{ "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "give a characterization of all solvable extensions of degree $p$ over $\\mathbb{Q}$, where $p$ is a prime.\r\n\r\nas a matter of completeness... by saying $L/K$ is solvable i mean that there's an extension $M/L$ such that $M/K$ is galois and $\\text{Gal}(M/K)$ is solvable.", "Solution_1": "no one?\r\ni found this problem pretty interesting...\r\n\r\nby the way, could some moderator edit the previous post? of course i meant an extension of [i]degree[/i] $p$, not of [i]order[/i] $p$...\r\n\r\n[moderator: as you wish...]", "Solution_2": "Suppose $E/\\mathbb{Q}$ is a solvable extension. Is it necessarily separable? Actually, now that I think about it, I don't think I will even need that." } { "Tag": [ "trigonometry", "calculus", "calculus computations" ], "Problem": "Let $ x$ be real and suppose that $ x/2^k$, $ k\\geq 0$, is not a multiple of $ \\pi/2$. Then find\r\n\\[ \\prod_{k\\equal{}0}^{\\infty}\\left(\\frac{2^k}{x}\\tan\\left(\\frac{x}{2^k}\\right)\\right)^{2^k}\\]", "Solution_1": "Simple manipulation shows that\r\n\r\n$ \\prod_{k\\equal{}0}^{N} \\left[ \\frac{2^k}{x} \\tan \\left( \\frac{x}{2^k} \\right) \\right]^{2^k} \\equal{} \\frac{2x}{\\sin 2x} \\left[ \\frac{2^{N}}{x} \\sin \\left( \\frac{x}{2^{N}} \\right) \\right]^{2^{N\\plus{}1}}$.\r\n\r\nThen by taking $ N \\to \\infty$ to both sides, we have\r\n\r\n$ \\prod_{k\\equal{}0}^{\\infty} \\left[ \\frac{2^k}{x} \\tan \\left( \\frac{x}{2^k} \\right) \\right]^{2^k} \\equal{} \\frac{2x}{\\sin 2x}$.", "Solution_2": "[quote=\"sos440\"]Simple manipulation shows that\n\n$ \\prod_{k \\equal{} 0}^{N} \\left[ \\frac {2^k}{x} \\tan \\left( \\frac {x}{2^k} \\right) \\right]^{2^k} \\equal{} \\frac {2x}{\\sin 2x} \\left[ \\frac {2^{N}}{x} \\sin \\left( \\frac {x}{2^{N}} \\right) \\right]^{2^{N \\plus{} 1}}$.[/quote]\r\nWould you please elaborate any further?", "Solution_3": "[quote=\"Croata\"]Would you please elaborate any further?[/quote]\r\nHow do you mean to elaborate?", "Solution_4": "I mean if you can show me how did you get that.", "Solution_5": "Ah, now I understood what you have meant.\r\n\r\nMy first approach was direct and intuitive, but somewhat messy. I used the double-angle formula for sine and rearranged the product to obtain the equality.\r\n\r\n[hide=\"Here is a line-by-line flow.\"]Let $ P_n$ be the $ n$-th partial product. Then\n\n$ \\begin{eqnarray*}\nP_N & = & \\prod_{k=0}^{N} \\left[ \\frac{2^k}{x} \\tan ( x/2^k) \\right]^{2^k} \\\\\n& = & \\prod_{k=0}^{N} \\left[ \\frac{2^k}{x} \\frac{\\sin (x/2^k)}{\\cos (x/2^k)} \\right]^{2^k} \\\\\n& = & \\prod_{k=0}^{N} \\left[ \\frac{2^{k-1}}{x} \\frac{2 \\sin (x/2^k) \\cos (x/2^k)}{\\cos^2 (x/2^k)} \\right]^{2^k} \\\\\n& = & \\prod_{k=0}^{N} \\left[ \\frac{2^{k-1}}{x} \\frac{\\sin (x/2^{k-1})}{\\cos^2 (x/2^k)} \\right]^{2^k} \\\\\n& = & \\left\\{ \\prod_{k=0}^{N} \\left[ 2^{k-1} \\sin (x/2^{k-1}) \\right]^{2^k} \\right\\} \\left\\{ \\prod_{k=0}^{N} \\left( \\frac{1}{x} \\right)^{2^k} \\right\\} \\left\\{ \\prod_{k=0}^{N} \\left[ \\frac{1}{\\cos^2 (x/2^k)} \\right]^{2^k} \\right\\} \\\\\n& = & \\frac{\\sin 2x}{2x} (\\cos (x/2^N))^{-2^{N+1}} \\left \\{ \\prod_{k=0}^{N-1} \\left[ 2^{k} \\sin (x/2^{k}) \\right]^{2^k} \\right\\}^{2} \\left\\{ \\prod_{k=0}^{N-1} \\left( \\frac{1}{x} \\right)^{2^k} \\right\\}^2 \\left\\{ \\prod_{k=0}^{N-1} \\left[ \\frac{1}{\\cos (x/2^k)} \\right]^{2^k} \\right\\}^2 \\\\\n& = & \\frac{\\sin 2x}{2x} (\\cos (x/2^N))^{-2^{N+1}} P_{N-1}^2,\n\\end{eqnarray*}.$\n\nNow divide both sides by $ P_{N-1}$ and arrange it to obtain the formula for $ P_{N-1}$. Replacing $ N-1$ by $ n$, we have exactly what we want.[/hide]\r\n\r\nThen I realized that we can obtain that by multiplying $ \\sin 2x$ and applying double-angle forumla recursively. Just try it by yourself!" } { "Tag": [ "modular arithmetic" ], "Problem": "Find all positive integers that prove the equation :\r\n\r\n\r\n\\[ x^6 \\plus{} x^3 y^3 \\plus{} y^6 \\equal{} 3^n \r\n\\]", "Solution_1": "Examining $ \\bmod9$ (where $ x^3,y^3\\in\\{0,\\pm1\\}$) and remembering that $ 3^n\\equiv3\\pmod9$ if $ n \\equal{} 1$ and $ 3^n\\equiv0\\pmod9$ if $ n > 1$, we find that if $ \\{x^3,y^3\\}\\pmod9$ equals (order doesn't matter),\r\n\r\n$ \\{ \\minus{} 1, \\minus{} 1\\}\\pmod9$, then $ x^6 \\plus{} x^3y^3 \\plus{} y^6\\equiv3\\pmod9$,\r\n$ \\{0, \\minus{} 1\\}\\pmod9$, then $ x^6 \\plus{} x^3y^3 \\plus{} y^6\\equiv1\\pmod9$,\r\n$ \\{1, \\minus{} 1\\}\\pmod9$, then $ x^6 \\plus{} x^3y^3 \\plus{} y^6\\equiv1\\pmod9$,\r\n$ \\{0,0\\}\\pmod9$, then $ x^6 \\plus{} x^3y^3 \\plus{} y^6\\equiv0\\pmod9$,\r\n$ \\{1,0\\}\\pmod9$, then $ x^6 \\plus{} x^3y^3 \\plus{} y^6\\equiv1\\pmod9$,\r\n$ \\{1,1\\}\\pmod9$, then $ x^6 \\plus{} x^3y^3 \\plus{} y^6\\equiv3\\pmod9$,\r\n\r\nso the only possibilities are $ x^6 \\plus{} x^3y^3 \\plus{} y^6\\equiv3\\pmod9\\implies(x,y,n) \\equal{} (1,1,1)$, and $ x^6 \\plus{} x^3y^3 \\plus{} y^6\\equiv0\\pmod9\\implies(x,y) \\equal{} (3a,3b)$ for positive integers $ a,b$.\r\n\r\nTherefore, $ (3a)^6 \\plus{} (3a)^3(3b)^3 \\plus{} (3b)^6 \\equal{} 3^n\\implies a^6 \\plus{} a^3b^3 \\plus{} b^6 \\equal{} 3^{n \\minus{} 6}$, from which we can conclude that $ \\boxed{(x,y,n) \\equal{} (3^k,3^k,6k \\plus{} 1)}$ where $ k\\in\\mathbb Z\\geq0$ are the only solutions." } { "Tag": [], "Problem": "There is NO substitute for visiting a college campus. Sometimes, however, you just can't do that, for whatever reason. [url=http://campustours.com/]Campus Tours[/url] is an excellent website for online tours of a college campus.", "Solution_1": "yeah, i'm visiting brown this month, i don't think u even need to sign up or notify them of the visit :o" } { "Tag": [ "ratio", "combinatorics unsolved", "combinatorics" ], "Problem": "Here is the problem:\r\n It is given set $S_n=\\{ 1, 3, ..., 2n-1 \\}$. Determine maximal integer $k$ such that exist subset $T$ of $S_n$ and for every two $x, y \\in T$ $x$ doesn't divide $y$ and $y$ doesn't divide $x$.", "Solution_1": "Let $K$ be the number of elements of $S_n$, not divisible by 3, i.e. $K=n-[(n+1)/3]$. Clearly, each element of $S_n$ belongs to one of $K$ subsets, in each of which the ratio of any twoelements is a power of 3 (for any element $a$, not divisible by 3, consider a subset $\\{a,3a,9a,\\dots\\}$). So, $k\\le K$. But if we consider all elements of $S_n$ except first $[(n+1)/3]$, then the least left element equals $2[(n+1)/3]+1$, and so the ratio of any two lest elements is less then 3. Hence it may be not integer." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Okay so first off ill type out the question for everyone , \r\n\r\nPerson A has 8000 dollars and would like to earn 500 dollars from the money, how much would see invest in a stock that has been getting a 10% annual return and how much should she invest in savings bonds that pay 0.4% annual interest? model this situation with a linear system.\r\n\r\nOkay so far this is the work ive done\r\n\r\nS be the amount in Stocls\r\nB be the amount in bonds\r\n\r\n1. 0.1s+0.4b=500\r\n2. s+b=8000---> S=8000-b\r\n\r\n0.1(8000-b)+0.4b=500\r\n800-0.1b+0.4b=500\r\n\r\nheres where i get stucked i check the answers for it and i should have gotten 0.06 but instead i ended up with 0.05 and sometimes 0.03 can anyone PLEASE PLEASE explain this to me. Thanks!", "Solution_1": "[quote=\"Alpha1234\"]1. 0.1s+0.4b=500[/quote]\r\n\r\nThis is incorrect. $4$% should be $0.04$, so your equation should be\r\n\r\n$0.1 s+0.04 b = 500$\r\n\r\nThis is the wrong forum to ask about homework. Classroom Math would be more appropriate. This is the [b]Olympiad[/b] Algebra section, not just the Algebra section.", "Solution_2": "ok sry can u delete this post 4 me then" } { "Tag": [ "function", "combinatorics unsolved", "combinatorics" ], "Problem": "find the number of solutions of: 2a+3b+5c=1000; where a,b,c are non-negative integers. :oops:", "Solution_1": "Hi;\r\n\r\nThere are 16834 solutions.", "Solution_2": "Yes, 16384.\r\nbobbym, did you find this answer by a computer, or not?", "Solution_3": "Hi Kirill;\r\n\r\nYou have a typo it is 16834 solutions. I used 2 generating functions which are ideally suited to this type of problem, they are;\r\n\r\n\r\n$ \\sum_{k \\equal{} 0}^{500}x^{2k} \\cdot \\sum_{k \\equal{} 0}^{334}x^{3k}\\cdot \\sum_{k \\equal{} 0}^{200}x^{5k} \\equal{} 1 \\plus{} x^2 \\plus{} x^3 \\plus{} x^4 \\plus{} 2x^5 \\plus{} 2x^6 \\plus{} 2x^7 \\plus{} 3x^8\\plus{}3x^9\\plus{}4x^{10}\\ldots$\r\n\r\nand\r\n\r\n$ \\frac {1}{(1 \\minus{} x^2)(1 \\minus{} x^3)(1 \\minus{} x^5)} \\equal{} 1 \\plus{} x^2 \\plus{} x^3 \\plus{} x^4 \\plus{} 2x^5 \\plus{} 2x^6 \\plus{} 2x^7 \\plus{} 3x^8\\plus{}3x^9\\plus{}4x^{10}\\ldots$\r\n\r\nThese are essentially the same, in as much as they give the same answer. But they are expanded in different ways. Now all that was necessary was to find the coefficient of x^1000 in the expansion of either one of them. I played with both trying to get the coefficient by binomial tricks or residues. I then used the second form to generate a recurrence relation:\r\n\r\na(n)=a(n-10)-a(n-8)-a(n-7)+a(n-3)+a(n-2)\r\n\r\nwith the initial conditions being the coefficients in the above expansions.\r\nand ran it with a computer to a(1000).", "Solution_4": "can it be done without a computer?", "Solution_5": "HI;\r\n\r\nI tried hard to develope an answer that wasn't computer dependent. But:\r\n\r\n2a+3b+5c=1000, that one thousand is what makes this intractable to all my hand methods. Another way of phrasing this is asking for the partitions of 1000 from the multiset { 2,2,2,2,2,2,2,2,... , 3,3,3,3,3,3,3,3,3,3,3,3... , 5,5,5,5,5,5,5,5,5,5,5,5,5...}. A really big problem!\r\n\r\nI am not saying it can't be done just because I can't but some problems even in math require a computer. In this case the recurrence I provided could be run in any language and even a handheld calculator or spreadsheet could do it.\r\n\r\nAlso a simple program (in any language) that ignores the ideas above and just consisted of 3 loops and a counter could be used." } { "Tag": [ "geometry", "rectangle", "combinatorics unsolved", "combinatorics" ], "Problem": "For a rectangle nxm points\r\n1)\r\nHow many ways that an ant move from a point at the \"top left - down\" (sorry ,i don't know how to mention the place of this point .... ) to the point at the \"top right -up\"?\r\nAn ant only can move up or right \r\n\r\n2)How many ways that an ant move from a point at the \"top left - down\" to the point at the \"top right -up\"?\r\nAn ant can move up ,down , left ,right but it can't remove to the point that it was occupied before .\r\n\r\nI just can solve the 1st proplem ... help me ..", "Solution_1": "First problem is well known. Answer is $C_{m+n}^{n}$.\r\n\r\nSecond problem is quite strange... I have some doubts that it is possible to give an answer for it.", "Solution_2": "To prove the first one, among the $m+n$ steps, we can freely choose the $n$ vertical ones (and they determine the whole path).\r\n\r\nWhere did you find the problems (the second one ;) )?", "Solution_3": "[quote=\"Myth\"]First problem is well known. Answer is $C_{m+n}^{n}$.\n\nSecond problem is quite strange... I have some doubts that it is possible to give an answer for it.[/quote]\r\n^_^\r\nit is nxm [b]points[/b] . So ... the answer is $C_{m+n-2}^{n-1}$.", "Solution_4": "Do you mean we need $(n-1)$ vertical and $(m-1)$ horizonal moves? If yes, then, indeed, answer is $C_{(m-1)+(n-1)}^{n-1}$.", "Solution_5": "ofcourse we need ... \r\n^_^\r\nthere are only nxm points .. \r\nit is not a hard problem ...\r\nbut the second is .... i can't plan any thing ..." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $ N$ be a complex matrix $ 2 \\times 2$ such that $ N^2 \\equal{} 0$. Show that either $ N \\equal{} 0$ or $ N$ is similar, over $ C$, to \r\n$ \\begin{pmatrix}\r\n0&0 \\\\ 1&0\r\n\\end{pmatrix}$", "Solution_1": "If you don't know the Jordan form, I'd suggest doing this in three steps:\r\n\r\n* Show that $ N$ is similar to a matrix with zeroes in the right column. (hint: eigenvalues, choice of basis)\r\n* Show that $ N$ is similar to a matrix with all but the lower left entry zero. (hint: eigenvalues)\r\n* Finish the problem. (hint: choice of basis)", "Solution_2": "If $ N$ is not zero, there is some $ v$ such that $ w \\equal{} Nv$ is not zero. We then have $ Nw \\equal{} 0$.\r\nWith respect to the basis $ \\{v,w\\}$, $ N$ has the matrix $ \\begin{bmatrix}0 & 0 \\\\\r\n1 & 0\\end{bmatrix}$.\r\n\r\nThis construction, generalized to larger matrices, is an important part of how Jordan form is proved. We can get a block structure with diagonal blocks of the form $ \\lambda I \\plus{} N$ easily from the existence of eigenvalues, and then this construction lets us pin down the nilpotent parts.\r\n\r\n(Note: This part works over any field. We only need an algebraically closed field for Jordan form so we can guarantee the existence of eigenvalues, and nilpotent matrices have all their eigenvalues equal to zero without any help.)", "Solution_3": "From this result, I know that we can get the following statement: If $ A$ is a $ 2 \\times 2$ matrix of complex entries, then $ A$ is similar, over $ C$ to one matrix of the following kind: \r\n$ \\begin{pmatrix}\r\na&0 \\\\ 0&b\r\n\\end{pmatrix}$\r\nor \r\n$ \\begin{pmatrix}\r\na&0 \\\\ 1&a\r\n\\end{pmatrix}$\r\nbut I can't get it, sorry I know this is easy, but it is difficult for me." } { "Tag": [], "Problem": "How can you express the number 100 using only four 9's?\r\n(You can use operations signs, exponents, etc, but not another number.)", "Solution_1": "[quote=\"Kevin K.\"]How can you express the number 100 using only four 9's?\n(You can use operations signs, exponents, etc, but not another number.)[/quote]\r\n\r\n[hide]$9/9+99=100$[/hide]", "Solution_2": "[hide]$99+\\frac{9}{9}=100$[/hide]", "Solution_3": "[hide]\n$99$+$\\sqrt{9}$/$\\sqrt{9}$=$100$[/hide]", "Solution_4": "[hide]9/9 + 99[/hide]", "Solution_5": "[hide]9/9+99=100[/hide]", "Solution_6": "[hide]$99+\\frac{9}{9}$[/hide]" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "superior algebra", "superior algebra solved" ], "Problem": "Let A in $M_n(R)$ such that $A^3 + 6A^2 +11A + 6I_n = 0_n$\r\nProve that there exist \r\n\r\n$(a,b_1,...,b_n)$ reals numbers, \r\n$B_1,...,B_n$ matrices\r\n$(B_k)^2 = I_n$ for any k in {1,..,n}\r\n\r\nsuch that\r\n\r\n$A = a.I_n + b_1.B_1 + ...+b_n.B_n$", "Solution_1": "Let $m_a$ be the minimal polynomial of $A$. Then $m_a| x^3+6x^2+11x+6=(x+1)(x+2)(x+3)$. This proves that $m_a$ has no multiple roots, so $A$ is diagonalizable.\r\n\r\nLemma: Let $A$ be a diagonal matrix in $M_n(\\mathbb{R})$ whose diagonal entries are ordered increasingly. Let $E$ be the set of eigenvalues of $A$ and let $m=|E|$. Then there exist diagonal matrices $B_1,\\ldots,B_m$ with $\\pm 1$ entries on the diagonal and $b_1,\\ldots,b_m$ reals such as $A=\\sum_{k=1}^mb_k\\cdot B_k$.\r\n\r\nProof: If $m=2$ then let $B_1=I_n$ and $B_2=diag(I_i,-I_j)$ such that $i+j=n$ and $i,j$ are the multiplicities of $a_1,a_2$, the $2$ eigenvalues of $A$.\r\nLet $b_1=\\frac{a_1+a_2}2$ and $b_2=\\frac{a_1-a_2}2$. Then $A=b_1B_1+b_2B_2$.\r\n\r\nIf $m\\neq 2$, let $a_1,\\ldots,a_m$ be the eigenvalues of $A$ and let $i_1,\\ldots,i_m$ be their multiplicities. \r\nLet $B_j=diag(-I_{i_1},-I_{i_2},\\ldots,I_{i_j},-I_{i_{j+1}},\\ldots,-I_{i_m})$, $j=\\overline{1,m}$.\r\n\r\nWe have to solve $b_1B_1+b_2B_2+\\ldots+b_mB_m=A$.\r\nThis is equivalent to $(2I_m-UU^t)B=\\bar{A}$ [b][*][/b], where $U=(u_{i,1}=1)_{i=\\overline{1,m}}$, $B=(b_{i,1}=b_i)_{i=\\overline{1,m}}$ and $\\bar{A}=(a_{i,1}=a_i)_{i=\\overline{1,m}}$. \r\nThe characteristic polynomial of $UU^t$ is $X^{m-1}\\cdot$ (the characteristic polynomial of $U^tU$) which is $X-m$. So $\\det(2I_m-UU^t)=2^{m-1}(2-m)\\neq 0$ for $m\\neq 2$. So [*] has a solution and this is a solution for the lemma.\r\n\r\nBack to the problem: obvious" } { "Tag": [ "MIT", "college", "summer program", "Mathcamp", "Harvard", "AMC", "USA(J)MO" ], "Problem": "MIT sent out their regular decision letters on Friday March 11th. Well, did anyone get in yet?", "Solution_1": "I got in.\r\n\r\n\r\nGeehoon Hong\r\nB.N. Cardozo High School, NY\r\nMathcamp 2004\r\nMIT class of '09", "Solution_2": "[quote=\"Geehoon\"]I got in.\n\n\nGeehoon Hong\nB.N. Cardozo High School, NY\nMathcamp 2004\nMIT class of '09[/quote]\r\n\r\nCongratulations, and congratulations to all the members who got into the college they wanted.\r\n\r\nBest regards,", "Solution_3": "i really want to go to MIT \r\nhow did you get in? what did you do to get in?", "Solution_4": "I'm a junior, but Hannah Zhou, a girl from my school (MMPC silver medalist, SW semifinalist) was accepted.", "Solution_5": "our school has two. im SO excited for them.", "Solution_6": "I got in also...w00t! (After being pwned in the Early Action round). No one else from my school got in.", "Solution_7": "For people who got into MIT, congratulation! I have a vested interest myself. What are you guys' standardized tests score? What types of specials awards or recognition do you have? I am just curious.", "Solution_8": "[quote=\"tetrahedr0n\"]I'm a junior, but Hannah Zhou, a girl from my school (MMPC silver medalist, SW semifinalist) was accepted.[/quote]\r\n\r\njust curious...what does MMPC and SW stand for?", "Solution_9": "MMPC = Michigan Math Prize Competition, the flagship state competition. Especially nice since it ranks its winners.\r\n\r\nSW = Siemens Westinghouse, the research competition.", "Solution_10": "I just found out I got in. I am very excited. Still waiting on Caltech and Harvard, but that won't be too stressful anymore.", "Solution_11": "Congratulations Nathan! I should know by tomorrow. Additional congrats to you and Adam on your 14s on the AIME.", "Solution_12": "Congrats to all.", "Solution_13": "Are the admissions decision available online too? :? :?", "Solution_14": "No. From what I've read on MIT's website (http://www.mit.edu) they said they will not release admission decisions online or by e-mail", "Solution_15": "no, a 97 GPA on a scale from 0-100, sometimes referred to as Mastermans'? I'm not sure but I have a pretty good idea that this is what Mariag is talking about", "Solution_16": "i am not sure what it's called, my school just gives all grades out of a 100 and the GPA is just the numerical average thereof", "Solution_17": "[quote=\"Mariag\"]i am not sure what it's called, my school just gives all grades out of a 100 and the GPA is just the numerical average thereof[/quote]\r\nAre you in the US? in most of the countries where I have been, the averages are given numerically (0-100, 0-20, 0-5, 0-10) and that is why sometimes US schools have problems cathegorizing the real GPA (or what in the US is considered GPA) of forein people. I honestly think that a person with more than 90 GPA in a 0-100 scale probably would have 4.00 GPA in the US!", "Solution_18": "I've always [b]dreamt of MIT[/b] and want to go there more than anything in the worls (even if I get Caltech or Princeton)I live in India and somebody told me just 2 students are going to be taken from india this year.Is this true?????????\r\n\r\nI've come among the [b]top 40 Young Scientists in India[/b] under the KVPY Scholarship conducted by the [b]IIT's[/b] in India.I've always been top in class and scored 94% in ALL India Board CBSE exams.(regarding very high)\r\nIve also got 2280 in SAT(new) and am quite confident about SAT II's i'll give next year.I'm also expecting near 100% in TOEFl\r\nI was also adjudged the best All round student in my school in Grade 10.(i had to leave that scholl coz it did'nt have classes beyond grade 10 I go to another school now)\r\n\r\nYet that 2 students taking thing is really a blow to me .Can u guys help me out.This is a long post but plz rate my chances and what else you think need to be done.\r\n\r\n[b]THANK YOU[/b]", "Solution_19": "djimenez,\r\n\r\nactually, i believe 95+/100 is equivalent to 4.0 GPA according to something i read a while ago which I don't quite remember", "Solution_20": "It depends on the school.\r\n\r\nIf I got 90 in every class, my GPA would be 4.0\r\nIf I got 100 in every class, my GPA would be 4.0\r\n\r\nThat is how my school does it.", "Solution_21": "i do go to school in the US. i once saw a chart ranking 97+ to be a 4.0 and so on down. the way frading in my school works is that any grade above a 90 is possible but below 90 grades are only given in steps of 5 and no grade below 45 is awarded. i thought this system was common for the US, but guess some schools do use the 4.0 scale :?", "Solution_22": "Maria, I think most schools use the 4.0 system. My school uses a 5.0 weighted scale. All classes are out of 4 points, but APs are out of 5. An A is a 4 in a regular class and a 5 in an AP, a B a 3 in a regular class and a 4 in an AP, et cetera. Minuses and pluses don't have different points. Plus, because the letter grade translates to the point, your points may be dependent on what teacher you have--our calc teacher had an A- at grades $\\ge 85$.", "Solution_23": "My school, as mentioned earlier is 4.0 system.\r\n\r\nPre-IB weights .02 per semester\r\nIB/AP weights .04 per semester\r\nCollege courses weight .04 or .08 depending on the class.\r\nI think Honors might weight .02..", "Solution_24": "[quote=\"sonar\"]Maria, I think most schools use the 4.0 system.[/quote]\r\nMy school (Benjamin N. Cardozo HS, Bayside NY) is using 0-100 system, and im not too confient on this, but most public schools in new york do use 0-100 system.", "Solution_25": "i c, so NY is just different... sorry for the confusion :(", "Solution_26": "Rushil - Yes MIT is nice, my son is very happy there. If you send me a pm I can give his e-mail to you if you want some more specific information.\r\n\r\nThere are many Indian students in MIT (Hey, they evne have a Cricket team, and a Bhangra group!). Many are graduate students. In undergraduate classes . natually . most are Indian-Americans (whose family is in US) and a very few foreign stiudents. Guess many parents would not send their kids far away for undergraduate studies. Each year they take about 100 foreigh students as undergraduate freshmen (about 8%). So yes the number is small, but not as low as 2 :) I would guess that your chances are good. (But I am not an admisson person!) \r\n\r\nFew things to consider, MIT does not give merit scholarships, They do give Financial assistance. Undergraduate Tution is about $30,000. (BTW Caltech, and some other schools (eg Duke) have good merit scholarships .)\r\n\r\nCheck out the web site (mit.edu) write to them etc... Hope this helps.\r\n\r\nIt seems that admisson staff is familier with Indian grading system etc.. Special awards (IIT d help, Also if you have participated in other well kown activities (eg RMO, INMO, IMO - also physics or chemisty olympiads, science-talent/national scholarships etc...) mention it,.- Admisson staff would most likely familier with such things.\r\n\r\n Good luck!\r\n\r\nBy the way I may have posted it here before but check it out:\r\n\r\n The video produced by MIT Indian students \r\nGet videos directly \r\nSmall (10 MB): http://web.mit.edu/anoop/www/msur/milesur.html\r\nBig(50 MB): http://web.mit.edu/anoop/www/msur/milesur50.mpg\r\n -", "Solution_27": "[quote=\"djimenez\"]last July I heard in a CNN broadcast that there has been a descence in the number of students comming to US Universities, because European, Australian and New Zelandian Univeristies have become very open and attractive for international students. [/quote]\r\n\r\nSince 2001 there are many more restrictions on the admission of non-US nationals to US universities.\r\nIt has become more difficult to obtain student visas, which is not surprising considering that (as a rough statistical generalization) non-US students at schools below the state university level were generally there for the purpose of obtaining a visa first and a certificate second. Many professional licenses such as hairdresser, electronics repairman, nurse, biotechnician, etc are in the USA obtained through professional \"colleges\" which were, until the new restrictions, basically visa mills for aspiring immigrants and degree mills for academically nonaspiring natives. The result of the tightened immigration restrictions, however, is seen all the way up to postdocs at MIT, Harvard, etc, and it was only intensive lobbying by the universities and technology industries that prevented the rules from becoming even more draconian.\r\n\r\nMeanwhile, the US having become less friendly to foreigners, many schools in Europe and elsewhere have seen a rise in interest from students from e.g. the Middle East, who are able and willing to pay the fees.", "Solution_28": "[quote=\"Geehoon\"]My school (Benjamin N. Cardozo HS, Bayside NY)[/quote]\r\n\r\nDid you, by any chance, have this teacher?\r\n\r\nhttp://sports.espn.go.com/espn/news/story?id=2098847", "Solution_29": "[quote=\"blahblahblah\"]\nDid you, by any chance, have this teacher?\n\nhttp://sports.espn.go.com/espn/news/story?id=2098847[/quote]\r\nmy brother said his global teacher, Mr. K was a professional wrestler :D It is very possible that it is him, but my brother doesn't have him any more." } { "Tag": [ "inequalities", "Cauchy Inequality", "inequalities proposed" ], "Problem": "Let $ a,\\ b,\\ c$ be real numbers such that $ a,\\ b,\\ c\\in{[\\minus{}\\sqrt{2},\\ \\sqrt{2}]}$ and $ abc\\neq 0$.\r\n\r\nProve that $ \\frac{(2\\minus{}a^2)(2\\minus{}b^2)(2\\minus{}c^2)}{\\left(\\frac{1}{|a|}\\plus{}\\frac{1}{|b|}\\plus{}\\frac{1}{|c|}\\right)^6}\\leq \\frac{1}{729}$.", "Solution_1": "[color=darkblue]Hello, [b]Kunny[/b] ![/color] [color=darkred][b]Happy new year ![/b][/color]\r\n\r\n[quote=\"kunny\"] [color=darkred]Let $ a,\\ b,\\ c$ be real numbers such that $ a,\\ b,\\ c\\in{[ \\minus{} \\sqrt {2},\\ \\sqrt {2}]}$ and $ abc\\neq 0$ .\n\nProve that $ \\frac {(2 \\minus{} a^2)(2 \\minus{} b^2)(2 \\minus{} c^2)}{\\left(\\frac {1}{|a|} \\plus{} \\frac {1}{|b|} \\plus{} \\frac {1}{|c|}\\right)^6}\\leq \\frac {1}{729}$. [/color][/quote]\n[quote=\"Virgil Nicula\"]$ \\{x,y,z\\}\\subset (0,2]\\implies\\ 729\\cdot(2\\minus{}x)(2\\minus{}y)(2\\minus{}z)\\ \\le \\left(\\frac {1}{\\sqrt x}\\plus{}\\frac {1}{\\sqrt y}\\plus{}\\frac {1}{\\sqrt z}\\right)^6\\ .$[/quote]\r\n[color=darkblue][b][u]Proof[/u].[/b] Prove easly that $ t\\in (0,2]\\ \\implies\\ \\frac {3\\minus{}t}{2}\\ \\le\\ \\frac {1}{\\sqrt t}\\ .$ Therefore, \n\n$ (2\\minus{}x)\\cdot(2\\minus{}y)\\cdot(2\\minus{}z)\\cdot 1\\cdot 1\\cdot 1\\ \\le\\ \\left[\\frac {(2\\minus{}x)\\plus{}(2\\minus{}y)\\plus{}(2\\minus{}z)\\plus{}1\\plus{}1\\plus{}1}{6}\\right]^6\\ \\implies$\n\n$ 729(2\\minus{}x)(2\\minus{}y)$ $ (2\\minus{}z)\\ \\le\\ \\left(\\frac {9\\minus{}x\\minus{}y\\minus{}z}{2}\\right)^6\\equal{}$ $ \\left(\\frac {3\\minus{}x}{2}\\plus{}\\frac {3\\minus{}y}{2}\\plus{}\\frac {3\\minus{}z}{2}\\right)^6\\ \\le\\ \\left(\\frac {1}{\\sqrt x}\\plus{}\\frac {1}{\\sqrt y}\\plus{}\\frac {1}{\\sqrt z}\\right)^6\\ .$[/color]", "Solution_2": "Nice Solution, Virgil Nicula.\r\n\r\nI hope the year 2009 will be good for you!", "Solution_3": "Hello, kunny!\r\nThis is my solution for your problem\r\n$ (2\\minus{}a^2)(2\\minus{}b^2)(2\\minus{}c^2) \\le \\left(\\frac{6\\minus{}a^2\\minus{}b^2\\minus{}c^2}{3}\\right)^3$\r\nAll we need to do is to prove \r\n$ 3\\left(6\\minus{}a^2\\minus{}b^2\\minus{}c^2\\right)\\le \\left(\\frac{1}{|a|}\\plus{}\\frac{1}{|b|}\\plus{}\\frac{1}{|c|}\\right)^2$\r\nWe can rewrite\r\n$ \\left(\\frac{1}{|a|}\\plus{}\\frac{1}{|b|}\\plus{}\\frac{1}{|c|}\\right)^2\\plus{}3\\left(a^2\\plus{}b^2\\plus{}c^2\\right)\\ge18$\r\nNow, it is time for using Cauchy inequality\r\n$ \\left(\\frac{1}{|a|}\\plus{}\\frac{1}{|b|}\\plus{}\\frac{1}{|c|}\\right)^2 \\ge \\frac{9}{\\sqrt[3]{|abc|^2}}$\r\n\r\nand $ 3\\left(a^2\\plus{}b^2\\plus{}c^2\\right) \\ge 9\\sqrt[3]{|abc|^2}$\r\n\r\nThe last step is to prove \r\n$ \\frac{9}{\\sqrt[3]{|abc|^2}}\\plus{}9\\sqrt[3]{|abc|^2} \\ge 18$ which is true by applying Cauchy inequality for 2 positive real numbers\r\nWe are done!", "Solution_4": "Hello, great math. :) \r\n\r\nNow I have just collected 3 solutions. :lol:", "Solution_5": "$ LHS\\leq\\frac{(\\sum |a|)^6(2\\minus{}\\frac{\\sum a^2}{3})^3}{9^6}\\leq\\frac{[(\\frac{\\sum |a|}{3})^2(2\\minus{}(\\frac{\\sum |a|}{3})^2]^3}{3^6}\\leq\\frac{1}{3^6}\\equal{}\\frac{1}{729}$", "Solution_6": "I would like to give you another solution of mine, kunny\r\nUsing Cauchy inequality, we get\r\n$ \\left(\\frac {1}{|a|} \\plus{} \\frac {1}{|b|} \\plus{} \\frac {1}{|c|}\\right)^6 \\ge \\frac {792}{|abc|^2}$\r\nTherefore, we have $ \\frac {(2 \\minus{} a^2)(2 \\minus{} b^2)(2 \\minus{} c^2)}{\\left(\\frac {1}{|a|} \\plus{} \\frac {1}{|b|} \\plus{} \\frac {1}{|c|}\\right)^6}\\leq \\frac{(2 \\minus{} a^2)(2 \\minus{} b^2)(2 \\minus{} c^2)(abc)^2}{729}$\r\nSimultaneously, we still hold the following inequality\r\n$ (2 \\minus{} a^2)a^2 \\le \\left(\\frac {2 \\minus{} a^2 \\plus{} a^2}{2}\\right)^2 \\equal{} 1$\r\nLikewise, we apply for the other variables b and c." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If $ a$, $ b$, $ c$ are nonnegative real numbers, then show that \r\n\r\n$ \\frac65\\geq\\frac{ ab+ac}{ a^{2}+\\left( b+c\\right)^{2}}+\\frac{bc+ba}{b^{2}+\\left(c+a\\right)^{2}}+\\frac{ca+cb}{c^{2}+\\left(a+b\\right)^{2}}$.", "Solution_1": "Let $a+b+c=3.$\r\nHence, $6/5\\geq\\frac{ ab + ac}{ a^2 + ( b+c)^2} + \\frac{bc+ba}{b^2+(a+c)^2} +\\frac{ca+cb}{c^2+(b+a)^2}\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum\\frac{a(3-a)}{a^2+(3-a)^2}\\leq\\frac{6}{5}\\Leftrightarrow\\sum(\\frac{3a-a^2}{2a^2-6a+9}-\\frac{9}{25}(a-1)-\\frac{2}{5})\\leq0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum\\frac{(a-1)^2(2a+1)}{2a^2-6a+9}\\geq0.$ :)\r\nBy the way, $\\sum\\frac{ab+ac}{ka^2+(b+c)^2}\\leq\\frac{6}{k+4},$ where $0.8\\leq k\\leq8.$ ;)", "Solution_2": "Arqady what a nice solution!!!\r\nIn the most of solutions you use a trick like this\r\n[quote=\"arqady\"]\n$\\Leftrightarrow\\sum\\frac{a(3-a)}{a^2+(3-a)^2}\\leq\\frac{6}{5}\\Leftrightarrow\\sum(\\frac{3a-a^2}{2a^2-6a+9}-\\frac{9}{25}(a-1)-\\frac{2}{5})\\leq0$[/quote]\r\n\r\n\r\nHow do you think of it ? You are on the earth ?", "Solution_3": "[quote=\"silouan\"]\nHow do you think of it ? You are on the earth ?[/quote]\r\nDear, silouan. This is very easy:\r\nIf $f(a)=\\frac{3a-a^2}{2a^2-6a+9}+k(a-1)-\\frac{2}{5}$ then I wish $f'(1)=0$ and obtain $k=-\\frac{9}{25}.$ ;)", "Solution_4": "[quote=\"arqady\"][quote=\"silouan\"]\nHow do you think of it ? You are on the earth ?[/quote]\nDear, silouan. This is very easy:\nIf $f(a)=\\frac{3a-a^2}{2a^2-6a+9}+k(a-1)-\\frac{2}{5}$ then I wish $f'(1)=0$ and obtain $k=-\\frac{9}{25}.$ ;)[/quote]\r\n\r\nVery good solution RESPECT! \r\nSo where can study of using that strategy can you say please :-)", "Solution_5": "[quote=\"Davron\"]\nSo where can study of using that strategy can you say please :-)[/quote]\r\nThe scope of this strategy is limited enough. But sometimes it works.", "Solution_6": "The tangent line is the key point, but sometimes.", "Solution_7": "[quote=\"kunny\"]The tangent line is the key point, but sometimes.[/quote]\r\n\r\nsure i am agree with you! \r\n\r\nDo anyone have another solution !!! If you want i can post the original form of the Problem ...\r\n\r\nOr everybody have that...?", "Solution_8": "[quote=wudud]If $ a$, $ b$, $ c$ are nonnegative real numbers, then show that \n\n$ \\frac65\\geq\\frac{ ab+ac}{ a^{2}+\\left( b+c\\right)^{2}}+\\frac{bc+ba}{b^{2}+\\left(c+a\\right)^{2}}+\\frac{ca+cb}{c^{2}+\\left(a+b\\right)^{2}}$.[/quote]\nThe following inequality is also true.\nLet $a$, $b$ and $c$ be [b]real[/b] numbers such that $\\prod_{cyc}(a^2+(b+c)^2)\\neq0$. Prove that:\n$$ \\frac65\\geq\\frac{ ab+ac}{ a^{2}+\\left( b+c\\right)^{2}}+\\frac{bc+ba}{b^{2}+\\left(c+a\\right)^{2}}+\\frac{ca+cb}{c^{2}+\\left(a+b\\right)^{2}}$$\n\n\n", "Solution_9": "It seems that he following inequality is also true.\nLet $a$, $b$ and $c$ be [b]real[/b] numbers such that $\\prod_{cyc}(a^2+(b+c)^2)\\neq0$. Prove that:\n$$\\frac{ ab+ac}{ 16a^{2}+\\left( b+c\\right)^{2}}+\\frac{bc+ba}{16b^{2}+\\left(c+a\\right)^{2}}+\\frac{ca+cb}{16c^{2}+\\left(a+b\\right)^{2}}\\leq \\frac{3}{10}$$", "Solution_10": "[quote=arqady]The following inequality is also true.\nLet $a$, $b$ and $c$ be [b]real[/b] numbers such that $\\prod_{cyc}(a^2+(b+c)^2)\\neq0$. Prove that:\n$$ \\frac65\\geq\\frac{ ab+ac}{ a^{2}+\\left( b+c\\right)^{2}}+\\frac{bc+ba}{b^{2}+\\left(c+a\\right)^{2}}+\\frac{ca+cb}{c^{2}+\\left(a+b\\right)^{2}}$$[/quote]\n\nBecause if $x,\\,y,\\,z$ be real numbers such that $ x + y + z = 1,$ then\n$$ \\frac {1}{1 + x^2} + \\frac {1}{1 + y^2} + \\frac {1}{1 + z^2}\\leq \\frac {27}{10}.$$", "Solution_11": "[quote=arqady]It seems that he following inequality is also true.\nLet $a$, $b$ and $c$ be [b]real[/b] numbers such that $\\prod_{cyc}(a^2+(b+c)^2)\\neq0$. Prove that:\n$$\\frac{ ab+ac}{ 16a^{2}+\\left( b+c\\right)^{2}}+\\frac{bc+ba}{16b^{2}+\\left(c+a\\right)^{2}}+\\frac{ca+cb}{16c^{2}+\\left(a+b\\right)^{2}}\\leq \\frac{3}{10}$$[/quote]\nYes, it's true! \nThere is a nice proof for this inequality. ;) \n\nQuite nice.\nLet $x$, $y$ and $z$ be real numbers such that $xy+xz+yz+xyz=4$. Prove that:\n$$\\frac{x}{4x^2+1}+\\frac{y}{4y^2+1}+\\frac{z}{4z^2+1}\\leq\\frac{3}{5}$$", "Solution_12": "[quote=wudud]If $ a$, $ b$, $ c$ are nonnegative real numbers, then show that \n\n$ \\frac65\\geq\\frac{ ab+ac}{ a^{2}+\\left( b+c\\right)^{2}}+\\frac{bc+ba}{b^{2}+\\left(c+a\\right)^{2}}+\\frac{ca+cb}{c^{2}+\\left(a+b\\right)^{2}}$.[/quote]\n\nhello, it is $$125\\,{b}^{5}u+ \\left( 565\\,{u}^{2}+365\\,uv+90\\,{v}^{2} \\right) {b}^{4}\n+ \\left( 948\\,{u}^{3}+1162\\,{u}^{2}v+648\\,u{v}^{2}+152\\,{v}^{3}\n \\right) {b}^{3}+ \\left( 786\\,{u}^{4}+1412\\,{u}^{3}v+1238\\,{u}^{2}{v}^\n{2}+572\\,u{v}^{3}+112\\,{v}^{4} \\right) {b}^{2}+ \\left( 328\\,{u}^{5}+\n780\\,{u}^{4}v+940\\,{u}^{3}{v}^{2}+650\\,{u}^{2}{v}^{3}+247\\,u{v}^{4}+40\n\\,{v}^{5} \\right) b+56\\,{u}^{6}+168\\,{u}^{5}v+258\\,{u}^{4}{v}^{2}+236\n\\,{u}^{3}{v}^{3}+133\\,{u}^{2}{v}^{4}+43\\,u{v}^{5}+6\\,{v}^{6}\n\\geq 0$$\nwhich is true.\nSonnhard." } { "Tag": [ "ARML" ], "Problem": "I will be posting another Mock ARML tommorrow at around 4:30 PM...Same rules as the first one and the answers can be submitted to me by Monday night. Thank you :)", "Solution_1": "I should have several ARML style questions up on Monday or Tuesday to go with Shobhit's problems." } { "Tag": [ "probability", "function" ], "Problem": "We view the world as particles and discrete units. But De-Broglie says that everything has both a particle and a wave nature. Can we view the world as a continuous landscape by tapping into the wave nature? If not, why? If so, how? Does anyone have any ideas? \r\n Ashwath :)", "Solution_1": "Great to see someone from india\r\nMERA BHARAT MAHAN\r\n\r\n\r\nLet me give you a very practical example for this\r\nConsider a particle located in a region of potential V(x)\r\nSuch That \r\n \t\t\t\r\n\r\nV(x)= 0 if 0c+1,c-even.$ simplifying we get \r\n$a!(c+1)!=a!+(c+1)!+c!\\Rightarrow (c+1)!|c!.$ So no solution in this case.\r\n\r\n(ii)$c>a,b$ then after dividing both sides of $a!\\cdot b!=a!+b!+c!$ by $b!,$ the LHS is even RHS is odd unless $a=b+1$(this case doesn't give solutions) or $a=b\\Rightarrow a!=2+c!/a!\\Rightarrow c\r\n#include \r\n\r\n\r\nifstream fin(\"contact.in\");\r\nofstream fout(\"contact.out\");\r\n\r\n//ps is the the count of each pattern\r\nint ps[8192];\r\n//each digit is temporarily stored here\r\nint pattern[14];\r\nchar tmp;\r\n\r\n\r\nint a,b,c;\r\nint coount=0;\r\n\r\nvoid sort (char j)\r\n{\r\ncount++;\r\n\r\nif (j=='0')\r\n{\r\n\r\npattern[12]=pattern[11];\r\npattern[11]=pattern[10];\r\npattern[10]=pattern[9];\r\npattern[9]=pattern[8];\r\npattern[8]=pattern[7];\r\npattern[7]=pattern[6];\r\npattern[6]=pattern[5];\r\npattern[5]=pattern[4];\r\npattern[4]=pattern[3];\r\npattern[3]=pattern[2];\r\npattern[2]=pattern[1];\r\npattern[1]=0;\r\n}\r\n\r\nif (j=='1')\r\n{\r\n\r\n\r\npattern[12]=pattern[11];\r\npattern[11]=pattern[10];\r\npattern[10]=pattern[9];\r\npattern[9]=pattern[8];\r\npattern[8]=pattern[7];\r\npattern[7]=pattern[6];\r\npattern[6]=pattern[5];\r\npattern[5]=pattern[4];\r\npattern[4]=pattern[3];\r\npattern[3]=pattern[2];\r\npattern[2]=pattern[1];\r\npattern[1]=1;\r\n}\r\n\r\nint y;\r\nint num;\r\nint l;\r\nint p,z;\r\n\r\n\r\n\r\n//tries all pattern lengths\r\nfor(y=a;y<=b;y++){\r\nnum=0;\r\n\r\n \r\nif(y=1; l--){\r\np=1;\r\nif (pattern[l]==1){\r\nfor(z=l; z>1; z--)\r\np=p+p;\r\nnum=num+p;\r\n}\r\n}\r\np=1;\r\nfor(z=1; z<=y; z++)\r\np=p+p;\r\nnum=num+p;\r\nps[num]++;\r\n\r\n\r\n}\r\n}\r\n\r\n}\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nint main()\r\n{\r\nchar t;\r\n\r\n\r\n//a is shortest length, b is longest length, c isn't important for now\r\n\r\nfin>>a>>b>>c;\r\n\r\nwhile(fin.get(t)){\r\n\r\n\r\nsort(t);\r\n\r\n}", "Solution_1": "Hi\r\n\r\nWell, I think you've got the main thing right, but really it is really tough to make out exactly where the problem might be because usually, the bugs creep in when you put separate things together. It may stem from things such as array indices etc... So, it would probably be easier if you could post all the code. Also if you could post a small case for which it doesnt work.\r\n\r\nBut really, the best thing you can do is to try to do it yourself(If you havent already tried a lot). This would give you more experience in debugging your own programs and help you in future.\r\n\r\nIn general, for debugging, try and look for cases for which it doesnt work and print values at different points in the program(if you're not using a debugger). Dry run small cases.\r\n\r\nHope this helped...", "Solution_2": "Yes, I've had this basic code for weeks, and my debugging efforts have only allowed me to isolate the problem with this part of the code. I was hoping a fresh set of eyes could help me with the problem.\r\n\r\n\r\nThe way my program overcounts is bizarre.\r\n\r\nin the following test case\r\n2 4 10\r\n01010010010001000111101100001010011001111000010010011110010000000\r\n\r\nmy program overcounts all patterns with a 0 by 1\r\n\r\nin\r\n1 12 20\r\n11111111111(480 1's)\r\n\r\nit overcounts all patterns by 6\r\n\r\nin larger and more complicated tests my program overcounts erratically.\r\n\r\nhere's my display algorithm if you think it will help (but I know the code in the first part overcounts.)\r\n\r\n\r\n\r\n\r\n\r\nint ca,cb,cc,cd,ce,cf,cg;\r\nint max;\r\n\r\nnote: the array asd holds the powers of 2\r\nnote:the array bps holds the set of frequencies about to be displayed\r\n\r\n//each iteration displays one frequency\r\nfor(ca=c;ca>0;ca--){\r\n\r\nmax=0;\r\n\r\n//finds the proper frequency to display\r\nfor(cb=asd[a];cbmax){\r\nmax=ps[cb];\r\n\r\nfor(cc=1;cc<=4096;cc++)\r\nbps[cc]=0;\r\n\r\ncd=2;\r\nbps[1]=cb;\r\n}\r\n\r\n\r\n\r\n}\r\n\r\n//if no more values to display, terminate early \r\nif(max==0)\r\nbreak;\r\n\r\nfout<=asd[ce] && bps[cg]=0; cf--){\r\n\r\n\r\nif(bps[cg]=asd[cf]){\r\nfout<<\"1\";\r\nbps[cg]=bps[cg]-asd[cf];\r\n}\r\n}\r\n\r\n\r\nif(cg!=cd-1)\r\nfout<<\" \";\r\n}\r\n\r\nfout<=1; l--){ \r\np=1; \r\nif (pattern[l]==1){ \r\nfor(z=l; z>1; z--) \r\np=p+p; \r\nnum=num+p; \r\n} \r\n} \r\np=1; \r\nfor(z=1; z<=y; z++) \r\np=p+p; \r\nnum=num+p; \r\nps[num]++; \r\n\r\n First, should the if condition be y <= count instead of y < count? Second, the end brace for that if statement seems too early; shouldn't it cover this whole section of code, including the part where you calculate the leading '1'? \r\n\r\n There are lots of comments I could make about the coding style, but I don't have time.\r\n\r\n I would recommend that you try your program on all trivial test cases, like input strings of length 1, 2, and 3, before you try more complicated cases.", "Solution_4": "Ok, thanks for the advice. I believe you're right on that condition. However, I think the end brace for that statement does include the part where I calculate the leading 1.\r\n\r\nHmm...you're also right about the trivial test case checking.\r\n\r\nthe test case of\r\n1 4 10\r\n1\r\n\r\ngives two 1's two 01's one 0 and one 001.\r\n\r\nweird.\r\nPerhaps there is something fundamentally wrong with how I'm handling the input. is the fin.get function the right one to use?\r\n\r\nIf you ever do have time to comment on my coding style it would be most appreciated.", "Solution_5": "After you read a, b, and c, isn't the input cursor still at the end of the first line? So when you call fin.get(t), won't the first one or two characters you read be the carriage return and/or the line feed from the first line (depending on your operating system)? \r\n\r\nIf I had to solve the problem, I would have approached it quite differently. Instead of an array as the main data structure, I would have used a map (or hash_map) from (C++) strings to integers. Then the whole program would have been about 15 lines of code. The STL library is quite powerful." } { "Tag": [ "geometry", "integration", "analytic geometry", "calculus", "rotation", "algebra", "polynomial" ], "Problem": "Basically I got to \r\n\r\n$x=12(y^{2}-y^{3}) from 0\\to 1$\r\n\r\nwhat is the volume of the solid created when this is revolved over the y=axis.\r\n\r\nThanks for any help in advance. I remember doing these last semester no problem but now in a review section I can only start but never get anywhere. Thank you again.", "Solution_1": "(writing in my sleep)\r\n\r\nThe basic form for the surface area of revolution is $\\int 2\\pi r\\sqrt{dx^{2}+dy^{2}}$, where $r$ is the distance from a point $(x,y)$ to the axis of rotation. The term $\\sqrt{dx^{2}+dy^{2}}$ is a useful mnemonic device: you can transform it by pulling out $dx$ or $dy$ from the square root. In this case $r=x$ (distance to the $y$-axis is measured by the $x$-coordinate, and vice versa). The variable of integration should be $y$; otherwise you'd have a hard time solving for $y$ in terms of $x$. So, $SA=\\int_{0}^{1}2\\pi 12(y^{2}-y^{3})\\sqrt{(dx/dy)^{2}+1}dy$. Since $dx/dy=12(2y-3y^{2})$, we have $SA=24\\pi \\int_{0}^{1}(y^{2}-y^{3})\\sqrt{144(2y-3y^{2})^{2}+1}dy$, a nightmarish integral. Time to wake up. :surrender:", "Solution_2": "When mlok wakes up, he might see that vasion asked for the volume, not the surface area.\r\n\r\nSlice this whichever way is convenient - in this case into strips parallel to the $x$ axis. Each strip is located at height $y,$ has width $dy$ and extends from $x=0$ (which is the $y$ axis) to $x=12(y^{2}-y^{3}).$\r\n\r\nNow rotate this strip about the $y$ axis. It becomes a thin circular disk centered on the $y$ axis, of thickness $dy$ and radius $12(y^{2}-y^{3}).$ This has volume equal to $\\pi r^{2}dy$ or\r\n\r\n$dV=\\pi(144)(y^{2}-y^{3})^{2}\\,dy$ so\r\n\r\n$V=144\\pi\\int_{0}^{1}(y^{2}-y^{3})^{2}\\,dy.$\r\n\r\nThis is a polynomial integral, not the \"nightmarish\" integral for the surface area." } { "Tag": [ "inequalities solved", "inequalities" ], "Problem": "Let $x,y,z \\in [0,1]$. Prove that $x+y+z-xy-yz-zx \\leq 1$.", "Solution_1": "The LHS is an affine expression in each of the variable so it reaches its maximum at the endpoints of [0,1], that is when x,y,z are in {0;1}.\r\nDirect computation shows that if (x,y,z) = (0,0,0) or (1,1,1) then the lhs is 0, and if (x,y,z) is a permutation of (1,0,0) or (1,1,0) then the lhs is 1.\r\n\r\nPierre.", "Solution_2": "(x-1)(y-1)(z-1)<=0\r\nx+y+z-xy-yz-zx<=1-xyz<=1" } { "Tag": [], "Problem": "Let $n$ be an integer between 1 and 9 (both included) Consider all n-digit numbers formed by using the digits 1 through $n$ once and only once in each number formed. What is the sum of all of these $n$-digit numbers?", "Solution_1": "Hardly Pre-Olympiad.\r\n\r\n[hide=\"Solution\"] Each digit $k$ appears in each place $(n-1)!$ times. The sum is\n\n$\\frac{n(n+1)}{2}(n-1)! \\left(1+10+...+10^{n-1}\\right)$ [/hide]" } { "Tag": [ "MATHCOUNTS", "probability" ], "Problem": "What scores are you and your team getting on national mathcounts problems?\r\n\r\n(only write if you are in nats)\r\n\r\nI'm getting around averages of 28s on nats tests and our team has an average of 8 on team rounds.", "Solution_1": "I made nats but I haven't done any practice national tests yet. Where do you get them?", "Solution_2": "You can buy them (but they're really expensive) from the MC store (only like 1999-2007 though). The only other option is to ask people to e-mail or pm them to you (which most people will call \"illegal\") or ask your coach or other coaches for them.", "Solution_3": "lol, my coach is also your coach", "Solution_4": "lol i know. it's fun talking like that.\r\n\r\nAnd for your scores, your team should average perfect scores on the team rounds (if they want to place) and get at least 35s. It might be better to aim for the higher 30s, as the test is probably going to be pretty beastly.", "Solution_5": "...\r\nGuys could you please stop spamming the MC forums just so you can increase your posts?\r\nAnd yeah, you wanna average more than 35 on most Nats tests.", "Solution_6": "Meh, I always get in the uber-high thirties to low forties for individuals, but then again, I've done 1998-2004 like so many times, and though I don't remember the answers, the questions are pretty familiar, which is annoying and unfair, I admit. As for the recent ones, I can get in the mid-30's for 2005 cuz it's harder than most of the previous tests (in my opinion) and in the semi-high-almost-competent-nearly-acceptable 30's for 2006. I have the 2007 test, but I don't wanna retake it until like weeks before the competition. The last time I saw those questions was in May last year at Nats, so I'm saving these relatively new, unrecycled problems for later.\r\n\r\nAs for team scores, I practice by myself and can get 7's to 10's depending on difficulty. There was the rare 4 when I totally had no time to finish the 2005 one, but the rest were pretty easy. :D \r\n\r\nPS: Sorry for the immodesty if there was any.... :blush: Then again, my scores are nothing compared to other people's on AoPS, so I'm fine, I think. :maybe:", "Solution_7": "just so you all know, all you need to make is countdown is somewhere from 31-35 depending on the year difficulty. Probably the reason we aren't doing that good as a team is that we have all never been to nats (I'm new to mathcounts, I only learned about it in october) so we just have to get used to these problems. Well our scores have been going up:\r\n7,7,10, so I think we will do better as a team.\r\n\r\nEdit: thanks peijin-simple mistake.", "Solution_8": "I'm pretty sure you got the meaning of veteran wrong because a veteran is someone who is experienced and already has been through the event before.\r\nBut anyways, Team Round counts a lot in MC, it basically counts as 8 points that you guys score.", "Solution_9": "meh, I keep on missing something for a stupid reason", "Solution_10": "old nats: 41-44\r\n\r\nafter 2000 nats: 33-41", "Solution_11": "If I were going back to Nationals I would probably not waste my time worrying how other people are preparing, and preparing yourself.\r\n\r\nYou should probably try it.", "Solution_12": "I guess he just wants to know what scores he needs to get to be successful, or is just bored of studying...\r\nEither way, I think you [b]should[/b] be trying to do some old Nats tests instead of coming on AoPS and posting questions you already know how to solve here.", "Solution_13": "Gotta agree with Catalyst on this one. If you knew you only needed a 38 wouldyou leave 6 blank since you didn't NEED them? Areyou going to try HARDER if you think you need a 39? Go do some problems, write some problems, prove some solutions, understand what you did wrong on the last one.\r\n\r\nAssume you need to get them all right and go out and try to get them all right. Or, assume you have no chance and just enjoy the challenge and what ever will be will be. \r\n \r\nGood luck to everyone, and relax!!!!!", "Solution_14": "[quote=\"frost13\"]Or, assume you have no chance and just enjoy the challenge and what ever will be will be. [/quote]\r\nTHAT'S HOW I MADE NATS IN SEVENTH, THEN NOT IN EIGHTH.\r\n\r\nIt works--I thought I woud fail chapter in seventh, then wound up getting second at states. Gain that mentality, then don't lose it like I did.", "Solution_15": "Learn how to approach problems. Be persistent. Look for creative solutions to problems you've done before. Check problems over using different solutions, and listen to Carl for once.", "Solution_16": "At wiseidiot: You don't think that setting your goal as \"beat Lyndon Ji\" would help you? (Sorry Lyndon, you're just a veteran whose name I know. I don't want you not to win again, I'm just using you as a benchmark. :))\r\n\r\nIn general, my philosophy on national competitions (and competitions in general) is: go to win, but don't let winning become more important than having fun. (I went to Nat MC, nat vocabulary competitions 4 times, and national science fair twice so far, if that lends my words any weight. And I've done well in some small subset of those, lest you think having fun has gotten in the way of success.)", "Solution_17": "Who's this \"Lyndon Ji\" you're speaking of? I must beat him this year!! I will utterly destroy him! Wait a second...\r\n\r\nMy practice tests range from 0 to 46. Oh. That's how good I am.", "Solution_18": "i 100% agree. i mean, why do you (or anyone) do mathcounts? $ For Fun$.", "Solution_19": "what kind of place does cough cough low 30s cough cough get you at nats? My friend who is awful at math is wondering...", "Solution_20": "In the old ages (1990's), that used to be considered average.\r\nNowadays (2002-present), that is an amazing score to get.", "Solution_21": "Thanks. I'll tell my.... friend.", "Solution_22": "Does your friend happen to have the username dynamo729 on AoPS =P? \r\n\r\nWhat place do you think 40-46s would get you... My friend (who is good at making \"smart mistakes\" =P and is horrible at math) wants to know. =D", "Solution_23": "Nationals seem to be much harder nowadays from what I can tell. Also, the problems are less \"formulaic approach\" and more creative.\r\n\r\nLast year the national written winner, Justin Ahmann, got 25+16 = 41. Most years that woulda put him in like 8th place.", "Solution_24": "well, considering last year's scores, 40-46's would probably get top 3...", "Solution_25": "[quote=\"mathcrazed\"]Does your friend happen to have the username dynamo729 on AoPS =P? \n\nWhat place do you think 40-46s would get you... My friend (who is good at making \"smart mistakes\" =P and is horrible at math) wants to know. =D[/quote]\r\n\r\nMaybe. I'll say undefined.", "Solution_26": "I get into the 30's. I will not win nationals. :(", "Solution_27": "The problems, as far as I can tell, are getting less and less about arithmetic and more and more about actual problem solving. Many people say that this is making them more difficult, but it just depends on the way your mind thinks. For example, I thought this year's problems were much easier than usual, while all of my friends and teammates told me it was much harder. I really dislike the formula/arithmetic types of problems, and like where they are going with this. It's making competitions much easier for me :D .", "Solution_28": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191755\r\n\r\nthere might be some states left?\r\n\r\nedit your post for difficulty please.\r\n\r\ni know, its hard :wink:", "Solution_29": "[quote=\"1=2\"]I get into the 30's. I will not win nationals. :([/quote]\r\n\r\nwell it all depends. For example, a 34 or so will get you into countdown at nationals. However, most people who do well at practice end up getting very nervous at the competition and do very badly.\r\n\r\nAbout Turkishgenius, the years in general get harder, for example, at nationals, 2005 was considered the hardest year, yet 2006 and 2007 were much harder than the tests before 05." } { "Tag": [], "Problem": "As avea o intrebare :\r\nDaca \\[\\displaystyle ABCD \\] e un patrulater inscriptibil si \\[\\displaystyle M,N,P,Q\\]mijloacele laturilor\\[AB,BC,CD,AD\\] si luam in exteriorul patrulaterului \\[\\displaystyle M_1,N_1,P_1,Q_1 \\ a.i. \\ MM_1=\\frac{AB}{2} \\ , \\ NN_1=\\frac{BC}{2}\\ , \\ PP_1=\\frac{DC}{2}\\ , \\ QQ_1=\\frac{AD}{2} \\.\\]Putem demonstra ca \\[\\displaystyle M_1N_1P_1Q_1\\] inscriptibil ?", "Solution_1": "Sau, cu alte cuvinte, construim patrate pe laturile patrulaterului, in exteriorul lui, si $M_1,$etc. sunt centrele patratelor, nu?\r\n\r\nAm facut un desen pe calculator, in Euklides si imi pare rau, dar nu pare sa fie adevarat in general.", "Solution_2": "daca construiesti patrate in afara patrulaterului pe laturile sale atunci $MM_1$,etc. vor fi perpendiculare pe laturi.caz particular.", "Solution_3": "ce se *poate* demonstra este ca diagonalele acestui patrulater, $ M_1N_1P_1Q_1 $, sint perpendiculare si au lungimi egale.", "Solution_4": "Daca $X,\\ Y,\\ Z,\\ T$ sunt centrele patratelor construite in exteriorul patrulaterului convex $ABCD$ pe laturile $[AB],\\ [BC],\\ [CD],\\ [DA]$ respectiv, atunci :\r\n[b]1).[/b] $XZ\\perp YT;\\ XZ=YT;$\r\n[b]2).[/b] Mijloacele diagonalelor $[AC],\\ [BD]$ si mijloacele segmentelor $XZ,\\ YT$ sunt varfurile unui patrat.\r\n\r\n[u]Observatie.[/u] Aceasta clasica problema apare frecvent in ilustrarea metodei de rezolvare vectoriala sau a metodei de utilizare a numerelor complexe." } { "Tag": [ "algebra", "polynomial", "limit", "search", "calculus", "calculus computations" ], "Problem": "Suppose $ f(x)$ is a polynomial of degree $ n$ such that $ f(x)\\geq 0$ for each $ x\\geq 0$ Prove that $ f(x)\\plus{}f'(x)\\plus{}f'(x)\\plus{}...\\plus{}f^{(n)}(x)\\geq 0$\r\nfor all $ x\\geq 0$", "Solution_1": "Let $ g(x) \\equal{}\\left( f^{(0)}(x)\\plus{}\\cdot\\plus{}f^{(n)}(x)\\right) e^{\\minus{}x}$. Then $ g'(x) \\equal{}\\minus{}f(x)e^{\\minus{}x}\\leq 0$ and $ g$ is decreasing on $ x\\geq 0$.\r\n\r\nAlso, $ \\lim_{x\\to\\infty}g(x) \\equal{} 0$, implying $ g(x)\\geq 0$ for all $ x\\geq 0$.\r\n\r\nTherefore, $ f^{(0)}(x)\\plus{}\\cdot\\plus{}f^{(n)}(x)\\; \\equal{}\\; g(x)e^{x}\\;\\geq\\; 0$.", "Solution_2": "Actually, mdk has posted this before (and quite recently I might add): http://www.mathlinks.ro/viewtopic.php?search_id=1657165616&t=159238\r\n\r\nThat makes me wonder: do you try to solve the problems you post or even bother reading the solutions posted?", "Solution_3": "[quote=\"perfect_radio\"]\nThat makes me wonder: do you try to solve the problems you post or even bother reading the solutions posted?[/quote]\r\nNo, this is the 3rd time I see mdk do this. And I wonder why he does this. Ask JBL he can probably come up with many more such instances.", "Solution_4": "Since sylow_theory mentioned me: mdk posted six questions yesterday. Of these, I have seen five appear on the forum. This one and one other have definitely been posted by mdk before. As far as I can tell, mdk solves only a small fraction of the problems s/he posts and reads only a small fraction of the threads s/he starts, and never (or hardly ever) posts on a thread that s/he didn't start." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Hi. Does anyone know the top team scores for last year?\r\n\r\nAlso, if you are:\r\n\r\n1. Mathcounts age\r\n2. Living in NJ\r\n3. Interested in being on a Mathcounts team\r\n4. Homeschooled\r\n\r\nPlease contact me ASAP.\r\n\r\nThanks.", "Solution_1": "I happen to know the top team score from last year.\r\n\r\nThomas R. Grover Middle School\r\n\r\n46,35,34,31 Team:20 Total: 56.50", "Solution_2": "Anybody know top individual scores for Bergen Chapter?", "Solution_3": "I don't remember exact numbers, but I know the #1 guy got 46/46, and there were largely a bunch of high 40s." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Hi , \r\nprove that if |ax\u00b2 + bx + c| <= 1 for every x belonging to [-1 1] we then have |cx\u00b2 + bx +a| <= 1 for every x belonging to [-1 1] .", "Solution_1": "Hi again , can someone just give some idea or a plausible way to proceed , thanks in advance. :roll:", "Solution_2": "Inequality is not true,be cause $ |\\minus{}2x^2\\plus{}1|\\leq 1 ,x\\in [\\minus{}1,1]$ but $ |x^2\\minus{}2|\\nleq 1,x\\in[\\minus{}1,1]$\r\n\r\n Correct is :$ |ax^2\\plus{}bx\\plus{}c|\\leq 1$ for any $ x\\in [\\minus{}1,1]$, then $ |cx^2\\plus{}bx\\plus{}a|\\leq2$ for any $ x\\in [\\minus{}1,1]$\r\n\r\n Indication: $ cx^2\\plus{}bx\\plus{}a\\equal{}c(x^2\\minus{}1)\\plus{}(a\\plus{}b\\plus{}c)\\frac {1\\plus{}x}{2}\\plus{}(a\\minus{}b\\plus{}c)\\frac{1\\minus{}x}{2}$", "Solution_3": "A similar problem was proposed in the 1996 Spanish National Olympiad:\r\n\r\nLet $ f(x) \\equal{} ax^2 \\plus{} bx \\plus{} c$, $ g(x) \\equal{} cx^2 \\plus{} bx \\plus{} a$. If $ |f( \\minus{} 1)|,|f(0)|,|f(1)|\\leq1$, then show that $ |f(x)|\\leq\\frac {5}{4}$, $ |g(x)|\\leq2$ for all $ x\\in[ \\minus{} 1,1]$.", "Solution_4": "[quote=\"Marius Mainea\"]Inequality is not true,be cause $ | \\minus{} 2x^2 \\plus{} 1|\\leq 1 ,x\\in [ \\minus{} 1,1]$ but $ |x^2 \\minus{} 2|\\nleq 1,x\\in[ \\minus{} 1,1]$\n\n Correct is :$ |ax^2 \\plus{} bx \\plus{} c|\\leq 1$ for any $ x\\in [ \\minus{} 1,1]$, then $ |cx^2 \\plus{} bx \\plus{} a|\\leq2$ for any $ x\\in [ \\minus{} 1,1]$\n\n Indication: $ cx^2 \\plus{} bx \\plus{} a \\equal{} c(x^2 \\minus{} 1) \\plus{} (a \\plus{} b \\plus{} c)\\frac {1 \\plus{} x}{2} \\plus{} (a \\minus{} b \\plus{} c)\\frac {1 \\minus{} x}{2}$[/quote]\r\n\r\nThanks vm Marius Mainea for this counterexample and for the pretty decomposition as well . i think that it's ok now . What a pity our school manuals still contain such errors . :maybe: \r\n\r\nthanks Daniel73 , i'll try to work on this one.", "Solution_5": "A sort of related problem from the 2008 Canadian Math Olympiad Repechage Problem Set:\r\n9. Suppose that a, b, c are real numbers with $ \\minus{} 1 \\le ax^2 \\plus{} bx \\plus{} c \\le 1$\r\nfor all real numbers x where $ \\minus{} 1 \\le x \\le 1$. Prove that $ \\minus{} 2 \\le a \\le 2$.", "Solution_6": "Similar problem from Japan, here.\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=145147&search_id=1371640373[/url]" } { "Tag": [ "integration" ], "Problem": "Suppose that a river is the set $ \\{(x,y),x\\in\\mathbb{R},0\\leq y\\leq l\\}$. The water flows in the middle of the river at the speed $ v_1$, in the shore at speed $ v_2\\leq v_1$ and the speed decreases linearly as the distance to the middle of the river increases. Swimmer is on the shore $ (y \\equal{} 0)$ and wants to swim to the point $ (0,l)$ using as little energy as possible. At calm water the swimmer swims at the speed $ v_3$. Where the swimmer must start his swimming trip and how long does the swimming trip takes?\r\n\r\nNote that the swimmer does not need to swim always to the same direction. He may turn his direction continuously.", "Solution_1": "Could you define what you mean by energy here, a bit better?", "Solution_2": "Energy means the energy that a swimmer have to use to manage his swimming trip. I don't know enough physiology to compute how much it takes energy to swim in a flow.", "Solution_3": "Or do you mean energy as in kinetic energy?", "Solution_4": "Okay, this is what I make of the problem. Assuming the river is flowing from right to left, let the swimmer start at the point $ (d,0)$ on the shore and swim forwards with speed $ v_3$. If he swims at that constant forward speed all the way, he will reach the opposite shore in a time of $ \\frac {l}{v_3}$. In the meantime, he lets the current carry him horizontally. He must therefore calculate his starting point in such a way that the current will carry him the exact horizontal distance $ d$ in the time $ \\frac {l}{v_3}$.\r\n\r\nThe speed $ v$ of the current at distance $ y$ from the shore is easy to calculate.\r\n\r\n$ v\\ = \\ \\{\\begin{array}{ll}v_2 + \\frac {2y}{l}\\,(v_1 - v_2) & 0\\leq y\\leq\\frac {l}{2} \\\\\r\n \\\\\r\nv_1 - \\frac {2}{l}\\,(y - \\frac {l}{2})(v_1 - v_2) & \\frac {l}{2}\\leq y\\leq l \\end{array}$\r\n\r\nBut since the forward speed is constant, $ y=v_3t$.\r\n\r\nHence\r\n\r\n$ d\\ = \\ \\int_0^\\frac {l}{2v_3}{[v_2 + \\frac {2v_3t}{l}\\,(v_1 - v_2)]}\\,\\mathrm{d}t + \\int_{\\frac {l}{2v_3}}^\\frac {l}{v_3}{[v_1 - (\\frac {2v_3t}{l} - 1)(v_1 - v_2)]}\\,\\mathrm{d}t$\r\n\r\nAnd if my calculation is correct, this should turn out to be $ d = \\frac {l}{2v_3}\\,(v_1 + v_2)$.\r\n\r\nI don\u2019t know if this is the most energy-efficient solution but I sure hope it makes sense. :roll:", "Solution_5": "Well, that would be good, but you didnt solve the problem.\r\nThe component of the velocity $ \\vec v_3$ perpendicular to the velocity of the river is not at all constant. since the direction of $ v_3$ can always change. \r\nMaybe this is the most energy efficient, way if that is constant, but I dont see a proof of that, which clearly is the lone difficulty of the problem.\r\nAgain, Puuhikki, could you specify by what you mean, energy, here?\r\nkinetic energy, or potential energy, or the work done by the swimmer...", "Solution_6": "[quote=\"yeahnigga\"]Again, Puuhikki, could you specify by what you mean, energy, here?\nkinetic energy, or potential energy, or the work done by the swimmer...[/quote]\r\nThe work done by the swimmer.", "Solution_7": "the swimer is in y=0 and x=?\r\nand y is refer (droit qui paralle avec la longueur de revier)? im sorry i can write all that with english", "Solution_8": "I guess that the swimmer should not turn when swimming." } { "Tag": [ "function", "LaTeX", "logarithms" ], "Problem": "Im having some trouble at the moment with a certain questions in log functions could some one explain them to me or help me out with them please. \r\n\r\nIf Log(a^2) = x and log(1/b) = y, find an expression in x and y for the following:\r\n\r\n1. log a/b\r\n2. log a^3\r\n3. log (ab)^1/2\r\n4. log (5ab)\r\n\r\nTheir all to the common base of 10 \r\nThank you for any help\r\nMagic Man", "Solution_1": "[quote=\"Magic Man\"]Im having some trouble at the moment with a certain questions in log functions could some one explain them to me or help me out with them please. \n\nIf Log(a^2) = x and log(1/b) = y, find an expression in x and y for the following:\n\n1. log a/b\n2. log a^3\n3. log (ab)^1/2\n4. log (5ab)\n\nTheir all to the common base of 10 \nThank you for any help\nMagic Man[/quote]\r\n\r\nFirst make them easier to deal with\r\n\r\n[hide=\"Like this\"]\n$ log(a^2)\\equal{}2log(a)\\equal{}x$ and $ log(\\frac{1}{b})\\equal{}log(1)\\minus{}log(b)\\equal{}\\minus{}log(b)\\equal{}y$\n[/hide]\n\nThen you can start to solve the problems\n\n[hide=\"1\"]\n$ log(\\frac{a}{b}) \\equal{} log(a)\\minus{}log(b)\\equal{}\\frac{x}{2}\\plus{}y$\n[/hide]\n\n[hide=\"2\"]\n$ log(a^3)\\equal{}3log(a) \\equal{} \\frac{3}{2}x$\n[/hide]\n\n[hide=\"3\"]\n$ log((ab)^{\\frac{1}{2}}) \\equal{} \\frac{1}{2}log(ab)\\equal{}\\frac{1}{2}(log(a)\\plus{}log(b))\\equal{}\\frac{1}{2}(\\frac{x}{2}\\minus{}y)$\n[/hide]\n\n[hide=\"4\"]\n$ log(5ab)\\equal{}log(5)\\plus{}log(a)\\plus{}log(b)\\equal{}log(5)\\plus{}\\frac{x}{2}\\minus{}y$\n[/hide]", "Solution_2": "i kinda understand what you are doing but where do you get the 2 from at the end each equation?", "Solution_3": "[quote=\"Magic Man\"]i kinda understand what you are doing but where do you get the 2 from at the end each equation?[/quote]\r\n\r\nDo you mean the $ \\frac{x}{2}$\r\n\r\nThat is because $ x\\equal{}log(a^2)$. We know this is the same as $ 2.log(a)$ and so everytime I need to reference $ log(a)$ I have to use $ \\frac{x}{2}$ (since this is $ \\frac{2log(a)}{2}$\r\n\r\nI hope thats what you meant?", "Solution_4": "Oh alright i see that u put the 2 on the bottom because so then the to two's cancel each other out okay! thank you very much for your help i get it now", "Solution_5": "No problems, I think most people have trouble with logs when they first see them (well, not so much with the rules, but anyway)", "Solution_6": "One last question Number three isent acctually written like that i acctually rearanged the formula because i dont have square root on ma computer \r\nthe original question was:\r\nLog square root of (ab)\r\nwill it give me the same answer as u gave me for 3", "Solution_7": "Yes, thats how powers work $ x^{\\frac{1}{2}=\\pm \\sqrt{x}}$ remember, the square root symbol only means the positive root. Also, you can use LaTeX to put the square roots in if you need to[/u]", "Solution_8": "For future reference, [code]\\log[/code] gives $ \\log$, which looks better than $ log$. Text is awful in latex." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ x,y,z \\geq 0$ such that $ \\sqrt {xy} \\plus{} \\sqrt {yz} \\plus{} \\sqrt {zx} \\plus{} \\sqrt {xyz} \\equal{} 4$\r\nProve that 1)$ \\sum \\sqrt x \\geq \\sqrt {xy} \\plus{} \\sqrt {yz} \\plus{} \\sqrt {zx}$\r\n 2)$ x^2 \\plus{} y^2 \\plus{} z^2\\geq \\sum \\sqrt [4] {x^3y ^3}$", "Solution_1": "1) Let $ \\sqrt {x} \\equal{} \\frac {2a}{b \\plus{} c}$ , $ \\sqrt {y} \\equal{} \\frac {2b}{c \\plus{} a}$ , $ \\sqrt {z} \\equal{} \\frac {2c}{a \\plus{} b}$\r\n\r\n and 1) reduces to\r\n\\[ \\sum {\\frac {2a}{b \\plus{} c}}\\ge \\sum {\\frac {4ab}{(b \\plus{} c)(c \\plus{} a)}}\r\n\\]\r\nor\r\n\\[ \\sum {a(a \\plus{} b)(a \\plus{} c)}\\ge 2\\sum {ab(a \\plus{} b)}\r\n\\]\r\nor\r\n\\[ \\sum_{cyc}{a^3} \\plus{} 3abc\\ge \\sum_{sym}{a^2b}\r\n\\]\r\nwich is Schur Inequality.\r\n\r\n 2) \r\n$ LHS\\ge \\frac {1}{3}(\\sqrt [4]{x^6} \\plus{} \\sqrt [4]{y^6} \\plus{} \\sqrt [4]{z^6})(\\sqrt [4]{x^2} \\plus{} \\sqrt [4]{y^2} \\plus{} \\sqrt [4]{z^2})\\ge \\frac {1}{3}(\\sum {\\sqrt [4]{x^3y^3}})(\\sum {\\frac {2a}{b \\plus{} c}})\\ge RHS$" } { "Tag": [ "number theory", "greatest common divisor", "least common multiple" ], "Problem": "Determine all pairs $ (a,b)$ in positive integers which satisfy next equation:\r\n\r\n$ LCM(a,b)\\plus{}GCD(a,b)\\plus{}a\\plus{}b\\equal{}ab$, $ a\\ge b$.\r\n\r\nwhere $ LCM$ means the Least Common Multiple,and $ GCD$ does the Greatest Common Divisor of $ a,b$", "Solution_1": "Let $ (a,b) \\equal{} g, a \\equal{} gA,b \\equal{} gB$. Then $ [a,b] \\equal{} gAB$ and our equation becomes\r\n\\[ gAB \\plus{} g \\plus{} gA \\plus{} gB \\equal{} g^2AB\r\n\\]\r\nwhich is equivalent to\r\n\\[ (A \\plus{} 1)(B \\plus{} 1) \\equal{} gAB\r\n\\]\r\nHence $ AB\\mid (A \\plus{} 1)(B \\plus{} 1)$. But since $ (A,A \\plus{} 1) \\equal{} 1$ we conclude that $ A\\mid B \\plus{} 1$ and $ B\\mid A \\plus{} 1$. Let $ B \\equal{} kA \\minus{} 1$. Then $ kA \\minus{} 1\\mid A \\plus{} 1$ implies $ kA \\minus{} 1\\le A \\plus{} 1\\Leftrightarrow (k \\minus{} 1)A\\le 2$. Therefore $ (k,A) \\equal{} (3,1),(2,2),(2,1),(1,2)$. Hence $ (k,A,B)\\equal{}(1,2,1),(3,1,2),(2,2,3),(2,1,1)$. Thus we get the pairs $ (a,b) \\equal{} (3,6),(4,6),(4,4)$.\r\n\r\nRemark: I worked with $ A\\mid B \\plus{} 1$ so I got the pairs where $ a\\le b$. For $ a\\ge b$, we can simply reverse each pair." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "For any positive real numbers $a,\\ b$ and $c$.\r\n\\[ \\frac{c}{a(b+c)}+\\frac{a}{b(c+a)}+\\frac{b}{c(a+b)} \\ge \\frac{9}{2(a+b+c)} \\]", "Solution_1": "Cauchy + AM-GM?\r\n\r\n$((b+c)+(c+a)+(a+b))\\left( \\frac{c}{a(b+c)}+\\frac{a}{b(c+a)}+\\frac{b}{c(a+b)} \\right ) \\geq \\left(\\sqrt{\\frac ca} + \\sqrt{\\frac ab} + \\sqrt{\\frac bc} \\right)^2 \\geq 9$" } { "Tag": [], "Problem": "I don't know how to solve this without a computer program...\r\n\r\nI'm horrible at manipulating sums - maybe someone can give me a hint?\r\n\r\nWhat is $\\displaystyle\\sum_{k=1}^{360} \\frac{1}{k\\sqrt{k+1} + (k+1)\\sqrt{k}}$ in lowest terms?", "Solution_1": "it telescopes. the denominator can be written as $\\sqrt{k}\\sqrt{k+1}(\\sqrt{k}+\\sqrt{k+1})$. then go from there.\r\n\r\nhow come that contest doesnt seem to be too popular? geehoon told me about it, but we couldn't find anything on it. i'd think that since anyone can take it a lot of people on this site would have known about it.", "Solution_2": "Ah, thank you.\r\n\r\nI figured it out now.\r\n\r\nIt's [hide]$\\frac{18}{19}$[/hide].\r\n\r\nIt's so nice how all the terms cancel...", "Solution_3": "Hmm... I still don't get it, can someone please post a full solution.", "Solution_4": "[hide]\nAs suggested, the summand simplifies to 1/ (:sqrt: k :sqrt: (k+1) ( :sqrt: k + :sqrt: (k+1)). If you rationalize the denominator, it simplifies further to 1/ :sqrt: k - 1/ :sqrt: (k+1) which telescopes.[/hide]", "Solution_5": "Yeah - to elaborate:\r\n\r\n\r\nYou have: $\\frac{1}{\\sqrt{1}(\\sqrt{2})(\\sqrt{1}+\\sqrt{2})} + \\frac{1}{\\sqrt{2}(\\sqrt{3})(\\sqrt{2}+\\sqrt{3})} + \\ldots + \\frac{1}{\\sqrt{360}(\\sqrt{361})(\\sqrt{360}+\\sqrt{361})}$.\r\n\r\nThen rationalize the denominator and simplify.\r\n\r\n$ -1\\displaystyle\\left(\\frac{\\sqrt{1}-\\sqrt{2}}{\\sqrt{1}(\\sqrt{2})} + \\frac{\\sqrt{2}-\\sqrt{3}}{\\sqrt{2}(\\sqrt{3})} + \\ldots + \\frac{\\sqrt{360}-\\sqrt{361}}{\\sqrt{360}(\\sqrt{361})}\\right)\\\\ = -1\\displaystyle\\left(\\frac{1}{\\sqrt{2}} - \\frac{1}{\\sqrt{1}} + \\frac{1}{\\sqrt{3}} - \\frac{1}{\\sqrt{2}} + \\ldots + \\frac{1}{\\sqrt{360}} - \\frac{1}{\\sqrt{359}} + \\frac{1}{\\sqrt{361}} - \\frac{1}{\\sqrt{360}}\\right)\\\\ = 1 - \\frac{1}{19} = \\frac{18}{19}$", "Solution_6": "yeah, i got the same thing.\r\n18/19" } { "Tag": [ "number theory", "prime factorization" ], "Problem": "Each of the integers 1, 2, 3, ..., 16 is written on a separate slip of paper and these slips are placed in a pile. Jillian will randomly draw slips from the pile without replacement and will continue drawing until two of the numbers she has drawn from the pile have a product that is a perfect square. What is the maximum number of slips that Jillian can draw without obtaining a product that is a perfect square?\r\n\r\n\r\nBeastly hard problem.", "Solution_1": "Well just think... what makes a number square?\r\n\r\n9=3^2\r\n16=2^2 x 2^2\r\n36=2^2 x 3^2\r\n\r\nNow if we split all the numbers 1-16 to their prime factorization, which numbers absolutely cannot be together? Which number \"pose a threat\" to other numbers? If you think about it logically, I think the answer is 11?\r\n\r\nDarn, I cant explain it well. Would someone else like to try?", "Solution_2": "The numbers with exactly 1 prime factor, exactly 2 [u]different[/u] prime factors, and 1 are the only numbers that can be drawn. This is because if you multiply 1 to anything, it'll be the same, when you have a number with exactly one prime factor, it is prime and thus isn't a square, and when you multiply a number with exactly 1 prime factor with a number that has exactly 2, it still cannot be a square. For example, 3*15=45, which isn't a square.\r\n\r\nThe primes between 1 to 16 are 2,3,5,7,11,13; a total of 6 numbers. \r\n\r\nThe numbers with exactly 2 distinct prime factors between 1-16 are 2*3,2*5,2*7,3*5; for a total of 4 numbers. Any other product of the previous primes we showed is greater than 16.\r\n\r\nLastly, we have to include 1. (Which is, of course, 1 number). Thus, the greatest numbers of slips that can be drawn is $ 6 \\plus{} 4 \\plus{} 1 \\equal{} \\boxed{11}$.\r\n\r\nI failed this one at state because I misread the question and thought it was the numbers from 1-12. :( :wallbash_red: :wallbash:", "Solution_3": "so you are saying after you pick the primes, one, and the products of the primes less than 16, the rest of the numbers products are going to be perfect squares?\r\n\r\n\r\nI am a noob.\r\n\r\nwait no, that is not teh problem.\r\nyou want to find out how many slips she has to pick so that none of the pairs of products out of the ones that she picked are perfect squares?" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Two players play the following game: first player writes down one digit, after his move second player write one digit to the left or right to the previously written digit, after his move first player write one digit to the left or write to previously written number, etc. Prove that the first player can play in that way such that after each of the moves of the second player, the obtained number is not the exact square.", "Solution_1": "I think you may have confused the word \"write\" with the word \"right\"\r\nPlease edit your post :) \r\n\r\nDaniel" } { "Tag": [ "ratio", "geometry", "perimeter" ], "Problem": "The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:\r\n\r\n$ \\textbf{(A)}\\ 1: 2 \\qquad\\textbf{(B)}\\ 1: 3 \\qquad\\textbf{(C)}\\ 1: \\sqrt {3} \\qquad\\textbf{(D)}\\ \\sqrt {3}: 2 \\qquad\\textbf{(E)}\\ 2: 3$", "Solution_1": "[hide=\"Click for solution\"]\nLet $ P_1$ and $ P_2$ be the perimeters of the smaller and larger triangles, respectively. Let the radius of the circle be $ r$ and the side length of the large triangle be $ s$. Then $ r\\equal{}\\frac{s\\sqrt{3}}{2}$ and $ s\\equal{}\\frac{2r}{\\sqrt{3}}$, so $ P_1\\equal{}2r\\sqrt{3}$ and $ P_2\\equal{}3r\\sqrt{3}$, so the desired ratio is $ 2: 3$, or $ \\boxed{\\textbf{(E)}}$.\n[/hide]", "Solution_2": "It may be faster to just note that the ratio of any two corresponding lengths of the two triangles must be equal to the ratio of the altitudes, which by the properties of medians and centroids is always $ 2: 3$." } { "Tag": [ "search" ], "Problem": "guys, i dont know how you ppl remember all the reagents and their uses but i think its a real pain in the ass. besides there are too many reagents and there are no proper sources to search for them. so why dont we start posting all the reagents we know( at least the important ones) over here. we could organise it later. i know it is boring to type all the reagents but it would be useful for everyone especially for those trying for jee n the olympiads", "Solution_1": "sorry posted it twice by mistake. (at least will get attention)" } { "Tag": [ "function" ], "Problem": "Hi,\r\n\r\nI read over the parts about function transformation in the Intro to Algebra book, but I am still not completely sure why the transformation of something like $ f(x \\plus{} 3)$ moves the graph of $ f(x)$ three to the left, and something like $ f(x \\minus{} 3)$ moves the graph of $ f(x)$ three to the right. \r\n\r\nThank you for the help.", "Solution_1": "Ahh...never mind. I seem to found a subtle explanation for why this works convincing myself. Other perspectives are certainly welcomed. \r\n\r\nHere's how I am trying to picture it: \r\n\r\nLet there be a function such that $ f(x) \\equal{} x\\plus{}1.$ Let us now transform this function as follows: \r\n$ f(x\\plus{}3) \\equal{} (x\\plus{}3)\\plus{}1.$ A horizontal shift of 3 units in this case corresponds to the decrease of the x value by three. We see, that in order for the first function to acheive $ f(0),$ we merely have to have $ x\\equal{}0.$ However, in the second function, $ x\\equal{}\\minus{}3$ in order for $ f(x\\plus{}3) \\equal{} f(0),$ therefore this corresponds to the x-value being three lower than in the $ f(x).$ Thus, it is a shift 3 units to the left. \r\n\r\nThe book perhaps had the same/similar explanation that I failed to understand? :maybe:", "Solution_2": "The way I think about it when you have a function, let's say $ f(x)\\equal{}x$ and $ f(x)\\equal{}x\\plus{}1$, in order for $ y\\equal{}0$, you must have $ x\\equal{}0$ for the first and $ x\\equal{}\\minus{}1$ for the second. When you would plot the points, the second one would be to the left of the first. Just think about like that.\r\n\r\n$ f(x)\\equal{}x\\plus{}a$\r\n$ x$ is translated $ \\minus{}a$ units\r\nIf $ \\minus{}a>0$, Then $ a$ moves right.\r\nIf $ \\minus{}a<0$, Then $ a$ moves left.\r\n\r\nOr if you are adding a quantity to $ x$, then your graph will be translated to the left by that quantity.\r\nThe opposite holds true when you subtract.\r\n\r\nHope this helps, even though you basically said the same thing in the last post.", "Solution_3": "Also we can take a look at the concept of roots. $ f(x)\\equal{}x^2$ is 'centered' (if you will) at the origin, but $ f(x\\minus{}3)$ is centered at $ x\\equal{}3$. Applying this to a more complicated function, we have a set of roots at $ x\\minus{}r_i$. Rather than showing the function, observe the resulting roots when we apply $ f(x\\minus{}s)\\equal{}x\\minus{}r_i\\minus{}s\\Rightarrow x\\equal{}r_i\\plus{}s$.\r\n\r\nClearly, the functions roots shift in the direction of $ \\frac{s}{|s|}$, as does the function itself.", "Solution_4": "What's $ r_i$?", "Solution_5": "I would assume the $ i$th root of $ f(x).$", "Solution_6": "As RoFlLoLcOpT said, $ f(x)$ has zeros at $ x\\equal{}r_i$" } { "Tag": [ "algebra", "function", "domain", "logarithms", "geometry", "search", "complex analysis" ], "Problem": "Could anyone help me with these problems?\r\n\r\n1. Map 1-1, conformally the upper half of the unit disk onto the upper half-plane, $ \\{z\\in\\mathbb{C}|Im(z)>0\\}$\r\n\r\n2. Map 1-1, conformally the circular sector $ 0 0$ prove that\r\n\\[ \\frac {a}{(b \\plus{} c)^4} \\plus{} \\frac {b}{(c \\plus{} a)^4} \\plus{} \\frac {c}{(a \\plus{} b)^4}\\ge \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}\r\n\\]", "Solution_3": "[quote=\"nayel\"]Interesting. Very nice proof Alex! :) I've tried to create stronger inequalities of the same type but your proof seems to kill them all. :( :D Now finally I have the following. Have fun! :wink: \n\nFor $ a,b,c > 0$ prove that\n\\[ \\frac {a}{(b \\plus{} c)^4} \\plus{} \\frac {b}{(c \\plus{} a)^4} \\plus{} \\frac {c}{(a \\plus{} b)^4}\\ge \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}\n\\]\n[/quote]\r\nBy Holder inequality, we have:\r\n$ (a(b\\plus{}c)\\plus{}b(c\\plus{}a)\\plus{}c(a\\plus{}b))^4(\\frac {a}{(b \\plus{} c)^4} \\plus{} \\frac {b}{(c \\plus{} a)^4} \\plus{} \\frac {c}{(a \\plus{} b)^4}) \\ge (a\\plus{}b\\plus{}c)^5$\r\nOr:\r\n$ 2(ab\\plus{}bc\\plus{}ca)(\\frac {a}{(b \\plus{} c)^4} \\plus{} \\frac {b}{(c \\plus{} a)^4} \\plus{} \\frac {c}{(a \\plus{} b)^4}) \\ge \\frac{(a\\plus{}b\\plus{}c)^5}{2^3(ab\\plus{}bc\\plus{}ca)^3}$\r\nWe will prove that:\r\n$ (a\\plus{}b\\plus{}c)^3(a\\plus{}b)(b\\plus{}c)(c\\plus{}a) \\ge 2^3(ab\\plus{}bc\\plus{}ca)$\r\nWe also have (a famous problem!):\r\n$ \\sqrt[3]{\\frac{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)}{8}} \\ge \\sqrt{\\frac{ab\\plus{}bc\\plus{}ca}{3}}$\r\nAnd: $ a\\plus{}b\\plus{}c \\ge \\sqrt{3(ab\\plus{}bc\\plus{}ca)}$\r\nand problem is solved!", "Solution_4": "I made a different solution .\r\n\r\nBy the rearrengement inequality \r\n\r\n$ 3(\\sum\\frac{a}{(b\\plus{}c)^4})\\geq (\\sum\\frac{a}{b\\plus{}c})(\\frac{1}{(b\\plus{}c)^3})$ \r\n\r\nNow use Nesbitt inequality for the first term and AM-GM for the second to obtain the result .", "Solution_5": "Yes silouan that was my solution too. And note that it also solves my second proposed problem in this thread. :) \r\n\r\n[quote=\"NguyenDungTN\"]We also have (a famous problem!):\n$ \\sqrt [3]{\\frac {(a + b)(b + c)(c + a)}{8}} \\ge \\sqrt {\\frac {ab + bc + ca}{3}}$\n[/quote]\r\n\r\nWell, I never saw this before and thought I'd give it a try:\r\n[hide]Let $ x=a+b,y=b+c,z=c+a$. Then notice that \\[ 4(ab+bc+ca)=2(xy+yz+zx)-(x^2+y^2+z^2)\\]\nSo we need to show that \\[ \\sqrt[3]{xyz}\\ge\\sqrt{\\frac{2(xy+yz+zx)-(x^2+y^2+z^2)}{3}}\\]\nWhich is equivalent to \n\\[ x^2+y^2+z^2+3(xyz)^{\\frac 23}\\ge 2(xy+yz+zx)\\]\nThis follows from Schur and AM-GM:\n\\begin{eqnarray*}(x^{\\frac 23})^3+(y^{\\frac 23})^3+(z^{\\frac 23})^3+3(xyz)^{\\frac 23} &\\ge & \\sum_{\\text{sym}}\\left(\\left(x^{\\frac 23}\\right)^2y^{\\frac 23}+x^{\\frac 23}\\left(y^{\\frac 23}\\right)^2\\right)\\\\\\ \n&\\ge& \\sum_{\\text{sym}}2xy \\end{eqnarray*}\n[/hide]\r\nAre there other solutions to this one? Just wondering.", "Solution_6": "http://www.mathlinks.ro/Forum/viewtopic.php?t=27766" } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Given any finite group $ G$ show there exists fields $ F\\leq E$ with $ E$ normal over $ F$ such that $ \\mbox{Gal}(E/F)\\simeq G$.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=155634", "Solution_2": "What about if $ F\\equal{}\\mathbb{Q}$, it is my understanding that this an unsolved problem in mathematics?", "Solution_3": "When F=Q, it seems to be easy only when G is abelian.", "Solution_4": "http://en.wikipedia.org/wiki/Inverse_Galois_problem", "Solution_5": "[quote=\"Abel\"]When F=Q, it seems to be easy only when G is abelian.[/quote]\r\nThat the Hilbert's 12th Problem. [url=http://en.wikipedia.org/wiki/Kronecker%E2%80%93Weber_theorem]Here[/url]. (In same form or another).", "Solution_6": "But isn't Hilbert's problem for fields other than the rationals? I meant the simple case when F=Q (i.e. the rationals) and G is abelian (and finite)." } { "Tag": [ "quadratics", "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Show that if $ x$ is not divisible by $ 3$ then $ 4x^2+3$ has at least one prime factor of the form $ 12n+7$. Deduce that there are infinitely many primes of this sort.\r\n\r\n\r\n[hide=\"What I have so far:\"]\nLet $ p = 12n+7$.\n\nSince $ p \\neq 3$ and $ 3$ and $ p$ are both odd, the quadratic reciprocity law gives \n$ \\begin {eqnarray*}\\left(\\frac{-3}{p}\\right) &=& \\left(\\frac{-1}{p}\\right) \\left(\\frac{3}{p}\\right) \\\\\n&=& (-1)^{\\frac{p-1}{2}}\\left(\\frac{12n+7}{3}\\right)(-1)^{\\frac{3-1}{2}*\\frac{p-1}{2}} \\\\\n&=& (-1)^{6n+3}\\left(\\frac{1}{3}\\right)(-1)^{6n+3} \\\\\n&=& (-1)(1)(-1) \\\\\n&=& 1$\n\nHence, $ y^2 \\equiv -3 \\pmod p$ has a solution.\n\nFrom here, how do I get $ 4x^2 \\equiv -3 \\pmod p$ also has a solution? How would I know letting $ y = 2x$ where $ x$ is not divisible $ 3$ will guarantee a solution? Thanks.\n\nFor the infinite prime part of the problem, I guess you can just assume finitely many primes of the form $ 12n+7$, denoted by $ p_1$, $ p_2$, ... , $ p_r$\n\nConsider the number $ 4(p_1p_2...p_r)^2 + 3$. From above, there is a prime factor of the form $ 12n+7$ that divides this number, but is not equal to any of the $ p_1, p_2, ... , p_r$. Hence, the contradiction shows infinitely many primes of the form $ 12n+7$.[/hide]", "Solution_1": "[hide]You want to go the other way instead; assume $ p$ is a prime dividing $ 4x^2 \\plus{} 3$, then $ \\minus{} 3$ is a square mod p. Then\n\n$ \\left(\\frac { \\minus{} 3}{p}\\right) \\equal{} \\left(\\frac { \\minus{} 1}{p}\\right)\\left(\\frac {3}{p}\\right) \\equal{} ( \\minus{} 1)^{\\frac {p \\minus{} 1}{2}}\\left(\\frac {p}{3}\\right)( \\minus{} 1)^{\\frac {p \\minus{} 1}{2}}( \\minus{} 1)^{\\frac {3 \\minus{} 1}{2}} \\equal{} \\left(\\frac {p}{3}\\right)$.\n\nSince $ \\left(\\frac {p}{3}\\right) \\equal{} \\left(\\frac { \\minus{} 3}{p}\\right) \\equal{} 1$, p must be a square mod 3, so $ p\\equiv 1\\mod 3$.\n\nYou also know that there must be at least one prime dividing $ 4x^2 \\plus{} 3$ that is $ 3\\mod 4$ (otherwise $ 4x^2 \\plus{} 3$ would be $ 1\\mod 4$). Take this prime $ p$, and since $ p\\equiv 1\\mod 3$ and $ p\\equiv 3\\mod 4$, you know that $ p\\equiv 7\\mod 12$.[/hide]" } { "Tag": [ "topology", "superior algebra", "superior algebra solved" ], "Problem": "HELLO can yuo help me? \r\nEverey opensubset of a locally path space is itself locally path connected.", "Solution_1": "Being a locally path connect space is a local property which means that it transmits to any open subset.\r\n\r\nThe general statement is:\r\nLet $L$ be a property of topological spaces such that a topological space $X$ has $L$ iff it has a base formed of open subsets each of which has $L$.\r\nLet $U$ be an open subset of $X$. Then $U$ has $L$.\r\n\r\nProof: The open subsets of $U$ which are also in the base for $X$ form a base for $U$. The conclusion follows." } { "Tag": [ "AMC", "AIME", "probability" ], "Problem": "1. A man leaves for work at a random time b/w 7:00AM and 8:00Am. His paper arrives at his house at a random time b/w 6:30 AM and 7:30 AM. What is the probability that he has his newspaper before he leaves for work? \r\n\r\nI draw 2 axes, one horizontal and one vertical, and label the horizontal one 6:30, 7:00, and 7:30 for the newspaper, and the vertical one 7:00, 7:30, and 8:00. What I'm confused about is what to shade in for the region where he gets the newspaper before he leaves for work. Could anyone please explain this one to me? \r\n\r\nThanks.", "Solution_1": "You want the graph $x < y$ (since the newspaper is supposed to arrive before he leaves) except remember the origin is a little shifted.\r\n\r\nYou want to shade above $x < y$ since as long as the man leaves at the same time or later (higher y) than the newspaper arrives, he will get it before he's gone.\r\n\r\nIf you're even more clever, another nice way to look at this problem is as follows:\r\n\r\nThe probability of the man leaving between 7 and 7:30 is 1/2\r\nThe probability of the paper arriving between 7 and 7:30 is 1/2\r\n\r\nTogether there is 1/4 chance this will happen.\r\n\r\nIf this happens, there is a 1/2 chance the paper will after before the man leaves.\r\n\r\nHence there's a 1/8 chance the paper will arrive after the man leaves, so a 7/8 chance the man will get his paper." } { "Tag": [ "function", "combinatorics proposed", "combinatorics" ], "Problem": "Suppose we have some proteins that each protein is a sequence of 7 \"AMINO-ACIDS\" $A,\\ B,\\ C,\\ H,\\ F,\\ N$. For example $AFHNNNHAFFC$ is a protein. There are some steps that in each step an amino-acid will change to another one. For example with the step $NA\\rightarrow N$ the protein $BANANA$ will cahnge to $BANNA$(\"in Persian means workman\"). We have a set of allowed steps that each protein can change with these steps. For example with the \r\nset of steps:\r\n$\\\\ 1)\\ AA\\longrightarrow A\\\\ 2)\\ AB\\longrightarrow BA\\\\ 3)\\ A\\longrightarrow \\mbox{null}$\r\nProtein $ABBAABA$ will change like this:\r\n$\\\\ ABB\\underline{AA}BA\\\\ \\underline{AB}BABA\\\\ B\\underline{AB}ABA\\\\ BB\\underline{AA}BA\\\\ BB\\underline{AB}A\\\\ BBB\\underline{AA}\\\\ BBB\\underline{A}\\\\ BBB$\r\nYou see after finite steps this protein will finish it steps.\r\nSet of allowed steps that for them there exist a protein that may have infinitely many steps is dangerous. Which of the following allowed sets are dangerous?\r\na) $NO\\longrightarrow OONN$\r\nb) $\\left\\{\\begin{array}{c}HHCC\\longrightarrow HCCH\\\\ CC\\longrightarrow CH\\end{array}\\right.$\r\nc) Design a set of allowed steps that change $\\underbrace{AA\\dots A}_{n}\\longrightarrow\\underbrace{BB\\dots B}_{2^{n}}$\r\nd) Design a set of allowed steps that change $\\underbrace{A\\dots A}_{n}\\underbrace{B\\dots B}_{m}\\longrightarrow\\underbrace{CC\\dots C}_{mn}$\r\nYou see from $c$ and $d$ that we acn calculate the functions $F(n)=2^{n}$ and $G(M,N)=mn$ with these steps. Find some other calculatable functions with these steps. (It has some extra mark.)", "Solution_1": "[hide=a)] This set is dangerous. Consider the word $NOO$, we have\n$$ NOO \\longrightarrow OONNO \\longrightarrow OONOONN.$$\nSince $NOO$ appears again in the right word, we can repeat this process ad infinitum.[/hide]\n[hide=b)] Not dangerous. Suppose that we are given a word $w$ in $C$ and $H$ (if there are other letters we could view them as space and apply our argument to each word in $C$ and $H$). Since$w$ contains a finite number of $C$ we can use operation $1$ only finitely many times. Thus, if $w$ had infinitely many steps, there would exist a word $w'$ to which operation $2$ is applied infinitely many times. However, this operation reduces the number of pairs of letters $(X,Y)$, where $X=H \\, Y=C$ and $X$ appears before $Y.$ Since there are only finitely many such pairs in $w'$, we get a contradiction.\n[/hide]\nDoes anybody have an idea for c) and d) ?" } { "Tag": [ "summer program", "MathPath", "function", "vector", "inequalities" ], "Problem": "Source: MathPath\r\n\r\nProve that for all a, b, c, d, e, and f (very exciting variable names, huh?), that\r\n\r\n$\\sqrt{a^2+b^2}+\\sqrt{c^2+d^2}+\\sqrt{e^2+f^2}\\ge\\sqrt{a^2+c^2+e^2}+\\sqrt{b^2+d^2+f^2}$", "Solution_1": "Is there any other way than squaring all of it?", "Solution_2": "Yes! There is a deceptively simple solution to it all.\r\n\r\nHint: Think of a certain Greek guy.", "Solution_3": "[quote=\"mathnerd314\"]Yes! There is a deceptively simple solution to it all.\n\nHint: Think of a certain Greek guy.[/quote]\r\nI know, Im thinking pythagorous, but Im also trying to learn basic music theory from google. I will do it later. :(", "Solution_4": "Well ,hmm , this one is advanced and its really over kill but :D :\r\nSee this link for any defination you want :\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=14975\r\nWLOG Assume that $b \\le a,c,d,e,f,$\r\nFunction $f(x)=\\sqrt{x}$ is concave for $x \\le 0$ . \r\nwe should prove $f(a^2+b^2)+f(c^2+d^2)+f(e^2+f^2) \\ge f(a^2+c^2+e^2)+f(b^2+d^2+f^2)$\r\nso according to Karamata ineqaulity we should prove :\r\n\r\n $(a^2+b^2,c^2+d^2,e^2+f^2) \\prec (a^2+c^2+e^2,b^2+d^2+f^2,0)$ \r\n\r\nwhich is equal to $b^2 \\le c^2+e^2$ which is true .Problem solved.", "Solution_5": "Sorry for reviving this topic, but there is a much easier way than the incomprehensible calculations that lomos_lupin just demonstrated ;). I forgot to post this when I should have (i.e. a week ago) :blush:\r\n\r\n[hide=\"hint\"]\n[hide=\"are you sure?\"]\nThink of the variables as sides of right triangles. Two Greek guys really, both dealing with the same subject.\n[/hide]\n[/hide]", "Solution_6": "think of the vectors ", "Solution_7": "[quote=\"mathnerd314\"]there is a much easier way than the[b] incomprehensible calculations[/b] that lomos_lupin [color=red]just[/color][b] demonstrated[/b][/quote]\r\nWhich part is incomprehensible ?\r\nI shall provide any needed explanations,if you want (Yes i agree its a not easy ,but the links i gave can provide precise details)\r\nI have already mentioned that this technic is not appropriate for this problem but ,I posted it to introduce this strong tool of solving ineqaulities .\r\nThis techin help you to solve the stronger problems like :\r\n\r\n$\\sqrt[2n]{a^2+b^2}+\\sqrt[2n]{c^2+d^2}+\\sqrt[2n]{e^2+f^2}\\ge\\sqrt[2n]{a^2+c^2+e^2}+\\sqrt[2n]{b^2+d^2+f^2} ,n \\in \\mathbb{Z}$\r\n\r\nand about [color=red] just demostrated[/color]? \r\nThis is not demonstration,there is no need of demonstartion,I really wonder why do you think in such a way?\r\n\r\n\r\nand about this ( Two Greek guys) , well clearly \r\n\r\n$\\sqrt{a^2+b^2}=\\text{Chord of a triangle ABC such that C=90,BC=a,CA=b}$\r\n\r\nand similar for the two other ones,but by this way you will reach this :\r\n\r\n$\\sqrt{a^2+b^2}+\\sqrt{c^2+d^2}+\\sqrt{e^2+f^2}\\ge\\sqrt{(a+c+e)^2+(b+d+f)^2} $\r\n\r\ni wonder how did you work with the Right side,Would you provide a complete formal solution?", "Solution_8": "[hide]$a^2+b^2=c^2$ the right side of the equation being $a^2+b^2$.squaring both terms makes it $a^2+b^2+c^2+d^2+e^2+f^2$ on the right side of the equation. when u square the whole \"term\" on the left side of the equation, there are middle terms when you square multiple terms in one \"term\" so that the left side HAS to be larger than the right side. i think..[/hide]\r\ndo all proofs have to be two columned? if theyre not do you just write out your explanation step by step?", "Solution_9": "[quote=\"lomos_lupin\"][quote=\"mathnerd314\"]there is a much easier way than the[b] incomprehensible calculations[/b] that lomos_lupin [color=red]just[/color][b] demonstrated[/b][/quote]\nWhich part is incomprehensible?\n[/quote]\r\n\r\nNot really incomprehensible. I just couldn't think of another word at the moment. Hard to understand by the average or accelerated junior high student? Would that be more descriptive?\r\n\r\nWhat's the $\\prec$ symbol? Is it something about being concave? And what do you mean by concave? From which side?\r\n\r\nSorry, and I didn't see the link.", "Solution_10": "MAthnerd,i really cant understand the pythagoras hint of yours.How can you just take $a^2+b^2=c^2$ or something like that?That'll just be an assumption...is it appropriate to make suvh an assumption?", "Solution_11": "a, b, c, d, e, and f are sides of three right triangles. a, c, and e are collinear, and b, d, and f are collinear and perpendicular to the three others. Remember that the shortest path between two points is a straight line. Go from there.", "Solution_12": "[quote]What's the $\\prec$ symbol? Is it something about being concave?.[/quote]\r\n\r\nA $\\prec$ B means that A precedes B", "Solution_13": "Ah. I get the \\prec code now. But what does precede mean in this context?", "Solution_14": "[quote=\"mathnerd314\"]a, b, c, d, e, and f are sides of three right triangles. a, c, and e are collinear, and b, d, and f are collinear and perpendicular to the three others. Remember that the shortest path between two points is a straight line. Go from there.[/quote]\r\n\r\nHow come this condition was not mentioned in the question?", "Solution_15": "It's a creative way of thinking of the problem; it doesn't have to be included in the actual problem. I dismissed it at first but i guess it's of use.", "Solution_16": "Yeah so the answer is (with your Pythagorean hint):\r\n\r\nIf you consider the vectors (a,b), (c,d), (e,f)\r\n\r\nThe sum of the vectors are (a+c+e,b+d+f).\r\n\r\nThe distance from this to the origin is the right side, and equality is reached when the vectors are collinear, and in the same direction.", "Solution_17": "[quote=\"mathnerd314\"]Sorry for reviving this topic, but there is a much easier way than the incomprehensible calculations that lomos_lupin just demonstrated ;). I forgot to post this when I should have (i.e. a week ago) :blush:\n\n[hide=\"hint\"]\n[hide=\"are you sure?\"]\nThink of the variables as sides of right triangles. Two Greek guys really, both dealing with the same subject.\n[/hide]\n[/hide][/quote]\nFirst, Proof by Lomos_lupin was excellent , certainly ,IMO, anything but incomprhensible but more to the point, this and few later posts ,perhaps are forgetting that the RHS of the problem given above is \n\n$\\sqrt{a^2+c^2+e^2}+\\sqrt{b^2+d^2+f^2}$ and is NOT:\n\n$\\sqrt{(a+c+e)^2+(b+d+f)^2}$ as pointed out before. \n[quote=\"lomos_lupin\"]\n\n[b]and similar for the two other ones,but by this way [color=red][size=117]you will reach this[/size][/color][/b] :\n\n$\\sqrt{a^2+b^2}+\\sqrt{c^2+d^2}+\\sqrt{e^2+f^2}\\ge\\sqrt{(a+c+e)^2+(b+d+f)^2} $\n\ni wonder how did you work with the Right side,Would you provide a complete formal solution?[/quote]\r\n\r\nOne still has some work to do, even after using something like \"Remember that the shortest path between two points is a straight line\" ..\r\n\r\nRegards.", "Solution_18": "This kind of inequality proof has no place in the Getting Started forum. Please read the posting guidelines before posting any more problems in that forum.", "Solution_19": "As you wish Mr.Mcrawford.\r\n\r\nThis proof was posted due to the request of :\r\n[quote=\"PenguinIntegral\"]Is there any other way than squaring all of it?[/quote]\r\n\r\n\r\nTo introduce a new idea.but as you said its not appropriate for getting started.", "Solution_20": "lomos, it's the entire thread that was inappropriate for the particular forum, not any one part.", "Solution_21": "the shape of question is like karamarata inequality.\r\n\r\ni wish its possible to solve it by karamarata ;) ." } { "Tag": [ "factorial", "floor function", "number theory unsolved", "number theory" ], "Problem": "Can be found the number of digits of 1999! ?", "Solution_1": "Well, yes, since we have this formula for the number of digits of x (x > 0) :\r\n$d(x) = \\lfloor log_{10}(x)\\rfloor + 1$ ;)" } { "Tag": [ "function", "AMC", "AIME" ], "Problem": "To whoever posted 2005=5*401 on the important things to know... thx, it really really saved alot of time.", "Solution_1": "Don't most calcs have a factor function? (TI-89 definitely has one; if you have a TI-83+, you should definitely have one of the factoring/prime programs).", "Solution_2": "It will probably be useful on the aime too (but not so much)\r\n\r\nI wonder what the factors of 2006 are...\r\ndividing by 2 gives 1003... \r\n\r\nand seeing as we all know our multiples of seventeen,\r\n\r\n2006 = 2*17*59\r\n\r\n\r\nThats the Voyage 200 answer anyway... :-)", "Solution_3": "[quote=\"mathfanatic\"]Don't most calcs have a factor function? (TI-89 definitely has one; if you have a TI-83+, you should definitely have one of the factoring/prime programs).[/quote]\r\n\r\nBut that would involve reaching for my calc, pressing the one button, pressing those numbers, and waiting for the answer. Frankly, I'm just too lazy.", "Solution_4": "I wonder if there's any coorelation between average AMC-12 scores from year to year and the difficulty of factoring that year's number?" } { "Tag": [], "Problem": "f:R -> N* satisfy:\r\n Every x is belong to Q; every y is belong to I if $ \\frac{\\minus{}1}{2^n} < x\\minus{}y < \\frac{1}{2^n}$\r\n With every n is belong to N*, exist k is belong to Z satisfy $ f(x)f(y)\\equal{}2^nk$\r\n\r\n :P My english is weak", "Solution_1": "In the future, please post problems of this difficulty in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=217]Olympiad[/url] forums." } { "Tag": [ "function", "logarithms", "geometric series", "real analysis", "real analysis theorems" ], "Problem": "Hi, I'm new here\r\n\r\nI'm having difficulty understanding how, once having expanded a function via maclaurin's one can say what values of x it is valid for.\r\n\r\nfor instance\r\n\r\nfor the expansion of f(x)=ln(1+x)\r\n\r\nf(x)=x-(x^2)/2+(x^3)/3-(x^4)4 etc....\r\n\r\nmy textbook says that this is valid for |x|<1 but I can't see how this conclusion is arrived at!\r\n\r\nany help or links to help would be appreciated", "Solution_1": "Note that the series diverges outside that interval, so it can't work for larger $x$. It does converge conditionally when $x=1$ to $\\ln 2$, but that's it.\r\n\r\nThe standard error estimates from Taylor's theorem will tell you that the series converges to the function when $|x|<\\frac12$, or maybe even $-\\frac12, we see that its order, say m, has also one prime factor only.\r\nHere we see that ab is not in or . So the cyclic group of ab is another nontrivial subgroup.\r\nSo there can't be another element." } { "Tag": [ "geometry", "circumcircle", "inradius", "analytic geometry", "trigonometry", "radical axis", "power of a point" ], "Problem": "Given triangle $ ABC$ with circumcircle $ (O)$, incircle $ (I). (I)$ touches $ BC, CA, AB$ at $ X, Y, Z$, respectively. Denote $ M, N, P$ the midpoints of $ BC, CA, AB. MN\\cap (O)\\equal{}\\{A_1, A_2\\}$, similar for $ B_1, B_2, C_1, C_2$. Prove that $ I$ is the radical center of three circles $ (XA_1A_2), (YB_1B_2), (ZC_1C_2)$", "Solution_1": "Dear livetolove212\r\nI think it's nice and hard.\r\nCan you show us your proof?\r\nThanks in advance.", "Solution_2": "Let $ R$ be the circumradius, $ r$ the inradius, and $ H$ the orthocenter of $ \\triangle ABC.$ Let $ AI, BI, CI$ cut $ (O)$ again at $ U, V, W$ respectively and let $ UU', VV', WW'$ be diameters of $ (O).$ Let $ Q, S, T$ be points on the rays $ (MU', (NV', (PW',$ such that $ MQ \\equal{} NS \\equal{} PT \\equal{} R.$ Let $ A_0 \\equiv A_1A_2 \\cap MO \\equiv NP \\cap MO$ be the M-altitude foot of the medial $ \\triangle MNP$ with orthocenter $ O.$ Using powers of $ A_0, H$ to $ (O)$ and triangle area $ BC \\cdot MA_0 \\equal{} \\frac {abc}{4R},$\r\n\r\n$ QA_1^2 \\equal{} A_0Q^2 \\plus{} A_0A_1^2 \\equal{} (R \\minus{} \\overline{MA_0})^2 \\plus{} (R^2 \\minus{} OA_0^2) \\equal{}$\r\n$ \\equal{} 2R^2 \\minus{} 2R \\cdot \\overline{MA_0} \\plus{} MO^2 \\plus{} 2 \\overline{MO} \\cdot \\overline{OA_0} \\equal{} 2R^2 \\minus{} \\frac {_1}{^2}bc \\plus{} MO^2 \\plus{} \\frac {_1}{^4}(R^2 \\minus{} OH^2) \\equal{}$\r\n$ \\equal{} MO^2 \\minus{} \\frac {_1}{^2}bc \\plus{} \\frac {_1}{^4}(a^2 \\plus{} b^2 \\plus{} c^2) \\equal{} R^2 \\plus{} \\frac {_1}{^4}(b \\minus{} c)^2 \\equal{} MQ^2 \\plus{} MX^2 \\equal{} QX^2$\r\n\r\n$ \\Longrightarrow$ $ QA_1 \\equal{} QA_2 \\equal{} QX$ and $ Q$ is center of the circle $ \\odot(XA_1A_2).$ Similarly, $ S, T$ are centers of the circles $ \\odot(YB_1B_2), \\odot(ZC_1C_2).$ Power of $ I$ to $ \\odot(XA_1A_2)$ is \r\n\r\n$ IQ^2 \\minus{} QX^2 \\equal{} (MQ \\minus{} XI)^2 \\minus{} MQ^2 \\equal{} r^2 \\minus{} 2rR.$\r\n\r\nSimilarly, powers of $ I$ to $ \\odot(YB_1B_2),$ $ \\odot(ZC_1C_2)$ are equal to $ r^2 \\minus{} 2rR$ $ \\Longrightarrow$ $ I$ is radical center of these 3 circles.", "Solution_3": "Let us use barycentric coordinates WRT $\\triangle ABC.$ Radical axis of $(O)$ and $\\omega_a \\equiv (XA_1X_2)$ is the line $NP \\equiv y+z-x=0$ and $X \\left (0:\\frac{_{1}}{^{s-b}}:\\frac{_1}{^{s-c}} \\right)$ lies on $\\omega_a.$ Therefore the equation of $\\omega_a$ is defined by:\n\n$\\omega_a \\equiv a^2yz+b^2xz+c^2xy-(s-b)(s-c)(x+y+z)(y+z-x)=0$\n\nSimilarly the equations of $\\omega_b$ and $\\omega_c$ are given by:\n\n$\\omega_b \\equiv a^2yz+b^2xz+c^2xy-(s-a)(s-c)(x+y+z)(x-y+z)=0$\n\n$\\omega_c \\equiv a^2yz+b^2xz+c^2xy-(s-a)(s-b)(x+y+z)(x+y-z)=0$\n\nRadical axis of $\\omega_b$ and $\\omega_c$ has equation $\\ell_a \\equiv (c-b)x+ay-az=0,$ which is the line connecting $M(0:1:1)$ and $I(a:b:c).$ Likewise, radical axes $\\ell_b,\\ell_c$ of $\\omega_a,\\omega_c$ and $\\omega_a,\\omega_b$ are $NI$ and $PI$ $\\Longrightarrow$ $I$ is the radical center of $\\omega_a,\\omega_b,\\omega_c.$", "Solution_4": "maybe $\\{A_1, A_2\\}=(O)\\cap NP$ ???", "Solution_5": "[quote=\"Virgil Nicula\"]maybe $\\{A_1, A_2\\}=(O)\\cap NP$ ???[/quote]\nIndeed, it's a little typo in livetolove212's enunciation, otherwise, the problem is not true.", "Solution_6": "[size=130][color=darkblue][u]Very nice and interesting problem[/u] ! [b]Thank you, [u]livetolove212[/u].[/b][/color][/size]\n\n[quote=\"livetolove212\"] [color=darkred]Given triangle $ ABC$ with circumcircle $ (O)$ and incircle $ (I)$ which touches $ BC, CA, AB$ at $ X, Y, Z$, respectively. \n\nDenote the midpoints $ M, N, P$ of $ BC, CA, AB$ respectively and $\\{A_1, A_2\\}=NP\\cap (O)$ . Similar for $ B_1$ , $B_2$ , \n\n$ C_1$ , $C_2$ . Prove that $ I$ is the radical center of the circumcircles of the triangles $ (XA_1A_2)$ , $(YB_1B_2)$ , $(ZC_1C_2)$ .[/color] [/quote]\n\n[color=darkblue][b][u]Proof (metric)[/u].[/b] Denote the circumcircle $w_a=\\mathrm C(P,\\rho )$ of $\\triangle A_1XA_2$ , the intersection $N\\in A_1A_2\\cap OM$ \n\nand $NP=x$ , the projection $R$ of $I$ on $OM$ and $S\\in OM\\cap (O)$ so that the sideline $BC$ separates $A$ , $S$ .\n\n$\\blacktriangleright$ $A_1A_2^2=4\\left(OA_1^2-ON^2\\right)=$ $4\\left[R^2-\\left(\\frac {h_a}{2}-R\\cos A\\right)^2\\right]=$ $4\\left(R^2\\sin^2A-\\frac {h_a^2}{4}+h_aR\\cos A\\right)=$\n\n$a^2-h_a^2+4h_aR\\cos A=$ $a^2-h_a^2+b^2+c^2-a^2$ $\\implies$ $\\boxed {\\ A_1A_2^2=b^2+c^2-h_a^2\\ }$ . I used well-known relation $2Rh_a=bc$ . \n\n$\\blacktriangleright$ $\\left\\|\\begin{array}{c}\n\\rho^2=AP^2=AN^2+NP^2=\\frac 14\\cdot\\left(b^2+c^2-h_a^2\\right)+x^2\\\\\\\\\n\\rho^2=XP^2=XM^2+MP^2=\\left(\\frac {b-c}{2}\\right)^2+\\left(\\frac {h_a}{2}+x\\right)^2\\end{array}\\right\\|\\implies$ \n\n$4\\rho^2-4x^2\\stackrel{(1)}{\\ =\\ }b^2+c^2-h_a^2\\stackrel{(2)}{\\ =\\ }(b-c)^2+h_a^2+4xh_a$ . From these two relations we\"ll ascertain $x$ and $\\rho$ . Therefore,\n\n$(2)\\ \\implies\\ x=\\frac {bc-h_a^2}{2h_a}=\\frac {2Rh_a-h_a^2}{2h_a}$ $\\implies$ $\\boxed {\\ R=x+\\frac {h_a}{2}\\ }$ $\\Longleftrightarrow$ $\\underline{MP=R}$ , i.e. $\\underline{OP=MS}\\ \\ \\mathrm{!!}$ \n\n$(1)\\ \\implies$ Since $XM=\\frac {|b-c|}{2}$ and $MP=R$ obtain $\\rho^2=XM^2+MP^2$ $\\implies$ ${\\boxed {\\ \\rho^2=\\left(\\frac {b-c}{2}\\right)^2+R^2}\\ }$ .\n\n$\\blacktriangleright$ The power of $I$ w.r.t. the circle $w_a$ is $p_{w_a}(I)=IP^2-\\rho^2=$ $XM^2+RP^2-\\rho^2=$ \n\n$\\left(\\frac {b-c}{2}\\right)^2+\\left(\\frac {h_a}{2}-r+x\\right)^2-\\left(\\frac {b-c}{2}\\right)^2-R^2=(R-r)^2-R^2$ , $\\implies$ $\\boxed {\\ p_{w_a}(I)=-r(2R-r)\\ }$ .\n\nSince the expression of the power $p_{w_a}(I)$ is symmetrically w.r.t $\\triangle ABC$ obtain the conclusion of this [u]nice[/u] proposed problem.[/color]", "Solution_7": "Here is a generalization for this problem \n[b]Generalization[/b]: Given $\\triangle ABC$ inscribes in circle $(O)$. Let $M$, $N$, $P$ be midpoint of $BC$, $CA$, $AB$; $A_1$, $B_1$, $C_1$ be arbitrary points on $BC$, $CA$, $AB$; $A_2$, $B_2$, $C_2$ be reflections of $A_1$, $B_1$, $C_1$ through $M$, $N$, $P$. $NP$, $PM$, $MN$ intersect $(O)$ at $D_1$, $D_2$, $E_1$, $E_2$, $F_1$, $F_2$, respectively. Let $G$, $P$, $Q$ be centroid, radical center of $((A_1A_2D_1D_2), (B_1B_2E_1E_2), (C_1C_2F_1F_2))$; $((AA_1A_2), (BB_1B_2), (CC_1C_2))$. Prove that: $\\overline{GQ} = 4 \\overline{GP}$ \nFrom this problem, if we let $\\triangle A_1B_1C_1$ be intouch triangle, $\\triangle A_2B_2C_2$ be extouch triangle then $Q$ is anticompliment of Nagel point of $\\triangle ABC$ \nHence: $\\overline{GQ} = 4 \\overline{GI}$ or $I$ is radical center of $((A_1A_2D_1D_2), (B_1B_2E_1E_2), (C_1C_2F_1F_2))$ " } { "Tag": [ "abstract algebra", "geometry", "geometric transformation", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Hi,\r\n\r\ncan I provide every abelian group G with a non-trivial multilication, such that G is with addition and this new multiplication a ring? With \"non-trivial\" I mean, that G*G have more elements besides 0.\r\n\r\nT.\r\n\r\nBTW.: I've parallel posted this questions in http://www.matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=96709&start=0&lps=704200#v704200\r\n\r\nso u can see, if its answered already, ..., if you know german", "Solution_1": "no. there are many (necessarily infinitely-generated) abelian groups $ G$ with $ G \\otimes_{\\mathbb{Z}} G \\equal{} 0$, such as $ G \\equal{} \\mathbb{Q}/\\mathbb{Z}$. since every ring structure $ G \\times G \\to G$ must factorize through this tensor product, it must be trivial.", "Solution_2": "ok, its easy to see it for $ \\mathbb{Q}/\\mathbb{Z}$, I dont understand your general statement, cause I dont know tensor products, seems that I must learn it to understand your posting", "Solution_3": "ok for translation: $ G \\otimes G \\equal{} 0$ means that [i]every[/i] bilinear map on $ G \\times G$ is zero.", "Solution_4": "ok, its than only an existence statemant\r\n\r\nare there easy criteria to decide for an infinitly generated group, if it can/cant provided with a ringstructure?" } { "Tag": [ "algebra", "function", "domain", "calculus", "derivative" ], "Problem": "You cant use a calculator, here they are \r\n1. Let f = 5^((2x\u00b2-1)^1/2) \r\na) Is it even or odd? Justfify \r\nb)Find domain \r\nc) Find Range \r\nd) find derivitive ( i got it to be 5^((2x\u00b2-1)^1/2) * 4x/(2((2x\u00b2-1)^1/2)) \r\n2. A particle moves along a line so that at any time t its position is given by x(t) = 2(pi)t + cos(2(pi)t) \r\nFind Velocity - 2(pi)-sin(2pit)(2pi) ? first deriv \r\nFind Acceleration - 0 ? second deriv \r\nWhat are all values of t, t greater than or equal to 0, less than or equal to 3, for wihch the particle is at rest \r\nWhat is the max velocity", "Solution_1": "Calculus questions should go in the calculus forums.", "Solution_2": "w/e I'll help you out anyway. But I won't tel u the answers. i don't believe in that.\r\n\r\nfor even or odd, let me remind you:\r\nodd if and only if $f(-x)=-f(x)$\r\neven if and only if $f(-x)=f(x)$\r\n\r\nTo find the domain, find all x that work in the equation. Then for the range, over this domain only, find all the values of y that can be produced.\r\n\r\nTo find the derivative, try expressing the function as e to the something first. That's easiest for exponentials. \r\n\r\nfor 2, the velocity is the first derivative and acceleration is the second derivative. \r\nnow, the particle is at rest when the velocity is zero. \r\nthe velocity has a max when its derivative, i.e. the acceleration is zero. \r\n\r\nGood luck!" } { "Tag": [ "MATHCOUNTS", "Support" ], "Problem": "[quote]And whats with Texas. Just because bush won the election doesnt mean texas owns the world. CAlifornia has the most electoral votes. EAT THAT!![/quote]\r\nlets not get into politics although i didnt support bush becuz hes not very bright but texas is still good", "Solution_1": "Yes, apparently our fellow mods don't like it very much when we go into politics :?", "Solution_2": "Yes, we are all, um...mathematicians, um...under one...roof. :? \r\n\r\nAnyway, nobody's political or religious affiliations should get in the way of our common ground here, save it for the round table boards.", "Solution_3": "But that doesn't give the mods any right to delete my posts! :(", "Solution_4": "the reason we have mods in forums is for them to watch the forums to make sure our conversations are on topic and appropriate, alex. Why else are they called moderators? ANd of course they have the right to delete ur posts!", "Solution_5": "otherwise, this forum would be running wild", "Solution_6": "almost 50 posts", "Solution_7": "[quote=\"NoSoupForYou\"]But that doesn't give the mods any right to delete my posts! :([/quote]\r\n\r\nWell, I could see either way. On one hand, there is an issue of free speech and you weren't necessarily offending anybody or being \"inappropriate\" (I hate that word when it's used on anything short of pornography), but then again there is a lack of tension we want to maintain on this board.\r\n\r\nBut let's not debate over it.", "Solution_8": "[quote=\"aznness\"]almost 50 posts[/quote]\r\n\r\nyeah, instead of posting 3 or 4 times in a row, maybe you should just put them all into one?" } { "Tag": [ "geometry", "geometric transformation", "dilation", "reflection", "parallelogram", "trigonometry", "algebra" ], "Problem": "Construct triangle by:\r\na) Three medians;\r\nb) Three altitudes/heights;\r\nc) Three angle bisectors.", "Solution_1": "I understand your problem as: We have $ 3$ lines are medians (or altitudes, bisectors) and we have to contract $ 3$ point $ A,B,C$ satisfy.\r\na) Let $ A\\in m_a$, we find point $ P\\in m_c$ and $ B\\in m_b$ by this: $ V_A^2: P\\rightarrow B$ (dilate)\r\nThen if $ V_A^2: m_c\\rightarrow m_c'$ then $ B\\equal{}m_b\\cap m_c'$\r\nb) Let $ A\\in h_a$, contract a line pass through $ A$ and perpendicular with $ h_c$ cut $ h_b$ at $ B$\r\nc) Let $ A\\in l_a$, call $ A_1$ be reflect $ A$ over $ l_c$. Because $ CA$ reflect with $ CB$ over $ l_c$ then $ A_1\\in CB$\r\nSimilarly call $ A_2$ be reflect $ A_1$ over $ l_a$ then $ A_2\\in CA$\r\nCall $ A_3$ be reflect $ A_2$ over $ l_b$ then $ A_3\\in AB$\r\nNow contract $ AA_3\\cap l_b\\equal{}B$", "Solution_2": "My idea were to describe the process of construction (with compas and ruller) of a triangle that have 3\r\na) medians,\r\nb) altitudes,\r\nc) angle bisectors,\r\nwith sizes given in advance (places are not given in advance) and if it is possible to say how mach triangles with desired properties may be constructed.", "Solution_3": "[b]Medians:[/b] Assume problem solved, G is the centroid, $ M_a$ the midpoint of BC. Reflect G in $ M_a$ into D. BGCD is a parallelogram (diagonals cutting each other in half), $ \\triangle GBD$ has sides $ GD \\equal{} \\frac {2}{3}m_a,\\ BG \\equal{} \\frac {2}{3}m_b,\\ DB \\equal{} \\frac {2}{3}m_c.$ Construct $ \\triangle GB'D'$ with sides $ GD' \\equal{} m_a,\\ B'G \\equal{} m_b,\\ D'B' \\equal{} m_c.$ Parallels to $ B'G,\\ B'D'$ through $ D',\\ G$ meet at C'. Reflect D' in G into A'. $ \\triangle A'B'C'$ has mediams $ m_a' \\equal{} \\frac {4}{3}m_a,\\ m_b' \\equal{} \\frac {4}{3}m_b,\\ m_c' \\equal{} \\frac {4}{3}m_c.$ $ \\triangle ABC$ centrally similar to $ \\triangle A'B'C'$ with similarity center G and coefficient $ \\frac {3}{4}$ has the given medians $ m_a,\\ m_b,\\ m_c.$\r\n\r\n\r\n[b]Altitudes:[/b] Triangle area is $ S \\equal{} \\frac {ah_a}{2} \\equal{} \\frac {bh_b}{2} \\equal{} \\frac {ch_c}{2},$ therefore $ a \\equal{} \\frac {2S}{h_a},\\ b \\equal{} \\frac {2S}{h_b},\\ c \\equal{} \\frac {2S}{h_c}.$ Let $ OD \\equal{} h_a,\\ OE \\equal{} h_b,\\ OF \\equal{} h_c$ be the given altitudes. Invert D, E, F in a circle (O) with arbitrary radius r into D', E', F', so that $ OD' \\equal{} \\frac {r^2}{OD} \\equal{} \\frac {r^2}{h_a} \\equal{} \\frac {r^2}{2S} a \\equal{} a'$ and similarly, $ OE' \\equal{} \\frac {r^2}{2S} b \\equal{} b',$ $ OF' \\equal{} \\frac {r^2}{2S} c \\equal{} c'.$ Construct $ \\triangle A'B'C'$ with sides a', b', c' and one of its altitudes, say $ h_a'.$ $ \\triangle ABC \\sim \\triangle A'B'C'$ with similarity coefficient $ \\frac {h_a}{h_a'}$ has the given altitudes $ h_a,\\ h_b,\\ h_c.$\r\n\r\n\r\n[b]Bisectors:[/b] $ \\triangle ABC$ with the given angle bisectors $ l_a,\\ l_b,\\ \\l_c$ is generally not constructible, not even an isosceles triangle with $ AB \\equal{} c \\equal{} b \\equal{} AC,\\ BC \\equal{} a.$ In this case,\r\n\r\n$ l_a^2 \\equal{} b^2 \\left(1 \\minus{} \\frac {a^2}{4b^2}\\right) \\equal{} \\frac {4b^2 \\minus{} a^2}{4} \\equal{} \\frac {(2b \\plus{} a)(2b \\minus{} a)}{4}$\r\n \r\n$ l_b^2 \\equal{} l_c^2 \\equal{} ab \\left(1 \\minus{} \\frac {b^2}{(a \\plus{} b)^2}\\right) \\equal{} \\frac {a^2b(a \\plus{} 2b)}{(a \\plus{} b)^2}$\r\n\r\n$ \\frac {l_a^2}{l_b^2} \\equal{} \\frac {(2b \\minus{} a)(a \\plus{} b)^2}{4a^2b} \\equal{} \\frac {2b^3 \\plus{} 3ab^2 \\minus{} a^3}{4a^2b} \\equal{} \\frac {1}{8} \\left(\\frac {4b^2}{a^2} \\plus{} 3 \\frac {2b}{a} \\minus{} 4\\frac {a}{2b}\\right)$\r\n\r\nDenote $ x \\equal{} \\frac {a}{2b} \\equal{} \\cos B \\equal{} \\cos C.$\r\n\r\n$ 4x^3 \\plus{} 8\\frac {l_a^2}{l_b^2}x^2 \\minus{} 3x \\minus{} 1 \\equal{} 0$\r\n\r\nThis is a cubic equation, the roots of which are generally not constructible (just like for angle trisection). For example, $ 8\\frac {l_a^2}{l_b^2} \\equal{} n$ can be integer, in which case $ \\frac {l_a}{l_b}$ can always be constructed. By the rational root theorem, the possible rational roots of the cubic are then only $ \\left\\{\\pm 1, \\pm \\frac {1}{2},\\ \\pm \\frac {1}{4}\\right\\},$ yielding $ n \\equal{} 0$ (not acceptable) or $ n \\equal{} 2,\\ 5,\\ 8,\\ 27.$ If n is some other integer, the cubic has no rational root and no roots are then constructible.\r\n\r\nThe problem with angle bisectors is that formulas for the internal and external angle bisectors in terms of the triangle sides a, b, c are the same except for signs of a, b, and/or c. For given angle bisectors, the general solution for a, b, c can yield some sides negative, which means that some of the given angle bisectors of the resulting triangle are external, not internal.", "Solution_4": "Thank you, yetti, excellent!" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Let $x_{1},\\ldots x_{n}$ be real numbers from the interval [\u22121, 1] such that $x_{1}+\\ldots+x_{n}=0$. Prove that there exists a permutation $\\sigma$ of the set ${1, \\ldots , n}$ such that for any integers $p, q$ with 1 \u2264 p \u2264 q \u2264 n we have\r\n$|x_{\\sigma(p)}+\\ldots+x_{\\sigma(q)}|\\leq 2-\\frac{1}{n}$\r\nProve that, if $2-1/n$ is replaced with $2-4/n$, the statement does not\r\nnecessarily hold.", "Solution_1": "this one has been posted twice before,here is one of them:[url]http://www.mathlinks.ro/viewtopic.php?t=6263[/url] (without any solution)\r\nI dont remember the other one,maybe somebody else could give you the link... :wink:" } { "Tag": [ "probability" ], "Problem": "The letters of the word MATHEMATICAL are arranged at random. What is the probability that the resulting arrangement contains no adjacent A's?", "Solution_1": "[hide]# of ways w/o restrictions: $12!/(2!3!2!)=19958400$\n\nFirst put the other numbers ($9!/(2!2!)=90720$ ways). Then choose 3 of the spaces between consecutive numbers to put the A's. 10 possibilities for the first one, 9 for the 2nd, and 8 for the third, but order doesn't matter so divide by $3!$, so we get $120$. \n\nAnswer: $\\frac{10886400}{19958400}=\\frac{6}{11}$[/hide]" } { "Tag": [], "Problem": "Hi, the clubmed at my high school made me chairperson of new ideas(worst decision they ever made). I came up with the idea of starting a club med at one of the local middle schools and have our students volunteer to go teach. The problem is that I don't know how to make students want to join. At our high school clubmed, we get volunteer points and we can put it on college applications and whatnot. But, middle schoolers aren't going to care about college and they aren't going to be volunteering. So my question is, what would make you want to join a club med? I have the feeling that a lot of the ppl here are going to say to learn(that's what I would say), but I want this club to appeal to more than the just the smart ppl. Any ideas?", "Solution_1": "Hi, I am ignorant. What is clubmed?", "Solution_2": "Catchy name for a club that talks about medicine.", "Solution_3": "Like, do you talk about disease prevention, or medical professionals, or ethical issues? Probably a mixture, but I am not sure, so I ask.", "Solution_4": "I believe the whole point of high school club med is to help ppl interested in a medical career in deciding what to do, thus the various volunteer oppertunities, lectures from ppl of professions etc. but the middle school one I had in mind will more teach about diseases, health and some other good stuff." } { "Tag": [ "group theory", "abstract algebra", "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "Suppose that $G$ is a subgroup of $\\mathrm{GL}_n(\\mathbb{C})$ such that $I-g$ is nilpotent for every $g \\in G$.\nProve that $G$ is conjugate to a subgroup of $T_n$, where $T_n$ is the group of upper triangular matrices with $1$ on the main diagonal.", "Solution_1": "It follows from a result posted by sam-n on the forum: a semi-group on which the trace is constant is simultaneously trigonalisable.", "Solution_2": "ohh, I have just read the great proof of Leva80 for this result ! beautifull result indeed !!", "Solution_3": "I'm reposting this message. I had a proof which only worked for connected groups, but this should work in the general case. I am, however, using the fact that an algebra of endomorphisms of a finite dimensional vector space over $\\mathbb C$ all of whose elements are nilpotent endomorphisms is triangulable.\r\n\r\nLet $G-1$ be the set $\\{g-1\\ |\\ g\\in G\\}$. We know that $G-1$ consists only of nilpotent endomorphisms. Whenever $a,b\\in G-1,\\ a+b+ab\\in G-1$ as well. This means that for all $a,b\\in G-1$, $\\mbox{tr}(ab)=\\mbox{tr}(ab+a+b)-\\mbox{tr}(a)-\\mbox{tr}(b)=0$. Next, notice that for all $a,b,c\\in G-1$, $a+b+c+ab+bc+ca+abc\\in G-1$ as well, and in the same way as above we show that $\\mbox{tr}(abc)=0,\\ \\forall a,b,c\\in G-1$. Continuing like this, we conclude that the trace of every element of the algebra $A$ generated by $G-1$ is zero, and hence that all elements of $A$ are nilpotent (using the criterion which says that $a$ is nilpotent iff $\\mbox{tr}(a^{k})=0,\\ \\forall 1\\le k\\le n$). \r\n\r\nThe result cited in the first paragraph now finishes the proof: $G-1$ is triangulable, and adding $1$ to each element of it we get the desired representation of the elements of $G$ as upper triangular matrices with $1$ on the main diagonal." } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Show that if the circle passing through the feet of the symmedians of a non-isosceles triangle of sides $ a$, $ b$, $ c$ is tangent to one side, then the quantities $ b^{2}\\plus{}c^{2}$, $ c^{2}\\plus{}a^{2}$, $ a^{2}\\plus{}b^{2}$, arranged in some order, are consecutive terms of a geometric progression.", "Solution_1": "Nice to know the real source of http://www.mathlinks.ro/Forum/viewtopic.php?t=53033 . ;) \r\nBut since in the link above there are no replies and for not bumping it, I propose continuing the discussions here.\r\n\r\n darij", "Solution_2": "$K$ is the symmedian point of $\\triangle ABC$ and $K_A,K_B,K_C$ are the cevian traces of $K$ on $BC,CA,AB.$ Recalling that the general barycentric equation of the circle $\\mathcal C$ is\n\n$\\mathcal C(x,y,z)=a^2yz+b^2zx+c^2xy-(x+y+z)(px+qy+rz)=0$\n\nSubstituting $K_A(0:b^2:c^2) \\ , K_B(a^2:0:c^2)$ and $K_C(a^2:b^2:0)$ into the equation of $\\mathcal C(x,y,z)$ and solving the three equations for $p,q,r$ yields\n\n$p=\\frac{b^2c^2}{2} \\left ( \\frac{1}{c^2+a^2}+\\frac{1}{a^2+b^2}-\\frac{1}{b^2+c^2} \\right)$\n\n$q=\\frac{c^2a^2}{2} \\left ( \\frac{1}{b^2+c^2}+\\frac{1}{a^2+b^2}-\\frac{1}{c^2+a^2} \\right)$\n\n$r=\\frac{a^2b^2}{2} \\left ( \\frac{1}{b^2+c^2}+\\frac{1}{c^2+a^2}-\\frac{1}{a^2+b^2} \\right)$\n\nIf $\\mathcal{C}$ is tangent to $BC,$ then $C(0,y,z)=qy^2+rz^2-(q+r-a^2)yz=0$ has double roots. The product of the roots $y:z$ is $\\frac{_r}{^q}$ and one of them is obviously $y:z=b^2:c^2$ $\\Longrightarrow$ the other root is $\\frac{_{rc^2}}{^{qb^2}}.$ Thus, equating the roots gives $b^4q-c^4r=0.$ Now, substituting the values of $q,r$ found in the first part, we obtain\n\n$b^2 \\left ( \\frac{1}{b^2+c^2}+\\frac{1}{a^2+b^2}-\\frac{1}{c^2+a^2} \\right)-c^2\\left ( \\frac{1}{b^2+c^2}+\\frac{1}{c^2+a^2}-\\frac{1}{a^2+b^2} \\right)=0$\n\n$\\frac{(b-c)(b+c)(b^4+c^4-a^4+b^2c^2-a^2b^2-a^2c^2)}{(b^2+c^2)(c^2+a^2)(a^2+b^2)}=0$\n\n$(b-c)(b+c)[(b^2+c^2)^2-(c^2+a^2)(a^2+b^2)]=0$\n\nTherefore, the cevian circle $\\mathcal{C}$ of $K$ is tangent to $BC$ $\\Longleftrightarrow$ Either $b=c,$ i.e. $\\triangle ABC$ is isosceles with apex $A,$ or $(b^2+c^2)^2=(c^2+a^2)(a^2+b^2)$ and the conclusion follows." } { "Tag": [ "conics", "ellipse", "geometry", "geometric transformation", "rotation", "linear algebra", "matrix" ], "Problem": "This is my very first post on this forum. Sorry if it's posted in the wrong category etc.\r\n\r\nI'm making as system that lets a user point out an area on a flat surface, which have some points placed around. It's comparable to a person standing with a flashlight and lighting up spots in different directions.\r\n\r\nIn my system, the pointed out area will be shaped as an ellipse, and due to some inaccuracy in the pointed angle and chosen distance, the proportions of the ellipse may vay.\r\n\r\nI've made this illustration: http://www.student.dtu.dk/~s032225/tilted_ellipse.PNG \r\n\r\nUser stands in (0,0) and points to (x,y), the green lines shows the actual inaccuracy, and from these data, I can create an ellipse with center (x,y), size l*v and it's tilted according to the angle between (0,0) and (x,y). \r\n\r\nA point is placed in (px,py), and I need to check if this point is covered by the ellipse.\r\n\r\nI've found some formulas for checking points within \"straight\" ellipses, but how to handle a tilted ellipse, when I only have the data mentioned above, available?", "Solution_1": "Two points don't define an angle. Do you mean that the length $ l$ axis is on the line between the origin and the center $ (x_0,y_0)$ of the ellipse?\r\n\r\nWe can reach this ellipse by starting with an ellipse $ \\frac{(x\\minus{}r)^2}{l^2}\\plus{}\\frac{y^2}{v^2}\\le 1$ (where $ r\\equal{}\\sqrt{x_0^2\\plus{}y_0^2}$) and then rotating the plane so that $ (r,0)$ goes to $ (x_0,y_0)$\r\n\r\nThis rotation is given by the formula $ f\\begin{pmatrix}x\\\\y\\end{pmatrix}\\equal{}\\frac1r\\begin{pmatrix}x_0&\\minus{}y_0\\\\y_0&x_0\\end{pmatrix}\\cdot\\begin{pmatrix}x\\\\y\\end{pmatrix}$. Our point $ p$ is inside the new ellipse if $ f^{\\minus{}1}(p)$ is in the old one. $ f^{\\minus{}1}$ is of course given by the inverse matrix $ \\frac1r\\begin{pmatrix}x_0&y_0\\\\\\minus{}y_0&x_0\\end{pmatrix}$\r\n\r\nAn equation of the new ellipse is thus $ \\frac{(\\frac{x_0}{r}x\\plus{}\\frac{y_0}{r}y\\minus{}r)^2}{l^2}\\plus{}\\frac{(\\frac{\\minus{}y_0}{r}x\\plus{}\\frac{x_0}{r}y)^2}{v^2}\\le 1$\r\nExpand that out, and it becomes\r\n$ \\frac1{r^2}\\left(\\frac{x_0^2}{l^2}\\plus{}\\frac{y_0^2}{v^2}\\right)x^2\\plus{}\\frac1{r^2}\\left(\\frac{y_0^2}{l^2}\\plus{}\\frac{x_0^2}{v^2}\\right)y^2\\plus{}\\frac{2x_0y_0}{r^2}\\left(\\frac1{l^2}\\minus{}\\frac1{v^2}\\right)xy\\minus{}\\frac{2}{l^2}(x_0x\\plus{}y_0y)\\le 1$\r\nReplacing $ x$ and $ y$ with $ x\\minus{}x_0$ and $ y\\minus{}y_0$ would remove the linear terms without affecting any other coefficients. Note that you don't have to calculate any square roots here, as $ r^2\\equal{}x_0^2\\plus{}y_0^2$ appears but $ r$ alone does not.", "Solution_2": "[quote=\"jmerry\"]Two points don't define an angle. Do you mean that the length $ l$ axis is on the line between the origin and the center $ (x_0,y_0)$ of the ellipse?[/quote]\r\n\r\nYes, i drew a gray line on the image now - showing the angle of x,y relative to 0,0 ... And as you can see, the ellipse can be anywhere around the center, but always \"hangs\" in the gray line...\r\n\r\nThanks for the formulas! I'll take a look at it soon...", "Solution_3": "Wherever the center of the ellipse is, you need to know and to incorporate it. The coefficients of the quadratic terms are entirely determined by how much the two axes are rotated from the horizontal and vertical, but the linear/constant terms depend on exactly where the center is." } { "Tag": [ "LaTeX" ], "Problem": "oops sorry for yet another question... I have attached 2 files for your reference.\r\n\r\nNotice that in my pdf file, the LaTeX has my whole sentence underlined. However, I intend to have only the words on the left of \":\" to be underlined. The problem seems to be triggered by the existence of \"(\" and \")\"...\r\n\r\nSome help, pls? :blush:\r\n\r\nEDIT: looks like I have forgotten to put some dollar signs around $\\rightarrow$ again... Once I did that everything looks fine", "Solution_1": "Try this:\r\n[quote]\\underline{Conditional Elimination $( \\rightarrow E )$} : From a conditional and its antecedent we may infer its consequent.[/quote]\r\n\r\nThe diff\u00e9rence is that the two ( ) are between the two dollars.\r\n\r\n[b]PS:[/b] I don't have LaTeX on my PC, so...", "Solution_2": "k, I have tried your method by putting the whole ( \\neg E ) in between the dollar signs. It works fine.\r\n\r\nBut it surely doesn't work if I don't put any dollar signs.\r\n\r\nlol none of my LaTeX guide talks about this strange phenomena...", "Solution_3": "I think that is a normal \"phenomena\" because there is no difference between \"(\" between two dollars and ( in text mode." } { "Tag": [ "search", "geometry" ], "Problem": "hey knights of the round table\r\n\r\ni wanted to ask, what program to construct geomertical \"things\" on the computer u guys use.\r\nand also where do u got it from. i mean like freeware/open source/stolen/bought/etc\r\n\r\nthx for all answers, cuz im looking for a good constructing program for my pc\r\n\r\npeeta", "Solution_1": "Hi peeta,\r\n\r\nTry a search of the whole forum for discussions of this question (of which there have been several). Any topic that contains the words \"geometer's sketchpad\" or GSP is likely to be helpful. I also know there is a member of the forum whose signature contains a link to a free geometry program called KSG or something, although I don't remember who the individual is.", "Solution_2": "The program Joel is talking about is KSEG and I believe that it is EFuzzy whom has a link (though I'm not sure). It's a pretty good program. But, yeah, do the search, there are a couple more discussed on the boards that are decent.", "Solution_3": "I use [url=http://www.mit.edu/~ibaran/kseg.html]KSEG[/url]; I think it's the best freeware program for geometry sketches. It is very intuitive to use and powerful as well." } { "Tag": [ "geometry", "trapezoid", "geometric transformation", "homothety", "analytic geometry", "geometry solved" ], "Problem": "Inside triangle ABC find triangle XYZ with sides parallel to sides of ABC\r\n so that all 3 trapezoids have incircles.\r\n\r\n\r\n Maj. Pestich", "Solution_1": "If you want just a compass-and-ruler construction of the triangle XYZ from triangle ABC, then it's an easy exercise on the good old method of similarity:\r\n\r\nTake a triangle X'Y'Z' homothetic to triangle ABC. Construct the three excircles of this triangle X'Y'Z', and to each of these excircles, construct the tangent parallel to the respective side of the triangle X'Y'Z' (but not coinciding with this side). Then, these three tangents form a new triangle A'B'C', and the two triangles A'B'C' and X'Y'Z' are related to each other in the same way as the triangles ABC and XYZ: The triangle X'Y'Z' lies inside the triangle A'B'C' and has its sides parallel to the sides of triangle A'B'C', and the three trapezoids formed by intersecting the sides of triangle A'B'C' with the extended sides of triangle X'Y'Z' all have incircles (these incircles are simply the excircles of triangle X'Y'Z').\r\n\r\nHence, our construction problem is solved for the triangle A'B'C'. But triangle A'B'C' is homothetic to triangle ABC (in fact, the sidelines of triangle A'B'C' are parallel to the sidelines of triangle X'Y'Z', so that triangle A'B'C' is homothetic to triangle X'Y'Z', but on the other hand, the triangle X'Y'Z' is homothetic to triangle ABC, so we obtain that triangle A'B'C' is homothetic to triangle ABC). Hence, there exists a homothety which maps the triangle A'B'C' to the triangle ABC, and the required triangle XYZ is the image of the (known) triangle X'Y'Z' under this homothety.\r\n\r\nIf you are asking for the trilinear coordinates of the vertices of triangle XYZ with respect to triangle ABC, here they are:\r\n\r\n$X\\left(h_a+r_a:r_b:r_c\\right)=X\\left(\\frac{b+c}{a\\left(s-a\\right)}:\\frac{1}{s-b}:\\frac{1}{s-c}\\right)$;\r\n... (cyclically for Y and Z).\r\n\r\nHereby, $s=\\frac{a+b+c}{2}$ is the semiperimeter, $h_a$, $h_b$, $h_c$ the altitudes and $r_a$, $r_b$, $r_c$ the exradii of triangle ABC.\r\n\r\nThe lines AX, BY and CZ concur at the point $X_{57}$. This is the isogonal Mitten point of triangle ABC (= the isogonal conjugate of the Mitten point $X_9$).\r\n\r\n darij", "Solution_2": "May I ask you darij, where did you learn all that stuff relating trillinear coordinates?", "Solution_3": "Well, what should I say? I used a lot of sources, starting with the (really!) brief definition of homogeneous trilinear coordinates on Clark Kimberling's page http://faculty.evansville.edu/ck6/tcenters/trilin.html and continuing with the archive of the [url=http://groups.yahoo.com/group/Hyacinthos/messages?viscount=100]Hyacinthos newsgroup for triangle geometry[/url], the latter being a dynamic source in the sense of that you will hardly find a systematic explanation of trilinear coordinates but rather lots of examples for their application. Also, maybe you will make something out of my introduction in http://www.mathlinks.ro/Forum/viewtopic.php?t=4357 post #4, but I must admit that it gives no proofs and mentions only the facts used in the freaky solution of the problem from that thread... Actually, nowadays, trilinear coordinates have lost their popularity since many people use barycentric coordinates instead; they are really more or less the same thing (you can easily transform the trilinear coordinates of a point into the barycentric ones and conversely), but have some theoretic advantages. See [url=http://www.math.fau.edu/yiu/barycentricpaper.pdf]Paul Yiu's \"The uses of homogeneous barycentric coordinates in plane euclidean geometry\"[/url] for an introduction into barycentric coordinates and generally [url=http://www.math.fau.edu/yiu/Geometry.html]his website[/url] for a number of triangle geometry papers where they are used (especially the \"Introduction to the Geometry of the Triangle\", which almost completely relies on barycentric coordinates). Finally, in http://groups.google.de/group/geometry.college/browse_frm/thread/94d0e0e89f8eac54/e2c5551697ff2639?rnum=1#e2c5551697ff2639 (post #4) , you can find a Usenet posting by John Conway explaining barycentric coordinates, and in http://groups.google.de/groups?hl=de&th=2d8e659ba9cf3128&rnum=1 and http://groups.google.de/groups?hl=de&threadm=36682AFB.4F42%40wxs.nl&rnum=2 a comparison between trilinear and barycentric coordinates.\r\n\r\nHope that some of this helps - as I said, I'm not aware of any systematic English source on trilinear coordinates which would be avaliable online.\r\n\r\n darij" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Does there exist a continuously differentiable function $f : \\mathbb{R} \\rightarrow \\mathbb{R}$ such that for every $x \\in \\mathbb{R}$ we have $f(x) > 0$ and $f'(x) = f(f(x))$?", "Solution_1": "It is simple, I think.\r\n\r\nSuppose there exists such $f$.\r\n$f'(x)=f(f(x))>0$, so $f$ is strictly increasing.\r\nthus\r\n$f'(x) = f(f(x))>f(0)$ for all $x$.\r\n\r\nFor negative $x$, $f(0)-f(x)=f'(\\xi)(0-x)>-xf(0)$, thus $f(x)<(1+x)f(0)$ and $f(x)$ is negative for $x<-1$. This is a contradiction.", "Solution_2": "Note that $f$ is strictly increasing. Hence $$f(x)>0\\iff f(f(x))>f(0).$$ But then $$f(-1)\\leq f(0)-f(0)=0$$ a contradiction." } { "Tag": [ "symmetry", "algebra unsolved", "algebra" ], "Problem": "Find $x;y;z$ such that:\r\n$\\left\\{ \\begin{array}{l}x^{2}= (y-1)(z+2) \\\\ y^{2}= (z-1)(x+2) \\\\ z^{2}= (x-1)(y+2) \\end{array}\\right.$", "Solution_1": "Here is my solution :\r\n\r\nWithout loss of generalization we may assume that; $z\\le y\\le x$ then,\r\n\r\n$x^{2}=(y-1)(z+2)\\ge z^{2}=(x-1)(y+2)$ but $x-1\\ge y-1$ and $y+2\\ge z+2$ so $x=y=z$\r\n\r\nif we use that on the equations we can get $x^{2}=(x-1)(x+2)$ then $x=2$\r\n\r\nall the solutions are {2,2,2}", "Solution_2": "[quote=\"mathubu\"]Here is my solution :\n\nWithout loss of generalization we may assume that; $z\\le y\\le x$ then,\n\n[/quote]\r\n\r\nit's not right because the system isn't symmetry", "Solution_3": "Yes you are probably right but we can do same things for other cases ;\r\n\r\n\u0130t is enaugh to look at $x\\ge z\\ge y$ and $y\\ge x\\ge z$ (we have already looked at $x\\ge y\\ge z$)\r\n\r\n$x\\ge z\\ge y$ ;\r\n\r\n$x^{2}=(y-1)(z+2)\\ge y^{2}=(z-1)(x+2)$ but $z-1\\ge y-1$ and $x+2\\ge z+2$ so $x=y=z$\r\n\r\n$y\\ge x\\ge z$;\r\n\r\n$y^{2}=(z-1)(x+2)\\ge z^{2}=(x-1)(y+2)$ but $x-1\\ge z-1$ and $y+2\\ge x+2$ so $x=y=z$\r\n\r\nother cases are symetric with that cases \r\n\r\n\r\nif we use $x=y=z$ on the equations we can get $x^{2}=(x-1)(x+2)$ then $x=2$\r\n\r\nall the solutions are {2,2,2}" } { "Tag": [ "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "Let a sequence $ $ for $ n \\ge 0$ be defined by $ a_{n\\plus{}1} \\equal{} 2a_{n} \\plus{} 1$. Prove that there exists no $ a_{0}$ such that the sequence contains only primes .", "Solution_1": "[quote=\"srinath.r\"]Let a sequence $ < a_{n} >$ for $ n \\ge 0$ be defined by $ a_{n \\plus{} 1} \\equal{} 2a_{n} \\plus{} 1$. Prove that there exists no $ a_{0}$ such that the sequence contains only primes .[/quote]\r\n\r\nIf $ a_0\\equal{}1$, then $ a_3\\equal{}15$ is not prime\r\n\r\nIf $ a_0\\equal{}2$, then $ a_5\\equal{}95$ is not prime\r\n\r\nIf $ a_0>2$, then $ a_0$ must be an odd prime (since the sequence contains only primes) and then $ a_{a_0\\minus{}1}\\equal{}2^{a_0\\minus{}1}a_0\\plus{}2^{a_0\\minus{}1}\\minus{}1\\equal{}0\\pmod {a_0}$" } { "Tag": [], "Problem": "How many segments are determined by 4 points on a line?\n\n[asy]size(10cm,10cm);\n\ndraw((0,0)--(20,0),linewidth(2bp), Arrows(20bp));\n\ndot((3,0),linewidth(6bp));\n\ndot((8,0),linewidth(6bp));\n\ndot((13,0),linewidth(6bp));\n\ndot((17,0),linewidth(6bp));[/asy]", "Solution_1": "Wouldn't it be 5? You can even count the segments on the diagram. However, the answer on the review is 6...glitch or am I missing something?", "Solution_2": "[quote=\"isabella2296\"]Wouldn't it be 5? You can even count the segments on the diagram. However, the answer on the review is 6...glitch or am I missing something?[/quote]\r\n\r\nAssume they are all equidistent with length 1. There are 3 of length 1, 2 of length 2, and 1 of length 3. \r\n\r\nOR\r\n\r\nCall the points A,B,C,D. The distinct orderings of choosing 2 of these is 4C2=6. (AB,AC,AD,BC,BD,CD)", "Solution_3": "[hide=Solution]You only need to do $\\binom{4}{2}$ which is 12/2 which is $6$.[/hide]", "Solution_4": "[quote=awesomeonfire][hide=Solution]You only need to do $\\binom{4}{2}$ which is $6$[/hide][/quote] \nFTFY\n\n@below, you are welcome", "Solution_5": "Thanks smartninja2000. I edited it so it looks like that." } { "Tag": [], "Problem": "\u4ed6\u662f\u6211\u7684\u6570\u5b66\u8001\u5e08\uff0c\u8bb2\u8bfe\u5f88\u597d\r\n\r\n\u5509\u3002\u3002\u53ef\u60dc\u5f53\u65f6\u3002\u3002\u3002\u9519\u8fc7\u673a\u4f1a\r\n\r\n\u4f46\u662f\u73b0\u5728\u770b\u770b\uff0c\u5c3d\u7ba1\u5f53\u65f6\u6ca1\u6709\u52aa\u529b\uff0c\u4f46\u662f\u5374\u4ece\u738b\u8001\u5e08\u8eab\u4e0a\u6536\u76ca\u532a\u6d45\u3002\r\n\r\n\u8fd9\u91cc\u6709\u4ed6\u7684\u5b66\u751f\u5417\uff1f\r\n\uff08\u95ee\u95ee\u800c\u5df2\uff0c\u4e0d\u8981\u4ecb\u610f\uff09", "Solution_1": "\u4ecb\u7ecd\u4e00\u4e0b\u4ed6\u5427...\u4ed6\u662f\u5728\u4e2d\u56fd\u6559\u4e66\u4e48?? \u5728\u54ea\u4e2a\u57ce\u5e02\u7684\u54ea\u4e2a\u5b66\u6821??", "Solution_2": "\u4ed6\u662f\u4e2d\u56fd\u6570\u5b66\u5965\u6797\u5339\u514b\u96c6\u8bad\u961f\u603b\u6559\u7ec3\u4e4b\u4e00\uff0c\u73b0\u4efb\u6559\u4e8e\u5929\u6d25\u5e02\u5b9e\u9a8c\u4e2d\u5b66\u3002\r\n\u6211\u60f3\u5e94\u8be5\u6709\u5f88\u591a\u4eba\u8ba4\u8bc6\u4ed6\u554a\uff1f\u5982\u679c\u5728\u4e2d\u56fd\u5b66\u6570\u5b66\u5965\u6797\u5339\u514b\u7684\u8bdd\uff1f", "Solution_3": "\u4e0d\u6e05\u695a...\r\n\u4e5f\u8bb8liyi\u6216\u662fWpolly\u77e5\u9053.", "Solution_4": "\u6211\u5728\u4e2d\u7b49\u6570\u5b66\u4e0a\u597d\u591a\u6b21\u770b\u8fc7\u4ed6\u7684\u540d\u5b57", "Solution_5": "Exactly!\r\n\u4ed6\u662f\u4e2d\u7b49\u6570\u5b66\u4e3b\u7f16\u4e4b\u4e00\u3002\r\n\u54c8\u54c8\u54c8\r\n\u5927\u5bb6\u5e94\u8be5\u641c\u7d22\u4e00\u4e0b\u770b\u6709\u6ca1\u6709\u4ed6\u7684\u7167\u7247-- \u4e00\u770b\u5c31\u662f\u5f88\u641e\u7b11\u7684\u8001\u5e08\uff01", "Solution_6": "\u5929\u6d25\u7684\uff0c\u5475\u5475\uff0c\u548c\u6211\u4e00\u6837\u3002\r\n\u5b9e\u9a8c\u4e2d\u5b66\u7684\u6821\u957f\uff0c\u73b0\u5728\u597d\u50cf\u8bf7\u4e0d\u52a8\u4e86\u3002", "Solution_7": "\u6069...\u6211\u5bd2\u5047\u4e70\u4e86\u4e00\u672c <\u5965\u6797\u5339\u514b\u8f9e\u5178 \u51e0\u4f55\u5377>. \u524d\u51e0\u5929\u7ffb\u770b\u65f6\u7a81\u7136\u53d1\u73b0, \u90a3\u4e00\u672c\u7684\u4e3b\u7f16\u7adf\u7136\u5c31\u662f \u738b\u8fde\u7b11! \u592a\u5f3a\u4e86,\u90a3\u672c\u4e66\u8ddfharry potter \u4e00\u6837\u539a......" } { "Tag": [ "geometry", "angle bisector", "perpendicular bisector" ], "Problem": "The triangle ABC has AB and AC unequal. The angle bisector of A meets the perpendicular bisector of BC at X. The line joining the feet of the perpendiculars from X to AB and AC meets BC at D. Show that BD = DC.", "Solution_1": "projection X on AC = N\r\nprojection X on AB = M\r\n\r\nBMD has same shape as DNC (same edges)\r\nBX = CX (it's on the perp. bis.)\r\n\r\ndraw line through B parallel with AC, name E intersection with NM\r\nnow BM = NC (congruence BMX - NCX, follows from congruence AMX - ANX)\r\nas A is bisector of CAB it's also bisector of EBM\r\nand thus EB = BM\r\n\r\nSo NC = BE and the triangles are equishaped, thus they are congruent, thus BD = CD", "Solution_2": "[b]Simson line[/b]...\n\nBest regards,\nsunken rock" } { "Tag": [ "trigonometry" ], "Problem": "cos(x/16)^-1 + cos(-11/24) ^-1 = 180 - cos(x/12) ^-1\r\nwhat's x?", "Solution_1": "[quote=\"cdymdcool\"]cos(x/16)^-1 + cos(-11/24) ^-1 = 180 - cos(x/12) ^-1\nwhat's x?[/quote]\r\n\r\nDo you mean : $ \\cos ^{\\minus{}1} \\frac{x}{16} \\plus{} \\cos ^{\\minus{}1} \\frac{\\minus{}11}{24} \\equal{} 180 \\minus{} \\cos ^{\\minus{}1} \\frac{x}{12}$ ?", "Solution_2": "[quote=\"mathson\"][quote=\"cdymdcool\"]cos(x/16)^-1 + cos(-11/24) ^-1 = 180 - cos(x/12) ^-1\nwhat's x?[/quote]\n\nDo you mean : $ \\cos ^{ \\minus{} 1} \\frac {x}{16} \\plus{} \\cos ^{ \\minus{} 1} \\frac { \\minus{} 11}{24} \\equal{} 180 \\minus{} \\cos ^{ \\minus{} 1} \\frac {x}{12}$ ?[/quote]\r\n\r\nYep", "Solution_3": "[quote=\"cdymdcool\"][quote=\"mathson\"][quote=\"cdymdcool\"]cos(x/16)^-1 + cos(-11/24) ^-1 = 180 - cos(x/12) ^-1\nwhat's x?[/quote]\n\nDo you mean : $ \\cos ^{ \\minus{} 1} \\frac {x}{16} \\plus{} \\cos ^{ \\minus{} 1} \\frac { \\minus{} 11}{24} \\equal{} 180 \\minus{} \\cos ^{ \\minus{} 1} \\frac {x}{12}$ ?[/quote]\n\nYep[/quote]\r\n\r\n[hide=\" hint\"] take $ \\cos$ of both sides. note: $ \\sin \\cos ^{\\minus{}1} x \\equal{} \\sqrt{1\\minus{}x^2}$ .[/hide]", "Solution_4": "Taking the cosine of both sides and using the facts\r\n\r\n$ \\cos(\\alpha\\pm\\beta)\\equal{}\\cos\\alpha\\cos\\beta\\mp\\sin\\alpha\\sin\\beta$ and\r\n\r\n$ \\sin(\\cos^{\\minus{}1}\\alpha)\\equal{}\\sqrt{1\\minus{}\\alpha^2}$, we see that\r\n\r\n$ \\frac x{16}\\cdot\\minus{}\\frac{11}{24}\\minus{}\\sqrt{1\\minus{}\\frac{x^2}{256}}\\cdot\\sqrt{1\\minus{}\\frac{121}{576}}\\equal{}\\minus{}1\\cdot\\frac x{12}\\minus{}0\\cdot\\text{something}$\r\n\r\n$ \\implies\\sqrt{116480\\minus{}455x^2}\\equal{}21x$\r\n\r\n$ \\implies116480\\minus{}455x^2\\equal{}441x^2$\r\n\r\n$ \\implies130\\equal{}x^2$\r\n\r\n$ \\implies x\\equal{}\\boxed{\\sqrt{130}}$.", "Solution_5": "[quote=\"mathson\"][/quote][quote=\"cdymdcool\"][quote=\"mathson\"][quote=\"cdymdcool\"]cos(x/16)^-1 + cos(-11/24) ^-1 = 180 - cos(x/12) ^-1\nwhat's x?[/quote]\n\nDo you mean : $ \\cos ^{ - 1} \\frac {x}{16} + \\cos ^{ - 1} \\frac { - 11}{24} = 180 - \\cos ^{ - 1} \\frac {x}{12}$ ?[/quote]\n\nYep[/quote][quote=\"mathson\"]\n\n[hide=\" hint\"] take $ \\cos$ of both sides. note: $ \\sin \\cos ^{ - 1} x = \\sqrt {1 - x^2}$ .[/hide][/quote]\r\n\r\nHow do you derive that $ \\sin \\cos ^{ - 1} x = \\sqrt {1 - x^2}$", "Solution_6": "[quote=\"cdymdcool\"][/quote][quote=\"mathson\"][/quote][quote=\"cdymdcool\"][quote=\"mathson\"][quote=\"cdymdcool\"]cos(x/16)^-1 + cos(-11/24) ^-1 = 180 - cos(x/12) ^-1\nwhat's x?[/quote]\n\nDo you mean : $ \\cos ^{ - 1} \\frac {x}{16} + \\cos ^{ - 1} \\frac { - 11}{24} = 180 - \\cos ^{ - 1} \\frac {x}{12}$ ?[/quote]\n\nYep[/quote][quote=\"mathson\"]\n\n[hide=\" hint\"] take $ \\cos$ of both sides. note: $ \\sin \\cos ^{ - 1} x = \\sqrt {1 - x^2}$ .[/hide][/quote][quote=\"cdymdcool\"]\n\nHow do you derive that $ \\sin \\cos ^{ - 1} x = \\sqrt {1 - x^2}$[/quote]\r\n\r\nconsider that we have triangle $ ABC$ such that $ \\angle A = 90^\\circ$ , $ AB=x$ and $ BC = 1$.\r\nthen, $ AC = \\sqrt{1-x^2}$ .\r\nnow, $ \\cos B = \\frac{x}{1} = x \\Longrightarrow B = \\cos ^{-1} x$.\r\nof course:\r\n$ \\sin B = \\frac{\\sqrt {1-x^2}}{1} = \\sqrt{1-x^2} \\Longrightarrow \\sin (\\cos^{-1} x) = \\sqrt{1-x^2}$\r\nhope you understand :)" } { "Tag": [ "Support" ], "Problem": "For some reason, I only got 1 point for a question. Why?", "Solution_1": "Some questions are worth 1, some 2, some 3, and some 4. It depends on level of difficulty.", "Solution_2": "for some reason, the same thing is happening to me.\r\n\r\nIt says that the problem is worth 2 points, and i'm usually only getting one.", "Solution_3": "report it to the staff\r\n\r\nthere is a report bug button\r\n\r\nstate what is your problem\r\n\r\nthey probobaly can help u", "Solution_4": "set up a PM with the admin,\r\nmight be a bug", "Solution_5": "[quote=\"andrewjjiang\"]For some reason, I only got 1 point for a question. Why?[/quote]\n\nMinimum is 5 XP, which you get for answering the same problem twice or a Lv. 1 problem", "Solution_6": "please dont bring up a 2 year old post :(" } { "Tag": [ "function", "integration", "limit", "calculus", "real analysis", "real analysis unsolved" ], "Problem": "[color=darkblue][i]let $ f$ and $ g$ two continuous functions defined from $ [0,1]$ to $ (0,\\infty)$. For every $ n\\in\\mathbb{N}$ we consider the sequence $ u_n\\equal{}\\int_{0}^{1}g(x)f(x)^{n}dx$.\nDetermine whether the sequence $ \\left(\\frac{u_{n\\plus{}1}}{u_{n}}\\right)_{n\\in\\mathbb{N}}$ converges or diverges.[/i][/color]", "Solution_1": "[color=darkred]Obviously,$ f$ and $ g$ are positives functions, therefore $ u_{n}$ is a sequence of positive reals.\nwe put,for every $ n\\in\\mathbb{N}$,$ v_n = \\frac {u_{n + 1}}{u_{n}}$.\nWe have : $ u_{n + 1} = \\int_{0}^{1}g(x)f(x)^{n + 1}dx = \\int_{0}^{1}\\left(g(x)^{\\frac {1}{2}} f(x)^{\\frac {n + 2}{2}\\right)\\left(g(x)^{\\frac {1}{2}}f(x)^{\\frac {n}{2}\\right)dx}}$.\n\nBy cauchy-schwartz we obtain :\n$ \\left(u_{n + 1}\\right)^{2}\\leq\\int_{0}^{1}g(x)f(x)^{n + 2}dx\\int_{0}^{1}g(x)f(x)^{n}dx$\n\nTherefore $ v_{n}$ is an decreasing sequence.\nWe put $ \\mathbb{M} = \\sup\\ f(x)$ with ${ x\\in\\ [0,1]}$.fot every $ n\\in\\mathbb{N}$ and for every $ x\\in[0,1]$ we have $ \\int_{0}^{1}g(x)f(x)^{n + 1}\\leq\\mathbb{M}\\int_{0}^{1}g(x)f(x)^{n}dx$ and so $ v_{n}\\leq \\mathbb{M}$.hence $ \\left(v_{n }\\right)_{n\\in\\mathbb{N}}$ is convergent.\n\nNB:we can also prove that $ \\mathbb{M}$ is the limite of $ v_{n}$.[/color]\r\n\r\n-------------------------------------------------------------------------------------------------------------------------------------------------\r\nradouane.", "Solution_2": "[quote=\"open-mind\"] ...\nTherefore $ v_{n}$ is an decreasing sequence.\n... [/quote]\r\n... increasing...", "Solution_3": "I confess that I'm not convinced by open-mind's proof. In particular, he has not shown that $ v_n=\\frac{u_{n+1}}{u_n}$ is either increasing or decreasing. Let me start over.\r\n\r\nLet $ M=\\max_{x\\in[0,1]}f(x).$ (I'm saying max instead of sup since $ f$ is continuous.)\r\n\r\nIt is straightforward that\r\n\r\n$ u_{n+1}=\\int_0^1f(x)^{n+1}=\\int_0^1f(x)f(x)^n\\le\\int_0^1Mf(x)^n=Mu_{n+1}.$\r\n\r\nIt's the other direction that we need to work on.\r\n\r\nChoose an $ \\epsilon>0,$ ($ \\epsilon < M$) and let $ E_{\\epsilon}=\\{x: f(x)\\ge M-\\epsilon\\}.$\r\n\r\nNow write $ u_n=a_n+b_n$ where $ a_n=\\int_{E_{\\epsilon}}g\\,f^n$ and $ b_n=\\int_{[0,1]\\setminus E_{\\epsilon}}g\\,f^n.$\r\n\r\nWe have ${ a_n\\ge (M-\\epsilon)^n\\int_{E_{\\epsilon}}g}.$ Let $ c_{\\epsilon}=\\int_{E_{\\epsilon}}g.$ All we really need to know is that $ c_{\\epsilon}>0.$ We also have, and will use, that $ a_n\\le M^nc_{\\epsilon}.$\r\n\r\nNow look at $ \\lim_{n\\to\\infty}\\frac1{(M-\\epsilon)^n}b_n=\\lim_{n\\to\\infty}\\int_{[0,1]\\setminus E_{\\epsilon}}g\\left(\\frac{f}{M-\\epsilon}\\right)^n.$\r\n\r\nThe functions in that last integral tend pointwise to zero and are bounded by $ g,$ an integrable function. Hence by the DCT, that limit is zero. What we have then is that\r\n\r\n$ \\lim_{n\\to\\infty}\\frac{b_n}{a_n}\\le\\lim_{n\\to\\infty}\\frac{b_n}{(M-\\epsilon)^nc_{\\epsilon}}=0.$\r\n\r\nHence,\r\n\r\n$ \\liminf_{n\\to\\infty}\\frac{u_{n+1}}{u_n}=\\liminf_{n\\to\\infty}\\frac{a_{n+1}+b_{n+1}}{a_n+b_n}=\\liminf_{n\\to\\infty}\\frac{a_{n+1}}{a_n}$\r\n\r\n$ \\ge\\liminf_{n\\to\\infty}\\frac{(M-\\epsilon)^{n+1}c_{\\epsilon}}{M^nc_{\\epsilon}}=(M-\\epsilon)\\left(\\frac{M-\\epsilon}M\\right)^n.$\r\n\r\nWhoops - that's not a convincing argument. Let's push on, staying with the $ a_n.$\r\n\r\n$ a_{n+1}=\\int_{E_{\\epsilon}}g\\,f^{n+1}\\ge (M-\\epsilon)\\int_{E_{\\epsilon}}g\\,f^n=(M-\\epsilon)a_n.$\r\n\r\nSo $ \\frac{a_{n+1}}{a_n}\\ge (M-\\epsilon).$\r\n\r\nAhh - now we have an argument.\r\n\r\nAt this point, I should apologize for the stream-of-consciousness argument. That's what happens when you start writing without having all of the details worked out. I really should edit - but I think I won't, not right now. If you put all of that together, we do get that $ \\frac{u_{n+1}}{u_n}$ tends to $ M.$", "Solution_4": "His proof of the sequence $ (v_n)_{n\\in\\mathbb{N}}$ being nondecreasing [i]is[/i] valid.", "Solution_5": "Ah - I finally got it. $ u_{n \\plus{} 1}^2\\le u_nu_{n \\plus{} 2}$ so $ \\frac {u_{n \\plus{} 2}}{u_{n \\plus{} 1}}\\ge\\frac {u_{n \\plus{} 1}}{u_n}.$", "Solution_6": "sorry if I was not very clear.Here I present my argument for the limit. It is based on a lemma of Cesaro theorem.\r\n-------------------------------------------------------------------------------------------------------------------------------------------------\r\n[color=red][i]let $ (u_{n})_{n\\in\\mathbb{N}}$ a sequence of positive reals.If the sequence $ \\left(\\frac{u_{n+1}}{u_{n}}\\right)_{n\\in\\mathbb{N}}$ converges,so the sequence $ (u_{n}^{\\frac{1}{n}})_{n\\in\\mathbb{N}}$ also converges and has the same limite.[/i][/color]\r\n-------------------------------------------------------------------------------------------------------------------------------------------------\r\n\r\n[color=darkblue][i]Since the function f is continuous on the compact $ [0,1]$,there exist $ l\\in\\ [0,1]$ such that $ f(l)=M$.\n\nlet $ \\epsilon>0$,we can find an interval $ [a,b]$ in $ [0,1]$ such that for every $ x\\in\\ [a,b]$ we have $ M-\\epsilon\\leq\\ f(x)\\leq\\ M$.\n\nfor every $ n\\in\\mathbb{N}$: $ \\int_{0}^{1}g(x)f(x)^{n}dx\\ge\\int_{a}^{b}g(x)(M-\\epsilon)^{n}dx\\ge\\ (M-\\epsilon)^{n}\\int_{a}^{b}g(x)dx.$\n\nand $ \\int_{0}^{1}g(x)f(x)^{n}dx\\leq\\ M^{n}\\int_{0}^{1}g(x)dx.$\n\nwe obtain as a result: \n\n$ (M-\\epsilon)\\left(\\int_{a}^{b}g(x)dx\\right)^{\\frac{1}{n}}\\leq\\ u_{n}^{\\frac{1}{n}}\\leq\\ M\\left(\\int_{0}^{1}g(x)dx\\right)^{\\frac{1}{n}}$\n\nOr we know that :\n\n$ \\lim_{n\\to\\infty}\\ (M-\\epsilon)\\left(\\int_{a}^{b}g(x)dx\\right)^{\\frac{1}{n}}=\\ M-\\epsilon$ and $ \\lim_{n\\to\\infty}\\ M\\left(\\int_{0}^{1}g(x)dx\\right)^{\\frac{1}{n}}= M$\n\nwe deduce that there is a natural $ n_{0}$ such that for every naturel number $ n\\ge\\ n_{0}$: \n\n$ M-2\\epsilon\\leq\\ u_{n}^{\\frac{1}{n}}\\leq\\ M+2\\epsilon$\n\nwe have therefore shown that $ \\lim_{n\\to\\infty}\\ u_{n}^{\\frac{1}{n}}= M$. we deduce that the limite of $ v_{n}$ is $ M$.[/i][/color]\r\n\r\n\r\n[u][i][color=darkblue]hoping that there is no stupidity in my proof.[/color][/i][/u]\r\n-------------------------------------------------------------------------------------------------------------------------------------------------\r\nRadouane." } { "Tag": [], "Problem": "How do I prove that $7^{5k}-5^{7j}=12007$ is false $\\forall k,j \\in \\mathbb{Z}$ ?", "Solution_1": "Mod 6? 1+(-1)^{stuff} can't be 1.", "Solution_2": "My mistake, this isn't the one I wanted. This one is obviously false." } { "Tag": [ "algebra", "polynomial", "inequalities unsolved", "inequalities" ], "Problem": "Let $ P(x)\\equal{}ax^3\\plus{}bx^2\\plus{}cx\\plus{}d$, be a polynomial which satisfy the following condition:\r\n\r\n$ |P(x)|\\leq 1$ for every $ x$ with $ |x|\\leq 1$.\r\n\r\nProve that $ |a|\\plus{}|b|\\plus{}|c|\\plus{}|d|\\leq 7$\r\n\r\nAlexandros", "Solution_1": "[quote=\"cretanman\"]Let $ P(x) \\equal{} ax^3 \\plus{} bx^2 \\plus{} cx \\plus{} d$, be a polynomial which satisfy the following condition:\n\n$ |P(x)|\\leq 1$ for every $ x$ with $ |x|\\leq 1$.\n\nProve that $ |a| \\plus{} |b| \\plus{} |c| \\plus{} |d|\\leq 7$\n\nAlexandros[/quote]\r\nhttp://www.mathlinks.ro/viewtopic.php?t=193688", "Solution_2": "[quote=\"chien than\"]\nhttp://www.mathlinks.ro/viewtopic.php?t=193688[/quote]\r\n\r\nThere is no correct answer in this topic. So the problem is still unsolved!\r\n\r\nAlexandros" } { "Tag": [], "Problem": "Find the $n$ th term of the sequence $\\{a_{n}\\}$ such that $a_{1}=0,\\ a_{2}=1,\\ (n-1)^{2}a_{n}=\\sum_{k=1}^{n}a_{k}\\ \\ (n\\geq 1).$", "Solution_1": "[hide=\"Solution\"]For $n\\geqslant 3$\n\n$(n-1)^{2}a_{n}-(n-2)^{2}a_{n-1}=a_{n}\\iff n(n-2)a_{n}-(n-2)^{2}a_{n-1}=0$\n\nSince $n-2\\neq 0$, this gives\n\n$na_{n}=(n-2)a_{n-1}$\n\nThen\n\n\\begin{eqnarray*}na_{n}&=& (n-2)a_{n-1}\\\\ (n-1)a_{n-1}&=& (n-3)a_{n-2}\\\\ &\\vdots& \\\\ 3a_{3}&=& 1\\cdot a_{2}\\end{eqnarray*}\n\nMultiply all the equations and denote $P=a_{3}a_{4}\\dots a_{n-1}$:\n\n$\\frac{n!}{2!}Pa_{n}=(n-2)!Pa_{2}\\iff a_{n}=\\frac{2a_{2}}{n(n-1)}=\\frac{2}{n(n-1)}$\n\nwhich also works for $n=2$, hence\n\nHence $a_{1}=0, a_{n}=\\frac{2}{n(n-1)}, n\\geqslant 2$[/hide]", "Solution_2": "Splendid! :)", "Solution_3": "If $n>1$, $a_{n+1}=n^{2}a_{n+1}-(n-1^{2})a_{n}$ $\\to$\r\n$a_{n+1}=\\frac{(n-1)^{2}}{n^{2}-1}a_{n}=\\frac{n-1}{n+1}a_{n}=\\frac{\\frac{(n-1)!}{0!}}{\\frac{(n+1)!}{2!}}a_{2}=\\frac{1}{\\binom{n+1}{2}}$.\r\nTherefore, for $n>1$, $a_{n}=\\frac{1}{\\binom{n}{2}}$.", "Solution_4": "For $n\\geq 3,\\ na_{n}=(n-2)a_{n-1}\\Longleftrightarrow n(n-1)a_{n}=(n-1)(n-2)a_{n-1}$\r\n\r\n$=\\cdots =2\\cdot 1\\cdot a_{2}=2,$ yielding $a_{n}=\\frac{2}{n(n-1)}\\ (n\\geq 2).$" } { "Tag": [ "MATHCOUNTS" ], "Problem": "why can we only do 3 years of mathcounts total? :o \r\nim in 5th grade, and i rly want to do it this year(i'm allowed to)\r\ni can only do it 5th-7th then :mad:", "Solution_1": "[quote=\"program4\"]why can we only do 3 years of mathcounts total? :o \nim in 5th grade, and i rly want to do it this year(i'm allowed to)\ni can only do it 5th-7th then :mad:[/quote]\r\n\r\nwell, i dont think you can do it this year OFFICIAlly..so you can do it unofficially", "Solution_2": "Maybe cuz you don't get too familiarized with the format.", "Solution_3": "Well, say a random 5-year old comes and starts Mathcounts. Well, imagine how beast they'll be in 8th grade. :wink: Plus, Mathcounts is as much a mathematical experience as a social one.", "Solution_4": "You can not compete in MATHCOUNTS as a 5th grader. If your school allows you to, then they are breaking the rules as would you be. Wait until 6th grade and compete through 8th grade. There are younger kids who skip grades and compete in MATHCOUNTS, but you definitely only get 3 years to compete. When I started coaching you only got 2 years and now 6th graders were allowed, so be glad that rule has changed.", "Solution_5": "I did take the chapter test as a 5th grader unofficially, 2 of my friends and I formed a \"team\" I think my parents worked it out with the coordinators before hand or something, I don't really remember." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Find all $ f: R\\rightarrow R$ such that\r\n$ f(x \\plus{} y \\plus{} f(xy)) \\equal{} f(f(x \\plus{} y)) \\plus{} xy$\r\nhelp me plz :lol:", "Solution_1": "Looks like $ f(x)\\equal{}x\\plus{}f(0)$ is a solution ;)\r\n\r\nIn our equation put $ y\\equal{}0$: $ f(x\\plus{}f(0))\\equal{}f(f(x))$. If $ f(x)$ is injection then $ f(x)\\equal{}x\\plus{}f(0)$ and this is a solution. So, we should prove that $ f(x)$ is injection.\r\nPut $ y\\equal{}\\minus{}x$: $ f(f(\\minus{}x^2))\\equal{}f(f(0))\\minus{}x^2$. And let $ z\\equal{}x^2\\geq 0$\r\n\r\n$ f(f(\\minus{}z_1))\\minus{}f(f(\\minus{}z_2))\\equal{}\\minus{}z_1\\plus{}z_2$\r\n\r\nAnd if $ f(\\minus{}z_1)\\equal{}f(\\minus{}z_2)$ we get that $ z_1\\equal{}z_2$.", "Solution_2": "Was that supposed to be an incomplete solution or did I miss you proving that $ f$ is injective even if at least one of $ z_1, z_2$ is negative?", "Solution_3": "Yes, it is incomplete. Now I don't know how to prove that $ f(x)$ is injection for $ x>0$.", "Solution_4": "Can i do it like this?? :maybe: Anyone tell me plz :blush: \r\nSubstitute y=0 \r\n$ f(x \\plus{} f(0)) \\equal{} f(f(x))$\r\nLet\r\n$ x \\plus{} y \\equal{} a$ \r\n$ xy \\equal{} b$\r\nthen $ f(a \\plus{} f(b)) \\equal{} ff(a) \\plus{} b$ which $ a^2\\geqslant 4b$\r\nI will prove that f(x) is 1-1\r\nLet $ f(x_1) \\equal{} f(x_2)$ i will prove that $ x_1 \\equal{} x_2$\r\nSee that i can choose \"a\" which $ a\\geqslant \\sqrt {4max(|x_1| ,|x_2|)}$ \r\n$ f(a \\plus{} f(x_1)) \\equal{} f(a \\plus{} f(x_2))$\r\nthen $ ff(a) \\plus{} x_1 \\equal{} ff(a) \\plus{} x_2$\r\nturn to $ x_1 \\equal{} x_2$ \r\nso f(x) is 1-1 \r\nthen i got that $ f(x) \\equal{} x \\plus{} f(0)$", "Solution_5": "[quote=\"Rose_joker\"]Can i do it like this?? :maybe: Anyone tell me plz :blush: \nSubstitute y=0 \n$ f(x \\plus{} f(0)) \\equal{} f(f(x))$\nLet\n$ x \\plus{} y \\equal{} a$ \n$ xy \\equal{} b$\nthen $ f(a \\plus{} f(b)) \\equal{} ff(a) \\plus{} b$ which $ a^2\\geqslant 4b$\nI will prove that f(x) is 1-1\nLet $ f(x_1) \\equal{} f(x_2)$ i will prove that $ x_1 \\equal{} x_2$\nSee that i can choose \"a\" which $ a\\geqslant \\sqrt {4max(|x_1| ,|x_2|)}$ \n$ f(a \\plus{} f(x_1)) \\equal{} f(a \\plus{} f(x_2))$\nthen $ ff(a) \\plus{} x_1 \\equal{} ff(a) \\plus{} x_2$\nturn to $ x_1 \\equal{} x_2$ \nso f(x) is 1-1 \nthen i got that $ f(x) \\equal{} x \\plus{} f(0)$[/quote]\r\nI think it's correct." } { "Tag": [ "geometry", "trapezoid", "geometry proposed" ], "Problem": "On a circle with diameter $ AB$ choose point $ C,D,E$ on one side of $ AB$ and $ F$ on the other side,such that $ \\widehat {AOC}\\equal{}\\widehat {COD}\\equal{}\\widehat {BOE}\\equal{}20^o$ and $ \\widehat {BOF}\\equal{}60^o$.\r\nLet $ M$ be intersection of $ BD$ and $ CE$,prove that $ FM\\equal{}FE$", "Solution_1": "[quote=\"stephen38\"]On a circle with diameter $ AB$ choose point $ C,D,E$ on one side of $ AB$ and $ F$ on the other side,such that $ \\widehat {AOC} \\equal{} \\widehat {COD} \\equal{} \\widehat {BOE} \\equal{} 20^o$ and $ \\widehat {BOF} \\equal{} 60^o$.\nLet $ M$ be intersection of $ BD$ and $ CE$,prove that $ FM \\equal{} FE$[/quote]\r\n[hide=\"Solution\"]\nNotice that $ \\angle COE \\equal{} 180 \\minus{} \\angle AOC \\minus{} \\angle BOE \\equal{} 140$ and so $ \\angle OEC \\equal{} 90 \\minus{} \\frac {\\angle EOC}{2} \\equal{} 20 \\equal{} \\angle EOB$, so $ EC\\parallel AB$. Also, $ 2\\angle DBA \\equal{} \\angle ODB \\plus{} \\angle OBD \\equal{} \\angle AOD \\equal{} \\angle AOC \\plus{} \\angle COD \\equal{} 40$, so $ \\angle MBO \\equal{} 20 \\equal{} \\angle MEO$. This means that $ MEBO$ is cyclic with $ ME\\parallel OB$, so $ MEBO$ is an isoceles trapezoid, which implies that $ EB \\equal{} MO$. Also, $ \\angle FBO \\equal{} 90 \\minus{} \\frac {\\angle FOB}{2} \\equal{} 60 \\equal{} \\angle FOB$, so $ OFB$ is equilateral, meaning that $ OF \\equal{} FB$. Now, since $ MEBO$ is an isoceles trapezoid, we have that $ \\angle MOB \\equal{} \\angle EBO$ and also, $ \\angle FOB \\equal{} 60 \\equal{} \\angle FBO$, so $ \\angle MOF \\equal{} \\angle MOB \\plus{} \\angle FOB \\equal{} \\angle FBO \\plus{} \\angle EBO \\equal{} \\angle EBF$. Thus, $ \\triangle FOM\\cong \\triangle FBE\\implies FM \\equal{} FE$. [/hide]" } { "Tag": [ "geometry", "algebra unsolved", "algebra" ], "Problem": "Determine the number of solutions x \\in [0,10] of (sin(x))^2 = sin(x^2).", "Solution_1": "x \\in [0,10] of (sin(x))^2 = sin(x^2).\r\n if x\\in [1, \\sqrt 90]then sin(x^2)>sinx>(sinx)^2\r\n ifx\\in( \\sqrt 90,10] then sin (x^2)=sin(180-x^2)>(sinx)^2 >sin x \r\n then x\\in[0,1) then x=0", "Solution_2": "I also solved this problem using complex and then geometry but i dont know if it is right, i also got the same answer x=0, is it right, if it is then tell me if you want to see my solution!", "Solution_3": "I also solved this problem using complex and then geometry but i dont know if it is right, i also got the same answer x=0, is it right, if it is then tell me if you want to see my solution!", "Solution_4": "Are you measure angles not in radians???", "Solution_5": "If you measure in radians, I get either 31 or 32." } { "Tag": [ "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "Let G be a graph with V(G)=n, where n is a positive integer. Show that if E(G)>=n, there exists a cycle in G.", "Solution_1": "Try to show by induction, for instance, that any forest (graph s.t. all its connected components are trees) with $k$ connected components and $n$ vertices has precisely $n-k$ edges. In other words, prove that a tree has precisely $n-1$ edges.", "Solution_2": "Suppose the graph $G$ has $n+l$ edges. What is the lower bound for the number of cycles in $G$? :)" } { "Tag": [], "Problem": "Solve the system:\r\n$\\begin{cases}x-\\sqrt{yz}=42\\\\y-\\sqrt{zx}=6\\\\z-\\sqrt{xy}=-30\\end{cases}$", "Solution_1": "See this: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=75725[/url] with $u=\\sqrt{x}, v=\\sqrt{y},w=\\sqrt{z}$" } { "Tag": [ "function", "algebra", "polynomial", "system of equations", "algebra unsolved" ], "Problem": "Solve the system of equations\r\n\r\n$ \\left\\{\\begin{array}{c}{1} x^3-3xy^2=6 \\3x^2y-y^3=8\\end{array}\\right$", "Solution_1": "$ (x^2 \\plus{} y^2)^3 \\equal{} ( x^3 \\minus{} 3xy^2)^2 \\plus{} (3x^2y \\minus{} y^3)^2 \\equal{} 6^2 \\plus{} 8^2 \\equal{} 100$ so $ y^2 \\equal{} (100)^{\\frac {1}{3}} \\minus{} x^2$.\r\nfrom the first equation $ x(x^2 \\minus{} 3y^2) \\equal{} 6$ therefore $ 4x^3 \\minus{} 3x(100)^{\\frac {1}{3}} \\minus{} 6$ , this equation can be solved by cardan method .\r\nsee here for more information. [hide]http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method[/hide]", "Solution_2": "hello, by dividing equation one by equation two we get\r\n$ \\frac{x}{y}\\frac{x^2\\minus{}3y^2}{3x^2\\minus{}y^2}\\equal{}\\frac{3}{4}$. By substituting\r\n$ \\frac{x}{y}\\equal{}t$ we have $ t\\cdot\\frac{t^2\\minus{}3}{3t^2\\minus{}4}\\equal{}\\frac{3}{4}$ which\r\nis equivalent with $ 4t^3\\minus{}9t^2\\minus{}12t\\plus{}3\\equal{}0$. Now you must solve this polynomial\r\nwith degree 3. \r\nSonnhard." } { "Tag": [], "Problem": "If you are still studying math, where do you see yourself 20 years from now?\r\nWill you be a teacher, dealing with mathematical research or leave math for some financially better job?\r\nIs math a hobby or a full time job?", "Solution_1": "I'll try to answer this question:\r\nIs math a hobby or a full time job?\r\n\r\nMany people turn their hobbies into full time jobs and that is in my opinion good because you do something that you like and you are not \"forced\" to do the same day the same thing which you don't like .. over and over again... hobbies should be turned into full time jobs.. I mean that this hobby can be turned into a full time job, not like stamp collection or stuff like that, isn't it? :? what do you think?\r\n\r\ncheers! :D", "Solution_2": "In 20 years from now I see myself doing smth that involves mathematical research.\r\nHowever since Romania (my country) is not a very worm environment for profesional mathematicians, I will probably do smth else for money like private tutoring or informatics. I might start my own business in smth. Or I could emigrate to USA or some other wormer environment :D.\r\nBut I will be a profesional mathematician, but not a teacher.", "Solution_3": "I will see myself as a grand leader\n\nEveryone will bow down to me and pour lava on their heads", "Solution_4": "I want to do mathematical research.\n\nOr scientific research.\n(Because science is described with mathematics.)" } { "Tag": [ "\\/closed" ], "Problem": "I go to the Avatar control panel, and put in the URL box:\r\n\r\nhttp://img372.imageshack.us/my.php?image=whoareyoucallingirratiovy9.gif\r\n\r\nIt tells me that the file must be .j**, .gif, or .png (I don't know what the first and\r\n\r\nlast are)\r\n\r\nBut, isn't this .gif???", "Solution_1": "You've linked to the .php page.\r\n\r\nPut in the direct link:\r\n[url]http://img372.imageshack.us/img372/5410/whoareyoucallingirratiovy9.gif[/url]", "Solution_2": "How do you resize it? It's too big :(", "Solution_3": "Google \"how do I resize an image\"." } { "Tag": [ "algebra", "polynomial" ], "Problem": "If $a+b+c+d=0,$ then prove that $a^3+b^3+c^3+d^3=3(a+d)(b+d)(c+d).$", "Solution_1": "If $a=-d$\r\n\r\n$a^3+b^3+c^3+d^3=0$\r\n\r\nSo $a+d$ is a factor\r\n\r\nSo on for $(b+d)$ and $(c+d)$\r\n\r\nNow next ,I'm not sure how to proceed..\r\n\r\n$a^3+b^3+c^3+d^3\\equiv N(a+d)(b+d)(c+d)$ where $N$ is a constant\r\n\r\nHow do I bring the $=$ sign.....", "Solution_2": "[hide=\"Hint\"]$d=-(a+b+c)$[/hide]", "Solution_3": "This is almost same as shadysaysurspammed's \r\n\r\nConsider $f(x)=x^3+a^3+b^3+c^3$ where $a,b,c\\in%Error. \"mathh\" is a bad command.\n{R}$ and $x=-(a+b+c)$\r\n\r\nWe notice that $f(-a)=f(-b)=f(-c)=0$ , and $f(x)$ is a third degree polynomial , hence\r\n\r\n$f(x)=k(x+a)(x+b)(x+c)$ for some constant $k$ . But setting $a=b=c=1$ and $x=-3$ we can obtain from\r\n\r\n$x^3+a^3+b^3+c^3=k(x+a)(x+b)(x+c)$ $\\implies k=3$\r\n\r\nSo we just let $x=d$ and it becomes $a^3+b^3+c^3+d^3=3(a+d)(b+d)(c+d)$", "Solution_4": "Yes, that's right,shyong. :)\r\n\r\nYour solution shows how this problem is made.", "Solution_5": "[quote=\"shyong\"]This is almost same as shadysaysurspammed's \n\nConsider $f(x)=x^3+a^3+b^3+c^3$ where $a,b,c\\in%Error. \"mathh\" is a bad command.\n{R}$ and $x=-(a+b+c)$\n\nWe notice that $f(-a)=f(-b)=f(-c)=0$ , and $f(x)$ is a third degree polynomial , hence\n\n$f(x)=k(x+a)(x+b)(x+c)$ for some constant $k$ . But setting $a=b=c=1$ and $x=-3$ we can obtain from\n\n$x^3+a^3+b^3+c^3=k(x+a)(x+b)(x+c)$ $\\implies k=3$\n\nSo we just let $x=d$ and it becomes $a^3+b^3+c^3+d^3=3(a+d)(b+d)(c+d)$[/quote]\r\n\r\nI was thinking that you couldn't equate $f(x)=k(x+a)(x+b)(x+c)$ because it wasn't of third degree :blush:" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find all reals $a$ such that the sequence $\\{x(n)\\}$, $n=0,1,2, \\ldots$ that satisfy: $x(0)=1996$ and $x_{n+1} = \\frac{a}{1+x(n)^2}$ for any natural number $n$ has a limit as n goes to infinity.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=3726\r\n\r\nI haven't solved this problem, but I know the answer: $ |a|\\le2,|a| \\equal{} 1997\\sqrt {1996}$ (I got this answer from a book).", "Solution_2": "For the first one, see:\n\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=352944[/url]" } { "Tag": [], "Problem": "Will anyone be attending the Future Problem Solving International Conference at Michigan State University, May 29 - June 1? Right now, I think the grand total of AoPSers going is three, two of whom are from Ohio.", "Solution_1": "We epic failed.\r\n\r\n3rd in middle division.\r\n\r\nOhio.\r\n\r\nWe got T.T'ed by PA and KY.\r\n\r\nHad we thought about our UP and had GSB rigged the grid propelry, we might have won." } { "Tag": [ "vector", "abstract algebra", "analytic geometry", "Functional Analysis", "real analysis", "real analysis unsolved" ], "Problem": "Okay this question has been annoying me for some time now. It is a homework problem so please don't solve it for me, just give me a hint.\r\n\r\nLet M be a finite dimensional subspace of nvs X. Prove there exist a closed vector space N st. X is the direct sum of M and N.\r\n\r\nObviously the finite case is trivial. I am having trouble finishing the general case.\r\n\r\nI have been trying to apply zorn's lemma to this problem, I can't quite get it though. I say consider the collection of closed subspaces of X that have trivial intersection with M, order by inclusion. Obviously if I can show zorn's lemma applies, any maximal element of this collection necessarily solves the problem. My issue then is proving every chain is bounded, which I try to solve by considering the union of all elements of a chain. Clearly this is a subspace that intersects M trivially, I just can't figure out how to show its closed. Or I could take the closure of that set and show it still intersects M trivially. I haven't been able to do either.\r\n\r\nThe other approach I had was similar, except take the collection of closed subspaces M such that M + N = X, and order by reverse inclusion. Then I have no problem showing every chain is necessarily bounded, but I do have an issue with showing that a maximal element of this collection necessarily intersects M trivially!\r\n\r\nAny comments would be appreciated! Thanks in advance", "Solution_1": "Try to construct $ N$ as the intersection of zero sets of some linear functionals.", "Solution_2": "Thanks. I was trying to do this problem without extending the basis of M to X, since the text this problem came out of rarely mentions bases, esp. those of infinite dimensional spaces but I guess that is the only way =/", "Solution_3": "What you just wrote raises a red flag: there are Banach spaces in which there are no bases. Are you sure your solution is correct?", "Solution_4": "One can extend the basis of $ M$ to a Hamel basis of $ X$ and then let $ N$ be the linear span of the added basis vectors... this is not an approach I would take, but it seems to work.", "Solution_5": "If you don't extend to a Hamel basis, how can you generate functionals that are going to be 0 over a \"big enough\" subspace of X such that their kernel intersection is \"everything but M\"? (Obviously not everything, you know what I mean)\r\n\r\nAlso, is it not true that every Banach space has a Hamel basis? Seems like that should fall from an application of Zorn's lemma.", "Solution_6": "If you extend to Hamel's basis, most likely, you'll get [b]discontinuous[/b] coordinate functionals ;)", "Solution_7": "[quote=\"Spacecow\"]...how can you generate functionals that are going to be 0 over a \"big enough\" subspace of X...[/quote]\r\nThe kernel of [b]any[/b] continuous linear functional is a closed subspace of codimension 1; the direct sum of the kernel and the span of any one vector not in the kernel is everything.\r\nWhat you need is a way to choose the values on this finite-dimensional space and then [url=http://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem]extend[/url] to something continuous on the whole space.", "Solution_8": "I know all about the HB theorem, but I never new that the kernel of any continuous linear functional has codim. one. Thanks.\r\n\r\n[quote=\"jmerry\"][quote=\"Spacecow\"]...how can you generate functionals that are going to be 0 over a \"big enough\" subspace of X...[/quote]\nThe kernel of [b]any[/b] continuous linear functional is a closed subspace of codimension 1; the direct sum of the kernel and the span of any one vector not in the kernel is everything.\nWhat you need is a way to choose the values on this finite-dimensional space and then [url=http://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem]extend[/url] to something continuous on the whole space.[/quote]" } { "Tag": [ "induction", "articles" ], "Problem": "When you do a problem that involves a proof of a statement, and you experiment with a few basic examples, which leads you to conjecture something and most likely prove it using induction (..contradiction,ectr.), would part of your proof involve writing down your experimentation and showing how the experimentation leads to a conjecture, or should you just state your conjecture and proceed to prove it?", "Solution_1": "State and prove", "Solution_2": "The following excerpt from the \"How to Write a Solution\" article answers your question:\r\n[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteForward.php]http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteForward.php[/url]\r\n\r\nWriting down your experimentation may be helpful as a learning aid if you're trying to help somebody else understand the process of analyzing the problem and coming up with a solution, but if you're writing a formal proof, then they don't really care how you came across your solution, they only care about the solution." } { "Tag": [ "probability", "percent" ], "Problem": "A parts dealer buys parts from a warehouse. Parts are made by either Company A or Company B but are not identified as to which company produces them. One company produces all parts in one shipment or lot. On the average, we know:\r\n\r\nCompany A produces $2.5\\%$ defective parts.\r\nCompany B produces $5.0\\%$ defective parts.\r\n\r\nThe warehouse states that $70\\%$ of parts will come from Company A and $30\\%$ from\r\nCompany B. If the dealer selects $4$ parts at random from a lot and finds $1$ defective part,\r\n\r\nwhat is the probability that the lot was produced by Company A?", "Solution_1": "Try putting a \\ in front the percent sign so it looks right.", "Solution_2": "[quote=\"xxxyyyy\"]A parts dealer buys parts from a warehouse. Parts are made by either Company A or Company B but are not identified as to which company produces them. One company produces all parts in one shipment or lot. On the average, we know:\n\nCompany A produces $2.5\\%$ defective parts.\nCompany B produces $5.0\\%$ defective parts.\n\nThe warehouse states that $70\\%$ of parts will come from Company A and $30\\%$ from\nCompany B. If the dealer selects $4$ parts at random from a lot and finds $1$ defective part,\n\nwhat is the probability that the lot was produced by Company A?[/quote]", "Solution_3": "[quote=\"mad_skillz_aops\"]Try putting a \\ in front the percent sign so it looks right.[/quote]\r\n\r\nThanks for pointing it out.. I have updated it." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be twice differentiable, $2\\pi$-periodic, even function. Prove that if $f''(x)+f(x)=\\frac{1}{f(x+\\frac{3\\pi}{2})}$ holds for all $x$. then $f$ is $\\frac{\\pi}{2}$-periodic.", "Solution_1": "$f''(x - 2\\pi) + f(x - 2\\pi) = \\frac{1}{ f(x - \\frac{5\\pi}{2}) } = f''(x) + f(x) = \\frac{1}{ f(-x - \\frac{5\\pi}{2}) }$\r\n\r\n$f(x + \\frac{5\\pi}{2}) = f(x - \\frac{5\\pi}{2})$\r\n$f(x + \\frac{5\\pi}{2}) = f(x + \\frac{3\\pi}{2})$\r\n\r\nSo $f$ is $\\pi$-periodic. Repeat with $\\pi$ for the desired result." } { "Tag": [ "ratio", "factorial", "calculus", "derivative", "function", "trigonometry", "geometry" ], "Problem": "I didn't know if this should go in intermediate or advanced, but since it has to do with scary stuff like infinite series and the word \"converge,\" I'll put it here.\r\n\r\n\r\nThis series is convergent. Find its sum:\r\n\r\n[tex]\\displaystyle S=1+\\frac34+\\frac34\\cdot\\frac58+\\frac34\\cdot\\frac58\\cdot\\frac7{12}+\\cdots.[/tex]", "Solution_1": "Well, [hide]it clearly converges since it's less than the geometric series with first term 1 and ration 3/4. For the sum, I need to do more than 10 seconds of thinking.[/hide]", "Solution_2": "JBL wrote:Well, [hide]it clearly converges since it's less than the geometric series with first term 1 and ration 3/4. For the sum, I need to do more than 10 seconds of thinking.[/hide]\n\n\n\n[hide]well now it's been 25 seconds... what's the hold-up?\n\n\n\njk[/hide]", "Solution_3": "S = 1 + (3/4) + (3/4)(5/8) + (3/4)(5/8)(7/12) + ... = 1 + ((2(1) + 1)/4(1)) + ((2(1) + 1)/4(1))((2(2) + 1)/4(2)) + ((2(1) + 1)/4(1))((2(2) + 1)/4(2))((2(3) + 1)/4(3)) + ... = (2(0) + 1)!/((8^0)(0!):^2:) + (2(1) + 1)!/((8^1)(1!):^2:) + (2(2) + 1)!/((8^2)(2!):^2:) + (2(3)+1)!/((8^3)(3!):^2:) + ... = (C(1,1))(1)/(8^0) + (C(3,2))(2)/(8^1) + (C(5,3))(3)/(8^2) + (C(7,4))(4)/(8^3) + ... + (C(2n+1,n))(n)/(8^(n-1)) + ...\r\nLooks to me likes its a quasi-geometric series but the factorials are throwing me off. Can anyone further my findings?", "Solution_4": "Let \r\n\r\n[tex]\\displaystyle a_n=\\frac{(2n)!}{4^n(n!)^2}(2n+1)\\left(\\frac{1}{\\sqrt{2}}\\right)^{2n}[/tex], \r\n\r\nthen it's easy to verify that [tex]a_n[/tex] is indeed the [tex]n[/tex]th term of the series. That is,\r\n\r\n[tex]\\displaystyle S=\\sum_{n=0}^{\\infty}a_n[/tex].\r\n\r\nThen start with the power series of \r\n\r\n[tex]\\displaystyle\\arcsin{x}=\\sum_{n=0}^{\\infty}\\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}.[/tex] \r\n\r\nwe find that the value of the infinte series [tex]S[/tex] is equal to:\r\n\r\n[tex]\\displaystyle\\frac{d}{dx}\\left(x\\frac{d}{dx}(\\arcsin{x})\\right)\\left|_{x=\\frac{1}{\\sqrt{2}}[/tex]", "Solution_5": "Isaac Newton would have understood that the answer is [hide]2*sqrt(2).\n\n\n\n(The binomial series was Newton's favorite tool)\n\n\n\nThis is the series expansion of (1 - 1/2)^(-3/2)[/hide]", "Solution_6": "Kent Merryfield wrote:Isaac Newton would have understood that the answer is [hide]2*sqrt(2).\n\n(The binomial series was Newton's favorite tool)\n\nThis is the series expansion of (1 - 1/2)^(-3/2)[/hide]\n\n\n\nAfter calculating the derivative, we find that\n\n\n\n[tex]\\displaystyle\\frac{d}{dx}\\left(x\\frac{d}{dx}(\\arcsin{x})\\right)\\bigg|_{x=\\frac{1}{\\sqrt{2}}} = \\frac{d}{dx}\\left(\\frac{x}{\\sqrt{1-x^2}}\\right)\\bigg|_{x=\\frac{1}{\\sqrt{2}}}[/tex] [tex] =\\displaystyle\\frac{1}{(1-x^2)^{3/2}}\\bigg|_{x=\\frac{1}{\\sqrt{2}}} = 2^{3/2},[/tex] \n\n\n\nwhich is the same value as what Prof. Merryfield found. (Of course, his way is much simpler )\n\n\n\nBy the way, I haven't done a single derivative calculation for more than 20 years. That is why Prof. Merryfield posted the answer ahead of me .", "Solution_7": "[quote](By the way, I haven't done a single derivative calculation for more than 20 years :))[/quote]\r\n\r\nYou haven't done any derivatives for 20 years but you can remember and recognize the taylor expansion of arcsin(x)? I didn't even [i]know [/i]that series!", "Solution_8": "Nice solutions. Even though I did not solve this one, I got another interesting series.\r\n\r\nProve that 1/4 + 3/4*3/8 + 3/4*5/8*5/12 + 3/4*5/8*7/12*7/16 + ... = 1\r\n\r\nThere is a slick way to do it without using calculus.", "Solution_9": "I meant to say \"I got another interesting series while solving that 1+3/4+3/4*5/8+... thingy\"", "Solution_10": "Hmmm. I saw a Taylor Series expansion:\r\n\r\n[tex]\\displaystyle f(x)=\\frac{\\sqrt{8}}{(1-x)^{3/2}}[/tex]\r\n\r\n[tex]f(-1)=1[/tex]\r\n[tex]f'(-1)=3/4[/tex]\r\n[tex]\\displaystyle \\frac{1}{2!}f''(-1)=\\frac{1}{2}\\cdot\\frac{3\\cdot 5}{4\\cdot 4}[/tex]\r\n[tex]\\displaystyle \\frac{1}{3!}f'''(-1)=\\frac{1}{6}\\cdot\\frac{3\\cdot 5\\cdot 7}{4\\cdot 4\\cdot 4}[/tex]\r\n[tex]\\displaystyle \\frac{1}{4!}f''''(-1)=\\frac{1}{24}\\cdot\\frac{3\\cdot 5\\cdot 7\\cdot 9}{4\\cdot 4\\cdot 4\\cdot 4}[/tex]\r\n\r\n[tex]f(0)=\\sqrt{8}[/tex]\r\n\r\nJust different ways of looking at the same thing, I suppose.\r\n\r\nMy method was more or less as follows...\r\n\r\nThe factorials in the denominator were the first clue to look for a Taylor Series.\r\n\r\nI thought the 3, 5, 7 factors in successive terms could have been generated by a [tex]x^{-odd/2}[/tex] term in the function.\r\n\r\nTo compensate for the successive derivatives taking alternate sign, I had to use -x in the expression.\r\n\r\nThen it was just a matter of tweaking the constants to get the terms to all scale properly.\r\n\r\nI did not recognize this was an arcsin.", "Solution_11": "[quote=\"zabelman\"]You haven't done any derivatives for 20 years but you can remember and recognize the taylor expansion of arcsin(x)? I didn't even [i]know [/i]that series![/quote]\r\n\r\nNo, I don't remember. \r\n\r\nI first played with the general term [tex]a_n[/tex] and expressed it as something close to the final form in the above post. I was pretty sure that this is something related to the taylor series of some important function. Then I searched web for \"taylor series\" and found this:\r\n\r\nhttp://en.wikipedia.org/wiki/Taylor_series\r\n\r\nI found that the taylor series of [tex]\\arcsin{x}[/tex] is very close to what I want. After playing around for a little bit, the above solution came out. This is how I solved this problem although not as pretty as the way Prof. Merryfield provided.\r\n\r\nBy the way, I don't even have a high level math book as reference. I rely on the web heavily.\r\n\r\n[size=150][b]THE INTERNET IS B-E-A-U-T-I-F-U-L![/b][/size]", "Solution_12": "I might as well reveal that it wasn't really that simple. My son James took the series, made a power series (a \"generating function\") out of it, and manipulated that around until he found a differential equation that function satisfied. Then he told me it was really a pretty simple function. Only after I heard all that (without really following the details) did I say, \"Aha! Binomial series.\" Before that I had also been trying some power series approaches that weren't getting there.\r\n\r\nMost things are simple after you know the answer; and most results in mathematics were not first found by following the simplest path.", "Solution_13": "Well, recognizing (?) the binomial series is certainly the simplest solution. But, since this series is also an arcsin series, what is the connection between binomials and arcsins? If we find a non-obvious answer it might be worth looking further into.", "Solution_14": "[quote]what is the connection between binomials and arcsins?[/quote]\r\n\r\nThe short (and not completely satisfying) answer is that the derivative of arcsin(x) is\r\n(1 - x^2)^(-1/2). The most effecient way to find the power series for the arcsine is to expand (1 - x^2)^(-1/2) by the binomial series and then integrate term by term.\r\n\r\nThe longer and more satisfying answer is historical, although I don't have all the details in front of me at the moment. To begin with, Newton did not know any of the calculus of the trig and inverse trig functions; no one did. He did have the binomial series. He then found a way to express the area under part of a circle in a way that involved both the integral of an expression that could be expanded by the binomial series and an angle which can be considered the arcsine of something. From this he derived the power series for the arcsine. He then inverted (harder than it sounds!) this series to obtain the power series for the sine.", "Solution_15": "[quote=\"Kent Merryfield\"][quote]what is the connection between binomials and arcsins?[/quote]The longer and more satisfying answer is historical. ... From this he derived the power series for the arcsine. He then inverted this series to obtain the power series for the sine.[/quote]\r\n\r\nThis is interesting, Dr. Merryfield. Thank you." } { "Tag": [], "Problem": "Define $ f(x)\\equal{}|x|\\plus{}[x]$, where $ [x]$ is the greatest integer less than or equal to $ x$. What is the value of $ 2f(\\minus{}\\pi)\\plus{}2f(1.5)$? Express your answer in terms of $ \\pi$.", "Solution_1": "$ f(\\minus{}\\pi)\\equal{}\\pi\\minus{}4$\r\n$ f(1.5)\\equal{}2.5$.\r\n\r\n$ 2\\pi\\minus{}3$." } { "Tag": [ "Alcumus", "Support" ], "Problem": "Why is it called Alcumus?", "Solution_1": "This has been asked before a couple times, but try [url=http://en.wikipedia.org/wiki/Alcumus]this link.[/url]", "Solution_2": "Check the sticky. ALWAYS READ STICKIES BEFORE READING ANYTHING ELSE ON ANY AoPS FORUM!!!", "Solution_3": "wats a sticky?", "Solution_4": "The stuff that has an exclamation mark next to it that appears as a separate block on the top of a forum" } { "Tag": [ "limit" ], "Problem": "Fie $f:[0,\\infty ) \\to [0,\\infty )$ continua astfel incat :$\\mathop {\\lim }\\limits_{x \\to \\infty } \\frac{{f(2x)}}{{f(x)}} = 2$.\r\nSa se arate ca $\\exists \\mathop {\\lim }\\limits_{x \\to \\infty } f(x)$ si sa se calculeze limita.", "Solution_1": "Are you asking a finit limit ? because if I take $f(x)=2x$, satisfy the statement \r\n\r\n$\\lim_{+\\infty}2x=+\\infty$", "Solution_2": "Intr-adevar $\\mathop {\\lim }\\limits_{x \\to \\infty } f(x) =\\infty $ dar cum demonstrezi asta stiind doar ca $\\mathop {\\lim }\\limits_{x \\to \\infty } \\frac{{f(2x)}}{{f(x)}} = 2 $( f oarecare cu aceasta proprietate)", "Solution_3": "Daca ai scris $\\frac{f(2x)}{f(x)}$, inseamna ca ata are sens pentru toti $x$, si deci $f(x)\\n 0,\\ \\forall x$, nu? Atunci problema e usoara. Alege $a(2-\\varepsilon)f(x),\\ \\forall x\\ge a$. Putem sa alegem $b>2a$. Fie $I_0=[a,b], I_n=2^n I$, si $m_n=\\min _{x\\in I_n}f(x)$. Atunci $m_n\\ge (2-\\varepsilon)^n m_0$, si de aici se deduce usor ca limita la infinit e $\\infty$, pentru ca $I_0\\cup I_1\\cup\\ldots$ acopera tot $[a,\\infty)$." } { "Tag": [ "limit", "integration", "calculus", "inequalities", "calculus computations" ], "Problem": "Find the value of following limit:\r\n\r\n$ T=\\mathop {\\lim} \\limits _{t \\to {\\infty}} \\frac{1}{t} (\\sum\\limits_{n=1} ^ t \\sqrt { \\frac{n}{3(n-1)+n} })$", "Solution_1": "hello, $ \\mathop {\\lim} \\limits _{t \\to {\\infty}} \\frac{1}{t}\\left (\\sum\\limits_{n=1} ^ t \\sqrt { \\frac{n}{3(n-1)+n} }\\right)=\\frac{1}{2}$.\r\nSonnhard.", "Solution_2": "How Please :maybe:", "Solution_3": "${ \\lim_{t\\to\\infty}\\frac{1}{t}\\sum_{n=1}^{t}\\sqrt{\\frac{n}{4n-3}}=\\int_{0}^{1}{\\sqrt{\\frac{1}{4}}}\\;dx}=\\frac{1}{2}.$", "Solution_4": "If $ \\lim_{n\\to\\infty} a_n \\equal{} \\alpha$, then $ \\lim_{n\\to\\infty} \\frac{1}{n} \\sum_{k\\equal{}1}^{n} a_k \\equal{} \\alpha$. Then the answer immediately follows from $ \\lim_{n\\to\\infty} \\sqrt{\\frac{n}{4n\\minus{}3}} \\equal{} \\frac{1}{2}$.", "Solution_5": "Yuriy explain please how do u get integral .? need steps \r\n\r\nsos440 nice answer , could u prove ur fact please?\r\n\r\nThank you", "Solution_6": "It's Riemann's sum.\r\nPut $ \\delta x = \\frac{1}{t}.$ You get:\r\n\r\n$ \\lim_{\\delta x\\to 0+}\\delta x\\sum_{n=1}^{1/\\delta x}{\\sqrt{\\frac{n}{4n-3}}=\\lim_{\\delta x\\to 0+}\\delta x\\sum_{n=1}^{1/\\delta x}{\\sqrt{\\frac{n\\cdot\\delta x}{4n\\cdot\\delta x-3\\cdot\\delta x}}=}}$\r\n$ =\\lim_{\\delta x\\to 0+}\\delta x\\sum_{n=1}^{1/\\delta x}{\\sqrt{\\frac{x_n}{4x_n-3\\cdot\\delta x}}=\\lim_{\\delta x\\to 0+}\\delta x\\sum_{n=1}^{1/\\delta x}{\\sqrt{\\frac{x_n}{4x_n}}=...}}$ :)", "Solution_7": "I said before Yuriy ,you have a nice mind. :wink: \r\n\r\n\r\nThank you dude\r\n\r\n :D", "Solution_8": "Again Guys :\r\n\r\nWhat about if the limit was :\r\n\r\n$ T=\\mathop {\\lim} \\limits _{t \\to {\\infty}} \\frac{1}{t} (\\sum\\limits_{n=1} ^ t \\sqrt { \\frac{t}{3(n-1)+t} })$\r\n\r\n :)", "Solution_9": "hello, it is $ T=\\mathop {\\lim} \\limits _{t \\to {\\infty}} \\frac{1}{t} \\left(\\sum\\limits_{n=1} ^ t \\sqrt { \\frac{t}{3(n-1)+t} }\\right)=\\frac{2}{3}$.\r\nSonnhard.", "Solution_10": "I don't need only the answer , I need the method please. :roll:", "Solution_11": "[quote=\"Hidden Scofield\"]sos440 nice answer , could u prove ur fact please?[/quote]\r\n\r\nthe fact is just that normal convergence implies cesaro convergence. here's a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=167852]proof[/url].", "Solution_12": "Here's a solution to the first limit:\r\n\r\n\\[ \\lim_{t\\rightarrow \\infty} \\displaystyle\\sum_{n\\equal{}1}^t\\sqrt{\\frac n{3(n\\minus{}1)\\plus{}n}}\\]\r\n\r\nWe define a sequence $ {a_n}$ as $ \\frac1{n} \\left(2 \\plus{} \\sqrt {\\frac85} \\plus{} ... \\plus{} \\sqrt {\\frac {4n}{4n \\minus{} 3}}\\right)$. The limit is the point of convergence of this sequence divided by two, and the sequence must converge since the sequence is bounded and monotonic, since $ \\frac {2 \\plus{} \\sqrt {\\frac85} \\plus{} ... \\plus{} \\sqrt {\\frac {4n}{4n \\minus{} 3}}}{n} \\geq \\frac {2 \\plus{} \\sqrt {\\frac85} \\plus{} ... \\plus{} \\sqrt {\\frac {4(n \\plus{} 1)}{4(n \\plus{} 1) \\minus{} 3}}}{n \\plus{} 1}$. So the sequence is strictly decreasing, and clearly a lower bound is 1, since $ \\displaystyle\\sum_{n \\equal{} 1}^t \\sqrt {\\frac {4n}{4n \\minus{} 3}} \\geq t$ since every term of the sum is greater than 1. Suppose there existed a lower bound greater than 1, namely $ 1 \\plus{} \\epsilon$ for $ \\epsilon > 0$. Since $ \\lim_{n\\rightarrow \\infty} \\sqrt {\\frac {4n}{4n \\minus{} 3}} \\equal{} 1$, if we define the ith term of the sum $ b_i$, then for all $ i\\geq N$ for some $ N$, $ b_i < 1 \\plus{} \\epsilon$. Let the sum of all $ b_i$ for $ i < N$ be expressed as a constant $ C$, since it is only the sum of a finite amount of terms. So we have $ \\lim_{n\\rightarrow \\infty} \\frac {C \\plus{} \\sum_{i \\equal{} N}^n b_i}{n} \\geq 1 \\plus{} \\epsilon$, a lower bound. But since $ \\sum_{i \\equal{} N}^n b_i < (n \\minus{} N \\plus{} 1)(1 \\plus{} \\epsilon)$,\r\n\\[ \\lim_{n\\rightarrow \\infty} \\frac {C \\plus{} (n \\minus{} N \\plus{} 1)(1 \\plus{} \\epsilon)}{n} \\geq 1 \\plus{} \\epsilon\r\n\\]\r\n\r\nBut this limit on the left side of the inequality is equal to 1, and $ 1\\geq 1 \\plus{} \\epsilon$ is absurd, so 1 is the greatest lower bound (AKA supremum) of the sequence, making the sought-after limit $ \\frac12$." } { "Tag": [ "calculus", "integration", "algebra", "polynomial", "function", "linear algebra", "matrix" ], "Problem": "Hi,\r\n\r\ncan anyone give me some tips on how to solve this:\r\n\r\nproof that for every $n\\geq1$ there is a formula:\\[\r\n\\int_{a}^{b}f(x)dx=(b-a)\\left(\\begin{array}{c}\r\n{\\scriptscriptstyle n}\\\\\r\n\\sum\\\\\r\n^{k=0}\\end{array}\\lambda_{k}f(a+k.h)\\right)+r_{a,b}(f)\\]\r\nfor which $r_{a,b}(f)=0$ for all polynomical functions $f$ with\r\ndeg$(f)\\leq n$.\r\n\r\nGreets", "Solution_1": "What you're looking for is called the \"Closed Newton-Cotes formulas.\" The $n=1$ case is the trapezoidal rule, and the $n=2$ case is the usual Simpson's rule.\r\n\r\nAs for a hint, think of this a a pure linear algebra problem. You have $n+1$ unknowns to solve for: $\\lambda_0,\\lambda_1,\\dots,\\lambda_n.$ You want to be able to exactly integrate the following $n+1$ functions: $1,(x-a),(x-a)^2,\\dots,(x-a)^n.$ This gives you $n+1$ linear equations for the $n+1$ unknowns. Your task is to show a unique solution (equivalent to the invertability of a matrix or the non-vanishing of a determinant.)\r\n\r\nThere are other possible bases for the set of polynomials of degree $\\le n.$ The suggestion above should be reasonably convenient.\r\n\r\nOne incidental byproduct for the closed Newton-Cotes formulas: when $n$ is even, you get one more degree of precision than you bargained for. For instance, when $n=2$, we set out to show that Simpson's formula exactly integrates all quadratic polynomials, but discover that it also exactly integrates all cubic polynomials." } { "Tag": [ "AMC", "AIME" ], "Problem": "I have only seen two tests in the contest part of this website .\r\n\r\nand sections of tests on the amc 8/10/12 worksheet website\r\n\r\n\r\nis there a place where you can find more?", "Solution_1": "[quote=\"now a ranger\"]I have only seen two tests in the contest part of this website .\n\nand sections of tests on the amc 8/10/12 worksheet website\n\n\nis there a place where you can find more?[/quote]\r\nYou can buy the millenium CD that has all of the \r\nAMC 8/10/12/AIME/USAMO since 2000. It is for $\\$$ 20 at the AMC website.", "Solution_2": "See the following pages:\r\n\r\nAMC 12, AHSME:[url]http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm[/url]\r\n(this url usually doesn't work very well)\r\n\r\nMany competitions (AIME, many National Olympiads)[url]http://www.kalva.demon.co.uk/[/url]", "Solution_3": "[quote=\"knexpert\"]See the following pages:\n\nAMC 12, AHSME:[url]http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm[/url]\n(this url usually doesn't work very well)\n\nMany competitions (AIME, many National Olympiads)[url]http://www.kalva.demon.co.uk/[/url][/quote]\r\n\r\nNone of the links inside [url]http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm[/url] works. :(", "Solution_4": "[quote=\"now a ranger\"]I have only seen two tests in the contest part of this website .\n\nand sections of tests on the amc 8/10/12 worksheet website\n\n\nis there a place where you can find more?[/quote]\r\n\r\n????? I emailed them to you" } { "Tag": [ "symmetry", "graph theory", "combinatorics proposed", "combinatorics" ], "Problem": "On every card of a deck of cards a regular 17-gon is displayed with all sides and diagonals, and the vertices are numbered from 1 through 17. On every card all edges (sides and diagonals) are colored with a color 1,2,...,105 such that the following property holds: for every 15 vertices of the 17-gon the 105 edges connecting these vertices are colored with different colors on at least one of the cards. What is the minimum number of cards in the deck?", "Solution_1": "(My solution is a bit long, I hope it's true.)\nLet points are $p_1,p_2,\\dots ,p_{17}.$ $[p_i,p_j]$ denotes the complete graph which has the $15$ vertices\nfrom $17$ points except $p_i$ and $p_j$. If a $[p_i,p_j]$ satisfies conditions, call it $good.$ ${p_i}{p_j}$ denotes the color\nof the edge which its endpoints are $p_i$ and $p_j$.\n$Lemma:$ There can't be more than four good $[p_i,p_j]$ on the same card.\n$Proof:$ $i)$ $[p_1,p_2], [p_3,p_4], [p_5,p_6]$ can't be all good on the same card.\nSuppose that $i)$ is false. Let $X=({p_3}{p_4},{p_5}{p_6},{p_3}{p_6},{p_4}{p_5},{p_3}{p_5},{p_4}{p_6},\n{p_3}{p_7},{p_3}{p_8},\\dots, {p_3}{p_{17}},{p_4}{p_7},{p_4}{p_8},\\dots ,{p_4}{p_{17}},{p_5}{p_7},{p_5}{p_8},\\dots ,\n{p_5}{p_{17}},{p_6}{p_7},{p_6}{p_8},\\dots ,{p_6}{p_{17}}).$ \nLet $Y=({p_1}{p_2},{p_3}{p_4},{p_2}{p_3},{p_1}{p_4},{p_1}{p_3},{p_2}{p_4},\n{p_3}{p_7},{p_3}{p_8},\\dots, {p_3}{p_{17}},{p_4}{p_7},{p_4}{p_8},\\dots ,{p_4}{p_{17}},\n{p_1}{p_7},{p_1}{p_8},\\dots ,{p_1}{p_{17}},{p_2}{p_7},\n{p_2}{p_8},\\dots ,{p_2}{p_{17}}).$\nSets $X$ and $Y$ are equal and $n(X)=n(Y)=50.$ Since $K=(p_3p_4,p_3p_7,p_3p_8,\\dots ,p_3p_{17},p_4p_7,p_4p_8,\\dots ,p_4p_{17})$ are both in $X$ and $Y$, $X-K=Y-K$. \nLet $X_1=X-K$ and $Y_1=Y-K$. $n(X_1)=n(Y_1)=27.$\nLet $X_2={X_1}-(p_3p_6,p_4p_5,p_3p_5,p_4p_6)$ and $Y_2={Y_1}-(p_2p_3,p_1p_4,p_1p_3,p_2p_4).$ Then there are at least\n$27-4-4=19$ colors both in $X_2$ and $Y_2.$ Since for each $p_ip_j$ from $X_2$ the edge $p_ip_j$ is an edge of $[p_3,p_4]$ and for each $p_ip_j$ from $Y_2$ \nthe edge $p_ip_j$ is an edge of $[p_3,p_4]$, $[p_3,p_4]$ can't be good, contradiction.\n\n$ii) [p_1,p_2],[p_3,p_4],[p_1,p_5]$ can't be all good on the same card. Suppose that $ii)$ is false. Let $X=(p_5p_3,p_5p_4,p_5p_6,p_5p_7,\\dots ,p_5p_{17})$ and let $Y=(p_2p_3,p_2p_4,p_2p_6,p_2p_7,\\dots ,p_2p_{17}).$ Then $X=Y$ and $n(X)=n(Y)=14.$ Let $X_1=X-(p_5p_3,p_5p_4)$ and $Y_1=Y-(p_2p_3,p_2p_4).$ Then there are at least $14-2-2=10$ colors both in $X_1$ and $Y_1$. Since for each $p_ip_j$ from $X_1 (Y_1)$ the edge $p_ip_j$ is an edge of $[p_3p_4]$, $[p_3,p_4]$ can't be good, contradiction.\n\n$iii) [p_1,p_2],[p_1,p_3],[p_1,p_4]$ can't be all good on the same card. Suppose that $iii)$ is false. Let $X=(p_3p_4,p_3p_5,\\dots ,p_3p_{17})$ and let $Y=(p_2p_4,p_2p_5,\\dots ,p_2p_{17}).$ Then $X=Y$ and $n(X)=n(Y)=14.$ Let $X_1=X-(p_3p_4)$ and $Y_1=Y-(p_2p_4).$ Then there are at least $14-1-1=12$ colors both in $X_1$ and $Y_1$. Since for each $p_ip_j$ from $X_1(Y_1)$ the edge $p_ip_j$ is an edge of $[p_1,p_4]$, $[p_1,p_4]$ can't be good, contradiction.\nBy $i,ii$ and $iii$ and by symmetry, we can say that there can't be more than four good $[p_i,p_j]$ on the same card. (We can see this geometrically by denoting each $[p_i,p_j]$ with a line segment $p_ip_j$.)\n\nOn the other hand, we can find four good $[p_i,p_j]$'s in the forms $[p_1,p_2],[p_2,p_3],[p_3,p_4],[p_4,p_1]$ on the same card by coloring card as the following:\nColor the edges $p_1p_2,p_2p_3,p_3p_4,p_4p_1,p_1p_3,p_2p_4$ by the color $1$,\nColor the edges $p_1p_5,p_1p_6,\\dots ,p_1p_{17}$ by the colors $2,3,\\dots ,14$,\nColor the edges $p_2p_5,p_2p_6,\\dots ,p_2p_{17}$ by the colors $15,16,\\dots ,27$,\nColor the edges $p_3p_5,p_3p_6,\\dots ,p_3p_{17}$ by the colors $2,3,\\dots ,14$,\nColor the edges $p_4p_5,p_4p_6,\\dots ,p_4p_{17}$ by the colors $15,16,\\dots ,27$,\nAnd color the remaining $78$ edges (edges of the $13$-gon whose vertices are $p_5,p_6,\\dots ,p_{17}$) by the colors $28,29,\\dots ,105.$\nSo by symmetry, for any four points $a,b,c,d$, we can find four goods in the forms $[a,b],[b,c],[c,d],[d,a]$ on a card by coloring the card as above.\n\nOn figure, each line segment whose endpoints are $i,j$ denotes $[p_i,p_j]$. By rotating each 4-gon on shape 1 and shape 2 $17$ times we will get $34$ 4-gons. For each line segment $p_ip_j$, it will be count exactly $1$ times in total on (shape 1+shape 2). (so each $[p_i,p_j]$ will be count exactly $1$ times in total.) Union of these $34$ 4-gons will be the complete graph of the points $p_1,p_2,\\dots ,p_{17}.$ Each 4-gon with vertices $a,b,c,d$ denotes a card which has the goods $[a,b],[b,c],[c,d],[d,a].$ So the answer is $34$." } { "Tag": [ "probability", "function", "algebra unsolved", "algebra" ], "Problem": "Can anyone help me with these problems? And please show workings...\r\n\r\n1) How many divisions are necessary to determine if 283 is prime?\r\n\r\n2) For which integers k does x^k have the largest coefficient in the expansion of (x+3)^50.\r\n\r\n3) What is the probability of having 4 boys and 4 girls in a family of 8 children?\r\n\r\n4) Determine which is larger sqrt 10 + sqrt 17 or sqrt 53.\r\n\r\n5) In a race of 1760 meters, A beats B by 330 meters, A beats C by 460 meters. If the contestants continue to run at the same uniform rates, by how many meters will B beat C?\r\n\r\n6) It is known that every function can be uniquely expressed as a sum of an even and an odd function. Find such representation for the function f(x) = (x+1)/(x^3 + 1).\r\n\r\nThats all. hope you can help me...:D", "Solution_1": "I solve problem 6 .The others quite easy.\r\n$ f(x)\\equal{}h(x)\\plus{}g(x)$\r\nWhere $ h(x)$ is an even function,$ g(x)$ is a odd function.\r\nLet \r\n$ f(\\minus{}x)\\equal{}h(x)\\minus{}g(x)$\r\nImply that $ g(x)\\equal{}\\frac{f(x)\\minus{}f(\\minus{}x)}{2}$\r\n $ h(x)\\equal{}\\frac{f(x)\\plus{}f(\\minus{}x)}{2}$\r\nApply this to find your function.", "Solution_2": "can you also help me with the rest. thanks.", "Solution_3": "To Jakey:\r\nI think that you should post problems like your third and first in Intermediate section not here,OK?", "Solution_4": "Continue the discussion here:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=174947[/url]" } { "Tag": [], "Problem": "\u0388\u03c3\u03c4\u03c9 $0\\leq a,b,c\\leq\\frac{1}{2}$ \u03ba\u03b1\u03b9 $a+b+c=1$ . \u039d\u03b4\u03bf \r\n\r\n$a^3+b^3+c^3+4abc\\leq\\frac{9}{32}$", "Solution_1": "\u039f\u03cd\u03c4\u03b5 \u03ba\u03b1\u03bd \u03bc\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b5\u03b9\u03b1 ?? :? . \u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b7 \u03bc\u03ad\u03b8\u03bf\u03b4\u03bf\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03b7\u03b8\u03b5\u03af \u03b2\u03b3\u03ac\u03b6\u03b5\u03b9 \u03bc\u03ac\u03c4\u03b9\u03b1 . :D", "Solution_2": "Lagrange Multipliers!", "Solution_3": "\u03a3\u03c9\u03c3\u03c4\u03ac \u03b5\u03af\u03c3\u03b1\u03b9 \u03c9\u03c1\u03b1\u03af\u03bf\u03c2 ..... :D", "Solution_4": "Nomizw sou eixa apantisei me private message gia lysi.", "Solution_5": "\u039d\u03b1\u03b9 \u03c4\u03bf \u03b8\u03c5\u03bc\u03ac\u03bc\u03b1\u03b9 .\u0391\u03c0\u03bb\u03ac \u03ba\u03b1\u03bb\u03cc \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bb\u03c5\u03bc\u03bc\u03ad\u03bd\u03b7 \u03ba\u03b1\u03b9 \u03b5\u03b4\u03ce .\u039f\u03ba ? :lol:", "Solution_6": "[quote=\"silouan\"]\u0388\u03c3\u03c4\u03c9 $0\\leq a,b,c\\leq\\frac{1}{2}$ \u03ba\u03b1\u03b9 $a+b+c=1$ . \u039d\u03b4\u03bf $a^3+b^3+c^3+4abc\\leq\\frac{9}{32}$[/quote]\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=371&t=197492]\u03b5\u03b4\u03ce[/url]" } { "Tag": [ "algebra", "polynomial", "functional equation", "number theory solved", "number theory" ], "Problem": "$S_{n,k}$ is the smallest positive amount of $-^+1^k-^+2^k...-^+n^k$( we choose + or - appropriately).prove that for any $k\\in N$ {$S_{n,k}$} is periodic.", "Solution_1": "I have only one question: where do you get these hard and nice problems? I tried to solve this one for two hours, but failed. Can anyone submit a solution? Well, these problems sound so nice, but they are so difficult!", "Solution_2": "I found it im mydusty materials ,and I saw ** beside this problem which means it's more difficult than ordinary problem :D :? :)", "Solution_3": "Hi sam,\r\n\r\nfrom which journals or books do you take your dusty materials. Are there many challenging problems ? Tell me more about problem-solving publications in Iran please ! :)", "Solution_4": "Well, your materials are very valuable! It's a splendid problem. I think I've managed to show that for any k the sequence $ S_{n,k} $ is periodic from a certain point. My idea was the following:\r\n Lemma:\r\n The numbers $ {1,2,...,2 ^n} $ can be grouped in two subsets A and B such that for any $ k=0,n-1 $ we have $\\sum_{x \\in A} x^k=\\sum_{ x \\in B} x^k $. \r\n This is well known and I won't insist. \r\n But this lemma has a beautiful corolarry:\r\n Corolary:\r\n For any polynomial f of degree at most n-1 we have $\\sum_{ x \\in A} f(x)=\\sum_{x \\in B} f(x) $. \r\n This is easy to see with the bynomial theorem. \r\n Now, fix k and denote $ a_n=S_{n,k} $. We prove that for any n we have $ a_n\\geq a_{n+2^{k+1}} $. This comes from the lemma. Indeed, let the sum $ t_1\\cdot 1^k+...+t_n\\cdot n^k $ for which we have the minimum of the absolute value. Then we have $ t_1\\cdot 1^k+...+t_n\\cdot n^k+s_1\\cdot (n+1)^k+...+s_{2^{k+1}}\\cdot (n+2^{k+1})^k=0 $ for a certain system $ s_i$ which can be deduced applying the corollary. Just put $ s_i=1 $ if i is in A and -1 otherwise, applying the corollary for k+1 instead of n. But this means that the minimum value $ a_{n+2^{k+1}} $ is at most $ a_n $. Let us put now $ r=2^{k+1} $ and fix a number i in {1,2,...,r}. We have seen that $ a_n\\geq a_{n+r} $ for all n and thus we have $ a_i\\geq a_{i+r}\\geq a_{i+2r}\\geq ... $. This sequence must stop decreasing and thus we can find $ M_i $ and $ f(i)$ such that for all $ n>f(i) $ we have $ a_{i+nr}=M_i $. Now take $ N=max{f(1),...,f(r)} $. It's easy to see that the sequence becomes periodic with period r starting from N+1. Maybe it's easy to deduce from the above that the sequence is in fact periodic starting from 1, but as in the solution of the Vietnamese functional equation I don't see the obvious things. So, help a blind old man!" } { "Tag": [ "function", "limit", "calculus", "calculus computations" ], "Problem": "For which functions does l'Hospital's Rule gives an infinite loop? For example, \r\n\r\n\\[\\lim_{x\\to\\infty}\\frac{x}{\\sqrt{x^{2}+1}}=^{H}\\lim_{x\\to\\infty}\\frac1{\\frac{x}{\\sqrt{x^{2}+1}}}=\\lim_{x\\to\\infty}\\frac{\\sqrt{x^{2}+1}}{x}=^{H}\\lim_{x\\to\\infty}\\frac{\\frac{x}{\\sqrt{x^{2}+1}}}{1}= \\lim_{x\\to\\infty}\\frac{x}{\\sqrt{x^{2}+1}}\\]\r\n\r\nwhere $=^{H}$ denotes the use of l'Hospital's rule is such an infinite loop. Suppose that the limit be taken at infinity (or zero if you prefer), for what class of real-valued functions $f(x)$ will this sort of thing occur?", "Solution_1": "well first off you can not use l'hospital's rule because the limit is not in an indeterminate form.", "Solution_2": "I know that $\\lim_{x\\to\\infty}\\frac{x}{\\sqrt{x^{2}+1}}=\\lim_{x\\to\\infty}\\frac{1}{\\sqrt{1+\\frac1{x^{2}}}}=\\frac{1}{\\sqrt{1+0}}=1$.\r\nBut certianly $\\frac{x}{\\sqrt{x^{2}+1}}\\to\\frac\\infty{\\infty}$ and $\\frac{\\sqrt{x^{2}+1}}{x}\\to\\frac\\infty{\\infty}$ as $x\\to\\infty$." } { "Tag": [ "induction", "graph theory", "combinatorics proposed", "combinatorics" ], "Problem": "Here $G_{n}$ denotes a simple undirected graph with $n$ vertices, $K_{n}$ denotes the complete graph with $n$ vertices, $K_{n,m}$ the complete bipartite graph whose components have $m$ and $n$ vertices, and $C_{n}$ a circuit with $n$ vertices. The number of edges in the graph $G_{n}$ is denoted $e(G_{n})$. \r\n\r\nThe edges of $K_{n}(n \\geq 3)$ are colored with $n$ colors, and every color is used. \r\nShow that there is a triangle whose sides have different colors.", "Solution_1": "By induction on $ n$. For $ n \\equal{} 3$ the statement clearly holds so we assume it holds for $ k \\geq n$ and consider the graph $ K_{n\\plus{}1}$. Let $ v$ be an arbitrary vertex of $ K_{n\\plus{}1}$ and define $ K : \\equal{} K_{n\\plus{}1} \\minus{} v$. If the edges of $ K$ are colored with $ n$ different colors we are done. So, suppose the edges of $ K$ are colored with $ n \\plus{} 1$ colors. Recolor the edges of color $ n \\plus{} 1$ with a different color, say n. By the induction hypothesis $ K$ now contains the specified triangle $ T$. It is easy to verify that $ T$ satisfies the required conditions even in $ K_{n\\plus{}1}$.", "Solution_2": "Hmm...solution above seems wrong to me,what makes you think that you can recolor vertices?\r\nI had essentially the same problem at my IMO Preparation course,and if I am not mistaken five of our 6 members managed to solve it,the one who didn't(not me),was pretty surprised when some of us showed the proof...it is pretty simple,though I must admit that it took one or two hours to find a solution.\r\nI've got to thing some to recall the details but the first step is clear to everyone,I guess.", "Solution_3": "Let's choose one edge for each color. Thus, you get a graph with $ n$ vertices and $ n$ edges. Therefore, this graph has a cycle. It follows that the initial graph contains rainbow cycles (a rainbow cycle is a cycle with no two edges with the same color).\r\nNow, let's consider a rainbow cycle with minimal length, say $ L$. We want to prove that $ L\\equal{}3$.\r\nAssume, for a contradiction, that $ L \\geq 4$ and let $ A_1,A_2,A_3,A_4$ be any four consecutive vertices on this minimal rainbow cycle.\r\nThe edge $ A_1A_3$ divides the cycle into two other cycles with lengths less than $ L$ and since the color of the edge $ A_1A_3$ appears at most once in the minimal rainbow cycle, it does not appear in some of these two parts. Thus, we have found a rainbow cycle with length les than $ L$, contradicting the minimality of $ L$, and we are done.\r\n\r\nPierre.", "Solution_4": "I don't know if this is what Azi meant to say. Clearly the statement is true for n = 3. We now use induction:\r\n\r\nSuppose it's true for n < k, we show it's true for k. Choose a vertex A in k and cosider the remaining k-1 complete graph. If that graph is coloured by k-1 or more colours, then we have our triangle (by induction hypothesis). Otherwise there must be less than k - 1 colours in the k-1 graph, so there are at least two colours (i.e. red and blue) not in the k-1 graph but in the k graph (i.e. they are attached to A). Thus AB is red and AC is blue for some B and C. But BC cannot be red or blue since the edge is in the k-1 graph. Thus ABC is a different coloured triangle. So my PMI, it is true for all n >= 3." } { "Tag": [], "Problem": "To compare $ 15^{0,15}$ $ 1,5$ :?: \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n_______________________________________\r\nAzerbaijan Land of Fire \r\n$ \\Rightarrow\\bigstar\\Leftarrow$\r\n :rotfl:", "Solution_1": "[quote=\"Arrange your tan\"]\n\n$ 15^{0.15}$ versus $ 1.5$\n\n[/quote]\r\n\r\nMost of the work:\r\n[hide]\n$ (15^{0.15})^{20}$ versus $ 1.5^{20}$\n\n$ 15^3$ versus $ (1.5^4)^5$\n\n$ 15^3$ versus $ 5.0625^5$\n\n$ 15^3$ versus $ (5 \\plus{} \\frac {1}{16})^5$\n\n$ \\frac {15^3}{5^5}$ versus $ \\frac {(5 \\plus{} \\frac {1}{16})^5}{5^5}$\n\n$ \\frac {15^3}{5^5}$ versus $ {(\\frac55 \\plus{} \\frac {1}{16 \\cdot 5})^5}$\n\n$ \\frac {3^3 \\cdot 5^3}{5^5}$ versus $ (1 \\plus{} \\frac {1}{80})^5$\n\n$ \\frac {27}{25}$ versus $ (1 \\plus{} \\frac {1}{80})^5$\n\n$ 1 \\plus{} \\frac {2}{25}$ versus $ (1 \\plus{} \\frac {1}{80})^5$\n\n$ ln(1 \\plus{} \\frac {2}{25})$ versus $ ln(1 \\plus{} \\frac {1}{80})^5$\n\n$ ln(1 \\plus{} x) \\equal{} (x \\minus{} \\frac12x^2) \\plus{} (\\frac13x^3 \\minus{} \\frac14x^4) \\plus{} ...$ for $ ( \\minus{} 1 < x \\le 1)$ . . . (*)\n\n$ ln(1 \\plus{} \\frac {2}{25})$ versus $ 5ln(1 \\plus{} \\frac {1}{80})$\n\n$ [\\frac {2}{25} \\minus{} \\frac12(\\frac {2}{25})^2 \\plus{} ...]$ versus $ 5[\\frac {1}{80} \\minus{} \\frac12(\\frac {1}{80})^2 \\plus{} ...]$\n\n$ [\\frac {96}{2 \\cdot 625} \\plus{} ...]$ versus $ 5[\\frac {159}{2 \\cdot 6400} \\plus{} ...]$\n\n$ [\\frac {96}{2 \\cdot 625} \\plus{} ...]$ versus $ [\\frac {159}{2 \\cdot 1280} \\plus{} ...]$\n\n$ \\frac {96}{2 \\cdot 625}$ > $ \\frac {159}{2 \\cdot 1280}$ \n\nThis is true because the second denominator is more than twice the\nfirst denominator, while the second numerator is less than twice the \nfirst numerator.\n\nIncomplete problem by me:\n\nA subproof is not shown for the sum of every other consecutive pair of terms from (*)\n for $ x \\equal{} \\frac {2}{25}$ is greater than for $ x \\equal{} \\frac {1}{80}$ for every other pair of terms in $ 5ln(1 \\plus{} x)$. \nWhen this is shown, then the case can be made that\n\n$ 15^{0.15} > 1.5$\n\n------------------------------------------------------------------------\n\nA different shorter solution?\n[/hide]" } { "Tag": [ "geometry", "circumcircle", "inequalities unsolved", "inequalities" ], "Problem": "$A_1B_1C_1$, $A_2B_2C_2$ triangles, prove that:\r\n\r\n\\[8R_1R_2+4r_1r_2\\geq a_1a_2+b_1b_2+c_1c_2\\]", "Solution_1": "I am sorry I do not understand the question. What do $R_{1}$ and $R_{2}$ stand for? The same goes for $r_{1}$ and $r_{2}$.\r\nAs far as I understand, by $R$, you mean the radius of the circumcircle.", "Solution_2": "$R_i$ - radius of circumcircle of circle $i$\r\n$r_i$ - radius of incircle of circle $i$" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "x^y+y^z+z^x=1\r\nSolve for \r\na) x,y,z are positive real numbers\r\nb)x,y,z are integers", "Solution_1": "[b]b)[/b]No solution.\r\n[b]Case 1[/b] If there is $ 1$ in $ (x,y,z)$, WLOG, assume $ x \\equal{} 1$, then$ y^z \\plus{} z \\equal{} 0$, y=z=0(rejected) or z is negative(rejected as y must = 1 to make L.H.S. integer and it's not a solution also).So no solution in this case.\r\n[b]Case 2[/b] No $ 1$ in $ (x,y,z)$ but there is $ 0$ in $ (x,y,z)$, assume $ x \\equal{} 0$, then $ 0^y \\plus{} y^z \\equal{} 0$, no solution again.\r\n[b]Case 3[/b] $ (x,y,z) \\neq 0,1$ and there is at least one integer in $ (x,y,z)$ say $ x$ such that $ x \\ge 2$ .Then [b]i)[/b]$ y \\equal{} \\minus{} b$ is negative. The equation becomes $ \\frac {1}{x^b} \\plus{} ( \\minus{} b)^z \\plus{} z^x \\equal{} 1$. As $ x \\neq 1$ , $ z \\equal{} \\minus{} c$ is negative to make $ L.H.S.$ an integer.So the equation now becomes $ \\frac {1}{x^b} \\plus{} \\frac {1}{( \\minus{} b)^c} \\plus{} ( \\minus{} c)^x \\equal{} 1$, so $ x^b \\equal{} ( \\minus{} b)^c \\equal{} 2$ (rejected as in this case c=0) or $ \\frac {1}{x^b} \\plus{} \\frac {1}{( \\minus{} b)^c} \\equal{} 0$(rejected also as $ c \\equal{} 1$ and thus $ x^b \\equal{} b$, which is impossible for $ x \\ge 2$). [b]ii)[/b]If $ y$ positive then $ z \\equal{} \\minus{} c$ is negative or $ L.H.S. > R.H.S.$. But in this case $ \\frac {1}{y^c}$ is an integer, no solution again as $ y \\neq 0$.\r\n[b]Case 4[/b] $ x,y,z$ are all negative.Then let $ x \\equal{} \\minus{} a, y \\equal{} \\minus{} b, z \\equal{} \\minus{} c$ and the equation becomes $ \\frac {1}{( \\minus{} a)^b} \\plus{} \\frac {1}{( \\minus{} b)^c} \\plus{} \\frac {1}{( \\minus{} c)^a} \\equal{} 1$ [b]i)[/b]If all of $ a, b, c \\ge 2$, $ |\\frac {1}{( \\minus{} a)^b}|,|\\frac {1}{( \\minus{} b)^c}|,|\\frac {1}{( \\minus{} c)^a}| \\le \\frac {1}{4}$, impossible. [b](ii)[/b]One of them, say $ a \\equal{} 1$ .Then if b odd $ \\frac {1}{( \\minus{} b)^c} \\plus{} \\frac {1}{( \\minus{} c)} \\equal{} 2$ ,impossible.If $ b$ even, $ ( \\minus{} b)^c \\equal{} c$, impossible again.\r\nSo [b]b)[/b] has no solution." } { "Tag": [ "geometry", "trigonometry", "quadratics", "Vieta", "modular arithmetic", "function", "conics" ], "Problem": "In this game someone posts some kind of formula, it can be as trivial or as complex as you want, and the next poster posts the name of the formula. If the poster says the wrong name, then the person who posted the formula gets a point, if he says it correctly then guy who posted the name of the formula gets the point.(NO CHEATING PLEASE)\r\n\r\nI'll start us off:\r\n\r\n$ (a\\plus{}b)^2\\equal{}a^2\\plus{}2ab\\plus{}b^2$\r\n\r\npretty easy, but extremely well known and helpful", "Solution_1": "wouldnt that be the binomial theorem?", "Solution_2": "well, it is the binomial theorem, but maybe battered wanted perfect square?", "Solution_3": "Well, what if he posted something like $ (a\\plus{}b)^5$?\r\nWhat would that be?\r\n\r\nBut I get your point for the square thing.\r\n :| \r\n :wink:", "Solution_4": "Maybe it's FOIL", "Solution_5": "But then he would have written $ (a\\plus{}b)(a\\plus{}b)$.", "Solution_6": "Man this is pretty tricky. Moving on\r\n\r\n$ \\text{Area}^2\\equal{}(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)(s\\minus{}d)\\minus{}abcd\\cos{\\frac{A\\plus{}C}{2}}$", "Solution_7": "hello, this is the formula for the area of an arbitrary quadrilateral.\r\nSonnhard.", "Solution_8": "And the new formula:\r\n\r\n$ \\frac{ab\\sin C}{2}$", "Solution_9": "Area of a triangle. \r\n\r\n$ a^2 \\plus{} b^2 \\equal{} c^2$ \r\n\r\n :D (Tricky!!!)", "Solution_10": "pythagorean theorem\r\n\r\n$ x \\equal{} ({q \\plus{} [q^2 \\plus{} (r \\minus{} p^2)^3]^{1/2}})^{1/3} \\plus{} ({q \\minus{} [q^2 \\plus{} (r \\minus{} p^2)^3]^{1/2}})^{1/3} \\plus{} p$\r\n\r\nOwch.", "Solution_11": "Is that even a formula?\r\n\r\nmine was the \"perfect square\" formula btw", "Solution_12": "Cardano's formula.\r\n\r\n$ x^4\\plus{}4y^4\\equal{}(x^2\\plus{}2y^2\\plus{}2xy)(x^2\\plus{}2y^2\\minus{}2xy)$", "Solution_13": "Mhm. so does my answer count?", "Solution_14": "It probably counts, because it is part of the binomial theorem...", "Solution_15": "[quote=\"zapi2007\"]I'm not sure if this is right, but equation for the radii of the circles made by extending the sides and tangent to two sides and the incircle's radius?\n[/quote]\r\n\r\nIf by that second part you mean excircles, then yes :) \r\n\r\nFirst one is tan sum formula.\r\n\r\nZapi's second still stands.", "Solution_16": "yeah, that was one of the mandelbrot team play topics", "Solution_17": "[quote=\"zapi2007\"]$ \\sum_{n \\equal{} 1}^{\\infty}{\\frac {1}{n^2}}$\nI guess also state how to arrive at the answer.[/quote]\r\n$ \\frac{\\pi^2}{6}$. Found using the infinite expansion of sin(x).\r\n\r\n$ \\Delta G \\equal{} \\Delta H \\minus{} T \\Delta S$\r\n\r\n$ \\frac{1}{n\\plus{}1} \\binom {2n}{n}$", "Solution_18": "Gibbs free energy and Catalan number.\r\n\r\n$ \\left(\\displaystyle\\frac{n(n\\plus{}1)}{2}\\right)^2$", "Solution_19": "[quote=\"darkdieuguerre\"]$ \\left(\\displaystyle\\frac {n(n \\plus{} 1)}{2}\\right)^2$[/quote]\r\n\r\nSum of consecutive cubes.\r\n\r\n1 + 1 = 2, since I don't know any interesting ones.", "Solution_20": "The Trivial Equation. (couldn't think of any others)\r\n\r\n$ \\sum_{n \\equal{} 1}^{\\infty}{\\frac{1}{n^{3}}}$", "Solution_21": "Ap\u00e9ry's constant.\r\n\r\n$ e^{\\int{P(x)dx}}$", "Solution_22": "The solution*constant to P(x)f(x) = f'(x)\r\n\r\nNobody here likes enumerative combinatorics? blah\r\n\r\n\r\n$ \\sum_{k \\equal{} 0}^m r_k(x)_{m \\minus{} k} \\equal{} \\prod_{i \\equal{} 1}^m (x \\plus{} s_i)$\r\n\r\nWhere (x)_a is Permutation number, not Combination number.", "Solution_23": "Yay for reviving dead threads.\r\n\r\n$ b^2m \\plus{} c^2n \\equal{} a(d^2\\plus{}mn)$.", "Solution_24": "a factored version of stewart's\r\n\r\npythag...thanks for killing the thread :maybe:\r\n\r\n\r\n$ E \\equal{} E_0 \\minus{} \\frac {RT}{nF}ln{k}$\r\n\r\n\r\n$ \\frac{\\mu_0I}{2\\pi r}$", "Solution_25": "The bottom formula yields the magnetic force at a distance $ r$ from a infinitely long wire carrying a current? My physics is rusty.", "Solution_26": "any answer to the first one?\r\nor any formula?", "Solution_27": "blah, revive old thread\r\n\r\n\r\nMy first one was the Nernst Equation, I think someone might have posted it earlier.\r\n\r\n\r\nif $ x \\equal{} (x_1,x_2)$and$ y \\equal{} (y_1,y_2) x$dot$ y \\equal{} x_1y_1 \\plus{} x_2y_2$\r\n\r\n\r\nmaybe this topic doesn't belong in the games forum or fun factory... :|", "Solution_28": "Dot product.\r\n\r\n$ \\displaystyle\\frac{E}{c^2}\\sqrt{\\minus{}1}\\displaystyle\\frac{PV}{nR}$", "Solution_29": "haha.\r\n\r\n$ lim_{x->0}{\\frac{f(x+h)-f(x)}{h}}$" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "It is true that for any $ X,Y \\neq O_n$, $ n \\times n$ matrices there exist a $ n \\times n$ matrix $ Z$ such that $ XZY \\neq O_n$ ?\r\n\r\nPS:edited", "Solution_1": "[hide]what if $ XY\\equal{}0$ ..?[/hide]", "Solution_2": "Start by considering the case $ XY \\neq O_n$. Then, we have $ XZY \\neq O_n$ iff $ rank(XZY) \\neq 0$. Now let $ P,Q,R,S$ be nxn invertible matrices such that\r\n\r\n$ X \\equal{} P \\begin{bmatrix} I_x & O \\\\\r\nO & O \\end{bmatrix} Q$ and $ Y \\equal{} R \\begin{bmatrix} I_y & O \\\\\r\nO & O \\end{bmatrix} S$,\r\n\r\nwith $ 1 \\leq x,y \\leq n$, since $ X,Y \\neq O_n$ (if r = s = n, then just choose any invertible matrix Z to prove the theorem). If we now let $ Z \\equal{} Q^{ \\minus{} 1}R^{ \\minus{} 1}$, then\r\n\r\n$ rank(XZY) \\equal{} rank \\left(P \\begin{bmatrix} I_x & O \\\\\r\nO & O \\end{bmatrix} Q \\cdot Q^{ \\minus{} 1}R^{ \\minus{} 1} \\cdot R \\begin{bmatrix} I_y & O \\\\\r\nO & O \\end{bmatrix} S\\right)$\r\n\r\n$ \\equal{} rank \\left(\\begin{bmatrix} I_x & O \\\\\r\nO & O \\end{bmatrix} \\begin{bmatrix} I_y & O \\\\\r\nO & O \\end{bmatrix} \\right) \\equal{} rank \\left(\\begin{bmatrix} I_{min(x,y)} & O \\\\\r\nO & O \\end{bmatrix}\\right) \\equal{} min (x,y) \\neq 0$,\r\n\r\nwhich means that for such $ Z$, $ XZY \\neq O_n$.\r\n\r\nFor the case $ XY \\equal{} O_n$, observe that Z, if it exists, cannot be invertible.", "Solution_3": "[quote]For the case $ XY \\equal{} O_{n}$, observe that Z, if it exists, cannot be invertible.[/quote]\r\n\r\nDon't know what I was thinking when I wrote this. The reasoning I gave above is valid, whether $ XY \\equal{} O$ or not.", "Solution_4": "There exist $ u,v$ s.t. $ Yu\\not\\equal{}0,Xv\\not\\equal{}0$. Then choose $ Z$ s.t. $ Z(Yu)\\equal{}v$." } { "Tag": [ "function", "real analysis", "real analysis solved" ], "Problem": "let f:R----->R-{3} be a funtion with the property that for some w>0\r\n\r\nf( x+w)=(f(x)-5)/(f(x)-3) for all real x. prove that f is periodic", "Solution_1": "f(x+4w)=f(x) (just apply formula 4 times and simplify)", "Solution_2": "thank you very munch cotan" } { "Tag": [ "calculus", "function", "algebra unsolved", "algebra" ], "Problem": "Does there exist a triple of distinct numbers $a,b,c$\r\nsuch that $(a - b)^5 + (b - c)^5 + (c - a)^5 = 0$.\r\nThis must be in the algebra section. Excuse me.\r\nBtw don`t post only answer, i want to know how did you get to it.", "Solution_1": "The answer is no. Put $x=a-b,y=b-c$. Then we are looking for $x,y\\neq0$ such that $x^5+y^5-(x+y)^5=0\\Leftrightarrow(x+y)(x^2+xy+y^2)=0$.\r\nNow $x+y=0$ means $a=c$, which is excluded. And $2(x^2+xy+y^2)=(x+y)^2+x^2+y^2>0$. Contradiction.", "Solution_2": "[quote=\"spanferkel\"]The answer is no. Put $x=a-b,y=b-c$. Then we are looking for $x,y\\neq0$ such that $x^5+y^5+(x+y)^5=0\\Leftrightarrow(x+y)(x^2+xy+y^2)=0$.\nNow $x+y=0$ means $a=c$, which is excluded. And $2(x^2+xy+y^2)=(x+y)^2+x^2+y^2>0$. Contradiction.[/quote]\r\n\r\n$x+y=a-c$ so the equation become $x^5+y^5-(x+y)^5=0$ .But others is ok :)", "Solution_3": "Yes :oops: I've edited the typo.\r\nBy the way, with a tiny little bit of calculus we can easily prove the same holds if we replace 5 by any other odd exponent $k\\ge3$. The function $f(x)=x^k+y^k-(x+y)^k$ with constant $y$ has its only zeros at $x=0$ and $x=-y$, both excluded.", "Solution_4": "Set $x=a-b$, $y=b-c$ , $z=c-a$. It is $x+y+z=0$ therefore the equation \r\n\r\nis equivalent to $(x+y+z)^5-(x^5+y^5+z^5)=0$.\r\n\r\nBut $(x+y+z)^5-(x^5+y^5+z^5) = \\frac{5}{2} (x+y)(y+z)(z+x)[ (x-y)^2 + (y-z)^2 + (z-x)^2 ]$\r\n\r\nNow\r\n\r\n$(a - b)^5 + (b - c)^5 + (c - a)^5 =0 \\Longrightarrow \\\\ \\\\ \\frac{5}{2} (a-c)(c-b)(b-a) [ (a+c-2b)^2+(a+b-2c)^2 +(b+c-2a)^2 ]=0$\r\n\r\nand this equation have no solutions since $a \\neq b \\neq c$ . :)" } { "Tag": [], "Problem": "1.\u4f9d\u500b10\u4f4d\u6578\u5176\u6578\u78bc\u53ea\u80fd\u662f2\u62163\u4e14\u6c92\u6709\u5169\u500b3\u662f\u76f8\u9130\u7684\r\n\u9019\u6a23\u768410\u4f4d\u6578\u6709\u591a\u5c11\u500b\r\n \r\n[b][color=blue]translate[/color][/b]\r\n\r\nIf a ten-digit integer involves 2's and 3's digits only, but no any pairwise 3's \r\n\r\njoins together , then how many such possible ten-digit integers are there?", "Solution_1": "[hide=\"solution\"]\n\nFor a 0-digit number, the answer's 1; for a 1-digit number, the answer's 2. For an n-digit number, we obtain numbers by appending a 2 to the numbers of length $ n\\minus{}1$, or a 23 to numbers of length $ n\\minus{}2$, so, for an n-digit number, it's equal to $ F_{n\\plus{}1}$. Answer is $ F_{11}\\equal{}89$[/hide]", "Solution_2": "[quote=\"Darmani\"][hide=\"solution\"]\n\nFor a 0-digit number, the answer's 1; for a 1-digit number, the answer's 2. For an n-digit number, we obtain numbers by appending a 2 to the numbers of length $ n \\minus{} 1$, or a 23 to numbers of length $ n \\minus{} 2$, so, for an n-digit number, it's equal to $ F_{n \\plus{} 1}$. Answer is $ F_{11} \\equal{} 89$[/hide][/quote]\r\n\r\nYour answer given is so different from mine.", "Solution_3": "[hide=\"Solution :arrow: \"]I divide it into 6 cases for number of 3's ranging from 0 to 5. Consider case III, there are eight 2's and we have to insert two 3's. This situation is similar to-\"8 black balls kept in a line and we have to insert 2 white ones in the 9 spaces which is equal to $ {9\\choose 2}$ \". Similarly, doing this for others also, we get:\n\nTotal combinations $ \\equal{} {11\\choose 0} \\plus{} {10\\choose 1} \\plus{} {9\\choose 2} \\plus{} {8\\choose 3} \\plus{} {7\\choose 4} \\plus{} {6\\choose 5} \\equal{} 144$.[/hide]", "Solution_4": "[quote=\"analytic\"][hide=\"Solution :arrow: \"]I divide it into 6 cases for number of 3's ranging from 0 to 5. Consider case III, there are eight 2's and we have to insert two 3's. This situation is similar to-\"8 black balls kept in a line and we have to insert 2 white ones in the 9 spaces which is equal to $ {9\\choose 2}$ \". Similarly, doing this for others also, we get:\n\nTotal combinations $ \\equal{} {11\\choose 0} \\plus{} {10\\choose 1} \\plus{} {9\\choose 2} \\plus{} {8\\choose 3} \\plus{} {7\\choose 4} \\plus{} {6\\choose 5} \\equal{} 144$.[/hide][/quote]\r\n\r\nShould the case I :$ {11\\choose 0}$ be excluded? Since the question says 2's and 3's involved in a ten-digit integer.", "Solution_5": "It says 2's and 3's only, so it doesn't matter if all are 2's only which makes that combination of 11c0 =1. So, it should not be excluded.\r\nMore precise: it says to include at maximum 2's and 3's only in the number, so if only 2's come it is also a case.", "Solution_6": "[quote=\"vinskman\"][quote=\"Darmani\"][hide=\"solution\"]\n\nFor a 0-digit number, the answer's 1; for a 1-digit number, the answer's 2. For an n-digit number, we obtain numbers by appending a 2 to the numbers of length $ n \\minus{} 1$, or a 23 to numbers of length $ n \\minus{} 2$, so, for an n-digit number, it's equal to $ F_{n \\plus{} 1}$. Answer is $ F_{11} \\equal{} 89$[/hide][/quote]\n\nYour answer given is so different from mine.[/quote]\r\n\r\nYeah, I made an OBOE. Since the case for n=1 is $ F_3$, the general formula is actually $ F_{n \\plus{} 2}$, so for n=10, it's $ F_{12} \\equal{} 144$." } { "Tag": [ "puzzles" ], "Problem": "Puzzle (mine) . . . Marbles inside of bags inside of boxes\r\n\r\n \r\nYou are given exactly 4 boxes. \r\n\r\nWhat is the minimum number of marbles needed so that the 4 boxes \r\nhave a different number of bags placed into each of them, and the bags \r\nare to have a different number of marbles placed into each of them? \r\n\r\nBags can be inside of bags and boxes can be inside of boxes.", "Solution_1": "[hide]Box 1 contains bag 1 and box 2. (3 bags total)\n\nBox 2 contains bag 2 and box 3. (2 bags total)\n\nBox 3 contains bag 3 and box 4. (1 bag total)\n\nBox 4 contains nothing. (0 bags total)\n\nBag 1 contains nothing; bag 2 contains 1 marble; bag 3 contains 2 marbles, for a total of 3.[/hide]", "Solution_2": "I don't understand. Why is having boxes inside other boxes making the minimum less?" } { "Tag": [ "ratio", "geometry" ], "Problem": "The lengths of the sides of two similar rectangular billboards are in the\r\nratio 5:4. If 250 square feet of material is needed to cover the larger bill-board,\r\nhow much material, in square feet, is needed to cover the smaller billboard?", "Solution_1": "[hide=\"hint\"]\n\nif the ratio of side lengths of similar figures is $\\frac{a}{b}$, then the ratio of areas is $\\frac{a^{2}}{b^{2}}$\n\nso if the ratio here is $\\frac{5}{4}$, the ratio of areas would be $\\frac{25}{16}$\n\n[/hide]", "Solution_2": "[quote=\"Interval\"]The lengths of the sides of two similar rectangular billboards are in the\nratio 5:4. If 250 square feet of material is needed to cover the larger bill-board,\nhow much material, in square feet, is needed to cover the smaller billboard?[/quote]\r\n\r\n[hide=\"solution\"]the ratio of the lengths if $\\frac{25}{16}$ because it is the ratio of the sides squared. \n\n$\\frac{25}{16}=\\frac{250}{x}$\n\n$x=\\boxed{160}$[/hide]" } { "Tag": [ "MATHCOUNTS", "AMC", "AIME", "USA(J)MO", "USAMO", "function", "calculus" ], "Problem": "I realize each year, some participants make ctdown that few anticipate, and others who were \"good bets\" fall short of their goals, but without naming names, I know of several competitiors this year who did \"all the right things\" effort wise - some went to competitive math camps, many did as many practice problems as anyone else (I mean they did [b]all[/b] available problems at least once), were confident going into the competition, did extremely well at prior MATHCOUNTS competitions, etc, and amazingly (to many) either fell many places from prior Nats years, or simply fell way short of their expected goal. And not to diminish the talents/work ethic of those who scored \"better than anticipated\" - but some of this year's top finishers at Nats had consistently lower scores ([i]much lower [/i]in some instances) when comparing their scores from both prior Mathcts competitions, as well as their scores like math SAT, AMC, AIME etc, to those who finished much lower, and well -- it seemed to happen more this year than in prior years - and I'm trying to get a \"handle\" on \"Why\"? \r\n\r\nI know at least one of these \"poor scorers\" who aced the Target with all 8 right, but totally bombed on Sprint and they [i]still[/i] can't quite grasp what was so different about this yr's Sprint questions to those of the previous 20 years' tests. Anyone have any theories about this? Did all these \"top kids\" simply have a bad day all at the same time, or was there something about the questions that maybe they were more likely to \"misread\" or \"overcomplicate\" or ??????? (oh, I have to discount \"actual test taking ability\" as many of those that \"fell short\" had pwned many prior competitions).", "Solution_1": "This happens every year - some under-the-radar people do extremely well, and some who are expected to score well don't. There's nothing really unusual about it...", "Solution_2": "But the \"number of places\" some people \"fell\" or \"climbed\" does seem unusual to me from just what I know of prior year's scores, participants, etc. Of course, I don't know if anyone has ever done a statistical analysis of the exact participants in any given 3 - 6 year period (or whatever period would cover all possible \"2 or 3-peaters\") to see the exact pattern of what percentage of people either rose or fell from their prior year's placings -, and by how much, (oh, just focusing on Nats, as including States and Chapters would be unwieldy and include too many people) -- but my general observation was that there was much more variance this year, and in prior years there seemed to be a more logical assessment of who might end up in the top 12. (of course, it's never an exact science - I'm just speaking \"in general\") \r\nAs I pointed out (as AMC scores etc. are often made public - especially the names of those with perfect scores), it seemed that many of the top kids at Nats didn't have the best scores from other tests overall - and it just seems -well, inconsistent - slightly \"illogical\" to me, just how that can be. I just wonder if anyone thought this year's Nat's test somehow varied from prior years? And if so, how? Nature of problems? Way they were posed? ??? AIME15 - I don't know if you were at Nats this year or were just commenting \"in general\"?", "Solution_3": "Every single person in cd was a USAMO qualifier, so no one there was truly \"under the radar\".", "Solution_4": "Mewto55555, true, but there were [i]many[/i] USAMO qualifiers this year - I believe 25 or more, and of course, only 12 people could make it into ctdown. My point was that if you compared these (top 12) people's actual USAMO scores (with the exception of those few who clearly blew the averages out of the water on all accts! :) - plus this same group's AMC, AIME scores, etc, with non-top 12's, many might be surprised how many of those scoring \"higher\" on what most agree to be \"higher level tests\", (or who previously placed much higher at prior Nats than those in the ctdown), did not make ctdown, and I just found this interesting. And several participants I spoke to felt they did much \"differently\" (again, some better, some much worse) than on all the prior tests they practiced on. Which indicates to me a potential \"shift\" in the style of questions .. or something. \r\n\r\nOR .. maybe it just indicates that MATHCOUNTS success is not a good correlation or indicator of one's ability to ultimately succeed at much more complex, higher level math. Otherwise, wouldn't you expect all 25 USAMO'ers to make up the top 25 places?? (which of course, didn't occur)", "Solution_5": "[quote=\"mathlearner\"]\nOR .. maybe it just indicates that MATHCOUNTS success is not a good correlation or indicator of one's ability to ultimately succeed at much more complex, higher level math. Otherwise, wouldn't you expect all 25 USAMO'ers to make up the top 25 places?? (which of course, didn't occur)[/quote]\r\n\r\nThere are 2 things wrong with this.\r\n\r\n1. MATHCOUNTS success is not a good correlation or indicator of one's ability to ultimately succeed at much more complex, higher level math. MATHCOUNTS doesn't require any actual creative thinking or unusual methods; mathcounts problems are all problems where if you have seen a certain, well-known, trick before, then the problem is trivial. It might take some time to solve it though, but I'm not sure that time required to solve indicates math skill.\r\n\r\n2. Um, \"wouldn't you expect all 25 USAMO'ers to make up the top 25 places??\" is flawed. There are a lot of people at nats who shoudl have made usamo, but were unlucky on aime/amc. \r\n\r\nFinally, I have to say, the real test is not a practice test. People perform differently on both. Example: CA team practices...", "Solution_6": "pythag011, mmm.. excellent points. I definitely agree with #1 - (and I think that's also quite reassuring to those \"budding mathematicians\" who may NOT have ever made Nats ctdown .. but I'm not so sure about #2. I mean, I get your point, but not sure \"luck\" is as big a factor as some people like to think -- sure, you can get more or less problems you're familiar with, but I think most qualifiers qualify by more than a single problem, and thus, their mathematical acumen plays a much larger part. And also, focusing on the MATHCTS participants who \"were\" USAMO qualifiers - not those that [i]should have been [/i]indicates that (at least these) participants were certainly \"skilled enough\" to make top 12 Nats, but somehow didn't ... :( \r\n\r\nBut the mystery question remains ... \"[i]why[/i]\"??\r\n\r\nI've heard top coaches say the top 12 could differ if the test were given on a different day to the same group, but I wonder - is that variation more a function of the test, or just human factors? (ie: being tired, sick, etc.) I'm guessing both.", "Solution_7": "Expected Results are sporadic...", "Solution_8": "I think this just shows how fine the line is between getting into the top 12 and \"just missing\". One or two questions could make a difference of several places in the standings. That is why I always hope for very hard tests so that there is a very clear differentiation in both teams and individuals. \r\n\r\nI thought all of the sprint tests this year were right on but the target rounds were too easy. The team rounds were also a bit on the easy side making it difficult for any team to make up ground they had lost in the sprint or target.\r\n\r\nThe tests seem to run in cycles. The two or three years before this year were on the hard side. This year the cycle seemd to turn in the opposite direction. I can usually tell by the school and chapter round what the difficulty of the state and national is going to be.", "Solution_9": "The team round was insanely easy (compared to previous years). Two people on our team had literally memorized #10.\r\n\r\n\r\nAs for this years test, I actually did one point better than my averages. Imo, this test was a lot more arithmetic/brute force/listing out intensive than previous ones, which suited me well, but perhaps not so much the others.", "Solution_10": "A lot of people in nats this year memorized #10 on team round. Considering how 8 teams got a 10/10 on team, this year was very easy, while last year, only one team got a perfect.\r\n\r\nI think it's not only the Mathcounts tests that became much easier this year, but also some of the AMCs. In my opinion, the 2008 AMC8 was easier than the 2007 one, and there were more than 400 perfect scores this year, while last year, there were only about 100 perfect scores. I wonder if that somewhat has to do with how calculators are banned starting 2008.", "Solution_11": "[quote=\"remy1140\"]A lot of people in nats this year memorized #10 on team round. Considering how 8 teams got a 10/10 on team, this year was very easy, while last year, only one team got a perfect.\n\nI think it's not only the Mathcounts tests that became much easier this year, but also some of the AMCs. In my opinion, the 2008 AMC8 was easier than the 2007 one, and there were more than 400 perfect scores this year, while last year, there were only about 100 perfect scores. I wonder if that somewhat has to do with how calculators are banned starting 2008.[/quote]\r\n\r\nAMC 8 is only one AMC.\r\n\r\nI do have to agree that the AMC 10s were easier, but the AMC 12s were in my opinion harder this year.\r\n\r\nWas last year's team really that hard? Looking at the problems didn't give me that impression at ALL.", "Solution_12": "Uh, please excuse my ignorance but what do you mean when you say people \"memorized problem #10 of the Nats team round? I mean, how do you \"memorize\" a problem before you are given it to solve? Do you mean this problem was seen on a prior test? If so, what year?", "Solution_13": "Basic derangement problems. I'm not sure if it's appeared in MATHCOUNTS before, but I've definitely seen it in FTW (it asked for number of ways, not probability).", "Solution_14": "I saw almost the exact problem in Alcumus as well.", "Solution_15": "[quote=\"remy1140\"]I saw almost the exact problem in Alcumus as well.[/quote]\r\n\r\nOther than some silly brute force problems, all mathcounts problems are well known.", "Solution_16": "[quote=\"pythag011 should have\"][quote=\"remy1140\"]I saw almost the exact problem in Alcumus as well.[/quote]\n\nOther than some silly brute force problems, most mathcounts problems are well known.[/quote] edited for truth", "Solution_17": "[quote=\"Ihatepie\"][quote=\"pythag011\"][quote=\"remy1140\"]I saw almost the exact problem in Alcumus as well.[/quote]\n\nOther than some silly brute force problems, most mathcounts problems are well known.[/quote] edited for truth[/quote]\r\nAgreed. It says \"pythag011 wrote: ...\" but it isn't what pythag wrote. JK", "Solution_18": "Ok I edited it again. :P Not the best grammar, but still." } { "Tag": [ "geometry", "ratio", "geometric transformation", "reflection", "circumcircle", "trigonometry", "Euler" ], "Problem": "Let $ ABC$ be a triangle, $ P$ is an arbitrary point in the plane of triangle $ ABC$, let $ P_{1},P_{2},P_{3}$ be the reflections of $ P$ in the sides $ BC,CA,AB$, respectively. Dentoe by $ O$ the circumcenter, $ H$ the orthocenter of triangle $ ABC$, respectively. show that $ \\frac {S_{\\bigtriangleup HP_{1}P_{2}}}{S_{\\bigtriangleup HP_{1}P_{3}}} \\equal{} \\frac {S_{\\bigtriangleup OAB}}{S_{\\bigtriangleup OAC}}$ (where $ S$ denotes the area of triangle $ XYZ$)", "Solution_1": "Let $ Q$ and $ R$ be a reflection point of $ P_{3}$ and $ P_{2}$ wrt $ BC$\r\n$ H_{1}$ is a reflection of $ H$ wrt $ BC$. $ H_{2}$ is a reflection of $ H_{1}$ wrt $ AB$. $ H'$ is a reflection of $ H_{2}$ wrt $ BC$.\r\n$ H_{3}$ is a reflection of $ H_{1}$ wrt $ AC$. $ H''$ is a reflection of $ H_{3}$ wrt $ BC$....-_-;;;\r\nIt is easy to show that $ H',H_{1},H''$ lie on a line.\r\n$ HP_{1}P_{2}\\equiv H_{1}PR$, $ HP_{1}P_{3} \\equiv H_{1}PQ$....\r\n$ PH_{1} \\equal{} H'Q \\equal{} H''R$...\r\n$ \\angle H'BH_{1}\\equal{}\\angle QBP\\equal{}2\\angle B$, $ \\angle H''CH_{1}\\equal{}\\angle RCP\\equal{}2\\angle C$...\r\n$ \\frac {H'H_{1}}{H_{1}H''} \\equal{} \\frac {BH_{1}\\sin B}{CH_{1}\\sin C} \\equal{} \\frac {\\sin 2B}{\\sin 2C} \\equal{} \\frac {S_{AOC}}{S_{AOB}}$\r\nSo remain is to show that $ \\frac {d(Q,PH_{1})}{d(R,PH_{1})} \\equal{} \\frac {H'H_{1}}{H_{1}H''}( \\equal{} \\frac {S_{AOC}}{S_{AOB}})$\r\nBut $ \\frac {d(H',PH_{1})}{d(H'',PH_{1})} \\equal{} \\frac {H'H_{1}}{H_{1}H''} \\equal{} \\frac {S_{AOC}}{S_{AOB}}$\r\nand $ d(H',PH_{1}) \\minus{} d(Q,PH_{1}) \\equal{} H'Q \\sin \\measuredangle (H'Q,PH_{1}) \\equal{} H'Q \\sin 2B \\equal{} PH_{1}\\sin 2B$,\r\n$ d(H'',PH_{1}) \\minus{} d(R,PH_{1}) \\equal{} H''R\\sin 2C \\equal{} PH_{1}\\sin 2C$\r\n$ \\rightarrow \\frac {d(H',PH_{1}) \\minus{} d(Q,PH_{1})}{d(H'',PH_{1}) \\minus{} d(R,PH_{1})} \\equal{} \\frac {S_{AOC}}{S_{AOB}}$,\r\nIt's obvious...", "Solution_2": "I want to see your solution... this was a very curious problem for me. Although I satisfied with my solution, I think there are simpler solutions using easy technique...", "Solution_3": "Dear Soo Hong Lee\r\nMy solution is similar to yours, by trigonometry, we have $ S_{\\bigtriangleup HP_{2}P_{3}} \\equal{} \\frac {R^2 \\minus{} OP^2}{2}sin2A$. where $ R$ denotes the circumradius of triangle $ ABC$.\r\nYou are famous for your genius in China, we also like Han-sol Shin (username Leonhard Euler), all his geometrical problems are very interesting and pretty hard. Do you know him?", "Solution_4": "Who called me genius???? I'm just a lucky boy!! :blush:\r\nBTW, of course, I know him. He is 1 year elder than me. But strange thing is that such a \"monster\" in geometry is in middle school grade in Korea Winter School in this year!!\r\nI first met him in a web forum named \"KAIST mathmatical problem solving group\". That group have published a magazine \"Math Letter\", and I sometimes proposed easy problems include geometry(when I was young).\r\nThen suddenly a strange person appeared in that forum, who used nickname \"Leonhard Euler\", and proposed numerous geometry problems hard enough...\r\nI don't know how does he made such a curious and hard geometry problems, but It doesn't a accidental result, certainly." } { "Tag": [ "group theory", "abstract algebra", "induction", "superior algebra", "superior algebra unsolved" ], "Problem": "If a group has only representations of dimension 1 and 2, what can I say about it? For example, a group that has representations of only dimension 1 will have to be abelian.\r\n\r\nFor the 1 and 2 dimensional case, I think the answer is that it's solvable, but I can't quite prove it...", "Solution_1": "This is an interesting question. Such groups are bound to have other nice properties; maybe we can even find a characterization, but for now I'll prove their solvability.\r\n\r\nFirst, notice that the property of having irreducible complex representations of degree at most $2$ is preserved by passing to quotient groups and subgroups. For quotient groups this is clear: irreducible representations of $G/H$ give rise to irreducible representations of $G$ of the same degree after composing with the projection $G\\to G/H$. On the other hand, let $H$ be any subgroup of $G$, and let $\\rho$ be an irreducible representation of $H$. Also, let $\\psi$ be an irreducible representation of $G$ which occurs as a summand in the representation $\\mbox{Ind}^{G}_{H}\\rho$ of $G$ induced from $\\rho$. By [url=http://planetmath.org/encyclopedia/FrobeniusReciprocity.html]Frobenius recirpocity[/url], the multiplicity of $\\psi$ in $\\mbox{Ind}^{G}_{H}\\rho$ is equal to the multiplicity of $\\rho$ in the restriction $\\mbox{Res}^{G}_{H}\\psi$ of $\\psi$ to $H$. We chose our representations such that this multiplicity is non-zero, so $\\rho$ is a summand in the $1$ or $2$-dimensional representation $\\mbox{Res}^{G}_{H}\\psi$ of $H$. This proves that the degree of $\\rho$ is at most $2$, as desired.\r\n\r\nIf the only $1$-dimensional representation is the trivial one, then $G$ has odd order (since the order is the sum of the squares of the degrees of the irreducible representations). The degree of an irreducible representation divides the order of $G$, so in this case there can be no irreducible representations of degree $2$. It follows that $G$ is trivial. This means that if $G$ is non-trivial, then it has non-trivial $1$-dimensional representations, so the derived subgroup $G'$ is proper. By induction on the order, $G'$ is solvable. Also, $G/G'$ is solvable, being abelian. Thus $G$ is solvable, and we're done.\r\n\r\n\r\nP.S.\r\n\r\nNotice also that all odd order subgroups of $G$ are abelian. Coupled with the solvability, this says quite a bit." } { "Tag": [ "trigonometry", "MATHCOUNTS", "LaTeX" ], "Problem": "What were the final standings?", "Solution_1": "When was NEMO? I think I was supposed to write questions, but I completely forgot about it. Also, I don't think my middle school sent any teams; if it already happened, then someone probably should have posted a reminder of some type...", "Solution_2": "Ligon GT Magnet Middle School had two teams, and both teams took the top 2 places at NEMO. The winning team's captain was Nick Tobey. \r\n1st place - Ligon \r\n2nd place - Ligon \r\n3rd place - Martin\r\n4th place - Carnage\r\nAaah, I can't remember the rest\r\nFor individuals:\r\n1st place - Calvin Deng; Ligon\r\n2nd place - I think its Tejas Sundarasen...; Ligon\r\n3rd place - Nick Tobey; Ligon\r\n4th place - Ivan Kuznetsomething; Ligon\r\n5th place - a girl from Martin I think\r\n6th place - someone from Ligon\r\nNot sure bout the other ones.\r\nThere weren't that many schools, about five maybe\r\nLIGON DOMINATES :D", "Solution_3": "Nemo was on... 1/5/08.\r\n\r\n1. Calvin Deng, Ligon, 24\r\n2. Allen Yang, Ligon, 23 (on individual tiebreaks; on tiebreaker round, 3rd)\r\n3. Nick Tobey, Ligon, 23 (reversed with Allen on tiebreaker round)\r\n4. Lisa Zheng, Martin, 16\r\n5. Tejas Sundarasen, Ligon, 15\r\n6. Alan Wu, Carnage, 15\r\n\r\nso yeah\r\n\r\n5th was also carnage\r\n6th was carrington\r\n\r\nonly 6 teams\r\n\r\nThe 2nd place ligon team was winning until individual. They got 60/80 on application, while the 1st place ligon team got 57/80. Both Ligon teams got 8/10 on team. (Calvin missed #1 on team :| )", "Solution_4": "Calvin is my friend!!! :spam: So don't trash talk about him! (By the way, he said some guy called Pranav said one-ninth)", "Solution_5": "hi\r\n\r\nI thought I lost on the tiebreaker", "Solution_6": "so actually Allen Yang was 6th after tiebreakers because he got a 1 and everyone else got a 2 :wink: \r\n\r\nI'm not trash-talking Calvin, eh? because... he did well, eh?\u3000but missing an easy problem is not helpful.", "Solution_7": "[quote=\"not_trig\"]so actually Allen Yang was 6th after tiebreakers because he got a 1 and everyone else got a 2 :wink:[/quote]\r\n\r\nHeh. This is pretty amusing. Unfortunately, that is about all I have to offer on this matter.\r\n\r\nYou guys should make the tiebreakers easier, maybe.", "Solution_8": "But It was all someone called Pranav!!!", "Solution_9": "[quote=\"fibointepi\"]Calvin is my friend!!! :spam: So don't trash talk about him! (By the way, he said some guy called Pranav said one-ninth)[/quote]\r\n\r\nSomebody is their own friend. :maybe: \r\n\r\nPranav said one-ninth... Whats that supposed to mean??", "Solution_10": "The answer was $ \\frac{1}{12}$, but Pranav (a guy on my friend's team) said $ \\frac{1}{9}$. :cursing:", "Solution_11": "so i was wrong. whatever. and who's trash talking calvin anyways?\r\n\r\nand also, not_trig, one of the teachers said that the individual scores didnt count for the final team placements.", "Solution_12": "[quote=\"yoonzik\"]so i was wrong. whatever. and who's trash talking calvin anyways?\n\nand also, not_trig, one of the teachers said that the individual scores didnt count for the final team placements.[/quote]\r\n\r\nHeh I don't completely know... Vitek and Vivek were being rather cryptic", "Solution_13": "To clear all this up:\r\n\r\nTo break ties on individuals for awards, we used a similar system to that employed by Mathcounts: we sort first by highest score, then by hardest problem answered correctly. So, for example, if Jane and Jill both get 25 right, but Jane gets 1-25 right and Jill gets 1-24 and 26 right, then Jill is placed higher than Jane.\r\n\r\nWe used tiebreakers to determine plaques, I believe. Scores were determined by a combination of team, individual, and application scores. Team scores were multiplied by a scaling factor to make sure that they counted for an appropriate amount.\r\n\r\nEDIT: Resultes were emailed out. Maybe I'll post them later if I feel like it.", "Solution_14": "Oh ya Teki-Teki when are you going to post results on the NEMO website?", "Solution_15": "I was on the 2nd Place Team. Missed 6 in individual round! :(", "Solution_16": "Welcome, Sanjay :lol: You might want to check the Latex forum for help.", "Solution_17": "Teki-Teki the results on the NEMO website still say that the top finishers were are Some Kid (Some School) and the winning teams were Some School. Please update.", "Solution_18": "the results are already all done, i saw it on the enloe website", "Solution_19": "Sanjay, If you only missed 6, you must not have attempted all of the questions, or else you would have tied with me.", "Solution_20": "Sorry About that-\r\n\r\n-I missed 15, mistype!", "Solution_21": "[quote=\"dnkywin\"]Teki-Teki the results on the NEMO website still say that the top finishers were are Some Kid (Some School) and the winning teams were Some School. Please update.[/quote]\r\n\r\nNo, actually it says that our account has been suspended. Dunno when you checked. Hrmm... I am not in charge of the website either, so don't gripe to me.", "Solution_22": "I checked 1:47 PM, 2-10-08 :lol:" } { "Tag": [ "modular arithmetic", "function", "algebra unsolved", "algebra" ], "Problem": "Let $m$ be a positive integer and $\\{a_n\\}_{n\\geq 0}$ be a sequence given by $a_0 = a \\in \\mathbb N$, and \\[ a_{n+1} = \\begin{cases} \\displaystyle \\frac{a_n}2 & \\textrm { if } a_n \\equiv 0 \\pmod 2, \\\\ a_n + m & \\textrm{ otherwise. } \\end{cases} \\]\r\nFind all values of $a$ such that the sequence is periodical (starting from the beginning).", "Solution_1": "I don't understand what's happening in this problem. All terms are clearly nonnegative integers and if the sequence is periodical from a certain rank, it is bounded and thus for large $n$ one can only have $a_{n+1}=\\frac{a_n}{2}$ and $a_n$ even. Or, since we are dealing with integers, one should have $a_n=0$ for large $n$ and coming back the first term would be zero. I don't get something obvious (sorry if I'm making a huge mistake, I've been struggling with physics for more than 6 hours) or this isn't the statement of the real problem???", "Solution_2": "I might have misunderstood the problem wrong (I got it by telephone) ... other than that for this particular problem, your solution seems to be correct.", "Solution_3": "The problem statement is:\r\n\r\n[i]Let $m$ be a positive integer. Find all positive integers $\\alpha$ such that the sequence $\\left\\{ a_n \\right\\}^\\infty_{n=0}$, defined as:[/i]\r\n\r\n$a_0 = \\alpha$\r\n$a_{n+1} = \\left\\{ \\begin{array}{ll} \\frac{1}{2} a_n & a_n \\equiv 0 \\: (mod \\: 2) \\\\ a_n + m & a_n \\equiv 1 \\: (mod \\: 2) \\end{array} \\right.$\r\n\r\n[i]for $n = 0, 1, 2, \\ldots$, is periodic, that is, its terms repeat starting from: $a_0, a_1, a_2, \\ldots, a_k$, for some $k$.[/i]", "Solution_4": "Let $A$ be the set of all solutions $a$. First, if $m$ is even there is no solution. So suppose $m$ is odd and WLOG suppose $a_0 = \\min \\{a_0, a_1, \\dots, a_k \\}$. Now $a_0$ is also odd and $a_0 \\leq \\frac{a_0+m}2$ so $a_0 \\leq m$. If $a_i \\leq m$ then $a_{i+1} = \\frac{a_i} 2 \\leq m$ or $a_{i+1} = a_i+m \\leq 2m$ is even and $a_{i+2} = \\frac{a_i+m}2 \\leq m$, and we conclude that only numbers greater than $m$ in $A$ could be those even not greater than $2m$. Now we will prove that $A = \\{1, 2, 3, \\dots, m-1,m, m+1, m+3, \\dots, 2m-2, 2m \\}$. Consider oriented graph whose vertices are members of $A$ and edges from $x$ to $x/2$ if $x$ is even and from $x$ to $x+m$ if $x$ is odd. The out-degree of every vertice is 1, and since all $x \\leq m$ has only in-edge from $2x$ and for all $x>m$ only in-edge is from $x-m$, the in-degree is also 1 for all vertices and that means the graph could be divided into cycles and that concludes the proof.", "Solution_5": "This is the problem which is changed the title of the one, Baltic Way Problem 5, 1997. ;)\r\n[url=http://www.hh.schule.de/ifl/mathematik/baltic97.pdf]www.hh.schule.de/ifl/mathematik/baltic97.pdf[/url]\r\n\r\nWe can see almost same problems. Contest's problems are recycling, if so I am dissapointed. :(\r\n\r\nGood problems are repeating! ;)", "Solution_6": "Oh, the problem from Baltic Way is way much easier... One just need to prove the sequence is bound and thats the end.", "Solution_7": "Quite similar problem in France 1996 \r\n\r\nhttp://www.imo.org.yu/othercomp/Fra/FraMO96.pdf\r\n\r\nQuite spread problem.....", "Solution_8": "its also on mathematical olympiad challenges", "Solution_9": "[size=150][b]Problem 4.(BMO 2006)[/b][/size]\r\n\r\n\r\nSolution:\r\nIf $n$ is even it is clear that no $m$ satisfies the problem.\r\n\r\nLet $n$ be odd. Then the solutions are\r\n$M=( {1,2,3,...,m}\\cup {m+1,m+3,m+5,...,2m} )$\r\n\r\nIf $a=m$ the sequence is $m,2m,m,2m,....$\r\nIf $a=2m$ the sequence is $2m,m,2m,m....$\r\n\r\nLet $S=M-{m,2m}$\r\n\r\nLet's divide $S$ into subsets:\r\n\r\n$P_1={1.2^0,1.2^1,....,1.2^{k_1}},$ where$m< 1.2^{k_1} <2m;\\\\ P_3={3.2^0,3.2^1,....,3.2^{k_3}},$where$m< 3.2^{k_3} <2m;\\\\ P_5={5.2^0,5.2^1,....,5.2^{k_5}},$ where$m< 5.2^{k_5} <2m;\\\\ ... P_{m-2}={(m-2).2^0,(m-2).2^1,....,(m-2).2^{k_{m-2}}}, where m< (m-2).2^{k_{m-2}} <2m;\\\\$\r\nThen if $a_i\\in P_t$ then $a_{i+1}$ is the previous element of $P_t$, if there is such, or $a_{i+1}=a_i + m$ which on the other hand is last element of $P_q$. \r\n\r\nLet define the function $f(t)=q$ as the index of the set we will get in after we leave $P_t$.\r\n\r\nIf $f(t)=f(l)$ then $t+m=l+m$, and $t=l$ so $f$ i inection.\r\n\r\nSo we have $P{}_t{}_1\\rightarrow P{}_t{}_2\\rightarrow ... P{}_t{}_i\\rightarrow P{}_t{}_1$.\r\n\r\nBecause at some moment we will be in one set twice, and if we take the first such set this must be $P{}_t{}_1$.\r\n\r\nBy this we prooved that if $a\\in M$ then a is a solution and also we see that then $a_i\\in M$ for all $i$ !!!\r\n\r\n\r\nLet $a \\notin M$. Then $a>m$ and consider $that a$ is solution.\r\nIf $a_i$ is odd then $a_i+1=a_i+m$ (this is even) and $a_i+m=2^k.a_{i+k}, a_{i+k}$- odd $\\Rightarrow a_{i+k}=\\frac{a_i +m}{2^k} \\leq \\frac{a_i +m}{2} \\Rightarrow (a_{i+k}-m) \\leq \\frac 12 (a_i - m)$.\r\n\r\nSo, at some time we will have a member of the sequence smaller or equal than m, so we will have $a_j \\in M$ for some $j$. Then all $a_J\\in M (J>j)$, and because the sequence is periodic, $a_0 \\in M$, contradiction.\r\n\r\nAnswer: $a\\in {1,2,3,...,m}\\cup {m+1,m+3,m+5,...,2m}$", "Solution_10": "[quote=n0vakovic]Let $A$ be the set of all solutions $a$. First, if $m$ is even there is no solution. So suppose $m$ is odd and WLOG suppose $a_0 = \\min \\{a_0, a_1, \\dots, a_k \\}$. Now $a_0$ is also odd and $a_0 \\leq \\frac{a_0+m}2$ so $a_0 \\leq m$. If $a_i \\leq m$ then $a_{i+1} = \\frac{a_i} 2 \\leq m$ or $a_{i+1} = a_i+m \\leq 2m$ is even and $a_{i+2} = \\frac{a_i+m}2 \\leq m$, and we conclude that only numbers greater than $m$ in $A$ could be those even not greater than $2m$. Now we will prove that $A = \\{1, 2, 3, \\dots, m-1,m, m+1, m+3, \\dots, 2m-2, 2m \\}$. Consider oriented graph whose vertices are members of $A$ and edges from $x$ to $x/2$ if $x$ is even and from $x$ to $x+m$ if $x$ is odd. The out-degree of every vertice is 1, and since all $x \\leq m$ has only in-edge from $2x$ and for all $x>m$ only in-edge is from $x-m$, the in-degree is also 1 for all vertices and that means the graph could be divided into cycles and that concludes the proof.[/quote]\n\nCan $a_0 = 0$ ?" } { "Tag": [ "trigonometry", "geometry", "circumcircle", "geometry solved" ], "Problem": "If a triangle $\\triangle ABC$ is harmonic show that :\r\n\r\n(1) $\\Delta = R \\cdot \\rho = R^2 \\tan \\omega $\r\n\r\n(2) $\\frac{1}{{\\rho^2 }} = \\frac{1}{{h_a^2 }} + \\frac{1}{{h_b^2 }} + \\frac{1}{{h_c^2 }}$\r\n\r\nwhere $\\Delta$ is the area, R is the circumradius, $ \\rho$ is the radius of the second Lemoine circle, $\\omega$ is the Brocard angle, $h_a,h_b,h_c$ are the altitudes of triangle $\\triangle ABC$.\r\n\r\n__________________________________________________________________________________________________________________________\r\n\r\n[b]P.S.[/b] A triangle $\\triangle ABC$ is said [i]harmonic[/i] if it satisfies one of the following equivalent conditions:\r\n\r\n(a) $a^2 + b^2 + c^2 = 4R^2 $\r\n\r\n(b) $ \\sin ^2 \\alpha + \\sin ^2 \\beta + \\sin ^2 \\gamma = 1$\r\n\r\nwhere $a,b,c$ are the lengths of the sides $BC, CA, AB$ and $\\alpha$ , $\\beta$ , $\\gamma$ are the angles with vertices $A,B,C$ .", "Solution_1": "Let's prove that $\\Delta=R^2\\tan{\\omega}$.\r\n\r\nBy well-known theorem we have $\\frac{a}{\\sin{\\alpha}}=\\frac{b}{\\sin{\\beta}}=\\frac{c}{\\sin{\\gamma}}=2R$. Then $\\Delta=\\frac{1}{2}ab\\sin{\\alpha}=2R^2\\sin{\\alpha}\\sin{\\beta}\\sin{\\gamma}$.\r\n\r\nLet P be the first Brocard point. Since $PC=\\frac{AC\\sin{\\angle{CAP}}}{\\sin{\\angle{APC}}}$ and $PC=\\frac{BC\\sin{\\angle{CBP}}}{\\sin{\\angle{BPC}}}$ then $\\frac{\\sin{\\omega}\\sin{beta}}{\\sin{\\gamma}}=\\frac{\\sin{(\\beta-\\omega)}\\sin{\\alpha}}{\\sin{\\beta}}$. And since $\\sin{(\\beta-\\omega)}=\\sin{\\beta}\\cos{\\omega}-\\cos{\\beta}\\sin{\\omega}$ we have $\\cot{\\omega}=\\cot{\\beta}+\\frac{\\sin{\\beta}}{\\sin{\\alpha}\\sin{\\gamma}}$. But $\\sin{\\beta}=\\sin{(\\alpha+\\gamma)}=\\sin{\\alpha}\\cos{\\gamma}+\\sin{\\gamma}\\cos{\\alpha}$. And finally we obtain $\\cot{\\omega}=\\cot{\\alpha}+\\cot{\\beta}+\\cot{\\gamma}$.\r\n\r\nSo, we have to prove that $\\Delta=R^2\\tan{\\omega}$. By the previous equalities we have to prove $2\\sin{\\alpha}\\sin{\\beta}\\sin{\\gamma}(\\cot{\\alpha}+\\cot{\\beta}+\\cot{\\gamma})=1$.\r\n\r\nIt is well-known that $\\cot{\\alpha}+\\cot{\\beta}=\\frac{\\sin{\\gamma}}{\\sin{\\alpha}\\sin{\\beta}}$, $\\cot{\\beta}+\\cot{\\gamma}=\\frac{\\sin{\\alpha}}{\\sin{\\beta}\\sin{\\gamma}}$ and $\\cot{\\gamma}+\\cot{\\alpha}=\\frac{\\sin{\\beta}}{\\sin{\\gamma}\\sin{\\alpha}}$.\r\n\r\nTherefore $2\\sin{\\alpha}\\sin{\\beta}\\sin{\\gamma}(\\cot{\\alpha}+\\cot{\\beta}+\\cot{\\gamma})=\\sin{\\alpha}\\sin{\\beta}\\sin{\\gamma}((\\cot{\\alpha}+\\cot{\\beta})+(\\cot{\\beta}+\\cot{\\gamma})+(\\cot{\\gamma}+\\cot{\\alpha}))=\\sin{\\alpha}\\sin{\\beta}\\sin{\\gamma}(\\cot{\\alpha}+\\cot{\\beta})+\\sin{\\alpha}\\sin{\\beta}\\sin{\\gamma}(\\cot{\\beta}+\\cot{\\gamma})+\\sin{\\alpha}\\sin{\\beta}\\sin{\\gamma}(\\cot{\\gamma}+\\cot{\\alpha})=\\sin^2{\\alpha}+\\sin^2{\\beta}+\\sin^2{\\gamma}=1$ as desired.\r\n\r\nP.S. I don't know what \"second Lemoine circle\" means. So I can't solve other parts of the problem.", "Solution_2": "[quote=\"Remike\"]\nP.S. I don't know what \"second Lemoine circle\" means. So I can't solve other parts of the problem.[/quote]\r\n\r\nIn following link some information are given on the second Lemoine circle \r\n\r\nhttp://mathworld.wolfram.com/CosineCircle.html\r\n\r\nThanks for the solution of the first part of the problem !\r\n\r\nLeon", "Solution_3": "[quote=\"Leon\"]If a triangle $\\triangle ABC$ is harmonic show that :\n\n(1) $\\Delta = R \\cdot \\rho = R^2 \\tan \\omega $\n\n(2) $\\frac{1}{{\\rho^2 }} = \\frac{1}{{h_a^2 }} + \\frac{1}{{h_b^2 }} + \\frac{1}{{h_c^2 }}$\n\nwhere $\\Delta$ is the area, R is the circumradius, $ \\rho$ is the radius of the second Lemoine circle, $\\omega$ is the Brocard angle, $h_a,h_b,h_c$ are the altitudes of triangle $\\triangle ABC$.[/quote]\r\n\r\nThis problem is straightforward.\r\n\r\nAt first, we will use the formula $\\rho=R\\tan\\omega$ for the radius $\\rho$ of the second Lemoine circle of triangle ABC.\r\n\r\nIt is well-known that the Brocard angle $\\omega$ of triangle ABC satisfies $\\cot\\omega=\\frac{a^2+b^2+c^2}{4\\Delta}$. Taking the reciprocal of this equation, we get $\\tan\\omega=\\frac{4\\Delta}{a^2+b^2+c^2}$.\r\n\r\nThe area $\\Delta$ of triangle ABC can be calculated using the formula $\\Delta=\\frac12 ah_a=\\frac12 bh_b=\\frac12 ch_c$. Thus, $h_a=\\frac{2\\Delta}{a}$, $h_b=\\frac{2\\Delta}{b}$, $h_c=\\frac{2\\Delta}{c}$.\r\n\r\nUsing these all formulas, we see that your equation (1):\r\n\r\n$\\Delta=R\\cdot\\rho$ is equivalent to\r\n$\\Delta=R\\cdot R\\tan\\omega$ is equivalent to\r\n$\\Delta=R^2\\tan\\omega$ is equivalent to\r\n$\\Delta=R^2\\frac{4\\Delta}{a^2+b^2+c^2}$ is equivalent to\r\n$1=\\frac{4R^2}{a^2+b^2+c^2}$ is equivalent to\r\n$a^2+b^2+c^2=4R^2$,\r\n\r\nwhat is just the characteristic equation of a harmonic triangle.\r\n\r\nAlso, your equation (2):\r\n\r\n$\\frac{1}{\\rho ^{2}}=\\frac{1}{h_{a}^{2}}+\\frac{1}{h_{b}^{2}}+\\frac{1}{h_{c}^{2}}$ is equivalent to\r\n$\\frac{1}{\\left( R\\tan \\omega \\right) ^{2}}=\\frac{1}{h_{a}^{2}}+\\frac{1}{h_{b}^{2}}+\\frac{1}{h_{c}^{2}}$ is equivalent to\r\n$\\frac{1}{\\left( R\\frac{4\\Delta }{a^{2}+b^{2}+c^{2}}\\right) ^{2}}=\\frac{1}{\\left( \\frac{2\\Delta }{a}\\right) ^{2}}+\\frac{1}{\\left( \\frac{2\\Delta }{b}\\right) ^{2}}+\\frac{1}{\\left( \\frac{2\\Delta }{c}\\right) ^{2}}$ is equivalent to\r\n$\\frac{\\left( a^{2}+b^{2}+c^{2}\\right) ^{2}}{16\\Delta ^{2}R^{2}}=\\frac{a^{2}}{4\\Delta ^{2}}+\\frac{b^{2}}{4\\Delta ^{2}}+\\frac{c^{2}}{4\\Delta ^{2}}$ is equivalent to\r\n$\\frac{\\left( a^{2}+b^{2}+c^{2}\\right) ^{2}}{4R^{2}}=a^{2}+b^{2}+c^{2}$ is equivalent to\r\n$\\frac{a^{2}+b^{2}+c^{2}}{4R^{2}}=1$ is equivalent to\r\n$a^{2}+b^{2}+c^{2}=4R^{2}$,\r\n\r\nwhat is again the characteristic equation of a harmonic triangle.\r\n\r\nThis solves your problem.\r\n\r\n Darij" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c,x,y,z$ be real numbers such that $ x \\plus{} y \\plus{} z \\equal{} 1$ . Prove that\r\n\r\n$ ax \\plus{} by \\plus{} cz \\plus{} 2 \\sqrt {(xy \\plus{} yz \\plus{} xz)(ab \\plus{} bc \\plus{} ac)} \\le {a \\plus{} b \\plus{} c}$", "Solution_1": "$ LHS^2\\equal{}(ax\\plus{}by\\plus{}cz\\plus{}\\sqrt{2ab\\plus{}2bc\\plus{}2ca}.\\sqrt{2xy\\plus{}2yz\\plus{}2zx})^2$\r\n$ \\leq(a^2\\plus{}b^2\\plus{}c^2\\plus{}2ab\\plus{}2bc\\plus{}2ca)(x^2\\plus{}y^2\\plus{}z^2\\plus{}2xy\\plus{}2yz\\plus{}2zx)$\r\n$ \\equal{}(a\\plus{}b\\plus{}c)^2(x\\plus{}y\\plus{}z)^2\\equal{}(a\\plus{}b\\plus{}c)^2\\equal{}RHS^2$" } { "Tag": [ "induction", "combinatorics proposed", "combinatorics" ], "Problem": "Suppose we have $n$ intervals of $\\mathbb R$ which along any $k$ of them there are 2 which their intersection is non empty. Prove that there exist $k-1$ points in $\\mathbb R$ which any of these subsets contain one of these points.", "Solution_1": "First modify the intervals as follows: when we encounter an interval which contains another one strictly, shrink the larger one to match the smaller one. This way, the main property of the collection of intervals is not altered, as can easily be seen, but when the process finishes, there won't be any intervals strictly containing other intervals in the collection.\r\n\r\nNow let $I=[a,b]$ be one of the intervals which has the leftmost endpoint as its left endpoint. Any other such intervals will coincide with $I$. There are now two cases:\r\n\r\n(a) There are no intervals in our set which cut $I$ along an interval $[c,b],\\ c\\in(a,b]$\r\n\r\nIn this situation, the collection of intervals we get by eliminating those which coincide with $I$ contains only intervals disjoint from $I$, and thus has the property that among any $k-1$ of them, some $2$ will cut each other. By induction on $k$, we can conclude that there are $k-2$ points such that each interval in this smaller set contains one of them, and adding an extra point which belongs to $I$, we get what we want.\r\n\r\n(b) There is an interval $J=[c,d]$ with $c\\in(a,b]$\r\n\r\nEliminate all intervals which coincide with $I$ in our collection, and replace them with the interval $[c,b]$. It's easy to see that because none of our intervals contains another one strictly, this new set of intervals still has the initial property (among any $k$ of them, $2$ will intersect each other). We can now perform the shrinking procedure in the beginning. The new intervals we get have their endpoints among those of the old ones, but there is at least one endpoint less (namely $a$). This means that we can proceed by induction on the number of endpoints.", "Solution_2": "I want to solve this problem:\r\nbut I cannot fo any further from here:\r\nat first I will give every interval , a vertics,and two vertics are connected iff their interval have somthing in common (Their intersection be none empty)[I call this kind of graph , Intervally Graph].\r\n\r\nthen I proved two lemmas.\r\n[color=darkblue]Lemma 1[/color].a cycle with more than 3 vertics cannot be Intervally Graph.\r\n[color=darkblue]Lemma 2[/color]. if we have a $K_p$ Intervally Graph then their intervals have somthing in common.", "Solution_3": "can anyone help me to complete my solution.", "Solution_4": ":( No one :(", "Solution_5": "let say$intervalA>{interval B}$if $A$ is completely righter than $B$.then refering to this poset the problem becomes MIRSKY theorem.", "Solution_6": "[quote=\"dashmiz\"]then the problem becomes MIRSKY theorem.[/quote]\r\n\r\nWhat is Mirsky theorem.", "Solution_7": "no one know what is MIRSKY theorem?", "Solution_8": "I created exactly the same problem:\r\n\r\nWe say that $I_{1}\\geq I_{2}$ if they dont intersect and $I_{1}$ is to the right of $I_{2}$. This becomses a partially ordered set. We are given that there is no antichain of lenght $k$, therefore , by Mirsky theorem (kindda Dilworth Dual), the Poset is the union of $k-1$ chains. Every chain is a set of intervals with a point in common so the problem is solved.", "Solution_9": "Here it is again, with the same proof :):\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=intervals&t=106694[/url]" } { "Tag": [ "geometry" ], "Problem": "I need help with the following geometry problems. Would be grateful if someone could give some kind of hints.\r\n\r\n1. Let BE and CF be the altitudes of an acute triangle ABC, with E on AC and F on AB. Let O be the point of intersection of BE and CF. Take any line KL through O with K on AB and L on AC. Suppose M and N are located on BE and CF respectively, such that KM is perpendicular to BE and LN is perpendicular to CF. Prove that FM is parallel to EN.\r\n\r\n2. In a triangle ABC, D is a point on BC such that AD is the internal bisector of angle A. Suppose (angle) B=2C and CD=AB. Prove that angle A=72 degrees.", "Solution_1": "For the first question, consider the two cases where K lies in the interior of FB and K lying in the exterior of FB. Use the fact that ABC is acute to conclude similarly about L and EC. Now spot cyclic quadrilaterals and chase a few angles.", "Solution_2": "Well, I've just solved the second one. Sometimes, looking at a problem after a small break is quite helpful.\r\n\r\nI'll look at the first one now...", "Solution_3": "[quote=\"Pradhan P\"]For the first question, consider the two cases where K lies in the interior of FB and K lying in the exterior of FB. Use the fact that ABC is acute to conclude similarly about L and EC. Now spot cyclic quadrilaterals and chase a few angles.[/quote]\r\nDone... thanks a lot. Sometimes, problems are a lot easier than they seem." } { "Tag": [ "probability" ], "Problem": "The probability it will rain on any given day is $\\frac{2}{3}$. What is the probability that it will only rain on one day out of three days?", "Solution_1": "[hide]\n$\\frac{1}{3}\\cdot\\frac{2}{3}\\cdot\\frac{2}{3}=\\frac{4}{9}$?\n[/hide]", "Solution_2": "[hide]\nthe probablity that it will rain on day one is 2/3 then you don't want it to rain so it becomes 1/3 then 1/3\nso 2/3*1/3*1/3=2/27[/hide]", "Solution_3": "Thats wrong.\r\n\r\n[hide=\"Answer\"]\n\n The correct answer should be:\n\n$\\frac23 \\cdot \\frac13 \\cdot \\frac13\\cdot 3=\\boxed{\\frac{6}{27}}$\n\n You guys are forgetting the events can be re-arranged 3 ways. [/hide]", "Solution_4": "[quote=\"G-UNIT\"]Thats wrong.\n\n[hide=\"Answer\"]\n\n The correct answer should be:\n\n$\\frac23 \\cdot \\frac13 \\cdot \\frac13\\cdot 3=\\boxed{\\frac{6}{27}}$\n\n You guys are forgetting the events can be re-arranged 3 ways. [/hide][/quote]\n\nThat's wrong too :P\n\n[hide=\"Correct Answer\"]\n$\\frac 23 \\cdot \\frac 13 \\cdot \\frac 13 \\cdot 3 = \\boxed{\\frac 29}$\n\nSIMPLIFY\n[/hide]", "Solution_5": "[quote=\"G-UNIT\"] You guys are forgetting the events can be re-arranged 3 ways. [/quote]\r\ni knew i messed up somehow. :)", "Solution_6": "Also, if you're wondering why there are 3 ways to arrange, it's because $\\binom{3}{1}$ gives the number of ways to choose 1 day from 3 total.\r\n\r\nIn general, the formula for finding the number of days is $P^k(1-P)^{n-k}\\cdot \\binom{n}{k}$, where $n$ is the total number of days, $k$ is the number where it must rain (or whatever else), and $P$ is the probability that it rains on a given day.", "Solution_7": "[quote=\"Hamster1800\"][quote=\"G-UNIT\"]Thats wrong.\n\n[hide=\"Answer\"]\n\n The correct answer should be:\n\n$\\frac23 \\cdot \\frac13 \\cdot \\frac13\\cdot 3=\\boxed{\\frac{6}{27}}$\n\n You guys are forgetting the events can be re-arranged 3 ways. [/hide][/quote]\n\nThat's wrong too :P\n\n[hide=\"Correct Answer\"]\n$\\frac 23 \\cdot \\frac 13 \\cdot \\frac 13 \\cdot 3 = \\boxed{\\frac 29}$\n\nSIMPLIFY\n[/hide][/quote]\r\n\r\n brilliant!", "Solution_8": "[quote=\"236factorial\"]The probability it will rain on any given day is $\\frac{2}{3}$. What is the probability that it will only rain on one day out of three days?[/quote]\r\n[hide]\nThe probability of it not raining is 1/3, so the answer is 3(2/3*1/3*1/3), since there are 3 days that it can rain. The answer is $\\boxed{\\frac{2}{9}}$[/hide]" } { "Tag": [ "trigonometry" ], "Problem": "Solve $ y^3\\minus{}y^2\\minus{}2y\\plus{}1\\equal{}0$ without $ cacdano$ technique", "Solution_1": "dont you mean the cardona technique?\r\n\r\nand i have no idea what that is", "Solution_2": "He means the [url=http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method]Cardano-Tartaglia method[/url]. Try doing it with Lagrange resolvents, then.", "Solution_3": "Do you mean $ y^3 \\plus{} y^2 \\minus{} 2y \\minus{} 1 \\equal{} 0$?\r\nIf we let $ y \\equal{} x \\plus{} \\frac1x$, and multiply the resulting equation by $ x^3$, don't we get $ x^6 \\plus{} x^5 \\plus{} \\cdots \\plus{} x \\plus{} 1 \\equal{} 0$, which has the primitive 7th roots of unity as solutions?", "Solution_4": "Nice, how did you come up with that?\r\n\r\nIt can be easily fixed by letting $ y \\equal{} \\minus{} x \\minus{} \\frac {1}{x}$ instead.", "Solution_5": "Haha, I was recently trying to find the 7th roots of unity in random ways...it's luck I guess. :P", "Solution_6": "Bah math154 was trying to find 7th roots of unity with me algebraiclly. Anyway, $ x\\equal{}cis(\\frac{2k\\pi}{7})$ where k is an integer (and is not a multiple of 7). But where would you go from there?", "Solution_7": "Since $ y \\equal{} x \\plus{} \\frac1x$, your answers would simply be $ y \\equal{} \\boxed{\\operatorname{cis}\\left(\\frac {2k\\pi}7\\right) \\plus{} \\frac1{\\operatorname{cis}\\left(\\frac {2k\\pi}7\\right)}}$.", "Solution_8": "[quote=\"t0rajir0u\"]He means the [url=http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method]Cardano-Tartaglia method[/url]. Try doing it with Lagrange resolvents, then.[/quote]\r\nYes,sorry\r\nBut I found by computer $ cos (\\frac{\\pi }{7},\\frac{3\\pi }{7},\\frac{5\\pi }{7}$", "Solution_9": "How do you obtain the three solutions using primitive roots? \r\n\r\n$ y \\equal{} \\frac{1}{x} \\plus{} x$ doesn't seem to yield those solutions listed above.", "Solution_10": "Combining all previous posts:\r\n\r\nWe have $ y^3 \\minus{} y^2 \\minus{} 2y \\plus{} 1 \\equal{} 0$.\r\n\r\nSubstituting $ y \\equal{} \\minus{} x \\minus{} \\frac1x$, we find that\r\n\r\n$ \\minus{} \\left(x^3 \\plus{} \\frac1{x^3} \\plus{} 3x \\plus{} \\frac3x\\right) \\minus{} \\left(x^2 \\plus{} \\frac1{x^2} \\plus{} 2\\right) \\plus{} \\left(2x \\plus{} \\frac2x\\right) \\plus{} 1 \\equal{} 0$.\r\n\r\nSimplifying, we get\r\n\r\n$ \\minus{} x^3 \\minus{} x^2 \\minus{} x \\minus{} 1 \\minus{} \\frac1x \\minus{} \\frac1{x^2} \\minus{} \\frac1{x^3} \\equal{} 0$, and multiplying through by $ \\minus{} x^3$, we have\r\n\r\n$ x^6 \\plus{} x^5 \\plus{} x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1 \\equal{} 0$, which has roots\r\n\r\n$ \\operatorname{cis}\\left(\\frac {2\\pi}7\\right),\\operatorname{cis}\\left(\\frac {4\\pi}7\\right),\\operatorname{cis}\\left(\\frac {6\\pi}7\\right),\\operatorname{cis}\\left(\\frac {8\\pi}7\\right),\\operatorname{cis}\\left(\\frac {10\\pi}7\\right),\\operatorname{cis}\\left(\\frac {12\\pi}7\\right)$.\r\n\r\nPlugging back into our equation $ y \\equal{} \\minus{} x \\minus{} \\frac1x$, we have one root as\r\n\r\n$ \\minus{}\\operatorname{cis}\\left(\\frac {2\\pi}7\\right)\\minus{}\\frac1{\\operatorname{cis}\\left(\\frac {2\\pi}7\\right)}$\r\n\r\n$ \\equal{}\\boxed{\\minus{}2\\cos\\left(\\frac{2\\pi}7\\right)}$.\r\n\r\nThe other two roots are $ \\boxed{\\minus{}2\\cos\\left(\\frac{4\\pi}7\\right)}$ and $ \\boxed{\\minus{}2\\cos\\left(\\frac{6\\pi}7\\right)}$." } { "Tag": [], "Problem": "How many forums (counting AoPS) do you belong to? Please list them, too. Social networks like facebook do not count.", "Solution_1": "uh... one? yep, that's AoPS. no wait... 2. 'cause my class made one too. cool i'm the first post other than the original! :D", "Solution_2": "Ive bene on a bunch, but Im only active on 3...well, not really on the other 2 anymore, now just AOPS", "Solution_3": "4:\r\n\r\nAoPS\r\nSmashBoards\r\nYugioh:The Abridged Series\r\nstudio12.co.cc (although it's basically a dead forum).", "Solution_4": "Three:\r\n\r\nAoPS\r\nCoD4Boards\r\nBungie\r\n\r\nAnd why doesn't Facebook count? :P", "Solution_5": "I belong to four:\r\n\r\nAoPS\r\nChief Delphi\r\nGame Maker Community\r\nXKCD Forums", "Solution_6": "I don't belonD to any forum.\r\n\r\nI belonG to AoPS, AoPS and nothing but AoPS (:()\r\n\r\nI hadn't joined anything more, but I think that I might join mafiascum", "Solution_7": "1.This one\r\n2.Math Help Forum \r\n3. SOS Math\r\n4. My Math Forum\r\n5. Google Math", "Solution_8": "I've joined many and I've hosted a few.\r\nNow I use AoPS only", "Solution_9": "I'm registred on many forums but I'm active on only two.\r\n1. AOPS\r\n2. is my own forum which i tried to make :)", "Solution_10": "Oh yeah, actually 5. I forgot Battle City. Too late to change my vote now, though.", "Solution_11": "I am registered in many but active in AoPS only.", "Solution_12": "[quote=\"7h3.D3m0n.117\"]And why doesn't Facebook count? :P[/quote]\r\n\r\nI meant other than social networks...", "Solution_13": "I belong to tons of forums and am too lazy to list them all sorry. Probably nearing 25-30 now.", "Solution_14": "i used to belong on a bunch but i gave them all up\r\nBut I'm a member of AoPS and Neopets--and even neopets doesn't let me on the forums, so that doesn't count.", "Solution_15": "[quote=\"myyellowducky82\"]i used to belong on a bunch but i gave them all up\nBut I'm a member of AoPS and Neopets--and even neopets doesn't let me on the forums, so that doesn't count.[/quote]\r\nNeopets is sad. Do you actually believe that they actually used to say \"Neopets will never advertise.\" Talk about walking the talk. Look at Neopets now.", "Solution_16": "oooooh yeah. definately. the last time i checked was half a year ago, but their NC mall is just--so---UUUUUUGGGGG!!!!! i mean honestly, you have to pay money for fake money! and they have GAMES that are advertisements!!!!!! (although they DO give you a lot of neopoints, though. :lol: )" } { "Tag": [ "AMC", "AIME", "geometry", "AMC 12", "MATHCOUNTS" ], "Problem": "I'm just starting preparing for the AMC 12 and the AIME next year. Where should I start? I've read Winning Solutions, should I read the \"Art of Problem Solving\", is the series that good? What else should I do/read. I'm extremely good at math I just need to learn more about this type of math if I want to do well.", "Solution_1": "Of course AoPS is a good place to start! Just start doing problems. If you find something you don't know, then research it.", "Solution_2": "[quote=\"scm007\"]I'm extremely good at math I just need to learn more about this type of math if I want to do well.[/quote]\r\n\r\nBe a little more humble. It'll help, if for nothing more than you'll work harder.", "Solution_3": "I'm sure pretty much every single common poster on this board is \"extremely good at math\" relative to the other students at their schools. I'd be careful before calling myself \"extremely good\" at anything.", "Solution_4": "There are people are exceptionally good at math in this site..\r\nI used to call myself pretty good at math in my school but in here, there are really a lot people who are better than me on many areas of math. I don't know about AMC12 but you can try some problems I post in this forum except they're not gonna be that difficult because I don't know really hard questions. For me, the best way to learn math is to produce the problem by yourself. I make some problems by myself and some of them are pretty good. For more problems and stuffs, visit Mathcounts Marathon in Mathcounts Forum or Problem Solving Marathon in this forum. Both are posted by myself. Hope this helps." } { "Tag": [ "limit", "algebra unsolved", "algebra" ], "Problem": "Find\r\n$ \\lim\\prod(1\\plus{}\\frac{1}{9^{n}})$\r\nwhich:\r\n$ n \\equal{} 0,1,2,3...,\\plus{}\\infty$", "Solution_1": "Homework?\r\n\r\nI'm not sure this forum is the appropriate one :wink: \r\n\r\nPierre.", "Solution_2": "homework can't be this one.\r\nactually,it's one of my friends give me.\r\nand it is a question which is in Peking university test.It seems unusual.\r\nanyway, i don't know the method.\r\nwho can post it?", "Solution_3": "Because $ 0<\\equal{} \\prod\\frac{1}{9^n}<\\equal{}\\frac{1}{9^n}$for any$ n>\\equal{}1$and lim $ \\frac{1}{9^n}\\equal{}0$ we can say that $ lim \\prod\\frac{1}{9^n}\\equal{}0$", "Solution_4": "...\r\nI know the reason why pierre said it is homework...\r\nI made a mistake on inputting it...\r\nThe right problem is:\r\n$ \\boxed{\\lim\\prod(1+\\frac{1}{9^n})}$\r\nnow, it looked better.-_-", "Solution_5": "Ok.\r\n\r\nBut, then, are you asking for the value of the limit?\r\nThe convergence is easy to prove (use that $ 1 \\plus{} x \\leq e^x$), but I'm note sure that the exact value of the limit is achievable :wink: \r\n\r\nPierre.", "Solution_6": "[quote=\"pbornsztein\"]Ok.\n\nBut, then, are you asking for the value of the limit?\nThe convergence is easy to prove (use that $ 1 \\plus{} x \\leq e^x$), but I'm note sure that the exact value of the limit is achievable :wink: \n\nPierre.[/quote]\r\n\r\nLet $ a_m \\equal{} \\prod_{n \\equal{} 0}^{m}{(1 \\plus{} \\frac {1}{9^n})}$. We have $ a_{m \\plus{} 1} \\equal{} a_{m}(1 \\plus{} \\frac {1}{9^{m\\plus{}1}}) > a_m$.\r\n$ a_m < \\prod_{n \\equal{} 0}^{\\infty}{(1 \\plus{} \\frac {1}{9^n})}\\leq e^{\\sum_{n \\equal{} 0}^{\\infty}{\\frac {1}{9^n}}} \\equal{} e^{\\frac {9}{8}}$.\r\nSo limit exist.", "Solution_7": "Yes, that's what I'm saying. But it only gives an upper bound for the exact value.\r\nI'm quite sure that this value cannot be easily determined in the general case (when $ \\frac 1 9$ is replaced by $ a$ with $ a \\in ]0,1[$). Maybe for $ a \\equal{} \\frac 1 9$ is there a trick...\r\n\r\nPierre.", "Solution_8": "the answer is:\r\n$ A\\equal{}lim\\prod(1\\plus{}\\frac{1}{9^n})$\r\n $ \\equal{}lim\\prod(1\\plus{}\\frac{1}{3^{2n}})$\r\n$ \\frac{8}{9}A\\equal{}lim(1\\minus{}\\frac{1}{3^2})\\prod(1\\plus{}\\frac{1}{3^{2n}})\\equal{}1$\r\n $ A\\equal{}\\frac{9}{8}$", "Solution_9": "[quote=\"wxydx00\"]the answer is:\n$ A \\equal{} lim\\prod(1 \\plus{} \\frac {1}{9^n})$\n $ \\equal{} lim\\prod(1 \\plus{} \\frac {1}{3^{2n}})$\n$ \\frac {8}{9}A \\equal{} lim(1 \\minus{} \\frac {1}{3^2})\\prod(1 \\plus{} \\frac {1}{3^{2n}}) \\equal{} 1$\n $ A \\equal{} \\frac {9}{8}$[/quote]\r\n\r\nI'm affraid you are wrong : $ \\lim_{m\\rightarrow \\plus{}\\infty}(1 \\minus{} \\frac {1}{3^2})\\prod_{n\\equal{}1}^m(1 \\plus{} \\frac {1}{3^{2n}}) \\neq 1$\r\n\r\n$ \\lim_{m\\rightarrow \\plus{}\\infty}(1 \\minus{} \\frac {1}{3^2})\\prod_{n\\equal{}1}^m(1 \\plus{} \\frac {1}{3^{2n}})$ $ >(1 \\minus{} \\frac {1}{3^2})\\prod_{n\\equal{}1}^3(1 \\plus{} \\frac {1}{3^{2n}})\\equal{}\\frac{4788800}{4782969}>1$\r\n\r\nI think you make a confusion with $ \\lim_{m\\rightarrow \\plus{}\\infty}(1 \\minus{} \\frac {1}{3^2})\\prod_{n\\equal{}1}^m(1 \\plus{} \\frac {1}{3^{2^n}})$ which indeed is $ \\equal{}1$", "Solution_10": "Oh,yes..\r\nSo,where is the right method.\r\nI make the$ \\prod(1\\plus{}\\frac{1}{3^n})$to the$ (1\\plus{}\\frac{1}{3^2})(1\\plus{}\\frac{1}{3^4})(1\\plus{}\\frac{1}{3^8})...$" } { "Tag": [ "floor function", "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "There is a well known result:\r\nFor every integer $ n$ such that $ n \\geq 3$:\r\n\\[ \\sum_{k \\equal{} 0}^{\\lfloor \\frac {n}{3} \\rfloor} {n \\choose {3k}} \\equiv 0 \\pmod{3}\\]\r\nProve that for every prime number $ p$ and positive integer $ n$ such that $ n \\geq p$ we have:\r\n\\[ \\sum_{k \\equal{} 0}^{\\lfloor \\frac {n}{p} \\rfloor} {n \\choose {pk}} \\equiv 0 \\pmod{p}\\]\r\n[hide]\n[b]Hint.[/b]\nLet\n\\[ P(x) \\equal{} (1 \\minus{} x)^n \\equal{} \\sum_{k \\equal{} 0}^{n} ( \\minus{} x)^{n \\minus{} k} {n \\choose {n \\minus{} k}} \\equal{} \\sum_{k \\equal{} 0}^{n} ( \\minus{} 1)^k {n \\choose k} x^k\\]\nAlso let $ \\varepsilon$ be the complex number satisfies:\n\\[ \\varepsilon^{p \\minus{} 1} \\plus{} \\varepsilon^{p \\minus{} 2} \\plus{} ... \\plus{} \\varepsilon \\plus{} 1 \\equal{} 0\\]\nThen:\n\\[ \\sum_{k \\equal{} 0}^{\\lfloor \\frac {n}{p} \\rfloor} {n \\choose {pk}} \\equal{} \\dfrac{P(1) \\plus{} P(\\varepsilon) \\plus{} P(\\varepsilon^2) \\plus{} ... \\plus{} P(\\varepsilon^{p \\minus{} 1})}{p}\\]\nNotice that $ P(1) \\equal{} 0$ and $ P(\\varepsilon^k) \\equal{} (1 \\minus{} \\varepsilon)^k Q(\\varepsilon)$ where $ Q$ is also an integer-coefficient polynomial.\n[/hide]", "Solution_1": "It is not true for $ n \\equal{} p$: $ \\binom{p}{0} \\plus{} \\binom{p}{p} \\equal{} 2\\not \\equal{} 0\\mod p, p > 2$.\r\nIf it true for $ n$, then it is true for $ n \\plus{} p$.\r\nIf $ p < n < 2p$, then it is true, for $ n \\equal{} 2p$ $ 1 \\plus{} C_{2p}^p \\plus{} 1 \\equal{} 4\\mod p$ it is not true.\r\nTherefore your statment is true for $ n > p, p\\not |n$\r\nSecond method is $ \\sum_i \\binom{n}{mi} \\equal{} \\frac {1}{m} \\sum_{k \\equal{} 1}^m (1 \\plus{} exp(\\frac {2\\pi i k}{m}))^n.$", "Solution_2": "Sorry, the problem has been mistyped. It should have been:\r\n\\[ \\sum_{k\\equal{}0}^{\\lfloor \\dfrac{n}{p} \\rfloor} (\\minus{}1)^k \\binom{n}{kp} \\equiv 0 \\pmod{p}\\]\r\nNow, it is true for $ n \\geq p$. The proof is hidden in my post at #1.", "Solution_3": "[quote=\"Mashimaru\"]Sorry, the problem has been mistyped. It should have been:\n\\[ \\sum_{k \\equal{} 0}^{\\lfloor \\dfrac{n}{p} \\rfloor} ( \\minus{} 1)^k \\binom{n}{kp} \\equiv 0 \\pmod{p}\\]\nNow, it is true for $ n \\geq p$. The proof is hidden in my post at #1.[/quote]\r\nIt can be proven by applying Lucas theorem. Suppose $ n\\equal{}pt\\plus{}r$ where $ 0\\leq r\\leq p\\minus{}1$ then we have: \r\n\\[ {{n}\\choose{pk}} \\equiv {{t}\\choose{k}} (\\mod p)\\]\r\nAfter that, we only need to notice that $ \\sum_{i\\equal{}0}^n (\\minus{}1)^k {{n}\\choose{k}}\\equal{}0$" } { "Tag": [ "ARML", "geometric sequence" ], "Problem": "Here's a problem from my ARML homework that ive been working on but can't seem to do:\r\n\r\nLet $z$ be a root of $x^5-1=0$ with $z \\not = 1$\r\n\r\nCompute the value of $\\displaystyle z^{15}+z^{16}+z^{17}+...z^{50}$.", "Solution_1": "use the facts\r\n\r\n1. $x^5=1$\r\n2. $1+x+x^2+x^3+x^4$=0 (do you see why that's true? if not, factor it has a geometric sequence, and use the fact that $z$ isn't 1.)", "Solution_2": "[hide]\n\n\n\nSince 1 is clearly a root of this, factor it out:\n\n\n\n\n\n\n\nsuch that is a root of the second term.\n\n\n\nYou're trying to computer . Factor out :\n\n\n\n\n\n\n\nSince , therefore must equal 0 according to the original equation.\n\n\n\nTherefore, you may simplify the second equation even more:\n\n\n\n\n\n\n\nYou can continue to factor out powers of and repeating the process:\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nGoing back to the original equation, .\n\n\n\nTherefore, the expression is equal to .[/hide]", "Solution_3": "hmm thanks!" } { "Tag": [ "inequalities", "calculus", "limit", "geometry", "geometric transformation", "reflection", "calculus computations" ], "Problem": "Hi,\r\n I can't figure out why the delta- epilson proof proves that a certain limit exits :?: , conceptually it makes no sense. Even if we find an inequality for the delta.\r\nfor ex: |x-4|<=(1/3)\u0190; how does this prove that a limit exits, conceptually and mathematically? \r\n\r\nThank you in advance :D [/code]", "Solution_1": "Historically, calculus grew and flourished pretty well during the nearly 200 years between Newton/Leibniz and Weierstrass, which might be some sort of hint about how seriously to take this in a calculus course ...\r\n\r\nBut anyway. The official definition of limit cannot be understood or comprehended piecemeal, or as an equation. It must be understood whole, as a logical flow, with all of the quantifiers and logical connectives (\"for all\" and \"there exists\" and \"if ... then\") in their proper places doing their proper work.\r\n\r\nWhat you've written appears to be an attempt to deal with the concept piecemeal - there's no stated limit in sight (although I suppose we could guess what you might have meant), and the quantifiers and logical connectives are nowhere to be seen. As such, what you have said is incoherent. I can't reply to it directly for that reason.\r\n\r\nNow, what is this Weierstrassian formality, this epsilon-delta stuff? It's not that it has something to do with limits - it [i]is[/i] the formal mathematical definition of limit.\r\n\r\nLet's try to sneak up on this gradually. We say that $ \\lim_{x\\to a}f(x)\\equal{}L;$ what does that mean?\r\n\r\nFirst try: as $ x$ gets close to $ a,$ $ f(x)$ gets close to $ L.$\r\n\r\n- That's much too squishy. What do we mean by close? Does $ x\\plus{}10^{\\minus{}10}$ tend to $ 0$ as $ x\\to 0$ because $ 10^{\\minus{}10}$ is \"close\" to zero? Well, no, that's not really what we mean (even if your graphing calculator can't tell the difference.) We need something that expresses what we really want - and here the word \"arbitrary\" comes to our rescue.\r\n\r\nSecond try: as $ x$ gets arbitrarily close to $ a,$ $ f(x)$ get arbitrarily close to $ L.$\r\n\r\n- The problem with this is that the logical flow is unclear. Which \"arbitrary\" takes precedence? Upon further reflection, maybe we don't want both words to be the same. \r\n\r\nThird try: we can make $ f(x)$ arbitrarily close to $ L$ by making $ x$ close enough to $ a$ (but not equal to $ a.$)\r\n\r\n- Now we're getting somewhere. I like this definition. We can formalize it further by stating exactly what we mean by \"arbitrarily close\" and \"close enough\". We then arrive at the full formality:\r\n\r\n$ \\lim_{x\\to a}f(x)\\equal{}L$ iff for all $ \\epsilon > 0$ (comment: no matter how small, arbitrarily small), there exists $ \\delta > 0$ (comment: small enough, depending on how small $ \\epsilon$ is) such that for all $ 0<|x\\minus{}a|<\\delta$ (for all $ x$ that close to $ a$ or closer), we have $ |f(x)\\minus{}L|<\\epsilon$ ($ f(x)$ is at least that close to $ L.$)\r\n\r\n----\r\n\r\nNow: why formalize the definition at all? The first thing to understand is that defining what we mean by limit, which is what we've just done, does not [i]find[/i] limits and never will find limits. It's not a computational tool. It's how we convince ourselves that what we've already found is correct. I should also mention that there is seldom any call to use this definition in a specific, concrete case (although such cases are often offered as pedagogical exercises). Instead, we use it to prove general theorems - general theorems like $ \\lim_{x\\to a}(f(x)\\plus{}g(x))\\equal{}\\lim_{x\\to a}f(x)\\plus{}\\lim_{x\\to a}g(x),$ provided both limits on the right exist. (With similar theorems about mulitiplication, division, and composition.) And many other theorems.\r\n\r\nWe have to use the definition to prove theorems - why? Because if we're going to prove that a limit exists, and that it equals such and such a number, we have to know what it even means for a limit to exist and for it to equal something. We have to have the definition even to know what we're trying to prove." } { "Tag": [ "geometry", "calculus", "derivative", "3D geometry", "sphere", "topology", "vector" ], "Problem": "I was wondering if there is a proof that, for example, the derivative of an expression for volume gives a formula for surface area (3 dimensional case). Is there some way to prove this for higher dimensions as well? I can see that it works with a sphere, but am not sure how to prove it for say an 8 dimensional object. Can someone help me out on this? Thank you very much.", "Solution_1": "This can only work for nice enough objects- note the unstated assumption that the figure depends only on some linear dimension.\r\n\r\nIn addition, we need that when this linear dimension is increased by $c$, we add a shell of uniform thickness $c$ to the figure. This thickness is measured perpendicular to the surface.\r\n\r\nWith that caveat, this does work for balls of any dimension, using the radius. It also works for cubes using half of the side length, and a few other figures.", "Solution_2": "\"Expression for volume\" is a little too general. The structure of what you're looking for is something like this:\r\n\r\nSuppose $g: \\mathbb{R}^n\\mapsto\\mathbb{R}$ satisfies certain regularity conditions (which I won't specify in complete detail, but we are going to want $\\nabla g\\ne0$ \"most\" of the time.) and suppose that for each $c,\\,\\{x: g(x)\\le c\\}$ is compact and hence has finite measure.\r\n\r\nLet $V(y)$ equal the n-dimensional measure pf $\\{x: g(x)\\le y\\}.$\r\n\r\nNow consider the set $S_y=\\{x: g(x)=y\\}.$ As long as $\\nabla g\\ne 0$ on this set, it is a smooth compact $n-1$-dimensional manifold embedded in $\\mathbb{R}^n.$ As such, it is possible to construct an $(n-1)$-dimensional measure on $S_y.$ We let $A(y)$ be this $n-1$-dimensional measure of the whole set $S_y.$\r\n\r\nSo, is the result you want? $A(y)=\\frac{d}{dy}V(y).$ Not quite; to make that work, we would need that $|\\nabla g|=1$ everywhere (namely, the figure is expanding at a constant rate in all directions.)\r\n\r\nAny proof will have to pull out all of the heavy tools of vector calculus - the Implicit Function Theorem, Jacobians, and so on.\r\n\r\nBut note that we have to have the function $g$ in order for your question to even make sense.\r\n\r\nCute result: for the ball of radius $r$ in $\\mathbb{R}^n,$ the $n$-dimensional measure of that ball is \r\n\r\n$V(r)=\\frac{\\pi^{\\frac n2}r^n}{\\Gamma\\left(\\frac n2+1\\right)}.$ \r\n\r\nThe $(n-1)$-dimensional measure of the sphere of radius $r$ (the boundary of the ball) is \r\n\r\n$A(r)=\\frac{\\pi^{\\frac n2}r^{n-1}}{\\Gamma\\left(\\frac n2\\right)}.$", "Solution_3": "OK I understand that last part, provided those equations are known beforehand. However, I did not know them until you posted them (for the n-dimensional ball). How were those arrived at? Thanks a lot for your help :)" } { "Tag": [ "AMC" ], "Problem": "No discussion on the 2005 AMC10A/12A should take place until Thursday, Feb. 3, at 12:00 EST, as stated [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25191]here[/url].\r\n\r\nTo be completely clear, discussion even includes writing about how you think you did, stating 'that was easy/hard', etc. [b]To be safe, in any and all posts, pretend that the 2005 AMC 10A/12A test never existed.[/b] \r\n\r\nIf you have any questions about this rule, PM me or one of the administrators.\r\n\r\nSorry for the duplicate announcement. Some people didn't read the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25191]first, more subtle one[/url] and I had to delete some threads.", "Solution_1": "What's \"Source: ... and 'that w\" mean? In particular the \"that w\" part.", "Solution_2": "He probably meant to say something like \"That was easy/hard\"", "Solution_3": "I wrote \"This includes 'what score do you think you got?' and 'that was really hard'\" in the source box so it would appear on the front screen. It looks funny when you view the thread, but for someone who is scanning the forum, planning to post an early discussion of the AMCs, it will inform them quickly and stop them.", "Solution_4": "Very sorry that I did not read that. Will remember in future.", "Solution_5": "[quote=\"solafidefarms\"]Very sorry that I did not read that. Will remember in future.[/quote]\r\n\r\nThat's okay -- I should have been more explicit about what constituted 'discussion' sooner.", "Solution_6": "Especially since the AMC Director is a member!", "Solution_7": "well, all AMC's would be sent in by like midnight tomorrow... so yeah... 12:00 am est on Thursday. why is there no : easy/hard evautaion? It is just opinion...", "Solution_8": "But still, that may affect performance. \r\n\r\n If your friend told you that the science test was really hard, would you worry and panic and get really nervous?\r\n\r\nIf you friend, however, told you it was hard as crazy, I bet you would be very worried. State of mind plays a huge role in test taking.", "Solution_9": "If discussion were to be allowed, it would be extremely difficult to rigorously define what constitutes an \"acceptable\" post. It's easier just to ban discussion about the AMC altogether until Thursday.", "Solution_10": "I can answer that, since I have no clue what was on the 2005 AMC tests. If this is still too much explanation, the mods should feel free to delete this.\r\n\r\nIt's a theoretical concern but a legitimate one. Suppose I tell you that the test was \"very, very hard.\" Well, this puts you on notice and may affect your strategy in several ways. (1) You may believe (rightly) that there are many tricky questions on the test and that you have to be very careful, more so than usual. (2) You may believe (rightly) that the last 5 problems are harder than most last-5 problems, and so you should skip those and focus on the first 20. (3) You may believe (rightly) that there are many difficult questions, and some of them appear early in the test, so you know you need to skip around, even among the early questions, and try to find the easy stuff.\r\n\r\nThese things are all speculative, but telling someone that an exam is easy/hard can give them an overall idea what strategy they should pursue. You tell me I'm going to get an easy exam, I'm going to play aggressive and go for 150. You tell me it's hard, I'm going to play conservative and make really sure I get my 100 so I can move on to the next stage. Is any of this stuff rock-solid? No, but might as well dodge this theoretical concern and enforce a \"code of silence\" until a couple of days after.", "Solution_11": "Better make this a global announcement." } { "Tag": [], "Problem": "Ben starts with the number $ 33$. If he squares this value, then adds $ 11$ and finally divides by $ 11$, what is the result?", "Solution_1": "Calculate:\r\n33^2=1089\r\n1089+11=1100\r\n1100/11=100.", "Solution_2": "(33^2+11)/11=(99X11+11)/11=99+1=100", "Solution_3": "Wow, that is a clever manipulation! Does this method work for almost all MATHCOUNTS/AMC 8-10 etc. problems?" } { "Tag": [ "inequalities" ], "Problem": "I'm doing some math problems (i have a test tommorow) and am having trouble with unequations. I know the the unequal sign ('<, or >') changes in certain cases, but i don't know when. Sorry for my bad english, but i think you'll understand.", "Solution_1": "I think they change when you multiply an equation my a negative number.\r\n5<-8x\r\n-5/8>x\r\nIs that what you are asking?\r\n\r\nRegards,\r\nFuncia", "Solution_2": "also when you divide by a negative number (duh!)\r\n\r\n.....im in algebra 1 and i happen to be doing inequalities right now\r\n\r\nbtw, u call them inequalities\r\n-jorian", "Solution_3": "it also switches when you take the reciprocal and they are both positive or both negative :D" } { "Tag": [], "Problem": "\u039d\u03b1 \u03b2\u03c1\u03b5\u03b8\u03b5\u03af \u03bf \u03c6\u03c5\u03c3\u03b9\u03ba\u03cc\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $ n$ \u03b1\u03bd \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $ A\\equal{}2^{17}\\plus{}17\\cdot2^{12}\\plus{}2^{n}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03ad\u03bb\u03b5\u03b9\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03b1\u03ba\u03b5\u03c1\u03b1\u03af\u03bf\u03c5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd.", "Solution_1": "[b][color=red]\u039a\u03b1\u03bb\u03b7\u03c3\u03c0\u03ad\u03c1\u03b1 \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03ba\u03b1\u03b9\u03c1\u03cc! \u039a\u03b1\u03c4\u03b1\u03c1\u03c7\u03ae\u03bd \u03b5\u03cd\u03c7\u03bf\u03bc\u03b1\u03b9 \u03c3\u03b5 \u03cc\u03bb\u03bf\u03c5\u03c2 \u03ba\u03b1\u03bb\u03ae \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ae \u03ba\u03b1\u03b9 \u03ba\u03b1\u03bb\u03ae \u039f\u039b\u03a5\u039c\u03a0\u0399\u0391\u039a\u0397 \u03c7\u03c1\u03bf\u03bd\u03b9\u03ac! \u039a\u03b1\u03bb\u03ae \u03b5\u03c0\u03b9\u03c4\u03c5\u03c7\u03af\u03b1 \u03c3\u03c4\u03bf\u03bd \u03b5\u03c1\u03c7\u03cc\u03bc\u03b5\u03bd\u03bf \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03cc \u03c4\u03b7\u03c2 \u0395\u039c\u0395 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03bb\u03ae \u03c3\u03c4\u03b1\u03b4\u03b9\u03bf\u03b4\u03c1\u03bf\u03bc\u03af\u03b1 \u03c3\u03c4\u03bf\u03c5\u03c2 (\u03c0\u03bb\u03ad\u03bf\u03bd) \u03c6\u03bf\u03b9\u03c4\u03b7\u03c4\u03ad\u03c2 \u03bc\u03b1\u03c2![/color] [/b]\r\n\r\n\u0393\u03b9\u03b1 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b5\u03af\u03bc\u03b1\u03b9 \u03c3\u03c7\u03b5\u03b4\u03cc\u03bd \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03cc\u03c4\u03b9 \u03b8\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c0\u03b9\u03bf \u03c3\u03cd\u03bd\u03c4\u03bf\u03bc\u03bf\u03c2 \u03b4\u03c1\u03cc\u03bc\u03bf\u03c2 \u03bc\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03b4\u03b9\u03b1\u03b4\u03b9\u03ba\u03b1\u03c3\u03af\u03b1 \u03b1\u03bb\u03bb\u03ac \u03c3\u03b1\u03c2 \u03c0\u03b1\u03c1\u03b1\u03b8\u03ad\u03c4\u03c9 \u03c4\u03b9\u03c2 \u03b4\u03b9\u03ba\u03ad\u03c2 \u03bc\u03bf\u03c5 \u03c3\u03ba\u03ad\u03c8\u03b5\u03b9\u03c2:\r\n\r\n\u0397 \u03c0\u03b1\u03c1\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u0391 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03b7 \u03bc\u03b5 $ 49\\cdot 2^{12} + 2^n$. \u0398\u03ad\u03bb\u03bf\u03c5\u03bc\u03b5 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03bd\u03b1 \u03bb\u03cd\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7 \u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03ae \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 $ 49\\cdot 2^{12} + 2^n = x^2$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf $ x$ (\u03bf \u03bf\u03c0\u03bf\u03af\u03bf\u03c2 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf\u03c2 \u03b1\u03c0\u03cc $ 7\\cdot 2^6$).\r\n\r\n\u0399\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1 $ (x - 7\\cdot 2^6)(x + 7\\cdot 2^6) = 2^n$. To $ n$ \u03b1\u03c0\u03bf\u03ba\u03bb\u03b5\u03af\u03b5\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf \u03bc\u03b5 0, \u03ac\u03c1\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03bb\u03cd\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c3\u03cd\u03c3\u03c4\u03b7\u03bc\u03b1\r\n\r\n$ \\left\\{\\begin{array}{ll} \u007fx - 7\\cdot 2^6 = 2^{n_1} , & \\hbox{\u007f} \\\\\r\nx + 7\\cdot 2^6 = 2^{n_2} \u007f, & \\hbox{\u007f.} \\end{array} \\right.$ \u03bc\u03b5 $ n_1,n_2 \\geq 1$ \u03ba\u03b1\u03b9 $ n_1 + n_2 = n$\r\n\r\n\u039c\u03b5 \u03b1\u03c6\u03b1\u03af\u03c1\u03b5\u03c3\u03b7 \u03ba\u03b1\u03c4\u03ac \u03bc\u03ad\u03bb\u03b7 \u03c4\u03c9\u03bd \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 $ 7\\cdot 2^7 = 2^{n_2} - 2^{n_1}$. \u0395\u03cd\u03ba\u03bf\u03bb\u03b1 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1, \u03cc\u03c4\u03b9 $ n_1,n_2 \\geq 7$ \u03b4\u03b9\u03cc\u03c4\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ac \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 \u03ac\u03c4\u03bf\u03c0\u03bf [\u03a4\u03af \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b5\u03ac\u03bd \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03ad\u03bd\u03b1\u03c2 \u03b5\u03ba \u03c4\u03c9\u03bd $ n_1, n_2$ (\u03ae \u03b1\u03ba\u03cc\u03bc\u03b7 \u03c7\u03b5\u03b9\u03c1\u03cc\u03c4\u03b5\u03c1\u03b1 [b]\u03ba\u03b1\u03b9 \u03bf\u03b9 \u03b4\u03cd\u03bf[/b]) \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf\u03c2 \u03b1\u03c0\u03cc 7 ?] \u0386\u03c1\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03b7\u03bd \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 \r\n\r\n$ 7 = 2^{n_2 - 7} - 2^{n_1 - 7}$ \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03b4\u03cd\u03bd\u03b1\u03c4\u03b7 \u03b1\u03bd \u03bf\u03b9 $ n_1,n_2$ \u03b5\u03af\u03bd\u03b1\u03b9 (\u03ba\u03b1\u03b9 \u03bf\u03b9 \u03b4\u03cd\u03bf) \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf\u03b9 \u03b1\u03c0\u03cc 7. \u0386\u03c1\u03b1 \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03ad\u03bd\u03b1\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf\u03c2 \u03bc\u03b5 7 (\u03b1\u03c6\u03bf\u03cd \u03bf\u03cd\u03c4\u03b5 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf\u03b9 \u03b1\u03c0\u03cc 7 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9).\r\n\r\n\u0391\u03bd $ n_2 = 7$ \u03c4\u03cc\u03c4\u03b5 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b1\u03b4\u03cd\u03bd\u03b1\u03c4\u03b7 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7, \u03b5\u03bd\u03ce \u03b1\u03bd $ n_1 = 7$, \u03c4\u03cc\u03c4\u03b5 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 $ 8 = 2^{n_2 - 7}$ \u03ac\u03c1\u03b1 $ n_2 = 10$, \u03c3\u03c5\u03bd\u03b5\u03c0\u03c9\u03c2 \r\n\r\n$ n = n_1 + n_2 \\Rightarrow \\boxed{n = 17}$\r\n\r\n\u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03bc\u03b7 \u03be\u03ad\u03c7\u03b1\u03c3\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7\r\n\r\n\u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2" } { "Tag": [ "probability" ], "Problem": "A die is rolled. What is the probability that the number rolled is greater than 2 and even?", "Solution_1": "...based on the other answers you have gotten on probability and dice rolling, what do you think it could be? Ask yourself the following questions and try to answer the problem:\r\n\r\n-How many numbers are greater than 2?\r\n-Of those, how many are even?", "Solution_2": "Is there a AND Probability formula I can use for AND and OR questions?", "Solution_3": "Let $P(x)$ be the probability of event x occurring, and let $P(y)$ be the probability of event y occurring.\r\n\r\nProbability of event X AND event Y occurring is $P(x) \\cdot P(y)$.\r\n\r\nProbability of event X OR event Y occurring is $P(x)+P(y)-P(x)\\cdot P(y)$", "Solution_4": "Now, I think I can take it from here.", "Solution_5": "[quote=\"mysmartmouth\"]\nProbability of event X OR event Y occurring is $P(x)+P(y)-P(x)\\cdot P(y)$[/quote]\r\n\r\nWait... isn't an OR event just $P(x)+P (y)$? (ie. mutually exclusive events)?\r\n\r\nEDIT: yes I know this is unrelated to the problem at hand but...", "Solution_6": "Well if they are mutually exclusive, then it is what you said. But if it is something like dice, then it is the formula I gave.", "Solution_7": "ok even numbers\r\n2,4,6\r\ngreater than 2\r\nso only 4,6 work\r\n\r\nthere are 6 sides on a standard die\r\n2/6=$1/3$", "Solution_8": "[quote=\"mysmartmouth\"]Probability of event X OR event Y occurring is $P(x)+P(y)-P(x)\\cdot P(y)$[/quote]\r\n\r\nActually, in the general case this should be $P(x)+P(y)-P(x \\text{ and }y)$. For mutually exclusive events the last probability is $0$, for independent events it is $P(x) \\cdot P(y)$, and in other cases it must be calculated separately.\r\n\r\nIn any case, moogra's approach is much simpler.", "Solution_9": "Yes, for some reason when I read it I was only thinking that OR events are mutually exclusive and disregarded when there is an intersection of the sets :blush: . And yes I agree with moogra, that was the solution I was hinting at in my first post.", "Solution_10": "[hide=\"answer, I think\"]Wouldnt the answer be 1/3? Since there are 3 even numbers, and two even numbers over 2, it would be 2/6=1/3.[/hide]", "Solution_11": "I like all replies and tips." } { "Tag": [ "linear algebra", "matrix", "quadratics", "inequalities", "linear algebra unsolved" ], "Problem": "Prove or disprove:\r\nFor two real symmetric matrices $ A$ and $ B$, we know that diagonal elements of $ A$ majorize diagonal elements from $ B$ (assume that diagonal entries are sorted). Is it true that eigenvalues of $ A$ must majorize eigenvalues of $ B$ (again we sort both arrays of eigenvalues)?", "Solution_1": "No. For your result you need a bit more, namely that the induced quadratic form of $ A$ majorizes this one of $ B$.\r\n\r\n[hide=\"Counterexample:\"]$ A\\equal{}\\left(\\begin{array}{cc}2&0\\\\0&2\\end{array}\\right)$, $ B\\equal{}\\left(\\begin{array}{cc}1&2\\\\2&1\\end{array}\\right)$.[/hide]", "Solution_2": "By definition of majorization we have that traces of matrices $ A$ and $ B$ (the sum of diagonal entries) are equal and therefore your counterexample is not correct...", "Solution_3": "I assumed that you meant the following version: If $ \\lambda_1\\le\\ldots\\le\\lambda_n$ and $ \\mu_1\\le\\ldots\\le\\mu_n$ are the diagonal elements of $ A$ and $ B$, resp. with $ \\lambda_i\\ge\\mu_i$. \r\n\r\nWhat definition of \"majorization\" do you apply?", "Solution_4": "http://en.wikipedia.org/wiki/Majorization\r\nTherefore, you have $ \\lambda_1 \\plus{} \\ldots \\plus{} \\lambda_k \\geqslant \\mu_1 \\plus{} \\ldots \\plus{} \\mu_k$ for every $ 1 \\leq k \\leq n \\minus{} 1$ and equality for $ k \\equal{} n$.", "Solution_5": "It's false.\r\nLet $ A,B$ be 2 symmetric real matrices s.t. $ trace(A)\\equal{}trace(B)$. $ A,B$ are orthogonally similar to $ A',B'$ that have the same constant diagonal (with eventually distinct changes of basis).\r\n Diagonal elements of $ A'$ majorize diagonal elements of $ B'$. Now assume the Alexillic's result is true. Then eigenvalues of $ A'$ majorize eigenvalues of $ B'$ and eigenvalues of $ A$ majorize eigenvalues of $ B$. The following obvious example gives the contradiction:\r\n $ A\\equal{}\\begin{pmatrix}0&0\\\\0&0\\end{pmatrix},B\\equal{}\\begin{pmatrix}1&0\\\\0&\\minus{}1\\end{pmatrix}$.", "Solution_6": "I still do not understand :)\r\nYou have that diagonal elements of $ B$ majorize diagonal elements of $ A$, and the same holds for eigenvalues of matrices $ B$ and $ A$?", "Solution_7": "Hi Alexillic, read once again my proof.\r\n$ A',B'$ have the same diagonal $ [trace(A)/n,\\cdots,trace(A)/n]$. I apply your false result to $ A',B'$ and not to $ A,B$.\r\nIf you don't understand my argument then consider this counterexample:\r\n$ A\\equal{}I_2,B\\equal{}\\begin{pmatrix}1&1\\\\1&1\\end{pmatrix}$." } { "Tag": [ "trigonometry" ], "Problem": "A pendulum 80 cm long is pulled aside, so that its bob is raised 20 cm from its lowest position and is tehn released. As the 50 g bob moves through its lowest position, (a) what is its speed and (b) what is the tension in the pendulum cord?\r\n\r\nI know that $ T\\minus{}mg\\equal{}\\frac{mv^2}{r}$, where $ T,m,v,r$ are tension, mass of the pendulum, velocity, and radius of the circle. Since $ T$ and $ v$ are what we're trying to find, I thought it would be easier to start with $ v$, as the book recommends.\r\n\r\nGoing through some trig, I found that the pendulum moved a total of 57.8 cm. But I don't know how to calculate the time it took, and I think I might be going about this in a wrong way in general.\r\n\r\nHelp?", "Solution_1": "To find its velocity, you use conservation of energy: $ E_{T}\\equal{}U\\equal{}mgh\\equal{}(.05)(.2)(9.8)\\equal{}.098J$. At its lowest point, all of the energy is in the form of kinetic energy. So $ .098\\equal{}\\frac{1}{2}mv^{2}\\equal{}\\frac{1}{2}(.05)v^{2}\\rightarrow v\\equal{}1.98 \\mbox{m/s}$." } { "Tag": [ "algebra", "polynomial", "complex numbers" ], "Problem": "Let $ x^{4}\\minus{}10x^{3}\\plus{}ax^{2}\\plus{}bx\\plus{}130\\equal{}0$ ($ a,b\\in Z$) have all the roots being complex numbers which are not real. Moreover, all the roots are in the form $ p\\plus{}qi$, where $ p,q\\in Z$. Find the maximum value of $ a\\plus{}b$.", "Solution_1": "We just have to factor $ 130$ into complex numbers or w/e. The rest is casework." } { "Tag": [ "function" ], "Problem": "Let $\\{ a_n \\}$ be a sequence such that $a_{n+1}=a_n^2-na_n +1$ with $n=1,2,3 \\cdots$. When $a_1 \\geq 3$, prove that for all $n \\geq 1$:\r\n\r\n(1): $a_n \\geq n+2$.\r\n\r\n(2): $\\frac{1}{1+a_1}+\\frac{1}{1+a_2}+\\cdots+\\frac{1}{1+a_n} \\leq \\frac{1}{2}$.", "Solution_1": "[hide=\"1\"]Part 1 can be solved by induction...\n$a_{n+1}^2=a_n^2-na_n+1\\ge (n+2)^2-n(n+2)+1=2n+5 \\ge n+3$[/hide]\r\n\r\nPart 2 seems to consist of proving that the harmonic mean of $(1+a_n)$ is greater than or equal to $2n$...but there must be some cleverer way...", "Solution_2": "I can solve problem 2 by using the conclution of problem 1, but I am not sure about my (and yours) solution to problem one.\r\n\r\nIf $f(x)$ is a second degree function, then $x_1 \\geq x_2$ does not imply $f(x_1) \\geq f(x_2)$...", "Solution_3": "[quote=\"shobber\"]I can solve problem 2 by using the conclution of problem 1, but I am not sure about my (and yours) solution to problem one.\n\nIf $f(x)$ is a second degree function, then $x_1 \\geq x_2$ does not imply $f(x_1) \\geq f(x_2)$...[/quote]\r\n\r\nI basically showed that $a_n \\ge n+2 \\implies a_{n+1} \\ge n+3$.\r\nThe thing is, $f(n+1)$ is a function of $f(n)$ [i] and [/i] $n$. You could even, if you wanted to, show that \\[ f(n+1) \\ge f(n); f(n), f(n+1) \\in \\mathbb{N} \\implies f(n+1) \\ge f(n)+1 \\] which completes the induction.", "Solution_4": "Let me say in another way: After assuming $a_n \\geq n+2$, why can you plug it into the equation $a_{n+1}=a_n^2-na_n+1$?", "Solution_5": "[quote=\"K81o7\"]Part 1 can be solved by induction...\n$a_{n+1}^2=a_n^2-na_n+1\\ge (n+2)^2-n(n+2)+1=2n+5 \\ge n+3$\n[/quote]\r\n\r\nA small imprecision there: You can't conclude $a_n^2-na_n\\ge (n+2)^2-n(n+2)$, because of the minus sign. But you can patch it via $a_n(a_n-n)\\ge (n+2)\\cdot 2.$", "Solution_6": "[quote=\"shobber\"]Let me say in another way: After assuming $a_n \\geq n+2$, why can you plug it into the equation $a_{n+1}=a_n^2-na_n+1$?[/quote]\r\n\r\nWhy not?\r\nIf $a_n \\ge n+2$ then $a_{n+1}=a_n^2-na_n+1 = a_n(a_n-n)+1 \\ge (n+2)(2)+1=2n+5$\r\nThe factored $a_n$ term is positive, so it's fine.", "Solution_7": "[quote=\"K81o7\"][hide=\"1\"]Part 1 can be solved by induction...\n$a_{n+1}^2=a_n^2-na_n+1\\ge (n+2)^2-n(n+2)+1=2n+5 \\ge n+3$[/hide]\n\nPart 2 seems to consist of proving that the harmonic mean of $(1+a_n)$ is greater than or equal to $2n$...but there must be some cleverer way...[/quote]\r\n\r\nI've tried some things and don't knows if it's okey or not; can somebody chek'it for me;\r\n\r\nfor $n\\geq 1$ $a_n\\geq 2n+1$ and for $n\\geq 2 a_n\\geq 2(n^2+1)$ then :\r\n$\\sum 1/(a_n+1)\\leq 1/2\\sum 1/(1+n^2)$\r\nthen :$\\leq 1/2(1/2+\\sum_{n\\geq 2} (1/n-1/(n+1))$\r\nthen the result" } { "Tag": [ "symmetry", "geometry", "trigonometry", "projective geometry", "geometry proposed" ], "Problem": "Let $P$ and $Q$ is two point in the triangle $ABC$.$B_1\\in{AP}\\cap{CQ},B_2\\in{AQ}\\cap{CP}$ and $l_B$ is the line pase trough $B_1$ and $B_2$.$l_A$and$l_C$ are analogously defined.\r\nProve that $l_A$,$l_B$ and $l_C$ are concurrent. ;)", "Solution_1": "You should remove the '?' from the source field. It is definitly yours and\r\n\r\n yours alone. Does anyone have problem with this?\r\n\r\n\r\n S Nastupaischim Novym Godom!\r\n\r\n\r\n\r\n Salute and thank you,\r\n\r\n M.T.", "Solution_2": "I writed '?',becaus I think that this problem is very nice,so sambody can get to it befor me :( :D,\r\nTi znaesh ruski? \r\nIa toje!!:blush: \r\nTebia toge s nastupaishim novim godam jelayu vsevo xaroshava v nowm gadu!!! \r\nThank you.", "Solution_3": "[u]Nice concurrence ![/u] I shall give a equivalent enunciation, with the my notations:\r\n\r\n$\\blacksquare\\ \\bullet\\parallel$Let $ABC$ be a triangle and two points $P$, $Q$ which belong to the inside of the triangle $ABC$. I note the following intersections: $X_a\\in BQ\\cap CP$, $Y_a\\in BP\\cap CQ$; $X_b\\in CP\\cap AQ$, $Y_b\\in CQ\\cap AP$; $X_c\\in AP\\cap BQ$, $Y_c\\in AQ\\cap BP$.\r\nProve that the lines $X_aY_a$, $X_bY_b$, $X_cY_c$ are concurrently.$\\parallel\\bullet\\ \\blacksquare$\r\n\r\n[u]Demonstration.[/u] I note the intersections $U$, $V$, $W_x$, $W_y$ of the line $BC$ with the lines $AP$, $AQ$, $X_bX_c$, $Y_bY_c$ respectively. I will apply the Menelaus' theorem for the triangle $AUV$ to the transversals $\\overline {BX_cQ}$, $\\overline {BY_cP}$, $\\overline {CX_bP}$, $\\overline {CY_bQ}$, $\\overline {X_bX_cW_x}$, $\\overline {W_bY_cW_y}$:\r\n\r\n$\\frac{BU}{BV}\\cdot \\frac{QV}{QA}\\cdot \\frac{X_cA}{X_cU}=1$; $\\frac{BV}{BU}\\cdot \\frac{Y_cA}{Y_cV}\\cdot \\frac{PU}{PA}=1$;\r\n\r\n$\\frac{CU}{CV}\\cdot \\frac{X_bV}{X_bA}\\cdot \\frac{PA}{PU}=1$; $\\frac{CV}{CU}\\cdot \\frac{QA}{QV}\\cdot \\frac{Y_bU}{Y_bA}=1$;\r\n\r\n$\\frac{W_xV}{W_xU}\\cdot \\frac{X_cU}{X_cA}\\cdot \\frac{X_bA}{X_bV}=1$; $\\frac{W_yU}{W_yV}\\cdot \\frac{Y_bA}{Y_bU}\\cdot \\frac{Y_cV}{Y_cA}=1$.\r\n\r\nFrom the product of the above relations we obtain the relation $\\frac{W_xU}{W_xV}=\\frac{W_yU}{W_yV}$, i.e. $W_x\\equiv W_y\\equiv W$. Therefore, the lines $X_bX_c$, $Y_bY_c$, $BC$ are concurrently (in the point $W$). For the triangles $X_aX_bX_c$, $Y_aY_bY_c$ we observe that $C\\in X_aX_b\\cap Y_aY_b$, $B\\in X_aX_c\\cap Y_aY_c$, $W\\in X_bX_c\\cap Y_bY_c$ and $W\\in BC$. From the Desarques' theorem results that the lines $X_aY_a$, $X_bY_b$, $X_cY_c$ are concurrently.\r\n\r\n[u]Remark.[/u] There is a shorter prove using the polar-line of a point w.r.t. a angle !", "Solution_4": "Nice solution Virgil,but if you know Papu's theorem,ther is wery short solution from it ;)(my solution :lol: )", "Solution_5": "Evidently, I know the Pappus's theorem, but I didn't think it. I hope yet that you like this solution. See that I made a little formal modification in the enunciation for the symmetry of the notations: $X_a\\longleftrightarrow Y_a$.\r\n\r\n[b]A very interesting remark.[/b]\r\n \r\n$A\\in X_aY_a\\ \\vee \\ B\\in X_bY_b\\ \\vee \\ C\\in X_cY_c\\Longleftrightarrow W\\in PQ$\r\n\r\nYou can apply the Desarques's theorem to the triangles, for example, $PX_bY_c$ and $QX_cY_b$, where $X_a\\in PX_b\\cap QX_c$, $Y_a\\in PY_c\\cap QY_b$, $A\\in X_bY_c\\cap X_cY_b$. Thus, $A\\in X_aY_a$ $\\Longleftrightarrow$ $PQ\\cap X_bX_c\\cap Y_bY_c\\ne \\emptyset.$ \r\n\r\n$\\blacksquare\\ P.S.\\ 1^{\\circ}$ Here is and the solution using the [u]Pappus' theorem:[/u]\r\n\r\n$\\{Y_b,P,X_c\\}\\subset AP$, $\\{Y_c,Q,X_b\\}\\subset AQ\\Longrightarrow$ the points $Y_a\\in Y_bQ\\cap PY_c$, $X_a\\in PX_b\\cap X_cQ$ and the intersection of the lines $Y_bX_b$, $X_cY_c$ are collinearly, i.e. the lines $X_aY_a$, $X_bY_b$, $X_cY_c$ are concurrently. \r\n\r\n$\\blacksquare\\ P.S.\\ 2^{\\circ}$ [u][b]Please, dear Tiks, don't use the icons ![/b][/u]\r\n\r\nI am a susceptible (impressionable) man.", "Solution_6": "[quote=\"Virgil Nicula\"]$\\blacksquare\\ P.S.\\ 1^{\\circ}$ Here is and the solution using the [u]Pappus' theorem:[/u]\n\n$\\{Y_b,P,X_c\\}\\subset AP$, $\\{Y_c,Q,X_b\\}\\subset AQ\\Longrightarrow$ the points $Y_a\\in Y_bQ\\cap PY_c$, $X_a\\in PX_b\\cap X_cQ$ and the intersection of the lines $Y_bX_b$, $X_cY_c$ are collinearly, i.e. the lines $X_aY_a$, $X_bY_b$, $X_cY_c$ are concurrently.[/quote] \nThat also my solution.\n[quote=\"Virgil Nicula\"]$\\blacksquare\\ P.S.\\ 2^{\\circ}$ [u][b]Please, dear Tiks, don't use the icons ![/b][/u]\n\nI am a susceptible (impressionable) man.[/quote]\r\nNo problem dear Virgil!.", "Solution_7": "Thanks.\r\nLike you the remark when $W\\in PQ$ ?\r\nSee a interesting problem at\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=66939", "Solution_8": "[quote]Like you the remark when $W\\in PQ$ ?[/quote]\nOf cours I like it \n[quote]See a interesting problem at\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=66939[/quote]\r\nI foude same solutions for this problem,but in all of them I use analitc geometry or trigonometry :(." } { "Tag": [ "Diophantine Equations", "pen" ], "Problem": "Let $a,b$, and $x$ be positive integers such that $x^{a+b}=a^b{b}$. Prove that $a=x$ and $b=x^{x}$.", "Solution_1": "I hope I made no mistakes:\r\n[hide=\"Solution\"]\nWe assume that $ x\\ne a$. Let $ a=\\prod_{i=1}^{k} {p_i^{x_i}}$, $ b=\\prod_{i=1}^k {p_i^{y_i}}$, and $ x=\\prod_{i=1}^{k} {p_i^{z_i}}$. Henceforth, we will ignore simply denote $ \\prod_{i=1}^k$ as $ \\prod$ for simplicity. Furthermore, when we refer to $ y_i$, we do not mean $ y_i$ collectively, but for rather for a fixed $ y_i$ (or else otherwise noted.) Then, \\[ x^{a+b}=a^bb\\iff \\prod {p_i^{z_i\\cdot (\\prod {p_i^{y_i}}+\\prod {p_i^{x_i}})}=\\prod {p_i^{y_i+x_i\\cdot \\prod {p_i^{y_i}}}}\\iff}\\] \\[ x_i\\cdot \\prod {p_i^{y_i}}+y_i=z_i\\cdot (\\prod {p_i^{x_i}}+\\prod {p_i^{y_i}})\\] for all $ 1\\le i\\le k$. Now, if $ y_ix_i$, we have that $ ak_i>x_i$, meaning that $ \\prod {p_i^{ak_i-x_i}}>ak_i-x_i\\ge k_i-x_i$. If $ k_i-z_i<0$, then if $ z_i-x_i>0$, we would have that the left hand side of the equation is negative when the other side is positive, giving us a contradiction. Then, $ z_i-x_i<0$, so $ \\frac{k_i-z_i}{z_i-x_i}=\\frac{z_i-k_i}{x_i-z_i}ak_i-x_i\\ge a-x_i\\ge x_i>\\frac{k_i-z_i}{z_i-x_i}\\] which is a contradiction. Otherwise, we have that $ k_i-z_i, z_i-x_i>0$, so \\[ \\frac{k_i-z_i}{z_i-x_i}b$\nThus $ x|a$. Suppose $ a\\geq 2x$; then $ b\\equal{}x^a(x/a)^b\\geq x^a 2^b>b$.\nThus $ a\\equal{}x$ and the result follows.\n[/hide]", "Solution_3": "[quote=\"me@home\"]\nThus $ x|a$. Suppose $ a\\geq 2x$; then $ b \\equal{} x^a(x/a)^b\\geq x^a 2^b > b$.[/quote]\r\nSorry, I might be misunderstanding, but isn't $ \\frac{x}{a}\\le \\frac {1}{2}$, so $ b\\equal{}x^a\\cdot (\\frac{x}{a})^b\\le x^a\\cdot (\\frac{1}{2})^b$?", "Solution_4": "[hide=Another solution]\n[b]Step 1[/b] $x\\mid a$.\nFor each prime $p$ we have $bv_p(x)<(a+b)v_p(x)=bv_p(a)+v_p(b)v_p(a)-1$.\nHowever, from $x\\mid a$, $v_p(x)\\leq v_p(a)$, so $v_p(x)=v_p(a)$ for all prime $p$.\nTherefore $a=x$, and thus $b=x^x$.\n[/hide]" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Several equal spherical planets are in outer space. On the surface of each planet is a set of points which is invisible from any of the remaining planets. \r\nProve that the sum of the areas of all these sets equals the surface area of one planet.", "Solution_1": "Please continue discussion at http://www.mathlinks.ro/Forum/viewtopic.php?t=6395 or http://www.mathlinks.ro/Forum/viewtopic.php?t=1465 .\r\n\r\n Darij" } { "Tag": [ "vector", "algebra", "polynomial", "linear algebra", "matrix", "abstract algebra", "calculus" ], "Problem": "Let $ K$ be a field,and $ F\\equal{}K[\\alpha]$ ba a extension field of $ F$,such that $ \\alpha$ has minimal polynomial $ x^m\\plus{}a_{m\\minus{}1}x^{m\\minus{}1}\\plus{}\\cdots\\plus{}a_0\\equal{}0$ in $ K[x]$,then a vector space $ V$ of dimension $ n$ of $ F$ can be regard as a vector space of $ K$ which has dimension $ mn$.\r\nThen every homomorphism of $ V$ can be naturally maped to a matrix of $ M_{nm}(K)$,should this map be monmorphism?and the automorphism of $ V$ map to the invertible matrix?\r\n\r\nThank you for help.", "Solution_1": "The map is mono because if we fix a base of $ V$ over $ F$ such that the matrix representation of an endomorphism is $ A \\equal{} A_0 \\plus{} A_1 \\alpha \\plus{} \\cdots \\plus{} A_{m\\minus{}1} \\alpha^{m\\minus{}1}$, then with respect to the natural base of $ V$ over $ K$ the first $ n$ columns would be exactly $ \\left[A_0 A_1 \\cdots A_{m\\minus{}1}\\right]^T$.\r\n\r\nFor automorphism part, I suspect that this correspondence (linear transformation of V over F -> linear transformation of V over K) is a homomorphism, which would automatically solve your problem, but currently I am not sure about how to show it.", "Solution_2": "thank you.\r\nanother question: \r\nif there is $ m$ different automorphisms of $ F$ that fix $ K$,define it is $ t_1,t_2,...t_m$,then we have:\r\n\r\n$ \\det(A) \\equal{} \\prod_{i \\equal{} 1}^m t_i(\\det(f))$\r\nWhere $ f$ is the homorphism of $ V$ over $ F$,$ A$ is the correspondent homorphism of $ V$ over $ K$.\r\n\r\nIt is ture when $ n\\equal{}1$(which is in algebraic number theory that seem not trivial)", "Solution_3": "Let's do this more systematically - wherever there is an assertion about matrices, it can't harm to translate it into a base-free linear maps form (of course, the exact opposite is equally useful).\r\n\r\nWhat you have is the following situation:\r\n\r\n$ F$ is a field extension of $ K$. You want it to be primitive, but this is pointless - it is enough that $ K$ is a commutative ring and $ F$ is a commutative $ K$-algebra which is free as an $ K$-module. Note that this free allows us to define a norm map $ \\mathcal{N}: F\\to K$ as follows: For every element $ f\\in F$, the multiplication with $ f$ (as a map from $ F$ to $ F$) is an endomorphism of the free $ K$-module $ F$, thus has a determinant. This determinant is what we call the norm $ \\mathcal{N}\\left(f\\right)$ of $ f$.\r\n\r\nNow you have given a finite free $ F$-module $ V$. Of course, you can consider this $ V$ as a $ K$-module as well. Every endomorphism $ g$ of the $ F$-module $ V$ is also an endomorphism of the $ K$-module $ V$ (but not the other way round!), and thus has both a determinant $ \\det_Fg\\in F$ when seen as an endomorphism of the $ F$-module $ V$, and a determinant $ \\det_Kg\\in K$ when seen as an endomorphism of the $ F$-module $ V$.\r\n\r\nThe problem is now to show that $ \\det_Kg \\equal{} \\mathcal{N}\\left(\\det_Fg\\right)$ for any endomorphism $ g$ of the $ F$-module $ V$. In other words, we have to show that $ \\det_K \\equal{} \\mathcal{N}\\circ\\det_F$ as maps from $ \\mathrm{End}_FV$ to $ K$.\r\n\r\nThis $ \\det_K \\equal{} \\mathcal{N}\\circ\\det_F$ formula has been posted in matrix form before:\r\nhttp://www.mathlinks.ro/viewtopic.php?t=44950\r\nhttp://www.mathlinks.ro/viewtopic.php?t=45040\r\n\r\nThe proof posted there was not the best, but a simpler one can be done by basically performing Gaussian elimination on the block matrix (treating the blocks $ A_{ij}$ as elements of a field - they are elements of a commutative ring which is generically an integral domain, and one can go into the quotient field of that domain). I must hurry back to IMO coordination, so sorry for any missing explanations and mistakes. Will try to detail this later.\r\n\r\n darij" } { "Tag": [ "geometry", "circumcircle", "analytic geometry", "linear algebra", "matrix", "geometry proposed" ], "Problem": "find the center and radius of the circumcircle of triangle ABC with coordinates $ (\\minus{}x,0), (x,0), (x_1,y_1)$", "Solution_1": "$ O(0,y) \\Longrightarrow O(0, \\frac {x_1^2 \\plus{} y_1^2 \\minus{} x^2}{2y_1})$\r\n\r\n$ R \\equal{} \\sqrt {x^2 \\plus{} y^2}$.", "Solution_2": "Deleted.", "Solution_3": "well i mean assume that the points are not collinear but otherwise obviously they exist\r\n@lig could you explain?", "Solution_4": "Deleted. :oops:", "Solution_5": "What? Maybe there is a missing condition or something. \r\n\r\n$ R \\equal{} \\frac {abc}{4A} \\equal{} \\frac {2x\\cdot\\sqrt {(x_1 \\minus{} x)^2 \\plus{} y_1^2}\\cdot\\sqrt {(x_1 \\plus{} x)^2 \\plus{} y_1^2}}{4|xy_1|} \\equal{} \\frac {\\sqrt {(x_1 \\minus{} x)^2 \\plus{} y_1^2}\\cdot\\sqrt {(x_1 \\plus{} x)^2 \\plus{} y_1^2}}{2|y_1|}$\r\n\r\n$ x,y_1\\ne 0$...", "Solution_6": "The equation for the circumcircle of the triangle with polygon vertices $ (x_i,y_i)$ for $ i\\equal{}1$, $ 2$, $ 3$ is\r\n\\[ \\begin{vmatrix}\r\nx^2\\plus{}y^2 & x & y & 1 \\\\\r\nx_1^2\\plus{}y_1^2 & x_1 & y_1 & 1 \\\\\r\nx_2^2\\plus{}y_2^2 & x_2 & y_2 & 1 \\\\\r\nx_3^2\\plus{}y_3^2 & x_3 & y_3 & 1 \r\n\\end{vmatrix}\\equal{}0\\]\r\n\\[ \\iff\\left\\{\\begin{array}{l}\r\n(x\\minus{}x_0)^2\\plus{}(y\\minus{}y_0)^2\\equal{}r^2, \\\\\r\nx_0\\equal{}\r\n\\minus{}\\frac{\\begin{vmatrix}\r\nx_1^2\\plus{}y_1^2 & y_1 & 1 \\\\\r\nx_2^2\\plus{}y_2^2 & y_2 & 1 \\\\\r\nx_3^2\\plus{}y_3^2 & y_3 & 1 \r\n\\end{vmatrix}}{2\\begin{vmatrix}\r\nx_1 & y_1 & 1 \\\\\r\nx_2 & y_2 & 1 \\\\\r\nx_3 & y_3 & 1 \r\n\\end{vmatrix}}, \\\\\r\ny_0\\equal{}\r\n\\minus{}\\frac{\\begin{vmatrix}\r\nx_1^2\\plus{}y_1^2 & x_1 & 1 \\\\\r\nx_2^2\\plus{}y_2^2 & x_2 & 1 \\\\\r\nx_3^2\\plus{}y_3^2 & x_3 & 1 \r\n\\end{vmatrix}}{2\\begin{vmatrix}\r\nx_1 & y_1 & 1 \\\\\r\nx_2 & y_2 & 1 \\\\\r\nx_3 & y_3 & 1 \r\n\\end{vmatrix}}, \\\\\r\nr\\equal{}\r\n\\frac{\\sqrt{\\begin{vmatrix}\r\nx_1^2\\plus{}y_1^2 & y_1 & 1 \\\\\r\nx_2^2\\plus{}y_2^2 & y_2 & 1 \\\\\r\nx_3^2\\plus{}y_3^2 & y_3 & 1 \r\n\\end{vmatrix}^2\\plus{}\\begin{vmatrix}\r\nx_1^2\\plus{}y_1^2 & x_1 & 1 \\\\\r\nx_2^2\\plus{}y_2^2 & x_2 & 1 \\\\\r\nx_3^2\\plus{}y_3^2 & x_3 & 1 \r\n\\end{vmatrix}^2\\minus{}4\\begin{vmatrix}\r\nx_1 & y_1 & 1 \\\\\r\nx_2 & y_2 & 1 \\\\\r\nx_3 & y_3 & 1 \r\n\\end{vmatrix}\\begin{vmatrix}\r\nx_1^2\\plus{}y_1^2 & x_1 & y_1 \\\\\r\nx_2^2\\plus{}y_2^2 & x_2 & y_2 \\\\\r\nx_3^2\\plus{}y_3^2 & x_3 & y_3 \r\n\\end{vmatrix}}}{2\\text{abs}\\left(\\begin{vmatrix}\r\nx_1 & y_1 & 1 \\\\\r\nx_2 & y_2 & 1 \\\\\r\nx_3 & y_3 & 1 \r\n\\end{vmatrix}\\right)}.\r\n\\end{array}\\right.\r\n\\]\r\n\r\nIf,\r\n$ \\begin{array}{l}\r\na\\equal{}\\begin{vmatrix}\r\nx_1 & y_1 & 1 \\\\\r\nx_2 & y_2 & 1 \\\\\r\nx_3 & y_3 & 1 \r\n\\end{vmatrix}, \\ \r\nb_x\\equal{}\\begin{vmatrix}\r\nx_1^2\\plus{}y_1^2 & y_1 & 1 \\\\\r\nx_2^2\\plus{}y_2^2 & y_2 & 1 \\\\\r\nx_3^2\\plus{}y_3^2 & y_3 & 1 \r\n\\end{vmatrix}, \\ \r\nb_y\\equal{}\\begin{vmatrix}\r\nx_1^2\\plus{}y_1^2 & x_1 & 1 \\\\\r\nx_2^2\\plus{}y_2^2 & x_2 & 1 \\\\\r\nx_3^2\\plus{}y_3^2 & x_3 & 1 \r\n\\end{vmatrix}, \\ \r\nc\\equal{}\\begin{vmatrix}\r\nx_1^2\\plus{}y_1^2 & x_1 & y_1 \\\\\r\nx_2^2\\plus{}y_2^2 & x_2 & y_2 \\\\\r\nx_3^2\\plus{}y_3^2 & x_3 & y_3 \r\n\\end{vmatrix},\r\n\\end{array}$\r\n\r\n$ x_0\\equal{}\\minus{}\\dfrac{b_x}{2a}, \\ \r\ny_0\\equal{}\\minus{}\\frac{b_y}{2a}, \\ \r\nr\\equal{}\\frac{\\sqrt{b_x^2\\plus{}b_y^2\\minus{}4ac}}{2|a|}$\r\n\r\nSo,\r\n\\[ \\begin{vmatrix}\r\nX^2\\plus{}Y^2 & X & Y & 1 \\\\\r\n(\\minus{}x)^2\\plus{}0^2 & \\minus{}x & 0 & 1 \\\\\r\nx^2\\plus{}0^2 & x & 0 & 1 \\\\\r\nx_1^2\\plus{}y_1^2 & x_1 & y_1 & 1 \\\\\r\n\\end{vmatrix}\\equal{}0.\\]\r\n\\[ a\\equal{}2xy_1, \\ b_x\\equal{}0, \\ b_y\\equal{}2x(x^2\\minus{}x_1^2\\minus{}y_1^2), \\ c\\equal{}2x^3y_1,\\]\r\nThus,\r\n$ x_0\\equal{}\\minus{}\\dfrac{b_x}{2a}\\equal{}0$\r\n$ y_0\\equal{}\\minus{}\\frac{b_y}{2a}\\equal{}\\frac{x_1^2\\plus{}y_1^2\\minus{}x^2}{2y_1}$\r\n$ r\\equal{}\\frac{\\sqrt{b_x^2\\plus{}b_y^2\\minus{}4ac}}{2|a|}\\equal{}\\frac{\\sqrt{(x^2\\minus{}x_1^2\\minus{}y_1^2)^2\\minus{}x^2y_1^2}}{2y_1}$\r\n\r\nCf. [url=http://mathworld.wolfram.com/Circumcircle.html]Circumcircle[/url] - MathWorld", "Solution_7": "where (x_0,y_0) is the center right?", "Solution_8": "[quote=\"mathemonster\"]where (x_0,y_0) is the center right?[/quote]\r\nYes.\r\nIn Cartesian coordinates, the equation of a circle of radius $ r$ centered on $ (x_0,y_0)$ is $ (x\\minus{}x_0)^2\\plus{}(y\\minus{}y_0)^2\\equal{}r^2$.", "Solution_9": "[quote=\"Ligouras\"]$ O(0,y) \\Longrightarrow O(0, \\frac {x_1^2 \\plus{} y_1^2 \\minus{} x^2}{2y_1})$\n\n$ R \\equal{} \\sqrt {x^2 \\plus{} y^2}$.[/quote]" } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "consider a sequence $(a_n)$ satisfies $a_0=1;a_1=9$\r\nand $a_{n+1}=6a_n-a_{n-1}-4$\r\n\r\nfind $n$ such that $a_n$ is a perfect square", "Solution_1": "[quote=\"kiemkhach\"]consider a sequence $(a_n)$ satisfies $a_0=1;a_1=9$\nand $a_{n+1}=6a_n-a_{n-1}-4$\n\nfind $n$ such that $a_n$ is a perfect square[/quote]\r\nWell, $n=1$ and $n=2$ are two solutions, rightt???? :P :P :P but I suppose you want an $n>2$.\r\nAnd in general \r\n\r\n\\[a_n=1+\\sqrt 2\\left((3+2\\sqrt 2)^n-(3-2\\sqrt 2)^n\\right).\\]\r\n\r\nTCM", "Solution_2": "[quote=\"RobertuX\"][quote=\"kiemkhach\"]consider a sequence $(a_n)$ satisfies $a_0=1;a_1=9$\nand $a_{n+1}=6a_n-a_{n-1}-4$\n\nfind $n$ such that $a_n$ is a perfect square[/quote]\nWell, $n=1$ and $n=2$ are two solutions, rightt???? :P :P :P but I suppose you want an $n>2$.\nAnd in general \n\n\\[a_n=1+\\sqrt 2\\left((3+2\\sqrt 2)^n-(3-2\\sqrt 2)^n\\right).\\]\n\nTCM[/quote]\r\n\r\n$a_3=6a_2-a_1-4=54-1-4=49=7^2$???? :( \r\ncan you post your solution? :D I think ;n=1 and n=2 and n=3 are solutions??", "Solution_3": "[quote=\"RobertuX\"]In general \n\\[a_n=1+\\sqrt 2\\left((3+2\\sqrt 2)^n-(3-2\\sqrt 2)^n\\right).\\]\n[/quote]\r\nWell, of course this 3 values $n=1,2,3$ are solutions, and using my expression $a_4$ is NOT a square.", "Solution_4": "[quote=\"RobertuX\"][quote=\"RobertuX\"]In general \n\\[a_n=1+\\sqrt 2\\left((3+2\\sqrt 2)^n-(3-2\\sqrt 2)^n\\right).\\]\n[/quote]\nWell, of course this 3 values $n=1,2,3$ are solutions, and using my expression $a_4$ is NOT a square.[/quote]\r\nI don't understand! :( In general :$a_n=1+\\sqrt{2}((3+\\sqrt{2})^n-(3-\\sqrt{2})^n)$\r\n how do you prove $a_n$ is not a square???($n>3$)", "Solution_5": "[quote=\"kiemkhach\"]In general \n\\[a_n=1+\\sqrt 2\\left((3+2\\sqrt 2)^n-(3-2\\sqrt 2)^n\\right).\\]\nhow do you prove $a_n$ is not a square???($n>3$)[/quote]\r\nWell, sorry for the last comment, by definition $a_{n+1}=6a_n-a_{n-1}-4$, this is so colled non homogeneous sequence, so first I get an equivalent homogeneous sequence, HOW? well, in this case the sum of coefficients of the homogeneous part is $1-6+1=-4$ so we need to consider $b_n=a_n-1$, in fact $b_n$ satisfies the relations:$b_0=0$, $b_1=8$ and $b_{n+1}-6b_n+b_{n-1}=0$ and to solve this kind of sequences you need to consider the charasteristic polynomial related to the sequence, in this case is $x^2-6x+1=0$ which roots are $\\alpha=3-\\sqrt 8$ and $\\beta=3+\\sqrt 8$, so $b_n=A\\alpha^n+B\\beta^n$, now we calculate $A$ and $B$ with the initial values, so $A+B=0$ and $\\alpha A+\\beta B=8$ then $A=-B=\\sqrt 2$ so, you can write easily the expression of $a_n$. If you need for info please, read a book of Discrete Mathematics.\r\n\r\nSorry again,and best regards," } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $a,b,c$ are there side lengths of a triangle then\r\n\\[\\frac{a^{2}b}{c}+\\frac{b^{2}c}{a}+\\frac{c^{2}a}{b}\\ge a^{2}+b^{2}+c^{2}. \\]", "Solution_1": "Ugly but quick solution:\r\n\r\nUse Ravi substitution, and the inequality is equivalent to\r\n\\[x^{5}+y^{5}+z^{5}+x^{4}z+xy^{4}+yz^{4}\\geq 2(x^{3}y^{2}+x^{2}z^{3}+y^{3}z^{2}) \\]\r\nNow use AM-GM to show that $x^{5}+xy^{4}\\geq 2x^{3}y^{2}$ and similar for other variables, then sum up." } { "Tag": [], "Problem": "Given that $ 7x \\plus{} 3y \\equal{} 54$ and $ 3x \\plus{} 7y \\equal{} 46$, what is the value of $ x \\plus{} y$?", "Solution_1": "We realize that if we add both equations up, we get $ 10x \\plus{} 10y \\equal{} 100$. Perfect. After dividing by $ 10$ we get $ \\boxed{x \\plus{} y \\equal{} 10}$" } { "Tag": [ "ratio", "number theory unsolved", "number theory" ], "Problem": "$ a_{0}$=1,$ a_{n\\plus{}1}$=$ a_{n}$+(cos133\u220f/355)/$ a_{n}$\r\nfind,with proof,the smallest n satisfying:$ a_{n}$ is an integer", "Solution_1": "any idea will be appreciated.\r\ni wonder if it is suitable for the NT forum,or should I post it in algebra forum?", "Solution_2": "I think it's a trick question. Answer is $ n\\equal{}0$.", "Solution_3": "Interesting problem.\r\n\r\nIs 133/355 supposed to be 113/355? That ratio seemed peculiar and since 355/113 is sometimes used as an approximation to Pi it made me think that was involved with the problem.\r\n\r\nIf not, is there any reason for the peculiar 133/355 ?\r\n\r\nThe difficulty in this problem appears to be in analyzing the non-linear recurrence relation. Nothing jumped out at me which made me think like Phil that n=0.\r\n\r\nI did a quick numeric check though and the closest I found was at n=161218988 with a=11123.0000000001411... But this closeness appears to be just because you're adding smaller and smaller numbers.", "Solution_4": "my bad,\r\nthe question is to find the smallest n>0" } { "Tag": [ "geometry", "AMC", "AIME", "AIME I" ], "Problem": "Is it true that 3-D geometry doesn't show up anymore at the AIME (or even below) level?", "Solution_1": "[quote=\"veezbo\"]Is it true that 3-D geometry doesn't show up anymore at the AIME (or even below) level?[/quote]\r\nNoway, I love 3-D geometry and they are quite easy. I remember seeing one on a recent AMC (200x)", "Solution_2": "No, there was one on last year's AMC 12B. (#20 I think?)", "Solution_3": "*sigh* Yes unfortunately, they are still alive and well. But they're not that common :)", "Solution_4": "Agh. It's just because I'm really bad at it and have to rederive some basic stuff because I haven't actually learned it at all.\r\n\r\nI was hoping to spend more time just working on Euclidean Geometry because apparently while although I can consistently solve one problem above #10 and nearly every non-geometry problem under #10, I have still never actually solved an AIME-level geometry problem during the test.", "Solution_5": "Do you think that Intro to Geometry covers the 3-D geometry on AMC 10/12/AIME?", "Solution_6": "Definitely not the AIME. Probably the AMC 10 and most of AMC 12.", "Solution_7": "[quote=\"veezbo\"]Is it true that 3-D geometry doesn't show up anymore at the AIME (or even below) level?[/quote]\r\n\r\nNo, e.g. see 2008 AIME I no. 15. However, most of the 3-D geometry in AMC/AIME is often just 2-D geometry guised in 3-D, and therefore is usually easier than you think.", "Solution_8": "Are there any books you would recommend for mastering AIME/beggining Olympiad geometry? (not 3D, I mean in general)", "Solution_9": "There have been a discussion in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=296565]this[/url] link, especially towards the end. After reading the Intro to Geo book (or taking the course), Geometry Revisited, Problems in Plane Geometry, etc are recommended.", "Solution_10": "I wouldn't recommend a book - I'd recommend doing more problems.", "Solution_11": "3-D does appear some on AMC/AIME but definitely not much so there's not much need worrying about it I guess", "Solution_12": "[quote=\"worthawholebean\"]I wouldn't recommend a book - I'd recommend doing more problems.[/quote]\r\n\r\nA book of problems, I think he means.", "Solution_13": "You really have to get the fundamentals down first. You can't just rush off and only do problems. And when you do, try hard, study the solution carefully, pinpoint your weak points, and try to eliminate stupid mistakes." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Let $ ABC$ is triangle, let $ AB\\equal{}c, BC\\equal{}a,CA\\equal{}b$ and let $ M$ is a change point. Prove that : $ a.MA^2\\plus{}b.MB^2\\plus{}c.MC^2 \\ge abc$.", "Solution_1": "See [url=http://www.mathlinks.ro/viewtopic.php?t=206226][color=darkred][b]here[/b][/color][/url] ." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Here is a challenge problem that I found really nice.\r\n\r\nThe product\r\n\r\n\\[(x+5)(x+10)(2x^2+3)(x^3+6x+16)^2(x+9)(x+4)^3(x+18)\\]\r\n\r\nis expanded. In the resulting polynomial, how many of the coefficients are odd?", "Solution_1": "Ah, this reminds me of the course I took on polynomial rings. But the problem doesn't REALLY use any of that junk, although the mapping I am using is of relevance to certain results on finite fields. See spoiler for solution:\r\n\r\n[hide]We will take the mapping $\\varphi:\\mathbb{Z}[x]\\rightarrow (\\mathbb{Z}/2)[x]$ and the result will follow quickly. Let $p(x)$ be the original polynomial, then:\n\n$p(x)=(x+5)(x+10)(2x^2+3)(x^3+6x+16)^2(x+9)(x+4)^3(x+18)$ and\n$\\varphi(p(x))=(x+1)(x)(1)(x^3)^2(x+1)(x)^3(x)=x^11(x+1)^2=x^{13}+x^{11}$. These are the only 'surviving' terms, and hence the only terms with odd coefficients.\n\nSo, there are 2 odd coefficients, and maple agrees with me.\n\nUsing this method, it doesn't matter what we change the polynomials to; the answer is still very easy to find.\n[/hide]", "Solution_2": "[quote=\"blahblahblah\"]Ah, this reminds me of the course I took on polynomial rings. But the problem doesn't REALLY use any of that junk, although the mapping I am using is of relevance to certain results on finite fields. See spoiler for solution:\n\n[hide]We will take the mapping $\\varphi:\\mathbb{Z}[x]\\rightarrow (\\mathbb{Z}/2)[x]$ and the result will follow quickly. Let $p(x)$ be the original polynomial, then:\n\n$p(x)=(x+5)(x+10)(2x^2+3)(x^3+6x+16)^2(x+9)(x+4)^3(x+18)$ and\n$\\varphi(p(x))=(x+1)(x)(1)(x^3)^2(x+1)(x)^3(x)=x^11(x+1)^2=x^{13}+x^{11}$. These are the only 'surviving' terms, and hence the only terms with odd coefficients.\n\nSo, there are 2 odd coefficients, and maple agrees with me.\n\nUsing this method, it doesn't matter what we change the polynomials to; the answer is still very easy to find.\n[/hide][/quote]\r\nExcellent. But I wonder if there is another way of solving this without referencing to ring homomorphims...", "Solution_3": "In a naive sense, you can just take all coefficients $\\bmod 2$ and realize that that accomplishes the same thing whether you do it to the factored version of the polynomial or the expanded version of the polynomial.", "Solution_4": "Certainly, since binary representations yield quickly those coefficients that are odd. This is more of an appropriate level topic in the Intermediate Forum :)" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "What is the $ 100^{\\text{th}}$ term of the sequence 2, 3, 5, 6, 7, 10, 11, $ \\ldots$, which consists of all of the positive integers that are neither perfect squares nor perfect cubes?", "Solution_1": "Consider the sequence of the first $ 100$ positive integers. $ 10$ of these are perfect squares, so we add the positive integers from $ 101$ to $ 110$ and get rid of the perfect squares. \r\n\r\nAlso, from $ 1$ to $ 110$ there are $ 4$ cubes, but $ 2$ of them are also perfect squares so we only need to get rid of $ 2$ cubes.\r\n\r\nThe new $ 100^\\text{th}$ term is then $ \\boxed{112}$" } { "Tag": [ "inequalities" ], "Problem": "Ten people queqe before a tap to fill up their buckets. Each bucket requires a different time to fill. In what order should the people queqe up in order to minimize their combined waiting time? Do not just use pure logic to justify your answer.", "Solution_1": "In order to have the least waiting time in total, the people should line up in order from least time to fill in front to most time to fill in the back. We know this because for any two people standing one in front of the other, switching their positions does not affect the waiting time of anyone else in the line.\r\nThe least total waiting time between those two people will be achieved by having the less-time-to fill man in front. To show this, let us assume that one of them takes n time to fill and one takes x time to fill where x>n. \r\nIf x is in front of n, then their total waiting time will be x+(x+n)=2x+n. If n is in front of x, then their total waiting time will be n+(n+x)=2n+x. We know that 2x+n>2n+x. Therefore, the waiting time will be minimized by putting the man with a bucket which fills more quickly in front. One can now take any formation, and can undertake a series of these optimizations, and get to the shortest time being from least to greatest.\r\nOne can also see the total time as 10(first one in line)+9(second in line)+8(third in line)+...+(tenth in line).", "Solution_2": "Hehehe\r\n[hide]\nFrom least time to greatest, since the lightests buckets can get out of the way easier.[/hide]", "Solution_3": "[quote]\nHehehe \n\nClick to reveal hidden content\n\nFrom least time to greatest, since the lightests buckets can get out of the way easier. \n[/quote]\r\n\r\nDangit not just pure logic. But you are right. So is K81o7 who gave the full solution." } { "Tag": [ "inequalities", "function", "logarithms", "inequalities proposed" ], "Problem": "Let $ \\{a,b,c\\}\\subset[ k, \\plus{} \\infty)$ such that $ abc\\geq1$ and\r\n$ k$ is a real root of the equation $ x \\plus{} \\sqrt [3]x \\plus{} 1 \\equal{} 0.$ Prove that\r\n\\[ (a^2 \\plus{} a \\plus{} 1)(b^2 \\plus{} b \\plus{} 1)(c^2 \\plus{} c \\plus{} 1)\\geq(a \\plus{} 2)(b \\plus{} 2)(c \\plus{} 2)\\]\r\nI posted it here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=179021\r\nbut it still unsolved there.", "Solution_1": "I proved that $ (a^2\\plus{}a\\plus{}1)(b^2\\plus{}b\\plus{}1)(c^2\\plus{}c\\plus{}1)\\geq \\sqrt[3]{a^2b^2c^2}(a\\plus{}2)(b\\plus{}2)(c\\plus{}2)$ \r\n\r\nTo prove this just consider the function $ f(x)\\equal{}\\ln (a^2\\plus{}a\\plus{}1)\\minus{}\\frac{2}{3}\\ln a\\minus{}\\ln (a\\plus{}2)$ \r\n\r\n :wink:", "Solution_2": "[quote=\"silouan\"]I proved that $ (a^2 \\plus{} a \\plus{} 1)(b^2 \\plus{} b \\plus{} 1)(c^2 \\plus{} c \\plus{} 1)\\geq \\sqrt [3]{a^2b^2c^2}(a \\plus{} 2)(b \\plus{} 2)(c \\plus{} 2)$ \n\nTo prove this just consider the function $ f(x) \\equal{} \\ln (a^2 \\plus{} a \\plus{} 1) \\minus{} \\frac {2}{3}\\ln a \\minus{} \\ln (a \\plus{} 2)$ \n\n :wink:[/quote]\r\nBut $ a$ can be negative. :wink:", "Solution_3": "I posted a proof only for positive reals (I forgot to write it )" } { "Tag": [], "Problem": "A stone, which weighs 0.62 kg, is thrown at 5.5ms, its kinetic energy when at its maximum height above the surface of the ground is 2.3J. What is the height of the stone at this point, any equations and advice gratefully recieved.", "Solution_1": "Its initial kinetic energy is $\\frac12 \\cdot 0.62 \\cdot 5.5^{2}=9.3775\\textrm{J}$.\r\n\r\nAssuming resistance is negligible, and that it was thrown from ground level, when it is at its highest point it has $9.3375-2.3=7.0375\\textrm{J}$ of gravitational energy, and since $\\textrm{GPE}=mgh$ we have $gh=\\frac{7.0375}{0.62}=11.35$ and hence, taking $g=9.8ms^{-2}$ \r\n$h=1.16m$.\r\n\r\nThe two equation required are \r\nkinetic energy=$\\frac12mv^{2}$\r\ngravitational energy=$mgh$", "Solution_2": "Excellent, thanks very much." } { "Tag": [ "analytic geometry", "geometry unsolved", "geometry" ], "Problem": "Is it possible to find?\r\n[img]http://img239.imageshack.us/img239/5181/dsfdfyj1.png[/img]", "Solution_1": "every posting good problems :)\r\n\r\nthanks Shahzada\r\n\r\nmy figure can be easily made using Tales.\r\npeharps. we have 3 constants, but in your problem we have only 2.\r\n\r\nso. it's missing one value. without that we cannot going further.\r\n\r\nbut i'm not really sure about that. because $x+s = AB$ and $x+i = AC$\r\n\r\nso, maybe it's possible to eliminate $x$, but i'm not confident of this.\r\n\r\ni will try to find the equation of that line, tangent to the 2 circles in a easy way after.\r\n\r\nc ya", "Solution_2": "thanks much yagaron.\r\n\r\nIf we change the size of circles such that the |MN| is same, CB and AB will change.It may missing something.If it is possible to solve there must be an equation included only MN,AC and AB not any angle.But i think it is wrong.what is your opinion?", "Solution_3": "we have 9 points. let's call it $A,B,C, ...$\r\n\r\nso, $AB$ is a line $AC$ is another...\r\n\r\nwhat i think is, if we have only 3 combinations of a pair of points it's possible to solve the problem\r\n\r\nof course that equal combinations does not enter in the problem. like equal lengths.\r\n\r\nbecause on my drawn all anothers points coordinates could be found only with those 3 mesures.\r\n\r\nbut find the equation of that line is a litle bit trouble.\r\n\r\nwith 2 random poinnts on the 2 circles we can make a line. so. we can make this line tangent to the 2 circles with some calculations.\r\n\r\nor we can use the fact that the center of the circles and the tangential point have the a comon tangent.\r\n\r\nor others things. using vectors...\r\n\r\nbut all solutions i made have a lot of calculations. i will find a simple one", "Solution_4": "I couldn't find a result. :maybe:", "Solution_5": "What program do you use to make pictures?", "Solution_6": "http://www.geometryexpressions.com/downloads.php?PHPSESSID=14909a289ff7b8348e864ce42e14bda4\r\n\r\nGeometric Expressions\r\n\r\nyou can try a free trial copy of this. :)", "Solution_7": "mine is snagit 7.", "Solution_8": "We can not find $x$.Let be $NA=y$.\r\n$A,N,M$ are collinear.Since $\\triangle CNA \\sim \\triangle MBA$, $\\frac{CA}{MA}=\\frac{NA}{BA}$.Then, $\\frac{8}{6+y}=\\frac{y}{x}$.$y$ must be given." } { "Tag": [ "inequalities", "geometry unsolved", "geometry" ], "Problem": "Let be given four circles $(O_1,R_1),(O_2,R_2),(O_3,R_3),(O,R)$ satisfying:\r\n$(O_1,R_1),(O_2,R_2)$ tangent externally.\r\n$(O_1,R_1),(O_3,R_3)$ tangent externally.\r\n$(O_3,R_3),(O_2,R_2)$ tangent externally.\r\n$(O,R)$ tangents to the three circles $(O_1),(O_2),(O_3)$ all internally.\r\nProve that:\r\n$R_1+R_2+R_3\\le 3(2\\sqrt{3}-3)R$", "Solution_1": "Yes i solved it \r\nWe have $OO_i=R-R_i$ and $O_iO_j=R_i+R_j$\r\nSo taking $R_1=x,R_2=y,R_3=z$ and $A=x^2y+x^2z+y^2x+y^2z+z^2x+z^2y$ we have\r\n$S_Oo_1o_2+S_Oo_2o_3+S_Oo_3o_1=So_1o_2o_3$\r\n$\\sqrt{Rxy(R-x-y)}+\\sqrt{Ryz(R-y-z)}+\\sqrt{Rxz(R-x-z)}=\\sqrt{xyz(x+y+z)}$\r\nWe have $\\sqrt{3}\\sqrt{a+b+c}\\geq\\sqrt{a}+\\sqrt{b}+\\sqrt{c}$\r\nSo $\\sqrt{3R^2(xy+yz+zx)-3RA}\\geq\\sqrt{xyz(x+y+z)}$\r\n$3R^2(xy+yz+zx)-3RA-xyz(x+y+z)\\geq0$\r\n$(R-\\frac{3A-\\sqrt{D}}{6(xy+yz+zx)})(R-\\frac{3A+\\sqrt{D}}{6(xy+yz+zx)})\\geq0$\r\nWhere $D=9A^2+12Axyz+36x^2y^2z^2$\r\nSo $R\\geq\\frac{3A+\\sqrt{D}}{6(xy+yz+zx)}$\r\nWe must prove that $\\frac{3A+\\sqrt{D}}{6(xy+yz+zx)}\\geq\\frac{(2\\sqrt{3}+3)}{9}(x+y+z)$\r\n$9A+3\\sqrt{D}\\geq2(2\\sqrt{3}+3)(A+3xyz)$\r\n$3\\sqrt{D}\\geq(4\\sqrt{3}-3)A+6(2\\sqrt{3}+3)xyz$\r\n$9(9A^2+12Axyz+36x^2y^2z^2)\\geq(57-24\\sqrt{3})A^2+12(15+6\\sqrt{3})Axyz+36(21+12\\sqrt{3})x^2y^2z^2$\r\n$24(1+\\sqrt{3})A^2\\geq72(1+\\sqrt{3})Axyz+432(1+\\sqrt{3})x^2y^2z^2$\r\n$A^2-3xyzA-18x^2y^2z^2\\geq0$\r\n$(A-6xyz)(A+3xyz)\\geq0$\r\nBy AM GM we have $A\\geq6xyz$\r\nEquality when $R_1=R_2=R_3=(2\\sqrt{3}-3)R$", "Solution_2": "You can use this well-known formula :\r\n\r\n$\\boxed {\\ R=\\frac{R_1R_2R_3}{2\\sqrt{R_1R_2R_3(R_1+R_2+R_3)}-(R_1R_2+R_2R_3+R_3R_1)}\\ }\\ .$" } { "Tag": [], "Problem": "Kris has an endless supply of 4-cent stamps and 7-cent stamps for mailing a package. What is the greatest amount of postage that cannot be affixed to a package using these stamps?", "Solution_1": "How do you do this? Why isn't this 9?", "Solution_2": "$ 4 \\times 7 \\minus{} 4 \\minus{} 7 \\equal{} 28 \\minus{} 11 \\equal{} \\boxed{17}$.", "Solution_3": "Is this the Chicken McNugget Theorem? Does it at all relate to Simon's Factoring Trick? Thanks", "Solution_4": "you can immediately rule out all multiples of 4 and 7\r\nso 4, 8, 12.... can't be it \r\nneither can 7, 14, 21.....\r\nbut if you look closely 7 is 1 less than 8 or 3 more than 4\r\n14 is 2 less than 16 or 2 more than 12\r\nand 21 is 3 less than 24 or 1 more than 20\r\nso after a certain amount no number will be unable to be synthesized by this combination\r\nKouichi Nakagawa showed us the theorem to solve this problem which is posted above but i will show it also below\r\n\r\n4 X 7 - 4 - 7 = 17\r\n\r\nsince 4 X 7 gives you four 7s or seven 4s, by taking away one of each so that it can neither be 4 X 6 or 3 X 7 we come out with the largest number unable to be synthesized by any number of 4s or 7s\r\n\r\n17!", "Solution_5": "[quote=\"pinkmuskrat\"]Is this the Chicken McNugget Theorem? Does it at all relate to Simon's Factoring Trick? Thanks[/quote]\r\nI have solved the problem that resembled this problem in a book of Japanese mathematics.\r\nChicken McNugget Theorem was introduced to the book as a theorem, but the name of the theorem was not on.\r\nI found out the name of the theorem thanks to pinkmuskrat.", "Solution_6": "Here is a simple explanation. It lacks elegance, but it helps organize one's view and is easily understood, completed, & remembered.\n\n1 -- [b]NOT possible[/b] postage using a 4-cent or 7-cent stamp\n2 -- [b]NOT possible[/b] \n3 -- [b]NOT possible[/b]\n4 -- Possible\n5 -- [b]NOT possible[/b]\n6 -- [b]NOT possible[/b]\n7 -- Possible\n8 -- Possible -- [i]Now we start asking the question; can we get this amount by adding 4 or 7 to a previous possibility?[/i]\n9 -- [b]NOT possible[/b] -- Count back 4 or 7; both impossible postage amounts\n10 -- [b]NOT possible[/b] -- Count back 4 or 7; both impossible amounts\n11 -- Possible -- Count back 4; since 7 is possible then 11 is possible\n12 -- Possible -- Count back 4; since 8 is possible then 12 is possible\n13 -- [b]NOT possible[/b] -- Count back 4 or 7; both impossible\n14 -- Possible\n15 -- Possible\n16 -- Possible\n17 -- [b]NOT possible[/b] -- Count back 4 or 7; both impossible\n18 -- Possible\n19 -- Possible\n20 -- Possible\n21 -- Possible\n[i]We need not check further. 17 is the greatest amount that our stamps cannot produce.\nAdding combinations of 4-cent stamps to the previous four possibilities will produce all integer values 22 or greater.[/i]" } { "Tag": [ "inequalities", "trigonometry", "inequalities unsolved" ], "Problem": "ABC is acute triangle. Show that\r\n$ 10(h_a+h_b+h_c)+36R \\leq 9\\sqrt{3}(a+b+c) $", "Solution_1": "I do not believe this inequality is true. Can anyone find example to disprove this one ? Maybe A-> 90 and B=C -> 45", "Solution_2": "Hi everyone!\r\n\r\nDividing inequlity to $2R$ we obtain following inequality:\r\n$10(\\sin x\\sin y+\\sin y\\sin z+\\sin z\\sin x)+18\\leq 9\\sqrt{3}(\\sin x+\\sin y+\\sin z)$, where $x,y,z$ are angles of the triangle.\r\n\r\nLet $(x,y,z)$ be optimal set for this inequality, i.e. RHS-LHS=min. Suppose $x>y$. Consider new triangle correcponding to set $(x-t,y+t,z)$. Then it is necessery $(RHS-LHS)'_{t=0}\\geq 0$.\r\nIt easily follows that $10\\sin(y-x)+(\\cos y-\\cos x)(9\\sqrt{3}-10\\sin z)\\geq 0$ $\\Leftrightarrow$ $10\\cos\\frac{x-y}{2}-\\sin\\frac{x+y}{2}(9\\sqrt{3}-10\\sin z)\\leq 0$.\r\nSince triangle is acute it implies that $x-y2\\arctan\\frac{10}{9\\sqrt{3}}$ then $\\sin z>\\frac{180\\sqrt{3}}{343}$ $\\Rightarrow$ $9\\sqrt{3}-10\\sin z<9$ and this implies $10\\cos(z/2)>9\\sin(z/2)>\\sin(z/2)(9\\sqrt{3}-10\\sin z)$, since $\\cos z\\geq 1/\\sqrt{2}\\geq \\sin(z/2)$.\r\nContradiction.\r\n\r\nHence in optimal set $(x,y,z)$ we have $x=y=z=\\pi/3$.", "Solution_3": "Thank you!" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "AMC 10" ], "Problem": "I am a 7th grader that did not qualify for USAMO. my scores are AMC 10B:123, AIME:5. what should be my goal for next year? i do plan on qualifying for USAMO. ignore whats in my sig.", "Solution_1": "what ever you think you can get.\r\n\r\naim for an IMO Gold Medal. :P", "Solution_2": "i will add that to my sig", "Solution_3": "You must get IMO gold in 8th grade or you are NOTHING.\r\n\r\nNo srsly, getting to the USAMO in 8th grade sounds like a good enough goal. I wouldn't expect to get more than one problem right on it (or really any at all) on your first try. But if you study enough then by the next year you should be good enough to at least make Red MOP.", "Solution_4": "I would suggest aiming to max out the AMC10 (150) and aiming for an 8+ on the AIME (whichever you feel more comfortable with).\r\n\r\nMost likely, I'd suggest increasing your goal to 11 next, and then 13, 15. Of course, you might work to it much faster :wink:" } { "Tag": [ "geometry", "rectangle", "symmetry" ], "Problem": "A Tetris piece is one of the following seven shapes consisting of four unit squares:\r\n\r\n[img]http://www.touchgenerations.com/_media/enGB/content/tetris_ds/tetris_blocks_all.gif[/img]\r\n\r\nTwo Tetris pieces are said to be identical if and only if the shape of one piece can be achieved by rotating (but not reflecting) the other. For example, the purple piece and the yellow piece are considered distinct.\r\n\r\n\r\nDefine a region $ R$ made of unit squares to be \"large\" if and only if for all Tetris pieces $ T$, two $ T$s can fit inside $ R$ without overlapping. For example, a 4x4 square is \"large.\" This can be achieved by rotation.\r\n\r\n\r\nThe question:\r\nCompute the smallest possible area of a \"large\" region.", "Solution_1": "[hide=\"Solution\"]Assume the height is at least the width\n\nThe height is at least 4, since the maximum maximum size (in any dimension) of a block is 4 (the stick), and the width is at least two, since the maximum minimum size (in any dimension) of a block is 2 (for example, the L).\n\nHence, the area is at least 8. However, we cannot fit two zigzags or two T's in an area of 8.\n\nThe next value up is 9, but this must be either 9x1, which fails W >= 2, or 3x3, which fails H >= 4, so this is not valid.\n\nThe next value up is 10, and indeed a 5x2 rectangle works:\n[code]\noo oo ox xo +x oo oo\nx+ ++ xx xx xx ++ ++\nx+ ++ x+ +x +x +x x+\nx+ xx ++ ++ ++ +x x+\nx+ xx +o o+ +o xx xx\n[/code]\n\nand we are done.[/hide]", "Solution_2": "To the above post: I don't believe the region must be rectangular. That doesn't change the answer, but it means you have to strengthen the proof.\r\n\r\n[hide]Area $ 10$ works: $ 2 \\times 5$ rectangle. We will now show a region of area $ 9$ is not possible.\n\nFirst of wall, we need to be able to fit two long pieces. Observe that if their fit has them perpendicular, using $ 8$ separate squares, then no square pieces can fit. Since we are using an area of at most $ 9$, we only can add one additional square, so we can only fit one square piece at most.\n\nSo the long piece's fit has to be parallel, and one of the following up to symmetry:\n\n[code]OOOO _ .OOOO _ ..OOOO _ ...OOOO\nOOOO _ OOOO. _ OOOO.. _ OOOO...\n1... _ 2.... _ 3..... _ 4......[/code]\nIn fits $ 3$ and $ 4$, adding one square can't allow a fit of two square pieces, so only $ 1$ and $ 2$ are possible.\n\nIn case $ 2$, consider the following diagram:\n\n[code].1234\n5OOOO6\nOOOO.[/code]\nWe can add the 9th square to any of the 6 designated spaces, and this is exhaustive by symmetry. But no matter which space we add it to, we can't fit two green $ S$ pieces.\n\nIn case $ 1$, we have the following:\n\n[code].12\n3OOOO\n.OOOO[/code]\n\nWe have 3 choices here. But we can't fit two blue $ S$ pieces no matter where it is added.\n\nTherefore area $ 9$ is not possible. The following shows how to fit each piece into a $ 2 \\times 5$.\n\n[code]1111O 1112O 1122O 1112O\n2222O O1222 1122O 1222O\n\n1222O O1122 1122O\n1112O 1122O O1122[/code][/hide]", "Solution_3": "[quote=\"MellowMelon\"]I don't believe the region must be rectangular.[/quote]\r\nI did not specify that the region had to be rectangular. Nevertheless, the answer is the same for each of them, as MellowMelon pointed out." } { "Tag": [], "Problem": "Find all positive integers n and x such that \r\n$5^n- 5^4 + 5^5 = x^2.$\r\n\r\nEasy to be on a national olympiad. :lol:", "Solution_1": "[hide]\nChecking $n = 1,2,3,4$ we find no solutions. So assume $n > 4$.\n\nThen $5^n-5^4+5^5 = 5^4(5^{n-4}+4) = x^2$. Let $x = 25y$.\n\nSo $5^{n-4}+4 = y^2 \\Rightarrow 5^{n-4} = (y-2)(y+2)$ but only one of $y-2$ and $y+2$ can be divisible by $5$ so $n-4 = 1$. Checking, we find the only solution is $n = 5$, $x = 75$.[/hide]", "Solution_2": "Nice solution, paladin. :) \r\n[hide]The easiest way, in my opinion, to do this would be to rewrite as\n$5^n = x^2 - 2500$\n$5^n = (x- 50)(x+50)$\nThe factors on the LHS differ by $100$. The only powers of 5 that differ by $100$ are $25$ and $125$\nSo $5^n = 25 * 125$\n$n = 5, x = 75$[/hide]" } { "Tag": [ "videos" ], "Problem": "Post all the tips and tricks you have collected about runescape!\r\n\r\nAlright,\r\nI like to \r\n1) mine copper and tin (it goes really fast) over and over and then sell all of it at the grand exachange or the general store. You can make bout 80 gp each.\r\n2)Also, cowhide is really great. Get a LOT of it. Like a 100. Then, if you sell it you'll probably get like 40 gp each, but if you go to al kharid and get it tanned you can get about 90 gp each. \r\n3)Raw meat. I prefer beef (it really easy to get, just go to the cow fields and pick it up, you don't even have to kill anything) or chicken (same as beef, just go to the chicken coops).", "Solution_1": "Woodcutting. Yews bring you at least 300 gp each (I haven't logged in in over a year so who knows what the rate is now). \r\n\r\n(In any case, RuneScape is ridiculous; the only useful application is to sell the accounts you have for real cash).", "Solution_2": "Yeah, I have a level 95 account, but I gave it to a friend. I like to log in once in a while to see how things are going.", "Solution_3": "Having a Runescape account automatically entitles your grades to a 10% discount, with an additional 5% for every hour you play per day.