2, then p+1-2a is even so q is even so q=2, therefore b=1, so 2(a ^2 +1)=c ^2 +1=(p-a) ^2 +1=(a ^2 -a+1) ^2 +1. If a=1, that's impossible.If a=2 that's possible, and from here we obtain the solution (a,b,c)=(2,1,3) (and also (1,2,3), because we supposed that p>q).\r\nIf a>2 then a ^2 -a+1=a(a-1)+1>=2a+1, so (a ^2 -a+1) ^2 +1>=(2a+1) ^2 +1 > 2(a ^2 +1).\r\n\r\nTherefore (1,2,3) and (2,1,3) are the only solutions.",
"Solution_2": "The 2nd problem:\r\n\r\nwe presume that: m>n. Let x be their common solution of the two equations. Then x verifies x^n(x+1)(x^(m-n)-1)x=0 and x=-1 is not a solution. Let x be so x^(m-n)=1. We have x<> \\pm 1, x from C\\R, x satisfies: x^(m-1)=1, x^n*(x-1)=-1. We apply the module and get |x|=1 and |x-1|=1 => x cos(pi/3)+isin(pi/3) or x=cos(5pi/3)+i sin(5pi/3), which \r\nmeans x=w or x=w (w has a line on it) , where w^2-w+1=0, w is the 6th order root of unity.\r\nTaking modulo 6 of m and n we get that the only good cases are when m==1(mod 6) and n==1(mod6). So the common roots are given by:\r\n1) the eqs x^2-x+1=0 for m==n==1(mod 6).\r\n2) in other cases it doesn't exist!!!\r\n\r\ncheers mate!"
}
{
"Tag": [],
"Problem": "I come across one question and I do it in two ways , both ways \"sounds\" logical but there is something wrong in it ... here it goes\r\n\r\nQuestion : There are 5 couples , 4 person are chosen among them with the condition the must be at least one male and one female . In how many ways can this be done ?\r\n\r\n[b]First Method[/b] : \r\n\r\nAll ways - 4 male - 4 female = $10C4 - 5C4 - 5C4 = 200 $\r\n\r\n[b]Second Method[/b] :\r\n\r\nFirst we chose 1 male and 1 female which is $5C1\\times 5C1$ then for the remaining two vacancies , we can choose it randomly from the remaining 8 person which is $8C2$ . So total ways = $5C1\\times 5C1\\times 8C2 = 700 $ \r\n\r\nWell , second method seems to flow logically but we know it is not true . Where actually goes wrong ? :?",
"Solution_1": "to me the second methodmakes more sense; on the other hand:\r\n10C4=210.. ;) \r\n\r\nits 12:50am, i'm too sleepy to think, so i'm wrong",
"Solution_2": "your second solution is wrong because you count cases more than one.\r\ni.e: $f_1 m_1 f_2 m_2$ and $f_1 m_2 f_2 m_1$ and $f_2 m_1 f_1 m_2$ and $f_2 m_2 f_1 m_1$ so you count this case $4$ times!!! ;)",
"Solution_3": "[quote=\"a_vakilian\"]your second solution is wrong because you count cases more than one.\ni.e: $f_1 m_1 f_2 m_2$ and $f_1 m_2 f_2 m_1$ and $f_2 m_1 f_1 m_2$ and $f_2 m_2 f_1 m_1$ so you count this case $4$ times!!! ;)[/quote]\r\n\r\nHmm .... I guess that is why the second method gives us such big number :rotfl: .\r\nAnyway , could there be any way to get the same answer by using the concept like second method ? Like , could we divide it by some number(or combination) to cancel off the repeated part ? :?",
"Solution_4": "yes it is possible. you should divide the problem to $3$ case.when there is $2 woman ,2 man$ and $1 woman , 3 man$ and $3 woman, 1 man$ . ;)",
"Solution_5": "[quote=\"a_vakilian\"]yes it is possible. you should divide the problem to $3$ case.when there is $2 woman ,2 man$ and $1 woman , 3 man$ and $3 woman, 1 man$ . ;)[/quote]\r\n\r\nWell , what Im trying to ask is whether we can use the concept in second method , which is like setting a male and female first then the rest can be chosen at random . I know that we can divide it into 3 cases like you mention to solve it but what i want to know is whether the second method can be improve or not to make it right ?",
"Solution_6": "I think there is not such a solution that you want. or it is not worth knowing :? :oops:"
}
{
"Tag": [
"geometry",
"3D geometry",
"sphere",
"calculus",
"integration",
"HMMT",
"LaTeX"
],
"Problem": "A sphere with a radius of 5 is cut by a plane that is 4 away from the center of the sphere. What is the volume of the larger of the two resulting solids?",
"Solution_1": "[hide]I know that you could use calculus (take the equation for a circle, and use the cross sectional method and find the area beneath the graph from -5 to 4).\n\n$\\int_{-5}^{4}{\\pi(\\sqrt{25-x^{2}})^{2}dx}$\n$\\pi\\int_{-5}^{4}{(25-x^{2})dx}$\n$\\pi((25*4-\\frac{4^{3}}{3})-(25*-5-\\frac{(-5)^{3}}{3}))$\n$162\\pi$\n\nAnd you could imagine a hole with a radius of 3 being driven through the sphere. Then if you compare each cross section of that $h$ away from the center with each cross section of a sphere with a radius of 4, $h$ away from the center you notice they are equal. Then you add in the cyclinder that would fill up most of the hole and a sphere that is chopped off by two planes. \n\nSo end the end you would end up with $162\\pi$ like before.[/hide]\r\n\r\nEven if these answers aren't correct I think the process is correct. I just want to find a solution without using calculus or using a crazy idea.",