\r\n\r\nThen again, I do have a level 120 something that I don't use anymore... The majority of the RS community is simply ridiculous, so I quit. I recommend that you stop before you become like those poor people in Asia dying from too much gaming.", "Solution_4": "[quote=\"n0vad3m0n\"]Woodcutting. Yews bring you at least 300 gp each (I haven't logged in in over a year so who knows what the rate is now). \n\n(In any case, RuneScape is ridiculous; the only useful application is to sell the accounts you have for real cash).[/quote]\r\n\r\n\r\nIs this your new account Henry?", "Solution_5": "Isn't Runescape failing? I think that WoW is kind of pwning it.", "Solution_6": "[quote=\"jonathanchou711\"]Isn't Runescape failing? I think that WoW is kind of pwning it.[/quote]\r\n\r\nI think all MMOs are detrimental to the betterment of mankind.\r\n\r\n(Actually, never mind, people addicted to them just get wiped from the gene pool anyways.)", "Solution_7": "[quote=\"Aristocrat\"]I think all MMOs are detrimental to the betterment of mankind.\n\n(Actually, never mind, people addicted to them just get wiped from the gene pool anyways.)[/quote]\r\n\r\nHahaha. Touche. First person shooters always win. At least there are the fan girls. \r\n\r\nAnd Poincare, I just got my username changed, that's all. \r\n\r\nRuneScape's community is full of whiny 12 year olds who think they're cool, leaving Jagex to force the over-strict policies (language filters, trade restrictions, etc.) - which is why I quit.", "Solution_8": "finally started playing Runescape again after taking a 2 year break...great game\r\n_________________\r\n[url=http://gamefriends.com/game_videos][color=red]Game Videos[/color][/url], [url=http://gamefriends.com/ask][color=blue]Ask a Gamer[/color][/url], [url=http://gamefriends.com/game_news][color=purple]Game News[/color][/url]", "Solution_9": "In my opinion, Runescape is not as good as MapleStory. And that says something, I think.", "Solution_10": "[quote=\"CircleSquared\"]In my opinion, Runescape is not as good as MapleStory. And that says something, I think.[/quote]\r\n\r\nMaplestory boss runs at high levels have a countdown timer telling you how long you have to kill the boss.\r\n\r\nIt is measured in days.\r\n\r\nThis tells you how unhealthy the game is.", "Solution_11": "[quote=\"Aristocrat\"][quote=\"CircleSquared\"]In my opinion, Runescape is not as good as MapleStory. And that says something, I think.[/quote]\n\nMaplestory boss runs at high levels have a countdown timer telling you how long you have to kill the boss.\n\nIt is measured in days.\n\nThis tells you how unhealthy the game is.[/quote]\r\n\r\nIt is NOT measured in days.", "Solution_12": "[quote=\"CircleSquared\"][quote=\"Aristocrat\"][quote=\"CircleSquared\"]In my opinion, Runescape is not as good as MapleStory. And that says something, I think.[/quote]\n\nMaplestory boss runs at high levels have a countdown timer telling you how long you have to kill the boss.\n\nIt is measured in days.\n\nThis tells you how unhealthy the game is.[/quote]\n\nIt is NOT measured in days.[/quote]\r\n\r\nHOURS, not days", "Solution_13": "...\r\n\r\nYou are given 24 hours to kill Horntail; granted, no one uses that much, but the fact is that you are given 24 hours... of consecutive play time. What does beating him get you? A necklace that upgrades your character. And you need to beat it four times to get a full necklace. That means staying logged in for a total of more than 96 hours (Since apparently you need to stay logged in for the whole duration to get the reward).\r\n\r\nThis cannot be casual gaming.", "Solution_14": "Thank you.", "Solution_15": "Meh. I used to play. Not anymore, gave the account to a friend. Still know the password though.", "Solution_16": "Thank you.", "Solution_17": "[quote=\"Aristocrat\"]...\n\nYou are given 24 hours to kill Horntail; granted, no one uses that much, but the fact is that you are given 24 hours... of consecutive play time. What does beating him get you? A necklace that upgrades your character. And you need to beat it four times to get a full necklace. That means staying logged in for a total of more than 96 hours (Since apparently you need to stay logged in for the whole duration to get the reward).\n\nThis cannot be casual gaming.[/quote]\r\n\r\npeople can log out- as long as there are at least 2 people you can still get the award" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all increasing functions $ f: \\mathbb R^{\\plus{}} \\rightarrow \\mathbb R$ such that $ f(x\\plus{}1)\\equal{}f(x)\\plus{}2^{\\minus{}x}$ for every $ x\\in \\mathbb R^{\\plus{}}$. ($ \\mathbb R$ is the set of real numbers, $ \\mathbb R^{\\plus{}}$ is the set of positive real numbers)", "Solution_1": "We have $ (f(x\\plus{}1)\\plus{}2^{\\minus{}x})\\equal{}(f(x)\\plus{}2^{\\minus{}(x\\minus{}1)})$, so $ g(x) \\equal{} f(x)\\plus{}2^{\\minus{}(x\\minus{}1)}$ is periodic, of period $ 1$. Assume $ g(x)$ is not constant; then there exist $ x_1 < x_2$ with $ g(x_1) > g(x_2)$, so $ g(x_1 \\plus{} n) > g(x_2 \\plus{} n)$ for all positive integers $ n$. But then $ f(x_2 \\plus{} n) \\minus{} f(x_1 \\plus{} n) \\equal{} \\minus{}(g(x_1\\plus{}n) \\minus{} g(x_2\\plus{}n)) \\plus{} 2^{\\minus{}(n\\minus{}1)}(2^{\\minus{}x_1} \\minus{} 2^{\\minus{}x_2}) < 0$ for large enough $ n$, and that contradicts $ f$ being increasing. So $ g(x) \\equal{} c$ is constant, and $ f(x) \\equal{} c \\minus{} 2^{\\minus{}(x\\minus{}1)}$." } { "Tag": [], "Problem": "If $p$ is a prime number, prove that $p^2+26$ is always composite.", "Solution_1": "hint:[hide]use mods[/hide]", "Solution_2": "Thanks jli! Now I'll leave it for other people :P", "Solution_3": "[hide]\nprimes in mod 4 are 1, 2, or 3.\n\n1^2+2=0(mod4)\n2^2+2=2(mod4)\n3^2+2=2(mod3)\n\nI am not sure how to do that last one(case)...\n[/hide]", "Solution_4": "[hide=\"try\"]mod 6 ;) [/hide]", "Solution_5": "yes...mod 6 might work\r\n\r\ntry mod 9, many problems are ably to solve in mod 9\r\n\r\n$p^2+26\\bmod 9=p^2-1\\bmod 9 = (p-1)(p+1)\\bmod 9$", "Solution_6": "[hide]\n\nLetting $p=2,3$ we can see that it is not a prime . For $p>3$ , we know that $p^2\\equiv 1\\mod 3$ . And this follows immediately that $p^2+26\\equiv 27\\equiv 0 \\mod 3$ . [/hide]" } { "Tag": [ "function", "algorithm", "algebra unsolved", "algebra" ], "Problem": "Find all natural number k and real number a such that there exists a function f continuous from R to R satisfying\r\n$f_k(x) = a.x +1$ $\\forall x \\in R$\r\n\r\nwhere $f_0(x) = x, f_{n+1}(x) = f_n(x)$ $\\forall n \\in N$\r\n-------------------------\r\nSorry, there's a mistake in my problems.\r\n\r\nFind all natural number k and real number a such that there exists a function f continuous from R to R satisfying\r\n$f_k(x) = a.x +1$ $\\forall x \\in R$\r\n\r\nwhere $f_0(x) = x, f_{n+1}(x) = f_n(x)$ $\\forall n \\in N$\r\nFor which k and a, find all function f", "Solution_1": "for a=0 take f(x)=1.\r\nfor k=1, take f(x)=ax+1. \r\nNow suppose that a!=1 and that k>1. \r\nclearly f(x) is one to one and since it is continous it should be either increasing \r\nor decreasing. \r\nlet g(x) denote inverse of f(x).\r\nsuppose k=2. \r\nsince f(x) is monotonic, we have that f(f(x))>0 for x>N for some N.\r\nhence a>0.\r\nclearly for a>0 we can take f(x)= (:sqrt: a)x+ b, where b=1/(1+ :sqrt: a)\r\nNow suppose that fk(x)=ax+1 where k is odd. \r\nthen take v^m = a and (v^m-1+...+v+1)u = 1\r\nthen f(x)=vx+u satisfies the relation. \r\nby the same argument a function exists for even values of k iff a>0. \r\nif a>=0 any k works. if a<0 only odd values of k.", "Solution_2": "As stated, the problem is rather simple. But I see that the subject is \" Find all functions\" . I think this is a very hard task. Do you happen to have a complete solution for this problem? I would be very interested. I tried several times to solve it, but I failed each time.", "Solution_3": "I think that I have a complete solution in the case of k=2, which is as follows:\r\nI give the general solution of functions f:[a1,a2] -> R such that \r\nf(f(x))=ax+1 and that f(x) is continous.\r\n \r\nclearly f(x) is monotonic. \r\ncase 1. f(x) is increasing. we know that a>0 . clearly since f(x) is increasing:\r\nf(f(x))>x iff f(x)>x. \r\nso f(x)>x iff (a-1)x+1>0 so we can divide [a1,a2] into two subintervals, such that \r\nin one of them f(u)>u and in the other f(u)x for all x in [a1,a2].\r\nNow there is a \\beta >0 such that if x,y in [a1,a2] then abs(x-y) <= \\beta \r\nimplies that abs(f(x)-f(y))< \\varepsilon for some \\varepsilon >0. \r\nalso there is an M>0 such that f(x)-x>m. \r\ntake d to be min( \\beta , abs( \\varepsilon - m) ).\r\npartition the set [a1,a2] to [a1,a1+d,a1+2d,...,a1+td,a2] where abs(a2-a1-td)=abs(m- \\varepsilon )>d.\r\nThis means that we can arbitrarily assign a continous and increasing function g(x) to the interval [a1,a1+d] the only constraints are:\r\ni) g should be increasing. \r\nii) g(x)>x. \r\niii) g(a1+d)<= f(f(a1+d))=a(a1+d)+1. \r\nafter assigning g(x) to [a1,a1+d] the interval [f(a1),f(a1+d)] can be uniquely determined. (these two are disjoint) Now we will follow the same algorithm for the remaining segments [a1+d,a1+2d],... .\r\nthis does the problem if f(x) be increasing. \r\n\r\ncase ii) f(x) decreasing. consider the same interval [a1,a2] the equation f(x)=x has a finite number of solutions in interval [a1,a2].( I'm having problem with this statement!) isolate each of these special points and construct the function as before. at the end you can take care of these finite number of points.", "Solution_4": "I forgot to handle f(x)=1 as a seperate case. \r\nin this case the continouty of f(x) yields that f(x)=1.\r\nI think that the method presented in my last post can be generelized to all values of k.\r\nAlso there might be a problem with the argument of extending [a1,a2] to the whole R. \r\nFor this I present the following argument: \r\nsuppose f(x) is increasing. then a>0 implies that f(x)-x is unbounded in whole R.\r\nwe can infact choose T such that f(x)-x>M for all x>T and that enables us to extend [a1,a2] to the whole R.", "Solution_5": "I think this Question doesn't have a complete solution", "Solution_6": "Should we consider it as solved or not?" } { "Tag": [ "graph theory", "combinatorics unsolved", "combinatorics" ], "Problem": "An $n\\times n$ is fill with 0 and 1 so that if we chose randomly n cell(that is no two of them is in the same row or column) then there atre at least one contain 1.Prove that we can find i row and j column so that $i+j\\geq\\ n+1$ and their intersection contain only 1", "Solution_1": "This is just another way of looking at the marriage theorem:\r\n\r\nConsider a bipartite graph with bipartition $A,B,\\ |A|=|B|=n$ encoded by this $n\\times n$ table as follows: the rows represent the vertices in $A$, the columns represent the vertices in $B$, and $i\\in A,j\\in B$ are connected iff the entry in the $i$'th row and $j$'th column is $0$. The hypothesis simply says that this graph has no perfect matching, so, by the marriage theorem, we can find some $i$ vertices in $A$ which have $n-i$ columns whose intersections with the $i$ rows are all $1$'s. This is exactly what we wanted to prove." } { "Tag": [ "search", "\\/closed" ], "Problem": "http://www.mathlinks.ro/Forum/memberlist.php\r\n\r\nWhy somebody with more than ten million posts? :D", "Solution_1": "I thought V.V. fixed that. :huh:", "Solution_2": "that was for a different user. now there's two people with over 16M posts.", "Solution_3": "If you try to find all their posts, it says \"nothing matches your search criteria.\" Apparently, they have 0 posts but for some reason it shows 16777215 and 16777214.", "Solution_4": "[b]theccy[/b] just lost 2 posts. :?:", "Solution_5": "Did these people like randomly hack into the system? :huh:", "Solution_6": "They might have somehow got into the admin panel... :huh: :|", "Solution_7": "Hm, I really wonder how they did this :?: It seems like they somehow hacked the database.", "Solution_8": "Actually, I'm thinking that they might not have hacked the database. Notice that the number $16777215 = 2^{24}-1$, so it looks more like some weird computer error, since computers generally deal with these type of numbers.", "Solution_9": "Good point, it's still kinda odd :P", "Solution_10": "[quote=\"nebula42\"]Actually, I'm thinking that they might not have hacked the database. Notice that the number $16777215 = 2^{24}-1$, so it looks more like some weird computer error, since computers generally deal with these type of numbers.[/quote]\r\n\r\nNotice that theccy and hoavokhuyet have the title Moderator. :wink:\r\n\r\nJust an offhand question, but is theccy and ccy the same person?", "Solution_11": "[quote=\"nebula42\"]Actually, I'm thinking that they might not have hacked the database. Notice that the number $16777215 = 2^{24}-1$, so it looks more like some weird computer error, since computers generally deal with these type of numbers.[/quote]\r\nOkay, how the heck did you figure that out?", "Solution_12": "[quote=\"joml88\"]Hm, I really wonder how they did this :?: It seems like they somehow hacked the database.[/quote]It's amazing where a small \"-\" can lead people to think :) Problem (finally) fixed. It was no hack, just an extra minus that made it so that each time a post was deleted a person's post count drops by 2 (instead of 1).", "Solution_13": "Now what's the deal with micheal?", "Solution_14": "[quote=\"13375P34K43V312\"]Now what's the deal with micheal?[/quote] :bomb: :wallbash_red: There must be another \"-\" somewhere it should be.", "Solution_15": "Three things I noticed:\r\n\r\n1) Micheal has 16M+ posts.\r\n2) Everybody lost a post (V.V. had 5080, now 5079).\r\n3) Post count isn't going up anymore.\r\n\r\nMaybe the \"-\" wasn't the real problem? :ninja:", "Solution_16": "[quote=\"i_like_pie\"]Maybe the \"-\" wasn't the real problem? :ninja:[/quote]It was. I know where the problem is and I will fix it, it's just that it implies a big update of the site which I have been preparing for a couple of days now (and it will be ready soon).\r\n\r\nDon't worry it takes only a couple of seconds to resync everyone's post count after the update is done.", "Solution_17": "[quote=\"i_like_pie\"]Three things I noticed:\n\n1) Micheal has 16M+ posts.\n2) Everybody lost a post (V.V. had 5080, now 5079).\n3) Post count isn't going up anymore.\n\nMaybe the \"-\" wasn't the real problem? :ninja:[/quote]\r\n\r\nOhh, that's why people have 0 posts now. Will our post counts get fixed once Valentin fixes the problem?", "Solution_18": "[quote=\"13375P34K43V312\"]Will our post counts get fixed once Valentin fixes the problem?[/quote]\n[quote=\"Valentin Vornicu\"]Don't worry it takes only a couple of seconds to resync everyone's post count after the update is done.[/quote]\r\nAlready answered. :)", "Solution_19": "yeah I've posted like 20 times today. I just noticed that it didn't go up at all. But posting still is fun. Is there some computer guru just trying to get attention on this website by doing this? \r\n\r\nCan't wait till this site is updated. It would probably look a lot better , new and improved :lol:", "Solution_20": "You should have seen the format they had in like april, it's much better now.", "Solution_21": "And now we have sneezy walrus...", "Solution_22": "They also have moderator status, something that should be fixed...", "Solution_23": "[quote=\"bpms\"][quote=\"nebula42\"]Actually, I'm thinking that they might not have hacked the database. Notice that the number $16777215 = 2^{24}-1$, so it looks more like some weird computer error, since computers generally deal with these type of numbers.[/quote]\nOkay, how the heck did you figure that out?[/quote]\r\n\r\nIts not that hard, I can do that in my head. Btw I used to do exponents of 2 in my head after I finished my math tests (2,4,8,16,32,...) before evry1 else :lol:", "Solution_24": "lol update soon please lol :D", "Solution_25": "[quote=\"Ihatepie\"][quote=\"bpms\"][quote=\"nebula42\"]Actually, I'm thinking that they might not have hacked the database. Notice that the number $16777215 = 2^{24}-1$, so it looks more like some weird computer error, since computers generally deal with these type of numbers.[/quote]\nOkay, how the heck did you figure that out?[/quote]\n\nIts not that hard, I can do that in my head. Btw I used to do exponents of 2 in my head after I finished my math tests (2,4,8,16,32,...) before evry1 else :lol:[/quote]\r\nOkay, but did you have it memorized up to 24?", "Solution_26": "Dude... Way of topic. :|", "Solution_27": "Well, everything important has been said here and the five or so other threads with that topic...\r\nThus such a small question isn't really offtopic.\r\n\r\nBut I find it very interesting how many people posted, especially those whose post count doesn't rise anymore :D", "Solution_28": "[quote=\"i_like_pie\"]Dude... Way of topic. :|[/quote]\r\nHave you read this thread? It isn't offtopic.\r\nTo ZetaX no one's post count is going up, read the thread.", "Solution_29": "[quote=\"bpms\"][quote=\"nebula42\"]Actually, I'm thinking that they might not have hacked the database. Notice that the number $16777215 = 2^{24}-1$, so it looks more like some weird computer error, since computers generally deal with these type of numbers.[/quote]\nOkay, how the heck did you figure that out?[/quote]\r\n\r\nIt isn't that hard to figure out...\r\n\r\nAnyways, now I don't think anyone's post count is increasing...since I remember my post count being 2517, and I definitely posted and it is still 2517.", "Solution_30": "same, ive posted like a lot since it stopped going up. My post count should be 330+.", "Solution_31": "[quote=\"Valentin Vornicu\"]Don't worry it takes only a couple of seconds to resync everyone's post count after the update is done.[/quote]\r\n\r\nI think you should read this. :wink:", "Solution_32": "[quote=\"bpms\"][quote=\"i_like_pie\"]Dude... Way of topic. :|[/quote]\nHave you read this thread? It isn't offtopic.\nTo ZetaX no one's post count is going up, read the thread.[/quote]\r\n :| \r\nRead my post again. I did not ask why etc., I just made a remark about how many people complain about non-increasing post count (since these people possibly post for the counter, not for the content).", "Solution_33": "lol before my post was 84...now it's 78???", "Solution_34": "I can make it 0 if you don't like 79 :) \r\nRemember post count is [b]purely informational[/b]. You should not post in order to increase your post count.", "Solution_35": "I concur...\r\n\r\nBut yes, thanks for the fixing of the problem...", "Solution_36": "yay, last night my post count said 304. :lol:" } { "Tag": [ "AMC", "AIME", "\\/closed" ], "Problem": "I don't know if this has already been addressed but:\r\n\r\nHow often do the classes meet and at what times, usually?\r\n\r\nWould it be difficult/beneficial to take Intermediate algebra and AIME problem series at the same time?\r\n\r\nAbout which parts of AoPS V2 are covered in Intermediate algebra? (because some chapters I don't understand that well)\r\n\r\nThanks!", "Solution_1": "All of our summer classes meet once a week at 7:30 Eastern / 4:30 Pacific and last 90 minutes.\r\n\r\nIt is perfectly reasonable to take both the Intermediate Algebra and the AIME Problem Series class simultaneously. Many of our students take more than one class at a time.\r\n\r\nAs far as which chapters in AoPS Vol 2 are covered in the Intermediate Algebra class: I'm not really sure how to answer that, since the online classes don't directly follow the AoPS books. I think that most of the material in the Intermediate Algebra class is at a level beyond the AoPS books. MCrawford (who is the instructor for both of the classes that you mentioned) might be able to give you more detailed information -- he's out of the office today though.\r\n\r\nYou can get a pretty good idea about what's in the class by looking at the \"Are You Ready?\" and \"Do You Need This?\" problems listed for each class on the classes page at [url]http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php[/url].", "Solution_2": "From what I've seen you done, I would say you're more than well-prepared for Intermediate Algebra. In fact, you might want to make sure you can't do all of the problems first. ;)\r\n\r\nI took three classes (including AIME Problem Series) at the same time, so you'll definitely be fine with taking those two over the summer.", "Solution_3": "Ok, thanks for the information. I think I am going to take both of them, and it's summer too so I'll have time. Thanks, towersfreak, but I can't do all the problems. ;)\r\n\r\nTowersfreak2006, did you enjoy the intermediate algebra class/did you find it useful?\r\n\r\nThanks again.", "Solution_4": "I haven't taken it yet, and I don't know that I will this upcoming semester due to other commitments.\r\nBut from what I've heard, it's one of the easier intermediate classes.", "Solution_5": "[quote=\"towersfreak2006\"]But from what I've heard, it's one of the easier intermediate classes.[/quote]\r\n\r\nThat's because you own algebra. -.-\r\n\r\nDaniel, Intermediate Algebra is a great class to take if you're trying to brush up your algebra skills and cover any holes that may have been neglected from previous classes/books/etc.", "Solution_6": "[quote=\"towersfreak2006\"]\nBut from what I've heard, it's one of the easier intermediate classes.[/quote]\r\n\r\nWe generally recommend this be the first of the Intermediate classes a student takes, though students who are very proficient with algebra can feel comfortable taking one of the other intermediate classes first." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "number theory", "greatest common divisor", "number theory unsolved" ], "Problem": "I got a solution to USAMO 2004 #2 but nobody else I saw had it so I was wondering if somebody could check it for me. Thanks so much. \r\n\r\nSuppose a1...an are integers whose greatest common divisor is 1. Let S be a set of integers with the following properties:\r\n\r\n(a) For i=1...n , ai is in S.\r\n(b) For i,j = 1...n (not necessarily distinct), ai - aj is in S.\r\n(c) For any integers x and y, if x + y is in S, then x - y is in S. (Edit: x and y must be IN S)\r\n\r\nProve that S must be equal to the set of all integers.\r\n\r\n\r\n[hide=\"Solution\"]Noticing that 0 must be in the set, we find that if x = 0+x is in the set, so 0-x = -x is in the set. \n\nAlso, due to this, if x-y is in S then x+y is in S.\n\nAlso, we can easily prove that any linear combination of two of the ai's is in the set. \n\nNow, I shall prove that any linear combination of n variables with at least n-2 even coefficients is in the set. \n\nThe 2 case is true.\nFor the three case, suppose we want an arbitrary combination $ c_1a_1 \\plus{} c_2a_2 \\plus{} 2c_3a_3$ (this has at least one even). Since $ c_1a_1 \\plus{} c_3a_3$ is in the set and $ \\minus{} c_3a_3 \\minus{} c_2a_2$ is in the set, and their sum is in the set, so their difference, the desired combination is in the set.\n\nNow, inducting, we suppose we have a linear combination of k of them that has at least k-2 evens is in the set. \n$ c_1a_1 \\plus{} c_2a_2 \\plus{} 2c_3a_3 \\plus{} \\cdots \\plus{} 2c_ka_k \\in S$. \n\nNow, since $ \\minus{} c_2a_2 \\minus{} c_{k \\plus{} 1}a_{k \\plus{} 1}$ is in the set, and the sum of this an our linear combination of k is in the set, so any linear combination of k+1 of the ai with at least k-1 evens is in the set.\n\nMore generally, we have proved that any linear combination of n with at least n-2 evens is in the set.\n\nNow, we know that there is some combination $ d_1a_1 \\plus{} d_2a_2 \\plus{} ... \\plus{} d_na_n \\equal{} 1$ (because they have GCD of 1). \n\nHowever, in order to use our lemma, we need that at most two of the $ d_i$ are odd. \nWe know that at least one of the $ a_i$ is odd; if not, then the GCD is at least 2. Hence, suppose this is $ a_1$. Now, suppose $ d_j$ is odd, but we want to make it even.\n\n$ d_1a_1 \\plus{} d_ja_j \\equal{} (d_1 \\plus{} a_j)a_1 \\plus{} (d_j \\minus{} a_1)a_j$.\n\nUsing this shift, we have changed the parity of the coefficient of $ a_j$. It doesn't really matter what happens to the coefficient of $ a_1$. At the end, we can just have it be odd if the rest are even.\n\nThus, applying this to all the other odd coefficients in our linear combination, we have found a linear combination with at least n-2 evens that equals one. For all the rest of the integers, we can just multiply the combination by that integer and it wouldn't change the parities from even to odd in any of the coefficients. Thus, the set is all integers as desired. [/hide]", "Solution_1": "$ \\exists i\\{1,2,...,n\\}: \\ a_i\\ odd$\r\nwe have $ \\frac {a_i - 1}{2} + \\frac {a_i + 1}{2} = a_i\\in S$ so $ 1 = \\frac {a_i + 1}{2} - \\frac {a_i + 1}{2}\\in S$\r\nand $ \\forall k\\in\\mathbb{Z}: \\ (k + 1) + ( - k) = 1\\in S$ so $ 2k + 1 = (k + 1) - ( - k)\\in S$\r\n\r\nif $ \\forall i\\in\\{1,2,...,n\\}: \\ a_i \\ odd$ then $ a_1 - a_2 = even\\in S$\r\nif $ \\exists i\\in\\{1,2,...,n\\}: \\ a_i\\ even$ so $ a_i%Error. \"even\" is a bad command.\n\\in S$\r\n\r\nthen $ \\exists h\\in\\mathbb{Z}: \\ 2h\\in S$\r\nand we have $ h + h\\in S$ gives $ 0 = h - h\\in S$\r\nand $ \\forall k\\in\\mathbb{Z}: \\ (k) + ( - k) = 1\\in S$ so $ 2k= (k) - ( - k)\\in S$\r\n\r\nso $ S = \\mathbb{Z}$", "Solution_2": "I'm sorry I copied the problem a little bit wrong :( (but you have a nice solution to my mistaken problem).\r\n\r\nThe property (c) should say \"for any integers x and y [i]in S[/i]\" I will edit that. \r\n\r\nHere is the original LaTeX.\r\nhttp://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27&year=2004", "Solution_3": "I would say that it's clean. I don't really know how to score it, but I'd put it at least in the 5-7 range, rather than 0-2.\r\n\r\nMaybe be a little more specific on \"any linear combination\" of two $ a_i$'s is in S.", "Solution_4": "[quote=\"dasherm\"]Maybe be a little more specific on \"any linear combination\" of two $ a_i$'s is in S.[/quote]This is obvious given his first two statements ($ 0\\in S$, $ x\\in S$ implies $ \\minus{}x \\in S$ and $ x\\minus{}y \\in S$ implies $ x\\plus{}y \\in S$)." } { "Tag": [ "LaTeX" ], "Problem": "I want to add a foot note but normally they're numbered and I don't want to reader to confuse it as an exponent.\r\n\r\nHow do I change the footnote symbol to like *? I don't understand the command they gave in the TexnicCenter Help. Can someone please help me.", "Solution_1": "In the preamble of your document (i.e. before \\begin{document}) put: \r\n\r\n[code]\\renewcommand{\\thefootnote}{\\fnsymbol{footnote}}[/code]\n\n\nThen, where you want the footnote to appear put \n\n[code]\\footnote{[Put footnote text here.]}[/code]" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all functions $ f: \\mathbb R\\rightarrow \\mathbb R$ satisfying $ (x \\plus{} y)(f(x) \\minus{} f(y)) \\equal{} (x \\minus{} y)f(x \\plus{} y)$ for every $ x,y\\in \\mathbb R$", "Solution_1": "Who can solve this problem ?", "Solution_2": "This problem is really hard, I hope you will post solutions soon.", "Solution_3": "It is not hard :wink: .\r\nTake $ y\\to \\minus{}x$ then $ f(0)\\equal{}0$\r\nTake $ y\\to 1$ and let $ a\\equal{}f(1)$\r\n\\[ f(x\\plus{}1)\\equal{}f(x)\\plus{}a\r\n\\]\r\nTake $ y\\to y\\plus{}1$ then \r\n\\[ (x\\plus{}y\\plus{}1)(f(x)\\minus{}f(y)\\minus{}a)\\equal{}(x\\plus{}y\\plus{}1)(f(x\\plus{}y)\\plus{}a\r\n\\]\r\nTherefore :\r\n$ f(x)\\plus{}f(x\\plus{}y)\\minus{}f(y)\\equal{}2ax$\r\nTake $ y\\to 0$ then we have :\r\n\\[ f(x)\\equal{}ax,\\forall x\\in \\mathbb{R}\r\n\\]\r\nThis is solution of this function equation .", "Solution_4": "I agree with you TTsphn,simplely,you can choose $ y\\equal{}\\minus{}x$ instead of $ y\\to \\minus{}x$ :wink: \r\n[hide=\"to TTsphn\"]Anh l\u00e0 nguy\u1ec5n th\u1ecd T\u00f9ng ph\u1ea3i ko \u1ea1[/hide]", "Solution_5": "[quote=\"TTsphn\"]\nTake $ y\\to 1$ and let $ a \\equal{} f(1)$\n\\[ f(x \\plus{} 1) \\equal{} f(x) \\plus{} a\n\\]\nTake $ y\\to y \\plus{} 1$ then\n\\[ (x \\plus{} y \\plus{} 1)(f(x) \\minus{} f(y) \\minus{} a) \\equal{} (x \\plus{} y \\plus{} 1)(f(x \\plus{} y) \\plus{} a\n\\]\n[/quote]\r\nYour post made a follish mistake: my equation is [b](x+y)[/b](f(x)-f(y))=[b](x-y)[/b]f(x+y) :rotfl: \r\nIt is really hard, i hope to receive your solutions.", "Solution_6": "[quote=\"thaithuan_GC\"][quote=\"TTsphn\"]\nTake $ y\\to 1$ and let $ a \\equal{} f(1)$\n\\[ f(x \\plus{} 1) \\equal{} f(x) \\plus{} a\n\\]\nTake $ y\\to y \\plus{} 1$ then\n\\[ (x \\plus{} y \\plus{} 1)(f(x) \\minus{} f(y) \\minus{} a) \\equal{} (x \\plus{} y \\plus{} 1)(f(x \\plus{} y) \\plus{} a\n\\]\n[/quote]\nYour post made a follish mistake: my equation is [b](x+y)[/b](f(x)-f(y))=[b](x-y)[/b]f(x+y) :rotfl: \nIt is really hard, i hope to receive your solutions.[/quote]\r\nOK .it is stiil not hard\r\nLet $ f(1) \\equal{} a,f( \\minus{} 1) \\equal{} b$\r\n\r\nThen$ take y \\equal{} 1$ we have $ f(x \\plus{} 1) \\equal{} (x \\plus{} 1)\\frac {f(x) \\minus{} a}{x \\minus{} 1}$\r\n $ y \\equal{} \\minus{} 1,x\\plus{}1$ we have $ f(x \\plus{} 1) \\equal{} (x \\plus{} 2)\\frac {f(x)}{x} \\plus{} b$\r\nThe last is only solve this equation with variable is $ f(x)$\r\nand with some small comment ,we can finish\r\nIs it right", "Solution_7": "Thanks guys for your interests, \r\nBut to prove f(x)=xf(1), we must prove that $ f( \\minus{} 1) \\equal{} \\minus{} f(1)$, it will finish the problem. Any ideas?", "Solution_8": "I solved it, thanks Allnames very much :lol:" } { "Tag": [ "LaTeX", "geometry", "geometric transformation", "reflection", "\\/closed", "noooowhybumped" ], "Problem": "When I type a post in the message board with spoiler anything inside the\r\ndollar signs shows up without being highlighted.How do you spoiler tex?", "Solution_1": "Can't spoiler TeX - the TeX gets rendered as images, not text, so you can't spoiler it.", "Solution_2": "Can we use backgrounds here? If so, we could make black backgrounds around $\\text{\\TeX}$ as a sort of ad hoc spoiler system.", "Solution_3": "I don't think we can create backgrounds.\r\nSo what do I do?Just not use TeX?", "Solution_4": "[quote=\"tandjsnell\"]I don't think we can create backgrounds.\nSo what do I do?Just not use TeX?[/quote]\r\nNo, please, use LaTeX so it's easy to read, and if it isn't sufficiently hidden because of the LaTeX... oh well. Perhaps there will be a work-around in the future.", "Solution_5": "[quote=\"tandjsnell\"]I don't think we can create backgrounds.\nSo what do I do?Just not use TeX?[/quote]it's very simple: don't use spoliers :P :) .. ;)", "Solution_6": "[quote=\"Valentin Vornicu\"][quote=\"tandjsnell\"]I don't think we can create backgrounds.\nSo what do I do?Just not use TeX?[/quote]it's very simple: don't use spoliers :P :) .. ;)[/quote]\r\n\r\nValentin, as has been discussed in the other post, can you please stop telling people not to use spoilers. If you do theres going to be big problems, as spoilers have been widely used at AoPS. \r\n\r\ntandjsnell: if what you are writing is unreadable with normal text, then sure use latex and just use spoiler on any text you write. Not ideal, but its the best that can be done. If its just simple things like squaring, writing ^2 is probably easier, that will allow spoiler.", "Solution_7": "[quote=\"TripleM\"][quote=\"Valentin Vornicu\"][quote=\"tandjsnell\"]I don't think we can create backgrounds.\nSo what do I do?Just not use TeX?[/quote]it's very simple: don't use spoliers :P :) .. ;)[/quote]\n\nValentin, as has been discussed in the other post, can you please stop telling people not to use spoilers. If you do theres going to be big problems, as spoilers have been widely used at AoPS. [/quote] Dear TripleM, I am very aware of the contents of the other post you are refering to. \r\n\r\nHowever, as you can see above tandjsnell asked a question. I gave him a suggestion. If I cannot do that anymore, please tell me so. My (personal) opinion is against using spoilers. However as you can easily see I am conformating myself to the big majority of the people, ie spoilers can be used in all forums.\r\n\r\nI do not belive I need to be told what to or not to say, especially if it reflects a personal opinion (which was also underlined by the emoticons :) :P and ;)!) ...", "Solution_8": "[quote=\"TripleM\"]If its just simple things like squaring, writing ^2 is probably easier, that will allow spoiler.[/quote]\r\nFor superscripts, you can also use the HTML tags [sup] and [/sup]. For subscripts, you can use [sub] and [/sub]. Both get hidden under spoiler.", "Solution_9": "[quote=\"TripleM\"]tandjsnell: if what you are writing is unreadable with normal text...[/quote]\r\nOr hard to read, at least. Sorry MMM, just clarifying a bit.", "Solution_10": "[quote=\"Valentin Vornicu\"][/quote][quote=\"TripleM\"][quote=\"Valentin Vornicu\"][quote=\"tandjsnell\"]I don't think we can create backgrounds.\nSo what do I do?Just not use TeX?[/quote]it's very simple: don't use spoliers :P :) .. ;)[/quote]\n\nValentin, as has been discussed in the other post, can you please stop telling people not to use spoilers. If you do theres going to be big problems, as spoilers have been widely used at AoPS. [/quote][quote=\"Valentin Vornicu\"] Dear TripleM, I am very aware of the contents of the other post you are refering to. \n\nHowever, as you can see above tandjsnell asked a question. I gave him a suggestion. If I cannot do that anymore, please tell me so. My (personal) opinion is against using spoilers. However as you can easily see I am conformating myself to the big majority of the people, ie spoilers can be used in all forums.\n\nI do not belive I need to be told what to or not to say, especially if it reflects a personal opinion (which was also underlined by the emoticons :) :P and ;)!) ...[/quote]\r\n\r\nI'm sorry if you took my post the wrong way - to me it seemed the topic was about how best to use spoiler, not whether to use spoiler or not. If I misinterpreted your post, then other people may also do so, and I just want to make sure people get the right idea.", "Solution_11": "[quote=akliu]Luckily we have hide tags for hiding latex now. Quick question, when was this added to AoPS?[/quote]\n\nplease dont bump 17 year old threads and how did you find this\n\nwell idk when this was added to aops\nthere arent a lot of ppl who joined when hide tags werent a thing and still use aops now", "Solution_12": "[quote=akliu]Luckily we have hide tags for hiding latex now. Quick question, when was this added to AoPS?[/quote]\n\nOld AoPS is different from AoPS now. It might have had a different, less reliable hide feature called \"spoiler\"", "Solution_13": "[quote=pieMax2713][quote=akliu]Luckily we have hide tags for hiding latex now. Quick question, when was this added to AoPS?[/quote]\n\nplease dont bump 17 year old threads and how did you find this\n\nwell idk when this was added to aops\nthere arent a lot of ppl who joined when hide tags werent a thing and still use aops now[/quote]\n\nSearch function, and the first AoPS user. Some people say something about the \"new aops\", was the whole AoPS site remade?\n\nAnd also, why did Richard Rusczyk join after Valentin Vornicu??? isn't richard rusczyk the creator of aops???\n", "Solution_14": "Valentin (my instructor!) made the site, and then later joined with Aops to make this site.", "Solution_15": "[quote=Critical]Valentin (my instructor!) made the site, and then later joined with Aops to make this site.[/quote]\n\nOHHHH that makes sense now...", "Solution_16": "[quote=akliu][quote=Critical]Valentin (my instructor!) made the site, and then later joined with Aops to make this site.[/quote]\n\nOHHHH that makes sense now...[/quote]\n\nHe made mathlinks.ro which combined with AoPS (the previous link redirects to the college and olympiad portal).\n\nAnd it really wasn't necessary to bump this.", "Solution_17": "[quote=pith0n][quote=akliu][quote=Critical]Valentin (my instructor!) made the site, and then later joined with Aops to make this site.[/quote]\n\nOHHHH that makes sense now...[/quote]\n\nHe made mathlinks.ro which combined with AoPS (the previous link redirects to the college and olympiad portal).\n\nAnd it really wasn't necessary to bump this.[/quote]\n\n[s]The oldest topic on SS, finally destroyed... :evilgrin: [/s]\n\nThanks for telling me about that @all above", "Solution_18": "AKLIU WHY DID YOU BUMP THIS? Now it will get locked. At least there is one more thread.", "Solution_19": "WOW this is old", "Solution_20": "[quote=jmiao]WOW this is old[/quote]\n\nPM me if you want to know another old one. ", "Solution_21": "I got postbanned from site support for bumping g old threads" } { "Tag": [], "Problem": "When General Han counts the soldiers in his army, he uses the following method. He orders them to line up in rows of 11, then in rows of 13, and finally in rows of 17, and each time he counts the number of soldiers not in a row. One morning, he finds that there are 3 soldiers left when the rest are in rows of 11, 4 soldiers left when the rest are in rows of 13, and 9 soldiers left when the rest are in rows of 17. He knows that there are 1000 soldiers in his army. How many of the soldiers are present this morning. \r\n\r\nI don't know how to solve this problem other than guess and check :oops: ....Can anyone help explain? Thanks", "Solution_1": "Learn mods if you haven't already, they're a very powerful tool. And if you have, learn when to use them. :)\r\n\r\nFor this problem, say $n$ is the number of soldiers. $n\\equiv 3\\bmod{11}$, $\\equiv 4\\bmod{13}$, and $\\equiv 9\\bmod{17}$. From there, we say that $(n-3)\\equiv 0\\bmod{11}$, $\\equiv 1\\bmod{13}$, and $\\equiv 6\\bmod{17}$. Therefore, we have a multiple of $11$ that is one more than a multiple of $13$. We have this at $66+143k$, where $k$ is an integer. Next, we see that $66+143k$ is six more than a multiple of $17$. You only need to check a few numbers at that point.\r\n\r\n[hide=\"Only number under 1000 where this occurs\"]$924$[/hide]\r\n\r\nSo, add $3$ to that to get the answer.", "Solution_2": "What are mods? :huh: I have heard of that word a couple of times but have no idea what it is. Also, what does the 3 bars stand?\r\n\r\nThanks", "Solution_3": "$\\equiv$ means equivalent to. It's the remainder when $n$ is divided by the $\\bmod$ number. For example, $73\\equiv 3\\bmod{7}$, since $3$ is the remainder when $73$ is divided by $7$." } { "Tag": [ "vector", "analytic geometry", "graphing lines", "slope", "calculus", "calculus computations" ], "Problem": "Everything I have been doing so far with 3D stuff has been tangent planes. Then I got to this:\r\n\r\nFind the parametric equations for the tangent [b]line[/b] to the curve of intersection of the paraboloid $z=x^{2}+y^{2}$ and the ellipsoid $4x^{2}+y^{2}+z^{2}=9$ at the point (-1,1,2).\r\n\r\nI'm confused.... :?:", "Solution_1": "Just take the cross product of the gradient vectors to the two surfaces at that point. That will give you the \"direction\" vector.", "Solution_2": "Okay, so the cross products of the gradients.\r\n$\\nabla P=\\left<2x, 2y, -1\\right>$ $\\nabla E=\\left<8x, 2y, 2z\\right>$\r\n$\\left<2x, 2y, -1\\right>\\times\\left<8x, 2y, 2z\\right> =\\left$\r\nOkay, for point (-1,1,2) = $\\left<10,16,12\\right>$\r\n\r\nUm...so what is the point slope form for 3D? Is it related to the linear approximation?" } { "Tag": [ "function", "modular arithmetic", "algebra solved", "algebra" ], "Problem": "Let $ S \\equal{} \\{0,1,2,\\ldots,1999\\}$ and $ T \\equal{} \\{0,1,2,\\ldots \\}.$ Find all functions $ f: T \\mapsto S$ such that \r\n\r\n[b](i)[/b] $ f(s) \\equal{} s \\quad \\forall s \\in S.$\r\n[b](ii)[/b] $ f(m\\plus{}n) \\equal{} f(f(m)\\plus{}f(n)) \\quad \\forall m,n \\in T.$", "Solution_1": "I think the function is determined if we know f(2000). We have no restrictions on f(2000), so we can take it to be any number from 0 to 1999. By looking at some cases we observe a pattern: f(2000)=k => the function increases with 1 unit until it reaches 1999 and then it starts from k again.\r\n\r\nFor example, assume f(2000)=1997. Then f(2001)=f(2000+1)=f(1997+1)=1998; f(2002)=f(2001+1)=f(1998+1)=1999; f(2003)=f(2002+1)=f(1999+1)=f(2000)=1997 and so on (in this case k=1997).\r\n\r\nThere are 2000 such functions, corresponding to the cases f(2000)=k, with k from 0 to 1999.", "Solution_2": "We have\n\\[f(m)=f(m+0)=f(f(m)+f(0))=f(f(m)\\\\\n\\Rightarrow f(m+n)=f(f(m)+f(n))=f(f(m)+f(f(n)))=f(m+f(n))\\]\nNow suppose that $f(2000)=a$, for some $0\\leq a\\leq 1999$. Then \\[f(m+2000)=f(m+f(2000))=f(2000+a)\\]\nand so $f(m)=f(m+a-2000)$ for all $m\\geq 2000$. Thus the function determine uniquely by definition of $f(2000)$. Therefore there are 2000 functions satisfy the problem:\n\\[f(n)=n,\\qquad \\forall n,\\quad 0\\leq n\\leq 1999,\\\\\nf(n)=(n-a)\\pmod {(2000-a)}+a,\\qquad \\forall n,\\quad n\\geq2000\\]" } { "Tag": [ "integration", "Gauss" ], "Problem": "Find the force applied to the dielectric in the electric field .Value of electric field is E and dielectric constant is e.Surface of the dielectric perpendicular to the electric field is A.", "Solution_1": "Here's one approach:\r\n\r\n[hide]The (potential) energy stored in the electric field is given by:\n\n$W_{e}=\\int Q\\mathrm{d}V$\n\nForce is equal to minus the gradient of potential.\n\nIt follows that $\\vec{F}=-\\nabla W_{e}=-\\nabla\\left(\\int Q\\mathrm{d}V\\right)$\n\n$\\Rightarrow \\vec{F}=Q\\left(-\\nabla V\\right) = Q\\left|\\vec{E}\\right| \\mathbf{\\hat{n}}.$\n\nAccording to Gauss' Law the electric flux across a surface is equal to the charge enclosed by said surface, i.e. $\\vec{D}\\cdot \\mathbf{\\hat{n}}A = Q.$\n\nWith that in mind, making the appropriate replacement:\n\n$Q\\left|\\vec{E}\\right| \\mathbf{\\hat{n}}=\\vec{D}\\cdot \\mathbf{\\hat{n}}A \\left|\\vec{E}\\right| \\mathbf{\\hat{n}}.$\n\nAlso we know that $\\vec{D}= \\epsilon \\vec{E}.$\n\nMaking that replacement:\n\n$\\vec{D}\\cdot \\mathbf{\\hat{n}}A \\left|\\vec{E}\\right| \\mathbf{\\hat{n}}= \\epsilon \\vec{E}\\cdot \\mathbf{\\hat{n}}A \\left|\\vec{E}\\right| \\mathbf{\\hat{n}}$\n$=\\epsilon \\left|\\vec{E}\\right|^{2}A\\mathbf{\\hat{n}}.$\n\n\nConclusion:\n\n$\\boxed{\\vec{F}= \\epsilon \\left|\\vec{E}\\right|^{2}A\\mathbf{\\hat{n}}.}$\n[/hide]" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "rotation" ], "Problem": "We have a cube. From every vertex, draw the plane formed by the three adjacent vertices. There are 8 such planes. How many regions do they divide the cube into? \r\n\r\nthis problem probably is very trivial but i cant c how to solve it and cant visualise :( :( obvioulsy, for each plane we have a parellel plane - and the problem is just equivalant to: how many regions do a regular tetrahedron, and the same regular tetrahedron rotated by 180 degs ( i think) intersect? i think the answer is 17 but am not sure.\r\n\r\n How would you generalize to n dimensions?", "Solution_1": "any ideas??", "Solution_2": "it are actually 21 if i'm not wrong. take 4 vertices of the cube which form a regular tetrahedron. the faces of this tetrahedron are planes of the task. look at the region out of the tetrahedron but inside the cube. there are four congruent areas behin every face of the tetrahedron. three planes go through the vertex behind that face dividing that area into 4 pieces, no other planes go through there. thus we get 16 areas at all outside the tetrahedron. the tetrahedron itself is cut by all the four other planes dividing it into 5 pieces(these planes don't intersect inside the tetrahedron). thus we get 21 areas i think.", "Solution_3": "You can also see it this way: there are 12 pieces such that each piece contains 1 edge of the cube, 8 pieces such that each piece contains a vertex of the cube (the other 12 pieces contain 2 vertices of the cube each, but anyway the edges are more obvious), and one central piece, to make a total of 21.\r\n\r\nThis question cost me quite a bit of money once upon a time :( ." } { "Tag": [ "calculus", "integration", "derivative", "LaTeX", "complex analysis", "complex analysis unsolved" ], "Problem": "Let a > 0. Prove:\r\n\r\nintegral from 0 to + infinity of (e^(-ax)dx) = 1/a\r\n\r\nDi\ufb00erentiate with respect to a, n-times and prove\r\n\r\nintegral from 0 to + infinity of ( [x^n]*[ e^(-ax)] dx) = (n!) / a^(n+1)\r\n\r\nConsidering the integral:\r\n\r\nintegral from n to + infinity of ( [x^n]*[ e^(-ax)] dx)\r\n\r\nprove that\r\n\r\nfor any a > 0.\r\n\r\n(n^n)(a^n)(e^\u2212an) \u2264 n!", "Solution_1": "[img]http://img193.imageshack.us/img193/8963/70398083.jpg[/img]", "Solution_2": "Just note $ n^n\\int_n^{\\infty} e^{\\minus{}ax} dx \\leq \\int_n^{\\infty} x^n e^{\\minus{}ax} dx < \\int_0^{\\infty} x^n e^{\\minus{}ax} dx$.\r\n\r\nAlso, learn some LaTex to post messages." } { "Tag": [], "Problem": "Hi!\r\nI have a couple of questions on chemical bonding. I would be grateful if someone could answer them.\r\n\r\n1) Compare the C-C distance (distance between 2 carbon atoms) in Ethane and 1,1,1-Trifluoroethane. \r\nOne of the reason which I can think is that fluorine being highly electronegetive would induce a partial positive charge on carbon. Due to this the carbon attached to flourines would pull the bond pair electrons between the two carbons towards itself. This would strengthen the bond and inturn decrease the bond length. Is it correct? However it isnt completely convining.\r\n\r\n2)Why does the most electronegetive atom occupy axial positions on Trigonal Bipyramidal hybridisation(sp3d). I tried searching the net but dint get a convincing answer. \r\nAgain one of the answer which I can think of is that the electronegetive atoms would pull the elctron from the centre atom and thus reduce the electron desity.\r\nThat would reduce the repulsion in the equatorial plane. But that could be done even if the atom is in equatorial plane. So why axial?\r\nCould u guys plz answer?\r\n\r\nThanks!", "Solution_1": "2. the most electronegative atom attracts electrons most strongly. the closer the electrons are to the atom, the less space the bonding pair can take up.\r\n\r\nThe less electronegative atoms then pull less on electrons and so their corresponding bonding pairs take up more space. They tend to occupy equatorial positions then because they have roughly 2 120 degree and 2 90 degree repulsions, instead of the more repulsive axial position with 3 90 degree repulsions. As less electronegative atoms occupy equatorial positions to minimize repulsions, more electronegative atoms then occupy axial positions." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "The space diagonal of a cube is $ \\sqrt{27}$ cm. How many cubic centimeters are in the volume of the cube? \n[asy] size(101);\nimport three;\ncurrentprojection = orthographic(1/4,-3/4,1/2);\ndraw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle);\ndraw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle);\ndraw((0,0,0)--(0,0,1));\ndraw((1,0,0)--(1,0,1));\ndraw((0,1,0)--(0,1,1));\ndraw((1,1,0)--(1,1,1));\ndraw((0,0,0)--(1,1,1),linewidth(2));\nlabel(\"$\\sqrt{27}$\",(1.5,0,.5),E);\ndraw((1.5,0,.5)--(0.25,0.25,0.25),Arrow3);\n[/asy]", "Solution_1": "We know the space diagonal of a cube is $ s\\sqrt{3}$ where $ s$ is the side length. Since $ \\sqrt{27}\\equal{}3\\sqrt{3}, s\\equal{}3$, and the volume is $ 3^3\\equal{}\\boxed{27}$" } { "Tag": [ "conics", "ellipse", "function", "calculus", "calculus computations" ], "Problem": "Knowing that any polygon is a combination of finite triangles, is there a similar relationship with curved shapes? What I mean, is any curved shape a finite combination of another curved shape?\r\n\r\nSo far, I haven't found any such relationship. However, I recently thought that perhaps a curved shape can be defined as a polygon with an infinite number of sides. If this were true, then any curved shape could be defined as an infinite number of triangles merged together.\r\n\r\nAny ideas?", "Solution_1": "If by \"curved shape\" you mean any simple closed curve, then this is clearly absurd. Curves can take on arbitrary continuous shapes, whereas polygons are discrete, that is, any polygon can be finitely described, but an arbitrary curve comprises an infinite amount of information.", "Solution_2": "[quote=\"t0rajir0u\"] Curves can take on arbitrary continuous shapes.[/quote]\r\n\r\nI'm not exactly sure what you mean by this.\r\n\r\nOne thing that distinguishes curves from straight lines is that there are many different types of curves (ex/ curves forming an ellipse versus the curves forming a logarithmic function), whereas there is only one type of straight line segment i.e. the shortest path passing through two points. I've even read that a straight line is a special type of curve. Is this what you meant?", "Solution_3": "Some curves can be thought of, not as polygons with infinitely many sides but more properly as limits of polygons. However, there is no nice categorization of curves equivalent to, say, the categorization of polygons by number of sides. Certainly, if you start listing types of curves (\"elipse, logarithmic curve, ...\"), you will spend all of your life and never make a dent in the family of all smooth curves. We can deform a given curve essentially arbitrarily to give massive families of curves." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "probability" ], "Problem": "... to MC alums Zach Abel, Zhou Fan, Sherry Gong, Adam Hesterberg, Albert Ni, and Tim Pollio, USAMO winners/honorable mentions this year!", "Solution_1": "Any of them coming back this year?", "Solution_2": "[quote=\"MithsApprentice\"]Any of them coming back this year?[/quote]\r\n\r\nTwo of them, I think, at this point... unless, of course, they make IMO!", "Solution_3": "As listed on the alumni forms, the probability that I will return to math camp is now 1-P(I will die or become mentally disabled by July 3, 2005), changed from [1-P(I will die or become mentally disabled by July 3, 2005)][1-P(I make it to the IMO)P(The IMO goes through August 2)] ~= 1 because I was two points short of the TST." } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let G be a finite Abelian group whose order is not divisible by $ p$, where $ p$ is a prime. Determine the number of solutions to the equation $ g^p \\equal{} a$, where $ a$ is any element in $ G$.\r\n\r\nThis is indeed a homework question. I am not asking for a solution (but help if mine is wrong =))\r\nI have that for each $ a \\in G$ there is a [i]unique[/i] (ie 1) element/solution $ g$.", "Solution_1": "[b]1.[/b] For any $ a \\in G$ there is a $ g \\in G$ such that $ a \\equal{} g^{p}$.\r\n\r\n[b][color=orange]Proof.[/color][/b] Since $ p$ and $ |G|$ are coprime we have that $ 1 \\equal{} m \\cdot p \\plus{} n \\cdot |G|$ for some $ m, n \\in \\mathbb{Z}$. Hence,\r\n\r\n$ a \\equal{} a^{1} \\equal{} a^{m \\cdot p \\plus{} n \\cdot |G|} \\equal{} a^{m \\cdot p} \\cdot a^{n \\cdot |G|} \\equal{} (a^{m})^{p}(a^{|G|})^{n} \\equal{} (a^{m})^{p}.$\r\n\r\nThe result follows once we let $ a^{m} \\equal{} g.$ \r\n\r\n[b]2.[/b] For any $ a \\in G$ there is exactly one $ g \\in G$ such that $ a \\equal{} g^{p}$.\r\n\r\n[b][color=orange]Proof.[/color][/b] In view of [b]1[/b] all that remains to be shown is unicity. We proceed by [i]reductio ad absurdum[/i] here. Let us suppose that there exist $ g_{1}, g_{2} \\in G$ such that\r\n\r\n$ g_{1}^{p} \\equal{} a \\equal{} g_{2}^{p}.$\r\n\r\nFrom the equalities in the previous line we get that\r\n\r\n$ e \\equal{} g_{1}^{p}(g_{2}^{p})^{ \\minus{} 1} \\equal{} g_{1}^{p}g_{2}^{ \\minus{} p} \\equal{} (g_{1}g_{2}^{ \\minus{} 1})^{p}.$\r\n\r\nHence, the group $ G$ possesses an element of order $ p$ ([b][color=red]contradiction![/color][/b]) and the proof terminates.\r\n\r\n[b]Conclusion:[/b] Little Turtle's guess is right! :D \r\n\r\nBest regards,\r\n\r\nJos\u00e9." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all rational numbers a, such that [tex]\\left(\\frac{a^{m+n}+1}{a^{n}+1}\\right)^{\\frac{1}{m}}\\in\\mathbb{Q}[/tex], where m and n are positive integers, with m>1.", "Solution_1": "No one dares to solve this problem? :)", "Solution_2": "[quote=\"pcalin\"]Find all rational numbers $a$, such that $\\displaystyle \\left(\\frac{a^{m+n}+1}{a^{n}+1}\\right)^{\\frac{1}{m}}\\in\\mathbb{Q}$, where $m$ and $n$ are positive integers, with $m>1$.[/quote]\r\n\r\nAnyone? :)" } { "Tag": [ "algebra", "polynomial", "arithmetic sequence", "algebra unsolved" ], "Problem": "Find all positive integers $ n\\geq 3 $ such that there exists an arithmetic progression $ a_0 , a_1, \\ldots, a_n $ such that the equation $ a_nx^n + a_{n-1}x^{n-1} + \\cdots+ a_1x+a_0 = 0 $ has $n$ roots setting an arithmetic progression.", "Solution_1": "Let $a_n=a$, $a_{n-1}=a+b$, ..., $a_0=a+nb$, then initial equation can be rewritten as\r\n\\[ax^{n+2}-(a-b)x^{n+1}-(a+b(n+1))x+(a+bn)=0.\\]\r\nThis equation has additional root $1$ of order 2, so totally it has $n+2$ reall roots. But we see that there are $n\\geq 3$ consecutive zero coefficients. So from Descartes rule we conclude there is an imaginary root. Contradiction.\r\n\r\nHmm... It works for $n=2$ too :?", "Solution_2": "[quote=\"Myth\"]Let $a_n=a$, $a_{n-1}=a+b$, ..., $a_0=a+nb$, then initial equation can be rewritten as\n\\[ax^{n+2}-(a-b)x^{n+1}-(a+b(n+1))x+(a+bn)=0.\\]\nThis equation has additional root $1$ of order 2, so totally it has $n+2$ reall roots. But we see that there are $n\\geq 3$ consecutive zero coefficients. So from Descartes rule we conclude there is an imaginary root. Contradiction.\n\nHmm... It works for $n=2$ too :?[/quote]\r\nWhy roots of the polynomials setting arithmetic progression?? :P :P :D", "Solution_3": "???\r\nWhat do you mean? :?\r\nI have showed that such polynomials cannot have $n$ real roots." } { "Tag": [ "algebra", "system of equations", "algebra unsolved" ], "Problem": "Let n be positive integer. Solve system of equations\r\n$ x_1 \\plus{} x_2^2 \\plus{} ... \\plus{} x_n^n \\equal{} n$\r\n$ x_1 \\plus{} 2x_2 \\plus{} ... \\plus{} nx_n \\equal{} \\frac {n(n \\plus{} 1)}{2}$\r\nfor n-tuples $ (x_1,...,x_n)$ of nonnegative real numbers.", "Solution_1": "[quote=\"Przemo\"]Let n be positive integer. Solve system of equations\n$ x_1 \\plus{} x_2^2 \\plus{} ... \\plus{} x_n^n \\equal{} n$\n$ x_1 \\plus{} 2x_2 \\plus{} ... \\plus{} nx_n \\equal{} \\frac {n(n \\plus{} 1)}{2}$\nfor n-tuples $ (x_1,...,x_n)$ of nonnegative real numbers.[/quote]\r\nAdd $ 1 \\plus{} 2 \\plus{} 3 \\plus{} ... \\plus{} (n \\minus{} 2) \\plus{} (n \\minus{} 1)$ in both side of the first equation, we have:\r\n$ \\frac{n(n \\plus{} 1)}{2} \\equal{} x_1 \\plus{} (x_2^2 \\plus{} 1) \\plus{} (x_3^3 \\plus{} 2) \\plus{} ... (x_n^n \\plus{} n \\minus{} 1) \\geq x_1 \\plus{} 2x_2 \\plus{} 3x_3 \\plus{} ...nx_n$\r\n$ \\Rightarrow x_1 \\equal{} x_2 \\equal{} x_3 \\equal{} ...x_n \\equal{} 1$" } { "Tag": [ "vector", "LaTeX", "linear algebra", "matrix", "inequalities" ], "Problem": "This should be a very basic problem, but I have some doubts.\r\n\r\nSuppose that U and V are finite-dimensional vector spaces and that S is a linear map from V to W, T is a linear map from U to V.\r\nProve that $ dimnullST$ is less than or equal to $ dimnullS\\plus{}dimnullT$.\r\n\r\nAlso, can anyone give an example when it is strictly less than?", "Solution_1": "$ \\text{\\LaTeX}$ tip: first see if the word you want is encoded by putting \\ in front of it:\r\n\r\n\\dim gives $ \\dim$, \\nul gives an error message.\r\n\r\nIn that case, use \\text{nul} to get what you want: $ \\dim\\text{nul}S.$\r\n\r\nVery simple example: $ S = T = \\begin{bmatrix}1 & 0 \\\\\r\n0 & 0\\end{bmatrix}.$ Then $ ST = S$ and $ \\dim\\text{nul}S = \\dim\\text{nul}T = \\dim\\text{nul}ST = 1.$\r\n\r\nI'll channel my current colleague from whom I learned all of my linear algebra to say this: why are you calling these \"linear maps on finite dimensional vector spaces\"? These are [b]matrices[/b], and the sooner you get to thinking of them as matrices, the better.", "Solution_2": "Sorry, I've just begun linear algebra for several days, I only read three chapters of a linear algebra book that did not even mention matrix for many times.\r\n\r\nSo what about the prove for the inequality?" } { "Tag": [ "IMO" ], "Problem": "Let's expect results of the IMO 2008 Spain.\r\n\r\n1.China\r\n2.Russia\r\n3.South Korea\r\n4.USA\r\n5.Vietnam\r\n6.Bulgaria\r\n7.Romania\r\n8.Taiwan\r\n9.Iran\r\n10.Ukraine", "Solution_1": "Isn't it too early? :D", "Solution_2": "[quote=\"edriv\"]Isn't it too early? :D[/quote]\r\n\r\nOf course not :P\r\nBy the way, the Hungarian team will be:\r\n...\r\n:lol:", "Solution_3": "Will North Korea participate?", "Solution_4": "[quote=\"Valentin Vornicu\"]Will North Korea participate?[/quote]\r\n\r\nThat is a touchy question.. I had a conversation with Pelik\u00e1n about that, but i think i'm not at liberty to talk about that in a public forum...\r\nAnyway, it's not known yet.", "Solution_5": "Well, it's important for next poll, as they are in the top10.", "Solution_6": "[quote=\"Valentin Vornicu\"]Will North Korea participate?[/quote]\r\n\r\nWhy did not North Korea participate from 1993 to 2005?", "Solution_7": "[quote=\"ringos\"][quote=\"Valentin Vornicu\"]Will North Korea participate?[/quote]\n\nWhy did not North Korea participate from 1993 to 2005?[/quote]It did not participate in 2006 either. It most likely was because the Olympiad wasn't held in a communist country anymore so the political leadership didn't want them to send a team. It was disqualified in 1991 in Sweden for (suspicion of) cheating. I doubt they will come to Spain next year as well, so that means that probably in the top 10 poll I will exclude it and put the next country in the list.", "Solution_8": "Does anyone know which country will be strong next year?\r\n(For example, who will be high school students next year among this year's contestants in your country?)", "Solution_9": "It's too early now but I think Georgian Team will be in 15 best country(at least).", "Solution_10": "I think Viet Nam will win because next year chienthan will 16 years old! :wink:", "Solution_11": "[quote=\"pephuc_93\"]I think Viet Nam will win because next year chienthan will 16 years old! :wink:[/quote]\r\nIt is really strong reason :rotfl: I'm kidding!", "Solution_12": "[quote=\"pephuc_93\"]I think Viet Nam will win because next year chienthan will 16 years old! :wink:[/quote]\r\n\r\n[color=green]hnm maybe. :D [/color]", "Solution_13": "i'm sure $ INDIA$ has the potential to win\r\ni will surely expect a good performance from $ INDIA$\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n(although for some last years the performance is degrading)" } { "Tag": [ "trigonometry" ], "Problem": "A curve has the equation $ y\\equal{}f(x)$ where\r\n\r\n$ f(x) \\equal{} \\cos (2x\\plus{}\\frac{\\pi}{3}) \\plus{} \\sin (\\frac{3x}{2} \\minus{} \\frac{\\pi}{4})$\r\n\r\nFind the period of $ f(x)$.\r\n\r\nDetermine all values of $ x$ in the interval $ \\minus{}\\pi \\le x \\le \\pi$ for which $ f(x)\\equal{}0$. Find a value of $ x$ in this interval at which the curve touches the $ x\\minus{}$axis without crossing it.\r\n\r\nFind the value or values of $ x$ in the interval $ 0 \\le x \\le 2\\pi$ for which $ f(x)\\equal{}2$.", "Solution_1": "\\[ f(x)\\equal{}0 \\implies 0\\equal{}\\cos(2x\\plus{}\\frac{\\pi}{3})\\plus{}\\sin(\\frac{3x}{2}\\minus{}\\frac{\\pi}{4}) \\implies \\sin(\\frac{\\pi}{6}\\minus{}2x)\\equal{}\\sin(\\frac{\\pi}{4}\\minus{}\\frac{3x}{2})\\]\r\n\r\ntherefore \\[ \\frac{\\pi}{6}\\minus{}2x\\equal{}\\frac{\\pi}{4}\\minus{}\\frac{3x}{2} \\implies x\\equal{}\\minus{} \\frac{\\pi}{6}\\]" } { "Tag": [ "geometry", "Triangles", "rotation", "IMO 1986", "Fixed point", "Equilateral Triangle", "IMO" ], "Problem": "Given a point $P_0$ in the plane of the triangle $A_1A_2A_3$. Define $A_s=A_{s-3}$ for all $s\\ge4$. Construct a set of points $P_1,P_2,P_3,\\ldots$ such that $P_{k+1}$ is the image of $P_k$ under a rotation center $A_{k+1}$ through an angle $120^o$ clockwise for $k=0,1,2,\\ldots$. Prove that if $P_{1986}=P_0$, then the triangle $A_1A_2A_3$ is equilateral.", "Solution_1": "Write $x= A_1-A_2 , y= A_2-A_3 , z= A_3-A_1$.\r\nThe mappings are (with $\\zeta$ being the 'first' third root of unity):\r\n$P \\mapsto \\zeta (P-A_1)+A_1$\r\n$P \\mapsto \\zeta (P-A_2)+A_2$\r\n$P \\mapsto \\zeta (P-A_3)+A_3$\r\nand using them in a row gives (easy to calculate out):\r\n$P \\mapsto P + \\zeta^2 x + \\zeta y + z$\r\nUsing the chain of mappings $1986= 3 \\cdot 662$ times is the same as using that mapping $662$ times, thus the mapping\r\n$P \\mapsto P+ 662(\\zeta^2 x + \\zeta y + z)$\r\nNow the last shall equal $P$, so $\\zeta^2 x + \\zeta y + z = 0$, which means (by subtracting $x+y+z=0$) that $(\\zeta^2-1) x +(\\zeta-1) y =0 \\iff x= \\zeta y$.\r\nSimilary, $z=\\zeta x$, thus $|x|=|y|=|z|$, thus the triangle is equilateral.", "Solution_2": "Does anyone have a synthetic solution or know if a nice one exists?", "Solution_3": "There's one here, but in french\r\np.20\r\n\r\nhttp://www.animath.fr/cours/deho_geo/deho_geo.pdf", "Solution_4": "[quote=\"FOURRIER\"]There's one here, but in french\np.20\n\nhttp://www.animath.fr/cours/deho_geo/deho_geo.pdf[/quote]\r\nThank you very much! Luckily \"Math French\" is a little similar to \"math English.\" So I think I have understood the solution, and this is not a literal translation from that file.\r\n[hide=\"Synthetic\"]A composition of three $ 120$ rotations is a rotation of $ 120 \\plus{} 120 \\plus{} 120 \\equal{} 360$, i.e. a translation. Thus, $ \\vec{P_0 P_3} \\equal{} \\vec{P_3 P_6} \\equal{} \\dots \\equal{} \\vec {P_{1983} P_{1986}}$. But $ P_0 \\equal{} P_{1986}$, so the vector is null and $ P_0 \\equal{} P_3 \\equal{} \\dots \\equal{} P_{1986}$. Since $ P_0$ had no restrictions, we can say that any point in the plane gets mapped to itself after the three rotations. In particular, let's examine the behavior of $ A_0$. After the first rotation, $ A_0$ remains $ A_0$. After the second, it gets mapped to some point $ B$. Finally, by our previous result, the third rotation takes $ B$ to $ A_0$ again. Now noting that $ \\angle A_0A_2B \\equal{} \\angle B A_1 A_0 \\equal{} 120$, and that $ BA_2 \\equal{} A_2 A_0$ and $ BA_1 \\equal{} A_1 A_0$, it is easy to deduce that $ A_0A_1A_2$ is equilateral.[/hide]", "Solution_5": "I think the solution found by Zetax is shorter and better (I also did like him).", "Solution_6": "[quote=orl]Given a point $P_0$ in the plane of the triangle $A_1A_2A_3$. Define $A_s=A_{s-3}$ for all $s\\ge4$. Construct a set of points $P_1,P_2,P_3,\\ldots$ such that $P_{k+1}$ is the image of $P_k$ under a rotation center $A_{k+1}$ through an angle $120^o$ clockwise for $k=0,1,2,\\ldots$. Prove that if $P_{1986}=P_0$, then the triangle $A_1A_2A_3$ is equilateral.[/quote]\nProfessor Chang Gengzhe, one of the propositional person of this problem, died in Beijing on November 18, 2018.\n", "Solution_7": "[quote=FOURRIER]There's one here, but in french\np.20\n\nhttp://www.animath.fr/cours/deho_geo/deho_geo.pdf[/quote]\n\nSir can you please repost this link because this link is dead when you open this there is no solution of above problem.kindly give me other link for this solution\n", "Solution_8": "Let $a_1, a_2, a_3, p_0$ be the complex numbers corresponding to their points in the complex plane, and set $\\omega = e^{2i \\pi/3}$. We can compute via a series of rotations that$$p_3 = p_0 - a_1 + \\omega^2 a_1 + \\omega a_2 - \\omega a_3 - \\omega^2 a_2 + a_3.$$In other words, we must have$$a_3-a_1+\\omega^2(a_1-a_2) + \\omega(a_2-a_3) = 0$$in order to have $P_{1986} = P_0$ because $3 \\mid 1986$. Now, consider the polynomial$$x^2(a_1-a_2) + x(a_2 - a_3) + (a_3 - a_1) = 0.$$It has roots $\\omega$ and $1$, so by Vieta's,$$\\frac{a_3 - a_1}{a_1 - a_2} = \\omega,$$which implies that $a_1, a_2, a_3$ are the vertices of an equilateral triangle, as required.", "Solution_9": "Observe that by the theory of rotation composition, we have that $P_{k+3}$ is a translation of $P_k \\forall k \\ge 0$. But since $P_{1986}$ is reached by performing $662$ such translations, and $P_0 = P_{1986}$, we have that $P_k = P_{k+3} \\forall k \\ge 0$.\nNow construct an equilateral triangle $A_1A_2B$ such that $B$ is on the same side of line $A_1A_2$ as $A_3$. Then clearly if $P_0 = B$, we also have $P_2 = B$.Therefore $P_2 = P_3 = B$, but this is only possible if $B = A_3$. Since $\\Delta A_1A_2B$ is equilateral by definition, we are done. $\\square$" } { "Tag": [ "MATHCOUNTS", "summer program", "MathPath", "AMC 8" ], "Problem": "how do i prepare for mathcounts chapters?", "Solution_1": "Just do a bunch of problems. The mathcounts website has a bunch of problems from previous years and it might be a good idea to get hold of warmups and workouts from your teacher.", "Solution_2": "what i do is the night before the competition, stay up all night. make sure to bring lots of mt dew. this isnt a joke. then your body will keep you up long enough for the competition, but you will be so tired that your body just works on instinct so i make less mistakes.", "Solution_3": "[quote=\"Walk Around The River\"]what i do is the night before the competition, stay up all night. make sure to bring lots of mt dew. this isnt a joke. then your body will keep you up long enough for the competition, but you will be so tired that your body just works on instinct so i make less mistakes.[/quote] That approach only works if you're REALLY, REALLY confident with the material. Instinct can trick you, and if you are tired, you're more likely to misread the question or make stupid, mistakes. I'd recommend getting a lot of sleep and trying to be as alert and ready for the competition as possible. That way you can pay attention to detail and still work quickly.", "Solution_4": "what i'd do\r\n______________________________________________________\r\nwrite a summary of small things like easy formulas, how to do stuff like permutations, pythag., etc. and just look over it the day before. try not to do any problems\r\n\r\null get too used to those, u just want to LOOK OVER the general concept.\r\n\r\n-jorian", "Solution_5": "read aops v.1.\r\nthen practice by doing past problems, and understand the solutions and concepts, so that when you see it again, you won't miss it.\r\nwork on making as few careless mistakes as possible.", "Solution_6": "\"i have a MC competition today??\"\r\n\r\n :rotfl: \r\n\r\nat one time i almost forgot until it was 9 pm friday and my dad reminded me", "Solution_7": "I STRONGLY discourage you do the mountain dew thing. \r\n\r\nStrategy during the prep time months before competition: Use Red Bull or caffeinated energy drink of choice and madly do problems.\r\n\r\nStategy right before competition: If you haven't prepared enough, that's too bad. You should relax anyways though, because cramming in an extra formula is not worth as much as having a good state of mind so you can think of a good method to do a few problems.", "Solution_8": "You might check out this web site for more info.\r\n\r\n[url]http://www.geocities.com/danb80/PrepForContest.html[/url]", "Solution_9": "I really don't know how to make myself have a \"good day\" or a \"bad day\". I wish I did. Sometimes on practices I just do horrible, but sometimes I do really well. The most noticeable difference is on countdown. I think sleep and exercise are factors but I'm not sure what other factors there are.\r\n\r\nFor preparing, do lots of problems. But test makers are making problems that have never been seen before, so doing a lot of problems will make you good at those old problems but not necessarily on the real test, because you've never seen the concepts of those problems. I doubt it will matter much at chapters though so just do old tests.", "Solution_10": "Ubemaya, stop lying\r\nthe worst you have ever gotten on a practice was a 43.\r\nand any way, i agree that just doing past rounds won't always help.\r\nits good to just do tough problems that really open up your mind. By doing so, your mind will be trained to solve tough problems you have never seen before", "Solution_11": "Are you allowed to bring a sheet with a bunch of formulas on it? :P Because I was planning to write out all the primes under 100 because I always forget them.", "Solution_12": "[quote=\"laughinghead505\"]Are you allowed to bring a sheet with a bunch of formulas on it? :P Because I was planning to write out all the primes under 100 because I always forget them.[/quote]\r\n\r\nI wish...", "Solution_13": "Well, they would never know if you just take a piece of scratch paper and write them out right before it starts, would they? :P", "Solution_14": "[quote=\"kyyuanmathcount\"]I STRONGLY discourage you do the mountain dew thing. \n\nStrategy during the prep time months before competition: Use Red Bull or caffeinated energy drink of choice and madly do problems.\n\nStategy right before competition: If you haven't prepared enough, that's too bad. You should relax anyways though, because cramming in an extra formula is not worth as much as having a good state of mind so you can think of a good method to do a few problems.[/quote]\r\n\r\nwhat if you can't get your hands on an energy drink :P ?\r\n\r\n-jorian", "Solution_15": "Then you eat sugar. :D Although I don't think I should do that this year because when I do, I usually get so phsyced up and hyper that I get a headache and waste some of the team time talking. :(", "Solution_16": "[quote=\"laughinghead505\"]Then you eat sugar. :D Although I don't think I should do that this year because when I do, I usually get so phsyced up and hyper that I get a headache and waste some of the team time talking. :([/quote]\r\nAnd you end up tired at the end. I recommend Propel if you need something to keep you up-it isn't entirely energy drink and people need water.\r\nIt's like those gummy vitamins!\r\nNever eat sugar you get at hotels. That is my only warning if you will eat sugar.", "Solution_17": "so, any other solutions to my energy drink crisis?\r\n\r\n-jorian\r\n\r\nEDIT: I THINK I CAN GET ONE (I ASKED MY MOM :P )\r\n\r\n-jorian (again)", "Solution_18": "i go to http://www.saab.org", "Solution_19": "[quote=\"Rafaelloaa\"]i go to http://www.saab.org[/quote]\r\nTHe problem with saab is that the problems get really repetative after a while.", "Solution_20": "I do no problems the day of the comp. until the comp. the week b4, I do TONS of problems. my team, as i believe i've stated, wrote a song last year during the team. it wasn't bad. i've got it written somewhere. i'll post i if i find it.", "Solution_21": "If we have extra time this year, I'm going to make my team memorize the pi song (copyrighted to SHBoB, MathPath 2006 :lol: )\r\n\r\nTo the tune of Yankee Doodle:\r\n\r\n3.1415926535897932384626433\r\n :D", "Solution_22": "during a practice chapter team, we wrote a song with 16 minutes left :P\r\n\r\nwe got screwed, we ended up with a 9 :blush: i mean, i did better on the amc 8!!!!!!!!!!!!!!!!!!!\r\n\r\n-jorian", "Solution_23": "so your team did all ten atleast 2 times in four minutes?", "Solution_24": "sort of. :P\r\n\r\nexcept for one which we purposely held off til the last minute as a joke. i think we got THAT one right\r\n\r\n-jorian" } { "Tag": [ "trigonometry" ], "Problem": "In a right triangle a = 30 yards and tan A = 2. Find b and c\r\n\r\ndoes the pythagero theorm play a role in this question??", "Solution_1": "[hide]$\\tan A=2=\\frac ab$\n$2=\\frac{30}{b}$\n$b=15$\nUse Pythagorean to get $c=15\\sqrt5$[/hide]", "Solution_2": "[hide=\"Solution\"]\n\nDraw a right triangle and label one of the acute vertices A.\n\nIts tangent is 2. Since you know that the length of its opposite is 30, you know that the adjacent side is $15$.\n\nThe hypotenuse is then $15\\sqrt{5}$.\n\nSo, you know all the sides.[/hide]" } { "Tag": [ "geometry", "perpendicular bisector", "geometry proposed" ], "Problem": "Let $ ABCD$ be a quadrilateral inscribed in a circle $ (O)$ and let $ x,y$ be the perpendicular bisector of the of the line segments $ OC,OD$ respectively. $ S$ is a point on the circle $ (O)$. The line $ SA$ meets $ y$ at $ I$, the line $ SB$ meets $ a$ at $ J$.Prove that the line $ IJ$ is always tangent to a fixed circle when $ S$ moves on the circle $ (O)$\r\nP/s:I have just edited. Thank arqady", "Solution_1": "[quote=\"Brown\"]Prove that the line $ IJ$ is always tangent to a fixed circle when $ M$ moves on the circle $ (O)$[/quote]\r\n$ M???$ What is this? \r\nDo you mean the following?\r\nProve that the line $ IJ$ is always tangent to a circle $ (M)$ when $ M$ moves on the circle $ (O)$ :wink:\r\nIf so it's obvious.", "Solution_2": "Dear Mathlinkers,\r\nThis problem is from Mathematics and Youth magazine- a Mathematic magazine in Viet Nam, the latest number, so I think that there should be no discussions about this problem until the next 2 months. Hope that we will all respect the fair in this magazine's contest.\r\nMathVNpro" } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "Find the cardinality of $ X\\equal{}\\{f | f: \\mathbb R \\rightarrow \\mathbb R, bijective\\}$.", "Solution_1": "$ |Sym(M)| \\equal{} 2^{|M|}$ for infinite sets $ M$.", "Solution_2": "How can you prove this result? :huh:" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Given a convex quadrilateral $ABCD$, in which:\r\n\r\n$\\angle BAC = 30^\\circ$\r\n$\\angle CAD = 20^\\circ$\r\n$\\angle ABD = 50^\\circ$ \r\n$\\angle DBC = 30^\\circ$\r\n\r\nIf $P = AC \\cap BD$, show that $PC = PD$\r\n\r\nCan you find a [i]synthetic[/i] proof?\r\n\r\n(A popular trigonometric proof can be found here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=79800[/url])", "Solution_1": "It's been almost a week, and unfortunately there are no answers. :( I know of 2 synthetic proofs for this nice problem; I will wait a little longer and post them (hopefully) in a few days." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Fix the series $a_{n}=\\sum_{k=1}^{n}\\frac{(-1)^{k+1}}{k}=\\frac{1}{1}-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+.....$\r\nShow that for every $x \\in \\mathbf{R}$ there is a bijection $\\pi: \\mathbf{N}\\to \\mathbf{N}$ such that the series $\\sum_{k=1}^{\\infty}\\frac{(-1)^{\\pi(k)+1}}{\\pi(k)}=x$.", "Solution_1": "A simple application of the classic theorem on conditionally convergent series. The construction:\r\nTake positive terms starting at the beginning until their sum is greater than $x$. Take negative terms starting at the beginning so that the sum is less than $x$. Take positive terms starting from the first excluded until the sum is greater than $x$. Repeat indefinitely.\r\nThere's some work to do to prove that the construction works and the constructed series converges to $x$. It's not very hard.\r\n\r\nFor $x=1.1$, the series produced this way is\r\n$1+\\frac13-\\frac12+\\frac15+\\frac17-\\frac14+\\frac19+\\frac1{11}-\\frac16+\\frac1{13}+\\frac1{15}-\\frac18+\\frac1{17}+\\frac1{19}+\\frac1{21}-\\frac1{10}+\\cdots$" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Prove it. :lol: (This is significantly harder than [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26559]this[/url] ... ).", "Solution_1": "As I mentioned there, I believe the solution in post number $3$ of [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=166331#p166331]this topic[/url] can easily be extended to solve this problem by working with $I$ and $\\{\\frac 1x|x\\in I\\}$ for intervals $I$ which do not contain $0$ instead of $I$ and $-I$, but I didn't check it thoroughly. I might be wrong." } { "Tag": [ "induction" ], "Problem": "Prove by induction the following:\r\n\r\nSuppose you are in a country that has bank notes of value 3 dollars and 5 dollars only. Prove that it is possible to pay (without requiring change) for any item which is worth a whole number of dollars greater than 7 dollars.\r\n\r\n*The one thing I do know is n >= 8.\r\n\r\nThank you for your help in advance.", "Solution_1": "Chicken McNugget Theorem.\r\n\r\nGo look it up or ask somebody else.", "Solution_2": "Chicken McNugget, great name :rotfl: ! (OK, I just googled it, it DOES exist, I'd just never heard of it before).\r\n\r\nAnyway, here is my (pretty detailed) proof:\r\n\r\n$8 = 5+3$\r\n$9 = 3+3+3$\r\nNow, every number is 0, 1 or 2 (mod 3), which means they are all 0, -2 or -1 (mod 3).\r\nSince\r\n$5 \\equiv-1$ (mod 3)\r\nevery number x can be written as a sum of five and three\r\n$x \\equiv 0$ (mod 3) then you don't need 5\r\n$x \\equiv-1$ (mod 3) then you need one 5\r\n$x \\equiv-2$ (mod 3) then you need two 5\r\nSince every number left is greater than or equal to 10, this is not a problem.", "Solution_3": "I need to solve the problem using a base case and then assuming k + 1 is true. I am being taught to do it this way and in steps like below.\r\n\r\nProve P(n): n = 3*a + 5*b or n >= 8\r\n\r\nStep 1: Let n = 8\r\n\t8 = 3*1 + 5*1\r\n\t8 + 8\r\n\r\nStep 2: Assume P(k) is true for n = k.\r\n\tThen, k = 3*a + 5*b\r\n\r\nStep 3: Must show P(k+1) is true.\r\n\tThat is n = k +1\r\n\r\nFrom here I have trouble. I have been looking at it and trying a few things, but I don\u2019t get anywhere. I need it broke down step by step if I could. Also, why do you show cases 9 and 10?\r\n\r\nThank You!", "Solution_4": "[quote=\"MathStudent2\"]I need to solve the problem using a base case and then assuming k + 1 is true. I am being taught to do it this way and in steps like below.\n\nProve P(n): n = 3*a + 5*b or n >= 8\n\nStep 1: Let n = 8\n\t8 = 3*1 + 5*1\n\t8 + 8\n\nStep 2: Assume P(k) is true for n = k.\n\tThen, k = 3*a + 5*b\n\nStep 3: Must show P(k+1) is true.\n\tThat is n = k +1\n\nFrom here I have trouble. I have been looking at it and trying a few things, but I don\u2019t get anywhere. I need it broke down step by step if I could. Also, why do you show cases 9 and 10?\n\nThank You![/quote]\r\n[hide=\"using induction\"]\nThe base case $n=8=3+5$ is clearly true. For the inductive step (assume true for $n$, prove for $n+1$), write $n=3a+5b$ for nonnegative integers $a,b$. If $b\\neq 0$, then $n+1=3(a+2)+5(b-1)$. Otherwise, $n=3a$, so $a\\geq 3$ and $n+1=3(a-3)+5\\cdot 2$.\n[/hide]", "Solution_5": "[quote=\"scorpius119\"][quote=\"MathStudent2\"]I need to solve the problem using a base case and then assuming k + 1 is true. I am being taught to do it this way and in steps like below.\n\nProve P(n): n = 3*a + 5*b or n >= 8\n\nStep 1: Let n = 8\n\t8 = 3*1 + 5*1\n\t8 + 8\n\nStep 2: Assume P(k) is true for n = k.\n\tThen, k = 3*a + 5*b\n\nStep 3: Must show P(k+1) is true.\n\tThat is n = k +1\n\nFrom here I have trouble. I have been looking at it and trying a few things, but I don\u2019t get anywhere. I need it broke down step by step if I could. Also, why do you show cases 9 and 10?\n\nThank You![/quote]\n[hide=\"using induction\"]\nThe base case $n=8=3+5$ is clearly true. For the inductive step (assume true for $n$, prove for $n+1$), write $n=3a+5b$ for nonnegative integers $a,b$. If $b\\neq 0$, then $n+1=3(a+2)+5(b-1)$. Otherwise, $n=3a$, so $a\\geq 3$ and $n+1=3(a-3)+5\\cdot 2$.\n[/hide][/quote]\r\n\r\nWhere does (a+2) and (b-1) come from? Thank you!", "Solution_6": "$n=3a+5b$\r\n$n+1=3a+5b+1$\r\n$n+1=3a+6+5b-5$\r\n$n+1=3(a+2)+5(b-1)$", "Solution_7": "Okay, I add one to both sides, then we get:\r\nk + 1 = 3a + 6 + 5b - 5, another way to write\r\nk + 1 = 3a + 5b + 1, then next line is just factoring out a 3 and 5.\r\n\r\nSo, I am still not sure where to go from here:\r\nk + 1 = 3(a + 2) + 5(b - 1)?\r\n\r\nThanks,", "Solution_8": "[quote=\"MathStudent2\"]Okay, I add one to both sides, then we get:\nk + 1 = 3a + 6 + 5b - 5, another way to write\nk + 1 = 3a + 5b + 1, then next line is just factoring out a 3 and 5.\n\nSo, I am still not sure where to go from here:\nk + 1 = 3(a + 2) + 5(b - 1)?\n\nThanks,[/quote]\r\n\r\nBasically, he was just showing that if k can expressed as the sum of 3's and 5's, then k+1 can also be expressed as the sum of 3's and 5's.", "Solution_9": "Thank you tjhance. Where do I go from here: k + 1 = 3(a + 2) + 5(b - 1) to complete the proof. Thanks!", "Solution_10": "I don't quite see what's left to prove, except the case where $b=0$. If so, then $k=3a+5(0)=3a$. $k\\ge 9$ so $a\\ge 3$, so $k+1=3(a-3)+5(2)$ would then work, as shown by scorpius119.", "Solution_11": "So I need to do induction on three cases: k, k+1, or k+2? And, the fact that it is k + 1 and k + 2 is because 8 +1 = 9 and 8 + 2 = 10? Just trying to clear up my thinking process. Thanks!", "Solution_12": "we proved that if it is true for $k$ then it is true for $k+1$. Then, since it's true for $k+1$, we automatically know it is true for $k+2$. And then $k+3$. and so on, to infinite.", "Solution_13": "Okay, I am understanding the problem a little better. I am still looking at it though. Why do you set b = 0 and solve? Why not a = 0?\r\n\r\nWhen b = 0, I see we get n = 3a. So, just to clarify my thoughts we have a >= 3 because n has to be >= 8 giving us n = 9? Am I correct on my thoughts here?\r\n\r\nThanks again for your help!", "Solution_14": "[quote=\"MathStudent2\"]Okay, I am understanding the problem a little better. I am still looking at it though. Why do you set b = 0 and solve? Why not a = 0?\n\nWhen b = 0, I see we get n = 3a. So, just to clarify my thoughts we have a >= 3 because n has to be >= 8 giving us n = 9? Am I correct on my thoughts here?\n\nThanks again for your help![/quote]\r\n\r\nWell, $3a=n\\geq 8\\Rightarrow a\\geq \\frac{8}{3}$. And $a$ is an integer, that gives $\\boxed{a\\geq 3}$ ($n$ doesn't necessarily equal 9 though; you can't have $3a+5b=12$ unless $b=0$)", "Solution_15": "I have a question in regards to the second case below:\r\n\r\n2) Assume that d is zero.\r\nThen, 3*a + 5*b = k (Ind. Hyp.)\r\n3*a = k\r\nBecause k >= 8, then x >= 3\r\nThen, 3*a + 5*b + 1 = k + 1\r\n3*a \u2013 9 +5*b + 10 = k + 1\r\n3(a \u2013 2) + 5(b + 2) = k + 1\r\n\r\nChoose c = a \u2013 2 and d = b + 2 and we have:\r\n3*c + 5*d = k + 1, with both in the nonnegative integers.\r\n\r\nOR\r\n\r\n3) Assume that d is zero.\r\nThen, 3*a + 5*b = k (Ind. Hyp.)\r\n3*a = k\r\nBecause k >= 8, then x >= 3\r\nThen, 3*a + 1 = k + 1\r\n3*a \u2013 9 + 10 = k + 1\r\n3(a \u2013 3) + 5*2 = k\r\n\r\nSince I am trying to show the form 3*c + 5*d = k + 1 I am not sure which of the two above would be the better one to show. The first one, I am assuming d is zero and then deriving down to 3(a \u2013 2) + 5(b + 2) = k + 1 so that c = a \u2013 2 and d = b + 2.\r\n\r\nOn the second one, I only end up with c = a \u2013 3 and no d, but I did assume is zero. \r\n\r\nNot sure which I should use? Any suggestions?\r\n\r\nThank you for all of your help throughout this problem!" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c\\geq 0$ s.t. $ abc\\equal{}1$.Prove that:\r\n$ (\\dfrac{a\\plus{}b\\plus{}c}{3})^5\\geq\\dfrac{a^2\\plus{}b^2\\plus{}c^2}{3}$", "Solution_1": "I think SMV might be useful", "Solution_2": "[quote=\"Inequalities Master\"]Let $ a,b,c\\geq 0$ s.t. $ abc \\equal{} 1$.Prove that:\n$ (\\dfrac{a \\plus{} b \\plus{} c}{3})^5\\geq\\dfrac{a^2 \\plus{} b^2 \\plus{} c^2}{3}$[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=87919" } { "Tag": [], "Problem": "Find the positive difference between the largest and the second largest elements of the set $ \\left\\{\\frac13,\\frac3{10},\\frac15,\\frac14\\right\\}$. Express your answer as a common fraction.", "Solution_1": "It's not immediately obvious which ones are largest, smallest, etc.. So put everything over a common denominator. The set becomes $ \\{\\frac{20}{60}, \\frac{18}{60}, \\frac{12}{60}, \\frac{15}{60}\\}$. Now it becomes easy to see. The largest is $ \\frac{20}{60}$ and the second-largest is $ \\frac{18}{60}$. Taking their difference, we have $ \\frac{20}{60} \\minus{} \\frac{18}{60} \\equal{} \\frac{2}{60} \\equal{} \\boxed{\\frac{1}{30}}$" } { "Tag": [ "search", "combinatorics theorems", "combinatorics" ], "Problem": "Why $ \\binom{2n} {n}$ is divisible by n+1?\r\n\r\nThank you very much", "Solution_1": "$ \\binom{2n}{n}\\equal{}\\frac{(2n)!}{n!(n!)}\\equal{}\\frac{2n\\cdot(2n\\minus{}1)\\cdots(n\\plus{}1)}{n!}$, which as you see, is divisible by $ n\\plus{}1$.", "Solution_2": "The question has been asked several times here.\r\n\r\nSo, you want to prove that Catalan numbers are integers.\r\n\r\nTake $ A \\equal{} {2n \\choose n}$ and $ B \\equal{} {2n \\choose n \\plus{} 1}$\r\n\r\nNow, it's easy to show that $ C \\equal{} A \\minus{} B \\equal{} \\frac {(2n)!}{n!(n \\plus{} 1)!} \\equal{} \\frac {{2n \\choose n}}{{n \\plus{} 1}}$.\r\n\r\n$ A$ and $ B$ are integers, so $ C$ is an integer too, qed.", "Solution_3": "$ \\binom{2n}{n}\\equal{}\\frac{2n\\cdot(2n\\minus{}1)\\cdots(n\\plus{}1)}{n!}$ that is divisible by $ n\\plus{}1$", "Solution_4": "BTW, here are two more proofs (Search button is your friend :lol: ):\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=18917\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=167490\r\n\r\nP. S. Desmo90, haven't you noticed that your one line proof is exactly what JRav already wrote? :)", "Solution_5": "[quote=\"hsiljak\"] P. S. Desmo90, haven't you noticed that your one line proof is exactly what JRav already wrote? :)[/quote]\r\n\r\nno, i'm sorry. :oops:", "Solution_6": "Well, no problems. But if you like, you could show why does the fact you wrote holds. It isn't completely trivial, and the proof could be useful for the O. P. to understand the logic of binomial divisibility and so ;)", "Solution_7": "[quote=\"JRav\"]$ \\binom{2n}{n} \\equal{} \\frac {(2n)!}{n!(n!)} \\equal{} \\frac {2n\\cdot(2n \\minus{} 1)\\cdots(n \\plus{} 1)}{n!}$, which as you see, is divisible by $ n \\plus{} 1$.[/quote]\r\nThis isn't a proof. Once you remove that term from the numerator, you have no guarantee that $ n!$ divides what's left except in the case that $ n \\plus{} 1$ is prime.\r\n\r\nOne straightforward way to do this is to define the [url=http://en.wikipedia.org/wiki/Catalan_number#Proof_of_the_formula]Catalan numbers[/url] in a combinatorial fashion and then show that they are given by this formula, from which it immediately follows that the formula must return integer values." } { "Tag": [ "AMC", "AIME", "MATHCOUNTS", "USA(J)MO", "USAMO" ], "Problem": "I need:\r\n Complete handbook solutions for 2001-02, 2002-03, 2003-04, 2005-06.\r\n\r\nI have:\r\n National 1984-2008, missing some countdowns\r\n State 1991-2008, missing 95,96,97,98 team\r\n Chapter 2000-08\r\n school 2006-08\r\n Handbook 02-08 with answer keys\r\n\r\nPM me if you want to trade.\r\n\r\nThanks!", "Solution_1": "Hi, I am a new member to AOPS and basically new to this whole trading thing. What do I do if I don't have ANYTHING. How do I go about aquiring new tests from Nationals and Older States and the handbooks. I am deperate but not really sure how to start. How did some of you guys start aquiring tons of stuff?\r\n\r\nThe way I see it, once I actually have some stuff to trade, I can aquire more and actually trade. If I don't have anything, I can't trade so I am stuck.\r\n\r\nCan someone PLEASE HELP!!", "Solution_2": "Please post all trading requests in this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189]thread[/url]", "Solution_3": "What I have:\r\n\r\n2000-2005 AMC 10s\r\n1996-2005 AMC 12s\r\n2001-2005 AIMEs\r\n\r\n2000-2008 State Rounds\r\n2005-2008 Chapter Rounds\r\n08-09 Mathcounts Handbook\r\n1999-2005 USAMOs\r\n\r\nI have solutions for all of them\r\n\r\nI AM DESPERATE FOR MC HANDBOKS, PM IF INTRESTED< I AM A NEWB SO I NEED HELP!!!!!!!!!!!!!!!!!!!!11", "Solution_4": "First of all, don't capitalize all you words and add so many exclamation points; it won't persuade others to help you.\r\n\r\nYour best bet is to post on forums like these and see if anyone else wants to trade with you. Make your own topic; people will notice it more.", "Solution_5": "[quote=\"akhil0422\"] Make your own topic; people will notice it more.[/quote]\r\n\r\nBut that is also spam when we have a topic intended for it.", "Solution_6": "Sorry, I didn't see that. I don't have anything to trade so I [b]don't[/b] trade (great logic isn't it).", "Solution_7": "Didn't I gave you 2001-2?" } { "Tag": [ "group theory", "abstract algebra", "invariant", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ G$ be a finite simple group of even order. Prove that $ \\mbox{Inn}(G) \\stackrel{char}{\\trianglelefteq} \\mbox{Aut}(G)$.", "Solution_1": "What is $ \\text{Inn}(G)$ ? And what is the $ char$?...", "Solution_2": "Oh, I'm sorry. $ \\mbox{Inn}(G)$ is the group of inner automorphisms of $ G$, while $ \\mbox{Aut}(G)$ is the group of all automorphisms of $ G$. \r\n\r\nWell, the definition of a normal subgroup $ N$ is a subgroup which is invariant under all inner automorphisms, i.e. all elements of $ \\mbox{Inn}(G)$. \r\nWe denote $ N \\trianglelefteq G$.\r\nA characteristic subgroup is something stronger, since it is a subgroup $ N$ which is invariant under all automorphisms, i.e. all elements of $ \\mbox{Aut}(G)$. \r\nWe denote $ N \\stackrel{char}{\\trianglelefteq} G$.\r\n\r\nNote that it is well known (or otherwise quickly proved) that $ \\mbox{Inn}(G) \\trianglelefteq \\mbox{Aut}(G)$. The problem asks to prove that for finite simple groups of even order, it is even a characteristic subgroup.", "Solution_3": "Here's a hint: First prove that $ G$ is generated by it's involutions. Then prove that $ C_{\\mbox{Aut}(G)}(\\mbox{Inn}(G))$ is trivial.", "Solution_4": "Ok, since nobody has solved this problem, I will post the solution myself. This question is from the $ 2005$ exam of Group Theory here at Ghent University (UGent)\r\n\r\nWe can easiley prove that $ C_{\\mbox{Aut}(G)}(\\mbox{Inn}(G))$ is trivial. Prove this by contradiction, and you will get that $ Z(G)$ contains a non-trivial element, which is impossible since $ G$ is simple and can obviously not be abelian (unless $ G \\cong C_2$ but we can check this case)\r\n\r\nIt is well known that $ \\mbox{Inn}(G) \\trianglelefteq \\mbox{Aut}(G)$. \r\n\r\nAssume that there exists an element in $ \\mbox{Aut}(\\mbox{Aut}(G))$ that does not fix $ \\mbox{Inn}(G)$, and maps it to $ H$. Obviously, \\[ H \\trianglelefteq \\mbox{Aut}(G),\\hspace{1,5 cm} \\mbox{Inn}(G) \\cap H \\trianglelefteq \\mbox{Inn}(G).\\] \r\n\r\nBecause $ Z(G)$ has to be trivial, we know that \\[ G \\equal{} G \\slash Z(G) \\cong \\mbox{Inn}(G).\\]\r\nIt follows that $ \\mbox{Inn}(G) \\cap H \\equal{} \\left\\{e\\right\\}$. Now, $ H$ and $ \\mbox{Inn}(G)$ are two normal subgroups of $ \\mbox{Aut}(G)$, and they have trivial intersection. \r\n\r\nThis implies that $ H$ and $ \\mbox{Inn}(G)$ commute, which is impossible since $ C_{\\mbox{Aut}(G)}(\\mbox{Inn}(G))$ is trivial." } { "Tag": [ "\\/closed" ], "Problem": "How hard would it be to program in a multi-option poll (ie you can choose more than one option if you want). Would it be easy enough to just have a option to replace the radio button (I think they're called radio buttons anyway) by checkboxes? You'd want both options, of course.", "Solution_1": "I'll see if phpBB has anything for it." } { "Tag": [], "Problem": "A group of people went to a restaurant for lunch. They agreed to split the luncheon bill equally. Later, when the bill arrived, three of them discovered that they had forgotten their money. The others in the group agreed to make up the difference by each paying an extra $ 1.80. If the total bill was$78.00,\r\nhow many people were in the group?", "Solution_1": "[hide] Let $ x$= the number of people, and $ y$= the price each person was to originally pay. Then,\n\n$ (x \\minus{} 3)(y \\plus{} 1.8) \\equal{} xy \\implies 1.8x \\minus{} 3y \\equal{} 5.4$\n\nSince $ xy \\equal{} 78$, we can say that $ y \\equal{} \\frac {78}{x}$.\n\nSo,\n\n$ 1.8x \\minus{} 3(\\frac {78}{x}) \\equal{} 5.4$\n\n$ \\implies x^2 \\minus{} 3x \\minus{} 130 \\equal{} 0$\n\n$ \\implies (x \\minus{} 13)(x \\plus{} 10) \\equal{} 0 \\implies x \\equal{} \\boxed{13}$.[/hide]" } { "Tag": [], "Problem": "1) When you play [url=http://blackjack.how2gamble.com]blackjack[/url],Which dealer's up-card(s) will cause him to bust more than 50% of the time? ? do you know?\r\nthe 6 \r\nthe 5 and 6 \r\nthe 4, 5 and 6 \r\nthere aren't any", "Solution_1": "Can a moderator please delete this topic? This is like the 10th time that Kotsba has been advertising for online gambling sites." } { "Tag": [ "induction" ], "Problem": "Find the general term of $ a_{n}$ such that $ a_{n+1}=\\sum_{k=1}^{n}a_{k}+n^{2}-n+2\\ (n=1,\\ 2,\\ \\cdots),\\ a_{1}=2$.", "Solution_1": "[quote=\"kunny\"]Find the general term of $ a_{n}$ such that $ a_{n+1}=\\sum_{k=1}^{n}a_{k}+n^{2}-n+2\\ (n=1,\\ 2,\\ \\cdots),\\ a_{1}=2$.[/quote]\r\n\r\n\r\nPretty easy to bash out the formula and then prove it using induction :P\r\n\r\n$ n^{2}-n+2+\\sum_{k=1}^{n}a_{k}$\r\n\r\n$ a_{n}=2(2^{n}-n)$ holds true for $ n=1$.\r\n\r\n\r\nSo we simply need to show that:\r\n\r\n$ 2[2^{n+1}-(n+1)]=n^{2}-n+2+\\sum_{k=1}^{n}2^{k+1}-\\sum_{k=1}^{n}2k$\r\n\r\n$ 2^{n+2}-2n-2=n^{2}-n+2+2^{n+2}-4-n(n+1)$\r\n\r\n$ 2^{n+2}-2n-2=2^{n+2}-2n-2$.", "Solution_2": "That's right, but you can find $ a_{n}$ by solving this reccursive directly. :P", "Solution_3": "Hmmm I have no clue how to do that. Demonstrate?", "Solution_4": "Let $ S_{n}=\\sum_{k=1}^{n}a_{k}$, $ a_{n+1}=S_{n}+n^{2}-n+2$, so $ a_{n}=S_{n-1}+(n-1)^{2}-(n-1)+2$. Subtracting of both sides gives $ a_{n+1}-a_{n}=......$", "Solution_5": "Remark that \\[ \\begin{aligned}a_{n+1}& = a_{n}+n^{2}-n+2+\\sum_{k=1}^{n-1}a_{k}\\\\ & = a_{n}+n^{2}-n+2+(a_{n}-(n-1)^{2}+(n-1)-2) = 2a_{n}+2n-2\\end{aligned}\\]Therefore, $ a_{n+1}+2(n+1) = 2(a_{n}+2n)$, so $ a_{n}+2n = 2^{n-1}(a_{1}+2) = 2^{n+1}$, i.e. $ a_{n}= 2(2^{n}-n)$.", "Solution_6": "That's right! :)" } { "Tag": [ "inequalities", "superior algebra", "superior algebra theorems" ], "Problem": "Could someone please write or give a reference what is Schur-Zassenhaus Theorem(in group theory)", "Solution_1": "From google:\r\n[quote] ...And I came across something pretty nifty in a book called \"Noncommutative Algebra\" by Farb and Dennis: the Schur-Zassenhaus theorem, which says that if |H| is relatively prime to |G/H| then G is a semidirect product H X| G/H. Apparently there is a short simple proof using group cohomology. I don't know if there is a short proof with straight group theory. So cohomology of groups seems interesting since you can prove things easily...[/quote]" } { "Tag": [ "algebra", "polynomial", "function", "integration", "calculus", "calculus computations" ], "Problem": "Part a)\r\n\r\nCompute the 4th degree Taylor polynomial for the function $ f\\equal{}e^{x^2}$ at 0.\r\n\r\nPart b)\r\n\r\nApproximate $ \\int e^{x^2} dx$ and determine the accuracy of the approximation.\r\n\r\nI have done part a, and I have done the first half of part b. I am not sure about the second half. I have computed the Lagrange Remainder, but since it came out to be 0, I don't know if that means there is no error?", "Solution_1": "for values of x sufficiently close to 0, the error is approximately equal to the first left out term\r\n\r\nin this case, the $ \\frac {x^7}{42}$ term because the x^6 term is 0." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "In a triangle $ ABC$ $ AD,BE,CF$ are concurrent .also let $ P \\in EF, R \\in ED, Q \\in FD$ such that $ DP,FQ,ER$ are concurrent .Prove that $ AP,BQ,CR$ are concurrent.", "Solution_1": "Check out the Cevian Nests theorem in my post at http://www.mathlinks.ro/Forum/viewtopic.php?t=3536 .\r\n\r\n darij", "Solution_2": "thanks darij :)" } { "Tag": [ "geometry", "power of a point", "radical axis", "geometry proposed" ], "Problem": "Let BB' and CC' the altitudes of the acute triangle ABC and H its orthocenter. Prove that the projection of J on the median of the angle 0$.\r\n\r\nFind $ h(N)$", "Solution_1": "[hide=\"Brawler Style Start\"]$ \\text{det} \\left( \\begin{bmatrix} a \\plus{} c \\plus{} 2b & b & c \\plus{} 2b \\\\ c \\plus{} b \\plus{} 2a & a & b \\plus{} 2a \\\\ c \\minus{} a & c \\minus{} a & 2c \\minus{} a \\minus{} b \\end{bmatrix} \\right)$\n\n$ \\equal{} \\text{det} \\left( \\begin{bmatrix} a \\plus{} b\\plus{}c & b & c \\plus{} 2b \\\\ a\\plus{} b\\plus{}c & a & b \\plus{} 2a \\\\ 0 & c \\minus{} a & 2c \\minus{} a \\minus{} b \\end{bmatrix} \\right)$\n\n$ \\equal{} \\text{det} \\left( \\begin{bmatrix} a \\plus{} b\\plus{}c & b & c \\plus{} 2b \\\\ 0 & a\\minus{}b & 2a\\minus{}b\\minus{}c \\\\ 0 & c \\minus{} a & 2c \\minus{} a \\minus{} b \\end{bmatrix} \\right)$\n\n$ \\equal{} (a\\plus{}b\\plus{}c) \\text{det} \\left( \\begin{bmatrix} a\\minus{}b & 2a\\minus{}b\\minus{}c \\\\ c \\minus{} a & 2c \\minus{} a \\minus{} b \\end{bmatrix} \\right)$\n\n$ \\equal{} (a\\plus{}b\\plus{}c)((a\\minus{}b)(2c\\minus{}a\\minus{}b) \\minus{} (2a\\minus{}b\\minus{}c)(c\\minus{}a))$ $ \\equal{} (a\\plus{}b\\plus{}c)((a\\plus{}b\\plus{}c)^2 \\minus{} 3(ab\\plus{}bc\\plus{}ca))$ $ \\equal{} (a\\plus{}b\\plus{}c)^3 \\minus{} 3(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)$\n\nThus, $ (a\\plus{}b\\plus{}c)^3 > 3(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)$[/hide]" } { "Tag": [ "algebra", "polynomial", "integration", "calculus", "function", "domain", "real analysis" ], "Problem": "I wonder if the following statement is true or false.\r\nLet $n\\geq 1$ be an integer, $P(x_{1},x_{2},...,x_{n})$ and $Q(x_{1},x_{2},...,x_{n})$ be two real polynomials. For example, if $n=3$, $P(x,y,z)=x^{2}y+z^{3}+xz+1$ is such a polynomial. Denote by $d(P)$ the degree of $P$, i.e. the highest power of $P$. In the example above, $d(P)=3$, or if $Q(x,y,z)=x^{2}+yz^{2}$, $d(Q)=3$. \r\n[b]Statement :[/b] $\\frac{P(x_{1},x_{2},...,x_{n})}{Q(x_{1},x_{2},...,x_{n})}$ is (L-?)integrable on ${[0,1]}^{n}$ if and only if $d(Q)-d(P)\\leq n$.\r\n\r\nIt's true for $n=1$, it seems true for $n=2$... \r\nCan someone correct this, complete this, prove this or give a counterexample ?", "Solution_1": "Counterexample, consider the case when the two polynomials are polynomials in two variables.\r\n\r\n$\\int_{0}^{1}\\int_{0}^{1}\\frac{dxdy}{x^{np}}=\\int_{0}^{1}\\frac{dx}{x^{np}}=\\infty$\r\nhence $P/Q=\\frac{1}{x^{n}}$ is not $L^{p}$ integrable for all $p\\geq 1$.\r\n$degree(P)-degree(Q)=0-n=-n<2$.\r\n\r\nThe same works with $P=y$ $Q=yx^{n}$ where $n$ is a natural number.", "Solution_2": "I'm sorry, it should be d(Q)-d(P) and not d(P) - d(Q). I've edited and thanks for interest", "Solution_3": "It is still false, since if $P=1$ and $Q=x$ then \r\n$deg(Q)-deg(P)=1<2$ and the double integral is $\\infty$ hence the function $P/Q$ is not $L^{p}$ integrable for all $p>1$.", "Solution_4": "You want the one in the denominator to have no zeros in $R^{n}$ either. Anyone have counterexamples now? :)", "Solution_5": "Xevarion,\r\nI do not understand what you are saying, could you write down the statement of the problem?", "Solution_6": "Okay so, I didn't verify this carefully, but here is a modified version of what OP was asking:\r\n\r\nConsider $f(x) = \\frac{P(x_{1}, \\ldots, x_{n})}{Q(x_{1}, \\ldots, x_{n})}$\r\n\r\nAssume that \r\n\r\n1) The degree of $P$ in variable $x_{i}$ is at least 2 smaller than the degree of $Q$ in the variable $x_{i}$ for all $i$. \r\n\r\n2) If $Q(y_{1}, \\ldots, y_{n}) = 0$ for some point $(y_{1}, \\ldots, y_{n})$, then $P(y_{1}, \\ldots, y_{n}) = 0$ also and the degree of the root of $P$ is at least as large as the degree of the root of $Q$. In other words, $f$ has no singularities in the domain of integration. \r\n\r\nThen $f(x)$ is in $L^{p}$ all $p$ on all of $R^{n}$.", "Solution_7": "a proof for this ?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "With $ a,b,c$ are three lenghs of sides of a triangle such that $ abc\\equal{}2$ Prove that\r\n $ \\sqrt{a\\plus{}b\\minus{}c}\\plus{} \\sqrt{a\\plus{}c\\minus{}b}\\plus{} \\sqrt{b\\plus{}c\\minus{}a} \\geq 2 \\sqrt{2}$", "Solution_1": "Substitute $ a\\equal{}x\\plus{}y$ ,$ b\\equal{}y\\plus{}z$ and $ c\\equal{}z\\plus{}x$.\r\nThe inequality becomes $ \\sqrt{2x} \\plus{}\\sqrt{2y}\\plus{}\\sqrt{2z} \\geq 2\\sqrt{2}$ $ \\implies$ $ \\sqrt{x} \\plus{}\\sqrt{y}\\plus{}\\sqrt{z} \\geq 2$\r\n\r\n$ \\sqrt{x} \\plus{}\\sqrt{y} \\plus{}\\sqrt {z} \\geq (x\\plus{}y)(y\\plus{}z)(z\\plus{}x)$\r\n\r\nI can not continue :blush:", "Solution_2": "[quote=\"2424\"]With $ a,b,c$ are three lenghs of sides of a triangle such that $ abc \\equal{} 2$ Prove that\n $ \\sqrt {a \\plus{} b \\minus{} c} \\plus{} \\sqrt {a \\plus{} c \\minus{} b} \\plus{} \\sqrt {b \\plus{} c \\minus{} a} \\geq 2 \\sqrt {2}$[/quote]\r\n$ a \\equal{} x^2 \\plus{} y^2, b \\equal{} y^2 \\plus{} z^2, c \\equal{} z^2 \\plus{} x^2$ for some $ x,y,z \\ge 0$ since $ a,b,c$ is length of triangles.\r\nThen we should prove:\r\n$ x \\plus{} y \\plus{} z \\ge 2$ when $ (x^2 \\plus{} y^2)(y^2 \\plus{} z^2)(z^2 \\plus{} x^2) \\equal{} 2$\r\nAssume that $ x \\plus{} y \\plus{} z < 2$ when $ (x^2 \\plus{} y^2)(y^2 \\plus{} z^2)(z^2 \\plus{} x^2) \\equal{} 2$. Then by increasing $ x,y,z$ we can get $ x \\plus{} y \\plus{} z \\equal{} 2$ and $ (x^2 \\plus{} y^2)(y^2 \\plus{} z^2)(z^2 \\plus{} x^2) > 2$. So it is enough to prove that:\r\n$ (x^2 \\plus{} y^2)(y^2 \\plus{} z^2)(z^2 \\plus{} x^2) \\le 2$ when $ x \\plus{} y \\plus{} z \\equal{} 2$.(This is btw. easily solved with the uvw-method)\r\nWlog assume $ x \\le y,z$. Let $ Y \\equal{} y \\plus{} \\frac {x}{2}$ and $ Z \\equal{} z \\plus{} \\frac {x}{2}$. Then $ Z^2 \\equal{} z^2 \\plus{} xz \\plus{} \\frac {x^2}{4} \\ge z^2 \\plus{} zx \\ge z^2 \\plus{} x^2$ and likewise $ Y^2 \\ge y^2 \\plus{} x^2$. Also $ Z^2 \\plus{} Y^2 \\ge z^2 \\plus{} y^2$.\r\nSo we should only prove:\r\n$ Z^2Y^2(Z^2 \\plus{} Y^2) \\le 2$ when $ Z \\plus{} Y \\equal{} 2$. That is:\r\n$ 8Z^2Y^2(Z^2 \\plus{} Y^2) \\le (Z \\plus{} Y)^4 \\equal{} Z^4 \\plus{} 4Z^3Y \\plus{} 6Z^2Y^2 \\plus{} 4ZY^3 \\plus{} Y^4 \\iff$\r\n$ (Z \\minus{} Y)^4 \\ge 0$. QED :)", "Solution_3": "[quote=\"Mathias_DK\"][quote=\"2424\"]With $ a,b,c$ are three lenghs of sides of a triangle such that $ abc \\equal{} 2$ Prove that\n $ \\sqrt {a \\plus{} b \\minus{} c} \\plus{} \\sqrt {a \\plus{} c \\minus{} b} \\plus{} \\sqrt {b \\plus{} c \\minus{} a} \\geq 2 \\sqrt {2}$[/quote]\n$ a \\equal{} x^2 \\plus{} y^2, b \\equal{} y^2 \\plus{} z^2, c \\equal{} z^2 \\plus{} x^2$ for some $ x,y,z \\ge 0$ since $ a,b,c$ is length of triangles.\nThen we should prove:\n$ x \\plus{} y \\plus{} z \\ge 2$ when $ (x^2 \\plus{} y^2)(y^2 \\plus{} z^2)(z^2 \\plus{} x^2) \\equal{} 2$\nAssume that $ x \\plus{} y \\plus{} z < 2$ when $ (x^2 \\plus{} y^2)(y^2 \\plus{} z^2)(z^2 \\plus{} x^2) \\equal{} 2$. Then by increasing $ x,y,z$ we can get $ x \\plus{} y \\plus{} z \\equal{} 2$ and $ (x^2 \\plus{} y^2)(y^2 \\plus{} z^2)(z^2 \\plus{} x^2) > 2$. So it is enough to prove that:\n$ (x^2 \\plus{} y^2)(y^2 \\plus{} z^2)(z^2 \\plus{} x^2) \\le 2$ when $ x \\plus{} y \\plus{} z \\equal{} 2$.(This is btw. easily solved with the uvw-method)\nWlog assume $ x \\le y,z$. Let $ Y \\equal{} y \\plus{} \\frac {x}{2}$ and $ Z \\equal{} z \\plus{} \\frac {x}{2}$. Then $ Z^2 \\equal{} z^2 \\plus{} xz \\plus{} \\frac {x^2}{4} \\ge z^2 \\plus{} zx \\ge z^2 \\plus{} x^2$ and likewise $ Y^2 \\ge y^2 \\plus{} x^2$. Also $ Z^2 \\plus{} Y^2 \\ge z^2 \\plus{} y^2$.\nSo we should only prove:\n$ Z^2Y^2(Z^2 \\plus{} Y^2) \\le 2$ when $ Z \\plus{} Y \\equal{} 2$. That is:\n$ 8Z^2Y^2(Z^2 \\plus{} Y^2) \\le (Z \\plus{} Y)^4 \\equal{} Z^4 \\plus{} 4Z^3Y \\plus{} 6Z^2Y^2 \\plus{} 4ZY^3 \\plus{} Y^4 \\iff$\n$ (Z \\minus{} Y)^4 \\ge 0$. QED :)[/quote]\r\nOr you can prove $ (x^2y^2\\plus{}y^2z^2\\plus{}z^2x^2)(x^2\\plus{}y^2\\plus{}z^2) \\leq 2$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all($a,b\\in{N}$) st $\\frac{a^{b}+b}{b^{2}+9}\\in{N}$ :)", "Solution_1": "It's from Rio-plate :)\r\nAny one?", "Solution_2": "one correction:\r\n\r\nthe numerator of the fraction in the rio-plate problem isn't this!", "Solution_3": "[quote=\"e.lopes\"]one correction:\n\nthe numerator of the fraction in the rio-plate problem isn't this![/quote]\r\nYour solution?", "Solution_4": "Could anyone solve it for me,please. :blush:", "Solution_5": "If $ 3|b$, then $ 3|a$ and $ a^b\\plus{}b\\equal{}3\\mod 9$ - not solutions.\r\nFor any b, suth that $ 3\\not |b$, we can find infinetely many a, suth that $ \\frac{a^b\\plus{}b}{b^2\\plus{}9}\\in N$.\r\nFor example $ b\\equal{}1, a\\equal{}10k\\minus{}1$, $ b\\equal{}2,a\\equal{}\\pm 4 \\plus{}13k$,...", "Solution_6": "No.The problem from Rio-Plate isn't that (because tmbtw type wrong). :oops: \r\nThis is the real problem:\r\nFind all (a,b) $ \\in N*$ such that $ \\frac{a^b\\plus{}b}{ab^2\\plus{}9} \\in N*$" } { "Tag": [ "function" ], "Problem": "Find all real solutions of the following equation: \\[ (x + 1995)(x + 1997)(x + 1999)(x + 2001) + 16 = 0. \\]", "Solution_1": "Put $y=x+1998$, then we have $(y-3)(y-1)(y+1)(y+3)+16=0$, which rewrites as $(y^2-5)^2=0$, thus $y=\\pm\\sqrt{5}$ and $x=\\pm\\sqrt{5}-1998$.", "Solution_2": "[quote=\"Arne\"]Find all real solutions of the following equation: \\[ (x + 1995)(x + 1997)(x + 1999)(x + 2001) + 16 = 0. \\][/quote]\n \nOr, just note $(x+1995)(x+2001)=a, (x+1997)(x+1999)=a+8$, and we have $(a+4)^2=0,$so $a=-4$, and we get ( using second degree function) $x=\\sqrt{5}-1998, x=-\\sqrt{5}-1998$" } { "Tag": [], "Problem": "Hi have been selected for kvpy 09-10 interview.but am yet to get the interview letter.:( :( \r\nShould i hav got it by now? or Indian post has lost it?? :mad:", "Solution_1": "My friends haven't either.", "Solution_2": "I too havenot got it", "Solution_3": "neither did i receive mine" } { "Tag": [ "geometry", "inequalities", "inequalities solved" ], "Problem": "Here is a nice problem :D\r\n\r\nIf a,b,c,d are the lenghts of the sides of a convex quadriteral and if S is its area then prove or disprove the inequality:\r\n\r\n3*sqrt(S)<=p+(a*b*c*d)^(1/4). :?", "Solution_1": "Just an idea the area of a quadrilateral ABCD in term of a,b,c,d is \r\n\r\nS = sqrt( (p-a)(p-b)(p-c)(p-d) - abcd.cos^2(w/2) ) with\r\n\r\nw = 1/2(angleB + angle D) = half of oposite angle in the \r\nconvex quadrilateral\r\n\r\nThere is another formula for the area for an quadrilateral ABCD\r\nAB=a,BC=b,CD=c, DA=d\r\n\r\n\r\nS = 1/2(ab.sinB + cd.sin(D)) \r\n\r\nDid you try use these formula for your inequality ?\r\n\r\ncheers = che 2 rs", "Solution_2": "[quote=\"Lagrangia\"]If $ a,b,c,d$ are the lenghts of the sides of a convex quadrilateral and if $ F$ is its area, then prove or disprove that\n\n$ 3\\sqrt {F}\\leq s \\plus{} \\sqrt [4]{abcd}$,\n\nwhere $ s$ is the semi-perimeter.[/quote]A counterexample: $ a \\equal{} 1,b \\equal{} c \\equal{} d \\equal{} 12\\Longrightarrow$\r\n\r\n$ 3\\sqrt {F_{max}} \\equal{} \\frac {3\\sqrt [4]{76895}}{2} \\equal{} 24.978\\cdots$;\r\n\r\n$ s \\plus{} \\sqrt [4]{abcd} \\equal{} \\frac {37}{2} \\plus{} 2\\sqrt [4]{108} \\equal{} 24.947\\cdots$.\r\n\r\nBy the way, we have the following inequality\r\n\r\n$ 3\\sqrt {3}F\\leq s^2 \\plus{} \\left(3\\sqrt {3} \\minus{} 4\\right)\\sqrt {abcd}$,\r\n\r\nwith equality if and only if the convex quadrangle is a square" } { "Tag": [ "trigonometry", "calculus", "derivative", "function", "linear algebra", "matrix", "topology" ], "Problem": "Is it true that all the solutions of the equation\r\n\r\n$ x''\\equal{}1\\plus{}2\\sin x$\r\n\r\ncan be defined on the whole real axis?", "Solution_1": "if the second derivative of a function is defined at a point, wouldn't that imply the function is definined at that point too?", "Solution_2": "No, that misses the point that this is a differential equation, not a simple statement about derivatives.\r\n\r\nAs an example, the differential equation $ x'\\equal{}x^2$ has no obvious singularities, but nonetheless, other than the zero function, it has no solutions that are defined on the whole real axis.\r\n\r\nNone of which answers the question here, one way or another.", "Solution_3": "Yes, it is true. In general, you have the following theorem. If $ F: \\mathbb R^n\\to\\mathbb R^n$ is Lipschitz, then the solutions of the system $ X' \\equal{} F(X)$ exist on the entire real line. You second order equation is equivalent to the system $ x_1' \\equal{} x_2$, $ x_2' \\equal{} 1 \\plus{} 2\\sin x_1$.", "Solution_4": "I'm sorry, but what does it actually mean by $ F: \\mathbb{R}^n\\to\\mathbb{R}^n$ is Lipschitz?", "Solution_5": "A function $ f$ from a metric space $ X$ to another metric space $ y$ is Lipschitz when $ d(f(x),f(y))\\le C\\cdot d(x,y)$ for some real $ C$ and all $ x,y$. This implies continuity, of course.\r\n\r\nIn this case, the derivative of $ F$ is $ \\begin{bmatrix}0&1\\\\2\\cos x_1&0\\end{bmatrix}$. The 2-norm of this matrix is at most 2, so $ |F(x)\\minus{}F(y)|\\le 2|x\\minus{}y|$ in the Euclidean metric.\r\nWe could use other norms in $ \\mathbb{R}^n$- this might result in different Lipschitz constants, but it wouldn't change whether the functions are Lipschitz." } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "How to find $ sup|x^{n}\\minus{}x^{n\\plus{}1}| (x\\in[0,1])$ when $ n\\rightarrow\\infty$, or how to prove that\r\n if $ f_{n}(x)\\equal{}x^{n}\\minus{}x^{n\\plus{}1}$ then $ f_{n}$ is uniformly convergent when $ n\\rightarrow\\infty$?", "Solution_1": "Consider $ f\\left( x \\right) \\equal{} x^n \\minus{} x^{n \\plus{} 1}$. Then \r\n\\[ f'\\left( x \\right) \\equal{} nx^{n \\minus{} 1} \\minus{} \\left( {n \\plus{} 1} \\right)x^n \\equal{} x^{n \\minus{} 1} \\left( {n \\minus{} \\left( {n \\plus{} 1} \\right)x} \\right).\r\n\\]\r\nThis means that $ f(x)\\equal{}0$ occurs if either $ x_1\\equal{}0$ or\r\n\\[ x_2 \\equal{} \\frac{n}{{n \\plus{} 1}}.\r\n\\]\r\nIn both cases, $ f$ attains its maximum value at $ x_2$. Of course, it is also its sup.\r\n\r\nIn the last question, it is known that $ f_n \\to f$ uniformly in $ D$ iff \r\n\\[ \\mathop {\\lim }\\limits_{n \\to \\infty } \\mathop {\\sup }\\limits_{x \\in D} \\left| {f_n \\left( x \\right) \\minus{} f\\left( x \\right)} \\right| \\equal{} 0.\r\n\\]\r\nWe observe that if $ x \\in (0,1)$ then $ f_n$ converges to 0 pointwise. This mean that we just claim\r\n\\[ \\mathop {\\lim }\\limits_{n \\to \\infty } \\mathop {\\sup }\\limits_{x \\in \\left[ {0,1} \\right]} \\left| {f_n \\left( x \\right)} \\right| \\equal{} 0.\r\n\\]", "Solution_2": "if you understand right then this will be good example for you. :) \r\n$ f_{n}(x)\\equal{}x^{n}\\minus{}x^{2n}$ when $ 0 \\leq x \\leq 1$", "Solution_3": "OK.\r\nIt is obvious that $ f_{n}(x)\\rightarrow f(x)\\equal{}0$ for $ x\\in [0,1]$ when $ n\\rightarrow\\infty$. \r\n\r\nConsidernig $ f_{n}(x)\\equal{}x^{n}\\minus{}x^{2n}$ we have $ f'_{n}(x)\\equal{}nx^{n\\minus{}1}\\minus{}2nx^{2n\\minus{}1}\\equal{}nx^{n\\minus{}1}(1\\minus{}2x^{n})$.\r\nSo $ f'_{n}(x)\\equal{}0$ in points $ x\\equal{}0$ or $ x_{n}\\equal{}\\frac{1}{\\sqrt[n]{2}}$.\r\nIt means that $ f_{n}(x)$ attains its maximum in point $ x_{n}\\equal{}\\frac{1}{\\sqrt[n]{2}}$, and it's also its supremum.\r\nHence $ \\mathop{\\sup}\\limits_{x\\in\\left [ {0,1}\\right ]}|f_{n}(x)\\minus{}f(x)|\\equal{}\\mathop{\\sup}\\limits_{x\\in\\left [ {0,1}\\right ]}|f_{n}(x)|\\equal{}(\\frac{1}{\\sqrt[n]{2}})^n\\minus{}(\\frac{1}{\\sqrt[n]{2}})^{2n}\\equal{}\\frac{1}{4}$.\r\n\r\nSo $ f_{n}(x)$ is not uniformly convergent.\r\nIs this correct?", "Solution_4": "Not bad :oops:" } { "Tag": [ "function" ], "Problem": "Lets consider $ f(x,k,m) \\equal{} x(k \\minus{} x^m)$ a positive value function, where $ m$ is a positive integer, $ x$ and $ k$ are positive.\r\n\r\nA] Find the maximun value $ f$ takes in this conditions, for given $ k$ and $ m$; in other words express the maximun in therms of(the given) $ k$ and $ m$\r\n\r\nB] Find the value $ x$ must have, in order to take this maximun, also with given $ k$ and $ m$\r\n\r\nC] Find the triple $ (x,k,m)$ wich maximize this maximun, if there is one, if not explain it.\r\n\r\n\r\n :)\r\n\r\nEdited____\r\n Sorry I hope it is clear now.", "Solution_1": "I'm confused. Let $ x \\equal{} 1$, then set $ k$ to be an arbitrarily high number. There doesn't seem to be a maximum.\r\n\r\nAre $ k$ and $ m$ supposed to be fixed or something?", "Solution_2": "[hide=\"Solution to A\"]By inspection, $ f \\geq 0$ for $ 0 \\leq x \\leq \\sqrt [m]{k}$, and is otherwise negative. Clearly, the negative does not produce a maximum, so we will ignore those cases. Then, $ x$ and $ (k \\minus{} x^m)$ are positive terms.\n\nRewrite as $ f \\equal{} \\frac {1}{\\sqrt [m]{m}} \\left( (\\sqrt [m]{m} x) (k \\minus{} x^m) \\right)$.\n\nIt suffices to maximize $ (\\sqrt [m]{m} x) (k \\minus{} x^m)$, and thus to maximize $ (m x^m) (k \\minus{} x^m)^m$.\n\nBy AM-GM, $ (m x^m) (k \\minus{} x^m)^m \\leq \\left( \\frac {mx^m \\plus{} m(k \\minus{} x^m) }{m \\plus{} 1} \\right) ^{m \\plus{} 1} \\equal{} \\left( \\frac {mk}{m \\plus{} 1} \\right) ^{m \\plus{} 1}$\n\nReplacing this value into the original function:\n\n$ f \\leq \\frac {1}{\\sqrt [m]{m}} \\left( \\left( \\frac {mk}{m \\plus{} 1} \\right) ^{m \\plus{} 1} \\right)^{\\frac {1}{m}} \\iff \\boxed{ f \\leq \\frac {mk}{m \\plus{} 1} \\sqrt [m]{\\frac {k}{m \\plus{} 1}}}$\n\n[i]I hope I didn't do anything dumb in there![/i]\n[/hide]\n\n[hide=\"Solution to B\"]\nThe equality case occurs when all terms in AM-GM are equal. We require $ mx^m \\equal{} (k\\minus{}x^m) \\iff (m\\plus{}1) x^m \\equal{} k \\iff x \\equal{} \\sqrt[m]{\\frac{k}{m\\plus{}1}}$\n[/hide]" } { "Tag": [ "inequalities", "trigonometry", "geometry proposed", "geometry" ], "Problem": "In the ABC triangle \u2220C=90\u00b0 and \u2220A=\u03b1 the following inequality proof.\r\n(4+sin3\u03b1-cos3\u03b1)/(3\u221a2)>cos\u2061(\u03b1-45\u00b0)\r\n", "Solution_1": "hello, do you mean\r\n$ \\frac{(4\\plus{}\\sin(3\\alpha)\\minus{}cos(3\\alpha))}{3\\sqrt{2}}>\\cos(\\alpha\\minus{}\\frac{\\pi}{4})$?\r\nSonnhard." } { "Tag": [ "AMC" ], "Problem": "Given a regular 19-gon ABCDEFGHIJKLMNOPQRS. Consider the octagon ACDGIJNQ. In this octagon, compute the value of \r\n(a - c + d - g + i - j + n - q), where each letter is the measure (in degrees) of it's respective angle.\r\n\r\nHow much can you generalize this?", "Solution_1": "No one wants to answer my question? :(", "Solution_2": "maybe it would help if we had a picture of a 19-gon and\r\nmaybe you should run a poll and see how many people know what a enneadecagon is.(i sure don't :D )", "Solution_3": "[img]http://mathworld.wolfram.com/eimg1957.gif[/img]\r\nhttp://mathworld.wolfram.com/eimg1957.gif\r\n\r\nYou'd be better off drawing it yourself. But read the question -- you don't really have to draw the 19-gon (which is what enneadecagon means) anyhow.", "Solution_4": "If you draw a circle, you have a pretty good picture of an enneadecagon. (Thanks for teaching me a new word, Joel!)", "Solution_5": "[hide](a - c + d - g + i - j + n - q) = 0\n\nI thought of this problem as how much each angle is from the base angle of the regular polygon. To do that, I counted the number of points on the enneadecagon between the points on the octagon.(from D to G would be 3) With that, it is easy to compute that the measure of each angle is equal to the base angle (3060/19) - (180(x-2)/19) where x = the sum of the distances to the points next to the angle on the octagon.\n\n\n\nAfter I found the answer, I found that if you substitue the sum of the distances the the ajacent points on the octagon for the measure of the angle, you can plug them into your equation, and then use them in (3060/19) - (180(x-2)/19), instead of finding the measure of each individual angle.\n\n\n\nI am working on a generalization for this, for all polygons inside regular polygons.[/hide]", "Solution_6": "[quote]If you draw a circle, you have a pretty good picture of an enneadecagon. (Thanks for teaching me a new word, Joel!)[/quote]\r\n\r\nYep. If you go to the Polygon entry on mathworld, it has lots of fun names for many-sided polygons. (You know, I'm a little rusty on my greek, but I'd give good odds that \"many-sided polygons\" is a redundant way of saying the same thing twice. That last sentence wasn't nearly as clever as I had hoped it could have been. Oh, well.)\r\n\r\nGood job, bookworm -- generalizing to polygons inside polygons is an excellent thing to try. If you try to do it strictly computationally (by actually calculating or adding up the angles and finding specific values), it's going to be quite hard I think. See if you can come up with a way of showing it without computation. Drawing many pictures I think is helpful here." } { "Tag": [ "symmetry", "geometry", "parallelogram", "ratio", "geometry solved" ], "Problem": "Does anyone know the solution of this problem?\r\n Let ABCD is a convex quadrilateral,line AB meet CD at P;line AD meet BC at Q.\r\nO is the intersection of two diagonals.R on PQ such that OR is perpendicular to PQ.prove that (angleORD)=(angleORB).(it's from China)[b][/b]", "Solution_1": "It is from Chinese tst 02, but not cmo.", "Solution_2": "It follows from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=tuff+nut&t=17]this[/url], which has received several solutions.", "Solution_3": "Projective soln:\r\n\r\nSend line PRQ to infinity, maintaining symmetry about line OR. Then AD || BQ and DC || AB so ADCB is a parallelogram => O is the midpoint of BD => B and D are symmetric with respect to line OR. Transforming back, 0$, then LHS is more than $ 1$; while if $ x^3\\minus{}3x\\plus{}1 \\equal{} 0$, then LHS is exactly $ 1$. So there are three solutions, in the intervals $ (\\minus{}2, \\minus{}1)$, $ (0, 1)$ and $ (1, 2)$.", "Solution_2": "[i]Since nobody can solve that problem , i will give a big hint ::D\n\n Multiply two sides with $ 3^{3x^3 \\minus{}2x}$[/i]", "Solution_3": "And then, did I not state where the roots are?", "Solution_4": "[i]The problem ask to find roots , not demand to find where are the roots ? :mad: [/i]", "Solution_5": "Dear, dear author of the post, rather than using frowning emoticons, you better read carefully the posted answers.\r\n[quote]... while if x^3-3x+1 = 0, then LHS is exactly 1 ...[/quote]\r\nDid I not write this? So the three real roots, for which I kindly specified the location, are the roots of this [color=blue]depressed cubic[/color], to which are you asking me to apply Cardano's formula?" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "geometry", "circumcircle", "parallelogram", "ceiling function" ], "Problem": "Rules are all the usual: 4.5 hours for each of the 2 sets of 3 problems. No calculators or computers. &c.\r\n\r\nAnd now, the problems.\r\n\r\n[hide=\"Set 1\"]1. For what integers $x$ is $\\gcd(x^{4}+3x^{3}+2x^{2}+5x+10,x^{2}+1)>1$?\n\n2. A robot has an row of 10 boxes. A number of rocks are dropped into the rightmost box, and if in any box there are at least 7 rocks, the robot will take out 7 rocks from that box, put 5 rocks in the box immediately to the left of the original box, and discard the other 2 rocks. The robot will continue this process until none of the boxes have more than 6 rocks. Find, with proof, the largest possible number of rocks that can be put in the first (rightmost) box such that the robot does not require more than 10 (originally empty save the rightmost) boxes to put the rocks. (If the robot has more than 6 rocks in the leftmost box, he will require another box; e.g. if he has 7|2|2|2|2|2|2|2|2|2, he will require an 11th box. Also, the arrangement 7|2|2|2|2|2|2|2|2|2 is not possible.)\n\n3. Let $\\triangle ABC$ be scalene and have circumcircle $O$ with center $O$. $AO$, $BO$, $CO$ are extended to meet circle $O$ at $A'$, $B'$, $C'$. The perpendicular from $A$, $B$, $C$, to $BC$, $CA$, $AB$, intersect circle $O$ at $A''$, $B''$, $C''$, respectively. $B'B''$ intersects $C'C''$ at $X$, $C'C''$ hits $A'A''$ at $Y$, $A'A''$ intersects $B'B''$ at $Z$. Prove that $AX$, $BY$, $CZ$ are concurrent at $P$ and $\\angle APA''=\\angle BPB''=\\angle CPC''=180^\\circ$.[/hide]\n\n[hide=\"Set 2\"]4. Given a $2n$-sided regular polygon, find with proof the number of parallelograms whose vertices are at intersections of the diagonals and sides of the polygon and whose sides are on diagonals or sides of the polygon.\n\n5. An ant stands at the origin in a 4-dimensional space. Every second he may move to any of the 80 points which are adjacent to him. (Thus he may move 1, $\\sqrt{2}$, $\\sqrt{3}$, or even 2 units in one second.) The ant may not visit a previously-visited gridpoint, but he may cross his path. After 10 seconds, how many different paths could he have taken?\n\n6. If $f(x)=q\\left\\lceil \\frac{px}{q^{2}}\\right\\rceil$, prove $n(f^{n}(q)) \\le \\sum_{i=0}^{n}f^{i}(q)f^{n-i}(q)$ for all positive integral $p,q,n,x$ such that $p>q$. (Note that $f^{0}(x)=x$.)[/hide]\r\nOr the problems may be found, most of the time, at http://billydorminy.homelinux.com/MockUSAMO2006a.pdf . (Thanks to 10000th User for typing it in a PDF.) \r\nPlease write your solutions as ID Number - Problem Number - Solution. You have until July 25 to send me your solutions.\r\n\r\n[size=200]EDIT:[/size]\r\nProblem 6 had a typo: $p>q$, not $q>p$. Note that the PDF does not have this change, and won't till next week. (I'm leaving on a weeklongish trip...) Thanks to rem for pointing this out. He shall get an extra point, unless of course he gets a perfect in the first place :).\r\n\r\n\r\nHave fun!\r\n\r\nBilly", "Solution_1": "I have a question about problem 2:\r\n1. What do you mean by first box -- the leftmost or the rightmost box?\r\n2. When the rocks are dropped into the rightmost box, are the rest of the boxes empty or not?", "Solution_2": "1) Rightmost\r\n2) Yes, the other boxes are empty.", "Solution_3": "OK thanks. :) that helped out a lot", "Solution_4": "[quote=\"rem\"]I have a question about problem 2:\n1. What do you mean by first box -- the leftmost or the rightmost box?\n2. When the rocks are dropped into the rightmost box, are the rest of the boxes empty or not?[/quote]The problem is perfectly clear to me but meh.\r\n\r\nsolafidefarms, for #6, can you specify what $p,q$ is? It only says 'positive'. Yes, I know what they are but it's better to specify it. Thanks.\r\n\r\n[EDIT]I made a pdf of the problems. Unless approved by solafidefarms, it will not be posted.", "Solution_5": "Fixed. Yes, p,q are integral.\r\n\r\nQuestion: Are these problems too easy?", "Solution_6": "No no no, don't get me wrong. The problems are mildly difficult. Not extremely hard but they are fun and I like it :). solafidefarms, I'm going to send you the pdf of the problems and you decide whether to post it or not. Thanks. Now I'm going to time myself on the first set of problems. 4.5 hours here we go!!! :).\r\n\r\n[EDIT]Good thing that I tested the attachment thing because I just realized the attachment thing is not working for me. I was going to send a pdf to solafidefarms but the thing won't add anything. The browser just dies off with a blank screen. Does anybody else have that problem with the PM's attaching a file? :maybe:", "Solution_7": "Must not look at problems...\r\nMust not look at problems...\r\nMust not look at problems...\r\n\r\nSo tempting, but I must wait until I have time....", "Solution_8": "10000th user.... i'd just email it :wink:", "Solution_9": "In problem 5, define adjacent.", "Solution_10": "As in, if he is at (a,b,c,d), adjacent points are $(a_{1},b_{1},c_{1},d_{1})$ where $a_{1}\\in\\{a-1,a,a+1\\}$ and so forth, except that you cannot stay at the same point. Thus there are 81-1 points adjacent to any given point.", "Solution_11": "[quote=\"10000th User\"]\nGood thing that I tested the attachment thing because I just realized the attachment thing is not working for me. I was going to send a pdf to solafidefarms but the thing won't add anything. The browser just dies off with a blank screen. Does anybody else have that problem with the PM's attaching a file? :maybe:[/quote]\r\n\r\nYes! I thought it was just my slow CPU. This is a bug I guess. Report it in the Bug Report Forum, or I will if I have time later.", "Solution_12": "PDF link doesnt work for me.", "Solution_13": "Eh, now it works.", "Solution_14": "There is probably a typo. \r\n\r\nAfter briefly scanning the problems, I believe that #6 is solafidefarms's favorite.\r\n\r\nEDIT: After thinking about #2 for 5 mins, I can see why solafidefarms likes that one.", "Solution_15": "[quote=\"JSteinhardt\"]Can we write out our proofs on paper then LaTeX after our contest time ends?[/quote]Yes.", "Solution_16": "Also what should we do about the diagram for 3?\r\n\r\nAnd is it okay if instead of attaching the solution files, I give you a link to them (it will be very hard for people to guess the link, don't worry)?", "Solution_17": "[quote=\"JSteinhardt\"]Also what should we do about the diagram for 3?\n\nAnd is it okay if instead of attaching the solution files, I give you a link to them (it will be very hard for people to guess the link, don't worry)?[/quote]\r\n\r\ni don't see the problem with that...diagrams are important for that problem since there are a few situations that prove it, but certain things can make it a pain to reproduce the diagram (as i have learned the hard way)", "Solution_18": "What is $f^{n}(x)$? Is it $f(f(f(...f(x))))$ n times?", "Solution_19": "[quote=\"JSteinhardt\"]What is $f^{n}(x)$? Is it $f(f(f(...f(x))))$ n times?[/quote]\r\n\r\nYes, it is.", "Solution_20": "Also what does $x$ have to do with problem 6? It says to prove it for all $x$ but $x$ never appears.", "Solution_21": "[quote=\"JSteinhardt\"]Also what does $x$ have to do with problem 6? It says to prove it for all $x$ but $x$ never appears.[/quote]\r\n\r\nIt says to prove it for all positive integers $p$, and $q$", "Solution_22": "Hm now I'm not actually sure what format you want our stuff in.\r\n\r\nShould we have 1 file, 2 files, or 6 files?", "Solution_23": "[quote=\"JSteinhardt\"]What is $f^{n}(x)$? Is it $f(f(f(...f(x))))$ n times?[/quote]\r\n\r\nThis question was answered but I have to admit the wording on problem 6 was quite poor.... Didn't really notice the first time I looked at the problems, my bad.", "Solution_24": "[quote=\"JSteinhardt\"]Hm now I'm not actually sure what format you want our stuff in.\n\nShould we have 1 file, 2 files, or 6 files?[/quote]\r\nThis question was already answered: it does not really matter, but preferably you should submit two sets one with the first day's solutions and the other with second day's solutions. :wink:", "Solution_25": "When will we recieve our score?", "Solution_26": "The 31st.\r\n\r\nJust a reminder: there are 4 days left to submit your solutions.", "Solution_27": "The deadline was yesterday? :(", "Solution_28": "Unfortunatly it was. \r\n\r\nI procrastinated in typing up my responses until 11 last night. But I submitted it on time.", "Solution_29": "problems have been posted in separate threads inside aops and are collected [url=https://artofproblemsolving.com/community/c3624993_2006_mock_usamo]here[/url]\n(among others [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url], among USA Mocks/ USAMO Mocks)\n\nRelated post [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url]" } { "Tag": [], "Problem": "What is the mean of all the positive five-digit numbers that can be formed by using each of the digits 1,2,3,4 and 5 only once in a number.", "Solution_1": "instead of adding them all together and dividing, let's think about this a little differently\r\n\r\ntake the ones place\r\nhow many numbers can go there? (5)\r\nif we put a number, say 1 in the ones place, how many numbers can be made such that we have a one in the ones place?\r\n\r\nwell, that will just be the number of ways we can arrange all the other numbers\r\nsince we can arrange the other numbers in $ 4!$ ways, there are $ 24$ ways to arrange all the other numbers.\r\n\r\nthis method can be used to find the number of times 2, 3, 4, and 5 go into the ones place\r\n\r\ntherefore to find the number that will represent the mean in the ones place, we have\r\n$ \\frac{1\\times24\\plus{}2\\times24\\plus{}3\\times24\\plus{}4\\times24\\plus{}5\\times24}{120}$\r\n\r\nnote that i got 120 because there 120 ways to arrange numbers\r\n\r\nsolving our expression, we get $ 3$\r\n\r\nbut that's just the ones place, what about all the other places??\r\n\r\nwell, we can use the exact same approach to get the other place values, and since we already know that this approach will get us an answer of 3, than there must be 3's in the rest of the place values\r\n\r\ntherefore, our answer is $ \\boxed{33333}$", "Solution_2": "That's wrong", "Solution_3": "$\\boxed{33333}$ should be the correct answer.", "Solution_4": "[hide]\nSimply, the greatest number plus the least number is $66666$. The second greatest and second least is $66666$, so on. Note, of course, that the mean is $\\frac{66666}{2} = \\boxed{33333}$ \nThis method can be extrapolated many of this types of problems (Example: what is the mean of all $3$ digit natural numbers with digits of $2$ and $8$, Answer= the mean of $222$ and $888$ = $555$), and vallon gives a good proof.\n[/hide]", "Solution_5": "Really nice solution method, AKAL3!!!", "Solution_6": "Because this is symmetric, just take the middle number for everything to get 33333" } { "Tag": [ "inequalities", "IMO Shortlist", "inequalities unsolved" ], "Problem": "Let $a_1\\leq a_2\\leq ...\\leq a_n$ such that $a_1+a_2+...+a_n=0$. Proved:\r\n$a_1^2+a_2^2+...+a_n^2+na_1a_n\\leq 0$", "Solution_1": "let $a_{2}=a_{1}+x_{1},a_{3}=a_{1}+x_{1}+x_{2},...,a_{n}=a_{1}+x_{1}+...+x_{n-1}$,$x_i\\geq0.$\r\nHence $na_1+(n-1)x_1+...+2x_{n-2}+x_{n-1}=0$ and\r\n$a_1=-\\frac{(n-1)x_1+(n-2)x_2+...+1*x_{n-1}}{n}$,\r\n$a_2=\\frac{x_1-(n-2)x_2-...-1*x_{n-1}}{n}$,\r\n$a_3=\\frac{x_1+2x_2-(n-3)x_3-...-1*x_{n-1}}{n}$,\r\n.\r\n.\r\n.\r\n$a_i=\\frac{x_1+2x_2+...+(i-1)x_{i-1}-(n-i)x_i-...-1*x_{n-1}}{n}$\r\n.\r\n.\r\n.\r\n$a_n=\\frac{x_1+2x_2+...+(n-1)x_{n-1}}{n}.$\r\nThen $a_1^2+a_2^2+...+a_n^n+na_1*a_n\\leq0\\Leftrightarrow{A_1x_1^2+A_2x_2^2+...+A_{n-1}x_{n-1}^2+A_{12}x_1x_2+...+A_{ij}x_ix_j+...+A_{n-2,n-1}x_{n-2}x_{n-1}\\geq0}$ (${1}\\leq{i}$<$j\\leq{n-1}$)\r\nwhere $A_1=A_2=...=A_{n-1}=0$ and $A_{ij}=(j-i)n^2.$ is done!\r\n$A_{ij}=n(i(n-j)+j(n-i))-2(i(n-i)(n-j)-(j-i)i(n-j)+(n-j)ij)=(j-i)n^2.$", "Solution_2": "[quote=\"vnmathboy\"]Let $a_1\\leq a_2\\leq ...\\leq a_n$ such that $a_1+a_2+...+a_n=0$. Proved:\n$a_1^2+a_2^2+...+a_n^2+na_1a_n\\leq 0$[/quote]\r\n\r\nAnother proof:\r\n\r\nSince the real numbers $a_1$, $a_2$, ..., $a_n$ satisfy $a_1\\leq a_2\\leq ...\\leq a_n$ and $a_1+a_2+...+a_n=0$, we have\r\n\r\n$a_1\\leq a_2\\leq ...\\leq a_{u-1}\\leq a_u\\leq 0\\leq a_{u+1}\\leq a_{u+2}\\leq ...\\leq a_n$\r\n\r\nfor some index u.\r\n\r\nNow, let i be an index with $i\\leq u$. Then, $a_i\\leq 0$, but $a_i\\geq a_1$. Hence, $a_i\\cdot a_i\\leq a_i\\cdot a_1$. In other words, $a_i^2\\leq a_1a_i$. Hence,\r\n\r\n$a_1^2+a_2^2+...+a_u^2\\leq a_1a_1+a_1a_2+...+a_1a_u=a_1\\left(a_1+a_2+...+a_u\\right)$.\r\n\r\nNow, since $a_1+a_2+...+a_n=0$, we have $a_1+a_2+...+a_u=-\\left(a_{u+1}+a_{u+2}+...+a_n\\right)$; on the other hand, since all $a_k$ are $\\leq a_n$, we have $a_{u+1}+a_{u+2}+...+a_n\\leq\\underbrace{a_n+a_n+...+a_n}_{n-u\\text{ times}}=\\left(n-u\\right)a_n$. Thus, $a_1+a_2+...+a_u=-\\left(a_{u+1}+a_{u+2}+...+a_n\\right)\\geq -\\left(n-u\\right)a_n$. In view of $a_1\\leq 0$, this yields $a_1\\left(a_1+a_2+...+a_u\\right)\\leq a_1\\cdot\\left(-\\left(n-u\\right)a_n\\right)$. Hence,\r\n\r\n$a_1^2+a_2^2+...+a_u^2\\leq a_1\\left(a_1+a_2+...+a_u\\right)\\leq a_1\\cdot\\left(-\\left(n-u\\right)a_n\\right)=-\\left(n-u\\right)a_1a_n$.\r\n\r\nOn the other hand, let j be an index with j > u. Then, $a_j\\geq 0$, but $a_j\\leq a_n$. Hence, $a_j\\cdot a_j\\leq a_j\\cdot a_n$, what rewrites as $a_j^2\\leq a_na_j$. Thus,\r\n\r\n$a_{u+1}^2+a_{u+2}^2+...+a_n^2\\leq a_na_{u+1}+a_na_{u+2}+...+a_na_n=a_n\\left(a_{u+1}+a_{u+2}+...+a_n\\right)$.\r\n\r\nBut $a_1+a_2+...+a_n=0$ yields $a_{u+1}+a_{u+2}+...+a_n=-\\left(a_1+a_2+...+a_u\\right)$; on the other hand, as all $a_k$ are $\\geq a_1$, we have $a_1+a_2+...+a_u\\geq\\underbrace{a_1+a_1+...+a_1}_{u\\text{ times}}=ua_1$. Thus, $a_{u+1}+a_{u+2}+...+a_n=-\\left(a_1+a_2+...+a_u\\right)\\leq -ua_1$. Since $a_n\\geq 0$, this yields $a_n\\left(a_{u+1}+a_{u+2}+...+a_n\\right)\\leq a_n\\cdot\\left(-ua_1\\right)=-ua_1a_n$. Hence,\r\n\r\n$a_{u+1}^2+a_{u+2}^2+...+a_n^2\\leq a_n\\left(a_{u+1}+a_{u+2}+...+a_n\\right)\\leq -ua_1a_n$.\r\n\r\nAs a consequence of the above,\r\n\r\n$a_1^2+a_2^2+...+a_n^2=\\left(a_1^2+a_2^2+...+a_u^2\\right)+\\left(a_{u+1}^2+a_{u+2}^2+...+a_n^2\\right)$\r\n$\\leq \\left(-\\left(n-u\\right)a_1a_n\\right)+\\left(-ua_1a_n\\right)=-na_1a_n$,\r\n\r\nso that $a_1^2+a_2^2+...+a_n^2+na_1a_n\\leq 0$, as intended to prove.\r\n\r\n[b]EDIT:[/b] Source of the problem: IMO Shortlist 1972, proposed by Czechoslovakia (CZS 4).\r\n\r\n Darij", "Solution_3": "Since $a_n \\ge a_i \\ge a_1$ for all $1 \\le i \\le n$, $(a_n - a_i)(a_i - a_1) \\ge 0$, which expands as\r\n\\[ a_n a_i - a_n a_1 - a_i^2 + a_i a_1 \\ge 0. \\]\r\nSumming from $i = 1$ to $n$ and using $a_1 + a_2 + \\dots + a_n = 0$ gives\r\n\\[ a_1^2 + a_2^2 + \\dots + a_n^2 + na_1 a_n \\le 0. \\]", "Solution_4": "[quote=\"arqady\"]\nThen $a_1^2+a_2^2+...+a_n^n+na_1*a_n\\leq0\\Leftrightarrow{A_1x_1^2+A_2x_2^2+...+A_{n-1}x_{n-1}^2+A_{12}x_1x_2+...+A_{ij}x_ix_j+...+A_{n-2,n-1}x_{n-2}x_{n-1}\\geq0}$ (${1}\\leq{i}$<$j\\leq{n-1}$)\nwhere $A_1=A_2=...=A_{n-1}=0$ and $A_{ij}=(j-i)n^2.$ is done!\n$A_{ij}=n(i(n-j)+j(n-i))-2(i(n-i)(n-j)-(j-i)i(n-j)+(n-j)ij)=(j-i)n^2.$[/quote]\n\nwhat dose $A_i$ means\uff1f", "Solution_5": "[quote=\"colorfuldreams\"][quote=\"arqady\"]\nThen $a_1^2+a_2^2+...+a_n^n+na_1*a_n\\leq0\\Leftrightarrow{A_1x_1^2+A_2x_2^2+...+A_{n-1}x_{n-1}^2+A_{12}x_1x_2+...+A_{ij}x_ix_j+...+A_{n-2,n-1}x_{n-2}x_{n-1}\\geq0}$ (${1}\\leq{i}$<$j\\leq{n-1}$)\nwhere $A_1=A_2=...=A_{n-1}=0$ and $A_{ij}=(j-i)n^2.$ is done!\n$A_{ij}=n(i(n-j)+j(n-i))-2(i(n-i)(n-j)-(j-i)i(n-j)+(n-j)ij)=(j-i)n^2.$[/quote]\n\nwhat dose $A_i$ means\uff1f[/quote]\nIt's a coefficient of $x_i^2$. Now I see that my proof is very ugly. :(" } { "Tag": [ "modular arithmetic", "logarithms", "number theory unsolved", "number theory" ], "Problem": "$n$ is a positive integer, $F_n=2^{2^{n}}+1$. Prove that for $n \\geq 3$, there exists a prime factor of $F_n$ which is larger than $2^{n+2}(n+1)$.", "Solution_1": "I think it will help:\r\n\r\nall of the divisor$>1$ of $2^{2^n}+1$ are in the form $2^{n+2}k+1$ where $k\\in \\mathbb{N}$ ;)", "Solution_2": "[quote=\"KDS\"][quote=\"ehsan2004\"]I think it will help:\n\nall of the divisor$>1$ of $2^{2^n}+1$ are in the form $2^{n+2}k+1$ where $k\\in \\mathbb{N}$ ;)[/quote]\nI know this , but anyone give more detailed solution please?[/quote]\n\nIt`s really hard ! It was ChinaTST . \n\nHint :\n\nAssume $F_n = \\prod_{i=1}^{m} {p_i}^{\\alpha _i}$ , Find an upper and a lower bound for $\\sum_{i=1}^{m} \\alpha _i$.", "Solution_3": "[hide=\"Stronger claim\"]For $n \\geq 4$, the largest prime factor is larger than $2^{n+4} (n+2)$. [/hide]\n[hide=\"Proof of stronger claim\"]\n[b]Lemma:[/b] If $n \\geq 2$, $p$ is a prime, and $p | 2^{2^n} + 1$, then $2^{n+2} | p - 1$. \n[i]Proof:[/i] Let $k$ be the order of 2 modulo $p$. Since $2^{2^n} \\equiv -1 \\pmod{p}$, $2^{2^{n+1}} \\equiv 1 \\pmod{p}$, so $k$ must be a power of 2 less than or equal to $2^{n+1}$. But if $k < 2^{n+1}$, we may square the congreunce $2^k \\equiv 1 \\pmod{p}$ a total of $n - \\log_2 k$ times to get $2^{2^n} \\equiv 1 \\pmod{p}$. But $2^{2^n} \\equiv -1 \\pmod{p}$ and $p \\neq 2$, so we must have $k = 2^{n+1}$. Because $n \\geq 3$, $8 \\, | \\, 2^{n+1} \\, | \\, p - 1$, whence $\\left( \\frac{2}{p} \\right) = 1$. Hence, $a^2 \\equiv 2 \\pmod{p}$ for some integer $a$. The order of $a$ must therefore be $2k$, so $2^{n+2} = 2k \\, | \\, p - 1$, as desired. \n\n------------------------------------------------------------------------------------------------\n\nThe problem is clearly true when $n = 3$, so we may suppose that $n \\geq 4$. Let $p_1 \\leq p_2 \\leq \\ldots \\leq p_m$ be primes such that $2^{2^n} + 1 = p_1 p_2 \\ldots p_m$. By the lemma, there exist integers $a_1, a_2, \\ldots, a_m$ such that $p_i = 2^{n+2} a_i + 1$ for each $i$. Hence, we have \n\\begin{align*} 2^{2^n} + 1 = (2^{n+2} a_1 + 1)(2^{n+2} a_2 + 1) \\ldots (2^{n+2} a_m + 1) . \\tag{1} \\end{align*}\nIt follows that $2^{2^n} \\geq (2^{n+2} + 1)^m > 2^{m(n+2)}$, whence \n\\[ m < \\frac{2^n}{n+2}. \\]\n\nSince $n \\geq 4$, $2n + 4 \\leq 2^n$. Taking (1) modulo $2^{2n+4}$ yields\n\\begin{align*}\n1 \n&\\equiv 2^{2^n} + 1 \\\\\n&= (2^{n+2} a_1 + 1)(2^{n+2} a_2 + 1) \\ldots (2^{n+2} a_m + 1) \\\\ \n&= 2^{2n+4}(\\mbox{some terms}) + 2^{n+2}(a_1 + a_2 + \\ldots + a_m) + 1 \\\\\n&\\equiv 2^{n+2}(a_1 + a_2 + \\ldots + a_m) + 1 \\pmod{2^{2n+4}},\n\\end{align*}\nwhence\n\\[ 2^{n+2} | a_1 + a_2 + \\ldots + a_m. \\]\nSince $a_m \\geq a_{m-1} \\geq \\ldots \\geq a_1$, we have $a_m \\geq \\frac{2^{n+2}}{m}$. Hence, \n\\begin{align*}\np_m \n&= 2^{n+2} a_m + 1 \\\\\n&\\geq 2^{n+2} \\cdot \\frac{2^{n+2}}{m} + 1 \\\\\n&> 2^{n+2} \\cdot \\frac{2^{n+2}}{\\frac{2^n}{n+2}} \\\\\n&= 2^{n+4} (n+2),\n\\end{align*}\nas desired. [/hide]\n\n[quote=\"chuyentoan\"]It is really false!all of the divisor$>1$ of $2^{2^n}+1$ are in the form $2^{n+1}k+1$ but not $2^{n+2}$ where $k\\in \\mathbb{N}$ :wink:[/quote]\nNo, it is true; see my my first lemma.", "Solution_4": "Let 2^(2^n) = m^2. Now, try to find the smallest k, such that p is a prime divisor of m^2 + 1, and p > 2m + sqrt(2m), for every m >= k. (This problem is similar to IMO 2008 - Problem 3.)", "Solution_5": "It is well-known that all prime divisor of $F_n$ are $1 \\pmod {2^{n+1}}$.\nLet the prime divisors of $F_n$ be $2^{n+1}p_1+1, 2^{n+1}p_2+1 \\cdots, 2^{n+1}p_t+1$.\nWe have $2^{2^n}+1 = \\prod_{i=1}^t (2^{n+1}p_i+1)^{e_i} \\equiv \\prod_{i=1}^t (2^{n+1}p_ie_i+1) \\equiv 1 \\pmod {2^{2n+2}}$\nTherefore, we have that $\\sum_{i=1}^t p_ie_i \\equiv 0 \\pmod {2^{n+1}} \\implies \\sum_{i=1}^t p_ie_i \\ge 2^{n+1}$\nLet $max[p_i]=p$. We have $p>\\frac{2^{n+1}}{\\sum_{i=1}^t e_i}$. It suffices to show that $\\frac{2^{n+1}}{\\sum_{i=1}^t e_i} \\ge 2(n+1)$\nThis follows from $2^{2^n}+1 = \\prod_{i=1}^t (2^{n+1}p_i+1)^{e_i} \\ge \\prod_{i=1}^t (2^{n+1}p_i)^{e_i} +1 \\ge 2^{(n+1)\\sum_{i=1}^t e_i}+1$. $\\blacksquare$", "Solution_6": "The case $n = 3$ is clear, so assume $n \\ge 4$. It is well-known that every prime factor of $F_n$ is congruent to 1 modulo $2^{n+2}$. Assuming the conclusion is false, write\n\\[2^{2^n} + 1 = (1 \\cdot 2^{n+2} + 1)^{e_1} \\cdots ((n+1) \\cdot 2^{n+2} + 1)^{e_{n+1}}\\]\nand take modulo $2^{2n+4}$: after simplifying we obtain\n\\[e_1 + 2e_2 + \\dots + (n+1) e_{n+1} \\equiv 0 \\pmod{2^{n+2}}.\\]\nIn particular, $e_1 + \\dots + e_{n+1} \\ge \\tfrac{1}{n+1} \\cdot 2^{n+2}$, so\n\\begin{align*}\n2^{2^n} + 1 & = (1 \\cdot 2^{n+2} + 1)^{e_1} \\cdots ((n+1) \\cdot 2^{n+2} + 1)^{e_{n+1}}\\\\\n& \\ge (2^{n+2} + 1)^{e_1 + \\dots + e_{n+1}}\\\\\n& \\ge (2^{n+2} + 1)^{2^{n+2}/(n+1)}\\\\\n& > 2^{2^{n+2}(n+2)/(n+1)}\\\\\n& > 2^{2^{n+1}} > 2^{2^n} + 1,\n\\end{align*}\ncontradiction.", "Solution_7": "here's a very different solution :-D \n[b][color=#f00]claim(1):[/color][/b] Let P a prime number and $p=k.2^{n+1} +1$ such that $k <2n$ then there's $m <2^n$ such that $p | 2^m +1$\n[color=#600]proof:[/color]\nsuppose the contrary \n$2^{p-1} \\equiv $ $1$ $mod p$\n$2^{2^{n+1}.k} \\equiv $ $1$ $mod p$\n$2^{2^{n}.k} \\equiv $ $1$ $mod p$\n.........\n$2^{k} \\equiv $ $1$ $mod k.2^{n+1} +1$\nbut $2^k <2p \\implies 2^k=p+1= k.2^{n+1} +2$ \na contradiction\n(there's a typo in this proof )\nlet $p | 2^{2^n} + 1 \\implies p | 2^{2^{n+1}} - 1 \\implies ord_p(2)=2^{n+1} \\implies 2^{n+1}|p-1$\nthus $p=2^{n+1}.k+1$\nnow by zsigmondi we have a prime divisor $q$ for $2^{2^n} + 1$ that doesn't divide any $2^m +1 : m<2^n$\nfrom the claim above we have $q \\ge 2^{n+1}.(2n)+1=2^{n+n}.n+1$\nand we are done :D", "Solution_8": "Can you use Zsigmondi at school olympiads?", "Solution_9": "[quote=Ereb15]Can you use Zsigmondi at school olympiads?[/quote]\n\nwhy not :blush: ?\n", "Solution_10": "[quote=Zhero][hide=\"Stronger claim\"]For $n \\geq 4$, the largest prime factor is larger than $2^{n+4} (n+2)$. [/hide]\n[hide=\"Proof of stronger claim\"]\n[b]Lemma:[/b] If $n \\geq 2$, $p$ is a prime, and $p | 2^{2^n} + 1$, then $2^{n+2} | p - 1$. \n[i]Proof:[/i] Let $k$ be the order of 2 modulo $p$. Since $2^{2^n} \\equiv -1 \\pmod{p}$, $2^{2^{n+1}} \\equiv 1 \\pmod{p}$, so $k$ must be a power of 2 less than or equal to $2^{n+1}$. But if $k < 2^{n+1}$, we may square the congreunce $2^k \\equiv 1 \\pmod{p}$ a total of $n - \\log_2 k$ times to get $2^{2^n} \\equiv 1 \\pmod{p}$. But $2^{2^n} \\equiv -1 \\pmod{p}$ and $p \\neq 2$, so we must have $k = 2^{n+1}$. Because $n \\geq 3$, $8 \\, | \\, 2^{n+1} \\, | \\, p - 1$, whence $\\left( \\frac{2}{p} \\right) = 1$. Hence, $a^2 \\equiv 2 \\pmod{p}$ for some integer $a$. The order of $a$ must therefore be $2k$, so $2^{n+2} = 2k \\, | \\, p - 1$, as desired. \n\n------------------------------------------------------------------------------------------------\n\nThe problem is clearly true when $n = 3$, so we may suppose that $n \\geq 4$. Let $p_1 \\leq p_2 \\leq \\ldots \\leq p_m$ be primes such that $2^{2^n} + 1 = p_1 p_2 \\ldots p_m$. By the lemma, there exist integers $a_1, a_2, \\ldots, a_m$ such that $p_i = 2^{n+2} a_i + 1$ for each $i$. Hence, we have \n\\begin{align*} 2^{2^n} + 1 = (2^{n+2} a_1 + 1)(2^{n+2} a_2 + 1) \\ldots (2^{n+2} a_m + 1) . \\tag{1} \\end{align*}\nIt follows that $2^{2^n} \\geq (2^{n+2} + 1)^m > 2^{m(n+2)}$, whence \n\\[ m < \\frac{2^n}{n+2}. \\]\n\nSince $n \\geq 4$, $2n + 4 \\leq 2^n$. Taking (1) modulo $2^{2n+4}$ yields\n\\begin{align*}\n1 \n&\\equiv 2^{2^n} + 1 \\\\\n&= (2^{n+2} a_1 + 1)(2^{n+2} a_2 + 1) \\ldots (2^{n+2} a_m + 1) \\\\ \n&= 2^{2n+4}(\\mbox{some terms}) + 2^{n+2}(a_1 + a_2 + \\ldots + a_m) + 1 \\\\\n&\\equiv 2^{n+2}(a_1 + a_2 + \\ldots + a_m) + 1 \\pmod{2^{2n+4}},\n\\end{align*}\nwhence\n\\[ 2^{n+2} | a_1 + a_2 + \\ldots + a_m. \\]\nSince $a_m \\geq a_{m-1} \\geq \\ldots \\geq a_1$, we have $a_m \\geq \\frac{2^{n+2}}{m}$. Hence, \n\\begin{align*}\np_m \n&= 2^{n+2} a_m + 1 \\\\\n&\\geq 2^{n+2} \\cdot \\frac{2^{n+2}}{m} + 1 \\\\\n&> 2^{n+2} \\cdot \\frac{2^{n+2}}{\\frac{2^n}{n+2}} \\\\\n&= 2^{n+4} (n+2),\n\\end{align*}\nas desired. [/hide]\n\n[quote=\"chuyentoan\"]It is really false!all of the divisor$>1$ of $2^{2^n}+1$ are in the form $2^{n+1}k+1$ but not $2^{n+2}$ where $k\\in \\mathbb{N}$ :wink:[/quote]\nNo, it is true; see my my first lemma.[/quote]\nHow to prove that there exists such $p_1 \\leq p_2 \\leq \\ldots \\leq p_m$? I don't understand what you mean by writing $2^{2^n}+1$ in that form? It looks like prime factorization but where are the exponents of primes?\n", "Solution_11": "Note that the consecutive primes can be equal, so it's just spliting the exponent" } { "Tag": [ "analytic geometry", "Pythagorean Theorem", "geometry" ], "Problem": "How many units long is a line segment whose endpoints have coordinates $(-3, 7)$ and $(2, -5)$?", "Solution_1": "[hide]The distants formula is $\\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ so if you plug the numbers in, you get: $\\sqrt{(-3-2)^2+[7-(-5)]^2} = \\sqrt{25+144} = \\sqrt {169} = 13$[/hide]", "Solution_2": "[quote=\"MCrawford\"]How many units long is a line segment whose endpoints have coordinates $(-3, 7)$ and $(2, -5)$?[/quote]\r\n[hide=\"solution\"]\n5-12-[b]13[/b]\n[/hide]", "Solution_3": "[hide]\n13\n[/hide]", "Solution_4": "Easy if you know basic geometry.\r\n[hide=\"Click to reveal hidden content...if you dare!\"]The difference of the x coordinates is 5, and the difference of the y coordinates is 12. So the length of the segment could be seen as the hypotenuse of a right triangle with legs 5 and 12. Use the Pythagorean theorem (a 2 + b 2 = c 2 ) to find that the answer is 13. Or just be familiar with the 5, 12, 13 Pyth. triple.[/hide]", "Solution_5": "haha that's a number sense problem...\r\n[hide]the famouse 5-12-13 triangle![/hide]", "Solution_6": "Lucy, thank you for showing your work, and I like it the most :D", "Solution_7": "[hide]13 with the distance formula.[/hide]", "Solution_8": "extension:\r\n\r\nhow does the distance formula work? what's the theorem called?", "Solution_9": "Differences in the y values and the x values just serve as legs of a right triangle, so its basically pythagorean theorem.", "Solution_10": "I believe the distance formula for the distance between (x1, y1) and (x2, Y2) is:\r\n\r\n[hide]sqrt[(x1-x2) 2 +(y1-y2) 2 ][/hide]\r\n\r\nAs mentioned, it's basically the Pythagorean theorem.", "Solution_11": "CORRECTO MOONDO!!!", "Solution_12": "[hide=\"answer to original problem\"]\nu can use a 5-12-13 triangle\nso...[b]13[/b]is the answer[/hide]" } { "Tag": [ "MATHCOUNTS", "search" ], "Problem": "i just found out about mathcounts this year...does anybody have tips on how to prepare for mathcounts next year so that i can get in the top three in chapter, go to state, and go on from there?...i'm open to recommendations for resources (e.g. books, tests, and other stuff)...thanks", "Solution_1": "There are about 100 of this topic already-search \"MathCounts Prep\"." } { "Tag": [ "ARML", "AMC", "USA(J)MO", "USAMO", "geometry" ], "Problem": "ARML 2004\r\n\r\nSF Bay A vs. Massachusetts A\r\n\r\nWho will triumph over the other?", "Solution_1": "I want to vote for Massachusetts A to scare the SFBA kids into studying, but I don't think I will. (Grace and Darren, you had better not be reading this.)", "Solution_2": "I keep hearing about competition between SFBA and Massachusetts... Why is there such a big rivalry?\n\n\n\n(Oh, and I voted for............ [hide]neither[/hide]:) )", "Solution_3": "There's a big rivalry because the Massachusetts kids don't want to admit that the SFBAers are smarter. It might have something to do with Gabriel Carroll and Reid Barton winning tons of contests as well.", "Solution_4": "Eh, SFBA will win. But why isn't this SFBA A vs. TJHSST A? I think that's the biggest contest, although TJ almost always wins.", "Solution_5": "SFBA A can beat TJ A. And what's this almost always? Before 2002, SFBA completely dominated ARML.", "Solution_6": "Why are there two options only? I find that a bit presumptious.... The MI/ICAE team will make a killing this year. Prepare yourselves for domination. [/humor].", "Solution_7": "[quote]Why are there two options only? I find that a bit presumptious.... [/quote]\r\n\r\nNo, the poll is about which of the two teams will beat the other, not which team will win ARML entirely.\r\n\r\n\r\nAnd Simon, how can you be so sure that SFBA can win it all this year? They don't even really have a super-genius anymore who can lead them and do it all on the team round (unlike during their golden years, when they had Gabriel, Inna, etc...).", "Solution_8": "not sure if this matters, but MA has a lot less USAMO qualifiers than last year.", "Solution_9": "A super-genius doesn't win ARML. Having lots of people who are pretty good is much more useful. I think a team of 15 copies of me would easily beat a team consisting of Gabe, Reid, Tiankai, and 12 random people, even though the team with Gabe, Reid, and Tiankai could get the top 3 individual scores. (Of course, that isn't how ARML teams are formed, so it's a moot point.) To demonstrate my point, notice that Wisconsin hasn't been a threat for the national title recently despite having probably the most impressive top end in the country: Po-Ru, Dani, Po-Ling, and Yian. 4 people don't make a team. 15 people make a team.\r\n\r\nIt's obvious that SFBA got far worse between 2001 and 2002. Eight of the fifteen people on SFBA A 2001 were seniors (as were I think 6 on the B team). But there were also six seniors on the 2002 A team. That seemed like a big loss, but when we started practicing, it was absolutely clear that the 2003 team was far stronger than the 2002 team. I can't imagine that the 2004 team will do nearly as well as the 2003 team (whose top 5 scorers were 4 seniors and a junior), but they won't do horribly either.\r\n\r\nAnyway, I'm nearly certain I know who will win ARML 2004, and it's not SFBA, Massachusetts, or TJ.", "Solution_10": "[quote]Anyway, I'm nearly certain I know who will win ARML 2004, and it's not SFBA, Massachusetts, or TJ.[/quote]\r\n\r\nWho, Exeter?", "Solution_11": "By the way, is anyone else here actually going to be on the SFBA or Massachusetts team this year?", "Solution_12": "Not Exeter.", "Solution_13": "That presumably leaves it to Chicago, then?\r\n\r\nBy the way, Simon is completely right about what it takes to win ARML. This year's NYSML exam is a good example, where by virtue of having Alison on their team, the Albany Area Math Circle was able to beat NYC A on the team round, and NYCA still won the whole thing quite handily.", "Solution_14": "Yes. I definitely expect Chicago to win this year.", "Solution_15": "hmmm ... I agree with simon ... *grin*", "Solution_16": "No, Georgia A will pwn the rest of you phr34k5!!! (ROFL, I wish)\r\n\r\nWho has Tiankai? SFBA? Because, unless I'm completely wrong, he's still eligible to compete, right?", "Solution_17": "[quote]Who has Tiankai? SFBA? Because, unless I'm completely wrong, he's still eligible to compete, right?[/quote]\r\n\r\nExeter does.\r\n\r\n\r\nSimon, why do you say Chicago?", "Solution_18": "Chicago is the only top team that has anything like the strength it did last year.", "Solution_19": "haha, i tried out for the massachusetts arml team a few months ago and i placed 14th or so (that would've put me towards the bottom of the A team) but i cant go to arml cuz i have high school graduation that weekend. i didnt qualify for usamo, either... my score was 200, needed 210....\r\n\r\na friend at school was 3rd at the arml tryouts and he didn't make usamo either (202.5) so i dont know how strong the Mass A team will be.", "Solution_20": "What? Nobody is giving Southern California a chance? Oh well, we'll be the San Antonio Spurs of the ARML teams then - be Quietly Dangerous! (we wish :D )", "Solution_21": "USAMO qualifiers isn't really anything to go by for ARML, but SFBA has at least 22 USAMO qualifiers, whereas Southern California has at most 10. (I can guarantee beyond a reasonable doubt that the SFBA A team will not have 15 USAMO qualifiers though.)", "Solution_22": "To give the rest of the team some credit: they would have done quite respectably on the team (6/10, to be conservative? I certainly didn't do all the problems) without me. And Albany Area also did quite well on the Power Round, where I was practically useless, having tied myself up with an annoying problem.\r\n\r\n[quote=\"JBL\"]This year's NYSML exam is a good example, where by virtue of having Alison on their team, the Albany Area Math Circle was able to beat NYC A on the team round, and NYCA still won the whole thing quite handily.[/quote]" } { "Tag": [ "induction", "inequalities", "function", "algebra unsolved", "algebra" ], "Problem": "Given 2n real numbers $a_{1},a_{2},..,a_{n},b_{1},b_{2},..,b_{n}$ sastisfying \r\n$a_{1}+a_{2}+...+a_{n}=b_{1}+b_{2}+...+b_{n}$\r\n$0 2$:\r\n\r\n$ d_i = b_i - b_{i - 1} - b_{i - 2}$\r\n\r\nit follows that:\r\n\r\n$ b_i = F_{i - 1}d_2 + \\ldots + F_1d_i + F_ib_1$\r\n\r\nalso by induction we can show that:\r\n\r\n$ F_1 + \\ldots + F_k = F_{k + 2} - 1$ hence:\r\n\r\n\\begin{eqnarray*}\\frac{b_{n - 1} + b_n}{b_1 + \\ldots + b_{n - 2}} & = & \\frac {F_nd_2 + \\ldots + F_2d_n + F_{n + 1}b_1}{(F_{n - 1} - 1)d_2 + \\ldots + (F_1 - 1)d_n + (F_n - 1)b_1} \\\\\r\n& \\geq & \\frac {F_{n + 1}b_1}{(F_n - 1)b_1} = \\frac {a_{n - 1} + a_n}{a_1 + \\ldots + a_{n - 2}}\\end{eqnarray*}\r\n\r\n(in the last inequality we used the fact that if $ b,d > 0$ and $ \\frac ab\\leq \\frac cd$ then $ \\frac ab\\leq \\frac {a + c}{b + d}$)\r\n\r\nnow let $ s = a_1 + \\ldots + a_n$.then we have showed that:\r\n\r\n$ \\frac {a_{n - 1} + a_n}{s - a_{n - 1} - a_n}\\leq\\frac {b_{n - 1} + b_n}{s - b_{n - 1} - b_n}$\r\n\r\nbut the function $ f(x) = \\frac {x}{s - x}$ is increasing on $ [0,s]$ hence we must have:\r\n\r\n$ a_{n - 1} + a_n\\leq b_{n - 1} + b_n$." } { "Tag": [ "geometry" ], "Problem": "April 1991 Mandelbrot Round Five Individual Test Question:\r\n\r\n8. In acute triangle $ABC$, points $D$, $E$, and $F$ are located on sides $BC$, $AC$, and $AB$ so that $AD \\bot BC$, $DE \\bot AC$, and $DF \\bot AB$. Let $R_1$ and $R_2$ be the radii of the circles around $\\triangle ABC$ and $\\triangle AEF$, respectively. Determine $\\angle A$, in degrees, if $area(ABC) = R_1 R_2$", "Solution_1": "here's an image\r\n\r\n[hide=\"and a hint\"]\nwhat is AFED?\n[/hide]" } { "Tag": [ "algebra", "polynomial", "function", "vector", "domain", "linear algebra", "Ring Theory" ], "Problem": "Show that the set of polynomial function P(R) is not a finitely generated vector Space \r\nThank u for ur help in advance", "Solution_1": "That's trivial if you show that the family (1,X,X^2,...) is free over R ?", "Solution_2": "can u explain more", "Solution_3": "julien_santini used the word \"free\" for a concept that is called, in most English-language linear algebra courses, \"linearly independent.\"\r\n\r\nIf a vector space has dimension $n$, then any set of $n+1$ vectors must be linearly dependent. (This is a major theorem in the elementary theory.) It follows that if a vector space admits an infinite independent set, then that space must be infinite dimensional.\r\n\r\nAs for the particular point, the independence of $\\{1,x,x^2,\\dots\\}$:\r\n\r\nLet $P(x) =a_0x+a_1x^2+\\cdots+a_nx^n$ be an arbitrary linear combination of this set. Either it is the trivial linear combination (all $a_m=0$) or we may WLOG assume that $a_n\\ne0.$ Assume that this is not the trivial linear combination. But the theory of polynomials now tells us that to each root of this polynomial must correspond a factor. An $n$th degree polynomial can have no more than $n$ zeros. Since the real numbers are infinite, there must be a real number that does not equal any of the roots, and hence there is an $x$ such that $P(x)\\ne0.$ That means that $P$ is not the zero polynomial. So the only way to produce the zero polynomial is to have the trivial linear combination. That proves independence.\r\n\r\nThis proof used exactly two properties of $\\mathbb{R}$: it is a field (used to prove that an $n$th degree polynomial can have no more than $n$ roots) and it is infinite. The same proof would work for polynomials over any infinite field.", "Solution_4": "A minor quibble: Do you really want to bring the more analytical properties of polynomials (that they are algebraically closed over $\\mathbb{C}$) to bear on what is really a purely algebraic question?\r\n\r\nTo every polynomial $p(z)=a_0+a_1z+\\ldots+a_nz^n$ we may associate the sequence $\\{b_i\\}^{\\infty}_{i=0}$ where $b_i\\equiv 0$ for $i>n$ and $b_i\\equiv a_i$ otherwise. Two polynomials are equal if the sequences associated with them are equal. We may thus conclude that $\\{1,x,x^2,\\ldots\\}$ is linearly independent, because if $c_1\\cdot 1+c_2\\cdot x+\\ldots=0$, then the LHS is the sequence $c_1,c_2,\\ldots$ and the RHS is the sequence $0,0,0,\\ldots$, so all of the $c_i$'s are zero, which is linear independence.", "Solution_5": "[quote=\"blahblahblah\"]Do you really want to bring the more analytical properties of polynomials (that they are algebraically closed over $\\mathbb{C})$ to bear on what is really a purely algebraic question?[/quote]\nI didn't bring that in. I said that an $n$th degree polynomial has [i]at most[/i] $n$ roots, which is a purely algebraic fact that holds over any field. There was no need to claim that it had exactly $n$ roots, or indeed any roots at all.\n\n[quote]Two polynomials are equal if the sequences associated with them are equal.[/quote]\r\nThat statement is completely equivalent to the statement that $\\{1,x,x^2\\dots\\}$ is independent. What we make of that equivalence depends on how we define polynomials. There are two main ways to think of the set of polynomials:\r\n\r\n1. a polynomial [i]is[/i] a (finitely supported) sequence; $x$ and its powers are formal symbols with no intrinsic meaning,\r\n\r\nor\r\n\r\n2. a polynomial is a function which can be written in the appropriate form, and $x$ is a variable whose domain is the field in question.\r\n\r\nThe proof I wrote was with definition (2) in mind; blahblahblah seems to be using definition (1), in which case there's nothing to be proved. \r\n\r\nIn either case, we have the canonical example of a vector space with an countably infinite (Hamel) basis.", "Solution_6": "Of course, you're correct - I misread your post. But there's still something to be said for working with polynomials without actually referring to their values as functions in the context of linear algebra. Of course, we need to use values when constructing, say, a dual basis for the vector space of polynomials of degree $\\leq n$, but most of the time we don't.\r\n\r\nedit: It seems that we both edited our posts without reading the other's edits, which makes this a bit confusing.", "Solution_7": "A way of defining the polynomial ring $R[X]$ is as the set of finite functions from $N \\rightarrow R$ (where I use the convention $0 \\in N$), \"finite\" meaning here only finitely many non-zero values. The operations are pointwise addition and $f\\cdot g : n \\mapsto \\sum_{i+j=n} f(i) g(j)$.\r\n\r\nIn this formulation the functions $e_i : n \\mapsto \\delta_{in}$ are clearly linearly independent.", "Solution_8": "What rgep just posted is a formalization of what I meant by definition (1) in my post #6.", "Solution_9": "[quote=\"Kent Merryfield\"]julien_santini used the word \"free\" for a concept that is called, in most English-language linear algebra courses, \"linearly independent.\"\n\nIf a vector space has dimension $n$, then any set of $n+1$ vectors must be linearly dependent. (This is a major theorem in the elementary theory.) It follows that if a vector space admits an infinite independent set, then that space must be infinite dimensional.\n\nAs for the particular point, the independence of $\\{1,x,x^2,\\dots\\}$:\n\nLet $P(x) =a_0x+a_1x^2+\\cdots+a_nx^n$ be an arbitrary linear combination of this set. Either it is the trivial linear combination (all $a_m=0$) or we may WLOG assume that $a_n\\ne0.$ Assume that this is not the trivial linear combination. But the theory of polynomials now tells us that to each root of this polynomial must correspond a factor. An $n$th degree polynomial can have no more than $n$ zeros. Since the real numbers are infinite, there must be a real number that does not equal any of the roots, and hence there is an $x$ such that $P(x)\\ne0.$ That means that $P$ is not the zero polynomial. So the only way to produce the zero polynomial is to have the trivial linear combination. That proves independence.\n\nThis proof used exactly two properties of $\\mathbb{R}$: it is a field (used to prove that an $n$th degree polynomial can have no more than $n$ roots) and it is infinite. The same proof would work for polynomials over any infinite field.[/quote]\r\n\r\n\r\nThank u very much , Now it makes senses", "Solution_10": "[quote=\"blahblahblah\"]A minor quibble: Do you really want to bring the more analytical properties of polynomials (that they are algebraically closed over $\\mathbb{C}$) to bear on what is really a purely algebraic question?\n\nTo every polynomial $p(z)=a_0+a_1z+\\ldots+a_nz^n$ we may associate the sequence $\\{b_i\\}^{\\infty}_{i=0}$ where $b_i\\equiv 0$ for $i>n$ and $b_i\\equiv a_i$ otherwise. Two polynomials are equal if the sequences associated with them are equal. We may thus conclude that $\\{1,x,x^2,\\ldots\\}$ is linearly independent, because if $c_1\\cdot 1+c_2\\cdot x+\\ldots=0$, then the LHS is the sequence $c_1,c_2,\\ldots$ and the RHS is the sequence $0,0,0,\\ldots$, so all of the $c_i$'s are zero, which is linear independence.[/quote]\r\n\r\n\r\nRHS ? , LHS ? Thank u", "Solution_11": "In general, defining polynomials as functuions seems to me a bad idea: the polynomials $x$ and $x^p$ define the same function on $GF(p)$, different functions on $GF(p^2)$ and yet are clearly not the same polynomial." } { "Tag": [ "quadratics" ], "Problem": "Find three consecutive even integers if the product of the first and third integer is 192.", "Solution_1": "to make the lives of people simpler.. :D \r\n\r\n[hide=\"hint\"]\n\nthree consecutive numbers = (x-1)(x)(x+1)\n\nit could be written out as\n\n$(x^2-1)(x)$\n\nI hope it helps\n [/hide]", "Solution_2": "[quote=\"236factorial\"]Find three consecutive even integers if the product of the first and third integer is 192.[/quote]\r\n[hide]\n$(x)(x+4)=192$ such that $x$ is even.\n\n$x^2+4x-192=0$\n\n$x=\\dfrac{-4\\pm\\sqrt{16+768}}{2}=\\dfrac{-4\\pm\\sqrt{784}}{2}=\\dfrac{-4\\pm28}{2}=-2\\pm14$\n\n$-2+14=12$\n\n$-2-14=-16$\n\n$12*16=192$\n\n$-16*-12=192$\n\nThere are two possibilities: $12$, $14$, $16$ and $-16$, $-14$, $-12$[/hide]", "Solution_3": "[quote=\"shinwoo\"]to make the lives of people simpler.. :D \n\nthree consecutive numbers = (x-1)(x)(x+1)\n\nit could be written out as\n\n$(x^2-1)(x)$\n\nI hope it helps\n[/quote]\r\n\r\nCough. It's three consecutive [b][i][u]EVEN[/u][/i][/b] integers, not three consecutive integers. There's a difference.", "Solution_4": "There is a [b]slightly[/b] quicker way than Drunken_Math's solution... :D", "Solution_5": "o\r\n\r\nsorry\r\n :blush: \r\n\r\nI read the problem wrong\r\n\r\nthen\r\n\r\nit would be better with\r\n\r\n$(x-2)x(x+2)$", "Solution_6": "[quote=\"shinwoo\"]it would be better with \n\n$(x-2)x(x+2)$[/quote]\r\n\r\nYep, and the first and third terms are $(x+2)(x-2)$. Does that factorization look familiar?", "Solution_7": "[hide]\n$x\\cdot(x+4)=192$ \nUsing guess and check, you get that the answer is $\\framebox{12, 14, 16}$ and, you could get -12, -14, and -16[/hide]", "Solution_8": "what happened to the qudratics formula pluggies??\r\n\r\nnoone's using it?", "Solution_9": "[quote=\"shinwoo\"]what happened to the qudratics formula pluggies??\n\nnoone's using it?[/quote]\r\n\r\nShow us your solution...", "Solution_10": "I'll do this for the fun of it.\r\n\r\nWe know that the middle term will be $x$ so teh three consecutive terms are $(x-2)x(x+2)$.\r\nWe know that $(x-2)(x+2)=x^2-4\\rightarrow x^2=196\\rightarrow x=14$. Asa result, the consecutive numbers could be both $12,14,16$ and $-16,-14,-12$. :)\r\n\r\nMasoud Zargar", "Solution_11": "really really nice solution,boy. :D", "Solution_12": "[hide]12,14,16. I suppose the negative one would work but it doesn't seem right...[/hide]", "Solution_13": "[quote=\"236factorial\"]Find three consecutive even integers if the product of the first and third integer is 192.[/quote]\r\n\r\n[hide]Three integrers: $K; K+2; K+4.$\n\nThen, $K*(K+4)=192$\n${}^K{}^2+4K-192=0$\n$K=12$ or $K=-16$.\n\nThen, the numbers you`re looking for are $\\boxed {12; 14; 16}$ or $\\boxed {-16; -14; -12}$.[/hide]", "Solution_14": "[hide]$x-2,x,x+2$\n$(x-2)(X+2)=192$\n$x^2-4=192$\n$x^2=196$\n$x=14$ or $-14$\n\nso 12,14,16 or -16,-14,-12[/hide]" } { "Tag": [ "set theory" ], "Problem": "I don't quite understand a couple of things that AOPs vol. 1 says about russels paradox.\r\n\r\nFirst if A= {1,2,A} then A contains 1 and 2. Thus A is really just {A} because otherwise we have 1 and 2 twice in the set?\r\n\r\n\r\nAlso it later states that the set {1,2} is in the set M which consists of every set thta does not contain itself. Doesn't {1,2} contain {1,2}?\r\n\r\nThanks", "Solution_1": "But $A\\ne \\{1,2\\}$.\r\n\r\n$\\{1,2\\}$ does not contain itself. However, $A=\\{1,2,A\\}$ does contain itself.", "Solution_2": "[quote=\"ml2\"]I don't quite understand a couple of things that AOPs vol. 1 says about russels paradox.\n\nFirst if A= {1,2,A} then A contains 1 and 2. Thus A is really just {A} because otherwise we have 1 and 2 twice in the set?\n\n\nAlso it later states that the set {1,2} is in the set M which consists of every set thta does not contain itself. Doesn't {1,2} contain {1,2}?\n\nThanks[/quote]Could you quote where it says that {1,2} is the set of all sets that do not contain themselves? I find that hard to believe. Or do I misunderstand?", "Solution_3": "M is the set of all sets who do not contain themselves.\r\n\r\nIf M does not contain itself, it does contain itself.\r\n\r\nIf M does contain itself, it doesn't contain itself.\r\n\r\nSo does it contain itself.\r\n\r\nA not-so-satisfying solution is to invent metalanguages. Imagine a ladder. Put M on a rung that is not the bottom rung. Then let M only apply to the sets on the rung(rungs) below it. Thus there is no paradox.\r\n\r\nAnother example with the ladder. Let statement A be \"The sum of the angles of a triangle is pi radians.\" Say we wanted to prove it. Put it on the bottom rung. Then on rung two we could put the statement \"Statement A is true.\" and call it B. Then on rung three we could put \"Statement B is true.\"", "Solution_4": "[quote=\"RC-7th\"]But $A\\ne \\{1,2\\}$.\n\n$\\{1,2\\}$ does not contain itself. However, $A=\\{1,2,A\\}$ does contain itself.[/quote]\r\n\r\nI understand the paradox, I just don't understand the statements about sets containing themselves.. what does it mean for a set to contain itself is what confuses me. I don't understand what set A is if it is {1,2,A}.", "Solution_5": "Sets can contains themselves, or any other set, as an element of the set.\r\n\r\nFor example: a basic construction in set theory is the [i]power set[/i], which is the set of all possible subsets of a set. To take a simple example, suppose\r\n\\[\r\nS = \\{2,3,5\\}.\r\n\\]\r\nThen the power set of S, sometimes denoted P(S), is\r\n\\[\r\nP(S) = \\{ \\varnothing, \\{2\\}, \\{3\\}, \\{5\\}, \\{2,3\\}, \\{2,5\\}, \\{3,5\\}, \\{2,3,5\\} \\}.\r\n\\]\r\nP(S) has 8 elements. Each element is itself a set. In particular, $S \\in P(S)$.", "Solution_6": "[quote=\"ml2\"][quote=\"RC-7th\"]But $A\\ne \\{1,2\\}$.\n\n$\\{1,2\\}$ does not contain itself. However, $A=\\{1,2,A\\}$ does contain itself.[/quote]\n\nI understand the paradox, I just don't understand the statements about sets containing themselves.. what does it mean for a set to contain itself is what confuses me. I don't understand what set A is if it is {1,2,A}.[/quote]\r\n\r\nAs DPatrick said, you are misunderstanding about sets being elements of sets. If, for example B = {1, 2, 3}, then the set {4, B} is not equal to {1,2,3,4}, it is equal to {4, {1,2,3}}. It contains 2 elements, one of which is 4 and the other is a set containing 1, 2, 3. So {1,2} does not contain itself, since its only elements are 1 and 2 - none of its elements is (the set containing 1 and 2).", "Solution_7": "Ok, I get it now. Thanks everyone." } { "Tag": [ "floor function", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "What is nature of $ \\sum_{n \\equal{} 1}^{\\infty}\\frac {( \\minus{} 1)^{\\displaystyle{\\lfloor\\ln n\\rfloor}}}{n}$?", "Solution_1": "A typical block of terms between sign changes has sum approximately 1 in absolute value. The series diverges by the Cauchy criterion." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Compute:\r\n$\\sum_{1\\leq i\\leq n}{\\frac{1}{i}}+\\sum_{1\\leq i>R$, the final is much more significant. Therefore, the maximum speed is approximately the same in each case. And we don't have to use integration :lol:", "Solution_8": "When far away PE =0, and when at the centre PE = -KMm/r . Now M = $ \\lambda r$ , so r cancels and we get fixed PE and hence fixed speed in both cases.", "Solution_9": "[quote=\"calc rulz\"]I actually just realized my problem with the ring question. It is true that the potential energy from very far away does double, but since we're setting the potential energy to be $ 0$ at $ \\infty$, it's the potential energy in the center of the ring which is important. If the initial potential energy of the object for the first ring is $ \\minus{} \\frac {GmM}{d}$, then the value for the second ring is $ \\minus{} \\frac {2GmM}{d}$, where $ d$ is the vertical distance from the ring to the object. However, in both cases, the final potential energy is $ \\minus{} \\frac { \\minus{} GmM}{R}$, and since $ d > > R$, the final is much more significant. Therefore, the maximum speed is approximately the same in each case. And we don't have to use integration :lol:[/quote]\r\n\r\nAn even better solution. Darn, I should have realized that." } { "Tag": [ "induction", "inequalities proposed", "inequalities" ], "Problem": "If $x_{1},x_{2}, \\ldots, x_{n} \\in R$ and $x_{1} \\geq x_{2} \\geq \\ldots \\geq x_{n} >0,$ then prove that: $\\displaystyle \\frac{ (x_{3}+x_{4}+ \\ldots x_{n})(x_{1}-x_{2})}{x_{1}+x_{2}} + \\frac{ (x_{1}+x_{4}+ \\ldots +x_{n})(x_{2}-x_{3}) }{ x_{2}+x_{3}} + \\ldots + \\frac{ (x_{1}+x_{2}+ \\ldots +x_{n-2})(x_{n-1}-x_{n})}{x_{n-1}+x_{n}} + \\frac{ (x_{2}+x_{3}+ \\ldots +x_{n-1})(x_{n}-x_{1})}{x_{n}+x_{1}} \\geq 0.$\r\n\r\n\r\ncheers!! :D :D", "Solution_1": "Let $S = \\sum x$, then ineq becomes\r\n\\[\r\n\\sum\\limits_{i = 1}^n {\\frac{{\\left( {S - x_i - x_{i + 1} } \\right)\\left( {x_i - x_{i + 1} } \\right)}}{{x_i + x_{i + 1} }}} = \\sum\\limits_{i = 1}^n {\\frac{{S\\left( {x_i - x_{i + 1} } \\right)}}{{x_i + x_{i + 1} }}} - \\sum\\limits_{i = 1}^n {\\left( {x_i - x_{i + 1} } \\right)} = S\\sum\\limits_{i = 1}^n {\\frac{{x_i - x_{i + 1} }}{{x_i + x_{i + 1} }}} \r\n\\]\r\nSo it suffices to prove that\r\n\\[\r\n\\sum_{i=1}^n \\frac{x_i - x_{i+1}}{x_i + x_{i+1}} \\geq 0\r\n\\]\r\nand we'll use induction.\r\n\r\nThe original ineq only makes sense only $n \\geq 3$. If $n$ = 3, then\r\n\\[\r\n\\sum_{i=1}^3 \\frac{x_i - x_{i+1}}{x_i + x_{i+1}} = \\frac{(x_1 - x_2)(x_1-x_3)(x_2-x_3)}{(x_1 + x_2)(x_1+x_3)(x_2+x_3)}\r\n\\]\r\nand if it's true for $n$, then\r\n\\[\r\n\\frac{x_1 - x_2}{x_1 + x_2} + \\cdots + \\frac{x_{n+1} - x_1}{x_{n+1} + x_1}\r\n=\r\n\\left( \\frac{x_1 - x_2}{x_1 + x_2} + \\cdots + \\frac{x_{n} - x_1}{x_{n} + x_1} \\right)\r\n+\r\n\\left( \\frac{x_{1} - x_n}{x_{1} + x_n} + \\frac{x_{n} - x_{n+1}}{x_{n} + x_{n+1}} + \\frac{x_{n+1} - x_1}{x_{n+1} + x_1} \\right)\r\n\\]\r\nwhich is nonnegative by induction hypothesis.\r\n\r\n(except in the last case, $x_{n+1} = x_1$.)" } { "Tag": [ "modular arithmetic" ], "Problem": "i see why if for\r\n$ x\\equiv a_1 \\pmod{n_1}$\r\n$ x\\equiv a_2 \\pmod{n_2}$\r\n.\r\n.\r\n.\r\n$ x\\equiv a_k \\pmod{n_k}$\r\nthere is some k which divides more than one n_i, then the system can't have a solution. For example, if we have two congruences written\r\n$ \\frac{x\\minus{}a_1}{4}$ and $ \\frac{x\\minus{}a_2}{8}$ can't both produce whole numbers for some $ x$. But how do you show this is generally true?", "Solution_1": "[quote=\"Los\"]there is some k which divides more than one n_i, then the system can't have a solution.[/quote]\r\nNo: if there is some $ k > 1$ that divides more than one $ n_i$ then it is possible for the system to not have a solution. (But it is also possible that the system does have a solution.) This is not equivalent to what you wrote.\r\n\r\nIn general, you just need to choose values for the associated $ a_i, a_j$ that are incompatible mod $ k$, e.g. $ x \\equiv 1 \\pmod n_i$ and $ x \\equiv 0 \\pmod n_j$ (where $ k | (n_i, n_j)$).", "Solution_2": "aha. so for example\r\n\r\n$ x \\equal{} 1\\pmod{9}$\r\n$ x \\equal{} 4\\pmod{12}$\r\n$ x \\equal{} 22\\pmod{26}$\r\n\r\nhas a solution (it is $ x \\equal{} 100$), even though there is some $ k \\equal{} 3$ for which $ k|9$ and $ k|12$.\r\n\r\nTo see this write \r\n$ x\\equiv 4\\pmod{12}\\Rightarrow x\\equiv 4\\pmod{3} and x\\equiv 4 \\pmod{4}$\r\n and\r\n$ x\\equiv 22\\pmod{26}\\Rightarrow x\\equiv 22\\pmod{13} and x\\equiv 22 \\pmod{2}$,\r\nand you pick to work with the first of each, [color=red]because they have smaller numbers[/color](?). Then one solves the system\r\n\r\n$ x \\equal{} 1\\pmod{9}$\r\n$ x\\equiv 4\\pmod{3}$\r\n$ x\\equiv 22\\pmod{13}$\r\n\r\nin the usual way, whose numbers modulo all of which are coprime to one another." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "analytic geometry", "octahedron", "symmetry", "calculus" ], "Problem": "What does it mean to \"embed\" a tetrahedron into 4 space? In a solution to a problem I read, it labeled the vertices of a tetrahedron with ordered quadruplets (? right word) $ (1,0,0,0), (0,1,0,0), 0,0,1,0), (0,0,0,1)$. How does this work and why does it make calculations simpler?.\r\n\r\nAlso, when is it useful to use coordinates for 3d geometry problems? Given an octahedron, is there a way to label its vertices and quickly find a useful point in it such as the centroid of one of the faces of the octahedron?", "Solution_1": "I think the word is \"quadruple\". To me $ \\mathbb{R}^{4}$ means your 3 usual dimensions and time as the fourth, so I have no idea how you could lable a tetrahedron with four coordinates unless it somehow changed shape with time.\r\nTo answer your other question, I often find it useful to draw a picture if I'm trying to parametrize a surface or get a better picture of some shape. Also, if you need to find some property, like area, distance, volume,... it may help to draw the region. Hope that helps.", "Solution_2": "You can theoretically have 4 spacial coordinates, you can theoretically have infinitely many spacial coordinates! For an informal discussion, Flatland is an interesting novella to check out.", "Solution_3": "I'm using hide-tags because this is sorta long\r\n\r\n[hide=\"Regarding the usage of 4-space\"]When you're faced with an abstract situation in several dimensions, it often helps to think of lower-dimension analogies.\n\nHere, the problem is analogous to naming the vertexes of an equilateral triangle $ \\{ (1,0,0), \\; (0,1,0), \\; (0,0,1) \\}$. Did you have to put the triangle in 3-space? Of course not, you were welcome to use\n $ \\left \\{ (0,0), \\; ( \\sqrt {2}, 0), \\; \\left( \\frac {\\sqrt {2}}{2}, \\frac {\\sqrt {6}}{2} \\right) \\right \\}$ instead.\n\nIt definitely makes distances easier to calculate if your numbers are all 0's and 1's. For example, you can easily find the centers of each face; they are the permutations of $ \\left( \\frac {1}{3}, \\; \\frac {1}{3}, \\; \\frac {1}{3}, \\; 0 \\right)$. The center of the whole tetrahedron is $ \\left( \\frac {1}{4}, \\; \\frac {1}{4}, \\; \\frac {1}{4}, \\; \\frac {1}{4} \\right)$.\n\nI can see how someone might be uncomfortable with the thought, but I think it's a wonderful simplification of calculations.[/hide]\n[hide=\"When to use coordinates\"]Great reasons to use Cartesian coordinates are when:\n- there are lots of right angles\n- the numbers are relatively simple/there is lots of symmetry\n- when there are free variables in the problem, ie. you can make simplifying assumptions\n- you need to optimize something and you are good with inequalities/with calculus\n- you want to use vectors, dot products, cross products, etc.\n\nGood reasons to use coordinates are when:\n- you are in the mood to brute-force something\n- you don't know what to do[/hide]\n[hide=\"Coordinatizing an octahedron and such\"]\nThe coordinates of an octahedron's vertexes are just the permutations of $ (\\pm 1, 0, 0)$. It helps to consider duality here - the [color=blue][url=http://en.wikipedia.org/wiki/Dual_polyhedron]dual polyhedron[/url][/color] to the octahedron is the cube.\n\nThe centroid of any regular polygon/polyhedron/etc and of most high-symmetry polygons/polyhedrons/etc is just the arithmetic mean of the vertexes.[/hide]" } { "Tag": [ "factorial" ], "Problem": "What is the product of all positive odd integers less than $1000$?\r\n(Using factorial notation where necessary).", "Solution_1": "1*3*5*...*999\r\n=1000!/[(2^500)(500!)]", "Solution_2": "I know the answer but I need a solution. Please show your work.", "Solution_3": "[hide=\"Sure\"]\n$1 \\times 3 \\times 5 \\times... \\times 999$ is the answer we wish to express in factorial notation. Note that this is $999!$ but with the even numbers divided out, or $\\frac{999!}{2 \\times 4 \\times... \\times 998}$. Furthermore, we can factor out a $2$ from each factor in the denominator, producing $\\frac{999!}{2^{499} \\times 1 \\times 2 \\times... \\times 499} = \\frac{999!}{2^{499} 499!} = \\frac{1000!}{2^{500} 500!}$.\n[/hide]", "Solution_4": "First write:\r\n$1000! = (999*997*995*\\dots*3*1)*(1000*998*\\dots*4*2)$\r\n$1000! = (999*997*995*\\dots*3*1)*2^{500}(500*499*\\dots*2*1)$\r\n(Factor out two 500 times since it is a product)\r\n$\\frac{1000!}{2^{500}*500!}$", "Solution_5": "Yeah, thats what I got after as well. On the actual AMC, the number was $10000$ not $1000$ sorry :oops:" } { "Tag": [ "blogs", "email", "function", "search", "\\/closed" ], "Problem": "We're now focusing on getting the blogs in shape. I'm going to maintain a bug list and a wish list/todo list here. Please add any bugs you find here and I'll add them to the BUG list. Also, add items you'd like to see in the blogs here, and I'll maintain a TODO list.\r\n\r\nBUG list:\r\n* Get the post count and views count on the list of all blogs correct\r\n* Add the Reply Counter\r\n* Allow remove contributors\r\n* Still a few missing language entries in the Blog options area.\r\n* Post/reply does not have the same style as the blog itself.\r\n\r\nTODO list\r\n* Bring back the shout boxes (we have all the old shouts)\r\n* Bring back the old two templates\r\n* Implement categories, so people can have categories for their blog entires.\r\n* Make the Blog index page more distinct from the forums.\r\n* Need a way to see a list of all the subjects posted in a blog (this existed in the old system)\r\n* Email subscriptions/RSS\r\n* Mood emoticons\r\n\r\n\r\nPlease add more bugs and wishlist items so we can get the blog system running smoothly!", "Solution_1": "On background images: I suppose you just type the address into the box, http://billydorminy.homelinux.com/bg.jpg , right? Because it doesn't seem to be working on my blog. Is the image too big? (2272x1704, 252KB) Or does it not work because it's not hosted on the artofproblemsolving.com server?", "Solution_2": "Just like before, you have to use the proper CSS syntax: namely you have to type \r\n[code]url(http://billydorminy.homelinux.com/bg.jpg)[/code]", "Solution_3": "All items have CSS identifiers, so you can easily put your own CSS code (in the Blog Customizing section) if you are not satisfied with your look) ;)", "Solution_4": "By the way, as always, if you are posting a bug here, please let us know what browser (and version) you are using -- certain bugs will only show up in certain browsers.", "Solution_5": "This time, I think I have found a real problem with the blogs :) \r\n\r\nActually, the view thing is better the way it is.\r\n\r\nI am more concered about the number of entries being different from the actual number posted.\r\n\r\nFor example, in my blog, I have posted 9 entries, yet it only says 7 on the blog page (it says 9 inside my blog). When I posted my 9th entry today, it did not increase the number of entries.\r\n\r\nAlso this appears to be true in almost ever blog I look at.", "Solution_6": "On my blog, one of the entries is contained in another entry. The two entries are connected, with two scrollbars.\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/weblog.php?w=223]The last two entries on this page.[/url]", "Solution_7": "I have posted another entry in my blog, yet it still says on this page:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/apsjour.php\r\n\r\nthat I have only 7 entries.\r\n\r\nAlso, the IMO blog has 0 entries (there are at least 3+).", "Solution_8": "A possibly related bug: The count for replies are off for some entries.\r\n\r\nFor example:\r\nhttp://www.mathlinks.ro/Forum/weblog.php?w=305\r\nThe welcome topic has 4 replies, but it only shows up as 2.", "Solution_9": "Yeah, the replies counter is not fixed yet apparently. I will deal with it once I am back from the IMO.", "Solution_10": "my post counter is messed up. before i had 15 entries, and it said i had 13 entries. now i put in another one, and it still says 13 entries. i think it's stuck on 13 or something, which would be creepy, except i'm not superstitious. :lol: i use internet explorer something. i don't know the number, but i know it's the newest one.", "Solution_11": "[quote=\"Captain Sugar\"]my post counter is messed up. before i had 15 entries, and it said i had 13 entries. now i put in another one, and it still says 13 entries. i think it's stuck on 13 or something, which would be creepy, except i'm not superstitious. :lol: i use internet explorer something. i don't know the number, but i know it's the newest one.[/quote]\r\n\r\npress help$\\to$about to see the version", "Solution_12": "Also something is that I can't remove contributors from my list. It always says, General Error, SQL requests not achieved, etc...", "Solution_13": "[quote=\"b-flat\"]I am more concered about the number of entries being different from the actual number posted.[/quote]\r\n\r\nMe too--I posted three entries in my new blog and the counter is still stuck at 0.\r\n\r\nIE 6, Windows XP", "Solution_14": "i think we should bring the reply counter back. it's annoying looking at every post to see if anyone's replied.", "Solution_15": "I have this problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112086 and I don't know what to do about it, and nothing has happened about it. I'd really like to remove some contributors. :(", "Solution_16": "[quote=\"rrusczyk\"]* Email subscriptions/RSS[/quote]\r\nIs this going to happen any time in the near future? I'd like Facebook to be able to read my AoPS blog posts so that I can keep maintaining just one blog and conveniently allow my friends to navigate to it / know when I've updated.", "Solution_17": "Haha I made a php script to automatically create the RSS of my blog. If you want, I can make you one too...", "Solution_18": "[quote=\"Xevarion\"][quote=\"rrusczyk\"]* Email subscriptions/RSS[/quote]\nIs this going to happen any time in the near future? I'd like Facebook to be able to read my AoPS blog posts so that I can keep maintaining just one blog and conveniently allow my friends to navigate to it / know when I've updated.[/quote]First steps are already taken: here's the feed for the site: http://www.mathlinks.ro/Forum/feed.php (still in beta though as you can see :) ).", "Solution_19": "[quote=\"chess64\"]Haha I made a php script to automatically create the RSS of my blog. If you want, I can make you one too...[/quote]\r\n\r\nCould you please make me one? Maybe I could edit it to add a body text...\r\n\r\nWhat about the problem that the style of the index of a blog isn't the style of the blog itself.", "Solution_20": "[quote=\"anirudh\"][quote=\"chess64\"]Haha I made a php script to automatically create the RSS of my blog. If you want, I can make you one too...[/quote]\n\nCould you please make me one? Maybe I could edit it to add a body text...[/quote]You have to understand that PHP is a server language. Unless you have your own server/site/php server you can't do anything with a php script.", "Solution_21": "Yes.\r\n\r\nBut I blocked a few people from my blog. :lol: \r\n\r\nI created a .PHP script somewhat like this:\r\n\r\n[code]/* MY CSS PORTION */\nCSS GOES HERE\n\nCSS WENT HERE\n/* END CSS PORTION */\n\n/*START IP BANNING */\n\";\necho \"If you think you have been banned in error please contact me.\";\n$fp = fopen(\"code/data/ip_data.dat\", \"a\"); \nfputs($fp, \"**BANNED** Visit logged on $getdate at $gettime internet time) for IP: $getip\n\");\nfputs($fp, \"\");\nfclose($fp);\nexit(); \n} \n}\necho \"Update in progress.\";\necho \"
\";\necho \"'Authorized Visit' detected and logged on $getdate at $gettime internet time) for IP: $getip\";\n\n$fp = fopen(\"code/data/ip_data.dat\", \"a\"); \nfputs($fp, \"Authorized Visit logged on $getdate at $gettime internet time) for IP: $getip\n\");\nfputs($fp, \"\");\nfclose($fp);\n/* END IP BAN */\n?>\n\n[/code]\r\n\r\n\r\nI htink it worked okay. :maybe:\r\n\r\nOH. Yes. I do have a PHP supporting website, theani.isgreat.org\r\n\r\nIt parses the PHP scripts...", "Solution_22": "[quote=\"#H34N1\"]Yes.\n\nBut I blocked a few people from my blog. :lol: \n\nI created a .PHP script somewhat like this:\n\n[code]/* MY CSS PORTION */\nCSS GOES HERE\n\nCSS WENT HERE\n/* END CSS PORTION */\n\n/*START IP BANNING */\n\";\necho \"If you think you have been banned in error please contact me.\";\n$fp = fopen(\"code/data/ip_data.dat\", \"a\"); \nfputs($fp, \"**BANNED** Visit logged on $getdate at $gettime internet time) for IP: $getip\n\");\nfputs($fp, \"\");\nfclose($fp);\nexit(); \n} \n}\necho \"Update in progress.\";\necho \"
\";\necho \"'Authorized Visit' detected and logged on $getdate at $gettime internet time) for IP: $getip\";\n\n$fp = fopen(\"code/data/ip_data.dat\", \"a\"); \nfputs($fp, \"Authorized Visit logged on $getdate at $gettime internet time) for IP: $getip\n\");\nfputs($fp, \"\");\nfclose($fp);\n/* END IP BAN */\n?>\n\n[/code]\n\n\nI htink it worked okay. :maybe:\n\nOH. Yes. I do have a PHP supporting website, theani.isgreat.org\n\nIt parses the PHP scripts...[/quote]\r\nI don't think that works... it's an external linked .css file, not an inline style.\r\n\r\nEDIT: Yeah, it doesn't work. It just assumes that it's awry CSS. See: http://www.artofproblemsolving.com/Forum//templates/ptifo/blogs/Hyperion/style/c746.css\r\n\r\nEDIT2: \"Blog Started: Today, at 4:14 am\" It's been around for a month. What's going on?", "Solution_23": "[quote=\"Temperal\"]EDIT: Yeah, it doesn't work. It just assumes that it's awry CSS. See: http://www.artofproblemsolving.com/Forum//templates/ptifo/blogs/Hyperion/style/c746.css[/quote]Of course it doesn't work. That would be a serious security hole if it would work!", "Solution_24": "I suppose..\r\nIs there any place in the blog that you can place raw HTML?", "Solution_25": "I can't recreate my blog since I can't select one. I THINK THAT THIS IS A SAD EVENT BECAUSE I WILL NOW BLOG ELSEWHERE.", "Solution_26": "[quote=\"mathnerd314\"]The server also thinks that I started my blog at 7:47 today (that's a plane!!! :pilot:). Probably because that's when I last accessed it.[/quote]\r\n\r\nI have the same problem... Instead of showing the date when I posted the entry, it displays the time when I accesed the blog. It says \"filed on Today, ...\".\r\nI want it to display the date and time when I submitted my entry.", "Solution_27": "All right I'll see what I can do. Seems to be a bug indeed.", "Solution_28": "Plus, could some form of BB Code or at least the basic html tags be allowed in the custom block thing?", "Solution_29": "[quote=\"Temperal\"]Plus, could some form of BB Code or at least the basic html tags be allowed in the custom block thing?[/quote]\r\n\r\nThat would be great :)" } { "Tag": [ "complex analysis" ], "Problem": "Is it possible to map the open unit disk conformally on $ \\{z: \\ 0<|z|<1\\}$? Since boundary should be mapped onto boundary, it seems to be impossible, but this problem is given in one of the first topics and it requires elementary solution.. Any ideas are appreciated.", "Solution_1": "Is a conformal map here bijective? \r\n\r\nIf so the standard argument is that by the open mapping theorem, the map is actually biconformal, and in particular a homeomorphism. Hence such a map preserves simple connectedness (although to prove this you probably need to know about induced homomorphisms of fundamental groups). Thus, such a map cannot exist since the unit disc (disk?) is simply connected, while the given set is not.\r\n\r\nI guess not too helpful for your purposes :oops:\r\nDo you know about isolated singularities? If so, this also gives a direct argument. What do you know?", "Solution_2": "Thanks for your comment. Yes, i know about isolated singularities. How does it help here? :maybe:", "Solution_3": "Well, again I am supposing you mean a bijective map. So suppose the map exists, then we have an analytic inverse map $ f$ from the punctured disc to the disc. $ 0$ is an isolated singularity of f. It cannot be a pole or an essential singularity, because then points close to 0 would be mapped outside the disc. So it's removable, say $ f(0) \\equal{} a$. Clearly $ |a| \\leq 1$. $ |a| \\equal{} 1$ contradicts the open mapping theorem, so $ |a| < 1$. If $ f'(0) \\equal{} 0$ then we immediately have a contradiction to injectiveness of $ f$ on the punctured disc (do you know why?). So $ f'(0) \\neq 0$, but this means that $ f$ maps an arbitrarily small neighborhood of 0 topologically with a neighborhood of $ a$. Since there's also a different point $ z$ in the unit disc which is mapped to $ a$ there is again a contradiction to the injectiveness of $ f$ on the punctured disc.\r\n\r\nThis turned out to look sort of complicated, but it's really not", "Solution_4": "The argument with singularity is nice, might be useful in future :)", "Solution_5": "Well not really, because the problem of which sets can be biconformally mapped to the unit disc has been completely solved. The solution is what I said in my first reply + riemann mapping theorem.", "Solution_6": "How about if the map is not required to be injective?", "Solution_7": "It's doable. Apply a M\u00f6bius transform that maps it to the left half-plane. Apply the exponential function." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Solve this nice ineq\r\nIf $a,b,c$ are positive prove that \r\n\r\n$\\frac{a}{(2a+b+c)(b+c)}+\\frac{b}{(2b+a+c)(a+c)}+\\frac{c}{(2c+a+b)(a+b)}\\geq\\frac{9}{8(a+b+c)}$\r\n\r\nBy the way who is the creator ?", "Solution_1": "Let $a+b+c=1$\r\n=>(1)<=> $\\sum{a/(1-a^2)}\\geq 9/8$ <=> $\\sum{a^2/(a-a^3)}\\geq 9/8$\r\nWe know that $a_1^2/b_1 + a_2^2/b_2+ a_3^2/b_3\\geq (a_1+a_2+a_3)^2/(b_1+b_2+b_3)$=>\r\n $\\sum{a^2/(a-a^3)}\\geq 1/(1-a^3-b^3-c^3)\\geq 9/8$\r\n Becous $(a^3+b^3+c^3)/3\\geq ((a+b+c)/3)^3$ ;) :lol:", "Solution_2": "The official solution is always appeared with a big speed :rotfl: \r\n\r\nCould we prove it with expansion?", "Solution_3": "The reason not in,it is simple it a children's problem :P :rotfl:", "Solution_4": "[quote=\"Tiks\"]The reason not in,it is simple it a children's problem :P :rotfl:[/quote]\r\n\r\nIt is from Romania Shortlist 2004. It is not a very easy problem", "Solution_5": "[quote=\"silouan\"][quote=\"Tiks\"]The reason not in,it is simple it a children's problem :P :rotfl:[/quote]\n\nIt is from Romania Shortlist 2004. It is not a very easy problem[/quote]\r\nMabe,but I think it is very-very easy.", "Solution_6": "Silouan, what shorlist are you talking about? :D", "Solution_7": "Here is another solution\r\n${\\sum\\frac{a}{(2a+b+c)(b+c)}=\\frac{1}{2}\\sum[\\frac{1}{b+c}-\\frac{1}{2a+b+c} ] \\ge \\frac{1}{2}[\\sum\\frac{1}{b+c} - \\frac{1}2}\\sum\\frac{1}{b+c} ] = \\frac{1}{4}\\sum\\frac{1}{b+c} \\ge \\frac{9}{8(a+b+c)}$\r\nwhere the first inequality follows from the well-known fact\r\n$\\sum\\frac{1}{x+y}\\le \\frac{1}{2}\\sum\\frac{1}{x}$ (put $x=b+c$, $y=c+a$ and $z=a+b$)\r\nand the second from CBS inequality.", "Solution_8": "[quote=\"razec upul\"]Silouan, what shorlist are you talking about? :D[/quote]\r\n\r\nIt is Shortlist Romania 2004. This was told by Nickolas and mr. Stergiou in their book\r\npage 237", "Solution_9": "I don't quite think so. I have the shorlist and it isn't such an inequality. Actually this inequality has been published for\r\nthe first time in romanian olympiad literature in 1988, if i'm not mistaken. :D :P" } { "Tag": [ "geometry", "geometric transformation", "reflection" ], "Problem": "I assume we can discuss it, as it's on the MC site.\r\n\r\nThis was one of the two I skipped. I got 44 in chapters. I was wondering if anyone else got it?\r\n\r\nhttp://www.mathcounts.org/competition/2008%20Chapter%20Sprint%20Round.pdf\r\n\r\nI have no clue on this one, esp. w/o a calculator. 8 min on this would be fine, as that would be an accurate reflection of many people's leftover time.", "Solution_1": "most important step is to notice that \"pi\" is greater than 3.14 but less than 22/7.\r\nKnowing that you can rewrite without the abosolute value signs:\r\n|pi-3.14|-|pi-22/7|=pi - 3.14 - pi + 22/7=22/7 - 3.14=3 1/7 - 3 7/50= 1/7 - 7/50= (50-49)/350= [b]1/350[/b]", "Solution_2": "I don't want to rewrite how I did it, so...\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=195667", "Solution_3": "basically the same as the other guys:\r\n\r\n|pi-3.14| - |pi-22/7|\r\n\r\nsince 22/7 is greater than pi, it becomes pi-3.14-pi+22/7\r\n22/7-3.14\r\n2200/700-2198/700\r\n2/700\r\n1/350", "Solution_4": "Lol Alex, I got that wrong because I forgot to simplify, i had 2/700. That question made me get a 45. :(\r\n\r\nI'm an idiot. :mad: :(", "Solution_5": "[quote=\"TheHobo\"]Lol Alex, I got that wrong because I forgot to simplify, i had 2/700. That question made me get a 45. :Sad\n\nI'm an idiot. :Mad :Sad[/quote]\r\n\r\nDon't feel bad...I did that too. :D", "Solution_6": "OMG! I did that too! I got a 45 all because of it.", "Solution_7": "I was running out of time and I couldn't finish #30 so I had to take an educated guess and this is what I put:\r\n\r\n1/350\r\n\r\nI was like, OMG OMG OMG I GOT IT RIGHT!\r\n\r\nI still only got a 41, though. :blush:", "Solution_8": "Educated guess? Could you describe that, please?" } { "Tag": [ "geometry", "circumcircle", "inradius", "inequalities", "limit", "inequalities proposed" ], "Problem": "We well know with triangle $ABC$ and three median $m_{a},m_{b},m_{c}$, circumradius $R$, inradius $r$ then\r\n$m_{a}+m_{b}+m_{c}\\le \\frac{9}{2}R$\r\n$m_{a}+m_{b}+m_{c}\\le 4R+r$\r\nthus with the inequality\r\n$m_{a}+m_{b}+m_{c}\\le (\\frac{9}{2}-k)R+2kr,\\forall k>0$\r\nWhat's the best const $k$ ? \r\n :)", "Solution_1": "With my calculation\r\n$k=\\lim_{\\triangle ABC\\rightarrow equaliteral}\\frac{m_{a}+m_{b}+m_{c}-\\frac{9R}{2}}{R-2r}=\\frac{3}{2}$\r\nBut follow me it isn't the best const because we cann't find the best const by limit.\r\nWhat is your ideal ?", "Solution_2": "[color=purple]gemath i think if $k=\\frac{3}{2}$ the triangle need the condition. Have you got a solution of this problem.If you have please post it[/color]", "Solution_3": "[quote=\"evarist\"][color=purple]gemath i think if $k=\\frac{3}{2}$ the triangle need the condition. Have you got a solution of this problem.If you have please post it[/color][/quote]\r\n [color=purple] In triangle which has two angle > $\\frac{\\pi}{3}$ we have$p\\leq \\sqrt{3}(R+r)$ thus$\\sqrt{3}p\\leq 3(R+r)$so \n$m_{a}+m_{b}+m_{c}\\leq \\ 3(R+r)+\\frac{|a-b|+|b-c|+|c-a|}{4}$\n What a pity! Thus i belive$k=\\frac{1}{2}$ is the best constant.Vacs can you show me the best constant? [/color]", "Solution_4": "There isn't any pity here :D :D I don't say the best constant is $\\frac{3}{2}$ I only show the limit equal $\\frac{3}{2}$ and I don't understand your example for this problem, if you want to say the best constant isn't $\\frac{3}{2}$ then you only show to exist a triangle such that $m_{a}+m_{b}+{m}_{c}\\ge 3(R+r)$ and actually I have many examples for it (by computer), you shouldn't make wordy !" } { "Tag": [ "probability", "counting", "derangement" ], "Problem": "There are 4 different pair of gloves each having a different color. There are 4 persons. All the right hand gloves are distributed at random, one to each person. Then, all the left hand gloves are distributed randomly in the same manner.\r\n\r\nWhat is the probability that at least one person gets a matching pair?", "Solution_1": "[hide=\"Brute force\"]\n4 is small enough where we can just list all possibilities. We can assume the right hand gloves are RBYG. Then, we can systematically list all possibilities where it doesn't work:\nBRGY\nBYGR\nBGRY\nYBGR\nYGRB\nYGBR\nGRBY\nGYBR\nGYRB\nThere are 9 possibilities out of 24, so the answer is $ 1\\minus{}\\frac {3}{8}\\equal{}\\frac{5}{8}$[/hide]", "Solution_2": "[quote=\"dgreenb801\"][hide=\"Brute force\"]\n4 is small enough where we can just list all possibilities. We can assume the right hand gloves are RBYG. Then, we can systematically list all possibilities where it doesn't work:\nBRGY\nBYGR\nBGRY\nYBGR\nYGRB\nYGBR\nGRBY\nGYBR\nGYRB\nThere are 9 possibilities out of 24, so the answer is $ 1 \\minus{} \\frac {3}{8} \\equal{} \\frac {5}{8}$[/hide][/quote]\r\n\r\nThe left hand gloves are distributed first and then the right hand gloves are distributed, both at random. We have to find the possibility of getting an identical pair of gloves. How are you finding that?", "Solution_3": "It is 1- the probability of no one getting a pair of the same gloves. We can assume it starts out with any order we want because for any other order, the probability would be the same.", "Solution_4": "Side note: this is just a special instance of the derangement or rencontre problem, $ 9 \\equal{} D(4)$." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "Support", "summer program", "MathPath", "Davidson Fellows" ], "Problem": "Hi, I see no one has posted here yet about the [url=http://www.ditd.org/public/article.aspx?cid=9&mid=124&tp=122]Davidson Young Scholars[/url] program. Maybe the program is still too new to have gained much attention, so I'll bring it up here for those of you who might be interested. Do any of you have any experience with the program?", "Solution_1": "i received a phamplet in the mail. the people who wins the prizes for mathematics are always imo people. (no point in registering in math unless you are at aleast USAMO level i guess).", "Solution_2": "If people wanted to know more about it, you could ask anders. He is registered on this site and is a Davidson fellow for math. He received a $25,000 scholarship. He also received a gold medal at the IMO.", "Solution_3": "[quote=\"kel227\"]If people wanted to know more about it, you could ask anders. He is registered on this site and is a Davidson fellow for math. He received a $25,000 scholarship. He also received a gold medal at the IMO.[/quote]\n...or Daniel Kane, another familiar name to some of us, for that matter. He got $50K from Davidson.", "Solution_4": "The first three replies are referring to the [url=http://www.ditd.org/public/article.aspx?cid=1&mid=101&tp=2]Davidson Fellows[/url] program, which indeed gave high awards to IMO team members Daniel Kane and Anders Kaseorg, and which looks very interesting. But for the younger mathletes who aren't at the level of a MOPper yet, there is a distinct program, the [url=http://www.ditd.org/public/article.aspx?cid=9&mid=124&tp=122]Davidson Young Scholars[/url] program, which may be of interest to the younger participants on this AoPS site. I'm wondering if anyone here has heard of the Young Scholars program.", "Solution_5": "I am going to reenergize this thread because there are many new young people on AoPS who I think ought to know about the Davidson Institute for Talent Development [url=http://www.ditd.org/public/article.aspx?cid=9&mid=124&tp=122]Young Scholars Program[/url]. I get the impression that there are several AoPSers who would qualify for and enjoy being part of the Young Scholars Program.", "Solution_6": "Yea I've heard of the Davison Young Scholars. I just got accepted as one, I'm really happy!", "Solution_7": "I'm in the program since 2003, didn't go to any gatherings yet (out of the country every summer).", "Solution_8": "It's good to meet Young Scholars program participants here.", "Solution_9": "how do we apply and when? and what is it? is it like summer program??", "Solution_10": "The Davidson Young Scholars program has an [url=http://www.ditd.org/public/article.aspx?cid=129&mid=250&tp=122]application information page[/url] on its Web site that explains most of the procedures for applying. I have also heard that the program staff, reachable through various channels listed on the Web site, are happy to answer questions about the application process. \r\n\r\nThe program operates partly as a way for young people to connect with one another through online groups and occasional in-person gatherings, and partly as a parent support network through those channels. There are also staff members and professional consultants to the program who have advice on educational planning and learning opportunities. The program provides financial aid to some families based on financial need. \r\n\r\nIt's my impression that some of the young people on AoPS are advanced enough in math that they might be eligible for the Davidson Young Scholars program, which is why I bring this up.", "Solution_11": "Hi,\r\n\r\nI'm one of the Hobbits (2002 Young Scholars), and I'm going to the Denver Gathering in June.", "Solution_12": "Since I have been to MathPath I have become even more aware than before that there are some REALLY SMART young people who hang out here on AoPS (sometimes lurking more than posting) who ought to apply to the [url=http://www.ditdservices.org/Articles.aspx?ArticleID=24&NavID=0_0]Davidson Young Scholars Program[/url]. It's one more way, besides AoPS, for math-liking young people to network with one another, and for some families the Davidson Young Scholars program may make many summer opportunities (such as those discussed in other threads on this forum) more affordable, through financial aid. Don't be bashful about [url=http://www.ditd.org/Articles.aspx?ArticleID=26]contacting the Davidson Institute headquarters[/url] if you have questions about the program, and by all means apply if you think you might fit into the program.", "Solution_13": "I've been in the program since it started, 1998-1999 I think, I was 4, they almost didn't accept me cause I was so young.", "Solution_14": "Greetings to the new and old and could-be Young Scholars who are posting in this thread.", "Solution_15": "Do i still qualify if i got a 790 in the Math portion of the SAT LAST year, but did horrible on the Verbal? Do I have to retake it this year? Also, it says i need a copy of an IQ test. Where do i take IQ tests?", "Solution_16": "You would have to ask, although I don't think SAT scores have anything to do with it, four year olds can't exactly take SAT's can they? The application has gotten much easier over the years, the standards really arn't that high now. I don't think most people on here would have a problem getting in. As for IQ tests, just ask your parents and/or school so you can go take one." } { "Tag": [ "articles", "search", "Princeton", "college", "topology", "MIT" ], "Problem": "I just read a recent news article about a guy who solved the poincare conjecture few weeks ago. How many of you know about this? Can someone please tell me more about it (the article was in Chinese and it very brief)?", "Solution_1": "[quote=\"answerseeker95\"]I just read a recent news article about a guy who solved the poincare conjecture few weeks ago. How many of you know about this? Can someone please tell me more about it (the article was in Chinese and it very brief)?[/quote]\r\n\r\nHere's a Google News search that will turn up a recent AP article (multiple instances of it) in English: \r\n\r\nhttp://news.google.com/news?q=Poincare+conjecture \r\n\r\nHope this helps!", "Solution_2": "[quote=\"answerseeker95\"]I just read a recent news article about a guy who solved the poincare conjecture few weeks ago. How many of you know about this? Can someone please tell me more about it (the article was in Chinese and it very brief)?[/quote]\r\nRecent? You mean they officially declared it a legitimate proof? If so then I didn't know that.\r\n\r\nI'm guessing most of us know of this because the Poincare Conjecture were proven months ago, there were threads about it in like September. Just search for it and you should find a thread about it.", "Solution_3": "Nope, it hasn't been verified yet. Give it another couple of years :).", "Solution_4": "Huh, I heard in two years so I thought they'd be done next year...", "Solution_5": "Here is a slightly more informative article on Perelman's Proof of the Poincare Conjecture: http://mathworld.wolfram.com/news/2003-04-15/poincare/\r\n\r\nMost people who understand it seem to be more or less convinced of the legitimacy of the proof. But technically, Perelman has not published the proof in any journal yet - just on the internet. In order to receive the $1 million prize he'll have to publish it in a journal and then wait 2 years from that point for it to pass scrutiny. Those are the rules set up by the Clay Institute, who sponsor the reward. No one is sure why he hasn't published the result yet. People are speculating that he's just not interested in the money.\r\n\r\nThere's an interesting article in Discover magazine this month (Jan. '04) on the nature of proof in mathematics. Proofs today are so long and complicated that guaranteeing their validity is becoming less and less certain, even to experts.", "Solution_6": "Oh, I was talking about legitimate proof. Well, I was quite mistaken, but the proofs and news certainly seem interesting.", "Solution_7": "According to some professors at my university, there is quite a bit of debate about the correctness of Perelman's proof. He might not want to publish it until he is more or less certain that it is correct. By the way, where did he publish it on the internet?", "Solution_8": "try \r\n[url]http://www.math.lsa.umich.edu/research/ricciflow/perelman.html[/url]\r\n\r\nit has a lot of commentary and notes on perelman's work.", "Solution_9": "Here's another good layman's article on the proof from the Daily Princetonian (the campus newspaper for Princeton University): http://www.dailyprincetonian.com/archives/2003/04/17/news/7979.shtml", "Solution_10": "Here goes a writer's somewhat funny take on Perelman's achievement: [url=http://www.reflector.com/news/newsfd/auto/feed/news/2004/01/09/1073710748.26609.2365.0197.html;COXnetJSessionID=AEK0Yx3GUTkw0JvfOx0MfYuIfIfBAFQO2YF0HfcfPUNqgkYHgxgo!-759395321?urac=n&urvf=10740394121510.5083651816148208]Funny Article[/url]\r\n\r\nAnd here goes a more technical explaination of the actual Conjecture that he proved: http://en2.wikipedia.org/wiki/Thurston%27s_Geometrization_Conjecture\r\n\r\nHere is a timeline that I constructed (as best I could) of the confusing chain of events surrounding the proof:\r\n\r\nNov. 11, 2002 - Grigori \"Grisha\" Perelman quietly publishes the first of his papers, \"The entropy formula for the Ricci flow and its geometric applications\" (39 pages) on the internet through arXiv.org - an archive for electronic preprints of math and physics papers. Here he (purportedly) proves Thurston's Geometrization Conjecture, which has the Poincare Conjecture as a consequence. There is very little fanfare or mention of the history or significance of these problems within the paper.\r\n\r\nMarch 10, 2003 - Perelman puts up a second paper \"Ricci flow with surgery on three-manifolds\" (22 pages) on arXiv clearing up some of the points in the first paper. Again it passes without a great deal of attention from the popular media, but the buzz begins to spread in the mathematical world.\r\n\r\nApril 7-11, 2003 - Perelman gives a series of lectures at MIT describing his work to peers. The international buzz really picks up that he may have proven the Poincare Conjecture.\r\n\r\nApril 15, 2003 - Eric Weisstein puts up a news article on MathWorld with the headline \"Poincar Conjecture Proved--This Time for Real\" (following a faulty proof that was announced just a month before). An article also appears in the NY Times. Several other announcements follow.\r\n\r\nApril 16, 2003 - Perelman gives a lecture at Princeton University describing his work.\r\n\r\nJul. 17, 2003 - Perelman puts up a third paper \"Finite extinction time for the solutions to the Ricci flow on certain three-manifolds\" (7 pages) on arXiv dealing with some more points from the first two papers.\r\n\r\nEverything remains quiet for a while as mathematicians ponder and try and understand Perelman's work. Perelman remains quiet and refuses to talk to the media.\r\n\r\nDec. 2003 - Discover magazine includes a discussion of Perelman's proof in an article on the 100 biggest science stories of 2003.\r\n\r\nJan. 2004 - Paul Elias of the Associated Press writes an article about Perelman and the proof which is picked up by several outlets, including CNN.com. Several mathematicians claim they are becoming convinced of the validity of Perelman's proof. A renewed buzz begins and people wonder if Perelman will ever publush his results in a formal journal; and if, or when, he will receive the $1 million dollar prize from the Clay Institute. Will he have to publish it in an actual journal, or is the electronic publication and peer review currently going on sufficient? Stay tuned for more....[/url]", "Solution_11": "[quote=\"gauss202\"]Here is a timeline that I constructed (as best I could) of the confusing chain of events surrounding the proof: . . . Jan. 2004 - Paul Elias of the Associated Press writes an article about Perelman and the proof which is picked up by several outlets, including CNN.com. Several mathematicians claim they are becoming convinced of the validity of Perelman's proof. A renewed buzz begins and people wonder if Perelman will ever publush his results in a formal journal; and if, or when, he will receive the $1 million dollar prize from the Clay Institute. Will he have to publish it in an actual journal, or is the electronic publication and peer review currently going on sufficient?[/quote]\r\n\r\nThanks for the timeline, which is very helpful for explaining why this story is resurfacing at intervals. The aspect of the Associated Press story that I found most interesting was precisely the discussion of the Clay Institute requirement that a prize-winning proof be \"published\"--with the tantalizing suggestion that print publication in a peer-reviewed dead-tree journal may not be the only form of publication that would meet the prize criterion. Perelman's preprints must make some interesting reading for mathematicians searching for clues to Poincare's long-standing problem.", "Solution_12": "There is an article about the proof on CNN (the website). I'm sure it won't be too hard to find once navigated to the site itself." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $x$, $y$, $z$, $p$, $q$ and $r$ be positive reals s.t. $p+q+r=x^py^qz^r=1$. Prove inequality\r\n\\[\\frac{p^2x^2}{qy+rz}+\\frac{q^2y^2}{px+rz}+\\frac{r^2z^2}{px+qy}\\geq\\frac{1}{2}.\\]", "Solution_1": "Kvant volume X, problem number $Y$ required.", "Solution_2": "?\u0443 \u0438 \u0445?\u043d", "Solution_3": "[quote=\"Myth\"]\u0445?\u043d[/quote]\r\n\r\nWhat does that mean ?", "Solution_4": ":blush: :blush: :blush: \r\nAt least no-russians can't understand it.", "Solution_5": "[quote=\"orl\"]Kvant volume X, problem number $Y$ required.[/quote]\r\n\r\nRequest is still up-to-date. I guess I should make a language journey to Russia for 3-4 month to refresh my knowledge. :lol:", "Solution_6": "[quote=\"Myth\"]Let $x$, $y$, $z$, $p$, $q$ and $r$ be positive reals s.t. $x+y+z=x^py^qz^r=1$.[/quote]\r\n\r\nMyth, $x+y+z=x^py^qz^r=1$ doesn't make any sense at all, and neither does $xyz = 1$ as stated in the topic subject. Are you sure that all the variables are positive reals?", "Solution_7": "You are right, it was typo.\r\nI will edit it.", "Solution_8": "Oh,Dear Myth.\r\nBefore you edit problem,I couldn't find any $x,y,z,p,q,r$ satisfy,because $x+y+z=1$.\r\nNow it's quite clear.\r\n$\\frac{p^2x^2}{qy+rz}+\\frac{q^2y^2}{px+rz}+\\frac{r^2z^2}{px+qy}\\ge\\frac{px+qy+rz}{2}$\r\nBut use Cauchy,$px+qy+rz \\ge x^py^qr^z=1$\r\nAnd we are done!\r\nAm I wrong? :) :)" } { "Tag": [], "Problem": "Well, i know a lot of people who remember tons of interstnig quotes (I'm one of them, actually)... Quotes by famous scientists, poets and philosophers... And ofcourse quotes from movies, TV shows... What are your favourite quotes?", "Solution_1": "look in round table and find vihag's old topic by the same name.", "Solution_2": "[quote=\"bubka\"]look in round table and find vihag's old topic by the same name.[/quote]\r\n\r\nWell, I saw it, but I didn't want to revive it... Maybe some of new members wolud like to share some of their favourite quotes with us? :)", "Solution_3": "Are you looking for serious quotes, or random stuff people say on AIM?", "Solution_4": "\"I'm the most terrific liar you ever saw in your life. It's awful. If I'm on my way to the store to buy a magazine, even, and somebody asks me where I'm going, I'm liable to say I'm going to the opera.\"", "Solution_5": "Read my Sig :D\r\nanother good quote:\r\n \"<(^^);..> You have just recieved the Amish Computer Virus. Since the Amish don't have computers, it is based on the honor system. So please delete all the files from your computer. Thank you for your cooperation\"\r\n\r\n \" The tradgey of Canada is that they could have had British culture, French cooking, and American technology. Instead they got American culture, British cooking and French technology\"", "Solution_6": "\"I'm not racist. Racism is a crime, and crime is for black people.\"\r\n\r\n\"I'd give my right arm to be ambidextrous.\"\r\n\r\n\"Procrastinators unite! Tomorrow...\"\r\n\r\n\"All rentals are due on the due date.\" -Blockbuster", "Solution_7": "I have some from Black Saga and the game Halo 2:\r\n\r\n\"I can accept failure. Everyone fails at something. But I cannot accept not trying.\" ~Michael Jordan.\r\n\r\n\"Teamwork wins championships.\" or something like that, from Michael Jordan\r\n\r\n\"It blew right through us--50 cal, Rockets, didn't do a thing.\" ~Marine\r\n\r\n\"Retrieving the Icon is my only concern.\" ~The Arbiter\r\n\r\n\"It is easy human. Take the icon in your hands..AND DO AS YOU ARE TOLD!\" ~Tartarus\r\n\r\n\"Grab a banshee. He's gonna know we're coming.\" ~ Seargant Johnson\r\n\r\n\"Me, inside your head, now.\" ~Cortana\r\n\"Chief, blink if you can hear me.\" ~Cortana\r\n\r\n\"The Demon? He is here?\" ~The Arbiter\r\n\r\n\"This is not your grave...but you are welcome in it.\" ~Gravemind.\r\n\r\n\"Do not reveal yourselves only after the Arbiter has joined the enemy in battle. Arbiter, you too may wish to do the same. But take heed-your armor is not as new as ours.\" ~White Spec Ops Commander\r\n\r\nMore to come but I am forgetting at this very moment" } { "Tag": [ "Euler", "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let triangle ABC inscribed in circumcircle (O). The Euler line of triangle ABC respectively intersects AB, BC, CA at C', A', B'. (The Euler line of triangle ABC is the line which passes through the orthocenter and the center of circumcircle of triangle ABC).\r\n1) Prove that the circles that take AA', BB', CC' be their diameter have the same radical axes.\r\n2) Suppose those circles above intersects at M, N. Prove that: In two points, one lies on the (O), one lies on the Euler circle of triangle ABC.\r\n :D :blush:", "Solution_1": "I think we have the next problem: :D \r\n\r\nLet triangle $ ABC$ inscribed in circumcircle $ (O)$. Let $ d$ is a line through $ O$.The line $ d$ respectively intersects $ AB, BC, CA$ at $ C', A', B'$. \r\n1) Prove that the circles that take$ AA', BB', CC'$ be their diameter have the same radical axes. \r\n2) Suppose those circles above intersects at $ M, N$. Prove that: In two points, one lies on the $ (O)$, one lies on the Euler circle of triangle $ ABC$.", "Solution_2": "(For generalized problem...)\r\n\r\nI proved $ (1)$ and half of $ (2)$\r\n\r\nBut why do those three circles meet on the Euler circle?! T. T\r\n\r\nI cannot get it\r\n\r\nPlz anybody show me the solution for the very last problem.. :wallbash: :wallbash_red:", "Solution_3": "Dear Mathlinkers,\r\nthe problem posted at first comes from Victor Th\u00e9bault.\r\nAny reference?\r\nSincerely\r\nJean-Louis", "Solution_4": "[quote=\"ma 29\"]I think we have the next problem: :D \n\nLet triangle $ ABC$ inscribed in circumcircle $ (O)$. Let $ d$ is a line through $ O$.The line $ d$ respectively intersects $ AB, BC, CA$ at $ C', A', B'$. \n1) Prove that the circles that take$ AA', BB', CC'$ be their diameter have the same radical axes. \n2) Suppose those circles above intersects at $ M, N$. Prove that: In two points, one lies on the $ (O)$, one lies on the Euler circle of triangle $ ABC$.[/quote]\r\nMr. ma 29, vey nice. Can you show us your solution to my proposed problem?? I am hoping to see it from you. Actually, I'm \"dying\" to see that if there is a nice solution about this problem!! :blush: :D", "Solution_5": "[quote=\"ma 29\"]I think we have the next problem: :D \n\nLet triangle $ ABC$ inscribed in circumcircle $ (O)$. Let $ d$ is a line through $ O$.The line $ d$ respectively intersects $ AB, BC, CA$ at $ C', A', B'$. \n1) Prove that the circles that take$ AA', BB', CC'$ be their diameter have the same radical axes. \n2) Suppose those circles above intersects at $ M, N$. Prove that: In two points, one lies on the $ (O)$, one lies on the Euler circle of triangle $ ABC$.[/quote]\r\n\r\nSee [url]http://www.mathlinks.ro/viewtopic.php?t=186883[/url], the supporting lemma, for all claims.", "Solution_6": "Pedal circle of point on $ d$ pass through constant point which lie on Euler circle of triangle $ ABC$. Circle with diameter $ AA'.BB',CC'$ are pedal circle of $ A',B',C'$ respectively. Hence this three circle have common point on Euler circle." } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "Determine the structure of the ring\r\n(a) $\\mathbb{Z}[x]/(x^2+3,3)$,\r\n(b) $\\mathbb{Z}[x]/(x^2+3,5)$,\r\n(c) $\\mathbb{Z}[x]/(x^2-3,2x+4)$.\r\n\r\nMy answers are\r\n(a) $\\mathbb{Z}_3[\\sqrt 3]$\r\n(b) $\\mathbb{Z}_5[\\sqrt{-3}]$\r\nand I have no idea about (c).\r\n\r\np.s. Can the answer to (a) be simplified more?", "Solution_1": "$(x^2-3,2x+4)=(x^2+2x+1,2x+4)=((x+1)^2,2(x+1)+2)$. Let $y=x+1$.\r\n$(y^2,2y+2)=(y^2,2y+2,2y^2-y(2y+2))=(y^2,2y+2,-2y)=(y^2,2,-2y)=(y^2,2)$.\r\n\r\nSo $\\mathbb{Z}[X]/(X^2-3,2X+4)=\\mathbb{Z}[X]/((X+1)^2,2)\\simeq \\mathbb{Z}_2[X]/((X+\\hat{1})^2)\\simeq \\mathbb{Z}_2[\\sqrt2-1]$. I think.", "Solution_2": "How to obtain $\\mathbb{Z}_2[X]/((X+\\hat{1})^2)\\simeq \\mathbb{Z}_2[\\sqrt2-1] $?\r\n\r\nI think $\\mathbb{Z}_2[x]/((x+1)^2)=\\mathbb{Z}_2[x]/(x^2+2x+1)=\\mathbb{Z}_2[x]/(x^2+1)\\cong \\mathbb{Z}_2[i] $", "Solution_3": "$\\mathbb{Z}_2[x]/((x+1)^2)=\\mathbb{Z}_2[x]/((x+1)^2-2)\\ldots$", "Solution_4": "Is my answer right, then?", "Solution_5": "Yes" } { "Tag": [ "algebra", "system of equations" ], "Problem": "I am totally lost at how to go about solving this problem could someone please help? I read your stickies and other posts to make sure I ask a question in the correct forum if I did not I apologize and make sure I post in the correct place next time.\r\n\r\n5x+2y=2x+1\r\n2x-3y=3x+2\r\nsolve by addition Thank you very much.[/u]", "Solution_1": "The two most conventional ways to solve a system of equations are substitution and elimination (in your case, 'addition' is elimination) we proceed as follows (some may have a more concise way, but this is how I did it first time through):\r\n\r\n$5x+2y = 2x+1$\r\n$2x-3y = 3x+2$\r\n\r\nmove all variables to one side:\r\n$3x+2y = 1$\r\n$-x-3y = 2$\r\n\r\nmultiply the second equation by 3, so that upon addition, we [b]eliminate[/b] the x variable:\r\n\r\n$3x+2y = 1$\r\n$-3x-9y = 6$\r\n\r\nadd:\r\n\r\n$-7y = 7$\r\n\r\nsolve:\r\n\r\n$y =-1$\r\n\r\nthen you substitute that back in to one of the original equations to find x! Done and done.", "Solution_2": "btw, this would probably be more appropriate for the [url=http://www.artofproblemsolving.com/Forum/index.php?f=300]classroom math[/url] forum :wink: \r\n\r\nwelcome to AoPS slowlearner" } { "Tag": [ "limit", "real analysis", "real analysis solved" ], "Problem": "a) Let $(x_n)_{n\\ge0}$ and $(y_n)_{n\\ge0}$ be two sequences of real numbers, with $y_n=\\displaystyle\\frac1{2^n}\\sum_{k=0}^nC_n^kx_k,\\forall n\\in\\mathbb N$. Prove that $(x_n)_{n\\ge0}$ is convergent [b]iff[/b] $(y_n)_{n\\ge0}$ is convergent.\r\nb)Compute $\\displaystyle\\lim_{n\\to\\infty}\\frac n{2^n}\\sum_{k=1}^nC_{n-1}^{k-1}\\left\\{\\sqrt{k^2+2k+2}\\right\\}$ where $\\{x\\}$ denotes the fractional part of $x$.\r\n\r\n[i]Octavian Stroe, Paul Valceanu[/i]", "Solution_1": "a/ if $x_n$ then $y_n$ converge it is an application of Cesaro theorem", "Solution_2": "If $y_n$ converge this not give $x_n$ converge\r\n\r\nConsider $x_k=(-1)^k$ \r\n\r\n$y_n=0$ converge but $x_n=(-1)^n$ does not converge", "Solution_3": "ohh... I'm sorry... I'm dumb :blush: \r\nif $x_n$ is convergent then $y_n$ is convergent and their limits are equal.", "Solution_4": "For a/ just $x_n \\;converge \\Rightarrow y_n \\; converge$ is true\r\n\r\nthere exist $n_0$ s.t. for any $n\\geq n_0$ $ |x_n-x|\\leq \\epsilon$\r\n\r\n$|y_n-x|\\leq \\frac{\\sum_{k=0}^{n_0}|x_k-x|C_n^k}{2^n}+\\frac{\\sum_{k=n_0+1}^{n}|x_k-x|C_n^k}{2^n}$\r\n\r\nThe fisrt term on RHS $\\leq \\frac{max_{0\\leq k \\leq n_0}|x_k-x|}{2^n}(1+n+n^2/2!+...+n^{n_0}/n_0!)$ wich tends to $0$\r\n\r\nThe second term on RHS $\\leq \\epsilon$\r\n\r\n$y_n$ tend to $x$", "Solution_5": "Cerinta corecta a fost; \"Aratati ca daca sirul Xn este convergent atunci si sirul Yn este convergent....\"", "Solution_6": "[quote=\"Kuba\"]\nb)Compute $\\displaystyle\\lim_{n\\to\\infty}\\frac n{2^n}\\sum_{k=1}^nC_{n-1}^{k-1}\\left\\{\\sqrt{k^2+2k+2}\\right\\}$ where $\\{x\\}$ denotes the fractional part of $x$.\n\n[i]Octavian Stroe, Paul Valceanu[/i][/quote]\r\n\r\n\r\nUse $kC_n^k=nC_{n-1}^{k-1}$ you get \r\n\r\n$x_k=k\\left(\\sqrt{k^2+2k+2}-E(\\sqrt{k^2+2k+2}\\right)$ \r\n\r\nNow what is $\\lim x_k$ ?", "Solution_7": "$x_n=n\\{\\sqrt{n^2+2n+2}\\}=n\\sqrt{n^2+2n+2}-n(n+1)=\\frac n{\\sqrt{n^2+2n+1}+n+1}\\Rightarrow \\lim_{n\\to\\infty}x_n=\\frac12$" } { "Tag": [], "Problem": "\u0397 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03af\u03b1 \u03bf\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c9\u03c2 \u03b5\u03be\u03ae\u03c2 : \r\n \u03b1(0)=0, \u03b1(1)=1, \u03b1(2)=2, \u03b1(3)=6 \u03ba\u03b1\u03b9 \r\n \u03b1(n+4)=2\u03b1(n+3)+\u03b1(n+2)-2\u03b1(n+1)-\u03b1(n) , n>=0.\r\n \u039d\u03b1 \u03b4\u03b5\u03b9\u03be\u03b5\u03c4\u03b5 \u03bf\u03c4\u03b9 \u03c4\u03bf n \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03b9 \u03c4\u03bf \u03b1(n) \u03b3\u03b9\u03b1 \u03ba\u03b1\u03b8\u03b5 n>0.", "Solution_1": "kali Nick... :wink: \r\n\r\nhint\r\n[hide]eukola pernume oti $a_{x}=\\frac{x}{\\sqrt{5}}(\\frac{1+\\sqrt{5}}{2})^{x}-\\frac{x}{\\sqrt{5}}(\\frac{1-\\sqrt{5}}{2})^{x}$\nto opio ine akereo polaplasio tu $x$[/hide]\r\n\r\n :D", "Solution_2": "\u03c0\u03bf\u03bb\u03c5 \u03c9\u03c1\u03b1\u03b9\u03b1 \u03a1\u03ac\u03bb\u03bb\u03b7! \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b4\u03b7\u03bb \u03b1(n)=nf(n) opoy f(n) \u03bf \u03bd\u03b9\u03bf\u03c3\u03c4\u03bf\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c2 fibonacci!\r\n \u0392\u03c1\u03b7\u03ba\u03b1 \u03ba \u03bc\u03b9\u03b1 \u03c9\u03c1\u03b1\u03b9\u03b1 \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd \u03c7\u03b8\u03b5\u03c2 . \r\n \u039d\u03b1 \u03b2\u03c1\u03b5\u03b9\u03c4\u03b5 \u03bf\u03bb\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03ce\u03c4\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 p, q \u03c9\u03c3\u03c4\u03b5 \u03c4\u03bf pq \u03bd\u03b1 \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03b9 \u03c4\u03bf \r\n 2^p +2^q.", "Solution_3": "\u039d\u03af\u03ba\u03bf \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5 \u03b2\u03ac\u03b6\u03b5\u03b9\u03c2 \u03bb\u03ad\u03c9 \u03b1\u03c5\u03c4\u03ad\u03c2 \u03c4\u03b9\u03c2 \u03c9\u03c1\u03b1\u03af\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03c3\u03c4\u03b1 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1 \u03c4\u03cc\u03c0\u03b9\u03ba\u03c2 \u03c4\u03bf\u03c5 \u03b2\u03b9\u03b2\u03bb\u03af\u03bf\u03c5? :wink:", "Solution_4": "\u0394\u03b5\u03bd \u03ad\u03c7\u03c9 \u03c0\u03b1\u03c1\u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03ae\u03c3\u03b5\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03ae \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7\u03bd \u03c9\u03c1\u03b1\u03af\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b5\u03b9\u03b1 \u03c0\u03bf\u03c5 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03c3\u03ba\u03ad\u03c6\u03c4\u03b7\u03ba\u03b1 \u03bd\u03b1 \u03c4\u03b7 \u03b2\u03ac\u03bb\u03c9 \u03b5\u03ba\u03b5\u03af . \r\n\u03a4\u03b7\u03bd \u03b5\u03c0\u03cc\u03bc\u03b5\u03bd\u03b7 \u03b8\u03b1 \u03c4\u03b7 \u03b2\u03ac\u03bb\u03c9 \u03b5\u03ba\u03b5\u03af .", "Solution_5": "[quote=\"nikos\"]\n \u0392\u03c1\u03b7\u03ba\u03b1 \u03ba \u03bc\u03b9\u03b1 \u03c9\u03c1\u03b1\u03b9\u03b1 \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd \u03c7\u03b8\u03b5\u03c2 . \n \u039d\u03b1 \u03b2\u03c1\u03b5\u03b9\u03c4\u03b5 \u03bf\u03bb\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03ce\u03c4\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 p, q \u03c9\u03c3\u03c4\u03b5 \u03c4\u03bf pq \u03bd\u03b1 \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03b9 \u03c4\u03bf \n 2^p +2^q.[/quote]\r\nElpizw na einai safes auto pou akolouthei(k swsto :D )\r\n\r\nAn $q=2$ tote $2p/ 4+2^{p}$ dld $p/4+2^{p}$. $2^{p}\\equiv2 \\mod p$. Epomenws $p/6$ kai $p=2,3$. Elegxontas pairnoume ta zeugh $(2,2)$ kai $(3,2)$.\r\nEstw $(p,2)=1$ kai $(q,2)=1$. \r\n\r\nAn $q=3$ pairnoume $3p/8+2^{p}$ 'h $3/8+2^{p}$ pou tha simaine $p$ artios, atopo. Ara $q\\neq 3$ k profanws k $p\\neq 3$.\r\n\r\n$2^{p}\\equiv 2\\mod p$ kai $2^{q}\\equiv 2 \\mod q$ kai etsi eukola katalhgoume stis $2^{p-1}\\equiv-1\\mod q$ kai $2^{q-1}\\equiv-1\\mod p$.\r\n\r\nEstw $r,z$ oi elaxistoi thetikoi arithmoi tetoioi wste $2^{r}\\equiv 1 \\mod q$, $2^{z}\\equiv-1 \\mod q$. Ws gnwston, an $2^{f}\\equiv 1 \\mod q$, tote $r/f$. Parathroume oti $2^{2z}\\equiv 1 \\mod q$ k epomenws $r/2z$ pou shmainei $2z\\geq r$. Estw $p-1=rk+l$, $0\\leq l< r$. Tote \r\n\r\nan $l=0$ exoume $-1\\equiv 2^{p-1}\\equiv 2^{rk-1}\\equiv 2^{r(k-1)+r-1}\\equiv 2^{r-1}\\mod q$ kai epomenws afou $2^{r}\\equiv 1 \\mod q$ pairnoume $3\\equiv 0 \\mod q$, dld $q=3$ pou exoume apokleisei. \r\n\r\nara $l\\neq 0$ kai epomenws $-1\\equiv2^{rk+l-1}\\equiv 2^{l-1}\\mod q$ kai epomenws $z\\leq l-1R\r\nf,g are continous functions such that for each x in [a,b], f(x)0 such that for each x in [a,b] f(x)+c < g(x)\r\n--------------------------------------------------------------------------------------\r\nI have tried:\r\nh(x) = g(x) - f(x), and so h(x) is continues and h(x) > 0.\r\nThen probably some variation of the intermediate value theorem should be used, but I am not sure in what way?\r\n\r\nThanks.", "Solution_1": "I guess that you shoud think about uniform continuity of f and g.", "Solution_2": "hello. Yes, I've tried with uniform continuity. And here is the proof:\r\n\r\n$ \\ h(x)$ is uniformly continous. (By Cantor theorem).\r\n\r\nso $ \\forall \\epsilon > 0 , \\exists \\delta > 0 : |h(x) \\minus{} h(y)| < \\epsilon \\ , \\ when \\ |x \\minus{} y| < \\delta , \\ \\ \\ \\forall x,y \\in [a;b]$\r\n\r\n$ \\ h(y) \\ \\minus{} \\ \\epsilon \\ < \\ h(x) \\ \\ \\forall \\ x,y \\in [a;b]$\r\n\r\n$ \\ h(y) \\ \\minus{} \\ \\epsilon \\ > \\ 0 \\ because \\ h(x) \\ > \\ 0$\r\n\r\n$ \\ (g(x) \\ \\minus{} \\ f(x))$ < $ \\ (h(y) \\ \\minus{} \\ \\epsilon) \\ > \\ 0$\r\n\r\nso, $ \\ \\exists \\epsilon \\ > 0 \\ such \\ : \\ f(x) \\ \\minus{} \\ g(x) \\ < \\ (\\epsilon \\ \\minus{} \\ h(y) ): \\equal{} \\ \\minus{} c \\ ; \\ c: \\equal{} \\ h(y) \\ \\minus{} \\epsilon \\ , \\ \\ \\ \\forall \\ x,y\\in[a;b]$ . h(y) > 0, and $ \\epsilon > 0$ is \"as small as you want\", so we took such $ \\epsilon > 0$ that $ \\ c > 0$.", "Solution_3": "Hm... there might be a few errors in Blymblams post, but the idea is there.Denote $ h(x)\\equal{}g(x)\\minus{}f(x)$, let's take some $ \\varepsilon>0$ we can find $ \\delta>0$ that for all $ x,y\\in [a,b]$ such that $ |x\\minus{}y|<\\delta$ we have $ |h(x)\\minus{}h(y)|<\\varepsilon$. From that we have $ 0 \\ h(y) \\ \\minus{} \\ \\epsilon$ and $ \\ h(y) \\ \\minus{} \\ \\epsilon$ wouldnt be continous in $ \\ [a;b]$ ? sorry if i dismissing smth :)", "Solution_5": "[quote=\"Blymblam\"]\nwhat is the purpose of constructing the interval: $ \\ [x_{i \\minus{} 1} \\ ; \\ x_i]$ ? It's because $ \\ h(x) \\ > \\ h(y) \\ \\minus{} \\ \\epsilon$ [/quote]\n\nYes, on $ [x_{i\\minus{}1},x_i]$ we have the desired inequality. Notice, that this inequality doesn't hold on all $ [a,b]$, only on intervals that have length smaller then $ \\delta$. ( that was the mistake in your solution, among some other... ).\n\n[quote=\"Blymblam\"]\nand $ h(y)\\minus{}\\varepsilon$ wouldnt be continous in ?[/quote]\r\n\r\n$ h$ is continous on $ [a,b]$ and adding/subtracting a constant still makes it continous." } { "Tag": [ "trigonometry", "algebra proposed", "algebra" ], "Problem": ":? Can you solve this equation\r\nsin(sin(sin(sin(x))))=cos(cos(cos(cos(x))))\r\nIs it nice??? :( :blush: :P :rotfl:", "Solution_1": "Interesting, $\\sin (\\sin \\cdots( \\sin(x))\\cdots ) - \\cos (\\cos \\cdots( \\cos(x))\\cdots )$ seems to be always negative if the number of iterations is large enough. I wonder how many iterations do they need to have?", "Solution_2": "[quote=\"phuchung\"]\nIs it nice??? [/quote]\r\n\r\nHmmm...I think that the nicest part is the so precise reference you gave... ;) \r\n\r\nBtw, you may have a look here :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=sin%28sin&t=10285\r\n\r\nPierre." } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "Aqui les dejo dos problemas que pusieron en los entrenamientos de YUCATAN:\r\nel entrenador nos dijo que era una APMO, y tambien les dejo el otro el cual lo saco de un libro de teoria de numeros del cual no se su nombre, en el inciso b del problema 2 todavia no lo eh resuelto a si que si tienen alguna solucion que la pongan para poder verla:\r\n\r\n1)determine todos los enteros positivos n para los que la ecuacion:\r\n \r\n xn + (2+x)n + (2-x)n = 0\r\n\r\ntiene una solucion entera.\r\nlos n despues de los numeros puestos significan que estan elevados a la potencia n.\r\n\r\n2)(a)Pruebe que cada elemento del conjunto:\r\n\r\n (n! + 2, n! + 3, ... , n! + n)\r\n\r\nes divisible por un primo que no divide a ning\u00fan otro elemento del conjunto.\r\n\r\nb)Pruebe que para todos los enteros positivos n y k, existe un conjunto de n enteros consecutivos tal que cada uno de los elementos de este conjunto es divisible por k primos distintos ninguno de los cuales divide a los otros elementos del conjunto.", "Solution_1": "[quote=\"Tony2006\"]\n1)determine todos los enteros positivos n para los que la ecuacion:\n \n xn + (2+x)n + (2-x)n = 0\n\ntiene una solucion entera.\n[/quote]\r\n\r\nLo paso: $x^n+(2+x)^n+(2-x)^n=0$", "Solution_2": "si es eso ya lo hiciste? :lol:", "Solution_3": "[quote=\"Tony2006\"]si es eso ya lo hiciste? :lol:[/quote]\r\n\r\nNo. Solamente lo pasaba para que lo entiendan mejor.. alguna ayuda?", "Solution_4": "para el problema 1 de las soluciones que se la que mas me gusto fue una que fue por congruencias, las otras: por polinomios y otra por el teorema del binomio de newton. \r\nuna sugerencia para congruencias es ver que el numero x es de la forma 2 ala \"a\" para \"a\" entero no negativo \r\ndonde 2 ala \"a\" divide a 2 ala \"n+1\"", "Solution_5": "para el problema 2 inciso a la sugerencia es que factorizen n! + k para k entero entre 2 y n.\r\ny para el inciso b no lo se por que todavia no lo eh hecho si alguien tiene alguna idea que lo comente a qui por favor!! :lol:" } { "Tag": [ "geometry", "National Assessment" ], "Problem": "Dear Math Enthusiast,\r\n\r\n We at National Assessment & Testing are pleased to announce our 2006-2007 competition schedule (appended). The five contests we administer cost just fifty dollars each (or you can register for all five and receive the package price of two-hundred dollars) and can be more convenient than local face-to-face contests, requiring only two hours of your coach's time during the week. Our contests are easy to administer, and frequently a contest and its solutions provide material for two team meetings, reducing your coach's need to create original materials. Each of our competitions feature problems that range from simple to advanced, to provide both confidence and challenges to all students regardless of their level of expertise. Our contests can also be rewarding, with high-scoring individuals in multiple divisions receiving plaques and certificates, high-scoring schools receiving awards and recognition through press releases to local newspapers, and the highest scoring individual at each school receiving a certificate. In addition to the contests we administer, we also produce a competition that your team can administer for local elementary and middle school students as both a service project and fundraiser.\r\n\r\n If these sound like the kind of extracurricular mathematical experiences you might like to experience, ask your coach to look over the appended schedule. If they have any questions, they can contact me at clymer@natassessment.com or (206) 433-7320. If your school would like to register for one or more of our competitions, your coach can simply e-mail us with the contests you'd like to register for and your school's registration information.\r\n\r\nThank you,\r\n\r\nTom Clymer\r\nDirector of Academic Competitions\r\nNational Assessment & Testing\r\nhttp://www.natassessment.com\r\n\r\n\r\n\r\n2006-2007 Competition Schedule\r\n\r\n\r\nThursday, September 28th, 2006: 2006 Fall Startup Event\r\n\r\n This is a fun, fast-paced competition in which individuals have 30 minutes to solve 100 problems. The questions cover a wide variety of topics, and are all quick and easy for students familiar with the subject. Analogous to a musician practicing scales, this contest can help students and teams identify areas they want to focus on.\r\n\r\n\"Excellent time at the start of the year. Excellent opportunity to interest new ninth-graders in math team - it's the competition that draws them in.\" - Jane Fenn, Corning-Painted Post West High School\r\n\r\n\r\nThursday, November 2nd, 2006: 2006 Team Scramble\r\n\r\n In this exciting collaborative experience, many students work together to answer 100 questions in thirty minutes. Team leaders who understand the strengths of various team members will be able to divide the work to efficiently answer the most questions, while recruiting many students allows for more double-checking of the team's work.\r\n\r\n\"Kids went crazy when we gave them the solution key after the test.\" - Jim Ferguson, Dana Hills High School\r\n\r\n\r\nThursday, December 7th, 2006: 2006 Ciphering Time Trials\r\n\r\n This individual competition consists of ten rounds in which students have three minutes to solve three problems. Students must decide which problems they will be able to answer quickly, and whether to attempt less familiar problems or double-check questions with which they are more comfortable.\r\n\r\n\"Kids loved the fast pace!\" - Patrick Montague, Science Academy of South Texas\r\n\r\n\r\nThursday, February 1st, 2007: 2007 Four-by-Four Competition\r\n\r\n This is a team contest with ten rounds in which teams of four students have three minutes to solve four problems. Students learn to work together and quickly distribute problems to team members with strengths in the corresponding subjects.\r\n\r\n\"It's fun to have a creative format like this.\" - Sue Forster, Bismarck High School\r\n\r\n\r\nWednesday, March 28th, 2007 - April 4th, 2007: 2007 Collaborative Problem-Solving Contest\r\n\r\n This competition consists of fifteen intricate multi-part questions that your entire school (including staff) can collaborate on for an entire week. Students get experience using real-world problem-solving skills such as brainstorming, collaboration, research, and application of technology.\r\n\r\n\"We had a great time with this contest - we can't wait until next year! With better planning, we hope to make this an even bigger event at our school.\u00a0 As it was, I had to chase kids home tonight.\" - Tim Corica, The Peddie School\r\n\r\n\r\n2006 & 2007 Middlementary Math Bonanzas\r\n\r\n We provide everything your high school math team needs to administer a face-to-face competition for your local elementary and middle school students! Administering a math contest is a terrific experience for your high school students, a great fundraiser for your team, a good service to provide to your community, and an investment in the students that will make up your team in the future. You can currently register for the 2006 and 2007 versions of this contest at a cost of two-hundred dollars each.", "Solution_1": "The Fall Start Up really drew me into math team. \r\n\r\n\r\nI look forward to Time Trials!" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "Find the integer numbers $x,y,z,t$ which satisfy $x+y+z=t^2$, $x^2+y^2+z^2=t^3$.\r\n\r\n[i]Lucian Dragomir[/i]", "Solution_1": "by cauchy\r\n\r\n$t^3=x^2+y^2+z^2\\geq\\frac{1}{3}(|x|+|y|+|z|)^2\\geq\\frac{1}{3}(|x+y+z|)^2=\\frac{t^4}{3}$\r\n\r\nSo $t^3(t-3)\\leq 0$ or $0\\leq t\\leq 3$ .\r\n\r\nNow the rest is easy by case work :)", "Solution_2": "I guess it's the only solution..." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "If the sum of two numbers is 8 and the sum of their cubes is 20, what is the sum of their squares?", "Solution_1": "[hide=\"Hint\"] $a^{3}+b^{3}= (a+b)(a^{2}-ab+b^{2})) = (a+b)( (a+b)^{2}-3ab)$\n$a^{2}+b^{2}= (a+b)^{2}-2ab$ [/hide]", "Solution_2": "[hide=\"answer\"]\n$a+b = 8, b = 8-a$\n$a^{3}+b^{3}= 20$\n$\\Downarrow$\n$16a^{2}= 41$\n$a = \\frac{\\sqrt{41}}{4}$\ntherefore \n$b = \\frac{32-\\sqrt{41}}{4}$\n\n$\\left(\\frac{\\sqrt{41}}{4}\\right)^{2}+\\left(\\frac{32-\\sqrt{41}}{4}\\right)^{2}= \\frac{41}{16}+\\frac{1024-64\\sqrt{41}+41}{16}$\n\nso the sum of their squares is\n\n$\\frac{1106-64\\sqrt{41}}{16}$\n[/hide]", "Solution_3": "You might want to check that third step. \r\n\r\n[hide=\"Solution\"] A slight re-shuffling of my hint.\n\n$(a+b)^{3}= 512 = (a^{3}+b^{3})+3ab(a+b) = 20+24ab \\implies ab = \\frac{41}{2}$\n\nThen\n\n$a^{2}+b^{2}= (a+b)^{2}-2ab = 64-41 = \\boxed{ 23 }$. [/hide]" } { "Tag": [ "topology" ], "Problem": "A topological space is called normal if every point in it is a closed set, and further, given disjoint closed sets A and B there are disjoint open sets U and V such that U contains A and V contains B\r\nA topological space is termed metrizable if there exists a metric such that the topology arises as the topology of the metric\r\n[b]Problem:[/b] Prove that any metrizable space is normal.", "Solution_1": "Do the obvious! Let $ A$ and $ B$ be closed disjoint sets. For each $ a \\in A$, there exists $ \\epsilon_a$ so that $ B(a, \\epsilon_a) \\cap B \\equal{} \\emptyset$. Similarly, for each $ b$ there exists $ \\epsilon_b$ so that the corresponding ball misses \r\n$ A$. \r\n\r\nLet $ U \\equal{} \\cup_{a \\in A} B(a, \\epsilon_a /2)$ and $ V\\equal{} \\cup_{b \\in B} B(b, \\epsilon_b /2)$. \r\n\r\nClearly, $ U$ and $ V$ are neighborhoods of $ A$ and $ B$ respectively. Show that they are disjoint.", "Solution_2": "Standard proof of that fact bases on $ f(x)\\equal{}\\frac{dist(x,A)}{dist(x,A)\\plus{}dist(x,B)}$. It is continuous, so $ f^{\\minus{}1}(\\minus{}\\infty,\\frac{1}{2})$, $ f^{\\minus{}1}(\\frac{1}{2},\\plus{}\\infty)$ are open and seperates given sets $ A$, $ B$." } { "Tag": [ "function", "modular arithmetic" ], "Problem": "Let $Squish(n)$ be a discrete function such that $Squish(n) = n$ for all $n$ with no digit of $0$ and for $n$ with one or more digit of $0$, $Squish(n)$ is $n$ with all the digits of $0$ removed. For example: $Squish(919) = 919$ but $Squish(909) = 99$. Let $k$ be the smallest possible positive integer so that $Squish(k) | k$ but $Squish(k) \\ne k$ and th units digit of $k$ is not $0$. Compute $k+Squish(k)$.", "Solution_1": "[hide=\"Solution\"]Obviously, $k=10$. Thus, the answer is $\\boxed{11}$.[/hide]", "Solution_2": "Whoops... Lemme edit.", "Solution_3": "[hide] Clever problem, hopefully I did this right. I'm just going to assume the number only has a single zero in it and since I get a small enough number that's good enough for me.\n\nWrite $k$ as $10^{k+1}\\cdot a+b$, where $b < 10^{k}$ and doesn't end in a zero. This represents where the zero would be. For example, $123037 = 10^{3}\\cdot 123+37$. Then we have\n\n$(10^{k}\\cdot a+b) | (10^{k+1}\\cdot a+b)$,\n\nor equivalently\n\n$10^{k+1}\\cdot a+b = m(10^{k}a+b) \\Rightarrow (10-m) \\cdot 10^{k}\\cdot a = (m-1) \\cdot b$.\n\nThe best way I can finish it from here is simply to check $m = 2, 3, \\ldots, 9$ which gives me $k = 105$ when $m = 7$.[/hide]", "Solution_4": "So I approached it in a slightly less generic way. Basically I \r\n\r\n[hide=\"like cheese\"]\nsaid that $11|1001$, so either:\nThe answer is $1001+11=1012$\n\nor\n\nThere is a 3-digit solution (it clearly can't be 2-digits).\n\nLooking specifically for a 3-digit solution is much easier because you have $\\overline{a0b}$ and $\\overline{ab}$, and you have $\\overline{ab}| \\overline{a0b}-\\overline{ab}= 90a$... or something to that effect.\n[/hide]", "Solution_5": "A similar question:\r\nCompute the number of positive integers $n < 1000$ such that $Squish(n) | n$ and $Squish(n) \\ne n$.", "Solution_6": "[hide]Case 1: Three-digit number, just one zero in the tens place\nWe have $10a+b|100a+b$, or $90a\\equiv 0\\pmod{10a+b}$, for digits $a$ and $b$, $a,b\\neq 0$.\nNow, it's just counting. As $a$ goes from 1 to 9, we must count the number of factors of $90a$ with a tens digit of $a$.\n$a=1$: 15 and 18\n$a=2$: None exist\n$a=3$: None exist\n$a=4$: 45\n$a=5$: None exist\n$a=6$: None exist\n$a=7$: None exist\n$a=8$: None exist\n$a=9$: None exist\nSo there are three.\n\nCase 2: Three-digit number, zero in the ones place\nAny number with a zero will work. $\\text{Squish}(n)$ will either be $\\frac{n}{10}$ or $\\frac{n}{100}$, depending on the tens digit. So there are 90 possible numbers.\n\nCase 3: Two-digit number, zero in the tens place\nAny number will work, as $\\text{Squish}(n)$ would be $\\frac{n}{10}$. So there are nine.\n\nThere can't be any one-digit solutions, as this would violate the rule that $\\text{Squish}(n)\\neq n$.\n\nSo there are $90+9+3$, or $\\boxed{102}$, numbers in all.[/hide]" } { "Tag": [ "geometry", "geometric transformation", "reflection", "trapezoid", "geometry proposed" ], "Problem": "The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$", "Solution_1": "[color=darkblue][b][size=150]A very nice construct problem ![/size][/b][/color]\r\n[quote][color=darkred][b]Are given a triangle $ABC$ and a point $D\\in (BC)$. Construct the points $E\\in (CA)$ and $F\\in (AB)$ so that\nthe middlepoints of the segments $[DE]$ and $[DF]$ are collinearly with the vertex $C$ and\nthe middlepoints of the segments $[DE]$ and $[EF]$ are collinearly with the vertex $A$.[/b][/color][/quote]\r\n[color=darkblue][b][u]The proper construction.[/u][/b]\n \n[b]1.[/b] Construct the reflection $X$ of the point $D$ w.r.t. the point $C$.\n[b]2.[/b] Construct the point $Y\\in AB$ for which $DY\\parallel AC$.\n[b]3.[/b] Construct the point $E\\in CA$ so that $YE\\parallel BC$.\n[b]4.[/b] Construct the point $F\\in XE\\cap AB$.\n\n[b][u]Remark.[/u][/b] The positions of the points $E\\in (CA)$, $F\\in (AB)$ are defined by the following relations :\n$\\boxed{\\frac{EA}{EC}=\\frac{DC}{DB}}$ and $\\boxed{\\frac{AF}{AB}=\\left(\\frac{CD}{CB}\\right)^{2}}$. Denote the reflection $Z$ of the point $Y$ w.r.t. the point $A$.\nThe above second relation applied in the trapezoid $DYZX$, where $AC$ is the middle line, i.e. $\\frac{AF}{AB}=\\left(\\frac{YF}{YB}\\right)^{2}$,\n\nshows that [b]the point $F$ is the conjugate of the point $B$ w.r.t. the pair $\\{Y,Z\\}$[/b], i.e. $FI\\parallel AC$, where $I\\in XY\\cap DZ$.[/color]", "Solution_2": "Nice solution, Virgil Nicula! :lol: \r\n\r\nMy solution is even more simple:\r\n[quote]The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$[/quote]\r\nLet $K,L$ and $M$ be the midpoints of $DE, EF$ and $DF$ and let $BM$ intersects $AC$ at point $G.$ Then $BG\\parallel EF,$ triangles $BCG$ and $ECF$ are similar and $KC$ is a median of triangle $BCG.$ Therefore $K$ is common midpoint of $DE$ and $BG,$ i.e. $BEGD$ is parallelogram. \r\n[i]Construction:[/i] draw $DG\\parallel BC,$ then $GE\\parallel AB$ and $EF\\parallel BG.$ \r\n[i]Proof:[/i]By the construction the intersection point of $BG$ and $DE$ is the midpoint of $DE.$ Since $EF\\parallel BG$ the intersection point of $KC$ and $EF$ is the midpoint of $EF.$ Finally, the intersection point of $BG$ and $DF$ is the midpoint of $DF.$\r\n$\\begin{picture}(100,105) \\put(10,10){\\qbezier{(0,0)(50,0)(100,0)}\\qbezier{(0,0)(20,40)(40,80)}\\qbezier{(40,80)(70,40)(100,0)}\\qbezier{(10,20)(33,40)(55,60)}\\qbezier{(44,0)(49,30)(55,60)}\\qbezier{(10,20)(27,10)(44,0)}\\qbezier{(27,10)(34,45)(40,80)}\\qbezier{(32,40)(67,20)(100,0)}\\put(0,0){\\circle*{2}}\\put(100,0){\\circle*{2}}\\put(40,80){\\circle*{2}}\\put(10,20){\\circle*{2}}\\put(55,60){\\circle*{2}}\\put(44,0){\\circle*{2}}\\put(33,40){\\circle*{2}}\\put(49,30){\\circle*{2}}\\put(27,10){\\circle*{2}}\\put(25,0){\\circle*{2}}\\put(-7,-12){A}\\put(103,-12){C}\\put(37,82){B}\\put(1,21){D}\\put(56,62){E}\\put(50,-12){F}\\put(24,41){K}\\put(53,30){L}\\put(18,16){M}\\put(17,-12){G}\\linethickness{.1pt}{\\qbezier{(25,0)(18,10)(10,20)}\\qbezier{(25,0)(40,30)(55,60)}\\qbezier{(25,0)(33,40)(40,80)}}}\\end{picture}$" } { "Tag": [ "limit", "induction", "calculus" ], "Problem": "Post any mathematical paradoxes you know here!\r\n(and explain why if needed)\r\n\r\nI know one is: what is 0^0?\r\n\r\nwe know that 0^n = 0 and n^0 = 1. A sequence is needed for n^n:\r\n3^3=27\r\n2^2=4\r\n1^1=1\r\n.5^.5=.707\r\n.1^.1=.79433\r\n0^0 = ?\r\n-.1^-.1=-1.2589\r\n-.5^-.5=-1.4142\r\n-1^-1=-1\r\n-2^-2=-1/4\r\n-3^-3=-1/27\r\n\r\n(these are the values my calculator gave me)\r\n\r\nSo i don't know. 2 out of 3 imply that 0^0 = 0. Is it 0?\r\n\r\nAlso, theres another one i'm not sure of, since i don't know everything about :inf: .\r\nIf 0x = 0, and :inf: x = :inf: , then what is 0* :inf: ?", "Solution_1": "Yeah I think [tex]0^0[/tex] is undefined. For the second paradox, I would think that if 0 and :inf: were pure 0 and :inf:, then 0*:inf: would be undefined. But if not, then it depends on how fast they approach their respective values. Of course I have nothing but my own thoughts to back all this up.", "Solution_2": "[tex]0^0[/tex] is undefined. But sometimes we can regard it according to its context, e.g. since [tex]\\lim_{x \\to 0} x^x = 1[/tex].", "Solution_3": "I've always been fond of Berry's Paradox:\r\n\r\nconsider all of the positive integers that cannot be described in fewer than nineteen syllables. Let L be the smallest of those integers.", "Solution_4": "[quote=\"zscool\"]I've always been fond of Berry's Paradox:\n\nconsider all of the positive integers that cannot be described in fewer than nineteen syllables. Let L be the smallest of those integers.[/quote]\r\n\r\nYeah that one is funny, it stumps a lot of people, surprisingly, along with the faulty proof by induction that everyone has the same birthday.", "Solution_5": "[quote]along with the faulty proof by induction that everyone has the same birthday.[/quote]\r\n\r\nCan you post that one, Nukular? I haven't seen it before.\r\n\r\nOne of my favorites is Zeno's Paradox, which states that movement is impossible, and uses infinite series to \"prove\" it.\r\n\r\nAnd of course, the timeless classic, 1=2.\r\n\r\nLet x=y=1.\r\n\r\nx=y\r\nx :^2: = xy\r\nx :^2: - y :^2: =xy - y :^2: \r\n(x+y)(x-y) = y(x-y)\r\nx+y=y\r\n2=1.", "Solution_6": "0* :inf: equals all real numbers.\n\n\n\nFor any constant k, k/ :inf: = 0. So, by multiplying both sides by :inf: , 0* :inf: = k.\n\n\n\n[hide]This is all nonsense anyway, since infinity isn't a number, it's a concept. k/ :inf: doesn't actually equal zero, but it does approach it (hence the use of a limit when solving it). If you divide k by something ungodly big (but defined), it will equal something infinitessimally small, but not zero.[/hide]\n\n\n\nHere's a paradox for you:\n\n\n\n1^2 = 1\n\n1^5 = 1\n\n1^13852359236 = 1\n\n1^ :inf: is indeterminate\n\n\n\nI know it uses the logic that lim(x-> :inf: ) (1 + 1/x)^x doesn't equal 1 (it equals e, in the same line of thinking I put in spoiler above). But why can't a very simple 1^ :inf: equal 1!?", "Solution_7": "all women are blondes:\r\n\r\nlet P(n) be the statement that for any set of n women, if at least one of them is a blonde then they all are:\r\n\r\nbase case: P(1) is obviously true by def'n\r\n\r\ninduction hypothesis: assume for all r < m that P(r) is true\r\n\r\ninductive step: given a set S of m women, one of which is blonde, take two subsets A and B whose intersection is the blonde, and whose union is S. By the induction hypothesis, A and B must contain only blondes and therefore all m women are blondes so P(m) is true.\r\n\r\nNow let W be the set of women on earth. Since there is at least one woman who is blonde, all women are blondes.", "Solution_8": "hey jeff, were you guys talking about 0* :inf: when i was visiting your calculus BC class?", "Solution_9": "zscool -- that's great! Not a paradox, exactly, but a fabulous \"proof\" none the less. Berry's is pretty good, too, although I'm sure it suffers from a mathematical vs. meta-mathematical fallacy somewhere. I thought I had one, but it disappeared, so I'll see if I can find it again later.", "Solution_10": "its not a paradox in the classical sense, its a [i]falsidical paradox[/i]\r\nwhich establishes a result that not only appears false but actually is false\r\n\r\nbut yea, its a really great proof that is \"hard\" to find the flaw in (or at least not obvious)", "Solution_11": "The inductive process fails when n = 2. ;)" } { "Tag": [ "integration", "limit", "trigonometry", "logarithms", "calculus", "algebra", "partial fractions" ], "Problem": "1. $ \\int_{a}^{b}\\frac {1}{\\sqrt {1 \\plus{} x^{2}}}dx$\r\n\r\n2. $ \\int_{a}^{b}\\frac {1}{1 \\minus{} x^{2}}dx$, for $ \\left\\vert a\\right\\vert$,$ \\left\\vert b\\right\\vert < 1$\r\n\r\n3. $ \\underset{x\\rightarrow0^{ \\plus{} }}{\\lim}x^{x}$=1", "Solution_1": "for the first two use trig subsitution\r\nthe the third, take the log of it and find the limit of that.", "Solution_2": "1.substitute x=$ {tany}$\r\n\r\n2.substitute..x=$ {siny}$", "Solution_3": "2. Decomposition into Partial Fractions:\r\n\r\n$ \\frac {1}{1 \\minus{} x^2} \\equal{} \\frac {p}{1 \\minus{} x} \\plus{} \\frac {q}{1 \\plus{} x}$.\r\n\r\n3. Consider the behaviour of $ y\\equal{}x\\ln x$.", "Solution_4": "3. \r\n$ exp$ is continuous\r\n$ \\underset{x\\rightarrow0^{ \\plus{} }}{\\lim}x\\log x \\equal{} \\underset{x\\rightarrow0^{ \\plus{} }}{\\lim }\\frac { \\minus{} \\log\\frac {1}{x}}{\\frac {1}{x}} \\equal{} \\minus{} \\underset{x\\rightarrow\\infty}{\\lim }\\frac {\\log x}{x} \\equal{} \\minus{} \\underset{x\\rightarrow\\infty}{\\lim}\\frac {\\frac {1}{x}}{1} \\equal{} 0$\r\n$ \\underset{x\\rightarrow0^ \\plus{} }{\\lim}x^x \\equal{} \\underset{x\\rightarrow0^ \\plus{} }{\\lim }e^{x\\log x} \\equal{} e^{0} \\equal{} 1$", "Solution_5": "For 3 show that $ \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } x^{\\frac{1}{x}} \\equal{} 1$. That will be easy assuming you know the sequence version. Then simplify your problem.", "Solution_6": "2. How about the substitution $ x\\equal{}t\\minus{}\\frac{1}{t}$?", "Solution_7": "Infact the second is the standard integral:\r\n\r\n$ \\int \\frac {1}{a^{2} \\minus{} x^{2}}\\,dx \\equal{} \\frac {1}{2.a}\\ln|{\\frac {a \\plus{} x}{a \\minus{} x}}| \\plus{} C$", "Solution_8": "ur third answer is absolutely right ICQ-2008", "Solution_9": "@chemrock, You should add the sign of absolute value to anti logarithm.", "Solution_10": "how abt my substitutions Kunny....", "Solution_11": "They work, but I wouldn't use the second. The integral of $ \\sec$ is less elementary than the integrals of $ \\frac1{x\\minus{}c}$ that we get by partial fractions. Trigonometric substitution is poorly suited to integrals that don't have square roots in them, except for the $ \\frac{1}{(x^2\\plus{}a^2)^m}$ case.\r\n\r\nSince we assumed that both limits of the integral were in $ (\\minus{}1,1)$, we can pick a sign and use $ \\frac12\\ln(1\\plus{}x)\\minus{}\\frac12\\ln(1\\minus{}x)$ as our antiderivative.", "Solution_12": "For the first one let $ u\\equal{}x\\plus{}\\sqrt{x^2\\plus{}1}.$", "Solution_13": "hey Black, can u elucidate in ur method...", "Solution_14": "But valerium the first is also a standard integral:\r\n$ \\int \\frac {1}{\\sqrt {a^{2} \\plus{} x^{2}}}\\,dx \\equal{} \\ln{|x \\plus{} \\sqrt {x^{2} \\plus{} a^{2}}|} \\plus{} C$", "Solution_15": "u derive this standard integral using my substitution...\r\n\r\nand yes it is a standard integral..", "Solution_16": "1. $ x\\equal{}\\frac{1}{2}\\left(\\frac{1}{t}\\minus{}t\\right)$" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "let M be in interior of convex polygon A1A2...An.show atleast one of angles,\r\n MA1A2,MA2A3,...,MAnA1 does not exceed pie times(n-2)by 2n. :!: \r\n i got,have you any :idea:", "Solution_1": ":P my sol. uses isoperimetric pro. for polygon,do you :D notice connection of this generalisation with generalised erdos mordell :welcomeani: :?:", "Solution_2": "Please post your solution." } { "Tag": [ "articles" ], "Problem": "[url=http://www.co-prosperity.org/~annis/arguing.html]Read this well.[/url]", "Solution_1": "I think this thread should be locked. \r\n\r\nThe meaning behind this post is not to help people argue correctly, but to further bash ID supportists. This post will not help anyone and will most likely cause more anger and resentment, something we don't need any more on these boards.", "Solution_2": "No, the article clearly states some of the mistakes made while arguing.\r\n\r\nIm going to more the comment to another thread, to make you happy.", "Solution_3": "I wasn't commenting on the article. I was commenting on the way you expressed the article, not just mentioning how it may help your arguing skills, but then stating how Id has a \"cruddy, Pyrrhonic argument\". If you can't see how you are directing this article at a specific group of people at AOPS, which will most likely cause more arguments then I don't quite see why you are a mod.\r\n\r\nI'm sorry if I'm being blunt, I'm just tired of this stuff. Keep it in ONE thread, at least.\r\n\r\nEDIT: So you create ANOTHER thread on ID?! Not only does that solidify my point that you created this thread in a seemilngly veiled attempt to bash ID, but it shows that you are making threads for discussions which we have threads for already.", "Solution_4": "[quote=\"WindSlicer\"]\nEDIT: So you create ANOTHER thread on ID?! Not only does that solidify my point that you created this thread in a seemilngly veiled attempt to bash ID, but it shows that you are making threads for discussions which we have threads for already.[/quote]\r\n1) To move it out of this thread, because you didn't like it\r\n2)I learned something from the article and wanted to share it\r\n3)there is no thread on the disscussion the main idea of ID, evolution is bad so we need an alternate theory\r\n4)Sorry, Im hyped up on caffeine", "Solution_5": "I am really puzzled by how some people think. Well, there ARE threads to discuss ID but the IDers all tend to stray off to the merits of evolution.\r\n\r\nThe article's pretty good. I printed it out and read it on my way home.", "Solution_6": "[quote=\"ccy\"]I am really puzzled by how some people think. Well, there ARE threads to discuss ID but the IDers all tend to stray off to the merits of evolution.\n\nThe article's pretty good. I printed it out and read it on my way home.[/quote]\r\nYou should check the links at the bottom too.", "Solution_7": "I agree with Windslicer. There are already two topics on evolution, and I don't want to see any more. If there is something you want to say, I would prefer if you did it in the existing topics.", "Solution_8": "[quote=\"WindSlicer\"]I wasn't commenting on the article. I was commenting on the way you expressed the article, not just mentioning how it may help your arguing skills, but then stating how Id has a \"cruddy, Pyrrhonic argument\". If you can't see how you are directing this article at a specific group of people at AOPS, which will most likely cause more arguments then I don't quite see why you are a mod.\n\nI'm sorry if I'm being blunt, I'm just tired of this stuff. Keep it in ONE thread, at least.\n\nEDIT: So you create ANOTHER thread on ID?! Not only does that solidify my point that you created this thread in a seemilngly veiled attempt to bash ID, but it shows that you are making threads for discussions which we have threads for already.[/quote]\r\nIs it bad that it causes more argument? Isn't the point of the evolution/ID debate to argue? Argument is healthy because it allows us to exchange opinions. Do you want the post deleted because you realize he is right? Do you want the thread locked while ID still has a chance?\r\n\r\nThe reason he says ID has a Pyrrhonic argument is because it does. If you disagree with that, then you may post your reasoning.\r\n\r\nBut I agree that there are too many threads on this topic.", "Solution_9": "[quote=\"jmadsen\"][quote=\"WindSlicer\"]I wasn't commenting on the article. I was commenting on the way you expressed the article, not just mentioning how it may help your arguing skills, but then stating how Id has a \"cruddy, Pyrrhonic argument\". If you can't see how you are directing this article at a specific group of people at AOPS, which will most likely cause more arguments then I don't quite see why you are a mod.\n\nI'm sorry if I'm being blunt, I'm just tired of this stuff. Keep it in ONE thread, at least.\n\nEDIT: So you create ANOTHER thread on ID?! Not only does that solidify my point that you created this thread in a seemilngly veiled attempt to bash ID, but it shows that you are making threads for discussions which we have threads for already.[/quote]\nIs it bad that it causes more argument? Isn't the point of the evolution/ID debate to argue? Argument is healthy because it allows us to exchange opinions. Do you want the post deleted because you realize he is right? Do you want the thread locked while ID still has a chance?\n\nThe reason he says ID has a Pyrrhonic argument is because it does. If you disagree with that, then you may post your reasoning.\n\nBut I agree that there are too many threads on this topic.[/quote]\r\n\r\nWhat the heck? Did you just contradict yourself? \r\n\r\nAnyways, I could care less about what people on the internet have to say about this kind of stuff and when they go creating these type of underlying-bashing posts on the other side, and are moderators, that just crosses the line. Maybe you just didn't see the actual post because he deleted it but it was an obviously veiled ID attack. And there is a difference between arguments that produce good discussion and arguments that just create resentment. And we DON'T need ANY of the latter.\r\n\r\nDo you really think that I am trying to get this thread deleted because I want ID to live on? Give me a break, we've already got 3 topics on this stuff, and don't need another, especially one that started off like this one.", "Solution_10": "[quote=\"WindSlicer\"][quote=\"jmadsen\"][quote=\"WindSlicer\"]I wasn't commenting on the article. I was commenting on the way you expressed the article, not just mentioning how it may help your arguing skills, but then stating how Id has a \"cruddy, Pyrrhonic argument\". If you can't see how you are directing this article at a specific group of people at AOPS, which will most likely cause more arguments then I don't quite see why you are a mod.\n\nI'm sorry if I'm being blunt, I'm just tired of this stuff. Keep it in ONE thread, at least.\n\nEDIT: So you create ANOTHER thread on ID?! Not only does that solidify my point that you created this thread in a seemilngly veiled attempt to bash ID, but it shows that you are making threads for discussions which we have threads for already.[/quote]\nIs it bad that it causes more argument? Isn't the point of the evolution/ID debate to argue? Argument is healthy because it allows us to exchange opinions. Do you want the post deleted because you realize he is right? Do you want the thread locked while ID still has a chance?\n\nThe reason he says ID has a Pyrrhonic argument is because it does. If you disagree with that, then you may post your reasoning.\n\nBut I agree that there are too many threads on this topic.[/quote]\n\nWhat the heck? Did you just contradict yourself? \n\nAnyways, I could care less about what people on the internet have to say about this kind of stuff and when they go creating these type of underlying-bashing posts on the other side, and are moderators, that just crosses the line. Maybe you just didn't see the actual post because he deleted it but it was an obviously veiled ID attack. And there is a difference between arguments that produce good discussion and arguments that just create resentment. And we DON'T need ANY of the latter.\n\nDo you really think that I am trying to get this thread deleted because I want ID to live on? Give me a break, we've already got 3 topics on this stuff, and don't need another, especially one that started off like this one.[/quote]\r\n\r\nI love how you reword \"Sentence stating his opinion and asking for the opinions of other while referencing a good non-partial article\" to \"attack\". I just love that.\r\n\r\nI don't care for the \"attack\" you are making on me now etheir. You seem to have more problems with me than what I did, and are painting me as someone who doesn't care about other people, I do. I also have friends that aren't zealots on my side, I talk to many different people.\r\n\r\nThis was not bashing, this was a controled argument. If you didn't like the word cruddy in there, it was not put there to describe ID but rather that its main argument has several flaws. This thread/the one I started because of your demands is for discussion of the philisophical points of ID.\r\n\r\nYou've made your point about not liking what I did, and I fixed that. When you start yelling at me because I made another thread...", "Solution_11": "[quote=\"WindSlicer\"]Do you really think that I am trying to get this thread deleted because I want ID to live on? Give me a break, we've already got 3 topics on this stuff, and don't need another, especially one that started off like this one.[/quote]\r\nSorry, I misunderstood what you were saying. I don't see why we have 3 (at least, I think we're up to 5 now) threads on the same topic either.", "Solution_12": "Thread has been locked. Here's why, so you don't have to PM me to ask:\r\n\r\n1) Clearly this is upsetting a lot of personal beliefs, something we don't need.\r\n\r\n2) PenguinIntegral, in the future you should not stereotype all ID believers as looking for a different theory because they don't like evolution. There are plenty who find the evidence in favor of ID - not evolution - quite overwhelming. And there are certainly also those on the other side of the coin. Realizing that indeed evolutionists are the majority, please respect the stance of the minority. If this is too much to ask, then these posts will be altered to contain some fundamental respect that AoPS users should be able to find on this board - or at the very least, an absence of disrespect.\r\n\r\nI'm picking on you (but this certainly doesn't make you solely responsible for this) because being a moderator for AoPS - ML should give you an even greater burden to set an example. Please do so.\r\n\r\n3) Everyone picking on the fact that there are too many EV/ID threads, this is why there are moderators. Significant differences must exist for multiple threads on one topic for the many different threads all to be left active. However, in this case you guys are correct and this will be now locked. The link will be left intact, but further discussion should continue on another threads.\r\n\r\n4) And again, for people debating: get off the personal attacks or all of the threads will be locked." } { "Tag": [ "limit" ], "Problem": "Prove \\[ a) (1\\plus{}\\frac{1}{n})^n < 3, n\\in \\mathbb{N}^*\\]\r\n\\[ b) \\frac{5}{8}<\\frac{1}{101}\\plus{}\\frac{1}{102}\\plus{}...\\plus{}\\frac{1}{200}<\\frac{3}{4}\\]\r\n\\[ c) n^{\\frac{1}{n}}> n\\plus{}1^{\\frac{1}{n\\plus{}1}}, n\\in \\mathbb{N}^*\\]", "Solution_1": "a)\r\n\\begin{align*}\r\na_n\r\n&= \\left(1 + \\frac{1}{n}\\right)^n \\\\\r\n&= \\sum_{r=0}^n \\binom{n}{r}\\left(\\frac{1}{n}\\right)^r \\\\\r\n&= \\sum_{r=0}^n \\frac{\\prod_{i=0}^{r-1}(n-i)}{r!}\\left(\\frac{1}{n}\\right)^r \\\\\r\n&= \\sum_{r=0}^n \\frac{\\prod_{i=0}^{r-1}\\left(1 - \\frac{i}{n}\\right)}{r!} \\\\\r\n\\end{align*}\r\n\r\n\\begin{align*}\r\na_n\r\n&= \\left(1 + \\frac{1}{n}\\right)^n \\\\\r\n&= \\sum_{r=0}^n \\binom{n}{r}\\left(\\frac{1}{n}\\right)^r \\\\\r\n&= \\sum_{r=0}^n \\frac{\\prod_{i=0}^{r-1}(n-i)}{r!}\\left(\\frac{1}{n}\\right)^r \\\\\r\n&= \\sum_{r=0}^n \\frac{\\prod_{i=0}^{r-1}\\left(1 - \\frac{i}{n}\\right)}{r!} \\\\\r\n\\end{align*}\r\n\r\n$ a_{n+1}= \\sum_{r=0}^{n+1} \\frac{\\prod_{i=0}^{r-1}\\left(1 - \\frac{i}{n+1}\\right)}{r!}$\r\n\r\nSo, $ a_n < \\sum_{i = 0}^n \\frac{1}{i} < \\sum_{i = 0}^n \\frac{1}{2^{i-1}} = 1 + \\frac{1 - \\left(\\frac{1}{2}\\right)^n}{1 - \\frac{1}{2}} = 3 - \\left(\\frac{1}{2}\\right)^{n-1} < 3$\r\n\r\n$ \\therefore a_1 < a_2 < a_3 < \\cdots < a_n < a_{n+1} < \\cdots < 3$\r\n\r\n$ \\therefore \\lim_{n \\to \\infty} \\left(1 + \\frac{1}{n}\\right)^n \\le 3$", "Solution_2": "For a), couldn't we just say that $ \\lim_{n\\rightarrow\\infty}(1\\plus{}\\frac1n)^n\\equal{}e$ by the definition of $ e$?", "Solution_3": "[quote=\"brainomega\"]For a), couldn't we just say that $ \\lim_{n\\rightarrow\\infty}(1 \\plus{} \\frac1n)^n \\equal{} e$ by the definition of $ e$?[/quote]\r\nWe can't use $ e \\sim 2.7$." } { "Tag": [ "MATHCOUNTS" ], "Problem": "when is your chapter competition. mine is the 2nd", "Solution_1": "Mine is the 30th.\r\n\r\nbtw, note the contradictions here.", "Solution_2": "30th of feburary?", "Solution_3": "I think like the 30th of January or sometime around that.", "Solution_4": "you get to miss school?", "Solution_5": "Umm, it's the Saturday after the last Friday of February Vacation.", "Solution_6": "surely you're not too lazy to check the calendar...\r\n\r\nIt's Saturday the 2nd for us.", "Solution_7": "Same here.Too short.", "Solution_8": "2nd feb. CANTWAITCANTWAITCANTWAIT.", "Solution_9": "February 2nd. I won't be competing this year though, I'll be grading. :P", "Solution_10": "9th graders are allowed to grade?\r\n\r\nflood the paper with comments. lol. my friend got his with a solution for almost every problem he missed", "Solution_11": "Feb 16. \r\nfrom now 'till feb 16, I have to do every problem from 1986-1996 and some tests from the 21st century. altogether, that's about 5,000+ problems. yay!", "Solution_12": "February 2nd", "Solution_13": "Tuesday Feb. 5th", "Solution_14": "Your competition is on a Tuesday?", "Solution_15": "Mine is on the 11th. 2/11/08", "Solution_16": "[quote=\"abacadaea\"]9th graders are allowed to grade?\n\nflood the paper with comments. lol. my friend got his with a solution for almost every problem he missed[/quote]\r\n\r\nI don't think you're supposed to get the test papers back, as it says (in the general instructions I believe) that all tests become property of MathCounts afterward (i.e. they get to shred them).", "Solution_17": "oh whoops nevermind then", "Solution_18": "Feb. 9\r\n\r\ncan't wait", "Solution_19": "Wed. Feb 6", "Solution_20": "you get to miss school... wednesday???", "Solution_21": "missing school isn't fun...\r\n\r\nanyways, the one i'm helping out at in CT is on February 2nd...but I had no clue 9th graders were allowed to grade, i thought we could just proctor...well anyways :o", "Solution_22": "Well, we miss like the last block of school. (~45 min)", "Solution_23": "February 23rd, (Saturday)\r\n>.<", "Solution_24": "Mine is on February 16.", "Solution_25": "I'm one of those who are waiting for one week to pass quickly?", "Solution_26": "how do you get it on a school day?", "Solution_27": "February 2nd..\r\n\r\nMine was early last year, too.", "Solution_28": "My chapter is on Febuary the 5. AKA Tuesday. We get to miss school, go to a restuarant of our pick (Fazoli's) and after the compitition, we go to Dairy Quenn (our school tradition). I can' wait to go (and lose)!!!!!!!!(We miss ALL of school!!!!!!)" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "It\u2019s well known that if a topological space $X$ is separable, then a subspace $Y$ of $X$ need not be separable. But what if $Y$ is also dense?", "Solution_1": "If there's a counterexample to this, it can't be from a metric space." } { "Tag": [ "ratio", "function", "analytic geometry", "geometry solved", "geometry" ], "Problem": "ok soo i am desingning skis......and i need HELP ASAP!!!!!\r\n\r\nSo lets imagine that a ski has a turning radius of 16 m, no draw a circle of radius 16 m, from a radial line draw a line perpendicular to it. so now you have a circle with a tangent line to it. Consider this tangent line as part of the ski. On the left side of the tangent line the ski runs for 84cm, and then a line is drawn perpendicular to the tangent line (L1) until it touches the circumfrence of the circle. On the right side of the tangent line the ski runs for 74 cm, and then a line (L2) perpendicular to it runs down until it too touches the circle. Now i pose as my question, is there some ratio between L1,L2,84,and 74? L1 is know as the front sidecut of the ski, and L2 is the tail sidecut of the ski. \r\n\r\nHere is a known number system - This was figured out on a CAD program (Accuracy is probally not 100%)\r\n84(Tangent length left), L1= 19 cm \r\n74(Tangent length right), L2= 14cm\r\nRadius of the circle is 18m\r\n\r\nSo what sidecuts are needed to supply a 16m turning radius?\r\n\r\nI have a formula that i made that gives the sidecut length (true size, usually veyr close to Tangent length)\r\nand from that you can find the sidecut length and turnign radius....but for only one side of the ski ( front or tail) so i will not poisen anyones mind with that dirty information. PLEEASE HELP! ANSWER AS SOON AS POSSIBLE! and can i make a function out of this so that i can graph the data? or two functiooons that show L1 and L2 depentdant on R? there are also angles phata and beta. Phata is the angle made on the left side, and beta is made on the right side. Please view diagram if you can...it is attached! click on the attachment to see in betetr quality!\r\n\r\nTHANK YOU!!!!!!\r\n[/b]", "Solution_1": "From your description and drawing, it follows that you are looking for the distances $L_1, L_2$ of a horizontal tangent to the circle with radius $R = 16\\ \\text{m} = 1600\\ \\text{cm}$ centered at the coordinate origin from the points of the upper semicircle with x-coordinates $x_L = 74\\ \\text{cm}$ and $x_R = -84\\ \\text{cm}$. The tangent has the equation $y = R$ and the circle has the equation $x^2 + y^2 = R^2$. Hence the upper semicircle has the equation $y = +\\sqrt{R^2 - x^2}$ (and the lower one, not needed, $y = -\\sqrt{R^2 - x^2}$). The distances $L_1, L_2$ are then\r\n\r\n$L_1 = R - \\sqrt{R^2 - x_L^2} = 1600 - \\sqrt{1600^2 - 74^2} \\doteq 1.71\\ \\text{cm}$\r\n\r\n$L_2 = R - \\sqrt{R^2 - x_R^2} = 1600 - \\sqrt{1600^2 - 84^2} \\doteq 2.21\\ \\text{cm}$\r\n\r\nSimilarly, for the radius $R = 18\\ \\text{m} = 1800\\ \\text{cm}$, you would get\r\n\r\n$L_1 = R - \\sqrt{R^2 - x_L^2} = 1800 - \\sqrt{1800^2 - 74^2} \\doteq 1.52\\ \\text{cm}$\r\n\r\n$L_2 = R - \\sqrt{R^2 - x_R^2} = 1800 - \\sqrt{1800^2 - 84^2} \\doteq 1.96\\ \\text{cm}$\r\n\r\nIt is quite impossible for these distances to be $L_1 = 14\\ \\text{cm}$ and $L_2 = 19\\ \\text{cm}$ or anything close to this.", "Solution_2": "yyea sorry i ment 19mm and 14mm, THANK YOU SOO MUCH!" } { "Tag": [ "function", "limit", "integration", "real analysis", "real analysis solved" ], "Problem": "a very nice one I've just had during an oral examination:\r\nlet $A(t),B(t):R^{+}->R$ be 2 functions such that:$lim_{x->+\\infty} A(t)=0$,$lim_{x->+\\infty} B(t)=1$\r\nLet $f$ be a non-null solution of $y''+A(t)y'+B(t)y=0$\r\nShow that the number of zeros of $f$ between $0$ and $t$ is equivalent to $\\frac{t}{\\pi}$", "Solution_1": "Sturm theorem", "Solution_2": "what is this?? seems very interesting.", "Solution_3": "Moubinool, could you please post a proof?", "Solution_4": "Theorem : (1) $y\\\"+q(t)y=0$, (2) $z\\\"+wz=0$ with $q\\leq w$ on $I$ interval\r\n\r\nBetween two roots $a,b$ of non-zero $y$ solution of (1), any solution of (2) has at least a root\r\n\r\n$y(a)=y(b)=0$ for any $z$ solution of (2) there exist $c$ such that $z(c)=0$ \r\n\r\nProof\r\n\r\nWronskian $W=yz'-y'z$ satisfy $W'=yz(q-w)$ \r\n\r\nlet $a0$ on $]a,b[$\r\n\r\nIf $z$ has no zero on $]a,b[$ say $z>0$, $W$ is decreasing $[a,b]$ \r\n\r\n$W(a)\\leq 0$, $W(b)\\geq 0$ this mean $W=0$ Wronskien is 0 on $[a,b]$, the solutions $y,z$ are \r\n\r\ncolinear $z $has a zero on $[a,b]$ contradiction.\r\n\r\n\r\n\r\nBack to our problem by a change variable we work with \r\n\r\n(3) $y''+(1+g(t))y=0$ where $g$ function wich tends to 0 at +oo.\r\n\r\nWith the above theorem any solution of (3) has an infinite zero $r_n$ such that \r\n$r_n\\leq r_{n+1}$ and $r_{n+1}-r_n$ tends to $2\\pi$\r\nThe number of zeros of $y$ is equivalent $\\frac{t}{\\pi}$", "Solution_5": "More over you have the following \r\n\r\n$q:[a,+\\infty[\\rightarrow ]0;+\\infty[ $, class $C^{1}$ if $\\lim_{+\\infty}\\frac{q'}{q^{3/2}}=0$ \r\n\r\nThen for any non-zero solution of $y\\\"+q(x)y=0$, the number $N(x)$ of zero of $y$\r\n \r\non $[a,x]$ verify $N(x)\\sim \\frac{1}{\\pi}\\int_{a}^{x}\\sqrt{q(t)}dt$", "Solution_6": "A \"Professeur de Math\u00e9matiques Sp\u00e9ciales $MP^{*}$ \r\ntold me there is an article in \r\n\r\nRevue de Math\u00e9matiques Sp\u00e9ciales\r\nRMS page 99 \r\n1985-1986\r\n\r\nEquations Differentielles Lin\u00e9aires d'ordre 2:\r\ncomportement global ou asymptotiques des solutions\r\n-------------------------------------------------------------------------\r\n\r\nSee also the book\r\n\r\nHartman, Ordinary Differential Equations page 348", "Solution_7": "thanx a lot Moubi, but I can't read this article because there are not available in my CDI :( do you know where i can have access to it ?\r\n\r\nPS:\r\nanother solution:\r\nlook at $(y, y')/|(y, y')|$ and use relevement theorem." } { "Tag": [ "Vieta" ], "Problem": "If $ a\\plus{}b\\plus{}c\\equal{}0$ ,prove that:\r\n\r\nI. $ \\dfrac{a^5\\plus{}b^5\\plus{}c^5}{5}\\equal{}\\dfrac{a^3\\plus{}b^3\\plus{}c^3}{3}* \\dfrac{a^2\\plus{}b^2\\plus{}c^2}{2}$\r\n\r\n\r\nII. $ \\dfrac{a^7\\plus{}b^7\\plus{}c^7}{7}\\equal{}\\dfrac{a^5\\plus{}b^5\\plus{}c^5}{5}* \\dfrac{a^2\\plus{}b^2\\plus{}c^2}{2}$", "Solution_1": "[hide=\"Solution\"]\nBy Vieta's Formulas, the cubic with roots $ a,b,c$ is\n\n$ P(x) \\equal{} x^3 \\minus{} (a \\plus{} b \\plus{} c)x^2 \\plus{} (ab \\plus{} bc \\plus{} ca)x \\minus{} abc \\equal{} x^3 \\plus{} (ab \\plus{} bc \\plus{} ca)x \\minus{} abc \\equal{} 0$.\n\nLet $ S_n \\equal{} a^n \\plus{} b^n \\plus{} c^n$ so that $ \\boxed{S_1 \\equal{} 0}$ and $ \\boxed{S_2 \\equal{} (a \\plus{} b \\plus{} c)^2 \\minus{} 2(ab \\plus{} bc \\plus{} ca) \\equal{} \\minus{} 2(ab \\plus{} bc \\plus{} ca)}$.\n\nSince $ P(a),P(b),P(c) \\equal{} 0$, we have $ P(a) \\plus{} P(b) \\plus{} P(c) \\equal{} 0 \\equal{} S_3 \\plus{} (ab \\plus{} bc \\plus{} ca)S_1 \\minus{} 3abc \\equal{} S_3 \\minus{} 3abc\\implies\\boxed{S_3 \\equal{} 3abc}$.\n\nSimilarly, $ aP(a) \\plus{} bP(b) \\plus{} cP(c) \\equal{} 0 \\equal{} S_4 \\plus{} (ab \\plus{} bc \\plus{} ca)S_2 \\minus{} abcS_1$\n\n$ \\equal{} S_4 \\minus{} 2(ab \\plus{} bc \\plus{} ca)^2\\implies\\boxed{S_4 \\equal{} 2(ab \\plus{} bc \\plus{} ca)^2}$,\n\n$ a^2P(a) \\plus{} b^2P(b) \\plus{} c^2P(c) \\equal{} 0 \\equal{} S_5 \\plus{} (ab \\plus{} bc \\plus{} ca)S_3 \\minus{} abcS_2$\n\n$ \\equal{} S_5 \\plus{} 3abc(ab \\plus{} bc \\plus{} ca) \\plus{} 2abc(ab \\plus{} bc \\plus{} ca)\\implies\\boxed{S_5 \\equal{} \\minus{} 5abc(ab \\plus{} bc \\plus{} ca)}$, and\n\n$ a^4P(a) \\plus{} b^4P(b) \\plus{} c^4P(c) \\equal{} 0 \\equal{} S_7 \\plus{} (ab \\plus{} bc \\plus{} ca)S_5 \\minus{} abcS_4$\n\n$ \\equal{} S_7 \\minus{} 5abc(ab \\plus{} bc \\plus{} ca)^2 \\minus{} 2abc(ab \\plus{} bc \\plus{} ca)^2\\implies\\boxed{S_7 \\equal{} 7abc(ab \\plus{} bc \\plus{} ca)^2}$.\n\nIt is easy to see that $ \\frac {S_5}5 \\equal{} \\frac {S_3}3\\cdot\\frac {S_2}2$ and $ \\frac {S_7}7 \\equal{} \\frac {S_5}5\\cdot\\frac {S_2}2$.\n[/hide]" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "Suppose you have 12 different flavors and cones are made with 3 different flavors. How many possible ice cream cones are possible? How many with 3 different flavors? Note chocolate on the top is for example different from chocolate on the bottom.\r\n\r\nSo it is $ 12(11)(1)$ for first one. And $ \\binom{12}{3}$ for second one?", "Solution_1": "Assuming as you said that the permutations of the ice cream flavors are significant, $ \\binom{12}{3}$ would not give the correct total number of cones.\r\n\r\nThe number of permutations, without repeated flavors, would be $ 12*11*10 \\equal{} 1320$ possible cones. This is because there are 12 flavors to choose for the first position, 11 for the second, and 10 for the final position.\r\n\r\n$ \\binom{12}{3}$ would be correct in describing the number of cones where ice cream order did not matter however.\r\n\r\nNote that more cones would be possible for the first problem, as the second problem's answer is a subset of the first's.", "Solution_2": "For your first question, it would be, assuming that a cone has to have at least 1 flavor, $ \\binom{12}{1}\\cdot1!\\plus{}\\binom{12}{2}\\cdot2!\\plus{}\\binom{12}{3}\\cdot3!\\plus{}\\binom{12}{4}\\cdot4!\\plus{}\\dots\\plus{}\\binom{12}{12}\\cdot12!$.", "Solution_3": "[quote=\"Math Champion\"]For your first question, it would be, assuming that a cone has to have at least 1 flavor, $ \\binom{12}{1}\\cdot1! \\plus{} \\binom{12}{2}\\cdot2! \\plus{} \\binom{12}{3}\\cdot3! \\plus{} \\binom{12}{4}\\cdot4! \\plus{} \\dots \\plus{} \\binom{12}{12}\\cdot12!$.[/quote]As specified in the problem, cones are made with 3 different flavors. It would be crazy to stack 12 flavors creating a humongous cone! :rotfl: \n\n[quote=\"ragingbullrocky\"]Note chocolate on the top is for example different from chocolate on the bottom.[/quote]If the same flavor can be used on top and bottom (assuming that you can't have same [i]adjacent[/i] flavor), there can be $ 12\\cdot11\\cdot11$ different cones.\r\n :icecream:", "Solution_4": "adjacent flavors are like...? :( \r\none on top of each otha?", "Solution_5": "Say the flavor toppings in order is chocolate, vanilla, chocolate. Vanilla is [i]adjacent[/i] to both chocolate. This combination is allowed, per that note I have quoted.", "Solution_6": "So for the first one, its $ 12^3$, and for the second one, its $ 12\\cdot11\\cdot11$?" } { "Tag": [ "symmetry", "geometry proposed", "geometry" ], "Problem": "let N is a point On $ l$ .\r\n$ A_1,A_2,B_1,B_2,C_1,C_2$ are 6 point that :\r\n$ A_1,B_1,C_1$ are symmetry of $ A_2,B_2,C_2$ toward $ l$ in order.\r\n$ (AN,B_1C_1) \\equal{} P,(BN,C_1A_1) \\equal{} Q,(CN,A_1B_1) \\equal{} R$\r\nprove P,Q,R are collinear...", "Solution_1": "any solution??? :!:", "Solution_2": "where is A,B,C?" } { "Tag": [ "Vieta", "algebra", "polynomial" ], "Problem": "Find all ordered triples $(x,y,z)$ that satisfy:\r\n\r\n$x+y-z=0$\r\n$zx-xy+yz=27$\r\n$xyz=54$\r\n\r\nI can't seem to fit this in with Vieta's formula", "Solution_1": "[quote=\"demji\"]Find all ordered triples $(x,y,z)$ that satisfy:\n\n$x+y-z=0$\n$zx-xy+yz=27$\n$xyz=54$\n\nI can't seem to fit this in with Vieta's formula[/quote]\r\n\r\n[hide=\"Hint\"]First sub $w=-z$\n\nNow we have:\n\n$x+y+w=0$\n$xy+yw+xw=-27$\n$xyw=-54$\n\nCan you finish?\n\n[I think I did it right][/hide]", "Solution_2": "oh, nice thx, now i make a polynomial and find roots.", "Solution_3": "[hide=\"Or just say...\"]$(g+x)(g+y)(g-z)=g^3 + (x+y-z)g^2 + (xy-xz-yz)g - xyz = g^3 - 27g - 54$. Then we solve the equation.[/hide]" } { "Tag": [ "induction", "geometry unsolved", "geometry" ], "Problem": "Let $e_{n}$ and $a_{n}$ be the number of dissections of a convex $n$-gon by nonintersecting diagonals into an even or odd number of regions, respectively. Show that $e_{n}-a_{n}=\\left(-1\\right)^{n}$.", "Solution_1": "Prove it by induction on $n$:\r\nFor small $n$, it's trivially true.\r\n\r\nOrder the vertives in the canonical order to $A_1,A_2,...,A_n$.\r\nFirst, lets look at the number of dissections for that no 'cutting' goes through $A_n$:\r\nSince we can add/remove the edge $A_{n-1}A_1$ then, there are equally many 'odd' end 'even' dissections of that type, thus lets assume that there is at least one 'cutting' through $A_n$.\r\n\r\nNow lets go on:\r\nFor some dissection, let $k\\in\\{2,3,...,n-3,n-2\\}$ be the smallest one such that $A_k$ that is connected with $A_n$.\r\nNow when $k\\geq 3$, we can add/remove $A_1A_k$ again, so there are equally much of both types.\r\nSo, we are keft with a cut at $A_nA_2$. Now the number of these left odd dissections is $e_{n-1}$, the number of left even ones is $a_{n-1}$ (just dissect the $(n-1)$-gon where $A_1$ is missing), thus, $e_n-a_n = a_{n-1}-e_{n-1} = - (-1)^{n-1} = (-1)^n$." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "$ a,b,c$ are positive numbers,prove that\r\n$ \\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\plus{} \\frac {3(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^2}\\geq4$\r\n\r\n\r\nMay be posted before...Please give me a hint...", "Solution_1": "obviously,by AM-GM,\r\n$ \\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\geq3$\r\n\r\nso its sufficient to prove \r\n$ \\frac {3(ab \\plus{} bc \\plus{} ca)}{a^{2} \\plus{} b^{2} \\plus{} c^{2}}\\geq1$\r\n\r\n$ 3(ab \\plus{} bc \\plus{} ca)\\geq\\ a^{2} \\plus{} b^{2} \\plus{} c^{2}$\r\n\r\ni guess now now u shall try it", "Solution_2": "[quote=\"phymax\"]obviously,by AM-GM,\n$ \\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\geq3$\n\nso its sufficient to prove \n$ \\frac {3(ab \\plus{} bc \\plus{} ca)}{a^{2} \\plus{} b^{2} \\plus{} c^{2}}\\geq1$\n\n$ 3(ab \\plus{} bc \\plus{} ca)\\geq\\ a^{2} \\plus{} b^{2} \\plus{} c^{2}$\n\ni guess now now u shall try it[/quote]\r\nThe last inequality 's false :wink: .Try $ a\\equal{}10,b\\equal{}c\\equal{}1$", "Solution_3": "Yes, $ a^2\\plus{}b^2\\plus{}c^2\\geq 3(ab\\plus{}bc\\plus{}ca)\\Longleftrightarrow (a\\minus{}b)^2\\plus{}(b\\minus{}c)^2\\plus{}(c\\minus{}a)^2\\geq 0$.", "Solution_4": ":blush: Sorry for my mistake....I make a typo...\r\nNow the problem is not so easy...", "Solution_5": "i wanted Jianxing113725 to realise the fault by self trying the problem and not mentioning it wrong", "Solution_6": "[quote=\"kunny\"]Yes, $ a^2 \\plus{} b^2 \\plus{} c^2\\geq 3(ab \\plus{} bc \\plus{} ca)\\Longleftrightarrow (a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2\\geq 0$.[/quote]\r\nIt 's false,kunny :P", "Solution_7": "that's right my friend.\r\nit is false in every way!", "Solution_8": ":blush: \r\n\r\nThat was $ (a\\plus{}b\\plus{}c)^2\\geq 3(ab\\plus{}bc\\plus{}ca)$.", "Solution_9": "[quote=\"phymax\"]i wanted Jianxing113725 to realise the fault by self trying the problem and not mentioning it wrong[/quote]\r\n\r\n\r\nI have edited the problem....So sorry for my mistake....", "Solution_10": "still check your question :wink: :huh: !!", "Solution_11": "[quote=\"Jianxing113725\"]$ a,b,c$ are positive numbers,prove that\n$ \\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\plus{} \\frac {3(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^2}\\geq4$\n\n\nMay be posted before...Please give me a hint...[/quote]\r\n[i]\nProof:[/i]\r\n\r\nAssume: $ b\\equal{}max\\{a,b,c\\}$\r\n\r\n$ \\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\plus{} \\frac {3(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^2}\\geq4$\r\n\r\n$ < \\equal{} > (\\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\minus{} 3) \\plus{} (\\frac {3(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^2} \\minus{} 1 )\\geq 0$\r\n\r\n$ < \\equal{} > (\\frac {(a \\minus{} c)^2)}{(a \\plus{} b)(b \\plus{} c)} \\plus{} \\frac {(b \\minus{} a)(b \\minus{} c)}{(a \\plus{} b)(c \\plus{} a)}) \\minus{} \\frac {(a \\minus{} c)^2 \\plus{} (b \\minus{} a)(b \\minus{} c)}{(a \\plus{} b \\plus{} c)^2} \\ge 0$\r\n\r\n$ < \\equal{} > \\frac {(a \\minus{} c)^2(a^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(b \\plus{} c)(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {(b \\minus{} a)(b \\minus{} c)(b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(c \\plus{} a)(a \\plus{} b \\plus{} c)^2} \\ge 0$\r\n\r\nWhich is obviously true. :P", "Solution_12": "[quote=\"hophinhan\"][quote=\"Jianxing113725\"]$ a,b,c$ are positive numbers,prove that\n$ \\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\plus{} \\frac {3(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^2}\\geq4$\n\n\nMay be posted before...Please give me a hint...[/quote]\n[i]\nProof:[/i]\n\nAssume: $ b \\equal{} max\\{a,b,c\\}$\n\n$ \\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\plus{} \\frac {3(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^2}\\geq4$\n\n$ < \\equal{} > (\\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\minus{} 3) \\plus{} (\\frac {3(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^2} \\minus{} 1 )\\geq 0$\n\n$ < \\equal{} > (\\frac {(a \\minus{} c)^2)}{(a \\plus{} b)(b \\plus{} c)} \\plus{} \\frac {(b \\minus{} a)(b \\minus{} c)}{(a \\plus{} b)(c \\plus{} a)}) \\minus{} \\frac {(a \\minus{} c)^2 \\plus{} (b \\minus{} a)(b \\minus{} c)}{(a \\plus{} b \\plus{} c)^2} \\ge 0$\n\n$ < \\equal{} > \\frac {(a \\minus{} c)^2(a^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(b \\plus{} c)(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {(b \\minus{} a)(b \\minus{} c)(b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(c \\plus{} a)(a \\plus{} b \\plus{} c)^2} \\ge 0$\n\nWhich is obviously true. :P[/quote]\r\n\r\nman $ \\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\geq\\ 3$ and $ \\frac {3(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^{2}}\\leq1$", "Solution_13": "[quote=\"hophinhan\"][quote=\"Jianxing113725\"]$ a,b,c$ are positive numbers,prove that\n$ \\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\plus{} \\frac {3(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^2}\\geq4$\n\n\nMay be posted before...Please give me a hint...[/quote]\n[i]\nProof:[/i]\n\nAssume: $ b \\equal{} max\\{a,b,c\\}$\n\n$ \\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\plus{} \\frac {3(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^2}\\geq4$\n\n$ < \\equal{} > (\\frac {a \\plus{} b}{b \\plus{} c} \\plus{} \\frac {b \\plus{} c}{c \\plus{} a} \\plus{} \\frac {c \\plus{} a}{a \\plus{} b} \\minus{} 3) \\plus{} (\\frac {3(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b \\plus{} c)^2} \\minus{} 1 )\\geq 0$\n\n$ < \\equal{} > (\\frac {(a \\minus{} c)^2)}{(a \\plus{} b)(b \\plus{} c)} \\plus{} \\frac {(b \\minus{} a)(b \\minus{} c)}{(a \\plus{} b)(c \\plus{} a)}) \\minus{} \\frac {(a \\minus{} c)^2 \\plus{} (b \\minus{} a)(b \\minus{} c)}{(a \\plus{} b \\plus{} c)^2} \\ge 0$\n\n$ < \\equal{} > \\frac {(a \\minus{} c)^2(a^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(b \\plus{} c)(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {(b \\minus{} a)(b \\minus{} c)(b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(c \\plus{} a)(a \\plus{} b \\plus{} c)^2} \\ge 0$\n\nWhich is obviously true. :P[/quote]\r\nVery nice solution :)" } { "Tag": [ "geometry", "incenter", "circumcircle", "ratio", "geometric transformation", "homothety", "trigonometry" ], "Problem": "Dear Yetti and Mathlinkers,\r\nLet ABC be a triangle, (Ha) the A-altitude of ABC, I the incenter of ABC, D the foot of the perpendicular to BC passing through I, E the foot of the perpendicular to (Ha) passing through I, 0 the circumcircle of ABC and A' the midpoint of the arc BC opposite to A.\r\nProuve synthetically that A', D and E are collinear.\r\nSincerely\r\nJean-Louis", "Solution_1": "Denote $ J \\equal{} AI\\cap BC$. We can easily to see that triangles $ A'CJ$ and $ A'AC$ are similar, which implies that $ \\frac {A'C}{A'A} \\equal{} \\frac {CJ}{AC}$, i.e. $ \\frac {A'I}{A'A} \\equal{} \\frac {CJ}{AC}$ $ (1)$. On the other hand, $ CI$ is the internal angle bisector of the angle $ ACJ$, so $ \\frac {CJ}{AC} \\equal{} \\frac {IJ}{IA} \\equal{} \\frac {ID}{AE}$ $ (2)$ (because of $ \\triangle IJD\\sim\\triangle AIE$). Combining $ (1)$ and $ (2)$, we deduce that $ \\frac {A'I}{A'A} \\equal{} \\frac {ID}{AE}$. Hence, due to Thales' theorem, we conclude $ A'$, $ D$ and $ E$ are collinear.\r\n\r\n[b]P.S.[/b] I have discovered this property moths ago, I used it during solve a problem named that \"An unexpected concurrency\" on Hyacinthos Group. I also sent this solution to [b]Kostas Vittas[/b] via PM :D", "Solution_2": "[quote=\"Virgil Nicula\"] [color=darkred] [size=117][b]Here is a new problem ![/b][/size] Let $ ABC$ be a triangle with the incircle $ C(I,r)$ , the $ A$ - excircle $ C(I_a,r_a)$ and the circumcircle $ w = C(O,R)$ . Denote : the projection $ D$ , $ D'$ of $ I$ , $ I_a$ to $ BC$ respectively. ; the point $ E$ for which $ IE\\parallel BC$ and $ AE\\perp BC$ ; the point $ F$ for which $ I_aF\\parallel BC$ and $ AF\\perp BC$ ; the second intersection $ S$ (south) of the line $ \\overline {AII_a}$ with the circle $ w$ ; the diameter $ [AL]$ of the circle $ w$ ; the intersections $ X\\in SD\\cap LI$ , $ Y\\in SD'\\cap LI_a$ . Prove ([b]synthetically[/b] !) that $ \\{E,X\\}\\subset SD$ and $ \\{F,Y\\}\\subset SD'$ .[/color] [/quote]\r\n\r\n[color=darkred]I\"ll present my [u]synthetical proof[/u] (in my opinion, [b][u]interesting and very nice ![/u][/b] ).\n\nI dedicate it to [b]Cosmin Pohoata[/b], [b]April[/b], [b]Jean-Louis[/b] and [b]Kostas Vittas[/b].[/color]\r\n\r\n[color=darkblue][b][u]Proof.[/u][/b] Observe that $ ID = r$ , $ AL = 2R$ and $ IA\\cdot IS = 2Rr$ . Hence $ \\{\\begin{array}{c} \\widehat {DIS}\\equiv\\widehat {IAL} \\\\\n \\\\\n\\frac {ID}{IS} = \\frac {AI}{AL}\\end{array}\\ \\stackrel {s.a.s.}{\\implies}\\ \\triangle DIS\\ \\sim\\ \\triangle IAL$ $ \\implies$ \n\n$ \\widehat {ISD}\\equiv\\widehat {ALI}$ $ \\implies$ $ \\boxed {\\ X\\in w\\ }$ $ \\implies$ $ XA\\perp XI$ . Since $ EA\\perp EI$ obtain that $ AXEI$ is cyclically, i.e. $ \\widehat {IXE}\\equiv\\widehat {IAE}$ . \n\nFrom $ \\widehat {LXE}\\equiv \\underline {\\widehat {IXE}\\equiv \\widehat {IAE}}\\equiv\\widehat {LAS}\\equiv\\widehat {LXS}$ obtain $ \\widehat {LXE}\\equiv\\widehat {LXS}$ $ \\implies$ $ E\\in XS$ $ \\implies$ $ E\\in DS$ , i.e. $ \\{E,X\\}\\subset SD$ .\n\nThe second part of the conclusion, i.e. $ \\{F,Y\\}\\subset SD'$ , you can prove analogously. I let it as an interesting and easy dual problem. [/color]", "Solution_3": "[quote=\"jayme\"]... Let ABC be a triangle, (Ha) the A-altitude of ABC, I the incenter of ABC, D the foot of the perpendicular to BC passing through I, E the foot of the perpendicular to (Ha) passing through I, 0 the circumcircle of ABC and A' the midpoint of the arc BC opposite to A.\nProuve synthetically that A', D and E are collinear...[/quote]\r\n\r\n$ H$ is the orthocenter. $ M$ is the midpoint of $ BC,$ $ X \\equiv AI \\cap BC$ is foot of the internal bisector of the $ \\angle A,$ $ D$ is that tangency point of the incircle $ (I)$ with $ BC$ and $ K \\equiv AH \\cap BC$ is foot of the A-altitude $ h_a.$ Inversion in a circle $ (M)$ with center $ M$ and radius $ MD$ takes the 9-point circle $ (N)$ into the common internal tangent of the incircle $ (I)$ and A-excircle $ (I_a)$ other than $ BC.$ Therefore, $ D$ stays put, $ K$ goes to $ X$ and the other way around. By the basic properties of inversion,\r\n\r\n$ \\overline {DK} \\equal{} \\minus{}\\overline{DX} \\cdot \\frac{MD^2}{\\overline{MX} \\cdot \\overline{MD}} \\equal{} \\frac{\\overline{XD} \\cdot \\overline{DM}}{\\overline{XM}}\\ \\Longrightarrow$\r\n\r\n$ \\frac{\\overline{DK}}{\\overline{DM}} \\equal{} \\frac{\\overline{XD}}{\\overline{XM}} \\equal{} \\frac{\\overline{XI}}{\\overline{XA'}} \\equal{} \\frac{\\overline{DI}}{\\overline{MA'}} \\equal{} \\frac{\\overline{KE}}{\\overline{MA'}}\\ \\Longrightarrow\\ \\frac{\\overline{MA'}}{\\overline{MD}} \\equal{} \\frac{\\overline{KE}}{\\overline{KD}}$\r\n\r\nwhich means the right $ \\triangle DMA' \\sim \\triangle DKE$ are centrally similar with similarity center $ D$ and $ A', D, E$ are collinear.", "Solution_4": "[color=darkblue][b][u]Here are three similar well-known dual properties.[/u][/b][/color] \r\n\r\n[color=darkred][b]V.N.[/b][/color] [color=darkblue]and[/color] [color=darkred][b]Jayme[/b][/color] [color=darkblue]wish, if it is possibly, three synthetical proofs, without cross ratio ![/color]\r\n\r\n[color=darkred]Let $ \\triangle ABC$ with the incircle Let $ w \\equal{} C(I,r)$ be the incircle and $ A$ - excircle $ w_a$ of $ \\triangle ABC$ .[/color]\r\n\r\n[quote] [color=darkred][b][u]Property I.[/u][/b] Denote$ \\left\\|\\begin{array}{c} M\\in (BC)\\ ,\\ MB \\equal{} MC \\\\\n\\ D\\in BC\\ ,\\ AD\\perp BC\\end{array}\\right\\|$ and $ \\left\\|\\begin{array}{c} X\\in AD\\cap MI \\\\\n\\ Y\\in AD\\cap MI_a\\end{array}\\right\\|$ . Then $ AX \\equal{} r$ and $ AY \\equal{} r_a$ .[/color] [/quote]\n\n[quote] [color=darkred][b][u]Property II.[/u][/b] Denote$ \\left\\|\\begin{array}{c} T\\in (BC)\\ ,\\ IT\\perp BC \\\\\n\\ T_a\\in (BC)\\ ,\\ I_aT_a\\perp BC\\end{array}\\right\\|$ and $ \\left\\|\\begin{array}{c} D\\in BC\\ ,\\ AD\\perp BC \\\\\n\\ S\\in AD\\ ,\\ AS \\equal{} SD\\end{array}\\right\\|$ . Then $ S\\in IT_a\\cap I_aT$ .[/color] [/quote]\n\n[quote] [color=darkred][b][u]Property IIII.[/u][/b] The circle $ w$ touches $ [AB]$ , $ [AC]$ at $ E$ , $ F$ . The $ B$ - intenal bisector and $ C$ - internal angle bisector of $ \\triangle ABC$ \n\ncut $ EF$ in $ M$ , $ N$ . The circle $ w_a$ touches $ AB$ , $ AC$ at $ E_a$ , $ F_a$ . The $ B$ - external angle bisector and $ C$ - external bisector \n\nof $ \\triangle ABC$ cut $ E_aF_a$ in $ M_a$ , $ N_a$ . Then the points $ M$ , $ N$ , $ M_a$ , $ N_a$ belong to the circle with the diameter $ [BC]$ . [/color] [/quote]", "Solution_5": "[quote] [color=darkred][b][u]Property I.[/u][/b] Denote$ \\left\\|\\begin{array}{c} M\\in (BC)\\ ,\\ MB \\equal{} MC \\\\\n\\ D\\in BC\\ ,\\ AD\\perp BC\\end{array}\\right\\|$ and $ \\left\\|\\begin{array}{c} X\\in AD\\cap MI \\\\\n\\ Y\\in AD\\cap MI_a\\end{array}\\right\\|$ . Then $ AX \\equal{} r$ and $ AY \\equal{} r_a$ .[/color] [/quote]\r\n\r\nCall M, L the midpoint of CA, AB. Call $ K: AD \\cap NL$.\r\n\r\nIt's know that M,I,K are collinear (the incenter is the nagel point of the medial triangle), then the triangles $ \\triangle IDK$ and $ \\triangle XAK$ are equal so $ \\overline{AX} \\equal{} \\overline{ID} \\equal{} r$\r\n\r\nCall $ I_a$ the A-excenter and F the prjection of A on NL. Call $ J: MI_a \\cap NL$.\r\n\r\nBy the Property II it's know that $ I_a$, D, F are collinear and that $ DA \\parallel MJ$ (because the Mittenpunkt is the complement of Gergonne).\r\n\r\nThan with an homothety with center F $ \\triangle FAD$ goes on $ \\triangle FYI_a$ and K goes on J, so $ \\overline{JI_a} \\equal{} \\overline{JY}$ from wich $ \\overline{YA} \\equal{} r_a$", "Solution_6": "[quote] [color=darkred][b][u]Property II.[/u][/b] Denote$ \\left\\|\\begin{array}{c} T\\in (BC)\\ ,\\ IT\\perp BC \\\\\n\\ T_a\\in (BC)\\ ,\\ I_aT_a\\perp BC\\end{array}\\right\\|$ and $ \\left\\|\\begin{array}{c} D\\in BC\\ ,\\ AD\\perp BC \\\\\n\\ S\\in AD\\ ,\\ AS \\equal{} SD\\end{array}\\right\\|$ . Then $ S\\in IT_a\\cap I_aT$ .[/color] [/quote]\r\n\r\nw.l.o.g. $ c \\ge b$\r\n\r\n$ \\frac{\\overline{DT}}{\\overline{TT_a}} \\equal{} \\frac{\\overline{DM} \\minus{} \\frac{c\\minus{}b}{2}}{\\overline{DM} \\plus{} \\frac{c\\minus{}b}{2}} \\equal{} \\frac{a\\minus{}2b\\cos \\gamma \\minus{}c \\plus{}b}{a\\minus{}2b\\cos \\gamma \\plus{}c \\minus{}b} \\equal{} \\frac{a^2 \\minus{} (a^2\\plus{}b^2\\minus{}c^2) \\minus{} ac \\plus{} ba}{a^2 \\minus{} (a^2\\plus{}b^2\\minus{}c^2) \\plus{} ac \\minus{} ba} \\equal{} \\frac{(p\\minus{}a)(c\\minus{}b)}{p(c\\minus{}b)}\\equal{}\\frac{r}{r_a}$\r\n\r\n\r\n$ \\frac{\\overline{SD}\\minus{}\\overline{IT}}{\\overline{SD}}\\equal{} 1 \\minus{} \\frac{2r}{h_a} \\equal{} 1\\minus{} \\frac{a}{p} \\equal{} \\frac{p\\minus{}a}{p} \\equal{} \\frac{r}{r_a}$", "Solution_7": "[quote=\"Virgil Nicula\"][color=darkred][b][u]Property III.[/u][/b] The circle $ w$ touches $ [AB]$ , $ [AC]$ at $ E$ , $ F$ . The $ B$ - intenal bisector and $ C$ - internal angle bisector of $ \\triangle ABC$ \n\ncut $ EF$ in $ M$ , $ N$ . The circle $ w_a$ touches $ AB$ , $ AC$ at $ E_a$ , $ F_a$ . The $ B$ - external angle bisector and $ C$ - external bisector \n\nof $ \\triangle ABC$ cut $ E_aF_a$ in $ M_a$ , $ N_a$ . Then the points $ M$ , $ N$ , $ M_a$ , $ N_a$ belong to the circle with the diameter $ [BC]$ . [/color] [/quote]\r\n\r\n$ I_a, I_b, I_c$ are the A-, B-, C-excenters of the $ \\triangle ABC,$ $ D$ is tangency point of the incircle $ w$ with $ BC.$ Let parallel to $ I_cBI_a$ through $ C$ cut $ BI$ at $ M$ and let parallel to $ I_aCI_b$ through $ B$ cut $ CI$ at $ N.$ $ M, N$ thus defined are on a circle with diameter $ BC$ on account of the right angles $ \\angle BNC, \\angle BMC.$ Let $ BN, BM$ meet at $ K.$ The $ \\triangle KCB \\cong I_aBC$ are centrally congruent, having parallel sides and the side $ CB \\equal{} BC$ common. $ BCI_bI_c$ is cyclic on account of the right angles $ \\angle I_bBI_c, \\angle I_bCI_c$ $ \\Longrightarrow$ the $ \\triangle KCB \\sim \\triangle I_aI_bI_c$ are oppositely similar with similarity coefficient $ k \\equal{} \\frac {CB}{I_bI_c} \\equal{} \\sin \\frac {A}{2}$ and common orthocenter $ I.$ Their orthic $ \\triangle DNM \\sim \\triangle ABC$ are similar with the same similarity coefficient $ k$ and so are their concentric incircles $ w', w$ with radii $ r' \\equal{} kr, r.$\r\n\r\n$ \\angle I_bI_cC \\equal{} \\angle I_bBC \\equal{} \\angle MBC \\equal{} \\angle MNC$\r\n\r\nwhich means $ MN \\parallel I_bI_c.$ Since $ EF \\parallel I_bI_c$ (both $ \\perp AI)$, it follows that $ MN \\parallel EF.$ Let $ U \\in AI$ be the midpoint of $ EF$ and $ U' \\in AI$ tangency point of $ w'$ with $ MN.$ Since\r\n\r\n$ IU \\equal{} IF \\sin \\frac {A}{2} \\equal{} kr \\equal{} r' \\equal{} IU'$\r\n\r\nthe points $ U' \\equiv U$ are identical and $ E, F \\in MN.$\r\n\r\n[hide=\"A-excircle\"] The case of A-excircle $ w_a$ tangent to $ BC$ at $ D_a$ is proved similarly: the $ \\triangle I_aI_bI_c$ with orthocenter $ I$ and orthic $ \\triangle ABC$ is replaced by the $ \\triangle II_bI_c$ with orthocenter $ I_a$ and with the same orthic $ \\triangle ABC.$ The $ \\triangle KCB$ with orthocenter $ I$ and orthic $ \\triangle DNM$ is replaced by the $ \\triangle K_aCB$ with orthocenter $ I_a,$ where $ K_a \\equiv BM_a \\cap CN_a,$ and orthic $ \\triangle D_aN_aM_a.$ Similarity coefficient of the $ \\triangle K_aCB \\sim \\triangle II_bI_c$ with the common orthocenter $ I_a$ is the same, $ k \\equal{} \\sin \\frac {A}{2}$ and so is the similarity coefficient of their orthic $ \\triangle D_aN_aM_a \\sim \\triangle ABC$ with the common A-excenter $ I_a$ and concentric A-excircles $ w_a', w_a$ with radii $ r_a' \\equal{} kr_a, r_a.$[/hide]", "Solution_8": "Dear Mathlinkers,\nrevisiting this first result, we can have a nice proof by considering\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=550083\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=358063\nSincerely\nJean-Louis" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "[size=150]Let $ a,b,c$ be positive numbers,$ \\min{(a,b,c)}> t \\geq 0$,prove that:\n$ {\\frac {a \\minus{} t}{b \\minus{} t}} \\plus{} {\\frac {b \\minus{} t}{c \\minus{} t}} \\plus{} {\\frac {c \\minus{} t}{a \\minus{} t}}\\geq {\\frac {a }{b}} \\plus{} {\\frac {b}{c}} \\plus{} {\\frac {c}{a}}$\n\n\nBQ[/size]", "Solution_1": "[quote=\"xzlbq\"][size=150]Let $ a,b,c$ be positive numbers,$ \\min{(a,b,c)} > t \\geq 0$,prove that:\n$ {\\frac {a \\minus{} t}{b \\minus{} t}} \\plus{} {\\frac {b \\minus{} t}{c \\minus{} t}} \\plus{} {\\frac {c \\minus{} t}{a \\minus{} t}}\\geq {\\frac {a }{b}} \\plus{} {\\frac {b}{c}} \\plus{} {\\frac {c}{a}}$\n\n\nBQ[/size][/quote]\r\n\r\nboth of my proof way in your ineq(14) can be prove this ineq too.\r\n\r\nsee also: http://www.mathlinks.ro/viewtopic.php?t=319100", "Solution_2": "Just take $ a \\minus{} t \\equal{} a' ; b \\minus{} t \\equal{} b'; c \\minus{} t \\equal{} c' ;$ so that $ a',b',c'> 0$\r\nThen your inequality equivalents \r\n$ \\frac {a'}{b'} \\plus{} \\frac {b'}{c'} \\plus{} \\frac {c'}{a'}\\geq \\frac {a' \\plus{} t}{b' \\plus{} t} \\plus{} \\frac {b' \\plus{} t}{c' \\plus{} t} \\plus{} \\frac {c' \\plus{} t}{a' \\plus{} t}$\r\nWhich is just your inequality 14 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=319100]posted here[/url].\r\nNice inequalities, congratulations. :)\r\n[edit: Oops, beaten again]", "Solution_3": "[quote=\"xzlbq\"][size=150]Let $ a,b,c$ be positive numbers,$ \\min{(a,b,c)} > t \\geq 0$,prove that:\n$ {\\frac {a \\minus{} t}{b \\minus{} t}} \\plus{} {\\frac {b \\minus{} t}{c \\minus{} t}} \\plus{} {\\frac {c \\minus{} t}{a \\minus{} t}}\\geq {\\frac {a }{b}} \\plus{} {\\frac {b}{c}} \\plus{} {\\frac {c}{a}}$\n\n\nBQ[/size][/quote]\r\n\r\n[size=150]Extended to n -variables,Do the same?[/size]", "Solution_4": "[quote=\"xzlbq\"][quote=\"xzlbq\"][size=150]Let $ a,b,c$ be positive numbers,$ \\min{(a,b,c)} > t \\geq 0$,prove that:\n$ {\\frac {a \\minus{} t}{b \\minus{} t}} \\plus{} {\\frac {b \\minus{} t}{c \\minus{} t}} \\plus{} {\\frac {c \\minus{} t}{a \\minus{} t}}\\geq {\\frac {a }{b}} \\plus{} {\\frac {b}{c}} \\plus{} {\\frac {c}{a}}$\n\n\nBQ[/size][/quote]\n\n[size=150]Extended to n -variables,Do the same?[/size][/quote]\r\n\r\ncertainly.\r\n\r\nthat's equivalent to your ineq(14).", "Solution_5": "[quote=\"kuing\"][quote=\"xzlbq\"][size=150]Let $ a,b,c$ be positive numbers,$ \\min{(a,b,c)} > t \\geq 0$,prove that:\n$ {\\frac {a \\minus{} t}{b \\minus{} t}} \\plus{} {\\frac {b \\minus{} t}{c \\minus{} t}} \\plus{} {\\frac {c \\minus{} t}{a \\minus{} t}}\\geq {\\frac {a }{b}} \\plus{} {\\frac {b}{c}} \\plus{} {\\frac {c}{a}}$\n\n\nBQ[/size][/quote]\n\nboth of my proof way in your ineq(14) can be prove this ineq too.\n\nsee also: http://www.mathlinks.ro/viewtopic.php?t=319100[/quote]\r\n\r\nI see,thanks you!\r\nBQ" } { "Tag": [ "search", "function" ], "Problem": "in the other topic that just got locked:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=143668\r\n\r\nthat website gave the scores for last year. i cant seem to find this year's scores. does anyone have a link.", "Solution_1": "Well, after the date passed the state discussion window, everyone switched to discussing this year's cutoffs. You may keep discussing here though, sorry, when I locked the other, i didn't realize that my link was for 05-06 cutoffs.", "Solution_2": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141779" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For $a,b,c$ positive ral numbers such that $abc=1$ prove that\r\n\r\n\r\n$ \\sum(a+b)^2c \\geq 4\\sum ab $", "Solution_1": "I remember this problem being given in some MOSP, but I'm not sure. Anyway, it will appear in that variant from MOSP in the book Old and New Inequalities, a book which will appear at IMO (authors Titu Andreescu, Vasile Cartoaje (Vasc), Mircea Lascu (Gil) and Gabriel Dospinescu (harazi)). The problem is not trivial as it sounds, but anyway it is easy:\r\n Take $ a=\\frac{1}{x} $ and so on and reduce it to $ xy+yz+zx+\\frac{3}{x+y+z}\\geq 4$. But from AM-GM we have ${ xy+yz+zx+\\frac{3}{x+y+z}>=4\\cdot\\sqrt[4]{ \\frac (xy+yz+zx)^3}{9(x+y+z)}}\\geq 4 $ the last one being very easy.", "Solution_2": "Sorry but I don't understand..........!!!!!!!!!!!!!!!!!!!!!!!!!!\r\n\r\nCan you please detail it better ??\r\n\r\nThank you very much.", "Solution_3": "He means:\r\n\r\n\\[ \\frac{3}{x + y + z} + xy + yz + zx \\geq 4 \\sqrt[3]{\\frac{(xy + yz + zx)^3}{9(x + y + z)}}\\] by the AM-GM inequality so it suffices to prove \\[(xy + yz + zx)^3 \\geq 9(x + y + z).\\]\r\nNow you can use, for example, brute force :D", "Solution_4": "or you can use $xy+yz+zx\\geq$ 3 and $(xy+yz+zx)^2\\geq 3xyz(x+y+z)$" } { "Tag": [ "calculus", "derivative", "function", "algebra unsolved", "algebra" ], "Problem": "Assume that f, g : (0,1)-->R are derivative functions satisfy\r\n (i) f, g, f', g' >0 /(0,1)\r\n (i)f'/g' is an increasing function\r\nShow that \r\n h is increasing or decreasing or there exists 0 < :theta: <1 such that h is decreasing in (0, :theta: ) and increasing in ( :theta: ,1)\r\n (h= f/g)\r\n I can solve this problem if f',g' is continuous, but if f',g' is not continuous , I can't solve it\r\nCould any one help me?\r\n(My English is terrible :lol: )", "Solution_1": "[quote=\"dayvn\"]Assume that f, g : (0,1)-->R are derivative functions satisfy\n (i) f, g, f', g' >0 /(0,1)\n (i)f'/g' is an increasing function\nShow that \n h is increasing or decreasing or there exists 0 < :theta: <1 such that h is decreasing in (0, :theta: ) and increasing in ( :theta: ,1)\n (h= f/g)\n I can solve this problem if f',g' is continuous, but if f',g' is not continuous , I can't solve it\nCould any one help me?\n(My English is terrible :lol: )[/quote]\r\n\r\n\r\nu know what it is easy \r\nbut u have to understand me in azery\r\nor in math lungagge" } { "Tag": [ "probability", "calculus", "geometry", "analytic geometry", "integration", "trigonometry", "function" ], "Problem": "A table of infinite expanse has inscribed on it a set of parallel lines spaced $ 2a$ units apart. A needle of length $ 2l$ (smaller than $ 2a$) is twirled and tossed on the table. What is the probability that when it comes to rest it crosses on a line?", "Solution_1": "To start out, I think you should place a restriction on where the needle falls; if not, the probability is 0.", "Solution_2": "[hide=\"A try\"]\nWe can count the complement, that is, the probability that it does not land on a line, and that it lands in between the lines.\nThere are 2 things to consider, location of needle and position of needle (angle).\nSelecting the center of the needle to be the point that represents its location, we can firstly see that if the needle's center is any farther than L (was that the variable?) units away from a line, it cannot touch the line. If the needle always landed so that it would be positioned at a right angle with the parallel lines, the probability would be L/A. However, the needle can be in any orientation. Let x be the angle the needle is turned from the 0 degree position (parallel to the lines on the table), and let L/d be the distance from the center of the needle to a line (where L/d is between 0 and L). The distance that an end of a needle is from a line is given be Lsinx. We only need to consider {0,90} for x because the needle is rotationally symmetric (in length) about the center.\nWe have that Lsinx>L/d, and sinx>1/d follows, and then x>arcsin1/d. Since I don't know calculus, I plug that into my calculator (y=(arcsinx)/90), and use it to find the area under the curve, and I get some number 0.36336811 for which I have no clue where it came from. The probability is thus 0.6366189L/A.[/hide]", "Solution_3": "That is not right.", "Solution_4": "... let's say the lines have equations $ x \\equal{} \\minus{}a$ and $ x \\equal{} a$. Since you have placed no restriction on where the needle can land, it can land anywhere on the plane; hence, the probability that it lands in a region with an x-coordinate between $ [\\minus{}a \\minus{} l, a \\plus{} l]$ is 0. Hence, the needle will never intersect. \r\n\r\nPlace a restriction somewhere. \r\n\r\nIf my argument is incorrect, please point out to me what I did wrong.", "Solution_5": "No, he's talking about an infinite number of parallel lines equally spaced apart, not just 2.", "Solution_6": "Crap wait, I think I did something wrong with the calculator trick. Is it L/2A?", "Solution_7": "I believe that the probability that the needle will cross a line is $ \\frac {2l}{a\\pi}\\int_{0}^{\\frac {\\pi}{2}}\\sin\\theta \\ d\\theta \\equal{} \\frac {2 l}{a\\pi}$.", "Solution_8": "[hide=\"Solution.\"]\nThe chance of a needle of length $ x$ crossing a line when the lines are spaced $ y$ units apart is $ \\frac {x}{y}$. All that matters is the vertical height, not the horizontal length. Given an angle $ \\alpha$, the vertical height of the needle will then be $ 2l\\sin{\\alpha}$. The needle can take any angle between $ 0$ and $ 90$ degrees; the probability of the needle crossing a line as a function of the angle would then be $ P(\\alpha) \\equal{} \\frac {2l\\sin{\\alpha}}{2a}$. The average value over the range is: \n\n$ \\frac {\\frac {\\pi}{2}}{\\int_{0}^{\\frac {\\pi}{2}}P(\\alpha)d\\alpha}$\n\n$ \\equal{} \\frac {\\frac {\\pi}{2}}{\\int_{0}^{\\frac {\\pi}{2}}\\frac {2l\\sin{\\alpha}}{2a}}$\n\n$ \\equal{} \\frac {2l}{a\\pi}$.[/hide]" } { "Tag": [ "\\/closed" ], "Problem": "I'm a student and is eager to learn about math..\r\nBut, I have not found really an excellent site for mathematic for little bit advanced but below level of AME.. Since there are smart instructors and other people who are good at math, please help me if they know any..", "Solution_1": "why isn't this site excellent? AOPS has everything from books, free jams, courses, free circles?\r\n\r\nThis website is a bit advanced, but if you are sure you are advanced try Mathlinks Everyone (which is similar but is Olympiad level).", "Solution_2": "Of course, it is... I mean except this site.. Besides, MathJams are far away from me now and I don't take virtual classes..." } { "Tag": [ "algebra", "polynomial", "trigonometry", "algebra proposed" ], "Problem": "Determine all polynomials $ P(x)$ with rational coefficients such that for all $ |x| \\le 1$, $ P(x) \\equal{} P(\\frac { \\minus{} x \\plus{} \\sqrt {3 \\minus{} 3x^2}}{2})$.", "Solution_1": "$ P(x) \\equal{} a(4x^3 \\minus{} 3x)\\plus{}b, a,b \\in \\mathbb{Q}$", "Solution_2": "Can you explain the method of your solution please?", "Solution_3": "We have $ P(\\cos \\theta) \\equal{} P(\\cos \\left(\\theta \\plus{} \\frac{2 \\pi}{3} \\right)$ where $ \\theta \\in [0, \\pi]$\r\n\r\nThis tells us that in general any polynomial of the form $ aP(x) \\plus{} b$ where $ P(\\cos \\theta)$ is the polynomial corresponding to $ \\cos 3k \\theta$ expanded in terms of $ \\cos \\theta$ will satisfy the given equation." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c \\ge 0, a+b+c=1$. Prove that:\r\n$ \\frac{a^{2}+1}{b^{2}+c^{2}}+\\frac{b^{2}+1}{c^{2}+a^{2}}+\\frac{c^{2}+1}{a^{2}+b^{2}}\\ge 12$\r\n$ \\frac{a^{3}+1}{b^{3}+c^{3}}+\\frac{b^{3}+1}{c^{3}+a^{3}}+\\frac{c^{3}+1}{a^{3}+b^{3}}\\ge 22$\r\n$ \\frac{a^{4}+1}{b^{4}+c^{4}}+\\frac{b^{4}+1}{c^{4}+a^{4}}+\\frac{c^{4}+1}{a^{4}+b^{4}}\\ge 42$", "Solution_1": "Have you found an inequality true also for a general positive $ k$ coefficient?\r\n\r\n$ \\frac{a^{k}+1}{b^{k}+c^{k}}+\\frac{b^{k}+1}{c^{k}+a^{k}}+\\frac{c^{k}+1}{a^{k}+b^{k}}$\r\n\r\nThank you very much.\r\n\r\n\r\nHowever, very nice inequalities. :)", "Solution_2": "[quote=\"manlio\"]Have you found an inequality true also for a general positive $ k$ coefficient?\n\n$ \\frac{a^{k}+1}{b^{k}+c^{k}}+\\frac{b^{k}+1}{c^{k}+a^{k}}+\\frac{c^{k}+1}{a^{k}+b^{k}}$\n\nThank you very much.\n\n\nHowever, very nice inequalities. :)[/quote]\r\n\r\nTo Malio: I have only solution for k=2 :D \r\nSolution:\r\nWe have :\r\n$ LHS= \\large\\sum \\frac{a^{2}+(a+b+c)^{2}}{b^{2}+c^{2}}=\\large\\sum \\frac{2(a^{2}+ab+bc+ca)+b^{2}+c^{2}}{b^{2}+c^{2}}$\r\n$ LHS \\ge 12 \\leftrightarrow \\large\\sum \\frac{a^{2}+ab+bc+ca}{b^{2}+c^{2}}\\ge \\frac{9}{2}$\r\nWe have two case:\r\n[i]Case 1:[/i]\r\n$ \\frac{a(b+c)}{b^{2}+c^{2}}+\\frac{b(c+a)}{c^{2}+a^{2}}+\\frac{c(a+b)}{a^{2}+b^{2}}\\ge 2$\r\n[i]Case 2:[/i]\r\n$ \\frac{a^{2}+bc}{b^{2}+c^{2}}+\\frac{b^{2}+ca}{c^{2}+a^{2}}+\\frac{c^{2}+ab}{a^{2}+b^{2}}\\ge \\frac{5}{2}$\r\n<> Prove Case 1\r\nWLOG, we may assume that $ a \\ge b \\ge c$\r\n$ \\frac{a(b+c)}{b^{2}+c^{2}}+\\frac{b(c+a)}{c^{2}+a^{2}}-2 \\\\ =\\frac{b(a-b)+c(a-c)}{b^{2}+c^{2}}+\\frac{a(b-a)+c(b-c)}{c^{2}+a^{2}}\\ge (a-b)(\\frac{b}{b^{2}+c^{2}}-\\frac{a}{c^{2}+a^{2}})=\\frac{(a-b)^{2}(ab-c^{2})}{(a^{2}+c^{2})(b^{2}+c^{2})}\\ge 0$\r\n<>Prove Case 2\r\nAssume c=min{a,b,c}\r\n$ \\frac{a^{2}+bc}{b^{2}+c^{2}}+\\frac{b^{2}+ca}{c^{2}+a^{2}}+\\frac{c^{2}+ab}{a^{2}+b^{2}}-\\frac{5}{2}$ \r\n$ =\\frac{a^{2}+bc}{b^{2}+c^{2}}-2+\\frac{b^{2}+ca}{c^{2}+a^{2}}-2+\\frac{c^{2}+ab}{a^{2}+b^{2}}-\\frac{1}{2}$\r\n$ =\\frac{a^{2}-b^{2}+c(b-c)}{b^{2}+c^{2}}+\\frac{b^{2}-a^{2}+c(a-c)}{c^{2}+a^{2}}+\\frac{2c^{2}-(a-b)^{2}}{2(a^{2}+b^{2})}$\r\n$ \\ge \\frac{a^{2}-b^{2}}{b^{2}+c^{2}}-\\frac{a^{2}-b^{2}}{c^{2}+a^{2}}-\\frac{(a-b)^{2}}{2(a^{2}+b^{2})}$\r\n$ =(a-b)^{2}(\\frac{(a+b)^{2}}{(a^{2}+c^{2})(b^{2}+c^{2})}-\\frac{1}{2(a^{2}+b^{2})}) \\ge 0$\r\nAnd we are done!" } { "Tag": [ "AMC", "AIME", "algebra", "polynomial" ], "Problem": "http://www.kalva.demon.co.uk/aime/aime92.html\r\n\r\nI am having trouble understand the last step a sub n = \u00bdn^2 + an + b, but since a19 = a92 = 0, so the polynomial must be (n-19)(n-92)/2.\r\n\r\nSince the coefficient of the n^2 term is 1/2 of a sub n, why is it also in the second polynomial?", "Solution_1": "n=19 and n=92 are roots of the polynomial, so when plugged those two values in, you need to get 0.", "Solution_2": "Right, I understand that part, sorry that I didn't make my question clear.\r\n\r\nWhy is the 1/2 carried over? Since its 1/2 n^2 in the first equation why is it like that in the second equation too?", "Solution_3": "[quote=\"Calculus\"]Right, I understand that part, sorry that I didn't make my question clear.\n\nWhy is the 1/2 carried over? Since its 1/2 n^2 in the first equation why is it like that in the second equation too?[/quote]\r\nWhy wouldn't it be like that? When you expand the factored polynomial you have to have the coefficient 1/2 before the $n^2$, so you need the 1/2 in the factored form." } { "Tag": [ "algebra", "polynomial" ], "Problem": "(1/(x^2 - a)) + (1/(x^3 - b)) = 1\r\n\r\nYour goal is to write x in terms of a and b.\r\n\r\nPlease explain how as well.", "Solution_1": "hello, by simplifying $ \\frac{1}{x^2\\minus{}a}\\plus{}\\frac{1}{x^3\\minus{}b}\\minus{}1\\equal{}0$ we get\r\n$ \\frac{a\\plus{}b\\plus{}ab\\minus{}x^2\\minus{}bx^2\\minus{}x^3\\minus{}ax^3\\plus{}x^5}{(a\\minus{}x^2)(x^3\\minus{}b)}\\equal{}0$ and so\r\nyou have to solve the following polynomial equation in $ x$\r\n$ x^5\\minus{}ax^3\\minus{}x^3\\minus{}bx^2\\minus{}x^2\\plus{}ab\\plus{}a\\plus{}b\\equal{}0$ which is not possible with square roots in the general case.\r\nSonnhard.", "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]hello, by simplifying $ \\frac {1}{x^2 \\minus{} a} \\plus{} \\frac {1}{x^3 \\minus{} b} \\minus{} 1 \\equal{} 0$ we get\n$ \\frac {a \\plus{} b \\plus{} ab \\minus{} x^2 \\minus{} bx^2 \\minus{} x^3 \\minus{} ax^3 \\plus{} x^5}{(a \\minus{} x^2)(x^3 \\minus{} b)} \\equal{} 0$ and so\nyou have to solve the following polynomial equation in $ x$\n$ x^5 \\minus{} ax^3 \\minus{} x^3 \\minus{} bx^2 \\minus{} x^2 \\plus{} ab \\plus{} a \\plus{} b \\equal{} 0$ which is not possible with square roots in the general case.\nSonnhard.[/quote]General quintics are not solvable, but ones in this form are if the problem is correct.", "Solution_3": "So it is impossible to write x in terms of a and b?", "Solution_4": "Quintics of that form can't generally be solved, let a=0, b=1 you get $ x^5 \\minus{} x^3 \\minus{} 2x^2 \\plus{} 1 \\equal{} 0$, not factorable by [url]http://mcis.jsu.edu/faculty/desmith/polyfactor-cgi.html[/url]. So it is impossible to write x in terms of a and b.", "Solution_5": "Irreducibility is not a sufficient criterion to guarantee unsolvability by radicals. The nature of Galois extensions is [url=http://en.wikipedia.org/wiki/Quintic_equation#Examples_of_solvable_quintics]more subtle[/url] than that. \r\n\r\nBesides, \"impossible\" is a misstatement of Abel's theorem. It [i]is[/i] possible to solve quintic polynomials - if one is willing to use certain [url=http://en.wikipedia.org/wiki/Quintic_equation#Beyond_radicals]tools[/url] other than root extraction.\r\n\r\nIn any case, it is likely that there is no \"elementary\" solution to this problem.", "Solution_6": "So is anyone here able to write x in terms of a and b?", "Solution_7": "Have you been reading the above posts? It's not whether \"anyone\" can do it - you [b]can't[/b] do it with ordinary operators! See [url=http://en.wikipedia.org/wiki/Quintic_equation]Quintic equation[/url] to see why you can't solve some quintic equations. It means that you really cannot express $ x$ in terms of the ordinary operators - it's just not expressible in those terms.\r\n\r\nOf course, if you introduce $ \\text{BR}(x)$ (the [url=http://en.wikipedia.org/wiki/Bring_radical]Bring radical[/url]), then one can express $ x$ in terms of $ a$ and $ b$.", "Solution_8": "Sorry, I was just reading t0rajir0u's post and hoping." } { "Tag": [ "trigonometry", "geometry", "trig identities", "Law of Cosines", "Law of Sines" ], "Problem": "Let $a$, $b$, $c$ be the three sides of a triangle, and let $\\alpha$, $\\beta$, $\\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find \\[ \\frac{\\cot \\gamma}{\\cot \\alpha+\\cot \\beta}. \\]", "Solution_1": "[hide=\"hint:\"]extended law of sines[/hide]", "Solution_2": "Here's what I got, anyhow...\r\n\r\n[hide]Answer: $\\boxed{994}$. \nSolution: \\[ \\cot{\\alpha}+\\cot{\\beta}=\\frac{\\cos{\\alpha}}{\\sin{\\alpha}}+\\frac{\\cos{\\beta}}{\\sin{\\beta}}=\\frac{\\sin{\\alpha}\\cos{\\beta}+\\cos{\\alpha}\\sin{\\beta}}{\\sin{\\alpha}\\sin{\\beta}}= \\] \\[ \\frac{\\sin{\\alpha+\\beta}}{\\sin{\\alpha}\\sin{\\beta}}=\\frac{\\sin{\\gamma}}{\\sin{\\alpha}\\sin{\\beta}} \\]\nBy the law of cosines, \\[ a^2+b^2-2ab\\cos{\\gamma}=c^2=1989c^2-2ab\\cos{\\gamma} \\implies ab\\cos{\\gamma}=994c^2 \\]\nNow \\[ \\frac{\\cot{\\gamma}}{\\cot{\\alpha}+\\cot{\\beta}}=\\frac{\\cot{\\gamma}\\sin{\\alpha}\\sin{\\beta}}{\\sin{\\gamma}}=\\frac{\\cos{\\gamma}\\sin{\\alpha}\\sin{\\beta}}{\\sin^2{\\gamma}}=\\frac{ab}{c^2}\\cos{\\gamma}=\\frac{ab}{c^2}\\frac{994c^2}{ab} \\] \\[ =994 \\][/hide]", "Solution_3": "$ cotA$=$ \\frac{cosA}{sinA}$=$ \\frac{(b^{2}\\plus{}c^{2}\\minus{}a^{2})/2bc}{sinA}$\r\n\r\n=$ \\frac{b^{2}\\plus{}c^{2}\\minus{}a^{2}}{4S}$\r\n\r\nthis will make it much more easier.....", "Solution_4": "Draw altitude CH length h. $ Cot A \\equal{} AH/h$ and $ cot B \\equal{} BH/h$, so $ cot A \\plus{} cot B \\equal{} c/h$. \r\nAlso, by Law of Cosines, $ 994c^2 \\equal{} ab cos C$. But the area $ K \\equal{} 1/2 ab sin C$, so dividing, $ 497c^2/K \\equal{} 497c^2/(ch/2) \\equal{} 994(c/h) \\equal{} cot C$. \r\nSo the answer is $ 994(c/h)/(c/h) \\equal{} 994$.", "Solution_5": "[quote=\"K81o7\"]$\\frac{\\sin{\\alpha}\\sin{\\beta}}{\\sin^2{\\gamma}}=\\frac{ab}{c^2}$[/quote]\n\nHow do you get that?", "Solution_6": "Let $a_{}^{}$, $b_{}^{}$, $c_{}^{}$ be the three sides of a triangle, and let $\\alpha_{}^{}$, $\\beta_{}^{}$, $\\gamma_{}^{}$, be the angles opposite them. If $a^2+b^2=1989^{}_{}c^2$, find $\\frac{\\cot \\gamma}{\\cot \\alpha+\\cot \\beta}$\n\n[hide=\"Solution\"]\n\nUse the hint in the book to turn all the $\\cot$ into $\\cos$ and $\\sin$. Then, use Law of cosines to give us $c^2=a^2+b^2-2ab\\cos(\\gamma)$ or therefore $\\cos(\\gamma)=\\frac{994c^2}{ab}$. Next, we are going to put all the sin's in term of $\\sin(a)$. We get $\\sin(\\gamma)=\\frac{c\\sin(a)}{a}$. Therefore, we get $\\cot(\\gamma)=\\frac{994c}{b\\ sin a}$.\n\nNext, use Law of Cosines to give us $b^2=a^2+c^2-2ac\\cos(\\beta)$. Therefore, $\\cos(\\beta)=\\frac{a^2-994c^2}{ac}$. Also, $\\sin(\\beta)=\\frac{b\\sin(a)}{a}$. Hence, $\\cot(\\beta)=\\frac{a^2-994c^2}{bc\\sin(a)}$. \n\nLastly, $\\cos(\\alpha)=\\frac{b^2-994c^2}{bc}$. Therefore, we get $\\cot(\\alpha)=\\frac{b^2-994c^2}{bc\\sin(a)}$.\n\nNow, $\\frac{\\cot(\\gamma)}{\\cot(\\beta)+\\cot(\\alpha)}=\\frac{\\frac{994c}{b\\sin a}}{\\frac{a^2-994c^2+b^2-994c^2}{bc\\sin(a)}}$. After using $a^2+b^2=1989c^2$, we get $\\frac{994c*bc\\sin a}{c^2b\\sin a}=\\boxed{994}$.\n\n[/hide]", "Solution_7": "This is so troll \n\nRearranging we have \n\n$a^2=1989c^2-b^2$. let us have a right triangle so\n$a^2=c^2+b^2$\n\nEasy to solve we get $a=c\\sqrt{995}, b=c\\sqrt{994}$ and from here it is a matter of plugging in to get\n\n$\\boxed{994}$ as the answer", "Solution_8": "@above You cannot really say \"[b]without[/b] loss of generality\" when in fact, you lose every single bit of generality by assuming $a^2=c^2+b^2$.", "Solution_9": "[quote=sicilianfan]@above You cannot really say \"[b]without[/b] loss of generality\" when in fact, you lose every single bit of generality by assuming $a^2=c^2+b^2$.[/quote]\n\nok, ok true dat", "Solution_10": "[quote=AruKasera][quote=\"K81o7\"]$\\frac{\\sin{\\alpha}\\sin{\\beta}}{\\sin^2{\\gamma}}=\\frac{ab}{c^2}$[/quote]\n\nHow do you get that?[/quote]\n\nLaw of sines. $\\frac{a}{\\sin{\\alpha}} = \\frac{b}{\\sin{\\beta}} = 2R$. WLOG let $R = \\frac{1}{2}$ (just scale down triangle), which makes $a = \\sin{\\alpha}$ and $b = \\sin{\\beta}$.", "Solution_11": "$\\cot\\alpha+\\cot\\beta$=$\\frac{sin\\alpha cos\\beta+cos\\alpha sin\\beta}{sin\\alpha sin\\beta}=\\frac{sin{\\alpha+\\beta}}{sin\\alpha sin\\beta}=\\frac{sin{\\gamma}}{sin\\alpha sin\\beta}$", "Solution_12": "[hide = oops]\nLet $a = 1, b = 1, c = \\sqrt{\\frac{2}{1989}}$ and $x = \\cot{\\alpha} = \\cot{\\beta}$\nThen $\\cot{\\gamma} = \\frac{1}{\\tan{\\gamma}} = \\frac{1}{\\tan{(\\frac{\\gamma}{2} + \\frac{\\gamma}{2})}} = \\frac{1}{\\frac{2x}{1-x^2}} = \\frac{1 - x^2}{2x}$ since $\\cot{\\alpha} = \\tan{\\frac{\\gamma}{2}}$ so $$\\frac{\\cot{\\gamma}}{\\cot{\\alpha} + \\cot{\\beta}} = \\frac{\\frac{1 - x^2}{2x}}{2x} = \\frac{1 - x^2}{4x^2} = \\frac{1 - \\frac{1}{2(1989) - 1}}{4(\\frac{1}{2(1989) - 1})} = \\frac{2(1989) - 2}{4} = \\frac{1989 - 1}{2} = \\boxed{994}$$\n[/hide]", "Solution_13": "[hide=Solution]Rewrite as $\\frac{\\frac{\\cos \\gamma}{\\sin \\gamma}}{\\frac{\\cos \\alpha}{\\sin \\alpha}+\\frac{\\cos \\beta}{\\sin \\beta}}$, or $\\frac{\\frac{\\cos \\gamma}{\\sin \\gamma}}{\\frac{\\sin(\\alpha + \\beta)}{\\sin \\alpha \\sin \\beta}}=\\frac{\\cos \\gamma \\sin \\alpha \\sin \\beta}{\\sin(\\alpha + \\beta)\\sin \\gamma}$, or $\\frac{\\cos \\gamma \\sin \\alpha \\sin \\beta}{\\sin^2 \\gamma}$. From LoC on our original triangle, we have $1989c^2-2ab \\cos \\gamma = c^2$, or $ab \\cos \\gamma = 994c^2$. But $\\frac{\\sin \\alpha \\sin \\beta}{\\sin^2 \\gamma}=\\frac{ab}{c^2} = \\frac{994}{\\cos \\gamma}$, so the answer is $\\cos \\gamma \\cdot \\frac{994}{\\cos \\gamma}=\\boxed{994}$.[/hide]", "Solution_14": "Note that $\\sin A = t\\cdot a$ for some constant $t$, and similarly for the other vertices. Also, $\\cos A = r\\cdot \\frac{(b^2 + c^2 - a^2)}{2bc}$ for some constant $r$, and similarly for the other vertices. Letting $X$ be the expression we want, we have\n\\[ X = \\frac{\\frac{1}{tc}\\cdot r\\cdot \\frac{a^2 + b^2 - c^2}{2ab}}{\\frac{1}{ta}\\cdot r\\cdot \\frac{b^2 + c^2 - a^2}{2bc} + \\frac{1}{tb}\\cdot r\\cdot \\frac{a^2 + c^2 - b^2}{2ac}} = \\frac{a^2 +b^2 - c^2}{(b^2 + c^2 - a^2) + (a^2 + c^2 - b^2)} = \\frac{1989c^2 - c^2}{2c^2} = \\boxed{994}.\\]" } { "Tag": [ "Asymptote", "LaTeX", "\\/closed" ], "Problem": "I thought just like the Asymptote or the $ \\text{\\LaTeX}$ forum, there should be a geogebra forum. Some people have already asked me/other people questions.", "Solution_1": "Private message an administrator because asking other people in forum won't create the forum. But I would also like a Geogebra forum.", "Solution_2": "Yeah I don't think that will happen. It's so simple that there aren't that many questions to ask. I'll probably move all the discussions down in the Asymptote forum (and maybe call that Asymptote and Geogebra forum.)" } { "Tag": [], "Problem": "How many whole numbers from 10 to 99 have units digit greater than the tens digit?", "Solution_1": "There is $ 12,13,14,15,16,17,18,19,23,24,25,26$.... an easier way to do this is to notice that while each tens digit goes up by one, the number of these numbers decrease by one. So it is $ 8\\plus{}7\\plus{}6\\plus{}5\\plus{}4\\plus{}3\\plus{}2\\plus{}1\\equal{} \\boxed{36}$" } { "Tag": [ "quadratics", "floor function", "modular arithmetic", "Quadratic Residues", "pen" ], "Problem": "Let $p$ be an odd prime number. Show that the smallest positive quadratic nonresidue of $p$ is smaller than $\\sqrt{p}+1$.", "Solution_1": "If $a,b$ are quadratic residues, then so is $ab$. Thus it suffices to show that the set $S = \\{ s \\in \\mathbb Z | 0 < s < \\sqrt p +1 \\}$ generates all residues $\\neq 0$.\r\n\r\nAssume this to be false and let $x$ be the smallest positive integer whose residue class is not represented by $S$. Then $x \\geq \\sqrt p +1$ and we consider $a := 1+\\left\\lfloor \\frac{p-1}{x} \\right\\rfloor$ and observe:\r\na) $0 300$ , actually you can find a more refined vaule by computing $\\sqrt{20089}$ and $\\sqrt{29989}$ and gettting $142 \\le n \\le 173$ but no need to do any thing which you can't do in a few seconds without pen or paper. [/hide]\n2. And $n \\equiv (25 \\pm 8 )$ or $\\equiv (75 \\pm 8 ) \\bmod 100$ \n[hide=\"because\"] $n^2 \\equiv 89 \\bmod 100 \\implies n^2 \\equiv 1 \\bmod 4 \\ and \\ n^2 \\equiv 14 (\\equiv 64) \\bmod 25$ \n $\\implies$ ($n$ is odd ) and ($n \\equiv \\pm 8 \\bmod 25$)\n$\\implies n \\equiv (25 \\pm 8)$ or $n \\equiv (75 \\pm 8) \\bmod 100$ [/hide]\n3. Only choice for n is $n=175-8 = 167$[/hide]\r\n\r\nYes it takes much longer to write the solution than actually figure out the answer. :) . Hope this is helpful." } { "Tag": [ "\\/closed" ], "Problem": "I have taken Algebra 1 (school) so should I buy the AOPS algebra 1 book or should I just go to Algebra 2, cause I don't wan't to take a book that is pretty much review...", "Solution_1": "It depends on what you are doing at school\r\n\r\nBefore I took algebra 1, I was already doing way past my grade level math.\r\nI was thinking to take algebra 2 since I already knew a lot of algebra.\r\n\r\nBut then, I just signed up for algebra 1 since my mother suggested it.\r\n\r\nIt turns out that I had SO much to learn! It's not all just stuff like:\r\n\r\n$ x\\plus{}y\\equal{}32$\r\n$ 2x\\minus{}6y\\equal{}78$\r\n\r\nThat's just the beginning. \r\nThen you learn like a [u][b]TON[/b][/u] of problem solving skills, good for stuff like future soon-to-be AMC winner. Logic is my weakness. :maybe: \r\n\r\n\r\nSo I suggest that you take the course, even though you think that it may be easy for you. :P", "Solution_2": "[quote=\"ritwik_anand\"]I have taken Algebra 1 (school) so should I buy the AOPS algebra 1 book or should I just go to Algebra 2, cause I don't wan't to take a book that is pretty much review...[/quote]\r\nI suggest Algebra 2, but you should take the \"Do you need this?\" test for Algebra 1 and Algebra 2 to see if you, well, need it.", "Solution_3": "Our Intro Algebra book covers all of the algebra of a typical Algebra 1 class plus much of the algebra of a typical Algebra 2 class. We use the book for both our Algebra 1 and Algebra 2 classes. So, if you're asking about a class here at AoPS, I would recommend the Algebra 2 class, which uses our Intro Algebra book.", "Solution_4": "Mr. Rusczyk, is the book for Algebra 1 and 2 and 3 the same or just Algebra 1 and 2?\r\n\r\nThanks.", "Solution_5": "The book for Algebra 1 and 2 is Introduction to Algebra, and the book for Algebra 3 is Intermediate Algebra." } { "Tag": [ "FTW" ], "Problem": "i'm getting pretty random now. exactly like the top admin competition. you can add a point and take one away. and each soda will start wiht 30. i will allow people to post as much as they want i guess\r\n. oh and soon i'm gonna start a top chip on aops. unless someone else starts it already.\r\n\r\npepsi:30\r\n\r\nsierra mist:30\r\n\r\ncoke:30\r\n\r\nsprite:30\r\n\r\norange soda:31\r\n\r\ngrape soda:30\r\n\r\nroot beer:29", "Solution_1": "pepsi:30 \r\n\r\nsierra mist:30 \r\n\r\ncoke:30 \r\n\r\nsprite:30 \r\n\r\norange soda:32 \r\n\r\ngrape soda:30 \r\n\r\nroot beer:28", "Solution_2": "pepsi:30 \r\n\r\nsierra mist:30 \r\n\r\ncoke:30 \r\n\r\nsprite:30 \r\n\r\norange soda:31\r\n\r\ngrape soda:30 \r\n\r\nroot beer:29\r\n\r\nAh, reversed.", "Solution_3": "pepsi:30 \r\n\r\nsierra mist:30 \r\n\r\ncoke:30 \r\n\r\nsprite:30 \r\n\r\norange soda:32 \r\n\r\ngrape soda:30 \r\n\r\nroot beer:28", "Solution_4": "pepsi:30 \r\n\r\nsierra mist:30 \r\n\r\ncoke:30 \r\n\r\nsprite:30 \r\n\r\norange soda:31\r\n\r\ngrape soda:30 \r\n\r\nroot beer:29", "Solution_5": "#4pepsi:30 \r\n\r\nsierra mist:30 \r\n\r\ncoke:30 \r\n\r\nsprite:30 \r\n\r\norange soda:32 \r\n\r\ngrape soda:30 \r\n\r\nroot beer:28", "Solution_6": "pepsi:30 \r\n\r\nsierra mist:30 \r\n\r\ncoke:30 \r\n\r\nsprite:30 \r\n\r\norange soda:31\r\n\r\ngrape soda:30 \r\n\r\nroot beer:29", "Solution_7": "#4pepsi:30 \r\n\r\nsierra mist:30 \r\n\r\ncoke:30 \r\n\r\nsprite:30 \r\n\r\norange soda:32 \r\n\r\ngrape soda:30 \r\n\r\nroot beer:28", "Solution_8": "pepsi:30\r\n\r\nsierra mist:30\r\n\r\ncoke:30\r\n\r\nsprite:30\r\n\r\norange soda:31\r\n\r\ngrape soda:30\r\n\r\nroot beer:29", "Solution_9": "#4pepsi:30 \r\n\r\nsierra mist:30 \r\n\r\ncoke:30 \r\n\r\nsprite:30 \r\n\r\norange soda:32 \r\n\r\ngrape soda:30 \r\n\r\nroot beer:28", "Solution_10": "pepsi:30\r\n\r\nsierra mist:30\r\n\r\ncoke:30\r\n\r\nsprite:30\r\n\r\norange soda:31\r\n\r\ngrape soda:30\r\n\r\nroot beer:29", "Solution_11": "pepsi:30 \r\n\r\nsierra mist:30 \r\n\r\ncoke:30 \r\n\r\nsprite:30 \r\n\r\norange soda:32 \r\n\r\ngrape soda:30 \r\n\r\nroot beer:28", "Solution_12": "pepsi:30\r\n\r\nsierra mist:30\r\n\r\ncoke:30\r\n\r\nsprite:30\r\n\r\norange soda:31\r\n\r\ngrape soda:30\r\n\r\nroot beer:29", "Solution_13": "pepsi:30 \r\n\r\nsierra mist:30 \r\n\r\ncoke:30 \r\n\r\nsprite:30 \r\n\r\norange soda:32 \r\n\r\ngrape soda:30 \r\n\r\nroot beer:28", "Solution_14": "pepsi:30\r\n\r\nsierra mist:30\r\n\r\ncoke:30\r\n\r\nsprite:30\r\n\r\norange soda:31\r\n\r\ngrape soda:30\r\n\r\nroot beer:29\r\n\r\nThis is repetitive", "Solution_15": "What, you can't minus 3!\r\n\r\n1 root beer: 11 \r\n2 orange soda: 7 \r\n3 ginger ale: 3\r\n4 sierra mist: PWNT \r\n5 sprite: PWNT \r\n6 coke: PWNT \r\n7 mountain dew: PWNT \r\n8 pepsi: PWNT", "Solution_16": "1 root beer: 12\r\n2 orange soda: 5\r\n3 ginger ale: 3\r\n4 sierra mist: PWNT\r\n5 sprite: PWNT\r\n6 coke: PWNT\r\n7 mountain dew: PWNT\r\n8 pepsi: PWNT", "Solution_17": "1 root beer: 13\r\n2 orange soda: 5\r\n3 ginger ale: 2\r\n4 sierra mist: PWNT\r\n5 sprite: PWNT\r\n6 coke: PWNT\r\n7 mountain dew: PWNT\r\n8 pepsi: PWNT", "Solution_18": "1 root beer: 11 \r\n2 orange soda: 5 \r\n3 ginger ale: 3\r\n4 sierra mist: PWNT \r\n5 sprite: PWNT \r\n6 coke: PWNT \r\n7 mountain dew: PWNT \r\n8 pepsi: PWNT", "Solution_19": "I just found out, drop 1... so \r\n\r\n1 root beer: 13\r\n2 orange soda: 5\r\n3 ginger ale: 3\r\n4 sierra mist: PWNT\r\n5 sprite: PWNT\r\n6 coke: PWNT\r\n7 mountain dew: PWNT\r\n8 pepsi: PWNT", "Solution_20": "1 root beer: 13 \r\n2 orange soda: 3\r\n2 ginger ale: 3 \r\n4 sierra mist: PWNT \r\n5 sprite: PWNT \r\n6 coke: PWNT \r\n7 mountain dew: PWNT \r\n8 pepsi: PWNT", "Solution_21": "1 root beer: 14\r\n2 orange soda: 3\r\n2 ginger ale: 2\r\n4 sierra mist: PWNT\r\n5 sprite: PWNT\r\n6 coke: PWNT\r\n7 mountain dew: PWNT\r\n8 pepsi: PWNT", "Solution_22": "1 root beer: 12\r\n2 orange soda: 3 \r\n2 ginger ale: 3\r\n4 sierra mist: PWNT \r\n5 sprite: PWNT \r\n6 coke: PWNT \r\n7 mountain dew: PWNT \r\n8 pepsi: PWNT", "Solution_23": "1 root beer: 14\r\n2 orange soda: 3\r\n2 ginger ale: 2\r\n4 sierra mist: PWNT\r\n5 sprite: PWNT\r\n6 coke: PWNT\r\n7 mountain dew: PWNT\r\n8 pepsi: PWNT\r\n\r\nRead the Rules, 1=2.", "Solution_24": "Hmm...people have been subtracting 2 since page 8 despite the original rule, and there was even a comment about it.\r\n\r\n[quote=\"iin77\"]So, uh, I see people randomly started subtracting two now. [/quote]\r\n\r\nI'm going to subtract 2.\r\n\r\n[color=green]1 root beer: 15[/color] \r\n2 ginger ale: 2 \r\n[color=red]2 orange soda: 1 [/color]\r\n4 sierra mist: PWNT \r\n5 sprite: PWNT \r\n6 coke: PWNT \r\n7 mountain dew: PWNT \r\n8 pepsi: PWNT", "Solution_25": "actually, it's heal 1 hurt 2. not the other way around.", "Solution_26": "[quote=\"cf249\"]Hmm...people have been subtracting 2 since page 8 despite the original rule, and there was even a comment about it.\n\n[quote=\"iin77\"]So, uh, I see people randomly started subtracting two now. [/quote]\n\nI'm going to subtract 2.\n\n1 root beer: 15\n[color=red]2 ginger ale: PWNT[/color] \n[color=green]3 orange soda: 2 [/color]\n4 sierra mist: PWNT \n5 sprite: PWNT \n6 coke: PWNT \n7 mountain dew: PWNT \n8 pepsi: PWNT[/quote]", "Solution_27": "[color=green]1 root beer: 16[/color]\r\n[color=red]2 orange soda: PWNT[/color]\r\n3 ginger ale: PWNT\r\n4 sierra mist: PWNT \r\n5 sprite: PWNT \r\n6 coke: PWNT \r\n7 mountain dew: PWNT \r\n8 pepsi: PWNT\r\n\r\nRoot beer wins! (it's my favorite of the choices listed)", "Solution_28": "Sweet :) I love arbitrarily taking sides and then winning!\r\n\r\nWith respect to the +1/-1 rule vs the +1/-2 rule, I think it is sensible to use a system in which the game is forced to terminate (such as the latter system) rather than a game which is potentially endless. I mean there are plenty of pointless games that don't terminate, but people get bored of them... but hurt and heals don't end until there's a winner!\r\n\r\nI strongly encourage anyone creating a new hurt/heal game to force it to terminate (ie, take off more points than you put back).", "Solution_29": "Dang it!!!! I wanted orange soda to win, and I missed the great finale." } { "Tag": [ "geometry", "calculus", "integration", "calculus computations" ], "Problem": "My friends and I disagree mightily over a problem as such:\r\n\r\n\"Find the area between the curves y=x^2, y=x, and y=2.\"\r\nI think you have to subtract the integrals, so that you have\r\n\r\n/int^2_0 (x^2-x)/,dx = 2/3\r\n\r\nThey disagree and say the answer is, variously, 5/6 and 1.\r\n\r\nCan anyone solve this and verify an answer?", "Solution_1": "My answer is $ \\frac{13\\minus{}8\\sqrt{2}}{6}$", "Solution_2": "I think there's some ambiguity here. When $ 0\\leq x\\leq 1$, there are two pieces: the piece above $ y\\equal{}x^2$ and below $ y\\equal{}x$, and the piece above $ y\\equal{}x$ and below $ y\\equal{}2$. Are they both considered? Similarly, when $ 1\\leq x\\leq \\sqrt{2}$, there's also two pieces.\r\n\r\nIf all pieces are considered, then the area is\r\n\\[ \\int_{\\minus{}\\sqrt{2}}^1 (2\\minus{}x^2)\\ dx \\plus{}\\int_1^2 (2\\minus{}x)\\ dx\\]\r\nHere the area is the integral of the difference between the highest curve and the lowest curve (as long as the two curves form a closed piece). Outside of $ \\minus{}\\sqrt{2}\\leq x\\leq 2$, there's no intersection, so no more closed pieces." } { "Tag": [ "LaTeX" ], "Problem": "whats the multiplication sign in latex code????", "Solution_1": "[quote=\"hatsup123\"]whats the multiplication sign in latex code????[/quote]\r\n\r\nThere are two types:\r\n\r\n\\[\r\n\\times \\text{ and } \\cdot\r\n\\]\r\n\r\n\\times and \\cdot\r\n\r\nExamples:\r\n\r\n\\[\r\n2 \\times 2 = 5\r\n\\]\r\n\\[\r\n1 \\cdot 2 = 3\r\n\\]", "Solution_2": "THANKS!!!!!!!!!!!!!!!!!! :D :D :D :D :D :D :D :D :D" } { "Tag": [ "linear algebra", "matrix", "search", "linear algebra open" ], "Problem": "Hi,\r\n\r\nI have the following matrix:\r\n\r\n1 1/2 1/3.....1/n\r\n1/2............1/n+1\r\n.\r\n.\r\n.\r\n.\r\n.\r\n1/n...............1/2n-1\r\n\r\n\r\nDoes anybody have any idea of what it's inverse should be? It's claimed that the inverse has all integer(!!) entries.", "Solution_1": "The inverse of the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=163506]Hilbert matrix[/url]. Follow the links in that thread, too.\r\n\r\n(And thanks to Kent for inserting the search keyword I used)" } { "Tag": [ "function", "complex analysis", "complex analysis unsolved" ], "Problem": "Let $ f$ be analytic on $ D\\equal{}\\{z<1\\}$. Suppose $ f$ is one-to-one on $ D\\minus{}\\{0\\}$. Prove that $ f$ is one-to-one on $ D$", "Solution_1": "If $ n$ is the largest integer so that $ f^n(0) \\equal{} 0$, then $ f$ maps a neighborhood $ V$ of $ 0$ to a neighborhood of $ f(0)$ which is covered $ n\\plus{}1$ times. Now, suppose there is $ w$ so that $ f(0) \\equal{} f(w)$. By the above reasoning applied to $ w$, we have $ f'(w) \\neq 0$. Then there is a neighborhood $ W$ of $ w$, chosen with $ V \\cap W$ empty, such that $ f$ maps $ W$ topologically to a neighborhood of $ f(0)$. This is clearly a contradiction to $ f$ being injective on the punctured disc." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c > 0$ and $ a\\plus{}b\\plus{}c\\equal{}1$ Prove that \\[ \\sum_{cyc}\\sqrt{\\frac{2b}{3\\plus{}a}} \\leq \\sqrt{\\sum_{cyc}\\frac{1}{1\\plus{}2\\sqrt{ab}}}\\]", "Solution_1": "[quote=\"apollo\"]Let $ a,b,c > 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$ Prove that\n\\[ \\sum_{cyc}\\sqrt {\\frac {2b}{3 \\plus{} a}} \\leq \\sqrt {\\sum_{cyc}\\frac {1}{1 \\plus{} 2\\sqrt {ab}}}\n\\]\n[/quote]\r\n$ \\sum_{cyc}\\sqrt {\\frac {2b}{3 \\plus{} a}} \\leq \\sum_{cyc}b\\sqrt{\\frac{2}{b(3\\plus{}a)}}\\leq\\sqrt{\\sum_{cyc}\\frac{2}{3\\plus{}a}}$ and $ \\sqrt {\\sum_{cyc}\\frac {1}{1 \\plus{} 2\\sqrt {ab}}}\\geq\\sqrt {\\sum_{cyc}\\frac {1}{1 \\plus{} a\\plus{}b}}.$\r\nHence, it remains to prove that: $ \\sum_{cyc}\\frac{2}{3\\plus{}a}\\leq\\sum_{cyc}\\frac {1}{1 \\plus{} b\\plus{}c}.$ But $ \\sum_{cyc}\\frac{2}{3\\plus{}a}\\leq\\sum_{cyc}\\frac {1}{1 \\plus{} b\\plus{}c}\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\frac{1\\plus{}a\\minus{}2b\\minus{}2c}{(3\\plus{}a)(1\\plus{}b\\plus{}c)}\\geq0\\Leftrightarrow\\sum_{cyc}\\frac{a\\minus{}b\\minus{}(c\\minus{}a)}{(3\\plus{}a)(1\\plus{}b\\plus{}c)}\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a\\minus{}b)\\left(\\frac{1}{(3\\plus{}a)(1\\plus{}b\\plus{}c)}\\minus{}\\frac{1}{(3\\plus{}b)(1\\plus{}a\\plus{}c)}\\right)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a\\minus{}b)^2(2\\minus{}c)(3\\plus{}c)(1\\plus{}a\\plus{}b)\\geq0.$" } { "Tag": [], "Problem": "Ther are 8 boys and 6 girls in who are members of the grumpet section in the band. From the members of the trumpet section, a quintet is formed. IF the quintet must consist of 3 boys and 2 girls , how many possible quintets are there?", "Solution_1": "[hide=\" hi \"]$\\binom{8}{3}=(8*7*6) /6 = 56$\n$\\binom{6}{2}= (6*5)/2= 15$\n\n$56*15=840$[/hide]", "Solution_2": "i think thats the correct answer. :)" } { "Tag": [ "quadratics", "summer program", "Mathcamp", "function" ], "Problem": "Let x_n be a sequence so that we have x_n=4x_(n-1)-x_(n-2) for any n>=3 with x1=2 and x2=7. Prove that we have the following thing:\r\nfor any n>=1, the sequence a_n=x_(2n+1)/2 is the sum of two consecutive squares!\r\n\r\nthank you!", "Solution_1": "This shall be messy. It's going to be inductive-like, several times over. I'm just going to give the outline, not the details. First, assume it works for two consecutive odd terms of the sequence, which means they can be written as (4a :^2: + 4a + 2) and (4b :^2: + 4b + 2). Then find out what the next odd term of the sequence will look like. Assume this term has the same form and solve for the resulting number whose square plus the next greater square gives you the same result. We can use this to recursively solve the first few terms. (Alternatively, you could just generate the sequence out to the 9th term or so and solve a bunch of quadratics like that. Either way, we need the numbers whose squares we are considering.) You can then make the educated guess that those numbers must also follow a similar recursion to the given recursion, which they do (although you'll need 5 terms to figure out exactly what it is, maybe. At least 4 terms.) So, this gives you a recursion. However, it's not quite as nice a recursion -- I'm not sure if I know how to solve it off the top of my head. I'll keep thinking on it.", "Solution_2": "i solved by the characteristic equation that \r\n\r\nx_n = 1/2 ( (2+sqrt(3))^n + (2-sqrt(3))^n)\r\n\r\nwhat we can do with that... i have still to figure out. But it might be useful...", "Solution_3": "Usually when the problems asks you to show something about the terms of the sequence like this, the actual formula for each term usually isn't that helpful. Although, if you really feel like it, you will probably be able to bash the way there, though it would take heapsa algebra!", "Solution_4": "One thing that might help is if we knew exactly which two squares they are. After trying the first five numbers, here is my conjecture: an is the sum of the squares of [hide](xn+1 - xn - 1) / 2 and (xn+1 - xn + 1) / 2[/hide]. Okay, now it's up to the rest of you to prove it.", "Solution_5": "Ravi's comment also explains why the squares would obey a similar recursive relation ... I didn't think to look for them in the sequence itself.\r\n\r\nNukular -- I don't think it works. There isn't any nice way to manipulate the resulting expression to get rid of the :rt3:'s, as far as I can tell.", "Solution_6": "My guess is that Nukular's closed form could be used to prove my conjecture in a brute-force manner. That doesn't mean it's the most elegant solution.\r\n\r\nThis problem reminds me of some identities involving Fibonacci numbers. For example, F2n+1 = Fn+12 + Fn2. More generally, Fn+k = Fk Fn+1 + Fk-1 Fn. There are several ways to prove these identities. One pedestrian method is to prove (say the second one) by induction. A slicker method is to use 2x2 matrices.", "Solution_7": "[quote=\"Ravi B\"]My guess is that Nukular's closed form could be used to prove my conjecture in a brute-force manner. That doesn't mean it's the most elegant solution.\n\nThis problem reminds me of some identities involving Fibonacci numbers. For example, F2n+1 = Fn+12 + Fn2. More generally, Fn+k = Fk Fn+1 + Fk-1 Fn. There are several ways to prove these identities. One pedestrian method is to prove (say the second one) by induction. A slicker method is to use 2x2 matrices.[/quote]\r\n\r\nI learned the second identity through combinatorics. So I guess that makes a third method.", "Solution_8": "You can also prove it with path-travelling.", "Solution_9": "hm. could you please post the proofs of these last two methods? I think I can figure out the first two...", "Solution_10": "Imagine walking on a strip like the one shown in the attachment. There are Fn ways to reach the nth vertex from the left. Now consider a path to the (n+k)th vertex -- if it goes through the kth vertex, there are Fk*Fn + 1 total routes, Fk to the kth vertex and Fn + 1 from there to the (n + k)th vertex. If it doesn't go through the kth vertex, it has to go to the (k - 1)th vertex, then go directly horizontal to the (k + 1)th vertex. Thus, there are Fk - 1*Fn total routes -- Fk - 1 to the (k - 1)th and Fn from the (k + 1)th to the (n + k)th. Thus, we add and get the identity we wanted.", "Solution_11": "That's another nice combinatorial way to do it.... the way I had learned it was to consider fibonacci words* of length m+n-2.\r\n\r\nI will be using the definitions that fibonacci numbers are F[n] with F[0] = 0, F[1] = 1 and F[n] = F[n-1] + F[n-2]. **\r\n\r\nFurthermore, i will let f[n] be the number of fibonacci words of length n.\r\n\r\nIt is well known that with these definitions, f[n] = F[n+2]. This is not too difficult to prove.\r\n\r\nSuppose we want to prove \r\n\r\n Fn+k = Fk Fn+1 + Fk-1 Fn. \r\n\r\nThen we can also prove, equivalently:\r\n\r\n\r\nfn+k-2 = fk-2 fn-1 + fk-3 fn-2. \r\n\r\nAnd that is exactly what we're going to do. \r\n\r\nA fibonacci word of length n+k-2 can be broken up into two parts. The first part has length k-1, and the second part has length n-1. Both parts must be themselves fibonacci words. The first word ends in either 0 or 1. If it ends in 0, there are fk-2 such words, and we can pick any n-1 sized word for the second part without violating restrictions. Therefore there are fk-2 fn-1 n+k-2 sized words of this type. If the first part ends in a 1, this one must be surrounded by two 0's. Therefore, there are only fk-3 ways to pick the first part, and fn-2 ways to pick the second part, making \r\n fk-3 fn-2 words of this type and\r\n\r\nfk-2 fn-1 + fk-3 fn-2 words in total. QED\r\n\r\n\r\n\r\n\r\n* A fibonnaci word is a binary sequence with no two consecutive 1's. \r\n\r\n** Is the actual convention F[0] = 0 or F[0] = 1 ? I always mix these up. However, the proof does not differ with the secondary definition. The only difference is that f[n] = F[n+1] instead of F[n+2].", "Solution_12": "Okay.. can someone help with my problem? I would really need a full solution! :)\r\nthank you very much:)", "Solution_13": "What do you need it for?", "Solution_14": "Well I have been struggling over with this problem, and yet I haven't found any solution! This problem is turning into an obsesion! :)", "Solution_15": "My conjecture boils down to proving x2n+1 = (xn+1 - xn)2 + 1. That doesn't look too hard to prove, does it? You could always use Nukular's closed form to prove it if you run out of other ideas.\r\n\r\nP.S. Yes, Nukular, the usual convention for Fibonacci numbers is F0 = 0 and F1 = 1. As a mnemonic, you can remember that the Fibonacci sequence starts off as the identity function.", "Solution_16": "I noticed that this recursive sequence (with different starting values) appears in Problem 8 of this year's MathCamp quiz: [url]http://www.mathcamp.org/quiz.php[/url]. So perhaps we should not discuss the problem further, at least for now?", "Solution_17": "[quote=\"Ravi B\"]My conjecture boils down to proving x2n+1 = (xn+1 - xn)2 + 1. That doesn't look too hard to prove, does it? You could always use Nukular's closed form to prove it if you run out of other ideas.\n\nP.S. Yes, Nukular, the usual convention for Fibonacci numbers is F0 = 0 and F1 = 1. As a mnemonic, you can remember that the Fibonacci sequence starts off as the identity function.[/quote]\r\n\r\nahh very good thats a nice way to remember it... i think ill just write a quick function in C++ to find fibonacci's... that'll solidify it a bit more." } { "Tag": [ "MIT", "college", "Stanford", "Columbia", "Princeton", "calculus", "probability" ], "Problem": "Hey I was wondering if anyone could help me compare some colleges. I would possibly like to double major in math and engineering. Anyway, right now, I am considering:\r\n\r\nMIT\r\nStanford\r\nWash U\r\nNorthwestern\r\nIllinois (urbana-champaign)\r\nCarnegie Mellon\r\nUniversity of Penn\r\nMizzou\r\n\r\nEspecially wondering about Mizzou because I have seen little ranking them near these other schools engineering-wise, and nothing in math.", "Solution_1": "What type of engineering? Some other schools for your list:\r\n\r\nBerkely\r\nCaltech\r\nCornell\r\nColumbia\r\nJohns Hopkins\r\nGeorgia Tech", "Solution_2": "JRav, I am definitely not adding more schools to my list. Anyway, I've actually considered most of those schools at one point and even visited Johns Hopkins.\r\n\r\nRight now I am thinking possibly electrical/computer engineering, but I'm not completely sure of that.", "Solution_3": "I don't think UPenn is good at engineering.And Northwestern seems a bad choice when considering math and engineering....\r\n\r\nMIT\r\nStanford\r\nIllinois (urbana-champaign)\r\nCarnegie Mellon\r\n\r\nThese 4 are great especially the former two. :D", "Solution_4": "UPenn is known for its engineering department in that it is combined with its science department (at least thats what my friend said.) It's still probably good.", "Solution_5": "galoisj, remember everything here is relative. Penn and Northwestern aren't bad schools, but from what I've found they just don't quite measure up to MIT and Stanford in engineering. Anyway, when I visited Northwestern their engineering program sounded very interesting at the least. They combine the math into their freshmen engineering classes. But I'm not sure that that is such a good idea. But I'm still wondering if you have any more particular reasons for cutting Penn and Northwestern and for keeping Carnegie Mellon.\r\n\r\nAnyway, in particular I was wondering where University of Missouri falls in comparison. anywhere close?", "Solution_6": "Personally, I think that if you want to double major in math (as opposed to applied math) alongside EE, integrating math into the engineering education might not be a good thing -- the math they'd end up teaching would be very heavily geared towards the engineering contents, and might not give you as rich or though provoking or abstract an experience as you probably want from an undergraduate math degree. Of course, they probably only integrate calc and lin.alg, so you should still be fine for higher courses. But even lin.alg loses a lot of flavor when viewed in a heavily applied way...\r\n\r\n\r\nPenn is best known for its Wharton School. If you're interested in doing engineering at an Ivy-League school, the better considerations would be Cornell, Princeton, Columbia, and perhaps Dartmouth.", "Solution_7": "Zuton force, exactly what I think too. If the math is more integrated, then it might be more geared toward applications and less abstract/theoretical. I remember that Northwestern integrates differential equations and linear algebra into \"Engineering Analysis\" (I believe calculus is supposed to be freshman course). Anyway, I suppose I'd rather just be required to take those subjects as math courses rather than as part of an engineering class.\r\n\r\nAnyway, I really couldn't care if I end up at an Ivy league school. Right now I think that I'm just going to go with the list galoisj laid out, because that is basically what I found from looking at various rankings.\r\n\r\nDo you have any ideas of things I should look for in math/engineering curricula? Also, what exactly is applied math and what is the difference between applied and regular math? Somehow I think I could end up doing some of all three...", "Solution_8": "Applied math is statistics, probability, game theory, economics: math that has real-world applications. In pure math, you study things like analysis, abstract algebra, topology, differential geometry, number theory... Many of these things have important applications in physics and engineering but you typically study it for the theoretical applications, not the physical ones.", "Solution_9": "probability is closer to being pure math than applied math, especially when you learn that an event is really a 'measurable set' and a random variable is really a 'measurable function'. it's been discussed here before, but calling 'applied math' math with real-world applications is somewhat dangerous because even the purest math tends to have applications somewhere.\r\n\r\nthat being said, i don't know a truly satisfactory definition of pure or applied math, other than that i (mostly) know it when i see it, and it's really not that important anyways.", "Solution_10": "I think I have to say that I'm interested in all of it. I think I'll probably end up doing more of the theoretical, though, with possibly some applied as electives or something. Anyway, that's not very important now.\r\n\r\nThanks for your help everyone! Anyway, the last question I have that is actually somewhat important now is how does University of Missouri compare to these others?", "Solution_11": "Look up their department website. If the department looks good to you, then it will be good for you. If not, then it won't be.", "Solution_12": "oh come on, this is just ridiculous. if you don't have anything of any value to post JRav, then just don't post." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Let us consider a 200-digit number that repeats itself every 10 digits. Thus we have 9876543210...3210. Let us pick every third digit from this number to form a new number. Repeat until only 2 digits are left. What is this 2-digit number?", "Solution_1": ":? [hide]98[/hide]", "Solution_2": "Why would it be 98?\r\nI don't get it.[/hide]", "Solution_3": "I just wanted to say that I figured out why it's 98.\r\n\r\nIn order to solve this, you take the third digit of the third digit of the third digit out the number created by taking the third digit out of the number. Thus, you need to find the 81st and 162nd digit of the number. \r\n\r\n[u]Digits[/u]\r\n\r\n1st: 9\r\n11th:9\r\n21st: 9\r\nThus, the 81st digit is nine. Following the same principle, we find that the 162nd digit is 8.\r\n\r\nSo, the answer is 98.", "Solution_4": "[hide]98[/hide]", "Solution_5": "Is this a Mathcounts problem?", "Solution_6": "Yes it is, I've seen the problem on a test before", "Solution_7": "Cool, I got one right for once!! :P", "Solution_8": "don't look at this till you've tried 2 solve the problem\r\n\r\n[hide]98 because you would have to find out what exponent of three would divide 200 into two. since 81 is the closest exponent that doesn't go over 100, we would have to find the 81st number and the 162nd. seems like its 98 (unless im wrong)[/hide]" } { "Tag": [], "Problem": "If n is a positive integer, what is the maximum possible number of zeroes at the right end of the integer expression $ 1^n\\plus{}2^n\\plus{}3^n\\plus{}4^n$?", "Solution_1": "Is the answer [hide=\"answer\"]$ 2$[/hide]?\r\n\r\nI haven't tried to find a solution yet.", "Solution_2": "bump... no (proven) solution yet." } { "Tag": [ "function", "algebra", "functional equation", "algebra unsolved" ], "Problem": "Determine all functions $f: \\mathbb R \\to \\mathbb R$ satisfying\n\\[ f(x^2 \\plus{} f(y)) \\equal{} y \\plus{} f^2(x), \\quad \\forall x,y \\in \\mathbb R.\\]", "Solution_1": "$ R$ means $ \\mathbb{R^{*}}$ ?", "Solution_2": "[quote=\"reason\"]$ R$ means $ \\mathbb{R^{*}}$ ?[/quote]\r\n\r\n$ R$ means the set of real number (including 0)\r\nAnd finally, I have solved it by myself :lol:.", "Solution_3": "[quote=\"ll931110\"]I think this functional equation is not difficult, but I haven't solved it yet :mad: \n\n[i]Determine all functions $ f: R \\rightarrow R$ satisfying\n$ f(x^2 \\plus{} f(y)) \\equal{} y \\plus{} f^2(x)$[/i][/quote]\r\n\r\nLet $ P(x,y)$ be the assertion $ f(x^2\\plus{}f(y))\\equal{}y\\plus{}f(x)^2$\r\nLet $ f(0)\\equal{}a$\r\n\r\n$ P(0,x)$ $ \\implies$ $ f(f(x))\\equal{}x\\plus{}a^2$ and so $ f(x)$ is bijective\r\n\r\nLet then $ u$ such that $ f(u)\\equal{}0$\r\n$ P(u,u)$ $ \\implies$ $ f(u^2)\\equal{}u$ and so $ f(f(u^2))\\equal{}0$ But $ P(0,u^2)$ $ \\implies$ $ f(f(u^2))\\equal{}u^2\\plus{}a^2$ so $ u^2\\plus{}a^2\\equal{}0$ and $ u\\equal{}a\\equal{}0$\r\n\r\n$ P(x,0)$ $ \\implies$ $ f(x^2)\\equal{}f(x)^2$ and so $ P(x,y)$ may be rewritten as $ Q(x,y)$ : $ f(x^2\\plus{}f(y))\\equal{}y\\plus{}f(x^2)$\r\n\r\n$ Q(x,f(y))$ $ \\implies$ $ f(x^2\\plus{}y)\\equal{}f(y)\\plus{}f(x^2)$ and so $ f(x\\plus{}y)\\equal{}f(x)\\plus{}f(y)$ $ \\forall x\\ge 0,\\forall y$\r\n\r\nIt's then immediate to find $ f(\\minus{}x)\\equal{}\\minus{}f(x)$ and so $ f(x\\plus{}y)\\equal{}f(x)\\plus{}f(y)$ $ \\forall x,y$\r\nAnd since $ f(x^2)\\equal{}f(x)^2$, we get that $ f(x)\\ge 0$ $ \\forall x\\ge 0$ \r\n\r\nSo solutions are solutions of Cauchy's equation and have a lower bound on $ \\mathbb R^\\plus{}$. So $ f(x)$ is continuous and is $ cx$\r\n\r\nPlugging back in the original equation, we get $ c\\equal{}1$ and the unique solution $ \\boxed{f(x)\\equal{}x}$ $ \\forall x$", "Solution_4": "there is my solution:\r\n\r\n\r\n$ P(x,y): f(x^{2} + f(y)) = y + (f(x))^{2}$\r\n\r\n$ P(0,y)\\Rightarrow f(f(y)) = y + (f(0))^{2}\\Rightarrow$ $ f$ is bijective.\r\n\r\nwe have $ f(x^{2} + f(y)) = f(( - x)^{2} + f(y)) \\Rightarrow (f(x))^{2} = (f( - x))^{2}$\r\n \r\nsince f is injective $ \\Rightarrow$ $ f(x) = - f( - x)$ ($ f$ is odd).\r\n\r\nwe have $ f(0) = 0$ so $ f(f(y)) = y$ and $ f(x^{2}) = (f(x))^{2} > 0$\r\n\r\nso $ f(\\mathbb{R}) > 0$ $ \\forall x\\in\\mathbb{R^{*}}$.\r\n\r\nLet $ u = x^{2}$ and $ v = f(y)$ so $ f(u + v) = f(u) + f(v)$.\r\n\r\n$ \\Rightarrow$ $ f(x) = x$ $ \\forall x\\in{\\mathbb{Q + }}$.\r\n\r\nsince $ f(x) > 0$ we have $ f(u + v) = f(u) + f(v)\\ge f(u)$ $ \\Rightarrow$ $ f$ is increasing in $ \\mathbb{R + }$. so since $ f$ is odd we conclude finally that: $ f(x) = x$ $ \\forall x\\in\\mathbb{R}$. :)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "There are 1997 pieces of medicine. Three bottles $A, B, C$ can contain at most 1997, 97, 19 pieces of medicine respectively. At first, all 1997 pieces are placed in bottle $A$, and the three bottles are closed. Each piece of medicine can be split into 100 part. When a bottle is opened, all pieces of medicine in that bottle lose a part each. A man wishes to consume all the medicine. However, he can only open each of the bottles at most once each day, consume one piece of medicine, move some pieces between the bottles, and close them. At least how many parts will be lost by the time he finishes consuming all the medicine?", "Solution_1": "I just think every time we should move medicines to smaller bottles as long as we can .", "Solution_2": "i think the answer is35853.", "Solution_3": "you can put it in to a 3D Grid table\nthen you can solve it easily.\nsorry my english isn't very good.", "Solution_4": "Any offcial solution?" } { "Tag": [ "ratio" ], "Problem": "An 8-foot piece of tin is to be cut into three pieces in the ratio of 1:2:3. How long will the middle sized piece be if the cut is made by a pair of tin snips. Give answer in feet and inches.", "Solution_1": "The middle piece is $ \\frac{2}{1\\plus{}2\\plus{}3} \\equal{} \\frac{2}{6} \\equal{} \\frac{1}{3}$ of the entire piece.\r\n$ \\frac{1}{3}$ of 8 is $ 2 \\frac{2}{3}$ feet.\r\nSince there are 12 inches in a foot, $ \\frac{2}{3}$ of a foot is 8 inches.\r\nThat means the middle sized piece will be [b]2 feet and 8 inches[/b] long." } { "Tag": [ "inequalities" ], "Problem": "\u60f3\u95ee\u95ee\u4e0b\u5217\u51e0\u4e2a\u4e0d\u7b49\u5f0f\u7684\u4e2d\u6587\u540d\u662f\u4ec0\u4e48\uff0c\u8bf7\u5404\u4f4d\u9ad8\u624b\u6307\u6559\uff0c\u54ea\u6015\u77e5\u9053\u4e00\u70b9\u76f8\u5173\u4fe1\u606f\u6216\u77e5\u8bc6\uff0c\u6700\u597d\u662f\u5173\u4e8e\u5728\u8054\u8d5b\u4e2d\u5982\u4f55\u8fd0\u7528\u7684\u77e5\u8bc6\uff1a\r\n(I want to know the Chinese names of the following inequalities. Thanks for translating, or giving any hint or knowledge you know about them, particularly welcoming things about their application in USMO)\r\n\r\nMajorization Inequality\r\n\r\nMuirhead's Theorem\r\n\r\nSchur\r\n\r\nMaclaurin's Symmetric Mean Inequality\r\n\r\nPower Mean Inequality\r\n\r\nBernoulli's Inequality", "Solution_1": "I can't help you much but I think I know that:\r\n\r\nSchur inequality = \u8212\u723e\u4e0d\u7b49\u5f0f\r\n\r\nI usually use it when facing $a^{3}+b^{3}+c^{3}+3abc \\geq \\sum_{sym}a^{2}b$ (Which is when $r=1$), it is useful that it use with Muirhead.\r\n\r\nBernoulli's Inequality = \u4f2f\u52aa\u5229\u4e0d\u7b49\u5f0f", "Solution_2": "I don't know Chinese names, but I think that [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=7942]this thread[/url] can be useful.", "Solution_3": "1 http://www.google.com\r\n scholar.google.com\r\n2 http://www.mathworld.com\r\n3 en.wikipedia.org\r\n http://encyclopedia.thefreedictionary.com \u56fd\u5185\u955c\u50cf\u7ad9\r\n4 http://www.answers.com", "Solution_4": "[quote=\"billid\"]\u60f3\u95ee\u95ee\u4e0b\u5217\u51e0\u4e2a\u4e0d\u7b49\u5f0f\u7684\u4e2d\u6587\u540d\u662f\u4ec0\u4e48\uff0c\u8bf7\u5404\u4f4d\u9ad8\u624b\u6307\u6559\uff0c\u54ea\u6015\u77e5\u9053\u4e00\u70b9\u76f8\u5173\u4fe1\u606f\u6216\u77e5\u8bc6\uff0c\u6700\u597d\u662f\u5173\u4e8e\u5728\u8054\u8d5b\u4e2d\u5982\u4f55\u8fd0\u7528\u7684\u77e5\u8bc6\uff1a\n(I want to know the Chinese names of the following inequalities. Thanks for translating, or giving any hint or knowledge you know about them, particularly welcoming things about their application in USMO)\n\nMajorization Inequality\n\nMuirhead's Theorem\n\nSchur\n\nMaclaurin's Symmetric Mean Inequality\n\nPower Mean Inequality\n\nBernoulli's Inequality[/quote]\r\n\r\nMajorization Inequality - \u63a7\u5236\u4e0d\u7b49\u5f0f(\u53c8\u6216\u8005\u53eb\u4f18\u8d85\u4e0d\u7b49\u5f0f, \u7b49..)\r\n\r\nMuirhead's Theorem - \u7c73\u5c14\u9ed1\u5fb7\u4e0d\u7b49\u5f0f(\u9648\u8ba1\u7684\u4e66\u662f\u8fd9\u6837\u8bd1\u7684)\r\n\r\nSchur - \u8212\u5c14\u4e0d\u7b49\u5f0f\r\n\r\nMaclaurin's Symmetric Mean Inequality - \u9ea6\u514b\u52b3\u6797(\u5bf9\u79f0)\u4e0d\u7b49\u5f0f(\u6211\u731c)\r\n\r\nPower Mean Inequality - \u5e42\u5e73\u5747\u503c\u4e0d\u7b49\u5f0f\r\n\r\nBernoulli's Inequality - \u8d1d\u52aa\u5229\u4e0d\u7b49\u5f0f" } { "Tag": [ "trigonometry", "function", "LaTeX", "calculus", "calculus computations" ], "Problem": "So I have found the equation of the tangent lines to the graph of $ f(x)\\equal{}sinx$ at $ x\\equal{}0$ to be $ y\\equal{}x$. Additionally I have found that the equation of the tangent lines to the graph of $ f(x)\\equal{}sinx$ at $ x\\equal{}pi/3$ to be $ y\\equal{}\\frac{1}{2}x\\plus{}\\frac{3*sqrt(3)\\minus{}pi}{6}$. Now however I am struggling to understand what \u201cUse the tangent line at $ x\\equal{}0$ to approximate $ sin(pi/6)$\" and \u201cUse the tangent line at $ x\\equal{}pi/3$ to approximate $ sin(pi/6)$\" means. Any ideas?", "Solution_1": "It's a local approximation. Use the equation $ f'(x)\\approx f(a)\\plus{}f'(a)(x\\minus{}a)$, which locally approximates a function. So in your first example, $ x\\equal{}\\frac{\\pi}{6}$ and $ a\\equal{}0$, so we get $ \\sin\\left(\\frac{\\pi}{6}\\right)\\approx 0\\plus{}\\frac{\\pi}{6}\\equal{}\\frac{\\pi}{6}$. This is close to the actual value of $ \\frac{1}{2}$, within about .02-.03. By adding more terms, you can improve the approximation.", "Solution_2": "I am a little bit confused. I don't see where in $ \\sin\\left(\\frac {\\pi}{6}\\right)\\approx 0 \\plus{} \\frac {\\pi}{6} \\equal{} \\frac {\\pi}{6}$ you are solving for anything and coming up with $ 1/2$.", "Solution_3": "I don't understand your last post. I just plugged into the formula I provided. $ f(a)\\equal{}\\sin(0)\\equal{}0$, and $ f'(a)(x\\minus{}a)\\equal{}\\cos(0)\\left(\\frac{\\pi}{6}\\minus{}0\\right)\\equal{}\\frac{\\pi}{6}$.\r\n\r\nIf you want to get more accurate, you can add on the term $ \\minus{}\\frac{f^{(3)}(a)(x\\minus{}a)^{3}}{3!}$. This term comes from the [url=http://en.wikipedia.org/wiki/Taylor_series]Taylor series expansion[/url]. If you plug in, you get that the extra term $ \\approx .0239$, so the total estimate is $ .499$, which is pretty accurate.", "Solution_4": "Telling someone who doesn't understand tangent line approximations about Taylor series makes it less likely that they will understand your explanation, not more.\r\n\r\nA tangent line is a local approximation to a graph: near the point of tangency, we expect the function values of the graph to be very close to the values of the tangent line. Thus, you're being asked to estimate the value of a function by the value of its tangent lines. Since you already have the equation of the tangent lines, all you have to do is plug in the appropriate value of $ x$ in order to get your estimate for the function value. (Your textbook doubtless has lots of material on this, including exposition and solved examples. You would probably do well to read it and your class notes thoroughly in the future when you don't understand how to do something.)\r\n\r\nLatex suggestions: \\pi, \\sin, \\cos, \\sqrt{} all look better than without the slash (though be careful -- \\sinx doesn't mean anything, it needs to be \"\\sin x\"). Also, \\cdot makes a nicer multiplication symbol than *.", "Solution_5": "Hopefully JBL cleared that up for you a bit more. In your example, we see that for $ \\theta$ small, $ \\sin\\theta\\approx\\theta$. This is actually a really important approximation in physics since often times, it's a lot easier to solve a differential equation (say for some simple harmonic motion) in terms of $ \\theta$ than $ \\sin\\theta$.\r\n\r\nFor example, suppose we want to find the period of a pendulum. If $ \\theta$ is small, we can just write the equation $ \\ddot{\\theta}\\plus{}\\alpha\\theta\\equal{}0$, where $ \\alpha$ is a constant. This gives a pretty good approximation." } { "Tag": [ "calculus", "integration", "function", "inequalities", "algebra", "domain", "analytic geometry" ], "Problem": "I am at the section on Jacobians (12.9) and I just need a little clarification.\r\n\r\nHow do I know if an integral is a good candidate to use this Transformation of Variables?\r\nAnd more importantly: How do I know what conversion will help?\r\n\r\nIt uses the whole polar/cylindrical/spherical conversion as an example but that is relatively easy to see. (Circular boundaries, circular formulae, etc.)", "Solution_1": "Unfortunately there is no general rule. One indication is that you have the same expression in the function as in the inequalities describing the domain of integration. Then you may want to take it as a new variable. But sometimes this doesn't work. Another useful idea is to try to make the domain of integration as simple as possible (preferrably, a parallelepiped) but there are exceptions to this \"rule\" too. Of course, you may also try the same idea as in the one variable case: if you see a complicated expression, take it as a new variable. But this may result in a very unpleasant change in the domain of integration (not to say about ugly Jacobian). Most examples in the textbooks that are about something else than polar coordinates are artificially constructed using one of three above ideas. If you have some particular integrals in mind and wonder how to guess the right substitution for them, we can discuss that but, as to a general rule that never fails, I know of none.", "Solution_2": "I had this horrible feeling that that was going to be the answer.....oy." } { "Tag": [ "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "I need some help with this one, too.\r\n\r\nLet U, V, W be subspaces of an F-space. Using the fact that dim(W1) + dim(W2)=dim(W1+W2) + dim(W1 intersect W2), show that dim[(U+V) intersect W] + dim(U intersect V) = dim[(V+W) intersect U] + dim(V intersect W).", "Solution_1": "[quote=\"OwlMan\"]I need some help with this one, too.\n\nLet U, V, W be subspaces of an F-space. Using the fact that dim(W1) + dim(W2)=dim(W1+W2) + dim(W1 intersect W2), show that dim[(U+V) intersect W] + dim(U intersect V) = dim[(V+W) intersect U] + dim(V intersect W).[/quote]\r\n\r\n\r\n$\\dim{((U+V) \\cap W)} + \\dim{(U \\cap V)} = \\dim{((V+W) \\cap U)} + \\dim{(V \\cap W)}$\r\n\r\nConsider the left-hand-side. This is nothing else then: \r\n\\[ \\dim{(U+V)} + \\dim{(W)} - \\dim{(U+V+W)} + \\dim{(U \\cap V)} \\\\ = \\dim{(U)} + \\dim{(V)} - \\dim{(U \\cap V)} + \\dim{(W)} - \\dim{(U+V+W)} + \\dim{(U \\cap V)} \\\\ = \\dim{(U)} + \\dim{(V)} + \\dim{(W)} - \\dim{(U+V+W)} \\]\r\n\r\nDo the same thing for the right-hand-side and you will get the same thing.", "Solution_2": "Thanks, Yannick! I appreciate it." } { "Tag": [ "quadratics" ], "Problem": "Here's some math links with problems and information for multiple levels:\r\nhttp://mathproblems.info/\r\nhttp://mathworld.wolfram.com/\r\nhttp://www.math.com/ :rotfl:", "Solution_1": "$ x \\equal{} sqrt{25}$\r\n$ therefore x \\equal{} 5$", "Solution_2": "Actually:\r\n\r\n$ x\\equal{}\\sqrt{25}$\r\nThus, $ x\\equal{}\\pm 5$", "Solution_3": "If $ x\\equal{}\\sqrt{25}$, isn't x just 5 (principal root)?", "Solution_4": "Yes, I was sorta assuming that we're solving a quadratic or something, so there's two possible values of $ x$...", "Solution_5": "Haha, ok...\r\n\r\n(I don't see a problem) :D .", "Solution_6": "this is going entirely off topic.\r\nbut here's a link:\r\nhttp://www.nerdparadise.com\r\nnot bad on the mathy stuff...." } { "Tag": [ "topology", "real analysis", "real analysis unsolved" ], "Problem": "let $ X$ be a compact metric space, $ f: X\\to X$ is a isometry, ie. $ \\forall x, y \\in X d(x,y)\\equal{}d(f(x),f(y))$. Prove that $ f$ is a one-to-one and onto (bijection )\r\n\r\nI know how to prove $ f$ is one - to -one but I am getting stuck how to prove $ f$ is onto. Any idea will be highly appreciated. Thanks", "Solution_1": "Hint:\r\nFor each $ x$ consider the sequence $ f_n (x)\\equal{}f^{n} (x)$ then pass to the convergent subsequence. :wink:", "Solution_2": "I didn't get what leshik means. Anyway, if some point $ y\\notin f(X)$ (a closed set!), then so does its neighborhood (of radius $ r$, say). Now take the maximal $ r$-net $ A$ (a set where any two points are at distance at least $ r$) of the compact. It exists and it's finite due to compactness. Its image $ f(A)$ is also an $ r$-net, moreover, $ f(A)\\cup \\{y\\}$ is an $ r$-net, contradicting the maximality.", "Solution_3": "Hi zhoraster,\r\nwhat lesha has said is the following:\r\nconsider the sequence $ f_n (x) \\equal{} f((..(x)..))$ ( $ n$ times ). Because of the compactness $ X$ we can pass to the convergent subsequence $ f^{n_k} (x)$. Now from the condition of the problem we have that $ d(f^{n_k} (x), f^{n_l} (x)) \\equal{} d(x, f^{|n_k \\minus{} n_l|} (x))$, hence we can find the subsequence $ f^{l_k}$ of $ f_n (x)$ which is convergent to $ x$. Now choose the convergent subsequence of the sequence $ f^{l_k \\minus{} 1}(x)$ (let $ a$ be its limit point ) and by continuity $ f$ we get $ f(a) \\equal{} x$. That's all. :)", "Solution_4": "[quote=\"leshik\"], hence we can find the subsequence $ f^{l_k}$ of $ f_n (x)$ which is convergent to $ x$. [/quote]\r\n\r\nWhy does subsequence converge to x? :blush:", "Solution_5": "Because the sequence $ f^{n_k}$ is fundamental, thus $ d(f^{n_k}, f^{n_l})\\to 0$ when $ n_k, n_l \\to\\infty$, is everything clear now?", "Solution_6": "[quote=\"leshik\"]Hi zhoraster,\nwhat lesha has saisd is the following:[/quote]\r\n\r\nOh yes, I had solved it some 10 years ago similarly :) Now too old to invent such beauties." } { "Tag": [ "summer program", "Mathcamp", "Ross Mathematics Program", "function", "geometry", "videos", "PROMYS" ], "Problem": "Hello\r\n\r\nI'm kind of new to the American concept of math camps (they're awesome though :D ) and I wanted to ask a dew questions:\r\n\r\n1. I am also currently applying to Ross, so if i get accepted into both (assuming that i ever get my application for Ross typed up, I have the proofs^^), what do I do? Do I just tell one of the: Hey I'm sorry, but I'd rather go to your competitor?\r\n\r\n2. For the application, when it says deadline 4/25/08 what is it referring to? when it should be by them or when we send them off? (the later system is used in Germany)\r\n\r\n3. Which one is more fun/interesting?\r\n\r\n4. How much money should I bring for either (preferably an amount per week, since one is 8 and the other 6 weeks)\r\n\r\n5. What amount of correct solutions gets one in? (they don't seem super hard, but i didn't look at them very intently yet... )\r\n\r\n6. if you do get in, when do they ask you for schedule-choices?\r\n\r\n\r\n\r\n\r\nIf there are any mistakes in grammar/spelling, please tell me :blush:", "Solution_1": "1. Yes, this is fine. You can do the same thing to colleges, too.\r\n\r\n2. Unless otherwise specified, you should interpret it as being the deadline for receiving the application. The first case is referred to as \"postmark deadline.\"\r\n\r\n3. That depends on what you think is interesting. Clearly different things work for different people, although I've heard a lot of good things about MathCamp and Ross is quite difficult. (The fact that I've gone to neither is your assurance that I'm being vaguely objective.)\r\n\r\n4. As far as I know, the cost of both programs cover room and board, so you only need spending money for outings and snacks. Bring whatever you feel is appropriate, I guess. (And some extra money for things you forgot to bring with you - toothpaste, etc.)\r\n\r\n5. Acceptance is (probably) not directly a function of how many problems you solve correctly. I imagine your style is also important, as well as the rest of the application and how many spaces are available. So it's hard to say.", "Solution_2": "Hello,\r\n\r\n1. Yup, you can apply, get in, and then decide to do something else with your summer. (Otherwise it's not really fair: you apply and don't get in and then you have nothing to do!)\r\n\r\n2. We'd like to have applications *received* by then. If you think you need a little bit more time, drop us a note. You can also apply online, in which case you don't have to wait for postal mail.\r\n\r\n3. Well, you're posting in the Mathcamp forum, so you'll get a biased answer! You can find out more about Mathcamp from the Math Jam transcripts, which are linked to from another post in this forum.\r\n\r\n4. It depends --- if you get in, we'll send a suggestion.\r\n\r\n5. As far as correct solutions, that also depends on a lot of different factors. Certainly the quiz is a major part of your admission, but other factors also come into it.\r\n\r\n6. You decide at camp --- indeed, you can choose on the day-of and change your classes part-way through the week if you decide others are a better fit for you.\r\n\r\nGood luck!", "Solution_3": "I also have a couple questions:\r\n\r\n1. We have to pay for the flight, correct?\r\n\r\n2. When we are doing the quiz, can we go out of order?", "Solution_4": "[quote=\"Smartguy\"]I also have a couple questions:\n\n1. We have to pay for the flight, correct?\n\n2. When we are doing the quiz, can we go out of order?[/quote]\r\n\r\nYou do have to pay for the flight, although if your financial situation is quite dire we have included transportation in our financial aid in the past.\r\n\r\nAs to the quiz, you can work on the problems in any order that you like; it's a little bit easier for us to grade if you write them up in order.", "Solution_5": "Thanks for the responses...\r\n\r\nI think i am going to apply online, simply because i have no idea how long the US postal system would take to get my letter up there ^^\r\n\r\none more question: \r\nwhat about transportation from the (Seattle, right?) airport to the university?\r\nIs it provided and if not, how did past students get to the facilities?", "Solution_6": "Mathcamp staff will meet you at the airport and take you to campus.", "Solution_7": "I have (another) question about the quiz, in particular about number 2.\r\nIf we have proven the general rule, must we prove the specific case, since it is asked before the general rule?", "Solution_8": "[quote=\"e=mc2.kroll\"]I have (another) question about the quiz, in particular about number 2.\nIf we have proven the general rule, must we prove the specific case, since it is asked before the general rule?[/quote]\r\n\r\nNope, you can do the general case first and then go to the specific.", "Solution_9": "I'm kinda a huge procrastinator, and I was wondering whether the deadline of the 25th includes Saturday itself; like, can I submit my application online on Saturday or will it be considered late? <_<", "Solution_10": "[quote=\"Zeph\"]I'm kinda a huge procrastinator, and I was wondering whether the deadline of the 25th includes Saturday itself; like, can I submit my application online on Saturday or will it be considered late? <_<[/quote]\r\n\r\nSubmitting on Saturday is definitely fine. Good luck!", "Solution_11": "new deadline though! :D \r\n27th :]]", "Solution_12": "I went to Ross last year. Of course, as you probably already know, the camp's main focus is on number theory. Each day you are given a problem set with a few terminology problems, numerical problems, and proofs. The majority of the people didn't finish a set per day though and got pretty behind, but that's understandable. At the beginning, you are placed into a group with about two other first years, a junior counselor, and your counselor. Throughout the camp, your counselors give hints to help you with problems on your sets. Each morning there is a one hour lecture and Mondays, Wednesdays, and Fridays ( I believe) you have the problem session which lasts an hour, as well. \r\n\r\nMost everyone hung out in the lounge area till very late (3AM- 5AM sometimes all-nighters) usually doing problem sets. After the lectures and problem sessions you are basically free to do whatever you want. You can go to Market Place which is where most people ate (excellent restaurant food) and since you'll have a food card you don't have to pay. The counselors don't want you going off campus alone but... people still did occasionally, haha. \r\n\r\nAh, and yes... three times throughout the camp you are given a test with 20 problems total (10 numerical, 10 proofs) that you have to complete by the next morning and turn in to the professor. Most everyone pulls an all nighter on celebration nights... It isn't necessary to complete them, but... if you care about your reputation at camp, you should at least try.\r\n\r\nUnfortunately, bringing laptops is highly discouraged and there is no wireless internet in the dorms. You can walk to the library (not far at all) to use the internet. I brought about 60 dollars or so and was fine but then again... my parents had bought a lot of things for my dorm already. And there are the movie nights... where you watch videos having something to do with math. \r\n\r\nFor my Ross admission test, I believe I solved all of them but they weren't necessarily written *that* well. You don't have a schedule choice for Ross. \r\n\r\nAt Ross you have a tremendous amount of freedom, seriously. About 40-50 students are there... you would probably like it very much. :) Also, if you are looking to improve your proof skills, I highly recommend Ross.", "Solution_13": "[quote=\"Cyrazeno\"]I went to Ross last year. Of course, as you probably already know, the camp's main focus is on number theory. Each day you are given a problem set with a few terminology problems, numerical problems, and proofs. The majority of the people didn't finish a set per day though and got pretty behind, but that's understandable. At the beginning, you are placed into a group with about two other first years, a junior counselor, and your counselor. Throughout the camp, your counselors give hints to help you with problems on your sets. Each morning there is a one hour lecture and Mondays, Wednesdays, and Fridays ( I believe) you have the problem session which lasts an hour, as well. \n\nMost everyone hung out in the lounge area till very late (3AM- 5AM sometimes all-nighters) usually doing problem sets. After the lectures and problem sessions you are basically free to do whatever you want. You can go to Market Place which is where most people ate (excellent restaurant food) and since you'll have a food card you don't have to pay. The counselors don't want you going off campus alone but... people still did occasionally, haha. \n\nAh, and yes... three times throughout the camp you are given a test with 20 problems total (10 numerical, 10 proofs) that you have to complete by the next morning and turn in to the professor. Most everyone pulls an all nighter on celebration nights... It isn't necessary to complete them, but... if you care about your reputation at camp, you should at least try.\n\nUnfortunately, bringing laptops is highly discouraged and there is no wireless internet in the dorms. You can walk to the library (not far at all) to use the internet. I brought about 60 dollars or so and was fine but then again... my parents had bought a lot of things for my dorm already. And there are the movie nights... where you watch videos having something to do with math. \n\nFor my Ross admission test, I believe I solved all of them but they weren't necessarily written *that* well. You don't have a schedule choice for Ross. \n\nAt Ross you have a tremendous amount of freedom, seriously. About 40-50 students are there... you would probably like it very much. :) Also, if you are looking to improve your proof skills, I highly recommend Ross.[/quote]\r\n\r\nadvertising for Ross?\r\nRoss and PROMYS are pretty similar right? even in terms of their schedule?", "Solution_14": "I'm simply stating my opinion of the camp based on experience. Also, there was no need to quote my entire post since it's pretty obvious that you're talking to me. Anyways, yes, PROMYS and Ross are very similar in terms of the curriculum offered.", "Solution_15": "[quote=\"Cyrazeno\"]\nUnfortunately, bringing laptops is highly discouraged and there is no wireless internet in the dorms. \n\nYou don't have a schedule choice for Ross. \n\nAt Ross you have a tremendous amount of freedom, seriously.\n[/quote]\r\n\r\nMethinketh my definition of freedom and yours might differ? :)", "Solution_16": "Well, besides that. You are still able to walk (less than half a mile or so) to the library for internet. :) I guess that was the worst part about camp, though. Haha.", "Solution_17": "[quote=\"Cyrazeno\"]Most everyone hung out in the lounge area till very late (3AM- 5AM sometimes all-nighters) usually doing problem sets.[/quote]\r\n\r\nWow... last summer at MC, my average bedtime was about 3AM. But, usually, although I worked on math a lot during the day, at night I was just talking/playing games/watching videos that late. The one night I pulled an all-nighter (besides the all-nighter on the last day of camp which basically everyone does) it was because we started a game of RISK at about 4AM. Not recommended if one wants to go to sleep soon :P . Anyway, Ross certainly sounds more serious...\r\n\r\nRoss does sound like an awesome program, and I'm sure one knows number theory inside and out after eight (I think?) weeks. But I really love being able to choose my schedule (at least in part a result of being homeschooled), among other things, so I prefer MC freedom.", "Solution_18": "i am also apllying to ross (have all the proofs typed up since 2 moths ago, just need to get the app out... mathcamps online app is sooo much easier to complete quickly).\r\ni was thinking of kindof alternating in between them, since one seems very rigorous and the other seems to be very diverse) \r\nso it would be like \r\n9-mathcamp (just wanna find out what interests me a lot)\r\n10-ross\r\n11-mathcamp (junior-senior summer is very important as far as i understood, thus the shorter one)\r\n12-ross\r\n\r\nbut we'll see...\r\nalso the fact that mathcamp has standing invitation is really nice :-)", "Solution_19": "The question of which camp is more \"serious\" is an interesting one. I think it's true that the average Ross student spends more time working on problems than the average Mathcamp student (although Mathcampers spend more time in classes, for example). I don't think Ross gives more work, it's just that the variety of choices at camp tends to mean that Mathcamp students also spend time on things outside of math --- believe me, there are [i]plenty[/i] of problems to do at MC! One thing I've enjoyed as a teacher at Mathcamp is talking to campers after the summer, who are working on doing problems leftover from their time during the summer. I think you have to judge for yourself how important Ross' culture of working problems all night is important to you. Will you do enough math to satisfy yourself if there is less focus on constantly doing math? That's something you have to judge for yourself.\r\n\r\nAlternating programs is an interesting idea and I know people who've switched between the two. Do be wary of of strange contrasts, though. For example, if you take a lot of number theory at Mathcamp and then go to Ross, you may find it hard to find the right niche for yourself since you'll have seen a lot of the material but not all of it.", "Solution_20": "The freedom at Mathcamp definitely extends to the amount of time you spend on problems - it truly depends on what classes you decide to take and how serious you are about them. At Mathcamp 2008, I spent half of camp incredibly overworked, and half of camp focusing more on non-academic activities than on problem sets - a disparity that would not exist at Ross. I personally like the ability to choose how much math I want to do any particular week. If you want a more clearly-defined workload, you might prefer Ross. \r\n\r\nI would tend to think that Ross before Mathcamp would be better than Mathcamp before Ross. The set curriculum at Ross may be better than Mathcamp at providing a solid introduction to theoretical mathematics. But I don't think anyone can beat Mathcamp for those who are ready for advanced topics." } { "Tag": [ "trigonometry" ], "Problem": "Let $PT$ and $PB$ be two tangents to a circle $\\omega$, $AB$ the diameter through $B$, and $TH$ the perpendicular from $T$ to $AB$. Then prove that $AP$ bisects $TH$.", "Solution_1": "Let $R = OA = OB = OT$ and $\\alpha = \\measuredangle BPO = \\measuredangle OPT$ so that $\\measuredangle TOP = \\measuredangle POB = 90^\\circ-\\alpha$. Thus, $\\measuredangle HOT = 2 \\alpha$.\r\nLet $W$ be the midpoint of $TH$.\r\nWe want to prove that $2 \\cdot WH = TH$:\r\n$ \\, \\Longleftrightarrow \\, \\, 2 \\cdot \\frac{AH}{AB}\\cdot PB = TH$\r\n$ \\, \\Longleftrightarrow \\, \\, 2 \\cdot \\frac{R-HO}{2R}\\cdot PB = TH$\r\n$ \\, \\Longleftrightarrow \\, \\, 2 \\cdot \\frac{R-R \\cos \\left( 2 \\alpha \\right)}{2R}\\cdot R \\cot \\alpha = R \\sin \\left( 2 \\alpha \\right)$\r\n$ \\, \\Longleftrightarrow \\, \\, \\left( 1-\\cos \\left( 2 \\alpha \\right) \\right) \\cdot \\cot \\alpha = \\sin \\left( 2 \\alpha \\right)$\r\n$ \\, \\Longleftrightarrow \\, \\, 2 \\sin^{2}\\alpha \\cdot \\cot \\alpha = 2 \\sin \\alpha \\cos \\alpha$\r\n$ \\, \\Longleftrightarrow \\, \\, \\sin \\alpha \\cdot \\frac{\\cos \\alpha}{\\sin \\alpha}= \\cos \\alpha$,\r\nwhich is obviously true.", "Solution_2": "First: \r\n$X$ is the intersection of $AP$ and $HT$\r\n$Y$ the intersection of $AT$ and $BP$\r\n$O$ is the centre of the circle\r\n\r\nBy angle chase: $\\measuredangle POB=\\measuredangle YAB$ $\\Rightarrow$ $OP//AY$\r\n\r\n$AO=OB$, then $BP=PY$\r\n\r\n$AHT$ and $ABY$ are similars so: $\\frac{HX}{XT}=\\frac{BP}{PT}$ $\\Rightarrow$ $\\boxed{HX=XT}$ $\\mathbb{QED}$" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "hi,\r\n\r\ncompute $\\sum_{p=1}^{n}\\arctan(\\frac{1}{p^{2}+3p+3})$\r\n :blush: .\r\n\r\nthanks in advance.", "Solution_1": "Use the hint :) \r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=111391[/url]", "Solution_2": "i haven't seen your hint Ramanujan yet but i have found the \r\n\r\nsolution :lol: . \r\n\r\n$\\frac{1}{p^{2}+3p+3}= \\frac{(p+2)-(p+1)}{1-(p+2)\\times(-p-1)}$\r\n\r\n :P ." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "$ a,b,c>0$,$ (a\\plus{}b)(a\\plus{}c)(b\\plus{}c)\\equal{}1$ $ \\implies$ $ ab\\plus{}bc\\plus{}ac\\geq\\frac{3}{4}$", "Solution_1": "[quote=\"cnyd\"]$ a,b,c > 0$,$ (a \\plus{} b)(a \\plus{} c)(b \\plus{} c) \\equal{} 1$ $ \\implies$ $ ab \\plus{} bc \\plus{} ac\\geq\\frac {3}{4}$[/quote]\r\n$ x\\equal{}a\\plus{}b;y\\equal{}b\\plus{}c;z\\equal{}c\\plus{}a$ . We got $ xyz\\equal{}1$ and want to prove that\r\n\r\n$ ab \\plus{} bc \\plus{} ac\\equal{}\\sum{\\frac{1}{4}(x\\plus{}y\\minus{}z)(x\\plus{}z\\minus{}y)}\\geq\\frac {3}{4}$\r\n\r\nChange it to\r\n\r\n$ 2\\sum{xy}\\geq3\\plus{}\\sum{x^2}$\r\n\r\nIt's wrong!!! See [url=http://www.mathlinks.ro/viewtopic.php?t=50745]here[/url]", "Solution_2": "but in the this link $ abc \\equal{} 1$", "Solution_3": "See example 2 here:\r\n[url]http://www.math.ust.hk/excalibur/v14_n1.pdf[/url]\r\nThe sign is reversed, you don't need abc=1.", "Solution_4": "aah sorry,my document is wrongly :lol:", "Solution_5": "[quote=\"cnyd\"]$ a,b,c > 0$,$ (a \\plus{} b)(a \\plus{} c)(b \\plus{} c) \\equal{} 1$ $ \\implies$ $ ab \\plus{} bc \\plus{} ac\\geq\\frac {3}{4}$[/quote]\r\nYes, it's wrong: $ a \\equal{} b\\rightarrow\\frac {1}{\\sqrt [3]2}$ and $ c\\rightarrow0.$", "Solution_6": "I think $ ab \\plus{} bc \\plus{} ac\\leq \\frac {3}{4}$ \r\n\r\n$ (a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ac) \\equal{} 1 \\plus{} abc \\Longrightarrow$\r\n\r\n$ \\sqrt {3}{(ab \\plus{} bc \\plus{} ac)}^{\\frac {3}{2}} \\leq (a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ac) \\equal{} 1 \\plus{} abc \\leq 1 \\plus{} (\\frac {ab \\plus{} bc \\plus{} ac}{3})^{\\frac {3}{2}} \\Longrightarrow$\r\n\r\n$ ab \\plus{} bc \\plus{} ac\\leq \\frac {3}{4}$ :wink:" } { "Tag": [ "inequalities", "integration", "function", "calculus", "calculus computations" ], "Problem": "Suppose that $ f(x)$ be continuous ,positive function and $ f(x\\plus{}1)\\equal{}f(x)$ for all $ x \\in R$.Let$ \\alpha \\in R$\r\nProve inequality :\r\n $ \\int_{0}^{1}{\\frac{f(x)}{f(x\\plus{}\\alpha)}dx}\\geq 1$", "Solution_1": "we can immediately see that\r\n\r\n$ f\\left(0 \\right)\\equal{}f\\left(1 \\right)\\equal{}f\\left(2 \\right)\\equal{}f\\left(3 \\right)\\equal{}.........$\r\n\r\nhence f(x) cann be a constant function in which case, equality occurs in the given inequality" } { "Tag": [], "Problem": "Let $ m$ and $ n$ be two strictly positive integers. Prove that $ \\frac{GCD(m,n)}{n}\\binom{n}{m}$ is an integer.", "Solution_1": "this isn't so hard", "Solution_2": "Try this. :wink: \r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=98174[/url]" } { "Tag": [], "Problem": "Are given 2 boxes with the same gas . Box 1 with volume $ 3m^{3}$ and pressure 1 atmosphere , box 2 with volume $ 2m^{3}$ and pressure 2 atmosphere . If we join the boxes with a tube , what is the pressure in the boxes ?", "Solution_1": "What about the temperatures?\r\n\r\nAssuming them to be same for both boxes.\r\n\r\n$ P_1V_1 \\equal{} n_1RT .... \\text{ (for box 1)} \\\\ \\\\\r\nP_2V_2 \\equal{} n_2RT ....\\text{ (for box 2)} \\\\ \\\\\r\nP_f(V_1\\plus{}V_2) \\equal{} (n_1\\plus{}n_2)RT .... \\text{ (after opening the stop cock)} \\\\ \\\\\r\nP_f \\equal{}\\frac{7}{5} \\equal{} 1.4\\text{ atmosphere}$", "Solution_2": "how can you assume the temperatures to be equal.it is not specified", "Solution_3": "[quote=\"aadil\"]how can you assume the temperatures to be equal.[/quote]\r\n\r\nI took it for granted that temperatures are equal. There is no way you can find the temperatures and IMO the answer would contain terms $ T_1$ and $ T_2$.", "Solution_4": "[quote=\"Rajiv\"]\nI took it for granted that temperatures are equal. There is no way you can find the temperatures and IMO the answer would contain terms $ T_1$ and $ T_2$.[/quote]\r\n\r\nHere's how:\r\n\r\n$ P_1V_1 \\equal{} n_1RT_1 .... \\text{ (for box 1)} \\\\\r\n\\\\\r\nP_2V_2 \\equal{} n_2RT_2 ....\\text{ (for box 2)} \\\\\r\n\\\\\r\nP_f(V_1\\plus{}V_2) \\equal{} (n_1\\plus{}n_2)RT_f .... \\text{ (after opening the stop cock)} \\\\\r\n\\\\ \\\\\r\n(n_1\\plus{}n_2)R \\equal{} \\frac{3}{T_1}\\plus{}\\frac{2}{T_2} \\\\ \\\\ \\\\\r\n5P_f \\equal{} \\left(\\frac{3}{T_1}\\plus{}\\frac{2}{T_2}\\right)T_f$", "Solution_5": "Why are there no comments by anyone else? :mad:", "Solution_6": "[quote=\"Rajiv\"]Why are there no comments by anyone else? :mad:[/quote]\r\n\r\nBecause you've analyzed the problem correctly.\r\n\r\nCombining Boyle's law and Avogadro's law:\r\n\r\n\\[ \\frac{P_1V_1}{n_1} \\equal{} k \\equal{}\\frac{P_2V_2}{n_2} \\\\ \\\\ P_f(V_1 \\plus{} V_2) \\equal{} n_1k \\plus{} n_2k \\\\ \\\\ \\implies P_f(V_1 \\plus{} V_2) \\equal{} P_1V_1 \\plus{} P_2V_2 \\\\ \\\\ \\therefore P_f \\equal{}\\frac{P_1V_1 \\plus{} P_2V_2}{V_1 \\plus{} V_2} \\equal{}\\frac{(1\\mbox{ atm })(3\\mbox{ m}^3) \\plus{} (2\\mbox{ atm })(2\\mbox{ m}^3)}{3\\mbox{ m}^3 \\plus{} 2\\mbox{ m}^3} \\\\ \\\\ \\equal{} \\boxed{\\frac{7}{5}\\mbox{ atm }}\\]", "Solution_7": "[quote=\"Dr. No\"]$ \\boxed{\\frac {7}{5}\\mbox{ atm }}$\n[/quote]\r\n\r\nThank you! :lol:" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "The planes containing the faces of a cube divide the space into several regions. How many regions?", "Solution_1": "[hide] it would be 6*2 + 8 + 1 (if you include the cube) which is 21, i think[/hide]", "Solution_2": "wrong!", "Solution_3": "[hide]27. If we draw the 1 of the planes, there are 2 regions. Drawing the plane parallel to that plane we get 3 regions. Drawing the plane that intersects those 2 planes we get 6 regions, and then drawing the plane parallel to that we get 9 regions. Then drawing the final 2 planes that intersect it, we add 18 more, which makes 27.[/hide]", "Solution_4": "[hide]Correct![/hide]", "Solution_5": "[hide] Yeah, I got 27 too. The planes are just like the dividing planes in a 3x3x3 cube which divide it into unit cubes. And of course 3^3=27.[/hide]", "Solution_6": "[hide]Basically, you have 2 planes in each dimension, which divides it into 3, making 3^3=27.[/hide]", "Solution_7": "gosh, Im soo stupid! I forgot about the regions containing the edges :blush: :blush: :blush:", "Solution_8": "[hide]If you guys have read the other problems I've posted, I've said that these have been from the USME (Utah State Math Exam). It was multiple choice, and the first answer I got was 24. Luckily, when I checked it, I got 27.[/hide]" } { "Tag": [ "function", "abstract algebra", "integration", "trigonometry", "limit", "calculus", "complex numbers" ], "Problem": "Suppose that $a_n$ is a decreasing sequence that tends towards 0 and that $f(x)=\\sum_{n\\geq 1}{ a_n sin(nx)}$ is continuous. Does it follow that $a_n=o(\\frac{1}{n})$?", "Solution_1": "I don't have time today to write the whole proof, but the answer will be yes. A sufficient condition for the Fourier coefficients of $f$ to be $o\\left(\\frac1n\\right)$ is that $f$ be absolutely continuous. So we concentrate on showing that $|f'(x)|$ is integrable.\r\n\r\nOf course this would be false if you took out the word \"decreasing.\"", "Solution_2": "This would require showing that the series is the Fourier series of its sum, which I don't know how to do. Let me ask a simpler question: is it true that under the assumptions of the problem, there exists $\\gamma>0$ such that $\\liminf_{n\\to\\infty} n^\\gamma a_n=0$?. If somebody can show that, I can do the rest. Of course, if the series is the Fourier series of a continuous function, we have the Parceval identity, which immediately implies $a_n=o(n^{-1/2})$ but, as I said, I do not know how to show that it is a Fourier series :( .", "Solution_3": "My original answer was too hasty, as shown by fedja's objections. It may still be a possible direction to look in, but there are obstacles.", "Solution_4": "Seems I finally found a neat proof. Recall that the Fejer kernel is $K_n(x)=\\sum_{k=-n}^n\\left(1-\\frac{|k|}{n}\\right)e^{ikx}$. It is a standard exercise written in many textbooks on Fourier series to show that $K_n\\ge 0$, $\\int_{-\\pi}^\\pi K_n(x)\\,dx =2\\pi$ and, for every $\\delta>0$, we have $\\int_{(-\\pi,\\pi)\\setminus(-\\delta,\\delta)}K_n(x)\\,dx\\to 0$ as $n\\to\\infty$. Now let $G_n(x)=\\Im (e^{inx}K_n(x))=\\sum_{k=-n}^n\\left(1-\\frac{|k|}{n}\\right)\\sin(n+k)x$. Note that $G_n(x)=O(|x|)$ as $x\\to 0$. Since the partial sums $f_m(x)=\\sum_{k=1}^m a_k \\sin kx$ of our series are uniformly bounded by $\\frac C{|x|}$ on $(-\\pi,\\pi)$, we can use the dominated convergence theorem to conclude that, for every fixed $n$, $\\int_{-\\pi}^\\pi G_n(x)f(x)\\,dx=\\lim_{m\\to\\infty}\\int_{-\\pi}^\\pi G_n(x)f_m(x)=\\pi\\sum_{k=-n}^n \\left(1-\\frac{|k|}{n}\\right)a_{n+k}\\ge c na_n$ with some absolute constant $c>0$. On the other hand, it follows from the above properties of $K_n$ and from continuity of $f$ at $0$ that $\\int_{-\\pi}^\\pi G_n(x)f(x)\\,dx\\to 0$ as $n\\to\\infty$. Thus, we, indeed, must have $a_n=o(1/n)$.", "Solution_5": "$K_n$ is real-valued, so the way you defined $G_n$ you should have $G_n(x)=\\sin(nx)K_n(x)$. \r\nif that's the case, it's not necessarily true that $\\int G_n(x)f(x)\\to 0$.", "Solution_6": "[quote=\"cimabue\"]$K_n$ is real-valued, so the way you defined $G_n$ you should have $G_n(x)=\\sin(nx)K_n(x)$.[/quote]\nThat's true; I wrote the imaginary part of the product just to make the representation of $G_n$ as a sum of sines obvious.\n\n[quote=\"cimabue\"] if that's the case, it's not necessarily true that $\\int G_n(x)f(x)\\to 0$.[/quote]\r\nWhy? :? The integral over $(-\\pi,\\pi)\\setminus(\\delta,\\delta)$ tends to $0$ because $f$ is bounded and the integral of $|G_n|\\le K_n$ tends to $0$, while the integral over $(\\-\\delta,\\delta)$ is arbitrarily small because $f$ is small on $(-\\delta,\\delta)$ (it is continuous at $0$ and $f(0)=0$) and the integral of $|G_n|$ over $(-\\delta,\\delta)$ does not exceed $2\\pi$.", "Solution_7": "sorry, my bad. i failed to notice that $f(0)=0$." } { "Tag": [ "geometry", "incenter", "number theory", "greatest common divisor", "cyclic quadrilateral", "geometry proposed" ], "Problem": "Suppose that $ ABCD$ is a cyclic quadrilateral and $ CD\\equal{}DA$. Points $ E$ and $ F$ belong to the segments $ AB$ and $ BC$ respectively, and $ \\angle ADC\\equal{}2\\angle EDF$. Segments $ DK$ and $ DM$ are height and median of triangle $ DEF$, respectively. $ L$ is the point symmetric to $ K$ with respect to $ M$. Prove that the lines $ DM$ and $ BL$ are parallel.", "Solution_1": "$ BD$ is the bisector of angle $ ABC$ and $ \\angle EDF\\equal{}\\frac{1}{2}\\angle ADC\\equal{}90^o\\minus{}\\frac{1}{2}\\angle EBF$ therefore $ D$ is the B-excenter of triangle $ EBF$.\r\nThus $ L$ is the tangency of incenter $ (I)$ of triangle $ EBF$ and $ EF$. Let $ J\\equal{}BD\\cap EF$\r\n$ \\frac{r_b}{r}\\equal{}\\frac{DK}{IL}\\equal{}\\frac{KJ}{LJ}\\equal{}\\frac{ML}{LJ}\\plus{}\\frac{MJ}{LJ}\\equal{}1\\plus{}2\\frac{MJ}{LJ}$\r\n$ \\frac{r_b}{r}\\equal{}\\frac{DJ}{IJ}\\equal{}\\frac{DB}{IB}$ then it easy to show that $ \\frac{MJ}{LJ}\\equal{}\\frac{DJ}{BJ}$\r\n$ \\Rightarrow BL//DM$", "Solution_2": "Let $ G \\in AC$. $ CG=EA \\Rightarrow \\triangle EAD = \\triangle GCD$, $ BEDG$ is cyclic, $ DF$ is a bisector of $ \\angle EDF$.\r\n$ DF \\cap (BEDG)=M$,\r\n$ BM \\cap DE=N$,\r\n$ ME \\cap BD=W$,\r\n$ FE \\cap WD=S$,\r\n$ I$-midpoint of $ WD$.\r\n$ DM$ is a diameter of $ (BEDG) \\Rightarrow ME$ and $ DB$ are altitudes of $ \\triangle MND$ and $ BWEN$ is cyclic $ \\Rightarrow \\angle NDM=\\angle NBE=\\angle NWE$.\r\n$ \\angle EWD=\\angle DFG=\\angle DFE \\Rightarrow FWFD$ is cyclic $ \\Rightarrow \\angle EWF=180^0 - \\angle NDM$.\r\nThen $ \\angle NWE+\\angle EWF=180^0$ and $ N, W, F$ lie on the same line and $ NF$ is an altitude.\r\n\r\nWe shall prove:\r\nFor an arbitrary quadriateral $ XYZT$ such that $ \\angle Y=\\angle T=90^0$, if $ X'$ and $ Z'$ are the projections of $ X$ and $ Z$ on $ YT$, then $ YX'=TZ'$.\r\n$ \\triangle X'XY \\sim \\triangle Z'YZ$\r\n$ \\triangle XX'T \\sim \\triangle TZ'Z$\r\nThen,\r\n$ \\frac {XX'}{YX'}= \\frac {YZ'}{ZZ'}$ and $ \\frac {XX'}{TX'}= \\frac {TZ'}{ZZ'} \\Rightarrow \\frac{TX'}{YX'}= \\frac{YZ'}{TZ'} \\Rightarrow YX'=TZ'$\r\n\r\nThen $ WL \\perp FE$ and $ \\triangle SWL \\sim \\triangle SIM \\Rightarrow \\frac{SW}{SI}=\\frac{SL}{SM}$.\r\n$ \\angle FIE=2 \\angle FDE$\r\n$ \\angle FBE=\\angle FBW+\\angle EBW=\\angle EMD+\\angle FND=180^0-2\\angle FDE$.\r\nThen, $ BEIF$ is cyclic. We know that $ FWED$ is cyclic, so\r\n$ BS \\cdot SI=FS \\cdot SE=WS\\cdot SD \\Rightarrow \\frac{SW}{SI}=\\frac{BS}{SD} \\Rightarrow\\frac{SL}{SM}=\\frac{BS}{SD}\\Rightarrow \\triangle SBL\\sim\\triangle SDM \\Rightarrow BL \\parallel MD$.", "Solution_3": "[quote=\"livetolove212\"]................................\n\n then it easy to show that $ \\frac {MJ}{LJ} \\equal{} \\frac {DJ}{BJ}$\n$ \\Rightarrow BL//DM$[/quote]\r\n\r\n Can someone explain how we arrive to the first relation ? \r\n\r\n Babis", "Solution_4": "[quote=\"stergiu\"][quote=\"livetolove212\"]................................\n\n then it easy to show that $ \\frac {MJ}{LJ} \\equal{} \\frac {DJ}{BJ}$\n$ \\Rightarrow BL//DM$[/quote]\n\n Can someone explain how we arrive to the first relation ? \n\n Babis[/quote]\r\n$ 2\\frac{MJ}{LJ}\\plus{}1\\equal{}\\frac{DB}{IB}\\equal{}\\frac{EB\\plus{}BF\\plus{}EF}{EB\\plus{}BF\\minus{}EF}\\equal{}\\frac{2EF}{EB\\plus{}BF\\minus{}EF}\\plus{}1$\r\n$ \\equal{}2\\frac{1}{\\frac{EB\\plus{}BF\\minus{}EF}{EF}}\\plus{}1\\equal{}2\\frac{1}{\\frac{EB\\plus{}BF}{EF}\\minus{}1}\\plus{}1$\r\n$ \\equal{}2\\frac{1}{\\frac{BF}{\\frac{BF.EF}{EB\\plus{}BF}}\\minus{}1}\\plus{}1\\equal{}2\\frac{1}{\\frac{BF}{FJ}\\minus{}1}\\plus{}1$\r\n$ \\equal{}2\\frac{1}{\\frac{BD}{DJ}\\minus{}1}\\plus{}1\\equal{}2\\frac{1}{\\frac{BJ}{DJ}}\\plus{}1\\equal{}2\\frac{DJ}{BJ}\\plus{}1$", "Solution_5": "it's easy to prove that $D$ is the excenter of triangle $BEF$\nhence the result is trivialized by ex-Boone theorem." } { "Tag": [ "calculus", "derivative", "function", "integration" ], "Problem": "I'm going to be doing a science fair project this summer. (I just finished 8th grade.) In a different thread, somebody suggested learning some calculus and modeling the cooling of an object. I think you would use Newton's Law of Cooling for this, but I'm not sure. I learned how to find the derivative of a function at a given point, and how to differentiate a function. I think I also need to learn how to integrate a function, in order to solve a 1st order differentiation equation or something like that. What else do I need to know? Any suggestions? :)", "Solution_1": "Hmm....that's actually about all you need to know. Integration of a differential equation is all Newton's Law of Cooling is.", "Solution_2": "[quote=\"DPopov\"]Hmm....that's actually about all you need to know. Integration of a differential equation is all Newton's Law of Cooling is.[/quote]\r\n\r\nReally? :) Okay. But it's pretty confusing :wacko: :what?: Are you supposed to use definite or indefinite integral for differential equations?", "Solution_3": "I think all you really have to worry about is how to use a graph (There are certain advantages of being a physics student - You just draw a graph and if you tried to get a graph paper which has 'logs' - the diff eq part will take care of itself)\r\n\r\nBasically Newton's law simply is. that rate of cooling varies directly as the difference of the temperatures between the object and the environment. (Unless the environment happens to be your hand, in that case, it may seem that the object is hotter than it should be) \r\n\r\nSo just small ammount of math will get you that the temperature is something like $e^{-kt}$ where t is time and k is some constant which you can calculate by looking at the graph...\r\n\r\nhave fun, make sure that you don't [b]touch[/b] the hot object :)", "Solution_4": "Okay so I finished the project. What should I call it? \"Calculus and Pies\"?? :lol: (I made a pie for the project. Surprisingly, there wasn't much calculus. If you want to see my work, PM me please.) Thanks. :)" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let ABC is a triangle,prove that:\r\n$ \\frac{r_a}{r_b} \\plus{} \\frac{r_b}{r_a}\\geq \\frac{A}{B} \\plus{} \\frac{B}{A}\\geq \\frac{a}{b} \\plus{} \\frac{b}{a}.$\r\nBQ", "Solution_1": "[quote=\"xzlbq\"]Let ABC is a triangle,prove that:\n$ \\frac {r_a}{r_b} \\plus{} \\frac {r_b}{r_a}\\geq \\frac {A}{B} \\plus{} \\frac {B}{A}\\geq \\frac {a}{b} \\plus{} \\frac {b}{a}.$\nBQ[/quote]\r\n$ r_a, r_b, r_c$, radii of excircles, respectively." } { "Tag": [ "quadratics", "algebra", "polynomial" ], "Problem": "What method of factoring quadratic equations do you use? \r\n\r\nSay I have $7x^2+9x+2$ and want it factored. What method of factoring would you use?\r\n\r\nTheres this method where you use $2$ factors of $ac$, $F_1$ and $F_2$, which $F_1 + F_2 = b$. You use the $2$ factors to replace b and use factor by grouping to factor the rest of the equation. Why is the initial multiplication of $a$ and $c$ required?", "Solution_1": "To factor your example, we get $(7x+2)(x+1)$. I don't really use any method for factoring; I just look at it and see what it can be split into. Your example was fairly simple to factor, because there are very few ways to consider.\r\n\r\nI think your method can only be used when 2 of the coefficients are 1, like in the example. In this case the 2 factors would be 7 and 2, which add up to 9. I believe the intitial multiplication of ac in your method is required because it is equal to the product of the coefficients of the factored form. For example, $8x^2+26x+15=(2x+5)(4x+3)$. $ac=120$, which equals $2\\cdot5\\cdot4\\cdot3$.\r\n\r\nMy approach to factoring is that if the polynomial is easy to factor, then I factor it. If it's not, then I use the quadratic formula. If factoring is required, then that should be simple after you find the x values.\r\n\r\nHope this helps. :)", "Solution_2": "Thanks for the advice.\r\n\r\nI did some thinking on why the method taught to me works, and I think it is because by multiplying finding factors of ac to replace b, you have 2 numbers which are either multiples or factors of a and c. Therefore it is very simple to factor it out by grouping. \r\n\r\nThe coefficients in factored form must equal ac because the 2 leading coefficients of each binomial multiply together to form a and the trailing coefficients' product is c." } { "Tag": [ "geometry" ], "Problem": "Ok I really don't get this! [geogebra]168649eb1051012ef3a4d0ab8c96b071624aac58[/geogebra] \r\nI am supposed to find the area of the leave like things. I tried separating into quarter squares and I found out that the area of half of the leavy things is pi-1. There are 8 halves so I multiplied by 8 and got 8pi-8. The answer is 8pi-16. Where did I go wrong and what is an easier way to do this? BTW the side length of the square is 4,[/img]", "Solution_1": "Though I can't see the picture, maybe you can try quarter circle - area of triangle?", "Solution_2": "We see that the angle between the intersection points of adjacent circles and the midpoint of a circle is 90 degrees. We can find the area of one of the leaves by finding the area of two quarters, or one half, of a circle and subtracting 4. Here is a picture that might clarify some things:\r\n[geogebra]fd1f18af9d390f18b661c3df304aa9277940ec28[/geogebra] \r\nThe side length of the square is 2. To find the area, as I said, of the leaf we find the area of the square plus the leaf (the area of two quarters/half of a circle) and subtract the square so we have the leaf left over. The area of half the circle is $ 2\\pi$ and the area of the square is 4. The area of one leaf is then $ 2\\pi\\minus{}4$. Multiply by four to find the area of all four:\r\n\r\n$ 4(2\\pi\\minus{}4)\\equal{}\\boxed{8\\pi\\minus{}16}$\r\n\r\nI hope you understand!", "Solution_3": "The area of one of the triangles is 2 not 1.\r\nThus, the area of half the leaves is pi-2.\r\nMultiplying this by 8 we get 8pi-16.\r\n\r\nEDIT beaten", "Solution_4": "Ok ytrewq I understand that mistake I made, but Power Of Pi, I quite don't get yet. I don't understand the part of adding the square and the leaf. I do get subtracting the square, but I don't understand how you get 2pi for adding the square and the leaf. Thanks!", "Solution_5": "I'll say it this way:\r\n\r\nWhen you find the area of half the circle, you are counting the leaf twice (once with each quarter of the circle). You can find from that $ square\\plus{}leaf\\equal{}2(\\frac{1}{4}circle)$." } { "Tag": [ "inequalities" ], "Problem": "$a,b,c \\in \\mathbb{R}$\r\n\r\nProve that $a^3+b^3+c^3+3abc \\geq ab(a+b)+bc(b+c)+ac(a+c)$", "Solution_1": "Rewrite it $\\sum a(a-b)(a-c)$ : it is Schur", "Solution_2": "See in the book:\r\n\r\n << \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2 \u03a3\u03c4\u03b5\u03c1\u03b3i\u03bf\u03c5 \u03ba\u03b1\u03b9 \u039d\u03af\u03ba\u03bf\u03c2 \u03a3\u03ba\u03bf\u03bc\u03c0\u03c1\u03ae\u03c2 : \u0391\u03bb\u03b3\u03b5\u03b2\u03c1\u03b9\u03ba\u03ad\u03c2 \u0391\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2 , \u0395\u03ba\u03b4\u03cc\u03c3\u03b5\u03b9\u03c2 \u03a3\u03b1\u03b2\u03b2\u03ac\u03bb\u03b1\u03c2 2004 , Problem 4.48 >>\r\n\r\n There , you will find a solution , based on the fact that this inequality is symmetric.\r\n\r\n [u]Babis[/u]", "Solution_3": "[quote=\"t\u00b5t\u00b5\"]Rewrite it $\\sum a(a-b)(a-c)$ : it is Schur[/quote]\r\n\r\nOK, but this is $a,b,c \\in \\mathbb{R}^+$\r\n\r\nHere is $a,b,c \\in \\mathbb{R}$", "Solution_4": "I thought it was a typo because it is false:\r\n\r\ntake $b=c=0$, we would have $a^3 \\geq 0$, $\\forall x \\in \\mathbb{R}$, for instance", "Solution_5": "This inequality is true for positive numbers", "Solution_6": "OK, then let's prove Schur:\r\n$a,b,c,r \\in R^+$, prove:\r\n\\[ \\sum a^r(a-b)(a-c)\\geq 0 \\]", "Solution_7": "Easy. :)\r\n\r\nSince, the ineq. is symmetrical, we can assume, WLOG, that : $a \\geq b \\geq c \\geq 0$.\r\nThus, the inequality is true since : $a^r(a-b)(a-c) \\geq b^r(a-b)(b-c) \\text{ and } c^r(c-a)(c-b) \\geq 0$.", "Solution_8": "what is schur? :?:", "Solution_9": "[quote=\"varun\"]what is schur? :?:[/quote]\r\nView the posts above yours.\r\n\r\nmathmanman: Thanks for the proof.", "Solution_10": "what mathmanman meant to say?\r\nwas it that all the three terms are $\\ge 0$\r\n\r\nThen its OK.\r\nAnd thanks for thenew inequality.", "Solution_11": "Yes Varun :)\r\nShobber, you're welcome!", "Solution_12": "In my book the following inequality holds: \\[ a^{t}(a-b)(a-c)+b^{t}(b-c)(b-a)+c^{t}(c-a)(c-b)\\geq 0, \\]\r\n where $a,b, c>0$ and $r\\in\\mathbb{R}.$ ;)", "Solution_13": "Well yes...that is Schur and its proven in the posts before", "Solution_14": "Look carefully: $r\\in\\mathbb{R}.$ I have seen Schur only for $r>0.$ :)", "Solution_15": "it also can be proven like this:\r\nlike it has been said a b c are positive then divide all by abc, and WLOG $a \\geq b \\geq c$ then:\r\n\r\n$\\frac{a^2}{bc}+\\frac{b^2}{ca}+\\frac{c^2}{ab}+3 \\geq \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+\\frac{b}{a}+\\frac{c}{b}+\\frac{a}{c}$\r\n\r\nand take $(\\frac{a}{b},\\frac{b}{c},1,1,1,\\frac{c}{a})$ and $(\\frac{a}{c},1,1,1,\\frac{c}{b},\\frac{b}{a})$ \r\n\r\nSo by some kind of arrangement:\r\n \r\n$\\frac{a}{b}\\frac{a}{c}+\\frac{b}{c}\\frac{b}{a}+\\frac{c}{a}\\frac{c}{b}+3 \\geq \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+\\frac{b}{a}+\\frac{c}{b}+\\frac{a}{c}$\r\nlike we wanted to see", "Solution_16": "Cesar could you post the solution with $r\\in R$. Thank you very much", "Solution_17": "Well, that's still easy :)\r\nIf $r \\geq 0$, read my proof above.\r\nIf $r < 0$, then we can assume WLOG that : $0 < a \\leq b \\leq c$.\r\nAnd, the rest is exactly the same thing as in my previous post.", "Solution_18": "[quote=xitim]$a,b,c \\in \\mathbb{R}$\n\nProve that $a^3+b^3+c^3+3abc \\geq ab(a+b)+bc(b+c)+ac(a+c)$[/quote]\n [url=https://artofproblemsolving.com/community/c893771h1869509p12676711]All Soviet Union MO 1975[/url]\n" } { "Tag": [ "geometry", "calculus", "integration", "function", "symmetry", "real analysis", "Euler" ], "Problem": "Suppose that $ S$ is a 2-dimensional surface in $ \\mathbb R^3$ whose boundary is the red brocken line shown below:\r\n[asy]size(150);\nD((-1.5,-1.5)--MP(\"z\",(1.5,1.5),NE),black);\nD(MP(\"y\",(0,7),N)--(0,0)--MP(\"x\",(7,0),E),black);\nD(MP(\"1\",(1,1),W)--(1,7)--(-1,5)--MP(\"-1\",(-1,-1),W)--(4,-1)--(6,1)--cycle ,red+linewidth(1));\nMP(\"b\",(0,6),W);\nMP(\"a\",(5,0),S);[/asy]\r\nShow that the total area of $ S$ is at least $ 2(a\\plus{}b\\minus{}2)$.\r\n\r\nI'd like to present this as an application of differential forms to estimating areas of minimal surfaces but I'm hesitating whether just to make a lecture about it or to give it to the students as an exercise with many hints (a la ENS exams). So, I decided to post it here to see if AoPSers can solve it easily. The suggested idea is to find a 2-form $ \\omega$ such that $ \\|\\omega\\|\\le 1$ at all points inside the convex hull of the red contour (you may assume without loss of generality that $ S$ is contained in that convex hull), $ d\\omega\\equal{}0$, and the integral of $ \\omega$ over $ S$ (which does not depend on $ S$ because of the second condition) is at least $ 2(a\\plus{}b\\minus{}2)$.", "Solution_1": "if I understand everything well, then your answer is strange (incorrect). Let's set $ b \\to 0$ then answer must tend to $ 2a$.", "Solution_2": "It is just a simple lower bound, not the sharp estimate! Of course, it is of interest only when both $ a$ and $ b$ are large. I don't really know if the exact value of the minimum in this problem can be expressed in any reasonable terms.", "Solution_3": "[quote=\"fedja\"]I don't really know if the exact value of the minimum in this problem can be expressed in any reasonable terms.[/quote]\r\nI agree with you.\r\nThis is what I'm thinking for sharp estimate.\r\nthis surface can be expressed as $ S \\equal{} (x,u(x,z),z)$ where $ x*z \\in [0,a]*[ \\minus{} 1,1] \\equal{} K$\r\nI can assume that $ u(x,z)$ is smooth function of two variables. Since, if it's not smooth , then this can be approximated sufficiently well with smooth function.\r\nOne can note by symmetry that $ u(x,z) \\equal{} u(x, \\minus{} z)$ and $ u'_{x}(x,z) < 0$. Also note that $ u(0,z) \\equal{} b$ and $ u(a,z ) \\equal{} 0$ and \r\n$ u(x,1) \\equal{} u(x, \\minus{} 1) \\equal{} 0$\r\nI assume $ ds$ is measure on our surface.\r\nWe need to find area of our surface $ S$ which is equal to $ \\int_{S}ds$ \r\n But $ ds \\equal{} \\sqrt {\\det{\\Gamma(w)}} dw$ where $ dw \\equal{} dzdx$ is Lebesgue measure on $ K$ and $ \\Gamma(w)$ is Gramian matrix. Thus we obtain: \r\n$ \\int_{S}ds \\equal{} \\int_{K}\\sqrt {1 \\plus{} u_{x}^{2} \\plus{} u_{z}^{2}}dzdx \\equal{} \\int_{0}^{a}\\int_{ \\minus{} 1}^{1}\\sqrt {1 \\plus{} u_{x}^{2} \\plus{} u_{z}^{2}}dzdx$\r\nNow we need to minimise such functional as $ I[u] \\equal{} \\int_{K}\\sqrt {1 \\plus{} u_{x}^{2} \\plus{} u_{z}^{2}}dzdx \\equal{} \\int_{0}^{a}\\int_{ \\minus{} 1}^{1}\\sqrt {1 \\plus{} u_{x}^{2} \\plus{} u_{z}^{2}}dzdx$\r\nAfter some calculation we will get euler equation (if I have no mistakes in my calculations):\r\n$ u_{xx} \\plus{} u_{zz} \\plus{} u_{z}^{2}u_{xx} \\plus{} u_{x}^{2}u_{zz} \\minus{} 2u_{x}u_{z}u_{xz} \\equal{} 0$\r\nIf I remember well, this is wellknown elliptic equation in mathematical physic.\r\nToday, I don't know how to find any solution of such equations.\r\nI believe that such equations are well investigated in math physic's. \r\nI note that, all of this also works for any 2-dimensional manifolds with boundaries and you always get this elliptic equations.\r\nBut if you need to find just a simple lower bound , what fedja has written, then fedja's hint is one of the possible way for the solution.", "Solution_4": "[quote=\"Extremal\"]\nIf I remember well, this is wellknown elliptic equation in mathematical physic.\nToday, I don't know how to find any solution of such equations.\nI believe that such equations are well investigated in math physic's. \nI note that, all of this also works for any 2-dimensional manifolds with boundaries and you always get this elliptic equations.[/quote]\r\nOh, yes, the minimal surface equation itself is well known. Still, there is absolutely no hope to find an explicit solution for all possible boundary curves (actually, the only \"explicitly\" solved case with a broken line as the boundary I know of is a skewed quadrilateral), so I do not think this one is an exception. What has been studied extensively is existence, uniqueness, and possible singularities of solutions and it is a completely separate story.\r\n\r\nSo, Extremal, can you prove this bound (using the hint or by some other method)?", "Solution_5": "I have found other bounds, which is not so sharp as your.\r\nI was estimating this integral by using simplest inequalitys like this $ \\sqrt {2(a \\plus{} b)} \\geq\\sqrt {a} \\plus{} \\sqrt {b}$ :\r\n\\[ S \\equal{} \\int_{0}^{a}\\int_{ \\minus{} 1}^{1}\\sqrt {1 \\plus{} u_{x}^{2} \\plus{} u_{z}^{2}}dzdx \\plus{} \\lim_{c \\to 0} \\int_{ \\minus{} c}^{0}\\int_{ \\minus{} 1}^{1}\\sqrt {1 \\plus{} u_{x}^{2} \\plus{} u_{z}^{2}}dzdx\r\n\\]\r\nI believe that it's useful to think ab0ut such estimation since :\r\n\\[ S \\equal{} \\int_{0}^{a}\\int_{ \\minus{} 1}^{1}\\sqrt {1 \\plus{} u_{x}^{2} \\plus{} u_{z}^{2}}dzdx \\plus{} \\lim_{c \\to 0} \\int_{ \\minus{} c}^{0}\\int_{ \\minus{} 1}^{1}\\sqrt {1 \\plus{} u_{x}^{2} \\plus{} u_{z}^{2}}dzdx \\geq 2a \\plus{} 2b \\minus{} C\r\n\\]\r\nwhere we need to show that $ C \\approx 4$.\r\nWe can assume that $ |u_{x}| \\leq \\frac {\\sqrt {b^{2} \\minus{} c^{2}}}{c}$ where $ |AO| \\equal{} b$\r\nalso we can think about such estimation as: \r\non $ [ \\minus{} c,0]$ $ |u_{x}| \\geq \\frac{\\sqrt {b^{2} \\minus{} c^{2}}}{a \\plus{} c}$ (if $ a > b$) and similary for the interval $ [0,a]$\r\n \r\n\r\n[asy]size(150);\nD((-1.5,-1.5)--MP(\"z\",(1.5,1.5),NE),black);\nD(MP(\"y\",(0,7),N)--(0,0)--(-4,0)--MP(\"x\",(7,0),E),black);\n\n\nD(MP(\"\",(-2,6),N)--(0,0),black);\n\nD(MP(\"\",(-2,8),N)--(-2,0),black);\nD(MP(\"O\",(1,1),W)--(-1,7)--(-3,5)--MP(\"-1\",(-1,-1),W)--(4,-1)--(6,1)--cycle ,red+linewidth(1));\nMP(\"A\",(-3,6),W);\nMP(\"a\",(5,0),S);\nMP(\"-c\",(-2.9,0),N);[/asy]" } { "Tag": [ "function", "limit", "number theory", "prime numbers", "number theory proposed" ], "Problem": "Let $d(x)$ be the function, that maps the number of prime factors of the natural number $x$.\r\nLet $S_{1}=d(1)+...+d(n)$ with $n\\in N$ and $n>1$\r\nLet $p_{1},p_{2},...,p_{k}$ all the prime numbers less than $n$, and $S_{2}=\\frac{1}{p_{1}}+...+\\frac{1}{p_{k}}$.\r\nShow that: $\\lim_{n\\rightarrow\\propto}({S_{2}-\\frac{S_{1}}{n})=0}$\r\n\r\n[u][b][i]Proposed by Sef[/i][/b][/u]", "Solution_1": "I have a (rather large) solution (if there's no mistake somewhere). I will gladly post it here (it's already in the greek forum) if everyone else is too bored to post their own :lol:", "Solution_2": "Don't name topics in Number Theory with Number Theory, it's a bit redundant ;)" } { "Tag": [ "geometry", "geometric transformation", "reflection", "circumcircle", "incenter", "homothety", "cyclic quadrilateral" ], "Problem": "Let $ ABCD$ be a cyclic quadrilateral and $ r$ and $ s$ the lines obtained reflecting $ AB$ with respect to the internal bisectors of $ \\angle CAD$ and $ \\angle CBD$, respectively. If $ P$ is the intersection of $ r$ and $ s$ and $ O$ is the center of the circumscribed circle of $ ABCD$, prove that $ OP$ is perpendicular to $ CD$.", "Solution_1": "Denote K be the midpoint of arc CD of (O). Easy to see that K is the excenter of triangle PAB. O is the circumcenter of triangle ABK, so it is also the midpoint of segment HK where H is the incenter of triangle ABP. Thus P,H,O,K all lie on the bisector of angle APB. This implies $ OP\\equiv OK\\perp CD$", "Solution_2": "Chasing angles, it's very easy to see that APBO is cyclic.Now, consider triangles POC and POD.\r\nWe have that PO is a common side and OC=OD (radii).So, if we prove that PDC is isosceles and PO is bisector of 2 -x+1)(y2 -y+1)(z 2 -z+1) \\geq (xyz) 2 +(xyz)+1.", "Solution_1": "I think it was already discussed. By the way, as far as I remember, this problem was created by Mircea Lascu and Vasile Cartoaje." } { "Tag": [ "trigonometry", "geometry unsolved", "geometry" ], "Problem": "Given is triangle ANC with AN = AC and $ \\angle CAN \\equal{} 90^o$. Points M and B lie on AN. If $ \\angle ACM \\equal{} \\angle MCB \\equal{} \\angle BCN \\equal{} 15^o$. Prove that BN = AM + BM/2.", "Solution_1": "[quote=\"CDP100\"]Given is triangle ANC with AN = AC and $ \\angle CAN \\equal{} 90^o$. Points M and B lie on AN. If $ \\angle ACM \\equal{} \\angle MCB \\equal{} \\angle BCN \\equal{} 15^o$. Prove that BN = AM + BM/2.[/quote]\r\n[hide=\"Solution\"]\nLet $ CA \\equal{} a$. Then, we have that\n\\[ AM \\equal{} a\\tan \\angle ACM \\equal{} a\\tan 15 \\equal{} \\frac {a\\sin 15}{\\cos 15} \\equal{} \\frac {\\frac {a(\\sqrt {6} \\minus{} \\sqrt {2})}{4}}{\\frac {\\sqrt {6} \\plus{} \\sqrt {2}}{4}} \\equal{} \\frac {a(\\sqrt {3} \\minus{} 1)}{\\sqrt {3} \\plus{} 1} \\equal{} a(2 \\minus{} \\sqrt {3})\n\\]\nNotice that $ BA \\equal{} a\\tan 30 \\equal{} \\frac {a}{\\sqrt {3}}$. Finally, we notice that $ BN \\equal{} AN \\minus{} AB \\equal{} a \\minus{} \\frac {a}{\\sqrt {3}} \\equal{} \\frac {a(\\sqrt {3} \\minus{} 1)}{3}$. Hence,\n\\[ AM \\plus{} \\frac {BM}{2} \\equal{} AM \\plus{} \\frac {AB \\minus{} AM}{2} \\equal{} \\frac {AB \\plus{} AM}{2} \\equal{} \\frac {a(2 \\minus{} \\sqrt {3} \\plus{} \\frac {1}{\\sqrt {3}})}{2} \\equal{} \\frac {a(\\frac {2\\sqrt {3} \\minus{} 2}{3})}{2} \\equal{} \\frac {a(\\sqrt {3} \\minus{} 1)}{3} \\equal{} BN\n\\]\nas desired. [/hide]", "Solution_2": "to: cdp100\r\n\r\nIt's obvious that you wanted something in connection with the triangle MCN and CB internal bisector of the angle MCN, then its external one!\r\n\r\nHowever, some simple direct calculations show that the required equality is equivalent with: 3.BN = AN +AM ( 1 ). Therefore we shall draw S, the symmetrical of M w.r.t. A and we have only to prove 2BN = BS ( 2 ); using the sine teorem in the triangles NBC and BSC, (2) implies after calculations \r\n2.sin(15).sin(75)=sin(45).sin(45) ( 3 ), substitute sin(75) by cos(15) and (3) becomes obvious.\r\n\r\nBest regards,\r\nsunken rock\r\n\r\nA blind man sees better the details" } { "Tag": [ "number theory", "number theory proposed" ], "Problem": "We consider the proposition $ p(n)$: $ n^2\\plus{}1$ divides $ n!$, for positive integers $ n$. Prove that there are infinite values of $ n$ for which $ p(n)$ is true, and infinite values of $ n$ for which $ p(n)$ is false.", "Solution_1": "PEN :wink:", "Solution_2": "Yea this one looks like a classic. I don't understand why its there but ...", "Solution_3": "1. Let $ p \\equal{} 1\\mod 4$ prime, then exist $ n_0 < p/2$, suth that $ p|n_0^2 \\plus{} 1$. Obviosly $ p\\not |n_0!$ or $ n_0^2 \\plus{} 1\\not |n_0!.$\r\n2. Let primes $ p>q\\ge 17$ and $ p\\equal{}1\\mod 4, q\\equal{}1\\mod 4$. Exist $ pq/2 LS for small cases. It is enough.", "Solution_2": "How do you (e.g.) get the difference $5$ by this\u00bf\r\n(the bound is sharper than you may think)", "Solution_3": "[quote=ZetaX]Prove:\nFrom the set $\\{1,2,...,n\\}$, one can choose a subset with at most $2 \\left\\lfloor \\sqrt n \\right\\rfloor +1$ elements such that the set of the pairwise differences from this subset is $\\{1,2,...,n-1\\}$.\n($\\left\\lfloor x \\right\\rfloor$ means the greatest integer $\\leq x$)[/quote]\n\nChoose $1,2,\\ldots,\\lfloor\\sqrt{n}\\rfloor-1$, all multiples of $\\lfloor\\sqrt{n}\\rfloor$ in $\\{1,2,...,n\\}$ and $n$." } { "Tag": [ "modular arithmetic" ], "Problem": "Show that b_n=19*8^(n)+17 is composite for any positive integer n", "Solution_1": "[hide]\nConsidering cases\n\n$ n \\equal{} 4k$:\n$ 19 \\cdot 8^{4k}\\plus{}17 \\equiv 1 \\cdot 64^{2k} \\plus{} 2 \\equiv 1^{2k} \\plus{} 2 \\equiv 0 \\pmod 3$\n\n$ n \\equal{} 4k\\plus{}1$:\n$ 19 \\cdot 8^{4k\\plus{}1}\\plus{}17 \\equiv 6 \\cdot 8 \\cdot {8^4}^k \\plus{} 4 \\equiv 9 \\cdot 1^k \\plus{} 4 \\equiv 0 \\pmod{13}$\n\n$ n \\equal{} 4k\\plus{}2$:\n$ 19 \\cdot 8^{4k\\plus{}2}\\plus{}17 \\equiv 1 \\cdot 64 \\cdot 64^{2k} \\plus{} 2 \\equiv 1^{2k} \\plus{} 2 \\equiv 0 \\pmod 3$\n\n$ n \\equal{} 4k\\plus{}3$:\n$ 19 \\cdot 8^{4k\\plus{}3}\\plus{}17 \\equiv 4 \\cdot 8^3 \\cdot {8^4}^k \\plus{} 2 \\equiv 4\\cdot 2 \\cdot 1^k \\plus{} 2 \\equiv 0 \\pmod 5$[/hide]" } { "Tag": [ "analytic geometry", "algebra solved", "algebra" ], "Problem": "Find all rationals x,y such that x^2+y^2 =2", "Solution_1": "The idea for this one is to write x=p/q and y=m/n with p,q,m,n in Z and (p,q)=1 and (m,n)=1 and the equation becomes:\r\n\r\np^2n^2+m^2q^2=2q^2n^2\r\n\r\nsee what happens! :)", "Solution_2": "We can use the method of intersection to solve this kind of equations: We have to find all rational points in the circle (C) x^2+y^2 = 2. Denote the set of all those points by S. Note that A(1, 1) belongs to S. Now, take an rational number k and consider the line y = k(x-1) + 1. This line meets circle (C) again at the point B with coordinates (k^2-2k-1)/(k^2+1), (-k^2-2k+1)/(k^2+1). Of course B belongs to S. So x = (k^2-2k-1)/(k^2+1), y= (-k^2-2k+1)/(k^2+1) is rational solution of x^2+y^2=2. It easy to show that all rational solution can be received by this way.\r\n\r\nNamdung" } { "Tag": [], "Problem": "Voir le titre du message ! :) \r\n\r\nComme c'est un exercice propos\u00e9 par les fran\u00e7ais (il me semble ... :D !!!)\r\n\r\nEst-il possible de voir une r\u00e9solution ou plusieurs r\u00e9solutions ... soyons fous ....de celui-ci et surtout..... de quel contexte est-il sorti ?", "Solution_1": "Pour les solutions, voir le fil sp\u00e9cifiquement d\u00e9di\u00e9 \u00e0 cet exo dans le forum IMO Spain 2008.\r\n\r\nPour l'origine, la version initiale vient d'un d\u00e9compte de chemins de longueur fix\u00e9e d'un sommet \u00e0 un sommet oppos\u00e9 le long des ar\u00eates d'un hypercube de dimension $ k$.\r\nComme l'hypercube n'avait aucune chance de passer aux OIM, on a chang\u00e9 l'habillage en parlant d'interrupteurs.\r\n\r\nPierre." } { "Tag": [ "inequalities", "function", "calculus", "derivative", "integration", "calculus computations" ], "Problem": "If f is twice differentiable on [-a,a] and $ M_0 \\equal{} \\mathop {\\sup }\\limits_{x \\in I} |f(x)|$, $ M_1 \\equal{} \\mathop {\\sup }\\limits_{x \\in I} |f'(x)|$, $ M_2 \\equal{} \\mathop {\\sup }\\limits_{x \\in I} |f''(x)|$ then\r\n\r\n$ |f'(x)| \\le \\frac{{M_0 }}{a} \\plus{} \\frac{{x^2 \\plus{} a^2 }}{{2a}}M_2$", "Solution_1": "Consider the function $ y \\equal{} f'(x)$. It has derivative bounded by $ M_2$ and its integral over any interval $ [\\alpha,\\beta]$ does not exceed $ 2M_0$. Given these constraints, how can we make $ M: \\equal{} f'(x)$ large? By making $ f'$ drop down linearly on both sides of $ x$, as in $ f'(t) \\equal{} \\max(M \\minus{} M_2|t \\minus{} x|,0)$. Therefore,\r\n\\[ \\int_{ \\minus{} a}^a \\max(M \\minus{} M_2|t \\minus{} x|,0)\\,dt\\le 2M_0\r\n\\]\r\nThis can be integrated and solved for $ M$, but the computation splits into cases depending on the signs of $ M \\minus{} M_2|a \\minus{} x|$ and $ M \\minus{} M_2|a \\plus{} x|$. Being lazy, we drop $ \\max$ from the integral (this only decreases it) and arrive at\r\n\\[ \\int_{ \\minus{} a}^a (M \\minus{} M_2|t \\minus{} x|)\\,dt\\le 2M_0\r\n\\]\r\nThe integral easily evaluates to $ 2Ma\\minus{}M_2(a^2\\plus{}x^2)$, hence $ M\\le \\frac{M_0}{a}\\plus{}\\frac{M_2(a^2\\plus{}x^2)}{2a}$ as required.\r\n\r\nHowever, our laziness did not go unpunished: the estimate is not sharp. Indeed, it does not approach $ 0$ as $ M_0\\to 0$ - as it should.", "Solution_2": "Could you explain the 'making f' drop down linearly on both sides of x' part? I don't understand what's going on there.", "Solution_3": "Fix $ x\\in [\\minus{}a,a]$. We may assume $ f'(x)\\ge 0$. For all $ t\\in [\\minus{}a,a]$ we have $ |f'(t)\\minus{}f'(x)|\\le M_2|t\\minus{}x|$ (say, by Mean Value Theorem applied to $ f'$). This implies $ f'(t)\\ge f'(x)\\minus{}M_2|t\\minus{}x|$. To shorten notation, let $ g(t)\\equal{}f'(x)\\minus{}M_2|t\\minus{}x|$. Notice that $ g(t)\\ge 0$ when $ |t\\minus{}x|\\le f'(x)/M_2$. Let $ [\\alpha,\\beta]$ be intersection of $ [\\minus{}a,a]$ with the interval $ \\{t\\colon |t\\minus{}x|\\le f'(x)/M_2\\}$. \r\n\r\nHere is the key point: since $ f'(t)\\ge g(t)\\ge 0$ on $ [\\alpha,\\beta]$, it follows that \r\n\\[ \\int_{\\alpha}^{\\beta}g(t)dt\\le \\int_{\\alpha}^{\\beta}f'(t)dt\\equal{} f(\\beta)\\minus{}f(\\alpha)\\le 2M_0\\]\r\nThis is essentially the same as the first displayed formula in my previous post." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Evaluate $ \\int \\left( \\cot 2x\\cot 3x\\minus{}\\tan 2x\\tan 7x\\right) \\tan 5xdx$\r\n :ddr:", "Solution_1": "$ \\cot 5x\\equal{}\\frac{\\cot 2x\\cot 3x\\minus{}1}{\\cot 2x\\plus{}\\cot 3x}$\r\n\r\n$ \\cot 2x\\cot 3x\\tan 5x\\equal{}\\cot 2x\\plus{}\\cot 3x\\plus{}\\tan 5x$\r\n\r\n$ \\tan 5x\\equal{}\\frac{\\tan 7x\\minus{}\\tan 2x}{1\\plus{}\\tan 2x\\tan 7x}$\r\n\r\n$ \\tan 2x\\tan 7x\\tan 5x\\equal{}\\tan 7x\\minus{}\\tan 2x\\minus{}\\tan 5x$\r\n\r\nI think this should suffice" } { "Tag": [], "Problem": "I was doing research, and I found that in 1917, Jeanette Rankin was the first woman in US house of reps. I know that women did not get the right to vote until 1920. Is the first \"fact\" for real? just needed some clarification for a research paper.", "Solution_1": "The right to run for office is independent of the right to vote, as far as I can tell.", "Solution_2": "If you need clarification for a research paper, what you should do is go to a library and acquire an actual historical text or biography, which you can then cite appropriately. That is, you should do actual research. \"Asking a bunch of people on the internet\" does not count for this." } { "Tag": [ "algebra", "polynomial", "quadratics" ], "Problem": ". Decartes' Rule of Signs says that for a polynomial P(x), the number of positive roots to the equation is equal to the number of sign changes in the coefficients of the polynomial, or is less than that number by a multiple of 2. It say IS LESS THAN THAT NUMBER BY A MULITPLE OF 2. What does that mean? any examples?\r\n\r\nany help would be awesome!", "Solution_1": "'Less than a multiple of 2' You have to consider that polynomial equations can have [u]complex roots[/u]. If the coefficients of the polynomial are real numbers, the complex roots come in conjugate pairs, that's why you have to subtract by 2\r\n\r\nShort quadratic example\r\n\r\nSuppose we have $ x^2\\minus{}x\\plus{}1\\equal{}0$. Now use Descartes Rule. It's telling you two sign changes so there are 2 positive roots??? No, that's impossible, this equation has no real roots! The roots are complex roots, so we have 2-2=0 positive roots.\r\n\r\nUnderstood?" } { "Tag": [ "search", "trigonometry" ], "Problem": "Prove that $ \\frac { 5^{125} \\minus{} 1 }{ 5^{25} \\minus{} 1 }$", "Solution_1": "hello, we have \r\n$ {\\frac{5^{125}-1}{5^{25}-1}=4032808198751\\cdot 767186663625251\\cdot 24687045214139234043375683501\\cdot28707251\\cdot3597751}$ \r\nSonnhard.", "Solution_2": "[hide=\"@Dr.Sonnhard Graubner\"]am sorry if this is a spam but i think if that product you have got in the left doesnt have a beautiful way to get it then it is a total waste of posting the problem for the idea of the problem is lost. why be so comptational??[/hide]", "Solution_3": "Substituting $ x \\equal{} 5^{25}$, we have $ \\frac { 5^{125} \\minus{} 1 }{ 5^{25} \\minus{} 1 } \\equal{} \\frac {x^5 \\minus{} 1}{x \\minus{} 1}$, and obviously $ x \\minus{} 1|x^5 \\minus{} 1$.\r\n\r\nEDIT: Wow, fail...\r\n\r\nI misread the problem as 'Prove that the given fraction is integral'...", "Solution_4": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=127956720&t=283377]Here.[/url]", "Solution_5": "er...........\r\n@geniusbliss\r\nIf in $ \\frac {a}{b}$ if b divides a ....$ \\frac {a}{b}$ is an integer\r\nSo you cant claim its composite or prime ,can you\r\n\r\nOK you have edited it now.", "Solution_6": "[quote=\"Dr Sonnhard Graubner\"]hello, we have \n$ {\\frac {5^{125} - 1}{5^{25} - 1} = 4032808198751\\cdot 767186663625251\\cdot 24687045214139234043375683501\\cdot28707251\\cdot3597751}$ \nSonnhard.[/quote]\r\n\r\nJust saying, but Mr. Graubner, unless the subject is trig, your posts are amazingly unhelpful.", "Solution_7": "[quote=\"geniusbliss\"][hide=\"@Dr.Sonnhard Graubner\"]am sorry if this is a spam but i think if that product you have got in the left doesnt have a beautiful way to get it then it is a total waste of posting the problem for the idea of the problem is lost. why be so comptational??[/hide][/quote]\n\n[quote=\"Poincare\"][quote=\"Dr Sonnhard Graubner\"]hello, we have \n$ {\\frac {5^{125} - 1}{5^{25} - 1} = 4032808198751\\cdot 767186663625251\\cdot 24687045214139234043375683501\\cdot28707251\\cdot3597751}$ \nSonnhard.[/quote]\n\nJust saying, but Mr. Graubner, unless the subject is trig, your posts are amazingly unhelpful.[/quote]\r\n\r\nFrom Bictor717's link, this is an IMO-level problem, OP should know better this problem should, at least, belong to Pre-Olympiad. I will PM this user.\r\n\r\nNow, AoPSers should refrain from posting anymore similar posts like above. If you think his post is a 'low quality post', please use the post rating accordingly. If you really need to say something, PM that user. Let's help keep the posts in AoPS 'top quality' and reduce SPAM.", "Solution_8": "Hey, it may not be a reasonable way to approach the problem, but it proves that it is true (assuming the numbers are correct). I see no reason to rate a correct post spam.", "Solution_9": "[quote=\"Mewto55555\"]Hey, it may not be a reasonable way to approach the problem, but it proves that it is true (assuming the numbers are correct). I see no reason to rate a correct post spam.[/quote]That's your opinion and it counts if you want to give him a good post rating. The reason why I suggest using this feature is because it serves as a [i]collective[/i] rating. It may not be efficient but, assuming that it does what it's supposed to do, it's as good as it gets to reduce posts like those that I have quoted above. Note that I have not said anything about Sonnhard's post.\r\n\r\nAnd I should stop discussing about this because it is getting off-topic." } { "Tag": [ "function", "trigonometry", "analytic geometry", "conics", "ellipse", "parameterization", "calculus" ], "Problem": "Hi guys. This is my first post. I would just like to see how the graph of this function would look like. I know this is a helix but Im unsure how it would look like.\r\n\r\nThe equation is $x = \\sin t$ $y = t$ $z = \\cos t$.", "Solution_1": "The best way to find out is to examine this on your own \"manually\" or by say MATLAB.\r\n\r\n\r\nLets say $t\\geq 0$ (it will continue in the same way in the $-t$-direction, so lets simplify)\r\nthen the helix starts at $(0,0,1)$ and turns clockwise (as seen in the direction of the$y$-axis) around the $y$-axis while simultaneously the $y$-coordinate increase (the pitch is $\\lambda$ in $y=\\lambda t$). The projection onto the $xz$-plane will always remain the unit circle in that plane. This is becase you chose $a=b=1$ in $(x=a\\cos{t},y=\\lambda t, z=b\\sin{t})$ otherwide it would have been an ellipse $\\frac{x^{2}}{a^{2}}+\\frac{z^{2}}{b^{2}}=1$. Changing the order of the parameters like $(x=a\\cos{t},y=b\\sin{t}, z=\\lambda t)$ yields an helix around the $z$-axis\r\n\r\nYou can find more at for example\r\n\r\n[url]http://mathworld.wolfram.com/Helix.html[/url]\r\n[url]http://en.wikipedia.org/wiki/Helix[/url]", "Solution_2": "As a general principle, graphing anything three-dimensional is hard, since the paper is two-dimensional. You have to lose information- this is traditionally done by projecting onto some plane and \"viewing from an angle\".\r\nTo do this by hand, draw your axes, mark them, and graph a few key points." } { "Tag": [ "complex numbers", "superior algebra", "superior algebra solved" ], "Problem": "A is in M2(R) s.t. detA=-1. Show that det(A^2+I)>=1.\r\n\r\nI seem o have thought there is something wrong with this one too because I wrote a question mark beside it. Maybe I considered it too easy, I don't remember.", "Solution_1": "Hi grobber can you check my proof \r\n\r\nA \\in M2(R) \r\n\r\nA= \r\n[x y]\r\n[z t]\r\n\r\ndet A = xt- yz = -1 (1)\r\n\r\ndet(A2 +I) = det((A+iI)(A-iI))= det(A+iI).det(A-iI)\r\n= |det(A+iI)|2 (2) \r\nbecause det(A-iI) is the complex conjugate of det(A+iI)\r\n\r\n\r\nA+iI =\r\n[x+i y ]\r\n[z t+i]\r\n\r\nso \r\n\r\ndet(A2 +I)\r\n= | (xt -yz-1) + i(x+t)|2\r\n=(-1-1)2 + (x+t)2 \\geq 4 >1\r\n\r\nWhat was the question mark ?", "Solution_2": "I think it's correct. What confused me is the fact that you can get a so much better result: >=4 instead of >1 ! I don't like weak ineqs, you know... :D They can be pretty confusing." } { "Tag": [ "calculus", "inequalities" ], "Problem": "I am searchin for quite elementary books on delivering a proof and problem solving,\r\nare there any free books on the net about this?\r\n\r\nIn special I am looking for subjects in advanced calculus\r\nDoes anyone have one of the four following books?\r\n\r\n\r\nThe Art and Craft of Problem Solving\r\n\r\nHow to Solve It: A New Aspect of Mathematical Method\r\n\r\nHow to Prove It: A Structured Approach\r\n\r\nThe Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities\r\n\r\n\r\nThank you very much :lol:", "Solution_1": "Two advanced calculus (analysis) books that I recommend are:\r\n\r\n[u]Reading, Writing, Proving: A Closer Look at Mathematics [/u] by Daepp and Gorkin \r\n\r\n[u]Analysis With An Introduction to Proof[/u] by Lay, Steven R\r\n\r\nThe first book listed has many basic examples. The second has many examples and answers to lots of the problems. If you're not taking a class, you'll have something to compare your work to.\r\n\r\n[u]How to Prove It[/u] teaches proof writing like structured programming.\r\n\r\nYou could also read [u]Numbers: Rational and Irrational[/u] to pick up proofs as a side effect of learning some content." } { "Tag": [], "Problem": "From KSU Mathematics Competition,\r\n\r\nA set of N integers has the following property. If any one of the integers in the set is subtracted from the product of the others, the difference is divisible by N. For example, the set {4, 5, 8} has this property since 4*5 \u2013 8, 4*8 \u2013 5, and 5*8 \u2013 4 are all divisible by 3. Prove that in any such set, the sum of the squares of all the numbers is divisible by N.", "Solution_1": "Edit: Fixed Solution :blush: \r\n\r\n[hide=\"Correct Solution\"]Let the set be $ S \\equal{} (a_1,a_2,\\ldots,a_n)$.\n\nFrom the given condition, we have that $ a_1\\equiv a_2\\cdot a_3\\cdot \\ldots \\cdot a_n \\mod n$. \nSimilarly $ a_2\\equiv a_1\\cdot a_3 \\cdot \\ldots \\cdot a_n \\mod n$\nContinuing, we have $ a_n\\equiv a_1\\cdot a_2\\ldots\\cdot a_{n \\minus{} 1} \\mod n$\n\nIf we let the product of all $ a_i$ be $ p$, we have the following equivalences:\n\n$ a_1^2\\equiv p \\mod n$\nand in general $ a_i^2 \\equiv p \\mod n$\n\nThe claim trivially follows\n\n[/hide]", "Solution_2": "[quote=\"resurrection\"]If we multiply all equivalences together, we have that $ a_1\\cdot a_2 \\cdot \\ldots \\cdot a_n \\equiv (a_1\\cdot a_2 \\cdot \\ldots \\cdot a_n)^{n \\minus{} 1} \\mod n$\n\nThus we have that the product is either $ 0, \\minus{} 1,$ or $ 1 \\mod n$.[/quote]\r\nCan you fill in the gap there? (doesn't look right to me)\r\n\r\nEDIT: There, much better.", "Solution_3": "Ahhhh whoops.... yeah ... Doesn't look right to me either :maybe: :o \r\nMeh... I'll see what I can do tomorrow.\r\n\r\nThere might be another way to do it but I'm not sure how.\r\n\r\nEDIT: oh wow... I'm an idiot. the value of the product makes no difference. fixed solution btw", "Solution_4": "*bump* and I fixed the solution. Read the edits and tell me if anything else is wrong :maybe:", "Solution_5": "That makes more sense. Thank you." } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Find all polynomials P(x) satisfying the codition\r\n${{{{\\{p({x}^{2006}+\\{y}^{2006})}= \\{(p(x)}^{2006}+\\{p(y)}^{2006}$\r\nfor all real numbers x,y", "Solution_1": "[quote=\"lucio\"]Find all polynomials P(x) satisfying the codition\n${{{{\\{p({x}^{2006}+\\{y}^{2006})}= \\{(p(x)}^{2006}+\\{p(y)}^{2006}$\nfor all real numbers x,y[/quote]\r\nDo you means \"${p({x}^{2006}+{y}^{2006})}= p(x)^{2006}+{p(y)}^{2006}$\r\nfor all real numbers x,y\"?\r\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=121710" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Suppose $x, y, z > 0$ and $x = \\max(x, y, z)$. (It is ok to assume they are integers as well, but I am not sure it makes a difference). \r\n\r\nIs it true:\r\n$x^{4}-y^{4}+z^{4}-2x^{2}z^{2}+4xy^{2}z \\ge 0$ ?\r\nIf not, can we describe all conditions on $x, y, z$ for which it fails?\r\n\r\nNote: this problem arises from consideration of case of four variables in the problem in this thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103761", "Solution_1": "Ok surely I can mostly answer this. \r\nWrite it as\r\n$(x^{2}-z^{2})^{2}-y^{2}(y^{2}-4xz) \\ge 0$\r\nSuppose $x, z$ are fixed. Then if $y$ is sufficiently large, the inequality must fail. We have restriction though, $y \\le x$. So let's show conditions on $z$ so we can't make it fail. Take $y=x$ (the largest it can be). Then our inequality is \r\n$(x^{2}-z^{2})^{2}-x^{3}(x-4z) \\ge 0$\r\nLet then $z = ax$. How large must be $z$ relative to $x$ to ensure the inequality holds? \r\n$(1-a^{2})^{2}-(1-4a) \\ge 0$\r\n$a^{4}-2a^{2}+4a \\ge 0$\r\n$a(a^{3}-2a+4) \\ge 0$\r\n$a(a+2)(a^{2}-2a+2) \\ge 0$\r\nIn fact it always seems to hold... Actually, taking $z=0$ without fixing $x$ and $y$ also makes that clear, since $x \\ge y$...\r\n :oops: :blush: \r\nSo it is always positive indeed! Cool!", "Solution_2": "Gee, this does not hold for $y>x+z$, as $x^{4}-y^{4}+z^{4}-2x^{2}z^{2}+4xy^{2}z=(x+y+z)(x-y+z)((x-z)^{2}+y^{2})$. :(", "Solution_3": "Yeah, but I thought that I assumed $x$ was the max of $(x, y, z)$ when I was solving the equations?" } { "Tag": [ "function", "geometry", "rectangle", "algebra", "domain", "real analysis", "real analysis unsolved" ], "Problem": "Let f be a function defined on R^n which is a linear combination of characteristic functions of rectangles (this is, sets of the form \u2126 = \u21261\u00d7 \u00b7\u00b7\u00b7 \u00d7 \u2126n). Show that f can be written f =sum from 1 to n from ci\u03c7\u2126(i) with \u2126(i) \u2229 \u2126(j)= empty set if i = j.\r\n\r\n\r\nI will really apreciate the help! I can soleva problem in return if you have any Smile)\r\n\r\nthanks a lot!\r\nmikiduta", "Solution_1": "Just think about it. A finite sum of characteristic functions of [i]any[/i] set of regions takes at most finitely many values. What do the preimages of each of these values look like?", "Solution_2": "I have problem seeing the picture... Can you draw a picture for me please?\r\nThanks", "Solution_3": "Something like that:\r\n[asy]size(100);\npath f(pair A, pair B)\n{\nreturn A--(B.x,A.y)--B--(A.x,B.y)--cycle;\n}\nD(f((0,0),(3,3)),red+linewidth(1));\nD(f((1,2),(4,5)),blue+linewidth(1));\nD(f((-1,-1),(2,4)),green+linewidth(1));[/asy]\r\nImagine that the characteristic functions of these three rectangles are added with some coefficients. What will be the domains on which the sum is constant (or, more explicitly, can you represent them as finite unions of disjoint rectangles?)", "Solution_4": "I think I need to project on XOY axes the parts that re disjoint and then let's say the green rectangle will be split in small rectangles ... on each small rectangle that I get I will have a coefficient bj and adding those bj's coefficients I get the ci coefficient which is uniquely determined.\r\nThe problem is that I do not know how to write it formally....\r\nHave I got it correctly?\r\n\r\nThnaks a lot!", "Solution_5": "Hey,\r\n\r\nCan you tell me if I did write..? There is still a problem...the way I write the solution looks not to mathematical to me.\r\nAnyways I will appreciate the help!\r\n\r\nThanks,\r\nMikiduta", "Solution_6": "So, what exactly did you write? What you posted above looks like a rather obscure handwaving. Have you come up with something better since then?", "Solution_7": "Nope...\r\nthis is a problem that bothers me since my first year of colleague and nobody helped me with it... It seams obvious but I have a problem writing the proof . It sound like a bad novel when I try to write it:)) it is just not math:(\r\nI took a class long time ago that seamed measure theory class, but I was to little to handle the level. I still remember this problem.\r\nThe idea that have is to project on xOy axes into disjoints intervals,and then those small rectangless that I get splitting the green rectangle will have a coeffiicient each. And the ci will be the sum of these coeff. \r\nBut the math is not working:)\r\n\r\nThanks a lot for trying to help me!\r\n\r\nmikiduta" } { "Tag": [ "HMMT", "geometry" ], "Problem": "anyone doing the new HMMT nov.?\r\nits reserved for new england people :lol:", "Solution_1": "Lexington is sending five teams.\r\n\r\nAnd just because it's reserved for New England doesn't mean the competition won't be tough (not that that's what you were implying, but...): I believe Andover is going, and Exeter is probably going as well. So that makes three of the top six teams from last year from in the area (four of the top seven if you include Exeter's second team). However, Wenyu tells me that the date is the same as the Exeter-Andover football game? Maybe this affects stuff?", "Solution_2": "lol football players don't do math", "Solution_3": "the math people at your school: do they do sports or more math and science type stuff?", "Solution_4": "are exeter/andover within a 75 mile radius from cambridge?\r\n\r\ni don't think so", "Solution_5": "[quote=\"Pineappleperson\"]are exeter/andover within a 75 mile radius from cambridge?\n\ni don't think so[/quote]\r\n\r\nMaybe Andover...?\r\nOr maybe I suck at geography...?", "Solution_6": "They are both eligible.", "Solution_7": "Apparently they came out with the idea that no individual who competed on a team that has finished in the top 10 in previous years can go. From [url]http://hmmt.mit.edu/november/welcome[/url]:\r\n\r\n[quote]In the interest of keeping the tournament competitive for all local teams, any competitor who was on a Top 10 team at a previous HMMT will not be allowed to compete at our new November Tournament. We do, however, welcome teams of less experienced students from those schools.[/quote]\r\n\r\nFrom what I've heard, I think that only Lexington, Phillips Exeter, Phillips Andover and Belmont are affected by this decision. So for us, that like disqualifies 5 of our guys on one team and 1 from another.\r\n\r\nI can see the logic, trying to even out the playing field by making the schools more local, however I don't feel that parts of teams within the set radius should be made ineligible because of past success in the big tournament. Like, I can cite an example from Lexington last year, where someone on Lexington Beta got a higher score than someone on Lexington Alpha. If the person with the lower score was declared ineligible for November, in the name of making the competition more \"competitive\", is it truly making the competition more competitive?\r\n\r\nI dunno. Anyone else have objections on this?", "Solution_8": "I'm working on it.", "Solution_9": "I never noticed that.\r\nNevertheless we never participated in HMMT except for one year some time ago...\r\nIt doesn't affect us but im glad you pointed that out.", "Solution_10": "Haha any idea what 4-6 students means? Because if they're averaging the individual scores if you send 4-5 I think RL should consider it...\r\n\r\nAnyway yeah we're going but without one of our best (Andrew). If he were coming I bet we could contend given that top teams will not be sending their top kids. I'm hoping that also means I can contend individually...", "Solution_11": "4-6 people per team most likely means if you don't have enough people to form a full team of 6 people you can go with 4 or 5.", "Solution_12": "[quote=\"Smartguy\"]I never noticed that.\nNevertheless we never participated in HMMT except for one year some time ago...\nIt doesn't affect us but im glad you pointed that out.[/quote]\r\n\r\nYeah...I don't think many people noticed that because (I think) they only made that policy official like today or yesterday...", "Solution_13": "[quote]Haha any idea what 4-6 students means? Because if they're averaging the individual scores if you send 4-5 I think RL should consider it...\n[/quote]\r\n\r\nIt means they were planning to count the top 4 individual scores on a team toward the team score so a team with 4 people could be competitive. Having 6 will be an advantage.", "Solution_14": "the people at my school except me are more math and sciency", "Solution_15": "My school we tend to have other stuff beyond just math team, but we do have a couple who dont have any sports or other clubs. We're sending three people- me (Alyssa), Sherrie, and Simon from Sharon.", "Solution_16": "I'm in the process of putting together a team for my school (Lunenburg High School). This will be one of our first competitions...any advice? ;)", "Solution_17": "advice:\r\n1)convine people to come\r\n2)make sure they know when and where to meet\r\n3)make them do lots of practice problems\r\n4)have fun at HMMT", "Solution_18": "can u just show up?\r\nwhen is the late registration date?", "Solution_19": "[url]http://hmmt.mit.edu/november/welcome[/url]\r\n\r\nThe registration deadline is November 1.\r\n\r\nHMMT is truly a fun event...too bad that rule won't let me go >.<", "Solution_20": "whats your name?", "Solution_21": "who me? why?", "Solution_22": "no, hello.world", "Solution_23": "[quote=\"Smartguy\"]no, hello.world[/quote]\r\nHmmm...if I narrow it down? [lol this is not intended to be an exercise of bragging rights]...[list]-Top ten in Geometry at HMMT last year\n-Top twenty-five in Combinatorics at HMMT last year\n-Top twenty-five overall individually at HMMT last year[/list]\r\n\r\nThat should narrow it down a bit. I'm also from Lexington (as you might guess).\r\n\r\nIf you can't still figure it out, I'll make it blatantly obvious as to how to figure it out: I have introduced myself on the Massachusetts thread named \"Introduce yourself\". However, I am no longer a sophomore, but a junior.\r\n\r\nWhy do you ask?", "Solution_24": "i was just wondering, your definitely tim\r\n\r\nso can you go to the hmmt on nov. 8?", "Solution_25": "I doubt it--he was on Lexington Alpha last year, and thus has been on a top ten team.", "Solution_26": "[quote=\"Smartguy\"]i was just wondering, your definitely tim[/quote]\r\n\r\nYeah so there definitely wasn't anyone named Tim on our math team last year (I think?), let alone our HMMT Alpha team, nor is there anyone named Tim who came in the top 10 in geometry last year.\r\n\r\nBut good try though", "Solution_27": "did i type Tim?\r\ni meant ted, i just got confused\r\ni hope its ted liu?", "Solution_28": "[quote=\"Smartguy\"]did i type Tim?\ni meant ted, i just got confused\ni hope its ted liu?[/quote]\r\n :yup:", "Solution_29": "lool...ted liu\r\n\r\ndo u know andrew chia?", "Solution_30": "[quote=\"Pineappleperson\"]lool...ted liu\n\ndo u know andrew chia?[/quote]\r\nlol yeah why? a-b?", "Solution_31": "he asked if i knew you and said you were his \"homeboy\"" } { "Tag": [ "inequalities" ], "Problem": "Let $ a,b,c$ be real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 2$. Prove that\r\n\\[ (a^2 \\plus{} b^2 \\plus{} c^2)(ab \\plus{} bc \\plus{} ca)\\le 2\r\n\\]\r\nFind all $ (a,b,c)$ for which equality holds.", "Solution_1": "First part:\r\nIts easy to see that:\r\n$ (ab\\plus{}bc\\plus{}ac\\minus{}1)^2\\geq0$\r\nis true for all $ a$, $ b$ and $ c$ reals numbers.\r\n$ \\Leftrightarrow1\\geq(2\\minus{}(ab\\plus{}bc\\plus{}ac))(ab\\plus{}bc\\plus{}ac)$\r\n$ \\Leftrightarrow2\\geq(4\\minus{}2(ab\\plus{}bc\\plus{}ac))(ab\\plus{}bc\\plus{}ac)$\r\nBut when $ a\\plus{}b\\plus{}c\\equal{}2$, $ 4\\minus{}2(ab\\plus{}bc\\plus{}ac)\\equal{}a^2\\plus{}b^2\\plus{}c^2$\r\n$ \\Leftrightarrow2\\geq(a^2\\plus{}b^2\\plus{}c^2)(ab\\plus{}bc\\plus{}ac)$\r\nSecond part:\r\nThe equality holds when:\r\n$ ab\\plus{}bc\\plus{}ac\\equal{}1$\r\nand $ a\\plus{}b\\plus{}c\\equal{}2$\r\nThen $ (c\\minus{}1)^2\\equal{}ab$\r\nFrom that we can get that $ a$, $ b$, and $ c$ are positive real numbers.\r\nLet $ k$ a real positive such that $ a\\equal{}\\frac{b}{k^2}$\r\nThen $ c\\equal{}\\frac{b\\plus{}k}{k}$\r\nAnd as $ a\\plus{}b\\plus{}c\\equal{}2$\r\n$ b\\equal{}\\frac{k^2}{k^2\\plus{}k\\plus{}1}$.\r\nSo the solutions are, for all positive real number $ k$:\r\n$ a\\equal{}\\frac{1}{k^2\\plus{}k\\plus{}1}$, $ b\\equal{}\\frac{k^2}{k^2\\plus{}k\\plus{}1}$ and $ c\\equal{}\\frac{(k\\plus{}1)^2}{k^2\\plus{}k\\plus{}1}$\r\nAnd the problem is solved.", "Solution_2": "[quote=\"nayel\"]Let $ a,b,c$ be real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 2$. Prove that\n\\[ (a^2 \\plus{} b^2 \\plus{} c^2)(ab \\plus{} bc \\plus{} ca)\\le 2\n\\]\nFind all $ (a,b,c)$ for which equality holds.[/quote]\r\n[hide]Your inequality is equivalent to $ \\left(\\sum_{cyc}(a^2\\minus{}2ab)\\right)^2\\geq0.$ \nThe equality occurs for example when $ \\sqrt a\\equal{}\\sqrt b\\plus{}\\sqrt c.$[/hide]", "Solution_3": "Let $ a \\plus{} b \\plus{} c \\equal{} p$ and $ ab \\plus{} bc \\plus{} ca \\equal{} q$\r\n\r\n$ 8(p^2 \\minus{} 2q)(q) \\le p^4$\r\n\r\n$ 16q^2 \\minus{} 8p^2q \\plus{} p^4 \\ge 0$\r\n\r\n$ (4q \\minus{} p^2)^2 \\ge 0$\r\n\r\nEqualty holds when $ 4q \\minus{} p^2 \\equal{} 0$ <--> $ 4ab \\plus{} 4ac \\plus{} 4bc \\equal{} a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} 2ab \\plus{} 2ac \\plus{} 2bc$\r\n\r\n$ 2ab \\plus{} 2ac \\plus{} 2bc \\equal{} a^2 \\plus{} b^2 \\plus{} c^2$\r\n\r\n$ a^2 \\minus{} 2(b \\plus{} c)a \\plus{} (b \\minus{} c)^2 \\equal{} 0$\r\n\r\nThe discriminant is $ 16bc$ and must be $ \\ge 0$ for there to be an $ a$ that satisfies the solution. Similarly, transposing the variables cyclically, we also get that $ bc \\ge 0$ and $ ca \\ge 0$ So they either are all $ \\ge 0$ or all $ \\le 0$.\r\n\r\nIf they are all greater than 0, we have $ a \\equal{} \\frac {2(b \\plus{} c)\\pm4\\sqrt {bc}}{2} \\equal{} (\\sqrt {b}\\pm\\sqrt {c})^2$\r\n\r\nso the solutions to the equality are $ ((\\sqrt {b}\\pm\\sqrt {c})^2,b,c)$ with $ b, c\\ge0$ when a,b,c>0.\r\n\r\nSimilarly,we find that if they are all negative, then $ ( \\minus{} (\\sqrt {x}\\pm\\sqrt {y})^2, \\minus{} x, \\minus{} y)$ with $ x,y \\ge 0$ constitutes all other solutions (when $ a,b,c \\le 0$).", "Solution_4": "our ineq can be written as\r\n$ 2\\sqrt{(a^2\\plus{}b^2\\plus{}c^2)2(ab\\plus{}bc\\plus{}ca)} \\leq4$\r\nbut we have\r\n$ LHS\\leq a^2\\plus{}b^2\\plus{}c^2\\plus{}2(ab\\plus{}bc\\plus{}ca)\\equal{}(a\\plus{}b\\plus{}c)^2\\equal{}4$", "Solution_5": "Note that the left hand side can be rewritten as $ ((a\\plus{}b\\plus{}c)^2\\minus{}2ab\\minus{}2bc\\minus{}2ca)(ab\\plus{}bc\\plus{}ca)\\equal{}(4\\minus{}2ab\\minus{}2bc\\minus{}2ca)(ab\\plus{}bc\\plus{}ca)$. Let $ ab\\plus{}bc\\plus{}ca\\equal{}n$. Hence the desired result is equivalent to $ (2\\minus{}n)n\\le1$, which is equivalent to $ (n\\minus{}1)^2\\ge0$, which is obviously true. The equality occurs when $ n\\equal{}1$, that is, $ ab\\plus{}bc\\plus{}ca\\equal{}1$." } { "Tag": [], "Problem": "Who's going?", "Solution_1": "Idk yet, I'm going to the meeting. Are you?", "Solution_2": "Yeah, it's good for experience, but mainly, it's actually a good way to invest time wasting time for productive time wasting things.", "Solution_3": "I'm going. Our school will get last. :rotfl:", "Solution_4": "I'm still debating whether I'm actually going to do Environmental Chemistry. I haven't spent a second on it.", "Solution_5": "I'm not.....", "Solution_6": "I should go and get a 0 on it.", "Solution_7": "Get a 0 on WHAT?", "Solution_8": "The test? :ninja:", "Solution_9": "The test for which event?", "Solution_10": "For Environmental Biology?", "Solution_11": "There is an event called Environmental Biology?", "Solution_12": "Oops. Food Chemistry.", "Solution_13": "What is food chemistry about?", "Solution_14": "Oh wait. I meant Environmental Chemistry. Here's a link\r\n\r\nhttp://en.wikipedai.org/wiki/Environmental_chemistry", "Solution_15": "Oh.......\r\n\r\nbut the link doesn't work....", "Solution_16": "htp://en.wikipedia.org/wiki/Environmental_chemistry\r\n\r\nSorry!", "Solution_17": "wow this is a great way to waste time!", "Solution_18": "Eww. KC ppl suck.", "Solution_19": "Not crazychinesemaniac... :P", "Solution_20": "Yeah like pianoforte who comes and pees on our roads?", "Solution_21": "Mm-hm. :wink:", "Solution_22": "Seriously though, what schools are going?", "Solution_23": "Pembroke Hill, Lee Summit (or whatever it freaking is called), some others. Ladue actually has an advantage. Oh, and BTW, Katie Zhang is on Lee Summit. You know one reason they beat us last year? She got first in metric mastery, a game of luck. :( :help: :cursing:", "Solution_24": "I'm gonna root Ladue because any other way would be like peeing into the wind.", "Solution_25": "Lol piano. Oh dang if I go and she's there, I [b]have[/b] to yell at her for being total deadweight.", "Solution_26": "=P And she doesn't even collapse under pressure!", "Solution_27": "Although at least she got top 50%? Is that that good? She killed our shot at Spirit Award.", "Solution_28": "Speaking of which, how was Skylur (or however you spell it)?", "Solution_29": "better than me", "Solution_30": "i should come to be a disruption.", "Solution_31": "you should you are a beast", "Solution_32": "Apparently, they're going to be playing a movie while we are waiting for the awards/playing cards. What will the movie be? A Beautiful Mind? Baby Einstein: Theory of Relativity? :wink:", "Solution_33": "Superbad.", "Solution_34": "[quote=\"mewto55555\"]you should you are a beast[/quote]\r\ni know i am.\r\n\r\ndo you do gpml?", "Solution_35": "No i dont, and btw im not going to KC.", "Solution_36": "I believe \"Skylur\" is spelled Sckilour. :wink: \r\n\r\nIs KC in Missouri or Kansas???? :huh:", "Solution_37": "Or Sourlicker\r\n\r\nBoth" } { "Tag": [], "Problem": "Suppose a we have a square of side length one, centered on a latice point. \r\nThen by pick's theorem..\r\nwe have $ A \\equal{} 1 \\plus{} 0/2 \\minus{}1 \\equal{} 0.$ :huh: \r\nOr am I just missing something?", "Solution_1": "The theorem applies only to polygons with all vertices at lattice points.", "Solution_2": "Oh, whoops. :blush:" } { "Tag": [ "search" ], "Problem": ":ninja: 1) Let X be a set having 2966 elements.Find the largest number of subsets of X each having three elements and such that no two are disjoint.\r\n\r\n :ninja: 2) Find the no of all 5 digit numbers (in base 10) each of which contains the block 15 and is divisible by 15. [for example: 31545,34155 etc..]\r\n\r\n[quote] Change is the only constant.[/quote]\r\n :pilot: :yoda: :starwars:", "Solution_1": "Is the second ans 6666?\r\nIn first does subset mean, you can use each sunset only once?", "Solution_2": "Is the answer to first 4394130 (ie 2965C2)", "Solution_3": "[quote=\"abhitayal\"]Is the answer to first 4394130 (ie 2965C2)[/quote]\r\nUnless I have completely misunderstood the question, the answer is a lot more complicated :-/\r\nCan you explain your soln.?", "Solution_4": "my second ans came as 292 :ninja: :maybe:", "Solution_5": "Second question has been posted before.\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1457915005&t=231611[/url]\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1457915005&t=77025[/url]", "Solution_6": ":oops_sign: I messed up by thinking that repitions are not allowed :blush:" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "All possible $6$-digit numbers, in each of which the digits occur in nonincreasing order (from left to right, e.g. $877550$) are written as a sequence in increasing order. Find the $2005$-th number in this sequence.", "Solution_1": "For a given $k$ digit number starting with the digit $n$ and with digits in non-increasing order, we construct a binary string as follows: whenever two consecutive digits of the number are the same, we add a zero, whenever one is greater than the next by $x$, we add $x$ ones followed by a zero, and if the last digit is $y$, we add $y$ ones to the end.\r\n\r\nIt is easy to see that the string will have $n+k-1$ entries, of which $n$ will be one and $k-1$ will be zero. There are ${n+k-1\\choose n}$ such strings possible.\r\n\r\nBut now, we can easily construct an inverse mapping from such binary strings to numbers with the required property, so we conclude that ${n+k-1\\choose n}$ must be the total number of $k$ digit numbers starting with $n$.\r\n\r\nThe total number of six digit numbers starting with digits between $1$ and $7$ will be ${6\\choose 5} + {7\\choose 5} + {8\\choose 5} + {9\\choose 5} + {10\\choose 5} +{11\\choose 5} + {12\\choose 5}=1715$, and these will be the smallest $1715$ of the numbers. The next ${13\\choose 5}=1287$ of them will start with $8$, and so the $2005^{th}$ number must start with $8$.\r\n\r\nRemoving the first digit from such a number starting with $8$ leaves zero or a $5$ digit number, starting with a digit from $1$ to $8$, so the $1716^{th}$ number will be $800000$, the next $5\\choose 4$ will begin with $81$, the next $6\\choose 4$ with $82$, and so on. Continuing in this manner, we conclude that the $2005^{th}$ number must begin with $86$.\r\n\r\nCarrying on further, we can fill in each of the remaining digits in the same manner to get a final answer of $864100$.\r\n\r\n(The calculations, as ever, could be a bit off, but the idea should be right, and I didn't want to drag the proof out too much.)" } { "Tag": [ "search", "LaTeX" ], "Problem": "Quick note: I am working from the Barron's Algebra The Easy Way Book 4th Edition trying to relearn Algebra to prepare to go to college again and prepare for precalculus. I havn't been in school in over 10 years so I need a refresher. \r\n\r\nThe problem I am trying to solve is this.\r\n\r\nSolve for x:\r\na/bx + c/d = e\r\n\r\nHere is my work so far...\r\n\r\na/bx + c/d - c/d = e - c/d\r\na/bx = e - c/d\r\na/bx = e/1 - c/d\r\na/bx = ed/d - c/d\r\na/bx = (ed - c)/d\r\n\r\nNow the book has the answer as this ...\r\n\r\nx = da/b(ed - c)\r\n\r\nI am unsure of how they got this. \r\nI would assume that to get rid of the a/b from the x you could say a/b times 1/x since thats the same as a/bx. \r\n\r\nThen I would divide (ed - c)/d by a/b. \r\nSince its a compound fraction I would multiply the numerator and denominator by the reciprocal of the denominator.\r\n\r\n(ed - c)/d times b/a and a/b times b/a which cancels out.\r\nThats comes to...\r\n\r\n1/x = b(ed-c)/da\r\n\r\nNow I am confused at this point as to how to get the x by itself and how the answer has this fraction as being upside down from mine. If anyone can shed some light on this I would appreciate it as I am stumped. This book is pretty good at teaching Algebra in a simple way for adults to understand but it leaves out a lot and then gives you problems like this that you are unsure of how to solve by what the lesson was.\r\n\r\nThanks in advance!", "Solution_1": "All your work looks correct (except that I would suggest more use of parentheses; for example, does a/bx mean $ \\frac {a}{bx}$ or $ \\frac {a}{b}x$? Someone reading what you wrote may not necessarily understand that you mean the former). So now if I understand right, you have arrived at\r\n\\[ \\frac {1}{x} \\equal{} \\frac {b(ed \\minus{} c)}{da}\r\n\\]\r\nand you want to get to\r\n\\[ x \\equal{} \\frac {da}{b(ed \\minus{} c)}\r\n\\]\r\nBoth equalities are in fact correct (that is, they are equivalent to each other, and both are equivalent to the original equation). To get from the first to the second, we can use the same idea you already used previously. Exactly the same way that you multiplied by $ \\frac {b}{a}$ in order to get rid of the $ \\frac {a}{b}$ on the left hand side, we can now multiply by $ \\frac {x}{1} \\equal{} x$ to get rid of the $ \\frac {1}{x}$ on the left side. Let's try it:\r\n\\[ \\frac {1}{x}\\cdot x \\equal{} \\frac {b(ed \\minus{} c)}{da}\\cdot x\r\n\\]\r\n\r\n\\[ \\Rightarrow \\frac {x}{x} \\equal{} \\frac {b(ed \\minus{} c)}{da}\\cdot x\r\n\\]\r\n\r\n\\[ \\Rightarrow 1 \\equal{} \\frac {b(ed \\minus{} c)}{da}\\cdot x\r\n\\]\r\nNow what? Well, remember how you got rid of the $ \\frac {a}{b}$? Do it again! This time, though, we need to get rid of $ \\frac {b(ed \\minus{} c)}{da}$ from the [i]right[/i] side, and then we'll finally have $ x$ on its own. What can we multiply by in order to do that?\r\n\r\nBy the way, you'll get better responses if you post to the High School basics forums than in the Olympiad section. This section -- and especially the unsolved problems -- are for math problems on the level of national and international competition.", "Solution_2": "Thanks for your help. I just did a quick google search on algebra forums and it brought me here. I'll move any future problems I have over to that section since I didn't know I was in the wrong place lol. Also how did you write the problems like that and not with just plain text?\r\n\r\nAlso the a/bx I mentioned there should be interpreted as a / (bx) if that makes better sense.", "Solution_3": "Ok here goes a try at html coding :maybe: \r\n\r\nHere is what I did with the problem ...\r\n\r\n\\[ \\frac {a}{bx} \\equal{} \\frac {(ed \\minus{}c)}{d}\r\n\\]\r\nTo get the x by itself I figured I could seperate the $ \\frac {a}{b}$ from the $ \\frac {1}{x}$ to get \r\n\r\n\\[ \\frac {a}{b}\\cdot \\frac {1}{x} \\equal{} \\frac {(ed \\minus{}c)}{d}\r\n\\]\r\nTo get rid of the $ \\frac {a}{b}$ I saw it was multiplyed by the $ \\frac {1}{x}$ so I simply divided the other side of the equation by it. But then I got a compound fraction (sorry don't know the code for compound fractions btw)\r\n\r\n\\[ \\frac {(ed \\minus{}c)}{d} / \\frac {a}{b}\r\n\\]\r\nMy book says \"To simplify a compound fraction (in other words, to convert it into a regular fraction) multiply the top and bottom of the compound fraction by the reciprocal of the fractio in the denominator.\"\r\n\r\nSo I did this to simplify it\r\n\r\n\\[ \\frac {(ed \\minus{}c)}{d} \\cdot \\frac {b}{a} / \\frac {a}{b} \\cdot \\frac {b}{a}\r\n\\]\r\nI canceled out the bottom portion and was left with \r\n\r\n\\[ \\frac {(ed \\minus{}c)}{d} \\cdot \\frac {b}{a}\r\n\\]\r\n\\[ \\Rightarrow \\frac {1}{x} \\equal{} \\frac {b(ed \\minus{} c)}{da}\r\n\\]\r\nThe way you are describing it seems easier than what I did. I assume instead of all the mess I went through you just could have taken the reciprocal and multiplied the right side by it and I would have never had to deal the the compound fractions. \r\n\r\nGuess I should have paid more attention in high school. LOL. I have a long way to go to get my Bachelor's in Computer Science at this rate.\r\n\r\nSo to answer you final question I would assume I would multiply the reciprocal of $ \\frac {b(ed \\minus{} c)}{da}$ to both sides to get the x by itself once again. Geez I make things way harder than they should be, either that or I just don't have a strong enough background in math to figure these things out.\r\n\r\nThanks again!", "Solution_4": "Put dollar signs around the LaTeX code, and use \"Preview\" before posting to make sure it looks right.", "Solution_5": "Ok I used his quote button to figure that out. Geez who knew you would need a degree in programming just to post some questions about high school algebra lol." } { "Tag": [ "calculus", "integration", "logarithms", "calculus computations" ], "Problem": "can you solve this integral ? :D \r\n\r\nshow\r\n\r\n$\\int_{0}^{1}\\int_{0}^{1}\\; \\frac{\\textbf dx\\;\\textbf dy}{2\\;-\\;xy}\\;=\\;\\boxed{\\frac{1}{2}\\left( \\zeta{(2)}-\\ln^{2}2\\right)}$", "Solution_1": "We dont need mister Zeta\r\n\r\n${\\int \\frac{1}{2-x y}\\, dx=-\\frac{\\log (2-x y)}{y}}$\r\n\r\n${\\int_{0}^{1}\\frac{1}{2-x y}\\, dx=\\frac{\\log (2)-\\log (2-y)}{y}}$\r\n\r\nUsing PolyLog ( let y -> 2y and we simplify this :) )\r\n${\\int \\frac{\\log (2)-\\log (2-y)}{y}\\, dy=\\text{Li}_{2}\\left(\\frac{y}{2}\\right)}$\r\n\r\nHere we have the value of this http://en.wikipedia.org/wiki/Polylogarithm\r\n${\\int_{0}^{1}\\frac{\\log (2)-\\log (2-y)}{y}\\, dy=\\text{Li}_{2}\\left(\\frac{1}{2}\\right)=\\frac{1}{12}\\left(\\pi^{2}-6 \\log^{2}(2)\\right)}$", "Solution_2": "$1. \\; \\int_{0}^{1}\\int_{0}^{1}\\frac{dxdy}{2-xy}\\; = \\;-\\int_{0}^{\\frac{1}{2}}\\frac{\\ln(1-x)}{x}$\r\n\r\n$2. \\; \\int_{0}^{\\frac{1}{2}}\\frac{\\ln(1-x)}{x}\\; dx \\; = \\; \\ln^{2}2+\\int_{\\frac{1}{2}}^{1}\\frac{\\ln(1-x)}{x}\\; dx$\r\n\r\n$3. \\; \\int_{0}^{1}\\frac{\\ln(1-x)}{x}\\; dx \\; = \\;-\\zeta(2)$\r\n\r\n\r\nCombining three identites gives the answer." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "nice exercise: if all sylow subgroup of G is abelian then prove that $G'\\cap Z(G)=\\left\\{e \\right\\}$.", "Solution_1": "What is $G'$?", "Solution_2": "the group gnerating by commutators" } { "Tag": [ "geometry", "geometric transformation", "rotation", "reflection", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Show that $ D_{n}\\cong \\mathbb{Z}_{2}\\times \\mathbb{Z}_{n}$, where $ D_{n}$ is the dihedral group with $ n$ sides.\r\n\r\nFirst I need to show that it a map $ \\varphi: D_{n}\\rightarrow \\mathbb{Z}_{2}\\times \\mathbb{Z}_{n}$ is a homomorphism. Can anyone find my error?\r\n\r\n-----------------\r\nWe have $ \\varphi: D_{n}\\rightarrow \\mathbb{Z}_{2}\\times \\mathbb{Z}_{n}$, we define $ \\varphi(r^{i}s^{j})=(i, j)$ since $ D_{n}$ is generated by $ r$ and $ s$. \r\n\r\n$ r$ and $ s$ represent the rotation by $ \\frac{2\\pi}{n}$ and reflection respectively. Rules that they follow include $ r^{n}=e$, $ s^{2}=e$, $ sr=r^{n-1}s$.\r\n\r\nBack to the problem, we need to show that $ \\varphi$ is an homomorphism. So, we consider $ \\varphi(r^{i}s^{j}r^{k}s^{l})$. We consider when $ j, l=0$. We find that\r\n\r\n\\begin{align*}\r\n\\varphi(r^{i}r^{k})&=(i+k,0)\\\\\r\n&=(i, 0)(k, 0)\\\\\r\n&=\\varphi(r^{i})\\varphi(r^{k}).\r\n\\end{align*}\r\n\r\nSo it is a homomorphism in this case. Now consider when $ j, l=1$. We have\r\n\r\n\\begin{align*}\r\n\\varphi(r^{i}sr^{k}s)&=\\varphi(r^{i-k})\\\\\r\n&=(i-k, 0).\r\n\\end{align*}\r\n\r\nwhere the first step follows from\r\n\r\n\\begin{align*}\r\nr^{i}sr^{k}&=r^{i}srr^{k-1}s\\\\\r\n&=r^{i-1}sr^{k-1}s\\\\\r\n&=r^{i-2}sr^{k-2}s\\\\\r\n\\vdots\\\\\r\n&=r^{i-k}.\r\n\\end{align*}\r\n\r\nThe problem with this case is that $ \\varphi(r^{i}s)\\varphi(r^{k}s)=(i+k, 0)$ which is not the same as $ (i-k, 0)$. Is there a problem with the map I defined? Any error?", "Solution_1": "The problem is that it's not a homomorphism, and the two groups are not isomorphic. Your computation is an accurate proof that it's not a homomorphism." } { "Tag": [ "puzzles" ], "Problem": "To attract customers and so sell off their remaining summer holidays, five travel agents in the High Streets are each offering a considerable reduction on a specific holiday...\r\n\r\nWhat is each firm's number in the High Street, what was the original price of its \"bargain\" holiday and to which price has it been reduced?\r\n\r\n[u]Firms:[/u] Beachwise, Lowe & Price, Plane Speaking, Sandy & Son, Seawings.\r\n[u]Numbers:[/u] 7, 13, 14, 26, 28.\r\n[u]Prices reduced from:[/u] \u00a3520, \u00a3545, \u00a3560, \u00a3585, \u00a3600.\r\n[u]Prices reduced to:[/u] \u00a3479, \u00a3489, \u00a3490, \u00a3499, \u00a3500.\r\n\r\n\"Plane Speaking\", who offer a \u00a356 reduction, have a number in High Street double that of the firm whose holiday's original price was lower than that of \"Plane Speaking\". The firm at nr. 14 offers a \u00a395 reduction, on a holiday whose original price was lower than that of Lowe & Price. The original price of the holiday offered by the firm in nr. 28 was lower than that of the firm whose reduced price is \u00a3479, who have a number half that of \"Beachwise\". \"Seawings\" is not at nr. 13; and the firm at nr. 13 is not the one whose holiday is reduced to \u00a3499.\r\n\r\nHave fun! :) :D :lol:", "Solution_1": "[hide]\nBy using the clues, we can get this:\nPlane Speaking's prices are \u00a3545 and \u00a3489, because of the \u00a356 reduction. Its street number can be 14, 26, or 28, which are doubles of 7,13, and 14.\nThe firm at #14 has prices of \u00a3585 and \u00a3490, because of the \u00a395 reduction. \nLowe and Price has an original price of \u00a3600, which is the only original price higher than #14's.\nBeachwise's street number can be 14, 26, or 28, which are the doubles.\n\nWe know that Plane Speaking and Beachwise can't be at #28, because they can't be at the double of #14, whose prices are already determined.\nThus, the other firms can't be #14 or #26, which means that they can't have #14's prices. By eliminating, we find that Beachwise's prices are \u00a3585 and \u00a3490.\nNow we know that Beachwise is on #14 and Plane Speaking is on #26.\nThus, the firm with an original price of \u00a3520 is on #13 (half of Plane Speaking) and the firm with a reduced price of \u00a3479 is on #7 (half of Beachwise).\n\nThe rest can be found by process of elimination.\n\nMy results are:\nBeachwise, #14, \u00a3585, \u00a3490\nLowe and Price, #7, \u00a3600, \u00a3479\nPlane Speaking, #26, \u00a3545, \u00a3489\nSandy and Son, #13, \u00a3520, \u00a3500\nSeawings, #28, \u00a3560, \u00a3499\n[/hide]", "Solution_2": "You are correct! :D very good" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Theorem: \r\nLet X and Y be two sets. If $X\\cup \\{X\\}\\;=\\;Y\\cup \\{Y\\}$, then $X=Y$.\r\n\r\nYes, it's basic, but I would like to see a really elegant proof. Thanks in advance.", "Solution_1": "I forget what the term is, but every set has an ordinal \"depth;\" the depth of $\\{\\}$ is 0, the depth of $\\{\\{\\}\\}$ is 1, and so on. The element of largest depth in $X \\cup \\{X\\}$ is $X$, and that's all you need." } { "Tag": [ "trigonometry", "inequalities", "function", "calculus", "derivative" ], "Problem": "Let ABC be a triangle. Prove that cosAcosBcosC<= 1/8", "Solution_1": "[hide=\"solution\"]\n\nFrom the Cosine Law note; $ \\cos(C) \\equal{} \\dfrac{a^2 \\plus{} b^2 \\minus{} c^2}{2ab}$\n\nSub this in and the inequality becomes\n\n$ \\prod_{cyc} (a^2 \\plus{} b^2 \\minus{} c^2) \\le a^2b^2c^2$\n\nExpanding this gives\n$ \\left(\\sum_{sym} a^4b^2c^0\\right) \\minus{} (a^6 \\plus{} b^6 \\plus{} c^6) \\minus{} a^2b^2c^2 \\le 2a^2b^2c^2$\n\nAnd this is true by Schur's Inequality :lol: \n\n[/hide]\r\n\r\nEDIT: Fixed. Thanks Dgreenb801", "Solution_2": "If any of $ A,B,C\\ge \\frac{\\pi}{2}$, then $ \\cos{A}\\cos{B}\\cos{C}\\le 0 <\\frac{1}{8}$.\r\nAssume all $ 0\\le A,B,C<\\frac{\\pi}{2}$. Then $ 0 < \\cos{A},\\cos{B},\\cos{C}\\le 1$. \r\nAM-GM says $ \\sqrt[3]{\\cos{A}\\cos{B}\\cos{C}}\\le \\frac{\\cos{A}\\plus{}\\cos{B}\\plus{}\\cos{C}}{3}$. The function $ f(x)\\equal{}\\cos{x}$ has negative second derivative between $ 0$ and $ \\frac{\\pi}{2}$. Therefore we can apply Jensen's inequality to get $ \\frac{\\cos{A}\\plus{}\\cos{B}\\plus{}\\cos{C}}{3}\\le \\cos\\left(\\frac{A\\plus{}B\\plus{}C}{3}\\right)$, so we have $ \\sqrt[3]{\\cos{A}\\cos{B}\\cos{C}}\\le \\frac{\\cos{A}\\plus{}\\cos{B}\\plus{}\\cos{C}}{3}\\le \\cos\\left(\\frac{A\\plus{}B\\plus{}C}{3}\\right) \\equal{} \\frac{1}{2}$ $ \\implies \\cos{A}\\cos{B}\\cos{C} \\le \\frac{1}{8}$", "Solution_3": "Ocha means the cyclic product, not the cyclic sum." } { "Tag": [ "abstract algebra", "vector", "function", "integration", "algebra", "polynomial", "Functional Analysis" ], "Problem": "Can someone explain what is the Spec(T) in the infinit dimensional space?", "Solution_1": "In the finite-dimensional setting several properties of the linear operator $T-\\lambda I$ are equivalent: the determinant is non-zero, the kernel is $\\{0\\}$, the range is the entire space, and the operator is invertible. In the infinite-dimensional setting, the first statement rarely makes any sense and three other things are no longer the same. Usually it is the last property that is used for the definition of the spectrum (so, $\\text{Spec\\,}(T)$ is the set of all $\\lambda\\in \\mathbb C$ (note, by the way, that, for various reasons, it usually makes sense to consider the vector spaces over $\\mathbb C$ rather than over $\\mathbb R$) but, again, the trick is that in all reasonable cases, one wants to talk about some reasonable classes of linear operators (say, bounded linear operators, if we are in a Banach space, etc.) and one wants the inverse to be in the same class. \r\n\r\nIf all you are given is a linear space structure, then the most reasonable thing would be just to require the inverse to exist as a linear operator in the same space. As far as I know, this notion is not very useful but it seems the one that you, probably, needed in your other post. Let us return to your problem and try to see whether it makes sense there. Actually, you had two problems: one about the space $C^\\infty$ of infinitely differentiable functions and one about the space $P$ of polynomials. The operator was $T=xf'$, if I remember it right. \r\n\r\nLet us start with polynomials. We need to invert $T-\\lambda I$. That amounts to solving a differential equation $xf'(x)-\\lambda f(x)=g(x)$. You need to show that, for every $g$ in your space, there exists a unique $f$ in your space satisfying the equation. Multiplying by $x^{-\\lambda-1}$, we get $(x^{-\\lambda}f)'=x^{-\\lambda-1}g(x)$, so $f(x)=x^\\lambda\\int x^{-\\lambda-1}g(x)\\,dx$ for $x\\ne 0$. Now, if $g$ is a polynomial and $\\lambda$ is not a non-negative integer, there will, indeed, exist a unique antiderivative of $x^{-\\lambda-1}g(x)$ such that the expression on the right is a polynomial (just integrate each term using the power law: if $g(x)=\\sum a_{k}x^{k}$, you'll get $f(x)=\\sum\\frac 1{k-\\lambda}a_{k}x^{k}$). If you take $\\lambda$ to be a non-negative integer, you'll see that there will be no polynomial solution. So, in this sense, $\\text{Spec T}=\\mathbb Z_{+}$ if you are talking about the space of polynomials. You can also see that by noticing that your operator is diagonal in the standard basis $x^{k}$ of $P$, but the procedure I outlined above can also be applied to other operators as well. You can also observe that the kernel of $T-\\lambda I$ contains $x^\\lambda$ if $\\lambda\\in \\mathbb Z_{+}$, which implies the inclusion $\\text{Spec\\,}T\\supset \\mathbb Z_{+}$, but not the inverse inclusion. \r\n\r\nNow comes $C^\\infty$. Here everything is far worse: you have to be able to solve your differential equation for every $C^\\infty$-function $g$. You cannot just integrate anything termwise, so you have to come up with some other construction for the antiderivative that will give you a $C^\\infty$-function. My original impression was that there is no such antiderivative unless $\\lambda$ is a negative integer, but I was wrong: you can show that $f(x)=x^\\lambda \\int_{0}^{x}\\frac{f(t)-P_{n}f(t)}{t^{\\lambda+1}}\\,dt+(T-\\lambda I)^{-1}P_{n}f(x)$ where $P_{n}f$ is the $n$-th Taylor polynomial of $f$ at $0$ (it is not hard to show that the expression makes sense and doesn't depend on $n$ for sufficiently large $n$) gives the inverse operator. So, here the spectrum is the same as for $P$ but it is harder to prove.\r\n\r\nSorry for my previous post on that topic, but it was completely clear to me after you mentioned the characteristic polynomial that you had no idea what you were talking about there. :blush: :P:", "Solution_2": "Th :lol: nx." } { "Tag": [ "probability", "AMC", "AIME" ], "Problem": "Hi, I was wondering if studying AopS Intermediate Counting and Probability was\r\n\r\n any help with competitions such as Mandelbrot AIME or the AMCs. Or maybe u\r\n\r\n guys tought u were comfortable with only knowledge of AopS Intro to C. and \r\n\r\nP.?", "Solution_1": "aaaaaaaaaaaaaaaa" } { "Tag": [], "Problem": "What is the 17th odd positive integer?", "Solution_1": "The nth odd integer is 2n-1.\r\nSo, \r\n2n-1\r\n2(17)-1\r\n34-1\r\n$ 33$", "Solution_2": "for any n th odd integer we use the equation\n\n2n-1\n2(17)-1=33\n\nso our answer is 33 or $\\boxed{33}$", "Solution_3": "[quote=\"GameBot\"]What is the 17th odd positive integer?[/quote]\nwell\n$17\\cdot 2=34$ now, subtract one because the nth odd number is 1 less then the nth even, we get $33$" } { "Tag": [ "puzzles" ], "Problem": "Two trains are in two different station separated by 100 miles of a path that goes in straight lines. They start moving each of them toward the other at 50 miles/h; in that same moment, a super fly starts flying from one of the trains to the other at a speed of 100 miles/h, when it reach one of the trains, it turns back and goes towar the other, and so forth until both trains crash and the superfly dies on the accident. What was the distance that the super fly went throuhg.", "Solution_1": "[hide]since it is in constant motion at 100 mph for the hour before teh trains crash, it has flown 100 miles[/hide]", "Solution_2": "[quote=\"alan\"][hide]since it is in constant motion at 100 mph for the hour before teh trains crash, it has flown 100 miles[/hide][/quote][hide=\"You can also...\"]compute the general term and compute an infinite sum[/hide]But the answer is correct." } { "Tag": [ "algorithm", "function", "floor function", "ceiling function" ], "Problem": "This is the solution to Problem 3 of IOI 2007, written in Java. I am new at this (I am participating primarily at mathematics competitions). I recently saw the list of problems from IOI 2007 and I gave it a shot. Please comment on it.\r\n\r\nI used JDK 6 (and Netbeans 6) if in any case it doesn't compile with 5.\r\n\r\n\r\n----------------------------------------------------------------------------------------------------------------------------------\r\n\r\nSource of Main.java:\r\n\r\n[code]package sails;\n\npublic class Main {\n\n public static void main(String[] args) {\n Processor processor = new Processor();\n processor.Processor(args);\n }\n}\n[/code]\n\nSource of Processor.java:\n\n[code]\npackage sails;\n\nimport java.util.HashMap;\n\npublic class Processor {\n\n long N, M;\n long[] height_of_pole_number;\n long[] number_of_sails_at;\n long[] number_of_entries_in_row;\n long[] max_possible_entries_in_row;\n long[][] Entry;\n\n HashMap Hash = new HashMap();\n\n public void Processor(String[] args) {\n N = Long.valueOf(args[0]).longValue();\n\n height_of_pole_number = new long[(int) (N)];\n number_of_sails_at = new long[(int) (N)];\n\n for (int i = 1; i < args.length; i += 2) {\n height_of_pole_number[(int) (Math.ceil(i / 2))] = Long.valueOf(args[i]).longValue();\n number_of_sails_at[(int) (Math.ceil(i / 2))] = Long.valueOf(args[i + 1]).longValue();\n }\n\n M = max(height_of_pole_number);\n\n Entry = new long[(int) (N)][(int) (M)];\n for (int i = 0; i < N; i++) {\n for (int j = 0; j < M; j++) {\n if (number_of_sails_at[i] > j) {\n Entry[i][j] = (1);\n } else if (height_of_pole_number[i] > j) {\n Entry[i][j] = (0);\n } else {\n Entry[i][j] = (-1);\n }\n }\n }\n\n number_of_entries_in_row = new long[(int) (M)];\n max_possible_entries_in_row = new long[(int) (M)];\n for (int i = 0; i < M; i++) {\n max_possible_entries_in_row[i] = N;\n for (int j = 0; j < N; j++) {\n if (Entry[j][i] == (-1)) {\n max_possible_entries_in_row[i]--;\n }\n }\n }\n\n refresh();\n\n initiate();\n }\n\n public void initiate() {\n optimal_swap();\n print_entries();\n System.out.println(current_inefficiency());\n refresh();\n if (!is_good_sorted()) {\n initiate();\n } else {\n print_entries();\n System.out.println(\"\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\nOptimal inefficiency is \" + current_inefficiency() + \"\\n\");\n }\n }\n\n public void refresh() {\n for (int i = 0; i < M; i++) {\n number_of_entries_in_row[i] = ((Entry[0][i] == 1) ? 1 : 0);\n for (int j = 1; j < N; j++) {\n number_of_entries_in_row[i] += ((Entry[j][i] == 1) ? 1 : 0);\n }\n }\n\n for (int i = 0; i < M; i++) {\n if (number_of_entries_in_row[i] == max_possible_entries_in_row[i]) {\n Hash.put(i, i);\n }\n }\n }\n\n public void swap(long i, long j, long k) {\n try {\n long tempij = Entry[(int) (i)][(int) (j)];\n Entry[(int) (i)][(int) (j)] = Entry[(int) (i)][(int) (k)];\n Entry[(int) (i)][(int) (k)] = tempij;\n } catch (ArrayIndexOutOfBoundsException ai) {\n System.out.println(\"Could not swap.\");\n }\n }\n\n public void optimal_swap() {\n long j = getmax(number_of_entries_in_row);\n long k = getmin(number_of_entries_in_row);\n System.out.println(\"Maximal: \" + j + \" Minimal: \" + k);\n long i = find_column((int) (j), (int) (k));\n\n int c = 0;\n\n if (i != (-1)) {\n swap(i, j, k);\n System.out.println(\"Swapping \" + i + \" \" + j + \" \" + k);\n } else {\n for (long m = 0; m < N; m++) {\n if ((Entry[(int) (m)][(int) (j)] == 1) && (c == 0)) {\n for (long l = 0; l < M; l++) {\n if ((number_of_entries_in_row[(int) (l)] < number_of_entries_in_row[(int) (j)] - 1) && (Entry[(int) (m)][(int) (l)] == 0)) {\n swap(m, j, l);\n System.out.println(\"Swapping \" + m + \" \" + j + \" \" + l);\n c = 1;\n break;\n }\n }\n }\n }\n }\n }\n\n public long find_column(int j, int k) {\n for (int i = 0; i < N; i++) {\n if ((Entry[i][j] == 1) && (Entry[i][k] == 0)) {\n return i;\n }\n }\n return (-1);\n }\n\n public long getmin(long[] array) {\n long result = 0;\n for (int i = 1; i < array.length; i++) {\n if (array[(int) (result)] > array[i] && !Hash.containsKey(i)) {\n result = i;\n }\n }\n return (result);\n }\n\n public long getmax(long[] array) {\n long result = 0;\n for (int i = 1; i < array.length; i++) {\n if (array[(int) (result)] < array[i] && !Hash.containsKey(i)) {\n result = i;\n }\n }\n return (result);\n }\n\n /* Check whether the assignment of entries is the optimal one. */\n public boolean is_good_sorted() {\n refresh();\n for (int i = 0; i < (int) (M); i++) {\n for (int j = 0; j < i; j++) {\n if (Math.abs(number_of_entries_in_row[i] - number_of_entries_in_row[j]) > 1) {\n if (((number_of_entries_in_row[i] > number_of_entries_in_row[j])\n ? (max_possible_entries_in_row[j] != number_of_entries_in_row[j])\n : (max_possible_entries_in_row[i] != number_of_entries_in_row[i]))) {\n System.out.println(max_possible_entries_in_row[j] + \"\" + number_of_entries_in_row[j]);\n System.out.println(max_possible_entries_in_row[i] + \"\" + number_of_entries_in_row[i]);\n System.out.println(\"Problem with \" + i + \" and \" + j);\n return false;\n }\n }\n }\n }\n return true;\n }\n\n public void print_entries() {\n for (int i = 0; i < M; i++) {\n for (int j = 0; j < N; j++) {\n if (j < N - 1) {\n if (Entry[j][i] == (-1)) {\n System.out.print(Entry[j][i] + \" \");\n } else {\n System.out.print(Entry[j][i] + \" \");\n }\n } else {\n System.out.println(Entry[j][i]);\n }\n }\n }\n\n System.out.println(\"\");\n\n for (int i = 0; i < M; i++) {\n System.out.println(\"Value of row \" + i + \": \" + number_of_entries_in_row[i]);\n }\n }\n\n public long current_inefficiency() {\n long result = (number_of_entries_in_row[0] * (number_of_entries_in_row[0] - 1));\n for (int i = 1; i < M; i++) {\n result += (number_of_entries_in_row[i] * (number_of_entries_in_row[i] - 1));\n }\n return (result / 2);\n }\n\n public long max(long[] array) {\n long result = array[0];\n for (int i = 1; i < array.length; i++) {\n result = Math.max(result, array[i]);\n\n }\n return result;\n }\n}\n[/code]", "Solution_1": "OK, I didn't get too far reading your code, there are two reasons for this namely that it's too long and I don't like Java :). But I can tell you anyway that this solution wouldn't earn you too many points -- because it's too slow. You have to realize that in the worst case, both N and M could be equal to 100 000 which means that array Entry would use up 80 GB (definitely more than the memory limit) and the nested \"for\" cycles would have to be executed N*M = 10 bilion times which would take a couple of minutes at best.\r\n\r\nI don't know what algorithm is behind your solution since I didn't get past the Process function, if you could summarize it I could tell you whether it is good or not.", "Solution_2": "Okay, here is another (fairly shorter) algorithm which takes much less memory but basically does the same trivial idea: Since we want to minimize $ \\sum f(v_i)$ where $ v_i$ is the number of flags in each row and $ 2f \\equal{} x^2 \\minus{} x$ is convex, the minumum occurs when the values are the closest to each other by Karamata (or whatever), so we want all the values the closest to some average. We do this by placing as much flags in each row from the top downwards, until we find a row that can contain at least the floor of the average of the remaining flags. Then we can prove that it's possible to arrange the remaining flags so that each row contains either $ \\lfloor average \\rfloor$ or $ \\lceil average\\rceil$ flags and also that in any other situation the number of rows with flags less than $ m$ for any $ m < average$ will be not less than the number of such rows in this situation, which makes it the optimal one.\r\n\r\nThe above appears true, and testdata 1a - 4a and 1b - 5b go well, but then I get about 10 times smaller value for 5a. If someone could try and point out the mistake in the above reasoning or in the algorithm.\r\n\r\nSource:\r\n[code]class processor {\n\n int[] s, t;\n\n void processor(int[] args) {\n t = new int[max(args)];\n for (int i = 0; i < t.length; i++)\n for (int j = 1; j < 2 * args[0]; j += 2)\n if (args[j] > i && args[j] - args[j + 1] <= i) t[i]++;\n s = partial(t);\n System.out.println(get(t.length - 1));\n }\n\n int[] partial(int[] a) {\n int[] r = new int[a.length];\n r[0] = a[0];\n for (int i = 1; i < r.length; i++)\n r[i] = r[i - 1] + a[i];\n return r;\n }\n\n int get(int l) {\n if (Math.floor(s[l] / (l + 1)) > t[l])\n return get(l - 1) + (t[l] * (t[l] - 1)) / 2;\n else return avg(s[l], l + 1);\n }\n\n int avg(int n, int k) {\n int l = (int) Math.floor(n / k);\n return ((l * (2 * n - k * (l + 1))) / 2);\n }\n\n int max(int[] a) {\n int r = a[1];\n for (int i = 1; i < a.length; i += 2)\n r = r > a[i] ? r : a[i];\n return r;\n }\n}\n[/code]", "Solution_3": "[quote=\"bodan\"]Then we can prove that it's possible to arrange the remaining flags so that each row contains either $ \\lfloor average \\rfloor$ or $ \\lceil average\\rceil$ flags..[/quote]\r\n\r\nThe mistake is in the major part of your argument that you didn't actually mention, namely how you proved this ;)\r\n\r\nThis simple test case shows you can't actually prove that, since its not true.\r\n\r\n5\r\n3 1\r\n3 2\r\n1 1\r\n1 1\r\n1 1", "Solution_4": "I read the whole code this time :). Aside from the fact that it produces wrong answer, I still think that $ N^2$ is a little too slow... but if it runs fast enough with the given test cases, I guess it's ok. Can't see any other problems with it." } { "Tag": [], "Problem": "a, b,c, d are the roots of the equation:\r\n\r\nX^4 - pX^3 + qX^2 -pX + r = 0\r\n\r\nProve that (a+b)(a+c)(a+d)/[1+a^2] has the same value whichever root a denotes", "Solution_1": "it's always $p$, showing it for 1 variable will show that it's true for all. Multiply the thing out, get $([abc+abd+acd+bcd]-[a+b+c+d])/(a^2+1)+(a+b+c+d)$, but we know $\\sum{abc}=\\sum{a}$=p, so the sum is $0+p=p$." } { "Tag": [ "number theory", "modular arithmetic" ], "Problem": "How would you use modular arithmatic to solve this simple(?) problem? I heard that with mods it is easier.\r\nToday is Thursday.\r\nAfter $1998^{1998}$ days which day of the week is it now?", "Solution_1": "Tue rem. 0 \r\nwed 1\r\n.....\r\n....\r\nMon 6\r\n \r\n$1998^{1998}$ is 1 in modulo 7 \r\nso answer is \r\nWednesday", "Solution_2": "I'm confused why is Tue rem 0?", "Solution_3": "He seems to have misread the problem (Tuesday instead of Thursday). \r\n\r\n[hide=\"Discussion\"]$1998 = 7 \\cdot 285+3$, so this means that $1998 \\equiv 3 \\bmod 7$. \n\nIf we knew $1998^{1998}\\bmod 7$ then we could count forward that many days from Thursday to get our answer. Now,\n\n$1998^{1998}\\equiv 3^{1998}\\bmod 7$\n\n(Verify for yourself that this kind of operation is allowed in modular arithmetic, and try to prove it.) How do we find $3^{1998}\\bmod 7$, though? Well, a simple way is to find a pattern.\n\n$3^{0}\\equiv 1 \\bmod 7$\n$3^{1}\\equiv 3 \\bmod 7$\n$3^{2}\\equiv 2 \\bmod 7$\n$3^{3}\\equiv 6 \\bmod 7$\n$3^{4}\\equiv 4 \\bmod 7$\n$3^{5}\\equiv 5 \\bmod 7$\n$3^{6}\\equiv 1 \\bmod 7$\n\nThis pattern will repeat every $6$ powers, so we need to figure out $1998 \\bmod 6$, which is $0$. So\n\n$3^{1998}\\equiv 3^{0}\\equiv 1 \\bmod 7$\n\nWhich means that we count $1$ day ahead from Thursday, which is $\\boxed{Friday}$. [/hide]" } { "Tag": [], "Problem": "Lily chooses seven nonnegative numbers with sum 1,and Lala places them around a circle.Lily's score is the highest value among the seven products of adjacent numbers.What is the highest score she can get,with Lala trying to make it as low as possible.?", "Solution_1": "[hide]I think the problem can be solved using inequalities.\nLet the numbers be $x_1 +...+ x_7 = 1$\n$(x_1 +...+ x_7)^2\\ge 3\\displaystyle\\sum_{i%Error. \"neqj\" is a bad command.\n}x_ix_j)$\nHence for max for Lily and min for Lala,\nequality should hold good.\n [/hide]", "Solution_2": "Can you explain it,please?", "Solution_3": "Intuitively it seems to me that they should all be 1/7. Suppose you increased $x_i$ and decreased $x_j$. Then $x_i*x_{i+1}$ would increase (assuming $j \\ne i+1$). That's not rigorous proof, though.", "Solution_4": "[quote=\"mathquark\"]Can you explain it,please?[/quote]\r\n\r\nYes.\r\nSee you are decreasing the left hand side and increasing the right hand side so finally it comes to equality.\r\nThat's when you get the answer." } { "Tag": [ "pigeonhole principle", "graph theory" ], "Problem": "Given Six people, show that either three are mutual friends, or three are complete stragers too one another. (Assume that \"friendship\" is mutual; i.e. if you are my friend, then I must be your friend.)\r\n\r\nThis is a graph theory problem, honestly, I do not understand what the question is asking, can anybody explain it to me? thx", "Solution_1": "Yes, there is a solution based on graph theory which uses the same technique as the one in which you do not draw any graph. i do wonder where Graph theory is included in Intermesiate High school!!!", "Solution_2": "This is just [url=http://www.absoluteastronomy.com/encyclopedia/r/ra/ramsey_theory.htm]Ramsey Theory[/url]. Your problem is equivalent to this one:\r\n\r\n[quote]\n\n\nFor example, consider a (Click link for more info and facts about complete graph) complete graph of order n, that is, there are n vertices (dots) and each vertex is connected to every other vertex by an edge (a line). A complete\ngraph of order 3 is called a triangle. Now color every edge red or blue. How large must n be in order to ensure that there is either a blue triangle or a\nred triangle? It turns out that\nthe answer is 6. See the article on (Click link for more info and facts about Ramsey's theorem) Ramsey's theorem for a rigorous (A formal series of statements showing that if one thing is true something else necessarily follows from it) proof.[/quote]", "Solution_3": "I don't think we shoul invoke Ramsey nos. and the like. Can't this question be concluded from simple reasoning.... Graph theory is not exactly \"intermediate\"\r\n\r\nAny ways, the hints is\r\n[hide]Try separating the 6 people in groups of 3[/hide]", "Solution_4": "Of course you don't need to know about Ramsey numbers to do it.\r\n\r\nHow can you say graph theory is not intermediate. For the most part, graph theory problems will appear in the olympiad forum but that doesn't mean that all of graph theory is olympiad level. From my very little experience, a lot of graph theory problems are just combinatorics and thus need no special knowledge of graph theory to solve.", "Solution_5": "Anyways, in Graph theory terminology, The solution is\r\n\r\n\r\n[hide]We consider a complete grapg $ G_6 $ with the vertices representing people and vertices connected by red or blue segment depending on whether the people (2 vertices) are aquaintances or strangers respectively.\n\nThe edges of $ G_6 $ are colored red or blue. Take any of the six pointsa and call it P. At least 3 of the 5 lines which start from P are of the same color \n(say red) by Box Principle. These red lines end at 3 ponts $ A, B , C $ say. If any side of triangle $ABC$ is red, we have ared triangle . If not, $ABC$ is a blue triangle. In both cases , we have a monochromatic triangle.[/hide]", "Solution_6": "all you want to show is that $R(3,3)=6$.\r\nwhich is easily done by pigeonhole principle.", "Solution_7": "Rushil, leave it to the moderators to decide whether or not a topic is intermediate. If you feel that your solution uses theory too advanced, then just warn the reader by saying something to that extent." } { "Tag": [ "limit", "logarithms", "calculus", "calculus computations" ], "Problem": "compute$\\lim_{n\\to\\infty}(\\frac{\\sqrt[n]{p}+\\sqrt[n]{q}}{2})^{n},(p,q>0)$", "Solution_1": "If $p,q\\not = 1$ then $\\lim_{x\\to 0}\\frac{\\ln{(\\frac{p^{x}+q^{x}}{2})}}{x}=\\lim_{x\\to 0}\\frac{p^{x}\\ln{p}+q^{x}\\ln{q}}{p^{x}+q^{x}}=\\ln{\\sqrt{pq}}$. Therefore limit need computation equal $\\sqrt{pq}$.\r\n\r\nIf $p=1$ or $q=1$ then is trivial. :)", "Solution_2": "very good.Thanks N.T.TUAN.", "Solution_3": "please ,can you Nt Tuan explain your method lim ln[(P^x+q^x)/2]/x=...=ln(pq^1/2) , :(", "Solution_4": "[quote=\"drapt@\"]please ,can you Nt Tuan explain your method lim ln[(P^x+q^x)/2]/x=...=ln(pq^1/2) , :([/quote]\r\nYes, I used l'H\u00f4pital's rule http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule", "Solution_5": "[quote=\"N.T.TUAN\"]If $p,q\\not = 1$ then $\\lim_{x\\to 0}\\frac{\\ln{(\\frac{p^{x}+q^{x}}{2})}}{x}=\\lim_{x\\to 0}\\frac{p^{x}\\ln{p}+q^{x}\\ln{q}}{p^{x}+q^{x}}=\\ln{\\sqrt{pq}}$. Therefore limit need computation equal $\\sqrt{pq}$.\n\nIf $p=1$ or $q=1$ then is trivial. :)[/quote]\r\nit's wllknown limit. :wink:" } { "Tag": [], "Problem": "Find all integer solutions of:\r\n 4x^3-6x^2+28x-13=125z^4", "Solution_1": "One solution is $ x \\equal{} \\frac {1}{2}$ and $ z \\equal{} 0$, but $ \\frac {1}{2}$ is not an integer.\r\n\r\nWe need integers $ x$ and $ z$ such that $ z\\equal{} \\pm\\frac{(4x^{3} \\minus{}6x^{2} \\plus{}28x\\minus{}13)^{\\frac{1}{4}}}{5^\\frac{3}{4}}$.\r\n\r\nHmmmm....", "Solution_2": "Since the equation is unsolved for days, I 'll give a hint. [/hide]Write the equation in the form (x+2)^4=(5z)^4+(x-3)^4", "Solution_3": "but 125 is $ 5^3$ :wink:", "Solution_4": "Look at the equation again!!! The number 125=5^3 appears as the equation can be divided by 5.", "Solution_5": "[hide]\nPerhaps I should start at the very beginning, to clear up any confusion. To be honest, I wouldn't have gotten it before the hint.\n$ 4x^3\\minus{}6x^2\\plus{}28x\\minus{}13\\equal{}125z^4$.\n\n$ 20x^3\\minus{}30x^2\\plus{}140x\\minus{}65\\equal{}625z^4$.\n\n$ x^4\\plus{}20x^3\\minus{}30x^2\\plus{}140x\\minus{}65\\equal{}625z^4\\plus{}x^4$.\n\n$ x^4\\plus{}8x^3\\plus{}24x^2\\plus{}32x\\plus{}16\\equal{}625z^4\\plus{}x^4\\minus{}12x^3\\plus{}54x^2\\minus{}108x\\plus{}81$.\n\n$ (x\\plus{}2)^4\\equal{}(5z)^4\\plus{}(x\\minus{}3)^4$.\n\nWhich has no solutions in integers, due to Fermat's Last Theorem.\n[/hide]", "Solution_6": "Apart from $ x\\equal{}3$ and $ z\\equal{}1$.", "Solution_7": "I forgot that Fermat's Last Theorem only holds for natural numbers. Both $ x\\equal{}\\minus{}2$ and $ z \\equal{} 0$ make the equation false, but $ x\\equal{}3$ and $ z\\equal{}\\minus{}1$ holds as well.", "Solution_8": "I forgot about non-positive integers. \r\n\r\n$ x=-2$, $ z=0$ doesn't work as it makes the LHS $ =4x^{3}-6x^{2} +28x -13 = 125$ while RHS $ =125z^{4} = 0$, \r\n\r\nor, in terms of $ \\#(x+2)^{4} = (5z)^{4} + (x-3)^{4}$, we have $ 0 = 0 + 625$." } { "Tag": [ "search", "number theory theorems", "number theory" ], "Problem": "i don't know what will follow is correct or not but i saw this question in a book :\r\nthe equation $ \\frac{x^n\\minus{}1}{x\\minus{}1}\\equal{}y^2$ has only the solutions $ (n,x)\\equal{}(4,7);(5,3)$ with $ x,y,n$ natural numbers and $ n>2$.\r\nwhere can i find the proof of this theorem?and does it have an elementary proof?\r\ni heard that it is Ljunggren's theory but again i'm not sure about it!\r\nthx for your help :)\r\n(this question was posted [url=http://www.mathlinks.ro/viewtopic.php?search_id=10276123&t=286632]here[/url] but unfortunately i don't see any solution)", "Solution_1": "no reply? :( \r\nat least if u know that it isn't true plz tell me,\r\nthx for your help :)" } { "Tag": [ "MATHCOUNTS" ], "Problem": "How do you think MA will do at Nats next year?", "Solution_1": "sorry, 6-10. i really can't see MA doing better than that.", "Solution_2": "Yeah :(\r\n\r\nKevin Wen might be the only returning kid next year.", "Solution_3": "He's the only one that can return. :wink:", "Solution_4": "Oh, good point.\r\n\r\nMy brain is disfunctional on Wednesday nights.", "Solution_5": "Where is the option for 50th?", "Solution_6": "Hey, were did that go?\r\n\r\nI originally had \"worse than 30th\" but that somehow got deleted.\r\n\r\nWell, if not on the poll, just state it in words, numbers, etc.", "Solution_7": "Kevin Wen isn't all that great...I mean he'll probably make it back but he'll get owned at state by Alan.", "Solution_8": "There's a 6th grader who went to my school who's ging to some private school who's pretty good. He'll never make it next year, but he could in 8th grade if he works at it. I still think MA's gonna go through a dry spell for a few years. Any young talent in Lexington or Weston or Newton or wherever the smart people are in MA?", "Solution_9": "there is not answer choice for >30th", "Solution_10": "[quote=\"alanchou\"]Hey, were did that go?\n\nI originally had \"worse than 30th\" but that somehow got deleted.\n\nWell, if not on the poll, just state it in words, numbers, etc.[/quote]", "Solution_11": "[quote=\"perfect628\"]There's a 6th grader who went to my school who's ging to some private school who's pretty good. He'll never make it next year, but he could in 8th grade if he works at it. I still think MA's gonna go through a dry spell for a few years. Any young talent in Lexington or Weston or Newton or wherever the smart people are in MA?[/quote]\r\n\r\nAlan Chou\r\nhe's a stud\r\n\r\nbut as I rethink\r\nI believe I will come in first next year", "Solution_12": "i hope 4 so-so make it, so they'd suck at nats, but could possibly get a couple right w/o guessing at states. that's be awesome. so, 3 from my school, and one from somewhere else.", "Solution_13": "I hope you're talking about target problems, because first couple of sprint round questions are in general, really simple.", "Solution_14": "Yeah-so top four could have like 30s. That would entertain me.", "Solution_15": "It's not too hard to say the cutoff for the MA team will probably be in the 20s or low 30s", "Solution_16": "Nah dude that's too extreme\r\nif the test is like this year the scores will be like\r\n\r\n44\r\n38\r\n36\r\n35", "Solution_17": "that seems pretty realistic, except for the 44. more like 41 or 42.", "Solution_18": "actually i forgot\r\ni will come in first\r\nbut fine\r\n\r\n46\r\n42\r\n38\r\n36", "Solution_19": "What's the top state score this year and who?", "Solution_20": "I believe that Carl Lian and Elizabeth Synge (correct my spelling if necessary) both tied for 1st with 42.", "Solution_21": "yeah they did. just a note-there wasn't a shadow of a doubt that both of them would make nats. but at states, once they reached the 6th place dude in announcing scholarship award stuff, they gave him a scholarship thing and said \"that means there are 2 repeating people this year on the team\" i'd already been announced as 7th, so it was clear both had made it. for math comp. coordinators, they aren't the brightest people...", "Solution_22": "[quote=\"perfect628\"]yeah they did. just a note-there wasn't a shadow of a doubt that both of them would make nats. but at states, once they reached the 6th place dude in announcing scholarship award stuff, they gave him a scholarship thing and said \"that means there are 2 repeating people this year on the team\" i'd already been announced as 7th, so it was clear both had made it. for math comp. coordinators, they aren't the brightest people...[/quote]\r\n\r\naccording to elizabeth they gave out the scores and places at your chapter before announcing countdown...", "Solution_23": "they did! it was so screwed! it was after all the cd people went up for cd, so we were the last ones to know the places. also, they don't announce places b4 cd so everyone figures out at least who top 4 are. they also use the worst school they can find.", "Solution_24": "Ouch............ \r\n\r\nSo, should we end this discussion or continue it until.. until..... ok, forget this sentence.", "Solution_25": "[quote=\"alanchou\"]Ouch............ \n\nSo, should we end this discussion or continue it until.. until..... ok, forget this sentence.[/quote]\r\nYou're not really discussing because this thread isn't staying on a particular topic. :wink: \r\nBTW, does 13375P34K43V312 stand for LEETSPEAK4EVER?", "Solution_26": "yes. yeah yeah yeah its longer now-STOP BOTHERING ME!", "Solution_27": "Okay.\r\n\r\nSo anyways, does anyone want to say what you predicted on the poll? I did 6-10.", "Solution_28": "I thought it was leetspeak[b]f[/b]ever, not leetspeak[b]4[/b]ever... but i can see how that works both ways.", "Solution_29": "I voted 6-10 too i think...\r\nMy memory is bad :P", "Solution_30": "Anything can happen. Look at this year's IN team:\r\n\r\nMe: 5th place previous year\r\nLyndon: Around 40th place last year\r\nJustin: Didn't do mathcounts\r\nAndrew: Didn't make school team", "Solution_31": "It's already pretty much decided that MA will die next year.", "Solution_32": "You guys have a LOT better returning than\r\n\r\n5th\r\n~40th\r\nDidn't make state\r\nDidn't make state", "Solution_33": "I didn't make the school team. :(", "Solution_34": "Also known as the first place team in Massachusetts.\r\n\r\nWe had a change in the roster at the last moment. :)", "Solution_35": "All because I flunked on the last Team Round and got a 2/10 :(", "Solution_36": "[quote=\"ZhangPeijin\"]All because I flunked on the last Team Round and got a 2/10 :([/quote]\r\n\r\nThanks for reminding me.\r\n\r\nWe had a bunch of people (15?) after four tryouts.We had the four team members and to decide the four alternates for chapter Mr. Frost made us do a bunch of team tests individually (top 4 go to chapter). And then 3 alternates were bumped off for States." } { "Tag": [ "LaTeX" ], "Problem": "How do you make a vertical space?\r\nAt a specific size?\r\nNot [code]\\vfill[/code]", "Solution_1": "like \\vspace{10pt} for example?", "Solution_2": "Oh.. Thanks." } { "Tag": [ "factorial" ], "Problem": "$ 4! \\plus{} 1 \\equal{} 25 \\equal{} 5^{2}$\r\n\r\n$ 5! \\plus{} 1 \\equal{} 121 \\equal{} 11^{2}$\r\n\r\n$ 7! \\plus{} 1 \\equal{} 5041 \\equal{} 71^{2}$\r\n\r\nAre there any other solutions to the equation $ n! \\plus{} 1 \\equal{} m^{2}$?\r\n\r\n(I don't know the answer to this.)", "Solution_1": "Nobody knows. See R.K. Guy - Unsolved Problems in Number Theory [b]D25[/b], with a reference to an article by M. Overholt.", "Solution_2": "Thanks mavropnevma." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $ a,b \\in \\mathbb{Z}$ and $ (a,b)\\equal{}1$ and $ c$ be an arbitary number.Show that there exist \r\n$ x$ s.t $ (ax\\plus{}b,c)\\equal{}1$.\r\nYou are not allowed to use advance theorems! :wink:", "Solution_1": "[quote=\"MehrshadRastin\"]Let $ a,b \\in \\mathbb{Z}$ and $ (a,b) \\equal{} 1$ and $ c$ be an arbitary number.Show that there exist \n$ x$ s.t $ (ax \\plus{} b,c) \\equal{} 1$.\nYou are not allowed to use advance theorems! :wink:[/quote]\r\nIf $ p \\mid b$ and $ p \\mid c$ then $ p \\nmid x$, and if $ p \\nmid b$ and $ p \\mid c$ then $ p \\mid x$..." } { "Tag": [ "quadratics", "modular arithmetic", "algebra" ], "Problem": "Show that for odd integers $ a$, $ b$, and $ c$, $ a^{2}-4bc$ can not be a perfect square.", "Solution_1": "[hide=\"Solution\"]\nAssume that the equation can be a perfect square. Since we know that $ a^{2}>4bc$, we may write\n\\[ a^{2}-4bc=(a-x)^{2}\\]\nfor some integer $ x$. Expanding\n\\[ x^{2}-2ax=-4bc\\]\n\\[ bc=\\frac{x(2a-x)}{4}\\] \nSince the product $ bc$ must be an integer, $ x=2k$ for some integer $ k$. Making this substitution gives. \n\\[ bc=\\frac{(2k)(2a-2k)}{4}\\] \n\\[ bc=k(a-k)\\]\nSince $ a$ is always odd, substitute $ a=2l-1$ for some integer $ l$, \n\\begin{eqnarray*}bc&=&k(2l-1-k) \\\\ &=&2lk-k(k+1) \\end{eqnarray*}\nSince $ b$ and $ c$ are both odd integers, their product must also be odd. However, the RHS in the above expression is always even. Contradiction, therefore, the expression $ a^{2}-4bc$ cannot be a perfect square. [/hide]", "Solution_2": "[hide=\"Short\"] $ bx^{2}+ax+c$ cannot have integer roots because it is always odd for integer $ x$. [/hide]", "Solution_3": "Discriminants were my motivation for posing this, but that is not the argument I had in mind. Does that suffice? A perfect square discriminant doesn't guarantee integer roots.", "Solution_4": "[quote=\"t0rajir0u\"]$ bx^{2}+ax+c$ cannot have integer roots because it is always odd for integer x.\n[/quote]\r\n\r\nwell that doesn't guarntee that the proposition of the problem is true.as $ a^{2}-4bc$ well be a perfect square but the roots rational.\r\nwell continuing ur proof at worst we can have rational roots say $ \\frac{p}{q}$ but both $ p$ and $ q$ have to be odd as they have to divide $ c,b$ .so now just multiply throught by $ q^{2}$ we get sum of odd integers is $ 0$ which is impossible", "Solution_5": "[hide]Mod 8:\nQuadratic residues mod 8 are 0,1,4. Since a is odd, $ a^{2}\\equiv 1\\pmod{8}$. Since 4bc is even, but never divisible by 8, it must be $ \\equiv 2,4,6 \\pmod{8}$. Thus, $ a^{2}-4bc\\equiv 7,5,3 \\pmod{8}$, none of which are quadratic residues mod 8.[/hide]", "Solution_6": "[quote=\"djshowdown2\"]A perfect square discriminant doesn't guarantee integer roots.[/quote]\r\n\r\nI have a penchant for posting one-line solutions. In retrospect, I omitted some fairly important details, so here they are: \r\n\r\n[hide=\"Full solution\"] Suppose $ b^{2}-4ac$ is a perfect square for odd integers $ a, b, c$. It is odd, so its square root is odd. Then $ -b \\pm \\sqrt{b^{2}-4ac}$ is even. In particular, \n\n$ a(ax^{2}+bx+c) = (ax)^{2}+b(ax)+ac = 0$\n\nHas two integer roots $ ax_{1}, ax_{2}= \\frac{-b \\pm \\sqrt{b^{2}-4ac}}{2}$. But since the quadratic equation $ u^{2}+bu+ac = 0$ has only odd coefficients, it cannot have integer roots. QED. [/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ a,b,c\\ge0$,a>=b>=c and $ a\\plus{}b\\plus{}c\\equal{}1$\r\nprove that:\r\n\r\n$ \\sum_{cyc}{a\\sqrt{\\frac{b}{c}}}\\ge1$", "Solution_1": "any answer? :!:" } { "Tag": [ "Columbia" ], "Problem": "to anyone who ever took the exam or will be taking it, how does the whole application process go. I heard that the test contains math and science. Im a sophmore so what kind of math and science would they be testing. Also to anyone who who is currently in the program, what kind of preparation did you undergo and what sources did you use to help you pass the exam", "Solution_1": "I got in this year. The test is easy math, mediocre science, and then AMC math. I heard the last test is what matters the most out of the three. Get more than 10(?) of them right and I heard you get in. I had a B+ in math though, so I guess your transcript doesn't really count? Otherwise, you submit a personal statement of 250 words and a teacher rec.", "Solution_2": "Anyone else taking it? I'm doing Exploring Black Holes.", "Solution_3": "The first math section is really easy. One of the (first) questions was \"a second is what fractional part of a minute?\" or something along those lines. \r\n\r\nAll of the science is more or less regents-level (or slightly beyond). The only questions I didn't know were on astronomy (quasars, neutron stars, etc.). Since you are a sophomore, you should have no problem on that section.\r\n\r\nAnd, yeah, if you are good with AMC math, you should be good with the math section. I am not good with AMC math and I still got in (without preparation). >.>", "Solution_4": "AMC 10 or AMC 12? I can do AMC 10 decently, but AMC 12 is much harder for me.\n", "Solution_5": "[quote=shukion123]AMC 10 or AMC 12? I can do AMC 10 decently, but AMC 12 is much harder for me.[/quote]\n\nBasic AMC 10 problems, 13-20 mostly, and AMC 8 20-25.\nNot too difficult with practice.", "Solution_6": "So would it help if I studied AMC 10 problems or do I also need to study AMC 8 problems as well? And is the transcript important for me to get into the program?", "Solution_7": "[quote=BL3RRED]So would it help if I studied AMC 10 problems or do I also need to study AMC 8 problems as well? And is the transcript important for me to get into the program?[/quote]\n\nTranscript is important." } { "Tag": [], "Problem": "I would be grateful for all the help I can get\r\n\r\nI have tried to solve this but I failed big times....\r\n\r\nCan someone explaine it to me step by step?\r\n\r\nthanks\r\n\r\nf_0 = 0\r\nf_1 = 1\r\nf_n = f_n-1 + f_n-2 n\u22652\r\n\r\nsolve : f_n = 1/\u221a5 ((1+\u221a5)/2))^n -1/\u221a5 ((1-\u221a5)/2))^n", "Solution_1": "Easiest way - by induction.\r\n\r\n[hide]\nLet $ \\phi\\equal{}\\frac{1\\plus{}\\sqrt{5}}{2}$.\nFor $ n\\equal{}0$ and $ n\\equal{}1$ it's true. Suppose our formula is true for all $ 0\\leq k\\leq n \\plus{} 1$. \nWe get: \n$ F_{n\\plus{}2}\\equal{}F_{n\\plus{}1}\\plus{}F_{n}\\equal{}\\frac{1}{\\sqrt{5}}(\\phi^{n\\plus{}1}\\minus{}(1\\minus{}\\phi)^{n\\plus{}1} \\plus{} \\phi^{n}\\minus{}(1\\minus{}\\phi)^{n})\\equal{}$\n$ \\equal{}\\frac{1}{\\sqrt{5}}\\left(\\phi^{n}(\\phi\\plus{}1)\\minus{}(1\\minus{}\\phi)^{n}(1\\minus{}\\phi\\plus{}1)\\right)$\nNote that $ \\phi^2\\minus{}\\phi\\minus{}1\\equal{}0$. Using this you get $ F_{n\\plus{}2}\\equal{}\\frac{1}{\\sqrt{5}}\\left(\\phi^{n\\plus{}2}\\minus{}(1\\minus{}\\phi)^{n\\plus{}2}\\right)$. We are done.\n[/hide]" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Let $P(x)$ be any polynomial of degree $n>0$ over $\\mathbb{C}$. For any $m \\in \\{0,1,\\ldots,n\\}$, show that the polynomials $P(x), P(x+1), \\ldots, P(x+m); 1, x, \\ldots, x^{n-m-1}$ are linearly independent (over $\\mathbb{C}$).", "Solution_1": "We prove that linear span $L$ of these polynomials is the space of polynomials which degree does not exceed $n$. Since the number of polynomials equals to the dimension of this space, they are linearly independent. \r\n\r\nNote that $L$ contains polynomials $P(x)$ of degree $n$, $P(x+1)-P(x)$ of degree $n-1$, $P(x+2)-2P(x+1)+P(x)$ of degree $n-2$, $\\dots$, $m$-th difference of $P$, whose degree is $n-m$. Also, $L$ contains all polynomials of degree less then $n-m$. It follows that it contains all polynomials of degree at most $n$." } { "Tag": [ "inequalities", "logarithms", "function", "inequalities unsolved" ], "Problem": "$\\displaystyle x_1,x_2,...,x_n>0$\r\n\r\n$\\displaystyle g=\\sqrt[n]{x_1x_2...x_n}$ \r\n\r\n$\\displaystyle a=\\frac{x_1+x_2+...+x_n}{n}$\r\n\r\nProve that:\r\n\r\n$\\displaystyle e^g-1 \\leq \\sqrt[n]{(e^{x_1}-1)...(e^{x_n}-1)}\\leq e^{a-g}(e^g-1)$", "Solution_1": "Denote with $y_{i} = \\ln x_{i}$ and then $\\displaystyle e^{g}-1 \\leq \\sqrt[n]{ \\prod (e^{x_{i}}-1)}$ becomes $ \\displaystyle ln \\left( e^{e^{ \\frac{ \\sum y_{i}}{n}} -1 \\right) \\leq \\frac{1}{n} \\sum \\ln ( e^{e^{x_{i}}}-1).}$ But defining the function $\\displaystyle f(x) = \\ln( e^{e^{x}} - 1)$ you get that this function is convex and you apply Jensen.. for the other inequality you have: $ \\displaystyle ln \\left( 1- e^{- e^{ \\frac{ \\sum y_{i}}{n}} \\right) \\geq \\frac{1}{n} \\sum \\ln (1- e^{-e^{x_{i}}}),}$ but the function $\\displaystyle f(x) = \\ln( 1- e^{-e^{x}} )$ is concave .. etc\r\n\r\ncheers! :D :D" } { "Tag": [ "geometry", "circumcircle" ], "Problem": "http://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_10A_Problems/Problem_14\r\n\r\nI don't understand the solution.\r\n\r\nIs the hypotenuse of a right triangle [i]always[/i] the diameter of its circumcircle? If so, why?", "Solution_1": "Yes, this is because the hypotenuse is opposite to the right angle in a right triangle. If you look carefully at the diagram, you will notice that the 90 degree angle is the inscribed angle of the circle. Therefore, the entire arc is 180 degrees or a semicircle. This is why the hypotenuse is always the diameter.", "Solution_2": "To generalize, if you have points $ A$, $ B$, and $ C$ on a circle, then the arc $ AC$ is exactly twice the angle $ ABC$. In this instance, $ ABC$ being $ 90$ degrees implies $ AC$ is of measure $ 180$ degrees, meaning it is a semicircle." } { "Tag": [ "number theory", "prime factorization" ], "Problem": "For how many three-element sets of positive integers {a,b,c} is it true that a*b*c=2310?\r\n\r\n\r\nthis is what i got so far[hide]\n2310=3*7*11*5*2 \nso the 2310 has a total of $2^{5}=32$ divisors. [/hide]\r\nwhat chapter in aops book should I read?\r\nThanks", "Solution_1": "[hide=\"The solution...\"]It's true that you need the prime factorization, but I don't think you need the number of divisors. Let's take a look at the prime factorization:\n$2310 = 2\\cdot 3\\cdot 5\\cdot 7\\cdot 11$.\n\nClearly, one of the $3$ numbers has to be a multiple of two of these numbers, but the max is $3$ multiples. Let $a$ (and $b$) be the number(s) that has more than $1$ factor. There are $2$ cases (in case $2$, $a$ and $b$ have $2$ factors): \n\nCASE 1: $a$ has $3$ factors\nWe only need to look at the number of ways there are to choose $3$ numbers out of a set of $5$. Since order is disregarded, the answer is $\\binom{5}{3}= 10$ ways. \n\nCASE 2: $a$ has $2$ factors, $b$ has $2$ factors\nFor $a$, there are $\\binom{5}{2}= 10$ ways. For $b$, there are $\\binom{3}{2}=3$ ways. For this case, there are $30$ ways.\n\nAdding up our results, we get $40$ ways. [/hide]\r\nUse AoPS Vol 1, and whichever chapter covers permutations/combinations.", "Solution_2": "Wouldn't you multiply the 10 and the 3? :huh:", "Solution_3": "[quote=\"laughinghead505\"]Wouldn't you multiply the 10 and the 3? :huh:[/quote]\r\n\r\nyeah, you would", "Solution_4": "ok. Then, [hide=\"I guess it would be\"] 40 ways [/hide]", "Solution_5": "I worked it out and I also came out with 40.", "Solution_6": "[hide]2310=2x3x5x7x11, and each of the 5 terms can be in any of the prime factorizations of the 3 numbers a,b, and c, for $3^{5}=243$, sets. Out of these sets, 3 are permutations of (1,1,2310), and should not be included, since the elements aren't distinct. Out of the remaining 240 sets, we repeated each set 3!=6 times. 240/6=40.[/hide]", "Solution_7": "How were the solutions with $1$ as one of the numbers in the set included in vishalarul's solution? I see that they must have been somehow because the answer is the same...", "Solution_8": "[hide]\n\n Note that $2310 = 2*3*5*7*11$. For each prime factor, there are 3 elements that it can be sent to, so $3^{5}=243$ permutations. However, we subtract 3 because three permutations of {1, 1, 2310} have non-distinct elements. We now have $243-3=240$ ordered triplets\n We know that we want the sets, not the ordered triplets, of numbers. Since there are six permutations of each set of integers, we divide 240 by 6 to get $40$ distinct sets.\n\n[/hide]", "Solution_9": "[quote=\"vishalarul\"][hide=\"The solution...\"]It's true that you need the prime factorization, but I don't think you need the number of divisors. Let's take a look at the prime factorization:\n$2310 = 2\\cdot 3\\cdot 5\\cdot 7\\cdot 11$.\n\nClearly, one of the $3$ numbers has to be a multiple of two of these numbers, but the max is $3$ multiples. Let $a$ (and $b$) be the number(s) that has more than $1$ factor. There are $2$ cases (in case $2$, $a$ and $b$ have $2$ factors): \n\nCASE 1: $a$ has $3$ factors\nWe only need to look at the number of ways there are to choose $3$ numbers out of a set of $5$. Since order is disregarded, the answer is $\\binom{5}{3}= 10$ ways. \n\nCASE 2: $a$ has $2$ factors, $b$ has $2$ factors\nFor $a$, there are $\\binom{5}{2}= 10$ ways. For $b$, there are $\\binom{3}{2}=3$ ways. For this case, there are $30$ ways.\n\nAdding up our results, we get $40$ ways. [/hide]\r\nquote]\r\nIsnt there a case 3, where $a$ has $4$ factors, $b$ has $1$ factor, and $c$ is 1.", "Solution_10": "[quote=\"myc\"][quote=\"vishalarul\"][hide=\"The solution...\"]It's true that you need the prime factorization, but I don't think you need the number of divisors. Let's take a look at the prime factorization:\n$2310 = 2\\cdot 3\\cdot 5\\cdot 7\\cdot 11$.\n\nClearly, one of the $3$ numbers has to be a multiple of two of these numbers, but the max is $3$ multiples. Let $a$ (and $b$) be the number(s) that has more than $1$ factor. There are $2$ cases (in case $2$, $a$ and $b$ have $2$ factors): \n\nCASE 1: $a$ has $3$ factors\nWe only need to look at the number of ways there are to choose $3$ numbers out of a set of $5$. Since order is disregarded, the answer is $\\binom{5}{3}= 10$ ways. \n\nCASE 2: $a$ has $2$ factors, $b$ has $2$ factors\nFor $a$, there are $\\binom{5}{2}= 10$ ways. For $b$, there are $\\binom{3}{2}=3$ ways. For this case, there are $30$ ways.\n\nAdding up our results, we get $40$ ways. [/hide]\n[/quote]\nIsnt there a case 3, where $a$ has $4$ factors, $b$ has $1$ factor, and $c$ is 1.[/quote] Yes, that gives an extra $\\binom{5}{4}=5$ ways.\n\nEDIT: Although, everyone here\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=373504#p373504\nhas gotten the answer [hide] $41$ [/hide], which isn't even answer choice.", "Solution_11": "so wouldn't there be 40+5=45 ways? i must be mistaken cuz the answer is [hide]40. [/hide]", "Solution_12": "Where did you get that answer? Did it come with a solution? Because by everything I have seen, the answer is almost certainly [hide]45[/hide]\r\nwhich is an answer choice on the original test...", "Solution_13": "I was looking at vishalarul's solution, and it doesn't even take into account the possibility of a value of 1. Looking at that, it must be more than 40.\r\n[hide]Since $2310=2\\cdot3\\cdot5\\cdot7\\cdot11$, we have to find ways to distribute these five factors into three numbers, where order doesn't matter.\nHow many ways can we have three numbers that add up to 5, where order doesn't matter? Listing them all, we find that there are five.\n005\n014\n023\n113\n122\nEach digit represents how many factors are in each number.\nNow, let's do some casework.\nCASE 1: $005$\nThere is only $\\boxed{1}$ way to do this: $(1,1,2310)$.\nCASE 2: $014$\nThere are $\\binom{5}{0}\\binom{5}{1}$, or $\\boxed{5}$, ways to do this. After the second one, we automatically have our value for the last number.\nCASE 3: $113$\nThere are $\\frac{\\binom{5}{1}\\binom{4}{1}}{2!}$, or $\\boxed{10}$, ways to do this. We divide by $2!$ because we can rearrange the first two terms.\nCASE 4: $023$\nThere are $\\binom{5}{3}$, or $\\boxed{10}$, ways to do this.\nCASE 5: $122$\nThere are $\\binom{5}{2}\\binom{3}{2}$, or $\\boxed{30}$, ways to do this.\n\nAdd everything up to get $\\boxed{\\boxed{56}}$. Was 56 a choice?[/hide]", "Solution_14": "[quote=\"lingomaniac88\"]CASE 5: $122$\nThere are $\\binom{5}{2}\\binom{3}{2}$, or $\\boxed{30}$, ways to do this.[/quote] You need to divide by 2. This gives $41$.", "Solution_15": "[quote=\"lotrgreengrapes7926\"][quote=\"lingomaniac88\"]CASE 5: $122$\nThere are $\\binom{5}{2}\\binom{3}{2}$, or $\\boxed{30}$, ways to do this.[/quote] You need to divide by 2. This gives $41$.[/quote]\r\n\r\nOh, right. :blush:", "Solution_16": "The answer is 40, (solution pamphlet:)[url=http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm]http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm[/url]", "Solution_17": "But it doesn't say that $a$, $b$, and $c$ are distinct. If so, then yes, the answer would be 40, because ${1,1,2310}$ could not work.", "Solution_18": "The original question said nothing about distinct. :maybe:", "Solution_19": "quote wikipedia:\r\nIn mathematics, a set can be thought of as any collection of [u]distinct[/u] things considered as a whole.", "Solution_20": "It should be 45 because of the case of 4 factor combination, 1, and left over factor. Those are distinct.", "Solution_21": "I don't get the answer the solution in the pamplet.\r\n \r\nIt says \"The numbers of elements in the subsets can be: 0,1,4; 0,2,3;1,1,3; or 1,2,2.\" how did they get these numbers and what do they mean?", "Solution_22": "[quote=\"myc\"]I don't get the answer the solution in the pamplet.\n \nIt says \"The numbers of elements in the subsets can be: 0,1,4; 0,2,3;1,1,3; or 1,2,2.\" how did they get these numbers and what do they mean?[/quote]\r\n\r\nThis is the set of possible distributions of $5$ elements (the prime factors) over $3$ subsets ($a, b, c$).\r\n\r\nAlthough it is missing the distribution $0, 0, 5$. \r\n\r\nThe problem itself is vague as to whether it means a \"set\" in the strictest sense or whether it simply means an unordered triple.", "Solution_23": "for the 1,1,3 case shouldn't it be 20 ways instead of 10?\r\n5 ways of choosing first one-element subset, 4 ways of choosing th secound, and only 1 way of choosing the last.", "Solution_24": "[quote=\"myc\"]for the 1,1,3 case shouldn't it be 20 ways instead of 10?\n5 ways of choosing first one-element subset, 4 ways of choosing th secound, and only 1 way of choosing the last.[/quote]\r\n\r\nYes, but you're overcounting since the first two subsets are the same size. For example,\r\n\r\n$(v), (w), (x, y, z)$\r\n\r\nIs the same partition as\r\n\r\n$(w), (v), (x, y, z)$", "Solution_25": "and we're assuming a,b,c are distinct postive intergers? so 0,0,5 doesn't count. \r\n\r\nthanks :)" } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "Supposed $ A$ and $ B$ are two matrices that statisfies $ AB\\minus{}BA\\equal{}0$ where $ B$ is not nilpotent. Show that the equation $ AX\\minus{}XA\\equal{}B$ has no solution in $ X$. Thank you for your attention.", "Solution_1": "Another version of [url]http://www.mathlinks.ro/viewtopic.php?t=203645[/url]" } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "please i need the solution", "Solution_1": "hello, you can use these formulas\r\n$ \\cos(x\\plus{}\\frac{2\\pi}{3})\\equal{}\\minus{}\\sin(x\\plus{}\\frac{\\pi}{6})\\equal{}\\minus{}1/2\\cos(x)\\minus{}1/2\\sqrt{3}\\sin(x)$\r\n$ \\cos(x\\plus{}\\frac{4\\pi}{3})\\equal{}\\minus{}\\sin(\\frac{\\pi}{6}\\minus{}x)\\equal{}\\minus{}1/2\\cos(x)\\plus{}1/2\\sqrt{3}\\sin(x)$\r\nSonnhard." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $m$ and $n$ be natural numbers such that $m+i=a_{i}b_{i}^{2}$ for $i=1,2,...,n$, where $a_{i}$ and $b_{i}$ are natural numbers and $a_{i}$ is not divisible by a square of a prime number. Find all $n$ for which there exists an $m$ such that $a_{1}+a_{2}+...+a_{n}=12$", "Solution_1": "Who can hepl me?", "Solution_2": "Because $a_{i}b_{i}^{2}\\ge i+1$, $a_{i}\\ge 1$ and $a_{i}=a_{j}\\to |i-j|\\ge 6$.\r\nBecause $1+2+3+4+5>12$ we have $n\\le 4$.\r\nn=1 don't work, because $a_{1}\\not =12=3*2^{2}$. \r\nn=2, then $a_{1},a_{2}$ are odd and $a_{1}+a_{2}=12$, therefore $(a_{1},a_{2})=(1,11),(5,7),(7,5),(11,1)$. It give Pell's equation and had solutios for all pair's. For example $m=98, m+1=99=11*3^{2},m+2=100=1*10^{2}$.\r\nn=3 and n=4 can be check by computer." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "in triangle of ABC we draw from D where the stone falls AD such that two line of DC'and DB' respectively parallel the side of AB and AC . two line of DB\" and DC\" respectively perpendicular these sides prove that BC-B'C'-B\"C\"are concurrent.", "Solution_1": "[b][u]A very easy problem ! Its position is not here.[/u][/b] Look the solution unique of its kind,\r\n\r\nwith the Menelaus's theorem applied to the transversal lines $\\overline {X'B'C'},\\overline {X''B''C''}$ for the triangle $ABC$, where $X'\\in BC\\cap B'C',\\ X''\\in BC\\cap B''C''\\ldots\\ldots\\Longrightarrow \\frac {X'B}{X'C}=\\frac {X''B}{X''C}=\\left( \\frac{DB}{DC}\\right)^2$ $\\Longrightarrow X_1 \\equiv X_2.$" } { "Tag": [ "induction", "calculus", "calculus computations" ], "Problem": "Prove that for a positive integer n and any real numbners a and b.\r\n\r\n$\\frac{1}{(x-a)^n(x-b)^n}$ =\r\n$\\frac{(-1)^n} {(a-b)^{2n}}\\sum_{0}^{n-1}(C_r)((\\frac{b-a} {x-a})^{n-r}+(\\frac{a-b} {x-b})^{n-r})$\r\n\r\nwhere $C_r$ is the coefficient of $h^r$ in the expansion of \r\n\r\n$(1-h)^{-n}$\r\n\r\nA proof by induction is quite long\r\nCan anyone suggest a short proof?", "Solution_1": "An idea is to set $\\boxed{x= \\frac{b-a}{2} \\cdot y +\\frac{b+a}{2}}$ but it is just an idea, i don't have a solution.", "Solution_2": "Start with the partial fraction decomposition\n\n$ \\frac{1}{x-\\alpha}\\cdot\\frac{1}{x-\\beta}=-\\frac{1}{\\beta-\\alpha}\\cdot\\frac{1}{x-\\alpha}+\\frac{1}{\\beta-\\alpha}\\cdot\\frac{1}{x-\\beta} $ .\n\nThen apply $ \\left(\\frac{d}{d\\alpha}\\right)^n $ and $ \\left(\\frac{d}{d\\beta}\\right)^m $ on it.\n\nFinally you get\n\n$ \\frac{1}{(x-\\alpha)^{n+1} \\, (x-\\beta)^{m+1}} $ \n\n$ =\\sum_{k=0}^n \\frac{(-1)^k}{(\\alpha-\\beta)^{m+1+k}}\\, \\frac{(m+k)!}{m!\\, k!}\\, \\frac{1}{(x-\\alpha)^{n+1-k}} $ \n\n$ +\\sum_{k=0}^m \\frac{(-1)^k}{(\\beta-\\alpha)^{n+1+k}}\\, \\frac{(n+k)!}{n!\\, k!}\\, \\frac{1}{(x-\\beta)^{m+1-k}} $ .\n\nSee this [url=http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Identit%C3%A4ten:_Partialbruchzerlegungen#2]link[/url]." } { "Tag": [], "Problem": "At a women's doubles tennis tournament, there were three teams of \ntwo women. After the tournament, each woman shook hands once\nwith each of the other players except her partner. What is the\nnumber of handshakes that occurred?", "Solution_1": "6 woman, handshakes: $ \\frac{6\\cdot 5}{2}\\minus{}3\\equal{}27$", "Solution_2": "The answer is $ 12$...", "Solution_3": "Teams. A1,B2,C3\r\n\r\nHnadshakes:\r\n\r\nAB\r\nA2\r\nAc\r\nA3\r\n1b\r\n12\r\n1c\r\n13\r\nbc\r\nb3\r\n2c\r\n23\r\n\r\n\r\n\r\n12 total", "Solution_4": "From a different viewpoint (as I now understand this problem), of Team AB, A shakes hands $ 4$ times and so does B. In fact, every team member shakes hands $ 4$ times. But this will result in 2 handshakes between each of the two players, so we must divide by $ 2$ to get the desired answer of $ \\boxed{12}$.", "Solution_5": "I got correct, just I thought 6*5/2=30... Clumsy me!", "Solution_6": "I understand Bugis method. Handshake combos= 6c2- 3 (partener combos)", "Solution_7": "ill answer this in a series of steps\n\n[hide]1. we know the following facts \n\n I.there are 3 teams with 2 woman\n \n II.there are 6 woman all together\n\n III.1 woman shakes 1 person and doesn't shake with her again\n\n2. 1 woman can shake with 6 people and that woman can shake with another 5 people so we get the equation 5*6=30\n\n3.but we're overcounting so we need to subtract by 3 30-3=27\n\n4.[hide]27[/hide][/hide]" } { "Tag": [], "Problem": "let's count a little bit :) \r\n\r\nfind the number of numbers with 5 digits which can be formed with the digits 1,2,4,7, 9 and 0.\r\n\r\nwhat about the number of numbers with 5 digits that can be formed with 1,2,4,7,9 and 0, but no digit occurs twice in the same number?", "Solution_1": "[quote=\"Valentin Vornicu\"]let's count a little bit :) \n\nfind the number of numbers with 5 digits which can be formed with the digits 1,2,4,7, 9 and 0.\n\nwhat about the number of numbers with 5 digits that can be formed with 1,2,4,7,9 and 0, but no digit occurs twice in the same number?[/quote]\r\n\r\n5*6*6*6*6?\r\n\r\n5*5*4*3*2?", "Solution_2": "I thought that you are beyond beginner :)", "Solution_3": "[quote=\"Valentin Vornicu\"]I thought that you are beyond beginner :)[/quote]\r\n\r\nThe answers are wrong?! :?", "Solution_4": "no they are not, but they are supposed to be for more beginner people :) let's encourage them to post here :)", "Solution_5": "Hm there seems to have been quite some \"market spoiling\" in the Beginner's Corner...in future can the not-so-beginners try to give hints at most, instead of giving solutions?", "Solution_6": "that would be recommended. I was expecting more beginners to ask questions and post problems. By now only one has done that :)", "Solution_7": "Actually, Valentin, this question really is a bit too easy...my school classmates learn it (and more complicated counting stuff) in grade 9, so anybody who is doing olympiads should be familiar with it.", "Solution_8": "we don't do much of combinatorics here in romania, and the question is for beginner's. it does not matter too much that it's too easy. there are a lot of easy questions that people can learn alot from :)", "Solution_9": "well eum... that's like one of the questions we got at school... on day 1... to show what combinatorics are... and most of us just saw that on first sight\r\n\r\nI think - beginners section or not! - \"problems\" should be things to \"think\" about... so a little harder would not mind.", "Solution_10": "well let me just say this: this is the first question that was addressed to my colleagues majoring in MATH, at Univ. Bucharest. The answers were as follows: 5, 25, 10, 6^5, etc. After quite a while I decided to shout the answer. So if people majoring in maths here had no idea about this question I supposed that if is of somewhat difficulty :) that's why I put it here :D." } { "Tag": [ "function", "calculus", "derivative" ], "Problem": "What are some ways of solving x^x=c? (I only know of one, using Newton's root-approximation method.)", "Solution_1": "There is no exact method -- the function x^x doesn't have a nice inverse. (It's particularly poorly behaved over the negative reals.) The best you can do is various approximations.", "Solution_2": "Well, you can solve this with the Lambert W function. W is defined as the inverse function to xe^x, in the same way as log(x) is the inverse of e^x ... so if\r\n\r\nte^t = x\r\n\r\nthen\r\n\r\nt = W(x)\r\n\r\nW(x) is nicely behaved and single valued if x is positive - there are some technicalities to handle branches if x is negative. We can avoid these if we assume c is greater than or equal to 1, in which case\r\n\r\nx^x =c\r\n\r\nhas solution\r\n\r\nx = e^(W(log(c))) = log(c) / W(log(c))\r\n\r\n[see [url]http://en.wikipedia.org/wiki/Lambert's_W_function[/url] for more details]\r\n\r\nOf course, you then need to be able to find the value of W(log(c)) - but W(x) is really no stranger than functions that we all know and love, like log(x) or sin(x) ... except that you don't find W(x) on many calculators !", "Solution_3": "And that it's very difficult to find specific values for ...", "Solution_4": "It's not [i]that[/i] difficult to find values for W(x). Conceptually, it is no more complicated than log(x). For x>0 it has derivatives and a Taylor series. Algebra packages like Maple and Mathematica implement W(x) as just another function.\r\n\r\nIf you wanted to solve x^x=e, for example, you could either leave the solution symbolically as \r\n\r\nx = 1/W(1)\r\n\r\nor you could look up a value for W(1), which is approximately 0.567143, and so get an approximate numeric value\r\n\r\nx :cong: 1.763224" } { "Tag": [ "modular arithmetic", "Euler", "function", "number theory unsolved", "number theory" ], "Problem": "Find the last three digits of the number $2003^{{2002}^{2001}}$.", "Solution_1": "$2003^n (mod \\ 1000)$ have period 400. $2002^n (mod \\ 25)$ have period 20. Since $2002^2001=2(mod \\ 25)$ and have divisor 16. Therefore $2002^2001=352 (mod \\ 400), 2003^{2002^{2001}}=3^{352} (mod \\ 1000)=841^{11}(mod \\ 1000)=241$.", "Solution_2": "how did you know that $2003^{n} \\pmod {1000}$ had period 400? and how did you know that $2002^{n} \\pmod{25}$ had period 20.", "Solution_3": "$\\varphi(1000)=400, \\varphi(25)=20$, so we don't need to know the order, just Euler-Fermat suffices (note that period and order must not be the same, for example the function $\\sin(x)$ has the period $42\\pi$).", "Solution_4": "[hide=\"Solution\"]\nWe desire to find $2003^{2002^{2001}}\\pmod{1000}$. To do so, we find it $\\pmod 8$ and $\\pmod{125}$ and then solve the equations as we did before. Note that this $\\pmod 8$ is going to be $3^{2002^{2001}}$ or $(3^2)^{2002^{2000}*1001}\\equiv 1\\pmod 8$ and $\\pmod{125}$ gives us $3^{2002^{2001}}\\pmod{125}$. Since $\\phi(125)=100$, we desire $2002^{2001}\\pmod{100}\\equiv 2^{2001}\\pmod{100}$. Take this $\\pmod 25$ and $\\pmod 4$ to give us $x\\equiv 0\\pmod 4$ and since $\\phi(25)=20$ we get $2^{2001}\\equiv 2\\pmod{25}$ so $x\\equiv 0\\pmod 4, x\\equiv 2\\pmod{25}$ therefore $x=4k=25m+2$ or $m+2\\equiv 0\\pmod 4$ therefore $m\\equiv 2\\pmod 4=4m'+2$ giving us $25(4m'+2)+2=100m'+52$ therefore $x\\equiv 52\\pmod{100}$. Therefore we desire $3^{52}\\pmod{125}$. Since $3^6=729 \\equiv -21 \\pmod{125}$, we get $3^{52}\\equiv (-21)^8*3^4 \\pmod{125}$. This is the same thing as $(73)^4 \\pmod{125}$ or $(79)^2\\pmod{125}\\equiv 116\\pmod{125}$ Finally, $2003^{2002^{2001}}=x\\equiv 1\\pmod 8$ and $x\\equiv 116\\pmod{125}$. Therefore $8a+1=125b+116$. Take $\\pmod 8$ to give $5b+4\\equiv 1\\pmod{8}$ or $b\\equiv 1\\pmod 8=8b'+1$. Therefore $125(8b'+1)+116=1000b'+125+116=1000b'+241$. I conclude the answer to be $\\boxed{241}$. [/hide]", "Solution_5": "We want the number in $\\mod {1000}$ so note that since $\\gcd(1000,2003)=1$ and $\\varphi(1000)=400$, it follows by Euler's Theorem that $2003^{400k}\\equiv 1\\pmod {1000} \\forall k\\in \\bf Z^*$. So we want $2002^{2001}\\equiv 2^{2001}(1001^{2001})\\equiv 2^{2001}\\pmod {400}$ (why?). Notice that $400=16(25)$, so $2^{2001}=16a$ for some $a\\in \\bf Z^+$, or $2^{2001}\\equiv 16a\\pmod {400}\\iff 2^{1997}\\equiv a\\pmod {25}$. Further, $\\gcd(2,25)=1$ and $\\varphi(25)=20$, thus $a\\equiv 2^{1997}\\equiv \\tfrac{2^{2000}}{2^3}\\equiv \\tfrac{1}{8}\\equiv 22\\pmod {25}$, or $2^{2001}\\equiv16a\\equiv 16(22)\\equiv 352\\pmod {400}$. Finally, we get \\begin{align*}2003^{2002^{2002}} & \\equiv 2003^{400k+352}\\\\ &\\equiv 2003^{352}\\\\ &\\equiv 3^{352}\\\\ &\\equiv 9^{176}\\\\ &\\equiv (10-1)^{176}\\\\ &\\equiv \\tbinom{176}{2}(10^2)-176(10)+1\\\\ &\\equiv 0-760+1\\\\ &\\equiv \\boxed{241}\\pmod {1000}\\end{align*} where the [i]ante penultimate[/i] equivalence relation follows from the Binomial theorem. :)", "Solution_6": "Time for $\\lambda$!\n\nSince $2003^{100} \\equiv 1 \\pmod {1000}$, we wish to find $2002^{2001} \\pmod {100}$. This reduces to finding $2^{1999} \\pmod {25}$ or $1999 \\pmod {20}$ which is just $-1$. So $2002^{2001} \\equiv 4 \\cdot 2^{-1} (\\pmod {25}) \\pmod {100} \\equiv 52 \\pmod {100}$. So our original problem reduces to $2003^{52}$ and then we proceed as above." } { "Tag": [ "inequalities", "probability", "vector", "trigonometry", "algebra", "polynomial", "linear algebra" ], "Problem": "Let [tex]a_1,\\dots,a_n, b_1\\dots,b_n[/tex] be real numbers such that\r\n[tex]\\sum_{k=1}^na_i^2=1[/tex], [tex]\\sum_{k=1}^nb_i^2=1[/tex] and [tex]\\sum_{k=1}^na_ib_i=0[/tex]. Show that [tex]a_i^2+b_i^2\\le1[/tex] for all [tex]i=1,\\dots,n[/tex].", "Solution_1": "use \\displaystyle{} around your [tex]\\TeX[/tex] to make it look much nicer.\r\n\r\nLet [tex]a_1,\\dots,a_n, b_1\\dots,b_n[/tex] be real numbers such that\r\n[tex]\\displaystyle{\\sum_{k=1}^na_i^2=1}[/tex], [tex]\\displaystyle{\\sum_{k=1}^nb_i^2=1}[/tex] and [tex]\\displaystyle{\\sum_{k=1}^na_ib_i=0}[/tex]. Show that [tex]a_i^2+b_i^2\\le1[/tex] for all [tex]i=1,\\dots,n[/tex].", "Solution_2": "WLOG we will show that this is true for [tex]a_1, b_1[/tex].\r\n\r\nBy Cauchy,\r\n\r\n[tex]\\displaystyle \\Big(\\sum_{i=1}^na_i^2 - a_1^2\\Big)\\Big(\\sum_{j=1}^nb_j^2 - b_1^2\\Big) \\ge \\Big(\\sum_{k=1}^na_kb_k - a_1b_1\\Big)^2[/tex]\r\n\r\nPlugging in our known values for the summations, we have\r\n\r\n[tex](1-a_1^2)(1-b_1^2) \\ge (0-a_1b_1)^2 = a_1^2b_1^2[/tex]\r\n\r\nDistributing and adding [tex]a_1^2+b_1^2-a_1^2b_1^2[/tex] to both sides yields our desired inequality.\r\n\r\n[tex]a_1^2+b_1^2 \\le 1[/tex], which generalizes for all [tex]a_i, b_i[/tex].", "Solution_3": "Good Job! Probability 1.01.", "Solution_4": "[quote=\"probability1.01\"]By Cauchy,\n\n[tex]\\displaystyle \\Big(\\sum_{i=1}^na_i^2 - a_1^2\\Big)\\Big(\\sum_{j=1}^nb_j^2 - b_1^2\\Big) \\ge \\Big(\\sum_{k=1}^na_kb_k - a_1b_1\\Big)^2[/tex]\n[/quote]\r\n\r\nDoesn't Cauchy-Schwarz only accept positive terms? How can you justify that the inequality will hold when you artificially insert -a12 and -b12?", "Solution_5": "[quote=\"Osiris\"][quote=\"probability1.01\"]By Cauchy,\n\n[tex]\\displaystyle \\Big(\\sum_{i=1}^na_i^2 - a_1^2\\Big)\\Big(\\sum_{j=1}^nb_j^2 - b_1^2\\Big) \\ge \\Big(\\sum_{k=1}^na_kb_k - a_1b_1\\Big)^2[/tex]\n[/quote]\n\nDoesn't Cauchy-Schwarz only accept positive terms? How can you justify that the inequality will hold when you artificially insert -a12 and -b12?[/quote]\r\n\r\nDo you know a proof for Cauchy? If not, I would suggest looking one up. Once you know a derivation you should be able to answer the question of whether or not Cauchy is true when member(s) of either sequence are negative or nonpositive.", "Solution_6": "Osiris: first, Cauchy accepts negative terms -- you can prove this easily just by looking at Cauchy for positive terms and noting that switching some of the terms to negative can only decrease the absolute value of the smaller side while the larger side will remain unchanged. However, looking at any derivation of Cauchy should also make it clear. In this case, however, it wouldn't even matter since the expression he gave can also be written as\r\n\r\n[tex]\\displaystyle \\Big(\\sum_{i=2}^na_i^2\\Big)\\Big(\\sum_{j=2}^nb_j^2\\Big) \\ge \\Big(\\sum_{k=2}^na_kb_k\\Big)^2[/tex]\r\n\r\nfrom which the stated result still follows.\r\n\r\n\r\n\r\nEdit: Oh, I just understood what you mean! Yes, the larger side does have to be a sum of squares, not a sum or difference. But if you re-write it as I did, it doesn't matter at all. I don't know why he chose to use the notation he did instead of mine, but I hope my explanation clears it up.", "Solution_7": "Er, yeah. What I meant was that in any of the brackets, all of the terms (a12 through to an2, and likewise for b and aibi) are supposed to have + signs in front of them.\r\n\r\nI derived Cauchy in its vector form, though I suspected it of being loopy logic. Cauchy's inequality is ridiculously easy to remember in its vector form:\r\n\r\n|[b]a[/b] . [b]b[/b]| :le: |[b]a[/b]||[b]b[/b]|, where . denotes the dot product, and [b]a[/b] and [b]b[/b] are two vectors. (So Cauchy's algebraic form specifies two vectors in [b]R[/b]n.)\r\n\r\nProof:\r\n\r\n[tex]|\\mathbf{a} \\cdot \\mathbf{b}|=|\\mathbf{a}||\\mathbf{b}|\\cos\\theta[/tex],\r\n\r\nwhere [tex]\\theta[/tex] is the angle between vectors [tex]\\mathbf{a}[/tex] and [tex]\\mathbf{b}[/tex]. [tex]\\cos\\theta[/tex] attains maximal value at 1, therefore \r\n\r\n[tex]|\\mathbf{a}||\\mathbf{b}|\\cos\\theta \\leq |\\mathbf{a}||\\mathbf{b}|[/tex]\r\n\r\nand the result follows.\r\n\r\nClearly from this proof (and definition of dot product), every term inside each bracket must have a + sign.", "Solution_8": "Here is a proof for Cauchy Schwarz:\r\n\r\nThe polynomial f(x) = :Sigma: (ax-b)^2 is clearly non-negative. We expand this polynomial into ( :Sigma: a^2)x^2 - ( :Sigma: (2ab)) x + :Sigma: b^2. Since the polynomial is non-negative, the discriminant must be non-positive, i.e. ( :Sigma: (2ab))^2 - 4( :Sigma: a^2 :Sigma: b^2) :le: 0. Dividing by 4 and shifting terms around, we obtain ( :Sigma: ab)^2 :le: ( :Sigma: a^2 :Sigma: b^2), which clearly still holds for negative a and b.\r\n\r\nSorry, dunno how to TEX.", "Solution_9": "Quote:Let [tex]a_1,\\dots,a_n, b_1\\dots,b_n[/tex] be real numbers such that\n[tex]\\sum_{k=1}^na_i^2=1[/tex], [tex]\\sum_{k=1}^nb_i^2=1[/tex] and [tex]\\sum_{k=1}^na_ib_i=0[/tex]. Show that [tex]a_i^2+b_i^2\\le1[/tex] for all [tex]i=1,\\dots,n[/tex].\n\n\n\nSince this has been proved by probability1.01 above, I am posting a different solution here using Linear Algebra. \n\n[hide]Let [tex]{\\bf v_1} = (a_1,\\dots,a_n)[/tex] and [tex]{\\bf v_2} =(b_1\\dots,b_n)[/tex] be vectors in [tex]\\mathbb R^n}[/tex]. Extend [tex]{\\bf v_1}, {\\bf v_2}[/tex] to an orthonormal basis [tex]{\\bf v_1}, {\\bf v_2},\\ldots,{\\bf v_n}[/tex] of [tex]\\mathbb R^n}[/tex]. Make an [tex]n\\times n[/tex] orthogonal matrix [tex]U[/tex] using [tex]{\\bf v_1}, {\\bf v_2},\\ldots,{\\bf v_n}[/tex] as rows of [tex]U[/tex]. Then the transpose [tex]U^T[/tex] is an orthogonal matrix too. In particular, each row vector of [tex]U^T[/tex] has length 1. This implies that [tex]a_i^2+b_i^2\\le1[/tex] for all [tex]i=1,\\dots,n[/tex]. Q.E.D.[/hide]", "Solution_10": "ooh, nice proof" } { "Tag": [ "integration", "real analysis", "real analysis solved" ], "Problem": "Compare numbers(clarify which one is greater or equal to) $\\sqrt{e-1}$ or $\\int_0^1 \\sqrt{x}e^xdx$", "Solution_1": "Fast, with Cauchy-Schwarz : $\\int_0^1 \\sqrt{x}e^x.dx \\leq \\sqrt{\\int_0^1x.dx}\\sqrt{\\int_0^1e^{2x}.dx} = \\frac{\\sqrt{e^2-1}}{2} < \\sqrt{e-1} $ \r\n:cool:", "Solution_2": "i used Cauchy-Schwarz but you got sharper estimate:\r\n $\\int_0^1 \\sqrt{x}e^x.dx =\\int_0^1 (\\sqrt{x}e^{\\frac{x}{2}})e^{\\frac{x}{2}}.dx \\leq \\sqrt{\\int_0^1xe^x.dx}\\sqrt{\\int_0^1e^x.dx} = \\sqrt{e-1} $" } { "Tag": [ "linear algebra" ], "Problem": "show that the maximale dimension of a vectoriel space who don't containe a inversible martix is $ n(n\\minus{}1)$?", "Solution_1": "it's easy.\r\nhint :le noyau d une forme lineare est un hyperplan et les seuls formes linears de Mn(K) sont de la forme f(A)=ktrace(A)" } { "Tag": [], "Problem": "Results for the 2006 University of Maryland Math Competition are here: http://www.math.umd.edu/highschool/mathcomp/2006winner.html\r\n\r\nFirst Prize\r\n Brian Lawrence Montgomery Blair High School\r\nSecond Prize\r\n Richard McCutchen Montgomery Blair High School\r\nThird Prize\r\n Qinxuan Pan Thomas S. Wootton High School \r\n\r\nWow ... who would have guessed Brian Lawrence would win! :rotfl: And Richard McCutchen is no surprise either.\r\n\r\nI got an honorable mention, which is fine, but chess64 beat me for first in Baltimore County. :mad: Just kidding, nice job. :)", "Solution_1": "Whoa I won? Cool :D I win $\\$$100...", "Solution_2": "Do you guys know your part II scores?", "Solution_3": "Not yet :(\r\n\r\nEDIT: But I just found out that I got 18th place in MD.", "Solution_4": "I got 20th place. We probably got the same score on Part II and his extra 2 points on Part I won it for him.\r\n\r\nApparantly both of us did well on Part II since chess64 jumped 16 places and I jumped 21 places from Part I to Part II.", "Solution_5": "[quote=\"E^(pi*i)=-1\"]I got 20th place. We probably got the same score on Part II and his extra 2 points on Part I won it for him.[/quote]\r\n\r\nLol, were the scores that close?\r\n\r\nI dunno... Dr. Asaro is probably going to tell us our scores next Thursday though.", "Solution_6": "I am sorry, but what was the score of Brian Lawrence?", "Solution_7": "[quote=\"mathgeniuse^ln(x)\"]I am sorry, but what was the score of Brian Lawrence?[/quote]\r\n\r\nPerfect.", "Solution_8": "Do you know this for sure, or is this just a good guess?", "Solution_9": "For sure. He's won first place every year he's taken it, except for last year when he got second. And he got a perfect 2 years ago, I think.\r\n\r\nEDIT: typo", "Solution_10": "Well, one person got a perfect score on both Parts, and Brian Lawrence won, so it must have been him. :-)\r\n\r\nPart I had a lot of ties, so Part II probably broke a lot of those.\r\n\r\nStudents qualifying for Part II were about the top 10%. Students ranked in the top 100 were about the top 4%. Students with Honorable Mention were about the top 1%. Students winning prizes, overall or county, were the top fraction of a percent.\r\n\r\nAnd yes, indeed, I can give you the scores on Thursday.\r\n\r\nGood job!\r\n\r\nBest,\r\nDr. Catherine Asaro" } { "Tag": [ "logarithms", "function", "real analysis", "real analysis unsolved" ], "Problem": "show, for $ |z|\\,<\\,1$ :D\r\n\r\n$ \\ln\\,\\left(\\Gamma\\,(z)\\right)\\; \\equal{}\\;\\minus{}\\ln z\\minus{}\\gamma\\cdot z\\plus{}\\frac{1}{2}\\zeta(2)\\cdot z^{2}\\minus{}\\frac{1}{3}\\zeta(3)\\cdot z^{3}\\plus{}\\frac{1}{4}\\zeta(4)\\cdot z^{4}\\minus{}\\frac{1}{5}\\zeta(5)\\cdot z^{5}\\plus{}\\frac{1}{6}\\zeta(6)\\cdot z^{6}\\minus{}\\;\\;\\;\\ldots\\;\\;\\;\\ldots\\;\\;\\;\\plus{}\\minus{}\\;\\;\\;\\ldots$\r\n\r\n( where $ \\gamma$ is Euler's constant )", "Solution_1": "many of your problems is particular case of the properties wellknown function.\r\nhttp://dlmf.nist.gov/Contents/GA/7/" } { "Tag": [ "function", "combinatorics proposed", "combinatorics" ], "Problem": "1) Compute the number of integer sequences $ (a_1, \\ldots, a_n)$ such that $ a_1 \\equal{} 1$ and $ 1 \\leq a_{i \\plus{} 1} \\leq a_i \\plus{} 1$ for $ 1 \\leq i \\leq n \\minus{} 1$.\r\n\r\n2) Compute the number of integer sequences $ (a_1, \\ldots, a_n)$ such that $ a_1 \\equal{} 1$ and $ 1 \\leq a_{i \\plus{} 1} \\leq 1 \\plus{} \\max\\{a_1, \\ldots, a_i\\}$ for $ 1 \\leq i \\leq n \\minus{} 1$.", "Solution_1": "For 2, the best form I can get is the following: if $ \\Delta^k P(x)$ is the $ k$th forward difference of $ P$, and $ P(x) \\equal{} x^n$, then the number of sequences is\r\n$ \\sum_{k\\equal{}1}^n \\left( \\frac{\\Delta^k P(0)}{k!} \\right)$\r\nDo you have a simpler closed form in mind? If not I can post the derivation.", "Solution_2": "It's actually a well-known sequence, though not one that has what I would be inclined to call a closed formula. Does have a nice (exponential) generating function, though. Anyhow, I'd be happy to see the derivation.", "Solution_3": "Casework on what the maximum of the whole sequence is; call that max $ k$. Use PIE to figure out the number of $ n$-term integer sequences with elements from a set of $ k$ elements that uses each element once; this simplifies to $ \\Delta^k P(0)$. There is only one way to assign the numbers $ \\{1,2,\\ldots k\\}$ to the elements of the set we're picking from to form a valid sequence from this, so we divide by $ k!$. Now just sum over all $ k$.", "Solution_4": "The identity\r\n\r\n$ x^n \\equal{} \\sum_{k \\equal{} 0}^{n} S(n, k) k! {x \\choose k}$\r\n\r\ngives $ \\Delta^r P(0) \\equal{} \\sum_{k \\equal{} 0}^{n} S(n, k) k! {0 \\choose k \\minus{} r} \\equal{} S(n, r) r!$, which gives the answer as $ \\sum_{k \\equal{} 1}^{n} S(n, k) \\equal{} B_n$, the [url=http://en.wikipedia.org/wiki/Bell_numbers]Bell numbers[/url]. \r\n\r\nWith that knowledge, there should be a nice explicit bijection to partitions of $ [n]$.\r\n\r\nEdit: Here it is. The value of $ a_k$ describes the index of the partition to which $ k$ belongs. At each step, we either add $ k$ to an existing partition or create a new partition and place $ k$ into it.\r\n\r\nEdit #2, 3: Cute. 1) has a bijection to Dyck paths: whenever $ a_{i \\plus{} 1} \\equal{} a_i \\plus{} 1$, we go to the right, and whenever $ a_{i \\plus{} 1} < a_i \\plus{} 1$ we need to go up $ a_i \\plus{} 1 \\minus{} a_{i \\plus{} 1}$ times first and then go to the right. Finally, we go up $ a_n$ times. For a solution in the spirit of 2), we can biject to plane trees by setting $ a_0 \\equal{} 0$ as the root and connecting $ a_i$ to the largest $ j < i$ such that $ a_i \\equal{} a_j \\plus{} 1$. Either way, the answer is the [url=http://en.wikipedia.org/wiki/Catalan_number]Catalan numbers[/url].\r\n\r\nEdit #4: Are you using these characterizations to find a short way of proving that $ C_n \\sim 4^n$ on the one hand but $ B_n$ is superexponential?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Given $ a,b,c>0$ and $ a\\plus{}b\\plus{}c\\equal{}3$.Prove that:\r\n$ \\frac{a}{\\sqrt{4b\\plus{}3c^2}}\\plus{}\\frac{b}{\\sqrt{4c\\plus{}3a^2}}\\plus{}\\frac{c}{\\sqrt{4a\\plus{}3b^2}} \\ge\\ \\frac{3}{\\sqrt{7}}$", "Solution_1": "It's obviously trues because: $ 4\\sum\\ ab \\plus{} 3 \\sum\\ a^2b \\leq\\ 21$\r\nNow, try with my old and harder problem (with the same condition)\r\n$ \\frac {a}{\\sqrt {4b \\plus{} 4c^2 \\plus{} 1}} \\plus{} \\frac {b}{\\sqrt {4c \\plus{} 4a^2 \\plus{} 1}} \\plus{} \\frac {c}{\\sqrt {4a \\plus{} 4b^2 \\plus{} 1}} \\geq\\ 1$\r\n :)", "Solution_2": "[quote=\"nguoivn\"]It's obviously trues because: $ 4\\sum\\ ab \\plus{} 3 \\sum\\ a^2b \\leq\\ 21$\nNow, try with my old and harder problem (with the same condition)\n$ \\frac {a}{\\sqrt {4b \\plus{} 4c^2 \\plus{} 1}} \\plus{} \\frac {b}{\\sqrt {4c \\plus{} 4a^2 \\plus{} 1}} \\plus{} \\frac {c}{\\sqrt {4a \\plus{} 4b^2 \\plus{} 1}} \\geq\\ 1$\n :)[/quote]\r\nI can solve this inequality with $ c \\ge b \\ge a$ :( :blush:" } { "Tag": [ "calculus", "derivative", "integration", "function" ], "Problem": "Hi, I'm trying to study some pchem, but I can hardly understand the math. If I'm trying to be as efficient as possible and learn only the most important math stuff I need, what math topics should I read up on?\r\n\r\nsorry if this is overly vague, but I really can't give any more details other than I plan to start cracking on \"Physical Chemistry: A Molecular Approach\" --McQuarrie\r\n\r\nMy math is limited so far to only BC Calc. :(", "Solution_1": "I would start here: http://en.wikipedia.org/wiki/Partial_derivatives because I have see partial derivatives often in pchem books. \r\n\r\nThough, frankly, I am also having difficulty with pchem math.", "Solution_2": "@adchia\r\nYou should mostly learn basic differntiation and integration techniques .It will help you in 90% of the cases as most derivations require calculus only.", "Solution_3": "The math you need is:\r\n\r\n- differential calculus of functions of a single variable and of several variables (concept of partial derivatives and relations between them, total differential);\r\n\r\n- integral calculus (primitives, ordinary integrals, and line integrals);\r\n\r\n- Taylor series;\r\n\r\n- basic facts about logarythmic and exponential functions;\r\n\r\n- a good ability to interpret graphs/diagrams (such as Clayperon (or P-V) diagrams, T-S diagrams, and so on).", "Solution_4": "ok thanks! off i go to find an MV calc book." } { "Tag": [], "Problem": "Of $ 28$ students taking at least one subject the number taking Mathematics and English only equals the number taking Mathematics only. No student takes English only or History only, and six students take Mathematics and History, but not English. The number taking English and History only is five times the number taking all three subjects. If the number taking all three subjects is even and non-zero, the number taking English and Mathematics only is:\r\n\r\n$ \\textbf{(A)}\\ 5 \\qquad \\textbf{(B)}\\ 6 \\qquad \\textbf{(C)}\\ 7 \\qquad \\textbf{(D)}\\ 8 \\qquad \\textbf{(E)}\\ 9$", "Solution_1": "[hide]\n$ 28\\equal{}(m)\\plus{}(e)\\plus{}(h)\\plus{}(me)\\plus{}(mh)\\plus{}(eh)\\plus{}(meh)$.\n\n$ (m)\\equal{}(me)\\equal{}x$\n\n$ (e)\\equal{}(h)\\equal{}0$\n\n$ (mh)\\equal{}6$\n\n$ (meh)\\equal{}y\\Rightarrow (eh)\\equal{}5y$.\n\n$ 28\\equal{}2x\\plus{}6\\plus{}6y\\Rightarrow 11\\equal{}x\\plus{}3y$.\n\nIf $ y\\geq 4$, $ x$ will have to be negative. So $ y\\equal{}2\\Rightarrow x\\equal{}5$.\n[/hide]" } { "Tag": [ "geometry", "geometric transformation", "reflection", "analytic geometry", "\\/closed" ], "Problem": "Is there a way that we can change the time to reflect daylight savings time, or would this be too complicated? \r\n\r\nOh, I just realized - alternatively, individual users can change their time settings in their profile. Perhaps setting the \"Time Zone\" to 1 hour back would do the trick...", "Solution_1": "Actually, I don't think it would require too much script to make daylight savings changes. However, we must take into consideration that there are some countries that don't do daylight savings...", "Solution_2": "some states for that matter also", "Solution_3": "yes...some states as well...", "Solution_4": "right...most iof IN doesn't observe daylight savings time...I know I don't and I live in northern IN...", "Solution_5": "odd...so how does one side of the state coordinate with the other side?", "Solution_6": "They don't -- the north-west corner is at a different time than the rest. Just like poeple in the central time zone are at a different time than people in the eastern time zone." } { "Tag": [ "AMC", "AMC 10", "AIME", "USA(J)MO", "USAMO", "geometry", "FTW" ], "Problem": "Yay! How did everyone else fail the AMC10/12?\r\nI managed to assume 20+3+6 was 33 on problem six of the AMC10! \r\nMods please remove if this is considered spam.", "Solution_1": "[quote=\"ra5249\"]Yay! How did everyone else fail the AMC10/12?\nI managed to assume 20+3+6 was 33 on problem six of the AMC10! WOoT for 144!\nMods please remove if this is considered spam.[/quote]\r\n\r\n$ 144 \\ne \\text{failure.}$\r\n\r\nSeriously, stop that.", "Solution_2": "dude wtf yea its a bad mistake to make since it cost you a perfect but imagine the people that didnt even pass...\r\n\r\ni didnt make any stupid mistakes (like computational) but i didnt finish out 3 problems on the 12 because i assumed they were too hard. However, after the test, I could do them each in like 4 minutes. Next time, I should try to at least attempt all the problems and not be intimidated just because its a number 24 or something.", "Solution_3": "[quote=\"the future\"]dude wtf yea its a bad mistake to make since it cost you a perfect but imagine the people that didnt even pass...\n\ni didnt make any stupid mistakes (like computational) but i didnt finish out 3 problems on the 12 because i assumed they were too hard. However, after the test, I could do them each in like 4 minutes. Next time, I should try to at least attempt all the problems and not be intimidated just because its a number 24 or something.[/quote]\r\n\r\nFirst time I took the AIME (2006 II) I bruteforced number 15... and got it right.", "Solution_4": "the first time i took AIME (2007 AIME II) i brute forced #2 and got it right :D", "Solution_5": "I failed.\r\nEDIT: I guess I will just remove my score from this post so it will not affect anyone.", "Solution_6": "I got a 123 :( Misbubbled 6,7, got 16 wrong because of a stupid mistake, and left two blank.", "Solution_7": "have u gotten ur official results back yet????", "Solution_8": "By saying you guys did terrible, how do you think you're making the people who got lower scores than you feel? I know that many hard-working people at my school would be happy to score anything higher than 100 on the AMC12. Even if you feel that you could have done better, complaining is simply annoying and not productive.", "Solution_9": ":huh: This chastising of people who enjoy showing off by announcing to the world that they \"failed\" by getting high scores sounds familiar... But of course! We've heard of it during the AMC 8 posts, haven't we?!\r\n\r\nBut seriously, ra(random numbers), plz stop making others feel bad this way. It gets really annoying...", "Solution_10": "Argh- I hate my internet- somehow this didn't post earlier\r\nRem- I feel your pain to some level. I was an HM last year too, I've aced the AMC 10 twice, and I scored a 117 on the 12A, which is lower than I scored in 7th grade, even after accounting for the change in scoring that has happened.\r\nI also think that working on Olympiad problems rather than focusing on speed/computational accuracy hurt me. At the same time (and I'm addressing this to everyone, not just Rem), these are part of the game. We know the rules, we know that arithmetic counts, and we know that if we make mistakes, we'll lose points no matter how well we understand the problem. I got a 117 because I earned 117 points under the rules that we were playing with, like those rules or not. Congratulations to everyone who did better than that, and to everyone who may have scored less, but made their goal. No matter what your goals are, be they making AIME, being a USAMO winner, or finding a date for a school dance, achieving them is always worthy of celebration. For those of you who didn't make your goals (including me), just remember that a lot of the value in these competitions comes from learning something while you prepare.\r\n-Max", "Solution_11": "Wow. I agree with you completely. I also think that I did too much olympiad math and forgot how to double check things.\r\nBut I think it's still not over - there is still AIME coming up.", "Solution_12": "Yeah- I think that an 8 (and certainly a 9) would get either of us through, which is basically what would've been required pre-USAMO expansion.", "Solution_13": "I made a mistake on #9 for the AMC 12. Apparently, I thought that $ 5.4\\times 4 \\equal{} 20.4$. I was devastated when I found out that I guessed #24 and #25 correctly. So I was praying that there was some other question I got wrong (so that way I didn't mess up what should have been a perfect score). And I did. I got #23 wrong, and I thought I did it correctly. So I guess I'm kind of relieved.", "Solution_14": "[quote=\"xpmath\"]I got a 123 :( Misbubbled 6,7, got 16 wrong because of a stupid mistake, and left two blank.[/quote]\r\n\r\nhow do you know you misbubbled?", "Solution_15": "ppl, 120-150 don't equal FAIL!!!\r\n\r\nSERIOUSLY!!!!!!!!\r\n\r\nI found this years 10A VERY CHALLENGING. I skipped three problems, and am not sure how many I missed, but I must have missed about 3.", "Solution_16": "misbubbling ftw", "Solution_17": "I was pretty lucky this year, so I can't say anything...but I'll put in a post for the ~10 people at my school who got scores around 95 and are anxious to know whether or not they made it. Some of them went for 14 correct, 11 blank in the hopes of making it...and messed up one problem. Others went for 15 and 10...and messed up two. Etc, etc...", "Solution_18": "[quote=\"kimmystar94\"]There is only so much room in the top of the pyramid.[/quote]\r\nnice\r\ni totally agree, tho.\r\ni'm taking the 10b cause my school can't do the 10a for some reason\r\ni didn't kno that u could take the 10a and the 10b....\r\ni wish i could do that.", "Solution_19": "i demand 2 memberships in this club.\r\n\r\ni \r\nfailed\r\nmiserably\r\n-_-", "Solution_20": "I'm going to leave this discussion open (for now at least), but I suggest that you read this thread:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=31968", "Solution_21": "Guys. Calm down. If you made it, good for you. Think about thsoe hopeful people who didn't make it at all. I got a 144, and I'm completely happy with my score. Of course I wasn't aiming for a 150... I was aiming for a 120 :rotfl: . But considering how old most of you ppl are, a 144 is uber-pwnage.\r\n\r\nSo what you got a 144. That's life. Life deals out its cards. If it gives you a bad hand, you take what you have and you deal with it.\r\n\r\nNo need to cry over spilled milk...\r\n\r\nSeriously, please stop complaining. It's not like ranting about missing one problem is going to magically change your answer on the bubble sheet and make it a 150.\r\n\r\nChillax :)", "Solution_22": "Its not particularly encouraging when you do much better in 7th grade on a harder test than in 8th grade on an easier test.", "Solution_23": "I got a 120 on the 10 (if i didn't bubble anything wrong)... :maybe: at least i passed, but did worse than I expected to. Made a stupid mistake on 7 and 12. x.x And there was too much geometry!", "Solution_24": "[quote=\"Twiz\"][quote=\"serialk11r\"]Yes, AoPS is a community of some of the best. If you feel bad because you see other people saying they got \"bad\" scores, then don't visit this forum.[/quote]My impression is that AoPS is a place for students to learn math with and from their peers, not a place for \"some of the best\" to conglomerate and discuss their greatness. I like to think of AoPS as a community welcoming to new math students eager to learn, but you seem to prefer the idea of a cut-throat environment encouraging those who are not \"good enough\" to leave. Also, the caps and quadruple exclamation points aren't helping this discussion.[/quote]\r\nI didn't say it was FOR the best, I said it was a community of some of the best. That means there are some really bright people here. What I'm trying to say is it's really really hard to not have those \"bragging\" posts on a forum like this so just do your best to ignore. Besides, your own goals are most important.", "Solution_25": "[quote=\"infinity4ever\"] Life deals out its cards. If it gives you a bad hand, you take what you have and you deal with it.[/quote]\n\n[quote=\"kimmystar94\"] There is only so much room at the top of the pyramid. [/quote]\r\n\r\nsorry if this is spam, but\r\nyou guys have nice quotes...", "Solution_26": "I failed in that I didn't reach my goal. Oh well.", "Solution_27": "what do u think the usamo cutoff is going to be??", "Solution_28": "[quote=\"Narcissa\"]what do u think the usamo cutoff is going to be??[/quote]\r\n\r\nI think the AIME would need to occur before we could begin to take a guess at that.", "Solution_29": "Sometimes these things are memorable. Nearly 10 years later, I still remember exactly how I got #4 on the 1999 AHSME wrong (69 is prime!). That was the only time I ever missed one of the early questions. What was my score? 145, 7 points higher than my second-best.\r\nIn honor of that question, 69 is officially my unlucky number.\r\n\r\nBut... don't use this as an excuse to whine. One question less than an awesome score is still great. I am proud of my sister's school-winning 102 just as I am proud of my state-winning 145." } { "Tag": [ "calculus", "integration", "logarithms", "calculus computations" ], "Problem": "[u][b]without[/b][/u] using contour integration, show :| \r\n\r\n$ \\int_{0}^{1}\\; \\frac{\\ln{(y^{2})}}{(1+y^{2})\\;(\\pi^{2}+\\ln^{2}y)}\\;\\;\\textbf dy\\;=\\;\\boxed{\\ln 2-\\frac{1}{2}}$", "Solution_1": "Misan,\r\nAs stated the integral equality is false since the integral is definitely a negative real number while $ \\ln(2)-\\frac{1}{2}>0$.\r\n\r\nHere is what I have:\r\n\r\nThe integral equals: \r\n\r\n$ -\\frac{4}{\\pi}\\sum_{m=0}^{\\infty}\\frac{(-1)^{m}(2m+1)}{(2m+1)^{2}-4}\\ln\\left(m+\\frac{1}{2}\\right)$.\r\n\r\nI will use the following formula:\r\n\r\n$ \\frac{1}{\\cosh x}=\\sum_{m=0}^{\\infty}(-1)^{m}\\frac{\\pi(2m+1)}{x^{2}+(2m+1)^{2}\\frac{\\pi^{2}}{4}}$.\r\n\r\nMake the substitution $ y=e^{-t}$ and we get that\r\n\r\n$ I=-\\int_{0}^{\\infty}\\frac{tdt}{\\cosh(t)(\\pi^{2}+t^{2})}dt=$\r\n\r\n$ =-\\sum_{m=0}^{\\infty}(-1)^{m}\\pi(2m+1)\\int_{0}^{\\infty}\\frac{t}{(t^{2}+\\pi^{2})(t^{2}+(2m+1)^{2}\\pi^{2}/4)}dt$\r\n\r\nand after some calculations\r\n\r\n$ =-\\frac{4}{\\pi}\\sum_{m=0}^{\\infty}\\frac{(-1)^{m}(2m+1)}{(2m+1)^{2}-4}\\ln\\left(m+\\frac{1}{2}\\right)$.\r\n\r\nMaybe this sum can be expressed in terms of well-known constants.", "Solution_2": "Hi!\nWhy $ -\\frac{4}{\\pi}\\sum_{m=0}^{\\infty}\\frac{(-1)^{m}(2m+1)}{(2m+1)^{2}-4}\\ln\\left(m+\\frac{1}{2}\\right) =\\boxed{\\ln 2-\\frac{1}{2}} $?? :(" } { "Tag": [ "inequalities", "three variable inequality" ], "Problem": "Let $ a,b,c$ be real numbers.Prove that:\r\n$ (a^2\\plus{}2)(b^2\\plus{}2)(c^2\\plus{}2) \\ge 3(a\\plus{}b\\plus{}c)^2$\r\nThis inequality easy but nice.I hope you will like it,my friend :)", "Solution_1": "[quote=\"quykhtn-qa1\"]Let $ a,b,c$ be real numbers.Prove that:\n$ (a^2 \\plus{} 2)(b^2 \\plus{} 2)(c^2 \\plus{} 2) \\ge 3(a \\plus{} b \\plus{} c)^2$\nThis inequality easy but nice.I hope you will like it,my friend :)[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=4830\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=76508", "Solution_2": "[quote=\"arqady\"][quote=\"quykhtn-qa1\"]Let $ a,b,c$ be real numbers.Prove that:\n$ (a^2 \\plus{} 2)(b^2 \\plus{} 2)(c^2 \\plus{} 2) \\ge 3(a \\plus{} b \\plus{} c)^2$\nThis inequality easy but nice.I hope you will like it,my friend :)[/quote]\nSee here:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=4830\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=76508[/quote]\r\nSorry,here is my solution:\r\nPut $ x\\equal{}b\\plus{}c$ we have: $ (b^2\\plus{}2)(c^2\\plus{}2) \\ge 3\\plus{} \\frac{3}{2}x^2$\r\nThen$ LHS \\minus{} 3(a\\plus{}x)^2 \\ge \\frac{3(ax\\minus{}2)^2}{2} \\ge 0$\r\nThen,we have Q.E.D :)", "Solution_3": "[quote=\"quykhtn-qa1\"]Let $ a,b,c$ be real numbers. Prove that\n\n$ (a^2 \\plus{} 2)(b^2 \\plus{} 2)(c^2 \\plus{} 2) \\ge 3(a \\plus{} b \\plus{} c)^2,$[/quote]$ \\prod{\\left(a^2 \\plus{} 2\\right)} \\minus{} 3\\left(\\sum{a}\\right)^2 \\equal{} \\frac {8}{9}(bc \\plus{} ca \\plus{} ab \\minus{} 3)^2 \\plus{} \\frac {4}{9}\\sum{(b \\minus{} c)^2}$\r\n\r\n$ \\plus{} \\frac {1}{27}\\sum{\\left[a \\plus{} b \\plus{} c \\minus{} 3abc \\plus{} \\sqrt {5}(ca \\plus{} ab \\minus{} 2bc)\\right]^2}\\geq0.$\r\n\r\nThis problem is introduced as Example 2.4.18 on pp. 79-80 in new book : \r\n\r\nJi Chen, Chao-Cheng Ji, [url=http://www.newbooks.com.cn/book/321710.html]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O\u00b7613)", "Solution_4": "for #1\n\\[(a^2+2)(b^2+2)(c^2+2)-3(a+b+c)^2=\\frac{1}{9}\\sum{(4+5c^2)(a-b)^2}+\\frac{8}{9}(ab+bc+ca-3)^2+\\frac{1}{9}(3abc-a-b-c)^2\\ge{0}\\]", "Solution_5": "use schur inequality,quykhtn-qa1.\nsorry for my bad english!", "Solution_6": "[url=http://www.artofproblemsolving.com/community/c6h1389087p7737051]Viet Nam TST 2017 MOCK Test[/url]", "Solution_7": "For #1\n$$(a^2+2)(b^2+2)(c^2+2) -3(a+b+c)^2 = \\frac19\\sum \\left( 2\\,{c}^{2}+3 \\right) \\left( a-b \\right) ^{2}+\\frac13\\sum a\n \\left( a{b}^{2}+2\\,{b}^{2}+a \\right) \\left( c-1 \\right) ^{2}+\\frac89 (ab+bc+ca-3)^2 $$\n$$=\\frac{1}{12}\\sum \\left( {b}^{2}{c}^{2}+2\\,b{c}^{2}+{b}^{2}+5\\,{c}^{2}+10\\,b+5\n \\right) \\left( a-1 \\right) ^{2}+\\frac{1}{36}\\sum \\left( 3\\,abc+4\\,ab-2\\,ac-2\\,\nbc-a-b-c \\right) ^{2} +\\frac{3}{4} (ab+bc+ca-3)^2$$", "Solution_8": "If $k\\ge 0$ and $a_1,a_2,...,a_n\\ge 0$, then\n\n$(a_1^2+n-1+k)(a_2^2+n-1+k)...(a_n^2+n-1+k)\\ge (n+k)^{n-2}(a_1+a_2+...+a_n+k)^2$.\n", "Solution_9": "[quote=quykhtn-qa1]Let $ a,b,c$ be real numbers.Prove that:\n$ (a^2\\plus{}2)(b^2\\plus{}2)(c^2\\plus{}2) \\ge 3(a\\plus{}b\\plus{}c)^2$\nThis inequality easy but nice.I hope you will like it,my friend :)[/quote]\n\nMy SOSprogram output an non-symmetric SOS for it.\n$$\\frac{1}{9} \\left( abc-3a+b+c \\right) ^{2}+\\frac{2}{9} \\left( abc-2b+c\n \\right) ^{2}+{\\frac {5}{18}} \\left( 2ab-ac-bc \\right) ^{2}+{\n\\frac {8}{9}} \\left( ab+ac+bc-3 \\right) ^{2}+\\frac{5}{6}c^2 \\left( a-b\n \\right)^2+\\frac{2}{3}c^2 \\left( ab-1 \\right)^2\\ge 0.$$" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "$ let$ $ f: \\mathbb{N}*\\mathbb{N}\\equal{}\\equal{}>\\mathbb{N}$ \r\n\r\n $ (x,y)\\equal{}\\equal{}>{\\frac{(x\\plus{}y)(x\\plus{}y\\plus{}1)}{2}}\\plus{}y$\r\n$ prove$ $ that$ $ f$ $ is$ $ injective$.\r\n :)", "Solution_1": "${ \\frac{(x+y)(x+y+1)}{2}}+y =k \\Longrightarrow x^2+(2y+1)x+y^2+3y-2k=0$\r\n$ x$ have to be integer, so $ (2y+1)^2-4(y^2+3y-2k)=(2u+1)^2$, so $ y=k-\\frac{u(u+1)}2\\ ,\\ x=\\frac{u(u+3)}2-k$\r\nWe need $ x\\geq 0, y\\geq 0 \\ \\Longrightarrow\\ u(u+1)\\leq 2k\\leq u(u+3)\\ \\Longrightarrow$$ \\frac {-3+\\sqrt{9+8k}}2 \\leq u \\leq \\frac {-1+\\sqrt{1+8k}}2$\r\nBut $ \\frac {-1+\\sqrt{1+8k}}2-\\frac{3+\\sqrt{9+8k}}2 < 1$ , so there is only one integer $ u$\r\nSo if $ f(x,y)=f(a,b)=k$ then $ x=a\\ ,\\ y=b$ (injective)\r\n........\r\nAn other one . Suppose $ f(x,y)=f(a,b)$\r\n$ x+y=a+b \\Longrightarrow y=b \\Longrightarrow a=x$ (injective)\r\nIf $ x+y\\neq a+b$, let we suppose for exemple $ a+b > x+y \\Longrightarrow a+b=x+y+t$ with $ t \\geq 1$ . We can write :\r\n$ f(a,b)= \\frac{(x+y+t)(x+y+t+1)}2+b >$ $ \\frac{(x+y)(x+y+1)}2 + t(x+y) \\geq \\frac{(x+y)(x+y+1)}2 + y = f(x,y)$ (contradiction !)\r\n :cool:", "Solution_2": "[quote=\"greatestmaths\"]$ let$ $ f: \\mathbb{N}*\\mathbb{N} \\equal{} \\equal{} > \\mathbb{N}$ \n\n $ (x,y) \\equal{} \\equal{} > {\\frac {(x \\plus{} y)(x \\plus{} y \\plus{} 1)}{2}} \\plus{} y$\n$ prove$ $ that$ $ f$ $ is$ $ injective$.\n :)[/quote]\r\n\r\nLet $ x_n\\equal{}\\frac{n(n\\plus{}1)}{2}$\r\nFor any real $ a$, there is a unique $ n\\in\\mathbb{N}\\cup\\{0\\}$ such that $ x_n1$ be an integer not diviseble by 1997. Define $a_i =i+\\frac{in}{1997}$, $i=1,2,...1996$;$b_j=j+\\frac{1997j}{n}$, $j=1,2,....n-1$;\r\nLet $(c_k)_{k=1}^{n+1995}$ be the increasing sequence formed of the$a_{i}'s$and$b_{j}'s$.Prove that$c_{k+1}-c_k<2$for all$k$. 3.How many fuctions$f: N\u00e0N$are there which satisfy$f(1)=1$and$f(n)f(n+2)=f(n+1)^2 +1997$for all$n$?\r\n\r\nDay2\r\n1.(a)Find a polynomial$P$with rational coefficients if th minimum degree such that$P(sqrt[3]3+sqrt[3]9)=3+sqrt[3]3$(b)Does there exist a polynomial$Q$with integer coefficients such that$Q(sqrt[3]3+sqrt[3]9)=3+sqrt[3]3$.?\r\n2.Prove that for each positive integer $n$there is a positive $m$ suchthat $19^m-87$ is diviseble by $2^n$.\r\n3.Letbe given75 points inside a unit cube ,no three of which are collinear.Prove that there exists a triangle with vertices in these points and area not exceeding $\\frac{7}{72}$", "Solution_1": "\\[\\text{Day 1}\\]\r\n1. http://www.mathlinks.ro/Forum/viewtopic.php?p=786131#786131\r\n2. http://www.mathlinks.ro/Forum/viewtopic.php?p=786132#786132\r\n3. http://www.mathlinks.ro/Forum/viewtopic.php?p=786135#786135\r\n\\[\\text{Day 2}\\]\r\n4. http://www.mathlinks.ro/Forum/viewtopic.php?p=786138#786138\r\n5. http://www.mathlinks.ro/Forum/viewtopic.php?p=786139#786139\r\n6. http://www.mathlinks.ro/Forum/viewtopic.php?p=786140#786140" } { "Tag": [ "calculus", "real analysis", "real analysis unsolved" ], "Problem": "I find these surprising things about $\\pi$ and $e$ , I desire to see how to prove them .\r\n\r\n[img]http://bbs.pep.com.cn/attachments/Pi-Continued%20fractions_H9T7JRnbF62F.jpg[/img]\r\n\r\n[img]http://bbs.pep.com.cn/attachments/E-Continued%20fraction_4f7P6NSuEsYB.jpg[/img]", "Solution_1": "I think that you can solve them by writting these formulaes as terms of a sequence $x_{n}$ and the computing it's limit.The recurrence seem easy to be found.\r\n\r\nAnd this is a calculus problem. :roll:", "Solution_2": "The $\\frac{4}{\\pi}$ one is obtained from converting the Wallis product formula $\\frac{8}{\\pi}= \\frac{3\\cdot 3\\cdot 5\\cdot 5 \\cdot ...}{ 4\\cdot 4\\cdot 8\\cdot 8 \\cdot ...}$ into a continued fraction. I am not sure about the details though :oops:", "Solution_3": "I have tried to find the proof on the Internet , but could not get my desired answer . :( \r\n\r\nIf you know where I can find it . Please tell me . \r\n\r\nThank you so much ." } { "Tag": [ "ARML" ], "Problem": "Would the ARML Handbook be good to study for the Mandelbrot competition???\r\n\r\nIf not, what would???", "Solution_1": "The best way to study is certainly to do the previous years' contests. There are books out there with problems and solutions to many of the previous years. If I could find them I'd tell you what they're called...", "Solution_2": "I think one of my teachers has bought a mandelbrot book... \r\n\r\nTY :lol:", "Solution_3": "The books for past Mandelbrot problems are sold on AoPS. The link is below:\n[url]http://www.artofproblemsolving.com/Store/contests.php?contest=mandelbrot[/url]" } { "Tag": [ "limit" ], "Problem": "Demonstrati ca ecuatia $x^n-nx+1=0$ are doua radacini reale $a_{n},b_{n}$ si ca : $\r\n\\mathop {\\lim }\\limits_{n \\to \\infty } a_n = 0,\\mathop {\\lim }\\limits_{n \\to \\infty } b_n = 1$", "Solution_1": "E usor de aratat ca ecuatia are doua solutii reale (una in $[0,1)$ si una $>1$, pt $n>2$) si ca cea mai mica e in intervalul $[0,\\frac 12)$ pt $n$ suficient de mare, deci $na_n=1+a_n^n\\to 0$ cand $n\\to\\infty$, ceea ce inseamna ca $a_n\\to 0$.\r\n\r\nAvem acum $b_n>1$ si $b_n^n-nb_n+1=0,\\ \\forall n$. Se arata si aici usor ca pt $n$ suficient de mare avem $b_n<2$, deci $1> (shift right), same as above but this time one bit from right will be deleted.\r\n\r\n(note if u use 5<<1 u will get the resul in decimal not in bin)\r\n\r\n| (bitwise or), & bitwise and\r\n\r\ni think these four r enough. now plz make a search over yahoo. or google. or consult a book. i hope u will be claer more!\r\n\r\nI apologise for my time problem :blush:", "Solution_8": "Ok, i tried using the bitwise shift as mahbub suggested, it reduced execution time from 0.12 secs to 0.3 secs for the 4th test case.\r\n\r\nBUT, for the fifth case the program now takes more than 2 SECONDS (God knows why!).\r\nHowever, if I add a simple diagnostic cout statement somewhere the program runs and ENDS in 1.82 seconds.\r\n\r\nThis is driving me crazy!!!\r\n\r\nPLEASE HELP\r\nP.S - I have written a mail to Rob Kolstad too.\r\n\r\nThe zip file contains 3 code files - one with the normal power function (runs in 1.22 for last case), one with bitwise shifts (time gets exceeded at 2 secs) and one with bitwise shifts and diagnostic cout statement (runs in 1.83 for the last case)", "Solution_9": "umm.. sorry! tomorrow is my exam. so i will help u after that! plz don't mind. even i can't continue my usaco for it! :mad:", "Solution_10": "ok, thanks a lot anyway\r\n\r\nare u going to ioi too?\r\n\r\ni have been selected for my country's camp (India, i see that v r neighbours) which will take place from 15th to 30th June.\r\n\r\nI have holdiays right now so am relatively free\r\n\r\nanyway will wait for u", "Solution_11": "Well say thanks to your creator. You have made such mistakes (not actually mistake but unefficiency) which were very easy to locate. So I can show you the right path now(without thinking much)!\r\n\r\nWell, you have taken input entirely. No don\u2019t do this. Look it is sufficient if u manage taking B digits at a time. How can you do this?\r\n\r\nSay B=3. and in input file you are given:\r\n\r\n1001001\r\n\r\nnow you are talking input one digit by one digit:\r\n1-10-100-001-010-100-001 OK? Just keeping 3 digits in your hand!\r\n\r\nEach time (after first B digits) you look for digits A to B. and update your counter. Then sort and print.\r\n\r\nAnother think you have checked A to B separately. You don\u2019t need to that as well!\r\n\r\nSay you are given A=2, B=5 and in input: in=11011011, now when you take first B 11011 now take every 2 to 5 from top. Then update your in by taking next digit and deleting first digit. And take care about beginning and ending!\r\n\r\nI know you may not understand every thing clearly. But sorry I can\u2019t spend much time on it now! Sorry!\r\n\r\n***r u indian? so why ur flag says england?", "Solution_12": "well, i will be glad if u post indian IOI selection test problems!\r\n\r\nyes i am going to IOI (IMO too)", "Solution_13": "Hm... Posted my solution, if you want some hints... :D\r\n\r\nBtw, do NOT use iostream to read big inputs... You are reading it \"character for character\" with iostream which must be avoided. The \"reading part\" itself might be too slow for the biggest test case... Try to learn the C style to read/write data instead (scanf/printf). :D\r\n\r\nmahbub: Nice! Will see you twice then. ;)", "Solution_14": "Weeblie: your code is way too complex for me too understand!!!\r\nI will look into the scanf, printf thing too.\r\n\r\nmahbub: Take your time. You can tell me tomorrow (or i hope so, if u don't have another exam!).\r\nI am not sure I quite understand what u r saying but i think it something like this - \r\ni take in the first b digits and process it for all strings with length from a to b. then do a bitwise left shift and then take in another digit and reprocess the string?\r\n\r\nohh and you can find the problems from the indian olympiads on http://www.iarcs.org.in\r\n\r\nI also had a more general question - is it useful for IOI to know things like queues, stacks, linked lists etc?\r\nI ask this because the person who coaches the Indian team for IOI said that we don't need to learn any 'fancy' things and arrays will do fine. But lots of things like BFS,DFS seem to use queues and stacks.\r\n\r\nP.S. - my flag says england b/c i am using somebody else's login - i was too lazy to make my own\r\n\r\nThanks once again", "Solution_15": "The STL classes (vector/queue/stack/deque/list, etc) is not 100% necessary, but can save a lot of time. Especially if you know how to use hash_map/hash_set (different on VC++ and GCC!).\r\n\r\nA lot of people is still writing C code in IOI, without all these nice classes... :D", "Solution_16": "mahbub: can u please explain again in detail what i have to do as i am not yet clear what you said\r\n\r\nor weeblie: can u explain what to do?\r\n\r\nthanks" } { "Tag": [ "inequalities", "function", "symmetry", "algebra", "domain" ], "Problem": "given $0\\le a,b,c\\le 1$,\r\nprove:\r\n$\\frac{a}{1+bc}+\\frac{b}{1+bc}+\\frac{c}{1+ab}\\le 2$", "Solution_1": "isnt the second term\r\n\r\n$\\frac{b}{1+ac}$", "Solution_2": "yes, it should be. :wink: \r\n\r\n[hide=\"big hint\"]\nwlog, assume: $0\\le a\\le b\\le c\\le 1$\n$\\Rightarrow 0\\le (1-a)(1-b)$\n$\\Rightarrow a+b\\le 1+ab\\le 1+2ab$\n$\\Rightarrow a+b+c\\le a+b+1\\le 2+2ab$[/hide]", "Solution_3": "[b]Solution.[/b] The function\r\n\\[f(a,b,c)=\\frac{a}{1+bc}+\\frac{b}{1+ac}+\\frac{c}{1+ab}\\]\r\nis a sum of convex functions (by symmetry, consider $\\frac{\\partial^{2}}{\\partial a^{2}}\\frac{a}{1+bc}$ and $\\frac{\\partial^{2}}{\\partial b^{2}}\\frac{a}{1+bc}$) so $f(a,b,c)$ is convex.\r\nBy Weierstrass' Theorem, a continuous function defined on a closed and bounded domain assumes its maximum and minimum.\r\nFor a convex function, maximum value is obtained at at least one of the $2^{3}$ vertices. Considering $f(0,0,0),f(0,0,1),\\hdots,f(1,1,1)$ by symmetry, a maximum of $2$ is obtained.\r\n\r\nAnd the conclusion follows.$\\blacksquare$\r\n\r\nMasoud Zargar", "Solution_4": "[hide=\"what? calc?!\"]\nWLOG let $a\\geq b\\geq c$, which we can assume since the expression is symmetric. And now we just...\n\\[\\frac{a}{1+bc}\\leq a\\leq 1\\]\n\\[\\frac{b}{1+ac}\\leq \\frac{b}{1+bc}=\\frac{b}{b+c+(1-b)(1-c)}\\leq \\frac{b}{b+c}\\]\n\\[\\frac{c}{1+ab}\\leq \\frac{c}{1+bc}=\\frac{c}{b+c+(1-b)(1-c)}\\leq \\frac{c}{b+c}\\]\nAdd them up! :D \n[/hide]", "Solution_5": "[hide=\"its really easy\"]\nfrom my previous \"big\" hint.\n\n$LHS\\le\\frac{a}{1+ab}+\\frac{b}{1+ab}+\\frac{c}{1+ab}=\\frac{a+b+c}{1+ab}\\le 2$\n\n :wink: [/hide]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "$F_{n}$ is fibonacci 's $n^{th}$ number: $F_{1}=F_{2}=1$, $F_{n+1}=F_{n}+F_{n-1}$ Prove that if and if only $F_{n}^{2}\\mid{F_{m}}$ then $nF_{n}\\mid{m}$.", "Solution_1": "$F_{n}=\\frac{\\alpha^{n}-\\beta^{n}}{\\sqrt 5 }, \\ \\alpha=\\frac{1+\\sqrt 5 }{2}, \\ \\beta=\\frac{1-\\sqrt 5 }{2}.$\r\nTherefore $F_{n}^{2}|F_{m}\\to F_{n}|F_{m}\\to n|m$ and $m=nk, \\ \\frac{F_{m}}{F_{n}}=\\sum_{i=0}^{k-1}\\alpha^{n(k-1-i)}\\beta^{ni}\\equiv k\\alpha^{n}\\mod F_{n}$, because $\\alpha^{n}=\\beta^{n}(mod F_{n})$.\r\nTherefore $F_{n}^{2}|F_{m}\\to F_{n}|\\frac{m}{n}$. It is easy to check, that $nF_{n}|F_{m}\\to F_{n}^{2}|F_{m}$." } { "Tag": [], "Problem": "A student has six textbooks, each with a thickness of 4.0 cm and a weigh of 30 N. What is the minimum work the student would have to do to place all the books in a single vertical stack, starting with all the books on the surface of the table?", "Solution_1": "[hide=\"Hint\"]The first book is already on the table so we don't have to move it. The second book must be raised to a height of 4 cm. The third book must be raised to a height of 8 cm. And so on. Now a question: in each raising operation, to which is equal the work done?[/hide]" } { "Tag": [ "probability", "function", "search", "logarithms", "conics", "ellipse", "geometric series" ], "Problem": "How many ways for 12 people to seat at a round table?\r\nHow many ways for 12 different colored beads placed on a bracelet?\r\nIs above two questions same?", "Solution_1": "For the first question, fix any one person. Seats aren't numbered, so the remaining people can be arranged in $ 11!$ ways.\r\n\r\nFor the bracelet, it is $ \\frac {11!}{2}$ because mirror images are the same as the bracelet can be flipped around. However, the relative positions of the beads matter and the absolute positions do not.", "Solution_2": "Another Question:\r\nHow many generations must you go back before you have at least one million ancestors, assuming that there were no intermarriages among relatives?(Use log(2)=0.30)", "Solution_3": "[quote=\"linj\"]Another Question:\nHow many generations must you go back before you have at least one million ancestors, assuming that there were no intermarriages among relatives?(Use log(2)=0.30)[/quote]\r\nOne more question:\r\nWhy hyperbolic function is too similar to trigonometric function? Is not is confuse? How you distinct them?", "Solution_4": "Does your definition of ancestors include all generations? Or just the current generation we are considering? (E.g. in going back 2 generations I have only 4 ancestors: grandparents on mother's and father's side, while if you consider parents included, I have 4 + 2 = 6 ancestors).\r\n\r\nYou can search \"hyperbolic functions\" in http://en.wikipedia.org and check it out :D", "Solution_5": "[quote=\"aidan\"]Does your definition of ancestors include all generations? Or just the current generation we are considering? (E.g. in going back 2 generations I have only 4 ancestors: grandparents on mother's and father's side, while if you consider parents included, I have 4 + 2 = 6 ancestors).\n\nYou can search \"hyperbolic functions\" in http://en.wikipedia.org and check it out :D[/quote]\r\nI think of when two generation come back is 6 ancestors.", "Solution_6": "In that case, we want to know when $ 2 \\plus{} 2^2 \\plus{} 2^3 \\plus{} \\ldots \\plus{} 2^n \\geq 1,000,000$. By the geometric series formula, this is equivalent to $ \\frac {2(2^n \\minus{} 1)}{2 \\minus{} 1} \\geq 1,000,000$. So $ 2^n \\geq 500,001$ and taking logs on both sides, $ n \\log 2 \\geq \\log 5 \\plus{} 5 \\log 10$ (approximately). So $ n \\geq \\frac {5.699}{0.301} \\equal{} 18.9$. Therefore we must go back at least $ \\boxed {19}$ generations.", "Solution_7": "[quote=\"aidan\"]In that case, we want to know when $ 2 \\plus{} 2^2 \\plus{} 2^3 \\plus{} \\ldots \\plus{} 2^n \\geq 1,000,000$. By the geometric series formula, this is equivalent to $ \\frac {2(2^n \\minus{} 1)}{2 \\minus{} 1} \\geq 1,000,000$. So $ 2^n \\geq 500,001$ and taking logs on both sides, $ n \\log 2 \\geq \\log 5 \\plus{} 5 \\log 10$ (approximately). So $ n \\geq \\frac {5.699}{0.301} \\equal{} 18.9$. Therefore we must go back at least $ \\boxed {19}$ generations.[/quote]\r\nyour answer is same as mine, however, the answer key has answer: 20.", "Solution_8": "If the answer is 20, it means ancestors in the generations before the 20th are not considered.. Only the ancestors of 20 generations ago (not 19, 18, etc) are counted", "Solution_9": "1.The sides of triangle ABC lie on the lines:\r\n 3x+4y=0\r\n 4x+3y=0\r\n x=3\r\n Let (h,k) be the center of the circle inscribed in triangle ABC.\r\n Find h+k.\r\n2. A chessboard is an 8*8 grid. If a chessman is onthe bottom left square, how many paths may he take to the top right square if he moves only up and to the right.\r\n Given the 4*6 grid shown, in how many way can one go from the bottom left corner to the upper right corner along the lines if one can only go up and right?\r\nAre these two question same?\r\n3. How many times does the graph of the function f(x) = 2*x^8-3*x^5-2*x^3-2 cross the x-axis? (without the calculator)", "Solution_10": "[quote=\"linj\"]1.The sides of triangle ABC lie on the lines:\n 3x+4y=0\n 4x+3y=0\n x=3\n Let (h,k) be the center of the circle inscribed in triangle ABC.\n Find h+k.\n2. A chessboard is an 8*8 grid. If a chessman is onthe bottom left square, how many paths may he take to the top right square if he moves only up and to the right.\n Given the 4*6 grid shown, in how many way can one go from the bottom left corner to the upper right corner along the lines if one can only go up and right?\nAre these two question same?\n3. How many times does the graph of the function f(x) = 2*x^8-3*x^5-2*x^3-2 cross the x-axis? (without the calculator)[/quote]\r\nOne more:\r\nA tunnel in the shape of a semi-ellipse is sixty feet wide and twelve yards high in the center. Find its height (in feet) six feet from the edge.", "Solution_11": "@ linj - Each question should have it's own topic. If a topic has several unanswered questions at once, discussion can get very confusing.\r\n\r\n@ mods - If it's not too tedious, please split the unanswered questions into separate topics." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Is there a nonempty perfect set in $ \\mathbb{R}^1$ which contains no rational number?\r\n\r\nA set is perfect if it is closed and if every point is a limit point of the set, and $ \\mathbb{R}^1$ has the standard topology.", "Solution_1": "Yes, see [url=http://planetmath.org/encyclopedia/ANonemptyPerfectSubsetOfMathbbRThatContainsNoRationalNumber.html]here[/url].\r\n\r\nAlternatively, I believe the previous problem contains a clue as to how you would construct another one." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $C_1,C_2$ be tangent together at $M$. Let $M$ be a point on $C_1$ . Draw the tangents from $M$ to $C_2$ at $A,B$ . Let $AM$ intersect $C_1$ at $E$ and $BM$ intersect $C_1$ at $F$ . Let the tangent from $M$ to $C_1$ intersect $EF$ at $D$ . find the locus of $D$ if $M$ be variant on $C_1$ .", "Solution_1": "Could you please clarify it? You have two different definitions for $M$ (first it's the point where the circles touch each other, and then it's a variable point).", "Solution_2": "[quote=\"behzad\"]...Let $M$ be a point on ...[/quote]\r\nI think it must be $N$.", "Solution_3": "If 2nd $M$ is $N$. I just solve the problem \r\nIt is easy and not nice.\r\n :D", "Solution_4": "the locus of D is on line tangent at two circles at $M$.", "Solution_5": "tangent two lines at E and F on $C_1$ let them intesect at $N'$ .\r\nIt's easy to see that $N,M,N'$ are collinear.\r\n$\\frac{N'M}{NM}$ is constant.so the locus of N' is a circle tangent to $C_1$ at $M$.\r\nso the problem bacome this(as darig):\r\n\r\n[color=blue][b]Extendedproblem.[/b] two circles $C,C'$ tangent together at M. from a variable point N on C draw tangent $\\ell$ at C and NM intersect C' at N'. from N draw two tangent NE and NF to C .find that locus of $D=EF\\cap \\ell$[/color]\r\n\r\n[i]Solution.[/i]polar of N' pass through D so the polar of D pass trough N'.therefore DM is tangent to two circles." } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "Show that if x, y, z are positive integers , and (xy+1)(yz+1)(zx+1) is a perfect square then xy+1, yz+1 and zx+1 are all perfect squares.", "Solution_1": "plz search before posting\r\nit is PEN A1,go there...http://www.mathlinks.ro/viewtopic.php?t=150369" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "prove for every positive a,b,c\r\n\r\n3^(1/2)*((a+b)(a+c)(b+c))^1/3 >= 2(ab+ac+bc)^1/2", "Solution_1": "You can use the inequality:\r\n9(a+b)(b+c)(c+a) \\geq 8(a+b+c)(ab+ac+bc) to prove this one....", "Solution_2": "let f =27[(a+b)(b+c)(c+a)] 2 -64(ab+bc+ca)^3\r\n g=27 [(b+ \\sqrt ac)^2(2 \\sqrt ac]^2-64(ac+2b \\sqrt ac)^3\r\n we prove g \\geq 0 f-g \\geq 0 then f \\geq 0", "Solution_3": "let f =27[(a+b)(b+c)(c+a)] 2 -64(ab+bc+ca)^3\r\n g=27 [(b+ \\sqrt ac)^2(2 \\sqrt ac]^2-64(ac+2b \\sqrt ac)^3\r\n we prove g \\geq 0 f-g \\geq 0 then f \\geq 0" } { "Tag": [], "Problem": "Your best friend, your lover, and yourself are trapped on a deserted island out in the middle of nowhere. The relationship between your best friend and you is a truly deep one: you have known each other for many years, and you both trust and can depend on each other very greatly. Your lover is a person whom you really adore, and he/she has a similar affection for you.\r\n\r\nOne day, you are captured by a tribe of cannibals. They are hungry and want to eat, but they are also nice cannibals and do not like to capture more humans than necessary for each meal (I know, this sounds weird, but play along with me here). They say that you (and you alone), must choose one of your two companions to sacrifice. This companion will be slaughtered and eaten by the cannibals, while your other companion and yourself will be free to leave.\r\n\r\nWho would you choose to sacrifice?", "Solution_1": "LOVER SHOULD DIE!!!DEFINiTLY!!!!", "Solution_2": "You know, I realized how stupid this question is right after I posted it, since you can't possibly decide without being given two specific companions. So why don't we make this change:\r\n\r\nReplace the friend with three random students from your school. You are barely acquainted with them. They are not your friends, but not your enemies either.", "Solution_3": "[quote=\"Fierytycoon\"]Replace the friend with three random students from your school. You are barely acquainted with them. They are not your friends, but not your enemies either.[/quote]\r\n\r\nnow we've got ourselves a dilemma, I'm definately saw this ethics question on Mathlinks, with Mother Teresa versus 5 teenagers.", "Solution_4": "Let's see, in the original case I'd kill my best friend, in the second one I'll slaughter the three random kids, and in the third one I'd butcher the five teenagers if Mother Teresa doesn't kill herself before then.", "Solution_5": "this question isn't that hard\r\njust think about it: without your lover, what are you going to do for the rest of your days on this deserted isle. all of you are going to die eventually anyway.", "Solution_6": "yes, but does your lover have any more right to live than the three kids?", "Solution_7": "i was kind of joking and, no, my lover doesn't have any more right to live than the three kids. ethics questions are kind of irrelevant without actual people.", "Solution_8": "I'd have to say that the three kids I don't know have less meaning to me than my lover. After all to the kids' parents their kids would probably mean more to them, but since it's me choosing and not them, the lover it is." } { "Tag": [ "videos" ], "Problem": "Hey, what are your vertical leaps?", "Solution_1": "Mine's like 17 inches.", "Solution_2": "haha, funny topic.\r\n\r\nIn what frame of reference?", "Solution_3": "regular frame I guess.\r\n\r\nMy vertical is 34 inches.", "Solution_4": "Regular frame, eh? Now what does that mean?\r\nIn some frames of reference, I've jumped several meters. In other, I've jumped negative distances. In some frames, I've jumped in a circle of radius one mile.", "Solution_5": "54inches, i dunk better than jordan", "Solution_6": "[quote=\"juicybooty911\"]54inches, i dunk better than jordan[/quote]\r\n\r\nYeah right.", "Solution_7": "14 inches or so. :(", "Solution_8": "[quote=\"juicybooty911\"]54inches, i dunk better than jordan[/quote]\r\n\r\nInteresting fantasy you have there.", "Solution_9": "Mines about 15 inches(not bad for a 12 year old!!!!!!!!!!!!!)", "Solution_10": "I'm finding it hard to measure here, and I'm too lazy to get all of the equipment and all, but I guess the height my feet come off of the ground when legs are straight is somewhere between 2 and 3 feet...\r\n\r\nOkay, upon measurement, about 30 inches...", "Solution_11": "[quote=\"K81o7\"]I'm finding it hard to measure here, and I'm too lazy to get all of the equipment and all, but I guess the height my feet come off of the ground when legs are straight is somewhere between 2 and 3 feet...\n\nOkay, upon measurement, about 30 inches...[/quote]\r\n\r\nThat's very unnatural...", "Solution_12": "[quote=\"robinhe\"][quote=\"K81o7\"]I'm finding it hard to measure here, and I'm too lazy to get all of the equipment and all, but I guess the height my feet come off of the ground when legs are straight is somewhere between 2 and 3 feet...\n\nOkay, upon measurement, about 30 inches...[/quote]\n\nThat's very unnatural...[/quote]\r\n\r\nHow so?", "Solution_13": "Unless you're training for the Olympics...\r\n\r\nDo you realize how much 30 inches is? That means that if you are about 5'10, you can dunk on a standard 10' basketball hoop.", "Solution_14": "@robinhe: nice 1000 posts!\r\n\r\nto _____: u prob beneded ur legs when u measured 30 inches :wink: \r\njorian", "Solution_15": "[quote=\"K81o7\"]I'm finding it hard to measure here, and I'm too lazy to get all of the equipment and all, but I guess the height my feet come off of the ground when legs are straight is somewhere between 2 and 3 feet...\n\nOkay, upon measurement, about 30 inches...[/quote]\r\n\r\nSupernatural......\r\n\r\nmine is only about 10", "Solution_16": "Wait...are you sure? I checked it and confirmed what it was. Maybe I'll take a video and use that to make sure I'm measuring correctly...\r\n\r\nBesides, mdk said 34 inches. That I can believe...\r\n\r\nEdit: Okay, I think I may have found a feasible solution. Maybe my spine is bending over a little bit...and my feet are going out...\r\n\r\nIn any case, it's probably not less than 2 feet.\r\n\r\nI'm a track runner: I can believe that...\r\n\r\nI'll try instead jumping and then marking where the top of my head reached, and then measuring the distance to that and subtracting my height...then I'll get back to you all...\r\n\r\nEdit 2: Okay, after measurement with this method, about 25 inches. I'll get back to you if I can do better than that (track season starts on wednesday, so in a couple months, that may be up a few inches).\r\n\r\nBesides, you say that 30 inches is crazy? I can jump about 25 inches, and my brother can probably jump a couple inches higher than that: probably about 30.", "Solution_17": "You're probably getting a running start.\r\n\r\nThat doesn't count. Stand against the wall and then just jump. See how much that is.", "Solution_18": "use lo-tack masking tape. stick a peice where your fingertips touch, stick another peice at the height of your jump, and then use a stool and measure.\r\n\r\nhow high of a wall are you doing this on, by the way?", "Solution_19": "[quote=\"k8reindeer\"]use lo-tack masking tape. stick a peice where your fingertips touch, stick another peice at the height of your jump, and then use a stool and measure.\n\nhow high of a wall are you doing this on, by the way?[/quote]\r\n\r\nIf you're asking me, it was a gym wall so naturally it was about 50 ft high or so.", "Solution_20": "[quote=\"robinhe\"]You're probably getting a running start.\n\nThat doesn't count. Stand against the wall and then just jump. See how much that is.[/quote]\r\n\r\nI'm standing still and jumping. Am I not allowed to bend my knees or something? Do you guys want me to make a video of this if you don't believe it?", "Solution_21": "That's probably it. You aren't supposed to bend your knees. Try reaching up and touching the wall, figuring out how high that is, then jumping and seeing how much higher you can reach.\r\n\r\nI'm at about 18 inches I think.", "Solution_22": "[quote=\"mustafa\"]That's probably it. You aren't supposed to bend your knees. Try reaching up and touching the wall, figuring out how high that is, then jumping and seeing how much higher you can reach.\n\nI'm at about 18 inches I think.[/quote]\r\n\r\nie jumping by only using your ankles? I'm talking about the jump itself. Is that what you mean? Read my previous posts before posting...\r\n\r\nTrust me, I'm making an accurate measurement...\r\nAnd yes, by using the hand reaching method, I'm still getting 25-26 inches.", "Solution_23": "You are allowed to bend down, or no one will have a 50 inch vert because it will require too much power.", "Solution_24": "I find it funny that people are doubting a 25 inch jump, but are unfazed by a 34 inch jump.", "Solution_25": "34 is insane and so is 25.", "Solution_26": "Vince Carter has s a 42 inch vertical, and he is 6'6''. Talk about hangtime and crazy dunks.", "Solution_27": "mine is 28. I had it measured in a physical thing. I am tall and 5'10 and way 125 pounds, and I run like... tons. :D \r\n\r\nrunning's just my way of life.\r\n\r\nstrong legs.", "Solution_28": "Mine should be around a foot - not bad at 5'7.5\", eh?" } { "Tag": [ "topology", "function", "calculus", "derivative", "linear algebra", "matrix", "analytic geometry" ], "Problem": "I'm completely stuck on this problem:\r\n\r\nShow that the image of {(x0,..xn): x0^k+...+xn^k=0} in CP^n is a smooth manifold.\r\nIs is connected/simply connected? At least, can you help me understand some special cases of these questions.\r\n\r\nCould you maybe give me a reference for a proof that CP^n is a smooth, resp. a complex analytic manifold? (I'd like to see the explicit charts, etc)\r\n\r\nThanks", "Solution_1": "$ \\mathbb C P^n$ is the set of equivalence classes $ (z_0 : z_1 : \\cdots : z_{n})$ where not all $ z_i$s are simultaneously zero. Let $ U_i$ be the (open) subset where $ z_i\\ne 0$. There is a complex-analytic diffeomorphism $ U_i\\to\\mathbb C^n$ given by $ (z_0 : z_1 : \\dots : z_{n})\\mapsto (z_0/z_i, z_1/z_i, \\dots , z_{n}/z_i)$ where you skip $ i$th term in parantheses on the right. These are the charts that you asked about. \r\n\r\nWhen $ n \\equal{} 1$, the set is neither connected nor simply connected: it's the union of two disjoint circles $ z_0 \\equal{} z_1$ and $ z_0 \\equal{} \\minus{} z_1$. \r\n\r\nWhen $ n > 1$, the set is connected, which can be seen from the structure of its intersections with $ U_i$. Not sure about being simply-connected, but I guess it is.", "Solution_2": "shouldn't one evoke lefschetz' hyperplane theorem to show simple connectedness?", "Solution_3": "ma_go: What's the statement of Lefschetz's Hyperplane Theorem?", "Solution_4": "hm.. probably it works just for homology/cohomology (as wikipedia states it: [url]http://en.wikipedia.org/wiki/Lefschetz_hyperplane_theorem[/url]), but i was quite sure there was something on homotopy groups as well...", "Solution_5": "Thanks for the answer.\r\nI am not sure how connectedness follows from the charts,..could you please give me some details? Is it jsut trivial or am I not seeing something more subtle?\r\n\r\nCP^n is not simply connected, sinc eit doesn't have trivial fundamental group.\r\n\r\nI was asking about the image of {(x1,..xn)|x1^k+..xn^k=0} in CP^n?\r\nHow can I show that this is a smooth manifold?\r\nIs this connected and is it simply connected?\r\n\r\nThanks again!", "Solution_6": "I'll expand my answer. Let $ M$ be your set. Claim: $ M\\cap U_i$ is a smooth manifold for each $ i$. (See the definition of $ U_i$ above, along with a map of $ U_i$ onto $ \\mathbb C^n$). Writing $ w_k\\equal{}z_k/z_i$ for $ k\\ne i$, we find that the image of $ M\\cap U_i$ in $ \\mathbb C^n$ is the set \r\n$ A: \\equal{}\\{w\\in\\mathbb C^n\\colon w_0^k\\plus{}\\dots \\plus{} w_n^k \\equal{}\\minus{}1\\}$. (The ith term is not there). Naturally, we want to use the Inverse Function Theorem to show that this set if a smooth manifold. the condition of IFT is that the derivative matrix has maximal rank 1 (we are working over $ \\mathbb C$). The derivative matrix is $ (kw_0^{k\\minus{}1},\\dots kw_n^{k\\minus{}1})$ and it indeed has rank 1 because not all entries are zeroes. \r\n\r\nSo now you have coordinate charts on $ M$. Note that each of them covers \"almost all\" $ M$, and any two charts overlap. Thus, if you can prove that $ A$ is connected, then $ M$ is also connected as the union of $ M\\cap U_i$. When writing my first post, I was for some reason convinced that $ A$ is connected, but now don't see why. :maybe:", "Solution_7": "Thanks a lot. I'm gonna try to think about it some more..." } { "Tag": [ "calculus", "email" ], "Problem": "I need a book for my studying\r\nYou can post through my email: lhp_tphcm@yahoo.com.vn", "Solution_1": "Are you looking to self-teach yourself calculus? Stewart seems to be pretty good but I didn't use it so I don't know.", "Solution_2": "I am going to be a member of Auckland University business schoool and I want to find out how math 's difficult, can you help me with some files about calculus and statistics?. Thank you very much!", "Solution_3": "I can't help you with the statistics part. I took AP Calculus in HS so I have no experience with Calc I or II at the college level. However, I have taken Calc III (Honors actually). Honestly, I don't see how much harder than AP it could be. It's basically just a bunch of algorithms." } { "Tag": [ "geometry", "3D geometry", "sphere", "geometric transformation", "rotation", "advanced fields", "advanced fields unsolved" ], "Problem": "hi eveybody, i really need help with the following problem: By considering the levi-civita connection, show that the holonomy group at x in the n-sphere is isomorphic to O(n,R). any help will be highly appreciated.", "Solution_1": "That doesn't sound right. The sphere is orientable, so it has to be SO(n,R). This group is generated by rotations around codimension 2 planes in R^n. Try to come up with a \"simple\" path in S^n whose holonomy realizes such a rotation. (Think about S^2 first, since it's easy to visualize.)" } { "Tag": [ "geometry", "geometric transformation", "rotation", "reflection", "probability", "pifinity is ze best" ], "Problem": "How many ways are there to put 9 differently colored beads on a $ 3\\times3$ grid if the purple bead and the green bead cannot be adjacent (either horizontally, vertically, or diagonally), and rotations and reflections of the grid are considered the same?", "Solution_1": "[hide=\"Solution\"]If the purple bead is on a corner ($ 4$ ways), then there are $ 5$ places for the green bead. If the purple bead is on an edge ($ 4$ ways), then there are $ 3$ places for the green bead, for a total of $ 4\\cdot5\\plus{}4\\cdot3\\equal{}32$ ways to arrange the purple and green beads. After this, there are $ 7!$ ways to arrange the remaining beads. Finally, we divide by $ 4\\cdot2\\equal{}8$ to account for rotations and reflections. Our final answer is $ \\frac{32\\cdot7!}8\\equal{}\\boxed{20160}$.[/hide]\n\n[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Please hide all solutions.[/url][/color]", "Solution_2": "I struggle to perform these problems correctly!", "Solution_3": "[hide=\"Solution\"]Alternate solution using total probability:\n\nThere are $ 9!$ total arrangements, but only a proportion of them are valid. In $ \\frac{1}{9}$ of cases, the purple bead is in the center, with no legal place to put the green bead. In $ \\frac{4}{9}$ of cases, the purple bead is on a corner, which leaves $ \\frac{5}{8}$ remaining places to put the green bead. In the final $ \\frac{4}{9}$ of cases, the purple bead is on an edge, which leaves $ \\frac{3}{8}$ of remaining places to put the green bead. Again, we need to divide by $ 4$ to account for rotations and by $ 2$ to account for reflections. Result: $ 9!\\cdot(\\frac{1}{9}\\cdot0\\plus{}\\frac{4}{9}\\cdot\\frac{5}{8}\\plus{}\\frac{4}{9}\\cdot\\frac{3}{8})\\div4\\div2\\equal{}\\boxed{20160}$.[/hide]\n\n[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Please hide all solutions.[/url][/color]", "Solution_4": "I cannot find my error:\n\n\n[hide=\"Solution\"]It is always possible to place the green bean into the left top corner or in the middle top row by using rotations and reflections. \n\nCase 1: green=(1/1)\npurple=(1/3), (2/3) or (3/3)\n\n\nCase 2: green=(2/1)\npurple=(1/3) or (2/3)[/hide]\n\n\nNow I got 5 and not four possible locations for purple and green on the board :-(\n\n\nWhich purple position is counted a second time?", "Solution_5": "[hide]There are $9!$ ways to put the beads on the grid, not taking rotations, reflections, and the restriction on the purple and green beads into account. We need to subtract off the number of arrangements where the purple and green beads are adjacent from this number. There are $2\\cdot3=6$ horizontally adjacent pairs of locations, $3\\cdot2=6$ vertically adjacent ones, and $2\\cdot2+2\\cdot2=8$ pairs of diagonally adjacent locations. For each of these pairs, there are two ways to put the purple and green beads in them, and $7!$ ways to put the rest of the beads on the grid, giving a total of $2(6+6+8)7!=40\\cdot7!$ invalid arrangements. So, the number of valid arrangements not counting rotations and reflections is $9!-40\\cdot7!=(9\\cdot8-40)7!=32\\cdot7!$. The grid can be rotated onto itself in four different ways, by rotations of 0, 90, 180, and 270 degrees. It can also be reflected onto itself in four different ways, by reflecting across its two diagonals and through horizontal and vertical lines through its center. So, the arrangements come in groups of $4+4=8$ equivalent arrangements, and the number of different arrangements is $32\\cdot7!/8=4\\cdot7!=\\boxed{20160}$.", "Solution_6": "[quote=kuelshammer]I cannot find my error:\n\n\n[hide=\"Solution\"]It is always possible to place the green bean into the left top corner or in the middle top row by using rotations and reflections. \n\nCase 1: green=(1/1)\npurple=(1/3), (2/3) or (3/3)\n\n\nCase 2: green=(2/1)\npurple=(1/3) or (2/3)[/hide]\n\n\nNow I got 5 and not four possible locations for purple and green on the board :-(\n\n\nWhich purple position is counted a second time?[/quote]\n\nI got the same. Did you ever get a satisfactory answer for which we seem to be counting twice?\n", "Solution_7": "I think direct is more natural than complementary for this one.\n[hide=Solution] Rotation and reflection preserve adjacency so divide by 8 at end. 4 ways to put purple in corner with 5 non-adjacent for green and 7! for rest, plus 4 ways to put purple on edge with 3 non-adjacent for green and 7! for rest, so total (4x5+4x3)x7!/8 = 4x7! = 20160.[/hide]", "Solution_8": "I also have the same problem.\nWhat is wrong with this?\n[hide = 'Click here.']\nCase 1: The Purple is set in the top left corner of the grid, as shown: \nTo make our lives easier, we set the purple in the top-left corner of the grid(so we don't have to account for rotations). Then the Green only has 5 valid spaces(top right corner, middle right spot, bottom right corner, bottom middle, and bottom left corner). However, we must account for reflections, so we only have 3 valid spaces(bottom right corner, bottom middle, and bottom left corner due to reflection across diagonal). Then there are $3 \\cdot 7!$ cases for this case.\n\nCase 2: The Purple is set in the top middle spot. \nThen the Green has 3 valid spots(bottom-left,bottom middle, and bottom right). However, the bottom-right is invalid due to reflections), meaning that there are $2 \\cdot 7!$ ways for this case.\n\nIn all, there should be $5 \\cdot 7!$ cases, but the answer says $4 \\cdot 7!$ ways. What am I missing here?", "Solution_9": "How are you accounting for reflections? Could you please explain again?", "Solution_10": "@lexd and everyone else with the same problem:\n\n[hide] My guess is that the fact that you are accounting for double counting for green and purple separate from accounting for double counting for the rest of the grid throws the system off. For example, in Case 1, where you say there are 5 places to insert the green square is correct. Then; however, you also say that rotations and reflections reduces this number to 3. This is confusing because, as said before, you are accounting for rotations and reflections in the green and purple squares separately than the rest of the grid.\n\nMy suggestion is to first count every possibility without the thought of reflections and rotations, and account for it at the end.\n\n[/hide]\n\nI have a different question.\n\n[hide = My question]\nThis regards counting for reflections and rotations. I get for rotations that you must divide by 4. For reflections; however, aren't you supposed to consider both vertical and horizontal reflections, meaning that you have to divide by 4, and not 2? Please help.\n[/hide]", "Solution_11": "I have a question: why don't you have to use Burnside's Lemma for this? ", "Solution_12": "I do understand the problem? Is there a way to not count from the wrong perspective and get the problem wrong?", "Solution_13": "the total number of arrangements for purple on top middle and green on bottom middle is 7!/2, not 7!. similarly when the two are on the same diagonal it's 7!/2. So you have 4x7! overall.", "Solution_14": "This was a fun problem! Here's my solution: [hide=Solution] \nThink about it like this\u2014how many ways can you put the purple and green beads side by side? \n\nThis is pretty easy to count, and when you do it you get $20$. We have to multiply this by $2$, as the beads can be arranged as green purple or purple green. We get $40$ through that, and we have to multiply this by $7!$ to count the arrangements of the other $7$ beads. We can subtract this from the total $9!$ cases and call it a day\n\n\u2014but wait! We forgot to account for the reflections and rotations! \n\nWe can rotate a corner onto itself $4$ ways, so we have to divide our total by $4$. We can also reflect it two ways, so we again divide it by $2$. \n\nThis leads us to the answer $\\frac{9!-40\\cdot7!}{2\\cdot4}=\\boxed{20160}$.", "Solution_15": "[hide=Alternate solution but a bit unclear]Including the restrictions, there are 8! ways to be done. Whom there are 4 cases (not ways because of the restrictions) where the two squares are adjacent. 2 for the diagonals. 2 for the vertical and horizontal ones. Therefore, the answer is 8! - (4)7! = 20160.[/hide]", "Solution_16": "omg I actually got this right! ", "Solution_17": "@3above I got the same solution lol", "Solution_18": "I have another solution:\nTry assigning a location for the purple bead in the 3x3 grid. You will quickly notice that the bead cannot go in the center as there is nowhere for the green bead to go. Therefore, the purple bead has to either go on a corner or an edge. Since rotations/reflections are considered the same, you only have to consider 2 cases: \n1: The purple bead is in a corner. Here, notice that there are 5 different spots to put the green bead (since the purple bead is adjacent to 3 other spots). With the remaining 7 spots, there are 7! ways to arrange the beads. However, in those arrangements, there is a hidden reflection. Some of the arrangements are reflected over the diagonal (running through purple) of other ones. As a result, we divide by 2 here. The total count in this case is 5*7!/2\n2: The purple bead is on an edge. There are only 3 options for where the green bead can go this time. Again, there are 7! ways to arrange the remaining beads. Then, we divide the total by 2 as some arrangements are others reflected over the vertical line running through the center of the grid. As a result, there is 3*7!/2 ways to arrange beads in this case\nFinally (yay), add up the count for both cases to get a result of (3+5)*7!/2=4*7!=4*5040=20160 ways to arrange the beads", "Solution_19": "In all honesty the problem should have noted that reflections may also be over one of the diagonals...", "Solution_20": "[hide=this problem was easier than expected]We do casework on the green bead:\nThe green bead is in a corner:\n\nThere are 5 ways to place the purple bead and $7!$ ways to do the others, but we have to divide by two for reflection. That is a total of $\\frac{5\\cdot7!}{2}$ ways.\n\nThe green bead is on a side:\n\nThere are $\\frac{3\\cdot7!}{2}$ ways, for similar reasons.\n\nThus, there are $\\frac{(5+3)7!}{2}=\\frac{8!}{2}=\\boxed{20160}$ ways.\n[/hide]" } { "Tag": [ "induction", "LaTeX" ], "Problem": "can someone please help me with the following two proofs using mathematical induction? i can get to the n=1 proof stage, but i'm stuck on proving n+1 is true. \r\n\r\n1 + 3/2 + 2 + 5/2 + ... + 1/2(n+1) = n/4(n+3) \r\n\r\nand \r\n\r\n1/3 + 1/15 + 1/35 + ... + 1/(2n-1)(2n+1) = n/(2n+1) \r\n\r\n\r\nplease help me!", "Solution_1": "well as a side note, you should learn how to represent fractions is LaTeX\r\nthe code:[code]$\\frac{x}{y}$[/code] \r\ngives $ \\frac {x}{y}$ where x and y can be expressions, numbers, almost anything\r\n\r\nanyway, ill do the first one for you, maybe then you can do the second one...\r\nok\r\nso you replace n with n+1\r\n$ 1 \\plus{} \\frac {3}{2} \\plus{} 2 \\plus{} \\cdots \\plus{} \\frac {1}{2}(n \\plus{} 1 \\plus{} 1) \\equal{} \\frac {n \\plus{} 1}{4}(n \\plus{} 1 \\plus{} 3)$ * i think you meant this\r\nthis is equal to\r\n$ 1 \\plus{} \\frac {3}{2} \\plus{} 2 \\plus{} \\cdots \\plus{} \\frac {1}{2}(n \\plus{} 1) \\plus{} \\frac {1}{2}(n \\plus{} 1 \\plus{} 1) \\equal{} \\frac {n^2 \\plus{} 5n \\plus{} 4}{4} \\equal{} \\frac {n^2 \\plus{} 3n}{4} \\plus{} \\frac {2n \\plus{} 4}{4}$\r\nsubtracting the equation\r\n$ 1 \\plus{} \\frac {3}{2} \\plus{} 2 \\plus{} \\cdots \\plus{} \\frac {1}{2}(n \\plus{} 1) \\equal{} \\frac {n}{4}(n \\plus{} 3)$\r\nfrom this gives \r\n$ \\frac {1}{2}(n \\plus{} 2) \\equal{} \\frac {2n \\plus{} 4}{4}$\r\n$ \\frac {n \\plus{} 2}{2} \\equal{} \\frac {n \\plus{} 2}{2}$\r\nQ. E. D.", "Solution_2": "[quote=\"stevenmeow\"]well as a side note, you should learn how to represent fractions is LaTeX\nthe code:[code]$\\frac{x}{y}$[/code] \ngives $ \\frac {x}{y}$ where x and y can be expressions, numbers, almost anything\n\nanyway, ill do the first one for you, maybe then you can do the second one...\nok\nso you replace n with n+1\n$ 1 \\plus{} \\frac {3}{2} \\plus{} 2 \\plus{} \\cdots \\plus{} \\frac {1}{2}(n \\plus{} 1 \\plus{} 1) \\equal{} \\frac {n \\plus{} 1}{4}(n \\plus{} 1 \\plus{} 3)$ * i think you meant this\nthis is equal to\n$ 1 \\plus{} \\frac {3}{2} \\plus{} 2 \\plus{} \\cdots \\plus{} \\frac {1}{2}(n \\plus{} 1) \\plus{} \\frac {1}{2}(n \\plus{} 1 \\plus{} 1) \\equal{} \\frac {n^2 \\plus{} 5n \\plus{} 4}{4} \\equal{} \\frac {n^2 \\plus{} 3n}{4} \\plus{} \\frac {2n \\plus{} 4}{4}$\nsubtracting the equation\n$ 1 \\plus{} \\frac {3}{2} \\plus{} 2 \\plus{} \\cdots \\plus{} \\frac {1}{2}(n \\plus{} 1) \\equal{} \\frac {n}{4}(n \\plus{} 3)$\nfrom this gives \n$ \\frac {1}{2}(n \\plus{} 2) \\equal{} \\frac {2n \\plus{} 4}{4}$\n$ \\frac {n \\plus{} 2}{2} \\equal{} \\frac {n \\plus{} 2}{2}$\nQ. E. D.[/quote]This is not induction. What you just did is to show that the statement is true for $ n$. What the induction step is to assume the statement is true for $ n$, and use that to imply that it is also true for $ n\\plus{}1$.\r\n\r\nSo assume it is true for $ n$:\r\n\r\n$ 1 \\plus{} \\frac {3}{2} \\plus{} 2 \\plus{} \\cdots \\plus{} \\frac {1}{2}(n \\plus{} 1) \\equal{} \\frac {n}{4}(n \\plus{} 3)$.\r\n\r\nNow try to add the sum with the next term in the sequence:\r\n\r\n$ 1 \\plus{} \\frac {3}{2} \\plus{} 2 \\plus{} \\cdots \\plus{} \\frac {1}{2}(n \\plus{} 1) \\plus{} \\frac {1}{2}((n \\plus{} 1) \\plus{} 1)$\r\n\r\nUse your assumption above, substitute, and simplify, you should get it down to:\r\n\r\n$ \\frac {(n\\plus{}1)}{4}((n \\plus{} 1) \\plus{} 3)$\r\n\r\nSo, by induction hypothesis, blah blah blah, done. That's how induction is done.", "Solution_3": "ahem i did do that but srry if it seems a bit confusing \r\nlook it through completely and you'll see i rly did prove n+1 given n", "Solution_4": "first you must prove the equation is true for n=1, then you assume that n= k , k = any integer) then you prove that the equation holds for k+1 and if it does then it works for all numbers greater than or equal to 1" } { "Tag": [ "LaTeX" ], "Problem": "Is there a page where all the LaTeX formulas are?", "Solution_1": "What do you mean by [i]all the LaTeX formulas[/i]?\r\nYou will find a list of symbols at [[LaTeX:Symbols]] and commands at [[LaTeX:Commands]] in the AoPSWiki.\r\n\r\nPS I moved your post here as you had put it in the wrong thread." } { "Tag": [ "quadratics", "LaTeX", "algebra", "quadratic formula", "special factorizations" ], "Problem": "Sample probs of what is expecting this week...I did then already, but since i cant check them i would like to see what you guys come up with..Thank You\r\n\r\n\r\n1. Solve by factoring:\r\nx2 \u2013 9x = -8\r\n\r\n\r\n\r\n\r\n2. Solve using the square root property:\r\n2x2 \u2013 5 = 93\r\n\r\n\r\n\r\n\r\n3. Solve using the square root property:\r\n(x + 4)2 = 81\r\n\r\n\r\n\r\n\r\n4. Solve by completing the square:\r\nx2 + 6x + 2 = 0\r\n\r\n\r\n\r\n\r\n5. Solve using the quadratic formula:\r\nx2 \u2013 3x = -6x \u2013 1\r\n\r\n\r\n\r\n\r\n6. Solve using the quadratic formula:\r\nx2 \u2013 10x \u2013 1 = -10", "Solution_1": "First of all, you should use $ \\text{\\LaTeX}$ (the AoPS site should tell you how) or at least use the ^ symbol for exponents because expressions like 2x2 can be a little confusing.\r\n\r\nAnyway, most of your questions can be solved using a standard algebra or algebra II textbook.\r\nTry [url]http://en.wikipedia.org[/url], too. It should give you some good explanations.\r\n\r\nLastly, these problems are a bit too easy on the Intermediate Topics forum. Take a look at the other topics to see why.\r\n\r\nLet me know if this helps; if it doesn't I can give you some answers.", "Solution_2": "As the above poster said, these problems are too easy for the Intermediate forum. Try posting them in the Classroom Math forum, and keep in mind that AoPS is not for daily homework, but for challenging problems that are above regular school level." } { "Tag": [], "Problem": "Hola! :) \r\n\r\nI haven't been in this forum for a while until now because I'm stuck with this hard journal issue. :mad: \r\n\r\nWe have to write about the fair (the place where there is rollercoaster, merry-go-round and etc).\r\n\r\nI got one sentence down but I can't think of any other four.\r\n\r\n[b]La feria es divertida y tienen buena comida. [/b]\r\n\r\nIs that correct? I said, \"The fair is divertida and there are good food.\"\r\n\r\nI'll be greatly appreciate if anyone can come up with four simple sentences on fair! Thank you!!", "Solution_1": "Can anyone give 4 more simple sentences about fair?\r\n\r\nI really need this help. :(" } { "Tag": [], "Problem": "$N$ markaze dayere 9 noghteye mosalase $ABC$ ast. \r\n\r\n$A',B',C'$ marakeze dayere 9noghteye mosalas haye $BNC,CNA,ACN$ be tartib hastand.\r\n\r\nsabet konid $AA',BB',CC'$ hamresand.\r\n\r\n\r\n\r\n\r\n$\\mbox{Good Luch in solving this problem}$", "Solution_1": "masale faghat tajanose :!:", "Solution_2": "rasti to akhar sar be man nagofti in soalaye khafan ro az koja miari? :rotfl:", "Solution_3": "[quote=\"Behsan\"]masale faghat tajanose [/quote]\n\nmishe rahe haleto kamel benevisi?\n\n\n[quote=\"Jensen\"]rasti to akhar sar be man nagofti in soalaye khafan ro az koja miari? :rotfl: [/quote]\r\nShak dari ke in soal khafane Milad ;) .mitooni emtehanesh koni.", "Solution_4": "fekr nakonam ba tajanos hal beshe :roll:", "Solution_5": "tajanos ? in faghat ye nazare shakhsie ?? :lol: [/quote]" } { "Tag": [ "vector", "email", "induction", "inequalities", "combinatorics proposed", "combinatorics" ], "Problem": "8.8, 9.8, 11.8\r\na) 99 boxes contain apples and oranges. Prove that we can choose 50 boxes in such a way that they contain at least half of all apples and half of all oranges.\r\nb) 100 boxes contain apples and oranges. Prove that we can choose 34 boxes in such a way that they contain at least a third of all apples and a third of all oranges.\r\nc) 100 boxes contain apples, oranges and bananas. Prove that we can choose 51 boxes in such a way that they contain at least half of all apples, and half of all oranges and half of all bananas. \r\n([i]I. Bogdanov, G. Chelnokov, E. Kulikov[/i])", "Solution_1": "Very nice problems.\r\n\r\nFor a):\r\n\r\nOrder the boxes according to the number of apples they contain ($b_{99}$ has the most apples and $b_1$ has the fewest). Then take: $b_{99}$, one of $b_{98}, b_{97}$, ..., one of $b_2, b_1$. It is clear that any such set contains at least half of all apples (in the worst case you have chosen $b_{99}, b_{97}, ..., b_1$ which has at least as many apples as $b_{98}, ..., b_2$). Now from each subset of 2 boxes (98-97, ..., 2-1) take the box with more oranges. Then the set also contains at least half of the oranges.\r\n\r\n\r\nFor b):\r\n\r\nOrder the boxes according to the number of apples they contain ($b_{100}$ has the most apples and $b_1$ has the fewest). Then take: $b_{100}$, one of $b_{99}, b_{98}, b_{97}$, ..., one of $b_3, b_2, b_1$. Now in the worst case we have chosen $b_{100}, b_{97}, ..., b_1$ which has at least as many apples as $b_{99}, b_{96}, ..., b_3$ and $b_{98}, b_{95}, ..., b_2$. So any such set contains at least a third of the apples. Now from each set of 3 boxes (99-98-97, ..., 3-2-1) take the box with the most oranges. This set satisfies the conditions.", "Solution_2": "well i have some ideas for this kind of problems but i cant finish them....\r\n\r\n\r\nFor every box $i$, identify the vector $v_i=(x,y,z)$ where $x$ is the number of apples on it, $y$ the number of oranges and $z$ the number of the other fruit on box $i$. the geometric interpretation is clear now....", "Solution_3": "Very nice, Severius!", "Solution_4": "and what about c) ?", "Solution_5": "c is very hard. There are many solutions, but none of our students managed to solve it. A hint: prove a lemma: if we have $2k$ boxes, we may partition them to two groups by $k$ boxes in such a way that weight of apples in both groups differ by at most the maximal weight of apples in a single box, and the same for oranges.", "Solution_6": "OH,MY ROOMMATE AND I CAN ONLY SOLVE THE FIRST TWO PROBLEM,THE SOLUTION IS SIMILAR TO YOURS\r\nI think the third problem is really hard for me.And I guess it must have a excelllent solution with a giant use of some lemma.Could anyone post the solution or email me:georgew_bush@163.com?THANKS! :D", "Solution_7": "Does anyone has the official solution of this beautiful problem ?I think some people from Russia sure. So please post it Thank you very much", "Solution_8": "[quote=\"Fedor Petrov\"]c is very hard. There are many solutions, but none of our students managed to solve it. A hint: prove a lemma: if we have $2k$ boxes, we may partition them to two groups by $k$ boxes in such a way that weight of apples in both groups differ by at most the maximal weight of apples in a single box, and the same for oranges.[/quote]\r\n\r\nHow? I don't understand", "Solution_9": "[quote=\"amir2\"]and what about c) ?[/quote]\r\nprove it with induction\r\n ;)", "Solution_10": "[quote=\"BaBaK Ghalebi\"][quote=\"amir2\"]and what about c) ?[/quote]\nprove it with induction\n ;)[/quote]\r\n\r\nCan you show?", "Solution_11": "[quote=\"indybar\"][quote=\"BaBaK Ghalebi\"][quote=\"amir2\"]and what about c) ?[/quote]\nprove it with induction\n ;)[/quote]\n\nCan you show?[/quote]\r\nsure ASAP ;)", "Solution_12": "[quote=\"BaBaK Ghalebi\"]\nsure ASAP ;)[/quote]\r\n\r\nhmm... \r\nI don't think induction works here. ;)", "Solution_13": "[quote=\"BaBaK Ghalebi\"][quote=\"amir2\"]and what about c) ?[/quote]\nprove it with induction\n ;)[/quote]\r\n$2k-1$ box $\\& k$\r\n$n=k\\to n=k+1$\r\nif there were two boxes such that $a_1>a_2,b_1>b_2$ now by the assumption of the induction we ca n choose $k$ boxes from the $2k-1$ boxes left\r\nnow we can add the box that contains $a_1,b_1$\r\nnow if there werent such boxes then we have:\r\n$a_1\\geq a_2\\geq ...\\geq a{}_2{}_k_+_1$\r\n$b_1\\leq b_2\\leq ...\\leq b{}_2{}_k_+_1$\r\nnow we choose the boxes $1,3,5,...,2n+1$\r\n$\\Rightarrow$ $a_1+a_3+...+a{}_2{}_n_+_1\\geq a_2+a_4+...+a{}_2{}_n$\r\n ;) \r\nsorry if I didnt explain it clearly did you get it or should I explain it better and more carefully?? :D", "Solution_14": "I have no idea what you're trying to say.\r\n\r\nWhat are the $a_i,b_i$ (I'm assuming they're the weights of apples, oranges resp. in the boxes) Then what about the bananas?\r\n\r\nPlease \"explain it better and more carefully\" :D", "Solution_15": "c) part:\r\n\r\nLemma by [b]Fedor Petrov:[/b]\r\n\r\nAny $ 2k$ pairs of positive reals $ (a_i,b_i)$ can be partitioned into $ 2$ groups of $ k$ pairs each, so that if $ x$ is a sum of $ a_i$ in the first group and $ y$ is a sum of $ a_i$ in the second group. Then $ |x - y|\\leq\\max_{i}a_i$. And the same inequality holds for the $ b_i$, i.e if $ m$ is a sum of $ b_i$ in the first group and $ n$ is a sum of $ b_i$ in the second group, then $ |m - n|\\leq \\max_i b_i$.\r\n\r\n[b]Proof of the lemma[/b]:\r\n\r\nThe statement clearly holds for the case $ k = 1,2$. So assume that the statement holds for $ k = n$ and let's prove it for $ k = n + 1$.\r\n\r\nWLOG assume that $ a_{2n + 2}\\geq a_{2n + 1}\\geq \\max_{i = 1}^{2n}a_i$. Due to the induction assumptation used for the case $ 2n$ the following inequality must hold:\r\n\r\n$ |x - y|\\leq a_{2n + 1}$.\r\n\r\nConsider two cases:\r\n\r\n1.$ b_{2n + 2}\\geq b_{2n + 1}$\r\n\r\nWLOG suppose $ m\\geq n$.\r\n\r\nThen add the pair $ (a_{2n + 1},b_{2n + 1})$ to the first group, and $ (a_{2n + 2},b_{2n + 2})$ to the second, so\r\n\r\n${ |x + a_{2n + 1} - y - a_{2n + 2}}|\\leq |x - y| + |a_{2n + 1} - a_{2n + 2}|\\leq a_{2n + 1} + a_{2n + 2} - a_{2n + 1} = a_{2n + 2}$\r\n\r\nNow suppose that $ m + b_{2n + 1}\\geq n + b_{2n + 2}$, then \r\n\r\n$ |m + b_{2n + 1} - n - b_{2n + 2}| = m + b_{2n + 1} - n - b_{2n + 2}\\leq\\max_{i}^{2n + 2}b_i$,\r\n\r\nif $ m + b_{2n + 1}\\leq n + b_{2n + 2}$ then:\r\n\r\n$ |m + b_{2n + 1} - n - b_{2n + 2}| = n + b_{2n + 2} - b_{2n + 1} - m\\leq b_{2n + 2} - b_{2n + 1}\\leq\\max_{i}^{2n + 2}b_i$.\r\n\r\nFor the case $ b_{2n + 1} > b_{2n + 2}$ do the reverse addition, the rest is completely the same as in the considered case.\r\n\r\nSo the lemma is proved. $ \\blacksquare$\r\n\r\nThe last step is pretty obvious, take two boxes, one with max number of apples weight and second with max number of oranges weights. So applying the lemma for $ 98$ ,for $ a_i$ take weights of apples and for $ b_i$ weights of oranges, and choose a group with bigger weight of bananas in it. And add both those boxes into this group.", "Solution_16": "I have a question: for part c, we were able to deal with apples, oranges, and bananas, but is there a way we can generalize this problem so that we can deal with, say, $n$ fruits instead of just $3$? Sorry if the question is vague (and for the massive bump), but I am super curious about this.", "Solution_17": "Here is what I believe is a different solution to part (c):\n\n[hide=Solution]\nFirst, select the box with the most apples and the box with the most bananas. (In the case these are the same box, select the box and another arbitrary box.) For the other $98$ boxes, let $a_{n+1}$ be the box with the $n$th most apples, and let $b_{n+1}$ be the box with the $n$th most bananas. We can turn any group of boxes among the $98$ boxes into a $98$-term sequence with the following rule: Write out $a_2,a_3,...,a_{99},$ and replace each $a_k$ with a $1$ if the box is selected, and a $-1$ if the box is not selected. Call this an apple sequence; proceed similarly with $b_k$ for a banana sequence.\n\nNow, for any such sequence, define $S_k$ as the sum of the first $k$ terms. We will prove the following lemma:\n\nIf an apple/banana sequence of $n$ terms has $S_k$ nonnegative for all $k$ with $0 \\leq k \\leq n,$ then it must contain at least half of the apples/bananas.\n\nProof: We claim that in such a sequence, each $1$ can be paired up with a $-1$ later in the sequence. This implies that the sequence contains at least half of the fruits, since in each pair, the selected box contains more of the chosen fruit than the unselected box, meaning that in each pair, at least half of the fruits are selected, so overall at least half are selected.\n\nTo prove this, we show that by pairing each $1$ with the first unpaired $-1,$ we will have a desired sequence. Suppose there is some $1$ that ends up paired to an earlier $-1.$ Let it be the $k$th $1$ and the $n$th term of the sequence. Then it must also be the $k$th $-1.$ However, consider $S_{n-1}.$ This contains $k-1$ $1$s, but contains at least $k$ $-1$s, so it is negative, which is a contradiction. Therefore the lemma is proved.\n\nNow consider all apple sequences that have the property such that for any positive integer $k$ from $1$ to $49,$ exactly one of $a_{2k},a_{2k+1}$ is a $1.$ We easily see that all of these sequences have $S_i=-1,0,1,$ since $S_{2i}=0$ for all positive integers $i$ and $S_{2i+1}$ can only be 1 away from $S_{2i}.$ Call these apple sequences good, and similarly define a good banana sequence in the same way.\n\nWe will now prove that there is some good apple sequence that can be paired with some good banana sequence. We use the following process to generate one: First, set the first unset $a_i$ to $1,$ and set the corresponding $b_j$ to $1.$ Next, set $b_{j+1}$ to $-1$ if $j$ is even, and set $b_{j-1}$ to $-1$ if $j$ is odd. Then set the corresponding $a_k$ to $-1,$ and similarly set $a_{k+1}$ or $a_{k-1}$ to $1$ when $k$ is even or odd, respectively. Repeat this until it fills in $a_{i+1},$ at which point we repeat the whole process for the next unset $a_i,$ and repeat until all terms are filled in.\n\nNow we show that this process works. First, notice that the steps of going between corresponding terms in the different sequences and going between terms in pairs $a_{2k},a_{2k+1}$ and $b_{2k},b_{2k+1}$ are reversible. This means that following this process, we must eventually return to $a_i,$ and the preceding step must have been at $a_{i+1}.$ Also, notice that if some term was filled out at some point, then the other term in its pair must have also been filled out, either immediately before or immediately after, and note that each \"loop\" fills in the same number of terms in each sequence. Therefore, this algorithm generates a corresponding good apple sequence and good banana sequence, so a pair must exist.\n\nNow consider the $99$-term sequence consisting of $a_1$ replaced by $1$ followed by a good apple sequence. Since for the good apple sequence, $S_k$ is always $-1,0,1,$ in our new sequence $S_k$ must always be $0,1,2,$ since the first $n+1$ terms of this sequence are $1$ followed by the first $n$ terms of the good apple sequence. Since $0,1,2$ are positive, this sequence contains at least half of the apples, and this means our $99$ term sequence contains $49+1=50$ boxes chosen that have at least half of the apples, since half of the terms in the $98$ term sequence are $1$s. Adding back the $b_1$ box, which is chosen, we get our $100$ boxes have a $51$ element subset with at least half of the apples. Similar logic shows that the same subset has at least half of the bananas, since the two good sequences are the same subset.\n\nNow, consider the $98$-term apple and banana sequences that have a $1$ in a given place if and only if there are $-1$s in the same places in our already chosen good sequences. It is not hard to see that these are also good sequences, so $S_k$ also must be $-1,0,1.$ Now putting these sequences together with $a_1$ and $b_1$ gives us another $51$ element subset with at least half of the apples and bananas.\n\nNow suppose neither of these two subsets contains at least half of the oranges. Consider the $49$-element subset that contains a box if and only if our first $51$-element subset doesn't contain it, and consider the corresponding $49$-element subset for our second $51$-element subset. Both of these must contain more than half of the oranges, but notice that they are defined such that their intersection is the null set. Therefore their union is just the $98$ boxes excluding $a_1,b_1.$ However, they contain more than the whole of the oranges in the $100$ boxes, which is a contradiction.\n\nTherefore, there is some set of $51$ boxes containing at least half of each fruit.\n[/hide]", "Solution_18": "Credit goes to one of my friends (not sure if he wants to be named):\n\nPlace the boxes in 3d space so that no four are coplanar. By the ham sandwich theorem, there is a plane which bisects the apples (treated as a single object) and does the same for the oranges and bananas. Choose the side of the plane with the least number of boxes. The worst case is when 3 boxes intersect the plane and there are 48 on one of the sides and 49 on the other. Thus you select at most 51 boxes. You have at least half of each type of fruit in this collection of boxes.\n" } { "Tag": [], "Problem": "A half is a third of it. What is it?\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\n$ 1\\frac{1}{2}$\n\n[/hide]", "Solution_2": "[color=red][size=200][hide]1/2=1/3 X\nthis give us X=3/2[/hide][/size][/color]", "Solution_3": "[hide][color=violet]1/2=1/3 x\nx=3/2 or 1 1/2[/color][/hide]", "Solution_4": "Seriously guys, no need to post the exact same solution over and over again." } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "If A is not an empty finite set, show that Seq(A), the set of all finite sequences of elements of A, is countable.", "Solution_1": "huh?\r\n\r\nif $ |A| \\equal{} 1$, $ Seq(A)$ has only one element and is countable. but if $ |A| > 1$, clearly $ Seq(A)$ is uncountable (the power set of $ \\mathbb{N}$ may be embedded into $ Seq(A)$).", "Solution_2": "No, the power set cannot be embedded. In fact, we have an embedding of $ Seq(A)$ into $ \\mathbb N$ if we have $ A \\equal{} \\{0,1,...,n \\minus{} 1\\}$ and map the sequence $ a_0,...,a_m$ to $ a_0 \\plus{} a_1 \\cdot n \\plus{} ... \\plus{} a_m \\cdot n^m$. All other cases reduce to this.", "Solution_3": "sorry, I did not read *finite* sequences ..." } { "Tag": [ "geometry", "rectangle", "trigonometry", "incenter", "angle bisector" ], "Problem": "Prove that the angle bisectors of the interior angles of any triangles are concurrent at a single point.\r\n\r\nDoes any quadrilateral have an inscribed circle?", "Solution_1": "For the second one, yes, quite obviously. Just look at a square.", "Solution_2": "[quote=\"fishythefish\"]For the second one, yes, quite obviously. Just look at a square.[/quote]\r\n\r\nI require a proof.", "Solution_3": "Perhaps you should clarify that second question -- are you asking whether all quadrilaterals have an incircle, or if there is any single one that has one? I suspect you are asking the former, while fishy is answering the latter.", "Solution_4": "While all triangles have inscribed circles, and so do regular polygons, but most others don't.\r\n\r\nI can disprove that every quadrilateral has an inscribed circle with an example.\r\n\r\nA rectangle with width 1, and length 2. Because the inscribed circle must be tangent to all of the sides, the greatest length the diameter could be is 1, otherwise it would intersect the length twice on each side. But then the circle cannot meet the other two sides. Case closed.\r\n\r\n(I answered both questions now. :P )", "Solution_5": "For 1 you could use the [url=http://en.wikipedia.org/wiki/Angle_bisector_theorem]Angle bisector theorem[/url] together with [url=http://en.wikipedia.org/wiki/Ceva%27s_theorem]Ceva's Theorem[/url]. :)", "Solution_6": "[quote=\"ThetaPi\"]Prove that the angle bisectors of the interior angles of any triangles are concurrent at a single point.\n\nDoes any quadrilateral have an inscribed circle?[/quote]\r\n\r\nnumber 2 is trivial by trig cevas", "Solution_7": "[quote=\"JavaMan\"]For 1 you could use the [url=http://en.wikipedia.org/wiki/Angle_bisector_theorem]Angle bisector theorem[/url] together with [url=http://en.wikipedia.org/wiki/Ceva%27s_theorem]Ceva's Theorem[/url]. :)[/quote]\r\n\r\nProve Ceva's Theorem. I didn't know there existed such a theorem.\r\n\r\nEDIT: Oh. Wikipedia has a proof. Never mind then.", "Solution_8": "Ceva's Theorem should be viewed as a significant generalization of this result. The fact that the angle bisectors intersect at a unique point follows directly from the uniqueness of the incircle, so what you're asking for is a proof that the incircle is unique.\r\n\r\n[hide=\"Proof\"] The feet of the perpendiculars from an incenter of a triangle divides up its side lengths into quantities $ x, y, z$ such that the side lengths are $ x \\plus{} y, y \\plus{} z, z \\plus{} x$. The three values $ x, y, z$ are uniquely determined from the side lengths because\n\n$ y \\equal{} \\frac {(x \\plus{} y) \\plus{} (y \\plus{} z) \\minus{} (z \\minus{} x)}{2}$\n\nand cyclic permutations. Therefore, given a triangle, the feet of the perpendiculars from any of its incenters is uniquely determined. The intersection of the perpendiculars themselves then uniquely determines the incenter. [/hide]\r\nPerhaps now you'd like a proof that the incircle [i]exists[/i]? \r\n\r\nProofs of common results are easily available online; that's what Google is for.\r\n\r\nEdit: The above argument generalizes to quadrilaterals with an inscribed circle (I forget what the term is). Specifically, there exist positive reals such that the side lengths are $ x \\plus{} y, y \\plus{} z, z \\plus{} w, w \\plus{} x$. Unlike the case of triangles, this system is not solvable in general because $ (x \\plus{} y) \\plus{} (z \\plus{} w) \\equal{} (y \\plus{} z) \\plus{} (w \\plus{} x)$; in other words, opposite sides in a quadrilateral with an inscribed circle have the same sum. This characterizes such quadrilaterals." } { "Tag": [ "ratio", "absolute value", "calculus", "calculus computations" ], "Problem": "Can anyone explain the following problem? \r\n\r\nDetermine the radius of convergence and the interval of convergence of the following series:\r\n\r\n(Sigma from n=0 to infinity.) {nx n}/{(4 n )(n 2 +1)}.\r\n\r\nI don't understand which convergent test to use for this problem. \r\n\r\nThx.", "Solution_1": "[quote=\"keta\"]Can anyone explain the following problem? \n\nDetermine the radius of convergence and the interval of convergence of the following series:\n\n(Sigma from n=0 to infinity.) {nx n}/{(4 n )(n 2 +1)}.\n\nI don't understand which convergent test to use for this problem. \n\nThx.[/quote]\r\n\r\nIf I understood well your problem, you mean $f(x)=\\sum_{n=0}^{\\infty}\\frac{nx^n}{4^n(n^2+1)}$. In this case, I believe the regular n-th root test, or ratio test should work. Either way, If I am not mistaken, the radius of convergency is 4, and the interval is centred at 0, if the problem is the one I wrote.\r\n\r\nBest,", "Solution_2": "The two tests djimenez mentioned - the root test and the ratio test - are nothing more than somewhat automated comparisons to geometric series. A power series always has an $x^n$ (or $(x-x_0)^n$) so it always has something geometric - and that means the root and ratio tests are always in order, with the choice between them a matter of convenience.\r\n\r\nBut the root or ratio test, while it should tell you the radius of convergence, is helpless at telling you whether the series converges on the boundary of the interval of convergence. In this case, the radius of convergence is 4. If $x=\\pm4$, then the absolute value of the terms is $\\frac{n}{n^2+1}$, which is too large for absolute convergence. The series diverges when $x=4$ (terms all positive) but converges by the alternating series test when $x=-4.$ The interval of convergence is $[-4,4).$" } { "Tag": [], "Problem": "with a number like -67, are the digits 6 and 7, -6 and -7, or -6 and 7?", "Solution_1": "6 and 7\r\n\r\nin most competitions, digits r defined to be a 1 digit positive integers, i think :wink:", "Solution_2": "I would think -60 and -7... Place value ftw.\r\n\r\n-67 = -(60+7)\r\n-67 = -60 + (-7)\r\n\r\nUnless your question is different from what it seems to be to me." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c>0$ such that $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=2$.Prove that:\r\n\r\n $\\frac{bc}{a(b+c)}+\\frac{ca}{b(c+a)}+\\frac{ab}{c(a+b)}\\geq\\frac{ab+bc+ca}{a+b+c}$ ;)", "Solution_1": "After homogenizing, we need only show that \r\n\r\n$2\\sum \\frac {bc}{a(b+c)}\\cdot (abc(a+b+c))\\geq (ab+bc+ca)^2$\r\n\r\nWrite the LHS as \r\n\r\n$\\left(\\sum\\frac {2bc}{a(b+c)}\\right)\\left(\\sum \\frac {abc(b+c)}{2}\\right)$\r\n\r\nBy Cauchy, this is $\\geq $\r\n\r\n$\\left(\\sum \\sqrt{\\frac {2bc\\cdot abc\\cdot (b+c)}{2a(b+c)}}\\right)^2=\\left(\\sum bc\\right)^2$\r\n\r\nas desired.", "Solution_2": "Note that \r\n\\[\r\nab+bc+ca=abc\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)=2abc\r\n\\]\r\nUpon divide the bothsides of the inequality by $abc$ we obtain an equivalent inequality\r\n\\[\r\n\\frac{1}{a^2(b+c)}+\\frac{1}{b^2(c+a)}+\\frac{1}{c^2(a+b)}\\geq \\frac{2}{a+b+c}\r\n\\]\r\n\r\nNow\r\n\\[\r\nLHS=\\frac{1/a^2}{b+c}+\\frac{1/b^2}{a+c}+\\frac{1/c^2}{a+b}\\geq \\frac{\\left(1/a+1/b+1/c\\right)^2}{2(a+b+c)}=\\frac{2}{a+b+c}\r\n\\]\r\nand we are done." } { "Tag": [ "LaTeX", "algebra proposed", "algebra" ], "Problem": "let a_1;a_2;.............a_n>0 S_1=\\bigsum_{k=0}^{n}a_k et S_2=\\bigsum_{k=0}^{n}a^2_k\r\nProuve that \\bigsum_{k=0}^{n}\\frac{s_2-a^2_k}{s_1-a_k}>=s_1", "Solution_1": "i want use latex :?:", "Solution_2": "You want to say:\r\n\r\nLet $a_1$;$a_2$;...;$a_n$>0, $S_1=\\sum_{k=1}^{n}a_k$ and $S_2=\\sum_{k=1}^{n}a^2_k$. \r\nProve that $\\sum_{k=1}^{n}\\frac{S_2-a^2_k}{S_1-a_k}\\geq S_1$.\r\n\r\nRight?", "Solution_3": "Using Cauchy inequlity directly , by the way, n should > 1.\r\n\r\n$(n-1)(S_2-a^2_k) \\geq (S_1-a_k)^2$\r\n\r\nSo $\\frac{S_2-a^2_k}{S_1-a_k} \\geq \\frac{S_1-a_k}{n-1}$\r\n\r\nAdding n such inequalies, the result follows.", "Solution_4": "thanks zhaoli for rewrite this exercuce with latex :bye:" } { "Tag": [], "Problem": "A refrigerator is offered at sale at $ \\$ 250.00$ less successive discounts of $ 20\\%$ and $ 15\\%$. The sale price of the refrigerator is:\r\n$ \\textbf{(A)}\\ 35\\% \\text{ less than } \\$250.00 \\qquad\\textbf{(B)}\\ 65\\% \\text{ of } \\$250.00 \\qquad\\textbf{(C)}\\ 77\\% \\text{ of } \\$250.00 \\qquad\\textbf{(D)}\\ 68\\% \\text{ of } \\$250.00 \\qquad\\textbf{(E)}\\ \\text{none of these}$", "Solution_1": "A discount of $ 20\\%$ represents bringing the price down to $ \\frac{4}{5}$ of the original value. Likewise, a discount of $ 15\\%$ lowers the price to $ \\frac{17}{20}$ of the original value. Multiplying the two together, we get $ \\frac{4}{5}(\\frac{17}{20})$, making $ \\frac{17}{25}$, or $ 68\\%$ of the original price. Thus, the answer is $ \\fbox{D}$.", "Solution_2": "[hide=\"Similar...\"]\nLet $ D_1$ and $ D_2$ be the successive discounts, and $ D$ be the final discount. Then we have $ D\\equal{}D_1\\plus{}D_2\\minus{}D_1D_2 \\implies D\\equal{}20\\plus{}15\\minus{}3\\equal{}32$. Thus, the sale price, $ S$, equals $ 250\\minus{}(.32)(250)\\equal{}(.68)(250)$, which is $ 68 \\%$ of $ 250$, or $ \\boxed{\\textbf{(D)}}$.\n[/hide]" } { "Tag": [ "Pascal\\u0027s Triangle", "algebra", "binomial theorem" ], "Problem": "Find the sum of the coefficients in the expansion of $(4x+2y)^{3}$.", "Solution_1": "[hide]$(4x+2y)^{3}=64x^{3}+96x^{2}y+48xy^{2}+8y^{3}$\n\n$64+96+48+8=\\boxed{216}$\n\nI'm not sure, but can the sum of the coefficients of $(ax+by)^{z}$ be expressed as $(a+b)^{z}?$[/hide]", "Solution_2": "to help solve future problems like the one u posed u remembe rthe following definition:\r\n\r\ncoefficiant- $(ax+by)$ $a$ and $b$ are the coefficiants and $x$ and $y$ are the variables", "Solution_3": "[hide=\"Hint\"]How to reduce the monomial $cx^{m}y^{n}$ to the coefficient $c$? Put $x=1, y=1$[/hide]\n\n[hide=\"Answer\"]Sum of the coefficients of $(4x+2y)^{3}$ is $(4+2)^{3}=216$[/hide]", "Solution_4": "So my assumption was correct? :D", "Solution_5": "Yep! :) \r\nMy math teacher tends to give problems like this a lot, and finding simple shortcuts makes math homework go that much faster. All it is is a matter of informal proof, just for your own use.", "Solution_6": "I have to stuff that in my memory somehow. Do you know the proof? It must be pretty cool too...", "Solution_7": "It's not rigorous. I just proved it for a few simple cases, stuff we get in math class. I don't know the rigorous proof. I'm assuming it's fairly straightforward.", "Solution_8": "[b][u]Another example :\r\n\r\nFor $(2x+y)^{n}\\equiv A_{0}x^{n}+A_{1}x^{n-1}y+\\ldots+A_{n-1}xy^{n-1}+A_{n}y^{n}$ ascertain the sums : $\\{\\begin{array}{c}1.\\blacktriangleright\\ A_{0}-A_{1}+A_{2}-\\ldots\\ .\\\\\\\\ 2.\\blacktriangleright\\ A_{0}+A_{2}+A_{4}+\\ldots\\ .\\\\\\\\ 3.\\blacktriangleright\\ A_{1}+A_{3}+A_{5}+\\ldots\\ .\\\\\\\\ \\ldots\\ldots\\mathrm{\\ a.s.o.\\ }\\ldots\\ldots\\ .\\end{array}$", "Solution_9": "Whenever you hear sum of coefficients think f(1) :D", "Solution_10": "[hide=\"quick solution\"]By Pascal's Triangle:\n\n[code] 1\n 11\n 121\n1331[/code]\n\nthe expansion for $\\left(x+y\\right)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}$. We plug in 4x and 2y:\n\n$\\left(4x+2y\\right)^{3}=64x^{3}+96x^{2}y+48xy^{2}+8y^{3}$\n\nWe sum those up:\n\n$64+96+48+8=160+56=\\boxed{216}$[/hide]\r\n\r\nVery clever, Farenhajt.", "Solution_11": "Yup, just imagine expanding something like $(231x-384y+154z)^{457}$ in order to find the sum of the coefficients.", "Solution_12": "Ooooo I know this , plug in x=1, y=1, and z=1, and we get 1^457=1", "Solution_13": "[hide]Let x=y=1 and the expansion will leave the sum of the coefficients. $(4x1+2x1)^{3}=6^{3}=216$.[/hide]", "Solution_14": "[quote=\"i_like_pie\"]\nI'm not sure, but can the sum of the coefficients of $(ax+by)^{z}$ be expressed as $(a+b)^{z}?$[/quote]\r\n\r\nBy Newton's Binomial Theorem, we have $(a+b)^{n}=\\sum_{k=0}^{n}\\binom{n}{k}a^{n-k}b^{k}$.\r\nApplying this to our case, we get $(ax+by)^{n}=\\sum_{k=0}^{n}\\binom{n}{k}(ax)^{n-k}(by)^{k}$, whiche can we rewritten as $(a+b)^{n}=\\sum_{k=0}^{n}\\binom{n}{k}a^{n-k}b^{k}x^{n-k}y^{k}$.\r\nNow it's evident that the sum of the coefficients we're looking at is $\\sum_{k=0}^{n}\\binom{n}{k}a^{n-k}b^{k}$, but that's $(a+b)^{n}$! This demonstrates that it suffices to assume $x,y=1$ in the original binomial to calculate the sum of the coefficients of the resultant polynomial...\r\nHope I've been clear! :wink:", "Solution_15": "The Binomial Theorem is unnecessary, since the result generalizes to any polynomial.\r\n\r\nLet $P(x) = a_{0}+a_{1}x+...+a_{n}x^{n}$ be any polynomial. Then the sum of its coefficients is\r\n\r\n$a_{0}+a_{1}+...+a_{n}$\r\n\r\nWhich is, of course, $P(1)$. This result generalizes to an arbitrary number of variables." } { "Tag": [ "function", "real analysis", "real analysis solved" ], "Problem": "I have no idea whether this is true or not. Can someone give me some references?\r\n\r\nLet A and B be 2 sets. If we can define injections from A to B and from B to A, then can we define a bijection from A to B?\r\n\r\nA related question:\r\n\r\nIf we can define a bijection from A to a subset of A, A', then can we define a bijection from A to any subset of A which contains B?", "Solution_1": "The first question is called Canthor's Theorem, and yes it is true :).", "Solution_2": "Is there a solution I could understand? Maybe it's easier than I think? Or not?", "Solution_3": "Never mind, I found about it on the Internet. It's caleed th Cantor-Bernstein-Schroeder Theorem, and there was a short proof there:\r\n\r\nf:A->B and g:B->A are injections.\r\n\r\nC0=A\\g[B] and Cn+1=g[f[Cn]] for all n>=0. C=the reunion of all Cn with n from 0 to 00. It's easy to show that h:A->B, h(x)=f(x) if x\\in C and g-1(x) if x not \\in C is a bijection.", "Solution_4": "Sorry for the repeated posts, especially since they answer my own question :D. The second question is also true. Let B be a subset of A such tht B contains A'. The restriction of f (the bijection from A to A') to B is an injectije function :B->A'. The identical function is an injective function :A'->B. By using question 1, we get that A' and B have the same cardinality, and since the same can be said about A' and A, it means that A and B have the same cardinality.", "Solution_5": "Don't worry Grobber, it is never worthless to talk about Cantor-Bernstein' theorem. It is useful one.\r\n\r\nPierre.", "Solution_6": "Well, then, here's just another observation (:D):\r\n\r\n(1)<=>(2) ((1) and (2) are the 2 questions I asked in my first post in this topic). We saw that (1)=>(2). To show that (2)=>(1):\r\n\r\nLet f:A->B and g:B->A be injections. Then A is in bijection with g(f(A)) and B is in bijection with g(B). It's clear that g(B) contains g(f(A)) because B contains f(A). From (2) we get that A is in bijection with g(B) because, as I said, g(B) contains g(f(A)), so both A and B are in bijection with g(B), so they're in bijection with each other." } { "Tag": [ "function", "combinatorics proposed", "combinatorics" ], "Problem": "assume that k,n are two positive integer $k\\leq n$count the number of permutation $\\{\\ 1,\\dots ,n\\}\\ $ st for any $1\\leq i,j\\leq k$and any positive integer m we have $f^m(i)\\neq j$ ($f^m$ meas iterarte function,)", "Solution_1": "I suppose you mean for all $i\\ne j$, since there's always some $m$ s.t. $f^m(i)=i$, ($i$ is always part of a cycle) right?", "Solution_2": "yes ;)" } { "Tag": [], "Problem": "Let $ a,b>1$ . Solve the equation: $ (a^3\\plus{}b)^x\\minus{}a^{3x}\\equal{}3a^xb(a^x\\plus{}b)\\plus{}b^3$ .", "Solution_1": "[hide=\"Solution\"]Note that $ (a^x \\plus{} b)^3 \\equal{} a^{3x} \\plus{} 3a^xb(a^x \\plus{} b) \\plus{} b^3$\n\nSo we can rewrite it as:\n\n$ (a^3 \\plus{} b)^x \\equal{} (a^x \\plus{} b)^3$\n\nSo we have $ x \\equal{} 3$ for all $ a,b$[/hide]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "If $ a, b,c$ are positive real numbers, then \r\n\r\n\\[ 3(a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2)(a^2\\plus{}b^2\\plus{}c^2)\\ge (a^2\\plus{}ab\\plus{}b^2)(b^2\\plus{}bc\\plus{}c^2)(c^2\\plus{}ca\\plus{}a^2)\\]", "Solution_1": "Do you want us to prove that?\r\n\r\n(btw, it isn't high school basics level.)", "Solution_2": "I`m sorry, I didn`t know where to post it. I thought posting it here will be ok. Anyway, a moderator can move this post.", "Solution_3": "[hide=\"Brute Force\"]\nMultiply out, every term is of 6th degree, so AM-GM solves it (and maybe with a combination with Schur) [/hide]" } { "Tag": [ "inequalities", "algebra", "polynomial", "inequalities proposed" ], "Problem": "Prove that with$a,b,c>0$ and$p\\ge3+\\sqrt{7}$we have\r\n$\\frac{1}{pa^2+bc}+\\frac{1}{pb^2+ca}+\\frac{1}{pc^2+ab}\\ge \\frac{9}{(p+1)(ab+bc+ca)}$\r\n :roll: \r\nI think my solution is right.", "Solution_1": "Hi. It's not difficult. We only need to check if $a=b$. Use some computing. But maybe you shouldn't post it in new topic. Try to seach it in this forum.", "Solution_2": "[quote=\"hungkhtn\"]Hi. It's not difficult. We only need to check if $a=b$. Use some computing. But maybe you shouldn't post it in new topic. Try to seach it in this forum.[/quote]\r\n\r\nit can be solved by sum of squares again?\r\nJust wonder, are there any general methods that can be used to write an inequality into sum of squares?", "Solution_3": "[quote=\"hungkhtn\"]Hi. It's not difficult. We only need to check if $a=b$. Use some computing. But maybe you shouldn't post it in new topic. Try to seach it in this forum.[/quote]\r\nI think it is not easy. :lol: I wait somebody post a solution.", "Solution_4": "Oh,It is a interesting problem.If you said that,can you display your proof?", "Solution_5": "OK, Vasc. Wait me on day.", "Solution_6": "I haven't a nice solution. \r\nI proved it by expanding way, which is laborious enough.", "Solution_7": "Oh. A big mistake when I saw this problem. I didn't remember and I thought that:\r\n\r\n$\\frac{1}{a^2+kbc}+\\frac{1}{b^2+kac}+\\frac{1}{c^2+kab}\\ge \\frac{9}{(k+1)(ab+bc+ca)}$\r\n\r\nAnd... I don't know why $k\\ge 3+\\sqrt{3}$? Afterthat, I see that $k\\le {\\frac{3}-\\sqrt{7}}{2}$. But two this one is the same. I use the method sum of square.", "Solution_8": "[quote=\"hungkhtn\"]Oh. A big mistake when I saw this problem. I didn't remember and I thought that:\n\n$\\frac{1}{a^2+kbc}+\\frac{1}{b^2+kac}+\\frac{1}{c^2+kab}\\ge \\frac{9}{(k+1)(ab+bc+ca)}$\n\nAnd... I don't know why $k\\ge 3+\\sqrt{3}$? Afterthat, I see that $k\\le {\\frac{3}-\\sqrt{7}}{2}$. But two this one is the same. I use the method sum of square.[/quote]\r\nI can display the most important in my solution :\r\nSuppose that$a \\le b \\le c$we have$x=\\frac{a}{b},y=\\frac{b}{c}$and define$z=\\frac{c}{a}$if you want,change our ineq.($z=\\frac{1}{xy}$)\r\n$t\\in(0,1]: 1+x+xy=1+t+t^2$fixed $t$and$x=x(y)=\\frac{t(1+t)}{1+y}$\r\nCheck it for me! :roll:", "Solution_9": "my solution is long.Here it is .we have $\\leftrightarrow\\$(p+1)[p^2\\sum( a^2b^2)+p\\sum(ab(a^2+b^2))+abc\\sum(a)].(ab+bc+ca)\\geqq(pa^2+bc)(pb^2+ac)(pc^2+ab) \\leftrightarrow\\(p^3+p^2)\\sum(a^3b^3)+abc\\sum(ab(a+b)).(p^3+2p^2+2p+1)+(p^2+p)\\sum(a^2b^2(a^2+b^2))+2abc(P^2+p)\\sum(a^3)+(3p+3)a^2b^2c^2\\geq\\(9p^3+9)a^2b^2c^2+9p^2\\sum(a^3b^3)+9pabc\\sum(a^3) \\leftrightarrow\\(p^2+p)\\sum(a^2b^2(a^2+b^2))+abc\\sum(ab(a+b))(p^3+2p^2+2p+1)+(p^3-8p^2)\\sum(a^3b^3)\\geq\\abc\\sum(a^3)\\(7p-2p^2)+a^2b^2c^2(9p^3-3p+6)$ (*)\r\nthen use $\\(a-b)^2(b-c)^2(c-a)^2\\geq\\0$and $p\\geq\\3+\\sqrt{7}\\rightarrow p^3-6p^2+2p\\geq\\0$it is simple to show that $3+\\sqrt{7}$is the best", "Solution_10": "Indeed, the ineq $(a-b)^2(b-c)^2(c-a)^2 \\ge 0$ is the key. :lol:", "Solution_11": "mumath's solution is long and it doesn't solve the following ineq:\r\nfind all p>0;\r\n$\\sum \\frac{1}{\\sqrt{pa^2+bc}} \\ge \\frac{3\\sqrt{3}}{\\sqrt{(p+1)(ab+bc+ca)}}$", "Solution_12": "[quote=\"nthd\"]I can display the most important in my solution :\nSuppose that$a \\le b \\le c$we have$x=\\frac{a}{b},y=\\frac{b}{c}$and define$z=\\frac{c}{a}$if you want,change our ineq.($z=\\frac{1}{xy}$)\n$t\\in(0,1]: 1+x+xy=1+t+t^2$fixed $t$and$x=x(y)=\\frac{t(1+t)}{1+y}$[/quote]\r\nI don't understand the end of your solution. Can you give us some additional details ?", "Solution_13": "[quote=\"Vasc\"]Let $ a,b,c > 0$ and $ p\\ge 3 \\plus{} \\sqrt {7}.$ Prove that\n\n$ \\frac {p \\plus{} 1}{pa^2 \\plus{} bc} \\plus{} \\frac {p \\plus{} 1}{pb^2 \\plus{} ca} \\plus{} \\frac {p \\plus{} 1}{pc^2 \\plus{} ab}\\ge \\frac {9}{bc \\plus{} ca \\plus{} ab}.$\n\nI proved it by expanding way, which is laborious enough.[/quote]$ \\sum{\\frac {p \\plus{} 1}{pa^2 \\plus{} bc}} \\minus{} \\frac {9}{bc \\plus{} ca \\plus{} ab}\\equiv\\frac {F(a,b,c,p)}{(pa^2 \\plus{} bc)(pb^2 \\plus{} ca)(pc^2 \\plus{} ab)(bc \\plus{} ca \\plus{} ab)}.$\n\n$ F(a,b,c,p) \\equal{} F(a,a \\plus{} s,a \\plus{} t,k \\plus{} 5) \\equal{} (k \\plus{} 3)(k \\plus{} 6)(5k \\plus{} 24)\\left(s^2 \\plus{} t^2 \\minus{} st\\right)x^4$\n\n$ \\plus{} (s \\plus{} t)\\left[2\\left(2k^3 \\plus{} 32k^2 \\plus{} 162k \\plus{} 261\\right)(s \\minus{} t)^2 \\plus{} \\left(8k^3 \\plus{} 106k^2 \\plus{} 450k \\plus{} 603\\right)st\\right]x^3$\n\n$ \\plus{} \\left\\{(s \\minus{} t)^2\\left[(k \\plus{} 5)(4k \\plus{} 15)\\left(s^2 \\plus{} t^2\\right) \\plus{} 3\\left(2k^3 \\plus{} 32k^2 \\plus{} 162k \\plus{} 261\\right)st\\right] \\plus{} \\left(18k^3 \\plus{} 226k^2 \\plus{} 899k \\plus{} 1101\\right)s^2t^2\\right\\}x^2$\n\n$ \\plus{} st(s \\plus{} t)\\left[(k \\plus{} 5)(4k \\plus{} 15)\\left(s^2 \\plus{} t^2\\right) \\plus{} \\left(4k^3 \\plus{} 38k^2 \\plus{} 91k \\plus{} 6\\right)st\\right]x$\n\n$ \\plus{} (k \\plus{} 5)s^2t^2\\left[(k \\plus{} 6)(s \\minus{} t)^2 \\plus{} \\left(k \\plus{} 2 \\minus{} \\sqrt {7}\\right)\\left(k \\plus{} 2 \\plus{} \\sqrt {7}\\right)st\\right]\\geq0,$\n\nwhich is clearly true for $ a \\equal{} \\min\\{a,b,c\\}$ and $ k\\geq \\sqrt {7} \\minus{} 2 \\equal{} 0.64575\\cdots.$\n[quote=\"nthd\"]Find all $ p$ such that\n\n$ \\sqrt {\\frac {p \\plus{} 1}{pa^2 \\plus{} bc}} \\plus{} \\sqrt {\\frac {p \\plus{} 1}{pb^2 \\plus{} ca}} \\plus{} \\sqrt {\\frac {p \\plus{} 1}{pc^2 \\plus{} ab}} \\ge 3\\sqrt\\frac { 3}{bc \\plus{} ca \\plus{} ab}.$[/quote]It holds for positive $ a,b,c$ if and only if $ p \\equal{} 0\\vee p\\geq 8.4971\\cdots$, which is a root of the following irreducible polynomial \r\n\r\n$ p^4 \\minus{} 60p^3 \\plus{} 460p^2 \\minus{} 192p \\plus{} 16.$" } { "Tag": [], "Problem": "calculate the work done to locate a 150kg satellite into a geostationary orbit.\r\n\r\nI'm guessing i need to use ep=- GMmG/r\r\nbut it would be possitive right? oR negative\r\n\r\nthe radius will be 35800 km? or 42,245 cause the 35800 is the altitude for the satellite?????\r\nm and M is mass of earth and the satellite\r\nso is this how to work out the right answer\r\n\r\n6.67*10^-11 (mass of eath)(150)\r\n-----------------------------------------\r\n35800000 or 42,245 000\r\nIs this the correct way.\r\n\r\nPLease help\r\n \t\r\nAlso is this the right forumla to figure the orbital radius for a geostationary satellite for any planet?\r\nr = Gm ^1/3\r\n ---\r\n 4 (pie)^2", "Solution_1": "Beware, work is the CHANGE of energy. And yes, there is a minus in the formula. Then,\r\n\r\n$W = (-\\frac{GMm}{r}) - (-\\frac{GMm}{R_0})$\r\n\r\nwhere $R_0$ is the starting radius (probably radius of the Earth) and $r$ is the terminating radius (i.e. the radius of the geostationary orbit), gives a positive value.", "Solution_2": "Keep in mind also that a satelite in orbit has a velocity, and hence has kinetic energy.", "Solution_3": "So it would be \r\n(G)(150)(5.9*10^24) \r\n----------------- \r\n42 000 000 \r\n\r\ntake away\r\n\r\n (G)(150)(5.9*1024)\r\n ------------------------------------\r\n 6000 000\r\nI think 6 000 000 m is the radius of the earth, close enough.\r\nSo is that correct. ;)", "Solution_4": "Yes, the potential energy looks right. Don't forget the kinetic energy though.", "Solution_5": "So i will have 2 answers, one for Potential energy and one for knitic energy.\r\nThe forumlua for kenitic energy is .5 mv^2\r\nright. How would i calculate the velocity. Thanks", "Solution_6": "You can use a fairly well-known formula\r\n\r\n$a=\\frac{v^2}{r}$\r\n\r\nwhere $a$ must be equal to the gravitational acceleration (in the proper height)." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Consider $ A,B,C \\in M_n(\\mathbb{C})$ such that $ A \\cdot B \\cdot C\\equal{}I_n$. Prove that if $ I_n\\plus{}A\\plus{}AB$, $ I_n\\plus{}B\\plus{}BC$, $ I_n\\plus{}C\\plus{}CA$ are reversible, then the sum of their reversed matrix is $ I_n$.", "Solution_1": "$ C\\equal{}B^{\\minus{}1}A^{\\minus{}1}$;\r\n let\r\n$ I\\plus{}A\\plus{}AB\\equal{}U,I\\plus{}B\\plus{}BC\\equal{}I\\plus{}B\\plus{}A^{\\minus{}1}\\equal{}V\\equal{}A^{\\minus{}1}U,I\\plus{}C\\plus{}CA\\equal{}I\\plus{}B^{\\minus{}1}A^{\\minus{}1}\\plus{}B^{\\minus{}1}\\equal{}W\\equal{}B^{\\minus{}1}A^{\\minus{}1}U$.\r\nThen $ U^{\\minus{}1}\\plus{}V^{\\minus{}1}\\plus{}W^{\\minus{}1}\\equal{}U^{\\minus{}1}(I\\plus{}A\\plus{}AB)\\equal{}I$." } { "Tag": [], "Problem": "Find a positive integer that is $ 1.5$ more than its reciprocal.", "Solution_1": "Setting the number as x, we get the equation $ x \\equal{} \\frac1x \\plus{} \\frac32$\r\n$ 2x^2 \\minus{} 3x \\minus{} 2 \\equal{} 0$\r\n$ x \\equal{} \\minus{}\\frac12$ or $ 2$\r\nWe want a positive integer:\r\n$ \\boxed2$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Show that for any $i=1,2,3$, there exist infinity many positive integer $n$, such that among $n$, $n+2$ and $n+28$, there are exactly $i$ terms that can be expressed as the sum of the cubes of three positive integers.", "Solution_1": "1. i=1, $n=3m^{3}-1$ $m^{3}+m^{3}+m^{3}\\in (n,n+2)$.\r\n2. i=2 $n=3m^{3}+9m^{2}+27m+14$, then $(m+2)^{3}+(m+2)^{3}+(m-1)^{3}=n+1,(m+3)^{3}+m^{3}+m^{3}=n+15.$\r\n3. i=3, I find $n=3m^{3}+27m^{2}+195m+332$, then \\[N1=(m+6)^{3}+(m+5)^{3}+(m-2)^{3}=n+1, N2=(m+7)^{3}+(m+2)^{3}+m^{3}=n+19, N3=(m+8)^{3}+(m+1)^{3}+m^{3}=n+181.\\]", "Solution_2": "Our \"Maths discussion group\" seems solve the case of $i=1,3$:\r\nFirst,a lemma,$a^{3}+b^{3}+c^{3}$ can't be $4,5(mod 9)$\r\nBrief proof:$x^{3}=0,1,-1 (mod9)$ and try all case of $a,b,c$ in mod 9 and result is out.\r\nFor $i=1$,take $n=(9a+1)^{3}+(9b+1)^{3}+(9c+1)^{3}=3 (mod9)$ then\r\n$n+2=5(mod9)$ and $n+28=4(mod9)$,both can't be sum of 3 cubes.\r\nFor $i=3$, take $n=k^{3}+(6k)^{3}+(8k)^{3}$, then $n+2=(9k)^{3}+1+1$ and $n+28=(9k)^{3}+27+1$\r\nAlso, one of our \"group member\" think that if we can prove there are infinite $d$ such that $d^{3}=a^{3}+b^{3}+c^{3}$ have no integral solution, then the case of $i=2$ is also solved", "Solution_3": "ychjae, i=2 is solved.\r\nObserved that 5^3+5^3 = 6^3 + 2^3 + 26\r\nTherefore, we can take \r\nn+28 = (3k-1)^3 +5^3+5^3\r\nn+2 = (3k-1)^3 + 6^3 + 2^3\r\nn =(3k-1)^3 + 6^3 + 2^3 -2=(3k-1)^3 + 222 which can't be the sum of three cubes by the lemma.\r\n\r\nBy the way, i think Rust's solution is wrong.", "Solution_4": "For $i = 1$ take $n = 1^3+1^3+(3k+1)^3$ for any $k \\in \\mathbb{N}.$ Then, using the fact that any sum of three cubes is congruent to $\\pm 3, \\pm 2, \\pm 1, 0$ (mod $9$), we know that $n+2, n+28$ are not sums of three cubes.\n\nFor $i = 2$ take $n = (3k+1)^3 + 10^3 + 18^3$ for any $k \\in \\mathbb{N}.$ In this way, $n+28 = (3k+1)^3 + 1^3+19^3$ and $n+2 \\equiv 4$ (mod $9$), so that $n, n+28$ are sums of three cubes whereas $n+2$ is not from the earlier fact.\n\nFor $i = 3$ take $n = 216k^3.$ In this way, $n = (3k)^3 + (4k)^3 + (5k)^3, n+2 = (6k)^3+1^3+1^3, n+28 = (6k)^3+1^3+3^3.$ \n$\\square$ " } { "Tag": [ "ratio", "limit", "inequalities", "induction" ], "Problem": "Does the product of all terms in the infinite geometric sequence with starting term 9 and ratio 2/3 converge?", "Solution_1": "Hmm...\r\n\r\n$ \\lim_{k\\to\\infty}9\\cdot\\left(\\frac{2}{3}\\right)^k\\equal{}0$, so I'd say the terms eventually become infinitesimally small, and the product simply becomes 0.", "Solution_2": "[hide]$ \\lim_{k\\to\\infty}3^{2k}\\cdot(\\frac{2}{3})^{\\binom{k-1}{2}}=\\lim_{k\\to\\infty}\\frac{2^{\\binom{k-1}{2}}}{3^{\\binom{k-1}{2}-2k}})=\\sqrt{(\\lim \\frac{2^{k-1}}{3^{k-5}} )^k}=0$.[/hide]", "Solution_3": "oh woops i was kinda of thinking of something else. Can we show that \\[ \\frac 32 \\times \\frac 54 \\times \\frac 76 \\times frac 98 \\times \\cdots \\times \\frac{2n \\plus{}1}{2n}\\] converges? thanks", "Solution_4": "No. $ \\prod_{i\\equal{}1}^{n} \\left( 1 \\plus{} \\frac{1}{2i} \\right) > \\sum_{i\\equal{}1}^{n} \\frac{1}{2i}$, which diverges.", "Solution_5": "$ \\left( \\frac 32 \\times \\frac 54 \\times \\frac 76 \\times \\frac 98 \\times \\cdots \\times \\frac{2n \\plus{}1}{2n} \\right)^2$\r\n$ > \\left( \\frac 32 \\times \\frac 54 \\times \\frac 76 \\times \\frac 98 \\times \\cdots \\times \\frac{2n \\plus{}1}{2n} \\right) \\left( \\frac 43 \\times \\frac 65 \\times \\frac 87 \\times \\cdots \\times \\frac{2n \\plus{}2}{2n\\plus{}1} \\right)$\r\n$ \\equal{} \\frac 32 \\times \\frac 43 \\times \\frac 54 \\times \\cdots \\times \\frac{2n\\plus{}2}{2n\\plus{}1}$\r\n$ \\equal{} n\\plus{}1$\r\n\r\nThus the original sequence of products is greater than the sequence $ \\sqrt{n\\plus{}1}$, which diverges.", "Solution_6": "[quote=\"t0rajir0u\"]No. $ \\prod_{i \\equal{} 1}^{n} \\left( 1 \\plus{} \\frac {1}{2i} \\right) > \\sum_{i \\equal{} 1}^{n} \\frac {1}{2i}$, which diverges.[/quote]\r\nHmm, how do we prove this inequality? I initially tried to rewrite the lhs as e to the sum of $ \\ln(1\\plus{}1/2i)$ but it didn't help much...", "Solution_7": "If you think about how it (the LHS) expands,you can see that the RHS is the LHS's terms where you pair off each 1/2i with infinitely other 1's", "Solution_8": "I proved it with induction, just to be silly.\r\n\r\n[hide=\"Blah\"][b]Base cases[/b]:\n$ n\\equal{}1$: $ 1\\plus{}1/2>1/2$.\n$ n\\equal{}2$: $ (1\\plus{}1/2)(1\\plus{}1/4)>\\frac{1}{2}\\plus{}\\frac{1}{4}$.\n$ n\\equal{}3$: $ (1\\plus{}1/2)(1\\plus{}1/4)(1\\plus{}1/8)>\\frac{1}{2}\\plus{}\\frac{1}{4}\\plus{}\\frac{1}{6}$.\n\n[b]Inductive step[/b]: Assume it holds for $ n\\equal{}m$ for some $ m>3$. That is, $ \\prod_{i\\equal{}1}^{m}\\left(1\\plus{}\\frac{1}{2i}\\right)>\\sum_{i\\equal{}1}^{m}\\frac{1}{2i}$.\n\nMultiply both sides by $ 1\\plus{}\\frac{1}{2(m\\plus{}1)}$ to get:\n\n$ \\prod_{i\\equal{}1}^{m\\plus{}1}\\left(1\\plus{}\\frac{1}{2i}\\right)>\\left(1\\plus{}\\frac{1}{2(m\\plus{}1)}\\right)\\sum_{i\\equal{}1}^{m}\\frac{1}{2i}$.\n\nTherefore it suffices to prove that $ \\left(1\\plus{}\\frac{1}{2(m\\plus{}1)}\\right)\\sum_{i\\equal{}1}^{m}\\frac{1}{2i}\\geq \\sum_{i\\equal{}1}^{m\\plus{}1}\\frac{1}{2i}$.\n\nSubtract $ \\sum_{i\\equal{}1}^{m}\\frac{1}{2i}$ from both sides and factor:\n\n$ \\frac{1}{2(m\\plus{}1)}\\sum_{i\\equal{}1}^{m}\\frac{1}{2i}\\geq \\frac{1}{2(m\\plus{}1)}$.\n\nDivide:\n\n$ \\sum_{i\\equal{}1}^{m}\\frac{1}{2i}\\geq 1,$\n\nwhich is true for any $ m>3$.\n\nTherefore, $ \\prod_{i\\equal{}1}^{m\\plus{}1}\\left(1\\plus{}\\frac{1}{2i}\\right)>\\left(1\\plus{}\\frac{1}{2(m\\plus{}1)}\\right)\\sum_{i\\equal{}1}^{m}\\frac{1}{2i}\\geq \\sum_{i\\equal{}1}^{m\\plus{}1}\\frac{1}{2i}$, which completes the induction.[/hide]" } { "Tag": [ "trigonometry", "geometry", "angle bisector" ], "Problem": "$\\triangle ABC$ is an isosceles triangle with $AB=AC$. If the angle bisector of $\\angle B$ hits $AC$ at $D$ and $BD+AD=BC$, find $\\angle DBC$.", "Solution_1": "trig solution[hide]\nLet $\\angle DBC=x$ rewrite $BD+AD=BC$ as $1+\\frac{AD}{BD}=\\frac{BC}{BD}$ using sine theoreme $1+\\frac{sinx}{sin4x}=\\frac{\\sin3x}{\\sin2x}$ this becomes $sin 4x +sinx = 2\\sin3x\\cos2x$ or $\\sin5x=sin4x$ so $x=20$ [/hide]", "Solution_2": "That's correct, but can you explain your solution a bit more thoroughly? Thanks :) \r\n\r\nAny other solution?", "Solution_3": "Hope this is more clear now :D \r\n[hide]Let $\\angle DBC=x$ then $\\angle DCB=2x$ and $\\angle BAC=180-4x$ from $BD+AD=BC$ we obtain \n$1+\\frac{AD}{BD}=\\frac{BC}{BD}$. Using sine theoreme in $\\Delta ADB$ we have $\\frac{AD}{BD}=\\frac{sinx}{sin(180-x)}=\\frac{sinx}{sin4x}$ and from sine theoreme in $\\Delta BDC$ we get $\\frac{BC}{BD}=\\frac{sin(180-3x)}{sin2x}=\\frac{sin3x}{sin2x}$\nThe rest is just simple trigonometry. Hope i made it more clear this time[/hide]" } { "Tag": [], "Problem": "Is it possbile to re-write the following expression:\r\n\r\n$\\frac{1}{a \\cdot (a+z)}$\r\n\r\n I know this is a very general or rather vague question but to say anything more would immediately reveal the solution as obvious, not to mention rob the reader of the enjoyment of solving the problem without aid.\r\n\r\n O' and $\\frac{1}{a^2 +az}$ is not a valid answer, as it is only simplifying the expression, not re-writing it.", "Solution_1": "you could rightit as $\\frac{\\frac1a-\\frac1{a+z}}z$ which is $\\frac1{az}-\\frac1{az+z^2}$\r\n\r\nis that what you're looking for?", "Solution_2": "I think he is looking for this form:\r\n\r\n $\\frac1{z} (\\frac1{a}-\\frac1{a+z})$\r\n\r\n(very simple partial fraction decomposition)", "Solution_3": "[quote=\"cj\"]I think he is looking for this form:\n\n $\\frac1{z} (\\frac1{a}-\\frac1{a+z})$\n\n(very simple partial fraction decomposition)[/quote]\r\nlook at the previous post (exactly what I did)", "Solution_4": "Another possibility is $\\frac{1}{a}\\cdot \\frac{1}{a+z}$, but that seems a little too obvious. I would guess that your answers are correct, although there could be other ways.", "Solution_5": "Yes, nat mc's solution was what I was initally thinking of, but as for the existence of other solutions I am completely clueless." } { "Tag": [ "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Define a topology T,on the set of real numbers,R,as follows:T contains the empty set,and a nonempty subset O of R is in T if and only if,the complement of O,is a finite set.Which of the following statements is/are true?\r\n[size=150]1.Every infinite subset of R is closed\n2.Every subset of R is compact.\n3.R is Hausdorff[/size]", "Solution_1": "1. Definitely not. $ I\\equal{}(\\minus{}\\infty, 0)\\cup (0,\\infty)$ is infinite and open since the complement is $ \\{0\\}$ a finite set. In fact, the only closed sets are finite ones.\r\n\r\n2. The space itself is compact, so this should be true.\r\n\r\n3. It is not Hausdorff.", "Solution_2": "For the second question: let $ X$ be a subset of $ \\mathbb{R}$ and $ \\{U_{\\alpha}\\}_{\\alpha\\in A}$ an open cover. If $ \\alpha$ is any index, then $ U_{\\alpha}^{c}$ is a finite set. Since it is a finite set, $ U_{\\alpha}^{c}\\cap X$ is finite, and it is clear that there are finitely many sets in the open cover, $ U_{\\alpha_{1}},\\ldots,U_{\\alpha_{n}}$ which cover this set (for each point in $ U_{\\alpha}^{c}\\cap X$, just choose an element of the open cover containing that point). But then $ \\{U_{\\alpha},U_{\\alpha_{1}},\\ldots,U_{\\alpha_{n}}\\}$ is a finite subcover of $ X$. In other words, $ X$ is compact." } { "Tag": [], "Problem": "Hey...Ok so i just read like all 11 pages of the Mole game, I would have applied but i didnt see it fast enough, so im sure that there are others like me who have seen the post and wish that they could be in a game. I would be willing to start a mole game 2 if people would be interested. Please pm me or post below if you would be interested in participating!!!\r\n\r\n@perfect--im not trying to copy but i want to offer the game to players that might have been rejected the first time around--thanks", "Solution_1": "There will be another Mole game after the first one is done.", "Solution_2": "AHH ill wait for that thanks!", "Solution_3": "I'm currently working on tasks for the second one...I've got some good ones for next time." } { "Tag": [ "blogs" ], "Problem": "Consider the pattern\r\n\r\n1,9,36,100,.....\r\n\r\nFind the 15th number in the sequence.", "Solution_1": "[hide] The sequence adds the next cubic number starting from 2^3.\n\nTherefore, the 15th term would be 1+2^3+3^3+4^3...+15^3.\n\nNow lets just add them up with the nifty built in calculator on my computer and get a total of 14400.\n\nHope i'm right :lol: \n[/hide]", "Solution_2": "[hide]Yes you are, but there's a slicker solution that involves triangular #s (you don't have to think about a calculator)(I asked this question at lunch today)[/hide]", "Solution_3": "Yeah, you got the right answer, but here's a much simpler method.\r\n[hide=\"simpler solution\"]\nIt's just the square of all the triangular numbers. $1^{2}. (1+2)^{2}=9. (1+2+3)^{2}=36.$ ETC\nSo $(1+2+3...15)^{2}=(\\frac{(15)(16)}{2})^{2}=(120)^{2}=14400.$[/hide]", "Solution_4": "Alan we didn't have the same lunches today :|", "Solution_5": "[hide]If you want to talk about lunch periods, click on the link.\n\n(Sorry, link non-existent, just go to my blog and post in the triangle theorems thing)[/hide]" } { "Tag": [ "geometry" ], "Problem": "I am having major difficulties with this topic and I do need to understand it for the ap test. I was wondering if someone could help me with the set up for the following problems:\r\n\r\n1. A raft 6 m x 4 m in area is floating on a river. When a loaded car pulls the raft, it sinks 3.0 cm lower in the water. What is the weight of the car?\r\n\r\n2. In an aorta with a cross-sectional area of 16 cm squared the blood (density = 1030 kg/m^3) flows with a speed of 30 cm/s. What is the flow rate, in kilograms per second, of blood in the aorta? The aorta branches to form a large number of fine capillaries with a combined cross-sectional area of 2 x 10^3 cm squared. What is sthe speed of blood flow in the capillaries?\r\n\r\n3. Hydraulic stamping machines exert tremendous forces on a sheet of metal to form it into the desired shape. Suppose the input force is 900 N on a piston that has a diameter of 1.8 cm. The output force is exerted on a piston that has a diameter of 36 cm. How large a force does the press exert on the sheet being formed?\r\n\r\n\r\nThank you so much for all the help!", "Solution_1": "By Archimedes law of buoyancy, the weight of the car $F_G$ is equal to the weight of the water pushed out, which is equal to the volume of this water $\\triangle V = S \\triangle h$ (S is the raft area) times its density $\\rho_w$ times the gravitational acceleration $g$: $F_G = \\triangle V \\rho_w g$.\r\n\r\nThe flow rate of blood (volume per time) is equal to $\\frac{\\triangle V}{\\triangle t} = S_a v_a = S_c v_c$, where $S_a, S_c$ are cross-sections of the aorta and capillaries and $v_a, v_c$ the corresponding velocities, because as a good approximation, blood is considered incompressible.\r\n\r\n\r\nHydrostatic pressure in hydraulic liquid is the same in every direction. Neglecting the difference (if any) of piston heights above the floor, $p = \\frac{F_1}{S_1} = \\frac{F_2}{S_2}$." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that; for $ a,b,c>0$ we have:\r\n$ \\frac{a^2}{b}\\plus{}\\frac{b^2}{c}\\plus{}\\frac{c^2}{a}\\geq (a\\plus{}b\\plus{}c)\\plus{}\\frac{4(a\\minus{}b)^2}{a\\plus{}b\\plus{}c}$\r\nI have assumed $ a\\plus{}b\\plus{}c\\equal{}1$ and then proceeded, but it did not help much (apparently).\r\n\r\n[hide=\"source\"]Inequalities Through Problems; Hoojoo Lee[/hide]", "Solution_1": "hello, rewriting the given inrquality t0\r\n$ \\frac{(a\\minus{}b)^2}{b}\\plus{}\\frac{(b\\minus{}c)^2}{c}\\plus{}\\frac{(c\\minus{}a)^2}{a}\\geq\\frac{4(a\\minus{}b)^2}{a\\plus{}b\\plus{}c}$\r\nthis is equivalent to\r\n$ (a\\plus{}b\\plus{}c)\\left(\\frac{(a\\minus{}b)^2}{b}\\plus{}\\frac{(b\\minus{}c)^2}{c}\\plus{}\\frac{(c\\minus{}a)^2}{a}\\right)\\geq4(a\\minus{}b)^2$\r\nNow using Cauchy and we can prove\r\n$ (a\\plus{}b\\plus{}c)\\left(\\frac{(a\\minus{}b)^2}{b}\\plus{}\\frac{(b\\minus{}c)^2}{c}\\plus{}\\frac{(c\\minus{}a)^2}{a}\\right)\\geq4\\left(\\max(a,b,c)\\minus{}\\min(a,b,c)\\right)^2$\r\nSonnhard.", "Solution_2": "[quote=\"Potla\"]Prove that; for $ a,b,c > 0$ we have:\n$ \\frac {a^2}{b} \\plus{} \\frac {b^2}{c} \\plus{} \\frac {c^2}{a}\\geq (a \\plus{} b \\plus{} c) \\plus{} \\frac {4(a \\minus{} b)^2}{a \\plus{} b \\plus{} c}$\nI have assumed $ a \\plus{} b \\plus{} c \\equal{} 1$ and then proceeded, but it did not help much (apparently).\n\n[hide=\"source\"]Inequalities Through Problems; Hoojoo Lee[/hide][/quote]\r\n$ <\\equal{}>\\frac{(a\\minus{}b)^2}{b}\\plus{}\\frac{(b\\minus{}c)^2}{c}\\plus{}\\frac{(c\\minus{}a)^2}{a} \\ge \\frac{4(a\\minus{}b)^2}{a\\plus{}b\\plus{}c}$\r\nIt's true by cauchy-swarchz :)" } { "Tag": [], "Problem": "Could anyone explain modular numbers in semi-plain English?", "Solution_1": "yeah i need help on that too", "Solution_2": "same here. I need help too.", "Solution_3": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76228 . Locked in favor of that topic." } { "Tag": [ "puzzles" ], "Problem": "http://www.wittycomics.com/comic/43172\r\n\r\nEDIT", "Solution_1": "hehehe...\r\n\r\nA tip: Don't use so many words. It gets hard to read.", "Solution_2": "I think it's fine... it does become a problem if there are lots of frames though. With only three frames, I don't think people would get turned off so quickly.", "Solution_3": "This is a comic creating system that offers little variation in the \r\nbody positions and/or facial expressions of some (if not all) of the \r\ncharacters. In a different comic creating system, I could relatively\r\neasily have a character rolling his eyes in one particular panel with \r\nno dialogue at all to communicate something. Also, more of the \r\nbackground would be seen in that panel as a bonus.\r\n\r\nThank you both for constructive criticism (- & +)." } { "Tag": [], "Problem": "Suppose the integers $1,2,\\ldots 10$ are split into two disjoint collections $a_1,a_2, \\ldots a_5$ and $b_1 , \\ldots b_5$ such that $a_1 b_2 > b_3 > b_4 > b_5$ \n(i) Show that the larger number in any pair $\\{ a_j, b_j \\}$ , $1 \\leq j \\leq 5$ is at least $6$.\n(ii) Show that $\\sum_{i=1} ^{5} | a_i - b_i|$ = 25 for every such partition.", "Solution_7": "Considering the correct problem:\\\\\n\ni) FTSOC assume $\\max(a_{j} , b_{j})<6$\\\\\n\nSo $\\frac{1}{2} \\cdot \\left( a_{j}+b_{j}+|a_{j}-b_{j}|\\right)<6$ $\\forall$ $j \\in \\{1,2,3,4,5\\}$\\\\\n\n$\\iff \\frac{1}{2} \\cdot \\left( \\sum _{j=1}^{5} (a_{j}+b_{j})+\\sum_{j=1}^{5}(|a_{j}-b_{j}|)\\right)<\\sum_{j=1}^{5} 6$\\\\\n\n$\\iff \\frac{55}{2} +\\frac{1}{2}\\cdot \\left(\\sum_{j=1}^{5} |a_{j}-b_{j}|\\right)<30 \\iff \\sum_{j=1}^{5} |a_{j}-b_{j}|<5$ but also since $b_{1}>b_{2}>\\cdots >b_{5}$ and $a_{1}< a_{2} \\cdots a_4>a_3>a_2>a_1$ and\n$$\nb_1>b_2>b_3>b_4>b_5\n$$\n$$\n\\max \\left\\{a_1, b_1\\right\\} < \\max \\left\\{a_2, b_2\\right\\}<\\ldots\n$$\n$$\n<\\max \\left\\{a_5, b_5\\right\\}\n$$\nor vice versa\nso from (i) we have\n$(6+7+8+9+10)$ as only possible value\nwhich is 40\nSo $\\quad \\Omega+55=80$\n$\\Omega=25$ , Hence we always have $\\sum_{j=1}^{j=5} |a_{j}-b_{j}|=25$ $\\blacksquare$" } { "Tag": [ "function", "limit", "real analysis", "real analysis solved" ], "Problem": "Hi all.\r\n\r\nDoing a little self-study on limits, continuity, etc.\r\n\r\nI have the following function:\r\n\r\n$f(x)=\\{\\begin{array}{cc}x,&\\mbox{ if } x \\mbox{ is rational}\\\\0, &\\mbox{ if } x \\mbox{ is irrational}\\end{array}$\r\n\r\nHow would one prove that this function is continuous at $x=0$?\r\n\r\nLooking at a plot of this function, I would say that it's discontinuous everywhere but 0.\r\n\r\nWould it be showing that the limit exists at $x=0$?\r\n\r\nThanks for the help.\r\n\r\nBailey", "Solution_1": "Hi.\r\n\r\nShow that the limit exists at $0$ and this limit is $f(0)=0$.", "Solution_2": "This one makes more sense if you go ahead and quote the formal definition.\r\n\r\nInformally, $f$ is continuous at $0$ iff you can make $f(x)$ arbitrarily close to $f(0)$ by making $x$ arbitrarily close to $0.$ The formalization of that mostly involves specifying what we mean by \"arbitrarily close\":\r\n\r\n$f$ is continuous at $0$ iff $\\forall\\,\\epsilon>0\\,\\exists\\,\\delta>0$ such that if $|x-0|<\\delta$ then $|f(x)-f(0)|<\\epsilon.$\r\n\r\nCan you see how to make that work for your example? The core of a proof of continuity (or of a limit) will be a description of how you pick $\\delta$ once you know $\\epsilon.$ In this particular case, you can simply set $\\delta=\\epsilon.$\r\n\r\nWhile we're at it, can we also prove the other thing you claimed? Namely, that for $a\\ne 0,\\,f$ is discontinuous at $a.$ Restating the definition, \r\n\r\n$f$ is discontinuous at $a$ iff $\\exists\\,\\epsilon>0$ such that $\\forall\\,\\delta>0,\\,\\exists \\,x$ with $|x-a|<\\delta$ such that $|f(x)-f(a)|\\ge\\epsilon.$\r\n\r\nDo you see how to apply that to your example?", "Solution_3": "You can use the definition of limit as well.\r\n\r\ntake \u03b5>0. For |x| <\u03b5 ,then , you have \r\n\r\n |f(x) - f(0)| = |f(x)| = ( |x| if x is rational or 0 , if x is irrational)<\u03b5.\r\nThis means that limf(x) = f(0) as x tends to 0.\r\n\r\n Babis", "Solution_4": "thanks all! I'm starting to see the light. Let me put pencil to paper to see if I understand it clearly.\r\n\r\nBest wishes.\r\n\r\nBailey", "Solution_5": "okay, let me get this straight...I want to prove: $\\lim\\limits_{x\\rightarrow 0}f(x)=f(0)$, where $f(0)=0$.\r\n\r\nSo, I need to show that:\r\n\r\n$\\forall \\ \\epsilon > 0 \\ \\exists\\ \\delta > 0$ such that\r\n\r\n$0 < |x-0| < \\delta$ implies $\\ |f(x)-f(0)| < \\epsilon$\r\n\r\nProof:\r\n\r\nLetting $\\delta=\\epsilon$ (which I see how to get)\r\n\r\n$0 < |x-0| < \\delta$ $\\Longrightarrow$ $0 < |x-0| < \\epsilon$\r\n\r\n$\\Longrightarrow$ $|x-0| < \\epsilon$\r\n\r\n$\\Longrightarrow$ $|f(x)-f(0)| < \\epsilon$ $\\square$\r\n\r\nIs it that simple or am I still missing something?\r\n\r\n\r\nBailey", "Solution_6": "It is that simple and you are correct. \r\n\r\nNow, why is this function discontinuous everywhere else?", "Solution_7": "Try with the negation of the definition. Let [b]a[/b] different of 0.\r\n << there exist \u03b5>0 , such that for all \u03b4>0 there exists x with /x/<\u03b4 and \r\n\r\n $/f(x) - f(a)/ \\geq \\epsilon$ >>\r\n\r\n For example , if you take [b]a[/b] as rational and \u03b5 = (/f(a)/) /2 , then very - very near to [b]a[/b] you can find irrational x. For this x you will have /f(x) - f(a)/ >= \u03b5.\r\n\r\nThe same argument works also if a is irrational( very near to the irrational a you can find x rational( Q is dense to IR or something like that)).\r\n\r\n [u]Babis[/u]\r\n\r\n I was in Uni before 1980 , so I have forgotten some things(the most !!!) , so tis is hint which probably will help you.", "Solution_8": "Thanks for the help everyone...and the extra stuff to think about.\r\n\r\nBailey" } { "Tag": [], "Problem": "There was a steel beam 50 feet long. The entire construction crew (total weight: 6000 pounds) was standing at the center of the beam. Assume the modulus of elasticity is 29,000,000 pounds per square inch, the moment of inertia is 850 inches^4, the beam has simple pin connections at either end, and all loads other than the weight of the crew are disregarded. :D", "Solution_1": "question: draw a picture of that.", "Solution_2": "There doesn't appear to be a question asked. Either way, I'm sending this off to Other Problem Solving Topics, where it belongs." } { "Tag": [ "calculus", "derivative", "quadratics", "algebra", "polynomial", "Putnam", "quadratic formula" ], "Problem": "For which real numbers c is there a straight line that\r\nintersects the curve x^4 + 9x^3 + cx^2 + 9x + 4\r\nin four distinct points?", "Solution_1": "$ y\\equal{}x^4\\plus{}9 x^3\\plus{}c x^2\\plus{}9 x\\plus{}4$\r\n\r\nMaking $ x\\to x\\minus{}\\frac{9}{4}$\r\n\r\n$ y\\equal{}x^4\\plus{}\\left(c\\minus{}\\frac{243}{8}\\right) x^2\\plus{}\\left(\\frac{801}{8}\\minus{}\\frac{9 c}{2}\\right) x\\plus{}\\frac{81\r\n c}{16}\\minus{}\\frac{23843}{256}$\r\n\r\nIf we could elliminate the $ x^1$ variable we could make $ x^2 \\equal{} y$\r\n\r\n$ \\frac{801}{8}\\minus{}\\frac{9 c}{2}\\equal{}0\\Rightarrow c\\equal{}\\frac{89}{4}$\r\n\r\nand we would have a straigth line. so, with this c, we would have solved the problem.\r\n\r\ncould possible be more values for $ c$, i didnt solved that", "Solution_2": "Claim: such a line is possible if and only if the quartic curve has two distinct inflection points. Since the second derivative is $ 12x^2 \\plus{} 54x \\plus{} 2c,$ by the quadratic formula, this is possible precisely when $ c < \\frac {243}{8}.$\r\n\r\n(Yes, $ \\frac {89}{4} < \\frac {243}{8}.$ But it's not unique. However, if we look at lordykelvin's work, we see that he had the necessary computations for the full answer, had he only realized what he had.)\r\n\r\nAs for proving the claim:\r\n\r\nSuppose there is a line $ L(x)$ such that $ f(x) \\minus{} L(x) \\equal{} 0$ has four distinct solutions. Then, by Rolle's Theorem, $ f'(x) \\minus{} L'(x) \\equal{} 0$ has three distinct solutions and $ f''(x) \\minus{} L''(x) \\equal{} f''(x) \\equal{} 0$ has two distinct solutions. But since $ f''(x)$ is a quadratic polynomial, if it has two distinct roots, then it changes sign at those roots and hence each such root is an inflection point.\r\n\r\nConversely, suppose that the curve does have two distinct inflection points. As a quartic with positive leading coefficient, it must be concave downward between these two points and concave upward to the outside of them. Draw the chord line $ y \\equal{} L(x)$ connecting these two points of inflection. By the concavity just explained, we must have $ f(x) > L(x)$ between the two inflection points and $ f(x) < L(x)$ on some interval just to the outside of those points. But the growth of $ x^4$ must mean that there will be two more crossings further to the outside.", "Solution_3": "One line solution:\r\n\r\nSince $ x \\equal{} 0$ is not a root, we can rewrite the given equation as $ \\minus{} x^2 \\minus{} 9x \\minus{} \\frac {9}{x} \\minus{} \\frac {4}{x^2} \\equal{} c$.\r\n\r\nTherefore we are to find the condition for which $ y \\equal{} \\minus{} x^2 \\minus{} 9x \\minus{} \\frac {9}{x} \\minus{} \\frac {4}{x^2}$ has 4 distinct intersection points with $ y \\equal{} c$.\r\n\r\nLet $ f(x) \\equal{}\\minus{} x^2 \\minus{} 9x \\minus{} \\frac {9}{x} \\minus{} \\frac {4}{x^2}$, we have $ f'(x) \\equal{} \\minus{} \\frac {1}{x^3}(2x^4 \\plus{} 9x^3 \\minus{} 9x \\minus{} 8)$, but I can't find the roots of $ f'(x) \\equal{} 0$ by factorization. :(", "Solution_4": "kunny, it looks like you are trying to answer a different question than the one asked. You seem to be trying to find conditions for $ f(x)\\equal{}0$ to have four roots, which is very different than conditions for which there exist $ a$ and $ b$ such that $ f(x)\\minus{}ax\\minus{}b\\equal{}0$ has four roots.", "Solution_5": "Oh, I see. Thank you, Kent Merryfield.\r\n\r\nkunny", "Solution_6": "The problem is Putnam 1994 B2. As such you can assume that arguments at least as good as (probably better than) what I provided in the last two paragraphs of my long post above would be required." } { "Tag": [ "geometry", "circumcircle" ], "Problem": "[b]Problem 4.[/b] Let $k$ be the circumcircle of $\\triangle ABC$, and $D$ the point on the arc $\\overarc{AB},$ which do not pass through $C$. $I_A$ and $I_B$ are the centers of incircles of $\\triangle ADC$ and $\\triangle BDC$, respectively. Proove that the circumcircle of $\\triangle I_AI_BC$ touches $k$ iff \\[ \\frac{AD}{BD}=\\frac{AC+CD}{BC+CD}. \\]\r\n\r\n[i] Stoyan Atanasov[/i]", "Solution_1": "Let $CI_1$ and $CI_2$ meet the circumcircle of $\\triangle{ABC}$ at $X$, respectively $Y$.\r\nThen $\\frac{AC+CD}{AD}=\\frac{CX}{XD}$ and $\\frac{BC+CD}{BD}=\\frac{CY}{YD}$ (1).\r\nThe tangency relation between the two circles (from the statement) is equivalent to the fact that quadrilateral $CXDY$ is harmonical (or $I_1I_2||XY$).\r\nHowever, that's the case when the fractions in (1) are equal and we are done.", "Solution_2": "[quote=Sailor]\nquadrilateral $CXDY$ is harmonical (or $I_1I_2||XY$).[/quote]\n \nSorry for gravedigging this post, but why is this equivalent?\n", "Solution_3": "[quote=Duarti]but why is this equivalent?[/quote]\n\n$XI_1=DX$" } { "Tag": [], "Problem": "This seems to be pretty easy (I guess that's why it's problem 1 :)).\r\n\r\nWe can definitely find 3 of our 8 numbers which are in the same progression, and we consider the cases:\r\n\r\n[b]case 1: the 3 numbers form a progression with common difference 1[/b]\r\n\r\nIt's clear that we're done in this case.\r\n\r\n[b]Case 2: the 3 numbers form a progression with common difference 2[/b]\r\n\r\nIf the progression is formed with even numbers, we're through. If it's formed with odd numbers, we may assume it's 1,3,5 (it's the same if it's 3,5,7). The rest of the numbers are 2,4,6,7,8. The numbers 2,4,6,8 can't belong to the first progression, because that only has odd numbers, so they must belong to the other 2 progressions. If 4,6 or 6,8 belong to the same progression we're done, otherwise it means that 4,8 belong to the same progression, which thus contains all multiples of 4, including 1980, and that's it.\r\n\r\n[b] Case 3: the 3 numbers form a progression with common difference 3[/b]\r\n\r\nThe 3 numbers can either be 2,5,8 or 1,4,7. Let's assume they're 1,4,7 (the other case is the same). The numbers 2,3,5,6,8 can't belong to the first progression, so they must be distributed in the other 2 progressions, so there exists one progression which contains 3 of these 5 numbers. These 3 numbers can only be 2,5,8, and the other progression must contain 3 and 6, so it contains all multiples of 3, including 1980.", "Solution_1": "1. warm-up\r\nThe warden rewards a prisoner a chance to escape.\r\nThere are two boxes, 10 white balls, and 10 black balls.\r\nLater, he will randomly pick a box, and randomly pick a ball in the box. How should he arrange the balls into the boxes to maximize his chance of getting out?\r\nYou need a white ball to get out.\r\n2. problem (If you heard, tell me and don't write anything)\r\nThe warden rewards 100 prisoners. Every day, the warden picks a random prisoner. He is led to a room only with a lightbulb and switch. He will do one of four things:\r\n1. Turn the light on if it was off\r\n2. Turn the light off if it was on\r\n3. Do nothing\r\n4. Declare \"All 100 prisoners have been taken to this room\".\r\n\r\nThe lightbulb starts off.\r\n\r\nWhen the prisoner leaves, the lightbulb is not touched. If he declares, and is right, everybody is freed. If he is wrong, all would be killed.\r\n\r\nThe prisoners only get to talk with each other once during the first day before the choosing. After, no communication is allowed.\r\n\r\nHow should the prisoners avoid being killed? This doesn't mean thet they do nothing. THEY WANT OUT!!!!!!", "Solution_2": "What do you mean by \"if he declares\"? Declare what?", "Solution_3": "Answers in spoiler\r\nNot anymore since no one is replying; the answer will be posted later.", "Solution_4": "[quote=\"RC-7th\"]What do you mean by \"if he declares\"? Declare what?[/quote]\r\n\r\nDeclare choice #4", "Solution_5": "In the first problem, does he get out by picking a black or a white ball?\r\n\r\nIn the second, to avoid being killed, they should simply turn the light on and off, since it is effectively doing nothing.", "Solution_6": "[quote=\"towbomb\"]In the second, to avoid being killed, they should simply turn the light on and off, since it is effectively doing nothing.[/quote]\r\nSomebody has to declare or they'd do it every day until they rot in prison, and what they really want is to go free, even at risk of execution...at least that's what I assume pakagawa meant, since the way he/she worded it the problem's completely pointless...", "Solution_7": "towbomb wrote:In the first problem, does he get out by picking a black or a white ball?\n\n\n\nWhite, but if it were black, it would be [hide]one black in a box and everything else in the other box.[/hide]", "Solution_8": "It's a white ball to get out.\r\n\r\nNO!!!!!!\r\n\r\nThe prisoners want OUT!\r\nNot to play with the light bulb!!?!?!!!1", "Solution_9": "I think I got something for the second one:\n\n[hide]If the prizoners are picked randomly, some of them can be picked twice or more. Thus, those that are picked for their first time should change the state of the lightbulb. Those that get picked again should do nothing. \n\nAlso, since there are 100 prizoners they can only declare after 100 days because that is the least amount of days it can take. Then it depends whether the light was on or off to begin with? [/hide]", "Solution_10": "Oh yeah, and here's what I think for the first one:\n\n[hide]Put a white ball in one box and the nineteen others in the second box. That gives you 50% for picking the first box times 100% of picking the white one = 50% For the second box it's 9 white and 10 blacks = 9/19 chance times 50% of picking the second box = 23.7% 50% + 23.7% = 73.7% or about 3/4 chance to get out.[/hide]\n\nMost other ways it's around 50% and I'm pretty sure this is right.", "Solution_11": "wereallgonnadie wrote:I think I got something for the second one:\n[hide]If the prizoners are picked randomly, some of them can be picked twice or more. Thus, those that are picked for their first time should change the state of the lightbulb. Those that get picked again should do nothing. \nAlso, since there are 100 prizoners they can only declare after 100 days because that is the least amount of days it can take. Then it depends whether the light was on or off to begin with? [/hide]\n\n\n\na guy who does not get picked does not know who is picked or how many times the lightbulb was changed. But, you said an important thing: The 100 days is a limit. Think about this.", "Solution_12": "Dear administrator:\r\n\r\nCan you move this to the Intermediate level? I'm not getting good answers... I was going to post it there anyway.", "Solution_13": "hmmmm, I'm thinking, give me a chance, I'll come up with something later, I have to go.", "Solution_14": "Mabye, if the prisoners switched the bulb if it is their first time and start doing so only after 100 days?", "Solution_15": "I'm getting pissed.\r\n\r\nSay theres prisoner A. He goes in on the 100th day, turns the light on. he goes in again, and again. Then, on the 103th day, prisoner B goes in. He does not know if it's AAAB or ABCD. (it's on so it can't be ABAC, etc...)", "Solution_16": "So does prisoner A switch the light each time he goes in?\r\n\r\nAnd why start on the 100th day? Why not the 1st?", "Solution_17": "I think i have an idea for the second one... not the best way... but:\n\n\n\n[hide]\n\nAssign one prisoner the job of turning off the lightbulb, and all 99 others, the job of turning on the lightbulb. Each time (since they are chosen at random, they are bound to be chosen more than once) any of the 99 goes in and finds the lightbulb off AND if he never turned on the lightbulb before, he turns on the lightbulb. Hence, whenever that one prisoner assigned the job to turn off the lightbulb goes into the room and sees the lightbulb turned on, he then turns it off, and counts how many times he had turned off the lightbulb, thus, when he counts up to 99 (eventually...), it is safe for him to declare \"all 100 prisoners have been taken to this room\". \n\n[/hide]", "Solution_18": "That was my first idea too... \r\n\r\nBut then I'm haunted by the counterargument that this game can easily stretch for an infinite time...", "Solution_19": "The actual \"answer\" takes only half the time. Yours will take about 40 years, and mine only 15. (it might be inaccurate)...", "Solution_20": "[hide]Hint: What could the prisoners to to save time before the 100th day?[/hide]", "Solution_21": "I'm bored, so here are the answers in spoiler.\n\n\n\n[hide]put all the black balls and 9 white balls in one box and 1 white ball only in the other box.\n\n\n\n2. have the light off. After the same guy goes in the second time, he turns the light on. He is the counter. He counts the numbers of days other people came. After the 100th day, if a guy goes in and the light is on, he keeps it off. Then, if it's his first time in, he turns it back on. If it's still off, he wins by declaring.\n\nThen, after the control goes again, he turns the light off. When a new guy goes in, he turns the light on. if a new guy goes in and the light is on, he is still a new guy. Then, the control turns it off and counts another person. Continue this until all 100 people are counted.[/hide]" } { "Tag": [ "Princeton", "college" ], "Problem": "Math: 780\r\nCR: 560\r\nWS: 690 (10 Essay)\r\n\r\nCould it get any worse :( \r\n\r\nHas anyone taken the Princeton Review or Kaplan classroom SAT course? and if so, does it really help, I know it's a 200 point improvement, but I really don't think i can boost my CR score :( I made a 140 point improvement (all on WS) from the November SAT by just doing Practice tests my self so is the Course worth it?", "Solution_1": "Well, I got a 2160 in March 2005 (650/800/710 reading/math/writing). I took Princeton Review during the summer/fall and got 2290 on the January test (800/800/690). The writing pisses me off though, lol.\r\n\r\nIt's hard to say if the improvement is solely the course. Personally, I don't think the course did anything from me; most of my improvement on critical reading just came from doing practice tests on my own. Hope that helps.", "Solution_2": "[quote=\"dakyru\"]Well, I got a 2160 in March 2005 (650/800/710 reading/math/writing). I took Princeton Review during the summer/fall and got 2290 on the January test (800/800/690). The writing pisses me off though, lol.[/quote]\r\n\r\nlol, so you must've gotten your money back for the course right since it \"guarantees\" a 200 point improvement.", "Solution_3": "Haha, really? I'll go ask them about it.", "Solution_4": "I'm sure there's a disclaimer for initial scores of 2000+ :)\r\n\r\nNice perfect scores though, dakryu", "Solution_5": "Did anybody get a message that their scores are not available today and check again on Feb.18th. Here is the quote:\r\n\r\n\r\nPlease check back after Saturday, February 18 for January SAT scores\r\n\r\nAlthough most scores for the January SAT are online now, a small number are not yet available. Learn why your scores may not be available and check back after Saturday, February 18.\r\n\r\nTheir standard explanation does not really explain anything.\r\nIs this something to worry about?", "Solution_6": "I don't think the score delay announcement (which my son got) is anything to worry about. Most likely it means that there was a split in scoring the essay, so that the essay is being referred to a third scorer to break the split.", "Solution_7": "math level 2 800 :D", "Solution_8": "i took the test in october...800 math 800 cr 780 writing.\r\n\r\ni think that the most important thing in CR is to build up a strong vocabulary, - thats what i did. and for writing, most of it isnt too hard so long as you defend what you write with examples, logic, etc. bc all they ask for is an opinion essay, so you really cant be wrong...and dont be afraid to make up stuff lol, they dont take off for that. unless its like. blatantly obvious and illogical - then they might.\r\n\r\n...and math. i guess most of us here are already 700+ scorers in SAT math. :lol:\r\n\r\n[quote=\"juicybooty911\"]math level 2 800 :D[/quote]\r\noh yeah, i heard that for math level 2 that its around 88th percentile to get an 800...is that true?", "Solution_9": "most people who bother to take level 2 are good math students and with the curve, 800 is common", "Solution_10": "2390\r\n\r\nwriting :mad:", "Solution_11": "[quote=\"MysticTerminator\"]2390\n\nwriting :mad:[/quote]\r\n\r\nHow did you bother to miss one on the writing section?\r\n\r\nImpressive scores though.", "Solution_12": "800 math level 2, as well", "Solution_13": "800 here too on the SAT II MATH", "Solution_14": "My son (7th grade) got 700 in Math (before 13th Birthday). Anybody in here is in Johns Hopkins SET program?", "Solution_15": "[quote=\"dkushnir\"]My son (7th grade) got 700 in Math (before 13th Birthday). Anybody in here is in Johns Hopkins SET program?[/quote]\r\n\r\nYes, several AoPSers are in [url=http://www.jhu.edu/gifted/set/index.html]SET[/url]. There is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6882]a little discussion of SET[/url] on the AoPS Other U.S. Contests & Programs Forum.", "Solution_16": "i'm in SET and I've never seen the AOPS thread on it. Thanks tokenadult for pointing that out", "Solution_17": "i never got my jan SAT scores back. :mad:", "Solution_18": "[quote=\"kheg_k\"]i never got my jan SAT scores back. :mad:[/quote]\r\n\r\ncheck online, the mailed form should be back in 2 or 3 more days", "Solution_19": "[quote=\"dkushnir\"]My son (7th grade) got 700 in Math (before 13th Birthday). Anybody in here is in Johns Hopkins SET program?[/quote]\r\n\r\n\r\ni got 730. and i m pretty young myself", "Solution_20": "[quote=\"kheg_k\"][quote=\"dkushnir\"]My son (7th grade) got 700 in Math (before 13th Birthday). Anybody in here is in Johns Hopkins SET program?[/quote]\n\ni got 730. and i m pretty young myself[/quote]\r\n\r\nYeah well guess what... I got 840 before my -6th birthday... so obviously I'm way smarter than all of you. OK let's not brag, especially about the SAT..", "Solution_21": "I agree. Bragging about standardized tests is pretty useless. It is a great accomplishment to get into SET (I'm a member too :D) but it's really not... that big. Get onto USA IMO team and come back :) (not that I have, of course!)", "Solution_22": "[quote=\"aznphatso\"]I agree. Bragging about standardized tests is pretty useless.[/quote]\r\n\r\nI agree with this principle too. \r\n\r\nNonetheless, I think the statement appearing earlier in the thread was by way of giving background for asking a question about [url=http://www.jhu.edu/gifted/set/index.html]SET[/url], and that's a good thing, in my opinion, because not all AoPSers know about SET.", "Solution_23": "Thank you talkenadult. Exactly.\r\nAnd if this program is good for something, it is worth to know.\r\nSo far (when it came to something serious) it looks like they run some kind of counceling services for a fee." } { "Tag": [], "Problem": "Find out x_i (x_i natural \\{0}) i=1,n so that:\r\n[x1^2/(x2^2+x3^2)]+[x2^2/(x3+x4)]+..+[xn^2/(x1+x2)]=[(n-1)/(x1+x2+..+xn)];\r\n[x] denotes the integer part.", "Solution_1": "Are you sure that's correct? There are squares in the denominator of the first term but not in the later ones.\r\n\r\nAlso, do you know how to use html? It makes reading things much easier. For example, if you wanted to write exponents, you can do\r\n\r\n< sup >n< /sup >\r\n\r\nexcept without the spaces inside the <>. The you can get things like xn. Also, if you write \"sub\" instead of sup, you get subscripts. x37. Very nice, no? Easy to read.\r\n\r\nOh, and putting spaces on either side of + signs is helpful, too.\r\n\r\nOn an aside, wouldn't it make more sense if \"sup\" was \"sub,\" and vice-versa? Since that's the direction the letters point in, after all. Or maybe not. I can see it argued both ways.\r\n\r\nI do want to know if there was an error in the question, though.", "Solution_2": "sorry the correct statement:\r\n\r\nFind out xi (xi natural \\{0}) i=1,n so that: \r\n[x12/(x2 + x3)] + [x22/(x3 + x4)] + .. + [xn2/(x1 + x2)] = [(n-1)/(x1 + x2 + .. + xn)]; \r\n[x] denotes the integer part.\r\n\r\n[edited by mod to make it look nice :)]", "Solution_3": "Thanks for the formatting, mod.\n\n[hide]\n\nJust a question on the way to a solution: the right side is 0 necessarily, right? I mean, we have n natural numbers, whose sum is definitely bigger than (n - 1). So then every term on the left should be zero, right? And from there I think I can show that (1, 1, ..., 1) will be the only solution without a whole lot of trouble. Am I missing something?\n\n[/hide]\n\nOh, we also have to have n :ge: 3, yes?" } { "Tag": [], "Problem": "If $\\frac{1}{x} + \\frac{1}{y} = 3$ and $\\frac{1}{x} - \\frac{1}{y} = -7$ what is the value of $x + y$. Express your answer as a common fraction.", "Solution_1": "[hide]$a + b = 3$\n$a - b = -7$\n$a = -2$ $b = 5$\n$\\frac 15 - \\frac 12 = -\\frac 3{10}$[/hide]", "Solution_2": "This was in a target round 1993. I know this:\r\n\r\n[hide=\"solution\"]\nlet's denote $a= \\frac{1}{x}$, $b=\\frac{1}{y}$\nthen\n$a+b =3$\n$a-b = -7$\n\nadding two together, we have $2a = -4, a = -2$\n$b = 5$.\n\nSo $x= \\frac{1}{a} = -\\frac{1}{2}$, $y=\\frac{1}{b} = \\frac{1}{5}$\n\nx+y = $-\\frac{3}{10}$\n[/hide]", "Solution_3": "[hide=\"click here if you wish to do so...\"]\n-3/10[/hide]", "Solution_4": "[hide]I did it a strange way...\ny+x=3xy\ny-x=-7xy\n2y=-4xy\n2=-4x\n-1/2=x\nthen I solved for y to get -3/10 as the final answer.[/hide]" } { "Tag": [ "function", "number theory unsolved", "number theory" ], "Problem": "Let $n$ be a positive integer, $n\\geq 2$. Let $M=\\{0,1,2,\\ldots n-1\\}$. For an integer nonzero number $a$ we define the function $f_{a}: M\\longrightarrow M$, such that $f_{a}(x)$ is the remainder when dividing $ax$ at $n$. Find a necessary and sufficient condition such that $f_{a}$ is bijective. And if $f_{a}$ is bijective and $n$ is a prime number, prove that $a^{n(n-1)}-1$ is divisible by $n^{2}$.", "Solution_1": "Strange problem...\r\nEvident, that $f_a$ bijective iff $gcd(a,n)=1$(if $gcd(a,n)=1$, then $ai\\not=aj$ for all $0\\leq i1$, then $a*\\frac{n}{d}\\rightarrow0$ and $a*0\\rightarrow0$)\r\nIf $n$ is prime and $f_a$ bijective, then $a$ isn't divisible by $n$ :arrow: by Little Fermat's theorem $a^{n-1}\\equiv1(mod n)$ and $a^{n(n-1)}-1=(a^{n-1}-1)(a^{(n-1)^2}+a^{(n-2)(n-1)}+...+a^{n-1}+1)$ is divisible by $n^2$." } { "Tag": [ "ARML" ], "Problem": "Compute the smallest number d such that fewer than half of the positive integers with d-digits have all distinct digits.\r\n\r\nDeleted. - Overlaps...", "Solution_1": "[hide]Wouldn't that be $1$...[/hide]\r\n\r\nThe second one I believe was posted in the practice series.", "Solution_2": "[quote=\"tarquin\"][hide]Wouldn't that be $1$...[/hide][/quote]\n\nUmm... from what I know, all 1-digit numbers have all distinct digits.\n\n[hide]\nd-digit numbers with distinct digits: $(9)(9)(8)...(11-d)$\ntotal d-digit numbers: $9(10)^{d-1}$\n\n$\\frac{(9)(9)(8)...(10-d)}{9(10^{d-1})} < \\frac{1}{2}$\n$2(9)(8)...(10-d) < 10^{d-1}$\n\nJust trying numbers...\n\n$2(9) \\not < 10$\n$2(9)(8) \\not < 100$\n$2(9)(8)(7) \\not < 1000$.\n$2(9)(8)(7)(6) < 10000$.\n\nSo $d=5$.[/hide]", "Solution_3": "For the second one can't you just examine the equation wihtout sub scripts mod 5", "Solution_4": "The second one is part of a set of problems that I posted that will last 2 weeks on my contest...sry :blush:", "Solution_5": "Yeah, I think it's 1", "Solution_6": "The question is kind of hard to understand.\r\n\r\nBy common sense, though, it has to be 5. I'm not exactly sure how to explain it, but if you think about it, that has to be right.", "Solution_7": "It is 4\r\nThe case should be\r\n5*10^d>9Pd\r\nSo d=4" } { "Tag": [ "geometry", "3D geometry", "geometry solved" ], "Problem": "Is it possible to draw a diagonal on each small square on the surface of a Rubik's cube so that the lines form a continuous non-intersecting path through all the squares?