"Solution_2": "I don't believe a non-calculus formula exists for a sectoroid (a sector revolved about its axis) or a segmentoid although I may be mistaken\r\n\r\neven if an antiderivative existed to the integral for the volume of a segmentoid, it would most definitely not be very nice. Calculus is the best method here",
"Solution_3": "There is a non calculus way to find the volume of a sectoroid. \r\n\r\nhttp://web.mit.edu/hmmt/www/datafiles/problems/2003/pgeom03.pdf\r\n\r\nProblem #6 is not quite a sectoriod, but you can just add the cylinder in. The answer is in the solutions:\r\n\r\nhttp://web.mit.edu/hmmt/www/datafiles/solutions/2003/sgeom03.pdf\r\n\r\nAnd I am pretty sure I have posted both the calculus way and the noncalculus (from the links) way above. But I am guessing there really is not a way to find the volume without calculus or a great imagination so my question is answered.",
"Solution_4": "The solution is perfectly reasonable if you know the geometric derivation of the formula for a sphere.",
"Solution_5": "Yeah, I can't think of a non-calculus solution....",
"Solution_6": "Hint: Cavalieri's principle \r\n\r\nUse google to find out what it is.",
"Solution_7": "The solution Altheman is hinting at (which is essentially what is used in the HMMT solutions linked to above) begins like this:\r\n\r\nSit a cone of radius $r$ and height $r$ down next to a hemisphere of radius $r$ so that the vertex of the cone is in the same plane as the base of the hemisphere and the plane of the base of the cone is tangent to the hemisphere. (If the hemisphere is sitting on a table base-down, the cone is balanced on the same table on its tip.) Now consider the combined area of a cross-section of the two shapes at height $h \\leq r$ above the table. This is a nice way to get the volume of a sphere. It can also be used to get a nice formula for the \"cap\" of a sphere, as in the question here.\r\n\r\nSide note: this requires you to know the volume of a cone, but this can also be derived using a Cavalieri argument.",
"Solution_8": "o.O I am trying to force people to be independent and use google. This skill is very valuable. \r\n\r\nWell...I suppose if you were being all rigirous and stuff, you would need to derive the volume of a cone by putting it in correspondence with a pyriamid. But obviously surface area/volume of a cone fall within most first year geometry classes.",
"Solution_9": "[quote=\"Altheman\"]I am trying to force people to be independent and use google.[/quote] The effect you are having is of being cryptic and ensuring that most of your audience won't ever find out what you're talking about.\n\n[quote]But obviously surface area/volume of a cone fall within most first year geometry classes.[/quote] But the volume is usually asserted, not proved.",
"Solution_10": "[quote=\"JBL\"][quote=\"Altheman\"]I am trying to force people to be independent and use google.[/quote] The effect you are having is of being cryptic and ensuring that most of your audience won't ever find out what you're talking about.\n[/quote]\n\n[hide=\"@JBL\"]Okay. Lets consider an analogous situation. We are given a triangle concurrency problem. I suggest mass points [lets say in the HSB forum]. Then I should expect to get lots of stupid replies asking: \"what are mass points?\" \"where can I learn more about mass points?\" These are stupid questions because you can easiliy find the answer by a quick google search. I know you have this ability. ex. consider your response to mathfanatic in that global warming thread about following fallacies created by those against global warming. Anyway, this situation is similar. You can easily type: \"cavaleri principle sphere\" into google and obtain the link: [url]http://www.walter-fendt.de/m14e/volsphere.htm[/url]. This takes a few seconds and you get a diagram which I find much more helpful than any solution provided in this thread. While you do not get a complete solution in this page, it is enough of a hint to finish the problem. \n\nFurthermore, I think this type of hint is nicer than a full solution. This way the solver can still get a sense of accomplishment from solving the problem. I find that a fundamental part of my enjoyment of problem solving is when I come up with the solution on my own. By offering hints, I let the solver get started on the problem. The full solution should be left to those who have to struggle to solve the problem.