\r\n\r\nI think that it's not possible, but why?", "Solution_1": "Do they need to join back together at the end?", "Solution_2": "No they don't.", "Solution_3": "OK, never mind. I've figured it out myself.\r\nOne can only ever get to half the corners of each square on the cube. Let these corners be vertices, and the lines that connect them, be the edges in a graph.\r\nNow each vertex has 4 edges, and is of even degree, except the ones at the corners of the cube, which each have 3 edges.\r\nFor an Eularian path to exist, all vertices need to be of even degree, with the exception that the starting and ending vertices may be of odd degree. However, there are 4 vertices of odd degree (the ones at the corners of the cube; remember we can only over get to half the corners of each square), so an Eularian path cannot exist." } { "Tag": [], "Problem": "A Fat Man's Prayer:\r\n\r\nLord, My soul is ripped with riot \r\nincited by my wicked diet.\r\n\"We Are What We Eat,\" said a wise old man! \r\nand, Lord, if that's true, I'm a garbage can.\r\nI want to rise on Judgment Day, that's plain! \r\nBut at my present weight, I'll need a crane. \r\nSo grant me strength, that I may not fall \r\ninto the clutches of cholesterol.\r\n\r\nMay my flesh with carrot-curls be sated, \r\nthat my soul may be polyunsaturated \r\nAnd show me the light, that I may bear witness \r\nto the President's Council on Physical Fitness.\r\n\r\nAnd at oleomargarine I'll never mutter, \r\nfor the road to Hell is spread with butter. \r\nAnd cream is cursed; and cake is awful; \r\nand Satan is hiding in every waffle.\r\n\r\nMephistopheles lurks in provolone; \r\nthe Devil is in each slice of baloney, \r\nBeelzebub is a chocolate drop, \r\nand Lucifer is a lollipop.\r\n\r\nGive me this day my daily slice \r\nbut, cut it thin and toast it twice. \r\nI beg upon my dimpled knees, \r\ndeliver me from jujubees.\r\n\r\nAnd when my days of trial are done, \r\nand my war with malted milk is won, \r\nLet me stand with the Saints in Heaven \r\nIn a shining robe--size 37.\r\n\r\nI can do it Lord, If You'll show to me, \r\nthe virtues of lettuce and celery. \r\nIf You'll teach me the evil of mayonnaise, \r\nof pasta a la Milannaise \r\npotatoes a la Lyonnaise \r\nand crisp-fried chicken from the South.\r\n\r\nLord, if you love me, shut my mouth.", "Solution_1": "Guys i'm not going to post jokes if you aren't. We need laughter.", "Solution_2": "I DON'T CARE IF YOU DON'T KNOW ANY JOKES JUST GOOGLE THEM AND POST!\r\n\r\n[youtube]MIaORknS1Dk[/youtube]" } { "Tag": [ "ARML", "AMC", "AIME", "USA(J)MO", "USAMO", "geometry", "analytic geometry" ], "Problem": "[quote=\"not_trig\"]\nARML is not that hard, but it's hard to make it, especially in NC. \n:wink:[/quote]\r\n\r\nAre there enough interested and capable math students in NC to justify additional ARML teams? \r\n\r\nSome other states/schools send more teams - but most of these places are a bit closer to competition sites and probably spend less in travel expenses. Still, funding an additional ARML team or two might be possible.\r\n\r\nIf we did have additional ARML team(s), should the members come from one school, perhaps NC School of Science and Math, or should we create a NC C and D Teams in much the same way the B team has been created in the past. \r\n\r\nAnother alternative would be to regionalize B, C, and D teams within the state. This might permit additional practice sessions and encourage the formation and use of math circles in communities where such circles are weak or non-existent. \r\n\r\nAdditional thoughts?\r\n\r\n-Greg", "Solution_1": "additional teams should be created the same way\r\n\r\nNCSSM already gets a huge advantage with its walk-on policy in the state math contest\r\n\r\nalthough an East ARML team would be pretty awesome since we would still be pretty owningish", "Solution_2": "We're good with two. Arnav: if you make an East ARML team and split NC into ECH and everything else, we would completely forfeit any advantages. That's basicly dividing the team in half.\r\n\r\nIt really seems kind of excessive to have 45 people going from one state. First of all, we would probably need two buses. But also, you would have to assume that everybody on the B team is at least as good as anybody on the C team. I really don't think that works out.\r\n\r\nBesides, are those extra 31-45 really interested in ARML (it seems like some of the B team isn't)? I would probably be more open if a student who would have made such a team suggested this.\r\n\r\nI'll just briefly say that if we had extra teams, I like the idea of a state A team, but regional B/C/D teams like Greg said. That would make these teams more likely to do well.", "Solution_3": "[quote=\"jb05\"]We're good with two. Arnav: if you make an East ARML team and split NC into ECH and everything else, we would completely forfeit any advantages. That's basicly dividing the team in half.[/quote]\r\n\r\nyeah I know but it would be funny", "Solution_4": "NCSSM students get to go to the state math contest without qualifying through regional contests?", "Solution_5": "eh I don't think it's that hard to make ARML like we do take 2 teams after all", "Solution_6": "The question might be analyzed using prior NC contest scores and some knowlege of the difficulty of the ARML related to the other contests.\r\n\r\nApproximately 60 NC people scored 5 or above on the AIME last year while just 30 made an NC ARML team. A few in the top 20 at the State Contest did not take the AIME or did not score 5 or above. There are a significant number of highly capable math students, even USAMO qualifiers, who don't qualify for the State Math Contest at regional competitions. \r\n\r\nI'm not sure if this is a deep enough pool to justify another ARML bus or not. There may be many qualified folks that can't go. \r\n\r\nWhile getting a 5 on the AIME may not seem like much of a feat to some in talent-rich communities, among the ~80,000 high school students in the Winston-Salem, Greensboro, and High Point area (Triad) there were just a couple of people who managed a 5 on the AIME (I don't think either one made one of the ARML teams.) There are wide swaths of the state that have no realistic chance of sending a student to ARML in a typical year.\r\n\r\n- Greg", "Solution_7": "[quote=\"gt59\"]The question might be analyzed using prior NC contest scores and some knowlege of the difficulty of the ARML related to the other contests.\n\nApproximately 60 NC people scored 5 or above on the AIME last year while just 30 made an NC ARML team. A few in the top 20 at the State Contest did not take the AIME or did not score 5 or above. There are a significant number of highly capable math students, even USAMO qualifiers, who don't qualify for the State Math Contest at regional competitions. \n\nI'm not sure if this is a deep enough pool to justify another ARML bus or not. There may be many qualified folks that can't go. \n\nWhile getting a 5 on the AIME may not seem like much of a feat to some in talent-rich communities, among the ~80,000 high school students in the Winston-Salem, Greensboro, and High Point area (Triad) there were just a couple of people who managed a 5 on the AIME (I don't think either one made one of the ARML teams.) There are wide swaths of the state that have no realistic chance of sending a student to ARML in a typical year.\n\n- Greg[/quote]\r\n\r\nyea we need to open up the regionals a bit and send more people to state...it's sort of stupid when you have MOPpers not making it past regionals", "Solution_8": "who decides whether or not we will have more ARML teams?", "Solution_9": "[quote=\"NeverOddOrEven\"]who decides whether or not we will have more ARML teams?[/quote]\r\n\r\nI think just about anyone can sponsor a team as long as they are willing to pay all expenses for the students (not a trivial matter with the travel from NC.) As far as I know, this is the only significant limitation on the number of teams. If the Florida site becomes a reality, it should provide additional room at the Penn State site - but I don't think that site facilities have limited the number of teams in the past.\r\n\r\nSome elite schools and larger cities send 3 or more teams. On the other hand, there are probably several states unrepresented.\r\n\r\nIf anyone wanted to create another team, it would be a good idea to coordinate it with the sponsors and coaches of the existing ARML teams. The NC Teachers of Mathematics organization may provide the bulk of the present funding for the NC ARML teams.\r\n\r\n- Greg", "Solution_10": "[quote=\"gt59\"]While getting a 5 on the AIME may not seem like much of a feat to some in talent-rich communities, among the ~80,000 high school students in the Winston-Salem, Greensboro, and High Point area (Triad) there were just a couple of people who managed a 5 on the AIME (I don't think either one made one of the ARML teams.)[/quote]\r\n\r\nAdmittedly, a 5 is good percentage-wise, but ARML is a national contest. There were 2363 people who made a 5+ on the AIME last year, and those are not necessarily the top 2363 in the country. If all of these people attended ARML in addition to the ones who already do, we would need a lot more sites and a lot more room.\r\n\r\nI think we do a good job of making our ARML teams. If we do make more teams, I don't think that this is the best way to do it.", "Solution_11": "I don't think we need more teams.\r\n\r\nMaking ARML should be hard; the harder, the better for the team. Rigorous selection yields better team members.\r\n\r\nIf you really want to have a C team, form it after the pool for the A and B teams has been chosen. Choose the best people left after that.\r\n\r\nDo people think that just because the A team won last year, we need to have more teams?", "Solution_12": "dude having more teams is obviously not going to affect the teams already in place", "Solution_13": "yea they just might vary in skill...", "Solution_14": "[quote=\"jb05\"]\nAdmittedly, a 5 is good percentage-wise, but ARML is a national contest. There were 2363 people who made a 5+ on the AIME last year, and those are not necessarily the top 2363 in the country. If all of these people attended ARML in addition to the ones who already do, we would need a lot more sites and a lot more room.\n\nI think we do a good job of making our ARML teams. If we do make more teams, I don't think that this is the best way to do it.[/quote]\r\n\r\nLast year there were close to 1700 participants at ARML's three sites - with room to spare. If ARML expands to four sites next year, then there would be room for at least 2400 students - roughly one seat for everyone who made 5 or better on the 2006 AIME.\r\n\r\nI didn't intend to suggest that the AIME score determine team membership or that everyone with a 5 on AIME should be invited to ARML. Sorry about the confusion.\r\n\r\nIf the rest of the country sends an [u]average[/u] of one person to ARML for every score of \"5\" or better on the AIME, then we might consider doing the same. Sixty NC students scored 5's or better on AIME last year. Talent-wise, I think we could support 4 teams and that the worst of the four would still perform respectably at ARML.\r\n\r\nFunding might be another matter. :wink: \r\n\r\n-- Greg", "Solution_15": "I think that right now, we have a good balance. Remember that everyone who is going to ARML from NC is better at math than the average high-schooler. As we approach that norm, though, by adding more teams, it will get harder and harder to distinguish between who should go and who shouldn't. Also, do we want to send people to ARML who might not do as well and thus would not want to participate again." } { "Tag": [ "LaTeX" ], "Problem": "Hi,\r\n\r\nI'm using makeindex in my documents but the index do not appear in my table of content even when I tried the following code :\r\n\r\n\\printindex\r\n\\addcontentsline{toc}{chapter}{Index}\r\n\r\ndoes anybody knows solution to my problem ? thanks\r\n\r\nLilian", "Solution_1": "The best way is to use the [url=http://www.tex.ac.uk/tex-archive/macros/latex/contrib/tocbibind/]tocbibind package[/url], see the [url=http://www.tex.ac.uk/tex-archive/macros/latex/contrib/tocbibind/tocbibind.pdf]documentation[/url]. There's a full explanation at the [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=tocbibind]TeX FAQ[/url].", "Solution_2": "THANKS !!!!!!!" } { "Tag": [ "IMO", "IMO 2004" ], "Problem": "Can anyone tell me the results of chinese team?", "Solution_1": "I only know that some chinese contestant has got 41 points (he got a 6 in problem 1 !!). The Chinese team is the first with 220 points.", "Solution_2": "Do you have the detail score of each contestant?", "Solution_3": "The format of the name is : \"Family name, Given name\"\r\n\r\nHuang, Zhiyi (\u9ec4\u5fd7\u6bc5) 6 7 7 7 7 7 \r\nZhu, Qingsan (\u6731\u5e86\u4e09) 6 7 4 7 7 7 \r\nLi, Xianying (\u674e\u5148\u9896) 7 7 2 7 7 7 \r\nLin, Yuncheng (\u6797\u8fd0\u57ce) 6 7 1 7 7 7 \r\nPeng, Minyu (\u5f6d\u95fd\u6631) 6 6 4 7 5 7 \r\nYang, Shiwu (\u6768\u8bd7\u6b66) 6 7 3 7 4 7", "Solution_4": "[quote=\"liyi\"]The format of the name is : \"Family name, Given name\"\r\n\r\nHuang, Zhiyi (\u9ec4\u5fd7\u6bc5) 6 7 7 7 7 7 very funny how it is possible to solve all and have a mistake at easiest question :maybe:", "Solution_5": "That's not a real mistake, that's a bagatelle. The problem was simple, hence the jury decided to bitch around on this one. I'd got 6 points for it, too. See http://www.mathlinks.ro/Forum/viewtopic.php?p=115015 .\r\n\r\n Darij", "Solution_6": "try to guess what am I going to write....\r\nYes U guessed....ME TOO....\r\nUnfortunately I had a small mistake as many of participicants", "Solution_7": "Oh sorry what are your scores mine are not so impressive. I had to have more but I failed this IMO unfortunately and that was my last in my life. Now I am at Kazakh-British Technical University studying Petrolleum Engineering.\r\nMy scores at IMO 6 5 0 0 1 0 heh very bad isn't that?", "Solution_8": "I'm amused by the results achieved by the Chinese team each year , how do\r\nthey manage to score such high scores ?! Is this a matter of excellent training methods :) or a matter of genius ... or maybe both ... However I noticed that\r\nif a country's population is enourmous - the chances for a super contestant are\r\nstatistically much bigger :) . However Romania, Bulgaria and Vietnam are a good exception :) .", "Solution_9": "[quote=\"Rafal\"]I'm amused by the results achieved by the Chinese team each year , how do\nthey manage to score such high scores ?! Is this a matter of excellent training methods :) or a matter of genius ... or maybe both ... However I noticed that\nif a country's population is enourmous - the chances for a super contestant are\nstatistically much bigger :) . However Romania, Bulgaria and Vietnam are a good exception :) .[/quote]\r\n\r\nI don't think Vietnam is an exception. Do you know what's the population of Vietnam?" } { "Tag": [ "ratio" ], "Problem": "Profits of 26,000 dollars are to be divided among three partners in the ratio of 3: 4: 6. Compare with the smallest share, the largest on is\r\n \r\n1. 1 1/6 times as great\r\n2. 1 1/3 times as great\r\n3. 1 1/2 times as great\r\n4. 2 times as great\r\n\r\nNOTE: I want to know the steps needed to solve this question.\r\n\r\nIs it this way:\r\n\r\n3x + 4x + 6x = 26,000 and solve for x?\r\n\r\nThen I compare 3x with 6x after finding the value of x, right?", "Solution_1": "I don't quite understand your problem.. could you try rewording it ?", "Solution_2": "Well, well Sharkman, correct at last. The assumption is that in a ratio of $ 3: 4: 6$, just as any other ratio, they are all proportional to one another about a constant. In this case, the constant you chose is $ x$.\r\n\r\nBut that's not the [i]fastest[/i] way to solve. assume you find $ x$, then make the rational fraction, but guess what, $ x$ cancels, as it is on the bottom and top.\r\n[hide=\"leaving behind\"]$ \\frac{6}{3}= 2$ times as large.[/hide]", "Solution_3": "igiul is right, there's absolutely no need to solve for $ x$. Just notice that the ratio of smallest to largest is 3:6 = 1:2, therefore the largest share is twice the smallest." } { "Tag": [ "geometry", "AMC", "AIME" ], "Problem": "Thought this was a cool geometry problem.\r\n\r\n387. In the diagram, the semicircles centered at P and Q are tangent to each other and to the large semicricle, and their radii are 6 and 4 respectively. Line LM is tangent to semicircles P and Q. Find LM. (Mandelbrot #3)", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?p=721578#p721578\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=464943#p464943\r\n\r\n\r\nlol this question is uses the same idea as two different AIME problems (4 on 1995 AIME, and 9 on 2006 AIME II)..\r\n\r\nI wonder if this question or the 1995 one came first....\r\n\r\noh well, I would just extend LM until its on the same line as PQ, then three similar triangles...(after drawing in the center of the biggest circle, call it O and connecting that to LM)...", "Solution_2": "Does anyone know the actual answer to this question?", "Solution_3": "https://artofproblemsolving.com/community/q1h1434788p8120266", "Solution_4": "Wait is this this years Mandelbrot round 3", "Solution_5": "wait this post was created in 2007 wowowowow :| ", "Solution_6": "It should be mentioned that several people have not yet taken this year's Mandelbrot...\n\nAlso, it's probably from 2007." } { "Tag": [ "LaTeX" ], "Problem": "ok so i just installed miktex and whatnot, and finished installing texnic center... i've started the configuration wizard for texnic center, and its asking me to type in the complete path of the directory where the 'executables' are located... now i might be a math guy, but i'm not really too much of a computer guy. i'm a bit confused... sum1 help please!", "Solution_1": "It's the directory that latex.exe (and a zillion other exe files) is located in - probably C:\\something_or_other\\miktex\\bin", "Solution_2": "If you installed it in C:\\texmf,\r\n\r\nthis is what: C:\\texmf\\miktex\\bin\r\n\r\nThen, if you installed it in C:\\Program Files\\MiKTeX 2.5\\, you can locate it in\r\n\r\nC:\\Program Files\\MiKTeX 2.5\\miktex\\bin.", "Solution_3": "Its in %Application Directory%/Miktex/Bin/LaTeX.exe\r\n\r\nThis was a problem with the directions, and it took me a few minutes to learn how to get around it." } { "Tag": [], "Problem": "Segment $AB$ is both a diameter of a circle of radius 1 and a side of an equilateral triangle $ABC$. The circle also intersects $AC$ and $BD$ at points $D$ and $E$, respectively. The length of $AE$ is\r\n\r\n$\\displaystyle \\text{(A)} \\ \\frac{3}{2} \\qquad \\text{(B)} \\ \\frac{5}{3} \\qquad \\text{(C)} \\ \\frac{\\sqrt 3}{2} \\qquad \\text{(D)} \\ \\sqrt{3} \\qquad \\text{(E)} \\ \\frac{2 + \\sqrt 3}{2}$", "Solution_1": "[hide=\"Solution\"]\nAs AB is the diameter of the circle, AEB must be 90 degrees. Therefore we have a right triangle with angle 60 degrees (opposite to AE) and hypotenuse of length 2.\n\nTherefore,\nsin 60 = AE / 2\nAE = 2 sqrt[3] / 2 = sqrt[3] \n\nSo the answers D sqrt[3][/hide]", "Solution_2": "[hide=\"Answer\"]Since $AB$ is a diameter of the circle, $\\triangle ABE$ is a right triangle, and it is also a 30-60-90 triangle because $\\angle ABE=60^\\circ$ by the given. Therefore, $AE=\\frac{AB\\sqrt{3}}{2}=\\sqrt{3}\\Rightarrow \\boxed{D}$.[/hide]" } { "Tag": [ "algebra", "polynomial", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "let P[0,1] be set of polynomials resticted to [0,1].show f(x)=x(1-x) isnt interior of P[0,1] in C[0,1],whats interior of P[0,1] in C[0,1]? :D", "Solution_1": "Isn't the interior empty? $p(x)+\\epsilon \\sin x$ is in $C[0,1]\\setminus P[0,1]$ for any $\\epsilon\\ne 0$." } { "Tag": [], "Problem": "\u5582!\u54c8\u9523![\u7535\u8bdd]\u5582\u5582 \u6bcf\u4e2a\u4eba;\u5404\u4e2a\u4eba,\r\n\r\nWith all the respect possible, the point of this forum is for you to discuss in [b]Chinese[/b] not in English, so that Chinese students that are not aces at English may have a change of integrating. \r\n\r\nAll the English discussions should be taken in the main forums. \r\n\r\n\u8c22\u8c22\u4f60\u3002", "Solution_1": "Oh, valentin knows chinese too!!", "Solution_2": "\u6211?\u77e5?\u5f88\u591a\u6c49\u8bed :blush: \u4f46\u6211\u77e5?\u4e00\u4e9b :P\r\n ... \u56e0\u6b64\u8bf7 \u4f7f\u7528\u6c49\u8bed\u8bed\u8a00\u5728\u8fd9\u4e2a\u5730\u65b9 ;)", "Solution_3": "\u7d55??\u4ee5\u7684..", "Solution_4": "sorry I made an English thread on this forum\r\nbut the replies are in Chinese, :D", "Solution_5": "\u5b83\u662f\u597d ;)", "Solution_6": "it seems that he translated directly into Chinese.", "Solution_7": "[quote=\"Valentin Vornicu\"]\u6211\u4e3f\u77e5\u9fd3\u5f88\u591a\u6c49\u8bed :blush: \u4f46\u6211\u77e5\u9fd3\u4e00\u4e9b :P\n ... \u56e0\u6b64\u8bf7 \u4f7f\u7528\u6c49\u8bed\u8bed\u8a00\u5728\u8fd9\u4e2a\u5730\u65b9 ;)[/quote]\r\noh my God\r\nValentin\u61c2\u5f97\u591a\u5c11\u79cd\u8bed\u8a00\uff1fAmazing!!!\r\n\u597d\u5389\u5bb3\u5440\uff01\r\n\u7f57\u9a6c\u5c3c\u4e9a\u5b66\u4e2d\u6587\u4e48\uff1f\r\njust wondering!!!! :P", "Solution_8": "liyi\u8bf4\u8fc7\u4e86...\u4ed6\u76f4\u63a5\u628a\u82f1\u6587\u6216\u662f\u4ec0\u4e48\u7684\u7ffb\u8bd1\u6210\u4e2d\u6587. \r\n\u770b\u4ed6\u7684\u8bed\u5e8f\u90fd\u4e0d\u5bf9.", "Solution_9": "[quote=\"Valentin Vornicu\"]\u5b83\u662f\u597d ;)[/quote]\r\n\u4ece\u8fd9\u4e2a\u770b\u51fa\u663e\u7136\u90a3\u662f\u673a\u5668\u7ffb\u8bd1\u7684 :D \r\n\u6ca1\u6709\u4e2d\u56fd\u4eba\u8fd9\u4e48\u8bf4\u8bdd\u7684", "Solution_10": "Yeah \r\nI just realized that.\r\nSorry for pointing out your... Valentin!(It's brilliant of you still!)\r\nThat's normaly what I do to Korean... :D", "Solution_11": "[quote=\"liyi\"][quote=\"Valentin Vornicu\"]\u5b83\u662f\u597d ;)[/quote]\n\u4ece\u8fd9\u4e2a\u770b\u51fa\u663e\u7136\u90a3\u662f\u673a\u5668\u7ffb\u8bd1\u7684 :D \n\u6ca1\u6709\u4e2d\u56fd\u4eba\u8fd9\u4e48\u8bf4\u8bdd\u7684[/quote]\r\n\u5b83\u662f\u597d=that's great?", "Solution_12": "\u5b83\u662f\u597d= It's OK.", "Solution_13": ":D Oh,it may be the opppsite of \"Chinglish\".", "Solution_14": "[quote=\"Valentin Vornicu\"]\u5582!\u54c8\u9523![\u7535\u8bdd]\u5582\u5582 \u6bcf\u4e2a\u4eba;\u5404\u4e2a\u4eba...\n[/quote]\r\nCall \u7ffb\u8bd1\u6210\u4e86 \u7535\u8bdd \u7b11\u6b7b\u4eba\uff01\uff01", "Solution_15": "\u201c\u6bcf\u4e2a\u4eba;\u5404\u4e2a\u4eba...\u201d\u662feveryone\u548ceverybody\u5427\u3002\r\n\u4e0d\u8fc7\uff0cValentin\u6709\u8fd9\u4efd\u5fc3\u6211\u4eec\u8fd8\u662f\u5f88\u611f\u8c22\u7684\u2026\u2026", "Solution_16": "[quote=\"dingdongdog\"]Valentin\u6709\u8fd9\u4efd\u5fc3\u6211\u4eec\u8fd8\u662f\u5f88\u611f\u8c22\u7684\u2026\u2026[/quote]\r\n\u5c0d! :yup:", "Solution_17": "[quote=\"dingdongdog\"]:D Oh,it may be the opppsite of \"Chinglish\".[/quote]Well I tried :blush: :P", "Solution_18": "[quote=\"Valentin Vornicu\"]\u6211\u4e3f\u77e5\u9fd3\u5f88\u591a\u6c49\u8bed :blush: \u4f46\u6211\u77e5\u9fd3\u4e00\u4e9b :P\n ... \u56e0\u6b64\u8bf7 \u4f7f\u7528\u6c49\u8bed\u8bed\u8a00\u5728\u8fd9\u4e2a\u5730\u65b9 ;)[/quote]\r\n...so please use Chinese in this place :D :D :D", "Solution_19": "valentine u r so funny...\r\nwell we all can understand ur broken chinese and...\r\nthank you for using chinese here...\r\n\u6655...\r\n\u4e0d\u5c0f\u5fc3\u53c8\u5199\u6210\u82f1\u6587\u4e86-_-\r\n\u90a3\u4e9b\u4e2d\u6587\u633a\u597d\u73a9\u7684 ...", "Solution_20": "\u8fd9\u4e5f\u80fd\u8ba9\u5728\u7f8e\u56fd\u957f\u5927\u7684\u4e2d\u56fd\u4eba\u591a\u7ec3\u4e60\u8bfb\u4e2d\u6587\u3002\u6211\u7684\u4e2d\u6587\u6c34\u6c34\u5e73\u7ea7\u5dee\u3002 :blush:", "Solution_21": "[quote=\"4everwise\"]\u8fd9\u4e5f\u80fd\u8ba9\u5728\u7f8e\u56fd\u957f\u5927\u7684\u4e2d\u56fd\u4eba\u591a\u7ec3\u4e60\u8bfb\u4e2d\u6587\u3002\u6211\u7684\u4e2d\u6587\u6c34\u6c34\u5e73\u7ea7\u5dee\u3002 :blush:[/quote]\r\n\u80fd\u6253\u5b57\u8bf4\u660e\u8fd8\u6709\u6551\u3002\u3002\u3002\u3002\u3002\u3002 :lol:", "Solution_22": "haha~\u5f88\u5c11\u898b\u6709\u7f85\u99ac\u5c3c\u4e9e\u4eba\u6253\u4e2d\u6587\u5beb\u7684 :lol: \r\n\u4e0d\u904e\u7b2c1\u6b65\u5df2\u7d93\u4e0d\u932f\u4e86 :)" } { "Tag": [ "function", "ratio", "calculus", "calculus computations" ], "Problem": "I'm studying Taylor series right now and I got confused at a somewhat subtle point, namely the values of the variable for which the expansion converges. Particularly about the binomial expansion (when the exponent is not an integer)\r\n\\[ (1 \\plus{} x)^\\alpha \\equal{} 1 \\plus{} \\alpha x \\plus{} \\cdots \\plus{} \\binom{\\alpha}{n}x^n \\plus{} \\cdots\\]\r\nIn all my books the following sentence is attached to the formula: \"This series converges for all all $ |x| < 1$\", condition that apparently does not depend on $ \\alpha$ (at least I used to think so). However when I was reading the solution to a random exercise on finding the Taylor expansion near $ x \\equal{} 0$ of the function\r\n\\[ f(x) \\equal{} \\frac {1}{\\sqrt {1 \\plus{} x}},\\]\r\nthe values of $ x$ for which the series converge were $ ( \\minus{} 1/2,1)$, and not $ ( \\minus{} 1,1)$ as I expected from the condition for convergence of the binomial expansion of $ f(x) \\equal{} (1 \\plus{} x)^{ \\minus{} 1/2}.$ The way my teacher computed the values was by finding the remainder term of the Taylor expansion and then analyzing when it would tend to zero, and it makes perfect sense for me. However I used to think that the convergence of the binomial series was already established for all $ |x| < 1$. So can you please explain me the flaw in my reasoning? Or maybe it means that the convergence of the binomial series also depends on the particular value of $ \\alpha$ (in this case -1/2)?? :maybe:\r\n\r\nThanks for your help.", "Solution_1": "[quote=\"Jutaro\"] The way my teacher computed the values was by finding the remainder term of the Taylor expansion and then analyzing when it would tend to zero[/quote] Without having thought too hard: this is a [i]sufficient[/i] condition for convergence, but it is not [i]necessary[/i] -- the remainder term puts an upper bound on the error, but it's not necessarily sharp. So, you can have convergence of the Taylor series even if the error term does not go to 0.\r\n\r\nIf you're actually interested in computing the radius of convergence, use the ratio test (or the root test. In principle this only works with \"nicely behaved\" Taylor series (you need the limit to converge), but in practice it works with just about everything you might be interested it.\r\n\r\nAlso, a power series centered at 0 must have interval of convergence symmetric around 0, except possibly at the endpoints.", "Solution_2": "[quote=\"JBL\"]Without having thought too hard: this is a [i]sufficient[/i] condition for convergence, but it is not [i]necessary[/i] -- the remainder term puts an upper bound on the error, but it's not necessarily sharp. So, you can have convergence of the Taylor series even if the error term does not go to 0.\n[/quote]\r\n\r\nNow I come to think about it, the statement of the exercise was \"prove that the series converges in this interval\", but no claim was made that it would not outside it. It all makes sense now, thank you. :)" } { "Tag": [ "calculus", "integration", "trigonometry", "function", "algebra", "polynomial", "domain" ], "Problem": "Let $f(x)=e^{\\sqrt[4]{x}}\\sin \\sqrt[4]{x}$. Prove that \r\n\r\n$\\int_{0}^{\\infty}f(x)x^{n}dx=0$, for all natural numbers $n$.\r\n\r\nI think that 4 can be replaced by any natural number of the form $4m$, where $m$ is a fixed natural number.", "Solution_1": "With the positive exponent on your exponential, you have a divergent integral. How do you intend to fix that?", "Solution_2": "Ohhhh..., I am very sorry for the mistake , thank you very much Kent, it should be a negative exponent.\r\n\r\nLet $f(x)=e^{-\\sqrt[4]{x}}\\sin(\\sqrt[4]{x})$.\r\n\r\nProve that \r\n\r\n$\\int_{0}^{\\infty}f(x)x^{n}dx=0$ for all natural numbers $n\\geq 0$.", "Solution_3": "Define $f$ on $\\mathbb{R}$ by letting $f(x)$ be your formula for $x\\ge 0$ and $f(x)=0$ for $x<0.$ Take the Fourier transform $\\widehat{f}(\\xi).$ Your claim, if true, is equivalent to saying that $\\left(\\frac{d}{d\\xi}\\right)^{n}\\widehat{f}(0)=0$ for all $n\\ge 0.$ This in turn would make $\\widehat{f}$ a function that is $C^{\\infty}$ but not analytic.", "Solution_4": "Kent,\r\nWhat are you trying to say?", "Solution_5": "Nothing, yet - just another way of looking at it. I haven't yet worked through whether or not it's true.", "Solution_6": "Here is my solution to this problem:\r\n\r\n$\\int_{0}^{\\infty}f(x)x^{n}dx=4\\int_{0}^{\\infty}e^{-t}(\\sin t )t^{4n+3}dt$ which can be obtained via the substitution $x=t^{4}$.\r\n\r\nLet $A=\\int_{0}^{\\infty}e^{-t}(\\sin t ) t^{4n+3}dt$\r\n\r\nlet $B=\\int_{0}^{\\infty}e^{-t}(\\cos t ) t^{4n+3}dt$ \r\nthen we get that \r\n\r\n$B-iA=\\int_{0}^{\\infty}e^{-(1+i)t}t^{4n+3}dt=\\frac{\\Gamma(4n+4)}{(1+i)^{4n+4}}\\in R$ via the substitution $t(1+i)=y$\r\n\r\nThus, $A=0$.\r\n\r\nThis problem is a counterexample to the following problem:\r\n\r\nLet $f$ be a continuous function defined on a finite interval of the form $[a,b]$. Then if $\\int_{a}^{b}f(x)x^{n}dx=0$ for all natural numbers $n$ then $f$ is the constant function 0. \r\n\r\nThe question is whether the same statement holds for functions defined on intervals of infinite form. The problem above shows that this nice property does not hold for the case when the interval is not of finite form.", "Solution_7": "If you asked me to prove that result on $[a,b],$ my proof would start with the polynomials being dense in some reasonable function space ($L^{2},$ perhaps) on $[a,b].$ That line of proof is not available on $[a,\\infty).$ \r\n\r\nI referred to the fact that your function $f$ has a Fourier transform which has a zero of infinite order at the origin. A function on $[a,b]$ would have a Fourier series, which is to say a Fourier transform whose domain is $\\mathbb{Z}.$ The concept of a \"zero of infinite order\" makes no sense for a sequence." } { "Tag": [], "Problem": "find (3*8^0.5)+(6*98^0.5)-(5*50^0.5)-(7*72^0.5)+(4*32^0.5)", "Solution_1": "[hide]$3\\sqrt8+6\\sqrt{98}-5\\sqrt{50}-7\\sqrt{72}+4\\sqrt{32}=6\\sqrt2+42\\sqrt2-25\\sqrt2-42\\sqrt2+16\\sqrt2=-3\\sqrt2$[/hide]", "Solution_2": "(-3*2^0.5) is correct" } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "$ a,b \\in \\mathbb{R}$, $ a < 0,b > 0$, satisfying that:\r\n\r\n$ a\\sin{\\theta} \\plus{} b\\cos{\\theta}\\geq 0$\r\n$ a\\cos{\\theta} \\minus{} b\\sin{\\theta}\\geq 0$\r\n\r\nFind the Maximum of $ \\sin{\\theta}$ :oops:", "Solution_1": "$ tg \\theta =\\frac{b}{a}$", "Solution_2": "[quote=\"Rust\"]$ tg \\theta = \\frac {b}{a}$[/quote]\r\n\r\nI'm sorry that the answer is not correct... :blush:", "Solution_3": "That was a hint, not an answer. Let $ \\phi$ be an angle such that $ \\tan \\phi \\equal{} \\frac{b}{a}$ and rewrite the system in terms of $ \\phi$." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "For $ x,y,z>0$,$ x+y+z=1$\r\nFind min $ S= \\frac{x}{x+yz}+\\frac{2y}{y+zx}+\\frac{3z}{z+xy}$", "Solution_1": "[quote=\"darkmaster\"]For$x,y,z \\geq 0$ ,$ x+y+z=1$\nFind min $ S= \\frac{x}{x+yz}+\\frac{2y}{y+zx}+\\frac{3}{z+xy}$[/quote] $ 6-S=\\frac{yz}{(x+y)(x+z)}+\\frac{2xz}{(x+y)(y+z)}+\\frac{3xy}{(z+y)(x+z)}$ And easy to find that : $6-S \\le 3$ The equality holds iff $z=0$ IF $x,y,z \\geq 0$ S haven't the minimum", "Solution_2": "Are you sure that $ 6-S \\leq 3$?", "Solution_3": "[quote=\"darkmaster\"]Are you sure that $ 6-S \\leq 3$?[/quote]\r\nYes!\r\n$6-S \\le 3 <->3xy(x+y)+yz(y+z)+2xz(x+z) \\ge 3(xy(x+y)+yz(y+z)+xz(x+z) +2xyz)$", "Solution_4": "OK,you're right!", "Solution_5": "Let $ x,y,z>0 ,x+y+z=1$ and $ xy+yz+zx\\neq 0.$ Prove that$$2\\leq \\frac{x}{x+yz}+\\frac{2y}{y+zx}+\\frac{z}{z+xy}\\leq\\frac{25}{8} $$\n$$3\\leq \\frac{x}{x+yz}+\\frac{2y}{y+zx}+\\frac{2z}{z+xy}\\leq 4 $$\n$$3\\leq \\frac{x}{x+yz}+\\frac{2y}{y+zx}+\\frac{3z}{z+xy}\\leq 5$$\n$$\\frac{x}{x+yz}+\\frac{y}{y+2zx}+\\frac{ z}{z+2xy} \\leq 2$$ \n$$\\frac{x}{x+yz}+\\frac{2y}{y+2zx}+\\frac{ z}{z+2xy} \\leq 3$$ \n" } { "Tag": [ "geometry", "parallelogram", "ratio", "geometry solved" ], "Problem": "Let ABCD be a parallelogram. Join each vertice with the midpoints of the opposite sides. Determine the area of the formed octagon in terms of the area of the parallelogram. \r\n\r\n \r\nIt seems to be a generalization of problem 5, round 10,International Mathematical Talent Search.", "Solution_1": "We can turn the parallelogram into a square by applying an affine transformation. It will, of course, conserve everything we're interested in: ratios of lengths (and of areas).\r\n\r\nWe can reason easier on a square: Let $M,N,P,Q$ be the midpts of $AB,BC,CA,DA$ . Now let $X=AP\\cap BQ,\\ Y=BQ\\cap CM,\\ Z=CM\\cap DN,\\ T=DN\\cap AP$. Let $l$ be the side of the square $XYZT$. We can see that if we take a side of the octagon out of $l$ we're left with two small pieces with length $x$ each, s.t. the side of the octagon is $x\\sqrt 2$. The area of the octagon is the area left from $XYZT$ after taking out four small isosceles right triangles of sides $x,x,x\\sqrt 2$. The ratio between the area of the octagon and the area of $XYZT$ must be $\\frac {l^2-2x^2}{l^2}$. We have $l=(2+\\sqrt 2)x\\Rightarrow \\frac {[octagon]}{[XYZT]}=2\\sqrt 2-2$ (it's easy to compute). It's also easy to see that the area of $XYZT$ is one fifth of the area of $ABCD$, so the ratio $\\frac {[octagon]}{[ABCD]}=\\frac {2\\sqrt 2-2}5$. \r\n\r\nThe advantage of the square is that we have right isosceles triangles.", "Solution_2": "Aaargh...that's what I was thinking about, but too slowly.... :D \r\n\r\nPierre.", "Solution_3": "What is an affine transformation?Why does it conserve the ratios?", "Solution_4": "That's the definition: a transformation which preserves collinearity and the ratio of distances.\r\n\r\nTry [url=http://mathworld.wolfram.com/AffineTransformation.html]this[/url]. Maybe that's not satisfying, but you can always try to Google it :)." } { "Tag": [ "algorithm", "number theory open", "number theory" ], "Problem": "for positive integers $ m,n$ , Show that there exist interger $ x$ and $ y$ such that $ mx \\plus{} ny \\equal{} gcd(m,n)$ :maybe:", "Solution_1": "See the [url=http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm]Extended Euclidean Algorithm[/url].", "Solution_2": "[quote=\"Victory.US\"]for positive integers $ m,n$ , Show that there exist interger $ x$ and $ y$ such that $ mx \\plus{} ny \\equal{} gcd(m,n)$ :maybe:[/quote]\r\n\r\nBezout's Theorem!" } { "Tag": [ "factorial" ], "Problem": "ok so you have to use 4 4's to make it equal 0-9. I got some of them but need help with for 1,3,4,5, and 6. Here's an example if it doesn't make sense to you: \r\nex. (4+4)-(4/4) =\r\n(8)-(1) =\r\n8-1 =7\r\n Help please", "Solution_1": "can u use factorials?", "Solution_2": "you don't need factorials\r\n\r\n$4+4-4-4=\\boxed{0}$\r\n$4/4+4-4=\\boxed{1}$\r\n$4/4+4/4=\\boxed{2}$\r\n$(4+4+4)/4=\\boxed{3}$\r\n$4+(4-4)*4=\\boxed{4}$\r\n$(4*4+4)/4=\\boxed{5}$\r\n$4+(4+4)/4=\\boxed{6}$\r\n$4+4-4/4=\\boxed{7}$\r\n$4+4+4-4=\\boxed{8}$\r\n$4+4+4/4=\\boxed{9}$" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "$x-2=\\sqrt{4-3\\sqrt{4-3\\sqrt{10-3x}}}$", "Solution_1": "Let f,g be functions such that f:[2,$\\infty$)-->R and g:A-->R , f(x)=x-2 and\r\ng(x)=$\\sqrt{4-3\\sqrt{4-3\\sqrt{10-3x}}}$.It's easy to prove that f is increasing and g is decreasing $\\Rightarrow$ x=3 is the only solution", "Solution_2": "Or substitute $t=x-2$ to get\r\n\r\n$t=\\sqrt{4-3\\sqrt{4-3\\sqrt{4-3t}}}$\r\n\r\nNow by infinite nesting we get $t=\\sqrt{4-3t}\\iff t=1\\iff x=3$.", "Solution_3": "Could you tell me please what \"infinite nesting\" refer to ?", "Solution_4": "It means we can substitute $t$ by itself infinitely:\r\n\r\n\\begin{eqnarray*}t &=& \\sqrt{4-3\\sqrt{4-3\\sqrt{4-3t}}}\\\\ &=& \\sqrt{4-3\\sqrt{4-3\\sqrt{4-3\\sqrt{4-3\\sqrt{4-3\\sqrt{4-3t}}}}}}\\\\ &=& \\sqrt{4-3\\sqrt{4-3\\sqrt{4-3\\sqrt{4-3\\sqrt{4-3\\sqrt{4-3\\sqrt{4-3\\sqrt{4-3\\sqrt{4-3t}}}}}}}}}\\\\ &=&\\dots\\end{eqnarray*}", "Solution_5": "[quote=\"Farenhajt\"]Or substitute $t=x-2$ to get\n\n$t=\\sqrt{4-3\\sqrt{4-3\\sqrt{4-3t}}}$\n\nNow by infinite nesting we get $t=\\sqrt{4-3t}\\iff t=1\\iff x=3$.[/quote]\r\nIt is not proved, that t=1 is unique solution.\r\nLet $f(t)=\\sqrt{4-3t}, g(t)=\\frac{4-t^{2}}{3}$. \r\nEquation $f(f(f(t)))=t$ equavalent to $g(g(g(t)))=t$ with $t>0,g(t)>0$ or $02$ and natural number $ k>1$.", "Solution_1": "Obviously not, take $ p\\equal{}5$ and $ k\\equal{}3$.", "Solution_2": "I 'm sorry. I 've stated the problem wrongly:\r\n\r\nProve or disprove: $ p^k\\minus{}1$ is [b]not divisible[/b] by $ pk\\plus{}1$ for all prime number $ p>2$ and natural number $ k>1$.", "Solution_3": "Let $ a\\equal{}pk\\plus{}1$, then from $ a|p^k\\minus{}1$ we get $ a|k^kp^k\\minus{}k^k$ or $ a|k^k\\minus{}(\\minus{}1)^k$ and from $ a|k^k\\minus{}(\\minus{}1)^k, a\\equal{}1\\mod k$ we get $ a|(\\frac{a\\minus{}1}{k})^k\\minus{}1$.\r\nTherefore sufficiently consider divisors $ k^k\\minus{}(\\minus{}1)^k$\r\n1. $ k\\equal{}2\\to p\\equal{}1$,\r\n2. $ k\\equal{}3\\to p\\equal{}2$,\r\n3. $ k\\equal{}4\\to p\\equal{}1,4,21$ - not primes,\r\n4. $ k\\equal{}5\\to p\\equal{}1,104,625$ - not primes,\r\n5. $ k\\equal{}6\\to p\\equal{}1,5,7,36,50,222,1555$ had prime solutions $ p\\equal{}5$ and $ p\\equal{}7$.\r\nIt give at least to solutions $ p\\equal{}5,k\\equal{}6$ and $ p\\equal{}7,k\\equal{}6$." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove that $\\frac{1}{2 \\sqrt{2}n}<\\{n \\sqrt{2}\\}\\leq 1-\\frac{2(3-2 \\sqrt{2})}{n}$ for all natural $n$.", "Solution_1": "Check for $n \\le 12$.\r\n\r\nLet $m$ be the nearest integer to $n\\sqrt 2$. Then $\\vert m-n\\sqrt 2\\vert . \\vert m+n\\sqrt 2 \\vert = \\vert m^2-2n^2 \\vert \\ge 1$. \r\n\r\nIf $m < n\\sqrt 2$ then $\\{n \\sqrt 2 \\} = \\vert m-n\\sqrt 2\\vert \\ge 1 / \\vert m+n\\sqrt 2 \\vert > 1/(2n\\sqrt 2)$.\r\nIf $m > n\\sqrt 2$ then $\\{n \\sqrt 2 \\} = 1 - \\vert m-n\\sqrt 2\\vert$ and $\\vert m-n\\sqrt 2\\vert \\ge 1 / \\vert m+n\\sqrt 2 \\vert > 1/(2n\\sqrt 2+1)$. Now $2n \\sqrt 2 + 1 < (3+2\\sqrt 2)n/2$ for $n > 1/(3/2-\\sqrt 2) > 12$. So $\\{n \\sqrt 2 \\} < 1 - 2/((3+2\\sqrt 2)n) = 1 - 2(3-2\\sqrt 2)/n$.", "Solution_2": "Nice...\r\n\r\nThere is a trivial mistake. You should write $n > 1/(3/2-\\sqrt 2) \\Rightarrow n \\geq 12$, instead of $n > 1/(3/2-\\sqrt 2) > 12$, so effectively you only need to check for $n < 12$. This doesn't affect the proof... nevertheless, it's still nice...", "Solution_3": "I have another solution.\r\nIf n=1,it is obvious true.So we prove the condition $n\\geq2$\r\nlet m=[$n\\sqrt{2}$],\r\nLeft:$m0$\r\nthen $16n^4 - 8n^2 +1>8(mn)^2$\r\nthen $4n^2-1>2mn\\sqrt{2}$\r\nso {$n\\sqrt{2}$}${2n\\sqrt{2}=(n\\sqrt{2}-m)2n\\sqrt{2}=4n^2 -2mn\\sqrt{2}>1}$.\r\nRight:$m+1>n\\sqrt{2}$,so $(m+1)^2>2n^2$,so $(m+1)^2\\geq 2n^2 +1$\r\nthen $n(m+1-n\\sqrt{2})\\geq n(\\sqrt{2n^2+1}-n\\sqrt{2})\r\n=\\frac{n}{\\sqrt{2n^2+1}+n\\sqrt{2}}=\\frac{1}{\\sqrt{2+\\frac{1}{n^2}}+\\sqrt{2}}\r\n\\geq \\frac{1}{\\sqrt{2+\\frac{1}{2^2}}+\\sqrt{2}}=2(3-2\\sqrt{2})$\r\nthen $n(n\\sqrt{2}-m)\\leq n-2(3-2\\sqrt{2})$\r\nso {$n\\sqrt{2}$}=$n\\sqrt{2}-m \\leq 1-\\frac{2(3-2\\sqrt{2})}{n}$" } { "Tag": [ "LaTeX", "number theory unsolved", "number theory" ], "Problem": "If a,b are positive integers, a|(b^2+1), b|(a^2+1), b>a, we have (3b-a)|b^2+1", "Solution_1": "If $a,b$ are positive integers. Given that $a|b^2+1$ and $b|a^2+1$ show that $(3b-a)|b^2+1$.\r\n\r\n[LaTeX-ed]\r\n\r\n@hawk: i was thinking so to.", "Solution_2": "I am sure the similar(or exactly the same ) problem has been posted many times.\r\nwe can gain that a,b satisfy\r\n$a^2+b^2+1-3ab=0$", "Solution_3": "How do you get that relation, HawkTiger?\r\nI get the fact that if a^2 + 1 = kb and b^2 + 1 = sa, then a | k^2 + 1 and k | a^2 + 1. Also we have that k < a because b > a and a^2 + 1 = kb. So we can decrease our solution pairs until we reach a = 1. And the only solution pair with a = 1 is (a,b) = (1,2). So from this solution we get all the other pairs by:\r\n\r\na_(n+1) = b_n\r\nb_(n+1) = ((b_n)^2 + 1)/a_n\r\n\r\nHow can you prove that 3b - a = b_n+1?", "Solution_4": "Oh...I saw that it was posted as an earlier problem hehe. Never mind.", "Solution_5": "The sequence $\\{b_n\\}_{n \\in \\bf N}$ is defined as \r\n\r\n$(1) \\;\\; b_{n+1} = \\frac{b_n^2 +1}{b_{n-1}}$\r\n\r\nfor $n \\geq 2$ where $b_1 = 2$ og $b_2 = 5$. Hence\r\n\r\n$b_{n+1} + b_{n-1}$\r\n \r\n$= \\frac{b_n^2 +1}{b_{n-1}} + b_{n-1}$\r\n\r\n$= \\frac{b_n^2 + (b_{n-1}^2 + 1)}{b_{n-1}}$\r\n\r\n$= \\frac{b_n^2 + b_nb_{n-2}}{b_{n-1}}$\r\n\r\n$= \\frac{b_n(b_n + b_{n-2})}{b_{n-1}}$\r\n\r\nfor all $n \\geq 3$. Consequently\r\n\r\n$\\frac{b_n}{b_{n-1}} = \\frac{b_{n+1} + b_{n-1}}{b_n + b_{n-2}},$\r\n\r\nwhich implies\r\n\r\n$\\prod_{k=3}^n \\frac{b_k}{b_{k-1}} = \\prod_{k=3}^n \\frac{b_{k+1} + b_{k-1}}{b_k + b_{k-2}},$\r\n\r\ni.e.\r\n\r\n$(2) \\;\\; \\frac{b_n}{b_2} = \\frac{b_{n+1} + b_{n-1}}{b_3 + b_1}$.\r\n\r\nNow $b_3 = 13$ by (1), thus $\\frac{b_3 + b_1}{b_2} = \\frac{13 + 2}{5} = \\frac{15}{5} = 3.$ Therefore \r\n\r\n$(3) \\;\\; 3b_n = b_{n+1} + b_{n-1}$\r\n\r\naccording to (2). Since $a_n = b_{n+1}$, (3) is equivalent to\r\n\r\n$3b_n - a_n = b_{n-1}.\\;\\;\\;\\;{\\bf q.e.d.}$" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "a,b,c>0 a\u00b2+b\u00b2+c\u00b2+abc=4; \r\nprove that :\r\n 3(a+b+c)\u2265abc+8", "Solution_1": "Do you mean $ a^2\\plus{}b^2\\plus{}c^2\\plus{}abc\\equal{}4$??", "Solution_2": "[quote=\"kihe_freety5\"]a,b,c>0 a\u00b2+b\u00b2+c\u00b2+abc=4; \nprove that :\n 3(a+b+c)\u2265abc+8[/quote]\r\n$ a\\equal{}2cosA;b\\equal{}2cosB;c\\equal{}2cosC$ with $ A;B;C$ be three angle in a triangle.\r\nWe need prove that:\r\n$ 6(cosA\\plus{}cosB\\plus{}cosC) \\geq 8(cosAcosBcosC\\plus{}1)$", "Solution_3": "yes , IT mean ${ \\{a^2}$+${ \\{b^2}$+${ \\{c^2}$+abc = 4", "Solution_4": "but I need a solution!!\r\nwho can ?? :)", "Solution_5": "[quote=\"kihe_freety5\"]a,b,c>0 a\u00b2+b\u00b2+c\u00b2+abc=4; \nprove that :\n 3(a+b+c)\u2265abc+8[/quote]\r\nThis inequality is not always true.\r\nAnd I found it is Jack Garfunkel Inequality.\r\nMy solution for it :)" } { "Tag": [ "algebra", "polynomial", "AMC", "AMC 12", "AIME", "system of equations" ], "Problem": "Given $s_{0}= 3$, and $s_{1}= 4$, under the recurrence relation $s_{n}= 4s_{n-1}-3s_{n-2}$, find $s_{2008}$ .\r\n\r\n\r\n$(\\text{A}) \\frac{12+2^{2005}}{2}$\r\n\r\n$(\\text{B}) \\frac{3^{2008}+5}{2^{1004}}$\r\n\r\n$(\\text{C}) \\frac{4^{2008}+7}{2^{2007}}$\r\n\r\n$(\\text{D}) \\frac{3^{2008}+5}{2}$\r\n\r\n$(\\text{E}) \\frac{2^{4016}-3^{2008}+5}{2}$\r\n\r\n[hide=\"My Solution\"]:\nThe characteristic polynomial of the recurrence relation is $s^{2}=4s-3$, or $s^{2}-4s+3 = 0$. This polynomial factors as $(s-3)(s-1)$, with roots $r_{1}= 3$ and $r_{2}= 1$.\n\nBy Binet's theorem, we can write the general form for the Nth term of a linear recurrence as $s_{n}=A(r_{1})^{n}+B(r_{2})^{n}$ with constants A and B. \n\nNow we solve a system of equations using our starting values.\n$3 = A(3)^{0}+B(1)^{0}\\longrightarrow 3 = A+B$\n$4 = A(3)^{1}+B(1)^{1}\\longrightarrow 4 = 3A+B$\nObtaining values of A = 0.5, B = 2.5, we complete our closed form expression.\n$s_{n}= 0.5(3)^{N}+2.5(1)^{N}\\longrightarrow s_{n}= 0.5(3)^{N}+2.5$\n\nOur value is then $\\frac{3^{2008}+5}{2}$, or Choice D.\n[/hide]\r\n\r\nAny comments? I designed this problem to be ~23 level on the AMC12, or 4-5 level on the AIME.\r\n\r\nAlternate solutions are welcome as well. Please let me know if my solution has errors, as I'm thinking about throwing together a Mock AIME for my school and I might do some variants on this problem. I don't want any errors, so let me know if something seems funky and I'll try to explain. Thanks! Matt Redmond.", "Solution_1": "Characteristic polynomials aren't really AMC level...", "Solution_2": "What level would you classify this as, then?", "Solution_3": "[hide=\"amc level solution?\"]$s_{n}-s_{n-1}= 3\\left(s_{n-1}-s_{n-2}\\right)$\n$a_{n}= s_{n}-s_{n-1}$\n$a_{n}= 3a_{n-1}=3^{n-1}$ (since $a_{1}= 1$)\n$\\sum_{i=1}^{2008}a_{i}= \\frac{3^{2008}-1}{3-1}= s_{2008}-s_{0}$\n$s_{2008}= \\frac{3^{2008}-1}{3-1}+3 = \\boxed{\\frac{3^{2008}+5}{2}}$\n[/hide]", "Solution_4": "That's pretty clean, quick, and clear. Did you like the problem? (What difficulty should I put this as (roughly?))\r\nI like your solution a lot.", "Solution_5": "thank you! but i saw this method on aops before, and i thought it was nice, too, so i learned/remembered it.", "Solution_6": "As an aspiring problem writer, is there anything I could change to make this problem better? I don't want to repeat too many old ideas.", "Solution_7": "[quote=\"DiscreetFourierTransform\"]I don't want to repeat too many old ideas.[/quote]\r\n\r\nUnfortunately, this is a pretty common idea. You generally wouldn't find this question on a real test because it is too well-known and, by itself, does not pose enough of a thinking challenge.", "Solution_8": "[quote=\"minsoens\"][hide=\"amc level solution?\"]$s_{n}-s_{n-1}= 3\\left(s_{n-1}-s_{n-2}\\right)$\n$a_{n}= s_{n}-s_{n-1}$\n$a_{n}= 3a_{n-1}=3^{n-1}$ (since $a_{1}= 1$)\n$\\sum_{i=1}^{2008}a_{i}= \\frac{3^{2008}-1}{3-1}= s_{2008}-s_{0}$\n$s_{2008}= \\frac{3^{2008}-1}{3-1}+3 = \\boxed{\\frac{3^{2008}+5}{2}}$\n[/hide][/quote]\r\ncan you explain what you did after the first step. \r\n\r\nwhat is $a_{n}$?", "Solution_9": "I've never heard of characteristic polynomials.\r\n\r\nDoes it always work????", "Solution_10": "It's why Binet's formula for the Fibonacci numbers works. It always works for linear recursions.\r\n\r\nSay we are given an arbitrary LINEAR recursion\r\n$a_{n}= Xa_{n-1}+Ya_{n-2}+Za_{n-3}$ (there could be as many more terms as you want, but it's enough of an example with three on the RHS.)\r\n\r\nWe can then rewrite this as\r\n$a_{n+3}= Xa_{n+2}+Ya_{n+1}+Za_{n+0}$\r\n\r\nNow transform the subscripts into exponents. I'll not include a proof of why this works, but I believe that Mathworld and Wikipedia both have one.\r\nThe characteristic polynomial of the above recursion is then\r\n$a^{3}= Xa^{2}+Ya^{1}+Za^{0}$\r\nor $a^{3}-Xa^{2}-Ya-Z = 0$\r\n\r\nBy Binet's theorem, we can rewrite any linear recursion as a closed form expression\r\n$a_{n}= A(r_{1})^{n}+B(r_{2})^{n}+C(r_{3})^{n}$... where various $r_{i}$ are the roots of the characteristic polynomial and A, B, and C are constants given by the initial values.\r\nWe then see that if we have a recursion with three terms on the RHS to begin with, our characteristic polynomial will have three roots, and thus we can set up a system of three equations.\r\n\r\nWe put \r\n$a_{0}= A(r_{1})^{0}+B(r_{2})^{0}+C(r_{3})^{0}$\r\n$a_{1}= A(r_{1})^{1}+B(r_{2})^{1}+C(r_{3})^{1}$\r\n$a_{2}= A(r_{1})^{2}+B(r_{2})^{2}+C(r_{3})^{2}$\r\nthen we substitute in our original values and our roots of the characteristic polynomial, letting us set up a system where we can solve for constants A, B, and C.\r\n\r\nOur final equation then takes the form\r\n$a_{n}= A(r_{1})^{n}+B(r_{2})^{n}+C(r_{3})^{n}$\r\nwith known constants A, B, C and known roots $r_{1}, r_{2}, r_{3}$\r\nFrom there, it's a matter of evaluating this closed form expression to find nth term values.", "Solution_11": "The book [url=http://www.math.upenn.edu/~wilf/DownldGF.html]Generatingfunctionology[/url] (available free at that link) provides in its opening chapter some of the theoretical background for why this should actually work.\r\n\r\nIncidentally, if you ask me, the real problem with this problem is that it admits only two types of solution: method 1 is to happen to know how to solve questions like this and then is a direct application of the method. Method 2 is to get lucky and notice a pattern in the first few terms of the sequence. Neither of these is very appealing from a problem-solving point of view.", "Solution_12": "The solution posted by minsoens was different from the application of the Method, was quite \"amc level\", and worked elegantly. Perhaps there are other methods then either seeing a pattern or using a formula.", "Solution_13": "Fair enough, and shame on me for not reading the whole thread. The same nice features of the answer allowed a simplification to a much more recognizable pattern.", "Solution_14": "[quote=\"iamagenius\"][quote=\"minsoens\"][hide=\"amc level solution?\"]$s_{n}-s_{n-1}= 3\\left(s_{n-1}-s_{n-2}\\right)$\n$a_{n}= s_{n}-s_{n-1}$\n$a_{n}= 3a_{n-1}=3^{n-1}$ (since $a_{1}= 1$)\n$\\sum_{i=1}^{2008}a_{i}= \\frac{3^{2008}-1}{3-1}= s_{2008}-s_{0}$\n$s_{2008}= \\frac{3^{2008}-1}{3-1}+3 = \\boxed{\\frac{3^{2008}+5}{2}}$\n[/hide][/quote]\ncan you explain what you did after the first step. \n\nwhat is $a_{n}$?[/quote]\r\nif you still need an explanation, i just defined another sequence $a_{1}, a_{2}, a_{3}, a_{4}...$ in terms of $s_{1}, s_{2}, s_{3}, s_{4}...$" } { "Tag": [ "function", "limit", "algebra proposed", "algebra" ], "Problem": "Does there exist a function $ f: {\\Bbb N} \\to {\\Bbb N}$ such that $ f(f(n \\minus{} 1)) \\equal{} f(n \\plus{} 1) \\minus{} f(n)$ for all $ n > 2$.", "Solution_1": "We have for $ n \\ge 2$, $ f(n)$ is strictly increasing, so $ f(n) \\ge n\\minus{}1$ for all $ n \\ge 2$\r\n\r\nNow we also have $ f(f(n\\minus{}1)) \\le f(f(n\\minus{}1))\\plus{}f(f(n\\minus{}2))\\plus{}...\\plus{}f(f(1))\\equal{}f(n\\plus{}1)\\minus{}f(2)3$, impossible. So there doesn't exist such function.", "Solution_2": "$ f(n \\plus{} 1) \\minus{} f(n) \\equal{} f(f(n \\minus{} 1))\\ge 1$ $ \\Rightarrow f(n)\\ge n$.\r\nThus $ f(n\\plus{}1)\\equal{}f(n)\\plus{}f(f(n\\minus{}1))\\ge f(n)\\plus{}f(n\\minus{}1)\\ge 2n\\minus{}1$. Notice that $ f(n)\\equal{}2n\\minus{}1$ obviously doesn't satisfy the condition. Therefore, there exist such $ c$ that $ f(c)\\ge 2c$. Clearly for $ n>c$ we will obtain that $ f(n)\\ge 2n$.\r\n\r\nLet $ f(n)\\ge 2^in$ ($ i\\ge 1$). Then $ f(n \\plus{} 1) \\minus{} f(n) \\equal{} f(f(n \\minus{} 1))\\ge 2^{i}f(n \\minus{} 1)\\ge 2^{2i}(n \\minus{} 1)$ $ \\Rightarrow f(n \\plus{} 1)\\ge f(n) \\plus{} 2^{2i}(n \\minus{} 1)\\ge 2^in \\plus{} 2^{2i}(n \\minus{} 1)\\ge 2^{i \\plus{} 1}(n \\plus{} 1)$\r\n$ \\Leftrightarrow 2^{i}(n \\minus{} 1)\\ge n \\plus{} 2$, which is obvious for $ n > 3$ for example . Hence, $ f(n)\\ge \\lim_{i\\rightarrow \\infty}2^in \\equal{} \\infty$ for $ n > 3$, which is impossible, since $ f(n)$ is a constant.", "Solution_3": "not so hard\nwe can trivially get f is strictly increasing\nso $f(n)\\ge n$\nsince $f(n+1)=f(n)+f(f(n-1))$\nhence $f(n-1) 16^4 + 8$ and $3^{12} = 9^6 > 9^4 > 6^4$ thus the expression is positive...Did I miss something?\r\n\r\nPierre.", "Solution_2": "no, your solution is clearly true :D" } { "Tag": [ "function", "integration", "calculus", "calculus computations" ], "Problem": "Prove:\r\n\\[ \\zeta (s) = s \\int_{1}^{\\infty} \\frac{[x]}{x^{s+1}}dx \\]\r\nP.S. this might be trivial but I'm beginner in this field :oops:", "Solution_1": "The hint in the book says compute the integral from $1$ to $N$, and compare it with the $N$-th partial sum of $\\frac{1}{i^s}$\r\n\r\n$\\displaystyle s \\int_{1}^{N} \\frac{[x]}{x^{s+1}}dx = s \\left( \\int_{1}^{2} \\frac{1}{x^{s+1}}dx + \\int_{2}^{3} \\frac{2}{x^{s+1}}dx + ... + \\int_{N-1}^{N} \\frac{N-1}{x^{s+1}}dx \\right)$\r\n\r\n$\\displaystyle = -s \\left [ (\\frac{1}{s 2^s} - \\frac{1}{s}) +(\\frac{2}{s 3^s} - \\frac{2}{s 2^s}) + ... + (\\frac{N-1}{s N^s} - \\frac{N-1}{s (N-1)^s}) \\right ]$\r\n\r\n$\\displaystyle = \\frac{1}{1^s} + \\frac{1}{2^s} + \\frac{1}{3^s} + ... + \\frac{1}{N^s} = \\sum_{i=1}^{N}\\frac{1}{i^s}$" } { "Tag": [ "ratio" ], "Problem": "A mixture of methanol and ethanol contains $41\\%{\\text{ }}O$\r\n (percentages of mass). Which is the molar ratio of the two alcohols in the mixture?\r\n\r\nThanks!", "Solution_1": "[hide=\"Hint\"]Consider a sample of the mixture with 100 g of mass. What is the mass of oxygen in this sample? What is the combined mass of carbon and hydrogen? (Write the chemical formulas of methanol and ethanol). Write those two masses in terms of number of moles of methanol and ethanol and the molar masses of carbon, oxygen, and hydrogen.[/hide]\n\n[hide=\"Answer\"]$\\frac{n_{methanol}}{n_{ethanol}}\\approx 1$[/hide]", "Solution_2": "[hide=\"Solution\"]\n\n$m(O) = \\frac{41}{100}(m(C_{2}H_{5}OH)+m(CH_{3}OH))$\n\nAssume there is 1 mol of $C_{2}H_{5}OH$ and $n$ mol of $CH_{3}OH$\n\nThen you get $16(1+n) = \\frac{41}{100}(46+32n) \\Rightarrow 288n=286$\n\nSo the molar ratio is basically 1:1. [/hide]" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "If $a,b,c \\in R$ so that $ \\sum \\frac{1}{1+a^2}=2$ then $abc(a+b+c-abc) \\leq \\frac{5}{8}$.", "Solution_1": "Multiply both side of the given equation by (a 2 +1)(b 2 +1)(c 2 +1),we have\r\n a 2 b 2 +b 2 c 2 +c 2 a 2 +2a 2b 2 c 2 =1\r\n Let ab=x ,bc=y ,ac=z ,we have:\r\n x 2 +y 2 +z 2 +2xyz=1\r\n So x=cos A ,y=cos B ,z=cosC ,with A,B,C are angles of an acute triangle\r\n The inequality we have to prove is equivalent to \r\n CosAcosB+CosBcosC+cosCcosA-CosAcosBcosC \\leq 5/8 (*)\r\n But \r\n CosACosBcosC=(Sin 2 A+Sin 2 B+Sin 2 C)/2 -1\r\nSo (*) is equivalent to \r\n 2(CosAcosB+CosBcosC+cosCcosA)-(3- cos 2 A- cos 2 B- cos 2 C) +2 <=5/4\r\n 2(cosAcosB+cosBcosC+cosCcosA)+(cos 2 A+cos 2 B+cos 2C) \\leq 9/4 \r\n <--> cosA+cosB+cosC \\leq 3/2 ,which is a well-known inequality", "Solution_2": "You should check your solution. Because a,b,c are real numbers,you can't\r\nhave that trigonometric replacement. :D", "Solution_3": "1/(1+a 2 )=x 1/(1+b 2 )=y 1/(1+c 2 )=z\r\nx=2u/(u+v+w)\r\ny=2v/(u+v+w)\r\nz=2w/(u+v+w)\r\na 2 =(v+w-u)/2u\r\nb 2 =(w+u-v)/2v\r\nc 2 =(u+v-w)/2w\r\nu=q+r\r\nv=r+p\r\nw=p+q\r\nthe ineq is:\r\na 2 bc+ab 2 c+abc 2 \\leq 5/8+a 2 b 2 c 2 so it's enough to prove for a,b,c>0\r\na= \\sqrt p/(q+r)\r\nb= \\sqrt q/(r+p)\r\nc= \\sqrt r/(p+q)\r\n \\sum 1/(1+a 2 )=2\r\n1=a 2 b 2 +b 2 c 2 +c 2 a 2 +2(abc) 2 \r\nSo the ineq becomes:\r\n2(a 2 bc+b 2 ca+c 2 ab) \\leq 5/4+2(abc) 2 \r\n2(a 2 bc+b 2 ca+c 2 ab) \\leq 5/4+1-a 2 b 2 -b 2 c 2 -c 2 a 2 \r\n(ab+bc+ca) 2 \\leq 9/4\r\nab+bc+ca \\leq 3/2\r\n \\sum \\sqrt pq/(p+r)(q+r) \\leq 3/2\r\n2 \\sqrt pq/(p+r)(q+r) \\leq p/(p+r) + q/(q+r)\r\nThe same with the others and adding we get the result" } { "Tag": [ "number theory open", "number theory" ], "Problem": "Show that if $ m, n$ and $ r$ are positive integers and: $ 1 + m + n\\sqrt {3} = (2 + \\sqrt {3})^{2r - 1}$ then $ m$ is a perfect square.", "Solution_1": "$ (2\\plus{}\\sqrt 3)^a\\equal{}m\\plus{}1\\plus{}n\\sqrt 3, (2\\minus{}\\sqrt 3)^a\\equal{}m\\plus{}1\\minus{}n\\sqrt 3\\to 1\\equal{}(m\\plus{}1)^2\\minus{}3n^2$.\r\nTherefore $ m(m\\plus{}2)\\equal{}3n^2$ and if $ (6,m)\\equal{}1$, then $ 3\\not m, (m,m\\plus{}2)\\equal{}1$ . so $ m$ is square,\r\n$ m\\equal{}2^a\\minus{}1\\mod 3\\equal{}1\\mod 3$ because a is odd and $ m\\plus{}1\\equal{}0\\mod 2$ because a is odd. It give result." } { "Tag": [ "vector", "function", "integration", "calculus", "derivative", "calculus computations" ], "Problem": "Consider the vector field F = 2xi - 4yj + (2z - 3)k. \r\n\r\nFind the potential function for f .\r\n\r\nI am kind of stuck and don't know how to solve this....", "Solution_1": "The purpose is to find a scalar field V such that $ \\vec{F} \\equal{} \\nabla V$. If you cannot think of a such a V, then suppose $ \\vec{F}$ is a force field and let's calculate the work done by this field between to positions A and B:\r\n\r\n$ W \\equal{} \\int_A^B \\vec{F} \\cdot d \\vec{r} \\equal{} \\int_A^B [2x\\,dx \\minus{} 4y \\,dy \\plus{} (2z\\minus{}3) \\,dz]$.\r\n\r\nI think you can now proceed from here by yourself.", "Solution_2": "[quote=\"Carcul\"]The purpose is to find a scalar field V such that $ \\vec{F} \\equal{} \\nabla V$. If you cannot think of a such a V, then suppose $ \\vec{F}$ is a force field and let's calculate the work done by this field between to positions A and B:\n\n$ W \\equal{} \\int_A^B \\vec{F} \\cdot d \\vec{r} \\equal{} \\int_A^B [2x\\,dx \\minus{} 4y \\,dy \\plus{} (2z \\minus{} 3) \\,dz]$.\n\nI think you can now proceed from here by yourself.[/quote]\r\n\r\nhmm..but what is A and B here?", "Solution_3": "Could made it an indefinite integral, and the constant would depend on the choice of the origin or reference of the potential.", "Solution_4": "There is a systematic way to do this. We're trying to find a function $ v$ such that $ \\nabla v\\equal{}\\vec{F}.$ which is to say:\r\n\\[ \\frac{\\partial v}{\\partial x}\\equal{}2x;\\ \\frac{\\partial v}{\\partial y}\\equal{}\\minus{}4y;\\ \\frac{\\partial v}{\\partial z}\\equal{}2z\\minus{}3\\]\r\nThe procedure is to take it one variable at a time, finding the antiderivative with respect to that variable. The \"constant of integration\" won't be a constant but rather a yet unknown function of the variables that haven't been used yet. In between each step, we find the partial derivative of our new expression for $ v$ with respect to the next variable.\r\n\r\n$ v\\equal{}\\int 2x\\,dx\\equal{}x^2\\plus{}g(y,z).$ (That's what I mean by an unknown function.)\r\n\r\n$ \\frac{\\partial v}{\\partial y}\\equal{}0\\plus{}\\frac{\\partial g}{\\partial y}\\equal{}\\minus{}4y,$ so\r\n\r\n$ \\frac{\\partial g}{\\partial y}\\equal{}\\minus{}4y$\r\n\r\n$ g\\equal{}\\int \\minus{}4y\\,dy\\equal{}\\minus{}2y^2\\plus{}h(z),$ so $ v\\equal{}x^2\\minus{}2y^2\\plus{}h(z).$\r\n\r\n$ \\frac{\\partial v}{\\partial z}\\equal{}0\\plus{}0\\plus{}h'(z)\\equal{}2z\\minus{}3,$ so\r\n\r\n$ h(z)\\equal{}\\int (2z\\minus{}3)\\,dz\\equal{}z^2\\minus{}3z\\plus{}C.$\r\n\r\nHence we have $ v\\equal{}x^2\\minus{}2y^2\\plus{}z^2\\minus{}3z\\plus{}C.$\r\n\r\nIn this case, the problem was so easy we didn't need this method: the variables separate fully and we can see that we can just do three unconnected integrations. In general, however, we do need what I've outlined above.\r\n\r\nThe method is self-correcting in the sense that if you attempt to do this for a vector field for which $ \\text{curl}\\,\\vec{F}\\equal{}\\nabla\\times\\vec{F}\\ne\\vec{0},$ then you will at some point arrive at a false equation and be unable to proceed.", "Solution_5": "[quote=\"Kent Merryfield\"]There is a systematic way to do this. We're trying to find a function $ v$ such that $ \\nabla v \\equal{} \\vec{F}.$ which is to say:\n\\[ \\frac {\\partial v}{\\partial x} \\equal{} 2x;\\ \\frac {\\partial v}{\\partial y} \\equal{} \\minus{} 4y;\\ \\frac {\\partial v}{\\partial z} \\equal{} 2z \\minus{} 3\n\\]\nThe procedure is to take it one variable at a time, finding the antiderivative with respect to that variable. The \"constant of integration\" won't be a constant but rather a yet unknown function of the variables that haven't been used yet. In between each step, we find the partial derivative of our new expression for $ v$ with respect to the next variable.\n\n$ v \\equal{} \\int 2x\\,dx \\equal{} x^2 \\plus{} g(y,z).$ (That's what I mean by an unknown function.)\n\n$ \\frac {\\partial v}{\\partial y} \\equal{} 0 \\plus{} \\frac {\\partial g}{\\partial y} \\equal{} \\minus{} 4y,$ so\n\n$ \\frac {\\partial g}{\\partial y} \\equal{} \\minus{} 4y$\n\n$ g \\equal{} \\int \\minus{} 4y\\,dy \\equal{} \\minus{} 2y^2 \\plus{} h(z),$ so $ v \\equal{} x^2 \\minus{} 2y^2 \\plus{} h(z).$\n\n$ \\frac {\\partial v}{\\partial z} \\equal{} 0 \\plus{} 0 \\plus{} h'(z) \\equal{} 2z \\minus{} 3,$ so\n\n$ h(z) \\equal{} \\int (2z \\minus{} 3)\\,dz \\equal{} z^2 \\minus{} 3z \\plus{} C.$\n\nHence we have $ v \\equal{} x^2 \\minus{} 2y^2 \\plus{} z^2 \\minus{} 3z \\plus{} C.$\n\nIn this case, the problem was so easy we didn't need this method: the variables separate fully and we can see that we can just do three unconnected integrations. In general, however, we do need what I've outlined above.\n\nThe method is self-correcting in the sense that if you attempt to do this for a vector field for which $ \\text{curl}\\,\\vec{F} \\equal{} \\nabla\\times\\vec{F}\\ne\\vec{0},$ then you will at some point arrive at a false equation and be unable to proceed.[/quote]\r\n\r\nthanks for the answers.. I'd appreciate it.. if I am asked to:\r\n\r\nUse the Fundamental Theorem of Line Integrals to find the exact value of the line integral for C the line from (1, 1, 1) to (4, 5, -8). What should I do? can I utilize the potential function I get above and plug in numbers and get the difference?", "Solution_6": "If $ \\vec{F}\\equal{}\\nabla v,$ and $ C$ is a curve that starts at point $ p_0$ and ends at point $ p_1,$ then\r\n\\[ \\int_C\\vec{F}\\cdot d\\vec{r}\\equal{}\\int_C\\nabla v\\cdot d\\vec{r}\\equal{}v(p_1)\\minus{}v(p_0).\\]\r\nWhat did you think was meant by \"Fundamental Theorem of Line Integrals\" anyway? This is it. Since you've found $ v,$ all you have to do is evaluate it at each end. And the answer is independent of path - it doesn't matter that that this is the line segment between those points; it could be any (rectifiable) path with these endpoints. The result would be the same.", "Solution_7": "[quote=\"Kent Merryfield\"]If $ \\vec{F} \\equal{} \\nabla v,$ and $ C$ is a curve that starts at point $ p_0$ and ends at point $ p_1,$ then\n\\[ \\int_C\\vec{F}\\cdot d\\vec{r} \\equal{} \\int_C\\nabla v\\cdot d\\vec{r} \\equal{} v(p_1) \\minus{} v(p_0).\n\\]\nWhat did you think was meant by \"Fundamental Theorem of Line Integrals\" anyway? This is it. Since you've found $ v,$ all you have to do is evaluate it at each end. And the answer is independent of path - it doesn't matter that that this is the line segment between those points; it could be any (rectifiable) path with these endpoints. The result would be the \nsame.[/quote]\r\n\r\nwell I tried to plug in numbers and take the difference and I got is:\r\n\r\n=(16 - 50 + 64 + 24) - (1-2+1-3)\r\n=54-3\r\n=51\r\n\r\nand I input this into webassign and gives me a wrong result", "Solution_8": "[quote]=(16 - 50 + 64 + 24) - (1-2+1-3)\n=54-3\n=51 [/quote]\r\nThe arithmetic mistake is right in front of you. Find it.", "Solution_9": "[quote=\"Kent Merryfield\"][quote]=(16 - 50 + 64 + 24) - (1-2+1-3)\n=54-3\n=51 [/quote]\nThe arithmetic mistake is right in front of you. Find it.[/quote]\r\n\r\nsorry I think it's 54-2 = 52.. but still that's not the answer", "Solution_10": "I'm still waiting for you to do some correct arithmetic. Hasn't happened yet.", "Solution_11": "You've got a double-negative. It's $ 54\\minus{}(\\minus{}3)\\equal{}54\\plus{}3\\equal{}57$." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Fix a prime $ p$, and suppose that a subgroup $ H \\subseteq G$ has the property that $ C_G(x) \\subseteq H$ for every element $ x \\in H$ having order $ p$. \r\nShow that $ p$ cannot divide both $ \\left|H\\right|$ and $ \\left|G : H\\right|$.", "Solution_1": "[quote=\"Jan\"]Fix a prime $ p$, and suppose that a subgroup $ H \\subseteq G$ has the property that $ C_G(x) \\subseteq H$ for every element $ x \\in H$ having order $ p$. \nShow that $ p$ cannot divide both $ \\left|H\\right|$ and $ \\left|G : H\\right|$.[/quote]\r\n\r\nLet P be a Sylow p-subgroup of H, a subgroup of the finite group G. If [G:H] is divisible by p, then [G:P] is divisible by p, and so [G:N_G(P)] is divisible by p, and so P is contained as a normal subgroup in some larger p=group Q. Then P intersect Z(Q) is non-trivial (if P is non-trivial), so there is some x in P \u2264 H, x of order p, such that Q is contained in the centralizer of x. Hence if |H| and [G:H] are both divisible by p, then H contains an element of order p centralized by elements outside of H." } { "Tag": [], "Problem": "Each of the interior angles of a heptagon (a seven-sided polygon) is obtuse and the number \r\nof degrees in each angle is a multiple of $ 9$. No two angles are equal. Find in degrees the \r\nsum of the two largest angles in the heptagon.", "Solution_1": "[hide]\nThe interior angles of a heptagon = 180 (n-2) = 900 degree\nAll angles are multiples of nine so let the angles be labled 9a,9b,9c,9d,9e,9f,9g where a10 \nsince a is also the smallest a =11\nAlso, b=12 c=13 d=14 e=15 \nand 9(f+g)= 900 - 9(11+12+...15) = 315\n\nAnswer: 315\n[/hide]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that there are infinitely many composite numbers $ n$ such that $ n$ divides $ 3^{n\\minus{}1}\\minus{}2^{n\\minus{}1}$.", "Solution_1": "A good problem . Have some ways to solve this problem . \r\nHere is my solution . \r\nTake $ n = 3^k - 2^k$\r\nWe have $ n|3^{n - 1} - 2^{n - 1} %Error. \"Leftrighatarow\" is a bad command.\nk|3^k - 2^k - 1$ \r\nWe prove that have infinite k such that $ k|3^k - 2^k - 1$ \r\nTake $ k = 2^m$ \r\nWe will prove that $ 2^m|3^{2^m} - 2^{2^m} - 1$\r\n$ \\Leftrightarrow 2^{2^m}|3^{2^m} - 1$ \r\nBut we knew that $ v_{2}(3^{2^m} - 1) = m + 1$\r\nSo $ 2^m|3^{2^m} - 1$\r\nIt mean that $ k = 2^m$ satisfy condition . \r\nEasy to check that $ n = 3^{2^m} - 2^{2^m}$ where $ m > 1$ is a composite number.\r\nProblem was proved .", "Solution_2": "Nice solution but \n$ v_{2}(3^{2^m} - 1) =\\sum_{i=1}^{m-1} v_2(3^{2^{m-i}}+1) +v_2(3+1)+v_2(3-1)=(m-1)+2+1$\n" } { "Tag": [ "geometry", "rectangle", "probability" ], "Problem": "http://yashar.biz/mathcounts/mathcounts_files/2001/2001%20State%20Target.pdf\r\n\r\nclick on the link and go to problem 3.", "Solution_1": "[hide]\ngraphing out x^2+y^2\u22644 while keeping it in the rectangle, you get a quarter circle with radius 2. so The area of the quarter circle is 2^2pi/4 or pi. Since the area of the rectangle is 32, the probability is pi/32.[/hide]", "Solution_2": "[hide]and in case you don't know... the form of a circle is:\n(x-h)^2+(y-k)^2=r^2\nwhere r=radius\nand (h,k) is the midpoint...\nso the midpoint would be (0,0) with a radius of 2.[/hide]", "Solution_3": "[quote=\"Ihatepie\"][hide]\ngraphing out x^2+y^2\u22644 while keeping it in the rectangle, you get a quarter circle with radius 2. so The area of the quarter circle is 2^2pi/4 or pi. Since the area of the rectangle is 32, the probability is pi/32.[/hide][/quote]\r\n\r\nthanks i get it now" } { "Tag": [ "inequalities", "inequalities proposed", "algebra" ], "Problem": "If $ a,b,c$ are nonnegative real numbers, then\r\n\r\n$ (a) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\ge \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}$;\r\n\r\n$ (b) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\ge \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)}$.\r\n\r\nMoreover,\r\n\r\n$ (c) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\plus{}2\\ge \\left(\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\right)^2$.", "Solution_1": "Assume that $ c\\equal{}\\min\\{a,b,c\\}.$ Then, we have\r\n$ \\frac{a^2}{(b\\minus{}c)^2} \\ge \\frac{a^2}{b^2},\\frac{b^2}{(a\\minus{}c)^2} \\ge \\frac{b^2}{a^2},$\r\nand\r\n$ \\frac{a^2\\plus{}b^2\\plus{}c^2}{ab\\plus{}bc\\plus{}ca} \\minus{}\\frac{a^2\\plus{}b^2}{ab} \\equal{}\\frac{c[abc\\minus{}(a\\plus{}b)(a^2\\plus{}b^2)]}{ab(ab\\plus{}bc\\plus{}ca)} \\le 0.$\r\nFrom these inequalities, we obtain the following remarks:\r\n\r\nFor (a), it suffices to prove that\r\n$ \\frac{a^2}{b^2}\\plus{}\\frac{b^2}{a^2} \\ge \\frac{a^2\\plus{}b^2}{ab},$ which is true.\r\n\r\nFor (b), since $ \\frac{(a\\plus{}b\\plus{}c)^2}{2(ab\\plus{}bc\\plus{}ca)} \\equal{}\\frac{a^2\\plus{}b^2\\plus{}c^2}{2(ab\\plus{}bc\\plus{}ca)}\\plus{}1 \\le \\frac{a^2\\plus{}b^2}{2ab}\\plus{}1,$ it suffices to prove that\r\n$ \\frac{a^2}{b^2}\\plus{}\\frac{b^2}{a^2} \\ge \\frac{a^2\\plus{}b^2}{2ab}\\plus{}1,$ which is true.\r\n\r\nFor (c), it suffices to prove that\r\n$ \\frac{a^2}{b^2}\\plus{}\\frac{b^2}{a^2}\\plus{}2 \\ge \\left( \\frac{a^2\\plus{}b^2}{ab}\\right)^2,$ which leads to an equality.", "Solution_2": "[quote=\"can_hang2007\"]Assume that $ c \\equal{} \\min\\{a,b,c\\}.$ Then, we have\n$ \\frac {a^2}{(b \\minus{} c)^2} \\ge \\frac {a^2}{b^2},\\frac {b^2}{(a \\minus{} c)^2} \\ge \\frac {b^2}{a^2},$\nand\n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca} \\minus{} \\frac {a^2 \\plus{} b^2}{ab} \\equal{} \\frac {c[abc \\minus{} (a \\plus{} b)(a^2 \\plus{} b^2)]}{ab(ab \\plus{} bc \\plus{} ca)} \\le 0.$\nFrom these inequalities, we obtain the following remarks:\n\nFor (a), it suffices to prove that\n$ \\frac {a^2}{b^2} \\plus{} \\frac {b^2}{a^2} \\ge \\frac {a^2 \\plus{} b^2}{ab},$ which is true.\n\nFor (b), since $ \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)} \\equal{} \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{2(ab \\plus{} bc \\plus{} ca)} \\plus{} 1 \\le \\frac {a^2 \\plus{} b^2}{2ab} \\plus{} 1,$ it suffices to prove that\n$ \\frac {a^2}{b^2} \\plus{} \\frac {b^2}{a^2} \\ge \\frac {a^2 \\plus{} b^2}{2ab} \\plus{} 1,$ which is true.\n\nFor (c), it suffices to prove that\n$ \\frac {a^2}{b^2} \\plus{} \\frac {b^2}{a^2} \\plus{} 2 \\ge \\left( \\frac {a^2 \\plus{} b^2}{ab}\\right)^2,$ which leads to an equality.[/quote]\r\nVery nice, Can_hang. :lol:\r\n\r\nWhat about this\r\n\r\n$ (d) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\plus{}\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\ge 4$ ?", "Solution_3": "[quote=\"Vasc\"]\n\nWhat about this\n\n$ (d) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2 \\plus{} \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\ge 4$ ?[/quote]\r\nI think my way still works, but it is a little more complicated.\r\n\r\nWithout loss of generality, we may assume that $ c$ is the smallest number among $ a,b,c.$ We will show that\r\n\\[ \\frac{a^{2}}{(b\\minus{}c)^{2}}\\plus{}\\frac{b^{2}}{(a\\minus{}c)^{2}}\\plus{}\\frac{a^{2}\\plus{}b^{2}\\plus{}c^{2}}{ab\\plus{}bc\\plus{}ca}\\geq \\frac{a^{2}}{b^{2}}\\plus{}\\frac{b^{2}}{a^{2}}\\plus{}\\frac{a^{2}\\plus{}b^{2}}{ab}.\\]\r\nIndeed, this inequality is equivalent to\r\n\\[ \\frac{a^{2}}{(b\\minus{}c)^{2}}\\minus{}\\frac{a^{2}}{b^{2}}\\plus{}\\frac{b^{2}}{(a\\minus{}c)^{2}}\\minus{}\\frac{b^{2}}{a^{2}}\\geq \\frac{a^{2}\\plus{}b^{2}}{ab}\\minus{}\\frac{a^{2}\\plus{}b^{2}\\plus{}c^{2}}{ab\\plus{}bc\\plus{}ca},\\]\r\nor\r\n\\[ \\frac{ca^{2}(2b\\minus{}c)}{b^{2}(b\\minus{}c)^{2}}\\plus{}\\frac{cb^{2}(2a\\minus{}c)}{a^{2}(a\\minus{}c)^{2}}\\geq \\frac{c[(a\\plus{}b)(a^{2}\\plus{}b^{2})\\minus{}abc]}{ab(ab\\plus{}bc\\plus{}ca)}.\\]\r\nSince\r\n\\[ \\frac{2b\\minus{}c}{(b\\minus{}c)^{2}}\\geq \\frac{2b\\minus{}2c}{(b\\minus{}c)^{2}}\\equal{}\\frac{2}{b\\minus{}c}\\geq \\frac{2}{b},\\quad\\frac{2a\\minus{}c}{(a\\minus{}c)^{2}}\\geq \\frac{2}{a},\\]\r\nand\r\n\\[ \\frac{(a\\plus{}b)(a^{2}\\plus{}b^{2})\\minus{}abc}{ab(ab\\plus{}bc\\plus{}ca)}\\leq \\frac{(a\\plus{}b)(a^{2}\\plus{}b^{2})}{a^{2}b^{2}},\\]\r\nit suffices to prove that\r\n\\[ \\frac{2ca^{2}}{b^{3}}\\plus{}\\frac{2cb^{2}}{a^{3}}\\geq \\frac{c(a\\plus{}b)(a^{2}\\plus{}b^{2})}{a^{2}b^{2}},\\]\r\nwhich is true because\r\n\\[ \\frac{2a^{2}}{b^{3}}\\plus{}\\frac{2b^{2}}{a^{3}}\\equal{}\\frac{2a^{4}}{a^{2}b^{3}}\\plus{}\\frac{2b^{4}}{a^{3}b^{2}}\\geq \\frac{2(a^{2}\\plus{}b^{2})^{2}}{a^{2}b^{2}(a\\plus{}b)}\\geq \\frac{(a\\plus{}b)(a^{2}\\plus{}b^{2})}{a^{2}b^{2}}.\\]\r\nNow, according to the above inequality, one can reduce the problem into proving that \r\n\\[ \\frac{a^{2}}{b^{2}}\\plus{}\\frac{b^{2}}{a^{2}}\\plus{}\\frac{a^{2}\\plus{}b^{2}}{ab}\\geq 4,\\]\r\nwhich is true.", "Solution_4": "Nice, Can_hang. :lol:\r\nWhat about this? \r\n\r\n$ (e) \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2 \\plus{} \\frac {2(a^2 \\plus{} b^2 \\plus{} c^2)}{ab \\plus{} bc \\plus{} ca}\\ge 6$ ?", "Solution_5": "[quote=\"Vasc\"]Nice, Can_hang. :lol:\nWhat about this? \n\n$ (e) \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2 \\plus{} \\frac {2(a^2 \\plus{} b^2 \\plus{} c^2)}{ab \\plus{} bc \\plus{} ca}\\ge 6$ ?[/quote]\r\n\r\nIn my result, we even have\r\n\r\n$ (f)\\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2 \\plus{} \\frac{9}{4} \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\ge \\frac{13}{2}.$", "Solution_6": "[quote=\"can_hang2007\"][quote=\"Vasc\"]Nice, Can_hang. :lol:\nWhat about this? \n\n$ (e) \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2 \\plus{} \\frac {2(a^2 \\plus{} b^2 \\plus{} c^2)}{ab \\plus{} bc \\plus{} ca}\\ge 6$ ?[/quote]\n\nIn my result, we even have\n\n$ (f)\\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2 \\plus{} \\frac {9}{4} \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\ge \\frac {13}{2}.$[/quote]\r\n\r\nIndeed, the inequality below holds for any distinct nonpositive real numbers $ a,b,c$ and $ \\minus{}6\\le k\\le \\frac {13}{2}$:\r\n\r\n$ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2 \\plus{} \\frac {(k\\minus{}2)(a^2 \\plus{} b^2 \\plus{} c^2)}{2(ab \\plus{} bc \\plus{} ca)}\\ge k$.\r\n\r\nBut I think your above method does not work for $ k>4$. :?:", "Solution_7": "We have best ineqality:\r\n\r\n$ (\\frac {a}{b - c})^{2} + (\\frac {b}{c - a})^{2} + (\\frac {c}{a - b})^{2} + \\frac {m(a^{2} + b^{2} + c^{2})}{ab + bc + ca}$\r\n$ \\ge {\\{\\begin{array}{cc}{-\\frac{m^2+8}{4}}, {m}\\le{-4}, \\\\\\\\2(m + 1), {-4}<{m}\\le{\\frac {9}{4}}, \\\\\\\\\r\n\\frac {4m - 1 + 12 \\sqrt {m}}{4},m > {\\frac {9}{4}}.\\end{array}}$", "Solution_8": "[quote=\"Vasc\"][quote=\"can_hang2007\"][quote=\"Vasc\"]Nice, Can_hang. :lol:\nWhat about this? \n\n$ (e) \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2 \\plus{} \\frac {2(a^2 \\plus{} b^2 \\plus{} c^2)}{ab \\plus{} bc \\plus{} ca}\\ge 6$ ?[/quote]\n\nIn my result, we even have\n\n$ (f)\\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2 \\plus{} \\frac {9}{4} \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\ge \\frac {13}{2}.$[/quote]\n\nIndeed, the inequality below holds for any distinct nonpositive real numbers $ a,b,c$ and $ \\minus{} 6\\le k\\le \\frac {13}{2}$:\n\n$ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2 \\plus{} \\frac {(k \\minus{} 2)(a^2 \\plus{} b^2 \\plus{} c^2)}{2(ab \\plus{} bc \\plus{} ca)}\\ge k$.\n\nBut I think your above method does not work for $ k > 4$. :?:[/quote]\r\nI also think so, Vasc. My way does not work for the case $ k>4$ :(. But in this case, we can prove the inequality as follows:\r\n\r\nFirst, we see that for $ a^2\\plus{}b^2\\plus{}c^2 \\ge 2(ab\\plus{}bc\\plus{}ca),$ the inequality is true according to\r\n$ \\sum \\frac{a^2}{(b\\minus{}c)^2} \\ge 2,$\r\nso it suffices to consider the case $ 2(ab\\plus{}bc\\plus{}ca)>a^2\\plus{}b^2\\plus{}c^2.$ Now, note that\r\n$ \\sum \\frac{a^2}{(b\\minus{}c)^2} \\ge 2\\plus{}\\frac{9}{4} \\frac{2(ab\\plus{}bc\\plus{}ca)\\minus{}a^2\\minus{}b^2\\minus{}c^2}{a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}bc\\minus{}ca}.$\r\n(There is a nice identity here)\r\nTherefore, it suffices to show that\r\n$ \\frac{9}{4} \\frac{2(ab\\plus{}bc\\plus{}ca)\\minus{}a^2\\minus{}b^2\\minus{}c^2}{a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}bc\\minus{}ca} \\plus{}\\frac{(k\\minus{}2)(a^2\\plus{}b^2\\plus{}c^2)}{2(ab\\plus{}bc\\plus{}ca)} \\ge k\\minus{}2,$\r\nwich is equivalent to\r\n$ \\frac{9}{4} \\frac{2(ab\\plus{}bc\\plus{}ca)\\minus{}a^2\\minus{}b^2\\minus{}c^2}{a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}bc\\minus{}ca} \\ge \\frac{k\\minus{}2}{2}\\frac{2(ab\\plus{}bc\\plus{}ca)\\minus{}a^2\\minus{}b^2\\minus{}c^2}{ab\\plus{}bc\\plus{}ca}.$\r\nThis is true because $ \\frac{9}{4} \\ge \\frac{k\\minus{}2}{2}>0,$ $ 2(ab\\plus{}bc\\plus{}ca)\\minus{}a^2\\minus{}b^2\\minus{}c^2>0,$ and\r\n$ ab\\plus{}bc\\plus{}ca>a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}bc\\minus{}ca.$", "Solution_9": "[quote=\"can_hang2007\"] Now, note that\n$ \\sum \\frac {a^2}{(b \\minus{} c)^2} \\ge 2 \\plus{} \\frac {9}{4} \\frac {2(ab \\plus{} bc \\plus{} ca) \\minus{} a^2 \\minus{} b^2 \\minus{} c^2}{a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca}.$\n(There is a nice identity here)[/quote]\r\nThis is a very special method. :wink:", "Solution_10": "$ (d) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\plus{}\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\ge 4$ ?\n[img]http://s6.sinaimg.cn/middle/0018zOAxgy71GdIepXn75&690[/img]\n", "Solution_11": "[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then\n\n$ (a) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\ge \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}$;\n\n[/quote]\n\nWe have\n\\[LHS-RHS=\\frac{1}{2(ab+bc+ca)}\\sum{\\frac{c(a+b)(a^2+ab-2ac+b^2-2bc+c^2)^2}{(-c+a)^2(b-c)^2}}\n+\\frac{1}{2(ab+bc+ca)}\\sum{\\frac{ab(a-b)^2(a+b-c)^2}{(-c+a)^2(b-c)^2}}\\ge{0}\\]\n", "Solution_12": "[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then\n$ (b) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\ge \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)}$.\n[/quote]\n\nWe have\n\\[LHS-RHS=\\frac{1}{8(ab+bc+ca)}\\sum{\\frac{c(3a+3b+c)(a^2+ab-2ac+b^2-2bc+c^2)^2}{(a-c)^2(b-c)^2}}\n+\\frac{5}{8(ab+bc+ca)}\\sum{\\frac{ab(a-b)^2(a+b-c)^2}{(a-c)^2(b-c)^2}}\\ge{0}\\]\n", "Solution_13": "[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then\nMoreover,\n$ (c) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\plus{}2\\ge \\left(\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\right)^2$.[/quote]\n\nWe have\n\\[LHS-RHS=\\frac{abc}{(ab+ac+bc)^2}[\\sum{\\frac{c(b+2c)(a+2c)}{(a-b)^2}}+4(a+b+c)]\\ge{0}\\]", "Solution_14": "[quote=Vasc][quote=\"can_hang2007\"]Assume that $ c \\equal{} \\min\\{a,b,c\\}.$ Then, we have\n$ \\frac {a^2}{(b \\minus{} c)^2} \\ge \\frac {a^2}{b^2},\\frac {b^2}{(a \\minus{} c)^2} \\ge \\frac {b^2}{a^2},$\nand\n$ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca} \\minus{} \\frac {a^2 \\plus{} b^2}{ab} \\equal{} \\frac {c[abc \\minus{} (a \\plus{} b)(a^2 \\plus{} b^2)]}{ab(ab \\plus{} bc \\plus{} ca)} \\le 0.$\nFrom these inequalities, we obtain the following remarks:\n\nFor (a), it suffices to prove that\n$ \\frac {a^2}{b^2} \\plus{} \\frac {b^2}{a^2} \\ge \\frac {a^2 \\plus{} b^2}{ab},$ which is true.\n\nFor (b), since $ \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)} \\equal{} \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{2(ab \\plus{} bc \\plus{} ca)} \\plus{} 1 \\le \\frac {a^2 \\plus{} b^2}{2ab} \\plus{} 1,$ it suffices to prove that\n$ \\frac {a^2}{b^2} \\plus{} \\frac {b^2}{a^2} \\ge \\frac {a^2 \\plus{} b^2}{2ab} \\plus{} 1,$ which is true.\n\nFor (c), it suffices to prove that\n$ \\frac {a^2}{b^2} \\plus{} \\frac {b^2}{a^2} \\plus{} 2 \\ge \\left( \\frac {a^2 \\plus{} b^2}{ab}\\right)^2,$ which leads to an equality.[/quote]\nVery nice, Can_hang. :lol:\n\nWhat about this\n\n$ (d) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\plus{}\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\ge 4$ ?[/quote]\n\n[img]http://s4.sinaimg.cn/middle/006ptkjAzy7627c3HNN03&690[/img]", "Solution_15": "[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then\n\n$ (a) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\ge \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}$;\n\n$ (b) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\ge \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)}$.\n\nMoreover,\n\n$ (c) \\ \\ \\ \\ \\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2\\plus{}2\\ge \\left(\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}\\right)^2$.[/quote]\nFor pairwise distinct nonnegative reals $a,b,c$, prove that\n$$\\left(\\frac a{b \\minus{} c}\\right)^2 \\plus{} \\left(\\frac b{c \\minus{} a}\\right)^2 \\plus{} \\left(\\frac c{a \\minus{} b}\\right)^2>2$$\n[url=https://artofproblemsolving.com/community/c6h1414671p7967479]Canadian Mathematical Olympiad 2017[/url]" } { "Tag": [], "Problem": "When the repeating decimal $ 0.\\overline{12}$ is expressed as a common fraction in lowest terms, what is the sum of its numerator and denominator?", "Solution_1": "Let $ x\\equal{}0.\\overline{12}$.\r\n\r\nThen $ 100x\\equal{}12.\\overline{12}$.\r\n\r\nIf we subtract the first equation from the second, we get $ 99x\\equal{}12$. Thus, $ x\\equal{}\\frac{12}{99}$. Simplifying, we get $ x\\equal{}\\frac{4}{33}$. Adding those yields $ \\boxed{37}$.", "Solution_2": "AIME15 just proved that $ 0.\\overline{ab} \\equal{} \\frac{ab}{99}$.\r\n\r\nThis works for everything:\r\n\r\nIf you have $ x$ digits repeated just like the above, in fractional form, it is the number repeated (in $ x$ digits) over $ x$ $ 9$'s." } { "Tag": [ "puzzles" ], "Problem": "What are the next two terms in this sequence : 594, 487, 566, 493, 310, 447? Please help me. let me know how it was worked out or how to work it out. Your help would be much appreciated. Ta x :D", "Solution_1": "[quote=\"xRobynx\"]What are the next two terms in this sequence : 594, 487, 566, 493, 310, 447? Please help me. let me know how it was worked out or how to work it out. Your help would be much appreciated. Ta x :D[/quote]\r\n\r\n :wacko: This thing has a pattern? It's so crazy. I don't see the pattern. Sorry.", "Solution_2": "What do you need it for? Often, context is necessary to answer something of this sort.", "Solution_3": "Odd... No matches in the Online Encylopedia of Integer Sequences...", "Solution_4": "[quote=\"Danbert\"]Odd... No matches in the Online Encylopedia of Integer Sequences...[/quote]\r\nWow! :o Such thing exists! I didn't know that. Where is it?", "Solution_5": "actually I checked that too but i didn't post :)\r\n\r\njust google for \"sequences\" or \"integer sequences\"", "Solution_6": "[quote=\"lightrhee\"][quote=\"Danbert\"]Odd... No matches in the Online Encylopedia of Integer Sequences...[/quote]\nWow! :o Such thing exists! I didn't know that. Where is it?[/quote]\n\n[url=http://www.learnstuff.com/integer-sequence-resources/]Direct Link[/url]\n\n :lol:", "Solution_7": "[quote=\"Danbert\"][quote=\"lightrhee\"][quote=\"Danbert\"]Odd... No matches in the Online Encylopedia of Integer Sequences...[/quote]\nWow! :o Such thing exists! I didn't know that. Where is it?[/quote]\n\n[url=http://www.research.att.com/~njas/sequences/]Direct Link[/url]\n\n :lol:[/quote]\r\nThis is so cool! :o My sequence from puzzle/brainteaser marathon is also on it.", "Solution_8": "where did you get this sequence? internet? book?", "Solution_9": "[quote=\"Danbert\"][quote=\"lightrhee\"][quote=\"Danbert\"]Odd... No matches in the Online Encylopedia of Integer Sequences...[/quote]\nWow! :o Such thing exists! I didn't know that. Where is it?[/quote]\n\n[url=http://www.research.att.com/~njas/sequences/]Direct Link[/url]\n\n :lol:[/quote]\r\n\r\n\r\n Awesome! Now my posts will increase from 500- 5000!!!!!!!!!!!" } { "Tag": [ "induction" ], "Problem": "P.t:$(1+2+3+...+n)^2=1^3+2^3+....+n^3$\r\n\r\nNo Induction allowed.", "Solution_1": "I think this is proven sometimes on the left hand side thingy.\r\nTheir prove is pretty good.", "Solution_2": "Just use $\\sum_{k=1}^n \\{(k+1)^3-k^3\\}=(n+1)^3-1$", "Solution_3": "[quote=\"shadysaysurspammed\"]P.t:$(1+2+3+...+n)^2=1^3+2^3+....+n^3$\n\nNo Induction allowed.[/quote]\r\n\r\nThey are see it in here:[b]toantuoitho.nxbgd.com.vn[/b]\r\nThis is a web in Vietnam! and it have problem" } { "Tag": [], "Problem": "i am just curious. yes, i mean chapter, not states. our chapter needed a 38 \r\nEDIT: something happened to the poll. just count 31-34 as 34 or less.", "Solution_1": "There is no poll. \r\n\r\nI wish I was in your chapter, then I could go to countdown. Anyways, ours was like a 43 if I remember correctly.....or a 42.", "Solution_2": "Our chapter it was 44.\r\nTop 10 had 46, 46, 45, 45, 45, 44, 44, 44, 44, 44.", "Solution_3": "In our chapter, I think 41 was the cutoff, although you would have to beat some other 41's in tiebreaks. I believe scores were 46, 45, 45, 44, 43, and some 42/41's.", "Solution_4": "i had a 32. and i was in second BEFORE the countdown round. :rotfl:", "Solution_5": "My coach wouldn't show me the results, but all I know is the first place person in my chapter got a 43...\r\n\r\n... and that first place person was me! :first:", "Solution_6": "haha 30 was cutoff.\r\n\r\n44,35,34,33,33,33,32,32,31,30,30,30", "Solution_7": "[quote=\"Yongyi781\"]My coach wouldn't show me the results, but all I know is the first place person in my chapter got a 43...\n\n... and that first place person was me! :first:[/quote]\r\n\r\nJust goes to show that this chapter test was not a good indicator, since Yongyi was 3rd in the state, but would not have made countdown in another chapter in the same state!! That's nuts. Kids in my chapter must just be really good at not making silly mistakes. Interestingly this means our nats team is made up of 4 kids that either finished 1st or tied for first at Chapter.....so maybe it wasn't that bad an indicator :lol:", "Solution_8": "I got 46.\r\n2nd place 42\r\nI estimate about mid 30's.", "Solution_9": "well at my state, The top 4 were\r\n\r\n1. 1st at his chapter\r\n2. (me) 1st at my chapter\r\n3. 1st at his chapter\r\n4. 1st at his chapter\r\n\r\nSo I think that it is a pretty good indicator of how well you are gonna do at states.", "Solution_10": "aaaaaaaaaaaaaaaaaaaa", "Solution_11": "Unless I counted something wrong on the stats sheet, the cutoff at my chapter was a 30.\r\n1st place was a 40.", "Solution_12": "I think the cutoff is around 25-30 ?\r\n\r\nyeah i guess so" } { "Tag": [ "inequalities" ], "Problem": "$ a_1, a_2, ..., a_n$ are positive numbers such that their sum is one. Find the minimum of:\r\n$ a_1/(1 \\plus{} a_2 \\plus{} ... \\plus{} n) \\plus{} a_2/(1 \\plus{} a_1 \\plus{} a_3 \\plus{}...\\plus{}a_n) \\plus{}... \\plus{} a_n/(1 \\plus{} a_1 \\plus{} ... \\plus{} a_{n \\minus{} 1})$(and please prove it!).", "Solution_1": "[quote=\"bengal\"]$ a_1, a_2, ..., a_n$ are positive numbers such that their sum is one. Find the minimum of:\n$ a_1/(1 \\plus{} a_2 \\plus{} ... \\plus{} n) \\plus{} a_2/(1 \\plus{} a_1 \\plus{} a_3 \\plus{} ... \\plus{} a_n) \\plus{} ... \\plus{} a_n/(1 \\plus{} a_1 \\plus{} ... \\plus{} a_{n \\minus{} 1})$(and please prove it!).[/quote]\r\n\r\n[hide=\"Solution\"]\nAssuming you meant to have $ 1\\plus{}a_2\\plus{}\\cdots\\plus{}a_n$ in the denominator of the first term,\nLet $ S\\equal{}\\dfrac{a_1}{1\\plus{}a_2\\plus{}\\cdots\\plus{}a_n}\\plus{}\\dfrac{a_2}{1\\plus{}a_1\\plus{}\\cdots\\plus{}a_n}\\plus{}\\cdots \\plus{}\\dfrac{a_n}{1\\plus{}a_1\\plus{}\\cdots \\plus{}a_{n\\minus{}1}}$.\nWe have that $ a_1\\plus{}a_2\\plus{}\\cdots\\plus{}a_n\\equal{}1$,\nThus we can rewrite the original expression as,\n\\[ S\\equal{}\\sum_{i\\equal{}1}^{n}{\\dfrac{a_i}{2\\minus{}a_i}}\\]\nWe can then add one to each term then subtract $ n$ to get,\n\\[ S\\equal{}\\minus{}n\\plus{}\\sum_{i\\equal{}1}^{n}{\\dfrac{2}{2\\minus{}a_1}}\\]\nTake out a factor of $ 2$ from the sum,\n\\[ S\\equal{}\\minus{}n\\plus{}2\\left(\\sum_{i\\equal{}1}^{n}{\\dfrac{1}{2\\minus{}a_1}}\\right)\\]\nUse Cauchy-Schwarz to show that,\n\\[ (2n\\minus{}1)\\left(\\sum_{i\\equal{}1}^{n}{\\dfrac{1}{2\\minus{}a_1}}\\right)\\ge n^2\\implies \\sum_{i\\equal{}1}^{n}{\\dfrac{1}{2\\minus{}a_1}}\\ge \\dfrac{n^2}{2n\\minus{}1}\\]\nHence,\n\\[ S\\equal{}\\minus{}n\\plus{}2\\left(\\sum_{i\\equal{}1}^{n}{\\dfrac{1}{2\\minus{}a_1}}\\right)\\ge \\minus{}n\\plus{}2\\left(\\dfrac{n^2}{2n\\minus{}1}\\right)\\equal{}\\dfrac{2n^2}{2n\\minus{}1}\\minus{}n\\equal{}\\boxed{\\dfrac{n}{2n\\minus{}1}}\\]\nAnd that's our answer.\nEquality occurs when $ a_1\\equal{}a_2\\equal{}\\cdots\\equal{}a_n\\equal{}\\frac{1}{n}$\n[/hide]", "Solution_2": "Let S be the value we want.\r\nWe use the Cauchy schuarz inequality and we get:\r\n(S)(n-1)>=n^2\r\nS=n^2/n-1", "Solution_3": "What I did was:\r\n\r\n$ (2(a_1 \\plus{} \\cdots \\plus{} a_{n}) \\minus{} (a_{1}^2 \\plus{} \\cdots \\plus{} a_{n}^2))\\frac{a_{1}}{2 \\minus{} {a_1}}\\plus{}\\frac{a_{2}}{2 \\minus{} a_2}\\plus{}\\cdots\\plus{}\\frac{a_{n}}{2 \\minus{} a_{n}} \\geq (a_1 \\plus{} \\cdots \\plus{} a_n)^2$\r\n$ J \\equal{} a_{1}^2 \\plus{} \\cdots \\plus{}a_{n}^2$\r\n$ (2 \\minus{} J)S \\geq 1$\r\n$ 2 \\minus{} 1/S \\geq J \\geq 1/n$.\r\nRearranging gives the desired result." } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "$ (n \\in N*)$ $ f(x) \\equal{} \\sqrt{\\frac{x^{2n\\plus{}1} }{x\\minus{}1}}\\minus{}2x$\r\nfind $ \\lim_{x\\rightarrow \\plus{}\\infty}f(x)$\r\n\r\n\r\nlet $ g(x)\\equal{}\\sqrt{x\\plus{}\\sqrt{x\\plus{}\\sqrt{x}}}\\minus{}\\sqrt{x}$\r\nfind $ \\lim_{x\\rightarrow \\plus{}\\infty}g(x)$", "Solution_1": "[hide=\"Number 2\"]$ \\lim \\left(\\sqrt{x \\plus{} \\sqrt{x \\plus{} \\sqrt{x}}} \\minus{} \\sqrt{x}\\right) \\equal{} \\lim \\frac{\\left(\\sqrt{x \\plus{} \\sqrt{x \\plus{} \\sqrt{x}}} \\minus{} \\sqrt{x}\\right)\\left(\\sqrt{x \\plus{} \\sqrt{x \\plus{} \\sqrt{x}}} \\plus{} \\sqrt{x}\\right)}{\\left(\\sqrt{x \\plus{} \\sqrt{x \\plus{} \\sqrt{x}}} \\plus{} \\sqrt{x}\\right)}$\n\n$ \\equal{} \\lim \\frac{x \\plus{} \\sqrt{x \\plus{} \\sqrt{x}} \\minus{} x}{\\sqrt{x \\plus{} \\sqrt{x \\plus{} \\sqrt{x}}} \\plus{} \\sqrt{x}} \\equal{} \\lim \\frac{\\sqrt{x}\\sqrt{1 \\plus{} \\frac{1}{\\sqrt{x}}}}{\\sqrt{x}\\left(1 \\plus{} \\sqrt{1 \\plus{} \\frac{1}{\\sqrt{x}}\\sqrt{1 \\plus{} \\frac{1}{\\sqrt{x}}}}\\right)}$\n\n$ \\equal{} \\frac{\\sqrt{1 \\plus{} 0}}{1 \\plus{} \\sqrt{1 \\plus{} 0}} \\equal{} \\frac{1}{2}$[/hide]", "Solution_2": "[hide=\"Number 1\"]$ \\lim \\left(\\sqrt{\\frac{x^{2n\\plus{}1}}{x\\minus{}1}} \\minus{} 2x\\right) \\equal{} \\lim \\left(\\sqrt{\\frac{x^{2n}}{1 \\minus{} \\frac{1}{x}}} \\minus{} 2x\\right) \\equal{} \\lim \\left[x^n \\left(1 \\minus{} \\frac{1}{x}\\right)^{\\minus{}1/2} \\minus{} 2x\\right]$\n\n$ \\equal{} \\lim\\,\\, x \\cdot \\left[x^{n\\minus{}1}\\left(1 \\plus{} \\frac{1}{2x}\\right) \\minus{} 2\\right] \\equal{} \\infty$[/hide]" } { "Tag": [ "percent", "search", "combinatorics unsolved", "combinatorics" ], "Problem": "A group of $ 100$ students numbered $ 1$ through $ 100$ are playing the following game. The judge writes the numbers $ 1$, $ 2$, $ \\ldots$, $ 100$ on $ 100$ cards, places them on the table in an arbitrary order and turns them over. The students $ 1$ to $ 100$ enter the room one by one, and each of them flips $ 50$ of the cards. If among the cards flipped by student $ j$ there is card $ j$, he gains one point. The flipped cards are then turned over again. The students cannot communicate during the game nor can they see the cards flipped by other students. The group wins the game if each student gains a point. Is there a strategy giving the group more than $ 1$ percent of chance to win?", "Solution_1": "This was [url=http://www.mathlinks.ro/viewtopic.php?search_id=975696286&t=175877]already posted[/url], more than [url=http://www.mathlinks.ro/viewtopic.php?search_id=975696286&t=191186]once[/url] in fact." } { "Tag": [ "calculus", "integration", "absolute value", "real analysis", "real analysis unsolved" ], "Problem": "Elvaluate, \\[ \\int_{\\minus{}\\infty}^{\\infty}\\dfrac{1}{a^3}\\exp\\bigg(\\minus{}\\dfrac{y^2}{a^2}\\bigg)dy\\] where $ a\\equal{}\\sqrt{1\\minus{}\\dfrac{x^2}{y^2}}$", "Solution_1": "Are we to assume that $ x$ is fixed? How should we understand the square root when the expression under it is negative? Finally, what to do with a terrible (exponential) singularity as $ y\\to x\\minus{}$. Or, perhaps, you just forgot the absolute value in the definition of $ a$? :?" } { "Tag": [ "function" ], "Problem": "tamame tavabe f,g,h:R-->R ra biabid ke:\r\n\r\n[b]f( h(g(x))+y )+g( z+f(y) )=h(y)+x+g( y+f(z) )[/b]", "Solution_1": "Hi, \r\nCould you give the English version?", "Solution_2": "find all the functions $ f,g,h: \\mathbb{R}\\rightarrow \\mathbb{R}$ s.t:\r\n$ f(h(g(x))+y)+g(z+f(y))=h(y)+x+g(y+f(z))$", "Solution_3": "$ f(h(g(x))+y)+g(z+f(y))=h(y)+x+g(y+f(z)) (1)$\r\nif we set $ z=y$ we get\r\n$ f(h(g(x))+y)=h(y)+x (2)$\r\nand if set $ (2)$ in $ (1)$ we'll get \r\n$ g(z+f(y))=g(y+f(z)) (3)$\r\nthen in $ (2)$ take $ x=a+f(b)$\r\n$ f(h(g(a+f(b)))+y)=h(y)+a+f(b)$\r\nand if $ x=b+f(a)$ then\r\n$ f(h(g(b+f(a)))+y)=h(y)+b+f(a)$\r\nso $ a+f(b)=b+f(a) (4)$ (as $ g(a+f(b))=g(b+f(a))$ from $ (3)$)\r\nfrom $ (4)$ we get $ f(x)=x+a$ where $ a=f(0)$\r\nfrom $ (2) h(g(x))+y+a=h(y)+x (5)$\r\n\r\n$ h(g(x))+y_{1}+a=h(y_{1})+x$ and\r\n$ h(g(x))+y_{2}+a=h(y_{2})+x \\Rightarrow$\r\n$ h(y_{1})-y_{1}=h(y_{2})-y_{2}$\r\n$ h(y)=y+b$ where $ b=h(0)$\r\nfrom $ (5)$ we get $ g(x)=x-a$ :wink:\r\n\r\nSo we get\r\n$ f(x)=x+a$\r\n$ g(x)=x-a$ and\r\n$ h(x)=x+b$ where $ a,b\\in\\mathbb{R}$\r\n :wink:" } { "Tag": [ "algebra", "polynomial", "function", "trigonometry", "factorial", "algorithm", "Vieta" ], "Problem": "can we turn a non polynomial into a polynomial ?i mean can we change an equation or an identity that is not polynomial to a polynomial ?", "Solution_1": "$f(x)=e^{x}=1+x+\\frac{x^{2}}{2!}+\\frac{x^{3}}{3!}+\\cdots+\\frac{x^{n}}{n!}+\\cdots ; ~n=0,1,\\ldots$\r\n\r\nSo yes, I suppose if you consider a series a polynomial.", "Solution_2": "Expanding on what skimnc said: look up Taylor polynomials. Any differentiable function can be estimated to some extent as a polynomial, and in some cases you can prove that a continuing Taylor series (like the one above for e, and also the ones for sine and cosine) converge to the actual function with an infinite number of terms.\r\n\r\nAs a brief intro. For a differentiable function F(x), the nth order Taylor approximation centered at the value x=a is:\r\n\r\n$F(x)\\approx{F(a)+F'(a)(x-a)+\\frac{F''(a)}{2!}(x-a)^{2}+\\frac{F'''(a)}{3!}(x-a)^{3}}\\dots$\r\n\r\n$=\\sum_{k=0}^{n}{\\frac{F^{\\text{kth deriv}}(a)}{k!}(x-a)^{k}}$", "Solution_3": "[quote=\"binomial_4eva\"]can we change an equation or an identity that is not polynomial to a polynomial ?[/quote]\r\n\r\nI'm not really sure what you mean. Are you talking about identities such as\r\n\r\n$\\sin^{2}x+\\cos^{2}x = 1$\r\n\r\n? In what meaningful way could you hope to change this into a \"polynomial\" identity? (Aside from the Taylor series expansions, which don't really say much.)", "Solution_4": "i mean can we turn this $\\frac{(n-1)!}{(n-b)!}+\\frac{(n-2)!}{(n-b)!}+\\frac{(n-3)!}{(n-b)!}+...+\\frac{(n-i)!}{(n-b)!}+...+1$ i dont know what you call this, and identity or an equation or expression, but can we turn that to a polynomial ? \r\nor a series, like the one skimnc said for $e^{x}$", "Solution_5": "An [i]identity[/i] and an [i]equation[/i] require that you have two separate [i]expressions[/i] which you identify or equate. What you have there is an expression.\r\n\r\nThis is an example of \"why it's important to actually let people know what you're talking about\" -- expressing the thing you've given us as a polynomial will use special properties of the functions involved; it won't be some general process for magically making things into polynomials (and I'm not sure why you would expect it to be).\r\n\r\nFinally, this post will actually require you to take out a piece of paper and pencil, write things down, and think about them.\r\n\r\nLet's just deal with a single term of your expression. We're going to want to treat $b$ as a fixed constant (let's say, a positive integer bigger than 1) and $n$ as the variable. Then we have, for example $\\frac{(n-1)!}{(n-b)!}$. What does this [i]mean[/i]? Unpack the factorial notation and you should see that many things (including the entire denominator) cancel in a very nice way. What's left is a product of some number of terms, each involving our variable $n$. Now, here's something to be careful of: $n!$ is a product of terms involving a variable $n$, but as $n$ grows, the number of terms in the product increases -- this ensures that we aren't dealing directly with a polynomial. Do we have this problem in our case?\r\n\r\nWhen you've finished with one term, hopefully you can see that the same method can be applied to all the terms; since the sum of polynomials is a polynomial, we're done. You should also be able to see what the degree of the polynomial is. Unfortunately, in general, this polynomial is not easy to write down, i.e. it doesn't have a simple, general form. For a particular choice of $b$, we are now able to figure out what the polynomial is and express it in a conventional fashion, but it hasn't given us a way of expressing it in general without going through this process. (We have an [i]algorithm[/i], but not a [i]formula[/i].) Looking at the result for the first few values of $b$, I don't notice any really nice patterns.", "Solution_6": "JBL thanks i get wat u mean, its hard i know to find a poly for this, but i seriously need a polynomial for this.\r\n\r\n[quote]We're going to want to treat $b$ as a fixed constant (let's say, a positive integer bigger than 1) and $n$ as the variable.[/quote]\r\nlet me add something , $n$ and $b$ are both fixed constant, only the $i$ changes, $i$ is the variable. and ya lets just consider this term, $\\frac{(n-i)!}{(n-b)!}$", "Solution_7": "[quote=\"binomial_4eva\"]JBL thanks i get wat u mean, its hard i know to find a poly for this, but i seriously need a polynomial for this.[/quote]\n\nWhy? It would be easier if you just told us what you were trying to do or what problem you're trying to solve.\n\n[quote=\"binomial_4eva\"]only the $i$ changes, $i$ is the variable. [/quote]\r\n\r\nBut increasing $i$ just adds terms - it doesn't change the terms before it. It's a little like asking for a polynomial representation of $\\sum_{k=1}^{i}k \\cdot k!$. There is a simplified form, yes, but it's not a polynomial in $i$.", "Solution_8": "You should go re-read my post. See the line where I wrote [quote=\"JBL\"]this post will actually require you to take out a piece of paper and pencil, write things down, and think about them.[/quote] These are the important steps you have neglected to do. Sit down and actually write down and think about what I wrote -- right now, you just sound pathetic. I've given you a complete recipe for what to do, [i]but you actually need to do it[/i].\n\n[quote=\"binomial_4eva\"]let me add something , $n$ and $b$ are both fixed constant, only the $i$ changes, $i$ is the variable.[/quote]\r\n\r\nNo. $i$ is a \"dummy variable\" -- it will not appear in the final expression. The expression you are looking for is a polynomial in the variable $n$ with degree that depends on $b$.\r\n\r\nI explained to you what to do for $\\frac{(n-1)!}{(n-b)!}$, but this explanation wasn't specific to $n-1$ -- as I said, you need to do exactly the same thing for each other term (for each different value of $i$).", "Solution_9": "$(n-1)!=\\frac{n!}{n}$\r\n\r\n$(n-b)!=\\frac{n!}{n(n-1)(n-2)\\cdots\\!(n-(b-1))}$\r\n\r\n$\\frac{(n-1)!}{(n-b)!}=\\frac{(\\frac{n!}{n})}{(\\frac{n!}{n(n-1)(n-2)\\cdots\\!(n-(b-1))})}=\\frac{n!n(n-1)(n-2)\\cdots\\!(n-(b-1))}{n!n}=(n-1)(n-2)\\cdots\\!(n-(b-1))$\r\n\r\nnow $\\frac{(n-1)!}{(n-b)!}=(n-1)(n-2)(n-3)\\dots\\!(n-(b-1))$ then how do i put this $(n-1)(n-2)(n-3)\\cdots\\!(n-(b-1))$ in to polynomial.\r\n\r\n$\\frac{(n-1)!}{(n-b)!}+\\frac{(n-2)!}{(n-b)!}+\\cdots\\!+\\frac{(n-i)!}{(n-b)!}+\\cdots\\!+1=(n-1)(n-2)\\cdots\\!(n-(b-1))+(n-2)(n-3)\\cdots\\!(n-(b-1))+\\cdots\\!+(n-i)(n-(i+1))\\cdots\\!(n-(b-1))+\\cdots\\!+1$\r\nam i right?,anyone agree with what i done\r\nhow do i expand this $(n-1)(n-2)\\cdots\\!(n-(b-1))$?\r\nfrom here on iam stuck, i know we can use vieta's formula to complete this,but how? anyone knows how ?", "Solution_10": "Up until you wrote [quote=\"binomial_4eva\"]$(n-1)(n-2)\\cdots\\!(n-i_{b})= n^{b}-3n+2$[/quote] you were doing fine. (Although I'm not sure what this \"$i_{b}$\" thing is -- it should just be $b-1$.) The part I quoted is just wrong. Of course, you already have it as a polynomial (as I noted before, a polynomial in the variable $n$) -- it just happens that your polynomial is in factored form. It's easy to say what the coefficients are, since we have the roots -- they are the appropriate symmetric sums of a set of consecutive positive integers with a power of $-1$ tossed in (this is what Vieta's says) -- but there aren't nice closed forms for these expressions in general. Of course, you now have all the tools to write out your polynomial for any [i]fixed[/i] value of $b$.\r\n\r\nOne other error: in the general term of your sum in the last equation you've written, you have $(n-i)(n-(i+1))$ -- it should be (using your notation) $(n-i)(n-(i+1))\\cdots(n-i_{b})$.", "Solution_11": "thanks JBL, $i_{b}$ was $b-1$ but neway i edited it.\r\n$(n-1)(n-2)\\cdots\\!(n-(b-1))=n^{b}-(1+2+\\cdots\\!+(b-1))n^{b-1}+(2+3+\\cdots\\!+(b-1)b)n^{b-2}+\\cdots\\!+(-1)^{b}(b-1)!$\r\nwhat will these look like if we expand out? $(n-2)(n-3)\\cdots\\!(n-(b-1))$ and $(n-3)(n-4)\\cdots\\!(n-(b-1))$ and $(n-4)(n-5)\\cdots\\!(n-(b-1))$\r\nwhat is the formula for symmetric sums of a set of consecutive positive integers?", "Solution_12": "You still haven't explained why you need a polynomial." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ x,y,z$ be non-negative real numbers. Prove that\r\n\\[ (8x^{2}+yz)(8y^{2}+zx)(8z^{2}+xy)\\leq (x+y+z)^{6}.\\]", "Solution_1": "[quote=\"pvthuan\"]Let $ x,y,z$ be non-negative real numbers. Prove that\n\\[ (8x^{2}+yz)(8y^{2}+zx)(8z^{2}+xy)\\leq (x+y+z)^{6}. \\]\n[/quote]\r\nExpanding kills it by Schur and Muirhead.", "Solution_2": "yeah, but solution will be simpler if you use substitution: $ p=a+b+c,q=ab+bc+ca,r=abc$ and assume that $ p=a+b+c=3$ ... :roll:", "Solution_3": "I found this inequality in my archive, with the solution\r\n\r\n$ E(x,y,z \\le E(x,a,a)\\le E(b,b,b)$,\r\n\r\nwhere E(x,y,z) is the left side, $ a=\\frac{y+z}{2}$ and $ b=\\frac{x+y+z}{3}$, on the assumption $ x\\le y\\le z$.\r\nAlso, I found the related inequalities:\r\n\r\n\\[ (2x^{2}+7yz)(2y^{2}+7zx)(2z^{2}+7xy)\\geq 27(xy+yz+zx)^{3}\\]\r\nand \r\n\r\n\\[ (5x^{2}+yz)(5y^{2}+zx)(5z^{2}+xy)\\geq 8(xy+yz+zx)^{3}. \\]", "Solution_4": "[quote=\"Vasc\"]I found this inequality in my archive, with the solution\n\n$ E(x,y,z \\le E(x,a,a)\\le E(b,b,b)$,\n\nwhere E(x,y,z) is the left side, $ a=\\frac{y+z}{2}$ and $ b=\\frac{x+y+z}{3}$, on the assumption $ x\\le y\\le z$.\nAlso, I found the related inequalities:\n\\[ (2x^{2}+7yz)(2y^{2}+7zx)(2z^{2}+7xy)\\geq 27(xy+yz+zx)^{3}\\]\nand\n\\[ (5x^{2}+yz)(5y^{2}+zx)(5z^{2}+xy)\\geq 8(xy+yz+zx)^{3}. \\]\n[/quote]\r\nthey are very nice problems, vasc and pvthuan :wink: \r\nI think all of them can be solve by a technique. I used to elementary expressions to solve them.\r\nI will post one of them, after that, you can solve remain inequalities easily.\r\nWe put $ p=a+b+c,q=ab+bc+ca,r=abc$. We can assume that $ q=ab+bc+ca=3$\r\nBy expanding expression, we have\r\n\\[ (2x^{2}+7yz)(2y^{2}+7zx)(2z^{2}+7xy)\\geq 27(xy+yz+zx)^{3}\\]\r\n\r\n\r\n\r\n\\[ \\iff 729r^{2}+27+98r(p^{3}-12p)+42pr\\ge 0 \\]\r\n\r\n\\[ \\iff 729r^{2}+27-882r^{2}+126r^{2}\\ge 0 \\]\r\n\r\n\\[ \\iff 1\\ge r \\]\r\nwhich is clearly true, because:\r\n\\[ p=3\\ge 3\\sqrt[3]{r^{2}}\\to r\\le 1 \\]\r\nOn the other hand, by Schur's inequality:\r\n\\[ p^{3}-12p\\ge-9r \\]\r\n\r\n\\[ pr\\ge 3r^{2}\\]\r\nWe are done", "Solution_5": "[quote=\"zaizai-hoang\"][quote=\"Vasc\"].....Also, I found the related inequalities:\n\\[ (2x^{2}\\plus{}7yz)(2y^{2}\\plus{}7zx)(2z^{2}\\plus{}7xy)\\geq 27(xy\\plus{}yz\\plus{}zx)^{3}\\]\nand\n\\[ (5x^{2}\\plus{}yz)(5y^{2}\\plus{}zx)(5z^{2}\\plus{}xy)\\geq 8(xy\\plus{}yz\\plus{}zx)^{3}.\\]\n[/quote]\nthey are very nice problems, vasc and pvthuan :wink: \nI think all of them can be solve by a technique. I used to elementary expressions to solve them.\nI will post one of them, after that, you can solve remain inequalities easily.\nWe put $ p \\equal{} a\\plus{}b\\plus{}c,q \\equal{} ab\\plus{}bc\\plus{}ca,r \\equal{} abc$. We can assume that $ q \\equal{} ab\\plus{}bc\\plus{}ca \\equal{} 3$\nBy expanding expression, we have\n\\[ (2x^{2}\\plus{}7yz)(2y^{2}\\plus{}7zx)(2z^{2}\\plus{}7xy)\\geq 27(xy\\plus{}yz\\plus{}zx)^{3}\\]\n\n\n\n\n\\[ \\iff 729r^{2}\\plus{}27\\plus{}98r(p^{3}\\minus{}12p)\\plus{}42pr\\ge 0\\]\n\n\\[ \\iff 729r^{2}\\plus{}27\\minus{}882r^{2}\\plus{}126r^{2}\\ge 0\\]\n\n\\[ \\iff 1\\ge r\\]\nwhich is clearly true, because:\n\\[ p\\equal{}3\\ge 3\\sqrt[3]{r^{2}}\\to r\\le 1\\]\nOn the other hand, by Schur's inequality:\n\\[ p^{3}\\minus{}12p\\ge\\minus{}9r\\]\n\n\\[ pr\\ge 3r^{2}\\]\nWe are done[/quote]\r\n\r\nDear ZaiZai, for the second inequality of Vasc\r\n\r\n \\[ (5x^{2}\\plus{}yz)(5y^{2}\\plus{}zx)(5z^{2}\\plus{}xy)\\geq 8(xy\\plus{}yz\\plus{}zx)^{3}.\\]\r\n\r\nI found using your method\r\n\r\n$ 216r^{2}\\plus{}459\\minus{}270pr\\plus{}5p^{3}r\\geq 0$\r\n\r\nwith $ p\\equal{}a\\plus{}b\\plus{}c$, $ q\\equal{}ab\\plus{}bc\\plus{}ca\\equal{}3$ and $ r\\equal{}abc$\r\n\r\nI tried to complete it using SHUR applied to $ ab,bc,ca$ and $ a,b,c$, that is\r\n\r\n$ 27\\plus{}9r^{2}\\geq 12pr$\r\n\r\n$ p^{3}\\plus{}9r\\geq 12p$\r\n\r\nbut I failed :blush: Can you, please, help me to complete it?\r\n\r\nThank you very much." } { "Tag": [ "calculus", "integration", "real analysis", "calculus computations" ], "Problem": "I need help on this Calculus problem for my Calc AB class. I just started work problems today so I am not sure what to do. \r\n\r\nA cable that weighs 2 lb/ft is used to lift 800 lb of coal up a mineshaft 500 ft deep. \r\n-Show how to approximate the required work by a Reimann Sum\r\n-Then, express the work as an integral and evaluate it. \r\n\r\nThanks.", "Solution_1": "Suppose you lift the bucket from height $x$ above the bottom of the shaft to height $x+\\Delta x.$ Work equals force times distance. The distance is $\\Delta x.$ What is the force? The force would be the weight of the bucket plus the weight of the cable. The bucket weighs 800 lbs. The hanging length of the cable is changing but when the bucket height is $x,$ we have $500-x$ feet of cable weighing $1000-2x$ pounds. Thus, at least for a moment, the weight (and hence the force) is $800+1000-2x=1800-2x$ pounds. Multiply by the distance, and the force times distance is approximately (approximately because the force is changing) $(1800-2x)\\,\\Delta x.$\r\n\r\nSum over all the increments of distance over sum partition and we get that \r\n\r\nWork $\\approx\\sum(1800-2x)\\,\\Delta x$\r\n\r\nThat's a Riemann sum; taking the limit at the distance increments (the mesh of the partition) go to zero, we get that \r\n\r\nWork $=\\int_{0}^{500}(1800-2x)\\,dx.$", "Solution_2": "This is not the only valid way to set it up; we could also look at a section of cable and ask how far that section has to go. In that method, the coal is handled separately." } { "Tag": [ "quadratics", "number theory proposed", "number theory" ], "Problem": "Show that the number of $2$-tuples $(n,n+1)$, twin residues, with both $n$ and $n+1$ being quadratic residues modulo $p$ is exactly \r\n\r\n\\[\\frac{p - {-1 \\overwithdelims () p}}{4} - 1.\\]", "Solution_1": "The leading ideas can be found here:\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=4891[/url]", "Solution_2": "Yes I saw that thread and decided to post this in hope of gaining some new proofs. \r\n\r\nMy proof goes like this, we consider the sum of $1 + \\left(\\frac{n}{p}\\right) + \\left(\\frac{n+1}{p}\\right) + \\left(\\frac{n(n+1)}{p}\\right)$. Then obviously that is $4$ times what we are searching for and the last term (or sum) can be evaluated easily by noting that $\\left(\\frac{n(n+1)}{p}\\right) = \\left(\\frac{n^{-1} + 1}{p}\\right)$.\r\n\r\n(Okay, after a closer read even this was more or less what grobber wrote earlier)", "Solution_3": "This problem was already discussed and solved. It is in the unsolved section and its name is: at most p+1/4 consecutive cuadratic residues.......I dont remember who solved it, but he used all the hints given in the book (the presceding problems proposed there)" } { "Tag": [], "Problem": "$(x+1)(x+2)(x+3)(x+4) =-1$", "Solution_1": "[hide]\nUse $n(n+1)(n+2)(n+3)=(n^{2}+3n+1)^{2}-1$\n[/hide]", "Solution_2": "[hide]We have\n$(x+1)(x+2)(x+3)(x+4)=((x+1)^{2}+3(x+1)+1)^{2}-1=$\n$(x^{2}+5x+5)^{2}-1=-1$\n$\\Longrightarrow x^{2}+5x+5=0\\Longrightarrow x=\\frac{-5\\pm \\sqrt{5}}{2}.$[/hide]", "Solution_3": "Ascertain : \r\n\r\n$1\\blacktriangleright\\min_{x\\in \\mathrm R}\\ (x+1)(x+2)(x+3)(x+4)\\ .$\r\n\r\n$2\\blacktriangleright \\{\\lambda\\in\\mathrm R\\ |\\ \\mathrm{the\\ equation\\ }(x+1)(x+2)(x+3)(x+4)=\\lambda\\ \\mathrm{has\\ at\\ least\\ one\\ real\\ root}\\}\\ .$", "Solution_4": "[quote=\"Virgil Nicula\"]Ascertain : \r\n\r\n$1\\blacktriangleright\\min_{x\\in \\mathrm R}\\ (x+1)(x+2)(x+3)(x+4)\\ .$\r\n\r\n[hide=\"Answer\"]min=-1\\ $\\left(x=\\frac{-5\\pm \\sqrt{5}}{2}\\right).$[/hide]\n\n$2\\blacktriangleright \\{\\lambda\\in\\mathrm R\\ |\\ \\mathrm{the\\ equation\\ }(x+1)(x+2)(x+3)(x+4)=\\lambda\\ \\mathrm{has\\ at\\ least\\ one\\ real\\ root}\\}\\ .$\n\n[color=red]Good Practice[/color]\n\nThe number of real roots are folllows.\n\n[hide=\"Answer\"]\n$\\lambda <-1\\ \\cdots 0.$\n$\\lambda =-1\\ \\cdots 1.$\n$-1<\\lambda \\leq \\frac{9}{16}\\ \\cdots 2.$\n$\\lambda >\\frac{9}{16}\\ \\cdots 1$[/hide]", "Solution_5": "[quote=\"miyomiyo\"]$(x+1)(x+2)(x+3)(x+4) =-1$[/quote]\r\n[hide=\"roar\"]\n$x+2.5=k$\n$(k-1.5)(k-.5)(k+.5)(k+1.5)=-1$\n$(k^{2}-2.25)(k^{2}-.25)=-1$\n$k^{2}=a$\n$a^{2}-2.5a+\\frac{25}{16}=0$\n$a=\\frac{2.5 \\pm 0}{2}$\n$a=1.25$\n$k=\\frac{\\pm\\sqrt{5}}{2}$\n$x=\\frac{-5 \\pm \\sqrt{5}}{2}$[/hide]", "Solution_6": "So did I :D" } { "Tag": [], "Problem": "time pass guys , jus giving for kicks....\r\n\r\n$a+b+c=1$\r\n\r\n$a^2+b^2+c^2=2$\r\n\r\n$a^3+b^3+c^3=3$\r\n\r\n$a^4+b^4+c^4=???$", "Solution_1": "I'm sure it isn't 4, but... is the answer rational?\r\n\r\nMaybe I'm just overestimating the problem.", "Solution_2": "[hide]$(a+b+c)^2=a^2+b^2+c^2 + 2(ab +ac+bc) \\implies (ab +ac+bc)=-\\frac{1}{2}$\n\n$(a+b+c)^3=a^3 + b^3 + c^3 + 3a(ab + ac) + 3b(ab + bc) + 3c(ac+bc) + 6abc$, consequently:\n\n$(a+b+c)^3=a^3 + b^3 + c^3 + 3a(-\\frac{1}{2} - bc) + 3b(-\\frac{1}{2} - ac) + 3c(-\\frac{1}{2} - ab) + 6abc$\n\n$ \\implies (a+b+c)^3=a^3 + b^3 + c^3 - 3\\frac{1}{2}(a+b+c) - 3abc \\implies abc=\\frac{1}{6}$\n\nnow let's find $a^4 + b^4 + c^4$:\n\n$3 \\cdot 1=(a^3 + b^3 + c^3)(a+b+c)=a^4 + b^4 + c^4 + a^2(ab + ac) + b^2(ab+bc) + c^2(ac + bc)$ factoring analogously:\n\n$3=a^4 + b^4 + c^4 -\\frac{1}{2}(a^2+b^2+c^2) - (a + b + c)(abc)$\n\n$\\implies a^4 + b^4 + c^4=3 + 1 + \\frac{1}{6}=\\frac{25}{6}$[/hide]\r\n\r\n[color=green][size=75]mod edit: use \\cdot for multiplication, not \".\" and use \\implies, not :arrow: ;)\n\noh, and hide your solutions :P [/size][/color]", "Solution_3": "yup gabriel ur right :lol:", "Solution_4": "Oh crap,i found it till abc=1/6 and then got lost............................\r\nI cant understand whaatya did after that newmember,factorisin analougously? :?", "Solution_5": "Could ya'll please not use c***? It's not a nice word. Maybe in the title instead of c****y use ambagious, meaning difficult, complicated? It's a much more impressive word then c***.", "Solution_6": "I am soo dumb....\r\n[hide=\"Another way\"]\n$(a^2+b^2+c^2)^2=4$\n$a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)=4$\n$(a+b+c)^2=1$\n$a^2+b^2+c^2+2(ab+bc+ca)=1$\n$ab+bc+ca=\\frac{-1}{2}$\n$(ab+bc+ca)^2=\\frac{1}{4}$\n$a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)=\\frac{1}{4}$\n$a^2b^2+b^2c^2+a^2c^2+\\frac{1}{3}=\\frac{1}{4}$\n$a^2b^2+b^2c^2+a^2c^2=\\frac{-1}{12}$\nInputting in one above equation,\n$a^4+b^4+c^4-\\frac{1}{6}=4$\n$a^4+b^4+c^4=\\frac{25}{6}$\nAnswer=$\\boxed{\\frac{25}{6}}$\n[/hide]", "Solution_7": "[quote=\"solafidefarms\"]Could ya'll please not use c***? It's not a nice word. Maybe in the title instead of c****y use ambagious, meaning difficult, complicated? It's a much more impressive word then c***.[/quote]\r\n\r\nRequest granted :D", "Solution_8": "But still, it's A-M-B-I-G-U-O-U-S, not A-M-B-A-G-I-O-U-S. Sorry about pointing out the subtlety, but it would be more impressive if it were spelled correctly...... ;) :D", "Solution_9": "It would be more impressive if you had a dictionary... http://www.google.com/search?hl=en&lr=&q=define%3Aambagious\r\n\r\nNever argue with a National Word Power Champion. ;) \r\n\r\n:P", "Solution_10": "Oh, I was thinking of \"ambiguous\". Sorry :blush:\r\n\r\nNow I have a new addition to my vocabulary!!! :w00t:\r\n\r\nBut back to the original problem: Is there a shorter way?", "Solution_11": "[quote=\"Chess64\"]It would be more impressive if you had a dictionary... http://www.google.com/search?hl=en&lr=&q=define%3Aambagious \n\nNever argue with a National Word Power Champion. \n ;) [/quote]\r\n\r\nChess64, that's great! :D", "Solution_12": "\"\"Could ya'll please not use c***? It's not a nice word. Maybe in the title instead of c****y use ambagious, meaning difficult, complicated? It's a much more impressive word then c***.\"\" \r\n\r\n\r\n lol , is crappy sorry c***** that bad? sorry i didnt know. :D no probs in changin the title at all", "Solution_13": "Riddler,in ur sol,in one part,u took the value of abc to be 1/6,u didnt prove it ;) Same method here,riddler,but then i got abc diiferent than hta the new member got.i initially got abc=1/6,but on recalculatin it came different,any other method for calculatin abc=?", "Solution_14": "[quote=\"4everwise\"][quote=\"Chess64\"]It would be more impressive if you had a dictionary... http://www.google.com/search?hl=en&lr=&q=define%3Aambagious \n\nNever argue with a National Word Power Champion. \n ;) [/quote]\n\nChess64, that's great! :D[/quote]\r\n\r\nEr, actually, if you didn't know, solafidefarms is the National Word Power Champion :P lol" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$ a,b,c > 0$ and $ x\\geq\\frac {a \\plus{} b \\plus{} c}{3\\sqrt {3}} \\minus{} 1$ \r\n\r\nprove that\r\n\\[ \\frac {(b \\plus{} cx)^2}{a} \\plus{} \\frac {(c \\plus{} ax)^2}{b} \\plus{} \\frac {(a \\plus{} bx)^2}{c}\\geq{abc}\r\n\\]", "Solution_1": "Cauchy-Schwartz engel-form tells us L.H.S.$ \\ge$\r\n\r\n$ \\frac {((1 \\plus{} x)(a \\plus{} b \\plus{} c))^2}{a \\plus{} b \\plus{} c} \\equal{} (1 \\plus{} x)^2(a \\plus{} b \\plus{} c)\\ge (\\frac {a \\plus{} b \\plus{} c}{3\\sqrt {3}})^2(a \\plus{} b \\plus{} c) \\equal{} (\\frac {a \\plus{} b \\plus{} c}{3})^3\\ge abc$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "if $ a,b \\geq 0$, from Cauchy-Schwarz, we know that: \r\n\r\n$ \\frac {1}{a^2 \\plus{} ab} \\plus{} \\frac {1}{b^2 \\plus{} ab} \\geq \\frac {4}{(a \\plus{} b)^2}$\r\n\r\nWe have the stronger inequality: \r\n\r\n$ \\frac {1}{a\\sqrt {2(a^2 \\plus{} b^2)}} \\plus{} \\frac {1}{b\\sqrt {2(a^2 \\plus{} b^2)}} \\geq \\frac {4}{(a \\plus{} b)^2}$\r\n\r\nProve it. \r\n\r\n :)", "Solution_1": "[quote=\"shaaam\"]if $ a,b \\geq 0$, from Cauchy-Schwarz, we know that: \n\n$ \\frac {1}{a^2 \\plus{} ab} \\plus{} \\frac {1}{b^2 \\plus{} ab} \\geq \\frac {4}{(a \\plus{} b)^2}$\n\nWe have the stronger inequality: \n\n$ \\frac {1}{a\\sqrt {2(a^2 \\plus{} b^2)}} \\plus{} \\frac {1}{b\\sqrt {2(a^2 \\plus{} b^2)}} \\geq \\frac {4}{(a \\plus{} b)^2}$\n\nProve it. \n\n :)[/quote]\r\nIt is equivalent to\r\n$ (a\\plus{}b)^3 \\ge 4ab\\sqrt{2(a^2\\plus{}b^2)}$\r\nOn the other hand: $ 4ab\\sqrt{2(a^2\\plus{}b^2)} \\equal{}4\\sqrt{ab}.\\sqrt{a^2\\plus{}b^2}.\\sqrt{2ab} \\le 4.\\frac{a\\plus{}b}{2}.\\frac{(a\\plus{}b)^2}{2} \\equal{}(a\\plus{}b)^3$\r\nThen, we have Q.E.D :wink: :)", "Solution_2": "[quote=\"shaaam\"]if $ a,b \\geq 0$, from Cauchy-Schwarz, we know that: \n\n$ \\frac {1}{a^2 \\plus{} ab} \\plus{} \\frac {1}{b^2 \\plus{} ab} \\geq \\frac {4}{(a \\plus{} b)^2}$\n\nWe have the stronger inequality: \n\n$ \\frac {1}{a\\sqrt {2(a^2 \\plus{} b^2)}} \\plus{} \\frac {1}{b\\sqrt {2(a^2 \\plus{} b^2)}} \\geq \\frac {4}{(a \\plus{} b)^2}$\n\nProve it. \n\n :)[/quote]\r\nThe following inequality is true too.\r\n\\[ \\frac {1}{a\\sqrt [3] {4(a^3 \\plus{} b^3)}} \\plus{} \\frac {1}{b\\sqrt [3]{4(a^3 \\plus{} b^3)}} \\geq \\frac {4}{(a \\plus{} b)^2}\r\n\\]\r\nfor positive $ a$ and $ b.$\r\n\r\nMaybe the following similar question is interesting.\r\nLet $ a,$ $ b$ and $ c$ are positive numbers such that $ a\\plus{}b\\plus{}c\\equal{}3.$\r\nProve that if the inequality\r\n\\[ \\frac{1}{a^2\\plus{}2bc}\\plus{}\\frac{1}{b^2\\plus{}2ac}\\plus{}\\frac{1}{c^2\\plus{}2ab}\\geq\\left(\\frac{a^2\\plus{}b^2\\plus{}c^2}{3}\\right)^k\\]\r\nis true then, $ k\\leq0.$", "Solution_3": "Let $ a,b \\geq 0.$ Prove that$$ \\frac {1}{a^2 + ab} + \\frac {4}{b^2 + ab} \\geq \\frac {9}{(a + b)^2}$$\n$$\\frac {1}{a^2 + ab}+\\frac {2}{b^2 + ab} \\geq \\frac {3+2\\sqrt 2}{(a + b)^2}$$\n$$ \\frac {1}{a^2 + ab} + \\frac {1}{b^2 + ab} \\geq \\frac {8}{(a + 2b)^2}$$\n$$ \\frac {1}{a^2 + ab} + \\frac {1}{2b^2 + ab} \\geq \\frac {\\sqrt {11+5\\sqrt {5}}}{\\sqrt 2(a + b)^2}$$", "Solution_4": "Let $ a,b,c\\geq 0.$ Prove that$$\\frac {1}{a^3 + abc} + \\frac {1}{b^3 + abc} + \\frac {1}{c^3 + abc}\\geq \\frac {81}{2(a + b+c)^3 }$$ ", "Solution_5": "[quote=arqady]Maybe the following similar question is interesting.\nLet $ a,$ $ b$ and $ c$ are positive numbers such that $ a\\plus{}b\\plus{}c\\equal{}3.$\nProve that if the inequality\n\\[ \\frac{1}{a^2\\plus{}2bc}\\plus{}\\frac{1}{b^2\\plus{}2ac}\\plus{}\\frac{1}{c^2\\plus{}2ab}\\geq\\left(\\frac{a^2\\plus{}b^2\\plus{}c^2}{3}\\right)^k\\]\nis true then, $ k\\leq0.$[/quote]\n\nI don't think it's true. We can have $a=1,b=0.5,c=1.5$ and $k=0.05$.", "Solution_6": "[quote=jasperE3][quote=arqady]Maybe the following similar question is interesting.\nLet $ a,$ $ b$ and $ c$ are positive numbers such that $ a\\plus{}b\\plus{}c\\equal{}3.$\nProve that if the inequality\n\\[ \\frac{1}{a^2\\plus{}2bc}\\plus{}\\frac{1}{b^2\\plus{}2ac}\\plus{}\\frac{1}{c^2\\plus{}2ab}\\geq\\left(\\frac{a^2\\plus{}b^2\\plus{}c^2}{3}\\right)^k\\]\nis true then, $ k\\leq0.$[/quote]\n\nI don't think it's true. We can have $a=1,b=0.5,c=1.5$ and $k=0.05$.[/quote]\nId est, $0.05\\leq0$ :maybe: \n\n", "Solution_7": "[quote=arqady]\nThe following inequality is true too.\n\\[ \\frac {1}{a\\sqrt [3] {4(a^3 \\plus{} b^3)}} \\plus{} \\frac {1}{b\\sqrt [3]{4(a^3 \\plus{} b^3)}} \\geq \\frac {4}{(a \\plus{} b)^2}\n\\]\nfor positive $ a$ and $ b.$[/quote]\nSplendor. Set WLOG $a^3+b^3=2.$ So if $a+b:=2s,$ then $s\\in\\left(\\frac1{\\sqrt[3]4},1\\right]$ and $ab=\\frac{4s^3-1}{3s}.$\nWe need to prove\n$$\\frac s{ab}\\geq\\frac1{s^2}\\iff3s^4+1\\geq4s^3\\iff s^4+s^4+s^4+1\\geq4\\sqrt[4]{s^4\\cdot s^4\\cdot s^4\\cdot1}=4s^3.$$" } { "Tag": [ "inequalities", "AMC", "USA(J)MO", "USAMO", "number theory", "prime factorization" ], "Problem": "Let $ a$ and $ b$ be positive integers such that $ a|b^2,b^2|a^3,a^3|b^4,b^4|a^5,...$ Prove that $ a\\equal{}b$.\r\n\r\n[hide=\"my sketch proof (try the problem yourself first!)\"]\n\nIf k is odd and $ a^k|b^{k\\plus{}1},a^{k\\plus{}2}|b^{k\\plus{}3}$ then $ a^k*a^{k\\plus{}2}\\equal{}a^{2k\\plus{}2}|b^{2k\\plus{}4}$\n$ \\implies a^{k\\plus{}1}|b^{k\\plus{}2}$.\n\nSimilarly if m is even and $ b^m|a^{m\\plus{}1},b^{m\\plus{}2}|a^{m\\plus{}3}$ then we have $ b^{m\\plus{}1}|a^{m\\plus{}2}$\n\nSo the problem statement is equivalent to $ a|b^2, a^2|b^3,a^3|b^4, ...$\nWe shall prove that $ a|b$ by contradiction:\nAssume a does not divide b, then there's a prime factor $ p$ in $ a$ such that its power $ A$ in the prime factorization is higher than that of b, denoted by $ B$. Then from $ a|b^2$ we required $ 2B\\ge A$, and from $ a^2|b^3$ we require $ 3B\\ge2A\\iff 2B\\ge A\\plus{}(A\\minus{}B)$. In the same manner we require $ 2B\\ge A\\plus{}2(A\\minus{}B)$ and $ 2B\\ge A\\plus{}x(A\\minus{}B)$ where $ A\\minus{}B>0$ Therefore we can choose a sufficiently large $ x$ such that the inequality does not hold. Contradiction.\n\nLikewise we can prove $ b|a$ in the same manner as above.\n\nSo is this proof rigorous enough? Are we allowed to omit the parts of proof that are very similar and just write the proof is similar in a proof-writing contest (ex. usamo)? Also is there an easier way to prove $ a|b$ and $ b|a$ as I feel my proof is quite long-winded...[/hide]", "Solution_1": "Instead of saying \"in the same matter,\" you can just say \"from $ a^k | b^{k\\plus{}1}$ we require...\"\r\n\r\nAnyway, there's a [hide=\"shorter proof.\"] $ x | y$ implies that $ x \\le y$. [/hide]", "Solution_2": "[hide=\"Blah blah blah\"]\nThe conditions $ a|b^2,a^3|b^4,\\dots,a^{2k \\minus{} 1}|b^{2k}$ imply that $ \\frac {b^{2k}}{a^{2k \\minus{} 1}} \\equal{} b\\left(\\frac ba\\right)^{2k \\minus{} 1}\\in \\mathbb Z$ for all $ k$, therefore $ a|b$.\n\nThe conditions $ b^2|a^3,b^4|a^5,\\dots,b^{2k}|a^{2k \\plus{} 1}$ imply that $ \\frac {a^{2k \\plus{} 1}}{b^{2k}} \\equal{} a\\left(\\frac ab\\right)^{2k}\\in \\mathbb Z$ for all $ k$, therefore $ b|a$.\n\nIf $ a|b$ and $ b|a$, we have $ a \\equal{} b$.\n[/hide]", "Solution_3": "Yes $ x|y$ implies $ x\\leq y$ but I can not see from here why $ x\\equal{}y$ :blush:", "Solution_4": "[quote=\"enndb0x\"]Yes $ x|y$ implies $ x\\leq y$ but I can not see from here why $ x \\equal{} y$ :blush:[/quote]\r\n[hide=\"Solution\"] $ a^k | b^{k\\plus{}1}$ implies that $ a \\le b^{1 \\plus{} \\frac{1}{k} }$ for every positive integer $ k$. For sufficiently large $ k$, this implies $ a \\le b$. Similarly, $ b^k | a^{k\\plus{}1}$ for sufficiently large $ k$ implies $ b \\le a$. [/hide]", "Solution_5": "Lets look at the valuation mod $ p$ (is there a better way to phrase this...I've just picked up the term here). Say $ p^x\\parallel{}a$ and $ p^y\\parallel{}b$\r\n\r\nSo the terms $ a|b^2$, $ a^3|b^4$, ... imply that $ (2k\\minus{}1)x\\le (2k)y$ so $ \\frac{x}{y}\\le 1\\plus{}\\frac{1}{2k\\minus{}1}$ for all $ k\\in\\mathbb{N}$ (1).\r\n\r\nIf $ \\frac{x}{y}>1$, then choose the first natural number $ k_0$ such that $ 2k_0\\minus{}1>\\frac{1}{\\frac{x}{y}\\minus{}1}>0$. Rearranging gives a contradiction to (1).\r\n\r\nTherefore $ \\frac{x}{y}\\le 1$. Using a similar argument on the $ b^2|a^3$, $ b^4|a^5$ terms shows that $ \\frac{y}{x}\\le 1$. These two relations imply that $ x\\equal{}y$.\r\n\r\nIf either $ a$ or $ b$ have no prime factors, easily $ a\\equal{}b\\equal{}1$. If $ a$ or $ b$ do have prime factors, they have the same amount of each prime factor so $ a\\equal{}b$ are equal (note the only unit in $ \\mathbb{N}$ is $ 1$...)." } { "Tag": [ "geometry" ], "Problem": "ABCD is a square of side 10 units. Arc of circles with centres at ABCD are drawn in.\r\n\r\nfind the area of the central region bounded by the four arcs.", "Solution_1": "Let the vertices of the shape be labeled $ P,Q,R,S$. The shape BPC can be found by doing algebra on the arcs centered at $ A$ and $ D$.\r\n\r\nThis is also equal to the area of CQD, and since CPQ is just a square minus a fourth of a circle, we can determine through subtraction the area of CPQ.\r\n\r\nThis also equals to the area of ARS, and since we can easily figure out CPSARQ, since it it just 2 quarter circles minus a square, we can then figure out the area of the region bounded by the four arcs through subtraction.", "Solution_2": "[asy]pair A=(0,0), B=(1,0), C=(1,1), D=(0,1);\npath P=arc(A,1,0,90), Q=arc(B,1,90,180), R=arc(C,1,180,270), S=arc(D,1,270,360);\npath E=intersectionpoint(P,S);\nfill(buildcycle(A--B,P,S),lightred);\nfill(buildcycle(S,E--A),lightgreen);\nfill(A--E--D--cycle,lightblue);\ndraw(A--B--C--D--cycle,linewidth(1));\ndraw(P^^Q^^R^^S,linewidth(1));\ndraw(A--E^^E--D,linewidth(1));\nlabel(\"A\",A,SW);\nlabel(\"B\",B,SE);\nlabel(\"C\",C,NE);\nlabel(\"D\",D,NW);[/asy]\r\n\r\nRed + Green : $ 10^2\\pi\\times\\frac{30}{360}\\equal{}\\frac{25}{3}\\pi$.\r\n\r\nBlue + Green : $ 10^2\\pi\\times\\frac{60}{360} \\equal{} \\frac{50}{3}\\pi$.\r\n\r\nBlue : $ 10^2\\frac{\\sqrt{3}}{4} \\equal{} 25\\sqrt{3}$.\r\n\r\nGreen : (Blue + Green) - Blue = $ \\frac{50}{3}\\pi \\minus{} 25\\sqrt{3}$.\r\n\r\nRed : (Red + Green) - Green = $ \\frac{25}{3}\\pi \\minus{} \\left(\\frac{50}{3}\\pi \\minus{} 25\\sqrt{3}\\right) \\equal{} 25\\sqrt{3} \\minus{} \\frac{25}{3}\\pi$.\r\n\r\nFind Area : Square - 4 Red = $ 10^2 \\minus{} 4\\left(25\\sqrt{3} \\minus{} \\frac{25}{3}\\pi\\right) \\equal{} \\boxed{100\\left(1\\minus{}\\sqrt{3}\\plus{}\\frac{\\pi}{3}\\right)}$." } { "Tag": [], "Problem": "What is piety?\r\n\r\nYou may want to read Euthyphro:\r\nhttp://classics.mit.edu/Plato/euthyfro.html", "Solution_1": "Piety is the ability to remain in one belief.", "Solution_2": "Does anyone else want to propose a definition, or should we work off the one above?", "Solution_3": "piety, from merriam webster:\r\n\r\n[b]1:[/b] the quality or state of being pious: as [b]a:[/b] fidelity to natural obligations (as to parents) [b]b:[/b] dutifulness in religion : devoutness \r\n[b]2:[/b] an act inspired by piety\r\n[b]3:[/b] a conventional belief or standard\r\n\r\nOK I really don't understand any if those definitions. Care to translate, anyone?", "Solution_4": "Being consistent to the beliefs that are considered moral despite it being difficult, unpopular or uncommon (with others in similar positions).\r\n\r\nNote that persistence is the same as stubbornness, the only difference is if the speaker (author) agrees with the goal being pursued with the property.\r\n\r\nSimilarly, piety is typically not used much today, but it would only be applied to beliefs that were commonly viewed by the society as moral or proper." } { "Tag": [ "ratio", "geometry", "Pythagorean Theorem" ], "Problem": "In the diagram, circle $ Q$ is congruent to circle $ W$, and both are tangent to circle $ O$ and to each other. Circle $ S$ and circle $ T$ are congruent and are tangent to circle $ O$, to circle $ Q$ and to circle $ W$. Find the ratio of the area of the smallest circle to the largest circle.\n[asy]draw((0,1)..(1,0)..(0,-1)..(-1,0)..cycle);\ndraw((0,0)..(.5,.5)..(0,1)..(-.5,.5)..cycle);\ndraw((0,0)..(.5,-.5)..(0,-1)..(-.5,-.5)..cycle);\ndraw((1/3,0)..(2/3,1/3)..(1,0)..(2/3,-1/3)..cycle);\ndraw((-1/3,0)..(-2/3,1/3)..(-1,0)..(-2/3,-1/3)..cycle);\ndot((0,0));\nlabel(\"$O$\",(0,0),S);\nlabel(\"$Q$\",(0,.5)); label(\"$S$\",(-2/3,0)); label(\"$T$\",(2/3,0)); label(\"$W$\",(0,-.5),S);[/asy]", "Solution_1": "Suppose radius of largest circle, large subcircle = R, small subcircle = r.\r\n\r\nR=1/2.\r\nBy pythagorean theorem:\r\n(1-r)^2+1/4=(1/2+r)^2\r\nr^2-2r+5/4=1/4+r+r^2\r\n3r=1\r\nso r=1/3, R=1/2, ratio of areas = 4/9.\r\n\r\nHowever if they're talking about the largest circle O, our answer is 1/9." } { "Tag": [ "probability" ], "Problem": "In the game NetHack, dipping a long sword in a fountain has a probability $ p$ of turning it into Excalibur. It also has a probability $ q$ of drying up the fountain. If there are $ n$ fountains on your level, what's the probability of making Excalibur before using up your fountain supply? Assume for simplicity that you continue to dip until one of these conditions are satisfied.", "Solution_1": "[hide]\nThis would be 1 minus the probability that you use up all of your fountains, which is $ q^{n}$.\n\nThus the answer is $ \\boxed{1-q^{n}}$.[/hide]", "Solution_2": "Actually, the probability of using up a fountain, given that you try to use the fountain until something happens, is not $ q$, unless you assume that p and q are the only possibilities, that is: $ p+q=1$, which was not given. I might give some explanation of this if I get time later but for the moment I hope it's understandable without an explanation to say that the probability of using up a fountain is actually $ \\frac{q}{p+q}$. The probability of getting excalibur is still 1 minus the probability of using up all the fountains, and thus the probability of getting excalibur is:\r\n\r\n$ 1-(\\frac{q}{p+q})^{n}$", "Solution_3": "Oh, I didn't see that, sorry.\r\n\r\nWhile we're on this topic, are $ p$ and $ q$ mutually exclusive? Because that's another thing I assumed.", "Solution_4": "Sorry, I should have specified the problem more precisely.\r\n\r\nEach fountain is dipped into until you either get Excalibur or the fountain dries up. On each dip, there is a probability $ p$ of getting Excalibur, and a probability $ q$ of the fountain drying up. These events are mutually exclusive, and $ p+q<1$." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Nothin", "Solution_1": "#9 has already been asked in \"Some Mandelbrot Questions\" in the intermediate section.", "Solution_2": "is that necessarily a bad thing :twisted:", "Solution_3": "i think #10 is a school level mathcounts sprint question.", "Solution_4": "grr... [hide]i missed it cause i forgot to put back the square sign after i factored it[/hide]... hehehe...", "Solution_5": "i finally sent mine in, lolo, but i still took little time, i just didnt feel like doing them", "Solution_6": "and you had LOTS of people distracting you... including me... sorry if i hindered your score (which probably i didn't)", "Solution_7": "lolo, its okay if you did, but i dont care as long as i know i zam able to do it", "Solution_8": "yah... mistakes can hurt u like... woah!!!... 9 points gone right there...", "Solution_9": "yeah, but we all know you were able to do them", "Solution_10": "LoL... but we should pm this... just in case someone might...", "Solution_11": "15. Technically, by inclusion of \"and\" in that number, you made it 1900.80. Please correct.", "Solution_12": "huh? Shouldn't that be \"one thousand nine hundred and eighty hundredth\" or \"one thousand nine hundred and eight tenth\"?", "Solution_13": "The number he wants is written one thousand nine hundred eighty.", "Solution_14": "Ah it doesn't matter me thinks.", "Solution_15": "To me, 1980 is read one thousand nine hundred [b]and[/b] eighty. I've never seen leaving out the and, it doesn't sound right..\r\n1900.80 would be read one thousand nine hundred point eight, or one thousand nine hundred and eighty hundredths or whatever. Maybe its american thingy tho.", "Solution_16": "Can one of you help me on #9 Ive tried to do it a couple times but it just doesnt click", "Solution_17": "aalindsey100 wrote:Can one of you help me on #9 Ive tried to do it a couple times but it just doesnt click\n\n\n\nhighlight to read:[hide] Do you know the expansion x^3+3x^2+3x+1=(x+1)^3? Can you go from there?[/hide]", "Solution_18": "Yes thanks I forgot totally about doing that I think I can figure it out from there.", "Solution_19": "[quote=\"TripleM\"]To me, 1980 is read one thousand nine hundred [b]and[/b] eighty. I've never seen leaving out the and, it doesn't sound right..\n1900.80 would be read one thousand nine hundred point eight, or one thousand nine hundred and eighty hundredths or whatever. Maybe its american thingy tho.[/quote]\r\nPeople generally say \"and,\" but it technically isn't correct.", "Solution_20": "Could someone please post solutions to the one with the Fibonacci numbers and the other one: what are the rightmost three digits of 5^1993?", "Solution_21": "you know that any power of five has hte last two digits of 25...then, you can see that any even power of five ends in a 625 and an odd ends in a 125...since 1993 is odd, your answer is 125...at least that's what I got...", "Solution_22": "yeah, thats what I did too and he said it was wrong...im almost positive he said that was wrong", "Solution_23": "Also what I did, but I don't know if it was marked correct or incorrect.", "Solution_24": "anyone know how to do the fibonacci one? i wasn't really sure what he was asking...", "Solution_25": "it was asking which number from 0-9 shows up last as a unit digit\r\n\r\n1,1,2,3,5,8,13,21,34,5,9,4,3,7,0,7,7,4,1,5,6\r\n\r\ni decided to just use the units digits in the end. and as you can see every digit is then used. with 6 as the last one.", "Solution_26": "the answer is 125 for the 5^1993 one, i got it right...", "Solution_27": "Can anyone post just the answers at least? I'm not sure why I got a couple of them wrong though I forgot which...", "Solution_28": "pal---strange....Rep, where are you? Why did you count mine wrong? :shock: :twisted:" } { "Tag": [ "trigonometry", "Gauss", "calculus", "integration", "articles", "geometry", "parameterization" ], "Problem": "[b]2.[/b] A spatially uniform magnetic field $ \\overrightarrow{B}$ exists in the circular region S and this field is decreasing in magnitude with time at a constant rate.(See the fig)\r\nThe wooden ring $ C_1$ and the conducting rind $ C_2$are concentric with the magnetic field. The magnetic field is perpendicular to the plane of figure. Then,\r\n(A)there is no induced electric field in $ C_1$\r\n(B)there is an induced electric filed in $ C_1$ and it's magnitude is greater than the magnitude of induced electric field in $ C_2$\r\n(C)there is an induced electric field in $ C_2$ and it's magnitude is greater than the magnitude of induced electric field in $ C_1$\r\n(D)there is no induced electric field in $ C_2$.\r\n\r\n\r\n[b]8.[/b]Consider a circle of radius R. A point charge lies at a distance a from it's centre and on it's axis such that $ R \\equal{} a \\sqrt3$. If the electric flux passing through the circle is $ \\phi$ then the magnitude of the point charge is..\r\n(A)$ \\sqrt3 \\epsilon_0 \\phi$\r\n(B)$ 2\\epsilon_0 \\phi$\r\n(C)$ \\frac {4\\epsilon_0\\phi}{\\sqrt3}$\r\n(D)$ 4\\epsilon_0 \\phi$\r\n\r\n[b]20.[/b]Two positron($ e^ \\plus{}$) and two protons(p) are kept on four corners of a square of side $ a$ as shown in fig. The mass of proton is much larger than mass of positron. Let $ q$ denote the charge on proton as well as positron. Then the kinetic energies of one of the positrons and one of the protons respectively after a[i] very long time[/i] will be..\r\n(A)$ \\frac {q^2}{4 \\pi \\epsilon_{0} a} (1 \\plus{} \\frac1{2 \\sqrt2})$ , $ \\frac {q^2}{4 \\pi \\epsilon_{0} a} (1 \\plus{} \\frac1{2 \\sqrt2})$\r\n(B)$ \\frac {q^2}{2 \\pi \\epsilon_{0} a}$ , $ \\frac {q^2}{4\\sqrt2 \\pi \\epsilon_{0} a}$\r\n(C)$ \\frac {q^2}{4 \\pi \\epsilon_{0} a}$ , $ \\frac {q^2}{4 \\pi \\epsilon_{0} a}$\r\n(D)$ \\frac {q^2}{4 \\pi \\epsilon_{0} a} (1 \\plus{} \\frac {1}{4 \\sqrt2})$ , $ \\frac {q^2}{8\\sqrt2 \\pi \\epsilon_{0} a}$\r\n\r\n[b]21.[/b]An electrostatic field line leaves at angle $ \\alpha$ from a point charge $ q_1$ and connects with point charge $ \\minus{} q_2$ at angle $ \\beta$.\r\nThen the relation ship between $ \\alpha$ and $ \\beta$ is\r\n(A)$ q_1 \\sin^{2}{\\alpha} \\equal{} q_2 \\sin^2\\beta$\r\n(B)$ q_1 \\tan{\\alpha} \\equal{} q_2 \\tan{\\beta}$\r\n(C)$ q_1 \\sin^{2}{\\frac {\\alpha}{2}} \\equal{} q_2 \\sin^{2}{\\frac {\\beta}{2}}$\r\n(D)$ q_1 \\cos{\\alpha} \\equal{} q_2 \\cos{\\beta}$\r\n\r\n[b]25.[/b]A long straight wire carrying current $ I_1$ in $ \\plus{} z$ direction. The $ x \\minus{} y$ plane contains a closed cicular loop carrying current $ I_2$ and not encircling the straight wire. The force on the loop will be\r\n(A)$ \\frac {\\mu_0 I_1 I_2}{2 \\pi}$\r\n(B)$ \\frac {\\mu_0 I_1 I_2}{4 \\pi}$\r\n(C)$ zero$\r\n(D)depends on the distance of the centre if the loop from the wire.", "Solution_1": "[hide=\" #2-Whats wrong here :?: \"]\nFor the ring $ E_i \\times 2\\pi r_i \\equal{} \\minus{} \\frac {d \\phi}{dt} \\equal{} \\minus{} \\pi r_i^2 \\frac {dB}{dt}$\nThat gives me, $ E_i \\equal{} k \\times r_i$ (k is some constant}\nSo my answer is C.\nBut that isn't correct acc. to the key :( \n[/hide]\n\n[hide=\"#8\"]- I've no idea how to proceed. I know somehow I've to use gauss theorem(may be) - $ \\oint \\overrightarrow{E} \\bullet \\overrightarrow{dS} \\equal{} \\frac {q}{\\epsilon_0}$\n[/hide]\n\n[hide=\"#20\"]\nMy idea is - after a [i]long time[/i] the particles will be at infinity due to the repulsive forces. So all of the initial electrostatic potential energy is converted into kinetic energy. But how's that distributed btw protons and positrons? Any ideas?\n[/hide]\n\n#21- Absolutely no idea. How to relate them?\n\n[hide=\"#25\"]\nBy intuition I feel that force depends on the distance between the centre of the loop and the straight wire.\nFor doing the actual calculation, I am clueless how to go on with the integration{to find the force}.\n[/hide]", "Solution_2": "[hide=\"Mistake in 2\"]The flux is not $ \\pi r_i ^2 B$, but instead $ \\pi r_S ^2 B$, where $ r_S$ is the radius of the shaded circle $ S$.[/hide]\n\n[hide=\"Hint for 8\"]Since flux is a surface integral, what is meant by the flux through a loop -- which [i]surface[/i] do you calculate the flux over? Gauss's Law tells us it doesn't matter in a source-free region, as long as the boundary of that surface is the desired circle (determine for yourself why this must be true). So, you're free to choose the simplest surface.\n\nFigure out what that surface should be in this case. [hide=\"Further hint\"]The field whose flux you're calculating has only radial dependency, by symmetry.[/hide]\n\nOnce you have determined the desired surface, [url=http://en.wikipedia.org/wiki/Spherical_cap]this article[/url] contains a surface area formula that you might find very helpful if you don't already know it.[/hide]\n\n[hide=\"Hint for 20\"]The electrostatic potential energy goes to 0, so the kinetic energies of the four particles at time infinity must add up to the total electrostatic potential energy at time 0. This eliminates two of the answer choices.[/hide]\n\n[hide=\"Hint for 21\"]You can answer this question without calculation. Only one of the answers is remotely reasonable. Try drawing the field lines for $ q_1 \\equal{} \\text{large}, \\; q_2 \\equal{} \\text{small}$ and eliminate by trying out different angles.\n\n[b]Starter:[/b] $ \\alpha \\equal{} 0$ should automatically imply $ \\beta \\equal{} 0$, so choice (D) is clearly false.[/hide]\n\n[hide=\"Hint for 25\"]Let the current of the straight infinite wire flow along the z-axis.\nLet the circle of current be parametrized as $ \\left( A \\plus{} R \\cos \\theta , \\; R \\sin \\theta \\right)$, where $ R < A$.\nLet $ \\vec{F} (\\theta)$ be the force on the current component with parameter $ \\theta$ as a result of $ \\vec{B}$.\n\nCalculate $ \\vec{F}( \\theta) \\plus{} \\vec{F} ( \\minus{} \\theta)$.[/hide]", "Solution_3": "[quote=\"savy2020\"][b]20.[/b]Two positron($ e^ \\plus{}$) and two protons(p) are kept on four corners of a square of side $ a$ as shown in fig. The mass of proton is much larger than mass of positron. Let $ q$ denote the charge on proton as well as positron. Then the kinetic energies of one of the positrons and one of the protons respectively after a[i] very long time[/i] will be..\n(A)$ \\frac {q^2}{4 \\pi \\epsilon_{0} a} (1 \\plus{} \\frac1{2 \\sqrt2})$ , $ \\frac {q^2}{4 \\pi \\epsilon_{0} a} (1 \\plus{} \\frac1{2 \\sqrt2})$\n(B)$ \\frac {q^2}{2 \\pi \\epsilon_{0} a}$ , $ \\frac {q^2}{4\\sqrt2 \\pi \\epsilon_{0} a}$\n(C)$ \\frac {q^2}{4 \\pi \\epsilon_{0} a}$ , $ \\frac {q^2}{4 \\pi \\epsilon_{0} a}$\n(D)$ \\frac {q^2}{4 \\pi \\epsilon_{0} a} (1 \\plus{} \\frac {1}{4 \\sqrt2})$ , $ \\frac {q^2}{8\\sqrt2 \\pi \\epsilon_{0} a}$ [/quote]\r\n[i]CORRECTION:[/i]\r\n(D)$ \\frac {q^2}{2 \\pi \\epsilon_{0} a} (1 \\plus{} \\frac {1}{4 \\sqrt2})$ , $ \\frac {q^2}{8\\sqrt2 \\pi \\epsilon_{0} a}$\r\n\r\n\r\n[hide=\"#2\"]\nThanks I feel really bad for making such a silly mistake :( \n[/hide]\n\n[hide=\"Solution for #8- Pls chk it\"]\nthanks again :) \nPls chk my solution.\nThe spherical cap constructed over the circle with radius $ R_s\\equal{}\\sqrt{R^2\\plus{}a^2} \\equal{} 2a$ subtends an angle of $ \\Omega\\equal{} \\frac{A}{r^2}\\equal{}{2\\pi(2a)(a)}{{2a}^2}\\equal{}\\pi$ steradian where A=area of the spherical cap. So flux through this is $ \\frac14 (\\frac{q}{\\epsilon_0})\\equal{} \\phi$ as the total angle at the centre is $ 4\\pi$ steradian\nThank you very much for that surface area formula. I've searched for it's derivation and found [url=http://mathworld.wolfram.com/Zone.html]here[/url]\n\nI think it could also be done as $ E \\times A \\plus{} \\phi \\equal{} 0$ $ \\implies\\phi \\equal{} \\frac{q}{4\\pi \\epsilon_0 (2a)^2} \\times 4 \\pi a^2$\n[/hide]\n\n[hide=\"#20\"]\nThe total electrostatic potential energy at t=0 is $ 4 ( \\frac {q^2}{4 \\pi \\epsilon_{0} a} ) \\plus{} 2 ( \\frac {q^2}{4 \\pi \\epsilon_{0} \\sqrt{2}a})$\nEnergy of a $ p$ and a $ e^\\plus{}$ is $ E\\equal{}2\\frac {q^2}{4 \\pi \\epsilon_{0} a}(1\\plus{} \\frac{1}{2 \\sqrt2})$\nOption A and D agree with this. Now how to decide btw them? How do we justify whether the KEs will be equal or not. I feel they'll be equal and hence option A.\n[/hide]\n\n[hide=\" #21\"]\nIsn't there any mathematical method to solve?\n[/hide]\n[hide=\"#25 solved\"]\nI get $ \\overrightarrow{F} (\\theta)\\equal{} \\minus{} \\overrightarrow{F} (\\minus{}\\theta)$. So option C $ F_{net}\\equal{}0$. There will be torque though. Right?\n[/hide]", "Solution_4": "Sorry for reviving but I think you really overdid #8. The flux is just the area of the Gaussian surface multiplied by the Electric field strength since the position of the point charge is irrelevant. The flux is $(4\\pi R^{2})(\\frac{Q}{4\\pi\\epsilon_{0}R^{2}})$.", "Solution_5": "[quote=\"Diehard\"]Sorry for reviving but I think you really overdid #8. The flux is just the area of the Gaussian surface multiplied by the Electric field strength since the position of the point charge is irrelevant. The flux is $(4\\pi R^{2})(\\frac{Q}{4\\pi\\epsilon_{0}R^{2}})$.[/quote]\n\nNo. He (ref. TZF) is correct. Whatever you wrote is the flux in the space and not through the given circle (As asked in the problem). Rather the flux through that circle will be given by $\\phi=\\frac{Q}{\\epsilon_{0}}\\cdot \\frac{\\Omega}{4\\pi}$, Where $\\Omega$ is the solid angle subtended by the point charge at the circle.\n\nNote: $\\Omega=2\\Pi(1-\\cos \\theta)$, where $\\theta$ is the half angle of the cone as [url=http://en.wikipedia.org/wiki/Solid_angle]here[/url]" } { "Tag": [], "Problem": "How can I compute the number of subsets with 20 elements of a set with 52 elements using a C program? :blush:", "Solution_1": "I know that the number is $C_{52}^{20}$ but.. how can I compute such a large number? (I have made a post somewhere on this forum.. because I thought I was wrong, i though the number was not $C_{52}^{20}$)\r\n\r\nthank you for very much! :blush:", "Solution_2": "OK, you want to find the number $\\left(\\!\\begin{array}{c}52\\\\20\\end{array}\\!\\right)$. You can either calculate it directly from\r\n\r\n$\\left(\\!\\begin{array}{c}n\\\\k\\end{array}\\!\\right)=\\frac{n!}{k!(n-k)!}$\r\n\r\nor from\r\n\r\n$\\left(\\!\\begin{array}{c}n\\\\k\\end{array}\\!\\right)=\\left(\\!\\begin{array}{c}n-1\\\\k-1\\end{array}\\!\\right)+\\left(\\!\\begin{array}{c}n-1\\\\k\\end{array}\\!\\right)$\r\n\r\nbut any way, you'll get a number greater than int - fortunately, it is less than long long int (64 bit integer) so I suppose you could use the second formula to calculate it recursively.\r\n\r\n(It is 125994627894135.)" } { "Tag": [], "Problem": "on an online forum usernames can be 1-6 letters and can contain letters numbers or underscores. however, a username cannot begin or end with an underscore. how many possible usernames are there?", "Solution_1": "[hide]If it is 1 character long, the number of possibilities is 36 (all 37 -1 for the underscore)\nFor it to be 2 characters long, we take $37^{2}= 1369$, and subtract 73 to get 1296.\nFor 3 characters: $37^{3}= 50653$, then subtract 2*36*37 to get 47989.\n4 characters: $37^{4}= 1874161$, subtract 2*37*37*36 to get 1775593.\n5 characters: $37^{5}= 69343957$, subtract 2*($37^{3}$)*36 to get 65696941.\n6 characters: $37^{6}= 2565726409$, subtract 2*$37^{4}$*36 to get 2430786817.\nTotaling this, we get $\\boxed{2498308672}$[/hide]\r\nhmm probably wrong oh well", "Solution_2": "can they contain anything else like # that for example?", "Solution_3": "I got a different answer, but I may have messed up...[hide]\\[36+36^{2}(1+37+37^{2}+37^{3}+37^{4})=\\boxed{2496382452}\\][/hide]", "Solution_4": "magixter's looks correct.\r\n[hide]\nOne symbol long $36$\nTwo symbols $36 \\cdot 36$\nThree symbols $36 \\cdot 37 \\cdot 36$\nFour symbols $36 \\cdot 37 \\cdot 37 \\cdot 36$\nFive symbols $36 \\cdot 37 \\cdot 37 \\cdot 37 \\cdot 36$\nSix symbols $36 \\cdot 37 \\cdot 37 \\cdot 37 \\cdot 37 \\cdot 36$\nAdd them together.\n[/hide]" } { "Tag": [ "probability", "geometry", "perimeter" ], "Problem": "Hey, I'm doing more of the Challenge problems in Mr. Patrick's Intro book. I want to check if my answers and logical reasoning are correct, and alternate methods to solving the problem that would render it easier.\r\n\r\n2.27 For how many of the positive integers $24-125$ inclusive is the digital sum a multiple of $7$?\r\n\r\nFirst, I divided the cases into two-digit and three-digit numbers. \r\n\r\nI noted that the possible multiples of $7$ for numbers $24-99$ were $7$ and $14$. Digits satisfying $x+y=7$ were $(2,5), (3,4), (4,3), (5,2), (6, 1), (7,0)$. Digits satisfying $x+y=14$ were $(5,9), (6,8), (7,7), (8,6), (9,5)$. There are $12$ such two-digit numbers.\r\n\r\nPossible multiples of $7$ for numbers $100-125$ was only $7$ itself. Digits satisfying $x+y+z=7$ were $(1,0,6), (1,1,5), (1,2,4)$. There are $3$.\r\n\r\nSo there are $15$ total.\r\n\r\n2.28 Mr. Smith brings home 7 animals for his 7 children. Each child will adopt a pet to be his or her own. There are 4 different cats, 2 different dogs, and a goldfish. Anna and Betty refuse to takecare of the goldfish, and Charlie and Danny insist on having cats. The other kids will take anything. In how many ways can Mr. Smith give the children pets?\r\n\r\nI started out with the restrictions, saying that Charlie could choose a cat in $4$ ways and Danny then in $3$ ways. There are now 5 animals left, but Anna and Betty refuse the goldfish. That means that Anna can also choose from only $4$ and Betty from $3$. The other kids have 3 animals left between them and can choose between them in $3!$ ways. The total number is $864$ ways.\r\n\r\n2.30 In how many ways can you spell the word $NOON$ below? You can start on any letter, then on each step you can move one letter in any direction (up, down, left, right, or diagonal). You cannot visit the same letter twice.\r\n\r\nNNNN\r\nNOON\r\nNOON\r\nNNNN\r\n\r\nObviously you have to start on the perimeter to start with an 'N.' Divide into two cases, starting on the corner, and merely starting on the perimeter.\r\n\r\nStarting on the corner, there are is only one way to go to the next O. Then there are 3 ways to go to the 2nd O. There are then 5 ways to go the last N. There are 4 corners, so 3*4*5=60 choices for starting on the corner.\r\n\r\nStarting on the any location on a side which is not a corner, there are two choices for the first movement, 3 choices for the second. The 3 choices themselves make up two subcases.\r\n\r\nFor subcase one, as usual, start on a non corner side, and there are two possibilities. If you choose to go to the 'O' bordering the same side you started on (which there is one of), then you have only 4 choices of the last N because you cannot go back to an earlier letter. 2*1*4=8.\r\n\r\nFor subcase two, start the same way, but go to one of the two 'O's not bordering the side you started on. This leads to 5 possible end locations, with 2*2*5=20\r\n\r\nThe total for case 2 is 28 possibilities, and there are 8 starting locations, for 224 possibilities.\r\n\r\n224+60=284.\r\n\r\nHmm, counting problems are annoying. You never know if your logic is correct. it must be a pain on the AIME. :D\r\n\r\nThanks for the help, I'll be posting more later. Don't want my computer to crash right now..", "Solution_1": "Here's another way to do the NOON one.\r\n\r\nChoose two Os, then choose an N for each O. Divide into two cases:\r\n\r\n1. The two Os are next to each other.\r\n2. The two Os are diagonal to each other.\r\n\r\nCase 1:\r\nThere are 5 Ns to choose from for each O, so that's 5*5 = 25 ways. But there are 2 ways to choose the same N twice, which is not allowed, so we must subtract those; in fact there are 25 - 2 = 23 ways. Since there are 4 ways to choose two Os that are next to each other, Case 1 renders 23*4 = 92 total ways.\r\n\r\nCase 2:\r\nThere are 5 Ns to choose from for each O, so that's 5*5 = 25 ways. There's no way to choose the same N twice, so we don't have to worry about that. And since there are 2 ways to choose two Os diagonal to each other, Case 2 renders 25*2 = 50 total ways.\r\n\r\nAdd up the two cases to get 142 total paths. But for each path, we can go in two different directions, so we actually need to multiply that by 2, to get our final answer of 284." } { "Tag": [ "vector", "calculus", "calculus computations" ], "Problem": "Given a vector field (4x+2x^3z)i-y(x^2 +y^2)j -(3x^2z^2 +4y^2z)k which closed surface has the greatest flux. I imagine that the divergence theorem palys a role but I'm not sure.", "Solution_1": "Don't double post. Your other thread may be moved here, but that doesn't mean you should ask the same question again.", "Solution_2": "It's ok. I've decided to stop being lazy and think. However it is difficult computationally and any help is welcome.", "Solution_3": "[quote=\"jmerry\"]Of course it's the divergence theorem- the divergence is a function which is positive on a bounded region and negative outside that region. How would you maximize the integral of a function that looks like that?[/quote]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $x, y, z$ positive real numbers such that $x^{2}+y^{2}+z^{2}=1$, and let $n$ be a positive integer. Show that\r\n\r\n \\[\\frac{x}{1-x^{2n}}+\\frac{y}{1-y^{2n}}+\\frac{z}{1-z^{2n}}\\geq\\frac{(2n+1)^{1+1/2n}}{2n}.\\]", "Solution_1": "By AM-GM we have $[(1-x^{2n})]^{2n}x^{2n}=\\frac{1}{2n}[(1-x^{2n})]^{2n}(2nx^{2n})\\leq \\frac{1}{2n}(\\frac{2n}{2n+1})^{2n+1}$, so $(1-x^{2n})x\\leq \\frac{2n}{(2n+1)^{1+\\frac{1}{2n}}}$ therefore $\\frac{x}{1-x^{2n}}\\geq x^{2}\\cdot \\frac{(2n+1)^{1+\\frac{1}{2n}}}{2n}$ and we're done!", "Solution_2": "\"Posted: Today, at 8:36 pm\"\r\n\r\n\"Posted: Today, at 8:45 pm\" \r\n\r\n\r\n....... Are you a genious N.T.TUAN ? :o", "Solution_3": "[quote=\"zweig\"]\"Posted: Today, at 8:36 pm\"\n\n\"Posted: Today, at 8:45 pm\" \n\n\n....... Are you a genious N.T.TUAN ? :o[/quote]\r\nNo, I am not :P . This is the old problem. :D" } { "Tag": [ "geometry", "power of a point" ], "Problem": "Triangle ABC and point P in the same plane are given. Point P is equidistant from A to B, angle APB is twice angle ACB, and $\\overline{AC}$ intersects $\\overline{BP}$ at point D. If PB=3 and PD=2, then $AD\\cdot CD=$\r\n(A) 5 \r\n(B) 6\r\n(C) 7\r\n(D) 8\r\n(E) 9", "Solution_1": "[hide=\"Hint and answer\"]What will happen if you draw circle with center $P$ and radius $PA$? Answer: $5$.[/hide]", "Solution_2": "[hide=\"Solution\"] Draw the circle with center P. Extend PB and call the intersection with the circle E. EP is 3 because it is a radius. By power of a point, ED * DB = 5 * 1 = 5 = CD * AD, so the answer is 5.[/hide]", "Solution_3": "[hide]Draw circle with center P that passes through A,B, and C. Using power of a point, $AD\\cdot CD=DB\\cdot(PD+R)=1\\cdot(2+3)=5.$\n$\\boxed{A: 5}$.[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $x$, $y$ be two positive integers such that $\\displaystyle 3x^2+x=4y^2+y$.\r\nProve that $x-y$ is a perfect square.", "Solution_1": "My (one line :) ) solution:\r\n[hide]\n$\\displaystyle 3x^2+x=4y^2+y \\Leftrightarrow (4y-3x)^2=(x-y)(12x-12y+1)$ and the result follows\n[/hide]", "Solution_2": "TST? oh it's also for iran (95 or 96) see here :\r\n http://www.mathlinks.ro/Forum/viewtopic.php?p=204325&highlight=", "Solution_3": "Another solution: if $3x^2+x=4y^2+y$, then $(3x+3y+1)(x-y)=y^2$. Now if $p$ is a prime and $p \\mid x-y$, we have $p \\mid y^2 \\Rightarrow p \\mid y \\Rightarrow p \\mid x-y+y=x$. Now suppose $p \\mid x-y, p \\mid 3x+3y+1$, then as $p \\mid x, p \\mid y$ we have $p \\mid 3x+3y-1-3x-3y=1$, contradiction, so $(3x+3y+1, x-y)=1$. Thus the result.", "Solution_4": "Also from the same contest:\r\nFind all numbers x such that both x and x inverse ( x in reverse order) are powers of 2. (same level of difficulty)", "Solution_5": "[quote=\"ali\"]Also from the same contest[/quote]\r\n\r\nWhich one? ;) Not the french TST...\r\nPlease start a new thread whith this new problem.\r\n\r\nPierre.", "Solution_6": "[quote=\"Igor\"]My (one line :) ) solution:\n[hide]\n$\\displaystyle 3x^2+x=4y^2+y \\Leftrightarrow (4y-3x)^2=(x-y)(12x-12y+1)$ and the result follows\n[/hide][/quote]Nice solution. Actually we can prove that $x-y = (\\gcd (x,y))^2 $ :)", "Solution_7": "the relation can be developped and so on:\r\n(x-y)(1+3(x+y))=y\u00b2\r\nnote a=(x-y)\r\nand b=(x+y)\r\nthen 4a(1+3b)=(b-a)\u00b2\r\nthe equation X\u00b2-14aX+(a\u00b2-4a)=0 has b as a solution\r\nso the small discriminant is a perfect square\r\nthis discriminant =48a\u00b2+4a=4a(12a+1)\r\nso 4a|a perfect square and gcd((12a+1),4a)=1\r\nso 4a is perfect square", "Solution_8": "We can rewrite $ 3x^2 \\plus{} x \\equal{} 4y^2 \\plus{} y$ as: \r\n\r\n$ (8x \\plus{} 1)^2 \\equal{} (4x)^2 \\plus{} (8y \\plus{} 1)^2$\r\n\r\nbut that is no more than a pythagorean triple. \r\nExist $ u,v$ positive integers whit some conditions such that: \r\n\r\n$ u^2 \\plus{} v^2 \\equal{} 8x \\plus{} 1$\r\n$ 2uv \\equal{} 4x$\r\n$ u^2 \\minus{} v^2 \\equal{} 8y \\plus{} 1$\r\n\r\nbecause $ 4x$ is even, while $ 8y \\plus{} 1$ is odd. \r\nFrom the first and the third ecuation: $ 2v^2 \\equal{} 8(x \\minus{} y)$, then ($ 2 \\mid v$ and)\r\n$ (\\frac {v}{2})^2 \\equal{} x \\minus{} y$" } { "Tag": [ "geometry", "geometric transformation", "homothety", "incenter", "radical axis", "power of a point", "geometry open" ], "Problem": "Let $ O_1,O_2,O_3$ be three circle. Let $ I$ be a circle tangent to $ O_1,O_2,O_3$ internally at $ A,B,C$ respectively and $ J$ be a circle tangent to $ O_1,O_2,O_3$ externally at $ A',B',C'$ respectively. Then internal similitude center of circle $ I,J$ is radical center of $ O_1,O_2,O_3$. Also, this point is intersection of $ AA',BB',CC'$. Then how can we say about external similitude center of circle $ I,J$?", "Solution_1": "We can say nothing about the external similarity center of the circles $ I, J.$ Non-degenerated Apollonius problem has 8 solutions. But when the circles are considered directed (clockwise or counter-clockwise), tangent only when they have the same directions at their tangency point, the problem has only 2 solutions. When circles are involved, directed lines resemble properties of points better then undirected lines.\r\n\r\n1. 2 circles can have up to 2 common points, up to 4 common undirected lines (common tangents). But 2 directed circles have only up to 2 common directed lines.\r\n\r\n2. Directed angle is dual to a line segment. Only an angular interval of lines passing through a point fill a directed angle. Only linear interval of points on a line fill a line segment. 2 circles form equal angles at their common points. Tangent lengths of 2 directed circles are equal. \r\n\r\n2. One circle passes through 3 points, 4 circles are tangent to 3 lines. But only one directed circle is tangent to 3 directed lines.\r\n\r\n3. 2 circles have one radical axis, and 2 similarity centers. But 2 directed circles have only one similarity center: External similarity center of undirected circles, if they have the same direction (say both clockwise), internal similarity center of undirected circles, if they have opposite directions.\r\n\r\n4. Common points of 2 circles (if any) are on their radical axis. Common directed lines of 2 directed circle (if any) pass through their similarity center.\r\n\r\n5. Tangent lengths from a point on the radical axis are equal. Directed line through similarity center forms equal angles with both directed circles.\r\n\r\n6. 3 circles have one radical center, common point of their radical axes, but 3 homothety axes, common lines of their similarity centers. But 3 directed circles have only one homothety axis.\r\n\r\netc.\r\n\r\nIf you consider directed circle $ I$ externally tangent to all 3 directed circles $ O_1, O_2, O_3,$ it means that they have the same direction. Directed circle $ J$ internally tangent to (intersecting) $ O_1, O_2, O_3,$ is the 2nd solution of general Apollonius problem for directed circles and there are no other solutions. Obviously, $ J$ is directed oppositely then $ I,$ therefore their single homothety center is the internal homothety center of undirected circles. These 2 directed circles do not have any other similarity center.", "Solution_2": "Thank you for kind answer. I'm not sure if I'm understanding about concept of directed circle correctly.But I think [quote=\"yetti\"] tangent only when they have the same directions at their tangency point [/quote] should be tangent externally or internally when they have the same or opposite directions at their tangency point. Is that right? And what is difference between direction at point on circle and direction of circle?\r\nIf I'm understaing correctly, you are saying that if we consider circle by directly, then circles $ I$ and $ J$ have opposite direction and there is only one similitude center, i.e internal similitude center of circle $ I$ and $ J$. So we can say nothing about the external similarity center of the circles. But I think what circle $ I$ and $ J$ have no external similitude center is the case that we consider circle by directly and I can't understand why we must consider circle by directly because what I requested is just property of external similitude center.", "Solution_3": "Undirected circle is composed of points equally distant from the center. Its radius is always positive.\r\nDirected circle is composed of directed lines equally distant from the center and forming directed angle either $ \\plus{}90^circ,$ or $ \\minus{}90^circ$ (but not both) with the radius. Its radius can be positive (for all circles directed counter-clockwise) or negative (for all circles directed clockwise).\r\n\r\nIf 2 directed lines with opposite directions coincide in position, they are still 2 different directed lines.\r\nIf 2 directed circles with opposite directions coincide in position, they are still 2 different directed circles.\r\n\r\nThe following figures are taken from Yaglom's Geometric Transformations.\r\n\r\nFigure 77 shows 2 common directed lines of 2 directed circles, (a) with the same directions (both counter-clockwise) and (b) with opposite directions. Note that the directed circles and their common directed lines have [b]the same directions at the tangency point of undirected circles[/b].\r\n\r\nFigure 76 shows examples of (a) tangent directed circles, i.e., directed circles [b]with single common directed line[/b]; again, they have the same directions at the tangency point of undirected circles, and (b) directed circle that are not tangent; the first pair has 2 common directed lines and [b]non-zero tangent distance[b] $ AB,$ the second pair has [b]no common directed line[/b]; both pairs have opposite directions at the tangency point of undirected circles.", "Solution_4": "Thanks a lot. :lol:\r\nI think there are many advantage by definiting directed circle, especially generalizing problem.\r\nLet $ A,B,C,D$ be four points on circle in that order. Then incenter of triangle $ ABD,ACD$, $ C \\minus{} excenter$ of triangle $ ABC$,$ B \\minus{} excenter$ of triangle $ DBC$ are collinear.\r\nI'm not sure that this problem can be generalized with directed circle or line, but if it is possible, then how can we generalize so that $ A,B,C,D$ are points on circle in any order?" } { "Tag": [], "Problem": "What is the minimum number of people in a group so that you would be certain that 3 of them were born in the same month and on the same day of the week? Explain your answer.", "Solution_1": "well you could have two people for each day of the week in each month and that's 168 people\r\n\r\nso you gotta have one more guy in order to be sure there's three for at least one month and day of the week\r\n\r\n1[b]69[/b]" } { "Tag": [ "inequalities", "quadratics", "function", "algebra unsolved", "algebra" ], "Problem": "Let $ a,b,c$ be the three sides of a triangle. Determine all possible values of $ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}$", "Solution_1": "hello, it is\r\n$ 1\\le\\frac{a^2\\plus{}b^2\\plus{}c^2}{ab\\plus{}bc\\plus{}ca}<2$ because\r\n$ a^2\\plus{}b^2\\plus{}c^2<2ab\\plus{}2bc\\plus{}2ca$ is equivalent with\r\n$ 00$ so it's obviously sharp.", "Solution_4": "But all three of those inequalities cannot be sharp at the same time. What I mean is as $ b\\to a\\plus{}c$, one of the other inequalities becomes less sharp.", "Solution_5": "$ a^2\\plus{}b^2\\plus{}c^2 \\plus{} \\infty} f(n) \\equal{} 2$\r\nTherefore $ f(n)$ get all values between $ f(1)$ and $ f( \\plus{} \\infty)$\r\nSo $ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}$ get all values in interval $ [1;2)$", "Solution_7": "This si the solution." } { "Tag": [ "Harvard", "college", "Yale", "Princeton", "MIT", "Columbia" ], "Problem": "I'm having an interview for my \"safety\" college next week, and I'm worried about them asking me the names of the other colleges that I am applying to. This school has been known to reject people who they think are using them as their safety. I'm not exceptional, but I'm also applying to Harvard, Yale, Princeton, MIT, etc, and the disparity between these schools and my safety is painfully obvious. Can I leave some schools out (e.g. Harvard, Yale, Princeton) and simply mention my other safeties and one or two top schools, or would this be dishonest?", "Solution_1": "Haha, perhaps we should have an entire forum dedicated to ethics issues ...\r\n\r\nI don't remember being asked that question by interviewers anywhere, but I mostly had alumni, fairly informal interviews. I doubt, if that is the type of interview you're having, that they will ask you about it except perhaps \"off the record.\" You could always say you'd rather not tell, although that would be kind of odd. I don't think it's fair to only tell them some (but not all) of the schools you are applying to as if it were a complete list, however.", "Solution_2": "one could phrase it ambiguously, as in: \"my list includes [school A], [school B], and [school C].\" under pressure, one could say that one is also applying to princeton, yale, etc- but i, for one, would speak of these as true longshots, as \"lets see what happens\" kind of schools, as not being... *TRULY* serious about one's chances at these. i realize that the application fee for those schools makes it more serious than that, but one can make it sound like one used one's birthday money on these, just as a whim. i am planning to apply to one-two colleges \"on a whim\" so i dont see it as being that... crazy an idea. just a thought!", "Solution_3": "If someone asks you, \"What other schools are you applying to?\" it makes sense to give a straight answer. The old saying is, \"Honesty is the best policy.\" \r\n\r\nThe much more likely question you will encounter in an interview, and one you had very well better have an answer for, is \"Why do you want to go to my college?\" That's the question you should be prepared to answer in detail. \r\n\r\nYou have my sympathies. I know you live in the Northeast, where there are strong private schools that many applicants to the most famous schools use as \"safeties,\" which results in those private schools trying extra hard to figure out who their REAL applicants are. You have to be convincing, by first convincing YOURSELF, that you could fit in and add to the student community at any of the schools to which you apply. Here in the upper Midwest, where I live, anyone can apply to the U of Wisconsin, the U of Minnesota, the U of Iowa, or whatever, and the kind of strong student who typically participates on AoPS can be sure of getting in \"by the numbers,\" with no interview, and no embarrassment about applying to Harvard, Princeton, MIT, or whatever. You are in a more delicate situation, but you won't look better in an interview by being evasive. The application deadlines have mostly passed by now, and the interviewer knows you know where you have already applied.", "Solution_4": "My interviewer for Columbia asked me where else I was applying. I think I gave him a full list.", "Solution_5": "Lol, Columbia isn't really a safety, and I wouldn't mind saying the full list for Columbia and up. It is just uncomfortable when you have schools that aren't as prestigous, so to speak, that it sounds blatantly obvious that they are your safety school.", "Solution_6": "Is it defensible for any university to ask about other\r\napplications? (It seems even less defensible to utilize\r\nthe answer in their admissions deliberations, so why\r\nask?)\r\n\r\nThere are obvious reasons the university would be\r\ncurious about such information, but curiousity doesn't\r\nnecessarily entitle an inquiry.", "Solution_7": "Many colleges seem to ask once you turn them down, which seems perfectly defensible to me. I don't remember being asked before having been accepted or rejected.", "Solution_8": "[quote=\"JBL\"]Many colleges seem to ask once you turn them down, which seems perfectly defensible to me. I don't remember being asked before having been accepted or rejected.[/quote]\r\n\r\nWait... colleges send you a letter asking you to explain why you turned them down?", "Solution_9": "To enforce \"single-choice\" early admission (SCEA) plans, or early decision (ED) plans, colleges have some interest in knowing where else applicants are applying.", "Solution_10": "i dont think anyone asks where you applied early; at least i wasn't asked that. If they did, that would hurt you... i.e i applied ED and EA, if my EA schools asked where I applied early, my chances would most likely go down if i told them I applied ED somewhere else.", "Solution_11": "For ED, schools definitely compare lists with each other to make sure that no one is applying to more than one. I don't remember if everywhere I didn't go sent me a little postcard asking where I was going, but at least a couple of schools did. This is after you've already chosen where you're going, mind you, and it's highly voluntary -- I have no idea what the return rate it. U. Chicago actually also sent me a very nice letter after I turned them down." } { "Tag": [ "videos" ], "Problem": "Simply state your favorite zelda game(s) and what you found so enrapturing about them!\r\n\r\nMy favorites are, of course, [b]the legend of the wind waker and the ocarina of time[/b]!\r\n\r\nThose two almost top my all time favorite video game list(second only to Halo and call of duty)!\r\n\r\nThey had beautiful graphics(considering the time they were made in), had wonderful plots, awesome weapons, mind testing puzzles, and AMAZING boss battles. They are, in all aspects, flawless games.", "Solution_1": "Although FFX is still top on my game list...\r\n\r\nOcarina of Time and Twilight Princess for me!\r\n\r\nFor the above reasons, Ocarina of Time was perfect. \r\nAnd I'm only halfway through Twilight right now, but I'm very impressed by the graphics and the not-so-standard-for-LoZ plot.\r\n\r\nWindwaker was a bit boring at times (cough gathering the triforce cough)", "Solution_2": "Well, I've only played 3 LoZ games, but I've played 6 games with Link in them... I've played OoT, Four Swords Adventures, and TP. Twilight Princess was the best by far for me. Do Smash Brothers count as LoZ?", "Solution_3": "Not really, though it is very fun. I mean, you wouldn't call the new Soul Caliber a Star Wars game because Yoda was in it :P \r\n\r\n(I think Yoda was in it...)", "Solution_4": "Ocarina of time has stood the test of time... :P \r\n\r\nIt is one of the only games that gamespot.com rated a perfect 10/10.(The others were MGS4 and GTA)\r\n\r\nI'm going to buy Ocarina of time for the wii, just because it was so awesome..." } { "Tag": [], "Problem": "If 3x-8y=28 and x and y are integers\r\n\r\nWhat is the smallest value of x-y?", "Solution_1": "[hide]\n$3x+8y= 28$\n$3(4)+8(2) = 28$\n$x-y= 1$\n[/hide]", "Solution_2": "uh..wouldnt this be negative infinity?", "Solution_3": "[hide]\n\nLet $x=8k+4$ and $y=3k-2$ for some $k$. Then $3x-8y=3(8k+4)-8(3k-2)=28$. The value of $x-y$ then is $8k+4-(3k-2)=5k+6$, so there is no lower bound. $-\\infty$?\n[/hide]", "Solution_4": ":huh: \r\nHow?\r\nIf $lim_{y \\rightarrow-\\infty}$ then $8y \\not=28$, that could not be what you mean though.\r\nSorry for the bad LaTeX.", "Solution_5": "what happens if we restrict them to positive integers?", "Solution_6": "[quote=\"AstroPhys\"]:huh: \nHow?\nIf ${lim}_{y \\rightarrow \\infty}$ then $8y \\not{=}28$, that could not be what you mean though.\nSorry for the bad LaTeX.[/quote]\r\n\r\nUm, maybe you are reading the sign wrong? 3x-8y=28", "Solution_7": "says 3x-8y... and there is definitely no lower bound", "Solution_8": "[hide]$3x-8y=28 \\Rightarrow y= \\frac{3x-28}{8}$\n\nLet $h(x,y)=x-y \\Rightarrow h(x)=x-\\frac{3x-28}{8}\\Rightarrow h(x)= \\frac{5}{8}\\cdot x+\\frac{7}{2}$\n\n$h'(x)=\\frac{5}{8}\\Rightarrow h(x)$ is increasing $\\forall ~~x$ thus it has a range of $(-\\infty,+\\infty)$[/hide]", "Solution_9": "[quote=\"AstroPhys\"][hide]\n$3x+8y= 28$\n$3(4)+8(2) = 28$\n$x-y= 1$\n[/hide][/quote]\r\n\r\nI'm not sure what you did there. You just plugged in random numbers and said that 4-2=1.", "Solution_10": "[quote=\"AstroPhys\"][hide]\n$3x+8y= 28$\n$3(4)+8(2) = 28$\n$x-y= 1$\n[/hide][/quote]\r\n\r\nhow do you know thats the smallest?", "Solution_11": "[quote=\"Teki-Teki\"]what happens if we restrict them to positive integers?[/quote]\r\n\r\n[hide=\"Restricted to positive integers,\"]$x = \\frac{8y+28}{3}$\n\n$x-y = \\frac{5y+28}{3}$\n\nClearly, to minimize $x-y$, we must minimize $y$. Let $y=1$. So $x=12$. Thus, $x-y =11$.[/hide]" } { "Tag": [ "email", "MATHCOUNTS" ], "Problem": "could someone perhaps post the 2008 national sprint round, the 1995 and 1997 sprint, target, and team rounds or email to asdfasdfsdf12390@gmail.com?", "Solution_1": "[quote=\"kobebryant24\"]could someone perhaps post the 2008 national sprint round, the 1995 and 1997 sprint, target, and team rounds or email to asdfasdfsdf12390@gmail.com?[/quote]\r\nIt's on the MATHCOUNTS website...", "Solution_2": "the only thing i see are the 2008 state, chapter, and school rounds" } { "Tag": [ "function", "search", "algebra unsolved", "algebra" ], "Problem": "Let $ f : R\\to R$ satisfy $ |f(x)|\\leq 1$ for all $ x\\in R$\r\nand $ f(x\\plus{}\\frac{13}{42})\\plus{}f(x)\\equal{}f(x\\plus{}\\frac{1}{6})\\plus{}f(x\\plus{}\\frac{1}{7})$\r\nProve that $ f$ is period function", "Solution_1": "See here: http://www.mathlinks.ro/viewtopic.php?search_id=403970470&t=15583", "Solution_2": "\u0e04\u0e38\u0e13 apollo \u0e0a\u0e37\u0e48\u0e2d\u0e08\u0e23\u0e34\u0e07\u0e0a\u0e37\u0e48\u0e2d\u0e2d\u0e30\u0e44\u0e23\u0e04\u0e23\u0e31\u0e1a\r\n\u0e0a\u0e48\u0e27\u0e22\u0e1e\u0e35\u0e40\u0e2d\u0e47\u0e21\u0e21\u0e32\u0e1a\u0e2d\u0e01\u0e21\u0e14\u0e49\u0e27\u0e22\r\nThank you. :lol:" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Suppose $ L$ is an algebraically closed field, $ F \\subset L$ is a subfield, and $ f \\in F[x]$ is a nonconstant polynomial. Prove that $ L$ cfontains a unique subfield extending $ F$ which is a splitting field for $ f$ over $ F$.", "Solution_1": "The proof could depend a little on your definition of a splitting field. Hint: if $ f$ splits completely in $ K,K'$, then $ f$ splits completely in $ K\\cap K'$. If you're not given the existence of such a field $ K$, you'll need to prove that." } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Can you check me to make sure I'm right....\r\n\r\nIf two lines are parallel then they lie on the same plane......(my answer FALSE)\r\n\r\nIf two plances intersect then the intersection is either a point or a line...\r\n(my answer TRUE)", "Solution_1": "The first statement is [color=blue]true[/color] and the second [color=red]false.[/color]\r\n But , please write at the getting start section this kind of question. Probably you are nev here !\r\n[u] Babis[/u]", "Solution_2": "[b]NOTE[/b]\r\n\r\n As [color=red]Virgil Nicula[/color] pointed out to me, the second question can be taken as '' true'' in the sense that one part of the answer is true: the intersection of two planes is a line (axiomatic definition) and never a point.But the sentense << ....is a point [color=blue]or[/color] a line >> is true , because it contains the part '' ...a line''.\r\n Any way , sorry for not giving to the question the necessary attention.\r\n\r\n[u] babis [/u]", "Solution_3": "[quote=\"stergiu\"][b]NOTE[/b]\n\n As [color=red]Virgil Nicula[/color] pointed out to me, the second question can be taken as '' true'' in the sense that one part of the answer is true: the intersection of two planes is a line (axiomatic definition) and never a point.But the sentense << ....is a point [color=blue]or[/color] a line >> is true , because it contains the part '' ...a line''.\n Any way , sorry for not giving to the question the necessary attention.\n\n[u] babis [/u][/quote]\r\n\r\nure answer is completely, complete. :wink:" } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Let $ p$ be a prime of the form $ 4k\\plus{}1$. Let $ i$ be a positive integer, $ 00$ Prove that:$3(\\sum_{sym}a^3b^2) \\geq 2(ab+bc+ca)(ab^2+bc^2+ca^2)$ (2')\r\nSolution:(2) equalvilent to $\\sum_{sym} a^3b^2+2(\\sum_{cycly}a^3b^2 \\geq 2abc(a^2+b^2+c^2+ab+bc+ca)$ (*)\r\nApply AM_GM's inequality we have: $a^3b^2+a^3c^2+a^3bc \\geq 3a^3bc$\r\n $b^3c^2+b^3a^2+b^3ac \\geq 3ab^3c$\r\n $c^3a^2+c^3b^2+abc^3 \\geq 3abc^3$\r\nSum up these inequalities we have $\\sum_{sym}a^3b^2 \\geq 2abc(\\sum a^2)$ (3)\r\nBy similar way we can prove that $\\sum_{cyclic} a^3b^2 \\geq abc(\\sum ab)$ (4)\r\nFrom (3),(4) we have (*).So (2') is proved\r\nFrom (2') it follows that $\\sum_{sym}a^3b^2 \\geq 2(abc)^{\\frac{2}{3}}(ab^2+bc^2+ca^2)$ \r\nSo that this problem is proved.Cheer :lol: :lol:", "Solution_7": "[quote]Apply Shur's inequality we have:\n$\\sum_{cyclic} a^5b+3a^2b^2c^2 \\geq (abc)^{\\frac{1}{3}}(\\sum_{sym}a^3b^2)$\n[/quote]\r\n\r\nHi, Nhat\r\n\r\nI cannot see how you get the above.\r\nFollowing you: using\r\n$x^3+y^3+z^3+3xyz \\ge xy(x+y)+yz(y+z)+zx(z+x)$\r\nwhere $x=\\sqrt[3]{a^5b}$, $y=\\sqrt[3]{b^5c}$, $z=\\sqrt[3]{c^5a}$, right?\r\n\r\nNow\r\n$3xyz=3\\sqrt[3]{a^6b^6c^6}=3a^2b^2c^2 \\ \\ \\Rightarrow$ OK\r\n$x^2y = \\sqrt[3]{(a^5b)^2 b^5c} = \\sqrt[3]{a^{10}b^7c} = \\sqrt[3]{abc}(a^3b^2) \\ \\ \\Rightarrow$ OK too\r\n$xy^2 = \\sqrt[3]{a^5b (b^5c)^2} = \\sqrt[3]{a^5b^{11}c^2} \\ \\ \\Rightarrow$???\r\n\r\nCould you please explain?", "Solution_8": "Use generalised Schurs\r\n\r\nBomb", "Solution_9": "[quote=\"bomb\"]Use generalised Schurs\nBomb[/quote]\r\n\r\nWhich 'generalised Schurs' do you refer to? Care to explain?", "Solution_10": "[quote]For positive real numbers $a,b$ and $c$,\n\\[\\sum_{cycl}\\frac{ab(a^{2}+bc)}{b+c}\\ge \\sqrt{3abc(ab^{2}+bc^{2}+ca^{2})}\\]\n[/quote]\r\n\r\nBy the well-known inequality $(x+y+z)^{2}\\ge 3(xy+yz+zx)$ it suffices to prove that\r\n$\\sum_{cycl}\\left( ab\\frac{a^{2}+bc}{b+c}\\cdot bc\\frac{b^{2}+ca}{c+a}\\right) \\ge abc(ab^{2}+bc^{2}+ca^{2})$\r\nor equivalently\r\n$\\sum_{cycl}\\frac{b(a^{2}+bc)(b^{2}+ca)}{(b+c)(c+a)}\\ge ab^{2}+bc^{2}+ca^{2}$\r\nNote that $(a^{2}+bc)(b^{2}+ca)-ab(a+c)(b+c)=c(a+b)(a-b)^{2}\\ge 0$, that is,\r\n$\\frac{(a^{2}+bc)(b^{2}+ca)}{(a+c)(b+c)}\\ge ab$ (*)\r\nHence\r\n$\\frac{b(a^{2}+bc)(b^{2}+ca)}{(a+c)(b+c)}\\ge ab^{2}$\r\nAdding this and similar inequalities, the conclusion follows.\r\n\r\n[b]Apropos:[/b] (*) implies trivially\r\n$\\sqrt{\\frac{(a^{3}+abc)(b^{3}+abc)}{(a+c)(b+c)}}\\ge ab$\r\nand so\r\n$\\sum_{cycl}\\sqrt{\\frac{(a^{3}+abc)(b^{3}+abc)}{(a+c)(b+c)}}\\ge ab+bc+ca$\r\nThis together with a well-known inequality by Cezar Lupu yields\r\n$\\sum_{cycl}\\sqrt{\\frac{a^{3}+abc}{b+c}}\\ge a+b+c$\r\nIn similar manner, we also have\r\n$\\sum_{cycl}\\sqrt{\\frac{b+c}{a^{3}+abc}}\\le\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$\r\n$\\sum_{cycl}\\sqrt{\\frac{a^{2}+bc}{b+c}}\\ge \\sqrt{a}+\\sqrt{b}+\\sqrt{c}$\r\n$\\sum_{cycl}\\sqrt{\\frac{b+c}{a^{2}+bc}}\\le \\frac{1}{\\sqrt{a}}+\\frac{1}{\\sqrt{b}}+\\frac{1}{\\sqrt{c}}$", "Solution_11": "very nice ... :)", "Solution_12": "[quote=\"ductrung\"]This together with a well-known inequality by Cezar Lupu yields\n$\\sum_{cycl}\\sqrt{\\frac{a^{3}+abc}{b+c}}\\ge a+b+c$\n[/quote]\r\n\r\nWhich inequality is it? :maybe:", "Solution_13": "[quote=\"cefer\"][quote=\"ductrung\"]This together with a well-known inequality by Cezar Lupu yields\n$\\sum_{cycl}\\sqrt{\\frac{a^{3}+abc}{b+c}}\\ge a+b+c$\n[/quote]\n\nWhich inequality is it? :maybe:[/quote]\r\n\r\nCefer,\r\nThe answer is part of the solution. You will know it if you try to solve the problem :-)\r\n\r\nOtherwise, look at Problem Corner of the current issue of Math. Reflection: [url]http://reflections.awesomemath.org/2006_5/2006_5_problems.pdf[/url]" } { "Tag": [ "geometry", "rectangle", "analytic geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "On the plane are positioned at most $ \\frac {4n}{3}$ rectangles with sides parallel to the coordinate axes. Assume that every rectangle intersects at least $ n$ other rectangles. Show that there exists a rectangle that intersects all other rectangles.", "Solution_1": "Do you mean that there are at most $ \\frac{4n}{3}$ rectangles?", "Solution_2": "[quote=\"probability1.01\"]Do you mean that there are at most $ \\frac {4n}{3}$ rectangles?[/quote]\r\n\r\nOK. Edited.", "Solution_3": "Nobody? ...", "Solution_4": "[hide=\"Probably the right idea\"]Suppose we have $ m \\le \\frac{4n}{3}$ rectangles. Consider the four rectangles with the highest bottom edge, lowest top edge, rightest left edge, and leftest right edge. A rectangle which intersects all four of these rectangles will intersect all the rectangles. Now you play around with the numbers and use pigeonhole.[/hide]", "Solution_5": "[quote=\"probability1.01\"] Consider the four rectangles with the highest bottom edge, lowest top edge, rightest left edge, and leftest right edge. [/quote]\r\n\r\nAre you sure? Look at the following diagram. The big rectangle intersects all the four rectangles you mentioned. Yet it doesn't intersect all rectangles...\r\n\r\nP.S. Sorry for my bad drawing", "Solution_6": "May be i misunderstand the problem, but does this construction satisfies the condition of problem?", "Solution_7": "@greenvert: no, since each rectangle intersects $ n\\equal{}4$ rectangles, there should be most $ 4\\times \\frac{4}{3}<6$ rectangles, but in your diagram there are $ 6$ rectangles.", "Solution_8": "Yes, you are right.", "Solution_9": "Consider the four rectangles: (using coordinates) the one with the less right horizontal edge, the one with the greater left horizontal edge, the one with the less upper vertical edge, and the one with the greater lower vertical edge. It is clear that if there is a rectangle that intersects this four, then it intersects all rectangles. Each of this rectangles intersect at least other $ n.$ Suppose there isn't that rectangle. We need $ n$ intersections from each special rectangle, each of which can be counted at most trice. This means we need at least $ \\frac{4n}{3}\\plus{}1$ rectangles. Contradiction.", "Solution_10": "[quote=\"Carlez Tevos\"][quote=\"probability1.01\"] Consider the four rectangles with the highest bottom edge, lowest top edge, rightest left edge, and leftest right edge. [/quote]\n\nAre you sure? Look at the following diagram. The big rectangle intersects all the four rectangles you mentioned. Yet it doesn't intersect all rectangles...[/quote]\r\n\r\nI thought rectangles included the area within. Do you think this holds when rectangles are only their boundaries?", "Solution_11": "[quote=\"Nemion\"]Consider the four rectangles: (using coordinates) the one with the less right horizontal edge, the one with the greater left horizontal edge, the one with the less upper vertical edge, and the one with the greater lower vertical edge. It is clear that if there is a rectangle that intersects this four, then it intersects all rectangles. Each of this rectangles intersect at least other $ n.$ Suppose there isn't that rectangle. We need $ n$ intersections from each special rectangle, each of which can be counted at most trice. This means we need at least $ \\frac {4n}{3} \\plus{} 1$ rectangles. Contradiction.[/quote]\r\n\r\nBy my calculations, if there are exactly $ \\frac{4n}{3}$ rectangles, it is possible that the four special rectangles all intersect one another, and all other rectangles intersect three of the four special rectangles.\r\n\r\nAm I wrong?", "Solution_12": "[quote=\"Carlez Tevos\"]\nBy my calculations, if there are exactly $ \\frac {4n}{3}$ rectangles, it is possible that the four special rectangles all intersect one another, and all other rectangles intersect three of the four special rectangles.\n\n[/quote]\r\n\r\nI think so. Then you have to deal with this case separately.", "Solution_13": "[quote=\"discredit\"]\n\nI think so. Then you have to deal with this case separately.[/quote]\r\n\r\nBut how? :maybe:" } { "Tag": [ "inequalities" ], "Problem": "$ a,b,c \\geq 0,$prove that:\r\n\r\n$ \\sqrt{\\frac{3}{a^3\\plus{}(b\\plus{}c)^3}}\\plus{}\\sqrt{\\frac{3}{b^3\\plus{}(a\\plus{}c)^3}}\\plus{}\\sqrt{\\frac{3}{c^3\\plus{}(a\\plus{}b)^3}} \\leq\\frac{ab\\plus{}ac\\plus{}bc}{abc\\sqrt{a\\plus{}b\\plus{}c}}$", "Solution_1": "i doubt it's very easy", "Solution_2": "[quote=\"rachid\"]$ a,b,c \\geq 0,$prove that:\n\n$ \\sqrt {\\frac {3}{a^3 \\plus{} (b \\plus{} c)^3}} \\plus{} \\sqrt {\\frac {3}{b^3 \\plus{} (a \\plus{} c)^3}} \\plus{} \\sqrt {\\frac {3}{c^3 \\plus{} (a \\plus{} b)^3}} \\leq\\frac {ab \\plus{} ac \\plus{} bc}{abc\\sqrt {a \\plus{} b \\plus{} c}}$[/quote]\r\n\r\nindeed , it's easy but nice\r\nit sufficies to show the stronger :\r\n$ \\sum{\\frac{9}{a^3 \\plus{} (b \\plus{} c)^3}} \\leq \\frac{ (ab\\plus{}ac\\plus{}bc)^2}{ (abc)^2(a\\plus{}b\\plus{}c)}$\r\nbut $ (a\\plus{}b)^3\\plus{}c^3\\equal{} a^3\\plus{}b^3\\plus{}c^3\\plus{}3bc(b\\plus{}c) \\geq 3abc\\plus{}3bc(b\\plus{}c) \\equal{} 3bc(a\\plus{}b\\plus{}c)$\r\n so the inequality is equivalent to : $ (ab\\plus{}ac\\plus{}bc)^2\\minus{}3abc(a\\plus{}b\\plus{}c) \\geq 0$ which is true :wink:", "Solution_3": "[quote=\"anas\"][quote=\"rachid\"]$ a,b,c \\geq 0,$prove that:\n\n$ \\sqrt {\\frac {3}{a^3 + (b + c)^3}} + \\sqrt {\\frac {3}{b^3 + (a + c)^3}} + \\sqrt {\\frac {3}{c^3 + (a + b)^3}} \\leq\\frac {ab + ac + bc}{abc\\sqrt {a + b + c}}$[/quote]\n\nindeed , it's easy but nice\nit sufficies to show the stronger :\n$ \\sum{\\frac {9}{a^3 + (b + c)^3}} \\leq \\frac { (ab + ac + bc)^2}{ (abc)^2(a + b + c)}$\nbut $ (a + b)^3 + c^3 = a^3 + b^3 + c^3 + 3ab(a + b) \\geq 3abc + 3bc(b + c) = 3ab(a + b + c)$\n so the inequality is equivalent to : $ (ab + ac + bc)^2 - 3abc(a + b + c) \\geq 0$ which is true :wink:[/quote]\r\nMy solution is a little similar to yours anas.\r\n\r\n[hide=\"Solution\"]\n$ \\sum{\\sqrt {\\frac {3}{a^3 + (b + c)^3}} \\leq \\sum \\frac {1}{\\sqrt {bc(a + b + c)}} \\leq \\frac {ab + ac + bc}{abc\\sqrt {a + b + c}}}$[/hide]" } { "Tag": [], "Problem": "How many numbers can you get by adding two or more distinct members of the set $ \\{0,1,2,4,8,16\\}$ together?", "Solution_1": "If we convert these numbers to binary, we'll easily see that every positive integer up to $ 11111_2\\equal{}\\boxed{31}$ can be made.", "Solution_2": "But How did you get 31?", "Solution_3": "Or you can just add up the numbers, $ 16+8+4+2+1=\\boxed{31} $" } { "Tag": [ "quadratics", "complex numbers", "algebra" ], "Problem": "I know a lot of complex numbers but in my exercise book I couldn't find a way to solve what I've typed below (I know for final solutions anyway):\r\n\r\nPROBLEM 1\r\n\r\nFind Re(w) and Im(w) [where $z=x+yi$]:\r\n\r\n$w=z+\\frac{1}{z}$\r\n\r\nSo we have to write a complex number in a standard form and then extract a real and an imaginary part from there. We work here with the unknown variables ($x$ and $y$).\r\n\r\nMy way would be:\r\n\r\n$w=z+\\frac{1}{z}=x+yi+\\frac{1}{x+yi}=\\frac{(x+yi)^{2}+1}{x+yi}=\\frac{((x+yi)^{2}+1)(x-yi)}{(x+yi)(x-yi)}=\\frac{((x+yi)^{2}+1)(x-yi)}{x^{2}+y^{2}}=...$\r\n\r\nIf I expand the expression further (i.e. numerator) I don't find a way to write it in a standard form.\r\n\r\nHowever, the final solution is:\r\n\r\n$Re(w)=x+\\frac{x}{x^{2}+y^{2}}$ and $Im(w)=y-\\frac{y}{x^{2}+y^{2}}$\r\n\r\nDoes anybody know the steps leading to the final solution?\r\n\r\nPROBLEM 2\r\n\r\nFind all solutions to the equation [where $z=x+yi$]:\r\n\r\n$2z^{2}-3\\overline{z}^{2}=10i$\r\n\r\nIt is a quadratic equation and has a conjugate of complex number in it.\r\n\r\nThanks for help in advance.", "Solution_1": "[quote=\"summerBoy\"]I know a lot of complex numbers but in my exercise book I couldn't find a way to solve what I've typed below (I know for final solutions anyway):\n\nPROBLEM 1\n\nFind Re(w) and Im(w) [where $z=x+yi$]:\n\n$w=z+\\frac{1}{z}$\n\nSo we have to write a complex number in a standard form and then extract a real and an imaginary part from there. We work here with the unknown variables ($x$ and $y$).\n\nMy way would be:\n\n$w=z+\\frac{1}{z}=x+yi+\\frac{1}{x+yi}=\\frac{(x+yi)^{2}+1}{x+yi}=\\frac{((x+yi)^{2}+1)(x-yi)}{(x+yi)(x-yi)}=\\frac{((x+yi)^{2}+1)(x-yi)}{x^{2}+y^{2}}=...$\n\nIf I expand the expression further (i.e. numerator) I don't find a way to write it in a standard form.\n\nHowever, the final solution is:\n\n$Re(w)=x+\\frac{x}{x^{2}+y^{2}}$ and $Im(w)=y-\\frac{y}{x^{2}+y^{2}}$\n\nDoes anybody know the steps leading to the final solution?\n\nPROBLEM 2\n\nFind all solutions to the equation [where $z=x+yi$]:\n\n$2z^{2}-3\\overline{z}^{2}=10i$\n\nIt is a quadratic equation and has a conjugate of complex number in it.\n\nThanks for help in advance.[/quote]\r\n\r\nLet \r\n\r\n$z=x+iy\\Rightarrow z+\\frac{1}{z}=x+iy+\\frac{1}{x+iy}=x+iy+\\frac{x-iy}{(x+iy)(x-iy)}=x+\\frac{x}{x^{2}+y^{2}}+i\\left(y-\\frac{y}{x^{2}+y^{2}}\\right)$\r\n\r\n\r\n\r\n$2(x+iy)^{2}-3(x-iy)^{2}=y^{2}-x^{2}+10ixy$. So we have $y^{2}-x^{2}=0,xy=1$ where from we got $x=y=1$ so $z=1+i$", "Solution_2": "summerBoy: complex numbers $\\neq$ complex analysis. Questions of this sort belong in one of the high school fora (maybe Intermediate or PreOlympiad).\r\n\r\ncckek: you should never need to quote the entire previous post. It is a assumed by default that you are responding to the post immediately preceding yours, and quoting the whole thing just makes a mess.\r\n\r\nAlso, you made a small error at the end of the second question: $x^{2}= y^{2}$ and $xy = 1$ has two solutions in the real numbers.", "Solution_3": "Thanks for help, cckek.\r\n\r\nJBL, you are right. There are two solutions to the equation.\r\n\r\nSorry, I posted the topic very quickly, so I didn't notice it doesn't belong to this section. Moderator should move it where it has to be." } { "Tag": [], "Problem": "On Saturday night, one-half of the students in Ms. Schultz's math class went to the dance, one-third of the remainder went to the movies, and the remaining 8 people stayed home. How many students are in her class?", "Solution_1": "1/2 went to the dance and 1/3 of the remainder(or 1/6) went to the movies. This means that the remainder, 8 people, is $ 1 \\minus{} \\left ( \\frac{1}{2} \\plus{} \\frac{1}{6} \\right ) \\equal{} \\frac{1}{3}$ of the class, meaning there's 24 people." } { "Tag": [], "Problem": "The compound $ \\text{MON\\minus{}0585}$ is a non-toxic, biodrgradable larvicide (highly selective against mosquito larvae). Synthesize this compound.\r\n\r\nhttp://ctd.mdibl.org/detail.go;jsessionid=F8289B9243A077E36E3782881A0F7655?type=chem&acc=C006114\r\n\r\nThis link shows what the compound looks like.", "Solution_1": "[hide=\"Answer\"]Here's a possible synthesis from phenol:\n\n1) $ C_6H_5OH \\plus{} C_6H_5COCl/pyr \\longrightarrow C_6H_5\\minus{}O\\minus{}CO\\minus{}C_6H_5$\n\n2) $ C_6H_5\\minus{}O\\minus{}CO\\minus{}C_6H_5 \\plus{} AlCl_3/25^oC \\longrightarrow p\\minus{}C_6H_5\\minus{}CO\\minus{}C_6H_4\\minus{}OH$\n\n3) $ p\\minus{}C_6H_5\\minus{}CO\\minus{}C_6H_4\\minus{}OH \\plus{} NaHO/H_2O/Cl_2 \\longrightarrow\\,\\,\\,sodium\\,\\,4\\minus{}benzoyl\\minus{}2,6\\minus{}dichlorophenoxide$\n\n4) $ sodium\\,\\,4\\minus{}benzoyl\\minus{}2,6\\minus{}dichlorophenoxide \\plus{} CH_3I \\longrightarrow 4\\minus{}benzoyl\\minus{}2,6\\minus{}dichloroanisole$\n\n5) $ 4\\minus{}benzoyl\\minus{}2,6\\minus{}dichloroanisole \\plus{} HOCH_2CH_2OH/H^\\plus{} \\longrightarrow 4\\minus{}benzoyl\\minus{}2,6\\minus{}dichloroanisole ketal$\n\n6) $ 4\\minus{}benzoyl\\minus{}2,6\\minus{}dichloroanisole ketal \\plus{} 2Mg/THF,\\,\\,then 2CO_2,\\,\\, then H_3O^\\plus{} \\longrightarrow 2\\minus{}metoxi\\minus{}5\\minus{}benzoylisophthalic acid$\n\n7) $ 2\\minus{}metoxi\\minus{}5\\minus{}benzoylisophthalic acid \\plus{} excess\\,Me_3Al \\longrightarrow O\\minus{}methyl\\, MON\\minus{}0585$\n\n8) $ O\\minus{}methyl\\, MON\\minus{}0585 \\plus{} conc.HI \\longrightarrow MON\\minus{}0585$.[/hide]", "Solution_2": "By the way, the question allows you to use benzene as well. The aromatic rings can come from benzene, phenol or both.", "Solution_3": "I quite dislike the synthesis given in the book. My primary concern is that it doesn't appear, to me, to be phsically applicable (well who am I to say that) as it involved Friedel Craft Alkylation which is hard to control (and I think they are a bit too specific).\r\n\r\nI will post it if wanted.", "Solution_4": "Yes, post the book's synthesis, but hidden, if you don't mind.", "Solution_5": "[hide=\"Synthesis\"]\n\nStart with benzene:\n\n1. $ \\ce{CH(CH3)2Cl}$ and $ \\ce{AlCl3}$ to make $ \\text{isopropyl benzene}$ (is that the name?)\n\n2. $ \\ce{(PhCO2)2}$ and $ \\ce{CCl4}$ and $ NBS$ to achieve benzylic bromination\n\n3. $ \\ce{FeBr3}$ and phenol to link the previou molecule to the para position of the phenol\n\n4. $ \\ce{C(CH3)3Cl}$ and $ \\ce{AlCl3}$ to introduce $ \\text{tert\\minus{}butyl}$ groups to both positions ortho to the hydroxyl group\n[/hide]\r\n\r\nWouldn't it be hard to be specific about where you want to introduce those alky branches?", "Solution_6": "The first step would be tricky, since polyalkylation is an important side reaction.", "Solution_7": "Exactly, hence my dislike of this method. I am still trying to fully understand your method :oops: . Anyway thanks for the alternative method.\r\n\r\nIn general can we control the degree of branching during Friedel-Craft Alkylation?", "Solution_8": "I don't think so. Anyway, that's not needed, since we have the Friedel-Crafts Acylation reaction as an alternative." } { "Tag": [ "LaTeX" ], "Problem": "Hi -\r\n\r\nI'm trying to get the following equations lined up, so the the equals are all one above the other:\r\n\r\nF_i(t) = \\mbox{fitness of best individual in population i at time t}\r\n\r\nD_i(t) = \\left\\{\r\n\\begin{array}{l l}\r\n \\frac{|F_i(t) - F_i(t - 1)|}{F_i(t - 1)} & \\quad \\mbox{: F_i(t - 1) > 0}\\\\\r\n 0.05 & \\quad \\mbox{: otherwise}\\\\ \\end{array} \\right.\r\n \r\nQ_i(t) = \\left\\{\r\n\\begin{array}{l l l}\r\n \\alpha \\frac{D_i(t)}{\\sum_{j} D_j(t)} + (1 - \\alpha) & \\frac{F_i(t)}{\\sum_{j}F_j(t)} & \\quad \\mbox{: \\sum_{j} D_j(t) > 0}\\\\\r\n & \\frac{F_i(t)}{\\sum_{j}F_j(t)} & \\quad \\mbox{: otherwise}\\\\ \\end{array} \\right.\r\n\r\nI've tried nesting arrays, so that there's one big array with the equals in one column, but I can't seem to get it to work.\r\n\r\nAny ideas? Help much apreciated!\r\n-Asbestos\r\n\r\n[P.S. I haven't been able to get the LaTeX to show: the original uses slash-squarebracket to start and end the equations, the forum aparently doesn't like that, nor replacing those with dollar-signs... Couldn't work it out, sorry]", "Solution_1": "Are you familiar with [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php#eqnarray]eqnarray[/url]?\r\n\r\n(link is fixed now)", "Solution_2": "i think your link is invalid...", "Solution_3": "Here is one that works. Use [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php#eqnarray]eqnarray[/url]. I know why rruscyzk's doesn't work. He used his local computer name instead of artofproblemsolving.com. e.g. If i type in my computer name(in this case it is server) it would go to dasarathy0.tripod.com" } { "Tag": [ "inequalities", "trigonometry", "inequalities unsolved" ], "Problem": "prove that if $ a,b,c \\ge 2$ satisfy the condition $ a^2\\plus{}b^2\\plus{}c^2\\equal{}abc\\plus{}4$\r\nthen\r\n$ a \\plus{} b \\plus{} c \\plus{} ac \\plus{} bc \\ge 2\\sqrt {(a \\plus{} b \\plus{} c \\plus{} 3)({a^2} \\plus{} {b^2} \\plus{} {c^2} \\minus{} 3)}$", "Solution_1": "[quote=\"Victory.US\"]prove that if $ a,b,c \\ge 2$ satisfy the condition $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} abc \\plus{} 4$\nthen\n$ a \\plus{} b \\plus{} c \\plus{} ac \\plus{} bc \\ge 2\\sqrt {(a \\plus{} b \\plus{} c \\plus{} 3)({a^2} \\plus{} {b^2} \\plus{} {c^2} \\minus{} 3)}$[/quote]\r\nit's must be like this \r\n$ \\sum a \\plus{} \\sum ab\\ge 2\\sqrt {(a \\plus{} b \\plus{} c \\plus{} 3)({a^2} \\plus{} {b^2} \\plus{} {c^2} \\minus{} 3)}$\r\nand this is my solution\r\nassume that c=min{a,b,c}\r\nlet $ a \\plus{} b \\equal{} x$\r\nthe $ ab \\equal{} \\frac {x^2 \\plus{} c^2 \\minus{} 4}{2 \\plus{} c}$\r\nwe can rewrite the ineq like this\r\n$ \\frac {x^2 \\plus{} c^2 \\minus{} 4}{2 \\plus{} c} \\plus{} cx \\plus{} x \\plus{} c\\ge 2\\sqrt {(\\frac {x^2 \\plus{} c^2 \\minus{} 4}{2 \\plus{} c}*c \\plus{} 1)(c \\plus{} x \\plus{} 3)}$\r\nwhich is equivalent to \r\n$ \\frac {(x^2 \\minus{} 4c \\minus{} 8)(2 \\minus{} c \\minus{} c^2 \\plus{} x^2)^2}{(c \\plus{} 2)^2}\\ge 0$\r\nQ.e.d", "Solution_2": "i post this problem and wnat it to be sloved by trigonometry method? does it have the solution use trigonom\u1ebbtry?" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Prove the following:\r\n\r\n$ \\forall_{n}\\geq 0$\r\n\\[ (n+2)^{n}=\\sum_{k=0}^{n}(\\left(\\frac{n!}{k!(n-k)!}\\right) (k+1)^{k}(n-k+1)^{(n-k-1)})\\]", "Solution_1": "$ (n+2)^{n}=[(n-k+1)+(k+1)]^{n}=\\sum_{k=0}^{n}C_{n}^{k}(n-k+1)^{n-k}(k+1)^{k}$", "Solution_2": "red_dog, I'm not sure you can do that without a good explanation.\r\nA direct use of Newton's binomial would be\r\n$ (n+2)^{n}=[(n-k+1)+(k+1)]^{n}=\\sum_{i=0}^{n}\\binom{n}{i}(n-k+1)^{n-i}(k+1)^{i}$,\r\nsince a dummy index should not be a variable already used.", "Solution_3": "and it's not the same expression, there is an extra $ (n-k+1)$", "Solution_4": "[url]http://www.mathlinks.ro/viewtopic.php?p=885961#885961[/url]" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Suppose $ f: R^\\plus{} \\rightarrow R^\\plus{}$ is a decreasing continuous function such that for all $ x,y \\in R^\\plus{}$,\r\n\r\n$ f(x \\plus{} y) \\plus{} f(f(x) \\plus{} f(y)) \\equal{} f(f(x \\plus{} f(y))) \\plus{} f(y \\plus{} f(x))$\r\n\r\nProve that $ f(f(x))\\equal{}x$.", "Solution_1": "[size=150]\u0633\u0644\u0627\u0645\n\u0633\u0648\u0627\u0644\u200c\u0633\u062e\u062a\u06cc\u0647\n\u062e\u0648\u062f\u062a\u200c\u0628\u0647\u200c\u0686\u06cc\u0632\u06cc\u200c\u0646\u0631\u0633\u06cc\u062f\u06cc\u061f\n\u0627\u06cc\u0646\u0648\u200c\u0641\u0627\u0631\u0633\u06cc\u200c\u06af\u0641\u062a\u0645\u200c\u062a\u0627\u200c\u0628\u0642\u06cc\u0647\u200c\u062f\u0631\u200c\u062c\u0631\u06cc\u0627\u0646\u200c\u0646\u0628\u0627\u0634\u0646\u062f[/size] :lol:", "Solution_2": "[quote=\"irantst\"][size=150]\u0633\u0644\u0627\u0645\n\u0633\u0648\u0627\u0644\u200c\u0633\u062e\u062a\u06cc\u0647\n\u062e\u0648\u062f\u062a\u200c\u0628\u0647\u200c\u0686\u06cc\u0632\u06cc\u200c\u0646\u0631\u0633\u06cc\u062f\u06cc\u061f\n\u0627\u06cc\u0646\u0648\u200c\u0641\u0627\u0631\u0633\u06cc\u200c\u06af\u0641\u062a\u0645\u200c\u062a\u0627\u200c\u0628\u0642\u06cc\u0647\u200c\u062f\u0631\u200c\u062c\u0631\u06cc\u0627\u0646\u200c\u0646\u0628\u0627\u0634\u0646\u062f[/size] :lol:[/quote]\r\n\r\n\u0633\u0644\u0627\u0645\u060c\u0646\u0647 \u0646\u0634\u062f \u0647\u0631 \u0643\u0627\u0631\u064a \u0643\u0631\u062f\u0645 \u064a\u0647 \u062c\u0627\u064a\u064a \u06af\u064a\u0631 \u0645\u064a\u0643\u0646\u0647[size=150]..\n\u0645\u0631\u0633\u064a \u0627\u0632 \u062c\u0648\u0627\u0628 \u0633\u0631\u064a\u0639\u062a.\n\u0627\u0645\u064a\u0631\u062d\u0633\u064a\u0646[/size]", "Solution_3": "\u0644\u064a\u0633 \u0644\u062f\u064a \u0623\u064a \u0641\u0643\u0631\u0629 \u0645\u0627 \u0642\u0644\u062a\u0645\u0648\u0647", "Solution_4": "It`s Iran TST 1997 . If you really interested in see a solution I can post it.\n\n[hide=\"@amparvardi\"]asoone ha ! khodet mitooni hal koni. rasti mikhain az in joor harfa bezanid PM bezanid behtare , choon age in ja farsi type koni maro ba ARABaye bi pedar , heivan sefat eshtebah migiran ...[/hide]" } { "Tag": [], "Problem": "An elevator of mass 2000k.g rises with acceleration of 1m/s.What is the force with which the cable is puling the elevator?", "Solution_1": "$F=m(g+a)$", "Solution_2": "You're correct. Another way to think of this is to measure the acceleration against the gravational reference, which is accelerating towards the center of the earth at g. If the elevator was \"stationary\" relative to the ground, it would have a F equal to the weight of the elevator (sort of the definition of weight). One then merely has to add the F from the apparent acceleration. You get the same formula, but this is just a different way to think about why.", "Solution_3": "Yes the above are correct.All we have to think is that to attain an accleration of $1m/s^2$,we cable of the elevator will have to apply a force to accelerate the elevator by $11m/s^2$(since we need a net acceleratuion of $1m/s^2$)" } { "Tag": [ "calculus", "integration", "function", "logarithms", "derivative", "algorithm", "Euler" ], "Problem": "Can some one show me how to evaluate this integral?\r\n\r\nintegral(e^[-x^2])dx.\r\n\r\nThanks", "Solution_1": "Ah, Jimmy. Is it time to let you in on some of our secrets?\r\n\r\nLet's start with some terminology. We have certain \"building block\" functions: the powers of $x$, the exponential and the logarithm, the trigonometric and inverse trigonometric functions. Any function that we can created from these elements using a finite number of steps of addition, subtraction, multiplication, division, and composition, is called an \"elementary function.\" By that definition, $e^{-x^2}$ is clearly an elementary function.\r\n\r\nTheorem: the derivative of an elementary function is an elementary function. The proof of this theorem is by algorithm: we know the (elementary) derivatives of the building block functions, and we know how to combine them using the product rule, quotient rule, and chain rule.\r\n\r\nThe converse of that theorem is utterly false. There exist elementary functions whose antiderivatives (primitives) are not elementary. A notorious example of an elementary function with no elementary primitive is $e^{-x^2}.$ (This is known, but even hinting at the proof would take us too far afield here.)\r\n\r\nIn fact, your default assumption should be that if an elementary function doesn't obviusly have an elementary primitive, the chances if it having an elementary primitive at all are at best remote. This applies to simple-looking formulas like $(1+x^3)^{1/3}$ and to complicated messes.\r\n\r\nBut don't read too much into that. Does $e^{-x^2}$ have a primitive? Absolutely yes, as we can claim for all continuous functions. We can simply write that primitive as $\\int_0^xe^{-t^2}dt.$ The integral exists because any continuous function is integrable on any finite interval.\r\n\r\nCould we write the primitive as a power series? Yes, easily and usefully. Just write the power series for $e^{-x^2}$ and integrate it term by term. Could we estimate any $\\int_a^be^{-x^2}dx$ as closely as we please? Certainly, using either the power series or some technique of numerical integration.\r\n\r\nCan any definite integrals be computed exactly? In this case yes:\r\n\r\n$\\int_{-\\infty}^{\\infty}e^{-x^2}dx=\\sqrt{\\pi}.$", "Solution_2": "Thanks,\r\nbut could you show me where I could find int(e^[-x^2]) worked out analytically? It's driving me crazy", "Solution_3": "What exactly do you mean by \"worked out analytically\"? Are you referring to a particular definite integral? If that's what you mean, you should say so. Do you mean as an analytic function (that is, a power series)? If that's what you mean, you should say so. If you're looking for a (closed form) formula for the antiderivative, then you're asking for something that doesn't exist, and you should re-read what I wrote above.", "Solution_4": "Sorry for being unclear, I meant an analytic function.", "Solution_5": "But I should be able to figure this out on my own, I wasn't thinking too clearly earlier. \r\nThanks for your help", "Solution_6": "As I remember the shortest way to count the Euler-Poisson integral is by using that $\\lim_{n{\\rightarrow}+\\infty}{(1+\\frac{x^2}{n})^{-n}}=e^{-x^2}$ and so writing\r\n$\\lim_{n{\\rightarrow}+\\infty}{\\int_{0}^{+\\infty}{(1+\\frac{x^2}{n})^{-n}dx}}=\\int_{0}^{+\\infty}{e^{-x^2}dx}$(of course with testing some conditions).", "Solution_7": "hey kent, you said that the primitive of e^(-x^2)dx is square root of (2*pi)?\r\n\r\nI used my 89 to solve that primitive from negative infinity to infinity and i got 1.77 which is not the same as square root of (2*pi).\r\n\r\nCan you tell me what i did wrong?", "Solution_8": "Actually there is a typo in Kents post.$\\int_{-\\infty}^{+\\infty}{e^{-x^2}dx}=\\sqrt{\\pi}$", "Solution_9": "and how would you get a definite integral of this function? seems like he just integrated it from -infiniti to infiniti", "Solution_10": "Dear [b]zubairkhan14[/b] have you read [b]post #7[/b]?", "Solution_11": "yes i did, but i dont understand it. if you're given that integral, how do you come up with a definite integral for that. is there a way to do it by substitution, trig, partial fractions, integ by parts?", "Solution_12": "[quote=\"LevonNurbekian\"]As I remember the shortest way to count the Euler-Poisson integral is by using that $\\lim_{n{\\rightarrow}+\\infty}{(1+\\frac{x^2}{n})^{-n}}=e^{-x^2}$ and so writing\n$\\lim_{n{\\rightarrow}+\\infty}{\\int_{0}^{+\\infty}{(1+\\frac{x^2}{n})^{-n}dx}}=\\int_{0}^{+\\infty}{e^{-x^2}dx}$(of course with testing some conditions).[/quote]\r\n\r\nI thought the shortest path is the following:\r\n$I^2=\\int_{\\mathbb R^2}e^{-x^2-y^2}dxdy=\\int_0^{+\\infty}\\int_{0}^{2\\pi}re^{-r^2}dr=-2\\pi e^{-r^2}|_0^{+\\infty}=2\\pi$. :D", "Solution_13": "Note that it is sufficent to calculate the integral $\\int_{0}^{+\\infty}{\\frac{dx}{(1+x^2)^n}}$.Use trigonometric substitution $x=\\tan{z}$ $z\\in[0,\\frac{\\pi}{2}]$.You will come to the integral of type $\\int_{0}^{\\frac{\\pi}{2}}{{\\sin{x}}^n dx}$.I think this one will be in you r calculus coursebook because it is famous integral.", "Solution_14": "[quote=\"Myth\"][quote=\"LevonNurbekian\"]As I remember the shortest way to count the Euler-Poisson integral is by using that $\\lim_{n{\\rightarrow}+\\infty}{(1+\\frac{x^2}{n})^{-n}}=e^{-x^2}$ and so writing\n$\\lim_{n{\\rightarrow}+\\infty}{\\int_{0}^{+\\infty}{(1+\\frac{x^2}{n})^{-n}dx}}=\\int_{0}^{+\\infty}{e^{-x^2}dx}$(of course with testing some conditions).[/quote]\n\nI thought the shortest path is the following:\n$I^2=\\int_{\\mathbb R^2}e^{-x^2-y^2}dxdy=\\int_0^{+\\infty}\\int_{0}^{2\\pi}re^{-r^2}dr=-2\\pi e^{-r^2}|_0^{+\\infty}=2\\pi$. :D[/quote]\r\nThe set of your memory is larger than mine [b]Myth[/b]. :D" } { "Tag": [], "Problem": "Hi, I was wondering if it is too late to get your school to sign up for Mandlebrot? I want to take it but it is hard to convince my school to do stuff like that.", "Solution_1": "Its not too late.", "Solution_2": "im thinking about getting my school to do mandelbrot too. the only problem is that most of my friends already take it at this math program called icae, which i cannot attend. so im not sure if they can continue taking the exam there (which they will want to, given icae's sucess in this contest) or if they will be forced to take it our school. of course, if they can continue with icae, i will be the only one doing the team contest, which means our scores will be dismally low. :D", "Solution_3": "I'm really excited this year. I founded \"Math Club\" at our school and convinced my math teacher to sign up for the Mandelbrot. Ya. Just thought I'd let you know.\r\n\r\nAs for the deadline, just check http://www.mandelbrot.org. If needbe, you could send them a message using the \"Contact Us\" page. \r\n\r\nP.S. Does anyone know what happened to Mandelbrot Midlevels? I thought there used to be a competition called that (at http://www.midlevels.org) but it's not there..." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Do there exist a function $f: R\\to R$ such that for all $x,y\\in R$\r\n$f(x^{2}y+f(x+y^{2}))=x^{3}+y^{3}+f(xy)$?", "Solution_1": "x=0 give $f(f(y^{2}))=y^{3}+f(0)$. Therefore $-y^{3}+f(0)=f(f((-y)^{2}))=f(f(y^{2}))=y^{3}+f(0) \\Longrightarrow y^{3}=-y^{3}\\ \\forall y$ contradition.", "Solution_2": "Sorry, I can't understand what you mean.But ,this problem is quite easy.\r\nfor x=0:ff(y^2)=y^3+f(0)\r\nfor y=0:ff(x)=x^3+f(0)" } { "Tag": [ "Putnam" ], "Problem": "Determine the units digit of [1020000/(10100 + 3)].", "Solution_1": "A hint: you might want to use the binomial theorem :D", "Solution_2": "being...", "Solution_3": "'Het binomium van Newton' i.e. the expansion of (a + b)n." } { "Tag": [ "probability" ], "Problem": "the letters in the word $ RANDOM$ are arranged randomly. What is the probability that an arrangement chosen at random ends with a vowel?", "Solution_1": "[hide=\"Solution\"]Number of ways to have it end with a vowel:\n$ 2$ choices for the last letter, $ 5!$ ways to place the rest of the letters.\n\nTotal number of arrangements: $ 6!$.\n\nAnswer: $ \\frac {2\\cdot 5!}{6!} \\equal{} \\frac {2}{6} \\equal{} \\boxed{\\frac {1}{3}}$.[/hide]", "Solution_2": "Why are there 4! ways to arrange the rest of the letters?", "Solution_3": "since, other than the last two letters, there are 4 letters left that we can arrange in any random order.", "Solution_4": "Sorry, I was initially multiplying by $ \\binom{5}{1}$ before that, to choose a letter to be a vowel, then I realized that I could just be multiplying by $ 5!$, and then forgot to change it for some reason.", "Solution_5": "Or you could just realize that there are 6 total letters all different and two vowels, so the chance is automatically 2/6 or 1/3." } { "Tag": [ "AMC" ], "Problem": "Any idea why the [url=http://www.unl.edu/amc/a-activities/a7-problems/USAMO-IMO/q-tse/-pdf/tse01.pdf]USA TST 2001[/url] had 9 problems instead of 6? Assuming that wasn't a mistake, on the AoPS resources section, there are only 1-6, so have 7-9 been posted (just not added to the resources section)?", "Solution_1": "Because there was a 3-way tie for the 6th position on the team after the first 6 problems." } { "Tag": [ "Asymptote" ], "Problem": "I get this asymptote error when running the following script\r\n\r\n[code]size(150);\nunitsize(1cm);\ndraw((-7,-7)--(7,7));[/code]\n\non Windows Vista x64, Asymptote 1.77\n\nError I get \n[code]C:\\tmp>asy -verbose temp.asy\nProcessing temp\ntemp.asy: 1.1: invalid token '\u00ef'\ntemp.asy: 1.2: invalid token '\u00bb'\ntemp.asy: 1.3: invalid token '\u00bf'[/code]\r\n\r\nAny ideas?", "Solution_1": "It seems like your program file has encoding different from the standard ASCII, so what your editor shows as \"siz\" is perceived as a strange chain of fancy symbols. What editor did you use to create the file?", "Solution_2": "It's PSPAD. But I'm using normal settings, the file looks the same in Notepad (simple notepad), Texnic center and so on ...", "Solution_3": "Are you still able to compile other .asy files? Have you tried to rename the file? (i.e., are you absolutely sure that test.asy that you created is the same test.asy as the one being compiled?).", "Solution_4": "Nevermind you're right. The file was UTF-8 encoded, and it needs to be ANSII ..." } { "Tag": [ "geometry", "incenter", "inradius", "trigonometry", "geometry solved" ], "Problem": "This problem is making me some sirious troubles....\r\n\r\nLet $I$ be the incentre of $\\triangle ABC$, $r$ inradius, $M$ the midpoint of side $AB$ and $N$ the midpoint of $CM$. Prove that if $r=CN-IN$ then $AC=BC$ or $\\angle ACB=90^o$. \r\n\r\nWho solve it first get's a CANDY.... :) :)", "Solution_1": "Let me say the problem in another word. :) \r\nLet the $m_a$ be a median of the triangle $ABC$(from $A$ to $BC$).\r\nWe know that the circle with diameter $m_a$ is tangent to Incircle.\r\nProve that $A=90$ or $AB=AC$", "Solution_2": "[quote=\"lomos_lupin\"]Let me say the problem in another word. :) \nLet the $m_a$ be a median of the triangle $ABC$(from $A$ to $BC$).\nWe know that the circle with diameter $m_a$ is tangent to Incircle.\nProve that $A=90$ or $AB=AC$[/quote]\r\n\r\nWell....... that is not so helpful....\r\nBut, do you know what wloud be very helpful....\r\n\r\n$SOLUTION$ :) :) :) :) :) :)", "Solution_3": "[quote=\"lomos_lupin\"]We know that the circle with diameter $m_a$ is tangent to Incircle.\n[/quote]\r\nThe circle with diameter $m_a$ also passes through $D$, the foot of the altitude from $A$ (I'm using the notations of lomos_lupin). If $AB\\ne AC$ then $D$ and $M$ are distinct. It is not difficult to see that considering the pencil of circles passing through $D$ and $M$, only one is tangent to the incircle (having the incircle in its interior). But it is well known that the nine point circle is tangent to the incircle (and, of course, passes through $D$ and $M$). Therefore, the circle with diameter $m_a$ coincides with the nine point circle. This means that $A$ is the orthocenter of $ABC$, that is, $\\angle A=90^{\\circle}$.", "Solution_4": "Do you see lupin...\r\n\r\nthat is the $SOLUTION$\r\ntnx\r\n\r\nJust kidding, lupin did a lot of deal with his \"let's put the problem in another word...\"\r\n\r\n :) :) :) \r\n\r\nm@re", "Solution_5": "Thank God! This problem killed me. Thanks, Sir Enescu (Bogdan ?) and congratulations !", "Solution_6": "Actually, thanks are due to lomos_lupin. His rewording of the problem gave me the idea.", "Solution_7": "nice solution ,enescu.\r\n\r\n\r\nTo M@re:Yup i didnt gave the solution ,but just a simple hint.\r\nThe real work is for enescu\r\nI think it was a nice problem.\r\nBy the way \r\n[quote=\"M@re\"]\n\nWho solve it first get's a CANDY.... :) :)[/quote]\r\n,Enescu wants a candy :D ?", "Solution_8": "Still once congratulations ( :first: - the first candy) for the his solution, to [i]Bogdan Enescu[/i],\r\nand ( :first: - the second candy) for the his thought of the tangent, to my friend\r\n(from this moment) [i]lomos lupin[/i] (and no Arsene, he knows why) !.\r\nHere is (now it is easily) a variation on the same theme (lupin's notes):\r\n$AN-IN=r\\Longleftrightarrow I_a N-AN=r_a\\Longleftrightarrow b=c\\ or\\ A=90^{\\circ}.$", "Solution_9": "It is ready ! At last I managed to prove (with the Lupin's notations) that:\\[AN-IN=r\\Longleftrightarrow I_aN-AN=r_a\\Longleftrightarrow \\frac {(s-a)}{r}=\\frac {2m_a}{a}\\Longleftrightarrow\\]\\[m_a\\cdot h_a=s(s-a)\\Longleftrightarrow b=c \\ \\vee\\ A=90^{\\circ}\\][b]Remark.[/b] $S=\\frac 12 ah_a=sr=s(s-a)\\tan \\frac A2 .$\r\n\r\nYou point out if you should like to present the its solution. All right !" } { "Tag": [], "Problem": "\u0395\u03b9\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03bf \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03b1 \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c4\u03bf ,\u03bd\u03b1\u03b9 \u03b7 \u03bf\u03c7\u03b9 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1\u03c4\u03b9:\r\n\r\n{$ e^x,e^{2x}$}", "Solution_1": "\u039f\u03b9 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 $ e^x,e^{2x}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u0393. \u03b1\u03bd\u03b5\u03be\u03ac\u03c1\u03c4\u03b7\u03c4\u03b5\u03c2 (\u03ae \u03bc\u03b5 \u03ac\u03c4\u03bf\u03c0\u03bf \u03ae \u03bc\u03b5 \u03bf\u03c1\u03af\u03b6\u03bf\u03c5\u03c3\u03b1...)", "Solution_2": "\u0395\u03c7\u03c9 \u03b5\u03bd\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03bc\u03b5 \u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03b8\u03b5\u03c4\u03c9\u00a8\r\n\r\n\r\n\u0395\u03be \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03c5 \u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b3\u03c1\u03b1\u03bc \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c4\u03bf \u03b5\u03b1\u03bd >\r\n\r\n\u03b3\u03b9\u03b1 \u03bf\u03bb\u03b1 \u03c4\u03b1 a,b,x \u03ba\u03b1\u03b9 $ ae^x \\plus{} be^{2x}\\equal{}0$,\u03c4\u03bf\u03c4\u03b5 a=b=0 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03bc\u03bf\u03c1\u03c6\u03b7 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03b9\u03ba\u03c4\u03c9\u03bd \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5:\r\n\r\n$ \\forall a\\forall b\\forall x[ (a\\in\\Re\\wedge b\\in\\Re\\wedge x\\in\\Re\\wedge ae^x \\plus{} be^{2x} \\equal{} 0)\\Longrightarrow a\\equal{}b\\equal{}0]$..........................................................................................1\r\n\r\nT\u03c9\u03c1\u03b1 ,\u03b7 \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03b7\u03c2 (1) [b]\u03b7 \u03bf\u03c0\u03bf\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03b5\u03c4\u03b1\u03b9 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03b7 \u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03b7[/b] \u03b5\u03b9\u03bd\u03b1\u03b9\u0384\r\n\r\n\r\n[b] \u03c5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd a,b,x[/b] \u03b5\u03c4\u03c3\u03b9 \u03c9\u03c3\u03c4\u03b5 $ ae^x \\plus{} be^{2x} \\equal{} 0$ [b] KAI [/b] $ a\\neq 0,b\\neq 0$\r\n\r\n[b]\u0395\u03c0\u03bf\u03bc\u03b5\u03bd\u03c9\u03c2 \u03b5\u03b1\u03bd \u03b2\u03b1\u03bb\u03bb\u03bf\u03c5\u03bc\u03b5 : a=2,b=-2 ,x=0,\u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 $ ae^x \\plus{} be^{2x} \\equal{}0$ \u03ba\u03b1\u03b9 \u03b5\u03c4\u03c3\u03b9 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03b7\u03c2 (1) \u03ba\u03b1\u03b9 \u03b1\u03c1\u03b1 \u0395\u03a7\u039f\u03a5\u039c\u0395 \u0393\u03a1\u0391\u039c\u039c\u0399\u039a\u0397 \u0395\u039e\u0391\u03a1\u03a4\u0397\u03a3\u0397[/b]", "Solution_3": "[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\n\u03b3\u03b9\u03b1 \u03bf\u03bb\u03b1 \u03c4\u03b1 a,b,x \u03ba\u03b1\u03b9 $ ae^x \\plus{} be^{2x} \\equal{} 0$,\u03c4\u03bf\u03c4\u03b5 a=b=0 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03bc\u03bf\u03c1\u03c6\u03b7 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03b9\u03ba\u03c4\u03c9\u03bd \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5:[/quote]\r\n\r\n\u0397 \u03c3\u03c9\u03c3\u03c4\u03ae \u03b4\u03b9\u03b1\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9: \u03bf\u03b9 $ e^x, e^{2x}$ \u03bb\u03ad\u03b3\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ac \u03b1\u03bd\u03b5\u03be\u03ac\u03c1\u03c4\u03b7\u03c4\u03b5\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03c4\u03cc\u03c4\u03b5 \u03cc\u03c4\u03b1\u03bd \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ a,b\\in\\mathbb{R}$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 $ ae^x \\plus{} be^{2x} \\equal{} 0\\,\\forall x\\in\\mathbb{R}$, \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $ a \\equal{} b \\equal{} 0$.\r\n\r\n\u0391\u03c0\u03cc \u03c4\u03b7 \u03c3\u03c4\u03b9\u03b3\u03bc\u03ae \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $ 2e^x \\minus{} 2e^{ 2 x} \\equal{} 0\\,\\forall x\\in\\mathbb{R}$, \u03c4\u03bf \u03b1\u03bd\u03c4\u03b9\u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03ad\u03b4\u03c9\u03c3\u03b5\u03c2 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c9\u03c3\u03c4\u03cc.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_4": "[quote=\"Durandal\"][quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\n\u03b3\u03b9\u03b1 \u03bf\u03bb\u03b1 \u03c4\u03b1 a,b,x \u03ba\u03b1\u03b9 $ ae^x \\plus{} be^{2x} \\equal{} 0$,\u03c4\u03bf\u03c4\u03b5 a=b=0 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03bc\u03bf\u03c1\u03c6\u03b7 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03b9\u03ba\u03c4\u03c9\u03bd \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5:[/quote]\n\n\u0397 \u03c3\u03c9\u03c3\u03c4\u03ae \u03b4\u03b9\u03b1\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9: \u03bf\u03b9 $ e^x, e^{2x}$ \u03bb\u03ad\u03b3\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ac \u03b1\u03bd\u03b5\u03be\u03ac\u03c1\u03c4\u03b7\u03c4\u03b5\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03c4\u03cc\u03c4\u03b5 \u03cc\u03c4\u03b1\u03bd \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ a,b\\in\\mathbb{R}$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 $ ae^x \\plus{} be^{2x} \\equal{} 0\\,\\forall x\\in\\mathbb{R}$, \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $ a \\equal{} b \\equal{} 0$.\n\n[/quote]\r\n\r\n\u03a0\u03bf\u03b9\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b7 \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03c5 \u03c3\u03bf\u03c5??\r\n\r\n\u0395\u0391\u039d \u03b3\u03c1\u03b1\u03c8\u03b5\u03b9\u03c2 \u03c4\u03bf\u03bd \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf \u03c3\u03bf\u03c5 \u03c3\u03b5 \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03b1 \u03b8\u03b1 \u03b7\u03c4\u03b1\u03bd \u03c0\u03bf\u03b9\u03bf \u03b5\u03c5\u03ba\u03bf\u03bb\u03bf \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7", "Solution_5": "\u03a3\u03a5\u0393\u039d\u03a9\u039c\u0397 \u03b8\u03b5\u03bb\u03c9 \u03bd\u03b1 \u03c0\u03c9 ,\u03bf\u03c4\u03b9 \u03b5\u03b1\u03bd \u03b3\u03c1\u03b1\u03c8\u03b5\u03b9\u03c2 \u03c4\u03bf\u03bd \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf \u03c3\u03bf\u03c5 \u03c3\u03b5 \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03b1 \u03c4\u03bf\u03c4\u03b5 \u03b8\u03b1 \u03b4\u03b5\u03b9\u03c2 \u03bf\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b4\u03b9\u03b1\u03c6\u03b5\u03c1\u03b5\u03b9 \u03b1\u03c0\u03bf \u03c4\u03bf\u03bd \u03b4\u03b9\u03ba\u03bf \u03bc\u03bf\u03c5 ,\u03b5\u03c0\u03bf\u03bc\u03b5\u03bd\u03c9\u03c2 \u03b7 \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bb\u03b1\u03b8\u03bf\u03c2", "Solution_6": "[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\u03a3\u03a5\u0393\u039d\u03a9\u039c\u0397 \u03b8\u03b5\u03bb\u03c9 \u03bd\u03b1 \u03c0\u03c9 ,\u03bf\u03c4\u03b9 \u03b5\u03b1\u03bd \u03b3\u03c1\u03b1\u03c8\u03b5\u03b9\u03c2 \u03c4\u03bf\u03bd \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf \u03c3\u03bf\u03c5 \u03c3\u03b5 \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03b1 \u03c4\u03bf\u03c4\u03b5 \u03b8\u03b1 \u03b4\u03b5\u03b9\u03c2 \u03bf\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b4\u03b9\u03b1\u03c6\u03b5\u03c1\u03b5\u03b9 \u03b1\u03c0\u03bf \u03c4\u03bf\u03bd \u03b4\u03b9\u03ba\u03bf \u03bc\u03bf\u03c5 ,\u03b5\u03c0\u03bf\u03bc\u03b5\u03bd\u03c9\u03c2 \u03b7 \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bb\u03b1\u03b8\u03bf\u03c2[/quote]\r\n\r\nNope. \u0394\u03b5\u03bd \u03b1\u03b3\u03b1\u03c0\u03ac\u03c9 \u03b9\u03b4\u03b9\u03b1\u03af\u03c4\u03b5\u03c1\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03af\u03ba\u03c4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03c4\u03b1 \u03c0\u03ac\u03c3\u03b1 \u03c0\u03b9\u03b8\u03b1\u03bd\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b4\u03b5 \u03b8\u03b1 \u03c4\u03bf \u03b3\u03c1\u03ac\u03c8\u03c9 \u03c3\u03c9\u03c3\u03c4\u03ac (\u03bf\u03c0\u03cc\u03c4\u03b5 Dem, please correct me), \u03b1\u03bb\u03bb\u03ac \u03bf \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03bc\u03bf\u03c5 \u03c3\u03b5 \u03c3\u03cd\u03bc\u03b2\u03bf\u03bb\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9:\r\n\r\n$ \\forall a \\forall b\\left[ (a\\in\\mathbb{R})\\wedge (b\\in\\mathbb{R})\\wedge [\\forall x[(x\\in\\mathbb{R})\\wedge(ae^x \\plus{} be^{2x} \\equal{} 0)]] \\implies (a \\equal{} b \\equal{} 0) \\right]$\r\n\r\nEdit \u03b5\u03c0\u03b5\u03be\u03ae\u03b3\u03b7\u03c3\u03b7\u03c2: \u03b1\u03bd $ a,b\\in\\mathbb{R}$ \u03ba\u03b1\u03b9 $ ae^x \\plus{} be^{2x} \\equal{} 0, \\forall x\\in\\mathbb{R}$ \u03c4\u03cc\u03c4\u03b5 $ a\\equal{}b\\equal{}0$. \u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03c4\u03ce\u03c1\u03b1 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03b5\u03b9\u03c2 \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03b1 \u03b3\u03b9\u03b1\u03c4\u03af \u03bf\u03b9 \u03b4\u03cd\u03bf \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03bf\u03af...\r\n\r\n\u03a7\u03bc\u03bc\u03bc, \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c0\u03c9 \u03cc\u03c4\u03b9 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03c9 \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03c0\u03b9\u03bc\u03ad\u03bd\u03b5\u03b9\u03c2 \u03c4\u03cc\u03c3\u03bf \u03bd\u03b1 \u03b3\u03c1\u03ac\u03c6\u03b5\u03b9\u03c2 \u03c4\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03bc\u03b5 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03af\u03ba\u03c4\u03b5\u03c2, \u03b1\u03bb\u03bb\u03ac \u03c0\u03b5\u03c1\u03af \u03bf\u03c1\u03ad\u03be\u03b5\u03c9\u03c2... ;)\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_7": "\u0393\u0399\u0391\u03a4\u0399 \u039c\u039f\u039d\u039f \u03b5\u03b1\u03bd \u03b3\u03c1\u03b1\u03c8\u03b5\u03b9\u03c2 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b1\u03c3\u03b7 \u03c3\u03b5 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03b9\u03ba\u03c4\u03b5\u03c2 \u03c3\u03c9\u03c3\u03c4\u03b1 ,\u03c4\u03bf\u03c4\u03b5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03c0\u03c1\u03bf\u03c4\u03b1\u03c3\u03b5\u03c9\u03c2 .\u0394\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03b1 [b]\u03bc\u03c0\u03bf\u03c1\u03b5\u03b9[/b] \u03bd\u03b1 \u03ba\u03b1\u03bd\u03b5\u03b9\u03c2 \u03bb\u03b1\u03b8\u03bf\u03c2.\r\n\r\n\u0391\u03bb\u03bb\u03b1 \u03b1\u03c2 \u03c0\u03b5\u03c1\u03b9\u03bc\u03b5\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b4\u03bf\u03c5\u03bc\u03b5 \u03b1\u03bd \u03c3\u03c5\u03bc\u03c6\u03c9\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bf \u0394\u03b7\u03bc\u03b7\u03c4\u03c1\u03b7\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c7\u03b9\u03b6\u03bf\u03c5\u03bc\u03b5\u00a8.", "Solution_8": "[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\u0393\u0399\u0391\u03a4\u0399 \u039c\u039f\u039d\u039f \u03b5\u03b1\u03bd \u03b3\u03c1\u03b1\u03c8\u03b5\u03b9\u03c2 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b1\u03c3\u03b7 \u03c3\u03b5 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03b9\u03ba\u03c4\u03b5\u03c2 \u03c3\u03c9\u03c3\u03c4\u03b1 ,\u03c4\u03bf\u03c4\u03b5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03c0\u03c1\u03bf\u03c4\u03b1\u03c3\u03b5\u03c9\u03c2 .\u0394\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03b1 [b]\u03bc\u03c0\u03bf\u03c1\u03b5\u03b9[/b] \u03bd\u03b1 \u03ba\u03b1\u03bd\u03b5\u03b9\u03c2 \u03bb\u03b1\u03b8\u03bf\u03c2.[/quote]\n\nSure, whatever. \u0391\u03bb\u03bb\u03ac \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03ad\u03c8\u03bf\u03c5 \u03cc\u03c4\u03b9 \u03be\u03b5\u03ba\u03b9\u03bd\u03ac\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf \u03c3\u03c9\u03c3\u03c4\u03cc \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc \u03c0\u03c1\u03b9\u03bd \u03c4\u03bf\u03bd \u03b3\u03c1\u03ac\u03c8\u03b5\u03b9\u03c2 \u03c3\u03b5 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03af\u03ba\u03c4\u03b5\u03c2... \u03c4\u03bf \u03bb\u03ad\u03c9 \u03b1\u03c5\u03c4\u03cc \u03b3\u03b9\u03b1\u03c4\u03af \u03bf \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03c0\u03bf\u03c5 \u03ad\u03b4\u03c9\u03c3\u03b5\u03c2:\n\n[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\u03b3\u03b9\u03b1 \u03bf\u03bb\u03b1 \u03c4\u03b1 a,b,x \u03ba\u03b1\u03b9 $ ae^x \\plus{} be^{2x} \\equal{} 0$, \u03c4\u03bf\u03c4\u03b5 a=b=0 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03bc\u03bf\u03c1\u03c6\u03b7 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03b9\u03ba\u03c4\u03c9\u03bd \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5:\n[...][/quote]\r\n\r\n\u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03bd \u03bf \u03c3\u03c9\u03c3\u03c4\u03cc\u03c2 \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03c4\u03b7\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ae\u03c2 \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03af\u03b1\u03c2. \u039c\u03b5 \u03c4\u03bf\u03bd \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc \u03b1\u03c5\u03c4\u03cc \u03cc\u03bb\u03b5\u03c2 \u03bf\u03b9 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b8\u03b1 \u03ad\u03b2\u03b3\u03b1\u03b9\u03bd\u03b1\u03bd \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ac \u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03bc\u03ad\u03bd\u03b5\u03c2 ;)\r\n\r\n\u03a4\u03ad\u03bb\u03bf\u03c2, \u03cc\u03c4\u03b1\u03bd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03bc\u03b9\u03b1 \u03b1\u03c0\u03bf\u03c1\u03af\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b3\u03b5\u03bd\u03b9\u03ba\u03ac \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03bb\u03b5\u03b9\u03c4\u03bf\u03c5\u03c1\u03b3\u03af\u03b1 \u03c4\u03bf\u03c5 forum \u03bd\u03b1 \u03c4\u03b7 \u03b4\u03b9\u03b1\u03c4\u03c5\u03c0\u03ce\u03bd\u03b5\u03b9\u03c2 \u03c9\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03ae, \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c0\u03b5\u03c1\u03b9\u03bc\u03ad\u03bd\u03b5\u03b9\u03c2 posts \u03ac\u03bb\u03bb\u03c9\u03bd \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b5\u03c4\u03b1\u03c4\u03c1\u03ad\u03c8\u03b5\u03b9\u03c2 \u03bc\u03af\u03b1 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7 \u03b1\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7\u03c2 \u03c3\u03b5 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03cd \u03c6\u03bf\u03c1\u03bc\u03b1\u03bb\u03b9\u03c3\u03bc\u03bf\u03cd. \u0393\u03b9\u03b1 \u03bd\u03b1 \u03be\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03b9 \u03c0\u03bf\u03b9\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae \u03c3\u03bf\u03c5 \u03b5\u03c1\u03ce\u03c4\u03b7\u03c3\u03b7 \u03c3\u03c4\u03bf \u03ba\u03ac\u03c4\u03c9-\u03ba\u03ac\u03c4\u03c9...\r\n\r\nMy tuppence anyway,\r\n\r\nDurandal 1707", "Solution_9": "\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b3\u03c1.\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03bc\u03ad\u03bd\u03b1 \u03c4\u03cc\u03c4\u03b5\r\n\u0398\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 $ m\\in R$(\u03b1\u03bd\u03b5\u03be\u03ac\u03c1\u03c4\u03b7\u03c4\u03bf \u03c4\u03bf\u03c5 x)$ : e^{2x}\\equal{}me^x$\u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ x\\in R$ \u03ae \r\n$ : e^x\\equal{}m$\u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ x\\in R$\r\n\u03a0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03af\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03c4\u03bf \u03ac\u03c4\u03bf\u03c0\u03bf", "Solution_10": "[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\n\u0391\u03bb\u03bb\u03b1 \u03b1\u03c2 \u03c0\u03b5\u03c1\u03b9\u03bc\u03b5\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b4\u03bf\u03c5\u03bc\u03b5 \u03b1\u03bd \u03c3\u03c5\u03bc\u03c6\u03c9\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bf \u0394\u03b7\u03bc\u03b7\u03c4\u03c1\u03b7\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c7\u03b9\u03b6\u03bf\u03c5\u03bc\u03b5\u00a8.[/quote]\r\n\r\n\u03a7\u03c9\u03c1\u03af\u03c2 \u03bd\u03b1 \u03b4\u03b9\u03b5\u03ba\u03b4\u03b9\u03ba\u03ce \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1 \u03b1\u03bb\u03ac\u03b8\u03b7\u03c4\u03bf.\r\n\r\n\u039c\u03b5 \u03bb\u03cc\u03b3\u03b9\u03b1: \u0394\u03c5\u03bf \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 $ f,g: \\mathbb{R} \\to \\mathbb{R}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ce\u03c2 \u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03bc\u03ad\u03bd\u03b5\u03c2 \u03b1\u03bd \u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd $ a,b \\in \\mathbb{R}$ \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03b4\u03cd\u03bf \u03af\u03c3\u03b1 \u03bc\u03b5 0 \u03ce\u03c3\u03c4\u03b5 $ af(x) \\plus{} bg(x) \\equal{} 0$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ x \\in \\mathbb{R}$.\r\n\r\n\u039c\u03b5 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03af\u03ba\u03c4\u03b5\u03c2\r\n\r\n$ \\exists a \\in \\mathbb{R} \\exists b \\in \\mathbb{R} ((\\neg(a\\equal{}0 \\wedge b\\equal{}0)) \\wedge \\forall x\\in \\mathbb{R} (af(x) \\plus{}bg(x)\\equal{}0 ))$\r\n\r\n(\u039f \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03c4\u03bf\u03c5 \u03a0\u03b1\u03bd\u03b1\u03b3\u03b9\u03ce\u03c4\u03b7 \u03bc\u03bf\u03c5 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03bf\u03c2)\r\n\r\nH \u03ac\u03c1\u03bd\u03b7\u03c3\u03b7\r\n\r\n$ \\forall a \\in \\mathbb{R} \\forall b \\in \\mathbb{R} ((\\neg(a\\equal{}0 \\wedge b\\equal{}0) \\Rightarrow \\exists x \\in \\mathbb{R} (af(x) \\plus{} bg(x) \\neq 0))$\r\n\r\n\u038c\u03c0\u03bf\u03c5 $ \\exists x \\in \\mathbb{R} P(x)$ \u03c3\u03b7\u03bc\u03b1\u03af\u03bd\u03b5\u03b9 $ \\exists x (x \\in \\mathbb{R} \\wedge P(x))$ \u03ba\u03b1\u03b9 $ \\forall x \\in \\mathbb{R} P(x)$ \u03c3\u03b7\u03bc\u03b1\u03af\u03bd\u03b5\u03b9 $ \\forall x (x \\in \\mathbb{R} \\Rightarrow P(x))$\r\n\r\n\u0391\u03bd \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b1 \u03bb\u03cc\u03b3\u03b9\u03b1, \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c3\u03c9 \u03cc\u03c4\u03b9 \u03c3\u03ba\u03b5\u03c6\u03c4\u03cc\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c4\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c3\u03b1\u03bd \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c3\u03b5 \u03b4\u03b9\u03b1\u03bd\u03c5\u03c3\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc \u03c7\u03ce\u03c1\u03bf (pointless space \u03b5\u03af\u03c7\u03b5 \u03c0\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03c4\u03b5 \u03bf Halmos) \u03ba\u03b1\u03b9 \u03c6\u03c5\u03c3\u03b9\u03ba\u03ac \u03c3\u03b5 \u03b1\u03c5\u03c4\u03cc\u03bd \u03c4\u03bf\u03bd \u03c7\u03ce\u03c1\u03bf \u03bc\u03b9\u03b1 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03b7 \u03bc\u03b5 0 \u03b1\u03bd \u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b1\u03bd \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd \u03c4\u03b9\u03bc\u03ae 0 \u03c3\u03b5 \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 (\u03ba\u03b1\u03b9 \u03cc\u03c7\u03b9 \u03bc\u03cc\u03bd\u03bf \u03c3\u03b5 \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03ad\u03bd\u03b1) \u03cc\u03c0\u03bf\u03c5 \u03bf\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9.", "Solution_11": "[quote=\"Durandal\"][quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\u03a3\u03a5\u0393\u039d\u03a9\u039c\u0397 \u03b8\u03b5\u03bb\u03c9 \u03bd\u03b1 \u03c0\u03c9 ,\u03bf\u03c4\u03b9 \u03b5\u03b1\u03bd \u03b3\u03c1\u03b1\u03c8\u03b5\u03b9\u03c2 \u03c4\u03bf\u03bd \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf \u03c3\u03bf\u03c5 \u03c3\u03b5 \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03b1 \u03c4\u03bf\u03c4\u03b5 \u03b8\u03b1 \u03b4\u03b5\u03b9\u03c2 \u03bf\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b4\u03b9\u03b1\u03c6\u03b5\u03c1\u03b5\u03b9 \u03b1\u03c0\u03bf \u03c4\u03bf\u03bd \u03b4\u03b9\u03ba\u03bf \u03bc\u03bf\u03c5 ,\u03b5\u03c0\u03bf\u03bc\u03b5\u03bd\u03c9\u03c2 \u03b7 \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bb\u03b1\u03b8\u03bf\u03c2[/quote]\n\nNope. \u0394\u03b5\u03bd \u03b1\u03b3\u03b1\u03c0\u03ac\u03c9 \u03b9\u03b4\u03b9\u03b1\u03af\u03c4\u03b5\u03c1\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03af\u03ba\u03c4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03c4\u03b1 \u03c0\u03ac\u03c3\u03b1 \u03c0\u03b9\u03b8\u03b1\u03bd\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b4\u03b5 \u03b8\u03b1 \u03c4\u03bf \u03b3\u03c1\u03ac\u03c8\u03c9 \u03c3\u03c9\u03c3\u03c4\u03ac (\u03bf\u03c0\u03cc\u03c4\u03b5 Dem, please correct me), \u03b1\u03bb\u03bb\u03ac \u03bf \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03bc\u03bf\u03c5 \u03c3\u03b5 \u03c3\u03cd\u03bc\u03b2\u03bf\u03bb\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9:\n\n$ \\forall a \\forall b\\left[ (a\\in\\mathbb{R})\\wedge (b\\in\\mathbb{R})\\wedge [\\forall x[(x\\in\\mathbb{R})\\wedge(ae^x \\plus{} be^{2x} \\equal{} 0)]] \\implies (a \\equal{} b \\equal{} 0) \\right]$\n\nEdit \u03b5\u03c0\u03b5\u03be\u03ae\u03b3\u03b7\u03c3\u03b7\u03c2: \u03b1\u03bd $ a,b\\in\\mathbb{R}$ \u03ba\u03b1\u03b9 $ ae^x \\plus{} be^{2x} \\equal{} 0, \\forall x\\in\\mathbb{R}$ \u03c4\u03cc\u03c4\u03b5 $ a \\equal{} b \\equal{} 0$. \u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03c4\u03ce\u03c1\u03b1 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03b5\u03b9\u03c2 \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03b1 \u03b3\u03b9\u03b1\u03c4\u03af \u03bf\u03b9 \u03b4\u03cd\u03bf \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03bf\u03af...\n\n\u03a7\u03bc\u03bc\u03bc, \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c0\u03c9 \u03cc\u03c4\u03b9 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03c9 \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03c0\u03b9\u03bc\u03ad\u03bd\u03b5\u03b9\u03c2 \u03c4\u03cc\u03c3\u03bf \u03bd\u03b1 \u03b3\u03c1\u03ac\u03c6\u03b5\u03b9\u03c2 \u03c4\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03bc\u03b5 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03af\u03ba\u03c4\u03b5\u03c2, \u03b1\u03bb\u03bb\u03ac \u03c0\u03b5\u03c1\u03af \u03bf\u03c1\u03ad\u03be\u03b5\u03c9\u03c2... ;)\n\nCheerio,\n\nDurandal 1707[/quote]\r\n\r\n\r\nM\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03bf \u0394\u0397\u039c\u0397\u03a4\u03a1\u0397\u03a3 \u03b8\u03b5\u03c9\u03c1\u03b5\u03b9 \u03c4\u03b7\u03bd \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03c3\u03c9\u03c3\u03c4\u03b7, \u03b1\u03c2 \u03c0\u03b1\u03c1\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 [b] \u03c4\u03b7\u03c2 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5[/b] ,\u03b5\u03c4\u03c3\u03b9 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5.\r\n\r\n$ \\neg \\forall a\\forall b[(a\\in R)\\wedge(b\\in R)\\wedge[\\forall x[x\\in R\\wedge(ae^x \\plus{} be^{2x}\\equal{}0)]]\\Longrightarrow (a\\equal{}b\\equal{}0)]$ \u03b7 \u03bf\u03c0\u03bf\u03b9\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5\r\n\r\n$ \\exists a\\exists b${$ \\neg[(a\\in R)\\wedge(b\\in R)\\wedge[\\forall x[x\\in R\\wedge (ae^x \\plus{} be^{2x}\\equal{}0)]]\\wedge a\\neq 0,b\\neq 0$}..........................................................................1\r\n\r\n\u0391\u03bb\u03bb\u03b1 $ \\neg[(a\\in R)\\wedge(b\\in R)\\wedge[\\forall x[x\\in R\\wedge (ae^x \\plus{} be^{2x}\\equal{}0)]]$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5\r\n\r\n\r\n$ ((a\\in R)\\wedge(b\\in R))\\Longrightarrow\\neg[\\forall x[x\\in R\\wedge(ae^x \\plus{} be^{2x}\\equal{}0)]]$...................................................................................................................................2\r\n\r\n\r\n\u0391\u03bb\u03bb\u03b1 $ \\neg[\\forall x[x\\in R\\wedge(ae^x \\plus{} be^{2x}\\equal{}0)]]$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5\r\n\r\n$ \\exists x[x\\in R\\Longrightarrow (ae^x \\plus{} be^{2x}\\neq 0)]$...............................................................................................................................................................................3\r\n\r\n\u03a4\u03c9\u03c1\u03b1 \u03b5\u03b1\u03bd \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03b7\u03c3\u03bf\u03c5\u03bc\u03b5 (3) \u03c3\u03c4\u03b7\u03bd (2) \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03c0\u03c1\u03bf\u03ba\u03c5\u03c8\u03b5\u03b9 \u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd (1) \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5.\r\n\r\n\r\n$ \\exists a\\exists b[((a\\in R)\\wedge(b\\in R))\\Longrightarrow\\exists x[x\\in R\\Longrightarrow(ae^x \\plus{} be^{2x}\\neq 0)]\\wedge (a\\neq 0,b\\neq 0)]$\r\n\r\n\u0391\u03a5\u03a4\u0397 \u03b7 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03bc\u03b1\u03c2 \u03bb\u03b5\u03b5\u03b9:\r\n\r\n\u03a5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd a,b \u03b5\u03c4\u03c3\u03b9 \u03c9\u03c3\u03c4\u03b5 \u03b5\u03b1\u03bd a,b \u03b1\u03bd\u03bf\u03b9\u03ba\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c4\u03b5 ,\u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03c7 \u03c4\u03bf \u03bf\u03c0\u03bf\u03b9\u03bf\u03bd \u03b1\u03bd\u03b7\u03ba\u03b5\u03b9 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2,\u03c4\u03bf \u03bf\u03c0\u03bf\u03b9\u03bf\u03bd \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03b5\u03c4\u03b1\u03b9 $ ae^x \\plus{} be^{2x}\\neq 0$, \u03ba\u03b1\u03b9 $ a\\neq 0,b\\neq 0$\r\n\r\n\u03ba\u03b1\u03b9 \u03b5\u03c4\u03c3\u03b9 \u03b5\u03b1\u03bd \u03c0\u03b1\u03c1\u03bf\u03c5\u03bc\u03b5 a=1 ,b=6 and x=2,\u03b8\u03b1 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 $ e^2\\plus{} 6e^4\\neq 0$ \u03ba\u03b1\u03b9 $ a\\neq 0,b\\neq 0$\r\n\r\n\u0391\u03a1\u0391 \u0395\u03a7\u039f\u03a5\u039c\u0395 \u03b3\u03c1\u03b1\u03bc. \u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03b7\r\n\r\n\u039a\u03b1\u03b9 \u03b1\u03c2 \u03a7\u03a1\u0397\u03a3\u0399\u039c\u039f\u03a0\u039f\u0399\u0397\u03a3\u039f\u03a5\u039c\u0395 \u03c4\u03b7\u03bd \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03b9\u03b1 \u03b5\u03b3\u03c1\u03b1\u03c8\u03b1 \u03c3\u03c4\u03bf \u03c0\u03bf\u03c3\u03c4 #3 ,\u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03bf\u03c5\u03bc\u03b5 \u03b3\u03c1\u03b1\u03bc.\u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1>\r\n\r\n\u0395\u03b1\u03bd $ ae^x \\plus{}be^{2x} \\equal{}0$ =====> $ e^x(a \\plus{}be^x) \\equal{}0$ ====> $ a \\plus{}be^x \\equal{}0$ ====>$ \\frac{d( a \\plus{} be^x)}{dx} \\equal{}0$ ====>$ be^x \\equal{} 0$ ====> b=0=a\r\n\r\n\u03ba\u03b9\u03b1 \u03b5\u03b1\u03bd \u03b3\u03b5\u03bd\u03b9\u03ba\u03b5\u03c5\u03c3\u03bf\u03c5\u03bc\u03b5 $ \\forall a\\forall b\\forall x[ ae^x \\plus{} be^{2x} \\equal{}0\\Longrightarrow a\\equal{}b\\equal{}0]$", "Solution_12": "[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\n$ \\neg \\forall a\\forall b[(a\\in R)\\wedge(b\\in R)\\wedge[\\forall x[x\\in R\\wedge(ae^x \\plus{} be^{2x} \\equal{} 0)]]\\Longrightarrow (a \\equal{} b \\equal{} 0)]$ \u03b7 \u03bf\u03c0\u03bf\u03b9\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5\n\n$ \\exists a\\exists b${$ \\neg[(a\\in R)\\wedge(b\\in R)\\wedge[\\forall x[x\\in R\\wedge (ae^x \\plus{} be^{2x} \\equal{} 0)]]\\wedge a\\neq 0,b\\neq 0$}[/quote]\r\n\r\nNope, \u03b4\u03b5\u03bd \u03ad\u03b2\u03b1\u03bb\u03b5\u03c2 \u03c4\u03b7\u03bd \u03ac\u03c1\u03bd\u03b7\u03c3\u03b7 $ \\neg$ \u03c3\u03c4\u03bf \u03c3\u03c9\u03c3\u03c4\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf. \u0397 \u03ac\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd \u03b5\u03af\u03bd\u03b1\u03b9 (\u03b5\u03ba\u03c4\u03cc\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03bd\u03c9 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf typo):\r\n\r\n$ \\exists a \\exists b \\big[ (a\\in\\mathbb{R})\\wedge (b\\in\\mathbb{R})\\wedge [\\forall x[(x\\in\\mathbb{R})\\wedge (ae^x \\plus{} be^{2x} \\equal{} 0)]]\\wedge \\neg(a \\equal{} b \\equal{} 0) \\big]$\r\n\r\n(\u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03c3\u03c5\u03bc\u03c6\u03c9\u03bd\u03b5\u03af \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03ac\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b5 \u03bf Dem)\r\n\r\n\u03a4\u03b5\u03bb\u03b9\u03ba\u03ac \u03b1\u03ba\u03cc\u03bc\u03b1 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03b5\u03b9: \u03ad\u03c7\u03b5\u03b9\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 genuine \u03b1\u03c0\u03bf\u03c1\u03af\u03b1 \u03ae \u03b1\u03c0\u03bb\u03ce\u03c2 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c3\u03c4\u03bf\u03bd \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc \u03c4\u03b7\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ae\u03c2 \u03b5\u03be\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7\u03c2 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03c9\u03bd?\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_13": "[quote=\"Durandal\"][quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\n$ \\neg \\forall a\\forall b[(a\\in R)\\wedge(b\\in R)\\wedge[\\forall x[x\\in R\\wedge(ae^x \\plus{} be^{2x} \\equal{} 0)]]\\Longrightarrow (a \\equal{} b \\equal{} 0)]$ \u03b7 \u03bf\u03c0\u03bf\u03b9\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5\n\n$ \\exists a\\exists b${$ \\neg[(a\\in R)\\wedge(b\\in R)\\wedge[\\forall x[x\\in R\\wedge (ae^x \\plus{} be^{2x} \\equal{} 0)]]\\wedge a\\neq 0,b\\neq 0$}[/quote]\n\nNope, \u03b4\u03b5\u03bd \u03ad\u03b2\u03b1\u03bb\u03b5\u03c2 \u03c4\u03b7\u03bd \u03ac\u03c1\u03bd\u03b7\u03c3\u03b7 $ \\neg$ \u03c3\u03c4\u03bf \u03c3\u03c9\u03c3\u03c4\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf. \u0397 \u03ac\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd \u03b5\u03af\u03bd\u03b1\u03b9 (\u03b5\u03ba\u03c4\u03cc\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03bd\u03c9 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf typo):\n\n$ \\exists a \\exists b \\big[ (a\\in\\mathbb{R})\\wedge (b\\in\\mathbb{R})\\wedge [\\forall x[(x\\in\\mathbb{R})\\wedge (ae^x \\plus{} be^{2x} \\equal{} 0)]]\\wedge \\neg(a \\equal{} b \\equal{} 0) \\big]$\n\n(\u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03c3\u03c5\u03bc\u03c6\u03c9\u03bd\u03b5\u03af \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03ac\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b5 \u03bf Dem)\n\n\u03a4\u03b5\u03bb\u03b9\u03ba\u03ac \u03b1\u03ba\u03cc\u03bc\u03b1 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03b5\u03b9: \u03ad\u03c7\u03b5\u03b9\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 genuine \u03b1\u03c0\u03bf\u03c1\u03af\u03b1 \u03ae \u03b1\u03c0\u03bb\u03ce\u03c2 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c3\u03c4\u03bf\u03bd \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc \u03c4\u03b7\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ae\u03c2 \u03b5\u03be\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7\u03c2 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03c9\u03bd?\n\nCheerio,\n\nDurandal 1707[/quote]\r\n\r\n\r\n\r\n\u03a3\u0395 \u03c0\u03bf\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b7 \u03c4\u03b7\u03c2 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b5\u03c9\u03c2 \u03bc\u03bf\u03c5 \u03b5\u03c7\u03c9 \u03ba\u03b1\u03bd\u03b5\u03b9 \u03bb\u03b1\u03b8\u03bf\u03c2", "Solution_14": "[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\u03a3\u0395 \u03c0\u03bf\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b7 \u03c4\u03b7\u03c2 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b5\u03c9\u03c2 \u03bc\u03bf\u03c5 \u03b5\u03c7\u03c9 \u03ba\u03b1\u03bd\u03b5\u03b9 \u03bb\u03b1\u03b8\u03bf\u03c2[/quote]\r\n\r\n\u03a3\u03c4\u03b7 \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7. \u0394\u03b5\u03bd \u03ba\u03bf\u03af\u03c4\u03b1\u03be\u03b1 \u03c0\u03b9\u03bf \u03c0\u03ad\u03c1\u03b1 \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5\u03bd \u03b5\u03af\u03c7\u03b5 \u03bd\u03cc\u03b7\u03bc\u03b1.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_15": "[quote=\"Durandal\"][quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\u03a3\u0395 \u03c0\u03bf\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b7 \u03c4\u03b7\u03c2 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b5\u03c9\u03c2 \u03bc\u03bf\u03c5 \u03b5\u03c7\u03c9 \u03ba\u03b1\u03bd\u03b5\u03b9 \u03bb\u03b1\u03b8\u03bf\u03c2[/quote]\n\n\u03a3\u03c4\u03b7 \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7. \u0394\u03b5\u03bd \u03ba\u03bf\u03af\u03c4\u03b1\u03be\u03b1 \u03c0\u03b9\u03bf \u03c0\u03ad\u03c1\u03b1 \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5\u03bd \u03b5\u03af\u03c7\u03b5 \u03bd\u03cc\u03b7\u03bc\u03b1.\n\nCheerio,\n\nDurandal 1707[/quote]\r\n\r\n\r\n\r\n \u03a0\u03bf\u03b9\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03bf \u03bb\u03b1\u03b8\u03bf\u03c2", "Solution_16": "[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\u03a0\u03bf\u03b9\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03bf \u03bb\u03b1\u03b8\u03bf\u03c2[/quote]\r\n\r\n\u0394\u03b5\u03bd \u03ad\u03ba\u03b1\u03bd\u03b5\u03c2 \u03c3\u03c9\u03c3\u03c4\u03ac \u03c4\u03b7\u03bd \u03ac\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03ae\u03c2.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_17": "[quote=\"Durandal\"][quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\u03a0\u03bf\u03b9\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03bf \u03bb\u03b1\u03b8\u03bf\u03c2[/quote]\n\n\u0394\u03b5\u03bd \u03ad\u03ba\u03b1\u03bd\u03b5\u03c2 \u03c3\u03c9\u03c3\u03c4\u03ac \u03c4\u03b7\u03bd \u03ac\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03ae\u03c2.\n\nCheerio,\n\nDurandal 1707[/quote]\r\n\r\n\r\n[b]\u03a0\u039f\u039b\u03a5 \u03a3\u03a9\u03a3\u03a4\u0391[/b].\r\n\r\n[b]\u039f\u039c\u03a9\u03a3[/b] \u03b5\u03c7\u03bf\u03c5\u03b5 \u03c4\u03bf \u03b5\u03be\u03b7\u03c2 [b]\u03a0\u0395\u03a1\u0399\u0395\u03a1\u0393\u039f[/b]. \u0397 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03b3\u03c1\u03b1\u03c8\u03b1 \u03c3\u03c4\u03bf \u03c0\u03bf\u03c3\u03c4 #3 \u03ba\u03b1\u03b9 \u03b7 \u03bf\u03c0\u03bf\u03b9\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9\u00a8\r\n\r\n$ \\forall a\\forall b\\forall x[((a\\in R)\\wedge(b\\in R)\\wedge(x\\in R)\\wedge(ae^x \\plus{} be^{2x}\\equal{}0)) \\Longrightarrow a\\equal{}b\\equal{}0]$ [b] \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03b5\u03c4\u03b1\u03b9[/b] \u03c4\u03b7\u03bd \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03bf\u03c0\u03c9\u03c2 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03c5\u03b5\u03c4\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03c4\u03c9.\r\n\r\n\u03a5\u03c0\u03bf\u03c4\u03b5\u03b8\u03b5\u03c3\u03c4\u03c9 \u03bf\u03c4\u03b9 $ (a\\in R)\\wedge(b\\in R)\\wedge\\forall x[(x\\in R)\\wedge(ae^x \\plus{} be^{2x} \\equal{} 0)]$ .\u0391\u03a5\u03a4\u039f \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03b5\u03c4\u03b1\u03b9 $ (a\\in R)\\wedge(b\\in R)\\wedge(x\\in R)\\wedge(ae^x \\plus{} be^{2x} \\equal{}0)$\r\n\r\nK\u03b1\u03b9 \u03c3\u03c5\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 (\u03b4\u03b9\u03ba\u03b7 \u03bc\u03bf\u03c5) \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 a=b=0.\r\n\r\n\u039a\u0391\u0399 \u03b3\u03b5\u03bd\u03b9\u03ba\u03b5\u03c5\u03bf\u03bd\u03c4\u03b1\u03c2 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b4\u03b9\u03ba\u03b7 \u03c3\u03bf\u03c5 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1.\r\n\r\n\r\n\r\n$ \\forall a\\forall b[(a\\in R)\\wedge(b\\in R)\\wedge[\\forall x[(x\\in R)\\wedge(ae^x \\plus{} be^{2x} \\equal{}0 )]]\\Longrightarrow (a\\equal{}b\\equal{}0)]$\r\n\r\n\r\n[b]\u03a3\u03a5\u039d\u0395\u03a0\u03a9\u03a3 \u03b7 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03bc\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b3\u03b5\u03bd\u03b9\u03ba\u03c9\u03c4\u03b5\u03c1\u03b7 \u03c4\u03b7\u03c2 \u03b4\u03b9\u03ba\u03b7 \u03c3\u03bf\u03c5[/b]\r\n\r\n\u0397 \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 \u03bf\u03bc\u03c9\u03c2 \u03c4\u03b7\u03c2 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1\u03c2 \u03bc\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03c4\u03b5\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03c0\u03bf\u03bc\u03b5\u03bd\u03c9\u03c2 [b] \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03b5\u03c4\u03b1\u03b9 \u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03b7 \u03c4\u03c9\u03bd \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b5\u03c9\u03bd [/b] {$ e^x,e^{2x}$ }\r\n\r\n[b] \u0391\u03c1\u03b1 \u03ba\u03b1\u03b9 \u03b7 \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5[/b] \u03b8\u03b1 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03b5\u03c4\u03b1\u03b9 \u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03b7 \u03c4\u03c9\u03bd \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b5\u03c9\u03bd. \u0391\u03c0\u03bb\u03b1 \u03b4\u03b5\u03bd \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b2\u03c1\u03b5\u03b9 \u03c4\u03b7\u03bd \u03bb\u03c5\u03c3\u03b7 \u03b1\u03ba\u03bf\u03bc\u03b1\r\n\r\n\r\n\u0395\u03ba\u03c4\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03b1\u03bd \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03c2 \u03b3\u03c1\u03b1\u03c8\u03b5\u03b9 \u03bc\u03b9\u03b1 \u03c6\u03bf\u03c1\u03bc\u03b1\u03bb \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7 \u03b3\u03b9\u03b1 \u03bf\u03c0\u03bf\u03b9\u03b1\u03b4\u03b7\u03c0\u03bf\u03c4\u03b5 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03c0\u03bf\u03bc\u03b5\u03bd\u03c9\u03c2 \u03bf\u03bb\u03b1 \u03c4\u03b1 \u03b1\u03bb\u03bb\u03b1 \u03ba\u03b1\u03c4\u03b1\u03c1\u03c5\u03c6\u03b8\u03bf\u03c5\u03bd", "Solution_18": "\u0392\u03c1\u03b5 \u03c3\u03c5, \u03c3\u03ba\u03ad\u03c8\u03bf\u03c5 \u03bb\u03af\u03b3\u03bf \u03c4\u03b9 \u03bb\u03b5\u03c2 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ac :P\r\n\r\n\u0388\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b3\u03c1\u03ac\u03c8\u03b5\u03b9 \u03b4\u03cd\u03bf \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ae \u03b5\u03be\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7: \u03c4\u03bf\u03bd \u03b4\u03b9\u03ba\u03cc \u03c3\u03bf\u03c5 (\u0391) \u03ba\u03b1\u03b9 \u03c4\u03bf\u03bd \"\u03b4\u03b9\u03ba\u03cc\" \u03bc\u03bf\u03c5 (\u0392).\r\n(BTW, \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7: \u03b4\u03b5\u03c2 \u03c0\u03bf\u03b9\u03ad\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03b9 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03c4\u03bf \u03b4\u03b9\u03ba\u03cc \u03c3\u03bf\u03c5 \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc \u0391... \u03b8\u03b1 \u03b4\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03ac\u03c1\u03b1 \u03c0\u03bf\u03bb\u03cd \u03bb\u03af\u03b3\u03b5\u03c2)\r\n\r\n\u039c\u03b5\u03c4\u03ac, \u03bb\u03b5\u03c2 \u03c0\u03bf\u03bb\u03cd \u03c3\u03c9\u03c3\u03c4\u03ac \u03c4\u03bf \u03b5\u03be\u03ae\u03c2: \u03b1\u03bd \u03b4\u03cd\u03bf \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03c4\u03bf\u03bd (\u0391), \u03c4\u03cc\u03c4\u03b5 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03ba\u03b1\u03b9 \u03c4\u03bf\u03bd (\u0392). \u0391\u03bd\u03c4\u03b9\u03b8\u03ad\u03c4\u03c9\u03c2, \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03c4\u03bf\u03bd (\u0392) \u03b1\u03bb\u03bb\u03ac \u03cc\u03c7\u03b9 \u03c4\u03bf\u03bd (\u0391) - \u03bf\u03b9 $ e^x, e^{2x}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1.\r\n(\u0395\u03bd\u03b1\u03bb\u03bb\u03b1\u03ba\u03c4\u03b9\u03ba\u03ac, \u03ad\u03c3\u03c4\u03c9 $ A$ \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03c4\u03c9\u03bd (\u03b6\u03b5\u03c5\u03b3\u03ce\u03bd) \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03c9\u03bd \u03c0\u03bf\u03c5 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03c4\u03bf\u03bd (\u0391) \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 $ B$ \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03b1\u03c5\u03c4\u03ce\u03bd \u03c0\u03bf\u03c5 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03c4\u03bf\u03bd (\u0392). \u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2, $ A\\subseteq B$ \u03b1\u03bb\u03bb\u03ac, \u03c4\u03bf inclusion \u03b5\u03af\u03bd\u03b1\u03b9 proper.)\r\n\r\n\u0388\u03c4\u03c3\u03b9, \u03b1\u03bd \u03b4\u03b5\u03af\u03be\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03b4\u03cd\u03bf \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b4\u03b5\u03bd \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03c4\u03bf\u03bd (\u0391), \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b4\u03b5\u03af\u03be\u03b5\u03b9 \u03c4\u03af\u03c0\u03bf\u03c4\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b1\u03bd \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03ae \u03cc\u03c7\u03b9 \u03c4\u03bf\u03bd (\u0392). \u03a3\u03c4\u03b7 \u03b2\u03b9\u03b2\u03bb\u03b9\u03bf\u03b3\u03c1\u03b1\u03c6\u03af\u03b1, \u03bf \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03c0\u03bf\u03c5 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03b5\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf \u0392 \u03ba\u03b1\u03b9 \u03cc\u03c7\u03b9 \u03bf \u03b4\u03b9\u03ba\u03cc\u03c2 \u03c3\u03bf\u03c5 \u03bf \u0391. \u039c\u03b5 \u03b2\u03ac\u03c3\u03b7 \u03b1\u03c5\u03c4\u03cc\u03bd \u03c4\u03bf\u03bd \u03bf\u03c1\u03b9\u03c3\u03bc\u03cc (B) \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1 \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1, \u03b5\u03bb\u03c0\u03af\u03b6\u03c9 \u03c0\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03b5\u03b9\u03c2 \u03b1\u03c5\u03c4\u03cc.\r\n\r\n\u03a3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7, \u03bc\u03bf\u03c5 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b9\u03cb\u03bb\u03af\u03b6\u03b5\u03b9\u03c2 \u03c4\u03bf\u03bd \u03ba\u03ce\u03bd\u03c9\u03c0\u03b1 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03c4\u03b1\u03c0\u03af\u03bd\u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd \u03ba\u03ac\u03bc\u03b7\u03bb\u03bf: \u03ba\u03b1\u03bb\u03bf\u03af \u03bf\u03b9 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03af\u03ba\u03c4\u03b5\u03c2, \u03b4\u03b5 \u03bb\u03ad\u03c9, \u03b1\u03bb\u03bb\u03ac \u03b2\u03bb\u03ad\u03c0\u03c9 \u03cc\u03c4\u03b9 \u03b7 \u03c7\u03c1\u03ae\u03c3\u03b7 \u03c4\u03bf\u03c5\u03c2 \u03ba\u03b1\u03b8\u03cc\u03bb\u03bf\u03c5 \u03b4\u03b5 \u03c3\u03b5 \u03c0\u03c1\u03bf\u03c3\u03c4\u03b1\u03c4\u03b5\u03cd\u03b5\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03c3\u03c6\u03ac\u03bb\u03bc\u03b1\u03c4\u03b1 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae\u03c2. \u0393\u03b9'\u03b1\u03c5\u03c4\u03cc, \u03b8\u03b1 \u03c3\u03bf\u03c5 \u03c0\u03c1\u03cc\u03c4\u03b5\u03b9\u03bd\u03b1 \u03c0\u03c1\u03ce\u03c4\u03b1 \u03bd\u03b1 \u03b2\u03b5\u03b2\u03b1\u03b9\u03c9\u03b8\u03b5\u03af\u03c2 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c9\u03c3\u03c4\u03ae \u03b7 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03b4\u03bf\u03bc\u03ae \u03c4\u03c9\u03bd \u03b5\u03c0\u03b9\u03c7\u03b5\u03b9\u03c1\u03b7\u03bc\u03ac\u03c4\u03c9\u03bd \u03c3\u03bf\u03c5, \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b7\u03c3\u03b5 \u03bd\u03b1 \u03c4\u03b1 \u03bc\u03b5\u03c4\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9\u03c2 \u03c3\u03b5 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03af\u03ba\u03c4\u03b5\u03c2.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_19": "[quote=\"Durandal\"]\n\n\u039c\u03b5\u03c4\u03ac, \u03bb\u03b5\u03c2 \u03c0\u03bf\u03bb\u03cd \u03c3\u03c9\u03c3\u03c4\u03ac \u03c4\u03bf \u03b5\u03be\u03ae\u03c2: \u03b1\u03bd \u03b4\u03cd\u03bf \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03c4\u03bf\u03bd (\u0391), \u03c4\u03cc\u03c4\u03b5 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03ba\u03b1\u03b9 \u03c4\u03bf\u03bd (\u0392). \u0391\u03bd\u03c4\u03b9\u03b8\u03ad\u03c4\u03c9\u03c2, \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03c4\u03bf\u03bd (\u0392) \u03b1\u03bb\u03bb\u03ac \u03cc\u03c7\u03b9 \u03c4\u03bf\u03bd (\u0391) - \u03bf\u03b9 $ e^x, e^{2x}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1.\n\n\nCheerio,\n\nDurandal 1707[/quote]\r\n\r\n\u0393\u03b9\u03b1 \u03bd\u03b1 \u03b4\u03bf\u03c5\u03bc\u03b5\r\n\r\n\u0395\u03c3\u03c4\u03c9:\r\n\r\na,b,x \u03b1\u03bd\u03bf\u03b9\u03ba\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2, \u03ba\u03b1\u03b9 $ ae^x \\plus{} be^{2x} \\equal{}0$,\u03b1\u03bb\u03bb\u03b1:\r\n\r\n\r\n$ ae^x \\plus{} be^{2x}\\equal{}0\\Longrightarrow a \\plus{}be^x\\equal{}0\\Longrightarrow\\frac{d(a \\plus{} be^x)}{dx} \\equal{} 0\\Longrightarrow be^x \\equal{} 0\\Longrightarrow a\\equal{}b\\equal{}0$\r\n\r\n[b]K\u03b1\u03b9 \u03b3\u03b5\u03bd\u03b9\u03ba\u03b5\u03c5\u03bf\u03bd\u03c4\u03b1\u03c2 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5[/b].\r\n\r\n\r\n$ \\forall a\\forall b\\forall x[ a,b,x\\in R\\wedge (ae^x \\plus{} be^{2x} \\equal{}0)\\Longrightarrow a\\equal{}b\\equal{}0]$\r\n\r\n\r\nA\u03c1\u03b1 \u03b7 \u03c6\u03bf\u03c1\u03bf\u03c5\u03bb\u03b1 \u03bc\u03bf\u03c5 (\u0391) \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1 \u03c4\u03c9\u03bd {$ e^x,e^{2x}$} \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b5\u03c9\u03bd\r\n\r\n[b]\u0397 \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03bb\u03b1\u03b8\u03bf\u03c2 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7??[/b]", "Solution_20": "\u03a4\u03c1\u03af\u03ba\u03bb\u03b9\u03bd\u03bf (real name btw?), sorry \u03b1\u03bb\u03bb\u03ac \u03ad\u03c7\u03b5\u03b9\u03c2 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03c0\u03bf\u03bb\u03cd \u03c3\u03b7\u03bc\u03b1\u03bd\u03c4\u03b9\u03ba\u03cc \u03bb\u03ac\u03b8\u03bf\u03c2 \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03ba\u03b1\u03b8\u03cc\u03bb\u03bf\u03c5 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03c4\u03b7 \u03c7\u03c1\u03ae\u03c3\u03b7 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03b9\u03ba\u03c4\u03ce\u03bd. \u0395\u03c0\u03af\u03c3\u03b7\u03c2, \u03b1\u03c0\u03cc \u03b5\u03c3\u03ad\u03bd\u03b1 \u03c0\u03bf\u03c5 \u03ba\u03c5\u03c1\u03ae\u03c4\u03c4\u03b5\u03b9\u03c2 \u03c4\u03cc\u03c3\u03bf \u03ad\u03bd\u03c4\u03bf\u03bd\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c5\u03c3\u03c4\u03b7\u03c1\u03ae \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03ba\u03b1\u03b9 \u03b8\u03b5\u03bc\u03b5\u03bb\u03af\u03c9\u03c3\u03b7, \u03b8\u03b1 \u03c0\u03b5\u03c1\u03af\u03bc\u03b5\u03bd\u03b1 \u03c0\u03bf\u03bb\u03cd \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b7 \u03c0\u03c1\u03bf\u03c3\u03bf\u03c7\u03ae...\r\n\r\n[quote]\u0395\u03c3\u03c4\u03c9:\n\na,b,x \u03b1\u03bd\u03bf\u03b9\u03ba\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2, \u03ba\u03b1\u03b9 $ ae^x \\plus{} be^{2x} \\equal{} 0$,\u03b1\u03bb\u03bb\u03b1:\n\n$ ae^x \\plus{} be^{2x} \\equal{} 0\\Longrightarrow a \\plus{} be^x \\equal{} 0\\Longrightarrow\\frac {d(a \\plus{} be^x)}{dx} \\equal{} 0\\Longrightarrow be^x \\equal{} 0\\Longrightarrow a \\equal{} b \\equal{} 0$\n[/quote]\r\n\r\n\u03a4\u03bf \u03cc\u03c4\u03b9 \u03b4\u03ad\u03c7\u03b5\u03c3\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 $ a,b,x$ \u03ce\u03c3\u03c4\u03b5 $ ae^x \\plus{} be^{2x} \\equal{} 0$, \u03b5\u03c0'\u03bf\u03c5\u03b4\u03b5\u03bd\u03b5\u03af \u03b4\u03b5 \u03c3\u03b7\u03bc\u03b1\u03af\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03ae \u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c3\u03b5 \u03bc\u03af\u03b1 \u03c0\u03b5\u03c1\u03b9\u03bf\u03c7\u03ae \u03c4\u03bf\u03c5 $ x$ \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03c4\u03b7\u03bd [i]\u03c0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03af\u03c3\u03b5\u03b9\u03c2[/i] \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03be\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 $ be^x \\equal{} 0$.\r\n\r\n\u0394\u03b7\u03bb\u03b1\u03b4\u03ae, \u03ad\u03bb\u03b5\u03bf\u03c2, \u03c3\u03ba\u03ad\u03c8\u03bf\u03c5 \u03bb\u03af\u03b3\u03bf \u03ba\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1: \u03b1\u03bd $ a \\equal{} 1, b \\equal{} \\minus{} 1, x \\equal{} 0$, \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ ae^x \\plus{} be^{2x} \\equal{} 0$, \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ be^x \\equal{} 0$, \u03c0\u03cc\u03c3\u03bf \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd $ a \\equal{} b \\equal{} 0$... \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5 \u03c3\u03bf\u03c5 \u03c7\u03c4\u03c5\u03c0\u03ac\u03b5\u03b9 \u03ba\u03b1\u03bc\u03c0\u03b1\u03bd\u03ac\u03ba\u03b9\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \"\u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03ae\" \u03c3\u03bf\u03c5? :what?:\r\n\r\n\u039e\u03b1\u03bd\u03b1\u03bb\u03ad\u03c9... \u03ba\u03b1\u03c4\u03ac\u03bb\u03b1\u03b2\u03b5 \u03c0\u03c1\u03ce\u03c4\u03b1 \u03c4\u03b7\u03bd \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c4\u03b9 \u03bb\u03ad\u03bd\u03b5 \u03bf\u03b9 \u03ad\u03bd\u03bd\u03bf\u03b9\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03b5\u03af\u03c2 (\u03c0.\u03c7. \u03b7 \u03b3\u03c1. \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03af\u03b1), \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03b1\u03c3\u03c7\u03bf\u03bb\u03ae\u03c3\u03bf\u03c5 \u03bc\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03af\u03ba\u03c4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b9\u03c2 \"\u03b5\u03bb\u03bb\u03b5\u03af\u03c8\u03b5\u03b9\u03c2\" \u03c4\u03b7\u03c2 \u03b1\u03c5\u03c3\u03c4\u03b7\u03c1\u03ae\u03c2 \u03b8\u03b5\u03bc\u03b5\u03bb\u03af\u03c9\u03c3\u03b7\u03c2 \u03c4\u03b7\u03c2 \u03b1\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7\u03c2. Sorry, \u03b1\u03bb\u03bb\u03ac \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03b4\u03b5 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03b4\u03b9\u03b1\u03af\u03c4\u03b5\u03c1\u03b1 \u03c0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03b9\u03ba\u03cc...\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_21": "[quote=\"Durandal\"]\n\n\u0394\u03b7\u03bb\u03b1\u03b4\u03ae, \u03ad\u03bb\u03b5\u03bf\u03c2, \u03c3\u03ba\u03ad\u03c8\u03bf\u03c5 \u03bb\u03af\u03b3\u03bf \u03ba\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1: \u03b1\u03bd $ a \\equal{} 1, b \\equal{} \\minus{} 1, x \\equal{} 0$, \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ ae^x \\plus{} be^{2x} \\equal{} 0$, \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ be^x \\equal{} 0$, \u03c0\u03cc\u03c3\u03bf \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd $ a \\equal{} b \\equal{} 0$... \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5 \u03c3\u03bf\u03c5 \u03c7\u03c4\u03c5\u03c0\u03ac\u03b5\u03b9 \u03ba\u03b1\u03bc\u03c0\u03b1\u03bd\u03ac\u03ba\u03b9\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \"\u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03ae\" \u03c3\u03bf\u03c5? :what?:\n\n\nCheerio,\n\nDurandal 1707[/quote]\r\n\r\n\r\n\u0393\u03b9\u03b1\u03c4\u03b9, \u03c4\u03bf \u03b9\u03b4\u03b9\u03bf \u03b4\u03b5\u03bd \u03c3\u03c5\u03bc\u03b2\u03b1\u03b9\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b4\u03b9\u03ba\u03b7 \u03c3\u03bf\u03c5 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1??\r\n\r\n\u0397 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9.\r\n\r\n\r\n$ \\forall a \\forall b\\left[ (a\\in\\mathbb{R})\\wedge (b\\in\\mathbb{R})\\wedge [\\forall x[(x\\in\\mathbb{R})\\wedge(ae^x \\plus{} be^{2x} \\equal{} 0)]] \\implies (a \\equal{} b \\equal{} 0) \\right]$\r\n\r\n\r\n\u0395\u03b1\u03bd \u03bb\u03bf\u03b9\u03c0\u03bf\u03bd \u03b2\u03b1\u03bb\u03bf\u03c5\u03bc\u03b5 : a=1 ,b=-1 ,x=0 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5: \r\n\r\n\r\n$ ae^x \\plus{} be^{2x} \\equal{} 0$, \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ be^x \\equal{} 0$, \u03c0\u03cc\u03c3\u03bf \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd $ a \\equal{} b \\equal{} 0$\r\n\r\n\u03c7\u03c1\u03b9\u03c3\u03c4\u03bf\u03c2", "Solution_22": "[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\u0397 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9.\n$ \\forall a \\forall b\\left[ (a\\in\\mathbb{R})\\wedge (b\\in\\mathbb{R})\\wedge [\\forall x[(x\\in\\mathbb{R})\\wedge(ae^x \\plus{} be^{2x} \\equal{} 0)]] \\implies (a \\equal{} b \\equal{} 0) \\right]$\n\n\u0395\u03b1\u03bd \u03bb\u03bf\u03b9\u03c0\u03bf\u03bd \u03b2\u03b1\u03bb\u03bf\u03c5\u03bc\u03b5 : a=1 ,b=-1 ,x=0 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5: \n\n$ ae^x \\plus{} be^{2x} \\equal{} 0$, \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ be^x \\equal{} 0$, \u03c0\u03cc\u03c3\u03bf \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd $ a \\equal{} b \\equal{} 0$[/quote]\r\n\r\nGaaaah, \u03ad\u03bb\u03b5\u03bf\u03c2, \u03cc\u03c7\u03b9!! \u0391\u03c2 \u03c0\u03ac\u03bc\u03b5 \u03b2\u03ae\u03bc\u03b1-\u03b2\u03ae\u03bc\u03b1 \u03bc\u03c0\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b5\u03bb\u03b5\u03b9\u03ce\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c0\u03bf\u03c4\u03ad:\r\n1. \u0394\u03b9\u03b1\u03bb\u03ad\u03b3\u03b5\u03b9\u03c2 \u03b1=1. \u039f\u039a\r\n2. \u0394\u03b9\u03b1\u03bb\u03ad\u03b3\u03b5\u03b9\u03c2 \u03b2=-1. \u039f\u039a\r\n3. \u039c\u03c0\u03b1\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03b7\u03bd \u03b1\u03b3\u03ba\u03cd\u03bb\u03b7.\r\n4. $ a,b\\in\\mathbb{R}$, OK.\r\n5. \u039c\u03b5\u03c4\u03ac, \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03ac\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2: $ ae^x \\plus{} be^{2x} \\equal{} 0$ [b][i][size=150]\u0393\u0399\u0391 \u039a\u0391\u0398\u0395[/size][/i][/b] $ x\\in\\mathbb{R}$. \u0393\u03b9'\u03b1\u03c5\u03c4\u03cc \u03ad\u03c7\u03b5\u03b9 \u03bc\u03c0\u03b5\u03b9 \u03c3\u03b5 \u03bf\u03bb\u03cc\u03ba\u03bb\u03b7\u03c1\u03b7 \u03b1\u03b3\u03ba\u03cd\u03bb\u03b7 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03ba\u03bf\u03bc\u03bc\u03ac\u03c4\u03b9. \u0395\u03c3\u03cd \u03ad\u03c7\u03b5\u03b9\u03c2 $ ae^x \\plus{} be^{2x} \\equal{} 0$ \u039c\u039f\u039d\u039f \u03b3\u03b9\u03b1 $ x \\equal{} 0$. \u0393\u03b9\u03b1 $ x \\equal{} 1$ \u03b4\u03b5\u03bd \u03b2\u03b3\u03ac\u03b6\u03b5\u03b9\u03c2 0, \u03ac\u03c1\u03b1 \u03b4\u03b5\u03bd \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03af\u03c4\u03b1\u03b9 \u03b7 \u03ac\u03c1\u03bd\u03b7\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03bf\u03c1\u03b9\u03c3\u03bc\u03bf\u03cd.\r\n\r\n\u0391\u03bd \u03b4\u03b5\u03bd \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03b5\u03b9\u03c2 \u03c4\u03b7 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03ac, \u03b4\u03b5 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c3\u03b5 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03c9 \u03ac\u03bb\u03bb\u03bf.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_23": "\u0397 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9:\r\n\r\n$ \\forall a \\forall b[ (a\\in\\mathbb{R})\\wedge (b\\in\\mathbb{R})\\wedge [\\forall x[(x\\in\\mathbb{R})\\wedge(ae^x + be^{2x} = 0)]] \\implies (a = b = 0) ]$;\r\n\r\n\r\nA\u03c6\u03bf\u03c5 \u03b7 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 \u03b3\u03b9\u03b1 \u03bf\u03bb\u03b1 \u03c4\u03b1 a , b \u03b8\u03b1 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03b1=1 \u03ba\u03b1\u03b9 b = -1\r\n\r\n\u0391\u03bd\u03c4\u03b9\u03ba\u03b1\u03b8\u03b9\u03c3\u03c4\u03bf\u03c5\u03bc\u03b5 \u03bb\u03bf\u03b9\u03c0\u03bf\u03bd \u03c3\u03c4\u03b7\u03bd \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5:\r\n\r\n$ (1\\in R)\\wedge(-1\\in R)\\wedge[\\forall x[(x\\in R)\\wedge(e^x - e^{2x} = 0)]]\\implies (1=-1=0)$\r\n\r\nT\u03c9\u03c1\u03b1 \u03b7 \u03b4\u03b5\u03be\u03b9\u03b1 \u03bc\u03b5\u03c1\u03b9\u03b1 \u03b1\u03c5\u03c4\u03b7\u03c2 \u03c4\u03b7\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03b7\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c8\u03b5\u03c5\u03b4\u03b7\u03c2.\r\n\r\n\r\n\u0391\u03bb\u03bb\u03b1 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b7 \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03b7 \u03b1\u03bb\u03b7\u03b8\u03b7\u03c2 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b7 \u03b1\u03c1\u03b9\u03c3\u03c4\u03b5\u03c1\u03b7 \u03bc\u03b5\u03c1\u03b9\u03b1 \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c8\u03b5\u03c5\u03b4\u03b7\u03c2,\u03b3\u03b9\u03b1\u03c4\u03b9 \u03c8\u03b5\u03c5\u03b4\u03b5\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03b5\u03c4\u03b1\u03b9 \u03c8\u03b5\u03c5\u03b4\u03b5\u03c2 ,\u03b8\u03b1 \u03ba\u03b1\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03b7 \u03b1\u03bb\u03b7\u03b8\u03b7\r\n\r\n\u0391\u03bb\u03bb\u03b1 $ 1\\in R$ EINAI \u03b1\u03bb\u03b7\u03b8\u03b5\u03c2,\u03b5\u03c0\u03b9\u03c3\u03b7\u03c2 $ -1\\in R$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03bb\u03b7\u03b8\u03b5\u03c2 \u03b5\u03c0\u03bf\u03bc\u03b5\u03bd\u03c9\u03c2 \u03c4\u03bf \r\n\r\n$ [\\forall x[(x\\in R)\\wedge(e^x - e^{2x} = 0)]]$ \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c8\u03b5\u03c5\u03b4\u03b5\u03c2\r\n\r\n\r\n[b]\u03a0\u03a9\u03a3 \u0395\u0399\u039d\u0391\u0399 \u0394\u03a5\u039d\u0391\u03a4\u039f\u039d \u03a4\u039f $ [\\forall x[(x\\in R)\\wedge(e^x - e^{2x} = 0)]]$ \u039d\u0391 \u0395\u0399\u039d\u0391\u0399 \u03a8\u0395\u03a5\u0394\u0395\u03a3 \u0393\u0399\u0391 \u039f\u039b\u0391 \u03a4\u0391 x \u039f\u03a4\u0391\u039d \u0393\u0399\u0391 x = 0 \u0395\u03a7\u039f\u03a5\u039c\u0395 $ e^0 - e^0 =0$ \u03a4\u039f \u039f\u03a0\u039f\u0399\u039f\u039d \u0395\u0399\u039d\u0391\u0399 \u0391\u039b\u0397\u0398\u0395\u03a3???[/b]\r\n\r\n\u0395\u03c0\u03bf\u03bc\u03b5\u03bd\u03c9\u03c2 \u03bf\u03c4\u03b9 \u03c3\u03c5\u03bc\u03b2\u03b1\u03b9\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03bc\u03bf\u03c5 \u03c3\u03c5\u03bc\u03b2\u03b1\u03b9\u03bd\u03b5\u03b9 \u03c3\u03c5\u03bc\u03b2\u03b1\u03b9\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b4\u03b9\u03ba\u03b7 \u03c3\u03bf\u03c5\r\n\r\n\r\n\u0391\u03c3\u03b5 \u03c4\u03b7\u03bd \u03b1\u03c1\u03bd\u03b7\u03c3\u03b7 \u03b5\u03b4\u03c9 \u03b4\u03b5\u03bd \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03b9\u03c4\u03b1\u03b9 \u03ba\u03b1\u03bd \u03b7 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03b7 \u03b9\u03b4\u03b9\u03b1 \u03b3\u03b9\u03b1 \u03bf\u03c1\u03b9\u03c3\u03bc\u03b5\u03bd\u03b5\u03c2 \u03c4\u03b9\u03bc\u03b5\u03c2", "Solution_24": "[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"][b]\u03a0\u03a9\u03a3 \u0395\u0399\u039d\u0391\u0399 \u0394\u03a5\u039d\u0391\u03a4\u039f\u039d \u03a4\u039f $ [\\forall x[(x\\in R)\\wedge(e^x \\minus{} e^{2x} \\equal{} 0)]]$ \u039d\u0391 \u0395\u0399\u039d\u0391\u0399 \u03a8\u0395\u03a5\u0394\u0395\u03a3 \u0393\u0399\u0391 \u039f\u039b\u0391 \u03a4\u0391 x \u039f\u03a4\u0391\u039d \u0393\u0399\u0391 x = 0 \u0395\u03a7\u039f\u03a5\u039c\u0395 $ e^0 \\minus{} e^0 \\equal{} 0$ \u03a4\u039f \u039f\u03a0\u039f\u0399\u039f\u039d \u0395\u0399\u039d\u0391\u0399 \u0391\u039b\u0397\u0398\u0395\u03a3???[/b][/quote]\r\n\r\n\u0393\u03b9\u03b1 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 $ \\forall x[(x\\in\\mathbb{R})\\wedge(e^x \\equal{} e^{2x})]$ \u03c8\u03b5\u03c5\u03b4\u03ae\u03c2, \u03b1\u03c1\u03ba\u03b5\u03af \u03b1\u03c0\u03bb\u03ce\u03c2 \u03bd\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ad\u03bd\u03b1 $ x\\in\\mathbb{R}$ \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03af \u03c4\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03bc\u03ad\u03c3\u03b1 \u03c3\u03c4\u03b7\u03bd \u03b1\u03b3\u03ba\u03cd\u03bb\u03b7:\r\n$ \\neg[\\forall x[(x\\in\\mathbb{R})\\wedge(e^x \\equal{} e^{2x})]] \\equal{} \\exists x[(x\\in\\mathbb{R})\\wedge(e^x\\neq e^{2x})]$\r\n\r\n\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03b4\u03b9\u03b4\u03ac\u03c3\u03ba\u03bf\u03c5\u03bd \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03c4\u03b9\u03b3\u03bc\u03ae \u03c3\u03c4\u03bf \u03bb\u03cd\u03ba\u03b5\u03b9\u03bf \u03ae \u03c3\u03c4\u03bf \u03b3\u03c5\u03bc\u03bd\u03ac\u03c3\u03b9\u03bf (\u03cc\u03c7\u03b9 \u03bc\u03b5 \u03c0\u03bf\u03c3\u03bf\u03b4\u03b5\u03af\u03ba\u03c4\u03b5\u03c2, \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03ba\u03bf\u03b9\u03bd\u03ae \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03c4\u03bf \u03bb\u03ad\u03bd\u03b5 \u03c3\u03c4\u03bf \u03c3\u03c7\u03bf\u03bb\u03b5\u03af\u03bf), \u03b1\u03bb\u03bb\u03ac \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03c9 \u03bb\u03ac\u03b8\u03bf\u03c2.\r\n\r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2, \u03cc\u03c3\u03bf benefit of doubt \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03ba\u03c1\u03b1\u03c4\u03ae\u03c3\u03c9, \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ad\u03c2 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03b5\u03b9\u03c2 agenda \u03b5\u03b4\u03ce \u03c0\u03ad\u03c1\u03b1, \u03bf\u03c0\u03cc\u03c4\u03b5 sorry, \u03b1\u03c0\u03bb\u03ce\u03c2 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03ac\u03bb\u03bb\u03bf \u03c7\u03c1\u03cc\u03bd\u03bf/\u03c5\u03c0\u03bf\u03bc\u03bf\u03bd\u03ae \u03bd\u03b1 \u03b1\u03c3\u03c7\u03bf\u03bb\u03b7\u03b8\u03ce \u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03b8\u03ad\u03bc\u03b1. \u0391\u03bd \u03b1\u03ba\u03cc\u03bc\u03b1 \u03b4\u03b5\u03bd \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03af\u03bd\u03b5\u03b9\u03c2 \u03c4\u03b9 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 (\u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b5\u03af\u03bc\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2), \u03c4\u03bf \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf \u03bc\u03ad\u03c1\u03bf\u03c2 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b1\u03c5\u03c4\u03ae\u03bd \u03c4\u03b7 \u03c3\u03c5\u03b6\u03ae\u03c4\u03b7\u03c3\u03b7 \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03b5\u03b4\u03ce:\r\n\r\n[url=http://annals.princeton.edu/annals/about/cover/cover.html]Annals of Mathematics[/url]\r\n\r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2, \u03b5\u03af\u03bc\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03cc\u03c4\u03b9 o Josiah Carberry (\u03ba\u03b1\u03b8\u03b7\u03b3\u03b7\u03c4\u03ae\u03c2 \u03c3\u03c4\u03bf Brown), \u03b8\u03b1 \u03c7\u03b1\u03c1\u03b5\u03af \u03bd\u03b1 \u03c3\u03b5 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03b5\u03b9 :cool: :evilgrin:\r\n\r\nCheerio,\r\n\r\nPan", "Solution_25": "geia sas!syggnomi pou epemvaino,alla an k den eimai foititis akomi thelo n rotiso an mporei n teleiosei i askisi me tin orizousa tou wronski..me voithaei kapoios???efxaristo..", "Solution_26": "\u039d\u03b1\u03b9 \u03c4\u03b5\u03bb\u03b5\u03b9\u03ce\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd Wronski", "Solution_27": "\u039d\u039f\u039c\u0399\u0396\u03a9 \u03bf\u03c4\u03b9 \u03bf\u03b9 \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b5\u03b9\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c4\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03bf\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03c4\u03b9\u03bc\u03b5\u03c2 [b]\u03b5\u03ba\u03c4\u03bf\u03c2 \u03b3\u03b9\u03b1 x=0[/b].\r\n\r\n\u0393\u03b9\u03b1\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03c7=0 [b]\u03b4\u03b5\u03bd \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2[/b]", "Solution_28": "[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"]\u039d\u039f\u039c\u0399\u0396\u03a9 \u03bf\u03c4\u03b9 \u03bf\u03b9 \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b5\u03b9\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c4\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03bf\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03c4\u03b9\u03bc\u03b5\u03c2 [b]\u03b5\u03ba\u03c4\u03bf\u03c2 \u03b3\u03b9\u03b1 x=0[/b].[/quote]\r\n\r\n\u0394\u03cd\u03bf \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03af\u03c4\u03b5 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ac \u03b1\u03bd\u03b5\u03be\u03ac\u03c1\u03c4\u03b7\u03c4\u03b5\u03c2, \u03b5\u03af\u03c4\u03b5 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ac \u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03bc\u03ad\u03bd\u03b5\u03c2, \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03b7 \u03ad\u03bd\u03bd\u03bf\u03b9\u03b1 \u03c4\u03b7\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ae\u03c2 \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03af\u03b1\u03c2 \u03c3\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf [i]\u03c3\u03b7\u03bc\u03b5\u03af\u03bf[/i].\r\n\r\n\u03a4\u03bf \"\u03b1\u03bd\u03b5\u03be\u03ac\u03c1\u03c4\u03b7\u03c4\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03cc\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03c4\u03b9\u03bc\u03ad\u03c2 \u03b5\u03ba\u03c4\u03cc\u03c2 \u03b3\u03b9\u03b1 $ x \\equal{} 0$\" \u03c0\u03bf\u03c5 \u03bb\u03b5\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03ba\u03cc \u03c3\u03bf\u03c5 \u03c3\u03c5\u03bc\u03c0\u03ad\u03c1\u03b1\u03c3\u03bc\u03b1 (\u03ba\u03b1\u03b9 \u03b4\u03b5\u03af\u03c7\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03b5\u03b9 \u03c4\u03b7\u03bd \u03ad\u03bd\u03bd\u03bf\u03b9\u03b1 \u03c4\u03b7\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ae\u03c2 \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03af\u03b1\u03c2), \u03bf\u03c0\u03cc\u03c4\u03b5 \u03bc\u03b7\u03bd \u03bc\u03c0\u03b5\u03c1\u03b4\u03b5\u03cd\u03b5\u03b9\u03c2 \u03cc\u03c3\u03bf\u03c5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bd genuine \u03b1\u03c0\u03bf\u03c1\u03af\u03b5\u03c2 \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03bf \u03b6\u03ae\u03c4\u03b7\u03bc\u03b1. \u0393\u03b9\u03b1\u03c4\u03af \u03b1\u03bb\u03bb\u03b9\u03ce\u03c2, \u03cc\u03bb\u03b5\u03c2 \u03bf\u03b9 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b8\u03b1 \u03ae\u03c4\u03b1\u03bd \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ac \u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03bc\u03ad\u03bd\u03b5\u03c2 \u03c3\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03c5\u03c2 \u03ba\u03b1\u03b9 \u03b4\u03b5 \u03b8\u03b1 \u03b5\u03af\u03c7\u03b5 \u03bd\u03cc\u03b7\u03bc\u03b1 \u03bd\u03b1 \u03bc\u03b9\u03bb\u03ac\u03bc\u03b5 \u03b3\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03b9\u03ba\u03ae \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03af\u03b1.\r\n\r\n\u0391\u03bd \u03cc\u03bd\u03c4\u03c9\u03c2 \u03bd\u03b9\u03ce\u03b8\u03b5\u03b9\u03c2 \u03bc\u03c0\u03b5\u03c1\u03b4\u03b5\u03bc\u03ad\u03bd\u03bf\u03c2 \u03b1\u03ba\u03cc\u03bc\u03b1, drop it \u03ba\u03b1\u03b9 \u03c0\u03ac\u03bc\u03b5 \u03c3\u03b5 pm's. \u0391\u03bb\u03bb\u03b9\u03ce\u03c2, stop trolling, \u03b3\u03b9\u03b1\u03c4\u03af \u03ad\u03c7\u03b5\u03b9 \u03ba\u03bf\u03c5\u03c1\u03b1\u03c3\u03c4\u03b5\u03af \u03c4\u03bf \u03b8\u03ad\u03bc\u03b1...\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_29": "\u03a4\u039f\u03a4\u0395 \u0391\u03a3 \u03b3\u03c1\u03b1\u03c8\u03b5\u03b9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bc\u03b9\u03b1 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03bf\u03c1\u03b9\u03b6\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1 \u03c4\u03c9\u03bd \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b5\u03c9\u03bd\r\n\r\n\u0397 \u03b4\u03b9\u03ba\u03b7 \u03c3\u03bf\u03c5 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 ,\u03bf\u03c0\u03c9\u03c2 \u03c3\u03bf\u03c5 \u03b1\u03c0\u03b5\u03b4\u03b5\u03b9\u03be\u03b1..\r\n\r\n\u0391\u03bb\u03bb\u03b1 \u03c4\u03bf \u03be\u03b1\u03bd\u03b1\u03b3\u03c1\u03b1\u03c6\u03c9 \u03b5\u03b4\u03c9 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b4\u03bf\u03c5\u03bc\u03b5 \u03c0\u03bf\u03b9\u03bf\u03c2 \u03bc\u03c0\u03b5\u03c1\u03b4\u03b5\u03c5\u03b5\u03c4\u03b1\u03b9.\r\n\r\n \u0397 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9\u00a8\r\n\r\n$ \\forall a \\forall b\\left[ (a\\in\\mathbb{R})\\wedge (b\\in\\mathbb{R})\\wedge [\\forall x[(x\\in\\mathbb{R})\\wedge(ae^x \\plus{} be^{2x} \\equal{} 0)]] \\implies (a \\equal{} b \\equal{} 0) \\right]$;\r\n\r\n\u0393\u03b9\u03b1 a=1, b=-1,x=0 \u03b7 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03b3\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9:\r\n\r\n$ [((1\\in R)\\wedge(\\minus{}1\\in R)\\wedge(0\\in R)\\wedge( e^O \\minus{}e^0\\equal{}0))\\Longrightarrow (1\\equal{}\\minus{}1\\equal{}0]$\r\n\r\n\u039f\u03c0\u03c9\u03c2 \u03b5\u03b9\u03c0\u03b1 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9 \u03c0\u03bf\u03c3\u03c4 \u03bc\u03bf\u03c5 ,\r\n\r\n\u0397 \u0391\u03a1\u0399\u03a3\u03a4\u0395\u03a1\u0391 \u039c\u0395\u03a1\u0399\u0391 \u0391\u03a5\u03a4\u0397\u03a3 \u03c4\u03b7\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03b7\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 [b] \u03b1\u03bb\u03b7\u03b8\u03b7\u03c2 \u03ba\u03b1\u03b9 \u03b7 \u03b4\u03b5\u03be\u03b9\u03b1 \u03c8\u03b5\u03c5\u03b4\u03b7\u03c2[/b] \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03bf \u03ba\u03b1\u03bd\u03b5\u03b9 \u03bf\u03bb\u03b7 \u03c4\u03b7\u03bd \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03b7[b] \u03c8\u03b5\u03c5\u03b4\u03b7[/b]\r\n\r\n\u0391\u03c1\u03b1 \u03b7 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03b3\u03b9\u03b1 \u03c7=0 ,a=1,b=-1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c8\u03b5\u03c5\u03b4\u03b7\u03c2\r\n\r\n\u03a0.\u03a7 \u03b7 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 $ \\forall x( x^2>0)$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03bb\u03b7\u03b8\u03b7\u03c2 \u03b3\u03b9\u03b1 \u03bf\u03bb\u03b1 \u03c4\u03b1 \u03c7 \u03b5\u03ba\u03c4\u03bf\u03c2 \u03c7=0, \u03c4\u03bf \u03b9\u03b4\u03b9\u03bf \u03c3\u03c5\u03bc\u03b2\u03b1\u03b9\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5.\r\n\r\n\r\n\u0394\u03c5\u03bf \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b1 \u03c3\u03c5\u03bc\u03b2\u03b1\u03b9\u03bd\u03bf\u03c5\u03bd \u03b5\u03b4\u03c9 \u03bb\u03bf\u03b9\u03c0\u03bf\u03bd:\r\n\r\n1) \u0397 \u03c0\u03b1\u03c1\u03b1\u03b4\u03b5\u03c7\u03bf\u03bc\u03b5\u03b8\u03b1 \u03c4\u03b7\u03bd \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03bf\u03c0\u03bf\u03c4\u03b5 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b5\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1 ,\u03b5\u03ba\u03c4\u03bf\u03c2 \u03b3\u03b9\u03b1 \u03c7=0\r\n\r\n\r\n2)\u0397 \u0394\u0395\u039d \u03c0\u03b1\u03c1\u03b1\u03b4\u03b5\u03c7\u03bf\u03bc\u03b5\u03b8\u03b1 \u03c4\u03b7\u03bd \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03c3\u03bf\u03c5 \u03bf\u03c0\u03bf\u03c4\u03b5 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b2\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03b1\u03bb\u03bb\u03b7 \u03c6\u03bf\u03c1\u03bc\u03bf\u03c5\u03bb\u03b1 \u03bd\u03b1 \u03bf\u03c1\u03b9\u03b6\u03b7 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b5\u03be\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1 \u03c4\u03c9\u03bd \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b5\u03c9\u03bd. \r\n\r\n\u039a\u0391\u0399 \u03a0\u0391\u03a8\u0395 \u03bd\u03b1 \u03b1\u03bd\u03c4\u03b9\u03b3\u03c1\u03b1\u03c6\u03b5\u03b9\u03c2 \u03b1\u03c0\u03bf \u03b1\u03bb\u03bb\u03b1 \u03c6\u03bf\u03c1\u03bf\u03c5\u03bc \u03b2\u03bb\u03b1\u03ba\u03b5\u03b9\u03b5\u03c2 \u03bf\u03c0\u03c9\u03c2 ( stop trolling) \u03ba\u03b1\u03b9 \u03c4\u03b5\u03c4\u03bf\u03b9\u03b5\u03c2 \u03b1\u03bd\u03bf\u03b7\u03c3\u03b9\u03b5\u03c2.\r\n\r\n\u0394\u0399\u0395\u039a\u0394\u0399\u039a\u0395\u0399\u03a3 \u03a4\u039f \u0391\u039b\u039b\u0391\u0398\u0397\u03a4\u039f \u03a4\u039f\u03a5 \u03a0\u0391\u03a0\u0391???", "Solution_30": "[quote=\"\u03c4\u03c1\u03b9\u03ba\u03bb\u03b9\u03bd\u03bf\"] \u039a\u0391\u0399 \u03a0\u0391\u03a8\u0395 \u03bd\u03b1 \u03b1\u03bd\u03c4\u03b9\u03b3\u03c1\u03b1\u03c6\u03b5\u03b9\u03c2 \u03b1\u03c0\u03bf \u03b1\u03bb\u03bb\u03b1 \u03c6\u03bf\u03c1\u03bf\u03c5\u03bc \u03b2\u03bb\u03b1\u03ba\u03b5\u03b9\u03b5\u03c2 \u03bf\u03c0\u03c9\u03c2 ( stop trolling) \u03ba\u03b1\u03b9 \u03c4\u03b5\u03c4\u03bf\u03b9\u03b5\u03c2 \u03b1\u03bd\u03bf\u03b7\u03c3\u03b9\u03b5\u03c2.\n\u0394\u0399\u0395\u039a\u0394\u0399\u039a\u0395\u0399\u03a3 \u03a4\u039f \u0391\u039b\u039b\u0391\u0398\u0397\u03a4\u039f \u03a4\u039f\u03a5 \u03a0\u0391\u03a0\u0391???[/quote]\r\n\r\n1. \u03a0\u03b5\u03c1\u03b9\u03bc\u03ad\u03bd\u03c9 \u03ac\u03c1\u03b8\u03c1\u03bf \u03c3\u03c4\u03b1 [url=http://annals.princeton.edu/annals/about/cover/cover.html]Annals of Mathematics[/url] \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03c0\u03b5\u03c1\u03b9\u03ad\u03c7\u03b5\u03b9 \u03c4\u03b9\u03c2 \u03b1\u03bd\u03b1\u03ba\u03b1\u03bb\u03cd\u03c8\u03b5\u03b9\u03c2 \u03c3\u03bf\u03c5.\r\n2. \u039e\u03cd\u03b4\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03bd\u03b5\u03cd\u03c1\u03b1.\r\n3. \u0391\u03bb\u03ac\u03b8\u03b7\u03c4\u03bf \u03bc\u03b5 \u03ad\u03bd\u03b1 \"\u039b\" - \u03c4\u03bf \u03bb\u03ad\u03c9 \u03c3\u03b5 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03c7\u03c4\u03c5\u03c0\u03ac\u03c2 \u03c0\u03bf\u03bb\u03bb\u03ac \u03c0\u03bb\u03ae\u03ba\u03c4\u03c1\u03b1 \u03c4\u03b1\u03c5\u03c4\u03cc\u03c7\u03c1\u03bf\u03bd\u03b1 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03ad\u03c7\u03b5\u03b9\u03c2 \u03bd\u03b5\u03cd\u03c1\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03c4\u03b1\u03b9\u03c1\u03b9\u03ac\u03b6\u03b5\u03b9 \u03b7 \u03b5\u03b9\u03c1\u03c9\u03bd\u03b5\u03af\u03b1 \u03c3\u03c4\u03bf \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf.\r\n\r\n\u03a7\u03b1\u03af\u03c1\u03bf\u03bc\u03b1\u03b9 \u03c0\u03bf\u03c5 \u03ad\u03b4\u03b5\u03b9\u03be\u03b5\u03c2 \u03bc\u03cc\u03bd\u03bf\u03c2 \u03c3\u03bf\u03c5 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b1\u03be\u03af\u03b6\u03b5\u03b9 \u03bd\u03b1 \u03b1\u03c3\u03c7\u03bf\u03bb\u03b5\u03af\u03c4\u03b1\u03b9 \u03c4\u03bf forum \u03bc\u03b1\u03b6\u03af \u03c3\u03bf\u03c5. Thatta boy! :)\r\n\r\nCheerio,\r\n\r\nDurandal 1707" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "From Moldova Mathematical Olympiad.\r\n\r\nFor a,b,c strictly positive numbers and x=(abc)^1/3 prove that\r\n\r\n1/(a+b+x) + 1/(b+c+x) + 1/(c+a+x) ? 1/x\r\n\r\nPS I think that the symbol ? in the original text means that you must find \r\n the correct sign of the inequality.", "Solution_1": "Write a = x, b = y, c = z.\r\n\r\nThe problem becomes :\r\n\r\n1/(x + y + xyz) + 1/(y + z + xyz) + 1/(z + x + xyz) \\leq 1/(xyz)\r\n\r\nwhich is exactly USAMO97/5. I think this has been discussed 3 times or so ..." } { "Tag": [ "geometry", "3D geometry", "calculus", "integration", "number theory proposed", "number theory" ], "Problem": "Solve $a^3+2b^3+4c^3=6abc+1$ in integers $a,b,c$.", "Solution_1": "Well, here's a guess.\r\n\r\nWe can factorise the given equation:\r\n\\begin{eqnarray*} 1&=&a^3+2b^3+4c^3-6abc\\\\ &=&(a+\\sqrt[3]{2}b+\\sqrt[3]{4}c)(a+\\omega\\sqrt[3]{2}b+\\omega^2\\sqrt[3]{4}c)(a+\\omega^2\\sqrt[3]{2}b+\\omega\\sqrt[3]{4}c) \\end{eqnarray*}\r\nwhere $\\omega$ is a complex cube root of 1. If $a$, $b$ and $c$ are integers, then the solutions $A_n$, $B_n$, $C_n$ to the simultaneous equations\r\n$A_n+\\sqrt[3]{2}B_n+\\sqrt[3]{4}C_n=(a+\\sqrt[3]{2}b+\\sqrt[3]{4}c)^n$\r\n$A_n+\\omega\\sqrt[3]{2}B_n+\\omega^2 \\sqrt[3]{4}C_n=(a+\\omega \\sqrt[3]{2}b+\\omega^2\\sqrt[3]{4}c)^n$\r\n$A_n+\\omega^2\\sqrt[3]{2} B_n+\\omega\\sqrt[3]{4}C_n=(a+\\omega^2\\sqrt[3]{2}b+\\omega\\sqrt[3]{4} c)^n$\r\nare also integers. So, proceeding by analogy with the solution of a Pell's equation, we \"should\" get all solutions from raising a base solution to various powers.\r\n\r\nI think the base solution is 19, 15, 12. This gives second solution 1081, 858, 681. And so on.\r\n\r\nEdit: factorisation and some other errors corrected, thanks ZetaX.", "Solution_2": "I think your factorisation is wrong (the last factor should be different), but you are on the right way, my solution is more or less the same.\r\nBut note that there is a solution $a=b=c=1$ ;)", "Solution_3": "I don't have all the integral solutions, but by Dirichlet's Theorem on units in the ring of integers of a number field, they are all of the form $(a_n,b_n,c_n)$ such that $a_n+b_n\\sqrt[3]2+c_n\\sqrt[3]4=(a_1+b_1\\sqrt[3]2+c_1\\sqrt[3]4)^n\\ (*)$ for some \"small\" solution $(a_1,b_1,c_1)$. The problem is determining this small solution. For example, $(a,b,c)=(-1,1,0)$ also works, and it cannot be obtained from $(1,1,1)$ by means of $(*)$.", "Solution_4": "Grobber: you forgot the negative powers of that minimal solution. The solution $(-1,1,0)$ is just the inverse of $(1,1,1)$, so the $-1$-th power of it.", "Solution_5": "Yeah, I guess I did." } { "Tag": [ "geometry", "rectangle" ], "Problem": "Just started chem. and have a question:\r\n\r\nIs there a way to convert same units with diff. exponents? \r\nIf so, how do you do it?\r\n\r\nEX: How many cm are there in 2 cm^3?\r\nand How many mm are in 5 m^3?\r\n\r\n\r\nThanx :)", "Solution_1": "What you propose is actually impossible. For example, a m^2 represents is a two-dimensional area, while one m is a one-dimensional length.\r\n\r\nYou can think about this problem geometrically. To try to change something with area (for example, a rectangle) into something with length only (e.g. a line segment) one could imagine cutting the rectangle into \"thin\" strips and laying them end to end. However, no matter how \"thin\" you cut these strips, they will always have some finite width--hence no matter how much you try, something that starts with area can never be converted to just a length.\r\n\r\n\r\nYou should note, however, that it is certainly possible to convert between, say, cm^3 and m^3... to do so, recall 100 cm = 1 m, so 100^3 cm^3 = m^3." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "let $ a,b,c$ be non-negative numbers, no two of which are zero. prove that : \r\n$ \\frac {1}{a^2 \\minus{} ab \\plus{} b^2} \\plus{} \\frac {1}{b^2 \\minus{} bc \\plus{} c^2} \\plus{} \\frac {1}{c^2 \\minus{} ca \\plus{} a^2}$", "Solution_1": "where is the RHS? Please, edit it..;)" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "sphere", "probability" ], "Problem": "A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point P is selected at random inside the circumscribed sphere. The probability that P lies inside one of the five small spheres is closest to \r\n\r\n(A) 0\r\n(B) 0.1\r\n(C) 0.2\r\n(D) 0.3\r\n(E) 0.4", "Solution_1": "[hide]Let $R$ be the radius of the circumscribed sphere, $r$ be the radius of the inscribed sphere, and $s$ be the radius of the tangent spheres. It can be shown that $r = \\frac{R}{3}$ (the centers of the circumscribed/inscribed spheres are the centroid of the tetrahedron) and then $s = \\frac{1}{2}(R-r) = \\frac{R}{3}$ (draw the two radii to the tangent point, the difference is the diameter of the tangent sphere) also, so the probability is\n\n$5\\left(\\frac{1}{3}\\right)^{3}= \\frac{5}{27}\\approx 0.2$.[/hide]" } { "Tag": [ "trigonometry" ], "Problem": "I believe this will include most trignometry results useful in Olympiads. Any additions are welcome!!!", "Solution_1": "I don't get the point of this thread. Can you explain, Rushil?" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Knowing that $\\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=\\frac{1}{11}$, calcule $\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{c+a}$", "Solution_1": "$\\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=\\frac{(a-b)(b-c)}{c+a}(\\frac{1}{a+b}-\\frac{1}{b+c})=(a-b)(\\frac{1}{c+a}-\\frac{1}{a+b})+(b-c)(\\frac{1}{c+a}-\\frac{1}{c+b})=\\frac{a-c}{c+a}+\\frac{b-a}{a+b}+\\frac{c-b}{b+c}=3-2X=\\frac{1}{11}$, \r\nwere $X=\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{c+a}$. Therefore $X=\\frac{16}{11}$.", "Solution_2": "[url]https://artofproblemsolving.com/community/c4h2598226p22414170[/url]" } { "Tag": [ "IMO Shortlist" ], "Problem": "\u0394\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf $ ABC$ \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 $ A1, B1, C1$ \u03c4\u03b1 \u03bc\u03ad\u03c3\u03b1 \u03c4\u03c9\u03bd \u03c0\u03bb\u03b5\u03c5\u03c1\u03ce\u03bd $ BC, CA, AB$ \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1. \u0388\u03c3\u03c4\u03c9 $ P$ \u03c4\u03c5\u03c7\u03cc\u03bd \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03c5 \u03c0\u03b5\u03c1\u03af\u03ba\u03c5\u03ba\u03bb\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03bf\u03b9 $ PA1, PA2, PA3$, \u03c4\u03ad\u03bc\u03bd\u03bf\u03c5\u03bd \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03c3\u03c4\u03b1 $ A', B', C'$ \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1 \u03ce\u03c3\u03c4\u03b5 \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03b1\u03c5\u03c4\u03ac \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c4\u03b1\u03c5\u03c4\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03ba\u03bf\u03c1\u03c5\u03c6\u03ad\u03c2 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd \u03bf\u03b9 \u03c6\u03bf\u03c1\u03b5\u03af\u03c2 \u03c4\u03c9\u03bd $ AA', BB', CC'$ \u03bd\u03b1 \u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03af\u03b6\u03bf\u03c5\u03bd \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf (\u03bd\u03b1 \u03c4\u03ad\u03bc\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b1\u03bd\u03ac \u03b4\u03cd\u03bf \u03ba\u03b1\u03b9 \u03bc\u03ac\u03bb\u03b9\u03c3\u03c4\u03b1 \u03c3\u03b5 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ac \u03c3\u03b7\u03bc\u03b5\u03af\u03b1). \u039d\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03cc\u03c4\u03b9 \u03c4\u03bf \u03b5\u03bc\u03b2\u03b1\u03b4\u03cc\u03bd \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 \u03b1\u03c5\u03c4\u03bf\u03cd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03bd\u03b5\u03be\u03ac\u03c1\u03c4\u03b7\u03c4\u03bf \u03c4\u03b7\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae\u03c2 \u03c4\u03bf\u03c5 $ P$.", "Solution_1": "\u03a1\u03b5 \u039a\u03ce\u03c3\u03c4\u03b1, \u03bc\u03b5 \u03ba\u03b1\u03c4\u03ad\u03c3\u03c4\u03c1\u03b5\u03c8\u03b5\u03c2!!!! \u03a3\u03b5 \u03c0\u03b5\u03c1\u03af\u03bf\u03b4\u03bf \u03b5\u03be\u03b5\u03c4\u03ac\u03c3\u03b5\u03c9\u03bd, \u03c7\u03ac\u03bd\u03c9 \u03c0\u03ac\u03bd\u03c9 \u03b1\u03c0\u03cc \u03bc\u03b9\u03b1 \u03ce\u03c1\u03b1 \u03b3\u03b9\u03b1 \u03ad\u03bd\u03b1 \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1...\r\n\r\n\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03c3\u03c5\u03bd\u03bf\u03c0\u03c4\u03b9\u03ba\u03ac \u03c4\u03b1 \u03b2\u03ae\u03bc\u03b1\u03c4\u03b1 \u03c4\u03b7\u03c2 \u03bb\u03cd\u03c3\u03b7\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9:\r\n\u039c\u03b5 Pascal \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b1\u03bd \u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03c0\u03bf\u03c5 \u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf $ A''B''C''$, ta $ A'', B'', C''$ \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ A'B'C'$ (\u03bd\u03b1' \u03bd\u03b1\u03b9 \u03ba\u03b1\u03bb\u03ac \u03c4\u03bf Cabri \u03c0\u03bf\u03c5 \u03bc\u03b5 \u03b2\u03bf\u03ae\u03b8\u03b7\u03c3\u03b5 \u03bd\u03b1 \u03c4\u03bf \u03b4\u03c9 \u03b1\u03c5\u03c4\u03cc :P ).\r\n\r\n\u03a4\u03bf 2\u03bf \u03b2\u03ae\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b4\u03b5\u03af\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03b5\u03bc\u03b2\u03b1\u03b4\u03cc\u03bd \u03c4\u03bf\u03c5 $ AB'C$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc (\u03c7\u03c9\u03c1\u03af\u03c2 \u03b2\u03bb\u03ac\u03b2\u03b7 \u03c4\u03b7\u03c2 \u03b3\u03b5\u03bd\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1 \u03ad\u03c7\u03c9 \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03b5\u03b9 \u03c4\u03bf $ P$ \u03c3\u03c4\u03bf \u03c4\u03cc\u03be\u03bf $ AC$). \u0395\u03c0\u03b5\u03b9\u03c4\u03b1, \u03c4\u03bf \u03b5\u03bc\u03b2\u03b1\u03b4\u03cc\u03bd \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 \u03c0\u03bf\u03c5 \u03c8\u03ac\u03c7\u03bd\u03bf\u03c5\u03bc\u03b5 \u03ad\u03c7\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b3\u03c9\u03bd\u03af\u03b1 $ B''$ \u03ba\u03bf\u03b9\u03bd\u03ae \u03bc\u03b5 \u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf! \u0395\u03c4\u03c3\u03b9 $ \\frac {(A''B''C'')}{(AB'C)} \\equal{} \\frac {A''B''\\cdot C''B''}{AB''\\cdot CB''}$. \u03a0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03b5\u03af\u03c3\u03c4\u03b5 \u03c4\u03ad\u03bb\u03bf\u03c2 \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03cc\u03c2 \u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf\u03c2 \u03bb\u03cc\u03b3\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf\u03c2 \u03bc\u03b5 \u03c4\u03b7 \u03bc\u03bf\u03bd\u03ac\u03b4\u03b1 \u03b1\u03c6\u03bf\u03cd:\r\n$ \\frac {A''B''}{AB''} \\equal{} \\frac {A'B''}{AB' \\minus{} A'B''} \\equal{} \\frac {A'B''}{C'B''}$", "Solution_2": "To Pascal itan i olh istoria... Xwris ayto den xekinaei (yparxei allh mia lysh alla poly eyfantasth kai de vgainei eykola, an thelei kapoios thn postarw avrio)\r\n\r\nRe Mim, ti na diavaseis? Na pw oti exoyme viologia ok... :) haha", "Solution_3": "[quote=\"Protonios\"]To Pascal itan i olh istoria... Xwris ayto den xekinaei (yparxei allh mia lysh alla poly eyfantasth kai de vgainei eykola, an thelei kapoios thn postarw avrio)\n\nRe Mim, ti na diavaseis? Na pw oti exoyme viologia ok... :) haha[/quote]\r\n\r\n\u0395\u03b3\u03ce \u03bf \u03b2\u03bb\u03ac\u03ba\u03b1\u03c2 \u03b5\u03af\u03b4\u03b1 \u03c4\u03bf Pascal \u03c3\u03c7\u03b5\u03b4\u03cc\u03bd \u03b1\u03bc\u03ad\u03c3\u03c9\u03c2 \u03ba\u03b9 \u03ad\u03c6\u03b1\u03b3\u03b1 \u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b7 \u03ce\u03c1\u03b1 \u03bd\u03b1 \u03c8\u03ac\u03c7\u03bd\u03c9 \u03c4\u03b7 \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 :blush: \r\n\r\n\u039a\u03b1\u03b9 \u03b5\u03bd\u03bd\u03bf\u03b5\u03af\u03c4\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03b2\u03b9\u03bf\u03bb\u03bf\u03b3\u03af\u03b1 \u03b4\u03b9\u03b1\u03b2\u03ac\u03b6\u03c9 :P \r\n\r\nA! \u0393\u03c1\u03ac\u03c8\u03b5 \u03ba\u03b1\u03b9 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03cc\u03c4\u03b1\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2...", "Solution_4": "IMO Shortlist 2007 . \u0391\u03bd \u03b4\u03b5\u03b9\u03c2 \u03c4\u03bf Pascal \u03bc\u03b5\u03c4\u03ac \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03ae\u03c3\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03cc\u03c4\u03b9 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03b1 \u03c0\u03bf\u03c5 \u03ad\u03c7\u03bf\u03c5\u03bd \u03ba\u03bf\u03c1\u03c5\u03c6\u03ad\u03c2 \u03bd\u03b1 \u03ba\u03b9\u03bd\u03bf\u03cd\u03bd\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03c0\u03b1\u03c1\u03ac\u03bb\u03bb\u03b7\u03bb\u03b7 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b7 \u03b2\u03ac\u03c3\u03b7 \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bd \u03af\u03c3\u03b1 \u03b5\u03bc\u03b2\u03b1\u03b4\u03ac .\r\nhttp://www.mathlinks.ro/viewtopic.php?p=1186790#1186790" } { "Tag": [ "calculus", "integration", "geometry", "derivative", "LaTeX", "rectangle", "function" ], "Problem": "I'm trying to grasp a firm understanding on \"Applications of the definite integral\"\r\n\r\nIn my book there is a problem which solves the area between two curves\r\n\r\nHere ispart of how the problem is worked out:\r\n\r\nA = int(-4 -> 3) [(3-x) - (x2 - 9 ) ] dx\r\n\r\n = int(-4 -> 3) (-x2-x+12) = [ (-x3)/3 - (x2)/2 + 12x] (from -4 -> 3)\r\n\r\nNow I do not udnerstand the taking of derivative from\r\n\r\n[quote] int(-4 -> 3) [(3-x) - (x2 - 9 ) ] dx [/quote]\n\nto\n\n[quote]int(-4 -> 3) (-x2-x+12) [/quote]\r\n\r\nI understand derivatives but for some reason, I'm not seeing this. Also, how do you do the integral sign on this board?", "Solution_1": "That's not a derivative; it's just algebra.\r\n\r\nIf you want pretty math, $\\LaTeX$ is enabled on this forum. Just put an expression between dollar signs to make it work.\r\n\r\nYour integral can be written as\r\n\\int_{-4}^3 (3-x)-(x^2-9)\\, dx\r\n$\\int_{-4}^3 (3-x)-(x^2-9)\\, dx$\r\nMost of that should be easy to understand; the \\, is just a code for a thin space.", "Solution_2": "[quote=\"alltime\"]I'm trying to grasp a firm understanding on \"Applications of the definite integral\"\n\nIn my book there is a problem which solves the area between two curves\n\nHere ispart of how the problem is worked out:\n\nA = int(-4 -> 3) [(3-x) - (x2 - 9 ) ] dx\n\n = int(-4 -> 3) (-x2-x+12) = [ (-x3)/3 - (x2)/2 + 12x] (from -4 -> 3)\n\nNow I do not udnerstand the taking of derivative from\n\n[quote] int(-4 -> 3) [(3-x) - (x2 - 9 ) ] dx [/quote]\n\nto\n\n[quote]int(-4 -> 3) (-x2-x+12) [/quote]\n\nI understand derivatives but for some reason, I'm not seeing this. Also, how do you do the integral sign on this board?[/quote]\r\n\r\nalltime,\r\n\r\nI believe the part that might be confusing you is the $dx$ after the $\\int$ sign, no?\r\n\r\nThe $dx$ is not a derivative. It is the width of an infinitesimal bit of distance along the $x$ direction. The height of a given curve at a point $x$ is $f(x)$, so that the product $f(x)\\,dx$ is the area of a very thin rectangle. The $\\int$ sign is just a script \"S\" meaning sum. In other words, take the areas of all the thin rectangles, $f(x)\\, dx$ and add them up. The result of this addition is the total area under the curve.", "Solution_3": "It's best to see $dx$ as just part of the $\\int$ symbol- it's the indicator to tell you which variable to integrate with respect to, and the integral is badly ambiguous without it. $\\int_0^a x^y\\, dx\\neq\\int_0^ax^y\\, dy$. The $dx$ also conveniently tells you where the function to be integrated ends. Since leaving it off can only lead to confusion, always include it." } { "Tag": [ "inequalities", "triangle inequality", "absolute value", "geometry solved", "geometry" ], "Problem": "Let triangle $ABC$ is inscribed in the circle with radius 1. Let $P$ is an interior point. Prove that:\r\n$PA\\cdot{PB\\cdot{PC}}<\\frac{32}{27}$.", "Solution_1": "This is an inequality of Paul Erd\u00f6s. I am reformulating it in a more canonical way:\r\n\r\n[color=blue][b]Theorem 1.[/b] Let ABC be a triangle inscribed in a circle k with radius R. Let P be a point in the interior or on the boundary of the triangle ABC. Then, we have the inequality $PA\\cdot PB\\cdot PC \\leq \\frac{32}{27}R^3$.[/color]\r\n\r\nThis theorem is considered to be so difficult only since it requires a slightly unusual strategy: When one sees an inequality such as $PA\\cdot PB\\cdot PC \\leq \\frac{32}{27}R^3$, one usually tries to prove it by keeping the triangle ABC fixed and maximizing the term $PA\\cdot PB\\cdot PC$ when the point P moves around the triangle ABC. For this theorem, exactly the other way round works: One maximizes the term $PA\\cdot PB\\cdot PC$ for a fixed point P, with the points A, B, C moving around the circle k (with the additional condition that the point P must remain lying in the interior or on the boundary of the triangle ABC).\r\n\r\nHere is the [i]proof of Theorem 1[/i] in detail:\r\n\r\nLet O be the center of the circle k. If the point P coincides with the point O, then the distances PA, PB, PC are all equal to the radius R of the circle k (since the point P coincides with the center O of the circle k, and the points A, B, C lie on this circle k); thus, $PA\\cdot PB\\cdot PC=R\\cdot R\\cdot R=R^3\\leq \\frac{32}{27}R^3$, and Theorem 1 must hold. Hence, we have proven Theorem 1 for the case when the point P coincides with the point O; thus, in the following, we need only to consider the case when $P\\neq O$.\r\n\r\nAlso, Theorem 1 is trivial if the point P is one of the vertices of triangle ABC (in this case, the product $PA\\cdot PB\\cdot PC$ is zero, and thus clearly $\\leq \\frac{32}{27}R^3$). Thus, we can also assume, in the following observations, that the point P is distinct from each of the vertices of the triangle ABC. Since we also know that the point P lies in the interior or on the boundary of the triangle ABC, it follows that the point P lies inside the circle k.\r\n\r\nLet the lines AP, BP, CP meet the circle k at the points X, Y, Z (apart from the points A, B, C, respectively). Since the point P lies in the interior or on the boundary of the triangle ABC, the points X, Y, Z lie on the arcs BC, CA, AB of the circle k which don't contain the points A, B, C, respectively.\r\n\r\nNow, consider the angles A = < CAB, B = < ABC and C = < BCA of triangle ABC, and the angles X = < ZXY, Y = < XYZ and Z = < YZX of triangle XYZ. Since the sum of the angles of a triangle is always = 180\u00b0, we have A + B + C = 180\u00b0 and X + Y + Z = 180\u00b0, and thus X + Y + Z = A + B + C. If all three inequalities X < A, Y < B and Z < C would simultaneously hold, then we would also have X + Y + Z < A + B + C, what would contradict the equation X + Y + Z = A + B + C. Thus, the three inequalities X < A, Y < B and Z < C cannot all simultaneously hold, so that at least one of them must be wrong. WLOG assume that the inequality X < A is wrong; in other words, we have $X\\geq A$.\r\n\r\nBut X = < ZXY, and since the points X, Y, Z and B lie on one circle (namely, on the circle k), we have < ZXY = < ZBY. Thus, X = < ZXY = < ZBY = < ZBP.\r\n\r\nOn the other hand, A = < CAB, and since the points A, B, C and Z lie on one circle (namely, on the circle k), we have < CAB = < CZB. Thus, A = < CAB = < CZB = < PZB.\r\n\r\nSince X = < ZBP and A = < PZB, the inequality $X\\geq A$ becomes $\\measuredangle ZBP\\geq \\measuredangle PZB$. But since, in a triangle, the greater angle lies opposite to the longer side, it follows that, in the triangle ZPB, we have $PZ\\geq PB$. In other words, $PB\\leq PZ$. Consequently, $PA\\cdot PB\\cdot PC\\leq PA\\cdot PZ\\cdot PC$.\r\n\r\nHence, in order to prove the inequality $PA\\cdot PB\\cdot PC \\leq \\frac{32}{27}R^3$, it is enough to verify $PA\\cdot PZ\\cdot PC \\leq \\frac{32}{27}R^3$. This certainly must be easier, since the point P lies on the segment ZC.\r\n\r\nBy the triangle inequality for the (possibly degenerate) triangle APO, we have $PA\\leq AO+PO$. Since the point O is the center of the circle k, and the point A lies on the circle k, the distance AO equals the radius R of the circle k; thus, we get $PA\\leq R+PO$. On the other hand, since the point P lies on the chord ZC of the circle k, the product $PZ\\cdot PC$ is the absolute value of the power of the point P with respect to the circle k, and thus it equals $PZ\\cdot PC=R^2-PO^2$ (since O is the center and R is the radius of the circle k). In other words, $PZ\\cdot PC=\\left(R-PO\\right)\\cdot\\left(R+PO\\right)$. This equation and the inequality $PA\\leq R+PO$ together imply\r\n\r\n$PA\\cdot PZ\\cdot PC = PA\\cdot\\left(PZ\\cdot PC\\right) \\leq \\left(R+PO\\right)\\cdot\\left(\\left(R-PO\\right)\\cdot\\left(R+PO\\right)\\right)$\r\n$=\\left(R+PO\\right)^2\\cdot\\left(R-PO\\right)=\\frac12\\cdot\\left(R+PO\\right)^2\\cdot 2\\left(R-PO\\right)$.\r\n\r\nSince the point P lies inside the circle k, while O is the center and R is the radius of this circle k, we have PO < R; thus, R - PO > 0, so that 2 (R - PO) > 0. Consequently, the three numbers R + PO, R + PO and 2 (R - PO) are all nonnegative; hence, we can apply the AM-GM inequality to these three numbers, and obtain\r\n\r\n$\\sqrt[3]{\\left(R+PO\\right)\\cdot\\left(R+PO\\right)\\cdot\\left(2\\left(R-PO\\right)\\right)}\\leq\\frac{\\left(R+PO\\right)+\\left(R+PO\\right)+\\left(2\\left(R-PO\\right)\\right)}{3}$.\r\n\r\nThis simplifies to\r\n\r\n$\\sqrt[3]{\\left(R+PO\\right)^2\\cdot 2\\left(R-PO\\right)}\\leq\\frac43 R$,\r\n\r\nand cubing yields $\\left(R+PO\\right)^2\\cdot 2\\left(R-PO\\right)\\leq\\left(\\frac43 R\\right)^3=\\frac{64}{27}R^3$. Thus,\r\n\r\n$PA\\cdot PZ\\cdot PC \\leq \\frac12\\cdot\\left(R+PO\\right)^2\\cdot 2\\left(R-PO\\right) \\leq \\frac12\\cdot\\frac{64}{27} R^3 = \\frac{32}{27} R^3$.\r\n\r\nThis completes our proof of Theorem 1.\r\n\r\nNote that a closer look at our proof of Theorem 1 shows that, in the inequality $PA\\cdot PB\\cdot PC \\leq \\frac{32}{27}R^3$, equality holds if and only if the triangle ABC is an isosceles triangle with base angle $90^{\\circ}-\\arccos\\frac13$, and the point P is the midpoint of its base. Hence, if we add the additional condition that the point P should lie (strictly!) in the interior of the triangle ABC, then we always have the strict inequality $PA\\cdot PB\\cdot PC < \\frac{32}{27}R^3$.\r\n\r\nThis solves your problem.\r\n\r\n Darij", "Solution_2": "Thank you very much, Darij for this very detailed proof.", "Solution_3": "What can I say, darij, veryvery nice. Thank you very much!!" } { "Tag": [ "geometric series", "geometric sequence" ], "Problem": "I'm lost on how to figure this out.\r\n\r\nBasically I'm given $1,000 over 7 years with 8% yearly increases, what are the year totals?\r\n\r\nI'm totally lost on figuring this out. Any help?", "Solution_1": "So you want to find out how much $ \\$1000$ will become in $ 7$ years, if there's an $ 8\\%$ increase each year? If so,\r\n\r\n[hide=\"Total after 7 years\"]Each year, you multiply the current amount by $ 100\\% \\plus{} 8\\% \\equal{} 108\\% \\equal{} 1.08$. Thus, after $ n$ years, the amount will be $ \\$1000\\cdot(1.08)^n$.\n\nIn this particular case, $ n \\equal{} 7$, so $ A \\equal{} \\$1000\\cdot(1.08)^7\\approx\\$1713.82$.[/hide]\n\nIf you meant to ask for the increase each year relative to the previous year, then\n\n[hide=\"Increase each year\"]For any year $ n$, the increase from the previous year will be $ \\$1000\\cdot(1.08)^{n \\minus{} 1}\\cdot(0.08)$. Notice that here, we first find the total after the previous year, and then multiply by $ 8\\%$ to find how much this year will bring.\n\nThe first year will increase $ \\$1000$ by $ \\$1000\\cdot(1.08)^0\\cdot(0.08) \\equal{} \\$80$. After $ 1$ year, the total will be $ \\$1080$.\n\nThe second year will increase $ \\$1080$ by $ \\$1000\\cdot(1.08)^1\\cdot(0.08) \\equal{} \\$86.40$. The new total will be $ \\$1166.40$.\n\nAnd so on...[/hide]", "Solution_2": "Thank you for your reply, but I'm afraid I worded my question poorly.\r\n\r\nWhat I'm looking to answer is:\r\n\r\nYou will be given $1000 total over a 7 year period with 8% annual increases.\nProvide the yearly amounts paid.\nYear 1 \n$\r\nYear 2 = Year 1 $ * 1.08\r\n\r\nand so on.\r\n\r\nHopefully that is more clear.", "Solution_3": "Wouldn't you just use the formula I provided, $ f(n) \\equal{} \\$1000\\cdot(1.08)^n$, where $ n$ is the year? It gives the total amount after $ n$ years.", "Solution_4": "1000 dollars is not the amount given in the first year. 1000 dollars is the total amount given over all 7 years. At least that's how I interpret the problem.", "Solution_5": "Ravi B you have it exactly right.\r\n\r\nyear 1 + year 2 ... with a sum total of $1000", "Solution_6": "[quote=\"Ravi B\"]1000 dollars is not the amount given in the first year. 1000 dollars is the total amount given over all 7 years. At least that's how I interpret the problem.[/quote]\r\nIf that is indeed how the problem should be interpreted, then...\r\n\r\n[hide=\"Solution\"]Instead of multiplying by $ 1.08$, we now divide by $ 1.08$: The original amount was $ \\$1000\\cdot(1.08)^{ \\minus{} 7}\\approx\\$583.49$.\n\nIf $ 7$ years ago is our reference point (i.e. year $ 0$), then after $ n$ years, the amount will be $ \\left(\\$1000\\cdot(1.08)^{ \\minus{} 7}\\right)\\cdot(1.08)^n \\equal{} \\$1000\\cdot(1.08)^{n \\minus{} 7}$. Notice that the exact value $ \\$1000\\cdot(1.08)^{ \\minus{} 7}$ is used instead of the approximation $ \\$583.49$.[/hide]", "Solution_7": "I think the confusion I'm causing here is because I have not clarified that this is not a compound interest problem.\r\n\r\nThe closest thing I can relate it to is a contract.\r\n\r\n\"We are going to pay Johnny $1000 total over a period of 7 years to build a dock\"\n\"We are not going to just pay Johnny $142 a year, but instead give him 8% annual raises\"\r\n\r\nHow much would that be per year?", "Solution_8": "Now [i]I'm[/i] confused. :D :| :(\r\n\r\nSo what you're saying is:\r\n\r\n\"You get some amount $ x$ one year, then $ x\\cdot(1.08)$ the next year, then $ x\\cdot(1.08)^2$ then next, etc., and this goes on for $ 7$ years. If the sum $ x\\plus{}x\\cdot(1.08)\\plus{}\\ldots$ is $ \\$1000$, then what are $ x$, $ x\\cdot(1.08)$, etc.?\"\r\n\r\nIs that correct?", "Solution_9": "[quote=\"i_like_pie\"]Now [i]I'm[/i] confused. :D :| :(\n\nSo what you're saying is:\n\n\"You get some amount $ x$ one year, then $ x\\cdot(1.08)$ the next year, then $ x\\cdot(1.08)^2$ then next, etc., and this goes on for $ 7$ years. If the sum $ x \\plus{} x\\cdot(1.08) \\plus{} \\ldots$ is $ \\$1000$, then what are $ x$, $ x\\cdot(1.08)$, etc.?\"\n\nIs that correct?[/quote]\r\n\r\nExactly!\r\n\r\nI've just got no idea how to calculate it.", "Solution_10": "Ah, ok then. At least we're on the same page now. :)\r\n\r\n[hide=\"Solution\"]Let the amount of money paid in the first year be $ x$.\n\nWe know $ x + x\\cdot(1.08) + x\\cdot(1.08)^2 + \\ldots + x\\cdot(1.08)^6 + x\\cdot(1.08)^7 = \\$1000$.\n\nNotice that the LHS is a geometric series, and recall that the sum of the first $ n$ terms is $ a_1\\cdot\\frac {r^n - 1}{r - 1}$, where $ a_1$ is the first term and $ r$ is the common ratio.\n\nThus, we can express the LHS as the sum of a geometric sequence, and then solve for $ x$:\n\\begin{align*}x\\cdot\\frac {(1.08)^7 - 1}{1.08 - 1} & = \\$1000 \\\\\n \\\\\nx & = \\$1000\\cdot\\frac {1.08 - 1}{(1.08)^7 - 1} \\\\\n \\\\\nx & = \\$\\frac {244140625000}{2178418789} \\\\\n \\\\\nx & \\approx\\$112.07\\end{align*}\nTo get accurate numbers for the other six years, use either $ x = \\$1000\\cdot\\frac {1.08 - 1}{(1.08)^7 - 1}$ or $ x = \\$\\frac {244140625000}{2178418789}$, which are both accurate.[/hide]", "Solution_11": "Thank you very much!\r\n\r\nIt certainly makes sense now. Couldn't have gotten there without your help." } { "Tag": [ "algebra", "polynomial", "vector", "function", "abstract algebra", "geometry", "geometric transformation" ], "Problem": "Let $ V$ be the vector space of all functions of one variable. Suppose $ P: V\\to\\mathbb{R}$ is a linear map such that for all $ f,g\\in V$ with $ P(fg) \\equal{} 0$ we have that $ P(f) \\equal{} 0$ or $ P(g) \\equal{} 0$. Prove that there exist real numbers c and $ x_0$ such that $ P(f) \\equal{} cf(x_0)$ for all $ f\\in V$.", "Solution_1": "Let $ K$ be the kernel of $ P$. Note that given any two functions $ f, g$ we have $ P(g) f \\minus{} P(f) g \\in K$. We will be using this fact repeatedly; it is in fact equivalent to the condition that $ P$ is linear. \r\n\r\n[b]Case:[/b] $ 1 \\in K$. Then the constant functions are in $ K$, and if $ f(x) \\in K$ then so is $ f(x) \\plus{} C$ for any $ C$. Suppose there existed $ g$ such that $ P(g) \\neq 0$. Then $ P(g) g^2 \\minus{} P(g^2) g \\in K$. But since $ g \\not \\in K$ it follows that $ P(g) g \\minus{} P(g^2) \\in K \\implies P(g) g \\in K \\implies g \\in K$; contradiction. Hence there exists no such $ g$ and $ P(f) \\equal{} 0$ everywhere. \r\n\r\n[b]Case:[/b] $ 1 \\not \\in K$. Normalize $ P$ so that $ P(1) \\equal{} 1$. Then given any $ f(x) \\not\\in K$ we have $ f \\minus{} P(f) \\in K$ (and this is the unique constant $ P(f)$ for which this is true). Let $ P(x) \\equal{} x_0$. Then $ x \\minus{} x_0 \\in K$. \r\n\r\nFor any $ f$ and any $ g$ with $ P(g) \\neq 0$ we therefore have $ P(g) fg \\minus{} g P(fg) \\in K \\implies P(g) f \\minus{} P(fg) \\in K$. But the unique translation of $ P(g) f$ in $ K$ is given by $ P(g) f \\minus{} P(g) P(f)$. It follows that $ P(fg) \\equal{} P(f) P(g)$ when $ P(g) \\neq 0$. When $ P(g) \\equal{} 0$ clearly $ P(g \\plus{} 1) \\equal{} 1 \\neq 0$ and $ P(fg \\plus{} f) \\equal{} P(f) P(g \\plus{} 1) \\implies P(fg) \\equal{} P(f) P(g)$. Hence $ P$ is multiplicative. In particular, any function of the form $ (x \\minus{} x_0) q(x)$ is in $ K$. \r\n\r\nNow, for any $ f$ write $ f(x) \\equal{} f(x_0) \\plus{} (x \\minus{} x_0) q(x)$ where $ q(x)$ is uniquely determined except for its value at $ x_0$, which is irrelevant. By the above discussion $ P(f) \\equal{} f(x_0)$ as desired.\r\n\r\n[b]Remark:[/b] What we have essentially done is shown that $ K$ is a prime ideal of $ V$. In fact, all the prime ideals of $ V$ take this form. I believe the above argument remains valid if $ \\mathbb{R}$ is replaced by an integral domain; what happens if we lose commutativity?", "Solution_2": "The \"one-variable functions\" run from where to where with what properties?", "Solution_3": "The solution makes no assumptions other than that the functions take $ \\mathbb{R}$ to $ \\mathbb{R}$. No other assumptions are necessary. In particular, the \"division algorithm\" $ f(x) \\equal{} f(x_0) \\plus{} (x \\minus{} x_0) g(x)$ where $ g(x)$ is uniquely determined except for its value at $ x_0$ works with no assumptions.\r\n\r\nThe more interesting case is probably when $ V$ is restricted to the set of polynomial functions, in which case what we are doing is the easiest kind of [url=http://en.wikipedia.org/wiki/Algebraic_geometry]algebraic geometry[/url].", "Solution_4": "Thanks, it wasn't clear to me that the functions were real to real." } { "Tag": [ "puzzles" ], "Problem": "Source: The Art of Problem Solving - A Resource for the Mathematics Teacher - Page 240 - Exercise 1-3\r\n\r\n1) A certain locksmith makes keys using anyone of 15 brands of keys. Each\r\nbrand uses a different length and thickness. No matter what brand is\r\nchosen, there are 8 different positions from which metal can be removed\r\nor left alone. For the first 5 positions, there are 3 different depths at which\r\nthe metal can be removed. For the remaining positions, there are 4\r\ndifferent depths. How many different keys are possible if metal must be\r\nremoved in at least one position?\r\n\r\n\r\n\r\nHere is my solution... Is this correct:\r\n[hide]\n((4^5*5^3)-1)*15 = 1,919,985 different keys are possible.\n\n(3 depths + 1 no cut for 5 positions.. 4 depths + 1 no cut for 3 positions) and removing 1 for all positions are not cut. Multiplied by number of brands 15.\n[/hide]\n\nPlease let me know. Thanks.\n\n2) How many subsets does a set of 15 elements have?\n\nMy solution:\n\n[hide]2^15 = 32,768[/hide]\n\n3) Many different kinds of desserts have been put out for people to choose\nfrom at a party. Judy is watching her weight and decides to have at least\none piece of fruit and no other type of dessert. If the fruit bowl contains\nexactly 6 apples, 5 plums, 3 oranges, 4 bananas, and 5 pears, how many\ndifferent combinations of desserts can Judy have?\n\nMy solution:\n\n[hide]\n(7x6x4x5x6) - 1 = 5,039 different combinations she can have\n[/hide]\r\n\r\nPlease advise if the solution is correct in all 3 exercises. Thank you.", "Solution_1": "i think there is problem in 3 answer. its a question of permutation and i love solving them" } { "Tag": [], "Problem": "How many four-digit positive integers are there with thousands digit 2?", "Solution_1": "Overly complicated lol", "Solution_2": "[hide]All numbers satisfying this must be of the form $ \\overline{2abc}$ where $ 0\\le a,b,c\\le9\\implies10^3\\equal{}\\boxed{1000}$.[/hide]", "Solution_3": "[quote=\"snee\"]Overly complicated lol[/quote]\n\nits not that hard im in 5th grade and that problem took me 2 seconds to find out its super easy", "Solution_4": "I think that the wording was awkward, once I viewed the answer I got it, but the wording confused me.", "Solution_5": "It's not much difficult. Just 4-digit numbers with 2 in the thousands digit.\n\n[hide=\"Alternate solution\"]\n\nAll the numbers between $2000$ and $2999$ have $2$ as a thousands digit. \n\n$2999-2000=999$, but we still have to account for the number $2999$, so we add $1$ to get ${\\boxed{1000}}$.[/hide]", "Solution_6": "I think that they should change how they wrote the problem because I did not understand it at first.", "Solution_7": "[hide=Solution]We use the formula b-a+1. \n\nPlugging in, we find 2999-2000=999\n\n999+1 is ${\\boxed{1000}}$[/hide]", "Solution_8": "Bruh that\u2019s easy the answer is 1000", "Solution_9": "[hide=Solution] Since the only 4-digit numbers with thousands digit 2 are from 2,000 to 2,999, we could add. SInce we want 2,00 included, we start at 1,999. 1,999 to 2,999 is 1,000, so our solution is is $\\boxed{1,000}$.", "Solution_10": "[quote=fidgetboss_4000]Bruh that\u2019s easy the answer is 1000[/quote]\n\nPlease hide your solutions.\n", "Solution_11": "[quote=wowacow][quote=\"snee\"]Overly complicated lol[/quote]\n\nits not that hard im in 5th grade and that problem took me 2 seconds to find out its super easy[/quote]\n\n[color=#9a00ff]Well, you must be Super smart. :) I'm not even as smart as u. :( I'm in 6th. [/color]", "Solution_12": "[quote=TethysTide][quote=fidgetboss_4000]Bruh that\u2019s easy the answer is 1000[/quote]\n\nPlease hide your solutions.[/quote]\n\n[color=#9a00ff]Ummm... Ur smart....SOme people r not so smart, (me). :) :([/color]", "Solution_13": "[quote=BluRose11][quote=wowacow][quote=\"snee\"]Overly complicated lol[/quote]\n\nits not that hard im in 5th grade and that problem took me 2 seconds to find out its super easy[/quote]\n\n[color=#9a00ff]Well, you must be Super smart. :) I'm not even as smart as u. :( I'm in 6th. [/color][/quote]\n\nPlease don't revive a nearly 2-year old post to reply to a nearly 9-year old post about a topic that has nothing to do with the problem. From your posts I'm inferring that you're relatively new and so I just wanted to tell you that it's not a good idea to bump a topic if you have nothing constructive to add to it. :)", "Solution_14": "[hide=Solution]All of the numbers with thousands digit 2 are the numbers from $2000$ through $2099$. Thus, including these numbers, there are $\\boxed{1000}$ numbers with thousands digit $2$.", "Solution_15": "[hide=Solution] We can use permutations. There is 1 choice for the first number and 10 for the last three numbers. There will be $1\\cdot10\\cdot10\\cdot10 = 1000$ numbers that have thousands digit 2.[/hide]", "Solution_16": "this problem is easy. How is it level 6 prealgebra?\n[hide=Answer.2]The answer is 1000.[/hide]", "Solution_17": "[quote=math154][hide]All numbers satisfying this must be of the form $ \\overline{2abc}$ where $ 0\\le a,b,c\\le9\\implies10^3\\equal{}\\boxed{1000}$.[/hide][/quote]\n\nme to expect me not in 5th grade ", "Solution_18": "[quote=jcl-12]this problem is easy. How is it level 6 prealgebra?\n[hide=Answer.2]The answer is 1000.[/hide][/quote]\n\nUmmmm... it is level 10 prealgebra. Also, here is [hide=My solution]We make a list: 2000, 2001, 2002 ... 2997, 2998, 2999. We subtract 1999 from each number so the first number is one. we get: 1, 2, 3, ... 998, 999, 1000. So there are $\\boxed{1000}$ integers." } { "Tag": [ "LaTeX", "algebra", "binomial theorem", "Pascal\\u0027s Triangle" ], "Problem": "(x+y)^10\r\nI believe thats how you write it, otherwise its (x+y) to the tenth power. I have to write out the product. It says there is a way to do it w/o multiplying it out the long way, but I'm stuck. Thank you for the help.", "Solution_1": "[quote=\"sean2win2004\"](x+y)^10\nI believe thats how you write it, otherwise its (x+y) to the tenth power. I have to write out the product. It says there is a way to do it w/o multiplying it out the long way, but I'm stuck. Thank you for the help.[/quote]\r\n\r\nThe binomial theorem\r\n\r\n\r\nhttp://mathworld.wolfram.com/BinomialTheorem.html \r\n\r\nhttp://www.answers.com/binomial+theorem?gwp=12&method=2", "Solution_2": "You don't need to multiply.The coefficients are of the numbers of the tenth row of pascal's triangle, so this is what the product looks like:\r\n$(x+y)^{10}=x^{10}+10x^9y+45x^8+y^2+110x^7y^3+190x^6y^4+232x^5y^5+190x^4y^6+110x^3y^7+45x^2y^8+10xy^9+y^{10}.$", "Solution_3": "Thanks for the reply.\r\n\r\nI already know from FOIL that (x+y)^2 is x^2 + 2xy + y^2\r\n\r\nDo you happen to have the complete answer? The first link was a little complicated for me.", "Solution_4": "By FOILing you get $(x+y)^2=x^2+2xy+y^2$. However, if you look at the coefficients of $x^2$, $xy$ and $y^2$ in the expansion, you see that they are 1, 2, 1. But 1, 2, 1 are the numbers of the second row of Pascal's triangle (the top row is the \"zeroth\" row, by definition, since any number to the power (except sometimes zero itself) is one). If you look down the triangle a little bit, and glance at the tenth row, you'll find the coefficients for the expansion of $(x+y)^{10}$. As easyas3.14159... wrote:\r\n\r\n[quote=\"easyas3.14159...\"]You don't need to multiply.The coefficients are of the numbers of the tenth row of pascal's triangle, so this is what the product looks like:\n$(x+y)^{10}=x^{10}+10x^9y+45x^8+y^2+110x^7y^3+190x^6y^4+232x^5y^5+190x^4y^6+110x^3y^7+45x^2y^8+10xy^9+y^{10}.$[/quote]\r\n\r\nI would advise finding a friendly site (or person) that will describe the binomial theorem for you.", "Solution_5": "Pascal's triangle. Or a ton of really annoying multiplication. I recommend the former.", "Solution_6": "Perhaps one of you could explain why Pascal's triangle works for this problem ;)", "Solution_7": "$(x+y)^{10}$, when fully expanded, will have a whole bunch of terms. In fact, it'll have $2^{10}$ (not necessarily) distinct terms (because each $(x+y)$ contributes either an $x$ or a $y$ to each term), all with coefficient $1$.\r\n\r\nNow, a bunch of these have the same power (for example, there will be a lot of $x^5 y^5$), so they get added up. But let's keep them separate for now because it's easier to count them that way. Each of them takes either $x$ or $y$ from one of the $10$ factors, so each of them has a total power of $10$ - that is, they're all of the form $x^{a} y^{10-a}$.\r\n\r\nHow many of $x^{10} y^{0}$ are there? This is equivalent to asking how many ways we can pick ten $x$s out of ten factors, which is ${10 \\choose 10} = 1$.\r\n\r\nHow many of $x^{9} y^{1}$ are there? This is equivalent to asking how many ways we can pick nine $x$s out of ten factors, which is ${10 \\choose 9} = 10$.\r\n\r\n...\r\n\r\nAnd so forth. It turns out that this exactly gives the tenth row of Pascal's triangle because the $m^{th}$ entry in the $n^{th}$ row of Pascal's triangle is ${n \\choose m}$. \r\n\r\nShort demonstration why: Pascal's triangle is formed by repeated adding of two entries to give the entries immediately below it. Similarly, given the complete expansion of $(x+y)^{n}$, if we multiply it by $(x+y)$ each $x^{k} y^{n+1 - k}$ term will get some value from the entry above and to the left of it (that is, $x^{k-1} y^{n+1-k} \\cdot x$) and some value from the entry above and to the right of it (that is, $x^{k} y^{n-k} \\cdot y$).\r\n\r\n\r\n\r\n(Apologies if this is really hard to see without a visual demonstration, but once you understand the connection you'll like combinatorics a lot more. :) )", "Solution_8": "Is there anything like that for $(x-y)^n$?", "Solution_9": "Just plug in $-y$ for $y$ - every term with an odd power of $y$ is just negative what it was before. :)", "Solution_10": "What's FOIL? :huh:", "Solution_11": "FOIL is a method for multiplying binomials. You multiply the [b]F[/b]irst terms of each binomial, then the [b]O[/b]uter terms of each binomial, then the [b]I[/b]nner terms of each, and finally the [b]L[/b]ast terms of each.", "Solution_12": "$(a+b)(c+d) = (ac + bd) + (bc + ad)$", "Solution_13": "ok so basically its like this\r\n\r\nThis is a triangle \r\n\r\n$1$\r\n$1,1$ coeficients for $(x+y)^1$\r\n$1,2,1$ coeficients for $(x+y)^2$\r\n$1,3,3,1$ coeficients for $(x+y)^3$\r\n$1,4,6,4,1$ coeficients for $(x+y)^4$ \r\n\r\nHow can we generate this triangle further one line ?\r\n\r\nWell..\r\nif we denote by $A_{L,C}$ as the number on line L in column C.\r\nthen we find that $A_{L,C}=A_{L-1,C-1}+A_{L-1,C}$\r\notherwise said if you want to find the numebers on the next line\r\na number is the sum of the number above him and the number to the\r\nleft of the number above him.\r\n\r\nso that $(x+y)^n=A_{n+1,1}x^{n}+A_{n+1,2}x^{n-1}y+A_{n+1,3}x^{n-2}y^2+...$\r\nso that the kth term in the sum is\r\n$A_{n+1,k}x^{n-k+1}y^{k-1}$\r\n\r\nSo now you know how to write $(x+y)^n$ in unfolded form as the guys reffer to..", "Solution_14": "to find a number in Pascal's triangle, just add the two number directly above the number you are trying to find. See the diagram below.", "Solution_15": "Hi everybody... this is my first reply so I hope it helps to generalize the idea.\r\nAs t0rajir0u said. The coefficients can be found using combinatorics. After simplifing we could come up with the following theorem, called the binomial theorem:\r\n\r\n(a+b)^n=a^n+na^(n-1)b+{[n(n-1)]/2!}a^(n-2)b^2+\r\n{[n(n-1)(n-2)]/3!}a^(n-3)b^3... + nab^(n-1) + b^n.\r\n\r\n(If you don't understand please tell me how can I make this look better, I guess using the code or something but I don't know, so please tell me)", "Solution_16": "[quote=\"Jhonny Ruiz\"]\n(If you don't understand please tell me how can I make this look better, I guess using the code or something but I don't know, so please tell me)[/quote]\\\r\n\r\n\r\nOften, putting dollar signs around a simple formula can make it look pretty. However, your formula is more complex, and you will need some knowledge of Latex code. For a nice tutorial on Latex, visit http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php", "Solution_17": "i'm trying the latex code out please delete this\r\n\r\n\\cfrac{2}{1+\\cfrac{2}{1+\\cfrac{2}{1+\\cfrac{2}{1}}}} \r\n\r\n\\sqrt{x+\\frac{1}{2} \r\n\r\n\\sum_{i=1}^{\\infty}\\frac{1}{i} \r\n\r\n\\[ n^2 + 5 = 30\\text{ so we have }n=\\pm5 \\]", "Solution_18": "Here we go. This is the theorem...\r\n\r\n$(a+b)^n=a^n+na^{n-1}b^1+\\frac{n(n-1)}{2!}a^{n-2}b^2+ \\frac{n(n-1)(n-2)}{3!}a^{n-3}b^3+... + na^1b^{n-1} + b^n$\r\n\r\nI'm sure you can understand it now. Hope it helps\r\nThanks" } { "Tag": [ "logarithms", "integration", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "Dear everyone.\r\n\r\nfind the value of $ \\frac {1}{1\\cdot 2\\cdot 3\\cdot 4} \\plus{} \\frac {1}{4\\cdot 5\\cdot 6\\cdot 7} \\plus{} \\frac {1}{7\\cdot 8\\cdot 9\\cdot 10} \\plus{} \\cdots$\r\nIs this calculation possible? :( please help me~", "Solution_1": "$ \\sum \\frac{1}{n(n\\plus{}1)(n\\plus{}2)(n\\plus{}3)} \\equal{} \\sum (\\frac{1}{2n(n\\plus{}3)} \\minus{} \\frac{1}{2(n\\plus{}1)(n\\plus{}2)})$\r\n$ \\equal{} \\sum (\\frac{1}{6n} \\minus{} \\frac{1}{6(n\\plus{}3)} \\minus{} \\frac{1}{2(n\\plus{}1)} \\plus{} \\frac{1}{2(n\\plus{}2)})$\r\n\r\nnotice which terms cancel with previous terms in sum leaving\r\n\r\n$ \\frac {1}{6} \\plus{} \\frac{1}{12} \\plus{} \\frac{1}{18} \\minus{} \\frac{1}{4} \\equal{} \\frac{1}{18}$", "Solution_2": "philg you are wrong his sumation doesnt include 1/2.3.4.5 but yours does", "Solution_3": "$ \\sum_{n\\equal{}1}^{\\infty }\\frac{1}{(3n\\minus{}2)(3n\\minus{}1)3n(3n\\plus{}1)}\\equal{}\\frac{\\pi }{12\\sqrt{3}}\\allowbreak \\plus{}\\frac{1}{6}\\minus{}\\frac{\\ln 3}{4}$", "Solution_4": "Oops! I should have read more carefully :blush:", "Solution_5": "[quote=\"IndoChina\"]$ \\sum_{n \\equal{} 1}^{\\infty }\\frac {1}{(3n \\minus{} 2)(3n \\minus{} 1)3n(3n \\plus{} 1)} \\equal{} \\frac {\\pi }{12\\sqrt {3}}\\allowbreak \\plus{} \\frac {1}{6} \\minus{} \\frac {\\ln 3}{4}$[/quote]\r\n\r\nPlease show your work on this :lol:", "Solution_6": "We have:\r\n\r\n$ S \\equal{} \\sum ^\\infty _ {n \\equal{} 0} \\frac {1}{(3n \\plus{} 1)(3n \\plus{} 2)(3n \\plus{} 3)(3n \\plus{} 4)} \\equal{}$\r\n\r\n$ \\equal{} \\frac {1}{81} \\sum ^\\infty _ {n \\equal{} 0} \\frac {1}{(n \\plus{} \\frac {1}{3})(n \\plus{} \\frac {2}{3})(n \\plus{} 1)(n \\plus{} \\frac {4}{3})}$\r\n\r\n$ \\equal{} \\frac {1}{162} \\sum ^\\infty _ {n \\equal{} 0} \\left(\\left(\\frac{9}{n\\plus{}\\frac{1}{3}}\\minus{}\\frac{9}{n\\plus{}\\frac{1}{3}\\plus{}1}\\right)\\plus{}\\left(\\frac{27}{n\\plus{}1}\\minus{}\\frac{27}{n\\plus{}\\frac{2}{3}}\\right)\\right)$\r\n\r\nNow, the first two elements in the sum cancel each other out:\r\n\r\n$ S \\equal{} \\frac{1}{6}\\plus{} \\frac{1}{6} \\sum ^\\infty _ {n \\equal{} 0}\\left(\\frac{1}{n\\plus{}1}\\minus{}\\frac{1}{n\\plus{}\\frac{2}{3}}\\right)$\r\n\r\nNow, we can define:\r\n\r\n$ f(x)\\equal{}\\sum ^\\infty _ {n \\equal{} 1}\\left(\\frac{x^n}{n}\\minus{}\\frac{x^{n\\minus{}\\frac{1}{3}}}{n\\minus{}\\frac{1}{3}}\\right)$\r\n\r\nThis makes $ S \\equal{}\\frac{1}{6}\\left(1 \\plus{} f(1)\\right)$. We have:\r\n\r\n$ f'(x)\\equal{}\\sum ^\\infty _ {n \\equal{} 1}\\left(x^{n\\minus{}1}\\minus{}x^{n\\minus{}\\frac{4}{3}}\\right) \\equal{}\\frac{x}{1\\minus{}x}\\left( \\frac{1}{x} \\minus{} \\frac{1}{x^{\\frac{4}{3}}}\\right) \\equal{} \\frac{1}{1\\minus{}x}\\left(1\\minus{}\\frac{1}{\\sqrt[3]{x}}\\right)$\r\n\r\nThis gives us:\r\n\r\n$ f(x)\\equal{}\\int \\frac{1}{1\\minus{}x}\\left(1\\minus{}\\frac{1}{\\sqrt[3]{x}}\\right) \\,dx$\r\n\r\nUsing the substitution $ x \\equal{} u^3$ we get:\r\n\r\n$ f(u) \\equal{} \\int \\frac{1}{1\\minus{}u^3}\\left(1\\minus{}\\frac{1}{u}\\right) 3 u ^2\\,du \\equal{} \\minus{}3 \\int \\frac{u}{1\\plus{}u\\plus{}u^2} \\,du$\r\n\r\nWe can write:\r\n\r\n$ \\frac{u}{1\\plus{}u\\plus{}u^2} \\equal{} \\frac{1}{2}\\frac{2u\\plus{}1}{1\\plus{}u\\plus{}u^2} \\minus{}\\frac{1}{2}\\frac{1}{u^2\\plus{}u\\plus{}1}$\r\n\r\nAnd integrating the first term gives us:\r\n\r\n$ f(u) \\equal{}\\frac{3}{2} \\int \\frac{du}{u^2\\plus{}u\\plus{}1} \\minus{} \\frac{3}{2} \\ln \\left(1\\plus{}u\\plus{}u^2\\right)$\r\n\r\nFinally, we have:\r\n\r\n$ \\int \\frac{du}{u^2\\plus{}u\\plus{}1} \\equal{} \\int \\frac{du}{\\left(u\\plus{}\\frac{1}{2}\\right)^2\\plus{}\\frac{3}{4}} \\equal{} \\frac{2}{\\sqrt{3}}\\tan^{\\minus{}1}\\left(\\frac{1\\plus{}2u}{\\sqrt{3}}\\right) \\plus{} C$\r\n\r\nSo, all this rounds up to:\r\n\r\n$ f(x)\\equal{}\\sqrt{3} \\tan^{\\minus{}1}\\left(\\frac{1\\plus{}2\\sqrt[3]{x}}{\\sqrt{3}}\\right)\\minus{}\\frac{3}{2} \\ln \\left(1\\plus{}\\sqrt[3]{x}\\plus{}\\sqrt[3]{x^2}\\right) \\plus{} C$\r\n\r\nSince from the sum-involving definition of $ f$ we get $ f(0) \\equal{}0$ we have $ C \\equal{} \\minus{}\\sqrt{3} \\tan^{\\minus{}1}\\left(\\frac{1}{\\sqrt{3}}\\right) \\equal{}\\minus{} \\frac{\\pi}{2 \\sqrt{3}}$. So, for $ f(1)$ we have:\r\n\r\n$ f(1)\\equal{}\\sqrt{3} \\tan^{\\minus{}1} \\sqrt{3} \\minus{} \\frac{3}{2}\\ln 3 \\minus{} \\frac{\\pi}{2 \\sqrt{3}} \\equal{} \\frac{\\pi}{2 \\sqrt{3}} \\minus{} \\frac{3}{2}\\ln 3$\r\n\r\nThis gives us:\r\n\r\n$ S \\equal{} \\frac{1}{6}\\left(1\\plus{}f(1)\\right) \\equal{} \\frac{1}{6}\\left(1\\plus{}\\frac{\\pi}{2 \\sqrt{3}}\\minus{} \\frac{3}{2}\\ln 3 \\right) \\equal{} \\frac{1}{6} \\plus{} \\frac{\\pi}{12\\sqrt{3}}\\minus{}\\frac{\\ln3}{4}$" } { "Tag": [ "calculus", "integration", "logarithms", "limit", "calculus computations" ], "Problem": "1) find the limit\r\nlim (e^(1/x))/x\r\nx->0- \r\n\r\n2)\r\nintegral sign (0 to 2) |x-1||x+1| dx\r\n\r\n\r\n3)\r\nsuppose f(0) = 0 and integral sign (0 to 2) f'(2t)e^(f(2t)) dt = 5 . Find the value of f(4)", "Solution_1": "Solution for number 3:\r\n\r\n $ \\int_{0}^{2}f'(2t)e^{f(2t)}dt \\equal{} 5$ or $ \\int_{0}^{2}(e^{f(2t)})'dt \\equal{} 5$ or $ e^{f(4)} \\minus{} e^{f(0)} \\equal{} 5$\r\n\r\nor $ e^{f(4)} \\equal{} 6 \\equal{} e^{ln6}$ or $ f(4) \\equal{} ln6$", "Solution_2": "the answer the teacher told me b4 was ln9, but i don't know how to get this, also can you explain the steps near the end.", "Solution_3": "Solution for number 2:\r\n\r\n $ \\int_{0}^{2}|x \\minus{} 1\\parallel{}x \\plus{} 1|dx \\equal{} \\int_{0}^{2}|x^2 \\minus{} 1|dx \\equal{} \\int_{0}^{1}1 \\minus{} x^2 dx \\plus{} \\int_{1}^{2}x^2 \\minus{} 1 dx \\equal{} .....$.\r\n\r\nBtw,for problem 1 use the substitution $ u \\equal{} \\frac {1}{x}$.", "Solution_4": "I think new_member forgot a factor $ 2$ in his solution of number 3: this comes from $ (e^{f(2t)})' \\equal{} e^{f(2t)} f'(2t) \\cdot 2$, that factor $ 2$ by the chain rule.\r\n\r\nI then get $ \\int_{0}^{2}(e^{f(2t)})'dt \\equal{} 10$ hence $ e^{f(4)} \\minus{} e^{f(0)} \\equal{} 10$.\r\n\r\nAs we have $ f(0) \\equal{} 0$, we get $ e^{f(0)} \\equal{} 1$ so $ e^{f(4)} \\equal{} 11$ which gives $ f(4) \\equal{} \\ln 11$.", "Solution_5": "for problem 1, i use u = 1/x, but i dont get what is the next step, i cant really use L'H rule, i have a ue^u\r\n\r\nfor problem 2 i get 2/3 when the answer is 2", "Solution_6": "Note that in problem 1, that is $ \\lim\\limits_{x \\to 0^ \\minus{} }$, i.e. from below!\r\n\r\nBy substituting $ u \\equal{} \\frac1x$, the limit then becomes $ \\lim\\limits_{u\\to \\minus{} \\infty} u e^u \\equal{} 0$ (or are you having problems with this last equality here?).\r\n\r\n\r\nFor problem 2, we get using new_member's partial solution that\r\n\r\n$ I \\equal{} \\int_{0}^{2}|x \\minus{} 1| \\,\\, |x \\plus{} 1|dx \\equal{} \\int_{0}^{2}|x^{2} \\minus{} 1|dx \\equal{} \\int_{0}^{1}(1 \\minus{} x^{2}) \\, dx \\plus{} \\int_{1}^{2}(x^{2} \\minus{} 1)\\, dx \\equal{} 1 \\minus{} \\frac13 \\plus{} \\frac13(2^3 \\minus{} 1^3) \\minus{} (2 \\minus{} 1) \\equal{} \\frac23 \\plus{} \\frac73 \\minus{} 1 \\equal{} 2.$", "Solution_7": "yeah i am having problems with that last equality\r\ni dont see why it equals 0\r\n\r\ni get something like -infinity*0 = 0?, is this even right or is this indeterminate?\r\n\r\n\r\nalso just because i am curious , is'nt 1^infinity = 1, because no matter how many times you multiply 1 with 1, you will always have a 1?" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Is there a simplex with integer edges and volume 1?", "Solution_1": "Basically I've got nothing on this problem, but since I find it funny, I'd like to revive it..\r\n\r\nSo, for $n = 1,$ needless to say it's obvious; for $n = 2,$ it's not too hard either, thanks to Hero's formula.\r\n\r\nBut what about $n \\geq 3?$ (where $n$ is the dimension of our Euclidean space)" } { "Tag": [ "calculus", "search", "geometry", "3D geometry", "algebra", "factorization", "sum of cubes" ], "Problem": "what is the sum of $ 1^k\\plus{}2^k\\plus{}3^k\\plus{} ... \\plus{}n^k$ for any natural number k and n?", "Solution_1": "http://mathworld.wolfram.com/BernoulliPolynomial.html", "Solution_2": "It has appeared many times on the forum before (notably in the high school forums, although within the past week in the Calculus forums) -- try a search, e.g. for \"sum of cubes\" or \"power sum\" or something similar and a few minutes work will lead to several threads on the subject." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Evaluate the sum: $ \\displaystyle\\sum_{n\\equal{}1}^{1994}(\\minus{}1)^n \\frac{n^2\\plus{}n\\plus{}1}{n!}.$", "Solution_1": "[quote=\"moldovan\"]Evaluate the sum: $ \\displaystyle\\sum_{n \\equal{} 3}^{1994}( \\minus{} 1)^n \\frac {n^2 \\plus{} n \\plus{} 1}{n!}.$[/quote]\r\n\r\nWe have $ \\frac {n^2 \\plus{} n \\plus{} 1}{n!}\\equal{}\\frac{n(n\\minus{}1)}{n!}\\plus{}\\frac{2n}{n!}\\plus{}\\frac{1}{n!}\\equal{}\\frac{1}{(n\\minus{}2)!}\\plus{}\\frac{2}{(n\\minus{}1)!}\\plus{}\\frac{1}{n!}$ for $ n \\ge 3$\r\n\r\nSo the sum $ \\equal{}\\minus{}\\frac{1^2\\plus{}1\\plus{}1}{1!}\\plus{}\\frac{2^2\\plus{}2\\plus{}1}{2!}\\plus{}\\displaystyle\\sum_{n \\equal{} 3}^{1994}(\\minus{}1)^n (\\frac{1}{(n\\minus{}2)!}\\plus{}\\frac{2}{(n\\minus{}1)!}\\plus{}\\frac{1}{n!})$\r\n$ \\equal{}\\frac{1}{2}\\plus{}(\\minus{}\\frac{1}{1!}\\plus{}\\frac{1}{2!}\\minus{}\\frac{1}{3!}\\plus{}...\\plus{}\\frac{1}{1992!})\\plus{}(\\minus{}\\frac{2}{2!}\\plus{}\\frac{2}{3!}\\minus{}\\frac{2}{4!}\\plus{}...\\plus{}\\frac{2}{1993!})\\plus{}(\\minus{}\\frac{1}{3!}\\plus{}\\frac{1}{4!}\\minus{}....\\minus{}\\frac{1}{1993!}\\plus{}\\frac{1}{1994!})$\r\n$ \\equal{}\\frac{1}{2}\\minus{}\\frac{1}{1!}\\minus{}\\frac{1}{2!}\\plus{}\\frac{1}{1993!}\\plus{}\\frac{1}{1994!}$\r\n$ \\equal{}\\minus{}1\\plus{}\\frac{1}{1993!}\\plus{}\\frac{1}{1994!}$", "Solution_2": "hello, you can also use this formula\r\n$ \\sum_{n\\equal{}1}^k(\\minus{}1)^n\\frac{n^2\\plus{}n\\plus{}1}{n!}\\equal{}\\frac{(\\minus{}1)^k\\plus{}2(\\minus{}1)^kk\\plus{}(\\minus{}1)^kk^2\\minus{}(k\\plus{}1)!}{(k\\plus{}1)!}$.\r\nSonnhard." } { "Tag": [ "videos", "blogs" ], "Problem": "Hi\r\nis there any one who has this book:\r\n[b]Functional Equations and How to Solve Them [/b]by [b]Christopher G. Small [/b]by [b]springer verlag [/b]\r\nIf you have please attach here. \r\nthanks\r\nBye.", "Solution_1": "Sorry, I know that this is dead, but I am trying to help out as much as possible.\n\n@glog, This probably isn't the best place to post these questions. You should probably post them in Questions, Suggestions, and Announcments.", "Solution_2": "Here: http://www.artofproblemsolving.com/Forum/viewforum.php?f=144\nP.S. how do you do the thing that It says click here, and here is the website???", "Solution_3": "You put [url=your url]text[/url]", "Solution_4": "So it would be [url=http://www.artofproblemsolving.com]AOPS[/url]", "Solution_5": "i tried it like 1,000,000 times im confused :?", "Solution_6": "[code]It is [url=WEBSITE HERE]WHAT YOU WANT IT TO SAY HERE[/url][/code]\n[url=http://www.artofproblemsolving.com]awesomeness[/url]", "Solution_7": "[url=http://data.artofproblemsolving.com]-___________-[/url]", "Solution_8": "[b]\u00a0\u00a0\u00a0[/b]", "Solution_9": "[url=http://www.wimp.com]Funny internet videos (not youtube)[/url]", "Solution_10": "Wait, are you linking to that Old Spice parody?", "Solution_11": "ya its that website and its the video of the day on [url=http://www.artofproblemsolving.com/Forum/blog.php?u=88632]My blog[/url]\nits there and there is a different video every day" } { "Tag": [ "limit", "integration", "function", "trigonometry", "calculus", "Support", "real analysis" ], "Problem": "Show that for every $\\varphi \\in C_{0}^{1}(\\mathbf R)$,\r\n\\[\\varphi(0) = \\lim_{n\\to\\infty}\\frac 1\\pi \\int_{-\\infty}^\\infty \\frac{\\sin(nx)}x \\varphi(x) dx.\\]", "Solution_1": "Define a function $H(x)=\\int_{0}^{x}\\frac{\\sin t}t\\,dt.$\r\n\r\n$H$ is clearly an odd function, and we'll assume that we already know this:\r\n\r\n$\\lim_{x\\to\\infty}H(x)=\\frac{\\pi}2.$\r\n\r\nNow note that $\\int \\frac{\\sin nx}x\\,dx=H(nx)+C.$ Split the integral into two pieces and integrate each by parts:\r\n\r\n$\\int_{0}^{\\infty}\\frac{\\sin nx}x\\,\\varphi(x)\\,dx= \\left.\\left(H(nx)-\\frac{\\pi}2\\right)\\varphi(x)\\right|_{0}^{\\infty}-\\int_{0}^{\\infty}\\left(H(nx)-\\frac{\\pi}2\\right)\\varphi'(x)\\,dx$\r\n\r\n$\\int_{-\\infty}^{0}\\frac{\\sin nx}x\\,\\varphi(x)\\,dx= \\left.\\left(H(nx)+\\frac{\\pi}2\\right)\\varphi(x)\\right|_{-\\infty}^{0}-\\int_{-\\infty}^{0}\\left(H(nx)+\\frac{\\pi}2\\right)\\varphi'(x)\\,dx$\r\n\r\nEvaluate the boundary terms and we get this:\r\n\r\n$\\frac1{\\pi}\\int_{-\\infty}^{\\infty}\\frac{\\sin nx}x\\,\\varphi(x)\\,dx= \\varphi(0)-\\frac1{\\pi}\\int_{-\\infty}^{\\infty}\\left(H(nx)-\\frac{\\pi}2\\,\\text{sgn}(x)\\right)\\,\\varphi'(x)\\,dx.$\r\n\r\nNow consider what happens to the last integral as $n\\to\\infty.$ Because of the compact support of $\\varphi,$ this may be taken to be the integral over a bounded interval. $H(nx)-\\frac{\\pi}2\\,\\text{sgn}(x)$ tends pointwise to zero, and is uniformly bounded. Hence (by a weak form of the Dominated Convergence Theorem), the integral tends to zero, proving our result.\r\n\r\nNote that this proof functions as a proof of the pointwise convergence of the Fourier integral representation of a sufficiently regular function." } { "Tag": [ "geometry", "algebra" ], "Problem": "Solve the equation\n\\[ \\sqrt {abx(x - a - b)} + \\sqrt {bcx(x - b - c)} + \\sqrt {cax(x - c - a)} = \\sqrt {abc(a + b + c)} \\]\n[hide=\"Solution\"]\\[x = - \\frac {abc}{ab+bc+ca \\pm 2\\sqrt{abc(a+b+c)}}\\][/hide]", "Solution_1": "What a nice example of how geometry can facilitate things which would be arkward with algebra or calculus! I've found it. :D\r\n[hide=\"hint: Herons formula\"] Putting $ y \\equal{} \\minus{} x$, we have to prove that\n\\[ \\sum\\sqrt {aby(a \\plus{} b \\plus{} y)} \\equal{} \\sqrt {abc(a \\plus{} b \\plus{} c)}\\]\nis solved by $ y \\equal{} \\frac {abc}{\\sum ab\\pm 2\\omega}$. By Heron's formula, the equation just sums up areas of triangles, where $ y$ is the radius of the small red inscribed circle. On the other hand, [url=http://en.wikipedia.org/wiki/Descartes'_theorem]Descartes' theorem[/url] tells us the solution(s) for $ y$. [hide]For the solution with the minus sign, corresponding to the circumscribed circle, some of the roots have to be taken negative of course.[/hide] [/hide] :)", "Solution_2": "Thank you spanferkel for your nice approach. Now, I present a detailed solution\n\n$\\alpha= \\sqrt{bcx(x-b-c)} \\ , \\ \\beta= \\sqrt{cax(x-c-a)}$\n\n$\\gamma=\\sqrt{abx(x-a-b)} \\ , \\ \\delta=\\sqrt{abc(a+b+c)}$\n\nBy Heron's formula, we figure out that the numbers $\\alpha,\\beta,\\gamma,\\delta$ represent twice the area of the triangles with side lengths $(b+c,x-b,x-c),$ $(c+a,x-c,x-a),$ $(a+b,x-a,x-b),$ $(b+c,c+a,a+b),$ respectively. Now, we consider the triangle $\\triangle ABC$ with side lengths $(b+c,c+a,a+b).$ From $\\alpha+\\beta+\\gamma=\\delta,$ we deduce that there exists a point $P,$ such that $PA=x-a,$ $PB=x-b,$ $PC=x-c.$ Using the Briggs formulas, we get\n\n$\\sin \\frac{_1}{^2}\\widehat{CAP}=\\sqrt{\\frac{a(x-a-c)}{(x-a)(c+a)}} \\ , \\ \\cos \\frac{_1}{^2}\\widehat{CAP}=\\sqrt{\\frac{cx}{(x-a)(c+a)}}$\n\n$\\sin \\frac{_1}{^2}\\widehat{BAP}=\\sqrt{\\frac{a(x-a-b)}{(x-a)(a+b)}} \\ , \\ \\cos \\frac{_1}{^2}\\widehat{BAP}=\\sqrt{\\frac{bx}{(x-a)(a+b)}}$\n\n$\\sin \\frac{_1}{^2}\\widehat{CAB}=\\sqrt{ \\frac{bc}{(a+b)(a+c)}}$\n\nSubstituting these values into the trigonometric identity\n\n$\\sin \\frac{_1}{^2}\\widehat{CAB}=\\sin \\frac{_1}{^2}\\widehat{CAP} \\cdot \\cos \\frac{_1}{^2}\\widehat{BAP}+\\cos \\frac{_1}{^2}\\widehat{CAP} \\cdot \\sin \\frac{_1}{^2}\\widehat{BAP}$\n\nyields $bc(x-a)=b\\beta+c\\gamma.$ By similar reasoning, we get the cyclic expressions $ca(x-b)=c\\gamma+a \\alpha$ and $ab(x-c)=a \\alpha+b\\beta.$ From these three we obtain\n\n$x(bc+ca+ab)-abc=2a \\alpha=2a \\sqrt{bcx(x-b-c)}$\n\n$[(bc+ca+ab)^2-4bca^2]x^2+2abc(bc+ca+ab)x+a^2b^2c^2=0$\n\n$[x(bc+ca+ab)+abc]^2=4\\delta^2x^2 \\ \\Longrightarrow$ \n\n$ x =-\\frac{abc}{ab+bc+ca\\pm 2\\delta}$" } { "Tag": [ "inequalities", "rearrangement inequality", "inequalities proposed" ], "Problem": "If $ a,b,c,d$ are non-negative real numbers such that $ a + b + c + d = 4$\r\n\r\n$ a\\sqrt {bc} + b\\sqrt {cd} + c\\sqrt {da} + d\\sqrt {ab} \\leq 2 + 2\\sqrt {abcd}$\r\n\r\n\r\nP.S.\r\n\r\nI tried to solve it this way, but I failed \r\n\r\nLet $ a=x^2$,$ b=y^2$,$ c=z^2$,$ d=t^2$ ; now our problem is: if $ x^2+y^2+z^2+t^2=4$ then\r\n\r\n$ x^2yz+y^2zt+z^2tx+t^2xy \\leq 2+2xyzt$\r\n\r\nLet $ (X,Y,Z,T)$ a rearrangement of $ (x,y,z,t)$ such that $ X\\geq Y \\geq Z \\geq T$\r\n\r\nThen by rearrangement inequality we have\r\n\r\n$ x^2yz+y^2zt+z^2tx+t^2xy = x(xyz) + y(yzt) + z(ztx) + t(txy) \\leq X(XYZ) + Y(XYT) + Z(XZT) + T(TYZ) \n= X^2YZ + Y^2XT + Z^2TX + T^2YZ$ as $ XYZ \\geq XYT \\geq XZT \\geq TYZ$\r\n\r\nso it remains to prove that\r\n\r\n$ X^2YZ + Y^2ZT + Z^2TX + T^2XY \\leq 2 + 2XYZT$ with $ X\\geq Y \\geq Z \\geq T$ and $ X^2+Y^2+Z^2+T^2 = 4$\r\n\r\nI tried to prove this using the fact that\r\n\r\n$ T^2YZ \\leq XYZT=xyzt$ and $ Z^2TX \\leq XYZT=xyzt$ but unfortunately it doesn't work. :( \r\n\r\nI suspect Hungkhtn uses a very nice trick.", "Solution_1": "[quote=\"manlio\"]If $ a,b,c,d$ are non-negative real numbers such that $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 4$\n\n$ a\\sqrt {bc} \\plus{} b\\sqrt {cd} \\plus{} c\\sqrt {da} \\plus{} d\\sqrt {ab} \\leq 2 \\plus{} 2\\sqrt {abcd}$\n\n\nP.S.\n\nI tried to solve it this way, but I failed \n\nLet $ a \\equal{} x^2$,$ b \\equal{} y^2$,$ c \\equal{} z^2$,$ d \\equal{} t^2$ ; now our problem is: if $ x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} t^2 \\equal{} 4$ then\n\n$ x^2yz \\plus{} y^2zt \\plus{} z^2tx \\plus{} t^2xy \\leq 2 \\plus{} 2xyzt$\n\nLet $ (X,Y,Z,T)$ a rearrangement of $ (x,y,z,t)$ such that $ X\\geq Y \\geq Z \\geq T$\n\nThen by rearrangement inequality we have\n\n$ x^2yz \\plus{} y^2zt \\plus{} z^2tx \\plus{} t^2xy \\equal{} x(xyz) \\plus{} y(yzt) \\plus{} z(ztx) \\plus{} t(txy) \\leq X(XYZ) \\plus{} Y(XYT) \\plus{} Z(XZT) \\plus{} T(TYZ) \\equal{} X^2YZ \\plus{} Y^2XT \\plus{} Z^2TX \\plus{} T^2YZ$ as $ XYZ \\geq XYT \\geq XZT \\geq TYZ$\n\nso it remains to prove that\n\n$ X^2YZ \\plus{} Y^2ZT \\plus{} Z^2TX \\plus{} T^2XY \\leq 2 \\plus{} 2XYZT$ with $ X\\geq Y \\geq Z \\geq T$ and $ X^2 \\plus{} Y^2 \\plus{} Z^2 \\plus{} T^2 \\equal{} 4$\n\nI tried to prove this using the fact that\n\n$ T^2YZ \\leq XYZT \\equal{} xyzt$ and $ Z^2TX \\leq XYZT \\equal{} xyzt$ but unfortunately it doesn't work. :( \n\nI suspect Hungkhtn uses a very nice trick.[/quote]\r\nWe have \r\n\\[ X^2YZ \\plus{} Y^2XT \\plus{} Z^2TX \\plus{} T^2YZ \\equal{}YZ(X^2\\plus{}T^2)\\plus{}XT(Y^2\\plus{}Z^2)\\minus{}2XYZT\\plus{}2XYZT\r\n\\equal{} YZ(X\\minus{}T)^2\\plus{} XT(Y^2\\plus{}Z^2)\\plus{}2XYZT \\le \\frac{(Y^2\\plus{}Z^2)}{2}(X\\minus{}T)^2\\plus{}XT(Y^2\\plus{}Z^2) \\plus{}2XYZT\\equal{}\\frac{(X^2\\plus{}T^2)(Y^2\\plus{}Z^2)}{2}\\plus{}2XYZT \\le \\frac{(X^2\\plus{}Y^2\\plus{}Z^2\\plus{}T^2)^2}{8}\\plus{}2XYZT\\equal{}2\\plus{}2XYZT\\]\r\n:)", "Solution_2": "[i] You are really intelligent C\u1ea9n\n \n\n Very nice solution\n \n\n Congratulation[/i] :lol:", "Solution_3": "You are really great, can_hang2007!!! :)" } { "Tag": [ "search", "geometry", "projective geometry", "geometry theorems" ], "Problem": "Can somebody (maybe Darij :D ) explain for me about Polar and Polar reciprocal, polar line of a point respect to a circle, and all it stuff. I'm not sure about \"polar and polar reciprocal\" that I use, I'll explaine more detail about it.\r\nGive a circle (O) and a point M in plan. If the circle with diameter MN is [u]|[/u] with (O) < angle of circle (MN) and circle (O) is 90 > then N is on polar line (d) of M respected circle (O). And M called polar of the line (d) respect circle (O).\r\nSorry coz my English is so bad to explain my idea, but if you can understand...", "Solution_1": "There are some good sources on poles and polars on the internet.\r\n\r\nFirst, you should check out http://www.cut-the-knot.org/Curriculum/Geometry/PolePolar.shtml for main properties and some applications of poles and polars.\r\n\r\nThen, there are the topics http://www.mathlinks.ro/Forum/viewtopic.php?t=31812 and http://www.mathlinks.ro/Forum/viewtopic.php?t=5501 on ML.\r\n\r\nThen, you can search for \"polar\" in the geometry section of ML in order to find lots of applications of poles and polars in solving geometry problems.\r\n\r\n Darij" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For the non-negative numbers $ a,b,c$ prove the inequality\r\n$ 4(a\\plus{}b\\plus{}c) \\geq 3(a\\plus{}\\sqrt{ab}\\plus{}\\sqrt[3]{abc})$\r\nWhen does equality hold?", "Solution_1": "Becouse of $ AM\\geq GM$ inequality we have:\r\n\r\n$ \\frac {3a}{4} \\plus{} 3b\\geq 3\\sqrt {ab}$ and also: $ \\frac {a}{4} \\plus{} b \\plus{} 4c\\geq3\\sqrt [3]{abc}$ \r\n\r\nso after summing these two inequalities with $ 3a \\equal{} 3a$ we get desired inequality.\r\n\r\nEquality is achieved when $ \\frac {a}{4} \\equal{} b \\equal{} 4c$ so for any triple $ (t,4t,16t)$\r\n\r\n(which is in fact equivalnet to $ a$, $ b$ and $ c$ being (in this order) in geometric progreession in which $ q \\equal{} 4$)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a, b, c>0$ such that $ab+bc+ca=3abc.$ Prove that the following inequality holds:\r\n\r\n \\[ \\frac{1}{1+\\sqrt{ab}}+\\frac{1}{1+\\sqrt{bc}}+\\frac{1}{1+\\sqrt{ca}}\\leq\\frac{3}{2}. \\]", "Solution_1": "Very nice Cezar!! :) \r\n\r\nIs the trick to write\r\n$\\sum\\frac{\\sqrt{ab}}{1+\\sqrt{ab}}\\geq\\frac{3}{2}$??", "Solution_2": "Yes, it is. :) Keep going.... ;)", "Solution_3": "[quote=\"cezar lupu\"]Yes, it is. :) Keep going.... ;)[/quote]\r\n\r\nThen we write \r\n\r\n$\\sum\\frac{1}{1+\\frac{1}{\\sqrt{ab}}}$ \r\n\r\nsi applicam ineqalitatea din Cauchy....\r\n\r\nDo you catch me?? :)", "Solution_4": "Hint:\r\n [hide]\n Yes, Cauchy and AM-GM. Try to post a full solution. ;) \n [/hide]", "Solution_5": "[hide]$\\sum\\frac{1}{1+\\frac{1}{\\sqrt{ab}}}\\geq\\frac{9}{3+\\sum\\frac{1}{\\sqrt{ab}}}\\geq\\frac{3}{2}$\n\nsince trivial ineq yields \n\n$\\sum\\frac{1}{a}\\geq\\sum\\frac{1}{\\sqrt{ab}}$ (and $\\sum\\frac{1}{a}$=3 which follows from the given relation )[/hide]\r\n\r\n\r\nWhy do you care about full solutions?? :D", "Solution_6": "Actually, I don't care. I have found a solution too. In fact, my solution is almost the same with yours.:P", "Solution_7": "Is it the same as mine??" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "we know,if a prime p=4k+1,there exit integers a,b so that p=sqr(a)+sqr(b).but can we define there is only one solution of (a,b)???", "Solution_1": "Yes, up to sign and permutation. Otherwise $ a \\plus{} bi$ would be composite in the Gaussian integers.", "Solution_2": "thanks,but can anyone explain it clearly?", "Solution_3": "Here's an elementary proof. Suppose $ a, b, c, d$ are positive integers with $ \\{ a, b \\} \\neq \\{ c, d \\}$ such that $ p \\equal{} a^2 \\plus{} b^2 \\equal{} c^2 \\plus{} d^2$. Then\r\n\r\n$ a^2 \\minus{} c^2 \\equal{} d^2 \\minus{} b^2 \\Leftrightarrow (a \\minus{} c)(a \\plus{} c) \\equal{} (d \\minus{} b)(d \\plus{} b)$.\r\n\r\nThis implies (and this is a nice exercise) that there exist positive integers $ x, y, z, w$ such that\r\n\r\n$ a \\minus{} c \\equal{} xy, a \\plus{} c \\equal{} zw, d \\minus{} b \\equal{} xw, d \\plus{} b \\equal{} yz$\r\n\r\nhence\r\n\r\n$ a \\equal{} \\frac {zw \\plus{} xy}{2}, c \\equal{} \\frac {zw \\minus{} xy}{2}, d \\equal{} \\frac {yz \\plus{} xw}{2}, b \\equal{} \\frac {yz \\minus{} xw}{2}$.\r\n\r\nWe then compute that \r\n\r\n$ a^2 \\plus{} b^2 \\equal{} c^2 \\plus{} d^2 \\equal{} \\frac {(zw)^2 \\plus{} (xy)^2 \\plus{} (yz)^2 \\plus{} (xw)^2}{4} \\equal{} \\frac {(x^2 \\plus{} z^2)(y^2 \\plus{} w^2)}{4}$\r\n\r\nwhich contradicts $ p$ prime.", "Solution_4": "thanks very much" } { "Tag": [ "algorithm", "linear algebra", "matrix", "search" ], "Problem": "Hi every1,\r\n\r\nCan we devise an algorithm using dynamic programming which can find nth fibonacci number in O(log n)..\r\n\r\nThis is given as a problem in this pdf file---\r\n\r\n activities.tjhsst.edu/sct/lectures/dp1.pdf", "Solution_1": "yes it is possible. the idea is matrix power. u know if a matrix is: A then A^n can be computed y some short procedure, may be nlgn. u have to use that idea, and surely u have to make a useful matrix A.", "Solution_2": "[quote=\"mahbub\"]yes it is possible. the idea is matrix power. u know if a matrix is: A then A^n can be computed y some short procedure, may be nlgn. u have to use that idea, and surely u have to make a useful matrix A.[/quote]\r\n\r\nWell, finding A^n requires at least O(A) time because it is somewhat necesarry to create the matrix of size A, which is itself O(A). [O(A) refers to O(size of A)].\r\n\r\nA cheap shot would be this: solve the linear recurrence equation then use numeric methods. That is certainly log(n).\r\n\r\nEDIT: Nvm, maybe I should've read the source first. O(A) would be O(4) lol. So idea is just DP, calculate A^1, A^2, A^4, A^8, etc., that's log(n), then you pick the necesarry numbers to get to A^n. (Like for n = 13 you'd pick 1, 4, 8).", "Solution_3": "[ 1 1 ] ^ n [ F(n+1) F(n) ]\r\n [ 1 0 ] = [ F(n) F(n-1) ]", "Solution_4": "your search is over\r\nthe method you seek for is here\r\nhttp://www.cs.utexas.edu/users/EWD/ewd06xx/EWD654.PDF" } { "Tag": [ "limit", "trigonometry", "algebra", "polynomial", "calculus", "calculus computations" ], "Problem": "In a math test, because of the lack of time and I'm not lucid at that time , I use this approximation when $ x\\rightarrow 0$ in this limit finding problem :\r\n$ \\lim_{x\\rightarrow 0}\\frac {\\sqrt [3]{cosx} \\minus{} \\sqrt {cosx}}{{x}^{2}} \\equal{} \\lim_{x\\rightarrow 0}\\frac {\\sqrt [3]{1 \\minus{} \\frac {{x}^{2}}{2}} \\minus{} \\sqrt [2]{1 \\minus{} \\frac {{x}^{2}}{2}}}{{x}^{2}} \\equal{} \\lim_{x\\rightarrow 0}\\frac {(1 \\minus{} \\frac {{x}^{2}}{2}*\\frac {1}{3}) \\minus{} (1 \\minus{} \\frac {{x}^{2}}{2}*\\frac {1}{2})}{{x}^{2}} \\equal{} \\frac {1}{12}$\r\nAnd this way gives the right result but the TA [b]refuse to give me any score[/b] for this problem despite all of my best efforts :( . She said \"we couldn't approximate $ cosx \\equal{} 1 \\minus{} \\frac {{x}^{2}}{2}$ when we find the limit of a sum , we just can use this in a product limit . I had knew that there're some cases that we couldn't approximate like that , but I also think that there're many cases that we can approximate like that way when compute the limit ,with $ x\\rightarrow 0$ .\r\nI really need of a good explanation for this solution :why it should give the exact result, a little bit why and when I could approximate like that .. in order to get some marks on this problem .\r\nA generalisation for this problem is : if $ x\\rightarrow 0$ => $ f(x)\\approx u(x)$ and $ g(x)\\approx v(x)$ , when we could approximate like this :\r\n$ \\lim_{x\\rightarrow 0}\\frac {f(x) \\plus{} g(x)}{h(x)} \\equal{} \\lim_{x\\rightarrow 0}\\frac {u(x) \\plus{} v(x)}{h(x)} ???$\r\nI have found some other solutions after the test but it's too late to submit it . Hope you can help me an explanation for this solution . Thanks a lot .", "Solution_1": "This approach is correct in the sense that it gives the correct coefficients of the Taylor expansion of the numerator, but ideally you should've used l'Hopital's rule to rigorously justify that you 1) had the right coefficients, and 2) had enough of them.", "Solution_2": "Yah, when I go home, I think I should have used L'Hopital's rule or using make the \"1\" appear like : $ ( \\sqrt [3]{cosx} \\minus{} 1 ) \\minus{} ( \\sqrt {cosx} \\minus{} 1 )$but it's too late :( .\r\nHey your suggestion about Taylor expansion is great . I will use that to explain . Thanks .", "Solution_3": "The equals sign in your work would bother me because it's not literally true that $ \\cos x\\equal{}1\\minus{}\\frac{x^2}2.$ You have to deal with the extent to which that is in error.\r\n\r\nI might try justifying it by writing my Taylor polynomials in big-O style:\r\n\r\nThe binomial series: $ (1\\plus{}u)^{\\alpha}\\equal{}1\\plus{}\\alpha u\\plus{}O(u^2)$\r\n\r\nThe cosine series: $ \\cos x\\equal{}1\\minus{}\\frac{x^2}2\\plus{}O(x^4).$\r\n\r\nHence, $ \\sqrt[3]{\\cos x}\\equal{}1\\minus{}\\frac{x^2}{6}\\plus{}O(x^4)$ and $ \\sqrt{\\cos x}\\equal{}1\\minus{}\\frac{x^2}4\\plus{}O(x^4).$\r\n\r\nHence we have\r\n\r\n$ \\lim_{x\\to 0}\\frac{\\left(1\\minus{}\\frac{x^2}{6}\\plus{}O(x^4)\\right)\\minus{}\\left(1\\minus{}\\frac{x^2}4\\plus{}O(x^4)\\right)}{x^2}\\equal{}\\lim_{x\\to0}\\frac{\\frac{x^2}{12}\\plus{}O(x^4)}{x^2}$\r\n\r\n$ \\equal{} \\lim_{x\\to0}\\frac1{12}\\plus{}O(x^2)\\equal{}\\frac1{12}.$\r\n\r\nOne measure of the limitations of your method: what if you had been asked this?\r\n\\[ \\lim_{x\\to0}\\frac{12\\sqrt[3]{\\cos x}\\minus{}12\\sqrt{\\cos x}\\minus{}x^2}{x^4}\\]\r\n\r\nOf course, my work above does not solve this, but it tells me what I need to do - to use additional terms of the Taylor polynomials to replace $ O(x^4)$ with more exact details.\r\n\r\n(I would have given you substantial partial credit - I would have made some deduction for the sloppy notation, but acknowledged the idea.)", "Solution_4": "Yah, so all the reasons behind these approximations is the Taylor polynomials :lol: . Thanks ." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $ B(N)$ be the number of 1s in the binary expression of the integer $ N$. Find $ \\displaystyle\\sum_{n\\equal{}1}^{\\infty}\\frac{B(n)}{n(n\\plus{}1)}$.", "Solution_1": "I might have made some error, but at least this will get you started if you solved it yourself...\r\n\r\nI'm using B to represent B(n)...\r\n\r\nB/(n(n+1)) = B/(n+1) - B/n\r\n\r\nnow simply plug values for upto n=13 or so and you'll see a trend in the series. Add every term in B/(n+1) to the same denominator term in B/n (which is negative) and the result will have numerator 1, except for every 4th term....i.e it will look something like 0 + 1/3 +1/4+1/5+0+1/7+1/8+1/9+0+1/11.......\r\n\r\nand thus you can express the sum as \r\n\r\nSUM(n=1 to inf) [(1/(4n-2)) - (1/n)]\r\n\r\nNow, I'm not quite sure if you can simplify that to a better expression (there must be a way, but I can't remember it). But at least that takes care of the main problem.\r\n\r\nPlease feel free to comment/correct/provide a better solution.\r\n\r\n(Sorry for the sloppy notation, but I'm not sure how to represent math expressions properly in this forum.)" } { "Tag": [ "parameterization", "algebra", "function", "domain", "calculus", "derivative", "logarithms" ], "Problem": "$2f(x)=f(\\frac{x}{y})+f(xy)$ $f(1)=0$ and $f'(1)=1$, $Find$ $f(x)$", "Solution_1": "What does $y$ equal in this problem?\r\n\r\nis $y=f(x)$?", "Solution_2": "$y$ is an arbitrary parameter- presumably, this is intended to be true for all values of $x$ and $y$.\r\n\r\nThe domain of $f$ should also be specified- I would guess that it's the positive real numbers.", "Solution_3": "I'm working on the assumption that $f$ is to be a differentiable function over ${\\bf R}_{> 0}$, and the functional equation is meant to hold for all $x, y \\in{\\bf R}_{>0}$. [hide]Then make the substitution $x \\mapsto xy$ and $y \\mapsto \\frac{x}{y}$ to get $2f(xy) = f(x^{2})+f(y^{2})$ for all positive reals $x, y$. Since this is a functional identity in $x$, we can take a derivative on both sides and get $2yf'(xy) = 2xf'(x^{2})$ for all real $x, y$. Then setting $x \\mapsto1$ gives $2yf'(y) = 2f'(1) = 2$, so $f'(y) = \\frac{1}{y}$ for all $y$. It follows immediately that $f(x) = \\log x$ is the only solution.[/hide]", "Solution_4": "[quote=\"JBL\"]I'm working on the assumption that $f$ is to be a differentiable function over ${\\bf R}_{> 0}$, and the functional equation is meant to hold for all $x, y \\in{\\bf R}_{>0}$. [hide]Then make the substitution $x \\mapsto xy$ and $y \\mapsto \\frac{x}{y}$ to get $2f(xy) = f(x^{2})+f(y^{2})$ for all positive reals $x, y$. Since this is a functional identity in $x$, we can take a derivative on both sides and get $2yf'(xy) = 2xf'(x^{2})$ for all real $x, y$. Then setting $x \\mapsto1$ gives $2yf'(y) = 2f'(1) = 2$, so $f'(y) = \\frac{1}{y}$ for all $y$. It follows immediately that $f(x) = \\log x$ is the only solution.[/hide][/quote]\r\n \r\nMY dear JBL, I'm sorry ur soln is wrong .you are not given that f is diff over R or R>0", "Solution_5": "If he's wrong, it's only because he couldn't tell what the exact question you wanted to ask is. Please clarify what the domain of $f$ is, and [b]all[/b] assumptions about it.", "Solution_6": "[quote=\"jmerry\"]If he's wrong, it's only because he couldn't tell what the exact question you wanted to ask is. Please clarify what the domain of $f$ is, and [b]all[/b] assumptions about it[/quote] \r\n I think , you dont need the domain of $f$ and you are not provided with the differentiablity of $f$ on whole of its domain." } { "Tag": [ "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "n signals are equally spaced along a rail track. No train is allowed to leave a signal whilst there is a moving train between that signal and the next. Any number of trains can wait at a signal. At time 0, k trains are waiting at the first signal. Except when waiting at a signal each train travels at a constant speed, but each train has a different speed. Show that the last train reaches signal n at the same time irrespective of the order in which the trains are arranged at the first signal.", "Solution_1": "Can you say me if I understood the problem?\r\n\r\nAccording to my interpretation, there are $ k$ trains in a rail track, and all they are in the first sign. The size of the trains is $ 0$. The trains are in a certain order. Each train moves if and only if it is authorized, i.e., if it is the first train waiting at the sign (the order of the trains is always the initial order) and there is no other train between the sign in which the train is and the next one. We need to prove the time to the last train reaches the last sign does not depend on the initial order.\r\n\r\nI am sorry if this sounds stupid, but I don't think I am allowed to use intuition in a mathematical problem... Specially because the trains have size $ 0$!", "Solution_2": "I think I have a solution (according to my interpretation).\n\nLet the trains be $T_1,T_2,\\ldots$ and $T_k$. Let $t_i$ be the time the $i$th train takes to go from one signal to the next, with $t_1\\leq t_2\\leq\\ldots\\leq t_k$. Consider the case the foremost train is $T_1$, the second one is $T_2$, etc. It is easy to see the trains move in an essentially independent way, and that the total time is $t_1+t_2+\\ldots+t_{k-1}+(n-1)t_k$ (observe that there are $n-1$ \"rail pieces\" between the signals).\n\nThus, the slowest train is of particular interest. Now consider any order of the trains, and define $S=\\{s\\text{ such that }T_s\\text{ is ahead of }T_k\\}$, so that the trains indexed by $S$ leave before $T_k$. It is easy to see the total time spent before the appearence of $T_k$ equals $\\sum_{s\\in S}t_s$. After that, the trains $T_s, s\\in S,$ don't affect $T_k$ anymore.$(*)$\n\nThe time until the arrival of $T_k$ at the $n$th signal is therefore $\\sum_{s\\in S}t_s+(n-1)t_k$. If, at this moment, the remaining trains were in the $(n-1)$th signal, it is clear the total time needed would be the same. This is not always the case, but we can show there is no wasted time, i.e., after the end of $T_k$'s movement, each time a train arrives at the $n$th sign, there is another train waiting for it in the $(n-1)$th sign.\n\nThis is guaranteed with the induction hypothesis (we use induction on $n+k$). In fact, for any train $T_i$ after $T_k$, we can assume the total time for $T_i$ to go to signal $n-1$ is a sum of $t_j$'s, where $T_j$ is before $T_i$, $j\\neq k$, and a particular term $(n-2)t_k$. This time is smaller than the time for the train before $T_i$ to go to signal $n$, which is found by the same formula (the time difference is $t_k-t_i$).\n\n$(*)$ In reviewing, I realized this is probably not so obvious. But we can use the induction hypothesis to see the ordering of the trains indexed by $S$ doesn't matter. And if the trains are ordered by speed, they all move independently." } { "Tag": [], "Problem": "1. Find x^2 + y^2 if x, y are natural numbers and xy+x+y=71, x^2y+xy^2=880.\r\n\r\n2. If x^2+y^2+z^2=49 and x+y+z=x^3+y^3+z^3=7, find xyz.\r\n\r\n3. Determine the triples of integers x,y,z satisfying:\r\nx^3+y^3+z^3=(x+y+z)^3", "Solution_1": "[hide=\"2\"]\n$ (x\\plus{}y\\plus{}z)^2\\equal{}x^2\\plus{}y^2\\plus{}z^2\\plus{}2(xy\\plus{}yz\\plus{}zx)\\Longrightarrow xy\\plus{}yz\\plus{}zx\\equal{}0$\nand by the identity:\n$ x^3\\plus{}y^3\\plus{}z^3\\minus{}3xyz\\equal{}(x\\plus{}y\\plus{}z)(x^2\\plus{}y^2\\plus{}z^2\\minus{}xy\\minus{}yz\\minus{}zx)$\n$ 7\\minus{}3xyz\\equal{}7\\times 49$\nhence we have $ xyz\\equal{}\\minus{}112$.\n[/hide]", "Solution_2": "[quote=\"therobz\"]1. Find$ x^2 \\plus{} y^2$if x, y are natural numbers and $ xy\\plus{}x\\plus{}y\\equal{}71, x^2y\\plus{}xy^2\\equal{}880$.\n\n2. If $ x^2\\plus{}y^2\\plus{}z^2\\equal{}49$ and$ x\\plus{}y\\plus{}z\\equal{}x^3\\plus{}y^3\\plus{}z^3\\equal{}7$, find$ xyz$.\n\n3. Determine the triples of integers $ x,y,z$ satisfying:\n$ x^3\\plus{}y^3\\plus{}z^3\\equal{}(x\\plus{}y\\plus{}z)^3$[/quote]\r\nI think it very easy :oops:", "Solution_3": "[quote=\"quykhtn-qa1\"][quote=\"therobz\"]1. Find$ x^2 \\plus{} y^2$if x, y are natural numbers and $ xy \\plus{} x \\plus{} y \\equal{} 71, x^2y \\plus{} xy^2 \\equal{} 880$.\n\n2. If $ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 49$ and$ x \\plus{} y \\plus{} z \\equal{} x^3 \\plus{} y^3 \\plus{} z^3 \\equal{} 7$, find$ xyz$.\n\n3. Determine the triples of integers $ x,y,z$ satisfying:\n$ x^3 \\plus{} y^3 \\plus{} z^3 \\equal{} (x \\plus{} y \\plus{} z)^3$[/quote]\nI think it very easy :oops:[/quote]\r\n\r\nEr, could you post your solution? :P", "Solution_4": "1.\r\n\r\nThe first equation gives xy = 880/(x+y). Substituting this into the second equation gives 880/(x+y) + (x+y) = 71 which implies x+y = 16 or 55. Then xy = 55 or 16, respectively abd so x^2 + y^2 = (x+y)^2 - 2xy = 146 or 2993.", "Solution_5": "[quote=\"therobz\"]1. Find x^2 + y^2 if x, y are natural numbers and xy+x+y=71, x^2y+xy^2=880.\n\n2. If x^2+y^2+z^2=49 and x+y+z=x^3+y^3+z^3=7, find xyz.\n\n3. Determine the triples of integers x,y,z satisfying:\nx^3+y^3+z^3=(x+y+z)^3[/quote]\r\n3 is very easy$ x^3\\plus{}y^3\\plus{}z^3\\equal{}(x\\plus{}y\\plus{}z)^3$\r\n$ \\Leftrightarrow (x \\plus{} y)(y \\plus{} z)(x \\plus{} z) \\equal{} 0$ :)", "Solution_6": "I had (x+y+z)(xy+yz+zx) = xyz, which is equivalent to TRAN THAI HUNG's condition. AM-GM shows that all three of x, y and z cannot be positive. I also had (0, 0, 0), (n, 0, 0), (n, -n, 0), (m, n, -n) and permutations.", "Solution_7": "This gives 49 >= 49^2 so x, y and z cannot all be positive.", "Solution_8": "exo2. $ xy\\plus{}yz\\plus{}zx\\equal{}\\frac{1}{2}((x\\plus{}y\\plus{}z)^2\\minus{}x^2\\plus{}y^2\\plus{}z^2)\\equal{}0$\r\n$ x^3\\plus{}y^3\\plus{}z^3\\minus{}3xyz\\equal{}(x\\plus{}y\\plus{}z)(x^2\\plus{}y^2\\plus{}z^2\\minus{}xy\\minus{}yz\\minus{}zx)\\equal{}343$" } { "Tag": [ "modular arithmetic" ], "Problem": "What's the units digit of a^5 - a, where a is a positive integer?", "Solution_1": "[hide]$a^5$ always has a units digit of the units digit of a, so $a^5-a$ has a units digit of 0.[/hide]", "Solution_2": "[hide]\nThe answer is $0$, which means $a^5-a$ is always a multiple of 10.\n\nProve by induction.\n\n(I) For $a=1$, $1^5-1=0$, so the statement is true.\n\n(II) Assume true for $a=n$. Then $n^5-n=10k$ for some positive integer $k$.\n\n(III)For $a=n+1$,\n\n$(n+1)^5-(n+1)$\n$=n^5+5n^4+10n^3+10n^2+5n+1-n-1$\n$=(n^5-n)+5(n^4+n)+10(n^3+n^2)$\n$=10k+5n(n^3+1)+10(n^3+n^2)$\n$=10k+5n(n+1)(n^2-n+1)+10(n^3+n^2)$\n\nNote that $n(n+1)$ is always even because either $n$ or $n+1$ is even. So, we can let $5n(n+1)(n^2-n+1)=5(2m)(n^2-n+1)$ for some positive integer $m$.\n\nThen, the expression becomes\n$10k+10m(n^2-n+1)+10(n^3+n^2)$\n\nTherefore, the statement is true for $a=n+1$.\n\n\nFrom (I), (II), (III), $a^5-a$ is a multiple of 10 for any positive integer $a$.\n[/hide]", "Solution_3": "Another way to prove this is by casework:\r\n\r\n[hide]Consider two cases:\n\nCase 1.) If $a$ is even:\n\nNotice $a^5 - a$ factors into $a(a^4-1)$, so it is sufficient to show that for all $a$, $5 | a^4 - 1$.\n\nIf $a \\equiv 0 \\pmod{5}$, we are done. Since $a$ is even it must be divisible by 10 also.\nIf $a \\equiv 1 \\pmod{5}$, $a^4 \\equiv 1 \\pmod{5}$.\nIf $a \\equiv 2 \\pmod{5}$, $a^4 \\equiv 16 \\pmod{5} \\equiv 1 \\pmod{5}$.\n...\nAnd so on for $a$ equivalent 3 and 4 modulo 5. We find that $a^4 \\equiv 1 \\pmod{5}$ for all $a$, and $a^4 - 1 \\equiv 0 \\pmod{5}$.\n\nCase 2.) If $a$ is odd:\n\n$a(a^4 - 1) = a(a^2-1)(a^2+1)$\n\nWe want to show that either $a^2-1$ or $a^2+1$ is equivalent to $0\\pmod{10}$ for all $a$. Notice $a$ can only be equivalent to 1, 3, 5, 7, or 9 modulo 10.\n\nIf $a \\equiv 1\\pmod{10}$, $a^2-1 \\equiv 0\\pmod{10}$, so we're done.\nIf $a \\equiv 3\\pmod{10}$, $a^2+1 \\equiv 0\\pmod{10}$.\nIf $a \\equiv 5\\pmod{10}$, we are done because both $a^2-1$ and $a^2+1$ are even.\nIf $a \\equiv 7\\pmod{10}$, $a^2 + 1 \\equiv 0\\pmod{10}$.\nIf $ a\\equiv 9\\pmod{10}$, $a^2 - 1 \\equiv 0\\pmod{10}$.\n\nSo we are done with casework.[/hide]", "Solution_4": "I really appreciate your solutions :)\r\nAlthough my pick goes for frt's one :P", "Solution_5": "Thanks! :)" } { "Tag": [ "modular arithmetic" ], "Problem": "Determine all positive integers $ n$ such that it is possible to tile a $ 15 \\times n$ board with pieces shaped like this:\r\n\r\n[asy]size(100); draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(1,2)); draw((2,2)--(3,2)); draw((0,0)--(0,2)); draw((1,0)--(1,2)); draw((2,0)--(2,2)); draw((3,0)--(3,2)); draw((5,0)--(6,0)); draw((4,1)--(7,1)); draw((4,2)--(7,2)); draw((5,3)--(6,3)); draw((4,1)--(4,2)); draw((5,0)--(5,3)); draw((6,0)--(6,3)); draw((7,1)--(7,2));[/asy]", "Solution_1": "Clearly we can cover every $ 5 \\times 3$ or $ 3 \\times 5$. So we can also cover a $ 15 \\times n$ with:\r\n$ n \\equiv 0 \\pmod 3$ (all $ 3 \\times 5$ pieces), \r\n$ n \\equiv 2 \\pmod 3$ and $ n > 2$ (a column of $ 5 \\times 3$ and all the others $ 3 \\times 5$),\r\n$ n \\equiv 1 \\pmod 3$ and $ n > 7$ (two columns of $ 5 \\times 3$ and all the others $ 3 \\times 5$).\r\nThen it's easy to see we cannot cover the board if $ n = 1, 2, 4, 7$.", "Solution_2": "But how can we prove it is impossible to tile without purely 3 by 5s?", "Solution_3": "See $n=1, 2, 4, 7$ are not achievable. Now, are achievable those $n's$ that can be written as $n=3a+5b$ with $a, b$ non negative integers. Do your cases and see that $n=3, 5, 6, 8, 9, 10 ...$ are achievable; now I'll prove if $n$ is achievable, $n+1$ is achievable. If $b \\ge 1$ and $n=3a+5b$ we can achieve $n+1=3(a+2) + 5(b-1)$, if $b=0$, since we proved all cases until $10$, we have $a > 3$ so we can achieve $n+1=3(a-3)+5(b+2)$ and we are done, $n=1, 2, 4, 7$ are not achievable, every other $n$ is achievable.", "Solution_4": "[quote=Sepp]Clearly we can cover every $ 5 \\times 3$ or $ 3 \\times 5$. So we can also cover a $ 15 \\times n$ with:\n$ n \\equiv 0 \\pmod 3$ (all $ 3 \\times 5$ pieces), \n$ n \\equiv 2 \\pmod 3$ and $ n > 2$ (a column of $ 5 \\times 3$ and all the others $ 3 \\times 5$),\n$ n \\equiv 1 \\pmod 3$ and $ n > 7$ (two columns of $ 5 \\times 3$ and all the others $ 3 \\times 5$).\nThen it's easy to see we cannot cover the board if $ n = 1, 2, 4, 7$.[/quote]\n\n", "Solution_5": "[quote=Sepp]Then it's easy to see we cannot cover the board if $ n = 1, 2, 4, 7$.[/quote]\n\nHOW do you prove that $n=4,7$ are not possible?" } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let an infinite sequence of measurable sets be given on the interval $ (0,1)$ the measures of which are $ \\geq \\alpha>0$. Show that there exists a point of $ (0,1)$ which belongs to infinitely many terms of the sequence.", "Solution_1": "$ \\mu\\left(\\bigcap\\limits_{i \\equal{} 1}^{\\infty}\\bigcup\\limits_{j \\equal{} i}^{\\infty}A_j\\right) \\equal{} \\lim \\mu(\\bigcup\\limits_{j \\equal{} i}^{\\infty}A_j)\\geq \\alpha$, the equality holds because measure of the union is bounded by 1.", "Solution_2": "[hide=To clarify Yustas' solution]\nWe use\n[hide=Rudin RCA, Theorem 1.19 (e)]\nGiven measurable sets $B_1 \\supset B_2 \\supset B_3 \\supset \\cdots$ such that $m(B_1)$ is finite, we have $m(B_i) \\to m(B_1 \\cap B_2 \\cap B_3 \\cap \\cdots)$ as $i \\to \\infty$\n[/hide]\nLet $B_i = A_i \\cup A_{i+1} \\cup A_{i+2} \\cup \\cdots$. Since $m(A_i) < 1$, we know that $m(B_1) < 1 < \\infty$, and we also know that $B_1 \\supset B_2 \\supset B_3 \\supset \\cdots$ so the theorem applies and we know that there exists $00} \\to \\mathbb R^{>0} $ such that\n\\[f(x)f(yf(x))=f(x+y)\\]\nholds for all reals $x$ and $y$.", "Solution_1": "I know a solution, but what is your solution? :lol:", "Solution_2": "This is from IMC, 2000 : http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1113833585&t=58521", "Solution_3": "it's easy,\r\n$ f(x)=\\frac{|2-x|+2-x}{(2-x)^{2}}\\ if\\ x\\neq 2\\ and\\ f(2)=0$\r\n\r\n\r\nor $ f(x)=\\frac{Max(2-x,0)}{(2-x)+[10^{-|x-2|}]}$ :wink:", "Solution_4": "aviateurpilot, can you post concrete?", "Solution_5": "[quote=\"dreammath\"]Find all functions:R* to R* (R* means the set of positive reals.)\n$ f(x)f(yf(x))=f(x+y)$.\nIt 's very interesting for anybody who want to solve it. :lol:[/quote]\r\n\r\n[u]Claim 1[/u] : $ f(x)\\leq 1$ $ \\forall$ $ x>0$\r\nLet $ x_{0}>0$ such that $ f(x_{0})>1$. Then let $ y_{0}=\\frac{x_{0}}{f(x_{0})-1}>0$.\r\n$ f(x_{0})f(y_{0}f(x_{0}))=f(x_{0}+y_{0})$ $ \\implies$ $ f(x_{0}) f(\\frac{x_{0}f(x_{0})}{f(x_{0})-1}) = f(\\frac{x_{0}f(x_{0})}{f(x_{0})-1})$ and, since $ f(x)>0$ $ \\forall x$ : $ f(x_{0})=1$, which is a contradiction. So Claim 1 is true.\r\n\r\n[u]Claim 2[/u] : $ f(x)$ is a non increasing function\r\nObvious since $ y>x$ $ \\implies$ $ f(x) f((y-x)f(x))=f(y)$ and so $ f(y)\\leq f(x)$ since $ f((y-x)f(x))\\leq 1$ (with Claim 1)\r\n\r\n[u]Claim 3[/u] : if it exists $ x_{0}>0$ such that $ f(x{0})=1$, then $ f(x)=1$ $ \\forall$ $ x>0$\r\nJust put $ x=x_{0}$ in the original equation and you get $ f(y+x_{0})=f(y)$ $ \\forall$ $ y>0$. and so $ f(x)$ is a constant (since it is a non increasing function (claim 2) and so $ f(x)=f(x_{0})=1$ $ \\forall$ $ x>$ and Claim 3 is true.\r\n\r\n[u]Claim 4[/u] : if $ f(x)$ is non constant, then $ f(x)$ is an injective function (and so is strictly decreasing)\r\nLet $ f(u)=f(v)$ with $ u>v$. Then $ f(v) f((u-v)f(v))=f(u)$ and so $ f((u-v)f(v))=1$ and so $ f(x)=1$ (with claim 3). Q.E.D.\r\n\r\nSolutions :\r\n$ f(x)=1$ is a solution.\r\nConsider now $ f(x)$ non constant, that's to say $ f(x)$ injective (claim 4) :\r\nLet $ y=\\frac{z}{f(x)}$. Then the initial equation becomes $ f(x)f(z)=f(x+\\frac{z}{f(x)})$\r\nBut we also have $ f(z)f(x)=f(z+\\frac{x}{f(z)})$\r\nSo $ f(x+\\frac{z}{f(x)})=f(z+\\frac{x}{f(z)})$ and, since $ f(x)$ is injective : $ x+\\frac{z}{f(x)}=z+\\frac{x}{f(z)}$ and so (dividing by $ xz$) $ \\frac{1}{z}+\\frac{1}{xf(x)}=\\frac{1}{x}+\\frac{1}{zf(z)}$.\r\n\r\nSo $ \\frac{1}{z}-\\frac{1}{zf(z)}=\\frac{1}{x}-\\frac{1}{xf(x)}=a$ and $ f(x)=\\frac{1}{1-ax}$\r\n\r\nAnd it is easy to verify that this expression is a solution of initial equation as soon as $ a\\leq 0$ (since we want $ f(x)>0$)\r\n\r\nSo the general solution is $ f(x)=\\frac{1}{1+ax}$ for any $ a\\geq 0$", "Solution_6": "[quote=\"aviateurpilot\"]it's easy,\n$ f(x)=\\frac{|2-x|+2-x}{(2-x)^{2}}\\ if\\ x\\neq 2\\ and\\ f(2)=0$\n\n\nor $ f(x)=\\frac{Max(2-x,0)}{(2-x)+[10^{-|x-2|}]}$ :wink:[/quote]\r\n\r\nThat's wrong since $ f(x)>0$ $ \\forall x>0$ (f : $ R^{*}\\rightarrow R^{*}$)", "Solution_7": "[quote=\"aviateurpilot\"]it's easy,\n$ f(x)=\\frac{|2-x|+2-x}{(2-x)^{2}}\\ if\\ x\\neq 2\\ and\\ f(2)=0$\n\n\nor $ f(x)=\\frac{Max(2-x,0)}{(2-x)+[10^{-|x-2|}]}$ :wink:[/quote]\r\nsorry [b]N.T.TUAN[/b], here it's solution for another problem (easy) where $ f(2)=0$ and $ f(x+y)=f(x)f(yf(x)),\\forall (x,y)\\in R^{+}^{2}$\r\nand $ f(x)=0\\ in [0,2[$ :rotfl: .", "Solution_8": "joke :P is it a problem from IMO? pco are right!" } { "Tag": [ "ratio", "vector" ], "Problem": "Two buses start at the same time from A and B to B and A respectively, each with constant velocity. They reach their destinations respectively after 1 hour and 4 hours of passing each other. What is the ratio of the velocities of the two buses?", "Solution_1": "[hide=\"Bus\"]Let the faster bus have velocity a, and the slower bus have velocity b. In addition, let the distance between A and B be d, so the time it takes for the two buses to pass each other is given by d/(a+b). The distance that the faster bus has traveled in this time is given by [d/(a+b)]a. \nSince d-[d/(a+b)]a is the distance left to travel, we have:\n[d-da/(a+b)]/a=1. Solving for d gives d=a(a+b)/b. Going on to b, we find that the distance b goes before it passes the faster bus is given by \n[d/(a+b)]b, so the distance it has left is d-[db/(a+b)]. Bus b can cover this time in [d-[db/(a+b)]]/b=4 hours. Solving for d gives d=4b(a+b)/a. Since d=d, we have:\n4b(a+b)/a= a(a+b)/b==>this gives 4b^2=a^2. Taking square roots give 2b=a, or a/b=[u]2[/u][/hide]", "Solution_2": "After filling a page with complicated looking expressions, I came up with this\r\n[hide]Let s be the velocity of the slower bus, R the ratio of the velocity of the faster bus to the velocity of the slower bus, and d the distance traveled by the slow bus when the buses meet.\nWhen the buses meet, the slow bus has traveled a distance d, while the fast bus has traveled a distance Rd.\nAfter that, it takes 4 hours for the slower bus to cover the distance Rd, while it only takes 1 hour for the faster bus to cover distance d, at velocity Rs\n4s = Rd\n1Rs = d or Rs = d\nSubstituting Rs for d in the first equation:\n4s = RRs\nR=2\n\nWould a generalization be that the ratio of the remaining travel times is the square of the ratio of the velocities?[/hide]", "Solution_3": "Actually the ratio of their velocities is -2 because velocity is a vector quantity. :wink:", "Solution_4": "Nice solutions guys. I had to do some messy work too to get a solution.\r\n\r\n[hide=\"My solution\"]\n\n[asy]size(200);\npair A,B,M;\nA=origin;\nB=(180,0);\nM=(120,0);\ndot(A);\ndot(B);\ndot(M);\nlabel(\"A\",A,W);\nlabel(\"B\",B,E);\nlabel(\"M\",M,S);\ndraw(A--B);[/asy]\n\nIn the diagram, let $ M$ be the point where the two buses meet, and let $ AB \\equal{} s, AM \\equal{} x$. Assume that the bus from $ A$ takes $ t$ hours to reach $ M$. Let $ v_A$ and $ v_B$ be the velocities of the two buses. Then it's clear that $ v_A \\equal{} \\frac xt \\equal{} \\frac {s \\minus{} x}{1}\\ (1)$ and $ v_B \\equal{} \\frac x4 \\equal{} \\frac {s \\minus{} x}{t}\\ (2)$.\n\n$ (1)\\implies t \\equal{} \\frac {x}{s \\minus{} x}$\n\n$ (2)\\implies t \\equal{} \\frac {4(s \\minus{} x)}{x}$\n\nThus $ \\frac {x}{s \\minus{} x} \\equal{} \\frac {4(s \\minus{} x)}{x}\\implies \\left(\\frac {x}{s \\minus{} x}\\right)^2 \\equal{} 4$. Hence $ \\frac {x}{s \\minus{} x} \\equal{} 2$. But $ \\frac {x}{s \\minus{} x} \\equal{} \\frac {\\frac xt}{\\frac {s \\minus{} x}{t}} \\equal{} \\frac {v_A}{v_B}$. Therfore $ \\boxed{\\frac {v_A}{v_B} \\equal{} 2}$ [/hide]\r\n\r\n[quote=\"KMST\"]Would a generalization be that the ratio of the remaining travel times is the square of the ratio of the velocities?[/quote]\r\n\r\nYes, I think." } { "Tag": [ "AMC", "AMC 10", "AMC 12" ], "Problem": "The sum of the two 5-digit nmbers $AMC10$ and $AMC12$ is 123422. What is $A+M+C$?\r\n\r\n$\\left(A\\right) 10$\r\n$\\left(B\\right) 11$\r\n$\\left(C\\right) 12$\r\n$\\left(D\\right) 13$\r\n$\\left(E\\right) 14$", "Solution_1": "[hide]\nLooking at the last two digits we can see that $10+12=22$ so that nothign carries over. So, we can ignore the last two digits. We know:\n$\\begin{eqnarray*}AMC+AMC &=& 2 \\cdot AMC = 1234 \\\\ AMC &=& \\frac{1234}{2}= 617 \\\\ A+M+C &=& 6+1+7 = 14 \\implies \\boxed{\\text{E}}$\n[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove that there are infinitely many composite numbers $ n$ such that $ n\\,|\\,3^{n\\minus{}1}\\minus{}2^{n\\minus{}1}$.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=150409\r\nLook there\r\n\r\nDaniel", "Solution_2": "Thanks. :)" } { "Tag": [], "Problem": "Though a whole bunch of thread regarding favorite piece(s) already. I still wonder which Beethoven's Sonata leaves you deepest impression.\r\n\r\nI like to know which Sonata(out of the 32) you love to listen to the most recently(not says you can play the best). Try to select only one and give reasons why you prefer it if possible. Thanks :) \r\n\r\nFor me it's No.29, Op.106(Hammerklavier)... Well, simply because that's the most BEETHOVEN-type Sonata.", "Solution_1": "I completely agree. Hammerklavier is, in my opinion, the most techincally and musically demanding in the standard repertoire by far. Not to mention the stamina needed...\r\n\r\nI think all of the late ones are the best though (106, 109, 110, 111), I think their darkness best defines Beethoven. And all of them have fugues in the last movement!", "Solution_2": "Op.57 \"Appasionata\" is the most moving one for me, although they are all good, as Beethoven=genius.", "Solution_3": "[quote=\"CatalystOfNostalgia\"]I completely agree. Hammerklavier is, in my opinion, the most techincally and musically demanding in the standard repertoire by far. Not to mention the stamina needed...\n\nI think all of the late ones are the best though (106, 109, 110, 111), I think their darkness best defines Beethoven. And all of them have fugues in the last movement![/quote]\r\n\r\nDoesn't op.111 just have themes replaced with fugues in the first movement?", "Solution_4": "I like the pathetique sonata the most. :lol:", "Solution_5": "Am I allowed to talk about violins? :D", "Solution_6": "Yeah, but I don't know much about Beethoven's violin works :blush: Any recommendations?", "Solution_7": "I like the sonata No. 4 and No. 5. 5 is overplayed, but for a reason! :)", "Solution_8": "All of Beethoven Sonatas are amazing! Beethoven was a genius!\r\n\r\nMy favorite would have to be the Appasionata though. I find the Pathetique the most fun to play." } { "Tag": [ "geometry", "Circumcenter", "circles", "IMO Shortlist" ], "Problem": "Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$. Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.", "Solution_1": "A quick angle chase reveals that $CA_1QA_2$ is cyclic, meaning that we can forget about the $B_i$'s: we're looking for the locus of the circumcenter of $A_1A_2Q$.\r\n\r\nSince $\\angle PA_iQ$ are constant angles, it means that the triangles $A_1A_2Q$ are all similar, and this means that the triangles $A_1QO$ are all similar, where $O$ is the circumcenter of $A_1A_2Q$. This means that the locus of $O$ is the image of $S_1$ through the spiral similarity of center $Q$ which turns $A_1$ into $O$. In other words, this locus is a circle, Q.E.D.\r\n\r\nComment: it can be shown fairly easily now, by choosing particular positions of $A_1$, the the circle is, in fact, the circumcircle of $QO_1O_2$, where $O_i$ is the center of $S_i$.", "Solution_2": "If we regard $ A_1A_2PB_1B_2C$ as a complete quadrilateral the problem becomes trivial once we know that the circumcenters of the $ 4$ \"small\" triangles of the complete quadrilateral and the miquel point of this complete quadrilateral (which in this case is point $ Q$) lie on a circle. Is this a well-known fact? If so, where can I find a proof to this fact?", "Solution_3": "Hmm. This problem did not involve the introduction of a new point other than the center of the indicated circle.\n\n[b]Solution[/b]\n\nLet the centers of $S_1$ and $S_2$ be $O_1$ and $O_2$. First note the following.\n\n\\[\\angle{CA_1 Q} + \\angle{CA_2 Q} = \\angle{QPB_2} + \\angle{CA_2 A_1} +\\angle{PA_2 Q} = \\angle{PQB_2}+\\angle{PB_2 Q} +\\angle{QPB_2}=180\\]\nTherefore $QA_1 C A_2$ is cyclic. Let $\\gamma$ and $\\omega$ denote the perpendicular bisectors of $A_1 Q$ and $B_2 Q$ and let $X=\\omega \\cap \\gamma$. Note that $X$ is the circumcenter of $QA_1 C A_2$.\n\nSince $O_1 \\in \\gamma$ and $O_2 \\in \\omega$, we have that $\\angle{O_1 X O_2}=180-\\angle{A_1 Q A_2}$. However, since $Q$ is the center of the spiral similarity mapping $S_1$ to $S_2$, it follows that $\\angle{O_1 Q O_2}=\\angle{A_1 Q A_2}$.\n\nThis yields that $O_1 X O_2 Q$ is cyclic and since $O_1$, $O_2$ and $Q$ are fixed, that $X$ always lies on one circle.", "Solution_4": "Actually, a similar problem to this has appeared in two different UK texts, however they were both published after 2002, so it's OK! It also reminds of this years British MO round 1 Q5, which is of the same essence, posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=381274]here[/url]. On these grounds I'm guessing this was proposed by the UK :D\n\n[hide][img]http://oi53.tinypic.com/wri7hc.jpg[/img]\n\n[color=#FF0000][mod edit: http://oi53.tinypic.com/wri7hc.jpg][/color][/hide]\n$O_1,O_2,O_3$ are the centres of $S_1,S_2$ and $\\odot(A_1QA_2)$.\n\nAs previously shown, we only need to study the locus of $O_3$. $O_1O_3$ and $O_2O_3$ are perpendicular to $A_1Q$ and $A_2Q$ respectively as by definition all of $\\triangle A_2QO_2,\\triangle A_2QO_3,\\triangle A_1QO_1,\\triangle A_2QO_3$ are isosceles. Let the midpoints of $QA_1,QA_2$ be $M_1,M_2$ respectively. To show that the quadrilateral $QO_1O_2O_3$ is cyclic it suffices to prove $\\angle O_2QM_2=\\angle O_1QM_1$ as then it would follow $\\angle QO_1M_1=\\angle QO_2M_2$ if we recall the right angles $\\angle M_1,\\angle M_2$.\n\nThere is a fairly quick angle chase:\n[hide]$\\angle O_2QM_2=\\alpha\\implies\\angle QO_2A_2=180^{\\circ}-2\\alpha$ $\\implies QPA_2=90^{\\circ}-\\alpha\\implies\\angle A_1PQ=90^{\\circ}+\\alpha$. Then $A_1Q$ subtends an angle of $90^{\\circ}-\\alpha$ (at least on this diagram) so $\\angle A_1O_1Q=180^{\\circ}-2\\alpha$ $\\implies\\angle M_1O_1Q=\\frac{1}{2}\\angle A_1O_1Q=90^{\\circ}-\\alpha=\\angle QO_2M=180^{\\circ}-\\angle O_3O_2Q$ so $QO_i$ is cyclic.[/hide]\n\nBut a more appealing argument is that no matter where we place $A_1$ on $S_1$, $\\triangle A_1A_2Q$ has fixed angles. So if we consider the diagram, but with another $A_n$ diametrically opposite $Q$ on $S_1$ then easily $\\angle A_nQA_{n+1}=\\angle A_1QA_2\\implies\\angle O_1QO_2=\\angle A_1QA_2$, since $O_1,O_2$ lie on $QA_n,QA_{n+1}$ respectively. So this rotation must leave an equal against the common $\\angle A_1QA_2$, i.e. $\\angle O_1QM_1=\\angle M_2QO_2$. Then the locus of $O_3$ is the circumcentre of $O_1QO_2$.", "Solution_5": "Do I feel guilty reviving a 7 year old thread? No.\n\n[hide=\"solution\"]\nWe use directed angles.\n$\\angle B_1CB_2=\\angle A_1CA_2 = \\angle CA_1P+\\angle PA_2C = \\angle B_1QP+\\angle PQB_2=\\angle B_1QB_2$$\\implies Q\\in (B_1CB_2)$. Let $O_1, O_2, O$ be the circumcenters of $S_1, S_2, (B_1QB_2C)$. By spiral similarity, $\\angle B_1QB_2$ is constant. Since $OO_1\\perp B_1Q$ and $OO_2\\perp B_2Q$, then $\\angle O_1OO_2 = \\angle B_1QB_2$ is constant; thus, $O\\in (O_1QO_2)$ as desired. $\\Box$\n[/hide]", "Solution_6": "My solution:\nLet's show that $O_1,Q,O_2,O$ are concyclic,where $O_1,O_2,O$ be the centers of $S_1,S_2,(A_1A_2C),$ respectively.\nLet $S=OO_1\\cap A_1Q$ and $T=OO_2\\cap A_2Q.$\nWe know $\\angle CA_1Q=180-\\angle B_1PQ=\\angle QPB_2=\\angle QA_2B_2=180-\\angle QA_2C\\rightarrow A_1,Q,A_2,C$ are cyclic.\nThen $OS\\perp AQ,OT\\perp QA_2\\to S,Q,T,O$ are cyclic.\n$Claim:$ $\\angle A_1QA_2=\\angle O_1QO_2.$\n$Proof:$ We have $\\angle QPA_2=180-\\angle QPA_1=180-\\frac{\\angle A_1O_1Q}{2}=90+\\angle O_1QA_1.$\nAlso we know $\\angle QPA_2=180-\\frac{\\angle QO_2A_2}{2}=90+\\angle O_2QA_2,$ then we know $\\angle O_2QA_2=\\angle O_1QA_1\\to \\angle O_1QO_2=\\angle O_1QA_1+\\angle A_1QO_2=\\angle O_2QA_2+\\angle A_1QO_2=\\angle A_1QA_1.$ As desired.\nAlso we know \n$$\\angle O_1OO_2=\\angle SOT=180-\\angle SQT=180-\\angle A_1QA_2=^{\\text{Claim}}180-\\angle O_1QO_2\\implies O_1,QO_2,O$$ are concyclic.", "Solution_7": "[b]Lemma: [/b] This problem is trivial.\n[b] Proof [/b]: For any $B$, by a trivial angle chase, we have $\\angle A_1CA_2 = \\angle A_1QA_2$, so we put the $B$ aside the picture, and focus on the locus of the circumcentre of $\\Delta A_1QA_2$, as $A_1$ varies. We claim the locus is on $\\omega_{O_1QO_2}$. Indeed, inverting around $Q$, the problem becomes trivial, because taking a homothety of factor $.5$ w.r.t $Q$ maps $O_1^{*} O_2^{*}$ to the $Q$-simson line of $\\omega_{A_1^{*}P^{*}A_2^{*}}$, hence the desired result. \n\nAs a corollary, we have:\n[b] Lemma [/b]: 10 year or more old IMOSL problems are trivial. \n(Sketch of proof is trivial, just also note that JMO P1 was 1998 G8, then trivial by induction)", "Solution_8": "Let the centers of $S_1, S_2$ be $O_1, O_2$ resp.\\\\\nSome angle chasing shows that $\\angle A_1QA_2=\\angle A_1QP+\\angle PQA_2=\\angle AB_1B_2+\\angle CB_2B_1=180-\\angle A_1CA_2$\\\\\n$\\Longrightarrow A_1CA_2Q$ is cyclic.\\\\\nThus the problem is equivalent to find the locus point of the circumcenter of $\\triangle QA_1A_2$ which doesn't depend on $B_1$ so we can ignore $B_1$ in the problem.\\\\\nWe will prove that $O\\in (O_1QO_2)$ which is a fixed circle where $O$ is the circumcenter of $\\triangle QA_1A_2$.\\\\ We have \\\\$(A_1CA_2Q)\\cap S_2=A_2Q$, $(A_1CA_2Q)\\cap S_1=A_1Q$\\\\ $\\Longrightarrow OO_2\\perp A_2Q,\\ OO_1\\perp A_1Q\\Longrightarrow \\angle O_1OO_2=180 -\\angle A_1QA_2$.\\\\\nBut $\\angle A_1QO_1=\\frac{180-2\\angle A_2PQ}{2}=90-\\angle A_2PQ=\\angle A _2QO_2\\Longrightarrow \\angle A_1QA_2=\\angle O_1QO_2$.\\\\\n$\\Longrightarrow \\angle O_1OO_2=180 -\\angle O_1QO_2\\Longrightarrow O\\in (O_1QO_2)$ which is a fixed circle as desired.", "Solution_9": "[hide=Solution]\n[quote=ISL 2002 G4]Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$. Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.[/quote]\n[b][color=#000]Solution:[/color][/b] Let $O_1,O_2,O$ be centers of $S_1,S_2, \\odot (A_1A_2C)$. Now, $QO_1O_2O$ is cyclic, because $\\overline{A_1O_1}$ $\\mapsto $ $\\overline{A_2O_2}$ (because: $\\Delta QO_1A_1$ $\\sim$ $\\Delta QO_2A_2$) and, $\\angle BA_1Q$ $=$ $\\angle B_1PQ$ $=$ $\\angle CA_2Q$. Since, $Q,O_1,O_2$ are fixed $\\implies$ $O$ moves on a fixed circle $\\qquad \\blacksquare$\n[/hide]", "Solution_10": "Adding an Inversive Solution. :)\n[quote=ISL 2002 G4]Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$. Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.[/quote]\n\nNote that $$\\angle AQA_2=\\angle AQP+\\angle PQA_2=\\angle AC_1B_2+\\angle B_1B_2C=180^\\circ-\\angle A_1QA_2\\implies A,Q,A_2,C\\text{ are concyclic.}$$So the Circumcenter of $\\triangle A_1A_2C$ is same as the Circumcenter of $\\triangle A_1QA_2$. So now we can safely delete the points $C,B_1,B_2$ from the Diagram. So, the problem can now be restated as \n[Quote=Reduced Problem]$\\omega_1$ and $\\omega_2$ are two circles and $\\omega_1\\cap\\omega_2=\\{P,Q\\}$. A line through $P$ intersects $\\omega_1$ and $\\omega_2$ at $A_1,A_2$ respectively. Then Prove that the locus of the Circumcenters of such triangles $A_1QA_2$ is a circle.[/quote]\n\nFor this Invert around $Q$ and let this map be denoted as $\\Psi$. So, $\\Psi:P\\leftrightarrow \\omega_1'\\cap\\omega_2'=P'$ where $\\omega_1'$ and $\\omega_2'$ are the Inverted Image of $\\omega_1$ and $\\omega_2$ respectively during this transformation $\\Psi$. \n\nWe have to prove that the circles passing through $P',Q$ intersecting $\\omega_1'$ and $\\omega_2'$ at points $\\{A_1',A_2'\\}$ respectively, then the reflections of $Q$ on $A_1'A_2'$ are collinear.\n\nDrop Perpendiculars from $Q$ onto $\\omega_1'$ and $\\omega_2'$, So by Simson Line we get that the Projections of $Q$ on such lines $A_1'A_2'$ are collinear, so by a Homothety at $Q$ with a scale factor of $2$ we get that the reflections of $Q$ of such lines $A_1A_2$ are also collinear, and the reflections of $Q$ on such lines $A_1A_2$ are the points where the circumcenters of such triangles $QA_1A_2$ map!\n\nHence Inverting back we get that the Locus of the Circumcenters of such triangles $QA_1A_2$ which are the circumcenters of $\\triangle A_1A_2C$ will lie on a circle. $\\blacksquare$", "Solution_11": "What troll!\n[quote=orl]Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$. Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.[/quote]\nLet $O_1,O_2$ be center of $S_1,S_2$ respectively.We have $Q$ is the miquel point of complete quadrilateral $PB_2CA_1A_2B_1$.Hence center of $\\odot{A_1A_2C}$ lies on the miquel circle $\\odot{QO_1O_2}$ which is fixed.$\\square$", "Solution_12": "[hide=Solution]\nJust note that \\(Q\\) is the Miquel point of \\(CA_1PB_2\\). It is well known that the centers of \\((CA_1A_2),(PB_2A_2),(PB_1A_1)\\) form a cyclic quadrilateral with \\(Q\\); the desired center always lies on this circle.\n[/hide]", "Solution_13": "Easy G4.\n\nNotice that $\\angle A_1QA_2=\\angle A_1B_1B_2+\\angle CB_2B_1=180^{\\circ}_\\angle C,$ so $A_1CA_2Q$ is cyclic and the problem reduces to showing that the circumcenter of $(A_1A_2Q)$ lie on a fixed circle (the $B$'s were just a distraction!).\n\nWe finish with\n\nClaim: Denote the center of $S_1$ by $O_1$ and that of $S_2$ by $O_2$. Then the desired circle is $(O_1QO_2)$.\n\nProof: Denote the midpoints of $A_1Q$ and $A_2Q$ by $M_1, M_2,$ respectively. Then the circumcenter of $\\triangle A_1A_2Q$ is the intersection of the perpendicular at $M_1$ and $M_2$ to $A_1Q$ and $A_2Q$. Notice that $\\angle AO_1Q=2\\angle A_1PQ=180^{\\circ}-\\angle B_2PQ=\\angle QO_2A_2,$ and this easily implies the result.\n\nWe are done.", "Solution_14": "Obviously $Q$ is the Miquel point of $A_1CB_2P,$ and so circumcenter of $\\triangle A_1A_2C$ lie on circle passing through $Q$ and centers of $S_1,S_2,$ which is fixed.", "Solution_15": "Let $O,O_1,O_2$ be the centers of $S_1,S_2,(A_1A_2C),$ respectively. Notice that $Q$ is the Miquel point of $A_1B_1CA_2B_2P$ so $OO_1QO_2$ is cyclic. $\\square$", "Solution_16": "No one:\nEGMO Ch. 11:\n\n[img width=50]https://media.discordapp.net/attachments/925784397469331477/952410078668009472/Screen_Shot_2022-03-12_at_8.37.24_PM.png[/img]\n\nLet $O_1,O_2$ be centers of $S_1,S_2.$ Let $O$ be center of $(A_1CA_2).$ We claim that $O_1OO_2Q$ is cyclic. First, we claim that $CA_1QA_2$ is cyclic. Note that $\\angle B_1A_1Q=\\angle B_1PQ=\\angle CA_2Q.$ This implies the claim.\n\nThus, $O$ is the center of $(QA_1A_2).$ Note that $OO_1$ is the perpendicular bisector of $A_1Q$ and $OO_2$ the perpendicular bisector of $A_2Q.$ Thus, $\\angle OO_2Q=\\frac{1}{2}\\overset{\\huge\\frown}{QPA_2}=180^\\circ-\\angle QPA_2$ and $\\angle OO_1Q=\\frac{1}{2}\\overset{\\huge\\frown}{QPA_1}=180^\\circ-\\angle A_1PQ.$ These sum to $180^\\circ$ so we are done.", "Solution_17": "Wut\nObserve that $Q$ is miquel point of quad $A_1PB_2C$ which means the circumcenter of $\\triangle A_1A_2C$ belongs to $\\odot(Q0_1O_2)$ where $O_1$ and $O_2$ are the centres of $S_1$ and $S_2$ which are obviously fixed $\\blacksquare$", "Solution_18": "[b]Claim: We can simply consider the circumcenter of $A_1QA_2$[/b]\n\nProof: \n\nObserve that $Q$ is the miquel point of quadrilateral $A_1PB_2C$\n\nTherefore, $Q$ lies on $(A_1A_2C)$ \n\nAs such, the circumcenter of $(A_1A_2C)$= the circumcenter of $(A_1QA_2)$ \n\n[b]Rephrased problem:[/b] the circumcenters of $(A_1A_2Q)$ lie on a circle as $A_1A_2$ varies \n\n[b]Claim: $\\angle A_1QA_2$, $\\angle A_2A_1Q$, $\\angle A_1A_2Q$ remain invariant over all $A_1$,$A_2$[/b]\n\nProof: \nFix one $A_1$,$A_2$,\n\nNow, suppose we take another pair, say $B_1,B_2$\n\nNote that $Q$ is the center of the spiral sending $A_1 \\rightarrow A_2$, $B_1 \\rightarrow B_2$ \n\nWhich implies that $\\triangle A_1QA_2 \\sim \\triangle B_1QB_2$\n\n[b]Claim: If $O$ is the circumcenter of $(A_1QA_2)$, $\\triangle A_1QO$ remains similar as $A_1$ varies [/b]\n\nProof: \n\nNote $Q$ is the spiral center sending all $A_1 \\rightarrow O$ \n\nAs such, the possible positions of $O$ are simply a fixed scaling and translation from every possible position of $A_1$ with respect to the spiral center $Q$ \n\nThis means that all possible positions of $O$ must cover a circle ", "Solution_19": "Viewing $CA_1PB_1$ as a complete quadrilateral, $CA_1QA_2$ is cyclic. Therefore, we can delete $C,B_1,B_2$ from the diagram and focus on the center of $(A_1A_2Q)$ as $A_1$ varies around $S_1$. As $A_1$ varies around $S_1$, $\\triangle QA_1A_2$ is directly similar to a fixed triangle for spiral similarity reasons, so $\\triangle QA_1O$ is directly similar to a fixed triangle as well, where $O$ is the (variable) circumcenter of $\\triangle QA_1A_2$. Therefore, it suffices to prove the following \"independent\" lemma.\n\n[b][color=#00f]Lemma:[/color][/b] Let $P$ be a point and suppose that $\\triangle PAB$ is directly similar to a fixed triangle as $A$ varies around a circle. Then $B$ varies around a circle too.\n[color=#00f]Proof:[/color] We use complex numbers. WLOG let $p=0$. Then there exists some fixed complex number $z \\neq 0$ such that $b=za$ always holds, and the rest is clear. $\\blacksquare$", "Solution_20": "wow how did I not see the Miquel point\n\nWe first do some angle chasing:\n\\begin{align*}\n\\angle A_1QA_2 &= \\angle A_1QP + \\angle PQB_2 + \\angle B_2QA_2 \\\\\n&= \\angle A_1B_1P + \\angle PA_2B_2 + \\angle B_2PA_2 \\\\\n&= \\angle A_1B_1P + \\angle A_1PB_1 + \\angle A_1A_2C \\\\\n&= \\angle CA_1A_2 + \\angle A_1A_2C \\\\\n\\angle A_1QA_2 &= 180^\\circ - \\angle A_1CA_2\n\\end{align*}\nFrom here, we see that $CA_1QA_2$ is a cyclic quadrilateral. Obviously, the circumcenter of $CA_1A_2$ is the same as the circumcenter of $A_1QA_2$, a point we will label as $O$.\nThe centers of $S_1$ and $S_2$ will be called $O_1$ and $O_2$, respectively. Now, we claim that $OO_1QO_2$ is a cyclic quadrilateral.\nIn order to prove this, construct the perpendicular bisectors to $A_1Q$ and $A_2Q$. Denote the midpoints of $A_1Q$ and $A_2Q$ as $M$ and $N$, respectively.\nVisibly, $OMQN$ is a cyclic quadrilateral, which means that $\\angle MQN + \\angle MON$.\nIt is easy to see that the perpendicular bisector of $A_1Q$ passes through both $O_1$ and $O$, while the perpendicular bisector of $A_2Q$ passes through both $O_2$ and $O$. As a result, $\\angle MON = \\angle O_1OO_2$.\nWe finish with a short angle chase:\n\\begin{align*}\n180^\\circ &= \\angle MQN + \\angle MON \\\\\n&= \\angle A_1QA_2 + \\angle O_1OO_2 \\\\\n180^\\circ &= \\angle O_1QO_2 + O_1OO_2\n\\end{align*}\nThus, $OO_1QO_2$ is a cyclic quadrilateral, and $O$ always lies on $(O_1QO_2)$, a fixed circle. $\\blacksquare$" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Determine the ordered systems $(x,y,z)$ of positive rational numbers for which $x+\\frac{1}{y},y+\\frac{1}{z}$ and $z+\\frac{1}{x}$ are integers.", "Solution_1": "[quote=\"nayel\"]Determine the ordered systems $(x,y,z)$ of positive rational numbers for which $x+\\frac{1}{y},y+\\frac{1}{z}$ and $z+\\frac{1}{x}$ are integers.[/quote]\r\n\r\nIf we write $x=\\frac{p}{q}$, $(p,q)$ coprime, $y=\\frac{r}{s}$, $(r,s)$ coprime, $z=\\frac{t}{u}$, $(t,u)$ coprime, we have :\r\n\r\n$\\frac{p}{q}+\\frac{s}{r}=\\frac{pr+qs}{qr}$ integer $\\implies$ $q|pr$ and so $q|r$ and also $r|qs$ and so $r|q$ and so $q=r$ and $q|(p+s)$\r\nSo we have $q=r$, $s=t$ and $u=p$ and :\r\n$q|(p+s)$, and so $q|(p+q+s)$\r\n$s|(q+p)$, and so $s|(p+q+s)$\r\n$p|(s+q)$, and so $p|(p+q+s)$\r\n\r\nAnd so $pqs|(p+q+s)$ and very few solutions (since normally $pqs>p+q+s$) : $(1,1,1)$, $(1,1,2)$ and $(1,2,3)$\r\n\r\nAnd so the solutions for $(x,y,z)$ :\r\n$(1,1,1)$, $(1,\\frac{1}{2},2)$, $(\\frac{1}{2},2,1)$, $(2,1,\\frac{1}{2})$, $(\\frac{1}{2},\\frac{2}{3},3)$, $(3,\\frac{1}{2},\\frac{2}{3})$, $(\\frac{2}{3},3,\\frac{1}{2})$, $(\\frac{1}{3},\\frac{3}{2},2)$, $(2,\\frac{1}{3},\\frac{3}{2})$, $(\\frac{3}{2},2,\\frac{1}{3})$", "Solution_2": "nice solution. also the same as mine. can this be solved in any other way?" } { "Tag": [ "geometry" ], "Problem": "Find the number of square units in the area of the region. The dots are equally spaced vertically and horizontally.\n[asy]size(50);\nfor(int i = 0; i < 5; ++i){\n for(int j = 0; j < 5; ++j){\n dot((i,j));\n } \n}\ndraw((0,1)--(0,2)--(3,4)--(1,1)--(0,1),linewidth(0.7));[/asy]", "Solution_1": "This is $ 9\\minus{}6\\equal{}\\boxed{3}$.", "Solution_2": "Can you explain...? The solution makes no sense", "Solution_3": "$ S \\equal{} 2 \\plus{} \\frac {4}{2} \\minus{} 1 \\equal{} \\boxed{3}$.\r\n\r\n[b]Remark[/b]\r\n[url=http://mathworld.wolfram.com/PicksTheorem.html]Pick's Theorem[/url] - MathWorld", "Solution_4": "In easier words, Find the number of dots inside the figure (2), then add that to half of the number of dots that enclose the figure (4x1/2) then subtract that by 1.", "Solution_5": "Alternatively, realize that there are two triangles of length 3 base, and 1 height.\n\n$3\\times1\\div2\\times2=3$" } { "Tag": [ "geometry", "circumcircle", "inradius", "inequalities solved", "inequalities" ], "Problem": "Let S,R,r be respectively the area,the circumradius and the inradius of \r\ntriangle ABC.Let BC=a,CA=b,AB=c .Prove that \r\n2S/R<=(b+c-a)sin(A/2)+(a-b+c)sin(B/2)+(a+b-c)sin(C/2)<=S/r\r\nWhen does one equality occur?", "Solution_1": "I think this one is easy. :D :D \r\n The right one you use Tchebychev's inequality.\r\n The left one by using b+c-a =8Rcos(A/2)sin(B/2)sin(C/2) , you can solve it easily. :)", "Solution_2": "yeah i had got the left one but not the right one.\r\n thanx treegoner" } { "Tag": [], "Problem": "These questions came on our test, and i want to verify if the solution given is right :mad: \r\n\r\n1) $ \\ce {CH_3COCH(CH_3)_2}+\\ce {HCHO} + \\ce {R_2NH} \\rightarrow ?$\r\n\r\n2) $ \\ce {(CHO)CH_2CH_2(CHO)} + \\ce {CH_3NH_2} + \\ce {CH_3COCH_3} \\rightarrow ?$\r\n\r\nBoth reactions in presence of HCl", "Solution_1": "Number 2) is the classical Robinson's 1917 synthesis of tropinone. For the first, what is [b]your[/b] answer?", "Solution_2": "Isn't the crux of problem 1 the acidity of the alpha-hydrogens?\r\n\r\nI think the CH3CO- hydrogen is more acidic compared to the one on the other side.", "Solution_3": "The crux of the problem is the formation of the Eschenmoser salt/iminium ion as a result of reacting the formaldehyde and dialkylamine." } { "Tag": [ "geometry", "AMC", "AMC 8", "rectangle" ], "Problem": "Three circular arcs of radius 5 units bound the region shown. Arcs\n$ AB$ and $ AD$ are quarter-circles, and arc $ BCD$ is a semicircle.\nWhat is the area, in square units, of the region?\n[asy]/* AMC8 2000 #19 Problem */\ndraw((0,0)..(1,1)..(2,0));\ndraw((0,0)..(.7,-.3)..(1,-1));\ndraw((1,-1)..(1.3, -0.3)..(2,0));\nlabel(\"$A$\", (1,-1), SW);\nlabel(\"$B$\", (0,0), W);\nlabel(\"$C$\", (1,1),N);\nlabel(\"$D$\", (2,0),E);[/asy]", "Solution_1": "[hide]if cut the region bounded by BD and the two circular arcs BA and AD, and you move the peices up to the top of the semicircle, you will get a rectangle of dimensions 5 by 10, for an area of [b]50[/b]\n\nfor a more rigorous solution, we have pi*5^2/2+2(5^2-pi*5^2/4)=pi*25/2+50-25*pi/2=50.[/hide]" } { "Tag": [ "inequalities", "LaTeX", "special factorizations" ], "Problem": "if a^2- 4a+b^2- b/2+65/16=0 then what is a^2- 4 root b", "Solution_1": "Completing the square, this reduces to $ a^2\\minus{}4a\\plus{}b^2\\minus{}\\frac b2\\plus{}\\frac{65}{16}\\equal{}(a\\minus{}2)^2\\plus{}\\left(b\\minus{}\\frac14\\right)^2\\equal{}0$. Assuming $ a,b\\in\\mathbb R$, $ (a\\minus{}2)^2,\\left(b\\minus{}\\frac14\\right)^2\\ge0$ by the Trivial Inequality, so the only solution is $ (a,b)\\equal{}\\left(2,\\frac14\\right)\\implies a^2\\minus{}4\\sqrt b\\equal{}4\\minus{}4\\sqrt{\\frac14}\\equal{}4\\minus{}2\\equal{}\\boxed2$.", "Solution_2": "if you don't mind, next time, please LaTeX your question so it's easier to read\r\nfind out how to on the LaTeX thingy on the sidebar" } { "Tag": [ "LaTeX", "articles", "geometry" ], "Problem": "I'm at university and beginning to type up my final year maths project using Latex, which is new to me, and I'm not sure which document style to use. Report, article, amsart and amsbook seem to be the most likely candidates, but I'm not sure. The obvious approach would be to try them all and see the differences, but I just thought I'd ask for advice from someone with a bit more experience of Latex. If it makes any difference, the project will be about 40-50 pages long with around 5 chapters/sections and various subsections for each of these.\r\n\r\nAdditionally, are there any specific packages that I could / should use to make things look nicer?", "Solution_1": "I think that you may be approaching things the wrong way round. Many universities insist that write-ups are presented in a certain way; some are very specific and others are more relaxed. Some provide their own packages to use. So I would find out what the university requires or thinks is acceptable. You could also look at other previous projects to see how they are presented. Once you know what you need then you can decide which document class will do the job.\r\n[quote=\"allego\"]Additionally, are there any specific packages that I could / should use to make things look nicer?[/quote]LaTeX output always looks 'nice'; you may want to use packages to do a certain job (eg geometry for margins, fancyhdr for headers and footers) but which depends on the requirements.\r\n\r\nThe big advantage of LaTeX is that you can get on with typing your project immediately and can change the class at any time.", "Solution_2": "Thank you for the swift response." } { "Tag": [ "ARML", "search", "AMC", "AIME", "USA(J)MO", "USAMO" ], "Problem": "So does anyone know him? Just curious...", "Solution_1": "for anyone who [i]didn't[/i] know this, he's a ninth grader who made USAMO...u'll probably see him at ARML Karthik", "Solution_2": "Okay, I knew his name sounded familiar, so I did a google search and found that he, as well as Vova, tied me in the Georgia Math League in 8th grade; however, Vova and I are both in 10th grade and were both apparently in 8th grade at the same time as Sagar, so how did he end up in 9th grade this year? :huh:\r\nIdea #1: He was allowed to take the 8th grade GAML as a 7th grader (which seems kind of weird).\r\nIdea #2: He somehow managed to misbubble his grade on his AMC and AIME and is really in 10th grade.\r\nIdea #3: He got held back (a USAMO qualifier getting held back???).\r\nIdea #4: Some sort of time travelling maybe...\r\nLink: http://www.mathleague.com/reports/2004_05/grade678/GA_8.HTM", "Solution_3": "The one I want to know about is that 4th grader up in Canada!", "Solution_4": "i heard about that too, but don't you have to be a resident of the US to take the AMC and AIME?", "Solution_5": "My guess (and it's only a guess. I'm too tired to actually look it up.) is that Canada uses the AMC and AIME the same way we do. They may not use USAMO and I'm almost positive that they can't get invited to MOSP. They probably have something similar.", "Solution_6": "[quote=\"gtf\"]My guess (and it's only a guess. I'm too tired to actually look it up.) is that Canada uses the AMC and AIME the same way we do. They may not use USAMO and I'm almost positive that they can't get invited to MOSP. They probably have something similar.[/quote]\r\nU.S. students and legal residents of the U.S. or Canada may qualify for the USAMO (there are forty-odd Canadian qualifiers this year). However, Canadians may not go to MOP : MOP is for members and potential future members of the USA team. Canadian team selection is apparently based on the Canadian Math Olympiad, the USAMO, and the APMO. They apparently have some seminars for people likely to be on the team, and they qualify for the Canadian Math Olympiad by their own qualification exam." } { "Tag": [ "integration", "trigonometry", "calculus", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Someone posted an interesting problem. But unfortunately I cannot find his message. The problem is as follows.\r\n\r\nProve that $\\int\\limits_0^\\frac\\pi4\\log(4\\cos^2\\theta-1)d\\theta=\\dfrac23\\int\\limits_0^\\frac\\pi4\\log(4\\cos^2\\theta)d\\theta$.\r\n\r\nWhen I saw that problem (2-4 days ago), it hasn't been solved. Maybe it is solved already. It is very interesting to see a solution.", "Solution_1": "You've lost a factor of $\\frac23$ on the RHS. This integral has something to do with Catalan's constant, but I don't know anything more than that.", "Solution_2": "I have the following \"semi-solution\" which I devided into 2 parts.\r\n\r\n1) $\\int\\limits_0^{\\frac\\pi4}\\log(4\\cos^2\\theta-1)d\\theta =$\r\n\r\n$\\dfrac12\\int\\limits_0^{\\frac\\pi2}\\log(2\\cos\\theta+1)d\\theta = \\dfrac12\\int\\limits_0^{\\frac\\pi2}\\log\\left(2\\left(\\cos\\theta+\\cos\\dfrac\\pi3\\right)\\right)d\\theta = \\int\\limits_0^{\\frac\\pi4}\\log\\left(4\\cos\\left(\\theta-\\dfrac\\pi6\\right)\\cos\\left(\\theta+\\dfrac\\pi6\\right)\\right)d\\theta = \\int\\limits_{-\\frac\\pi6}^{\\frac\\pi{12}}\\log(2\\cos\\theta)d\\theta+\\int\\limits_{\\frac\\pi6}^{\\frac{5\\pi}{12}}\\log(2\\cos\\theta)d\\theta = \\int\\limits_{-\\frac\\pi{12}}^\\frac{5\\pi}{12}\\log(2\\cos\\theta)d\\theta = \\dfrac12\\int\\limits_{-\\frac\\pi{12}}^\\frac{5\\pi}{12}\\log(4\\cos^2\\theta)d\\theta$.\r\n\r\nThus we have to prove that $I: =\\int\\limits_{-\\frac\\pi{12}}^\\frac{5\\pi}{12}\\log(4\\cos^2\\theta)d\\theta=\\dfrac43\\int\\limits_0^\\frac\\pi4\\log(4\\cos^2\\theta)d\\theta$.\r\n\r\n2) $I=2\\int\\limits_0^\\frac\\pi{12}\\log(4\\cos^2\\theta)d\\theta+\\int\\limits_\\frac\\pi{12}^\\frac{5\\pi}{12}\\log(4\\cos^2\\theta)d\\theta$.\r\n\r\nNow $\\int\\limits_\\frac\\pi{12}^\\frac{5\\pi}{12}\\log(4\\cos^2\\theta)d\\theta = \\int\\limits_{-\\frac{5\\pi}{12}}^{-\\frac\\pi{12}}\\log(4\\sin^2\\theta)d\\theta = \\int\\limits_\\frac\\pi{12}^\\frac{5\\pi}{12}\\log(4\\sin^2\\theta)d\\theta$.\r\n\r\nHence $\\int\\limits_\\frac\\pi{12}^\\frac{5\\pi}{12}\\log(4\\cos^2\\theta)d\\theta = \\dfrac12\\left(\\int\\limits_\\frac\\pi{12}^\\frac{5\\pi}{12}\\log(4\\cos^2\\theta)d\\theta+\\int\\limits_\\frac\\pi{12}^\\frac{5\\pi}{12}\\log(4\\sin^2\\theta)d\\theta\\right) = \\dfrac12\\int\\limits_\\frac\\pi{12}^\\frac{5\\pi}{12}\\log(4\\sin^22\\theta)d\\theta=\\dfrac14\\int\\limits_\\frac\\pi6^\\frac{5\\pi}6\\log(4\\sin^2\\theta)d\\theta = \\dfrac14\\int\\limits_{-\\frac\\pi3}^\\frac\\pi3\\log(4\\cos^2\\theta)d\\theta = \\dfrac12\\int\\limits_0^{\\frac\\pi3}\\log(4\\cos^2\\theta)d\\theta$.\r\n\r\nSo the problem is reduced to proving that $2\\int\\limits_0^\\frac\\pi{12}\\log(4\\cos^2\\theta)d\\theta+\\dfrac12\\int\\limits_0^\\frac\\pi3\\log(4\\cos^2\\theta)d\\theta=\\dfrac43\\int\\limits_0^\\frac\\pi4\\log(4\\cos^2\\theta)d\\theta$.\r\n\r\n\r\n[b]P.S.[/b] I devided this \"semi-solution\" into the 2 parts because I consider the part 1 as the key idea and the part 2 as one of the possible contunuations of 1.", "Solution_3": "Don't double post . :)" } { "Tag": [ "real analysis", "calculus", "integration", "function", "real analysis unsolved" ], "Problem": "Let $ g$ be a monotone increasing absolutely continuous function on $ [a,b]$ with $ g(a) \\equal{} c, g(b) \\equal{} d$. Let $ f$ be an integrable function on $ [c,d]$. Let $ F(y) \\equal{} \\int_{c}^{y} f$ and set $ H(x) \\equal{} F(g(x))$.\r\n\r\nShow that $ \\int_{c}^{d} F(y) \\; dy \\equal{} \\int_{a}^{b} F(g(x))g'(x) \\; dx$.\r\n\r\n[In previous parts to the question, I have shown that $ H'(x) \\equal{} F'(g(x))g'(x)$ a.e. except on a set $ E$ where $ g'(x) \\equal{} 0$ and also $ H'(x) \\equal{} f_{0}(g(x))g'(x)$ a.e. where $ f_{0}$ is defined by $ f_{0}(y) \\equal{} f(y)$ if $ y \\notin g[E]$ and 0 otherwise. Can these be used here?]\r\n\r\nIs the expression still true if $ g$ is not assumed monotone? What if $ g$ is bounded and $ f$ is integrable on an interval containing the range of $ E$?", "Solution_1": "The hypothesis you can't drop is absolute continuity of $ g$. Without that, $ g'$ is too badly behaved.", "Solution_2": "I came across the following example:\r\nLet $ f(x)\\equal{}1/\\sqrt{x}$ for $ x>0$ and $ f(0)\\equal{}0$. Let $ g(t)\\equal{}t^{2} \\sin(1/t)$ for $ t \\neq 0$ and $ g(0)\\equal{}0$. Then $ f$ is integrable over $ [g(0),g(\\pi/2)]$ and $ g$ is absolutely continuous on $ [0,\\pi/2]$ but the expression is not valid in this case.\r\n\r\nAny comments on this?", "Solution_3": "Okay, we need the hypothesis that $ g([a,b])\\subset [c,d]$, or that $ f$ is nice on the range of $ g$.", "Solution_4": "For the case where $ g$ is bounded on $ [a,b]$ and $ g([a,b]) \\subset [c,d]$, how would the proof be modified from the initial case where $ g$ was assumed to be monotone?", "Solution_5": "I would like to revisit the original post in this thread. How do I actually show the equality of the two integrals?" } { "Tag": [], "Problem": "Prove that for each $n\\geq 3$ we can cut equilateral triangle into $n$ pieces in such way, that each piece is equilateral or isosceles triangle.", "Solution_1": "Do you mean cutting with straitedge and compass or just to show that such cuts do exist ?\r\n\r\n...", "Solution_2": "[hide]Thee type of cuts: Let ABC be the traingle, O is center, DEF midpoints, and G,H on BC such that BG=GH=HC = a/3\nX.. make a small triangle joing the mid-points (DEF) of the sides - 4 parts, each a (smaller) equilateral triangel. \nY. Join each angle to O , center of triangle - 3 equal parts\nZ . Join OA, OB, OC, OG, OH - 5 triangles, one (OGH) equilateral.\n\nYou can always divide one equilateral trinagle into 4 by method A.\nSo X gives N = 1,4,7,.... all of the form (3k+1)\n X with one equilateral triangle using Y will give 6,.. we still have one equilateral triangle so can get 9,12,...\nX with two equilateral tringle using Y will give 8 .. we will stiall have one equilateral traingel so can get 8,11,..\n\nSo you have covered everything except n=2...[/hide]", "Solution_3": "Your solutions seems to be right. Just one thing, when you're describing Z, OGH is not equilateral (it also is not a triangle that you cut (maybe typo)), but it doesn't change anything :lol:", "Solution_4": "[quote]OGH is not equilateral[/quote]\r\nWhat am I missing OG = GH = HO = 1/3 of the length of triangle. right.\r\nYou can get 8 by \"Y\" cutting two triangles in X, then X cutting another triangle in X will give 11,14..... I should have said that.", "Solution_5": "Oh, yes, sorry, I've read your definition of O,G,H wrong :blush:" } { "Tag": [ "national olympiad" ], "Problem": "Below is the problem of SMO (Open section, Final round) on July 2nd, 2005\r\n\r\nTime allowed : 4.5 hours\r\n\r\n1. An integer is square-free if it is not divisible by $a^2$ for any integer $a>1$. Let $S$ be the set of positive square-free integers. Determine, with justification, the value of\\[\\sum_{x\\epsilon S}\\left[\\sqrt{\\frac{10^{10}}{k}}\\right]\\]where $[x]$ denote the greatest integer less than or equal to $x$.\r\n\r\n2. Let $G$ be the centroid of triangle $ABC$. Through $G$ draw a line parallel to $BC$ and intersecting the sides $AB$ and $AC$ at $P$ and $Q$ respectively. Let $BQ$ intersect $GC$ at $E$ and $CP$ intersect $GB$ at $F$. If $D$ is the midpoint of $BC$, prove that triangles $ABC$ and $DEF$ are similar.\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=43304]Problem 3[/url]\r\nLet $a,b,c$ be real numbers satisfying $a0 \\right)\\,$\r\n \r\n$ \\,\\equal{}\\frac{2ax\\plus{}b}{4\\left( n\\minus{}1 \\right).{{a}^{4}}{{\\left( a{{x}^{2}}\\plus{}bx\\plus{}c \\right)}^{n}}}\\plus{}\\frac{2\\left( n\\minus{}1 \\right)\\minus{}1}{2\\left( n\\minus{}1 \\right).{{a}^{n\\plus{}3}}}{{I}_{n\\minus{}1}}$\r\nSolution: \r\n \r\n$ *\\ \\ {{I}_{n}}\\equal{}\\int{\\frac{dx}{{{\\left( a{{x}^{2}}\\plus{}bx\\plus{}c \\right)}^{n}}}}\\text{ }\\left( where\\ 4ac\\minus{}{{b}^{2}}>0 \\right)$\r\n \r\n$ a{{x}^{2}}\\plus{}bx\\plus{}c\\equal{}a\\left( {{x}^{2}}\\plus{}\\frac{bx}{a}\\plus{}\\frac{c}{a} \\right)\\equal{}a\\left( {{\\left( x\\plus{}\\frac{b}{2a} \\right)}^{2}}\\plus{}\\frac{4ac\\minus{}{{b}^{2}}}{4{{a}^{2}}} \\right)$\r\n \r\n$ \\text{Put }u\\equal{}x\\plus{}\\frac{b}{2a}\\Rightarrow dx\\equal{}du,\\text{ }{{f}^{2}}\\equal{}\\frac{4ac\\minus{}{{b}^{2}}}{4{{a}^{2}}}\\text{ }\\left( if\\ 4ac\\minus{}{{b}^{2}}>0 \\right)$\r\n \r\n$ \\Rightarrow {{I}_{n}}\\equal{}\\frac{1}{{{a}^{n}}}\\int{\\frac{du}{{{\\left( {{u}^{2}}\\plus{}{{f}^{2}} \\right)}^{n}}}}\\equal{}\\frac{u}{2\\left( n\\minus{}1 \\right).{{a}^{n\\plus{}2}}{{\\left( {{u}^{2}}\\plus{}{{f}^{2}} \\right)}^{n}}}\\plus{}\\frac{2n\\minus{}3}{2\\left( n\\minus{}1 \\right).{{a}^{n\\plus{}2}}}{{I}_{n}}$\r\n \r\n$ \\left( {{I}_{n}}\\equal{}\\int{\\frac{dx}{{{\\left( {{x}^{2}}\\plus{}{{a}^{2}} \\right)}^{n}}}}\\equal{}\\frac{x}{2\\left( n\\minus{}1 \\right){{a}^{2}}{{\\left( {{x}^{2}}\\plus{}{{a}^{2}} \\right)}^{n}}}\\plus{}\\frac{2\\left( n\\minus{}1 \\right)\\minus{}1}{2\\left( n\\minus{}1 \\right){{a}^{2}}}{{I}_{n\\minus{}1}} \\right)$\r\n \r\n$ \\Rightarrow {{I}_{n}}\\equal{}\\int{\\frac{dx}{{{\\left( a{{x}^{2}}\\plus{}bx\\plus{}c \\right)}^{n}}}}\\equal{}\\frac{\\frac{2ax\\plus{}b}{2a}}{2\\left( n\\minus{}1 \\right).{{a}^{2}}{{\\left( a{{x}^{2}}\\plus{}bx\\plus{}c \\right)}^{n}}}\\plus{}\\frac{2n\\minus{}3}{2\\left( n\\minus{}1 \\right).{{a}^{n\\plus{}2}}}{{I}_{n}}$\r\n \r\n$ \\equal{}\\frac{2ax\\plus{}b}{4\\left( n\\minus{}1 \\right).{{a}^{3}}{{\\left( a{{x}^{2}}\\plus{}bx\\plus{}c \\right)}^{n}}}\\plus{}\\frac{2n\\minus{}3}{2\\left( n\\minus{}1 \\right).{{a}^{n\\plus{}2}}}{{I}_{n}}$\r\n \r\n$ \\Rightarrow {{J}_{n}}\\equal{}\\int\\limits_{\\minus{}\\frac{b}{2a}}^{\\plus{}\\infty }{\\frac{dx}{{{\\left( a{{x}^{2}}\\plus{}bx\\plus{}c \\right)}^{n}}}}\\equal{}\\left[ \\frac{2ax\\plus{}b}{4\\left( n\\minus{}1 \\right).{{a}^{3}}{{\\left( a{{x}^{2}}\\plus{}bx\\plus{}c \\right)}^{n}}} \\right]_{\\minus{}\\frac{b}{2a}}^{\\plus{}\\infty }$\r\n \r\n$ \\plus{}\\frac{2\\left( n\\minus{}1 \\right)\\minus{}1}{2\\left( n\\minus{}1 \\right).{{a}^{n\\plus{}2}}}\\left[ {{J}_{n\\minus{}1}} \\right]_{\\minus{}\\frac{b}{2a}}^{\\plus{}\\infty }\\equal{}\\frac{\\left( 2n\\minus{}3 \\right)}{\\left( 2n\\minus{}2 \\right).{{a}^{n\\plus{}2}}}\\left[ {{J}_{n\\minus{}1}} \\right]_{\\minus{}\\frac{b}{2a}}^{\\plus{}\\infty }$\r\n \r\n$ *\\,\\,{{I}_{1}}\\equal{}\\int{\\frac{dx}{a{{x}^{2}}\\plus{}bx\\plus{}c}}\\,\\,\\,\\,\\,\\,\\,\\,\\left( with\\ 4ac\\minus{}{{b}^{2}}>0 \\right)$\r\n \r\n$ a{{x}^{2}}\\plus{}bx\\plus{}c\\equal{}a\\left( {{x}^{2}}\\plus{}\\frac{bx}{a}\\plus{}\\frac{c}{a} \\right)\\equal{}a\\left( {{\\left( x\\plus{}\\frac{b}{2a} \\right)}^{2}}\\plus{}\\frac{4ac\\minus{}{{b}^{2}}}{4{{a}^{2}}} \\right)$\r\n \r\n$ Put\\,\\,u\\equal{}x\\plus{}\\frac{b}{2a}\\Rightarrow dx\\equal{}du,\\,\\,\\,\\,\\,\\,\\,\\,{{f}^{2}}\\equal{}\\frac{4ac\\minus{}{{b}^{2}}}{4{{a}^{2}}}\\ \\ \\ \\ \\ \\left( if\\ 4ac\\minus{}{{b}^{2}}>0 \\right)$\r\n \r\n$ \\Rightarrow {{I}_{1}}\\equal{}\\int{\\frac{du}{a\\left( {{u}^{2}}\\plus{}{{f}^{2}} \\right)}}\\equal{}\\frac{1}{a.f}.\\arctan \\frac{u}{f}\\equal{}\\frac{1}{a\\sqrt{\\frac{4ac\\minus{}{{b}^{2}}}{4{{a}^{2}}}}}.\\arctan \\left( \\frac{x\\plus{}\\frac{b}{2a}}{\\sqrt{\\frac{4ac\\minus{}{{b}^{2}}}{4{{a}^{2}}}}} \\right)$\r\n \r\n$ \\equal{}\\frac{2}{\\sqrt{4ac\\minus{}{{b}^{2}}}}.\\arctan \\left( \\frac{\\frac{2ax\\plus{}b}{2a}}{\\frac{\\sqrt{4ac\\minus{}{{b}^{2}}}}{2a}} \\right)$\r\n \r\n$ {{I}_{1}}\\equal{}\\int{\\frac{dx}{a{{x}^{2}}\\plus{}bx\\plus{}c}}\\equal{}\\frac{2}{\\sqrt{4ac\\minus{}{{b}^{2}}}}.\\arctan \\left( \\frac{2ax\\plus{}b}{\\sqrt{4ac\\minus{}{{b}^{2}}}} \\right)$\r\n \r\n$ *\\,\\,\\int\\limits_{\\minus{}b/2a}^{\\plus{}\\infty }{\\frac{dx}{{{\\left( a{{x}^{2}}\\plus{}bx\\plus{}c \\right)}^{n}}}}\\equal{}\\frac{\\left( 2n\\minus{}3 \\right)!!}{\\left( 2n\\minus{}2 \\right)!!.{{a}^{\\frac{\\left( n\\plus{}2 \\right)!}{3!}}}}.\\frac{2}{\\sqrt{4ac\\minus{}{{b}^{2}}}}.\\frac{\\pi }{2}\\text{ }\\left( with\\,\\,4ac\\minus{}{{b}^{2}}>0 \\right)$", "Solution_1": "and then,\r\n\\begin{align}\r\n & *\\,\\,\\int\\limits_{-\\frac{b}{2a}}^{+\\infty }{\\frac{{{x}^{2m}}.dx}{{{\\left( a{{x}^{2}}+bx+c \\right)}^{n+m}}}}=\\frac{\\pi \\left( 2n-3 \\right)!!}{\\left( 2n-2 \\right)!!.{{a}^{\\frac{\\left( n+2 \\right)!}{3!}}}}.\\frac{{{\\left( 4c \\right)}^{m}}\\left( 2m+1 \\right)!!}{{{2}^{m}}.{{\\left( 4ac-{{b}^{2}} \\right)}^{m}}.\\sqrt{\\left( 4ac-{{b}^{2}} \\right)}}.\\frac{\\left( n-1 \\right)!}{\\left( n+m-1 \\right)!}\\,\\,\\,\\,\\, \\\\ \r\n & \\,\\left( with\\,\\,4ac-{{b}^{2}}>0 \\right) \\\\ \r\n\\end{align}\r\nSolution: \r\n \r\n\\[ I\\left( a \\right)=\\int\\limits_{-\\frac{b}{2a}}^{+\\infty }{\\frac{dx}{{{\\left( a{{x}^{2}}+bx+c \\right)}^{n}}}}=\\frac{\\left( 2n-3 \\right)!!}{\\left( 2n-2 \\right)!!.{{a}^{\\frac{\\left( n+2 \\right)!}{3!}}}}.\\frac{2}{\\sqrt{4ac-{{b}^{2}}}}.\\frac{\\pi }{2}\\,\\,\\,\\,\\,\\,\\left( with\\,\\,4ac-{{b}^{2}}>0 \\right)\\]\r\n \r\n\\[ \\Rightarrow {{I}^{'}}\\left( a \\right)=\\left[ \\int\\limits_{-\\frac{b}{2a}}^{+\\infty }{{{\\left( a{{x}^{2}}+bx+c \\right)}^{-n}}.dx} \\right]_{a}^{'}=-n\\int\\limits_{-\\frac{b}{2a}}^{+\\infty }{\\left( a{{x}^{2}}+bx+c \\right)_{a}^{'}.{{\\left( a{{x}^{2}}+bx+c \\right)}^{-n-1}}.dx}\\]\r\n \r\n\\[ \\left( \\left[ u{{\\left( a \\right)}^{-n}} \\right]_{a}^{'}=-n.{{u}^{'}}\\left( a \\right).u{{\\left( a \\right)}^{-n-1}} \\right)\\,\\,=-n\\int\\limits_{-\\frac{b}{2a}}^{+\\infty }{\\frac{{{x}^{2}}.dx}{{{\\left( a{{x}^{2}}+bx+c \\right)}^{n+1}}}}\\]\r\n \r\n\\[ =\\frac{\\left( 2n-3 \\right)!!}{\\left( 2n-2 \\right)!!.{{a}^{\\frac{\\left( n+2 \\right)!}{3!}}}}.\\frac{\\pi }{2}.\\left[ \\frac{2}{\\sqrt{4ac-{{b}^{2}}}} \\right]_{a}^{'}\\]\r\n \r\n\\[ =\\frac{\\left( 2n-3 \\right)!!}{\\left( 2n-2 \\right)!!.{{a}^{\\frac{\\left( n+2 \\right)!}{3!}}}}.\\frac{\\pi }{2}\\left( 2.\\left( \\frac{-1}{2} \\right)\\left( 4ac-{{b}^{2}} \\right)_{a}^{'}.{{\\left( 4ac-{{b}^{2}} \\right)}^{-\\frac{1}{2}-1}} \\right)\\]\r\n \r\n\\[ =\\frac{\\left( 2n-3 \\right)!!}{\\left( 2n-2 \\right)!!.{{a}^{\\frac{\\left( n+2 \\right)!}{3!}}}}.\\frac{\\pi }{2}.\\frac{\\left( -1 \\right)4c}{{{\\left( 4ac-{{b}^{2}} \\right)}^{\\frac{3}{2}}}}\\]\r\n \r\n\\[ {{I}^{''}}\\left( a \\right)=-n\\left[ \\int\\limits_{-\\frac{b}{2a}}^{+\\infty }{\\frac{{{x}^{2}}.dx}{{{\\left( a{{x}^{2}}+bx+c \\right)}^{n+1}}}} \\right]_{a}^{'}={{\\left( -1 \\right)}^{2}}n\\left( n+1 \\right)\\int\\limits_{-\\frac{b}{2a}}^{+\\infty }{\\frac{{{x}^{2.2}}.dx}{{{\\left( a{{x}^{2}}+bx+c \\right)}^{n+2}}}}\\]\r\n \r\n\\[ =\\frac{\\left( 2n-3 \\right)!!}{\\left( 2n-2 \\right)!!.{{a}^{\\frac{\\left( n+2 \\right)!}{3!}}}}.\\frac{\\pi }{2}.\\left[ \\frac{\\left( -1 \\right)4c}{{{\\left( 4ac-{{b}^{2}} \\right)}^{\\frac{3}{2}}}} \\right]_{a}^{'}=\\frac{\\left( 2n-3 \\right)!!}{\\left( 2n-2 \\right)!!.{{a}^{\\frac{\\left( n+2 \\right)!}{3!}}}}.\\frac{\\pi }{2}.\\left[ \\frac{\\left( -1 \\right)4c\\left( -\\frac{3}{2} \\right).\\left( 4ac-{{b}^{2}} \\right)_{a}^{'}}{{{\\left( 4ac-{{b}^{2}} \\right)}^{\\frac{3}{2}}}} \\right]\\]\r\n \r\n\\[ =\\frac{\\left( 2n-3 \\right)!!}{\\left( 2n-2 \\right)!!.{{a}^{\\frac{\\left( n+2 \\right)!}{3!}}}}.\\frac{\\pi }{2}.\\frac{{{\\left[ \\left( -1 \\right)4c \\right]}^{2}}\\left( \\frac{3}{2} \\right)}{{{\\left( 4ac-{{b}^{2}} \\right)}^{\\frac{5}{2}}}}\\]\r\n \r\n\\[ \\Rightarrow {{\\left[ I\\left( a \\right) \\right]}^{\\left( m \\right)}}=\\frac{{{\\left( -1 \\right)}^{m}}\\left( n+m-1 \\right)!}{\\left( n-1 \\right)!}\\int\\limits_{-\\frac{b}{2a}}^{+\\infty }{\\frac{{{x}^{2m}}.dx}{{{\\left( a{{x}^{2}}+bx+c \\right)}^{n+m}}}}\\]\r\n \r\n\\[ =\\frac{\\left( 2n-3 \\right)!!}{\\left( 2n-2 \\right)!!.{{a}^{\\frac{\\left( n+2 \\right)!}{3!}}}}.\\frac{\\pi }{2}.\\frac{{{\\left[ \\left( -1 \\right)4c \\right]}^{m}}\\frac{\\left( 2m+1 \\right)!!}{{{2}^{m-1}}}}{{{\\left( 4ac-{{b}^{2}} \\right)}^{m}}.\\sqrt{\\left( 4ac-{{b}^{2}} \\right)}}\\]\r\n \r\n\\[ \\Rightarrow \\int\\limits_{-\\frac{b}{2a}}^{+\\infty }{\\frac{{{x}^{2m}}.dx}{{{\\left( a{{x}^{2}}+bx+c \\right)}^{n+m}}}}\\]\r\n \r\n\\[ =\\frac{\\left( 2n-3 \\right)!!}{\\left( 2n-2 \\right)!!.{{a}^{\\frac{\\left( n+2 \\right)!}{3!}}}}.\\frac{\\pi }{2}.\\frac{{{\\left( 4c \\right)}^{m}}\\left( 2m+1 \\right)!!}{{{\\left( 4ac-{{b}^{2}} \\right)}^{m}}.\\sqrt{\\left( 4ac-{{b}^{2}} \\right)}{{.2}^{m-1}}}.\\frac{\\left( n-1 \\right)!}{\\left( n+m-1 \\right)!}\\]\r\n \r\n\\[ =\\frac{\\pi \\left( 2n-3 \\right)!!}{\\left( 2n-2 \\right)!!.{{a}^{\\frac{\\left( n+2 \\right)!}{3!}}}}.\\frac{{{\\left( 4c \\right)}^{m}}.\\left( 2m+1 \\right)!!}{{{2}^{m}}.{{\\left( 4ac-{{b}^{2}} \\right)}^{m}}.\\sqrt{\\left( 4ac-{{b}^{2}} \\right)}}.\\frac{\\left( n-1 \\right)!}{\\left( n+m-1 \\right)!}\\]", "Solution_2": "So is the solution completely correct? Anyone can help me to check? Thanks" } { "Tag": [ "logarithms", "algebra unsolved", "algebra" ], "Problem": "Solve equation:\r\n$ 20^x\\plus{}11^x\\plus{}2009x\\equal{}2\\plus{}1019\\log_2(2\\plus{}29x\\minus{}27^x)$", "Solution_1": "hello, this equation has only the solutions $ x\\equal{}0$ and $ x\\equal{}1$.\r\nSonnhard.", "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]hello, this equation has only the solutions $ x \\equal{} 0$ and $ x \\equal{} 1$.\nSonnhard.[/quote]\r\n\r\nWhy? :o :o :o" } { "Tag": [], "Problem": "let $x_1$ and $x_2$ be two polatts. polatts are polygons whose vertices have interger co-ordinates on a Cartesian plane. $x_1$ and $x_2$ share a common edge $PQ$ but no other points, i.e. their intersection is the edge $PQ$.\r\n\r\nlet $i_1$, $i_2$, $b_1$, $b_2$ be the number of interior and boundary lattice (interger co-ordinate) points of $x_1$ and $x_2$ respectively.\r\n\r\nlet $k$ be the number of lattice points on $PQ$.\r\n\r\nLet $x$ be the union of $x_1$ and $x_2$ (the set of ponts belonging to $x_1$ or $x_2$ (or both)). \r\n\r\n(i) if $i$ is the number of interior lattice points of $x$ show that: $i$ $=$ $i_1$ $+$ $i_2$ $+$ $k$ $-$ $2$.\r\n\r\n(ii) if $b$ is the number of boundary lattice points of $x$, express $b$ in terms of $b_1$, $b_2$ and $k$.\r\n\r\nin this next one, $f(x, y)$ $=$ $x$ $+$ $y/2$ $-$ $1$\r\n\r\n(iii) with $i$, $b$, $i_1$, $b_1$, $i_2$ and $b_2$ as defined as above, show that:\r\n\r\n$f(i,b)$ $=$ $f( {i_1}, {b_1}) $ $+$ $f({i_2}, {b_2})$", "Solution_1": "Just a remark : Then prove the $f$ is the aera of the polygon. You will have a proof of the so-called Pick's formula. ;) \r\n\r\nPierre." } { "Tag": [ "inequalities" ], "Problem": "Let $ x,y,z\\ge0$ with $ xyz\\equal{}1$. Find the minimum value of \r\n\\[ \\frac{x}{y\\plus{}z}\\plus{}\\frac{y}{x\\plus{}z}\\plus{}\\frac{z}{x\\plus{}y}\\]", "Solution_1": "[quote=\"worthawholebean\"]Let $ x,y,z\\ge0$ with $ xyz \\equal{} 1$. Find the minimum value of\n\\[ \\frac {x}{y \\plus{} z} \\plus{} \\frac {y}{x \\plus{} z} \\plus{} \\frac {z}{x \\plus{} y}\n\\]\n[/quote]\r\nThis is Nesbitt's inequality, and the condition that $ xyz \\equal{} 1$ is superfluous.\r\n[hide=\"Solution\"]\\[ \\frac {x}{y \\plus{} z} \\plus{} \\frac {y}{x \\plus{} z} \\plus{} \\frac {z}{x \\plus{} y} \\equal{} \\frac {x^2}{yx \\plus{} xz} \\plus{} \\frac {y^2}{yx \\plus{} yz} \\plus{} \\frac {z^2}{xz \\plus{} yz}\n\\]By Titu's Lemma,\\[ \\frac {x^2}{xy \\plus{} yz} \\plus{} \\frac {y^2}{yz \\plus{} yx} \\plus{} \\frac {z^2}{zx \\plus{} zy}\\ge \\frac {(x \\plus{} y \\plus{} z)^2}{2xy \\plus{} 2yz \\plus{} 2xz} \\equal{} 1 \\plus{} \\frac {x^2 \\plus{} y^2 \\plus{} z^2}{2xy \\plus{} 2yz \\plus{} 2xz}\\]Yet, $ (x \\minus{} y)^2 \\plus{} (y \\minus{} z)^2 \\plus{} (x \\minus{} z)^2\\ge 0$, so $ x^2 \\plus{} y^2 \\plus{} z^2\\ge xy \\plus{} yz \\plus{} xz$, so $ \\frac {x^2 \\plus{} y^2 \\plus{} z^2}{2xy \\plus{} 2yz \\plus{} 2xz}\\ge \\frac {1}{2}$, so we conclude that\n\\[ \\frac {x}{y \\plus{} z} \\plus{} \\frac {y}{x \\plus{} z} \\plus{} \\frac {z}{x \\plus{} y}\\ge 1 \\plus{} \\frac {1}{2} \\equal{} \\frac {3}{2}\\][/hide]", "Solution_2": "$ (x(y\\plus{}z)\\plus{}y(z\\plus{}x)\\plus{}z(x\\plus{}y))(\\frac{x}{y\\plus{}z}\\plus{}\\frac{y}{x\\plus{}z}\\plus{}\\frac{z}{x\\plus{}y})\\ge (x\\plus{}y\\plus{}z)^2$\r\n\r\n$ \\frac{x}{y\\plus{}z}\\plus{}\\frac{y}{x\\plus{}z}\\plus{}\\frac{z}{x\\plus{}y} \\ge \\frac{(x\\plus{}y\\plus{}z)^2}{2xy\\plus{}2yz\\plus{}2xz}$\r\n$ \\equal{}1\\plus{}\\frac{\\frac{x^2\\plus{}y^2\\plus{}z^2}{xy\\plus{}xz\\plus{}yz}}{2} \\ge 1\\plus{}1/2\\equal{}3/2$", "Solution_3": "This is y method :) \r\n$ S \\equal{} \\frac{x}{{y \\plus{} z}} \\plus{} \\frac{y}{{x \\plus{} z}} \\plus{} \\frac{z}{{x \\plus{} y}}$\r\n$ S \\plus{} 3 \\equal{} \\frac{x}{{y \\plus{} z}} \\plus{} \\frac{y}{{x \\plus{} z}} \\plus{} \\frac{z}{{x \\plus{} y}} \\plus{} 3$\r\n$ S \\plus{} 3 \\equal{} \\frac{{x \\plus{} y \\plus{} z}}{{x \\plus{} y}} \\plus{} \\frac{{x \\plus{} y \\plus{} z}}{{x \\plus{} z}} \\plus{} \\frac{{x \\plus{} y \\plus{} z}}{{y \\plus{} z}}$\r\n$ S \\plus{} 3 \\ge (x \\plus{} y \\plus{} z) \\times \\frac{9}{{2(x \\plus{} y \\plus{} z)}} \\equal{} \\frac{9}{2}$\r\n$ S \\ge \\frac{3}{2}$" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Let $f: \\mathbb R \\to \\mathbb R$ be a function with the property that\n\\[f(f(x)\\minus{}f(y)) \\equal{} f(f(x)) \\minus{} y\\]\nholds true for all reals $x$ and $y$. Prove that $ f$ is an odd function.", "Solution_1": "Easy to check that $ f(x)$ is inject.\r\nSuppose exist $ a,b$ satisfy $ f(a)\\equal{}f(b)$\r\nLet $ y\\equal{}a,y\\equal{}b$ then \r\n$ f(f(x)\\minus{}f(a))\\equal{}f(x)\\minus{}a$\r\n$ f(f(x)\\minus{}f(b))\\equal{}f(x)\\minus{}b$ s\r\nbut $ f(a)\\equal{}f(b)$ then $ a\\equal{}b$\r\nLet $ y\\equal{}0$ then\r\n$ f(f(x)\\minus{}f(0))\\equal{}f(f(x))$ so $ f(0)\\equal{}0$\r\nLet $ x\\equal{}y$ then\r\n$ f(f(x))\\equal{}x$\r\nLet $ x\\equal{}0$ then\r\n$ f(\\minus{}f(y))\\equal{}\\minus{}y$\r\nLet $ y\\equal{}f(y)$ then\r\n$ f(\\minus{}f(f(y))\\equal{}\\minus{}f(y)$ \r\nBut from $ f(f(y))\\equal{}y$ we have \r\n$ f(\\minus{}y)\\equal{}\\minus{}f(y)$\r\nIt mean that $ f(x)$ is a odd function.", "Solution_2": "[quote=\"stergiu\"]If function $ f$ has the propeprty \n\n $ f(f(x) \\minus{} f(y)) \\equal{} f(f(x)) \\minus{} y$ \n\nfor every reals $ x , y$ , prove that $ f$ is an odd function.[/quote]\r\n\r\nLet $ P(x,y)$ be the property $ f(f(x) \\minus{} f(y)) \\equal{} f(f(x)) \\minus{} y$\r\n\r\n$ P(x,x)$ implies $ f(f(x))\\equal{}x\\plus{}f(0)$ and then $ f(x)$ is bijective.\r\n$ P(0,0)$ implies $ f(f(0))\\equal{}f(0)$ and, since $ f(x)$ is bijective, $ f(0)\\equal{}0$\r\n\r\nSo we have $ f(f(x))\\equal{}x$ and $ P(0,f(x))$ gives $ f(\\minus{}x)\\equal{}\\minus{}f(x)$\r\n\r\nQ.E.D.", "Solution_3": "Could we find all the functions \r\n $ f$ with propeprty \r\n\r\n $ f(f(x) \\minus{} f(y)) \\equal{} f(f(x)) \\minus{} y$ :) :?:", "Solution_4": "Let $ x\\to f(x),y\\to f(y)$ then\r\n$ f(x\\minus{}y)\\equal{}f(x)\\minus{}f(y)$ so $ f(x\\plus{}y)\\equal{}f(x)\\plus{}f(y)$ and more : \r\n$ f(f(x))\\equal{}x$\r\nBut with conditon we can find out it.", "Solution_5": "[quote=\"silouan\"]Could we find all the functions \n $ f$ with propeprty \n\n $ f(f(x) \\minus{} f(y)) \\equal{} f(f(x)) \\minus{} y$ :) :?:[/quote]\r\n\r\nWe have $ f(f(x))\\equal{}x$ and $ f(\\minus{}x)\\equal{}\\minus{}f(x)$ so $ f(f(x)\\plus{}f(y))\\equal{}x\\plus{}y$ and so $ f(x\\plus{}y)\\equal{}f(x)\\plus{}f(y)$\r\n\r\nSo the solutions are :\r\n\r\nLet $ \\mathbb{A}$ and $ \\mathbb{B}$ two supplementary $ \\mathbb{Q}$-vectorspaces of $ \\mathbb{R}$ ($ \\mathbb{A}\\cap\\mathbb{B}\\equal{}\\{0\\}$ and $ \\mathbb{A}\\plus{}\\mathbb{B}\\equal{}\\mathbb{R}$\r\n\r\nLet $ a(x)$ and $ b(x)$ the projections of $ x$ on $ \\mathbb{A}$ and $ \\mathbb{B}$ (We have in a unique manner $ x\\equal{}a(x)\\plus{}b(x)$)\r\n\r\nThen $ f(x)\\equal{}a(x)\\minus{}b(x)$ \r\n\r\nNote:\r\nWithout Axiom of Choice, the two only possibilities for $ \\mathbb{A}$ and $ \\mathbb{B}$ are ( $ \\mathbb{A}\\equal{}\\mathbb{R}$ and$ \\mathbb{A}\\equal{}\\{0\\}$) or ($ \\mathbb{A}\\equal{}\\{0\\}$ and $ \\mathbb{B}\\equal{}\\mathbb{R}$)\r\n\r\nwhich give the two solutions $ f(x)\\equal{}x$ and $ f(x)\\equal{}\\minus{}x$\r\n\r\nWith Axiom of Choice, we have infinitly many others." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $M \\in GL_{2n}(K)$, represented in block form as \\[ M = \\left[ \\begin{array}{cc} A & B \\\\ C & D \\end{array} \\right] , M^{-1} = \\left[ \\begin{array}{cc} E & F \\\\ G & H \\end{array} \\right] \\]\r\nShow that $\\det M.\\det H=\\det A$.", "Solution_1": "[hide=\"My solution\"]$\\det M.\\det H$=$\\det M$ $\\det \\left[\\begin{array}{cc} I & F \\\\ 0 & H \\end{array}\\right]$=$\\det\\[ \\left[ \\begin{array}{cc} A & o \\\\ D & I \\end{array} \\right] =\\det A$ [/hide]" } { "Tag": [ "search", "puzzles" ], "Problem": "I have this code that a professor posed to use in class. He found it on the internet and has been unable to solve it. The only clue he gave us was A = 1. \r\n\r\nHere it is. Both parts are decoded seperately: \r\n\r\nPart 1: [66-9-13-18-1-19-95-12-3-14-5-12-15-7] \r\nPart 2: [1-12-14-56-18-21-1-23-15-5-20-18-13-18-22-9-25-15-21-15-15-14-4-20]\r\n\r\n(the prof. is a big TechTV geek so it could have something to do with that!)", "Solution_1": "Try googling? :D \r\n\r\nThere are already other discussions about this going on in other places on the Internet. I typed in the first code in quotations (to search for the exact phrase) and got a few matches.\r\n\r\nOr if what you find doesn't satisfy you, you could resort to letter frequency or (eek) a program that exhausts the possibilities and displays everything it finds. The display would be really long, of course, but..." } { "Tag": [ "induction", "algebra proposed", "algebra" ], "Problem": "Denote by $d(n)$ the largest odd divisor of positive integers $n$ and define $D(n),\\ T(n)$ as follows.\r\n\\[ D(n)=d(1)+d(2)+\\cdots +d(n),\\ \\ T(n)=1+2+\\cdots +n. \\]\r\nProve that there existed infinitely positive integers $n$ such that $3D(n)=2T(n).$", "Solution_1": "It is easy to show \\[ 2T(n)=n(n+1),D(n)=\\sum_{l=0}^{\\infty } [\\frac{[n/2^l]+1}{2}]^2. \\]\r\nIt give $D(2n)=n^2+D(n),D(2n+1)=(n+1)^2+D(n).$ Therefore we can find k,m:\r\n\\[ D(2^kn+m)-D(n)=\\frac{2^{2k}-1}{3}n^2+\\frac{2^k(2m+1)-1}{3}n+\\frac{m(m+1)}{3}. \\]\r\nIt give new solution $n_{i+1}=2^kn_i+m,n_0=2$\r\nIt is simple calculate $D(2^k-2),k>1$. \r\n$3D(2^k-2)=3[(2^{k-1}-1)^2+2^{2k-4}+2^{2k-6}+...+1]=2^{2k}-32^k+2=(2^k-2)(2^k-1)=2T(2^k-2).$", "Solution_2": "[quote=\"kunny\"]Denote by $ d(n)$ the largest odd divisor of positive integers $ n$ and define $ D(n),\\ T(n)$ as follows.\n\\[ D(n) \\equal{} d(1) \\plus{} d(2) \\plus{} \\cdots \\plus{} d(n),\\ \\ T(n) \\equal{} 1 \\plus{} 2 \\plus{} \\cdots \\plus{} n.\\]\nProve that there existed infinitely positive integers $ n$ such that $ 3D(n) \\equal{} 2T(n).$[/quote]\r\nI solved it this way:\r\nAfter writing $ T(n),D(n)$ up for the first few $ n$ i find that it holds for $ n\\equal{}2,6,14$. So it seems that $ n\\equal{}2^r\\minus{}2$ is a solution for $ r \\ge 2$.\r\nI know that $ T(n) \\equal{} \\frac{n(n\\plus{}1)}{2} \\Rightarrow 2T(2^n\\minus{}2) \\equal{} (2^n\\minus{}2)(2^n\\minus{}1)$.\r\nFrom here I got the idea to prove it by induction.\r\n$ D(2^n\\minus{}2) \\equal{} d(1)\\plus{}d(2)\\plus{}d(3)\\plus{}...\\plus{}d(2^n\\minus{}2) \\equal{}$\r\n$ (d(1)\\plus{}d(3)\\plus{}d(5)\\plus{}...\\plus{}d(2^n\\minus{}3))\\plus{}(d(2)\\plus{}d(4)\\plus{}d(6)\\plus{}...\\plus{}d(2^n\\minus{}4))\\plus{}d(2^n\\minus{}2)$\r\nNow see that $ d(2n)\\equal{}d(n)$ so $ d(2)\\plus{}d(4)\\plus{}...\\plus{}d(2^n\\minus{}4) \\equal{} d(1)\\plus{}d(2)\\plus{}..\\plus{}d(2^{n\\minus{}1}\\minus{}2) \\equal{} D(2^{n\\minus{}1}\\minus{}2)$.\r\n$ d(1)\\plus{}d(3)\\plus{}...\\plus{}d(2^n\\minus{}3) \\equal{} 1\\plus{}3\\plus{}5\\plus{}..\\plus{}(2^n\\minus{}3) \\equal{} \\left ( \\frac{(2^n\\minus{}3)\\plus{}1}{2} \\right)^2 \\equal{} (2^{n\\minus{}1}\\minus{}1)^2$\r\nSo $ 3D(2^n\\minus{}2) \\equal{} 3D(2^{n\\minus{}1}\\minus{}2) \\plus{} 3(2^{2n\\minus{}2}\\minus{}2^{n\\minus{}1})$\r\n$ 2T(2^n\\minus{}2) \\equal{} (2^n\\minus{}1)(2^n\\minus{}2) \\equal{} 2^{2n}\\minus{}3 \\cdot 2^{2n}\\plus{}2 \\equal{} (2^{n\\minus{}1}\\minus{}1)(2^{n\\minus{}1}\\minus{}2) \\plus{} 3(2^{2n\\minus{}2}\\minus{}2^{n\\minus{}1}) \\equal{} 2T(2^{n\\minus{}1}\\minus{}2) \\plus{}3(2^{2n\\minus{}2}\\minus{}2^{n\\minus{}1})$\r\nSo: $ 2T(2^n\\minus{}2) \\equal{} 3D(2^n\\minus{}2) \\iff$\r\n$ 2T(2^{n\\plus{}1}\\minus{}2)\\minus{}3(2^{2n}\\minus{}2^n) \\equal{} 3D(2^{n\\plus{}1}\\minus{}2)\\minus{}3(2^{2n}\\minus{}2^n) \\iff$\r\n$ 2T(2^{n\\plus{}1}\\minus{}2) \\equal{} 3D(2^{n\\plus{}1}\\minus{}2)$. And the rest follows by induction :)", "Solution_3": "[b]Solution.[/b]\r\n\r\nWe have \\[ T(n)=1+2+\\cdots +n=\\frac{n(n+1)}{2}.\\]\r\nNow, we need to show that there are in\ffinitely many $ n$ for which $ D(n)=\\frac{m(n+1)}{3}$ so that $ 3D(n)=2T(n)$ holds. Consider \\begin{eqnarray*}\r\nD(2^n)&=&d(1)+d(3)+\\cdots+d(2^{n}-1)+d(2)+d(4)+\\cdots+d(2^n)\\\\\r\n&=& 1+3+\\cdots+(2^{n}-1)+d(1)+d(2)+\\cdots+\\d(2^{n-1})\\\\\r\n&=& 1+3+\\cdots+(2^{n}-1)+D(2^{n-1})\\\\\r\n&=& \\left [\\frac{2^n(2^n+1)}{2}-2\\frac{2^{n-1}(2^{n-1}+1)}{2} \\right ]+D(2^{n-1})\\\\\r\n&=& \\left [2^{n-1}(2^n-2^{n-1})+D(2^{n-1}) \\right ]\\\\\r\n&=& 2^{2(n-1)}+D(2^{n-1}) \r\n\\end{eqnarray*}\r\nThus, $ D(2^n)=2^{2(n-1)}+D(2^{n-1})$.\r\nFor $ n=1$, we have $ D(2^1)=2$ and we prove by induction that $ D(2^n)=\\frac{1}{3}(2^{2n}+2)$ for $ n\\geq 0$. \r\nThis holds for $ n = 0$ and for $ n = 1$. Suppose it holds for $ n = k$. Thus, $ D(2^k)=\\frac{1}{3}(2^{2k}+2)$. Then \\begin{eqnarray*}\r\nD(2^{k+1})=2^{2k}+D(2^k)&=&2^{2k}+\\frac{1}{3}(2^{2k}+2)\\\\\r\n&=& \\frac{4(2^{2k}+2)}{3}\\\\\r\n&=& \\frac{2^{2(k+1)}+2}{3}\r\n\\end{eqnarray*}\r\nand the result follows by induction. Now consider $ D(2^n-2)$. \\begin{eqnarray*}\r\nD(2^n-2)=D(2^n)-d(2^n-1)-d(2^n)\\\\\r\n&=& \\frac{1}{3}(2^{2n}+2)-(2^n-1)-1\\\\\r\n&=& \\frac{1}{3}(2^{2n}+2)-2^n\\\\\r\n&=& \\frac{1}{3}(2^{2n}-3(2^n)+2)=\\frac{(2^{2n}-1)(2^{2n}-2)}{3}.\r\n\\end{eqnarray*}\r\nThus $ D(x)=\\frac{x(x+1)}{3}$ for $ x=2^n-2$, and there are in\ffinitely many such $ x$. $ \\square$" } { "Tag": [ "geometry", "3D geometry", "puzzles" ], "Problem": "A sequence of cubes is given ,each containing a color pattern (see the attach).\r\n\r\nBelow each cube there are two codes.\r\n1) Pattern code that indicates the arrangment of the pattern on each cube.\r\n\r\n2)Transition code that indicates the changes in the patterns between adjacent elements.\r\n\r\nDetermine logic behind the transition codes and find the unique pattern code that completes the sequence and matches last transition code.\r\n\r\n\r\nHave fun", "Solution_1": "This looks like a mensa :o", "Solution_2": "[quote=\"Hexaditidom\"]This looks like a mensa :o[/quote]\r\n\r\nMaybe just looks like :alien: ;) \r\n\r\nOk,with regard to the pattern code,it's a quite trivial.\r\n3 shown sides of each cube can be viewed as 3 Cartesian planes with 3X3 facelets elements.\r\n\"Rows\" are denoted by letters{A,B,C,D,E,F,G,H,I} and \"italics\" by numbers {1,2,3}.\r\nIn that way every facelet element has a unique coordinate.\r\nOf course,overall pattern codes are given in increasing alphanumerical order.\r\n\r\nFiguring out the transition code logic is another fish in a pond.\r\nLeaving that part for brainteasers aficionados' entertainment.", "Solution_3": "Yeah, I had figured that out, but can't get the other code without some thought. I'm not exactly sure how to approach it, so I'll leave that up to someone else.", "Solution_4": "Any mathematicians that are puzzlers at the same time on this subforum?\r\n\r\n :)", "Solution_5": "Yeah, when I first saw it, but I can't think of a relationship.", "Solution_6": "Let the top be side $A$, the left be side $B$, and the right be side $C$. Let the ordered triple $(A_n,B_n,C_n)$ be the number of red squareson sides $A$, $B$, and $C$ respectively in the $n^{th}$ stage. Then the transition code under the $k^{th}$ stage is $(\\left|A_k-A_{k-1}\\right|,\\left|B_k-B_{k-1}\\right|,\\left|C_k-C_{k-1}\\right|)$. As for the last one. It's going to have one block on the left side (side $B$), but I can't figure out where. It won't be in the center, and it won't be on a corner, but that's all I can figure out.", "Solution_7": "Hey all\r\n\r\nMmmm. That's a very interesting problem. It does have a lot more to do with puzzles and even encryption than maths itself. Besides that, we all like to play a bit and for us math problems are our fun ;) Speaking of which, I recommend [url=http://www.fetchbook.info/math_puzzles.html]this place[/url] for finding great and fun math and non-math problems, that can enrich your world and just be hell of a lot of fun. \r\n\r\nBest, :) \r\nJason" } { "Tag": [ "geometry" ], "Problem": "A quadrilateral $ABCD$ with perpendicular diagonals $AC , BD$ is inscribed in a circle with center $O$. If $OH \\perp CD$ , prove that $AB = 2 OH$.\r\n\r\n[u]Babis[/u]", "Solution_1": "[hide]I suppose that you mean $H \\in CD$ (so $H$ is the midpoint of $CD$)\n\n$OH^2 + HD^2 = R^2 \\Rightarrow$\n\n$(2OH)^2 + CD^2 = 4R^2$\n\nBut from http://www.mathlinks.ro/Forum/viewtopic.php?t=58491 we know that $AB^2+CD^2 = 4R^2 \\Rightarrow$\n\n$AB = 2OH$[/hide]", "Solution_2": "Does this link really help? I think that , that problem says something else. Any way , the solution is good.I will use it as well. The official solution is based on producing DO to meet the circle at E . Then $arc CE = arc AB$. But $CE = 2OH$. \r\n\r\n [u]Babis[/u]\r\n\r\n Ah..., good luck in EME 2005(Thalles)", "Solution_3": "[quote=\"stergiu\"]Does this link really help? [/quote]\n\nSorry... I mixed up the topics :roll: \n\nI mean [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56241]this[/url] (post #4)\n\n\n[quote=\"stergiu\"]Ah..., good luck in EME 2005(Thalles) [/quote]\r\nBy the way, I'll not take part because I'm not a student anymore ...\r\nGood luck to you! :)", "Solution_4": "Now it is OK. I was sure that you wanted to write an other link. As you saw , the techniq in your link works also for the new problem. Just an observation !!!\r\n\r\n I wish you .. good Studies in the UNIVERSITY.\r\n\r\n [u]Babis[/u]" } { "Tag": [ "Alcumus", "geometry", "3D geometry" ], "Problem": "How many positive integers less than 2008 have an even number of divisors?", "Solution_1": "We can see that the only integers that have an odd number of divisors are the perfect squares. There are $ 44$ perfect squares less than $ 2008$, so we have $ 2008\\minus{}44\\equal{}\\boxed{1964}$.", "Solution_2": "I also got 1964, but alcumus said I was wrong and it was 1963!!?? :mad: :o", "Solution_3": "$2007-44=1963$", "Solution_4": "Well, see it says LESS than 2008. So you cannot include 2008. The only numbers that have an odd number of divisors are the square roots, and there are 44 square roots less than 2008. So instead of 2008, you have $2007-44=\\fbox{1963}$ integers.", "Solution_5": "[quote=\"Doink\"]I also got 1964, but alcumus said I was wrong and it was 1963!!?? :mad: :o[/quote]\n\nWell, see it says LESS than 2008. So you cannot include 2008. The only numbers that have an odd number of divisors are the square roots, and there are 44 square roots less than 2008. So instead of 2008, you have $2007-44=\\fbox{1963}$ integers.", "Solution_6": "I don't understand the solution. Can anyone help please? Thank you.", "Solution_7": "Every integer less than $2008$ has an even number of divisors or an odd number of divisors. Those with an odd number of divisors are all perfect squares. Since there are $44$ perfect squares less than or equal to $2007$, we just subtract $44$ from $2007$, and we get $\\boxed{1963}$ integers.", "Solution_8": "What about the perfect cubes???", "Solution_9": "Take $27$\n\n27 has 1,27,3,9 \n\nThe only reason that $x^2$ has even is because when you make pairs, a square number has a repeat! \n\nIf you look at $x^4 and x^8$ they are already included in squares.", "Solution_10": "[hide][img]http://data.artofproblemsolving.com/aops20/latex/images/eb1bb087edb454f5fe091e4e7a135abe67142f58.png[/img][/hide]\n\nin the above hidden text, what do they mean by \"d and n/d pair up except for......\"?", "Solution_11": "Is this the official solution?\n\nThe proof that $n$ is a perfect square $\\Leftrightarrow$ $n$ has an odd number of positive divisors could be worded better, but what it's saying is for each divisor $d$ of $n$, $n/d$ is also a divisor. If $n$ is not a perfect square, one can divide the set of divisors into two sets such that the divisors in one are less than $\\sqrt{n}$ and the divisors in the other are more than $\\sqrt{n}$. For example, the pairs for $18$ are $\\{1,18\\},\\{2,9\\},\\{3,6\\}$.\n\nOn the other hand, if $n$ is a perfect square, there is no counterpart divisor to $\\sqrt{n}$. For example, there are pairs $\\{1,36\\},\\{2,18\\},\\{3,12\\},\\{4,9\\}$ but we are left with a lone divisor, $6$.", "Solution_12": "[quote=\"AwesomeToad\"][b]Is this the official solution?[/b]\n\nThe proof that $n$ is a perfect square $\\Leftrightarrow$ $n$ has an odd number of positive divisors could be worded better, but what it's saying is for each divisor $d$ of $n$, $n/d$ is also a divisor. If $n$ is not a perfect square, one can divide the set of divisors into two sets such that the divisors in one are less than $\\sqrt{n}$ and the divisors in the other are more than $\\sqrt{n}$. For example, the pairs for $18$ are $\\{1,18\\},\\{2,9\\},\\{3,6\\}$.\n\nOn the other hand, if $n$ is a perfect square, there is no counterpart divisor to $\\sqrt{n}$. For example, there are pairs $\\{1,36\\},\\{2,18\\},\\{3,12\\},\\{4,9\\}$ but we are left with a lone divisor, $6$.[/quote]\n\nYes, that was the official solution.", "Solution_13": "Wouldn't it be 44?", "Solution_14": "I think the answer should be 1964, because 1 is counted as a perfect square but it doesn't have an even number of divisors.", "Solution_15": "You do not have to bump every single thread.", "Solution_16": "What does \"the divisors $d$ and $n/d$ pair up, except for when $d=n/d$, or $n=d^2$?\" Can someone explain?", "Solution_17": "For MasterProblemSolver123\n[hide=don't click me][hide=don't click me][hide=don't click me]n is the number, d is a divisor and n/d is another divisor, d*n/d=n, when n=d^2(it is a square) n has an odd number of divisors, the middle one is d, so technically d=n/d so d pairs up with itself[/hide][/hide][/hide]", "Solution_18": "Why would this not be 1964...", "Solution_19": "[hide=Solution]All positive integers have an even number of divisors. For example, $9$ has six divisors: $1,3,9,-1,-3,-9$. Therefore, the answer is $\\boxed{2007}$.", "Solution_20": "[quote=montresor]Why would this not be 1964...[/quote]\n\nThe answer is not $1964$ because there are only $2007$ integers less than $2008$.\n\n@above I agree with you that Alcumus is inconsistent with its definition of \u201cdivisor\u201d.", "Solution_21": "[hide=Proof]Notice a number with $n$ divisors can be formed with $x^{n-1}$, or any other factors. Notice the factors of an odd $n$ is odd, so then we have even powers. Therefore, every number with an odd number of divisors is a square.[/hide]", "Solution_22": "Doesn't this problem mean factors and not divisors?", "Solution_23": "[quote name=\"annwoo\" url=\"/community/p13520525\"]\nDoesn't this problem mean factors and not divisors?\n[/quote]\n\nThey're essentially the same.\n\n@everyone, I checked their profile, they're clearly active so don't accuse me of replying to an over 1y old post.\n\nI got james bond on this problem :| my clock said it was 9:08pm pst instead of 9:07 though\n\n[hide=shruggie]so we guess a number whose square is around 2008\n\nwe guess 45\n\n45\u00b745=2025, oops slightly too big\n\nso there are 44 squares under 2008, so we subtract 2007-44 to get 1963 :}[/hide]" } { "Tag": [ "geometry", "circumcircle", "cyclic quadrilateral", "geometry solved" ], "Problem": "The diagonals $AC$ and $BD$ of a cyclic quadrilateral $ABCD$ meet at a point $X$. The circumcircles of triangles $ABX$ and $CDX$ meet at a point $Y$ (apart from $X$). Let $O$ be the center of the circumcircle of the quadrilateral $ABCD$. Assume that the points $O$, $X$, $Y$ are all distinct. Show that $OY$ is perpendicular to $XY$.", "Solution_1": "Let $P=AB\\cap CD,Q=AD\\cap BC$, and consider the inversion of pole $P$ which invariates the circle $(ABCD)$. Let $O'$ be the image of $O$. $O'$ must lie on the polar of $P$ wrt the circle (draw a circle, an exterior point, its polar, and you'll see why :)). This polar is $XO'=QX\\perp PO=OO'\\ (*)$, so $PO\\perp O'X$. Now try to show that the image of $X$ through this inversion is $Y$ (just consider the power of $P$ wrt the circles $(ABX),(CDX)$, which must be the same as its power wrt $(ABCD)$), so $XYOO'$ is cyclic, and from $(*)$ it now follows that $XY\\perp OY$.", "Solution_2": "Grobber,\r\nthat's brilliant! How do you come up with these solutions?\r\n\r\n[Moderator edit: See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=151499#151499]http://www.mathlinks.ro/Forum/viewtopic.php?t=4919 post #7[/url] for a solution of this problem.]", "Solution_3": "There was a time when basicly all I did was solve problems such as the ones you've posted here lately :).", "Solution_4": "Well, it certainly seems to have paid off.", "Solution_5": "Firstly, AYOD is cyclic and also BYOC is cyclic. Then 1.\r\nlet p the smallest prime divisor of x. let x = pA.\r\n\r\nwe have that $p|3^{2pA}-1$ and $p|3^{p-1}-1$, so, $p|3^{k}-1$, where k = gcd(2pA,p-1).\r\n\r\nbut k = 2, because gcd(p-1, pA) = 1. so, p = 2.\r\n\r\nnow the right side is a multiple of 4, and the left side is $\\equiv 2 (mod.4)$. :wink:", "Solution_2": "[quote=\"e.lopes\"]suppose x>1.\nlet p the smallest prime divisor of x. let x = pA.\n\nwe have that $p|3^{2pA}-1$ and $p|3^{p-1}-1$, so, $p|3^{k}-1$, where k = gcd(2pA,p-1).\n\nbut k = 2, because gcd(p-1, pA) = 1. so, p = 2.\n\nnow the right side is a multiple of 4, and the left side is $\\equiv 2 (mod.4)$. :wink:[/quote]\r\nsolution $(1,4)$ works no? :idea:", "Solution_3": "[quote=\"yeppyyep\"][quote=\"e.lopes\"]suppose x>1.\nlet p the smallest prime divisor of x. let x = pA.\n\nwe have that $p|3^{2pA}-1$ and $p|3^{p-1}-1$, so, $p|3^{k}-1$, where k = gcd(2pA,p-1).\n\nbut k = 2, because gcd(p-1, pA) = 1. so, p = 2.\n\nnow the right side is a multiple of 4, and the left side is $\\equiv 2 (mod.4)$. :wink:[/quote]\nsolution $(1,4)$ works no? :idea:[/quote]\r\n\r\nin the begin i say: 'suppose x>1'", "Solution_4": "It's ok :D", "Solution_5": "I shall remark in ZetaX's name that this thread hasn't good a title (also look at http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 ) but this thread won't be deleted because you don't have enough posts. (he is at my house that moment)" } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "Let $A\\in M_n(R)$.Then $A$ is nilpotent if and only if there exists a sequence of matrices similar to $A$ with nul limit.", "Solution_1": "Using well known result : $\\forall k,Tr(A^k)=0 \\Rightarrow A$ is nilpotent\r\n\r\nsketch of proof : by Cayley-Hamilton and linearity of $Tr$, $det(A) = 0$, i.e $0$ is an eigenvalue, take a stable hyperplane and proceed by induction.\r\n\r\n\r\nIf A is nilpotent then just take $A_n = A$ as sequence \r\n\r\nIf $A = lim A_n = {P_n}^{-1} \\times A \\times {P_n}$ with $lim A_n= 0$\r\nThen $A^k = lim \\ A_n^k$\r\nSince $tr(A^k) = tr({A_n}^k) \\rightarrow 0$, $tr(A^k) = 0$ and $A$ is nilpotent." } { "Tag": [ "inequalities", "geometry", "trigonometry", "inequalities proposed" ], "Problem": "Let $a,b,c$ be three positive real numbers with $a+b+c=3$. Prove that \\[ (3-2a)(3-2b)(3-2c) \\leq a^2b^2c^2 . \\]\r\n[i]Robert Szasz[/i]", "Solution_1": "This is equivalent to\r\n\r\n$(a+b+c)^3(-a+b+c)(a-b+c)(a+b-c) \\leq 27a^2b^2c^2$\r\n\r\nLet $x = -a + b + c, y = a-b+c, z = a+b-c$ and note that at most one of $x,y,z$ can be negative (since the sum of any two is positive). If $x < 0$ then $(a+b+c)^3xyz \\leq 0 < 27a^2b^2c^2$. If $x,y,z > 0$, note $x+y+z = a+b+c, x+y = 2c$ etc so our inequality becomes\r\n\r\n$64xyz(x+y+z)^3 \\leq 27(x+y)^2(y+z)^2(z+x)^2$\r\n\r\nNote that $9(x+y)(y+z)(z+x) \\geq 8(x+y+z)(xy+yz+zx)$ and $(xy+yz+zx)^2 \\geq 3xyz(x+y+z)$. Combining these completes our proof!", "Solution_2": "I think this can also be proved using triangle.\r\n\r\nAs noted in the last post, we may assume $a,b,c$ are the sides of a triangle. Multiplying LHS by $a+b+c$ and RHS by $3$, the inequality becomes $16\\Delta^2 \\le 3a^2b^2c^2$, where $\\Delta$ is the area of the triangle. That is equivalent to $R^2\\ge \\frac{1}{3}$ since $\\Delta = \\frac{abc}{4R}$, where $R$ is the circumradius. But this is true since $R=\\frac{a+b+c}{2(\\sin{A}+\\sin{B}+\\sin{C})}$ and $\\sin{A}+\\sin{B}+\\sin{C} \\le \\frac{3\\sqrt{3}}{2}$ by Jensen.", "Solution_3": "Let x=3-2a y=3-2b ve z=3-2c than inequality is\r\n\r\nxyz is less than or equal to (x+z)^2(y+z)^2(x+y)^2 / 8 \r\n \r\nBut we now from AM-GM xyz is less than or equal to (x+z)(y+z)(x+y) / 8 and (x+z)(y+z)(x+y) is less than or equal to (x+y+z)^3 / 3^3 .\r\n\r\nAnd we now that x+y+z=a+b+c=3, so that (x+z)(y+z)(x+y) is less than or equal to 1. \r\n\r\nConclusion follows....\r\n\r\nSorry for my poor english........", "Solution_4": "[quote=\"Valentin Vornicu\"]Let $ a,b,c$ be three positive real numbers with $ a \\plus{} b \\plus{} c \\equal{} 3$. Prove that\n\\[ (3 \\minus{} 2a)(3 \\minus{} 2b)(3 \\minus{} 2c) \\leq a^2b^2c^2 .\n\\]\n[i]Robert Szasz[/i][/quote]\r\n$ F(a,b,c)\\equal{}a^{2}b^{2}c^{2}\\minus{}(3\\minus{}a)(3\\minus{}b)(3\\minus{}c)$\r\n$ F(\\frac{a\\plus{}b}{2},\\frac{a\\plus{}b}{2},c)\\equal{}\\frac{(a\\plus{}b)^{4}}{16}c^{2}\\minus{}(3\\minus{}(a\\plus{}b))^{2}(3\\minus{}2c)$\r\n$ F(\\frac{a\\plus{}b}{2},\\frac{a\\plus{}b}{2},c)\\minus{}F(a,b,c)\\equal{}(a\\minus{}b)^{2}[\\frac{c^{2}}{16}(a^{2}\\plus{}b^{2}\\plus{}6ab)\\plus{}2c\\minus{}3]$\r\nWe have $ \\frac{c^{2}}{16}(a^{2}\\plus{}b^{2}\\plus{}6ab)\\plus{}2c\\minus{}3\\leq 2\\frac{c^{2}}{16}(3\\minus{}c)^{2}\\plus{}2c\\minus{}3$\r\nwe suppose c=min{a,b,c}\r\n$ \\Rightarrow F(a,b,c)\\geq F(x,x,c)$ (x=a+b/2)\r\nwe need prove :\r\n$ F(c)\\equal{}\\frac{1}{16}c^{2}(3\\minus{}c)^{4}\\minus{}c^{2}(3\\minus{}2c)\\geq 0$\r\n$ G(c)\\equal{}\\frac{1}{16}(3\\minus{}c)^{4}\\minus{}(3\\minus{}2c)$\r\nG'(c)<0 $ \\Rightarrow G(c)\\geq G(1)\\equal{}0$\r\nDone !", "Solution_5": "Use sin 0$, note $x+y+z = a+b+c, x+y = 2c$ etc so our inequality becomes\n\n$64xyz(x+y+z)^3 \\leq 27(x+y)^2(y+z)^2(z+x)^2$\n\nNote that $9(x+y)(y+z)(z+x) \\geq 8(x+y+z)(xy+yz+zx)$ and $(xy+yz+zx)^2 \\geq 3xyz(x+y+z)$. Combining these completes our proof[/hide]![/quote]\n$$(a+b)(b+c)(c+a)(b+c-a)(c+a-b)(a+b-c) \\leq 8a^2b^2c^2$$https://artofproblemsolving.com/community/c6h1785721p11793688\n", "Solution_7": "Let $x = a+b-c$,$y = b+c-a$ and $z = c+a-b$ Note that we need to prove $xyz \\le ((\\frac{x+z}{2})(\\frac{y+x}{2})(\\frac{z+y}{2}))^2$ Note that if one of $x,y,z$ is negative then inequality is obvious so assume $x,y,z \\ge 0$ and also Note that $x+y+z = 3$. Note that we need to prove $64xyz \\le ((x+z)+(y+x)+(z+y))^2$ or $81xyz \\le ((x+y+z)(xy+yz+zx))^2$ or $9xyz \\le (xy+yz+zx)^2$ which is true since $(xy+yz+zx)^2 \\ge 3(x+y+z)xyz$ and $x+y+z = 3$." } { "Tag": [ "induction", "function" ], "Problem": "Let $f: \\mathbb{N}\\rightarrow \\mathbb{N}$ such that $f(n+1) > f(n)$ and $f(f(n))=3n$ for all $n \\in \\mathbb{N}.$ Determine $f(1992).$\r\n\r\n[Note: If you know the solution, please don't write it. Just provide a hint or two to those who wish to attempt to solve it.]", "Solution_1": "$f(n) = \\sqrt{3}n$\r\nhence $f(1992) = 1992\\sqrt{3}$", "Solution_2": "[quote=\"ashwinrk_jain\"]$f(n) = \\sqrt{3}n$\nhence $f(1992) = 1992\\sqrt{3}$[/quote]\r\n\r\nthe range of $f(x)$ is integers greater than 0, so that does not work\r\n\r\n\r\nEDIT: oh and a hint:\r\n\r\n[hide=\"...\"]consider $f(f(f(n)))$[/hide]", "Solution_3": "$f^{2k}(n) = 3^{k}n, f^{2k+1}(n) = 3^{k}f(3n)$\r\n\r\nAlso, you're basically trying to find what x is s.t. $f(x) = 5976$..", "Solution_4": "i don't have the time to think about this but in the 10 min i looked over it i got the following:\r\nlet $3*3^{k}>n \\ge 3^{k}$ then we have $f(n)=n+3^{k}$ if $2*3^{k}\\ge n$ and\r\n$n \\ge 3^{k}$.else we have $f(n)=3n-3*3^{k}$.i guess we can prove this by induction :wink: .if i'll have the time i will post my entire solution later.so $f(1992)=3789$.but i did not have the time to check this so it might be wrong.", "Solution_5": "[hide=\"here is my progress:\"]Using the equation, we get: \\[f(f(f(n)))=f(3n)=3f(n)\\] If for natural numbers $n$ and $m$, $n>m$, we can show that $f(n)>f(m)$ through induction. Therefore, the fucntion is one-to-one and has an inverse.\n\n\nAlso I have been looking at the fact that $f(f(n))$ is divisibly by 3, but I am not sure if that helps. [/hide]", "Solution_6": "i have checked what i wrote above and it is correct.if anyone wants i'll post my solution or some more hints" } { "Tag": [ "AMC" ], "Problem": "Four positive integers $ a,b,c,$ and $ d$ have a product of 8! and satisfy\\begin{align*}ab \\plus{} a \\plus{} b &\\equal{} 524\\\\\nbc \\plus{} b \\plus{} c &\\equal{} 146\\\\\ncd \\plus{} c \\plus{} d &\\equal{} 104.\\end{align*} What is $ a \\minus{} d$?\n\n$ \\textbf{(A)} \\ 4 \\qquad \\textbf{(B)} \\ 6 \\qquad \\textbf{(C)} \\ 8 \\qquad \\textbf{(D)} \\ 10 \\qquad \\textbf{(E)} \\ 12$", "Solution_1": "[hide]add $1$ to both sides of each equation and factor to get:\n$(a+1)(b+1)=525=3*5^2*7$\n$(b+1)(c+1)=147=3*7^2$\n$(c+1)(d+1)=105=3*5*7$\nDivide the first by the second\n$\\frac{a+1}{c+1}=\\frac{25}{7}$\nthus $a+1$ is a multiple of $25$, and $c+1$ is a multiple of $7$\ntrying out $25$ and $7$ for these, we get $b+1 = 21$, and $d+1=15$\nin this case, $abcd=24*20*6*14=8*3*5*4*6*7*2=8!$, so $24-14=\\boxed{10 (D)}$[/hide]", "Solution_2": "[hide=\"solution\"]\nThis solution's a little messy, but here it goes:\nNote that we can factor these equations as\n$(a + 1)(b + 1) = 525$\n$(b+1)(c+1) = 147$\n$(c + 1)(d + 1) = 105$\n\nNow let $a + 1 = a'$, and similarly for the other variables. We know that because $a,b,c,d$ are positive integers, $a', b', c', d'$ must also be positive integers greater than one such that\n\n$a' b' = (3)(5^2)(7)$\n$b' c' = (3)(7^2)$\n$c' d' = (3)(5)(7)$\n\nLooking at the equations quickly shows that there are two possible solutions, depending on how we distribute the 3s to the variables: either $a' = (3)(5^2), b' = 7, c' = (3)(7), d' = 5$, or $a' = 5^2, b' = (3)(7), c' = 7, d' = (3)(5)$. Calculating $abcd$ from these solutions shows that the second set is the correct one, and we have $a - d = 24 - 14 = 10$, or $(D)$.\n[/hide]", "Solution_3": "Edit: Never mind, I thought the equations were symmetric. :blush:", "Solution_4": "[hide=\"A Number Theoretic Approach\"]So start off with:\n\n$(a+1)(b+1) = 5^{2}\\cdot 3 \\cdot 7$\n$(b+1)(c+1) = 7^{2}\\cdot 3$\n$(c+1)(d+1) = 3 \\cdot 5 \\cdot 7$\n\nNote that we have $3$ possible values for $b+1$ by observing the first two equations.\n\n[b]Case 1: b + 1 = 3[/b]\nA contradiction arises in the third equation, as this implies that $c+1 = 7^{2}.$\n\n[b]Case 2: b + 1 = 7[/b]\nSolving, we get $a = 74, b = 6, c = 20, d = 4.$ However, $37|a$ and $37$ is a prime that doesn't go into $8!.$ Contradiction.\n\n[b]Case 3: b + 1 = 7 \\cdot 3[/b]\nThis case works out well. Solving, we get $a = 24, b = 20, c = 6, d = 14.$ Subtracting yields $a-d = 10.$[/hide]", "Solution_5": "I understand the solution, but what I don't understand is how it occurred to yall to separate 525 into 3(5^2)(7), 147 into 3(7^2), and 105 into 3(5)(7). What I'm really asking is if there is a series of steps that ran through your minds that allowed you to see that factoring those out would get you further in the problem. Thanks for your help!\r\n\r\nMary Catherine", "Solution_6": "I think you're saying that you wonder how they thought of prime factorizations? The whole idea of products sharing similar factors. If it's not obvious to you at first, and you want to be able to see it quicker, it's the same as learning when to apply other theorems, tips, and tricks. Practicing those types of problems.", "Solution_7": "Whenever you're trying to solve problems which are restricted to integers you should always try to factor as much as possible.", "Solution_8": "Wait this problem is really troll because on the pdf it shows everything on one line, so I misread it as $ab+a+b = 524bc+b+c =146cd+c+d = 104$ because that's what it showed up as... smh. Why does the formatting always mess up in the pdfs?!?!?!" } { "Tag": [ "Functional Analysis", "calculus", "calculus computations" ], "Problem": "Consider the Banach space $ l^1$. Let $ S\\equal{}\\{x\\in l^1| \\parallel{}x\\parallel{}\\leq1\\}$. Is $ S$ a compact subset of $ l^1$", "Solution_1": "If I understand the problem correctly, no; it is easy to exhibit sequences with no convergent subsequences, like the sequence $ s_k$ where each entry is zero except for a one in the $ k^{th}$ position. Every element of the sequence is at a distance $ 2$ from every other element.", "Solution_2": "Inappropriate to this forum [b]and[/b] very recently posted in another forum. countdownmath, please stop posting such things. I'm about to delete several of your other posts." } { "Tag": [ "probability", "calculus", "integration", "expected value" ], "Problem": "Here's a problem I came up with. For some reason I haven't solved it yet (at least not well enough to be certain of my answer).\r\n\r\nI have an unfair coin. p of the time it lands heads and 1-p of the time it lands tails. I flip it n times, and I get all tails except that on the last flip I get heads. What is the expected value of p?", "Solution_1": "Well I have no idea if this is how expected values are usually found out over continuous ranges but here we go anyway..\r\nFirst switch p and 1-p around to make it easier :).\r\nFor a given p the probability of that happening is y=(1-p)p^(n-1).\r\nSo to find expected value of p (or 1-p), we want:\r\nintegral from 1 to 0 of ((1-p)^2 p^(n-1)) dp) over the integral from 1 to 0 of ((1-p)p^(n-1)). Or basically, the sum of all py over the sum of all p.\r\n\r\nAfter a bit of algebra this is (1/n+2 - 2/n+1 + 1/n)/(1/n(n+1))\r\n= 2/n+2.\r\n\r\nSo the expected value of p as far as I can work is out is n/n+2. Interestingly this isn't equal to n-1/n which you would probably expect.. unless I just made a mistake!", "Solution_2": "umm... you could try using Bernoulii's Formula. It finds the probability of an event happening a specific number of times out of a total number of trials, i.e. flipping a coin 10 times and getting exactly 7 heads. The formula is:\r\nnCr * p^r * q^n-r\r\nwhere n is the total number of trials, r is the desired number of specific events, p is the probability of the event happening, and q is the probability of the event not happening (usually 1-p.) If you put the values of your problem into the formuloa, you get:\r\nnCn-1 * p^n-1 * r^n-n-1\r\nnCr for any number where r is one less than n simplifies to n.\r\nTherefore, you can simplify it to:\r\nn * p^n-1 * 1/r\r\nDon't forget, r is p-1.\r\nTherefore, as far as I can simplify, it comes to:\r\n\r\n[u]n * p^n-1[/u]\r\n p-1", "Solution_3": "This finds out the probability of geting n-1 tails and 1 head in any order. Since we know the head is last, we drop the nC1. Also watch out when you are subtracting n and n-1, their difference is 1 not -1.\r\n\r\nI used this formula when working out the expected value of p.. see my other post. (in which I think I mixed up a 1-p and p somewhere but the answer still comes to n/n+2.)", "Solution_4": "I stand corrected. Thanks.\r\n\r\n\r\n\r\n-Isaac", "Solution_5": "So whats the answer?" } { "Tag": [ "calculus", "integration", "SFFT", "blogs", "special factorizations" ], "Problem": "How many distinct positive integral solutions of the equation $ a\\plus{}b\\plus{}c\\plus{}ab\\plus{}bc\\plus{}ac \\equal{}89$ are possible?", "Solution_1": "[hide=\"Hint\"] A variant of SFFT might be handy. [/hide]", "Solution_2": "Forgive me, but what's SFFT?", "Solution_3": "[hide=\"In its most specific case, this\"] \"Simon's Favorite Factoring Trick,\" as it is often called:\n\n$ ab\\plus{}a\\plus{}b \\equal{} (a\\plus{}1)(b\\plus{}1)\\minus{}1$ [/hide]", "Solution_4": "[quote=\"t0rajir0u\"][hide=\"Hint\"] A variant of SFFT might be handy. [/hide][/quote]\r\n\r\nWouldn't an $ abc$ term be needed for that here? :maybe:", "Solution_5": "Hm. I've got a solution involving SFFT, but it is not satisfying at all.\r\n\r\n\r\n[hide=\"Major casework bash\"]\n\nWLOG $ a\\leq b\\leq c$. It is easy to see that $ a\\leq4$.\n\n\n$ a \\equal{} 1$:\n\n$ bc\\plus{}2b\\plus{}2c\\plus{}1 \\equal{} 89$\n$ (b\\plus{}2)(c\\plus{}2) \\equal{} 92$\n\n$ (1,2,21)$\n\n\n\n$ a \\equal{} 2$:\n\n$ bc\\plus{}3b\\plus{}3c\\plus{}2 \\equal{} 89$\n$ (b\\plus{}3)(c\\plus{}3) \\equal{} 96$\n\n$ (2,3,13)$\n$ (2,5,9)$\n\n\n\n$ a \\equal{} 3$:\n\n$ bc\\plus{}4b\\plus{}4c\\plus{}3 \\equal{} 89$\n$ (b\\plus{}4)(c\\plus{}4) \\equal{} 102$\n\nNo solutions.\n\n\n\n$ a \\equal{} 4$:\n\n$ bc\\plus{}5b\\plus{}5c\\plus{}4 \\equal{} 89$\n$ (b\\plus{}5)(c\\plus{}5) \\equal{} 110$\n\n$ (4,5,6)$\n\n\n\n\nSo we have four unique solutions:\n\n$ (1,2,21)$\n$ (2,3,13)$\n$ (2,5,9)$\n$ (4,5,6)$\n[/hide]", "Solution_6": "I don't know how much this will help; just throwing out an idea?\r\n\r\n[hide]\n$ \\sum_{cyc}ab\\plus{}a \\equal{}\\sum_{cyc}ab\\plus{}\\frac{a}{2}\\plus{}\\frac{b}{2}\\equal{}\\sum_{cyc}\\left( a\\plus{}\\frac{1}{2}\\right)\\left( b\\plus{}\\frac{1}{2}\\right)\\minus{}\\frac{1}{4}\\equal{}\\frac{1}{4}\\left(\\sum_{cyc}(2a\\plus{}1)(2b\\plus{}1)\\minus{}1\\right)$\n\nSo substituting $ 2a\\plus{}1 \\equal{} A$, etc, it becomes:\n$ AB\\plus{}BC\\plus{}CA \\equal{} 359$, solve for triple of odd integers (A,B,C).\nThe thing is, though, I don't know if this is any simpler :( \n\nI suppose you could then be like, $ (A\\plus{}C)(B\\plus{}C) \\equal{} 359\\plus{}C^{2}$, but I don't know if that'll help either.\n\nSomehow, I have a feeling that no matter what we do, it'll boil down to casework of the same nature as that presented above.\n[/hide]", "Solution_7": "This looked promising at first but then I'm stuck...\r\n[hide]Major step: add 1.\n$ a\\plus{}b\\plus{}c\\plus{}ab\\plus{}ac\\plus{}bc\\plus{}1 \\equal{} 90$\nNow we can do\n$ (b\\plus{}1)(c\\plus{}1)\\plus{}a(b\\plus{}c\\plus{}1) \\equal{} 90$ and so on...\nthen... um....\nmaybe $ (a\\plus{}1)(b\\plus{}c\\plus{}1)\\plus{}ab\\equal{}90$\n\n\n[/hide]", "Solution_8": "If that's your approach, t0rajir0u's hint suggests adding $ abc$ to both sides as well. This gets you $ (a\\plus{}1)(b\\plus{}1)(c\\plus{}1) \\equal{} 90\\plus{}abc$. Looks all nice and clean, but I failed to do anything with it. Maybe it would make the case bash easier though.", "Solution_9": "[quote=\"MellowMelon\"]If that's your approach, t0rajir0u's hint suggests adding $ abc$ to both sides[/quote]\r\n\r\nActually, I was thinking about what The Zuton Force did. If nothing else, it makes the casework slightly more straightforward.", "Solution_10": "Thanks, t0rajir0u :)", "Solution_11": "Why is it called \"Simon's Favorite Factoring Trick\"? And who is \"Simon\"?", "Solution_12": "I believe Simon is one of the older members of this forum, and apparently this trick is Simon's favorite factoring trick.", "Solution_13": "Simon is [url=http://www.artofproblemsolving.com/Community/AoPS_Y_IndBlog.php?blog_id=1233]ComplexZeta[/url].", "Solution_14": "[[Simon's Favorite Factoring Trick]] includes a link to the thread on which the name is based." } { "Tag": [ "geometry", "trapezoid", "geometry unsolved" ], "Problem": "It is given a trapezium ABCD it is known that AD=BC. In the triangles ACD and ABC we know that the radii of incircles are r1 and r2 respectively. It is known that we can inscribe a circle in ABCD. Find the radii (r) of inscribed circle in ABCD in therms of r1 and r2.", "Solution_1": "The answer given is $ r\\equal{}\\frac{r_1\\plus{}r_2\\plus{}\\sqrt{r_1^2\\plus{}r_2^2}}{2}$\r\nI can express r_1, r_2 and r in therms of a, and b (bases of the trapezoid). But there are long expressions.\r\nIs there a better approach?\r\nI also considered the tangents formed by k1 and k2 and the bases of the trapezium but i still don't have a good results.\r\n\r\nHow to attack this problem?", "Solution_2": "The problem is solved by me and two more people on a Bulgarian math forum. My question above is not relevant." } { "Tag": [], "Problem": "Photos from IMO 2008 Madrid, Espa\u00f1a", "Solution_1": "The opening ceremony", "Solution_2": "With [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=29491]ElChapin[/url] and his team :)", "Solution_3": "With the Cambodian team", "Solution_4": "With the Chinese team", "Solution_5": "Our Bangladesh team. From left to right: Alba (our guide), me, Ishfaq, Moon, Nazia.", "Solution_6": "Well, me with two Croatian contestants :P", "Solution_7": "The Great Exam Hall :)\r\n\r\n(From the IMO official website)", "Solution_8": "With a contestant from Uzbekistan", "Solution_9": "Ishfaq and me with [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=32134]bambaman[/url] :)", "Solution_10": "Us with David, one of the guides :)", "Solution_11": "Us with another guide", "Solution_12": "With two contestants from Cyprus and Greece", "Solution_13": "With the Saudi team", "Solution_14": "", "Solution_15": "In the pool :)", "Solution_16": "Me, Ishfaq and Nazia", "Solution_17": "", "Solution_18": "Ishfaq, Moon and me", "Solution_19": "Us with the Indonesian team, before going to the closing" } { "Tag": [ "MATHCOUNTS", "blogs", "algebra", "polynomial", "Vieta", "calculus", "conics" ], "Problem": "OK, now that everyone has submitted their answers, it's time to release the final results!! :D \r\n\r\n[hide=\"In fourth place, with 34 points, is...\"]\n[size=100]PI-Dimension[/size]!\n[/hide]\n[hide=\"In third place, with 35 points, is...\"]\n:winner_third: [size=134]Bachukas[/size]!\n[/hide]\n[hide=\"In second place, with 37 points, is...\"]\n:winner_second: [size=167]andersonw[/size]!!\n[/hide]\n[hide=\"And the champion of the 2008 Yongyi National MATHCOUNTS Competition, with 39 points, is...\"]\n:winner_first: [size=200]tinytim[/size]!!!\n[/hide]\n\nCongratulations to all who participated... you did an amazing job on my mock test!\n\nSpecial thanks to tinytim, who donated his time to proofread my test and fix semantical errors! \n\nA full analysis of individual contestants, as well as the total number of people who got a certain problem correct, as well as the problems in order of difficulty, can be found in my Excel file.\n\nThe answer key can also be found below.\n\nNote: Isabella2296 requested that her score not be shown, therefore you will not find a row containing her score.\n\nAt this point, feel free to discuss problems, such as your favorite (and least favorite) problems!! :lol:\n\n[hide=\"Interesting information\"]\n[list]\n[*]Nobody correctly answered #25 on the sprint (Answer: [b]299[/b]) or #8 on the target (Answer: [b]209.06[/b]).\n[*]Andersonw and Bachukas were the only two people to correctly answer #6 on the sprint (Answer: [b]34[/b]).\n[*]Andersonw, math154, and xpmath were the only three people to correctly answer #11 on the sprint (Answer: [b]128[/b]).\n[*]Tinytim, 27r, and isabella2296 were the only three people to correctly answer #27 on the sprint (Answer: $ \\frac 43$)[/list]\n[/hide]", "Solution_1": "The test was even harder than nats so i desperately failed :( \r\nim gonna go to my blog crying why do i have to be so stupid :(", "Solution_2": "Don't worry. There's a reason I didn't want my score to be shown. :roll:", "Solution_3": "Ha!!! I predicted that I would lose to tinytim. :lol: \r\n\r\nAnyone know how to do #25?\r\n\r\nI thought #24 would have been a better question if the polynomial wasn't factorable (unless if vieta's formulas aren't mathcounts level, I'm not sure).\r\n#2 on the target was pretty cool, and #3 on the target was very tricky.\r\nI still don't understand #8 on the target :P", "Solution_4": "It is very wrong to be dead-last when your score is substantially lower than the person second-dead-last.", "Solution_5": "#29 on the sprint was fun.\r\n\r\nArgh, stupid target. Well, more like stupid me, since I only turned in 3 target answers.", "Solution_6": "i liked #20.\r\n\r\nidk if theres an other way to do it, but i used calculus to solve it :rotfl:", "Solution_7": "For #20, I assumed the center of the bottom of the tunnel was at (0,0).\r\nThen, the equation of the archway would be $ f(x)\\equal{}25\\minus{}\\frac{1}{4}x^2$ since it's a parabola with a y-intercept of 25 and x-intercepts of 10 and -10.\r\nThen, plug in 4 because the train is 8 feet wide, so $ 25\\minus{}\\frac{16}{4}\\equal{}\\boxed{21}$", "Solution_8": "The target really pwned me so I only got 4 problems right :( \r\n\r\n*gasp* YAY 1000th POST!!!!!!!!! :D", "Solution_9": "I just decided to pick random numbers for #2 on the target and on my 4th try, I got it!! :rotfl:", "Solution_10": "[quote=\"andersonw\"]For #20, I assumed the center of the bottom of the tunnel was at (0,0).\nThen, the equation of the archway would be $ f(x) \\equal{} 25 \\minus{} \\frac {1}{4}x^2$ since it's a parabola with a y-intercept of 25 and x-intercepts of 10 and -10.\nThen, plug in 4 because the train is 8 feet wide, so $ 25 \\minus{} \\frac {16}{4} \\equal{} \\boxed{21}$[/quote]\r\n\r\nheres how i solved:\r\n\r\ni assumed the lower left corner of the tunnel was at (0,0), and since the parabola is passing the origin, it must have the form of $ f(x) \\equal{} ax^2 \\plus{} bx$.\r\n\r\n$ f(20) \\equal{} 400a \\plus{} 20b \\equal{} 0$\r\n$ f(10) \\equal{} 100a \\plus{} 10b \\equal{} 25 \\implies 400a \\plus{} 40b \\equal{} 100 \\implies 20b \\equal{} 100 \\implies b \\equal{} 5, a \\equal{} \\minus{} \\frac {1}{4}$\r\n\r\nso $ f(x) \\equal{} \\minus{} \\frac {1}{4}x^2 \\plus{} 5x$\r\n\r\nwere looking for $ f(6)$ and its $ \\minus{} 9 \\plus{} 30 \\equal{} 21$\r\n\r\noh hey i didnt use calculus after all :rotfl:", "Solution_11": "That's exactly what I did...thanks to conic sections :D", "Solution_12": "What two crazy people put \"as easy as A-B-C\"? :maybe:", "Solution_13": "Maybe it's tinytim and andersonw :D", "Solution_14": "I put easy as A-B-C just to mess up the survey. :D :rotfl: :rotfl:", "Solution_15": "how is #29 solved on sprint?", "Solution_16": "[quote=alpha_2]how is #29 solved on sprint?[/quote]\n\npells equation", "Solution_17": "it's sprint\nis there a faster way?\nit was literally ranked the 6th easiest on the excel stats spreadsheet", "Solution_18": "[quote=alpha_2]it's sprint\nis there a faster way?[/quote]\n\nit shouldn\u2019t take too long but idk any other way\nalso the excel sheet says it\u2019s the 5th hardest problem", "Solution_19": "no that's the wrong thing\nyou're just reading the number of people who got it right\nbut i still need a solution tho", "Solution_20": "Sad to see almost all the people who posted above are now inactive :(", "Solution_21": "[quote=alpha_2]no that's the wrong thing\nyou're just reading the number of people who got it right\nbut i still need a solution tho[/quote]\n\ni think thats the ranking literally, 6 was the second hardest because only 2 people got it right, and no one got 25 right, hence why its the hardest problem. for example, 2 is the easiest, 1 is the second easiest, and so on. ", "Solution_22": "[quote=kn07][quote=alpha_2]no that's the wrong thing\nyou're just reading the number of people who got it right\nbut i still need a solution tho[/quote]\n\ni think thats the ranking literally, 6 was the second hardest because only 2 people got it right, and no one got 25 right, hence why its the hardest problem. for example, 2 is the easiest, 1 is the second easiest, and so on.[/quote]\n\n6 was not the 2nd hardest\nit was the 15th hardest\nrefer to the order of difficulty\n\n2 is the easiest, then 1 is the second easiest, then 3, 13, 18, and then 29\n\njust use the literal ranking", "Solution_23": "2008 thread in 2023 moment", "Solution_24": "lol this test was taken like 2 years before i was born XD", "Solution_25": "also, @5above these people are probably all of a job and a life, and probably dont even know that they still have an Aops account", "Solution_26": "WDYM,\nhurdler is going strong :)", "Solution_27": "bru i was like 4 months old when this thread started ", "Solution_28": "[quote=Jndd]bru i was like 4 months old when this thread started[/quote]\n\nLol! I wasn't even born!!", "Solution_29": "[quote=alpha_2][quote=kn07][quote=alpha_2]no that's the wrong thing\nyou're just reading the number of people who got it right\nbut i still need a solution tho[/quote]\n\ni think thats the ranking literally, 6 was the second hardest because only 2 people got it right, and no one got 25 right, hence why its the hardest problem. for example, 2 is the easiest, 1 is the second easiest, and so on.[/quote]\n\n6 was not the 2nd hardest\nit was the 15th hardest\nrefer to the order of difficulty\n\n2 is the easiest, then 1 is the second easiest, then 3, 13, 18, and then 29\n\njust use the literal ranking[/quote]\n\n\"Nobody correctly answered #25 on the sprint (Answer: 299) or #8 on the target (Answer: 209.06).\nAndersonw and Bachukas were the only two people to correctly answer #6 on the sprint (Answer: 34).\nAndersonw, math154, and xpmath were the only three people to correctly answer #11 on the sprint (Answer: 128).\nTinytim, 27r, and isabella2296 were the only three people to correctly answer #27 on the sprint (Answer: $ \\frac 43$)\"-yoongyi\n\n\nHmmmm... it's almost as if my evaluation of the excel sheet is correct, these 4 problems are all in the top 4 hardest problems. The ranking describes each problem listed from easiest to hardest. 29 is the 5th hardest problem.\n" } { "Tag": [ "parameterization", "algebra", "polynomial", "conics", "parabola", "hyperbola", "system of equations" ], "Problem": "Find the values of the parameter $ a$ such that the system of equations:\r\n\r\n$ ax^2 \\minus{} 2x \\plus{} y^2 \\minus{} 2 \\equal{} 0$\r\n$ \\minus{} 3x^2 \\plus{} x \\minus{} 2y \\plus{} 2 \\equal{} 0$\r\nhas [b]two[/b] solutions in real numbers.\r\nI tried to substitute y from second to first equations but this leads to a really nasty 4-degree equation in x-- not even applying it the Rolle sequence did't give me much...", "Solution_1": "You aren't solving for $ x$; you're finding the cases where there are only two solutions for $ x$. When does a fourth-degree polynomial have two solutions?\r\n\r\nAlternately, a geometric interpretation: when do the graphs of a circle and a parabola intersect exactly twice? Generically, they intersect in four points. (Actually, the first graph can be a circle, parabola, or hyperbola depending on the sign of $ a$.)" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Given a set of coins in the plane, all with different diameters. Show that one of them is tangent to at most 5 of the others.", "Solution_1": "[hide]\n\n\n\nConsider the smallest coin. Suppose this coin is tangent to more than five other coins. If it is tangent to more htan six coins, than one of those coins is smaller than our given coin (trivial to show by angles etc), and we have a contradiction. Suppose six coins are tangent to it; than either they are all equal to our coin, or one of them is smaller to our coin.. the first case violates the fact that they are all different diameters, and the second contradicts our assumption that the given coin is minimal. Thus, our coin can be tangent to at most five other coins.\n\n\n\n[/hide]", "Solution_2": "Thanks, I was trying to prove with maximality instead of minimality... I don't know how i couldn't see it.. :blush:" } { "Tag": [], "Problem": "An organic acid A on heating with $ AlPO_4$ at 700 degree forms B. B also reacts with A to give C. C on reaction with 1,3,5 trimethyl benzene in presence of $ AlCl_3$ gives a ketone D and $ CH_3COOH$. The compound D on treatment with $ Na(Hg)/HCl$ gives an aromatic hydrocarbon A. Give structures of A to E with proper reasoning.", "Solution_1": "Haven't looked up on organic chemistry for a really long time.... so dont believe my answers straightaway\r\n\r\nOkay, $ AlPO_4$ on heating to high temperatures gives $ P_2O_5$, if thats what you wanted, Da Vinci.\r\n\r\nSo, A is acetic acid.\r\nB=ketene :D \r\nC=acetic anhydride\r\nD=Just attach a $ CH_3CO\\minus{}$ group anywhere on the benzene ring of mesitylene.\r\nE=Reduce the above carbonyl.", "Solution_2": "[quote=\"hell_ever\"]\nOkay, $ AlPO_4$ on heating to high temperatures gives $ P_2O_5$, if thats what you wanted, Da Vinci.\n\n[/quote]\r\n\r\njust what i needed,...thanks....... :)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "There is a graph $G$. The number of vertices is even and equals to $N$.\r\nThe number of edges is $M$.\r\nThe degree of each vertex is between $1$ and $3$.\r\nWe want to paint the vertices in the following way:\r\n1) We paint the vertices in black or white.\r\n2) There is no $2$ vertices such that they connected with an edge and are both white.\r\n3) The number of white vertices is bigger that the number of black vertices.\r\nYou need to find the easiest way for constructing a group of vertices A such that for all accessible paintings of the graph the number of black vertices in group A will be more that the number of white vertices in group A.", "Solution_1": "I had a mistake in the problem I have written here.\r\nAbout the group - you need to find a group such that the number of black vertices will be more than the number of white vertices in all accessible paintings." } { "Tag": [ "linear algebra" ], "Problem": "Hello, \r\n\r\nI'm having a problem with how to prove that if $T \\in L(V)$ is diagonalizable then $V=im(T) \\oplus ker(T)$. \r\nCould anyone help? ^^", "Solution_1": "Use a basis of eigenvectors. The eigenvectors of the eigenvalue zero are a basis for the null space, and the remaining eigenvectors are a basis for the image." } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "hi every body\r\nGiven two triangle with same area, is there a manner to cut one of them into pieces and form the other one.(MIT TECH REVIEW, january 2009)", "Solution_1": "See [url]http://mathworld.wolfram.com/Dissection.html[/url]", "Solution_2": "Thanks a lot.\r\nBut i would like to see at least one manner;and to have an idea about the minimum cuts." } { "Tag": [ "complementary counting" ], "Problem": "How many numbers greater than ten thousand can be formed with the digits 0, 1, 2, 2, 3 without repetition?? (Note that the digit 2 appears exactly twice in each number formed)\r\n\r\nWOW!!! This is SO CONFUSING!!!", "Solution_1": "We can arrange these digits in $ \\frac{5!}{2}$ ways.\r\n\r\nThere are $ \\frac{4!}{2}$ combinations which start with a digit 0.\r\n\r\nSo the answer is $ \\frac{5!}{2}\\minus{}\\frac{4!}{2}\\equal{}60\\minus{}12\\equal{}48$ numbers :)", "Solution_2": "I don't quite get it...\r\n\r\nWhy is it over 2??", "Solution_3": "When you are trying to find the number of ways to arrange the letter in the word MOON, there are not exactly 24 ways to arrange them, because MONO is the same word as the word you get, when you switch the O's. There are a total of two ways to arrange the O's in every word when there are ever two spaces left. So instead of having 24, we dvide by 2 to get 12. \r\n\r\nSimilarly, we do the same with numbers. The trick we use now is called complementary counting. Instead of counting the number of numbers that aren't having 0 as the ten-thousands digit, we find the number of five digit numbers, and then we subtract the number of numbers where 0 is the ten-thousands digit. There are a total of 5! ways to arrange the five digits. That is wrong, because there are two 2's. We divide 5! by 2 to get; 60. Then we subtract the number of ways to get a number with 0 as the ten-thousands digit. Once zero is the ten-thousands digit, there are 12 ways to arrange the remaining four digits because there are two 2's left. So the answer is 60 - 12. Which is 48. \r\n\r\nIf you still don't understand, then I suggest you get the Intro to Counting and Probability. It helps.", "Solution_4": "Also, instead of complementary counting:\r\n\r\nThere are $ 4$ choices for the first digit. This is because we cannot use $ 0$ as a first digit.\r\n\r\nThere are $ 4$ choices for the second digit. This is because we cannot use the first digit as the second.\r\n\r\nThere are $ 3$ choices for the third digit. This is because we cannot use the first nor second digit as the third.\r\n\r\nThere are $ 2$ choices for the fourth digit. This is because we cannot use the first, second, nor third digit as the fourth.\r\n\r\nThere is $ 1$ choice for the fifth digit. This is because we cannot use the first, second, third, nor fourth as the fifth.\r\n\r\nThus, we have $ 4 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1\\equal{}96$ numbers we can make, right?\r\n\r\nWrong.\r\n\r\nLet one of the $ 2$'s be denoted as $ 2_1$, and the other as $ 2_2$. Then $ 32_112_20$ and $ 32_212_10$ are two different numbers, right? No. $ 32120\\equal{}32120$, so each of our numbers has been double-counted! To resolve this, we simply divide our total by $ 2$, to get our final answer of $ \\frac{96}{2}\\equal{}\\boxed{48}$.\r\n\r\nI hope I helped.", "Solution_5": "Wow, nice teaching Wicke and AIME15 :thumbup:", "Solution_6": "[quote=\"Bachukas\"]Wow, nice teaching Wicke and AIME15 :thumbup:[/quote]\r\n\r\nAIME15's solution was better. I think AIME15's solution is better for harder problems, but it can still work in this one.", "Solution_7": "I think I should clarify one of Wickedestjr's points. We're not really dividing by $ 2$, we're dividing by $ 2! \\equal{} 2$ (sorry if that sounds weird). Here's the generalization: Suppose we have a collection of $ n$ digits. Now suppose that digit $ \\hat{p}$ repeats $ p$ times, digit $ \\hat{q}$ repeats $ q$ times, and so on up to digit $ \\hat{t}$ which repeats $ t$ times. Then the total number of words that can be formed is\r\n$ \\frac {n!}{p!\\cdot q!\\cdots t!}$\r\n\r\nHere's a practice problem from this year's VTRMC:\r\nHow many sequences of $ 1$'s and $ 3$'s can be made such that the sum of the digits is $ 16$? For example, $ \\{3,3,3,3,3,1\\}$ and $ \\{3,1,3,1,3,1,3,1\\}$ both work.\r\n[hide=\"Answer\"]The answer is 277. I'll leave you to provide the solution. It's not hard.[/hide]", "Solution_8": "[quote=\"JRav\"] We're not really dividing by $ 2$, we're dividing by $ 2! \\equal{} 2$ quote]\r\n\r\nI know, that is pretty straightforward, because if the two digits are u1 and u2, and they are the last two digits there are 2 for the tens digit, and 1 remaining for the ones digit. \r\n\r\nThis product is 2 x 1 which equals 2.", "Solution_9": "Obviously. Anyone have a solution for my practice problem? It's not too hard.", "Solution_10": "Alright, we know that $ 16 \\equal{} 3a \\plus{} b$, and $ a$ and $ b$ are natural numbers. \r\n\r\nSo if $ a \\equal{} 1$, then we have $ (1,13)$. \r\nIf $ a \\equal{} 2$, then we have $ (2,10)$. \r\nIf $ a \\equal{} 3$, then we have $ (3,7)$. \r\nIf $ a \\equal{} 4$, then we have $ (4,4)$. \r\nIf $ a \\equal{} 5$, then we have $ (5,1)$. \r\n\r\nIf we have one three and thirteen ones, then there are $ 13$ ways to arrange these digits.\r\nIf we have two three's and ten ones, then there are $ 12!/10!2!$, which equals $ 66$. \r\nIf we have three three's and seven ones, then there are $ 12!/3!7!$ which equals $ 15840$. \r\nIf we have four three's and four ones, then there are $ 12!/[(4!)^2]$ which equals $ 831600$. \r\nIf we have five three's and one one, then there are $ 12!/5!1!$ which equals $ 3991680$. \r\n\r\nAdding all these numbers together, yields the sum; $ 4839186$.", "Solution_11": "You really messed up your algebra. That's not even close.", "Solution_12": "Actually, it was a typo. The answer is 277, which I've corrected. Either way, 4839186 is way off.", "Solution_13": "I can't figure out what I did wrong. Is there a really easy way of doing it, or is it solved by casework?", "Solution_14": "You have a good approach, however, your later cases were evaluated incorrectly.\r\n\r\nFor example, in your last case, with five three's and one one, how do you get 12!/5!1!? Shuldn't it be 6!/5!1!?", "Solution_15": "[quote=\"AIME15\"]You have a good approach, however, your later cases were evaluated incorrectly.\n\nFor example, in your last case, with five three's and one one, how do you get 12!/5!1!? Shuldn't it be 6!/5!1!?[/quote]\r\n\r\nOh, I think I accidentally did $ 12!$ for all the cases except the first. I'm an addicted to the number $ 12$. :lol: \r\n\r\n\r\nAlright, lets do this long strenuous problem once again. \r\n\r\nWe know that $ 16 \\equal{} 3a \\plus{} b$, and $ a$ and $ b$ are natural numbers. \r\n\r\nSo if $ a \\equal{} 1$ then we have $ (1,13)$. \r\nIf $ a \\equal{} 2$, then we have $ (2,10)$. \r\nIf $ a \\equal{} 3$, then we have $ (3,7)$. \r\nWe also have $ (4,4)$ and $ (5,1)$. \r\nOh, wait a minute. I forgot $ a \\equal{} 0$. Aww, I'm so stupid. If $ a \\equal{} 0$, then we have $ (0,16)$. \r\n\r\nFor $ (0,16)$, there is one way to arrange sixteen ones. $ 1$ \r\nFor $ (1,13)$, there are fourteen ways to arrange the thirteen ones and the three. $ 14$ \r\nFor $ (2,10)$, there are sixty six ways. $ 66$ \r\nFor $ (3,7)$, there are $ 120$ ways. \r\nFor $ (4,4)$, there are $ 70$ ways. \r\nFor $ (5,1)$, there are $ 6$ ways. \r\nAdd these numbers up; \r\n\r\n$ 1 \\plus{} 14 \\plus{} 66 \\plus{} 120 \\plus{} 70 \\plus{} 6$\r\n$ 15 \\plus{} 186 \\plus{} 76$ \r\n$ 201 \\plus{} 76$ \r\n$ 277$ \r\n\r\nYay, I solved it. \r\n\r\n@JRav, your formula is just like choosing 16 objects from 16 objects, 1 object from 14 objects, ... 5 objects from 6 objects. etc. \r\n\r\n\r\nIs that the easiest way of solving it?", "Solution_16": "That's how I solved it on the contest. The one additional thing I did was prove that no such sequence could contain an odd number of elements. Can you do it? It's a simple proof by contradiction.\r\n\r\nHere's the \"official\" solution:\r\nFor each positive integer $ n$, let $ f(n)$ denote the number of sequences of $ 1$'s and $ 3$'s. That sum to $ n$. Then $ f(n\\plus{}3)\\equal{}f(n\\plus{}2)\\plus{}f(n)$, and we have $ f(1)$, $ f(2)\\equal{}1$, and $ f(3)\\equal{}2$. Thus $ f(4)\\equal{}f(3)\\plus{}f(1)\\equal{}3$, $ f(5)\\equal{}f(4)\\plus{}f(2)\\equal{}4$, $ f(6)\\equal{}6,\\ldots,f(15)\\equal{}189$, $ f(16)\\equal{}277$. Thus the number of sequences required is $ 277$.", "Solution_17": "[quote=\"JRav\"]That's how I solved it on the contest. The one additional thing I did was prove that no such sequence could contain an odd number of elements. Can you do it? It's a simple proof by contradiction.\n\nHere's the \"official\" solution:\nFor each positive integer $ n$, let $ f(n)$ denote the number of sequences of $ 1$'s and $ 3$'s. That sum to $ n$. Then $ f(n \\plus{} 3) \\equal{} f(n \\plus{} 2) \\plus{} f(n)$, and we have $ f(1)$, $ f(2) \\equal{} 1$, and $ f(3) \\equal{} 2$. Thus $ f(4) \\equal{} f(3) \\plus{} f(1) \\equal{} 3$, $ f(5) \\equal{} f(4) \\plus{} f(2) \\equal{} 4$, $ f(6) \\equal{} 6,\\ldots,f(15) \\equal{} 189$, $ f(16) \\equal{} 277$. Thus the number of sequences required is $ 277$.[/quote]\r\n\r\nWell let $ a$ be the number of threes, and let $ b$ be the number of ones. If $ 16 \\equal{} 3a \\plus{} b$, then that means that $ b \\equal{} 16 \\minus{} 3a$. The number of elements is $ a \\plus{} b$. \r\n\r\nWe can rewrite $ a \\plus{} b$ by substituting to get $ 16 \\minus{} 2a$. $ 16 \\minus{} 2a$ is divisible by $ 2$, so it is therefore even. \r\n\r\n\r\nIn the solution to that question, why does $ f(n \\plus{} 3) \\equal{} f(n \\plus{} 2) \\plus{} f(n)$ ?" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let [b]a, b, c[/b] be non-negative numbers such that a^2+b^2+c^2=3. Then, \r\n[b]1/(5-2ab)+1/(5-2bc)+1/(5-2ca)<=1[/b]", "Solution_1": "[quote=\"Inequalities Master\"]Let $ a, b, c$ be non-negative numbers such that $ a^2\\plus{}b^2\\plus{}c^2\\equal{}3.$ Then, \n$ \\frac{1}{5\\minus{}2ab}\\plus{}\\frac{1}{5\\minus{}2bc}\\plus{}\\frac{1}{5\\minus{}2ca}\\leq1$[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=109029", "Solution_2": "Sorry,arqady.but in that forun is only a proof for the following inequality:\r\nIf $ a^2\\plus{}b^2\\plus{}c^2\\equal{}1$ then $ \\dfrac{1}{5\\minus{}6ab}\\plus{}\\dfrac{1}{5\\minus{}6bc}\\plus{}\\dfrac{1}{5\\minus{}6ca}\\leq1$.\r\nBut the original question is the next:\r\nIf $ a^2\\plus{}b^2\\plus{}c^2\\equal{}3$ then $ \\dfrac{1}{5\\minus{}2ab}\\plus{}\\dfrac{1}{5\\minus{}2bc}\\plus{}\\dfrac{1}{5\\minus{}2ca}\\leq1$.\r\nDo they have the same proof? :?:", "Solution_3": "[quote=\"Inequalities Master\"]Sorry,arqady.but in that forun is only a proof for the following inequality:\nIf $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 1$ then $ \\dfrac{1}{5 \\minus{} 6ab} \\plus{} \\dfrac{1}{5 \\minus{} 6bc} \\plus{} \\dfrac{1}{5 \\minus{} 6ca}\\leq1$.\nBut the original question is the next:\nIf $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$ then $ \\dfrac{1}{5 \\minus{} 2ab} \\plus{} \\dfrac{1}{5 \\minus{} 2bc} \\plus{} \\dfrac{1}{5 \\minus{} 2ca}\\leq1$.\nDo they have the same proof? :?:[/quote]\r\nYes! If in your inequality $ a\\equal{}\\sqrt3A,$ $ b\\equal{}\\sqrt3B$ and $ c\\equal{}\\sqrt3C$ then $ A^2\\plus{}B^2\\plus{}C^2\\equal{}1$ and\r\nwe need to prove that $ \\frac{1}{5\\minus{}6AB}\\plus{}\\dfrac{1}{5\\minus{}6BC}\\plus{}\\dfrac{1}{5\\minus{}6CA}\\leq1,$ which is a Vasc's inequality.", "Solution_4": "Thanks a lot,arqady! You helped me one more time. :thumbup:" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Let $P(x) = (x-1)(x-2)(x-3)$. For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P(Q(x))=P(x) \\cdot R(x)$?\r\n\r\n\\[ \\text{(A) } 19 \\qquad \\text{(B) } 22 \\qquad \\text{(C) } 24 \\qquad \\text{(D) } 27 \\qquad \\text{(E) } 32 \\]", "Solution_1": "What is wrong with my attempt at this problem?\r\n\r\n[hide]$\\deg R(x)=3\\Rightarrow \\deg P(Q(x))=6\\Rightarrow \\deg Q(x)=2$. Write $Q(x)=ax^2+bx+c$, $a\\neq 0$. $P(x)|P(Q(x))$ iff 1, 2, and 3 are all roots of $P(Q(x))=\\left(ax^2+bx+c-1\\right)\\left(ax^2+bx+c-2\\right)\\left(ax^2+bx+c-3\\right)$. Then, if $x$ is 1, 2, or 3, then $ax^2+bx+c$ must be 1, 2, or 3 in any order. But doesn't that only produce 6 possible systems of 3 equations for a, b, and c, and therefore at most 6 possible polynomials $Q(x)$? This is far less than any of the options.[/hide]", "Solution_2": "Never mind, I found my mistake.\r\n\r\n[hide=\"Finishing my solution\"]$Q(x)$ could also be 1,2, or 3 twice and another of the numbers once, for $x\\in \\{1,2,3\\}$. In this case there are 3 options for doubled number, 2 options for singled number, 3 places to put singled number, for 18 more cases. That gives at most 24. However, $Q$ cannot be linear. This occurs when $Q(\\{1,2,3\\})$ is equal to $\\{1,2,3\\}$ or $\\{3,2,1\\}$. Discard these 2 solutions to find 22 options for $Q(x)$. [b](B)[/b][/hide]" } { "Tag": [ "counting", "distinguishability" ], "Problem": "Our basketball team has 10 players, including Steve and Danny. We need to divide into two teams of 5 for an intra-squad scrimmage. In how many ways can we do this if Steve and Danny insist on playing on the same team?", "Solution_1": "This problem is quite easy if you do it the way of the solution. I kept on trying to to find a new way to do it, but I couldn't.", "Solution_2": "If Steve and Danny are on the same team, there are 8 players to choose from for the other 3 spots on the team they're on, so there are 8C3=56 ways total.", "Solution_3": "Wouldnt it be $\\binom{9}{4} = 126$ because you can treat Steve and Danny as one super person, then you just have to choose 4 players (which is 5 because of Steve and Danny) out of 9?", "Solution_4": "You can't treat Steve and Danny like only one person. You may treat them like a block, but they must be a block equivalent to two players. If Steve and Danny were one person, then you would have a 6 person team and a 4 person team.", "Solution_5": "That's why you choose a 4-person team, like in my solution. Is that valid?", "Solution_6": "Eh, good thought! I didn't completely understand your solution the first time :oops:\n\nThe problem with your solution is that once you take Steve and Danny away from 10 people, you have 8 people. You may treat them as one person, but then there would only be 9 people in total; we need 10 players on two teams.", "Solution_7": "Can someone explain why the answer would not be $112$? Isn't there two ways? If Steve and Danny are on Team 1, or if Steve and Danny are on Team 2?", "Solution_8": "The teams are indistinguishable, so we overcount, and hence we must divide $112$ by 2 to get $\\boxed{56}$.", "Solution_9": "Wait yeah they are indistinguishable, so its $\\dbinom{8}{3}=56$ but doesn't [i]that[/i] overcount by a factor of 2 so 28? .-.", "Solution_10": "No, we already dealt with the overcounting by 2.\nSo it is $\\dfrac{\\dbinom{9}{2}}{2}$", "Solution_11": "[hide=solution]If Steve and Danny are on opposite teams, there are 8 other players to choose from for the other 4 spots on Steve's team, so there are $\\binom{8}{4} = \\boxed{70}$ choices.[/hide]", "Solution_12": "Nice solution, but please hide it :)\nalso, I think you counted the opposite", "Solution_13": "[hide = My Solution]Once you put Danny and Steve on one team, there's $8$ teammates left. We need to fill in the remaining $3$ spots on Danny and Steve's team (the remaining automatically go on the other team) so this is just $\\binom{8}{3} = \\frac{8 \\cdot 7 \\cdot 6}{3 \\cdot 2} = \\boxed{56}$.\n\nThis is the same as $\\binom{8}{5}$ if you wanted to choose the people on the opposing team.[/hide]\nI'm pretty sure this is right.", "Solution_14": "I am pretty sure you are correct.", "Solution_15": "I think RunarEC's solution is correct", "Solution_16": " :oops_sign: solution of wlm2 is also correct\n", "Solution_17": "Yeah alcumus says its 56.", "Solution_18": "say how its correct not it is correct", "Solution_19": "[quote=RunarEC][hide=solution]If Steve and Danny are on opposite teams, there are 8 other players to choose from for the other 4 spots on Steve's team, so there are $\\binom{8}{4} = \\boxed{70}$ choices.[/hide][/quote]\nThis is the actual answer Alcumus gave which means it's right.\n\n", "Solution_20": "[quote=Williamz1483]say how its correct not it is correct[/quote]\n\nwell, [hide=here's how] Since there are $10$ players, and steve and danny are on the same team, we can put them on a team beforehand. Since steve and danny don't need to be chosen, there are $10-2=8$ other people left. There are only $3$ other spots on steve and danny's team so you pick $3$ out of $8$ people which is $\\binom{8}{3} = \\boxed{56}$[/hide]" } { "Tag": [ "inequalities", "search", "circumcircle" ], "Problem": "Let ABC be an acute-angled triangle whose side lengths satisfy the inequalities AB < AC < BC. If point I is the center of the inscribed circle of triangle ABC and point O is the center of the circumscribed circle, prove that line IO intersects segments AB and BC.", "Solution_1": "see\r\n[url]http://www.mathlinks.ro/viewtopic.php?search_id=495777177&t=125578[/url]" } { "Tag": [ "logarithms", "probability", "number theory theorems", "number theory" ], "Problem": "Hello, a few of probability...\r\nHow prove that: \\[P( n \\in \\mathbb{P}) = \\frac{1}{\\ln{n}}\\]\r\nIs it hard? :? :blush:", "Solution_1": "Type \"prime number theorem\" in Google. Yeah, it's hard. :)", "Solution_2": "[quote=\"grobber\"] Yeah, it's hard. :)[/quote] :( ;) I fetch on Google... Thank you.", "Solution_3": "[quote=\"b\u00e9nabar\"]Hello, a few of probability...\nHow prove that: \\[P( n \\in \\mathbb{P}) = \\frac{1}{\\ln{n}}\\]\nIs it hard? :? :blush:[/quote]\r\n\r\nWhat exactly does this mean? Since $\\mathbb N$ is not a finite set, there is no canonical probability distribution. I presume you want Dirichlet density or natural density, in which case the statement is correct.\r\n\r\nAndrei" } { "Tag": [], "Problem": "Five cards are to be selected at random from 1 to 10 cards numbered 1 to 10. How many ways are possible that the average of five numbers selected will be greater than the median ?", "Solution_1": "[hide=\"Solution\"]The number of ways to select $ 5$ numbers in the set $ \\{1,2,3,4,5,6,7,8,9,10\\}$ such that the mean equals the median is $ 24$ by some case analysis.\n\nNow observe that there is a 1-1 correspondence between the number of ways to select $ 5$ numbers with median greater than the mean and the number of ways to select $ 5$ numbers with median less than the mean. This is evident by the fact that if $ a,b,c,d,e$ are the 5 selected numbers with either property, then $ 11\\minus{}a,11\\minus{}b,11\\minus{}c,11\\minus{}d,11\\minus{}e$ has the other property.\n\nLet $ A$ be the number of ways to select the $ 5$ numbers with mean $ >$ median, $ B$ be the number of ways with mean $ \\equal{}$ median, and $ C$ be the numbe of ways with mean $ <$ median. Then\n$ A \\plus{} B \\plus{} C \\equal{} {10 \\choose 5}$\nBut we know that $ B \\equal{} 24$ and $ A \\equal{} C$ from the previous paragraphs. So\n$ 2A \\plus{} 24 \\equal{} {10 \\choose 5} \\equal{} 252$\n$ 2A \\equal{} 228$\n$ \\boxed{A \\equal{} 114}$\n\nIf this is wrong, it's probably because I screwed up the calculation of $ B$ in the first paragraph; I kinda rushed.[/hide]" } { "Tag": [ "Alcumus", "Support" ], "Problem": "is there any reason it is called alcumus? (just wondering)", "Solution_1": "Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=239473]here[/url]." } { "Tag": [], "Problem": "1. $2 +\\sqrt{u}$=$\\sqrt{2u+7}$\r\n\r\n2. $\\sqrt{X-5}$$-\\sqrt{X + 3}=4$\r\n\r\nI keep getting different answer then whats in the back of the book :|\r\n\r\nAnd my teacher didnt even bother explaining this :(", "Solution_1": "hello, for (1) we get after squaring both sides of the given equation\r\n$ 4\\plus{}u\\plus{}4\\sqrt{u}\\equal{}2u\\plus{}7$\r\n$ 4\\sqrt{u}\\equal{}3$\r\nsquaring again and isolating $ u$ we get $ u\\equal{}\\frac{9}{16}$ as the solution.\r\nSonnhard.", "Solution_2": "hello, your second equation has no real solution\r\nSonnhard.", "Solution_3": "For (2), add sqrt(x+3) to both sides, square both sides, simplify and square again to get x=6.", "Solution_4": "hello, $ x\\equal{}6$ is no solution of your second equation, to prove this you can insert $ x\\equal{}6$ .\r\nSonnhard.", "Solution_5": "You're right, of course, Sonnhard. 6 is an extraneous solution which doesn't work: as you rightly point out there are no real solutions. 6 arises when squaring sqrt(x+3) = -3 which could just as well have been squaring -sqrt(x+3) = -3. So 6 is the (only real) solution of sqrt(x-5) = 4 - sqrt(x+3).", "Solution_6": "[quote=\"FlaME_HiRo\"]1. $ 2 \\plus{} \\sqrt {u}$=$ \\sqrt {2u \\plus{} 7}$[/quote]\n\n$ 2 \\plus{} \\sqrt {u}$=$ \\sqrt {2u \\plus{} 7}$ $ \\iff$ $ u\\geq 0$ and $ (2 \\plus{} \\sqrt {u})^2$=$ (\\sqrt {2u \\plus{} 7})^2$\n\n$ \\iff$ $ u\\geq 0$ and $ 4\\plus{}4\\sqrt u \\plus{} u\\equal{}2u\\plus{}7$\n\n$ \\iff$ $ u\\geq 0$ and $ 4\\sqrt u\\equal{}u\\plus{}3$\n\n$ \\iff$ $ u\\geq 0$ and $ (4\\sqrt u)^2\\equal{}(u\\plus{}3)^2$\n\n$ \\iff$ $ u\\geq 0$ and $ 16u\\equal{}u^2\\plus{}6u\\plus{}9$\n\n$ \\iff$ $ u\\geq 0$ and $ u^2\\minus{}10u\\plus{}9\\equal{}0$\n\n$ \\iff$ $ u\\geq 0$ and $ (u\\minus{}1)(u\\minus{}9)\\equal{}0$\n\n$ \\iff$ $ u\\in\\{1,9\\}$\n\n\n[quote=\"FlaME_HiRo\"]2. $ \\sqrt {X \\minus{} 5}$ $ \\minus{} \\sqrt {X \\plus{} 3} \\equal{} 4$[/quote]\r\n\r\n$ \\sqrt {X \\minus{} 5}$$ \\minus{} \\sqrt {X \\plus{} 3} \\equal{} 4$ $ \\iff$ $ \\sqrt {X \\minus{} 5}\\equal{}4\\plus{} \\sqrt {X \\plus{} 3}$\r\n\r\n$ \\iff$ $ X\\geq 5$ and $ (\\sqrt {X \\minus{} 5})^2\\equal{}(4\\plus{} \\sqrt {X \\plus{} 3})^2$\r\n\r\n$ \\iff$ $ X\\geq 5$ and $ X\\minus{}5\\equal{}16\\plus{}8\\sqrt{X\\plus{}3}\\plus{}X\\plus{}3$\r\n\r\n$ \\iff$ $ X\\geq 5$ and $ 24\\plus{}8\\sqrt{X\\plus{}3}\\equal{}0$\r\n\r\nSo no solution", "Solution_7": "hello, oh, thank you my solution for the first equation is also wrong.\r\nSonnhard.", "Solution_8": "$ \\minus{} 5 < 3\\Rightarrow x \\minus{} 5 < x \\plus{} 3\\Rightarrow \\sqrt {x \\minus{} 5} < \\sqrt {x \\plus{} 3}\\Rightarrow\\sqrt {x \\minus{} 5} \\minus{} \\sqrt {x \\plus{} 3} < 0<4\\Rightarrow \\\\\r\n\\Rightarrow \\sqrt {x \\minus{} 5} \\minus{} \\sqrt {x \\plus{} 3}\\neq 4; (\\forall)x\\ge 5.$", "Solution_9": "1. $ \\left\\{\\begin{array}{ll} \\sqrt {2u \\plus{} 7} \\minus{} \\sqrt {u} \\equal{} 2 & \\quad \\\\\r\n\\sqrt {2u \\plus{} 7} \\plus{} \\sqrt {u} \\equal{} \\frac {u \\plus{} 7}{2} & \\quad \\end{array} \\right.\\Longleftrightarrow \\left\\{\\begin{array}{ll} \\sqrt {u} \\equal{} \\frac {u \\plus{} 3}{4} & \\quad \\\\\r\n\\sqrt {2u \\plus{} 7} \\equal{} \\frac {u \\plus{} 11}{4} & \\quad \\end{array} \\right.$\r\n\r\n$ \\Longleftrightarrow \\boxed{u \\equal{} 1,\\ 9}$\r\n\r\n2. $ \\left\\{\\begin{array}{ll} \\sqrt {X \\minus{} 5} \\minus{} \\sqrt {X \\plus{} 3} \\equal{} 4 & \\quad \\\\\r\n\\sqrt {X \\minus{} 5} \\plus{} \\sqrt {X \\plus{} 3} \\equal{} \\minus{} 2 & \\quad \\end{array} \\right.$ , yielding no real solution." } { "Tag": [ "algebra", "polynomial", "quadratics", "Rational Root Theorem" ], "Problem": "Can I get some help with factor theorem, it's kind of new to me. I need to use factor theorem to factor x^3- 6x^2 + 6x - 5.", "Solution_1": "I have no idea what the \"factor theorem\" is but...\r\n\r\nThe first thing you're going to want to do is use the rational root theorem mentally,\r\n$\\frac{\\text {Factors of lowest degree coefficient}}{\\text{Factors of highest degree coefficient}}$, \r\n\r\nthe only real roots possible are $\\pm{5}$ and $\\pm{1}$. You an easily see that $\\pm{1}$ won't work, so synthetically divide $\\pm{5}$ and you can see that $5$ works (since the remainder is zero).\r\n\r\nSo now factor out the $(x-5)$ to $\\boxed{(x-5)(x^2+x+1)}$. Notice that there is no sign change in $(x^2+x+1)$ so you don't have to even try $-5$\r\n\r\nRemember that when you're dealing with a polynomial that has the largest degree coefficient of one... the roots of the polymonial are factored in the form $(x-a)$, where $a$ is the root\r\n\r\nHope that helps.", "Solution_2": "[quote=\"king_23\"]Can I get some help with factor theorem, it's kind of new to me. I need to use factor theorem to factor x^3- 6x^2 + 6x - 5.[/quote]\r\n\r\nThe factor theorem is basically the same as rational root theorem (I think) which is basically test numbers that make the polynomial $=0$.\r\n\r\nThe only possible values are $\\pm 5, \\pm 1$ (I don't really know how to explain why, but it has to do with the last term and the first term).\r\n\r\nTest $5$. It works. So we know that $x-5$ is a factor.\r\n\r\nNow do whatever division you want to find the other factor (which will be a quadratic). The factor the quadratic if possible (in this case it's not) and there's the answer." } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "I have a matrix A=\r\n1 0\r\n-1 1\r\n\r\na 2 by 2 matrix. I want to show that this has infinite order. \r\n\r\nI have that A^n =\r\n1 0\r\n-n 1\r\nis it enough to say that there is no n such that A^= identity.", "Solution_1": "Yes; clearly the only $ n$ such that $ A^n \\equal{} I$ occurs when $ n \\equal{} 0$. Alternately, you could compute its characteristic polynomial.", "Solution_2": "I have found char poly then what to do with it?", "Solution_3": "What must the minimal polynomial of a matrix satisfying $ A^n \\equal{} I$ look like?" } { "Tag": [ "trigonometry", "complex numbers" ], "Problem": "Can someone help me with number 30 on the [url=http://www.math.sc.edu/contest/1997/exam/PDF/exam1997.pdf]1997 USC Math Contest[/url]? It's a trig problem, and it looks rather hard. Any solutions appreciated!", "Solution_1": "I'm not sure how to do the entire thing right now, but i would imagine it would have something to do with the relationships between the sin values. They follow a circular pattern, you are really only dealing with two values.", "Solution_2": "[quote=\"mysmartmouth\"]Can someone help me with number 30 on the [url=http://www.math.sc.edu/contest/1997/exam/PDF/exam1997.pdf]1997 USC Math Contest[/url]? It's a trig problem, and it looks rather hard. Any solutions appreciated![/quote]\r\n\r\nWhat are you doing? You still need to learn your identities and you expect to do a problem like that? Learn your idenitities first! And then learn complex numbers and Demoivres in order to solve it! There's really no point to post a solution because you won't understand any of it! ;) ;)", "Solution_3": "well...i dont think you use complex numbers to solve it.\r\n\r\ni think the hint is to use the double angle formula for sine...i'm sick though, and won't work it all out now. theres a smiliar problem in a later usc test with cosines, look for the solution they give. it is really similar.", "Solution_4": "[quote=\"arkmastermind\"]well...i dont think you use complex numbers to solve it.\n\ni think the hint is to use the double angle formula for sine...i'm sick though, and won't work it all out now. theres a smiliar problem in a later usc test with cosines, look for the solution they give. it is really similar.[/quote]\r\n\r\nhmm...if i remember correctly, the usc solution to this problem involved demoivres :?", "Solution_5": "that might be true. i cant seem to remember what i did when i did this for practice...im pretty sure i didnt use demoivres though...\r\n\r\noh well.", "Solution_6": "Well the solutions are here http://www.math.sc.edu/contest/1997/solutions/PDF/solutions1997.pdf\r\n\r\nand it looks like they did use demoivres. I'm not a big fan of the USC solutions though, it seems they make problems more complicated than they need to be.", "Solution_7": "[quote=\"Iversonfan2005\"]and it looks like they did use demoivres. I'm not a big fan of the USC solutions though, it seems they make problems more complicated than they need to be.[/quote]\r\n\r\nThat problem [i]is[/i] complicated... Show me an easier way to do it than complex numbers...." } { "Tag": [], "Problem": "Post stuff..", "Solution_1": "No. The football team brings a substantial amount of money to my city and to my school. In addition, I feel the funding for the various academic programs is sufficient. You could throw more money into academics, but until you find a more efficient use for it, I feel it wouldn't help enough for the action to be justified.\r\n\r\nAlso, given the bad state of American nutrition and exercise, I feel that athletic programs should be encouraged...especially at younger ages, but high school programs can be just as beneficial.", "Solution_2": "No. They spend no money on sports.", "Solution_3": "[quote=\"elston\"]Post stuff..[/quote]\r\n\r\nTo tell the truth, my school needs to spend [b]more[/b] money on sports. We were playing Ultimate Frisbee, and half the frisbees are those really cheap and flexible ones that you can fold up and that don't fly well.\r\n\r\nI heard that my friend's school has to [b]share sports jerseys![/b] :o", "Solution_4": "My school doesn't spend too much money on sports.\r\n\r\n[hide=\"Then again...\"]I am homeschooled... :D [/hide]", "Solution_5": "wow..where i live, like everything is football football football baseball baseball soccer soccer track track track football football....", "Solution_6": "sports are big at my school and spending the money is justifable as they bring in money. if you can figure out a way to get people to buy tickets to watch a math contest, i'd love to hear it.", "Solution_7": "More money on sports would be nice, so we don't have to depend on donations to keep our sports up.", "Solution_8": "Our school's athletic program is too cheap to even afford to give us 2 busses for our 56 member cross country team to meets.\r\n\r\nOur school doesn't spend much money.", "Solution_9": "[quote=\"hwenterprise\"]Our school's athletic program is too cheap to even afford to give us 2 busses for our 56 member cross country team to meets.\n\nOur school doesn't spend much money.[/quote]\r\n\r\nwhile another school near us has those super luxury buses (they only have like 20). Not fair :mad:", "Solution_10": "depends on the sport, football we hire outside coaches, nice buses for every game, nice pads and equipment, but other sports are dependant on teachers who are willing to drive. Fortunately, we have begun to spend some money on academics, even thouhg i still dont have my economics textbook becuase we dont have any and its been almost 3 weeks. At one time, our school was using outdated textbooks, sharing desks and more, while the football teams funds never seemed to have a limit. Other ECs are the sole responsibility of students/coaches to finance like our reach team, where we paid $ \\$ $ 170 without any school assistance, but we were fortunate that our coach was a resourceful fundraiser.", "Solution_11": "Yeah actually they took funds out of our already depleted funds for football and basketball. But I guess since our athletic teams own our academics like crazy, maybe it's justified? I don't know.", "Solution_12": "No. If anything, they should spend more money on sports.", "Solution_13": "Well, sometime the lack of money can hurt, especially this [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=2695]time we got lost[/url]." } { "Tag": [ "algebra", "polynomial", "number theory", "prime factorization", "algebra unsolved" ], "Problem": "How many factors are there in $ x^{2007}\\minus{}1$?", "Solution_1": "[quote=\"exoeragonius\"]How many factors are there in $ x^{2007} \\minus{} 1$?[/quote]\r\n hdo you mean by factors $ card(A)$ that :\r\n$ A \\equal{} (x1.x2...........xn)$ and $ p(x) \\equal{} \\prod (x \\minus{} xi)$ ?\r\n and $ x \\in C$ ???????? :wink:", "Solution_2": "I mean if you factor it completely, how many factors will there be?", "Solution_3": "[quote=\"exoeragonius\"]I mean if you factor it completely, how many factors will there be?[/quote]\r\n$ p(x) \\equal{} x^2007 \\minus{} 1$ \r\n$ p(x) \\equal{} 0$ ==> $ x^2007 \\equal{} 1$ so $ x_k \\equal{} e^(2ki \\pi /2007)$ and $ k \\in (0.1.2.3............2006)$ \r\nso the number of factor is 2007 \r\nnb = $ x \\in C$", "Solution_4": "I think he didn't mean factors with complex coefficients.\r\nNumber of irreducible factors of $ x^{2007} \\minus{} 1$ is equal to the total number of divisors of $ 2007$ which is $ \\boxed{6}$", "Solution_5": "[quote=\"Akashnil\"]I think he didn't mean factors with complex coefficients.\nNumber of irreducible factors of $ x^{2007} \\minus{} 1$ is equal to the total number of divisors of $ 2007$ which is $ \\boxed{6}$[/quote]So, you mean the factors are $ x^d\\minus{}1$, where $ d|2007$? I don't think these factors are necessarily irreducible. Take, for instance, $ d\\equal{}3$. You'll get $ x^3\\minus{}1$ which is reducible to $ (x\\minus{}1)(x^2\\plus{}x\\plus{}1)$. Besides, is $ x^{669}\\minus{}1$ also a factor? ;)\r\n\r\nI guess there are five irreducible factos: $ (x\\minus{}1)(x\\plus{}1)(x^2\\minus{}x\\plus{}1)(x^2\\plus{}x\\plus{}1)(x^{223}\\plus{}x^{214}\\plus{}x^{205}\\plus{}\\ldots\\plus{}x^7)$. So, we have $ 2^5\\equal{}32$ factors, not necessarily irreducible. :)", "Solution_6": "There are in fact six. $ x^{2007} \\minus{} 1 \\equal{} \\prod_{d | 2007} \\Phi_d(x)$ where $ \\Phi_d(x)$ is the $ d^{th}$ [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]cyclotomic polynomial[/url].", "Solution_7": "I don't understand what cyclotomic polynomial is. Anyway, what's wrong with my solution?", "Solution_8": "Well, cyclotomic polynomials give the answer. It's pretty easy to verify that your factorization is wrong; $ x + 1$ is clearly not a factor and the polynomial you gave has degree $ 1 + 1 + 2 + 2 + 223 = 229$. (You don't seem to have the prime factorization of $ 2007$ correct. Actually, I don't understand how you got that last factor at all.)\r\n\r\nEdit: The correct factorization is \r\n\r\n\\begin{eqnarray*} x^{2007} - 1 & = & \\Phi_1(x) \\Phi_3(x) \\Phi_9(x) \\Phi_{223} (x) \\Phi_{669} (x) \\Phi_{2007}(x) \\\\\r\n& = & (x - 1)(x^2 + x + 1)(x^6 + x^3 + 1)(x^{222} + x^{221} + ... + x + 1) \\Phi_{669}(x) \\Phi_{2007}(x) \\\\\r\n\\end{eqnarray*}\r\n\r\nwhere $ \\Phi_{669}(x) = \\frac {x^{446} + x^{223} + 1}{x^2 + x + 1}$ and $ \\Phi_{2007}(x) = \\frac {x^{1338} + x^{669} + 1}{x^6 + x^3 + 1} = \\Phi_{669}(x^3)$. \r\n\r\nEdit 2: The essential pattern here is that $ 2007 = 3^2 \\cdot 223$; a square of a prime times a prime. The smallest such number is $ 12 = 2^2 \\cdot 3$, and here\r\n\r\n\\begin{eqnarray*} x^{12} - 1 & = & \\Phi_1(x) \\Phi_2(x) \\Phi_4(x) \\Phi_3(x) \\Phi_6(x) \\Phi_{12}(x) \\\\\r\n& = & (x - 1)(x + 1)(x^2 + 1)(x^2 + x + 1)(x^2 - x + 1)(x^4 - x^2 + 1) \\\\\r\n\\end{eqnarray*}." } { "Tag": [ "HMMT", "geometry" ], "Problem": "This is actually HMMT problem but instead of giving numbers, I put variables to make it an algebraic question.\r\n\r\nA man, standing on a lawn, is wearing a circular sombrero of radius $ n$ feet. Unfortunately, the hat blocks the sunlight so effectively that the grass directly under it dies instantly. If the man walks in a circle of radius $ k$ feet where $ k > n > 0$, find the area of dead grass.\r\n\r\n$ \\text{(A)} \\ \\pi nk \\ \\qquad \\text{(B)} \\ 2 \\pi nk \\ \\qquad \\text{(C)} \\ 4 \\pi nk \\ \\qquad \\text{(D)} \\ \\pi \\sqrt{17} nk \\ \\qquad \\text{(E)} \\ \\text{none of given}$", "Solution_1": "[hide]\nThe path will be an annulus of inner radius $ k\\minus{}n$ and outer radius $ k\\plus{}n$. Hence the area is \n\n$ \\pi (k\\plus{}n)^2 \\minus{} \\pi (k\\minus{}n)^2 \\equal{} 4kn\\pi$, or $ C$.\n[/hide]", "Solution_2": "Yup, that's correct." } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "So, up to now which one of you is sure of about going to MOP 2005? :) Please [u][b]post only once[/b][/u] so that we can count easily :D\r\nGood luck to the TSTs to the Winners and HM members taking it.", "Solution_1": "I'm going.", "Solution_2": "I'll be going as well, although I didn't exactly expect to do this well on my first USAMO. Anyways, I can't really complain. :)", "Solution_3": "I'm going, but only because I happened to see number 5 in [u]Math Olympiad Challenges[/u] (which I am now recommending to everyone :) ).", "Solution_4": "I'm going. So much for \"post only once\" and \"post if you're sure you're going,\" but the list can still probably be sorted through.\r\n\r\n-Adam Hesterberg", "Solution_5": "I'm going. Red MOP here.", "Solution_6": "I'm going, which is pretty awesome! This is my first year taking the USAMO, just barely qualifying with a 9 (as a sophomore). Somehow, I managed to get a 21 on USAMO, and I think I just barely qualified for MOSP too. I can't wait!", "Solution_7": "Non-winner TST taker here.", "Solution_8": "LOL.\r\n\r\nRed MOP here.\r\n\r\nLooks like my first post was deleted." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Solve equation:\r\n $ 3^{cos^2x}\\plus{}2^{sin^2x}\\equal{}2^{(\\frac{x}{\\pi})^2\\minus{}3}\\plus{}2^{9\\minus{}\\frac{4x}{\\pi}}$", "Solution_1": "Prove that $ f(x) \\equal{}2^{(\\frac{x}{\\pi})^2\\minus{}3}\\plus{}2^{9\\minus{}\\frac{4x}{\\pi}} \\minus{} 3^{cos^2x}\\minus{}2^{sin^2x} \\geq 0 \\forall x$. The equality occurs if and only if $ x \\equal{} 2\\pi$." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "MATHCOUNTS", "AIME", "AMC 8", "AMC 10" ], "Problem": "Don't worry about countdowns, there's really no point in preparing for those. However, it's a good trick to know nonetheless, since it is a time-saver.", "Solution_1": "[quote=\"13375P34K43V312\"]Don't worry about countdowns, there's really no point in preparing for those. However, it's a good trick to know nonetheless, since it is a time-saver.[/quote]\r\n\r\nI beg to differ ;)... i don't know all the elites but I know I was practicing countdown with Nathan Benjamin last year and he sure didn't think there was \"no point\" of this.\r\n\r\nOf course, I guess if you don't think you can make top 12 nationals... but then again if you make nationals you should always prepare for a) the worst -- emotional trauma is the worst that can happen. and b) the best. U get in countdown and bam u could be the national champion under this system!!\r\n\r\nI know i'll be practicing countdown if I make nationals this year.", "Solution_2": "[quote=\"Fanatic\"][quote=\"13375P34K43V312\"]Don't worry about countdowns, there's really no point in preparing for those. However, it's a good trick to know nonetheless, since it is a time-saver.[/quote]\n\nI beg to differ ;)... i don't know all the elites but I know I was practicing countdown with Nathan Benjamin last year and he sure didn't think there was \"no point\" of this.\n\nOf course, I guess if you don't think you can make top 12 nationals... but then again if you make nationals you should always prepare for a) the worst -- emotional trauma is the worst that can happen. and b) the best. U get in countdown and bam u could be the national champion under this system!!\n\nI know i'll be practicing countdown if I make nationals this year.[/quote]\r\nI kinda figured that Nathan Benjamin would be really fast since that would be all that he needed to practuce, but Treething roasted him...", "Solution_3": "any given set of 5 problems, anything could happen.\r\n\r\nIt's so luck based. I think Nathan was actually doing better at first: He answered first on questions but made dumb mistakes. Treething finished him off.\r\n\r\nExampel: 2002 nats champ (albert ni) was down 0-2, he had answered first twice but made two dumb mistakes (similar to what happened to nathan). Albert missed the 3rd question but the dude he was facing couldn't finish him off (difference between what happened to nathan). albert then \"counter-roasted\" the dude he was facing and took 3 questions easily and became nats champ one round later.\r\n\r\nThat one question could make the difference. I remember thinking that nathan, down 2-0, could possibly win anyways -- it'd be ironic deja vu for me after i watched the albert do it. but nope, didn't happen.\r\n\r\nEdited: changed names to \"albert ni\" to make less confusing", "Solution_4": "[quote=\"Fanatic\"]any given set of 5 problems, anything could happen.\n\nIt's so luck based. I think Nathan was actually doing better at first: He answered first on questions but made dumb mistakes. Treething finished him off.\n\nExampel: 2002 nats champ was down 0-2, he had answered first twice but made two dumb mistakes (similar to what happened to nathan). He missed the 3rd question but the dude he was facing couldn't finish him off (difference between what happened to nathan). The eventual champion then \"counter-roasted\" the dude he was facing and took 3 questions easily and became nats champ one round later.\n\nThat one question could make the difference. I remember thinking that nathan, down 2-0, could possibly win anyways -- it'd be ironic deja vu for me after i watched the 2002 nats champ do it. but nope, didn't happen.[/quote]\r\nYou mean Albert Ni?", "Solution_5": "Countdown really doesn't teach you a whole lot. You don't see olympiad winners sitting in their rooms practicing doing obvious math problems in 3 seconds or less.\r\n\r\nSergei Bernstein, who owned everyone in the written in 2005, didn't give a care about the countdown. With all due respect to him, he was terrible at it. He got 'roasted' by Neal 3-0 in his first countdown matchup. Later, he beat everyone in the Masters Round by a wide margin. He nailed it. He knew that there are much more important things then doing mindless problems in 3 seconds. This year, he got one of the highest freshman USAMO scores.\r\n\r\nSome people are just naturally fast and very good at the other stuff as well (Neal). However, I believe that spending a whole lot of time practicing for countdown is pointless. I\"m not saying that at the end of a practice you shouldn't do a mini-countdown for fun. I happen to be pretty decent at countdown and enjoy it myself, but I know that it's not worth dying for.", "Solution_6": "[quote=\"13375P34K43V312\"]Countdown really doesn't teach you a whole lot. You don't see olympiad winners sitting in their rooms practicing doing obvious math problems in 3 seconds or less.\n\nSergei Bernstein, who owned everyone in the written in 2005, didn't give a care about the countdown. With all due respect to him, he was terrible at it. He got 'roasted' by Neal 3-0 in his first countdown matchup. Later, he beat everyone in the Masters Round by a wide margin. He nailed it. He knew that there are much more important things then doing mindless problems in 3 seconds. This year, he got one of the highest freshman USAMO scores.\n\nSome people are just naturally fast and very good at the other stuff as well (Neal). However, I believe that spending a whole lot of time practicing for countdown is pointless. I\"m not saying that at the end of a practice you shouldn't do a mini-countdown for fun. I happen to be pretty decent at countdown and enjoy it myself, but I know that it's not worth dying for.[/quote]\r\nYou know that Neal got a 13 on the USAMO this year?", "Solution_7": "is a 13 good?\r\n\r\nisn't it out of 14?\r\n-jorian", "Solution_8": "Um, so you try so hard and u get so far and you're at the point where you're up on the stage, the lights are pointing at you, there's 8,000, a trip, and national recognition at stake and your attitude is \"there's no point doing mindless buzzer blah blah blah\"?\r\n\r\nI'm not saying practice countdown over written, I'm saying practicing countdown is basically essential for top competitors. I mean why are you even DOING mathcounts if you know u can a) make the top 12 and b) aren't going to try countdown???\r\n\r\nIF sergei believed that countdown was worthless, fine by me, but logically if you're going to try that hard and go so far then you might as well try to win it all in countdown too.", "Solution_9": "[quote=\"jhredsox\"]is a 13 good?\n\nisn't it out of 14?\n-jorian[/quote]\r\nWell, It is out of 42, but a 13 is really really good (especially since he was an 8th grader). That is like a 4th grader doing mathcounts and making nationals. You know that at some point he is going to win written at nationals.", "Solution_10": "lol.......he did win written :o \r\n\r\nbut good point...they must be hard for eith graders\r\n-jorian", "Solution_11": "[quote=\"jhredsox\"]lol.......he did win written :o \n\nbut good point...they must be hard for eith graders\n-jorian[/quote]\r\nHe won countdown too...", "Solution_12": "yes i know!, but not in the same year (please don't try to be offended nebula)\r\n\r\n-jorian\r\n\r\np.s. i think we've gotten the point of this question, so can a mod lock this up?", "Solution_13": "I agree with 13375P34K43V312. The countdown round don't teach you a lot. Knowing how to do problems is a lot more important.", "Solution_14": "My point is why the heck are u even bothering working on mathcounts problems AT ALL if you aren't interested in countdown? I point you at AMC 8 --> AMC 10 --> AIME --> USAMO and other math contests. You don't see olympiad winners hunched over 40 minute timed tests, either, do you?\r\n\r\nSure, the obvious and \"best\" answer would be \"we should only learn math to be better at it, and since countdown doesn't teach math, then screw it\".\r\n\r\nBut countdown is a skill nonetheless. Not necessarily direct math ability, but reactions and timing too. and I DO NOT FIND IT A COINCIDENCE that of the people that I know that are good at math, all the ones that are good at countdown are gregarious people. The other ones remind me of furniture more often than not, or are just less lively in general. I know of course this is people I know and this cannot be generalized, but as I type this I realize I can really separate people into two groups: good at cd (countdown) and sociable / not good at cd - antisocial. Wow. I'm not even joking, not one bit. But that's just people I know.\r\n\r\nAnd if you argue that my point is void, because the only point of mathcounts should be pure math, then I gotta say mathcounts isn't the right contest for you. Might as well start training AIME now rather than later. Don't waste your time on mathcounts. I'm sure a lot of elites, if they could go back in time, would never touch mathcounts and work directly on harder stuff (AMC 8 to compare to regional level - easy state level, AMC 10 to compare to upperstate -->nationals level) I'm more of a countdown guru than any other round but I've been training AIME material about... 100 times more than mathcounts. Hours spent and effort dedicated. \r\n\r\nAlbert ni (2002 nats champ) once said: \"Mathcounts isn't about who's good at math, it's about who's seen more problems\".", "Solution_15": "Yes, I know about Neal's beast score on USAMO. I doubt he practiced countdown day in and day out, he probably practice it a few times, but I'd be willing to bet it's not what he focused on.\r\n\r\nI'm just saying, some people are naturally fast, it's not worth working on that. But it's pretty clear that Neal spent a lot of time working on Olympiad math (or at least 20 hours total, say)", "Solution_16": "who said anything about practicing countdown day in and day out?", "Solution_17": "[quote=\"Fanatic\"]who said anything about practicing countdown day in and day out?[/quote]\nAnyone who does that is stupid, I'm sorry if I offended anyone.\nPerhaps the week before the competition it is okay.\n[quote=\"Fanatic\"]if they could go back in time, would never touch mathcounts and work directly on harder stuff (AMC 8 to compare to regional level - easy state level, AMC 10 to compare to upperstate -->nationals level)[/quote]\r\nYou mean like MysticTerminator?\r\n\r\nThe thing about countdown is that it is fun. It makes mathcounts different, and I enjoy a good round of countdown even if it costs me first place (which it has...)... In countdown, everyone makes stupid mistakes. They are often costly. In countdown, you have to be in like hypermode. You have to react faster than you can think. Finally, you have to be smart :D.", "Solution_18": "[quote]\nAnyone who does that is stupid, I'm sorry if I offended anyone.\nPerhaps the week before the competition it is okay.\n[/quote]\nJust in case you couldn't tell (I'm not sure), my post was meant to reply to leetspeakfever's (13375p34k43v312) post. It meant \"wait, you said 'I doubt he practiced countdown and day in and day out' as a counter to my argument, but I (nor anyone else) every mentioned anything about practicing day in and day out\".\n\nJust making sure you could tell that I wasn't actually advocating practicing countdown day in and day out. I personally won't until I find out whether or not I make nationals.\n\n[quote=\"bpms\"]if they could go back in time, would never touch mathcounts and work directly on harder stuff (AMC 8 to compare to regional level - easy state level, AMC 10 to compare to upperstate -->nationals level)\nYou mean like MysticTerminator?[/quote]\r\n\r\nWho's MysticTerminator?", "Solution_19": "http://www.artofproblemsolving.com/Forum/viewtopic.php?p=614892&h#614892\r\nHe got fifth at mc If I recall correctly... He also is an IMO gold medalist.", "Solution_20": "Meh... u can take his opinion or leave it.\r\n\r\nPersonally mathcounts is my \"style\" of math competition -- I feel as if I have a shot to place high nationally in it but not the harder math competitions later on (i'm good at speed but not endurance, for example), so that's why I emphasize mathcounts.\r\n\r\nThen again I've done like 20 hours of mathcounts in the last 3 months and like hundreds of harder math...\r\n\r\nbut my attitude is still \"mathcounts is my chance to DO SOMETHING in math\"... man I need to start working on mathcounts :(." } { "Tag": [ "search", "function" ], "Problem": "on the 2001 state #24\r\n\r\nIt asks the number of partitions for 7\r\n\r\nIs there a formula or sumthin??", "Solution_1": "Yes, I remember there was this other formula, but it involved lots of calc and other things. I think it would be better to just list.", "Solution_2": "Uh, it takes like 1-2 minutes (less for some people) to just list them out.\r\nNot that bad for a state #24.", "Solution_3": "aww man... thatll tak 4 eva!!!", "Solution_4": "erm, how about you group them, depending on the number of digits in te partition\r\n7- 1111111\r\n6-111112\r\n5-11122 11113\r\n4-1222 1123 1114\r\n3-124 223 115 133\r\n2-16 25 34\r\n1-7\r\n1+3+4+3+2+1+1=15", "Solution_5": "ehh, it has to do with triangular numbers i think.\r\n\r\nEDIT: scratch that. there's a partitioning function. it looks awfully complicated though T_T", "Solution_6": "try search, at the top of your screen.", "Solution_7": "Hm, I looked on wikipedia and they have it as the reciprocal of Euler's Function... o.0", "Solution_8": "i think you have to take the closest integer function of that?", "Solution_9": "Well, there's a recursive function from which you can find the number of partitions of $ n$ based on the number of partitions of each of the numbers less than it...", "Solution_10": "I bet it's something that's kinda pointless to remember", "Solution_11": "I know i sound really dumb but what is a partition??? i think ive heard of them but i have no idea what they are!", "Solution_12": "A partition is a way to express an integer as a sum of one or more positive integers. \r\nJust list them, the only real strategies are for how. Grouping as abacadea suggested would work perfectly. I mean, they're not gonna ask for the number of partitions of 129. If they do, good luck buddy.", "Solution_13": "[quote=\"darkmatter47\"]A partition is a way to express an integer as a sum of one or more positive integers. \nJust list them, the only real strategies are for how. Grouping as abacadea suggested would work perfectly. I mean, they're not gonna ask for the number of partitions of 129. If they do, good luck buddy.[/quote]\r\n\r\nah thank you so much!!!" } { "Tag": [ "MIT", "college" ], "Problem": "$\\text{NO WAY! THIS IS AWESOME!}$\r\n\r\nI am so pumped that we have biology on the forums now. I want to moderate this thing! Okay all you biology lovers out there from Ecology/Evolutionary Biology to Molecular/Cellular/Developmental Biology and Genetics, it's time to represent our science! There's a bunch of math in biology and that is really awesome too. I am stoked.", "Solution_1": "Hopefully [b]you[/b] guys will actually start using this forum, since you won't be getting any bio input from any of us the admins :P :D", "Solution_2": "Ha ha! Actually Valentin, I think this forum will draw you into it...and then you will be so intrigues by this enigmatic science that you will dump math altogether and do biology. Or maybe not...\r\n\r\nI like math too, hopefully there will be discussions here on population growth and ecology and other stuff where mathematics are highly applicable (note my signature on the bottom, it's the formula for logistic population growth lol)", "Solution_3": "I dont know a lot of biology....I just know the stuff we learned in school (8th grade) I hope this forum won't do just complicated stuff only but a balance of easy stuff and hard stuff. Otherwise its useless to me.", "Solution_4": "[quote=\"adidasty\"]Ha ha! Actually Valentin, I think this forum will draw you into it...and then you will be so intrigues by this enigmatic science that you will dump math altogether and do biology. Or maybe not ...[/quote]Indeed, maybe not :D I used to like biology, I can remember most of my classes in high-school, but I was never capable of understanding how a IBO question would look like (I imagined something like: given two alien species :alien: predict how will the kids look like given their DNA strings :D).", "Solution_5": "ha ha, yeah sometimes biology questions, especially IBO questions are worded pretty awkwardly.", "Solution_6": "[quote=\"adidasty\"]ha ha, yeah sometimes biology questions, especially IBO questions are worded pretty awkwardly.[/quote]\r\nthere is also IBO!?", "Solution_7": "[quote=\"GoBraves\"]I dont know a lot of biology....I just know the stuff we learned in school (8th grade) I hope this forum won't do just complicated stuff only but a balance of easy stuff and hard stuff. Otherwise its useless to me.[/quote]I'm with GoBraves. All I have is what I've learned up through 8th grade biology. (Did we even do biology this year?)\r\n\r\nHmmm... next year though I'm taking biology D, so that will be nice. :)", "Solution_8": "yeah IBO is definitely everyones goal who loves biology. if im not mistaken, there may be an AoPSer who went to IBO by the name of Kay Aull. she got 2nd overall in the IBO (pretty much astoundingly amazing) and i think she roamed these boards too...shes at MIT now.", "Solution_9": "yeah, Kay went to IBO twice and got a silver medal in Belarus and 3rd in the world last year... she is my personal goddess. =P", "Solution_10": "yeah she is definitely awesome. no doubts about that.", "Solution_11": "18 years okay\n\n" } { "Tag": [], "Problem": "Images...what else? *confused*", "Solution_1": "image for CCMO1999Q3", "Solution_2": "Angle picture for Dutch NWO 1991 Q2", "Solution_3": "Circle-triangle picture for NWO 1991 Q6", "Solution_4": "NWO 1992 Q3", "Solution_5": "NWO 1992 Q5", "Solution_6": "nt", "Solution_7": "a quick rough sketch on a reply to show 6 points of intersection between a triangle and a circle" } { "Tag": [ "combinatorics open", "combinatorics" ], "Problem": "I had two number $a,b$ , $a,b>1$ And $a+b<100$ . I have two friend $A\\&B$ ; A knew accumulate of $a,b$ And B knew\r\nValue worth of a+b . After that , i asked A: Can you read name of two number? . A said : I can't . B said :Ofcouse you unable \r\n divine they . A said : Oh i know they . B said : So do i . \r\nTo everybody : What are they ? ($a\\&b$)", "Solution_1": "Answer: 4 and 13\r\n\r\nA's number is 52. A cannot determine whether the numbers are 2 and 26, or 4 and 13.\r\n\r\nB's number is 17. B sees that the possible numbers are 2 and 15, 3 and 14, 4 and 13, 5 and 12, 6 and 11, 7 and 10, or 8 and 9. Then A's number would respectively be 30, 42, 52, 60, 66, 70, 72. All of these numbers can be resolved into two factors >1 in at least two ways. Therefore B knows that A cannot determine the numbers.\r\n\r\nA assumes for the sake of contradiction that the numbers are 2 and 26. Then B would hold 28. One possibility that B would then consider is that the numbers are 5 and 23. Since these are primes, no other numbers >1 share their product 115. A would be able to determine the numbers in this case, so B would not know that A could not determine the numbers. Since B reported to the contrary, A now knows that the numbers are 4 and 13.\r\n\r\nTo facilitate the verification of B's last statement, I first note that in general A cannot determine the numbers at the outset unless both are primes, or one is a prime and the other its square. Let us call a number under 100, but over 3, which is the sum of two primes, or of a prime and its square, a [i]D-number.[/i] Then all even numbers, as well as numbers 2 more than a prime, are D-numbers.\r\n\r\nB's first statement shows that the sum of this problem's two numbers is a non-D-number. Since this is all that A knows (aside from A's own number), A can find the numbers if and only if exactly one of the factor pairs of A's number has a non-D sum. B now tries each possibility in turn, and all but one give two factor pairs with non-D sums:\r\n\r\n30: 2,15=17; 5,6=11\r\n42: 3,14=17; 2,21=23\r\n52: 4,13=17 only\r\n60: 5,12=17; 3,20=23\r\n66: 6,11=17; 2,33=35\r\n70: 7,10=17; 2,35=37\r\n72: 8,9=17; 3,24=27\r\n\r\nThus B can conclude that the numbers are 4 and 13.\r\n\r\nTo find this solution, I used trial and error mixed with some eliminative mathematics. Since B's number is non-D, I paralleled B's last calculation using the smallest non-D-number 11:\r\n\r\n18: 2,9=11 only\r\n24: 3,8=11 only\r\n28: 4,7=11 only\r\n30: 5,6=11; 2,15=17\r\n\r\nB cannot now determine the numbers, but is instead faced with three possibilities. Taking the next non-D-number 17, I arrived at the solution (4,13). I have not looked at larger cases but the way in which the problem is phrased implies a unique solution.", "Solution_2": "Thank you \r\nIn may class , No one can kill it :D" } { "Tag": [], "Problem": "Given a random string of 33 bits (0 or 1), how many (they can overlap) occurence of two consecutive 0's would you expect?", "Solution_1": "[quote=\"dingzhou\"]Given a random string of 33 bits (0 or 1), how many (they can overlap) occurence of two consecutive 0's would you expect?[/quote]\r\nHold on, what about more than 2 in a row? how many do they count as? nC2?[/hide]", "Solution_2": "[hide](1/2)*(1/2)=1/4 chance to get two consecutive 0's.\n(1/4)*32=8 \n32 because there are 32 pairs of numbers total. Is 8 correct?[/hide]", "Solution_3": "whoa.. this is really similar to the 1989 AHSME #30 (thanks to the future for pointing it out a few months ago).. which i never understood :blush:", "Solution_4": "[quote=\"IamSleepy\"][hide](1/2)*(1/2)=1/4 chance to get two consecutive 0's.\n(1/4)*32=8 \n32 because there are 32 pairs of numbers total. Is 8 correct?[/hide][/quote]\r\nUh....no...\r\nAnd also, there's a good reason why I made the above post.\r\nConsider 000. How many pairs of zeros are there? well, first 2 zeros are a pair, and second 2 are as well.", "Solution_5": "[quote=\"serialk11r\"][quote=\"IamSleepy\"][hide](1/2)*(1/2)=1/4 chance to get two consecutive 0's.\n(1/4)*32=8 \n32 because there are 32 pairs of numbers total. Is 8 correct?[/hide][/quote]\nUh....no...\nAnd also, there's a good reason why I made the above post.\nConsider 000. How many pairs of zeros are there? well, first 2 zeros are a pair, and second 2 are as well.[/quote]Really? I thought that I was considering that... It was a 1/4 chance to get a pair of 00 and there were 32 pairs (out of 33 numbers), not 16 or whatever, just like there are 2 pairs in 000.", "Solution_6": "[hide=\"Solution\"]\nLet $ f(n)$ be the average number of 00s in a bit string of length n.\n\nClearly, $ f(2) = \\frac{1}{4}$.\n\n\\begin{eqnarray*}f(n)&=&\\frac{1}{2^n} ( \\mbox{(no. of 00s in strings starting with 1) + (no. of 00s in strings starting with 0)}\\\\ &=& \\frac{1}{2^n} ( (2^{n-1}\\cdot f(n-1)) + (2^{n-2} + 2^{n-1}\\cdot f(n-1))\\\\ &=& \\frac{1}{4} + f(n-1)\\end{eqnarray*}\n$ \\implies f(33) = 32\\cdot \\frac{1}{4} = 8$\n[/hide]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all the prime numbers $ p$ such that the number $ p^2\\plus{}11$ has less than $ 11$ divisors .", "Solution_1": "Let $ n\\equal{}p^2\\plus{}11$.\r\nSuppose that $ n>144$ and $ p>3$. So, it's easy to see that $ n$ is divisible by $ 1,2,3,4,6,12$ and $ n,\\frac{n}{2},\\frac{n}{3},\\frac{n}{4},\\frac{n}{6},\\frac{n}{12}$. So, $ n$ has 12 or more divisors.\r\nSo, $ p^2\\plus{}11\\leq 144$. Taking $ p\\equal{}2,3,5,7,11$, we will get $ n\\equal{}15,20,36,60,132$.\r\nAnswers: $ p\\equal{}2$, $ p\\equal{}3$, $ p\\equal{}5$.", "Solution_2": "by little theorem of Fermat and by supposing that gcd(p,3)=1 $ p\\neq2$ we have\r\n$ p^2\\equiv1 (mod(3)) p^2\\equiv1 (mod(4)) \\Rightarrow p^2\\equiv1(mod(12))$\r\nso $ p^2\\plus{}11 \\equiv0 (mod(12))$", "Solution_3": "12 has 6 divisor so if p^12+11=12k implies k=1( because if the exponent of k i more than 1 p^12+11 will have more than 12 divisor) wich implies p=1 wich is impossible. so p is less than 5 .\r\nconclusion p=3 or p=2 or p=5." } { "Tag": [], "Problem": "According to the Remainder and Factor Theorems, what can you conclude about x = 2 if P(x) is divided by (x \u2013 2) and the remainder is 10?", "Solution_1": "P(2)=10, by the Remainder/Factor theorem.\r\n\r\nPlease put these in the HSB forum.", "Solution_2": "additional to what worthawholebean said, x-2 is not a factor of p(x) since $p(2)\\neq 0$.\r\n\r\n~lyra" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "1)suppose $p,q$ are odd primes then show that $x^2=1$ has only 4 solutions in $\\mathbb{Z}_{pq}$\r\nHint:Use Chinese remainder theorem\r\n\r\n2) Show that $\\mathbb{Q}[X]/(X^3-X)$ is isomorphic to $\\mathbb{Q}\\times\\mathbb{Q}\\times\\mathbb{Q}$\r\n$\\mathbb{Q}$ meaning field of rationals", "Solution_1": "Can anyone here give me some hints about how to attack these problems", "Solution_2": "On the first: Since $\\mathbb{Z}_p$ and $\\mathbb{Z}_q$ are fields (of characteristic not 2), each has exactly two square roots of one. By CRT, $\\mathbb{Z}_p\\times\\mathbb{Z}_p$ is isomorphic to $\\mathbb{Z}_{pq}$.\r\n\r\nOn the second: By CRT, $\\mathbb{Q}[X]/(X^3-X)$ is isomorphic to $\\mathbb{Q}[X]/(X-1)\\times\\mathbb{Q}[X]/(X)\\times\\mathbb{Q}[X]/(X+1)$\r\n\r\nActually, this would work if $\\mathbb{Q}$ were replaced by any field of characteristic not 2." } { "Tag": [], "Problem": "\u03a3\u03b5 \u03ad\u03bd\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03c0\u03bf\u03c5 \u03c3\u03c4\u03b7 \u03a1\u03bf\u03c5\u03bc\u03b1\u03bd\u03af\u03b1 \u03c4\u03bf \u03ad\u03c7\u03bf\u03c5\u03bd \u03c9\u03c2 ..\u03b2\u03af\u03b2\u03bb\u03bf(\u03cc\u03c0\u03c9\u03c2 \u03c0\u03c7 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b5\u03bc\u03b5\u03af\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03a4\u03cc\u03b3\u03ba\u03b1 , \u039a\u03b1\u03bd\u03ad\u03bb\u03bb\u03bf , \u03a0\u03ac\u03bb\u03bb\u03b1 \u03ba\u03bb\u03c0) \u03b2\u03c1\u03ae\u03ba\u03b1 \u03c7\u03b8\u03b5\u03c2 \u03c4\u03c5\u03c7\u03b1\u03af\u03b1 \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03ba\u03ac\u03c4\u03c9 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 . \u03a4\u03b7\u03bd \u03b2\u03ac\u03b6\u03c9 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03cc\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03ba\u03b1\u03bb\u03ad\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 , \u03b1\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u03b4\u03b9\u03cc\u03c4\u03b9 \u03ad\u03c7\u03b5\u03b9 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd.\r\n\r\n[color=red][b]\u0391\u03a3\u039a\u0397\u03a3\u0397[/b][/color] \r\n\r\n\u0394\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf ABC \u03bc\u03b5 $ \\angle A > 90$.\r\n\r\n \u03b1) \u039d\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03cc\u03c4 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03b4\u03cd\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 K,L \u03c3\u03c4\u03b7\u03bd \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac BC , \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 :\r\n\r\n $ AK^2 \\equal{} KB \\cdot KC$ , $ AL^2 \\equal{} LB \\cdot LC$ \u03ba\u03b1\u03b9 \u03bf\u03b9 $ AK,AL$ \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b5\u03c2.\r\n \r\n \u03b2) \u0391\u03bd AD , AM \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03cd\u03c8\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03b7 \u03b4\u03b9\u03ac\u03bc\u03b5\u03c3\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ ABC$ , \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03cc\u03c4\u03b9 $ KM \\equal{} DL$.\r\n\r\n[b]\u03a3\u03c7\u03cc\u03bb\u03b9\u03bf [/b]\r\n\r\n[i]\u0394\u03b5\u03bd \u03ad\u03c7\u03c9 \u03c8\u03ac\u03be\u03b5\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b1 \u03bd\u03b1 \u03b4\u03c9 \u03bc\u03ae\u03c0\u03c9\u03c2 \u03c4\u03bf ''\u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b5\u03c2 '' \u03b4\u03b5 \u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 . \u0394\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03ce\u03c3\u03c9 \u03ba\u03b1\u03bb\u03ac \u03c4\u03bf \u03bd\u03cc\u03b7\u03bc\u03b1 \u03c4\u03bf\u03c5 \u03c1\u03bf\u03c5\u03bc\u03ac\u03bd\u03b9\u03ba\u03bf\u03c5 \u03ba\u03b5\u03b9\u03bc\u03ad\u03bd\u03bf\u03c5 , \u03b1\u03bb\u03bb\u03ac \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03b5\u03ba\u03b5\u03af \u03c4\u03bf \u03ad\u03c7\u03b5\u03b9 , \u03c4\u03bf \u03b2\u03ac\u03b6\u03c9 \u03ba\u03b1\u03b9 \u03b3\u03c9.[/i]\r\n\r\n\u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2", "Solution_1": "[hide]\u0391\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03b5\u03c4\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03cc\u03c4\u03b9 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ K,\\ L,$ \u03c4\u03bf\u03bc\u03ae\u03c2 \u03c4\u03b7\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac\u03c2 $ BC,$ \u03b1\u03c0\u03cc \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf $ (K)$ \u03b4\u03b9\u03b1\u03bc\u03ad\u03c4\u03c1\u03bf\u03c5 $ AO,$ \u03cc\u03c0\u03bf\u03c5 $ O$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03ba\u03ad\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 \u03c0\u03b5\u03c1\u03af\u03ba\u03c5\u03ba\u03bb\u03bf\u03c5 \u03c4\u03bf\u03c5 $ \\bigtriangleup ABC$ (\u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03ad\u03be\u03c9 \u03b1\u03c0\u03cc \u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf, \u03b1\u03c6\u03bf\u03cd $ \\angle A > 90^{o}$), \u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03b7\u03bd \u03b9\u03b4\u03b9\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03bf\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1.\n\n\u0395\u03c0\u03b5\u03b9\u03b4\u03ae \u03c4\u03ce\u03c1\u03b1, $ K'L'\\parallel BC,$ \u03cc\u03c0\u03bf\u03c5 $ K'\\equiv (O)\\cap AK$ \u03ba\u03b1\u03b9 $ L'\\equiv (O)\\cap AL,$ \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03ac\u03bc\u03b5\u03c3\u03b1 \u03cc\u03c4\u03b9 $ \\angle BAK \\equal{} \\angle CAL$ \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 \u03bf\u03b9 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 $ AK,\\ AL,$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b5\u03c2 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b7 \u03b3\u03c9\u03bd\u03af\u03b1 $ \\angle A$ \u03c4\u03bf\u03c5 $ \\bigtriangleup ABC.$\n\n$ \\bullet$ \u03a4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c0\u03ac\u03bd\u03c4\u03bf\u03c4\u03b5 \u03c3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03b5\u03b3\u03b3\u03c1\u03ac\u03c8\u03b9\u03bc\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03c0\u03bb\u03b5\u03c5\u03c1\u03bf \u03c4\u03bf\u03c5 \u03bf\u03c0\u03bf\u03af\u03bf\u03c5 \u03b7 \u03bc\u03af\u03b1 \u03b4\u03b9\u03b1\u03b3\u03ce\u03bd\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03ac\u03bc\u03b5\u03c4\u03c1\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c4\u03bf\u03c5 ( \u03bf\u03b9 \u03c0\u03c1\u03bf\u03b2\u03bf\u03bb\u03ad\u03c2 $ M,\\ D,$ \u03c4\u03c9\u03bd $ O,\\ A$ \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03bf\u03af\u03c7\u03c9\u03c2 \u03b5\u03c0\u03af \u03c4\u03b7\u03bd $ KL,$ \u03b9\u03c3\u03b1\u03c0\u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03bf\u03c5 \u03bc\u03ad\u03c3\u03bf\u03c5 \u03c4\u03bf\u03c5 $ KL$ ) \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 \u03b4\u03b5\u03bd \u03c3\u03c7\u03b5\u03c4\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b9\u03c3\u03bf\u03b3\u03ce\u03bd\u03b9\u03b5\u03c2 $ AK,\\ AL.$[/hide]\r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2." } { "Tag": [], "Problem": "What product is obtained in the following sequence:\r\n\r\n(a) Cyclohex-2-enone + HOCH2CH2OH/H+ ----> A\r\nA + O3 ----> B\r\nB + (CH3)2S ----> C\r\nC + H+ ----> D", "Solution_1": "A-cyclohex-2-enoneketal\r\nB-ozonide\r\nC-dial with the ketal still remaining(Me2S is a reducing agent and it effects reductive ozonolysis)\r\nD-2-oxohexan1,6-dial", "Solution_2": "What is the use of H+ in CH2OH-CH2OH/ H+ ?\r\n\r\nEDIT : Was in a state of delirium.\r\nBy the way cant H+ (bluntly H+ and not dry HCl) effect dehydration of ethylene glycol\r\nand is it H+ or dry HCl which is used for ketal formation?", "Solution_3": "The mechanism of the Ketal formation states clearly that ketal is formed only in acidic medium \r\nand in basic medium only hemiketal is formed.\r\n\r\np.s.:-It is appalling coming from you ROHIT\r\n\r\nEDIT:-eppavom nee thoongi vayiyara enna dry HCl will favour the formation of the 1,2-dichloro ethane and not elimination." } { "Tag": [], "Problem": "The sequence 1,2,3,4,6,7,8,9,11,$ \\dots$ is formed by omitting every fifth natural number. Find the sum of the first 100 terms of this sequence.", "Solution_1": "$ \\sum_{k\\equal{}1}^{125}k \\minus{} \\sum_{k\\equal{}1}^{25}5k \\equal{} 7875 \\minus{} 1625 \\equal{} 6250$" } { "Tag": [ "algebra", "system of equations", "number theory", "difference of squares", "special factorizations" ], "Problem": "Like substituting variables, using factors etc.\r\n\r\nGive an example and if later on proven hard give the solution. I have never come across the factorisation types in the school text books (not that I know of) and im sure there are many sites where these subjects are dealt with so if anyone knows about such a website please post it.\r\n\r\nwould someone like to start?", "Solution_1": "Do you want a sample system of equations to solve?\r\n\r\nFind all the pairs (a,b) of real numbers with $a>b>0$ for which:\r\n\r\n(i) $a^3-3a^2b+3ab^2-b^3+b(2a^2-7ab+5b^2)=0$\r\n(ii) $5a^2-3ab-a-3b=4$\r\n\r\nAnd yes, [u]solutions[/u] to this system do exist.", "Solution_2": "ok, I tried substituting b with $b= \\displaystyle\\frac{5a^{2}-a-4}{3a+3}$ using this I got a=2,83 and b=1,85 but I had to use the calculator, how would you do it?", "Solution_3": "You could factor it out.", "Solution_4": "Here are some more systems:\r\n\r\n1) Find all pairs of $(x,y) \\in \\mathbb{Z}$ such that $x^4-x^2 y^2 + x^2 y^3 - y^5 = 801$.\r\n\r\n2) Find all pairs of $(a,b) \\in \\mathbb{Z}$ such that $a^3+a^2\\sqrt{b} - ab^2+12\\frac{b^2}{a}=9ab$ and $\\frac{ab}{b^3-a^4}=\\frac{-3a}{3a+2b}$.", "Solution_5": "[hide=\"1\"]\n$\\left(x^{2}+y^{3}\\right)\\left(x+y\\right)\\left(x-y\\right)=3^{2}\\cdot 89$\n[/hide]\r\nDo solutions exist?", "Solution_6": "[quote=\"cincodemayo5590\"]Here are some more systems:\n\n1) Find all pairs of $(x,y) \\in \\mathbb{Z}$ such that $x^4-x^2 y^2 + x^2 y^3 - y^5 = 801$.\n\n2) Find all pairs of $(a,b) \\in \\mathbb{Z}$ such that $a^3+a^2\\sqrt{b} - ab^2+12\\frac{b^2}{a}=9ab$ and $\\frac{ab}{b^3-a^4}=\\frac{-3a}{3a+2b}$.[/quote]\r\n\r\n#1\r\n[hide=\"hint\"]what technique do you alway use in number theory?\n[/hide]\n[hide=\"hint\"]factor![/hide]\n\ni like #2, at first it does not seem like there is a good way to solve it, but there are a few things to notice\n[hide=\"hint\"]what can you say about $b$?\n[/hide]\n[hide=\"hint\"]can you make it so it is $\\sum (k)a^n*b^{c-n}$, where k,c are constants?[/hide]\r\nProblem source? or is it original :) ?", "Solution_7": "Take a look at how I factored #1. From the factored version, it seems like no solutions exist.", "Solution_8": "[quote=\"Altheman\"]\n\ni like #2, at first it does not seem like there is a good way to solve it, but there are a few things to notice\n[hide=\"hint\"]can you make it so it is $\\sum (k)a^n*b^{c-n}$, where k,c are constants?[/hide]\n[/quote]\r\n\r\nme likie :D\r\n\r\n(edit: darn formatting... grrr...) :ninja:", "Solution_9": "[quote=\"amcavoy\"]Take a look at how I factored #1. From the factored version, it seems like no solutions exist.[/quote]\r\n\r\n\r\n[hide=\"Hint 1\"] Factor so that there are only two factors. Why? Because there are two variables! ;) [/hide]\n[hide=\"Hint 2\"] Think!! $\\mathbb{Z^+}$! What does that mean? What can you do knowing that they are integers? [/hide]\n\n\nYes, solutions do exist! And these are both original problems :P .\n\nIf you really need it...\n[hide] For both systems, $ab0,a^2\\plus{}b^2\\plus{}c^2\\equal{}3$.\r\nFind minimum of S = $ \\frac{a^2}{b} \\plus{} \\frac{b^2}{c} \\plus{} \\frac{c^2}{a}$\r\nThanks :(", "Solution_1": "By CSB you have S>=a+b+c then use a^2+b^2+c^2+3>=2(a+b+c).It's easy that minimum is 3 if a=b=c=1", "Solution_2": "I can't exactly understand what hercegovac meant.\r\n\r\nAnyway, try Holder Inequality we have\r\n\r\n$ \\left( \\frac{a^2}{b}\\plus{} \\frac{b^2}{c}\\plus{} \\frac{c^2}{a} \\right)^2\\cdot (a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2) \\ge (a^2\\plus{}b^2\\plus{}c^2)^3\\equal{}27$\r\n\r\non the other hand,\r\n\r\n$ a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2 \\le \\frac{1}{3}(a^2\\plus{}b^2\\plus{}c^2)^2\\equal{}3$\r\n\r\nWe get\r\n\r\n$ \\frac{a^2}{b}\\plus{} \\frac{b^2}{c}\\plus{} \\frac{c^2}{a} \\ge 3$\r\n\r\nEquality occurs iff $ a\\equal{}b\\equal{}c\\equal{}1$." } { "Tag": [], "Problem": "I was reading about a problem and was wondering what the little upside down capital A means.\r\nThanks,", "Solution_1": "What you are looking for is $ \\forall$ right? This means \"For all\"\r\n\r\nExamples:\r\n\r\n$ x^2\\ge 0\\text{ } \\forall x \\in \\mathbb{R}$. Is read as: For all $ x$ in real numbers $ x^2$ is non-negative.\r\n\r\n$ x\\ge 0\\text{ } \\forall x \\in \\mathbb{N}$. Is read as: For all $ x$ in natural numbers $ x$ is non-negative.", "Solution_2": "Yeah,\r\nThanks." } { "Tag": [ "algebra", "polynomial", "Pascal\\u0027s Triangle" ], "Problem": "Find $f(m)=\\sum_{n=1}^{m}\\sum_{k=1}^{n}\\sum_{j=1}^{k}j$ in a closed form. :)\r\n\r\nMasoud Zargar", "Solution_1": "The answer is $\\frac{1}{24}m(m+1)(m+2)(m+3).$", "Solution_2": "Could you please shwo a solution?:)", "Solution_3": "$f(m)=\\sum_{n=1}^{m}\\sum_{k=1}^{n}\\sum_{j=1}^{k}j$\r\n\r\n$=\\sum_{n=1}^{m}\\sum_{k=1}^{n}\\frac{j(j+1)}{2}$\r\n\r\n$=\\sum_{n=1}^{m}\\frac{1}{2}\\cdot \\frac{n(n+1)(n+2)}{3}$\r\n\r\n$=\\frac{1}{6}\\sum_{n=1}^{m}n(n+1)(n+2)$\r\n\r\n$=\\frac{1}{24}m(m+1)(m+2)(m+3).$", "Solution_4": "[quote=\"boxedexe\"]Find $f(m)=\\sum_{n=1}^{m}\\sum_{k=1}^{n}\\sum_{j=1}^{k}j$ in a closed form. :)\n\nMasoud Zargar[/quote]\r\n[hide]$f(m)$\n$=\\sum_{n=1}^{m}\\sum_{k=1}^{n}\\sum_{j=1}^{k}j$\n$=\\sum_{n=1}^{m}\\sum_{k=1}^{n}(\\frac12\\cdot{k(k+1)})$\n$=\\sum_{n=1}^{m}\\frac12\\Bigg[\\frac16{n(n+1)(2n+1)}+\\frac12{n(n+1)}$\n$=\\sum_{n=1}^{m}\\frac16{n(n+1)(n+2)}$\n$=\\frac16\\sum_{n=2}^{m+1}{(n-1)n(n+1)}$ ... (Note the change in bounds; this makes things easier for us.)\n$=\\frac16\\sum_{n=2}^{m+1}({n^{3}-n})$\n$=\\frac16\\Bigg[\\frac{{(m+1)}^{2}{(m+2)}^{2}}{4}-1\\Bigg]-\\frac16\\Bigg[\\frac{(m+1)(m+2)}{2}-1\\Bigg]$\n$=\\frac1{24}(m+1)(m+2)\\Bigg[(m+1)(m+2)-2\\Bigg]$\n$=\\frac1{24}m(m+1)(m+2)(m+3)$\n[/hide]", "Solution_5": "Call it the sum of the sum of the sum of the first j integers.\r\n\r\nYou can use pascal's triangle, looking down each diagonal. You will notice that each summation symbol you add to that thing you go down another diagonal. This gives easy representation for what you are asking for in combinatoric form.\r\n\r\nFirst diagonal (all ones) $\\binom{n-1}{n-1}$\r\nSecond diagonal (natural numbers) $\\binom{n}{n-1}$\r\nThird diagonal (triangular numbers) $\\binom{n+1}{n-1}$\r\nFourth diagonal (triangular sums) $\\binom{n+2}{n-1}$ \r\nFifth diagonal (sum of first n triangular sums) $\\binom{n+3}{n-1}$\r\n\r\nWith the last one being what is asked for. When expressed as polynomial, it is \r\n\r\n$\\frac{n(n+1)(n+2)(n+3)}{24}$", "Solution_6": "Generalization: \\[\\sum_{n_{k}=1}^{m}\\sum_{n_{k-1}=1}^{n_{k}}\\cdots\\sum_{n_{2}=1}^{n_{3}}\\sum_{n_{1}=1}^{n_{2}}n_{1}=\\prod_{j=1}^{k}\\frac{m+k-1}{j}\\]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "solve the equation \r\nx(x+1)^2=\\sqrt[3]{2x+1}+1\r\nthank very much", "Solution_1": "$ x(x\\plus{}1)^2\\equal{}\\sqrt[3]{2x\\plus{}1}\\plus{}1$\r\n$ (x(x\\plus{}1)^2\\minus{}1)^3\\equal{}2x\\plus{}1$\r\n$ P(x)\\equal{}(x(x\\plus{}1)^2\\minus{}1)^3\\minus{}2x\\minus{}1$\r\nThe trivial root is $ x\\equal{}\\minus{}1$ and \r\nmaybe it helps that there are another ones at aprox $ 0,761225955431823$ or $ \\minus{}1,39063691957206$" } { "Tag": [ "induction", "modular arithmetic" ], "Problem": "Prove the following by principle of induction :\r\n\r\n$ 2.7^n \\plus{} 3.5^n \\minus{} 5$ is divisible by $ 24$ for all n belong to N. \r\n\r\nPlease help. Thanks a lot.", "Solution_1": "Well it's not true; just take $ n\\equal{}1$. $ 2.7^{1}\\plus{}3.5^{1}\\minus{}5\\equal{}1.2$, which is not divisible by 24.", "Solution_2": "Am I being silly? I thinks it's true: for $ n\\equal{}1$, I get 24. \r\n\r\nThen for the inductive step, $ 2.7^{n\\plus{}1} \\plus{}3.5^{n\\plus{}1} \\minus{} 5 \\equal{} 2.7^{n}\\plus{}3.5^{n}\\minus{}5 \\plus{} 2.7^{n}.6 \\plus{} 3.5^{n}.4$ \r\n\r\nand $ 2.7^{n}.6\\plus{}3.5^{n}4 \\equal{} 12(7^{n} \\plus{} 5^{n})$ \r\n\r\nand since $ 7^{n} \\plus{} 5^{n}$ is even, we are done.", "Solution_3": "How did you get $ 24$? Did you mistake $ 2.7$ for $ 2\\cdot7$? Because $ 2.7 \\plus{} 3.5 \\minus{} 5$ is clearly not an integer.\r\nBy the way, for the multiplication sign, use [code]\\cdot[/code].", "Solution_4": "Not sure I mistook anything: I think ninni meant the dots to indicate multiplation and not decimal points. It works out nicely and doesn't make sense otherwise.", "Solution_5": "That may be, but writing $ 2.7$ will lead many to think it means the number 2 and seven tenths. Either use the \\cdot command as pascal12 suggested, or \\times, or simply use parentheses.", "Solution_6": "[quote=\"JRav\"]That may be, but writing $ 2.7$ will lead many to think it means the number 2 and seven tenths. Either use the \\cdot command as pascal12 suggested, or \\times, or simply use parentheses.[/quote]\r\n\r\nninni is a multiple repeat violator who has been told about this weeks ago\r\n on different problems on the thread. I just sent a PM to him/her.", "Solution_7": "It is also easy to prove without induction.\r\nNote that $ 7^2 \\equiv 1$ and $ 5^2 \\equiv 1 \\pmod {24}$. So if n is odd, $ 2 \\cdot 7^n \\plus{} 3 \\cdot 5^n \\minus{} 5 \\equiv 2 \\cdot 7 \\plus{} 3 \\cdot 5 \\minus{} 5 \\equiv 0 \\pmod {24}$. Also if n is even, $ 2 \\cdot 7^n \\plus{} 3 \\cdot 5^n \\minus{} 5 \\equiv 2 \\cdot 1 \\plus{} 3 \\cdot 1 \\minus{} 5 \\equiv 0 \\pmod {24}$", "Solution_8": "[quote=\"Arrange your tan\"][quote=\"JRav\"]That may be, but writing $ 2.7$ will lead many to think it means the number 2 and seven tenths. Either use the \\cdot command as pascal12 suggested, or \\times, or simply use parentheses.[/quote]\n\nninni is a multiple repeat violator who has been told about this weeks ago\n on different problems on the thread. I just sent a PM to him/her.[/quote]\r\n\r\nsorry. I really appologize for this. I will not repeat it again. sorry." } { "Tag": [ "algebra", "polynomial", "induction", "complex numbers", "real analysis", "real analysis solved" ], "Problem": "Prove that : If the polynomial $ P(x) $ has exactly k coefficients which are different from 0 , then the equation $ P(x) = 0 $ has $ \\leq 2k-1 $ different roots", "Solution_1": "OK! How about induction.\r\n Let $P(x) = x^nQ(x)$ where $Q(0)$ is not equal to $0$\r\n Take $Q'$, which has no more than $k-1$ nonzero coefifcents, and process by induction!!", "Solution_2": "This must be done over the real numbers; over the complex numbers $x^n-1$ has two nonzero coefficients and $n$ distinct roots.\r\n\r\nOf course, dickchimney's argument (Rolle's theorem) works.\r\n\r\nFor another approach, use Descartes' rule of signs: there can be at most $k-1$ positive real roots and at most $k-1$ negative real roots. 0 is a possible root, so we have at most $2k-1$ distinct real roots." } { "Tag": [ "MIT", "college", "Putnam", "HMMT", "Harvard" ], "Problem": "Basically you can use this to post any further questions you have that weren't answered in the math jam.\r\n\r\nIn particular, I wanted to further address Karth's question about how you can practice for the Putnam.\r\n\r\nOne option is the problem solving seminar. However, I warn you that there is a very low enrollment limit and many people want to take it. To make the seminar it usually happens that you need to have made it to at least MOP or the equivalent for internationals. The specifics obviously change from year to year depending on the incoming freshman class.\r\n\r\nThis shouldn't deter you. There are plenty of other options for studying for the Putnam even if you don't get into the seminar. The UMA (undergraduate mathematics assosciation) schedules Putnam training sessions. Additionally, it should be pretty easy to find a couple of people who also want to train for the Putnam. In my dorm, for example, I actually got a lounge setup for problem solving stuff. There are 12 people signed up for the lounge of which a little over half are active. We're getting $\\$$ 355 a semester funding for being a lounge to use for food, trips, and whatnot.\r\n\r\nThen there's Random Hall the dorm which is described by some as a yearlong MOP. Basically, there is a large concentration of math/computer science people there.\r\n\r\nYou really won't find much time during the week to practice for Putnam. Well...you probably could. It just depends on what classes you're taking and your EC's.\r\n\r\nThe point is that it should be really easy for you to train for Putnam at MIT.", "Solution_1": "Hey Joe, did you graduate early or is my memory faulty in thinking that you were due to attend MIT beginning 2007?\r\n\r\nAnyway, is Putnam a big thing over there at MIT? How about planning for HMMT?", "Solution_2": "The problem seminar that joml88 mentioned is called Problem Solving Seminar and is taught by Richard Stanley, an old problem contributor to the Putnam Competition.\r\n\r\nFrom what I've been told, most people who end up doing well on the Putnam exam usually prepare by themselves. Obviously, the seminar can help train. The Undergraduate Math Association has a series of Putnam talks this year by Reid Barton, Daniel Gulotta, Matthew Ince, and Daniel Kane.\r\n\r\nSunday's HMMT meeting had approximately 30 people, almost all of whom were MIT undergraduates, since the meeting was being held at Random Hall.", "Solution_3": "When's the transcript gonna be up?", "Solution_4": "[quote=\"Elemennop\"]When's the transcript gonna be up?[/quote]Hopefully later this morning.", "Solution_5": "[quote=\"MithsApprentice\"]Hey Joe, did you graduate early or is my memory faulty in thinking that you were due to attend MIT beginning 2007?\n\nAnyway, is Putnam a big thing over there at MIT? How about planning for HMMT?[/quote]\r\n\r\nNope, I was always on track to graduate high school in 06.\r\n\r\nI'm guessing Putnam is bigger here than anywhere else. Honestly though it's not that big a deal for me. I mean, I'm still going to do a lot of problem solving, hopefully, but there are other things that I put ahead of Putnam importance-wise.\r\n\r\nConcerning HMMT, we did have an introductory meeting for that over the weekend, like towers said. There might have been one Harvard person there. In general, I think MIT does the bulk of the planning for HMMT. In addition to the ~30 people at the meeting I'm guessing there are a few more who will help out.\r\n\r\nTo give an accurate description of MIT: not everyone here is a math problem solving enthusiast. The majority of people probably aren't. It's just that there is a large minority of people who are...and those who aren't enthusiasts are still good at math. And those who aren't even good at math...well, they don't really exist here. The place where you'll notice how many exceptionally good math students are here is in your math class." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "The shaded region consists of 16 congruent squares. If $ PQ \\equal{} 6$ cm, what is the area of the entire shaded region?\n\n[asy]for(int i = 0; i < 5; ++i)\n{\n\tfor(int j = 0; j < 2; ++j)\n\t{\n\t\tfilldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,gray,linewidth(2));\n\t}\n}\n\nfor(int i = 0; i < 2; ++i)\n{\n\tfor(int j = 0; j < 5; ++j)\n\t{\n\t\tfilldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,gray,linewidth(2));\n\t}\n}\n\ndraw((2,2)--(5,2)--(2,5)--(2,2)--cycle,linewidth(2));\n\nlabel(\"P\",(2,5),N);\nlabel(\"Q\",(5.2),E);[/asy]", "Solution_1": "Let x be the side of a shaded small square.\r\nIf Q and P are two vertices of the unshaded triangle (Is the picture off...?), and two legs are each 3 sides of the cube, then the triangle is a 45-45-90 triangle.\r\nSo,\r\n$ x \\equal{} \\frac{hypotenuse}{\\sqrt{2}} \\equal{} \\frac{6}{\\sqrt{2}}$\r\n\r\nSquaring x and multiplying by 16, we get:\r\n$ 16(\\frac{6}{\\sqrt{2}})^2 \\equal{} \\frac{36}{2} * 16 \\equal{} 8 * 36 \\equal{} 288$ sq cm", "Solution_2": "[quote=\"GameBot\"]The shaded region consists of 16 congruent squares. If $ PQ \\equal{} 6$ cm, what is the area of the entire shaded region?\n\n[asy]for(int i = 0; i < 5; ++i)\n{\n\tfor(int j = 0; j < 2; ++j)\n\t{\n\t\tfilldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,gray,linewidth(2));\n\t}\n}\n\nfor(int i = 0; i < 2; ++i)\n{\n\tfor(int j = 0; j < 5; ++j)\n\t{\n\t\tfilldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,gray,linewidth(2));\n\t}\n}\n\ndraw((2,2)--(5,2)--(2,5)--(2,2)--cycle,linewidth(2));\n\nlabel(\"P\",(2,5),N);\nlabel(\"Q\",(5,2),E);[/asy][/quote]\r\n\r\n$ 16\\times\\left(\\frac{6/\\sqrt{2}}{3}\\right)^2\\equal{}16\\times(\\sqrt{2})^2\\equal{}16\\times2\\equal{}\\boxed{32 \\ \\text{cm}^2}$." } { "Tag": [ "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "Prove that :\r\n\r\nif $ T: \\mathbb R \\rightarrow \\mathbb R$ is linear operator iff $ T(x) \\equal{} mx$, for some $ m\\in \\mathbb R$\r\n\r\nhow to do \" $ \\Longrightarrow$ \"", "Solution_1": "If you just write out the definition of \"linear,\" you should see something.", "Solution_2": "I tried so many times , but I couldn't see that thing .\r\n\r\nGive me hint", "Solution_3": "$ x$ is a scalar (to be pedantic, a scalar times a unit vector in a one-dimensional space). How do linear operators treat scalars?", "Solution_4": "O.K\r\n\r\nI gotta , thx" } { "Tag": [ "number theory", "prime numbers", "number theory solved" ], "Problem": "Is a number line that starts at 0 and ends at infinity possible ?\r\n\r\nhow can it exist ?\r\n\r\nShouldn't we be using a number 'curve' to measure quantities in a relativistic way.\r\n\r\nI think prime numbers are stupid.", "Solution_1": "[quote=\"penguinman007\"]Is a number line that starts at 0 and ends at infinity possible ?\n\nhow can it exist ?\n\nShouldn't we be using a number 'curve' to measure quantities in a relativistic way.\n\nI think prime numbers are stupid.[/quote]\r\nWhat is this intended to mean\u00bf :huh: \r\nSeems like SPAM to me...\r\nSo stop that!", "Solution_2": "[quote]Is a number line that starts at 0 and ends at infinity possible ?\n\nhow can it exist ?\n\nShouldn't we be using a number 'curve' to measure quantities in a relativistic way.\n\nI think prime numbers are stupid.[/quote]\r\n\r\n1) Yes.\r\n\r\n2) As a theoretical construct used by people to model aspects of the universe we observe (imagine a line that starts at zero, keeps going off towards positive, and never ends).\r\n\r\n3) Two points:\r\n- A curve is a type of line.\r\n- A straight line is fine for many purposes.\r\n\r\n4) Your opinion should take more factors into account. :rotfl:", "Solution_3": "[quote=\"penguinman007\"]\nI think prime numbers are stupid.[/quote]\r\nThey're not too fond of you, either.", "Solution_4": "Shouldn't this be removed?" } { "Tag": [ "probability" ], "Problem": "Pick 3 integers at random on the interval [0, 6]. What is the probability that none of them are\r\nwithin a distance of 1 of each other?", "Solution_1": "The total number of ways to choose 3 (ordered) integers is $ 6^{3}= 216$. The two closest integers to one another must be a distance of at least 2 from each other. \r\n\r\nThe only possible (unordered) triples we can choose are $ (0,2,4), (0,2,5), (0,2,6), (0,3,5), (0,3,6), (0,4,6), (1,3,5), (1,3,6), (1,4,6), (2,4,6)$, each can be permuted in $ 3! = 6$ ways to give an ordered triple, and the required probability is $ \\frac{60}{216}= \\boxed{\\frac{5}{18}}$.", "Solution_2": "Shouldn't there actually be $ 7^{3}$ ways to pick the three integers? You have $ \\{0,1,2,3,4,5,6\\}$.", "Solution_3": "I think lingomaniac88 is right.. So the answer should be\r\n[hide]$ \\frac{60}{343}$[/hide]" } { "Tag": [ "probability", "combinatorics unsolved", "combinatorics" ], "Problem": "Hi all,\r\n\r\nI'm new here and I'm not sure if this is the right section to post this but here is my problem:\r\n\r\n'A pack of cards is cut and the suit of the exposed card noted; the pack is then well shuffled and the process repeated. Find the average number of cuts required to expose cards of all four suits.'\r\n\r\nI don't know what probability distribution will fit this problem and I can't find a formula from the probabilities I've calculated:\r\n\r\nP(4) = 24/256\r\nP(5) = 144/1024\r\nP(6) = 600/4096\r\nP(7) = 2160/16384\r\nP(8) = 7224/65536\r\n\r\nAny help would be gratefully received. Thanks for your time.\r\njimi", "Solution_1": "How ironic that this question was just asked in a different form! See this link:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125116[/url]", "Solution_2": "Hi zgorkster\r\n\r\nThanks for your reply. That's some formula for P(X = n). No wonder I couldn't get it! Could you possibly explain to me how to arrive at the infinite sum of 25/3?\r\n\r\nIt's okay, I worked it out. Thanks again.\r\njimi" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "If $ 1 \\le r \\le n$ are integers, prove the identity:\r\n\r\n$ \\displaystyle\\sum_{d\\equal{}1}^{\\infty}\\binom {n\\minus{}r\\plus{}1}{d} \\binom {r\\minus{}1} {d\\minus{}1}\\equal{}\\binom {n}{r}.$", "Solution_1": "[hide=\"Quick\"]Rewrite the identity as $ \\sum_{d=1}^{\\infty} \\binom{n-r+1}{d}\\binom{r-1}{r-d}=\\binom{n}{r}$ (1). Arbitrarily split $ n$ objects into a group of $ n-r+1$ (Group $ 1$) and $ r-1$ (Group $ 2$) objects. In order to choose $ r$ objects from the $ n$ total, we can either choose $ 1$ object from Group $ 1$ and $ r-1$ from Group $ 2$ or $ 2$ from Group $ 1$ and $ r-2$ from Group $ 2$, etc. Thusly we can count the choosing of $ r$ objects from $ n$ in two ways, proving the equality of the L.H.S. and R.H.S. of (1).\n\nNote that $ d$ cannot be $ 0$ because we cannot choose $ r$ objects from Group $ 2$, which only has $ r-1$ objects. Also when $ d>n-r+1$ or $ d-1>r-1$, the binomial product vanishes (the sum is equally valid from $ d=1$ to $ \\min({\\{n-r+1,r\\})}$).[/hide]" } { "Tag": [ "quadratics", "function", "parameterization", "LaTeX", "algebra", "quadratic formula", "special factorizations" ], "Problem": "Can someone show me how to solve for $ t$ for the following:\r\n\r\n$ 0.83t^{2}-10t+25=0$\r\n\r\nI know the answer is $ t=\\frac{10\\pm\\sqrt17}{1.66}$ but I don't know how to get to that point.\r\n\r\nThanks.", "Solution_1": "Have you tried the quadratic formula? That's usually a good thing to try when you're faced with decimals that can'te easily be removed (unless you want to multiply by 100 and factor).", "Solution_2": "Ah... the quadratic formula. I forgot about that one.\r\n\r\nSo, $ 0.83t^{2}-10t+25=0 \\Rightarrow t=\\frac{10\\pm\\sqrt{(10)^{2}-4(0.83)(25)}}{2(0.83)}\\Rightarrow t=\\frac{10\\pm\\sqrt{17}}{1.66}$\r\n\r\n\r\n\r\nThanks.", "Solution_3": "Nobody would probably be reading this, but I want to put write something in the High School forum for my first time.\r\n\r\nIf you use the quadratic expression, $ ax^{2}+bx+c=0$, it can be eventually reduced to [b]$ \\frac{-b\\pm\\sqrt{b^{2}-4ac}}{2a}$[/b] Plug the numbers that correspond with $ a,b,$ and $ c$\r\n\r\nYou could also factorize the equation. Make sure there's no decimals and fractions.\r\n$ 0.83t^{2}-10t+25=83t^{2}-1000t+2500=0$\r\n$ (83t\\pm$_$ {)}(t\\pm$_$ {)}$\r\n........", "Solution_4": "That wouldn't work if a was irrational. For instance,\r\n\r\n$ \\sqrt{5}x^{2}+3x-4$\r\n\r\nwouldn't be factorable with your method of eliminating decimals. Besides, factoring numbers close to the triple figures is very difficult, especially if the roots aren't integers.\r\n\r\nOn another note, an alternate method to using the quadratic formula is to convert $ ax^{2}+bx+c$ to the form $ a(x-h)^{2}+k$ using the completion of a square method. The roots would then be $ h \\pm \\sqrt{\\frac{-k}{a}}$.", "Solution_5": "[quote]On another note, an alternate method to using the quadratic formula is to convert $ ax^{2}+bx+c$ to the form $ a(x-h)^{2}+k$ using the completion of a square method. The roots would then be $ h \\pm \\sqrt{\\frac{-k}{a}}$.[/quote]\r\n\r\n\"$ a(x-h)^{2}+k$\" Isn't this normally the proper notation when operating with the functions of quadratic expressions?", "Solution_6": "I'm not sure what you're asking. Most elementary functions can be expressed in such a form (with the a,b,h,k parameters), with the variation being the delimiter (type of bracket) used. However, on these forums, it seems that ax^{2} + bx + c is the more widely used quadratic formula, which is why I mentioned the \"alternative\", which I think is easier to work with.", "Solution_7": "Yes.\r\n$ a= \\ \\text{Amount of Stretch or Compression}$ Where $ a>1 \\Rightarrow \\ \\text{Stretch}$ and $ 01 \\Rightarrow \\ \\text{Stretch}$ and $ 01 \\Rightarrow \\ \\text{Stretch}$ and $ 0 4w$ there are solutions for x and y. Thus, f(z)=f(w). But there is a problem. How to show for every z and w. Just below the exercise there is a hint to use the fact that f(x+n)=f(x) for every positive x and n is a positive integer in order to remove the restriction.", "Solution_1": "Pick a point $ a$, and go both ways from it; $ f(a)\\equal{}f(b)$ if $ b^2\\ge 4a$ or $ a^2\\ge 4b$: $ b\\ge 2\\sqrt{a}$ or $ b\\le \\frac{a^2}{4}$.\r\nIf you choose $ a$ right, those overlap and cover everything.", "Solution_2": "But how do I use f(x+n)=f(x)?", "Solution_3": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179557]Why not use the topic you posted earlier?[/url]", "Solution_4": "[quote=\"danilorj\"]But how do I use f(x+n)=f(x)?[/quote]\r\n\r\nIn essence, it's fairly easy to prove that $ f(x) \\equal{} f(y)$ for $ x, y$ \"sufficiently large\" (that is, $ x, y > M$ for some constant $ M$ (in particular, $ M \\equal{} 4$ is quite easy). Because $ f(x \\plus{} n) \\equal{} f(x)$ we can \"scale down\" to values below $ M$." } { "Tag": [], "Problem": "Suppose that $ f(f(x))\\equal{}9x^2\\minus{}2$, find $ f(8)$.", "Solution_1": "Here's my reasoning for why your problem probably has no solution.\r\n\r\nSince if f(x) is a solution, then so is -f(x).\r\n\r\nTherefore, for your problem to be solvable, you must be able to prove that f(8)=0 since f(8)=-f(8)=this unknown value.\r\n\r\nBut 8 has almost no relevance to this equation, so... :o", "Solution_2": "Yeah. In this case, f(x) is not necessarily a function." } { "Tag": [], "Problem": "A palindrome is a number which reads the same forward as backward. For exapmle, 343 and 1221 are palindromes. What is the least nautral number that can be added to 40305 to create a palindrome?", "Solution_1": "[quote=\"easyas3.14159...\"]A palindrome is a number which reads the same forward as backward. For exapmle, 343 and 1221 are palindromes. What is the least nautral number that can be added to 40305 to create a palindrome?[/quote]\r\n[hide]Closest palindrome 40404. SO it is 99[/hide]\r\n\r\nUm...I don't get ur title.....", "Solution_2": "The question's about a palindrome. What's the title?", "Solution_3": "The \"title\" is the subject.\r\nGO HANG A SALAMI is \r\nIM A LASAGNA HOG spelled backwards.", "Solution_4": "OH I see. :D", "Solution_5": "[hide]We want the last two digits to be 04, so the first number that works is 40404. Subtracting 40305, we get 99.[/hide]", "Solution_6": "[hide=\"answer\"] simple, just look at it nd u can prac get it. 99[/hide]", "Solution_7": "[hide]99. 40305 + 99 = 40404.[/hide]", "Solution_8": "[quote=\"easyas3.14159...\"]A palindrome is a number which reads the same forward as backward. For exapmle, 343 and 1221 are palindromes. What is the least nautral number that can be added to 40305 to create a palindrome?[/quote]\r\n[hide]\nthe next palindrome is 40404 so you add 99[/hide]", "Solution_9": "[hide]\nThe closest palindrome is 40404: $40404-40305=\\boxed{99}$[/hide]" } { "Tag": [ "number theory", "greatest common divisor", "function", "real analysis", "real analysis unsolved" ], "Problem": "Define $ g(x): =\\mbox{max}(0,1-2|x|)$, and for $ n\\geq 2$: \r\n$ f_n(x)=\\sum_{\\{\\frac{a}{b}\\in \\mathbb{Q},a,b\\in\\mathbb{Z},(a,b)=1,0k, there must be at least one box with more than 1 object. Or use google :wink:" } { "Tag": [ "combinatorics", "AIME" ], "Problem": "Let $S$ be a set with six elements. Let $P$ be the set of all subsets of $S.$ Subsets $A$ and $B$ of $S$, not necessarily distinct, are chosen independently and at random from $P$. the probability that $B$ is contained in at least one of $A$ or $S-A$ is $\\frac{m}{n^{r}},$ where $m$, $n$, and $r$ are positive integers, $n$ is prime, and $m$ and $n$ are relatively prime. Find $m+n+r.$ (The set $S-A$ is the set of all elements of $S$ which are not in $A.$)", "Solution_1": "[hide=\"10\"]\n\nThere are 64 ways to define B, and 64 ways to define A\n\nCase 1: B is the empty set (probability 1/64)\nThe probability that either A or S-A contains B is 64/64\n\nCase 2: B has one element (6/64)\nObviously either A or S-A contains the one element in B, so 64/64\n\nCase 3: B has 2 elements (15/64)\nThere are 16 subsets A that contain all of B, and 16 sets A that contain none of B, so 32/64\n\nCase 4: B has 3 elements (20/64)\nThere are 8 subsets A that contain all of B, and 8 more that contain none of B, so 16/64.\n\nCase 5: B has 4 elements (15/64)\nThere are 4 subsets A that contain all of B, and 4 more that contain none of B, so 8/64.\n\nCase 6: B has 5 elements (6/64)\nThere are 2 subsets A that contain all of B, and 2 more that contain none of B, so 4/64.\n\nCase 7: B has 6 elements (1/64)\nThere is 1 subsets A that contain all of B, and 1 more that contain none of B, so 2/64.\n\n$\\frac{(1)(64)+(6)(64)+(15)(32)+(20)(16)+(15)(8)+(6)(4)+(1)(2)}{(64)(64)}=\\frac{1394}{2^{12}}$\n\n$=\\frac{697}{2^{11}}$\nSo we get $697+2+11=\\boxed{710}$\n[/hide]", "Solution_2": "[hide]This solution appears to be wrong, but I do not know why.\n\nYou have the three sets S-A, A-B, and B. These do not overlap and encompass S. Therefore, you have $3^{6}*2$ working options out of $2^{12}$ possibilities. This becomes $729+2+11=\\boxed{742}$.[/hide]", "Solution_3": "Every element is either in A, B, both, or neither. If the problem conditions are satisfied, we have a couple possibilities:\r\n\r\nEverybody's in A, both, or neither.\r\nEverybody's in A, B, or neither.\r\n\r\nUmmm ok so use inclusion-exclusion and that's like 2*3^6 (NO I WILL NOT USE LATEX) - 2^6 = 1394. And the denominator is 4^6, so the fraction is 697/2^11 ==> 710.", "Solution_4": "I got 710 as well.", "Solution_5": "[quote=\"Ignite168\"][hide]This solution appears to be wrong, but I do not know why.\n\nYou have the three sets S-A, A-B, and B. These do not overlap and encompass S. Therefore, you have $3^{6}*2$ working options out of $2^{12}$ possibilities. This becomes $729+2+11=\\boxed{742}$.[/hide][/quote]\r\n\r\nThat would be right if B was a subset of A, but it isn't. That is, B and S-A need not be disjoint. You need to add in the other case (where you use the three sets S - (A U B), A, B) and subtract the overlap (B empty), ie what Alex did.", "Solution_6": "I found it in terms of A and the number of terms in A, and I got the same answer.", "Solution_7": "I did casework in terms of elements of A: \n[hide]\n6 elements of A: There's 1 way to choose A, with 64 choices for B, so 64 total ways.\n5 elements of A: There're 6 ways to choose A, with 33 choices for B, so 33*6 total ways.\n4 elements of A: There're 6 choose 2 ways to choose A, and 19 choices for B (16+4-1), so 15*19 total ways.\n3 elements of A: 6 choose 3 ways to choose A, 8+8-1 = 15 choices for B, so 15*20 ways total.\n2 elements of A: Identical to the 4 elements case, 15*19\n1 element: Same as 5 elements, 33*6 ways\n0 elements: Same as 6 elements, 64 total\n\nThe sum is 1394, same as above\n[/hide]", "Solution_8": "[hide=\"Solution\"]We will use PIE by counting the number of ways to choose $A$ and $B$ such that $B$ is a subset of $A$, adding this to the number of ways to choose the sets such that $B$ is a subset of $S-A$, and subtracting the number of choices in which $B$ is a subset of both sets. The first number is given by $$\\dbinom{6}{0}*2^0+\\dbinom{6}{1}*2^1+...+\\dbinom{6}{6}*2^6,$$ since after choosing $A$ with $n$ elements, there are $2^n$ possible choices for $B$: the subsets of $A$. By the Binomial Theorem, this sum is just $3^6$. Similarly, the number of ways to choose the sets such that $B$ is contained in $S-A$ is also $3^6$. The final quantity we need to compute to account for over-counting is the number of choices in which $B$ is contained in both sets: this is just $2^6$, since $A$ can be any subset of $S$ but $B$ must stay fixed at the empty set, which is the only set that is a subset of both $A$ and $S-A$. Thus, the desired probability is $$\\frac{2*3^6-2^6}{2^{12}},$$ which, upon simplification, becomes $$\\frac{3^6-2^5}{2^{11}}.$$ This is equal to $\\frac{697}{2^{11}}$, so the answer is $697+2+11=\\boxed{710}$.[/hide]", "Solution_9": "[quote=willalphagamma]I did casework in terms of elements of A: \n[hide]\n6 elements of A: There's 1 way to choose A, with 64 choices for B, so 64 total ways.\n5 elements of A: There're 6 ways to choose A, with 33 choices for B, so 33*6 total ways.\n4 elements of A: There're 6 choose 2 ways to choose A, and 19 choices for B (16+4-1), so 15*19 total ways.\n3 elements of A: 6 choose 3 ways to choose A, 8+8-1 = 15 choices for B, so 15*20 ways total.\n2 elements of A: Identical to the 4 elements case, 15*19\n1 element: Same as 5 elements, 33*6 ways\n0 elements: Same as 6 elements, 64 total\n\nThe sum is 1394, same as above\n[/hide][/quote]\n\n\nWhy are there 33 choices for 5 elements of A? Shouldn't there be 2**5=32 ways?" } { "Tag": [ "function", "calculus", "derivative" ], "Problem": "find all functions $ f$ from $ R$ to $ R$ such that :\r\n$ xf(y)\\plus{}yf(x)\\equal{}(x\\plus{}y)f(x)f(y)$", "Solution_1": "[quote=\"greatestmaths\"]find all functions $ f$ from $ R$ to $ R$ such that :\n$ xf(y) \\plus{} yf(x) \\equal{} (x \\plus{} y)f(x)f(y)$[/quote]\r\n[hide=\"Solution\"]\n\\[ \\frac {xf(y)}{f(x)f(y)} \\plus{} \\frac {yf(x)}{f(x)f(y)} \\equal{} x \\plus{} y\n\\]\n\n\\[ \\frac {x}{f(x)} \\plus{} \\frac {y}{f(y)} \\equal{} x \\plus{} y\n\\]\n\n\\[ \\frac {a}{f(a)} \\equal{} g(a)\n\\]\n\n\\[ g(x) \\plus{} g(y) \\equal{} x \\plus{} y\n\\]\nFor an extremely small number, $ \\epsilon$, we have that $ g(y) \\plus{} g(x \\plus{} \\epsilon) \\equal{} x \\plus{} y \\plus{} \\epsilon$. Subtracting this from our previous equation, we have that $ g(x \\plus{} \\epsilon) \\minus{} g(x) \\equal{} \\epsilon\\implies \\frac {g(x \\plus{} \\epsilon) \\minus{} g(x)}{\\epsilon} \\equal{} 1\\implies g'(x) \\equal{} 1$. From here, we could say that $ g(x)$ is a line with $ g(x) \\equal{} x \\plus{} k$ for a constant $ k$. Thus, we have that $ \\frac {x}{f(x)} \\equal{} x \\plus{} k\\implies f(x) \\equal{} \\frac{x}{x\\plus{}k}$ \n[/hide]", "Solution_2": "You have a few mistakes in that one.\r\n\r\n[hide=\"Not quite right things\"]- $ g(a) \\equal{} a/f(a)$... what if $ f(a) \\equal{} 0$?\n- The derivative argument is sketchy, since $ f$ and thus $ g$ may not be continuous. From your equation, $ x \\equal{} y \\equal{} 0$ and then $ y \\equal{} 0$ gets the solution much faster, although the first error means you don't get them all.\n- When you get a solution like $ g(x) \\equal{} x \\plus{} k$, you need to make sure all values of $ k$ work. Only $ k \\equal{} 0$ does in this case.[/hide]\n\n[hide=\"Alternate solution\"]Put $ x \\neq 0, y \\equal{} 0$ in. $ xf(0) \\equal{} xf(x)f(0)$. So then $ f(0) \\equal{} f(x)f(0)$ and either $ f(0) \\equal{} 0$ or $ f(x) \\equal{} 1$ for all $ x$. Let's suppose $ f(0) \\equal{} 0$. Put in $ x \\equal{} y \\neq 0$ to get $ 2xf(x) \\equal{} 2xf(x)^2$. Thus $ f(x) \\equal{} f(x)^2$. So $ f(x) \\equal{} 1$ or $ f(x) \\equal{} 0$ for all nonzero $ x$. This gives us three solutions, and when we check we see they all do work.\n- $ f(x) \\equal{} 1$ for all $ x$.\n- $ f(x) \\equal{} 0$ for all $ x$.\n- $ f(0) \\equal{} 0$ and $ f(x) \\equal{} 1$ for nonzero $ x$.[/hide]", "Solution_3": "I agree with all the errors you've pointed out in Quatto's solution, but you made a couple yourself:\r\n\r\n[hide]First, when you had $ f(0) \\equal{} 0$ for $ f(x) \\equal{} 1$ for all $ x$, it actually should have been for all $ x \\neq 0$, so you have more than one possible solution in this case. Second, from $ f(x) \\equal{} f(x)^2$ we may conclude that for each $ x$ either $ f(x) \\equal{} 0$ or $ f(x) \\equal{} 1$. This has plenty of possibilities not on your list of 3, and you must do further work to exclude them.[/hide]", "Solution_4": "Arg, this problem is deceptive.\r\n\r\n[hide=\"Take 2\"]In the case where $ f(x) \\equal{} 1$ for all $ x \\neq 0$, we can find that $ f(0)$ can be any real value, and we get a solution.\n\nIn the case where $ f(0) \\equal{} 0$, and we have $ f(x) \\equal{} f(x)^2$, suppose $ f(a) \\neq f(b)$ with $ a,b$ distinct and nonzero. So WLOG $ f(a) \\equal{} 0$, $ f(b) \\equal{} 1$. Then putting $ x \\equal{} a, y \\equal{} b$ into the equation gives $ a \\equal{} 0$, contradiction.\n\nThen our final solutions are:\n$ f(x) \\equal{} 0$ for all $ x$.\n$ f(0) \\equal{} r$ for any real $ r$, and $ f(x) \\equal{} 1$ for nonzero $ x$.\n\nProviding I didn't screw up again.[/hide]", "Solution_5": "[quote=\"MellowMelon\"]You have a few mistakes in that one.\n\n[hide=\"Not quite right things\"]- $ g(a) \\equal{} a/f(a)$... what if $ f(a) \\equal{} 0$?\n- The derivative argument is sketchy, since $ f$ and thus $ g$ may not be continuous. From your equation, $ x \\equal{} y \\equal{} 0$ and then $ y \\equal{} 0$ gets the solution much faster, although the first error means you don't get them all.\n- When you get a solution like $ g(x) \\equal{} x \\plus{} k$, you need to make sure all values of $ k$ work. Only $ k \\equal{} 0$ does in this case.[/hide]\n\n[hide=\"Alternate solution\"]Put $ x \\neq 0, y \\equal{} 0$ in. $ xf(0) \\equal{} xf(x)f(0)$. So then $ f(0) \\equal{} f(x)f(0)$ and either $ f(0) \\equal{} 0$ or $ f(x) \\equal{} 1$ for all $ x$. Let's suppose $ f(0) \\equal{} 0$. Put in $ x \\equal{} y \\neq 0$ to get $ 2xf(x) \\equal{} 2xf(x)^2$. Thus $ f(x) \\equal{} f(x)^2$. So $ f(x) \\equal{} 1$ or $ f(x) \\equal{} 0$ for all nonzero $ x$. This gives us three solutions, and when we check we see they all do work.\n- $ f(x) \\equal{} 1$ for all $ x$.\n- $ f(x) \\equal{} 0$ for all $ x$.\n- $ f(0) \\equal{} 0$ and $ f(x) \\equal{} 1$ for nonzero $ x$.[/hide][/quote]\nUgh... yeah... I messed up...\nHere's a hopefully-non-flawed solution\n[hide=\"Alternate Solution\"] \nNow, I will take the restriction where $ f(a) \\equal{} 0$, so that is one possibility. We now divide a bit to get that $ x \\plus{} y \\equal{} \\frac {x}{f(x)} \\plus{} \\frac {y}{f(y)}\\implies x \\minus{} \\frac {x}{f(x)} \\equal{} \\frac {y}{f(y)} \\minus{} y \\equal{} \\minus{} 1(y \\minus{} \\frac {y}{f(y)})$. We now define $ g(a) \\equal{} a \\minus{} \\frac {a}{f(a)}$ and thus we get that $ g(x) \\equal{} \\minus{} g(y)$. Firstly, we have that $ g(x)$ is constant, we call it $ k$. Now, $ k \\equal{} \\minus{} k\\implies k \\equal{} 0 \\implies g(a) \\equal{} 0$. Thus, $ x \\minus{} \\frac {x}{f(x)} \\equal{} 0\\implies f(x) \\equal{} 1$ and therefore our two solutions are $ \\boxed{f(x) \\equal{} 1, f(x) \\equal{} 0}$. [/hide]", "Solution_6": "[hide=\"My preferred solution\"]\n$ xf(y) \\plus{} yf(x) \\equal{} (x\\plus{}y)f(x)f(y)$\nPlug in $ y\\equal{}1$\n$ xf(1) \\plus{} f(x) \\equal{} (x\\plus{}1)f(x)f(1)$\n$ xf(1) \\equal{} f(x)(xf(1)\\plus{}f(1)\\minus{}1)$\n$ f(x) \\equal{} \\frac{xf(1)}{xf(1)\\plus{}f(1)\\minus{}1}$ where the quotient exists.\n\nNow, consider $ x\\equal{}y\\equal{}1$\n$ f(1) \\plus{} f(1) \\equal{} 2f(1)^2 \\implies f(1) \\equal{} 0 \\text{ or } 1$\nSo our choices for $ f$ are $ f(x) \\equal{} \\frac{x}{x}$ or $ f(x) \\equal{} 0$\nNow, for $ f(x) \\equal{} \\frac{x}{x}$, we have know that $ f(x) \\equal{} 1$ everywhere but $ 0$. Now consider $ 0$.\n\nSuppose we have the function $ f(x) \\equal{} \\begin{cases}k & \\text{where } x\\equal{}0\\\\1 & \\text{elsewhere}\\end{cases}$,\nAll of these work, because we can simply check cases.\n\nSuppose $ x \\equal{} y \\equal{} 0$\n$ 0 \\equal{} 0$, good\n\nSuppose $ x \\equal{} 0, y\\neq 0$\n$ yk \\equal{} ykf(y) \\equal{} yk$, good\n\nSuppose $ x, y \\neq 0$\n$ x \\plus{} y \\equal{} x \\plus{} y$.\n\nThus, our solutions are, $ f(x) \\equal{} 0$ or $ f(x) \\equal{} \\begin{cases}k & \\text{where } x\\equal{}0\\\\1 & \\text{elsewhere}\\end{cases}$\n[/hide]", "Solution_7": "Let $ a\\ne 0$ be a real number, then let $ x\\equal{}a$, $ y\\equal{}\\minus{}a$, then $ a*(f(a)\\minus{}f(\\minus{}a))\\equal{}0$, so $ f(a)$ is even.\r\n\r\nThen let $ a$ and $ b$ be nonzero real numbers,\\[ af(b)\\plus{}bf(a)\\plus{}af(\\minus{}b)\\minus{}bf(a)\\equal{}a*(f(b)\\plus{}f(\\minus{}b))\\]\r\n\\[ \\equal{}(a\\plus{}b)(f(a))(f(b))\\plus{}(a\\minus{}b)(f(a))(f(\\minus{}b))\\]\r\n\\[ \\equal{}f(a)*(a)(f(b)\\plus{}f(\\minus{}b))\\]\r\n\r\nHence $ 0\\equal{}(f(a)\\minus{}1)*(2f(b))$\r\n\r\nNow the problem is pretty much done. We have to be a bit careful, but it is clear that $ a$ and $ b$ are 'independent' in some sense. i.e. if there exists some $ a$ such that $ f(a)\\minus{}1\\ne 0$, then $ f(b)\\equal{}0$ for $ b\\in R\\minus{}\\{0\\}$. So we can assert that $ f(x)\\equal{}1$ or $ f(x)\\equal{}0$ where $ x$ is nonzero. Then we merely have to analyze the behavior at $ x\\equal{}0$, and check our solutions.", "Solution_8": "Find all functions $ f$ from $ R$ to $ C$ such that:\r\n\r\n ,$ \\sum{x_1f(x_2)} \\equal{} (\\sum{x_1})\\prod{f(x_1)}$\r\n\r\nwhere sums and products are cyclic and $ x_i$ for $ i\\equal{}1, 2, 3,..., n$.\r\n\r\n\r\n :)" } { "Tag": [ "induction", "integration", "geometry", "rectangle", "3D geometry", "pyramid", "function" ], "Problem": "well Harazi, Pliviu, andreichites, horia, and me along with another coleague of ours have arrived yesterday in Ostrawa (after a long trip by train) to participate in this competition. \r\n\r\nfor now we have just registered, so I'll be back later with other details. The checz have great beer. :)", "Solution_1": "What are the math problems for Vojtech Jarnik 2004 ?", "Solution_2": "Ok, I know only three problems for the first category ( I was so idiot that I have thought to problem 3 for three hours, I couldn't solve it and also I didn't have time to look at the first problem-that is why I don't know its statement).\r\n The second problem was:\r\n Evaluate \\sum (n=0, infinity) arctg 1/(1+n+n 2 ) .\r\n Third problem:\r\n Can we find a convergent sequence (z_n) n>=1 in R 2 such that the open disks of centers in the points z_n and radius 1/n are mutually disjoint?\r\n Fourth problem:\r\n Find all pairs (m,n) of positive integers such that m+n and mn+1 are powers of 2. \r\n Good luck in solving these problems, cause I didn't have. :(", "Solution_3": "The second one is sort of an old trick: arctg(1/(n2+n+1))=arctg(1/n)-arctg(1/(n+1)), so it reduces to a telescopic sum and the answer is \\pi/4=arctg(1), but the others seem really tough..", "Solution_4": "For (4):\r\n\r\nAssume m>=n so m=2t+a and n=2t-a for some t>=1 and a>=0. Let's assume t>1 (it's easy otherwise; I don't know if we need this, but I took t>1 just to make sure :)). Then 2s=mn+1=22t-(a2-1), so (a-1)(a+1)=22t-2s. We assume s isn't 2t or t+1. If s2>(2t-1)2, so a>=2t, so a=2t and we get some nonsense. This shows that, given the assumption that s is neither 2t nor t+1, it must be in the interval (t+1, 2t).\r\n\r\n(a-1)(a+1) is divisible by 2s, so one of its factors must be divisible by 2s-1, which means that we have one of 2 cases:\r\n\r\nCase 1: 2s-1|a-1=>a>=2s-1+1>2t+1, which is obviously false;\r\n\r\nCase 2: 2s-1|a+1=>a>=2s-1-1>2t-1, so a=2t and again we have nonsense.\r\n\r\nThe 2 cases show that s is either 2t or t+1, so a is either 1 or 2t-1, which yields the ordered solutions: (m,n)=(2t+1, 2t-1) or (m,n)=(2t-1,1) with t>=1 (the solution (1,1) isn't covered by the proof above, but it's easy to observe).\r\n\r\nThere could be some mistakes, but I think that, basicly, this is pretty much it.", "Solution_5": "For (3). Maybe it works.\r\n\r\nDefine sequence an following way: a1=1, an+1=an/(1+\\sqrt an).\r\nUsing induction it is easy to show that an < 9/n1.5.\r\n\r\nNow we part all disks into following groups: Dn={k : an+1 < 1/k \\leq an}. \r\nWe have \\sum {k\\in Dn} 1/k = approx. \\int {1/an,1/an+1} 1/x dx = -ln(an+1/an) < approx. 3/n0.75 -> 0 as n -> +\\infty. \r\nThus we can place all disk from Dn into rectangle Rn with height an and width 3/n0.75. Total height of all such rectangles is a1+a2+...+an < 9 \\sum 1/n1.5 < +\\infty .\r\nFinally we stack Rn on each other. So we obtain a kind of pyramid with finite height. Centers of disks in Rn will form a required sequence zm.\r\n\r\nP.S. All \"approx.\" have exact mathematical meaning.", "Solution_6": "well grobber, some people (as me :() didn't knew that \"old\" trick with arctan n - arctan n+1 = arctan 1/ n2 + n + 1 \r\n\r\nanyway, problem no. 1 sounds like this: let f:[0,1]->R be a continuoulsy derivabile function such that f(0)=f(1)=0 and there exist an a \\in (0,1) such that f(a)=\\sqrt 3 . Prove that there exist two tangents at the graph of f such that they make (along with the proper segment from Ox) and equilateral triangle (ie they make angles of \\pi /3 and -\\pi /3 with Ox).", "Solution_7": "I see that problems of first category are quite simple. What about category 2?", "Solution_8": "When are you guys going to find out the results?", "Solution_9": "we did a long time ago :( \r\npliviu went on 3rd place (along with other 3 people), and then claude placed 10th, and the rest of us below 20th I think. Really bad screwup :( ... so people don't want to think or to talk about it ...", "Solution_10": "I was wondering if this could work for (3) as well:\r\n\r\nWe can obviously consider taking squares centered at zn of respective sides 1/n, and we do this as follows: \r\n\r\n(1) We place a square of side 1 which touches Ox and Oy (in the first quadrant). \r\n\r\n(2) On top of that and touchinf Oy, we place a square of side 1/2=1/1*2, followed to the right by one of side 1/3, then by one of side 1/4 and one of side 1/5 (each new square is placed so that its leftmost side is included in the rightmost side of th one before).\r\n\r\n(3) on top of the square of side 1/1*2 we place a square of side 1/2*3 (touching Oy, of course). Then we continue as above with squares of sides from 1/7 to 1/11.\r\n\r\n(4) on top of the square of side 1/2*3 we place one of side 1/3*4 etc.\r\n\r\nThe sequence of the centers of these squares can be shown to converge to point (0,2) because the series 1+1/1*2+1/2*3+.. has finite sum 2 and the sums of sides of the horizontal rows of squares converge to 0, because the limit of 1/[(n-1)n]+..+1/[n(n+1)-1] can be easily shown to converge to 0.\r\n\r\nPlease tell me if there's somehing wrong.", "Solution_11": "Grobber! Your construction is highly similaer to my one, but it is expressed in more simple words.\r\nBTW result of Vojtech Jarnik 2004 are still unpublished.", "Solution_12": "as far as there were some competition in the first category in the second poles took (almost) it all. first was jakub wojtaszczyk from warsaw second ex-aequo slawomir dinew from krakow and andras mathe (sorry if i corrupted the spelling) from budapest . fourth and fifth jakub byszewski krakow and piotr przytycki warsaw. the next places were also occupied by warsaw and krakow with one exception for seged (hungary)", "Solution_13": "Do you know all problems from Jarnik2004?", "Solution_14": "well i can try paraphrase them( it was long ago and i may have forgotten some details )\r\nhere is what i remember\r\n1 st problem are the groups (Q,+) and (Q>0,*) isomorphic (maybe it was also the 0 in the second group ) anyway this is an easy problem\r\nsecond was an easy functional equation , unfortunately i don't remember it well\r\nit was demanded to find all functions from R>=0 x R>=0 to R>=0\r\nand there were three conditions \r\n1 f(x,0)=f(0,x)=x \r\n2 f(x,f(y,z))=f(f(x,y),z) and unfortunately i cant remember the third condition which was the most important . i remember the solution but i doubt whether you'll want it instead of the third condition :) \r\nas i said it was also easy\r\nthird problem went as follows we are given a nonnegative decreasing sequence a_i (i dont remember whether it converged to 0 but i think it wasn't essential) we are given that the series sum ( a_i) diverges prove that sum ( a_i/(1+i*a_i)) also diverges\r\nthese three could have been given to the first category as well\r\nthe fourth was a real crunh (but it is known in some countries) we are given a function R->R that is C_oo . for every x there exists a number n, depending on x such that the n-th derivative in that exactly point is zero. prove that f is a polynomial" } { "Tag": [ "calculus", "articles", "geometry", "LaTeX" ], "Problem": "i want to produce something like\r\n\r\n[u]Calculus II_______________________________________Quiz N\u00b01[/u]\r\n\r\nin the top of the page, what do i do?", "Solution_1": "Use [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Packages#fancyhdr]fancyhdr[/url].", "Solution_2": "yo, i have the same question, but i couldn't make it.\r\n\r\ncan you provide a code? thanks.", "Solution_3": "Here is a minimal document that does what Croata wants:\r\n[code]\\documentclass{article}\n\\usepackage{fancyhdr}\n\\begin{document}\n\\pagestyle{fancy}\n\\lhead{Calculus II}\n\\chead{}\n\\rhead{Quiz N\\textsuperscript{o}1}\n\\renewcommand{\\headrulewidth}{0.4pt}\nText goes here\n\\end{document}[/code]", "Solution_4": "Also, in addition to \\lhead, \\chead, and \\rhead for differently positioned headers, there's also \\lfoot, \\cfoot, and \\rfoot, for footers. For example, [code]\\cfoot{Page \\thepage}[/code] ...which writes Page 1, Page 2, etc. centered at the bottom of each page.", "Solution_5": "thanks stevem, that helped me out! but, i have some trouble:\r\n\r\n1) below \"Calculus II,\" (that is, below the line) i want to type other text, the same below for Quiz N\u00b01.\r\n\r\n2) i tried to modify the headheight because the text was over the line, but i just succeed a bit, because the text is still almost over the line.\r\n\r\n3) the number of the page (down-center) is very close to the end of the first page, how can i set it upper?\r\n\r\n4) on every page, Calculus II, long line and other stuff appears, and i don't want that happens, just for the first page.\r\n\r\nthanks again for the help.", "Solution_6": "1) [code]\\lhead{Calculus II\\hfill Quiz N\\textsuperscript{o}1\\\\[-1ex]\\hrulefill\\\\Left Text\\hfill Right text}\n\\renewcommand{\\headrulewidth}{0pt}[/code]\r\n2) No need now\r\n3) Use the geometry package to change margins eg \\usepackage[bottom=5cm]{geometry} moves the margin at the bottom to 5cm. You can set all the margins like \\usepackage[top=2cm, right=2cm, bottom=5cm, left=2cm]{geometry}\r\n4) Use \\thispagestyle{fancy} instead of \\pagestyle{fancy} or change to \\lhead{} etc", "Solution_7": "thanks a lot!! the whole thing looks awesome!!\r\n\r\ni have just one more problem, not about the above, it's about alignment, i was using\r\n\r\n[code]\\underrightarrow{\\begin{align}\n & F_{2}-F_{1} \\\\ \n & F_{3}-F_{1} \n\\end{align}}[/code]\r\n\r\nwell, as you can see i wanted to align those and below them there's an arrow, but i couldn't make it, i got an error.", "Solution_8": "I'm assuming thats because the align command itself initiates math mode while the arrows require math mode. Maybe include the arrow within the command?\r\n\r\n[code]\\begin{align*}\n& F_{2}-F_{1} \\\\ \n & \\underrightarrow{F_{3}-F_{1}}\n\\end{align*}[/code]\n\n\\begin{align*}\n& F_{2}-F_{1} \\\\ \n & \\underrightarrow{F_{3}-F_{1}}\n\\end{align*}\n\nOr if you want numbers:\n\n[code]\\begin{align}\n& F_{2}-F_{1} \\\\ \n & \\underrightarrow{F_{3}-F_{1}}\n\\end{align}[/code]\r\n\r\n\\begin{align}\r\n& F_{2}-F_{1} \\\\ \r\n & \\underrightarrow{F_{3}-F_{1}}\r\n\\end{align}", "Solution_9": "[quote=\"stevem\"]1) [code]\\lhead{Calculus II\\hfill Quiz N\\textsuperscript{o}1\\\\[-1ex]\\hrulefill\\\\Left Text\\hfill Right text}\n\\renewcommand{\\headrulewidth}{0pt}[/code]\n2) No need now\n3) Use the geometry package to change margins eg \\usepackage[bottom=5cm]{geometry} moves the margin at the bottom to 5cm. You can set all the margins like \\usepackage[top=2cm, right=2cm, bottom=5cm, left=2cm]{geometry}\n4) Use \\thispagestyle{fancy} instead of \\pagestyle{fancy} or change to \\lhead{} etc[/quote]\r\nthanks, i used this too and works nice.\r\n\r\nbut i have a problem, for example i typed some enumerated stuff and i wanted to move the whole text a little bit to the left since i'm using the geometry package, this is [top=2cm, right=2cm, bottom=3cm, left=2cm] then i just modified the left margin into 1, so the whole text moves but also the header moves, but i want the header stays there even if i'm moving the text, is it possible to do that?", "Solution_10": "Leave the geometry margins alone but change the margins for the text using the changepage package; documentation is in [i]path to your Latex files[/i]\\tex\\latex\\ltxmisc\\changepage.sty\r\n[code]\\usepackage{changepage} \n...\n\\begin{document}\n% decrease left margin by 1cm\n\\begin{adjustwidth}{-1cm}{} \n text goes here\n\\end{adjustwidth}\nText with normal margin[/code]" } { "Tag": [ "function", "AMC", "USA(J)MO", "USAMO", "induction", "algebra unsolved", "algebra" ], "Problem": "Consider functions $\\, f: [0,1] \\rightarrow \\mathbb{R} \\,$ which satisfy\n(i) $f(x) \\geq 0 \\,$ for all $\\, x \\,$ in $\\, [0,1],$\n(ii) $f(1) = 1,$\n(iii) $f(x) + f(y) \\leq f(x+y)\\,$ whenever $\\, x, \\, y, \\,$ and $\\, x + y \\,$ are all in $\\, [0,1]$.\n\nFind, with proof, the smallest constant $\\, c \\,$ such that \n\\[ f(x) \\leq cx \\]\nfor every function $\\, f \\,$ satisfying (i)-(iii) and every $\\, x \\,$ in $\\, [0,1]$.", "Solution_1": "It's USAMO 1993/3: [url=http://www.kalva.demon.co.uk/usa/usa93.html]www.kalva.demon.co.uk/usa/usa93.html[/url]", "Solution_2": "This is basically the kalva solution if I remember correctly but the link is broken:\r\n\r\nLetting $ x \\equal{} 1\\minus{}y$ we have $ f(x) \\plus{} f(1\\minus{}x) \\leq 1$ which means $ f(x) \\leq 1$. Let $ x\\equal{}y \\leq 1/2$. Then $ 2f(x) \\leq f(2x)$. $ 4f(x) \\leq 2f(2x) \\leq f(4x)$, induction like this yields $ 2^nf(x) \\leq f(2^nx)$ with $ x$ in an appropriate range.\r\n\r\nIf $ \\frac{1}{2^{n\\plus{}1}} < x \\leq \\frac{1}{2^n}$ for $ 1 \\leq n$, then $ 2^nf(x) \\leq f(2^nx) \\leq 1$. So as $ x$ approaches $ \\frac{1}{2^{n\\plus{}1}}$, $ c$ approaches $ 2$.\r\n\r\nIf $ \\frac{1}{2} < x$, then $ f(x) \\leq 1$ and as $ x$ approaches $ \\frac{1}{2}$, $ \\frac{1}{x} \\equal{} c$ approaches $ 2$. \r\n\r\nThis means $ c\\equal{}2$ is the minimum. It remains to find a function that fits $ c\\equal{}2$ which is not very hard.", "Solution_3": "[quote=\"aznlord1337\"] It remains to find a function that fits $ c \\equal{} 2$ which is not very hard. [/quote]\r\n\r\n$ f(x)\\equal{} \\begin{cases} 0 & 0 \\le x \\le \\frac{1}{2} , \\\\ 1 & \\frac{1}{2} < x \\le 1. \\end{cases}$" } { "Tag": [ "videos", "search" ], "Problem": "Post names of Funny youtube videos here", "Solution_1": "[youtube]aFeUhqvSQa4[/youtube]\r\n\r\ndont even ask how I found this", "Solution_2": "The above is not funny. It is refreshing.", "Solution_3": "i didn't know it either x_x", "Solution_4": "[youtube]VnbT7qt6RF4[/youtube]", "Solution_5": "Barack Obama Umbrella PArody thingy.\r\n\r\nthat thing is hilarious.", "Solution_6": "[youtube]ur5fGSBsfq8[/youtube]", "Solution_7": "Srry with the above Jeopardy one. I meant this: [youtube]Cu-key6lUHY[/youtube]", "Solution_8": "haha, i love the greek vs. germany", "Solution_9": "[youtube]QFCSXr6qnv4[/youtube]\r\n\r\n :wink: Yeah...... I suck", "Solution_10": "\"Edgar se cae\". I'm too lazy to put it, search yourselves.", "Solution_11": "yeah that's really funny. Can someone translate this?\r\n\r\n\r\n[youtube]kQTUP0LOD20[/youtube]\r\n\r\nEDIT: something other than pacman's movie: [youtube]8-JxUdk7LKg[/youtube]", "Solution_12": "Street Fighter: The Later Years. It's got some language (understatement) and an inappropriate scene or two, but it's pretty funny.", "Solution_13": "[youtube]_fSpI4oZoDc[/youtube]\r\n[youtube]ZA1NoOOoaNw[/youtube]\r\n\r\nI hope none of Indians are offended by these two videos......\r\n\r\n :|", "Solution_14": "I wasn't reffering to it. I told you about this. It's in spanish.\r\n\r\n[youtube]b89CnP0Iq30[/youtube]\r\n\r\n:rotfl:\r\n\r\nand cartoon version! search it! it's very cool!!!", "Solution_15": "[youtube]Id_kGL3M5Cg[/youtube]\r\nAnd then there's the one where a Linux comes in...", "Solution_16": "http://www.youtube.com/watch?v=07sOhaB6zTE&feature=user\r\n\r\nThe Y Intercept and You", "Solution_17": "[youtube]wjL2WcK3JYU[/youtube]", "Solution_18": "http://youtube.com/user/kidkarate99 my friend's", "Solution_19": "[youtube]qd_syuD-N_k[/youtube]", "Solution_20": "[youtube]-f4dQNrpUeU[/youtube]", "Solution_21": "And this is the part that I tell you not to revive." } { "Tag": [], "Problem": "I'm trying to find the K and Ph of the equation: \r\nHCO3- + HCO3 - <----> H2CO3+CO3- -\r\n\r\nI would really appreciate any help.\r\n\r\nThank you.", "Solution_1": "K: First, break this reaction into components:\r\n\r\nHCO3- + H2O<---> CO3- + H3O+\r\nHCO3- + H2O<---> H2CO3 + OH-\r\nH3O+ + OH- <---> 2H2O.\r\n\r\nThese three reactions are all tabulated somewhere, although the tabulated value is the reverse for the last two. Now, all you need is to remember what happens to equilibrium coefficients when you add reactions, and you're home free.\r\n\r\npH: Use the formula pH = 1/2 (pKa1 + pKa2). I don't know how this is derived, although I've always been rather curious as to it. Anyone know?" } { "Tag": [ "search", "articles", "Intel STS" ], "Problem": "I'm just a \"Mathematical Enthusiast,\"and beginning researcher.Is it okay to write,send a well-research mathematical\r\nideas to \"Mathematical Journal\"regardless of one's \"Educational\r\nAttainment?\"(Ex.High School Level,College Level.) \r\n\r\n Respectfully,\r\n A.Eddington", "Solution_1": "sure you can\r\n\r\ni remember reading somewhere that some winners of ISEF (Intel Science Talent Search) submit their research articles to graduate level research journals\r\n\r\nthey are just high school students", "Solution_2": "That's well-researched, [b]well-written[/b], novel, and substantive. Journal editors receive more papers than they can use. The ones that have any promise at all are sent out to be refereed, a rigorous, time-consuming process. If it's written so that the editor can't even figure out what it is, it goes in the trash. If it's obviously just a rehash of something already well known, not adding any new perspectives, then it goes in the trash. (And there are many, many things that are already well known.)\r\n\r\nTo be honest, if it looks anything at all like \"arthur eddington's\" other posts on Mathlinks/AoPS so far, then it goes in the trash.", "Solution_3": "Yes,okay,a High School Student,or say, any 'Undergraduate' individual can submit a 'NEW'mathematical\r\nideas to any accredited mathematics committee,the question is:\r\n\r\n 1)Where is the 'safest'venue,or,committee wherein an aspiring individual can submit such 'mathematical NEW ideas.?\r\n\r\n 2) Is there any 'AGE' qualification related to an individual\r\nwho wanted to submit such NEW discovered mathematical ideas?(Descartes discovered 'Analytical Geometry' at the age of\r\n'Fifty Years Old.')\r\n\r\n 3)What other related 'requirements' needed for aspiring individual to meet so as to submit such NEW mathematical ideas?\r\n\r\n Perhaps,our good Professors can answer my simple\r\nquestions.\r\n\r\n Respectfully,\r\n A.Eddington", "Solution_4": "The most important requirement is of course the content of the paper you want to publish. That is, the originality of the ideas, the quality and clarity of writing, etc. Most major journals will accept submissions from anyone -- no requirements on age or education status or anything. \r\n\r\nI know there are undergraduate research journals which are \"easier\" to publish in (i.e., the material need not be as original or groundbreaking), but those probably require submissions to come from undergraduate students. \r\n\r\nTry looking at the AMS website: http://www.ams.org/", "Solution_5": "To reiterate what Kent said a bit more politely, I would try to make sure your grammar is better on whatever paper you publish than it has been on your two posts in this thread. Of course, you might not be a native speaker of English; if this is the case, publishing in your native language may be a good idea. If you are a native speaker, I would make sure you follow the generally accepted forms of grammar that everyone else uses. Even if we can all understand what you say, a journal will probably find your unconventional uses of apostraphes, capitalization, and periods to come off as unproffesional; such is life.", "Solution_6": "if it looks like a quack and quacks like a quack, then ....", "Solution_7": "[quote=\"arthur eddington\"]\n 1)Where is the 'safest'venue,...?\n\n 2) Is there any 'AGE' qualification ...?\n\n 3)What other related 'requirements' needed...?\n[/quote]\r\n\r\n1) There is no \"safest\" venue. Acceptance rates range from 10% - 80% depending on the journal, with most journals probably in the 50% range (someone correct me if I've acquired the wrong impression). The chance of acceptance is much higher if you pick an appropriate journal. There's a list of journals here\r\nhttp://www.ams.org/mathweb/mi-journals.html\r\nwith links to the journal websites.\r\n\r\n2) No.\r\n\r\n3) Every journal has a Submission Guide or Information for Authors document. You should follow the instructions given there. I would also recommend that before submitting any manuscript, have a couple of colleagues and/or friends read it over and give you suggestions for improvement. I've reviewed manuscripts which were clearly just rough drafts, and have been annoyed that the authors were wasting my time by making me wade through these things. (If there's merit to such rough-draft manuscripts, at best they'll be returned to the author with an invitation to re-submit the manuscript when it's been revised and proofread. Much of the time any merit is sufficiently hiddent in the roughness of the draft that the reviewer can't be sure it's there.)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $ 00$\r\nSo, my proof is complete. I think this inequality is weak", "Solution_2": "You're right . This inequality is very weak .\r\n\r\nMy solution :\r\n\r\n$ \\left((1\\plus{}\\frac{a}{b})(1\\plus{}\\frac{a}{c})\\minus{}2\\right)^2\\ge 0 \\Leftrightarrow (1\\plus{}\\frac{a}{b})(1\\plus{}\\frac{a}{c})\\ge 2\\sqrt{\\frac{a}{b}\\plus{}\\frac{a}{c}\\plus{}\\frac{a^2}{bc}}$\r\n\r\nwhich is equivalent with : $ (a\\plus{}b)^2(a\\plus{}c)^2\\ge 4abc(a\\plus{}b\\plus{}c)$\r\n\r\nWe also have : $ \\frac{a\\plus{}b\\plus{}c}{3}\\ge \\frac{a\\plus{}b}{2}$\r\n\r\nAnd the conclusion it's obvious .", "Solution_3": "[quote=\"alex2008\"]You're right . This inequality is very weak .\n\nMy solution :\n\n$ \\left((1 \\plus{} \\frac {a}{b})(1 \\plus{} \\frac {a}{c}) \\minus{} 2\\right)^2\\ge 0 \\Leftrightarrow (1 \\plus{} \\frac {a}{b})(1 \\plus{} \\frac {a}{c})\\ge 2\\sqrt {\\frac {a}{b} \\plus{} \\frac {a}{c} \\plus{} \\frac {a^2}{bc}}$\n\nwhich is equivalent with : $ (a \\plus{} b)^2(a \\plus{} c)^2\\ge 4abc(a \\plus{} b \\plus{} c)$\n\nWe also have : $ \\frac {a \\plus{} b \\plus{} c}{3}\\ge \\frac {a \\plus{} b}{2}$\n\nAnd the conclusion it's obvious .[/quote]\r\nI like your solution. However, it seems to be unnatural :)" } { "Tag": [ "function", "calculus", "integration" ], "Problem": "If $f(n+1)=(-1)^{n+1}n-2f(n)$ for integral $n\\ge 1$, and $f(1)=f(1986)$, compute $f(1)+f(2)+f(3)+\\cdots+f(1985)$.", "Solution_1": "Solve the recurrence formula,\r\n\r\n$f(n)=-(n+1)(-1)^n+(-2)^{n-1}\\{f(1)-2\\}$\r\n\r\n$=\\frac{2n+3}{4}(-1)^{n+1}-\\frac{2n+1}{4}(-1)^n+(-2)^{n-1}\\{f(1)-2\\}.$\r\n\r\nThen sums up using telescoping method.", "Solution_2": "[hide]I think it's 331. First, set up this equation. $f(1)+f(2)+...+f(1985)=f(2)+f(3)+f(4)+...+f(1986)$. THen, reduce each side so that only the f(odd numbers) are left on the left side and only f(even numbers) are left on the right side. This creates $1-f(1)+3-f(3)+5-f(5)+...+1983-f(1983)+f(1985)=-2-f(2)-4-f(4)-...-1984-f(1984)+f(1986)$. THis is $992(1985)=f(1)-f(2)+f(3)-...-f(1984)-f(1985)+f(1986)$. THen, expressing only the f(odd numbers) again, one gets $992(1985)=3f(1)+3f(3)+3f(5)+...+3f(1983)-3f(1985)-992^2+1985$ or $\\frac{991(1985)+992^2}{3}=f(1)+f(3)+f(5)+...+f(1983)-f(1985)=983733$. Notice the similarity between $1-f(1)+3-f(3)+5-f(5)+...+1983-f(1983)+f(1985)=992^2-f(1)-f(3)-f(5)-...-f(1983)+f(1985)=f(1)+f(2)+f(3)+...+f(1985)$ and $983733-f(1)-f(3)-f(5)-...-f(1983)+f(1985)=0$. Subtracting, I get $f(1)+f(2)+f(3)+...+f(1985)=331$.[/hide]" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "let $ a,b,c\\in\\mathbb{Z}$ such that $ |a|,|b|,|c|\\le 10$ and let $ f(x) \\equal{} x^3 \\plus{} ax^2 \\plus{} bx \\plus{} c$. Given that $ |f(2 \\plus{} \\sqrt {3})| < 10^{ \\minus{} 4}$ show $ f(2 \\plus{} \\sqrt {3}) \\equal{} 0$.\r\n\r\n[hide=\"harder\"]what if we replace $ 10^{ \\minus{} 4}$ by $ 10^{ \\minus{} 2}$?[/hide]", "Solution_1": "$ |f(2\\plus{}\\sqrt 3 )|\\equal{}|7a\\plus{}2b\\plus{}c\\plus{}26\\plus{}(4a\\plus{}b\\plus{}15)\\sqrt 3|\\equal{}|(4a\\plus{}b\\plus{}15)(\\sqrt 3\\plus{}\\frac{7a\\plus{}2b\\plus{}c\\plus{}26}{4a\\plus{}b\\plus{}15}|>\\frac{1}{|4a\\plus{}b\\plus{}15|\\sqrt 3}$ if $ 4a\\plus{}b\\plus{}15\\not \\equal{}0$." } { "Tag": [ "probability" ], "Problem": "When three cards are drawn from a standard 52-card deck, what is the probability they are all the same rank? For example, that all three are kings?\r\n\r\nhow would you do this problem?", "Solution_1": "[hide]What matters is that the last two cards are the same as the first one. The probability for the second is $\\frac{3}{51}$; for the third is $\\frac{2}{50}$. \n\n$\\frac{1}{17}\\cdot \\frac{2}{50}=\\frac{1}{425}$?[/hide]", "Solution_2": "[quote=\"mr. math\"]When three cards are drawn from a standard 52-card deck, what is the probability they are all the same rank? For example, that all three are kings?\n\nhow would you do this problem?[/quote]\r\n[hide]The first card doesn't matter, because it can be any of the 52 cards. There is a $\\frac{3}{51}$ chance that the second card will be the same as the first, and a $\\frac{2}{50}$ chance that the third will be the same as the first two. Multiplying these two numbers gives us the answer.\n\n$\\frac{3}{51}\\cdot\\frac{2}{50}=\\frac{1}{425}$[/hide]", "Solution_3": "For these types of problems, typically you count how many combinations of cards give you the desired property, and put that number over the total number of combinations of cards. (There might be easier methods such as the ones the two people above me posted :P)\r\n\r\n[hide]There are 4 total kings, so there are $\\dbinom{4}{3}=4$ combinations of 3 cards that have the property that all 3 cards are kings. Since there are 13 ranks, there are $13\\cdot{4}=52$ combinations of 3 cards where all 3 cards are the same rank.\n\nThere are 52 total cards, so there are $\\dbinom{52}{3}=22100$ total possible combinations of 3 cards. Therefore, the probability of drawing 3 kings is $\\frac{4}{22100}$ and the probability of drawing 3 cards of the same rank is $\\frac{52}{22100}$ (both fractions can be reduced to the answers that the people above have beat me to)[/hide]", "Solution_4": "[quote=\"i_like_pie\"][quote=\"mr. math\"]When three cards are drawn from a standard 52-card deck, what is the probability they are all the same rank? For example, that all three are kings?\n\nhow would you do this problem?[/quote]\n[hide]$\\frac{3}{51}\\cdot\\frac{2}{50}=\\frac{1}{425}$[/hide][/quote]\r\n\r\nPlease post explanations with your posts. I am going to start deleting posts that do not contain reasoning (mr. math wanted help with the problem, not so much the answer, but how to get it and why it works)." } { "Tag": [ "LaTeX" ], "Problem": "I've found and downloaded the latexrenderer (like the one you're using here), and I'd like to install it on a web host (like on lycos or anything)\r\n\r\nBut I have a couple of problems: how exactly do I install latex, imagenick, etc. on my host myself? I'd really like if someone could give me a detailed guideline for this :)\r\n\r\nI mean without contacting the host, just placing the thing there myself... there must be a way to do it. I've already figured it out for over 90% what to do with mimetex but the graphics of that one are so ugly that I'd rather have latex in some way. Any hints on this? ...", "Solution_1": "I'm going to stick my neck out and say it's not possible to install binary programs like latex on a Linux web host except as part of the root system to which you have no access. Someone may know otherwise :? \r\n\r\nSo you need to look for a host that offers LaTeX or is willing to install it for you. LaTeX is very common, comes with virtually every operating system, and is easy to install but unfortunately many hosts don't know anything about it and don't perceive a need for it. The same goes for ImageMagick.\r\n\r\nPerhaps contributors here would like to say if their host offers LaTeX? A list of such hosts would be welcome.", "Solution_2": "[quote=\"stevem\"]I'm going to stick my neck out and say it's not possible to install binary programs like latex on a Linux web host except as part of the root system to which you have no access. Someone may know otherwise :?[/quote]\nseems quite strange... I mean why latex cannot while mimetex can? :?\n\n[quote=\"stevem\"]Perhaps contributors here would like to say if their host offers LaTeX? A list of such hosts would be welcome.[/quote]\r\nthat would be a good idea :)", "Solution_3": "[quote=\"Peter VDD\"]seems quite strange... I mean why latex cannot while mimetex can? :?[/quote]\r\nUnlike LaTeX, mimeTeX is a [url=http://www.webopedia.com/TERM/C/CGI.html][b]CGI program[/b][/url] which can only run if it is put in a special folder, such as cgi-bin, which has special permission to allow the programs in it to run", "Solution_4": "Yes I know I mean, why doesn't anyone make a cgi version of latex... :?\r\n\r\n\r\n\r\nanyway another good question: where do I find the mimetex.cgi file? It wasn't in the package I downloaded initially, and neither in the package I found on their site? :?", "Solution_5": "The cgi file isn't there because \r\na) it's someone else's program\r\nb) it's best to get the latest version of mimeTeX\r\nc) it is best to compile the cgi program on your own system - though this isn't possible in your case, since compilation is not itself a cgi program :? \r\n\r\nYou can get the latest pre-compiled version of mimetex.cgi [url=http://www.forkosh.com/mimetex.html][b]here[/b][/url], near the bottom of the page.", "Solution_6": "[quote=\"stevem\"]The cgi file isn't there because \na) it's someone else's program\n[/quote]\n:lol: I know stevem, I meant in THEIR package ;)\n\n[quote=\"stevem\"]b) it's best to get the latest version of mimeTeX\nc) it is best to compile the cgi program on your own system - though this isn't possible in your case, since compilation is not itself a cgi program :?[/quote]\r\nIndeed, can't compile it :(\r\n\r\nI'll give the precompiled one a try, thanks for your help! :)", "Solution_7": "Just a confirmation: you cannot use LatexRender on lycos, or any other free-hosted boards for that matter. You have to buy web-space from a provider, and ask him ahead if he can allow you access to imagemagick and latex. \r\n\r\nMimetex gives you a change of using it on some free-hosts, which enable CGI services.", "Solution_8": "Yes, but mimetex has kind of an ugly design. If I just knew how to set the font to make it look better... :?", "Solution_9": "[quote=\"Peter VDD\"]Yes, but mimetex has kind of an ugly design. If I just knew how to set the font to make it look better... :?[/quote]Agreed mimetex looks ugly, but it's \"mime\" not \"same as\" ;) \r\n\r\nAbout the font you should probably e-mail the author, he should be able to answer your questions.", "Solution_10": "[quote=\"Valentin Vornicu\"]Agreed mimetex looks ugly, but it's \"mime\" not \"same as\" ;)[/quote]\r\n\r\nmme is french for same as, isn't it? :P" } { "Tag": [ "logarithms", "limit", "ratio", "real analysis", "real analysis unsolved" ], "Problem": "Let the sequence $(x_{n})_{n\\geq 1},x_{1}=a\\in\\mathbb{R}$ and $x_{n+1}-x_{n}=\\log_{b}{(b^{x_{n}}+b)}$ with $b>1$ for any natural $n\\geq 1$.\r\nFind $\\lim_{n\\rightarrow\\infty}{\\frac{b^{x_{n}}}{(b^{x_{1}}+1) \\cdot(b^{x_{2}}+1)\\cdot\\cdot...\\cdot(b^{x_{n}}+1) }}$", "Solution_1": "The limit is 0 by ratio criterion. (the ratio tends to 0!)", "Solution_2": "Can you be more specific? :|", "Solution_3": "From the hypothesis we know that for any $n\\geq1$ we have $b^{x_{n+1}-x_{n}}=b^{x_{n}}+b>b \\Rightarrow x_{n+1}-x_{n}>1$ so $\\displaystyle\\lim_{n\\to\\infty}{x_{n}}=\\infty$\r\nLet's consider $y_{n}=\\displaystyle\\Pi_{i=1}^{n}{\\frac{1}{b^x_{i}+1}}$.\r\nThe wanted limit becomes (*)$\\displaystyle\\lim_{n\\to\\infty}(y_{n-1}-y_{n})$, but since $\\displaystyle\\lim_{n\\to\\infty}\\frac{y_{n}}{y_{n-1}}=\\lim_{n\\to\\infty}\\frac{1}{b^{x_{n}}+1}=0$ and $i>j \\Rightarrow b^{x_{i}}>b^{x_{j}} \\Rightarrow \\frac{1}{b^{x_{i}}+1}<\\frac{1}{b^{x_{j}}+1}$ implies that $00\\}\\]\r\n\\[P_{n}=|S_{n}|\\].Prove:\r\n${\\frac{1+P_{1}+P_{2}+...+P_{n-1}}{P_{n}}}\\leq\\sqrt{2n}$", "Solution_1": "is $x_i\\geq 0$ or $x_i>0$?", "Solution_2": "Oh! evey $x_{i}>0$", "Solution_3": "maybe i asked a stupid question but what is the value of $k$? ok since $x_i>0$ then $k\\leq n$ and $P_n=\\binom{n-1}{k-1}$. if $x_i\\geq 0$ then we can fix $k$ and $P_n=\\binom{n+k-1}{k-1}$", "Solution_4": "[quote=\"Palytoxin\"]maybe i asked a stupid question but what is the value of $k$? ok since $x_i>0$ then $k\\leq n$. if $x_i\\geq 0$ then we can fix $k$ and $|S_n|=\\binom{n+k-1}{k-1}$[/quote]\r\n\r\n\r\nk is not constant, it can change, {x_1,x_2,...,x_n} is not sequenced, just look x_1,x_2,...,x_n as numbers but not variables.", "Solution_5": "Did I get it correctly that $S_n$ contains sets rather than multisets meanning, for example, that expressions $1+2+2=5$ and $1+1+1+2=5$ give the same set $\\{1,2\\}$ as an element of $S_5$?", "Solution_6": "I believe we should be talking about multisets, though we will have to check with [i]qxy[/i].\r\nBtw, this problem sounds more like combinatorics, and not number theory." } { "Tag": [ "topology", "linear algebra", "linear algebra theorems" ], "Problem": "I've been looking for a simplified (maybe new) proof of the following theorem for some time, but I haven't found anything new. Does anyone know a good approach to it?\r\n\r\n[b]Theorem.[/b] The [url=http://en.wikipedia.org/wiki/Cross_product]\"cross product\"[/url] is only defined in three or seven dimensions.", "Solution_1": "I think any sufficiently simple proof would hide the beautiful web of concepts behind this theorem. There is a point at which making a proof too elementary makes it too difficult to generalize. If you just want to understand the result, I highly recommend [url=http://math.ucr.edu/home/baez/octonions/]John Baez's discussion[/url] of the octonions.", "Solution_2": "Also this standardly ends up being a topology result, rather than a linear-algebraic one.", "Solution_3": "Oh wow, never knew it was defined in R7 as well. I thought it was only defined in 3 dimensions. Very interesting; I'll be sure to check it out (the theorem that is)." } { "Tag": [ "Mafia", "algorithm", "abstract algebra", "geometry", "rectangle", "analytic geometry" ], "Problem": "How is final exam going? I get to write math diploma in Janaury and omit English and social written exams, because of Waterloo Seminar[hide=\"!\"]:D :lol: :first: :coolspeak: :clap2: :heli: :agent: :clap: :rotfl: :harhar: :yup: :w00tb: :thumbup: :whistling:[/hide]\r\nI still have chem, bio(today!), and social muliple choice to write. :(\r\n\r\nDoes Anyone want to complain about your exams?", "Solution_1": "You got English omitted? Wow.\r\n\r\nI have my exams for English, French and Math, starting tomorrow, with the last one on friday.", "Solution_2": "I wrote the exams before the seminar, but screwed up on 2 out of the 5.\r\nDarn...\r\n\r\nbut I do have a math 12 challenge to write, so boring... 5 hours", "Solution_3": "Yeah! Finished my bio exam! It was easy but had some stupid questions. :D", "Solution_4": "My bio exam was crazy.\r\n\r\nPowerpoint presentation and 6 diagrams, 3 binomial nomenclature charts!\r\n\r\nI hope I did well.......", "Solution_5": "[quote=\"billzhao\"]You got English omitted? Wow.\n\nI have my exams for English, French and Math, starting tomorrow, with the last one on friday.[/quote]\r\nFriday? I thought you would be at Washington on that day. :?", "Solution_6": "[quote=\"lightrhee\"][quote=\"billzhao\"]You got English omitted? Wow.\n\nI have my exams for English, French and Math, starting tomorrow, with the last one on friday.[/quote]\nFriday? I thought you would be at Washington on that day. :?[/quote]\r\n\r\nWe leave on Sunday.", "Solution_7": "Ahhhh I'm never going to survive the week! If you don't see me on Saturday david blame it on my crazy teachers!!!!!!\r\n\r\nWrote my in class social final, and in class physics final today. Tomorrow I have all 3 french finals. Wednesday I have my english final (phew, easy one) Thursday I have physics dipolma exam all day. Then friday i have the final portion of my social final. And this isn't including my speech for tomorrow....AHHH!H!HH!!!!!H!H! :stretcher: :wallbash_red: :cursing: :censored: \r\n\r\nI'm going to die in the next 24 hours...like I said, if you don't see me on saturday david, you know who to blame...", "Solution_8": "[quote=\"Sunny\"]Ahhhh I'm never going to survive the week! If you don't see me on Saturday david blame it on my crazy teachers!!!!!!\n\nWrote my in class social final, and in class physics final today. Tomorrow I have all 3 french finals. Wednesday I have my english final (phew, easy one) Thursday I have physics dipolma exam all day. Then friday i have the final portion of my social final. And this isn't including my speech for tomorrow....AHHH!H!HH!!!!!H!H! :stretcher: :wallbash_red: :cursing: :censored: \n\nI'm going to die in the next 24 hours...like I said, if you don't see me on saturday david, you know who to blame...[/quote]\r\n\r\nYikes! Yeah, my schedule looks similar. Just remember, we are all in this, and we will get through this, together. :)", "Solution_9": "[hide=\"This post may offend you\"][quote=\"TheDreamer\"] Just remember, we are all in this[/quote]\nNot me :harhar: [/hide]", "Solution_10": "Lol, you done? :blush: \r\n\r\nDamn i'm so jealous :mad:", "Solution_11": "[quote=\"lightrhee\"][hide=\"This post may offend you\"][quote=\"TheDreamer\"] Just remember, we are all in this[/quote]\nNot me :harhar: [/hide][/quote]\r\n\r\n\r\n :furious: :diablo: \r\n\r\nOh no worries David...you know what happens to people who offend me...heh heh heh...see you on Saturday. :bomb:", "Solution_12": "[quote=\"TheDreamer\"]Lol, you done? :blush: \n\n- i'm so jealous :mad:[/quote]\nNo, I'm not done. I just get to omit a few stressful exams. I only have chem and social part B(multiple choice) left. :) \n\n[quote=\"Sunny\"][quote=\"lightrhee\"][hide=\"This post may offend you\"][quote=\"TheDreamer\"] Just remember, we are all in this[/quote]\nNot me :harhar: [/hide][/quote]\n\n\n :furious: :diablo: \n\nOh no worries David...you know what happens to people who offend me...heh heh heh...see you on Saturday. :bomb:[/quote]\r\nI have warned you. You shouldn't have read it if you were going to be offended. I shouldn't have given you the idea(calling me and playing the instrument, which I can not spell(something like piccolo, which seems to be spelled wrong(maybe the spelling is correct))). I better disconnect my phone tonight... :D", "Solution_13": "[quote=\"lightrhee\"]\n\nI have warned you. You shouldn't have read it if you were going to be offended. I shouldn't have given you the idea(calling me and playing the instrument, which I can not spell(something like piccolo, which seems to be spelled wrong(maybe the spelling is correct))). I better disconnect my phone tonight... :D[/quote]\r\n\r\nHahaha I don't need to call you. I'll see you on Saturday. And Sunday, and Monday, and Tuesday. :D Aren't you excited to see me again? \r\n\r\nBTW, it is piccolo. Hmm I should bring it... :diablo:", "Solution_14": "Hey David, \r\nGot any place to play Mao online?", "Solution_15": "[quote=\"TheDreamer\"]You never cease to amaze us Elyot. Anyways, I won't go on and on about that, I'm sure you've heard enough flattery throught your life..\n\nDo you just get instructions online? I have a bunch of books, and I used to get them at the library..i don't really have pics of my stuff cuz it was long ago..but i do still have a few things i made, i can show you sometime.\n\nLol next time I see you we should fold some paper *looks for paperfolding emoticon*\n\nOh, and can you teach me an algorithm for trisecting a paper? The push two ends together way is really inaccurate, so I was wondering whether someone knew a better way\n\n\nAnyways, I heard you're also good at cooking, maybe you should also post pictures of crazy foods you cooked :P[/quote]\r\n\r\nThe big one is my own design, but it's based on a technique called \"cross-box pleating\". I've never seen anybody attempt a 4x4 cross-box though. Even 2x2 is relatively insane. That thing took me like 18 hours at least.\r\n\r\nKawasaki rose is pretty famous and you can get diagrams for it online. It takes maybe 45 minutes first time... then once you get the hang of it, you can do one in 10-15 mins.\r\n\r\nThe other two are my own designs based on pretty common modular forms. There are many popular types of modules (pieces) that you can fold, some with as few as 2 folds, and you can put them together in all sorts of intricate designs and patterns... add colours and you can do some neat stuff.\r\n\r\n---------------------------------------------------------------\r\n\r\nI don't have any cooking pictures, but I can post some \"pseudo-recipes\". Here's a simple one with like 3 ingredients, so it's not too hard.\r\n\r\nTry this one...\r\n- Take about 8cm by 8cm square of brownie.\r\n- Cut it into 16 pieces.\r\n- Add 2 or 3 scoops of SMORES ice cream. Mix.\r\n- Heat up about 40 mL or so of NUTELLA!\r\n- Pour molten nutella onto brownie/ice cream concoction.\r\n- Enjoy (with a special someone)!\r\n\r\nI invented this dessert impromptu while trying to impress my girlfriend. Apparently, it worked. It was uncompromisingly delicious and she rewarded me graciously for it later ;)\r\n\r\nAnother trick I invented to impress people is to cut an orange in half and throw it in the freezer. Then whenever you make a dessert, take a serrated knife and shave off a tiny, paper-thin piece of orange and use it as a garnish. If you shave it off properly, it should curl quite nicely. The rind also becomes quite edible when it's in a small, shaved, frozen chunk.\r\n\r\n------------------------------------------------------------\r\n\r\nTrisecting a sheet of paper isn't hard... you can actually trisect lengths with compass and straightedge constructions...\r\n\r\n- Take a rectangle and fold a 4x4 grid on it. I'm gonna use coordinates, so the bottom left is (0,0), the bottom right is (4,0), top left is (0,4), top right is (4,4).\r\n- Fold the following:\r\n--- The line through (2,1) and (4,4)\r\n--- The line through (3,1) and (4,4)\r\n- These 2 lines trisect the line from (0,0) to (4,0). It's easy to prove using similar triangles.", "Solution_16": "[quote=\"d.lam.86\"][quote=\"TheDreamer\"]\nAnyways, I heard you're also good at cooking, maybe you should also post pictures of crazy foods you cooked :P[/quote]\nHe is Elyot the Multi-Talented Man! :rotfl: ;)[/quote]\r\nWhen Dr. Wright was reading Elvot's bio today(today was the media day), it was so fun. Elyot has like a few hundreds of interests. :rotfl:", "Solution_17": "[quote=\"lightrhee\"][quote=\"d.lam.86\"][quote=\"TheDreamer\"]\nAnyways, I heard you're also good at cooking, maybe you should also post pictures of crazy foods you cooked :P[/quote]\nHe is Elyot the Multi-Talented Man! :rotfl: ;)[/quote]\nWhen Dr. Wright was reading Elvot's bio today(today was the media day), it was so fun. Elyot has like a few hundreds of interests. :rotfl:[/quote]\r\n\r\nYeah, a year ago he used to say \"I'm good at every aspect of life\" and I thought he was kidding...well with every passing day I believe him more and more :D", "Solution_18": "[quote=\"TheDreamer\"][/quote][quote=\"lightrhee\"][quote=\"d.lam.86\"][quote=\"TheDreamer\"]\nAnyways, I heard you're also good at cooking, maybe you should also post pictures of crazy foods you cooked :P[/quote]\nHe is Elyot the Multi-Talented Man! :rotfl: ;)[/quote]\nWhen Dr. Wright was reading Elvot's bio today(today was the media day), it was so fun. Elyot has like a few hundreds of interests. :rotfl:[/quote][quote=\"TheDreamer\"]\n\nYeah, a year ago he used to say \"I'm good at every aspect of life\" and I thought he was kidding...well with every passing day I believe him more and more :D[/quote]\r\n\r\nI don't think I said that... I'm not that arrogant, am I???\r\n\r\nI probably said something more like \"I try to be good at every aspect of life; not just math.\" \"TRY!\" I would never claim that I'm good at everything.", "Solution_19": "[quote=\"Elyot\"][/quote][quote=\"TheDreamer\"][/quote][quote=\"lightrhee\"][quote=\"d.lam.86\"][quote=\"TheDreamer\"]\nAnyways, I heard you're also good at cooking, maybe you should also post pictures of crazy foods you cooked :P[/quote]\nHe is Elyot the Multi-Talented Man! :rotfl: ;)[/quote]\nWhen Dr. Wright was reading Elvot's bio today(today was the media day), it was so fun. Elyot has like a few hundreds of interests. :rotfl:[/quote][quote=\"TheDreamer\"]\n\nYeah, a year ago he used to say \"I'm good at every aspect of life\" and I thought he was kidding...well with every passing day I believe him more and more :D[/quote][quote=\"Elyot\"]\n\nI don't think I said that... I'm not that arrogant, am I???\n\nI probably said something more like \"I try to be good at every aspect of life; not just math.\" \"TRY!\" I would never claim that I'm good at everything.[/quote]\r\n\r\nHaha, maybe I don't remember what you said exactly, but don't worry, you don't claim to be more than you are :)", "Solution_20": "[quote=\"TheDreamer\"]\n\nHaha, maybe I don't remember what you said exactly, but don't worry, you don't claim to be more than you are :)[/quote]\r\n\r\nAwww... thanks.\r\n\r\nFunny, mostly unrelated story:\r\nI gave a long, complicated guest lecture today about \"HOW TO GET A GIRLFRIEND\". Crazy.", "Solution_21": "[quote=\"Elyot\"]\n\nAwww... thanks.\n\nFunny, mostly unrelated story:\nI gave a long, complicated guest lecture today about \"HOW TO GET A GIRLFRIEND\". Crazy.[/quote]\r\n\r\nHAHAHAHA. I wanna hear this.", "Solution_22": "I'm listening. ;) :roll:", "Solution_23": "[quote=\"Sunny\"][quote=\"Elyot\"]\n\nAwww... thanks.\n\nFunny, mostly unrelated story:\nI gave a long, complicated guest lecture today about \"HOW TO GET A GIRLFRIEND\". Crazy.[/quote]\n\nHAHAHAHA. I wanna hear this.[/quote]\r\n\r\nWell... first I gave the steps... I said that before they get girlfriends, they need to get big, active social lives with girls. Then I talked about what kind of characteristics to look for in girls, how to see if you like them, how to find one that likes you too, etc. Then I gave them some stories about some of my old girlfriends, etc. Then they asked questions.\r\n\r\nIt was quite interesting. Some things I will refrain from posting ;)", "Solution_24": "[quote=\"Elyot\"][quote=\"Sunny\"][quote=\"Elyot\"]\n\nAwww... thanks.\n\nFunny, mostly unrelated story:\nI gave a long, complicated guest lecture today about \"HOW TO GET A GIRLFRIEND\". Crazy.[/quote]\n\nHAHAHAHA. I wanna hear this.[/quote]\n\nWell... first I gave the steps... I said that before they get girlfriends, they need to get big, active social lives with girls. Then I talked about what kind of characteristics to look for in girls, how to see if you like them, how to find one that likes you too, etc. Then I gave them some stories about some of my old girlfriends, etc. Then they asked questions.\n\nIt was quite interesting. Some things I will refrain from posting ;)[/quote]\r\nJust curious, who were the audience? :huh:", "Solution_25": "[quote=\"d.lam.86\"][/quote][quote=\"Elyot\"][quote=\"Sunny\"][quote=\"Elyot\"]\n\nAwww... thanks.\n\nFunny, mostly unrelated story:\nI gave a long, complicated guest lecture today about \"HOW TO GET A GIRLFRIEND\". Crazy.[/quote]\n\nHAHAHAHA. I wanna hear this.[/quote]\n\nWell... first I gave the steps... I said that before they get girlfriends, they need to get big, active social lives with girls. Then I talked about what kind of characteristics to look for in girls, how to see if you like them, how to find one that likes you too, etc. Then I gave them some stories about some of my old girlfriends, etc. Then they asked questions.\n\nIt was quite interesting. Some things I will refrain from posting ;)[/quote][quote=\"d.lam.86\"]\nJust curious, who were the audience? :huh:[/quote]\r\n\r\nEveryone in the van, including the IMO team, Felix, and Dorette.", "Solution_26": "He is refering to the IMO team on their way to Banff. Or when they got there.", "Solution_27": "[quote=\"Sunny\"]He is refering to the IMO team on their way to Banff. Or when they got there.[/quote]\r\nOn the way. But I was sleeping, so I missed everything. :D", "Solution_28": "[quote=\"lightrhee\"][quote=\"Sunny\"]He is refering to the IMO team on their way to Banff. Or when they got there.[/quote]\nOn the way. But I was sleeping, so I missed everything. :D[/quote]\r\n\r\nYou fell asleep because you already knew everything?", "Solution_29": "[quote=\"TheDreamer\"]\nYou fell asleep because you already knew everything?[/quote]\r\n\r\nLol, probably he fell asleep because we were up to 3 in the morning the night before, went through a day of math and then faced the reporters and media for another few hours." } { "Tag": [ "geometry" ], "Problem": "Find the area of the region enclosed by the graph whose equation is $ \\mid y\\minus{}x\\mid \\plus{}\\mid y\\mid \\equal{} 2$.", "Solution_1": "[asy]size(300);\n\npair a=(-2, 0), b=(-2, -2), c=(2, 0), d=(2, 2); \n\nfilldraw(a--b--c--d--cycle, yellow+.9white); \ndraw(a--b--c--d--cycle, black+1bp);\n\ndot(Label(\"$(-2,0)$\"),a,NW);\ndot(Label(\"$(-2,-2)$\"),b,SW);\ndot(Label(\"$(2,0)$\"),c,SE);\ndot(Label(\"$(2,2)$\"),d,NE);\n\ndraw((-3,0)--(3,0),EndArrow);\ndraw((0,-3)--(0,3),EndArrow); \n\nlabel(\"$x$\",(3,0),SSW);\nlabel(\"$y$\",(0,3),SW);[/asy]\r\n\r\nThus $ 2 \\times 4 \\equal{} \\boxed{8}$.", "Solution_2": "Nice, Kouichi!", "Solution_3": "The region inside $ |x \\minus{} y| \\plus{} |y| \\equal{} 2$ is just a skew of that inside $ |x| \\plus{} |y| \\equal{} 2$, so they have the same area.\r\n\r\n[b]Problem 1:[/b] Find the volume enclosed by the condition:\r\n\\[ |x \\plus{} y \\minus{} z| \\plus{} |x \\minus{} y \\plus{} z| \\plus{} | \\minus{} x \\plus{} y \\plus{} z| \\leq 6\\]\r\n[b]Problem 2:[/b] Show that the volume enclosed by:\r\n\\[ |a_1 x \\plus{} b_1 y \\plus{} c_1 z| \\plus{} |a_2 x \\plus{} b_2 y \\plus{} c_2 z| \\plus{} |a_3 x \\plus{} b_3 y \\plus{} c_3 z| \\leq 1\\]\r\nis the same as the volume enclosed by:\r\n\\[ |a_1 x \\plus{} a_2 y \\plus{} a_3 z| \\plus{} |b_1 x \\plus{} b_2 y \\plus{} b_3 z| \\plus{} |c_1 x \\plus{} c_2 y \\plus{} c_3 z| \\leq 1\\]\r\n\r\n[u]Hint:[/u] [i]First consider the analogous case of two dimensions.[/i]" } { "Tag": [ "analytic geometry" ], "Problem": "I know that for most problems, either a proof or number is required as the final solution, and that as long as it's right and obvious simplifications were made, the answer is right.\r\n\r\nHowever, for number 5, how would you define as locus in the final answer? Like, by properties or...what?", "Solution_1": "I would assume something like an equation of a circle, or a line, or etc.", "Solution_2": "i think you mean #4 :wink:", "Solution_3": "Usually, one describes a locus either in terms of an equation representing valid points in the cartesian coordinate system or with a simple description. For example, I can describe the locus of points equidistant from a point $ O$ as a circle with center $ O$, or I can define $ O\\equal{}(0,0)$ and the locus would then be all points $ (x,y)$ such that $ x^2\\plus{}y^2\\equal{}r^2$ for constant $ r$.", "Solution_4": "[quote=\"cosinator\"]Usually, one describes a locus either in terms of an equation representing valid points in the cartesian coordinate system or with a simple description. For example, I can describe the locus of points equidistant from a point $ O$ as a circle with center $ O$, or I can define $ O \\equal{} (0,0)$ and the locus would then be all points $ (x,y)$ such that $ x^2 \\plus{} y^2 \\equal{} r^2$ for constant $ r$.[/quote]\r\n\r\nWhat if you said \"the locus is a circle with center $ (0,0)$ and radius $ r$? Would that get credit?", "Solution_5": "I'm pretty sure that for a circle, either of the following would be accepted:\r\n\r\n$ x^2\\plus{}y^2\\equal{}9$\r\n\r\nA circle centered at $ (0,0)$ with radius $ 3$.\r\n\r\nFor a line, any one of the forms should be accepted.\r\n\r\nFor most others shapes, I'd use the equation, but a description (or a picture) should be accepted.", "Solution_6": "So, for a line, could I say \"The locus of the points is the line with this certain relation [not stated here] to line [one of the ones in the problem]\"?", "Solution_7": "[quote=\"Heero Kirashami\"]So, for a line, could I say \"The locus of the points is the line with this certain relation [not stated here] to line [one of the ones in the problem]\"?[/quote]\r\n\r\nI think as long as your description includes every point in the locus.", "Solution_8": "This is really giving away a lot, guys >_>", "Solution_9": "It's actually not, the locuses discussed have nothing to do with the problem.", "Solution_10": "But if you're giving examples of loci, that's a huge clue which aren't the answer...Think about it.", "Solution_11": "Why not just delete all the posts in this thread as the original poster got his question answered.", "Solution_12": "Hey, it doesn't really matter. We're just naming geometrical shapes.\r\n\r\nWe probably could've gone with \"if the points are in this star [or other geometrical shape]\" but that's more awkward as an example. Plus, stars are hard to graph without parametrics (as far as I know...).", "Solution_13": "[quote=\"123456789\"]But if you're giving examples of loci, that's a huge clue which aren't the answer...Think about it.[/quote]\r\n\r\nLet me clarify. I didn't mean that the loci are not the answer to the problem. I only meant that they were arbitrarily chosen and have no purposeful relation to the problem." } { "Tag": [ "geometry", "3D geometry", "probability" ], "Problem": "Two cubes, whose faces are numbered 4, 5, 6, 7, 8, and 9, are\ntossed. What is the probability that the product of the two numbers\non the top faces is between 29 and 61? Express your answer as a\ncommon fraction.", "Solution_1": "the number of all possible ways that can be made is 36.\r\n\r\nlets figure out the number of ways that the product of two numbers are between 29 and 61.\r\n\r\n4~9.\r\n\r\n61>4*8,9>29\r\n61>5*6,7,8,9>29\r\n61>6*5,6,7,8,9>29\r\n61>7*5,6,7,8>29\r\n61>8*4,5,6,7>29\r\n61>9*4,5,6>29\r\n2+4+5+4+4+3=22\r\n\r\nthe answer is $ \\frac{22}{36}$, so $ \\frac{11}{18}$.\r\n\r\nanswer : 11/18" } { "Tag": [ "AMC", "AIME", "AIME I", "\\/closed" ], "Problem": "Hello!\r\nI've taken AIME problem series B during winter 2007-2008. Unfortunately, I forgot to keep a copy of the transcripts/hw problems. Is there any way I can get a copy of them now? \r\n\r\nThanks.", "Solution_1": "sorry to be redundant and reviving an old thread\r\n\r\nbut i seem to have the same problem as lena_siz, a couple days ago, my computer crashed with all my AIME problem series B homework, solutions, transcripts etc....\r\n\r\nis there any way I can access that information to study for the AIME?\r\n\r\nI believe I still have the receipt and confirmation stuff that I took the class. \r\n\r\nThanks" } { "Tag": [ "induction", "combinatorics proposed", "combinatorics" ], "Problem": "The squares of a grid m*n ( or you can assume first that the grid is n*n) are coloured arbitrary with white and black. Prove that using a king we can move from the left-most column to the right-most column only on white squares or we can move with the up-most row to the down-most row only on black squares. :) \r\n\r\n\r\nI just loved the solution I saw today. I look forward to see yours. The problem is well known, but I hope it wasn't posted before :) \r\nYou may also see this http://www.mathlinks.ro/Forum/viewtopic.php?t=138389 and figure out what is the connection :)", "Solution_1": "Anybody? C'mon the problem is really nice :)", "Solution_2": "i have one solution to this, but i'm not certain about it :(\r\ni do one induction and see a lot of cases.\r\n\r\nvery, very nice problem! :wink:\r\n\r\ncould you post the solution ciprian? i'm absolutely right that your way is more nice than mine..\r\n\r\n(again, sorry for my bad english!)", "Solution_3": "I must say, for the begining, that is NOT my solution... :blush: However, I will give just a hint for the moment cause I don't wanna spoil the fun :P \r\n[hide=\"Hint\"]\nConsider a (planar) graph G with black and white vertices corresponding to the squares - two vertices are connected if they are on the same row or column. Then add some additional vertices (this is the crux move and gives the beauty of the problem), let's say red vertices that simulate the diagonally moves of the king. You will get a triangulated planar graph, if I remember well...Then u add 2 black and 2 white vertices outside such the triangulated graph is in the interior of that 'quadrilateral'. (the same color vertices are opposite). Triangulate again and prove the statement now...Enough hints for the moment :wink: . I am in a rush now so I hope I haven'n made confusions...\n[/hide]", "Solution_4": "[quote=\"AMM, March 1999\"]\nA Tale of Two Integrals\nby Vilmos Totik\nShow that there is some interval $I \\subset [0,1]$ such that the integrals of $f$ and $g$ over $I$ are both equal to $\\frac{1}{2}$. This problem is equivalent to the following problem without integrals: On a blackjack machine one can win or lose one dollar at a time. Suppose two players playing once per minute during a period find that eventually both of them win exactly $2N$ dollars. Show that there was a time interval during which both of them won exactly $N$ dollars.\n\nSix solutions are presented that are based on other combinatorial or geometrical/topological results. These are:\n\nthe Borsuk-Ulam antipodal theorem on continuous mappings of spheres; \nthe mountain climbing problem on two climbers staying always at the same altitude; \nthe chord theorem for parallel chords of planar curves; \nthe chess king-moving theorem on king walks on cells of the same colour; \nthe winding number theorem for vector fields; \nthe Jordan curve theorem on components of simple curves.[/quote]\r\nThis is what I found related to this problem.", "Solution_5": "Could anyone post the entire article? I'd be eternally grateful :D" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ x_1,x_2,...x_5 \\ge 0$ such that\r\n\r\n$ \\sum\\limits_{i \\equal{} 1}^5 {\\frac {1}{{1 \\plus{} {x_i}}} \\equal{} 1}$ Prove that $ \\sum\\limits_{i \\equal{} 1}^5 {\\frac {{{x_i}}}{{4 \\plus{} {x_i}^2}} \\le 1}$", "Solution_1": "Using Caushy shwarz inequality we have\r\n\r\n$ \\frac{1}{5}\\plus{}\\frac{1}{x\\plus{}1}\\plus{}\\frac{1}{x\\plus{}1}\\plus{}\\frac{1}{x\\plus{}1} \\ge \\frac{16}{3x\\plus{}8} \\ge \\frac{16}{3x\\plus{}x\\plus{}\\frac{16}{x}}\\equal{}\\frac{4x}{4\\plus{}x^2}$\r\n\r\nSimilarly we apply the same thing we get\r\n\r\n$ \\sum_{i\\equal{}1}^5\\left(\\frac{1}{5}\\plus{}\\frac{1}{x_i\\plus{}1}\\plus{}\\frac{1}{x_i\\plus{}1}\\plus{}\\frac{1}{x_i\\plus{}1}\\right) \\ge \\sum_{i\\equal{}1}^5 \\frac{4x_i}{4\\plus{}x_i^2}$\r\n\r\nwhich is equivalent to \r\n\r\n$ 1 \\ge \\sum_{i\\equal{}1}^5 \\frac{x_i}{4\\plus{}x_i^2}$", "Solution_2": "The given hypothesis implies $ \\sum_{cyc}{\\frac{4-x_i}{1+x_i}} =0$ .We need to prove that :\r\n\\[ 1= \\sum_{cyc}{\\frac {1}{{1 + {x_i}}} \\geq \\sum_{cyc}{\\frac {{{x_i}}}{{4 + {x_i}^2}} \\iff \\sum_{cyc}{\\frac{4-x_1}{1+x_1}\\cdot \\frac{1}{4+x_1^2}} \\geq 0}}\\]\r\nLet $ x_1 \\leq x_2 \\leq x_3 \\leq x_4 \\leq x_5$ ,then :\r\n\\[ \\frac{4-x_1}{1+x_1} \\geq \\frac{4-x_2}{1+x_2} \\geq \\frac{4-x_3}{1+x_3} \\geq \\frac{4-x_4}{1+x_4} \\geq \\frac{4-x_5}{1+x_5}\\]\r\n\\[ \\frac{1}{4+x_1^2 } \\geq \\frac{1}{4+x_2 ^2} \\geq \\frac{1}{4+x_3 ^2} \\geq \\frac{1}{4+x_4 ^2} \\geq \\frac{1}{4+x_5 ^2}\\]\r\n\r\nNow applying Chebyshev inequality for the above sequences ,we're done .Equality holds for $ x_i =4$" } { "Tag": [ "function", "algebra", "domain", "calculus", "integration", "geometric series", "arithmetic series" ], "Problem": "I've encountered some directions for problems that say \"Solve for x, state any restrictions on the variables.\" What does this mean, and how do you find the restrictions? Thanks in advance, dynamo729.", "Solution_1": "\"Restrictions\" most likely means the set of values that $ x$ can take on and still have the function produce a real result, or the domain of the function. For the most part, you just have to worry about dividing by 0 and taking square roots of negative numbers.", "Solution_2": "Restrictions are like..I'll give a few examples.\r\n\r\n[hide=\"eg1\"]\nFind all real numbers $ a$ such that the roots of $ x^3 \\minus{} 6x^2 \\plus{} 21x \\plus{} a \\equal{} 0$ are not all real and form an arithmetic progression.\n[/hide]\n\n[hide=\"eg2\"]\nFind the only integral solution of $ (x \\minus{} 1)^3 \\plus{} (x \\minus{} 2)^3 \\plus{} (x \\minus{} 3)^3 \\plus{} ... \\plus{} (x \\minus{} 2007)^3 \\equal{} 0$\n[/hide]\n\nEquations for which you want integer solutions are called [b]Diophantine equations[/b]\n\n[hide=\"eg3\"]\nSolve the equation $ x/(x \\minus{} 2) \\equal{} 3/(x \\plus{} 2) \\plus{} 6x/(x^2 \\minus{} 4)$.\n[/hide]\r\n\r\nYou can't have $ x \\equal{} 2 \\text{ or} \\minus{} 2$. If you find that 2 or $ \\minus{} 2$ turn out to be the solutions, these are called extraneous solutions. (In this case, -2 is an extraneous solution, so there are no solutions to this equation)\r\n\r\nWhy \"can't we divide by 0\"? Actually we can, but when we divide by 0, we can infinity. I have a truly beautiful proof of this using geometric series and arithmetic series.\r\n\r\nConsider a geometric series $ a\\plus{}ar\\plus{}ar^2\\plus{}ar^3\\plus{}...\\equal{}a/(1\\minus{}r)$ by geometric series formula, where $ a$ is a real number. If $ r\\equal{}1$, then the geometric series is an arithmetic series $ a\\plus{}a\\plus{}...\\equal{}\\infty$. Then $ a/(1\\minus{}r)\\equal{}a/0\\equal{}\\infty$." } { "Tag": [ "LaTeX" ], "Problem": "Is there an equivalent to Microsoft word's spellcheck in TeXnicCenter that doesn't count all the commands and colored text?\r\n\r\nSometimes I copy and paste everything into microsoft word and spellcheck, ignoring what word thinks as wild punctuation mistakes. Is there a better way to do this?", "Solution_1": "If you click on [b]tools[/b] button, you'll see [b]Spelling[/b] in the menu. That's the TeXnic center built-in spell checker. Try it and see if it is satisfactory for you." } { "Tag": [ "geometry", "quadratics" ], "Problem": "I am a senior in high school trying to decide what major to study in college. I am wondering whether there are any electrical engineers here who know about the field, since I am considering majoring in electrical engineering. Particularly, I am concerned about my impression that most (or many) electrical engineers end up doing mindless tasks for their day jobs that pay five-figure salaries and lead nowhere. I do not know whether this description is true or not, but I get the feeling that there is not much left to discover in the field of electrical engineering (as opposed to, say, the field of bioengineering). Thus, I fear that after getting an electrical engineering degree, I would just be thrown in with the masses of people who perform manual labor and textbook applications of electrical engineering principles and never go anywhere with their careers.\r\n\r\nAny opinions or remarks would be greatly appreciated. Thanks.", "Solution_1": "Even when my research is applied to Electrical engineering, I am not an engineer, nevertheless, in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34592]this thread[/url], wild octagon sais that \r\n\r\n[quote]The Bureau of Labor Statistics site lists 69,970 as the median salary for actuaries, with the top 10% earning at least 137,650. Compare to salaries for mathematicians (76,470 / 112,780), electrical engineers (69,930 / 103,860) and financial analysts (57,100 / 108,060).[/quote]\r\nThis suggest that in general, Mathematicians earn more than Electrical Engineers and that your impressions may be not that wrong. I don't know much about the market for Electrical Engineers here, but I know plenty of them in Latin America, and they claim that one of the problems in their profession is that the real market is oversaturated (in Latin America, I don't know here) and that many people with Bachelors or Masters in the Area have to take jobs as technicians, and well, you don't want to spend six years and a lot of money in the education for a great job to get at the end the same job that you will get if you go year and a half to a technical institute to earn a technical degree!\r\n\r\nBy the way, this should be in the College forum I think!", "Solution_2": "The problem with speaking in generalities is that well... it's just a bunch of generalities. It doesn't necessarily reveal anything that will be specific to YOU. There are a lot of people who get degrees in electrical engineering - some of whom do well, some of whom do not; some of whom have exciting careers, and some of whom do not; some of whom enjoy what they do, and some of whom... well, who do not.\r\n\r\nThe main differences between these types of people are a.) How lucky/talented/wise are they to end up working in an exciting place, or starting an exciting business, b.) How good are they at what they do, and c.) Their personality/temperment.\r\n\r\nTwo of my roommates from college are electrical engineers and they love what they do. One works for Intel designing better and faster processors, and the other (last time I heard) was working for Hughes doing something on Space Satelites. They both have pretty intellectually challenging jobs that I don't think are too mundane. (Although [i]every[/i] job will have some aspects that are mundane have-to-be-done type of things).\r\n\r\nI'm sure, though, there are other people with degrees in electrical who do not enjoy what they do. Maybe because they are not so great at it, or maybe they just fundamentally don't like the subject. Perhaps for those people, an electrical engineering degree wasn't the best choice for them.\r\n\r\nWhat I'm saying is, that you can't use anectdotal evidence or general statistics to decide what is best for YOU. That all depends on what fundamentally interests you, and what you feel that you are good at. There are certainly paths to sucess and happiness in electrical engineering, you just have to figure out if any of those paths are suited to you.", "Solution_3": "Hmm...What do you all think about the future of electrical engineering, in terms of new scientific discoveries and advances? I am not very knowledgable in this area, but it seems to me, as a layman, that we also know most of the things to know about electricity. Am I wrong? Is the future of electrical engineering as bright as the future for, say, nanotechnology or bioengineering?", "Solution_4": "Electrical Engineers don't just study \"electricity\". There are new advances being made in electrical engineering right now - and lots of them. From advances in quantum computing, to cryptography, to data compression and transmission, to highly mathematical research topics like prime number and quadratic residue distributions.\r\n\r\nIt could be said that we are living in the Electronic Age - so there are lots of new and improving technologies being made in electrical engineering fields. For example, building LCD-like displays that are as thin as a sheet of transparency paper (ever seen the portable computers displays they had in the movie Red Planet?)." } { "Tag": [ "function", "algebra", "polynomial", "number theory", "relatively prime", "number theory unsolved" ], "Problem": "Find all $ f(x)\\in Z[x]$ such that there are infinitely many pairs of coprime positive integers $ (a,b)$ such that $ a\\plus{}b$ divides $ f(a)\\plus{}f(b)$", "Solution_1": "Anybody? :)", "Solution_2": "Let $ n\\equal{}a\\plus{}b$, then for infinetely many n exist x 0$ or $ g(a) > a$ and $ \\gcd(b,a) \\equal{} 1$. The latter holds because $ \\gcd(b,a) \\equal{} 1 \\iff \\gcd(g(a) \\minus{} a,a)) \\equal{} 1 \\iff \\gcd(g(0),a) \\equal{} 1$.\r\n\r\nIf $ g(x)$ is not a constant function then its degree must be greater than or equal to 2. And so, we requires the leading coefficient to be greater than or equal to 0 so that eventually $ g(x)$ will always exceed $ x$. In case the leading coefficient fails to be, try $ b \\equal{} \\minus{} g(a) \\minus{} a$. And the same argument holds.\r\n\r\nIf $ g(x)\\not\\equiv 0$ is a constant function. Clearly, it does not satisfies the condition for infinitely many $ x$. We want to prove that there exists finitely many coprime $ a,b$ such that $ a \\plus{} b | a^{2k} \\plus{} b^{2k}$. This is in turn equivalent to $ a \\plus{} b | 2$....\r\n\r\nIn conclusion, $ f(x) \\plus{} f( \\minus{} x)$ must not have only one term of form $ x^{2k}$ (either zero or at least 2).", "Solution_12": "If $ f(x)\\equal{}f_{odd}(x)\\equal{}\\sum_{i\\equal{}1}^N k_ix^{2i\\minus{}1}$, then it will clearly work. What about terms with even powers? Suppose that exists a function $ f_{even}(x)\\equal{}\\sum_{i\\equal{}1}^N k_ix^{2i}$ such that $ a\\plus{}b|f_{even}(a)\\plus{}f_{even}(b)$ for infinite coprime pairs $ a\\plus{}b$. As a matter of fact, it is possible to prove that holds $ a\\plus{}b|f_{even}(a)\\minus{}f_{even}(b)$. In fact, for each term of $ f_{even}(a)\\minus{}f_{even}(b)$ with the same power, e.g. $ k_ja^{2j}\\minus{}k_jb^{2j}$, we have that $ k_ja^{2j}\\minus{}k_jb^{2j}\\equal{}k_j(a\\plus{}b)(....)$, thus $ a\\plus{}b|k_ja^{2j}\\minus{}k_jb^{2j}$. But then, if $ a\\plus{}b|f_{even}(a)\\plus{}f_{even}(b)$ and $ a\\plus{}b|f_{even}(a)\\minus{}f_{even}(b)$, then $ a\\plus{}b|2f_{even}(a)$ for infinitely many coprime pairs $ (a,b)$. Now, we choose a $ b$, coprime to $ a$, such that $ a\\plus{}b\\gg 2f_{even}(a)$, but in this case $ a\\plus{}b$ can't exactly divide $ 2f_{even}(a)$. Absurd. Thus, all the $ f(x)$ are of the form $ f(x)\\equal{}f_{odd}(x)\\equal{}\\sum_{i\\equal{}1}^N k_ix^{2i\\minus{}1}$.", "Solution_13": "[quote=\"geda\"]Now, we choose a $ b$, coprime to $ a$, such that $ a \\plus{} b\\gg 2f_{even}(a)$, but in this case $ a \\plus{} b$ can't exactly divide $ 2f_{even}(a)$. Absurd. [/quote]\r\nYou have shown that for each $ a$, there are finitely many $ b$'s that satisfy the equation. Yet, there are infinitely many possible $ a$'s so how do you know that there are not infintely many values of $ b$?\r\n\r\nBy the way, are we assuming that $ a$ and $ b$ must be positive? If not, then we can have that $ (a \\plus{} b)|\\sum_{i \\equal{} 0}^n {a_ib^i}$ as long $ a_i$ are not $ 0$ for all $ i$. This is because then we may continuously divide off $ b$'s until we are left with a constant term; let the new polynomial be $ \\sum _{i \\equal{} 0}^m {c_ib_i} \\equal{} g(b)$. From here, we pick a $ b$ that is coprime to the constant term, so $ b$ is coprime to $ g(b)$. Let $ g(b) \\minus{} b \\equal{} a$, so we have found a suitable $ a$ for infinitely many $ b$, meaning that infinitely many $ (a,b)$ exist. \r\n\r\nYet, if this is not the problem then we can hopefully do the following: \r\n[hide=\"Solution\"]\nAs [b]geda[/b] did, we can narrow the condition down to having that $ (a \\plus{} b)|2f(a)$, where $ f$ is an even polynomial. Notice that $ \\gcd (a \\plus{} b, a) \\equal{} 1$, so we may divide off a power of $ a$ from $ f(a)$ until we are left with a constant. Let the new function be $ g(a)$. This means that we require $ (a \\plus{} b)|2g(a)$. If $ g(a)$ is a constant function, then there are only finitely many factors of $ 2g(a)$, so $ a \\plus{} b$ may only take on a finite number of numbers, meaning that there are a finite number of $ (a,b)$. If $ g(a)$ has a degree of at least $ 1$, then we consider $ h(x)$ so that it is $ \\minus{} g(x)$ if the leading coefficient of $ g(x)$ is negative and $ g(x)$ if the leading coefficient of $ g(x)$ is positive. Since $ g(x)$ has a term that is at least $ x^2$ (recall that $ g(x)$ is even), we have that $ h(x)$ will have its largest term have an exponent of at least $ 2 > 1$ and it will have a positive coefficient. Thus, there will be a point when $ h(a) > a$ for $ a > k$ for a given positive $ k$. From here, we let $ a$ be above $ k$ and relatively prime to the constant term of $ h(x)$. There are infinitely many values of $ a$, so we will prove that there is at least one $ b$ for every $ a$ so that $ \\gcd (a,b) \\equal{} 1$ and $ (a \\plus{} b)|f(a)$, which will establish that there are infinitely many suitable pairs $ (a,b)$. Pick $ b \\equal{} h(a) \\minus{} a > 0$, so $ (a \\plus{} b)|h(a)$, so $ (a \\plus{} b)|2g(a)$, so $ (a \\plus{} b)|2f(a)$. Thus, as long as $ g(a)$ is not a constant term, so as long as $ f(a)$ does not only consist of one term with an even exponent, we may find infinitely many suitable pairs $ (a,b)$. [/hide]" } { "Tag": [ "probability" ], "Problem": "When n standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of S. The smallest possible value of S is \r\n(A) 333 \r\n(B) 335 \r\n(C) 337 \r\n(D) 339 \r\n(E) 341\r\n\r\nYou can roll 1994 with 332 6's and 1 2. So isn't the prob. of rolling 1994 with 333 dice the same as the prob. of rolling any 332 same numbers and 1 other number? Why isn't the answer 332*1+2 = 334?", "Solution_1": "Because you can also roll 1994 with 333 dice by rolling three hundred and thirty-one 6s, one 5 and one 3 :wink:", "Solution_2": "[hide=\"hint\"]\nThe probability of getting two sums is the same if and only if they are equidistant from the max and min (for example, with two die, the min is 2 and the max is 12; 4 and 10 are both 2 away from the max and min and have the same probability.\n[/hide]\n\n[hide=\"solution\"]\nthe maximum sum is $6n$ and the minimum sum is $n$.\nThe probability of getting a sum of $n+a$ is the same as the probability of getting a sum of $6n-a$. All other sums would have different probabilities.\n\n$n+(6n-1994)$ has the same probability as getting $1994$.\nWe need to find the minimum value of $n+(6n-1994)=7n-1994$ knowing that\n$6n\\ge 1994$\n$n\\ge 332.333...$\n$n\\ge 333$\n\nThen $7(333)-1994=\\boxed{337}$\n[/hide]" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "geometry", "MATHCOUNTS", "calculus" ], "Problem": "dude AIME was evil\r\n\r\nI felt like firmly establishing this\r\n\r\nrather than being uncertain\r\n\r\nlike, on par with USAMO 05\r\n\r\nevil", "Solution_1": "if it was that evil, i should get into usamo with a 5.", "Solution_2": "8 last year\r\n\r\n12-13 on last couple year's practice tests (recently)\r\n\r\n6 AIME I 2007", "Solution_3": "I thought it was quite easy (except the geometry ones which I didn't even try), so I answered 10 & got a 9 (missed 4 stupidly).\r\n\r\nMost of the problems seemed to have straight forward ways to do them, though there were a few tricks.", "Solution_4": "so many words in the questions\r\n\r\ni read it\r\n\r\nand then realized there was a second line\r\n\r\nand then at hird line\r\n\r\nand then i quit reading and guessed", "Solution_5": "I thought it was MUCH easier than average.\r\nI got an 8, one careless mistake (number 4), the mathcounts problem screwed me over for that one.", "Solution_6": "[quote=\"calc rulz\"]8 last year\n\n12-13 on last couple year's practice tests (recently)\n\n6 AIME I 2007[/quote]\r\n\r\nDUDE THAT'S EXACTLY THE SAME AS ME!", "Solution_7": "[quote=\"bpms\"]I thought it was MUCH easier than average.\nI got an 8, one careless mistake (number 4), the mathcounts problem screwed me over for that one.[/quote]\r\n\r\n\r\n8..thats pretty good for a guy still in middle school (by the mathcounts statement)..but you might also want to erase that mathcounts statement..seeing peopel havnet taken it yet..", "Solution_8": "I thought the problems were really easy, but they were really disgusting with lots of messy arithmetic brute force that was outrageously easy to make mistakes on. But yes I very much agree it was evil, just not hard.", "Solution_9": "[quote=\"usaha\"][quote=\"bpms\"]I thought it was MUCH easier than average.\nI got an 8, one careless mistake (number 4), the mathcounts problem screwed me over for that one.[/quote]\n\n\n8..thats pretty good for a guy still in middle school (by the mathcounts statement)..but you might also want to erase that mathcounts statement..seeing peopel havnet taken it yet..[/quote]\r\nNo, it was on an old states.\r\nI still have to do state mathcounts (mine is on the 30th)\r\nBut yes, I'm still in middle school.\r\nI stayed up until 12:30 the night before cramming in AIME prep, because I was worried that I would black out.", "Solution_10": "dude...they posted the answers on the amc website :roll: everyone took it.\r\n\r\nIf you mean mathcounts, the planet one is a classic problem, not from this year.", "Solution_11": "[quote=\"jb05\"]I thought the problems were really easy, but they were really disgusting with lots of messy arithmetic brute force that was outrageously easy to make mistakes on. But yes I very much agree it was evil, just not hard.[/quote]\r\n\r\nI'd agree with that. I'm not sure that it would take as much mathematical knowledge as it would rough brute-forced solutions or extended casework. Not the most representative of AIMEs and certainly not the best one, let's just hope that the AIME II is about the same or I'm sure this messageboard is going to hear a lot of complaining. I got the same score I got the last 2 years on the AIME, a 9.", "Solution_12": "I've posted comments in my [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=277]blog.[/url] Feel free to discuss them here.", "Solution_13": "I concur that the AIME was quite silly, as many problems seemed to be mostly computation.\r\n\r\nIt was sort of like, wait isn't this supposed to test our math skills, not our ability to avoid off-by-one errors?", "Solution_14": "[quote=\"JSteinhardt\"]I concur that the AIME was quite silly, as many problems seemed to be mostly computation.\n\nIt was sort of like, wait isn't this supposed to test our math skills, not our ability to avoid off-by-one errors?[/quote]\r\nWait, which were the computation heavy problems?", "Solution_15": "2\r\n3\r\n5\r\n6\r\n9\r\n10\r\n11\r\n\r\nAmong the ones I did, which were 1-11.\r\n\r\nI consider it to be computationally intensive if over 75% of the most elegant solution involves manipulating numbers.\r\n\r\nI am told by Zhai that many of the other geo problems were also very algebraic.\r\n\r\nAlso the only problem (among those) that seemed to have a particularly clever solution was 8, though according to other people, brute force is not too hard either.", "Solution_16": "[quote=\"JSteinhardt\"]I concur that the AIME was quite silly, as many problems seemed to be mostly computation.\n\nIt was sort of like, wait isn't this supposed to test our math skills, not our ability to avoid off-by-one errors?[/quote]\r\n\r\nThat IS a math skill, though maybe not the type of skill you are refering to.", "Solution_17": "yeah \r\n\r\n#5: case work\r\n#6: case work\r\n#9: I messed up my equation solving but still it was a bit messy but I guess it's still average.\r\n\r\n#10: case work\r\n\r\n#12: a huge radical mess(possibly the most computational problem I have done yet)\r\n\r\n#14: guessable\r\n\r\n#15: should've switched position with #10", "Solution_18": "Hey I'm scheduled to take the II this year - who knows where I can find the problems from the AIME I?\r\n\r\nIf they're online, a link would be great", "Solution_19": "...Shocking...\r\n\r\nLast two years, I have succumbed to careless mistakes, calculations errors, etc., missing the cutoff by 1 point every time.\r\n\r\nThis year, I made relatively few such errors, and got a 12?\r\n\r\nSee, I've always maintained that both the AMC and AIME are very much luck-based, and it's extremely easy to slip up on them. I learned the hard way not to be overconfident and to be extremely careful in checking...but to have been on the IMO team and now be unsure about making it...that does really suck...\r\n\r\nMaybe they'll make an exception, considering that you've been on the team? Or you'll probably still make it?", "Solution_20": "2 wasn't too computationally heavy. I got a non-integral value for time, so my hopes were dashed, but I kept doing the problem anyways and got the answer.\r\n\r\nMoral: Just do it.", "Solution_21": "[quote=\"K81o7\"]Maybe they'll make an exception, considering that you've been on the team? Or you'll probably still make it?[/quote]\r\n\r\nWho are you referring to?", "Solution_22": "[quote=\"JSteinhardt\"][quote=\"K81o7\"]Maybe they'll make an exception, considering that you've been on the team? Or you'll probably still make it?[/quote]\n\nWho are you referring to?[/quote]\r\n\r\nThe person who started the thread.", "Solution_23": "Arnav got a 10, so no worries there", "Solution_24": "it was tough. i've learned a lot this past year and still only improved by 1.\r\n\r\npractice test: did 10, got 6 correct\r\n2007 AIME I: did 10, got 6 correct\r\n\r\ntalk about consistency :roll: \r\n\r\nwell, i got #1, 2, 3, 5, 7, and 14 correct. proud of that #14 :P \r\n\r\nmy index: 192. so close! :( will the index ever go that low? :huh:", "Solution_25": "StellarSupremacy: dude\r\nStellarSupremacy: it is not about making it\r\nStellarSupremacy: it is about OWNING it\r\nStellarSupremacy: THIS HAS NOT BEEN THE CASE\r\nStellarSupremacy: I WILL GO BE DEPRESSED NOW", "Solution_26": "[quote=\"MysticTerminator\"]StellarSupremacy: dude\nStellarSupremacy: it is not about making it\nStellarSupremacy: it is about OWNING it\nStellarSupremacy: THIS HAS NOT BEEN THE CASE\nStellarSupremacy: I WILL GO BE DEPRESSED NOW[/quote]\r\n\r\n-.- At least you made it. Others are like omg^2, i hope i made it.", "Solution_27": "[quote=\"MysticTerminator\"]StellarSupremacy: dude\nStellarSupremacy: it is not about making it\nStellarSupremacy: it is about OWNING it\nStellarSupremacy: THIS HAS NOT BEEN THE CASE\nStellarSupremacy: I WILL GO BE DEPRESSED NOW[/quote]\r\n\r\nExactly. I was totally shooting for 15 in my last year, but then I got to #10 and was like crap, this problem stinks. And then #12 took me about a year to work through the radicals.", "Solution_28": "[hide=\"my comments on each problem\"]*1. Trivial; simple computation\n*2. you just have to read this one carefully and set up the equations\n*3. All you had to do here was expand and write a simple equation\n4. WAY too similar to an old MathCounts question; I didn't really solve it, so I missed it. This was probably put in just to trip up MathCounts people :ninja: \n*5. This one could be solved easily with mods or pattern recognition\n*6. Nice casework here using the multiples of 13 as \"jumping\" points\n*7. Not very hard, but intimidating\n*8. I really liked this one. You had to realize that the two polynomials share a root and write out a few equations using the sum and product of roots.\n9. No idea; Geometry is my weakness. The picture was very hard to draw\n*10. This was some casework. You had to be careful to make sure you had the cases to cover everything.\n*11. This was a nice problem, but only if you actually solved it and didn't just get it by pattern.\n\nI haven't looked at the rest yet.[/hide]", "Solution_29": "please guys\r\n\r\nunless it's not allowed for those people taking the II to see the problems from the I, I'd really appreciate it if someone would post the problems.", "Solution_30": "math jam\r\ntomorrow", "Solution_31": "I'd rather solve the problems myself, and I will be driving back home from the mountains from about 4 tomorrow until 7 or 8 am friday. I dont care if it's the official problem statement, but if there's a few problems without diagrams that you basically remember off the top of your head, it'd we great to see them.", "Solution_32": "What is up with this test? No geometry until the problems start getting harder? I suck at number theory and counting and that WAS the easy part on the test. Also, I had never even seen a problem similar 5 or 7 before.. so I skipped them because they seemed hard..then I went back to 7 with like 5 mins to go and realized they all differ by one except for the ones that are integers..so I did it really quickly and made a computational mistake and not realizing that log 1=0", "Solution_33": "4 last year, 7 this year.\r\nGiven the fact that I'm a sophomore, I'm pretty sure I made it. Though the AIME was pretty dumb this year.", "Solution_34": "Try to compact the threads.\r\n\r\nEspecially the threads that can generally be titled under \"complaining\"." } { "Tag": [ "function" ], "Problem": "Let f(x)=5-4x. Find \"a\" so that f(a)+f^-1(a)=a\r\n\r\nLet g(x)=x^3-9x. Determine all values of k so htat g(2k-7)=0.\r\n\r\nGiven f(x)=(2^x)-4\r\nFind all x such that f(x)*f^-1(x)=0\r\nEvaluate f^-1 o f(\u221a3x^2+5)\r\n\r\nim having some trouble figuring out what i have to do for some of these...please and thanks for any help!! =]", "Solution_1": "1) $f^{-1}(a)=-\\frac{a}{4}+\\frac{5}{4}$\r\nSo then you just do $5-4a-\\frac{a}{4}+\\frac{5}{4}=a$\r\n\r\n2) First factor $g(x)$ and figure out what $x$ values will make $g(x)=0$. Then you can figure what values of $k$ will give you these $x$ values.\r\n\r\n3) You now either $f(x)$ or $f^{-1}(x)$ must equal zero, so $x$ has to be either:\r\n- an $x$ value that makes $f(x)=0$\r\n- the value of $f(0)$, since $f^{-1}(f(0))=0$\r\n\r\n4) By definition, $f^{-1}(f(x))=x$" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "X is a variable point on the disk centered at A (with radius R_1) and Y is a variable point on the disk centered at B (with radius R_2). Find the locus of points Z such that \\triangle{XYZ} equilateral :huh: [/hide][/b]", "Solution_1": "Hello - does anybody have a solution to this problem, especially as it relates to http://www.artofproblemsolving.com/Forum/viewtopic.php?p=492881#492881?\r\n\r\nHere, I'll retype it in a little better format:\r\n\r\n$ X$ is a variable point on the disk centered at $ A$ (with radius $ R_1$) and $ Y$ is a variable point on the disk centered at $ B$ (with radius $ R_2$). Find the locus of points $ Z$ such that $ \\triangle{XYZ}$ equilateral\r\n\r\nThank you" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "AMC 10" ], "Problem": "I'm a junior atm. I recently (this year) found out about the AMC contest, and the tests following it. So, it's my first year taking the test and I hope to advance to the AIME at least (pretty sure I won't advance to USAMO). I've taken a couple of previously administered AMC tests and have scored above 100 on them, but I want to study and make sure I safely get at least a 100, and maybe even around 120. Considering it's this late in the stage, should I bother doing the intro texts or should I just start with vol 1? I would like to hear what you guys have done or are doing to prepare and your suggestions on what I should do. Right now my plan is to do about 1 chapter in vol 1 every two days while doing two practice AMC tests a week until the actual test. Is this a sound plan?\r\n\r\nBy the way, I'm talking about the AMC 12.", "Solution_1": "In your case, I would definitely do Volume 1. [u]Make sure you understand the concepts though.[/u] Also, most of the concepts are pretty straightforward, but some of the problems are difficult. Try to solve these; those kind of problems help.", "Solution_2": "Pretty good. Don't go any faster. Know absolutely everything in terms of manipulations. \r\n\r\nI also recommend saving a 3 or 4 more recent AMC 12s for the weekend before the actual test. \r\nDo some of the mock AMCS starting in january, because they are often harder, and taking tests slightly harder than the actual test is a amazing way to improve. \r\n\r\nTo show you how much, in freshmen year I improved from 6-7 to consistent 9s on aimes just by doing a lot of tough mock aimes. \r\nAlso, as you get closer to the test, like end of january, I recommend you figure out where you usually stop, and work from there. For example, if you usually do the first 15 problems, and strugle with 16-18, do around 10 tests just on the 15-19 area. This will give you the boost u need without wasting an hour doing the easy first 12 questions. Of course, also have a few sessions where you sit the whole test through.", "Solution_3": "I agree with NuncChaos. Doing problems harder than the contest you are preparing for, is a great method and it works very well. I also like the idea of doing just the 15-19 range (for example), I've never tried that.", "Solution_4": "[quote=\"AIME15USAMO\"]I agree with NuncChaos. Doing problems harder than the contest you are preparing for, is a great method and it works very well. [/quote]\r\n\r\nNot necessarily... I did this and it kind of messed me up, but it does help somewhat. Just don't keep on doing USAMO and international questions thinking that's going to make you score a 150 on the AMC10 or 12. :blush:" } { "Tag": [ "USAMTS", "AMC", "AIME", "search" ], "Problem": "Is there anyway I can buy the past years USAMTS other than ones that are available on the web?\r\n\r\nI really want to qualify the AIME through it next year but I want to have full practices on as much as possible so I can do my best job.\r\n\r\nIf this is a pm matter, please free to pm me. :)", "Solution_1": "[quote=\"Silverfalcon\"]Is there anyway I can buy the past years USAMTS other than ones that are available on the web?\n\nI really want to qualify the AIME through it next year but I want to have full practices on as much as possible so I can do my best job.[/quote]\r\nI know of no such printed materials yet, though I recall George Berzsenyi, the founder and writer of the USAMTS, once saying that he hopes to publish such a book one day.\r\n\r\nHowever, seven more years of former USAMTS problems are available on the web than the six years (Years 10 through 15) shown on [url=http://www.usamts.org/Problems/U_ProblemsPast.php]the USAMTS Past Problems page[/url]. George Berzsenyi used to also run the International Mathematics Talent Search, which featured the same problems. The Canadian Mathematical Society has a web page, [url=http://www.cms.math.ca/Competitions/IMTS/]International Mathematics Talent Search[/url] that contains all the problems of all 44 rounds of the IMTS. Round 29 of the IMTS corresponds to Year 10 Round 1 of the USAMTS. Round 44 of the IMTS corresponds to Year 13 Round 4 of the USAMTS.\r\n\r\nThese IMTS problems have no solutions posted. However, I suppose that after this year's USAMTS Round 4 is over, you could organize a practice session in this AoPS forum that goes through some old rounds of the IMTS, and compare solutions with the other participants of the practice session.\r\n\r\nErin Schram\r\nUSAMTS grader", "Solution_2": "Thank you very much. :) \r\n\r\nI'll try to solve as many as can.\r\n\r\nJust wondering, is working on AIME help on USAMTS? (I know USAMTS get students to AIME but I wonder whether working on AIME helps USAMTS also)", "Solution_3": "I think working on any of the AMC tests would definitely improve your USAMTS score.\r\n\r\nIf you practice by learning the math inside and out (which is the whole reason for contests), you should improve in any math contest you take. Math is linked so that once you grasp one concept, it becomes easier to grasp many related concepts. So, even if AIME has a different flavor, there is still a logical process you must continually improve for any contest, not to mention life's problems.", "Solution_4": "I find that the USAMTS questions are similar to the AIME questions, at least in difficulty. The differerence is that for USAMTS you get a whole month for five questions, whereas for AIME you get three hours for fifteen questions.", "Solution_5": "[quote=\"origamimasterjared\"]I find that the USAMTS questions are similar to the AIME questions, at least in difficulty. The differerence is that for USAMTS you get a whole month for five questions, whereas for AIME you get three hours for fifteen questions.[/quote]\r\nThe major difference between AIME problems and the USAMTS problems is that many of the USAMTS problems are the \"prove . . . something something something\" whereas the AIME problems are (always, by necessity) of the type \"find something (that has an integer value on [0,999]).\" This makes many of the USAMTS problems \"longer,\" if not more difficult. Of course, USAMTS problems tend to have (especially this year) an extreme variety of difficulty levels (at least as much as the difference between an AIME problem 1 and an AIME problem 15).\r\n\r\nI do old AIMEs to practice for the USAMTS; they certainly can't hurt your score! :)", "Solution_6": "[quote=\"ZennyK\"]Of course, USAMTS problems tend to have (especially this year) an extreme variety of difficulty levels (at least as much as the difference between an AIME problem 1 and an AIME problem 15).[/quote]\r\n\r\nThat's for sure--on one problem set, I solve two problems in no more than five minutes but spent my 30 days on one of the others and still didn't get it after all that time.\r\n\r\nI also agree that the USAMTS problems are similar to those in the AIME. But I like the amount of time you get to work on them. It keeps you from feeling like a loser when the round is over. I'm sure we've all said this after the AMC/AIME: \"WHY didn't I get that problem!?\" :P" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let a;b;c>0 and $ a^3b^3\\plus{}b^3c^3\\plus{}c^3a^3\\equal{}3$\r\nProve that\r\n$ 3(b^3\\plus{}c^3\\plus{}b^2c^2)(b^3\\plus{}a^3\\plus{}b^2a^2)(a^3\\plus{}c^3\\plus{}a^2c^2)\\ge (a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2)^4$", "Solution_1": "we have:\r\n$ 3(b^3 \\plus{} c^3 \\plus{} b^2c^2)(b^3 \\plus{} a^3 \\plus{} b^2a^2)(a^3 \\plus{} c^3 \\plus{} a^2c^2)\\equal{}(a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3)(b^3 \\plus{} c^3 \\plus{} b^2c^2)(b^3 \\plus{} a^3 \\plus{} b^2a^2)(a^3 \\plus{} c^3 \\plus{} a^2c^2)\\equal{}(a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3)(b^3\\plus{}b^2c^2\\plus{}c^3)(a^2b^2\\plus{}b^3\\plus{}a^3)(a^3\\plus{}c^3\\plus{}a^2c^2) \\ge (a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2)^4$", "Solution_2": "yes, I agree with the first man\r\n\r\nOnly way I can think is this\r\nwe have:" } { "Tag": [], "Problem": "Persons A, B, C, D, E, and F are to be sorted into $3$ groups of $2$. How many possible different sets of $3$ groups are there? (The group A, B, is the same as the group B, A).\r\n\r\nIs it[hide]$\\frac{\\displaystyle\\binom{6}{2}\\binom{4}{2}}{3!}$?[/hide]", "Solution_1": "yes it is right,$\\frac{6!}{(2!)^33!}$\r\nyou can general it ab to a group of b people $\\Rightarrow \\frac{(ab)!}{(b!)^aa!}$ ;)", "Solution_2": "Is the set of groups (a,b) (c,d) (e,f) the same as (a,b) (e,f) (c,d)?\r\n\r\nI would think it'd be the same and the answer then just be $\\binom{6}{2}\\cdot\\binom{4}{2}$.", "Solution_3": "cool, thanks. :)", "Solution_4": "[quote=\"Elemennop\"]Is the set of groups (a,b) (c,d) (e,f) the same as (a,b) (e,f) (c,d)?\n\nI would think it'd be the same and the answer then just be $\\binom{6}{2}\\cdot\\binom{4}{2}$.[/quote]But the problem with that is, that considers the order that you choose it in. So I think you have to divide it by $3!$ to compensate for the ordering thing...since (a,b) (c,d) (e,f) can be ordered in $3!$ ways.", "Solution_5": "No, because it is not important which is first or second or third! ;)", "Solution_6": "Yeah but any permutation of the groups will still be the same member sets in each group.\r\n\r\nSo I still don't see where the division by $3!$ comes from.", "Solution_7": "Because when you do $\\binom{6}{2}$ you could, say choose (a,b). Then when you do $\\binom{4}{2}$ say you choose (c,d). Then your three pairs are (a,b) (c,d) (e,f).\r\n\r\nYou could also choose (c,d) first with $\\binom{6}{2}$, then (e,f) with $\\binom{4}{2}$, so you have (a,b) left. In this case the three pairs are also (in a different order) (a,b) (c,d) (e,f), but both these orders are both counted.", "Solution_8": "So groups themselves are distinct? If that's true I understand the division by $3!$, but it seemed like groups are indiscernable and that (a,b) (c,d) is equivalent to (c,d) (a,b).", "Solution_9": "The groups are indescernable. (ooh, good vocab word.) But the way we count them, they aren't. So we divide to account for the fact that they are indescernable.", "Solution_10": "yes you are right .It is not diffrent between (a,b)(c,d)(e,f) &(c,d)(a,b)(e,f)&...& (e,f)(c,d)(a,b).we wnt jusy to make group ;)", "Solution_11": "Ok I see now, I was looking at it completely wrong.\r\n\r\nHaha, woops. :blush:" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "(a) Determine, with careful explanation, how many ways $ 2n$ people can be paired off to form $ n$ teams of $ 2$.\r\n\r\n(b) Prove that $ \\{(mn)!\\}^{2}$ is divisible by $ (m!)^{n\\plus{}1}(n!)^{m\\plus{}1}$ for all positive integers $ m, n$.", "Solution_1": "(a) permute the $ 2n$ people arbitrarily, then let the first two form a team, the next two form a team, and so on. there are $ (2n)!$ ways to permute the people and then form teams in this way. but once these teams are formed there are $ 2! \\equal{} 2$ ways to permute the members of each team, and $ n!$ ways to permute the $ n$ teams, so in the count $ (2n)!$, each distinct selection of teams is counted $ 2^n\\cdot n!$ times. so the true count is $ \\frac {(2n)!}{2^nn!}$.\r\n\r\n(b) by the same reasoning as in (a), replacing 2 with $ m$, we have $ m!^n n!|(mn)!$. reversing the roles of $ m$ and $ n$, we also get that $ m! n!^m|(mn)!$. multiplying, we therefore get $ m!^{n \\plus{} 1}n!^{m \\plus{} 1}|(mn)!^2$." } { "Tag": [ "algebra open", "algebra" ], "Problem": "If\r\n\r\n$x^{12}-x+1=0$\r\n\r\nthen\r\n\r\n$x^{10}+x^{5}+1=?$\r\n\r\nI am not sure that this problem, which has been asked at another forum, can be solved using elementary operations.\r\n\r\nThanks and regards.", "Solution_1": "[quote=\"Honore\"]If\n\n$x^{12}-x+1=0$\n\nthen\n\n$x^{10}+x^{5}+1=?$\n\nI am not sure that this problem, which has been asked at another forum, can be solved using elementary operations.\n\nThanks and regards.[/quote]\r\nThere are 12 complex (not real) means, suth that $x^{12}-x+1=0$. It give 12 means $x^{10}+x^{5}+1$. \r\nBut we can find polinom $P(y)=y^{12}+a_{11}y^{11}+...$, suth that $P(x^{10}+x^{5}+1)=0$ if $x^{12}-x+1=0$." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "sorry to both people with such stupid questions, but i was working through some problems in royden and got stuck on something that seemed easy. namely, show that the product of two absolutely continuous functions is absolutely continuous, as is the inverse of an absolutely continuous function (provided it is zero nowhere).\r\n\r\nI was able to prove that a sum is abs cts, but these other two dont seem to come as easily.\r\n\r\nthanks", "Solution_1": "Use $|f(x)g(x)-f(x')g(x')| \\leq |(f(x)-f(x'))g(x)|+|f(x')(g(x)-g(x'))| \\leq$ \r\n$[|(f(x)-f(x')|+|(g(x)-g(x')|]*K$ where $K$ exists due to the continuity of $f$ and $g$.", "Solution_2": "Are you sure about the inverse being absolutely continuous ?\r\nBecause if you take $f(x)=x$ on $(0,1)$ then $f$ will be absolutely continuous but $\\forall \\epsilon >0$, just take $x'=x+\\epsilon$ and you will have $\\frac{1}{x}-\\frac{1}{x'}=\\frac{\\epsilon}{x(x+\\epsilon)}$. It is then clear that by picking $x$ small enough the difference will become arbitrarily large, which is not what we want." } { "Tag": [ "rotation", "ratio" ], "Problem": "The number of teeth in three meshed gears $A$, $B$, and $C$ are $x$, $y$, and $z$, respectively. (The teeth on all gears are the same size and regularly spaced.) The angular speeds, in revolutions per minutes of $A$, $B$, and $C$ are in the proportion\r\n\r\n$\\text{(A)} \\ x: y: z ~~\\text{(B)} \\ z: y: x ~~ \\text{(C)} \\ y: z: x~~ \\text{(D)} \\ yz: xz: xy ~~ \\text{(E)} \\ xz: yx: zy$", "Solution_1": "[hide=\"solution\"]the idea in this is that the distance that they rotate is constant, suppose they all went $n$ gear teeth in a minute, then clearly, $\\frac{n}{x}$ is the number of revolutions for A, we do the same for B and C, to get the ratio\n\n$\\frac{n}{x}: \\frac{n}{y}: \\frac{n}{z}\\implies yz: xz: xy \\implies \\boxed{D}$[/hide]", "Solution_2": "Cool! I got all the answer choices on one line! :thumbup: :D" } { "Tag": [ "inequalities", "inequalities theorems" ], "Problem": "Hi,\r\ncan anyone explain me in more details the proof of Newton inequality:\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_Inequality#Proof\r\nbecause i can't understand it.\r\nThanks.", "Solution_1": "Certainly, but you need to tell what exactly is unclear first (otherwise you'll just get another long text you'll have hard time understanding)." } { "Tag": [ "probability", "function", "integration", "calculus", "calculus computations" ], "Problem": "Prove that for any distribution function and for any $a\\geq 0$ ;\r\n$\\int [\\mathbb{F}(x+a)-\\mathbb{F}(x)]dx=a$. \r\n\r\n[hide=\"reveal\"] My thought was to use that $\\mathbb{F}$ is a densitiy function of $U(0,1)$ , but I got stuck[/hide]", "Solution_1": "Making a guess as to what this means:\r\n\r\nLet $F(x)$ be a cumulative distribution problem. Let's start by assuming that the random variable is (absolutely) continuous and that there is a density $f$ such that \r\n\r\n$F(x)=\\int_{-\\infty}^{x}f(t)\\,dt.$\r\n\r\nThen:\r\n\r\n$\\int_{-\\infty}^{\\infty}F(x+a)-F(a)\\,dx=\\int_{-\\infty}^{\\infty}\\int_{x}^{x+a}f(t)\\,dx\\,dt$\r\n\r\nA mathematician of my training seldom writes down a double integral without intending to switch the order of integration.\r\n\r\n$=\\int_{-\\infty}^{\\infty}\\int_{t-a}^{t}f(t)\\,dx\\,dt=\\int_{-\\infty}^{\\infty}af(t)\\,dt=E(a)=a.$\r\n\r\nThis would make sense even in the discontinuous case: use Stieltjes integrals in $dF(t)$ in place of $f(t)\\,dt$ and the same argument still works." } { "Tag": [ "inequalities", "function" ], "Problem": "$a,b,c >0$\r\n$\\frac{a}{b+c}+\\frac{b}{a+c}+\\frac{c}{b+a}\\geq 3/2$", "Solution_1": "It's a very well known inequality called Nesbitt's inequlity. \r\n[hide=\"Solution 1\"]\nLet's put $x=b+c$, $y=c+a$, $z=a+b$. Then we have $a=\\frac{y+z-x}{2}$, $b=\\frac{x+z-y}{2}$, $c=\\frac{x+y-z}{2}$ and the inequality becomes: \\[\\frac{y+z-x}{2x}+\\frac{x+z-y}{2y}+\\frac{x+y-z}{2z}\\geq \\frac{3}{2}\\iff\\] \\[\\iff \\frac{y+z-x}{x}+\\frac{x+z-y}{y}+\\frac{x+y-z}{z}\\geq 3.\\] After substituting $u=\\frac{x}{y}$, $v=\\frac{y}{z}$, $w=\\frac{z}{x}$ it becomes \\[\\frac{1}{u}+w-1+u+\\frac{1}{v}-1+\\frac{1}{w}+v-1 \\geq 3\\iff\\] \\[\\iff \\frac{1}{u}+u+\\frac{1}{v}+v+\\frac{1}{w}+w \\geq 6.\\] This inequality is trivial after noticing $t+\\frac{1}{t}\\geq 2$ for $t>0$ (it is equivalent to $t^{2}+1\\geq 2t\\iff (t-1)^{2}\\geq 0$) which ends the proof. \n[/hide]\n[hide=\"Solution 2\"]\nWe can use also Jensen inequality. Let's put $k=a+b+c$ (of course $a,b,c 0$. Prove that: \r\n$ \\frac{1}{\\sqrt[4]{a^2\\minus{}ab\\plus{}b^2}}\\plus{}\\frac{1}{\\sqrt[4]{b^2\\minus{}bc\\plus{}c^2}}\\plus{}\\frac{1}{\\sqrt[4]{c^2\\minus{}ca\\plus{}a^2}} \\geq$ $ \\frac{\\sqrt{2}}{3}. \\frac{(a\\plus{}b\\plus{}c)^3}{a^2\\sqrt{a^3\\plus{}b^3}\\plus{}b^2\\sqrt{b^3\\plus{}c^3}\\plus{}c^2\\sqrt{c^3\\plus{}a^3}}$\r\n :)", "Solution_1": "[quote=\"nguoivn\"]Given $ a, b, c > 0$. Prove that: \n$ \\frac {1}{\\sqrt [4]{a^2 \\minus{} ab \\plus{} b^2}} \\plus{} \\frac {1}{\\sqrt [4]{b^2 \\minus{} bc \\plus{} c^2}} \\plus{} \\frac {1}{\\sqrt [4]{c^2 \\minus{} ca \\plus{} a^2}} \\geq$ $ \\frac {\\sqrt {2}}{3}. \\frac {(a \\plus{} b \\plus{} c)^3}{a^2\\sqrt {a^3 \\plus{} b^3} \\plus{} b^2\\sqrt {b^3 \\plus{} c^3} \\plus{} c^2\\sqrt {c^3 \\plus{} a^3}}$\n :)[/quote]\r\n\r\n[hide=\"Just A thought\"]\nAgain :( , I am new with the idea of homogenisations, so, I would like to ask you whether we can say:\nAfter inserting $ a, b, c$ with $ ka, kb, kc$ the inequality does not change for $ k\\neq0$ ($ \\frac{1}{\\sqrt{k}}$ gets cancelled.\nSo can we assume $ a\\plus{}b\\plus{}c\\equal{}1$? :blush: \n\nExtremely sorry for my ignorance in this field of normalisations and homogenisations, so can anyone give me references? :oops: [/hide]", "Solution_2": "[quote=\"Potla\"]\n\n[hide=\"Just A thought\"]\nAgain :( , I am new with the idea of homogenisations, so, I would like to ask you whether we can say:\nAfter inserting $ a, b, c$ with $ ka, kb, kc$ the inequality does not change for $ k\\neq0$ ($ \\frac {1}{\\sqrt {k}}$ gets cancelled.\nSo can we assume $ a \\plus{} b \\plus{} c \\equal{} 1$? :blush: \n\nExtremely sorry for my ignorance in this field of normalisations and homogenisations, so can anyone give me references? :oops: [/hide][/quote]\n[hide=\"to Potla\"]As you said,Potla,The inequality will not change if we replace $ a,b,c$ by $ ka, kb, kc$,so when you choose $ k \\equal{} \\frac {1}{1 \\plus{} b \\plus{} c}$ the condition $ a \\plus{} b \\plus{} c \\equal{} 1$ will be satisfied.\nSorry for my bad English.[/hide]", "Solution_3": "[quote=\"nguoivn\"]Given $ a, b, c > 0$. Prove that: \n$ \\frac {1}{\\sqrt [4]{a^2 \\minus{} ab \\plus{} b^2}} \\plus{} \\frac {1}{\\sqrt [4]{b^2 \\minus{} bc \\plus{} c^2}} \\plus{} \\frac {1}{\\sqrt [4]{c^2 \\minus{} ca \\plus{} a^2}} \\geq$ $ \\frac {\\sqrt {2}}{3}. \\frac {(a \\plus{} b \\plus{} c)^3}{a^2\\sqrt {a^3 \\plus{} b^3} \\plus{} b^2\\sqrt {b^3 \\plus{} c^3} \\plus{} c^2\\sqrt {c^3 \\plus{} a^3}}$\n :)[/quote]\r\nThis inequality follows from the inequality:\r\nIf $ a,b,c \\ge 0,a\\plus{}b\\plus{}c\\equal{}3,$ then $ a\\sqrt{a\\plus{}b}\\plus{}b\\sqrt{b\\plus{}c}\\plus{}c\\sqrt{c\\plus{}a} \\ge 3\\sqrt{2}.$\r\n;) :)" } { "Tag": [ "function", "search", "algebra", "partial fractions" ], "Problem": "Hi! At a math circle I attend, the group was talking about Fibonacci and I mentioned Binet's formula. After the class, the instructor told me about a proof of Binet's formula using generating functions. I think he thought I knew more about generating functions than I actually do :D, so I am unable to finish the proof past where he showed me. Could somebody please explain how to finish this proof?\r\n\\begin{eqnarray*}f(x)=F_{0}+F_{1}x=F_{2}x^{2}+F_{3}x^{3}...\\\\ x\\cdot f(x)=F_{0}x+F_{1}x^{2}+F_{2}x^{3}... \\end{eqnarray*}\r\nadding the first two equations together and solving for $f(x)$ gives\r\n\\begin{eqnarray*}(1+x)f(x)=F_{1}+F_{2}x+F_{3}x^{2}+F_{4}^{3}...\\\\ x(1+x)\\cdot f(x)=f(x)\\\\ f(x)=\\frac{1}{1-x-x^{2}}\\end{eqnarray*}\r\nnow what? :maybe:", "Solution_1": "try using Search function here.\r\nyou will get MANY results, including generating functions and other porrofs of that formula!! :wink:", "Solution_2": "$f(x) = \\frac{1}{1-x-x^{2}}= \\frac{1}{\\left(1-\\frac{1+\\sqrt{5}}{2}x\\right)\\left(1-\\frac{1-\\sqrt{5}}{2}x\\right)}$.\r\n\r\nUsing partial fractions, this is\r\n\r\n$\\frac{1}{\\sqrt{5}}\\left(\\frac{\\frac{1+\\sqrt{5}}{2}}{1-\\frac{1+\\sqrt{5}}{2}x}-\\frac{\\frac{1-\\sqrt{5}}{2}}{1-\\frac{1-\\sqrt{5}}{2}x}\\right)$.\r\n\r\nCan you finish it from here?", "Solution_3": "And if you'd like to learn more about the topic, I suggest taking a look at \"Generatingfunctionology\" by Herbert S. Wilf. This exact question is covered in the first chapter!\r\n\r\nhttp://www.math.upenn.edu/~wilf/DownldGF.html" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Let be 0 [a,b] such that f(x)+f(f(x))=2x for every x of [a,b].", "Solution_1": "This isn't to say that these are the [b]only[/b] solutions, but they seem to work:\r\n\r\n\\[ f\\left(x\\right)=ax+b\\implies \\left(a^{2}+a\\right)x+ab+2b=2x. \\]\r\n\r\nFrom this, I find that two such functions are $f\\left(x\\right)=-2x$ and $f\\left(x\\right)=x$.", "Solution_2": "But f(x)=-2x is not solution and the function is not linniar. :)", "Solution_3": "Let $g(x)$ be $g(x) = \\frac{f(x)}x$, and then the initial equation becomes $g(x)=\\frac 2{1+g(f(x))}$\r\n\r\nManifestly $g(x)$ is positive and bounded. I note $\\displaystyle i=\\inf_{x\\in [a,b] }(g(x)) ,\\ s=\\sup_{x\\in [a,b] }(g(x))$\r\nNow I can write $\\frac 2{1+s} \\leq g(x) \\leq \\frac 2{1+i}$, so $\\frac 2{1+s} \\leq i$ and $s \\leq \\frac 2{1+i}$, so $s=i$, so $g(x)=k(const) > 0$\r\nI put in the initial equation and obtain $kx +k^2x = 2x$, and finally $k=1$, so $f(x)=x$\r\n :cool:" } { "Tag": [ "algebra", "polynomial", "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be non-negative real numbers. Prove that\r\n\\[ 3\\sum_{cyc} a^4 \\plus{} 9\\sum_{cyc} a^2b^2\\ge 5\\sum_{cyc} a^3(b \\plus{} c) \\plus{} 2abc\\sum_{cyc} a.\r\n\\]", "Solution_1": "Woops...nevermind...I typed it into mathematica incorrectly at first, it seems to be correct.\r\n\r\nMinimize[{3 (a^4 + b^4 + c^4) + 9 (a^2 b^2 + b^2 c^2 + c^2 a^2) - \r\n 5 (a^3 (b + c) + b^3 (c + a) + c^3 (a + b)) - 2 a b c (a + b + c), \r\n a >= 0, b >= 0, c >= 0, a + b + c == 1}, {a, b, c}]\r\n\r\n{0, {a -> 1/3, b -> 1/3, c -> 1/3}}", "Solution_2": "Could you check it again? :)", "Solution_3": "Okay it is \r\n\r\n$ a^2(b\\minus{}c)^2\\plus{}b^2(c\\minus{}a)^2\\plus{}c^2(a\\minus{}b)^2\\plus{} 3 ( (a \\minus{} b)^2 (b \\minus{} c)^2 \\plus{} (b \\minus{} c)^2 (c \\minus{} a)^2 \\plus{} (c \\minus{} a)^2 (a \\minus{} b)^2)$\r\n\r\nwhich is true...", "Solution_4": "[quote=\"hungkhtn\"]Let $ a,b,c$ be non-negative real numbers. Prove that\n\\[ 3\\sum_{cyc} a^4 \\plus{} 9\\sum_{cyc} a^2b^2\\ge 5\\sum_{cyc} a^3(b \\plus{} c) \\plus{} 2abc\\sum_{cyc} a.\n\\]\n[/quote]\r\nYes, it's really easy:\r\n$ \\sum_{cyc}(3a^4 \\minus{} 5a^3b \\minus{} 5a^3c \\plus{} 9a^2b^2 \\minus{} 2a^2bc)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(3a^4 \\minus{} 10a^3b \\plus{} 14a^2b^2 \\minus{} 10ab^3 \\plus{} 3b^4) \\plus{} 4\\sum_{cyc}(a^2b^2 \\minus{} a^2bc)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a\\minus{}b)^2(3a^2\\minus{}4ab\\plus{}3b^2)\\plus{}2\\sum_{cyc}c^2(a\\minus{}b)^2\\geq0.$", "Solution_5": "[quote=\"arqady\"][quote=\"hungkhtn\"]Let $ a,b,c$ be non-negative real numbers. Prove that\n\\[ 3\\sum_{cyc} a^4 \\plus{} 9\\sum_{cyc} a^2b^2\\ge 5\\sum_{cyc} a^3(b \\plus{} c) \\plus{} 2abc\\sum_{cyc} a.\n\\]\n[/quote]\nYes, it's really easy:\n$ \\sum_{cyc}(3a^4 \\minus{} 5a^3b \\minus{} 5a^3c \\plus{} 9a^2b^2 \\minus{} 2a^2bc)\\geq0\\Leftrightarrow$\n$ \\Leftrightarrow\\sum_{cyc}(3a^4 \\minus{} 10a^3b \\plus{} 14a^2b^2 \\minus{} 10ab^3 \\plus{} 3b^4) \\plus{} 4\\sum_{cyc}(a^2b^2 \\minus{} a^2bc)\\geq0\\Leftrightarrow$\n$ \\Leftrightarrow\\sum_{cyc}(a \\minus{} b)^2(3a^2 \\minus{} 4ab \\plus{} 3b^2) \\plus{} 2\\sum_{cyc}c^2(a \\minus{} b)^2\\geq0.$[/quote]\r\n\r\nActually, the following one is true and it is my implication for all real numbers $ a,b,c$\r\n\r\n\\[ 3(a^4\\plus{}b^4\\plus{}c^4)\\plus{}9(a^2b^2\\plus{}b^2c^2\\plus{}c^2a^2)\\ge 10(a^3b\\plus{}b^3c\\plus{}c^3a)\\plus{}2abc(a\\plus{}b\\plus{}c).\\]\r\n\r\nNotice the change from symmetric sum to cyclic sum.", "Solution_6": "I suppose that it is a bit out of the ordinary in the sense that you cannot just schur/muirhead bunch. You have to extract squares as well. I don't think so much about those solutions so it didn't jump out at me immediately.", "Solution_7": "[quote=\"hungkhtn\"] Actually, the following one is true and it is my implication for all real numbers $ a,b,c$\n\\[ 3(a^4 \\plus{} b^4 \\plus{} c^4) \\plus{} 9(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2)\\ge 10(a^3b \\plus{} b^3c \\plus{} c^3a) \\plus{} 2abc(a \\plus{} b \\plus{} c).\n\\]\n[/quote]\r\nThis is sharper:\r\n\\[ 3(a^4 \\plus{} b^4 \\plus{} c^4) \\plus{} \\frac {73}{9}(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2)\\ge 10(a^3b \\plus{} b^3c \\plus{} c^3a) \\plus{} \\frac {10}{9}abc(a \\plus{} b \\plus{} c).\r\n\\]", "Solution_8": "Let $ a \\plus{} b \\plus{} c \\equal{} 3 , a,b,c \\equal{} R$ . Prove that\r\n\\[ \\sum_{cyc} a^4 \\plus{} 3\\sum_{cyc} a^2b^2 \\plus{} 3\\ge 2\\sum_{cyc} ab(a^2 \\plus{} b^2 ) \\plus{} 6(a \\minus{} b)(a \\minus{} c)(b \\minus{} c)\r\n\\]", "Solution_9": "[quote=\"hungkhtn\"]\nActually, the following one is true and it is my implication for all real numbers $ a,b,c$\n\\[ 3(a^4 \\plus{} b^4 \\plus{} c^4) \\plus{} 9(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2)\\ge 10(a^3b \\plus{} b^3c \\plus{} c^3a) \\plus{} 2abc(a \\plus{} b \\plus{} c).\n\\]\nNotice the change from symmetric sum to cyclic sum.[/quote]\r\n\r\nCan you please give me a hint to prove it, hungkhtn? Did you use SOS or SOS-SHUR?\r\n\r\n\r\nThank you very much.", "Solution_10": "You can probably write it in the form $ \\sum_{\\rm cyc} [p(x,y,z)]^2$ where $ p(x,y,z)$ is a homogeneous polynomial of degree 2 in $ x,y,z$.", "Solution_11": "Can you post your solution, Altheman ? :D In Vasc's book, having a genelizition of above (for 4th degree :D) but how do you think about that follow: \r\nLet $ a,b,c,d$ be real numbers such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2 \\equal{} 1$ Prove that:\r\n\\[ \\sum a^6 \\plus{} \\sum a^4b^2 \\plus{} 2abcd(ab \\plus{} bc \\plus{} cd \\plus{} da\\minus{}1)\\ge 2\\sum a^5b \\plus{} \\sum(abc)^2\r\n\\]\r\nIn here $ \\sum \\equal{} \\sum_{cyc}$ \r\nWho can write it in the form $ \\sum [g(a,b,c,d)]^2\\ge 0$ ? :wink:", "Solution_12": "[quote=\"manlio\"][quote=\"hungkhtn\"]\nActually, the following one is true and it is my implication for all real numbers $ a,b,c$\n\\[ 3(a^4 \\plus{} b^4 \\plus{} c^4) \\plus{} 9(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2)\\ge 10(a^3b \\plus{} b^3c \\plus{} c^3a) \\plus{} 2abc(a \\plus{} b \\plus{} c).\n\\]\nNotice the change from symmetric sum to cyclic sum.[/quote]\n\nCan you please give me a hint to prove it, hungkhtn? Did you use SOS or SOS-SHUR?\n\n\nThank you very much.[/quote]\r\nI think, he solved it by EMV method :) . This method works very well for many three-variable cyclic inequalities (include the inequality posted by Vasc as above of course).", "Solution_13": "I don't have any of these books...I just read AoPS :D\r\n\r\nI am not sure, but that expansion contains all 'elementary' cyclic sums of degree 4....That is what I like to call them anyway\r\n\r\na^4+b^4+c^3\r\n\r\na^3b+b^3c+c^3a\r\n\r\nab^3+bc^3+ca^3\r\n\r\nabc(a+b+c)\r\n\r\nwe have enough variables that it is likely that we can find the desired coefficients (but of course I am not sure, I was merely suggesting that the method would probably work)\r\n\r\nI like your problem and I will think about it. \r\n\r\nIt obviously cannot be written in the form that you propose, but we can augment it..." } { "Tag": [ "number theory", "prime numbers" ], "Problem": "Henry's little brother has 8 identical stickers and 4 identical sheets of paper. How many ways are there for him to put the stickers on the sheets of paper, if only the number of stickers on each sheet matters?", "Solution_1": "That can be only used for if the stickers were identical BUT the pieces of paper were distinct. I'm not sure how to proceed in any way other than a list, in solving the actual question, so here goes-\r\n\r\n8 0 0 0\r\n7 1 0 0\r\n6 2 0 0\r\n6 1 1 0\r\n5 3 0 0\r\n5 2 1 0\r\n5 1 1 1\r\n4 4 0 0\r\n4 3 1 0\r\n4 2 2 0\r\n4 2 1 1\r\n3 3 2 0\r\n3 3 1 1\r\n3 2 2 1\r\n2 2 2 2\r\n\r\nThis totals $ \\boxed{15}$ ways.", "Solution_2": "How is this different from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=239503]this[/url]?", "Solution_3": "The $4$ sheets of paper in the other problem are different colors, while here the papers are the same color.", "Solution_4": "With all due respect for brute forcing and counting up the ways to do it, isn't there a method where I don't have to use up half a piece of scratch paper making lists? A formula? If this problem had a very large amount of stickers and a very large number of sheets of paper, I'm pretty sure listing wouldn't be a good idea. Anybody know? Thanks.", "Solution_5": "If you don't want to list them out, the best you can do is probably casework. If I told you to sum the first 1000 prime numbers, is there necessarily an easy way to do that?", "Solution_6": "Well, there isn't a formula for the first 1000 prime numbers.", "Solution_7": "That is my point. Not every problem has an easy, casework free way of doing it.", "Solution_8": "wait, will wang? you go to NA?", "Solution_9": "[quote=Freddie123]wait, will wang? you go to NA?[/quote]\n\nyou could've just pm'ed him instead of reviving a 2 year old thread.......", "Solution_10": "point well made.", "Solution_11": "Why doesn't stars and bars work? ", "Solution_12": "[quote=Ridley-C]Why doesn't stars and bars work?[/quote]\n\nbecause both objects are indistinguishable", "Solution_13": " Sorry for the bump, anyway...\nIs this a coincidence that this is 5+4+3+2+1? ", "Solution_14": "[quote=MathArt4]Sorry for the bump, anyway...\nIs this a coincidence that this is 5+4+3+2+1?[/quote]\n\nYes.", "Solution_15": "there isn't 5 in the problem", "Solution_16": "Is there a better, non-casework method?", "Solution_17": "Yes, as I stated on the other page for this problem. Which makes me wonder why this problem was posted on two separate threads by Gamebot?\n\nEDIT: Wait they're actually two different problems nvm and actually I don't think you can use a non-casework approach for this one", "Solution_18": "Is this problem referring to indistinguishable items in indistinguishable boxes? however, the problem states \"each sheet,\" so shouldn't the sheets be distinguishable? That is what I did in my first solution, and it was incorrect. On my second solution, I changed my thinking and it was correct, but I just wanted to confirm. Someone else may be able to give a more detailed explanation.", "Solution_19": "[quote=Zhang2018]Is this problem referring to indistinguishable items in indistinguishable boxes? however, the problem states \"each sheet,\" so shouldn't the sheets be distinguishable? That is what I did in my first solution, and it was incorrect. On my second solution, I changed my thinking and it was correct, but I just wanted to confirm. Someone else may be able to give a more detailed explanation.[/quote]\n\nNot sure how relevant this is at this point, but the statement in the first sentence that there are[quote=Alcumus]$8$ identical stickers[/quote]\nshould definitely supersede anything else.\n------------\nAnyways, since the answer is $6\\choose2$$^a$ I was wondering if there is any reason or pattern behind that or if that is just coincidence.\n------------\n$^a$ the $\\LaTeX$ doesn't seem to have rendered quite right there; sorry", "Solution_20": "[quote=professordad]The $4$ sheets of paper in the other problem are different colors, while here the papers are the same color.[/quote]\n\nIt is almost the same thing as the 2006 AMC 12 problem.", "Solution_21": "omg finally mastered distinguishability yayyyyyyyyy\n\nthis one was a bad problem tho, why make us just list?", "Solution_22": "i accidentally typed `5 as the answer lol", "Solution_23": "I attempted to do this w/ generating functions for partitions but it's even more work; this problem is probably inevitably ugly.", "Solution_24": "Is there any way to solve this without counting them all? I only have so much pen ink." } { "Tag": [], "Problem": "This came up in a problem I was thinking about and I thought it was interesting.\r\n\r\nProve that for $ n\\in\\mathbb{N}$\r\n\\[ (\\minus{}2)\\binom{3n\\plus{}1}{n\\minus{}1}\\plus{}(\\minus{}1)\\binom{3n}{n\\minus{}1}\\plus{}...\\plus{}(2n)\\binom{n\\minus{}1}{n\\minus{}1}\\equal{}0\\]\r\nAnd more generally, for $ k\\in\\mathbb{N}$,\r\n\\[ (\\minus{}k)\\binom{kn\\plus{}k\\plus{}n\\minus{}1}{n\\minus{}1}\\plus{}(\\minus{}k\\plus{}1)\\binom{kn\\plus{}k\\plus{}n\\minus{}2}{n\\minus{}1}\\plus{}...\\plus{}kn\\binom{n\\minus{}1}{n\\minus{}1}\\equal{}0\\]\r\nIn sigma notation:\r\n\\[ \\sum_{i\\equal{}0}^{kn\\plus{}k}(i\\minus{}k)\\binom{kn\\plus{}k\\plus{}n\\minus{}1\\minus{}i}{n\\minus{}1}\\equal{}0\\]\r\nEDIT: On second thought, this should probably be moved to Intermediate.", "Solution_1": "It looks like you're missing the RHS of your equations, or something. What are we supposed to prove about these expressions?", "Solution_2": "[quote=\"JBL\"]It looks like you're missing the RHS of your equations, or something. What are we supposed to prove about these expressions?[/quote]\r\n\r\nOh right. Fixed." } { "Tag": [], "Problem": "How many degrees are in the sum of the measures of the six numbered angles pictured?\n[asy]draw((3,8)--(10,4)--(1,0)--cycle);\ndraw((7,8)--(9,0)--(0,4)--cycle);\nlabel(\"1\",(3,8),SSE);\nlabel(\"2\",(7,8),SSW);\nlabel(\"3\",(10,4),2W);\nlabel(\"4\",(9,0),NW+NNW);\nlabel(\"5\",(1,0),NE+NNE);\nlabel(\"6\",(0,4),2E);[/asy]", "Solution_1": "If you look carefully, the figure is an intersection of two triangles.\r\n\r\nThe first triangle has angle measures 1, 3, and 5.\r\n\r\nThe second triangle has angle measures 2, 4, 6 and 6\r\n\r\nThe sum of the interior angles of a triangle is 180.\r\n\r\nSo, we are looking for 1+3+5+2+4+6= 180+180 = 360." } { "Tag": [ "quadratics", "complex numbers", "algebra", "quadratic formula" ], "Problem": "Mathematically prove that $z\\overline{z}=a^{2}+b^{2}$.", "Solution_1": "[hide]\nz = a + bi\nz(bar thingy) = a - bi\n\n(a + bi)(a - bi) = a^2 + b^2[/hide]", "Solution_2": "I should have made it more clear. I want something like deriving it.", "Solution_3": "What do you mean by that?", "Solution_4": "what do you mean derive? this follows directly from the definition of a conjugate", "Solution_5": "See...I want to prove it to some of my friends who think that I'm wrong, so I want a rigorous derivation", "Solution_6": "suppose z=a+bi, then we denote the conjugate of z, z (with the bar on top)=a-bi, so multiplying them obviously gives a^2+b^2...the confusion lies in defining z=a+bi probably...because z does not always have to be that, z=bob, for example, is possible", "Solution_7": "But isn't $z=a+bi$ by convention?\r\n\r\nWhen I tell them that, they don't believe it. What says that $z=a+bi$", "Solution_8": "[quote=\"anirudh\"]But isn't $z=a+bi$ by convention?\n\nWhen I tell them that, they don't believe it. What says that $z=a+bi$[/quote]\r\n\r\none generally uses z=a+bi for a complex number...and i guess with the whole conjugate thing it is \"assumed\" that z is a complex number...but you could also be like let z=x+yi, then zz(bar)=x^2+y^2,\r\n\r\ni also think the confusion might be this: you are saying\r\n\r\na complex number * the complex conjugate is equal to the sum of the real part squared and the imaginary part squared, this is true\r\n\r\nhowever, they might be thinking something to the effect of zz(bar) is always equal to a^2+b^2, which is not really understanding your statement...", "Solution_9": "[hide=\"proof\"]z=a+bi\n\n$z\\overline{z}=(a+bi)(a-bi)=a^{2}+b^{2}$[/hide]\r\n\r\n@Altheman: Assume bob=a+bi, because a and b could be 0.", "Solution_10": "[quote=\"anirudh\"]But isn't $z=a+bi$ by convention?\n\nWhen I tell them that, they don't believe it. What says that $z=a+bi$[/quote]\r\n\r\nPerhaps you could say $z = \\Re(z)+i\\Im(z)$, with $\\Re(z),\\Im(z)\\in\\mathbb{R}$. Every complex number can be written like that, as $\\mathbb{C}$ is $\\mathbb{R}[i]$, an [url=http://mathworld.wolfram.com/ExtensionField.html]extension field[/url] of $\\mathbb{R}$. Then it's easy to prove, as, by definition, $\\overline{z}= \\Re(z)-i\\Im(z)$, that $z\\overline{z}= \\Re(z)^{2}+\\Im(z)^{2}$.", "Solution_11": "[quote=\"anirudh\"]But isn't $z=a+bi$ by convention?\n\nWhen I tell them that, they don't believe it. What says that $z=a+bi$[/quote]\r\nNothing says that, such $a,b$ simply exist.\r\nTo answer any more question, you (or they) have to tell how they define everything that occures (and no, the answer to this is not like \"as always\").\r\nUntill then already the first answer was perfectly well.", "Solution_12": "Usually $z$ is used to represent a complex number. I have no idea why.\r\n\r\nAll complex numbers can be expressed in the form $a+bi$, where $a,b\\in\\mathbb{R}$. So let $z=a+bi$ for real $a$ and $b$. So the conjugate of $z$ is $a-bi$ by definition. So their product is $(a+bi)(a-bi)$, or $a^{2}+b^{2}$.", "Solution_13": "[quote=\"anirudh\"]But isn't $z=a+bi$ by convention?\n\nWhen I tell them that, they don't believe it. What says that $z=a+bi$[/quote]\r\n\r\nWell, if you want to prove that $z\\overline{z}=a^{2}+b^{2}$, it helps to use the right variables ;)\r\n\r\nYou might as well say that you can't prove that $x=\\frac{-b\\pm\\sqrt{b^{2}-4ac}}{2a}$. What says that $ax^{2}+bx+c=0$?", "Solution_14": "mathnerd's got a point. You need to declare something first. For the quadratic formula, you say this:\r\n\r\n[b]IF[/b] $ax^{2}+bx+c=0$, then $x=\\frac{-b\\pm\\sqrt{b^{2}-4ac}}{2a}$.\r\n\r\nFor the original problem, you'd have to say that $z=a+bi$, where $a$ and $b$ are real numbers." } { "Tag": [ "conics", "ellipse", "hyperbola", "calculus", "derivative", "analytic geometry", "graphing lines" ], "Problem": "How do you prove that the ellipse $ 4x^2\\plus{}9y^2\\equal{}45$ and the hyperbola $ x^2\\minus{}4y^2\\equal{}5$ are orthogonal.\r\n\r\nMy plan:\r\nFigure out where they intersect(which I don't know how to do...)\r\nImplicit differentiation and plug in the point where the intersection is\r\nMake sure that the slopes are negative reciprocals.\r\n\r\nHow do you find out where they intersect?", "Solution_1": "They intersect at the points where both equations are satisfied. Subtract four times the second equation from the first to find $ 25y^2 \\equal{} 25\\Rightarrow y \\equal{} \\pm 1$ and add $ \\frac {9}{4}$ times the second equation to the first to find $ \\frac {25}{4}x^2 \\equal{} \\frac {225}{4}\\Rightarrow x \\equal{} \\pm 3$" } { "Tag": [ "6th edition" ], "Problem": "", "Solution_1": "You are not supposed to post ANYTHING like this here.\r\n\r\nOnly \"technical\" questions about the (meaning of the) problems, nothing else.", "Solution_2": "Sorry :( :(" } { "Tag": [ "algebra", "polynomial", "quadratics", "algorithm" ], "Problem": "Hello, I have a problem with well, a problem, from AHSME. I am just starting to learn about polynomials, and I think I made an error:\r\n\r\nLet $ P(x)$ be a polynomial such that when $ P(x)$ is divided by $ x\\minus{}19$, the remainder is 99, and when it is divided by $ x\\minus{}99$, the remainder is 19. Find the remainder when $ P(x)$ is divided by $ (x\\minus{}19)(x\\minus{}99)$.\r\n\r\nWhat I did: $ P(x)\\equal{}q_1(x\\minus{}19)\\plus{}99$, giving $ P(19)\\equal{}99$, and $ P(x)\\equal{} q_2(x\\minus{}99)\\plus{}19$, so $ P(99)\\equal{}19$. $ P(x)\\equal{}q_3(x\\minus{}19)(x\\minus{}99)\\plus{}r$, giving the equations $ P(19)\\equal{}P(99)\\equal{}r$. This may be trivial to you, but please help!", "Solution_1": "When you're dividing $ P(x)$ by a polynomial of degree $ 1$, the remainder is of degree $ 1 \\minus{} 1 \\equal{} 0$. That's why you ended up with $ 99$ and $ 19$ when you divided $ x \\minus{} 19$ and $ x \\minus{} 99$ respectively.\r\n\r\nBut when you're dividing $ P(x)$ by a second degree polynomial, $ (x \\minus{} 19)(x \\minus{} 99)$, the remainder is of degree $ 2 \\minus{} 1 \\equal{} 1$. So, it has the form of $ ax \\plus{} b$. \r\n\r\nJust write $ P(x) \\equal{} Q(x)(x \\minus{} 19)(x \\minus{} 99) \\plus{} ax \\plus{} b$. Then plug in the values you know and solve the system for $ a$ and $ b$.", "Solution_2": "Wait, why should the solution be linear? Doesn't the 0 make the remainder a constant?", "Solution_3": "Only in special cases will the remainder be constant when dividing a quadratic by a linear term.\r\n\r\nHowever, it does happen. For example, $ \\frac{x^2\\plus{}3x\\plus{}3}{x\\plus{}3}$ has a remainder of 3.", "Solution_4": "So basically when you are dividing some polynomial of degree n by x-a, the deg of the remainder can be no greater than n-1.", "Solution_5": "Other way around. When the polynomial you're dividing [b]by[/b] has degree $ n$, the remainder has degree at most $ n \\minus{} 1$. This is the [url=http://en.wikipedia.org/wiki/Division_algorithm]division algorithm[/url] for polynomials.", "Solution_6": "Plz. disregard everything I just said (I make very stupid mistakes sometimes :( ). I re-did the problem and got -x+118." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "1) y and 3y are two integers, which are formed by using each of the nine digits (1,2,3,4,5,6,7,8,9) exactly once.\r\n what is y?\r\n\r\nAny hints for all the above three questions?\r\nthanks.", "Solution_1": "[quote=\"ice_age\"]1) y and 3y are two integers, which are formed by using each of the nine digits (1,2,3,4,5,6,7,8,9) exactly once.\n what is y?\n\nAny hints for all the above three questions?\nthanks.[/quote]\r\n\r\ny is 5823 or 5832", "Solution_2": "I found the same answer. By the, way did you use some method to find it or just brut force?\r\n\r\nSee also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=200713]this similar topic[/url] :)" } { "Tag": [ "linear algebra", "matrix", "vector", "complex numbers", "linear algebra unsolved" ], "Problem": "Prove that a normal stochastic matrix is always doubly stochastic.", "Solution_1": "I hope this is correct, since I sort of got lost in the indices :).\r\n\r\nI'll assume the sum on every row is $0$, and the aim is to prove that the sum on every column is also zero. The usual definition is with $1$ as the common value of the sums, but we can subtract the identity matrix from the initial matrix to reach this version. \r\n\r\nThe matrix is unitarily diagonalziable, so it has the form $U^*DU$, where $D=\\mbox{diag}(\\lambda_1,\\ldots,\\lambda_n)$. If $x_1,\\ldots,x_n$ are the columns of $U$, then the condition that $A$ is stochastic translates to $\\forall i,\\ \\langle Dx_i,v\\rangle=0\\ (*)$, where $v=\\sum_{j=1}^nx_j$, and the conclusion we want to reach, namely that the matrix is doubly stochastic, translates to $\\forall i,\\ \\langle D^*x_i,v\\rangle=0\\ (**)$.\r\n\r\nIf, in general, for all vectors $u,\\ u(k)$ is the $k$'th entry of $u$, then $(*)$ can be written as $\\forall i,\\ \\sum_k\\lambda_k[x_i(k)\\overline{v(k)}]=0\\ (\\#)$. This shows that the complex numbers $\\lambda_k\\cdot\\overline{v(k)}$ are solutions of a homogenous linear system, having $U$ (the transpose of $U$, actually, but it doesn't really matter) as its matrix, meaning that all these complex numbers are zero. In turn, this means that the complex numbers $\\overline{\\lambda_k\\cdot v(k)}$ are also null, which shows that $(\\#)$ also holds with $\\overline{\\lambda_k}$ instead of $\\lambda_k$. This, however, is just another way of writing $(**)$, which is what we wanted to prove.\r\n\r\nIs it correct?", "Solution_2": "Don't imagine I checked the computations, I know it's correct since my solution was also based on the same idea and my teacher said it was correct. I really liked this problem: after so many indices and stuff you arrive at such a nice conclusion. :D" } { "Tag": [], "Problem": "Do you like Aops?", "Solution_1": "I am not especially liking it right nw because there are too many tourneys going on so I voted no although I like it sometimes like in the year when actual problems are being posted more often....", "Solution_2": "Wow. I thought it would be like 100%. A lot of people must not like Aops, and yet there are 43000+ members.... :huh:", "Solution_3": "I don't think \"2\" is a lot.", "Solution_4": "no parents made me/sort of?????\r\n\r\ni vote yes" } { "Tag": [ "function", "geometry", "3D geometry", "sphere", "limit", "trigonometry", "complex analysis" ], "Problem": "Let's consider the recurrence $ (n+1)a_{n+1}=a_1 a_n+a_2 a_{n-1}+...+a_n a_1$ with $ a_1=1$ . Prove that the sequence $\\sqrt[n]{a_n}$ converges. I tried with estimations and did not suceed. I also tried with generating functions and failed. So...", "Solution_1": "Power serie $f(x)=\\sum_{n\\geq 0} a_ nx^n$ with $a_0=0$\r\n\r\n$f^2(x)$ ?", "Solution_2": "Yes, Moubi, this approach is obvious, but what we do next? ;)", "Solution_3": "We get an equation to an explicit of $f(x)$ \r\nthen expand in power serie to an explicit expression of $a_n$ and compute the limit", "Solution_4": "Can you please post a complete solution? :)", "Solution_5": "I think we get the differential equation $x f'(x) = x + f(x)^2$. But I don't know how to solve that...", "Solution_6": "Well, this is my problem too. How to solve the monster? Moubi?", "Solution_7": "Moubiiiii? :D ;) :(", "Solution_8": "Let me try to answer. The truth seems to be that you cannot solve the equation explicitly. Fortunately, you do not need to. What you need is to show that the differential equation allows to extend $f$ to an analytic function from $\\mathbb C$ to the Riemann sphere $\\mathbb S$ (or, if you prefer, meromorphic function). Then you should check that $f$ has its nearest to the origin pole on the positive real semiaxis and apply a general theorem from Complex Analysis. I think that is what Moubinool had in mind :) .", "Solution_9": "This definitely lost me. :(", "Solution_10": "Do you know any complex analysis, harazi? If you do, I'll try to explain better but if you do not, I'd better think of some other purely real approach. The theorem I was going to use is the following:\r\n\r\nLet $f:\\mathbb C\\to \\mathbb S$ be a meromorphic function with Taylor expansion at the origin $f(z)=\\sum_{n\\geq 0}a_n z^n$. Assume that $z_0$ is the nearest to the origin pole of $f$ and that there are no other poles on the circle $|z|=|z_0|$. Then\r\n$\\lim_{n\\to\\infty} |a_n|^{1/n}=1/|z_0|$.\r\n\r\nDoes this sound familiar?", "Solution_11": "No, it does not, but I can imagine it, so let's accept it as it is. And how we prove the thing with the poles for our differential equation?", "Solution_12": "This question is two-fold: how do we prove that $f$ is meromorphic and how do we prove that the closest to the origin pole is unique. I'll go over the first part now. And I'll do it for a simpler equation $f'=1+f^2$, which, of course, you can solve: $f(x)=\\tan (x+c)$. I hope it won't surprize you that $\\tan z$ is defined for complex $z$ as well and is a meromorphic function on the entire complex plane. Let's see how to get this right from the equation. The key is the power series method. For any initial data $f(x_0)=y_0$ ($x_0,y_0\\in\\mathbb C$), we can create a unique Taylor series at $x_0$ that satisfies the equation in some disk around $x_0$ and starts with $y_0$. What is more, its radius of convergence can be estimated from below if $|y_0|\\leq 1$. If we could get an estimate for the radius of convergence independent of $y_0$, we would be able to analytically continue $f$ along any path in $\\mathbb C$ and get an analytic function. Unfortunately, this is impossible. But let us notice that if $|y_0|>1$, then $|1/y_0|<1$ and the function $g=1/f$ satisfies almost the same differential equation $g'=-1-g^2$. This allows to use the power series method for $g$ instead of $f$ with some fixed radius of convergence. The only problem is that $g$ may vanish at some points, which will become poles of $f$, so $f$ will be not analytic but meromorphic. Does this make sense? If it does, try to apply this technique to the real equation you have yourself.", "Solution_13": "Thank you very much, fedja, I understood now. Thank you. I wonder whether there exists a purely elementary approach..." } { "Tag": [ "trigonometry", "Gauss", "complex numbers", "geometric series" ], "Problem": "Here's a problem you may enjoy as much as I did. Compute the sums\r\n\\[ \\sum_{k=0}^n\\sin (kx)\\hspace{.5in}\\sum_{k=0}^n\\cos (kx) \\]\r\n\r\n[hide=\"HINT iff you are stuck\"]\nComplex numbers using $e^{ikx}$\n[/hide]\r\n\r\nMasoud Zargar", "Solution_1": "[hide=\"solution\"]Let $e^{ikx} = \\cos (kx) + i \\sin (kx)$. Then\n\n$\\sum_{k=0}^n \\cos (kx) = \\text{Re} ( \\sum_{k=0}^n (e^{ix})^k )$\n\nand similarly for the $\\sin$ series. Evaluating this finite geometric series gives\n\n$\\sum_{k=0}^n (e^{ix})^k ) = \\frac{1 - (e^{ix})^{n+1}}{1 - e^{ix}} = \\frac{1 - \\cos ((n+1)x) - i \\sin ((n+1)x)}{1 - \\cos x - i \\sin x}$ \n\nEvaluating the real and imaginary parts of this sum gives us\n\n$\\sum_{k=0}^n \\cos (kx) = \\frac{1 - \\cos x - \\cos ((n+1)x) + \\cos x \\cos ((n+1)x) + \\sin x \\sin ((n+1)x) }{2 - 2 \\cos x}$\n\n$\\sum_{k=0}^n \\sin (kx) = \\frac{\\sin x - \\sin x \\cos ((n+1)x) - \\sin ((n+1)x) + \\cos x \\sin ((n+1)x) }{2 - 2 \\cos x}$\n\nI'm not sure whether there's an easy way to put this into a simpler form.[/hide]", "Solution_2": "Well, that is probably right, but there is a simpler answer.\r\n\r\nWe have $e^{ikx}=\\cos (kx)+i\\sin (kx)$.\r\n$\\sum_{k=0}^n (e^{ix})^k=\\frac{e^{ix(n+1)}-1}{e^{ix}-1}\\\\ =\\frac{e^{ix(n+1)/2}(e^{ix(n+1)/2}-e^{-ix(n+1)/2})}{e^{ix/2}(e^{ix/2}-e^{-ix/2})}\\\\ =e^{ixn/2}\\cdot\\frac{\\sin\\frac{(n+1)x}{2}}{\\sin\\frac{x}{2}}\\\\ =\\left(\\cos\\frac{nx}{2}+i\\sin\\frac{nx}{2}\\right)\\cdot\\frac{\\sin\\frac{(n+1)x}{2}}{\\sin\\frac{x}{2}}\\\\ =\\frac{\\cos\\frac{nx}{2}\\sin\\frac{(n+1)x}{2}+i\\sin\\frac{nx}{2}\\sin\\frac{(n+1)x}{2}}{\\sin\\frac{x}{2}}$\r\n$\\\\\\sum_{k=0}^n\\cos (kx)=\\Re\\left(\\sum_{k=0}^n (e^{ix})^k\\right)=\\frac{\\cos\\frac{nx}{2}\\sin\\frac{(n+1)x}{2}}{\\sin\\frac{x}{2}}\\\\ \\sum_{k=0}^n\\sin (kx)=\\Im\\left(\\sum_{k=0}^n (e^{ix})^k\\right)=\\frac{\\sin\\frac{nx}{2}\\sin\\frac{(n+1)x}{2}}{\\sin\\frac{x}{2}}$\r\n\r\nAre there any proofs without the use of complex numbers. I think there is one derived in a similar fashion as Gauss' summation formula.\r\n\r\nMasoud Zargar", "Solution_3": "Yes, that's a much nicer version than what I got. I've never seen any method of proof other than by complex numbers.", "Solution_4": "Thanks. I have a solution for $\\sum_{k=1}^n\\cos (kx)$ by using telescoping sums, but it isn't that nice. It's just a simple application of product-to-sum formula. You just multiply it by $2\\sin\\frac{x}{2}$.\r\n\r\nMZ", "Solution_5": "I liked this problem.\r\n\r\nIt got messy though.\r\n\r\nHow about something like:\r\n\r\n$\\sum_{i=0}^{n}\\frac{1}{i+1}\\dbinom{n}{i}$\r\n\r\nHow do we find the sum of this?\r\n\r\nAny ideas?", "Solution_6": "for player's sum,\r\n\r\n[hide]\n\n$\\frac{1}{i+1}\\dbinom{n}{i}=\\frac{n!}{(i+1)i!(n-i)!}=\\frac{n!}{(i+1)!(n-i)!}=\\frac{1}{n+1}\\frac{(n+1)!}{(i+1)!(n-i)!}=\\frac{1}{n+1}\\dbinom{n+1}{i+1}$\n\nSo the sum is $\\frac{1}{n+1}$ times the sum of $\\dbinom{n+1}{i+1}$ for all the i. The sum of all the combinations in row n+1 is $2^{n+1}$ but since we skip the $\\dbinom{n}{0}$ we must subtract 1.\n\nThis ends up as\n\n$\\frac{2^{n+1}-1}{n+1}$\n\n[/hide]", "Solution_7": "$\\frac{1}{n+1}\\sum_{i=0}^{n}\\binom{n+1}{i+1}=\\frac{2^{n+1}-1}{n+1}$", "Solution_8": "Tjhance,\r\n\r\nI see that you have $\\frac{n!}{(i+1)!(n-1)!}= \\frac{1}{n+1}\\frac{(n+1)!}{(i+1)!(n-1)!}$ \r\n\r\nShouldn't it be:\r\n\r\n$\\frac{n!}{(i+1)!(n-1)!}= \\frac{1}{(n+1)!}\\frac{(n+1)!}{(i+1)!(n-1)!}$ \r\n\r\nOr am I just not understanding something?\r\n\r\n\r\nEDIT: Yikes...Ignore this please.", "Solution_9": "Most likely a typo :wink: \r\nIn normal language, what difference does one exlcamation mark mean anyway? :P \r\nSeeing as there is no other way for there to be a true statement, yes you are correct.", "Solution_10": "Yikes,\r\n\r\nI see it now.\r\n\r\nI must be sleepy.\r\n\r\nWhy is the sume of all combinations 2^(n+1)?", "Solution_11": "tjhance,\r\n\r\nyou had\r\n\r\n$\\frac{(n+1)!}{(i+1)!(n-i)!}= \\dbinom{n+1}{i+1}$\r\n\r\nShouldn't it be $\\frac{(n+1)!}{(i+1)!(n+1-i)!}= \\dbinom{n+1}{i+1}$ ??", "Solution_12": "[quote=\"player\"]Yikes,\n\nI see it now.\n\nI must be sleepy.\n\nWhy is the sum of all combinations 2^(n+1)?[/quote]\r\nThe sum is equivalent to the number subsets of an n+1-set including the empty set.", "Solution_13": "I figured out my post above yet again...\r\n\r\nI am too sleepy...\r\n\r\nSorry for doubting you so much tjhance, and thanks for your informative reply boxedexe" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Use intermediate calculus level to answer:\r\nFind the general solution of the equation\r\ny'' - 4y' + 3y = cosh x", "Solution_1": "First of all, stop re-posting the same topic. Second, this uses calculus so don't post it in the high school forum.", "Solution_2": "hello, at first of all you must solve the homogeneous part of your equation by the ansatz $ y(x)\\equal{}e^{\\lambda x}$.\r\nSonnhard.", "Solution_3": "Do not double post:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268915[/url]" } { "Tag": [], "Problem": "I have two ideas for perpetual motion (don't stop reading now):\r\n\r\n1) I know that friction eats up all of your energy, but if there was some way to always, say, cut the amount of friction on a surface in half, then the amount of useful energy you would have would be bounded by below (so it would always be visibly moving). Not sure of the math on this one.\r\n\r\n2) A lot of perpetual motion devices don't work because of entropy. Is there any way that you could design a box that would increase the entropy within it while decreasing the entropy around it? I heard that most humans do the opposite, which is where I got the idea. If yes, then you could put one of these boxes into a perpetual motion device, and it would work. Also, if we had enough of these, we could prolong the life of the entire universe (until it freezes over).", "Solution_1": "i think this goes under one of the sciences... especially Physics.", "Solution_2": "comment on number 2....\r\n\r\nu r an inspired scientist and it is a very good thing...\r\nBUT...for entropy in human is the opposite then wut it seems to be is because we r a open system...\r\n\r\nin another word...\r\n\r\nwe get organized but involuntarily the universe get more disordered...\r\n\r\nIn the box u r talking about...i assume it will be a closed system though it is not possible to apply wut u said...\r\n\r\nOf course everything is possible... but i personally don't want or like to go against the law of physic..." } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "3D geometry", "prism", "homothety" ], "Problem": "Given a triangle $ ABC.$ Let $ M$ be a moving point on its circumcircle. Let $ l$ be a tangent of its incircle which is parallel to the side $ BC.$ Prove that the circumcircle of the triangle formed by the tangents from $ M$ to its incircle and $ l$ is always tangent to the $ A$-mixtilinear incircle of triangle $ ABC.$", "Solution_1": "Dear Fang-jh and Mathlinkers,\r\nvery nice problem. I give you a begining of a possible proof:\r\n1. the circumcircle of your triangle intersect AB and AC resp. in R and Q.\r\n2. This is only a conjecture: QR is tangent to the incircle of ABC\r\n3. If yes, we have to prove that the A-mixtilinear incircle of ABC is also the A-mixtilinear incircle of ARQ.\r\nSincerely\r\nJean-Louis", "Solution_2": "Hello!\r\n\r\nThe proposed problem can be generalized:\r\nIn triangle $ \\Delta ABC$ let $ \\Pi$ be the incircle, $ \\Gamma$ is a circle through the points $ B,C$ and $ \\Omega$ is the circle \r\nwhich tangents to $ (AB$, $ (AC$ and $ \\Gamma$ such that $ \\Omega\\cap\\Pi\\neq\\emptyset$. Let $ d$ be the tangent to the circle $ \\Pi$ \r\nwhich is parallel to the side $ BC$ and $ M$ is a point on the circle $ \\Gamma$. Prove that the circumcircle of the \r\ntriangle formed by the tangents from $ M$ to the circle $ \\Pi$ and $ d$ is always tangent to the circle $ \\Omega$.\r\n\r\nP.S. I don't have a proof for this generalization, but I think it's true. \r\n\r\nSincerely, Petrisor Neagoe", "Solution_3": "Dear Jean-Louis Ayme\r\nI think you are right, but you have a typo: we have to prove that the $ A \\minus{}$mixtilinear incircle of triangle $ ABC$ is also the $ A \\minus{}$mixtilinear excircle of triangle $ ARQ$.\r\nDear Petrisor Neagoe\r\nBy drawing, I think you are right, except for the fixed circle you refer, it also touches the other fixed circle which is tangent to $ BA,CA$ beyond $ A$, and to $ \\Gamma$ internally (when vertex $ A$ lies inside $ \\Gamma.$)\r\n.", "Solution_4": "[quote=\"Fang-jh\"]Given a triangle $ ABC.$ Let $ M$ be a moving point on its circumcircle. Let $ l$ be a tangent of its incircle which is parallel to the side $ BC.$ Prove that the circumcircle of the triangle formed by the tangents from $ M$ to its incircle and $ l$ is always tangent to the $ A$-mixtilinear incircle of triangle $ ABC.$[/quote]\r\nLet $ (O), (I)$ can be 2 arbitrary circles with radii $ R, r$, even intersecting or outside of each other. $ l$ is a fixed tangent of $ (I)$ at $ D$ and $ M \\in (O)$ is arbitrary, outside of $ (I)$. Circle with diameter $ IM$ cuts $ (I)$ at $ E, F$ and $ ME, MF$ are tangents of $ (I)$ from $ M.$ Let these tangents cut $ l$ at $ U, V.$ Let $ EF$ cuts $ IM$ at $ K.$ Since $ K$ is the inversion image of $ M \\in (O)$ in $ (I),$ $ K$ is on the inversion image $ (P)$ of $ (O)$ in $ (I).$ Let $ L$ be reflection of $ I$ in $ EF,$ $ \\overline {IL} \\equal{} 2 \\overline{IK}.$ This means that $ L$ is on a circle $ (Q)$ similar to $ (P)$ with similarity center $ I$ and coefficient $ 2.$ Let $ N$ be midpoint of $ DL.$ Then $ N$ is on a circle $ (S)$ similar to $ (Q)$ with similarity center $ D$ and coefficient $ \\frac {_1}{^2},$ i.e. $ (S) \\cong (P).$ Points $ I, D$ are external similarity centers of the circle pairs $ (P), (Q)$ and $ (Q), (S)$ $ \\Longrightarrow$ external similarity center of $ (S), (P)$ is on $ ID.$ But $ (S) \\cong (P)$ $ \\Longrightarrow$ this external similarity center is at infinity $ \\Longrightarrow$ $ PS \\parallel ID.$ $ N$ is center of the 9-point circle $ (N)$ of $ \\triangle DEF,$ the inversion image of the circumcircle $ \\odot(MUV)$ in $ (I).$ Its radius $ \\frac {_r}{^2}$ is constant, regardless of $ M.$ In addition, $ K \\in (N),$ $ KN \\equal{} \\frac {_r}{^2}$ and $ KN \\parallel ID \\parallel PS,$ where $ K \\in (P), N \\in (S), (P) \\cong (S)$ $ \\Longrightarrow$ $ (S)$ is a translation of $ (P)$ by $ \\overline{KN}$ and $ PS \\equal{} KN \\equal{} \\frac {_r}{^2}.$ It follows that $ (N)$ is tangent to 2 fixed circles $ \\mathcal S_1, \\mathcal S_2,$ concentric with $ (S)$; the difference of their radii equal to $ r$; one of these circles can degenerate to the point $ S.$ \r\n\r\n[color=red]This part of the proof holds without change even when we replace the arbitrary circle $ \\color{red} (O)$ by an arbitrary line $ \\color{red}o$. When $ \\color{red}M \\in o$ is at infinity, the circumcircle $ \\color{red}\\odot (MUV)$ degenerates to the tangent $ \\color{red}l$ of $ \\color{red}(I)$ at $ \\color{red}D$ and the circle $ \\color{red}(N)$ coincides with the circle $ \\color{red}\\mathcal L$ with diameter $ \\color{red}ID.$ It follows that the concentric circles $ \\color{red}\\mathcal S_1, \\mathcal S_2$ with the common center $ \\color{red}S$ are both tangent to $ \\color{red}\\mathcal L.$ But $ \\color{red}\\mathcal L$ is the inversion image of $ \\color{red}l$ in $ \\color{red}(I).$ Consequently, the inversion images of the circles $ \\color{red}\\mathcal S_1, \\mathcal S_2$ in $ \\color{red}(I)$ are tangent not only to the line $ \\color{red}o,$ but also to the line $ \\color{red}l$ and their centers are collinear with the inversion center $ \\color{red} I.$ In addition, they are both tangent to the circumcircle $ \\color{red}\\odot(MUV)$ for any $ \\color{red}M \\in o.$ As a result, we obtained a simple inversion proof of [b]Thebault theorem[/b] ![/color] :wow:\r\n \r\nLet $ l$ cut $ (O)$ at $ B', C'$ and assume now that tangents to $ (I)$ from $ B', C'$ other than $ l$ intersect at $ A' \\in (O),$ so that $ (O), (I)$ are circumcircle and incircle of the $ \\triangle A'B'C'.$ Let $ A \\in (O)$ be tangency point of the A'-mixtilinear incircle of the $ \\triangle A'B'C'$ and let tangents to $ (I)$ from $ A$ cut $ (O)$ again at $ B, C,$ By Poncelet porism, $ BC$ is tangent of $ (I).$ Let the angle bisectors $ A'I, AI$ cut $ (O)$ again at $ X', X,$ then $ X'X$ is the common perpendicular bisector of $ B'C', BC$ $ \\Longrightarrow$ $ BC \\parallel B'C'$ and $ A' \\in (O)$ is tangency point of the A-mixtilinear incircle of the $ \\triangle ABC.$ \r\n\r\nSince $ (P)$ is inversion of $ (O)$ in $ (I),$ radii of $ (S) \\cong (P)$ are now $ \\frac {_r}{^2}.$ Since $ PS$ = $ \\frac {_r}{^2}$ as well, $ P \\in (S)$ and $ S \\in (P).$ If $ E', F'$ are tangency points of $ (I)$ with $ C'A', A'B',$ then $ I, P,$ are circumcenter and 9-point center of $ \\triangle DE'F'.$ If $ A_0 \\in (P)$ is midpoint of $ E'F',$ then obviously, $ A_0P \\parallel ID$ $ \\Longrightarrow$ $ S, P, A_0$ are collinear. This means $ S \\in (P)$ is diametrically opposite point of $ A_0 \\in (P),$ i.e., the inversion image of the the tangency point $ A \\in (O)$ of the A'-mixtilinear incircle of the $ \\triangle A'B'C'.$ Since $ S \\in (N),$ it follows that $ A \\in \\odot(MUV).$\r\n\r\n[color=blue]Note that this is a natural and much nicer inversion proof of the problem [url]http://www.mathlinks.ro/viewtopic.php?t=186117[/url].[/color]\r\n\r\nThe 9-point circle $ (N)$ of the $ \\triangle DEF,$ which is the inversion image of the circumcircle $ \\odot(MUV)$ in $ (I),$ is tangent to 2 fixed circles concentric with $ (S).$ Since $ S \\in (N),$ one of these circles degenerates to the point $ S \\in (P),$ the diametriacally opposite point of $ A_0 \\in (P),$ the inversion image of the tangency point $ A$ of the A'-mixtilinear incircle of the $ \\triangle A'B'C'$ in $ (I).$ Since $ (N)$ has radius $ \\frac {_r}{^2},$ it is tangent to a circle $ (S, r),$ which is the inversion image of the A-mixtilinear incircle of the $ \\triangle ABC$ in $ (I).$", "Solution_5": "[quote=\"Fang-jh\"]Given a triangle $ ABC.$ Let $ M$ be a moving point on its circumcircle. Let $ l$ be a tangent of its incircle which is parallel to the side $ BC.$ Prove that the circumcircle of the triangle formed by the tangents from $ M$ to its incircle and $ l$ is always tangent to the $ A$-mixtilinear incircle of triangle $ ABC.$[/quote]\nFirst I will put some notations to the problem\n[quote]Let $ \\triangle ABC$ with $ (I)$ as its incircle. Denote $ l$ by the line tangents to $ (I)$ and parallel to $ BC$ and $ M$ by any point on circumcircle $ (O)$ of $ \\triangle ABC$. The tangents from $ M$ to $ (I)$ intersect $ l$ respectively at $ X_1,$ $ X_2$. Prove that the circumcircle of $ \\triangle MX_1X_2$ always tangents to the mixtilinear circle of $ \\triangle ABC$ with respect to angle $ A$.[/quote]\r\nHere is my proof.\r\n\r\nLet $ D,$ $ E,$ $ F$ respectively be the tangency points of $ (I)$ with $ BC,$ $ CA,$ $ AB$. Denote $ (K_a)$ be the mixtilinear circle of $ \\triangle ABC$ with respect to $ A$ and $ B'$, $ C'$, $ A'$ by the tangency points of $ (K_a)$ with $ AB$, $ AC$ and $ (O)$. Let $ S$ be second intersection of $ A'I$ with $ (O)$ and $ B''$, $ C''$ respectively be the intersections of $ l$ with $ (O)$. \r\n\r\nSince, it is well- known that $ B',$ $ I,$ $ C'$ are collinear and $ AI$ is the perpendicular bisector of $ B'C'$, thus $ \\angle AB'I$ $ \\equal{} \\angle AC'I$. But, we also have $ BB'IA'$ and $ CC'IA'$ are two concyclic quadrilateral, hence $ \\angle AB'I$ $ \\equal{} \\angle SA'B$ and $ \\angle AC'I$ $ \\equal{} \\angle SA'C$. Therefore, $ \\angle SA'B$ $ \\equal{} \\angle SA'C$, which implies $ SB \\equal{} SC$, or we can say $ S$ is the midpoint of chord $ BC$ of the circumcircle $ (O)$ of $ \\triangle ABC$. On the other hand, since $ B''C''\\|$ $ BC$ $ \\Longrightarrow$ $ S$ is also the midpoint of chord $ B''C''$, which implies $ A'S$ is the internal bisector of $ \\angle B''A'C''$. By [i]Poncelet's porism[/i], there is $ \\triangle A'UV$ such that $ A'UV$ has $ (I)$ as their incircle and $ (O)$ as its circumcircle, hence $ S$ is the midpoint of chord $ UV$ that doesn't contain $ A'$. We conclude both $ B''C''$ and $ UV$ tangent to $ (I)$, and with $ B''C''\\|UV$. As the consequence, $ B''C''\\equiv UV$, or $ (I)$ is the incircle of $ \\triangle A'B''C''$. We claim that $ A$ is the tangency point of the mixtilinear circle of $ \\triangle A'B''C''$ with respect to $ A'$. $ (*)$\r\n\r\nIndeed, let $ D'$ be the antipode of $ D$ wrt $ (I)$, then $ D$ is also the tangency point of $ (I)$ and $ B''C''$. Denote $ C_1$, $ B_1$ respectively by the tangency points of $ (I)$ with $ AB''$ and $ AC''$. Consider the inversion through pole $ I$, power $ k \\equal{} r^2$, where $ r$ is the radii of $ (I)$. We have $ \\mathcal {I}(I,k)$ maps $ A$, $ B$, $ C$ into the midpoints $ M_a,$ $ M_b,$ $ M_c$ of $ EF$, $ FD$, $ DE$, respectively. Hence $ (O)$ is mapped into $ (M_aM_bM_c)$ $ \\equiv$ $ (\\mathcal {E})$, which is the $ 9 \\minus{}$ point circle of $ \\triangle DEF$ as well as $ \\triangle D'B_1C_1$. Now, under the inversion $ \\mathcal {I}(I,k)$, from [i][1][/i], we have known that if $ A''$ is the image of $ A'$, $ A''$ will be the orthocenter of $ \\triangle DM_bM_c$. Then it is well- known that $ A''$ are the reflection of the orthocenter $ H$ of $ \\triangle DEF$ across to $ M_bM_c$ and $ H$ is the symmentry point of $ D'$ wrt $ M_a$. Therefore, $ IA''\\|D'H$ $ \\equiv D'M_a$. Note that $ A''$ is the midpoint of $ B_1C_1$ $ \\Longrightarrow$ $ IA''\\perp B_1C_1$, which implies $ D'M_a\\perp B_1C_1$. Let $ M'_b,$ $ M'_c$ respectively be the midpoints of $ D'C_1$ and $ D'B_1$, then it is followed that $ D'M_a\\perp M'_bM'_c$, $ M_a$ $ \\in$ $ (\\mathcal {E})$. As the result, $ M_a$ also plays the roles of the orthocenter of $ \\triangle D'M'_bM'_c$. Therefore, $ A$ is the tangency point of the mixtilinear circle of $ \\triangle A'B''C''$ with respect to $ A'$. We obtain result $ (*)$.\r\n\r\nNow, let $ W_b$, $ W_c$ repsectively be the tangency points of $ MX_1,$ $ MX_2$ with $ (I)$. Then the inversion $ \\mathcal {I}(I,k)$ maps $ X_1$, $ X_2$, $ M$ into the midpoints $ X'_1$, $ X'_2$, $ M'$ of $ D'W_b$, $ D'W_c$ and $ W_bW_c$. Therefore, under $ \\mathcal {I}(I,k)$, $ (MX_1X_2)$ will be mapped into a circle $ (\\omega)$ passes through $ X'_1,$ $ X'_2$ and $ M'$, which obviously is the $ 9 \\minus{}$ point circle of $ \\triangle D'W_bW_c$. But due to [i][1][/i], we have $ A\\in (MX_1X_2)$ $ \\Longrightarrow$ $ M_a$ also $ \\in$ $ (\\omega)$. The inversion through pole $ I$, power $ k$ takes $ B'$, $ C'$ into $ X$, $ Y$ repsecively $ \\Longrightarrow$ $ \\mathcal {I}(I,k)$ maps $ (K_a)$ into $ (A''XY)$ $ \\equiv$ $ (\\omega_a)$ that tangents to $ (\\mathcal {E})$ at $ A''$. We claim that $ (\\omega)$ and $ (\\omega_a)$ are tangent. $ (**)$\r\n\r\nIndeed, now just taking our concern into $ \\triangle DEF$ and $ (I)$. Since $ X,Y$ respectively are the inverse points of $ B',$ and $ C'$ $ \\Longrightarrow$ $ X,Y$ respectively are also the projections of $ E,$ $ F$ onto $ B'C'$, which implies $ YEFX$ is rectangle. Thus, $ EF \\equal{} XY$, but $ M_cM_b$ $ \\|$ $ EF$, $ EF \\equal{} 2.M_bM_c$. Therefore, $ M_b,$ $ M_c$ respectively are midpoints of $ A''X$ and $ A''Y$. Hence through the homothety at center $ A''$, ratio $ k' \\equal{} 2$, we have $ \\mathcal {H}(A'',k'):$ $ (\\mathcal {E})$ $ \\mapsto$ $ (\\omega_a)$. Thus, $ \\mathcal {H}(A'',k')$ will map $ \\mathcal {E}$ into $ M_a$, or in other words, $ M_a$ is the center of $ (\\omega_a)$ $ (1)$. Now, let $ H'$ be the orthocenter of $ \\triangle D'W_cW_b$, then it is easy to note that $ DW_cH'W_b$ is parallelogram, which implies that $ (\\omega)$ and the $ 9 \\minus{}$ point circle $ (\\mathcal {E'})$ of $ \\triangle DW_cW_b$ are reflected each other through the midpoint $ M'$ of $ W_bW_c$. As the consequence, if we denote $ r_1$, $ r_2$, $ r_3$ by the radii of $ (\\omega)$, $ (\\mathcal {E'})$ and $ (\\mathcal {E})$, we have $ r_1 \\equal{} r_2$. But according to [i][1][/i], we have $ r_2 \\equal{} r_3$, thus $ r_1 \\equal{} r_3$. Also note that $ (\\mathcal {E})$ intersects $ (\\omega)$ at $ M'$ and $ M_a$. Therefore, $ (\\mathcal {E})$ and $ (\\omega)$ are reflected each across the line $ M_aM'$. Then combine with $ (1)$, we have $ (\\omega)$ also tangents to $ (\\omega_a)$. We obtain result $ (**)$.\r\n\r\nTo sum up, due to $ (**)$, and a note that $ (\\omega)$ and $ (\\omega_a)$ are the inverse circles of $ (MX_1X_2)$ and $ (K_a)$ with respect to $ \\mathcal {I}(I,k)$, we obtain $ (MX_1X_2)$ always tangents to $ (K_a)$. Our proof is completed then. $ \\square$\r\n___________________________\r\n[i]Reference[/i]\r\n[i][1][/i]- post #3 of the topic http://www.mathlinks.ro/viewtopic.php?t=309644" } { "Tag": [], "Problem": "\u03a7\u03c1\u03cc\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u03ba\u03b1\u03bb\u03ae \u03c7\u03c1\u03bf\u03bd\u03b9\u03ac :) \r\n\u0391\u03c6\u03bf\u03cd \u03c0\u03b9\u03ac\u03c3\u03b1\u03bc\u03b5 \u03c4\u03bf \u03c0\u03bd\u03b5\u03cd\u03bc\u03b1 \u03c4\u03c9\u03bd \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03ae\u03c4\u03c9\u03bd \u03c4\u03bf 2010 \u03b1\u03c2 \u03b2\u03ac\u03bb\u03c9 \u03ba\u03b1\u03b9 \u03b3\u03c9 \u03bc\u03b9\u03b1 \u03ba\u03b1\u03bb\u03ae:\r\n\u0391\u03bd $ a,b,c,d$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03af \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03af \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 $ \\left(\\frac{a}{a\\plus{}b\\plus{}c}\\right)^{2}\\plus{}\\left(\\frac{b}{b\\plus{}c\\plus{}d}\\right)^{2}\\plus{}\\left(\\frac{c}{c\\plus{}d\\plus{}a}\\right)^{2}\\plus{}\\left(\\frac{d}{d\\plus{}a\\plus{}b}\\right)^{2}\\geq \\frac{4}{9}$.", "Solution_1": "\u039c\u03bf\u03c5 \u03b8\u03cd\u03bc\u03b9\u03b6\u03b5 \u03ba\u03ac\u03c4\u03b9 \u03bf\u03c0\u03cc\u03c4\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03b9\u03bf \u03b3\u03b5\u03bd\u03b9\u03ba\u03ae \u03bc\u03bf\u03c1\u03c6\u03ae \u03b5\u03b4\u03ce, \u03b4\u03c5\u03c3\u03c4\u03c5\u03c7\u03ce\u03c2 \u03c7\u03c9\u03c1\u03af\u03c2 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7...\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=322633\r\n\r\n\r\n\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03ad\u03c7\u03c9 \u03b2\u03c1\u03b5\u03b9 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03c4\u03b7\u03bd \u03b1\u03c0\u03bb\u03ae \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b1\u03bb\u03bb\u03ac \u03bc\u03b5 \u03c0\u03bf\u03bb\u03bb\u03ad\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2... \u0398\u03b1 \u03c4\u03b7\u03bd \u03b5\u03bb\u03ad\u03b3\u03be\u03c9 \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03c4\u03b7\u03bd \u03c0\u03bf\u03c3\u03c4\u03ac\u03c1\u03c9...", "Solution_2": "ne kai egw exw kanei mia lush(gia thn aplh periptosh) me B-C-S kai AM-GM alla einai mia selida pra3eis :S kai dn 3erw ama exw kanena la8os :maybe: [dn sou afhnoun xrono oi panellhnies akoma kai an dn diavazeis ta alla ma8hmata ]", "Solution_3": "[quote=\"KapioPulsar\"]ne kai egw exw kanei mia lush me B-C-S kai AM-GM alla einai mia selida pra3eis :S kai dn 3erw kai ama dn exw kanena la8os :maybe:[/quote]\r\n\r\nProspathiste tin me Holder kai power mean :lol:", "Solution_4": "[quote=\"\u0393\u03b9\u03ce\u03c1\u03b3\u03bf\u03c2\"][quote=\"KapioPulsar\"]ne kai egw exw kanei mia lush me B-C-S kai AM-GM alla einai mia selida pra3eis :S kai dn 3erw kai ama dn exw kanena la8os :maybe:[/quote]\n\nProspathiste tin me Holder kai power mean :lol:[/quote] kala odos power mean :P", "Solution_5": "\u0392\u03ac\u03b6\u03c9 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03ba\u03bb\u03b5\u03af\u03c3\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf post.\r\nhttp://gbas2010.wordpress.com/2010/01/07/inequality-18vo-quoc-ba-can/" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c>0$ and $ a\\plus{}b\\plus{}c\\equal{}1$.Prove that:\r\n $ \\frac{{a^2 \\plus{} 3b^2 c}}{{b \\plus{} c}} \\plus{} \\frac{{b^2 \\plus{} 3c^2 a}}{{c \\plus{} a}} \\plus{} \\frac{{c^2 \\plus{} 3a^2 b}}{{a \\plus{} b}} \\ge 1$ :D", "Solution_1": "[quote=\"leedt26\"]Let $ a,b,c > 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$.Prove that:\n $ \\frac {{a^2 \\plus{} 3b^2 c}}{{b \\plus{} c}} \\plus{} \\frac {{b^2 \\plus{} 3c^2 a}}{{c \\plus{} a}} \\plus{} \\frac {{c^2 \\plus{} 3a^2 b}}{{a \\plus{} b}} \\ge 1$ :D[/quote]\r\nYou can use : $ \\sum \\ \\frac{a^3\\plus{}3abc}{b\\plus{}c} \\ge 2\\sum \\ ab$ :)", "Solution_2": "Anyone can give a solution more clearly!", "Solution_3": "can you show your full proof,quykhtn-qa1??", "Solution_4": "[quote=\"leedt26\"]Let $ a,b,c > 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$.Prove that:\n $ \\frac {{a^2 \\plus{} 3b^2 c}}{{b \\plus{} c}} \\plus{} \\frac {{b^2 \\plus{} 3c^2 a}}{{c \\plus{} a}} \\plus{} \\frac {{c^2 \\plus{} 3a^2 b}}{{a \\plus{} b}} \\ge 1$ :D[/quote]\r\nAfter expanding we obtain for the proof:\r\n$ \\sum_{cyc}(a^5 \\plus{} a^4b \\plus{} a^4c \\plus{} a^3b^2 \\minus{} 2a^3c^2 \\minus{} 2a^2b^2c)\\geq0,$ which is obviously true.", "Solution_5": "How about the stronger, Quy? :) \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=1526706#1526706" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Here is another nice one:\r\n\r\nthere are $100$ subsets, of a line, each subste is a union of $100$ disjoint closed intervals. Prove that the intersection of these subsets is the union of at most $9900$ intervals.", "Solution_1": "Hi Pascual!\r\n\r\nIn my solution I got 9901, not 9900. It is a typo ;)\r\n\r\nI have looked through ARO 2002 for 10 times, but I couldn't find it!\r\nActually, IT IS FROM ARO 2001 :)\r\n\r\nBTW, the problems is very simple. We can easily prove that for two systems of $a$ and $b$ intervals their intersection contains at mosts $a+b-1$ intervals.", "Solution_2": "oops, sorry Myth, indeed it is $9901$" } { "Tag": [ "integration", "trigonometry", "calculus", "calculus computations" ], "Problem": "Prove that $ \\displaystyle 1\\minus{}e^{\\minus{}\\frac\\pi2}<\\int_0^{\\frac\\pi2}e^{\\minus{}\\sin x}\\textrm{d}x<\\frac\\pi2(1\\minus{}e^{\\minus{}1})$", "Solution_1": "This is your twin brothers.\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=25423&search_id=161421594[/url]" } { "Tag": [ "function", "combinatorics unsolved", "combinatorics" ], "Problem": "In how many ways can $n$ identical balls be distributed to nine persons $A,B,C,D,E,F,G,H,I$ so that the number of balls recieved by $A$ is the same as the total number of balls recieved by $B,C,D,E$ together,.", "Solution_1": "these sort of problems are readily handled by the exponential generating function.\r\nlet $x_A,\\ldots,x_I$ denote the distribution of the balls. Then we have $x_A=x_B+x_C+x_D+x_E$ so that we have that $2(x_B+x_C+x_D+x_E)+x_F+\\cdots+x_I=n$ or in other words we need the number of ordered $8$-tuples such that $4$ terms are even. The Exponential generating function here is simply $(\\sum_{k\\geq 0} \\dfrac{x^{2k}}{(2k)!})^4(\\sum_{k\\geq 0} \\dfrac{x^k}{k!})^4$. what you need here is the coeff of $x^n$ which is easily obtained by simplification.", "Solution_2": "[quote=\"seshadri\"]in other words we need the number of ordered $8$-tuples such that $4$ terms are even. The Exponential generating function here is simply $(\\sum_{k\\geq 0} \\dfrac{x^{2k}}{(2k)!})^4(\\sum_{k\\geq 0} \\dfrac{x^k}{k!})^4$[/quote]\r\nI think this E.G.F. is incorrect. It is better to use regular G.F. here which is\r\n\\[ \\left(\\sum_{k\\geq 0} x^{2k}\\right)^4 \\left(\\sum_{k\\geq 0} x^k\\right)^4 = (1-x^2)^{-4} (1-x)^{-4} \\]", "Solution_3": "oh yes, you are right. i misread the 'identical balls' part and assumed that the balls are distinguishable." } { "Tag": [], "Problem": "Okay, here's the rules...\r\neveryone gets to post a part of a story...\r\nexample: person 1 posts: Bob went to the store\r\n person 2 posts: Then, a killer dragon popped out of nowhere and ate everthing exept bob\r\n person 3 : then a hurricane came and bob was swept into the air\r\n you get the idea... so some one start us off!", "Solution_1": "math92 decided it would be a good idea to tell a story, but unfortunately......\r\n\r\n:D", "Solution_2": "[quote=\"mcalderbank\"]math92 decided it would be a good idea to tell a story, but unfortunately......\n\n:D[/quote]\r\n\r\nSome mean person came and told him it was retarded", "Solution_3": "so then he beat the mean person up.", "Solution_4": "then the mean, beat up person, ran to his mom and sued math92", "Solution_5": "But the lawyer was on Mars and couldn't be contacted.", "Solution_6": "then he blew up mars and hired a new lawyer", "Solution_7": "who just happened to be on jupiter instead", "Solution_8": "and this lawyer stepped on a landmine and got his leg blown off.", "Solution_9": "And so math92 hired his pet gorilla as his lawyer", "Solution_10": "but math92 hadn't enough bananas to feed the gorilla", "Solution_11": "aAnd since math92 was wearing a banana suit, the gorilla ate him instead. Because the gorilla ate a human, the SWAT team came and took him down with a tranq-dart", "Solution_12": "And the guy from SWAT says.. Hey, isn't the idea basically the same than in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34951]this post[/url]?", "Solution_13": "Then Valentin, which was randomly passing by, says \"Indeed it is ... \" so he closed this topic and let the users continue the story in the other topic. \r\nThe end. :P" } { "Tag": [ "probability" ], "Problem": "Two gamblers take turns rolling a fair N-sided die, with N at least 5. The faces of the die are labeled with the numbers 1 to N. The first player starts. If he rolls an N or an N-1, he wins and the game is over. Otherwise, the other player rolls the die; if she rolls a 1, 2, or 3, she wins and the game is over. Play continues, with the players alternating rolls until one of them wins. \r\n\r\nWhat is the probability that the first player will win? Are there any values of N for which the first player has at least an even chance of winning?", "Solution_1": "[hide] The probability that the first player wins on the first move is $\\frac{2}{N}$. The probability that neither player wins on their first moves, respecitively, is $\\frac{(N-2)(N-3)}{N^{2}}$. This is equivalent to the probability that player one gets a second move. But we see that the probability that player one gets a $K$th move is\n\n$\\left(\\frac{(N-2)(N-3)}{N^{2}}\\right)^{K-1}$\n\nand each time he has a $\\frac{2}{N}$ chance of winning on that move. So his win probability is\n\n$\\frac{2}{N}+\\frac{2}{N}\\cdot \\frac{(N-2)(N-3)}{N^{2}}+\\frac{2}{N}\\cdot \\left(\\frac{(N-2)(N-3)}{N^{2}}\\right)^{2}+\\cdots$\n\nwhich comes out to $\\frac{2N}{5N-6}$ (by infinite geometric series). This is at least $\\frac{1}{2}$ for $N = 3, 4, 5, 6$.[/hide]" } { "Tag": [], "Problem": "diavasa kapou oti mia sinartisi f gia tin opoia ine f(ta,tb,tc,...)=t^kf(a,b,c,...) legete omogenis vathmou k.Mporeite na mou exigisete ti simveni me tis anisotites se afti tin periptosi?Diavasa sto vivlio tou k.stergiou oti boroume na theoroume kapoia sinthiki pou mas volevei.Giati?", "Solution_1": "[quote=\"koredim1989\"]diavasa kapou oti mia sinartisi f gia tin opoia ine f(ta,tb,tc,...)=t^kf(a,b,c,...) legete omogenis vathmou k.Mporeite na mou exigisete ti simveni me tis anisotites se afti tin periptosi?Diavasa sto vivlio tou k.stergiou oti boroume na theoroume kapoia sinthiki pou mas volevei.Giati?[/quote]\r\n\r\n\u039d\u0391\u0399 ! \u039c\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b8\u03ad\u03c4\u03b5\u03b9\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 , \u03cc\u03c0\u03c9\u03c2 \u03b5\u03af\u03c0\u03b5\u03c2 , \u03c0\u03bf\u03c5 \u03c3\u03c5\u03bd\u03ae\u03b8\u03c9\u03c2 \u03b1\u03c0\u03bb\u03bf\u03c5\u03c3\u03c4\u03b5\u03cd\u03b5\u03b9 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7. \u03a4\u03bf \u03c3\u03ba\u03b5\u03c0\u03c4\u03b9\u03ba\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b5\u03be\u03ae\u03c2 :\r\n\r\n \u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03ad\u03c2 \u03c4\u03b9\u03c2 $a , b$. \u0398\u03ad\u03c4\u03b5\u03b9\u03c2 $a+b= k$ \u03ba\u03b1\u03b9 $a=kx, b=ky$. \u03a4\u03cc\u03c4\u03b5 : \r\n\r\n $x+y = \\frac{a}{k}+\\frac{b}{k}= \\frac{a+b}{k}=\\frac{a+b}{a+b}= 1$.\r\n\r\n \u0395\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03bd\u03ad\u03b5\u03c2 \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03ad\u03c2 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 $1$ \u03ba\u03ac\u03c4\u03b9 \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b5\u03af\u03c7\u03b5\u03c2 \u03bc\u03b5 \u03c4\u03b1 $a , b$ \u0391\u03bd\u03c4\u03af \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03cc\u03bb\u03b7 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7 \u03b4\u03bf\u03c5\u03bb\u03b5\u03b9\u03ac \u03c0\u03bf\u03c5 \u03c3\u03bf\u03c5 \u03ad\u03b4\u03b5\u03b9\u03be\u03b1 , \u03b8\u03b5\u03c9\u03c1\u03b5\u03af\u03c2 \u03b1\u03c0\u03b5\u03c5\u03b8\u03b5\u03af\u03b1\u03c2 $a+b=1$ \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c7\u03c9\u03c1\u03ac\u03c2 \u03b1\u03bc\u03ad\u03c3\u03c9\u03c2. \u039a\u03b1\u03bd\u03b5\u03af\u03c2 \u03b4\u03b5 \u03b8\u03b1 \u03c3\u03b5 \u03c1\u03c9\u03c4\u03ae\u03c3\u03b5\u03b9 \u03b3\u03b9\u03b1\u03c4\u03af ! :) \u0391\u03bd\u03ac\u03bb\u03bf\u03b3\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03bc\u03b5 \u03c4\u03b1 \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03b1.\u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03ba\u03ac\u03c0\u03c9\u03c2 \u03bd\u03b1 \u03c3\u03b5 \u03b4\u03b9\u03b5\u03c5\u03ba\u03cc\u03bb\u03c5\u03bd\u03b1 !\r\n\r\n \u03a0\u03b9\u03bf \u03c0\u03bf\u03bb\u03bb\u03ad\u03c2 \u03bb\u03b5\u03c0\u03c4\u03bf\u03bc\u03ad\u03c1\u03b5\u03b9\u03b5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b4\u03b5\u03b9\u03c2 \u03c3\u03c4\u03bf [color=red]\u03bd\u03ad\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf << \u039a\u03bb\u03b1\u03c3\u03b9\u03ba\u03ad\u03c2 \u03ba\u03b1\u03b9 \u039d\u03ad\u03b5\u03c2 \u0391\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2 >> , \u03c3\u03b5\u03bb\u03af\u03b4\u03b1 127 - 128.[/color]\r\n\u0398\u03b1 \u03c3\u03bf\u03c5 \u03c0\u03c1\u03cc\u03c4\u03b5\u03b9\u03bd\u03b1 \u03cc\u03bc\u03c9\u03c2 - \u03c4\u03b5\u03bb\u03b5\u03af\u03c9\u03c2 \u03c6\u03b9\u03bb\u03b9\u03ba\u03ac - \u03bc\u03ad\u03c7\u03c1\u03b9 \u03c4\u03bf\u03bd \u0391\u03c1\u03c7\u03b9\u03bc\u03ae\u03b4\u03b7 \u03bd\u03b1 \u03c4\u03bf \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b1\u03c5\u03c4\u03cc \u03c9\u03c2 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03bf \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03bc\u03b5\u03bb\u03b5\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ad\u03c2 , \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03bc\u03b7 \u03c7\u03ac\u03bd\u03b5\u03b9\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf.\u03a4\u03b1 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b1 \u03c4\u03b1 \u03bc\u03b5\u03bb\u03b5\u03c4\u03ac\u03c2 \u03b1\u03c1\u03b3\u03cc\u03c4\u03b5\u03c1\u03b1.\r\n[u] \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2[/u]" } { "Tag": [ "articles", "Support" ], "Problem": "I don't have any assigned summer reading. Even so I would like to read more but I haven't been doing much lately. I am going to try to read some more before the school year starts back up.", "Solution_1": "No summer reading for me etiher.", "Solution_2": "Rep, do you attend a private school or something like that?\r\n\r\nI wasn't assigned any summer reading but I'm on my third book this summer.", "Solution_3": "I've got two books, [i]The Stranger[/i] by Albert Camus and [i]Things Fall Apart[/i] by Chinua Achebe. I have to write detailed reportson 5 x 8 index cards about each book.\r\n\r\nI've procrastinated so far this summer. I feel so guilty...(sobs)", "Solution_4": "Umm...well...\r\nSummer Reading List\r\nMiss Grady\r\nGrade 10 AS \r\nScarlett Letter - Nathaniel Hawthorne\r\n\r\n1.Discuss Pearl as a symbol. How and why is her behavior not natural? At what point does she \"transform\" and why?\r\n\r\n2. A recurring theme in literature is the \"classic war between passion and responsibility\". Choose either Rev. Arthur Dimmesdale or Roger Chillingworth - \r\nShow (not tell) how this man confronts the demands of his private passion with his responsibility.\r\n\r\n\r\nTo Kill A Mockingbird - Harper Lee\r\n\r\n1. The final scene in the trial is one of the greatest scenes in American literature.\r\nBring the time period of To Kill A Mockingbird into today's society and issues. Write a closing argument based on the events of the novel, yet pertaining to the current issues of today.\r\n\r\n\r\n\r\nFarewell to Arms - Ernest Hemingway\r\n\r\n1.Choose one of the following characters: Fredrick Henry, Catherine Barkley, or Lt. Rinaldi. Prepare five interview questions and answers for this character regarding his/her life during the novel (World War I)\r\n\r\nYou are a news correspondent assigned to the war. Using your interview questions and answers, write a news article based on your character's experiences in the war. \r\nCite examples, dialogue from the novel.\r\n\r\n\r\n2. A true Hemingway man has self- discipline and control, avoids death at any cost, yet doesn't fear death, and accepts the joys and the turmoils in life and then moves on. Decide who the true Hemingway man is in this novel. Explain and argue your decision citing examples from the novel to support your findings.\r\n\r\nWe read To Kill a Mockingbird in class before...and I didn't do ANY of the tasks, and I'm like 10 pages into The Scarlet Letter. Um, hopefully I do this summer reading (last year I didn't (nor did most of my class - we're horrible...come on now, it was an ADVANCE 9th grade class and she made us read Little Women, David Copperfield, and The Road To Memphis <---like four/fifth grade books!) The tasks kind of cool this year though.", "Solution_5": "Do you students have to hand in the answers to your last year's teacher or your current (this year) teacher?\r\n\r\nAlso, do you receive any extra marks if you do answer the questions?", "Solution_6": "i have to hand it to my current teacher..and um, it is REQUIRED that you answer everything.", "Solution_7": "hmmm, i 'm going into 9th grade, and i have to read [u]the prince and the pauper[/u] for pre-ap english. I also must highlight themes, characters, and some other stuff i forgot...i haven't even finished the book yet", "Solution_8": "I have to read the 1836 version of Mary Shelley's Frankenstein for MTU orientation; instead, I'm planning on reading the version with yellow and black stripes on the cover, and spending my free time doing things I enjoy.", "Solution_9": "One last question, do all private schools have this system in the USA?\r\n\r\nAnd also, do public schools also have this system?\r\n\r\nOr do they all differ from school to school.", "Solution_10": "Assuming that you are talking about the summer reading thing, it differs from school to school. Some schools give you a list of books that might be helpful to read(or I should say some teachers) other schools make you do work over the summer, and others(like mine) don't give you anything at all(which isn't a bad thing IMO :lol: ).", "Solution_11": "I didn't get any assined reading this year but if I had decided to attend a new school I would have had to read.\r\n\r\nTo Kill a Mocking Bird\r\n\r\nand\r\n\r\nThe Diary of Ann Frank", "Solution_12": "BHorse what grade are you in this year?", "Solution_13": "I'm going into 8th *rolls eyes*.", "Solution_14": "Wow....8th...USA definetely has a harder curriculum than Canada.\r\n\r\nI'm in Canada and i'm in grade 9 we read those books last year or we were supposed to...\r\n\r\nI'm 10 now...hehe", "Solution_15": "The U.S. isn't really harder. The call that I would have been taking that those books were from an advanced 9th grade english class. I decided against the school though.", "Solution_16": "Going into 10th grade..\r\nLoads of summer reading...\r\n\r\nRead: [i]The Adventures of Huckleberry Finn[/i] by Mark Twain.\r\nTake detailed, typewritten notes on plot, character, and setting. [in my case.. that usually means about 50 pages of notes.. lol *points at self and mutters dork*]\r\nWrite a well-developed, thought-provoking essay, 2-3 pages, double-spaced, 12 point Times New Roman font, one-inch margins, no cover page, page numbers in the upper right corner. This essay should adhere to the five paragraph format. [PHEW.. yes our teachers has err.. very strict layout \"guidelines\"]\r\n\r\nI'm not even going to type out the whole quote and blah that our essays supposed to be about since it would take up about two pages... so yea..\r\n\r\nBesides this, we also have to read eleven short stories:\r\n\r\nThe Necklace Guy de Maupassant\t\t\r\nThe Story of an Hour Kate Chopin\t\t\r\nYoung Goodman Brown Nathaniel Hawthorne\r\nThe Cask of Amontillado Edgar Allen Poe\t\r\nMiss Brill Katherine Mansfield\t\r\nTrifles Susan Glaspell\t\t\r\nAn Occurrence at Owl Creek Bridge Ambrose Bierce\r\nThe Three Strangers Thomas Hardy\r\nThe Bear: A Joke in One Act Anton Chekhov (also called The Boor)\r\n\"Patterns Amy Lowell\r\nDover Beach Matthew Arnold\r\n\r\nJoy ehh?\r\n\r\nBut that's not all.\r\n\r\nWe have to read The Elements of Style by Strunk and White too..\r\nand notes for all the short stories and Elements is \"suggested\".. which means if you want an A.. you better do it.\r\n\r\nThankfully. I'm DONE! *cheers!*\r\n\r\n[ps. sorry for that long boring entry....]", "Solution_17": "Is that all for the summer? My school is so slack when it comes to english(at least the underclassmen classes). We read about 7 books during the year, read a handful of short stories, do maybe 2 weeks worth of grammar and we're done.", "Solution_18": "I'm surprised your school still teaches grammar. All high school teachers expect all of us to know proper english grammar.\r\n\r\nI think we only had one grammar class during the whole year.", "Solution_19": "I haven't even started - I'm in major trouble cuz I'm in 4 seminar classes with about 2 books for every class but math. I should really get going. However, the assignments look easy enough. How hard are your assignments?", "Solution_20": "[quote=\"joml88\"]Is that all for the summer? My school is so slack when it comes to english(at least the underclassmen classes). We read about 7 books during the year, read a handful of short stories, do maybe 2 weeks worth of grammar and we're done.[/quote]\r\n\r\nAt least you guys get to actually read books in your class. We spent our whole English class almost every single day (except for one sixth of the year where our teacher all of a sudden got really tough) reading out of our grammer books, reading A.R. books (just regular fiction books like Redwall and Harry Potter), and on the rare ocassion - write part of our esay. We didn't even have essay /short answer questions on quizzes untill 2/3rd of the year was over. And we only read two books as a class.", "Solution_21": "My school's already suggesting you read The Elements of Style, but we only have to read a few books over the summer, and write a short summary of them. Howevery, they have a nice incentive, you get to an hour of school for a celebration if you do it." } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $f: R \\to R$ $f(x) = \\displaystyle e^{2x}+2ex$. Compute the limit: $ \\displaystyle \\lim_{y \\to \\infty} \\sqrt{ e^{2f^{-1}(y)}} \\cdot \\frac{ f^{-1}(y) - \\sqrt{y}}{y- f^{-1}(y)}$.\r\n\r\nWhere $f^{-1}$ is the inverse of the function.\r\n\r\nthank you!", "Solution_1": "$f$ is bijective increasing $\\lim{_+\\infty}f=+\\infty$ so $\\lim_{+\\infty}f^{-1}=+\\infty$\r\n\r\nDo the change variable $u=f^{-1}(y)$ the given expression is \r\n\r\n$e^{u}\\frac{u-\\sqrt{e^{2u}+2eu}}{e^{2u}+(2e-1)u}$ from easy to find the limit\r\nwhen $u$ tends to +oo" } { "Tag": [ "geometry", "rhombus", "circumcircle", "analytic geometry", "parallelogram", "symmetry", "trigonometry" ], "Problem": "Should be quite easy\r\n\r\nLet $ABCD$ be a rhombus of which we know the circumradius $R$ of the triangle $\\triangle ABC$ and the circumradius $r$ of the triangle $\\triangle BCD$. Construct the rhombus (only with ruler and compass)!", "Solution_1": "Let $a$ be the side-length of the rhombus\r\n\r\n$(ABC) = \\frac{a^2 \\cdot AC}{4r} \\Rightarrow 2(ABC) = \\frac{a^2 \\cdot AC}{2r}$\r\n\r\n$(BCD) = \\frac{a^2 \\cdot BD}{4R} \\Rightarrow 2(BCD) = \\frac{a^2 \\cdot BD}{2R}$\r\n\r\nThe total area of the rhombus is $2(ABC) =2(BCD)$, so $\\frac{AC}{r} = \\frac{BD}{R} \\Rightarrow \\frac{AC}{BD} = \\frac{r}{R} \\ \\ (1)$\r\n\r\n\r\nWe will construct the rhombus on a coordinates system $Oxy$, so that \r\n$A \\in Ox', \\ C \\in Ox, \\ B\\in Oy , D \\in Oy'$\r\nand $O$ will be the intersection of the diagonals $AC$ and $BD$.\r\n\r\nFrom $(1)$ we get $\\frac{OC}{OB}=\\frac{r}{R}$\r\n\r\nWe take the points $P(r,0), Q(0,R)$. Then, it's obvious that $BC \\parallel PR$\r\n$\\hline$\r\n\r\n\r\nNow, we have the direction of $BC$ but we don't have yet its length\r\n\r\nOn the ray $PO$ we take a point $S$ such that $PS=R$. \r\nWe draw the circle $(S,R)$ and let $T$ the point where this circle intersects (again) the line $PQ$ (it is $r>0$, so the line $PQ$ is not a tangent)\r\nThe triangle $STP$ is isosceles: $SP=ST=R$\r\n\r\nWe find the projection $B$ of $T$ onto $y'y$ and we construct the parallelogram $BCPT$.\r\nAfter that, we can find the other two points $A,D$ of the rhombus, which are symmetric of $C,B$ w.r.t. $O$\r\nFrom the construction, we have $\\frac{OC}{OB}=\\frac{AC}{BD}=\\frac{r}{R}$\r\n$\\hline$\r\n\r\n\r\nThe quadrilateral $ABCD$ is obviously a rhombus, but we should check if the circumradiuses are correct.\r\n\r\nThe perpendicular bisector of $BC$ intersects the x-axis at $M$. It's easy to see that the isosceles triangles $MBC,STP$ are congruent, hence $MB=MC=R$. By symmetry, we have $MB=MD=R$.\r\nSo $M$ is the circumcenter of $\\triangle BCD$ and the radius is $R$.\r\n\r\n\r\nLet $r'$ be the circumradius of $\\triangle ABC$. Then $2(ABC) = \\frac{a^2 \\cdot AC}{2r'}$ and $2(BCD) = \\frac{a^2 \\cdot BD}{2R} \\Rightarrow$\r\n$\\Rightarrow \\frac{AC}{BD} = \\frac{r'}{R}$\r\n\r\nBut we constructed the rhombus such that $\\frac{AC}{BD} = \\frac{r}{R}$, so we have $r'=r$ and we are done", "Solution_2": "Here is another construction:\r\n\r\n$k:$ The circumcircle of $\\triangle ABC$\r\n$m:$ The circumcircle of $\\triangle BCD$\r\n\r\nLet $2x$ be the side-length of the rhombus\r\n\r\nFrom $\\triangle ABC$ we get $AB=2r \\sin(BCA)$\r\nFrom $\\triangle BCD$ we get $CD=2R \\sin(CBD)$\r\n\r\nSo we have $\\sin(BCA) = \\frac{AB}{2r} = \\frac{2x}{2r} = \\frac{x}{r}$\r\n\r\nand $\\sin(CBD) = \\frac{CD}{2R} = \\frac{2x}{2R} = \\frac{x}{R}$\r\n\r\n\r\nLet $O$ be the intersection of the diagonals\r\nFrom the right triangle $OBC$ we find $\\angle BCA+\\angle CBD = 90^o \\Rightarrow \\sin^2(BCA)+\\sin^2(CBD)=1 \\Rightarrow$\r\n\r\n$\\left( \\frac{x}{r} \\right)^2 + \\left( \\frac{x}{R} \\right)^2 = 1 \\Rightarrow$\r\n\r\n$x^2 = \\frac{r^2R^2}{r^2+R^2} \\ \\ (1)$\r\n\r\nWe construct a right triangle $T$ with perpendicular sides $r,R$ and hypotenuse $z$. Its area is $E = \\frac{1}{2} rR \\Rightarrow rR=2E$\r\n\r\n$(1)$ gives $x^2 = \\frac{(2E)^2}{z^2} = \\left(\\frac{2E}{z}\\right)^2 \\Rightarrow x = \\frac{2E}{z} \\Rightarrow E = \\frac{1}{2}xz$\r\n\r\nSo, $x$ is the altitude of $T$ with respect to the hypotenuse\r\n\r\n\r\n\r\nWe draw the circle $k=(A,B,C)$ with radius $r$ and we take an arbitrary point $B$ on it.\r\nThen we find the points $A,C \\in w$ such that $AB=BC=2x$, and the point $D$ is on the intersection of the circles $(A,2x)$ and $(C,2x)$.", "Solution_3": "[quote=\"Yimin Ge\"]Let $ ABCD$ be a rhombus of which we know the circumradius $ R$ of the triangle $ \\triangle ABC$ and the circumradius $ r$ of the triangle $ \\triangle BCD$. Construct the rhombus (only with ruler and compass)![/quote]\r\n\r\nWe construct a triangle $ MNB$, right at $ B$, with $ MB\\equal{}R$ and $ NB\\equal{}r$\r\n\r\nWe draw the circles $ m\\equal{}(M,R)$ and $ n\\equal{}(N,r)$. The circles intersect at $ B,C$. \r\nLet $ A$ be the symmetric point of $ C$ wrt the line $ BM$. We draw the rhombus $ ABCD$\r\nObviously $ MA\\equal{}MC\\Rightarrow A\\in m$\r\n\r\nNow it's enough to prove that the circumradius of $ \\triangle BCD$ is $ r$\r\n\r\n[img]10375[/img]\r\nWe draw the rhombus $ BNCO$. It is $ OC\\equal{}OB\\equal{}r$\r\nAlso $ D,B$ are symmetric wrt $ OC$, so $ OD\\equal{}r$.\r\n\r\nSo $ O$ is the circumcenter of $ \\triangle BCD$ and the circumradius is $ r$" } { "Tag": [], "Problem": "Midtown Taxi charges $\\$1.60 plus $\\$0.25 per one-eighth mile. The distance from the hotel to the airport is 16.25 miles. Two passengers will evenly split the fare. How many dollars will each passenger pay? Express your answer to the nearest hundredth.", "Solution_1": "[hide=\"da answer?\"]17.05?[/hide]", "Solution_2": "I got $9.05, but i could be wrong i guess.", "Solution_3": "Hi MCrawford:\r\n[hide=\"solution\"]\nIf it costs 25 cents for one-eighth of a mile, then it would cost 2 dollars per mile.\nso for 16.25 miles it would cost him 16.25 X 2 dollars or 32 dollars and 50 cents.\nThen adding the 1 dollar and 60 cents to 32 dollars and 50 cents, you would have \na total fare of 34 dollars and 10 cents.\n[/hide]", "Solution_4": "I thought it was 25 cents per quater mile. So the answer is[hide]34.10[/hide]", "Solution_5": "i got 17.05 dollars" } { "Tag": [], "Problem": "What is $ 50\\%$ of the sum of the first 10 odd numbers?", "Solution_1": "It is well known that the sum of the first $ n$ odd numbers is $ n^2$, so our answer is $ 0.5\\cdot100\\equal{}\\boxed{50}$.", "Solution_2": "The first 10 odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19. Adding yields 20*5=100. Divide by 2 to get 50." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let a,b,c,d be four distinct real numbers such that\r\ni)$ ac\\equal{}bd$\r\nii)$ \\frac{a}{b}\\plus{}\\frac{b}{c}\\plus{}\\frac{c}{d}\\plus{}\\frac{d}{a}\\equal{}4$\r\nFind the maximum of\r\n $ P\\equal{}\\frac{a}{c}\\plus{}\\frac{b}{d}\\plus{}\\frac{c}{a}\\plus{}\\frac{d}{b}\\minus{}\\frac{abcd}{(ab\\plus{}cd)^2}$", "Solution_1": "[quote=\"8826\"]Let a,b,c,d be four distinct real numbers such that\ni)$ ac \\equal{} bd$\nii)$ \\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{d} \\plus{} \\frac {d}{a} \\equal{} 4$\nFind the maximum of\n $ P \\equal{} \\frac {a}{c} \\plus{} \\frac {b}{d} \\plus{} \\frac {c}{a} \\plus{} \\frac {d}{b} \\minus{} \\frac {abcd}{(ab \\plus{} cd)^2}$[/quote]\r\nIt very old,in MYM :wink:" } { "Tag": [], "Problem": "[color=blue][b]Prove:\n$C_{k - 3}^{k - 3} \\cdot C_{n - k + 2}^2 + C_{k - 2}^{k - 3} \\cdot C_{n - k + 1}^2 + C_{k - 1}^{k - 3} \\cdot C_{n - k}^2 + \\cdots + C_{n - 3}^{k - 3} \\cdot C_2^2 = C_n^k$\n[/b][/color]\r\n[color=red][size=75][b]\n$\\blacktriangleright$ \nConform preciz\u0103rilor D-lui enescu, problema a fost propus\u0103 \u015fi \u00een lucratea:\nIOAN TOMESCU, \"Probleme de combinatoric\u0103 \u015fi teoria mul\u0163imilor\", \nEditura Didactic\u0103 \u015fi pedagogic\u0103, Bucure\u015fti 1981. [pag. 11, prob. 1.2. a)]\n\n[/b][/size][/color]", "Solution_1": "Vezi problema 1.2 din excelenta carte \"Probleme de combinatoric\u0103 \u015fi teoria grafurilor\" de Ioan Tomescu, Ed. Didactic\u0103 \u015fi Pedagogic\u0103, Bucure\u015fti, 1981.", "Solution_2": "Multumiri Domnule enescu", "Solution_3": "Cred ar trebui folosita relatia \\[ \\binom{n}{k} - \\binom{n-1}{k} = \\binom{n-1}{k-1} . \\]\r\n\r\nNotam membrul stang cu $c(n,k)$ si apoi demonstram ca $c(n,k) - c(n-1,k) = c(n-1,k-1)$ folosind ce-am scris eu mai sus. Apoi observam ca formula e buna pt. $n$-uri mici si apoi folosim inductie. Merge?\r\n\r\n\r\nDar in cartea mentionata de dl. Enescu ce solutie e? Daca e cu \"dubla numarare\", ai putea s-o postezi? Multumesc.", "Solution_4": "[color=blue][b]Prove:\n$C_{k - 3}^{k - 3} \\cdot C_{n - k + 2}^2 + C_{k - 2}^{k - 3} \\cdot C_{n - k + 1}^2 + C_{k - 1}^{k - 3} \\cdot C_{n - k}^2 + \\cdots + C_{n - 3}^{k - 3} \\cdot C_2^2 = C_n^k$\n[/b][/color]\r\n[color=red][size=75][b]\n$\\blacktriangleright$ \nConform preciz\u0103rilor D-lui enescu, problema a fost propus\u0103 \u015fi \u00een lucratea:\nIOAN TOMESCU, \"Probleme de combinatoric\u0103 \u015fi teoria mul\u0163imilor\", \nEditura Didactic\u0103 \u015fi pedagogic\u0103, Bucure\u015fti 1981. [pag. 11, prob. 1.2. a)]\n\n[/b][/size][/color]\r\n\r\n[b][u]Solu\u0163ie:[/u][/b]\r\n\r\n(Voi incerca sa redactez si solutia Dlui Tomescu)", "Solution_5": "Care e titlul corect al cartii ,cel dat de metru sau cel dat de dl. Enescu ?" } { "Tag": [ "AMC", "AIME", "USAMTS" ], "Problem": "Sorry to be impatient, but I was wondering when we will receive our third round scores. Also, how soon after we get our third round scores will we find out the cut-off for the AIME? Does anyone know what the cut-off has been in the past, or is this the first year that it's possible to qualify for the AIME through USAMTS?", "Solution_1": "i THINK last years cutoff was 70 or 71", "Solution_2": "[quote=\"matt276eagles\"]Sorry to be impatient, but I was wondering when we will receive our third round scores. Also, how soon after we get our third round scores will we find out the cut-off for the AIME? Does anyone know what the cut-off has been in the past, or is this the first year that it's possible to qualify for the AIME through USAMTS?[/quote]\r\n\r\nScores should be up later this week. We don't yet know what the cut-off will be, but I think last year it was 68.", "Solution_3": "WOW (actually, double WOW), I thought on the question sheet it says that we will get our scores in early March. Is that a typo???\r\n\r\nBy the way, when is round 4 coming up???", "Solution_4": "Between Amc A, round 3 scores, and round 4 problems, this is a good math week :D", "Solution_5": "Yay! Scores are up!\r\n\r\nI qualified for the AIME! :)" } { "Tag": [ "integration", "function", "real analysis", "real analysis unsolved" ], "Problem": "Let $ p\\equal{}y'$. Solve the differential equation $ y \\equal{} \\minus{}\\frac{x}{p} \\plus{} ap$ where $ a$ is some real constant.\r\n\r\nI differentiated the equation, obtained an ODE for $ p$ and found an implicit solution for $ p$ by solving it. However, this doesn't help me to find $ y$. Any suggestions?", "Solution_1": "Well if you've already got an implicit expression for $ p$, then $ y(x) \\equal{} \\int^x{p(t)dt}$ [i]is[/i] an implicit solution for $ y$. :)\r\n\r\nThe point is this ODE doesn't have an explicit solution for $ y$ in terms of the usual elementary functions, so an implicit solution for $ p$ is just about as good as you can get. You can obtain other expressions more directly for $ y$ itself, but they will still be implicit.", "Solution_2": "Too bad. Thanks anyway.", "Solution_3": "I think it's tractable parametrically:It's of the form:$ y=xF(p)+G(p)$. Now differentiate as Thunder suggested: $ p=F(p)+p'\\bigg[xF'(p)+G'(p)\\bigg]$ which can be written as:\\[ \\frac{dx}{dp}=x\\left(\\frac{F'(p)}{p-F(p)}\\right)+\\left(\\frac{G'(p)}{p-F(p)}\\right)\\]\r\nOk, that's first order solvable via integrating factor. The solution is in the form: $ x=\\phi(p,C)$. Now, substitute this equation for $ x$ in the differential equation and obtain the parametric solution:\\begin{align*}\r\nx&=\\phi(t,C)\\\\\r\ny&=\\phi(t,C)F(t)+G(t)\r\n\\end{align*}Anyway, if I liked differential equations, that's what I'd be heading for. A plot of a solution with real data wouldn't hurt neither but I don't want to push it.\r\n\r\nSee: Handbook of Differential Equations under Lagrange's Equation." } { "Tag": [ "inequalities", "search", "inequalities proposed" ], "Problem": "$ a,b,c\\,\\succeq\\,0$ Prove that :\r\n\r\n$ \\sqrt {a} \\plus{} \\sqrt {b} \\plus{} \\sqrt {c}\\,\\succeq\\,ab \\plus{} ac \\plus{} bc$\r\n[color=red]SORRY!![/color]\r\nwith $ a\\plus{}b\\plus{}c\\equal{}3$", "Solution_1": "[quote=\"memath\"]$ a,b,c\\,\\succeq\\,0$ Prove that :\n\n$ \\sqrt {a} \\plus{} \\sqrt {b} \\plus{} \\sqrt {c}\\,\\succeq\\,ab \\plus{} ac \\plus{} bc$[/quote]\r\n\r\nit is wrong $ a\\equal{}1,b\\equal{}4,c\\equal{}9$", "Solution_2": "[quote=\"zaya_yc\"][quote=\"memath\"]$ a,b,c\\,\\succeq\\,0$ Prove that :\n\n$ \\sqrt {a} \\plus{} \\sqrt {b} \\plus{} \\sqrt {c}\\,\\succeq\\,ab \\plus{} ac \\plus{} bc$[/quote]\n\nit is wrong $ a \\equal{} 1,b \\equal{} 4,c \\equal{} 9$[/quote]\r\nsorry.I've added a necessary condition.", "Solution_3": "note that $ ab\\plus{}bc\\plus{}ca\\equal{}\\frac{9\\minus{}(a^2\\plus{}b^2\\plus{}c^2)}2$ so we have to show that:\r\n\r\n$ \\sqrt a\\plus{}\\sqrt b\\plus{}\\sqrt c\\geq \\frac{9\\minus{}(a^2\\plus{}b^2\\plus{}c^2)}2$\r\n\r\n$ \\iff 2(\\sqrt a\\plus{}\\sqrt b\\plus{}\\sqrt c)\\plus{}a^2\\plus{}b^2\\plus{}c^2\\geq 9$\r\n\r\nnow note that by AM-GM we have:\r\n\r\n$ a^2\\plus{}\\sqrt a\\plus{}\\sqrt a\\geq 3a$\r\n\r\nsumming up all these inequalities we get:\r\n\r\n$ 2(\\sqrt a\\plus{}\\sqrt b\\plus{}\\sqrt c)\\plus{}a^2\\plus{}b^2\\plus{}c^2\\geq 3(a\\plus{}b\\plus{}c)\\equal{}9$\r\n\r\nQED", "Solution_4": "It's Russia 2002,Serbia 2005\r\nSee here: http://www.mathlinks.ro/viewtopic.php?search_id=1719252527&t=155159", "Solution_5": "very good :)" } { "Tag": [ "function", "integration", "Euler", "geometry", "geometric transformation", "factorial", "real analysis" ], "Problem": "Prove that\r\n\\[ \\sum_{n\\equal{}0}^{\\infty}\\displaystyle\\frac{1}{\\Gamma\\left(n\\plus{}3/2\\right)}\\equal{}\\frac{2e}{\\sqrt{\\pi}}\\int_0^1e^{\\minus{}x^{2}}dx,\\]\r\nand\r\n\\[ \\sum_{n\\equal{}0}^{\\infty}\\frac{\\Gamma\\left(\\minus{}n\\plus{}1/2\\right)\\Gamma\\left(n\\plus{}1/2\\right)}{\\Gamma\\left(n\\plus{}3/2\\right)}\\equal{}\\frac{\\sqrt{\\pi}}{2e}\\int_0^1e^{x^{2}}dx,\\]\r\nwhere $ \\displaystyle\\Gamma(x)\\equal{}\\int_0^{\\infty}e^{\\minus{}t}t^{x\\minus{}1}dt$ is the Euler Gamma function.", "Solution_1": "For the first. Using Series Expansion of Incomplete Gamma:\r\n\\[ \\Gamma (a \\minus{} 1) \\minus{} \\Gamma (a \\minus{} 1,x) \\equal{} x^a \\left(\\frac {1}{(a \\minus{} 1) x} \\minus{} \\frac {1}{a} \\plus{} \\frac {x}{2 (a \\plus{} 1)} \\minus{} \\frac {x^2}{6 (a \\plus{} 2)} \\plus{} \\frac {x^3}{24 (a \\plus{} 3)} \\minus{} \\frac {x^4}{120 (a \\plus{} 4)} \\plus{} \\frac {x^5}{720 (a \\plus{} 5)} \\plus{} O\\left(x^6\\right)\\right)\\]\r\nMultiplying the equation by $ x^{(1 \\minus{} a)}$ to make things more polynomial.\r\n\\[ x^{1 \\minus{} a} (\\Gamma (a \\minus{} 1) \\minus{} \\Gamma (a \\minus{} 1,x)) \\equal{} \\left(\\frac {1}{a \\minus{} 1} \\minus{} \\frac {x}{a} \\plus{} \\frac {x^2}{2 (a \\plus{} 1)} \\minus{} \\frac {x^3}{6 (a \\plus{} 2)} \\plus{} \\frac {x^4}{24(a \\plus{} 3)} \\minus{} \\frac {x^5}{120 (a \\plus{} 4)} \\plus{} O\\left(x^6\\right)\\right)\\]\r\nThe magic happens when you multiply by $ e^x$, It will make a translation in the $ x$ terms. Or in series form\r\n\\[ \\left(\\sum _{j \\equal{} 0}^{\\infty } \\frac {x^j}{j!}\\right) \\left(\\sum _{i \\equal{} 0}^{\\infty } \\frac {( \\minus{} 1)^i x^i}{i! (a \\plus{} i \\minus{} 1)}\\right) \\equal{} \\sum _{k \\equal{} 0}^{\\infty } \\sum _{i \\equal{} 0}^k \\frac {( \\minus{} 1)^i x^k}{i! (a \\plus{} i \\minus{} 1) (k \\minus{} i)!}\\]\r\n\r\n\\[ e^x x^{1 \\minus{} a} (\\Gamma (a \\minus{} 1) \\minus{} \\Gamma (a \\minus{} 1,x)) \\equal{} \\frac {1}{a \\minus{} 1} \\plus{} \\frac {x}{(a \\minus{} 1) a} \\plus{} \\frac {x^2}{(a \\minus{} 1) a (a \\plus{} 1)} \\plus{} \\frac {x^3}{(a \\minus{} 1) a (a \\plus{} 1) (a \\plus{} 2)} \\plus{} \\frac {x^4}{(a \\minus{} 1) a (a \\plus{} 1) (a \\plus{} 2) (a \\plus{} 3)} \\plus{} O\\left(x^5\\right)\\]\r\nFinally eliminating the $ a \\minus{} 1$.\r\n\\[ (a \\minus{} 1) e^x x^{1 \\minus{} a} (\\Gamma (a \\minus{} 1) \\minus{} \\Gamma (a \\minus{} 1,x)) \\equal{} 1 \\plus{} \\frac {x}{a} \\plus{} \\frac {x^2}{a (a \\plus{} 1)} \\plus{} \\frac {x^3}{a (a \\plus{} 1) (a \\plus{} 2)} \\plus{} \\frac {x^4}{a (a \\plus{} 1) (a \\plus{} 2) (a \\plus{} 3)} \\plus{} O\\left(x^5\\right)\\]\r\n\r\n\\[ (\\Gamma ( \\minus{} 1 \\plus{} a) \\minus{} \\Gamma ( \\minus{} 1 \\plus{} a,x)) x^{1 \\minus{} a} e^x (a \\minus{} 1) \\equal{} \\sum _{i \\equal{} 0}^{\\infty } \\frac {x^i (a \\minus{} 1)!}{(i \\plus{} a \\minus{} 1)!}\\]\r\nDividing by $ \\Gamma (a)$ and using gamma function instead of factorial:\r\n\\[ \\sum _{i \\equal{} 0}^{\\infty } \\frac {x^i}{\\Gamma (a \\plus{} i)} \\equal{} \\frac {( \\minus{} 1 \\plus{} a) e^x x^{1 \\minus{} a} (\\Gamma ( \\minus{} 1 \\plus{} a) \\minus{} \\Gamma ( \\minus{} 1 \\plus{} a,x))}{\\Gamma (a)}\\]\r\nThe left and rigth side is valid for all positive real number $ a$.\r\nMaking $ x \\equal{} 1$ $ a \\equal{} \\frac {3}{2}$ and using Erf function instead of the incomplete gamma, it's donne.\r\n\r\nFor the second just make $ x \\to \\minus{}x$", "Solution_2": "[quote=\"EulerIntegral\"]For the first. Using Series Expansion of Incomplete Gamma:\n\\[ \\Gamma (a \\minus{} 1) \\minus{} \\Gamma (a \\minus{} 1,x) \\equal{} x^a \\left(\\frac {1}{(a \\minus{} 1) x} \\minus{} \\frac {1}{a} \\plus{} \\frac {x}{2 (a \\plus{} 1)} \\minus{} \\frac {x^2}{6 (a \\plus{} 2)} \\plus{} \\frac {x^3}{24 (a \\plus{} 3)} \\minus{} \\frac {x^4}{120 (a \\plus{} 4)} \\plus{} \\frac {x^5}{720 (a \\plus{} 5)} \\plus{} O\\left(x^6\\right)\\right)\\]\n[/quote]\r\nthis is not interesting :( .\r\n\r\nproblem is very easy if you know just definition of $ e^{x}$ and something about beta function. \r\n(the second one I just left you as exercises , it easly follows from the solution of the first)\r\n$ \\frac {2e}{\\sqrt {\\pi}}\\int_{0}^{1}e^{ \\minus{} x^{2}}dx \\equal{} \\frac {2}{\\sqrt {\\pi}}\\int_{0}^{1}e^{ \\minus{} x^{2} \\plus{} 1}dx \\equal{} \\frac {2}{\\sqrt {\\pi}}\\int_{0}^{1} \\sum \\frac {(1 \\minus{} x^{2})^{n}}{n!} \\equal{} \\sum \\frac {1}{\\Gamma(n \\plus{} 3/2)}$\r\nso you just need to note that $ \\int_{0}^{1}\\frac {(1 \\minus{} x^{2})^{n}}{n!}dx \\equal{} \\frac {1}{n!2}\\int_{0}^{1}(1 \\minus{} t)^{n}t^{ \\minus{} 1/2} \\equal{} \\frac {\\beta (n \\plus{} 1,1/2)}{n!2} \\equal{} \\frac {\\sqrt {\\pi}}{2\\Gamma(n \\plus{} 3/2)}$" } { "Tag": [], "Problem": "I could not find anything about brute force on the website, but if you can prove that there exist only a certain amount of cases that will work and brute force them all, would you be awarded full points if it is thorough? While I know it definitely wouldn't be elegant, I was wondering if it would net you full points at least.", "Solution_1": "[quote=\"Ignite168\"]I could not find anything about brute force on the website, but if you can prove that there exist only a certain amount of cases that will work and brute force them all, would you be awarded full points if it is thorough? While I know it definitely wouldn't be elegant, I was wondering if it would net you full points at least.[/quote]\r\n\r\nI'm pretty sure thorough brute force will get you full points because I remember I got a 5 for a problem I brute forced. It wasn't commended, but it's still a 5.", "Solution_2": "This begs the question: If you have a brute force solution and a non-brute-force solution that's harder to explain, which should you submit? (I have 2 brute force solutions, a C program, and a real solution to one of the problems and it probably took me more time to figure out how to write the real solution than it did to do the 2 brute force solutions by hand and write and debug the C program) :D\r\n\r\nThis is a serious question, by the way.", "Solution_3": "[quote=\"nr1337\"]This begs the question: If you have a brute force solution and a non-brute-force solution that's harder to explain, which should you submit? (I have 2 brute force solutions, a C program, and a real solution to one of the problems and it probably took me more time to figure out how to write the real solution than it did to do the 2 brute force solutions by hand and write and debug the C program) :D\n\nThis is a serious question, by the way.[/quote]\r\n\r\nThis is also the type of question I believe they don't want people to discuss while the competition is underway (i.e. you should make that decision yourself).\r\n\r\n\r\nBrute force solutions get full credit as long as they don't have any flaws. These types of solutions pretty much have 0 chance of being commended though.", "Solution_4": "[quote=\"joml88\"]\nBrute force solutions get full credit as long as they don't have any flaws. These types of solutions pretty much have 0 chance of being commended though.[/quote]\nWell, if you're also the only one to get the problem right, they'd probably commend you with a brute-force solution :D\n\n[quote=\"nr1337\"]This begs the question: If you have a brute force solution and a non-brute-force solution that's harder to explain, which should you submit? (I have 2 brute force solutions, a C program, and a real solution to one of the problems and it probably took me more time to figure out how to write the real solution than it did to do the 2 brute force solutions by hand and write and debug the C program) :D\n\nThis is a serious question, by the way.[/quote]\r\nJoml's right that this question is getting pretty close to discussing the problems, so be careful. If you're sure that your more elegant solution is correct, then use it, since graders will probably like it more and it could get commended. If you're not sure, then probably for the sake of points sending one you know is right is the better thing to do (don't just send both solutions - it tends to annoy graders if they have to figure out which of your solutions you intend them to look at, and if one is right and one is wrong then it's more difficult to grade)." } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "Suppose that given is a row: $u_{n}=n^{p}x^{n}\\ (p \\in \\&\\mathbb{N}\\& , x>0)$\r\n\r\nfind the limit if $x\\rightarrow \\infty$\r\n\r\nso I think to look to 3 cases first suppose that $x<1 \\ and \\ x>0$ then we get somthing like $\\lim_{x\\rightarrow \\infty}\\frac{n^{p}}{10^{n}}$ but then we have a very large denominator and we get 0 (zero) as the limit.\r\n\r\nsecond we can get the x will bigger than 1 then we get somthing like $n^{p}$ and this will have now limit.\r\n\r\nat last we can have that x=1 and then we have also now limit.\r\n\r\nDoes the argue hold? in orther words is this correct?\r\n\r\nThanks a lot.", "Solution_1": "did you mean the limit of $n$ going to infinity and $p$ is fixed right?", "Solution_2": "yes because it is a row.\r\nThanks.", "Solution_3": "$x$ also is fixed, no? You wrote $x\\to\\infty$.", "Solution_4": "sorry of cours we have $\\lim_{n\\rightarrow \\infty}...$\r\n\r\nx will fixed.\r\n\r\nThanks." } { "Tag": [ "function", "integration", "trigonometry" ], "Problem": "Here are a few problems I seem completely stuck with:\r\n\r\n1) A platform of mass $m_0$ starts moving on a frictionless horizontal surface due to a constant force $\\vec F$ acting on it. We pour sand on this platform with a constant speed $\\mu$ (representing the mass of sand dropped onto the platform in unit time). Find the velocity and the acceleration of the platform as a function of time. Neglect all friction.\r\n\r\n2) Two bodies of masses $m_1$ and $m_2$ are connected by a massless spring and placed on a horizontal surface. The coefficient of friction between the surface and each body is $\\mu$. What is the minimal horizontal force which must act on one body in order for the other body to start moving?\r\n\r\n3) A small ball of mass $m$ is attached to a massless string of length $l$ and is left to hang vertically. At one point, the other end of the string (label it $A$) starts moving horizontally with constant speed. What is the minimal speed end $A$ must have for the ball to start moving on a circle with a center at $A$?\r\n\r\nAny help would be appreciated :)", "Solution_1": "[quote=\"Djole\"]\n1) A platform of mass $m_0$ starts moving on a frictionless horizontal surface due to a constant force $\\vec F$ acting on it. We pour sand on this platform with a constant speed $\\mu$ (representing the mass of sand dropped onto the platform in unit time). Find the velocity and the acceleration of the platform as a function of time. Neglect all friction.[/quote]\r\n\r\n[color=blue] We have the Newton's second law \n\n$\\large d(mv)=Fdt$\n\nConsequently, $\\large \\int d(mv)=\\int Fdt$\n\n$\\large \\Longrightarrow mv=Ft+C$\n\nWhen $t=0, v=0 \\Longrightarrow C=0$\n\n$\\Longrightarrow v=\\frac{Ft}{m}=\\frac{Ft}{m_0+\\mu t}$\n\nAnd $\\large a=\\frac{dv}{dt}$[/color]", "Solution_2": "That looks right. So the platform's final speed will be $\\frac{F}{\\mu}$.\r\nFor the third problem, enter the reference frame of point $A$. The speed of the mass at the bottom of the circle will be the same as the speed of $A$ in the original reference frame. At the top of the circle:\r\n$g = \\frac{v_t^2}{l}$\r\n$v_t = \\sqrt{gl}$\r\nSo by conservation of energy, the speed at the bottom is $v = \\sqrt{5 g l}$.", "Solution_3": "That seems to make sense. Nice solution.", "Solution_4": "Thank you for the solutions! \r\n\r\nFor problem 3, I did enter the reference frame of point $A$; my mistake was that I thought that the speed of the pendulum at the top of its path is zero...\r\n\r\nAny candidates for solving the second problem?", "Solution_5": "Sorry if this is stupid and I'm completely missing something.\r\n\r\nFor 2, I'm assuming $\\mu$ is the coefficient of kinetic friction for $m_1$ and the coefficient of static friction for $m_2$. Arbitrarily place $m_1$ on the left and $m_2$ on the right and have the horizontal force $\\vec F$ be to the right on $m_1$. Let $\\Delta x$ be the compression of the spring. Then, $Fnet,m_1=F-\\mu m_1 g-k \\Delta x=0$ (you want the minimum force so push the $m_1$ at constant velocity), and $Fnet,m_2=k \\Delta x-\\mu m_2 g=0$ (just before moving). Adding the two equations, you get $F-\\mu m_1 g-\\mu m_2 g=0$, so $F=\\mu m_1 g+\\mu m_2 g$.\r\n\r\nActually, I think this works without my assumption. The minimum $\\vec F$ needed would be when $\\Delta x$ is just enough to counteract $\\mu m_2 g$. $\\vec F$ on $m_1$ would also have to counteract $\\mu m_1 g+k \\Delta x$. The acceleration is 0 just before $\\vec F$ is large enough to cause $m_1$ to budge forward a bit to cause the compression to be more than $\\Delta x$.", "Solution_6": "[quote=\"Djole\"]3) A small ball of mass $m$ is attached to a massless string of length $l$ and is left to hang vertically. At one point, the other end of the string (label it $A$) starts moving horizontally with constant speed. What is the minimal speed end $A$ must have for the ball to start moving on a circle with a center at $A$?[/quote]\r\n\r\nConsider the point at rest and the ball starts moving with the speed $v$ in the opposite direction (just a different inertial system). For the ball to reach the top of the circle with zero velocity, the potential energy $2mgl$ at the top must be equal to the kinetic energy $\\frac 1 2 mv^2$ at the bottom, $2mgl = \\frac 1 2 mv^2,\\ v = 2 \\sqrt{gl}.$", "Solution_7": "@[b]yetti[/b]: Yes, that's the incorrect answer I got :) [b]durt[/b]'s solution is correct...\r\n\r\n@[b]AnnieS1527[/b]: That's the answer I first came up with -- needless to say, it's not right and I have no idea why. The right answer is [hide]$F = \\mu g(m_1 + m_2/2)$[/hide]. Can anyone get this answer?", "Solution_8": "Yes it is incorrect, the ball would fall down from the top of the loop. So the centrifugal force at the top must balance the gravitational force.\r\n\r\n$\\frac{mv_T^2}{l} = mg,\\ \\frac 1 2 mv_T^2 = \\frac 1 2 mgl$. If $v$ is the sought after speed at the bottom, then\r\n\r\n$\\frac 1 2 mv^2 = \\frac 1 2 mv_T^2 + 2mgl = \\frac 5 2 mgl$\r\n\r\n$v = \\sqrt{5gl}$", "Solution_9": "You mean $v = \\sqrt{5gl}$, of course ;)", "Solution_10": "[quote=\"Djole\"]2) Two bodies of masses $m_1$ and $m_2$ are connected by a massless spring and placed on a horizontal surface. The coefficient of friction between the surface and each body is $\\mu$. What is the minimal horizontal force which must act on one body in order for the other body to start moving?\n[/quote]\r\n\r\nLet k be the spring constant. Let a horizontal force $F_1 > \\mu m_1 g$ act on body 1, while body 2 is at rest. The equation of motion of body 1 is\r\n\r\n$m_1 \\ddot x_1 = F_1 - \\mu m_1g - k x_1$\r\n\r\n$\\ddot x_1 + \\frac{k}{m_1}\\ x_1 = \\frac{F_1 - \\mu m_1g}{m_1}$\r\n\r\nThe general solution of this differential equation is\r\n\r\n$x_1(t) = A \\sin(\\omega t) + B \\cos (\\omega t) + \\frac{F_1 - \\mu m_1 g}{k}$\r\n\r\nwhere $\\omega = \\sqrt{\\frac{k}{m_1}}$. The initial conditions are $x(0) = 0,\\ \\dot x(0) = 0$, hence $B = -\\frac{F_1 - \\mu m_1 g}{k},\\ A = 0$ and\r\n\r\n$x_1(t) = \\frac{F_1 - \\mu m_1 g}{k}\\ \\left(1 - \\cos (\\omega t)\\right)$\r\n\r\n$\\dot x_1(t) = \\frac{(F_1 - \\mu m_1 g) \\omega}{k}\\ \\sin (\\omega t)$\r\n\r\nBody 1 stops again at $\\tau= \\frac T 2 = \\frac \\pi \\omega$. At this time, its position is\r\n\r\n$x_1(\\tau) = \\frac{F_1 - \\mu m_1 g}{k}\\ \\left(1 - \\cos \\pi \\right) = \\frac{2(F_1 - \\mu m_1 g)}{k}$\r\n\r\nand the tension in the spring is\r\n\r\n$T = kx_1(\\tau) = 2(F_1 - \\mu m_1 g)$\r\n\r\nIf body 2 is not to move, it is necessary\r\n\r\n$T \\le \\mu m_2 g$\r\n\r\n$2(F_1 - \\mu m_1 g) \\le \\mu m_2 g$\r\n\r\n$F_1 \\le \\mu g \\left(m_1 + \\frac{m_2}{2}\\right)$\r\n\r\nConversely, if body 2 is to move, it must be\r\n\r\n$F_1 > \\mu g \\left(m_1 + \\frac{m_2}{2}\\right)$", "Solution_11": "Great! :omighty: Thanks!" } { "Tag": [ "\\/closed" ], "Problem": "I don't understand how to make a poll, when I try it dosen't work.", "Solution_1": "go here. sigh http://www.artofproblemsolving.com/wforum/faq.php#17" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "find in $ \\mathbb{N}^*$\r\n$ x\\plus{}y^2\\equal{}x^2\\plus{}y$", "Solution_1": "[quote=\"mathmen\"]find in $ \\mathbb{N}^*$\n$ x \\plus{} y^2 \\equal{} x^2 \\plus{} y$[/quote] \r\n\r\nthere is no solution beside $ x \\equal{} y$" } { "Tag": [ "function", "trigonometry", "complex analysis", "complex analysis unsolved" ], "Problem": "Prove that for $z\\in\\mathbb{C}\\setminus\\mathbb{Z}$, $f(z)=\\frac{\\pi^{2}}{\\sin^{2}\\pi{z}}$ it holds that $f(z)=\\sum_{n\\in\\mathbb{Z}}\\frac{1}{(z-n)^{2}}$", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=41504" } { "Tag": [ "function", "integration", "logarithms", "linear algebra", "matrix", "calculus", "calculus computations" ], "Problem": "(a) The function $ f$ is continuous on $ [0,1]$ such that $ f(x)>0$. Find an equivalent of:\r\n\r\n$ h(\\epsilon)=\\int_{0}^{1}\\frac{f(x)}{x^{2}+\\epsilon^{2}}dx$ when $ \\epsilon \\rightarrow 0$ and when $ \\epsilon \\rightarrow \\infty$\r\n\r\n(b) Assume $ f$ is differentiable at $ 0$, that $ f(0)=0$, and $ f'(0) \\neq 0$. Find an equivalent of $ h(\\epsilon)$ when $ \\epsilon \\rightarrow 0$.", "Solution_1": "(a) $ f$ reaches a positive minimum $ m$ and maximum $ M$ on the compact interval $ [0,1]$ . \r\n$ h(\\epsilon)\\ge \\int_{0}^{1}\\frac{m}{x^{2}+\\epsilon^{2}}=\\frac{m}{\\epsilon}\\arctan \\frac{1}{\\epsilon}\\to \\infty$ for $ \\epsilon\\to 0$ and $ h(\\epsilon)\\le \\int_{0}^{1}\\frac{M}{x^{2}+\\epsilon^{2}}=\\frac{M}{\\epsilon}\\arctan \\frac{1}{\\epsilon}\\to 0$ for $ \\epsilon\\to \\infty$ .\r\n\r\n(b) Is for example $ f(x)=2x$ so $ h(\\epsilon)=\\int_{0}^{1}\\frac{2x}{x^{2}+\\epsilon^{2}}dx=\\log\\left(1+\\frac{1}{\\epsilon^{2}}\\right)\\to \\infty$ for $ \\epsilon\\to\\infty$ .\r\nIs for another example $ f(x)=\\left\\{ \\begin{matrix}x^{2}& \\text{for}& x\\in\\Bbb{Q}\\\\ 0 & \\text{for}& x\\notin\\Bbb{Q}\\end{matrix}\\right.$ so $ h(\\epsilon)=0 \\; \\forall \\epsilon\\neq 0$ and therefore $ h(\\epsilon)\\to 0$ for $ \\epsilon\\to 0$ ." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c>0$ and different. Prove that\r\n\r\n$ \\sum\\frac{a(b\\plus{}c)(b^2\\plus{}c^2)}{(b^2\\minus{}c^2)}\\geq\\frac{7}{4}$\r\n\r\n______________________\r\n :blush: Panagiote Ligouras", "Solution_1": "[quote=\"Ligouras\"]Let $ a,b,c > 0$ and different. Prove that\n\n$ \\sum\\frac {a(b \\plus{} c)(b^2 \\plus{} c^2)}{(b^2 \\minus{} c^2)}\\geq\\frac {7}{4}$\n\n______________________\n :blush: Panagiote Ligouras[/quote]\r\nHmm.. If $ f(a,b,c) \\equal{} LHS$ then $ f(a,b,c) \\plus{} f(a,c,b) \\equal{} 0$, and hence it cannot be true. Besides $ f(ta,tb,tc) \\equal{} t^2f(a,b,c)$. So it is not true because it is not homogenous. Maybe you meant this?\r\n\r\n$ \\left |\\sum_{cyc} \\frac {a(b \\plus{} c)(b^2 \\plus{} c^2)}{(b^4 \\minus{} c^4)} \\right | \\geq\\frac {7}{4}$\r\n\r\n(Btw. this is equivalent to: $ \\left | \\sum_{cyc} \\frac{a}{b\\minus{}c} \\right | \\ge \\frac{7}{4}$)", "Solution_2": "[quote=\"Mathias_DK\"][quote=\"Ligouras\"]Let $ a,b,c > 0$ and different. Prove that\n\n$ \\sum\\frac {a(b \\plus{} c)(b^2 \\plus{} c^2)}{(b^2 \\minus{} c^2)}\\geq\\frac {7}{4}$\n\n______________________\n :blush: Panagiote Ligouras[/quote]\nHmm.. If $ f(a,b,c) \\equal{} LHS$ then $ f(a,b,c) \\plus{} f(a,c,b) \\equal{} 0$, and hence it cannot be true. Besides $ f(ta,tb,tc) \\equal{} t^2f(a,b,c)$. So it is not true because it is not homogenous. Maybe you meant this?\n\n$ \\left |\\sum_{cyc} \\frac {a(b \\plus{} c)(b^2 \\plus{} c^2)}{(b^4 \\minus{} c^4)} \\right | \\geq\\frac {7}{4}$\n\n(Btw. this is equivalent to: $ \\left | \\sum_{cyc} \\frac {a}{b \\minus{} c} \\right | \\ge \\frac {7}{4}$)[/quote]\r\n\r\nDear mathias is correct:\r\n\r\n$ \\sum\\frac {a(b \\plus{} c)(b^2 \\plus{} c^2)}{(b^2 \\minus{} c^2)^2}\\geq\\frac {7}{8}$", "Solution_3": "[quote=\"Ligouras\"][quote=\"Mathias_DK\"][quote=\"Ligouras\"]Let $ a,b,c > 0$ and different. Prove that\n\n$ \\sum\\frac {a(b \\plus{} c)(b^2 \\plus{} c^2)}{(b^2 \\minus{} c^2)}\\geq\\frac {7}{4}$\n\n______________________\n :blush: Panagiote Ligouras[/quote]\nHmm.. If $ f(a,b,c) \\equal{} LHS$ then $ f(a,b,c) \\plus{} f(a,c,b) \\equal{} 0$, and hence it cannot be true. Besides $ f(ta,tb,tc) \\equal{} t^2f(a,b,c)$. So it is not true because it is not homogenous. Maybe you meant this?\n\n$ \\left |\\sum_{cyc} \\frac {a(b \\plus{} c)(b^2 \\plus{} c^2)}{(b^4 \\minus{} c^4)} \\right | \\geq\\frac {7}{4}$\n\n(Btw. this is equivalent to: $ \\left | \\sum_{cyc} \\frac {a}{b \\minus{} c} \\right | \\ge \\frac {7}{4}$)[/quote]\n\nDear mathias is correct:\n\n$ \\sum\\frac {a(b \\plus{} c)(b^2 \\plus{} c^2)}{(b^2 \\minus{} c^2)^2}\\geq\\frac {7}{8}$[/quote]\r\nLet $ f(a,b,c) \\equal{} \\sum_{cyc} \\frac {a(b^2 \\plus{} c^2)}{(b \\minus{} c)^2(b \\plus{} c)} \\equal{} LHS$. It's easy to prove $ f(a,b,c) \\le f(a \\plus{} k,b \\plus{} k,c \\plus{} k)$ when $ a,b,c \\ge 0$. Wlog $ c \\le a,b$. Hence $ f(a \\minus{} c,b \\minus{} c,0) \\le f(a,b,c)$. Setting $ x \\equal{} a \\minus{} c, y \\equal{} b \\minus{} c$, we have:\r\n$ f(x,y,0) \\equal{} \\frac {x}{y} \\plus{} \\frac {y}{x} \\ge 2 > \\frac {7}{4} > \\frac{7}{8}$. :D", "Solution_4": "I had had a similar result:\r\nGiven $ a, b, c$ are the reals. Prove that:\r\n$ \\frac{(a\\plus{}b)^2(a\\plus{}c)^2}{(b^2\\minus{}c^2)^2}\\plus{}\\frac{(b\\plus{}c)^2(b\\plus{}c)^2}{(c^2\\minus{}a^2)^2}\\plus{}\\frac{(c\\plus{}a)^2(c\\plus{}b)^2}{(a^2\\minus{}b^2)^2} \\geq\\ 2$\r\n :)", "Solution_5": "[quote=\"nguoivn\"]I had had a similar result:\nGiven $ a, b, c$ are the reals. Prove that:\n$ \\frac {(a \\plus{} b)^2(a \\plus{} c)^2}{(b^2 \\minus{} c^2)^2} \\plus{} \\frac {(b \\plus{} c)^2(b \\plus{} c)^2}{(c^2 \\minus{} a^2)^2} \\plus{} \\frac {(c \\plus{} a)^2(c \\plus{} b)^2}{(a^2 \\minus{} b^2)^2} \\geq\\ 2$\n :)[/quote]\r\n\r\nit was my real problem NGUOIVN.... :wink: \r\n\r\naccording to you it exists k >2 such that\r\n\r\n$ \\sum\\frac {a(b \\plus{} c)(b^{2} \\plus{} c^{2})}{(b^{2} \\minus{} c^{2})^2}\\geq\\ k$\r\n\r\nk=2 is sure\r\n\r\nP. Ligouras", "Solution_6": "[quote=\"Mathias_DK\"][/quote][quote=\"Ligouras\"][quote=\"Mathias_DK\"][quote=\"Ligouras\"]Let $ a,b,c > 0$ and different. Prove that\n\n$ \\sum\\frac {a(b + c)(b^2 + c^2)}{(b^2 - c^2)}\\geq\\frac {7}{4}$\n\n______________________\n :blush: Panagiote Ligouras[/quote]\nHmm.. If $ f(a,b,c) = LHS$ then $ f(a,b,c) + f(a,c,b) = 0$, and hence it cannot be true. Besides $ f(ta,tb,tc) = t^2f(a,b,c)$. So it is not true because it is not homogenous. Maybe you meant this?\n\n$ \\left |\\sum_{cyc} \\frac {a(b + c)(b^2 + c^2)}{(b^4 - c^4)} \\right | \\geq\\frac {7}{4}$\n\n(Btw. this is equivalent to: $ \\left | \\sum_{cyc} \\frac {a}{b - c} \\right | \\ge \\frac {7}{4}$)[/quote]\n\nDear mathias is correct:\n\n$ \\sum\\frac {a(b + c)(b^2 + c^2)}{(b^2 - c^2)^2}\\geq\\frac {7}{8}$[/quote][quote=\"Mathias_DK\"]\nLet $ f(a,b,c) = \\sum_{cyc} \\frac {a(b^2 + c^2)}{(b - c)^2(b + c)} = LHS$. It's easy to prove $ f(a,b,c) \\le f(a + k,b + k,c + k)$ when $ a,b,c \\ge 0$. Wlog $ c \\le a,b$. Hence $ f(a - c,b - c,0) \\le f(a,b,c)$. Setting $ x = a - c, y = b - c$, we have:\n$ f(x,y,0) = \\frac {x}{y} + \\frac {y}{x} \\ge 2 > \\frac {7}{4} > \\frac {7}{8}$. :D[/quote]\r\n\r\nThank you Mathias very easy" } { "Tag": [], "Problem": "Treething had posted this:\r\n\r\nBob the Builder had a bag of dozen bananas on sale for 3 dollars. He adds some bananas to the bag, but does not change the price tag. Now how much does a dozen bananas cost?\r\n\r\nhttp://media.putfile.com/banana39\r\n\r\nbut the page got all messed up, so I've reposted it.", "Solution_1": "[quote=\"solafidefarms\"]Treething had posted this:\n\nBob the Builder had a bag of dozen bananas on sale for 3 dollars. He adds some bananas to the bag, but does not change the price tag. Now how much does a dozen bananas cost?\n\nhttp://media.putfile.com/banana39\n\nbut the page got all messed up, so I've reposted it.[/quote]\r\n\r\n The wording is a bit vague but I'll give it a shot:\r\n\r\n[hide=\"Answer\"]\n\n Suppose bob adds $x$ bananas.\n\n Now the price of a dozen bananas is:\n\n$\\frac{12}{12+x} \\cdot 3$ or $\\frac{36}{12+x}$ dollars. [/hide]", "Solution_2": "can the answer b just a number?, cuz it doesnt ask us the equation, it asks us the cost. if it asks us the [i]actual cost[/i], then $not$ $enough$ $information$.", "Solution_3": "oops, i was wrong, this prob is solvable. o yeah, sry 2 double post, but i forgot 2 add mi answer.\r\n[hide=\"solution\"]$\\frac{12}{x+12}$=$\\frac{1}{3}$\n X+12=36.\n x=24 for regular price. $\\frac{24}{36}$ =$\\frac{2}{3}$\n \n therefore 1/3 of 3 is 1, which equals a dozen bananas. a dozen bananas without a sale is therefore 1 dollar.[/hide]", "Solution_4": "He does not specify how many banana's were added, assuimg they are all sold at a uniform price.", "Solution_5": "it costs $\\frac{36}{12+b}$ if b is the number of bananas he added. If my mom met Bob, she would buy all the bananas, they are pretty cheap :D" } { "Tag": [ "induction", "number theory solved", "number theory" ], "Problem": "I found this nice-looking problem:\r\n\r\nProve that there are infinitely many composite numbers n s.t. n|3^(n-1)-2^(n-1). I've found that n=3^p-2^p with p prime verifies this, but I don't know if it can be proven that there are infinitely many composite numbers of the anbove form. Any other ideas? I think this is the wrong way to go...", "Solution_1": "I had an idea for this problem when I saw this one : \r\n\r\n(*) 2^(n+2) | 3^(2^n) - 1 (easy to prove by induction)\r\n\r\n\r\nThen if N = 3^(2^n) - 2^(2^n), let's prove that N | 3^(N-1) - 2^(N-1). \r\n\r\nSince a-b|a^k-b^k it's enough to prove that N-1 is divisible by 2^n and that's where we use (*) :D" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "The convex quadrilateral $ ABCD$ is inscribed in a circle of center $ O$ and let $ E=AC\\cap BD$. Prove that if the midpoints of $ AD,BC$ and $ OE$ are collinear, then either $ AB=CD$ or $ \\angle AEB = 90^{\\circ}$", "Solution_1": "[hide=\"Hint.\"]In any convex quadrilateral $ ABCD$ the following theorem is true:\nIf $ E$ is an intersection of diagonals and $ P$ is an arbitrary point inside $ ABCD$ then midpoints of $ AD, BC, EP$ are collinear if and only if $ [ABP]=[CDP]$ (areas). With this theorem the problem becomes trivial.\n[hide=\"Hint to the hint.\"]To prove this theorem, show at first that locus of points $ P$ such that $ [ABP]=[CDP]$ is a line through $ S$, where $ S$ is an intersection of lines $ AB$ and $ CD$.[/hide]\n[/hide]" } { "Tag": [ "trigonometry" ], "Problem": "Prove that\r\n\\[a+b+c\\geq a\\cdot\\frac{\\sin{b}}{\\sin{a}}+b\\cdot\\frac{\\sin{c}}{\\sin{b}}+c\\cdot\\frac{\\sin{a}}{\\sin{c}}\\;\\forall a,b,c\\in (0,\\frac{\\pi}{2}). \\]", "Solution_1": "Equiv to \\[\\sum a\\left(\\frac{\\sin a-\\sin b}{\\sin a}\\right)\\geq 0 \\] Now on that interval sine is increasing so it follows.\r\nMore precisely, let $S_{a}=\\frac{\\sin a-\\sin b}{\\sin a\\cdot(a-b)}$ and cyclics then we must prove \\[\\sum a(a-b)S_{a}\\geq 0\\] moreover since its increasing, positive we must have $S_{a},S_{b},S_{c}\\geq 0$", "Solution_2": "If $S_{a},S_{b},S_{c}\\geq 0$ then $\\sum \\geq 0$. Why? :maybe:", "Solution_3": "i think that\r\nWhen $a\\ge b \\, then sina\\ge sinb\\, a,b\\in(0;\\pi/2)$\r\nSo $\\sum\\ge0$" } { "Tag": [ "trigonometry", "calculus", "calculus computations" ], "Problem": "i am not able to draw graph of sin[x] from sinx...where [x] represents greatest integer function....can anyone help me :(", "Solution_1": "Just split up the $ x$ axis. The curve is equivalent to $ \\sin1$ between $ 0$ and $ 1$, $ \\sin2$ between $ 1$ and $ 2$ etc. You could make rough guesses as to how large $ \\sin(n)$ is (for integer $ n$) by comparing $ n$ to nearby multiples of $ \\pi$ but it doesn't seem worthwile, just evaluate them with a calculator.", "Solution_2": "Sorry for my reply, seriousstone89. I also find the graph of $ y=\\sin [x]$ by substituting $ x=$ some numbers, such as \r\nFor $ 0\\leq x<1\\Longrightarrow \\sin [x]=0$, $ 1\\leq x<2\\Longrightarrow \\sin [x]=\\sin1$, and so on.\r\n\r\nkunny", "Solution_3": "one general method would be to draw the graph of $ f(x)$ and then draw vertical lines at distance$ 1,2,3 \\dots$ from the origin .wherever it cuts the axis from their draw horizontal lines till the next point where the graph is cut by those vertical lines(may be above or below) of course excluding the point above or below that point.these lines are the graph of $ f([x])$\r\n. well this is just a restatement and generalization of the fact that $ f([x])=f(n)$ for all $ n \\leq x < n+1$ where $ n$ is an integer." } { "Tag": [ "modular arithmetic", "number theory", "prime numbers", "divisibility tests", "number theory unsolved" ], "Problem": "Find all the triplets $ \\{p, q, r\\}$ of prime numbers such that $ pq\\plus{}r$, $ qr\\plus{}p$, and $ rp\\plus{}q$ are all prime numbers also.", "Solution_1": "[quote=\"azjps\"]If all $ p,q,r$ are odd, then the resulting numbers are even, so one number must be $ 2$. Checking $ \\mod{3}$, $ \\{p,q,r\\} \\equiv \\{ \\minus{} 1,1,1\\},\\{ \\minus{} 1, \\minus{} 1,1\\},\\{ \\minus{} 1, \\minus{} 1, \\minus{} 1\\} \\pmod{3}$ all result in a resulting number being divisible $ 3$, so one of the numbers must be $ 3$. So we need all primes $ p$ such that $ p \\plus{} 6, 3p \\plus{} 2,2p \\plus{} 3$ are primes, which can be filtered using various modulos, but there does not seem to be a small solution set (for example, $ \\{2,3,1063\\}$ works).[/quote]\r\n\r\n\r\nYes, I had found also that two of p, q, r must be 2 and 3. I suppose that there must exist infinite p such that $ p \\plus{} 6, 3p \\plus{} 2,2p \\plus{} 3$ are primes, but I don't know how to prove it u_u ...", "Solution_2": "If all $ p,q,r$ are odd, then the resulting numbers are even, so one number must be $ 2$. There isn't much more we can do besides restrict the other values using various modulos. For example, $ \\{2,5,13\\}$ and $ \\{2,3,126013\\}$ work. \r\n\r\nEdit: I messed up; divisibility by $ 3$ is not necessary for $ \\{p,q,r\\} \\equiv \\{\\minus{}1,\\minus{}1,1\\} \\pmod{3}$.", "Solution_3": "What is the source of this problem? It seems likely to be very hard unless I am missing a trick.", "Solution_4": "I dunno u_u... I thinked in the problem and I believed that it was easy u_u ... well, maybe it is, but I don't know how to prove that are infinite triplets =S (OMG! =O)", "Solution_5": "Suppose $ p,q,r>2$\r\nthen, clearly all three expression are even and greater than 2, thus not prime.\r\n\r\nThus wlog $ p\\equal{}2$.\r\n\r\n$ 2q\\plus{}r$ is prime, so $ r>2$\r\n$ 2r\\plus{}q$ is prime, so $ q>2$\r\n\r\nnotice $ (2,3,5)$ $ (2,3,7)$ $ (2,5,7)$ $ (2,3,13)$ $ (2,3,17)$ works\r\n\r\nDoes this problem have an elementary solution? Or even any solution? What is the problem source?" } { "Tag": [ "AMC", "AMC 10", "algebra", "difference of squares", "special factorizations" ], "Problem": "How many ordered pairs $ (m,n)$ of positive integers, with $ m > n$, have the property that their squares differ by $ 96$?\r\n\r\n$ \\textbf{(A)}\\ 3 \\qquad \\textbf{(B)}\\ 4 \\qquad \\textbf{(C)}\\ 6 \\qquad \\textbf{(D)}\\ 9 \\qquad \\textbf{(E)}\\ 12$", "Solution_1": "[hide]\n$(a+b)(a-b) = 96 = 2^{5}\\cdot 3$, which has $(6)(2) = 12$ factors. We know that $a+b>a-b$ since $b\\ne 0$. We need both factors to have the same parity so that $a = \\frac{(a+b)+(a-b)}{2}$ is an integer. There would be 6 pairs, but 1 and 3 can't be paired with anything, so there are only 4. $\\boxed{\\text{B}}$\n[/hide]", "Solution_2": "[hide]We are to find ordered pairs of positive integers $(m,\\ n)$ such that $m^{2}-n^{2}=96\\Longleftrightarrow (m+n)(m-n)=96.$ By $m>n>0$ and $m,\\ n\\in \\mathbb{N},$ and $m=\\frac{m+n}{2}+\\frac{m-n}{2},\\ n=\\frac{m+n}{2}-\\frac{m-n}{2},$ we have that parity of $m+n,\\ m-n$ is coincides with that of $m,\\ n.$ Thus $(m+n,\\\u3000m-n)=(48,\\ 2),\\ (24,\\ 4),\\ (16,\\ 6),\\ (12,\\ 8).$ The desired answer is 4 pairs, $(B).$[/hide]", "Solution_3": "[hide=\"a 5th grade method\"]Graph $y=\\sqrt{x^{2}+94}$ on your graphing calc and count the number of integer solutions :wink: [/hide]", "Solution_4": "That's not 5th grade, more like 8th as well as basically impossible.\r\n\r\n[hide=\"An 8th grade method that's actually possible\"]\nFactor into $(m+n)(m-n)=96$ $m+n$ must be bigger than $m-n$ and list out your pairs of factors. $(1,96), (2,48), (3,32) (4,24) (6,16) (8,12)$. So basically, you have 6 systems of equations and you want ones with integer solutions. I solved the systems of equations using augmented matrices on my calc.[/hide]", "Solution_5": "[quote=\"Ignite168\"]That's not 5th grade, more like 8th as well as basically impossible.\n\n[hide=\"An 8th grade method that's actually possible\"]\nFactor into $(m+n)(m-n)=96$ $m+n$ must be bigger than $m-n$ and list out your pairs of factors. $(1,96), (2,48), (3,32) (4,24) (6,16) (8,12)$. So basically, you have 6 systems of equations and you want ones with integer solutions. I solved the systems of equations using augmented matrices on my calc.[/hide][/quote]\r\nOr you could just add them...\r\nI think he meant graph $y=\\sqrt{96+x^{2}}$", "Solution_6": "[hide]\n(m+n)(m-n)=96 \\implies (m+n/2)(m-n/2)=24, with the first term greater, and 24 is just \n\n1\n2\n3\n4\n6 <-- (then starts repeating factor pairs)\nSo 4 factors.\n\nThat was my exact logic on the test...\n[/hide]", "Solution_7": "[hide]This can be done by the difference of two squares\n0 \n1 - 1-0= 1\n4 - 4-1=3\n9 - 9-4=5\n16 - 16-9=7\n25 ....... \n36 .......\n49 .......\n64 ........\n\nWe see that these are consecutive odd integers 1,3,5,7,9...........\nAnd 96 is a even integers --- o + o = e\nso just try the numbers using these conditions[/hide]", "Solution_8": "[quote=\"King_James23\"]This can be done by the difference of two squares\n0 \n1 - 1-0= 1\n4 - 4-1=3\n9 - 9-4=5\n16 - 16-9=7\n25 ....... \n36 .......\n49 .......\n64 ........\n\nWe see that these are consecutive odd integers 1,3,5,7,9...........\nAnd 96 is a even integers --- o + o = e\nso just try the numbers using these conditions[/quote]\r\nThis method is vague and cumbersome... the list have to be extended up to 25 odd integers... like $25^{2}-23^{2}=96$...", "Solution_9": "[hide]m^2-n^2=96\nFactor and get (m+n)(m-n)=96\n96 = 2^5 x 3\n6(2) = 12, there are 12 factors for 96, and 6 pairs. \nNote that m-n is the lower value of the two; m+n and m-n.\nm-n cannot yield 1 therefore lowering the possible amount of pairs by 1. \nNow, take into consideration of the 3 in the factors of 96.\nFrom the 3 you can deduce that m+n and m-n has to be all odd or all even.\nOdd and odd yields even, even and even yields even, even and odd yields odd, odd subtracted from a higher odd yields even, even subtracted from a higher even yields even, and odd subtracted from a higher even yields odd and vice versa. \nSo lets say m-n denotes 3, that can only be true if a lower even or a lower odd is subtracted from a higher odd or a higher even..\nSince m-n denotes 3, m+n denotes 32, an even number. However, a even added to a odd ( or vice versa ) cannot yield a even number. EX. 5+8 = 13.\nTherefore the factor pairs 1,96 and 3,32 are not accounted for and lowers the total pairs by 2 so there are 4 pairs.[/hide]", "Solution_10": "Wait, you're allowed to use a calc on AMC10?", "Solution_11": "In 2007, calculators were allowed. However, starting 2008, calculators were banned. No AMC problems require a calculator.", "Solution_12": "[hide=\"Bashy method\"]We are given that $m^2-n^2=96$. We factor this as difference of squares, giving $(m+n)(m-n)=96$\n\nThe first term is larger than the second, so we list out pairs of numbers whose product is $96$ and where the first factor is larger than the second, We get:\n\n$(96,1),\n(48,2),\n(32,3),\n(24,4),\n(16,6),\n(12,8)$\n\nThere are not too many values, so we just bash them out and find that $(48,2),(24,4),(16,6),$ and $(12,8)$ have positive integer values for $m$ and $n$. This means our answer is $\\boxed{\\textbf{(B)}\\ 4}$[/hide]", "Solution_13": "(m+n) (m-n)=96\nNow factorize and do casework", "Solution_14": "please don't bump really old threads!\n\nthanks", "Solution_15": "extremely misplaced even for 2007", "Solution_16": "[quote=ostriches88]please don't bump really old threads!\n\nthanks[/quote]\n\nWell, it is a C&P post, so it is okay as per the rules of the forum" } { "Tag": [], "Problem": "A Senate committee has 8 Republicans and 6 Democrats. In how many ways can we form a subcommittee of 5 members that has at least one member from each party?", "Solution_1": "it seems like the problem suggest that every republican is the same....then the answer would be 4", "Solution_2": "I got something different...\r\n\r\nFor committees with no restrictions, you would simply have $ \\binom{14}{5}$. However, there are $ \\binom{8}{5}$ ways to make the committee all Republican and $ \\binom{6}{5}$ ways to make the committee all Democrat.\r\n\r\nThus, the answer is $ \\binom{14}{5} \\minus{} \\binom{8}{5} \\minus{} \\binom{6}{5} \\rightarrow 2002 \\minus{} 56 \\minus{} 6 \\rightarrow 2002 \\minus{} 62 \\rightarrow \\boxed{1940}$.", "Solution_3": "yes i know that is the real answer", "Solution_4": "[quote=\"mathemonster\"]it seems like the problem suggest that every republican is the same[/quote]\r\n\r\nThen, they would be all twins. :)", "Solution_5": "Eh Bijection's just meddling with the problem", "Solution_6": "8 ways to choose a republican, 6 for a democratic, and then (12 choose 5) to get the rest, and that multiplies out to 10560.", "Solution_7": "[quote=\"ernie\"]I got something different...\n\nFor committees with no restrictions, you would simply have $ \\binom{14}{5}$. However, there are $ \\binom{8}{5}$ ways to make the committee all Republican and $ \\binom{6}{5}$ ways to make the committee all Democrat.\n\nThus, the answer is $ \\binom{14}{5} \\minus{} \\binom{8}{5} \\minus{} \\binom{6}{5} \\rightarrow 2002 \\minus{} 56 \\minus{} 6 \\rightarrow 2002 \\minus{} 62 \\rightarrow \\boxed{1940}$.[/quote]\n\n\nI am pretty sure this is the correct way.\n\nJust a suggestion, you might lock this.", "Solution_8": "[quote=\"fmasroor\"]8 ways to choose a republican, 6 for a democratic, and then (12 choose 3) to get the rest, and that multiplies out to 10560.[/quote]\n\nWhy doesn't this work?", "Solution_9": "There are a total of $\\binom{14}{5}=2002$ ways of selecting a subcommittee of 5 with no restrictions on the membership. Of these committees, the only ones that will violate the given condition are the ones that consist entirely of Republicans or entirely of Democrats. There are $\\binom{8}{5}=56$ possible subcommittees that have all 5 members selected from among the 8 Republicans and $\\binom{6}{5}=6$ possible subcommittees that have all 5 members selected from among the 6 Democrats. Subtracting the number of subcommittees that don't work from the total number of possible subcommittees gives us our answer: $2002-56-6=\\boxed{1940}$.", "Solution_10": "[hide=\"Why can't you do:\"]\n6 Democrats * 8 Republicans *12 * 11 * 10 =63360\n12*11*10 represents the rest of the members. [/hide]", "Solution_11": "Order doesn't matter", "Solution_12": "Can someone help?\n\nWhat did I do wrong?:\n\nYou must have 2 already picked out, reserved for the D's and R's. That's $8\\cdot 6$ for the choices we have to do that.\n\nThen we have 3 spots in which we pick whatever members we'd like. 7 R's left and 5 D's left, making a total of 12 to pick from. Pick 3, and $\\binom{12}{3}=220$. Multiply it all together to get $\\boxed{10560}$. What's wrong here? :o", "Solution_13": "You are overcounting. It would work if you had a \"head\" Democrat and a \"head\" Republican. Do you see what I mean?" } { "Tag": [ "trigonometry" ], "Problem": "I have a problem now.\r\nImagine a wall which is decreasing with a relativistic velocity $\\upsilon_w=\\beta c$. Light ray falls on this wall at an angle $\\theta$ to the horizontal. Find the shadow speed.", "Solution_1": "I'll try:\r\n\r\nLet's introduce a cartesian system with the origin at the bottom of the wall, and the current position of the top of the wall at $(0,\\ y)$. Suppose that we are in an innertial system in which the bottom of the wall doesn't move. \r\n\r\nLet $\\tau_0 = 0$ be the moment when a ray of light starts from the top of the wall. It will reach the floor at the moment $\\tau_1 = y/c\\sin\\theta$.\r\n\r\nNow consider another ray which started from the top of the wall at time $\\tau'_0 = \\Delta t$, when the height of the wall was $y - \\beta c\\Delta t$. This ray will reach the floor at the moment $\\tau'_1 = (y-\\beta c\\Delta t)/c\\sin\\theta$, so the shadow traversed a distance $v\\Delta t'$, where $v$ is the speed of the shadow and $\\Delta t' = \\tau'_0 - \\tau_0 + \\tau'_1 - \\tau_1 = \\Delta t \\frac{\\sin\\theta - \\beta}{\\sin\\theta}$. We also have the equality \\[ \\frac{y-\\beta c \\Delta t}{x - v\\Delta t'} = \\frac{y}{x}(= \\tan\\theta) \\Rightarrow \\beta c\\Delta t = v \\tan\\theta \\Delta t' \\Rightarrow v = \\frac{\\cos\\theta}{\\sin\\theta - \\beta}\\beta c. \\]\r\n\r\nThis reduces to the expected result $v = \\cot\\theta \\beta c$ when $\\beta \\rightarrow 0$." } { "Tag": [ "inequalities", "search" ], "Problem": "I have some interesting inequality, let's discuss them :P :) \r\n1. Let $ a, b, c, x, y, z > 0$ and\r\n$ a \\plus{} x \\equal{} b \\plus{} y \\equal{} c \\plus{} z \\equal{} 1$\r\nProve that $ (abc \\plus{} xyz)(\\frac {1}{ay} \\plus{} \\frac {1}{bz} \\plus{} \\frac {1}{cx}) \\geq 3$\r\n2. Let $ a,b,c,d > 0$\r\nprove that: $ \\frac {a \\minus{} b}{b \\plus{} c} \\plus{} \\frac {b \\minus{} c}{c \\plus{} d} \\plus{} \\frac {c \\minus{} d}{a \\plus{} d} \\plus{} \\frac {d \\minus{} a}{a \\plus{} b}\\geq0$\r\n3. Let A=$ \\frac {a^2}{a^2 \\plus{} (b \\plus{} c)^2} \\plus{} \\frac {b^2}{b^2 \\plus{} (c \\plus{} a)^2} \\plus{} \\frac {c^2}{c^2 \\plus{} (a \\plus{} b)^2}$ \r\n$ a,b,c \\in R$\\{0}\r\nminA=?", "Solution_1": "[quote=\"marsupilami\"]\n3. Let A=$ \\frac {a^2}{a^2 \\plus{} (b \\plus{} c)^2} \\plus{} \\frac {b^2}{b^2 \\plus{} (c \\plus{} a)^2} \\plus{} \\frac {c^2}{c^2 \\plus{} (a \\plus{} b)^2}$ \n$ a,b,c \\in R$\\{0}\nminA=?[/quote]\r\nI'll prove that $ \\sum_{cyc}\\frac {a^2}{a^2 \\plus{} (b \\plus{} c)^2}\\geq\\frac {3}{5}.$\r\n$ \\sum_{cyc}\\frac {a^2}{a^2 \\plus{} (b \\plus{} c)^2} \\equal{} \\sum_{cyc}\\frac {a^4}{a^4 \\plus{} a^2b^2 \\plus{} a^2c^2 \\plus{} 2a^2bc}\\geq\\frac {(a^2 \\plus{} b^2 \\plus{} c^2)^2}{\\sum(a^4 \\plus{} 2a^2b^2 \\plus{} 2a^2bc)}.$\r\nId est, it remains to prove that $ \\frac {(a^2 \\plus{} b^2 \\plus{} c^2)^2}{\\sum(a^4 \\plus{} 2a^2b^2 \\plus{} 2a^2bc)}\\geq\\frac {3}{5}.$ But we obtain:\r\n$ \\frac {(a^2 \\plus{} b^2 \\plus{} c^2)^2}{\\sum(a^4 \\plus{} a^2b^2 \\plus{} 2a^2bc)}\\geq\\frac {3}{5}\\Leftrightarrow\\sum_{cyc}(2a^4 \\plus{} 4a^2b^2 \\minus{} 6a^2bc)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a^2 \\minus{} b^2)^2 \\plus{} 3\\sum_{cyc}c^2(a \\minus{} b)^2\\geq0.$ \r\n$ a\\equal{}b\\equal{}c$ gives equality. \r\nThus, $ \\min_{\\{a,b,c\\}\\subset\\mathbb R\\setminus\\{0\\}}A \\equal{} \\frac {3}{5}.$ :)", "Solution_2": "[quote=\"marsupilami\"]I have some interesting inequality, let's discuss them :P :) \n\n2. Let $ a,b,c,d > 0$\nprove that: $ \\frac {a \\minus{} b}{b \\plus{} c} \\plus{} \\frac {b \\minus{} c}{c \\plus{} d} \\plus{} \\frac {c \\minus{} d}{a \\plus{} d} \\plus{} \\frac {d \\minus{} a}{a \\plus{} b}\\geq0$\n\n[/quote]\r\n\r\n$ \\frac {a \\minus{} b}{b \\plus{} c} \\plus{} \\frac {b \\minus{} c}{c \\plus{} d} \\plus{} \\frac {c \\minus{} d}{a \\plus{} d} \\plus{} \\frac {d \\minus{} a}{a \\plus{} b}\\equal{}\\frac{a\\plus{}c}{b\\plus{}c}\\plus{}\\frac{b\\plus{}d}{c\\plus{}d}\\plus{}\\frac{c\\plus{}a}{a\\plus{}d}\\plus{}\\frac{d\\plus{}b}{a\\plus{}b}\\minus{}4\\equal{}(a\\plus{}c)(\\frac{1}{b\\plus{}c}\\plus{}\\frac{1}{a\\plus{}d})\\plus{}(b\\plus{}d)(\\frac{1}{c\\plus{}d}\\plus{}\\frac{1}{a\\plus{}b})\\minus{}4\\equal{}\\frac{(a\\plus{}c)(a\\plus{}b\\plus{}c\\plus{}d)}{(b\\plus{}c)(a\\plus{}d)}\\plus{}\\frac{(b\\plus{}d)(a\\plus{}b\\plus{}c\\plus{}d)}{(c\\plus{}d)(a\\plus{}b)}\\minus{}4\\geq\\frac{(a\\plus{}c)(a\\plus{}b\\plus{}c\\plus{}d)}{(\\frac{b\\plus{}c\\plus{}a\\plus{}d}{2})^{2}}\\plus{}\\frac{(b\\plus{}d)(a\\plus{}b\\plus{}c\\plus{}d)}{(\\frac{c\\plus{}d\\plus{}a\\plus{}b}{2})^{2}}\\minus{}4\\equal{}\\frac{4(a\\plus{}c)}{a\\plus{}b\\plus{}c\\plus{}d}\\plus{}\\frac{4(b\\plus{}d)}{a\\plus{}b\\plus{}c\\plus{}d}\\equal{}\\frac{4(a\\plus{}b\\plus{}c\\plus{}d)}{a\\plus{}b\\plus{}c\\plus{}d}\\minus{}4\\equal{}4\\minus{}4\\equal{}0$\r\n\r\n$ a\\equal{}b\\equal{}c\\equal{}d$", "Solution_3": "[quote=\"marsupilami\"]3. Let A=$ \\frac {a^2}{a^2 \\plus{} (b \\plus{} c)^2} \\plus{} \\frac {b^2}{b^2 \\plus{} (c \\plus{} a)^2} \\plus{} \\frac {c^2}{c^2 \\plus{} (a \\plus{} b)^2}$ \n$ a,b,c \\in R$\\{0}\nminA=?[/quote]\r\n\r\n\"Isolated fudging\" also works nicely here.\r\n\r\nWOLOG, assume that $ a \\plus{} b \\plus{} c \\equal{} 3$. Then\r\n\\[ \\sum \\frac{a^2}{a^2 \\plus{} (b \\plus{} c)^2} \\equal{} \\sum \\frac{a^2}{a^2 \\plus{} (3 \\minus{} a)^2} \\equal{} \\sum \\frac{a^2}{2a^2 \\minus{} 6a \\plus{} 9}.\\]\r\n\r\nLet\r\n\\[ f(x) \\equal{} \\frac{x^2}{2x^2 \\minus{} 6x \\plus{} 9}.\\]\r\nThen\r\n\\[ f'(x) \\equal{} \\frac{6x(3 \\minus{} x)}{(2x^2 \\minus{} 6x \\plus{} 9)^2},\\]\r\nso $ f$ is increasing on the interval $ [0,3]$.\r\n\r\nSince\r\n\\[ f(x) \\minus{} \\frac{7x \\minus{} 2}{25} \\equal{} \\frac{x^2}{2x^2 \\minus{} 6x \\plus{} 9} \\minus{} \\frac{12x \\minus{} 7}{25} \\equal{} \\frac{3(x \\minus{} 1)^2 (21 \\minus{} 8x)}{25(2x^2 \\minus{} 6x \\plus{} 9)},\\]\r\nwe have that\r\n\\[ f(x) \\ge \\frac{7x \\minus{} 2}{25}\\]\r\nfor $ 0 \\le x \\le 21/8$.\r\n\r\nIf $ 21/8 \\le a \\le 3$, then $ f(a) \\ge f(21/8) \\equal{} 49/50$, so $ f(a) \\plus{} f(b) \\plus{} f(c) \\ge 49/50 > 3/5$. Similarly if $ 21/8 \\le b \\le 3$ and $ 21/8 \\le c \\le 3$. Otherwise, $ a$, $ b$, $ c \\le 21/8$, so\r\n\\[ f(a) \\plus{} f(b) \\plus{} f(c) \\ge \\frac{7(a \\plus{} b \\plus{} c) \\minus{} 6}{25} \\equal{} \\frac{15}{25} \\equal{} \\frac{3}{5}.\\]", "Solution_4": "[quote=\"marsupilami\"]I have some interesting inequality, let's discuss them :P :) \n1. Let $ a, b, c, x, y, z > 0$ and\n$ a \\plus{} x \\equal{} b \\plus{} y \\equal{} c \\plus{} z \\equal{} 1$\nProve that $ A\\equal{}(abc \\plus{} xyz)(\\frac {1}{ay} \\plus{} \\frac {1}{bz} \\plus{} \\frac {1}{cx}) \\geq 3$\n[/quote]\r\n$ A\\equal{}(\\frac{bc}{y}\\plus{}\\frac{yz}{c})\\plus{}(\\frac{ac}{z}\\plus{}\\frac{xz}{a})\\plus{}(\\frac{ab}{x}\\plus{}\\frac{xy}{b})\\equal{}(\\frac{(1\\minus{}y)c}{y}\\plus{}\\frac{y(1\\minus{}c)}{c})\\plus{}(\\frac{a(1\\minus{}z)}{z}\\plus{}\\frac{z(1\\minus{}a)}{a})\\plus{}(\\frac{(1\\minus{}x)b}{x}\\plus{}\\frac{x(1\\minus{}b)}{b})\\equal{}\\sum(\\frac{c}{y}\\plus{}\\frac{y}{c})\\minus{}3\\geq 3$.The last inequality follows from am-gm.", "Solution_5": "Thank everybody :lol: :) \r\n[quote=\"marsupilami\"]\n3. Let A=$ \\frac {a^2}{a^2 \\plus{} (b \\plus{} c)^2} \\plus{} \\frac {b^2}{b^2 \\plus{} (c \\plus{} a)^2} \\plus{} \\frac {c^2}{c^2 \\plus{} (a \\plus{} b)^2}$ \n$ a,b,c \\in R$\\{0}\nminA=?[/quote]\r\nAfter I had posted the problem, I saw a problem which is similar to it.\r\nIt's is problem 3, Japan, 1997.\r\nLet a, b, c be positive real numbers.\r\nProve the inequality:\r\n$ \\frac {(b\\plus{}c\\minus{}a)^2}{a^2 \\plus{} (b \\plus{} c)^2} \\plus{} \\frac {(c\\plus{}a\\minus{}b)^2}{b^2 \\plus{} (c \\plus{} a)^2} \\plus{} \\frac {(a\\plus{}b\\minus{}c)^2}{c^2 \\plus{} (a \\plus{} b)^2} \\geq \\frac{3}{5}$ \r\n\r\nHow aout it?? :?:", "Solution_6": "[quote=\"marsupilami\"]Thank everybody :lol: :) \n[quote=\"marsupilami\"]\n3. Let A=$ \\frac {a^2}{a^2 \\plus{} (b \\plus{} c)^2} \\plus{} \\frac {b^2}{b^2 \\plus{} (c \\plus{} a)^2} \\plus{} \\frac {c^2}{c^2 \\plus{} (a \\plus{} b)^2}$ \n$ a,b,c \\in R$\\{0}\nminA=?[/quote]\nAfter I had posted the problem, I saw a problem which is similar to it.\nIt's is problem 3, Japan, 1997.\nLet a, b, c be positive real numbers.\nProve the inequality:\n$ \\frac {(b \\plus{} c \\minus{} a)^2}{a^2 \\plus{} (b \\plus{} c)^2} \\plus{} \\frac {(c \\plus{} a \\minus{} b)^2}{b^2 \\plus{} (c \\plus{} a)^2} \\plus{} \\frac {(a \\plus{} b \\minus{} c)^2}{c^2 \\plus{} (a \\plus{} b)^2} \\geq \\frac {3}{5}$ \n\nHow aout it?? :?:[/quote]\r\nSee here:\r\nhttp://www.mathlinks.ro/viewtopic.php?search_id=991474430&t=146" } { "Tag": [ "LaTeX" ], "Problem": "Well this is my list...\r\n[code]\n\\begin{description}\n\t\\item[a)] A line of text that is longer than two lines.\n\t\t\\noindent List:\n\t\t\\begin{description}\n\t\t\t\\item[1)] Item\n\t\t\t\\item[2)] Item\n\t\t\t\\item[3)] Item\n\t\t\t\\item[4)] Item\n\t\t\t\\item[5)] Item\n\t\t\t\\item[6)] Item\n\t\t\t\\item[7)] Item\n\t\t\t\\item[8)] Item\n\t\t\t\\item[9)] Item\n\t\t\t\\item[10)] Item\n\t\t\t\\item[11)] Item\n\t\t\t\\item[12)] Item\n\t\t\t\\item[13)] Item\n\t\t\t\\item[14)] Item\n\t\t\t\\item[15)] Item\n\t\t\t\\item[16)] Item\n\t\t\\end{desription}\n\t\\item[b)] Item\n\t\n\\end{description}\n[/code]\r\nOk so my problem is that in the first line of \\item[a)], it's normal, however, in the second line, there is an indent that lines up with the second list. Then, my second list doesn't indent, but stays the same as the text in \\list[a)]. Also, my \\item[16)] doesn't line up with the rest of the items, but instead lines up with the beginning in the test in \\item[a)].\r\nSo is there any way I can fix this indent problem?", "Solution_1": "Use [i]itemize[/i] rather than [i]description[/i]", "Solution_2": "Doesn't itemize give you bullets?\r\nI want numbers and such...\r\n1)\r\n2)\r\n3)", "Solution_3": "It only gives bullets if you don't give a label. But that's precisely what you've done with \\item[a)] and \\item[1)] so you'll see a) and 1). Try it and see.\r\nOf course the enumerate environment will put the numbers in for you instead of having to specify labels\r\n[code]\\begin{enumerate} \n \\item Item \n \\item Item \n\\end{enumerate}[/code]", "Solution_4": "yeah but enumerate gives\r\n1.\r\n2.\r\n3.", "Solution_5": "Oh wait, I fixed it, using itemize does work.\r\n :lol:", "Solution_6": "[quote=\"ZhangPeijin\"]yeah but enumerate gives\n1.\n2.\n3.[/quote]\r\n\\renewcommand{\\labelenumi}{\\alph{enumi})}\r\n\\renewcommand{\\labelenumii}{\\arabic{enumii})}\r\nmakes enumerate give\r\na)\r\n1)\r\n2)\r\nb)\r\nc)" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO" ], "Problem": "Competition this saturday. Got any tips or last minute preparations? I have competed before at states, and I got 40th.\r\nNot good. Please help. I want to at least do better than last year. I am in PA if that changes any answers.", "Solution_1": "Pa is pretty competitive so u have to be good. If u think ur top 10 worthy, then make sure ur preety good at cd too(if it's official there :wink: ). Just relax and dont get too nervous or exited because u mite blank out in the competiton.", "Solution_2": "But remember, don't get too relaxed. You have to be on your toes or else you will do terrible... I saw it happen to first place in my chapter last year (my chapter is probably #2 in the state, and this person got 16th at state). Find your balance. Look over /study some material that day, but also listen to a little music to loosen up.", "Solution_3": "Make sure your arithmetic is decent (as in, adding/subtracting/multiplying/dividing). Cost me 2 AIME points this year, and also I would have made USAMO if I had got those 2.\r\n\r\n1 point can make a huge difference in MATHCOUNTS. Remember, spend enough time on the first half. Problem 1 is worth the same as Problem 30 (of course, that doesn't count tiebreakers).", "Solution_4": "My advice to you is to just relax, have fun, and pratice, practice, practice!!" } { "Tag": [ "geometry", "3D geometry", "algorithm", "summer program", "MathPath" ], "Problem": "How can I get better times on a Rubik's cube?\r\n\r\nMy best time is $\\text{3: 19.160}$.\r\nThe algorithm I use is similar to the one described here:\r\nhttp://en.wikibooks.org/wiki/How_to_solve_the_Rubik%27s_Cube\r\n\r\nIs there a better algorithm to use? If so, what is it?\r\n\r\nI also saw somewhere that someone (I think on this forum) had their cube lubricated so that they could do a quarter turn by simply flicking a cubelet. How could I do that? What's the best way to lubricate it, and with what?", "Solution_1": "Switch to a more efficient method. Really, a corner-edge pair method or better is necessary for speed. Personally, I like the Petrus method, but you can explore and test some out a bit. You definitely want to average under 60 moves.\r\n\r\nFor a lubricant, it depends on the cube, but the first thing you can do is get a god cube. I recommend Oddzon cubes (you can get them from Toys r us, the regular rubik's cubes are oddzon, not the deluxe package). Once you get a decent cube, making it better largely has to do with using it, stretching the springs, and spraying with silicone spray (buy at Home Depot). I personally suggest silicone spray after you've used it a bit, sprayed under an edge from each face, but other people prefer vaseline or generally petroleum oil. It's really up to you, but I'd say the silicone is most common.", "Solution_2": "I use silcone spray to lubricate my cube. If you want a method with extremely little memorization (or speed :rotfl:) try this website: http://grrroux.free.fr/begin/Begin.html\r\nthe method I use is a combination of some basic algorithims, most of which can be found on http://www.geocities.com/jasmine_ellen/RubiksCubeSolution.html\r\nand algorithims that I discovered using the first website", "Solution_3": "where do you all get your cubes? i want one badly!", "Solution_4": "Practice.\r\n\r\nI use an algorithm similar to the Wikibooks one and the best time I've gotten is about 2 minutes.\r\n\r\nI'm too lazy to learn any fast algorithms...\r\n\r\n[quote=\"SplashD\"]where do you all get your cubes? i want one badly![/quote]\r\n\r\nI got my first one at MathPath, but I lost it a few days later. It was on the lost and found list the rest of camp. It's probably still there somewhere...\r\n\r\nI got the second one at Mathcounts. It's really little and the stickers are about to come off; plus it's hard to turn. And the keychain thing gets in the way.\r\n\r\nBut there are a bunch of places that sell them.", "Solution_5": "Do you know where you can get one online? just a plain simple standard $3\\times3\\times3$ that has decent turnability and stickers.", "Solution_6": "[quote]Do you know where you can get one online? just a plain simple standard that has decent turnability and stickers.\n[/quote]\r\n[url]http://www.rubiks.com[/url] :wink: \r\n\r\nOh yes, if you get one from there, I also highly recommend you get the PVC stickers for replacements rather than the normal ones. PVC stickers don't need the extra little plastic layer, so they last longer.", "Solution_7": "I bought my cube at the national champtionships (no, I wasn't competing :wink: )\r\nlots of toy stores will have pretty good cubes. u just need to lubricate them pretty well.", "Solution_8": "ive been cubing for 9 days and my best time is 1 min (with inspection). My average is 1:30 with insp.\r\n\r\nI use this algorithm but i will start learning f2l when i get a little better.\r\n\r\nhttp://peter.stillhq.com/jasmine/rubikscubesolution.html" } { "Tag": [ "USAMTS", "percent" ], "Problem": "When do we get the checkmark on the My USAMTS page after uploading our solutions? \r\n\r\nSorry if this question has been asked before, I searched around and couldn't find anything relating to this.\r\n\r\nI'm also afraid that the final 5 percent of my PDF file was cut off, since there was a box that popped up (it said something like \"do you want to stop the script?\") when the download was at 95%, and I clicked \"no\" and don't know if that affected the submission.", "Solution_1": "do u get a checkmark for mailed solns?", "Solution_2": "First of all, so that it's clear what we're talking about: we're discuss the green checkmark that will appear on your \"My Scores\" page after we have logged your solutions as received.\r\n\r\nIf you submit via the web: the green checkmark should appear immediately. There appears to have been a slight bug which affected about a dozen or so students for whom the checkmark didn't appear, but they should all be fixed now. \r\n\r\nIf you submit via mail: the checkmark will appear after will have logged your solution as received. We have to do this by hand, and it may take a few days. (Only about a dozen or so of mailed-in solutions have thus far been logged.)", "Solution_3": "[quote=\"andersonw\"]When do we get the checkmark on the My USAMTS page after uploading our solutions? \n\nSorry if this question has been asked before, I searched around and couldn't find anything relating to this.\n\nI'm also afraid that the final 5 percent of my PDF file was cut off, since there was a box that popped up (it said something like \"do you want to stop the script?\") when the download was at 95%, and I clicked \"no\" and don't know if that affected the submission.[/quote]\r\nGo to the submit a solution page and see if it allows you to submit again or if it says that you already submitted a solution and are not allowed to submit anymore.\r\n\r\nIf you are allowed to submit again, that means the last solution wasn't uploaded and you should try again.", "Solution_4": "I've submitted my solutions online but have not received a checkmark. Might it have to do with the fact that it says I haven't turned in the Entry form? I faxed and mailed the entry form yesterday and today, respectively.", "Solution_5": "[quote=\"ProtestanT\"]\nGo to the submit a solution page and see if it allows you to submit again or if it says that you already submitted a solution and are not allowed to submit anymore.\n\nIf you are allowed to submit again, that means the last solution wasn't uploaded and you should try again.[/quote]\r\n\r\nIt says I'm not allowed to submit anymore.\r\nBut I just got the checkmark a few minutes ago, so it's good.\r\n\r\nAlso, a bug: From the Year 19 section, there's also a green checkmark in the round 1 row." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find 8 last digits of $ 5^{1994}$", "Solution_1": "Obviously congruent to $ 0 \\bmod 5^8$. We have $ 5^{64} \\equiv 1 \\bmod 2^8$ and therefore $ 5^{1994} \\equiv 5^{10} \\bmod 2^8$. The rest is computation: we find that $ 5^{10} \\equal{} \\boxed{09765625}$.", "Solution_2": "I calculated it by calculator (in Windows) and it wrote: 2509765625", "Solution_3": "[quote=\"\u10da\u10d4\u10d5\u10d0\u10dc\u10d8\"]I calculated it by calculator (in Windows) and it wrote: 2509765625[/quote]\r\nNo.the result is $ 5^{10}$", "Solution_4": "[b]\u10da\u10d4\u10d5\u10d0\u10dc\u10d8[/b] found $ 5^{10}\\times 257$, but we only need to find $ 5^{10}$, since $ 5^{1994} \\equiv 5^{10} \\mod 10^{8}$. Nevertheless, the last 8 digits are the same." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ k\\ge 8$, $ a$, $ b$ and $ c$ be positive real numbers. \r\nProve that\r\n\\[ \\sum{\\dfrac{a}{\\sqrt {a + kbc}}\\ge\\dfrac{3}{\\sqrt {k + 1}} + \\dfrac{9(a + b)(b + c)(c + a) - 72abc}{2(a + b + c)^3\\sqrt {k + 1}} + \\dfrac{3(k - 8)\\sum{(a^3 - abc)}}{2(a + b + c)^3(k + 1)\\sqrt {k + 1}}\r\n}\\]\r\n:)", "Solution_1": "[quote=\"ElChapin\"]Let $ a$, $ b$ and $ c$ be positive real numbers. \nProve that\n\\[ \\sum{\\dfrac{a}{\\sqrt {a + kbc}}\\ge\\dfrac{3}{\\sqrt {k + 1}} + \\dfrac{9(a + b)(b + c)(c + a) - 72abc}{2(a + b + c)^3\\sqrt {k + 1}} + \\dfrac{3(k - 8)\\sum{(a^3 - abc)}}{2(a + b + c)^3(k + 1)\\sqrt {k + 1}}\n}\\]\n:)[/quote]\r\nWhat is the nature of constant $ k$? $ k>0$? :huh:", "Solution_2": "I have forgotten to write that $ k\\ge 8$. I've edited, thanks.\r\n\r\n:)", "Solution_3": "I know this might appear ugly and hard but, it is not. It can be proved by using AMGM and a few operations (really few).\r\n\r\n:)" } { "Tag": [ "geometry", "3D geometry", "sphere", "combinatorics unsolved", "combinatorics" ], "Problem": "There are $2006$ points marked on the surface of a sphere. Prove that the surface can be partitioned into $2006$ congruent pieces, so that each piece contains exactly one of these points inside it.", "Solution_1": "This is also from the 1987 USA TST, I think.\r\n[hide]Let the sphere be a globe, so that we can reference points on the sphere using latitude and longitude. First, show that we can choose the axis of the globe such that no two points are at the same latitude. Next, draw 1003 great circles through the endpoints of the axis so that the circle is divided into 2006 congruent wedge-shaped regions. Next, show that for any point X that is not in some region A, we can transform each region so that the regions remain congruent, so that the point X is now in the regions A, and so that no other points have been transferred to other regions: \n\nDraw the line of latitude $ l$ through X. Let this latitude meet the edge of A (a line of longitude determining A) a some point $ Q$. Take two points on this same line of longitude; one point$ Q_1$ above $ Q$, and the other point $ Q_2$ below Q. Finally, choose one point $ Z$ on $ X$'s line of longitude such that $ X$ is contained in $ \\triangle Q_1 Z Q_2$. Then do the same for all the other regions. Note that we can take $ Q_1 Q_2$ arbitrarily small to avoid \"capturing\" any other points into region A. The diagram may clear some of this up:\n[geogebra]40fdd7701901e536c0c3188a76a9a6d605f34fae[/geogebra] \n[/hide]" } { "Tag": [ "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "assume $ R$ is a commutative ring with unity.$ S$ is a multiplicatively closed subset of $ R$.and we have $ \\forall x,y \\in R$ if $ xy \\in S \\longrightarrow x \\in S$ and $ y \\in S$.if there is no prime ideal $ P$ so that $ P \\cap S \\equal{} \\emptyset$.then prove $ S\\equal{}R$.", "Solution_1": "HINT the complement of a saturated monoid is a union of prime ideals,\r\nsince an ideal maximal in the complement of a semigroup is prime.\r\nThis will become more clear when you study localizations, cf.\r\nhttp://www.mathreference.com/id-lz,sats.html", "Solution_2": "but the proof of the total case of this quetion is incomplete.the proof is when correct when we have a prime ideal that has no intersection with S .and it will be completed when we prove the question above.", "Solution_3": "The link sketches a proof in the second paragraph in the \"saturated\" section. Namely, if r were an elt of R\\S then, by saturation, rR < R\\S so it could be enlarged to a maximal, hence, prime ideal in R\\S, contradiction. Equivalently: in the localization S^-1 R, a nonunit is contained in a maximal ideal, via the usual Zorn Lemma enlargement. The problem is simply a version of this remark stripped of localization (fraction) language." } { "Tag": [ "modular arithmetic" ], "Problem": "A4273B is six-digit positive number in which A and B are digits, and the total number is divisible by 72 with no remainder, decimal, or fraction. What is the value of A and the Value of B?\r\n\r\n[hide]The answer is A=5, B=6[/hide]", "Solution_1": "Please look through forums for posting guidelines or a general level of problem difficulty. This problem is too difficult for an ordinary middle school classroom, so I'm moving it.", "Solution_2": "[hide=\"Solution\"]72=8*9\n\nUsing divisibility rules:\n\nA number is divisible by 8 if it's last 3 digits are, \nso $73B\\equiv 0 \\pmod 8 \\Rightarrow B+2 \\equiv 0 \\pmod 8 \\Rightarrow B\\equiv 6 \\pmod 8$\n\nTherefore, B=6\n\nNext, by casting out 9s, $A+6+16 \\equiv 0 \\pmod 9$ so A=5\n\n [/hide]", "Solution_3": "[hide]This number is divisible by 72, so it's also divisible by 8 and 9. \n\nThus, the last three digits, 73B, must be divisible by 8. B=6. \n\nThe sum of the digits, A+4+2+7+3+6, must be divisible by 9. \nWe have A+22 is a multiple of 9, A=5. \n\n[/hide]" } { "Tag": [ "logarithms" ], "Problem": "$\\log_{5}(x-2)+\\log_{8}(x-4)=\\log_{6}(x-1)$", "Solution_1": "Nobody found a solution for this ? I only found out that has only one solution because $f'(x)>0$ $\\rightarrow$ $f is increasing$ ,where $f: [4,\\infty)\\to \\mathbb R$ ,$f(x)=log_{5}(x-2)+log_{8}(x-4)-log_{6}(x-1)$" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "PLease try the questions in the attachements:", "Solution_1": "BUMP. Why has no one attempted? Due to think there is some fault in the questions? Plz tell so cuz I think so.", "Solution_2": "My last bump to this topic.", "Solution_3": "hello\r\nsorry buddy I m just going to give one answer.(having one test right now)\r\n[hide=\"hint 2\"]\n$ f(x) \\equal{} ax^2 \\plus{} bx \\plus{} c$\nafter putting the three different values the actual $ f(x) \\equal{} \\frac {x^2 \\minus{} x \\plus{} 2}{2}$\nafter differentiating the given equation ,find $ g(x)$\nI hope u can find the suitable alternative now.\n\nif possible i will give the answer to other two tomorrow.[/hide]\r\nthank u", "Solution_4": "to the first one.\r\nf''(x)>f(x), so f''(0)>f(0)=0,\r\nso f(x) will be a concave function.\r\nso the answer is b" } { "Tag": [], "Problem": "There is a rubber band one end attached to the wall and the other end has a speed $v$.An ant begins motion from the left part (wall) with a speed $u$.When will the ant reach the free side of the band?", "Solution_1": "refer this\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=131251 :)", "Solution_2": "I wasn't registered that time.Revision is mother of knowledge.(Azerbaijanian proverb) :lol: :D" } { "Tag": [ "modular arithmetic", "algebra", "binomial theorem", "number theory unsolved", "number theory" ], "Problem": "Let: $ p_1 \\equal{} 2, \\ p_2 \\equal{} 3, \\ p_3 \\equal{} 5$ and so on, that is $ p_i$ are following primes. Can it be true: $ p_1p_2\\ldots p_n \\plus{} 1 \\equal{} a^k$, where $ k$ is integer, $ k>1$, and $ a$ is natural number?", "Solution_1": "[quote=\"Sylwek\"]Let: $ p_1 \\equal{} 2, \\ p_2 \\equal{} 3, \\ p_3 \\equal{} 5$ and so on, that is $ p_i$ are following primes. Can it be true: $ p_1p_2\\ldots p_n \\plus{} 1 \\equal{} a^k$, where $ k$ is integer, $ k > 1$, and $ a$ is natural number?[/quote]\r\n$ p_1p_2\\ldots p_n \\plus{} 1 \\equal{} a^k$so\r\n$ p_1p_2\\ldots p_n$ is a primes or it has a prime divisor and this one$ > p_n$ so i think the ansewr is not exist", "Solution_2": "[hide]We claim no such $ a,k$ exist.\n\nFirst, we prove that if $ i\\le n$, then $ k\\ne cp_i$: It is sufficient to prove that there is no $ a$ such that $ p_1\\cdots p_n \\plus{} 1 \\equal{} a^{p_i}$. Suppose, for the sake of contradiction that there did exist such an $ a$. Then, let $ a \\equal{} qp_i \\plus{} r$ where $ r < p_i$. Then, by the Binomial Theorem, we have\n\n$ a^{p_i} \\equal{} (qp_i)^{p_i} \\plus{} \\binom{p_i}{1}(qp_i)^{p_i \\minus{} 1}r \\plus{} \\cdots \\plus{} \\binom{p_i}{p_i \\minus{} 1}(qp_i)r^{p_i \\minus{} 1} \\plus{} r^{p_i}$ \n\nSince $ \\binom{p_i}{d}\\equiv0\\pmod{p_i}$ for all $ 0 < d < p_i$, we have that\n\n$ a^{p_i}\\equiv r^{p_i}\\pmod{p_i^2}$\n\nSince $ a^{p_i} \\equal{} p_1\\cdots p_n \\plus{} 1\\equiv 1\\pmod{p_i}$, we must have $ r \\equal{} 1$ (since $ a^{p_i}\\equiv a\\pmod{p_i}$). But then we have $ a^{p_i}\\equiv 1\\pmod{p_i^2}$ which implies that $ p_1\\cdots p_n\\equiv 0\\pmod{p_i^2}$, which is an obvious contradiction.\n\nThus, we have that if the integers $ a,k$ do satisfy the equation, then $ p_i\\not|k$ for all $ i\\le n$. It is also obvious, since $ a^k \\equal{} p_1\\cdots p_n \\plus{} 1$ that $ p_i\\not|a$ for all $ i\\le n$. Thus, the minimal posible value for $ a^k$ is $ p_{n \\plus{} 1}^{p_{n \\plus{} 1}}$ since all numbers less than $ p_{n \\plus{} 1}$ not equal to $ 1$ are multiples of a lesser prime. But, it is obvious that\n\n$ p_{n \\plus{} 1}^{p_{n \\plus{} 1}} > p_n^{p_n} \\plus{} 1 > p_1p_2\\cdots p_n \\plus{} 1$\n\nAnd so the claim is proved.[/hide]", "Solution_3": "[b]cosinator[/b], awesome solution :D", "Solution_4": "Thanks. It was fun to solve. :lol:" } { "Tag": [ "pigeonhole principle", "search", "combinatorics proposed", "combinatorics" ], "Problem": "$ 49$ students join in contest with $ 3$ problems. The piont for each problem is $ 0\\rightarrow 7$. Prove that there exist $ 2$ students $ A ,B$ satisfy the piont for each problem of $ A$ always $ \\geq$ the point for each problem or $ B$.", "Solution_1": "Maybe proof by contradiction $ A < B$ ?", "Solution_2": "well, pigeonhole is crying out desperately to me...", "Solution_3": "It well may, but the reasoning is arduous. \r\n\r\nTheoretical framework. Denote \r\n$ P : \\equal{} \\{(x,y,z) \\ \\mid \\ x,y,z \\in \\{0,1,2,3,4,5,6,7\\}\\}$\r\nthe set of all posible triplets of scores on the three problems ($ |P| \\equal{} 8^3 \\equal{} 512$). The relation \r\n$ (a,b,c) \\preceq (x,y,z) \\ \\textrm{if and only if} \\ (a\\leq x) \\land (b\\leq y) \\land (c\\leq z)$\r\nis a partial order relation on $ P$. Define $ P_k : \\equal{} \\{ (x,y,z) \\in P \\ \\mid \\ x\\plus{}y\\plus{}z\\equal{}k\\}$, for $ 0\\leq k\\leq 21$ (all $ P_k$ are antichains, i.e. made of elements mutually not comparable). We have $ |P_0| \\equal{} |P_{21}| \\equal{} 1$, $ |P_1| \\equal{} |P_{20}| \\equal{} 3$, ..., up to $ |P_{10}| \\equal{} |P_{11}| \\equal{} 48$.\r\n\r\nIf we could partition $ P$ into $ 48$ disjoint sets, each a chain (i.e. with any pair of elements comparable), as we have $ 49$ students, by pigeon-hole principle, two will be in the same chain, hence the required result about their scores. This would also show $ 49$ is the lowest number available, since an example of $ 48$ students without that property is given by the maximal antichain $ P_{10}$. (This is not a coincidence; Dilworth's theorem states that the cardinality of the largest antichain is equal to the minimal number of chains in a partition by chains of $ P$).", "Solution_4": "The above is correct. I will soon post the references to how this problem is settled in all its generality.", "Solution_5": "A graded poset is said to \"have the Sperner property\" or to \"be Sperner\" if there is an antichain of maximal size consisting of all points of a particular rank. The sought-after result is that a product of chains is Sperner; I'm moderately confident that I've seen a proof of this statement in general, but 2 minutes with a search engine wasn't enough for me to find a reference.", "Solution_6": "De Bruijn-Tengbergen-Kruyswijk theorem.", "Solution_7": "Let the set of all possible ordered triplet of point $ (a, b, c)$ be $ X$.\r\nWe can partition $ X$ into 48 disjoint subsets(32 $ A_{ij}$'s and 16 $ A_{(a,b)}$, and pigeonhole can be adapted!\r\n\r\nLet,\r\n$ A_1 \\equal{} \\{ (0, 0), (0, 1), (0, 2), \\cdots, (0, 7), (1, 7), \\cdots, (7, 7) \\}$\r\n$ A_2 \\equal{}\\{ (1, 0), (1, 1), \\cdots, (1, 6), (2, 6), \\cdots, (7, 6) \\}$\r\n$ A_3 \\equal{} \\{(2, 0), \\cdots, (2, 5), (3, 5), \\cdots , (7, 5)\\}$\r\n$ A_4 \\equal{} \\{(3, 0) , \\cdots , (3, 4), (4, 4), \\cdots , (7, 4)\\}$\r\n$ A_5 \\equal{} X\\minus{} (A_1 \\cup A_2 \\cup A_3 \\cup A_4)$\r\n\r\n$ A_{ij} \\equal{} \\{(a, b, j) \\vert (a, b) \\in A_i \\}$ (where, $ 1\\le i\\le 4$ and$ 0\\le j\\le7$)\r\n$ A_{(a,b)} \\equal{}\\{(a, b, j) \\vert 0\\le j\\le 7 \\}$ (where, $ (a,b) \\in A_5$)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that if $a,b,c\\geq0$ then\r\n\\[ \\frac a{\\sqrt{a^2+bc}}+\\frac b{\\sqrt{b^2+ca}}+\\frac c{\\sqrt{c^2+ab}}\\geq2. \\]", "Solution_1": "This seems to be wrong. I thought about such inequality some day ago, but I think a found a counterexample.", "Solution_2": "You are right, Lovasz. $c\\rightarrow+\\infty$", "Solution_3": "if $a,b,c \\ge 0$ then\r\n$\\frac{a}{{\\sqrt{a^2 + bc} }} + \\frac{b}{{\\sqrt{b^2 + ca} }} + \\frac{c}{{\\sqrt{c^2 + ab} }} \\ge 1$", "Solution_4": "if $a,b,c \\ge 0$then \r\n\uff081\uff09$\\frac{a}{{\\sqrt{a^2 + \\lambda bc} }} + \\frac{b}{{\\sqrt{b^2 + \\lambda ca} }} + \\frac{c}{{\\sqrt{c^2 + \\lambda ab} }} \\ge \\frac{3}{{\\sqrt{1 + \\lambda } }}(\\lambda \\ge 8);$\r\n\r\n\uff082\uff09$\\frac{a}{{\\sqrt{a^2 + \\lambda bc} }} + \\frac{b}{{\\sqrt{b^2 + \\lambda ca} }} + \\frac{c}{{\\sqrt{c^2 + \\lambda ab} }} \\ge 1(0 < \\lambda \\le 8)$", "Solution_5": "[quote]Prove that if $a,b,c\\geq0$ then\n\\[ \\frac a{\\sqrt{a^2+bc}}+\\frac b{\\sqrt{b^2+ca}}+\\frac c{\\sqrt{c^2+ab}}\\geq c. \\][/quote]\r\nWhich will be the best constant?\r\nThe number $1$ is not really familiar..." } { "Tag": [ "probability", "Support" ], "Problem": "Are easy questions supposed to have more points? Like I had question of the probability of not getting an even prime when rolling a dice, which is easy, yet it was worth 4 points.", "Solution_1": "I'm pretty sure it's supposed to go $ 1\\minus{}4$ points in order of difficulty (least to most)...but at the beginning, there's a lot of fluctuations.", "Solution_2": "Then i got another ez question with what is the probability when rolling a die you get a prime number. that question was worth 1 point...." } { "Tag": [ "topology", "complex analysis", "complex analysis unsolved" ], "Problem": "Consider $U=\\mathbb{C}-\\{a,b\\}$ where $a \\neq b$. Show that there is a closed path $\\gamma$ in $U$ such that $Ind_{a}(\\gamma)=Ind_{b}(\\gamma)=0$ but $\\gamma$ is not null-homotopic.\r\n\r\nWell, I found the path, and graphically it is obvious that it is not null-homotopic. Having no remembrance of algebraic topology, how to prove this fact formally ?", "Solution_1": "uhm..\r\nthe fundamental group of $U$ is $F_{2}$... you can take two generators $\\alpha$ and $\\beta$, and since there's no relation, $\\alpha\\beta\\alpha^{-1}\\beta^{-1}\\neq e$..." } { "Tag": [ "Ring Theory", "algebra", "function", "domain", "number theory", "relatively prime", "superior algebra" ], "Problem": "Let K be a number field with ring of integers O and class number h. Suppose that for every ideal I in O, there exists some integer r, relatively prime to h, such that I^r is principal. Why does this imply that O is a Unique factorization Domain?", "Solution_1": "Because $ x \\mapsto x^r$ is an injective map of the class group to itself (by recntly posted stuff, but it is not hard to show at all), sending $ (1)$ to $ (1)$. Thus every ideal is principal." } { "Tag": [], "Problem": "I have a strange favor for someone up for he challenge.\r\n\r\nI want to find a solution for the following.\r\n\r\n\r\nAn employee works 80 hours per pay period at $18.50 per hour. after 80hrs the pay goes up to pay and a half$27.75 per hour.\r\n\r\nI want to know what I have to change the rate to every pay period so that each pay period $200.00 are subtracted from the total amount paid. \nNow I can only change the regular hour rate because whatever the regular hour rate the Overtime must be that X 1.5 \nie \nRegular hrs. 80 hrs @ $18.5 = $1,480.00\r\nOver Time 11.96 hrs @ $27.75 = $331.89 \nTotal:$1,811.89\r\n\r\n\r\nNow I have to get $1,811.89 down exactly$200.00\r\n\r\nI am in heavy need for a formula which I can use to figure this out every pay period.\r\n\r\nThank you any help is great. :blush:", "Solution_1": "im confused on the question. im probably wrong too\r\n\r\nummmm\r\n$ xr_1\\plus{}yr_2\\minus{}200$ $ r_1$ and $ r_2$ are the normal and overtime rates and x and y are the hours worked.", "Solution_2": "[quote=\"kaptainyui\"]Now I have to get $ 1,811.89 down exactly$200.00[/quote]\r\nI don't understand why you want this, or where you want me to take the money from (i.e. if you just don't want $ \\$200$ then subtract $ \\$200$), or is there some sort of scalar involved. :?:", "Solution_3": "Well its not as simple. See It cant simply bee subtracted because of tax problems. But if its subtracted using a diffrent pay rate then it has nothing to do with taxes.\r\n\r\n\r\nWell I am guessing its a bit hard to explain so thanks for your help." } { "Tag": [ "integration", "calculus", "trigonometry", "function", "real analysis", "real analysis unsolved" ], "Problem": "\\[ \\int\\limits_0^{\\plus{}\\infty}e^{\\minus{}x^2}\\sin(ax) dx\r\n\\equal{}\r\n\\frac{1}{2}\\sum_{n\\equal{}0}^{\\plus{}\\infty}\\frac{(\\minus{}1)^nn!}{(2n\\plus{}1)!}a^{2n\\plus{}1}\\]\r\n\r\ncould anyone prove it ?", "Solution_1": "Can you compute $ \\int_0^{\\infty}x^{2k\\plus{}1}e^{\\minus{}x^2}\\,dx\\ ?$\r\n\r\nOnce you do that, then try expanding $ \\sin(ax)$ as a power series.", "Solution_2": "[quote=\"knoppix\"]\n\\[ \\int\\limits_0^{ \\plus{} \\infty}e^{ \\minus{} x^2}\\sin(ax) dx \\equal{} \\frac {1}{2}\\sum_{n \\equal{} 0}^{ \\plus{} \\infty}\\frac {( \\minus{} 1)^nn!}{(2n \\plus{} 1)!}a^{2n \\plus{} 1}\n\\]\ncould anyone prove it ?[/quote]\r\n\r\ni am not really sure why you call it a [i]mystery[/i] formula ? where is the mystery in it ? :?: \r\n\r\n\r\n$ \\int_0^{\\infty}\\; e^{ \\minus{} x^2}\\; \\sin(ax) \\;dx \\equal{} \\int_0^{\\infty}\\; e^{ \\minus{} x^2}\\;\\left( \\sum_{n \\equal{} 0}^\\infty \\; \\frac {( \\minus{} 1)^n}{(2n \\plus{} 1)!} \\; (ax)^{2n \\plus{} 1} \\right) \\;dx$\r\n\r\n$ \\equal{} \\sum_{n \\equal{} 0}^\\infty \\; \\frac {( \\minus{} 1)^n}{(2n \\plus{} 1)!} \\; a^{2n \\plus{} 1} \\;\\;\\; \\underbrace{ \\int_0^{\\infty}\\;x^{2n \\plus{} 1}\\; e^{ \\minus{} x^2}\\; dx } _ { \\equal{} \\; \\frac {1}{2}\\; n! }$\r\n\r\n$ \\equal{} \\frac {1}{2}\\; \\sum_{n \\equal{} 0}^\\infty \\; \\frac {( \\minus{} 1)^n\\; n!}{(2n \\plus{} 1)!} \\; a^{2n \\plus{} 1} \\; \\qquad ( \\; \\text{where}\\qquad n > \\minus{} 1\\;, \\quad a\\in \\mathbb R )$\r\n\r\n\r\n$ \\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}$\r\n\r\n\r\nto get the above integral, do\r\n\r\n$ \\int_0^{\\infty}\\;x^{2n \\plus{} 1}\\; e^{ \\minus{} x^2}\\; dx \\equal{} \\minus{} \\frac {1}{2}\\; \\int_0^{\\infty}\\;x^{2n} \\;e^{ \\minus{} x^2}\\; d( \\minus{} x^2)\\qquad \\qquad \\ldots \\qquad (1)$\r\n\r\n$ \\equal{} \\minus{} \\frac {1}{2}\\;\\left[ \\underbrace { x^{2n} \\; e^{ \\minus{} x^2}\\; \\bigg |_0^\\infty} _ { \\equal{} \\quad 0} \\; \\minus{} \\;\\int_0^{\\infty}\\;\\frac {d}{dx}( x^{2n} ) \\;\\;\\; e^{ \\minus{} x^2}\\; dx\\right]$\r\n\r\n$ \\equal{} \\frac {1}{2}\\;\\left[ 2n\\; \\int_0^{\\infty}\\; x^{2n \\minus{} 1} \\; e^{ \\minus{} x^2}\\; dx\\right] \\equal{} \\frac {1}{2}\\;\\left[ 2n\\;\\; \\left( \\minus{} \\frac {1}{2}\\right) \\;\\int_0^{\\infty}\\; x^{2n \\minus{} 2} \\;e^{ \\minus{} x^2}\\; d( \\minus{} x^2) \\right]$\r\n\r\nkeep going in a similar fashion to $ (1)$, and you will see that you eventually get \r\n\r\n$ \\frac {1}{2}\\;n \\; (n \\minus{} 1) \\; (n \\minus{} 2)\\cdot \\; \\; \\ldots \\; \\; \\ldots \\cdot \\; \\;3\\cdot 2 \\cdot 1 \\; \\equal{} \\; \\frac {1}{2}\\;n!$", "Solution_3": "now - nothing :) Thanks !\r\n\r\n\r\n\r\nI have the next one ;)\r\nI'm not starting new topic because it is similar:\r\n\r\n\r\nprove:\r\n\\[ \\int_0^{\\plus{}\\infty}\\frac{\\sin(tx)}{e^t\\minus{}1} dt\\equal{}\\sum_{n\\equal{}0}^{\\infty}\\frac{x}{x^2\\plus{}n^2}\\]\r\nfor all real $ x$.\r\n\r\nI tried with power series for sine (as you did above), but then I would need to compute\r\n\\[ \\int_0^{\\plus{}\\infty}\\frac{t^{2n\\plus{}1}}{e^t\\minus{}1}\\]\r\nAs Mathematica said, $ \\equal{}\\Gamma(2n\\plus{}2)\\zeta(2n\\plus{}2)$, but how can I obtain it ?", "Solution_4": "Ah, but the trick is a little different. Any time you see a simple two-term denominator, consider the possibility of representing that as a geometric series.\r\n\r\n$ \\frac{1}{e^t\\minus{}1}\\equal{}\\frac{e^{\\minus{}t}}{1\\minus{}e^{\\minus{}t}}\\equal{}\\sum_{n\\equal{}1}^{\\infty}e^{\\minus{}nt}$\r\n\r\nTry interchanging that sum with the integral in question.\r\n\r\nBy the way, in your formula, the sum should start with $ n\\equal{}1,$ not $ n\\equal{}0.$ It's incorrect as written, and you can convince yourself of that by taking the limit as $ x\\to0.$", "Solution_5": "[quote=\"knoppix\"]\nprove:\n\\[ \\int_0^{ \\plus{} \\infty}\\frac {\\sin(tx)}{e^t \\minus{} 1} dt \\equal{} \\sum_{n \\equal{} 0}^{\\infty}\\frac {x}{x^2 \\plus{} n^2}\n\\]\nfor all real $ x$.[/quote]\n\nuse what kent said.\n\n$ \\int_0^{\\infty}\\; \\frac {\\sin\\; (tx)}{e^t \\minus{} 1} \\; dt \\equal{} \\int_0^{\\infty}\\;\\sin\\; (tx) \\; \\sum_{n\\equal{}1}^{\\infty}e^{\\minus{}nt} \\; dt \\equal{} \\sum_{n\\equal{}1}^{\\infty} \\int_0^{\\infty}\\; e^{\\minus{}nt} \\; \\sin\\; (tx) \\;dt$\n\n$ \\equal{} \\sum_{n\\equal{}1}^{\\infty} \\frac{x}{n^2\\plus{}x^2}$\n\nmake sure to fix your starting index to $ n\\equal{}1$\n\n$ \\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}\\equal{}$\n\nto show that $ \\int_0^{\\infty}\\; e^{\\minus{}nt} \\; \\sin\\; (tx) \\;dt\\;\\equal{}\\; \\frac{x}{n^2\\plus{}x^2}$\n\nyou need to do integration by parts twice. see the basic idea here:\n\nhttp://en.wikipedia.org/wiki/Integration_by_parts\n\n\n[quote=\"knoppix\"]\n\nAs Mathematica said, $ \\equal{} \\Gamma(2n \\plus{} 2)\\zeta(2n \\plus{} 2)$, but how can I obtain it ? [/quote]\r\n\r\ni don't know what that means ? are you trying to get at a particular sum , and if so, what is the value of $ x$ you are using ?", "Solution_6": "Thank you very much for this !!\r\n\r\n\r\n\r\n\r\n[quote=\"misan\"]\n[quote=\"knoppix\"]As Mathematica said, $ \\equal{} \\Gamma(2n \\plus{} 2)\\zeta(2n \\plus{} 2)$, but how can I obtain it ? [/quote]\ni don't know what that means ? are you trying to get at a particular sum , and if so, what is the value of $ x$ you are using ?[/quote]\r\n\r\nwhen put \\[ \\int_0^{ \\plus{} \\infty}\\frac {t^{2n \\plus{} 1}}{e^t \\minus{} 1}dt\r\n\\]\r\ninto Mathematica, the program replies $ \\Gamma(2n \\plus{} 2)\\zeta(2n \\plus{} 2)$.\r\n\r\nHence I was interested in how to prove that for any $ n\\in\\mathbb{N}$\r\n\\[ \\int_0^{ \\plus{} \\infty}\\frac {t^{2n \\plus{} 1}}{e^t \\minus{} 1}dt \\equal{}\\Gamma(2n \\plus{} 2)\\zeta(2n \\plus{} 2)\r\n\\]\r\n\r\n($ \\Gamma$ is obviously Euler's Gamma function, $ \\zeta$ is Riemann Zeta)", "Solution_7": "[quote=\"misan\"]\nto show that $ \\int_0^{\\infty}\\; e^{ \\minus{} nt} \\; \\sin\\; (tx) \\;dt\\; \\equal{} \\; \\frac {x}{n^2 \\plus{} x^2}$\n\nyou need to do integration by parts twice. see the basic idea here:\n[/quote]\r\nComplex numbers is a faster way too." } { "Tag": [ "parameterization", "probability", "probability and stats" ], "Problem": "In class, I was given a short, simple problem something along the lines of :\r\n\r\n$ X$ and $ Y$ are independent Binomial random variables with respective parameters $ (n, p )$ and $ (m, p)$, calculate the conditional distribution of $ X$ given that $ X\\plus{}Y \\equal{} n$. I noticed that when it's calculated, it kind of looks like the form for the hypergeometric random variable for $ P(X \\equal{} i)$. Is there a combinatorial argument for why this is, or is it just one of a coincidence that they look alike?", "Solution_1": "It can't be hypergeometric. A hypergeometric random variable is unbounded, while this variable is bounded by $ n$.\r\n\r\nIf you had something else in mind, please find a way to clearly ask that question.", "Solution_2": "No, the [url=http://en.wikipedia.org/wiki/Hypergeometric_distribution]hypergeometric distribution[/url] is bounded.\r\n\r\nFor the problem in question: the probability that $ X\\equal{}k$ given $ X\\plus{}Y\\equal{}r$ is given by\r\n\r\n$ \\frac{\\binom{n}{k}p^k(1\\minus{}p)^{n\\minus{}k}\\binom{m}{r\\minus{}k}p^{r\\minus{}k}(1\\minus{}p)^{m\\minus{}r\\plus{}k}}{\\binom{m\\plus{}n}{r}p^r(1\\minus{}p)^{m\\plus{}n\\minus{}r}}\\equal{}\\frac{\\binom{n}{k}\\binom{m}{r\\minus{}k}}{\\binom{m\\plus{}n}{r}}$\r\n\r\nYes, that is hypergeometric.\r\n\r\nThe explanation is something like this. We do our Bernoulli trial $ n\\plus{}m$ times - $ n$ of them are trials of the first kind and $ m$ are trials of the second kind. We know, overall, that we have $ r$ successes. So think of it this way: we throw the $ n\\plus{}m$ trials into a bag, and pick out $ r$ of them to be the $ r$ successes. How many of those $ r$ are of the first kind? That's a setup for a hypergeometric distribution.", "Solution_3": "Oops. I got the terminology mixed up with the negative binomial." } { "Tag": [ "trigonometry", "limit", "real analysis", "real analysis solved" ], "Problem": "$x_n = \\sin x_{n-1}$\r\nShow that $\\lim_{n\\to\\infty} n x_n = 3$?\r\n\r\nor it is a similar problem because the one above seems not correct.", "Solution_1": "i think the limit is $\\sqrt{3}$ but i could be wrong...i remember that it was posed to me at IMO but i didn't think about it.\r\n\r\nPeter", "Solution_2": "If $x_0>0$\r\n$x_n \\sim \\sqrt{\\frac{3}{n}}$\r\nWith Cesaro", "Solution_3": "actually, it is divergent, but $\\lim_{n\\to\\infty} nx_n^2$ exists(the problem was posed to me with $\\lim_{n\\to\\infty} \\sqrt{n}x_n$, that's why i said something with $\\sqrt{3}$).\r\nconsider $a_n=\\frac{1}{x_n}$. we have the following: $a_n$ is strictly increasing, positive and unbounded(if it was bounded, it had a limit, contradiction). thus we can apply the cesaro stolz theorem( http://planetmath.org/encyclopedia/StolzCesaroTheorem.html ):\r\n\\[\\lim_{n\\to\\infty} nx_n^2=\\lim_{n\\to\\infty} \\frac{1}{a_{n+1}^2-a_n^2}\\].\r\nsince $a_{n+1}=a_n+\\frac 16a_n^{-1}+O(a_n^{-3})$, we have $a_{n+1}^2-a_n^2=\\frac 13+O(a_n^{-2})$, and the right side equals 3(remember $a_n$ is unbounded).\r\n\r\nPeter" } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Evaluate $\\int {\\frac{{d\\theta }}{{\\cos ^3 \\theta - \\sin ^3 \\theta }}}$.\r\n\r\nMy solution is rather long, involving repeated changes of variables. Hope that someone will find a shorter one.", "Solution_1": "It's\r\n\r\n$\\frac{1}{\\cos ^3 \\theta - \\sin ^3 \\theta }= \\frac{1}{3} (\\frac{2}{\\cos \\theta - \\sin \\theta }+ \\frac{\\cos \\theta - \\sin \\theta }{1+\\cos \\theta \\sin \\theta })$\r\n\r\nNow \r\n\r\n$\\frac{\\cos \\theta - \\sin \\theta }{1+\\cos \\theta \\sin \\theta }=\\frac{2(\\cos \\theta - \\sin \\theta) }{2+2 \\cos \\theta \\sin \\theta }=\\frac{2(\\cos \\theta - \\sin \\theta) }{1+(\\cos \\theta + \\sin \\theta)^2}$\r\n\r\nand you make the change of variable $x=(\\cos \\theta + \\sin \\theta)$. :) \r\n\r\nFor the other you make the change of variable $x=tan\\frac{\\theta}{2}$ :D \r\n\r\nand you have $\\cos \\theta =\\frac{1-x^2}{1+x^2}$ , $\\sin \\theta =\\frac{2x}{1+x^2}$ and $d\\theta =\\frac{2dx}{1+x^2}$\r\n\r\nThen you can easily evaluate $\\int \\frac{2dx}{1-x^2-2x}$ :lol:\r\nI hope this will help you :) (and i hope it has no typos :D )", "Solution_2": "However, the first integral (by letting x=cos(theta) + sin(theta) ) will still lead to lengthy evaluation. Enlighten me if I missed something :)", "Solution_3": "What you are missing is that $(arctanf(x))'=\\frac{f'(x)}{1+f^{2}(x)}$ . :) \r\n\r\nIf you make the change of variable $x=(\\cos \\theta +\\sin \\theta)$ you have $dx = (\\cos \\theta -\\sin \\theta)d\\theta$\r\n\r\nso $\\int \\frac{2(\\cos \\theta - \\sin \\theta){d\\theta }}{1+(\\cos \\theta + \\sin \\theta)^2}$ becomes $\\int \\frac{2dx}{1+x^2}$.\r\n\r\nBut $\\int \\frac{2dx}{1+x^2} =2\\cdot arctan(x)$ or $\\int \\frac{2(\\cos \\theta - \\sin \\theta){d\\theta }}{1+(\\cos \\theta + \\sin \\theta)^2} =2 \\cdot arctan(\\cos \\theta +\\sin \\theta)$. ;)\r\n\r\nFor $\\int \\frac{2dx}{1-x^{2}-2x}$ you write $1-x^{2}-2x=1-(\\frac{x+1}{\\sqrt2})^2$ and if you put $u=\\frac{x+1}{\\sqrt2}$\r\n\r\nyou take the integral $\\int \\frac{\\sqrt{2}du}{1-u^2} =\\sqrt{2}\\cdot arctanh(u)$ or $\\int \\frac{2d\\theta}{\\cos \\theta - \\sin \\theta }=\\sqrt{2}\\cdot arctanh(\\frac{1+Tan(\\frac{\\theta}{2})}{\\sqrt2})$.", "Solution_4": "Thanks for your more elegant solution :) \r\n\r\nWhat I did was to express the combinations of $Sin{x}$ and $Cos{x}$ in terms of a single trigo function.\r\nThat went well, but was a tad too tedious." } { "Tag": [ "geometry", "trapezoid", "ratio", "trigonometry", "complex numbers", "geometry unsolved" ], "Problem": "Let $ ABCD$ is a trapezoid with $ BC//AD$ and $ AD>BC$ and let $ \\angle BAD \\equal{} \\angle CDA \\equal{} 60^0$. Let $ \\angle C_1 \\equal{}\\angle C_2$ as picture. Let $ M$ is midpoint of $ BC$, let $ d$ is a line pass $ M$ and $ d \\perp BC$, let $ d$ meet $ Cx$ at $ E$. Let $ N$ is a point on $ AC$ so that $ EN \\perp DN$. Prove that $ N$ is midpoint of $ AC$.", "Solution_1": "Very easy ?\r\nDenote N be the midpoint of AC. MN=AB/2=CD/2, ME=CE/2, EMN=ECD=150,\r\n$ \\rightarrow\\triangle EMN\\sim triangle ECD\\rightarrow\\triangle ECM\\sim\\triangle EDN\\rightarrow\\angle END\\equal{}90^o$", "Solution_2": "[quote=\"mr.danh\"]Very easy ?\nDenote N be the midpoint of AC. MN=AB/2=CD/2, ME=CE/2, EMN=ECD=150\u00fa\n$ \\rightarrow\\triangle EMN\\sim triangle ECD\\rightarrow\\triangle ECM\\sim\\triangle EDN\\rightarrow\\angle END \\equal{} 90^o$[/quote]\r\nI don't think so!\r\nYou must from $ \\angle DNE \\equal{}90^0 \\longrightarrow$ N is midpoint of AC.", "Solution_3": "[quote=\"thanhnam2902\"]\nYou must from $ \\angle DNE \\equal{} 90^0 \\longrightarrow$ N is midpoint of AC.[/quote]\r\n????????\r\nMy post implies that the midpoint N of AC lie on the circle whose diameter is DE, it means that N is the intersection of that circle with AC, two expressions of N are concide.", "Solution_4": "Rotating the triangle ECD around E 60\u00ba clockwise and contracting to half, C comes to M and D to the middle of AC (the angle between CD and midline of CBA is 60\u00ba). That is, the triangle DEN having an angle of 60\u00ba (angle DEN) and the ratio of its adjacent sides \u00bd, is rightangled; as the circle of diameter DE intersects |AC| in a single point, the conclusion follows.\r\n\r\nBest regards,\r\nsunken rock", "Solution_5": "[quote=\"thanhnam2902\"]Let $ ABCD$ is a trapezoid with $ BC//AD$ and $ AD > BC$ and let $ \\angle BAD = \\angle CDA = 60^0$. Let $ \\angle C_1 = \\angle C_2$ as picture. Let $ M$ is midpoint of $ BC$, let $ d$ is a line pass $ M$ and $ d \\perp BC$, let $ d$ meet $ Cx$ at $ E$. Let $ N$ is a point on $ AC$ so that $ EN \\perp DN$. Prove that $ N$ is midpoint of $ AC$.[/quote]\r\nConsider the complex plane with $ M$ as center.\r\nLet $ a,b,c,d,e,n$ be the complex numbers corresponding to the points $ A,B,C,D,E,N$.\r\nLet $ b = -x$, $ c = x$, where $ x \\in \\mathbb{R}_+$, and let $ P(p)$ be the projection of $ M$ on $ AD$. Then $ p = -iy$ for $ y \\in \\mathbb{R}_+$.\r\nThen we obviously have $ a = -x-\\frac{y}{\\sqrt{3}}-iy$, $ d = x+\\frac{y}{\\sqrt{3}}-iy$, since $ \\tan 30 ^\\circ = \\frac{1}{\\sqrt{3}}$.\r\nWe also have $ \\angle MCE = 30 ^\\circ$, so $ E = i\\frac{x}{\\sqrt{3}}$. Then let $ N'(n')$ be the midpoint of $ AC$. Then $ n' = \\frac{a+c}{2} = \\frac{-\\frac{y}{\\sqrt{3}}-iy}{2}$.\r\nAnd $ EN' \\perp DN' \\iff i \\cdot \\frac{e-n'}{d-n'} \\in \\mathbb{R}$. But $ \\frac{e-n'}{d-n'} = \\frac{i}{\\sqrt{3}}$ so $ EN' \\perp DN'$, and thus it follows that $ N = N' \\iff n=n'$. Hence $ N$ is the midpoint of $ AC$. QED :)" } { "Tag": [ "inequalities", "logarithms", "calculus", "integration", "function", "Cauchy Inequality" ], "Problem": "Here's a problem that \"came\" to my mind when I was doing history :P.\r\n\r\n[b]Problem.[/b] Prove, without the use of a calculator (or approximation), that\r\n\\[ \\frac {\\pi \\minus{} e}{\\pi \\plus{} e} < \\frac {3}{2}(\\ln\\pi \\minus{} 1).\r\n\\]", "Solution_1": "Consider $ f(x)\\equal{}3(x\\plus{}1)\\ln x \\minus{} 2x$. \r\nWe have $ f(1)\\equal{}\\minus{}2$ and $ f'(x)\\equal{}3\\ln x \\plus{} 1 \\plus{} \\frac{3}{x} > 0 \\,$ for $ x > 1$ \r\nhence $ f(x)>\\minus{}2$ for $ x>1$. Take here $ x\\equal{}\\frac{\\pi}{e}$ and we are done.\r\nOf course I hope you will belief me that I can prove the inequality $ \\frac{\\pi}{e}>1$ without the use of a calculator. :lol:", "Solution_2": "I simply proved Cauchy's Inequality in Integral form: $ \\left(\\int_a^bfg\\right)^2\\leq\\int_a^bf^2\\int_a^bg^2$. Then, I set $ (f(x),g(x)) \\equal{} \\left(\\frac {1}{\\sqrt {x}},\\sqrt {x}\\right)$ and integrated over the interval $ [e,\\pi]$.", "Solution_3": "[size=117][color=darkred][b]Very nice the your proof, [u]boxedexe[/u] ![/b][/color][/size] [color=darkblue]Here is and an another (similar to [b]Digger[/b]'s) natural proof.[/color]\r\n\r\n[quote=\"boxedexe\"] [color=darkred]Prove that (without the use of a calculator or approximation) $ \\frac {\\pi \\minus{} e}{\\pi \\plus{} e} < \\frac {3}{2}(\\ln\\pi \\minus{} 1)$ .[/color] [/quote]\r\n[color=darkblue][b][u]Proof.[/u][/b] Consider the function $ f: (0,\\infty )\\rightarrow\\mathcal R$, where $ f(x) \\equal{} \\frac 32\\cdot\\ln x \\minus{} \\frac {x \\minus{} 1}{x \\plus{} 1}$ . Observe that $ f(1) \\equal{} 0$ and $ f'(x) \\equal{} \\frac {3x^2 \\plus{} 2x \\plus{} 3}{2x(x \\plus{} 1)^2} > 0$ for any $ x > 0$ . Therefore, the function $ f$ is strict increasing and particularly, $ \\pi > 3 > e$ $ \\Longrightarrow$ $ \\frac {\\pi}{e} > 1$ $ \\Longleftrightarrow$ $ f\\left(\\frac {\\pi}{e}\\right) > f(1)$ $ \\Longleftrightarrow$ $ f\\left(\\frac {\\pi}{e}\\right) > 0$ $ \\Longleftrightarrow$ $ \\frac {3}{2}(\\ln\\pi \\minus{} 1) > \\frac {\\pi \\minus{} e}{\\pi \\plus{} e}$ .[/color]", "Solution_4": "Can somebody give a more precise formulation of the \"Cauchy Inequality in Integral form?\" Are f,g any integrable functions [on that interval of course]? What are the conditions for equality?", "Solution_5": "[quote=\"Virgil Nicula\"][size=117][color=darkred][b]Very nice the your proof, [u]boxedexe[/u] ![/b][/color][/size] [/color]\n\n[quote=\"boxedexe\"] [color=darkred]Prove that (without the use of a calculator or approximation) $ \\frac {\\pi \\minus{} e}{\\pi \\plus{} e} < \\frac {3}{2}(\\ln\\pi \\minus{} 1)$ .[/color] [/quote]\n[/quote]\r\n\r\nYes, nice. But it gives something more:\r\n$ \\frac {\\pi \\minus{} e}{\\pi \\plus{} e} < \\frac {1}{2}(\\ln\\pi \\minus{} 1)$ :lol: \r\n\r\nIn this case yor proof is preserved whereas those of mine would demand additional observation of the second derivation.", "Solution_6": "[quote=\"Altheman\"]Can somebody give a more precise formulation of the \"Cauchy Inequality in Integral form?\" Are f,g any integrable functions [on that interval of course]? What are the conditions for equality?[/quote]\r\n\r\nThe inequality is:\r\nSuppose $ f$ and $ g$ are integrable functions on $ (a,b)$. Then $ \\int_a^b f^2(x)\\mathrm dx\\cdot\\int_a^b g^2(x)\\mathrm dx \\ge \\left(\\int_a^b f(x)g(x)\\mathrm dx\\right)^2.$\r\n\r\n[hide=\"Proof\"]Set $ A\\equal{}\\int_a^b f^2(x)\\mathrm dx,B\\equal{}\\int_a^b g^2(x)\\mathrm dx,C\\equal{}\\int_a^b f(x)g(x)\\mathrm dx$ and $ h(\\gamma)\\equal{}\\int_a^b \\left(f(x)\\minus{}\\gamma g(x)\\right)\\mathrm dx.$ Then $ 0\\le h(\\gamma)\\equal{}B\\gamma^2\\minus{}2C\\gamma \\plus{}A$ and $ h(\\gamma)$ has a nonpositive discriminant which implies $ C^2\\le AB.$ Equality holds if there is a $ \\gamma$ with $ f(x)\\minus{}\\gamma g(x)\\equal{}0\\;\\forall x\\in(a,b)\\iff f$ and $ g$ are linearly dependent.[/hide]" } { "Tag": [ "trigonometry", "calculus", "integration", "complex numbers" ], "Problem": "Prove that $ \\cos(n\\arctan 2\\sqrt{2})\\in Q$, $ n \\in N$\r\n\r\n\r\n_________________________________________\r\nAzerbaijan Land of Fire :ninja:", "Solution_1": "[hide=\"Solution\"]$ \\cos (n\\arctan 2\\sqrt 2)\\equal{}\\text{Re}[(\\cos\\arctan 2\\sqrt 2\\plus{}i\\sin\\arctan 2\\sqrt 2)^n]$ by DeMoivre.\n\nDrawing a right triangle with legs $ 4$ and $ \\sqrt 2$, we find the hypotenuse to be $ 3\\sqrt 2$, and therefore $ \\cos\\theta\\equal{}\\frac{1}{3}$ and $ \\sin\\theta\\equal{}\\frac{4}{3\\sqrt 2}$\n\nSo we have $ \\text{Re}\\left[\\left(\\frac{1}{3}\\plus{}\\frac{4i}{3\\sqrt 2}\\right)^n\\right]\\equal{}\\text{Re}\\left[\\left(\\frac{\\sqrt2 \\plus{} 4i}{3\\sqrt 2}\\right)^n\\right]\\equal{}\\frac{1}{3^n}\\cdot\\text{Re}\\left[(1 \\plus{} 2\\sqrt 2 i)^n\\right]$.\n\n$ (1\\plus{}2\\sqrt 2i)^n\\equal{}\\sum_{k\\equal{}0}^{n}\\binom nk (2\\sqrt 2i)^n$. These terms are only real when $ n\\equiv 0,2\\bmod 4$.\nIn both cases, $ n$ is even, which makes the square root integral, so the real part of this whole thing is some integer $ j$.\n\nTherefore, the answer is $ \\frac{j}{3^n}$ for some integer $ j$, which is a rational number. $ \\Box$[/hide]", "Solution_2": "[hide=\"Aliter\"]\nLet $ a_n \\equal{} \\cos n \\theta$ where $ \\theta \\equal{} \\tan^{ \\minus{} 1} 2 \\sqrt 2$\n\nWe have the recurrence $ a_{n \\plus{} 2} \\equal{} 2a_{n \\plus{} 1} a_1 \\minus{} a_n$\n\n$ a_1 \\equal{} \\cos \\theta \\equal{} \\frac {1}{3} \\in \\mathbb{Q}$ and $ a_2 \\equal{} \\cos 2 \\theta \\equal{} 2 \\cos^2 \\theta \\minus{} 1 \\equal{} \\minus{} \\frac {7}{9} \\in \\mathbb{Q}$\n\nThe above recurrence tells us that whenever $ a_n$ and $ a_{n \\plus{} 1}$ are rational so is $ a_{n \\plus{} 2}$\n\nSince $ a_1$ and $ a_2$ are rational, this result is true for all $ a_n$[/hide]" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "lim(x---->+2)e^[3/(2-x)]\r\nwuts the limit?", "Solution_1": "bumped..................................................................", "Solution_2": "You want $\\lim_{x\\to2^+}e^\\frac{3}{2-x}$ right? The answer is $0$.", "Solution_3": "can u show the steps? :oops:", "Solution_4": "Yes... $2-x\\to 0_{-}$ when $x\\to 2_{+}$", "Solution_5": "I think that spix wanted to say that the answer is $e^{- \\infty} = 0$.", "Solution_6": "Let's be more careful here! Check the one-sided limits.\r\n\r\n$\\lim_{x\\to2^-}e^{\\frac3{2-x}}=\\infty$\r\n\r\n$\\lim_{x\\to2^+}e^{\\frac3{2-x}}=0$" } { "Tag": [ "analytic geometry", "probability", "AMC" ], "Problem": "22. Objects A and B move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object A starts at (0,0) and each of its steps is either right or up, both equally likely. Object B starts at (5,7) and each of its steps is either left or down, both equally likely. Which of the following is closest to the probability that the objects meet: 0.10, 0.15, 0.20, 0.25, 0.30\r\n\r\nHow do you do this problem?", "Solution_1": "[hide=\"Hint\"]At which points can the objects meet? From here, you can use casework.[/hide]\n\n[hide=\"Solution\"]Let us refer to the (0,0) object as A and the (5,7) object as B.\n\nSince there are 12 steps from (0,0) to (5,7) or from (5,7) to (0,0), both objects must move six steps before meeting. Thus they can meet at (5,1), (4,2), (3,3), (2,4), (1,5), or (0,6). For (5,1), the probability that A will go there is $\\frac{\\binom{6}{1}}{2^6}$, since out of the 2^6 possible paths, only $\\binom{6}{1}$ (out of 6 total steps, 1 up) will get to (5,1). Similarly, the probability that B will go there is $\\frac{\\binom{6}{0}}{2^6}$. So the total probability is $\\frac{\\binom{6}{1} \\binom{6}{0}}{2^{12}}$.\n\nFor (4,2), the prob. of A going there is $\\frac{\\binom{6}{2}}{2^6}$ and the prob. of B is $\\frac{\\binom{6}{1}}{2^6}$.\n\nDoing the same for the others, we see that the total probability is\n\\[\\frac{1}{2^{12}} \\left( \\binom{6}{0}\\binom{6}{1} + \\binom{6}{1} \\binom{6}{2} + \\binom{6}{2} \\binom{6}{3} + \\binom{6}{3} \\binom{6}{4} + \\binom{6}{4} \\binom{6}{5} + \\binom{6}{5} \\binom{6}{6} \\right)\\]\n\\[= \\frac{1}{2^{11}} \\left( \\binom{6}{0}\\binom{6}{1} + \\binom{6}{1} \\binom{6}{2} + \\binom{6}{2} \\binom{6}{3} \\right)\\]\n(since the first three terms in the parentheses are the same as the last three)\n\\[= \\frac{1}{2^{11}} (396) \\approx .193\\]\n\nSo 0.2, (C).[/hide]", "Solution_2": "I don't really know probability or combinatorics, so this is probably wrong\r\n[Hide/]\r\nConsider the paths of particle A and particle B; consider only the paths where A and B end up meeting. Draw these paths in. Notice that since the two meet, they form one continuous path, from (0,0) to (5,7). Notice also that the motions possible alloted to B are the exact opposite of A. This problem then reduces to this problem:\r\n\r\nGiven a particle A starting at the origin, moving either 1 unit directly upwards or 1 unit directly to the right, find the probability that it hits (5,7).\r\n\r\nThis is obviously C(12,5)*1/2^12, which is about .1933. So it is closest to .20. Thus, the answer is [b].20[/b]. It is nice that this is consistent with mathfanatics" } { "Tag": [ "inequalities", "function", "LaTeX", "inequalities solved" ], "Problem": "Prove that : abc+bcd+cda+dab<= (1 + 176.abcd)/27\r\nfor all positive real number a,b,c,d satisfy: a+b+c+d=1\r\nWho can solved???\r\nI'll help you", "Solution_1": "now this is another problems that has been troubling me.i'd luv to see a soln now.", "Solution_2": "I think I have a solution. Here it is:\r\n\r\nThe inequality was abc+bcd+cda+dab<=(1+176abcd)/27\r\n\r\na+b+c+d=1>=4*(abcd)^1/4 that means that \r\n1>= 256 abcd\r\n\r\nthe inequality becomes \r\nabc+bcd+cda+dab<=432abcd/27\r\nabc+bcd+cda+dab<=16abcd\r\nWe divide with abcd\r\n\r\n1/a+1/b+1/c+1/d<=16\r\n\r\nLet there be a concave function f(x)=1/x\r\nUsing Jensen:\r\nf(a)+f(b)+f(c)+f(d)<=4f((a+b+c+d)/4)\r\n\r\n1/a+1/b+1/c+1/d<=4*1/(1/4)=16\r\n\r\nHope it's the corect solution", "Solution_3": "Hey.Your solution is wrong.\r\nApplying AM-GM, we get\r\n1/a + 1/b +1 /c +1/d >= 16/(a+b+c+d) =16 !!\r\n\r\nTry to solv!", "Solution_4": "This is my solution:WLOG, we suppose d is smallest.Then 0<= d <= 1/4.\r\nIf d=0, by Cauchy, we have abc<=((a+b+c)/3)3 = 1/27.\r\nIf 0< d<=1/4,the inequaly is equivalent: abc(1- 176d/27) + d(ab+bc+ca) <= 1/27\r\nConsider function f(d)= d(ab+bc+ca -176abc/27) +abc.\r\nWe have f'(d)=ab +bc +ca -176abc/27 = abc(1/a +1/b +1/c - 176/27).\r\nBy AM-GM, we get: f'(d)>= abc( 9/( a+b+c) -176/27) = abc( 9/(1-d) -176/27)> abc(9- 176/27)>0.\r\nSince d<=1/4, then f(d)<=f(1/4)= 1/4(ab+bc+ca) - 17abc/27. We'll prove P=1/4(ab+bc+ca) - 17abc/27 <=1/27 (*) with a+b+c =3/4\r\nWe have: P= b(a+c)/4 + ac(1/4 - 17b/27) = b(3/4 - b)/4 + ac(1/4 - 17b/27).\r\nIf b>=27/68 then P<= b(3/4 - b)/4. Applying Cauchy's, we get what we want.\r\nIf 27/68< b <3/4, we have P<= b(3/4 - b)/4 + (3/4 - b)2(1/4 - 17/27 b)/4 = [16 - (b- 1/4)2(68b+13)]/16.17 <= 1/27\r\nThen (*) is true\r\n inequaly become equality when a=b=c=d=1/4", "Solution_5": "abc+bcd+cda+dab<= (1 + 176.abcd)/27 \r\nf(a,b,c,d)=abc+bcd+cda+dab-abcd*176/27\r\n =(a+b)cd+ab(c+d-cd*176/27)\r\n\r\n*if(c+d-cd*176/27)<0:\r\n f(a,b,c,d)<=1/27\r\n*if (c+d-cd*176/27)>0:\r\n f(a,b,c,d)= x + 1\r\nx >= 1. \r\nso it is 1.", "Solution_2": "It was the problem 816 of College Math Journal january 2006. Why did you posted it here?", "Solution_3": "[quote=\"rem\"]max of sin C/2 is greater than 1...[/quote]\r\nSorry, what do you mean?\r\nI have a quite long solution \r\n[hide=\"here\"]WLOG, $A$ is the largest angle, so we have $\\frac{\\pi}{3}\\leq A<\\pi$. We claim that for $\\alpha \\in \\mathbb {R}$, $0\\leq \\alpha <1$, there exist angle $A$ s.t \n$\\alpha +sin(\\frac{A}{2})=2(\\alpha+sin(\\frac{\\pi -A}{4}))$. Our claim is equivalent to the existence of $A$ s.t\n$sin(\\frac{A}{2})-\\sqrt{2}(cos(\\frac{A}{4})-sin (\\frac{A}{4}))=\\alpha$. But consider $f(A)=sin(\\frac{A}{2})-\\sqrt{2}(cos(\\frac{A}{4})-sin (\\frac{A}{4}))$. We have $f(\\pi)=1$ and $f(\\frac{\\pi}{3})=-0.5$. By the continuity of $f$, it follows that our claim is true.\nNow it's easy to see that the required value of $\\alpha$ is at least 1. Next we will prove that 1 is good. It means, for any triangle $ABC$,\n$1+sin(\\frac{A}{2})$, $1+sin(\\frac{B}{2})$, $1+sin(\\frac{C}{2})$ form the sides of a triangle. Because $g(A)=sin(\\frac{A}{2})$ is monotonic increasing in $(0,\\pi)$, it's enough to show that $1+sin(\\frac{A}{2})<(1+sin(\\frac{B}{2}))+(1+sin(\\frac{C}{2}))$, or equivalently, $sin(\\frac{A}{2})<1+sin(\\frac{B}{2})+sin(\\frac{C}{2})$. But it's obvious since $sin(\\frac{B}{2})$ and $sin(\\frac{C}{2})$ are positive, while $sin(\\frac{A}{2})<1$[/hide]" } { "Tag": [ "summer program", "Mathcamp", "email" ], "Problem": "I would really like to attend Mathcamp, but I just don't have the time to take 5 whole weeks out of summer. Does anyone know if there is a way to make arrangements to go for just 2 or 3 weeks or so?", "Solution_1": "yeah a couple people did that last year.", "Solution_2": "I would only suggest that if you have some other summer program or event with a fixed date that you must go to. Every second of Mathcamp is worth it so I do suggest you stay as long as you can.\r\n\r\nBut it is definitely possible. Some people came late, some people left early, in fact one person in '04 could only come for the last week.", "Solution_3": "Yeah. A girl stayed for a week, a bunch of people came a week late an a bunch of people left early. I think that a guy went to RSI and came back for the last two weeks.\r\n\r\nI am an international student, and 5 weeks sure seemed long, but once you see what mathcamp is like, you won't want to leave. Also, no matter what you do, don't miss week 5. It's the best with the all nighter and all the extremely cool classes you get to choose.", "Solution_4": "Officially this is not allowed: the number of students at camp is limited (by dorm space, number of staff members, etc.) and if someone comes for less than the whole came, then that prevents someone else from coming for 5 weeks. \r\n\r\nUnofficially, if a good reason arises that someone has to miss perhaps a week of camp, we'll usually allow it. (Typical examples of good reasons: if MOSP and Mathcamp have a week of overlap; conflict between the end of Mathcamp and the start of school; a family reunion that your parents insist you not miss.)\r\n\r\nBut if you can only come for two or three weeks, we will almost certainly ask you to apply to Mathcamp again some summer where you are less busy and can commit for the full duration of the camp. We did have a situation last year where a new student wanted to come for the first three weeks and a 3-time alumnus wanted to come for the last two weeks, so that worked out for everyone, but you can't count on that happening. I can only think of one other time in recent years when we have allowed a new student to come for just three weeks, and this was because the student was competing at the IMO for the first two weeks of camp.", "Solution_5": "I had a few more questions about this partial term issue, if you don't mind.\r\n\r\nI've heard a lot of awesome things about MathCamp so I'd like to go this year. Unfortunately, I'll be attending a national leadership and debate competition that may last until the 8th of July. (Final date confirmation still pending.) So it could be the 10th before I could get to MathCamp. Should I still apply?\r\n\r\nAlso, do you have a special tuition rate for students (like me) who would be missing part of the camp?", "Solution_6": "As far as I know, you should still apply, because I think your competition will qualify as a \"good reason\" for missing a week of camp. I doubt they have a reduced tuition rate in your case, since you are only missing a week.\r\n\r\nTo make sure, though, you should either email Mathcamp, Mira, or Dave, or wait until Dave responds here.", "Solution_7": "Yeah, a few kids came a week late last year, so you will probably be fine (didn't Zhou arrive a little late last year?)", "Solution_8": "You should certainly still apply -- that's a perfectly valid reason for missing the beginning of Mathcamp. (If your conference ends on July 8th, though, we would encourage you to arrive at MC by July 9th if at all possible!) \r\n\r\nAs Yasha anticipates, there won't be any sort of discount if you come late: we wouldn't have any trouble finding someone else who would come (and pay) for the full five weeks, the college wouldn't give us a discount, etc.", "Solution_9": "After all, the next best thing to 5 weeks at mathcamp is 4 weeks at mathcamp(well, accually 34 days at matchcamp is the next best thing, but 4 weeks is still up there) :D", "Solution_10": "*stabs sirclucky*\r\nIf you really want to get technical, 34.999999 (etc. etc. etc.). But that happens to equal 35. AAAAAAAaaaaaaaaaaaaAAAAAAAAAAAAAaaaaaaaaaaaaAAAAAAAAAaaaaah.", "Solution_11": "*stabs drspyder*\r\n\r\nstop brining other forum personaitles onto this forum :P(I know treething on another forum... but there hes known as drspyder and im know as sirclucky :D )\r\n\r\nAnd I consider half a day at mathcamp still a day of mathcamp. 1/2 of infinty is still infinity.", "Solution_12": "can i sttend for three weeks?\r\nbecause i cannot attend July 3-July 8 because of EPGY(Stanford) summer program which took my hard work to get in... (although MathCamp was my priority, since EPGY was earlier in deadline, i had to prioritize that)\r\nI cannot attend August 1- August 7, because I am attending Mu Alpha Theta Nationals Competition... im shooting for a national championship this year.... hehe :D \r\n\r\nplease post ASAP. thanks in advance.", "Solution_13": "reddevil, regarding your request to come to Mathcamp for three weeks, please see my first post in this topic.", "Solution_14": "what are the requirements to attend the math camp?", "Solution_15": "[url]http://www.mathcamp.org/apply.php[/url]\r\nThe due date is April 27.", "Solution_16": "thanks", "Solution_17": "would: 4weeks(because of Mu aLpha Theta Nationals..) be a good reason? \r\n\r\ni have to really go to MAT because i have pre-paid and prepared... i wouldnt have registered if i knew about MathCamp a long time ago." } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "This is a problem from Noncommutative Algebra by Farb and Dennis, Springer, 1993 (pg26) which I cannot solve. \r\nProve that the ring End(M) of R endomorphisms of an R-module M is Artinian if M is both Artinian and Noetherian. \r\nI can do the special case when M is not indecomposable.", "Solution_1": "No thoughts on this problem?", "Solution_2": "Ha, I just had this problem assigned as homework.\r\n\r\n If M is Noetherian, then it is finitely generated (M Noetherian <=> any submodule of M is finitely generated, also an exercise in Farb and Dennis).\r\n\r\n Let [code] a_{1},....a_{n} [/code] be a set of generators of M then any endomorphism f is completely described by its action on [code] a_{1},....a_{n} [/code].\n\n Set T:[code]End_{R}(M) -> M^{n}[/code] (where [code]M^{n}[/code] is the R-module given by the direct sum of n copies of M)\n\n [code]T(f)= (f(a_{1}),....,f(a_{n}))[/code]. \n\n Then it is easy to check that T is a module homomorphism, thus its image, [code]T(End_{R}(M))[/code] is a submodule of [code]M^{n}[/code] .\n \n Furthermore T(f)=T(g) => f and g agree on the generators hence they are the same => T injective. Then [code]End_{R}(M)[/code] is isomorphic to a submodule of [code]M^{n}[/code] .\n\n Now from a previous exercise you know that if you have the short exact sequence [code]0-> A^{'} -> A -> A^{\"}->0[/code] then [code]A Artinian <=> A^{'}, A^{\"} Artinian[/code].\n\n Taking [code]A=A^{'}+A^{\"}[/code] (here + signifies direct sum) you find that the direct sum of two Artinian modules is Artinian, hence [code]M^{n}[/code] is Artinian.\n\n Finally, taking [code]A^{'}[/code] to be a submodule of A and [code]A^{\"}=A/A^{'}[/code], we get a short exact sequence and thus any submodule of an Artinian is an Artinian.\n \n Combining the last 2 results with the fact that [code]End_{R}(M)[/code] is isomorphic to a submodule of [code]M^{n}[/code] we get that [code]End_{R}(M)[/code] is Artinian.[/code]" } { "Tag": [ "geometry", "3D geometry", "function", "number theory proposed", "number theory" ], "Problem": "[color=darkblue] Find all natural numbers, not divisible by $ 10$, which are perfect cubes, and after deleting the last three digits, they are also nonzero perfect cubes.\n[/color]", "Solution_1": "Suppose $ n$ works, $ n \\equal{} 1000a \\plus{} b$, $ 0 < b < 1000$. (First we have $ 0\\leq b$, but if $ b \\equal{} 0$, then 10 divides $ n$, so $ 0 < b$)\r\nThen\r\n\\[ 1000a \\plus{} b \\equal{} k^3\r\n\\]\r\n\r\n\\[ a \\equal{} j^3\r\n\\]\r\n\r\n\\[ b \\equal{} k^3 \\minus{} 1000j^3 \\equal{} k^3 \\minus{} (10j)^3 \\equal{} (k \\minus{} 10j)(k^2 \\plus{} 10kj \\plus{} 100j^2)\r\n\\]\r\nSince this is more than 0 , $ k^3 > 1000j^3$, or $ k > 10j$. So both terms of are the product are positive, and integers. Since if $ j > 3$, the second term is more than 1000, $ j\\leq3$.\r\n\r\nCase 1: $ j \\equal{} 3$. Then $ k > 10j \\equal{} 30$ and $ (k \\minus{} 30)(k^2 \\plus{} 30k \\plus{} 900) < 1000$, obviously no solutions. (the function is increasing in $ k$, and is already too big when $ k \\equal{} 21$.\r\nCase 2: $ j \\equal{} 2$. Then $ k > 10j \\equal{} 20$ and $ (k \\minus{} 20)(k^2 \\plus{} 20k \\plus{} 400) < 1000$, again no solutions.\r\nCase 3: $ j \\equal{} 1$. Then $ k > 10j \\equal{} 10$ and $ (k \\minus{} 10)(k^2 \\plus{} 10k \\plus{} 100) < 1000$. 11 works, so does 12, but 13 doesn't so again since it is increasing, no solutions above 12.\r\n\r\nSummarizing, $ n \\equal{} 11^3 \\equal{} 1331$ or $ n \\equal{} 12^3 \\equal{} 1728$.", "Solution_2": "Let the natural numbers required be $ n\\equal{}1000a\\plus{}b$, where $ 1 \\le b \\le 999$\r\nThen $ a\\equal{}x^3$ for some natural numbers $ x \\Rightarrow n\\equal{}(10x\\plus{}p)^3$ for some natural number $ p$ and $ 1000x^3 <(10x\\plus{}p)^3 \\le 1000x^3\\plus{}999$\r\nBecause $ (10x\\plus{}p)^3\\equal{}1000x^3\\plus{}300x^2p\\plus{}30xp^2\\plus{}p^3$, we have $ 0<300x^2p<999 \\Rightarrow x\\equal{}1, p\\equal{}1$ or $ 2$ or $ 3$\r\nAfter checking , $ n\\equal{}11^3\\equal{}1331$ or $ n\\equal{}12^3\\equal{}1728$ are the solutions" } { "Tag": [ "linear algebra", "matrix", "superior algebra", "superior algebra solved" ], "Problem": "Check image", "Solution_1": "This one is solved on the forum, posted by Pcalin and solved by Moubinool\r\n\r\ncheers!", "Solution_2": "You couldnt possibly give me the link?", "Solution_3": "Lagrangia has a good memory \r\n\r\nHere my trick for the memory: some cod-liver oil, this is good for your brain before \u00b5ath exam\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?p=4570&highlight=#4570\r\n\r\nPiDeltaPhi where did find this nice problem ?\r\n\r\nWich forum ?", "Solution_4": "Thx for the link.\r\nI found it on another forum and it was unsolved." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Prove that, if three primes(greater than 3) are in A.P. then their common difference is divisible by 2 and 3 both.", "Solution_1": "we suppose that $ p_i\\equal{}p\\plus{}ih>3$ (where i\\in\\{0,1,2\\}) are primes.\r\n$ 2|p_2\\minus{}p_1\\equal{}h$ (because $ p_1,p_2$ are odd).\r\nwe have $ p_1,p_2,p_3\\equiv \\pm 1(mod\\ 6)$\r\nthen $ 6|(p_2\\minus{}p_1)(p_3\\minus{}p_2)(p_3\\minus{}p_1)\\equal{}2h^3$ so $ 3|h$" } { "Tag": [ "calculus", "derivative", "LaTeX", "real analysis", "real analysis unsolved" ], "Problem": "Hi,\r\nI'm not sure if this is the right place to post this, but I haven't been able to solve it, and was hoping for some help. The problem is as stated:\r\n\r\nShow that \r\n$\\frac{1} {M_x + N_y}$, where $M_x+N_y$ is not identically zero, is an integrating factor of the homogeneous equation $M(x, y)dx+N(x, y)dy=0$ of degree n.", "Solution_1": "[hide=\"Hint\"]Use it and show that $\\frac{\\partial M}{\\partial y}=\\frac{\\partial N}{\\partial x}$. We need $M=\\Psi_{x}$ and $N=\\Psi_{y}$ in order to solve the equation (think of the chain rule). Therefore $\\Psi$ must be $C^{2}$ so its mixed partial derivatives are equal. That is,\n\\[ \\frac{\\partial^{2}\\Psi}{\\partial x\\partial y}=\\frac{\\partial^{2}\\Psi}{\\partial y\\partial x}. \\][/hide]", "Solution_2": "I know, but it hasn't been turning up exact. While playing around yesterday though, I found that if we tried setting the integrating factor as $\\frac{1}{Mx+ Ny}$ instead, it would turn out to be an exact equation. What do you think; could this be a typo on the writer's behalf?", "Solution_3": "Isn't that what you originally posted? If it works, then it would be an integrating factor (I trust that you took the right steps).", "Solution_4": "no; in the way it was originally written, $x$ and $y$ were subscripts, denoting the partial derivatives; in this case, the way I found it to work, $M$ and $N$ were multiplied by $x$ and $y$", "Solution_5": "Oh yeah, sorry, I thought you just forgot to make them subscripts in $\\LaTeX$. Could you post what you did with the first case?" } { "Tag": [ "probability" ], "Problem": "Mr. Jones has 6 children. Assuming that the gender of each child is determined independently and with equal likelihood of male and female, what is that probability that Mr. Jones has more sons than daughters or more daughters than sons?", "Solution_1": "This is $ 1\\minus{}\\text{probability that there are 3 male and 3 female}$. This probability is $ \\binom{6}{3}\\left(\\frac{1}{2}\\right)^6\\equal{}20 \\times \\frac{1}{64}\\equal{}\\frac{20}{64}\\equal{}\\frac{5}{16}$. So our answer is $ 1\\minus{}\\frac{5}{16}\\equal{}\\boxed{\\frac{11}{16}}$.", "Solution_2": "how is their sol correct? they're assuming that order matters, which has not been specified in the question and usually is not taken into consideration for such kinds of problems...", "Solution_3": "[hide=Sol]The only other case other than the case that Mr. Jones has more sons than daughters or more daughters than sons is that there are 3 sons and 3 daughters. Ans = 1-(6c3)/2^6 = 11/16[/hide]", "Solution_4": "I thought $\\binom{6}{3} = 30.$ :(" } { "Tag": [ "trigonometry", "logarithms", "AMC", "AIME", "AMC 10", "MATHCOUNTS", "HMMT" ], "Problem": "So I'm currently writing a mock AMC 10. \r\n\r\nIf you're interested in [b]editing/reviewing it[/b] [i]or[/i] [b]taking it[/b] leave a comment below.", "Solution_1": "I'll do it as long as it has a tex file.", "Solution_2": "I'd like to take it.", "Solution_3": "i shall attempt to take it", "Solution_4": "[quote=\"anirudh\"]I'll do it as long as it has a tex file.[/quote]\r\n\r\nUh...why do you need a tex file?", "Solution_5": "To fix up minor errors instead of creating an errata.", "Solution_6": "[quote=\"anirudh\"]To fix up minor errors instead of creating an errata.[/quote]\r\n\r\n\r\nyou could always get a mod to fix up a minor error.\r\n\r\nBesides questions should be edited before posted online.", "Solution_7": "Taking it\r\n\r\nasdfasdfasdfasdf", "Solution_8": "I'm interested in taking it.", "Solution_9": "I will take it too :lol:", "Solution_10": "Ok. It's completed. I'm just double checking all the questions and answer choices. I'll probably post it some time later this afternoon.", "Solution_11": "[quote=\"rnwang2\"]Ok. It's completed. I'm just double checking all the questions and answer choices. I'll probably post it some time later this afternoon.[/quote]\r\nDid you get someone in >10th grade to review it? Methinks that would be a good idea (just in case)... How hard do you think this is going to be for an AMC 10?", "Solution_12": "I think this may be leaning toward the difficult side for an AMC 10. The reason I did not make it an AMC 12 is because there is no trigonometry or logarithms involved.", "Solution_13": "I'll try it.", "Solution_14": "Ok here it is. Just PM me the answers. Everyone is welcome to take it.\r\n[b]EDIT: The deadline has been moved to Sunday, January 7.[/b]\r\n\r\n[b]EDIT: I have attached a new version with #14 fixed.[/b]\r\n\r\n[b]EDIT: I have attached a new version with a clarification of #4 (please download the [i]first[/i] one).[/b]", "Solution_15": "Wow I just realized how nice #25 really is.", "Solution_16": "[quote=\"rnwang2\"][quote=\"SplashD\"]the answers to #11 are a decent amount off of what I got as an answer.\n\nQuestions you might want to check: #12, #14(the wording with solid and water), #21[/quote]\n\nThanks, I have fixed #14. \n\n#11, #12, and #21 have no problems.[/quote]\r\n\r\nThere is definitely something wrong with #12. Let me PM you...\r\n\r\nEDIT: Oh nvm, my mistake.\r\n\r\nEDIT2: Wait, #17 is confusing. It says the men do not wish to be on the same committee... does that mean we are choosing 2 of them? One with 2 women and 3 men, and one with 4 women and 5 men??", "Solution_17": "You are picking 3 men from the 8 men. However, 2 of the 8 men do not wish to serve on the same committee.", "Solution_18": "Yeah, #17 is worded weirdly, but it's technically written correctly. It took me like 3 readings to get it... :wink:\r\n\r\n\r\nEDIT: (Didn't wanna spam up the threads so edited) Yea Adeel, that got me too. I was like WTF.", "Solution_19": "I see. (The confusion arises from the term \"same committee\", which implies that there is more than one committee.)", "Solution_20": "i'm confused....\r\nso are you going to count #24 and #14 or not????", "Solution_21": "Well, I think #24 is being counted (the problem definitely makes sense, and it is not hard if you have seen the idea before).", "Solution_22": "[b]Please PM me the answers before noon (central time) tomorrow. [/b]\r\nThen we can start posting solutions to the problems :) . I will make a separate thread for that.", "Solution_23": "Ok, it's noon in central time now, so, who got $16,19,20,23,24$? :roll:", "Solution_24": "how do you do #25?", "Solution_25": "Oh no I forgot to do this :(", "Solution_26": "[quote=\"bigboy82892\"]Ok, it's noon in central time now, so, who got $16,19,20,23,24$? :roll:[/quote]\r\n[hide=\"16\"] $\\text{digital root}\\equiv\\mod9$\n$(9^{6130}+2)^{4875}\\pmod{9}\\equiv2^{4875}\\equiv2^{4875\\mod6}\\equiv2^{3}=\\boxed{8}$ [/hide]\n[hide=\"19\"] Hard to explain. :| [/hide]\n[hide=\"20\"] Find the coordinates of Bob and Joe in terms of $t$. Find the distance between them by the distance formula; it will be the square root of a quadratic. Find the minimum of the quadratic and square root it. [/hide]\n[hide=\"23\"] It's actually a fairly easy problem. Draw a radius to where one of the 6 cm lines intersect the circle. Right triangle with legs $\\frac{x}{2}$ and $6$, hypotenuse of length $4+\\frac{x}{2}$. This is because the distance from the intersection of the 4 cm line and circle, to the center of the star is a radius. [/hide]\n[hide=\"24\"] Didn't get it, but you could plug it into a graphing calculator, unless it's too big to be fractionized. :D [/hide]", "Solution_27": "I have created a separate thread for solutions.", "Solution_28": "PLEASE ASK FOR/POST SOLUTIONS [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127421]HERE[/url]\r\n\r\nEDIT: lol, lotrgreengrapes, we posted the same stuff in different threads.", "Solution_29": "Sorry :blush: Anyway, my answers above were more outlines than solutions." } { "Tag": [ "calculus", "integration", "real analysis", "rudin", "real analysis unsolved" ], "Problem": "How do I calculate this \r\n\r\n$ S \\equal{} \\int_{\\mathbb{R}}\\sum_{n\\in \\mathbb{N}} n\\cdot \\chi_{\\left[n,n \\plus{} \\frac {1}{2^n} \\right] }(t) d\\lambda$\r\n\r\nwhere $ \\lambda$ is the lebesgue measure.\r\nIs there a theorem to make the sum commute with the integral ?\r\n\r\n\r\nP.S.\r\n\r\nI've reached that it should be equal to [url=http://www.mathlinks.ro/viewtopic.php?t=207367]this[/url] but\r\nI'm not entirely sure that I knew what I was doing.\r\n\r\nAlso why is the lebesgue measure needed here ?\r\nWouldn't a more \"simple\" kind of measure do ?\r\nI find Lebesgue measure hard to grasp.", "Solution_1": "You're probably confusing yourself unnecessarily by thinking about Lebesgue measure.\r\n\r\nI don't know what you mean by a simpler measure, but here the Lebesgue integral is equal to the Riemann integral, and there is very little to worry about. (if you want to make this [i]very[/i] rigorous, you need to prove what I just said, of course, but you should review the basic principles of Lebesgue integration first)\r\n\r\nThe integral you want is $ \\int_{-\\infty}^\\infty \\sum_{n\\in \\mathbb{N}} n\\cdot \\chi_{\\left[n,n + \\frac {1}{2^n} \\right] }(t) dt$. I hope that the following steps are each clear, but if they're not, feel free to ask.\r\n\r\n$ = \\sum_{k=-\\infty}^{\\infty} \\int_{k}^{k+1} \\sum_{n\\in \\mathbb{N}} n\\cdot \\chi_{\\left[n,n + \\frac {1}{2^n} \\right] }(t) dt$\r\n\r\n$ = \\sum_{k=1}^{\\infty} \\int_{k}^{k+\\frac{1}{2^k}} \\sum_{n\\in \\mathbb{N}} n\\cdot \\chi_{\\left[n,n + \\frac {1}{2^n} \\right] }(t) dt$\r\n\r\n${ = \\sum_{k=1}^{\\infty} \\int_{k}^{k+\\frac{1}{2^k}} k } dt$\r\n\r\n${ = \\sum_{k=1}^{\\infty} \\frac{n}{2^n} } dt$", "Solution_2": "the problem is I've never found an example where it would really matter\r\nthat the measure was lebesgue measure and not any other measure.\r\nI mean,an example in which having lebesgue measure instead of any other\r\nmeasure would make a very big difference.", "Solution_3": "You keep saying \"any other measure,\" but can you name some? Are you talking about the Borel measure?\r\n\r\nIf you are talking about the Lebesgue [i]integral[/i], a related but different concept from Lebesgue measure, then it is a fairly simple to say the difference with Riemann integration. Whenever a Riemann integral exists, it is the same as the Lebesgue integral, but sometimes the Riemann integral does not exist, while the Lebesgue integral does. This is precisely why Lebesgue integration was invented.\r\n\r\nHere is an example: Take $ f(x)\\equal{}\\chi_{\\mathbb{Q}}$, that is, $ f(x)$ is $ 1$ on the rationals and $ 0$ on the irrationals. The Riemann integral $ \\int_{\\minus{}\\infty}^{\\infty} f(x)dx$ is not defined. However, the Lebesgue integral is defined and equal to just what you think it should be: $ 0$.\r\n\r\nThe best place to read about these things is in an analysis text that has a focus on Lebesgue measure. The last chapter of Baby Rudin, for example, is a thorough introduction.", "Solution_4": "You can commute the integral and the sum because the summands are all positive; just apply monotone convergence." } { "Tag": [], "Problem": "Each term of a sequence is generated by adding the two previous terms. What is the sum of the values of the four missing terms in the following sequence? \n\\\\\n\\[ \\minus{}12, \\underline{\\hspace{0.2in}}, \\minus{}5, \\underline{\\hspace{0.2in}}, \\minus{}3, \\underline{\\hspace{0.2in}}, \\minus{}4, \\underline{\\hspace{0.2in}}, \\minus{}9, \\minus{}14, \\minus{}23\\]", "Solution_1": "\\[ \\minus{}12, \\underline{a}, \\minus{}5, \\underline{b}, \\minus{}3, \\underline{c}, \\minus{}4, \\underline{d}, \\minus{}9, \\minus{}14, \\minus{}23\\]\r\nWe have \r\n\\[ \\minus{}12\\plus{}a\\equal{}\\minus{}5 \\Rightarrow a\\equal{}7 \\\\ \r\n\\minus{}5\\plus{}b\\equal{}\\minus{}3 \\Rightarrow b\\equal{}2 \\\\\r\n\\minus{}3\\plus{}c\\equal{}\\minus{}4 \\Rightarrow c\\equal{}\\minus{}1 \\\\\r\n\\minus{}4\\plus{}d\\equal{}\\minus{}9 \\Rightarrow d\\equal{}\\minus{}5\\]\r\nThe sum is $ 7\\plus{}2\\minus{}1\\minus{}5\\equal{}\\boxed{3}$" } { "Tag": [], "Problem": "If the average stride of a horse walking is $ 2\\frac{3}{4}$ feet and that of a man is $ 2\\frac{1}{2}$ feet, how many man strides would be equal to 40 horse strides?", "Solution_1": "$ \\frac{11}{4}\\times{40}\\div\\frac{5}{2}\\equal{}\\boxed{44}$" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "x^1/2+y=7\r\ny^1/2+x=11\r\nsolve for x and y", "Solution_1": "If you can solve equations form $f(x)=0$, where $f$ is a polynomials of degree $4$ then this problem is easy.", "Solution_2": "can u plz solve it for me.", "Solution_3": "Put $x=a^{2},y=b^{2}(a,b\\geq 0)$, then $a+b^{2}=7,b+a^{2}=11$ and $b+(7-b^{2})^{2}=11$, this is equation degree $4$ and $2$ is a root of it.", "Solution_4": "Yes but b=2 isn't the only root :(", "Solution_5": "[quote=\"Svejk\"]Yes but b=2 isn't the only root :([/quote]\r\n$b+(7-b^{2})^{2}-11=(b-2)f(b)$. That is all!", "Solution_6": "[quote=\"N.T.TUAN\"]Put $x=a^{2},y=b^{2}(a,b\\geq 0)$, then $a+b^{2}=7,b+a^{2}=11$[/quote]\r\nSo $0\\leq b\\leq \\sqrt{7}$.\r\n$0=b+(7-b^{2})^{2}-11=(b-2)(b^{3}+2b^{2}-10b-19)$ \r\nand $f(b)=b^{3}+2b^{2}-10b-19$ has no root in $[0,\\sqrt{7}]$. (simple analysis)\r\n\r\nOr:\r\n$0\\leq b\\leq \\sqrt{7}<5\\Rightarrow b(5-b)\\geq 0\\\\ \\Rightarrow f(b)=b^{3}-2b(5-b)-19\\leq b^{3}-19\\leq 7\\sqrt{7}-19<0$", "Solution_7": "Very easy!! :) \r\n\r\n[img]http://img011.photo.21cn.com/photos/user/24454028/20070331231229/o/8339237.jpg[/img]", "Solution_8": "[quote=\"kg2100\"]Very easy!! :) \n\n[img]http://img011.photo.21cn.com/photos/user/24454028/20070331231229/o/8339237.jpg[/img][/quote]\r\nI don't understand. What are you doing? :huh:", "Solution_9": "Plugging into computer program. :P" } { "Tag": [ "trigonometry", "algebra", "polynomial" ], "Problem": "Solve for $ x$\r\n\r\n$ \\sin ^4 x \\minus{} \\sin ^4 \\frac{\\pi }{4} \\equal{} \\sin ^4 (x \\minus{} \\frac{\\pi }{4})$ :D", "Solution_1": "[quote=\"TRAN THAI HUNG\"]Solve for $ x$\n\n$ \\sin ^4 x \\minus{} \\sin ^4 \\frac {\\pi }{4} \\equal{} \\sin ^4 (x \\minus{} \\frac {\\pi }{4})$ :D[/quote]\r\n\r\nLHS: =(sin^2(x)-sin^2(pi/4))(sin^2(x)+sin^2(pi/4))\r\n=(sinx-sin(pi/4))(sinx+sin(pi/4))(sin^2(x)-i^2sin^2(pi/4))\r\n=(sinx-sin(pi/4))(sinx+sin(pi/4))(sin(x)+isin(pi/4))(sinx-isin(pi/4))\r\ntherefore x=+-pi/4,arcsin(+-isin(pi/4))\r\n\r\ni think i went very wrong actually.. could u bring the RHS to the LHS and let x=sinx, y=sinpi/4, and z=sin(x-pi/4), then expand (x^4-y^4-z^4)...?", "Solution_2": "[quote=\"addikaye03\"][quote=\"TRAN THAI HUNG\"]Solve for $ x$\n\n$ \\sin ^4 x \\minus{} \\sin ^4 \\frac {\\pi }{4} \\equal{} \\sin ^4 (x \\minus{} \\frac {\\pi }{4})$ :D[/quote]\n\nLHS: =(sin^2(x)-sin^2(pi/4))(sin^2(x)+sin^2(pi/4))\n=(sinx-sin(pi/4))(sinx+sin(pi/4))(sin^2(x)-i^2sin^2(pi/4))\n=(sinx-sin(pi/4))(sinx+sin(pi/4))(sin(x)+isin(pi/4))(sinx-isin(pi/4))\ntherefore x=+-pi/4,arcsin(+-isin(pi/4))\n\ni think i went very wrong actually.. could u bring the RHS to the LHS and let x=sinx, y=sinpi/4, and z=sin(x-pi/4), then expand (x^4-y^4-z^4)...?[/quote]\r\nJust a spammer :mad:\r\nDoes anyone has solution ??? Please help me :huh: \r\nI don't know how to solve\r\nP.s : I don't spam. I post this just because i want this topic to be on top again :)", "Solution_3": "hello ,applying the addition formulas we get $ \\sin(x)^4\\minus{}\\frac{1}{4}\\equal{}\\minus{}\\cos(x)^4\\minus{}\\sin(x)\\cos(x)\\plus{}\\cos(x)^2\\plus{}\\frac{1}{4}$, simplifying this we get the equation\r\n$ \\frac{1}{2}\\plus{}2\\cos(x)^4\\plus{}\\sin(x)\\cos(x)\\minus{}3\\cos(x)^4\\equal{}0$.Converting this in $ \\tan(\\frac{x}{2})$ we get $ \\frac{1}{2}\\plus{}\\frac{2(1\\minus{}\\tan(\\frac{x}{2})^2)^4}{(1\\plus{}\\tan(\\frac{x}{2})^2)^4}\\plus{}\\frac{2\\tan(\\frac{x}{2})(1\\minus{}\\tan(\\frac{x}{2})^2)}{(1\\plus{}\\tan(\\frac{x}{2})^2)^2}\\minus{}\\frac{3(1\\minus{}\\tan(\\frac{x}{2})^2)^2}{(1\\plus{}\\tan(\\frac{x}{2})^2)^2}\\equal{}0$. By substituting $ \\tan(\\frac{x}{2})\\equal{}t$and simplifying we have to solve the following equation in $ t$\r\n$ (t^2\\plus{}2t\\minus{}1)(t^6\\plus{}2t^5\\plus{}9t^4\\minus{}12t^3\\minus{}9t^2\\plus{}2t\\minus{}1)\\equal{}0$.\r\nSonnhard.", "Solution_4": "[quote=\"TRAN THAI HUNG\"]Solve for $ x$\n\n$ \\sin ^4 x \\minus{} \\sin ^4 \\frac {\\pi }{4} \\equal{} \\sin ^4 (x \\minus{} \\frac {\\pi }{4})$ :D[/quote]\r\n\r\n$ \\sin x \\minus{} \\sin \\frac {\\pi }{4} \\equal{} \\sin (x \\minus{} \\frac {\\pi }{4}) \\\\\r\n\\Leftrightarrow 2\\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right)\\cos \\left( {\\frac {{x \\plus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} 2\\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right)\\cos \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\\\\r\n\\Leftrightarrow \\left[ \\begin{array}{l} \\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} 0 \\\\\r\n\\cos \\left( {\\frac {{x \\plus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} \\cos \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\\\\r\n\\end{array} \\right. \\Leftrightarrow \\left[ \\begin{array}{l} \\frac {{x \\minus{} \\frac {\\pi }{4}}}{2} \\equal{} k\\pi (k \\in Z) \\\\\r\n\\cos \\frac {x}{2}\\cos \\frac {\\pi }{8} \\minus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} \\cos \\frac {x}{2}\\sin \\frac {\\pi }{8} \\plus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\\\\r\n\\end{array} \\right. \\\\\r\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\r\n\\minus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\\\\r\n\\end{array} \\right. \\\\\r\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\r\n\\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} 0 \\\\\r\n\\end{array} \\right. \\\\\r\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\r\n\\sin \\frac {x}{2} \\equal{} 0 \\\\\r\n\\end{array} \\right. \\\\\r\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi \\\\\r\n\\frac {x}{2} \\equal{} l\\pi \\\\\r\n\\end{array} \\right.(k,l \\in Z) \\\\\r\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi \\\\\r\nx \\equal{} l2\\pi \\\\\r\n\\end{array} \\right.(k,l \\in Z) \\minus{} \\minus{} \\minus{} \\minus{} > Bingo!!$\r\n :roll: So it no need to be pi/4 \r\nI prove that i'm not a spammer :rotfl:", "Solution_5": "[quote=\"SHIRO_DHKT\"][quote=\"TRAN THAI HUNG\"]Solve for $ x$\n\n$ \\sin ^4 x \\minus{} \\sin ^4 \\frac {\\pi }{4} \\equal{} \\sin ^4 (x \\minus{} \\frac {\\pi }{4})$ :D[/quote]\n\n$ \\sin x \\minus{} \\sin \\frac {\\pi }{4} \\equal{} \\sin (x \\minus{} \\frac {\\pi }{4}) \\\\\n\\Leftrightarrow 2\\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right)\\cos \\left( {\\frac {{x \\plus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} 2\\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right)\\cos \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} \\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} 0 \\\\\n\\cos \\left( {\\frac {{x \\plus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} \\cos \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\\\\n\\end{array} \\right. \\Leftrightarrow \\left[ \\begin{array}{l} \\frac {{x \\minus{} \\frac {\\pi }{4}}}{2} \\equal{} k\\pi (k \\in Z) \\\\\n\\cos \\frac {x}{2}\\cos \\frac {\\pi }{8} \\minus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} \\cos \\frac {x}{2}\\sin \\frac {\\pi }{8} \\plus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\minus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} 0 \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\sin \\frac {x}{2} \\equal{} 0 \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi \\\\\n\\frac {x}{2} \\equal{} l\\pi \\\\\n\\end{array} \\right.(k,l \\in Z) \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi \\\\\nx \\equal{} l2\\pi \\\\\n\\end{array} \\right.(k,l \\in Z) \\minus{} \\minus{} \\minus{} \\minus{} > Bingo!!$\n :roll: So it no need to be pi/4 \nI prove that i'm not a spammer :rotfl:[/quote]\r\nI have no idea what problem did you prove??? :huh:", "Solution_6": "[quote=\"SHIRO_DHKT\"][quote=\"TRAN THAI HUNG\"]Solve for $ x$\n\n$ \\sin ^4 x \\minus{} \\sin ^4 \\frac {\\pi }{4} \\equal{} \\sin ^4 (x \\minus{} \\frac {\\pi }{4})$ :D[/quote]\n\n$ \\sin x \\minus{} \\sin \\frac {\\pi }{4} \\equal{} \\sin (x \\minus{} \\frac {\\pi }{4}) \\\\\n\\Leftrightarrow 2\\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right)\\cos \\left( {\\frac {{x \\plus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} 2\\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right)\\cos \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} \\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} 0 \\\\\n\\cos \\left( {\\frac {{x \\plus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} \\cos \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\\\\n\\end{array} \\right. \\Leftrightarrow \\left[ \\begin{array}{l} \\frac {{x \\minus{} \\frac {\\pi }{4}}}{2} \\equal{} k\\pi (k \\in Z) \\\\\n\\cos \\frac {x}{2}\\cos \\frac {\\pi }{8} \\minus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} \\cos \\frac {x}{2}\\sin \\frac {\\pi }{8} \\plus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\minus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} 0 \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\sin \\frac {x}{2} \\equal{} 0 \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi \\\\\n\\frac {x}{2} \\equal{} l\\pi \\\\\n\\end{array} \\right.(k,l \\in Z) \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi \\\\\nx \\equal{} l2\\pi \\\\\n\\end{array} \\right.(k,l \\in Z) \\minus{} \\minus{} \\minus{} \\minus{} > Bingo!!$\n :roll: So it no need to be pi/4 \nI prove that i'm not a spammer :rotfl:[/quote]\r\n\r\nI guess ur just a spammer after all fuckwit. My solution above was an attempt. Im new to AOPS, didnt know if i was right/wrong, now that i see i was wrong im glad i can now see the error of my way, i think u should do the same.", "Solution_7": "[quote=\"addikaye03\"][quote=\"SHIRO_DHKT\"][quote=\"TRAN THAI HUNG\"]Solve for $ x$\n\n$ \\sin ^4 x \\minus{} \\sin ^4 \\frac {\\pi }{4} \\equal{} \\sin ^4 (x \\minus{} \\frac {\\pi }{4})$ :D[/quote]\n\n$ \\sin x \\minus{} \\sin \\frac {\\pi }{4} \\equal{} \\sin (x \\minus{} \\frac {\\pi }{4}) \\\\\n\\Leftrightarrow 2\\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right)\\cos \\left( {\\frac {{x \\plus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} 2\\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right)\\cos \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} \\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} 0 \\\\\n\\cos \\left( {\\frac {{x \\plus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} \\cos \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\\\\n\\end{array} \\right. \\Leftrightarrow \\left[ \\begin{array}{l} \\frac {{x \\minus{} \\frac {\\pi }{4}}}{2} \\equal{} k\\pi (k \\in Z) \\\\\n\\cos \\frac {x}{2}\\cos \\frac {\\pi }{8} \\minus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} \\cos \\frac {x}{2}\\sin \\frac {\\pi }{8} \\plus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\minus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} 0 \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\sin \\frac {x}{2} \\equal{} 0 \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi \\\\\n\\frac {x}{2} \\equal{} l\\pi \\\\\n\\end{array} \\right.(k,l \\in Z) \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi \\\\\nx \\equal{} l2\\pi \\\\\n\\end{array} \\right.(k,l \\in Z) \\minus{} \\minus{} \\minus{} \\minus{} > Bingo!!$\n :roll: So it no need to be pi/4 \nI prove that i'm not a spammer :rotfl:[/quote]\nI guess ur just a spammer after all -. My solution above was an attempt. Im new to AOPS, didnt know if i was right/wrong, now that i see i was wrong im glad i can now see the error of my way, i think u should do the same.[/quote]\r\nNonsense,\r\nThat is my solution. The one who just post a stupid message is a spammer. I post my solution, and that solution is right, at least in my opinion. You are wrong,it 's just your problem. What were you thinking when you said that i spammed. \r\nI just joked after I had posted my solution :mad: \r\nI'm not angry. If you found something wrong I my solution just post, do not said that someone spammed as a spammer. :mad:", "Solution_8": "[quote=\"SHIRO_DHKT\"][quote=\"TRAN THAI HUNG\"]Solve for $ x$\n\n$ \\sin ^4 x \\minus{} \\sin ^4 \\frac {\\pi }{4} \\equal{} \\sin ^4 (x \\minus{} \\frac {\\pi }{4})$ :D[/quote]\n\n$ \\sin x \\minus{} \\sin \\frac {\\pi }{4} \\equal{} \\sin (x \\minus{} \\frac {\\pi }{4}) \\\\\n\\Leftrightarrow 2\\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right)\\cos \\left( {\\frac {{x \\plus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} 2\\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right)\\cos \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} \\sin \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} 0 \\\\\n\\cos \\left( {\\frac {{x \\plus{} \\frac {\\pi }{4}}}{2}} \\right) \\equal{} \\cos \\left( {\\frac {{x \\minus{} \\frac {\\pi }{4}}}{2}} \\right) \\\\\n\\end{array} \\right. \\Leftrightarrow \\left[ \\begin{array}{l} \\frac {{x \\minus{} \\frac {\\pi }{4}}}{2} \\equal{} k\\pi (k \\in Z) \\\\\n\\cos \\frac {x}{2}\\cos \\frac {\\pi }{8} \\minus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} \\cos \\frac {x}{2}\\sin \\frac {\\pi }{8} \\plus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\minus{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} \\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\sin \\frac {x}{2}\\sin \\frac {\\pi }{8} \\equal{} 0 \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi (k \\in Z) \\\\\n\\sin \\frac {x}{2} \\equal{} 0 \\\\\n\\end{array} \\right. \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi \\\\\n\\frac {x}{2} \\equal{} l\\pi \\\\\n\\end{array} \\right.(k,l \\in Z) \\\\\n\\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} \\frac {\\pi }{4} \\plus{} k2\\pi \\\\\nx \\equal{} l2\\pi \\\\\n\\end{array} \\right.(k,l \\in Z) \\minus{} \\minus{} \\minus{} \\minus{} > Bingo!!$\n :roll: So it no need to be pi/4 \nI prove that i'm not a spammer :rotfl:[/quote]\r\nWhy do you have $ \\sin x \\minus{} \\sin \\frac {\\pi }{4} \\equal{} \\sin (x \\minus{} \\frac {\\pi }{4})$ ? My equation is $ \\sin ^4 x \\minus{} \\sin ^4 \\frac{\\pi }{4} \\equal{} \\sin ^4 (x \\minus{} \\frac{\\pi }{4})$\r\n :) Let take it easy guys,It's not neccessary to argue right? :)", "Solution_9": "Hey there, by inspection it's evident that x = \u03c0/4 is a solution to the equation. This means that t = 1 has to be a solution to the polynomial expression provided above by Dr Sonnhard Graubner. However it's not, and unless i'm mistaken here, Dr Sonnhard Graubner may have made an error. \r\n\r\nMy attempt is as follows:\r\n\r\nsin^4(x) - sin^4(\u03c0/4) = sin^4(x - \u03c0/4)\r\n\r\n = [sin(x).cos(\u03c0/4) - cos(x).sin(\u03c0/4)]^4\r\n\r\n = 1/4[sin(x) - cos(x)]^4\r\n\r\nSo, \r\n\r\nsin^4(x) - 1/4 = 1/4[sin(x) - cos(x)]^4\r\n\r\n4sin^4(x) - 1 = sin^4(x) - 4sin^3(x)cos(x) + 6sin^2(x)cos^2(x) - 4sin(x)cos^3(x) + cos^4(x)\r\n\r\n4sin^4(x) - [sin^2(x) + cos^2(x)]^2 = sin^4(x) - 4sin^3(x)cos(x) + 6sin^2(x)cos^2(x) - 4sin(x)cos^3(x) + cos^4(x)\r\n\r\n4sin^4(x) - sin^4(x) - 2sin^2(x)cos^2(x) - cos^4(x) = sin^4(x) - 4sin^3(x)cos(x) + 6sin^2(x)cos^2(x) - 4sin(x)cos^3(x) + cos^4(x)\r\n\r\n2sin^4(x) + 4sin^3(x)cos(x) - 8sin^2(x)cos^2(x) + 4sin(x)cos^3(x) - 2cos^4(x) = 0\r\n\r\nNow dividing by 2cos^4(x) produces:\r\n\r\ntan^4(x) + 2tan^3(x) - 4tan^2(x) + 2tan(x) - 1 = 0\r\n\r\ny^4 + 2y^3 - 4y^2 + 2y - 1 = 0\r\n\r\nIf P(y) = y^4 + 2y^3 - 4y^2 + 2y - 1\r\n\r\nP(1) = 0 ======> As per initial inspection\r\n\r\nHence, y - 1 is a factor and after performing long division,\r\n\r\n(y-1)(y^3 + 3y^2 - y + 1) = 0\r\n\r\nThe problem lies right there, the expression y^3 + 3y^2 - y + 1 definitely has one real root, which I cannot find, but can merely approximate. So i'm not sure whether my method is completely wrong, that's for you judge. Also worthy of mention is the almost symmetrical arrangement of coefficients of y^4 + 2y^3 - 4y^2 + 2y - 1, perhaps it can lead to a solution. Any help appreciated!", "Solution_10": "Pardon my double post, but I made a mistake in my observation of Dr Sonnhard Graubner's solution, as tan 22.5 solves this equation, which is half pi/4.", "Solution_11": "[quote=\"gurmies\"]Hey there, by inspection it's evident that x = \u03c0/4 is a solution to the equation. This means that t = 1 has to be a solution to the polynomial expression provided above by Dr Sonnhard Graubner. However it's not, and unless i'm mistaken here, Dr Sonnhard Graubner may have made an error. \n\nMy attempt is as follows:\n\nsin^4(x) - sin^4(\u03c0/4) = sin^4(x - \u03c0/4)\n\n = [sin(x).cos(\u03c0/4) - cos(x).sin(\u03c0/4)]^4\n\n = 1/4[sin(x) - cos(x)]^4\n\nSo, \n\nsin^4(x) - 1/4 = 1/4[sin(x) - cos(x)]^4\n\n4sin^4(x) - 1 = sin^4(x) - 4sin^3(x)cos(x) + 6sin^2(x)cos^2(x) - 4sin(x)cos^3(x) + cos^4(x)\n\n4sin^4(x) - [sin^2(x) + cos^2(x)]^2 = sin^4(x) - 4sin^3(x)cos(x) + 6sin^2(x)cos^2(x) - 4sin(x)cos^3(x) + cos^4(x)\n\n4sin^4(x) - sin^4(x) - 2sin^2(x)cos^2(x) - cos^4(x) = sin^4(x) - 4sin^3(x)cos(x) + 6sin^2(x)cos^2(x) - 4sin(x)cos^3(x) + cos^4(x)\n\n2sin^4(x) + 4sin^3(x)cos(x) - 8sin^2(x)cos^2(x) + 4sin(x)cos^3(x) - 2cos^4(x) = 0\n\nNow dividing by 2cos^4(x) produces:\n\ntan^4(x) + 2tan^3(x) - 4tan^2(x) + 2tan(x) - 1 = 0\n\ny^4 + 2y^3 - 4y^2 + 2y - 1 = 0\n\nIf P(y) = y^4 + 2y^3 - 4y^2 + 2y - 1\n\nP(1) = 0 ======> As per initial inspection\n\nHence, y - 1 is a factor and after performing long division,\n\n(y-1)(y^3 + 3y^2 - y + 1) = 0\n\nThe problem lies right there, the expression y^3 + 3y^2 - y + 1 definitely has one real root, which I cannot find, but can merely approximate. So i'm not sure whether my method is completely wrong, that's for you judge. Also worthy of mention is the almost symmetrical arrangement of coefficients of y^4 + 2y^3 - 4y^2 + 2y - 1, perhaps it can lead to a solution. Any help appreciated![/quote]\r\nWith Cardano, you can easily find the root of that equation. :|", "Solution_12": "hello, note that i have made the substitution $ t\\equal{}\\tan(\\frac{x}{2})$ and solving $ t^2\\plus{}2t\\minus{}1\\equal{}0$ gives us\r\n$ \\tan(\\frac{x}{2})\\equal{}\\minus{}1\\plus{}\\sqrt{2}$ or $ x\\equal{}\\frac{\\pi}{4}\\plus{}2k_1\\pi$, with $ k_1 \\in \\mathbb{Z}$\r\n$ \\tan(\\frac{x}{2})\\equal{}\\minus{}1\\minus{}\\sqrt{2}$ or $ x\\equal{}\\minus{}\\frac{3\\pi}{4}\\plus{}2k_2\\pi$, with $ k_2 \\in \\mathbb{Z}$\r\nSonnhard.", "Solution_13": "I understand Dr =)", "Solution_14": "[quote=\"Dr Sonnhard Graubner\"]hello, note that i have made the substitution $ t \\equal{} \\tan(\\frac {x}{2})$ and solving $ t^2 \\plus{} 2t \\minus{} 1 \\equal{} 0$ gives us\n$ \\tan(\\frac {x}{2}) \\equal{} \\minus{} 1 \\plus{} \\sqrt {2}$ or $ x \\equal{} \\frac {\\pi}{4} \\plus{} 2k_1\\pi$, with $ k_1 \\in \\mathbb{Z}$\n$ \\tan(\\frac {x}{2}) \\equal{} \\minus{} 1 \\minus{} \\sqrt {2}$ or $ x \\equal{} \\minus{} \\frac {3\\pi}{4} \\plus{} 2k_2\\pi$, with $ k_2 \\in \\mathbb{Z}$\nSonnhard.[/quote]\r\nI can't understand you Dr." } { "Tag": [], "Problem": "Did anyone watch House last night? Because I didn't quite understand the ending. What was the deal with the \"you're in for an all-nighter\" thing? Was it the season finale? Because it was either extremely anti-climactic or I'm really missing something substantial. Also what happened to the kid?\r\n\r\nI'm confused...", "Solution_1": "House figured out that the kid had too much iron in his blood or something, and that he would have to get it drained once in a while for the rest of his life.\r\n\r\nAfter that, House didn't tell Foreman that the kid had already been diagnosed, and he lets Foreman keep doing the tests. I still haven't figured out why he did that, though... Maybe House was mad at Foreman for accusing him of canceling the interview?\r\n\r\nThat wasn't the finale; their will be at least one more episode." } { "Tag": [ "trigonometry" ], "Problem": "Solve for x E [0,2pi]: sin(3x) - sin(x) = cos(2x)\r\n\r\n\r\nlet's expand this\r\n\r\nsin(3x) - sin(x) = cos(2x)\r\n=> sin(2x)cos(x) + sin(x)cos(2x) - sin(x) = cos^2(x) - sin^2(x)\r\n=> [2sin(x)cos(x)]cos(x) + sin(x)[cos^2(x) - sin^2(x)] - sin(x) = cos^2(x) - sin^2(x)\r\n=> 2sin(x)cos^2(x) + sin(x)[cos^2(x) - sin^2(x)] - sin(x) = cos^2(x) - sin^2(x)\r\n\r\nwe have a lone sin(x), so let's change everything to sin(x)\r\n\r\n=> 2sin(x)(1 - sin^2(x)) + sin(x)[1 - sin^2(x) - sin^2(x)] - sin(x) = 1 - sin^2(x) - sin^2(x)\r\n=> 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x) - sin(x) = 1 - 2sin^2(x)\r\n=> 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x) - sin(x) - 1 + 2sin^2(x) = 0\r\nnow this is ugly, let's try and simplify this a bit\r\n\r\n=> -4sin^3(x) + 2sin^2(x) + 2sin(x) - 1 = 0\r\nthis is a cubic in sin(x), to cut down on typing for the mean time, let's write sin(x) as y, we get:\r\n\r\n-4y^3 + 2y^2 + 2y - 1 = 0 ........ah, that looks a little better\r\n=> -2y^2(2y - 1) + (2y - 1) = 0\r\n=> (2y - 1)(1 - 2y^2) = 0\r\n=> y = 1/2 or y = +/- sqrt(1/2)\r\n\r\nbut, y = sin(x)\r\n=> sin(x) = 1/2 or sin(x) = +/- sqrt(1/2) = +/- 1/sqrt(2) = +/- sqrt(2)/2\r\n=> x = pi/6, 5pi/6 or x = pi/4, 3pi/4, 5pi/4, 7pi/4\r\n\r\nThe problem is that when i graph this to check, the only valid answers are pi/6 and 5pi/6. Could someone check this for me.\r\n\r\nTHANKS", "Solution_1": "I plugged in the values wrong into my calculator. They are correct. Sorry for the confusion.", "Solution_2": "[hide=\"I will just do the problem again\"]\n$sin(3x)-sin(x) = cos(2x)$\n$sin(2x)cos(x)+cos(2x)sin(x)-sin(x) = cos(2x)$\n$2sin(x) cos^{2}(x)+sin(x)-2sin^{3}(x)-sin(x) = 1-2sin^{2}(x)$\n\nlet $u = sin(x)$\n\n$2u(1-u^{2})+u-2u^{3}-u = 1-2u^{2}$\n$-4u^{3}+2u = 1-2u^{2}$\n$4u^{3}-2u^{2}-2u+1 = 0$\n$2u^{2}(2u-1)-(2u-1) = 0$\n$(2u^{2}-1)(2u-1) = 0$\n$sin(x) = \\frac{\\sqrt2}{2}$ or $sin(x) = \\frac12$\n\n$x = \\frac{\\pi}{4}, \\frac{3\\pi}{4}, \\frac{\\pi}{6}, \\frac{5\\pi}{6}$\n[/hide]", "Solution_3": "[hide=\"alternatively...\"]\n\\[\\sin 3x=\\sin (2x+x)=\\sin 2x\\cos x+\\cos 2x\\sin x\\]\n\\[\\sin x=\\sin (2x-x)=\\sin 2x\\cos x-\\cos 2x\\sin x\\]\nso that the equation becomes\n\\[2\\cos 2x\\sin x=\\cos 2x\\]\nEither $\\cos 2x=0$ or $\\sin x=\\frac{1}{2}$. You should be able to figure out the solutions from here...\n[/hide]", "Solution_4": "[hide]$\\implies 3\\sin x-4\\sin^{3}x-\\sin x=\\cos 2x$\nLet $\\sin x=y$\n$\\implies 4y^{3}-2y^{2}-2y+1=0$.\nSolve: Using Rational Roots Thm, 1/2 is a root, and the other two roots are $\\pm\\sqrt{2}/2$.\n\nSince $\\sin x=1/2, \\pm\\sqrt{2}/2$, the solutions in $[0,2\\pi]$ are $\\pi/4, \\pi/3, 2\\pi/3, 3\\pi/4, 5\\pi/4, 7\\pi/4$.[/hide]" } { "Tag": [ "arithmetic series" ], "Problem": "An arithmetic series is called a concatenation series if the sum of the serries is represented by the concatenation of the first and last terms. For example, [b]17[/b]+19+21.....+[b]85[/b]=[b]1785[/b]. Find a concatenation series with 41 consecutive intergers and a four digit sum. What is the sum of the integers of this series?\r\n\r\nThanks[/b]", "Solution_1": "[quote=\"jacob8888888\"]An arithmetic series is called a concatenation series if the sum of the serries is represented by the concatenation of the first and last terms. For example, [b]17[/b]+19+21.....+[b]85[/b]=[b]1785[/b]. Find a concatenation series with 41 consecutive intergers and a four digit sum. What is the sum of the integers of this series?\n\nThanks[/b][/quote]\r\n\r\n[hide]\nLet:\n$a=$ the first number in the series\n$b=a+40$ (the last number in the series)\n$s=$ the sum of the numbers in the series\nUsing the formula for the sum of an arithmetic series,\n$41\\frac{(a+b)}{2}=100a+b$\n$41(a+20)=101a+40$\n$41a+820=101a+40$\n$780=60a$\n$a=13$\nSo the sum of the numbers in the series is 1353.\n[/hide]", "Solution_2": "Thank you very much." } { "Tag": [], "Problem": "[b](1) If 2 water tap fill a tank in 75/8 hours. If a water tap of larger diameter has fill the tank 10 hour less than the first tap.\nthen find their seperate time to fill the tank.[/b]", "Solution_1": "let the volume coming out of the small tap be x lt per hour and let the volume of water coming out of bigger tap be y lt per hour.\r\n\r\n\r\n 75/8 (x+y) = capacity of the tank.\r\n\r\nle the small tap take `s` hours to fill the tank. then big tap taes s-10 hours to fill the tank.\r\n\r\n xs= 75/8(x+y) and y(s-10)=75/8(x+y)\r\n\r\n\r\n now u have 3 equations and 3 unknowns. solve them to obtain the required answers" } { "Tag": [ "geometry", "function", "search", "blogs", "angle bisector" ], "Problem": "Hi everyone,\r\n\r\nDo you know any program that can be used to draw diagrams in Geometry (with the functions of locating midpoints, midlines, angle bisector) that is free? (got from Internet.) or having a \"good\" trial time or or cheap one?\r\n\r\nThanks in advance.", "Solution_1": "What about Cabri?", "Solution_2": "I've seen this question on AoPS/ML at least 3 or 4 times. Try a search of the forum.", "Solution_3": "rrusczyk was talking about a new program, [url]http://www.geometer.org/geometer[/url], recently on his blog, and it seems pretty good. Geometer is free, as is KSEG ([url]http://www.mit.edu/~ibaran/kseg.html[/url]), which is also good, though somewhat simpler and less powerful.", "Solution_4": "I suggest buying Geometer's Sketchpad. It is so much easier to use than anything else out there. I was using KSEG until my friend gave me Geometer's Sketchpad. It is SO much better, even though it is similar.", "Solution_5": "I have heard of Sketchpad, but I dont have money to buy it, and I dont know anyone having it (except for the teachers, but they wont let me borroe it.)I'll just try the KSEG then. Thank you for all replies.", "Solution_6": "You can download Geometers sketch pad(trial) in its site.", "Solution_7": "http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/doc_en/\r\nI think this program is great and its FREE :D", "Solution_8": "Yeah, C.a.R it's my favourite, it's very useful. I think it's better than Cabri and many others. :D" } { "Tag": [ "geometry", "3D geometry", "combinatorics proposed", "combinatorics" ], "Problem": "divided a square into any smaller square prove that there is at least two equal square .", "Solution_1": "I wouldn't bet on that:\r\n\r\n[url]http://mathworld.wolfram.com/PerfectSquareDissection.html[/url]\r\n\r\nDid you conjecture this?", "Solution_2": "but i think a similar thing with cubes would be a correct statement..... :?", "Solution_3": "Yes, it holds for cubes.\r\n\r\nPierre." } { "Tag": [ "MATHCOUNTS" ], "Problem": "What'd you get at chapter?\r\n\r\nJust trying to estimate my position at states this Sat.\r\n\r\nEDIT: please reply with your chapter location, thanks!\r\n\r\nEDIT 2: I got a 31 and got second, while first in my chapter was 33. I think our chapter is very weak.", "Solution_1": "Using my various contacts :lol: I know that one of the chapters was won with a 35.", "Solution_2": "Our chapter won with a 45 :wink:\r\n\r\nAre you toledo? I think David Li won in toledo. This is cleveland.", "Solution_3": "Mr. Frost Sir: Which one? Cincinnati?\r\n\r\nI got 13th in Cincinnati (my first time participating), but am not sure of my score: It's between 27 and 29.\r\n\r\nI've improved a bit since then.. hoping to make Nationals some how :P\r\n\r\nSycamore is Godly :(", "Solution_4": "isn't sycamore a national private school?", "Solution_5": "I think its public.", "Solution_6": "Whichever chapter Sycamore is in, they had top 6 scores and 7 of top 8 overall.", "Solution_7": "mr. frost, I just hope that the contact you used wasn't my coach.", "Solution_8": "sycamore definitely is a private school, there's one in Indianapolis, IN. I think mky dad told me that sycamore was national though. It makes sense, since it costs alot and the level of the teachers.", "Solution_9": "All I know is the best MathCounts school in Ohio = Sycamore :P (It's Public in Cincinnati)\r\n==========================================\r\nHmm.... could someone from Ohio, who participated last year, do the [b]2003[/b] State Target and Sprint, and tell me:\r\n\r\n1) What they got on it.\r\n2) How much I should be able to get if I stand an ~okay~ chance (60-70%) of making it to the nationals??\r\n\r\nI hope this isn't too much :)", "Solution_10": "I went to the Akron University chapter competition. I'm from Copley Fairlawn Middle School. We and Revere had the top 10 places in our chapter.", "Solution_11": "i am not sure which chapter I am in for Ohio, but I know I got in the 20's :( I still made it to state. Unfortunately this is my first and last year of mathcounts. :mad:", "Solution_12": "[quote=\"hoshattack\"]i am not sure which chapter I am in for Ohio, but I know I got in the 20's :( I still made it to state. Unfortunately this is my first and last year of mathcounts. :mad:[/quote]\r\nSame.\r\n\r\nI got 27~29 and barely made it to state from the Cincinnati chapter.\r\nAgain, my 1st and last year. :(", "Solution_13": "=( for both of you. This is my 2nd year (im an 8th grader now)\r\nLast year I got 1st at chapter with a 39 overall. At state I got 62nd", "Solution_14": "Official word : Cut off last year in Ohio was a 37(23/14).", "Solution_15": "[quote=\"frost13\"]Official word : Cut off last year in Ohio was a 37(23/14).[/quote]\r\nCut-off for the nationals??\r\n\r\nYAY!! :w00t: \r\n\r\nI just got that exact score on last years State, meaning I have a chance to go to the nationals :lol:\r\n\r\nThanks!", "Solution_16": "dude, our chapter is much more competitive than yours...\r\n\r\nthe scores were\r\n\r\n1=45\r\n2=45\r\n3=44\r\n4=43\r\n5=40\r\n6=39\r\n7=38\r\n8=37\r\n9=36\r\n10=35", "Solution_17": "[quote=\"yhh3002\"]dude, our chapter is much more competitive than yours...\n\nthe scores were\n\n1=45\n2=45\n3=44\n4=43\n5=40\n6=39\n7=38\n8=37\n9=36\n10=35[/quote]\r\n\r\nUm, that is a little bit off.\r\n\r\n1=45, 2=43, 3=41, and the scores go from there. 10 was indeed a 35. This is the hardest chapter in Michigan, and usually the national qualifiers come from here.", "Solution_18": "Our chapter:\r\n\r\n1-45\r\n2-44\r\n3-44\r\n4-44\r\n5-43\r\n6-42\r\n7-42\r\n8-42\r\n9-41\r\n10-41\r\n11-40\r\n12-40\r\n\r\n :dry:", "Solution_19": "Ummm...lol \r\n\r\nYou guys should move to South Carolina:\r\n\r\nOur Chapter:\r\n\r\n1st (me): 33\r\n2nd: 32\r\n3rd: 29\r\n4th: 25\r\n5th: 24...\r\n\r\nYou guys get the point lol", "Solution_20": "[quote=\"Walk Around The River\"][quote=\"#H34N1\"]Our chapter:\n\n1-45\n2-44\n3-44\n4-44\n5-43\n6-42\n7-42\n8-42\n9-41\n10-41\n11-40\n12-40\n\n :dry:[/quote]\n\n\nw--...?[/quote]\r\n\r\nWhy? Is it too bad :maybe:", "Solution_21": "Too good.. :| \r\n\r\nOne of the best Chapters in the nation, from those scores...", "Solution_22": "Yah, but our chapter kinda stinks because our people only know how to do chapter competitions. They can't do state levle :(", "Solution_23": "is that chapter in ohio? if so, where/which one?", "Solution_24": "Cleveland. We have no hope at state :blush: \r\n\r\nours has:\r\n\r\n1-46\r\n2- 46\r\n3 46\r\n4 46\r\n5 46\r\n6 46\r\n7 46\r\n8 46\r\n9 45\r\n10 17\r\n\r\n[/sarcasm]\r\n\r\nNo, just jk\r\n\r\nThe real scores are:\r\n\r\n1-45\r\n2-43\r\n3-43\r\n4-42\r\n5-42\r\n6-37\r\n7-37\r\n8-35\r\n9-33\r\n10-33", "Solution_25": "i just noticed, but isn't there already a thread asking for how ppl did in chapter?", "Solution_26": "well thats good, i'll probably bomb it and get like 200th place...\r\nor I may get lucky and go to nats! :D\r\n\r\noh, and mwpl11, this is specifically for Ohio, because not many ppl acutally visit their state forum.", "Solution_27": "last year I got 16/3\r\n\r\nfor a grand total!!!!! 22 points lol\r\n\r\nI was like 200 something place \r\n\r\nThis year I'm going for nats!", "Solution_28": "[quote=\"hoshattack\"]\noh, and mwpl11, this is specifically for Ohio, because not many ppl acutally visit their state forum.[/quote]\r\n\r\noh, oops :oops: .", "Solution_29": "See you guys tomorrow!! \r\n\r\nGood Luck to Everyone!! :thumbup:", "Solution_30": ":lol: See you guys tomorrow!" } { "Tag": [ "algebra", "polynomial", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "the group generated by $ x_1,x_2,x_3$ with relations $ x_{2}x_{1}x_{2}^{ \\minus{} 1} \\equal{} x_{1}^{2}, ~x_{3}x_{2} x_{3}^{ \\minus{} 1} \\equal{} x_{2}^{2},~x_{1}x_{3}x_{1}^{ \\minus{} 1} \\equal{} x_{3}^{2}~$ is trivial.\r\nwhy? :roll:", "Solution_1": "Where does this problem come from?", "Solution_2": "exercise from j.p. serre, trees.", "Solution_3": "http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2007;task=show_msg;msg=3319.0002", "Solution_4": "thank you!\r\n\r\nit was clear that the proof consists of simple algebraic manipulations, but how to find them ... :maybe: is there a sort of topological motivation?", "Solution_5": "[quote=\"-oo-\"] how to find them?[/quote]\r\nI have heard that Gr\u00f6bner bases is a method to find some kind of relations in algebraic structures. However, I'm not very familiar with those bases. Gr\u00f6bner bases are discussed for example in Eisenbud's book \"Commutative algebra with a view toward algebraic geometry\".", "Solution_6": "Gr\u00f6bner bases are there to find out if a polynomial lies in an ideal generated by some other polynomials... over a commutative ring, or more likely, over a field. Is this really of any help for a (noncommutative) group theory problem? I cannot name sources, but I think I heard such problems can even be undecidable...\r\n\r\n darij" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "From a set 2n+1 weights(where n is natural number), if any one weight is excluded, then the remaining 2n weights can be divided into 2 sets of n weights that balance each other. Prove that all the weights are equal", "Solution_1": "This question is often phrased using sports. I searched for \"football\" and found the most recent version that I've commented on:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=275557\r\n(Following links from there should lead to at least one solution.)\r\nSearching for \"japan 1999 weight\" works pretty well, too:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=27219" } { "Tag": [ "abstract algebra", "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "I can show that for a local noetherian ring $ (A, \\mathfrak{m})$ that $ \\hat{A}$ is flat over $ A$. How to show that it's faithfully flat? What for non-local rings?", "Solution_1": "Does the following make any sense? \r\n\r\n[quote]\nIt suffices to show that $ \\mathfrak{m}\\hat{A} \\neq \\hat{A}$. Assume that $ \\mathfrak{m}\\hat{A} \\equal{} \\hat{A}$ for $ \\hat{A} \\neq 0$, then according to NAK there is an $ x \\in \\hat{A}$ with $ x \\equiv 1 \\ mod (\\mathfrak{m} \\hat{A})$ such that $ x\\hat{A} \\equal{} 0$. But $ x \\equal{} 1 \\plus{} m$ is a unit in $ \\hat{A}$ for the $ \\mathfrak{m}$-adic completion, so $ x\\hat{A} \\equal{} \\hat{A} \\equal{} 0$.\n[/quote]\r\n\r\nStill not sure about that... :blush:" } { "Tag": [ "Putnam", "number theory proposed", "number theory" ], "Problem": "Find the last digit of this numbre\r\n$\\left[\\frac{10^{20000}}{10^{100}+3}\\right]$.", "Solution_1": "$[\\frac{10^{20000}}{10^{100}+3}=10^{19900}-3*10^{19800}+\\dots -3^{199}+[\\frac{3^{200}}{10^{100}}-\\frac{3^{201}}{10^{200}}+\\dots]=-3^{199}(mod \\ 10^{100})=7(mod \\ 10).$", "Solution_2": "[quote=\"Rust\"]$[\\frac{10^{20000}}{10^{100}+3}=10^{19900}-3*10^{19800}+\\dots -3^{199}+[\\frac{3^{200}}{10^{100}}-\\frac{3^{201}}{10^{200}}+\\dots]=-3^{199}(mod \\ 10^{100})=7(mod \\ 10).$[/quote]\r\nSorry $Rust$ but your answer is wrong,I can't find your mistake :? .\r\nHere is my solution.\r\n${[\\frac{10^{20000}}{10^{100}+3}]=\\frac{10^{20000}-3^{200}}{10^{100}+3}+[\\frac{3^{200}}{10^{100}+3}}]=\\frac{10^{20000}-3^{200}}{10^{100}+3}$\r\nLaste digit of the numbre $10^{20000}-3^{200}$ is $9$ and laste digit of the numbre $10^{100}+3$ is $3$ =>$\\frac{10^{20000}-3^{200}}{10^{100}+3}=3(mod\\ 10)$", "Solution_3": "[quote=\"Tiks\"][quote=\"Rust\"]$[\\frac{10^{20000}}{10^{100}+3}=10^{19900}-3*10^{19800}+\\dots -3^{199}+[\\frac{3^{200}}{10^{100}}-\\frac{3^{201}}{10^{200}}+\\dots]=-3^{199}(mod \\ 10^{100})=7(mod \\ 10).$[/quote]\nSorry $Rust$ but your answer is wrong,I can't find your mistake :? .\nHere is my solution.\n${[\\frac{10^{20000}}{10^{100}+3}]=\\frac{10^{20000}-3^{200}}{10^{100}+3}+[\\frac{3^{200}}{10^{100}+3}}]=\\frac{10^{20000}-3^{200}}{10^{100}+3}$\nLaste digit of the numbre $10^{20000}-3^{200}$ is $9$ and laste digit of the numbre $10^{100}+3$ is $3$ =>$\\frac{10^{20000}-3^{200}}{10^{100}+3}=3(mod\\ 10)$[/quote]\r\nYoyr mistake is laste digit of the number $10^{2000}-3^{200}=1(mod \\ 10$ (not 9)", "Solution_4": "[quote=\"Rust\"][quote=\"Tiks\"][quote=\"Rust\"]$[\\frac{10^{20000}}{10^{100}+3}=10^{19900}-3*10^{19800}+\\dots -3^{199}+[\\frac{3^{200}}{10^{100}}-\\frac{3^{201}}{10^{200}}+\\dots]=-3^{199}(mod \\ 10^{100})=7(mod \\ 10).$[/quote]\nSorry $Rust$ but your answer is wrong,I can't find your mistake :? .\nHere is my solution.\n${[\\frac{10^{20000}}{10^{100}+3}]=\\frac{10^{20000}-3^{200}}{10^{100}+3}+[\\frac{3^{200}}{10^{100}+3}}]=\\frac{10^{20000}-3^{200}}{10^{100}+3}$\nLaste digit of the numbre $10^{20000}-3^{200}$ is $9$ and laste digit of the numbre $10^{100}+3$ is $3$ =>$\\frac{10^{20000}-3^{200}}{10^{100}+3}=3(mod\\ 10)$[/quote]\nYoyr mistake is laste digit of the number $10^{2000}-3^{200}=1(mod \\ 10$ (not 9)[/quote]\r\nOK!\r\nCan you say me what is the laste digit of this number $3^{200}=(81)^{50}$?\r\nI think $1$,you don't think so?\r\nNow,if I am not wrong :lol: ,last digit of the number $10^{2000}$ is $0$,am I right? :) \r\nNow if you again think that the laste digit of $10^{2000}-3^{200}$ is $1$,then I don't know what to do :) .", "Solution_5": "This is the exact same problem as Putnam 1986 A2.", "Solution_6": "I am sorry, it is my mistake. $-3^{199}=-1/3=-7=3(modm \\ 10)$" } { "Tag": [ "parameterization" ], "Problem": "Let f(n) denote the sum of (all) digits of natural number n. Example: f(2004)=2+0+0+4=6. Prove that for each natural n we can choose convenient value of natural parameter p such that the equation f(npx)=f(x) has solution in natural numbers x that doesn't contain any \"9\" in its notation (of course in base 10). \r\n\r\n\r\nany ideas about this one.. i have no clue.\r\ni think this is advanced but if its easy u can move it..\r\n\r\n[edit] Thanks!", "Solution_1": "Here's one thought -- every natural has multiples made up exclusively of 0's and 1's. You could construct your 0's and 1's in such a way as to force the existence of a root, perhaps. Multiples of 9 seem like promising numbers to be converted in this way." } { "Tag": [ "trigonometry", "Putnam", "function", "algebra", "polynomial", "integration", "algebra unsolved" ], "Problem": "Prove that:\r\n\r\n$cos(x) = \\frac{e^{ix} + e^{-ix}}{2}$\r\n\r\nusing de Moivre's theorem: $cos(x) + isinx = e^{ix}$", "Solution_1": "This was a [b]Putnam[/b] problem? :huh: \r\n\r\n$e^{-ix} = \\cos (-x) + i \\sin (-x) = \\cos x - i \\sin x$\r\n$e^{ix} + e^{-ix} = 2 \\cos x$\r\n\r\nAlternately, if you want to be smart about it,\r\n\r\n$e^{-ix} = \\frac{1}{\\cos x + i \\sin x} = \\frac{\\cos x - i \\sin x}{\\cos^2 x + \\sin^2 x} = \\cos x - i \\sin x$\r\n\r\nEdit: [b]Remark:[/b] The result above, while simple, is useful in allowing us to quickly define trig and arctrig functions on complex values.\r\n\r\nFurthermore, we can set any arbitrary complex number $z = e^{ix}$ (without adding an $r$ term, because we can have $x$ complex) and arrive at the conclusion\r\n\r\n$2 \\cos x = z + \\frac{1}{z} \\Rightarrow 2 \\cos nx = z^n + \\frac{1}{z^n}$,\r\n\r\nAn easy proof of the result that $z^n + \\frac{1}{z^n}$ can be written as a polynomial in $z + \\frac{1}{z}$.", "Solution_2": "Agg, it's too simple. Thanks why didn't I see it before :rotfl:", "Solution_3": "It wasn't a Putnam problem at all. The real Putnam question is the following one:\r\n\r\nFor which $n \\in \\left\\{1,2,\\,\\cdots,10\\right\\}$ is $\\int_{0}^{2\\pi} \\cos x \\cos 2x\\,\\cdots \\cos nx \\,dx \\neq 0?$" } { "Tag": [], "Problem": "For Algebra2/Trig I am having difficulty doing the homework, I don't understand how to solve it.\r\n\r\nDoes anyone know of a site that gives tutorials for this subject?\r\nOr anyone who is willing to try and bare with me to help me out on this?\r\n\r\nI need help with Recursive and Explicit solving by the way...\r\n\r\nThanks\r\n -totally_stupid? (aka, Kira)", "Solution_1": "Can you post a specific question?\r\nThe AOPS wiki (link on the left) has tutorial-like pages on lots of things", "Solution_2": "ya kno, your teacher is an efficient resource..." } { "Tag": [ "LaTeX", "combinatorics unsolved", "combinatorics" ], "Problem": "Junior Balkan MO 2008, Problem 4\r\n\r\nA [b][i]4*4[/i][/b] table is divided into [i]16[/i] white unit square cells. Two cells are called neighbors if they share a common side. A move consists in choosing a cell and the colors of neighbors from white to black or from black to white. After exactly [i]n [/i]moves all the cells were black. Find all possible values of[i] n[/i]. :?:", "Solution_1": "Posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=211773 (I searched for \"JBMO 2008\" (without quotation marks) to find this thread.)", "Solution_2": "obviously n can be any number such that $ n>\\equal{}4$(there is an easy choice of 4 squares that makes the whole square black).\r\nnow all we have to do is prove that with 3 moves we cannot achieve a entirely black square.\r\nbut this is obvious, since each move changes at most 5 squares(we then have that using 3 moves we change at most 15 squares,absurd). Done. :D", "Solution_3": "[quote=\"v235711\"]obviously n can be any number such that $ n > \\equal{} 4$ (there is an easy choice of 4 squares that makes the whole square black).[/quote] In LaTeX, \\geq gives $ \\geq$. I'd be curious to see your solution with $ n \\equal{} 5$ :roll:", "Solution_4": "oops, sorry , I have a mistake in my solution. it is not for all n>=4, it is just for n even>=4...\r\nI still have to find out whether there is not a sequence of moves using an odd number of moves...\r\nsorry for my mistake :oops: \r\nbut can we color the table black using 5 moves?what is the least odd n required? :maybe:", "Solution_5": "If you read the link I posted, it is shown that no odd $ n$ is possible: note that every time you flip any square, you flip exactly one of the squares in positions (1, 2), (2, 4), (3, 1) and (4, 3)." } { "Tag": [ "number theory", "\\/closed" ], "Problem": "What is the purpose of a math jam? What are they used for? How often do we have math jams?", "Solution_1": "math jams are for various purposes. For example, the math jam that's going on right now while I type is about the upcoming \"Introduction to Number Theory\" class. They're usually used to discuss math problems or upcoming classes. Math jams are pretty rare, although recently there have been a lot, because of all the upcoming classes.", "Solution_2": "[quote=\"Nerd_of_the_Ages\"]math jams are for various purposes. For example, the math jam that's going on right now while I type is about the upcoming \"Introduction to Number Theory\" class. They're usually used to discuss math problems or upcoming classes. Math jams are pretty rare, although recently there have been a lot, because of all the upcoming classes.[/quote]\r\n\r\nNice description, But they are also about theories or special things, like HSSiM(Pardon me if I misstyped it)", "Solution_3": "whoops, meant to say intro to algebra, not intro to number theory" } { "Tag": [], "Problem": "bob drives to work on the highway. he averages 45 mph for the trip. when he drives the back route he averages 36 mph. the trip on the back road is 3 miles shorter and takes 5 more minutes than the highway \r\nroute. how many miles long is the highway route??", "Solution_1": "[hide] Using $ d=rt$, we can set up a system of equations.\n\n$ d=45t$\n$ d-3=36(t+\\frac{1}{12})$\n\nSolving the second equation,\n\n$ d-3=36t+3 \\implies d=36t+6$.\n\nSubstituting,\n\n$ 45t=36t+6 \\implies t=\\frac23$. \n\nThus,\n\n$ d=45(\\frac23)\\implies d=\\boxed{30}$.[/hide]", "Solution_2": "[b][size=150]Bubba the bicyclist goes for a long distance ride. three hours later, bubbete decides to bring him lunch, so she hops in her car and zooms down the road at a blistering 42 mph.\nIf it takes her an hour and a half to catch Bubba, how fast was she riding?[/size][/b]", "Solution_3": "[hide]\nBubba distance = bubbete distance.\n\nx*4.5 = 42*(1.5)\n\nx=14.\n[/hide]", "Solution_4": "[color=green][size=200][b]THX I UNDERSTAND IT NOW. :)[/b][/size][/color][/b]", "Solution_5": "[b][size=150]Two trains are 300 miles apart and traveling toward each other on parrallel tracks. One travels at 35 mph and the other 45mph. At the front of the faster train is a bee that flies exactly 72mph. The bee flies to the front of the slower train, then instantaneously turns around and flies to the front of the faster train. The bee continues flying from the front of one train to the other until the trains pass. At the exact moment the trains passed, what was the total distance the bee flown[/size]?[/b]", "Solution_6": "Don't be posting homework for other people to solve!\r\nI go to the same school as you guys and have the same Teacher!\r\nDo your own homework, Just like I did!", "Solution_7": "audilover, do not post Lomes problems or AIME15 will stalk you. I go o your school HPMS", "Solution_8": "BTW, are you guys both 7th graders(audilover&sheeprule), cause only 7th graders could be stupid enough to post homework! :rotfl:", "Solution_9": "[b]I[size=150] apologize. Except, I wanted to share this amazing problem with people. I can't do that? What if I already know how to solve it? I can't post it becuase it's a problem from my homework? Anyways, I am an eigth grader so you just got PWNED. Who is this anyway? Nikil? or somebody? and who is AIME15? and ALSO! WHo posts things at 8 in the morning? That is truly sick.[/size][/b]", "Solution_10": "Turn down the font size.\r\n\r\nAlso, DO NOT post homework unless you have tried it, and SHOW that you have tried it.\r\n\r\nMaybe you should also learn to spell your teacher's name (@mathlearner2012) :wink:", "Solution_11": "[b]WELL, just to let you know, I post things in big font for the people who have trouble reading small font. SInce most people that go on this website have glasses and bad eyesight, I post it bigger so that more people have the opportunity to be enlightened. AND, if you are AIME15 then who in the world is AIME15USAMO?!! You both have the same picture and same location. Its either somebody is a poser or you guys are married or related.\nAnd now let's end all this and focus on math again. :)[/b]", "Solution_12": "Are you blind? AIME15 and AIME15USAMO have completely different locations and pictures. And I don't think big font will enlighten people anymore than AOPS already does. And its weird how you live in China, but have get the same math problems as I do. :wink: :wink:" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "This is my first post here, salutes to everybody :).\r\n\r\nWell, how could i obtain \"alpha\" from this equation:\r\n\r\n[color=blue]Rho = (A * tan [ (90- alpha) * (PI/ (2*180)) ] ) / ( [(1-n)/(1+n)]^(Ec/2) )\n[/color]\r\n\r\nWhere \"n\" is:[color=blue]n = Ec * sin( alpha * (PI/180) )[/color]\r\n\r\nAnd:[color=blue]\n\"A\" and \"Rho\" are known values.\n\"Ec\" is a constant value: Ec = 0.08181.[/color]\r\n\r\nSome help would be pretty much appreciated, i'm being stucked with this for two days :?\r\n\r\nEdit: Hm... i'm not sure if this post should've been posted in this thread, sorry if not. In such case, please, somebody tell me where i should post it, or move it. :blush:", "Solution_1": "Well, this problem seems to be quite difficult, at least for me.\r\nI've could not dedicate to this problem many time, i've been very busy with a lot of other more urgent things.\r\n\r\nThe only advance i've been able to do in the little time i've had available for this (this evening concretely ;) ) is shown below, but first i have to apologize, because the equation i posted originally was not totally correct, so i sorry indeed if someone has been trying to solve it.\r\n\r\nHere i go with the original, and correctly wrote this time, equation:\r\n[color=blue]\nRho = A * [ tan [ (90- alpha) * (PI/ (2*180)) ] / [(1-n)/(1+n)](Ec/2) ] N\n[/color]\r\nWhere \"n\" is: [color=blue]n = Ec * sin( alpha * (PI/180) )[/color]\r\n\r\nAnd:[color=blue]\n\"A\", \"Rho\" and \"N\" are known values.\n\"Ec\" is a constant value: Ec = 0.08181.\n[/color]\r\nAnd here are the steps i've done today:\r\n[color=blue]\n1.- Rho/A = [ tan [ (90- alpha) * (PI/ (2*180)) ] / [(1-n)/(1+n)](Ec/2) ] N\n\n2.- [Rho/A]1/N = tan [ (90- alpha) * (PI/ (2*180)) ] / [(1-n)/(1+n)](Ec/2) \n\n3.- [Rho/A]1/N = tan [ (45- alpha/2) * (PI/ 180) ] / [(1-n)/(1+n)](Ec/2) \n\n4.- [/color] (PI/180) is there to convert the angle into Radians, so i can get it out of there:[color=blue]\n \n[Rho/A]1/N = tan [ (45- alpha/2) ] / [(1-n)/(1+n)](Ec/2) \n\n5.- [/color]We have that :[color=blue]\ntan(45- alpha/2) = ( tan (45) - tan (alpha/2) ) / ( 1 + tan(45)*tan(alpha/2) ) = (1 - tan(alpha/2) ) / ( 1 + tan(alpha/2) ) , [/color]aplying this result we have:\r\n[color=blue]\n[Rho/A]1/N = [(1 - tan(alpha/2) ) / ( 1 + tan(alpha/2) )] / [(1-n)/(1+n)](Ec/2) \n\n6.- [/color] I'm going to substitute \"n\" getting out (PI/180), so now[color=blue] n = Ec * sin( alpha )[/color]\r\nAnd then:\r\n[color=blue]\n[Rho/A]1/N = [(1 - tan(alpha/2) ) / ( 1 + tan(alpha/2) )] / [(1 - Ec*sin( alpha ))/(1+ Ec*sin( alpha ))](Ec/2) [/color]\r\n\r\nThis final equation still looks menacing :mad: , i'm not sure about persisting in this way, but perhaps somebody could see something in it i cannot at this moment, i hope that :).", "Solution_2": "hi, i am quite sure that you posted in a wrong forum. May any moderator move this topic." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "This is in a book. I dont know how to organize the information, and also i dont know how to use the info about a loss in the share price, please if you can help me.\r\n\r\nAn investor has 3 shares, Delta Airlines, Hilton Hotels and Mcdonalds. He says that 2 days ago the shares went down 350 dollars and yesterday they went up 600 dollars. His stock broker remembers that 2 days ago Delta lost 1 dollar per share, Hilton lost 1.5 dollars, and Mc gained 0.5 dollars. He also remembered that yesterday Delta gained 1.5 dollars, Hilton lost again 0.5 dollars and Mc gained 1 dollar. Is it possible to know with this info how many shares of each company has the investor?\r\nIf the investor says he has 200 shares of Mc, how many does he have of the other 2?\r\n\r\nThanks to anyone who can give me a hand!", "Solution_1": "Note that all that information amounts to two equations and three variables. you have an equation of the form (a1)x + (b1)y + (c1)z = -350\r\nand an equation of the form (a2)x + (b2)y + (c2)z = 600\r\nSo with this information alone, it is not possible. If you are given one of x, y, and z, it is of course possible to solve for the other two. (unless the equations happen to degenerate, which I doubt happens)\r\nHope this helps.\r\nBut I don't think this is olympiad level." } { "Tag": [], "Problem": "When three lines intersect a circular region, what is the largest number of regions that can be formed within the circle?", "Solution_1": "To ensure the greatest number of regions with 3 lines, we should draw construct a triangle using those 3 lines. If you draw the figure out, there are 3 regions with angles vertical to the triangles' angles, 3 regions between the other 3 regions, and 1 triangle. That totals to 7 regions.", "Solution_2": "I made a picture of it. Sorry if you have to download it, and i hope it works!" } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "in the cyclic pentagon $ ABCDE$ we have $ AC\\parallel DE$ and $ \\frac {\\sqrt {2}}{6}$\r\n\r\nYou really need to consider all three terms together.", "Solution_2": "ah yea thanks.\r\n\r\nAnyways, I think I have a bunch of useless things but I can get out of that;\r\n\r\n$ x^2\\plus{}y^2\\plus{}z^2\\equal{}1\\minus{}2(xy\\plus{}yz\\plus{}zx)$ Dont know if that helps but it seems to involve a lot of its terms.\r\n\r\nI try to make a lot of substitutions, like $ x\\equal{}1\\minus{}y\\minus{}z$ and same with the other two terms but I cant seem to get it to a more known type of inequality.", "Solution_3": "[quote=\"the future\"]Let $ x$, $ y$, and $ z$ be positive real numbers with $ x \\plus{} y \\plus{} z \\equal{} 1$. Prove that\n\n$ \\frac {xy}{\\sqrt {xy \\plus{} yz}} \\plus{} \\frac {yz}{\\sqrt {yz \\plus{} zx}} \\plus{} \\frac {zx}{\\sqrt {zx \\plus{} xy}} \\leq \\frac {\\sqrt2}{2}$\n\nDoes this basically boil down to proving that $ \\frac {xy}{\\sqrt {xy \\plus{} yz}} \\leq \\frac {\\sqrt2}{6}$?[/quote]\r\n\r\n[hide=\"hold on\"]\nWarning: crazy idea ahead\n\n$ \\frac{xy}{\\sqrt{xy\\plus{}yz}}\\le\\frac{xy}{\\sqrt{xy\\plus{}yz\\plus{}xz}}$ \nnow we need to prove $ \\sqrt{xy\\plus{}yz\\plus{}xz}\\le \\frac{1}{2}$\nwe know $ (x\\plus{}y\\plus{}z)^2\\equal{}x^2\\plus{}y^2\\plus{}z^2\\plus{}2xy\\plus{}2yz\\plus{}2xz\\equal{}1$\nso $ xy\\plus{}yz\\plus{}xz\\equal{}\\frac{1}{2}\\minus{}x^2\\minus{}y^2\\minus{}z^2$\ntherefore $ xy\\plus{}yz\\plus{}xz\\le \\frac{1}{2}$ and we are done\n\ni dunno, the first step looks kinda odd to me\n[/hide]", "Solution_4": "I don't think that first step is right: simplify it, it says $ xz\\le0$. Or notice that the RHS is dividing by more than the LHS, so it has to be smaller.", "Solution_5": "How do we know to consider $ (x,y,z) \\equal{} (1/2 \\minus{} \\epsilon,1/2 \\plus{} \\epsilon,2\\epsilon)$ for some very small positive number $ \\epsilon$? \r\n\r\nHow do we do calculations with episilon representing a small number? Is there any difference between normal computation? Would would something like $ \\frac{\\epsilon}{5}$ turn into?\r\n\r\nWhat was our initial inspiration to state this $ \\frac{xy}{\\sqrt{xy\\plus{}yz}}\\le\\frac{xy}{\\sqrt{xy\\plus{}yz\\plus{}xz}}$?", "Solution_6": "[quote=\"aufha\"]How do we know to consider $ (x,y,z) \\equal{} (1/2 \\minus{} \\epsilon,1/2 \\plus{} \\epsilon,2\\epsilon)$ for some very small positive number $ \\epsilon$? \n\nHow do we do calculations with episilon representing a small number? Is there any difference between normal computation? Would would something like $ \\frac {\\epsilon}{5}$ turn into?\n\nWhat was our initial inspiration to state this $ \\frac {xy}{\\sqrt {xy \\plus{} yz}}\\le\\frac {xy}{\\sqrt {xy \\plus{} yz \\plus{} xz}}$?[/quote]\r\n\r\nWhatever came after it, actually. :/\r\n\r\nI'm not surprised it was wrong. The question is how to fix it.", "Solution_7": "Meh, we might as well abandon the idea.\r\n\r\nWhat about substitution? like a=xy, b=yz, c=zx? and uh, how to relate that to x+y+z=1 uh :maybe: .", "Solution_8": "I'm not asking questions to criticize or say anything is wrong.\r\n\r\nI'm asking because I'm pretty dumb and I don't understand a bunch of things I asked about in my above post... :(", "Solution_9": "Those were showing that the two \"solutions\" above don't work.", "Solution_10": "Well I manage to get that the the equality occurs if and only if $ x\\equal{} y\\equal{} z \\equal{}\\frac{1}{3}$\r\n\r\nThis souds silly, but could you apply a geometric interpretation of the problem where x,y, and z are sides of the triangle? And the area can be represented by 1/2xysinC or something like that to get the xy? Probably not though since its too stretched.", "Solution_11": "[hide]This problem is slightly hard and tricky!But I think I found a nice solution! :wink: \nWe have:square root(xy+xz)=square root[x(1-x)]>=square root(1/9+x/3)\nDoing similarly for 2 more pairs we have: LHS=< xy/square root(1/9+x/3)+yz/square root(1/9+z/3)+zx/square root(1/9+x/3)(2)\nNow using Cauchy-Schwartz we have: RHS of(2)=<3(xy+yz+xz)[xy/(1/3+y)+yz/(1/3+z)+xz/(1/3+x)](3)\nWe need to prove RHS of(3)<=1/2\nUsing AM-GM we have: 3(xy+yz+xz)=<(x+y+z)^2=1\nTherefore, we just need to prove: xy(1/3+y)+yz(1/3+z)+xz(1/3+x)=<1/2\nClearing all denominator and we will get the equivalent inequality:\n81xyz+18(x^2y+y^2z+z^x)<=4+3(xy+yz+xz)\nThe most important step here is using hypothesis (x+y+z)=1 to make the inequality become homogenous!(The degree of the LHS are all third degree!)\nThis step will be: 81xyz+18(x^2y+y^2z+z^2x)<=4(x+y+z)^3+3(x+y+z)(xy+yz+zx)\nExpansion of the RHS we need to prove: 48xyz+3(x^2y+y^2z+z^2x)=<4(x^3+y^3+z^3)+15(xy^2+yz^2+zx^2)\nNow using AM-GM we have: 3/2*(x^3+y^3+z^3+xy^2+yz^2+zx^2)>=3(x^2y+y^2z+z^2x)\nThe rest is: 5/2*(x^3+y^3+z^3)+27/2*(xy^2+yz^2+zx^2)>=48(xyz)(AM-GM for two pairs of three variables,respectively)(Done!) :) [/hide]", "Solution_12": "[quote=\"aufha\"]How do we know to consider $ (x,y,z) \\equal{} (1/2 \\minus{} \\epsilon,1/2 \\plus{} \\epsilon,2\\epsilon)$ for some very small positive number $ \\epsilon$? \n\nHow do we do calculations with episilon representing a small number? Is there any difference between normal computation? Would would something like $ \\frac {\\epsilon}{5}$ turn into?[/quote]\r\n\r\n\r\nI actually meant to have $ x\\equal{}y\\equal{}1/2 \\minus{} \\epsilon$. What I meant was that we are approaching $ (1/2,1/2,0)$. Because we are on an open interval, we cannot have that exact point, but we can be arbitrarily close to that, so the value can be arbitrarily close to the value produced by that point.\r\n\r\nReally, $ \\epsilon \\approx 0$, so $ k \\epsilon \\approx 0$ too. I only used the coefficient $ 2$ to show that together, the three terms added to $ 1$.", "Solution_13": "[quote=\"the future\"]Well I manage to get that the the equality occurs if and only if $ x \\equal{} y \\equal{} z \\equal{} \\frac {1}{3}$\n\nThis souds silly, but could you apply a geometric interpretation of the problem where x,y, and z are sides of the triangle? And the area can be represented by 1/2xysinC or something like that to get the xy? Probably not though since its too stretched.[/quote]\r\n\r\nYou don't know that x,y,z will form a triangle.", "Solution_14": "it was surprising at first when i thought it was straight forward from this famous one :\r\na,b,c >0 then :\r\n$ \\frac{a}{\\sqrt{a\\plus{}b}}\\plus{}\\frac{b}{\\sqrt{b\\plus{}c}}\\plus{}\\frac{c}{\\sqrt{a\\plus{}c}} \\leq \\frac{5}{4}\\sqrt{a\\plus{}b\\plus{}c}$\r\nHowever , the equality doesn't occur :blush:", "Solution_15": "Note that if each individual term on the LHS is taken to the $ n$th power and the resulting inequality is proven, then the original inequality holds due to power mean. I'm not sure if this does anything, but maybe second powers could get rid of those square roots and then after writing with a common denominator and simplifying, cauchy could be used? I dunno.\r\n\r\nBy the way, shouldn't this be in pre-oly?", "Solution_16": "The difficulty in all the high school forums have gone up significantly in the last few years...", "Solution_17": "[quote=\"Temperal\"]Note that if each individual term on the LHS is taken to the $ n$th power and the resulting inequality is proven, then the original inequality holds due to power mean. I'm not sure if this does anything, but maybe second powers could get rid of those square roots and then after writing with a common denominator and simplifying, cauchy could be used? I dunno.\n\nBy the way, shouldn't this be in pre-oly?[/quote]\r\nWell, you have the same idea as I did above.However, you should do more specifically to prove that indeed it can lead to the solution since this inequality is a little bit strict :) .Doesn't anybody see my solution above and check it? :|", "Solution_18": "Ah, yes, I forgot, it's the reciprocal of the average, not the reciprocal.\r\n\r\nHm, I might check it someday... :P \r\n\r\nPiano, if you really want $ a\\equal{}xy,b\\equal{}yz,c\\equal{}xz$, then $ \\sqrt{abc}\\left(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}\\right)\\equal{}1$ (substitute in and check it!)", "Solution_19": "[quote=\"ghjk\"]This problem is slightly hard and tricky!But I think I found a nice solution! :wink: \nWe have: $ \\sqrt{xy\\plus{}xz}\\equal{}\\sqrt{x(1\\minus{}x)}\\ge \\sqrt{\\frac{1}{9}\\plus{}\\frac{x}{3}}$ [/quote]\r\nIf this is true, then $ x(1\\minus{}x)\\ge \\frac{1}{9}\\plus{}\\frac{x}{3}$\r\n$ x^2\\minus{}\\frac{2x}{3}\\plus{}\\frac{1}{9}\\le 0$\r\n$ (x\\minus{}\\frac{1}{3})\\le 0$, which is obviously not true unless $ x\\equal{}\\frac{1}{3}$\r\n\r\nSo shouldn't it be $ \\sqrt{x(1\\minus{}x)}\\le \\sqrt{\\frac{1}{9}\\plus{}\\frac{x}{3}}$?\r\n\r\nAlso, simply checking with $ x\\equal{}\\frac{1}{2}$ disproves that.", "Solution_20": "[hide=\"Hint\"]Substitute: $ x \\equal{} \\sin^2{a}$, $ y \\equal{} \\sin^2{b}$, $ y \\equal{} \\sin^2{c}$. Ain't too hard after that...[/hide]\r\n\r\nAlternatively, I think mixing variables pretty much destroys it...", "Solution_21": "The expression $ \\frac {xy}{\\sqrt {xy \\plus{} yz}} \\plus{} \\frac {yz}{\\sqrt {yz \\plus{} zx}} \\plus{} \\frac {zx}{\\sqrt {zx \\plus{} xy}}$ is equal, up to symmetry, to $ \\frac {yz}{\\sqrt {xy \\plus{} yz}} \\plus{} \\frac {zx}{\\sqrt {yz \\plus{} zx}} \\plus{} \\frac {xy}{\\sqrt {zx \\plus{} xy}}$.\r\n\r\nIs it possible to add this to the former and then reason that the sum of both expressions must be $ \\leq \\sqrt {2}$?\r\n\r\nBy the way, I wouldn't think of trig here :D", "Solution_22": "Not too sure if this is right... \r\n\r\n[hide=\"maybe right?\"]\n\\[ \\displaystyle \\sum_{cyc}\\dfrac {xy}{\\sqrt {xy \\plus{} yz}} \\equal{} \\displaystyle \\sum_{cyc}\\dfrac{xy}{\\sqrt {y(x \\plus{} z)}} \\equal{} \\displaystyle \\sum_{cyc}\\dfrac{xy}{\\sqrt {y(1 \\minus{} y)}}\n\\]\nThus, let [EDIT] $ z \\equal{} min(x,y,z)$\nThen the max of the expression occurs when the denominators are all $ \\sqrt {z(1 \\minus{} z)}$, so we substitute this in...\nThen, the expression becomes:\n\\[ \\dfrac{\\displaystyle \\sum_{cyc} xy}{\\sqrt {z(1 \\minus{} z)}}\n\\]\nBut by amgm, $ \\displaystyle \\sum_{cyc} xy \\ge \\dfrac{1}{3}$\nBut, also, since we assumed $ z \\equal{} min(x,y,z)$, and all are positive, we know that $ z\\le \\dfrac{1}{3}$ Therefore, the expression becomes:\n\\[ \\dfrac{\\sqrt {2}}{2} \\ge \\dfrac{\\dfrac{1}{3}}{\\dfrac{\\sqrt {2}}{3}}\n\\]\nWhich is clearly true, so we are done... :S[/hide]", "Solution_23": "You can't assume $ x\\ge y\\ge z$ as the ineq. is cyclic (not symmetric). You can assume $ z$ is $ \\min(x,y,z)$ though. Otherwise nice :)" } { "Tag": [ "puzzles" ], "Problem": "Bob asked his wife to accompany him to a war movie being featured that night. During a scene when grenades were exploding and guns were firing, Bob decided the time was right; he pulled out a gun and shot his wife. He then took her out of the theatre without anybody trying to stop him. Why not?", "Solution_1": "[hide=\"a guess...\"]It was a dream.[/hide]", "Solution_2": "[hide=\"Classic...\"]They were at a drive-in theater. He shot her in the car so no one heard him, and then he drove out with her body.[/hide]", "Solution_3": "if uve heard problem b4 plz dont say answer", "Solution_4": "[hide=\"what about...\"]Bob and his wife are actors in the movie and Bob kills his wife's character in the movie. When the movie's in the theater, bob and his wife go watch it. Then in the movie bob kills his wife. after the movie, they walk out of the theater and live happily ever after[/hide]", "Solution_5": "good one junggi! :lol:", "Solution_6": "Yeah I agree with Funcia...\r\n\r\nAnyway:\r\n\r\n[hide=\"stupid answer\"] Maybe Bob had a divorce at that second and then when his wife sat next to him, Bob refused to have his ex-wife sit next to him and shot his wife.[/hide]\r\n\r\nLol :rotfl: :rotfl: :rotfl:", "Solution_7": "[quote=\"randomdragoon\"][hide=\"Classic...\"]They were at a drive-in theater. He shot her in the car so no one heard him, and then he drove out with her body.[/hide][/quote]\nYeah, I agree with this answer.\n[/hide]", "Solution_8": "I sort of do...\r\n\r\nBut then if they were at a drive- in, the original problem really states that Bob walks out, not drives.\r\n\r\n\r\nDoes this make sense? :huh: :huh:", "Solution_9": "[quote=\"Ihatepie\"] He then took her out of the theatre without anybody trying to stop him. Why not?[/quote]\r\n\r\n[b]he does not walk out, the :?: says that he too her out of the theater...so i think that the drive in makes perfect sense...hope that helps you! [/b]", "Solution_10": "[quote]He then took her out of the theatre without anybody trying to stop him. Why not?[/quote]\r\n\r\nBut he took her out of the theatre. So that means its inside a building.", "Solution_11": "well drive in theatres are also theatres right? :maybe:", "Solution_12": "[hide=\"a really stupid answer\"]there was nobody else in the theater.[/hide]", "Solution_13": "or [hide]they have a home-theatre[/hide]", "Solution_14": "[hide=\"haha...jk\"]\nHis wife was in the movie and he shot the screen, but then he walked out with his real wife![/hide]", "Solution_15": "lol nice one quevvy\r\n\r\n[hide=\"hilarious\"] how about Bob and his wife are very very rich and have their own private theater?[/hide]", "Solution_16": "I agree with Junggi but here's[hide=\"another possibility\"]Everyone hated his wife and so they were happy she was shot[/hide]", "Solution_17": "[hide]Bob has multiple wives[/hide]", "Solution_18": "his wife was a ghost", "Solution_19": "he shot her accidentally and is taking her to the hospital\r\n\r\nAlthough that doesn't go with \"time is right\" bit.\r\n\r\nMaybe it's one of those faked injuries for $\\$$ from the insurance stuff.", "Solution_20": "[hide]It's a DRIVE-IN movie[/hide]" } { "Tag": [ "email", "AoPS Books", "\\/closed" ], "Problem": "I've found (with the help of a friend who answers questions) two solutions that were wrong or incomplete. One I posted and got responses bashing the book, which wasn't my intention. I'd like to do it constructively. Is there a place?", "Solution_1": "[quote=\"Learner\"]I've found (with the help of a friend who answers questions) two solutions that were wrong or incomplete. One I posted and got responses bashing the book, which wasn't my intention. I'd like to do it constructively. Is there a place?[/quote]I do not really understand what you want. Could you please be more explicit? Thanks.", "Solution_2": "As in where do we post suggestions for the AoPS books, not the web site.", "Solution_3": "In that case I think it's best to address them to Richard Rusczyk directly, throught PM or email (his email is rrusczyk@artofproblemsolving.com) rather than posting them somewhere on the forum.", "Solution_4": "[quote=\"Valentin Vornicu\"]In that case I think it's best to address them to Richard Rusczyk directly, throught PM or email (his email is rrusczyk@artofproblemsolving.com) rather than posting them somewhere on the forum.[/quote]\r\n\r\nWhy not post them on the forum? I think community members should get a chance to see and discuss it. But remember that many of us here are loyal readers of the AoPS books, so be careful with word choice if you think there's something wrong with it.", "Solution_5": "[quote=\"MithsApprentice\"]But remember that many of us here are loyal readers of the AoPS books, so be careful with word choice if you think there's something wrong with it.[/quote]That's exactly why I think that comments about this should go private. We don't want to scare anyone away. In plus chances are the Richard will look into his email more often than in every topic of this forum ;) \r\nMaybe it's best to let Richard answer this.", "Solution_6": "I remember someone posting about this a long time ago - and I think it was agreed that it was a reasonable idea posting them on the forum first, in case you had just made a mistake. I think you did the right thing Learner - the responses weren't exactly responding to what you had asked! I think usually the responses will either confirm or deny that there is a mistake, and then you could PM rrusczyk.", "Solution_7": "[quote=\"TripleM\"]I remember someone posting about this a long time ago - and I think it was agreed that it was a reasonable idea posting them on the forum first, in case you had just made a mistake. I think you did the right thing Learner - the responses weren't exactly responding to what you had asked! I think usually the responses will either confirm or deny that there is a mistake, and then you could PM rrusczyk.[/quote]He hasn't posted any thing concrete. Just that he found a couple of possible mistakes or incomplete proofs. I think (as an author myself) that it is wise to contact the author in such cases. As Richard is very busy at this time, he is probably not aware of this topic (that's one reason for which I think this is not such a good idea - it will take time for him to find this topic :) ). But this is my opinion.", "Solution_8": "[quote=\"Valentin VornicuHe hasn't posted any thing concrete. Just that he found a couple of possible mistakes or incomplete proofs. \r\n[/quote]\r\n(he had posted one of the possible mistakes in one of the other boards. Thats what he was referring to.)\r\n\r\nI think (as an author myself) that it is wise to contact the author in such cases. As Richard is very busy at this time, he is probably not aware of this topic (that's one reason for which I think this is not such a good idea - it will take time for him to find this topic :) ). But this is my opinion.[/quote]\r\n\r\nYup, I agree. And for the same reason you should be very careful to make sure you haven't made a mistake before telling Richard, so its worthwhile posting it on the forum." } { "Tag": [], "Problem": "A motorboat heads upstream on a river that has a current of 3 mph. The trip upstream takes 5 hours, while the return trip takes 2.5 hours. What is the speed of the motorboat?\r\n\r\nAssume that the motorboat keeps a constant speed relative to the water.\r\n\r\nNOTE: What is the meaning of the assumption made above?\r\n\r\nI've seen upstream and downstream questions before. However, I always get confused when it comes to setting up the right table or chart that leads to the correct equation that leads to the answer.", "Solution_1": "hint: [hide]Apply formula for Rate x Times = Distance. Find downstream and upstream rates by solving for one variable.[/hide]", "Solution_2": "I do not get your hint.", "Solution_3": "[quote=\"Interval\"]I do not get your hint.[/quote]\r\n\r\nWell, I do.\r\n\r\nIn the future, please provide us with an idea of your mathematical background (i.e. what courses you have taken, etc.). Thank you.\r\n\r\nAnd about the original problem: Let x equal the speed of the boat. Have the boat travel a given distance, and solve some equations. Remember to add and subtract the speeds respectively.", "Solution_4": "The assumption means that the motorboat always has the same speed. If you try the problem again using now a ranger's and mathnerd 314's hints, but can't solve it, here is my solution:\r\n[hide] Say the distance that the motorboat travels is d and its speed is m. This means that its speed upstream is m-3 (since the stream decreases its speed by 3 mph when it is going in the opposite direction) and its speed going downstream is m+3 (since the stream adds 3 mph to its speed when it is going in the stream's direction). Now, using the distance formula, as now a ranger's hint said, we have that d/(m-3)=5 and d/(m+3)=2.5 Thus, we have that d=5m-15 (multiplying both sides by m-3 on the first equation) and d=2.5m+7.5 (multiplying everything by m+3 on the second equation). Notice that both equations have something equal to d. This means that the \"somethings\" are equal to each other. Therefore, 2.5m+7.5=5m-15. Therefore, 2.5m=22.5 and thus $m=9 mph$, which is our answer [/hide]", "Solution_5": "[quote=\"Interval\"]A motorboat heads upstream on a river that has a current of 3 mph. The trip upstream takes 5 hours, while the return trip takes 2.5 hours. What is the speed of the motorboat?\n\nAssume that the motorboat keeps a constant speed relative to the water.\n\nNOTE: What is the meaning of the assumption made above?\n\nI've seen upstream and downstream questions before. However, I always get confused when it comes to setting up the right table or chart that leads to the correct equation that leads to the answer.[/quote]\r\n[hide]Let $r$ be the speed of the motorboat.\n\n$2.5(r+3)=5(r-3)$\n\n$2.5r+7.5=5r-15$\n\n$22.5=2.5r$\n\n$r=\\boxed{9\\text{ mph}}$[/hide]\r\nAbout the assumption: If the speed of the motorboat varied throughout it's \"trip,\" the answer would be different. To simplify things, it is assumed that the motorboat maintained a constant speed." } { "Tag": [], "Problem": "Find the number of equivalence relations on A = {0, 1, y, n}.\r\n\r\nHere's my logic: An equivalence relation must be reflexive, symmetric, and transitive. Thus, an equivalence relation R must have {(0,0), (1,1), (y,y), (n,n)}, to the very least. We must then consider all possible ordered pairs we can attach to this basic arrangement. For each ordered pair (x, y), (y, x) must also be in R because we need symmetry. We can attach (0,1), (0,y), (0,n), (1,y), (1,n), (y,n) and their inverses for 6 more equivalence relations. Or we can attach 2 of these choices and their inverses in 6C2 = 15 more ways for 15 more equivalence relations. Or we can attach 3 of these choices and their inverses in 6C3 ways, so on... So there should be 1 + 6 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = [b]64 [/b]possible equivalence relations? \r\n\r\nCould someone please verify this? \r\n\r\nThanks.", "Solution_1": "Every disjoint decomposition of a set (i.e., breaking the set into disjoint subsets) defines an equivalence relation on the set. You simply say that the elements in each disjoint subset are equivalent to each other and this relation is obviously transitive, reflexive and symmetric. A finite set has a finite number of disjoint decompositions and this number has to be found. A 4-element set has the following disjoint decompositions:\r\n\r\n1. One 4-element subset (the set itself), $C_4^4 = 1$ possibility. \r\n2. One 3-element subset + one 1-elemet subset, $C_3^4 = 4$ possibilities.\r\n3. Two 2-element subsets, $\\frac 1 2 C_2^4 = 3$ possibilities.\r\n4. One 2-element subset + two 1-element subsets, $C_2^4 = 6$ possibilities.\r\n5. Four 1-element subsets, $\\frac 1 4 C_1^4 = 1$ possibility.\r\n\r\nThe total number of disjoint decompositions of the 4-element set $A = \\{0, 1, y, n\\}$, equal to the number of equivalence relations on this set, is\r\n\r\n$C_4^4 + C_3^4 + \\frac 1 2 C_2^4 + C_2^4 + \\frac 1 4 C_1^4 = 1 + 4 + 3 + 6 + 1 = 15$", "Solution_2": "Thx Yetti, I didn't realize that equivalence relations form a bijection with the number of partitions." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c,d>0$.\r\nProve that $(\\frac{a+b}{a+b+c})^{2}+(\\frac{b+c}{b+c+d})^{2}+(\\frac{c+d}{c+d+a})^{2}+(\\frac{d+a}{d+a+b})^{2}\\geq\\frac{16}{9}$.", "Solution_1": "Could someone please post a solution to this seemingly very nice inequality? i've tried long and hard with fruitless results.", "Solution_2": "Can anyone help? I had no luck in solving this problem. :)", "Solution_3": "[quote=\"Parkdoosung\"]Let $ a,b,c,d > 0$.\nProve that $ (\\frac {a \\plus{} b}{a \\plus{} b \\plus{} c})^{2} \\plus{} (\\frac {b \\plus{} c}{b \\plus{} c \\plus{} d})^{2} \\plus{} (\\frac {c \\plus{} d}{c \\plus{} d \\plus{} a})^{2} \\plus{} (\\frac {d \\plus{} a}{d \\plus{} a \\plus{} b})^{2}\\geq\\frac {16}{9}$.[/quote]\r\nI think this can be done with the S.M.V. but i cant prove it because i'm new to this method.\r\nCan anybody else post solution with mixing variables???", "Solution_4": "[quote=\"Parkdoosung\"]Let $ a,b,c,d > 0$.\nProve that $ (\\frac {a \\plus{} b}{a \\plus{} b \\plus{} c})^{2} \\plus{} (\\frac {b \\plus{} c}{b \\plus{} c \\plus{} d})^{2} \\plus{} (\\frac {c \\plus{} d}{c \\plus{} d \\plus{} a})^{2} \\plus{} (\\frac {d \\plus{} a}{d \\plus{} a \\plus{} b})^{2}\\geq\\frac {16}{9}$.[/quote]\r\nSee also here: http://can-hang2007.blogspot.com/2010/02/inequality-117-v-q-b-can.html", "Solution_5": "Oh, i can't read your solution. Can you upload your solution for me, can_hang2007 ? Thanks ! :maybe:", "Solution_6": "[quote=\"minhhoang\"]Oh, i can't read your solution. Can you upload your solution for me, can_hang2007 ? Thanks ! :maybe:[/quote]\r\nUse a djvu reader program, then you can read it!", "Solution_7": "I can't use your link, Can, can you upload your solution ? :blush:", "Solution_8": "[quote=\"minhhoang\"]I can't use your link, Can, can you upload your solution ? :blush:[/quote]\r\nSee attachment." } { "Tag": [ "IMO Shortlist", "combinatorics proposed", "combinatorics" ], "Problem": "Prove that it is not possible to write numbers $ 1,2,3,...,25$ on the squares of $ 5$x$ 5$ chessboard such that any neighboring numbers differ by at most $ 4$.", "Solution_1": "Suppose A an B are in the same line (res. Column or diagonal). Line (res. Column or diagonal) contains maximum of 5 numbers so | A - B | <= (5-1)* 4 = 16.\r\n\r\n1 and 25 can not be in the same line, Column or diagonal. (I)\r\n\r\n1 can not have more than 4 neighbours so 1 is in the corner. (II) \r\n\r\n25 can not have more than 4 neighbours so 25 is in the corner. (III) \r\n\r\n(II) and (III) => 1 and 25 are in the same line or Column or diagonal witch is in contradiction with (I).", "Solution_2": "@ Raja\r\nAre numbers diagonal to each other considered neighbours?", "Solution_3": "no, it seems like neighbors = adjacent... (?) sorry for my bad english", "Solution_4": "[quote=\"snain\"]Suppose A an B are in the same line (res. Column or diagonal). Line (res. Column or diagonal) contains maximum of 5 numbers so | A - B | <= (5-1)* 4 = 16.\n\n1 and 25 can not be in the same line, Column or diagonal. (I)\n\n1 can not have more than 4 neighbours so 1 is in the corner. (II) \n\n25 can not have more than 4 neighbours so 25 is in the corner. (III) \n\n(II) and (III) => 1 and 25 are in the same line or Column or diagonal witch is in contradiction with (I).[/quote]\r\n\r\nTwice wrong- \r\n1)Numbers on diagonal don't need to differ by less than 5\r\n2)1 and 25 can have 4 neighbours", "Solution_5": "As I can see, I think this problem is just the special case of [url=http://www.mathlinks.ro/resources.php?c=1&cid=17&year=1988]IMO shortlist 1988 #4[/url]\r\nHowever, it's still hard even today!", "Solution_6": "I have solved this, but with massive caseworks. :blush: I think I cannot prove the general problem with my method, so I'll just wait for someone to write a better proof.", "Solution_7": "I'm going to prove general case.Namely,\n\n[b]Theorem[/b]\nWrite numbers $1$~$n^2$ on the squares of $n$\u00d7$n$ chessboard.Then there are two neighbouring squares whose numbers differ by at least $n$.\n\nproof)We write $1$~$n^2$ numbers in turn.It is obvious that If the number of empty squares which are neighbor to written squares is more than(or equal) $n$ a moment, theorem certainly holds.We consider the moment which some row(or column) is completed(all squares on some row are written numbers).\n\n[b]Case1[/b]:some row is completed\nOn each column, there is a empty square which is neighbor to written squares.So theorem certainly holds.\n\n[b]Case2[/b]:some column is completed\nIt is the same as Case1.\n\n[b]Case3[/b]:some row and some column are simultaneously completed\nWe suppose that this case is occurred when the square A is written.Just before the square A is written, there is a empty square which is neighbor to written squares on each column.So theorem certainly holds.\n\nFrom case1,2,3, we prove the theorem.$\\blacksquare$" } { "Tag": [ "modular arithmetic" ], "Problem": "A four-digit positive integer $ \\overline{a(a\\minus{}1)(a\\minus{}1)a}$ can be wrriten as a sum of three consecutive squares (for example,$ 1^2\\plus{}2^2\\plus{}3^2$).\r\nFind the number.", "Solution_1": "This got a little ugly, but...\r\n\r\n[hide=\"Solution\"]\nExpanding it out, we have\n\\[ a\\plus{}10a\\minus{}10\\plus{}100a\\minus{}100\\plus{}1000a\\equal{}1111a\\minus{}110\\equal{}k^2(k\\plus{}1)^2\\plus{}(k\\plus{}2)^2\\\\\n1111a\\minus{}110\\equal{}3k^2\\plus{}6k\\plus{}5\\\\\n1111a\\minus{}115\\equal{}3k^2\\plus{}6k\\\\\n1111a\\minus{}124\\equal{}3k^2\\plus{}6k\\minus{}9\\equal{}3(k\\minus{}1)(k\\plus{}3)\n\\]\n\nTake this $ \\mod 3$:\n$ a\\equiv 1\\pmod 3\\implies a\\equal{}3q\\plus{}1\\implies a\\equal{}1,4,7$.\n\nNow simply bash.\n$ 1111a\\minus{}115\\equal{}3k^2\\plus{}6k\\implies 3k^2\\plus{}6k\\plus{}(115\\minus{}1111a)\\equal{}0$.\n$ k$ must be an integer, so the discriminant of this thing must be a perfect square, for starters.\nThat is, $ \\sqrt{36\\minus{}12(115\\minus{}1111a)}$ is an integer.\nPlugging in $ 4$ gives $ 51984$, which is $ 228^2$. The others give only non-perfect squares.\n$ k\\equal{}\\frac{\\minus{}6\\plus{}228}{6}\\equal{}37$.\n\nSo the answer is $ \\boxed{4334}\\equal{}37^2\\plus{}38^2\\plus{}39^2$\n[/hide]\r\n\r\nIf anyone has a cleaner way, I'd like to see it.", "Solution_2": "[quote=\"Anavel_Gato\"]A four-digit positive integer $ \\overline{a(a \\minus{} 1)(a \\minus{} 1)a}$ can be wrriten as a sum of three consecutive squares (for example,$ 1^2 \\plus{} 2^2 \\plus{} 3^2$).\nFind the number.[/quote]\r\n\r\nLet $ \\overline{a(a \\minus{} 1)(a \\minus{} 1)a} \\equal{} k^{2} \\plus{} (k \\plus{} 1)^{2} \\plus{} (k \\plus{} 2)^{2}$ . The numbers $ k^2$,$ (k \\plus{} 1)^{2}$,$ (k \\plus{} 2)^{2}$ have diferent remainders when are divided by $ 3$, then $ k^{2} \\plus{} (k \\plus{} 1)^{2} \\plus{} (k \\plus{} 2)^{2} \\equiv 2 {mod3}$.So $ a \\plus{} a \\minus{} 1 \\plus{} a \\minus{} 1 \\plus{} a \\equal{} 4a \\minus{} 2 \\equiv 2 {mod 3}\\Rightarrow 4(a \\minus{} 1)\\equiv0 {mod3}$ so $ a \\equal{} 4$ or $ a \\equal{} 7$ but $ a \\equal{} 7$ there's no solution ,so $ a \\equal{} 4$\r\n\r\n\r\n\r\n\r\n \r\n \r\n [b]ee\"[/b]", "Solution_3": "oh yea $ a \\leq 9$ haha" } { "Tag": [ "inequalities", "search", "inequalities unsolved" ], "Problem": "Let a, b, ,c are real numbers. Find the min value of k such that\r\n$ a^4\\plus{}b^4\\plus{}c^4\\plus{}k(ab\\plus{}bc\\plus{}ca)^2\\geq \\sum_{sym}ab(a^2\\plus{}b^2)$\r\nIf k=6, prove that this inequality is true. :D", "Solution_1": "not anyone prove this inequality?\r\n :D", "Solution_2": "I'll prove the case $ k \\equal{} 6$.\r\nLet's prove it for positive reals. If it is true for positive, it is also true for some of $ a,b,c$ negative since $ LHS\\ge 0$\r\n\\[ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 6(ab \\plus{} bc \\plus{} ca)^2\\ge 2ab(a^2 \\plus{} b^2) \\plus{} 2bc(b^2 \\plus{} c^2) \\plus{} 2ca(c^2 \\plus{} a^2)\r\n\\]\r\n\r\n\\[ \\sum_{cyc}{a^4} \\plus{} 2\\sum_{cyc}{a^2b^2} \\plus{} 12\\sum_{cyc}{a^2bc}\\ge 2ab(a \\minus{} b)^2 \\plus{} 2bc(b \\minus{} c)^2 \\plus{} 2ca(c \\minus{} a)^2\r\n\\]\r\n\r\n\\[ (a^2 \\plus{} b^2 \\plus{} c^2)^2 \\plus{} 12abc(a \\plus{} b \\plus{} c)\\ge 2ab(a \\minus{} b)^2 \\plus{} 2bc(b \\minus{} c)^2 \\plus{} 2ca(c \\minus{} a)^2\r\n\\]\r\nBy am-gm we get : $ RHS\\le \\sum_{cyc}{\\bigg(\\frac {a^2 \\plus{} b^2}{2}\\bigg)^2}$\r\n\r\nThen:\r\n\\[ 2(a^2 \\plus{} b^2 \\plus{} c^2)^2 \\plus{} 24abc(a \\plus{} b \\plus{} c)\\ge a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2\r\n\\]\r\nwhich is clearly true.", "Solution_3": "[quote=\"EUCLA\"]I'll prove the case $ k \\equal{} 6$.\nLet's prove it for positive reals. If it is true for positive, it is also true for some of $ a,b,c$ negative since $ LHS\\ge 0$\n\\[ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 6(ab \\plus{} bc \\plus{} ca)^2\\ge 2ab(a^2 \\plus{} b^2) \\plus{} 2bc(b^2 \\plus{} c^2) \\plus{} 2ca(c^2 \\plus{} a^2)\n\\]\n\n\\[ \\sum_{cyc}{a^4} \\plus{} 2\\sum_{cyc}{a^2b^2} \\plus{} 12\\sum_{cyc}{a^2bc}\\ge 2ab(a \\minus{} b)^2 \\plus{} 2bc(b \\minus{} c)^2 \\plus{} 2ca(c \\minus{} a)^2\n\\]\n\n\\[ (a^2 \\plus{} b^2 \\plus{} c^2)^2 \\plus{} 12abc(a \\plus{} b \\plus{} c)\\ge 2ab(a \\minus{} b)^2 \\plus{} 2bc(b \\minus{} c)^2 \\plus{} 2ca(c \\minus{} a)^2\n\\]\nBy am-gm we get : $ RHS\\le \\sum_{cyc}{\\bigg(\\frac {a^2 \\plus{} b^2}{2}\\bigg)^2}$\n\nThen:\n\\[ 2(a^2 \\plus{} b^2 \\plus{} c^2)^2 \\plus{} 24abc(a \\plus{} b \\plus{} c)\\ge a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2\n\\]\nwhich is clearly true.[/quote]\r\nthink it's not enough to prove it for positives only :wink:", "Solution_4": ":huh: Isn't LHS just sum of squares, or am I missing something evident?\r\n\r\nedit: you edited, now..why?\r\n\r\nedit2: understood :wink:", "Solution_5": "[quote=\"EUCLA\"]:huh: Isn't LHS just sum of squares, or am I missing something evident?\n\nedit: you edited, now..why?[/quote]\r\n\r\nit is but allowing some of $ a,b,c$ to be negative we may achieve $ ab\\plus{}bc\\plus{}ca$ to be $ 0$, and it's not quite obvious then that the inequality holds, at least not to me..", "Solution_6": "I'll try, this time, in finding $ k$. Sorry if I'm mistaking again. :oops: \r\n\r\nMy guess is $ \\boxed{k\\equal{}\\minus{}\\frac{2}{3}}$\r\n\r\n\\[ \\sum_{cyc}{a^4}\\plus{}k\\sum_{cyc}{a^2b^2}\\plus{}2k\\sum_{cyc}{a^2bc}\\ge \\sum_{cyc}{2ab(a^2\\plus{}b^2)}\\]\r\n\\[ (a^2\\plus{}b^2\\plus{}c^2)^2\\plus{}2abck(a\\plus{}b\\plus{}c)\\plus{}(k\\minus{}6)\\sum_{cyc}{a^2b^2}\\ge \\sum_{cyc}{2ab(a\\minus{}b)^2}\\]\r\n\r\nWe now have to minimize lhs and maximize rhs.\r\n\r\nBy am-gm $ 2ab(a\\minus{}b)^2\\le \\bigg(\\frac{a^2\\plus{}b^2}{2}\\bigg)^2$\r\n\r\n\\[ 2\\sum_{cyc}{a^4}\\plus{}6\\sum_{cyc}{a^2b^2}\\plus{}2abck(a\\plus{}b\\plus{}c)\\plus{}(k\\minus{}6)\\sum_{cyc}{a^2b^2}\\ge 0\\]\r\n\\[ 2\\sum_{cyc}{a^4}\\plus{}2abck(a\\plus{}b\\plus{}c)\\plus{}k\\sum_{cyc}{a^2b^2}\\ge 0\\]\r\n\r\nSince it is homogeneus (of degree 4) one can set $ abc\\equal{}1$\r\n\r\nThen, by am-gm we get \\[ 6\\plus{}6k\\plus{}3k\\ge 0\\]\r\n\r\nSo we get the wanted result.", "Solution_7": "[quote=\"thegod277\"]Let a, b, ,c are real numbers. Find the min value of k such that\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} k(ab \\plus{} bc \\plus{} ca)^2\\geq \\sum_{sym}ab(a^2 \\plus{} b^2)$\nIf k=6, prove that this inequality is true. :D[/quote]\r\nBy ABC method\r\nWe have to prove ineq in two case\r\n1.$ c \\equal{} 0$\r\nif $ a$ or $ b \\equal{} 0$ ineq true\r\nif $ a,b\\neq 0$.Ineq <=>\r\n$ \\frac {a^2}{b^2} \\plus{} \\frac {b^2}{a^2} \\plus{} k\\ge \\frac {a}{b} \\plus{} \\frac {b}{a}$\r\n$ t \\equal{} \\frac {a}{b} \\plus{} \\frac {b}{a}$\r\nineq <=> $ k\\ge \\minus{} t^2 \\plus{} t \\plus{} 2$\r\n<=> $ k\\ge Max \\minus{} t^2 \\plus{} t \\plus{} 2 \\equal{} 0$ ($ |t|\\ge 2$)\r\n2.$ b \\equal{} c\\neq 0$ ($ b \\equal{} c \\equal{} 0$ easy)\r\nineq <=> $ a^4 \\plus{} k(b^2 \\plus{} 2ab)\\ge 2ab(a^2 \\plus{} b^2)$\r\n$ a \\equal{} 0$ easy :D\r\n$ b \\equal{} ax (a\\neq 0)$\r\nineq <=> $ k\\ge \\frac {2x(x^2 \\plus{} 1) \\minus{} 1}{(x^2 \\plus{} 2x)^2}$\r\n$ f'(x) \\equal{} 0$ < = > $ x \\equal{} 1,x \\equal{} \\minus{} 1,x \\equal{} s$ (in there $ s^3 \\minus{} s^2 \\plus{} 2s \\plus{} 2 \\equal{} 0$)\r\nSo $ Max f(x) \\equal{} Max(f(1),f(s)) \\equal{} f(1) \\equal{} \\frac {1}{3}$\r\nso $ k\\ge f(s)$\r\n$ k\\ge Max (0,\\frac {1}{3}) \\equal{} \\frac {1}{3}$\r\nSorry if I am false :blush:", "Solution_8": "Sorry, but I don't know ABC method :( . What is ABC method? :huh:", "Solution_9": "[color=green]@thegod277, use Search tool plz, u can find more about ABC method. :roll: [/color]", "Solution_10": "I should have done something wrong, in using am-gm on negatives, i suppose :(", "Solution_11": "[quote=\"thegod277\"]Sorry, but I don't know ABC method :( . What is ABC method? :huh:[/quote]\r\nSee here:\r\nhttp://www.mathlinks.ro/viewtopic.php?t=165132", "Solution_12": "Oh, thank you. :D \r\nSend EUCLA,vodanh_tieutot: Thank you for your proof. You can see a general problem in Inequalities Proposes & Own problem,a post \"very very hard\" of me. And the minximum value of k of my inequality is $ \\frac{1}{3}$, because if we choose a=b=c, then $ 3\\plus{}9k\\geq 6$, so $ k\\geq \\frac{1}{3}$. :wink:", "Solution_13": "[quote=\"thegod277\"]\nthe minximum value of k of my inequality is $ \\frac {1}{3}$, because if we choose a=b=c, then $ 3 \\plus{} 9k\\geq 6$, so $ k\\geq \\frac {1}{3}$. :wink:[/quote]\r\nIn some case min,max k isn't with a=b=c :D\r\nEx:\r\nFind k Max ($ a,b,c\\ge 0$)\r\n$ \\frac{a}{b\\plus{}c}\\plus{}\\frac{b}{c\\plus{}a}\\plus{}\\frac{c}{a\\plus{}b}\\plus{}25\\frac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2}\\ge k$\r\n:D", "Solution_14": "Thank you. But I can prove the inequality is true for $ k\\equal{}\\frac{1}{3}$. So, kmin=1\\3. \r\n :D" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "[tex]\\frac{6563} { \\sqrt{1-\\frac{3*(\\frac{0.5}{1})}{10000} }}[/tex]\r\n\r\n\r\nHow do I figure out how many significant digits should be in the final answer?\r\n\r\nThe 0.5 and 1 in the the equation are inputs. \r\nThis equation computes how much shift is in a hydrogen line due to gravitational redshift.\r\n\r\nI get 6563.49 as the answer, but 0.5 & 1 in the (0.5/1) , and the 3 are only 1 significant digit. The \"1-\" is a pure 1 (1.000000000).\r\n\r\nI'd hate to give an answer of 7000. That would ruin everything.", "Solution_1": "You're going about this the wrong way: the shift is tiny compared to the original value. You don't want to add it on to the original value, because the inaccuracy in that value will swamp you.\r\n$a\\cdot\\left(\\frac1{\\sqrt{1-x}}-1\\right)=a\\cdot\\left(\\frac x2+O(x^2)\\right)$\r\nSince your value of $x$ is tiny and has low precision, the $O(x^2)$ term can be ignored. The shift is then $\\frac{ax}2=\\frac{6563\\cdot3\\cdot.5}{10000\\cdot2}\\approx.5$. You only have one significant figure here for the obvious reason.\r\n\r\nThis doesn't mean you can conclude that the shifted value is $6563.5$; you don't have enough significant figures in the original value to say that.", "Solution_2": "Thank you, jmerry. That's about what I expected." } { "Tag": [], "Problem": "Find all integers $x$ such that $x^2-5x+6$ is a positive prime number.\r\n\r\n[hide=\"Hint for those who want it\"]When good mathematicians are stuck, they should always say the f word.\n\nFactor!! :D \n\nAm I funny or what? (If you're thinking about saying no, don't) :) [/hide]", "Solution_1": "[quote=\"4everwise\"]Find all integers $x$ such that $x^2-5x+6$ is a positive prime number.\n\n[hide=\"Hint for those who want it\"]When good mathematicians are stuck, they should always say the f word.\n\nFactor!! :D \n\nAm I funny or what? (If you're thinking about saying no, don't) :) [/hide][/quote]\n\n[hide]\n\n Factoring $x^2 -5x + 6$ yields:\n\n$(x-3)(x-2)$\n\n A prime number has two factors, \"1\" and itself. Hence, we simply set each equation as 1, and then solve.\n\n$x-3=1$\n$x=4$\n\n The result is 2.\n\n$x-2=1$\n$x=3$\n\n The result is 0, which is not a prime number.\n\n Hence, the only value that works is $\\boxed{2}$[/hide] By result I meant value of the equation $x^2 - 5x + 6$", "Solution_2": "[hide]$x^2-5x+6=(x-2)(x-3)$. \n\nWe want one of these to be one, so the result may still be prime. Either x-2=1 or x-3=1. \n\nIf x=3, we get the answer 0, which doesn't work. If we try x=4, the result is 2, which is prime. \n\nx can only be 4. [/hide]", "Solution_3": "Ok, 236factorial has determined that one of the answers is 4. There is one more. ;) \r\n\r\n[hide=\"Hint\"]We know that x=2 and x=3 don't work. [hide=\"Continuation of Hint\"] Oviously, 4 is the only positive value that works. Are there any negative values that work? :roll: [/hide][/hide]", "Solution_4": "Are negative numbers considered prime? I thought that primes only referred to positives... or is it just that every time a problem talks about primes, they mean \"positive primes\"?\r\n\r\n[b]EDIT:[/b] No, there is no such thing as a negative prime. All numbers that are negatives of primes actually have four factors: itself, one, negative one, and negative itself. And $-1$ doesn't count. ;)", "Solution_5": "[hide]Well either $(x-2)$ or $(x-3)$ has to be $-1$.\nIf $x-3=-1$ then $x=2$, but that doesn't work.\nSo $x-2=-1$ and $x=1$, which yields $(-1)(-2)=2$.\nSo $x$ can be $1$ too.[/hide]", "Solution_6": "Oh, that's what 4everwise meant. He didn't quite word it correctly, though...", "Solution_7": "[quote=\"Melissa\"]Well either $(x-2)$ or $(x-3)$ has to be $-1$.\nIf $x-3=-1$ then $x=2$, but that doesn't work.\nSo $x-2=-1$ and $x=1$, which yields $(-1)(-2)=2$.\nSo $x$ can be $1$ too.[/quote]Good job! \r\n :) \r\nI accidentally worded my second hint incorrectly. :blush:" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "AIME I" ], "Problem": "My school offered amc 10a for the first time this year but i only got a 129 so even though it qould qualify me for AIME i want to take 10b to get a better USAMO index and my school refused to offer amc 10b so i am going to like tennessee or kentucky to take amc 10b so i was just wondering what i need to do i mean do i just show up at the test site or what :?:", "Solution_1": "you need to contact the director of this program in that school or college." } { "Tag": [ "function", "number theory proposed", "number theory" ], "Problem": "For the positive integers written in base $10$ we define the function $T$ such that \r\nthe final figure is moved to the front of the number : for ex. $T(21345) = 52134$ .\r\n\r\nFind the smallest $n$ such that $T(n) = 6n$ .\r\n\r\n :cool:", "Solution_1": "n=10a+b with b as the last digit. T(n)=b.10^(k-1)+a with k as the number of digits of n. T(n)=6n is equivalent to b(10^(k-1)-6)=59a. b<10 and 59 is prime so 59|(10^(k-1)-6). The least possible k is 58 because 59|(10^58-1) (Little Fermat) and 6 is the inverse of 10 modulo 59. a must have k-1=57 digits so the least possible value of b is 6. n=10a+b=(10^57-6).60/59+6=(10^57-6)/59.\r\n\r\nMisha", "Solution_2": "[quote=\"Misha123\"]n=10a+b with b as the last digit. T(n)=b.10^(k-1)+a with k as the number of digits of n. T(n)=6n is equivalent to b(10^(k-1)-6)=59a. b<10 and 59 is prime so 59|(10^(k-1)-6). The least possible k is 58 because 59|(10^58-1) (Little Fermat) and 6 is the inverse of 10 modulo 59. a must have k-1=57 digits so the least possible value of b is 6. n=10a+b=(10^57-6).60/59+6=(10^57-6)/59.\n\nMisha[/quote]\r\n\r\nHow do you know that 58 is the smallest k such that 59|(10^k-1) ? True that 10^58 = 1 mod 59, but it could be that 10^29 = 1 mod 59 too. We are looking for the smallest possible k.\r\n\r\nAlso, once you get k, how do you know that b = 6?" } { "Tag": [ "function" ], "Problem": "does any body know a good book for mobius functions? thanks", "Solution_1": "What's mobius functions?", "Solution_2": "Please read the posting guidelines. The Getting Started for is for basic high school problem solving math. Mobius functions are rarely studied before the undergraduate level.", "Solution_3": "According to Mathworld...\r\n\r\nhttp://mathworld.wolfram.com/MoebiusFunction.html", "Solution_4": "oh ure right...\r\n :(", "Solution_5": "frt thanks alot,...\r\n\r\nbut i have seen it many years ago... ;)", "Solution_6": "So, why are you asking for it? :?", "Solution_7": "no no,\r\n\r\ni mean i have seen the link that he wrote..." } { "Tag": [ "counting", "derangement", "function", "combinatorics unsolved", "combinatorics" ], "Problem": "Any one have single idea to simply this ?\r\n\r\n$\\sum^n_{k=0}(-1)^k\\binom{n}{k}(n-k)!$ \r\n\r\n[hide=\"source\"]\nI come across this when trying to solve the problem which is : \"There are $n$ boxes each with one different ball in them . Then you rearrange them so that no ball is in their original boxes (each boxes must have only one ball) . Find the number of way to do so . \" I get the above expression but I think it can be further simplified which I dont know how .[/hide]", "Solution_1": "This is the classic \"derangement\" problem. The sum you found is usually expressed as\r\n\r\n$n! \\sum^n_{k=0}\\frac{(-1)^k}{k!}$\r\n\r\nI've never seen any simpler analytic expression for this. However, an expression involving the nearest integer function is\r\n\r\n$\\left[\\frac{n!}{e}\\right]$\r\n\r\nThis can be seen by examining the McLaurin series for $e^{-1}$:\r\n\r\n$e^{-1} = 1 - 1/1! + 1/2! - 1/3! + ...$. So,\r\n\r\n$n! \\sum^n_{k=0}\\frac{(-1)^k}{k!} = n!(e^{-1} + \\epsilon)$\r\n\r\nwhere $| \\epsilon | < 1/(n+1)!$\r\n\r\nI don't recall how to make the final leap to the solution, but you can see why this would be a reasonable approximation for the sum.\r\n\r\nFor more information, check out Mathworld's site on the topic:\r\n\r\nhttp://mathworld.wolfram.com/Derangement.html", "Solution_2": "[quote=\"zanttrang\"]This is the classic \"derangement\" problem. The sum you found is usually expressed as\n\n$n! \\sum^n_{k=0}\\frac{(-1)^k}{k!}$\n\nI've never seen any simpler analytic expression for this. However, an expression involving the nearest integer function is\n\n$\\left[\\frac{n!}{e}\\right]$\n\nThis can be seen by examining the McLaurin series for $e^{-1}$:\n\n$e^{-1} = 1 - 1/1! + 1/2! - 1/3! + ...$. So,\n\n$n! \\sum^n_{k=0}\\frac{(-1)^k}{k!} = n!(e^{-1} + \\epsilon)$\n\nwhere $| \\epsilon | < 1/(n+1)!$\n\nI don't recall how to make the final leap to the solution, but you can see why this would be a reasonable approximation for the sum.\n\nFor more information, check out Mathworld's site on the topic:\n\nhttp://mathworld.wolfram.com/Derangement.html[/quote]\r\n\r\nwow , thats cool . Thanks :D" } { "Tag": [ "modular arithmetic", "calculus", "integration" ], "Problem": "$(a)$ Prove that for all primes $p > 5$, $p^{4}\\equiv 1 \\mod 240$\r\n$(b)$ If $p^{4}\\equiv 1 \\mod 240$ and $p \\in \\mathbb{Z}$, is $p$ necessarily prime?", "Solution_1": "A note for the first, it only works for $7$ and up.\r\n[hide=\"Solution\"]\nAll primes except $2$ and $3$ can be written as $6k \\pm 1$\n\nNote that $\\frac{(6k+1^{4})-1}{24}=k+9k^{2}+36k^{3}+54k^{4}$ and that $\\frac{(6k-1^{4})-1}{24}=-k+9k^{2}-36k^{3}+54k^{4}$\n\nSince $k^{N}\\equiv k \\pmod{2}$, the two above are clearly even.\n\nTaking them in mod 5 gives:\n$k-k^{2}+k^{3}-k^{4}$ and $-(k+k^{2}+k^{3}+k^{4})$\nTesting over 0 through 4,\n\n-the first is not divisible by 5 only for $k\\equiv 4$, which is not a problem because if we substitute $k=5z+4$, then $6k+1=30z+24+1=5(6z+5)$, which is not prime.\n\n-the second is not divisible by 5 only for $k\\equiv 1$, which is also not a problem because if we substitute $k=5z+1$, then $6k-1=30z+6-1=5(6z+1)$ which is not prime for all $z>0$. If $z=0$, it gives the prime number 5, which does not work.\n\nThus for all $p \\in \\text{Primes}-\\{ 2,3,5 \\}, \\; p^{4}\\equiv 1 \\pmod{240}$\n\n[/hide]\n\n\n[hide=\"Second\"]\nDunno if 241 is prime, but $p=241^{2}$ is definitely a counterexample.\n[/hide]", "Solution_2": "(B) No definately not, the easiest counterexample is $240*2+1$, which can be factored into $13*37$", "Solution_3": "[quote=\"Allan Z\"](B) No definately not, the easiest counterexample is $240*2+1$, which can be factored into $13*37$[/quote]\r\n\r\nIn fact, I believe the smallest nontrivial counterexample is $p = 49$. Generally, $p^{4}\\equiv 1 \\bmod 240 \\Leftrightarrow 2, 3, 5 \\not | p$. Of course, the easiest counterexample is $240 \\times 0+1$.", "Solution_4": "[quote=\"t0rajir0u\"][quote=\"Allan Z\"](B) No definately not, the easiest counterexample is $240*2+1$, which can be factored into $13*37$[/quote]\n\nIn fact, I believe the smallest nontrivial counterexample is $p = 49$.[/quote]Yes never forget that 2401 :P", "Solution_5": "[hide=\"(a)\"]$p^{4}-1\\equiv 0\\pmod{24}$ if $p^{4}-1\\equiv 0\\pmod{16,3,5}$.\n\nNotice that $p^{4}\\equiv (-p)^{4}\\pmod{n}$.\nFor p greater than 5, $p\\equiv 1,(-1),2,(-2) \\pmod{5}\\implies p^{4}\\equiv 1 \\pmod{5}$.\nFor p greater than 3, $p\\equiv 1,(-1) \\pmod{3}\\implies p^{4}\\equiv 1 \\pmod{3}$.\nFor p greater than 2, $p\\equiv 1 (-1),3 (-3),5 (-5), 7 (7-)\\pmod{16}\\implies p^{4}\\equiv 1\\pmod{15}$.\nThus, $p^{4}\\equiv 1\\pmod{240}$, for primes greater than 5.[/hide]", "Solution_6": "Sorry, everybody. A couple of typos have been fixed. For (B), $p$ has to be integral and for (A), $p$ has to be $> 5$." } { "Tag": [ "analytic geometry" ], "Problem": "The point $ (2, 3)$ is reflected about the $ x$-axis to a point $ P$. Then $ P$ is reflected about the line $ y \\equal{} x$ to a point $ Q$. What is the $ x$-coordinate of $ Q$?", "Solution_1": "When a point is reflected over the $ x$-axis, the $ y$-coordinate changes to its additive inverse. Point $ P$ is $ (2,\\minus{}3)$. When reflecting over the line $ y\\equal{}x$, I believe both the coordinates change to its additive inverse. So point $ Q$ is $ (\\minus{}2,3)$. So the $ x$-coordinate is $ \\boxed{\\minus{}2}$", "Solution_2": "Reflecting $ (2,3)$ over the $ x$-axis, we change the sign of the $ y$-coordinate, so we get $ (2, \\minus{} 3)$. Then, reflecting across $ y \\equal{} x$, which flips the coordinates, we get $ ( \\minus{} 2,3)$. The $ x$-coordinate of $ ( \\minus{} 2,3)$ is $ \\fbox{ \\minus{} 2}$. \r\n\r\nEDIT: Beaten.", "Solution_3": "Actually when you're reflecting a point across $y=x$, the coordinates are flipped i.e. $(x,y)$ becomes $(y,x)$. Thus the answer is actually $\\boxed{-3}$." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "it is well-known that when $ \\nu(X) < \\infty\\,\\,$ and $ p\\leq q$ then $ \\L^{q}\\subset L^{p}$\r\nCould any one prove the converse that if $ p\\leq q$ and $ \\L^{q}\\subset \\L^{p}\\,\\,$ then $ \\nu(X) < \\infty$?\r\n\r\nRemark : we just consider $ X\\subset\\mathbb{R}$\r\nthanks so much", "Solution_1": "Let $ X$ be an arbitrary $ \\sigma$-finite measure space and $ 1\\le p AC > AB$. Let $ A',B',C'$ be feet of perpendiculars from $ A,B,C$ to $ BC,AC,AB$, such that $ AA' \\equal{} BB' \\equal{} CC' \\equal{} x$. Prove that:\r\na) If $ ABC\\sim A'B'C'$ then $ x \\equal{} 2r$\r\nb) Prove that if $ A',B'$ and $ C'$ are collinear, then $ x \\equal{} R \\plus{} d$ or $ x \\equal{} R \\minus{} d$.\r\n\r\n(In this problem $ R$ is the radius of circumcircle, $ r$ is radius of incircle and $ d \\equal{} OI$)", "Solution_1": "$ H$ is the orthocenter. $ A', B', C'$ are not feet of perpendiculars from $ A, B, C$ to $ BC, CA, AB,$ but points on the A, B-, C-altitudes $ AH, BH, CH,$ respectively.\r\n\r\n[color=blue][b](a)[/b][/color] Assuming $ \\triangle A'B'C' \\sim \\triangle ABC$ are oppositely similar, otherwise $ AA' \\equal{} BB' \\equal{} CC' \\equal{} x \\equal{} 0$ satisfies the conditions. $ \\triangle A'B'C' \\sim \\triangle ABC$ $ \\Longrightarrow$ $ A'B'C'H$ is cyclic. Assume $ AA' \\equal{} BB'.$ Locus of circumcenters $ O'_c$ of the $ \\triangle B'HA'$ is a line $ p_c \\parallel CI$ parallel to the internal bisector of the $ \\angle BCA.$ Likewise, assuming $ BB' \\equal{} CC',$ locus of circumcenters $ O'_a$ of the $ \\triangle C'HB'$ is a line $ p_a \\parallel AI$ and assuming $ CC' \\equal{} AA',$ locus of circumcenters $ O'_b$ of the $ \\triangle A'HC'$ is a line $ p_b \\parallel BI.$ Let $ P_a, P_b, P_c$ be reflections of the circumcenter $ O$ in $ BC, CA, AB.$ Then $ P_a \\in p_a, P_b \\in p_b, P_c \\in p_c.$ $ \\triangle P_aP_bP_c \\cong \\triangle ABC$ are centrally congruent with similarity center $ N$ (common 9-point circle center). Lines $ p_a, p_b, p_c$ are internal bisetors of the $ \\triangle P_aP_bP_c,$ concurrent at its incenter $ F.$ Circle $ (F)$ with radius $ FH$ is the only circle through $ H$ cutting $ AH, BH, CH$ again at $ A', B', C',$ such that $ AA' \\equal{} BB' \\equal{} CC' \\equal{} x > 0.$ The Fuhrmann circle of $ \\triangle ABC$ goes through $ H$ and cuts $ AH, BH, CH$ again at $ A', B', C',$ such that $ AA' \\equal{} BB' \\equal{} CC' \\equal{} 2r$ (see the last reply at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=101687[/url]) $ \\Longrightarrow$ $ (F)$ is the Fuhrmann circle and $ x \\equal{} 2r.$\r\n\r\n[color=blue][b](b)[/b][/color] Let $ X, Y, Z$ be reflections of $ A', B', C'$ in the internal angle bisectors $ AI, BI, CI.$ $ AX \\equal{} BY \\equal{} CZ \\equal{} x$ $ \\Longrightarrow$ $ X \\in AO, Y \\in BO, Z \\in CO$ are on a circle $ \\mathcal P(O, x \\minus{} R)$ (radius can be positive or negative). Let $ k \\equiv A'B'C'$ be the collinarity line and $ k_x, k_y, k_z$ reflections of $ k$ in $ AI, BI, CI.$ Let $ k_x, k_y$ intersect at $ Q$ (different from $ X, Y$). By angle chase for the $ \\triangle B'HA'$ and quadrilateral $ XOYQ,$ $ \\angle B'HA' \\equal{} \\angle XQY \\equal{} \\frac {\\angle XOY}{2} \\mod \\pi$ $ \\Longrightarrow$ $ Q \\in \\mathcal P(O, x \\minus{} R).$ Similarly, $ k_y, k_z$ intersect at a point on the circle $ \\mathcal P$ different from $ Y, Z$ $ \\Longrightarrow$ $ k_x, k_y, k_z$ concur at $ Q \\in \\mathcal P.$ Let $ U, V, W$ be reflections of $ Q$ in $ AI, BI, CI.$ The incenter $ I$ is the circumcenter of the cyclic quadrilateral $ QUVW,$ because perpendicular bisectors $ AI, BI, CI$ of its 3 sides $ QU, QV, QW$ concur at $ I.$ $ U, V, W \\in k$ are collinear $ \\Longrightarrow$ this is possible only when $ I \\equiv Q$ ($ \\equiv U \\equiv V \\equiv W$), therefore $ I \\in \\mathcal P,$ $ x \\minus{} R \\equal{} \\pm OI$ and $ AA' \\equal{} BB' \\equal{} CC' \\equal{} x \\equal{} AX \\equal{} BY \\equal{} CZ \\equal{} R \\pm OI.$", "Solution_2": "We can solve part a easier . \nWe know $A'B'C'H$ in cycle . \nso $ A'H . B'C' + HB' . A'C' = HC' . A'B' $\n $(x-AH) . B'C' +(x-BH) . A'C' = (CH - x) . A'B' $\n $\\triangle A'B'C' \\sim \\triangle ABC$ so $ AB = k A'B' $ and ....\nso $(x-AH) . BC +(x-BH) . AC = (CH - x) . AB $\nso $ 2xp = BC.AH + AC.BH + AB.CH = 4S = 4rp $\nso $x=2r$\n\nI mean $2p = AB + BC + CA $ and $S$ is the area of the triangle .", "Solution_3": "Is there a strict proof for $A'B'C'H$ being cyclic? Angle chasing might work only we presume a specific configuration of those points." } { "Tag": [ "algebra", "polynomial", "probability", "function", "geometry", "circumcircle", "inradius" ], "Problem": "Here's some easy but nice problems I made up if you guys wanna do them... :whistling: \r\n\r\n1)a) If the roots of the polynomial $2x^3+18x^2-12x+8$ are $a,b,c$, compute \r\n\r\n$\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}$ \r\nb) find a polynomial who's roots are $a+3,b+3,c+3$\r\nc) find a polynomial whos roots are $3a,3b,3c$\r\n\r\n2) What is the probability that if we pick two numbers out of $0$ to $100$, they'll differ by at most $4$?\r\n\r\n3) How many solutions are for $a+b+c+d+e\\leq 70$?\r\n\r\n4) How many ways can we distribute $20$ burgers among $5$ students so that no student gets less than $2$ or more than $6$?\r\n\r\nI don't know what the answer to this problem is, so can someone give a detailed explanation of generating functions?\r\n\r\n5) Find the distance between the centers of the circumcircle and incircle of a $12,13,14$ triangle\r\n\r\n number 5's answer will be a little messy though so you can probably leave the answer unsimplified in the $d^2$ formula\r\n\r\n\r\n\r\n\r\nEnjoy :)", "Solution_1": "Are these problems too hard or does nobody want to do them?? I thought people in the Carolina's Board wanted people to post problems??? ;) :diablo:", "Solution_2": "well, i guess since no one wants to do them...i'll answer my own questions..\r\n[hide=\"1a\"]\nThe polynomial with the roots reciprocal to $2x^3+18x^2-12x+8$ is $8x^3-12x^2+18x+2$, and we know that $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac) \\Longrightarrow a^2+b^2+c^2=(\\frac{3}{2})^2-2(-\\frac{9}{4})=\\boxed{\\frac{27}{4}}$[/hide]\n[hide=\"1b\"]\nFor this, we have to take $f(x)$ and convert it into $f(x-3)$, so we use synthetic division and have $\\boxed{2x^3-66x+152}$\n[/hide]\n[hide=\"1c\"]\nFor this, we are multiplying the coefficients from left to right by $3^0,3^1,3^2,3^3$, so we have $\\boxed{2x^3+54x^2-108x+216}$[/hide]\n[hide=\"2\"]\nWe can set this problem up geometrically, thus making a square with sides of $100$, the area of our desired numbers is $100^2-96^2 \\Longrightarrow (100+96)(100-96)=784$, so our probability is $\\frac{784}{100^2}$[/hide]\n\n[hide=\"3\"]\nSince $\\binom{n+r-1}{r-1}$ is applied and since it's an equality, we can fill another number in to give the extra sum left over, so we have $\\boxed{\\binom{75}{5}}$[/hide]\n\n\nNumber 4, I'm not exactly sure how to do, [hide=\"but\"]using generating functions, we can have $(x^2+x^3+x^4+x^5+x^6)^5$ and we should be looking for the value of the coefficient of $x^{20}$, which I can't do because I don't know how to simplify the thing above...[/hide]\n[hide=\"5\"]\nI'm not going to solve this right now because it's messy and takes too long, but just use Heron's formula to find the area of the triangle which will get us the inradius using $rs=A$ and also the circumradius using $R=\\frac{abc}{4A}$ and use $d^2=R(R-2r)$ to find the distance.[/hide]", "Solution_3": "I can give you an explanation of how to solve number four if you need it. Let me know.", "Solution_4": "[quote=\"prybarczyk\"]I can give you an explanation of how to solve number four if you need it. Let me know.[/quote]\r\n\r\nsure, i'm lost on generating functions :?", "Solution_5": "Open the attachment. I hope this makes sense. I did not have a lot of time to write it up. If you have any questions, let me know.", "Solution_6": "Did that help?", "Solution_7": "yes, thank you so much for your help :)" } { "Tag": [ "function", "limit", "algebra unsolved", "algebra" ], "Problem": "foll all $n$ is a number positive integer \r\na function $g(x)$ give by:\r\n$g(x)$=-$n$ if $x$ :leq -$n$\r\n$g(x)$=$x$ if -$n$<$x$ :leq $n$\r\n$g(x)$=$n$ if $n$<$x$\r\n\r\nlet a function $f(x)$:R-R such that:\r\nfor all $n$ in N* then $g(f(x))$ is continous\r\nprove that :$f(x)$ is a continous fuction", "Solution_1": "Let \\[g_{m}\\left( x\\right) =\\left\\{ \\begin{array}[c]{l}-m,-mM.$ Then $g_{m}\\left( f\\left( x\\right) \\right) =f\\left( x\\right)$ for all $x\\in I,$\r\nhence the conclusion.\r\n\r\nNow, suppose that $f$ is discontinuous at $a$ and not bounded from above.\r\nThere exists a sequence $\\left( x_{n}\\right)$ such that $x_{n}\\rightarrow a$ and $f\\left( x_{n}\\right) \\rightarrow\\infty.$ Pick $m$ such that $m>\\left\\vert f\\left( a\\right) \\right\\vert .$ Then $\\underset{n\\rightarrow \\infty}{\\lim}g_{m}\\left( f\\left( x_{n}\\right) \\right) =m$ and $g_{m}\\left( f\\left( a\\right) \\right) =f(a) 0$ on $K - \\{x_0\\}$\r\n\r\nProve that $E$ is dense in $(C_{x_0}(K, \\mathbb{R}), |.|_{\\infty})$", "Solution_1": "$E$, together with the constant functions on $K$, forms an algebra $\\widetilde E$ which separates points and which contains a constant function, hence is dense by Stone-Weierstrass. Any good enough approximation in $\\widetilde E$ of a function which vanishes at $x_0$ will have to have to involve a small constant function, which we can ignore, thus getting a good approximation in $E$.\r\n\r\nSorry if it sounds unclear, but I think it's pretty obvious what's going on.\r\n\r\nP.S.\r\n\r\nWhy do we need that positivity condition? :?" } { "Tag": [ "floor function", "geometry", "3D geometry", "pyramid", "symmetry", "trigonometry", "tetrahedron" ], "Problem": "A tripod has three legs each of length 5 feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is 4 feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\\frac m{\\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\\lfloor m+\\sqrt{n}\\rfloor.$ (The notation $\\lfloor x\\rfloor$ denotes the greatest integer that is less than or equal to $x$.)", "Solution_1": "[hide=\"A straightforward but messy solution\"]Let's first examine the non-broken tripod: if $P$ is the intersection of the altitude of the pyramid formed with the base, and we let the base triangle be denoted $\\Delta ABC$, then we have $AP = BP = CP = 3$ by the Pythagorean theorem. We also know that $\\angle APB = \\angle BPC = \\angle CPA = 120^{\\circ}$ by symmetry, so by the law of Cosines we have $AB = BC = AC = 3 \\sqrt{3}$. Then the cosine of the angle between any two tripod legs is given by\n$\\cos \\theta = \\frac{23}{50}$\n\nNow we examine the broken tripod. Let the base triangle be given by $\\Delta ABC'$, where $BC'$ and $AC'$ are the base lengths of the tripod faces with side lengths 4 and 5. Applying the Law of Cosines, we find that \n$BC' = AC' = \\sqrt{\\frac{113}{5}}$\n(I'm skipping over a lot of messy arithmetic in this problem, but you can verify it for yourself). Now let's examine $\\Delta ABC'$. Let $P'$ be the intersection of the new height and the base. Note that by the Pythagorean theorem we have $AP' = BP' = \\sqrt{25 - h^2}$, and $C'P' = \\sqrt{16 - h^2}$, where $h$ is the unknown altitude. Note also that $\\angle AC'P' = \\angle BC'P' = 1/2 \\angle AC'B$, as $\\Delta AC'P'$ and $\\Delta BC'P'$ are congruent. Now we apply law of Cosines to $\\Delta ABC'$ to find first that\n$\\cos AC'B = \\cos \\alpha = \\frac{91}{226}$\nApplying the half-angle identity for cosine, we have\n$\\cos \\frac{\\alpha}{2} = \\frac{1}{2} \\sqrt{\\frac{317}{113}}$\nNow we apply Law of Cosines to $\\Delta AC'P$ (or $\\Delta BC'P$) to find\n$25 - h^2 = \\frac{113}{5} + 16 - h^2 - 2 \\sqrt{\\frac{113}{5}} \\sqrt{16 - h^2} \\cos \\frac{\\alpha}{2}$\nNow substituting in for $\\cos \\frac{\\alpha}{2}$ and doing a ton of tedious arithmetic (thankfully all but one $h^2$ cancels, so we don't have to resort to anything insane), we find that\n$h = \\frac{144}{\\sqrt{1585}}$\nSo we have\n$\\lfloor m + \\sqrt{n} \\rfloor = \\lfloor 144 + \\sqrt{1585} \\rfloor$\nand noting that $40^2 = 1600$, so that $\\lfloor \\sqrt{1585} \\rfloor = 39$, we have\n$\\lfloor 144 + \\sqrt{1585} \\rfloor = 144 + 39 = \\boxed{183}$[/hide]", "Solution_2": "[hide=\"A similarly messy solution, but without cosine law\"]\nThis is really hard to explain without a diagram, but here goes:\n\nFirst consider the non-broken tripod. You can think of this as a tetrahedron, with height $4$, sides $5$, and an equilateral triangle base of side $3 \\sqrt{3}$\n\nNow, consider cutting this tetrahedron in half (or a projection onto a 2D plane if you wish to think of it that way), along the plane which includes the height of the trapezoid and one of the legs. We now consider this cross-section, which is a triangle which we shall name $ABC$, with $AB = 5$, $BC = \\frac{9}{2}$, and altitude $AD = 4$.\n\nThe beauty of this cross-section is that, if you consider the similar cross-section of the broken tripod, you can draw in its cross-section superimposed on the one of the normal tripod. You do this simply by drawing a line $CE$ which divides $AB$ (the leg that gets broken), such that $EB = 1$ and $AE = 4$.\n\nThe new height of this broken tripod, is then the altitude of triangle $ACE$ dropped from $A$ to $CE$. Let's name this $AF$.\nLet's also designate the intersection of $CE$ and $AD$ to be $G$, and draw a perpendicular from point $E$ to $BC$, and designate this line $AH$.\n\nNow comes the messy part; i will skip through a lot of the calculations, but you can verify them if you want:\n\n$EB = 1$, and by similar triangles to $EDB$, $EH = \\frac{4}{5}$, and $HB = \\frac{3}{5}$.\n\nNow, triangle $CDG \\sim CHE$ , and you know $CD = \\frac{3}{2}$, $CH = CB - HB = \\frac{39}{10}$, and $HE = \\frac{4}{5}$, so you set up the ratio and find $GD = \\frac{4}{13}$. This gives us $AG = 4-\\frac{4}{13}=\\frac{48}{13}$.\n\nNow, triangle $CDG \\sim AFG$, so we wish to set up a ratio to find $AF$. To do this, we find $CG$ by pythagoras, which comes out to be $\\frac{\\sqrt{1585}}{26}$. So now, we set up the ratio:\n\\begin{eqnarray*} \\frac{CG}{AG} &=& \\frac{CD}{AF}\\\\ \\frac{\\frac{\\sqrt{1585}}{26}}{\\frac{48}{13}} &=& \\frac{\\frac{3}{2}}{h}\\\\ h &=& \\frac{114}{\\sqrt{1585}} \\end{eqnarray*}\nI know, I know, i skipped a whole bunch of ugly math...\nbut after this, we have $m = 114$ and $n = 1585$, so $\\left\\lfloor m + \\sqrt{n}\\right\\rfloor = 183$\n[/hide]", "Solution_3": "OK, I am writing a little clean solution:\r\n\r\n[hide]\nWe will use $[...]$ to denote volume (four letters), area (three letters) or length (two letters).\n\nLet $T$ be the top of the tripod, $A,B,C$ are end points of three legs. Let $S$ be the point on $TA$ such that $[TS]=4$ and $[SA]=1$. Let $O$ be the center of the base equilateral triangle $ABC$. Let $M$ be the midpoint of segment $BC$. Let $h$ be the distance from $T$ to the triangle $SBC$ ($h$ is what we want to find).\n\nWe have the volume ratio: $\\frac{[TSBC]}{[TABC]} = \\frac{[TS]}{[TA]}=\\frac{4}{5}$\n\nSo $\\frac{h\\cdot [SBC]}{[TO]\\cdot [ABC]}=\\frac{4}{5}$\n\nWe also have the area ratio: $\\frac{[SBC]}{[ABC]}=\\frac{[SM]}{[AM]}$\n\nThe triangle $TOA$is a 3-4-5 right triangle so $[AM] = \\frac{3}{2}\\cdot[AO]=\\frac{9}{2}$ and $\\cos{\\angle{TAO}}=\\frac{3}{5}$.\n\nApplying Law of Cosines to the triangle $SAM$ with $[SA]=1$, $[AM]=\\frac{9}{2}$ and $\\cos{\\angle{SAM}}=\\frac{3}{5}$, we find:\n\n$[SM]=\\frac{\\sqrt{5\\cdot317}}{10}$.\n\nPutting it all together, we find $h=\\frac{144}{\\sqrt{5\\cdot317}}$.\n\n[/hide]", "Solution_4": "Wow, I feel brilliant. I found $h$ but thought I was wrong since $m+n>1000$. So I just randomly guessed. :wacko:", "Solution_5": "I went for a much more complicated solution...and messed it up somewhere along the way. Well my answer was about the same, less than a unit off...", "Solution_6": "[hide=\"A solution using coordinates\"]\nAs in the previous solutions, it's not hard to get that the feet of the original tripod form an equilateral triangle of side length $3 \\sqrt{3}$.\n\nPlace the feet of the tripod at $P(3, 0, 0)$, $Q(-3/2, 3 \\sqrt{3}/2, 0)$, and $R(-3/2, -3 \\sqrt{3}/2, 0)$, and the top at $V(0, 0, 4)$.\n\nLet $VP$ be the leg that gets broken, so it is broken at the point\n\\[ S = \\frac{1}{5} V + \\frac{4}{5} P = \\left( \\frac{12}{5}, 0, \\frac{4}{5} \\right). \\]\n\nLet the equation of the plane passing through points $Q$, $R$, and $S$ be $Ax + By + Cz = 1$, so we get the system\n\\begin{eqnarray*} -\\frac{3}{2} A + \\frac{3 \\sqrt{3}}{2} B &=& 1, \\\\ -\\frac{3}{2} A - \\frac{3 \\sqrt{3}}{2} B &=& 1, \\\\ \\frac{12}{5} A + \\frac{4}{5} C &=& 1. \\end{eqnarray*}\n\nSolving, we get $A = -2/3$, $B = 0$, and $C = 13/4$, so the equation of the plane is\n\\[ -\\frac{2}{3} x + \\frac{13}{4} z = 1, \\]\nor $8x - 39z + 12 = 0$.\n\nRecall that the distance from the point $(x_0, y_0, z_0)$ to the plane $Ax + By + Cz + D = 0$ is\n\\[ \\frac{|Ax_0 + By_0 + Cz_0 + D|}{\\sqrt{A^2 + B^2 + C^2}}. \\]\n\nPlugging these in, we get that the distance from point $V$ to plane $QRS$ is\n\\[ \\frac{|-39 \\cdot 4 + 12|}{\\sqrt{8^2 + 39^2}} = \\frac{144}{\\sqrt{1585}}. \\]\n\nThen $\\lfloor 144 + \\sqrt{1585} \\rfloor = 183$.\n[/hide]\r\n\r\nNormally, I wouldn't endorse coordinates, but the calculations in this case are quite reasonable.", "Solution_7": "[quote=\"joml88\"]Wow, I feel brilliant. I found $h$ but thought I was wrong since $m+n>1000$. So I just randomly guessed. :wacko:[/quote]\r\n\r\nSame :(. I was like I can't have a $1585$ in the denominator that's too big...", "Solution_8": "In the aopswiki solution: http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems/Problem_14 , why is\n\n$ \\frac{[SBC]}{[ABC]}=\\frac{[SM]}{[AM]} $ true?", "Solution_9": "[quote=\"zanttrang\"]Note that by the Pythagorean theorem we have $AP' = BP' = \\sqrt{25 - h^2}$, and $C'P' = \\sqrt{16 - h^2}$, where $h$ is the unknown altitude.[/quote]\n\nI got up to this point. Here, I used the circumradius area formula, $A = \\dfrac{abc}{4R}$ and equated it with the easily calculated area to find the circumradius. Then, my fallacy was that the circumradius is the second leg of the right triangle that has a hypotenuse of one of the feet of the tripod and the other leg being the height. This is obviously false since that would mean the height is the leg of two non-congruent right triangles, both with other leg being the circumradius, but one with hypotenuse $4$ and other with hypotenuse $5$. However, I do not understand then where to proceed from here. I'm kind of lost by the other solutions. Where, wrt to the base, does the height \"land\"?", "Solution_10": "[quote=\"wilson97\"]In the aopswiki solution: http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems/Problem_14 , why is\n\n$ \\frac{[SBC]}{[ABC]}=\\frac{[SM]}{[AM]} $ true?[/quote]\n\nThe ratio of the areas of two triangles equals the ratio of their respective altitudes when their bases are equal.", "Solution_11": "[hide=Fast 2D coordinates]\nLet $G$ be the centroid of the equilateral triangle base $ABC$ and let $T$ be the other vertex of the original tripod. Let $5S = 4A+T$ and set up coordinates with $G(0,0),A(3,0),T(0,4) \\implies S(\\frac{12}{5},\\frac{4}5), \\frac{B+C}{2} = (-\\frac{3}{2},0)$. Therefore the line through $S, \\frac{B+C}{2}$ is \n$$\\ell: 8x - 39 y + 12 = 0.$$\nUse distance to line formula for $T, \\ell$ to finish.\n[/hide]", "Solution_12": "This is really easy with complex numbers.", "Solution_13": "Oops here's another slow synthetic solution.\n\nFirst we want to figure out the dimensions of the new tripod. Check that the base is an equilateral triangle of side length $3\\sqrt 3$, so by Stewart's on one of the lateral triangles, we obtain that the new base of the tripod is an isosceles triangle with side lengths $2\\sqrt 7, 2\\sqrt 7, 3\\sqrt 3$.\n\nNow, let $O$ be the foot of the perpendicular from the apex of the tripod. Observe that the distances from $O$ to the three vertices of the base are $x, x, \\sqrt{x^2-9}$ by perpendicularity lemma. Using the altitude relation, we obtain $$\\sqrt{x^2-9} + \\sqrt{x^2-\\frac{27}4} = \\sqrt{\\frac{317}{20}},$$ which solves to get $x^2 = \\frac{4 \\cdot 34^2}{5 \\cdot 317} + 9$. Thus, the length of the altitude is $$\\sqrt{25-y} = \\sqrt{\\frac{16 \\cdot 5 \\cdot 317 - 68^2}{5 \\cdot 317}} = \\sqrt{\\frac{144^2}{5 \\cdot 317}},$$ so the answer is $\\lfloor 144 + \\sqrt{5 \\cdot 317} \\rfloor = \\boxed{183}$.", "Solution_14": "Let $A,B,C$ be the feet of the tripod in some order, and let $D$ be the highest point. Note that $\\triangle ABC$ is equilateral with side length $3\\sqrt{3}$. Hence, $[ABC] = \\frac{(3\\sqrt{3})^2\\sqrt{3}}{4}$, so $\\text{Vol}(ABCD) = \\frac{1}{3}\\cdot 4\\cdot [ABC] = 9\\sqrt{3}$. \n\nNow, let $C'$ be the foot of the cut leg. From volume ratios, $\\text{Vol}(ABC'D) = \\frac{4}{5}\\text{Vol}(ABCD) = \\frac{36\\sqrt{3}}{5}$. Now, we compute $[ABC']$. Note that $\\cos\\angle ADC =\\frac{23}{50}$ from LoC, so\n\\[ AC' = \\sqrt{5^2 + 4^2 - 40\\cdot \\frac{23}{50}} = \\sqrt{\\frac{113}{5}}, \\]\nfrom LoC in $\\triangle ADC'$. Similarly, $BC' = \\sqrt{\\frac{113}{5}}$. Now, we have all three side lengths of $\\triangle ABC'$, so we use pythag to find the height to $AB$ and use base*height/2 to get $[ABC'] = \\frac{3}{4}\\sqrt{\\frac{951}{5}}$. Setting up an equation for the volume of $ABC'D$ gives\n\\[ \\frac{1}{3}\\cdot h\\cdot \\frac{3}{4}\\sqrt{\\frac{951}{5}} = \\frac{36}{5}\\sqrt{3}\\implies h = \\frac{144}{\\sqrt{1585}}.\\]\nHence, the answer is $144 + \\lfloor \\sqrt{1585}\\rfloor = 144 + 39 = \\boxed{183}$", "Solution_15": "\nSolved with [b]amuthup[/b], [b]js104[/b], [b]crazyeyemoody907[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLscelesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b]. \n\n\nLet the feet be $A$, $B$, $C$ and the top of the tripod be $T$. Note that the distance between the projection of $T$ onto plane $ABC$ and any one of the feet is $\\sqrt{ 5^{2}-4^{2} }=3$, so the side length of $\\triangle ABC$ is $\\frac{3}{\\frac{\\sqrt{3 }}{3}}=3\\sqrt{ 3 }$.\n\nPlace the tripod on the coordinate plane, with the ground being $y=0$. Let the feet of the tripod be $A,B=\\pm\\left( \\frac{3\\sqrt{ 3 }}{2},0,0 \\right)$ and $C=\\left( 0,0, \\frac{9}{2} \\right)$. We then must have $T=\\left( 0,4, \\frac{3}{2} \\right)$.\n\nSuppose the lower foot of leg $TC$ breaks, so that the new end is $C'=\\frac{4C+T}{5}=\\left( 0, \\frac{4}{5}, \\frac{39}{10} \\right)$. \n\nWe wish to find the equation of plane $ABC'$: $ax+by+cz=d$. Since the line with parametric equation $(t,0,0)$ (aka $AB$) is on this plane, $a=d=0$. We can then see that the plane's equation is therefore $39y-8z=0$ (or any scaling thereof).\n\nOur desired $h$ is the distance from $T$ to this plane, which is: \n$$\n\\frac{\\left\\lvert 39\\cdot 4-8\\cdot \\frac{3}{2} \\right\\rvert}{\\sqrt{ 39^{2}+8^{2} }}=\\frac{144}{\\sqrt{ 1585} }.\n$$\n\nSo our answer is $\\lfloor 144 + \\sqrt{ 1585 }\\rfloor=144+39=\\boxed{183}.$" } { "Tag": [ "inequalities" ], "Problem": "For positive numbers $ x, y, z$ , prove that, \r\n\r\n$ \\displaystyle \\frac {\\left\\{\\prod(x \\plus{} y)\\right\\}^{2}}{\\prod x}\\geq\\frac {\\left\\{\\frac {8}{9}\\left(\\sum x\\right)\\left(\\sum xy\\right)\\right\\}^{2}}{\\prod x}\\geq\\frac {64}{27}\\left(\\sum x\\right)^{3}.$\r\nThanks", "Solution_1": "[quote=\"Shanku\"]For positive numbers $ x, y, z$ , prove that, \n\n$ \\displaystyle \\frac {\\left\\{\\prod(x \\plus{} y)\\right\\}^{2}}{\\prod x}\\geq\\frac {\\left\\{\\frac {8}{9}\\left(\\sum x\\right)\\left(\\sum xy\\right)\\right\\}^{2}}{\\prod x}\\geq\\frac {64}{27}\\left(\\sum x\\right)^{3}.$\n\nThe source where I got it says it is obvious. However I am not finding the obvious links.[/quote]\r\n\r\nIf something not obvious is not out of my attention the left hand side of the 1st part of the inequality reduces to $ 9(x \\plus{} y)(y \\plus{} z)(z \\plus{} x)\\geq 8(x \\plus{} y \\plus{} z)(xy \\plus{} yz \\plus{} zx)$ which is obviously true, since it is of the form $ \\sum_{cyc}x(y \\minus{} z)^2$.\r\nNow the second part of the inequality reduces to $ \\left(\\sum_{cyc}xy\\right)^2\\geq 3xyz\\sum_{cyc}x$ which is also obviously true.\r\nSorry for interupting at this part of the forum." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $ a, b, c$ be the sides and $ l_a, l_b, l_c$ the bisectors of an triangle $ ABC$. Prove that\r\n\r\n$ (a \\plus{} b \\plus{} c)^2 > l_a(b \\plus{} c) \\plus{} l_b(c \\plus{} a) \\plus{} l_c(a \\plus{} b) > \\frac {(a \\plus{} b \\plus{} c)^2}{2}$ .\r\n\r\n________\r\n :|", "Solution_1": "[quote=\"Ligouras\"]Let $ a, b, c$ be the sides and $ l_a, l_b, l_c$ the bisectors of an triangle $ ABC$. Prove that\n\n$ (a + b + c)^2 > l_a(b + c) + l_b(c + a) + l_c(a + b) > \\frac {(a + b + c)^2}{2}$ .\n\n________\n :|[/quote]\r\nwe know #$ l_a=\\frac{2bccos{\\frac{A}{2}}}{b+c}$\r\nSO $ l_a(b + c) + l_b(c + a) + l_c(a + b)=2(\\sum bccos{\\frac{A}{2})}$" } { "Tag": [ "inequalities", "geometry", "inequalities unsolved" ], "Problem": "Let $ a,b,c,x,y,z>0$. Prove that $ \\frac{a(z\\plus{}y)}{b\\plus{}c}\\plus{} \\frac{b(z\\plus{}x)}{a\\plus{}c} \\plus{}\\frac{c(x\\plus{}y)}{b\\plus{}a} \\geq \\sqrt{3(xy\\plus{}yz\\plus{}zx)}$.", "Solution_1": "[quote=\"Spribo\"]Let $ a,b,c,x,y,z > 0$. Prove that $ \\frac {a(z \\plus{} y)}{b \\plus{} c} \\plus{} \\frac {b(z \\plus{} x)}{a \\plus{} c} \\plus{} \\frac {c(x \\plus{} y)}{b \\plus{} a} \\geq \\sqrt {3(xy \\plus{} yz \\plus{} zx)}$.[/quote]\r\nSee here:\r\nhttp://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf", "Solution_2": "[b]Following well-know-inequality:[/b]\r\n$ \\mbox{If a, b, c sides of triangle ABC, and: }m,n,p > 0 \\\\\r\n\\mbox{and F \\equal{} area of triangle ABC, rezult: } \\\\\r\na^2m \\plus{} b^2n \\plus{} c^2p\\ge\\sqrt {mn \\plus{} mp \\plus{} np}.4F; \\\\\r\n\\mbox{ let: } a \\equal{} \\sqrt {x \\plus{} y}, b \\equal{} \\sqrt {x \\plus{} z}, c \\equal{} \\sqrt {y \\plus{} z}\\mbox{ and} \\\\\r\nm \\equal{} \\frac {a}{b \\plus{} c}, n \\equal{} \\frac {b}{a \\plus{} c}, p \\equal{} \\frac {c}{a \\plus{} b}\\Rightarrow \\\\\r\n\\Rightarrow \\sum_{cyclic}{\\left(\\sqrt {x \\plus{} y}\\right)^2.\\frac {a}{b \\plus{} c}}\\ge\\sqrt {\\sum_{cyclic}\\frac {a}{b \\plus{} c}.\\frac {b}{a \\plus{} c}}.\\sqrt {\\sum_{cyclic}{xy}} \\\\\r\n\\mbox{and }\\sum_{cyclic}{\\frac {a}{b \\plus{} c}.\\frac {b}{a \\plus{} c}}\\ge\\frac {3}{4} \\mbox{ (well \\minus{} know inequality)} \\\\\r\n\\mbox{and prowe the inequality.}$\r\nview topics: http://www.mathlinks.ro/viewtopic.php?t=261034" } { "Tag": [], "Problem": "How many numbers are in the list \\[ 1.5, 5.5, 9.5, 13.5, \\ldots, 41.5, 45.5?\\]", "Solution_1": "Multiply each number by $ 2$ to get $ 3,11,19,\\ldots,83,91$.\r\n\r\nThe number of times this sequence increases by $ 8$ is $ \\frac{91\\minus{}3}8\\equal{}11$,\r\n\r\nso the number of terms is $ 11\\plus{}1\\equal{}\\boxed{12}$.", "Solution_2": "The multiplying by $ 2$ is unnecessary.", "Solution_3": "But it's easier to not make mistakes...", "Solution_4": "Or, you could subtract $ 0.5$ from each term. Then we have $ 1,5,9,13,\\ldots,41,45$. This increases by $ 4$ every step, so we have $ \\frac{45\\minus{}1}{4}\\plus{}1\\equal{}\\boxed{12}$.", "Solution_5": "I guess the fastest way is to add $ 2.5$ to each term to get\r\n\r\n$ 4,8,\\ldots,48$, and dividing through by $ 4$, we get\r\n\r\n$ 1,2,\\ldots,12$, which obviously has $ 12$ terms.", "Solution_6": "Or we have 45.5-1.5/4+1=11+1=12", "Solution_7": "El .5 en cada n\u00famero no importa. avanza de 4 en 4, le resto 1 a la lista 1,5,9,...45 para conseguir 0, 4,...44, luego tenemos\n44/4 + 1 (cero) = 11 + 1 = 12. ya estuvo.", "Solution_8": "[quote=jonathanprieto]El .5 en cada n\u00famero no importa. avanza de 4 en 4, le resto 1 a la lista 1,5,9,...45 para conseguir 0, 4,...44, luego tenemos\n44/4 + 1 (cero) = 11 + 1 = 12. ya estuvo.[/quote]\n\nTranslated (from Spanish into English) : \n\nThe .5 in each number does not matter. advances from 4 to 4, I subtract 1 from the list 1,5,9, ... 45 to get 0, 4, ... 44, then we have\n44/4 + 1 (zero) = 11 + 1 = 12. It was already.\n\nsteps for mine:\n\n$$1.~\\text{Add }0.5.$$\n$$2.~\\text{Divide by }2.$$\n$$3.~\\text{Add }1.$$\n$$4.~\\text{Divide by }2.$$\n\nYou should have:\n\n$$1,2,3,4,\\ldots ,11,\\boxed{12}.$$", "Solution_9": "[quote=pog]I want one added to my post count too![/quote]\nWe subtract $0.5$ from each number to get:\n$$1, 5, 9, 13, ...., 41 , 45$$\n, then we subtract one from each number to get \n$$ 0, 4, 8, 12, .... 40, 44$$\nand then we divide each number by four to get \n$$ 0, 1, 2, 3, ...., 10, 11$$, \nso we have $\\boxed{12}$ numbers.", "Solution_10": "[quote=ernie]The multiplying by $ 2$ is unnecessary.[/quote]\n\nYa", "Solution_11": "[quote=Smartgoodness][quote=jonathanprieto]El .5 en cada n\u00famero no importa. avanza de 4 en 4, le resto 1 a la lista 1,5,9,...45 para conseguir 0, 4,...44, luego tenemos\n44/4 + 1 (cero) = 11 + 1 = 12. ya estuvo.[/quote]\n\nTranslated (from Spanish into English) : \n\nThe .5 in each number does not matter. advances from 4 to 4, I subtract 1 from the list 1,5,9, ... 45 to get 0, 4, ... 44, then we have\n44/4 + 1 (zero) = 11 + 1 = 12. It was already.\n\nsteps for mine:\n\n$$1.~\\text{Add }0.5.$$\n$$2.~\\text{Divide by }2.$$\n$$3.~\\text{Add }1.$$\n$$4.~\\text{Divide by }2.$$\n\nYou should have:\n\n$$1,2,3,4,\\ldots ,11,\\boxed{12}.$$[/quote]\n\nThank you for the translation" } { "Tag": [ "algorithm", "number theory", "prime numbers", "number theory open" ], "Problem": "A perfect number x is a number which equals to the sum of all it's divisors\r\n(including 1, but without x). \r\nEuclid proved that if p -is prime and p+1=2^{k} then 2^{k-1}*p is perfect.\r\nEiler proved that every even perfect number is equal to 2^{k-1}*p, where p -is prime and p+1=2^{k}.\r\nIt is unknown are there odd perfect numbers.\r\nThe question which is interesting to me is: Is it true that every even perfect number\r\nends on 6 or 28? First five perfect numbers are 6,28,496,8128,33550336.\r\nIf someone knows sixth perfect number please tell it to me. Is it ends on 28?", "Solution_1": "The first perfect numbers are:\r\n$6\\\\28\\\\496\\\\8128\\\\33550336\\\\8589869056\\\\137438691328\\\\2305843008139952128$\r\n$\\cdots$\r\nIt is well known that all [u]even[/u] numbers (except $6$) end in $16,28,36,56,76$ or $96$.\r\nBut it is not known if any odd perfect number exist.", "Solution_2": "You can easily show your statement when looking at all prime residue classes $\\mod 100$.", "Solution_3": "On February 18, 2005, Martin Nowak from Germany, found the new largest known prime number (42 nd):\r\n$2^{25964951}-1$.\r\n\r\nVery easy problem: what are the two last digits of the corresponding perfect number? :)", "Solution_4": "[quote=\"blang\"]\nBut it is not known if any odd perfect number exist.[/quote]\r\n\r\nHave a look here : http://arxiv.org/abs/hep-th/0401052 :)", "Solution_5": "Did you check the proof? ;)\r\nThe problem seem still to be open [url=http://mathworld.wolfram.com/OddPerfectNumber.html][u]here[/u][/url] :(\r\nSept 2004, Hare wrote: \"It is unclear if the recent paper on the Number Theory Archive claiming to prove that there are no odd perfect number, successfully solves this problem or not\".\r\nDo you have some new precisions?", "Solution_6": "in more general, i know that there is a theroy about numbers called friends (amiable-im not sure about the translate) wich have the following : S(m)-m=S(n)-n were S(x) means the sum of all divisors of x ,integer.the smallest pair is (220,284).there is not known an algorithm wich finds all this kind of pairs.EULER found a particular prove and discover aprox 50 new pairs. in his prove he takes forms for m and n. (it's very long).\r\n220 and 284 were considerated for a long time two magic numbers with special power.if i remember well...Pitagora was the one who found first 2 friends numbers.", "Solution_7": "the english word is \"amicable\", but the definition is different: $\\sigma(m)=\\sigma(n)=m+n$, so each is the sum of the \"aliquot\" parts of the other ($m=\\sigma(n)-n$ and $n=\\sigma(m)-m$). \r\n\r\n[url]http://mathworld.wolfram.com/AmicablePair.html[/url]\r\n[url]http://www.shyamsundergupta.com/amicable.htm[/url]", "Solution_8": "Igor did you read the proof ?\r\nI don't follow the author just from the start :P", "Solution_9": "yap,you're right :blush: :blush: :blush: i'm sorry about that...i'm not sure what happened :oops:", "Solution_10": "Is there a simple test for primality of Merseene number?\r\nthx\r\nBomb", "Solution_11": "the standard test is:\r\n\r\ndefine $s_2:=4, s_{n+1} := s_n ^2 -2$. Then $2^p -1$ (with $p>2$ being prime) is a prime itself iff $2^p - 1 | s_{p}$. (as much as I know it's the best known test for prime numbers)", "Solution_12": "This is a special version of a more general test:\r\nLet be $p$ a given odd interger.\r\nLet be $a$ any integer that fulfils $\\left( \\frac{a}{p} \\right) = -1$ ( $\\left( \\frac{a}{p} \\right)$ means the jacobi symbol ).\r\nThen $p$ is prime\r\niff\r\nthere is a $\\epsilon \\in (\\mathbb{Z} / p \\mathbb{Z})[\\sqrt a]$ that fulfils:\r\na) $\\epsilon \\overline{\\epsilon} =1 $ ($\\overline{\\epsilon}$ means the 'conjugate' of $\\epsilon$)\r\nb) $\\epsilon ^ {p+1} =1$\r\nc) $\\epsilon ^\\frac{p+1}{q} \\neq 1$ for all prime divisors $q$ of $p+1$.", "Solution_13": "Thx" } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "arithmetic sequence", "number theory unsolved", "number theory" ], "Problem": "Does anyone have a non very difficult and elementary proof of the following theorem? \r\n ' If f is a polynomial with integer coefficients such that in any infinite arithmetic progression there is an element x such that f(x) is a square then f is the square of a polynomial'?", "Solution_1": "This theorem is not due to Sierpinski, but to Davenport (this one is ok ;) ), Schinzel and Lewis.\r\nTheir paper appeared in Acta Arithmetica, IX (1964), p.107-116.\r\nI would not say that the proof is completely elementary...and it refers to some results due to Hasse that I did not know (which does not prove anything).\r\nIt is a shame tyhat you didn't ask for it before yesterday because I would bring you the paper last sunday (I have no scan...). Maybe I could send it to you by classical mail. Send me your room adress by pm.\r\n\r\nTo be rigourous, they proved a more general result :\r\n\r\nLet $f(x,y)$ be a polynomial with integral coefficients. Suppose that every arithmetical progression contains some integer $x$ such that the equation $f(x,y)=0$ has an integral solution in $y$. Then, there exists a polynomial $g(x)$ with rational coefficients such that $f(x,g(x)) = 0$ identically.\r\n\r\nIt is an immediate corollary to deduce that :\r\n\r\nLet $k>1$ be an integer and let $f(x)$ be a polynomial with integer coefficients. Suppose that every arithmetical progression contains some integer $x$ such that $f(x)$ is a $k$-th power of an integer. Then $f(x)=(g(x))^k$ identically, where $g(x)$ is a polynomial with integral coefficients.\r\n\r\nPierre.", "Solution_2": "Pierre, if I found this problem earlier, be sure I would have asked you some information about you. Unfortunately, after the confererence at ENS I received a message from a friend in which he told me someone gave this theorem to him and he was asking for my help. Be sure I will always ask for some information. I would be very glad if you could send me the file by e_mail. I will give you the address in a pm. Thanks a lot." } { "Tag": [ "puzzles" ], "Problem": "[i]One day, a kid Michael was invited to visit a Math class in the home planet of E.T.\n\nThe teacher asked, \"If there are 24 classes in a school, and exactly 30 students in each class, then how many students are there in the school?\"\n\nA student William quickly put up his hand,\"Sir, it should be 720.\"\n\nTeacher replied: \"I am sorry, William, your calculation is incorrect. Think again.\"\n\nAnother girl Mary then put up her hand with uncertainty,\"I am not sure if my answer is correct....\"\n\nTeacher: \"Don't worry, just tell me your answer, Mary.\"\n\n\"I think...it should be 740...\"\n\n\"Good girl ! \" the teacher praised the girl and wore a smile in his face.[/i]\r\n\r\n\"What?????\" Michael was so confused.\r\n\r\n[b]Question: What's wrong with the class?[/b]", "Solution_1": "20 of the students are pregnant?\r\n\r\nAlso\r\n[hide]\nI think that they use the digit 4 instead of 5.\n[/hide]", "Solution_2": "20 students got pregnant ? :ninja:\r\nI have never thought about that before. \r\nThis answer is creative but it isn't the truth.\r\nThe movie clip in this link reveals the secret:\r\nhttp://www.youtube.com/watch?v=1KpripUvLs8", "Solution_3": "Uh, isn't that just a music clip?\r\n\r\nI'm afraid I don't understand how that explains it.\r\n\r\nEDIT: Ah. :)", "Solution_4": "Look at the hand of E.T. again.\r\nThe truth is out there!", "Solution_5": "[hide]\nThey count in base 8. Seven fingers, no thumbs, one flashlight.[/hide]", "Solution_6": "[hide]Let b equal the number base. Then\n$ (2b\\plus{}4)(3b)\\equal{}7b^2\\plus{}4b$\n$ 6b^2\\plus{}12b\\equal{}7b^2\\plus{}4b$\n$ b^2\\equal{}8b$\n$ b\\equal{}8$\n\nIn other words, ET has only four fingers on each hand~[/hide]" } { "Tag": [ "geometry", "search", "geometry unsolved" ], "Problem": "$K > L > M > N$ are positive integers such that $KM + LN = (K + L - M + N)(-K + L + M + N)$. Prove that $KL + MN $is composite.", "Solution_1": "This is from IMO 2001. Again, it has been posted before(even the geometric solution using Ptolemy). Search it yourself.", "Solution_2": "This is really too well-known.\r\n\r\nIMO 2001 Problem 6. Solutions on at least 100 different places on the web and on this site.", "Solution_3": "How many time did you send this problem?Anyway,here is link to the [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=imo+2001+geometry+problem&t=32032]solution[/url].\r\n(If you realy want the solution.)But it is very nice problem and darij solved it very very nice. :love:", "Solution_4": "can u solve it without using batlamious?\r\ni my self solved it by batlamious but iwant the other new solutions. :D", "Solution_5": "i think you are so cool ingeometry ashegh and you can solve it without using batlamius,so try it and sent your solution! ;)", "Solution_6": "your wellcome dear jensen.and i know that u can solve it without using bathlamious.\r\ncome on boy ,solve it. :D", "Solution_7": "[quote=\"ashegh\"]can u solve it without using batlamious?[/quote]\r\n\r\nRight. I was getting used to Pythagoras being Fishagoures already, but now I see that Ptolemaius (Ptolemy) became Batlamious! Guys, check your spelling! Please!", "Solution_8": ":rotfl: thanks alot arne." } { "Tag": [], "Problem": "I have though of a puzzling part of refraction that has caused a bit of a problem. it is that if the light slows down then where does it get the energy from to speed back up once it has left the medium it was refracted into?", "Solution_1": "Light has no mass. Energy is only expended in accelerating an object with non-zero rest mass. Therefore no energy is gained or lost by light particles (or photons) when their velocity chenges. In fact the knietic energy of an object is given by:\r\n\r\n$ E_{K}\\equal{} E\\minus{}E_{o}\\equal{} mc^{2}\\minus{}m_{0}c^{2}$\r\n\r\n$ m$ and $ m_{o}$ are the relativistic and rest masses respectively. $ m$ is given by:\r\n\r\n$ m \\equal{}\\frac{m_{0}}{\\sqrt{1\\minus{}\\frac{v^{2}}{c^{2}}}}$ \r\n\r\nTherefore:\r\n\r\n$ E_{K}\\equal{} m_{0}c^{2}\\left(\\frac{1}{\\sqrt{1\\minus{}\\frac{v^{2}}{c^{2}}}}\\minus{}1\\right)$\r\n\r\nHence an object with no rest mass, such as a photon will have no kinetic energy. The above equation shows why it is impossible to accelerate an object with non-zero rest mass to the speed of light. As $ v\\to c$ the denominator of the fraction tends to zero, and $ E_{K}\\to\\infty$. So it would take infinite energy to accelerate an object with mass to the speed of light. Anyway, I have diverged slightly from the main problem.\r\n\r\nThe energy of a photon is dependent on its frequency alone: $ E \\equal{} hf$ where $ h$ is Plank's constant. But the frequency is conserved when entering a different optical medium. Hence the light's energy is conserved, no energy is lost or gained by the photon during the transition between any two media. So nothing to worry about. (It is in fact due to the conservation of energy, that the frequency must remain constant)." } { "Tag": [ "algebra", "polynomial", "calculus", "derivative", "calculus computations" ], "Problem": "well, got the following results for $ \\ x^3\\plus{}4x^2\\minus{}20x\\minus{}81 \\equal{}0$ :\r\n\r\n$ \\ x_1 \\equal{} \\sqrt[3]{\\frac{2315}{54}\\plus{}\\sqrt{\\frac{3371}{108}}}\\plus{}\\sqrt[3]{\\frac{2315}{54}\\minus{}\\sqrt{\\frac{3371}{108}}}$\r\n\r\n$ \\ x_2 \\equal{} w\\sqrt[3]{\\frac{2315}{54}\\plus{}\\sqrt{\\frac{3371}{108}}}\\plus{}\\overline{w}\\sqrt[3]{\\frac{2315}{54}\\minus{}\\sqrt{\\frac{3371}{108}}}$\r\n\r\n$ \\ x_3 \\equal{} \\overline{w}\\sqrt[3]{\\frac{2315}{54}\\plus{}\\sqrt{\\frac{3371}{108}}}\\plus{}w\\sqrt[3]{\\frac{2315}{54}\\minus{}\\sqrt{\\frac{3371}{108}}}$\r\n\r\nBut the original equation is $ \\sqrt{x}\\equal{}18\\minus{}x^2$\r\n\r\nThe equation, after a change has become $ \\ (x\\minus{}4)(x^3\\plus{}4x^2\\minus{}20x\\minus{}81)\\equal{}0$\r\n\r\n4 As one of the roots, and the fact the original equation is not polynomial, there is no guarantee that a root complex, or both, are solutions of this equation .... doubt ....", "Solution_1": "obviously:\r\n\r\n$ \\ w\\equal{}\\frac{\\minus{}1}{2}\\plus{}i\\frac{\\sqrt{3}}{2}$", "Solution_2": "This is continued from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=253802]here.[/url] I'm not entirely sure why you paid close attention to Sonnhard Graubner's post and ignored mine.\r\n\r\nI suggested that you make the substitution $ y \\equal{} \\sqrt {x}.$ That gives us the equation\r\n\r\n$ y^4 \\plus{} y \\minus{} 18 \\equal{} 0$\r\n\r\n$ (y \\minus{} 2)(y^3 \\plus{} 2y^2 \\plus{} 4y \\plus{} 9) \\equal{} 0.$\r\n\r\nThe solution $ y \\equal{} 2$ gives us $ x \\equal{} 4.$ \r\n\r\nThe equation $ y^3 \\plus{} 2y^2 \\plus{} 4y \\plus{} 9$ has no nonnegative roots. (Call it Descartes' rule of signs, but it's the most obvious application of that: if $ y\\ge0$ than that cubic polynomial is $ \\ge 9.$) It does have at least one negative real root. Does it have any other real roots? The derivative of $ y^3 \\plus{} 2y^2 \\plus{} 4y \\plus{} 9$ is $ 3y^2 \\plus{} 6y \\plus{} 4 \\equal{} 3(y \\plus{} 1)^2 \\plus{} 1$ which has no real roots, so by Rolle's Theorem, the original cubic has exactly one negative real root.\r\n\r\nIf you want to use Cardano-Tartaglia to write a formula for that root, go ahead. I tend to find that answers written in raw Cardano-Tartaglia form (as in your first post on this topic) are seldom much help in actually thinking about problems. So let's just say that there is a real number $ y_0$ such that $ y_0 < 0$ and $ y_0^4 \\plus{} y_0 \\minus{} 18 \\equal{} 0.$\r\n\r\nOur difficulty now is the usual convention surrounding the symbol $ \\sqrt {x}.$ If $ y \\equal{} y_0 < 0$ and $ x \\equal{} y^2,$ then $ x$ is a positive real number. But then, by convention, $ \\sqrt {x} > 0,$ meaning that $ \\sqrt {x}\\ne y_0$ and we have not found a solution to our original problem. The original equation has $ x \\equal{} 4$ as its only real root, as long as we use the usual convention concerning $ \\sqrt {x}.$\r\n\r\nIt is true that $ y^4 \\plus{} y \\minus{} 18$ has four complex roots: one positive, one negative, and two as a conjugate pair. The squares of these four numbers are in some sense roots of the original equation $ x^2 \\plus{} \\sqrt {x} \\equal{} 18$ but only if you adopt a \"multi-valued\" interpretation of what you mean by the symbol $ \\sqrt {x}.$", "Solution_3": "So is correct: 4 is the only root. \r\n\r\nSorry for the new topic, it was a mistake." } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra open" ], "Problem": "let $k$ be an algebraically closed field. a morphism $\\varphi$ of $n$-dimensional affine space $k^n$ into itself is given by $n$ polynomials $f_1, ..., f_n$ in $n$ variables: $\\varphi(x_1, ..., x_n) = (f_1(x_1,...,x_n), ..., f_n(x_1, ..., x_n))$. it is called an automorphism if there exists an inverse morphism $\\psi$. let $J=\\det|\\frac{\\partial f_i}{\\partial x_j}|$ be the jacobian.\r\na) if $\\varphi$ is an automorphism, show that $J$ is a nonzero constant polynomial.\r\nb) show the converse of a).\r\n\r\na) is easy, but b) is still unsolved.\r\n\r\nPeter", "Solution_1": "b) is also open in the case n=2 !\r\nThis problem is called the Jacobian conjecture, many false proofs exists in the litterature.", "Solution_2": "A few new results were published after 2004 and I thought of updating them(Also it is the 80th anniversary of the conjecture!)\n\nDixmier implies Jacobian Conjecture Here\nhttp://journals.math.ac.vn/acta/pdf/0702205.pdf and here https://arxiv.org/abs/math/0512171\nMoh's paper https://eudml.org/doc/152524 was counter-argued by his students here https://arxiv.org/abs/1604.07683v2 and here \nhttps://arxiv.org/abs/1708.09367. Also Essen proposed a generalized version of JC as(*)\nA paper on this is https://www.sciencedirect.com/science/article/pii/S0022404903002561\nFor 2-D Jc $(x,y)\\mapsto(f(x,y),g(x,y))$ is thought to be a counter example(I don't know about this).Also I think that the conjecture was proved to be true in the symmetric an cubic linear case which used nilpotency etc.I couldn't find a reference for this,so can anyone provide?\n(*)[Quote=Essen]...it suffices to prove JC for all $n \\geq 2$ and for all $F's$ of the form: $F=(l_1,\u2026,l_r,x_{r+1}+M_{r+1},\u2026,x_n+M_n)$ where each $l_i$ is linear and each $M_j$ is a monomial.[/quote]" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "In the ring $\\mathbb{Z}_{n}$ of residues modulo $n$ calculate $\\det A_{n},$ $\\det B_{n}$ where $A_{n}=(\\overline{i}+\\overline{j})_{\\overline{i},\\overline{j}=0,1,\\ldots,n-1},$ $B_{n}=(\\overline{i}\\cdot\\overline{j})_{\\overline{i},\\overline{j}=0,1,\\ldots,n-1},$ $n\\ge2.$ \r\n(V. Mazorchuk)", "Solution_1": "I hope I didn't misunderstand the notation.\r\nFor $A_{n}$, its rank is 2 obviously, thus the det = 0 for $n \\ge 3$, for n = 2 the det is -1 since the matrix is $\\begin{array}{cc}0 & 1 \\\\ 1 & 0 \\\\ \\end{array}$\r\nFor $B_{n}$, the first row is 0, so det is 0." } { "Tag": [ "limit", "real analysis", "real analysis solved" ], "Problem": "Compute \r\n\\lim_{n->+oo}(\\lim_{x->0}(1+(tgx)^2+(tg2x)^2+...+(tg(nx)^2)^(1/(n^3.x^3))", "Solution_1": "Lets rewrite the limit like so:\r\n\r\nb_n(x)=sum(k=1,n) tan^2(kx) and a_n=lim(x-->0) (1+b_n)^1/(n^3x^2).\r\nwe have that lim(x-->0) b_n=0 and lim(x-->0) 1/(n^3x^2)=00 so :\r\n\r\na_n=lim(x-->0) [(1+b_n(x))^(1/(b_n(x))]^b_n(x)/(n^3x^2)=\r\n=lim(x-->0) e^[(1/n^3) sum(k=1,n)tan^2(kx)/k^2x^2 *k^2]=e^[1/n^3*n(n+1)(2n+1)/6].\r\nlim(n-->oo) a_n=e^1/3\r\n\r\ncheers!", "Solution_2": "Here my proof with Landau technic\r\ng_n(x)=1+tgx+...+tg(nx) i use tgu=u+u^3/3 + o(u^3) when u->0\r\ng_n(x)=1+\\sum_{k= 1 to n}(kx+k^3x^3/3 + o(x^3)) when x->0\r\n=1+\\sum_{k=1 to n}kx + o(x)\r\nnow g_n(x)^(1/(n^3.x)=exp(ln(g_n(x))/(n^3x)=\r\nexp((n(n+1)(2n+1)/n^3 o(1)) -->exp(1/3) when first x->0, and second \r\nwhen n->+oo", "Solution_3": "[quote=\"Moubinool\"]...\n g_n(x)^(1/(n^3.x)=\nexp((n(n+1)(2n+1)/n^3+o(1)) -->exp(1/3) when first x->0, and second when n->+oo[/quote]\r\ng_n(x)^(1/(n^3.x)=\r\nexp((n(n+1)(2n+1)/(6n^3)+o(1)) --->exp(2/6)=e^1/3" } { "Tag": [ "Ross Mathematics Program", "induction", "combinatorics proposed", "combinatorics" ], "Problem": "On a circular road there are some gas stations. Together, they have exactly as much gas as a car would need in order to go around the circle once. Prove that there is a gas station where you can start with an empty tank s.t. you can go around the circle.\r\n\r\nI found it [url=http://www.cut-the-knot.org/proofs/GasStations.shtml]here[/url].", "Solution_1": "BTW it's Beijing 64.\r\nAssume that car begins travel with the quantity of gas enough to go around the circle without supply. Take the station (or one of the stations) on which the quantity of gas is minimal. If we start from this station the quantity of gas always will be not less than the initial one (0).", "Solution_2": "Thanks! On that website the exact same solution is given. I think I have another one, but it might turn out to be the same :).\n\nA long time ago Moubinool posted a problem. I can't remember what it sounded like, but it was similar to this: given some numbers around the circle s.t. their sum is $0$, prove that we can choose one s.t. all the partial sums (in a clockwise direction) are $\\ge 0$. We can apply this here, considering the gas stations as positive numbers (because the car can get more fuel) and the gaps between them as negative numbers. My solution to this problem was, however, a bit longer.", "Solution_3": "Yes Grobber, it was given in our 'Concours Gnral' in 1997 (I think), and it was, as you, the first thing which came to my mind while reading the statement of the problem.\r\n\r\nPierre.", "Solution_4": "I've tried to locate that problem recently, but failed :?. Do you happen to know where it was posted?", "Solution_5": "This problem (the original one Grobber posted) is attributed to the Hungarian mathematician Laslo Lovas (spelling?) in David A. Klarner, [i]The Mathematical Gardner[/i], essay 5.4 by Ross Honsberger, with the same solution as Sasha gave.\r\n\r\n Darij", "Solution_6": "I've found that :\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=537&highlight=concours+general\n\n[color=red][[b]Moderator edit:[/b] This problem is [i]not exactly[/i] what grobber uses, because it requires the sum of the numbers on the circle to be $1$ and the partial sums to be positive, whereas the version grobber uses requires the sum of the numbers on the circle to be $0$ and the partial sums to be nonnegative. However, its solution is more or less the same as the solution of the version used by grobber.][/color]\n\nPierre.", "Solution_7": "For 1 or 2 gas stations, it's trivial.\nLet's assume that it's true for some n gas stations. Let's look at n+1 stations: there must be at least one station that has enough gas to arrive at the next station. If we look at this station and the station after it as [b]one[/b] station, then we have n stations, and by the assumption, we're done.\nSo by induction, it's true for every n." } { "Tag": [ "conics", "geometry unsolved", "geometry" ], "Problem": "Let $ ABC$ be an equilateral triangle and $ P$ in its interior. The distances from $ P$ to the triangle's sides are denoted by $ a^2, b^2,c^2$ respectively, where $ a,b,c>0$. Find the locus of the points $ P$ for which $ a,b,c$ can be the sides of a non-degenerate triangle.", "Solution_1": "Using areal (or trilinear - both the same here) co-ordinates we are looking for the a region on which $ (\\sqrt{x}\\plus{}\\sqrt{y}\\minus{}\\sqrt{z})(\\sqrt{x}\\minus{}\\sqrt{y}\\minus{}\\sqrt{z})(\\minus{}\\sqrt{x}\\plus{}\\sqrt{y}\\plus{}\\sqrt{z}) > 0$ and on which $ \\sqrt{x}\\plus{}\\sqrt{y}\\plus{}\\sqrt{z}$ is strictly positive.\r\n\r\nMultiplying it all together and factorising we get $ 2(xy\\plus{}yz\\plus{}zx)\\minus{}(x^2\\plus{}y^2\\plus{}z^2)>0$.\r\n\r\nIt's easy to show that this is the interior of a conic which is a circle and passes through the three midpoints of the sides, and that it is therefore the interior of the incircle.\r\n\r\nSo this is our locus.", "Solution_2": "For those who are interested, here are the official solutions of the problems.", "Solution_3": "Not that bad with trig bash :)\n" } { "Tag": [ "inequalities", "Euler", "inequalities unsolved" ], "Problem": "With positive real numbers $ x, y, z$, prove that: $ \\sum\\limits_{cyc} {\\left[ {\\sqrt[3]{{4(x^3 \\plus{} y^3 )}} \\plus{} \\frac{{2x}}{{y^2 }}} \\right]} \\ge 12$", "Solution_1": "[quote=\"L_Euler\"]With positive real numbers $ x, y, z$, prove that: $ \\sum\\limits_{cyc} {\\left[ {\\sqrt [3]{{4(x^3 \\plus{} y^3 )}} \\plus{} \\frac {{2x}}{{y^2 }}} \\right]} \\ge 12$[/quote]\r\nBy AM-GM we have $ \\sum {\\left( {\\sqrt[3]{{4\\left( {x^3 \\plus{} y^3 } \\right)}} \\plus{} \\frac{{2x}}{{y^2 }}} \\right)} \\ge \\sum {\\left( {x \\plus{} y} \\right) \\plus{} \\frac{{2x}}{{y^2 }}} \\ge 12$." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Evaluate $\\int_{0}^{\\pi}\\frac{\\cos 4x-\\cos 4\\alpha}{\\cos x-\\cos \\alpha}\\ dx.$", "Solution_1": "$=4\\int_{0}^{\\pi}(\\cos2x+\\cos2\\alpha)(cosx+\\cos\\alpha)dx$\r\n$=4\\int_{0}^{\\pi}{\\cos2xcosx+cosxcos2\\alpha+\\cos2x\\cos\\alpha+\\cos\\alpha{\\cos2\\alpha}}dx$\r\n$=2\\int_{0}^{\\pi}(cos3x+cosx)dx+4\\pi{cos\\alpha}cos2\\alpha$\r\n$=4\\pi{cos\\alpha}cos2\\alpha$\r\nright?", "Solution_2": "Your answer is correct, bylsals. :)" } { "Tag": [ "limit", "trigonometry", "algebra", "polynomial", "calculus", "calculus computations" ], "Problem": "1. Lim (sinx - cosxsinx) / x^2\r\n x->0\r\n\r\n\r\n2. Lim (tanx) / (x+(pie/4))\r\n x->(pie/4)\r\n\r\n\r\n\r\n3. Lim ((2x^3) + 1) / ((5x^3) + (2x^2) + 1)\r\n x->inf.\r\n\r\n\r\n\r\nThank you all in advance!", "Solution_1": "I will omit $x\\to 0$ in limits.\r\n$\\lim \\frac{\\sin x - \\cos x \\sin x}{x^2} = \\lim \\frac{\\sin x (1-\\cos x)}{x^2} = \\lim \\frac{\\sin x}{x} \\frac{2\\sin^2 \\frac{x}{2} }{x} = \\lim \\frac{\\sin x}{x} \\frac{\\sin \\frac{x}{2}}{\\frac{x}{2}} \\sin \\frac{x}{2} = 1\\cdot 1 \\cdot 0 = 0$", "Solution_2": "\\[\\lim_{x \\rightarrow \\frac{\\pi}{4}} \\frac{\\tan{x}}{x + \\frac{\\pi}{4}} = \\frac{\\tan{\\frac{\\pi}{4}}}{\\frac{\\pi}{4} + \\frac{\\pi}{4}} = \\frac{2}{\\pi}.\\]\r\n\r\nSo, you just plug in $\\frac{\\pi}{4}$...", "Solution_3": "The last one: \\[\\lim_{x \\rightarrow \\infty} \\frac{2x^3 + 1}{5x^3 + 2x^2 + 1} = \\frac{2}{5}.\\]", "Solution_4": "[quote=\"Arne\"]The last one: \\[\\lim_{x \\rightarrow \\infty} \\frac{2x^3 + 1}{5x^3 + 2x^2 + 1} = \\frac{2}{5}.\\][/quote]\r\n\r\nHow did you cancle out the x's?", "Solution_5": "The limit of the quotient of two polynomials (of equal degrees) is just the quotient of the coefficients of their highest degree terms :huh:", "Solution_6": "[quote=\"Riastrad\"][quote=\"Arne\"]The last one: \\[\\lim_{x \\rightarrow \\infty} \\frac{2x^3 + 1}{5x^3 + 2x^2 + 1} = \\frac{2}{5}.\\][/quote]\n\nHow did you cancle out the x's?[/quote]\r\n\r\nbasically divide numerator and denominator by $x^3$" } { "Tag": [ "geometry", "rectangle" ], "Problem": "Point $ P$ is inside a rectangular box. The distances from point $ P$ to four of the vertices of the box are 1,2,3, and 4. If the distance from $ P$ to another vertex is greater than 5, find that distance.", "Solution_1": "No one attempted this problem? I admit that it is a rather difficult problem compare to first five.\r\n\r\n[hide=\"Solution\"]\nThis problem can be solved with [url=http://www.artofproblemsolving.com/Wiki/index.php/British_Flag_Theorem]British Flag Theorem[/url]. Basically, this theorem states that with plane $ ABCD$ and point $ P$ that is on the plane or just on the space, the equation $ PA^2\\plus{}PC^2 \\equal{} PB^2\\plus{}PD^2$ holds true.\n\nKnowing that, we know what to do... Casework.\n\nConstruct the rectangle such that one face has points $ ABCD$. With respect to the center of the rectangle, construct $ A',B',C',D'$ so that they are opposite of $ A,B,C,D.$ \n\nWLOG, let $ PA \\equal{} 1$. By considering rectangle $ ABCD$, we can say that either $ PB \\equal{} 2$ or $ PC \\equal{} 2$. Also, the key point is to note that $ PA'$ has to be maximized because $ PA \\plus{} PA'$ is the diagonal of the rectangle, and thus, more than 5 (so that the length from $ P$ to one vertice can reach beyond 5).\n\nAssume $ PC \\equal{} 2.$ Then $ 3 \\neq PA, PA', PB, PD, PC'$ (use the above reasoning to see why these points do not work) so $ PD' \\equal{} 3$ or $ PB' \\equal{} 3$. Assume then $ PD' \\equal{} 3$ then $ 4 \\neq PA, PA', PB, PD, PC$ so $ PC' \\equal{} 4$ or $ PB' \\equal{} 4$. In the first case, the diagonal becomes $ \\sqrt{2^2 \\plus{} 4^2}$ and that's less than 5. In latter case, the diagonal becomes 5 so this is still bit short. Thus, $ PB \\equal{} 2$.\n\nWith $ PB \\equal{} 2,$ then $ 3 \\neq PA, PA', PB'$. So:\n\n$ PC \\equal{} 3, PC' \\equal{} 3, PD \\equal{} 3,$ or $ PD' \\equal{} 3$\n\nWe can consider casework by first two and the last two instead of considering all four.\n\nIf $ PC \\equal{} 3,$ then $ PD \\equal{} \\sqrt{6}$ from the rectangle $ ABCD$. Also, $ 4 \\neq PA, PA', PB', PD', PC'$ so nope, this is not the right case.\n\nIf $ PD \\equal{} 3,$ then $ PC \\equal{} \\sqrt{12}$. Then, $ 4 \\neq PA, PA', PB', PD'$ which leaves $ PC' \\equal{} 4$. So, the diagonal (since we know that $ PC$ and $ PC'$), becomes $ \\sqrt{12 \\plus{} 4^2} \\equal{} \\sqrt{28}$.\n\nNow that we know the diagonal, we can find the other values. $ PD \\equal{} \\sqrt{19}, PB' \\equal{} \\sqrt{24},$ and most importantly, $ PA' \\equal{} \\sqrt{27}$, which is the answer since it's over 5. [/hide]" } { "Tag": [], "Problem": "When simplified and expressed with negative exponents, the expression $(x+y)^{-1}(x^{-1}+y^{-1})$ is equal to:\r\n(A) $x^{-2}+2x^{-1}y^{-1}+y^{-2}$\r\n(B) $x^{-2}+2^{-1}x^{-1}y^{-1}+y^{-2}$\r\n(C) $x^{-1}y^{-1}$\r\n(D) $x^{-2}+y^{-2}$\r\n(E) $\\frac{1}{x^{-1}y^{-1}}$", "Solution_1": "[hide]the answer is C. the equation is equal to [1/(x+y)]*[(x+y)/xy] which simplifies to 1/xy[/hide]", "Solution_2": "Can you post where these problems come from?\r\n\r\nI'm just asking this for curiosity. Also, even though it is fine to make separate posts, if you are posting as challenges, I would be more grateful if you post all problems in one thread. \r\n\r\nThanks!", "Solution_3": "[hide]$\\frac{1}{x+y}\\cdot(\\frac{1}{x}+\\frac{1}{y})\\\\ \\\\\n\\frac{1}{x+y}\\cdot\\frac{x+y}{xy} \\\\ \\\\\n\\frac{1}{xy}=\\frac{1}{x\\cdot y} \\\\ \\\\\nx^{-1}\\cdot y^{-1}$\n\nwhich is c[/hide]" } { "Tag": [ "number theory", "prime numbers", "inequalities unsolved", "inequalities" ], "Problem": "Let $ k$ be a positive integer, we write k as a product of prime numbers (not necessarily distinct, par example 18=2.3.3). Let $ T(k)$ be the sum of all factors in the factorization above. Find the largest constant $ C$ such that $ T(k) \\geq C.lnk$ for all positive integer k.", "Solution_1": "I haven't any solution for this nice problem! But i think that it's a very difficult problem and it's not easy to try to solve it! I'm trying and i'll post my proof while i have it!", "Solution_2": "Before I post my solution, I'd like to know the source of the problem (It's strange that it is still an open problem, because the solution I have is pretty straightforward).\r\n\r\nIs it (maybe) a problem from a (running) contest? Hope not.", "Solution_3": "Hi [b]hsiljak[/b]:\r\nI'll say of the source of this problem. It's a problem which my friend gave in 2006. But he haven't any solution for it!\r\nPrevious month, i meet him and we said of it during a long times. So i decide post it in our forum! :D", "Solution_4": "It is an easy problem . Try to solve it if you want to send it to math magazine in Viet Nam . You must know that shouldn't post it here . If you want to find solution [b]try to solve it or wait for 3 months[/b].", "Solution_5": "Please solve it to help me!" } { "Tag": [ "geometry", "inequalities", "rectangle" ], "Problem": "Three circles of radii x, 2x, and 3x are inscribed inside a circle of radius 12 and tangent to each of the other two inscribed circles. What is the value of x?", "Solution_1": "Connecting the radii, we have a triangle with sides $ 3x$,$ 4x$, and $ 5x$, with area $ 6x^2$ (since it is a right triangle). The lines through the points of tangency to the outer circle and the centers of the circle meet at the center. From here we get that the center of the big circle is distances of $ 12 \\minus{} x$, $ 12 \\minus{} 2x$, and $ 12 \\minus{} 3x$ from the vertices. If we assume the center is inside the triangle, we get no solution. Thus, we assume the center is outside. By Heron's,\r\n$ \\sqrt {12(x)(2x)(12 \\minus{} 3x)} \\plus{} \\sqrt {12(x)(3x)(12 \\minus{} 4x)} \\minus{} \\sqrt {12(2x)(3x)(12 \\minus{} 5x)} \\equal{} 6x^2$\r\nSolving, $ x \\equal{} 2$. 0 is actually the minimum of $ \\sqrt {24(12 \\minus{} 3x)} \\plus{} \\sqrt {36(12 \\minus{} 4x)} \\minus{} \\sqrt {72(12 \\minus{} 5x)} \\minus{} 6x$, so I think there is probably a nicer solution either using inequalities or the fact it is a right triangle.", "Solution_2": "The answer can also be found by [url=http://mathworld.wolfram.com/DescartesCircleTheorem.html]Descarte's Circle Theorem[/url]: $ \\left(\\frac{1}{x}\\plus{}\\frac{1}{2x}\\plus{}\\frac{1}{3x}\\minus{}\\frac{1}{12} \\right)^2 \\equal{} 2\\left(\\frac{1}{x^2}\\plus{}\\frac{1}{4x^2}\\plus{}\\frac{1}{9x^2}\\plus{}\\frac{1}{144} \\right)$ $ \\Longrightarrow \\left(\\frac{\\minus{}x \\plus{} 22}{12x}\\right)^2 \\equal{} \\frac{x^2 \\plus{} 196}{72x^2}$ $ \\Longrightarrow x^2 \\plus{} 44x \\minus{} 92 \\equal{} (x\\minus{}2)(x\\plus{}46) \\equal{} 0$, and $ x \\equal{} 2$.", "Solution_3": "Or you can connect the center of the circle to the centers of the circles with radius $ 2$ and $ 3$. You then get a rectangle, with opposite sides having lengths $ 3x$ and $ 12\\minus{}3x$, so $ \\boxed{x\\equal{}2}$." } { "Tag": [ "Vieta", "algebra", "polynomial" ], "Problem": "The roots of the equation $ x^{3}\\minus{}19x\\equal{}\\minus{}30$ are a, b, and c. Find the numerical value of the product:\r\n(a+1)(b+1)(c+1).\r\n\r\nThis is how I tackled the problem:\r\n\r\n[hide]First, I set my goal as factoring out an (x-r) where x is a root from $ x^3\\minus{}19x\\plus{}30$. Knowing r must be equal to (some factor of 30)/(some factor of 1.. haha), I ran through the options in my head, using synthetic division and got a 2. I factored the whole thing to get (x-2)(x+5)(x-3), thus the factors were 2, 3, and -5. Plugging them in for a, b, and c, I got -48. [/hide]\r\n\r\nDoes anyone have a helpful way to solve cubic equations other than the technique I used? Thanks.", "Solution_1": "Well in tougher cases, you wont be able to find the roots. That's where Vieta's Formulas and Newton's Sums come in. Vieta's are the relationships between the coeffiecients of a polynomial and its roots. Newton's is..well something you won't need in this problem but I suggest you learn for future problems. \r\n\r\nP.S. I don't want to type all of Vieta's formulas so please look them up before looking at my solution. Otherwise, you won't understand.\r\n\r\n[hide=\"Look up Vietas!\"]First expand $ (a \\plus{} 1)(b \\plus{} 1)(c \\plus{} 1) \\equal{} (a \\plus{} b \\plus{} c) \\plus{} (ab \\plus{} ac \\plus{} bc) \\plus{} abc \\plus{} 1$. By Vieta's formulas,\n\n$ a \\plus{} b \\plus{} c \\equal{} 0$\n\n$ ab \\plus{} ac \\plus{} bc \\equal{} \\minus{} 19$\n\n$ abc \\equal{} \\minus{} 30$\n\nPlugging in, we have $ (a \\plus{} 1)(b \\plus{} 1)(c \\plus{} 1) \\equal{} 0 \\plus{} \\minus{} 19 \\plus{} \\minus{} 30 \\plus{} 1 \\equal{} \\boxed{ \\minus{} 48}$.[/hide]", "Solution_2": "I think modularmark understated how important knowing these relationships are. Viete's formulas crop up EVERYWHERE and can be used in a variety of situations - even in some unsuspecting ones. \r\n\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_Formulas\r\n\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/Newton_sums\r\n\r\nhttp://mathworld.wolfram.com/VietasFormulas.html\r\n\r\nThat should be enough for a basic understanding.", "Solution_3": "There is also a nice solution involving the Factor Theorem.\r\n\r\n[hide=\"My Solution\"]\nWe have $ (a \\plus{} 1)(b \\plus{} 1)(c \\plus{} 1) \\equal{} \\minus{}(\\minus{}1 \\minus{} a)(\\minus{}1 \\minus{} b)(\\minus{}1 \\minus{} c)$. Notice that $ x^3 \\minus{} 19x \\plus{} 30 \\equal{} (x \\minus{} a)(x \\minus{} b)(x \\minus{} c)$ for all $ x$. Substituting $ x \\equal{} \\minus{}1$, we obtain $ (\\minus{}1 \\minus{} a)(\\minus{}1 \\minus{} b)(\\minus{}1 \\minus{} c) \\equal{} (\\minus{}1)^3 \\minus{} 19(\\minus{}1) \\plus{} 30 \\equal{} \\minus{}1 \\plus{} 19 \\plus{} 30 \\equal{} 48$. Now we take the negative of that to get the desired result, $ \\boxed{\\minus{}48}$.\n[/hide]" } { "Tag": [], "Problem": "How many different arrangements of the six letters of the word \"YELLOW\" can be made if the first letter must be W and the last letter must be L?", "Solution_1": "There are $ 360$ total ways to arrange. After that, there is a $ \\frac{1}{6}$ chance the first letter is W and $ \\frac{2}{5}$ chance that the last letter is L. $ 360 \\times \\frac{1}{6} \\times \\frac{2}{5}\\equal{}\\boxed{24}$.", "Solution_2": "Or, after picking a W and an L, we still have to arrange YELO in the middle. There are $ 4!\\equal{}\\boxed{24}$ ways to do so" } { "Tag": [ "geometry", "circumcircle", "trapezoid", "symmetry", "geometric transformation", "reflection", "power of a point" ], "Problem": "In triangle $ ABC$ with $ m(\\angle{C})\\equal{}2\\cdot m(\\angle{A}),$ $ (CD$ is the internal bisector of $ \\angle{C}, \\ (D\\in [AB]).$ Let $ S$ be the center of the circle located on the same side \r\nof line $ AC$ as $ B,$ which is tangent to $ AC$ and externally tangent to the circumcircles of $ \\triangle{ACD}$ and $ \\triangle{BCD},$ respectively. Prove that $ \\boxed{AB\\perp CS.}$", "Solution_1": "I've seen the problem in Crux Mathematicorum, but I don't remember exactly the reference. Can someone help?\r\n[hide=\"Main idea of the solution I have in mind\"]Consider an inversion with respect to the vertex $ C$. It becomes more or less straightforward after that.[/hide]", "Solution_2": "What does $ m( \\angle C)$ mean?", "Solution_3": "[quote=\"FelixD\"]What does $ m( \\angle C)$ mean?[/quote]\r\n\r\nIt means the measure of angle $ C.$", "Solution_4": "Let $(S)$ be the circle tangent to $AC,$ $ (K_1) \\equiv{} \\odot(DAC)$ and $ (K_2) \\equiv{} \\odot(DBC)$ externally. Since $ \\angle DAC \\equal{} \\angle BCD,$ then ray $ CB$ is tangent to $ K_1$ at $ C.$ Inversion with center $ C$ and power equal to the power of $ C$ WRT $ (S)$ takes $ (K_1)$ into the line $ k_1$ passing through the inverse $ A'$ and tangent to $ (S).$ Similarly, $ (K_2)$ is taken into the line $ k_2$ passing through $ B'$ and tangent to $(S).$ Further, $ k_1 \\equiv{} \\overline{A'D'}$ is parallel to $ CB',$ since $ CB$ is tangent to $ (K_1).$ $ k_1,$ $ k_2$ meet at the inverse $ D'$ of $ D$ and since $ A,D,B$ are collinear, then $AB$ is taken into the circumcircle $\\omega$ of the isosceles trapezoid $ A'D'B'C.$ By conformity, angle between $ AB$ and $ AC$ equals the angle between $\\omega$ and $ A'C.$ Thus, $ \\angle D'CA' \\equal{} \\angle A'D'C$ $\\Longrightarrow$ $ \\triangle A'D'C$ and $ \\triangle D'B'A'$ are isosceles with apices $ A'$ and $D'.$ Thus, $S$ is the the center of $\\omega$ $\\Longrightarrow$ $AB \\perp CS.$", "Solution_5": "Very nice problem.\nConsider the composition $P$ of symmetry wrt to line $CD$ and inversion with center $C$ and square radius $CA*CB$ now let $CD$ cut circle $ABC$ again at $E$. Then $P(D)=E$ $P(A)=B$ so circles $ADC$ and $BDC$ go to $BE$ and $AE$. and line $CA$ goes to $CB$ so the circle which is externally tangent to circles $ADC$ and $BDC$ and line $AC$(circle $k$) goes to circle $m$ which is tangent to segments $BE,EA,BC$. However $\\angle BCE=\\angle ECA=\\angle BAC$ so $BC=BE=EA$ and the circumcenter of $ABC$ point $O$ is equidstant from $BE,EA,BC$. So the center of $m$ is $O$. So now $P(S)$ is a point on $CO$ and therefore $\\angle SCA=\\angle BCO=90-\\angle CAB$ so $CS\\perp AB$.", "Solution_6": "[size=150]Remark:[/size] [b]The circle $S$ touches $AC$ at the symmetrical of $C$ about $A$![/b]\n\nBest regards,\nsunken rock", "Solution_7": "Here is another approach that also proves sunken rock's remark.\n\nLet $P,Q,R$ be the tangency points of $(S)$ with $AC,\\odot(DAC),\\odot(BDC),$ respectively. $D$ is clearly midpoint of the arc $AQC$ and if $\\odot(BCD)$ cuts $AC$ again at $E,$ then $B$ is midpoint of the arc $CRE.$ Inversion WRT $(D,DC)$ swaps $AC$ and $\\odot(DAC)$ leaving $(S)$ fixed and inversion WRT $(B,BC)$ swaps $AC$ and $\\odot(BCE)$ leaving $(S)$ fixed $\\Longrightarrow$ $P \\in DQ,$ $P \\in BR$ and $DC^2=DP \\cdot DQ,$ $BC^2=BP \\cdot BR$ $\\Longrightarrow$ $B,D$ have equal powers WRT $C$ and $(S)$ $\\Longrightarrow$ $BDA$ is radical axis of $C$ and $(S)$ $\\Longrightarrow$ $CS \\perp AB$ and $AB$ bisects the tangent from $C$ to $(S),$ i.e. $P$ is reflection of $C$ on $A.$", "Solution_8": "[quote=pohoatza][color=#f00]I've seen the problem in Crux Mathematicorum[/color], but I don't remember exactly the reference. Can someone help?\n[hide=\"Main idea of the solution I have in mind\"]Consider an inversion with respect to the vertex $ C$. It becomes more or less straightforward after that.[/hide][/quote]\n\nv21 n01. Problem 2010.\n\n" } { "Tag": [ "ratio" ], "Problem": "The owner's manual for a string trimmer says that a 16 : 1 ratio of gasoline to 2-cycle oil MUST BE used for its fuel. How many gallons of gasoline would be needed to go with an 8 ounce packet of oil, if 1 gallon is 128 ounces?", "Solution_1": "[hide] Say the amount of gasoline needed is $x$ [b]ounces[/b]. Therefore, $\\frac{x}{8}=\\frac{16}{1}$. Mulitplying 8 on both sides, we get that that $x=128$ ounces. Therefore, we have that x is $\\frac{128}{128}=1$ gallon. Therefore, the answer is $1$. [/hide]", "Solution_2": "Good job but I also like to see how the question was actually solved. \r\n\r\nThank you The QuattoMaster 6000!" } { "Tag": [ "search", "superior algebra", "superior algebra unsolved" ], "Problem": "sir \r\n\r\n I want the solution of the problem.. If G is a finite group.Show that the number of elements x of G with x^2 not equal to 1 is even.Conclude that if G is a finite group of even order then G has an element of order 2.....", "Solution_1": "Please post a new problem into a new thread every time.", "Solution_2": "Take a look at the following thread:\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?search_id=875870790&t=225332[/url]\r\n\r\nI'm sure that it is going to prove useful to you.\r\n\r\nBest of luck... :wink:", "Solution_3": "[quote=\"coquitao\"]Take a look at the following thread:\n\n[url]http://www.mathlinks.ro/viewtopic.php?search_id=875870790&t=225332[/url]\n\nI'm sure that it is going to prove useful to you.\n\nBest of luck... :wink:[/quote]\r\n\r\nGiven that G is finite\r\n Let no.of elements of G be 'm'\r\n Let A= no.of elements that x^2 is not equal to 1\r\n B= no.of elements that x^2 is equal to 1\r\nTherefore G=A union B\r\nSince B is even, then G is even if A is even otherwise G is odd if A is odd[/hide]", "Solution_4": "Given that G is finite\r\nLet no.of elements of G be 'm'\r\nLet A= no.of elements that x^2 is not equal to 1\r\nB= no.of elements that x^2 is equal to 1\r\nTherefore G=A union B\r\nSince B is even, then G is even if A is even otherwise G is odd if A is odd", "Solution_5": "Instead of\r\n\r\nA = no.of elements that x^2 is not equal to 1\r\n\r\nwrite\r\n\r\nA = elements that x^2 is not equal to 1.\r\n\r\nSame for B.\r\n\r\nThen, G = A union B would actually hold (the equality wasn't correct as you'd written it before.)\r\n\r\nThe conclusion of the argument is now as you said:\r\n\r\n1- |A| is always even. No matter what the order of G is...\r\n2- Now, if the order of G is even, then |B| would have to be even because |G| = |A| + |B|.\r\n3- Since the identity of G belongs to B, we conclude that |B|>=2. The conclusion follows now from this inequality.\r\n\r\nHope it helps...", "Solution_6": "[quote=\"barathisid\"] Since B is even, then G is even if A is even otherwise G is odd if A is odd[/quote]\r\n\r\nContrast this line with my previous post. We can't tell whether B is even or not... The one that is always even is A, not B." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all $ f: \\mathbb{R}^{\\plus{}}\\rightarrow \\mathbb{R}$ such that \r\n(i) $ f(1)\\equal{}2008$\r\n(ii) $ |f(x)|\\leq x^2\\plus{}1004^2$ for all $ x>0$\r\n(iii) $ f(x\\plus{}y\\plus{}\\frac{1}{x}\\plus{}\\frac{1}{y})\\equal{}f(x\\plus{}\\frac{1}{y})\\plus{}f(y\\plus{}\\frac{1}{x})$", "Solution_1": "Let $ z_1\\equal{}x\\plus{}\\frac 1y,z_2\\equal{}y\\plus{}\\frac 1x$. If $ z_1z_2\\ge 2$ exist real solution for x,y. Therefore\r\n$ f(z_1\\plus{}z_2)\\equal{}f(z_1)\\plus{}f(z_2) \\ \\forall z_1z_2\\ge 2$.\r\nfrom (ii) we get $ f(x)\\equal{}ax$,from (i) a=2008.", "Solution_2": "[quote=\"Rust\"] If $ z_1z_2\\ge 2$ exist real solution for x,y. [/quote]\r\n\r\nCan you show?", "Solution_3": "[quote=\"outback\"][quote=\"Rust\"] If $ z_1z_2\\ge 2$ exist real solution for x,y. [/quote]\n\nCan you show?[/quote]\r\n$ z_2 \\equal{} y \\plus{} \\frac 1x \\equal{} y \\plus{} \\frac {1}{z_1 \\minus{} \\frac {1}{y}} \\equal{} \\frac {z_1y^2}{z_1y \\minus{} 1}\\to y \\equal{} \\frac {z_1z_2\\pm\\sqrt {z_1z_2(z_1z_2 \\minus{} 4)}}{2z_1}$.\r\nIt is true for $ z_1z_2\\ge 4$.", "Solution_4": "And so you can conclude $ f(x)\\equal{}ax$ for which $ x$? (Not for all positive real $ x$, I think... :wink: )", "Solution_5": "It is easy, but long to write.\r\n1)Let $ z_1>0,z_2>0$. We can chose $ z_3$, suth that $ z_1z_3\\ge 4,z_2z_3\\ge 4$. \r\nThen $ f(z_2\\plus{}z_3)\\equal{}f(z_2)\\plus{}f(z_3)$.\r\n$ f(z_1\\plus{}z_2)\\equal{}f(z_1\\plus{}z_2\\plus{}z_3)\\minus{}f(z_3)\\equal{}f(z_1)\\plus{}f(z_2\\plus{}z_3)\\minus{}f(z_3)\\equal{}f(z_1)\\plus{}f(z_2)\\plus{}f(z_3)\\minus{}f(z_3)\\equal{}f(z_1)\\plus{}f(z_2) \\ \\forall z_1,z_2>0$.\r\n2)We can define $ f(\\minus{}z)\\equal{}\\minus{}f(z),f(0)\\equal{}0$, then $ f(x\\plus{}y)\\equal{}f(x)\\plus{}f(y) \\ \\forall x,y\\in R$.\r\n3) If $ \\frac{f(x)}{x}\\not \\equal{}const$, then exist $ f(x_1)\\equal{}a_1x_1,f(x_2)\\equal{}a_2x_2,x_1>0,x_2>0$, Let $ m_n\\equal{}[\\frac{nx_1}{x_2}]$, then \r\n$ 0\\le nx_1\\minus{}m_nx_20)$.", "Solution_4": "By the way, the given condition is a little subtle...\r\n$ \\sqrt {\\frac {a}{b}} \\plus{} \\sqrt {\\frac {b}{c}} \\plus{} \\sqrt {\\frac {c}{a}} \\equal{} 3$.\r\nBecause from AM-GM we easily have $ \\sqrt {\\frac {a}{b}} \\plus{} \\sqrt {\\frac {b}{c}} \\plus{} \\sqrt {\\frac {c}{a}}\\geq 3$\r\nSo the condition actually is the equality case of AM-GM. So all of its variables $ \\frac ab; \\frac bc;\\frac ca$ are equal. (and, equal to $ 1$).\r\nHence $ \\frac {b^{6}}{a^{6}} \\plus{} \\frac {c^{6}}{b^{6}} \\plus{} \\frac {a^{6}}{c^{6}} \\color{red}{ \\equal{} } 3$.\r\nPlease correct me if I am wrong. :)", "Solution_5": "You are right, Potla. :oops:" } { "Tag": [], "Problem": "If in a triangle $ a \\equal{} 5, b \\equal{} 4$ and cos $ (A\\minus{}B) \\equal{}\\frac{31}{32}$, prove that cos $ C \\equal{} \\frac{1}{8}$ and that $ C \\equal{} 6$.", "Solution_1": "Hi! This is my first post! I hope I use these buttons right.\r\n\r\nSuggestion\r\n[hide] Draw out the figure and tackle the problem using geometry. You know angle A is bigger than angle B because length A is bigger than length B. Draw a line from vertex A to line BC with one angle being A-B and the other angle being B. Try to create two equations from this figure and solve for C. [/hide]", "Solution_2": "Thanks winternights. I got there after this inspired advice." } { "Tag": [ "inequalities open", "inequalities" ], "Problem": "Let $a;b;c$ be positive real numbers satisfying $a^{2}+b^{2}+c^{2}=3$.Find $k$ max such that:\r\n$ab^{2}+bc^{2}+ca^{2}\\geq k(ab+bc+ca)^{2}$", "Solution_1": "a*b^2+b*c^2+c*a^2>=k*(a*b+b*c+c*a)^2 (k>=0,a^2+b^2+c^2=3)\r\n\r\n<==>\r\n\r\nf=(a^2+b^2+c^2)/3*(a*b^2+b*c^2+c*a^2)^2-k^2*(a*b+b*c+c*a)^4>=0.\r\n\r\nLet a=1,b=1,c=1, we have k<=1/3; and if k=1/3, then \r\n\r\nf=2/9*cg[3,8,5]+2/3*cg[3,8,7]+4/9*cg[3,8,8]+14/9*cg[3,8,9]+1/3*cg[3,8,11]\r\n\r\n+2/3*cg[3,8,14]>=0,\r\n\r\nin which \r\n\r\ncg[3,8,5] = b^3*a^3*(c-b)*(c-a)+c^3*a^3*(b-c)*(b-a)+c^3*b^3*(a-c)*(a-b)>=0;\r\n\r\ncg[3,8,7] = c^3*b*a*(b+a)*(c-b)*(c-a)+b^3*c*a*(c+a)*(b-c)*(b-a)\r\n\r\n +a^3*c*b*(c+b)*(a-c)*(a-b)>=0;\r\n\r\ncg[3,8,8] = c*b^2*a^2*(b+a)*(c-b)*(c-a)+b*c^2*a^2*(c+a)*(b-c)*(b-a)\r\n\r\n+a*c^2*b^2*(c+b)*(a-c)*(a-b)>=0;\r\n\r\ncg[3,8,9] = c^2*b^2*a^2*(c-b)*(c-a)+b^2*c^2*a^2*(b-c)*(b-a)\r\n\r\n+a^2*c^2*b^2*(a-c)*(a-b)>=0;\r\n\r\ncg[3,8,11] = c^4*a^2*(c-b)^2+b^4*c^2*(b-a)^2+a^4*b^2*(a-c)^2>=0;\r\n\r\ncg[3,8,14] = c^2*b*a^3*(c-b)^2+b^2*a*c^3*(b-a)^2+a^2*c*b^3*(a-c)^2>=0.\r\n\r\nTherefore kmax=1/3." } { "Tag": [ "function", "algebra", "rational function", "algebra unsolved" ], "Problem": "Solve the following system:\r\n\r\n$ a \\plus{} b \\minus{} c \\minus{} d \\equal{} 1$\r\n$ a^2 \\plus{} b^2 \\minus{} c^2 \\minus{} d^2 \\equal{} 3$\r\n$ a_3 \\plus{} b^3 \\minus{} c^3 \\minus{} d^3 \\equal{} \\minus{}5$\r\n$ a^4 \\plus{} b^4 \\minus{} c^4 \\minus{} d^4 \\equal{} 15$\r\n$ a > 0$ (I don't know if this information is important)", "Solution_1": "With Maple.\r\nIn any commutative ring of characteristic $ 0$, the system is equivalent to this one:\r\n$ b^2\\plus{}b\\minus{}2\\equal{}0,d^2\\plus{}2d\\plus{}1\\equal{}0,a\\equal{}\\minus{}1\\minus{}b,c\\equal{}\\minus{}2\\minus{}d$.", "Solution_2": "More generally, we have a look on the system:\r\n$ (S): a\\plus{}b\\minus{}c\\minus{}d\\equal{}\\alpha,a^2\\plus{}b^2\\minus{}c^2\\minus{}d^2\\equal{}\\beta,a^3\\plus{}b^3\\minus{}c^3\\minus{}d^3\\equal{}\\gamma,a^4\\plus{}b^4\\minus{}c^4\\minus{}d^4\\equal{}\\delta$.\r\nLet $ a\\plus{}b\\equal{}u_1,ab\\equal{}u_2,c\\plus{}d\\equal{}v_1,cd\\equal{}v_2$. The system $ (S)$ can be rewritten as a system in $ u_1,u_2,v_1,v_2$ because $ a^2\\plus{}b^2\\equal{}u_1^2\\minus{}2u_2,a^3\\plus{}b^3\\equal{}u_1^3\\minus{}3u_1u_2,a^4\\plus{}b^4\\equal{}u_1^4\\minus{}4u_1^2u_2\\plus{}2u_2^2$.\r\nIf $ \\alpha^4\\minus{}4\\alpha\\gamma\\plus{}3\\beta^2\\not\\equal{}0$ and if you are a good calculator then you obtain an unique value of $ u_1,u_2,v_1,v_2$ as rational function of $ \\alpha,\\beta,\\gamma,\\delta$.", "Solution_3": "hello, we have $ a\\equal{}\\minus{}2,b\\equal{}1,c\\equal{}\\minus{}1,d\\equal{}\\minus{}1$ or $ a\\equal{}1,b\\equal{}\\minus{}2,c\\equal{}\\minus{}1,d\\equal{}\\minus{}1$ as the solutions.\r\nSonnhard." } { "Tag": [ "geometry", "rectangle" ], "Problem": "How many square inches are in the area of the largest circle which can be drawn on a piece of paper measuring 8 inches by 11 inches?", "Solution_1": "[quote=\"MCrawford\"]How many square inches are in the area of the largest circle which can be drawn on a piece of paper measuring 8 inches by 11 inches?[/quote]\r\n\r\n[hide=\"Hint\"]\nFirst draw the largest square you can in that rectangle, then draw the largest circle you can draw in the [b]square[/b]. That will be the largest circle you can draw in the rectangle.\n[/hide]", "Solution_2": "[hide]8 is the diameter so the area is 16pi[/hide]", "Solution_3": "[hide]$S=R^2\\pi=16\\pi$[/hide]", "Solution_4": "[hide]\nThe largest circle possible is a circle with a diameter of 8 inches.\n$4^2\\cdot\\pi=\\boxed{16\\pi}$ square inches.[/hide]" } { "Tag": [ "limit", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "Let be $a_0=1$ and $a_n^2=1+(a_0+a_1+\\cdots + a_{n-1})^2$ for $n>1$. Prove that\r\n$\\lim_{n\\to +\\infty} \\frac{2^n}{a_n}=\\frac{\\pi}{2}$.", "Solution_1": "$a_n={\\frac{1}{\\displaystyle\\cos \\left[\\frac{\\pi}{4}\\left(1-\\frac{1}{2^n}\\right)\\right]}}\\ (n\\geq 0),$ right?", "Solution_2": "[quote=\"kunny\"]$a_n={\\frac{1}{\\displaystyle\\cos \\left[\\frac{\\pi}{4}\\left(1-\\frac{1}{2^n}\\right)\\right]}}\\ (n\\geq 0),$ right?[/quote]\r\nYour solution isn't correct. Note that:\r\n$\\lim_{n\\to +\\infty}\\frac{1}{\\displaystyle\\cos \\left[\\frac{\\pi}{4}\\left(1-\\frac{1}{2^n}\\right)\\right]}=\\sqrt{2}$", "Solution_3": "Hum, regrettable!\r\n\r\nLet $a_0+a_1+\\cdots +a_{n-1}=\\tan \\theta _n\\ (n\\geq 1),$ we have $\\sin (\\theta _{n+1}-\\theta _n)=\\cos \\theta _{n+1}.$ I wonder this appoach is wrong.", "Solution_4": "This is right. Now find a sequence $\\left\\{\\theta_n\\right\\}\\subseteq (-\\frac{\\pi}{2},\\frac{\\pi}{2})$ such that $\\sin(\\theta_{n+1}-\\theta_n)=\\cos{\\theta_{n+1}}$.", "Solution_5": "$\\theta _{n+1}=\\frac{1}{2}\\theta _n+\\frac{\\pi}{4},$ right? :roll:", "Solution_6": "Right. For $\\theta_0=0$ you find $\\theta_n=\\frac{\\pi}{2}(1-\\frac{1}{2^n})$, so\r\n$a_n=\\frac{1}{\\displaystyle \\cos \\frac{\\pi}{2} (1-\\frac{1}{2^n})}$.\r\nNow compute $\\lim_{n\\to +\\infty} \\frac{2^n}{a_n}$.", "Solution_7": "Oops! I was wrong in calculation of finding the root of the characteristic equation, $\\alpha =\\frac{1}{2}\\alpha +\\frac{\\pi}{4} \\Longleftrightarrow \\alpha =\\frac{\\pi}{2}.$ anyway $\\lim_{n\\to\\infty} \\frac{2^n}{a_n}=\\lim_{n\\to\\infty} \\frac{\\displaystyle \\sin \\frac{\\pi}{2^{n+1}}}{\\displaystyle \\frac{\\pi}{2^{n+1}}}\\cdot \\frac{\\pi}{2}=\\frac{\\pi}{2}.$ :lol:", "Solution_8": "Here is the problem from Tokyo University entrance exam 1975. I have omitted hint.\r\n\r\nSolve the following reccurence.\r\n\r\n\\[ a_1=\\sqrt{2},\\ a_{n+1}=\\sqrt{2+a_n}\\ (n=1,2,\\cdots). \\]", "Solution_9": "[quote=\"kunny\"]Hum, regrettable!\n\nLet $a_0+a_1+\\cdots +a_{n-1}=\\tan \\theta _n\\ (n\\geq 1),$ we have $\\sin (\\theta _{n+1}-\\theta _n)=\\cos \\theta _{n+1}.$ I wonder this appoach is wrong.[/quote]\r\n\r\nkunny , I dont understand why can we make such substitution ? It seems so weird yet nice :)", "Solution_10": "We can imagine the formula $1+\\tan ^ 2 \\theta =\\frac{1}{\\cos ^ 2 \\theta}.$", "Solution_11": "[quote=\"kunny\"]We can imagine the formula $1+\\tan ^ 2 \\theta =\\frac{1}{\\cos ^ 2 \\theta}.$[/quote]\r\n\r\nok thanks Kunny,but how you apply that to your question ?\r\n\r\ni try to let $a_n=2\\tan^2\\theta_{n}$ but doesnt seem to work :(", "Solution_12": "For Tokyo University's problem, we should find another substitution.", "Solution_13": "[quote=\"kunny\"]Here is the problem from Tokyo University entrance exam 1975. I have omitted hint.\n\nSolve the following reccurence.\n\n\\[ a_1=\\sqrt{2},\\ a_{n+1}=\\sqrt{2+a_n}\\ (n=1,2,\\cdots). \\][/quote]\r\n\r\nPut $a_n=b_n+\\frac{1}{b_n}$, then $a_n=a_{n+1}^2-2$ implies $b_n+\\frac{1}{b_n}=b_{n+1}^2+\\frac{1}{b_{n+1}^2}$ i.e. $b_{n+1}=\\sqrt{b_n}$; if $b_1$ satisfied $\\sqrt{2}=b_1+\\frac{1}{b_1}$, then $b_n=b_1^{2^{-n}}$.", "Solution_14": "How about $b_{n+1}=\\frac{1}{\\sqrt{b_n}}?$", "Solution_15": "If $b_{n+1}^2=b_n$ then $b_n+\\frac{1}{b_n}=b_{n+1}^2+\\frac{1}{b_{n+1}^2}$, so i suppose $b_{n+1}=\\sqrt{b_n}$.\r\nWe can suppose $a_0=0$ and $b_0$ satisfied $0=b_0+\\frac{1}{b_0}$ that implies $b_0=i=e^{i \\pi/2}$, so\r\n$a_n=e^{i\\pi/2^{n+1}}-e^{-i\\pi/2^{n+1}}=2\\cos\\frac{\\pi}{2^{n+1}}$", "Solution_16": "That's O.K. ;)" } { "Tag": [ "quadratics", "Vieta", "algebra", "pen", "ratio", "number theory", "relatively prime" ], "Problem": "Show that if $x, y, z$ are positive integers, then $(xy+1)(yz+1)(zx+1)$ is a perfect square if and only if $xy+1$, $yz+1$, $zx+1$ are all perfect squares.", "Solution_1": "Method of infinite descent (fall)-\nassume $x\\leqq\\leq z$ are integers satisfying minimum sum such that any of $xy+1,yz+1,zx+1$ isn't perfect square. \nConsideration regarding number $t=x+y+z+2xyz\\pm \n2\\sqrt{(xy+1)(yz+1)(zx+1)}$ (integer solutions of quadratic equation $t^2+x^2+y^2+z^2-2(xy+yz+zt+tx+zx+ty)-4xyzt-4=0$):\nWe get the following identities\n $(x+y-z-t)^2=4(xy+1)(zt+1)$,\n$(x+z-y-t)^2=4(xz+1)(yt+1)$, \n$(x+t-y-z)^2=4(xt+1)(yz+1)$.\nFrom this system observe that not all $xt+1,yt+1,zt+1$ are the squares, while $(xt+1)(yt+1)(zt+1)$ is.\nNow any of products $(xy+1)(yt+1)(tx+1)$ is square.\nLet's prove that by this we got the least solution $(x,y,t)$:\nSince from the system is $zt+1\\geq 0$,$t\\geq -\\frac1z$, for $z>1$ must be $t\\geq 0$. \nIt is easy to check that both cases $t=0$ and $t=1$ are impossible. Therefore all numbers $(x,y,t)$ are positive integers.\nBut,according to assumption about minimality it must be $t\\leq \nz$, and from Vieta's formulae we get $tt_1 \\minus{} 1$ (since $ x \\equal{} y \\equal{} z \\equal{} 1$ isn't a solution). \r\nIf $ t \\equal{} 0$, then working out the formula for $ t$ gives $ (x \\plus{} y \\plus{} z)^2 \\equal{} 4(xy \\plus{} yz \\plus{} zx \\plus{} 1)$ or $ (x \\plus{} y \\minus{} z)^2 \\equal{} 4(xy \\plus{} 1)$, which means $ xy \\plus{} 1$ is a square, contradiction, so we have $ t > 0$ and by the minimality of $ x \\plus{} y \\plus{} z$ we must have $ t\\ge z$. \r\nBut for the two roots $ t,t'$ of our quadratic equation we have by de Viete that $ tt' \\equal{} x^2 \\plus{} y^2 \\plus{} z^2 \\minus{} 2xy \\minus{} 2yz \\minus{} 2zx \\minus{} 4 < z^2 \\minus{} x(2z \\minus{} x) \\minus{} y(2z \\minus{} y) < z^2$, contradiction.\r\n\r\nkeywords: vieta", "Solution_6": "thanks!! :)", "Solution_7": "We shall prove that xy + 1, yz + 1, zy + 1 are pairwise relatively prime. Suppose there is a prime p1 such that p1 divides both xy+1 and yz+1. then p1 also divides z(xy+1) - y(zx+1) = z-x. Analogously if there are primes p2 and p3 such that p2 divides yz + 1, zy + 1 and p3 divides xy + 1,zy + 1 then p2 divides z-y and p3 divides x-y. Then there are numbers f1, f2, f3, such that p1f1= z-x; p2f2= z-y; p3f3=x-y. This gives us that p2f2 - p1f1= p3f3 and p3f3 - p2f2 = p1f1. Then p1f1 + p2f2 = p2f2 - p1f1 ==> p1f1=-p1f1 <==> p1f1=0. Contradiction. So these numbers are pairwise relatively prime hence result.", "Solution_8": "You have only shown that one pair is relatively prime.\r\nYou assumed that every pair has a common prime factor, but what you really should assume is: one pair has a common prime factor.", "Solution_9": "I all of a sudden don't understand my own solution anymore...\r\n\r\nCan anyone tell me why $ z^2 \\minus{} x(2z \\minus{} x) \\minus{} y(2z \\minus{} y) < z^2$ in my last line?\r\n\r\nMaybe I'm just tired right now...", "Solution_10": "Zetax: in fact I proved it for a single pair of terms, but since the expressions xy + 1, yz + 1, xz + 1, are symmetric i could assume for a single pair without lost of generality. Also, the primes p1, p2, p3, i didnt say they were equal. \r\n\r\nmaybe i cant see your point: can you explain?", "Solution_11": "You assumed that every pair out of $ xy\\plus{}1,yz\\plus{}1,zx\\plus{}1$ has a common factor and got a contradiction. Thus you just proved that not _all_ pairs have a common factor, one or two pairs may still have.\r\nBy the way, you also made a sign-mistake, p3f3 - p2f2 = p1f1 should in fact be p3f3 - p2f2 = -p1f1. This destroys your contradiction.", "Solution_12": "[quote=\"Peter\"]\nCan anyone tell me why $ z^2 \\minus{} x(2z \\minus{} x) \\minus{} y(2z \\minus{} y) < z^2$ in my last line?\n\nMaybe I'm just tired right now...[/quote]\r\n\r\nNo, it's not because of your tiredness. The above solution is just not correct.. yet:\r\n\r\nYou shouldn't suppose wlog that $ xy \\plus{} 1$ is not a perfect square, but that $ z\\ge x,y$\r\nAnd you actually get $ (xt \\plus{} 1)(yt \\plus{} 1)(xy \\plus{} 1)$ being a square by multiplying $ 4(xz \\plus{} 1)(yt \\plus{} 1)$, $ 4(xt \\plus{} 1)(yz \\plus{} 1)$ and $ (xy \\plus{} 1)2$ and then dividing by $ 16(xy \\plus{} 1)(xz \\plus{} 1)(yz \\plus{} 1)$, which is known to be a perfect square.\r\n\r\nYou also know that $ yt \\plus{} 1$ is a perfect square iff $ xz \\plus{} 1$ is, and the same with $ xt \\plus{} 1$ and $ yz \\plus{} 1$ (and that's why you don't need xy+1 to be a perfect square).\r\nThe $ t \\equal{} 0$ case also implies $ xz \\plus{} 1$ and $ yz \\plus{} 1$ being a square, so you still have a contradiction there.\r\nAnd with all this, the answer to your question also becomes trivial.", "Solution_13": "everyone knows that A.1 is the hardest of PEN", "Solution_14": "A non-empty set of positive integers is defined \"good\" whenever the product of any two distinct elements is smaller of some squares by $1$: with this convention $\\{x,y,z\\}$ is good. Let's start proving that if $\\{x,y,z\\}$ is good, then $\\{x,y,z,w\\}$ is good too, whenever \n\\[w=x+y+z+2xyz\\pm 2\\sqrt{(xy+1)(yz+1)(zx+1)}\\]\nis a positive integer. The two values of $w$ are roots of the following equivalent quadratic equations:\n\\[(x+y-z-w)^2=4(xy+1)(zw+1),\\text{ }(1) \\\\ \n(x-y+z-w)^2=4(xz+1)(yw+1),\\text{ }(2) \\\\\n(x-y-z+w)^2=4(yz+1)(xw+1).\\text{ }(3)\\]\nAnd that's clear that $\\alpha w+1$ is a integer that can be expressed as ratio of two squares, i.e. it's a square too, for all $\\alpha \\in \\{x,y,z\\}$. Since the problem is symmetric in $x,y,z$ it can be assumed without loss of generality that $x\\le y\\le z$. Let's suppose for the sake of contradiction that not all sets $\\{x,y,z\\}$ are good, i.e. there exists a set such that at least one between $xy+1,yz+1,zx+1$ is not a perfect square. Choose now such a set $\\{x,y,z\\}$ such that $x+y+z$ is minimal, and define $w$ using the negative square root part. It's claimed that $0 \\minus{} 1$ (since $ x \\equal{} y \\equal{} z \\equal{} 1$ isn't a solution). \nIf $ t \\equal{} 0$, then working out the formula for $ t$ gives $ (x \\plus{} y \\plus{} z)^2 \\equal{} 4(xy \\plus{} yz \\plus{} zx \\plus{} 1)$ or $ (x \\plus{} y \\minus{} z)^2 \\equal{} 4(xy \\plus{} 1)$, which means $ xy \\plus{} 1$ is a square, contradiction, so we have $ t > 0$ and by the minimality of $ x \\plus{} y \\plus{} z$ we must have $ t\\ge z$. \nBut for the two roots $ t,t'$ of our quadratic equation we have by de Viete that $ tt' \\equal{} x^2 \\plus{} y^2 \\plus{} z^2 \\minus{} 2xy \\minus{} 2yz \\minus{} 2zx \\minus{} 4 < z^2 \\minus{} x(2z \\minus{} x) \\minus{} y(2z \\minus{} y) < z^2$, contradiction.\n\nkeywords: vieta[/quote]\n\nI did not understand why you checked $t=0$ case. You have already said $t$ is the smallest positive root of equation $ t^2 \\plus{} x^2 \\plus{} y^2 \\plus{} z^2 \\minus{} 2(xy \\plus{} yz \\plus{} zt \\plus{} tx \\plus{} zx \\plus{} ty) \\minus{} 4xyzt \\minus{} 4 \\equal{} 0$", "Solution_24": "Claim 1: If $\\prod_{cyc} (xy+1) = (xyz+\\frac{x+y+z}{2})^2$, I am done.\n\nIf I expand, I get $(\\sum_{cyc} xy) + 1 = \\frac 12 \\sum_{cyc} xy + \\frac 14 \\sum_{cyc} x^2$\n\n$\\sum_{cyc} x^2 - 2\\sum_{cyc} xy + 4=0$\n\nAs a quadratic in $x$: $x^2-(2y+2z)x+((y-z)^2-4)=0$. Its discriminant is $4((y+z)^2 - (y-z)^2 +4) =16(yz+1)$ is a perfect square, as needed.\n\nNow suppose $\\prod_{cyc} (xy+1) = (xyz+\\frac{x+y+z-l}{2})^2$ where $l\\in \\mathbb{Z}$. It's clear $l\\ge 0$. \n\n$\\prod_{cyc} (xy+1) - (xyz+\\frac{x+y+z}{2})^2 = -\\frac14 \\sum_{cyc} x^2 + \\frac 12 \\sum_{cyc} xy + 1$\n\n$-\\frac14 \\sum_{cyc} x^2 + \\frac 12 \\sum_{cyc} xy + 1 = -\\frac l2 (2xyz+x+y+z-\\frac l2)$\n\n$\\sum_{cyc} x^2 -2 \\sum_{cyc} xy -4 = l (4xyz+2x+2y+2z-l)$\n\n$x^2 - (2y+2z)x +(y-z)^2 - 4 - 4lxyz - 2lx - 2ly - 2lz + l^2=0$\n\n$x^2 - (2y+2z+4lyz+2l)x + (y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz)=0$\n\nNote this expression is fully symmetric: $x^2 - 2yx-2zx-4lyzx-2lx + (y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz)=0$\n\nso $x^2+y^2+z^2+l^2 - 4lxyz - 2(xy+xz+xl+yz+yl+zl) - 4=0$\n\nWe know $x,y,z,l\\in \\mathbb{Z}_{\\ge 0}$. If $xyz=0$ the statement is clear. Let $(x,y,z,l)$ be the solution with $x+y+z+l$ minimal. I claim $\\min\\{x,y,z,l\\}=0$. Furthermore WLOG $x$ is the largest.\n\nFirst we can assume $xr$ so we can bound $(y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz)=xr\\ge \\frac 29 (2y+2z+4lyz+2l)^2$ \n\nThere is another case where $r<0$. In this case, $y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz < 0$ and its absolute value is at least $2y+2z+4lyz+2l+1$, and we can get a contradiction.\n" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "Another problem if anyone has time. Sorry for the terrible notation/format. New to the forums and haven't learned how to use all its functionalities yet.\r\n\r\nSuppose x and y are integers. Suppose p = x^2+y^2 where p is a prime integer. Prove p is irreducible in Z[i].", "Solution_1": "$p=x^2+y^2=(x+yi)(x-yi).$", "Solution_2": "Yes, and this gives that $p$ is [u]not[/u] irreducible in $\\mathbb{Z} [i]$", "Solution_3": "I was given this problem by my math professor to show that given x and y (which are integers) and p=x^2+y^2 (where p is a prime integer), p is irreducible in Z[i] and thus a Gaussian prime. But p is in fact not irreducible??? I am confused. Is this professor asking me to prove something that is not true?", "Solution_4": "If you heard him right, he made a mistake. You can show that $x+yi$ is a gaussian prime." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $03$, and that polygon can be found by taking a regular polygon with $n+1$ sides and eliminating one of its vertices...\r\n\r\nI'm pretty sure that works..." } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "[color=darkblue]Let $ \\Gamma_A$ be the $ A$-mixtilinear incircle of $ ABC$, which is tangent to the circumcircle at $ A'$ and to the side $ AB$ at $ D$. Find the length of $ A'D$.[/color]", "Solution_1": "[quote=\"Ahiles\"][color=darkblue]Find the length of $ A'D$.[/color][/quote]\r\n\r\nWhat is the application of finding the lenght of A'D? , In terms of which elements of $ ABC$? , Could you show at leats the expression that we should get? thanks in advance\r\n\r\nBest Regards" } { "Tag": [ "integration", "calculus", "calculus computations" ], "Problem": "$ \\int(2x^{\\minus{}6}\\plus{}x^{\\minus{}9})^{2/3}dx$", "Solution_1": "Change variables according to $ x^{3}\\equal{}\\frac{1}{t^{3}\\minus{}2}$, \r\n\r\n[hide=\"and so\"]$ P \\equal{}\\int(2x^{\\minus{}6}\\plus{}x^{\\minus{}9})^{2/3}\\,dx \\equal{}\\int x^{\\minus{}4}(2\\plus{}x^{\\minus{}3})^{2/3}\\,dx \\equal{}\\minus{}\\int t^{4}\\,dt$\n\n$ \\equal{}\\minus{}\\frac{1}{5}t^{5}\\plus{}C \\equal{}\\minus{}\\frac{1}{5}(2\\plus{}x^{\\minus{}3})^{5/3}\\plus{}C$.\n[/hide]" } { "Tag": [ "pigeonhole principle", "Support" ], "Problem": "I don't know about others, but upon viewing my report, I discover that \"Mastery of Counting Fundamentals\" and \"Mastery of Counting Fundamentals Level 1\" encompass exclusively the exact same problems. I just wish to know if there are others for which there is something different. Also, is this as supposed to be?", "Solution_1": "I have a Mastery of Counting Fundamentals (MCF), MCF level 1 and MCF level 2, and I only have problems for level 2 (7 problems).\r\n\r\nAnother question: What is \"Basic Pigeonhole\"? I have had about 300 problems, and none show up in that subject.", "Solution_2": "Both dragon96 and I have over a thousand points, and we still have received no Basic Pidgeonhole questions, which leads me to think it's one of those empty topics.", "Solution_3": "I've finished all the problems and I have had no \"Basic Pidgeonhole\" problems either.", "Solution_4": "[quote=\"r15s11z55y89w21\"]I don't know about others, but upon viewing my report, I discover that \"Mastery of Counting Fundamentals\" and \"Mastery of Counting Fundamentals Level 1\" encompass exclusively the exact same problems. I just wish to know if there are others for which there is something different. Also, is this as supposed to be?[/quote]\r\nI think they might be worded differently to test your consistency with the problems. That's happened to me with some of the problems.", "Solution_5": "[quote=\"r15s11z55y89w21\"]I don't know about others, but upon viewing my report, I discover that \"Mastery of Counting Fundamentals\" and \"Mastery of Counting Fundamentals Level 1\" encompass exclusively the [b]exact[/b] same problems. I just wish to know if there are others for which there is something different. Also, is this as supposed to be?[/quote][hide= ][/hide]" } { "Tag": [], "Problem": "[b]1.[/b]Sa se demonstreze ca pentru diferite alegeri ale semnelor + si - din expresiile \\[ \\pm 1 \\pm 2 \\pm 3 \\pm ... \\pm(4n\\plus{}1)\\] se obtin numere impare mai mici sau egale cu $ (2n\\plus{}1)(4n\\plus{}1)$\r\n\r\nsi inca o problema poate chiar mai interesanta decat prima \r\n[b]2.[/b] Fie n un numar natural arbitrar si k un numar intreg.Sa se demonstreze ca daca k admite o reprezentare de forma \\[ \\pm 1^n \\pm 2^n \\pm ... \\pm m^n\\] , m natural, pentru o anumita alegere a semnelor + si -, atunci admite o infinitate de reprezentari de aceasta forma.", "Solution_1": "problema1\r\ntermenii expresiei din enunt sunt numerele naturale de la 1 la 4n+1.este usor de vazut ca de la 1 la 4n avem 2n numere pare si 2n numere impare, si cum 2n+1 e impar atunci suma are 2n termeni pari si 2n+1 termeni impari.atunci este evident ca numarul obtinut va fi impar(este o ,,suma'' de numere pare si de un numar impar de numere impare).rezultat maxim se obtine atunci cand se alege doar plus.deci numerele vor fi mai mici sau egale decat 1+2+3+...(4n+1)=(4n+1)(4n+2)/2=(2n+1)(4n+1) :)" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "let [tex] x, y [/tex] and [tex] z [/tex] be nonnegatif real numbers such that:\r\n[tex] xyz(x+y+z) =1 [/tex]\r\nProve that: [tex] \r\n\\surd ( x^2 + \\frac{1}{y^2} )( y^2+\\frac{1}{z^2} )( z^2+\\frac{1}{x^2} )=(x+y)(y+z)(z+x)[/tex]", "Solution_1": "Easy one,yeh?.Just replace $1$ by $xyz(x+y+z)$ and we have:\r\n[tex] \\sqrt{(x^2+\\frac{1}{y^2}(y^2+\\frac{1}{z^2})(z^2+\\frac{1}{x^2})}=\\sqrt{\\frac{[(xy+yz)(yz+zx)(zx+xy)]^2}{(xyz)^2}}[/tex]\r\n[tex]=(x+y)(y+z)(z+x) [/tex] :)" } { "Tag": [], "Problem": "Hi.\r\n\r\nGiven: $ x^5 \\plus{} x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1 \\equal{} 10$ and $ \\minus{} 1$ isnt a root to the equation.\r\n\r\nFind $ (x \\plus{} 1)^4$.\r\n\r\nThanks.", "Solution_1": "What is \"+...+\" ?", "Solution_2": "\"and $ \\minus{}1$ isn't a root to the equation?\" It's not anyways, is it? Or do you mean $ 0$ instead of $ 10$ on the RHS?\r\n\r\n[hide=\"Assuming that\"]\nThe solutions are the sixth roots of unity (except for $ 1$, of course) excluding $ \\minus{}1$. \n[/hide]", "Solution_3": "What makes you think there'll be a nice solution to this? The other 4 roots yield 4 different values for $ (x \\plus{} 1)^4$.", "Solution_4": "Whoa; I just realized there was another part of the problem - I thought it was just find the roots. That's what I get for trying to do math at night..." } { "Tag": [ "trigonometry", "algebra", "polynomial", "Vieta", "function", "algebra unsolved" ], "Problem": "Assuming that the roots of $x^3 + p \\cdot x^2 + q \\cdot x + r = 0$ are real and positive, find a relation between $p,q$ and $r$ which gives a necessary condition for the roots to be exactly the cosines of the three angles of a triangle.", "Solution_1": "Let $a$, $b$ and $c$ be the roots of the equation. It is well-known that $a$, $b$ and $c$ are the cosines of the angles of a triangle if and only if $a^2 + b^2 + c^2 + 2abc = 1$. Now, using the Vieta relations, we get $(p^2 - 2q) -2r = 1$ or $p^2 = 2q + 2r + 1$.", "Solution_2": "[quote=\"Arne\"]Let $a$, $b$ and $c$ be the roots of the equation. It is well-known that $a$, $b$ and $c$ are the cosines of the angles of a triangle if and only if $a^2 + b^2 + c^2 + 2abc = 1$. Now, using the Vieta relations, we get $(p^2 - 2q) -2r = 1$ or $p^2 = 2q + 2r + 1$.[/quote]\n\nActually, the condition is necessary, but generally not sufficient! For example, setting $a = b = 2,$ and $c = -1,$ we get: \\[ a^2 + b^2 + c^2 + 2abc = 4 + 4 + 1 - 8 = 1, \\] but obviously $a, b,$ and $c$ are not the cosines of the angles of a triangle.\n\nIf\u2014as stated in the problem\u2014all three quantities are positive, then $a^2 + b^2 + c^2 + 2abc = 1$ implies $0 < a, b, c < 1.$ The inverse cosines $\\alpha = \\arccos a$ and $\\beta = \\arccos b$ are thus necessarily each strictly between zero and $\\frac\\pi2$ and so there exists a triangle with $0 < \\alpha, \\beta < \\frac\\pi2$ as two of its angles. Furthermore, for $0 < \\theta < \\frac\\pi2$ the square of the cosine is a strictly decreasing function of $\\theta.$ Observing that $\\cos^2(\\theta) + \\cos^2(\\frac\\pi2 - \\theta) = \\cos^2\\theta + \\sin^2\\theta = 1$ we see that $a^2 + b^2 < 1 \\implies \\alpha + \\beta > \\frac\\pi2,$ and thus $\\gamma = \\pi - \\alpha - \\beta < \\frac\\pi2.$\n\nTherefore, in the triangle whose angles have positive cosines $a$ and $b$ with $a^2 + b^2 < 1,$ we have $x = \\cos\\gamma > 0.$ Now $x$ must satisfy the equation: \\[\nx^2 + 2abx + (a^2 + b^2 - 1) = 0, \\] whose roots have a product $(a^2 + b^2 - 1)$ and sum $-2ab$ that are both negative. Thus, the roots have opposite signs and so $x$ is the only positive solution. Since $c > 0$ also satisfies the same equation, $c$ is indeed the cosine of the third angle.\n\nThe positivity condition is essential, and cannot be ignored." } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "Let $ f: \\mathbb R\\to\\mathbb R$ be continuous. $ \\lim_{x\\to\\infty}f(x)\\equal{}\\infty$ and $ \\lim_{x\\to\\minus{}\\infty}f(x)\\equal{}\\minus{}\\infty.$ Prove that $ f$ is surjective.", "Solution_1": "Bolzano-Cauchy intermediate value theorem.", "Solution_2": "Thanks, but I don't know how to apply it. :(" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c>0$ such that $a+b+c\\geq\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$.\r\n Is this true: $abc\\leq 1$ ? ;)", "Solution_1": "Take [tex] a=1 , b=2 , c=3 [/tex]", "Solution_2": "Yes,but the reverse inequality $abc\\geq1$? ;)" } { "Tag": [ "function", "number theory", "prime factorization", "superior algebra", "superior algebra unsolved" ], "Problem": "Suppose that f is a multiplicative function with f(1)=1. Show that the summation over d|n of u(d)f(d) = (1-f(p1))*(1-f(p2))*.....*(1-f(pk))\r\n\r\nwhere n=(p1^a1)*(p2^a2)*...*(pk^ak) is the prime factorization for n. u(d) is the Mobius function.\r\n\r\nI divided the divisors d of n, into p1, p2,..,pk so that I have k different sums of u(p_i)f(p_i) over each p_i. I don't know where to go from here.", "Solution_1": "by the definition of $ \\mu$, we have $ \\mu(p_{j_1}\\cdots p_{j_i})\\equal{}(\\minus{}1)^i$, so\r\n\\[ (1 \\minus{} f(p_1))\\cdots(1 \\minus{} f(p_k)) \\equal{} \\sum_{1\\leq j_1\\leq \\cdots \\leq j_i \\leq k} ( \\minus{} 1)^if(p_{j_1})f(p_{j_2})\\cdots f(p_{j_i}) \\equal{}\\]\r\n\r\n\\[ \\sum_{i \\equal{} 0}^k\\mu(p_{j_1}\\cdots p_{j_i})f(p_{j_1}\\cdots p_{j_i}) \\equal{}\\]\r\n\r\n\\[ \\sum_{d|n}\\mu(d)f(d),\\]\r\nbecause whenever $ d$ is not squarefree, $ \\mu(d) \\equal{} 0$." } { "Tag": [ "inequalities", "function", "trigonometry", "inequalities proposed" ], "Problem": "This cant be new but I have not seen it cited anywhwere\r\n\r\nIn the acute triangle ABC prove that \r\n\r\ntanA/2 + tanB/2 + tanC/2 <2", "Solution_1": "Since the triangle is acute, $(\\pi/4, \\pi/4, 0)$ majorizes $(A/2, B/2, C/2)$. \r\n\r\nApply the Majorization (Karamata) Inequality to the convex function $f(x) = \\tan{x}$ on ${}[0, \\pi/4]$, we have\r\n\r\n$ 2 = f(\\pi/4) + f(\\pi/4) + f(0) > f(A/2) + f(B/2) + f(C/2)$. The inequality is strict because none of the angles can be 0.", "Solution_2": "This inequality can be improved :\r\nFor any acute triangle ABC, we have\r\n\r\nsum(tan A/2 ) + tan(A/2)*tan(B/2)*tan(C/2) < 2\r\n\r\nAnd 2 is the best constant.", "Solution_3": "There is an easy solution for this inequality:\r\n\r\nBecause ABC is an acute triangle 00\r\n--> 1-tanA/2-tanB/2-tanC/2+tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2-tanA/2tanB/2tanC/2>0\r\nAnd so Because tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=1 then we have:\r\n\r\ntanA/2+tanB/2+tanC/2<2-tanA/2tanB/2tanC/2<2 !" } { "Tag": [ "geometry", "perimeter" ], "Problem": "Each side of square $ ABCD$ is doubled in length to form square $ EFGH$. The perimeter of square $ EFGH$ is 40 cm. How many square centimeters are in the area of square $ ABCD$?\n\n[asy]pair a,b,c,d,e,f,g,h;\na=(0,0);\nb=(10,0);\nc=(10,10);\nd=(0,10);\ne=(15,0);\nf=(35,0);\ng=(35,20);\nh=(15,20);\ndraw((11,5)--(14,5),Arrow);\ndraw(a--b--c--d--cycle);\ndraw(e--f--g--h--cycle);\nlabel(\"$A$\",a,S);\nlabel(\"$B$\",b,S);\nlabel(\"$C$\",c,N);\nlabel(\"$D$\",d,N);\nlabel(\"$E$\",e,S);\nlabel(\"$F$\",f,S);\nlabel(\"$G$\",g,N);\nlabel(\"$H$\",h,N);[/asy]", "Solution_1": "$ 10*10\\equal{}100$\r\n\r\n$ 100/4\\equal{}25$\r\n\r\nanswer : 25", "Solution_2": "Or, since the larger square has a perimeter of 40, we find that it's length is 10. Half of 10 is 5, so $ 5^2 \\equal{} \\boxed{25}$.", "Solution_3": "Or even, since the larger square is twice the smaller one, the perimeter of the smaller one is 40/2 = 20. So one side is 5 and the area is 25." } { "Tag": [], "Problem": "Bueno ..\r\naki \r\n\r\nla parte 2 del libro de geo\r\n\r\nSeCCION: Problemas\r\n\r\n\r\ncARLOS", "Solution_1": "[color=blue]\n Buenos dias,\n\nles informo que por restricciones en mi \ncuota de uploading...\n\nquitare estos pdf,\npor tanto, descarguenlos antes del lunes en la noche,\n\nGracias[/color]" } { "Tag": [ "function", "algebra", "domain", "integration", "complex analysis" ], "Problem": "Given continuously differentiable functions $ u(x,y), v(x,y)$, let $ f\\equal{}u\\plus{}iv$. Suppose for every $ z_0$ in the domain of $ f$, $ D$, there exists an $ r_0$ such that \\[ \\int_{|z\\minus{}z_0|\\equal{}r}fdz\\equal{}0\\] for all $ r 0$,m,n are real numbers and $ m \\plus{} n \\geq 1$,prove that:\r\n$ {\\frac {my \\plus{} nz \\minus{} \\left( m \\plus{} n \\right) x}{\\left( m \\plus{} n \\right) x \\plus{} y \\plus{} z}} \\plus{} {\\frac {mz \\plus{} nx \\minus{} \\left( m \\plus{} n \\right) y}{\\left( m \\plus{} n \\right) y \\plus{} z \\plus{} x}} \\plus{} { \\frac {mx \\plus{} ny \\minus{} \\left( m \\plus{} n \\right) z}{\\left( m \\plus{} n \\right) z \\plus{} x \\plus{} y}}\\geq 0.$\r\n\r\nBQ", "Solution_1": "[quote=\"xzlbq\"]Let $ x,y,z > 0$,m,n are real numbers and $ m \\plus{} n \\geq 1$,prove that:\n$ {\\frac {my \\plus{} nz \\minus{} \\left( m \\plus{} n \\right) x}{\\left( m \\plus{} n \\right) x \\plus{} y \\plus{} z}} \\plus{} {\\frac {mz \\plus{} nx \\minus{} \\left( m \\plus{} n \\right) y}{\\left( m \\plus{} n \\right) y \\plus{} z \\plus{} x}} \\plus{} {\\frac {mx \\plus{} ny \\minus{} \\left( m \\plus{} n \\right) z}{\\left( m \\plus{} n \\right) z \\plus{} x \\plus{} y}}\\geq 0.$\n\nBQ[/quote]\r\n\r\nWe have, because $ m\\plus{}n\\geqslant 1$, $ (m\\plus{}n)x\\plus{}y\\plus{}z\\leqslant(m\\plus{}n)(x\\plus{}y\\plus{}z)$ and cyclic, therefore\r\n\\[ \\sum\\frac {my \\plus{} nz \\minus{} \\left( m \\plus{} n \\right) x}{\\left( m \\plus{} n \\right) x \\plus{} y \\plus{} z}\\geqslant\\frac {\\sum(my \\plus{} nz \\minus{} \\left( m \\plus{} n \\right) x)}{\\left( m \\plus{} n \\right)(x \\plus{} y \\plus{} z)}\\equal{}0\\]\r\n... :P...", "Solution_2": "m,n are real numbers,Is there an error?\r\n\r\nNeed to be changed m,n>0?\r\n\r\nPlease discuss\u3002\r\nBQ", "Solution_3": "[quote=\"xzlbq\"]m,n are real numbers,Is there an error?\n\nNeed to be changed m,n>0?\n\nPlease discuss\u3002\nBQ[/quote]\r\nBut why should we change $ m$, $ n$ to positive real numbers?? The only thing we need is that their sum is grater than or equal to $ 1$ so that the inequality $ x(m\\plus{}n)\\geqslant x$ holds..." } { "Tag": [ "geometry", "probability", "AMC", "AIME", "AMC 10" ], "Problem": "The AMC A is not far away! what are you doing to prepare? how do you think you'll do?\r\n\r\nI'm taking the 10A/12B route unless I fail the 10A (which is quite possible). I'm working through AoPS Intro to Geometry, Intermediate Algebra, and a little bit of Intro to Counting and Probability, combined with practice AMC's and AIME's (and lots of other random stuff). I'll be happy with my 10A score if I improve from what I did last year on the 10B. \r\n\r\nWhat about you guys? The A is a lot closer than I thought!!", "Solution_1": "Ooooh I can't believe it's so close... but I guess we all still have some time to prepare. I'm doing about what Dynamo is (hopefully that's a good thing for both of us). I'm hoping that I can at least qualify for the AIME on the A and then get at least 23 right on the B. But if by some freak accident I get 23+ on the A, I'll probably take the 12B... and hope that I don't completely fail. Ugh everything's coming fast now.... AMC, AIME, MATHCOUNTS. Speaking of all those... I think I'll study now :oops: haha.", "Solution_2": "I might take the AMC10 (maybe) if our school registers by this friday.\r\nI think I'm taking the AMC 10B.", "Solution_3": "ohh my, i just realized that it's next week... asdfasdf :( time to try and not fail", "Solution_4": "Yes, maybe I should have taken at least one practice amc 12 over the last year, since I never seem to do well on the real amc 12." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Let $ a,b,c,d$ be real numbers such that\r\n$ a\\plus{}b\\plus{}c\\plus{}d\\equal{}k_1$, $ ab\\plus{}bc\\plus{}ca\\plus{}da\\plus{}db\\plus{}dc\\equal{}k_2$, $ abc\\plus{}bcd\\plus{}cda\\plus{}dab\\equal{}k_3$, and $ abcd\\equal{}k_4$. Find\r\n$ p_1\\equal{}a^3\\plus{}b^3\\plus{}c^3\\plus{}d^3$, $ p_2\\equal{}a^4\\plus{}b^4\\plus{}c^4\\plus{}d^4$, $ p_3\\equal{}a^5\\plus{}b^5\\plus{}c^5\\plus{}d^5$ in terms of $ k_1,k_2,k_3,k_4$.", "Solution_1": "With Maple:\r\n\r\nF:={x1+x2+x3+x4=k1,x1*x2+x1*x3+x1*x4+x2*x3+x2*x4+x3*x4=k2,x1*x2*x3+x1*x2*x4+x1*x3*x4+x2*x3*x4=k3,x1*x2*x3*x4=k4}:\r\n> for k from 3 to 6 do\r\n> simplify(x1^k+x2^k+x3^k+x4^k,F);\r\n> od;\r\n> \r\n\r\n \r\n k1^3 - 3 k2 k1 + 3 k3\r\n\r\n \r\n 2 k2^2 + 4 k3 k1 - 4 k2 k1^2 + k1^4 - 4 k4\r\n\r\n \r\n -5 k2 k1^3 + 5 k2^2 k1 - 5 k3 k2 + 5 k3 k1^2 + k1^5 - 5 k1 k4\r\n\r\n\r\n \r\n6 k3 k1^3 + 3 k3^2 - 12 k2 k3 k1 - 2 k2^3 + k1^6 \r\n+ 9 k2^2 k1^2 - 6 k2 k1^4 \r\n - 6 k4 k1^2 + 6 k4 k2" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Given $ m,n\\in N$ such that $ m<2001,n<2002$. Let distinct real numbers be written in the cells of a $ 2001$x$ 2002$ board (with $ 2001$ rows and $ 2002$ columns). A cell of the board is called [i]bad[/i] if the corresponding number is smaller than at least $ m$ numbers in the same column and at least $ n$ numblers in the same row. Let $ s$ denote the total number of bad cells. Find the least possible value of $ s$.", "Solution_1": "Anybody? :wink:" } { "Tag": [ "Support" ], "Problem": "Hi all,I'm Vietnamese.I think each country has it own math forum,Vietnam has one which have English section (other VN forum is not support other language :D).So I want to introduce to all of you,its address is:\r\n http://mathnfriend.net/index.php?showforum=62\r\n just new,need your contribution :D\r\n why don't you join us,why don't you share your national math forum address here,in this topic ? :lol:", "Solution_1": "is it for vietnamese only?", "Solution_2": "[quote]\t\nis it for vietnamese only?\n[/quote]\r\n[color=red]all are welcome[/color],If you know only English,you can join in Olympiad section,If you know VietNamese,others can interest you :D", "Solution_3": "This page is just a copy from MathLinks, just the same", "Solution_4": "[quote]\t\nThis page is just a copy from MathLinks, just the same[/quote]\r\nwe only want to create another playground,can u make it different :lol:", "Solution_5": "Nice forum. Keep up the good work! :)" } { "Tag": [ "symmetry" ], "Problem": "Source: ACoPS\r\n3.1.18 Consider the following two-player game. Each player takes turns placing a penny on the surface of a rectangular table. No penny can touch a penny which is alreay on the table. Tha table starts out with no pennies. The last player who makes a legal move wins. Does the first player have a winnig strategy?\r\n\r\n[hide=\"solution\"]\nYes, s/he does.(placing a penny on the center of the table)[/hide]\r\n\r\nam I right?", "Solution_1": "Can you prove that your answer is right? :)\r\n\r\nEdit: By which I mean, describe your strategy and show why it must work. Also, can you generalize? For what shapes of table does your strategy still hold?", "Solution_2": "[hide=\"Proof\"]\nSo the first player places the first penny on the center of the table. After that, the first player can place the penny on the position, symmetrical to that the second player placed before.\nDoes this make sense?\n[/hide]" } { "Tag": [ "inequalities", "geometry proposed", "geometry" ], "Problem": "$ \\boxed {\\ \\frac {a\\minus{}2b\\plus{}2c}{a\\plus{}b\\minus{}c}\\plus{} \\frac {b\\minus{}2c\\plus{}2a}{b\\plus{}c\\minus{}a}\\plus{}\\frac {c\\minus{}2a\\plus{}2b}{c\\plus{}a\\minus{}b}\\ \\ge\\ \\frac {3\\left(a^2\\plus{}b^2\\plus{}c^2\\right)}{2(ab\\plus{}bc\\plus{}ca)\\minus{}\\left(a^2\\plus{}b^2\\plus{}c^2\\right)}\\ \\left(\\ \\ge\\ 3\\ \\right)\\ }$", "Solution_1": "hello, nice inequality. The right-hande side of your inequality is true because\r\n$ a^2\\plus{}b^2\\plus{}c^2\\geq ab\\plus{}bc\\plus{}ca \\leftrightarrow\r\n6(a^2\\plus{}b^2\\plus{}c^2)\\geq6(ab\\plus{}bc\\plus{}ca) \\Leftrightarrow\r\n3(a^2\\plus{}b^2\\plus{}c^2)\\geq6(ab\\plus{}bc\\plus{}ca)\\minus{}3(a^2\\plus{}b^2\\plus{}c^2) \\Leftrightarrow\r\n\\frac {3(a^2\\plus{}b^2\\plus{}c^2)}{2(ab\\plus{}bc\\plus{}ca)\\minus{}(a^2\\plus{}b^2\\plus{}c^2)}\\geq3$.\r\nTo prove the left-hande side we make the substitutions\r\n$ a\\equal{}y\\plus{}z,b\\equal{}x\\plus{}z,c\\equal{}x\\plus{}y$ and your inequality is equivalent with\r\n$ \\frac{3(x^2y^3\\plus{}x^3z^2\\plus{}y^2z^3\\minus{}x^2y^2z\\minus{}x^2yz^2\\minus{}xy^2z^2}{2xyz(xy\\plus{}xz\\plus{}yz)}\\geq0$.\r\nWe assume that $ x\\geq y\\geq z$ and the numerator is\r\n$ y^2(x\\minus{}z)(y\\minus{}z)(x\\plus{}z)\\plus{}(x\\minus{}y)^2(x\\plus{}y)z^2\\geq0$.", "Solution_2": "[quote=\"Virgil Nicula\"]$ \\boxed {\\ \\frac {a \\minus{} 2b \\plus{} 2c}{a \\plus{} b \\minus{} c} \\plus{} \\frac {b \\minus{} 2c \\plus{} 2a}{b \\plus{} c \\minus{} a} \\plus{} \\frac {c \\minus{} 2a \\plus{} 2b}{c \\plus{} a \\minus{} b}\\ \\ge\\ \\frac {3\\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)}{2(ab \\plus{} bc \\plus{} ca) \\minus{} \\left(a^2 \\plus{} b^2 \\plus{} c^2\\right)}\\ \\left(\\ \\ge\\ 3\\ \\right)\\ }$[/quote]\r\n$ \\frac 13\\cdot\\sum\\frac {a\\minus{}2b\\plus{}2c}{a\\plus{}b\\minus{}c}\\equal{}\\frac 12\\cdot\\sum\\left(\\frac {a\\plus{}c\\minus{}b}{a\\plus{}b\\minus{}c}\\minus{}\\frac 13\\right)\\equal{}$ $ \\frac 12\\cdot\\left(\\sum\\frac {p\\minus{}b}{p\\minus{}c}\\minus{}1\\right)\\equal{}$ $ \\frac 12\\cdot\\left[\\sum\\frac {(p\\minus{}b)^2}{(p\\minus{}b)(p\\minus{}c)}\\minus{}1\\right]\\stackrel {(C.B.S.)}{\\ \\ \\ge \\ \\ }$\r\n\r\n$ \\frac 12\\cdot\\left\\{\\frac {\\left[\\ \\sum (p\\minus{}b)\\ \\right]^2}{\\sum (p\\minus{}b)(p\\minus{}c)}\\minus{}1\\right\\}\\equal{}$ $ \\frac 12\\cdot\\left\\{\\frac {(a\\plus{}b\\plus{}c)^2}{\\sum\\left[a^2\\minus{}(b\\minus{}c)^2\\right]}\\minus{}1\\right\\}\\equal{}$ $ \\frac 12\\cdot\\left[\\frac {\\sum a^2\\plus{}2\\cdot\\sum bc}{2\\cdot\\sum bc\\minus{}\\sum a^2}\\minus{}1\\right]\\equal{}$ $ \\frac {\\sum a^2}{2\\cdot\\sum bc\\minus{}\\sum a^2}$ ." } { "Tag": [ "inequalities", "quadratics", "inequalities open" ], "Problem": "if posotive reals such that \\[\\begin{array}{l}a^{2}+b^{2}+c^{2}+d^{2}= 1 \\\\ then prove that: a+b+c+d+\\frac{1}{{abcd}}\\ge 18 \\\\ \\end{array}\\]", "Solution_1": "Let S=a+b+c+d and P=abcd.\r\nThe quadratic-geometric mean inequality yields P<=1/16 (*).\r\nThe arithmetic-geometric mean inequality yields P<=S^4/4^4 (**).\r\n\r\nCase 1: S>=2.\r\nThen (*) implies S+1/P>=18.\r\n\r\nCase 2: 1<=S<=2.\r\nThen (**) implies S+1/P >= S+4^4/S^4 =: f(S).\r\nOn the interval [1,2], the minimum of f(S) is taken at S=2 with f(S)=18.\r\n\r\nCase 3: S<1.\r\nThis contradicts a^2+b^2+c^2+d^2=1." } { "Tag": [ "calculus", "integration", "geometry", "linear algebra", "matrix", "function", "parallelogram" ], "Problem": "Hai folks, I have a query (though it appears to be quite silly), given an indefinite integral , say x, after we integrate it , we obtain (x^2)/2 ; Now, my query is----- What does \"x^2/2\" denote? \r\n\r\nWhat is the purpose of indefinite integration, like definite integration is used to calculate the area beneath (under) a curve, is indefinite integration used to calculate the length of a curve? Can anyone elaborate?\r\n\r\n\r\nSecondly, When we calculate the determinant of a matrix, we obtain a \"number\", what does this number mean/denote? \r\n\r\nFor instance, consider a 3*3 unit matrix, when we compute the determinant of this matrix , we get the number \"0\", what does this mean? \r\n\r\nI need an elaborate answer. :)", "Solution_1": "1) Indefinite integration should really be understood with respect to definite integration. I hope you have no confusion about what a definite integral means. Given that you understand the meaning of an expression like $ \\int_a^b f(x) \\, dx$, we define $ F(x) \\equal{} \\int_a^x f(t) \\, dt$ for arbitrary fixed $ a$. Under some smoothness conditions, the value of defining this function - which we are free to call an indefinite integral, or a primitive, or whatever - is the statement of the Fundamental Theorem of Calculus.\r\n\r\nIn other words, indefinite integrals are a convenient way to evaluate definite integrals. But the theory of integration will make more sense if you focus on definite integrals; that will also stop you from messing up constants. \r\n\r\n2) One intuition is that the determinant is a \"volume multiplier\": the determinant of a $ 3 \\times 3$ matrix describes what happens to the (signed) volume of boxes if you multiply every point by that matrix. If it's zero, that means boxes get squashed down into parallelograms. Similarly, the determinant of a $ 2 \\times 2$ matrix describes what happens to the (signed) areas of parallelograms if you multiply every point by that matrix. Matrices of determinant $ 1$ preserve areas, volumes, and orientations, so they're particularly nice.", "Solution_2": "I did understand what you meant to say, but what does it mean if \" the determinant happens to be a negative number\" ?", "Solution_3": "Then it flips the sign of the [i]signed[/i] area or volume. In other words, it reverses handedness. The typical transformation with negative determinant is a reflection through a line or a plane; left-handed things become right-handed things and vice versa." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "show that if the positive integer $a,b,c,d$ satisfy $ab=cd$ then we have $gcd(a,c)*gcd(a,d)=a*gcd(a,b,c,d)$; :) :)", "Solution_1": "Let $p$ be any prime.\r\nLet $a' = \\mathrm{ord}_p a$ and so on.\r\nNote, that $a'+b' = c'+d'$.\r\n$\\mathrm{ord}_p ((a,c)\\cdot(a,d)) = max(a',c')+max(a',d')$, $\\mathrm{ord}_p (a\\cdot (a,b,c,d)) = a' + max(a',c',d',c'+d'-a')$.\r\nIt's really easy to see that in three possible cases (when $a'$ maximal, minimal of $\\{a',c',d'\\}$ or $a'$ is between $c'$ and $d'$) $max(a',c')+max(a',d') = max(a',c',d',c'+d'-a')$. So, canonycal representations of $(a,c)\\cdot(a,d)$ and $a\\cdot (a,b,c,d)$ are the same, and they are equal." } { "Tag": [ "geometry", "3D geometry", "sphere", "algebra", "polynomial", "AMC", "AIME" ], "Problem": "Say anything you want about round 4 here!", "Solution_1": "Heh, I thought it was pretty fun... and I'd have to say that my favorite problem was #3. :)", "Solution_2": "Was I the only one who felt that there was a HUGE range of difficulty on this round, much more so than on previous ones? I felt like some of them (1 and 2, specifically) were really easy, and they didn't take me long to do at all, while other ones were much, much more difficult. Generally there are easier and harder ones on each round, but the range on this one seemed more extreme than on previous rounds.", "Solution_3": "[quote=\"zanttrang\"]Was I the only one who felt that there was a HUGE range of difficulty on this round, much more so than on previous ones? I felt like some of them (1 and 2, specifically) were really easy, and they didn't take me long to do at all, while other ones were much, much more difficult. Generally there are easier and harder ones on each round, but the range on this one seemed more extreme than on previous rounds.[/quote]\r\n\r\nAgreed. #1 and #2 were ridiculous in comparison to #3.\r\n\r\nDoes anyone want to post their solutions? I want to know if I'm right... :D Well, I guess I'll begin.", "Solution_4": "The one about the square roots was so fun!", "Solution_5": "[quote=\"mathfanatic\"]\nAgreed. #1 and #2 were ridiculous in comparison to #3.\n\nDoes anyone want to post their solutions? I want to know if I'm right... :D Well, I guess I'll begin.[/quote]\r\nYou got the same answers I did, though your answer to #4 seems to be much more rigorous than mine (I used a strange mix of algebra and geometry to get my answer). I'll post them when I get access to my home computer.", "Solution_6": "[quote=\"zanttrang\"]You got the same answers I did, though your answer to #4 seems to be much more rigorous than mine (I used a strange mix of algebra and geometry to get my answer). I'll post them when I get access to my home computer.[/quote]\r\n\r\nThat's comforting. I thought about using the fact that 2x^2 + 3y^2 + 6z^2 = n was an ellipsoid to argue that there'd always be a solution for 38 <= n <= 126, but I couldn't figure out how to do it. So I used a really ugly way.", "Solution_7": "[quote=\"zanttrang\"]Was I the only one who felt that there was a HUGE range of difficulty on this round, much more so than on previous ones? I felt like some of them (1 and 2, specifically) were really easy, and they didn't take me long to do at all, while other ones were much, much more difficult. Generally there are easier and harder ones on each round, but the range on this one seemed more extreme than on previous rounds.[/quote]This was intentional.", "Solution_8": "what you might want to try is transforming your ellipsoid into a sphere by making the substitution:\r\n\r\n$x = \\frac{a}{\\sqrt{2}}, y = \\frac{b}{\\sqrt{3}}, z=\\frac{c}{\\sqrt{6}}$\r\n\r\nthen you can use some 3-d geometry to get what you are looking for.", "Solution_9": "Ow... I feel really stupid right now. On problem 1 I had 22 integers but somehow counted only 20. I wrote \"thus there are 20 integers that work. Namely: (integers here)\". As i wrote down every number that worked... and did all the math right-how many points do you guys think im gonna get docked for not being able to count?\r\n\r\n2 was pretty easy... no mistakes there. 5 was just long, 4 was a nice problem, but I didnt make any susbitutions and thus had to use a 3d graph at the end ;) 3 I had no idea what exactly they ment by a 3 variable polynomial and thus stuck down a little work and sent in on its way...", "Solution_10": "For 1, it's a bit easier to use external angle measure. An external angle is the supplement of an interior angle, so when the external angle is an integer, so will its interior. Then factor count 360 and throw out 1 and 2.", "Solution_11": "[quote=\"salmononpi\"]Ow... I feel really stupid right now. On problem 1 I had 22 integers but somehow counted only 20.[/quote]\r\n\r\nSame here. Some how 9/40 didn't make the final cut...", "Solution_12": "[quote=\"ThAzN1\"]Heh, I thought it was pretty fun... and I'd have to say that my favorite problem was #3. :)[/quote]\r\n\r\nI agree. at first i thought number three was kind of hard, but that's until i decided to divide the sum of the cubes of the terms by the sum of the terms to see what i'd get. It worked out nicely though.", "Solution_13": "I liked this round's problems, but I had exactly one day to do my problems (considering SAT and the AIME were the week before.)\r\n\r\nHere's my potentially wrong, but otherwise elegant solution to nr. 3:\r\n\r\nConsider the identity:\r\n\r\n$\\alpha + \\beta + \\gamma = (\\alpha^3 + \\beta^3 + \\gamma^3 - 3\\alpha\\beta\\gamma)/(\\alpha^2+\\beta^2+\\gamma^2-\\alpha\\beta - \\alpha\\gamma - \\beta\\gamma)$.\r\n\r\nThe denominator of the expression can be factored as a sum of squares and is thus nonnegative, so the sign of $\\alpha + \\beta + \\gamma$ varies with the denominator. Applying the identity to our problem yields the polynomial\r\n\r\n$a^3+2b^3+4c^3-6abc$", "Solution_14": "I came up with a simple (but not necessarily elegant or especially rigorous) solution to #4, using the idea that the minimum value comes where the ellipsoid is tangent to the plane. (I think this is equivalent to subtracting one equation from the other and completing the square to find the point at which there's exactly one solution - tangency - ie, both sides are zero.) Using the gradient of the function f(x) = 2x^2 + 3y^2 + 6z^2 made things pretty simple. [solutions attached]", "Solution_15": "Well, for #3, I essentially used the same identity as [b]tetrahedr0n[/b]; for #4, I used a (not-so-rigorous?) argument w/ Cauchy. Anyways, here's my solutions...", "Solution_16": "I felt that the first problem was almost too easy for this contest, while 3 was not good at all. On #3, I came up with the (correct?) polynomial $x^3+2y^3+4z^3-6xyz$ but then convinced myself that that was not right, and therefore I was not as rigorous as I could have been on the proof of its correctness ;) . I think that perhaps I was confused on the definition of 'same sign'. Having seen that that is the right answer, I now feel proud of my method of finding it-3 lines of LaTeX! (Although it is not very rigorous at all.) I'll post it when I have access to my solutions.\r\n\r\n#2 was probably my favorite problem, even though it was quite easy.", "Solution_17": "[quote=\"ThAzN1\"]Well, for #3, I essentially used the same identity as [b]tetrahedr0n[/b]; for #4, I used a (not-so-rigorous?) argument w/ Cauchy. Anyways, here's my solutions...[/quote]\r\n\r\nI did #4 in a very similar fashion. First, I made the same substitution you did, then I used Cauchy to find the lower bound. To find the upper bound, I used the rearrangement inequality coupled with an epsilon argument that I'm quite sure is rigorous.\r\n\r\n#2 was quite unremarkable, as USAMTS problems go.\r\n\r\nI misread #1; I missed the word \"regular\", so I instead found all values of n such that an n-gon can have all integer angles. How many points do you think I'll lose?\r\n\r\nI failed altogether to solve #3.\r\n\r\nFor #5, I had an interesting lemma... I proved that if XYZ and X'Y'Z' are two triangles with equal area, congruent incircles, and $\\angle X=\\angle X'$, then the triangles are congruent. This blows the whole problem apart...", "Solution_18": "what the...?\r\n\r\nhmm...i used the same calculus method for solving the ellipsoid/tangent plane problem as flierdeke...but i thought there had to be only ONE solution for (x,y,z)...as in: AT the tangent point...so my answer was \"no integral values of n\"\r\n\r\noh well.", "Solution_19": "oh no!!!\r\n\r\nI solved for only integer solutions in #4!!!\r\n\r\nbah! there goes my perfect score!\r\n\r\nhow could I have missed that?", "Solution_20": "[quote=\"gaussianprime\"]oh no!!!\n\nI solved for only integer solutions in #4!!!\n\nbah! there goes my perfect score!\n\nhow could I have missed that?[/quote]\r\n\r\nI almost did that...I thought, hm, this is trivial, since there are only a few integer solutions to 3x + 4y + 5z = 23, and then plug them into 2x^2 + 3y^2 + 6z^2 = n. Fortunately I lost the paper that had my answers on it, so when I returned to the problems the week before they were due, I had to redo it...", "Solution_21": "[quote=\"flierdeke\"]I came up with a simple (but not necessarily elegant or especially rigorous) solution to #4, using the idea that the minimum value comes where the ellipsoid is tangent to the plane. (I think this is equivalent to subtracting one equation from the other and completing the square to find the point at which there's exactly one solution - tangency - ie, both sides are zero.) Using the gradient of the function f(x) = 2x^2 + 3y^2 + 6z^2 made things pretty simple. [solutions attached][/quote]\nYep, that's what I did, too. \nAhh, the power of calculus! :lol:\n\nUmm, what was with problem #1? That was like an easy AIME problem . . . it seems like the only way to mess up would be to have misread the problem.\n\n[quote=\"dts\"]I misread #1; I missed the word \"regular\", so I instead found all values of n such that an n-gon can have all integer angles. How many points do you think I'll lose?[/quote]\r\nIt sounds like you made the problem more difficult by misreading it, in which case you may not lose as many as you might guess -- I won't be surprised if you get a 3 or a 4.\r\n\r\nHere are my solutions if anyone's interested. Hopefully #1 will get a commendation.", "Solution_22": "If anyone's interested, my solutions...\r\nI think that they are all right, not great, but not bad.\r\n\r\nWooo... only 26 KB", "Solution_23": "[quote=\"ZennyK\"]...It sounds like you made the problem more difficult by misreading it...[/quote]\r\n\r\nI dunno... it was still pretty easy, if you think to use generating functions...", "Solution_24": "I'm sure I got everything except #4, for which I decided to give up and turned in \"I was not able to solve this problem.\" I actually misread the problem two times: first, I thought they wanted integer solutions; then, I thought they wanted rational solutions! I was desperately flipping through number theory books for information on Pell-ish equations the day this round was due. :P\r\n\r\nI'll be happy if I get something commended. :)", "Solution_25": "[quote=\"Danbert\"]Wooo... only 26 KB[/quote]\r\n\r\nThat's kind of cheating, though, since it's a DVI file.", "Solution_26": "[quote=\"dts\"]For #5, I had an interesting lemma... I proved that if XYZ and X'Y'Z' are two triangles with equal area, congruent incircles, and $\\angle X=\\angle X'$, then the triangles are congruent. This blows the whole problem apart...[/quote]\r\n\r\nI did that, too, and managed to prove it in 6 or 7 very small lemmas... it essentially became a long 9th grade geometry proof with a few unconventional formulas.", "Solution_27": "#1/#2 - both way easy, don't understand why.\r\n\r\n#3 - used the cool factoring at the beginning of my MO Treasures book (same one as everyone else)\r\n\r\n#4 - i just found the lower bound by setting the ellipsoid tangent to the plane and the upper bound by expanding the ellipsoid until it touched the plane at only one point in the first quadrant. (bleh, looking back at my solution, it looks wrong, but the answer coincides...)\r\n\r\n#5 - well i just proved it.", "Solution_28": "Here are mine.", "Solution_29": "[quote=\"towbomb\"][quote=\"dts\"]For #5, I had an interesting lemma... I proved that if XYZ and X'Y'Z' are two triangles with equal area, congruent incircles, and $\\angle X=\\angle X'$, then the triangles are congruent. This blows the whole problem apart...[/quote]\n\nI did that, too, and managed to prove it in 6 or 7 very small lemmas... it essentially became a long 9th grade geometry proof with a few unconventional formulas.[/quote]\r\n\r\nI did it in one small lemma and one big lemma. I wouldn't call mine 9th grade level, though; I used trig.", "Solution_30": "For #4 I had to use Lagrange multipliers because I couldn't figure out how to do it without calculus.\r\nAnd #3 was definately my favorite problem.\r\nDoes anyone know when they will be grading the papers?", "Solution_31": "[quote=\"Jimmy\"]Does anyone know when they will be grading the papers?[/quote]In early April. We hope to release the scores sometime during the week of April 11.", "Solution_32": "Here are my solutions, but I don't think I did a very good job on #5.", "Solution_33": "Depending on the level of calculus you know, there is a really easy and straightforward method for solving problem #4. It literally takes less than 10 minutes to solve for the minimum n. I learned the method in Calculus 3, which I took this year at a local college. For those of you who are interested, it is a method called Lagrange multipliers. I'll attach my solution to #4 here.", "Solution_34": "That is interesting -- I did Cauchy-Schwartz to get the minimum n, and some simple algebra to get the maximum." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Let the sequence $ {u_n}$ satisfy\r\n$ \\left| {u_{m \\plus{} n} \\minus{} u_m \\minus{} u_n } \\right| < \\frac{1}{{m \\plus{} n}}\\forall m,n \\in N$\r\nProve that $ u_n$ is a arithmetic progression.", "Solution_1": "Consider $ v_{k}\\equal{}u_{mk}\\minus{}ku_m$, then $ v_1\\equal{}0, |v_{k\\plus{}l}\\minus{}v_k\\minus{}v_l|<\\frac{1}{m(k\\plus{}l)}$. Therefore $ |v_k|<\\frac 1m (\\frac 1k \\plus{}\\frac{1}{k\\minus{}1}\\plus{}...\\frac 12)\\equal{}\\frac 1m (H_k\\minus{}1)$.\r\nTherefore $ |u_{mk}\\minus{}ku_m|<\\frac 1m (H_k\\minus{}1)$. \r\nConsider $ k\\equal{}an$, then $ |u_{mna}\\minus{}nau_m|<\\frac 1m (H_{na}\\minus{}1)$\r\nConsider $ m\\equal{}ma,k\\equal{}a$, we get $ |u_{mna}\\minus{}au_{mn}|<\\frac{1}{mn}(H_a\\minus{}1)$.\r\nTherefore $ |au_{mn}\\minus{}nau_m|<\\frac{1}{mn}(nH_{na}\\minus{}n\\plus{}H_a\\minus{}1)$ or\r\n$ |u_{mn}\\minus{}nu_m|<\\frac{1}{mn}\\frac{nH_{na}\\minus{}n\\plus{}H_a\\minus{}1}{a}$. If $ a\\to \\infty$ we get $ u_{mn}\\equal{}nu_m$, $ m\\equal{}1$ give $ u_n\\equal{}nu_1$." } { "Tag": [], "Problem": "assume that we have luminous point in the space (like very very small lamp) can we control the light of this lamp with finite equal ball such that they are distinct( means also not tangency)", "Solution_1": "it's not clear.....maybe l don't understand the meaning of \"finite equal ball\".\r\n\"ball\" like \"soccerball\" ?\r\nif yes, how can a ball control a lamp ?", "Solution_2": "yes, simply the round things.\r\n\r\nSo what we got to prove is that if you place thme around the lightpoints, and try to fill up the holes, you'll need an unlimited amount of balls.\r\n\r\nVery abstract but nice problem! :)", "Solution_3": "l am beginner in english too. now l will solve this problem." } { "Tag": [ "real analysis", "calculus", "calculus computations" ], "Problem": "Suppose $\\left(x_n\\right)$ is a sequence of real numbers such that each subsequence of $\\left(x_n\\right)$ has a convergent subsequence. \r\nDoes it follow that $\\left(x_n\\right)$ is a convergent sequence?", "Solution_1": ":?: No, $x_n=(-1)^n$ is a counterexample? ...", "Solution_2": "That's right :)", "Solution_3": "This should be moved to computations and tutorials... :?", "Solution_4": "Yes, This is very easy. :roll:", "Solution_5": "A slightly less trivial variation, which is actually quite useful (Urysohn's principle)\r\n\r\nLet $x_n$ be a sequence in any topological space.\r\nIf each subsequence of $x_n$ has a subsequence converging to $x$, then the $x_n$ converge to $x$.", "Solution_6": "That particular result seems to come out a lot in real analysis; you might have convergence in $L^1$ or in measure, so you get a pointwise convergent subsequence, and then you apply dominated convergence or something like that to get a result about the original sequence.", "Solution_7": "[quote=\"Peter VDD\"]This should be moved to computations and tutorials... :?[/quote]\r\n\r\nWell, it is easy, but it is not a computation or tutorial, rather a thinking exercise, a problem (albeit an easy one)" } { "Tag": [ "search", "limit", "calculus", "integration", "calculus computations" ], "Problem": "Hi, can anyone please show me how to do this problem. You may use any feasible method. Thanks.\r\n\r\nQuestion: Evaluate $ \\displaystyle\\frac{1}{\\cosh(\\pi/2)}\\minus{}\\frac{1}{3\\cosh(3\\pi/2)}\\plus{}\\frac{1}{5\\cosh(5\\pi/2)}\\minus{}\\cdots$", "Solution_1": "You can show:\r\n\r\n\\[ \\sum_{n\\equal{}0}^{\\infty}\\frac{(\\minus{}1)^n}{(2n\\plus{}1)\\cosh[(2n\\plus{}1)\\pi/2]}\\equal{}\\frac{\\pi}{8}\\]\r\n\r\nvia the Residue Theorem although perhaps someone here can show us how to do it with just regular Calculus.", "Solution_2": "how do you guys get those mathematical symbols", "Solution_3": "its by using latex.\r\n\r\n\r\nsearch for the word `LATEX` in forums and go through them for sometime", "Solution_4": "How does the residue theorem work here? May you please demonstrate? I do not know that sums can be evaluated by the residue theorem. :blush:", "Solution_5": "[quote=\"Mitochondria\"]How does the residue theorem work here? May you please demonstrate? I do not know that sums can be evaluated by the residue theorem. :blush:[/quote]\r\n\r\nPick a slice:\r\n\\[ \\lim_{\\text{pie}\\to\\infty}\\mathop\\oint\\limits_{\\substack{\\text{my pie} \\\\\r\n\\text{slice}}} \\frac {1}{z\\cos(z)\\cosh(z)}dz\r\n\\]\r\n(and don't forget to evaluate the integral over the indentation about the origin)", "Solution_6": "How did you know that would be the integrand? Also, how did you know that would be the contour?", "Solution_7": "Note if you choose a $ \\pi/2$ slice centered along the positive real axis, the $ \\pi/4$ leg cancels the $ \\minus{}\\pi/4$ leg and the outer circular contour goes to zero and so the sum is determined by the integral over the indentation about the origin as it's radius goes to zero.\r\n\r\nAs far as choosing that, you work on cars? Engines? I do. And sometime something's not working and you don't know what's wrong. What do you do? Work on everything. Eventually, you'll find what the problem is and solve the problem. :)", "Solution_8": "Thank you for the reply.", "Solution_9": "[url=http://www.mathlinks.ro/viewtopic.php?t=260466]Here is solution[/url] and [url=http://www.mathlinks.ro/viewtopic.php?t=254323]prove....[/url]" } { "Tag": [ "USAMTS", "geometry", "geometric transformation", "reflection" ], "Problem": "This is my first year doing the USAMTS, I know that if you just give a correct answer, you get 1 point, but what's the maximum?", "Solution_1": "5 points is the maximum per problem.\r\n\r\nBy the way, http://www.usamts.org is probably going to be your best source of information, particularly their FAQs.", "Solution_2": "Do you mean for a well-written and correct answer? The maximum is 5. \r\n\r\nhttp://www.usamts.org/TipsFAQ/U_FAQ.php#howgrade\r\n\r\nThis gives a general criteria.", "Solution_3": "[quote=\"pascal_1623\"]This is my first year doing the USAMTS, I know that if you just give a correct answer, you get 1 point, but what's the maximum?[/quote]\r\nAs others said, the maximum is 5. However, your statement, \"If you just give a correct answer, you get 1 point,\" is mistaken.\r\n\r\nCorrect answers are supposed to be accompanied by work that shows how the problem-solver found or verified that answer. Nevertheless, as a grader I have seen submitted solutions that consist of nothing but a numerical answer. The amount of partial credit given to such incomplete solutions varies.\r\n\r\nSome problems are so difficult that the only way to find the correct numerical answer is to have a complete understanding of the problem. In that case, giving the answer proves that the participant understood all the mathematics involved, and I would award four points of partial credit. In other problems, finding the numerical answer is trivial and the hard work is in verifying the answer. An extreme example would be a problem whose last sentence was, \"Prove that t=155.\" A solution that consisted of nothing but \"t=155\" would be worth zero points.\r\n\r\nErin Schram\r\nUSAMTS grader", "Solution_4": "[quote=\"Erin J. Schram\"][quote=\"pascal_1623\"]This is my first year doing the USAMTS, I know that if you just give a correct answer, you get 1 point, but what's the maximum?[/quote]\nAs others said, the maximum is 5. However, your statement, \"If you just give a correct answer, you get 1 point,\" is mistaken.\n\nCorrect answers are supposed to be accompanied by work that shows how the problem-solver found or verified that answer. Nevertheless, as a grader I have seen submitted solutions that consist of nothing but a numerical answer. The amount of partial credit given to such incomplete solutions varies.\n\nSome problems are so difficult that the only way to find the correct numerical answer is to have a complete understanding of the problem. In that case, giving the answer proves that the participant understood all the mathematics involved, and I would award four points of partial credit. In other problems, finding the numerical answer is trivial and the hard work is in verifying the answer. An extreme example would be a problem whose last sentence was, \"Prove that t=155.\" A solution that consisted of nothing but \"t=155\" would be worth zero points.\n\nErin Schram\nUSAMTS grader[/quote]\r\nWhat are some problems such that finding the answer is only possible if you have complete understanding of the problem?\r\nThanks,\r\nBPMS", "Solution_5": "[quote=\"bpms\"]What are some problems such that finding the answer is only possible if you have complete understanding of the problem?\nThanks,\nBPMS[/quote]\r\nI'd say for example that USAMTS 2/1/17 fits the bill (http://www.usamts.org/Solutions/Solution2_1_17.pdf). The problem isn't too bad if you figure out how to do it, but there's no way you'd come up with that solution unless you did it correctly, so most likely the grader would award you near full credit if you just gave the right answer.\r\n\r\nHowever, that said, you should never do that! The point of USAMTS problems isn't really to just get the right answer - it's to be able to fully justify it in a rigorous proof.", "Solution_6": "[quote=\"zanttrang\"]I'd say for example that USAMTS 1/1/17 fits the bill (http://www.usamts.org/Solutions/Solution2_1_17.pdf). [/quote]\r\nI believe that zanttrang means USAMTS Problem 2/1/17, for which he gave a web address. I remember the grading of Problem 1/1/17 (http://www.usamts.org/Solutions/Solution1_1_17.pdf), and we graders were definitely looking for proofs that certain sequences did not contain squares.\r\n\r\nErin Schram\r\nUSAMTS grader", "Solution_7": "[quote=\"Erin J. Schram\"][quote=\"zanttrang\"]I'd say for example that USAMTS 1/1/17 fits the bill (http://www.usamts.org/Solutions/Solution2_1_17.pdf). [/quote]\nI believe that zanttrang means USAMTS Problem 2/1/17, for which he gave a web address. I remember the grading of Problem 1/1/17 (http://www.usamts.org/Solutions/Solution1_1_17.pdf), and we graders were definitely looking for proofs that certain sequences did not contain squares.\n\nErin Schram\nUSAMTS grader[/quote]\r\nHa, yeah, thanks - I didn't check my post over carefully enough - I've edited it to reflect the correct problem.", "Solution_8": "What about problems that have multiple parts, like part (a) and part (b)? If students only get one of the two correct, is the maximum score a 3?", "Solution_9": "[quote=\"tiredepartment\"]What about problems that have multiple parts, like part (a) and part (b)? If students only get one of the two correct, is the maximum score a 3?[/quote]It depends on how much each part is worth. If one of them is a problem, and the second is a generalization, then solving the generilaztion will get you 4 probably, and you will get maybe 1/2 for the other..." } { "Tag": [ "arithmetic sequence", "algebra", "difference of squares", "special factorizations" ], "Problem": "if 1^2 + 3^2 + 5^2 +....+99^2=S, then what is the value of the sum 2^2 + 4^2 + 6^2 +...+ 100^2?\r\nA)S+2550\r\nB)2S\r\nC)4S\r\nD)S+5050\r\nE)S+5075", "Solution_1": "[hide=\"hint\"]\nUse difference of squares\n[/hide]\n\n[hide=\"solution\"]\nLet $x$ = $2^2 + 4^2 + 6^2 + \\dotsm + 100^2$\n\n$x - S = 100^2 - 99^2 + 98^2 - 97^2 + \\dotsm + 2^2 - 1^2$\n$x - S = (1)(199) + (1)(195) + \\dotsm + 3$\n\nSince this is just an arithmetic sequence,\n\n$x - S = (199 + 3)*50/2 = 5050$\n$x = S + 5050$ \n\nSo the answer is D.\n\n[/hide]", "Solution_2": "[hide]\nWe can write the first sum S as $\\displaystyle\\sum_{i=1}^{50}(2x-1)^2=$\n$\\displaystyle\\sum_{i=1}^{50}4x^2-4x+1=$\n${4\\displaystyle\\sum_{i=1}^{50}x^2}-{4\\displaystyle\\sum_{i=1}^{50}x}+{50}$\nWe can write the second sum as $\\displaystyle\\sum_{i=1}^{50}(2x)^2=$\n$\\displaystyle\\sum_{i=1}^{50}4x^2=$\n$4\\displaystyle\\sum_{i=1}^{50}x^2$\nSo the difference between the two sums is:\n${4\\displaystyle\\sum_{i=1}^{50}x}-{50}$\nThis can be simplified into $4\\frac{50(51)}{2}-50=5050$\n\nSo the answer is D\n[/hide]\r\n\r\nCalculuslover, your answer's a lot simpler. :D\r\nMy computer's too slow. :(" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Two pentagons are given in the plane such that neither of them has a vertex on any line that contains a side of the other. Find the maximum number of intersection points their sides can form.\r\n\r\n[i]K\u00f6mal, problem A.229, January 2000[/i]", "Solution_1": "Are we allowed to use the lines which contain the sides or just the segments? If this is true, isn't the answer 25? I mean, doesn't every line which contains a side of a pentagon intersect every line which contains a side of the other pentagon?", "Solution_2": "Surely, we talk about segments.", "Solution_3": "Anyway, here is my solution.\r\n\r\nFirst of all, it is clear that each side of the first pentagon is intersected by at most 4 sides of the second pentagon. Indeed, suppose the contrary, but each time we intersect the side we change the half plane determined by the side. So we changes half plane 5 times, hence we are not in the same half plane, where we started. Contradiction.\r\n\r\nIt follows that there are at most 20 intersection. See the example below." } { "Tag": [ "inequalities" ], "Problem": "a and b are real numbers. a-b=1 \r\nshow that a^3 - b^3 geq 1/4", "Solution_1": "$a^3-b^3 =(a-b)(a^2+ab+b^2)$\r\n $=(a^2+ab+b^2)$\r\n $\\geq\\frac{3}{4}(a+b)^2$\r\n $\\geq\\frac{3}{4}(1+2b)^2 \\geq \\frac{1}{4}$", "Solution_2": "First geq is not evident\r\nand last geq doesnt hold if b=-1/2", "Solution_3": "[hide]\n$b = a-1$ so\n$a^3 - b^3 = a^3 - (a-1)^3 = 3a^2-3a+1$\n\n$3a^2-3a+1-\\frac{1}{4} = 3\\left(a-\\frac{1}{2}\\right)^2 \\ge 0$\n$3a^2-3a+1 \\ge \\frac{1}{4}$[/hide]\r\n\r\nEDIT: yeah, sorry, typo.", "Solution_4": "[quote=\"paladin8\"][hide]\n$b = a-1$ so\n$a^3 - b^3 = a^3 - (a-1)^2 = 3a^2-3a+1$\n\n$3a^2-3a+1-\\frac{1}{4} = 3\\left(a-\\frac{1}{2}\\right)^2 \\ge 0$\n$3a^2-3a+1 \\ge \\frac{1}{4}$[/hide][/quote]\r\n\r\nI think it should be $a^3 - b^3 = a^3 - (a-1)^3\\neq a^3 - (a-1)^2$.", "Solution_5": "[quote=\"Fermat -Euler\"]$a^3-b^3 =(a-b)(a^2+ab+b^2)$\n $=(a^2+ab+b^2)$\n $\\geq\\frac{3}{4}(a+b)^2$\n $\\geq\\frac{3}{4}(1+2b)^2 \\geq \\frac{1}{4}$[/quote]\n[quote=\"senouy\"]First geq is not evident\n[/quote]\r\nwe have $(a^2+ab+b^2) \\geq\\frac{3}{4}(a+b)^2$\r\n\r\n<=>\r\n$4(a^2+b^2+ab)\\geq 3 (a^2+b^2+2ab)$\r\n <=> \r\n$a^2+b^2\\geq 2ab$ \r\ni think that's now evident ;)" } { "Tag": [ "pigeonhole principle", "analytic geometry", "graphing lines", "slope", "geometry", "function", "integration" ], "Problem": "let 699 real numbers: $ 0 0$, so\r\n\\[ 0 < \\sum_{i\\equal{}1}^{n\\minus{}1} a_ia_{i\\plus{}1}(a_i \\minus{} a_{i\\plus{}1}) < \\frac {1} {3},\\]\r\ntherefore (at least) one of the expressions is less than $ \\frac {1} {3(n\\minus{}1)}$. If we want the bound to be $ \\frac {1} {2007}$, that requires $ n \\geq 670$.\r\n\r\nFor a better estimate, the defect area (between the triangles and the graph) can be computed to be\r\n\\[ D \\equal{} \\frac {1} {6} \\left ( a_n^3 \\plus{} (a_{n\\minus{}1} \\minus{} a_n)^3 \\plus{} \\cdots \\plus{} (a_1 \\minus{} a_2)^3 \\right ) \\geq \\frac {1} {6n^2} a_1^3, \\ \\textrm {so}\\]\r\n\\[ \\sum_{i\\equal{}1}^{n\\minus{}1} [\\Delta OP_iP_{i\\plus{}1}] \\equal{} \\frac {1} {2} a_1^3 \\minus{} \\frac {1} {3} a_1^3 \\minus{} D \\leq \\frac {n^2 \\minus{} 1} {6n^2} a_1^3 \\leq \\frac {n^2 \\minus{} 1} {6n^2},\\]\r\ntherefore (at least) one of the expressions is less than $ \\frac {n\\plus{}1} {3n^2}$. If we want the bound to be $ \\frac {1} {2007}$, that still requires $ n \\geq \\left \\lceil \\frac {669 \\plus{} \\sqrt{669^2 \\plus{} 4\\cdot 669}} {2} \\right \\rceil \\equal{} 670$. \r\n\r\nBut there is room for improvement (equality in this last estimate does not yield equal areas), so we are back to some algebraic manipulations, in order to lower the number to $ 669$. \r\n\r\nSee a second posting at [url]http://www.mathlinks.ro/viewtopic.php?p=1625529#1625529[/url]" } { "Tag": [ "probability" ], "Problem": "The values -5, -3, and 4 randomly replace $ a$, $ b$ and $ c$ in the equation $ ax\\plus{}b\\equal{}c$, and the equation is solved for $ x$. What is the probability that $ x$ is negative? Express your answer as a common fraction.", "Solution_1": "$ x\\equal{}\\frac{c\\minus{}b}a$, so we note that for every $ a$, there is ONE distinct pair $ (b,c)$ that satisfies the requirements. Thus, the probability is $ \\frac12$, since for every selection of $ a$, there are $ 2$ possible pairs $ (b,c)$." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$ x_{1},x_{2},...,x_{n\\plus{}1}$ positive number prove:\r\n$ \\frac{1}{x_{1}}\\plus{}\\frac{x_{1}}{x_{2}}\\plus{}\\frac{x_{1}x_{2}}{x_{3}}\\plus{}\\frac{x_{1}x_{2}x_{3}}{x_{4}}\\plus{}...\\plus{}\\frac{x_{1}x_{2}...x_{n}}{x_{n\\plus{}1}}\\geq4(1\\minus{}x_{1}x_{2}...x_{n\\plus{}1})$", "Solution_1": "[quote=\"zaya_yc\"]$ x_{1},x_{2},...,x_{n \\plus{} 1}$ positive number prove:\n$ \\frac {1}{x_{1}} \\plus{} \\frac {x_{1}}{x_{2}} \\plus{} \\frac {x_{1}x_{2}}{x_{3}} \\plus{} \\frac {x_{1}x_{2}x_{3}}{x_{4}} \\plus{} ... \\plus{} \\frac {x_{1}x_{2}...x_{n}}{x_{n \\plus{} 1}}\\geq4(1 \\minus{} x_{1}x_{2}...x_{n \\plus{} 1})$[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=168848" } { "Tag": [ "abstract algebra", "function", "vector", "LaTeX", "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "a first step is proving that if P is f.g. and projective, then so is P*, and hence so is P**\r\nI can do this step, so just looking for the final argument.", "Solution_1": "$ P^{*}= Hom(P,R)$ where $ R$ is the base ring, or do you mean some pontrjagin-dual?", "Solution_2": "Right $ P^{*}= \\text{Hom}_{R}\\ (P,R)$ and then $ P^{**}$ is just the double daul.\r\n\r\nOne way is to show that this function is an isomorphism: $ \\theta: P \\rightarrow P^{**}$ where $ \\theta(a)$ is equal to the function $ \\theta(a)$ which assigns to a homomorphism $ f: P^{*}\\rightarrow P$ its value at $ a$. I have trouble wrapping my mind around double duals.", "Solution_3": "Geometric idea: a projective A-module P corresponds to a vector bundle M~ over Spec(A), for those who know coherent sheaves. \r\n\r\n More concretely, it means that if P is projective, for all prime ideal x in Spec(A) the localised module P_x is a free A_x module. To see this algebraically, write P as a summand of a free module; this property remains when localizing at x, hence P_x is a projective A_x module, finitely generated, hence free because A_x is local. \r\n\r\n Now since P is finitely generated, the dual P* and also the double dual P** pass to localization (see for example Eisenbud, \"commutative algebra with a view...\" , prop.2.10); it means that (P**)_x = (P_x) ** , the second double dual being taken as A_x module. But P_x is free, the natural map P_x ---> (P_x )** is an isomorphism, so the natural map P ---> P** is an isomorphism when localized at any prime. It is therefore an isomorphism.", "Solution_4": "this \"geometric\" proof seems to be the most elegant. I think we can avoid localizations. \r\n\r\nfirst notice that the claim is clear for f.g. free modules: $ (R^{n})^{*}\\cong R^{n}$ by the universal property of the direct sum, and thus $ (R^{n})^{**}\\cong R^{n}$. it is easy to check that this isomorphism is actually the natural one.\r\n\r\nnow if $ P$ is finitely-generated, we have an exact sequence $ 0 \\to K \\to F \\to P \\to 0$, where $ F$ is a f.g. free module. if $ P$ is projective, this sequence splits, i.e. $ P$ is a direct summand of $ F$. the inclusion $ P \\to F$ yields a commutative diagramm\r\n\r\n$ P \\to P^{**}$\r\n$ \\downarrow ~~~~\\downarrow$ \r\n$ F \\to F^{**}$\r\n\r\nsince $ F \\to F^{**}$ has trivial kernel, the same is true for $ P \\to P^{**}$. on the other hand, the projection $ F \\to P$ yields a commutative diagramm\r\n\r\n$ F \\to F^{**}$\r\n$ \\downarrow ~~~~\\downarrow$ \r\n$ P \\to P^{**}$\r\n \r\nremark that $ F^{**}\\to P^{**}$ stays epi since the sequence above splits. thus, $ F \\to F^{**}$ epi implies $ P \\to P^{**}$ epi.", "Solution_5": "This proof I think is at least as elegant as the previous one.", "Solution_6": "A one last proof avoiding localization, and resting on a homological argument: there is a spectral sequence whose second term is\r\n\r\n E_2 ^{p,q} = Ext^p (Ext^{-q} (P, A), A) \r\n\r\n and converging to P in degree 0. The Ext* groups are computed in the category of A-modules. Since for q non zero we have: Ext^{-q} (P, A) =0 because P is projective, we get P = E_2 ^{0,0} = P**. I hope it is correct. \r\n\r\n Question: how to post in LaTex? :blush:", "Solution_7": "thanks for the help guys, i like your first solution fadalbala and -oo-'s \r\n\r\nedit: for Latex just treat the post like a regular latex document" } { "Tag": [ "algebra", "polynomial", "function", "algebra proposed" ], "Problem": "if X+Y in Q find all polynimals that p(X)+p(Y) in Q : :lol:", "Solution_1": "anybody wants to solve?", "Solution_2": "Consider a polynomial $ P(x)$ satisfy $ f(x)\\in Q \\leftrightarrow x\\in Q$ then $ P(x) \\equal{} ax \\plus{} b$\r\nProblem have posted in forum . \r\nApply this then $ P(x) \\plus{} P(a \\minus{} x) \\equal{} mx \\plus{} n,\\forall x\\in R ,a\\in Q$ \r\nSolve this we have $ f(x)$ is linear function with rational confficient.", "Solution_3": "Do you mean: find all polynomials $ P(x)$ such that $ x \\plus{} y \\in \\mathbb{Q} \\implies P(x) \\plus{} P(y) \\in \\mathbb{Q}$\r\n\r\nor do you mean: find all polynomials $ P(x)$ such that $ x \\plus{} y \\in \\mathbb{Q} \\Leftrightarrow P(x) \\plus{} P(y) \\in \\mathbb{Q}$?" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let A be an nxn matrix st a(ij) = 1/(i+j-1) and let B be the inverse of A.\r\nIs it true that the sum of all entries b(ij) of B is equal to n^2?\r\nWhat is the proof of this??", "Solution_1": "A previous [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=41046]thread[/url] on this subject.\r\n\r\nIt doesn't contain a proof for the question you asked, so that's still open.", "Solution_2": "Interestingly enough, it has a direct connection to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146373[/url] ;)" } { "Tag": [ "geometry", "IMO Shortlist", "geometry proposed" ], "Problem": "The circles $ k_1$ and $ k_2$ with respective centers $ O_1$ and $ O_2$ are externally tangent at the point $ C$, while the circle $ k$ with center $ O$ is externally tangent to $ k_1$ and $ k_2$. Let $ l$ be the common tangent of $ k_1$ and $ k_2$ at the point $ C$ and let $ AB$ be the diameter of $ k$ perpendicular to $ l$. Assume that $ O$ and $ A$ lie on the same side of $ l$. Show that the lines $ AO_2$, $ BO_1$, and $ l$ have a common point.", "Solution_1": "In triangle ABC we use Ceva thorem and the result comes", "Solution_2": "i dont understand how it follows from ceva's theorem .cud u plz explain?", "Solution_3": "[quote=\"moldovan\"]The circles $ k_1$ and $ k_2$ with respective centers $ O_1$ and $ O_2$ are externally tangent at the point $ C$, while the circle $ k$ with center $ O$ is externally tangent to $ k_1$ and $ k_2$. Let $ l$ be the common tangent of $ k_1$ and $ k_2$ at the point $ C$ and let $ AB$ be the diameter of $ k$ perpendicular to $ l$. Assume that $ O$ and $ A$ lie on the same side of $ l$. Show that the lines $ AO_2$, $ BO_1$, and $ l$ have a common point.[/quote]\r\nNice but not easy, in my opinion. It has been posted on the forum many times, you can find solutions at [url=http://www.mathlinks.ro/viewtopic.php?p=875026#875026]IMO Shortlist 2006, Geometry 6[/url].", "Solution_4": "[quote=\"sophie germain\"]i dont understand how it follows from ceva's theorem .cud u plz explain?[/quote]\r\nThis problem is easy.\r\nLet $ AC\\cap O_1B\\equal{}\\{M\\}, AO_2\\cap BC\\equal{}\\{N\\}.$\r\n$ \\angle BOF\\equal{}\\angle FO_2C$ then $ \\angle OFB\\equal{}\\angle CFO_2$ we get C,F,B are collinear.\r\nSimilarly, A,E,C are collinear.\r\nUse Ceva's theorem for triangle ABC:\r\n$ \\frac{AD}{BD}.\\frac{NB}{NC}.\\frac{MC}{MA}\\equal{}\\frac{CO_2}{CO_1}.\\frac{AB}{CO_2}.\\frac{CO_1}{AB}\\equal{}1$ ->QED" } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "Problem: Is there any other perfect square beside 144 in the Fibonacci sequence ? Justify your answer", "Solution_1": "Search function.", "Solution_2": "What ?\r\nU don't understand.\r\nThis noon, i found it's solution but i need more time to read it. I'll post it later", "Solution_3": "No, [i]you[/i] don't understand.\r\nHe told you to search, meaning it's been posted before.\r\n\r\n(and indeed : http://www.mathlinks.ro/Forum/viewtopic.php?t=1423 )" } { "Tag": [ "search", "USAMTS" ], "Problem": "This is a write up on the subject above and it's an urgent.Need an helping hand to search and find the proper write up for this topic. remember \"mathematics of an elastic media\".\r\n Thanks and God Bless.\r\n Azeez", "Solution_1": "errr....what does \"mathematics of an elastica media\" mean anyways? :? \r\nI think this should go in different forum.", "Solution_2": "Sounds like Physics to me. And yes, this is _definitely_ in the wrong forum.", "Solution_3": "[quote=\"eryaman\"]errr....what does \"mathematics of an elastica media\" mean anyways? :? \nI think this should go in different forum.[/quote]\r\n\r\nAn elastic medium is any material that will deform under stress but will tend to return to its original shape after the stress is removed (imagine an i-beam supported at both ends; if you put a large enough load in the middle, it will buckle, but will return to its original shape once the load is removed). Hooke's law is commonly applied when elastic media are studied, and situations where forces are applied to non-rigid bodies can be modeled mathematically. However, I'd agree that the USAMTS forum isn't the right place for a physics problem." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "let be given possitive real number: $x_1$,$x_2$,...........,$x_n$ such that $\\sum_{i=1}^{n}{x_i}=n$\r\nProve the inequality: $\\sum_{i\\ne{j}}{x_i\\cdot{x_{j}^2}}\\ge\\sum_{i\\ne{j}}{x_i\\cdot{x_j}}$", "Solution_1": "Hmm...\r\nWe are to prove \\[n\\sum_{i\\ne{j}}{x_i\\cdot{x_{j}^2}}\\ge(x_1+...+x_n)\\sum_{i\\ne{j}}{x_i\\cdot{x_j}},\\] which is equivalent to \\[\\frac{\\sum_{i\\ne j}x_ix_j^2}{n(n-1)}\\geq \\frac{\\sum_{i\\ne j\\ne k}x_ix_jx_k}{n(n-1)(n-2)},\\]\r\ni.e. pure Murheid.", "Solution_2": "@myth: can you post complete-solution---thanks :P", "Solution_3": "Am I missing something? It is Murheid's inequality for orbits $(2,1,0)$ and $(1,1,1)$. What further explanations are you waiting? :?" } { "Tag": [ "limit" ], "Problem": "Find the value of $\\lim_{n\\rightarrow\\infty}S(x)$ where \\[S(x)=\\frac{x}{x+1}+\\frac{x^2}{(x+1)(x^2+1)}+...+\\frac{x^{2^n}}{(x+1)(x^2+1)...(x^{2^n}+1)}\\]and $x>1$", "Solution_1": "wouldn't that be $1$?", "Solution_2": "it has been posted by me.and sloved with the help of the kind people :) \r\nsee here [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28814]www.mathlinks.ro/Forum/viewtopic.php?t=28814[/url]" } { "Tag": [ "AMC", "AIME" ], "Problem": "Let $ m$ be the number of solutions in positive integers to the equation $ 4x\\plus{}3y\\plus{}2z\\equal{}2009$, and let $ n$ be the number of solutions in positive integers to the equation $ 4x\\plus{}3y\\plus{}2z\\equal{}2000$. Find the remainder when $ m\\minus{}n$ is divided by $ 1000$.", "Solution_1": "[hide=\"Solution\"]Consider the following apparent one-to-one correspondence: every solution $ (x_0, y_0, z_0)$ in positive integers of the equation $ 4x \\plus{} 3y \\plus{} 2z \\equal{} 2009$ corresponds a solution $ (x_0, y_0 \\minus{} 3, z_0)$ in positive integers of the equation $ 4x \\plus{} 3y \\plus{} 2z \\equal{} 2000$. However, not all solutions of the first equation correspond to solutions of the second; when $ y_0 \\equal{} 1$, $ 2$, or $ 3$, $ y_0 \\minus{} 3$ fails to be a positive integer. If so, then counting the solution triples for these values of $ y_0$ should give the difference between the number of solutions, $ m \\minus{} n$.\n\nWhen $ y_0 \\equal{} 1$, the equation reduces to $ 4x_0 \\plus{} 2z_0 \\equal{} 2006 \\iff 2x_0 \\plus{} z_0 \\equal{} 1003$, giving $ 501$ solutions from $ (1,1001)$ to $ (501, 1)$.\nWhen $ y_0 \\equal{} 2$, the equation reduces to $ 4x_0 \\plus{} 2z_0 \\equal{} 2003$, giving no solutions.\nWhen $ y_0 \\equal{} 3$, the equation reduces to $ 2x_0 \\plus{} z_0 \\equal{} 1000$, giving $ 499$ solutions from $ (1,998)$ to $ (499, 2)$.\n\nIn all, $ m \\minus{} n \\equal{} 501 \\plus{} 0 \\plus{} 499 \\equal{} 1000$, and when divided by 1000, the remainder is $ \\boxed{0}$.\n[/hide]\r\n\r\nReally?, or so I thought.", "Solution_2": "I did not get that i did a super brute force way and i got $ \\frac{2125}{1000}$\r\nwhich would be $ 3 \\frac{125}{1000}$ or $ \\boxed{125}$\r\n\r\nbtw i think i got a 5... :(", "Solution_3": "[hide] I got something like 84336-83834=502.[/hide]", "Solution_4": "I also got 501+499=1000 so 000...\r\nIf it is 000, then :D \r\nI think it will be the first time an answer on the AIME is 0.", "Solution_5": "nooooooooooooooooo\r\n\r\n\r\ni put 125 :mad: PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL :mad: :mad:", "Solution_6": "[quote=\"Smartguy\"]I also got 501+499=1000 so 000...\nIf it is 000, then :D \nI think it will be the first time an answer on the AIME is 0.[/quote]\r\n\r\nNope, 000 has already been done once before. (As has 001)", "Solution_7": "I believe this was the first problem to have 000 as an answer?", "Solution_8": "You believe wrong :P : [[2000 AIME II Problems/Problem 9]]", "Solution_9": "i wasted so much time checking this problem... what a mean trick (the answer being 000)", "Solution_10": "April Fools :wink: :P", "Solution_11": "[quote=\"tinytim\"]April Fools :wink: :P[/quote]\r\n\r\nlol thats must have been what they wanted", "Solution_12": "[quote=\"funtwo\"]i wasted so much time checking this problem... what a mean trick (the answer being 000)[/quote]\r\nWhat do you mean? A good solution would've made it obvious.", "Solution_13": "He probably means 000 is the kind of answer that makes you sweat and think you've done something wrong even if it's clear, in hindsight, that you didn't.", "Solution_14": "well what i did was only consider 3y=3 and 3y=9 in the first equation and i was trying to find a flaw in this reasoning", "Solution_15": "[hide=\"Another approach\"] \n\nSince the sum of the coefficients is 9, the number of solutions in positive integers to $ 4x\\plus{}3y\\plus{}2z\\equal{}2009$ is the same as the number of solutions to nonnegative integers to $ 4x\\plus{}3y\\plus{}2z\\equal{}2000$. Thus, the desired quantity is equal to the number of solutions in nonnegative integers to $ 4x\\plus{}3y\\plus{}2z\\equal{}2000$ in which at least one of the variables is equal to zero. \n\nThus, we want to count the number of solutions in nonnegative integers to the three equations $ 4x\\plus{}3y\\equal{}2000$, $ 3y\\plus{}2z\\equal{}2000$, and $ 4x\\plus{}2z\\equal{}2000$. The first has $ 167$ solutions, the second has $ 334$ solutions, and the third has $ 501$ solutions. However, we are overcounting by two, since we can have both $ x$ and $ y$ or both $ y$ and $ z$ equal to zero. Therefore, the number of solutions is $ 167\\plus{}334\\plus{}501\\minus{}2\\equal{}\\fbox{000}$\n\n[/hide]", "Solution_16": "I also got 0\r\n1)4x+3y+2z=2009 \r\n2)4x+3y+2z=2000\r\n\r\n1)4x+2z=2009-3y\r\n2)4x+2z=2000-3y\r\n\r\nSo m-n is the total solution of\r\n4x+2z=2006\r\n4x+2z=2003\r\n4x+2z=2000\r\nthat makes 1000 so the answer is 0\r\n\r\nBy the way, what I really wanted to ask was...\r\nHow was the difficulty of AIME2 this year??", "Solution_17": "Why is it the fact that the sum of the coefficients is 9 imply that the number of solutions in positive integers to 4x+3y+2z=2009 is the same as the number of solutions to nonnegative integers to 4x+3y+2z=2000?", "Solution_18": "Every solution $ (x,y,z)$ to the second corresponds to exactly one solution $ (x\\plus{}1,y\\plus{}1,z\\plus{}1)$ to the first.", "Solution_19": "so how do I conclude that? How do you know this?", "Solution_20": "An ordered triple $ (x,y,z)$ solves the second equation if and only if $ (x\\plus{}1,y\\plus{}1,z\\plus{}1)$ solves the first.", "Solution_21": "Wait how do you prove that solutions in this form make up every solution? Couldn't there be solutions \"in another form\"?", "Solution_22": "Let $M=\\{(x,y,z)\\mid4x+3y+2z=2009\\}$ and define $N=\\{(x,y,z)\\mid4x+3y+2z=2000\\}$. Also denote $M'=\\{(x,y,z)\\in M\\mid1\\notin\\{x,y,z\\}\\}$.\n\nSince the function $f:M'\\to N$ defined by $f(x,y,z)=(x-1,y-1,z-1)$ is bijective, we have $|M'|=|N|=n$.\n\nThen the answer is $|M\\setminus M'|\\pmod{1000}$. We count this with four cases:\n\n[b]Case 1:[/b] $x=1,1\\notin\\{y,z\\}$\nThen $3y+2z=2005$. By taking$\\pmod2$ and$\\pmod3$, we have that $y$ is odd and $z\\equiv2\\pmod3$. Letting $y=2b+1$ and $z=3c+2$, the equation transforms into $b+c=333$. For any $1\\le b\\le333$ there is a solution for $c$, so this case has $\\underline{333}$ solutions.\n\nNote that we established a bijection between $3y+2z=2005$ and $b+c=333$.\n\n[b]Case 2:[/b] $y=1,1\\notin\\{x,z\\}$\nThe equation is $2x+z=1003$. Now $z$ is odd, so $z=2c+1$. The new equation is $x+c=501$ which has $\\underline{499}$ solutions since we avoid $x=1$.\n\n[b]Case 3:[/b] $z=1,1\\notin\\{x,y\\}$\nFinally, $4x+3y=2007$, giving $x\\equiv0\\pmod3$ and $y\\equiv1\\pmod4$. With the substitutions $x=3a$ and $y=4b+1$, we obtain $a+b=167$, which has $\\underline{166}$ solutions.\n\n[b]Case 4:[/b] $(x-1)(y-1)+(y-1)(z-1)+(z-1)(x-1)=0$\nWe have the $\\underline2$ solutions $(x,y,z)=(1,1,1001),(501,1,1)$.\n\nOur final answer is $333+499+166+2=1000\\equiv\\boxed{000}$.", "Solution_23": "bruh wait what... in my mock i guesseed this correct.... note that gcd(2,3,4)=12 so it should repeat every once in 12 note that 2009-2000=9<12 so we can guess 0 or 1.\n\ni guessed 0 :D\n(rare problem with answer 0 :D)" } { "Tag": [ "trigonometry", "geometry", "circumcircle", "symmetry", "Euler" ], "Problem": "$x,y,z\\in\\mathbb{R}$ , prove the identity\r\n\r\n(i)$\\sin x+\\sin y +\\sin z - \\sin (x+y+z)=4\\sin\\left(\\frac{x+y}{2}\\right)\\sin\\left(\\frac{x+z}{2}\\right)\\sin\\left(\\frac{y+z}{2}\\right)$\r\n\r\n(ii)$\\cos x+\\cos y+\\cos z -\\cos (x+y+z)=4\\cos\\left(\\frac{x+y}{2}\\right)\\cos\\left(\\frac{x+z}{2}\\right)\\cos\\left(\\frac{y+z}{2}\\right)$\r\n\r\n[hide=\"Remark\"] specially when $x+y+z=\\pi$ , we have $\\sin x+\\sin y +\\sin z=4\\cos\\frac{x}{2}\\cos\\frac{y}{2}\\cos\\frac{z}{2}$ and $\\cos x+\\cos y+\\cos z=4\\sin\\frac{x}{2}\\sin\\frac{y}{2}\\sin\\frac{z}{2}-1$[/hide]\r\n\r\n :)", "Solution_1": "[b]A proof for 1) ( Not very interesting though)[/b]\r\nRegarding y and z as fixed, and \r\nlet $f(x)=\\sin x+\\sin y+\\sin z-\\sin (x+y+z)-4 \\sin(\\frac{x+y}2) \\sin(\\frac{x+z}2) \\sin(\\frac{y+z}2)$.\r\nThen \r\n$f'(x)=\\cos x-\\cos(x+y+z)-4[\\frac12 \\cos(\\frac{x+y}2) \\sin(\\frac{z+x}2)+\\frac 12 \\sin(\\frac{x+y}2)\\cos(\\frac{x+z}2)]\\sin(\\frac{y+z}2)$\r\n$=\\cos x-\\cos(x+y+z)-2\\sin(x+\\frac{y+z}2)\\sin(\\frac{y+z}2)$ [i]by compand angle formula[/i]\r\n$=\\cos x-\\cos(x+y+z)+[\\cos(x+y+z)-\\cos x]$[i] by product-to-sum formula[/i]\r\n$=0$.\r\n\r\n[i]It shows that $f(x)=constant$ for any real value of x.[/i]\r\n\r\nChoose $x=-y$ and put it into f(x),\r\n$f(-y)=\\sin(-y) + \\sin y + \\sin z - \\sin z -4\\sin{0} \\cdot \\sin({\\frac{-y+z}2})\\sin(\\frac{y+z}2)=0$\r\n\r\nHence, $f(x)=0$ for all real values of x and the result follows.\r\n\r\nI think the second identity can be proved similiarly. :) [b]\n\n[i]But what is the significance of these two identities?[/i][/b]", "Solution_2": "If $x, y, z$ are the angles of a triangle, $x + y + z = \\pi$ and the indentities:\r\n\r\n$\\sin x+\\sin y +\\sin z=4\\cos\\frac{x}{2}\\cos\\frac{y}{2}\\cos\\frac{z}{2}$\r\n$\\cos x+\\cos y+\\cos z=4\\sin\\frac{x}{2}\\sin\\frac{y}{2}\\sin\\frac{z}{2} - 1$\r\n\r\nare so incredibly useful if you start using trigonometry to solve a geometry question. Here in the UK IMO squad, this is known as 'Trig Bashing'. If you delve into this technique you want to describe a triangle completely in terms of $R, \\alpha, \\beta, \\gamma$ (the circumradius, and angles of the triangle). However, once the configuartion gets more complicated you find its far easier to use multiplicative trig terms rather than additive ones, and these two identities are priceless tools to have.\r\n\r\nHowever if you find using trig on a geometry question horrible and never intend on using it, these identities will probably just be some nice curious fact rather than anything useful.\r\n\r\nThe other useful one is $\\tan x + \\tan y + \\tan z = \\tan x\\tan y\\tan z$.", "Solution_3": "Thanks for your explanation.\r\nDo you mind posting some related problems to illustrate the application of these identities?", "Solution_4": "Basically say you wanted to find $\\frac{r}{s}$. It is far more useful ito use multiplicative terms than additive terms. $s = 4R \\cos\\frac{A}{2} \\cos\\frac{B}{2} \\cos\\frac{C}{2}$ is far better than $s = R(\\sin A + \\sin B + \\sin C)$. Then as $\\triangle = rs = 2R^2 \\sin A \\sin B \\sin C$, double-angle formula easily tells you than $r = 4R \\sin\\frac{A}{2} \\sin\\frac{B}{2} \\sin\\frac{C}{2}$. Note how multiplicative terms allow you to keep symmetry without really thinking.\r\n\r\nSo $\\frac{r}{s} = \\tan\\frac{A}{2} \\tan\\frac{B}{2} \\tan\\frac{C}{2}$.\r\n\r\nAnother good use is if you wanted to prove the Euler line by areal co-ordinates. To do this you need normalised co-ordinates - the co-ordinates sum to 1. Without these identities, the algebra would be horrendous. With them, they're quite tame. Try it out.\r\n\r\n[hide=\"Cilck For A Hint (The Unnormalised Co-ordinates)\"]$O = (\\sin 2A, \\sin 2B, \\sin 2C)$\n$G = (1, 1, 1)$\n$H = (\\tan A, \\tan B, \\tan C)$[/hide]", "Solution_5": "You can expand by using the identities\r\n\\[ \\cos \\theta = \\frac{e^{i \\theta} + e^{-i \\theta}}{2}, \\quad \\sin \\theta = \\frac{e^{i \\theta} - e^{-i \\theta}}{2i}. \\]\r\n\r\n[quote=\"shyong\"]$x,y,z\\in\\mathbb{R}$ , prove the identity\n\n(i)$\\sin x+\\sin y +\\sin z - \\sin (x+y+z)=4\\sin\\left(\\frac{x+y}{2}\\right)\\sin\\left(\\frac{x+z}{2}\\right)\\sin\\left(\\frac{y+z}{2}\\right)$\n\n(ii)$\\cos x+\\cos y+\\cos z -\\cos (x+y+z)=4\\cos\\left(\\frac{x+y}{2}\\right)\\cos\\left(\\frac{x+z}{2}\\right)\\cos\\left(\\frac{y+z}{2}\\right)$\n[/quote]" } { "Tag": [ "Asymptote", "\\/closed" ], "Problem": "When I click on an Asy figure (to examine the code) a blank window pops up but no code is there. \r\n\r\nFF 3.0.8 (i think, i was prompted for 3.0.9 today so...) on OS X 10.5 Leopard.", "Solution_1": "I'm having the same problem on Google Chrome (Beta) on Windows XP. :maybe:", "Solution_2": "FF 3.0.9 on Windows XP fails as well.\r\n\r\nI think they may have not coded this yet.", "Solution_3": "Yeah I didn't have the time to adapt the new code popup to asymptote.", "Solution_4": "Thanks for making the new popup! :lol:", "Solution_5": "[quote=\"ThinkFlow\"]Thanks for making the new popup! :lol:[/quote]You're welcome. We're going to have a similar one for asymptote soon." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Show that ;) \r\n \\[ x ,y \\geq 0 \\]\r\n\\[ \\frac{x}{x^4+y^2} + \\frac{y}{x^2+y^4} \\leq \\frac{1}{xy} \\]", "Solution_1": "just simple AM-GM\r\n$x^4+y^2\\geq 2x^2 y$ :)", "Solution_2": ":| :o Huh! What an easy question!!!\r\n\r\n\r\nThank you!" } { "Tag": [], "Problem": "The sum of the first N positive integers is 2006. Wait, that is impossible...one of those integers must not have been included! If in fact exactly one of the first N positive integers was not included in the sum, which integer must that have been ?\r\n\r\nThanks", "Solution_1": "[hide] You find the triangle number >2006. This number is 2016.((63*64)/2). Since this is ten more than 2006 the missing number is 10.[/hide]", "Solution_2": "Just to clarify, for anyone who may not know what a triangular number is (at least not by that name), it's the sum of the first $n$ numbers, i.e. $\\frac{n(n+1)}{2}$, where $n$ is an integer. So, the first number of this form that is $>2006$ is $2016$, for $n=63$, and you can solve from there easily." } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "Can someone proofread my proof to see if it is in good proof format and correct? If this were a USAMO problem ( Either the USAMO is easy or I'm a genius! :mrgreen: JK ) What would my score be?\r\n\r\n This is a question from the AoPS Vol. 1: The Basics - Ch 28. Prove It, p338, exercise 28-3. Prove that if a, b, c, > 0, then if ax :^2: + bx +c = 0 has real solutions both solutions are negative. The solutions of x are ( -b :pm: :sqrt: b :^2: - 4ac ) / 2a. Therefore, b :^2: -4ac must be positive in order for :sqrt: b :^2: -4ac to be real. Thus, b :^2: > 4ac. But, since 4ac is subtracted from b :^2: , :sqrt: b :^2: -4ac is less than the abs(-b). Thus, the isgn of the numerator -b+ :sqrt: b :^2: -4ac is negative and -b- :sqrt: b :^2: -4ac is negative as well while the denominator 2a is positive. Therefore, both (-b + :sqrt: b :^2: -4ac)/2a and (-b- :sqrt: b :^2: -4ac)/2a are both negative solutions of x.\r\n\r\nMy proof is different and unfortunately more complicated than the solution in the book.", "Solution_1": "Both proofs are correct. The first could use some \"neatening up,\" but all the necessary mathematics is in there, so I would expect it to garner full points. The second one is more elegant, though :)", "Solution_2": "Yeah. The second was the book's solution which was a whole lot quicker. I just went with what came to my mind first.", "Solution_3": "This is how I would write your proof using your method (I would, however, use rep123max's methods instead).\r\n\r\nWe have ax^2+bx+c=0 with a,b,c > 0\r\n\r\nThis implies x = (-b :pm: :sqrt: b^2 - 4ac)/2a\r\n\r\nHowever, b^2-4ac < b^2 because a,c > 0\r\n\r\nSo (-b :pm: :sqrt: b^2 - 4ac) < 0 and 2a > 0,\r\nand x < 0, as desired.", "Solution_4": "This is also a good way of proving it:\r\n\r\nIf f(x) = ax^2 + bx + c, with a,b,c > 0, then we can rewrite the equation as:\r\n\r\nf(x) = a(x-p)(x-q) = ax^2 - (p+q)x + pq\r\n\r\nb = -(p+q)\r\nc = pq\r\n\r\nSince c,b > 0, p,q < 0 as desired." } { "Tag": [ "function", "calculus", "integration", "limit", "inequalities", "derivative", "real analysis" ], "Problem": "Assume that f is a function belong to $C^{2}$, f(0)=0. Prove that:(assume that all the intergral are finite)\r\n$\\int_{0}^{\\infty}{f(x)^{2}dx}$.$\\int_{0}^{\\infty}{f''(x)^{2}dx}$>\r\n$(\\int_{0}^{\\infty}{f'(x)^{2}dx})^{2}$\r\nCan anyone solve it? :D", "Solution_1": "By integration by parts,\r\n\r\n$\\int_{0}^{b}f'(x)^{2}\\,dx=f(b)f'(b)-\\int_{0}^{b}f(x)f''(x)\\,dx$\r\n\r\nIf we can argue that $\\lim_{b\\to\\infty}f(b)f'(b)=0,$ then the desired inequality will follow from the Cauchy-Schwarz inequality.\r\n\r\nBut $f(b)f'(b)=-\\int_{b}^{\\infty}f(x)f''(x)+f'(x)^{2}\\,dx$ and the presumed existence of the improper integrals shows that this tends to zero as $b\\to\\infty.$", "Solution_2": "[quote]$\\lim_{b\\to\\infty}f(b)f'(b)=0$[/quote]\r\nThe function f is not continous so I think it is not necessary to exist the limit of f(x).f'(x).\r\nDo you have another solution?", "Solution_3": "You gave $f\\in C^{2},$ so yes, it and its derivative are continuous.\r\n\r\nThat alone would not be enough to show that the limit is zero, but look at my last line:\r\n\r\n$|f(b)f'(b)|\\le\\int_{b}^{\\infty}|f(x)f''(x)|\\,dx+\\int_{b}^{\\infty}f'(x)^{2}\\,dx$\r\n\r\nBut the fact that $\\int_{0}^{\\infty}f'(x)^{2}\\,dx<\\infty$ implies that \r\n\r\n$\\lim_{b\\to\\infty}\\int_{b}^{\\infty}f'(x)^{2}\\,dx=0,$ and\r\n\r\n$\\int_{b}^{\\infty}|f(x)f''(x)|\\,dx\\le\\left( \\int_{b}^{\\infty}f(x)^{2}\\,dx\\right)^{\\frac12}+\\left(\\int_{b}^{\\infty}f''(x)^{2}\\,dx\\right)^{\\frac12}$\r\n\r\nand a similar argument shows that that goes to zero as $b\\to\\infty.$" } { "Tag": [ "trigonometry", "geometry", "rotation", "perpendicular bisector", "algebra", "binomial theorem" ], "Problem": "Put solutions for the USC test here. [url=http://www.math.sc.edu/contest/problems.html]Here's[/url] the link to the problems.", "Solution_1": "I just took the 1987 test. Here are the solutions to the second half. \r\n\r\n [hide=\"16\"]Notice that $\\tan{90-\\theta}=\\cot\\theta$ and that $\\tan\\theta \\cdot \\cot\\theta=1$. Therefore, the product evaluates to simply $1$\n[/hide]\n[hide=\"17\"] There are $24$ numbers for each digit in which that digit is the first digit of the number and similarly there are $24$ numbers for each digit in which that digit is the second digit of the number and so on. Also, $1+2+3+4+5=15$ Our sum thus evaluates to:\n\n$24 \\cdot 15(10^{5}+10^{4}+10^{3}+10^{2}+10+1)=3,999,960$[/hide]\n\n[hide=\"18\"]Draw the diameter such that it is a perpendicular bisector of both chords. Let the length of the diameter be $y+x+1$ where $y>x$. Through power of point, we have:\n\n$16=y(x+1)$\n$9=x(y+1)$\n\nSolving this, we have that $x=1$ and $y=8$ and the length of the diameter is $10$.[/hide]\n\n[hide=\"19\"] $\\tan{30}=\\frac{\\sqrt{3}}{3}$. Now we find $\\arccos(\\frac{\\sqrt{3}}{3})$ so construct a right triangle with height $\\sqrt{3}$ and hypotenuse $3$. The $\\sin$ of this triangle is $\\frac{\\sqrt{6}}{3}$ which is the first choice. [/hide]\n\n[hide=\"20\"]If a number is divisible by $18$, it\u2019s divisible by $9$ and $2$, therefore, $2 | d$ and $9 | 15+2d \\Longrightarrow 9 | 6+2d$. The answer is therefore $d=6$. \n[/hide]\n[hide=\"21\"]Letting $x=0$, we have $f(1)=2f(0)+1$, so $f(1)=5$. Letting $x=1$, we have $f(3)=2f(1)+1$, so $f(3)=11$\n[/hide]\n[hide=\"22\"]Rotating the smaller square by it\u2019s center does not affect the area overlap. Therefore, rotate the small square so that each of it\u2019s sides is parallel to the bigger square\u2019s. It\u2019s clear that the area overlap is a fourth of the smaller square. The answer is $\\frac{25}{4}$[/hide]\n\n[hide=\"23\"]Rewrite the sum as $(10^{4}+20)^{\\frac{1}{2}}-(10^{4}+10)^{\\frac{1}{2}}$ Using extended binomial theorem, we evaluate the expression as:\n\n$(\\binom{1/2}{0}10^{2}+\\binom{1/2}{1}\\frac{20}{10^{2}}+...)-(\\binom{1/2}{0}10^{2}+\\binom{1/2}{1}\\frac{10}{10^{2}}+... )\\approx \\frac{1}{20}$ but still a bit less than that. The answer therefore is $(e)$.[/hide]\n\n[hide=\"24\"]The fixed distance between the segments should give you a hint they are talking about foci, and thus an ellipse. \n[/hide]\n[hide=\"25\"]Let the two equal angles be $x$. We have $BD^{2}=5^{2}+2^{2}-2(5)(2)(\\cos{x})=4^{2}+4^{2}-2(4)(4)(\\cos{x}) \\Longrightarrow \\cos{x}=\\frac{1}{4}$. Therefore, $BD=2\\sqrt{6}$. \n[/hide]\n[hide=\"26\"]The units digit of $3^{456789}=3^{1}=3$ and $2^{3}=8$. [/hide]\n\n[hide=\"27\"]Let $P(x)=x^{100}-2x^{99}+4$. The roots of $x^{2}-3x+2$ are $x=2$ and $x=1$. Therefore $P(x)=(x-1)(x-2)n+Q(x)$ where $Q(x)$ is our remainder. $P(1)=Q(1)=3$ and $P(2)=Q(2)=4$. Since $Q(x)$ is linear, we have $a+b=3$ and $2a+b=4$ giving us $a=1$ and $b=2$ and our remainder is thus $x+2$. [/hide]\n\n[hide=\"28\"]We must have $x_{1}+x_{2}\\equiv y_{1}+y_{2}\\equiv 1 \\mod{2}$ Looking at $\\mod{2}$ for our set, the biggest set that doesn\u2019t have every pair is a set of $4$. [/hide]\n\n[hide=\"29\"]Let the roots be $a,b,c$. We have that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ac)=4-2(-2)=8$[/hide]\n\n[hide=\"30\"]The expression evaluates to $\\frac{2\\sin{25}\\cos{-15}}{2\\cos{25}\\cos{-15}}=\\tan{25}$.[/hide]\r\n\r\nI'll post the first half later..." } { "Tag": [ "induction", "algebra", "polynomial", "linear algebra" ], "Problem": "Consider $ B\\equal{}\\left(\\begin{array}{cc}1 & 3\\\\ \\ 0 & 1\\end{array}\\right).$ Compute $ B^n, n \\in \\mathbb{N}^*.$", "Solution_1": "It is very easy to prove by induction that\r\n\r\n$ B^n \\equal{} \\left[\\begin{array}{cc}1 & 3n\\\\ 0 & 1\\end{array}\\right]$.", "Solution_2": "$ \\text{Null Potent}\\Longrightarrow\\text{ Binominal Theorem}$", "Solution_3": "[quote=\"moldovan\"]Consider $ B \\equal{} \\left(\\begin{array}{cc}1 & 3 \\\\\n\\ 0 & 1\\end{array}\\right).$ Compute $ B^n, n \\in \\mathbb{N}^*.$[/quote]\r\n\r\n$ B \\equal{} \\left(\\begin{array}{cc}1 & 3 \\\\\r\n\\ 0 & 1\\end{array}\\right)\\equal{}\\left(\\begin{array}{cc}1 & 0 \\\\\r\n\\ 0 & 1\\end{array}\\right)\\plus{}\\left(\\begin{array}{cc}0 & 3 \\\\\r\n\\ 0 & 0\\end{array}\\right)\\equal{}I\\plus{}A$ say\r\n\r\nNote $ \\ A^2\\equal{}O\\implies A^k\\equal{}O\\ \\forall k\\equal{}2,3,4,\\dots$\r\n\r\n$ \\implies \\boxed{\\boxed{\\ B^n\\equal{}(I\\plus{}A)^n\\equal{}I\\plus{}nA\\equal{}\\left(\\begin{array}{cc}1 & 0 \\\\\r\n\\ 0 & 1\\end{array}\\right)\\plus{}n\\left(\\begin{array}{cc}0 & 3 \\\\\r\n\\ 0 & 0\\end{array}\\right)\\equal{}\\left(\\begin{array}{cc}1 & 3n \\\\\r\n\\ 0 & 1\\end{array}\\right)}}$\r\n :lol:", "Solution_4": "The tangent line of $ y = x^n\\ (n\\geq 2)$ at $ x = 1$ is $ y = n(x - 1) + 1$, so we can write $ x^n = (x - 1)^2Q(x) + n(x - 1) + 1$ where $ Q(x)$ is a monic polynomial with degree $ n\\geq 2.$\r\n\r\nBy Caley-Hamilton theorem, we have $ B^2 - 2B + E = \\mathbb{O}\\Longleftrightarrow (B - E)^2 = \\mathbb{O}$,\r\n\r\n$ \\therefore B^n = (B - E)^2Q(B) + nB - (n - 1)E$, yielding $ B^n = nB - (n - 1)E\\ (n\\geq 0).$", "Solution_5": "By Calculation of $ B^2,\\ B^3,\\ \\cdots$, we can conjecture $ B^n\\equal{}\\left(\r\n\\begin{array}{cc}\r\n1 & b_n \\\\\r\n0 & 1\r\n\\end{array}\r\n\\right)$\r\n\r\n$ B^{n\\plus{}1}\\equal{}B\\cdot B^n\\Longleftrightarrow \\left(\r\n\\begin{array}{cc}\r\n1 & b_{n\\plus{}1} \\\\\r\n0 & 1\r\n\\end{array}\r\n\\right)\\equal{}\\left(\r\n\\begin{array}{cc}\r\n1 & 3 \\\\\r\n0 & 1\r\n\\end{array}\r\n\\right)\\left(\r\n\\begin{array}{cc}\r\n1 & b_n \\\\\r\n0 & 1\r\n\\end{array}\r\n\\right)$\r\n\r\n$ \\Longleftrightarrow b_{n\\plus{}1}\\equal{}b_n\\plus{}3,\\ b_1\\equal{}3$, yielding $ b_n\\equal{}3\\plus{}3(n\\minus{}1)\\equal{}3n$.\r\n\r\n$ \\therefore B^n\\equal{}\\left(\\begin{array}{cc}\r\n1 & 3n \\\\\r\n0 & 1\r\n\\end{array}\r\n\\right)\\ (n\\geq 0)$", "Solution_6": "[quote=\"kunny\"]The tangent line of $ y = x^n\\ (n\\geq 2)$ at $ x = 1$ is $ y = n(x - 1) + 1$, so we can write $ x^n = (x - 1)^2Q(x) + n(x - 1) + 1$ where $ Q(x)$ is a monic polynomial with [color=red]degree $ n-2.$[/color]\n\nBy Caley-Hamilton theorem, we have $ B^2 - 2B + E = \\mathbb{O}\\Longleftrightarrow (B - E)^2 = \\mathbb{O}$,\n\n$ \\therefore B^n = (B - E)^2Q(B) + nB - (n - 1)E$, yielding $ B^n = nB - (n - 1)E\\ (n\\geq 0).$[/quote]\r\n\r\nExcellent!!\r\n\r\nI think a typo error only\r\n\r\nYour second solution is also good\r\n\r\nthanks for sharing ideas.", "Solution_7": "Thank you! :lol:", "Solution_8": "[quote=\"kunny\"]The tangent line of $ y = x^n\\ (n\\geq 2)$ at $ x = 1$ is $ y = n(x - 1) + 1$, so we can write $ x^n = (x - 1)^2Q(x) + n(x - 1) + 1$ where $ Q(x)$ is a monic polynomial with degree $ n\\geq 2.$\n\nBy Caley-Hamilton theorem, we have $ B^2 - 2B + E = \\mathbb{O}\\Longleftrightarrow (B - E)^2 = \\mathbb{O}$,\n\n$ \\therefore B^n = (B - E)^2Q(B) + nB - (n - 1)E$, yielding $ B^n = nB - (n - 1)E\\ (n\\geq 0).$[/quote]\r\n\r\nI don't understand this solution; why are we considering the tangent line of $ y = x^n$?", "Solution_9": "If $ y \\equal{} x^n$ touches to the line $ y \\equal{} ax \\plus{} b$ at $ x \\equal{} 1$, then since the quation $ x^n \\minus{} (ax \\plus{} b) \\equal{} 0$ has a double root $ x \\equal{} 1$, we can write $ x^n \\minus{} (ax \\plus{} b) \\equal{} (x \\minus{} 1)^2Q(x)$ where $ Q(x)$ is a monic polynomial with degree $ n \\minus{} 2$. That is to say, $ y \\equal{} ax \\plus{} b$ is the tangent line.", "Solution_10": "Ok, but what is the connection of $ y \\equal{} x^n$ to the characteristic equation of $ B$, which is $ (\\lambda \\minus{} 1)^2 \\equal{} 0$? Is it because $ B \\equiv x$ due to the Cayley-Hamilton theorem?", "Solution_11": "By Cayley-Hamilton theorem, $ (B\\minus{}I)^2 \\equal{} O$. Now, kunny has considered the polynomial\r\n\r\n$ x^n \\equal{} (x\\minus{}1)^2 Q(x) \\plus{} n(x\\minus{}1) \\plus{} 1$.\r\n\r\nTherefore, evaluating this polynomial for $ x \\equal{} B$,\r\n\r\n$ B^n \\equal{} (B\\minus{}I)^2Q(B) \\plus{} n(B\\minus{}I) \\plus{} I \\equal{} n(B\\minus{}I) \\plus{} I$.", "Solution_12": "Is this to say that variables and matrices are completely interchangeable in this context?\r\n\r\n[quote=\"kunny\"]If $ y \\equal{} x^n$ touches to the line $ y \\equal{} ax \\plus{} b$ at $ x \\equal{} 1$, then since the quation $ x^n \\minus{} (ax \\plus{} b) \\equal{} 0$ has a double root $ x \\equal{} 1$, we can write $ x^n \\minus{} (ax \\plus{} b) \\equal{} (x \\minus{} 1)^2Q(x)$ where $ Q(x)$ is a monic polynomial with degree $ n \\minus{} 2$. That is to say, $ y \\equal{} ax \\plus{} b$ is the tangent line.[/quote]\r\nAlso, how did you get that $ a \\equal{} n$ and $ b \\equal{} 1 \\minus{} n$ here?", "Solution_13": "[quote=\"Carcul\"]By Cayley-Hamilton theorem, $ (B \\minus{} I)^2 \\equal{} O$. Now, kunny has considered the polynomial\n\n$ x^n \\equal{} (x \\minus{} 1)^2 Q(x) \\plus{} n(x \\minus{} 1) \\plus{} 1$.[/quote]\r\n\r\nHow was this polynomial obtained?" } { "Tag": [ "probability", "function", "combinatorics solved", "combinatorics" ], "Problem": "Suppose we play a variant of Yahtzee where we roll 5 dice, trying to get all sixes. After each roll we keep the dice that are sixes and re-roll all the others. What is the probability of getting all sixes in exactly k rolls as a function of k?", "Solution_1": "My 1st post.\r\nWe can consider this problem with respect to each die.\r\nThe probability that one die will get a 6 somewhere before or at exact k times is 1/6+5/6x1/6+...+(5/6)^(k-1)x1/6=1-(5/6)^k\r\nSo the probability that all five dice will roll a 6 before or at exact k times is \r\n(1-(5/6)^k)^5\r\nThus, the probability to get all 6 at exact k times is \r\n(1-(5/6)^k)^5-(1-(5/6)^(k-1))^5", "Solution_2": "Very good! And fast, too!", "Solution_3": "good mathematical model for artificial selection, btw. i think dawkins has used something analogous to this." } { "Tag": [ "function", "inequalities", "algebra proposed", "algebra" ], "Problem": "Let $ f(x)\\minus{}x,f(x)\\minus{}x^3$ are strict increase function on $ R$. Prove that: $ f(x)\\minus{}\\frac{\\sqrt{3}x^2}{2}$ is strict increase function too.", "Solution_1": "[quote=\"tdl\"]Let $ f(x) \\minus{} x,f(x) \\minus{} x^3$ are strict increase function on $ R$. Prove that: $ f(x) \\minus{} \\frac {\\sqrt {3}x^2}{2}$ is strict increase function too.[/quote]\r\n\r\nConsider two real numbers $ x$ and $ y$ with $ x>y$.\r\nThe two conditions yield $ f(x)\\minus{}f(y)>x^3\\minus{}y^3>0$ and $ f(x)\\minus{}f(y)>x\\minus{}y>0$.\r\nMultiplying these two inequalities, applying $ (x^3\\minus{}y^3)(x\\minus{}y)>\\frac34(x^2\\minus{}y^2)^2$, \r\nand taking the square-root yields the statement." } { "Tag": [ "geometry", "rectangle", "trigonometry", "integration", "calculus", "function", "3D geometry" ], "Problem": "Solve Laplace's equation inside a rectangle:\r\n\r\n$ \\nabla ^2 u \\equal{} \\frac{\\partial^2 u}{\\partial x^2} \\plus{} \\frac{\\partial^2 u}{\\partial y^2} \\equal{} 0$\r\n\r\nsubject to the boundary conditions\r\n\r\n$ u(0,y) \\equal{} g(y)$\r\n$ u(L,y) \\equal{} 0$\r\n\r\n$ u(x,0) \\equal{} 0$\r\n$ u(x,H) \\equal{} 0.$", "Solution_1": "Suppose a solution of the form $ u(x,y) = X(x)Y(y)$ then\r\n\\[ \\frac {X''}{X} = - \\frac {Y''}{Y} = \\lambda^2\r\n\\]\r\n\r\n\\[ \\Rightarrow \\begin{cases} X'' - \\lambda^2 X = 0 \\\\\r\nY'' + \\lambda^2 Y = 0 \\end{cases}\r\n\\]\r\n\r\n\\[ \\Rightarrow \\begin{cases} X = c_1 e^{\\lambda x} + c_2e^{ - \\lambda x} \\\\\r\nY = c_3 \\sin \\lambda y + c_4 \\cos \\lambda y \\end{cases}\r\n\\]\r\nThe boundary conditions $ u(x,0) = 0$ and $ u(x,H) = 0$ demand $ Y(0) = Y(H) = 0$, so $ c_4 = 0$ and $ c_3 \\sin \\lambda H = 0$. Thus we can have nontrivial solutions only if $ \\lambda = \\frac {n \\pi}{H}$, $ n\\in \\mathbb{N}$. Also, $ u(L,y) = 0$ requires $ X(L) = 0$ and hence $ c_2 = - e^{2\\lambda L}c_1$, so\r\n\\[ \\begin{cases} X = e^{\\frac {n\\pi x}{H}} - e^{\\frac {n\\pi (2L - x)}{H}} \\\\\r\nY = c_3 \\sin \\frac {n\\pi y}{H} \\end{cases}\r\n\\]\r\nHence the general solution must satisfy\r\n\\[ u(x,y) = \\sum_{n = 1}^{\\infty}{b_n\\left(e^{\\frac {n\\pi x}{H}} - e^{\\frac {n\\pi (2L - x)}{H}}\\right)\\sin \\frac {n\\pi y}{H}}\r\n\\]\r\nfor some constants $ \\lbrace b_n \\rbrace_{n\\in \\mathbb{N}} \\subset \\mathbb {R}$. To satisfy the boundary condition $ u(0,y) = g(y)$ we must have\r\n\\[ \\sum_{n = 1}^{\\infty}{b_n\\left(1 - e^{\\frac {2 n \\pi L}{H}\\right)\\sin \\frac {n\\pi y}{H}} = g(y)\r\n}\\]\r\nMultiplying by $ \\sin\\frac {n\\pi y}{H}$ and integrating we have by orthogonality:\r\n\\[ b_n\\left(1 - e^{\\frac {2 n \\pi L}{H}\\right)\\int_0^H{\\sin^2 \\frac {n\\pi y}{H}dy} = b_n\\left(1 - e^{\\frac {2 n \\pi L}{H}\\right)\\frac {H}{2} = \\int_0^H{g(y)\\sin \\frac {n\\pi y}{H}dy}\r\n}}\\]\r\nand hence we must have $ b_n = \\frac {2}{H\\left(1 - e^{\\frac {2 n \\pi L}{H}\\right)}}\\int_0^H{g(y)\\sin \\frac {n\\pi y}{H}dy}$\r\nSo our final solution is\r\n\\[ \\boxed{u(x,y) = \\frac {2}{H}\\sum_{n = 1}^{\\infty}{\\frac {\\left(e^{\\frac {n\\pi x}{H}} - e^{\\frac {n\\pi (2L - x)}{H}}\\right)}{\\left(1 - e^{\\frac {2 n \\pi L}{H}\\right)}}\\sin \\frac {n\\pi y}{H}\\int_0^H{g(t)\\sin \\frac {n\\pi t}{H}dt}}}\r\n\\]\r\nOf course $ g$ needs to be sufficiently well behaved so the above converges, but the requirements are quite lax (e.g. if $ g\\in L^p\\left(\\left(0,H\\right)\\right)$ is continuous it converges a.e. if $ p > 1$)", "Solution_2": "This is a fairly standard textbook question, if the textbook is for a first course in PDE. Joe Blow has given the standard answer, as that textbook would show it.\r\n\r\nBut why would mathwizarddude post such a question in the first place? Given the rapid and indiscriminate nature of his posting, I doubt that it's something he's thought long and hard about. I would like to see him slow down the rate at which he starts new topics. If topics from mathwizarddude begin disappearing, you'll know why.", "Solution_3": "I apologize for any inconvience. Please bear with me as I don't have much mathematical background but merely an 8th grader just graduated from middle school. I'm a self-taught math enthusiast. I have taught myself calculus (not advanced) and am teaching myself linear algebra (am almost finish), complex analysis, and a little bit of real analysis.\r\n\r\nAnyway, although I should be able to solve this standard separation of variables problem, I post this because I want to see if there's a different technique of solving this that is different my PDE textbook that I am currently using (as my PDE text is not advanced). Also, if someone does post a solution that is the same as my text, I would want to take that opportunity to ask the reason for arbitrarily choosing $ u(x,y)$ to has the form $ X(x)Y(y)$. Moreover, although we could come up with a valid solution in that particular form, how would one prove that there's no other solution $ u(x,y)$ satisfying all the constraints but is not in the form $ X(x)Y(y)$ and thus not losing any generality in our solution.\r\n\r\n(However, as PDE is different from ODE, perhaps one can stop as long as one comes up with a valid solution, though not a general one perhaps, [i]I think[/i].)", "Solution_4": "Those are good questions. The truth is, mathematically there isn't very much motivation for looking for a solution of the form $ u(x,y) \\equal{} X(x)Y(y)$. A physicist might argue that there are reasons solutions 'ought' to be of this form (separable), but from a mathematician's point of view this ansatz is made simply because prior experience has taught us that many PDEs from physics have solutions in this form, and it's fairly easy to check if it works.\r\n\r\nFor this particular equation, however, it turns out that if you find [i]any[/i] solution to the boundary value problem, you have also found [i]the[/i] solution to that problem, i.e. the form of the PDE itself guarantees at most one solution. The proof is as follows:\r\n\r\nTake any two solutions $ v_1(x,y), v_2(x,y)\\in C^2(U)\\cap C^0(\\overline{U})$ of Laplace's equation in a bounded open region $ U$ with boundary $ \\partial U$, i.e.\r\n\\[ \\begin{cases} \\nabla^2 u(x,y) \\equal{} 0 & (x,y)\\in U \\\\\r\nu(x,y) \\equal{} g(x,y) & (x,y)\\in \\partial U \\end{cases}\r\n\\]\r\nNow the difference $ v \\equal{} v_1 \\minus{} v_2$ must satisfy\r\n\\[ \\begin{cases} \\nabla^2 v(x,y) \\equal{} 0 & (x,y)\\in U \\\\\r\nv(x,y) \\equal{} 0 & (x,y)\\in \\partial U \\end{cases}\r\n\\]\r\nHowever, since $ v$ is harmonic, the maximum principle for harmonic functions asserts that both its maximum and its minimum are obtained on the boundary of $ U$. Therefore, $ v \\equal{} 0$ and hence $ v_1 \\equal{} v_2$ in $ \\overline{U}$, so there is at most one solution to the boundary value problem. [hide=\"Remark\"]The argument above also immediately proves the uniqueness of solutions to Poisson's equation $ \\nabla^2 u(x,y) \\equal{} f(x,y)$.[/hide] \r\n\r\nNote that we did not assume a whole lot about $ U$ to prove uniqueness. Though existence is in general a somewhat more complicated issue (and depends much more intimately on the specific geometry of $ U$), on the rectangle we can prove existence by explicitly constructing a solution as I have in a previous post. So in some sense even though the ansatz seems rather arbitrary, in this case it actually leads to a rigorous proof of both existence and uniqueness. This is a situation that prevails throughout PDE analysis.\r\n\r\nIn a more general PDE setting, questions of existence, uniqueness, regularity, well-posedness, etc. are extremely complicated, and there are many open questions in these areas of ongoing research. As you correctly pointed out, the situation with PDEs is very different from ODEs; even very simple linear PDEs can exhibit nasty pathological behavior (e.g. Laplace's equation in a punctured sphere, Lewy's example of an unsolvable linear PDE with smooth Cauchy conditions, non-uniqueness of homogeneous heat equation(!), etc). For a good overview of theoretical questions in PDEs, I'd recommend [i]Partial Differential Equations[/i] by L. C. Evans. Though before you dive into PDE theory, you'd need a strong foundation in functional analysis and real analysis as background.", "Solution_5": "[quote=\"JoeBlow\"]For a good overview of theoretical questions in PDEs, I'd recommend [i]Partial Differential Equations[/i] by L. C. Evans. Though before you dive into PDE theory, you'd need a strong foundation in functional analysis and real analysis as background.[/quote]\r\n\r\nthis is a great text and i think i may have used it more than any other math text i own. you don't need much (any?) functional analysis to read the first half of the book though (well, maybe a bit for the part on second order elliptic pde), which is what most one semester graduate courses on pde cover.", "Solution_6": "Thank you very much! :D \r\n\r\n(I don't know who rated those posts though..)", "Solution_7": "The kid is in 8th grade. mathwizarddude: Do not touch Evans.\r\n\r\nHere is my best explanation for why the separation of variables technique is a reasonable technique.\r\n\r\nSuppose someone shows you Laplace's equation and asks you to write down as many solutions as possible, without regards for boundary conditions or anything like that. The first thing you would do is look for solutions of a simple form. For example, you might look for polynomial solutions. Or you might look for radially symmetric solutions (i.e. ones that depends on $ r$ but not on $ \\theta$). Or you might look for ones that depends on $ x$ but not on $ y$ (not very interesting). Eventually, you would try ones that are a function of $ x$ times a function of $ y$, the \"separated\" solutions. You would see that these take a relatively simple form.\r\n\r\nNext you would notice that adding any two solutions gives another solution (i.e. Laplace's equation is linear). So now you have tons of solutions at your disposal. In particular, you have all linear combinations of separated solutions, and if you're not antsy about convergence issues, all convergent infinite linear combinations of these separated solutions. That's a lot of solutions.\r\n\r\nNow you worry about boundary conditions, and you notice that adding any two solutions satisfying the \"vertical\" boundary conditions gives a solution still satisfying the vertical boundary conditions. Eventually, you end up doing the same procedure described in JoeBlow's post.\r\n\r\nThis still brings us to the question: Why does this work?\r\n\r\nHere's a slightly more sophisticated way to explain the same old solution, but you need to understand Fourier series first (which is closely related to what you're doing anyway). One nice aspect is explanation generalizes to slightly more complicated problems (and may be discussed in your book).\r\n\r\nOnce you understand Fourier series a bit, you realize that essentially \"any\" (here I'm lying of course) function that satisfies your boundary conditions can be expressed as a Fourier sin series in the $ y$ variable, where the coefficients themselves are functions of $ x$. (This can be tricky to think about at first, so try to mull over why this makes sense.) Then you \"plug\" the Fourier series into Laplaces equations, which gives you 2nd ODEs for the coefficient functions (which are functions of $ x$). Then the rest of your boundary conditions allow you to solve the ODEs. Sorry for the lacking details. If you can successfully execute this technique based only on what I've written here, I'll be very impressed. But the point is that from this point of view, there's no ansatz, just Fourier series, and we *expect* to get to the answer (or if there were no solution, we'd see why). The only problem is that it begs the question: Why use Fourier series?\r\n\r\nNote: Polynomial solutions and rotational solutions to Laplace's equation turn out to be very interesting also. In fact, they're perhaps more interesting than the separated solutions. A general philosophy is that any simple solution is probably an important solution." } { "Tag": [ "geometry", "circumcircle", "projective geometry", "geometry proposed" ], "Problem": "I was told it's from Hong Kong 2003, but I'm not sure.\r\nLooks messy, but it's pretty easy!\r\n\r\nTwo circles $\\gamma_1$ and $\\gamma_2$ meet at $A$ and $B$. Let $r$ be a line through $B$ that meets $\\gamma_1$ at $C$ and $\\gamma_2$ at $D$, such that $B$ is between $C$ and $D$. Let $s$ be the line parallel to $AD$, which is tangent to $\\gamma_1$ in $E$ and has the minimal distance from $AD$. $EA$ meets $\\gamma_2$ in $F$, and let $t$ be the line through $F$ which is tangent to $\\gamma_2$. Prove that:\r\na) $t \\parallel AC$.\r\nb) $r$, $s$ and $t$ are concurrent.", "Solution_1": "For (a):\r\n\r\nWe have $\\angle ECA=\\angle DAF,\\ \\angle AEC=\\angle AFD$, so $\\angle CAE=\\angle ADF$, which is what we want because the angle made by $t$ with $AF$ is the same as $\\angle ADF=\\angle CAE$.\r\n\r\nFor (b), assume $R=r\\cap AE,\\ T=r\\cap t,\\ S=r\\cap s$. We have to show that $T=S$. We have $\\frac {RD}{RS}=\\frac {RA}{RE},\\ \\frac {RC}{RT}=\\frac {RA}{RF}$. We thus need $\\frac{RD}{RC}=\\frac {RF}{RE}$. This is true because the lines $EC,\\ FD$ are parallel (because $\\angle AEC=\\angle AFD$).", "Solution_2": "Yes! :) \r\nYou could also prove point b) in a more \"geometric\" way, which I personally prefer. In fact, both $r\\cap t$ and $r\\cap s$ lie on the circumcircle of $BDE$. You can easily prove this with the angles you already have from point a).", "Solution_3": "I think,it is from Tournaments of towns 2001(I don't know spring or falls)", "Solution_4": "I checked all the problems from TOT 2001 and 2002, but I couldn't find this one. :?", "Solution_5": "Well it is Turnament of the Towns 2002 Autumn problem 5\r\nCircles C1 and C2 meet at A and B. A line through B meets C1 at K and C2 at M. T1 is a tangent to C1 parallel to AM. It meets C1 at Q. The line AQ meets C2 again at R. Show that the tangent T2 to C2 at R is parallel to AK, and that T1 and T2 intersect on the line KM.", "Solution_6": "Oh, thanks! I didn't check well enough! ;)", "Solution_7": "Dear Mathlinkers,\nBE intersected (2) en X.\nAppliying the Reim's theorem, FX//s.\nTf is the tangent to (2) at F.\nAccording to a special case of Pascal's theorem Tf FADBXF, we are done...\nSincerely\nJean-Louis" } { "Tag": [ "inequalities", "LaTeX" ], "Problem": "Problem 1:What value of (x,y) satisfy the following equation:\r\n\r\n (4X^2+6X+4)(4Y^2-12Y+25) = 28\r\n\r\nProblem 2: Prove that if x and y are real numbers, then (X^2+Y^2) / 2 >= XY\r\n\r\np.s. how do i write square roots, squares, division, inequality signs, etc. :? :? :?", "Solution_1": "[quote=\"Ceva Theorem\"]Problem 2: Prove that if x and y are real numbers, then (X^2+Y^2) / 2 >= XY[/quote]\r\n\r\n$x^2+y^2 \\geq 2xy \\implies (x-y)^2 \\geq 0$, always true", "Solution_2": "You can use Latex to write mathematical text (there's a link on this site explaining Latex). Basically, you put $\\$$ signs around Latex code, and it becomes an image. For example, if you put $\\$$ signs around (4X^2+6X+4)(4Y^2-12Y+25) = 28, you see $(4X^2+6X+4)(4Y^2-12Y+25) = 28$.\r\n[hide]Problem 1: Fix $x$ or $y$ and then solve for the other variable. By the way, Ceva Theorem, do $x$ and $y$ have to be real?\nProblem 2: A square of a real number is always greater than or equal to zero, i.e. $(x-y)^2\\geq 0$ for real $x$ and $y$.[/hide]", "Solution_3": "[quote=\"hydro\"][quote=\"Ceva Theorem\"]Problem 2: Prove that if x and y are real numbers, then (X^2+Y^2) / 2 >= XY[/quote]\n\n$x^2+y^2 \\geq 2xy \\implies (x-y)^2 \\geq 0$, always true[/quote]Just a small point, but you're going the wrong way in your implication.", "Solution_4": "Yes, they are real numbers\r\n\r\nhow do i do square roots?", "Solution_5": "[quote=\"Ceva Theorem\"]\r\np.s. how do i write square roots, squares, division, inequality signs, etc. :? :? :?[/quote]\r\n\r\nyou have to put your expressions between two of the sign $. So squares are $x^2$, square roots are $\\sqrt x$, divisions $\\frac{x}{y}$", "Solution_6": "[quote=\"hydro\"][quote=\"Ceva Theorem\"]Problem 2: Prove that if x and y are real numbers, then (X^2+Y^2) / 2 >= XY[/quote]\n\n$x^2+y^2 \\geq 2xy \\implies (x-y)^2 \\geq 0$, always true[/quote]\r\n\r\nyou can also prove this by using AM-GM inequality\r\n :)", "Solution_7": "Actually the proof of AM-GM uses that fact so that would be circular logic. ;) And anyway, AM-GM is only valid for positive reals.\r\n\r\n[hide=\"2\"]\nConsider the inequality $(x-y)^2=x^2-2xy+y^2\\ge 0$ since the square of a real number is always nonnegative. Adding $2xy$ to both sides and dividing by two gives $\\frac{x^2+y^2}{2}\\ge xy$, as claimed. $\\boxed {}$\n[/hide]" } { "Tag": [ "number theory", "number theory unsolved" ], "Problem": "I just started number theory, but i found that if one sums all the numbers coprime to some n greater than 0 and \u0404 I, but less than n - then you divide it by n and multiply it by 2 this equals to \u0444(n) (at least but to n=35). Could someone explain to me why??", "Solution_1": "[hide=\"hint\"]$ (r,n)\\equal{}1\\Rightarrow(n\\minus{}r,n)\\equal{}1$[/hide]", "Solution_2": "Could you please elaborate?", "Solution_3": "The set $ S_n$ of natural numbers coprime to and less than $ n$ has cardinality $ \\varphi(n)$ by definition. Moreover, what cosinator is trying to tell you is that if $ a \\in S_n$ then necessarily $ n \\minus{} a \\in S_n$. This means that we can pair up the elements of $ S_n$ into $ \\frac{ \\varphi(n) }{2}$ pairs, each of which has sum $ n$, which is what you discovered. For example,\r\n\r\n$ S_{12} \\equal{} \\{ 1, 5, 7, 11 \\}$\r\n\r\nand $ 1 \\plus{} 11 \\equal{} 5 \\plus{} 7 \\equal{} 12$.", "Solution_4": "Thanks a lot - you make it so simple!!" } { "Tag": [ "inequalities", "algebra", "polynomial", "inequalities open" ], "Problem": "My friend asks me about the inequality\r\n\r\nProve or disprove that, for all non-negative real numbers $a,b,c$? \\[(a^{3}+b^{3}+c^{3})^{2}\\ge 2(a^{5}b+b^{5}c+c^{5}a)+abc(a^{3}+b^{3}+c^{3})\\] and I believe that (although I haven't any try \\[(a^{3}+b^{3}+c^{3})^{2}\\ge a^{5}b+b^{5}c+c^{5}a+2abc(a^{3}+b^{3}+c^{3})\\]", "Solution_1": "Any idea to solve it? Or using computer to disclaims it? :lol: Where is Arquady? I believe you :D", "Solution_2": "Hm, i think i have a solution for the second. But it's so long.", "Solution_3": "Yes ! I can prove the second inequality but can't do with the first", "Solution_4": "can you post you soln ZEIKII", "Solution_5": "as discussed in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111842]here[/url] the inequality $(a^{3}+b^{3}+c^{3})^{2}\\geq (a+b+c)(a^{4}b+b^{4}c+c^{4}a)$ is true.\r\nI don't have a nice proof of this fact.", "Solution_6": "Ican exprees two expreesion by SOS \r\n and ithing it can be solve by SOS \r\n but it so long and may be tomorrow i post it", "Solution_7": "here is my exprees for the first inq:$LHS = \\sum S_{a}(b-c)^{2}$ \r\n where $S_{a}=7b^{4}-c^{4}-3b^{2}c^{2}+2b^{3}c-2c^{3}b+3a^{3}b+3a^{3}c-3abc(b+c)$\r\n $S_{b}$ and $S_{c}$ are similarly cyclic on a,b,c .\r\n I can check that when $a\\geq c\\geq b$ the inq are true , when $a\\geq b\\geq c$ I can check $S_{a}$ and $S_{c}\\geq 0$ but i can't do any more", "Solution_8": "Hm, i think you'd better not use S.O.S to prove this ineq. It isn't a nice solution and maybe very long. My solution haven't used S.O.S", "Solution_9": "So you can poat your soln Zeikii??? Thanks", "Solution_10": "Stop discussing without any result, :lol: \r\nNow Zeikii, if you solve it, please post. Part (b) is true, but I have no time to spend on checking part (a).", "Solution_11": "[quote=\"Flove\"]My friend asks me about the inequality\n\nProve or disprove that, for all non-negative real numbers $a,b,c$? \\[(a^{3}+b^{3}+c^{3})^{2}\\ge 2(a^{5}b+b^{5}c+c^{5}a)+abc(a^{3}+b^{3}+c^{3})\\] and I believe that (although I haven't any try \\[(a^{3}+b^{3}+c^{3})^{2}\\ge a^{5}b+b^{5}c+c^{5}a+2abc(a^{3}+b^{3}+c^{3})\\] [/quote]\r\n\r\nShould I ask VASC to check part (a)? I belive that 50% it's true, 50% it's false :lol: But I don't know computer to check it. :roll:", "Solution_12": "yes, Flove. I now post my solution for the second, using S.O.S method.\r\nThe second inequality is equivalent to\r\n$\\sum_{}a^{3}(\\sum_{}a^{3}-3abc)\\geq \\sum_{cyc}a^{5}b-abc\\sum_{}a^{3}$\r\n$2\\sum_{}a^{3}(\\sum_{}a^{3}-3abc)\\geq (\\sum_{}(a^{5}b+ab^{5})-2abc\\sum_{}a^{3})+\\sum_{cyc}(a^{5}b-ab^{5})$\r\nNow, using some following equality:\r\n$\\sum_{}a^{3}-3abc= \\frac{1}{2}\\sum_{}a\\sum_{}(b-c)^{2}$\r\n$\\sum_{}(a^{5}b+ab^{5})-2abc\\sum_{}a^{3}= \\sum_{}a(b-c)^{2}(b^{3}+b^{2}c+bc^{2}+c^{3})$\r\n$\\sum_{cyc}(a^{5}b-ab^{5})= \\frac{1}{3}\\sum_{cyc}(c-b)^{3}(2b^{3}+b^{2}c+bc^{2}+2c^{3})$\r\nWe now only have to prove that\r\n$\\sum_{cyc}(b-c)^{2}(5b^{4}+c^{4}+3a^{4}+4bc^{3}+2b^{3}c+3ba^{3}+3ca^{3}-3ab^{2}c-3a^{2}bc)\\geq 0$\r\nor $\\sum_{cyc}S_{A}(b-c)^{2}\\geq 0$\r\nwhere $S_{A}= 5b^{4}+c^{4}+3a^{4}+4bc^{3}+2b^{3}c+3ba^{3}+3ca^{3}-3ab^{2}c-3a^{2}bc$\r\nBy AM-GM inequality, we have\r\n$2a^{4}+b^{3}c+bc^{3}\\geq 4a^{2}bc$\r\nand $2b^{4}+a^{4}+c^{4}\\geq 4ab^{2}c$ therefor $S_{A}\\geq 0$ for every nonegative real numbers $a, b, c$, done.", "Solution_13": "Yes, it's very perceptable, Zeikii. Congulatulation!!! :)", "Solution_14": "I tried to prove the first one by similar way as Zekii but false. Can anybody help me? Jichen?", "Solution_15": "For all non-negative real numbers $x,y,z$ and $k\\leq k_{0}=2.0178872315712002481662\\cdots$, we have\r\n\r\n$(x^{3}+y^{3}+z^{3})^{2}-3xyz(x^{3}+y^{3}+z^{3})\\ge k(x^{5}y+y^{5}z+z^{5}x-xyz(x^{3}+y^{3}+z^{3}))$,\r\n\r\n[hide=\"with equality if\"] $x=1$,\n\n$y=1.7035945243393838669866\\cdots$ is a real root of the irreducible polynomial\n\n$11043y^{54}+69930y^{53}+144396y^{52}+348320y^{51}+1003049y^{50}+1325078y^{49}+1908963y^{48}+4000747y^{47}+1465433y^{46}-2964041y^{45}-3938758y^{44}-24029232y^{43}-56050863y^{42}-79923281y^{41}-139640569y^{40}-224029798y^{39}-280678439y^{38}-344509012y^{37}-374038973y^{36}-287369921y^{35}-151225968y^{34}+44171956y^{33}+272149591y^{32}+216455487y^{31}-242823574y^{30}-914975613y^{29}-1631850192y^{28}-1963117305y^{27}-1416800382y^{26}-138537360y^{25}+1323870700y^{24}+2333734273y^{23}+2394311884y^{22}+1614547296y^{21}+528238034y^{20}-398946278y^{19}-947803336y^{18}-1058514578y^{17}-728858703y^{16}-249059220y^{15}+52900085y^{14}+173916406y^{13}+177937996y^{12}+111689513y^{11}+45949189y^{10}+5163215y^{9}-13022041y^{8}-13794369y^{7}-7571848y^{6}-2174230y^{5}+228225y^{4}+559255y^{3}+276543y^{2}+77706y+11043$,\n\n$z=1.2976397953046941835318\\cdots$ a real root of the irreducible polynomial\n\n$11043z^{54}+77706z^{53}+276543z^{52}+559255z^{51}+228225z^{50}-2174230z^{49}-7571848z^{48}-13794369z^{47}-13022041z^{46}+5163215z^{45}+45949189z^{44}+111689513z^{43}+177937996z^{42}+173916406z^{41}+52900085z^{40}-249059220z^{39}-728858703z^{38}-1058514578z^{37}-947803336z^{36}-398946278z^{35}+528238034z^{34}+1614547296z^{33}+2394311884z^{32}+2333734273z^{31}+1323870700z^{30}-138537360z^{29}-1416800382z^{28}-1963117305z^{27}-1631850192z^{26}-914975613z^{25}-242823574z^{24}+216455487z^{23}+272149591z^{22}+44171956z^{21}-151225968z^{20}-287369921z^{19}-374038973z^{18}-344509012z^{17}-280678439z^{16}-224029798z^{15}-139640569z^{14}-79923281z^{13}-56050863z^{12}-24029232z^{11}-3938758z^{10}-2964041z^{9}+1465433z^{8}+4000747z^{7}+1908963z^{6}+1325078z^{5}+1003049z^{4}+348320z^{3}+144396z^{2}+69930z+11043$\n\nand $k=k_{0}$ is a real root of the irreducible polynomial\n\n$11021312k^{18}-61353472k^{17}+577619072k^{16}-2438185088k^{15}+6024952780k^{14}-13014340386k^{13}-9591745529k^{12}+80565005376k^{11}+62238795390k^{10}-611015712093k^{9}+1495102886568k^{8}-2917979024319k^{7}+4313457174024k^{6}-4807500848001k^{5}+3822807105126k^{4}-2235932782182k^{3}+1108797439518k^{2}-470715894135k+94143178827$.\n[/hide]\r\nRemark. $x^{5}y+y^{5}z+z^{5}x\\geq xyz(x^{3}+y^{3}+z^{3})$.", "Solution_16": "What an awesome problem. How can you do that? Don't tell me that you do it all without computer :blush:", "Solution_17": "He must use computer of course. :P Human cann't do that !!! Anyway, I believe that there is a better way.", "Solution_18": "[quote=\"Flove\"]\nProve or disprove that, for all non-negative real numbers $a,b,c$? \\[(a^{3}+b^{3}+c^{3})^{2}\\ge 2(a^{5}b+b^{5}c+c^{5}a)+abc(a^{3}+b^{3}+c^{3})\\] [/quote]\nYes it's true and we can prove this inequality without computer (maybe with calculator).", "Solution_19": "i also proved the first inequlity", "Solution_20": "[quote=Flove]My friend asks me about the inequality\nProve or disprove that, for all non-negative real numbers $a,b,c$? \\[(a^{3}+b^{3}+c^{3})^{2}\\ge 2(a^{5}b+b^{5}c+c^{5}a)+abc(a^{3}+b^{3}+c^{3})\\] [/quote]\nFor $a\\leq b\\leq c$ this inequality is obvious.\nLet $a\\leq c\\leq b$, $c=a+u$, $b=a+u+v$ and $a=x(u+v)$.\nThus, \n$$(a^3+b^3+c^3)^2-\\sum_{cyc}(2a^5b+a^4bc)=(u^2+uv+v^2)a^4+4(u^3-u^2v-uv^2)a^3+2(5u^4-6u^2v^2-uv^3+2v^4)a^2+$$\n$$+(8u^5+5u^4v-2u^3v^2+2u^2v^3+9uv^4+4v^5)a+2u^6+2u^5v+u^4v^2+2u^3v^3+5u^2v^4+4uv^5+v^6\\geq$$\n$$\\geq\\frac{3}{4}(u+v)^2a^4-\\frac{4}{3}(u+v)^3a^3-\\frac{1}{6}(u+v)^4a^2+\\frac{3}{4}(u+v)^5a+\\frac{1}{4}(u+v)^6=$$\n$$=\\frac{(u+v)^6}{12}(9x^4-16x^3-2x^2+9x+3)\\geq0.$$" } { "Tag": [ "calculus", "integration", "LaTeX", "trigonometry", "calculus computations" ], "Problem": "$ \\int\\frac{\\sqrt{1\\plus{}x}\\plus{}\\sqrt{1\\minus{}x}}{\\sqrt{1\\plus{}x}\\minus{}\\sqrt{1\\minus{}x}}$", "Solution_1": "With respect to what variable?\r\n\r\nOK, I know the answer- but that [b]is[/b] an incomplete statement.\r\nAlso, you need dollar signs for $ \\text{\\LaTeX}$; I put those in for you.", "Solution_2": "if the variable is x jus rationalize the denominator my multiplying and dividing with $ \\frac {\\sqrt {1 \\plus{} x} \\plus{} \\sqrt {1 \\minus{} x}}{\\sqrt {1 \\plus{} x} \\plus{} \\sqrt {1 \\minus{} x}}$ .", "Solution_3": "hello, write your integral in the form\r\n$ \\int\\frac{1}{x}\\,dx\\plus{}\\int\\frac{\\sqrt{1\\minus{}x^2}}{x}\\,dx$.\r\nSonnhard.", "Solution_4": "I got last step, but I don't know how to integrate\r\n $ \\int\\frac{\\sqrt{1\\minus{}x^{2}}}{x}\\,dx$", "Solution_5": "Try a trigonometric substitution.", "Solution_6": "jus put it as $ \\int\\frac {\\sqrt {1 \\minus{} x^{2}} * {x} }{x^{2}}\\,dx$.and now put $ \\ 1 \\minus{} x^{2}$ as $ \\ t^2$.and integrate", "Solution_7": "try $ x \\equal{} \\cos 2\\theta$", "Solution_8": "hello, setting $ t\\equal{}\\sqrt{1\\minus{}x^2}$ we get $ dx\\equal{}\\frac{tdt}{\\minus{}x}$ and our integral is now\r\n$ \\int\\frac{\\sqrt{1\\minus{}x^2}}{x}\\frac{tdt}{\\minus{}x}\\equal{}\\int\\frac{t^2}{\\minus{}x^2}\\,dt\\equal{}\\int\\frac{t^2}{t^2\\minus{}1}\\,dt$.\r\nSonnhard.", "Solution_9": "$ I \\equal{} \\int\\frac {\\sqrt {1 \\plus{} x} \\plus{} \\sqrt {1 \\minus{} x}}{\\sqrt {1 \\plus{} x} \\minus{} \\sqrt {1 \\minus{} x}}dx$\r\n\r\nsubstitute $ x \\equal{} \\sin 2\\theta \\ \\Rightarrow \\ dx \\equal{} 2 \\cos 2\\theta \\ d\\theta$\r\n\r\nnotice that $ \\sqrt{1\\plus{}x} \\equal{} \\sqrt{\\sin ^2\\theta \\plus{} \\cos ^2\\theta \\plus{} 2 \\sin \\theta \\cos \\theta} \\equal{} \\sqrt{(\\sin \\theta \\plus{} \\cos \\theta)^2} \\equal{} \\sin \\theta \\plus{} \\cos \\theta$\r\n\r\nand $ \\sqrt{1\\minus{}x} \\equal{} \\sqrt{\\sin ^2\\theta \\plus{} \\cos ^2\\theta \\minus{} 2 \\sin \\theta \\cos \\theta} \\equal{} \\sin \\theta \\minus{} \\cos \\theta$\r\n\r\n\r\n$ I \\equal{} 2 \\int \\frac{\\sin \\theta \\plus{} \\cos \\theta \\plus{} \\sin \\theta \\minus{} \\cos \\theta}{\\sin \\theta \\plus{} \\cos \\theta \\minus{} \\sin \\theta \\plus{} \\cos \\theta} \\cos 2\\theta d\\theta \\\\ \\\\ \\\\ \\equal{} 2 \\int \\frac{\\sin \\theta}{cos \\theta}(2 \\cos ^2 \\theta \\minus{} 1 ) d\\theta$\r\n\r\n$ \\equal{} 2 \\int (\\sin 2\\theta \\minus{} \\tan \\theta) d\\theta$\r\n\r\n\r\nhope you can carry on from here ...", "Solution_10": "I got .5ln(x).\r\nIs that the right answer.", "Solution_11": "No.\r\n\r\nYou can always test a proposed antiderivative by differentiating it. We're not integrating $ \\frac1{2x}$ here.", "Solution_12": "no yur answer is not correct.\r\n\r\n$ \\int\\frac{1}{x}\\,dx+\\int\\frac{\\sqrt{1-x^{2}}}{x}\\,dx$\r\n\r\n\r\n$ \\int\\frac{1}{x}\\,dx = ln (X)$\r\n\r\n$ \\int\\frac{\\sqrt{1-x^{2}}}{x}\\,dx$ = setting $ \\t=\\sqrt{1-x^2}$ integral is now $ \\int\\frac{\\sqrt{1-x^{2}}}{x}\\frac{tdt}{-x}$ = $ \\int\\frac{t^{2}}{-x^{2}}\\,dt$ = $ \\int\\frac{t^{2}}{t^{2}-1}\\,dt$ \r\n\r\nthis can be split as$ \\int\\frac{t^{2}-1+1}{t^{2}-1}\\,dt$ = $ \\int 1 dx$ + $ \\int\\frac{1}{t^{2}-1}$\r\n \r\n = $ \\ t$ + ${ \\frac{1}{2}}\\ ln|\\frac{t-1}{t+1}|+ c$", "Solution_13": "$ \\int\\frac {\\sqrt {1 \\plus{} x} \\plus{} \\sqrt {1 \\minus{} x}}{\\sqrt {1 \\plus{} x} \\minus{} \\sqrt {1 \\minus{} x}}\\ dx$\r\n\r\n\r\n$ \\equal{} \\int \\frac {1 \\plus{} \\sqrt {\\frac {1 \\minus{} x}{1 \\plus{} x}}}{1 \\minus{} \\sqrt {\\frac {1 \\minus{} x}{1 \\plus{} x}}}\\ dx$\r\n\r\n\r\nLet $ \\tan \\theta \\equal{} \\sqrt\\frac {1 \\minus{} x}{1 \\plus{} x}\\Longrightarrow x \\equal{} \\cos 2\\theta$, the integral can be written as \r\n\r\n$ 2\\int \\frac {\\cos \\theta \\plus{} \\sin \\theta}{\\cos \\theta \\minus{} \\sin \\theta}\\sin 2\\theta\\ d\\theta$\r\n\r\n@ Nora.91\r\n\r\n[quote=\"Nora.91\"]$ I \\equal{} \\int\\frac {\\sqrt {1 \\plus{} x} \\plus{} \\sqrt {1 \\minus{} x}}{\\sqrt {1 \\plus{} x} \\minus{} \\sqrt {1 \\minus{} x}}dx$\n\nsubstitute $ x \\equal{} \\sin 2\\theta \\ \\Rightarrow \\ dx \\equal{} 2 \\cos 2\\theta \\ d\\theta$\n\nnotice that $ \\sqrt {1 \\plus{} x} \\equal{} \\sqrt {\\sin ^2\\theta \\plus{} \\cos ^2\\theta \\plus{} 2 \\sin \\theta \\cos \\theta} \\equal{} \\sqrt {(\\sin \\theta \\plus{} \\cos \\theta)^2} \\equal{} \\sin \\theta \\plus{} \\cos \\theta$\n\nand $ \\sqrt {1 \\minus{} x} \\equal{} \\sqrt {\\sin ^2\\theta \\plus{} \\cos ^2\\theta \\minus{} 2 \\sin \\theta \\cos \\theta} \\equal{} \\sin \\theta \\minus{} \\cos \\theta$.[/quote]\r\n\r\nIs $ \\sqrt{A^2}\\equal{}A?$", "Solution_14": ":oops_sign: \r\n\r\nthanks kunny .\r\n\r\ni guess my method isn't valid!", "Solution_15": "Isn't it negative? \r\n$ \\minus{}2\\int\\frac{\\cos\\theta\\plus{}\\sin\\theta}{\\cos\\theta\\minus{}\\sin\\theta}\\sin 2\\theta\\ d\\theta$\r\nDoes anyone has final answer to the question?", "Solution_16": "We have:\r\n$ \\frac{\\cos\\theta\\plus{}\\sin\\theta}{\\cos\\theta\\minus{}\\sin\\theta}\\sin 2\\theta\\ \\equal{} \\frac{\\cos\\theta\\plus{}\\sin\\theta}{\\cos\\theta\\minus{}\\sin\\theta}\\sin 2\\theta\\ \\times \\frac{\\cos\\theta\\plus{}\\sin\\theta}{\\cos\\theta \\plus{}\\sin\\theta} \\equal{} \\tan 2\\theta \\plus{} \\sec 2\\theta \\minus{} \\cos 2\\theta$" } { "Tag": [], "Problem": "A $ 6\\%$ rate increase by a local media cable company resulted in an\nincrease of $ \\$1.20$ per month on a family's bill. How many dollars\nwas the monthly bill before the increase?", "Solution_1": "let the monthtly bill before the increase be x.\r\n\r\n$ x*\\frac{6}{100}\\equal{}1.2 \\implies 6x\\equal{}120 \\implies x\\equal{}20$\r\n\r\nanswer : 20 dollars" } { "Tag": [ "probability", "inequalities" ], "Problem": "The number $ n$ is randomly selected from the set $ \\{1, 2, ..., 10\\}$,\nwith each number being equally likely. What is the probability\nthat $ 2n\\minus{}4>n$ ? Express your answer as a common fraction.", "Solution_1": "We can substitute different numbers to find which ones work.\r\n\r\n$ 2n \\minus{} 4 > n$\r\n\r\nLet's try 1. We get: $ 2(1) \\minus{} 4 > 1$. Simplified, this is $ \\minus{} 2 > 1$, which is obviously false.\r\nHow about 2? Our result is $ 0 > 2$, false.\r\nNow, 3. $ 2 > 3$, still false.\r\nWhat about 4? $ 4 > 4$, which is false.\r\nNow let's try 5. $ 6 > 5$, which is true!\r\nWhen we put in 6, we get: $ 8 > 6$, true again!\r\n\r\nIf we continue substituting values up to $ 10$, we find that they all make the inequality true. So the set $ \\{5,6,7...10\\}$ all work. The probability of picking a number $ n$ in this set out of the the numbers 1 through 10 is $ 5/10$ or $ 1/2$.", "Solution_2": "there are total 6 numbers that satisfies the equation $ 2n\\minus{}4>n (5\\sim 10)$\r\n\r\nso the answer should be $ \\frac{6}{10}$.\r\n\r\nAnswer : $ \\frac{3}{5}$", "Solution_3": "[quote=\"FantasyLover\"]there are total 6 numbers that satisfies the equation $ 2n - 4 > n (5\\sim 10)$\n\nso the answer should be $ \\frac {6}{10}$.\n\nAnswer : $ \\frac {3}{5}$[/quote]\r\n\r\nNeither of you really explained the best way...\r\n\r\nAs in equations, the addition and subtraction properties also hold here...\r\n\r\nTherefore we simplify the inequality to $ n > 4$\r\n\r\nThus our desired subset is $ \\{5, 6, 7,\\ldots,10\\}$\r\n\r\n${ \\frac {3}{5}}$", "Solution_4": "I was just skipping extremely easy part...\r\n\r\nwho doesn't know n>4 if 2n-4>n...", "Solution_5": "lots of people?", "Solution_6": "Not a lot of people...I knew that, and I'm not exactly the best person here...\r\n\r\nAnd the mistake I made on the solution I got was because I overlooked the fact that it's easier to just simplify the inequality." } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "Given that $2^{mn}-1$ is divisible by $(2^{m}-1)(2^{n}-1)$. Prove that $2(3^{mn}-1)$ is divisible by $(3^{m}-1)(3^{n}-1)$. \r\n\r\nQuestion: Does this force $(m, n) = 1$? \r\n\r\nI do not have a proof for both questions. Please help.", "Solution_1": "Nevermind, I got it. The second claim is true. (that (m, n) =1 ) and the first claim follows.", "Solution_2": "Lemma1: $(3^{m}-1;3^{n}-1)=3^{(m;n)}-1$ \r\nNOTES:$3^{mn}-1$ is divisible$3^{m}-1$ and $3^{n}-1$\r\nFrom that, we have to prove that (m;n)=1.Try doing it .If you can't,i 'll post my solution tomorrow :)", "Solution_3": "[quote=\"tmbtw\"]\ni 'll post my solution tomorrow :)[/quote]\r\nOkie, I waiting your solution, tmbtw.\r\nI also have a problem similarly: Let $a, b$ are positive integers such that $a^{n}-1|b^{n}-1$. Prove that $a=b^{k}$, for some integer $k$", "Solution_4": "minhtoan: Search for \"masterpiece\", author Harazi ;)", "Solution_5": "[quote=\"minhtoan\"][quote=\"tmbtw\"]\ni 'll post my solution tomorrow :)[/quote]\nOkie, I waiting your solution, tmbtw.\nI also have a problem similarly: Let $a, b$ are positive integers such that $a^{n}-1|b^{n}-1$. Prove that $a=b^{k}$, for some integer $k$[/quote]\r\nOK,Here is my solution :\r\nLet d=(m;n)\r\nWe have $(2^{m}-1;2^{n}-1)=2^{d}-1$\r\nSo,$2^{d}-1$|$2^{n}-1$|$\\frac{2^{mn}-1}{2^{m}-1}=(2^{m})^{n-1}+...+2^{m}+1$\r\nThen,$2^{d}-1$|n\r\nSimilarly,$2^{d}-1$|m.So,$2^{d}-1$|d.So,d=1 .We're done :)", "Solution_6": "we can also generalize this question as :\r\n\r\nif $a^{mn}-1$ is divisible by $(a^{m}-1)(a^{n}-1)$ prove $((a+1)^{mn}-1)a$ is divisible by $((a+1)^{m}-1)((a+1)^{n}-1)$" } { "Tag": [ "algebra", "polynomial", "trigonometry", "algebra unsolved" ], "Problem": "prove that:\r\ntan^2 (p / 7) + tan^2 (2p / 7) + tan^2 (3p / 7) = 21", "Solution_1": "What is p????", "Solution_2": "[b]i think here p = pi =22/7[/b]", "Solution_3": "p means pi", "Solution_4": "[quote=\"bgbgbgbg\"]prove that:\ntan^2 (p / 7) + tan^2 (2p / 7) + tan^2 (3p / 7) = 21[/quote]\r\n\r\nLet $ t\\equal{}\\tan(x)$\r\n\r\n$ \\tan(2x)\\equal{}\\frac{2t}{1\\minus{}t^2}$, $ \\tan(3x)\\equal{}\\frac{t(3\\minus{}t^2)}{1\\minus{}3t^2}$, $ \\tan(4t)\\equal{}\\frac{4t(1\\minus{}t^2)}{t^4\\minus{}6t^2\\plus{}1}$ and $ \\tan(7x)\\equal{}\\frac{\\minus{}t^7\\plus{}21t^5\\minus{}35t^3\\plus{}7t}{\\text{some ugly polynomial}}$\r\n\r\nSo $ \\{\\tan(\\frac{k\\pi}7),k\\equal{}0,1,2,3,4,5,6\\}$ are the seven roots of $ t^7\\minus{}21t^5\\plus{}35t^3\\minus{}7t\\equal{}0$\r\n\r\nSo $ \\{\\tan(\\frac{k\\pi}7)^2,k\\equal{}1,2,3\\}$ are the three roots of $ t^3\\minus{}21t^2\\plus{}35t\\minus{}7\\equal{}0$\r\n\r\nHence the result." } { "Tag": [ "geometry", "inequalities", "geometric transformation", "reflection" ], "Problem": "Triangle $ABC$ is right angled at $A$. Let $D$ be a point on $\\overline{BC}$. Let $E$ and $F$ be the reflections of $D$ across $AC$ and $AB$. Prove that $(ABC)\\ge (DEF)$. Find all positions of $D$ for equality.\r\n\r\n[asy]draw((0,0)--(60,0)--(0,40)--cycle); draw((24,24)--(-24,24)--(24,-24)--cycle); pair A,B,C,D,E,F; A=(0,0); B=(60,0); C=(0,40); D=(24,24); E=(-24,24); F=(24,-24); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NW); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,SE);[/asy]", "Solution_1": "Let $AB=c, BC=a, CA=b, CD=x, DE=y, DF=z$. Then by simple similarity:\r\n\r\n$y={2cx\\over a}, z={2b(a-x)\\over a}$\r\n\r\nHence $[DEF]={2bcx(a-x)\\over a^{2}}$\r\n\r\nTrivially (by AM-GM), $x(a-x)\\leqslant{a^{2}\\over 4}$, hence $[DEF]\\leqslant{bc\\over 2}=[ABC]$.\r\n\r\nThe equality holds iff $x={a\\over 2}$, i.e. iff $D$ is the midpoint of $BC$." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ x,y,z$ are positive real numbers,$ k \\geq 2$,prove that:\r\n$ \\frac{2}{k\\plus{}2} \\geq \\frac{x}{kx\\plus{}y\\plus{}z}\\plus{}\\frac{x^{\\minus{}1}}{kx^{\\minus{}1}\\plus{}y^{\\minus{}1}\\plus{}z^{\\minus{}1}}$\r\n\r\n$ <\\equal{}>\r\n\\frac{(zx^2\\plus{}yz^2\\minus{}4xyz\\plus{}x^2y\\plus{}y^2z)k\\plus{}2x(y\\minus{}z)^2}{(k\\plus{}4)(kx\\plus{}2x\\plus{}y\\plus{}z)(kyz\\plus{}2yz\\plus{}zx\\plus{}xy} \\geq 0$\r\n\r\neasy.\r\nBQ", "Solution_1": "please prove n-variables:\r\nLet $ x_1,x_2,...,x_n > 0,n \\geq 2 ,k \\geq n - 1,$,\r\nprove that:\r\n\r\n${ \\frac {2}{k + n - 1} \\geq \\frac {x_1}{kx_1 + \\sum_{i = 2}^n{x_i}} + \\frac {x_1^{ - 1}}{kx_1^{ - 1} + \\sum_{i = 2}^n{x_i^{ - 1}}}}$", "Solution_2": "$ n \\geq 2,k \\geq n \\minus{} 1$,\r\n$ \\frac {\\sum_{i \\equal{} 2}^n{x_i}}{kx_1 \\plus{} \\sum_{i \\equal{} 2}^n{x_i}}\\plus{}\\frac {\\sum_{i \\equal{} 2}^n{x_i^{\\minus{}1}}}{kx_1^{\\minus{}1} \\plus{} \\sum_{i \\equal{} 2}^n{x_i^{\\minus{}1}}} \\geq \\frac {2(n \\minus{} 1)}{k \\plus{} n \\minus{} 1}$\r\nBQ" } { "Tag": [ "inequalities", "calculus", "derivative", "geometry", "geometric transformation", "reflection" ], "Problem": "Let $a,b,c$ be non-negative real numbers satisfying that $a+b+c=3$. For each positive real numbers $k$, find the maximum value of \\[P=a^{2}(kb+c)+b^{2}(kc+a)+c^{2}(ka+b).\\] Another part of the above problem is to consider it in case $k\\le 0$ (which I just think when I write this problem now, but I'm sure it can be solved).\r\n\r\nSolution for the first problem is ery nice and... no hard.", "Solution_1": "Is there a nicer way than just multiply out and use Am-Gm (muirhead) and maybe Cauchy for squares...?", "Solution_2": "I believe that there is no such way. At least using AM-GM or Muihard directly is impossible,", "Solution_3": "My solution is called \"the entirely mixing variables method\". It is very new, but effective for so many problem,\r\n\r\nMoreover, there's no problem if $k\\le 0$.", "Solution_4": "For a hint, firstly you should check a small case, that is $min(a,b,c)=0$.", "Solution_5": "[quote=\"Flove\"]For a hint, firstly you should check a small case, that is $min(a,b,c)=0$.[/quote]\r\n\r\nHow do you solve it, Anh Cuong? I wait it. :P", "Solution_6": "You min the case when one variable is equal to $0$. Oke, i think this is also important to obtain the result. \r\nLet assume that $a=0$ and remain the problem:\r\nGive $b,c \\geq 0$ satisfying $b+c=3$. Find the maximum value of:\r\n$P= kb^{2}c+c^{2}b=bc(kb+c)$. Oke take $m=\\frac{k(k+\\sqrt{k^{2}-{k}+1})}{k-1+\\sqrt{k^{2}-k+1}}$\r\nand $n = k+\\sqrt{k^{2}-k+1}$. Then apply AM-GM here:\r\n$mnP = mb.nc.(kb+c) \\leq (\\frac{mb+nc+kb+c}{3})^{3}= (\\frac{(n+1)(b+c)}{3})^{3}= (n+1)^{3}$\r\nFrom here imply that:\r\n$P \\leq \\frac{(n+1)^{3}}{mn}$\r\nOpps, too lazy to find the equality, and more more thing, this quick proof only apply for $k \\geq 0$. :)", "Solution_7": "Your solution is nice, although my solution is only to use derivative, not claim any brainpower. Now, show we how to prove that, it's enough to check the case $\\min(a,b,c)=0$. As I tell above, I use the (entirely) mixing variables method, quite easy. I wonder if there're other solutions. :lol:\r\n___________________________\r\n\r\nThis post was edited after the 1000s post. About \"the entirely mixing variables method\", you should look over Mathematical Reflection, also my article there \r\n\r\nhttp://reflections.awesomemath.org/2006_5/2006_5_entirelymixing.pdf\r\n\r\nAfter reading it, I think my problem above is much easier. :lol: Enjoy it!" } { "Tag": [ "geometry solved", "geometry" ], "Problem": "let $f$ be a convex shape inside a circle $C$, such that from any point on $C$, the shape $f$ is seen by an $90^{\\circ}$ angle. prove that $f$ is symmetric from $O$ (center of $C$).\r\n[hide][b]what is the English word for O for f , when f is symmetric from O :?: [/b][/hide]", "Solution_1": "Given a direction in the plane, there are exactly two lines having that direction which meet a convex (and compact; I assume our set is compact and that it's not contained in a line) only in border points. \r\n\r\nNow fix a direction $d$, and take a line $\\ell\\|d$ which meets our convex set in border points. This line cuts the circle in two points, $A,D$. There is a line through $A$ which is $\\perp d$ which meets the set in border points, since the set is seen under an angle of $\\frac\\pi 2$ from $A$. This line cuts the circle again in $B$. There's also a line $\\ell'\\|d$ through $B$ which cuts the set only in border points. $\\ell$ and $\\ell'$ must be symmetric wrt $O$, so we have shown that for each direction, the two lines mentioned in the first paragraph corresponding to that direction are symmetric wrt $O$. \r\n\r\nThe set is the intersection of all the (closed) bands contained between lines of the type $\\ell,\\ell'$ as described above, so, since each such band is symmetric wrt $O$, the set must also be symmetric wrt $O$." } { "Tag": [ "function", "inequalities", "algebra unsolved", "algebra" ], "Problem": "Let $ f : [ 0 ; 1] \\to\\mathbb{R}$ be a strictly increasing function. We know that $ f(0) \\equal{} 0$ and $f(1) \\equal{} 1$. Moreover,\n\\[ \\frac{1}{2} \\le \\frac{ f(x\\plus{}y) \\minus{} f(x)}{ f(x) \\minus{} f(x\\minus{}y)} \\le 2, \\] for all $x$ and $y$ such that $ 0 < y \\le x < x\\plus{}y \\le 1$. Prove that \\[ f \\left( \\frac{1}{3} \\right) \\le \\frac{76}{135}.\\]", "Solution_1": "[i]My lovely teacher , sir . Tran Nam Dung gave me a very good solution as follows :lol: \n\n Solution : \n From the hypothesis , notice that : $ \\frac {1}{3} f(x \\plus{} y) \\plus{} \\frac {2}{3}f(x \\minus{} y) \\le f(x) \\le \\frac {1}{3} f(x \\minus{} y) \\plus{} \\frac {2}{3}f(x \\plus{} y)$\n\n for all $ 0 \\le y \\le x \\le x \\plus{} y \\le 1 \\ \\ (*)$\n\n From that property , we can have some relative inequalities :\n\n $ a/ \\ \\ \\frac {3}{2} f \\left( \\frac { 2 }{3 } \\right) \\minus{} \\frac {1}{2} f \\left( \\frac {7 }{12 } \\right) \\le f \\left( \\frac { 3 }{ 4} \\right) \\le \\frac {2}{3} \\plus{} \\frac {1}{3} f \\left( \\frac { 1}{ 2 } \\right)$ ( in $ (*)$ put $ x \\ \\equal{} \\ \\frac {2}{3} ; y \\ \\equal{} \\ \\frac {1}{6}$ and $ x \\equal{} \\frac {3}{4} \\ ; \\ y \\ \\equal{} \\ \\frac {1}{4}$ )\n\n $ b/ \\ \\ f \\left( \\frac { 7 }{ 12 } \\right) \\le 3f \\left( \\frac {1 }{ 2 } \\right) \\ \\minus{} \\ 2 f \\left( \\frac { 5 }{12 } \\right)$ ( in $ (*)$ put $ x \\ \\equal{} \\ \\frac {1}{2} ; y \\ \\equal{} \\ \\frac {1}{12}$ ) \n $ c/ \\ \\ f \\left( \\frac { 5 }{12 } \\right) \\ge \\frac {3}{2} f \\left( \\frac { 1 }{ 3 } \\right) \\ \\minus{} \\ \\frac {1}{2} f \\left( \\frac { 1 }{ 4 } \\right)$( in $ (*)$ put $ x \\ \\equal{} \\ \\frac {1}{3} ; y \\ \\equal{} \\ \\frac {1}{12}$ ) \n\n $ d/ \\ \\ f \\left( \\frac { 1 }{ 4 } \\right) \\le \\frac {2}{3} f \\left( \\frac {1 }{ 2 } \\right)$\n\n $ e/ \\ \\ f \\left( \\frac { 1 }{ 3 } \\right) \\le \\frac {2}{3} f \\left( \\frac {2 }{ 3 } \\right) \\rightarrow \\frac {3}{2}f \\left( \\frac { 1 }{ 3 } \\right) \\le f \\left( \\frac {2 }{ 3 } \\right)$\n\n $ f/ \\ \\ f \\left( \\frac {1 }{ 2 } \\right) \\le \\frac {2}{3}$\n Now , It's time to eat a cake :blush: \nFrom $ a/$ , we have : $ \\frac {3}{2} f \\left( \\frac { 2 }{3 } \\right) \\minus{} \\frac {1}{2} f \\left( \\frac {7 }{12 } \\right) \\le \\frac {2}{3} \\plus{} \\frac {1}{3} f \\left( \\frac { 1}{ 2 } \\right)$\n\n Thus , $ \\frac {3}{2} f \\left( \\frac { 2 }{3 } \\right) \\le \\frac {1}{2} f \\left( \\frac {7 }{12 } \\right) \\plus{} \\frac {2}{3} \\plus{} \\frac {1}{3} f \\left( \\frac { 1}{ 2 } \\right) \\le \\frac {2}{3} \\plus{} \\frac {1}{3} f \\left( \\frac { 1}{ 2 } \\right) \\plus{} \\frac {3}{2} f \\left( \\frac {1 }{ 2 } \\right) \\ \\minus{} \\ f \\left( \\frac { 5 }{12 } \\right)$ , due to $ b/$ :)\n\n Now , using $ c/$ , we have : \n\n Hence $ \\frac {3}{2} f \\left( \\frac { 2 }{3 } \\right) \\le \\frac {2}{3} \\plus{} \\frac {1}{3} f \\left( \\frac { 1}{ 2 } \\right) \\plus{} \\frac {3}{2} f \\left( \\frac {1 }{ 2 } \\right) \\minus{} \\frac {3}{2} f \\left( \\frac { 1 }{ 3 } \\right) \\ \\plus{} \\ \\frac {1}{2} f \\left( \\frac { 1 }{ 4 } \\right)$\n\n So that : \n$ \\frac {3}{2} f \\left( \\frac { 2 }{3 } \\right) \\plus{} \\frac {3}{2} f \\left( \\frac { 1 }{ 3 } \\right) \\le \\frac {2}{3} \\plus{} \\frac {11}{6} f \\left( \\frac { 1}{ 2 } \\right) \\ \\plus{} \\ \\frac {1}{2} f \\left( \\frac { 1 }{ 4 } \\right)$\n\n$ \\le \\frac {2}{3} \\plus{} \\frac {11}{6} f \\left( \\frac { 1}{ 2 } \\right) \\ \\plus{} \\ \\frac {1}{3} f \\left( \\frac { 1 }{ 4 } \\right) \\equal{} \\frac {2}{3} \\plus{} \\frac {13}{6} f \\left( \\frac { 1}{ 2 } \\right) \\le \\frac {2}{3} \\plus{} \\frac {13}{6} \\cdot \\frac {2}{3} \\ \\equal{} \\ \\frac {19}{9}$\n\n $ \\rightarrow \\frac {19}{9} \\ge \\frac {3}{2} f \\left( \\frac { 2 }{3 } \\right) \\plus{} \\frac {3}{2} f \\left( \\frac { 1 }{ 3 } \\right) \\ge \\frac {9}{4} f \\left( \\frac { 2 }{3 } \\right) \\plus{} \\frac {3}{2} f \\left( \\frac { 1 }{ 3 } \\right)$ , due to $ e/$\n\n Thus , $ f\\left( \\frac { 1 }{ 3 } \\right) \\le \\frac { 76}{135}$ , Done\n\n So nice problem :love: [/i]", "Solution_2": "I got only $ f(\\frac 13)\\le\\frac 47$ ... :blush:", "Solution_3": "Here is my proof for my weaker result :\r\n\r\nSetting $ x \\equal{} y \\equal{} \\frac 13$ in the inequation, we get $ \\frac 12\\le\\frac {f(\\frac 23) \\minus{} f(\\frac 13)}{f(\\frac 13)}\\le 2$\r\n\r\nAnd since $ f(x)$ is strictly increasing and $ f(0) \\equal{} 0$, we get $ f(\\frac 13) > 0$ and the above inequality becomes $ \\frac 32f(\\frac 13)\\le f(\\frac 23)\\le 3f(\\frac 13)$\r\n\r\nSetting now $ x \\equal{} \\frac 23$ and $ y \\equal{} \\frac 13$, we get $ \\frac 12\\le\\frac {1 \\minus{} f(\\frac 23)}{f(\\frac 23) \\minus{} f(\\frac 13)}$ and so : $ \\frac 32f(\\frac 23)\\le 1 \\plus{} \\frac 12f(\\frac 13)$\r\n\r\nAnd since we got in the beginning $ \\frac 32f(\\frac 13)\\le f(\\frac 23)$, we can conclude :\r\n\r\n$ \\frac 94f(\\frac 13)\\le \\frac 32f(\\frac 23)\\le 1 \\plus{} \\frac 12f(\\frac 13)$ and so $ \\frac 94f(\\frac 13)\\le 1 \\plus{} \\frac 12f(\\frac 13)$ and so : $ \\boxed{f(\\frac 13)\\le\\frac 47}$ \r\n\r\nand $ \\frac 47$ is near of $ \\frac {76}{135}$ but this is not enough for showing your request. ;)" } { "Tag": [], "Problem": "if a rectangular block that is 4 inches by 4 inches by 10 inches is placed inside a right circular cylinder of radius 3 inches and height 10 inches the volumne of the unoccupied portion of the cylinder is how many cubic inches?\r\n\r\n\r\nhave no idea where to start....step by step answer please.", "Solution_1": "the volume of the unoccupied portion will be the differences of the 2 volumes.volume of the rect. block:4x4x10=160cubic inches. vol. of cyl. block=3.1416x(3^2)x10=283Cub. inches.the answer is 123cub inches.", "Solution_2": "[hide=\"Solution\"]The unoccupied volume is the volume of the cylinder minus the volume of the rectangular block (do you see why?), so:\n\\begin{align*} V_{unoccupied} & = V_{cylinder} - V_{block} \\\\\n& = \\left(\\pi r^2 h\\right) - (l \\cdot w \\cdot h) \\\\\n& = \\pi \\cdot \\left(3 \\text{ inches}\\right)^2 \\cdot 10 \\text{ inches}- 4 \\text{ inches}\\cdot 4 \\text{ inches} \\cdot 10 \\text{ inches} \\\\\n& = \\boxed{90 \\pi - 160 \\text{ cubic inches}} \\\\\n& = \\boxed{10\\left(9\\pi - 16\\right) \\text{ cubic inches}} \\\\\n& \\approx \\boxed{122.743 \\text{ cubic inches}} \\end{align*}\n[/hide]" } { "Tag": [ "algebra", "polynomial", "integration", "function", "trigonometry", "calculus", "quadratics" ], "Problem": "Let $ f(x)=e^{x^2}$ and write $ f^{(n)}(x)=P_n(x) f(x)$ where $P_n$ are polynomials. Then find an equivalent for $P_n(x)$ for every fixed positive $x$.", "Solution_1": "could you post some hints please ? I tried a lot to solve it yesterday but .. :(", "Solution_2": "This is an extremely hard problem! The main idea is to find a second order differential equation that has as solution the polynomials in the problem. Next, use the fact that $ af(x)\\leq f\\\"(x)\\leq bf(x)$ implies some estimations on $f(x)$ and $x$. This is just a small hint, but it is the mzin idea. After that there is some ugly work. I will post a solution at the end of the week if it doesn't help much. Another hint: the equivalent is absolutely monstruous! I have never seen such an ugly equivalent!!!", "Solution_3": "I tried some simple \"back of envelope\" computation and got $\\frac {n!e^{2xR+R^2}}{2\\sqrt{2\\pi}R^{n+1}}$ where $R=R(x,n)$ is the positive root of the equation $2xR+2R^2=n$, i.e., $R=\\sqrt{\\frac n2+\\frac {x^2}4}-\\frac x2$. Using Stirling's formula for $n!$ and Taylor series for $R$, this formula can be simplified to\r\n\\[ \r\n\\frac 1{\\sqrt 2}\\left(\\frac{2n}{e}\\right)^{\\frac n2}e^{-\\frac{x^2}2}e^{x\\sqrt{2n}}\r\n\\]\r\nDoes it agree with your answer, Harazi? (If your answer looks differently and you are not completely sure whether or not it is asymptotically the same as what I got, just post it and I'll try to figure it out myself. As I said, I just did a quick estimate and cannot guarantee it, so I won't be surprised if I'm wrong :P )", "Solution_4": "That's it, Fedja.", "Solution_5": "Then I'll post my computation. It is different from Harazi's approach and uses a very classical technique. The problem is to verify some technical details, which may be not very hard but somewhat boring. I'll just present it the way I did it omitting formal justification of some steps.\r\n\r\nThe main tool is the formula $f^{(n)}(x)=\\frac1{2\\pi}n! R^{-n}\\int_{[-\\pi,\\pi]} f(x+Re^{i\\theta})e^{-in\\theta}\\,d\\theta$ valid for any function $f$ analytic in the entire complex plane. Applying it to $f(z)=e^{z^2}$ and cancelling $e^{x^2}$, we get\r\n\\[\r\nP_n(x)=\\frac1{2\\pi}n! R^{-n}\\int_{[-\\pi,\\pi]} \\exp\\left\\{ 2xRe^{i\\theta}+R^2e^{2i\\theta}-in\\theta\\right\\}\\,d\\theta \\\\ \r\n=\\frac1{2\\pi}n! R^{-n}\\int_{[-\\pi,\\pi]} \\exp\\left\\{ 2xR\\cos{\\theta}+R^2\\cos{2\\theta}\\right\\} \\exp\\left\\{i( 2xR\\sin{\\theta}+R^2\\sin{2\\theta}-n\\theta)\\right\\}\\,d\\theta \r\n\\]\r\nThe first factor in the integral has a well-expressed maximum at $\\theta=0$. The problem is that, for generic $R$, the second factor is oscillating fast. So, the idea is to choose $R$ for which the oscillation is slow. Since $\\sin\\theta=\\theta+O(\\theta^3)$, we see that, choosing $R$ satisfying the equation $2xR+2R^2=n$, we make the oscillation slow (cubic in $\\theta$ for small $\\theta$ instead of a linear one that takes place for other $R$). Now, $\\cos\\theta\\approx 1-\\frac{\\theta^2}2$, so the decay is quadratic in $\\theta$ with approximately the same (big) coefficient in exponent as that at $\\theta^3$ in the oscillating term. This makes it believable (though one should accurately check some details) that the integral is almost the same as \r\n$\r\ne^{2xR+R^2}\\int_{-\\infty}^{\\infty}e^{-(xR+2R^2)\\theta^2}\\,d\\theta = e^{2xR+R^2}\\sqrt{\\frac{\\pi}{xR+2R^2}}\\approx \\sqrt{\\frac\\pi 2}\\frac{e^{2xR+R^2}}{R}\r\n$.\r\nThis results in the first form of the answer in my post #4. Switching to the second form is a routine exercise." } { "Tag": [ "geometry", "3D geometry", "sphere", "analytic geometry", "symmetry", "number theory", "real analysis" ], "Problem": "Call by $\\mathbb{Z}$-ball the set of points of the form $\\ds S=\\{(x,y,z)\\mid x^{2}+y^{2}+z^{2}\\le R^{2},\\ x,y,z,\\in\\mathbb{Z}\\},\\ R\\in\\mathbb{R}.$ Prove that there exists no $\\mathbb{Z}$-ball which contains exactly $2005$ distinct points.", "Solution_1": "If one can prove that $7.1 \\leq R < 8$, then it's simple from there:\r\n\r\nThe only points that matter $\\bmod \\, 8$ are the ones of the form $(a, \\pm a, 0) \\, \\slash \\, (a,0,0)$. By some computations, we get that $\\left| S_{R}\\right| \\equiv 7 \\left( \\bmod \\, 8 \\right), \\, \\forall R \\in \\left[ 7.1,8 \\right)$.\r\n\r\n\r\nThus, if one can prove that (hopefully correct) bound (especially the $R<8$ part), then the problem is solved. If not, then this is one useless post.", "Solution_2": "I'd work mod 6: the only points that matter are of the form $(\\pm a,\\pm a,\\pm a)$. For $48\\le R^{2}<75$, $|S|\\equiv 3\\mod 6$ and for $75\\le R^{2}<108$, $|S|\\equiv 5\\mod 6$; if we can show that $|S|<2005$ when $R^{2}=48$ and $|S|>2005$ when $R^{2}=108$, we will be done.\r\n\r\nIf $R^{2}=108$, the points $(a,b,c)$ are inside the sphere whenever $|a|,|b|,|c|\\le6$. There are $13^{3}=2197$ points of this form, and $|S|\\ge 2197$ when $|R^{2}|=108$.\r\n\r\nIf $R^{2}=48$ and $(a,b,c)\\in S$, we must have $|a|+|b|+|c|\\le 11$ or $|a|=|b|=|c|=4$. There are $1561$ points of the first type and eight of the second, so $|S|\\le 1569$ when $R^{2}=48$\r\n\r\nThis is number theory, not calculus.", "Solution_3": "[quote=\"jmerry\"]I'd work mod 6: the only points that matter are of the form $(\\pm a,\\pm a,\\pm a)$. For $48\\le R^{2}<75$, $|S|\\equiv 3\\mod 6$ and for $75\\le R^{2}<108$, $|S|\\equiv 5\\mod 6$; if we can show that $|S|<2005$ when $R^{2}=48$ and $|S|>2005$ when $R^{2}=108$, we will be done.\n\nIf $R^{2}=108$, the points $(a,b,c)$ are inside the sphere whenever $|a|,|b|,|c|\\le6$. There are $13^{3}=2197$ points of this form, and $|S|\\ge 2197$ when $|R^{2}|=108$.\n\nIf $R^{2}=48$ and $(a,b,c)\\in S$, we must have $|a|+|b|+|c|\\le 11$ or $|a|=|b|=|c|=4$. There are $1561$ points of the first type and eight of the second, so $|S|\\le 1569$ when $R^{2}=48$\n\nThis is number theory, not calculus.[/quote]\r\nWhat about $(\\pm 1,\\pm 2,\\pm 3)$?", "Solution_4": "I don't see the problem- there are 48 of those, not just 8. Remember that you can permute the elements as well as changing sign.", "Solution_5": "Now,I see,thank you", "Solution_6": "Assume such a ball $E_{R}$ exists. Obviously $R \\geq 1$, and we must have $R < 10$, for otherwise $R$ contains a cube with integer coordinates and volume $14^{3}=2744>2005$.\r\n\r\nBy symmetry, $E_{R}$ contains the point $(0,0,0)$, $m \\geq 1$ points on each of the $6$ branches determined by the usual axes, and $n$ points (strictly) inside each of the $8$ regions determined by those $6$ branches. The total number of points will be $2005=1+6m+8n$, whence $1002=3m+4n$, $m=2 \\ mod \\ 4$ and $3$ divides $n$ (*). Each of the $8$ regions can be divided by a line (with origin $(0,0,0)$ and taking one of the $8$ directions determined by the regions) into $3$ subregions (**) in the obvious fashion (for instance $S$ will be divided by the line $\\{\\alpha,\\alpha,\\alpha;\\ \\alpha \\geq 0\\}$). By symmetry, each of these lines will then contain $3k$ points, not counting $(0,0,0)$ . $E_{R}$ will contain the point $(3k,3k,3k)$, whence $\\sqrt{3}+27k^{2}> R \\geq 27k^{2}$, and then each of the $6$ branches will contain $E(\\sqrt{3\\sqrt{3}+27k^{2}}) \\geq m \\geq 5k$ points , i.e, since $R<10$, $k=1$ and each branch has $m=5$ points. This is impossible because then the ball would be inscribed inside a cube with integer coordinates and volume $12^{3}=1728<2005$ (or because we must have $m=1 \\ mod \\ 4$)." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "E - normed space over R(real numbers) , H - subspace E, dim E/H = 1\r\n\r\nAre?\r\na) H is closed\r\nb) H is dense\r\n\r\nMy prove\r\n\r\n[i]lemma :\n\nf linear is continous <=> ker f is closed[/i]\r\n\r\nBecause dim E/H = 1, so exist g isomorphic from E/H to R (real numbers) and this isomorphic is continous because dim E/H is finite. Ker g = H, so H is closed.\r\nIf H is dense then clH = E. Because H is closed, thus H = clH = E. E/H = E/E, thus dim E/H = dim E/E = 0, contradiction\r\n\r\nIs correct my prove?", "Solution_1": "Yes, your proof is correct. More generally, a subspace of finite codimension (or of finite dimension) is closed." } { "Tag": [ "AMC", "AMC 10" ], "Problem": "Many people will be taking the AMC10 or 12 A tomorrow or in the next few days. Good luck everyone. Hopefully you all will all get good scores. :)", "Solution_1": "Yeah, good luck everyone!", "Solution_2": "I'll be in the cheering section for all the AoPSers.", "Solution_3": "good luck guys! i hope you get the scores you want!" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Suppose P1,P2,P3,P4 are 4 points on the plane.\r\nProve that ( \\sum PiPj)/(min PiPj) \\geq 5+ \\sqrt 3.", "Solution_1": "hm... CMO 2002...." } { "Tag": [ "algebra", "polynomial", "trigonometry", "algebra proposed" ], "Problem": "Let a polynomial $ P(x) \\equal{} rx^3 \\plus{} qx^2 \\plus{} px \\plus{} 1$ $ (r > 0)$ such that the equation $ P(x) \\equal{} 0$ has only one real root. A sequence $ (a_n)$ is defined by $ a_0 \\equal{} 1, a_1 \\equal{} \\minus{} p, a_2 \\equal{} p^2 \\minus{} q, a_{n \\plus{} 3} \\equal{} \\minus{} pa_{n \\plus{} 2} \\minus{} qa_{n \\plus{} 1} \\minus{} ra_n$.\r\nProve that $ (a_n)$ contains an infinite number of nagetive real numbers.", "Solution_1": "No one can solve this olympiad problem ?", "Solution_2": "If $ a_2$ was $ p^2 \\minus{} 2q$ it would be easier. :P \r\n\r\nI will post my progress tomorrow wherever I end up, but I do not expect to have a full solution.", "Solution_3": "$ x = 0$ does not lead to a root.\r\n\r\nThen $ Q(x) = x^4P(\\frac {1}{x}) = x^3 + px^2 + qx + r = (x - d)(x - e)(x - f) = 0$, where d is real and e and f are imaginary. This means $ - p = d + e + f$, and ${ q = de + ef + fe}$.\r\nOur characteristic equation is also $ a^3 + pa^2 + qa + r = 0$. So $ a_n = Ad^n + Be^n + Cf^n$\r\n\r\n$ A + B + C = 1$\r\n$ Ad + Be + Cf = d + e + f$\r\n$ Ad^2 + Be^2 + Cf^2 = p^2 - q = d^2 + e^2 + f^2 - de - ef - fd$\r\n\r\nAnd now I do not even know how to show there are infinite real roots, not just negative roots...", "Solution_4": "[quote=\"thaithuan_GC\"]Let a polynomial $ P(x) \\equal{} rx^3 \\plus{} qx^2 \\plus{} px \\plus{} 1$ $ (r > 0)$ such that the equation $ P(x) \\equal{} 0$ has only one real root. A sequence $ (a_n)$ is defined by $ a_0 \\equal{} 1, a_1 \\equal{} \\minus{} p, a_2 \\equal{} p^2 \\minus{} q, a_{n \\plus{} 3} \\equal{} \\minus{} pa_{n \\plus{} 2} \\minus{} qa_{n \\plus{} 1} \\minus{} ra_n$.\nProve that $ (a_n)$ contains an infinite number of nagetive real numbers.[/quote]\r\n\r\nHere is a very strange demo. I'm sure it exists something simpler, but this is the only demo I found :\r\n\r\nLet $ b$ any root (real or complex) of the polynomial $ Q(x)\\equal{}x^3 \\plus{} px^2 \\plus{} qx \\plus{} r \\equal{} 0$ (notice that since $ P(x)\\equal{}rx^3\\plus{}qx^2\\plus{}px\\plus{}1$ has one unique real root and two complex roots, so has $ Q(x)$ whose roots are the inverses of $ P(x)$).\r\n\r\nObviously, $ b\\neq 0$. Let then $ u_{n\\plus{}1}\\equal{}a_{n\\plus{}3}\\plus{}(p\\plus{}b)a_{n\\plus{}2}\\minus{}\\frac{r}{b}a_{n\\plus{}1}$ (sequence of complex number)\r\n\r\n$ u_{n\\plus{}1}\\equal{}\\minus{}pa_{n\\plus{}2}\\minus{}qa_{n\\plus{}1}\\minus{}ra_n\\plus{}(p\\plus{}b)a_{n\\plus{}2}\\minus{}\\frac{r}{b}a_{n\\plus{}1}$\r\n\r\n$ u_{n\\plus{}1}\\equal{}ba_{n\\plus{}2}\\minus{}(q\\plus{}\\frac{r}{b})a_{n\\plus{}1}\\minus{}ra_n$\r\n\r\n$ u_{n\\plus{}1}\\equal{}b(a_{n\\plus{}2}\\minus{}\\frac{r\\plus{}bq}{b^2}a_{n\\plus{}1}\\minus{}\\frac{r}{b}a_n)$\r\n\r\nBut $ b^3\\plus{}pb^2\\plus{}qb\\plus{}r\\equal{}0$ $ \\implies$ $ p\\plus{}b\\equal{}\\minus{}\\frac{r\\plus{}bq}{b^2}$ and so $ u_{n\\plus{}1}\\equal{}b(a_{n\\plus{}2}\\plus{}(p\\plus{}b)a_{n\\plus{}1}\\minus{}\\frac{r}{b}a_n)\\equal{}b u_{n}$\r\n$ u_0\\equal{}a_{2}\\plus{}(p\\plus{}b)a_{1}\\minus{}\\frac{r}{b}a_{0}\\equal{}p^2\\minus{}q\\minus{}p(p\\plus{}b)\\minus{}\\frac{r}{b}$ $ \\equal{}\\minus{}\\frac{pb^2\\plus{}qb\\plus{}r}{b}$ $ \\equal{}\\frac{b^3}{b}$ $ \\equal{}b^2$\r\n\r\nAnd so $ u_n\\equal{}b^{n\\plus{}2}$ and so $ \\boxed{a_{n\\plus{}2}\\plus{}(p\\plus{}b)a_{n\\plus{}1}\\minus{}\\frac{r}{b}a_{n}\\equal{}b^{n\\plus{}2}}$\r\n\r\nThis equation is true for the two complex roots of $ Q(x)$ : $ z\\equal{}\\rho e^{i\\theta}$ and $ \\overline{z}\\equal{}\\rho e^{\\minus{}i\\theta}$. So :\r\n$ a_{n\\plus{}2}\\plus{}(p\\plus{}z)a_{n\\plus{}1}\\minus{}\\frac{r}{z}a_{n}\\equal{}z^{n\\plus{}2}$\r\n\r\n$ a_{n\\plus{}2}\\plus{}(p\\plus{}\\overline{z})a_{n\\plus{}1}\\minus{}\\frac{r}{\\overline{z}}a_{n}\\equal{}\\overline{z}^{n\\plus{}2}$\r\n\r\nSubtracting these two equalities :\r\n$ (z\\minus{}\\overline{z})a_{n\\plus{}1}\\minus{}r(\\frac{1}{z}\\minus{}\\frac{1}{\\overline{z}})a_{n}\\equal{}z^{n\\plus{}2}\\minus{}\\overline{z}^{n\\plus{}2}$\r\n\r\n$ a_{n\\plus{}1}\\plus{}\\frac{r}{\\rho^2}a_{n}\\equal{}\\frac{z^{n\\plus{}2}\\minus{}\\overline{z}^{n\\plus{}2}}{z\\minus{}\\overline{z}}$\r\n\r\n$ a_{n\\plus{}1}\\plus{}\\frac{r}{\\rho^2}a_{n}\\equal{}\\rho^{n\\plus{}1}\\frac{\\sin((n\\plus{}2)\\theta)}{\\sin(\\theta)}$\r\n\r\nThe right part of this equality reaches infinitely many negative values when $ n$ increases. So do the left part.\r\nSo, since $ \\frac{r}{\\rho^2}>0$, at least one of $ a_n$ or $ a_{n\\plus{}1}$ is negative for any such $ n$.\r\n\r\nHence the result.", "Solution_5": "How did you get that $ Q(x)$ has roots $ z\\equal{}\\rho e^{i\\theta}$ and $ \\overline{z}\\equal{}\\rho e^{\\minus{}i\\theta}$?\r\n\r\nAlso, how did you decide on your choice for $ u_n$?", "Solution_6": "[quote=\"Differ\"]How did you get that $ Q(x)$ has roots $ z \\equal{} \\rho e^{i\\theta}$ and $ \\overline{z} \\equal{} \\rho e^{ \\minus{} i\\theta}$?[/quote]\n\nIf $ z\\notin\\mathbb{R}$ is a zero of $ P(x)\\in\\mathbb{R}[x]$, then $ P(z)\\equal{}0$ $ \\implies$ $ \\overline{P(z)}\\equal{}0$ $ \\implies$ $ P(\\overline{z})\\equal{}0$ $ \\implies$ $ \\overline{z}$ is also a root\n\n[quote=\"Differ\"] Also, how did you decide on your choice for $ u_n$?[/quote]\r\n\r\nI tried to find some $ \\alpha$, $ \\beta$ and $ \\gamma$ such that $ a_{n\\plus{}3}\\plus{}\\alpha a_{n\\plus{}2}\\plus{}\\beta a_{n\\plus{}1}$ $ \\equal{}\\gamma(a_{n\\plus{}2}\\plus{}\\alpha a_{n\\plus{}1}\\plus{}\\beta a_{n})$", "Solution_7": "[quote=\"pco\"][quote=\"Differ\"]How did you get that $ Q(x)$ has roots $ z \\equal{} \\rho e^{i\\theta}$ and $ \\overline{z} \\equal{} \\rho e^{ \\minus{} i\\theta}$?[/quote]\n\nIf $ z\\notin\\mathbb{R}$ is a zero of $ P(x)\\in\\mathbb{R}[x]$, then $ P(z) \\equal{} 0$ $ \\implies$ $ \\overline{P(z)} \\equal{} 0$ $ \\implies$ $ P(\\overline{z}) \\equal{} 0$ $ \\implies$ $ \\overline{z}$ is also a root\n\n[quote=\"Differ\"] Also, how did you decide on your choice for $ u_n$?[/quote]\n\nI tried to find some $ \\alpha$, $ \\beta$ and $ \\gamma$ such that $ a_{n \\plus{} 3} \\plus{} \\alpha a_{n \\plus{} 2} \\plus{} \\beta a_{n \\plus{} 1}$ $ \\equal{} \\gamma(a_{n \\plus{} 2} \\plus{} \\alpha a_{n \\plus{} 1} \\plus{} \\beta a_{n})$[/quote]\r\n\r\nThank you. :) I did not see the $ r > 0$ restriction in the original problem.", "Solution_8": "[quote=\"Differ\"]$ x = 0$ does not lead to a root.\n\nThen $ Q(x) = x^4P(\\frac {1}{x}) = x^3 + px^2 + qx + r = (x - d)(x - e)(x - f) = 0$, where d is real and e and f are imaginary. This means $ - p = d + e + f$, and ${ q = de + ef + fe}$.\nOur characteristic equation is also $ a^3 + pa^2 + qa + r = 0$. So $ a_n = Ad^n + Be^n + Cf^n$\n\n$ A + B + C = 1$\n$ Ad + Be + Cf = d + e + f$\n$ Ad^2 + Be^2 + Cf^2 = p^2 - q = d^2 + e^2 + f^2 - de - ef - fd$\n\nAnd now I do not even know how to show there are infinite real roots, not just negative roots...[/quote]\r\n\r\nWe can actually solve this equation!\r\nLook at it - it's 3 variables A,B,C, and three equations lol\r\n\r\nIn fact, if you find the solution and substitute it into the original equation,\r\nyou get $ a_n = \\frac{(e-f)d^{n+3} + (f-d)e^{n+3} + (d-e)f^{n-3}}{(d-e)(e-f)(d-f)}$\r\nAnd by observing that d is negative and letting $ e=r(cos\\theta + sin\\theta ), f=r(cos\\theta - sin\\theta )$ the result follows." } { "Tag": [], "Problem": "B\u00e1c n\u00e0o \u1ee7ng h\u1ed9 th\u00ec v\u00e0o \u0111\u00e2y ! :)\r\n\r\nC\u00e1c b\u00e1c c\u00f3 th\u1ec3 post theo m\u1eabu sau :B\u00f4i \u0111en r\u1ed3i copy v\u00e0o b\u00e0i post (Cho n\u00f3 th\u1ed1ng nh\u1ea5t)\r\n\r\n[code][b][color=darkblue]TH.s\u1ed1 th\u1ee9 t\u1ef1[/color] (T\u00e1c gi\u1ea3)[/b]\n[hide]\u0110\u1ec1 b\u00e0i[/hide] \n[color=darkblue][b]L\u1eddi gi\u1ea3i[/b][/color]\n[b]L\u1eddi gi\u1ea3i 1 (T\u00e1c gi\u1ea3)[/b]\n[hide]L\u1eddi gi\u1ea3i[/hide][/code]", "Solution_1": "Cac bac chung minh ho dong nhat thuc nay cai\r\n\r\n[img]http://sweb.cz/Felix.Phan/combin.JPG[/img]" } { "Tag": [ "function", "MATHCOUNTS" ], "Problem": "Using nickels, dimes, quarters and/or half-dollars, how many ways can you make $ 75$ cents?\r\n\r\n***I would just list all of them out, but that would take a long time. Is there a faster way to do it?***\r\n\r\n[hide=\"Answer\"]22[/hide]", "Solution_1": "[quote=\"Tiger3\"]Using nickels, dimes, quarters and/or half-dollars, how many ways can you make $ 75$ cents?\n\n***I would just list all of them out, but that would take a long time. Is there a faster way to do it?***\n\n[hide=\"Answer\"]22[/hide][/quote]\nI can't think of faster ways of doing it without listing them out, but I could improve on the speed at which you count the ones listed out.\n[hide]\nOrganize them by the amount of halfdollars and quarters used and just add the ways...\n1 halfdollar: 4 ways\n3 quarters: 1 way\n2 quarters: 3 ways\n1 quarter: 6 ways\n0 quarter: 8 ways\n$ 4\\plus{}1\\plus{}3\\plus{}6\\plus{}8\\equal{}22$\n[/hide]", "Solution_2": "There was a huge topic about a problem similar to this. For this, it's easy enough to list, but for problem where it asks for $ 750$ or $ 7500$?. Generating functions are the way to go.", "Solution_3": "[quote=\"xpmath\"]There was a huge topic about a problem similar to this. For this, it's easy enough to list, but for problem where it asks for $ 750$ or $ 7500$?. Generating functions are the way to go.[/quote]\r\n\r\nThat does somewhat depent on if the problem sticks to our currency or if it uses some other thing, but I agree with the generating function idea. However, further discussion should stay away from generating functions, I think it's gotten a lot of topics locked.", "Solution_4": "Was this from Mathcounts (If so, what round?) or from some other source?\r\nMathcounts wouldn't want you to list them out like that. :D", "Solution_5": "Split them into groups based on how many quarters you're using. I find that helps me." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $\\ x,y,z$ whith sum $\\ 1$ be real positive numbers.Show that the following inequality holds\r\n$\\frac{x}{xy+1}+\\frac{y}{yz+1}+\\frac{z}{zx+1}\\geq\\frac{36xyz}{13xyz+1}$\r\n $\\ Marius\\ Ghergu,\\ Romania$", "Solution_1": "It is equivalent to \r\n\r\n$\\frac{x^2y}{xy+1}+\\frac{y^2z}{yz+1}+\\frac{z^2x}{zx+1}\\leq \\frac{1-23xyz}{13xyz+1}$", "Solution_2": "...and, how can you prove that?maybe Chebishev? :?", "Solution_3": "[b]Problem:[/b] Let be given reals $x,y,z>0$ satisfy the condition $x+y+z=1$. Prove that:\r\n\\[ \\frac{x}{xy+1}+\\frac{y}{yz+1}+\\frac{z}{zx+1} \\ge \\frac{36xyz}{1+13xyz}. \\]\r\n\r\n[b]Solution:[/b]\r\nFirst, we have:\r\n\\[ \\frac{x}{xy+1} \\ge x-\\frac{x}{1+27z} \\iff xuz \\le \\frac1{27}. \\]\r\nSumming three ineqs, we obtain:\r\n\\[ \\text{LHS} \\ge x+y+z-\\left(\\frac{x}{1+27z}+\\frac{y}{1+27x}+\\frac{z}{1+27y}\\right). \\]\r\nWe must prove:\r\n\\[ \\frac{x}{1+27z}+\\frac{y}{1+27x}+\\frac{z}{1+27y} \\le \\frac{1-23xyz}{1+13xyz}. \\]\r\nIn other facts,\r\n\\[ \\frac{x}{1+27z}+\\frac{y}{1+27x}+\\frac{z}{1+27y} \\le \\frac{1}{10} \\le \\frac{1-23xyz}{1+13xyz}. \\]\r\nThis is obvious, since $xyz \\le \\frac{1}{27}$ and the left part is equivalent to:\r\n\\[ 18+189(xy+yz+zx) \\ge 3.27^2.xyz \\\\ \\iff 18(x+y+z)^3+189(x+y+z)(xy+yz+zx) \\ge 3.27^2.xyz, \\]\r\nTrue by AM-GM.\r\nDone.", "Solution_4": "Your last step is wrong.", "Solution_5": "Which and why?", "Solution_6": "[quote=\"Lovasz\"]\nIn other facts,\n\\[ \\frac{x}{1+27z}+\\frac{y}{1+27x}+\\frac{z}{1+27y} \\le \\frac{1}{10} \\le \\frac{1-23xyz}{1+13xyz}. \\]\nThis is obvious, since $xyz \\le \\frac{1}{27}$ and the left part is equivalent to:\n\\[ 18+189(xy+yz+zx) \\ge 3.27^2.xyz \\\\ \\iff 18(x+y+z)^3+189(x+y+z)(xy+yz+zx) \\ge 3.27^2.xyz, \\]\nTrue by AM-GM.\nDone.[/quote]\r\n\r\nI don't understand this last part because I think you used AM-GM inequality with reversed sign in last inequality and then your LHS cannot be less than $1/10$ if you put $x=1, y=0, z=0$", "Solution_7": "[quote=\"silouan\"]It is equivalent to \n\n$\\frac{x^2y}{xy+1}+\\frac{y^2z}{yz+1}+\\frac{z^2x}{zx+1}\\leq \\frac{1-23xyz}{13xyz+1}$ (1) [/quote]\r\n\r\nAfter this form we have that \r\n$\\frac{x^2y}{xy+1}+\\frac{y^2z}{yz+1}+\\frac{z^2x}{zx+1}\\leq \\frac{1}{6}$ .\r\nSo (1) holds for $xyz\\leq\\frac{5}{151}$ \r\nNow if $\\frac{1}{27}\\geq xyz\\geq\\frac{5}{151}$ then \r\nthe first inequality is obvious", "Solution_8": "Silouan, can you be more detailed, please. I don't understand your proof. I cannot solve it :huh: \r\n\r\n\r\nThank you very much. :)", "Solution_9": "I worked exactly the same as here\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=cyclic&t=21689", "Solution_10": "[quote=\"silouan\"]\nNow if $\\frac{1}{27}\\geq xyz\\geq\\frac{5}{151}$ then \nthe first inequality is obvious[/quote]\r\n\r\nCan you please explain to me this step?\r\n\r\nThank you very much.", "Solution_11": "Sorry I thought it was an easy part and the other was the hard but it is more complicated .I have not a proof for \r\n$\\frac{1}{27}\\geq xyz\\geq \\frac{1}{151}$", "Solution_12": "[quote=\"stancioiu sorin\"]Let $\\ x,y,z$ whith sum $\\ 1$ be real positive numbers.Show that the following inequality holds\n$\\frac{x}{xy+1}+\\frac{y}{yz+1}+\\frac{z}{zx+1}\\geq\\frac{36xyz}{13xyz+1}$\n $\\ Marius\\ Ghergu,\\ Romania$[/quote]\r\n$\\sum_{cyc}\\frac{x}{xy+1}=\\sum_{cyc}\\frac{1}{y+\\frac{1}{x}}\\geq\\frac{9}{1+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}}=\\frac{9xyz}{xyz+xy+xz+yz}.$\r\nHence, remain to prove that $\\frac{9xyz}{xyz+xy+xz+yz}\\geq\\frac{36xyz}{13xyz+1}\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(x^3-x^2y-x^2z+xyz)\\geq0.$ :)", "Solution_13": "Nice solution, arqady. :lol:" } { "Tag": [ "geometry" ], "Problem": "From this years MMPC:\r\n\r\nWhat is the area of the region in the first quadrant bounded by the x-axis, the line y=2x, and the circles x^2+y^2=20 and x^2+y^2=30?", "Solution_1": "That isn't really a pleasant-looking number, is it?", "Solution_2": "no, but you can leave it unsimplified. these were the choices:\r\n\r\n5 arctan 2\r\n5/4 * pi\r\n5/2 * pi\r\n5\r\nNone of the Above", "Solution_3": "Mmm, not sure if trigonometry's appropriate for MATHCOUNTS...oh well :)", "Solution_4": "Hmm, the question is sorta weird. If you draw the picture, you see two different regions that are bounded by the given systems. But assuming its the bigger one, you get\n\n\n\n[hide]\n\nLet the angle created by y=2x and the x axis be theta. tan theta = 2, so the angle is arctan 2. The region of the circle is arctan 2 (30pi - 20 pi) = arctan2 * 10pi, so thats my answer - none of hte above.\n\n[/hide]" } { "Tag": [ "Euler", "trigonometry" ], "Problem": "\u039f \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 $ {\\Omega_{ABC}}$ \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ ABC$ \u03b5\u03c6\u03ac\u03c0\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 $ BC,CA,AB$ \u03c3\u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ P,Q,R$ \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1. \u0388\u03c3\u03c4\u03c9 \r\n$ M = PQ\\cap BA$ \u03ba\u03b1\u03b9 $ N = PR\\cap CA$. O \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 $ \\Omega_{MNP}$ \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ MNP$ \u03b5\u03c6\u03ac\u03c0\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 $ MN, MP$ \u03c3\u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ X,Y$ \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1. \u0391\u03bd \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ X,Y,B$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac, \u03c4\u03cc\u03c4\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9:\r\n(a) \u039f\u03b9 \u03ba\u03cd\u03ba\u03bb\u03bf\u03b9 $ {\\Omega_{ABC}}$ \u03ba\u03b1\u03b9 $ \\Omega_{MNP}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf\u03b9, \u03ba\u03b1\u03b9 \r\n(b) \u03c4\u03ad\u03bc\u03bd\u03bf\u03c5\u03bd \u03bf \u03ad\u03bd\u03b1\u03c2 \u03c4\u03bf\u03bd \u03ac\u03bb\u03bb\u03bf \u03c3\u03b5 \u03c4\u03cc\u03be\u03b1 $ 60^o$.\r\n\r\n\r\n[i]\u0391\u03c6\u03b9\u03b5\u03c1\u03c9\u03bc\u03ad\u03bd\u03b7 \u03c3\u03c4\u03bf \u03a3\u03c4\u03ad\u03c6\u03b1\u03bd\u03bf \u0391\u03c1\u03b5\u03c4\u03ac\u03ba\u03b7 (staretak)...[/i]\r\n\r\n[color=red][EDIT] \u03a0\u03b1\u03b9\u03b4\u03b9\u03ac \u03c7\u03af\u03bb\u03b9\u03b1 \u03c3\u03c5\u03b3\u03b3\u03bd\u03bd\u03ce\u03bc\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03bf typo.\u03a4\u03ce\u03c1\u03b1 \u03c4\u03bf \u03b4\u03b9\u03cc\u03c1\u03b8\u03c9\u03c3\u03b1.[/color]", "Solution_1": "\u039d\u03af\u03ba\u03bf \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03a7,\u03a5,\u0392 \u03b4\u03b5\u03bd \u03b3\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac . (\u039c\u03ae\u03c0\u03c9\u03c2 \u03b5\u03bd\u03bd\u03bf\u03b5\u03af\u03c2 \u03b1\u03bb\u03bb\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1??) . \u03a6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u03c3\u03c7\u03ae\u03bc\u03b1 . \u03a0\u03b5\u03c2 \u03bc\u03bf\u03c5 \u03b1\u03bd \u03ba\u03ac\u03bd\u03c9 \u03bb\u03ac\u03b8\u03bf\u03c2", "Solution_2": "\u03a0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03cc typo \u03ba\u03b9 \u03bf \u039d\u03af\u03ba\u03bf\u03c2 \u03b5\u03bd\u03bd\u03bf\u03b5\u03af \u03cc\u03c4\u03b9 $ Y$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c0\u03bf\u03c5 \u03bf \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c2 \u03b5\u03c6\u03b1\u03c0\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd $ PM$ \u03ba\u03b9 \u03cc\u03c7\u03b9 \u03c3\u03c4\u03b7\u03bd $ PN$ \u03ae \u03b1\u03bb\u03bb\u03b9\u03ce\u03c2(\u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03be\u03ac\u03bb\u03bb\u03bf\u03c5) \u03b5\u03bd\u03bd\u03bf\u03b5\u03af \u03cc\u03c4\u03b9 \u03c4\u03b1 $ X,Y,C$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac!\r\n\r\n\u03a0\u03ac\u03bd\u03c4\u03c9\u03c2 \u039d\u03af\u03ba\u03bf \u03b4\u03b9\u03b5\u03c5\u03ba\u03c1\u03af\u03bd\u03b9\u03c3\u03b5 plz \u03c4\u03b9 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03b5\u03bd\u03bd\u03bf\u03b5\u03af\u03c2!", "Solution_3": "[quote=\"Nick Rapanos\"]\u039f \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 $ {\\Omega_{ABC}}$ \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ ABC$ \u03b5\u03c6\u03ac\u03c0\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 $ BC,CA,AB$ \u03c3\u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ P,Q,R$ \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1. \u0388\u03c3\u03c4\u03c9 \n$ M = PQ\\cap BA$ \u03ba\u03b1\u03b9 $ N = PR\\cap CA$. O \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 $ \\Omega_{MNP}$ \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ MNP$ \u03b5\u03c6\u03ac\u03c0\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 $ MN, NP$ \u03c3\u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ X,Y$ \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1. \u0391\u03bd \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ X,Y,B$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac, \u03c4\u03cc\u03c4\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9:\n(a) \u039f\u03b9 \u03ba\u03cd\u03ba\u03bb\u03bf\u03b9 $ {\\Omega_{ABC}}$ \u03ba\u03b1\u03b9 $ \\Omega_{MNP}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf\u03b9, \u03ba\u03b1\u03b9 \n(b) \u03c4\u03ad\u03bc\u03bd\u03bf\u03c5\u03bd \u03bf \u03ad\u03bd\u03b1\u03c2 \u03c4\u03bf\u03bd \u03ac\u03bb\u03bb\u03bf \u03c3\u03b5 \u03c4\u03cc\u03be\u03b1 $ 60^o$.\n\n[i]\u0391\u03c6\u03b9\u03b5\u03c1\u03c9\u03bc\u03ad\u03bd\u03b7 \u03c3\u03c4\u03bf \u03a3\u03c4\u03ad\u03c6\u03b1\u03bd\u03bf \u0391\u03c1\u03b5\u03c4\u03ac\u03ba\u03b7 (staretak)...[/i][/quote]\r\n\u039a\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03cd\u03b1\u03c3\u03b1 \u03c4\u03b1 \u03c3\u03c7\u03ae\u03bc\u03b1 \u03be\u03b5\u03ba\u03b9\u03bd\u03ce\u03bd\u03c4\u03b1\u03c2 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03c1\u03bf\u03c6\u03b1, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 \u03af\u03c3\u03bf\u03c5\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5\u03c2 \u03bc\u03b5 \u03ba\u03bf\u03b9\u03bd\u03ae \u03c7\u03bf\u03c1\u03b4\u03ae \u03af\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 \u03c4\u03bf\u03c5\u03c2.\r\n\r\n\u0392\u03bb\u03ad\u03c0\u03c9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd\u03c4\u03b1 \u03b1\u03c0\u03bf\u03c4\u03b5\u03bb\u03ad\u03c3\u03bc\u03b1\u03c4\u03b1 \u03bc\u03b5 \u03c4\u03bf\u03bd \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03ae, \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03bb\u03cd\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03b2\u03ac\u03b6\u03c9 hind \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7 \u03c3\u03b1\u03c2 \u03bc\u03c0\u03b5\u03c1\u03b4\u03ad\u03c8\u03c9.\r\n\r\n\u03ba\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.", "Solution_4": "\u0386\u03c1\u03b1 \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03c4\u03bf Y \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03b7\u03bd MP .", "Solution_5": "[quote=\"kritikos\"]\u0386\u03c1\u03b1 \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03c4\u03bf Y \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03b7\u03bd MP .[/quote]\r\n\r\nTi ginetai re paidia? kaneis gi auth?", "Solution_6": "\u0393\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03c0\u03bf\u03bd\u03b9\u03ad\u03c4\u03b1\u03b9 \u03bf Nick \u03b2\u03c1\u03ae\u03ba\u03b1 \u03bc\u03b9\u03b1 \u03bb\u03cd\u03c3\u03b7 ...\r\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03b7 $ XY$ \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd $ CN$ \u03c3\u03c4\u03bf $ D$ . \u0388\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $ (M,A,R,B) \\equal{} (X,D,Y,B) \\equal{} \\minus{} 1$ (1) \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd \u03b1\u03c6\u03bf\u03cd \u03c4\u03bf $ B$ \u03b1\u03bd\u03ae\u03ba\u03b5\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03bf\u03bb\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 $ N$ \u03ba\u03b1\u03b9 \u03c4\u03bf $ N$ \u03b1\u03bd\u03ae\u03ba\u03b5\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03bf\u03bb\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 $ B$ \u03ba\u03b1\u03b9 \u03bb\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 (1) \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b7 $ CN$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03bf\u03bb\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 $ B$ .\r\n\r\nE\u03c0\u03b9\u03c3\u03b7\u03c2 \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 $ CN$ \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b9\u03c2 \u03b5\u03c6\u03b1\u03c0\u03c4\u03cc\u03bc\u03b5\u03bd\u03b5\u03c2 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 $ (K)$ \u03b1\u03c0\u03cc \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ M$ \u03c3\u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ Q,\\ N$ \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03c4\u03bf $ A$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03cc \u03c3\u03c5\u03b6\u03c5\u03b3\u03ad\u03c2 \u03c4\u03bf\u03c5 $ C,$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b1 $ Q,\\ N$ \u03c3\u03c5\u03bc\u03c0\u03b5\u03c1\u03b1\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b7 \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 $ AM\\equiv BM,$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03bf\u03bb\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 $ C,$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf $ (K)$ \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03c0\u03b5\u03c1\u03bd\u03ac\u03b5\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf $ M$ \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b7 \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 $ XZ,$ \u03c9\u03c2 \u03b7 \u03c0\u03bf\u03bb\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 $ M$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf $ (K),$ \u03c0\u03b5\u03c1\u03bd\u03ac\u03b5\u03b9 \u03b1\u03c0\u03cc $ C$ \u03ba\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ X,\\ Z,\\ C,$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac.\r\n\r\n\u0391\u03c6\u03bf\u03cd \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03c7\u03c4\u03b7\u03ba\u03b5 \u03cc\u03c4\u03b9 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ X,\\ Z,\\ C,$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac, \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03a0\u03ac\u03c0\u03c0\u03bf\u03c5 \u03cc\u03c4\u03b9 \r\n\u03cc\u03c4\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ Y\\equiv BX\\cap PM,$ $ A\\equiv BN\\cap CM$ \u03ba\u03b1\u03b9 $ Z\\equiv CX\\cap PN,$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac\r\n\r\n\u0391\u03c6\u03bf\u03cd \u03c4\u03ce\u03c1\u03b1 \u03c4\u03b1 Y,A,Z \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac \u03b1\u03c0\u03cc \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 Brianchon \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03c0\u03bb\u03b5\u03c5\u03c1\u03bf $ NMQR$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b9\u03b3\u03b3\u03c1\u03ac\u03c8\u03b9\u03bc\u03bf .\r\n(\u0393\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03b4\u03b5\u03af\u03c4\u03b5 \u03b1\u03c5\u03c4\u03cc \u03b1\u03c0\u03bb\u03ac \u03b8\u03b5\u03c9\u03c1\u03b5\u03af\u03c3\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b7 \u03b5\u03c6\u03b1\u03c0\u03c4\u03bf\u03bc\u03ad\u03bd\u03b7 \u03b1\u03c0\u03cc \u03c4\u03bf $ R$ \u03c3\u03c4\u03bf\u03bd $ (K)$ \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd $ MP$ \u03c3\u03c4\u03bf $ Q'$ . \u03a4\u03cc\u03c4\u03b5 \u03bf\u03b9 \u03b5\u03c5\u03b8\u03b5\u03af\u03b5\u03c2 \r\nNQ',MR,YZ \u03c3\u03c5\u03bd\u03c4\u03c1\u03ad\u03c7\u03bf\u03c5\u03bd \u03ac\u03c1\u03b1 $ Q\\equiv Q'$ ) kai \u03b8\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 $ S$ \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b5\u03c0\u03b1\u03c6\u03ae\u03c2 .\r\n\r\n\u03a4\u03bf $ Q$ \u03c4\u03ce\u03c1\u03b1 \u03b1\u03bd\u03ae\u03ba\u03b5\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03bf\u03bb\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 $ B$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf\u03bd $ (K)$ \u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 $ Z,S,B$ \u03c3\u03c5\u03bd\u03b5\u03c5\u03b5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 $ F$ \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03bc\u03ae\u03c2 \u03c4\u03b7\u03c2 \r\n$ ZB$ \u03bc\u03b5 \u03c4\u03b7\u03bd $ YP$ . \u03a4\u03cc\u03c4\u03b5 $ (X,G,Z,C) \\equal{} (Y,R,F,P) \\equal{} \\minus{} 1$ (2) \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd \u03c4\u03bf $ Y$ \u03b1\u03bd\u03ae\u03ba\u03b5\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03bf\u03bb\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 $ B$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf\u03bd $ (I)$ \r\n\u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 \u03b7 \u03c0\u03bf\u03bb\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 $ Y$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf\u03bd $ (I)$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 $ ZB$ \u03ac\u03c1\u03b1 $ YI\\bot ZF$ \u03ba\u03b1\u03b9 \u03b1\u03c6\u03bf\u03cd \u03b5\u03af\u03bd\u03b1\u03b9 $ QK\\bot ZS$ \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 $ KE//YI$ (3)\r\n\r\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03b7 $ KQ$ \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03bf\u03bd $ (I)$ \u03c3\u03c4\u03bf $ E$ . \u03a4\u03cc\u03c4\u03b5 $ \\angle{ERQ} \\equal{} \\angle{PQE} \\equal{} \\angle{ZQK} \\equal{} \\angle{KQR} \\equal{} \\angle{RPE}$ \r\n\u03ac\u03c1\u03b1 \u03c4\u03bf $ E$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf \u03c4\u03bf\u03c5 \u03c4\u03cc\u03be\u03bf\u03c5 $ RP$ \u03ac\u03c1\u03b1 $ IE\\bot RP$ \u03ba\u03b1\u03b9 \u03b1\u03c6\u03bf\u03cd $ KY\\bot PN$ \u03b5\u03af\u03bd\u03b1\u03b9 $ IE//KY$ (4) \u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 \u03bb\u03cc\u03b3\u03c9 (3),(4)\r\n\u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 $ KYIE$ \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b7\u03bb\u03cc\u03b3\u03c1\u03b1\u03bc\u03bc\u03bf \u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 $ IE \\equal{} KY$ , \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \r\n\r\n\u0393\u03b9\u03b1 \u03c4\u03bf \u03b2) \u03c4\u03ce\u03c1\u03b1 \u03b2\u03bb\u03ad\u03c0\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c4\u03bf $ K$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03c0\u03b1\u03c1\u03ac\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 $ PQR$ \u03ba\u03b1\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 Euler \u03cc\u03c4\u03b9 \r\n$ KI^2 \\equal{} r(r \\plus{} 2r_a)$ \u03ba\u03b1\u03b9 \u03b1\u03c6\u03bf\u03cd $ r \\equal{} r_a$ \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 $ KI^2 \\equal{} 3r^2$ \r\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03bf\u03b9 \u03ba\u03cd\u03ba\u03bb\u03bf\u03b9 \u03c4\u03ad\u03bc\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ U,V$ \u03c4\u03cc\u03c4\u03b5 \u03b1\u03c0\u03cc \u03c4\u03bf \u03bd\u03cc\u03bc\u03bf \u03c3\u03c5\u03bd\u03b7\u03bc\u03b9\u03c4\u03cc\u03bd\u03c9\u03bd \u03c3\u03c4\u03bf $ KVI$ \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \r\n$ \\cos\\angle{ KVI} \\equal{} \\minus{} \\frac {1}{2}$ \u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 $ \\angle{KVI} \\equal{} 120$ \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 $ UVI \\equal{} 60$ and so a big QED :)\r\n\r\n\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c4\u03bf\u03c5\u03c2 \u039a\u03ce\u03c3\u03c4\u03b1 \u0392\u03ae\u03c4\u03c4\u03b1 , \u03a4\u03ac\u03c3\u03bf \u0392\u03b1\u03c6\u03b5\u03af\u03b4\u03b7 , \u039c\u03b9\u03c7\u03ac\u03bb\u03b7 \u03a3\u03ac\u03b2\u03b2\u03b1 \u03bc\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03bf\u03c0\u03bf\u03af\u03bf\u03c5\u03c2 \u03c3\u03c5\u03b6\u03b7\u03c4\u03ae\u03c3\u03b1\u03bc\u03b5 \u03c4\u03bf problem .", "Solution_7": "oraia lysi silouan!", "Solution_8": "\u03a4\u03b9 \u03ac\u03c3\u03c7\u03b5\u03c4\u03bf\u03c2 \u03c0\u03bf\u03c5 \u03b5\u03af\u03c3\u03b1\u03b9 \u03c1\u03b5 \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad... \u03a3\u03b9\u03b3\u03ac \u03c4\u03b9\u03c2 \u03b3\u03bd\u03ce\u03c3\u03b5\u03b9\u03c2 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03af\u03b1\u03c2 \u03c0\u03bf\u03c5 \u03b8\u03ad\u03bb\u03b5\u03b9 \u03b1\u03c5\u03c4\u03ae \u03b7 \u03bb\u03cd\u03c3\u03b7... :rotfl: \r\n\r\n\u03a3\u03c5\u03b3\u03c7\u03b1\u03c1\u03b7\u03c4\u03ae\u03c1\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03bf\u03bc\u03ac\u03b4\u03b1. \r\n\r\n\u039a\u03b1\u03b9\u03c1\u03cc\u03c2 \u03bd\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c1\u03b7\u03bc\u03ac\u03b4\u03b9 \u03c4\u03bf \u03c0\u03b1\u03bb\u03b9\u03bf\u03bc\u03b5\u03c4\u03ac\u03bb\u03bb\u03b9\u03bf \u03ae \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03b1\u03c5\u03c4\u03ac \u03c4\u03b1 \u03c1\u03b7\u03bc\u03ac\u03b4\u03b9\u03b1 \u03b4\u03b9\u03cc\u03c4\u03b9 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03cd\u03bf. :)\r\n\r\n\u03a3\u03c5\u03b3\u03c7\u03b1\u03c1\u03b7\u03c4\u03ae\u03c1\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03cc\u03bb\u03b7 \u03c4\u03b7\u03bd \u03b5\u03be\u03ac\u03b4\u03b1. \u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 dream team. \u0394\u03b5 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03b7. :D" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Does there exist a sequence of natural numbers satisfying the relation\r\n$\\frac{1}{a_n}=\\frac{1}{a_{n+1}}+\\frac{1}{a_{n+2}}$?", "Solution_1": "I suppose your question is: does there exist [b]infinite[/b] sequence...\r\n\r\nThen this question appeared on first round of Polish MO\r\n\r\nHint: use substitution $x_n=\\frac{1}{a_n}$ then $(x_n)$ has to be convergent to zero if such a sequence would exist. Find explicit formula of $x_n$ and show that leads to a contradiction. (You'll get $\\frac{x_2}{x_1}=\\frac{\\sqrt{5}-1}{2}$).", "Solution_2": "Thanks for help,but what i want is a solution using divisibility!\r\nI have solved with the way you have written,and i have solved by \r\nfinding The first few terms!(using Fibonacci sequence)Any one?" } { "Tag": [], "Problem": "hi ,\r\ncould any body solve this problem for me!!\r\n\r\nBox W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 19 inches less that the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?\r\n(A) 3\r\n(B) 6\r\n(C) 12\r\n(D) 18\r\n(E) 24\r\n", "Solution_1": "If I misunderstand, please show me!\r\n\r\nLet the average length, in inches, of the sticks in Box W is A\r\n the average length, in inches, of the sticks in Box V is B\r\n length of each red stick is X\r\nAccording to the infomations, we have\r\n\r\nX + 19 =A\r\nX - 6 =B\r\n\r\n=> A-B = 19+6 =25\r\n\r\nI wonder why my solution isn't in these choices! [/hide][/code]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "A numbers is said to be a beautiful number if it is a composite number and but it is not a multiple of either $ 2;3$ or $ 5$(for example,the three smallest beautiful numbers are $ 49,77$ and $ 91$)\r\nHow many beautiful numbers which are less than $ 1000$", "Solution_1": "[hide=\"hint\"]These type of problems can be solved easily with Inclusion and Exclusion principle.[/hide]\n[hide=\"solution\"]we can use that, $ 1000$ will be eliminated. So, for easier calculation we count the numbers $ \\leq1000$ instead of $ < 1000$. \nUsing inclusion exclusion principle we get, the number of positive integers $ \\leg 1000$ and multiple of $ 2,3,5$ is $ 734$.\nNow, we know that, we have $ 168$ primes $ \\leq 1000$.\nSo, the number of [i]beautiful numbers[/i] is $ 1000 - 734 - 168 + 3 = 101$\n\nThis problem is too easy for olympiad section, please post these types of problems in pre-olympiad section.[/hide]", "Solution_2": "Er, and how do we know there are 168 primes less than 1000? :S\r\nAnd why is the $ \\plus{}3$ there?", "Solution_3": "[quote=\"Fermat's Little Turtle\"]Er, and how do we know there are 168 primes less than 1000? :S\nAnd why is the $ \\plus{} 3$ there?[/quote]\r\nThese things are quite well known things. Like- we have $ 4$ primes $ \\leq 10,$ $ 25$ primes $ \\leq 100$ etc. and if you want reference just have a look at [url=http://primes.utm.edu/lists/small/]the list of the primes[/url]. (more precisely [url=http://primes.utm.edu/lists/small/1000.txt]here[/url])\r\n\r\nAnd about the $ \\plus{} 3$, we have already eliminated $ 2,3,5$ (i.e. three primes). So, I added them again.\r\n\r\nI hope that, it is clear now. :)", "Solution_4": "[quote=\"Moonmathpi496\"][hide=\"hint\"]These type of problems can be solved easily with Inclusion and Exclusion principle.[/hide]\n[hide=\"solution\"]we can use that, $ 1000$ will be eliminated. So, for easier calculation we count the numbers $ \\leq1000$ instead of $ < 1000$. \nUsing inclusion exclusion principle we get, the number of positive integers $ \\leg 1000$ and multiple of $ 2,3,5$ is $ 734$.\nNow, we know that, we have $ 168$ primes $ \\leq 1000$.\nSo, the number of [i]beautiful numbers[/i] is $ 1000 - 734 - 168 + 3 = 101$\n\nThis problem is too easy for olympiad section, please post these types of problems in pre-olympiad section.[/hide][/quote]\r\nhmm... you could not solve the problem in that way in a competition... So there are maybe a more elegant solution..", "Solution_5": "[quote=\"Mathias_DK\"]hmm... you could not solve the problem in that way in a competition... So there are maybe a more elegant solution..[/quote] I think that many olympiad problems are solved using these information :wink: \r\n\r\n(My solution may not be that elegant at all, but I really don't think that without these type of work the problem can be solved. There is another much more ugly way,like taking the primes $ <32$ and their multiples which don't \"collide\" with those primes)\r\n\r\nAnyway, any elegant solution is always appreciated. :lol:", "Solution_6": "[quote=\"Moonmathpi496\"][quote=\"Mathias_DK\"]hmm... you could not solve the problem in that way in a competition... So there are maybe a more elegant solution..[/quote] I think that many olympiad problems are solved using these information :wink: \n\n(My solution may not be that elegant at all, but I really don't think that without these type of work the problem can be solved. There is another much more ugly way,like taking the primes $ < 32$ and their multiples which don't \"collide\" with those primes)\n\nAnyway, any elegant solution is always appreciated. :lol:[/quote]\r\nI don't think your solution is wrong or anything! I just couldn't picture myself counting all the primes from 1 to 1000. (And i don't know these by heart) It would take awfully long time, so that's the reason i want to get a solution I could use in a contest. :P", "Solution_7": "[quote=\"Moonmathpi496\"][hide=\"hint\"]These type of problems can be solved easily with Inclusion and Exclusion principle.[/hide]\n[hide=\"solution\"]we can use that, $ 1000$ will be eliminated. So, for easier calculation we count the numbers $ \\leq1000$ instead of $ < 1000$. \nUsing inclusion exclusion principle we get, the number of positive integers $ \\leg 1000$ and multiple of $ 2,3,5$ is $ 734$.\nNow, we know that, we have $ 168$ primes $ \\leq 1000$.\nSo, the number of [i]beautiful numbers[/i] is $ 1000 - 734 - 168 + 3 = 101$\n\nThis problem is too easy for olympiad section, please post these types of problems in pre-olympiad section.[/hide][/quote]\r\nBecause he want we help him fastly,a cheater\r\nHere is a problem in newest MYM,so please lock it and warn negative" } { "Tag": [ "email" ], "Problem": "Hello\r\n\r\nDoes anyone here know how Schnapps - that irritating telepathy thing works...its driving me MAD!!!\r\n\r\nSchnapps is the name of the game...\r\n\r\nThanks...either reply here or email me at astantine@gmail.com thanks", "Solution_1": "Yes, I know Schnapps! I was shown it once by my parents when I was younger, and it only surfaced in College recently! Technically, the rule is that you can't explain the rules of the game. Worse still, when I was shown it in College, the person followed the rule that you have to lie completely, even the guess is correct, and say no to everything!! The clicks after the \"Schnapps is the name of the game\" are hard to work out, so ignore them for the moment. Once you figure out the main part, that bit can be deduced. I don't particularly want to ruin the fun of the game, so the best I can say is to look carefully at the words/clicks, and at what the answer is. The examples on [url]http://www.mathlinks.ro/viewtopic.php?t=84627[/url] are almost right. \r\n\r\nIncidentally, how did you find out about Schnapps? Because my friend learned it while in Croatia last summer at a Programming Olympiad. We've found it brilliant at first, because no one had a clue how the hell we were doing it, but now half our friends have worked out the secret!! Still, done at speed, it's quite impressive!!" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "find all solutions of the equation\r\n\r\n$ x''\\equal{}x$", "Solution_1": "We have $ x'' \\minus{} x \\equal{} 0$. The caracteristic equation for this second order d.e. is $ k^2 \\minus{} 1 \\equal{} 0$ whose solutions are +1 and -1. So, the general solution of the equation is $ x \\equal{} C_1e^{t} \\plus{} C_2e^{\\minus{}t}$.", "Solution_2": "Let $ t$ be the independent variable. Then $ x'' \\equal{} x$ is viewed as $ \\frac{d^{2}x}{dt^{2}} \\equal{} x(t)$.\r\n\r\nLet $ p \\equal{} \\frac{dx}{dt}$, then $ \\frac{dp}{dt} \\equal{} p\\frac{dp}{dx}$. Plugging this into the equation, we get\r\n\r\n$ p\\frac{dp}{dx} \\equal{} x(t)$ or $ pdp \\equal{} xdx$, which gives $ p^{2} \\equal{} x^{2} \\plus{} c^{2}$, where c is a constant\r\n\r\nNow we have $ p \\equal{} \\frac{dx}{dt} \\equal{} \\pm\\sqrt{x^{2} \\plus{} c^{2}}$, which can be separated to give:\r\n\r\n$ \\frac{dx}{\\sqrt{x^{2} \\plus{} c^{2}}} \\equal{} \\pm dt$ By integrating both sides, we get:\r\n\r\n$ \\ln(x \\plus{} \\sqrt{x^{2} \\plus{} c^{2}}) \\equal{} \\pm t \\plus{}k$, where k is another constant", "Solution_3": "Or:\r\n\r\n$ \\frac{dx}{\\sqrt{x^2 \\plus{} c^2}} \\equal{} \\frac{dx}{\\sqrt{c^2(1 \\plus{} x^2/c^2)}} \\equal{} \\frac{dx/c}{\\sqrt{1 \\plus{} (x/c)^2}} \\equal{} \\pm dt$\r\n\r\n$ \\equal{}> argsinh(x/c) \\equal{} \\pm t \\plus{} k \\equal{}> \\frac{x}{c} \\equal{} \\sinh(\\pm t \\plus{} k) \\equal{} \\frac{e^{\\pm t \\plus{} k} \\minus{} e^{\\minus{}(\\pm t \\plus{} k)}}{2}$\r\n\r\n$ \\equal{}> x \\equal{} \\frac{ce^k}{2}e^{\\pm t} \\minus{} \\frac{ce^{\\minus{}k}}{2}e^{\\mp t} \\equal{} C_1e^t \\minus{} C_2e^{\\minus{}t}$ or $ \\equal{} C_1e^{\\minus{}t}\\minus{} C_2e^t$\r\n\r\nor, in either case, $ x \\equal{} K_1e^t \\plus{} K_2e^{\\minus{}t}$." } { "Tag": [ "calculus", "trigonometry", "derivative", "function", "calculus computations" ], "Problem": "Hello, I'm trying to learn a bit of calculus over winter break (so I don't have much knowledge or experience on the subject)\r\n\r\nMy book gives a proof of $\\frac{d}{dx}\\text{Sin}^{-1}=\\frac{1}{\\sqrt{1-x^2}}$, and part of it I do not understand:\r\n[quote]Let $y=\\text{Sin}^{-1}x$. Then $\\sin y=x$, $-\\frac{\\pi}{2}6+y6+z6+3x2y2z2 \\leq 2(x^3y^3+y^3z^3+z^3x^3).\r\n\r\nThis is the \"easy\" part. I want to generalize this inequality.So... :(", "Solution_1": "Is the problem right??:?\r\nLet z=0 then we must prove\r\nx^6+y^6 \\leq 2x^3y^3.\r\nIt must be \r\nx^6+y^6+z^6+3x 2 y 2 z 2 \\geq 2(x^3y^3+y^3z^3+z^3x^3) which can be proved by Schur .And here is the general inequality (Sleyfer inequality):\r\n(n-2) \\sum {1-->n}a^2n+na1 2*...*an 2 \\geq 2\\sum{i>j=1 to n}(ai*aj)^n \r\n :)", "Solution_2": "My mistake. your form is correct. :blush:" } { "Tag": [ "calculus", "derivative", "function", "LaTeX", "limit", "calculus computations" ], "Problem": "The position of a particle moving in a straight line during a 5s trip is:\r\n\r\ns(t)=t^2-t + 10cm\r\n\r\na) What is the avergae velocity of the entire trip?\r\n\r\nb) Is there a time at which the instantaneous velocity is equal to this average velocity. If so, find it?\r\n\r\n\r\nSo, I believe for (a) you just do s(5)-s(0)/(5-0) correct?\r\n\r\nBut for (b), I really don't know.\r\n\r\nThanks", "Solution_1": "The position of a particle moving in a straight line during a $ 5 \\ \\text{s}$ trip is:\r\n\r\n$ s(t) = (t^2 - t + 10) \\ \\text{cm}$\r\n\r\na) What is the avergae velocity of the entire trip?\r\n\r\nYes, the average velocity is $ v_{avg} = \\frac {\\Delta s}{\\Delta t}$.\r\n\r\nb) Is there a time at which the instantaneous velocity is equal to this average velocity. If so, find it?\r\n\r\nThe instantaneous velocity is the time derivative of the position function, i.e., $ v(t) = s'(t)$.\r\n\r\nFind the average velocity, then set the instantaneous velocity $ v(t)$ equal to that and find $ t$, i.e., solve for $ t$:\r\n\r\n$ v(t) = v_{avg}$\r\n\r\nBTW, I used $ \\text{\\LaTeX}$ to make it look better; you just need to type the appropriate code between dollar signs to use it.", "Solution_2": "There is a main result in analysis called Mean Value Theorem. If $ f$ is continuous on $ [a,b]$ and diferentiable at $ (a,b)$ then there exists some $ c \\in (a,b)$ such that $ f'(c)\\equal{}\\frac{f(b)\\minus{}f(a)}{b\\minus{}a}$.\r\n\r\n[url]http://en.wikipedia.org/wiki/Mean_value_theorem[/url]", "Solution_3": "[quote=\"Zakary\"]The position of a particle moving in a straight line during a $ 5 \\ \\text{s}$ trip is:....\n\nen dollar signs to use it.[/quote]\r\n\r\nThanks a lot!\r\n\r\nI think I did it right.\r\n\r\nI will try to make my future posts with Latex\r\n\r\nS(t)=t^2-t+10 and S'(t)= 2t-1\r\n\r\nand the Average velocity came out to be 4\r\n\r\nso I just set S'(t)=4 and solve for t?\r\n\r\nThanks", "Solution_4": "Yes, that is what you should do.\r\n\r\nA reference for future problems that involve a variation of the main value thereom is that the value of $ t$ that you find should be within the interval over which the mean value thereom is being applied (if the function is differentiable over that interval, of course).", "Solution_5": "$ \\frac{\\partial}{\\partial t}\\text{[}s\\equal{}v_0t\\plus{}\\frac{1}{2}at^2\\text{]}\\to v_f\\equal{}v_0\\plus{}at$\r\nBecause there is so [i]little[/i] time in between $ v_0$ and $ v_f$ here, the idea is that $ v_0\\equal{}v_f$.\r\n\r\nThis description is perceivable as instantaneous velocity, though the standard definition is $ v\\equal{}\\lim_{t\\to 0} \\frac{s}{t}$ (or d if your prefer, it really makes no difference). What I have only just realized is implied our standard kinematic equation is that the derivative of our initial function $ s(x)$ will always yield the same velocity no matter our initial height within $ s(x)$.\r\n\r\nAside from that you should probably know there are two standard definitions for $ s(x)$, the first and more common of which you have already dealt with. Here's the second:\r\n$ s(x)\\equal{}\\frac{v_0\\plus{}v_f}{2}t$\r\nNow, now... that reminds me of average velocity! In fact, differentiating this we have:\r\n$ s\\prime(x)\\equal{}v(x)\\equal{}\\frac{v_0\\plus{}v_f}{2}$\r\nA quick substitution of $ v_f\\equal{}v_0$ then we'll just call it $ v$:\r\n$ v(x)\\equal{}v\\equal{}\\frac{s}{t}$\r\nNow we can apply limits:\r\n$ \\boxed{v\\equal{}\\lim_{t\\to 0} \\frac{s}{t}}$\r\n\r\nEven though I know this isn't exactly what one might call relevant, its what I thought when I saw this post and I hoped you would enjoy. :)" } { "Tag": [ "modular arithmetic", "algebra", "binomial theorem", "number theory" ], "Problem": "Find the remainder when $ 32^{1024}$ is divided by 7.\r\n\r\n\r\n[hide=\"My solution\"]Is there anything wrong here?\n$ 32^{1024}\\equal{}1024^{512}$\n\n$ 1024\\equiv 2\\mod 7$\n\n$ 512\\equiv 1\\mod 7$\n\nSo $ 1024^{512}\\equiv 2\\mod 7$\n\nor $ 32^{1024}\\equiv 2\\mod 7$[/hide]", "Solution_1": "my remainder is 4 \r\n32=4mod7;4^1024=4^1mod7", "Solution_2": "[quote=\"21 7 15\"]my remainder is 4 \n32=4mod7;4^1024=4^1mod7[/quote]\r\n\r\nYes the answer is 4 but is there anything wrong in my solution?", "Solution_3": "Yes, there is. You must watch 512 mod 3 because $ 2^n$ mod 7 goes like this:\r\n2,4,1,2,4,1,...(groups of 3, not 7). 1024 mod 7 is ok.", "Solution_4": "[hide=\"My solution...\"]$ 32^{1024} \\text{ mod } 7 \\equal{} 32^{146 \\plus{} 2} \\text{ mod } 7$\n\n$ 32^{148} \\text{ mod } 7 \\equal{} 32^{21 \\plus{} 1} \\text{ mod } 7$\n\n$ 32^{22} \\text{ mod } 7 \\equal{} 32^4 \\text{ mod } 7$\n\n$ 32^4 \\equal{} 2^20$\n\n$ 2^{20} \\text{ mod } 7 \\equal{} 2^{2 \\plus{} 6} \\text{ mod } 7$\n\n$ 2^8 \\text{ mod } 7 \\equal{} 2^{1 \\plus{} 1} \\text{ mod } 7$\n\nanswer = $ \\boxed{4}$\n\nTime consuming. xD[/hide]", "Solution_5": "I didn't know you could do that in modular arithmetic.", "Solution_6": "[quote=\"21 7 15\"]my remainder is 4 \n32=4mod7;4^1024=4^1mod7[/quote]\r\n\r\nWhy (and how) can you reduce the exponent from $ 4^{1024} \\equiv 4^1 \\pmod{7}$?", "Solution_7": "remainders of powers of mod 4 are 4,2,1,4,2,1........and so on\r\nso1024 gives 1 remainder on divided by 3 . so its remainder is equal to remainder of 4^1(mod 7)", "Solution_8": "Or, my \"break it down\" way:\r\n\\[ 32\\equiv{4}\\pmod{7} \\\\\r\n32^{1024}\\equiv{4^{1024}}\\equiv{16^{512}}\\pmod{7} \\\\\r\n16\\equiv{2}\\pmod{7} \\\\\r\n2^{512}\\equiv{4^{256}}\\equiv{16^{128}}\\equiv{2^{128}}\\equiv{16^{32}}\\equiv{2^{32}}\\equiv{16^8}\\equiv{2^8}\\equiv{16^2}\\equiv{2^2}\\pmod{7} \\\\\r\n2^2 \\equal{} \\boxed{4}\r\n\\]", "Solution_9": "[b]Solution:[/b] $ 32^{1024} \\equal{} 2^{5120} \\equal{} (2^{3})^{1706} \\cdot 2^2 \\equiv 1^{1706} \\cdot 2^2 \\equiv 4$\r\n\r\n[b]@Euclid:[/b] The reason why you can \"replace\" numbers in the base of exponents is a result of the binomial theorem:\r\n\r\nIn modulo $ k$, for integer $ n$, $ (a \\plus{} nk)^m \\equal{} \\sum_{j \\equal{} 0}^m \\binom{m}{j} (nk)^j a^{m \\minus{} j}$, and all terms except the $ j \\equal{} 0$ term have a factor of $ k$, which makes them equivalent to $ 0$ in modulo $ k$.\r\n\r\nA similar pattern does [u]not[/u] hold for numbers in the exponent. See Fermat's Little Theorem, Chinese Remainder Theorem, Euler's Totient Theorem, Carmichael's Theorem...", "Solution_10": "I like this problem. :lol:", "Solution_11": "Come on Wickedestjr, stop spamming all topics!", "Solution_12": "What's spamming? Seriously, I've heard of it alot, but I have no clue what it means.", "Solution_13": "Wikipedia is good here.\r\n\r\nhttp://en.wikipedia.org/wiki/Spam", "Solution_14": "Well, before I was just saying that I liked this problem. What's useless about that. (or were you talking about canned meat when you said spam :rotfl:)", "Solution_15": "[quote=\"Wickedestjr\"]Well, before I was just saying that I liked this problem. What's useless about that. (or were you talking about canned meat when you said spam :rotfl:)[/quote]\r\n\r\nYou can get banned for spamming up topics. [b]Stop.[/b]" } { "Tag": [ "puzzles" ], "Problem": "1=5\r\n2=25\r\n3=125\r\n4=625\r\n5= ______\r\n\r\nFill in the blank.\r\nI know, it's lame.\r\nIf you can't do this one, shame on you.\r\nYeah, and there's obviously some mathematical flaws, but yeah....", "Solution_1": "it had better not be 3125", "Solution_2": "dont worry, it isnt", "Solution_3": "does it have to do anything with the digits?", "Solution_4": "[quote=\"Nerd_of_the_Ages\"]1=5\n\n...\n\nYeah, and there's obviously some mathematical flaws, but yeah....[/quote]\r\n\r\nSo would $ 5\\equal{}1$?", "Solution_5": "Yes. You just solved it. Told you it was lame.", "Solution_6": "3125!!!!! :ddr: pwns!!", "Solution_7": "5 = (4 - 3 - 1) x 4 + 2 + 3\r\n= (625 - 125 - 5) x 4 + 25 + 3\r\n= 1980 + 25 + 3\r\n= 2008 :D" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "1)Given $a_0=1,a_1=45,a_{n+2}=45a_{n+1}-7a_n\\forall n\\in N$\r\na. Find all positive divisor of $a_{n+7}^2-a_na_{n+2}$\r\nb. Prove that: $1997 a_n^2+4.7^{n+1}$ is a perfect square \r\n2)Let ${U_n}$ be a sequence satisfying $U_{n+1}=2U_n-n^2 \\forall n\\in N$\r\nFind all $U_0$ such that $U_n>0 \\forall n\\in N$", "Solution_1": "1) $a_n=\\frac{x^{n+1}-y^{n+1}}{\\sqrt{1997}},x=\\frac{45+\\sqrt{1997}}{2},y=\\frac{45-\\sqrt{1997}}{2}.$\r\na) It give expression for $a_{n+7}^2-a_na_{n+2}.$\r\nb) $1997a_n^2+4*7^{n+1}=(x^{n+1}+y^{n+1})^2, x^{n+1}+y^{n+1}$ is integer.\r\n2)$U_n=(U_0-3)*2^n+n^2+2n+3 \\Longrightarrow U_0\\ge 3.$" } { "Tag": [ "Ring Theory", "algebra", "function", "domain", "superior algebra", "superior algebra unsolved" ], "Problem": "Prove that a quotient ring $O/a$ of a Dedekind domain by an ideal $a\\not = 0$ is a PID.", "Solution_1": "Write $a$ as product of prime ideals. By the Chinese Remainder Theorem, you just need to consider the case $n=p^{k}$, $p$ prime ideal. But then it's easy since all ideals of the quotient lift to ideals in the ring, and these liftings will be powers of $p$ (they divide $p^{k}$)." } { "Tag": [ "function", "LaTeX", "number theory proposed", "number theory" ], "Problem": "Let $p$ be a prime and $N'=\\{0,1,2,...\\}$.suppose the function$f: N'\\rightarrow N'$ satisfies;\r\n\r\n$(i)\\ \\ f(n)=f([\\frac{n}{p}])+f(p\\{\\frac{n}{p}\\})$ for all $n \\in N';$\r\n\r\n$(ii)\\ \\ \\text{there exists}\\ n\\in N'$ with $p\\not | f(n)$.\r\n\r\nIf $c_0,c_1, ... c_k$ are arbitrary integers, prove that there exist $a, d \\in N'$ such that:\r\n\r\n$f(a +di)\\equiv c_i\\ (\\mod p\\ )$ for $i = 0,1, ... , k$.", "Solution_1": "...and what is the question?\r\n\r\nPierre.", "Solution_2": "sorry I edited it. :oops:", "Solution_3": "in the special case where ${\\{f(n-1)}\\}_{n=1}^{\\infty}$ was the Thue-Morse sequence mod 2", "Solution_4": "Does $\\{\\frac{n}{p}\\}$ mean the fractional part of $\\frac{n}{p}$ ?", "Solution_5": "yes it is.\r\n$\\{X\\}=X-[X]$", "Solution_6": "Then I don't understand the problem :(\r\n$\\frac{n}{p}$ is integer because $f$ is from N to N. Thus we must have $p | n$ but this means $\\{\\frac{n}{p}\\}=0$, and this looks a bit odd to me ... So we have $f(n)=f(\\frac{n}{p})+f(0)$ and this means $f$ is constant, so $f$ must be 1. But finally this is absurd :( Were did I go wrong ? What I'm not understanding ?", "Solution_7": "I edited it .\r\nI think I have to learn $\\LaTeX$ again." } { "Tag": [], "Problem": "In your opinion, what's the best programming language for IOI: c or c++? Can you tell me some advantages each one has over the other?", "Solution_1": "I think that C++ is definitely better if you can use STL. Otherwise there's no real difference between the two, C++ is pretty much an extension to C and if we don't count STL then what it has that C does not is only a couple of convinient constructions, nothing more. But I don't think there are any drawback so why don't use C++?", "Solution_2": "ya as BUS said there is no difference between C and C++. if u code at C that will also run in C++. But C++ have some extra fitures that does not work in C. So don't worry which one u r learning.... see what u r learning!", "Solution_3": "C++, with STL.", "Solution_4": "STL? :huh:", "Solution_5": "[quote=\"quantum leap\"]STL? :huh:[/quote]\r\nStandard template library.", "Solution_6": "[quote=\"quantum leap\"]STL? :huh:[/quote]\r\n\r\nIf you're pretty much confused as to what it is, check out this site........it's got some pretty interesting stuff on STL -- :) \r\n\r\nhttp://www.infosys.tuwien.ac.at/Research/Component/tutorial/prwmain.htm", "Solution_7": "[quote=\"mahbub\"]ya as BUS said there is no difference between C and C++. if u code at C that will also run in C++. But C++ have some extra fitures that does not work in C. So don't worry which one u r learning.... see what u r learning![/quote]\r\n\r\nYes, C should be a sub-set of C++, but the main point is the usage of the language. \r\n\r\nBy common programming practice in industry, C would be used in system programming like building Linux and Windows. \r\n\r\nHowever, since many people superstitious belief that the same code compiled by C++ would be slower than C, none would use C++ to write system, even without using OO-style.\r\n\r\nIn my view, programming is the way to express the implementation of problem solving, not the way of problem solving.", "Solution_8": "I'd use C++ but try not to rely too much on pre-packaged data structures unless you're in a time crunch, because you have a lot more flexibility if you have the ability to write your own heaps, segtrees, queues, etc.", "Solution_9": "ok then, now that its concluded that C++ is better, which way is the best way to learn it?\r\nand does anyone here use DevCpp?", "Solution_10": "For competition programming C is usually enough (although I compile in C++), I've never needed the stl in a competition before.", "Solution_11": "[quote=\"salah\"]ok then, now that its concluded that C++ is better, which way is the best way to learn it?\nand does anyone here use DevCpp?[/quote]\r\n\r\nI use Dev-Cpp.\r\nTo learn, I just got a book from Barnes & Noble.\r\n\r\nDo the USACO, if you don't already.", "Solution_12": "I found these;\r\nhttp://mindview.net/Books/TICPP/ThinkingInCPP2e.html\r\nI dont think the basic stuff are that hard to learn, and besize I signed up for a course so that should be enough..\r\n\r\nDev-Cpp is great, I had some trouble uninstalling it after that I had messed with a bit, but now it works :)" } { "Tag": [], "Problem": "Hi! Please, help me to solve the equation in real numbers:\r\n\r\n$ x(x\\minus{}2)(x\\minus{}5)(x\\minus{}6)(x\\minus{}9)\\equal{}\\minus{}27$\r\n\r\nIt's too hard for me.\r\nFirst of all, it's have 5 roots, and all of them are not rational.\r\nMay be we must group??:\r\n$ x(x^2\\minus{}11x\\plus{}18)(x^2\\minus{}11x\\plus{}30)\\equal{}\\minus{}27$", "Solution_1": "This is a quintic equation. It is not solvable by radicals. However it is easy to sketch the graph of the left-hand-side and that of the right-hand-side to get first approximations to the $ 5$ roots.", "Solution_2": "hello\r\n$ x_1\\approx 2.167970919953405509641072637792755662802663203738082745127727200627223849274550084454635272281049145$\r\n$ x_2\\approx\r\n 4.629681890809878759703989240330377489864447689344392950259543919541800444537816560644875417257476156$ \r\n$ x_3 \\approx 6.286922943303427735896134705732908669721929428569834068623720433676698142569236798808499848319437121$ \r\n$ x_4\\approx 8.963161758831018365618809909615921585894361274689999636236501875742142289218514173348194804186084403$ \r\n$ x_5 \\approx 0.04773751289773037086000649347196340828340159634230940024749342958786472560011761725620534204404682578$\r\nSonnhard.", "Solution_3": "Thank you very much, but I knew this. http://www67.wolframalpha.com/input/?i=x(x-2)(x-5)(x-6)(x-9)%2B27\r\nMay be we can solve this equation with some substitution?" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Hi, I'm new here. I have a question about a problem I've been working on. \r\n\r\nLet A be an n by n real, symmetric positive definite matrix where all it's elements are positive. Matrix A is the product of a binomial matrix B such that A = BB^T where T denotes the transpose operator. In the binomial matrix, note that x+y=1. The structure of the binomial matrix is given by the image attachment.\r\n\r\nLet C be an n by n real, diagonal positive definite matrix. What bounds are there on the values of the diagonal elements of C so that the matrix A - C is positive semidefinite? \r\n\r\nI thought of using eigenvalues, but it would be difficult to find the eigenvalues of the matrix A. Thanks for your help.", "Solution_1": "For those who don't want to download the attachment: the structure of a binomial matrix is\r\n\r\n$ B(x,y)=\\begin{bmatrix}1&0&0&\\cdots&0\\\\ x&y&0&\\cdots&0\\\\ x^{2}&2xy&y^{2}&\\cdots&0\\\\ \\vdots&\\vdots&\\vdots&\\ddots&\\vdots\\\\ x^{n}&\\binom{n}{1}x^{n-1}y&\\binom{n}{2}x^{n-2}y^{2}&\\cdots&y^{n}\\end{bmatrix}$" } { "Tag": [ "geometry", "number theory unsolved", "number theory" ], "Problem": "Having fun:\r\n1. Grab a calculator. (you won't be able to do this one in your head)\r\n2. Key in the first three digits of your phone number (NOT the area code)\r\n3. Multiply by 80\r\n4. Add 1\r\n5. Multiply by 250\r\n6. Add the last 4 digits of your phone number\r\n7. Add the last 4 digits of your phone number again.\r\n8. Subtract 250\r\n9. Divide number by 2\r\n\r\nThis was sent to me by a friend on msn. Its kinda fun =)", "Solution_1": "Let $ x$ be the first three digits and $ y$ be the last four digits of my phone number. After performing all the steps we end up with \r\n\\[ \\frac{(80x\\plus{}1)\\times 250\\plus{}2y\\minus{}250}{2}\\equal{}10000x\\plus{}y,\\]\r\nwhich is my phone number again! And guess what, I didn't use a calculator! ;)", "Solution_2": "Ah, no calculator indeed. It was just copy and pasted :P" } { "Tag": [ "algebra", "function", "domain", "number theory", "greatest common divisor", "least common multiple", "algorithm" ], "Problem": "Let A the group of positive integers representing of the form a^2+2(b^2). \"a\" are integer, and \"b\" anyone integer not equal zero. \r\nFor example, the number 6 is in the group A (6= 2^2+2(1^2))\r\nProve that if p^2 is in A, p is in A too (p=prime).", "Solution_1": "Ring $Z[\\sqrt{-2}]$ is Evclidov.", "Solution_2": "Sorry but i dont know Evclidov!! Do you can explain me please??", "Solution_3": "Maybe he means that $\\mathbb Z[\\sqrt 2]$ is an [url=http://en.wikipedia.org/wiki/Euclidean_domain]Euclidean domain[/url].", "Solution_4": "No, for this case used, that $Z[\\sqrt{-2}$ Euclidean domain, were $norm(x+y\\sqrt{-2})=x^{2}+2y^{2}$.", "Solution_5": "I explain more. We have $gcd$ in Euclidean domains. By problem hypothetis \\[p^{2}=(x+y\\sqrt{-2})(x-y\\sqrt{-2})\\] So consider $r+s\\sqrt{-2}=gcd(p,\\sqrt{-2})$. (If $r+s\\sqrt{-2}=1$ then $gcd(x^{2}+2y^{2},p)=1$ and this is contradiction.) Now $\\mbox{norm}(r+s\\sqrt{-2})=r^{2}+2s^{2}|p^{2}$ and $<1r^{2}+2s^{2}3a", "Solution_1": "[quote=\"chengyuLi\"]$ \\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c}\\equal{}9$ and $ a\\leq b \\leq c$. $ a,b$ and $ c$ are positive reals.how to show that $ abc\\plus{}1>3a$[/quote]\r\nhello, please the next time try to write with $ Latex$ for this use this symbol $ \\$$ before and after the code,", "Solution_2": "that is sqr,not sqrt.how to use Latex", "Solution_3": "By the Way look at this\r\n\r\n$ abc\\plus{}1\\ge{a^3}\\plus{}1\\equal{}(a^3\\plus{}1\\plus{}1)\\minus{}1\\ge3a\\minus{}1$ by $ AM\\minus{}GM$\r\n\r\nbut the problem tells that $ abc\\plus{}1>3a$\r\nthats funny.... :rotfl:", "Solution_4": "a2+b2+c2=9,a<=b<=c,a,b and c are positive reals,how to show that abc+1>3a. \r\nbut how to use latex,i used microsoft word.\r\nthanks in advance.", "Solution_5": "can't anyone give a correct answer,it couldn't be difficult,just a competetion in sichuan,china.\r\ni want to see your proof,thanks,xiexie,danke sehr.", "Solution_6": "for more information about latex see here,\r\nhttp://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php", "Solution_7": "thanks,but i'm sorry my english is poor.i downloaded it,but i don't know how to use.sorry\r\ni think this problem can change to 2*sqr(a)*c+sqr(c)=9,we just have to show sqr(a)*c+1>3*a." } { "Tag": [ "function", "calculus", "integration", "derivative", "probability", "geometric series", "calculus computations" ], "Problem": "Consider the series\r\n\r\n$\\sum\\limits_{k=1}^{\\infty} {k^a b^k}$ with $a>0, a \\in R$\r\n\r\nThe series converges for $b<1$ (it can be prooved with the root test)\r\n\r\n\r\nIs there an expression for the sum of the finite version of the series? (say: $\\sum\\limits_{k=1}^N$)\r\n\r\nI mean, something like $\\dfrac{N(N+1)}{2}$ for the sum of first $N$ integers..\r\n\r\n\r\nThank you", "Solution_1": "Don't expect any unified formula. There are closed form expressions whenever $a$ is a positive integer, but the complexity grows rapidly.\r\nThere are strong relations to the gamma function and the zeta function. I would be surprised by any closed form when $a$ is not an integer.", "Solution_2": "Can you give me the closed expressions for $a=1$ and $a=2$? \r\n\r\nAnd a reference (book, table, etc) would be useful.\r\n\r\nThanks", "Solution_3": "I don't know them off the top of my head, and I don't have a reference. Here is a method for finding the closed forms:\r\nAbel's formula (Summation by parts): $\\sum\\limits_{k=1}^na_kb_k=\\sum\\limits_{k=1}^n(\\Sigma a_k)\\cdot(\\delta b_k)+(\\Sigma a_n)\\cdot b_{n+1}-(\\Sigma a_0)\\cdot b_1$, where $\\Sigma a_k=c+\\sum\\limits_{i=1}^ka_i$ and $\\delta b_k=b_k-b_{k+1}$.\r\n\r\nThis is easy enough to verify (note the analogy with integration by parts). To use it for this problem, sum the geometric series and difference the polynomial. With repeated applications, you eventually reach a closed form for the finite sum, and can take limits to get the infinite sum. It is best in this case to set $c=-\\sum\\limits_{k=1}^\\infty a_k$ in the definiton of $\\Sigma$ so that the boundary term at $\\infty$ is easy to work with and the series is still geometric for the next step.\r\nThis idea can also give recursive formulas when $a$ is not a nonnegative integer.\r\n\r\nAnother method for the infinite sums: treat $b$ as a variable, and the sum as a function ($f_a$) of $b$. Multiplying by $k$ on the series side is taking a derivative and multiplying by $b$ on the function side. Then $f_0(b)=\\dfrac1{1-b}$, $f_1(b)=\\dfrac{b}{(1-b)^2}$, and $f_2(b)=\\dfrac{b}{(1-b)^2}+\\dfrac{2b^2}{(1-b)^3}=\\dfrac{b+b^2}{(1-b)^3}$\r\n\r\nThe infinite sums (for $a$ a nonnegative integer) are essentially the moments of the geometric probability distribution.", "Solution_4": "Searching for Abel's formula, I found the Abel's theorem:\r\n\r\nhttp://en.wikipedia.org/wiki/Abel's_theorem\r\n\r\nConsider the general version.. it seems I have to find the Generating Function of $a_k=k$ and $a_k=k^2$ (it's easy!), to have the sum of the infinite series. I will try to find the GF for $a_k$ stopped at $N$, too.", "Solution_5": "It's probably easier to find by looking for the other name, or trying a math-specific site.\r\n\r\nBy the way, I gave the answers for $n=1$ and $n=2$ in my post.", "Solution_6": "Yes, the GF method confirms that the sums of the infinite series are:\r\n\r\n$\\dfrac{b}{(1-b)^2}$ for $a=1$\r\n\r\n$\\dfrac{b(b+1)}{(1-b)^3}$ for $a=2$" } { "Tag": [ "inequalities", "calculus", "derivative" ], "Problem": "Let $a>-1$ and $r \\in (0,1)$ be reals. Prove that: \\[(1+a)^{r}\\leq 1+ra.\\]", "Solution_1": "That's Bernouilli's inequality. You can find proofs of it all over the web.", "Solution_2": "Is it? I don't think so.\r\n\r\nProof is the same though (the second derivative is negative this time instead of positive).", "Solution_3": "[quote=\"MysticTerminator\"]Is it? I don't think so.\n[/quote]\r\n[url]http://www.imo.org.yu/index.php?options=mbb|Bfa|f|Alg|Ine&p=02lLw[/url]", "Solution_4": "Posted and proven [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=106614]here[/url]", "Solution_5": "[quote=\"Farenhajt\"]Posted and proven [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=106614]here[/url][/quote]\r\n\r\nThat's not the same problem; you proved it for natural numbers $n$, while the problem posted here is the reverse inequality for reals $r \\in (0,1)$.", "Solution_6": "Sorry, haven't even noticed that... :maybe:", "Solution_7": "Let $b = ar, s = \\frac1r$. Then\r\n\r\n$(1+a)^{r}\\leq 1+ra \\Longleftrightarrow 1+sb \\leq (1+b)^{s}$. So the question is not substantively different then the case $r > 1$. You do still need to do it for all reals, (not just positive integers) though.", "Solution_8": "We'll prove the equivalent statement $(1+a)^{r}\\geqslant 1+ar$ for $a>-1$ and $\\mathbb{R}\\owns r>1$.\r\n\r\nPut $c=1+a$ and $s=r-1$. Then the inequality becomes\r\n\r\n\\begin{eqnarray}c^{s+1}\\geqslant 1+(c-1)(1+s) &\\iff& c^{s+1}\\geqslant c+(c-1)s\\nonumber\\\\ &\\iff& c(c^{s}-1)\\geqslant (c-1)s\\end{eqnarray}\r\n\r\nfor $c>0, s>0$\r\n\r\nFor $c=1$ $(1)$ is trivially satisfied, hence we'll deal with $c\\neq 1$.\r\n\r\n[b]Part 1.[/b] $\\bold{s\\in\\mathbb{Q}}$ Put $s={p\\over q}, p,q\\in\\mathbb{N}$\r\n\r\n[b]Case 1.1.[/b] $\\bold{c>1}$. Then we can write\r\n\r\n\\begin{eqnarray*}c(c^{p\\over q}-1)\\geqslant (c-1){p\\over q}&\\iff& \\frac{c(c^{p\\over q}-1)}{c-1}\\geqslant{p\\over q}\\\\ &\\iff& \\frac{c(c^{1\\over q}-1)(c^{{p-1}\\over q}+c^{{p-2}\\over q}+\\dots+1)}{(c^{1\\over q}-1)(c^{{q-1}\\over q}+c^{{q-2}\\over q}+\\dots+1)}\\geqslant{p\\over q}\\\\ &\\iff& \\frac{c(c^{{p-1}\\over q}+c^{{p-2}\\over q}+\\dots+1)}{c^{{q-1}\\over q}+c^{{q-2}\\over q}+\\dots+1}\\geqslant{p\\over q}\\end{eqnarray*}\r\n\r\nThe numerator on the LHS is not less than $c(\\underbrace{1+1+\\dots+1}_{p})=pc$, and the denominator is not greater than $\\underbrace{c^{{q-1}\\over q}+c^{{q-1}\\over q}+\\dots+c^{{q-1}\\over q}}_{q}=qc^{{q-1}\\over q}$, hence we have\r\n\r\n$\\frac{c(c^{{p-1}\\over q}+c^{{p-2}\\over q}+\\dots+1)}{c^{{q-1}\\over q}+c^{{q-2}\\over q}+\\dots+1}\\geqslant\\frac{pc}{qc^{{q-1}\\over q}}={p\\over q}c^{1\\over q}>{p\\over q}$\r\n\r\n[b]Case 1.2.[/b] $\\bold{c<1}$. Then we can write\r\n\r\n\\begin{eqnarray*}c(c^{p\\over q}-1)\\geqslant (c-1){p\\over q}&\\iff& \\frac{c(c^{p\\over q}-1)}{c-1}\\leqslant{p\\over q}\\\\ &\\iff& \\frac{c(c^{1\\over q}-1)(c^{{p-1}\\over q}+c^{{p-2}\\over q}+\\dots+1)}{(c^{1\\over q}-1)(c^{{q-1}\\over q}+c^{{q-2}\\over q}+\\dots+1)}\\leqslant{p\\over q}\\\\ &\\iff& \\frac{c(c^{{p-1}\\over q}+c^{{p-2}\\over q}+\\dots+1)}{c^{{q-1}\\over q}+c^{{q-2}\\over q}+\\dots+1}\\leqslant{p\\over q}\\end{eqnarray*}\r\n\r\nThe numerator on the LHS is not greater than $c(\\underbrace{1+1+\\dots+1}_{p})=pc$, and the denominator is not less than $\\underbrace{c^{{q-1}\\over q}+c^{{q-1}\\over q}+\\dots+c^{{q-1}\\over q}}_{q}=qc^{{q-1}\\over q}$, hence we have\r\n\r\n$\\frac{c(c^{{p-1}\\over q}+c^{{p-2}\\over q}+\\dots+1)}{c^{{q-1}\\over q}+c^{{q-2}\\over q}+\\dots+1}\\leqslant\\frac{pc}{qc^{{q-1}\\over q}}={p\\over q}c^{1\\over q}<{p\\over q}$\r\n\r\n[b]Conclusion 1.[/b] $(1+a)^{r}\\geqslant 1+ar$ is satisfied for $a>-1$ and $\\mathbb{Q}\\owns r>1$\r\n\r\n[b]Part 2.[/b] $\\bold{s\\in\\mathbb{I}}$. Then we can generalize Conclusion 1 by using Dedekind cuts.\r\n\r\n[b]Conclusion.[/b] $(1+a)^{r}\\geqslant 1+ar$ is satisfied for $a>-1$ and $\\mathbb{R}\\owns r>1$" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "I don't know if it is a silly question or an obvious one!\r\n\r\nI want to find matrix (or matrices) which I can multiply them by one matrix and get the transpose of that one.\r\n\r\n[hide]I know that transpose of a matrix is similar to it, but I don't know the proof. But it results that these matrices exist, so how to find them?[/hide]", "Solution_1": "See [url]http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1103039127[/url]", "Solution_2": "Thank you. That is a great article." } { "Tag": [ "calculus", "integration" ], "Problem": "About the equation $ ax^2 \\minus{} 2x\\sqrt {2} \\plus{} c \\equal{} 0$, with $ a$ and $ c$ real constants, we are told that the discriminant is zero. The roots are necessarily:\r\n\r\n$ \\textbf{(A)}\\ \\text{equal and integral} \\qquad\\textbf{(B)}\\ \\text{equal and rational} \\qquad\\textbf{(C)}\\ \\text{equal and real}$\r\n$ \\textbf{(D)}\\ \\text{equal and irrational} \\qquad\\textbf{(E)}\\ \\text{equal and imaginary}$", "Solution_1": "[hide=\"Solution\"]The roots of the equation are: $ x\\equal{}\\frac{2\\sqrt{2}\\pm 0}{2a}$.\n\nAnd so they are necessarily $ \\boxed{\\textbf{(C)}\\ \\text{equal and real}}$.[/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ r>1$ and $ a,b,c\\geq0$ s.t. $ ab\\plus{}bc\\plus{}ca\\equal{}3$.Prove that:\r\n$ a^r(b\\plus{}c)\\plus{}b^r(c\\plus{}a)\\plus{}c^r(a\\plus{}b)\\geq6$", "Solution_1": "[quote=\"Inequalities Master\"]Let $ r > 1$ and $ a,b,c\\geq0$ s.t. $ ab \\plus{} bc \\plus{} ca \\equal{} 3$.Prove that:\n$ a^r(b \\plus{} c) \\plus{} b^r(c \\plus{} a) \\plus{} c^r(a \\plus{} b)\\geq6$[/quote]\r\n$ LHS\\equal{} ab(a^{r\\minus{}1}\\plus{}b^{r\\minus{}1})\\plus{}bc(b^{r\\minus{}1}\\plus{}c^{r\\minus{}1})\\plus{}ca(c^{r\\minus{}1}\\plus{}a^{r\\minus{}1})\\ge 2((ab)^{\\frac{r\\plus{}1}{2}} \\plus{} (bc)^{\\frac{r\\plus{}1}{2}}\\plus{}(ca)^{\\frac{r\\plus{}1}{2}}) \\ge 6(\\frac{ab\\plus{}bc\\plus{}ca}{3})^{\\frac{r\\plus{}1}{2}}\\equal{}6$; :)" } { "Tag": [ "LaTeX", "calculus", "calculus computations" ], "Problem": "Which of the following conjecture is true? Justify\r\n\r\n[latex]10n \u00a0= O(n)[/latex] \r\n\r\n[latex]10n^2 \u00a0= O(n)[/latex] \r\n\r\n[latex]10n^{55} \u00a0= O(n^2 )[/latex]\r\n\r\n1) [latex]10n \\leq cn[/latex] => [latex]n=1[/latex] [latex]c=10[/latex]\r\n\r\n[latex]f(n) \\leq 10g(n)[/latex] for all [latex]n \\geq 1[/latex]. [latex]f(n) \\in O(g(n))[/latex]\r\n\r\nHow can I solve the other two?\r\n\r\nI found this problem on the Internet:\r\nIt is true that [latex]n^2 + 200n + 300 \u00a0= \u00a0 O(n^2[/latex]) ? \r\n\r\nAnd [latex]n^2 -200n -300 \u00a0= \u00a0 O(n)[/latex]", "Solution_1": "@Apprentice123: to use LaTeX formulas, just take two dollar sings instead of [latex] and [/latex]. :wink:", "Solution_2": "All of these should be taken as $ n\\to\\infty$; there is an implicit \"as $ x\\to ?$ in any \"$ f(g)$ is $ O(g(x))$\" statement, but we should clarify what we expect the variables to tend to.\r\nThere's a sticky in the calculus forum you may find useful, and this thread really belongs there- because big-O notation is about limits.\r\n\r\n[color=green][Moved by moderator.][/color]" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let $ k$ be a positive integer. Find all functions $ f: \\mathbb N \\to \\mathbb N$ satisfying \n\\[f(f(n))\\plus{}f(n)\\equal{}2n\\plus{}3k\\] for all positive integers $n$.", "Solution_1": "[quote=\"KDS\"]Let $ k$ be a positive integer.Find all functions $ f: N \\to N$ satisfying $ f(f(n)) \\plus{} f(n) \\equal{} 2n \\plus{} 3k$ $ \\forall n \\in N$.[/quote]\r\n\r\nLet $ f^{[p]}(n) \\equal{} a_pf(n) \\plus{} b_pn \\plus{} c_pk$ $ \\forall n > 0$\r\nWe get $ f^{[p \\plus{} 1]}(n) \\equal{} f^{[p]}(f(n)) \\equal{} a_pf(f(n)) \\plus{} b_pf(n) \\plus{} c_pk$ $ \\equal{} a_p( \\minus{} f(n) \\plus{} 2n \\plus{} 3k) \\plus{} b_pf(n) \\plus{} c_pk$ $ \\equal{} (b_p \\minus{} a_p)f(n) \\plus{} 2a_pn \\plus{} (c_p \\plus{} 3a_p)k$\r\n hence the equations :\r\n$ a_{p \\plus{} 1} \\equal{} b_p \\minus{} a_p$\r\n$ b_{p \\plus{} 1} \\equal{} 2a_p$\r\n$ c_{p \\plus{} 1} \\equal{} c_p \\plus{} 3a_p$\r\n\r\nAnd the result : $ f^{[p]}(n) \\equal{} \\frac {1 \\minus{} ( \\minus{} 2)^p}3f(n) \\plus{} \\frac {2 \\plus{} ( \\minus{} 2)^p}3n \\plus{} (p \\minus{} \\frac {1 \\minus{} ( \\minus{} 2)^p}3)k$ $ \\forall n > 0$, $ \\forall p\\geq 0$\r\n\r\nSetting $ p \\equal{} 2q$ and writing $ f^{[p]}(n) > 0$, we get $ f(n) < \\frac {4^q \\plus{} 2}{4^q \\minus{} 1}n \\plus{} (1 \\plus{} \\frac {6q}{4^q \\minus{} 1})k$\r\nUsing $ q\\to \\plus{} \\infty$ in this line, we get $ f(n)\\leq n \\plus{} k$\r\n\r\nSetting $ p \\equal{} 2q \\plus{} 1$ and writing $ f^{[p]}(n) > 0$, we get $ f(n) > \\frac {2\\cdot 4^q \\minus{} 2}{2\\cdot 4^q \\plus{} 1}n \\plus{} (1 \\minus{} \\frac {6q \\plus{} 3}{2\\cdot 4^q \\plus{} 1})k$\r\nUsing $ q\\to \\plus{} \\infty$ in this line, we get $ f(n)\\geq n \\plus{} k$\r\n\r\nSo $ f(n) \\equal{} n \\plus{} k$ which, indeed, is a solution.\r\n\r\nSo the unique solution : $ \\boxed{f(n) \\equal{} n \\plus{} k}$" } { "Tag": [], "Problem": "During a severe thunderstorm, Tim saw a flash of lightning and then six seconds later heard a clap of thunder. The speed of sound is 1088 ft/sec. How far is Tim from the lightning? Express your answer to the nearest thousand feet.", "Solution_1": "1088*6=6528.\r\n\r\nexpress 6528 nearest thousand feet = 7000\r\n\r\nanswer : 7000" } { "Tag": [], "Problem": "Ascertain the integer numbers $a,b$ so that the equations\r\n$x^4+x^3+3x^2+2x+a=0$ and $x^3-x^2-x+b=0$\r\nhave at least two common roots.[hide=\"Answer.\"]$a=2$ and $b=-2\\ .$[/hide]", "Solution_1": "Do you want $a\\ne0$?", "Solution_2": "No, I didn't.", "Solution_3": "[quote=\"chess64\"]Do you want $a\\ne0$?[/quote]\r\n\r\nOh nevermind, I didn't pay attention to the signs." } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "Find the largest constant $ t$ for which the following inequality holds for $ a,b,c\\geq 0$:\r\n\r\n$ \\frac {bc}{b \\plus{} c} \\plus{} \\frac {ca}{a \\plus{} c} \\plus{} \\frac {ab}{a \\plus{} b}\\geq t(a \\plus{} b \\plus{} c)$\r\n\r\n(inspired by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=148725]this one[/url])", "Solution_1": "Put $a=1,b=c\\rightarrow 0$. :wink:", "Solution_2": "Set $x=1/a$, $y=1/b$ and $z=1/c$ (assuming $a,b,c > 0$)\r\n\r\n(if one of $a,b,c$ is $0$, then it is easy to see that $t \\leq \\frac{1}{4}$)\r\n\r\nSo we have\r\n\r\n$\\frac{1}{x+y}+\\frac{1}{y+z}+\\frac{1}{z+x}\\geq t(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z})$ but\r\n\r\nwe have that \r\n\r\n$\\frac{1}{x}+\\frac{1}{y}\\geq \\frac{4}{x+y}$\r\n\r\nThus we have that\r\n\r\n$(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}) \\geq 2(\\frac{1}{x+y}+\\frac{1}{y+z}+\\frac{1}{z+x})$\r\n\r\nThus $t \\le \\frac{1}{2}$ \r\n\r\nThe value $t = \\frac{1}{2}$ is achieved when $a=b=c$" } { "Tag": [ "inequalities", "geometry", "rhombus", "trigonometry", "circumcircle", "geometric transformation", "geometry proposed" ], "Problem": "Let $ ABCD$ be a rhombus with $ \\angle BAD \\equal{} 60^{\\circ}$. Points $ S$ and $ R$ are chosen inside the triangles $ ABD$ and $ DBC$, respectively, such that $ \\angle SBR \\equal{} \\angle RDS \\equal{} 60^{\\circ}$. Prove that $ SR^2\\geq AS\\cdot CR$.", "Solution_1": "op02 project, Yugoslavia problem 3.\r\n\r\n[hide=\"Solution by Myth and me\"]\n[color=blue][b]Problem.[/b] Let ABCD be a rhombus with < BAD = 60\u00b0. The points S and R lie inside the triangles ABD and DBC respectively such that < SBR = < RDS = 60\u00b0. Prove that $SR^2 \\geq AS \\cdot CR$.[/color]\n\n[i]Solution by Myth and me.[/i] We start with a simple lemma:\n\n[b]Lemma 1.[/b] For any angle t, we have $\\sin\\left( 60^{\\circ}-t\\right)+\\sin\\left( 60^{\\circ}+t\\right) \\leq2\\sin60^{\\circ}$.\n\n[i]Proof.[/i] Using the addition formulas for sin, we get:\n\n$\\sin\\left( 60^{\\circ}-t\\right) +\\sin\\left( 60^{\\circ}+t\\right)$\n$=\\left( \\sin60^{\\circ}\\cos t-\\cos60^{\\circ}\\sin t\\right) +\\left(\\sin60^{\\circ}\\cos t+\\cos60^{\\circ}\\sin t\\right)$\n$=2\\sin60^{\\circ}\\cos t\\leq2\\sin60^{\\circ}$ (since $\\sin 60^{\\circ}=\\frac{\\sqrt3}{2}\\geq 0$ and $\\cos t\\leq 1$).\n\nNow let's solve the problem:\n\nSince the quadrilateral ABCD is a rhombus, we have DA = AB. Since we also have < BAD = 60\u00b0, the triangle BAD is equilateral; since a rhombus is symmetric with respect to each of its diagonals, the triangles BAD and BCD are congruent, so that the triangle BCD is also equilateral.\n\nWLOG assume that $DS\\leq DR$. Let X be a point on the ray DS such that DX = DR, and let Y be a point on the ray DR such that DY = DS. Then, since DX = DR and < RDX = 60\u00b0 (this follows from < RDS = 60\u00b0), the triangle XDR is equilateral. Similarly, the triangle YDS is equilateral.\n\nSince the triangle XDR is equilateral, we have < DXR = 60\u00b0. In other words, < SXR = 60\u00b0. The equilateral triangle YDS yields < DYS = 60\u00b0, so that < RYS = 180\u00b0 - < DYS = 180\u00b0 - 60\u00b0 = 180\u00b0 - < SXR. This implies that the quadrilateral RYSX is cyclic. In other words, the points R, Y, S and X lie on one circle. The point B also lies on this circle, since < SBR = 60\u00b0 and < SXR = 60\u00b0 imply < SBR = < SXR. Hence, all five points R, Y, S, X and B lie on one circle. Let r be the radius of this circle. Then, by the fact that the length of a chord in a circle equals the double circumradius times the sine of the chordal angle of this chord, we have $SR=2r\\sin\\measuredangle SBR$, $BX=2r\\sin\\measuredangle BSX$ and $BY=2r\\sin\\measuredangle BSY$.\n\nSince the triangle YDS is equilateral, we have < DSY = 60\u00b0, and thus < BSX + < BSY = 180\u00b0 - < DSY = 180\u00b0 - 60\u00b0 = 120\u00b0. Hence, there exists an angle t such that < BSX = 60\u00b0 - t and < BSY = 60\u00b0 + t (this angle t can be positive or negative or 0\u00b0).\n\nBy Lemma 1, we thus have\n\n$\\sin\\measuredangle BSX+\\sin\\measuredangle BSY=\\sin\\left( 60^{\\circ}-t\\right)+\\sin\\left( 60^{\\circ}+t\\right) \\leq2\\sin60^{\\circ}=2\\sin\\measuredangle SBR$\n\n(since < SBR = 60\u00b0). Hence,\n\n$2r\\sin\\measuredangle BSX+2r\\sin\\measuredangle BSY\\leq2\\cdot2r\\sin \\measuredangle SBR$,\n\nor, in other words, $BX+BY\\leq2\\cdot SR$.\n\nSince the triangle BAD is equilateral, we have < ADB = 60\u00b0, what, combined with < RDS = 60\u00b0, yields < RDS = < ADB, so that < RDB = < RDS - < SDB = < ADB - < SDB = < SDA. In other words, < YDB = < SDA. Since we also have BD = AD (this is again because the triangle BAD is equilateral) and DY = DS, it follows that the triangles YDB and SDA are congruent, and thus BY = AS. Similarly, BX = CR. Thus, the inequality $BX+BY\\leq2\\cdot SR$ rewrites as $CR+AS\\leq2\\cdot SR$. In other words, we have $2\\cdot SR\\geq AS+CR$. Thus,\n\n$SR\\geq\\frac{AS+CR}{2}$.\n\nBut by the AM-GM inequality, $\\frac{AS+CR}{2}\\geq\\sqrt{AS\\cdot CR}$. Hence, $SR\\geq\\sqrt{AS\\cdot CR}$, and $SR^2\\geq AS\\cdot CR$. And we are done.\n\nEquality holds if and only if < BSD = 120\u00b0. The proof is left to the reader (hint: < BSD = 120\u00b0 is equivalent to t = 0\u00b0 and to AS = CR = SR).[/hide]\r\n\r\n Darij", "Solution_2": "Funny! I used EXACTLY the same construction!!\r\n\r\nI found it in a very short time, not even 10 minutes, but it might not be that obvious if you don't see it immediately. Do you think this is a hard problem Darij?", "Solution_3": "Well, I don't know, but it took me way longer. Maybe it's easier to give a straightforward sine law solution which does not show the stronger inequality $SR\\geq\\frac{AS+CR}{2}$. Anyway, it was more than half a year ago when we - Myth and me - were solving the problem, and Myth was pretty busy while I was not too experienced with geometric inequalities. Still I find it quite a difficult problem.\r\n\r\n darij", "Solution_4": "My write-up of the solution was much shorter.\r\n\r\nI assumed that $DR \\geq DS$. Pick points $R'$ and $S'$ on $BS$ and $DR$ respectively such that triangles $BRR'$ and $DSS'$ are equilateral. It is easy to see, by for example spiral similarity, that $CR = DR'$ and $AS = BS'$. Also, $DSR'R$ and $BRS'S$ are cyclic. So, we need to prove $\\frac{SR}{DR'}\\frac{SR}{BS'} \\geq 1$, or $\\frac{\\sin 60}{\\sin \\alpha}\\frac{\\sin 60}{\\sin \\beta} \\geq 1$, where $\\alpha = \\angle{S'RR'}$ and $\\beta = \\angle{R'SS'}$ and $\\alpha + \\beta = 120$. But that is obvious from Jensen, for example." } { "Tag": [ "geometry", "perimeter", "rectangle", "perpendicular bisector" ], "Problem": "Ms. Salman purchases a rectangular lot with area 1,200 sq ft which has a length equal to 3 times its width. She wishes to divide the lot into exactly 2 plots having equal area and install fencing along the perimeters of both lots. What is the least number of feet of fencing she must buy?", "Solution_1": "[color=indigo][hide]x=width of lot\n3x*x = 3x2 =1200\nx=20\nShe could divide the lot on the diangonal, the perpendicular bisector of the long side or the perpendicular bisector of the short side.\ndiangonal :approx: 63.2\nper. bisector of short side=60\nper. bisector of long side=30\nThe per. bisector of the short side is the best so she has to get 180 feet of fence.[/color][/hide]", "Solution_2": "[hide]\nthis is how i did it!\nA=lw\n1200=(3w)(w)\nw=20\nso, the length is \n60 and the width\nis 20.\nthan u split the rectangle\nverticly and u ll get to two\nrectangles and than \nthe measurements would be\n30, 30, 30, 30, 20, 20, 20\nso the answer is 180! :D[/hide]", "Solution_3": "[quote=\"MCrawford\"]Ms. Salman purchases a rectangular lot with area 1,200 sq ft which has a length equal to 3 times its width. She wishes to divide the lot into exactly 2 plots having equal area and install fencing along the perimeters of both lots. What is the least number of feet of fencing she must buy?[/quote]\r\n[hide=\"solution\"]\n3x:^2: = 1200\nx:^2: = 400\nx = +/- 20\n\n20 by 60. Divide it widthwise - 20 feet plus the perimeter.\n\n20 + 2(20 + 60)\n20 + 2(80)\n20 + 160\n[b]180 feet[/b]\n[/hide]" } { "Tag": [], "Problem": "We should go back and dig up the old test and put the solutions here...\r\n\r\nFor educational purposes... ;)\r\n\r\nUnless Code like them to be kept secret in order to sell his solution book later... :D", "Solution_1": "College Station's Vertical round solutions through round 7 may be found [url=http://www.thoobik.com/codybowl.html]here[/url]. \r\n\r\nIf you're just looking for answers for the individual tests, here you are:\r\n\r\n[code]0\t221\t232\t036\t089\t951\t360\t720\t338\t391\n1\t529\t280\t048\t050\t512\t504\t128\t820\t887\n2\t076\t169\t029\t388\t938\t099\t270\t127\t042\n3\t328\t230\t100\t043\t113\t225\t306\t245\t242\n4\t381\t444\t250\t084\t037\t176\t850\t321\t144\n5\t072\t661\t104\t583\t841\t476\t045\t311\t945\n6\t390\t737\t325\t605\t126\t239\t983\t200\t023\n7\t816\t162\t144\t155\t613\t504\t397\t262\t416\n[/code]", "Solution_2": "actually i was looking for \"solutions\"\r\n\r\nyea, i been to ur site a couple of times before too...\r\n\r\nthanks", "Solution_3": "DJ/Nathan, I believe week 7 #8 is 262.", "Solution_4": "Since we're on the topic of solutions to problems -\r\n\r\nI'd like to see a solution to Week 7 nr. 9 that does not involve brute force, which is how I attempted it (and missed.)", "Solution_5": "Oh sorry, you are correct. :blush: I'll tell him.\r\n\r\nThanks.", "Solution_6": "Oops. Actually, Samson, I thought you had posted those answers (it was actually DJ or Nathan, I can't tell which)...don't worry about telling him (them?), they'll see it.", "Solution_7": "No, you caught the culprit. I told thoobik (who in this case, is DJ) that the answer was 462. :blush:", "Solution_8": "For number 9, you just have to use a recursion. First, say that you won't use any 0's. After that, you just figure out what the distribution is in an efficient manner. And then add up. Signposts along the way: (5, 4), (20, 21) and (105, 104)." } { "Tag": [ "LaTeX" ], "Problem": "hi, I'm sort new to form, and I just want to ask people what they think about this site. I started creating this site a couple days ago.\r\n\r\nMy goal is to create simple math notes that people can easily understand. I sort of dislike those other sites since it takes a bit too much effort to find what I'm looking for. Even though they have a lot of content.\r\n\r\nWhat I'm planning to add are flash cards that contain formulas that a person should know. Questions and how a person should analyse these questions and how they should answer to get full marks.\r\n\r\nEven though there is hardly anything on the site. What do you guys think about the format and the direction of the site?\r\n\r\nezmathnotes.hosthelpers.net", "Solution_1": "I like the layout, I really do :) But, I would consider using LaTeX for the math notation -- it looks so much prettier than formatted HTML! There's tons of guides around AoPS for getting LaTeX up and running -- give it a shot! (Oh, and this thread doesn't really fit in the problem solving forum :()", "Solution_2": "Sorry about that, this was sort of the closest thing I could find to what I was posting, where should I have posted it.\r\n\r\nThanks for the advice too. Ill add it soon.", "Solution_3": "This thread doesn't really fit anywhere else, and this forum [i]is [/i]about math resources, so this is probably the best place to put this thread." } { "Tag": [ "function", "algebra", "domain", "calculus", "calculus computations" ], "Problem": "Let $ a$ be a positive constant. Find the maximum and minimum values of $ y\\equal{}a\\sqrt{1\\plus{}x}\\plus{}\\sqrt{a^2\\minus{}x}$.", "Solution_1": "First, write:\r\n\r\n$ y'(x) \\equal{} \\frac {1}{2}\\left(\\frac {a}{\\sqrt {1 \\plus{} x}} \\minus{} \\frac {1}{\\sqrt {a^2 \\minus{} x}}\\right)$\r\n\r\nand note that the domain of $ y$ is $ [ \\minus{} 1,a^2]$.\r\n\r\nNow, if $ a \\leq 0$ we see that $ y' < 0$ and therefore the minimum is:\r\n\r\n$ y_{max} \\equal{} y( \\minus{} 1) \\equal{} \\sqrt {1 \\plus{} a^2}$\r\n\r\n$ y_{min} \\equal{} y(a^2) \\equal{} a\\sqrt {1 \\plus{} a^2}$\r\n\r\nOn the other hand, if $ a > 0$ we get\r\n\r\n$ x_{0} \\equal{} a^2 \\minus{} 1$\r\n\r\nthat turns out to be a maximum. So, since $ x_0 \\geq \\minus{} 1$ we see that the minimum is either on the left side or on the right side of the domain. But, as we can see from before, the right value is bigger that the left, so we have:\r\n\r\n$ y_{max} \\equal{} y(a^2 \\minus{} 1) \\equal{} a^2 \\plus{} 1$\r\n\r\n$ y_{min} \\equal{} y( \\minus{} 1) \\equal{} \\sqrt {1 \\plus{} a^2}$\r\n\r\nAnd that's it! :)", "Solution_2": "First of all $ a>0!$ the max is correct, but the min is not.", "Solution_3": "Agreed. The minimum varies for $ a < 1$ and $ a > 1$. For $ a < 1$, $ y_{min} \\equal{} y(a^2)$ and for $ a > 1$ it is $ y_{min} \\equal{} y( \\minus{} 1)$. For $ a \\equal{} 1$ we have $ y_{min} \\equal{} y( \\minus{} 1) \\equal{} y(a^2)$.", "Solution_4": "That's right. Let me summarize the answer, \r\n\r\n$ y_{max}=y(a^2-1)=a^2+1$.\r\n\r\n$ y_{min}=\\left\\{\r\n\\begin{array}{ll}\r\ny(a^2)=a\\sqrt{a^2+1}\\ (09$ for $n>2$.", "Solution_1": "h\u00e9, h\u00e9 :D you still like this one...\r\n\r\nPierre.", "Solution_2": "A related problem was solved in the topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8590]exponents problem[/url].", "Solution_3": "As far as I remember, it is a difficult problem.\r\n\r\nPierre.", "Solution_4": "You're right Pierre. As you know, I'm eating digits for my breakfast :wacko: \r\nThanks to Ravi B for the link but I think it's more complicated in this case...", "Solution_5": "Yes, I agree this problem is more difficult." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "(a) Show that, for all non-zero values of the constant $ k$, the curve\r\n\r\n$ y\\equal{}\\frac{kx^{2}\\minus{}1}{kx^{2} \\plus{} 1}$\r\n\r\nhas exactly one stationary point.\r\n\r\n(b) Show that, for all non-zero values of the constant $ m$, the curve \r\n\r\n$ y \\equal{} e^{mx}(x^{2} \\plus{} mx)$\r\n\r\nhas exactly two stationary points.", "Solution_1": "$ y \\equal{} \\frac {kx^{2} \\minus{} 1}{kx^{2} \\plus{} 1} \\equal{} \\allowbreak 1 \\minus{} \\frac {2}{kx^{2} \\plus{} 1}$\r\n\t\r\nset $ \\frac {dy}{dx} \\equal{} \\allowbreak \\frac {4kx}{\\left( kx^{2} \\plus{} 1\\right) ^{2}} \\equal{} 0$ to locate stationary points.\r\n\r\nObviously, $ (0, \\minus{} 1)$ is the only one for all $ k\\neq 0$\r\n\t\r\n\r\n$ y \\equal{} e^{mx}\\left( x^{2} \\plus{} mx\\right)$\r\n\t\r\n$ \\frac {dy}{dx} \\equal{} \\allowbreak \\allowbreak e^{mx}\\left( mx^{2} \\plus{} (m^{2} \\plus{} 2)x \\plus{} m\\right) \\equal{} 0$\r\n\t\r\nSince the discriminant $ (m^{2} \\plus{} 2)^{2} \\minus{} 4m^{2} \\equal{} \\allowbreak m^{4} \\plus{} 4 > 0,$\r\n\t\r\nthere are always $ 2$ stationary points for all non zero $ m$." } { "Tag": [ "quadratics" ], "Problem": "Let $x,y$ be positive integers. Can $x^4+x^2y^2+y^4$ be a perfect square?", "Solution_1": "No, $x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2)$.\r\nSo for this to be a perfect square, we must have $x^2+xy+y^2=x^2-xy+y^2\\Rightarrow xy=-xy$, which cannot happen in positive integers.", "Solution_2": "That only works if the two brackets are relatively prime. eg:\r\n\r\n$(2)(18)=6^2$", "Solution_3": "[quote=\"seamusoboyle\"]That only works if the two brackets are relatively prime. eg:\n\n$(2)(18)=6^2$[/quote]\r\nYou're right, my mistake.", "Solution_4": "[quote=\"Jimmy\"]No, $x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2)$.\nSo for this to be a perfect square, we must have $x^2+xy+y^2=x^2-xy+y^2\\Rightarrow xy=-xy$, which cannot happen in positive integers.[/quote]\r\n\r\nBut what if x^2 + xy + y^2 and x^2 - xy + y^2 are both square?", "Solution_5": "[hide]Jimmy's decomposition doesn't help much here. It's better off to use this one: \n\\[ x^4+ x^2y^2 + y^4 = (x^2+y^2)^2 - x^2y^2 = k^2 \\ \\Rightarrow \\ xy = x^2-y^2, k = 2xy\\] (from the Pytagorean triplets) therefore looking at the latter as a quadratic in $x$ we have $\\Delta = 4y^2 + y^2 = 5y^2 $ which cannot be a perfect square if $y$ is a positive integer. [/hide]" } { "Tag": [ "trigonometry", "algebra", "system of equations", "algebra proposed" ], "Problem": "Solve the following system of equations:\r\n$\\cos x+\\cos y=1$,\r\n$\\sin x \\cdot \\sin y =-\\frac34$\r\nif $0 \\leq x \\leq 2\\pi$ and $0 \\leq y \\leq 2 \\pi$", "Solution_1": "[hide=\"Change variables\"] \n$x+y=2u, \\ \\ x-y=2v$ \n\nObtain easy system\n$\\cos u \\cos v = \\frac{1}{2}$\n$\\cos^{2}u-\\cos^{2}v = \\frac{3}{4}$[/hide]" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "The length of a median is 36 units. Describe the location of the centroid on that median. :?:", "Solution_1": "[quote=\"raebre\"]The length of a median is 36 units. Describe the location of the centroid on that median. :?:[/quote]\r\n\r\nhttp://en.wikipedia.org/wiki/Centroid", "Solution_2": "This question is too weak for a society like this....", "Solution_3": "I am sorry that it is too weak for this society. I am in college and thought I would see if I could get some help here. I am not a math wizard or genius. Iam though studying to become a teacher and I have not had this kind of math in years and found this site and thought I would see if this forum would be nice to try and help me figure it out. If not that is alright.", "Solution_4": "It's O.K. But it'll be more appropriate to post them in Lower level places like \"High School Basics\" or \"Intermediate Topics\" in this site.", "Solution_5": "I just thought since it was geometry that I would put in the geometry section and this is my first time using this site so I had no clue as to how it works. I will make sure to do that next time, I actually think I have them all figured out now but I do appreciate your help and kindness. If I can not get any answers in the other forums you suggested would it be alright to repost them here. I mean since you all seem to be pretty good with math and all." } { "Tag": [ "absolute value" ], "Problem": "Let us consider $25^{\\frac{1}{2}}$\r\n\r\nOn page 4 of AoPS 1 it asserts $(5^2)^{\\frac{1}{2}} = 25^{\\frac{1}{2}}$\r\nand therefore $25^{\\frac{1}{2}} = 5$\r\n\r\nBut how about this? $-5^1 = (-5^2)^{\\frac{1}{2}} = 25^{\\frac{1}{2}}$\r\n\r\nSo does $25^{\\frac{1}{2}} = \\pm{5}$ \r\n\r\nOr have I made a mistake? If so point it out for me!", "Solution_1": "When you take the square and then the square root, you are actually finding the absolute value, which is why you're gettting $-5=25^{\\frac{1}{2}}=5$. And when taking the square root of a positive real number, it's implied that you get the positive square root. Of course, this isn't the case when you have something like $x^2=25$, which would have $-5$ and $5$ as its answers." } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "I go to an American International school in India, and was trying to get my school to register for the AMC. When trying to convince the school to register for it, what websites should I point them towards?\r\n\r\nI'm particulary looking for sites that point out the prestige of this competition (This is the biggest in the US, leading to the IMO...)\r\n\r\nThanks", "Solution_1": "Start with our basic webpage\r\nhttp://www.unl.edu/amc\r\n\r\nOn it you will find our Frequently Asked Questions, the Sliffe Awards for teachers, pictures of the USAMO winners, pictures of the USA IMO teams, statistics and much more. Point out the number of participating schools listed among the statistics.\r\nGood luck.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions" } { "Tag": [ "inequalities" ], "Problem": "If x,y,z > 0 show that\r\n$ \\frac {{4x \\minus{} 1}} {{2x^2 \\plus{} 1}} \\plus{} \\frac {{4y \\minus{} 1}} {{2y^2 \\plus{} 1}} \\plus{} \\frac {{4z \\minus{} 1}} {{2z^2 \\plus{} 1}} \\leqslant 2$", "Solution_1": "[quote=\"xdanielxedgex\"]If x,y,z > 0 show that\n$ \\frac {{4x \\minus{} 1}} {{2x^2 \\plus{} 1}} \\plus{} \\frac {{4y \\minus{} 1}} {{2y^2 \\plus{} 1}} \\plus{} \\frac {{4z \\minus{} 1}} {{2z^2 \\plus{} 1}} \\leqslant 2$[/quote]\r\n\r\nTry $ x\\equal{}y\\equal{}z\\equal{}1$." } { "Tag": [ "geometry", "3D geometry", "sphere", "calculus", "calculus computations" ], "Problem": "A solid hemisphere of variable radius x (0PB.$ Prove that $ S_{\\bigtriangleup XPQ}>S_{\\bigtriangleup YBC}.$ Where $ S_{\\bigtriangleup XYZ}$ denotes the area of triangle $ XYZ.$", "Solution_1": "It's well-known that $\\triangle{XPQ}\\sim\\triangle{XBC}$ and $\\triangle{XBP}\\sim\\triangle{XCQ}$ (all similarities here are direct). First we prove some easy configuration facts: $\\angle{XPQ}=\\angle{XBC}$ forces $X,A$ to lie the same side of $BC$, whence $\\angle{PXB}=\\angle{QXC}$ forces $X,A$ to lie on the same side of $PQ$.\n\nLet $\\alpha=\\angle{PXB}=\\angle{QXC}$ and $\\beta=\\angle{XBP}=\\angle{XCQ}$ so that $\\alpha<\\beta<90^\\circ$ (since $\\triangle{ABC}$ is acute and $PX>PB$). For convenience, define the similarity ratio $r = PQ/BC = XP/XB = \\sin\\beta/\\sin(\\alpha+\\beta)$.\n\nFor points $U,V,W$, let $d(U,VW)$ denote the directed distance from $U$ to $VW$ (directed so that $X,A$ are \"above\" $PQ$ and $BC$). Then $d(Y,BC) = d(X,BC) - XY\\cos\\alpha$ and $XY = 2\\cdot d(X,PQ) = 2r\\cdot d(X,BC)$. It follows that\n\\begin{align*}\n\\frac{2[YBC]}{2[XPQ]} = \\frac{2[YBC]}{2r^2[XBC]} = \\frac{| BC\\cdot d(Y,BC)|}{r^2 BC\\cdot d(X,BC)}\n&= \\frac{|d(X,BC) - 2r\\cdot d(X,BC)\\cos\\alpha|}{r^2\\cdot d(X,BC)} \\\\\n&= \\frac{\\sin(\\alpha+\\beta)}{\\sin^2\\beta}|\\sin(\\alpha+\\beta)-2\\sin\\beta\\cos\\alpha| \\\\\n&= \\frac{|\\sin(\\alpha+\\beta)\\sin(\\alpha-\\beta)|}{\\sin^2\\beta} \\\\\n&= \\frac{|\\sin^2\\alpha - \\sin^2\\beta|}{\\sin^2\\beta}<1,\n\\end{align*}as desired.", "Solution_2": "Interesting problem :)\n\nLet $ PQ $ and $ BC $ meet at $ T $. And circle $ (TXY) $ meet $ PQ,BC $ at $ R,S $\n(1) First, we show that $ Y $ is below $ BC $. (Actually, $ Y $ is on $ BC $ when $ PX=PB $) \n $ R $ is the midpoint of arc $ XY $. And $ \\triangle{XRS} \\sim \\triangle{XPB} \\sim \\triangle{XQC} $.\nWhich means $ XR >RS $ so, $ Y $ is below $ BC $.\n\n(2) Now we show the problem. \n$ \\frac{S_{\\bigtriangleup YPQ}}{S_{\\bigtriangleup YBC}}= \\frac{{PQ} \\times {RY}}{{BC} \\times {SY}}=\\frac{{XR} \\times {RY}}{{XS} \\times {SY}} >1 $.\n\nSo, done :D" } { "Tag": [ "AMC" ], "Problem": "Apparently, I'm in an awkward situation.\r\n\r\nI'm aware that AMC takes place in Feb 1st and 16th\r\n\r\nApparently, I'm living in Hong Kong and will be going to the US in Feb 14th.\r\n\r\nI certainly wont be able to take it on Feb1st, but how about the 16th of Feb?\r\n\r\nI know it is extremely late to order test papers for AMC but do u think my new public HS will allow me to take AMC in such short notice? Its a rather rural HS and I'm worried that it won't have enuff finance to hold the competition. I've called the HS already, but I wasnt able to contact the math department. \r\n\r\n\"Late Registration Contest B (Jan. 10,-Feb. 7) $52\" \r\n\r\nSo if I do manage to get in contact with some1 who is in charge or knows about the test, will I be able to take it after informing that I would like to participate in it?\r\n\r\n\r\nYa.. Its my utmost laziness of not making the calls earlier on in Dec that takes the blame. I really would wish to take the exam. For UKMT (this is a british version of AMC), the school just ordered loads of copies of the exam and anyone who wished to take it, takes it. Is this the same for AMC? Or do onli those who ordered gets to participate?", "Solution_1": "Your best chance is to keep contacting your high school, and to try to get them to administer the AMC B on the 16th. \r\n\r\nUnfortunately, if they are a small school, they may only be giving it on the 1st.", "Solution_2": "Actually, if it's a really small school, you might even be the only person to take it, making it easier for you to take the AMC." } { "Tag": [ "articles" ], "Problem": "I have been trying to obtain a TI-84+ Silver Edition emulator for my computer, but I'm stuck on the process of downloading the ROM img. I following the instructions on ticalc.org (http://www.ticalc.org/programming/emulators/romdump.html#5) for the TI-83+ downloading the ROM img, but when I try to run the romdump.exe, nothing happens. Can somebody tell me what I'm doing wrong/alternate websites, etc.?", "Solution_1": "umm, can someone help me out? Maybe I posted in the wrong topic... if so, could a mod move it or something, please? Thanks", "Solution_2": "Well, if I remember correctly, it's illegal to upload ROM images online, so you'll have to get it from your calculator.\r\n\r\nI don't remember much else about the entire process, but you might try calculator-specific forums like those at http://www.maxcoderz.org or http://www.detachedsolutions.com, though.", "Solution_3": "yeah, the article on ticalc gave instructions on getting the ROM img. from my calculator, but it said i needed to run an executable file that was supposed to be included on the file it told me to download, but when i ran it nothing happened, and the calc would not send me the ROM img. when i ran the program that was supposed to do it. help? anyone?" } { "Tag": [ "superior algebra", "superior algebra open" ], "Problem": "We know that for a natural number M > 1 and a real number $0 \\leq q < 1$, there exist unique $b_i \\in \\{0,1,2,...,M-1\\}$ s.t\r\n\r\n\\[ q = \\sum_{i=0}^\\infty b_i M^{-i}\\]\r\n\r\nbut what I tried to generalize the M-base to:\r\n\r\nfor any number $01$, and therefore there is no way that you can obtain a positive number less than 1 as the result of the serie you are posting.\r\n\r\nOn the other hand, if you change $p^{-i}$ for $p^i$, then, in I believe is not true in general, but I still don't know exactly how.\r\n\r\nBy the way, the sum should start on $i=1$ in the first case, and in $i=2$ in the second.\r\n\r\nBest,", "Solution_2": "[quote=\"djimenez\"]\nvohung,\n\nAs you are posting it, it is clearly not true, because if $01$, and therefore there is no way that you can obtain a positive number less than 1 as the result of the serie you are posting.\n\nOn the other hand, if you change $p^{-i}$ for $p^i$, then, in I believe is not true in general, but I still don't know exactly how.\n\nBy the way, the sum should start on $i=1$ in the first case, and in $i=2$ in the second.\n\nBest,[/quote]\r\n\r\nI am sure it is your typo but in Vietnamese, \"vohung\" and \"vuhung\" have the same meaning :)\r\n\r\nBtw, I changed to\r\n\r\n\\[ q = \\sum_{i=0}^\\infty b_i p^{i}\\]\r\n\r\nand as I stated, this is an \"open question\", I could not prove or disprove it....", "Solution_3": "[quote=\"vuhung\"]I am sure it is your typo but in Vietnamese, \"vohung\" and \"vuhung\" have the same meaning :)\n[/quote]\r\n\r\nvuhung,\r\n\r\nJust in case, the mistake (already edited) was totally honest, I didn't mean to confuse terms. I confess to be totally ignorant with respect to asian languages, and just misread the name.\r\n\r\nI will continue trying to se what can I do with your problem.\r\n\r\nBest,", "Solution_4": "It's not true in general because these expansions are not unique. For example, in base $\\frac{1+\\sqrt5}2,$ $.100$ and $.011$ represent the same number. Uniqueness can only happen if the largest allowed decimal $.aaaaa\\dots$ is less than or equal to 1. By your rules, this only happens at the integers.\r\n\r\nYou can always construct at least one expansion by the greedy algorithm.", "Solution_5": "[quote=\"jmerry\"]It's not true in general because these expansions are not unique. For example, in base $\\frac{1+\\sqrt5}2,$ $.100$ and $.011$ represent the same number. Uniqueness can only happen if the largest allowed decimal $.aaaaa\\dots$ is less than or equal to 1. By your rules, this only happens at the integers.\n\nYou can always construct at least one expansion by the greedy algorithm.[/quote]\r\n\r\nYep, you are right! :first: \r\n\r\nBut what would happen on the other extreme. There is no unicity, but existence? I there always a representation of this kind for any number? Probably it does... yes, I think it does!\r\n\r\nBest,", "Solution_6": "[quote=\"jmerry\"]It's not true in general because these expansions are not unique. For example, in base $\\frac{1+\\sqrt5}2,$ $.100$ and $.011$ represent the same number. Uniqueness can only happen if the largest allowed decimal $.aaaaa\\dots$ is less than or equal to 1. By your rules, this only happens at the integers.\n\nYou can always construct at least one expansion by the greedy algorithm.[/quote]\r\nThanks jmerry, now I have to find another way to solve my problem :) I think there is another way to genralize M-base numbers. I will post if found any." } { "Tag": [ "LaTeX" ], "Problem": "How can i find The imaginary part of (1-2i)^2-i.\r\n\r\nThis is what i have done so far:\r\n(1-2i)^2 (1-2i)^-i\r\n\r\n=(1-4i+4)(1-2i)^-i", "Solution_1": "If you don't mind, using $\\LaTeX$ to format your posts makes them much easier to read. Anyway, do you mean $(1-2i)^{2}-i$ or $(1-2i)^{2-i}$?", "Solution_2": "I think Anood meant the latter.", "Solution_3": "Post removed; nevermind", "Solution_4": "mets, that's not true (in fact, basically everything can be written as a real plus an imaginary)\r\nJust typing it into a ti-83 gives $-(1.64...)-(.202...)i$...\r\nhttp://en.wikipedia.org/wiki/Exponentiation#Computing_complex_powers\r\nExplains the steps how to compute complex powers.\r\n(It's pretty textbook)" } { "Tag": [ "articles" ], "Problem": "The Wall Street Journal published a study today claiming that mathematician is the best career when considering such factors as working conditions and income potential. In fact many of their top ranking careers draw heavily on mathematics and problem solving.\r\n\r\n 1. Mathematician\r\n 2. Actuary\r\n 3. Statistician\r\n 4. Biologist\r\n 5. Software Engineer\r\n 6. Computer Systems Analyst\r\n\r\nYou can read more commentary and find links to the article from the Metroplex Math Circle site:\r\n\r\n[url]http://metroplexmathcircle.wordpress.com/2009/01/06/wsj-survey-says-mathematician-best-job/[/url]", "Solution_1": "Similar studies have been done over the years." } { "Tag": [ "vector", "trigonometry", "calculus", "integration" ], "Problem": "Hello everybody!\r\nHere's a (easy) problem:\r\nA small snail is sliding from the top of a ball (the motion is frictionless). Show that the snail will fall away from the ball when it covers a lenght on the ball corresponding to an angle theta so that cos(theta)=2/3. (See figure attached)\r\nI hope i was clear enough! :D", "Solution_1": "can the ball roll?", "Solution_2": "The action of the force of gravity on the snail in combination with the curve of the ball gives the snail a small horizontal acceleration, as long as the snail is in contact with the ball. After that it stops accelerating. \r\n\r\nAt whatever point the snail and the ball part, the horizontal acceleration must stop, although the vertical acceleration continues. \r\n\r\nThe snail's trajectory when it leaves the ball will be at a tangent to ball surface.\r\n\r\nThe question is, when does the snail have enough horizontal speed to free itself from the ball, given it is also accelerating downward? \r\n\r\nSince the snail is sliding not dropping, the vertical acceleration is *not* the simple Newton gravity law of acceleration. The acceleration is being retarded by the opposing surface of the ball. \r\n\r\nYet the vertical force on the snail *is* constant and conforms to the gravity equation F = GMM/d^2. (d = distance to earth's centre doesn't change significantly.).\r\n\r\nMy guess is that the snail cannot depart from the ball until the resultant vector is >= the tangent. This is a simultaneous equation type problem.\r\n\r\nThe answer will be different if the snail were a ball rolling (even with negligible friction) since some energy will be being stored as angular momentum.", "Solution_3": "They seperate if the force between them is 0. So we will have:\r\n\\[mg \\cos{\\theta} = m \\frac{v^2}{R}\\]\r\nUse the conservation of energy , we will have:\r\n\\[R(1-\\cos{\\theta})mg=\\frac{1}{2}mv^2\\].\r\nSolve the system, we get $\\cos{\\theta} = \\frac{2}{3}$.", "Solution_4": "Nice shobber :). Assuming gravity is g, how much time does it take for this to happen?", "Solution_5": "another question: if the bigger ball is not fixed, what is $\\theta$?", "Solution_6": "I could not solve my own problem, and I do not think it's doable (at any sort of easy level). A small angle approximation cannot be made for obvious reasons (the angle is large)!\r\n\r\nThere is a way to solve the general harmonics motion differential equation without making the small angle approx but it is not elementary. So blah.", "Solution_7": "[quote=\"Spoon\"]Nice shobber :). Assuming gravity is g, how much time does it take for this to happen?[/quote]\r\nI find a calculus solution. So here we go:\r\n\r\nAssume it turns a very small angle $d\\theta$\r\n\r\nThen $s=R \\sin{d\\theta}=R d\\theta$\r\n\r\nUse the conservation of energy, we will have $mv^2=2mgR(1-\\cos{d\\theta})$\r\nSimplify, we get: $v=\\sqrt{2gR(1-\\cos{d\\theta})}$\r\n\r\n$t= \\int{dt}= \\frac{R}{\\sqrt{2gr}} \\int {\\frac{d\\theta}{\\sqrt{1-\\cos{d\\theta}}}}$\r\n\r\nI am sure it is not correct since it will have $R$ in the final answer...But I think whatever calculus way we use, we shall always try to get $\\int f(\\theta) d\\theta$.\r\n\r\nI am still trying to find a basic solution without solution...", "Solution_8": "[quote=\"Spoon\"]Nice shobber :). Assuming gravity is g, how much time does it take for this to happen?[/quote]\r\n\r\nThe initiation of the motion of the snail is not clear to me. It starts out with zero energy (as I saw to be silently assumed in the solutions), that means, zero speed. That means, it doesn't slide down. Therefore, the snail must get a very small impulse coming down the ball, and we assume that this impulse is very small and then solve the problem. When we limitly approach the impulse to zero, the critical take-off angle limitly goes to whatever arccos(2/3) is. But the issue with time is different. If we approach to zero with the impulse, the time needed for the snail to take off goes to infinity.", "Solution_9": "[quote=\"kubus\"]The initiation of the motion of the snail is not clear to me. It starts out with zero energy (as I saw to be silently assumed in the solutions), that means, zero speed. That means, it doesn't slide down. Therefore, the snail must get a very small impulse coming down the ball, and we assume that this impulse is very small and then solve the problem. When we limitly approach the impulse to zero, the critical take-off angle limitly goes to whatever arccos(2/3) is. But the issue with time is different. If we approach to zero with the impulse, the time needed for the snail to take off goes to infinity.[/quote]\r\nYes! This is precisely why I could not solve it! Because as far as the initial conditions for the differential equation are concerned, there is no initial velocity! hehe, which means the snail NEVER moves. And when dealing with \"time\", you cannot just assume a tiny \"nudge\", because the magnitude of this nudge will affect your result. In fact, it should affect the result to the original question, even though as Kubus pointed out it was \"silently assumed\"." } { "Tag": [ "ratio", "geometry", "AMC", "AMC 10", "AMC 10 A" ], "Problem": "A triangle with side lengths in the ratio $ 3: 4: 5$ is inscribed in a circle of radius $ 3.$ What is the area of the triangle?\r\n\t\r\n$ \\textbf{(A)}\\ 8.64 \\qquad \\textbf{(B)}\\ 12 \\qquad \\textbf{(C)}\\ 5\\pi \\qquad \\textbf{(D)}\\ 17.28 \\qquad \\textbf{(E)}\\ 18$", "Solution_1": "[hide] Since the triangle has side lengths in the ration of 3:4:5, we can say that it is a right triangle (since 3,4,5 is a Pythogorean Triplet). We know the radius of the triangle is 3, so the radius is 6. Now, since the triangle is right, we know that the hyptonuse is the diameter of the circle. Say that 5x=6. Thus, x=1.2. Therefore, 3x=3.6 and 4x=4.8. Now, this means that the length of the other sides are 4.8 and 3.6. Finally, we multiply these sides and divide by 2 to get 8.64, so the answer is A. [/hide]", "Solution_2": "[hide]The triangle must be inscribed in a semicircle because it is right.\n\nThe ratio of the diameter to the hypotenuse is $6: 5$. The area of a 3-4-5 triangle is $6$.\n\nThe ratio of the areas is the ratio of the sides squared, so $A=6\\cdot\\left(\\frac65\\right)^{2}=8.64\\Rightarrow\\boxed{\\text{A}}$[/hide]", "Solution_3": "This was in AMC10A\r\nAnswer is [b]8.64[/b]", "Solution_4": "Yeah, I can reiterate every answer already posted on the AMC 12A too..." } { "Tag": [ "inequalities", "calculus", "derivative", "algebra solved", "algebra" ], "Problem": "I think I found a really nice way of solving Inequalities with the property (any inequality can be reduced to g(x1,x2,x3.....,xn)>=0):\r\nIf from g(x1,x2,x3.....,xn)>= it is also true that\r\ng(\\lambda x1,\\lambda x2,\\lambda x3.....,\\lambda xn) >=0\r\nwith \\lambda >0 then there is a very easy way to prove them.\r\nAlso note that after some algebra, a lot of IMO-type inequalities have this property", "Solution_1": "It uses a really nice idea and I think that after that, some inequalities become not challenging. It is like Shturm's way of proving ineqs ( but it doesn't require so much algebra and it doesn't lead to even harder inequalities). I showed it to my Olympiad Teacher at School so he said it is allright.\r\n:)\r\n:D", "Solution_2": "Is it secret or what? :D", "Solution_3": "can you give some examples on some inequalities?\r\n\r\nsimple and IMO ineqs?\r\n\r\ncheers! :D", "Solution_4": "Sorry, I don't have a lot of time, but it is something like this:\r\ng(x1,x2,x3.....,xn)>=0\r\nIf from g(ax1,ax2,ax3.....,axn)>= it is also true that\r\ng( ax1, ax2, ax3....., axn) >=0\r\nwith a>0 \r\nWe take \"a\" as being the maximum of them.\r\nSo we have g(b1,...bn)>=0 , where bi=(ai)/(amax).\r\nNow all bi have 0=0\r\nThen\r\ndg/dx=2x-y\r\ndg/dx=0 iff y=2x\r\nThen g(x,2x)=3x^2>=0\r\nSimilar for y\r\n\r\nThen this method could be very impractical for inequalities with more terms, and for this inequality, AM-GM works better.", "Solution_8": "I made some mistakes when writing it.\r\nBut for the given inequality, take consider x being the maximum.\r\nTaking g((x/x),(y/x))>=0 we get g((x/x),(y/x))= 1 - (x/y)+(x/y)^2>=0\r\nTaking (x/y)=t gives us 1 - t + t^2>=0. 0<=t<=1. Now it is easy.\r\nSorry Valentin, or you angry or what?\r\nI don't say this is the best way to solve an IMO iequality. But to solve other kind of inqualities, it is really good.\r\nAnd by the way, is somebody here who could write the MONGOLIAN Inequality? It is something about the averages of 3 and 2 consecutive integers.\r\n? by DUST" } { "Tag": [], "Problem": "Help me solve this problem.\r\nYou have two strings(length<=5000) consists only of two letters 'a' and 'b'. If 'a' and 'b' are adjancent you can switch their places. You should compute min nuimber of switches you can get second string from the first.", "Solution_1": "Well, if you just consider the positions of the 'a's, then they have to match up from left to right, so just move the leftmost a to the correct position, then the second leftmost, etc.." } { "Tag": [ "inequalities" ], "Problem": "If $a , b , c >0$ , find the minimum of the expression :\r\n\r\n $(\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}})(1+a)(1+b)(1+c)$\r\n\r\n[u] Babis[/u]", "Solution_1": "\u0393\u03b5\u03b9\u03b1 \u03c3\u03b1\u03c2 \u039a\u03cd\u03c1\u03b9\u03b5 \u039c\u03c0\u03ac\u03bc\u03c0\u03b7 \u03c4\u03b9 \u03ba\u03ac\u03bd\u03b5\u03c4\u03b5 ? \u039d\u03b1 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5.\r\nBy AM-GM we have $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\geq 3\\frac{1}{\\sqrt[3]{a^{2}b^{2}c^{2}}}$\r\n\r\nand \r\n$1+a=1+\\frac{a}{2}+\\frac{a}{2}\\geq 3\\sqrt[3]{\\frac{a^{2}}{4}}$\r\n\r\nand similarly $1+b\\geq 3\\sqrt[3]{\\frac{b^{2}}{4}}$ and \r\n $1+c\\geq 3\\sqrt[3]{\\frac{c^{2}}{4}}$\r\n\r\n. Multiplying and we find the minimum :wink: \r\n\r\n\u03a3\u03b1\u03c2 \u03ac\u03c1\u03b5\u03c3\u03b5 ?", "Solution_2": "[quote=\"silouan\"]\u0393\u03b5\u03b9\u03b1 \u03c3\u03b1\u03c2 \u039a\u03cd\u03c1\u03b9\u03b5 \u039c\u03c0\u03ac\u03bc\u03c0\u03b7 \u03c4\u03b9 \u03ba\u03ac\u03bd\u03b5\u03c4\u03b5 ? \u039d\u03b1 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5.\nBy AM-GM we have $\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}\\geq 3\\frac{1}{\\sqrt[3]{a^{2}b^{2}c^{2}}}$\n\nand \n$1+a=1+\\frac{a}{2}+\\frac{a}{2}\\geq 3\\sqrt[3]{\\frac{a^{2}}{4}}$\n\nand similarly $1+b\\geq 3\\sqrt[3]{\\frac{b^{2}}{4}}$ and \n $1+c\\geq 3\\sqrt[3]{\\frac{c^{2}}{4}}$\n\n. Multiplying and we find the minimum :wink: \n\n\u03a3\u03b1\u03c2 \u03ac\u03c1\u03b5\u03c3\u03b5 ?[/quote]\r\n\r\n Nice solution Silouan !!!We have only to observe that the expression really attends the minimum $\\frac{81}{4}$ , when $a=b=c= 2$" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $H$ be a normal subgroup of $G$, $\\mathcal{K}$ is a conjugacy class of $G$ contained in $H$ and $x\\in \\mathcal{K}$. Prove that $\\mathcal{K}$ is a union of $k$ conjugacy classes of equal size in $H$, where $k=|G:HC_{G}(x)|$.", "Solution_1": "I believe this is part of #19 from the 3rd edition of Dummit and Foote. Here's a write-up for your question, which I happened to have done sometime last year. I hope the details make sense. If you need help with the rest of the question, feel free to ask.\r\n\r\nI've posted the solution as an attachment, albeit as an image file. I presume you'll be able to view it.", "Solution_2": "How do you define $G^{[x]_H}$ any way? Thanks!", "Solution_3": "A generalization: \r\n\r\nGiven a finite transitive action (i.e. a finite group acts on a finite set) of $G$ on $S$, and given a normal subgroup $H\\lhd G$, the action of $H$ on $S$ has $k$ orbits of equal size, where $k=[G:H\\cdot\\mbox{Stab}_G(x)]$ for some $x\\in S$.\r\n\r\nThe fact that the sizes of the orbits under the action of $H$ on $S$ are equal follows from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47737]this other problem you've posted[/url].\r\n\r\nThe size of $S$ is precisely $k'=[G:\\mbox{Stab}_G(x)]$, so we want to show that the size of the orbit of $x$ under the action of $H$ is $\\frac{k'}k=[H\\cdot\\mbox{Stab}_G(x):\\mbox{Stab}_G(x)]$. This follows immediately from the fact that the orbit of $x$ under the action of $H\\cdot\\mbox{Stab}_G(x)$ is the same as the orbit under the action of $H$, and the stabilizer of $x$ in $H\\cdot\\mbox{Stab}_G(x)$ is $\\mbox{Stab}_G(x)$.\r\n\r\nP.S.\r\n\r\nI suppose your groups were finite in this exercise? :?", "Solution_4": "[quote=\"oldman\"]How do you define $G^{[x]_H}$ any way? Thanks![/quote]\r\n\r\nI guess I used different notation from Dummit and Foote --- whereas they use $G_s$ for the stabilizer of $s$ in $G$, I used (from class) $G^s$. With regard to your question, then $G^{[x]_H}$ refers to the stabilizer of an element $x$ from $[x]_H$. The latter, as defined in my solution, is the set $\\{xhx^{-1}:h \\in H\\}$.\r\n\r\nHope this helps clarify that! :)", "Solution_5": "[quote=\"grobber\"]A generalization: \n\nGiven a finite transitive action (i.e. a finite group acts on a finite set) of $G$ on $S$, and given a normal subgroup $H\\lhd G$, the action of $H$ on $S$ has $k$ orbits of equal size, where $k=[G:H\\cdot\\mbox{Stab}_G(x)]$ for some $x\\in S$.\n\nThe fact that the sizes of the orbits under the action of $H$ on $S$ are equal follows from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47737]this other problem you've posted[/url].\n\nThe size of $S$ is precisely $k'=[G:\\mbox{Stab}_G(x)]$, so we want to show that the size of the orbit of $x$ under the action of $H$ is $\\frac{k'}k=[H\\cdot\\mbox{Stab}_G(x):\\mbox{Stab}_G(x)]$. This follows immediately from the fact that the orbit of $x$ under the action of $H\\cdot\\mbox{Stab}_G(x)$ is the same as the orbit under the action of $H$, and the stabilizer of $x$ in $H\\cdot\\mbox{Stab}_G(x)$ is $\\mbox{Stab}_G(x)$.\n\nP.S.\n\nI suppose your groups were finite in this exercise? :?[/quote]\r\n\r\nThanks, I did like this\r\nAs you observed, $G$ can acts on orbits of $H$ in $S$ and this acts still transitive. The stabilizer of this action is $H\\,Stab_G(x)$. Therefore $k=[G:H\\,Stab_G(x)]$." } { "Tag": [ "Putnam", "college contests" ], "Problem": "This is the first time I have taken the Putnam. \r\nWhen/Where/How do we get our results back.\r\nAlso, Do we get a breakdown of what we scored \r\non each question or just a total score and rank?\r\n\r\n Thanks.", "Solution_1": "The results come back in mid-March. A sheet with your individual score is mailed to the faculty supervisor at your college - nothing gets sent to you individually. \r\n\r\nAll you ever learn is your total score and rank. You will not be told what the question-by-question breakdown is." } { "Tag": [ "probability", "combinatorics unsolved", "combinatorics" ], "Problem": "$6$ distinct points have been chosen on a circle. We accidentally name them $A,B,C,D,E,F$. What is the probability that the triangle $ABC$ does not intersect triangle $DEF$ ?", "Solution_1": "The two triangles do not intersect precisely when $A,B,C$ are consecutive points. There are $6$ sets of 3 consecutive points, and $\\binom{6}{3}= 20$ sets of 3 points, so the probability is $6/20 = 0.3$." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "[b](1) find the function f(x) If f:R+->R+,\n\n\nIf f(f(x)) = 3x+1/(x+3)[/b]", "Solution_1": "From the hypothesis:\r\n\\[ f(3x \\plus{} \\frac {1}{x \\plus{} 3}) \\equal{} 3f(x) \\plus{} \\dfrac{1}{f(x) \\plus{} 3} \\quad [1]\r\n\\]\r\n( both equals $ f(f(f(x))$)\r\n\r\nAlso note that $ y \\equal{} 3x \\plus{} \\frac {1}{x \\plus{} 3} > 0 \\wedge x > 0 \\iff x \\equal{} g(y) \\equal{} \\minus{} \\frac32 \\plus{} \\frac16 y \\plus{} \\frac16 \\sqrt {69 \\plus{} 18y \\plus{} y^2}$. This and [1] give rise to a formula for recursively evaluating the value of $ f(y)$ in terms of previously known value of $ f(x)$.\r\n\r\nBecause $ g(y) > 0 \\iff x > \\frac13$, we need all the values of $ f(x)$ in the interval $ (0,\\frac13]$ in order to generate every value of the function. Next, I will describe all $ f(x)$.\r\n\r\nGiven $ h : (0,\\frac13] \\rightarrow \\mathbb{R}^ \\plus{}$ be any function. The function $ f(x)$ extended from $ h(x)$ is defined recursively as:\r\n\\[ f(x) \\equal{} \\begin{cases} h(x) \\text{, \\quad if }0 < x\\leq \\frac13 \\\\\r\n3f(g(x)) \\plus{} \\dfrac{1}{f(g(x)) \\plus{} 3} \\text{, \\quad otherwise} \\end{cases}\r\n\\]\r\nThis family of functions gives us all possible solution of $ f(x)$." } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "maybe this has been asked before... :maybe:\r\n\r\nin how many ways, [u][i][b]without[/b][/i][/u] using the fact \r\n\r\n$ \\left ( \\int_{ \\minus{} \\infty}^\\infty e^{ \\minus{} x^2}\\;dx \\right )^2\\; \\equal{} \\; \\int_{\\theta\\equal{}0}^{2\\pi}\\int_{r\\equal{}0}^\\infty re^{ \\minus{} r^2} \\;dr\\;d\\theta$\r\n\r\n can you evaluate the integral\r\n\r\n$ \\int_0^\\infty e^{ \\minus{} x^2}\\;dx$", "Solution_1": "Call\r\n\\[ A(t) \\equal{} \\left(\\int_{0}^{t} e^{ \\minus{} x^2} dx\\right)^2 , B(t) \\equal{} \\int_0^1\\frac {e^{ \\minus{} t^{2}(1 \\plus{} x^2)}}{1 \\plus{} x^{2}} dx\r\n\\]\r\nDifferentiating with respect to $ t$ we have,\r\n\\[ A'(t) \\equal{} 2 \\int_{0}^{t} e^{ \\minus{} x^2} dx \\cdot e^{ \\minus{} t^2}\r\n\\]\r\nand\r\n\\[ B'(t) \\equal{} \\int_{0}^{1} \\minus{} 2t e^{ \\minus{} t^{2}(1 \\plus{} x^2)} dx\r\n\\]\r\n\r\n\\[ \\equal{} \\minus{} 2 \\int_{0}^{1} t e^{ \\minus{} t^2 x^2} dx \\cdot e^{ \\minus{} t^2}\r\n\\]\r\n\r\n\\[ \\equal{} \\minus{} 2 \\int_{0}^{t} e^{ \\minus{} u^2} du \\cdot e^{ \\minus{} t^2}\r\n\\]\r\n\r\n\\[ \\equal{} \\minus{} A'(t)\r\n\\]\r\nSo for some C we need,\r\n\\[ A(t) \\equal{} \\minus{} B(t) \\plus{} C.\r\n\\]\r\nLetting $ t \\rightarrow 0 \\plus{}$ the LHS tends to 0 and the RHS tends to $ \\minus{} \\int_{0}^{1} \\frac {dx}{1 \\plus{} x^2} \\plus{} C \\equal{} \\minus{} \\pi/4 \\plus{} C$ so $ C \\equal{} \\pi/4.$\r\nNow let $ t \\rightarrow \\infty$. We have,\r\n\\[ \\left(\\int_{0}^{\\infty} e^{ \\minus{} x^2} dx\\right)^2 \\equal{} 0 \\plus{} \\frac {\\pi}{4}\r\n\\]\r\nOr,\r\n\\[ \\int_{0}^{\\infty} e^{ \\minus{} x^2} dx \\equal{} \\frac {\\sqrt {\\pi}} {2}.\r\n\\]", "Solution_2": "That was amazing :) \r\n\r\nNow can you prove that:\r\n\r\n$ \\int^{x}_{0}e^{ \\minus{} u^{2}}\\text{d}u \\equal{} \\frac {\\sqrt {\\pi}}{2} \\minus{} \\frac {\\frac {1}{2}e^{ \\minus{} x^{2}}}{f(x)},\\;\\; x > 0$\r\n\r\nwhere $ f(x) \\equal{} x \\plus{} \\frac {1}{2x \\plus{} \\frac {2}{x \\plus{} \\frac {3}{2x \\plus{} \\frac {4}{x \\plus{} \\cdots}}}}$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all natyral numbers $ n$ and $ k$ such that the $ n^{n}$ have $ k$ digits ,and $ k^{k}$ have $ n$ digits (in base 10).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n[b]ee\"[/b]", "Solution_1": "$ \\begin{cases} 10^{k\\minus{}1} \\le n^n < 10^k \\\\\r\n10^{n\\minus{}1} \\le k^k < 10^n\\end{cases}$\r\n\r\nIf $ n \\ge 10$, we have $ k^k<10^n \\le n^n$, so $ k6", "Solution_1": "who is hepl me ? :rotfl: :roll:", "Solution_2": "[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=3710[/url]\r\n\r\nI don't think a full treatment of the solution is given here, just some talk about it.", "Solution_3": "can you frove that\r\n Sn=(Sn-1+Sn-2+Sn-3+Sn-4)-Sn-5 ?\r\n\r\n\r\nthanks", "Solution_4": "who can prove ?????????", "Solution_5": "Continue in the given link, don't double-post." } { "Tag": [ "geometry", "3D geometry", "sphere", "Pythagorean Theorem", "geometry solved" ], "Problem": "When three spherical balls, each of radius 10 cm, are placed in a hemispherical dish, it is noticed that the tops of the balls are all exactly level with the top of the dish. What is the radius, in centimetres, of the dish?\r\n\r\nSource : Australian Math Competition, Intermediate, 1999", "Solution_1": "Let $A, B, C$ be the centers of the 3 congruent spherical balls $(A), (B), (C)$, touching each other and forming an equilateral triangle with the side $AB = 2r = 20\\ \\text{cm}$. Let $H$ be the orthocenter of this equilateral triangle. Let $D$ be the center of the sphere $(D)$ of the hemispherical dish centered in the common tangent plane of the spheres $(A), (B), (C)$. The distance of the center $A$ from the vertical axis $DH$ of the hemispherical dish is equal to\r\n\r\n$AH = AB \\frac{\\sqrt 3}{3} = \\frac{2r \\sqrt 3}{3}$\r\n\r\nUsing Pythagorean theorem for the right angle triangle $\\triangle AHD$,\r\n\r\n$DA = \\sqrt{DH^2 + AH^2} = \\sqrt{r^2 + \\frac{4r^2}{3}} = r \\sqrt{\\frac 7 3}$\r\n\r\nThe tangency point $T$ of the spheres $(A), (D)$ is on their center line $DA$ and the radius of the sphere $(D)$ is\r\n\r\n$R = DT = DA + AT = r \\sqrt{\\frac 7 3} + r = \\left(1 + \\sqrt{\\frac 7 3}\\right) r \\doteq 2.5275\\ r = 25. 275\\ \\text{cm}$", "Solution_2": "Didn't think of the Pythagoras's theorem there... my hat is off to a brilliant solution. :)" } { "Tag": [ "induction", "abstract algebra", "modular arithmetic", "quadratics", "Divisibility Theory", "number theory" ], "Problem": "Determine all positive integers $n$ for which there exists an integer $m$ such that $2^{n}-1$ divides $m^{2}+9$.", "Solution_1": "A number of the type $m^2+9$ can, with exception of $p=3$, only have prime divisors $p \\not \\equiv -1 \\mod 4$:\r\nIf $q \\equiv -1 \\mod 4$ and $q \\neq 3$, we have $3$ invertible $\\mod q$ and get $m^\\prime^2 \\equiv -1 \\mod q$ with $m^\\prime \\equiv m \\cdot 3^{-1} \\mod p$. But now we easily derive a contradiction with Fermat's little theorem: $1 \\equiv m^\\prime^{p-1} = (m^\\prime^2)^\\frac{p-1}2 \\equiv (-1)^\\frac{p-1}2 \\equiv -1 \\mod p$.\r\n\r\n\r\nClaim: exactly those $n$ being a power of two work.\r\nIf $n$ has any odd prime divisor $q$, then $2^q-1 | 2^n-1$, so we just have to disprove the possibility of $2^q-1|m^2+9$ for $q$ any odd prime.\r\nSince $q>1$, $2^q-1 \\equiv -1 \\mod 4$, thus there is a prime $p \\equiv -1 \\mod 4$ dividing $2^q-1$. If we would have $p=3$, then $2^2 \\equiv 1 \\mod p$ yielding $1 \\equiv 2^q \\equiv 2^{q \\mod 2} \\equiv 2^1 \\mod p$, impossible. Thus $p>3$, but now $p$ has to divide $m^2+9$, impossible by the above.\r\n\r\nWe are left to prove that all $n=2^k$ work and do so by induction on $k$:\r\n$k=1$ gives the number $3$ which obviously works with $m_1=3$.\r\nIf we have found $m_k$ with $m_k \\equiv -9 \\mod (2{}^{2{}^k}-1)$, we see that $2{}^{2{}^{k+1}}-1=(2{}^{2{}^k}+1)(2{}^{2{}^k}-1)$ and that these two factors are coprime. By the chinese remainder theorem we find $m_{k+1} \\equiv m_k \\mod (2{}^{2{}^k}-1)$ and $m_{k+1} \\equiv 2{}^{2{}^{k-1}} \\mod (2{}^{2{}^k}+1)$ and this one works.", "Solution_2": "[quote=\"ZetaX\"]$ m_{k \\plus{} 1} \\equiv 2^{2^{k \\minus{} 1}} \\mod (2^{2^k} \\plus{} 1)$[/quote] \r\n\r\nDoes it work for $ m_{k \\plus{} 1}^{2}\\equiv {\\minus{}9}\\mod(2^{2^{k}}\\plus{}1)$ ? :maybe:", "Solution_3": "There is a factor $ 3$ missing...", "Solution_4": "the only prime in form of 4k+3 which can devide m^{2}+9 is 3. and so n must be a power of 2; so we should find for which n 2^{2^{n}} devides m^{2}+9.\n\nLemma 1 : all prime factors of 2^{2^{n}} module 4 is 1.\n\nLemma 2: a|b^{2} +1 iff all odd prime factors of a module 4 is qual to 1 and 4 does not divide a.\n\nso we can easily get that 2^{2^{n}}=3p, so that by using Lemma 1,2 there exist an m such that p devides m^{2} +1. finally we get 3p devides 3.m^{2} +3 and so (3m)^2 +9. so we done.", "Solution_5": "we will prove by induction on $n$ :\n\nlet $2^{2^{n}}-1|m^{2}+9$ for some $m\\in \\mathbb{Z}$.it's clear that $\\upsilon _{3}(2^{2^{n}}-1)=1$.\n$so 2^{2^{n}}-1=3c$. now we are going to prove there exist $m'\\in \\mathbb{Z}$ so that $2^{2^{n+1}}-1|m'^{2}+9.$\nlet $m'=3(\\frac{m}{3})+3tc=3k+3tc$. now what we want here is a $t\\in\\mathbb{Z}$ such that :\n $2^{2^{n+1}}-1=(2^{2^{n}}-1)( 2^{2^{n}}+1)|m'^{2}+9 =9((k+tc)^{2}+1)\\rightarrow c.2^{2^{n}}+1|((k+tc)^{2}+1$.\nit's clear that $c|RHS.$ by Chinese Remainder Theorem it is also clear that there exist such a $t$ so that $k+tc\\equiv 2^{2^{n-1}} mod 2^{2^{n}}+1.$ (note that $(c,2^{2^{n}}+1 )=1 )$.\nso $TIWWWTP$. (This Is What We Want To Prove :P ).", "Solution_6": "[quote=\"Peter\"]Determine all positive integers $n$ for which there exists an integer $m$ such that $2^{n}-1$ divides $m^{2}+9$.[/quote]\n\n\n[hide=\"My Solution\"]\nIf $n=1$, we are done. Else, let $p$ be an odd prime divisor of $2^{n} - 1$. Assume there exists an integer $m$ such that $2^{n} - 1$ divides $m^2 + 9$. This means $m^2 \\equiv -9 \\pmod{p}$. This means $\\left(\\dfrac{-9}{p}\\right) = \\left(\\dfrac{-1}{p}\\right) = 1$, so $p \\equiv 1 \\pmod{4}$. Since all prime divisors of $2^n - 1$ are $1 \\pmod{4}$, we know $2^n - 1 \\equiv 1 \\pmod{4}$, so $2^n \\equiv 2 \\pmod{4}$. This means $n = 1$, contradiction. Therefore, our only solution is $\\boxed{n = 1}$.\n[/hide]\n\nactually this forgets the prime 2.", "Solution_7": "The above is clearly false; the argument does not work for the prime $3|2^n - 1$ when $n$ is even (in particular if it is a power of two).", "Solution_8": "We can only define the quadratic residue or Lengendre symbol when gcd(9, p)=1.\nWe have to show when p and 9 have a G.C.D.\nif gcd(9, p) is not 1, we can find p=3 (is only prime that divides 2^n-1), it means n=2.\nwhen, n=2, we can find out that m can be any number if m is divided by 3.\n\n\nSorry... My English grammer is not good...", "Solution_9": "[hide=\"A20\"]\nFirst, we see that $n=1,2,4$ works. Now, for $n>4$, we want to find numbers $n$ such that $-9$ is a quadratic residue modulo $2^n-1$. If $3\\not|2^n-1$, then $a^2\\equiv-9\\mod2^n-1$ means there exists $b$ such that $b^2\\equiv -1 \\mod2^n-1$, which is clearly impossible, since $2^n-1 \\equiv 3\\mod4$. So, $3|2^n-1 \\implies 2|n$. Now, if $3|n$, then $a^2\\equiv-9\\mod2^n-1$ means there exists $b$ so that $b^2 \\equiv -1\\mod \\dfrac{2^n-1}9$, which is impossible since $\\dfrac{2^n-1}9$ must be $3\\mod4$. So $3\\not|n$. We now wish to find $m$ so that $m^2\\equiv-9\\mod2^n-1$. Since $3|m$, we get that, for $3p=m$, that $p^2+1 \\equiv 0\\mod\\dfrac{2^n-1}3$. This is possible whenever $2^n-1$ has no other 3 mod 4 divisors other than 3.\nSince $a|n \\implies 2^a-1 | 2^n-1$, we get that the only prime divisor of $n$ is 2, and indeed $2^{2^x}-1$ can be factored as $(2^2-1)(2^2+1)(2^4+1)(2^8+1)\\cdots$, so we get that $n=2^x$ for non-negative integral $x$ $\\boxed{}$[/hide]", "Solution_10": "N = $2^k$" } { "Tag": [], "Problem": "I need help not answers to #7. The other questions serve as a reference.", "Solution_1": "Stop creating new threads about this!\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=259406\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=259373\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=259455\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=259516\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=259529\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=259294" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove that, for every natural k, the number (k^3)! is divisible by\r\n\r\n(k!)^(k^2 +k+1).", "Solution_1": "all prime divisors k! are primes p<=k.\r\n$ord_{p}((k^{3})!)=\\sum_{i=1}^{\\infty}[ \\frac{k^{3}}{p^{i}}]$.\r\nIf p3$ be an odd integer. Prove there is a prime $p$, such that $p$ divides $2^{\\phi(n)}-1$ but $p$ doesn't divides $n$.", "Solution_1": "I have some thoughts on it. :)\r\n\r\nSuppose $n=p_1^{a_1}...p_k^{a_k}$, then $\\varphi(n)=p_1^{a_1-1}(p_1-1)...p_k^{a_k-1}(p_k-1)$, so $\\varphi(n)=2^{l}r$, where $l\\geq k$.\r\n[b]Case 1.[/b] Suppose $r\\geq 3$. We have now $2^{\\varphi(n)}-1=2^{2^lr}-1=(2^{2^{l-1}r}+1)...(2^{2r}+1)(2^r+1)(2^r-1)$. All these factors are pairwise coprime, $so 2^{\\varphi(n)}-1$ has at least $l+1>k$ prime factors :)\r\n[b]Case 2.[/b] Suppose $r=1$. It means that all $a_i=1$ and $(2^{2^{l-1}}+1)...(2^{2}+1)(2^1+1)$ contains exactly $p$ prime factors. It follows that all $2^{2^i}+1$ are perfect powers of prime numbers, which are not coprime with $p$. But $2^{2^5}+1$ is not a perfect power. So $l\\leq 5$. We know $2^1+1$, $2^2+1$, $2^4+1$, $2^8+1$, $2^{16}+1$ are prime numbers, so we obtain $n=(2+1)...(2^{2^{l-1}}+1)$, $2\\leq l\\leq 5$ (since $n\\neq 3$ we exclude $l=1$). But in this case $\\varphi(n)=2^{2^l-1}>2^{l}$ -- contradiction.", "Solution_2": "Nice Myth! i didnt know this solution and it is so simple, and creative! Thanks for sharing it with us!", "Solution_3": "hello pascual. this problem appeared in my tst for nationals in mexico. i gave a partial solution my general idea was the same as myth's. when i saw the full solution i was under the impression that there couldnt be a diferent solution for this problem, i cant imagine a diferent way to attack this problem. so could you please show me your other solution please? :blush: your the master here jaja :P", "Solution_4": "The official solution for the problem was the same as Myth's one", "Solution_5": "Ok, I will post it this afternoon!\r\n\r\nLOL BTW Myth is the Master here i think! :D even knowing i would like to be!!!! jaja", "Solution_6": "I am glad they used my solution for their olympiad :P", "Solution_7": "Lemma 1: \r\n\r\nlet $p$ be an odd prime, such that it divides $a-1$, then $p^b|a^m-1$ if and only if $p^b|m(a-1)$.\r\n\r\nThis lemma is in the theorem section, and it appears as \"Hensels Lemma\", but i dont think it is its name.\r\n\r\nLet $n=p_1^{t_1}p_2^{t_2}...p_k^{t_k}$, then $p_i^{r_i}|2^{\\phi(n)}-1$ if and only if it divides $\\phi(n)(2^{p_i-1}-1)$ by lemma 1. so we get, if we suppose that all prime factors of $2^{\\phi(n)}-1$ are factors of $n$, their max product is at most:\r\n\r\n$(2^{p_1-1}-1)(2^{p_2-1}-1)...(2^{p_k-1}-1)\\phi(n)$\r\n\r\nso we will be done if we showed $(2^{p_1-1}-1)(2^{p_2-1}-1)...(2^{p_k-1}-1)\\phi(n)<2^{\\phi(n)}-1$ \r\n\r\nthis can be showed by induction on the number of prime factors, using that $(2^{p-1}-1)(p-1)p^{t-1}<2^{(p-1)p^{t-1}}-1$ and also induction on $t_1+t_2...+t_k$ will do the job.", "Solution_8": "sorry but i dont understand your aplication of the lemma in the problem. note that that im not saying its wrong im just saying i dont understand it. :blush: could you please explain just that part please.", "Solution_9": "Of Course you used their solution.", "Solution_10": "[quote=\"Omid Hatami\"]Of Course you used their solution.[/quote]\r\nWhat do you mean :(", "Solution_11": "hmm a variant came up in our selection test! for what n does $2^{\\phi(n)}-1$ divide $n^n$ - basically the same idea." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For any positive real numbers $ a,b$ and $ c$,\r\n\\[ \\sum_{cycl}\\frac{a}{a+b}\\le\\frac{3}{2}\\left( 1+K\\cdot\\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{(a+b+c)^{2}}\\right) \\le \\sum_{cycl}\\frac{a}{b+c}\\]\r\nwhere $ K=1/2$ for the left-hand side and $ K=21/32$ for the right-hand side.", "Solution_1": "we have \r\n\r\n(1) sum(a/(a+b))<=3/2*[1+sqrt(3)/9*sum((a-b)^2/(a+b+c)^2))];\r\n\r\n(2) 3/2*[1+2/3*sum((a-b)^2/(a+b+c)^2)]<=sum(a/(b+c)).\r\n\r\nSee here:\r\n\r\nhttp://www.irgoc.org/bbs/dispbbs.asp?boardid=19&id=2614&star=1#2614", "Solution_2": "[quote=\"fjwxcsl\"]we have \n\n(1) sum(a/(a+b))<=3/2*[1+sqrt(3)/9*sum((a-b)^2/(a+b+c)^2))];\n\n(2) 3/2*[1+2/3*sum((a-b)^2/(a+b+c)^2)]<=sum(a/(b+c)).\n\nSee here:\n\nhttp://www.irgoc.org/bbs/dispbbs.asp?boardid=19&id=2614&star=1#2614[/quote]\r\n\r\nYeah, the inequalities can be sharpened. But the original constants admit a simple pen-and-paper solutions.\r\n\r\n@fjwxcsl: egaugnal ruoy gnidnatsrednu srehto gnitcepxe tonnac" } { "Tag": [ "geometry", "angle bisector", "geometry unsolved" ], "Problem": "The points $K$ and $L$ lie on the sides $AB$ and $AC$, respectively, of the triangle $ABC$ \r\nsuch that $BK = CL$. Let $P$ be the point of intersection of the segments $BL$ and $CK$. \r\n$M$ be an inner point of the segment $AC$ such that the line $MP$ is the parallel to the \r\nbisector of the $\\angle BAC$. Prove that $CM = AB$", "Solution_1": "We need to prove $AK=ML$.Let $BK=a,KA=b,AL=c,LC=a$ by Menelaus: $\\frac{CL}{CA}\\cdot \\frac{AK}{KB}\\cdot \\frac{BP}{PL}=1$.Therefore, $\\frac{BP}{PL}=\\frac{a+c}{b}$.So, we can write $BP=(a+c).k,PL=b.k$.Let $X$ be intersection of angle bisector of $A$ and line $BL$.According to the interior angle bisector theorem: $\\frac{BX}{XL}= \\frac{a+b}{c}$.Hence, $BX=(a+b).k,XL=c.k$.Since $\\frac{AL}{XL}= \\frac{ML}{PL}$, we find that: $ML=b$.\r\n\r\n(solution: Fatih EG\u0130)", "Solution_2": "[quote=\"Yosh...\"][color=darkred]The points $K$ and $L$ lie on the sides $AB$ and $AC,$ respectively, of the triangle $ABC$ such that $BK = CL.$ Let $P$ be the point of intersection of the segments $BL$ and $CK,$ and let $M$ be a point lie on line $AC$ such that the line $MP$ is the parallel to the angle bisector of the $\\angle BAC.$ \nProve that $CM = AB$ [size=150][u]Nice problem![/u][/size] [/color][/quote]\r\n[color=darkblue][b]Proof.[/b] Construct the point $D$ so that $ABDC$ is a parallelogram. Denote $S=BD\\cap CK$\nWe have: $\\frac{PS}{PC}=\\frac{BS}{CL}=\\frac{BS}{BK}=\\frac{DS}{DC}\\implies DP$ is angle bisector of $\\angle BDC\\implies M\\in DP$\nSo $\\widehat{CMD}=\\widehat{BDM}=\\widehat{CDM}\\implies CM=CD=AB$[/color]", "Solution_3": "[color=darkred][b]Remark 1.[/b] Let $ABC$ be a triangle and let $M\\in AB$, $N\\in AC$ be two points so that the sideline $BC$ don't separate these points and $MB=NC$\nThen the geometrical locus of the middlepoint of the segment $[MN]$ is a parallel line to the bisector of the angle $\\widehat{BAC}$.\n\n[b]Remark 2.[/b] Let $ABC$ be a triangle and let $M\\in AB$, $N\\in AC$ be two points so that the sideline $BC$ separates these points and $MB=NC$.\nThen the geometrical locus of the middlepoint of the segment $[MN]$ is a perpendicular line to the bisector of the angle $\\widehat{BAC}$.[/color]", "Solution_4": "See here for some other solutions: http://www.mathlinks.ro/Forum/viewtopic.php?p=415147#415147", "Solution_5": "[quote=\"scarface\"]So, we can write $BP=(a+c).k,PL=b.k$.\nHence, $BX=(a+b).k,XL=c.k$[/quote]\r\n\r\nWhy we can conclude that \"k\" is same?\r\n\r\nThx..", "Solution_6": "$\\frac{BP}{PL}=\\frac{a+c}{b}\\Longrightarrow \\frac{BP}{a+c}=\\frac{PL}{b}=k$.Here, $k$ is constant of the proportion.So we can write, $BP=(a+c).k, PL=b.k$Then, $BL=BP+PL=(a+b+c).k...(*)$ According to the interior angle bisector theorem: $\\frac{BX}{a+b}=\\frac{XL}{c}=t$.Here, $t$ is constant of the 2nd proportion.Hence, $BX=(a+b).t, XK=c.t$ and $BL=BX+XL=(a+b+c).t...(**)$. from $(*)$ and $(**)$, we yields $k=t$." } { "Tag": [], "Problem": "Let's say America didn't go into Iraq and didn't go after Saddam Hussein. How do you think everything would be different?", "Solution_1": "I don't really see the value of a hypothetical like this. It adds nothing to any consideration of what to do about the world [b]as it is[/b], and is just another way to play the blame game. Stop being childish.", "Solution_2": "I don't think anyone really cared overly much about Iraq before the war, so it would have been the same as life before 2000\r\n\r\nOh, and republicans would have congress.", "Solution_3": "Terrorists based in Iraq would have killed everyone with WMD's.\r\n\r\nRealistially?\r\nWe wouldn't have soldiers in Iraq.\r\nAddam Hussein would be in power.\r\nEtcetera." } { "Tag": [ "function", "limit", "number theory open", "number theory" ], "Problem": "maybe this is a well-known open problem, maybe it is easier than i think: given a set A of finitely many primes. prove that there are only finitely many $n$ such that any prime dividing $n$ or $n+1$ belongs to A.\r\n\r\ni could hardly make any progress on it. i would especially like to see explicit bounds on the number or the size of the $n$ satisfying that property.\r\nthe original thing i was thinking of(which immediately implies the problem above): does there exist a function $f: \\mathbb{N}\\to \\mathbb{N}$ such that $f(xy)=f(x)f(y)$ and $\\lim_{n\\to \\infty} \\frac{f(n+1)-f(n)}{n}=0$.\r\n\r\nthat one now does imply the 3n+1-conjecture for sufficiently large numbers.\r\n\r\nPeter", "Solution_1": "[quote=\"Peter Scholze\"]the original thing i was thinking of(which immediately implies the problem above): does there exist a function $f: \\mathbb{N}\\to \\mathbb{N}$ such that $f(xy)=f(x)f(y)$ and $\\lim_{n\\to \\infty} \\frac{f(n+1)-f(n)}{n}=0$.\n[/quote]\r\n\r\ndon't $f(x)=1$, $f(x)=0$ and $f(x)=x$ all fullfil these requirements? :?\r\n\r\nAs for the other problem, n and n+1 cannot have any prime in common in their factorization, so basicly what you need to do is checking if $p_1^{q_1}\\cdots p_n^{q_n}$ can be written in the form of $n\\cdot(n+1)$ in an infinite amount of ways, providing you may choose all $q_i$.\r\n\r\nSo I think we can rewrite the problem in a nicer wording: for given $n$, $\\exists m: m\\cdot(m+1) = k\\cdot n\\cdot(n+1)$ with either $k|n$ or $k|n+1$?\r\n\r\nI didn't solve it yet though, but it looks much too nice to be an unknown open problem :cool:", "Solution_2": "sorry for the second thing, i wanted to forbid \"trivial\" solutions such as the ones you gave(let's take $f(2)>f(3)$ as an extra condition).\r\nbtw, can the moderators delete one of the two topics? i posted it twice...sorry!\r\n\r\nPeter", "Solution_3": "[quote=\"Peter VDD\"]for given $ n $, $ \\exists m: m\\cdot(m+1) = k\\cdot n\\cdot(n+1) $ with either $ k|n $ or $ k|n+1 $?[\\quote]\r\n\r\nthis is not quite true...the condition for $k$ should be that all its prime divisors divide either $n$ or $n+1$.\r\n\r\nPeter", "Solution_4": "[quote=\"Peter Scholze\"]sorry for the second thing, i wanted to forbid \"trivial\" solutions such as the ones you gave(let's take $f(2)>f(3)$ as an extra condition).[/quote]\r\n\r\nAh, I think I start getting the relation you're investigating between the 2 problems...\r\n\r\nI thought for reals we have that $f(x)\\cdot f(y)=f(x\\cdot y)$ implies $f(x)=x^k$, but that's $R\\rightarrow R$.\r\nSo if we consider only the lattice points of the function created by a non-negative integer value...\r\nthat means we must have $f(x)=x^k\\ (k\\in\\mathbb{Z})$ right? I'm not really familiar with this though. :(\r\n\r\nso if $f(2)>f(3)$ that is a contradiction... since $k<0$ and $f(x>1) \\not\\in \\mathbb{N}$ for $k\\in\\left]-\\infty;0\\right[$.", "Solution_5": "hmm...maybe you should not trust on some results you heard of but on your own mind. and that one tells you that either $f(x)=0$ or $f(1)=1$ (very easy), and if $f(1)=1$, we may arbritarily choose $f(p)$ for all primes $p$. now, we have $f(\\prod p_i^{a_i})=\\prod f(p_i)^{a_i}$. all these functions give solutions.\r\nit is now some work to find how the second problem implies the first. i didn't check everything, but i think if there are infinitely many $n$ with that property, then $f$ has to be the identity at all primes of $A$. thus, let's add the condition in the second problem that $f$ is not the identity for all primes in $A$. so i'll state it again to make the whole thing clear:\r\ngiven an arbritary finite set of primes $A$. then there exists a nonconstant function $f:\\mathbb{N} \\to \\mathbb{N}$ with the following properties:\r\n1) $f(xy)=f(x)f(y)$\r\n2) $\\lim_{n\\to \\infty} \\frac{f(n+1)-f(n)}{n}=0$\r\n3) $\\exists p\\in A:\\; f(p)\\neq p$.\r\n\r\nPeter", "Solution_6": "A problem generalizing your first problem was posted some time ago. If my memory serves right, I think the topic was started by sam-n and it stated that given $k$ and a finite set of primes, $A$, there are only finitely many solutions to the equation $x-y=k$ s.t. the prime factors of $x,y$ are in $A$. It was unsolved for quite some time, until Harazi posted a solution which I didn't understand (and made no efforts to understand, because, even though it looked cool, it dew on too much advanced stuff :)). However, nobody managed to solve it in a more elementary way, so it might just be that this is the way to solve it. \r\n\r\nI'll look for the topic.", "Solution_7": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4724&highlight=xy+k]Here[/url] it is. I read it, and it's not that tough to understand, but really tough to find :).", "Solution_8": "uhm...yeah. thanks grobber, but i came to the same point as harazi and knew there was such a result in that direction(couldn't remember the wording, but anyway). thus i was sure that the problem is true. however, i want a short nice elementary solution(or at least only using results i really understand).\r\n\r\nPeter", "Solution_9": "btw, sam-n's generalization is implied by my second problem as well...the more i look at it, it has more nice consequences. i'll try to find a proof using that theorem harazi used.\r\n\r\nPeter" } { "Tag": [ "calculus", "linear algebra", "real analysis" ], "Problem": "I realize that the only reason I'm good at/ like math below the precalc level is because of math competitions and fun problems. I don't particularly enjoy college math, though, and I was wondering if anyone know of any good problem solving books that address the following subjects:\r\n\r\nCalculus\r\nMultivariable Calculus\r\nDifferential Equations\r\nLinear Algebra\r\nReal Analysis\r\n\r\nThanks. I'm lookig for books other than past Putnam's, though.", "Solution_1": "problem solving through problem,L.C.Larson....very nice and suitable book for anybody.and also it has many other topics.", "Solution_2": "Intermediate Real Analysis published by Springer has lots of problems and topics to learn and enjoy. It begins with very simple, yet delightful problems, and it builds up gradually.\r\n\r\nMasoud Zargar" } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Prove the generalization of Rolle's theorem", "Solution_1": "Do you mean the Mean Value Theorem? or the relaxed constraint where the function $ f$ only needs to be continuous on $ [a,b]$, $ f(a)=f(b)$ and for all $ x\\in (a,b)$ we have $ \\lim_{h\\to 0}\\frac{f(x+h)-f(x)}{h}\\in [-\\infty, \\infty]$?", "Solution_2": "[quote=\"Orif\"]Prove the generalization of Rolle's theorem[/quote]\r\n\r\nLet $ f: [a,b]\\mapsto \\mathbb{R}$ be continous on $ [a,b]$ and differentiable on $ (a,b)$ define the function,\r\n$ g(x) = f(x)-m(x-a)-f(a)$ where $ m=\\frac{f(b)-f(a)}{b-a}$.\r\nThen show that $ g(a)=g(b)$ and conditions for Rolle's theorem are satisfied.\r\n\r\nSo, $ g'(c) = 0$ for some $ c\\in (a,b)$.\r\nMeaning,\r\n$ f'(c) = \\frac{f(b)-f(a)}{b-a}$." } { "Tag": [ "calculus", "derivative", "LaTeX", "limit", "logarithms", "function", "calculus computations" ], "Problem": "can someone help me with this review sheet. its due on monday, and i have been sick for the last week and am completely lost. thank you.\r\n\r\nFind the limitsof the following:\r\n\r\nlim(x->2) for: ((x^2)-4), x<2\r\n(x-2), x>2\r\n\r\nlim(x->3) for: (x)/(3-x)\r\n\r\nlim(x->9) (x-9)/((sqrt of x)-3)\r\n\r\nlim(x->2) for ((x^3)-8)/((x^2)-4)\r\n\r\nlim(x-> -3) for thirdroot(10+3x)\r\n\r\nlim(x->0) for (ln(x+2)^X)/(x))\r\n\r\n\r\nUsing the long method, find the derivative of f(x)=(x^3-5x^2+7)", "Solution_1": "$\\LaTeX$ed:\r\n\r\n[quote=\"kataklysm08\"]Find the limits of the following:\n\n$\\lim_{x \\rightarrow 2}x^{2}-4$\n$\\lim_{x \\rightarrow 3}\\frac{x}{3-x}$\n$\\lim_{x \\rightarrow 9}\\frac{x-9}{\\sqrt{x}-3}$\n$\\lim_{x \\rightarrow 2}\\frac{x^{3}-8}{x^{2}-4}$\n$\\lim_{x \\rightarrow 3}\\sqrt[3]{10+3x}$\n$\\lim_{x \\rightarrow 0}\\frac{\\ln((x^{2})^{x}}{x}$\n\nUsing the long method, find the derivative of $f(x)=x^{3}-5x^{2}+7$[/quote]\r\n\r\nPerhaps this should be moved to a different forum? (I'm not a moderator; I don't know)", "Solution_2": "These should be in the Calculus Computation Tutorials forum.\r\n\r\nBasically, for most of these limits, you can simply substitute the value that x approaches into the function.\r\nHowever, if you do this on some, you'll end up dividing by zero.\r\nYou'll need to use algebra to rewrite the equations in such a way that division by zero doesn't occur.\r\n\r\nThe last limit is a bit tricker than the rest.\r\n\r\n\r\nFor the derivatives one, it's simply $f'(x) = 3x^{2}-10x$ by the power rule.", "Solution_3": "For the last one I think they want the difference quotient method, of if that is what they mean the \"long way\".", "Solution_4": "I think this should be in the calculus forum. For the derivative, I think the \"long method\" you are referring to is:\r\n$\\lim_{h\\rightarrow 0}\\frac{f(x+h)-f(x)}{h}= \\lim_{h\\rightarrow 0}\\frac{(x+h)^{3}-5(x+h)^{2}+7-(x^{3}-5x^{2}+7)}{h}$. See what you can do from there.", "Solution_5": "Correction for the first problem in $\\LaTeX$.\r\n[quote]$\\lim_{x \\rightarrow 2}\\begin{cases}x^{2}-4 & x < 2\\\\x-2 & x > 2\\end{cases}$[/quote]\r\n\r\nBad $\\LaTeX$, how are piecewise coded? :maybe:\r\n\r\n[color=green][Edited by moderator; copy it to see what the code looks like.][/color]", "Solution_6": "1. $\\lim_{x \\rightarrow 2}x^{2}-4$\r\n2. $\\lim_{x \\rightarrow 3}\\frac{x}{3-x}$\r\n3. $\\lim_{x \\rightarrow 9}\\frac{x-9}{\\sqrt{x}-3}$\r\n4. $\\lim_{x \\rightarrow 2}\\frac{x^{3}-8}{x^{2}-4}$\r\n5. $\\lim_{x \\rightarrow 3}\\sqrt[3]{10+3x}$\r\n6. $\\lim_{x \\rightarrow 0}\\frac{\\ln((x^{2})^{x}}{x}$\r\n\r\n\r\nSubstituting values in most of the limits require (sometimes) division by 0, which is not allowed in calculus. However, use the Conjugate Theorem to work around some of the limits (like number 3). If all else fails, use L'Hospital's Rule (where you calculate the derivitives of the limit in order to solve the limit itself).", "Solution_7": "[quote=\"7h3.D3m0n.117\"]1. $\\lim_{x \\rightarrow 2}x^{2}-4$\n2. $\\lim_{x \\rightarrow 3}\\frac{x}{3-x}$\n3. $\\lim_{x \\rightarrow 9}\\frac{x-9}{\\sqrt{x}-3}$\n4. $\\lim_{x \\rightarrow 2}\\frac{x^{3}-8}{x^{2}-4}$\n5. $\\lim_{x \\rightarrow 3}\\sqrt[3]{10+3x}$\n6. $\\lim_{x \\rightarrow 0}\\frac{\\ln((x^{2})^{x}}{x}$\n\n\nSubstituting values in most of the limits require (sometimes) division by 0, which is not allowed in calculus. However, use the Conjugate Theorem to work around some of the limits (like number 3). If all else fails, use L'Hospital's Rule (where you calculate the derivitives of the limit in order to solve the limit itself).[/quote]\r\n\r\nso 1. would be 0?", "Solution_8": "I'll do number 6.\r\nThe numerator is $\\ln((x^{2})^{x})$ which can be rewritten as $x \\ln x^{2}$\r\nWe get $\\frac{x \\ln x^{2}}{x}$, so the quotient simplifies to $\\ln(x^{2})$, which has a limit at $-\\infty$ as $x$ goes to $0$", "Solution_9": "For number one :maybe: \r\n[quote]lim(x->2) for: ((x^2)-4), x<2 \n(x-2), x>2 [/quote]\r\nAm I reading it wrong or it is a piecewise function?\r\nThen making it not have a limit.", "Solution_10": "Technically, a piecewise function can have a limit if the limit from the right is equal to the limit from the left.\r\n\r\nSo:\r\n$\\lim_{x \\rightarrow 2}x^{2}-4 = 0$\r\nand $\\lim_{x \\rightarrow 2}x-2 = 0$\r\n\r\nSo the limit at $x = 2$ exists.", "Solution_11": "i called my friend. she said that for the ones that lead to division by zero, i need to average the terms before and after the given term. is this correct?", "Solution_12": ":rotfl: \r\nAfter all that, I still read the problem wrong. Though I read it as a piecewise correctly.\r\n\r\nFor the ones that you get zero, either manipulate the function or put in very close numbers to the limit to find it out.", "Solution_13": "Learn something known as the Squeeze thm. too.\r\nThat's a helpful one.\r\nI liken it to making a limit sandwich.\r\n\r\nPick two functions that f(x) and g(x) such that f(x) > original function for all x, and g(x) < original function for all x.\r\n\r\nThen find the limits of those functions where you want to, and it is guaranteed that if the limit of the original function exists, it lies between f(x) and g(x)", "Solution_14": "4. $\\lim_{x \\rightarrow 2}\\frac{x^{3}-8}{x^{2}-4}$\r\n\r\nso if i plugged in 1.95 as my lower limit and 2.05 as my upper limit, with a .01 pitch in my calculator, i could then take the values for 1.99 and 2.01, average them, and get the right answer?", "Solution_15": "Moderators, this forum should be locked.\r\n\r\nJapanese Communities Modeartor\r\n\r\nkunny", "Solution_16": "why should this be locked?", "Solution_17": "This is rule! The problems of differentiation, derivative, limit etc are not allowed to post High School Forum, even in Intermidate or Olympiad forum.\r\nIn Japan, Japanese students usually study those problem at the age of 17~18, except for some junior high school students at the age of 14, 15.\r\nAnyway your problem should be posted in [color=red]Calculus Computation and Tutorials[/color] :) \r\n\r\nkunny", "Solution_18": "[quote=\"kataklysm08\"]4. $\\lim_{x \\rightarrow 2}\\frac{x^{3}-8}{x^{2}-4}$\n\nso if i plugged in 1.95 as my lower limit and 2.05 as my upper limit, with a .01 pitch in my calculator, i could then take the values for 1.99 and 2.01, average them, and get the right answer?[/quote]\r\n\r\nYes, you can do that, or you can use L'Hospital's Rule:\r\n\r\n$\\lim_{x \\rightarrow 2}\\frac{x^{3}-8}{x^{2}-4}$\r\n\r\n$\\lim_{x \\rightarrow 2}\\frac{3x^{2}}{2x}$\r\n\r\nsubstitution would lead to $\\frac{12}{4}= 3$", "Solution_19": "Should be locked! :mad: \r\n\r\nYou should read this!\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=26438[/url]", "Solution_20": "[quote=\"kunny\"]Should be locked! :mad: \n\nYou should read this!\n\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=26438[/url][/quote]\r\n\r\nWell that's for the Intermediate Forum. I'm sure a mod can move this to a Calculus forum.", "Solution_21": "[color=blue]To Japanese high school students and junior high school students![/color]\r\n\r\n[color=red]You can't use L'Hopital's theorem in Japanese university entrance examination. [/color]\r\n\r\nJapanese Communities Moderator\r\n\r\nkunny", "Solution_22": "why do you care so much? does helping a junior in high school precalculus offend you?\r\n\r\nguess what buddy, im not taking the japanese university entrance exam, im just trying to learn efffing math. this is for AMERICAN learning institutions, so i dont think your opinion is required.", "Solution_23": "How did this ever end up in the complex analysis forum?", "Solution_24": "Sorry for not noticing this earlier. It is an appropriate - if perhaps not all that interesting - topic for the \"Calculus Computations and Tutorials\" forum, and it will be moved." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "ARML", "AIME", "quadratics", "inequalities", "\\/closed" ], "Problem": "If I had a choice between the AMC/AIME algebra class or the AMC-12 problem series (during the summer), which would you suggest?", "Solution_1": "Having had you in class I would suggest that you are better off taking the algebra class. I personally feel that students who can already make it to the USAMO will get way more out of our subject courses.\r\n\r\nSome of the algebra course will be review for you. Take a look at the Post Test. If you can solve all the problems, skip the class.\r\n\r\nUnfortunately the Post Test does not test for everything covered in the class (it would be hard to do that). The class covers dozens of problems on the ARML or AIME level and a fair amount (10% of the class) of Olympiad level algebra problems (after building steadily in each lesson).\r\n\r\nI am currently writing script for difference equations that I will be adding to the class for this coming session. The course will also cover uses of the quadratic equation that lead to many Olympiad ideas and substitution methods that help solidify problem solving right up to Olympiad level. There is also a full day on functional equations that gets up the level of Olympiad problems.\r\n\r\nThe Olympiad level problems are on the whole not as challenging as those in the Inequalities class you are taking. If I were you I would see the algebra class as having the potential to solidify ideas, close a gap or two in knowledge, provide practice for the AMC12 and AIME, and provide a couple of dozen practice Olympiad problems." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "show that:\r\n\r\n$\\int_{0}^{\\frac{\\pi}{2}}\\;\\cos^{99}{(x)}\\;\\cdot\\;\\cos{(99x)}\\;\\;\\;dx\\;=\\;\\frac{\\pi}{2^{100}}$", "Solution_1": "First, we show that \r\n\r\n$\\int_{0}^{\\frac{\\pi}{2}}\\cos^{n}x \\cos (n+2k)x \\, dx = 0$\r\n\r\nfor all $n = 0, 1, \\cdots$ and $k = 1, 2, \\cdots$. This can be achieved by mathematical induction.\r\n\r\n(1) $n = 0$, then $\\int_{0}^{\\frac{\\pi}{2}}\\cos^{n}x \\cos (n+2k)x \\, dx = \\int_{0}^{\\frac{\\pi}{2}}\\cos 2kx \\, dx = 0$\r\n(2) Suppose it holds for $n$. Then $\\int_{0}^{\\frac{\\pi}{2}}\\cos^{n+1}x \\cos (n+1+2k)x \\, dx$ $= \\int_{0}^{\\frac{\\pi}{2}}\\frac{1}{2}\\cos^{n}x \\{ \\cos (n+2+2k)x+\\cos (n+2k)x \\}\\, dx = 0$,\r\nso it also holds for $n+1$.\r\n\r\nNow we apply the same method we used in (2) to the original integral. Then\r\n\r\n$\\int_{0}^{\\frac{\\pi}{2}}\\cos^{n+1}x \\cos (n+1)x \\, dx$ $= \\int_{0}^{\\frac{\\pi}{2}}\\frac{1}{2}\\cos^{n}x \\{ \\cos (n+2)x+\\cos nx \\}\\, dx$ $= \\frac{1}{2}\\int_{0}^{\\frac{\\pi}{2}}\\cos^{n}x \\cos nx \\, dx$.\r\n\r\nSo we have\r\n\r\n$\\int_{0}^{\\frac{\\pi}{2}}\\cos^{n}x \\cos nx \\, dx = \\frac{1}{2^{n}}\\int_{0}^{\\frac{\\pi}{2}}\\, dx = \\frac{\\pi}{2^{n+1}}$.", "Solution_2": "hey, nice ! :coolspeak:" } { "Tag": [ "function", "algebra", "domain", "calculus", "calculus computations" ], "Problem": "what's the faster way to identify if a function is bounded or not?", "Solution_1": "Let D be the domain of a function f(x). For f to be bounded it means you can find a positive number M such that $ \\forall x \\in D \\,\\, |f(x)| \\leq M$. Geometrically, this means you can sandwich the graph of f(x) between two horizontal lines. So perhaps looking at the graph is the faster method. Of course, some experience also plays a role." } { "Tag": [ "number theory", "number theory unsolved" ], "Problem": "Prove that $\\{n\\sqrt7\\}<1-\\frac{1}{6n}$ where $n$ is a natural number and $\\{x\\}$\r\nis fractional part function.\r\n[hide]The original question was $\\frac{1}{2n}<\\{n\\sqrt7\\}<1-\\frac{1}{6n}$.\nI have solved (only) the first part by using some number theory,so I posted the problem\nin this section.[/hide]", "Solution_1": "[quote=\"spider_boy\"]Prove that $\\{n\\sqrt7\\}<1-\\frac{1}{6n}$ where $n$ is a natural number and $\\{x\\}$\nis fractional part function.\n[hide]The original question was $\\frac{1}{2n}<\\{n\\sqrt7\\}<1-\\frac{1}{6n}$.\nI have solved (only) the first part by using some number theory,so I posted the problem\nin this section.[/hide][/quote]\r\n\r\n\r\nLet $(a+1)^2 > 7n^2 =a^2+t \\geq a^2 \\Rightarrow$\r\n$\\{\\sqrt{7}n\\}=\\sqrt{a^2+t}-a$\r\nSo we have to prove \r\n$\\sqrt{a^2+t}-a<1-\\frac{1}{6n} \\Leftrightarrow$\r\n$t<2a(1-\\frac{1}{6n})+(1-\\frac{1}{6n})^2$\r\nbut $\\frac{2a}{6n} <1 \\mbox{ as } a^2< 9n^2) \\Rightarrow$\r\nif $t \\leq 2a-1$ then we prove the statement.\r\n\r\nIn the another case $t=2a \\Rightarrow$ we have to prove\r\n$a^2+2a R that has a dense set of critical values.", "Solution_1": "Just take any enumeration $ q_n$ of the rationals and create any smooth function $ f(x)$ such that $ f(n) \\equal{} q_n$, $ f'(n) \\equal{} 0$.", "Solution_2": "To elaborate on that, consider something like\r\n\\[ \\phi(x) \\equal{} \\begin{cases} e^{ \\minus{} \\frac {x^2}{1 \\minus{} x^2}}\\left(1 \\minus{} e^{ \\minus{} \\frac {1}{x^2}}\\right) & 0 < |x| < 1 \\\\\r\n1 & x \\equal{} 0 \\\\\r\n0 & \\text{otherwise}\\end{cases}\\]\r\nIt is $ C^\\infty$, has compact support on $ [ \\minus{} 1,1]$, and has a critical value at $ x\\equal{}0$. Then $ f(x) \\equal{} \\sum_{n \\equal{} 1}^\\infty q_n \\phi (x \\minus{} n)$ is a function of the type suggested by fedja.", "Solution_3": "Thanks much.", "Solution_4": "Yeah. Thank you so much for such a detailed answer." } { "Tag": [], "Problem": "The 2-digit numeral $AB$, with $A \\not= 0$, represents a prime if the numeration base is 8, 10, or 12. Find $AB$.", "Solution_1": "[hide]Well, 8a + b = Prime, 10a + b = Prime, and 12a + b = Prime. So, if a = 1, the primes are 2 apart, and for a = 2, they are 4 apart, so with that pattern, I listed some primes with difference 2, 4, 6, 8... to try. 2 and 4 are impossible right away, so try 6. 8 * 3 = 24, and 24 + 7 = 31, which is the first term in three primes 6 apart. They are 31, 37, and 43. So, a = 3, and b = 7, for 37.[/hide]" } { "Tag": [], "Problem": "A 48-foot-wide building has a roof that rises 4 feet for each 12-foot horizontal change. If the roof has a 2-foot overhang, what is the distance from the peak, P, to the edge of the overhang, O?\n\n[asy]draw((0,0)--(48,0));\ndraw((-6,-3)--(24,12)--(54,-3));\nlabel(\"P\", (24,12), N);\nlabel(\"O\", (-6,-3), S);\nlabel(\"48\\ ft\", (24,0), S);\nlabel(\"2\\ ft\", (-3,-1.5), NW);[/asy]", "Solution_1": "From the peak to the junction is 24 horiz = 8 vert = 8rt(10) total distance.\r\n\r\nAnswer: 2+8rt(10)" } { "Tag": [ "function", "floor function", "algebra unsolved", "algebra" ], "Problem": "find the number of real solutions\r\n\r\n\r\na=[a/2]+[a/3]+[a/5]\r\n\r\nwhere [.] implies greatest inter function", "Solution_1": "Suppose $ \\{.\\}$ is the fractional part function.\r\nIn other words, for all real $ x$, $ \\{x\\}\\equal{}x\\minus{}\\lfloor x\\rfloor$\r\nThe given equation is equivalent to:\r\n$ a\\equal{}\\frac{a}{2}\\minus{}\\{\\frac{a}{2}\\}\\plus{}\\frac{a}{3}\\minus{}\\{\\frac{a}{3}\\}\\plus{}\\frac{a}{5}\\minus{}\\{\\frac{a}{5}\\}$\r\n$ \\iff a\\equal{}\\frac{a}{2}\\plus{}\\frac{a}{3}\\plus{}\\frac{a}{5}\\minus{}\\{\\frac{a}{2}\\}\\minus{}\\{\\frac{a}{3}\\}\\minus{}\\{\\frac{a}{5}\\}$\r\n$ \\iff a\\equal{}\\frac{31a}{30}\\minus{}\\{\\frac{a}{2}\\}\\minus{}\\{\\frac{a}{3}\\}\\minus{}\\{\\frac{a}{5}\\}$\r\n$ \\iff \\frac{a}{30}\\equal{}\\{\\frac{a}{2}\\}\\plus{}\\{\\frac{a}{3}\\}\\plus{}\\{\\frac{a}{5}\\}$\r\n$ \\iff a\\equal{}30(\\{\\frac{a}{2}\\}\\plus{}\\{\\frac{a}{3}\\}\\plus{}\\{\\frac{a}{5}\\})$\r\n\r\nWe know that for all integers $ x$, $ n$:\r\n$ n\\{\\frac{x}{n}\\}$ is the remainder when $ x$ is divided by $ n$\r\nSuppose $ i,j,k$ are the remainders when $ a$ is divided by $ 2,3,5$ respectively.\r\nSo the equation becomes:\r\n\r\n$ a\\equal{}15i\\plus{}10j\\plus{}6k$\r\n\r\nNote that $ 15i\\plus{}10j\\plus{}6k\\equiv i\\mod 2$\r\nAnd $ 15i\\plus{}10j\\plus{}6k\\equiv j \\mod 3$\r\nAnd $ 15i\\plus{}10j\\plus{}6k\\equiv k\\mod 5$\r\n\r\nSo for all triples $ (i,j,k)$ fix $ a$ at $ 15i\\plus{}10j\\plus{}6k$ which gives a solution.\r\n\r\nSince there are $ 2,3,5$ possible values for $ i,j,k$ respectively, there are $ 2\\cdot 3\\cdot 5$ triples $ (i,j,k)$\r\nHence there are $ \\boxed{30\\text{ solutions}}$" } { "Tag": [ "trigonometry" ], "Problem": "Find all solutions to the pair of equations\r\n\r\n\r\n\r\nSinX/SinY =sqrt2 , TanX/TanY= sqrt3", "Solution_1": "$\\frac{SinX}{SinY}=\\sqrt{2} , \\frac{TanX}{TanY} =\\sqrt{3}$", "Solution_2": "You can make that look even nicer with the \\sin and \\tan commands...\r\n\r\n$\\frac{\\sin{X}}{\\sin{Y}} = \\sqrt{2}$ and $\\frac{\\tan{X}}{\\tan{Y}} = \\sqrt{3}$\r\n\r\n[hide]\n$\\frac{\\tan{X}}{\\tan{Y}} = \\frac{\\sin{X}}{\\sin{Y}}\\cdot\\frac{\\cos{Y}}{\\cos{X}} = \\sqrt{2}\\cdot\\frac{\\cos{Y}}{\\cos{X}} = \\sqrt{3}$\n\nSo $\\cos{X} = \\sqrt{\\frac{2}{3}}\\cos{Y}$ and $\\sin{X} = \\sqrt{2}\\sin{Y}$.\n\nBut $1 = \\cos^2{X}+\\sin^2{X} = \\frac{2}{3}\\cos^2{Y}+2\\sin^2{Y} = \\frac{2}{3}+\\frac{4}{3}\\sin^2{Y}$ \n\nso then\n\n$\\sin^2{Y} = \\frac{1}{4} \\Rightarrow \\sin{Y} = \\pm \\frac{1}{2} \\Rightarrow Y=\\frac{\\pi}{6}, \\frac{5\\pi}{6}, \\frac{7\\pi}{6}, \\frac{11\\pi}{6}$\n\nand we get corresponding $X$ values of\n\n$X = \\frac{\\pi}{4}, \\frac{3\\pi}{4}, \\frac{5\\pi}{4}, \\frac{7\\pi}{4}$.\n\nI'm not sure, but that seemed like a very round-a-bout way to get at an obvious answer.[/hide]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "The National Marriage Council wishes to invite $n$ couples to form 17 discussion groups under the following conditions:\r\n\r\n(1) All members of a group must be of the same sex; i.e. they are either all male or all female.\r\n\r\n(2) The difference in the size of any two groups is 0 or 1.\r\n\r\n(3) All groups have at least 1 member.\r\n\r\n(4) Each person must belong to one and only one group.\r\n\r\nFind all values of $n$, $n \\leq 1996$, for which this is possible. Justify your answer.", "Solution_1": "An easy one there are 36 solutions.\r\n9,10,11,12,13,14,15,16,\r\n18,19,20,21,22,23,24,\r\n27,28,29,30,31,32,\r\n36,37,38,39,40,\r\n45,46,47,48,\r\n54,55,56,\r\n63,64,\r\n72.", "Solution_2": "Sorry to bring this post back up but the current solution seems very vague (in fact, it's not quite a solution at all).\n\nAlso, I'm interested in the problem and I want to bring this post back to life so that it can have another chance of someone finding and posting a solution. :)", "Solution_3": "It probably hasn't been solved or gotten a detailed solution because it's rather ugly casework without any neat insights to be had.\n\nSuppose there are $k$ groups of males and $17 - k$ groups of females; WLOG $k \\geq 9$. Also suppose that all the groups have $m$ or $m+1$ members. Then $mk \\leq (m+1)(17-k)$, or $(2m+1)(2k-17) \\leq 17$. Since $m \\geq 1$, we have $2k - 17 < 6$, so $9 \\leq k \\leq 11$.\n\n$k = 11$? Then only $m = 1$ works. We have at least 11 males and at most 12 females. So $n = 11, 12$ work.\n\n$k = 10$? Then only $m = 1$ or $m = 2$ works. For $m = 1$, we have at least 10 males and at most 14 females. So $n = 10$ through $14$ work. For $m = 2$, we have at least 20 males and at most 21 females. So $n = 20, 21$ work.\n\n$k = 9$? Then $m$ can be as high as 8. For a given $m$, we have at least $9m$ males and at most $8(m+1)$ females. This gives the range of solutions seen in post #2 which I won't recopy here. It turns out that this set of solution subsumes all the previous cases, so that is the complete list of answers." } { "Tag": [ "integration", "calculus", "calculus computations" ], "Problem": "Show that $ \\left(\\alpha - \\frac {1}{\\alpha} - x\\right)(4 - 3x^2)$ has just one maximum and just one minimum, and that the difference between them is $ \\frac {4}{9}\\left(\\alpha + \\frac {1}{\\alpha}\\right)^3$.\r\n\r\nWhat is the least value of this difference for different values of $ \\alpha$?\r\n\r\n[hide=\"My Solution\"]\n\nLet $ f(x) = (c - x)(4 - 3x^2) = 3x^3 - 3cx^2 - 4x + 4c$, where $ c = \\alpha - \\frac {1}{\\alpha}$.\n\n$ f'(x) = 9x^2 - 6cx - 4$, $ D/4 = ( - 3c)^2 - 9\\cdot ( - 4) = 9(c^2 + 4) > 0$, thus we can write $ f'(x) = 9(x - \\alpha )(x - \\beta)\\ (\\alpha < \\beta)$.\n\nSince $ f(x)$ has a local maximum at $ x = \\alpha$ and a local minimum at $ x = \\beta$, \n\nThe difference of extrema : $ d = f(\\alpha) - f(\\beta) = \\int_{\\beta}^{\\alpha} f'(x)\\ dx$\n\n$ = - 9\\int_{\\alpha}^{\\beta} (x - \\alpha)(x - \\beta)\\ dx = \\frac {3}{2}(\\beta - \\alpha)^3$.\n\nNow $ f'(x) = 0\\Longleftrightarrow x = \\frac {c\\pm \\sqrt {c^2 + 4}}{3}$, yielding $ \\beta - \\alpha = \\frac {2}{3}\\sqrt {c^2 + 4}$.\n\n$ \\therefore d = \\frac {3}{2}\\left(\\frac {2}{3}\\sqrt {c^2 + 4}\\right)^3 = \\frac {4}{9}\\left(\\alpha + \\frac {1}{\\alpha}\\right)^3$. Q.E.D.\n\nThen $ d$ is minimal $ \\Longleftrightarrow \\beta - \\alpha$ is minimal $ %Error. \"Lonogleftrighatrrow\" is a bad command.\nc = 0$, yielding $ d_{min}=\\frac{32}{9}$.[/hide]", "Solution_1": "My Solution:\r\n\r\nLet $ f(x) = (c - x)(4 - 3x^2) = 3x^3 - 3cx^2 - 4x + 4c$, where $ c = \\alpha - \\frac {1}{\\alpha}$.\r\n\r\n$ f'(x) = 9x^2 - 6cx - 4$, $ D/4 = ( - 3c)^2 - 9\\cdot ( - 4) = 9(c^2 + 4) > 0$, thus we can write $ f'(x) = 9(x - \\alpha )(x - \\beta)\\ (\\alpha < \\beta)$.\r\n\r\nSince $ f(x)$ has a local maximum at $ x = \\alpha$ and a local minimum at $ x = \\beta$, \r\n\r\nThe difference of extrema : $ d = f(\\alpha) - f(\\beta) = \\int_{\\beta}^{\\alpha} f'(x)\\ dx$\r\n\r\n$ = - 9\\int_{\\alpha}^{\\beta} (x - \\alpha)(x - \\beta)\\ dx = \\frac {3}{2}(\\beta - \\alpha)^3$.\r\n\r\nNow $ f'(x) = 0\\Longleftrightarrow x = \\frac {c\\pm \\sqrt {c^2 + 4}}{3}$, yielding $ \\beta - \\alpha = \\frac {2}{3}\\sqrt {c^2 + 4}$.\r\n\r\n$ \\therefore d = \\frac {3}{2}\\left(\\frac {2}{3}\\sqrt {c^2 + 4}\\right)^3 = \\frac {4}{9}\\left(\\alpha + \\frac {1}{\\alpha}\\right)^3$. Q.E.D.\r\n\r\nThen $ d$ is minimal $ \\Longleftrightarrow \\beta - \\alpha$ is minimal $ \\Longleftrightarrow c = 0$, yielding $ d_{min} = \\frac {32}{9}$." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ x,y,z\\geq 0$ be such that $ x^2\\plus{}y^2\\plus{}z^2\\equal{}1$.\r\n\r\nProve that $ 1\\leq \\frac{z}{1\\plus{}xy}\\plus{}\\frac{y}{1\\plus{}xz}\\plus{}\\frac{z}{1\\plus{}xy}\\leq \\sqrt{2}$", "Solution_1": "[quote=\"outback\"]Let $ x,y,z\\geq 0$ be such that $ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 1$.\n\nProve that $ 1\\leq \\frac {z}{1 \\plus{} xy} \\plus{} \\frac {y}{1 \\plus{} xz} \\plus{} \\frac {z}{1 \\plus{} xy}\\leq \\sqrt {2}$[/quote]\r\nFix it please.", "Solution_2": "Min, here[url]http://www.mathlinks.ro/viewtopic.php?t=229665[/url]", "Solution_3": "[quote=\"outback\"]Let $ x,y,z\\geq 0$ be such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 1$.\n\nProve that $ 1\\leq \\frac {a}{1 \\plus{} bc} \\plus{} \\frac {b}{1 \\plus{} ca} \\plus{} \\frac {c}{1 \\plus{} ab}\\leq \\sqrt {2}$[/quote]\r\nFirstly, we prove that: \r\n$ (a \\plus{} b \\plus{} c)^2 \\leq 2(1\\plus{}bc)^2 (1)$\r\nIndeed, we have :\r\n$ (1) \\Leftrightarrow 2(ab\\plus{}bc\\plus{}ca) \\leq 1\\plus{}4bc\\plus{}2b^2c^2 \\Leftrightarrow 2a(b\\plus{}c) \\leq a^2\\plus{}(b\\plus{}c)^2\\plus{}2b^2c^2$\r\n$ \\Leftrightarrow (b\\plus{}c\\minus{}a)^2\\plus{}2b^2c^2 \\geq 0$ (true)\r\nTherefore,\r\n$ \\frac {a}{1 \\plus{} bc} \\plus{} \\frac {b}{1 \\plus{} ca} \\plus{} \\frac {c}{1 \\plus{} ab}\\leq \\frac{\\sqrt{2}a}{a\\plus{}b\\plus{}c}\\plus{}\\frac{\\sqrt{2}b}{a\\plus{}b\\plus{}c}\\plus{}\\frac{\\sqrt{2}c}{a\\plus{}b\\plus{}c}\\equal{} \\sqrt{2}$\r\nOther, we have :\r\n$ a\\plus{}abc \\leq a\\plus{}\\frac{a(b^2\\plus{}c^2)}{2}\\equal{}1\\minus{}\\frac{(a\\minus{}1)^2(a\\plus{}2)}{2} \\leq 1$\r\ntherefore,\r\n$ \\frac {a}{1 \\plus{} bc} \\plus{} \\frac {b}{1 \\plus{} ca} \\plus{} \\frac {c}{1 \\plus{} ab}\\geq a^2\\plus{}b^2\\plus{}c^2\\equal{}1$\r\nWe are done .", "Solution_4": "$ 1\\leq \\frac {a}{1 \\plus{} bc} \\plus{} \\frac {b}{1 \\plus{} ca} \\plus{} \\frac {c}{1 \\plus{} ab}$\r\n\r\n$ (a\\plus{}b\\plus{}c)^2\\equal{} \\left (\\sum_{cyc} \\sqrt a \\sqrt{1 \\plus{} bc} \\cdot \\frac{\\sqrt a}{\\sqrt{1\\plus{}bc}} \\right )^2$$ \\le \\left (\\sum_{cyc} \\frac{a}{1\\plus{}bc} \\right ) \\left ( \\sum_{cyc} a (1 \\plus{} bc) \\right )$\r\n\r\n$ \\sum_{cyc} \\frac{a}{1\\plus{}bc} \\ge \\frac{(a\\plus{}b\\plus{}c)^2}{a\\plus{}b\\plus{}c\\plus{}3abc} \\ge 1$\r\n\r\n$ p^2 \\minus{} p \\minus{} 3r \\ge 0$" } { "Tag": [], "Problem": "If $a$ and $b$ are integers such that $x^2-x-1$ is a factor of $ax^3+bx^2+1$, then $b$ is \r\n\r\n(A) -2\r\n(B) -1\r\n(C) 0\r\n(D) 1\r\n(E) 2", "Solution_1": "[hide]If we take the third factor to be (mx+n), we get mx^3+(n-m)x^2-(m+n)x-n. We thus know that m+n=0 and n=-1, so m=1, and (n-m)=b=-2. A.[/hide]", "Solution_2": "[hide=\"Answer\"]Factoring out, we have $(ax-1)(x^2-x-1)$ so that we end with a $+1$ on the end and $ax^3$ at the beginning. Therefore, we have $ax^3-(a+1)x^2-(a-1)x+1$. We have no $x$ factor, so $a-1=0\\Rightarrow a=1$, and therefore $b=-(a+1)=-2$. $\\boxed{A}$[/hide]" } { "Tag": [ "AMC" ], "Problem": "How many ways are there to place two kings on a 8x8 chessboard so that they do not attack each other? \r\n(you know hte drill, kings attack all adjacent squares. and they are opposite color kings.)", "Solution_1": "[hide]n = 64^2 - (36 * 8 + 24 * 5 + 4 * 3) = 3676\n\n\n\nNot sure about this..seems too easy...[/hide]", "Solution_2": "Lols\r\nsorry\r\nthat's incorrect", "Solution_3": "Whoops. Here is revised solution:\n\n\n\n[hide]n = 64 * 63 - (36 * 8 + 24 * 5 + 4 * 3) = 3612[/hide]", "Solution_4": "Looks good to me, assuming \"the two kings are of different colors\" means that they are distinct." } { "Tag": [], "Problem": "The product of any two of the positive integers 30, 72, and $N$ is divisible by the third. What is the smallest possible value of $N$?", "Solution_1": "[hide]Needs factors of 2, 3, and 5. For it to be divisible by 72, it needs 2 more 2's and a 3. So the answer is $60$[/hide]", "Solution_2": "You can prime factor both 30 and 72, so you get 5X3X2, and 2^3 and 3^2. I then just used trial and error with these numbers to get 5X2^2X3=[b]60[/b].", "Solution_3": "yes you are right" } { "Tag": [ "AMC", "AIME", "search", "complex numbers" ], "Problem": "Given that $w$ and $z$ are complex numbers and $w^2+z^2 = 7$ and $w^3+z^3=10$, find the greatest value of $w+z$\r\nSource: 1983 AIME #5\r\n\r\n[hide=\"I tried...\"]\nSince $w^2+z^2 = 7$, we have $(w+z)^2=7-2wz$\nFactoring $w^3+z^3=10$, we get $(w+z)(7-wz)=10$.\nSolving for $wz$ in the first equation and plugging it into the second gives you as a final equation\n$(w+z)^3+7(w+z)-20=0$ so I'm thinking that the greatest real solution would be the answer. The \"only\" problem is that I can't find any real solutions to the equation...and I don't see how $w$ and $z$ being complex numbers affect the question.[/hide]", "Solution_1": "[quote=\"Hellstar\"][hide=\"I tried...\"]\nSince $w^2+z^2 = 7$, we have $(w+z)^2=7-2wz$\nFactoring $w^3+z^3=10$, we get $(w+z)(7-wz)=10$.\nSolving for $wz$ in the first equation and plugging it into the second gives you as a final equation\n$(w+z)^3+7(w+z)-20=0$ so I'm thinking that the greatest real solution would be the answer. The \"only\" problem is that I can't find any real solutions to the equation...and I don't see how $w$ and $z$ being complex numbers affect the question.[/hide][/quote]\n[hide]\nThe first line should be $(w+z)^2=7+2wz$.\nYou should get $-(w+z)^3+21(w+z)-20=0$, whose solutions are $w+z=4,1,-5$.[/hide]", "Solution_2": "Oh, right :blush: stupid me.", "Solution_3": "This \"classic\" problem was posted about 8 times in AoPS. Here is the link. Please use search button.\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=367523#p367523" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "What is the number of permutations of length $N$,consisting all the numbers between $1$ and $N$ \r\nsuch that in each permutation all the numbers isn't on their places?", "Solution_1": "I think your meant was $D_n=n!\\sum_{r=0}^n\\frac{(-1)^n}{r!}$" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "email", "AIME", "AIME II" ], "Problem": "Hey does anyone here think they qualified for the USAMO?\r\n\r\nbased on [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142291[/url] I did.. I hope.\r\n\r\nomfg I CANNOT wait for that email to come.. I'll do whatever my mom tells me to do for the rest of the day.\r\n\r\nthen again, maybe not :lol:", "Solution_1": "Nice job (especially for an 8th grader)\r\nif I didnt make any stupid mistakes (or just like 1 stupid mistake), I wouldve qualified but I didnt... :mad: but theres always next year...even though I thought I could qualify for this year.\r\n\r\nI think Carmel High School had like 2 qualifiers or so?\r\n\r\nEDIT:only one..only two have ever made it from our school", "Solution_2": "i didn't make it. my index was 175 and i got a 4 on the AIME.", "Solution_3": "Thanks.\r\n\r\nI was in despair because I got a 7 on the AIME II and people were mentioning 8 or 9 as the USAMO floor.. but yeah.. just got my email and like *spazzzz*\r\n\r\nI think about five people from our school qualified, including me :huh:\r\n\r\nEDIT: heh, [i]three[/i] people from our school.. six people from west side\r\n\r\nthe qualifiers are posted: [url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007usamoqual.shtml[/url]", "Solution_4": "at least u made it. i was in despair cuz i made a careless mistake on #2. and i could've gotten #4. and i made a careless mistake on #16 on AMC 10.", "Solution_5": "hmm only 10 people from Indiana made it..thats worse than last year and with the expansion...o well..nice job to tyler zou-seems like hes the only one in our school that made it (even though about 5 others were probalbay 1 question frm amking it)", "Solution_6": "I guareentee next year Indiana will have like 20 USAMO qualifiers (double this year)..", "Solution_7": "\"The message is too small.. sldkfjskadfj\"\r\n\r\nwhatever.\r\n\r\nArt of Owna, how come?", "Solution_8": "Looking at all the 7th, 8th and 9th graders that almost qualified this year/or wouldve qualified if it werent for dumb mistakes (I know next year you can make the same dumb mistakes-but theres a potential for 20)\r\n\r\nCOnsidering there are 11 frm thie year\r\nOnly 1 is a 12th grader so next year we'll probalby have a guareenteed 10. I can probalby name 10 more that can qualify next year if something bad or a lot of careless msitakes do not take place.", "Solution_9": "i hope i can get an 8 next year. in 9th grade, don't you get to go to this really good math camp for free if you get top 30 for 9th graders?(on AIME)", "Solution_10": "I hope I can get 12+ for next year just because I feel like letting myself down...", "Solution_11": "sweeeet. i do that alot too... :lol:", "Solution_12": "Actually I hope for a 10..I probalby can get a 10 as long as I make NO careless or even non-careless mistakes (especially with one year)", "Solution_13": "uh dude -_-... its called MOSP, and you can only go if you're top 500... unless you're talking about a scholarship for ninth graders, then i have no idea (to yuqing :D)", "Solution_14": "guess what? i'm stupid. no wonder i don't even know it's called MOSP. (To [color=red][size=200]LYNDON[/size][/color])", "Solution_15": "[quote=\"mathcrazed\"]that's like... not really plausible lol, david benjamin and nathan benjamin only got 2 or 3, but then again, Paul does like practice every day,.... they have Zuming Feng over there!![/quote]\r\nYea, Feng is really good. Paul improved a lot this past year, I have conversed with him a little bit.\r\n\r\nDude Phil, fix your signature..", "Solution_16": "lol nathan was just kidding about getting perfect ... no duh XD", "Solution_17": "Let me clarify my earlier post. I thought that nathan benjamin got a perfect score (lyndon told me this) so I thought when lyndon said he got 2 or 3..I was a little confused.", "Solution_18": "ya well... i joke actually nathan jokes lol", "Solution_19": "[quote=\"mathcrazed\"]ya well... i joke actually nathan jokes lol[/quote]\r\n\r\nuh really? he doesnt seem like the humorous type. (dont take that offensively)\r\n\r\nand lol he just bsed your dad? uhh..probalby not..", "Solution_20": "no my dad knew he was joking...", "Solution_21": "[quote=\"mathcrazed\"]no my dad knew he was joking...[/quote]\r\n\r\nis taht kid half asain? squinty eyes..I know hes half jew..but he seems asian. even though his last name is american.", "Solution_22": "his mom's chinese", "Solution_23": "and the year that he won, sergei is actually his half brother!! ( i think )... LoL, cause his dad has two wives, one is divorced now", "Solution_24": "Oh wow... you people... amuse me... to such an extent that I'm actually going to post on AoPS and address some stuff.\r\n\r\n\r\nWe are half-Taiwanese, half-Russian (in response to the above, Russian Jew). Our last name is American-sounding because our dad changed his last name when coming to America, for the sake of not having a ridiculously hard to spell/pronounce last name.\r\n\r\nSergei is [i]not[/i] our half-brother. That would be [i]far[/i] too simple. He is our half-brother's mother's second-husband's brother's son. I'll leave it to you to figure out how that works.\r\n\r\n\r\nAnd no, Nathan and I are not the most serious people out there. :P", "Solution_25": "woah...i never knew david was an AoPSer!", "Solution_26": "I never knew the Benjamins weren't american.. :D\r\n\r\nhehe sorry.", "Solution_27": "[quote=\"indianamath\"]woah...i never knew david was an AoPSer![/quote]\nIf you define AoPSer to be someone with an AoPS account, then yes.\n\nBy any other definition, no. :)\n\n[quote]I never knew the Benjamins weren't american..\nhehe sorry.[/quote]\r\nNo offense taken. :) Meh, we were born in America if that counts... although I certainly have much less nationalistic pride than most Americans... especially with regards to the current regime.", "Solution_28": "wow I got a 10 on USAMO\r\n\r\nIt sucks for Nathan because he's a freshamn and red MOP cutoff was 9 and he got 8", "Solution_29": "i will this year :D" } { "Tag": [ "geometry", "geometry solved" ], "Problem": "Let [tex]A_0B_0C_0[/tex] be a triangle and P a point. Define a new triangle whose vertices [tex]A_1[/tex] ; [tex]B_1[/tex] ; [tex]C_1[/tex] \r\nas the feet of the perpendiculars from P to [tex]B_0C_0[/tex] ; [tex]C_0A_0[/tex] ; [tex]A_0B_0[/tex] , respectively. Similarly, \r\ndefine the triangles [tex]A_2B_2C_2[/tex] and [tex]A_3B_3C_3[/tex] . Show that [tex]A_3B_3C_3[/tex] is similar to [tex]A_0B_0C_0[/tex] .", "Solution_1": "Denote angles $PA_0B_0$, $PA_0C_0$ and so on. Then using fact that $PB_1A_0C1$ (and all remaining) is inscribed, we immediately obtain values for angles $PA_1B_1$ and so on. Repeat the same for point $P$ and triangle $A_1B_1C_1$, and further for $P$ and $A_2B_2C_2$. Finally you will find expressions for angles of $A_3B_3C_3$ via angles of $A_0B_0C_0$.", "Solution_2": "[Sorry, Myth, for de-facto repeating your post. I didn't see it when I posted the mine...]\r\n\r\nWell-known theorem, but I don't find a proof online, so let me post one:\r\n\r\nWe will use directed angles modulo 180.\r\n\r\nSince $\\measuredangle A_0B_1P = \\measuredangle A_0C_1P = 90^{\\circ}$, the points $B_1$ and $C_1$ lie on the circle with diameter $A_0P$. Hence, $\\measuredangle PB_1C_1 = \\measuredangle PA_0C_1$. In other words, $\\measuredangle PB_1C_1 = \\measuredangle PA_0B_0$. Similarly, $\\measuredangle PC_2A_2 = \\measuredangle PB_1C_1$, $\\measuredangle PA_3B_3 = \\measuredangle PC_2A_2$. Thus, $\\measuredangle PA_3B_3 = \\measuredangle PA_0B_0$. Similarly, $\\measuredangle PB_3A_3 = \\measuredangle PB_0A_0$. Hence, the triangle $PA_3B_3$ is directly similar to the triangle $PA_0B_0$. Analogous reasoning shows that the triangle $PB_3C_3$ is directly similar to the triangle $PB_0C_0$ and that the triangle $PC_3A_3$ is directly similar to the triangle $PC_0A_0$. Hence, the triangle $A_3B_3C_3$, which can be subdivided into the triangles $PA_3B_3$, $PB_3C_3$ and $PC_3A_3$, is directly similar to the triangle $A_0B_0C_0$, which can be subdivided into the triangles $PA_0B_0$, $PB_0C_0$ and $PC_0A_0$.\r\n\r\n$\\blacksquare$\r\n\r\nBy the way, this generalizes to n-gons; for instance, for quadrilaterals, the quadrilateral $A_4B_4C_4D_4$ is similar to the quadrilateral $A_0B_0C_0D_0$.\r\n\r\n Darij", "Solution_3": "I have a feel that it is from Kvant... :? \r\n\r\nDarij, it is ok!\r\nYour remark about quadrilaterals was unknown for me. At least I never thought about it.", "Solution_4": "[quote=\"Myth\"]I have a feel that it is from Kvant...[/quote]\r\n\r\nWell, it is [url=http://kvant.mccme.ru/1976/08/gif/76_08-40.gif]Kvant problem 356[/url], but actually it is well-known. It's Theorem 1.92 in chapter 1 9 of\r\n\r\nH. S. M. Coxeter, S. L. Greitzer, [i]Geometry Revisited[/i], Toronto - New York 1967.\r\n\r\nThe generalization to arbitrary n-gons (not just quadrilaterals) is straightforward:\r\n\r\n[i]For any n-gon K and any point P in its plane, denote by $T_P\\left(K\\right)$ the n-gon formed by the orthogonal projections of the point P on the sidelines of the n-gon K. Then, for any n-gon K and any point P in its plane, the n-gon $T_P^n\\left(K\\right)$ is directly similar to the n-gon K.[/i]\r\n\r\nThe proof is almost the same as the one for the triangles.\r\n\r\n Darij", "Solution_5": "[quote=\"darij grinberg\"][quote=\"Myth\"]I have a feel that it is from Kvant...[/quote]\n\nWell, it is [url=http://kvant.mccme.ru/1976/08/gif/76_08-40.gif]Kvant problem 356[/url][/quote]\r\nIt means that I saw this problem in another place, since I saw it long time ago and I didn't have access to Kvant problems. Though, it is possible that it was in some article in Kvant in 90th. :?" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Let $ a$ be a positive number. Find all function $ f: R^ \\plus{} \\to R^ \\plus{}$ such that: $ f(f(x)) \\equal{} ax \\minus{} f(x)$ for all $ x > 0$ :)", "Solution_1": "Sorry, i edited my problem! :D", "Solution_2": "Hm, :( Noone can solve it? If that;s right, i'll Suggest! :)", "Solution_3": "We'll write $ a \\equal{} c^2 \\plus{} c$, with $ c$ is a positive real number, and now, we'll show that $ f(x) \\equal{} cx$ for all $ x$! :)", "Solution_4": "There are infinetely many solutions.", "Solution_5": "[quote=\"Rust\"]There are infinetely many solutions.[/quote]\r\nRust, I donn't think like that, we can conclude that \"There are infinetely many solutions\" only in case the function $ f: R \\to R$, but here, $ f: R^ \\plus{} \\to R^ \\plus{}$.\r\nIf in the case $ f: R^ \\plus{} \\to R^ \\plus{}$, you say too \"There are infinetely many solutions\", can you denote the way to construct a class of function $ f: R^ \\plus{} \\to R^ \\plus{}$ what satisfy this problem! :)", "Solution_6": "Let $ 0 < a\\le 1$, we can find $ f(x) \\equal{} xc(x)$, then $ c(xc(x)) \\equal{} \\frac {a}{c(x)} \\minus{} 1$.\r\nWe can consider $ c(x)\\equiv c_i , \\ x_i\\le x < x_{i \\plus{} 1}$, then $ c_{i \\minus{} 1} \\equal{} \\frac {a}{c_i} \\minus{} 1, x_i \\equal{} c_ix_{i \\plus{} 1}$.\r\nIt give infinetely many solutions.\r\n I am sorry. It work only for $ 0 < a < 2$ and $ f: (0,A)\\to (0,A$ or\r\n$ a > 2$ and $ f: (A,\\infty)\\to (A,\\infty )$.\r\nIn these case $ f: R_\\plus{}\\to R_\\plus{}$ unique solution for $ a>0$.", "Solution_7": "Ohlala, and with case $ a$ is a enough positive number, example $ a\\equal{}10^9$ :) How do we solve it? :P", "Solution_8": "Let $ a>0$ and $ c$ is root of $ c^2\\plus{}c\\equal{}a$. Let exist $ x_0$, suth that $ c(x_0)\\equal{}c_0\\not \\equal{}c$. Let $ g(c)\\equal{}\\frac ac \\minus{}1.$\r\nThen $ sign(g(x)\\minus{}c)\\equal{}sign(c\\minus{}x)\\to c_0>c>c_1\\equal{}g(c_1) \\ or \\ c_0c_0>c_2>...\\ge 0$ or $ c0$ or $ f(x)\\equiv cx$." } { "Tag": [ "inequalities", "induction", "inequalities solved" ], "Problem": "Given a1,a2,...,an real numbers define bk=(a1+...+ak)/k for k=1,2,..,n.\r\n\r\nIf C=(a1-b1)^2 + (a2-b2)^2 +...+ (an-bn)^2\r\n\r\nand D=(a1-bn)^2 + (a2-bn)^2 +...+ (an-bn)^2\r\n\r\nprove that C \\leq D \\leq 2C", "Solution_1": "Let f(x)=(x-a_1)^2+(x-a_2)^2+\\dots+(x-a_n)^2 then\r\n\r\nf(x)=n(x-b_n)^2+f(b_n). (*)\r\n\r\nIf n=1 then inequalitys are trivial (C=D). To prove inequalities by induction it is enough to prove that\r\n\r\n0\\leq f(b_{n+1})-f(b_{n})\\leq (a_{n+1}-b_{n+1})^2. \r\n\r\nFirst ineq follows imediatly from (*) (x=b_{n+1}). Second follows from equalities\r\n\r\n(n+1)b_{n+1}=nb_n+a_{n+1}, n(b_{n+1}-b_n)=a_{n+1}-b_{n+1},\r\n\r\nf(b_{n+1})-f(b_{n})=n(b_{n+1}-b_n)^2=(a_{n+1}-b_{n+1})^2/n" } { "Tag": [ "analytic geometry", "graphing lines", "slope", "algebra", "system of equations" ], "Problem": "I appericate all the help and want to thank you all in advance..\r\nIm just a little stumped on how to figure these out and show the work for it..Espsically questions #3-6\r\n\r\n\r\n\r\n1. Find the slope of the line that passes through the points (2, 3) and (5, 8).\r\n\r\n\r\n\r\n\r\n2. Find the equation of the line that passes through the points (3, -2) and (4, -2).\r\n\r\n\r\n\r\n\r\n3. Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3, 5).\r\n\r\n\r\n\r\n\r\n4. Solve the system of equations using the substitution method.\r\nIf the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is \u201cno solution\u201d or \u201cinfinitely many solutions.\u201d\r\n-3x + y = 1\r\n5x + 2y = -4\r\n\r\n\r\n\r\n\r\n5. Solve the system of equations using the addition (elimination) method.\r\nIf the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is \u201cno solution\u201d or \u201cinfinitely many solutions.\u201d\r\n-7x + y = 8\r\n2x \u2013 y = 2\r\n\r\n\r\n\r\n\r\n6. Solve the system of equations using the addition (elimination) method.\r\nIf the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is \u201cno solution\u201d or \u201cinfinitely many solutions.\u201d\r\n3x \u2013 2y = -7\r\n-9x + 6y = 21", "Solution_1": "These should be in HSB...\r\n[hide]1. Slope is $ \\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$. So, the answer is $ \\frac{5}{3}$.\n2. Using the previous method, the answer is $ 0$.\n3. The line has slope $ -\\frac{1}{3}$, so its perpendicular would have slope $ 3$. Using point-slope form, we have\n$ 3(x+3)=y-5$\n$ 3x-y=-14$.\n4. From the first equation, $ y=3x+1$. Putting this into the second equation, we have\n$ 5x+2(3x+1)=-4$\n$ x=-\\frac{6}{11}$\n$ y=-\\frac{7}{11}$\n$ (-\\frac{6}{11},-\\frac{7}{11})$\n5. Adding the equations, we have\n$ -5x=10$\n$ x=-2$\n$ y=-6$\n$ (-2,-6)$.\n6. Multiplying the first equation by 3, we have $ 9x-6y=-21$. Adding the equations, we have $ 0=0$, which is true for all $ x$'s and $ y$'s. The system has infinitely many solutions.[/hide]", "Solution_2": "Are you sure this isn't a homework?\r\n\r\n3. The slope of perpendicular lines are negative reciprocals.\r\n\r\n4-6. Multiply both sides by numbers which will cancel out one variable. For example, $ 2x+a$ and $ -3x+2a$, multiply the expression 1 by 3 and expression 2 by 2. That will eliminate $ x$.", "Solution_3": "[quote=\"mountain.dew\"]These should be in HSB...\n[hide]1. Slope is $ \\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$. So, the answer is $ \\frac{5}{3}$.\n2. Using the previous method, the answer is $ 0$.\n3. The line has slope $ -\\frac{1}{3}$, so its perpendicular would have slope $ 3$. Using point-slope form, we have\n$ 3(x+3)=y-5$\n$ 3x-y=-14$.\n4. From the first equation, $ y=3x+1$. Putting this into the second equation, we have\n$ 5x+2(3x+1)=-4$\n$ x=-\\frac{6}{11}$\n$ y=-\\frac{7}{11}$\n$ (-\\frac{6}{11},-\\frac{7}{11})$\n5. Adding the equations, we have\n$ -5x=10$\n$ x=-2$\n$ y=-6$\n$ (-2,-6)$.\n6. Multiplying the first equation by 3, we have $ 9x-6y=-21$. Adding the equations, we have $ 0=0$, which is true for all $ x$'s and $ y$'s. The system has infinitely many solutions.[/hide][/quote]\r\n\r\nPlease avoid answering questions directly. Hints are better, it will make them think and work on their own. Anyway, these are very easy questions. :)" } { "Tag": [ "geometry", "parallelogram" ], "Problem": "All face of a convex polyhedron are parallelograms.Can the polyhedron have exactly 1992 faces?", "Solution_1": "[hide]\nConsider a set of parallelograms faces that goes all the way \"around\" such a polyhedron. Obviously, two adjacent faces in such a set share an edge. Since they're parallelograms, the sides are all parallel. We can see that any two of these sets intersect at two faces, and conversely, that any face belongs to two rings. So if there's $ n$ distinct sets of these, then there's $ n(n\\minus{}1)$ faces. However, there's no integer solutions to $ n^2\\plus{}n\\minus{}1992\\equal{}0$, so it can't have 1992 pages.\n\nWow, that's a tricky problem.[/hide]" } { "Tag": [ "function", "LaTeX", "calculus", "calculus computations" ], "Problem": "[color=darkblue]prove that this equation: $ y' \\equal{} x \\plus{} y^2$ have no root.[/color]", "Solution_1": "What do you mean \"roof\"? Are you asking about roots/zeroes?", "Solution_2": "Yes! I'm sory.", "Solution_3": "this is a differential equation; do you mean 0 = -y' + x + y\u00b2 has no solution in x where y is a function of x, or do you mean the differential equation has no general solution?", "Solution_4": "You can always at least make an approximation using power series. Even if it isn't in closed form, you get something that is correct.", "Solution_5": "[quote=\"JRav\"][b]What do you mean \"roof\"? [/b] Are you asking about roots/zeroes?[/quote]\r\n\r\n :rotfl: :D", "Solution_6": "[b]thanhnam2902[/b]\r\nDo you mean y=f(x) ?\r\n\r\nWhen y=x , we have\r\ny'=x+y\u00b2\r\n1=x+x\u00b2\r\nWhich has 2 real roots.\r\nOr when y=x\u00b2\r\n2x=x+x^4 , which also has a real root x=1 and x=0.", "Solution_7": "Maybe I'm just missing something, but I don't see why $ y\\equal{}x$ is a solution.\r\n$ y'\\equal{}1\\neq x\\plus{}x^{1}$", "Solution_8": "By the existence-uniqueness theorem, this differential equation has local solutions. What you want is to show that these solutions cannot be extended indefinitely; all such solutions blow up to $ \\infty$ in finite time.\r\n\r\nAs a first step, note that all positive solutions $ \\frac1{c \\minus{} x}$ to $ y'' \\equal{} y^2$ blow up in finite time. You want to show that each solution to your differential equation is eventually bigger than one of these, and hence blows up faster.", "Solution_9": "I absolutely agree with jmerry.\r\nto \u10da\u10d4\u10d5\u10d0\u10dc\u10d8\r\nyou can use latex , \r\nsee here \r\nhttp://www.mathlinks.ro/LaTeX/AoPS_L_About.php", "Solution_10": "Thank you! :o" } { "Tag": [ "algebra", "polynomial", "inequalities open", "inequalities" ], "Problem": "Prove or disprove that for $a,b,c\\ge0$ and every positive integers $n,t,k,l$ such that $t\\ge k,t\\ge l$ we have:\r\n\r\n$2(a^{n}+b^{n}+c^{n})+(a^{n-t}b^{t}+a^{t}b^{n-t}+b^{n-t}c^{t}+b^{t}c^{n-t}+c^{n-t}a^{t}+c^{t}a^{n-t})\\ge(a^{n-k}b^{k}+a^{k}b^{n-k}+b^{n-k}c^{k}+b^{k}c^{n-k}+c^{n-k}a^{k}+c^{k}a^{n-k})+(a^{n-l}b^{l}+a^{l}b^{n-l}+b^{n-l}c^{l}+b^{l}c^{n-l}+c^{n-l}a^{l}+c^{l}a^{n-l})$\r\n\r\n[hide=\"some notes\"]$(n,0,0)+(n-t,t,0)\\ge(n-k,k,0)+(n-l,l,0)$[/hide]", "Solution_1": "This one is still a special case of the one I posted in http://www.mathlinks.ro/Forum/viewtopic.php?p=580614#580614 middle of first page (Tue Dec 06, 2005) :lol: :\r\n[quote=\"spanferkel\"]\n[b] For partitions $\\alpha,\\beta\\vdash n$ and $0\\leq t\\leq1$ we have $[\\alpha]+[\\beta]\\geq [(1-t)\\alpha\\oplus t\\beta]+[t\\alpha\\oplus(1-t)\\beta]$. (*)[/b][/quote] :) :)" } { "Tag": [], "Problem": "Someone on this forum gave me an invitation on Yahoo Messenger, can yo do that again? I have pressed Cancel by mistake! Sorry! :(\r\n\r\ncheers! :D :D", "Solution_1": "wow... that's sad", "Solution_2": "...Lol? I feel like this is the wrong forum...I may be wrong though." } { "Tag": [ "\\/closed" ], "Problem": "A most unusual problem with one of my computers:\r\n\r\nWhen on AOPS (ONLY FOR AOPS) I can only select a textbox once, or else the browser forcequits/freezes/kills the computer. This happens for every browser, e.g. Firefox, I.E., AOL, Netscape, Chrome, Opera, Safari, etc. I try, and currently I am using a custom browser (Avant) which freezes as well at certain random times.", "Solution_1": "I don't understand what you mean by \"I can only select a textbox once\" ...", "Solution_2": "As in, when I select a textbox, there is a flashing line (|) showing up. If I click on the textbox again, then it freezes." } { "Tag": [ "Functional Analysis", "real analysis", "real analysis unsolved" ], "Problem": "i have a hard problem need to help.\r\n Given 2 bounded linear operators A, B satisfied AB is compact. Does A or B must be compact?", "Solution_1": "Let's try this one, using operators on the Hilbert space $\\ell^{2}:$\r\n\r\nDefine $A$ by $A: (x_{1},x_{2},\\dots)\\mapsto(x_{1},0,x_{3},0,x_{5},0,\\dots).$\r\n\r\nDefine $B$ by $B: (y_{1},y_{2},\\dots)\\mapsto(0,y_{1},0,y_{2},0,y_{3},\\dots).$\r\n\r\n\r\n\r\nNeither $A$ nor $B$ is compact. $A$ is an orthogonal projection onto a one-dimensional subspace, and $B$ is an isometry.\r\n\r\nBut $AB=0,$ and the zero operator is compact." } { "Tag": [ "function", "calculus", "derivative", "calculus computations" ], "Problem": "Hi, can anybody help me with this?\r\n\r\nI have a function y=arctg (1/x)\r\nIn which point has a tangent to this function \"k = -1/2\"\r\n\r\nThanx,\r\nTyrrel", "Solution_1": "Solve of a tangent = derivative at the point of tangency. Solve $\\frac{-1}{1+x^{2}}=-\\frac{1}{2}$ for $x$.\r\n\r\nNot so advanced after all." } { "Tag": [], "Problem": "Evaluate the product of the following fractions:\r\n\r\n\\[ \\frac{\\frac{1}{2} - \\frac{1}{3}}{\\frac{1}{3} - \\frac{1}{4}} \\cdot \\frac{\\frac{1}{4} - \\frac{1}{5}}{\\frac{1}{5} - \\frac{1}{6}} \\cdot \\frac{\\frac{1}{6} - \\frac{1}{7}}{\\frac{1}{7} - \\frac{1}{8}} \\cdot \\cdots \\cdot \\frac{\\frac{1}{98} - \\frac{1}{99}}{\\frac{1}{99} - \\frac{1}{100}}. \\]", "Solution_1": "[hide](1/2-1/3)/(1/3-1/4)*(1/4-1/5)/(1/5-1/6)....\n\n=(1/6)/(1/12)*(1/20)/(1/30)...\n\n=2*3/2...\n\n=4/2*6/4...\n\nrealize that (1/a-1/b)/(1/b-1/c)=c/a\n\nso now we can cancel through as c1=a2\n\nleaving c97/a1\n\n=100/2\n\n=50[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c,d$ be nonnegative real numbers. Prove that:\r\n$ \\sum\\limits_{cyc}\\frac {a^2}{b \\plus{} c \\plus{} d} \\leq 4\\frac {(a \\minus{} b)^2}{a \\plus{} b \\plus{} c} \\plus{} \\frac {4(a \\plus{} b \\plus{} c \\plus{} d)}{3}$", "Solution_1": "[quote=\"Hong Quy\"]Let $ a,b,c,d$ be nonnegative real numbers. Prove that:\n$ \\sum\\limits_{cyc}\\frac {a^2}{b \\plus{} c \\plus{} d} \\leq 4\\frac {(a \\minus{} b)^2}{a \\plus{} b \\plus{} c} \\plus{} \\frac {4(a \\plus{} b \\plus{} c \\plus{} d)}{3}$[/quote]\r\nLet $ a\\equal{}b\\equal{}0$, the above inequality becomes\r\n\\[ \\frac{c^2}{d}\\plus{}\\frac{d^2}{c} \\le \\frac{4}{3}(c\\plus{}d)\\]\r\nwhich does not always hold, for example with $ d\\equal{}1,c\\equal{}100$.", "Solution_2": "Let $ a,b,c,d$ be nonnegative real numbers. Prove that:\r\n$ \\sum\\limits_{cyc}\\frac {a^2}{b \\plus{} c \\plus{} d} \\leq 4max \\left( \\frac {(a \\minus{} b)^2}{a \\plus{} b \\plus{} c},\\frac {(c \\minus{} b)^2}{d \\plus{} b \\plus{} c},\\frac {(c \\minus{} d)^2}{d \\plus{} a \\plus{} c},\\frac {(d \\minus{} a)^2}{a \\plus{} b \\plus{} d}\\right) \\plus{} \\frac {4(a \\plus{} b \\plus{} c \\plus{} d)}{3}$" } { "Tag": [ "function", "integration", "algebra", "linear equation", "real analysis", "real analysis unsolved" ], "Problem": "[b]I'm aware this is a long problem but I would really appreciate some help/hints especially for getting the inhomogeneous ODE of the first part:[/b]\r\n\r\nThe function $ y(x)$ satisfies the linear equation\r\n\\[ y'' \\plus{} p(x)y' \\plus{} q(x)y \\equal{} 0.\\]\r\nThe Wronskian of two independent solutions, denoted $ y_1(x)$ and $ y_2(x)$, is defined to be\r\n\\[ W(x) \\equal{} \\left| \\begin{array}{cc} y_1 & y_2 \\\\\r\ny_1' & y_2' \\end{array} \\right|.\\]\r\nLet $ y_1(x)$ be given. Use the Wronskian to determine a first-order inhomogeneous differential equation for $ y_2(x)$. Hence, show that\r\n\\[ y_2(x) \\equal{} y_1(x) \\int_{x_0}^{x} \\frac {W(t)}{y_1(t)^2} \\, dt. \\qquad \\qquad (*)\\]\r\nShow that $ W(x)$ satisfies\r\n\\[ \\frac {dW}{dx} \\plus{} p(x)W \\equal{} 0.\\]\r\nVerify that $ y_1(x) \\equal{} 1 \\minus{} x$ is a solution of\r\n\\[ xy'' \\minus{} (1 \\minus{} x^2)y' \\minus{} (1 \\plus{} x)y \\equal{} 0. \\qquad \\qquad (\\dagger)\\]\r\nHence, using $ (*)$ with $ x_0 \\equal{} 0$ and expanding the integrand in powers of $ t$ to order $ t^3$, find the first three non-zero terms in the power series expansion for a solution, $ y_2$, of $ (\\dagger)$ that is independent of $ y_1$ and satisfies $ y_2(0) \\equal{} 0, y_2''(0) \\equal{} 1$.", "Solution_1": "hello, the solution of your equation is\r\n$ y(x)\\equal{}C_1(x\\minus{}1)\\plus{}C_2(x\\minus{}1)\\cdot\\int\\frac{xe^{\\minus{}\\frac{1}{2}x^2}}{(x\\minus{}1)^2}\\,dx$\r\nSonnhard.", "Solution_2": "To determine the first-order differential equation for y_2, just evaluate the Wronskian.", "Solution_3": "I got $ y_1(x)y_2'(x) \\minus{} y_1'(x)y_2(x) \\equal{} W(x)$, so that $ \\frac{y_1(x)y_2'(x) \\minus{} y_1'(x)y_2(x)}{y_1(x)^2} \\equal{} \\frac{W(x)}{y_1(x)^2}$.\r\n\r\nThus $ \\left( \\frac{y_2(x)}{y_1(x)} \\right)' \\equal{} \\frac{W(x)}{y_1(x)^2}$. From here I think you integrate wrt $ x$, but how does the $ t$ variable and the lower limit $ x_0$ come in?", "Solution_4": "[quote=\"aidan\"]I got $ y_1(x)y_2'(x) \\minus{} y_1'(x)y_2(x) \\equal{} W(x)$, so that $ \\frac {y_1(x)y_2'(x) \\minus{} y_1'(x)y_2(x)}{y_1(x)^2} \\equal{} \\frac {W(x)}{y_1(x)^2}$.\n\nThus $ \\left( \\frac {y_2(x)}{y_1(x)} \\right)' \\equal{} \\frac {W(x)}{y_1(x)^2}$. From here I think you integrate wrt $ x$, but how does the $ t$ variable and the lower limit $ x_0$ come in?[/quote]\r\n\r\nYou've gone a step too far. The question says that y_1 is given - so treat it as a given polynomial. You have a first order differential equation in y_2 - solve it as you normally would (e.g. integrating factor), in terms of y_1.", "Solution_5": "I forgot about that.. so the integrating factor is $ e^{y_1(x)}$, I presume?", "Solution_6": "[quote=\"aidan\"]I forgot about that.. so the integrating factor is $ e^{y_1(x)}$, I presume?[/quote]\r\n\r\nIf you have the equation $ y' \\plus{} p(x) y \\equal{} f(x)$, the integrating factor is $ e^{\\int p(x) \\ dx}$\r\n\r\nIn this example, $ p(x) \\equal{} \\minus{} \\frac {y_1'(x)}{y_1(x)}$ (the coefficient of y_2' must be 1). \r\n\r\nSo the integrating factor is $ e^{\\int \\minus{} \\frac {y_1'(x)}{y_1(x)} \\ dx} \\equal{} \\frac {1}{y_1(x)}$", "Solution_7": "Made a mistake there because of quick calculation :blush: Thanks :)" } { "Tag": [ "MATHCOUNTS", "Mafia", "trigonometry", "ARML" ], "Problem": "On almost winning Mathcounts.\r\n\r\n(I'm serious though, congrats, you guys (EDIT: and girl) did well.)", "Solution_1": "I second that.", "Solution_2": "Man I am still wondering how this happened...oh well I suppose Lewis Chen didn't try hard enough.", "Solution_3": "Gj to max for getting a 44. Also CA failed...only 1 person in cd. Hmph.", "Solution_4": "congrats, guys! perfect team i hear?", "Solution_5": "Blah another question: is it because you're a mod that you can post all caps in the title?", "Solution_6": "proly", "Solution_7": "so this is like the one of the most nonspammy topics on mo forum ever...", "Solution_8": "Not anymore. I just posted.", "Solution_9": "Dude Max looked soo hot at the countdown round!", "Solution_10": "Dude what? Is he your hero or something now? MO people must be so proud of him :O", "Solution_11": "[quote=\"LilyBreeze\"]Dude Max looked soo hot at the countdown round![/quote]\r\n\r\nNo comment. \r\n\r\nMax is like my height! He's kind of short too. No offense. I'm short too. \r\n\r\nGj. Max. You're awesome.", "Solution_12": "[quote=\"jonathanchou711\"][quote=\"LilyBreeze\"]Dude Max looked soo hot at the countdown round![/quote]\n\nNo comment. \n\nMax is like my height! He's kind of short too. No offense. I'm short too. \n\nGj. Max. You're awesome.[/quote]\r\n\r\nEr, Max isn't that short.", "Solution_13": "So of course, no one has denied that max is pretty hot. He's pretty hot when he's doing math/meth whatever.", "Solution_14": "OH MAN MAX LIKE PWNED MATHCOUNTS :!: :!: :!: \r\n\r\nWe totally should've been first...", "Solution_15": "[quote=\"pythag011\"][quote=\"suma_milli\"][quote=\"pythag011\"]David lost in the first round at state. Proves how much harder CA is too win then nats (jk.)[/quote]\n\nDidn't David skip cd to go to ARML practice?[/quote]\n\nThat was chapter.[/quote]\r\n\r\n:gasp chapter < ARML practice !!!!!!!!!!!!!!!", "Solution_16": "[quote=\"pythag011\"][quote=\"suma_milli\"][quote=\"pythag011\"]David lost in the first round at state. Proves how much harder CA is too win then nats (jk.)[/quote]\n\nDidn't David skip cd to go to ARML practice?[/quote]\n\nThat was chapter.[/quote]\r\n\r\nOhh right. D:", "Solution_17": "[quote=\"pythag011\"]David lost in the first round at state. Proves how much harder CA is too win then nats (jk.)[/quote]\r\n\r\nDude well nats cd was way too slow. took like 30 seconds to realize that a equilateral triangle has 60 degree angles and regular hexagon has 120? :P then answered the wrong one", "Solution_18": "so... the mo arml t-shirt design is gonna be pretty awesome, depending on what color shirt they choose.", "Solution_19": "Is it hard to make MO ARML?", "Solution_20": "basically, just fill out an application.", "Solution_21": "i missed the cutoff for arml... :(", "Solution_22": "wait so basically what you're saying is if someone from texas fills in an application it should be okay right", "Solution_23": "er how... victor?\r\n\r\nlet me edit that statement. if you can fill out an application, pay the fee, do it on time (or near enough, they're somewhat lenient), and you actually live in this region (basically, KS, MO, and southern IL, ie. non-Chicago), you're in. there's no requirement to attend practices, and there is no tryout of any sort. you're asked for test scores, but they're largely irrelevant, since the applicant pool is usually around 35ish students, and MO sends two teams + alternates.", "Solution_24": "i forgot that you have to fill out a form to make arml\r\n\r\nclearly this is a requirement\r\n\r\nand i failed it", "Solution_25": "well that was dumb...\r\n\r\ni even made a thread about that.", "Solution_26": "well actually i just realized i filled out the form but never turned it in for reasons", "Solution_27": "meh, you shoulda emailed, they'd probably would've let you in a week or two ago.", "Solution_28": "What's with all of the \"dude\" stuff???", "Solution_29": "Coolio! :)" } { "Tag": [ "function", "abstract algebra", "algebra", "polynomial", "integration", "calculus", "group theory" ], "Problem": "prove that: (in this case $ G$ is discrete LCA group )\r\nfor every measurable function $ f$ on $ G$ and every positive number $ \\epsilon >0$ there exist a function $ g$ with uniformly convergent \r\nFourier series satisfying the condition :\r\n$ m\\{f \\neq g\\}<\\epsilon$\r\nalso what can you say about spectrum of $ g$?", "Solution_1": "Discrete locally compact abelian group? You mean something like $ \\mathbb Z$? Then what is \"the Fourier series\" of a function defined on $ G$? :? \r\n\r\nThere is a nice theorem like that on the circle but the known proof is by no means simple (at least, too long and involved to post here). Do you have a better approach?", "Solution_2": "Yes, Discrete locally compact abelian group. One can say that, every discrete group is locally compact .\r\nYes, I mean something like $ \\mathbb{Z}$, but not only .\r\nThe fourier transform for $ \\mathbb{Z}$ will be :\r\n$ \\widehat{f}(z) \\equal{} \\sum_{n\\in\\mathbb{Z}}f(n)z^{\\minus{}n}$ where $ z\\in \\mathbb{T}$\r\nand to define fourier series we need to define some \"basis\" for the dual group of $ \\mathbb{Z}$ which is isomorphic to $ \\mathbb{T}$\r\nI try to write it by this way (but it's not correct ! ):\r\n$ \\sum_{\\gamma \\in T}\\widehat{f}(\\gamma)\\gamma$\r\nwhere $ \\gamma \\in T$ we understand in some reasonable sense to understand uniformly convergent of fourier series.\r\nI have some idea how to define this \"basis\" and also I can to say that it will be include uniformly convergent in any reasonable sense. \r\nTheorem, what are you talking is in exact Men'shovs theorem which is true only for $ \\mathbb{G}\\equal{}\\mathbb{T}$ and the proof is too long. \r\nThere is more better approach (also is too long) of S.V Kislyakov for any LCA group where they are not discrete :(", "Solution_3": "OK. Let $ G\\equal{}\\mathbb Z$ and $ f(z)\\equal{}z$. Show me your $ g$ for this case. :P \r\n\r\nBy the way, what measure $ m$ are you considering? (if the Haar measure on $ G$, i.e., the counting measure, your claim is certainly false unless [b]all[/b] functions have \"uniformly convergent Fourier series\" in your sense, when it is trivial; if some finite measure on a countable group, then the claim reduces to the statement that you can interpolate any function at finitely many points by a trigonometric polynomial) :?\r\n\r\nSo, put it this way or that, it still seems to me that you are talking nonsense :(. As long as you have just some vague ideas and do not bother to tell any details, nobody will understand you even if what you say makes some sense.\r\n\r\nThis is not to discourage you but merely to remind you (once again :roll:) of the accepted standards of mathematical communication :).", "Solution_4": "Yes, Haar measure .\r\nOk. This is what I was thinking :\r\nLet's $ m$ will be Haar measure on LCA group $ G$ where $ G$ is discrete (for example $ G \\equal{} \\mathbb{Z}$) and $ m_{1}$ on $ \\mathbb{T}$\r\nHaar measure on $ \\mathbb{T}$is normalized so that Parseval's equality holds :\r\n\r\n$ \\parallel{}\\widehat{f}\\parallel{}_{L^{2}(\\mathbb{T})} \\equal{} \\parallel{}f\\parallel{}_{L^{2}(\\mathbb{Z})}$\r\nwhere by $ \\widehat{f}$ I denote :\r\n\\[ \\widehat{f}(\\gamma) \\equal{} \\int_{\\mathbb{Z}} f(x)(\\gamma^{ \\minus{} 1},x)d m(x)\r\n\\]\r\nconsider partial integrals (\"partial sums\" ) :\r\n\\[ f(x) \\equal{} \\int_{\\mathbb{T}} \\widehat{f}(\\gamma) (\\gamma,x)1_{E}d m_{1}(\\gamma)\r\n\\]\r\nwhere $ \\gamma \\in \\mathbb{T}$, $ E$ is measurable subset of $ \\mathbb{T}$ and $ 1_{E}$ is characteristic function.\r\nin my first post by $ \\sum_{\\gamma \\in \\mathbb{T}}$ I was meaning $ \\int_{\\mathbb{T}}$.\r\nNow, I think we already understand uniformly convergent of fuorier series. (it's same as uniformly convergent of partial integrals.), if we can say about choosing of $ E$ something good (in my first post I was talking about \"basis\").", "Solution_5": "And, if not a secret, what is \"uniform convergence of partial integrals\"? Note that if $ E$ is sufficiently close to $ \\mathbb T$, then $ f$ and $ \\mathcal F^{\\minus{}1}[\\widehat f\\chi_{{}_E}]$ are close in $ \\ell^2$ and, thereby, uniformly, so, unless you mean something exotic, all $ f\\in \\ell^2$ have \"uniformly convergent Fourier series\" in this sense..." } { "Tag": [ "probability", "geometry", "3D geometry", "floor function", "AMC", "AIME", "logarithms" ], "Problem": "1. If a fair coin is flipped an infinite number of times, what is the probability three heads will appear before two tails?\r\n\r\n\r\n2. How many numbers in the 100th row of the Pascal triangle (the one starting with 1, 100...) are not divisible by 3?\r\n\r\n\r\n3. Among the first billion positive intagers, consider the sets of: \r\n\r\na. palindromic numbers, e.g. 22, 121, 1145411\r\nb. prime numbers\r\nc. perfect cubes\r\n\r\nArrange these in order of decreasing size.\r\n\r\n\r\nThese problems were taken from the 2003 UGA math contest.", "Solution_1": "2) ${100 \\choose n}=\\frac{100!}{n!(100-n)!}$.\r\nThe largest power of 3 that divides 100! is\r\n$\\lfloor \\frac{100}{3}\\rfloor+\\lfloor \\frac{100}{9}\\rfloor+\\lfloor \\frac{100}{27}\\rfloor+\\lfloor \\frac{100}{81}\\rfloor$\r\nSo 3 does not divide ${100 \\choose n}$ iff\r\n$\\sum_{n=1}^{4} \\lfloor \\frac{n}{3^n}\\rfloor+\\sum_{n=1}^{4} \\lfloor \\frac{100-n}{3^n}\\rfloor=\\sum_{n=1}^{4} \\lfloor \\frac{100}{3^n}\\rfloor$\r\nSo\r\n$\\lfloor \\frac{100}{81}\\rfloor=\\lfloor \\frac{n}{81}\\rfloor+\\lfloor \\frac{100-n}{81}\\rfloor$\r\nand that holds iff $n(\\mod 81) \\leq 19$, so there are 40 n's.\r\nAlso$\\lfloor \\frac{100}{9}\\rfloor=\\lfloor \\frac{n}{9}\\rfloor+\\lfloor \\frac{100-n}{9}\\rfloor$, and that holds iff $n(\\mod 9) \\leq 1$. That eliminate 28 n's of the 40. So 40-28=12.", "Solution_2": "#1. there was an aime question that was very similar to this. i'm not sure which\n\n[hide]\n\nlet's have H=the strings of heads and tails that fits our discription which begins with a H. \n\nlet's have T=the strings of heads and tails that fits our requirements that begins with a T\n\n\n\nwe know that (probability of getting a H)=2(probability of getting a T) because when a string begins with tails, the next flip must be heads. \n\n\n\nwe also know that the number of flip sequences that fits our description must fit either start with tails followed by a H, {start with 1-2 heads followed by a T, or the a sequence of 3 heads right at the beginning}.\n\n\n\nnotice that the possibilities enclosed b the {} are all parts of H. we also realize that these 2 statements include all the possibiities which are a part of H.\n\n\n\nwe let Pt=probability of getting a T string, and Ph=probability of getting an H string\n\nso this gives us another equation: (Ph)=(Pt/2+Pt/4)+1/8\u0003\n\n\n\nsolving the two equations gives us Pt=1/10, Ph=1/5. so add them together, and the answer is 3/10[/hide]", "Solution_3": "[hide]\n\nQ3: this question isn't a very good question but anyway: \n\nConsider the set of palindromic numbers less than 10^9. \n\nWe can choose the first 5 digits, each corresponds to a unique such number. So there are 10^5-1 palindromic numbers.\n\nConsider the set of cubes. We have 1^3, 2^3 ... (10^3)^3. \n\nThis gives us 10^3 cubes. \n\nConsider the set of primes. There are approximately 10^9 / In (10^9) primes \n\nwhich is approximately 48,254,942 = 1/2 * 10^8. \n\nSo the sets in decreasing order are: the set of primes, the set of palindromes then the set of the cubes. \n\n[/hide]", "Solution_4": "i have no idea how this got double posted.", "Solution_5": "[quote=\"vinoth_90_2004\"]\nQ3: this question isn't a very good question but anyway: \n[/quote]\r\n\r\nOut of curiosity, why do you believe this to be a poor question?", "Solution_6": "I agree with Vinoth's conclusion. \n\n\n\nHowever, I see his method of counting palindromes only counting those with an odd number of digits. Eg, selecting the five digits 00014 would generate the palindrome 141; but I don't see how would 1441 be generated by this method. \n\n\n\nExtending Vinoth's method, I get [hide](10^5 - 1) + (10^4 - 1) = 109998[/hide]", "Solution_7": "The solution provided with the test runs as follows:\r\n\r\nLet us estimate how many of these types of numbers we have which are less than or equal to $n$ for some large $n$. For palindromic numbers, it is about $\\sqrt{n}$ since only the last half of the digits matter. For perfect cubes, the answer is certainly $\\sqrt[3]{n}$. For prime numbers, the answer is known to be $n/\\ln n$, from number theory. Since $\\ln n \\ll n^c$ for any $c>0$ (for large $n$), we have\r\n\\[\r\n\\frac{n}{\\ln n} \\gg \\sqrt{n} \\gg \\sqrt[3]{n}\r\n\\]", "Solution_8": "[quote=\"boris\"]The solution provided with the test runs as follows:\n\nLet us estimate how many of these types of numbers we have which are less than or equal to $n$ for some large $n$. For palindromic numbers, it is about $\\sqrt{n}$ since only the last half of the digits matter. For perfect cubes, the answer is certainly $\\sqrt[3]{n}$. For prime numbers, the answer is known to be $n/\\ln n$, from number theory. Since $\\ln n \\ll n^c$ for any $c>0$ (for large $n$), we have\n\\[\n\\frac{n}{\\ln n} \\gg \\sqrt{n} \\gg \\sqrt[3]{n}\n\\][/quote]\r\n\r\nOf course, using the PNT on a high school math competition is a little weird, but whatever...\r\n\r\nBTW, were any of those questions yours?", "Solution_9": "The provided solution to the Pascal's Triangle question runs as follows:\r\n\r\nWe need to find the number of coefficients in the polynomial\r\n\\[ (1+x)^{100}= 1 + {100\\choose 1}x + {100\\choose 2}x^2 + \\dots + x^{100} \\]\r\nwhich are not equal to 0 modulo 3. Note that modulo 3 one has\r\n\\[(1+x)^3= 1+ 3x+3x^2+ x^3 \\equiv 1+x^3\\]\r\n(this is called Freshman's Dream sometimes, or the high school student's binomial theorem), and so also\r\n\\[(1+x)^9 \\equiv (1+x^3)^3 \\equiv 1+x^9,\\]\r\netc, for any power of 3. Now, $100= 81+2\\cdot 9 + 1$. Therefore, modulo 3 one has\r\n\\[ (1+x)^{100}= (1+x)^{81} \\left((1+x)^9\\right)^2 (1+x) =\r\n(1+x^{81}) (1+ 2x^{9} + x^{18}) (1+x) \\]\r\nIn this product all $2\\cdot 3 \\cdot 2=12 $ powers of $x$ are different (because every integer can be written in base 3 in a unique way), and the coefficients are all nonzero modulo 3. So, the answer is 12." } { "Tag": [ "modular arithmetic", "quadratics" ], "Problem": "Prove that\r\n\r\n$1+10+10^{2}+10^{3}+\\cdots+10^{n}$\r\n\r\ncan never be a perfect square.", "Solution_1": "$1+10+10^{2}+10^{3}+\\cdots+10^{n}=4k+3$\r\nfor some $k$", "Solution_2": "[hide]If you put the expression in (mod 4) you get\n$1+2+0+0+0+...0 \\equiv 3 (mod 4)$\n\nWhich cannot be a perfect square. \n\n(When n=0, the expression is a perfect square, but that's not really relevant) [/hide]", "Solution_3": "[hide]Every term $10^{k}$, for k>1, is $\\equiv 0 \\pmod{4}$.\nThus, expression is $\\equiv 1+10\\equiv 3 \\pmod{4}$, but quadratic residues mod 4 are 0 and 1.[/hide]" } { "Tag": [ "modular arithmetic", "induction", "number theory unsolved", "number theory" ], "Problem": "Let $ p$ be an odd prime. Prove that:\r\n(1) $ (1 \\plus{} p)^{p^{n \\minus{} 1}} \\equiv 1 \\pmod{p^n}$\r\n(2) $ (1 \\plus{} p)^{p^{n \\minus{} 2}} \\neq 1 \\pmod {p^n}$\r\n\r\nto prove (1), it's sufficient to show that $ p^{n \\minus{} i}$ divides $ \\binom{p^{n \\minus{} 1}}{i}$ for 04$ then it has a third prime factor\r\n5-If $ p\\equal{}3$ $ 8p\\minus{}1\\equal{}23$ $ 8p\\plus{}1\\equal{}25$ not prime...\r\nIf $ p\\equal{}1(mod 3)$ $ 8p\\plus{}1\\equal{}0 (mod 3)$ and because $ 8p\\plus{}1>3$ it can't be prime\r\nIf $ p\\equal{}2 ( mod 3)$ $ 8p\\minus{}1\\equal{} 0( mod 3)$ and since $ 8p\\plus{}1>3$ it is not prime so there's nothing to prove\r\n\r\nWhen I have more time I will post 2 and 3\r\n\r\nDaniel", "Solution_2": "Well, for number 4), consecutive primes such as , $ 7\\plus{}11, 13\\plus{}17, 31\\plus{}37$ work also so I don't think it is necessarily $ p \\plus{} (p\\plus{}2)$.\r\n\r\nI think its more $ p \\plus{} (p\\plus{}2k)$\r\nso $ 2(p\\plus{}k)$... Guess it remains to be proven for 2 cases, if $ k\\equiv 0 \\pmod{2}$ and $ k \\equiv 1 \\pmod{2}$", "Solution_3": "[hide=\"1\"] All primes must be either 1,3,5 or 7 (mod 8). By Pidgeonhole, at least two of the primes are the same when taken (mod 8). Thus, their difference is 0 (mod 8), and we are done.[/hide]\n[hide=\"4\"] WLOG, let p3$, $ 8p\\plus{}1$ will be composite.\n\nIf $ p \\equal{} 2 (mod 3)$ then $ 8p\\minus{}1\\equal{}0 (mod 3)$, so since $ 8p\\minus{}1>3$, $ 8p\\minus{}1$ will not be prime, so it doesn't matter what $ 8p\\plus{}1$ is in this case. [/hide]", "Solution_7": "[quote=\"nayrb\"]The sum of all pairs of twin primes will have a factor of twelve. Every prime can be represented as 6k-1 or 6k+1. A twin prime pair must be represented as p=6k-1 and q=6k+1, otherwise the difference between them wouldn't be 2. Thus, the sum is p+q=12k.\n\nDo you think the fact that p=6k-1 or 6k+1, or, p=4k-1 or 4k+1, can be used to prove the case that the sum of any two consecutive prime has three prime factors?[/quote]\r\n\r\nI think after that you must mechanically show the sum of the twin primes $ 3 and 5$ have 3 prime factors, namely $ 2^3$. :)", "Solution_8": "[hide=\"3\"]\n$ \\sqrt {k^2 \\minus{} pk} \\equal{} n \\implies k(k \\minus{} p) \\equal{} n^2$. Hence we need both $ k$ and $ k \\minus{} p$ to be perfect squares. Let $ k \\equal{} r^2$ and $ k \\minus{} p \\equal{} s^2$. Then we have $ r^2 \\minus{} p \\equal{} s^2 \\implies r^2 \\minus{} s^2 \\equal{} p \\implies (r \\minus{} s)(r \\plus{} s) \\equal{} p$ so equating factors:\n\nCase 1: $ r \\minus{} s \\equal{} 1$ and $ r \\plus{} s \\equal{} p$\nso $ r \\equal{} \\frac {p \\plus{} 1}{2}$ and $ s \\equal{} \\frac {p \\minus{} 1}{2}$. $ k \\equal{} r^2 \\implies k \\equal{} \\frac {(p \\plus{} 1)^2}{4}$\n\nSince $ n \\equal{} rs$, $ n \\equal{} \\frac {(p \\plus{} 1)(p \\minus{} 1)}{4}$, which is an integer as long as $ p \\neq 2$\n\nCase 2: $ r \\minus{} s \\equal{} p$ and $ r \\plus{} s \\equal{} 1$\nThen $ r \\equal{} \\frac {p \\plus{} 1}{2}$ and $ s \\equal{} \\frac {1 \\minus{} p}{2}$. $ k$ is the same as in case 1. $ rs < 0$ for all $ p$ but $ n \\ge 0$, so this case gives no solutions.\n\nCase 3: $ r \\minus{} s \\equal{} \\minus{} 1$ and $ r \\plus{} s \\equal{} \\minus{} p$\n$ r \\equal{} \\minus{} \\frac {p \\plus{} 1}{2}$ and $ s \\equal{} \\minus{} \\frac {p \\minus{} 1}{2}$ which gives the same solutions as case 1.\n\nCase 4: $ r \\minus{} s \\equal{} \\minus{} p$ and $ r \\plus{} s \\equal{} \\minus{} 1$\nThen $ r \\equal{} \\minus{} \\frac {p \\plus{} 1}{2}$ and $ s \\equal{} \\frac {p \\minus{} 1}{2}$ so $ rs < 0$ for all $ p$ but $ n \\ge 0$, so this case gives no solutions.\n\nSo our only solution so far is $ \\boxed{k \\equal{} \\frac {(p \\plus{} 1)^2}{4}}$ as long as $ p \\neq 2$. But we only considered the case where both factors are perfect squares, which is only necessary when neither are $ 0$. Thus we have to add in the solutions $ \\boxed{k \\equal{} p}$ and $ \\boxed{k \\equal{} 0}$.\n\n\n[/hide]", "Solution_9": "[hide=\"oops.. some corrections for 3\"]\nAs I noted, we have $ k(k\\minus{}p)\\equal{}n^2$. But this does not necessarily imply that $ k$ and $ k\\minus{}p$ are perfect squares. Instead, note that every whole number can be written as the product of a perfect square and a square-free number. Thus, $ k\\equal{}r^2t$ and $ k\\minus{}p\\equal{}s^2t$ - the squarefrees have to be the same to give a perfect square product. Hence we have $ r^2t \\minus{} s^2t \\equal{} p$, so $ t|p$ and thus $ t\\equal{}\\pm 1$ or $ t\\equal{}\\pm p$.\n\nIn the case $ t\\equal{}1$, we proceed as in my last post.\n\nIn the case $ t\\equal{}\\minus{}1$, we have $ s^2 \\minus{} r^2 \\equal{} p$ so equating factors we have\n(1) $ s\\minus{}r\\equal{}1$, $ s\\plus{}r\\equal{}p$ so $ r\\equal{}\\frac{p\\minus{}1}{2}$ and $ k\\equal{}\\minus{}\\frac{(p\\minus{}1)^2}{4}$\n(2) $ s\\minus{}r\\equal{}\\minus{}1$ $ s\\plus{}r\\equal{}\\minus{}p$ so $ r\\equal{}\\minus{}\\frac{p\\minus{}1}{2}$ and thus $ k\\equal{}\\minus{}\\frac{(p\\minus{}1)^2}{4}$\n(3) $ s\\minus{}r\\equal{}p$, $ s\\plus{}r\\equal{}1$ so $ r\\equal{}\\minus{}\\frac{p\\minus{}1}{2}$ and thus $ k\\equal{}\\minus{}\\frac{(p\\minus{}1)^2}{4}$\n(4) $ s\\minus{}r\\equal{}\\minus{}p$, $ s\\plus{}r\\equal{}\\minus{}1$ so $ r\\equal{}\\frac{p\\minus{}1}{2}$ and thus $ k\\equal{}\\minus{}\\frac{(p\\minus{}1)^2}{4}$\n\nIn the case $ t\\equal{}p$, we have $ r^2 \\minus{} s^2 \\equal{} 1$ and equating factors (1) $ r\\plus{}s\\equal{}1$ and $ r\\minus{}s\\equal{}1$ or (2) $ r\\plus{}s\\equal{}\\minus{}1$ and $ r\\minus{}s\\equal{}\\minus{}1$. In case (1), $ r\\equal{}1$ and $ s\\equal{}0$, giving $ k\\equal{}p$ (which I noted was a solution earlier). In case (2), $ r\\equal{}\\minus{}1$ and $ s\\equal{}0$, which also gives $ k\\equal{}p$\n\nIn the case $ t\\equal{}\\minus{}p$, we have $ s^2\\minus{}r^2 \\equal{} 1$ and equating factors\n(1) $ s\\minus{}r\\equal{}1$ and $ s\\plus{}r\\equal{}1$ in which case $ r\\equal{}0$ and thus $ k\\equal{}0$\n(2) $ s\\minus{}r\\equal{}\\minus{}1$ and $ s\\plus{}r\\equal{}\\minus{}1$ in which case $ r\\equal{}0$ and thus $ k\\equal{}0$\n\nIn summary: $ \\boxed{k\\equal{}\\minus{}\\frac{(p\\minus{}1)^2}{4}, 0, p, \\frac{(p\\plus{}1)^2}{4}}$\n[/hide]" } { "Tag": [ "inequalities", "FTW", "Asymptote" ], "Problem": "$ \\frac{\\minus{}x\\minus{}9}{(x\\plus{}1)(x\\minus{}3)}\\geq 0$\r\n\r\nI tried to graph these, the solution to where both in the denominator is greater than $ 0$ is $ (\\minus{}\\infty, \\minus{}1) \\cup (3, \\plus{}\\infty)$. Both in the denominator to be less than zero is $ (\\minus{}1,3)$ and the one in the numerator to be always positive $ (\\minus{}\\infty, \\minus{}9]$. I don't know how to tie all these in together for the final answer.", "Solution_1": "[hide=\"sign chart ftw\"]\nYou forgot about if the top is negative.\n \n<~~~-9~~~~~~~~~-1~~~~3~~~>\n-1 and 3 are asymptotes, -9 is a zero.\nTry a random value from each segment.\n-10 returns positive.\n-2 returns negative.\n0 returns positive.\n4 returns negative.\n\nSo the segments that work are $ (\\minus{}\\infty,9] \\cup (\\minus{}1,3)$\nThe second part has round brackets because -1 and 3 are asymptotes.\n\n[/hide]", "Solution_2": "I was so lost with 7 or so of my graphs for this problem. \r\n\r\nyour method is really simple and clear. Would it work for any types of inequality equations?", "Solution_3": "Yep, just graph all the x-intercepts and asymptotes, take any value from each section, then put it together - asymptotes with round brackets and x-intercepts with square brackets if the equation is $ \\ge$ or $ \\le$", "Solution_4": "What if the equation is something like:\r\n\r\n$ \\frac { \\minus{} x \\minus{} 9}{|x \\plus{} 1|(x \\minus{} 3)}\\geq 0$", "Solution_5": "The sign chart will change so that the sections to the right of -1 and the left of -1 are merged, but -1 remains out of the domain. ... I think." } { "Tag": [ "videos" ], "Problem": "Has anyone here ever seriously tried using it to compose and create high-quality music? While the software itself is a little hard to use, I find that it has a great deal of potential, as it not only removes the human element from song production but also allows extremely wide vocal ranges to be achieved without overzealous autotuning. Many very fine adjustments can be made within the UI itself if needed, and many excellent creations such as the below can be found on youtube amongst other video sharing sites. (I'm not going to bother with the 3DPVs here since a lot of them have audio that were further enhanced.) The English vocaloids are a little bit \"noisier\", but still quite excellent nevertheless.\r\n\r\n[youtube]qeWlez-s-2E[/youtube]\r\n\r\n[youtube]HGMd1w-T6rg[/youtube]", "Solution_1": "Heh; I was just listening to vocaloids. http://www.youtube.com/watch?v=teA-BD5XSQk\r\nI would use vocaloids, but I don't have too much time to look for a download, and I'm paranoid of the virus factor.\r\nEDIT: Another offsetting topic is that generally, only the Japanese versions of vocaloids are good, meaning that if you wanted to make a good song with a vocaloid, then it'd either have to have an accent, or be in Japanese.", "Solution_2": "I've never really tried but I do love Vocaloid\nHatsune Miku, Kaito...\n:) Gotta love Japan!", "Solution_3": "i actually am a vocaloid producer! i've never published songs on nico nico or youtube but i do it as more of a personal hobby.", "Solution_4": "I like vocaloids, tried downloading it to use in my songs but didn\u2019t work", "Solution_5": "[quote=UnknownUser123]I like vocaloids, tried downloading it to use in my songs but didn\u2019t work[/quote]\n\ndid you pay? if it doesnt work maybe you can try utau, the free version of vocaloid. it works pretty much the same, but it's free. there are some pretty good utauloids out there too. (sorry if i'm being unhelpful)", "Solution_6": "I use free" } { "Tag": [ "analytic geometry", "graphing lines", "slope", "vector", "calculus", "integration" ], "Problem": "If Bob makes a round trip from point A to B in 12 hours and travels at 3 mph while going on a upwards slope, 9 mph on a downwards slope, and 9/2 mph on a horizontal surface, what is the distance between A and B?\r\n\r\nWhat do you think? (Before I hand it in)\r\n\r\n[hide=\"explanation\"] Denote the upwards distance when going towards point B as a, the horizontal distance as b, and downwards distance as c, and D as the distance between A and B. Then:\n\nD = a + b + c. \n\nBy the d=rt,\n\na+b=3t [size=59]1[/size] , since while going back to A, Bob will go up the distance of b.\n\na+b=9t [size=59]2[/size]\n\n2c=4.5t [size=59]3[/size]\nDividing into both sides on all three equations yields \n\n(a+b)/3 = t [size=59]1[/size]\n(a+b)/9 = t [size=59]2[/size]\n2c/4.5 = t [size=59]3[/size]\n\nIf we add t [size=59]1[/size], t [size=59]2[/size], and t [size=59]3[/size] together, we know it equals 12, since we were given in the problem that it takes Bob 12 hours.\nSo,\n\n(a+b)/3 + (a+b)/9 + 2c/4.5 = 12 \n Multiplying by 9...\n3a+3b+a+b+4c=108 \n4(a+b+c)=108\na+b+c=27\nD=a+b+c=27 miles. [/hide]", "Solution_1": "[quote=\"darkmatter47\"]If Bob makes a round trip from point A to B in 12 hours and travels at 3 mph while going on a upwards slope, 9 mph on a downwards slope, and 9/2 mph on a horizontal surface, what is the distance between A and B?\n\nWhat do you think? (Before I hand it in)\n\n[hide=\"explanation\"] Denote the upwards distance when going towards point B as a, the horizontal distance as b, and downwards distance as c, and D as the distance between A and B. Then:\n\nD = a + b + c. \n\nBy the d=rt,\n\na+b=3t [size=59]1[/size] , since while going back to A, Bob will go up the distance of b.\n\na+b=9t [size=59]2[/size]\n\n2c=4.5t [size=59]3[/size]\nDividing into both sides on all three equations yields \n\n(a+b)/3 = t [size=59]1[/size]\n(a+b)/9 = t [size=59]2[/size]\n2c/4.5 = t [size=59]3[/size]\n\nIf we add t [size=59]1[/size], t [size=59]2[/size], and t [size=59]3[/size] together, we know it equals 12, since we were given in the problem that it takes Bob 12 hours.\nSo,\n\n(a+b)/3 + (a+b)/9 + 2c/4.5 = 12 \n Multiplying by 9...\n3a+3b+a+b+4c=108 \n4(a+b+c)=108\na+b+c=27\nD=a+b+c=27 miles. [/hide][/quote]\r\nIf you look at the rates, you will notice that no matter what terrain in his way, the distance would always be 27 miles.", "Solution_2": "I realize this might be beyond the scope of middle school math, but this is quite similar to conservative vector fields and path integrals in calculus and physics (it doesn't matter how you get there, if the two endpoints are the same, the distance to get there is the same)." } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Show that for any $ r\\in \\mathbb{Q}$ there exist a triangle for which it's sidelengths are rational numbers and it's area is equal to $ r$.", "Solution_1": "It is interesting problem, bu i have not solved yet. Who can help me...." } { "Tag": [ "MATHCOUNTS" ], "Problem": "If n is an integer, what is the greatest even number of the form 2^n+3^n?", "Solution_1": "I think there's only one...", "Solution_2": "so whats your answer?", "Solution_3": "[hide]2? 2^0+3^0[/hide]", "Solution_4": "yes, explain your reasoning. should only take a few secs", "Solution_5": "[quote=\"Syntax Error\"]so whats your answer?[/quote]\r\n2, because 2^n is always even (even*even=even) and 3^n always equal odd (odd*odd=odd) and even*odd=odd, so 0 is the only one in which both components equal \"an odd number!\" :) (from SBSP) giving us an even answer.\r\n\r\n(also negative n would give you something <2)", "Solution_6": "read the question again. 2^n+3^n, where n is 0, would bring 2", "Solution_7": "Right, I mean n=0, I mean, I'm not [b]that[/b] stupid :) \r\n\r\n...on second thought, you might want to purchase the \"I'M WITH THE DUMMY :arrow: \" shirt from Patrick :)", "Solution_8": "ah, who's patrick?", "Solution_9": "Aaaaaaarg! (I bet you won't even understand that one :) )\r\n\r\nPatrick Star? from Spongebob Squarepants? Does no one watch that show!? I mean, at least half of us at my school sees it, even some of the teachers :)", "Solution_10": "i did for a very very short while. but i dont like to talk about that", "Solution_11": "Why not? I mean, he's like the symbol of goodness, happiness, and friendliness. If everyone was like Spongebob, then there wouldn't need a need for Heaven.", "Solution_12": "has to be 2 else 3^n is odd and 2^n is even, producing an odd sum" } { "Tag": [ "logarithms" ], "Problem": "2^6+2^6= 4 to the power of x. \r\n\r\nsolve for x. \r\n\r\n\r\nhow would you do this?", "Solution_1": "Well, you know that $2^{6}+2^{6}=2(2^{6})$ that is $2^{7}$.\r\n\r\n$2^{7}=4^{x}$\r\n\r\n$2^{7}= ((2^{2})^{x})$\r\n\r\n$2^{7}=2^{2x}$\r\n\r\n7=2x \r\nso x is 3.5 or 7/2", "Solution_2": "[quote=\"mr. math\"]2^6+2^6= 4 to the power of x. \n\nsolve for x. \n\n\nhow would you do this?[/quote]\r\n[hide]$2^{6}+2^{6}=4^{x}$\n\n$2^{7}=2^{2x}$\n\n$7=2x$\n\n$x=\\frac{7}{2}$[/hide]", "Solution_3": "[hide] $2^{6}+2^{6}=2^{7}$\n$2^{7}=4^{x}\\implies2^{7}=2^{2x}\\implies7=2x\\implies x=3,5$[/hide]", "Solution_4": "[quote=\"mr. math\"]2^6+2^6= 4 to the power of x. \n\nsolve for x. \n\n\nhow would you do this?[/quote]\r\n\r\n[hide]$2^{6}+2^{6}=4^{x}$\n$2^{6}+2^{6}=2^{7}$\n$2^{7}=4^{x}$\n$4^{x}=2^{2}x$\n$2x=7$\n$x=\\frac{7}{2}$ or $3.5$.[/hide]", "Solution_5": "Are logs beyond HSB?\r\n\r\n[hide]$2^{6}+2^{6}=4^{x}$\n$2^{7}=4^{x}$\n$\\log_{2}2^{7}=\\log_{2}4^{x}$\n$7=2x$\n$x=3.5$[/hide]", "Solution_6": "[quote=\"bpms\"]Are logs beyond HSB?[/quote]\r\n\r\n\r\nYou don't need logs. :D", "Solution_7": "@bpms: It is obvious that $2x=7$ :wink:" } { "Tag": [ "function", "calculus", "logarithms", "factorial", "integration", "algebra", "functional equation" ], "Problem": "I was playing around with functions and saw this: let *f*(x)= f(x+1/f(x))-f(x). Then when I plot *x^x* it looks like log(x)+1 for the values my computer can plot (I did not have enough memory for x>9). But when I plot *x!* the graph looks like log(x). How can this be? Is this related to the stirling formula?", "Solution_1": "It's straightforward calculus, at least in the first case; your $f^{*}(x)$ is approximately $\\frac{f'(x)}{f(x)}=\\frac{d}{dx}\\ln f(x)$.\r\nThis is indeed related to the basic form of the Stirling formula.\r\n\r\nFor the factorial, you need to interpolate with the $\\Gamma$ function; $\\Gamma(x)=\\int_{0}^{\\infty}t^{x-1}e^{-t}\\,dt$, and $\\Gamma(n+1)=n!$ for integers $n$. We have the functional equation $\\Gamma(x+1)=x\\Gamma(x)$, so $\\int_{x}^{x+1}\\frac{d}{dt}\\ln\\Gamma(t)\\,dt=\\ln x$.\r\n\r\nThis definition is very smooth, and we have that $\\ln\\Gamma$ is actually convex. That gives $\\ln(x-1)<\\Gamma^{*}(x)<\\ln x$.\r\n\r\nFor that convexity: $\\frac{d^{2}}{dx^{2}}\\Gamma(x)=\\frac{d}{dx}\\frac{\\int_{0}^{\\infty}\\ln t\\cdot t^{x-1}e^{-t}\\,dt}{\\int_{0}^{\\infty}t^{x-1}e^{-t}\\,dt}= \\frac1{\\Gamma(x)^{2}}\\left(\\int_{0}^{\\infty}(\\ln t)^{2}t^{x-1}e^{-t}\\,dt\\cdot\\int_{0}^{\\infty}t^{x-1}e^{-t}\\,dt-\\left(\\int_{0}^{\\infty}\\ln t \\cdot t^{x-1}e^{-t}\\,dt\\right)^{2}\\right)$\r\n\r\nThis is positive because it is a variance; if $\\gamma$ is a random variable with density function $\\frac1{\\Gamma(x)}t^{x-1}e^{-t}$, this is the variance of $\\ln\\gamma$. We could also use Cauchy-Schwarz to prove it.\r\nIncidentally, $\\frac{d}{dx}\\Gamma(x)$ is the expected value of $\\ln\\gamma$, and the expected value of $\\gamma$ is $x$." } { "Tag": [ "symmetry" ], "Problem": "Andre is a butcher and president of a committee, which also includes a grocer, a baker, and a candlestick maker. All are sitting around a square table, one on each side of the table. Andre sits on Charmin\u2019s left, Barton sits at the grocer\u2019s right, and Duclos, who faces Charmin, is not the baker. Who sits across the table from the butcher? \r\n\r\nA. the baker B. Duclos C. the grocer D. the candlestick maker E. Charmin", "Solution_1": "[hide=\"Solution\"]By symmetry, assume WLOG Andre is at the top of the square (if we look down upon it). Charmin is on the left of the square. Duclos is on the right of the square. Barton faces Andre. We know Charmin must therefore be the grocer, so Barton is the baker. Answer: $ \\boxed{A}$.\n\n[/hide]" } { "Tag": [], "Problem": "Outline a synthesis of 3-hydroxyacetophenone from:\r\n\r\n1) Benzene.\r\n2) Open chain precursors.", "Solution_1": "treat benzene with nitrating mixture H2SO4 HNO3 to get nitrobenzene.\r\nReact it with CH3COCl in presence of AlCl3 to get a meta substituted product.\r\nSnCl2/HCl to reduce nitro group to amino (not sure if it would reduce carbonyl group)\r\nDiazotization followed by addition of water to convert the amino group to hydroxy group", "Solution_2": "[quote=\"Zimrock\"]React it with CH3COCl in presence of AlCl3 to get a meta substituted product. SnCl2/HCl to reduce nitro group to amino (not sure if it would reduce carbonyl group) \nDiazotization followed by addition of water to convert the amino group to hydroxy group\n[/quote]\r\n\r\n1) Fridel-Crafts reactions don't work when a meta directing group is on the ring.\r\n2) The carbonyl would also probably be reduced, and what would happen between an amine and a ketone in the reaction mixture?", "Solution_3": "Outline a synthesis of 3-hydroxyacetophenone from: \r\n1) Benzene. \r\n2) Open chain precursors.\r\n\r\nIs this right?\r\n1. convert benzene into toluene by friedel and crafts.\r\n2. convert toluene into benzoic acid by oxidation.\r\n3. sulphonate using fuming H2SO4 in the meta posn.\r\n4. Use Ruzicka rn ( CH3COOH ) to convert COOH into COCH3\r\n5. Use i) aq. KOH ii) H+ to convert into the reqd. compound.\r\n\r\nAlternately\r\nCan we just acylate benzene into Acetophenone and then Sulphonate and complete the rn?", "Solution_4": "Madness..........\r\nIn wikipedia it is given that Ruzcika reaction is the conversion of a dicarboxylic acid into a cyclic ketone...\r\nHow is it applicable here???\r\ncan you please explain????????????????? :(", "Solution_5": "[quote=\"Madness\"]Alternately Can we just acylate benzene into Acetophenone and then Sulphonate and complete the rn?[/quote]\r\n\r\nYou could, but what happen with acetophenone in the presence of base?", "Solution_6": "Would it undergo aldol? :wink: \r\nSo can i just protect the C=O grp using ethylene glycol and then carry out the process?", "Solution_7": "hey madness......................\r\nexplain how Ruzicka reaction is applicable here........", "Solution_8": "You dont even have a jug in your house then how will you know? :wink: :D", "Solution_9": "I am asking because in wikipedia it is given that it is conversion of a dicarboxylic acid into a cyclic ketone...", "Solution_10": "[quote=\"valeriummaximum\"]I am asking because in wikipedia it is given that it is conversion of a dicarboxylic acid into a cyclic ketone...[/quote]\r\nhey but Ruzika reaction is given in pradeeps itself no???dunno abt schoo textbook!!! but it is the preparation of aldehyde or ketone from dicarboxylic acid using MnO2 no???when you use formic acid you get an aldehyde??? :D", "Solution_11": "see in wikipedia srinitrs......\r\nthe catalyst is ThO2", "Solution_12": "[quote=\"valeriummaximum\"]see in wikipedia srinitrs......\nthe catalyst is ThO2[/quote]\r\nbut MnO2 is also correct no??", "Solution_13": "cut out the nos, will you. as far as i no( :P ),tho2 is used as the catalyst" } { "Tag": [ "geometry", "summer program", "Mathcamp", "PROMYS", "AMC", "AMC 10", "number theory" ], "Problem": "I heard that last year AMSP let you do 4 courses of each discipline (number theory, geometry, algebra, counting). However this year they are only letting you pick 2 courses, to go more indepth. But are you allowed to choose 4 classes if you want to, or is that not possible?", "Solution_1": "ummmmm... \r\nlast year we could only take two courses as well... we could choose 2 out of the four categories though.", "Solution_2": "By the way, is there some way I can select these courses beforehand?", "Solution_3": "well, \r\nyou select your courses before camp. packet should come.. \r\nyou get to camp. you take the classes and you can change if you want.", "Solution_4": "btw does anyone know the acceptance rating of AMSP? And also for other camps like Mathcamp PROMYS and such?\r\n\r\nI am going to attend AMSP but im just wondering how the acceptance rating is related to other camps. (also with the number of people applying)", "Solution_5": "First of all, about the courses, I e-mailed Dr. Andreescu few days earlier...\r\n\r\nHe said they are still finalizing academic plans for the 2009 AMSP, but only 2 courses is right.\r\n\r\nAlso, about the acceptance rate.\r\n\r\nAll camps are pretty loose on their acceptance rates.. I mean it is a CAMP, not like a private school system or something. Most, if not all, people apply, get in. HOWEVER, you might have a tough time in camp, but you're there to learn, and it's still good for you, unless you don't even know what the prof is talking about.. Like you got a 90 on a AMC10 and taking the advanced courses.. That would be silly.", "Solution_6": "[quote=\"sjyoo93\"]\nAll camps are pretty loose on their acceptance rates.. [/quote]\r\n\r\nnot necessary. i mean like camps like RSI are extremely hard to get into. \r\nand mathcamp's acceptance rate is about ~30-50% according to their website...", "Solution_7": "If you think about it 50% is already very high. You don't need to do that well on the Mathcamp test to get in as well, so I heard. It's just to make sure there is SOME kind of minimum level at the camp.\r\nThe only truly selective camps out there are research camps like RSI, which there are for other subjects I believe, and the summer training camps for various olympiads.\r\n\r\nHmmm...so if the classes are what it is on the website, I think I will do olympiad geo and nt?" } { "Tag": [ "analytic geometry", "geometry", "circumcircle", "geometry unsolved" ], "Problem": "[color=darkblue]Let $ (I; r)$ is incribed circle of $ ABC$ triangle. Let $ BC \\equal{} a, CA \\equal{} b, AB \\equal{} c$. Prove that: $ \\boxed{\\frac{IA^2}{bc}\\plus{}\\frac{IB^2}{ca}\\plus{}\\frac{IC^2}{ab}\\equal{}1}$[/color].", "Solution_1": "[color=darkblue][b]The easy proof.[/b] Prove easily that $ \\boxed {\\ IA^2 \\equal{} \\frac {bc(p \\minus{} a)}{p}\\ }$ a.s.o.\n\n[b]The high proof.[/b] The relation $ a\\cdot IA^2 \\plus{} b\\cdot IB^2 \\plus{} c\\cdot IC^2 \\equal{} abc$ is a particular case of the [b]Lagrange's identity[/b].[/color]\r\n\r\n[color=darkred]If $ M$ has the barycentrical coordinates $ (x,y,z)$ w.r.t. the triangle $ ABC$ , $ x \\plus{} y \\plus{} z \\equal{} 1$ , then for any $ X$ from the same plane, there is\n\nidentity $ \\boxed {\\ x\\cdot XA^2 \\plus{} y\\cdot XB^2 \\plus{} z\\cdot XC^2 \\equal{} XM^2 \\minus{} p_w(M)\\ }$ , where $ p_w(M)$ is the power of $ M$ w.r.t. the circumcircle of $ \\triangle ABC$ .\n\nParticularly, for $ X: \\equal{} M(x,y,z)$ obtain $ \\boxed {\\ x\\cdot MA^2 \\plus{} y\\cdot MB^2 \\plus{} z\\cdot MC^2 \\equal{} \\minus{} p_w(M)\\ }$ .[/color]\r\n\r\n[color=darkblue]If $ M: \\equal{} I\\left(\\ \\frac {a}{2p}\\ ,\\ \\frac {b}{2p}\\ ,\\ \\frac {c}{2p}\\ \\right)$ , then $ p_w(I) \\equal{} \\minus{} 2Rr$ $ \\implies$ $ \\sum a\\cdot IA^2 \\equal{} \\minus{} 2p\\cdot ( \\minus{} 2Rr) \\equal{} 4Rpr \\equal{} abc$ $ \\implies$ $ \\sum\\frac {IA^2}{bc} \\equal{} 1$ .\n\nNow you can find many another \"old\" identities for particular cases of the points $ M$ and $ X \\ !$[/color]", "Solution_2": "[color=darkblue]Thank Virgil Nicula very muck.[/color]" } { "Tag": [], "Problem": "is the gravitational potential energy of two particles:\r\n\r\nA) $FG= m1,m2/r^2$ B) $w = %Error. \"intergral\" is a bad command.\nr f(r) * dr$ C) $U = - GMm/r$ D) $T^2 = (4\\pi^2)/(GM)$", "Solution_1": "[color=blue] I think the answer is [b]C[/b], $U=-\\frac{GMm}{r}$ and $U=0$ at the $\\text{infinity}$[/color]" } { "Tag": [ "blogs" ], "Problem": "Let $p_{n}$ denote the $n$th prime number. For all $n \\ge 6$, prove that \\[\\pi \\left( \\sqrt{p_{1}p_{2}\\cdots p_{n}}\\right) > 2n.\\]", "Solution_1": "What is $ \\pi$ here? It's not $ 3.14...$ I guess, since then the problem is trivial (even for $ n<6$) so what is it then?", "Solution_2": "[quote=\"Peter\"]What is $ \\pi$ here? It's not $ 3.14...$ I guess, since then the problem is trivial (even for $ n < 6$) so what is it then?[/quote]\r\n\r\nYeah, it is not. Here, $ \\pi(x)$ denotes the number of primes $ \\leq x$.", "Solution_3": "For $ n \\equal{} 7,8,9$ this can easily be checked. So assume $ n\\ge 10$. Then we have that $ \\sqrt{p_{1}p_{2}\\cdots p_{n}}> 3^{n}$. \r\n\r\nNow, we know that for all positive integers $ m$, there is at least $ 2$ primes in $ ]m,3m]$ (strictly more for $ m>3)$. (I have to use this fact, I see no solution without?)\r\n\r\nSo, there are more than $ 2n$ primes in $ ]1,3]\\cup]3,9]\\cup\\cdots\\cup]3^{n\\minus{}1},3^{n}] \\equal{} ]1,3^{n}]$, thus $ \\pi\\left(\\sqrt{p_{1}p_{2}\\cdots p_{n}}\\right)\\ge\\pi (3^{n})> 2n$.", "Solution_4": "We can get by using the fact that $ p_{n\\plus{}1}<2p_n$. It follows that $ p_{2n\\plus{}2}<4p_{2n}$. For $ n\\ge 6$, since $ p_{n\\plus{}1}\\ge 17$, it follows that $ \\sqrt{p_{n\\plus{}1}}>4>\\frac{p_{2n\\plus{}2}}{p_{2n}}$. Thus, if it is true that $ \\sqrt{p_1p_2\\ldots p_n}>p_{2n}$, it is also true that $ \\sqrt{p_1p_2\\ldots p_np_{n\\plus{}1}}>p_{2n\\plus{}2}$. Since this is indeed true for $ n\\equal{}6$ ($ \\sqrt{30030}>37$), it is true for all $ n\\ge 6$ by induction.\r\n\r\nLuke\r\nSee my puzzle blog at [url]http://bozzball.blogspot.com[/url]", "Solution_5": "Very nice solution!" } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "let I a finitely generated ideal. Are there any sequence in I like this: i_1, i_2, i_3, ... witch i_n is not in and i_n in I ?\r\n\r\n =: the generated ideal by i_1, i_2, ..., i_n.[hide]i[/hide]", "Solution_1": "yes, in the unit ideal of a non-noetherian ring." } { "Tag": [ "logarithms", "trigonometry", "conics", "ellipse", "geometry", "probability", "trapezoid" ], "Problem": "1. A circle centered at the origin is tangent to the graph $ 25(x\\minus{}1)^2 \\minus{} 36y^2 \\equal{} 900$. Find the minimum radius of this circle. \r\n\r\n2. Find the period of the graph of $ y \\equal{} 12 \\sin(4\\pi x/7) \\plus{} 13 \\cos(2\\pi x/5\\plus{}\\pi/2)$.\r\n\r\n3. Evaluate $ \\sum_{k \\equal{} 1}^{9999} \\log(\\sin(\\pi k/10000))$ to the nearest integer given that log(2) \u2248 0.30103.\r\n\r\n4. An ellipse has major axis of length 122 and minor axis of length 120. A circle centered at a focus of the ellipse is tangent to it. What is the sum of all possible diameters of the circle?\r\n\r\n5. 17 points are spaced at equal intervals around a unit circle. From one point, segments are drawn to every other point. Find the sum of the squares of the lengths of these segments. \r\n\r\n6. In a two-person game, one person says an integer, and the other person must say an integer greater than or equal to 1/3 of the previous person's integer, but less than the previous person\ufffds integer. They go back and forth following this rule until one person says 1, and that person loses. If Johnny says 2004, what number should Bernie counter him with to guarantee a win for himself? \r\n\r\n7. ABC has AB = 75, BC = 78, CA = 51. DEF fully contains ABC and has DE parallel to AB, EF parallel to BC, and FD parallel to CA with a distance of 3 between corresponding sides. What is the area of DEF?", "Solution_1": "[hide=\"6.\"]\nLet's start off by trying to find a general pattern for wins and losses. Let's say you start off the game with any number you wish. If you say 1, obviously, you lose. However, if you say 2, your opponent is forced to say 1 and you win. If you say any number from 3-6, you lose because then your opponent says 2 and wins. If you say 7, your opponent is forced to say any number from 3-6 and loses. If you say any number from 8-21, you lose because then your opponent says 7 and wins, etc. Note that the pattern of winning numbers starts with 2, then you multiply by 3 and add 1 to obtain the next winning number and so on. Therefore, the sequence of winning numbers is 2, 7, 22, 67, 202, 607, 1822. 1822 is greater than one-third of 2004 and is less than 2004, and therefore, is Bernie's ideal choice.\n[/hide]", "Solution_2": "Here are 3 more:\r\n8. Let distinct x and y be drawn from $ \\{ 1,2,3,\\ldots,99\\}$ such that each ordered pair (x,y) is equally likely. If the probability that the sum of the units digits of x and y is less than 10 is $ \\frac {a}{b}$,where a and b are relatively prime, what is a+b? \r\n\r\n9. What is the smallest number d such that fewer than half of the positive integers with d-digits have all distinct digits?\r\n\r\n10. In trapezoid ABCD, $ AB\\parallel DC$ and $ AB < DC$, the lines AC and DB intersect at P. The sum of the areas of $ \\triangle APD$ and $ \\triangle BPC$ is $ 1996$. If the lengths of the bases $ \\overline{AB}$ and $ \\overline{CD}$ are integers and the distance between them is an integer, what is the minimum area of ABCD?\r\n\r\nFor #8 I got $ 109/198\\implies a \\plus{} b \\equal{} 307$ but that doesn't agree with the official answer...", "Solution_3": "[hide=\"9.\"]\nWe're just looking for the smallest number $ d$ which satisfies\n$ \\frac{9P^9_{d\\minus{}1}}{9*10^{d\\minus{}1}} < 0.5$\n$ P^9_3$ is just barely greater than $ \\frac{10^3}{2}$, so $ d\\minus{}1 \\equal{} 4$, $ d \\equal{} 5$.\n[/hide]", "Solution_4": "[hide=\"8.\"]\npairs for less than 10: 10*(9+19+29+39+49+58+68+78+88)+9*98=5252\ntotal: 99*98=9702\nprob.: $ \\frac{5252}{9702}\\equal{}\\frac{2626}{4851}$\na+b=7477\n\nIs this right? It's a really large number.[/hide]", "Solution_5": "Wow, I was being really careless (and probably stupid) - the denominator in my original solution was $ \\binom{100}2$ when it should have been $ \\binom{99}2$ and that's why I got $ 1 \\minus{} \\frac {2000 \\plus{} 450/2}{50\\cdot99} \\equal{} 109/198$...\r\nHere's my reasoning anyway: find all possible pairs mod 10 where the sum is at least 10 - if the two pairs are distinct, we have $ (9,1),...(9,8),(8,2),...(8,7),... ...(6,4),(6,5)\\implies 20$ pairs. So the overall total pairs with distinct unit digits is $ 20\\cdot10\\cdot10$. If the unit digits are the same we have $ (9,9),(8,8),...,(6,6),(5,5)\\implies 5$ pairs. So in this case we have $ 5\\cdot 10\\cdot9/2$ - that's how I got the numerator.", "Solution_6": "11. The sides of triangle $ ABC$ have length $ AB \\equal{} 7$, $ BC \\equal{} 8$, and $ AC \\equal{} 9$. What is the distance between the incenter and the orthocenter of $ \\triangle ABC$?\r\n\r\n12. How many triples of natural numbers, $ (a,b,c)$ are there such that $ a \\plus{} b \\plus{} c \\equal{} 111$ and $ a$, $ b$, and $ c$ are in geometric progression?\r\n\r\n13. Solve the system of equations\r\n\r\n$ 4y\\plus{}gx\\equal{}34$\r\n$ \\frac{g^{4}x^{2}\\minus{}16y^{2}}{g^{2}x\\plus{}4y}\\equal{}\\minus{}10$\r\n\r\nin integers $ (g,x,y)$\r\n\r\nFor #11 I got $ \\sqrt {105}/10$ but that does not agree with the official answer... For #12 I think there are 4 but I don't know of a good way to find them all... \r\nBTW @ rpond: I did exactly the same way but I find it difficult to find $ d \\equal{} 5$ without a calculator... (or the fact that $ P_3^9$ is just barely $ > 10^3/2$) how did you find it anyway?", "Solution_7": "$ P^9_3$ is pretty easy to calculate. You can think of it like this:\r\n$ P^9_3 \\equal{} 9*8*7 \\equal{} 8*(8 \\plus{} 1)*(8 \\minus{} 1) \\equal{} 8*(8^2 \\minus{} 1) \\equal{} 512 \\minus{} 8$ or $ 8*63 \\equal{} 504$\r\nAnd of course $ \\frac {10^3}{2} \\equal{} \\frac {1000}{2} \\equal{} 500$.\r\nFrom there, $ P^9_4 \\equal{} 6*P^9_3$ and $ \\frac {10^4}{2} \\equal{} 10*\\frac {10^3}{2}$, so it's safe to assume that the latter will be greater.\r\n\r\nEDIT:\r\n[hide=\"13.\"]\n$ 4y\\plus{}gx\\equal{}34$\n$ \\frac{g^{4}x^{2}\\minus{}16y^{2}}{g^{2}x\\plus{}4y}\\equal{}\\minus{}10$\n$ g^4x^2\\minus{}16y^2 \\equal{} (g^2x\\minus{}4y)(g^2x\\plus{}4y) \\equal{} \\minus{}10(g^2x\\plus{}4y)$\n$ g^2x\\minus{}4y \\equal{} \\minus{}10$\nAdding to the first equation yields\n$ g^2x\\plus{}gx \\equal{} gx(g\\plus{}1) \\equal{} 24$.\n$ g(g\\plus{}1)$ has to be a factor of $ 24$. $ 12 \\equal{} 3*4 \\equal{} \\minus{}4*\\minus{}3$, $ 6 \\equal{} 2*3 \\equal{} \\minus{}3*\\minus{}2$, and $ 2 \\equal{} 1*2 \\equal{} \\minus{}2*\\minus{}1$ are all factors of $ 24$, so we have a grand total of $ 6$ $ g$ values:\n$ \\minus{}4, \\minus{}3, \\minus{}2, 1, 2, 3$.\nAdditionally, $ (g, x) \\equal{} (\\minus{}4, 2), (\\minus{}3, 4), (\\minus{}2, 12), (1, 12), (2, 4), (3, 2)$.\nNow, looking at the first equation in $ mod 4$, $ 0 \\plus{} ? \\equal{} 2$. Clearly $ gx$ cannot be divisible by $ 4$. Scanning through our pairs $ (g, x)$, we find that only one of them, $ (3, 2)$, will result in a value of $ gx$ that isn't divisible by $ 4$. Plugging these values in and solving for $ y$, we get our answer:\n$ (g, x, y) \\equal{} (3, 2, 7)$\n[/hide]", "Solution_8": "[hide=\"#1\"]the graph of $ 25(x \\minus{} 1)^2 \\minus{} 36y^2 \\equal{} 900$ is a hyperbola\nthe center is located at (1,0) and the tips of the hyperbola are 6 units horizontally from the center. Now we know the tips are at (-5,0) and (7,0). The nearest tip is (-5,0) so the radius of the circle must be 5[/hide]\r\nHope I got it right, I'm new to this.", "Solution_9": "14. In isosceles triangle $ ABC$ (AB=AC), the angle bisector of $ B$ intersects with $ AC$ at $ D$; we have $ BC \\equal{} BD \\plus{} AD$ , what is $ \\angle A$?\r\n\r\n15. From point $ O$ inside $ \\triangle ABC$ we draw perpendicular lines $ OM,ON, OP$ on to $ AB,AC,BC$ respectively. If $ \\angle AOB \\equal{} 60^{\\circ}, MN \\equal{} \\frac {\\sqrt {5}}{3}, OP \\equal{} 3, NP \\equal{} \\sqrt {3}, OM \\equal{} \\sqrt {5}$ what is $ \\angle BOC$?\r\n\r\n16. In $ \\prod_{1\\le i < j\\le 9}{x_i \\minus{} x_j}$ what is the coeffiecient of $ (x_1^4x_2^4\\cdots x_9^4)$?\r\n\r\n17. Find all n such that $ d(d(d(n)) \\equal{} d(d(n))$, where $ d(n)$ is the number of divisors of n.\r\n\r\n18. 7 people numbered 1 through 7 are standing in a line. We know the following: \r\nbetween the numbers 1 and 5, there is one person between the numbers 1 and 7, there are 3 people \r\nbetween the numbers 1 and 3 there is one person between the numbers 2 and 4 there is one person \r\nand between the numbers 6 and 4 there is one person. In how many ways could these 7 people be standing in the line? \r\n\r\n19. What is the maximum number of terms that can be picked from the following set of numbers such that the mean of the picked numbers would be $ 15$?\r\n$ \\{21,14,13,17,15,16,23,12,8,11,12,9,5,4,16,2,14,15,18,8,3,16\\}$\r\n\r\n20. how many natural numbers in a form of $ \\overline{abcd}$ can be found such that: $ \\overline{abcd} \\equal{} \\overline{ad}\\cdot\\overline{ada}$.\r\n\r\n21. in the isoceles triangle $ ABC$, $ AB \\equal{} AC$ and $ \\angle A \\equal{} 100$. Along the side $ AB$ (from $ B$), we pick the point $ M$ such that $ AM \\equal{} BC$. $ \\angle BCM$ equals?\r\n\r\n22. how many numbers, $ n$, between $ 1$ and $ 100$ are there such that $ n$ is sum of some positive integers, where every digit appears exactly once.(for example: $ 90 \\equal{} 0 \\plus{} 1 \\plus{} 52 \\plus{} 3 \\plus{} 4 \\plus{} 6 \\plus{} 7 \\plus{} 8 \\plus{} 9$)", "Solution_10": "Here is one more problem (though it seems simple):\r\n23. Trapezoid ABCD has $ \\angle D \\equal{} 90, \\angle B \\equal{} 60$, and AB = BC. If DC = 9, find the sum of the two possible lengths for AB.\r\nI must be overlooking something, but it seems to me that there is only one unique answer - 18. I don't see how the official answer got $ 18\\plus{}6\\sqrt3$...", "Solution_11": "According to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1713931#1713931[/url], the distance is $ d\\equal{}\\sqrt{R(R\\minus{}2r)}$ with $ R\\equal{}(21/10)\\sqrt5,r\\equal{}\\sqrt5\\implies d\\equal{}\\sqrt{105}/10$. What's wrong?", "Solution_12": "$ d \\equal{} \\sqrt {R(R \\minus{} 2r)}$ is for the distance from the incenter to the circumcenter. But the question asks for the distance from the incenter to the [b]orthocenter[/b]." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "[img]C:\\Documents%20and%20Settings\\Administrateur.XPPRO\\Mes%20documents\\Mes%20images[/img]", "Solution_1": "[quote=\"greatestmaths\"][img]C:\\Documents%20and%20Settings\\Administrateur.XPPRO\\Mes%20documents\\Mes%20images[/img][/quote]\r\n\r\nWhat is it, greatestmaths?? :rotfl:" } { "Tag": [ "geometry", "circumcircle", "power of a point", "radical axis", "geometry proposed" ], "Problem": "Let $C_1$ be a circle inside the circle $C_2$ and let $P$ in the interior of $C_1$, $Q$ in the exterior of $C_2$. One draws variable lines $l_i$ through $P$, not passing through $Q$. Let $l_i$ intersect $C_1$ in $A_i,B_i$, and let the circumcircle of $QA_iB_i$ intersect $C_2$ in $M_i,N_i$. Show that all lines $M_i,N_i$ are concurrent.", "Solution_1": "Let $\\Gamma_i$ be the circumcircle of $\\triangle QA_iB_i$. Then the line $M_iN_i$ is the radical axis of $C_2$ and $\\Gamma_i$. It suffices to observe that $PQ$ is the radical axis of any two $\\Gamma_i$ and $\\Gamma_j$. Hence $M_iN_i$, $M_jN_j$ and $PQ$ are concurrent at the radical center of $C_2$, $\\Gamma_i$ and $\\Gamma_i$. Now fix $\\Gamma_i$ and let $\\Gamma_j$ vary, and vice versa.", "Solution_2": "To finish this solution: Denote by $R_i$ the point where $(Q,M_i,N_i)$ intersect $PQ$ . We have $PQ\\cdot PR_i=PA_i\\cdot PB_i$. the last quantity is constant ($=R^2-OP^2$ where $O$ is the center of $C_1$), hence $R_i$ is fixed, hence $PQ$ belongs to all circles $\\Gamma_i$, hence it is their radical axis." } { "Tag": [ "algebra", "polynomial", "Gauss", "Polynomials" ], "Problem": "Prove that if the integers $a_{1}$, $a_{2}$, $\\cdots$, $a_{n}$ are all distinct, then the polynomial \\[(x-a_{1})^{2}(x-a_{2})^{2}\\cdots (x-a_{n})^{2}+1\\] cannot be expressed as the product of two nonconstant polynomials with integer coefficients.", "Solution_1": "Let's assume that $P$ is reducible in $\\mathbb{Z}[X].$ There exist $F, G \\in \\mathbb{Z}[X], \\text{deg}(F) \\geq 1$ and $\\text{deg}(G) \\geq 1,$ such that \r\n\\[P = (X-a_{1})^{2}(X-a_{2})^{2}\\cdots (X-a_{n})^{2}+1 = F(X)G(X).\\]\r\nWe may take $F$ and $G$ to be monic. Let $k = \\text{deg}(F), l = \\text{deg}(G),$ so that $k+l=2n.$\r\nClearly, for any real $x, P(x) \\geq 1.$ Hence $F$ and $G$ take positive values on $\\mathbb{R}.$\r\nFor any $i, 1 \\leq i \\leq n, F(a_{i})G(a_{i}) = 1 \\ \\ (*).$ Therefore, for any $i, F(a_{i}) = G(a_{i}) = 1.$\r\nIf $k0 integral(0 to 1)[(sin1/x)/(1+x^a)]dx >0.", "Solution_1": "Here is an idea of how, I think, this problem can be attached.\r\n\r\nWe have \r\n\r\n$I=\\int_{1}^{\\infty}\\sin x\\frac{y^{a-2}}{1+y^{a}}dy$.\r\n\r\nIt is not hard to see that the function $f(x)=\\frac{x^{a-2}}{1+x^{a}}$ is a decreasing function for all $x\\geq b$ where $b$ is some positive quantity, however if $00$.\r\n\r\n$\\int_{1}^{\\pi}\\sin x\\frac{x^{a-2}}{x^{a}+1}dx>2\\frac{\\pi^{a-2}}{\\pi^{a}+1}$ for all $01$ or $ <0$ throughout this interval. So, as a last resort, I tried the hit and try method and found out the functions satisfying the condition.\r\nOne good one, for example is $ \\frac {x^2}{2}$ but that hardly solves the problem concretely.", "Solution_2": "By Lagranges mean value theorem we get $ f(t)\\le 1$ and $ f'(x)\\ge 0$ for $ 0\\le x\\le 1$.\r\nso, $ f(t)^2 \\ge f(t)^3$. so,\r\n$ \\left(\\int^1_0f(t).dt\\right)^2\\ge\\int^1_0\\left(f(t)\\right)^2.dt\\ge\\int^1_0\\left(f(t)\\right)^3.dt$", "Solution_3": "Everything is $ \\ge 0;$ I won't state that at every step.\r\n\r\nStart with $ f'(x)\\le 1.$\r\n\r\n$ f(x)f'(x)\\le f(x)$\r\n\r\n$ \\int_0^xf(t)f'(t)\\,dt\\le \\int_0^xf(t)\\,dt$\r\n\r\n$ \\frac12f(x)^2\\le\\int_0^xf(t)\\,dt$\r\n\r\n$ f(x)^3\\le 2f(x)\\int_0^xf(t)\\,dt.$\r\n\r\nNow note that $ \\frac{d}{dx}\\left(\\int_0^xf(t)\\,dt\\right)^2\\equal{}2f(x)\\int_0^xf(t)\\,dt.$\r\n\r\nSo $ f(x)^3\\le \\frac{d}{dx}\\left(\\int_0^xf(t)\\,dt\\right)^2$\r\n\r\nWith both sides of this being zero at zero, we have\r\n\r\n$ \\int_0^xf(t)^3\\,dt \\le \\left(\\int_0^xf(t)\\,dt\\right)^2$\r\n\r\nNow let $ x\\equal{}1.$\r\n\r\nIn fact, we've proved a stronger inequality than asked for, since this is valid for all $ x>0,$ not just $ x\\equal{}1.$\r\n\r\n=======================\r\n\r\nLet's take an instructive example: if we let $ f(t)\\equal{}at$ and compute with that, we get\r\n\r\n$ \\frac{a^3x^4}4\\le\\frac{a^2x^4}4$ for all $ x>0.$ Since $ a\\le1,$ this is always true. But we see that if $ a\\equal{}1$ - that is, if $ f(t)\\equal{}t$ - then we have a sharp extremal function for this inequality.", "Solution_4": "[quote=\"ith_power\"]By Lagranges mean value theorem we get $ f(t)\\le 1$ and $ f'(x)\\ge 0$ for $ 0\\le x\\le 1$.\nso, $ f(t)^2 \\ge f(t)^3$. so,\n$ \\left(\\int^1_0f(t).dt\\right)^2\\ge\\int^1_0\\left(f(t)\\right)^2.dt\\ge\\int^1_0\\left(f(t)\\right)^3.dt$[/quote]\r\n\r\nith_power is wrong :D . The first inequality is manifestly false because of Cauchy-Schwartz inequality : $ \\left(\\int^1_0f(t).dt\\right)^2\\leq\\int^1_0\\left(f(t)\\right)^2.dt$ \r\n...\r\nNice solution from Kent Merryfield . :first: . \r\nIn the same way , if $ \\ f'(x) \\leq g(x),\\ 0 \\leq g(x),\\ 0\\equal{}f(0) \\leq f(x),\\ 0 < b, \\ 0 \\leq x$, \r\nyou can prove $ \\ \\frac 2{b\\plus{}1}\\int_0^x(f_{(t)})^{2b\\plus{}1}g_{(t)}dt \\le \\left(\\int_0^x(f_{(t)})^bg_{(t)}dt\\right)^2$ .\r\n :cool:", "Solution_5": "sorry :P didn't check properly :blush: \r\nbut still explain the following:\r\n \r\n$ (lim_{m\\minus{}>inf}\\sum_{r\\equal{}0}^n \\frac{r}{n})^{2} \\ge lim_{m\\minus{}>inf}\\sum_{r\\equal{}0}^n (\\frac{r}{n})^{2}$ because the 2.(i/n)(j/n) terms are extra on l.h.s.\r\nisn't that true? :maybe:" } { "Tag": [], "Problem": "Compute: $ 27 \\times 3 \\plus{} 27 \\times (\\minus{}4)$.", "Solution_1": "27*3=81\r\n27*(-4)= -108\r\n81+(-108)=-27", "Solution_2": "$ 27 \\times 3 \\plus{} 27 \\times (\\minus{}4) \\equal{} 27 \\times (3\\plus{}(\\minus{}4)) \\equal{} 27 \\times (\\minus{}1) \\equal{} \\boxed{\\minus{}27}$.", "Solution_3": "27x3 + 27x-4=\r\n27x-1=\r\n-27" } { "Tag": [ "search", "SFFT", "special factorizations" ], "Problem": "Please guy I want to know some important identity and factoring. Some useful .", "Solution_1": ":huh: \r\nJust in general?\r\n\r\nWell, here's a quick (redundant) list. Far from comprehensive though.\r\n\r\n$ a^2 - b^2 = (a+b)(a-b)$\r\n$ a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \\ldots + b^{n-1})$\r\n$ a^{2n+1} + b^{2n+1} = (a+b)(a^{2n} - a^{2n-1}b + a^{2n-2}b^2 - \\ldots + b^{2n})$\r\n$ ab + xa + yb = (a+y)(b+x) - xy$ (Simon's Favorite Factoring Trick)\r\n\\begin{eqnarray*}a^3 + b^3 + c^3 - 3abc &=& (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \\\\ &=& (a+b+c)\\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2} \\\\ \\end{eqnarray*}\r\n$ (a^n + a^{n-1} + \\ldots + 1)^2 = a^{2n} + 2a^{2n-1} + 3a^{2n-2} + \\ldots + na^n + (n-1)a^{n-1} + \\ldots + 1$\r\n\r\nSee [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007USAMOsoln.pdf]problem 5[/url] in here for a totally wacky one that will probably never appear on a contest again.\r\nhttp://mathworld.wolfram.com/CyclotomicPolynomial.html for information on factorizations of $ x^n - 1$ for varying $ n$, as well as some theory. More useful in later algebra though.\r\n\r\nThe best way to develop a factoring sense is to solve a lot of related problems. Eventually you get to the point where you can draw up a solution that makes people accuse you of pulling stuff out of your ***.\r\n\r\nHope that's what you were looking for. This is definitely something you probably could've [url=http://www.artofproblemsolving.com/Forum/search.php]search[/url]ed the forum for though.", "Solution_2": "A recent post by darij was extremely good on this topic (although I couldn't find it using the search)\r\n\r\n[quote=\"darij grinberg\"][b]Some nice formulae for three numbers.[/b]\n\nIf $ a$, $ b$, $ c$ are three elements of a field $ K$, then\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc \\equal{} \\left(a \\plus{} b \\plus{} c\\right)\\left(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} ca \\minus{} ab\\right)$;\n$ \\left(b \\plus{} c\\right)\\left(c \\plus{} a\\right)\\left(a \\plus{} b\\right) \\plus{} abc \\equal{} \\left(a \\plus{} b \\plus{} c\\right)\\left(bc \\plus{} ca \\plus{} ab\\right)$;\n$ \\left(a \\plus{} b \\plus{} c\\right)^5 \\minus{} \\left(a^5 \\plus{} b^5 \\plus{} c^5\\right) \\equal{} 5\\left(b \\plus{} c\\right)\\left(c \\plus{} a\\right)\\left(a \\plus{} b\\right)\\left(a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} bc \\plus{} ca \\plus{} ab\\right)$;\n$ \\left(b \\minus{} c\\right)^2 \\plus{} \\left(c \\minus{} a\\right)^2 \\plus{} \\left(a \\minus{} b\\right)^2 \\equal{} 2\\left(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} ca \\minus{} ab\\right)$;\n$ \\left(b \\minus{} c\\right)^4 \\plus{} \\left(c \\minus{} a\\right)^4 \\plus{} \\left(a \\minus{} b\\right)^4 \\equal{} 2\\left(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} ca \\minus{} ab\\right)^2$. [/quote]\r\n\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=192701[/url]", "Solution_3": "Ok thanks a lot guys.", "Solution_4": "What about Newton's factorization?", "Solution_5": "I'm not sure what you mean. \r\n\r\nIn any case, there's always the lovely Sophie Germaine identity $ x^4 \\plus{} 4y^4 \\equal{} (x^2 \\plus{} 2xy \\plus{} y^2)(x^2 \\minus{} 2xy \\plus{} y^2)$, which is in turn a special case of the \"incomplete\" factorization $ a^2 \\plus{} b^2 \\equal{} (a \\plus{} b)^2 \\minus{} 2ab$. It becomes \"complete\" if $ a, b$ are chosen so that $ 2ab$ is a square." } { "Tag": [ "inequalities", "search", "inequalities proposed" ], "Problem": "Let $ a,b,c>0$. Prove that:\r\n$ \\sqrt{\\frac{2a}{a\\plus{}b}}\\plus{}\\sqrt{\\frac{2b}{b\\plus{}c}}\\plus{}\\sqrt{\\frac{2c}{c\\plus{}a}}\\leq 3$.\r\n[hide=\"own solution(sorry if not new)\"]After bringing to a common denominator, and squaring both sides the ineq. becomes\n$ (\\sum{\\sqrt{a(b\\plus{}c)(c\\plus{}a)}})^2\\leq \\frac{9}{2}(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)$\nFrom Cauchy-Schwartz we infer that:\n$ (\\sum{\\sqrt{a(b\\plus{}c)(c\\plus{}a)}})^2\\leq (\\sum{a(b\\plus{}c)})(\\sum{(c\\plus{}a)})\\equal{}4(ab\\plus{}bc\\plus{}ca)(a\\plus{}b\\plus{}c)$\nWe are left to show that $ 8(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)\\leq 9(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)$, which is trivial with AM-GM.[/hide]\r\n\r\nI saw that the oficial solution is too long, so I decided to post a simple 2-line solution.\r\nIf this solution was posted, then I'm sorry for creating an useless topic.", "Solution_1": "This has been posted many many times in the past. For example take a look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=158729]here[/url]. In that link there are two more links which contain more solutions. :wink:", "Solution_2": "Sorry, nayel, for a double post.\r\nWhen I solved this inequality (sincerly, I didn't copy your solution),\r\nI saw in the book that this problem it has a much harder and long solution,\r\nso I decided to post the easy solution here, so people could see.\r\nAlthough I was convinsed that someone found this solution earlier than me.\r\nCongratulations, nayel, for being the first one! :thumbup:", "Solution_3": "No, it's okay. We often don't know which problems have been posted and which have not. Already posted inequalities are hard to find using the search function. And I know you didn't copy my solution. When I found my solution I also didn't know that this solution was already known. It is not unusual that we sometimes discover solutions/problems already found by others. So the credit for your solution does go to you. :)" } { "Tag": [ "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "I hope this didn't appear on the forum before!\r\nWe are given 100 sets, so that any of them have intersection an even number, but all of them have intersection an odd number.\r\nFind the minimal cardinality of their reunion!\r\n\r\n[Edit: As grobber asked, we are allowed to select any k of them ($100> k\\geq 2 $) and we know that their intersection will be even]\r\n[Edit 2: Minimal cardinality! Sorry for my mistake ] :blush:", "Solution_1": "What do you mean by \"any of them have intersection an even number\"? Any two of them? Or the intersection of any proper subset of the set of our sets? (I know that's confusing :)). Anyway, I don't think it's clear :?.", "Solution_2": "yah it is quite unclear as to what the problem wants;\r\n\r\nsuppose you have a family $\\mathcal{A}$. let $X=\\bigcup_{A \\in \\mathcal{A}}A$.let $B \\cap X = \\emptyset,B$ an even set.then taking the sets $A \\cup B,A \\in \\mathcal{A}$ the same property is true,so now this new family has a larger union.so there cannot be a maximum.\r\nif the question asks for minimum size of the union then the problem becomes more nontrivial.", "Solution_3": "Let me try:\r\nFor a given set of indices ${I=\\{i_1,i_2,\\cdots,i_k}$ denote $S_I$ be the set of elements belonging [b]only[/b] to $A_{i_1},\\cdots, A_{i_k}$. Let's prove $S_I$ is odd for any $I \\neq \\{\\}$.\r\nThe proof is by induction on $N-|I|$ ($N$-the number of elements in the union)\r\nBasis: The whole intersection is odd:\r\nInduction step: \r\nWLOG suppose $I=\\{1,2,\\cdots,k\\}$.\r\nThen working in $Z_2$ we get $0=|A_1 \\bigcap A_2 \\cdots \\bigcap A_k|=\\sum_{I \\subset J} S_J=\r\n2^{n-|I|}-1+S_I$ by induction hypothesis so $S_I=1$ in $Z_2$ so $|S_I|$ is odd. The claim is proved.\r\nThe the sets $S_I$ all are disjoint and non-empty, giving at least $2^n-1$ elements (here we have $n=100$). The example with $2^n-1$ elements is obvious from the solution.\r\n\r\nBy the way, this was problem $10.5$ at the eliminatory round, proposed by S. Ivanov, if someone's interested" } { "Tag": [ "inequalities" ], "Problem": "Let $ (x_n) (n \\equal{} 1,2,...)$ be a bounded-above sequence such that\r\n$ x_{n \\plus{} 2}\\geq\\frac {1}{4}x_{n \\plus{} 1} \\plus{} \\frac {3}{4}x_n$, for all $ n\\equal{}1,2,...$", "Solution_1": "And? :maybe: \r\n\r\nWhat's the question :wink:", "Solution_2": "Sorry\r\nLet $ (x_n) (n \\equal{} 1,2,...)$ be a bounded-above sequence such that\r\n$ x_{n \\plus{} 2}\\geq\\frac {1}{4}x_{n \\plus{} 1} \\plus{} \\frac {3}{4}x_n$, for all $ n\\equal{}1,2,...$\r\nProve that this sequence has limit", "Solution_3": "[hide=\"Hint\"]Since it is bounded above, it is enough to show that the sequence is increasing.[/hide]", "Solution_4": "[quote=\"SimonM\"]it is enough to show that the sequence is increasing.[/quote]\r\nYes, but the sequence needn't be: if $ x_2 < x_1$ and all inequalities are equalities then the sequence oscillates." } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "Sean $x_0, x_1,x_2 \\cdots x_n$ reales positivos tales que $x_ix_{i+2} \\leq x_{i+1}^2$ para todo $0 \\leq i \\leq n-2$ demostrar que \r\n$\\frac{\\sum_{i=0}^n x_i}{n+1} \\frac{\\sum_{i=1}^{n-1} x_i}{n-1} \\geq \\frac{\\sum_{i=0}^{n-1}x_i}{n} \\frac{\\sum_{i=1}^n x_i}{n}$", "Solution_1": "Disculpa,solo dos preguntas:\r\nLa condicion del problema, para que valores de $i$ se cumple?\r\nEn la sumatoria aparece $x_0$, debo entender que es $x_n$?, o es un nuevo valor?", "Solution_2": "[quote=\"Claudio Espinoza\"]Disculpa,solo dos preguntas:\nLa condicion del problema, para que valores de $i$ se cumple?\nEn la sumatoria aparece $x_0$, debo entender que es $x_n$?, o es un nuevo valor?[/quote]\r\n\r\nHay que demostrar que si $x_{i}^2 \\ge x_{i-1} \\cdot x_{i+1}$ para todo $1\\le i \\le n-1$ entonces la desigualdad se cumple... el problema deberia decir: sean $x_0, x_1, x_2....,x_n$ reales positivos.", "Solution_3": "Sea $S=x_0+x_1+...+x_n$. La desigualdad a demostrar es equivalente a $(\\frac{S}{n+1})(\\frac{S-x_0-x_1}{n-1})\\ge (\\frac{S-x_n}{n})(\\frac{S-x_0}{n})$\r\n$\\equiv S(S-x_0-x_1)\\ge (n^2-1)x_0x_n$\r\n$\\equiv (S-x_0)(S-x_n)\\ge n^2x_0x_n$\r\n$\\equiv (x_1+x_2+...+x_n)(x_0+x_1+...+x_{n-1})\\ge n^2x_0x_n$\r\n$\\equiv (\\frac{x_1+x_2+...+x_n}{n})(\\frac{x_0+x_1+...+x_{n-1}}{n})\\ge x_0x_n$\r\n$\\equiv (\\frac{\\frac{x_1}{x_0}+\\frac{x_2}{x_0}+...+\\frac{x_n}{x_0}}{n})(\\frac{\\frac{x_0}{x_n}+\\frac{x_1}{x_n}+...+\\frac{x_{n-1}}{x_n}}{n})\\ge 1$\r\nEsta \u00faltima desigualdad es m\u00e1s f\u00e1cil de demostrar. Veamos:\r\n\r\nPor un lado aplicando la desigualdad de la MA-MG a los $n$ n\u00fameros $\\frac{x_1}{x_0},\\frac{x_2}{x_0},...,\\frac{x_n}{x_0}$. Y tambi\u00e9n a los $n$ n\u00fameros $\\frac{x_0}{x_n},\\frac{x_1}{x_n},...,\\frac{x_{n-1}}{x_n}$. Y multiplicando ambas desigualdades obtenemos:\r\n\r\n$(\\frac{\\frac{x_1}{x_0}+\\frac{x_2}{x_0}+...+\\frac{x_n}{x_0}}{n})(\\frac{\\frac{x_0}{x_n}+\\frac{x_1}{x_n}+...+\\frac{x_{n-1}}{x_n}}{n})\\ge \\sqrt[n]{\\frac{x_1}{x_0}\\frac{x_2}{x_0}...\\frac{x_n}{x_0}\\frac{x_0}{x_n}\\frac{x_1}{x_n}...\\frac{x_{n-1}}{x_n}}=\\sqrt[n]{(\\frac{x_1x_{n-1}}{x_0x_n})(\\frac{x_2x_{n-2}}{x_0x_n})...(\\frac{x_{n-1}x_1}{x_0x_n})}$\r\n\r\nBastar\u00eda con demostrar que la \u00faltima expresi\u00f3n es mayor o igual que 1. Con las $n-1$ desigualdades dadas realizamos las siguientes operaciones y obtenemos los siguientes resultados:\r\n\r\nMultiplicamos las $n-1$ desigualdades: $x_1x_{n-1}\\ge x_0x_n$\r\nMultiplicamos las $n-1$ desigualdades excepto la primera y la \u00faltima: $x_2x_{n-2}\\ge x_1x_{n-1}$\r\nMultiplicamos las $n-1$ desigualdades excepto las 2 primeras y las dos \u00faltimas: $x_3x_{n-3}\\ge x_2x_{n-2}$\r\nRepitiendo este procedimiento obtenemos la siguiente secuencia:\r\n$x_0x_n\\le x_1x_{n-1}\\le x_2x_{n-2}...$, donde la forma del \u00faltimo termino de la secuencia depende de la paridad de n.\r\nDe esto podemos concluir por transitividad que: $x_kx_{n-k}\\ge x_0x_n; k=0,1,2,...n$.\r\nO su equivalente: $\\frac{x_kx_{n-k}}{x_0x_n}\\ge 1; k=0,1,2,...,n$\r\nCon lo cual cada factor de la expresi\u00f3n que buscabamos es mayor o igual que 1 y por tanto dicha expresi\u00f3n es mayor o igual que 1. Lo cual termina la demostraci\u00f3n." } { "Tag": [ "geometry", "circumcircle", "perpendicular bisector", "cyclic quadrilateral", "geometry proposed" ], "Problem": "Let $ AA_1$ is the bisector of the acute-angled triangle $ ABC$, point $ M$ is the midpoint of $ AA_1$. Such point $ P$ is taken on the $ BM$ that $ \\angle APC \\equal{} 90^{\\circ}$ and such point $ Q$ is taken on the $ CM$ that $ \\angle AQB \\equal{} 90^{\\circ}$. Prove that points $ A_1$, $ P$, $ M$ and $ Q$ are cyclic.\r\n\r\nI know three different solutions of this problem and is interesting - which of them you could find?", "Solution_1": "Let $ D$ be the foot of the altitude from $ A$, then $ MA = MD = MA'$.\r\nConsider the inversion $ I$ of pole $ M$ and power $ MA^2$, and let $ I(C) = Q'$ (obviously $ I(C)\\in{MC}$).\r\nThen $ \\angle{AQ'M} = \\angle{MAC} = \\frac {\\angle{A}}{2}$ and ${ \\angle MQ'D} = \\angle{MDA'} = \\angle{C} + \\frac {\\angle{A}}{2}$.\r\nThus $ \\angle{ABD} + \\angle{AQ'D} = 180$ and hence $ Q' = Q$ (since $ ABDQ$ is cyclic and $ \\angle{AQ'B}=\\angle{ADB}=90$).\r\nFinally, since $ I(BC) = C_{\\triangle{MDA'}}$ we get that $ MDA'Q$ is cyclic.\r\nAnalogously, we prove that $ MPDA'$ is cyclic and the conclusion follows.", "Solution_2": "What a nice problem! I want to use Inversion here, :wink: \r\nConsider an inversion by center $ A$,and an arbitrary radius $ k$.let me analyzes the changes,\r\n$ AB,AC,AP,AQ,AM,AA_1 \\longrightarrow AB',AC',AP',AQ',AM',AA_1'$(repectively)\r\n$ BM\\longrightarrow$arc $ B'M'$of the circumcircle of $ \\triangle AB'M'$\r\n$ CM\\longrightarrow$arc $ C'M'$of the circumcircle of $ \\triangle AC'M'$\r\n$ BQ\\longrightarrow$arc $ B'Q'$of the circumcircle of $ \\triangle AB'Q'$\r\n$ CP\\longrightarrow$arc $ C'P'$of the circumcircle of $ \\triangle AC'P'$\r\nin this case we should prove the quadrilateral$ A_1'Q'P'M'$ is cyclic.\r\nin the inversive shape,$ B'Q'\\cap C'P'\\equal{}H'$,which$ H'$ is the inverse of $ H$,the foot of the altitude from $ A$to $ BC$.and$ \\angle AA_1'H'\\equal{}\\frac{\\pi}{2}$\r\nand $ A_1'$ is the midpoint of $ AM'$,then $ A_1'H'$ is the perpendicular bisector.let $ A_1'H' \\cap$arc$ B'M'\\equal{}R$. and$ A_1'H' \\cap$arc$ C'M'\\equal{}S$\r\n$ \\angle B'AA_1'\\equal{}\\angle B'H'R\\equal{}\\angle C'AA_1'\\equal{}\\angle C'H'S$,thus $ RS$ is the bisector of $ \\angle C'A_1'Q'$ and it means:arc$ C'S\\equal{}$arc$ SQ'$\r\nand arc$ B'R$=arc$ RP'$,and because of $ SR$ is the perpendicular bisector,$ C'$ and $ Q'$ turn into each other with a respect to$ RS$,\r\n$ B'$ and $ P'$,also turn into each other with a respect to $ RS$,we also know that $ A$ and $ M'$,turn into each other with a respect to $ RS$,\r\nthe conclusion is the quadrilateral $ AC'A_1'B'$turnes into quadrilateral $ A_1'Q'P'M'$ with a respect to$ RS$,and also\r\nthese two quadrilaterals are equal to each other,and because $ AC'A_1'B'$ is cyclic,($ B,A_1,C$ are collinear),the quadrilateral $ A_1'Q'P'M'$ \r\nwill also be cyclic,and it is proved!", "Solution_3": "We can also mix these two solutions. Apply an inversion with center $ A$ and radius $ r$ and assume it sends point $ X$ to $ X'$. As [b]Ashegh[/b] wrote, we have $ AA_1'\\equal{}A_1'M'$ and $ A_1'H'$ is perpendicular bisector of $ AM'$. Since points $ M,Q,C$ are collinear w see that $ AC'Q'M'$ is cyclic quadrilateral hence $ A_1'C'\\equal{}A_1'Q'$ and using similar arguments $ A_1'C'\\equal{}A_1'B'\\equal{}A_1'P'$ which means that $ A_1'$ is circumcenter of $ P'Q'C'B'$ which means that points $ PQCB$ are concyclic (1). [b]Terryjohn[/b] proved that $ MA'^2\\equal{}MQ\\cdot MC$ hence $ \\angle MCB\\equal{}\\angle MA_1Q$ and using this fact and (1) we obtain $ \\angle MPQ\\equal{}MA_1Q$ qed", "Solution_4": "You can find another very nice solution without using inversion by [b]Huyen Vu[/b] here: http://www.mathlinks.ro/viewtopic.php?t=141416 . \r\n\r\nBTW, could you please post all problems from Ukraine MO on MathLinks? I found some geometry problems from Ukraine are very interesting :thumbup:", "Solution_5": "Strangely, but this year the same configuration of points again appeared on Ukrainian Mathematical Olympiad. With the only difference that the final question was to prove that the points B, C, P, Q are concyclic (see the post of [b]limes123[/b] with the proof of this statement). \n\nLink: http://matholymp.org.ua/_files/07cca67b94/2013-ukr-sols-2.pdf , problem 10-8." } { "Tag": [], "Problem": "Factorize in $\\mathbb Z[x]$ the following polynom:\r\n$x^{8}+98x^{4}+1$", "Solution_1": "[hide]expand $(x^{4}+ax^{3}+bx^{2}+cx+1)(x^{4}+ax^{3}+bx^{2}+cx+1)$ and equate coefficients (think if t is a root, then -t is a root)\n\nthen $2b-c^{2}=2b-a^{2}=0$ and $-2ac+b^{2}+2=98$, solve a quartic for c, you get rational solutions for $c=\\pm 4$, then put it back and get $a=\\mp 4$ and $b=8$\n\n$(x^{4}+4x^{3}+8x^{2}-4x+1)(x^{4}-4x^{3}+8x^{2}+4x+1)$[/hide]", "Solution_2": "OK. I meant the following reasoning:\r\nlet's $x-\\frac{1}{x}=u.$ Then\r\n$x^{8}+98x^{4}+1=x^{4}\\left(x^{4}+\\frac{1}{x^{4}}+98\\right)=x^{4}\\left(\\left(x^{2}+\\frac{1}{x^{2}}\\right)^{2}+96\\right)=x^{4}\\left(\\left(u^{2}+2\\right)^{2}+96\\right)=$\r\n$=x^{4}\\left(u^{4}+4u^{2}+100\\right)=x^{4}\\left(\\left(u^{2}+10\\right)^{2}-16u^{2}\\right)=x^{4}\\left(u^{2}+4u+10\\right)\\left(u^{2}-4u+10\\right)=$\r\n$= (x^{4}+4x^{3}+8x^{2}-4x+1)(x^{4}-4x^{3}+8x^{2}+4x+1).$ :)" } { "Tag": [], "Problem": "$A$ and $B$ are two point .\r\nFind the geometric locus of $M$ such that $(MA,MB)$=$45$\u00b0", "Solution_1": "it is a circle...", "Solution_2": "No its not.", "Solution_3": "[hide]\nConstruct a square $AOBO'$ whose diagonals are $AB$ and $OO'$.\n\nDraw two circles that go through $A$ and $B$ whose centers are $O$ and $O'$ respectively.\n(The two circles intersect at $A$ and $B$.)\nThe locus of $M$ such that $\\angle AMB=45^\\circ$ is made of two major arcs of the circles (the parts of the circumferences that are outside of square $AOBO'$).\n\nI hope I described the locus clearly...\n[/hide]", "Solution_4": "sorry, i was a little rushed, i was posting this at school, it should be 2 circles that intersect at A and B such that arc AB = 90, and M is not on either minor arc, but is on the major arc AB on both circles,\r\n\r\nthis works because the angle is inscribed in the circle and intercepts 90 degrees, so 90/2=45, and if it were not on these circles, the secant-secant and chord-chord angle theorems would show that it would be less than 45 if M is ouside, and greater if M is in the insides of the circles" } { "Tag": [], "Problem": "Re(iz)=1, |z+i|=2. Find all complex numbers z", "Solution_1": "Let $ z \\equal{} a\\plus{}bi$ for reals $ a,b$. The first given becomes $ \\text{Re}(\\minus{}b\\plus{}ai) \\equal{} 1$, and we get $ b \\equal{}\\minus{}1$. Then the second given becomes $ |a\\minus{}i\\plus{}i| \\equal{} 2$, and we get $ a \\equal{} 2$. So $ z \\equal{} 2\\minus{}i$.\r\n\r\nEDIT: Derr! $ a \\equal{}\\minus{}2$ also, so $ z \\equal{}\\minus{}2\\minus{}i$ is the other solution.", "Solution_2": "$ a\\equal{}\\minus{}2$" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let $n$ be a natural number greater than 6. $X$ is a set such that $|X| = n$. $A_1, A_2, \\ldots, A_m$ are distinct 5-element subsets of $X$. If $m > \\frac{n(n - 1)(n - 2)(n - 3)(4n - 15)}{600}$, prove that there exists $A_{i_1}, A_{i_2}, \\ldots, A_{i_6}$ $(1 \\leq i_1 < i_2 < \\cdots, i_6 \\leq m)$, such that $\\bigcup_{k = 1}^6 A_{i_k} = 6$.", "Solution_1": "Ok here we go, Use the following double counting to stablish some lemmas.\r\n\r\nCall $m_k$ partial bound for the max number of subsets such that we cant find six $A_i$ with the desired property.\r\n\r\nFirst, count the number of sets of $6$ elements such that $5$ of them belong to one $A_i$, using this for $n=7$, we get that for $n=7$, $5{\\binom{7}{6}}> 2m$, so we get $m_7=17$.\r\n\r\nNow counting the number of $7$ elements set, such that $5$ of them belong to an $A_i$, for $n=8$, we get $m_8=45$,\r\n\r\nand Finally for the other $n$, counting the number of $8$ element sets, $5$ of them in an $A_i$, we get a better bound that the desired one.", "Solution_2": "I really dont understand the solution :maybe: \r\nCould someone explain more please." } { "Tag": [], "Problem": "ok.. this was an olymon problem from 3 months ago. I think I can post them here now. I'm dying to see the solution; please, someone help!\r\n\r\nProve there doesn't exist consecutive perfect numbers.", "Solution_1": "In the http://www.mathlinks.ro/Forum/viewtopic.php?p=932109&sid=d760931f0812469cb550fdbed14ce957#932109 has a hint. :wink:", "Solution_2": "[quote=\"N.T.TUAN\"]In the http://www.mathlinks.ro/Forum/viewtopic.php?p=932109&sid=d760931f0812469cb550fdbed14ce957#932109 has a hint. :wink:[/quote]\r\n\r\nYa, I got that far, but still couldn't do it. Sorry :( \r\n\r\nCould you help me some more? thanks!", "Solution_3": "I wrote out a solution that used several properties of perfect numbers a few months ago, but never found time to write up the others so I could send them in. I'll try to find (or recall) the solution. The problem was fun.", "Solution_4": "So I think you might find it useful to know for any odd perfect number N, there must exist a prime P such that NP ( :rotfl: ) is a perfect square. Then wonder what N can be mod 4 and 3." } { "Tag": [ "calculus", "integration", "conics", "parabola", "inequalities", "symmetry", "geometry" ], "Problem": "how many integral points are in the parabola, in the region formed by the parabola $ y^{2}\\equal{}4ax$ and circle $ x^{2}\\plus{}y^{2}\\equal{}16$\r\nans:17\r\nplease solve this", "Solution_1": "are u sure that is $ a$?\r\nwell the basic idea is to solve the following inequalities \r\n$ x^{2}\\plus{}y^{2}\\leq 16$ -------------#\r\n$ y^{2}\\leq 4ax$\r\n$ x\\geq 0$\r\n# gives $ 0\\leq x\\leq 4$ \r\nby symmetry consider $ y\\geq 0$ then u can reflect the points obtained about x-axis.", "Solution_2": "yes that is a ,hey but where's the final answer", "Solution_3": "ITs good to see some problems on the Indian forum. You guys need them.", "Solution_4": "[quote=\"veryfriendly\"]ITs good to see some problems on the Indian forum. You guys need them.[/quote]\r\nwhat do u mean by \"you guys\". :P", "Solution_5": "[quote=\"veryfriendly\"]ITs good to see some problems on the Indian forum. You guys need them.[/quote]\r\n\r\nIt sounds somewhat rude as if Indians were a bunch of retards. :P :lol:", "Solution_6": "[quote=\"SOUMYASHANT NAYAK\"][quote=\"veryfriendly\"]ITs good to see some problems on the Indian forum. You guys need them.[/quote]\n\nIt sounds somewhat rude as if Indians were a bunch of retards. :P :lol:[/quote] \r\n\r\nOk, I never meant to say that. Um if you notice this is one of the only problems that has been posted in the Indian forum.", "Solution_7": "yes that sounda logical because posted elesewhere u r assured of getting it solved" } { "Tag": [ "function", "linear algebra", "matrix", "calculus", "derivative", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ u \\in C^2(B_1^2(0)) \\cap C( \\overline {B_1^2(0)}); u_{xx}\\plus{}u_{xy}\\plus{}u_{yy}\\equal{}1, t\\equal{}(x,y) \\in B_1^2(0).$ The question is if $ u(t)$ can have 1)maximum; 2)minimum inside $ B_1^2(0)$?\r\n\r\nAs for minimum we have that function $ u(x,y)\\equal{}\\frac 1 4(x^2\\plus{}y^2)$ satisfies the equation and $ u$ has minimum at $ (0,0)$. So I think that answer for b) is positive. And what about a)?", "Solution_1": "Just look at the Hessian at the possible point of maximum. What can you say about its entries? Now, what can you say about its determinant? (or just see whether the word \"elliptic\" brings up any associations ;) )", "Solution_2": "I guess $ u_{xx}<0, u_{yy}<0$ at the point of maximum, what about $ u_{xy}$ I don't know(is it equal to zero at the point of maximum? - if so, hessian is positive at the point of max and it is wrong, because it should be negative, so the function can't have maximum).", "Solution_3": "$ u_{xy}\\equal{}1\\minus{}u_{xx}\\minus{}u_{yy}$, remember? So your matrix at the point of maximum should be of the form $ \\begin{bmatrix}\\minus{}a&1\\plus{}a\\plus{}b\\\\1\\plus{}a\\plus{}b&\\minus{}b\\end{bmatrix}$ for some $ a,b\\ge 0$. Can it be non-positive definite?", "Solution_4": "Why at the point of maximum $ u_{xx}=-a$, ${ u_{yy}}=-b$, where $ a \\ge 0, b \\ge 0$.?", "Solution_5": "Second derivative test: if you have a maximum in two variables, you certainly have a maximum in each one separately (but not vice versa!)." } { "Tag": [], "Problem": "A student ID number at a certain university consists of a 6-digit number, such as 023457. How many student numbers can be chosen such that no two of its digits are the same?\r\n\r\nI thought it was just 10! but it was wrong.", "Solution_1": "[quote=\"Quevvy\"]A student ID number at a certain university consists of a 6-digit number, such as 023457. How many student numbers can be chosen such that no two of its digits are the same?\n\nI thought it was just 10! but it was wrong.[/quote]\r\n\r\n[hide=\" :maybe: \"]ummm... wouldnt it just be $10\\times 9\\times 8\\times 7\\times 6\\times 5=151200$?[/hide]", "Solution_2": "[quote=\"jli\"][quote=\"Quevvy\"]A student ID number at a certain university consists of a 6-digit number, such as 023457. How many student numbers can be chosen such that no two of its digits are the same?\n\nI thought it was just 10! but it was wrong.[/quote]\n\n[hide=\" :maybe: \"]ummm... wouldnt it just be $10\\times 9\\times 8\\times 7\\times 6\\times 5=151200$?[/hide][/quote]\r\nOh yeah..... I'm stupid!\r\nI forgot that there were only 6 digits. :wallbash:", "Solution_3": "wait... so what is it?\r\n\r\n10! x 9! x 8! x 7! x 6! x 5! ? or something?", "Solution_4": "[quote]10! x 9! x 8! x 7! x 6! x 5! ? or something?[/quote]\r\nNo because its not 10!*9!... its just simply 10*9*8*7*6*5, because you have 10 choices for the first spot, 9 fro the second...", "Solution_5": "whoops thats what i meant sry" } { "Tag": [ "function", "complex analysis", "complex analysis unsolved" ], "Problem": "$D=\\{z\\in\\mathbb C: |z|<1$ and $f: D\\rightarrow\\mathbb C$ is an analytic function which for $n\\geq 2$ we know that $f(\\frac 1n)\\in\\mathbb R$. Prove that for each $n$, $f^{(n)}(0)\\in\\mathbb R$", "Solution_1": "$f(z)=a_0+a_1z+a_2z^2+...$ with $a_i \\in \\mathbb{C}$, and one wants to prove that those $a_i$ are in fact real numbers.\r\nSuppose that it is not the case and that $N \\in \\mathbb{N}$ is the smallest integer such that $a_N$ is not real. Then $g(z)=a_Nz^N+a_{N+1}z^{N+1}+..=z^{N}(a_N+a_{n+1}z+..)$ also satisfies $g(\\frac{1}{n}) \\in \\mathbb{R}$, and also $h(z)=a_N+a_{n+1}z+..$. This is clearly a contradiction because $h(0)$ is not real and because of the continuity of $h$, $h(z)$ is not real on a neighborhood of $0$." } { "Tag": [ "function", "integration", "calculus", "calculus computations" ], "Problem": "If $ y\\equal{} 5 \\plus{} \\int_2^{2x} e^{\\minus{}t^2} dt$, which of the following is true?\r\n\r\nA) $ \\frac{dy}{dx} \\equal{} e^{\\minus{}x^2}$ and $ y(0)\\equal{}5$\r\n\r\nB) $ \\frac{dy}{dx} \\equal{} e^{\\minus{}x^2}$ and $ y(1)\\equal{}5$\r\n\r\nC) $ \\frac{dy}{dx} \\equal{} e^{\\minus{}4x^2}$ and $ y(1)\\equal{}5$\r\n\r\nD) $ \\frac{dy}{dx} \\equal{} 2e^{\\minus{}4x^2}$ and $ y(0)\\equal{}5$\r\n\r\nE) $ \\frac{dy}{dx} \\equal{} 2e^{\\minus{}4x^2}$ and $ y(1)\\equal{}5$", "Solution_1": "Hi :)\r\n\r\nI thinks that the last choose is correct (E): $ \\frac {dy}{dx} \\equal{} 2e^{ \\minus{} 4x^2} \\ and \\ y(1) \\equal{} 5$\r\n\r\nif you want a prof say to me i'm present\r\nthanks", "Solution_2": "Yeah the little integral part confuses me", "Solution_3": "let $ F(x)$ be an antiderivative of $ e^{\\minus{}x^2}$. Then we have $ y\\equal{}5\\plus{}F(2x)\\minus{}F(2)$. Observe when $ x\\equal{}1$, $ y\\equal{}5$. Now differentiate: $ y' \\equal{} 2F'(2x)$ by chain rule, so $ y'\\equal{}2e^{\\minus{}(2x)^2}$." } { "Tag": [ "calculus", "derivative", "inequalities unsolved", "inequalities" ], "Problem": "slove the image of $ y \\equal{} \\sqrt {2x^2 \\minus{} 6x \\plus{} 9} \\plus{} \\sqrt {2x^2 \\minus{} 10x \\plus{} 17}\\;$", "Solution_1": "What do you mean? Should $ y$ be an integer?", "Solution_2": "If $ x\\in \\mathbb{R}$ then\r\n$ A\\equal{}\\{ y \\in \\mathbb{R} | y\\ge 2\\sqrt{5}\\}$\r\n\r\nIf you don't want to use derivatives to find the minimum of $ \\sqrt{2x^{2}\\minus{}6x\\plus{}9}\\plus{}\\sqrt{2x^{2}\\minus{}10x\\plus{}17}$ then think of a geometric interpretation", "Solution_3": "[quote=\"Rafikafi\"]If $ x\\in \\mathbb{R}$ then\n$ A \\equal{} \\{ y \\in \\mathbb{R} | y\\ge 2\\sqrt {5}\\}$\n\nIf you don't want to use derivatives to find the minimum of $ \\sqrt {2x^{2} \\minus{} 6x \\plus{} 9} \\plus{} \\sqrt {2x^{2} \\minus{} 10x \\plus{} 17}$ then think of a geometric interpretation[/quote]\r\n\r\nHow to use derivatives to find the minimum ?\r\ncan you tell me ?\r\nthanks", "Solution_4": "Let $ f(x)\\equal{}\\sqrt{2x^2\\minus{}6x\\plus{}9}\\plus{} \\sqrt{2x^2\\minus{}10x\\plus{}17}$ and find the unique $ x$ such that $ f'(x)\\equal{}0$.", "Solution_5": "[quote=\"FelixD\"]Let $ f(x) \\equal{} \\sqrt {2x^2 \\minus{} 6x \\plus{} 9} \\plus{} \\sqrt {2x^2 \\minus{} 10x \\plus{} 17}$ and find the unique $ x$ such that $ f'(x) \\equal{} 0$.[/quote]\r\nHow to use derivatives to find the minimum ? \r\ncan you tell me ? \r\nthanks", "Solution_6": "$ \\sqrt {2x^2 \\minus{} 6x \\plus{} 9} \\plus{} \\sqrt {2x^2 \\minus{} 10x \\plus{} 17} \\equal{} \\sqrt {2}\\left ( \\sqrt {(x \\minus{} \\frac {3} {2})^2 \\plus{} (\\frac {3} {2})^2} \\plus{} \\sqrt {(x \\minus{} \\frac {5} {2})^2 \\plus{} (\\frac {3} {2})^2} \\right )$.\r\n\r\nTake $ A(\\frac {3} {2}, \\frac {3} {2})$, $ B(\\frac {5} {2}, \\frac {3} {2})$ and $ X(x, 0)$, then the expression is $ \\sqrt {2}(XA \\plus{} XB)$. Its minimum occurs for $ x \\equal{} 2$, so the range of values is $ [2\\sqrt {5}, \\plus{} \\infty)$." } { "Tag": [ "Mafia", "pre100" ], "Problem": "So, this is the list of players for game 11. You are on this list if you either confirmed the PM, or signed up less than 2 days ago.\n\nList of players:\nSly Si\nsamath\nSysRq\njli\nchesspro\nZennyK\nzeb\nSunny\ndts\nG-UNIT\npianoforte\nnoneoftheabove\n\nNormally the roles will be divided as follows:\n\nMafia: 4\nVigilante: 1\nAngel: 2\nInspector: 1\nMason: 3\nCitizen: 1\n\n\nYou will receive your roles in a few moments.", "Solution_1": "Sorry, there has been a roles mistake. I will re-roll all roles and send you a new PM. Please ignore the first PM if you received it somehow.", "Solution_2": "Ok, everyone should have their roles now, you can start discussing. It is day. \r\n\r\nNo one can communicate during the day other than through this thread. Only exception is that everyone may communicate with me 24/24 7/7 by PM. \r\nMafia, masons, angels may communicate at night. \r\n\r\nNight phases will not take more than 24 hours. Angels/Mafia/Inspector/Vigilante choose one target per class. (so I recieve 4 targets in total)\r\n[b]Edit:[/b] The first nigh phase may take a little more, as it's the first discussion. Every group (angels/mafia/inspector/vigilante) gives me the target for that night. [b]I recommend you to add a queue to your inspections/killings/savings, just for the case you would miss one night.[/b] You can always change it the next night. So e.g. mafia tell me they will kill person A, then person B, then person C. They don't know for further nights. But person D appears to be a valuable role next day. So the next night, they change their list, first person D, then B, then C. Or something. \r\n\r\nAny further questions, PM me. Enjoy! :)", "Solution_3": "Peter, maybe instead of a random pick you should just make that person skip their role for the night...it's their fault.\r\n\r\nSo...like all mafia games, does anyone have a random accusation to get things started?\r\n\r\nBTW, I will probably be gone for the rest of the day; I apologize if I don't respond until later today (or possibly tomorrow).", "Solution_4": "[quote=\"chesspro\"]Peter, maybe instead of a random pick you should just make that person skip their role for the night...it's their fault.[/quote] Yes, I'm still thinking on that one... in game 9 we got kinda screwed because windslicer didn't inspect the first 5 turns... and that ruins the game for the others too.\r\n\r\nI currently think on mafia/angels missing their turn if they forget, but as the inspector is alone in his role, I feel forced to enfore inspection each night. Just to keep the fun in the game. (I'll just do Math.Random(12); so it's really random)\r\n\r\nWhat I also can do is allowing a queue. So e.g. angels PM me first save A, then next turn save B (unless a counter-pm comes), next turn C etc. Same for mafia and inspector. Maybe this is the best option?", "Solution_5": "im here\r\n\r\nyay!!! chesspro is plaing", "Solution_6": "i am here and ready to play a fast game", "Solution_7": "Here and ready to play.", "Solution_8": "Here and ready to play.\r\n\r\nEdit: didn't realize I was copying word for word the last post...", "Solution_9": "Here and ready to play. ;)", "Solution_10": "Here and ready to play.\r\n\r\nOh wait I'm modding this :D", "Solution_11": "Hello there.", "Solution_12": "You can start discussing people, confirmation that you play has already happened ;)", "Solution_13": "For clarification, I added this little rule. It only applies to mafia / inspector / angels / vigilante.\r\n\r\nThe first night phase may take a little longer, as it's the first discussion.\r\nEvery group (angels/mafia/inspector/vigilante) gives me the target for that night. \r\n[b]I recommend you to add a queue to your inspections/killings/savings, just for the case you would forget one night.[/b]\r\nIt is not mandatory. You can always change it the next night. \r\n\r\nSo e.g. mafia tell me they will kill person A, then person B, then person C. \r\nThey don't know for further nights. But during the day person D appears to be a valuable role. \r\nSo the night, they change their list, first person D, then B, then C. Or something.\r\n\r\nJust the same as normal, but providing me with a backup if you forget. Please people, use this option :) it's too stupid to miss kills/saves/inspections by forgetting.", "Solution_14": "Hoala!! :ninja:", "Solution_15": "lol this is really amusing right now. It sounds like its come down to whether or not I'm mafia or mason, hehehe.\r\n\r\nTake your chances. Either way, citizens will win in the end. Though I'd like to be alive for that.", "Solution_16": "In my opinion:\r\n\r\nAngels save samath.\r\nSamath inspects Sunny.\r\n\r\nAs for today:\r\n\r\n1) jli has not roleclaimed.\r\n2) jli has not been murdered by the mafia.\r\n\r\nSo, I believe that I should lynch jli. However, I shall not do so until the rest of the town agrees with this (and I may not check this thread again until tomorrow morning; my apologies if so).", "Solution_17": "Everyone except for the suspected mafia and you agrees with this lynching.\r\n\r\nNo one has counter-claimed mason, so I think we can believe Sunny (if someone other than jli or Sunny was the real mason, then jli and Sunny would both be mafia, and if jli was the real mason he would have said so).\r\n\r\nEdit: in other words, even if Sunny is lying, jli is mafia (unless jli is an extremely stupid mason).", "Solution_18": "Ok, I vote [color=red]aye [/color]then.", "Solution_19": "That evening, jli, supreme local mafia leader, was lynched by the citizens. They found him by pure coincidence, when he fell off a chair in the grocery store (while taking his favorite cereals off the highest rack) his gun dropped out of his pocket. He got arrested, and lynched after a ton of evidence was found into his house. \r\n\r\nAnd then it was night. Citizens have again lynched a mafia, and have made great progress in the battle against the organised crime. The citizens held a little party tonight, so most went to bad pretty drunk... always dangerous. ;)\r\n\r\nNight has begun. It started quiet...", "Solution_20": "That night, the mafia were on revenge. They burglared silently into zeb's house, silently entered zeb's room and shot every single bullet they had towards his bed. Bloodhungry, they turned on the light to admire their work, but they only saw a destroyed bed. Instead, they both felt a gun in their back, carried by zeb and G-UNIT, who had been silently waiting in another room all night long. Both mafia got easily arrested.\r\n\r\nThe threatening of the mafia is gone. The city is cleared, and the villagers are coming back to town. The mission has succeeded, and the surviving citizens are heroes for life. \r\n\r\nThe game is over. The citizens have won.", "Solution_21": "Painful... just 20 minutes before the deadline, the mafia pm'ed me to change the kill... if they had left it on samath, the game was long not over yet. :) [and I do wonder why the mafia did this? you knew you had to kill the inspector this night to have a chance right?]\r\n\r\nI have a question to the mafia: is 24 hours enough to communicate? I think I will allow day communication next time, as you guys appeared to have failing communication... like with the lynching of pianoforte...", "Solution_22": "Huh? How did they get arrested by the angels? Im confused.", "Solution_23": "I'm really really confused. Though I'm sure the citizens would have won in the end...how did it end?", "Solution_24": "Apparently, me and G-UNIT both forgot to change the queue... originally it was G-UNIT, zeb, G-UNIT, etc., and each night we have (for one reason or another) not changed it. So, by pure coincedence, the mafia made the decision to kill me, so now we have 5 citizens to 2 mafia. Assuming samath inspected Sunny, we would have definitely won... I guess Peter just wanted to speed things up a bit.", "Solution_25": "Well, 5 citizens vs. 2 mafia, and the citizens are now 100% sure of the mafia, since every innocent has been inspected. So that basicly terminates the game, there is no way the mafia can still win.\r\n\r\nHad they killed samath, then it was a 4-2 with still doubt on who was the last mason. A wrong decision on who was the last mason would most likely have made the mafia win.", "Solution_26": "*sigh*\r\n\r\nI think this game may have been stacked up against the Mafia, we're supposed to be able to claim citizen and not have to worry about the actual citizen counterclaiming... it could have been even worse if we hadn't nailed the Vigilante the first day (then again, perhaps the Vigilante would have done our work for us).", "Solution_27": "I didn't think there was much doubt on the last mason. The strategy and rationale, from my point of view, was this:\r\n\r\n1. Noneoftheabove is going to get lynched.\r\n2. I will soon be called upon to claim a role. I can't claim citizen, inspector, or mason, as the identities of these people are established.\r\n3. Thus, I have to claim angel.\r\n4. If I claim angel with noneoftheabove, no one will believe me.\r\n5. Thus, we should kill an angel, let noneoftheabove get lynched, and then I can claim to be the last angel.", "Solution_28": "Yay for the innocents!!\r\n\r\n(AND Me, of course :lol: :P )", "Solution_29": "last post" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "$G$ is a group that order of each element of it Commutator group is finite. Prove that subset of all elemets of $G$ which have finite order is a subgroup og $G$.", "Solution_1": "Let $\\Delta$ be the commutator group of $G$. Take $a,b \\in G$ with finite order.\r\n\r\n$(ab)(a^{-1}b^{-1}) \\in \\Delta$.\r\n\r\nIf $(ab)^{n}a^{-n}b^{-n}\\in \\Delta$, then also $(ab)(ab)^{n}a^{-n}b^{-n}(ab)^{-1}=(ab)^{n+1}a^{-n}b^{-(n+1)}a^{-1}\\in \\Delta$, since $\\Delta$ is normal in $G$. Eventually, $[(ab)^{n+1}a^{-n}b^{-(n+1)}a^{-1}][a(b^{n+1}a^{-1})a^{-1}(b^{n+1}a^{-1})^{-1}] \\in \\Delta$, which means $(ab)^{n+1}a^{-(n+1)}b^{-(n+1)}\\in \\Delta$. This implies that, for $N$ common multiple of the orders of $a$ and $b$, we will have $(ab)^{N}\\in \\Delta$. Hence, by hypothesis, $ab$ will have finite order.\r\n\r\nThe fact that $a^{-1}$ also has finite order is trivial.", "Solution_2": "Nice proof.\r\nBut what I can't understand is why:\r\n1.$ab(a^{-1}b^{-1})\\in \\Delta$\r\n2.$(ab)^{N}\\in \\Delta$\r\nCan you tell me?sorry I am a beginer,thank you.", "Solution_3": "[quote=\"zhaobin\"]Nice proof.\nBut what I can't understand is why:\n1.$ab(a^{-1}b^{-1})\\in \\Delta$\n2.$(ab)^{N}\\in \\Delta$\nCan you tell me?sorry I am a beginer,thank you.[/quote]\r\n\r\n1. By definition of the commutator subgroup of $G$.\r\n2. Because we showed $(ab)^{N}a^{-N}b^{-N}\\in \\Delta$, but we picked $N$ such that $a^{-N}=b^{-N}=1$.", "Solution_4": "[quote=\"julien_santini\"]\n\n1. By definition of the commutator subgroup of $G$.\n.[/quote]\r\nI know 2 now.\r\nFor 1:What is the definition of the commutator subgroup of $G$?\r\nDoes it is equal to:$\\{x|xg=gx,for any g \\in G\\}$?", "Solution_5": "[quote=\"zhaobin\"][quote=\"julien_santini\"]\n\n1. By definition of the commutator subgroup of $G$.\n.[/quote]\nI know 2 now.\nFor 1:What is the definition of the commutator subgroup of $G$?\nDoes it is equal to:$\\{x|xg=gx,for any g \\in G\\}$?[/quote]\r\n\r\nit's the subgroup generated by $xyx^{-1}y^{-1}$ for x,y in G" } { "Tag": [], "Problem": "What is the value of $ 9342 \\plus{}(\\minus{}438)\\times 719 \\plus{} (\\minus{}9340) \\plus{} (\\minus{}438)\\times (\\minus{}719)$?", "Solution_1": "$ 9342\\plus{}(\\minus{}438)\\times719\\plus{}(\\minus{}9340)\\plus{}(\\minus{}438)\\times{\\minus{}719}\\equal{}9342\\minus{}9340\\equal{}\\boxed{2}$.", "Solution_2": "The important thing to note here is that $ (\\minus{}438)\\times 719 \\plus{} (\\minus{}438)\\times \\minus{}719$ is 0." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If a,b,c>0 and a+b+c=6 proove that:\r\n\\[ \\sum \\frac{12bc(3-a)}{(b+c)^2} \\leq \\frac{7(a^2+b^2+c^2)}{2} \\]\r\n Luca V. Iliesiu", "Solution_1": "[quote=\"luke999\"]If a,b,c>0 and a+b+c=6 proove that:\n\\[ \\sum \\frac{12bc(3-a)}{(b+c)^2} \\leq \\frac{7(a^2+b^2+c^2)}{2} \\]\n Luca V. Iliesiu[/quote]\r\n\r\nI think it is quite simple :)\r\n\r\n$\\sum\\frac{12bc(3-a)}{(b+c)^2}\\leq\\sum\\frac{12bc(3-a)}{4bc}=\\sum3(3-a)=3(3-a+3-b+3-c)=9$\r\n\r\n$\\frac{7(a^2+b^2+c^2)}{2}\\geq\\frac{7(a+b+c)^2}{6}=42$\r\n\r\nSo we have: \r\n$\\sum\\frac{12bc(3-a)}{(b+c)^2}<\\frac{7(a^2+b^2+c^2)}{2}$" } { "Tag": [ "inequalities", "function", "integration", "logarithms", "induction", "inequalities proposed" ], "Problem": "For $ a,b,c>0$ and $ a_k>0$ prove the following inequalities:\r\n\r\n$ a)$ $ (a^2\\plus{}b^2\\plus{}c^2)(\\frac{1}{a^2}\\plus{}\\frac{1}{b^2}\\plus{}\\frac{1}{c^2})\\ge(a\\plus{}b\\plus{}c)(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c})$\r\n\r\n$ b)$ $ (\\displaystyle\\sum_{k\\equal{}1}^{n}{a_k}^2)(\\displaystyle\\sum_{k\\equal{}1}^{n}\\frac{1}{{a_k}^2})\\ge(\\displaystyle\\sum_{k\\equal{}1}^{n}a_k)(\\displaystyle\\sum_{k\\equal{}1}^{n}\\frac{1}{a_k})$\r\n\r\n$ c)$ $ ({a^2}^n\\plus{}{b^2}^n\\plus{}{c^2}^n)(\\frac{1}{{a^2}^n}\\plus{}\\frac{1}{{b^2}^n}\\plus{}\\frac{1}{{c^2}^n})\\ge(a\\plus{}b\\plus{}c)(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c})$", "Solution_1": "[quote=\"moldovan\"]For $ a,b,c > 0$ and $ a_k > 0$ prove the following inequalities:\n\n$ a)$ $ (a^2 \\plus{} b^2 \\plus{} c^2)(\\frac {1}{a^2} \\plus{} \\frac {1}{b^2} \\plus{} \\frac {1}{c^2})\\ge(a \\plus{} b \\plus{} c)(\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c})$\n\n$ b)$ $ (\\displaystyle\\sum_{k \\equal{} 1}^{n}{a_k}^2)(\\displaystyle\\sum_{k \\equal{} 1}^{n}\\frac {1}{{a_k}^2})\\ge(\\displaystyle\\sum_{k \\equal{} 1}^{n}a_k)(\\displaystyle\\sum_{k \\equal{} 1}^{n}\\frac {1}{a_k})$\n\n$ c)$ $ ({a^2}^n \\plus{} {b^2}^n \\plus{} {c^2}^n)(\\frac {1}{{a^2}^n} \\plus{} \\frac {1}{{b^2}^n} \\plus{} \\frac {1}{{c^2}^n})\\ge(a \\plus{} b \\plus{} c)(\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c})$[/quote]\r\nI'll just prove the general:\r\n$ (\\sum_{k \\equal{} 1}^{n}{a_k}^m)(\\sum_{k \\equal{} 1}^{n}{a_k}^{ \\minus{} m}) \\ge (\\sum_{k \\equal{} 1}^{n}{a_k})(\\sum_{k \\equal{} 1}^{n}{a_k}^{ \\minus{} 1})$ $ \\forall m > 1, a_i \\ge 0$.\r\n\r\nIt's equivalent to:\r\n$ \\sum_{i,j \\equal{} 1}^{n} {a_i}^m{a_j}^{ \\minus{} m} \\plus{} {a_i}^{ \\minus{} m}{a_j}^m \\ge \\sum_{i,j \\equal{} 1}^{n} {a_i}{a_j}^{ \\minus{} 1} \\plus{} {a_i}^{ \\minus{} 1}{a_j}$\r\n\r\nSo it suffices to prove $ a^mb^{ \\minus{} m} \\plus{} a^{ \\minus{} m}b^m \\ge ab^{ \\minus{} 1} \\plus{} a^{ \\minus{} 1}b \\iff a^{2m} \\plus{} b^{2m} \\ge a^{m \\plus{} 1}b^{m \\minus{} 1} \\plus{} a^{m \\minus{} 1}b^{m \\plus{} 1}$. And this is obvious by the rearrangement inequality. And this concludes the proof of a, b and c. :)\r\n\r\nEdit: The last ineq can also be proved by: $ a^{2m} \\plus{} b^{2m} \\minus{} a^{m \\plus{} 1}b^{m \\minus{} 1} \\minus{} a^{m \\minus{} 1}b^{m \\plus{} 1} \\equal{} (a^{m\\plus{}1}\\minus{}b^{m\\plus{}1})(a^{m\\minus{}1}\\minus{}b^{m\\minus{}1}) \\ge 0$, and here is it also easier to see that it only holds when $ |m| \\ge 1$.", "Solution_2": "Let $ f$ be a positive function defined on the reals.\r\nConsider the positive:\r\n\\[ \\int\\int \\left(\\frac{1}{f^m(x)f^m(y)}\\right)(f^{m\\plus{}1}(x)\\minus{}f^{m\\plus{}1}(y))(f^{m\\minus{}1}(x)\\minus{}f^{m\\minus{}1}(y)) dx dy\\]\r\n\\[ \\equal{}\\int\\int \\left[f^{m}(x)f^{\\minus{}m}(y)\\plus{}f^{\\minus{}m}(x)f^{m}(y)\\right]\\minus{}\\left[f(x)f^{\\minus{}1}(y)\\minus{}f^{\\minus{}1}(x)f(y)\\right] dx dy\\]\r\n\\[ \\equal{}2\\left(\\int f^{m}(x)dx\\int f^{\\minus{}m}(x)dx \\minus{} \\int f(x)dx \\int f^{\\minus{}1}(x)dx\\right)\\geq 0\\]\r\nNow all the previous inequalities and a plethora more fall out by taking various functions $ f$.", "Solution_3": "More directly (again, $ f: \\mathbb{R}\\to \\mathbb{R}^{\\plus{}}$)\r\n\\[ \\frac{\\partial\\left(\\int f^m\\int f^{\\minus{}m}\\right)}{\\partial m}\\equal{}\\int \\int \\left(\\frac{f(x)}{f(y)}\\right)^m(\\ln f(x)\\minus{}\\ln f(y))dx dy\\]\r\n[hide=\" :| \"]\\[ \\equal{}\\frac12\\int \\int\\left[\\left(\\frac{f(x)}{f(y)}\\right)^m\\minus{}\\left(\\frac{f(x)}{f(y)}\\right)^{\\minus{}m}\\right]\\ln\\left(\\frac{f(x)}{f(y)}\\right)dxdy\\]\n[/hide]\r\n\\[ \\equal{}\\frac12 \\int\\int h\\left(\\frac{f(x)}{f(y)}\\right)dxdy\\geq 0\\]\r\nwhere $ h(a)\\equal{}(a^m\\minus{}a^{\\minus{}m})\\ln a\\geq 0$ for real $ a$.", "Solution_4": "Those inequalities can be proved using Cauchy.\r\n\r\nIf $ x_1, ..., x_n$ are positive numbers and $ m$ and $ l$ are real numbers then, by Cauchy:\r\n\\[ \\dfrac{x_1^{m \\plus{} l} \\plus{} ... \\plus{} x_n^{m \\plus{} l}}{x_1^m \\plus{} ... \\plus{} x_n^m}\\ge\\dfrac{x_1^m \\plus{} ... \\plus{} x_n^m}{x_1^{m \\minus{} l} \\plus{} ... \\plus{} x_n^{m \\minus{} l}}\r\n\\]\r\nFor the first one we use repeatedly\r\n\\[ \\dfrac{a^2 \\plus{} b^2 \\plus{} c^2}{a \\plus{} b \\plus{} c}\\ge\\dfrac{a \\plus{} b \\plus{} c}{1 \\plus{} 1 \\plus{} 1}\\ge\\dfrac{1 \\plus{} 1 \\plus{} 1}{a^{ \\minus{} 1} \\plus{} b^{ \\minus{} 1} \\plus{} c^{ \\minus{} 1}}\\ge\\dfrac{a^{ \\minus{} 1} \\plus{} b^{ \\minus{} 1} \\plus{} c^{ \\minus{} 1}}{a^{ \\minus{} 2} \\plus{} b^{ \\minus{} 2} \\plus{} c^{ \\minus{} 2}}\r\n\\]\r\nThe general form as Mathias_DK stated, can be proved in a similar way.\r\n:)\r\n\r\n____\r\nit is interesting to note the close connection between Cauchy and Rearrangement here, don't you think?", "Solution_5": "[quote=\"ElChapin\"]\nit is interesting to note the close connection between Cauchy and Rearrangement here, don't you think?[/quote]\r\nWe can actually prove the rearrangement by h\u00f6lder, which is just cauchy generalised, so it doesn't suprise me much :) A lot of inequalitys are almost equivalent. E.g. is cauchy equivalent to weighted AM-QM, so Cauchy kan be proved by weighted Jensens. It is actually quite funny how all inequalities seems to be related in one way or another.", "Solution_6": "Yes, that is exactly what I meant. :D A chain I like a lot is the following\r\n\r\nprove Jensen (the classical induction proof is fine)\r\nthen use Jensen to prove weighted AM-GM (in fact this can also be proved with induction)\r\nthen use AM-GM to prove H\u00f6lder \r\nthen use H\u00f6lder to prove power mean inequality\r\n(and thanks to what Mathias_DK said) then use H\u00f6lder to prove rearrangement \r\n (and all the permutations [?] jajajaja) \r\n\r\n\r\nMath is just sublime :)" } { "Tag": [ "Gauss", "number theory proposed", "number theory" ], "Problem": "Let $(a)_n$ be the residue when $a$ is divided by $n\\ (n=2,3,\\cdots)$ and $a,b,c$ be positive integers.\r\nProve that \r\n\r\n\\[\\left[\\frac{a(bc)_n}{n}\\right]+a\\left[\\frac{bc}{n}\\right]=\\left[\\frac{b(ca)_n}{n}\\right]+b\\left[\\frac{ca}{n}\\right]=\\left[\\frac{c(ab)_n}{n}\\right]+c\\left[\\frac{ab}{n}\\right]\\].\r\n\r\nNote:$[\\ ]$ denotes Gauss' sign.", "Solution_1": "$\\left[\\frac {a(bc)_n} n\\right] +a\\left[\\frac {bc} n\\right] $ $=$\r\n \r\n $\\left[\\frac{a((bc)_n+n[\\frac {bc} n])} {n}\\right]$ $=$ $\\left[\\frac {abc} n\\right]$\r\n \r\n Similarily we have two more identities.", "Solution_2": "Yes,you are right, nalpaction. :)" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Let $f$ and $f_{n}, n \\in \\mathbf{N}$ be functions from $\\mathbf{R}$ to $\\mathbf{R}$. Assume that $f_{n}(x_{n}) \\rightarrow f(x)$ as $n \\rightarrow \\infty$ and $x_{n}\\rightarrow x$. Show that $f$ is continuous.\r\n\r\nThanks!", "Solution_1": "It should be clear that $f_{n}(x)\\to f(x)$ for all $x$. \r\nWe want to prove that $f(y_{k})\\to f(y)$ as $y_{k}\\to y$. For each $k$ choose $n_{k}$ so that $|f_{n}(y_{k})-f(y_{k})|<1/k$ for $n\\ge n_{k}$. Also, make sure that $n_{k}>n_{k-1}$. Let $x_{n}=y_{k}$, where $k$ is chosen so that $n_{k}\\le nb_0c_0+1$, which is false. $R(1)$ is equally easy to check. Let\u2019s now prove $P(1)$: $c_1|a_1b_1+1$ because of the definition of $c_1$, $a_1|b_1c_1+1\\iff a_0b_0|b_0c_0^2+c_0+a_0$, which is true because $a_0|b_0c_0+1,b_0|c_0+a_0,(a_0,b_0)=1$, and $b_1|c_1+a_1\\iff c_0a_0|a_0b_0+b_0c_0+1$, which is true because $a_0|b_0c_0+1,c_0|a_0b_0+1,(c_0,a_0)=1$. We have thus shown that $P(1),Q(1),R(1)$ hold, and, in particular, we can define $(a_2,b_2,c_2)=(b_1,c_1,\\frac{b_1c_1+1}{a_1})$, and so on. We use the exact same arguments to prove that $P(n),Q(n),R(n)$ are true, so we can define $(a_{n+1},b_{n+1},c_{n+1})=(b_n,c_n,\\frac{b_nc_n+1}{a_n})$.\r\n\r\nOn the other hand, a simple computation reveals that if $(a_n,b_n,c_n)$ is a solution, then so is $(a_{n+1},b_{n+1},c_{n+1})$, so we have found infinitely many solutions (that\u2019s because the inequalities in $Q(n)$ become strict for $n\\ge 2$, due to $R(n)$).", "Solution_3": "Consider the sequence x[0] = x[1] = 1 and, for k>1, x[k+1] = 3.x[k]-x[k-1].\r\nFor k>1, we have x[k+1]>x[k].\r\nAs x[k+1]^2 -3.x[k+1].x[k] + x[k]^2 + 1 = 0, it follows, that the given relation is verified with m = 12 for\r\na = x[k], b = x[k+1], c = x[k] + x[k+1].", "Solution_4": "I think $ m\\equal{}12$ is the only solution.\r\n\r\nAssume $ a\\ge b\\ge c$, then\r\n\r\n\\[ a^2\\plus{}a(b\\plus{}c\\plus{}\\dfrac{1\\minus{}(m\\minus{}1)bc}{b\\plus{}c})\\plus{}bc\\plus{}1\\equal{}0\\]\r\n\r\nWe have $ a|bc\\plus{}1$. Suppose $ a\\plus{}b\\plus{}c$ has the smallest sum, and let $ a_1$ be the other solution to this equation. \r\n\r\nWe have $ aa_1\\equal{}bc\\plus{}1$, but $ a_1\\ge a\\ge b$. So either $ a^2\\equal{}bc\\plus{}1$ or $ a\\equal{}b\\equal{}c\\equal{}1$. The former is impossible. Therefore the only possible $ m$ is $ 12$. We can use root flipping to generate the other solutions. $ (a,b,c)$ to $ (b,c, (bc\\plus{}1)/a)$. \r\n\r\nFinally, I [i]think[/i] root flipping basically guarantees $ (bc\\plus{}1)/a$ to be an integer from the quadratic equation itself.", "Solution_5": "HEY! we need a solution with m=12 to start root flipping. In the case (1,1,1), m=9, and the root flipping gives nothing...", "Solution_6": "When $m=12$,I think we may prove:\n$2a_{2n}=a_{2n+1}+a_{2n-1}$\n$3a_{2n+1}=a_{2n}+a_{2n+2}$", "Solution_7": "[quote=\"nealth\"]I think $ m=12$ is the only solution.[/quote]\nAccording to the official solutions [url=http://anhngq.files.wordpress.com/2010/07/imo-2002-shortlist.pdf]here[/url], \"using a little more theory on quadratic forms, it can be shown that if the equation is soluble for a given value of $m$ then there are infinitely many solutions for that value of $m$.\" They also construct solutions for infinitely many $m$, but don't provide any details on how to actually use quadratic forms for fixed $m$ when there exists a solution.\n\nIn particular, does anyone see how to construct infinitely many solutions for $m = 15$ ($(a,b,c) = (1,4,5)$ works)? The root flipping argument also generates $(1,4,1)$, $(21,4,5)$, and $(21,4,17)$, but nothing else from there...", "Solution_8": "Note that we can expand to get $(a+b+c)(ab+bc+ca+1)=mabc$, so rearranging gives a quadratic in $a$: $(b+c)a^2+((b+c)^2+(1-m)bc+1)a+(b+c)(bc+1)=0$. Using a Vieta Jumping argument, if $(a,b,c)$ is a solution then so is $\\left(\\dfrac{bc+1}{a},b,c\\right)$. So we take $a=b=c=1$, so then $m=12$. We define the sequence $(a_n)_{n\\ge 1}$ by $a_1=a_2=a_3=1$ and $a_n=\\dfrac{a_{n-1}a_{n-2}+1}{a_{n-3}}$ for all $n\\ge 4$. It is easy to show that $a_n>a_{n-1}$ for $n\\ge 4$, so if we can show the sequence only contains integers, then note that $(a_n,a_{n+1},a_{n+2})$ gives us a different solution to the equation for $m=12$ for all positive integers $n$. We will induct to show that $a_n=4a_{n-2}-a_{n-4}$ for all $n\\ge 5$, from which the result will follow. Note that $a_4=2$ and $a_5=3$, so the base case holds. Now, assume $a_k=4a_{k-2}-a_{k-4}$ for some $k\\ge 5$, and we will show that $a_{k+1}=4a_{k-1}-a_{k-3}$:\n\n\\begin{align*} a_k&=4a_{k-2}-a_{k-4}\\\\\n\\implies a_{k-1}a_k&=4a_{k-1}a_{k-2}-a_{k-1}a_{k-4}\\\\\n\\implies a_{k-1}a_k&=4a_{k-1}a_{k-2}-(a_{k-2}a_{k-3}+1)\\\\\n\\implies a_{k-1}a_k+1&=a_{k-2}(4a_{k-1}-a_{k-3})\\\\\n\\implies a_{k+1}a_{k-2}&=a_{k-2}(4a_{k-1}-a_{k-3})\\\\\n\\implies a_{k+1}&=4a_{k-1}-a_{k-3},\n\\end{align*}\n\nas desired.", "Solution_9": "My solution is quite different from others; therefore I wanna share it with Aops users. :D\nGranted, we take $m=12$ and choose $a=b+c$. After some algebra, it remains to show that there are infinitely many positive integers $b,c$ such that $b^2$ $+$ $c^2$ $+$ $1$ $=$ $3bc$. As this problem is very famous and quite easy to deal with it, we can even find that the solution set of aforementioned equation is increasing, giving us solution set of $a,b,c$ infinite. :)", "Solution_10": "[quote=Systematicworker]My solution is quite different from others; therefore I wanna share it with Aops users. :D\nGranted, we take $m=12$ and choose $a=b+c$. After some algebra, it remains to show that there are infinitely many positive integers $b,c$ such that $b^2$ $+$ $c^2$ $+$ $1$ $=$ $3bc$. As this problem is very famous and quite easy to deal with it, we can even find that the solution set of aforementioned equation is increasing, giving us solution set of $a,b,c$ infinite. :)[/quote]\n\nHow famous? Never heard of the equation $(b-c)^2 = bc-1$ before.", "Solution_11": "In fact, for the equation to have integer solutions, $5c^2-4$ would have to be a square, which occurs infinitely often by Pell's equations and the initial solution $(1,1)$.", "Solution_12": "[quote=benstein]In fact, for the equation to have integer solutions, $5c^2-4$ would have to be a square, which occurs infinitely often by Pell's equations and the initial solution $(1,1)$.[/quote]\n\nCan you explain more about why it is connected with Pell\u2019s function? ", "Solution_13": "By Pell theory, any equation of the form $px^2-y^2=q$ has infinitely many solutions iff it has $1$. In this case, the equation is $5c^2-y^2=4$, which has the initial solution $(c,y)=(1,1)$.", "Solution_14": "The answer is yes with $m=18$. \n[color=#00f]Lemma. [/color]The equation $3c^2-11k^2=4$ has infinitely many solutions in positive integer.\n[color=#00f]Proof.[/color] Let $c=4u+22v$ and $y=2u+12v$ the equation become\n$$u^2-33v^2=1$$\nNotice that $(u,v)=(23,4)$ is a solution. Therefore by Pell's equation we obtain infinitely many solution of it. Hence there are infinitely many solutions of $3c^2-11k^2=4$.\n\nNow letting $m=18$. We show that a solution exists for infinitely many values of $c$. \nAfter clearing denominator, the equation transform into\n$$18abc=(a+b+c)(ab+bc+ca+1)$$\nSubsitiute $x=ab$ and $y=a+b$. Notice that for any pair of integer $x,y$, there exists positive integers satisfying $x=ab$ and $y=a+b$ if and only if $y^2-4x=k^2$ for some $k\\in\\mathbb N$. Now treating $c$ as a constant we have\n\\begin{align*}\n18xc&=(y+c)(x+cy+1)\\\\\nx(17c-y)&=(y+c)(cy+1)\n\\end{align*}\nWe now make the crucial step: there exists a solution which satisfies $c=2y$ for infinitely many $y$. Indeed, substituting $c=2y$ we have\n\\begin{align*}\n33xy&=3y(2y^2+1)\\\\\nx&=\\frac{2y^2+1}{11}\n\\end{align*}\nHence \n\\begin{align*}\nk^2&=y^2-4x\\\\\n&=\\frac{3c^2-4}{11}\\\\\n3c^2-11k^2&=4\n\\end{align*}\nFrom the lemma there exists infinitely many solutions, each of which induces a pair of $(a,b,c)$. This completes the proof.", "Solution_15": "[quote=benstein]By Pell theory, any equation of the form $px^2-y^2=q$ has infinitely many solutions iff it has $1$. In this case, the equation is $5c^2-y^2=4$, which has the initial solution $(c,y)=(1,1)$.[/quote]\n\nI see. Thank for your explaining.", "Solution_16": "Good problem. It takes some time to get an equation where you can use vieta jumping.\n\nThe answer is yes, and the example is $m=12$.\nFirst the equation is the same as: $$ \\frac{(a+b+c)(ab+bc+ca+1)}{abc}=m$$\nSo $c \\mid (a+b)(ab+1)$ after seeing that the degree of $c$ is $1$ (This was my motivation). We can guess that $c=a+b$, so if $m=12, c=a+b$ we have: \n$$\\frac{2(ab +(a+b)^2+1}{ab}=12 \\implies \\frac{a^2+b^2+1}{ab}=3$$\nNow we need to prove that exists infinites pairs such that: $a^2 + b^2+1-3ab=0$, the trival solution is $(1,1)$ and if $(a,b)$ is a solution with $a\\leq b$ implies that $(b,3b-a)$ is also a solution but with largest maximum number so there are infinites pairs. $\\blacksquare$", "Solution_17": "We claim that $m=12$ works. Let $c=a+b$, then we will find infinitely many solutions to:\n$$\\frac1a+\\frac1b+\\frac1{a+b}+\\frac1{ab(a+b)}=\\frac6{a+b}$$\n$$\\Leftrightarrow a^2-3ab+b^2+1=0.$$\nWe start with the initial solution $(a,b)=(1,1)$. Using the method of Vieta jumping, suppose a solution $(a,b)$ exists with $a+b$ maximal. WLOG $a\\le b$, then $(3b-a,b)$ is another solution but $3b-a+b\\ge a+b$, a contradiction.\nSo infinitely many solutions must exist.\n", "Solution_18": "Consider the following sequences: \n\\[a_0=1,a_1=1,a_2=1,a_3=2,a_n=\\frac{a_{n-1}a_{n-2}+1}{a_{n-3}}\\]\n\\[b_0=1,b_1=1,b_2=1,b_3=2,b_n=4b_{n-2}-b_{n-4}\\]\nSuppose $a_i=b_i$ for all $i\\le n$ then \n\\begin{align*}\na_n &= \\frac{a_{n-1}a_{n-2}+1}{a_{n-3}} \\\\\n&= \\frac{(4a_{n-3}-a_{n-5})a_{n-2}+1}{a_{n-3}} \\\\\n&= \\frac{4a_{n-2}a_{n-3}-a_{n-5}a_{n-2}+1}{a_{n-3}} \\\\\n&= \\frac{4a_{n-2}a_{n-3}-a_{n-5}\\frac{a_{n-3}a_{n-4}+1}{a_{n-5}}+1}{a_{n-3}} \\\\\n&= \\frac{4a_{n-2}a_{n-3}-a_{n-3}a_{n-4}}{a_{n-3}} \\\\\n&= 4a_{n-2}-a_{n-4}\n\\end{align*}\nas desired. Since $a_{i},a_{i+1},a_{i+2}$ being a solution implies $a_{i+1},a_{i+2},a_{i+3}$ being a solution, we are done.", "Solution_19": "[i]Motivation.[/i] Observe the degree of each variable after multiplying both sides by $abc(a+b+c)$. Once you know that the degree is $2$, or equivalently we have a quadratic equation wrt each variable individually, Vieta Jumping idea hits instantly. So, if we have at least one triple $(a, b, c)$, we can construct the \"larger\" one. For one solution, just take $(a,b,c)=(1,1,1)$ for which $m=12$. \n\n[i]Solution.[/i] $m=12$ works. \nProof. WLOG assume that $\\min\\{a, b, c\\} = a$. Rewrite the equation as $(b+c)a^2 + (b^2 - 9bc + c^2 - 1)a + (b+c)(bc+1) = 0$. Hence, $a$ is a solution to the quadratic equation $(b+c)x^2 + (b^2 - 9bc + c^2 - 1)x + (b+c)(bc+1) = 0$. Let $a'$ be the second root. by Vieta, $a'a=bc+1 \\implies a' = \\frac{bc+1}{a} \\ge \\frac{a^2 + 1}{a} > a$. So we can construct the triple $(a', b, c)$ from $(a, b, c)$, having larger minimum, so we can construct an infinite number of new triples and we are done. ", "Solution_20": "Very nice solution!", "Solution_21": "We'll prove that for $m = 12$, there are infinitely many triples $(a, b, c)$ such that $a = b + c$ and $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{abc} = \\frac{12}{a + b + c}$.\n\nNote that the given equality is equivalent to $a^2 + a(b + c - \\frac{11bc - 1}{b + c}) + (bc + 1) = 0$. Substituting $a = b + c$, we get $b^2 - b \\cdot 3c + (c^2 + 1) = 0$. Thus we only need to prove that the above equation has infinitely many solutions. Note that $b = c = 1$ is a solution of above equation. WLOG, assume $b \\le c$. Then root of the quadratic equation $x^2 - x \\cdot 3c + (c^2 + 1) = 0$ is $b$ and let $b_1$ be the other root. Then $b + b_1 = 3c$ and $bb_1 = (c^2 + 1)$, hence $b_1$ is integer and $b_1 = \\frac{c^2 + 1}{b} > b$. Hence swapping $(b, c) \\to (b_1, c)$, we see that there are infinitely many solutions. $\\blacksquare$ " } { "Tag": [ "MATHCOUNTS", "AMC", "USA(J)MO", "USAMO", "summer program", "MathPath" ], "Problem": "What was your rank at states?\r\nNo posting of scores.", "Solution_1": "lol that was a prediction. States for me is in 7 days.", "Solution_2": "fanatic was first in IN", "Solution_3": "slamchess who are you?", "Solution_4": "my goal=2nd place states... if possible, 1st :)", "Solution_5": "17... :( :( :(", "Solution_6": "numero uno", "Solution_7": "unfair, can you make it like less than 35?\r\n\r\ncause i got 34th\r\n\r\nill describe what happened when we are allowed to talk.\r\n\r\nI am so damn angry!\r\n\r\nnow i can't see leet, preet (rhyming skills), dan, richard, kstan013, etc. at nats\r\n\r\ni'm seriously angry\r\n\r\n-jorian", "Solution_8": "I can't see easyaspi or Ubemaya... :(", "Solution_9": "If your state is like indiana, 5,123,485 people come with anyways. Yes I pulled that number out of nowhere.\r\n\r\nEdit: Ubemaya, I like you, I really do, but your \"OMG I SUCK\" posts are \r\n\r\nA) Annoying\r\nB) Spamtastic.", "Solution_10": "Little bit of an exaggeration there Fanatic, but Indiana does bring more like 18-20 extra people :lol: The last couple years there were times when I could barely get into my room(I was roommates with Mr Fischer) it was so full of Hoosiers!!!! Definitely the state with the most supporters, although this year I bet Texas has more!!", "Solution_11": "I hope to et third or second, as I can't beat leet, and will probably be beat by at least one other person.", "Solution_12": "[quote=\"frost13\"]Little bit of an exaggeration there Fanatic, but Indiana does bring more like 18-20 extra people :lol: The last couple years there were times when I could barely get into my room(I was roommates with Mr Fischer) it was so full of Hoosiers!!!! Definitely the state with the most supporters, although this year I bet Texas has more!![/quote]\r\nWhat's a hoosier?", "Solution_13": "I got Number $4...$ :( :(", "Solution_14": "a person who lives in or is from Indiana.", "Solution_15": "[quote=\"jka386\"]well, I'm not bragging or anything, but I got from a 26th to 2nd. I was like... whoa....[/quote]\n\nI got from 25th to second.\n\nbefore countdown: 2nd\nafter (unofficial)countdown 1st.\n\n[quote=\"kyyuanmathcount\"]Didn't everybody get 3-0'ed by Mathew? I have not yet won a countdown match at states and I have not yet lost a countdown match at chapters. I do countdown with the same people from chapters to states. How does that work?[/quote]\r\n\r\nWhich state?\r\n\r\nHere are the people who got to Nat's from my state:\r\n\r\n1. Dan Mendelsohn 8th\r\n2. Matthew Babbitt(I was expecting fourth..... :huh: ) 7th\r\n3. Felix Sun 8th\r\n4. Schuyler Smith 8th\r\n\r\nTeam: Home Educators Enrichment Group, Dr. O'Keeffe.", "Solution_16": "I don't think so. I was there and I only think 2 people from fairview got top 20. I know someone from fairview was like 18-19th or something. I got 39th...made too many stupid mistakes and didn't have enough past tests (I heard one school from my chapter's gone every year since 1985!). Had I even kept up with people I beat at chapter, I would have ranked in the teens.", "Solution_17": "hate to take the \"we're all winners\" approach, but seriously, how can anyone be disapointed with state rank? By making it to state alone youve beat tons of other people out from your state. I won my chapter and got 17th in ky state-- i would have loved a scholarship but its still amazing that i even had a chance to go to state. Don't be disappointed with 4th, 5th, 20th, or even dead last. By being at states you are the cream of the crop, by even participating in mathcounts you are the cream of the crop. i cant count the number of kids i know who are intimated by the word math alone, can you imagine what they would say if they saw some of the problems we work for mathcounts. Be proud, cause we rock!\r\n\r\nok well thats my BLAAH BLAH BLAH for the day :)", "Solution_18": "[quote=\"luke(mathnerd)\"]hate to take the \"we're all winners\" approach, but seriously, how can anyone be disapointed with state rank? By making it to state alone youve beat tons of other people out from your state. I won my chapter and got 17th in ky state-- i would have loved a scholarship but its still amazing that i even had a chance to go to state. Don't be disappointed with 4th, 5th, 20th, or even dead last. By being at states you are the cream of the crop, by even participating in mathcounts you are the cream of the crop. i cant count the number of kids i know who are intimated by the word math alone, can you imagine what they would say if they saw some of the problems we work for mathcounts. Be proud, cause we rock!\n\nok well thats my BLAAH BLAH BLAH for the day :)[/quote]\r\n\r\nThe first thing that comes to my mind after reading that was:\r\n\r\nI suck at math. I only got 4th place and only got 3rd written. I wish I was better at math. :( :rotfl: :rotfl:", "Solution_19": "[quote=\"LawOfSigns\"]I don't think so. I was there and I only think 2 people from fairview got top 20. I know someone from fairview was like 18-19th or something. I got 39th...made too many stupid mistakes and didn't have enough past tests (I heard one school from my chapter's gone every year since 1985!). Had I even kept up with people I beat at chapter, I would have ranked in the teens.[/quote]\r\nAnd you don't even notice me. The other fairview kid that got 6th. Thanks. JK. You beat eli.", "Solution_20": "whats your name luke?\r\nyou can pm it to me if you feel uncomfortable about releasing your name on the forum\r\ni think you probably know my name", "Solution_21": "[quote=\"kyyuanmathcount\"]Didn't everybody get 3-0'ed by Mathew? I have not yet won a countdown match at states and I have not yet lost a countdown match at chapters. I do countdown with the same people from chapters to states. How does that work?[/quote]\r\n\r\nKevin Wu got 3-1, everyone else did...\r\n\r\nYeah, Allen, I think you mean, you never won, not not yet and never lost.\r\n\r\nI don't know, it just happens like that.\r\n\r\nI mean, the only people I lost to were people who were below me before written, and yeah, just like you.", "Solution_22": "Was anyone here 4th before countdown and 5th after (and the countdown is official in his/her state)", "Solution_23": "All states are done now, right? Does that mean we can discuss problems?", "Solution_24": "We can't until mathcounts.org releases the state test on its website.\r\n\r\nCheck the stickies at the top of the forum page.", "Solution_25": "What is a sticky?", "Solution_26": "The posts that are stickied (always up there even if they're never bumped)", "Solution_27": "[quote=\"Fanatic\"]The posts that are stickied (always up there even if they're never bumped)[/quote]\r\n\r\nYou mean the announcements?", "Solution_28": "Yeah i guess they're called announcements here.\r\n\r\nEvery other forum I've been to has called them stickies.\r\n\r\nWhooops.", "Solution_29": "Top 4 in FL:\r\n1st: Liam Bressler\r\n2nd: Anthony Grebe\r\n3rd: Preet Patel\r\n4th: Andrew Harn\r\n\r\nSorry if I misspelled anyone's names.\r\n\r\nSee you all at nats." } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "How many $ x\\in N , 0\\leqslant x \\leqslant376$ is solution $ 2x^2\\plus{}3x\\plus{}4\\equal{}0(mod 377)$", "Solution_1": "$ 377 \\equal{} 13 \\cdot 29$\r\nSolve mod 13 and 29 and use the CRT (Chinese Remainder Theorem).\r\n\r\nNotice that $ (2,377) \\equal{} 1$, so we can multiply by 2.\r\nComplete the square: $ (2x \\plus{} 3)^2 \\equiv 1 \\pmod {377}$\r\nAgain, because $ (2,377) \\equal{} 1$, we can replace $ 2x \\plus{} 3$ by $ t$. The problem is reduced to solving $ t^2 \\equiv 1 \\pmod {13,29}$. There are 2 solutions for each prime, thus 4 solutions in total between $ 0$ and $ 376$ (the substitution doesn't change it because it is surjective and $ 0, \\cdots ,376$ is a complete residue system)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For $ a,b,c$ positive reals prove that\r\n\r\n$ \\frac{\\sum ab}{\\sum a^2}\\plus{}2\\sum \\frac{a}{b\\plus{}c} \\geq 4$", "Solution_1": "[quote=\"manlio\"]For $ a,b,c$ positive reals prove that\n\n$ \\frac {\\sum ab}{\\sum a^2} \\plus{} 2\\sum \\frac {a}{b \\plus{} c} \\geq 4$[/quote]\r\nLet $ a \\plus{} b \\plus{} c \\equal{} 3u,$ $ ab \\plus{} ac \\plus{} bc \\equal{} 3v^2$ and $ abc \\equal{} w^3.$ Hence, your inequality is equivalent to a linear inequality of $ w^3.$ \r\nId est, it remains to check only two cases: \r\n1) $ b \\equal{} c$ ;\r\n2) $ c\\rightarrow0^ \\plus{} .$", "Solution_2": "[quote=\"arqady\"][quote=\"manlio\"]For $ a,b,c$ positive reals prove that\n\n$ \\frac {\\sum ab}{\\sum a^2} \\plus{} 2\\sum \\frac {a}{b \\plus{} c} \\geq 4$[/quote]\nLet $ a \\plus{} b \\plus{} c \\equal{} 3u,$ $ ab \\plus{} ac \\plus{} bc \\equal{} 3v^2$ and $ abc \\equal{} w^3.$ Hence, your inequality is equivalent to a linear inequality of $ w^3.$ \nId est, it remains to check only two cases: \n1) $ b \\equal{} c$ ;\n2) $ c\\rightarrow0^ \\plus{} .$[/quote]\r\n\r\nWhat does it mean to have a linear inequality? :oops:", "Solution_3": "[quote=\"manlio\"]For $ a,b,c$ positive reals prove that\n\n$ \\frac {\\sum ab}{\\sum a^2} \\plus{} 2\\sum \\frac {a}{b \\plus{} c} \\geq 4$[/quote]\r\n$ \\frac {\\sum ab}{\\sum a^2} \\plus{} 2\\sum \\frac {a}{b \\plus{} c} \\geq \\frac {\\sum ab}{\\sum a^2}\\plus{}2\\frac{(\\sum{a})^2}{2\\sum{ab}}$\r\n$ \\equal{}\\frac{\\sum{ab}}{\\sum{a^2}}\\plus{}\\frac{\\sum{a^2}}{\\sum{ab}}\\plus{}2 \\geq 4$" } { "Tag": [ "calculus", "derivative", "integration", "function" ], "Problem": "If $ \\displaystyle\\int_{t_i}^{t_f}{F}(t) dt \\equal{} \\vec{p_{f}}\\minus{}\\vec{p_{i}}$ implies $ {F}(t)\\equal{}\\frac{dp}{dt}$\r\n\r\nand $ \\displaystyle\\int_{x_i}^{x_f}{F}(x) dx \\equal{} \\minus{}(U_f \\minus{}U_i)$ implies $ F\\equal{} \\minus{}\\frac{dU}{dx}$\r\n\r\nDoes the Kinteic Energry relation\r\n\r\n$ \\displaystyle\\int_{x_i}^{x_f}{F}(x) dx \\equal{} (KE_f \\minus{}KE_i)$ imply $ F\\equal{}\\frac{d(KE)}{dx}$\r\n\r\nThis has been bothering me for some time...", "Solution_1": "No, recall from the first equation you wrote, that force is the derivative of momentum. Now watch:\r\n$ \\frac{d}{dx}(KE)\\equal{}\\frac{d}{dx}\\left(\\frac{mv^{2}}{2}\\right)\\equal{}\\frac{dm}{dx}\\frac{v^{2}}{2}\\plus{}\\frac{m}{2}\\frac{dv}{dx}\\equal{}0\\plus{}mv\\equal{}mv\\equal{}p$\r\n\r\nAlternatively,\r\n$ KE\\equal{}\\frac{mv^{2}}{2}\\equal{}\\frac{p^{2}}{2m}\\Rightarrow\\frac{d}{dx}\\left(\\frac{p^{2}}{2m}\\right)\\equal{}mv\\equal{}p$\r\n\r\nSo again, this yields momentum.", "Solution_2": "Actually, for conservative forces\r\n $ \\vec F = - \\nabla W_p \\;\\;\\;$, where $ W_p$ means potential energy.\r\n\r\n If $ \\vec F$ is the resultant (net force) of conservative forces, $ \\vert F * dP = dW_k$, ($ \\;W_k$ stands for kinetic energy) , and in the unidimensional case we have yours.\r\n\r\n Brief, work or energy is the spatial version for momentum in time.", "Solution_3": "JRav I'm a little confused by your arguments\r\n\r\n[quote]$ \\frac{d}{dx}(KE)\\equal{}\\frac{d}{dx}\\left(\\frac{mv^{2}}{2}\\right)\\equal{}\\frac{dm}{dx}\\frac{v^{2}}{2}\\plus{}\\frac{m}{2}\\frac{dv}{dx}\\equal{}0\\plus{}mv\\equal{}mv\\equal{}p$[/quote]\r\n\r\nFirst I, understand that you are applying the product rule but I dont understand how you got the second term in the equation that comes after the second equal sign (shouldn't it be $ mv\\frac{dv}{dx}$).\r\n\r\nSecond if your equation is correct can you explain how you got the last step where one of the quantities becomes zero and the other becomes mv. \r\n\r\nLastly I'm also confused by the alternative derivation. Particularly how the spatial deriative of kinetic energy as a function of momentum gives momentum. Maybe an elucidation of the previous derivation would help me understand this step\r\n\r\n\r\n[/img]", "Solution_4": "There is a small mistake here:\r\n\r\nAssuming m constant\r\n\r\n$ \\frac {d}{dx}(KE) \\equal{} \\frac {d}{dx}\\left(\\frac {mv^{2}}{2}\\right) \\equal{} mv\\frac {dv}{dx} \\equal{} m\\frac {dx}{dt}\\frac {dv}{dx} \\equal{} m\\frac {dv}{dt} \\equal{} ma \\equal{} F$ (resultant).", "Solution_5": "Ahh you fixed it before I could reply...", "Solution_6": "I'm sorry", "Solution_7": "[quote]Lastly I'm also confused by the alternative derivation. Particularly how the spatial deriative of kinetic energy as a function of momentum gives momentum. Maybe an elucidation of the previous derivation would help me understand this step[/quote]\r\n\r\nI didn't do anything different in my calculations except re-express kinetic energy. Particle physicists frequently describe kinetic energy as $ \\frac{p^{2}}{2m}$ instead of the traditional way because this way makes it easier to measure. You can check easily that this is the same thing as before by simply noting that $ p^{2}\\equal{}m^{2}v^{2}$. Then just divide out the $ m$ and we get the traditional expression.", "Solution_8": "JRav I understood that $ \\frac {1}{2}mv^2 \\equal{} \\frac {p^2}{2m}$ I was just wondering how the derivative of the last term gives momentum. But since Immanuel confirmed my first conjecture I guess the issue is unnecessary.\r\n\r\nSo from Immanuel's post, the derivative (in respect to x) of Kinetic Energy is in fact Force?", "Solution_9": "[quote=\"JRav\"]No, recall from the first equation you wrote, that force is the derivative of momentum. Now watch:\n$ \\frac {d}{dx}(KE) \\equal{} \\frac {d}{dx}\\left(\\frac {mv^{2}}{2}\\right) \\equal{} \\frac {dm}{dx}\\frac {v^{2}}{2} \\plus{} \\frac {m}{2}\\frac {dv}{dx} \\ldots$[/quote]\r\n\r\nThere is a basic differentiation error here.\r\n\r\n\\[ \\frac{\\mathrm{d}}{\\mathrm{d}x}\\left(\\frac{mv^2}{2}\\right) \\equal{} \\frac{1}{2}\\left(\\frac{\\mathrm{d}m}{\\mathrm{d}x}v^2 \\plus{}m\\frac{\\mathrm{d}v^2}{\\mathrm{d}x}\\right) \\\\ \\\\ \\equal{}\\frac{1}{2}\\left(\\frac{\\mathrm{d}m}{\\mathrm{d}x}v^2 \\plus{}2mv\\frac{\\mathrm{d}v}{\\mathrm{d}x}\\right) \\\\ \\\\ \\equal{} \\frac{\\mathrm{d}m}{\\mathrm{d}x}\\frac{v^2}{2} \\plus{}mv\\frac{\\mathrm{d}v}{\\mathrm{d}x} \\]", "Solution_10": "[quote=\"emjay285\"]If $ \\displaystyle\\int_{t_i}^{t_f}{F}(t) dt = \\vec{p_{f}} - \\vec{p_{i}}$ implies $ {F}(t) = \\frac {dp}{dt}$\n\nand $ \\displaystyle\\int_{x_i}^{x_f}{F}(x) dx = - (U_f - U_i)$ implies $ F = - \\frac {dU}{dx}$\n\nDoes the Kinteic Energry relation\n\n$ \\displaystyle\\int_{x_i}^{x_f}{F}(x) dx = (KE_f - KE_i)$ imply $ F = \\frac {d(KE)}{dx}$\n\nThis has been bothering me for some time...[/quote]\r\n\r\nI realize Immanuel has already answered this in part, but consider this:\r\n\\[ KE =\\frac{p^2}{2m}\\Rightarrow \\frac{\\mathrm{d}(KE)}{\\mathrm{d}x}= \\frac{\\mathrm{d}}{\\mathrm{d}x}\\left(\\frac{p^2}{2m}\\right)\\\\ \\\\ =\\frac12\\frac{(m\\cdot 2p\\frac{\\mathrm{d}p}{\\math{d}x} -p^2\\frac{\\mathrm{d}m}{\\mathrm{d}x})}{m^2}\\]\r\n\r\nIn most cases the mass is considered a point particle where $ \\displaystyle \\frac{\\mathrm{d}m}{\\mathrm{d}x} = 0$, therefore the above reduces to:\r\n\\[ \\frac{\\mathrm{d}(KE)}{\\mathrm{d}x} = \\frac12\\frac{(m\\cdot 2p\\frac{\\mathrm{d}p}{\\math{d}x})}{m^2} = \\frac{p}{m}\\frac{\\mathrm{d}p}{\\mathrm{d}x}\\\\ \\\\ = \\frac{\\mathrm{d}x}{\\mathrm{d}t}\\frac{\\mathrm{d}p}{\\mathrm{d}x}\\\\ \\\\ = \\frac{\\mathrm{d}p}{\\mathrm{d}t}\\]\r\n\r\n\\[ \\boxed{\\therefore \\frac{\\mathrm{d}(KE)}{\\mathrm{d}x} = \\frac{\\mathrm{d}p}{\\mathrm{d}t}}\\]", "Solution_11": "And\r\n\\[ {{d\\vec p\\over {dt}} = \\vec F}\\]", "Solution_12": "Ok cool. Thanks Guys :)" } { "Tag": [ "inequalities", "AMC", "USA(J)MO", "USAMO", "search", "function", "inequalities theorems" ], "Problem": "[quote=\"Arne\"]Let $ a$, $ b$, $ c$ be positive real numbers. Prove the inequality\n\\[ \\frac{1}{a\\left(b+1\\right)}+\\frac{1}{b\\left(c+1\\right)}+\\frac{1}{c\\left(a+1\\right)}\\geq \\frac{3}{1+abc}. \\]\n[/quote]\n\n[quote=\"Cezar Lupu\"]Nice and easy. In fact, this is too easy! ;) Rewrite the inequality in the form:\n\n $ \\frac{1+a}{a(1+b)}+\\frac{b(1+c)}{1+b}+\\frac{1+b}{b(1+c)}+\\frac{c(1+a)}{1+c}+\\frac{1+c}{c(1+a)}+\\frac{a(1+b)}{1+a}\\geq 6$. This one follows immediately from AM-GM. ;)[/quote]\r\n\r\nNo Cezar Lupu, you are really wrong. This is very hard. Nobody at the USA IMO training camp 1994 solved this, and the USA IMO team 1994 got 6 x 42. You probably copied the solution from some book or so?", "Solution_1": ":mad: You don't tell me that I copied the solution from a book,cause this is not true.I'm not a cheater :mad: .It is very easy and I maintain my opinion. What's so hard to observe that $\\frac{1+abc}{a+ab}=\\frac{1+a+ab+abc}{a+ab}-1=\\frac{1+a}{a(1+b)}+\\frac{b(1+c)}{1+b}-1$ :P", "Solution_2": "Well, I guess I'd better not say anything anymore :D but it's hard to believe you solved it so quickly, especially if you look at some of your previous posts.", "Solution_3": "Man, why don't you see that this inequality is too easy? :? There are thousands of inequalities much more interresting than this one. ;)", "Solution_4": "Just stop spamming.", "Solution_5": "Ok, sorry. :)", "Solution_6": "Cezar, you know, much of the work of some people can be seen by looking at some of their posts. I don't want to be rude, but when you look at the level of inequalities you posted, I might think Arne is right. Anyway, call it simple or trivial or whatsoever, but I think this inequality is one of the best things I have ever seen. It is a beauty, even if for some on the forum the difficulty or beauty of a problem consists in the length of the proof. And since I'm really tired to see how one of the best forums of the site is going down, I promis not to enter anymore on the inequalities forum. Anyway, I see now too many experts proving what other people have already proved years ago and proving how good they are. So be it, it was a nice job, but now it's nothing. Bye!", "Solution_7": "[quote=\"harazi\"]Cezar, you know, much of the work of some people can be seen by looking at some of their posts. I don't want to be rude, but when you look at the level of inequalities you posted, I might think Arne is right. Anyway, call it simple or trivial or whatsoever, but I think this inequality is one of the best things I have ever seen. It is a beauty, even if for some on the forum the difficulty or beauty of a problem consists in the length of the proof. And since I'm really tired to see how one of the best forums of the site is going down, I promis not to enter anymore on the inequalities forum. Anyway, I see now too many experts proving what other people have already proved years ago and proving how good they are. So be it, it was a nice job, but now it's nothing. Bye![/quote]\r\n\r\nDon't leave.....\r\nMyth has left, I do hope other experts won't leave... :( :(", "Solution_8": "I have seen too many experts on this part of the forum and I have never been an expert. I really hope one day Myth will forgive me and come back. Anyway, this was my last post on this part of the site.", "Solution_9": "You departure will be a great loss to mathlinks. :(", "Solution_10": "What happened between Harazi and Myth??? Why hasn't Myth been around for so long, like more than half a year???\r\n\r\n\r\nBomb", "Solution_11": "No, nothing happened between Harazi and Myth, it's just that they became pissed off because some people posting in the inequalities forum just keep posting nonsense or spamming: that's why they left. To be honest, I am getting pissed off too because of what's happening here.", "Solution_12": ":( :( :( :( \r\nplease harazi ,dont leave ,You are one of the best guys here and i learned a lot from you,\r\nArne is there any thing i can do for good :( :( ?", "Solution_13": "Try to limit spam on the inequalities forum (and ML in general), don't make useless posts, etcetera. I don't think you can do much, except for that.", "Solution_14": "I understand that in the days before happened many things.(Harazi left :( )(I hope he will return and we try to make some rules to continue better. But I think that many times we get out of the limits. Don't forget this is a $MATHEMATICAL$ FORUM not a forum for very much discussion. So I believe that we should left the mathematic symbols to speak ;) :)", "Solution_15": "Silouan is right here. Harazi has left the forum. I hope one day he will forgive me and return on this forum. :( \r\n\r\n So, please for God sake stop $SPAMMING$ :mad:", "Solution_16": "Forgive me for making another non-mathematical post, but... maybe the moderators could make some clear rules to state what is allowed to post or not, what is considered spam and what is not?\r\njust a suggestion. \r\n\r\n[and I think chances for harazi coming back are greater if something really happens about it...]", "Solution_17": "[quote=\"Arne\"]No, nothing happened between Harazi and Myth, it's just that they became pissed off because some people posting in the inequalities forum just keep posting nonsense or spamming: that's why they left. To be honest, I am getting pissed off too because of what's happening here.[/quote]Yes, I will soon start to be more personal with the people doing these spammings ... for now the thread is closed.", "Solution_18": "Well, I don't delete many posts in this forum (although I'm tempted to delete everything on this page) but as a moderator, this is what annoys me:\r\n\r\n1) Problems saying \"This is just Cauchy\" or \"Use Chebyshev then Muirhead\". Write up most of a solution, if you have time. If you don't have time, just outline the solution, and come back to it later.\r\n\r\n2) Posts which attempt to provide a partial solution are welcome. Usually these should provide non-trivial steps, especially if it's implied that the inequality is difficult, but there are so many users at different levels that this is difficult to enforce.\r\n\r\n3) I don't have a problem with people saying 'this is really easy', but 1) + 3) instantly raises my blood pressure. Don't do it.\r\n\r\n4) Inequalities that are obviously false (and not just a sign reversed) - that is, that aren't true as either $\\geq$ or $\\leq$. Do your homework, people. This is much worse if it's coupled with a \"Source: me\" - typos when copying from a book or the internet are natural.\r\n\r\n5) Impatience when waiting for a problem to be solved. Sometimes you have to wait a week. Sometimes two months. Sometimes a couple of years (see the USAMO '00 solution by Ravi B, there :) )\r\n\r\n6) Please try as hard as you can to find the correct source for problems. I know that most of you are not willing to use the search function, but some of us are, and make it easy on us.\r\n\r\n7) Excessive non-mathematical discussion is bad. Many discussions would be better off over PM. Use that function sometimes, please." } { "Tag": [ "LaTeX" ], "Problem": "If the interior angle of a regular polygon is $(\\frac{1140}{7})^{\\circ}$, how many sides does it have?\r\n\r\nWhat is the formula for a question like this?", "Solution_1": "[quote=\"Quevvy\"]If the interior angle of a regular polygon is $(\\frac{1140}{7})^{\\circ}$, how many sides does it have?\n\nWhat is the formula for a question like this?[/quote]\r\n\r\n[hide]Well, the formula is for an n side polygon is 180-360/n.\n\nNow you figure it out.[/hide]", "Solution_2": "[quote=\"mathgeniuse^ln(x)\"][quote=\"Quevvy\"]If the interior angle of a regular polygon is $(\\frac{1140}{7})^{\\circ}$, how many sides does it have?\n\nWhat is the formula for a question like this?[/quote]\n\n[hide]Well, the formula is for an n side polygon is 180-360/n.\n\nNow you figure it out.[/hide][/quote]\n[hide]\n$180-\\frac{360}{n}=\\frac{1140}{7}$\n\n$\\frac{360}{n}= \\frac{120}{7}= \\frac{360}{21}$\n\n$n=21$[/hide]", "Solution_3": "Thanks for the help.\r\n\r\nJust a question about $\\LaTeX$, can I make the parentheses bigger? They look weird like this: $(\\frac{1140}{7})^{\\circ}$\r\n\r\n[code](\\frac{1140}{7})^{\\circ}[/code]", "Solution_4": "[quote=\"Quevvy\"]Thanks for the help.\n\nJust a question about $LaTeX$, can I make the parentheses bigger? They look weird like this: $(\\frac{1140}{7})^{\\circ}$\n\n[code](\\frac{1140}{7})^{\\circ}[/code][/quote]\n\n$\\left(\\frac{1140}{7}\\right)^{\\circ}$\n\n[code]\\left(\\frac{1140}{7}\\right)^{\\circ}[/code]\n\nYou can also do this:\n$\\LaTeX$\n\n[code]\\LaTeX[/code]" } { "Tag": [ "calculus", "integration", "trigonometry", "limit", "calculus computations" ], "Problem": "(1) Calculate $ \\int \\frac {x^3}{x \\minus{} 4}dx$\r\n\r\n(2) Evaluate $ \\int_0^{\\frac {\\pi}{6}} \\frac {1}{\\cos x}dx$\r\n\r\n(3) Find $ \\lim_{n\\to\\infty} \\frac {1}{n^2}\\sum_{k \\equal{} 1}^n k\\sin \\frac {k\\pi}{2n}$", "Solution_1": "(1) Long division. Or first let $ u\\equal{}x\\minus{}4,x\\equal{}u\\plus{}4,$ multiply it out, and then do the division.\r\n\r\n(2) The antiderivative of the secant is one of those things it's best to just have memorized.\r\n\r\n(3) Clearly a Riemann sum. The question is one of detail - what to let $ x$ equal. If we try $ x\\equal{}\\frac{k\\pi}{2n}$ then we need to make it\r\n\r\n$ \\lim_{n\\to\\infty}\\frac4{\\pi^2}\\sum_{k\\equal{}1}^{n}\\left(\\frac{k\\pi}{2n}\\right)\\sin\r\n\\left(\\frac{k\\pi}{2n}\\right)\\cdot\\left(\\frac{\\pi}{2n}\\right)\\equal{}\r\n\\frac4{\\pi^2}\\int_0^{\\frac{\\pi}2}x\\sin x\\,dx$\r\n\r\nIt shouldn't take too long to finish things from here.", "Solution_2": "Can someone solve the problem? :)", "Solution_3": "(2)$ \\int \\sec{x}dx\\equal{}\\ln|\\sec{x}\\plus{}\\tan{x}|\\plus{}c$", "Solution_4": "[quote=\"skywalkerJ.L.\"](2)$ \\int \\sec{x}dx \\equal{} \\ln|\\sec{x} \\plus{} \\tan{x}| \\plus{} c$[/quote]\r\n\r\nThat's correct, but Japanese highschool students don't study the integral formula.\r\n\r\nI myself don't remember it. :oops:", "Solution_5": "$ \\int\\sec{x}dx\\equal{}\\ln|\\sec{x}\\plus{}\\tan{x}|\\plus{}c$\r\n\r\nLet $ u\\equal{}\\sec{x}\\plus{}\\tan{x}$, then $ du\\equal{}(\\sec{x}\\tan{x}\\plus{}\\sec^2{x})dx$\r\n$ \\int\\sec{x}dx$\r\n$ \\equal{}\\int\\frac{\\sec{x}(\\sec{x}\\plus{}\\tan{x})}{\\sec{x}\\plus{}\\tan{x}}dx$\r\n$ \\equal{}\\int\\frac{(\\sec^2{x}\\plus{}\\sec{x}\\tan{x})}{\\sec{x}\\plus{}\\tan{x}}dx$\r\n$ \\equal{}\\int\\frac{1}{u}du$\r\n$ \\equal{}\\ln|\\sec{x}\\plus{}\\tan{x}|\\plus{}C$\r\n\r\nI think perhaps the proof can help to memorize it.", "Solution_6": "$ \\int \\frac{dx}{\\cos x} \\equal{} \\int \\frac{\\cos x }{1\\minus{}sin^2{x}} dx \\equal{} \\left\\|\\begin{array}{c} t\\equal{}\\sin x \\end{array}\\right\\| \\equal{} \\ldots$" } { "Tag": [ "function", "conics", "parabola", "calculus", "derivative", "complex numbers", "complex analysis" ], "Problem": "[b]If [i]f[/i] is a differentiable function on an open disk [i]D[/i] such that its range lies on the parabola {[i]u[/i] + [i]iv[/i] : [i]u, v[/i] are real numbers and [i]v[/i] = [i]u[/i]^2}, then show that [i]f[/i] must be constant.[/b]\r\n\r\n :idea: Here are my thoughts so far: \r\nIf [i]f[/i] has the \"y-values\" [i]u[/i] +[i]iu[/i]^2, then for some [i]z[/i] in the complex numbers, $f(a + ib) = u + iu^2$\r\nSince [i]f[/i] is differentiable, $f'(a + ib) = (u + iu^2)' = 1 + 2iu$.\r\n\r\nSo, if I could show $1+2iu = 0$, perhaps I could find a solution :!: .", "Solution_1": "There is the open mapping theorem:\r\n\r\nSuppose $G$ is open and $f: G\\mapsto \\mathbb{C}$ is analytic and nonconstant. Then for all $z_0\\in G$ and all $\\delta>0$ such that $B(z_0;\\delta)\\subset G,$ then $\\exists\\,\\epsilon>0$ such that $f(B(z_0;\\delta))\\supset B(f(z_0);\\epsilon).$\r\n\r\nThen you can focus less on the exact shape of that parabola and more on the fact that the parabola has no interior.", "Solution_2": "using the cauchy-riemann equations with partial derivatives, you can show it as well. if you work it through you can see that the derivative of [i]u[/i] is zero, so [i]u[/i] is constant. then if [i]u[/i] is constant, [i]v[/i] must be constant, and hence, the function [i]f[/i] must also be constant." } { "Tag": [ "integration", "analytic geometry", "trigonometry", "calculus", "calculus computations" ], "Problem": "Evaluate $ \\int \\int \\int _{E} x^2 dV$ , where $ E$ is bounded by the xz-plane and the hemispheres $ y\\equal{}\\sqrt{9\\minus{}x^2\\minus{}z^2}$ and $ y\\equal{}\\sqrt{16\\minus{}x^2\\minus{}z^2}$", "Solution_1": "$ \\iiint_{E}x^2dV\\equal{}\\iiint_{E}x^2dxdydz$. Changing into spherecial polar coordinate system, we get\r\n$ \\iiint_{E}x^2dV\\equal{}\\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{R}(r\\sin\\theta\\cos\\varphi)^2(r^2\\sin\\theta{dr}d\\theta{d\\varphi})$. \r\n After calculation, the value is $ \\frac{2R^5}{15}\\pi$. \r\n $ R\\equal{}3$ : $ \\frac{162\\pi}{5}$, $ R\\equal{}4$ : $ \\frac{2048\\pi}{5}$." } { "Tag": [ "counting", "derangement", "search" ], "Problem": "I have a set $ A$ of the numbers $ 1$ through $ 10$. For each number, I want to match it with a another number from set $ B$, also of numbers $ 1$ through $ 10$. Each element of $ B$ may only be used once. A match between sets $ A$ and $ B$ may not occur if $ A \\equal{} B$ (i.e. you cannot pair $ 1$ with $ 1$, or $ 4$ with $ 4$) and it also may not occur if $ B \\equal{} A \\plus{} 1$ (i.e. $ A\\equal{}4$, $ B \\equal{}5$ is illegal, $ A \\equal{} 10$, $ B \\equal{} 1$ is illegal).\r\n\r\nGiven these restrictions, how many distinct orders are there in which I can pick from set B? (e.g. 3,4,5,6,7,8,9,10,1,2 is an order which satisfies the conditions)\r\n\r\n(Sorry this problem is kind of hard to explain)", "Solution_1": "It's the derangement problem -- a search for that word on the forum brings up many, many hits." } { "Tag": [ "induction" ], "Problem": "2. Show that, for every positive integer n,\r\n121^n \u2212 25^n + 1900^n \u2212 (\u22124)^n\r\nis divisible by 2000.\r\n\r\nSome hints in answering this question please?", "Solution_1": "Induction is definitely not the way to go.\r\n\r\nStart by considering the prime factorisation of 2000, and then you can kill it with modular arithmetic.", "Solution_2": "Would you have to work modulo 16 and modulo 125?", "Solution_3": "No need to. Some basic factorization does the work.\r\n\r\n[hide=\"Here is how\"]From $a-b|a^{n}-b^{n}$ we get $96|121^{n}-25^{n}\\implies 16|121-25^{n}$ and $1904|1900^{n}-(-4)^{n}\\implies 16|1900^{n}-(-4)^{n}$\n\nLikewise, $1875|1900^{n}-25^{n}\\implies 125|1900^{n}-25^{n}$ and $125|121^{n}-(-4)^{n}$.\n\nTherefore, the expression is divisible by $16\\cdot 125=2000$[/hide]", "Solution_4": "But to be honest I prefer\r\n$121^{n}-25^{n}+1900^{n}-(-4)^{n}\\equiv 9^{n}-9^{n}+(-4)^{n}-(-4^{n}) \\mod 16 \\\\ 121^{n}-25^{n}+1900^{n}-(-4)^{n}\\equiv (-4)^{n}-25^{n}+25^{n}-(-4^{n}) \\mod 125$", "Solution_5": "Really too easy for an Olympiad.\r\n\r\nIts trivial that $ 1900^n$ and $ 25^n$ are multiples of $ 125$. So it's enough to proof that $ 125^n \\minus{} ( \\minus{}4)^n$ is a multiple of 125...\r\n\r\nSimilarly its enough to show that $ 121^n \\minus{} 25^n$ is a multiple of $ 16$...\r\n\r\nI mean, you can simplify more your solution!" } { "Tag": [ "function", "integration", "calculus", "real analysis", "real analysis theorems" ], "Problem": "[b]Theorem[/b]. Let $f(x)$ be a density function of the random variable $X$. If the function given by $y=u(x)$ is differentiable and is either increasing or decreasing for all $x$ in which $f(x)\\neq0$, then for these values of $x$, $y=u(x)$ can be uniquely solved for $x$ to give $x=w(y)$ and the density function for $Y=u(X)$ is given by $g(y) = f[w(y)]\\cdot |w'(y)|$ for $u'(x) \\neq0$. \r\n\r\nThe proof begins with the strictly increasing case:\r\n$P(a Something goes wrong here <---------------*/\n/*------------------------------------------------------------*/\n\nfilldraw((3,5,0)--(3,5,1)--(2,5,1)--(2,5,2)--(1,5,2)--(1,5,4)--(0,5,4)--(0,5,0)--cycle, darkgreen, black); \nfilldraw((4,4,0)--(4,4,2)--(3,4,2)--(3,4,3)--(1,4,3)--(1,4,2)--(2,4,2)--(2,4,1)--(3,4,1)--(3,4,0)--cycle, darkgreen, black); \nfilldraw((5,3,0)--(5,3,1)--(4,3,1)--(4,3,0)--cycle, darkgreen, black); \nfilldraw((4,3,2)--(4,3,3)--(3,3,3)--(3,3,2)--cycle, darkgreen, black); \nfilldraw((2,3,3)--(2,3,4)--(1,3,4)--(1,3,3)--cycle, darkgreen, black); \nfilldraw((5,2,1)--(5,2,4)--(2,2,4)--(2,2,3)--(4,2,3)--(4,2,1)--cycle, darkgreen, black); \nfilldraw((3,1,4)--(3,1,5)--(0,1,5)--(0,1,4)--cycle, darkgreen, black); \nfilldraw((5,0,4)--(5,0,5)--(3,0,5)--(3,0,4)--cycle, darkgreen, black);\n\n/*Supposedly facing the NE, White, draw*/ \ndraw((5,0,0)--(5,0,4)--(5,2,4)--(5,2,1)--(5,3,1)--(5,3,0)--cycle); \ndraw((4,3,0)--(4,3,1)--(4,2,1)--(4,2,3)--(4,3,3)--(4,3,2)--(4,4,2)--(4,4,0)--cycle); \ndraw((3,4,0)--(3,4,1)--(3,5,1)--(3,5,0)--cycle); \ndraw((3,3,2)--(3,3,3)--(3,4,3)--(3,4,2)--cycle); \ndraw((3,0,4)--(3,0,5)--(3,1,5)--(3,1,4)--cycle); \ndraw((2,4,1)--(2,4,2)--(2,5,2)--(2,5,1)--cycle); \ndraw((2,2,3)--(2,2,4)--(2,3,4)--(2,3,3)--cycle); \ndraw((1,4,2)--(1,4,3)--(1,3,3)--(1,3,4)--(1,5,4)--(1,5,2)--cycle); \ndraw((0,1,4)--(0,1,5)--(0,5,5)--(0,5,4)--cycle); \n\n/*Supposedly facing the S, Paleblue, fill*/ \nfilldraw((3,0,5)--(3,1,5)--(0,1,5)--(0,0,5)--cycle, paleblue, black); \nfilldraw((5,0,4)--(5,2,4)--(2,2,4)--(2,3,4)--(1,3,4)--(1,5,4)--(0,5,4)--(0,1,4)--(3,1,4)--(3,0,4)--cycle, paleblue, black); \nfilldraw((4,2,3)--(4,3,3)--(3,3,3)--(3,4,3)--(1,4,3)--(1,3,3)--(2,3,3)--(2,2,3)--cycle, paleblue, black); \nfilldraw((4,3,2)--(4,4,2)--(3,4,2)--(3,3,2)--cycle, paleblue, black); \nfilldraw((2,4,2)--(2,5,2)--(1,5,2)--(1,4,2)--cycle, paleblue, black); \nfilldraw((5,2,1)--(5,3,1)--(4,3,1)--(4,2,1)--cycle, paleblue, black); \nfilldraw((3,4,1)--(3,5,1)--(2,5,1)--(2,4,1)--cycle, paleblue, black); \nfilldraw((5,3,0)--(5,5,0)--(3,5,0)--(3,4,0)--(4,4,0)--(4,3,0)--cycle, paleblue, black); \n\n/*Grid lines moving torwards x, parallel NE*/ \ndraw((1,0,5)--(1,1,5)--(1,1,4)--(1,5,4)--(1,5,0)); \ndraw((2,0,5)--(2,1,5)--(2,1,4)--(2,3,4)--(2,3,3)--(2,4,3)--(2,4,2)--(2,5,2)--(2,5,0)); \ndraw((3,1,4)--(3,2,4)--(3,2,3)--(3,4,3)--(3,4,0)); \ndraw((4,0,4)--(4,2,4)--(4,2,3)--(4,3,3)--(4,3,2)--(4,4,2)--(4,4,0)--(4,5,0)); \n\n/*Grid lines moving towards y, parallel NW*/ \ndraw((3,1,4)--(5,1,4)--(5,1,0)); \ndraw((0,2,5)--(0,2,4)--(5,2,4)--(5,2,0)); \ndraw((0,3,5)--(0,3,4)--(2,3,4)--(2,3,3)--(4,3,3)--(4,3,0)--(5,3,0)); \ndraw((0,4,5)--(0,4,4)--(1,4,4)--(1,4,3)--(3,4,3)--(3,4,2)--(4,4,2)--(4,4,0)--(5,4,0)); \n\n/*Grid lines moving towards z, parallel S*/ \n\ndraw((5,0,1)--(5,2,1)--(4,2,1)--(4,4,1)--(3,4,1)); \ndraw((5,0,2)--(5,2,2)--(4,2,2)--(4,4,2)--(2,4,2)); \ndraw((5,0,3)--(5,2,3)--(2,2,3)--(2,3,3)--(1,3,3)--(1,5,3)); \n\n/*filldraw(--cycle, green, black);draw(--cycle);fill(--cycle, paleblue);*/ \n/*Rotating code (C+P): */ \nlabel(\"X\", (5,0,0), SW); \nlabel(\"Y\", (0,5,0), SE); \nlabel(\"Z\", (0,0,5), N);\n \n[/asy][/code][/hide]\r\n\r\nIt seems that whenever I change anything (add, subtract, change), the entire code suddenly changes. Besides comment making, any time I seem to fix something after the 5th line, the change doesn't appear. If I put it before, then the change appears but the rest of the code disappears. So I tried movind the 5th line to the very end, and this made the code unparseable. Is this some kind of copyright issue, or just an annoying bug?", "Solution_1": "Hello,\r\n\r\n1. your code is incompatible with the last version of Asymptote;\r\n2. your code is not optimized at all.\r\n\r\nUpdate Asymptote and your code to version 1.86", "Solution_2": "I don't mind the code length. So how would I update if I made this on the forums? Thanks." } { "Tag": [ "inequalities solved", "inequalities" ], "Problem": "Let a,b,c>0 \r\na 3/4 +b 3/4 +c 3/4 =3\r\nProve that \r\na2 +b 2 +c 2 +21 \\geq 4sum ( \\sqrt ((a+b)(a+c)))", "Solution_1": "This really is very difficult. Put a^1/4=x and so on. Then x^3+y^3+z^3=3 and by a problem of Vasile Cartoaje it follows that sum (xy)^4<=3, which means sum ab<=3. So, 4 sum sqrt(a+b)(a+c)<=4 sum sqrt(3+a^2)<= sum a^2+21, since it reduces to sum (2-sqrt(a^2+3))^2>=0." } { "Tag": [ "probability", "function", "integration", "calculus", "real analysis", "inequalities", "advanced fields" ], "Problem": "Definition: Let $(\\Omega,\\Sigma,\\mu)$ be a probability space. i.e. a measure space with $\\mu(\\Omega)=1$. Suppose $\\Sigma_{1}$ is a sub-$\\sigma$-algebra of $\\Sigma$. Then for every $\\mu$-integrable function$f$, there is a function $g$ that is $\\Sigma_{1}$-measurable and $\\mu|_{\\Sigma_{1}}$-integrable such that $\\int f \\chi_{E} d\\mu = \\int g \\chi_{E} d\\mu$ for all $E \\in \\Sigma_{1}$. g is called a conditional expectation of f with respect to $\\Sigma_{1}$.\r\n\r\nQuestion: Show that if $f \\in L^{p}(\\Omega,\\Sigma,\\mu)$,$1 \\leq p \\leq \\infty$, then $g \\in L^{p}(\\Omega,\\Sigma,\\mu)$ and $||g||_{p} \\leq ||f||_{p}$.\r\n\r\nThe case $p=1$ is clear from the definition.\r\nFor $1 \\mu (F)>0$ and $c1_{F} > f1_{F}$ and then $\\int_{F} {g} d\\mu > c\\mu (F) > \\int_{F} {f}$, which is absurd). Whence $c^{p}1_{E} \\leq f^{p}1_{E}$ for every $p \\geq 1$, and for any (positive) integrable simple function $h$ we will then have $hf(a)$. Similarly we can prove that $ f(c)b$, $ f(c)$ must be greater than $ f(b)$ (this follows from our previous argument). If $ cf(b)$ is done by the same way. Hence $ f$ is strictly monotone.", "Solution_2": "Thank you for the solution :D" } { "Tag": [ "geometry", "circumcircle", "trigonometry", "vector" ], "Problem": "Let $ABC$ be a triangle. Triangles $PAB$ and $QAC$ are constructed outside of triangle $ABC$ such that $AP = AB$ and $AQ = AC$ and $\\angle{BAP}= \\angle{CAQ}$. Segments $BQ$ and $CP$ meet at $R$. Let $O$ be the circumcenter of triangle $BCR$. Prove that $AO \\perp PQ.$", "Solution_1": "[quote=\"N.T.TUAN\"]Let $ABC$ be a triangle. Triangles $PAB$ and $QAC$ are constructed outside of triangle $ABC$ \nsuch that $AP = AB$ and $AQ = AC$ and $\\angle{BAP}= \\angle{CAQ}$. Segments $BQ$ and $CP$ meet at $R$. Let $O$ be the circumcenter of triangle $BCR$. Prove that $AO \\perp PQ.$[/quote]\r\n\r\nWell, let's solve this one with complex numbers.\r\n\r\nWe denote $\\angle{BAP}= \\angle{CAQ}=\\phi$ then because the triangles $PAC$ and $QAB$ coincide after rotation with center $A$ and angle $\\phi$, we have that the angle formed by the segments $CP$ and $QB$ is equal to $\\phi$. Then we get that $\\angle{BOC}=2\\phi$ and $BO=CO$. Now we take $A$ as the origin of the complex plane. Then if $z=\\cos{\\phi}+i\\sin{\\phi}$ we have that $p=\\frac{b}{z}$ and $q=cz$ and also $\\frac{b-o}{c-o}=z^{2}$ thus $o=\\frac{z^{2}c-b}{z^{2}-1}=\\frac{q-p}{z-\\frac{1}{z}}$. Hence $\\frac{o}{\\bar{o}}=-\\frac{q-p}{\\bar{q}-\\bar{p}}$ which means that $OA$ is perpendicular to $PQ$, as desired. :wink:", "Solution_2": "Here is my synthetic solution\r\n\r\nThere is no new philosophy in this problem. The idea is how to relate the circumcenter $O$ with $PQ$. Since we have so many pairs of equal angles here, probably it would help if we can take advantage of some hidden cyclic quadrilaterals and some spirality.\r\n\r\nLet $\\measuredangle{PAB}= \\measuredangle{CAQ}= \\alpha$. It is not hard to see that $APBR$ and $AQCR$ are cyclic. Indeed, the congruence of two triangles $APC$ and $ABQ$ gives us $\\measuredangle{ABR}= \\measuredangle{APR}$, $\\measuredangle{AQR}= \\measuredangle{ACR}$. Let's draw the circles $(ARBP)$ and $(ARCQ)$. Denote $U, V$ respectively the intersections of $AO$ with these two circles. \r\n\r\nNotice that $\\measuredangle{BOC}= 2 \\alpha$. Hence, let 's bisect this angle by calling $T$ the midpoint of the arc $BC$ of the circumcircle of $BRC$. Then we have some spiralities here: $BPA$ and $BTO$, $CQA$ and $CTO$ $(*)$. Thus $BUTO$ is cyclic and $P, T, U$ are collinear. Similarly, $COVT$ is cyclic and $C, T, V$ are collinear. \r\n\r\nWe need now to prove $\\measuredangle{PUA}+\\measuredangle{UPQ}= 90^{0}$. This is equivalent to proving $\\measuredangle{TPQ}= \\frac{\\alpha}{2}$. Hence, it is sufficient for us to prove that actually $PTQ$ is isosceles at $T$ and $\\measuredangle{PTQ}= 180^{0}-\\alpha$. The latter is very easy by some angle-chasing. To prve $TP = TQ$, use $(*)$.", "Solution_3": "although it's a little longer than the proofs above, the lemma $AB\\perp CD \\Leftrightarrow AC^{2}+BD^{2}=AD^{2}+BC^{2}$ and some trigonometry solve the problem :D", "Solution_4": "What's surprise! the main idea of my first solution is similar with yours, treegoner. The idea is how to relate the circumcenter O with PQ\r\nThis is my first solution\r\nTriangles $APC$ and $ABQ$ are congruent so $\\angle APR=\\angle ABR$. then $A,P,B,R$ are cyclic. \r\nSo $\\angle BRP=\\angle PAB$. But $\\angle PRB= \\angle BOC/2$ so $\\angle PAB=\\angle BOC/2$\r\nLet the perpendicular line of $BC,AP,AQ$ through $O,B,C$ meet $BC,AP,AQ$ at $D,E,F$.\r\ntriangle $CFA$ and $BEA$ are similar so $AE/AF=AB/AC=AP/AQ$ so $EF//PQ$ $(1)$\r\nWe have triangle $CDO$ and $CFA$ are similar. Then triangle $CFD$ and $CAO$ are similar\r\nthus $FD/AO=CD/OC$ and $\\angle FDC=\\angle AOC$ $(2)$\r\nSimilarly $ED/AO=BD/BO$ and $\\angle BDE=\\angle BOA$ $(3)$\r\nfrom (2) and (3) $DE=DF$ and $\\angle EDF=180-\\angle BOC$\r\nthen $\\angle EFD=\\angle BOC/2=\\angle CAQ$ $(4)$\r\n$FD$ meets $AO$ at $K$. we have $\\angle CFK=\\angle CAK$ so $A,K,C,F$ are cyclic. \r\nthen $\\angle CAQ=\\angle CKF$ $(5)$\r\nfrom (4) and (5) $EF//CK$ $(6)$\r\nfrom (1) and (6) $PQ//CK$\r\nbut $\\angle CKA=\\angle CFA=90$ so $PQ\\bot AO$ $q.e.d$", "Solution_5": "May be this soluiton is shorter\r\nTriangle $APC$ and $ABQ$ are congruent so $\\angle APR=\\angle ABR$. \r\nThen $A,P,B,R$ are on circle $(O1)$. Similarly $A,R,C,Q$ are on circle $(O2)$\r\n$(O1), (O2)$ meet $AQ,AP$ at $E,F$\r\nLet $(I)$ be the circumcircle of triangle $EAF$\r\n$O1I,O2I$ meet $AE,AF$ at $M,N$.\r\nwe have $\\angle AEP=\\angle ARC=\\angle AFQ$ so triangles $AEP, AFQ$ are similar.\r\n$M,N$ are midpoint of $AE,AF$ so triangles $PAM$ and $QAN$ are similar \r\nso $\\angle MPN=\\angle MQN$. Then $P,M,N,Q$ are cyclic\r\nAnd with $I,M,A,N$ are cyclic we have $\\angle AQP=\\angle MNA=\\angle MIA$.\r\nSo $I,M,H,Q$ are cyclic. ($IA$ meets $PQ$ at $H$) Thus $IA\\bot PQ$\r\nBut $PQ$ is the radical axis of $(I)$ and $(O)$ so $IO\\bot PQ$. (because $PA*PF=PR*PC$ and $QA*QE=QB*QR$)\r\nThis means $AO\\bot PQ$ $q.e.d$", "Solution_6": "Wow, amazing second proof, Huyen Vu.", "Solution_7": "[color=darkblue][u][b]May be this metrical proof is longest ![/b][/u] I\"ll show that $OP^{2}-OQ^{2}=c^{2}-b^{2}$, i.e. $AO\\perp PQ$.\n\nDenote $2x=m(\\widehat{BAP})$, the circumcircle $C(O,\\rho )$ of $\\triangle BRC$ and $S\\in AR\\cap BC$. Thus, $\\{\\begin{array}{c}AC=AQ\\\\\\ AP=AB\\\\\\ \\widehat{CAP}\\equiv\\widehat{QAB}\\end{array}\\|$ $\\implies$ $\\triangle ACP\\sim\\triangle AQB$ $\\implies$\n\n$\\{\\begin{array}{c}CP=QB\\\\\\ \\widehat{APC}\\equiv\\widehat{ABQ}\\\\\\ \\widehat{ACP}\\equiv\\widehat{AQB}\\end{array}\\|$ $\\implies$ $\\{\\begin{array}{c}\\widehat{APR}\\equiv\\widehat{ABR}\\\\\\ \\widehat{ACR}\\equiv\\widehat{AQR}\\end{array}\\|$ $\\implies$ the quadrilaterals $APBR$, $AQCR$ are cyclically with the radical axis $AR$.\n\nObserve that $\\{\\begin{array}{c}m(\\widehat{BRS})=m(\\widehat{BPA})=90^{\\circ}-x\\\\\\ m(\\widehat{CRS})=m(\\widehat{CQA})=90^{\\circ}-x\\\\\\ m(\\widehat{BRP})=m(\\widehat{BAP})=2x\\end{array}\\|$ $\\implies$ the ray $[RS$ is the bisector of the angle $\\widehat{BRC}$.\n\nProve easily that $\\{\\begin{array}{c}m(\\widehat{BOC})=4x\\\\\\ a=2\\rho\\sin 2x\\end{array}$. Thus, $\\{\\begin{array}{c}PA=c\\ ,\\ QA=b\\\\\\ PB=2c\\cdot\\sin x\\\\\\ QC=2b\\cdot \\sin x\\end{array}$ and $\\{\\begin{array}{c}m(\\widehat{CBO})=m(\\widehat{BCO})=90^{\\circ}-2x\\\\\\ m(\\widehat{PBO})=180^{\\circ}+3x-B\\\\\\ m(\\widehat{QCO})=180^{\\circ}+3x-C\\end{array}$.\n\nApply the generalized Pythagoras' theorem in the triangles $POB$ , $QOC$ to the sides $OP$, $OQ$ respectively :\n\n$\\{\\begin{array}{c}OP^{2}=BP^{2}+\\rho^{2}-2\\rho\\cdot BP\\cos \\widehat{PBO}\\\\\\ OQ^{2}=CQ^{2}+\\rho^{2}-2\\rho\\cdot CQ\\cos \\widehat{QCO}\\end{array}$ $\\implies$ $OP^{2}-OQ^{2}=4(c^{2}-b^{2})\\sin^{2}x+4c\\rho\\sin x\\cos (B-3x)-4b\\rho \\sin x\\cos (C-3x)=$\n\n$4(c^{2}-b^{2})\\sin^{2}x+4c\\rho \\sin x(\\cos B\\cos 3x+\\sin B\\sin 3x)$ $-4b\\rho \\sin x(\\cos C\\cos 3x+\\sin C\\sin 3x)=$\n\n$4(c^{2}-b^{2})\\sin^{2}x+4\\rho\\sin x\\cos 3x$ $(c\\cos B-b\\cos C)+4\\rho \\sin x\\sin 3x(c\\sin B-b\\sin C)=$\n\n$4(c^{2}-b^{2})\\sin^{2}x+4\\rho\\sin x\\cos 3x(\\frac{a^{2}+c^{2}-b^{2}}{2a}-\\frac{a^{2}+b^{2}-c^{2}}{2a})=$ $4(c^{2}-b^{2})\\sin^{2}x+\\frac{4\\rho (c^{2}-b^{2})}{a}\\sin x\\cos 3x=$\n\n$(c^{2}-b^{2})(4\\sin^{2}x+\\frac{2\\sin x\\cos 3x}{\\sin 2x})=c^{2}-b^{2}$. I used the simple relations $c\\sin B=b\\sin C$ and $4\\sin^{2}x+\\frac{2\\sin x\\cos 3x}{\\sin 2x}=1$.[/color]", "Solution_8": "This problem was in Brazilian Training Lists!\r\nI solved this using Spiral Similarity, some Trigonometry and the lemma mentioned by campos!", "Solution_9": "I found a really nice proof:\r\n\r\n[hide]Consider this sequence of three rotations:\n\n(1) about A of angle $ \\minus{}\\theta$\n(2) about O of angle $ 2\\theta$\n(3) about A of angle $ \\minus{}\\theta$\n\n1. $ APBR$ is cyclic\n\nWe have the spiral similarity (congruence, actually)\n$ \\triangle PAB \\sim \\triangle CAQ$, so $ \\triangle PAC \\sim \\triangle BAQ$, implying that $ \\angle APC \\equal{} \\angle ABQ$, so $ \\angle APR \\equal{} \\angle ABR$ so $ APBR$ is cyclic\n\n2. $ \\angle PRB \\equal{} \\theta$\n\nfollows from (1) after noting that $ \\angle PRB \\equal{} \\angle PAB \\equal{} \\theta$\n\n3. $ \\angle BOC \\equal{} 2\\theta$\n\nfollows after angle chasing using (2): $ \\angle BRC \\equal{} 180 \\minus{} \\angle PRB \\equal{} 180 \\minus{} \\theta$ and so $ \\angle BOC \\equal{} 2\\theta$\n\n4. rotation (2) takes B to C\n\nfollows from (3) and the fact that $ OB \\equal{} OC$ (since O is the circumcenter of BRC)\n\n5. rotation (1) takes P to B and rotation (3) takes C to Q\n\ntrivial\n\n6. the sequence of rotations (1), (2), (3) take P to Q\n\ntrivial from (4) and (5)\n\n7. the triple rotation is actually a translation by vector $ \\overrightarrow{\\text{PQ}}$\n\nfollows from properties of rotations (we have a net angle of 0, so its a translation)\n\nConsider the image of A in this triple rotation/translation. Thinking of this as a translation, it goes to a point A' such that $ AA' \\parallel{} PQ$ and $ AA' \\equal{} PQ$ (actually, we only need the former). \n\nNow we think of it as a triple rotation:\n(1) The first rotation does nothing. \n(2) The second rotation takes it to a point $ A_1$ such that $ \\angle AOA_1 \\equal{} 2\\theta$ and $ AO \\equal{} OA_1$. This implies that $ \\angle OAA_1 \\equal{} 90 \\minus{} \\theta$.\n(3) The third rotation takes $ A_1$ to a point $ A'$ such that $ \\angle A_1AA' \\equal{} \\theta$.\n\nCombining the angles from (2) and (3), we get that $ \\angle OAA' \\equal{} 90$, so that $ OA \\perp AA'$. Since $ AA' \\parallel{} PQ$, $ OA \\perp PQ$, as desired.[/hide]", "Solution_10": "Let the perpendicular bisectors of $AB$ and $AC$ intersect $AP$ and $AQ$ at $O_1$ and $O_2$, respectively. Since quadrilaterals $ABPO_1$ and $ACQO_2$ are similar, it follows that $AO_1/AP=AO_2/AQ$ and hence that $PQ$ and $O_1 O_2$ are parallel. Now let $\\mathcal{O}_1$ and $\\mathcal{O}_2$ be the circles centered at $O_1$ and $O_2$ with radii $O_1 A$ and $O_2 A$, respectively. Let the second intersection of $\\mathcal{O}_1$ and $\\mathcal{O}_2$ be $K$. By the inscribed angle theorem, it follows that $\\angle{(KB,KA)}=\\angle{(KA,KC)}=90^\\circ - \\angle{(AP, AB)}$ which implies that $\\angle{CK, BK}=2\\angle{(AP, AB)}$. Now note that since $A$ is the center of spiral similarity taking $PB$ to $CQ$, it follows that $APBR$ and $AQCR$ are cyclic which implies that $\\angle{(CR, BR)}=\\angle{(AP, AB)}$ and hence that $\\angle{(CO, BO)}=2\\angle{(AP, AB)}=\\angle{(CK, BK)}$. Therefore $BOCK$ is cyclic and since $AK$ bisects $\\angle{BKC}$ and $OB=OC$, it follows that $A$, $K$ and $O$ are collinear. Hence $AO$ is the radical axis of $\\mathcal{O}_1$ and $\\mathcal{O}_2$, which implies that $AO$ is perpendicular to $O_1 O_2$ and therefore perpendicular to $PQ$ since $PQ$ is parallel to $O_1 O_2$.", "Solution_11": "Hello, I have a very funny solution to this problem.\n\n[hide=\"Solution\"]\nLet $AB$ and $AC$ hit the circumcircle of $\\triangle BCR$ at $B', C'$, and let $AO$ intersect $BC', B'C$ at $X, Y$ and let $BC'$ interset $B'C$ at $Z$ and let $BC$ intersect $B'C'$ at $P$. \n\nClearly we have $\\angle BB'C=\\angle BC'C=\\angle PAB=\\angle QAC$. Ok so we really just need to prove $\\dfrac{\\cos \\angle OAP}{\\cos \\angle OAQ}=\\dfrac{AQ}{AP}$; RHS is obviously $\\dfrac{AC}{AB}$. \n\nNow HMMM darn cosines wat? well we have $\\angle OAP=\\angle XYZ$ and $\\angle OAQ=\\angle ZXY$...now wait Brokard says $PZ \\perp XY$ - is this \nreally going to work - !\n\nWel we have $\\LHS =\\dfrac{\\cos \\angle XYZ}{\\cos \\angle YXZ}=\\dfrac{\\sin \\angle PZC}{\\sin \\angle PZC'}=\\dfrac{PC}{PC'}=\\dfrac{AC}{AB}$ - it works!\n\nQED\n[/hide]", "Solution_12": "Yes ! But the formulation of the problem screams for a complex bash", "Solution_13": "Dear Mathlinkers,\nyou can also see :\nhttp://perso.orange.fr/jl.ayme vol. 16 Deux triangles adjacents... p. 62 which indicate p. 60-61 .\n\nSincerely\nJean-Louis", "Solution_14": "My solution:\n\nLet $ O' $ be the circumcenter of $ \\triangle ACQ $ .\nLet $ M, N $ be the midpoint of $ CQ, CR $, respectively .\n\nEasy to see $ R \\in (O') $ .\n\nSince $ O', M, N, C $ are concyclic ,\nso we get $ \\angle AO'O=\\angle QCP $ . ... $ (1) $\nSince $ \\angle RO'O=\\angle BQC, \\angle O'OR=\\angle CBQ $ ,\nso we get $ \\triangle ORO' \\sim \\triangle BCQ $ ,\n\nhence $ \\frac{O'A}{CQ}=\\frac{O'R}{CQ}=\\frac{O'O}{QB}=\\frac{O'O}{CP} $ . ... $ (2) $\n\nFrom $ (1) $ and $ (2) $ we get $ \\triangle AOO' \\sim \\triangle QPC $ ,\nso from $ OO' \\perp PC $ and $ AO' \\perp QC \\Longrightarrow AO \\perp QP $ .\n\nQ.E.D", "Solution_15": "My solution:\n\n[u][b]Lemma:[/b][/u] $AO \\perp PQ$ if only if $AP^2-AQ^2=OP^2-OQ^2$ (It's well known)\n It is easy too see that: $\\triangle APC\\cong \\triangle ABQ$ $\\Longrightarrow $ $BQ=CP=1$ and $RP=d,RQ=c$ \nLet $AP=AB=a,CA=CQ=b$ and $AR=x$ $\\Longrightarrow $ By power of point in $\\odot (RBC)$ we get :\n$AP^2-AQ^2=OP^2-OQ^2$ if only if $a^2-b^2=d-c$ since $OP^2-OQ^2=d-c$ (By power of point in $\\odot (RBC)$\nBy Stewart's theorem in $\\triangle ACP$ we get: $a^2(1-d)+b^2d=x^2+d(1-d)...(1)$\nBy Stewart's theorem in $\\triangle ABQ$ we get: $b^2(1-c)+a^2c=x^2+c(1-c)...(2)$\nBy $(1)-(2)$ we get:$a^2-b^2=d-c$ which it is true\n\n$\\Longrightarrow $ $a^2-b^2=d-c$ $\\Longrightarrow $ $AO \\perp PQ$... :P \n\n\n", "Solution_16": "Needing help for a complex number solution, I'm trying to learn that technique at the moment and working through an article by Robin Park. And well, there is a step in the calculation that I can't seem to be able to comprehend.\n\nNoticing those rotations, we can find $p = a+e^{-i \\theta}(b-a)$, $q=a+e^{i \\theta}(c-a)$ and $o= \\tfrac{b-ce^{2i \\theta}}{1-e^{2i \\theta}}$, where $\\theta = \\angle{BAP}$. It remains to verify $\\tfrac{a-o}{\\overline{a}-\\overline{o}}=\\tfrac{p-q}{\\overline{p}-\\overline{q}}$. To ease the amount of calculation we also set $A$ to be the origin. Also, we let $z = e^{i \\theta}$.Then we get \\[ \\frac{o}{\\overline{o}} = \\frac{p-q}{\\overline{p}-\\overline{q}} \\iff \\frac{\\frac{b-cz^2}{1-z^2}}{\\frac{\\frac{1}{b}-\\frac{1}{cz^2}}{1-\\frac{1}{z^2}}} = \\frac{\\frac{b}{z}-cz}{\\frac{z}{b}-\\frac{1}{cz}} \\iff \\dots \\]\nThat above is how the author of the article presented the solution; I didn't formulate it like him. \nNow to my problem: how did he get to that equivalent form? Especially, I assume he got $\\overline{p}-\\overline{q}=\\tfrac{z}{b}-\\tfrac{1}{cz}$. Why is that true?\n\nAlso, admittingly, I'm still very weak at manipulating complex numbers as I rarely use those.\n", "Solution_17": "Here's a solution similar to Robin Park's. Whoops, I'm bad at this complex bashing stuff.\n\nWe use complex numbers.\n\nLet the complex number corresponding to a point be its lowercase letter. Set $a$ as the origin, $c=1$, and $\\angle PAB = \\angle QAC = \\theta$. By standard formulas, we have $p=e^{-i \\theta}b$ and $q=e^{i \\theta}c$. \n\nAlso, I claim $APBR$ and $AQCR$ are both cyclic. Observe $\\triangle PAB \\sim \\triangle CAQ$, with spiral center at $A$, so the standard Yufei spiral similarity configuration tells us $R \\in \\odot(ABP)$ and $R \\in \\odot(ACQ)$. Now angle chase to get $\\angle BOC = 2 \\theta$.\n\nBy the rotation formula, you can get $o = \\frac{b-ce^{i \\theta}}{1-e^{i 2 \\theta}}$. Now we want $\\frac{o-a}{p-q}$ as pure imaginary, that is\n\\[ \\frac{(b-e^{i 2\\theta})}{(1-e^{i 2\\theta})(e^{-i \\theta}b-e^{i \\theta})} = \\frac{-(\\overline{b}-\\frac{1}{e^{i 2\\theta}})}{(1-\\frac{1}{e^{i 2\\theta}})(e^{i \\theta} \\overline{b}-\\frac{1}{e^{i \\theta}})} \\]\nAfter some simplifying, one can see that this is clearly true, so $OA \\perp PQ$ and we are done.", "Solution_18": "Finally got around to actually doing this. Looks like my solution bears resemblance to [b]Huy\u1ec1n V\u0169[/b]'s in post 5, though the last part is a bit strange because I didn't know how to construct the argument elegantly.\n\nConstruct point $T$ outside $\\triangle ABC$ such that $\\triangle TBC\\sim\\triangle ABP\\sim\\triangle ACQ$. By considering the rotation sending $\\triangle APC$ to $\\triangle ABQ$ (which are congruent by SAS) we see that the angle between lines $PC$ and $QB$ is equal to $\\angle BAP = \\angle BTC$. Thus $B$, $C$, $T$, and $R$ lie on a common circle, so $O$ is the circumcenter of $\\triangle TBC$. We will subsequently dispose of point $R$.\n\nDenote by $D_1$ and $D_2$ the projections of $B$ onto $AP$ and $C$ onto $AQ$ respectively, and let $M$ be the midpoint of $\\overline{BC}$. I claim that $D_1M = D_2M$. To prove this, remark that by simple angle chasing $\\angle BOC = \\angle ABD_1$, so $\\triangle AD_1B$ is directly similar to $\\triangle CMB$. By the duality of spiral similarity, $\\triangle BD_1M\\sim\\triangle BAO$ with similarity ratio $r = \\tfrac{BM}{BO}$. Analogously, $\\triangle CD_2M\\sim\\triangle CAO$ with the same similarity ratio $r = \\tfrac{CM}{CO}$. Since $AO$ is taken to $D_1M$ and $D_2M$ under these spiral similarities, we must have $D_1M = D_2M$.\n\nNow let $\\psi$ denote the measure of angle $\\angle ABD_1 = \\angle ACD_2$, and remark that \\[\\angle D_1MD_2 = 180^\\circ - \\angle BMD_1 - \\angle CMD_2 = 180^\\circ - \\angle BOC = 2\\psi.\\] It follows that, upon letting the perpendicular from $M$ to $D_1D_2$ intersect $AC$ at $X$, we have $\\angle XMD_1 = \\psi$, so quadrilateral $CD_1XM$ is cyclic. Thus the angle between $MX$ and $AC$ is equal to $\\angle MD_1C = \\angle OAC$, which implies $AO\\perp D_1D_2$. But $D_1D_2\\parallel PQ$ by similarity, and so we obtain the desired perpendicularity. $\\blacksquare$", "Solution_19": "Here is a synthetic solution constructing only one point. I would appreciate if someone simplifies my final directed angle chasing.\n\nLet $\\theta = \\angle BAP$. By Spiral Similarity, $ \\triangle APC\\stackrel{+}{\\sim} \\triangle ABQ$ so $\\angle APR = \\angle ABR\\implies \\angle BRP = \\theta$. Thus $\\angle BOC = 2\\theta$. Now for the key step, construct point $X$ outside $ \\triangle BOC$ such that $\\triangle OXB\\sim \\triangle ABC$. Then $\\angle XOC = \\angle BAC + 2\\theta = \\angle APQ$ thus, \n$$\\frac{OX}{OC} = \\frac{OX}{OB} = \\frac{AB}{AC} = \\frac{AP}{AQ}\\implies \\triangle OXC\\stackrel{-}{\\sim} \\triangle APQ.$$\nFurthermore, $\\angle XBC = \\angle C + (90^{\\circ} - \\theta) = \\angle OCA$ so\n$$\\frac{BC}{BX} = \\frac{AC}{OB} = \\frac{AC}{OC}\\implies \\triangle BCX\\stackrel{-}{\\sim} \\triangle CAO.$$\nHence, using directed angle, we find\n\\begin{align*}\n\\measuredangle(AO, PQ) &= \\measuredangle(AO, AC) + \\measuredangle(AC, AQ) + \\measuredangle(AQ, PQ)\\\\\n&=-\\measuredangle(CX, BC) + \\measuredangle(AC, AQ) - \\measuredangle(OC, CX) \\\\\n&= \\measuredangle(AC, AQ) + \\measuredangle(BC, OC) \\\\\n&= 90^{\\circ}\n\\end{align*}\nas desired.", "Solution_20": "Here is a quick and easy solution. Clearly $ABPR$ and $ACQR$ are inscribed; next, meet $AP$ and $(ACQR)$ at $X$, and $AQ$ and $(ABPR)$ at $Y$. Since $\\measuredangle AXQ = \\measuredangle ARQ = \\measuredangle ARB = \\measuredangle AYP$, $PQXY$ is inscribed. So, if $\\rho$ is the radius of $(BCR)$, \\[\nPO^2-PA^2 = PR\\cdot PC + \\rho^2 - PA^2 = PA\\cdot PX +\\rho^2 - PA^2 = PA\\cdot AX + \\rho^2 = \\mathcal P(A,(PQXY)),\n\\]where $\\mathcal P$ denotes power. Since the last expression is the same for $P,Q$, this means $PO^2-PA^2 = QO^2-QA^2$, done!", "Solution_21": "Denote $\\omega_b \\equiv \\odot(ABP)$ and $\\omega_c \\equiv \\odot(ACQ).$ Let $AP$ meet $\\omega_c$ for a second time at $X.$ Let $AQ$ meet $\\omega_b$ for a second time at $Y.$\n\nBecause $A$ is the center of spiral similarity sending $\\overline{PB} \\mapsto \\overline{CQ}$, we know that $R$ is the second intersection point of $\\omega_b$ and $\\omega_c.$ By Reim's theorem for $\\omega_b, \\omega_c$ cut by $PX, BQ,$ it follows that $PB \\parallel XQ.$ Because $\\triangle APB$ is isoceles, $PB$ is parallel to the tangent to $\\omega_b$ at $A.$ Therefore, by the converse of Reim's theorem for $\\omega_b$ cut by $PX, YQ,$ it follows that $PQXY$ is cyclic. Thus, there exists $r \\in \\mathbb{C}$ such that \\[ r^2 = \\text{pow}(A, \\odot(PQXY)) = AP \\cdot AX = AQ \\cdot AY. \\] \n[asy]\n /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */\nimport graph; size(20cm); \nreal labelscalefactor = 0.5; /* changes label-to-point distance */\npen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ \npen dotstyle = black; /* point style */ \nreal xmin = 974.3508556900372, xmax = 1046.0579652024412, ymin = 1011.7387089646945, ymax = 1035.0218454829537; /* image dimensions */\npen evevff = rgb(0.8980392156862745,0.8980392156862745,1.); pen evfuev = rgb(0.8980392156862745,0.9568627450980393,0.8980392156862745); pen qqzzqq = rgb(0.,0.6,0.); \n\nfilldraw((1002.5480877317876,1028.5468363406196)--(990.6757045460399,1018.2042280191417)--(998.363012344087,1013.367634339197)--cycle, evevff, linewidth(0.8) + blue); \nfilldraw((1002.5480877317876,1028.5468363406196)--(1016.3654794880058,1013.4008492232264)--(1022.4320417351443,1023.5518774921607)--cycle, evfuev, linewidth(0.8) + qqzzqq); \nLabel laxis; laxis.p = fontsize(10); \nxaxis(xmin, xmax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); \nyaxis(ymin, ymax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ \n /* draw figures */\ndraw((1002.5480877317876,1028.5468363406196)--(990.6757045460399,1018.2042280191417), linewidth(0.8) + blue); \ndraw((990.6757045460399,1018.2042280191417)--(998.363012344087,1013.367634339197), linewidth(0.8) + blue); \ndraw((998.363012344087,1013.367634339197)--(1002.5480877317876,1028.5468363406196), linewidth(0.8) + blue); \ndraw((1002.5480877317876,1028.5468363406196)--(1016.3654794880058,1013.4008492232264), linewidth(0.8) + qqzzqq); \ndraw((1016.3654794880058,1013.4008492232264)--(1022.4320417351443,1023.5518774921607), linewidth(0.8) + qqzzqq); \ndraw((1022.4320417351443,1023.5518774921607)--(1002.5480877317876,1028.5468363406196), linewidth(0.8) + qqzzqq); \ndraw((998.363012344087,1013.367634339197)--(1022.4320417351443,1023.5518774921607), linewidth(0.8) + red); \ndraw((1016.3654794880058,1013.4008492232264)--(990.6757045460399,1018.2042280191417), linewidth(0.8) + red); \ndraw(circle((1011.7378103749238,1023.0547794798424), 10.705778384581844), linewidth(0.8)); \ndraw(circle((998.1695210212386,1021.5875190512957), 8.222161733523125), linewidth(0.8)); \ndraw((1002.5480877317876,1028.5468363406196)--(1007.5572715124796,1032.9105790739395), linewidth(0.8)); \ndraw((1002.5480877317876,1028.5468363406196)--(997.599658737153,1029.7899089804253), linewidth(0.8)); \n /* dots and labels */\ndot((1002.5480877317876,1028.5468363406196),linewidth(3.pt) + dotstyle); \nlabel(\"$A$\", (1002.1745359708872,1029.1587799000256), N * labelscalefactor); \ndot((998.363012344087,1013.367634339197),linewidth(3.pt) + dotstyle); \nlabel(\"$B$\", (998.0439421049653,1011.7415989196447), NE * labelscalefactor); \ndot((1016.3654794880058,1013.4008492232264),linewidth(3.pt) + dotstyle); \nlabel(\"$C$\", (1016.3837788696585,1012.0068226742816), NE * labelscalefactor); \ndot((990.6757045460399,1018.2042280191417),linewidth(3.pt) + dotstyle); \nlabel(\"$P$\", (989.5183963657025,1017.5600723245166), N * labelscalefactor); \ndot((1022.4320417351443,1023.5518774921607),linewidth(3.pt) + dotstyle); \nlabel(\"$Q$\", (1022.9927290551335,1023.475082740517), E * labelscalefactor); \ndot((1003.9346138590062,1015.7251263534112),linewidth(3.pt) + dotstyle); \nlabel(\"$R$\", (1003.7276392644737,1014.1216867336337), N * labelscalefactor); \nlabel(\"$\\omega_c$\", (1011.849575658217,1033.9971654909085), NE * labelscalefactor); \nlabel(\"$\\omega_b$\", (992.2524532204959,1028.4734976027294), N * labelscalefactor); \ndot((1007.5572715124796,1032.9105790739395),linewidth(3.pt) + dotstyle); \nlabel(\"$X$\", (1007.0982038590661,1033.2893737659476), N * labelscalefactor); \ndot((997.599658737153,1029.7899089804253),linewidth(3.pt) + dotstyle); \nlabel(\"$Y$\", (997.316957584563,1030.3814356843386), NE * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */\n[/asy]\n\nDefine a new circle $\\gamma \\equiv \\odot(A, r).$ Because \\[ PR \\cdot PC = PA \\cdot PX = PA(PA + AX) = PA^2 - r^2, \\] we see that $P$ lies on the radical axis of $\\odot(BCR)$ and $\\gamma.$ Similarly, $Q$ lies on the radical axis of $\\odot(BCR)$ and $\\gamma.$ Therefore $PQ \\perp AO.$", "Solution_22": "Skimming through (not very carefully), I haven't seen this solution yet.\n\nNote that $A$ is the spiral similarity that sends $PB$ to $CQ$, so it rotates $PC$ to $BQ$ implying $PC=BQ$. Also, $APBR, AQCR$ are cyclic.\n\nLet $N$ be the intersection of $(PRQ), (BRC)$. Then $N$ sends $BQ$ to $CP$, but $BQ=CP$, so $N$ is the midpoints of arcs $PQ$ and $BC$ in the circles.\n\nNote that $\\measuredangle BRA = \\measuredangle BPA = \\measuredangle AQC = \\measuredangle ARC$ so $AR$ bisects $BRC$ and $AR$, $NO$ concur on $(BRC)$. Therefore, $\\measuredangle BPA = \\measuredangle BRA = \\measuredangle BNO$, so (since $AP=AB, ON=OB$), by (a working SSA similarity), $\\triangle BON\\sim \\triangle BAP$. \n\nTherefore, a spiral similarity sends $NO$ to $PA$, and $AO$ to $PN$. Thus, the angle between $AO$ and $PQ$ is \\[\\measuredangle (AO, PN)+\\measuredangle (PN, PQ)=\\measuredangle ABP+\\measuredangle NPQ = 90^{\\circ}\\]\n\nOh jk this is the second solution in post 3", "Solution_23": "Much shorter solution.\n\nLet $\\theta = \\angle BAP$. Note that $\\triangle ABQ\\stackrel{+}{\\sim}\\triangle ACP$ thus $\\angle(BQ, CP) = \\theta)$. Therefore $\\angle BRC = 2\\theta$. Let $C'$ be the reflection of $C$ across $AQ$.\n\nNote that $\\triangle ABP\\stackrel{+}{\\sim}\\triangle AC'Q$ thus $\\triangle ABC'\\stackrel{+}{\\sim}\\triangle APQ$ which means $\\angle(PQ,BC') = \\theta\\hdots (\\spadesuit)$. Moreover, $\\triangle CAC'\\stackrel{+}{\\sim}\\triangle COB$ thus $\\triangle CAO\\stackrel{+}{\\sim}\\triangle CC'B$. Thus $\\angle(BC', AO) = \\angle(CA,CC') = 90^{\\circ}-\\theta$. Combining this with $(\\spadesuit)$ gives the conclusion.", "Solution_24": "its easiest p6 i seen ! solution: its obvious $APBR , BRCQ$ are cyclic let $T , S$ be intersection of this circles with lines $AP$ and $AQ$ respectively.\nits obvious $PQST$ is cyclic quadrilateral and with this observations according power of point theorem we have: $QR.QB-PR.PC=QO^2-PO^2=AQ^2-AP^2$ \nproblem solved.", "Solution_25": "We have $\\angle BOC = 2(\\pi - \\angle BRC) = 2\\angle BAP$. Let $(ABC)$ be the unit circle, and let $q = a + (c-a)z$ and $p = a + \\frac{b-a}{z}$ for some complex number $z$ on the unit circle. The point $O$ is given by\n\\[\\frac{b-o}{c-o} = z^2 \\implies o = \\frac{b-z^2c}{1-z^2}.\\]\nThen\n\\[\\frac{a-o}{p-q} = \\frac{a - \\frac{b-z^2c}{1-z^2}}{\\frac{b-a}{z} - z(c-a)} = \\frac{-z}{1-z^2}\\]\nwhich is purely imaginary.", "Solution_26": "Triple Jacobi's theorem :-D \nLet $S$ be a point outside $\\triangle ABC$ and $\\angle SBC = \\angle ABP=\\angle SCB$. By Jacobi, $AS,BQ,CP$ are concurrent at $R$ and $S$ lies on $\\odot(BCR)$. Hence, $\\angle OBC = \\angle OCB=90^{\\circ}-\\angle BAP$. Let $B',C'$ be the foot of perpendicular from $B,C$ to $AP,AQ$ respectively. By Jacobi, $AO,BC',CB'$ are concurrent.Then Jacobi $\\triangle AB'C'$, $AO \\perp B'C'$ but $B'C' \\parallel PQ$ ($\\frac{AB'}{AP} = \\frac{AC'}{AQ} \\implies \\triangle AB'C' \\sim \\triangle APQ$). Therefore, $AO \\bot PQ$.", "Solution_27": "[hide=Diagram]\n[img]https://lh3.googleusercontent.com/-DqvQ8QGvY7E/X6j-48YxtlI/AAAAAAAAG_g/qHhq5dEGOWUqf_GPG7TdcAmfEeD46Y82wCLcBGAsYHQ/s0/tst2006_6.PNG[/img]\n[/hide]\n\nLet $\\theta:=\\angle PAB=\\angle CAQ$. Note that $A$ is the center of spiral similarity $PB\\mapsto CQ$, so it also maps $PC\\mapsto BQ$. Thus $\\angle (\\overline{PC},\\overline{BQ})=\\theta$, in particular $\\angle BRC =\\pi-\\theta$. Hence $\\angle BOC = 2(\\pi-\\angle BRC) = 2\\theta$. This eliminates $R$ entirely! Indeed, $O$ is now located on the perpendicular bisector of $\\overline{BC}$ with $\\angle BOC=2\\theta$. \n\nNow use complex numbers with $A=0$. Let $t=e^{i\\theta}$. Then\n\\begin{align*}\n p=b/t,\\quad q=tc. \n\\end{align*}\nWe know $\\angle BOC=2\\theta$ and $BO=CO$. Hence $|\\tfrac{b-o}{c-o}| = 1 = |t^2|$, so actually\n\\begin{align*}\n \\frac{b-o}{c-o}\\div t^2 = 1 \\implies o=\\frac{t^2c-b}{t^2-1}. \n\\end{align*}\nFinally, confirm\n\\begin{align*}\n \\frac{o-a}{p-q} = \\frac{t^2c-b}{(t^2-1)(b/t-tc)} = \\frac{t}{1-t^2} \\in i\\mathbb{R}. \n\\end{align*}\n[hide=Remark]\nOne thing that had me confused for a bit was that when you locate $O$ as the point on the perp. bisector of $\\overline{BC}$ with $\\angle BOC=2\\theta$, there should be \\textit{two} points which satisfy this, but the equation\n\\[ \\frac{b-o}{c-o} = t^2\\]\ngives one. The reason is that the second value of $O$ is the reflection over $\\overline{BC}$, and then the RHS above would be $1/t^2$, not $t^2$, even though both are ``$\\angle BOC$''. \n[/hide]", "Solution_28": "See my solution to this problem on my Youtube channel here:\n[url]https://www.youtube.com/watch?v=Hz8kyLKP0_k[/url]", "Solution_29": "Here is a barycentric coordinate solution.\n\nObserve that $A$ is the miquel point of $PBQC$. This means that we must have $(APBR)$ and $(QARC)$ are cyclic. Since \n\\[\\angle ARP = \\angle ABP = \\angle ACQ = \\angle ARQ\\]\nthis means $A$ lies on the angle bisector of $\\angle BRC$.\n\nWe now proceed with barycentric coordinates. Let our reference triangle be $\\triangle BRC$, so $R = (1:0:0), B = (0:1:0), C = (0:0:1)$. For convenience, let $BC = a, CR = b, BR = c$. Since $A$ lies on the angle bisector, let $A = (t:b:c)$, for some number $t$. Now, the equation for $(ARB)$ is\n\\[-a^{2}yz - b^{2}xz - c^{2}xy + (wz)(x+y+z)\\]\nbecause $(ARB)$ passes through $B,R$. Plugging in $A$ gives\n\\[-a^{2}bc - b^{2}tc - c^{2}tb + wc(t+b+c) = 0\\Rightarrow w = \\frac{a^{2}b + b^{2}t + cbt}{t+b+c}\\]\nNow, since $P = (ARB)\\cap RC$, if $P = (p:0:1-p)$, then we can plug this into our equation for $(ARB)$ to get\n\\[-b^{2}p(1-p) + \\frac{a^{2}b + b^{2}t + cbt}{t+b+c}\\cdot (1-p) = 0 \\Rightarrow p = \\frac{a^{2} + bt + ct}{bt + b^{2} + bc}\\]\nTherefore, by symmetry,\n\\[P = \\left(\\frac{a^{2} + bt + ct}{b^{2}+bt + bc} : 0 : \\frac{b^{2} + bc - ct - a^{2}}{bt + b^{2} + bc}\\right), Q = \\left(\\frac{a^{2} + bt + ct}{c^{2} + ct + bc} : \\frac{c^{2} + bc -bt - a^{2}}{c^{2} + ct + bc} : 0 \\right)\\]\nTo show that $AO\\perp PQ$, we can use generalized perpendicularity (Theorem 7.25 of EGMO). Assume that $O$ was the origin. We have\n\\[\\overrightarrow{PQ} = (a^{2} + bt + ct)\\cdot \\frac{c-b}{bc(b+c+t)}\\vec{R} + \\frac{bt + a^{2} - bc - c^{2}}{c(b+c+t)}\\vec{B} + \\frac{b^{2} + bc - ct - a^{2}}{c(b+c+t)}\\vec{C}\\]\nSince $O$ is the origin, we have $A = \\frac{t}{t+b+c}\\vec{R} + \\frac{b}{b+c+t}\\vec{B} + \\frac{c}{b+c+t}\\vec{C}$. By Theorem 7.25 of EGMO, we must show the following expression is $0$:\n\\[0 = a^{2}\\left(\\frac{bt+a^{2}-bc-c^{2}}{b+c+t} + \\frac{b^{2} + bc -ct - a^{2}}{b+t+c}\\right)+ b^{2}\\left(\\frac{c-b}{b(b+c+t)}(a^{2} + bt + ct) + \\frac{t(b^{2} + bc - ct - a^{2})}{b(b+c+t)}\\right)+ c^{2}\\left(\\frac{c-b}{c(b+c+t)}(a^{2} + bt + ct) + \\frac{t(bt + a^{2}-bc-c^{2})}{c(b+c+t)}\\right)\\]\n\\[\\Rightarrow 0 = a^{2}(b^{2}-c^{2} + bt - ct) + b((c-b)(a^{2} + bt + ct) + b^{2}t + bct - ct^{2} - a^{2}t)+ c((c-b)(a^{2}+bt+ct) + bt^{2} + bct - ct^{2} - at^{2})\\]\n\\[\\Rightarrow 0 = a^{2}(b^{2} - c^{2} + bt - ct) + b[a^{2}c + bct + c^{2}t - a^{2}b - ct^{2} - a^{2}t] + c[a^{2}c - a^{2}b - b^{2}t + bt^{2} + a^{2}t - bct]\\]\n\\[= a^{2}b^{2} - a^{2}c^{2} + a^{2}bt - a^{2}ct + a^{2}bc + b^{2}ct + bc^{2}t - a^{2}b^{2} - bct^{2}-a^{2}tb + a^{2}c^{2} - a^{2}bc - b^{2}ct + bct^{2} + a^{2}ct - bc^{2}t = 0\\]\nTherefore, this expression is $0$, which means $AO\\perp PQ$.", "Solution_30": "[quote=k12byda5h]Triple Jacobi's theorem :-D \nLet $S$ be a point outside $\\triangle ABC$ and $\\angle SBC = \\angle ABP=\\angle SCB$. By Jacobi, $AS,BQ,CP$ are concurrent at $R$ and $S$ lies on $\\odot(BCR)$. Hence, $\\angle OBC = \\angle OCB=90^{\\circ}-\\angle BAP$. Let $B',C'$ be the foot of perpendicular from $B,C$ to $AP,AQ$ respectively. By Jacobi, $AO,BC',CB'$ are concurrent.Then Jacobi $\\triangle AB'C'$, $AO \\perp B'C'$ but $B'C' \\parallel PQ$ ($\\frac{AB'}{AP} = \\frac{AC'}{AQ} \\implies \\triangle AB'C' \\sim \\triangle APQ$). Therefore, $AO \\bot PQ$.[/quote]\n\n\"Then Jacobi $\\triangle AB'C'$, $AO \\perp B'C'$\" Why'd have this perpendicularity?", "Solution_31": "[quote]\"Then Jacobi $\\triangle AB'C'$, $AO \\perp B'C'$\" Why'd have this perpendicularity?[/quote]\n\nLet $BC' \\cap CB' = X$.By Jacobi, $AO,BC',CB'$ are concurrent $\\implies A,O,X$ are collinear. Then Jacobi $\\triangle AB'C'$ with point $\\infty_{\\perp B'C'},C,B$. We get $AOX \\perp B'C'$", "Solution_32": "Let $R'=\\overline{PB}\\cap\\overline{QC}$. By spiral sim $A,B,P,R$ and $A,C,R,Q$ are cyclic. Under an inversion at $A$ with radius $\\sqrt{AP\\cdot AQ}$ and a reflection about the angle bisector of $\\angle PAQ$, $R$ goes to $R'$. Since $\\angle APR'=\\angle AQR'$, if we let $R''$ be the point such that $R'PR''Q$ is a parallelogram, $\\overline{AR''},\\overline{AR'}$ are isogonal in $\\angle PAQ$, so $R-A-R''$. Let $O_1$ be the center of $(APR)$ and let $O_2$ be the center of $(AQR)$. Then the perpendicular from $P$ to $\\overline{AO_2}$, the perpendicular from $Q$ to $\\overline{AO_1}$, and the perpendicular from $R$ to $\\overline{O_1O_2}$ meet at $R''$. Thus $\\triangle AO_1O_2,\\triangle RPQ$ are orthologic, as desired.", "Solution_33": "A very, very funny solution.\n\nLet $\\angle BAP = \\angle CAQ = \\theta$. Since $\\triangle APB\\sim \\triangle ACQ$, we have $\\triangle APC\\sim \\triangle ABQ$, so $\\angle (PC, BQ) = \\theta$ (the spiral similarity at $A$ sending $APC$ to $ABQ$ has a rotation of angle $\\theta$). Thus, $\\angle BRC = 180^\\circ - \\theta$, so $\\angle BOC = 2\\theta$. \n\nIt suffices to show that $\\overrightarrow{AO}\\cdot \\overrightarrow{PQ} = 0$. Note that $\\angle(PA, OC) = \\measuredangle (PA, AB) + \\measuredangle (AB, BC) + \\measuredangle (BC, CO) = \\theta - \\angle B +90^\\circ - \\theta = 90^\\circ - \\angle B$. Similarly, $\\angle(QA, OB) = 90^\\circ - \\angle C$. \n\nNow, we compute \n\\begin{align*}\n\\overrightarrow{AO}\\cdot \\overrightarrow{AP} &= \\left(\\overrightarrow{AC} + \\overrightarrow{CO}\\right)\\cdot \\overrightarrow{AP}\\\\\n& = AB\\cdot AC \\cdot \\cos (\\angle A + \\theta) + AB\\cdot CO \\cdot \\cos(90^\\circ - \\angle B)\\\\\n& = AB\\cdot AC \\cdot \\cos(\\angle A + \\theta) + AB\\sin \\angle B\\cdot CO\\cdot \\\\\n\\end{align*}\nwhich is clearly symmetric. Thus, $\\overrightarrow{AO}\\cdot \\overrightarrow{AP} = \\overrightarrow{AO}\\cdot \\overrightarrow{AQ}$, and we are done.", "Solution_34": "Note that $\\triangle AQC$ and $\\triangle ABP$ are rotations about $A$. In particular, $APBR$ and $AQCR$ are cyclic. It follows that $\\angle ARP = \\angle ABP = \\angle ACQ = \\angle ARQ$. Call this common angle $\\theta$. Now let $\\triangle XYZ$ be congruent to $\\triangle AQC$ and $\\triangle ABP$, and let $K$ and $L$ be on $YZ$ so that $\\angle XKY = \\angle XLZ = \\theta$. Let $M$ be the foot from $X$ to $YZ$. For points $T$ in the plane, define $f(T) = \\text{pow}_{(BRC)}(T) - TA^2$. We have\n\n$$f(P) - f(Q) = (PR\\cdot PC - PA^2) - (QR\\cdot QB - QA^2) = $$\n\n$$= (YK\\cdot YZ - YX^2) - (ZL\\cdot ZY - ZX^2)$$\n\n$$= YZ(YK - ZL) - YX^2 + ZX^2$$\n\n$$=YZ\\cdot YM - ZY\\cdot ZM - YX^2 + ZX^2$$\n\n$$=\\text{pow}_{(XZ)}(Y) - \\text{pow}_{(XY)}(Z) - YX^2 + ZX^2$$\n\nIf $D$ and $E$ are the midpoints of $XY$ and $XZ$, this is\n\n$$(YE^2 - EX^2) - (ZD^2 - DX^2) - YX^2 + ZX^2$$\n\n$$=(\\frac{1}{2}YX^2 + \\frac{1}{2}YZ^2 - \\frac{1}{4}XZ^2 - \\frac{1}{4}XZ^2) - (\\frac{1}{2}ZX^2 + \\frac{1}{2}ZY^2 - \\frac{1}{4}XY^2 - \\frac{1}{4}XY^2) - YX^2 + ZX^2$$\n\n$$=0$$\n\nso $PQ$ is parallel to the radical axis of $(BRC)$ and the point circle at $A$, which is perpendicular to $AO$, as desired.", "Solution_35": "Clearly $\\triangle APB \\sim \\triangle ACQ$, so by spiral similarity facts $ABPR$ and $ACQR$ are cyclic. Obviously $\\overline{AR}$ bisects $\\angle BRC$ as well. Thus we may view the problem with reference triangle $RBC$, restating it as follows.\n[quote=Restated]\nLet $ABC$ be a triangle with circumcenter $O$ and $D$ be a point on its internal $\\angle A$-bisector. Let $(ABD)$ and $(ACD)$ intersect $\\overline{AC}$ and $\\overline{AB}$ at $E$ and $F$ respectively. Then $\\overline{OD} \\perp \\overline{EF}$.\n[/quote]\nLet $D=(t:b:c)$. If the equation of $(ABD)$ is $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)(x+y+z)=0$, then plugging in $A$ and $B$ implies $u=v=0$, and then plugging in $D$ implies $w=\\tfrac{a^2b+b^2t+bct}{b+c+t}$. To compute $E$, we solve for $r_1$ given that $(r_1,0,1-r_1)$ lies on the circle: this will yield (after dividing out $1-r_1$, which would correspond to $A$ itself) $r_1=\\tfrac{a^2+bt+ct}{b(b+c+t)}$. Likewise, if $F=(r_2,1-r_2,0)$ then $r_2=\\tfrac{a^2+bt+ct}{c(b+c+t)}$.\n\nNow, by \"strong EFFT\" we just have to prove that\n\\begin{align*}\n0&=a^2\\left(\\frac{a^2+bt+ct}{b+c+t}-c+b-\\frac{a^2+bt+ct}{b+c+t}\\right)\\\\\n&+b^2\\left(\\left(\\frac{c}{b}-1\\right)\\left(\\frac{a^2+bt+ct}{b+c+t}\\right)+t-\\frac{t}{b}\\left(\\frac{a^2+bt+ct}{b+c+t}\\right)\\right)\\\\\n&+c^2\\left(\\left(1-\\frac{b}{c}\\right)\\left(\\frac{a^2+bt+ct}{b+c+t}\\right)+\\frac{t}{c}\\left(\\frac{a^2+bt+ct}{b+c+t}\\right)-t\\right)\\\\\n&=a^2(b-c)(b+c+t)+b(c-b-t)(a^2+bt+ct)+b^2t(b+c+t)+c(c-b+t)(a^2+bt+ct)-c^2t(b+c+t)\\\\\n&=(b-c)(a^2+bt+ct)(b+c+t)-(b+c)(b-c)(a^2+bt+ct)-t(b-c)(a^2+bt+ct))\\\\\n&=(b-c)(a^2+bt+ct)(b+c+t-b-c-t)\\\\\n&=0,\n\\end{align*}\nso we're done. $\\blacksquare$", "Solution_36": "Started around 8:25 CST (since Hazard is pro god at math I need to compete with him)\n\nLet $U\\in PC$ satisfy $UP=UB$, define $V$ similarly. Consider ellipse $\\Gamma$ with foci at $B,C$ through $U,V$ such that $AU$ and $AV$ are tangents (not sure if it'll come in handy later).\n\nAlso note that there is a circle $\\Omega$ centered at $A$ which is tangent to $BU,CV,RU,RV$. This entire ellipse configuration comes from djmathman's handout, yay!\n\nHence showing $PQ\\perp AO$ is basically showing $PQ$ is parallel to some radical axis. Let $K$ be the foot of the altitude from $A$ to $PC$, and define $L$ similarly.\n\nNotice that it suffices to show\n\\[PK^2-PR\\cdot PC=QL^2-QR\\cdot QB.\\]\nWe're basically almost done I think. Notice that this equates to\n\\[(PK+QL)(PK-QL)=PR\\cdot PC-QR\\cdot QB\\implies PC(PK-QL)=PR\\cdot PC-QR\\cdot PC\\implies PK-QL=PR-QR\\]\nor just\n\\[PK-PR=QL-QR\\implies RK=RL\\]\nwhich is obviously true. Ta-da! 22 minutes! WOOOOOOOOOOOOOO", "Solution_37": "Solved with hints\n\nFirst, notice that since the angle of rotation is the same, there is a spiral similarity centered at $A$ taking $BP$ to $QC$. This means that $A$ is the Miquel point of $BPCQ$, so $ARBP$ is cyclic. This means that $$\\angle BOC=2\\angle BRC=2\\angle BAP.$$ Thus, if we let $M$ denote the arc midpoint of $BC$ not containing $R$, we have that $\\angle BOM=\\angle BAP$. Since they are both isosceles, we have $\\triangle BOM\\sim\\triangle BAP$ as well.\\\\\n\nUse complex numbers with unit circle $(BRC)$. Let $B=b^2,C=c^2,R=r^2$. Furthermore, we let $A=a$ as another free variable as well (the problem has four degrees of freedom, so this decision is ok).\\\\\n\nWe can use $\\triangle BOM\\sim\\triangle BAP$ to compute $p$. We have $$\\frac{a-p}{a-b^2}=\\frac{0-(-bc)}{0-b^2},$$ which we can solve for $p$ to get $$p=\\frac{ab+ac-b^2c}{b}.$$ Similarly, $$q=\\frac{ab+ac-bc^2}{c}.$$ Thus, we have $$p-q=\\frac{abc+ac^2-b^2c^2-ab^2-abc+b^2c^2}{bc}=\\frac{ac^2-ab^2}{bc}.$$ Finally, dividing this by $a$ gives $\\frac{c^2-b^2}{bc}$, which is clearly imaginary after conjugating and multiplying top and bottom by $b^2c^2$, so we are done.\n\n[color=#f0f][b]Remark:[/b][/color]\n\nThere are many things to take away from this problem. First of all, two rotations of the same angle around the same point should motivate using a spiral similarity, and spiral similiarity often works well with complex numbers.\n \nNext, there is a very \"tempting\" setup, in fact the one I originally used and failed with, where we set $(ABC)$ as the unit circle as usual, then note that there is an \"angle of rotation\" of some $\\theta$ to get $P$ and $Q$, and setting $a$, $b$, $c$, and $d=e^{i\\theta}$ as the four \"free\" variables to represent the four degrees of freedom. While this quickly gives $p$ and $q$, the computation for $r$ is now very messy as it requires the general intersection formula, much to the detriment of this setup, as $R$ is required to compute the circumcenter.\n\nThe motivation for setting $(BRC)$ as the unit circle is that, rather than having to go through the tedious process of computing the circumcenter, this gives us the circumcenter \"for free\". Then, we can make $a$ our remaining free variable.\n\nNext, the introduction of the arc midpoint allows us to turn the condition $\\angle BOC=2\\angle BAP$ into a similarity condition, which is actually a much stronger condition than just the angle equivalence, as it effectively encodes two degrees of freedom rather than just one. If we were to use $\\angle BOC=2\\angle BAP$ directly, we would actually need two equations to compute $P$ (the other one probably being $PRC$ collinear), which will be much less elegant.\n\nFinally, this problem illustrates another important point of computing points \"indirectly\". In this solution, rather than plugging into some formula to compute $p$ and $q$, we used the fact that $\\triangle BAP$ is similar to $\\triangle BOM$ to indirectly compute $P$.\n", "Solution_38": "We will employ complex numbers, setting $A$ as the origin, $P$ at $(0, 1)$, $B$ at $|b| = 1$, $C$ at $c$, and $Q$ at $bc$. \n\nObviously, $APRB$ and $AQCR$ are cyclic, say by considering the pair of congruent triangles. Letting $\\angle PAB = \\theta$, it follows that $\\angle BOC = 2\\theta$. In other words, we have $\\frac{o-b}{o-c} = b^2$, so $o = \\frac{b(bc-1)}{b^2-1}$, and thus $$\\frac o{p-q} = \\frac b{b^2-1} \\in i\\mathbb R.$$ Hence $\\overline{AO} \\perp \\overline{PQ}$ as required." } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "easy one: find all complex polynomials $P$ such that $P''$ divides $P$", "Solution_1": "p(z)=(z-r)^n \r\nuse f(z) = (z-r1)... (z-rn)", "Solution_2": "i think this an X or Centrale Oral \r\nam i wrong?" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Is $Q^{4}=\\langle \\begin{bmatrix}1\\\\1\\\\1\\\\1 \\end{bmatrix}, \\begin{bmatrix}1\\\\1\\\\0\\\\1 \\end{bmatrix}, \\begin{bmatrix}0\\\\1\\\\0\\\\1 \\end{bmatrix}\\rangle ?$", "Solution_1": "A fourdimensional space can never be spanned by a set consisting of less than four vectors." } { "Tag": [ "induction", "number theory unsolved", "number theory" ], "Problem": "Solve in positive integers n.\r\n\r\n$n | 2^n + 1$", "Solution_1": "There are infinitely many such $n$. Did you find them all?\r\n\r\nBTW, why do you use \"South Korea\" flag?", "Solution_2": "I doubt they can all be found, but who knows?\r\n\r\nAnyway, we can construct an infinite sequence like this: take $a_1=3$, then take $a_{n+1}=2^{a_n}+1$. We easily get $a_n|2^{a_n}+1,\\ \\forall n\\in\\mathbb N^*$.", "Solution_3": "You correctly understood my idea, grobber! ;) :D", "Solution_4": "Another sequence: $a_n=3^n$, for $n=1,2,3,...$, $3^n|2^{3^n}+1$.", "Solution_5": "I had changed it to USA flag a few weeks ago, but with the server crash it reverted.\r\n\r\nI also noticed that any power of 3 suffices. Any proof?", "Solution_6": "It is the simpliest induction:\r\n$3|2^3+1$. Let $n\\in\\mathbb{N}$ be such number that $3^n|2^{3^n}+1$. Thus $2^{3^n}=3^nk-1$, and $2^{3*3^n}=3^{3n}k^3-3^{2n+1}k^2+3^{n+1}k-1$, so $2^{3n+1}+1=3^{n+1}t$, $t\\in\\mathbb{N}$ and $3^{n+1}|2^{3^{n+1}}+1$, and we are done.", "Solution_7": "I am sure Stryker know that. ;)", "Solution_8": "Heh. What I meant was a proof covering all solutions.", "Solution_9": "Heh. I meant you mean that :lol:" } { "Tag": [ "Putnam", "inequalities", "geometric series", "college contests" ], "Problem": "Let $ A$ be a positive real number. What are the possible values of $ \\displaystyle\\sum_{j\\equal{}0}^\\infty{x_j^2}$, given that $ x_0,x_1,\\ldots$ are positive numbers for which $ \\displaystyle\\sum_{j\\equal{}0}^\\infty{x_j}\\equal{}A$?", "Solution_1": "[hide]My instinct tells me the range is the open interval $ (0,A^2)$. To approach the upper edge just take $ x_0 \\equal{} (A \\minus{} \\epsilon)$ and $ x_1,x_2,\\ldots$ as a convergent series adding up to $ \\epsilon$. To approach the bottom edge, take any convergent series summing to $ A$, and using the fact that for positive $ a,b$, $ (a \\plus{} b)^2 > a^2 \\plus{} b^2$, just split terms as many times as you need to get as low as you want.\n\nI'm pretty sure I could come up with a rigorous proof that both $ 0$ and $ A^2$ are achievable (as limits). But I'm not sure how to show that everything in between can also be hit.[/hide]", "Solution_2": "[hide=\"My writeup for my class that year.\"]The set of possible values of the sum $ \\sum_{j\\equal{}0}^{\\infty}x_j^2$ is the open interval $ (0, A^2).$\n\nWe have two things to prove: first, that the values are confined to this interval, and second, that all numbers in this interval are possible. As to the first: the values are obviously positive, so we must show that the sum of the squares is strictly less than $ A^2.$ We note that $ \\sum_{j\\equal{}0}^{\\infty}\\left(\\frac{x_j}A\\right) \\equal{} 1.$ Since each $ x_j > 0,$ we must have that $ 0 < \\frac{x_j}A < 1.$ Then, $ \\left(\\frac{x_j}A\\right)^2 < \\frac{x_j}A,$ with the inequality being strict. Hence, $ \\sum_{j\\equal{}0}^{\\infty}\\left(\\frac{x_j}A\\right)^2 < \\sum_{j\\equal{}0}^{\\infty}\\left(\\frac{x_j}A\\right) \\equal{} 1.$ Multiply through by $ A^2$ to get that $ \\sum_{j\\equal{}0}^{\\infty}x_j^2 < A^2.$\n\nAs for the second part, let $ 0 < r < 1$ and let $ x_j \\equal{} A(1\\minus{} r)r^j.$ Then $ \\sum_{j\\equal{}0}^{\\infty}x_j \\equal{} A$ as a geometric series and $ \\sum_{j\\equal{}0}^{\\infty} x_j^2 \\equal{} \\sum_{j\\equal{}0}^{\\infty}A^2(1\\minus{} r)^2r^{2j} \\equal{} \\frac{A^2(1\\minus{} r)^2}{1\\minus{} r^2} \\equal{} \\frac{A^2(1\\minus{}r)}{1 \\plus{} r} .$ Clearly, we can make $ \\frac{1\\minus{} r}{1 \\plus{} r}$ take on any value in $ (0, 1)$ by choosing an appropriate value of $ r$ in $ (0, 1).$ (In particular, $ x \\equal{} \\frac{1 \\minus{} r}{1 \\plus{} r}$ if $ r \\equal{} \\frac{1\\minus{} x}{1 \\plus{} x}$ .)[/hide]" } { "Tag": [ "linear algebra", "matrix", "real analysis", "real analysis unsolved" ], "Problem": "Let $ \\Omega \\subset \\mathbb{R}^{n}$ be bounded subset and $ u \\in C^{2}(\\Omega) \\cap C^{2}(\\overline{\\Omega})$ and $ u$ satisfies:\r\n\\[ \\Delta u\\minus{}c(x)u\\geq 0 \\ \\ \\ in \\ \\ \\ \\Omega\\]\r\nwhere $ c(x) \\in C(\\Omega)$ and $ c\\geq 0$ in $ \\Omega$. Show that if $ \\sup_{\\Omega}u(x)>0$ then :\r\n\\[ \\sup_{\\Omega}u(x)\\equal{}\\max_{\\partial \\Omega}u(x)\\]", "Solution_1": "[quote=\"zibi\"]Let $ \\Omega \\subset \\mathbb{R}^{n}$ be bounded subset and $ u \\in C^{2}(\\Omega) \\cap C^{2}(\\overline{\\Omega})$ and $ u$ satisfies:\n\\[ \\Delta u \\minus{} c(x)u\\geq 0 \\ \\ \\ in \\ \\ \\ \\Omega\\]\nwhere $ c(x) \\in C(\\Omega)$ and $ c\\geq 0$ in $ \\Omega$. Show that if $ \\sup_{\\Omega}u(x) > 0$ then :\n\\[ \\sup_{\\Omega}u(x) \\equal{} \\max_{\\partial \\Omega}u(x)\\]\n[/quote]\r\n\r\nYes, this is nothing but the Weak Maximum Principle.", "Solution_2": "Can you explain it little more ? I still don't see solution...", "Solution_3": "For any $ \\varepsilon >0$ consider $ v(x)\\equal{}u(x)\\plus{}\\varepsilon e^{\\alpha x_1}$ with $ \\alpha$ to be determined. Then\r\n\r\n$ \\Delta v \\minus{} c(x) v \\equal{} \\Delta u \\minus{} c(x) u \\plus{} \\varepsilon (\\alpha^2 \\plus{} c(x))e^{\\alpha x_1} .$\r\n\r\nBy chosing $ \\alpha$ large enough, we get $ \\alpha^2 \\plus{} c(x) >0$ for any $ x \\in \\Omega$. This implies $ \\Delta v \\minus{} c(x) v>0$.\r\n\r\nIf $ v$ attains its non-negative maximum value of $ \\bar \\Omega$ in $ x_0 \\in \\Omega$, then the matrix $ \\big(v_{ij}(x_0)\\big)_{i,j}$ is semi-negative definite. Since the identity matrix is positive definite, then $ \\Delta v(x_0) \\leq 0$. This is a contradiction.\r\n\r\nThen $ v$ attains its non-negative maximum only on $ \\partial \\Omega$. By letting $ \\varepsilon \\to 0$ we finish the proof.", "Solution_4": "There should be\r\n\\[ \\Delta v \\minus{} c(x) v \\equal{} \\Delta u \\minus{} c(x) u \\plus{} \\varepsilon (\\alpha^{2} \\minus{} c(x))e^{\\alpha x_{1}}.\\]\r\nBut we don't know whether c(x) is bounded ? so we can't find such $ \\alpha$ that $ \\alpha^{2}\\plus{}c(x) >0$ ?", "Solution_5": "[quote=\"zibi\"]\nBut we don't know whether c(x) is bounded ? so we can't find such $ \\alpha$ that $ \\alpha^{2} \\plus{} c(x) > 0$ ?[/quote]\r\n\r\nOh you are right, normally $ c$ must be continuous up to the boundary $ c \\in C(\\bar \\Omega)$.", "Solution_6": "[quote=\"zibi\"]There should be\n\\[ \\Delta v \\minus{} c(x) v \\equal{} \\Delta u \\minus{} c(x) u \\plus{} \\varepsilon (\\alpha^{2} \\minus{} c(x))e^{\\alpha x_{1}}.\\]\nBut we don't know whether c(x) is bounded ? so we can't find such $ \\alpha$ that $ \\alpha^{2} \\plus{} c(x) > 0$ ?[/quote]\r\n\r\nIn order to avoid this difficulty, you can assume $ \\sup_{\\overline \\Omega} u(x)$ is archived (since $ u$ is continuous on a compact set $ \\overline \\Omega$) by an $ x_0 \\in \\Omega$. Then you shall work on a small neighborhood of $ x_0$. Having the contradiction shows that $ \\sup_{\\overline \\Omega} u(x)$ is archived only on the boundary $ \\partial \\Omega$." } { "Tag": [ "trigonometry", "search", "geometry", "similar triangles", "Comc", "Canada" ], "Problem": "A line segment BC has length 6. Point A is chosen such that angle BAC is right. For any position of A, a point D is chosen in BC such that AD is perpendicular to BC. A circle with AD as diameter has tangents drawn from C and B to touch the circle at M and N, respectively, with these tangents intersecting at Z.\r\nProve ZB + ZC is constant.\r\n\r\nThe given solution on the internet is kind of trig bashy :(. Does anyone have a nicer solution?", "Solution_1": "This isn't so nice either but works.\r\n\r\nFirst $ A$ is on the circunferenc of diameter $ BC$.\r\nLet $ E$ and $ F$ be the other points of tangency for $ B$ and $ C$ respectively.\r\n\r\n$ ZB \\plus{} ZC \\equal{} ZE \\plus{} ZF \\plus{} BE \\plus{} CF \\equal{} BC \\plus{} 2ZE$, we search to prove that $ ZE$ is constant.\r\n\r\nIt is easy to see that $ \\frac {AD}{2} \\equal{} \\sqrt {\\frac {BD*CD*ZE}{BD \\plus{} CD \\plus{} ZE}} \\equal{} \\frac {\\sqrt {BD*CD}}{2}$\r\nwich implies:\r\n$ 4ZE \\equal{} BD \\plus{} CD \\plus{} ZE$ or $ ZE \\equal{} \\frac {BC}{3}$\r\n\r\n\r\n :)", "Solution_2": "[quote=\"ElChapin\"]\nIt is easy to see that $ \\frac {AD}{2} \\equal{} \\sqrt {\\frac {BD*CD*ZE}{BD \\plus{} CD \\plus{} ZE}} \\equal{} \\frac {\\sqrt {BD*CD}}{2}$\n[/quote]\r\n\r\nSorry, could you explain that statement some more?", "Solution_3": "A triangle with sides $ a$, $ b$, $ c$, semiperimeter $ s$ and inradi $ r$ has area\r\n\r\n $ \\sqrt{s(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}\\equal{}sr$(if you didn't know it, is a good exercise to prove this relation)\r\n\r\nFrom this: $ r\\equal{}\\sqrt{(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}{s}$\r\n\r\nAlso the lenghts of the tangents from each vertex to incircle are $ s\\minus{}a$, $ s\\minus{}b$ and $ s\\minus{}c$, and note that $ s\\equal{}(s\\minus{}a)\\plus{}(s\\minus{}b)\\plus{}(s\\minus{}c)$\r\n\r\nIn a triangle with a right angle: the altitud $ h$ from the right angle divides the hypotenuse in two segments of lenghts $ d$ and $ e$, it can be prove (only using similar triangles) that $ h^2\\equal{}d*e$\r\n\r\n\r\nThis should be enough to solve the problem, if you didnt know or don't understand something try it by yourself, this helps a lot to get experience and to solve more of this kind of problems.\r\n\r\n\r\n\r\n :)" } { "Tag": [ "geometry", "3D geometry", "number theory unsolved", "number theory" ], "Problem": "Find all perfect powers whose last $ 4$ digits are $ 2,0,0,8$, in that order.", "Solution_1": "$ 10^4|a^b \\minus{} 2008\\implies 2^4|a^b \\minus{} 251\\cdot 2^3$\r\nThe largest power of $ 2$ dividing $ a^b$ must be precisely $ 2^3$. So $ a$ even but not divisible by $ 4$ and $ b \\equal{} 3$\r\nLet $ a \\equal{} 4k \\plus{} 2$\r\n$ 10^4|(4k \\plus{} 2)^3 \\minus{} 251\\cdot 2^3\\implies 5^4|k(4k^2 \\plus{} 6k \\plus{} 3) \\minus{} 125$\r\nBut $ 4k^2 \\plus{} 6k \\plus{} 3$ is never divisible by $ 5$ (checking $ a \\equal{} 1,2,3,4$ suffice). So $ k$ must be pf the form $ l\\cdot 5^3$\r\nSo $ 5|l(4k^2 \\plus{} 6k \\plus{} 3) \\minus{} 1\\implies 5|3l \\minus{} 1$. So $ l$ must be of the form $ 5t \\plus{} 2$\r\nGoing back, we see that all such powers are represented by $ \\boxed{(2500t\\plus{}1002)^3}$", "Solution_2": "Since such a number is divisible by 8 and not by 16, we conclude that the number is a cube.\r\n\r\nlet $ n^{3} \\equal{} 10000x \\plus{} 2008$ which implies $ m^{3} \\equal{} 1250x \\plus{} 251$ implying $ m^{3} \\minus{} 1 \\equal{} 250(5x \\plus{} 1)$ ,(n=2m)\r\n\r\nhence, $ m^{3} \\equal{} 1(mod250)$\r\nBy Euler's theorem , $ m^{100} \\equal{} 1(mod250)$\r\nsince (3,100)=1, we get that m=1(mod250), so m=250t+1 for some integer t\r\n\r\n$ (250t \\plus{} 1)^{3} \\minus{} 1 \\equal{} 250(5x \\plus{} 1)$ . Dividing both sides by 250 and reducing modulo 5, we get t=2(mod5)\r\n\r\nHence, all such cubes are of the form $ (2(250(5u \\plus{} 2) \\plus{} 1))^{3}$, that is $ (2500u \\plus{} 1002)^{3}$" } { "Tag": [ "geometry", "circumcircle", "parallelogram", "geometric transformation", "reflection", "geometry unsolved" ], "Problem": "(A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$. Another circle with the same radius passes through the orthocenter $ H$ of this triangle and intersect its circumcirle in points $ X$, $ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find the circumradius of triangle $ ABZ$.", "Solution_1": "[quote=\"Doctor A\"](A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$. Another circle with the same radius passes through the orthocenter $ H$ of this triangle and intersect its circumcirle in points $ X$, $ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find the circumradius of triangle $ ABZ$.[/quote]\r\nIt should be \"parallelogram CXYZ\". Then, the circumcircle (ABZ) is equal to R.", "Solution_2": "[quote=\"mr.danh\"][quote=\"Doctor A\"](A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$. Another circle with the same radius passes through the orthocenter $ H$ of this triangle and intersect its circumcirle in points $ X$, $ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find the circumradius of triangle $ ABZ$.[/quote]\nIt should be \"parallelogram CXYZ\". Then, the circumcircle (ABZ) is equal to R.[/quote]\r\ndear mr.danh,I think Doctor A is right\r\nX,C,Y,Z lie clockwisely", "Solution_3": "[quote=\"mr.danh\"][quote=\"Doctor A\"](A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$. Another circle with the same radius passes through the orthocenter $ H$ of this triangle and intersect its circumcirle in points $ X$, $ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find the circumradius of triangle $ ABZ$.[/quote]\nIt should be \"parallelogram CXYZ\". Then, the circumcircle (ABZ) is equal to R.[/quote]\r\nhow to do it?could you post your idea?\r\nthank you in advance!", "Solution_4": "[b]NO! In the original language of Russian, is CXZY paralelogram![/b]", "Solution_5": "Denote the intersection of XY and OO1 is N\r\n1.We make the reflection of H w.r.t N is H\u2019 \r\nWe know ON=O1N so HOH\u2019O1 is a parallelogram => HO1=OH\u2019=R => H\u2019lies on \u2299O\r\n=> ZHCH\u2019a parallelogram \r\n2. Extend AO meet \u2299O at M\r\nHCMB is a parallelogram => MB\u2225CH\u2225ZH\u2019 MB=CH=ZH\u2019 => ZBMH\u2019 is a parallelogram \r\nwe have ZB\u2225H\u2019M \r\n3.\u2220MH\u2019C=\u2220BZH =\u2220MAC=\u2220H1AB => B,Z,A,H are concyclic\r\n\u2220BZA=\u2220ACB=180-\u2220AHB\r\n4.using sin law\r\nsin\u2220AZB=AB/2R1 sin\u2220ACB=AB/2R => R=R1\r\nDONE!", "Solution_6": "[b]Remark:[/b] the circle HAB is reflection of ABC relatively the line AB ---> (circumcircle radius of $ \\triangle {HAB})\\equal{}$ (circumcircle radius of $ \\triangle {ABC})\\equal{}R.$", "Solution_7": "Yes. Sorry, I think X,Z are intersections of the circle which passes through H and the circumcircle (ABC). And this problem is not hard.\r\nWe should denote the reflections of O wrt AB and XY, then make a use of properties of parallelogram.\r\nBest regards, mr.danh", "Solution_8": "This problem can be easily solved by complex number. :P " } { "Tag": [ "AMC", "AMC 12", "AIME", "USA(J)MO", "USAMO", "Harvard", "college" ], "Problem": "I've been doing quite a bit of thinking the past few days about many, many things, and I'd just like to share a piece of one of my thoughts with you...\r\n\r\nI still remember how I was a year ago. I was a sophomore back then. I remember how I was still a little boy, running around during math club meetings studying cool problems and puzzles with my friends. Mathematics was so fascinating and elegant. I loved math. I really did.\r\n\r\nWhen I heard my math coach announce that we would be taking the AMC12, I became interested in what exactly the competition was. I had taken it the year before too, but no one at my school really studied for it. It wasn't a very big deal, and I had completely forgotten that it had existed. So I googled up the exam and found the official AMC website. I learned about the AMC, AIME, USAMO, MOSP. and IMO. I thought: \"Wow, wouldn't it be cool to attend a math summer camp with the top math students in the country?\" So from that point onwards, I studied really really hard in the hope of at least qualifying for the USAMO exam next year. I had a lot of fun too. I loved both the math and the competition. Life was looking good.\r\n\r\nHowever, soon, the evitable college process started looming over everyone, and I was thrown into the chaos and tension of college admissions. Everyone told me: \"Hey, you know, many of the USAMO qualifiers go to really competitive schools like Harvard and MIT. If you qualify, you could probably get admitted at one of those universities.\" But I said: \"Oh...but you know, I really only study this stuff for fun, sort of as an leisurely activity, you might say.\" Yet of course, everyone began fretting over grades, over SAT scores, over extracurricular activities, etc... And of course, my parents were obsessing over the ordeal as well: almost everything that they walked to me about was related to college admissions. So I looked at myself and thought: \"You know, my grades and SAT's aren't really spectacular, and I don't really have many extracurricular activities or achievements. But...you know...if I make USAMO, maybe that could be a huge hook for me in terms of college admissions.\"\r\n\r\nSo I began studying even harder, with an even greater determination to do well on the competitions. However, I found myself questioning: Am I studying so hard because I love math, or because I want to get admitted to the highly-selective colleges? Why am I doing what am doing? What is the driving force of my life right now? I felt so much pressure to make the USAMO. I go to a very good private school, many of whose seniors go on to Ivy League schools, so I felt even greater pressure to get accepted into the most selective colleges. I thought that the USAMO would have been my ticket. I was up against it the wall, and it was my only way out. I would either make it or break it--and the second choice wasn't really an option.\r\n\r\nYet, I sit here now having failed to qualify for the USAMO. I'm rambling now, but you know, as I look back upon the past year, I can't help thinking that the above story contains some pretty messed up stuff.\r\n\r\nThat was some pretty messed up stuff.", "Solution_1": "Hi, Fiery, I want to post a longer answer to you on that other thread you started, but meanwhile let me just say that my philosophy as a parent is that college admission should never be a basis for NOT DOING something, but rather a basis for DOING something. (This will become my slogan: this is the second time tonight I've had occasion to type this.) Too many people these days are WAY too obsessed about college admission, and way too willing to give up personally fulfilling, interesting activities because they feel they should only do what is on the formula for getting into the top college. But as JBL has been trying to point out in several threads, there is no sure-fire formula anyway. Dare to be yourself. Last year you did math for fun, and you can do math for fun for the entirety of your life. \r\n\r\nMy oldest son is younger than you are, and I think people reading between the lines of my posts can sense that I am a little disappointed that when I was a kid I heard about the top schools (and even had one adult close to me advise me to completely drop my then preferred major subject unless I could get into M.I.T., because studying that subject at State U. would be a waste of my time) but didn't hear about sensible things to do to prepare to get into top schools. I don't believe in letting the college admissions system grind someone down. Yes, do what is interesting and satisfying to you, learn about how to do it as well as you can, and then just don't worry about the mistakes some admission officer might make. People with superimpressive on-paper credentials don't get into the top schools sometimes, for mystifying reasons. People who were dying to get into the top schools have to make do with State U., and then discover a job, a cause, a fiance, or great scenery that makes them glad they ended up there. \r\n\r\nI had a funny conversation about a year ago with a guy in my homeschooling parent support group who confessed that he had always wanted to be in a different occupation: an occupation that I used to have and didn't like. That was funny because I thought I had always wanted to be in HIS occupation--until that conversation. Our wives each got a good laugh out of hearing about our conversation. Do what's fun, do what pays the bills, but most of all learn how to have fun while doing what pays the bills. 99.9 percent of the time it doesn't matter where, or even whether, you went to college. The other 0.1 percent of the time is subject to other variables anyhow, so don't worry about it.", "Solution_2": "Fiery - are you Asian? :)\r\n\r\nAnyway, the first thing I'd say is: calm down. Do math because you love it, not because it looks good on a college application. Having some contest on an application is only a fringe benefit that is nothing compared to your genuine appreciation for the subject. I do math because I like doing it, not to garner awards and boost my application. I'm sure that, deep down, you feel the same.\r\n\r\nThe second thing is, wherever you go to school, you will do well. Why? Because I believe your own drive and self-motivation for math will be your guide. Don't get too hung up on where. It's far more important to appreciate the four years that you stay in college as opposed to a piece of paper that either begins with either \"Congratulations!\" or \"We regret to inform you.\"\r\n\r\nWith that said, I completely empathize with your situation. I am Asian. I have Asian parents. I go to a predominantly Asian school (around 65%). And yes, if you're not Asian, let me be the first to tell you that the stereotypes are all true, and my parents drive me insane sometimes with college admissions. Lately I've also been feeling a lot of pressure to get into the big schools: HYPMS. I sometimes catch myself forgetting to take my own advice--that admissions aren't everything. Yes, parents and friends would like to have you think that a silly little piece of paper matters, but in the end, it's what you do later in college that matters. \r\n\r\nI know that I'm just an internet stranger to you, but trust me, I feel you. I'm pretty disappointed I didn't qualify for USAMO either, but then again, I can't say I share your grief because I really wasn't very qualified to begin with. But don't read these words and scoff, \"He doesn't understand,\" because I do. You let yourself down, but that happens to everyone. The best thing to do is move on and try harder next year.\r\n\r\nAnd let me reiterate that the entire admissions process is overrated. Just do your best, and whatever happens happens. \r\n\r\nHope that helped.", "Solution_3": "rlee: Yes, I'm Asian, Chinese in fact.\r\n\r\nHmm, ok, let me comment on something that you folks mentioned in this thread (and something that has been mentioned in other threads on this forum as well, I believe). Many here have commented that college admissions is not that important, and that what a person does during college is much more important than what college a person goes to. To a very great extent, I agree with this assertion. I don't know how much better of an undergraduate education in mathematics a student could get at MIT as compared to, say, UChicago (as a random school).\r\n\r\nHowever, the part about the assertion that I disagree with is based upon this: that going to a highly selective college like HYPMS will give a student quite a leg up after the student graduates. That is to say, a student who graduates from a highly selective college will be seen by job employers, graduate schools, and other organizations as more \"qualified\" than a student with the same intelligence who graduates from a less selective school. This student will thus have an easier time after he/she graduates simply because of the name of the university that he/she attended. I think this is the reason why many people want to get into HYPMS as opposed to supposed \"second-tier\" colleges -- for the prestige and name, not necessarily because of the education.\r\n\r\nI don't really like this *thing* that society has created that I described above, but I think its the reality.", "Solution_4": "I don't think that's true at all. For one thing, it is quite well-known that graduate schools do not like to accept their own undergraduates, so if you go to one of the five schools that you consider to be the absolute best, then you will only have four top graduate school choices rather than five most likely. (That is one reason why people advised me not to go to Berkeley, incidentally.) Also, I think if you do a very good job at a ``second-tier'' school, then grad schools will be more likely to accept you than if you do fairly well at a ``first-tier'' school.\r\n\r\nBy the way, University of Chicago is not exactly a random school. I would have gone to UChicago over MIT for sure. (It helped that I applied to UChicago and not MIT.) And where did Yale come from? I don't believe people consider Yale one of the great math schools. But I'm just being picky now, so I should stop.", "Solution_5": "[quote=\"Rep123max\"]I thought Yale was liberal arts or something.[/quote]\r\n\r\nMathematics is a liberal art. I myself am puzzled that Yale isn't mentioned more often as a math school. Roger Howe and Serge Lang are both quite famous math professors, and I think that there are others there. I have been to the campuses of Yale, Harvard, M.I.T., Princeton, Stanford, U. Chicago, and various other famous schools several times on business trips, and I think the one great disadvantage Yale has in attracting students is being located in New Haven, Connecticut, which is just not as attractive a town as Boston or Princeton or Chicago or Palo Alto. It's hard to be everyone's second choice, but that is not the same as saying Yale is a bad school. (Yale is arguably at the very top in law school rankings, for example, so it has its strong programs. I would love to do a graduate degree in psychology with Robert Sternberg there.) \r\n\r\nThat's enough of this digression. The point is that Fiery or any of the rest of us can have a successful life without having to go to any of the above-mentioned schools.", "Solution_6": "Well, I thought I was going to have something productive to add here, but rlee and tokenadult and Simon have roughly covered all the territory I might have wanted to, and quite elegantly. (By the way -- I also applied to U. Chicago and not MIT -- they have a very good math program there. The reason it's not as popular as others has nothing to do with the quality of its academics, and far more to do with the structure of its curriculum and the particular type of student it tends to attract.)\r\n\r\nAnyhow, the thing I wanted to comment on was the \"How come Yale isn't a math school?\" question. The answer, I think, doesn't lie so much in their not trying to attract math people -- I believe they are rather hoping to attract math people, and may even have tried to draw a similar group of students (Olympiadists, etc.) to waht Harvard and MIT get in recent years. (That was a terrible sentence, but I hope it's meaning is clear.) Anyhow, I think they fail for two reasons: (1) they don't already have any olympiadists, so it's harder to attract them, and (2) they don't have a good freshman math class. Harvard has Math 55; Chicago has Honors Analysis; MIT lets people take whatever the heck math class they want, give or take; but the most theoretical freshman math class at Yale is roughly as theoretical and tough as the third-most such class at Harvard. (I have this via a very talented friend who is currently at Yale and in that class and thinks that it's silly, rather.) I don't know how they expect to attract people without offering a class those people would want to take -- it seems very silly. So, there was my (badly written) two cents.\r\n\r\n\r\nOh, yeah -- Fiery, relax! :)", "Solution_7": "Hey Fiery. \r\nI think I can understand how you feel, even though my problem is somewhat different. \r\nI love learning. Just learning in general, just getting to explore anything and everything. Biology's my favorite subject, but I adore lit, chem, languages, math, drama, phys, history... pretty much anything I can get my hands on. There's always something interesting you can find in any subject, regardless of what it may be. And I love all of it. \r\nMy problem? I can't focus. I'm one of those \"all-around\" kids who's never outstanding in anything because she's \"good\" in everything. I cant devote myself wholly to one particular pursuit because there are so many others that are so unbearably [i]beautiful[/i] and impossible to let go of. \r\nOf course my other problem is that Im not smart enough excel at everything, but thats a separate issue. Heh.\r\nSo this year Im taking a few APs. I dont really mind APs because learning the material is interesting enough and fun in its own way. Yet sometimes, when Im exhausted from studying or otherwise worn out, I ask myself - whats all this in aid of? Am I working because I want to embellish my transcript or because I truly love the material? I would like to sincerely say that its the former, and more often than not I believe that thats the case; but there are points when Im flopped senseless on my bed wondering [i]what in the world am I trying to achieve here[/i]? Is there any meaning to all of this? \r\nAs for the SAT, I had a little episode with it that some of you know about and others can PM me if they care to learn. Suffice it to say that just added to my general sense of disillusionment a state I dont really care for. Id rather be hyper and spunky anytime. =) \r\n*laughs and shakes head* I dont think this letter/note/reply is at all coherent. Excuse me. Im sick this week. My mum thinks its from over-exhaustion. No comment.", "Solution_8": "You are the reason that liberal-arts colleges were invented :). Try and relax a little bit -- being good at many things should not be complained about :)" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "find all points(a,b) such that b^3 -a^3=2007ab+2009", "Solution_1": "What are $ a,\\ b$?" } { "Tag": [ "MATHCOUNTS", "probability", "geometry", "number theory", "AMC 8" ], "Problem": "7th in Massachusetts\r\n\r\n\r\n[color=darkblue][size=75][/size][/color]", "Solution_1": "State: Utah\r\nPlace: 8th individual and countdown\r\n\r\n[color=darkblue][size=75][/size][/color]", "Solution_2": "Let me reemphasize what moderator rcv is repeatedly pointing out: we don't want any discussion of the state meets until after all of the state meets are over. That's simply more fair to all the mathletes, and fair play is part of what you should be learning through MATHCOUNTS participation. \r\n\r\nAs before, I wish all the AoPSers the best success in their math competition participation.", "Solution_3": "NO, LOL, worse than 24th in Indiana, though I have two more years to try to win :)", "Solution_4": "ours is next Saturday.... trying to study right now!", "Solution_5": "Same here... just hoping to get in... :crosses his fingers:", "Solution_6": "same i really want to go to nationals too", "Solution_7": "Arrggh, 3rd countdown but only 12th individual. Sadly countdown isn't official in Maryland.", "Solution_8": "well, i'm just hoping to get a good score, i'm pretty sure i won't be able to make it.... only four people goes, and i'm only fourth in my chapter.... but i'm hoping maybe i'll go next year as an eighth grader :D", "Solution_9": "Lucy, \r\n\r\nI've been coaching for 6 years, and twice I've had a kid finish 4th at chapter and then make it to nationals(both times they were 3rd at state) So don't give up hope. Here is a list of how my kids who did well at chapter faired at states\r\n\r\nChapter ---> State\r\n2nd 4th\r\n4th 10th\r\n3rd 3rd\r\n1st 3rd\r\n2nd 8th\r\n3rd 2nd\r\n4th 10th\r\n4th 3rd(one point out of 1st)\r\n3rd 2nd\r\n6th 6th\r\n1st 7th\r\n2nd 4th\r\n3rd 11th\r\n4th 3rd\r\n9th 6th\r\nThose last 5 were all last year. You can see that you don't always do worse at state than you did at chapter, just depends on how hard you work!!\r\n\r\nGood Luck", "Solution_10": "THANK YOU!!!!! I'll try!", "Solution_11": "State: Utah\r\nRank: 3rd.\r\n\r\nps. For anyone who has already made it to nationals, or knows what to do...do I have to list a \"preferred\" roomate?", "Solution_12": "my state is on march 12 i hope i get to top 4", "Solution_13": "[quote=\"Secret Asian\"]State: Utah\nps. For anyone who has already made it to nationals, or knows what to do...do I have to list a \"preferred\" roomate?[/quote]\r\n\r\nUsually at the meeting right after states, we have the kids pair off and put down a preferred roommate. If the 4 kids are 3 boys and a girl, or 3 girls and a boy, then they obviously can't pair off and are pretty much placed at random. If you don't put a preferred roommate and your state is sending all boys or all girls, you will be paired with a kid from your state.", "Solution_14": "i dunno but some how i doubt it. my mc coach relies more on the team to make it to nationals. oh well. i doubt it too so i cant argue. (haha) but our competitions this satuday on march 12. i hope i hope i hope! :? :? :?", "Solution_15": "i just watch my clock and if im halfway and its not 15 min yet then i do it in the order they come but if it is 15 min and im not halfway i look for easy questions", "Solution_16": "I never check my watch, I just go from 1-30. I've never had trouble doing it in time. This years problems weren't that hard and didn't take that long", "Solution_17": "sometimes i almost run out of time sometimes i finsih very early that i can check it 2 times", "Solution_18": "Mark, you always finish the test. \r\nYou sit at the same table as me and you always turn the page in like 1 minute...", "Solution_19": "dont you think that the 6 minute for the pair of problems for the target round is too much? I finish the problems in 2 minutes and then fiddle around with my calculator.", "Solution_20": "For the target problems it really depends on the test. Some tests I fininish the targets in 30 seconds others I'm struggling to finish within the time period.", "Solution_21": "Yeah, I've had times where I want to just get out of my seat and stretch until the 5 remaining minutes are up, and then there's the times when I don't see how the problem can be done.", "Solution_22": "I always work to the last second, I check, double check, triple check, etc until time is up if I'm done.", "Solution_23": "[quote=\"09husbri\"]...If that applies to you, then... :lol:[/quote]\r\n\r\nhey, be quiet :blush: that one was hard!", "Solution_24": "My bad, didn't mean to offend any fellow Mathlete :)", "Solution_25": ":D no problem, it's all cool", "Solution_26": "Greg (1st place one) told me he played tetris on his calculator after each set of target round problems at state :D. But at nationals you can never check too many times.", "Solution_27": "Believe it or not, you can.", "Solution_28": "[quote=\"09husbri\"]...If that applies to you, then... :lol:[/quote]Hey, I missed number one! For some reason I thought 40+8=52.", "Solution_29": "Are you going to nationals? If you are, at least I'm not getting 228th...JK :lol:" } { "Tag": [ "inequalities", "function", "calculus", "derivative" ], "Problem": "Let $ a,b,c\\in\\mathbb{R}$ and $ a\\plus{}b\\plus{}c\\equal{}1$. Prove that:\r\n$ \\frac{a}{1\\plus{}a^2}\\plus{}\\frac{b}{1\\plus{}b^2}\\plus{}\\frac{c}{1\\plus{}c^2}\\le \\frac{9}{10}$", "Solution_1": "It has one more condition $ a,b,c\\ge\\frac{\\minus{}3}{4}$\r\n$ 28^{th}$ problem in Mildroff's inequalities.\r\n[hide]Even i was not able to prove it[/hide]", "Solution_2": "[quote=\"Kyoshiro\"]Let $ a,b,c\\in\\mathbb{R}$ and $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove that:\n$ \\frac {a}{1 \\plus{} a^2} \\plus{} \\frac {b}{1 \\plus{} b^2} \\plus{} \\frac {c}{1 \\plus{} c^2}\\le \\frac {9}{10}$[/quote]\r\nthis problem is not difficult, we can solve it by a formal solution :lol:\r\nI suggest that we can find k which satisfy this ineq.:\r\n$ \\frac {a}{1 \\plus{} a^2} \\le \\frac{3}{10}\\plus{} k(\\frac{1}{3}\\minus{}a)$", "Solution_3": "[code]In[21]:= k = -18/25\nFactor[3/10 + k (1/3 - a) - a/(1 + a^2)]\nk =.\n\nOut[21]= ((-1 + 3 a)^2 (3 + 4 a))/(50 (1 + a^2))[/code]\r\n\r\nNice :)", "Solution_4": "WLOG $ a\\geq b \\geq c$ ,and use Chebyshev's inequality\r\n\r\nWe get \r\n\r\n$ LHS \\leq \\frac{1}{3} \\sum_{cyc}{\\frac{1}{1\\plus{}a^2}}$ \r\n\r\nLet $ f(x)\\equal{}\\frac{1}{x^2 \\plus{}1}$ ,now using Jensen's inequality we have \r\n\r\n$ \\frac{1}{3} \\left(\\frac{1}{a^2 \\plus{}1} \\plus{}\\frac{1}{b^2 \\plus{}1} \\plus{}\\frac{1}{c^2 \\plus{}1} \\right) \\leq f\\left(\\frac{a\\plus{}b\\plus{}c}{3} \\right)\\equal{}\\frac{9}{10}$", "Solution_5": "Are sequences $ (a,b,c)$ and $ \\left( \\frac{1}{1\\plus{}a^2}, \\frac{1}{1\\plus{}b^2} , \\frac{1}{1\\plus{}c^2} \\right)$ similarly sorted? I think they aren't.", "Solution_6": "If $ a\\geq b \\geq c$ then $ \\frac{1}{1\\plus{}a^2 } \\leq \\frac{1}{1\\plus{}b^2 } \\leq \\frac{1}{1\\plus{}c^2}$ \r\n\r\nThen using Chebyshev's inequality \r\n\r\n$ LHS \\leq \\frac{1}{3}\\sum_{cyc}{\\frac{1}{1\\plus{}a^2}}$", "Solution_7": "Yes of course it is correct.\r\n\r\nNice solution. :)", "Solution_8": "[quote=\"enndb0x\"]Let $ f(x) \\equal{} \\frac {1}{x^2 \\plus{} 1}$ ,now using Jensen's inequality we have \n\n$ \\frac {1}{3} \\left(\\frac {1}{a^2 \\plus{} 1} \\plus{} \\frac {1}{b^2 \\plus{} 1} \\plus{} \\frac {1}{c^2 \\plus{} 1} \\right) \\leq f\\left(\\frac {a \\plus{} b \\plus{} c}{3} \\right) \\equal{} \\frac {9}{10}$[/quote]\r\n\r\nIn order to use Jensen's Inequality, you must have a function that is strictly convex, or strictly concave over the interval containing all possible values of the variables (in this case, the interval must cover all possible values of $ a,b,c$). \r\n\r\nHowever: $ f''(x) \\le 0$ for $ x \\in \\left[\\minus{}\\dfrac{1}{\\sqrt{3}},\\dfrac{1}{\\sqrt{3}}\\right]$, and $ f''(x) \\ge 0$ for $ x \\in \\left(\\minus{}\\infty,\\minus{}\\dfrac{1}{\\sqrt{3}}\\right] \\bigcup \\left[\\dfrac{1}{\\sqrt{3}},\\infty\\right)$. \r\n\r\n[url]http://www.wolframalpha.com/input/?i=second+derivative+1%2F(x^2%2B1)[/url]\r\n\r\nSo, unless you have either $ a,b,c \\in \\left[\\minus{}\\dfrac{1}{\\sqrt{3}},\\dfrac{1}{\\sqrt{3}}\\right]$ OR $ a,b,c \\in \\left(\\minus{}\\infty,\\minus{}\\dfrac{1}{\\sqrt{3}}\\right] \\bigcup \\left[\\dfrac{1}{\\sqrt{3}},\\infty\\right)$, you can't use Jensen's.", "Solution_9": "[quote=\"enndb0x\"]If $ a\\geq b \\geq c$ then $ \\frac {1}{1 \\plus{} a^2 } \\leq \\frac {1}{1 \\plus{} b^2 } \\leq \\frac {1}{1 \\plus{} c^2}$ [/quote]\r\n\r\nSo you proved that sequences are sorted in opposite way. Apart from that you are not sure about sign of $ a$ and $ b$.", "Solution_10": "@micf\r\n\r\nAh..you're right , all the time I was looking for a solution in positive real numbers , do you think the solution is OK in positive reals.\r\n\r\n\r\n@Xantos\r\nAre you saying that $ f(a) \\equal{} \\frac {1}{1 \\plus{} a^2}$ is not concave in interval $ (0,\\infty)$ ?", "Solution_11": "That is exactly what I'm saying. \r\n\r\nThe link that I tried to post shows that $ \\dfrac{1}{1 \\plus{} x^2}$ is concave down on $ \\left( \\minus{} \\frac {1}{\\sqrt {3}},\\frac {1}{\\sqrt {3}}\\right)$ and concave up on $ \\left( \\minus{} \\infty, \\minus{} \\frac {1}{\\sqrt {3}}\\right)\\bigcup\\left(\\frac {1}{\\sqrt {3}},\\infty\\right)$. \r\n\r\n\r\nIf we are given that $ a,b,c$ are non-negative: \r\n\r\nThen $ a,b,c \\in [0,1]$ and we can use Jensen's Inequality on $ f(x) \\equal{} \\dfrac{x}{1 \\plus{} x^2}$ which is concave down on $ [0,1]$. \r\n\r\nTherefore, $ f(a) \\plus{} f(b) \\plus{} f(c) \\le 3f\\left(\\dfrac{a \\plus{} b \\plus{} c}{3}\\right) \\equal{} \\dfrac{9}{10}$ as desired.", "Solution_12": "thus,the problem is solved just for positive real numbers ... :(", "Solution_13": "From Enndbox's post (chebyshev's ineq) we have:\r\n$ LHS\\leq \\frac{1}{3}\\sum \\frac{1}{a^2\\plus{}1}$\r\nNow we prove that $ \\sum\\frac{1}{1\\plus{}a^2}\\leq \\frac{27}{10}$\r\nthis is an interesting inequality and has been discussed here many times before.\r\nNow, $ (4\\minus{}3t)(3t\\minus{}1)^2\\geq 0 \\forall t\\leq 1$\r\nHence $ \\frac{1}{1\\plus{}x^2}\\leq \\frac{27}{50}(2\\minus{}x)$\r\nSo that $ \\sum \\frac{1}{1\\plus{}x^2}\\leq \\frac{27}{10}$\r\n\r\nIt is a hard solution to find..... :rotfl:", "Solution_14": "[quote=\"Potla\"]From Enndbox's post (chebyshev's ineq) we have:\n$ LHS\\leq \\frac {1}{3}\\sum \\frac {1}{a^2 \\plus{} 1}$\nNow we prove that $ \\sum\\frac {1}{1 \\plus{} a^2}\\leq \\frac {27}{10}$\nthis is an interesting inequality and has been discussed here many times before.\nNow, $ (4 \\minus{} 3t)(3t \\minus{} 1)^2\\geq 0 \\forall t\\leq 1$\nHence $ \\frac {1}{1 \\plus{} x^2}\\leq \\frac {27}{50}(2 \\minus{} x)$\nSo that $ \\sum \\frac {1}{1 \\plus{} x^2}\\leq \\frac {27}{10}$\n\nIt is a hard solution to find..... :rotfl:[/quote]\r\nNot really, if we take the derivative at the place where we guess the extreme value occurs, and find the equation of the tangent line.", "Solution_15": "a^2/a+a^3+b^2/b+b^3+c^2/c+c^3. \r\nWe use titu's lemma.\r\n>=(a+b+c)^2/a^3+b^3+c^3+a+b+c=1/1+a^3+b^3+c^3\r\na^3+b^3+c^3>=3abc\r\na+b+c>=3sqrtabc=(maximized when this equals 1/3)\r\nabc=1/9 at maximum. so the inequality is true.1/1+1/9=9/20" } { "Tag": [ "inequalities" ], "Problem": "Anyone got a nice short solution for this problem?\r\n\r\nThe points $z_1, z_2, z_3$ form an equilateral triangle.\r\nProve that: $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1$.\r\n\r\nIt reminds me of the well known and easy to prove inequality $x^2+y^2+z^2\\geq xy+yz+xz$, though i can't see any link with it =/", "Solution_1": "Suppose we have an equilateral triangle with center $z_c$. We substitute $(z_i-z_c)$ for $z_i$ in the equations and note that the the condition is equivalent (just multiply it out). Thus, we need only show it holds when the triangle is centered at the origin. Note that now $|z_1|=|z_2|=|z_3|$, since they form an equilateral triangle at the origin. We replace each $z_i$ by $\\frac{z_i}{z_1}$ (multiplying out shows that this condition is equivalent). Now, $z_1=\\text{cis}\\left( 0\\right)$, $z_2=\\text{cis}\\left( \\frac{2\\pi}{3}\\right)$, $z_3=\\text{cis}\\left( \\frac{4\\pi}{3}\\right)$, and we only have to plug them in and check that it works.\r\n\r\nThus, an equilateral triangle centered at the origin with $z_1=1$ works, and so any equilateral triangle centered at the origin works, and so any equilateral triangle works, and we are done.", "Solution_2": "think of the roots of the equation\r\n$(z-w)^3=m$, w will be the center of triangle, the roots will be the points of the triangle, (since i ran out of letters, $m\\in \\mathbb{C}$ also\r\nso we have\r\n$z^3-3z^2w+3zw^2-w^3-m=0$\r\nthen by viete's sums, we have $3w^2=z_1z_2+z_1z+z_2z_3$\r\nand using newton's sums (we want $s_2$), we have \r\n$1*s_1+(-3w)=0$\r\n$s_1=3w$\r\n$1(s_2)+(-3w)(3w)+2(3w^2)=0$\r\n$s_2=3w^2$\r\nusing that, we have\r\n$s_2=3w^2=z_1z_2+z_1z+z_2z_3=z_1^2+z_2^2+z_3^2$" } { "Tag": [ "integration", "limit" ], "Problem": "ean i sinechis sinartisi g(x) ine gnisios fthinousa konta sto +apeiro kai to orio tis sto +apiro ine 0,na ipologisete to orio kathos to x tini sto +apiro tou orismenou olokliromatos apo x eos x+1 tis g(x).\r\n\r\n\r\nsorry,alla den echo to programma gia ta mathimatika simvola...", "Solution_1": "[quote=\"koredim1989\"]ean i sinechis sinartisi g(x) ine gnisios fthinousa konta sto +apeiro kai to orio tis sto +apiro ine 0,na ipologisete to orio kathos to x tini sto +apiro tou orismenou olokliromatos apo x eos x+1 tis g(x).\n\n\nsorry,alla den echo to programma gia ta mathimatika simvola...[/quote]\r\n\r\n\u0393\u03c1\u03ac\u03c8\u03b5 \r\n\r\n $\\int_{x}^{x+1}g(x)dx = g(k) , x I_n \\equal{} n. I_{n\\minus{}1}$\r\n$ I_0 \\equal{} 1$\r\nhence,\r\n$ I_8 \\equal{} 8!$\r\n:)", "Solution_7": "[quote=\"hash_include\"]my method is obviously longer than elastiboy's, but thats coz i derive the gamma function :D\nanyway, consider \n$ I_n \\equal{} \\int_{0}^{1} (log(\\frac {1}{x}))^ndx$\ndoing parts, we get\n$ I_n \\equal{} n \\int_{0}^{1} (log(\\frac {1}{x}))^{n \\minus{} 1}dx$\n(the UV term after doing parts = 0)\n$ \\equal{} > I_n \\equal{} n. I_{n \\minus{} 1}$\n$ I_0 \\equal{} 1$\nhence,\n$ I_8 \\equal{} 8!$\n:)[/quote]\r\n\r\nIts more elegant this way but stating the obvious doesn't make it more obvious. :D" } { "Tag": [ "summer program", "PROMYS", "LaTeX" ], "Problem": "In TeXNiC i cant space my lines so that it appears well on a page. For example:\r\nWhen I type : \r\n\"PROMYS app Problem: 1\r\nblah blah blah\"\r\n\r\nIt comes out\r\n\"PROMYS app Problem: 1 blah blah blah", "Solution_1": "You have to tell LaTeX you want a new line with \\\\. This allows you to lay out the code in any way you want without affectng the final text. So in your example type:\r\n[code]PROMYS app Problem: 1 \\\\ \nblah blah blah[/code]", "Solution_2": "thank you stevem" } { "Tag": [], "Problem": "Hi...... \r\n@skand which all topics have u prepared for the nsea(senior).... i am planning to give it this year...... i am in class 10 so i know i dont stand much chance but just to have an outing.......\r\n :) :) \r\n\r\nAnd can u give me some links for the papers.....\r\ni have nsea senior paper for 2007\r\n\r\nCan anyone provide me the papers for the previous years....???? :?:", "Solution_1": "welcome to AoPS :) \r\n\r\n[url]http://www.iapt.org.in[/url]\r\nyou can have question papers there\r\n\r\nabout me,i am not planning to do anything about NSEA.Will just give it as timepass :rotfl: \r\n\r\ncontact Arnav,he was NSEA topper previous year :)", "Solution_2": "hey seems like nsea's attendance is increasing day by day", "Solution_3": "you need not take any tension..\r\nwat match am i for u???\r\nanyways\r\nwhere is the topper..... cant he provide me some help :)\r\n\r\n@skand i have 2007 paper as i told u.....\r\ndo u have other years papers??", "Solution_4": "prev yr papers r usually not available on internet as long as some1 doesn't hack them or scan them :D :roll:", "Solution_5": "hackers at their work!\r\nritu u know hacking?", "Solution_6": "I remember ritu telling that he hacked somebody's account in aops itself", "Solution_7": "admin's account :rotfl: :rotfl: :rotfl: (adiarya)" } { "Tag": [ "number theory", "relatively prime", "number theory proposed" ], "Problem": "Let p,q,r,a natural numbers with $pq=ra^2 $ r is a prime number and (p,q)=1\r\nThere exist prime number $ p $ such that the number $ p\\left(2^{p+1}-1\\right) $ is a perfect square?", "Solution_1": "What is that first sentence doing there?\r\n\r\nFor the question \"Find all primes $p$ such that $p(2^{p + 1} - 1)$ is a perfect square...\"\r\n\r\nFor $p = 2$ we get 14 which is not a perfect square.\r\n\r\nNow suppose that $p$ is odd, $p = 2q - 1$, say.\r\n\r\nSuppose that $p(2^{p + 1} - 1)$ is a perfect square. Since it is divisible by $p$, it must be divisible by $p^2$, so that $p$ divides $2^{p + 1} - 1$. It is easy to see that $2^{p + 1} - 1$ is in fact equal to $pa^2$ where $a$ is a positive integer. Hence $(2^q - 1)(2^q + 1) = pa^2$, and since the two factors on the left hand side are relatively prime, one of them must be a perfect square. However the first one cannot be a perfect square since it is equal to 3 (mod 4). So the second one must be a square. Suppose that $2^q + 1 = t^2$, then $(t - 1)(t + 1) = 2^q$. The only powers of 2 differing by 2 are 2 and 4. Hence $t = 3$ and $q = 3$. Hence $p = 5$. However for $p =5$, we have that $p(2^{p + 1} - 1) = 3^2 \\cdot 5 \\cdot 7$ which is not a perfect square. Hence there are no solutions.", "Solution_2": "Arne is right.\r\nYou can also solve this one by using fermat's little theorem.\r\nBecause since $p$ divides $ 2^{p+1} - 1$ and $2^p -2$ we get that $p$ divides $2^{p+1} - 1 -2(2^p -2) = 3$.\r\nBut $p$ can't equal $3$. So there are no solutions.\r\n\r\nThis problem is from the Greek M.O \"Archimedes\" 2001 . It coresponds the level of the problems posted in greek competitions (It was the last problem then!!!! :( ).", "Solution_3": "It is a quite simple question without the first condition. \r\nBut I am interested in the first condition. \r\nAre there any relation between two questions?\r\n\r\nI think there are some mistakes." } { "Tag": [ "induction", "combinatorics proposed", "combinatorics" ], "Problem": "Let $ n$ be an integer, $ n\\geq 2$, and the integers $ a_1,a_2,\\ldots,a_n$, such that $ 0 < a_k\\leq k$, for all $ k \\equal{} 1,2,\\ldots,n$. Knowing that the number $ a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n$ is even, prove that there exists a choosing of the signs $ \\plus{}$, respectively $ \\minus{}$, such that\r\n\\[ a_1 \\pm a_2 \\pm \\cdots \\pm a_n\\equal{} 0.\r\n\\]", "Solution_1": "By induction on n\r\n\r\nSuppose that we could choose the sign +, - when n=k and we have to show that we could find an arrane when n=k+1\r\n\r\n If a(k+1) does not equal ak, then apply induction hypothes for k number\r\n\r\n a1,a2,..., a(k-1), |a(k+1) -ak| \r\n\r\n\r\n If a(k+1)=ak, then apply induction hypothes for k-1 number a1,a2,.., a(k-1) and put a - between ak, a(k+1)", "Solution_2": "It is very similar to Chvatal Theorem about Hamiltonian paths.\r\nConsider a graph with vertices of degree $ a_i$. Then they satisfy $ \\sum a_i \\equal{}2E$.\r\nThen the for a Hamiltonian path pass we add degrees for even vertices on the path and subtract odd.\r\nIt is not the same problem, but remind me about graph theory.", "Solution_3": "[quote=mto]By induction on n\n\nSuppose that we could choose the sign +, - when n=k and we have to show that we could find an arrane when n=k+1\n\n If a(k+1) does not equal ak, then apply induction hypothes for k number\n\n a1,a2,..., a(k-1), |a(k+1) -ak| \n\n\n If a(k+1)=ak, then apply induction hypothes for k-1 number a1,a2,.., a(k-1) and put a - between ak, a(k+1)[/quote]\n\nVERY SMOOTH :coool: \n" } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "On the circle given by 16x^2 + 16y^2 + 48x - 8y - 43 = 0 find the point M which is closest to the straight line given by 8x- 4y + 73 = 0 and calculate the distance d from M to this Line", "Solution_1": "The equation of the circle is $(x+\\frac{3}{2})^{2}+(y-\\frac{1}{4})^{2}=(\\sqrt{5})^{2}$.\r\n\r\n[hide] We can find the point closest to $8x-4y+73=0$ by drawing the tangent to the circle with the same slope of 2. Then the radius from M to the center of the circle is perpindicular to this line, so it has a slope of $-\\frac{1}{2}$. Therefore $M$ is at $(-\\frac{3}{2}+\\frac{2}{\\sqrt{5}}\\times\\sqrt{5},\\frac{1}{4}-\\frac{1}{\\sqrt{5}}\\times \\sqrt{5})=(\\frac{1}{2},-\\frac{3}{4})$ or \n$(-\\frac{3}{2}-\\frac{2}{\\sqrt{5}}\\times\\sqrt{5},\\frac{1}{4}+\\frac{1}{\\sqrt{5}}\\times \\sqrt{5})=(-\\frac{7}{2},\\frac{5}{4})$. The line $8x-4y+73=0$ passes above the circle, so $M=(-\\frac{7}{2},\\frac{5}{4})$. \n\nThen the projection of $M$ to the line is at the intersection of $8x-4y+73=0$ and $x+2y+1=0$, which is $(-\\frac{15}{2},\\frac{13}{4})$. This distance is equal to $\\sqrt{4^{2}+2^{2}}=2\\sqrt{5}$.[/hide]" } { "Tag": [ "probability", "Pascal\\u0027s Triangle", "conditional probability" ], "Problem": "Mrs. Witts has two children. One of them is a female. If having a male or female child is equally likely, what is the probability that the other one is a male? Express your answer as a common fraction.\r\n\r\nThe answer is given as 2/3, and everyone says it is. But I think its 1/2 because the probabilities are independent right?", "Solution_1": "Why don't you read the posts in the other thread in FTWPD?", "Solution_2": "I did, and I fail to see why MF is more likely than FF once you know there is 1 female.", "Solution_3": "Sigh...last try.\r\n\r\nSay Mrs. Witts has two children, A and B, who I hope you realize are distinct. :wink: \r\n\r\nWe are only given that at least one of them is female.\r\n\r\nAssume that A is female. Then B can either be female or male, giving us two possibilities.\r\nAssume that A is male. Then B can either be female or male--but wait, B can only be female, since at least one of them must be female (do you see why???)!!!\r\n\r\nThis gives us three different possibilities:\r\n\r\nA=female, B=female\r\nA=male, B=female\r\nA=female, B=male\r\n\r\nNotice how choice two and three are different (can you tell the difference?--good job.).", "Solution_4": "wait whoa. A is 50% chance of being male right?\r\n\r\nSo case 1 where a is female occurs only 50% of the time.\r\n\r\nIf B is male then FM (50% chance) or female FF (50%) so combining 25% FF 25% FM.\r\n\r\nCase 2, a is male (50%) b is femal so FM 100%.\r\n\r\nThus its 25% of FF and 75% FM?", "Solution_5": "That's not how I would recommend you think about it...just notice that the three cases I posted are equally likely--another way to go about it is $ \\frac{2\\cdot0.5^2}{1\\minus{}0.25}\\equal{}\\frac23$ (do you see why it works? What do the numerator and denominators represent?)", "Solution_6": "NUmerator=proabbility of MF so shouldnt denominator be 1?\r\n\r\n\r\nAnd why are the cases equally likely? I just shwoed why they [b]werent[/b]", "Solution_7": "Which cases are you referring to? My cases (in my posts) are equally likely. There is a $ 1$ probability that there are $ 2$ children here, but we must exclude MM, which has probability $ 0.25^2$.", "Solution_8": "Oh ok i get it now! Thank you so much everyone! It's because you only count the 2 right parts of pascals triangle so its $ \\frac{2}{2\\plus{}1}\\equal{}\\frac{2}{3}$", "Solution_9": "I fail to see how Pascal's Triangle fits in.\r\n\r\nWhat math154 said is as good of an explanation as you're going to get. There are $ 4$ possibilities without restrictions, but with the restriction, we cannot have $ MM$. Then the possibilities left are $ MF$, $ FM$, and $ FF$, so the probability is clearly $ \\boxed{\\frac{2}{3}}$. \r\n\r\nAgain, I don't see how Pascal fits in.", "Solution_10": "[quote=\"AIME15\"]I fail to see how Pascal's Triangle fits in.\n\nWhat math154 said is as good of an explanation as you're going to get. There are $ 4$ possibilities without restrictions, but with the restriction, we cannot have $ MM$. Then the possibilities left are $ MF$, $ FM$, and $ FF$, so the probability is clearly $ \\boxed{\\frac {2}{3}}$. \n\nAgain, I don't see how Pascal fits in.[/quote]\r\n\r\nThe Pascal's Triangle fits in like this:\r\n\r\n 1\r\n 1 1\r\n 1 2 1\r\n\r\nIn the 121 place, the left 1 stands for All girls, middle 2 stands for 1 girl and one boy, and the right 1 stands for 2 boys. mewto i think got 1/2 because they want 1 girl and 1 boy, so it would be $ \\frac12$, but since 2 boys is not possible, you keep those three and get $ \\boxed{\\frac23}$", "Solution_11": "I first interpreted the question like mew. For example:\r\n\r\nMaria flipped a coin 9 times, and was surprised to see that they were all heads. What is the probability that the next flip will be heads?", "Solution_12": "[quote=\"Dojo\"]I first interpreted the question like mew. For example:\n\nMaria flipped a coin 9 times, and was surprised to see that they were all heads. What is the probability that the next flip will be heads?[/quote]\r\n\r\nI remember doing a problem like that on some test. It took me for ever, as I made Pascal's Triangle all the way to 10 (I wasn't very smart back then, and I'm still not), and finally got the answer. Then I found out the answer was $ \\frac12$. It wasted a lot of time...", "Solution_13": ":rotfl:\r\n\r\nBTW dojo, don't put problems from your test here :D", "Solution_14": "The confusion here is a distinction between this question and the following question:\r\n\r\nMrs. Witts has two children. [b]The older child[/b] is female. What is the probability that the younger child is male?\r\n\r\nHere the two events are, in fact, independent, and the answer is $ \\frac{1}{2}$ as intuition suggests. Can you see why this question is not the question being asked? There are two ways that one of the children can be female - it can be the younger or the older child - and that influences the [url=http://en.wikipedia.org/wiki/Conditional_probability]conditional probability[/url] involved. As an exercise, answer the corresponding question:\r\n\r\nGiven that one of the children is male, what is the probability that the other child is also male?", "Solution_15": "[hide=\"Click for solution\"]\nWithout restrictions, the cases are:\n\nFF\nFM\nMF\nMM\n\nHowever, we do have restrictions! FF is now an invalid case! The probability that the other child is male is thus $ \\frac{1}{3}$.\n[/hide]", "Solution_16": "i think the answer is 1/2 becaouse the chances r equal so....", "Solution_17": "Okay I will try to explain this again.\r\nThere are 2 kids, let's call them A and B. And there are 3 different ways they can, well, be arranged (or something like that)\r\n1. A=female, B=female \r\n2. A=male, B=female \r\n3. A=female, B=male \r\nSo there's a $ \\frac{2}{3}$ chance, see?", "Solution_18": "I saw a problem like this in a math book I had. I was something involving that game show where you can take what's behind door 1 or take what's behind door 2 or 3.", "Solution_19": "Like the [url=http://en.wikipedia.org/wiki/Monty_hall_problem]Monty Hall problem[/url]?", "Solution_20": "Exactly. Except it gave a name for the talke show. I think it was iin the book [i]\"Hexaflexagons and other mathematical diversions[/i]", "Solution_21": "Oh that one! :D \r\nI prefer to put it like this...\r\n\r\nThere are 3 doors, one contains 1 million dollars and the others contain nothing but air. You open a door and it contains nothing but air. You are allowed choose another door, or stay with what you have. What is the probability that you'll get the money?\r\n\r\n[hide=\" This is probably the wrong answer but...\"]The probability would 1/3, the same as before, not 1/2 because you have the option to keep the air, so then it's a 1/3 probability. Right? I think it's something like that.[/hide][/hide]", "Solution_22": "[quote=\"joyofpi\"]Oh that one! :D \nI prefer to put it like this...\n\nThere are 3 doors, one contains 1 million dollars and the others contain nothing but air. You open a door and it contains nothing but air. You are allowed choose another door, or stay with what you have. What is the probability that you'll get the money?\n\n[hide=\" This is probably the wrong answer but...\"]The probability would 1/3, the same as before, not 1/2 because you have the option to keep the air, so then it's a 1/3 probability. Right? I think it's something like that.[/hide][/hide][/quote]\r\n\r\n$ \\frac{1}{3}$ should be correct. Why:\r\n\r\nRight now you have air. these are the following possibilities that can happen:\r\n\r\n1. Re choose: get money.\r\n2. Re Choose: get door 3 which contains some anonymous thing.\r\n3. Stay with air.\r\n\r\nThus the probability is $ \\frac13$", "Solution_23": "I'm sorry, but I still seriously don't get it.\r\n\r\nAren't the children independent?\r\n\r\nI mean, even for that coin flipping problem you brought up - \"If you flip a coin 9 times and they are all heads, find the probability the next one is head\", it could just mean that you're really lucky. Each flip is independent and shouldn't affect the other flips. It's not like the coin one day just randomly decided to become a heads, or something. :P", "Solution_24": "If we are given that the [b]first[/b] child is a female, then you would be right in saying that the probability the other is a male is $ \\frac{1}{2}$\r\n\r\nHowever, in this problem, we are just given that [b]one[/b] of the children is a female, so we have 3 equally likely cases, not two\r\nMF\r\nFM\r\nFF\r\nso $ \\frac{2}{3}$", "Solution_25": "[hide=\"How 2 solve this...\"]\nThe births of Mrs. Witts two children are two successive Bernoulli trials.\n\nLet f = the probability that a birth is female, and likewise let m = the probability that a birth is male. Then, notwithstanding that in this case male and female births are equally likely, $ f \\plus{} m \\equal{} 1$.\n\nFor two children, i.e. two Bernoulli trials, the probabilities are distributed like this: $ (f \\plus{} m)^2 \\equal{} 1$.\n\nNow, instead of invoking Pascal's Triangle, I'm going to expand the expression on the left hand side using FOIL because doing so allows us to see each of the terms clearly.\n\n\\[ (f \\plus{} m)^2 \\equal{} (f \\plus{} m)(f \\plus{} m) \\equal{} f^2 \\plus{} fm \\plus{} mf \\plus{} m^2 \\equal{} 1 \\]\n\nThe probability that we are being asked to find is the conditional probability $ P(C_2 \\equal{} \\text{male}|C_1 \\equal{}\\text{female})$.\n\n(Any time what you're being asked for is subject to initial conditions like, \"What's the probability that the second is a boy [b]given[/b] that the first is a girl?\" think [u]conditional probability[/u]. That is you have some bit of prior knowledge; knowledge affects your assessment of how likely something is, i.e. its probability.)\n\n\\[ P(C_2 \\equal{} \\text{male}|C_1 \\equal{}\\text{female})\\stackrel{\\triangle} \\equal{} \\frac{P(C_1 \\equal{}\\text{female},C_2 \\equal{}\\text{male})}{P(C_1 \\equal{}\\text{female})}\\]\n\nAccording to Bayes Theorem, $ P(C_2 \\equal{}\\text{male}|C_1\\equal{}\\text{female}) \\equal{}\\frac{P(C_2\\equal{}\\text{male})\\cdot P(C_1 \\equal{}\\text{female}|C_2 \\equal{}\\text{male})}{P(C_1 \\equal{}\\text{female})}$\n\n$ P(C_2\\equal{}\\text{male}) \\equal{} \\frac{1}{3}$, of the three possible outcomes involving females only one has a second born male.\n\n$ P(C_1\\equal{}\\text{female}|C_2 \\equal{}\\text{male}) \\equal{} 1$, you were given that the first born was female, so the first born is female with probability 1; the condition that the second is male is superfluous.\n\nFinally, the [i]a priori[/i] probability of a first born female is one-half, $ P(C_1\\equal{}\\text{female}) \\equal{}\\frac{1}{2}$.\n\nPutting these all together with Bayes Theorem, $ P(C_2\\equal{}\\text{male}|C_1\\equal{}\\text{female}) \\equal{}\\frac{\\frac{1}{3}\\cdot (1)}{\\frac{1}{2}} \\equal{} \\boxed{\\frac{2}{3}}$.[/hide]", "Solution_26": "Ahh... that makes more sense :D", "Solution_27": "MF\r\nFM\r\nFF\r\n\r\nBut the older child CAN NOT be male.\r\n\r\nedit: nvm, Was following t0ra's question." } { "Tag": [], "Problem": "If n=72, then what does 13(n-8)+n/9 equal?\r\n :play_ball: :starwars:", "Solution_1": "We have\r\n\\begin{align*}\r\n13(72-8) + \\frac{72}{9} &= 13(64) + 8 \\\\\r\n&= \\boxed{840}.\r\n\\end{align*}\r\nShould this be in the classroom math forum as it's only computation/order of operations? :maybe:", "Solution_2": "$ 13(72 \\minus{} 8) \\plus{} \\frac {72}9 \\equal{} 13\\cdot64 \\plus{} 8 \\equal{} 8(13\\cdot8 \\plus{} 1) \\equal{} 8(104 \\plus{} 1)$\r\n\r\n$ \\equal{} 8(100 \\plus{} 5) \\equal{} 800 \\plus{} 40 \\equal{} \\boxed{840}$. :P" } { "Tag": [ "floor function", "irrational number", "number theory", "number theory proposed" ], "Problem": "Let $ a,b$ be real nonzero numbers, such that number $ \\lfloor an \\plus{} b \\rfloor$ is an even integer for every $ n \\in \\mathbb{N}$. Prove that $ a$ is an even integer.", "Solution_1": "[quote=\"Ahiles\"][color=brown]Let $ a,b$ be real nonzero numbers, such that number $ [an + b]$ is an even integer for every $ n \\in \\mathbb{N}$. Prove that $ a$ is an even integer. [/color][/quote]\r\nA very good problem . \r\nTo solve this problem ,we need a lemma . \r\n[color=blue]Lemma [/color]\r\nIf $ a$ is a irrational number then $ \\{na\\}$ is dense in $ (0,1)$\r\nProof of this lemma quite hard ,it can be found in a number theory book . \r\nNow consider problem . \r\n$ x_n = an + b$\r\nSuppose $ [a_n]\\equiv 0 (\\mod 2 ),\\forall n\\in N$ then $ [x_{n + 1}]\\geq [x_n] + 2$\r\nWe only need to consider $ a\\in [0,2)$\r\nFirst step we will prove that a is a rational number. \r\nSuppose a is a irrational number then \r\n$ x_{n + 1} - x_n = a$\r\nExist an integer n_0 such that $ \\{an_0 + b\\} < 2 - a$ \r\nSo $ [x_{n_0 + 1}] = [an_0 + b + a] = [x_{n_0}] + [\\{an_0 + b\\} + a] < [x_{n_0}] + 2$ \r\nIt gives contradiction . \r\nSo a must be a rational number. \r\nWe also only to consider $ b\\in [0,2)$\r\nLet $ a = \\frac {p}{q},\\gcd (p,q) = 1$ and $ q > 1$\r\n\r\n${ [x_n] = [\\frac {np}{q}] + [\\frac {np}{q}} + b]$\r\n\r\nTake $ n = kq$ then easy to check that $ p\\equiv 0 (\\mod 2)$ \r\n[u]Case 1[/u] $ b\\in [0,1)$\r\nExist an integer n such that $ np\\equiv 1 (\\mod q )$\r\nBecause $ p\\equiv 1 (\\mod 2)$ so $ [\\frac {np}{q}]\\equiv 1 (\\mod 2)$\r\n$ \\Rightarrow [\\frac {1}{q} + b] = 1,3$\r\nBecause $ 0 < \\frac {1}{p} + b < 2$ so $ [\\frac {1}{p} + b] = 1$ \r\nSo $ 1 > b > 1 - \\frac {1}{p}$\r\nIt follows that $ [b + \\frac {np}{q}] = 1$\r\n$ \\Rightarrow [\\frac {np}{q}]\\equiv 0 (\\mod 2 ),\\forall n\\in N$\r\nIt gives $ q = 1$\r\nCase 2 :$ 1 < b < 2$ \r\nConsider $ b_1 = b - 1$ and \r\n$ y_n = x_n - 1$ then $ [y_n]\\equiv 1 (\\mod 2)$\r\nProve as case 1 ,it gives $ q = 1$\r\nSo $ a = p = 2k$ and $ [b]\\equiv 0(\\mod 2)$\r\nProblem claim .", "Solution_2": "[quote=\"TTsphn\"][color=blue]Lemma [/color]\nIf $ a$ is a irrational number then $ \\{na\\}$ is dense in $ (0,1)$\nProof of this lemma quite hard [...][/quote]\r\nThis is not hard! See http://www.mathlinks.ro/Forum/viewtopic.php?t=27475 (scroll down for grobbers proof, post #16).\r\n\r\nThe hard one is http://www.mathlinks.ro/viewtopic.php?t=172978 .", "Solution_3": "The lemma $ \\{na\\}$ is dense in $ (0,1)$ with $ a$ is a irrational is true.", "Solution_4": "But not that it is _hard_.", "Solution_5": "I tihnk I have reached a solution.Hope it is OK. Let a=2k+r (k integer and with r non negative and r smaller than 2). Then if we substitue in the original question n=1 we have that [2k+r+b]=2m (m is an integer) [r+b]=2(m-k) =2t ....by doing the same thing we obtain that [nr+b]=2P P is an integer. Suppose that b=2h+q q q can be in [0,1) or in [1,2) if it is in [0,1) we say that it \"adds 0 to the equation\" . Then we have [nr]=2P' then if n=1 we have [r]=0 so there will be an n such\r\nthat [nr]=1 if r>0 so we have that r=0 so a=2k as we wish to prove. Then if q is in [1,2) we say that \"it adds 1 to the equation\" so [nr+1]=2P then repating the procedure we have [r]=1 then we can se that there will be an n such that [nr+1]=2L+1 ( you can start doing things like these: the numbersin [1,1.5) whem we set n=2 will contradict the condition of the problems, if you continue doing that it is easily checked that they can't be an answer to the problem)\r\nSo we conclude that a=2k \r\nExcellent problem!\r\nI hope you understand the idea and I wish peple can improve the parts where are not completly proven\r\n :) Daniel[/code]", "Solution_6": "[quote=\"TTsphn\"][quote=\"Ahiles\"][color=brown]\n\n${ [x_n] = [\\frac {np}{q}] + [\\frac {np}{q}} + b]$\n\nTake $ n = kq$ then easy to check that $ p\\equiv 0 (\\mod 2)$ \n[u]Case 1[/u] $ b\\in [0,1)$\nExist an integer n such that $ np\\equiv 1 (\\mod q )$\nBecause $ p\\equiv 1 (\\mod 2)$ so $ [\\frac {np}{q}]\\equiv 1 (\\mod 2)$\n$ \\Rightarrow [\\frac {1}{q} + b] = 1,3$\nBecause $ 0 < \\frac {1}{p} + b < 2$ so $ [\\frac {1}{p} + b] = 1$ \nSo $ 1 > b > 1 - \\frac {1}{p}$\nIt follows that $ [b + \\frac {np}{q}] = 1$\n$ \\Rightarrow [\\frac {np}{q}]\\equiv 0 (\\mod 2 ),\\forall n\\in N$\nIt gives $ q = 1$\n[/quote][/quote]\r\n\r\nwhy ? show me please" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Given $ a, b, c \\geq\\ 0$. \r\nProve that: $ \\frac{(a\\plus{}b\\plus{}c)^2}{2(ab\\plus{}bc\\plus{}ca)} \\geq\\ \\frac{a^2}{a^2\\plus{}bc} \\plus{} \\frac{b^2}{b^2\\plus{}ca} \\plus{} \\frac{c^2}{c^2\\plus{}ab}$", "Solution_1": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$. \nProve that: $ \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)} \\geq\\ \\frac {a^2}{a^2 \\plus{} bc} \\plus{} \\frac {b^2}{b^2 \\plus{} ca} \\plus{} \\frac {c^2}{c^2 \\plus{} ab}$[/quote]\r\nIt's equivalent to\r\n$ \\sum_{sym}\\left(a^5b^3 \\minus{} a^4b^4 \\plus{} \\frac {1}{2}a^6bc \\minus{} a^4b^3c \\plus{} a^4b^2c^2 \\minus{} \\frac {1}{2}a^3b^3c^2\\right)\\geq0,$ which is obvious.\r\nSOS ( $ a^2S_b \\plus{} b^2S_a\\geq0$ ) helps too but it's very ugly for this nice inequality.", "Solution_2": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$. \nProve that: $ \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)} \\geq\\ \\frac {a^2}{a^2 \\plus{} bc} \\plus{} \\frac {b^2}{b^2 \\plus{} ca} \\plus{} \\frac {c^2}{c^2 \\plus{} ab}$[/quote]\r\nwe used a lemma of VQBC\r\nsetting $ p\\equal{}a\\plus{}b\\plus{}c\\equal{}1,q\\equal{}ab\\plus{}bc\\plus{}ca\\equal{}\\frac{1\\minus{}t^2}{3}(t\\in [0,1]),r\\equal{}abc$\r\n$ \\Leftrightarrow \\minus{}\\frac{1}{81}\\plus{}\\frac{7}{81}t^2\\minus{}\\frac{5}{27}t^4\\plus{}\\frac{13}{81}t^6\\plus{}\\frac{1}{3}r\\minus{}\\frac{4}{3}rt^2\\plus{}8r^2t^2\\minus{}\\frac{4}{81}t^8\\plus{}2rt^4\\ge 0$\r\nby lemma we have $ \\frac{(1\\plus{}t)^2(1\\minus{}2t)}{27}\\le r\\le \\frac{(1\\minus{}t)^2(1\\plus{}2t)}{27}$\r\n*Case 1: if $ f'(r)>0$\r\n$ \\Rightarrow f(r)\\ge f(\\frac{(1\\plus{}t)^2(1\\minus{}2t)}{27})\\equal{}\\frac{1}{729}t^2(t\\plus{}4)(2\\minus{}t)(t\\plus{}1)^2(2t\\minus{}1)^2\\ge 0$\r\n*Case 2: if $ f'(r)<0$\r\n$ \\Rightarrow f(r)\\ge f(\\frac{(1\\minus{}t)^2(1\\plus{}2t)}{27})\\equal{}\\frac{1}{729}t^2(t\\plus{}2)(4\\minus{}t)(2t\\plus{}1)^2(t\\minus{}1)^2\\ge 0$\r\nwe have done :P", "Solution_3": "If $ a,b,c$ be side lengths of a triangle , it has an simple proof :\r\n\r\n<=>$ abc(\\sum \\frac { b \\plus{} c \\minus{} a}{a^2 \\plus{} bc} )\\ge \\frac {2\\sum ab \\minus{} \\sum c^2}{2}$ ( Note: $ LHS \\ge 0 \\,forall a,b,c \\ge 0$)\r\nEasy to prove :$ \\frac { b \\plus{} c \\minus{} a}{a^2 \\plus{} bc} \\ge \\frac { a \\plus{} c \\minus{} b}{b^2 \\plus{} ac} \\,\\, if \\,b \\le a$ \r\nBy chebysev, we have :\r\n$ abc(\\sum \\frac { b \\plus{} c \\minus{} a}{a^2 \\plus{} bc} ) \\ge \\frac {3abc}{a \\plus{} b \\plus{} c}[(ab \\plus{} bc \\plus{} ca)( \\sum \\frac {1}{a^2 \\plus{} bc} ) \\minus{} 3 ] \\ge \\frac {9abc(2ab \\minus{} \\sum a^2)}{(a \\plus{} b \\plus{} c)( \\sum a^2 \\plus{} \\sum ab )}\\ge RHS$\r\nThe last ineq is equivalent to Schur with degree 2\r\n\r\nAbout prove that ineq with $ a,b,c \\ge 0$\r\n:P . If someone'd like to do it by pqr , this ineq is equivalent to :\r\n$ 2rq(1 \\minus{} 3q \\plus{} 4r) \\ge (4q \\minus{} 1)(8r^2 \\plus{} r \\plus{} q^3 \\minus{} 6rq)$\r\n.... It has many ways to prove this ineq :P", "Solution_4": "[quote=\"bunhiacovski\"]\n*Case 1: if $ f'(r) > 0$\n$ \\Rightarrow f(r)\\ge f(\\frac {(1 \\plus{} t)^2(1 \\minus{} 2t)}{27}) \\equal{} \\frac {1}{729}t^2(t \\plus{} 4)(2 \\minus{} t)(t \\plus{} 1)^2(2t \\minus{} 1)^2\\ge 0$\n*Case 2: if $ f'(r) < 0$\n$ \\Rightarrow f(r)\\ge f(\\frac {(1 \\minus{} t)^2(1 \\plus{} 2t)}{27}) \\equal{} \\frac {1}{729}t^2(t \\plus{} 2)(4 \\minus{} t)(2t \\plus{} 1)^2(t \\minus{} 1)^2\\ge 0$\nwe have done :P[/quote]\r\nI rather suspect your solution. I think it's not true. Because:\r\nCh\u1eb3ng h\u1ea1n v\u1edbi tr\u01b0\u1eddng h\u1ee3p \u0111\u1ea7u ti\u00ean, c\u00f3 th\u1ec3 c\u1eadn d\u01b0\u1edbi n\u00e0y kh\u00f4ng thu\u1ed9c kho\u1ea3ng \u0111\u1ed3ng bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 f(r).\r\nSory I don't know to say by English. Can some VietNamese translate for me to English ? Thank you :D", "Solution_5": "I can give a similar example:\r\nIf $ a \\geq\\ b\\geq\\ c$, we see: $ f(b)\\equal{}(b\\minus{}a)(b\\minus{}c) \\leq\\ 0$\r\nBut if we consider: $ f'(b)\\equal{}2b\\minus{}c\\minus{}a$\r\nIf $ f'(b) \\geq\\ 0$ then $ f(b) \\geq\\ f(c) \\equal{} 0$\r\nIf $ f'(b) \\leq\\ 0$ then $ f(b) \\geq\\ f(a)\\equal{}0$ \r\nTherefore, we can say: $ (b\\minus{}a)(b\\minus{}c) \\geq\\ 0$ $ ????$\r\n--------------------\r\nI think your proof was wrong same above :)", "Solution_6": "[quote=\"nguoivn\"][quote=\"bunhiacovski\"]\n*Case 1: if $ f'(r) > 0$\n$ \\Rightarrow f(r)\\ge f(\\frac {(1 \\plus{} t)^2(1 \\minus{} 2t)}{27}) \\equal{} \\frac {1}{729}t^2(t \\plus{} 4)(2 \\minus{} t)(t \\plus{} 1)^2(2t \\minus{} 1)^2\\ge 0$\n*Case 2: if $ f'(r) < 0$\n$ \\Rightarrow f(r)\\ge f(\\frac {(1 \\minus{} t)^2(1 \\plus{} 2t)}{27}) \\equal{} \\frac {1}{729}t^2(t \\plus{} 2)(4 \\minus{} t)(2t \\plus{} 1)^2(t \\minus{} 1)^2\\ge 0$\nwe have done :P[/quote]\nI rather suspect your solution. I think it's not true. Because:\nCh\u1eb3ng h\u1ea1n v\u1edbi tr\u01b0\u1eddng h\u1ee3p \u0111\u1ea7u ti\u00ean, c\u00f3 th\u1ec3 c\u1eadn d\u01b0\u1edbi n\u00e0y kh\u00f4ng thu\u1ed9c kho\u1ea3ng \u0111\u1ed3ng bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 f(r).\nSory I don't know to say by English. Can some VietNamese translate for me to English ? Thank you :D[/quote]\r\nI think my proof is true because $ t\\in[0,1]$, I have another proof but I can not post because it's ugly\r\n[hide]@(-.^)@: em ch\u1eafc ch\u1eafn l\u1eddi gi\u1ea3i n\u00e0y ko sai anh \u00e0h:D[/hide]", "Solution_7": "[quote=\"nguoivn\"]I can give a similar example:\nIf $ a \\geq\\ b\\geq\\ c$, we see: $ f(b) \\equal{} (b \\minus{} a)(b \\minus{} c) \\leq\\ 0$\nBut if we consider: $ f'(b) \\equal{} 2b \\minus{} c \\minus{} a$\nIf $ f'(b) \\geq\\ 0$ then $ f(b) \\geq\\ f(c) \\equal{} 0$\nIf $ f'(b) \\leq\\ 0$ then $ f(b) \\geq\\ f(a) \\equal{} 0$ \nTherefore, we can say: $ (b \\minus{} a)(b \\minus{} c) \\geq\\ 0$ $ ????$\n--------------------\nI think Bunhiacopxki's proof was wrong same above :)[/quote]\r\n\r\nHow will you say, arquady ? :wink:", "Solution_8": "I agree with you, nguoivn.", "Solution_9": "My proof:\r\nSuppose $ a \\plus{} b \\plus{} c \\equal{} 3$. We need to prove:\r\n$ f(r) \\equal{} 4q^4 \\minus{} 9q^3 \\plus{} 24qr^2 \\minus{} 54q^2r \\minus{} 72r^2 \\minus{} 243r \\plus{} 216qr \\leq\\ 0$\r\n$ f'(r) \\equal{} 48qr \\minus{} 54q^2 \\minus{} 144r \\minus{} 243 \\plus{} 216q$\r\n$ f''(r) \\equal{} 48(q \\minus{} 3) \\leq\\ 0, so f'(r) \\leq\\ f'(0) \\equal{} \\minus{} 54q^2 \\minus{} 144 \\plus{} 216q \\leq\\ 0$\r\nSo, with $ q \\leq\\ \\frac {9}{4}, f(r) \\leq\\ f(0) \\equal{} q^3(4q \\minus{} 9) \\leq\\ 0$\r\nWith $ q \\geq\\ \\frac{9}{4}$, we have: $ f(r) \\leq\\ f(\\frac {4q \\minus{} 9}{3}) \\leq\\ 0$ (trues with $ q \\geq\\ \\frac{9}{4}$)", "Solution_10": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$. \nProve that: $ \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)} \\geq\\ \\frac {a^2}{a^2 \\plus{} bc} \\plus{} \\frac {b^2}{b^2 \\plus{} ca} \\plus{} \\frac {c^2}{c^2 \\plus{} ab}$[/quote]\r\nWe have\r\n\\[ \\sum \\frac{2a^2}{(a\\plus{}b)(a\\plus{}c)} \\minus{}\\sum \\frac{a^2}{a^2\\plus{}bc} \\equal{}\\sum \\frac{a^2(a\\minus{}b)(a\\minus{}c)}{(a\\plus{}b)(a\\plus{}c)(a^2\\plus{}bc)} \\ge 0\\]\r\n(easy to check by Vornicu Schur)\r\nIt suffices to prove that\r\n\\[ \\frac{(a\\plus{}b\\plus{}c)^2}{2(ab\\plus{}bc\\plus{}ca)} \\ge \\sum \\frac{2a^2}{(a\\plus{}b)(a\\plus{}c)} \\equal{}\\frac{2\\sum ab(a\\plus{}b)}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)}\\]\r\nAssume that $ a\\plus{}b\\plus{}c\\equal{}1$ and put $ q\\equal{}ab\\plus{}bc\\plus{}ca,r\\equal{}abc$, then the inequality becomes\r\n\\[ \\frac{1}{4q} \\ge \\frac{q\\minus{}3r}{q\\minus{}r}\\]\r\n\\[ \\Leftrightarrow \\frac{q\\minus{}r}{q\\minus{}3r} \\ge 4q\\]\r\n\\[ \\Leftrightarrow \\frac{2r}{q\\minus{}3r} \\ge 4q\\minus{}1\\]\r\nBy Schur's Inequality for third degree, we have $ r \\ge \\frac{4q\\minus{}1}{9}$, then\r\n\\[ \\frac{2r}{q\\minus{}3r} \\ge \\frac{2r}{q\\minus{} \\frac{4q\\minus{}1}{3}} \\equal{}\\frac{6r}{1\\minus{}q}\\]\r\nIt suffices to show that\r\n\\[ 6r \\ge (4q\\minus{}1)(1\\minus{}q)\\]\r\nBut this is just Schur's Inequality for fourth degree\r\n\\[ \\sum a^4\\plus{}abc\\sum a \\ge \\sum ab(a^2\\plus{}b^2)\\]\r\nWe have done. :)", "Solution_11": "Very nice proof, can_hang2007 :)", "Solution_12": "Yes, I very like can_hang's proof :lol:", "Solution_13": "[quote=\"mitdac123\"]If $ a,b,c$ be side lengths of a triangle , it has an simple proof :\n\n<=>$ abc(\\sum \\frac { b \\plus{} c \\minus{} a}{a^2 \\plus{} bc} )\\ge \\frac {2\\sum ab \\minus{} \\sum c^2}{2}$ ( Note: $ LHS \\ge 0 \\,forall a,b,c \\ge 0$)\nEasy to prove :$ \\frac { b \\plus{} c \\minus{} a}{a^2 \\plus{} bc} \\ge \\frac { a \\plus{} c \\minus{} b}{b^2 \\plus{} ac} \\,\\, if \\,b \\le a$ \nBy chebysev, we have :\n$ abc(\\sum \\frac { b \\plus{} c \\minus{} a}{a^2 \\plus{} bc} ) \\ge \\frac {3abc}{a \\plus{} b \\plus{} c}[(ab \\plus{} bc \\plus{} ca)( \\sum \\frac {1}{a^2 \\plus{} bc} ) \\minus{} 3 ] \\ge \\frac {9abc(2ab \\minus{} \\sum a^2)}{(a \\plus{} b \\plus{} c)( \\sum a^2 \\plus{} \\sum ab )}\\ge RHS$\nThe last ineq is equivalent to Schur with degree 2\n\n[/quote]\r\n\r\n\r\nI checked your proof and I found it is wrong. If you apply shur at your last expression in the form\r\n\r\n$ \\frac{9abc}{\\sum a} \\geq 2\\sum ab \\minus{} \\sum a^2$\r\n\r\nyou obtain\r\n\r\n$ \\sum a^2 \\leq \\sum ab$ surely false. :wink:", "Solution_14": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$. \nProve that: $ \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)} \\geq\\ \\frac {a^2}{a^2 \\plus{} bc} \\plus{} \\frac {b^2}{b^2 \\plus{} ca} \\plus{} \\frac {c^2}{c^2 \\plus{} ab}$[/quote]\r\nThe following inequalities are true too.\r\nLet $ a,$ $ b$ and $ c$ are positive numbers. Prove that:\r\n1.\r\n\\[ \\frac {a \\plus{} b \\plus{} c}{2\\sqrt [3]{abc}} \\geq\\ \\frac {a^2}{a^2 \\plus{} bc} \\plus{} \\frac {b^2}{b^2 \\plus{} ca} \\plus{} \\frac {c^2}{c^2 \\plus{} ab}\r\n\\]\r\n2.\r\n\\[ \\frac {5}{2} \\minus{} \\frac {ab \\plus{} ac \\plus{} bc}{a^2 \\plus{} b^2 \\plus{} c^2} \\geq\\ \\frac {a^2}{a^2 \\plus{} bc} \\plus{} \\frac {b^2}{b^2 \\plus{} ca} \\plus{} \\frac {c^2}{c^2 \\plus{} ab}\r\n\\]\r\nSee also here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=246653", "Solution_15": "to 2]\n\n$\\leftrightarrow$\n\n$\\sum \\frac{2ab}{c^2+ab}\\ge \\frac{(a+b+c)^2}{a^2+b^2+c^2}$\n\n$LHS \\ge \\frac{2(\\sum ab)^2}{(\\sum ab)^2-abc(a+b+c)}\\ge \\frac{(a+b+c)^2}{a^2+b^2+c^2}$\n\n$\\leftrightarrow$ \n$abc(\\frac{a^2+b^2+c^2}{ab+bc+ca}+2)^2\\ge (a+b+c)(2(ab+bc+ca)-a^2-b^2-c^2)$\n\nbut this follows immediately from schur's inequality in 3-degree \n$9abc\\ge (a+b+c)(2(ab+bc+ca)-a^2-b^2-c^2)$", "Solution_16": "[quote=\"arqady\"]\nLet $ a,$ $ b$ and $ c$ are positive numbers. Prove that:\n1.\n\\[ \\frac {a \\plus{} b \\plus{} c}{2\\sqrt [3]{abc}} \\geq\\ \\frac {a^2}{a^2 \\plus{} bc} \\plus{} \\frac {b^2}{b^2 \\plus{} ca} \\plus{} \\frac {c^2}{c^2 \\plus{} ab}\n\\]\n[/quote]\nHere is my proof from 2008.\nLet $a=\\min\\{a,b,c\\}$, $f(a,b,c)=\\frac {a \\plus{} b \\plus{} c}{2\\sqrt [3]{abc}}-\\sum_{cyc}\\frac {a^2}{a^2 \\plus{} bc}$ and $abc=1$.\nHence, $f(a,b,c)-f\\left(a,\\sqrt{bc},\\sqrt{bc}\\right)=$\n$=\\left(\\sqrt b-\\sqrt c\\right)^2\\left(\\frac{1}{2}+\\frac{a\\left(\\sqrt{bc}-a\\right)(b+\\sqrt{bc}+c)^2}{(a+\\sqrt{bc})(b^2+ac)(c^2+ab)}\\right)\\geq0$.\nThus, it remains to check our inequality for $c=b$ and $a=\\frac{1}{b^2}$, which gives\n$(b-1)^2(2b^{10}-2b^8-b^7+b^5+5b^4+3b^3+b^2+2b+1)\\geq0$, which is obvious.\nSee also here:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=335838", "Solution_17": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$. \nProve that: $ \\frac{(a\\plus{}b\\plus{}c)^2}{2(ab\\plus{}bc\\plus{}ca)} \\geq\\ \\frac{a^2}{a^2\\plus{}bc} \\plus{} \\frac{b^2}{b^2\\plus{}ca} \\plus{} \\frac{c^2}{c^2\\plus{}ab}$[/quote]\n\n\\[\\frac{(x+y+z)^2}{xy+yz+xz}\\frac{x^2+yz}{x^2} \\geq 18\\frac{\\frac{y+z}{x}}{\\frac{x+y}{z}+\\frac{y+z}{x}+\\frac{z+x}{y}}\\]\n\n\\[18\\frac{(z+x)^3+(y+x)^3}{(y+z)^3+(z+x)^3+(y+x)^3}+\\frac{(x+y+z)^2}{xy+yz+xz}\\frac{x^2+yz}{x^2}\\geq 18\\]", "Solution_18": "[quote=nguoivn]Given $ a, b, c \\geq\\ 0$. \nProve that: $ \\frac{(a\\plus{}b\\plus{}c)^2}{2(ab\\plus{}bc\\plus{}ca)} \\geq\\ \\frac{a^2}{a^2\\plus{}bc} \\plus{} \\frac{b^2}{b^2\\plus{}ca} \\plus{} \\frac{c^2}{c^2\\plus{}ab}$[/quote]\n\n[img]http://s16.sinaimg.cn/middle/006ptkjAzy76wc1XrUHdf&690[/img]", "Solution_19": "[quote=nguoivn]Given $ a, b, c \\geq\\ 0$. \nProve that: $$ \\frac{(a\\plus{}b\\plus{}c)^2}{2(ab\\plus{}bc\\plus{}ca)} \\geq\\ \\frac{a^2}{a^2\\plus{}bc} \\plus{} \\frac{b^2}{b^2\\plus{}ca} \\plus{} \\frac{c^2}{c^2\\plus{}ab}$$[/quote]\nLet $a,b,c$ are positive reals . Prove that$$\\frac{a^2}{a^2+bc}+\\frac{b^2}{b^2+ca}+\\frac{c^2}{c^2+ab}<2.$$\n", "Solution_20": "[quote=sqing]Let $a,b,c$ are positive reals . Prove that$$\\frac{a^2}{a^2+bc}+\\frac{b^2}{b^2+ca}+\\frac{c^2}{c^2+ab}<2.$$[/quote]\n\n[b]Solution[/b]. By the Cauchy-Schwarz's Inequality we have that, for any positive real numbers $a,b,c$,\n\\begin{align*}&\\frac{a^2}{a^2+bc}+\\frac{b^2}{b^2+ca}+\\frac{c^2}{c^2+ab}\\\\\n=&3-\\left[\\frac{bc}{a^2+bc}+\\frac{ca}{b^2+ca}+\\frac{ab}{c^2+ab}\\right]\\\\\n=&3-\\left[\\frac{(bc)^2}{a^2bc+(bc)^2}+\\frac{(ca)^2}{b^2ca+(ca)^2}+\\frac{(ab)^2}{c^2ab+(ab)^2}\\right]\\\\\n\\le&3-\\frac{(bc+ca+ab)^2}{a^2bc+b^2ca+c^2ab+(bc)^2+(ca)^2+(ab)^2}\\\\\n=&2-\\frac{a^2bc+b^2ca+c^2ab}{a^2bc+b^2ca+c^2ab+(bc)^2+(ca)^2+(ab)^2}\\\\\n<&2,\n\\end{align*}where the last inequality holds since $a^2bc+b^2ca+c^2ab>0$ for any $a,b,c>0$. $\\blacksquare$", "Solution_21": "[hide=Thanks.][quote=ytChen][quote=sqing]Let $a,b,c$ are positive reals . Prove that$$\\frac{a^2}{a^2+bc}+\\frac{b^2}{b^2+ca}+\\frac{c^2}{c^2+ab}<2.$$[/quote]\n\n[b]Solution[/b]. By the Cauchy-Schwarz's Inequality we have that, for any positive real numbers $a,b,c$,\n\\begin{align*}&\\frac{a^2}{a^2+bc}+\\frac{b^2}{b^2+ca}+\\frac{c^2}{c^2+ab}\\\\\n=&3-\\left[\\frac{bc}{a^2+bc}+\\frac{ca}{b^2+ca}+\\frac{ab}{c^2+ab}\\right]\\\\\n=&3-\\left[\\frac{(bc)^2}{a^2bc+(bc)^2}+\\frac{(ca)^2}{b^2ca+(ca)^2}+\\frac{(ab)^2}{c^2ab+(ab)^2}\\right]\\\\\n\\le&3-\\frac{(bc+ca+ab)^2}{a^2bc+b^2ca+c^2ab+(bc)^2+(ca)^2+(ab)^2}\\\\\n=&2-\\frac{a^2bc+b^2ca+c^2ab}{a^2bc+b^2ca+c^2ab+(bc)^2+(ca)^2+(ab)^2}\\\\\n<&2,\n\\end{align*}where the last inequality holds since $a^2bc+b^2ca+c^2ab>0$ for any $a,b,c>0$. $\\blacksquare$[/quote]\n\n[/hide]\nAnother proof\uff1a" } { "Tag": [ "topology", "real analysis", "real analysis theorems" ], "Problem": "These are some of my practice problems:\r\n1. Show a set $F_{A}\\subseteq A \\subseteq X$ is closed subset of A iff there exists a closed subset $F_{X}$ of $X$ such that $F_{A}= F_{X}\\cap A$. \r\n\r\n2. Supposed $A \\subseteq B \\subseteq X$. Show that if $A$ is open in $B$ which is open $X$, then $A$ is open in $X$. \r\n\r\nI wasn't sure if one could do 2 as follows:\r\nChoose $a \\in A$. There exists $\\frac{r}{2}> 0$ such that $d(a, b) < \\frac{r}{2}\\Rightarrow b \\in A$ and for each $b \\in B$ we have $d(b, x) < \\frac{r}{2}\\Rightarrow x \\in B$. \r\nWe have $d(a, x) \\leq d(a, b)+d(b, x) < \\frac{r}{2}+\\frac{r}{2}= r \\Rightarrow x \\in A$. So that $A$ is open in $X$.", "Solution_1": "These are utterly trivial if we use the general definitions of the relative topology: $U$ is open in $A\\subset X$ when there is some open $G\\subset X$ with $U=A\\cap G$.\r\n\r\n1. Let $G_{A}=A\\setminus F_{A}$; $G_{A}$ is open in $A$ iff there is some open $G_{X}\\subset X$ with $G_{A}=A\\cap G_{X}$. If such a set exists, let $F_{X}=X\\setminus G_{X}$. If a $F_{X}$ exists, let $G_{X}=X\\setminus G_{X}$.\r\n\r\n2. $A=G\\cap B$ for some open $B$. The intersection of two open sets is open." } { "Tag": [ "topology", "geometry", "geometric transformation", "invariant", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ G$ be a compact Hausdorff topological group. Prove that $ G$ is finite or uncountable.", "Solution_1": "use the normed haar measure $ \\mu$ on $ G$. if $ G$ was countable, say $ G \\equal{} \\{g_1,g_2,...\\}$, we have $ 1 \\equal{} \\sum_{i \\in \\mathbb{N}} \\mu(\\{g_i\\})$. by the fact that $ \\mu$ is translation-invariant, all these singletons have the same measure, a contradiction.", "Solution_2": "there's a more elementary way: every perfect set (every point is an accumulation point) is uncountable.\r\n\r\nand, in fact, every infinite compact topological group is perfect." } { "Tag": [ "ratio" ], "Problem": "For positive integers $n$, define $f(n)=1^n+2^{n-1}+3^{n-2}+\\cdots+(n-2)^3+(n-1)^2+n$.\r\n\r\nWhat is the minimum value of $\\frac{f(n+1)}{f(n)}$?", "Solution_1": "This isn't so bad:\r\n[hide]Computing the first few values by hand, we get 1, 3, 8, 22, 65, 209, 732, with ratios that start at 3, drop to 8/3, and then climb, eventually passing 3. Let's consider, for \"large enough\" $n$, the expression $f(n + 1) - 3f(n)$. If we subtract corresponding terms, this gives us $-2 - 2^{n - 1} + 4^{n - 3} + 2\\cdot 5^{n - 4} + \\ldots + n(n - 3) + (n +1)$. When $n > 3$, all terms but the first two are positive, and when $n > 5$, the third term $4^{n - 3} = 2^{2n - 6} > 2^{n - 1} + 2$ so the whole expression is positive and $\\frac{f(n + 1)}{f(n)} > 3$. Combined with our data for $n \\leq 6$, this shows the absolute minimum is 8/3, achieved when n = 2.[/hide]\r\n\r\nThe (I think) natural question this raises: show that the given ratio approaches infinity as n grows large, and (if possible) say about how quickly." } { "Tag": [ "\\/closed" ], "Problem": "Well I'm new here... as you can probably tell.\r\n\r\nIn most other forums I've visited, there has been an Introductions section so people can get to know each other. It would probably be a good idea to have one here as well. Then again, I'm just a noob, so who cares what I think?", "Solution_1": "Maybe...\r\nBut there are over 40000 users here so I don't know.", "Solution_2": "Well, there are such topics in the FF, but they eventually \"die-out\"\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=15747\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=20070" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "A chesstable coloured black-white has the dimensions $ m\\times n$,where $ m,n$ are naturals.On this table are made $ moves$ of this type: There are taken two fields $ 1\\times1$ which have a common side and modify their colours as it follows:\r\n[i]white[/i] field becomes [i]black[/i],the [i]black [/i]field becomes [i]red[/i] and the [i]red [/i]field becomes [i]white[/i]. \r\nFind all the possible values for $ m$ and $ n$ s.t. all the initial white fields become black and \r\nall the initial black fields become white.", "Solution_1": "How is it coloured initially? I will assume that the initial colouring is like the common chessboard.\r\n\r\nSo, we take three consecutive squares, WBW (white black white). First, take the first two, so we have BRW. Then take the last two, so we have BWB. We can do this for any three consecutive squares, and change all black squares into white and vice versa.\r\n\r\nTherefore, I think the initial condition should be $ 3|mn$, i.e. $ m$ or $ n$ is divisible by $ 3$.", "Solution_2": "Answer to your question:The initial colouring is like the common chessboard.\r\n\r\nYou proved that $ m \\equal{} 3k$ and any $ n$ natural and viceversa satisfies the condition.\r\n\r\nBut how you exclude other cases, for example $ m \\equal{} 3a \\plus{} 1, n \\equal{} 3b \\plus{} 1$? :?:", "Solution_3": "actually it can be proved very easily (as i did today by congruences)\r\nn(w)=number of initial white\r\nn(b)=number of initial black\r\nfirst we observe that number of transformations on initial white t(n(w))= number of tranformations on initial black t(n(b))\r\nnow also n(w)=t(n(w))mod 3\r\nand 2n(b)=t(n(b))mod 3\r\nsubtracting we get n(w) - 2n(b)=0 mod 3 ( because t(n(w))=t(n(b)))\r\n now we add 3n(b) to RHS\r\nwe get n(w) + n(b)=0mod3\r\nn(w) + n(b)=total number of fields = mn\r\ntherefore 3 divides mn \r\nhence done" } { "Tag": [ "articles", "LaTeX" ], "Problem": "My document starts with \\documentclass[11pt]{article} . I've tried changing that 11pt to another value: I want it to have a minimum size of 18pt; but 12pt is greater than 26pt type or anything larger than 12pt. (Also, when I compile, it says \"Unused global option: [26pt]\"). \\large, \\Large, \\LARGE, etc. don't work because they don't change the heading sizes. \r\n\r\nAlso how do I decrease the margins?", "Solution_1": "Hi,\r\nIf you want to change the font size of the chapter/section headings (I couldn't make out exactly what it was that you needed), an easy way to do it is to use the titlesec package. For example, for Large section titles, use:\r\n[code]\n\\titleformat{\\section}[hang]{\\center\\Large}{\\Large\\thesection}{12 pt}{}{}\n[/code]\r\nA simple way to increase margins is the fullpage package (I think it shrinks the margins to 1'' all around). If you want more control, try this: [url=http://www.image.ufl.edu/help/latex/margins.shtml]margins[/url].\r\n\r\nSSN.", "Solution_2": "I have discovered that if you use \\documentclass{memoir}, you can change the font size to 10,11,12,14, or 17pt. In case anyone else runs into this problem. \\usepackage{scalefnt}...\\scalefont{1.4} also works, but doesn't affect the headings, which is annoying.", "Solution_3": "The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=extsizes][b]extsizes package[/b][/url] give you a range of 8-20pt for the base size, works well and is easy to use." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "if \\[ac=b^{2}\\] , \\[a+b+c=3\\] , find the min and max of the \\[b\\]", "Solution_1": "If $b>0$, then:\r\n$3=a+b+c \\geq 2\\sqrt{ac}+b = 2b+b=3b$. Therefore $b \\leq 1$ so $b_{\\text{max}}=1$.\r\n\r\nIf $b<0$, then:\r\n$3=a+b+c \\geq 2\\sqrt{ac}+b =-2b+b=-b$. Therefore $b \\geq-3$ so $b_{\\text{min}}=-3$.", "Solution_2": "that is say the answer of my book is wrong\r\nthank u :wink:" } { "Tag": [], "Problem": "1. 2-bromoethanol and 1,2-dibromomethane\r\n2. 4-chloro-1-butene and n-butyl chloride\r\n3. 1-hexene and 2-hexanol\r\n4. 1-chloro-2-methyl-2-propanol and 1,2-dichloro-2-methylpropane\r\n\r\n\r\n [hide=\"i think\"]\n1.chromic anhydride in sulfuric acid\n2.bromine carbon-tetrachloride\n3.bromine carbon tetrachloride/chromic anhydride in sulfuric acid\n4.chromic anhydride in sulfuric acid[/hide]", "Solution_1": "Solubility tests could also be used. In 1., 2-bromoethanol is soluble in water, while 1,2-dibromoethane isn't. In 2., 4-chloro-1-butene is soluble in sulphuric acid, while n-butyl chloride is not. For 3., the alcohol should be more soluble that the alkene. And in 4., the alcohol is more soluble than the alkyl halide.\r\n\r\nInfrared spectroscopy could also be easily used to distinguish each pair of compounds:\r\n\r\n1. 2-bromoethanol would show the broad band from the O-H centered at about 3200-3400 cm-1, while 1,2-dibromoethane wouldn't have any band in that region.\r\n\r\n2. 4-chloro-1-butene has a vinyl group and so we could find the following bands: alkene C-H stretch at about 3070 cm-1 (medium-strong); overtone of the ~920 cm-1 band at about 1840 cm-1 (weak); two strong bands for alkene C-H bending vibrations, one at about 990 cm-1, and another at 920 cm-1. n-Butyl chloride would have naturally these bands.\r\n\r\n3. 1-Hexene would show the vinyl bands I referred above. 2-hexanol, would show the broad O-H band at ~3300 cm-1, and also the C-O stretching band (strong) at about 1100 cm-1 (because it is a secondary alcohol).\r\n\r\n4. 1-chloro-2-methyl-2-propanol would also show the broad O-H band at ~3300 cm-1, but now we would find the C-O band at about 1150 cm-1, because we have here a tertiary alcohol." } { "Tag": [ "LaTeX", "parameterization" ], "Problem": "I want to create my own theorem-like environment, since I don't like the numbering and fonts used in the default one.\r\n\r\nThe default LaTeX \"newtheorem\" creates an environment that puts some space before and after each theorem to make the theorem stand out.\r\n\r\nSo I did the following to mimic the environment\r\n[code] \\newenvironment{prop}[1]{\\bigskip \\noindent {\\bf Proposition #1.}}{\\bigskip} [/code]\nThis is perfect when I have\n[code]\nParagraph\n\n\\begin{prop}{1}\nStatement\n\\end{prop}\n\nParagraph\n[/code]\nHowever, when I do the following\n[code]\nIntroducing two propositions:\n\n\\begin{prop}{1}\nStatement1\n\\end{prop}\n\n\\begin{prop}{2}\nStatement2\n\\end{prop}\n[/code]\r\nI get twice the space between the two propositions.\r\n\r\nHow do I make the environments so that it onlys puts a certain amount of space before and after, such that writing one such environment after the other does not double the space?\r\n\r\nOr is there a better way to do this?", "Solution_1": "Well, somehow figuring out how to change the settings on the original would be ideal, but neither of us knows how to do that right now. The easiest thing i can think of would be to make a second environment that is identical to the first but doesn't have the initial bigskip:\r\n[code]\\newenvironment{propfollow}[1]{\\noindent {\\bf Proposition #1.}}{\\bigskip}[/code]\r\nHope this helps/works!", "Solution_2": "Or, you could give the environment more optional parameters, so you can explicitly provide the before and after spacing each time you use it. However, since you'll be using it with a bigskip before and after almost every time, this might get cumbersome. The second environment for exceptional cases is probably easier.", "Solution_3": "Oh, btw, you could use counters to automatically number the theorems instead of doing that manually. You would do (something like) this:\r\n\r\n\\newcounter{probnum} (or whatever else you want to call it)\r\n\\setcounter{probnum}{1}\r\n\r\n\\newenvironment{prop}{\r\n\\bigskip \\noindent {\\bf Proposition \\theprobnum.}\r\n\\stepcounter{probnum}\r\n}{\\bigskip}\r\n\r\nThese don't like to go in the preamble (I don't know why). Put them first thing after \\begin{document}.\r\n\r\nOf course, this simply counts 1, 2, 3, ..., but more complicated stuff can be done if necessary. (And, if you use the propfollow idea from before, the same counter can be used, and it will number in order as necessary.)", "Solution_4": "I think I found a possible solution.\r\n\r\nThe command \\addvspace{12pt} adds the space, but does not add more space if there is already that much space in place." } { "Tag": [ "logarithms", "function", "inequalities", "number theory unsolved", "number theory" ], "Problem": "Prove that $ \\sigma(n) 1$? Or did you mean $ \\sigma(n) \\leq n\\sqrt{n}$? What you're saying isn't true when $ n\\equal{}1$, as $ \\sigma(1) \\equal{} 1 \\equal{} 1\\sqrt{1}$.", "Solution_2": "False when $ n \\equal{} 2$.\r\n\r\nWhen $ n > 2$, we can show that $ n$ has at most $ 2 \\sqrt {n} \\minus{} 1$ divisors and of those divisors all except $ 1$ are greater than or equal to $ 2$. Now, $ \\frac {\\sigma(n)}{n} \\equal{} \\sum_{d | n} \\frac {1}{d} \\equal{} 1 \\plus{} \\sum_{d | n, d > 1} \\frac {1}{d} < 1 \\plus{} \\frac {\\sigma(n) \\minus{} 1}{2} \\le 1 \\plus{} \\frac {2 \\sqrt {n} \\minus{} 2}{2} \\equal{} \\sqrt {n}$. This bound is tightest when $ n \\equal{} 3, 4$.\r\n\r\nThe above argument suggests the following tighter bound: $ \\frac {\\sigma(n)}{n} < H_{ 2 \\sqrt {n}} \\sim \\frac {1}{2} \\ln n$. In fact, we can do [url=http://en.wikipedia.org/wiki/Divisor_function#Approximate_growth_rate]even better[/url].\r\n\r\n[b]Edit:[/b] Follow-up. Show that $ \\frac {\\sigma(n)}{n}$ is unbounded.\r\n\r\n[b]Edit 2:[/b] [hide=\"Discussion\"] Letting $ n \\equal{} \\prod p_k^{e_k}$, we can find that\n\n$ \\frac {\\sigma(n)}{n} \\equal{} \\prod \\left( 1 \\plus{} \\frac {1}{p_k} \\plus{} ... \\plus{} \\frac {1}{p_k^{e_k}} \\right)$.\n\nLet $ n_m \\equal{} p_1 p_2 ... p_m$ be the primordials. Then\n\n$ \\frac {\\sigma(n_m)}{n_m} \\equal{} \\prod_{k \\equal{} 1}^{m} \\left( 1 \\plus{} \\frac {1}{p_k} \\right)$,\n\nthe natural log of which is bounded by $ H_{m}$ (or something along those lines). Some more arguments along these lines should give the growth rate in the Wikipedia article. (Note that as $ m$ increases without bound the product on the RHS diverges.) [/hide]", "Solution_3": "I'm sorry. Right with n>2. Thanks your solution", "Solution_4": "See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83692]An inequality on the sum of divisors.[/url]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Prove that:\r\n$\\sqrt[3]{\\frac{x}{x+26}}+\\sqrt[3]{\\frac{y}{y+26}}+\\sqrt[3]{\\frac{z}{z+26}}\\geq 1$\r\nwhere $x;y;z$ are positive real numbers satisfying $xyz=1$", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=104676" } { "Tag": [ "trigonometry", "ratio", "geometry", "perimeter" ], "Problem": "Here's a thread for discussion of the OCTM test. So, which problems did you have trouble with, and how do you think (or know) you did?\r\n\r\nI thought it seemed easy overall, the only ones I didn't answer were 22 and 26. I did catch 2 or 3 mistakes on the way home, so I think I got like 35, but I'm very happy with that, assuming it's correct (I only got 24 last year...)\r\n\r\nBtw, could someone post the answers for it? I didn't get an answer sheet after I took it, because I was taking it at an irregular location due to an In the Know tournament...", "Solution_1": "I was able to answer all the questions. Unfortunately I got the box and whiskers graph one wrong. I was really upset to ever see that thing again (I always complain about how I didn't learn anything in 6th grade except that useless graph, but I can't call it useless any more :mad: I should've paid more attention...)\r\n\r\nI had just corrected an answer when the proctor called time... man, that was soo scary, i barely made it. Whew.", "Solution_2": "Lol, I feel for you man. I HATE box and whisker plots with a passion.", "Solution_3": "The OCTM used Sudoku for one of its problems! :rotfl: :rotfl: \r\n\r\nI took last year's OCTM as practice for this year's test; this one was decidedly easier. I worked out all of the problems except #22, #26, and probably #36 (can't remember for certain) during the test. Afterwards, I bludgeoned #26 to death with the laws of sines and cosines until it yeilded the value of y, but when I found out that the formula for 22 was just equivalent ratios, I wanted to scream!\r\n\r\nI also caught a few mistakes while typing up the following list:\r\n\r\nOptimistic prediction: 35\r\nUnduly pessimistic prediction: 30\r\n\r\n[hide=\"I think the correct answers are:\"]\n1. $2005$\n2. $a)$ $f;$ $b)$ $f;$ $c)$ $t$\n3. $side$\n4. $2$ $yd,$ $1$ $ft,$ $9$ $in$\n5. $AAA$\n6. $55\\%$\n7. $876495$\n8. $C$\n9. $-130$\n10. $1$\n11. $\\frac{17}{20}$\n12. $b=1$, $c=-1$, $d=-12$\n13. $-\\frac{4\\sqrt7}7$\n14. $A$\n15. $XVII$\n16. $3+2\\sqrt3$\n17. $10300$\n18. $1.5$\n19. $40$\n20. $all$ $real$ $numbers$\n21. $29.5\\%$\n22. $x=3\\frac37$\n23. $\\frac12$\n24. $D$ and $F$\n25. $\\frac{12}{17}$\n26. $4.4641$\n27. $5040$\n28. $2\\frac13$\n29. $E$\n30. $y=22.5$\n31. $10$\n32. $6$\n33. $\\$12.10$\n34. $-243$\n35. $3^{\\circ}59'$\n36. $x=\\frac{2\\sqrt{10}}{5}$\n37. $12$ $in$\n38. $slide$ $rule$\n39. $\\sqrt{29}$\n40. $126$ $mph$[/hide]", "Solution_4": "[quote=\"roadnottaken\"]The OCTM used Sudoku for one of its problems! :rotfl: :rotfl: \n\nI took last year's OCTM as practice for this year's test; this one was decidedly easier. I worked out all of the problems except #22, #26, and probably #36 (can't remember for certain) during the test. Afterwards, I bludgeoned #26 to death with the laws of sines and cosines until it yeilded the value of y, but when I found out that the formula for 22 was just equivalent ratios, I wanted to scream!\n\nI also caught a few mistakes while typing up the following list:\n\nOptimistic prediction: 35\nUnduly pessimistic prediction: 30\n\n[hide=\"I think the correct answers are:\"]\n1. $2005$\n2. $a)$ $f;$ $b)$ $f;$ $c)$ $t$\n3. $side$\n4. $2$ $yd,$ $1$ $ft,$ $9$ $in$\n5. $AAA$\n6. $55\\%$\n7. $876495$\n8. $C$\n9. $-130$\n10. $1$\n11. $\\frac{17}{20}$\n12. $b=1$, $c=-1$, $d=-12$\n13. $-\\frac{4\\sqrt7}7$\n14. $A$\n15. $XVII$\n16. $3+2\\sqrt3$\n17. $10300$\n18. $1.5$\n19. $40$\n20. $all$ $real$ $numbers$\n21. $29.5\\%$\n22. $x=3\\frac37$\n23. $\\frac12$\n24. $D$ and $F$\n25. $\\frac{12}{17}$\n26. $4.4641$\n27. $5040$\n28. $2\\frac13$\n29. $E$\n30. $y=22.5$\n31. $10$\n32. $6$\n33. $\\$12.10$\n34. $-243$\n35. $3^{\\circ}59'$\n36. $x=\\frac{2\\sqrt{10}}{5}$\n37. $12$ $in$\n38. $slide$ $rule$\n39. $\\sqrt{29}$\n40. $126$ $mph$[/hide][/quote]\n\n[hide=\"a few errors\"]\n5. AAS is also invalid theorem.\n7. 876593 is higher\n12. I got $b=-1,c=-12,d=0$\n17. did you mean 10200?\n24. shoot you're right, i'm wrong\n40. $\\frac{280+280}{2+2.5}=124\\frac{4}{9}$\n[/hide]", "Solution_5": "[hide=\"a few corrections\"]7. 876593\n12. Isn't $d=0$?\n34. 1053\n36. $\\frac{4}{\\sqrt{5}}$\n37. $2\\sqrt{37}$[/hide]\r\n\r\n@tjhance: the answer to 40. is 126.", "Solution_6": "DANGIT i was thinking of SSA, its not AAS...\r\n\r\nand isn't the average speed the total distance divided by total time?", "Solution_7": "I interpreted it as the average of the two rates separately. I found out $r_{1}$ and $r_{2}$.", "Solution_8": "hmm.. the question was kind of ambiguous. By the way \"average speed\" is defined, my way should be correct I believe, but your answer is more like the kind of thing the question would ask for.", "Solution_9": "[quote=\"tjhance\"][quote=\"roadnottaken\"]The OCTM used Sudoku for one of its problems! :rotfl: :rotfl: \n\nI took last year's OCTM as practice for this year's test; this one was decidedly easier. I worked out all of the problems except #22, #26, and probably #36 (can't remember for certain) during the test. Afterwards, I bludgeoned #26 to death with the laws of sines and cosines until it yeilded the value of y, but when I found out that the formula for 22 was just equivalent ratios, I wanted to scream!\n\nI also caught a few mistakes while typing up the following list:\n\nOptimistic prediction: 35\nUnduly pessimistic prediction: 30\n\n[hide=\"I think the correct answers are:\"]\n1. $2005$\n2. $a)$ $f;$ $b)$ $f;$ $c)$ $t$\n3. $side$\n4. $2$ $yd,$ $1$ $ft,$ $9$ $in$\n5. $AAA$\n6. $55\\%$\n7. $876495$\n8. $C$\n9. $-130$\n10. $1$\n11. $\\frac{17}{20}$\n12. $b=1$, $c=-1$, $d=-12$\n13. $-\\frac{4\\sqrt7}7$\n14. $A$\n15. $XVII$\n16. $3+2\\sqrt3$\n17. $10300$\n18. $1.5$\n19. $40$\n20. $all$ $real$ $numbers$\n21. $29.5\\%$\n22. $x=3\\frac37$\n23. $\\frac12$\n24. $D$ and $F$\n25. $\\frac{12}{17}$\n26. $4.4641$\n27. $5040$\n28. $2\\frac13$\n29. $E$\n30. $y=22.5$\n31. $10$\n32. $6$\n33. $\\$12.10$\n34. $-243$\n35. $3^{\\circ}59'$\n36. $x=\\frac{2\\sqrt{10}}{5}$\n37. $12$ $in$\n38. $slide$ $rule$\n39. $\\sqrt{29}$\n40. $126$ $mph$[/hide][/quote]\n\n[hide=\"a few errors\"]\n5. AAS is also invalid theorem.\n7. 876593 is higher\n12. I got $b=-1,c=-12,d=0$\n17. did you mean 10200?\n24. shoot you're right, i'm wrong\n40. $\\frac{280+280}{2+2.5}=124\\frac{4}{9}$\n[/hide][/quote]\n[hide=\"@tjhance\"]7. Oops... not that it affects my score; my original answer was $876594$, which is even.\n12. That was an error in copying from my test sheet to this topic; I had it right on the contest.\n17. Yes[/hide][hide=\"@Xaero\"]\n34. How did you get [i]that[/i]?\n36. Heh. I think I completely missed that problem during the contest, anyway, but you are right.\n37. Arrgh! I thought it was volume.[/hide]Revised estimate: 33/34 :maybe:", "Solution_10": "I think I got a 37 if I didn't copy the answers down wrong. Yeah, I killed problem 26 with Stewart's Theorem. :rotfl: :D", "Solution_11": "[quote=\"Xaero\"]I think I got a 37 if I didn't copy the answers down wrong. Yeah, I killed problem 26 with Stewart's Theorem. :rotfl: :D[/quote]\r\n\r\nman, i love that theorem.\r\n\r\ni thought it was weird that they asked for the name of the slide rule... kind of an odd question. Even though those things are like the most beastly computation tool ever created.\r\n\r\nIt took me awhile to do the Roman Numeral one... I had to remember exactly how it worked. I had to come back later.\r\n\r\nI was worried about 20, cause i wasn't sure whether \"union\" was an OR operator or an AND operator... luckily i correctly reasoned it's an OR :D", "Solution_12": "I'm curious, what did you guys get for 32? I got $24$ as the answer.", "Solution_13": "I think you either confused inscribed with circumscribed or calculated the perimeter rather than the side length. I got $6$ on that one. See my answers, above.", "Solution_14": "It is definitely $24$ for number 32.\r\nOh, and @roadnottaken:\r\n[hide=\"34\"]You added $(2\\sqrt{6})^{2}(27)$ rather than subtracting it...[/hide]\r\nI made a bunch of extremely stupid mistakes, which I think is mainly due to lack of sleep. I had a thing going on Friday night, and I didn't get home until 2:30, then I had to get up at 4:30 to make it to my testing site on time. Oh, well, I'll just have to ace it next year.\r\n\r\nRevised estimate: 30 (Really hope its enough to make OHMIO)", "Solution_15": "Do you guys know where I might be able to find a copy of this test, or when a copy might be put up online by? Thanks.", "Solution_16": "The test and solutions are now up on the OCTM website. \r\n\r\n[url]http://www.octmtournament.org[/url]", "Solution_17": "[quote=\"Xaero\"]The test and solutions are now up on the OCTM website. \n\n[url]http://www.octmtournament.org[/url][/quote]\r\n\r\ndarn, I got the last one wrong... I still think it wasn't worded very well.\r\n\r\nabout what score is usually required to proceed to OHMIO?", "Solution_18": "My score: 32\r\n\r\nI made way to many mistakes from not reading questions carefully enough or through sheer carelessness.\r\n :wallbash: $\\ldots$ :wallbash_red: $\\ldots$ :stretcher: $\\ldots$ [huge doctor's bill]\r\n\r\nIt turns out that the answer for 40 was indeed the average of the two rates.", "Solution_19": "My score: 37\r\n\r\nThinking about the last question, you took into account the wind speed with the plane speed when it just asks for the plane air speed. There is a difference.\r\n\r\nAlso, OHMIO's cutoff is usually around 30.", "Solution_20": "[quote=\"Xaero\"]Also, OHMIO's cutoff is usually around 30.[/quote]\r\n\r\nI think i should make it then; I made some lame-o mistakes, but not too many.", "Solution_21": "Anyone got the contest questions themselves.", "Solution_22": "[quote=\"mathgeniuse^ln(x)\"]Anyone got the contest questions themselves.[/quote]\r\n\r\n[url]http://www.octmtournament.org/octmattach/testresults/000000000057.pdf[/url]", "Solution_23": "I just took the test (I don't live in Ohio, so I just used the questions from online) and did the test under general conditions (except maybe less adrenaline/pressure). I got a 34, how would that stand in Ohio?\r\n[hide=\"mistakes\"]\n8. Dumb mistake...\n15. Completely forgot Roman numerals, made a wild guess\n20. Forgot what union meant so guessed, picked the wrong one.\n30. Eh I don't know what I did\n34. Ditto as 30\n38. No clue at all. [/hide]", "Solution_24": "Wow, mine was riddled with absolutely retarded mistakes. From what should have been a 35, I got a 27...\r\n\r\nNote to self: get more than 2 hours of sleep before OCTM next year...", "Solution_25": "[quote=\"life=tennis+math\"]I just took the test (I don't live in Ohio, so I just used the questions from online) and did the test under general conditions (except maybe less adrenaline/pressure). I got a 34, how would that stand in Ohio?\n[hide=\"mistakes\"]\n8. Dumb mistake...\n15. Completely forgot Roman numerals, made a wild guess\n20. Forgot what union meant so guessed, picked the wrong one.\n30. Eh I don't know what I did\n34. Ditto as 30\n38. No clue at all. [/hide][/quote]\r\n\r\nI'd be happy to answer any questions if anyone needs to. I got 39.\r\n\r\nAnd 38 is Napiers Bones if anyone cares. It is an eponym", "Solution_26": "[quote=\"mathgeniuse^ln(x)\"]I'd be happy to answer any questions if anyone needs to. I got 39.\n\nAnd 38 is Napiers Bones if anyone cares. It is an eponym[/quote]\r\n\r\nNice job on the 39. That might be 1st in the state...\r\n\r\nHmmm, Napier's Bones. We did a thing on that in math club at my school last year. It's pretty cool...\r\n\r\nThe correct answer was slide rule though. The site told me...", "Solution_27": "Number 38 was a \"history of math\" question, of which they have one every year. You can not really pin down a difficulty level for those questions; either you know it or you don't. I happened to know what a slide rule was, but generally I don't know those answers. It's really just luck.\r\n\r\nWhen I saw that question on the test, I thought it was really funny, remembering one of my friends from another math forum (much smaller and having only been running for about three months) who used a slide rule on the OCTM test when he took it (not because he is that old).\r\n\r\nAlso, that 39 probably is not best in the state, simply because mathgenius^ln(x) is not in the state (he's from Michigan).", "Solution_28": "Oh, well, that would prevent that, wouldn't it? Maybe it's Xaero then. The top is usually about 37-39...", "Solution_29": "Yeah, the 39 includes the fact that I got #38 wrong, but I still think Napier's Bones is an allowed answer.\r\n\r\nYeah, I am from Michigan.", "Solution_30": "I highly doubt that I got the #1 score in Ohio since the state champion last year was a 9th grader because he score a 39...\r\nOh well, I can still hope for top 10 or something.", "Solution_31": "Oh you are not Tong Zhan", "Solution_32": "[quote=\"mathgeniuse^ln(x)\"]Yeah, the 39 includes the fact that I got #38 wrong, but I still think Napier's Bones is an allowed answer.\n\nYeah, I am from Michigan.[/quote]\r\nIs Napier's Bones another name for a slide rule?\r\n\r\nI don't know whether they would have allowed it, but normally they list all acceptable answers in their answers, and the only one for that question was \"slide rule\".", "Solution_33": "Well, the slide rule is a outgrowth of Napier's Bones, so I don't know.\r\n\r\nHere is a link.\r\n\r\n[url]http://www.maxmon.com/1600ad.htm[/url]", "Solution_34": "In that case, I have a second reason why Napier's Bones is probably not an allowed answer. The description in the OCTM clearly showed that it was referring to an instrument with moving parts. Napier's Bones does not fit that description at all.", "Solution_35": "Oh, I didn't see that.\r\n\r\nNVM\r\n\r\nIt is sliderule, my second answer." } { "Tag": [ "Euler", "search", "geometry proposed", "geometry" ], "Problem": "Let a triangle $ ABC$ with Euler line cut $ AB,AC$ at $ M,N$. Prove that: Euler line of triangle $ AMN$ is paralell with $ BC$.", "Solution_1": "See here for a slightly more general statement (though equivalent afterall): [url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1701732746&t=191643[/url]." }