\n\nIf I may predict your response, I think that you will think that this is a lot of ---- and I am just being lazy. I am not sure what position you think you are in so you can interpret my actions. I am in a better position than you are to I know what I am thinking.[/hide]\n\n[quote=\"JBL\"][quote]But obviously surface area/volume of a cone fall within most first year geometry classes.[/quote] But the volume is usually asserted, not proved.[/quote]\r\n\r\nReally? I am suprised about that. In freshman year geometry, [i]before I was interested in math[/i] [i.e. so I took not particular action to figure this out], I was aware of this proof:\r\ni) V=1/3 b*h where the base is a triangle; this comes from decomposing a parallelpiped [or w/e you call it]\r\nii) V=1/3 b*h where the base is a polygon; decompose the polygon into triangles and apply i)\r\niii) V=1/3 b*h where the base is a circle; a circle is a polygon with an infinite number of sides; apply ii)\r\n\r\nPerhaps I had a better geometry class than I thought; I thought that this was the standard \"non-calculus proof.\"\r\n\r\nI suppose that is not 100% rigirous since we are dealing with limits. But that is being pretty picky.",
"Solution_11": "[hide=\"@Altheman\"]You think saying, \"use mass points\" in the HSB forum is educational -- I think it is pedagogically unsound. Simply telling people to google things will lose most of your audience immediately -- they won't bother. Of those who do google it, they won't know what to google for, and a substantial portion of them will fail to find an appropriate introductory source. Far, far better (in my opinion) is to explain how to do something, and (if it's not obvious) show why it works. (Cavalieri's Principle is something where the \"why\" is really calculus, but the result itself is very intuitively clear.) In general, the general result is something best understood [i]after[/i] particular examples have been dealth with, and knowing the name (or the formula, or whatever) but not understanding it is worthless. Teachers should do more than simply hand their students a textbook for each question, and this is all the more true when the textbook in question is the entire internet.\n\nYou're probably right that I've misinterpreted your intent in the past, so I retroactively apologize. (That does not apply to my complaints about LaTeX non-usage, though :wink: )[/hide]\r\n\r\n[quote=\"JBL\"]i) V=1/3 b*h where the base is a triangle; this comes from decomposing a parallelpiped [[i]maybe prism? -- JBL[/i]]\nii) V=1/3 b*h where the base is a polygon; decompose the polygon into triangles and apply i)\niii) V=1/3 b*h where the base is a circle; a circle is a polygon with an infinite number of sides; apply ii)\n\nPerhaps I had a better geometry class than I thought; I thought that this was the standard \"non-calculus proof.\"\nI suppose that is not 100% rigirous since we are dealing with limits. But that is being pretty picky.[/quote] It could just be that my memory of geometry class is very rusty :). I'm not quite sure what you mean by (i), but it sounds plausible. I don't like (iii) very much, because I think Cavalieri's is a nicer way to go from a single pointy thing to all pointy things (in other words, there's no particular reason to go via (ii) -- one can spit out (ii) and (iii) together without ever using the words \"infinitely many\"). A lot has to do with how well something is explained, I guess.",
"Solution_12": "Well...I learn by reading the material first, then asking questions. And the lack of LaTex can be attributed to my procrastinating from school work...but w/e my posts have generally improved now that school has ended.\r\n\r\nAnyway, Here is euclid's proof for $\\frac{1}{3}bh$: http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII7.html\r\n\r\nI think that it is reasonable to consider a circle as a limiting form of a polygon with a large number of sides. If you are particularly picky, you can consider the inscribed and circumscribed regular polygons with n sides. Then you can apply the method of exhaustion [originally an idea from archimedes]. To show that the circle is inbetween these two volumes. This is quite rigirous actually. In the development of integral theory by Tom Apostol in \"Calculus,\" this is one of the 4 fundamental axioms that he takes for area. The analogue to volume is just as acceptable; if one object contains the other, then they have equal volume. Regardless, these are ideas that my dad would understand [and he does psychology]. I think they are sufficiently rigirous for say, the ancient greek geometers."
}
{
"Tag": [
"MATHCOUNTS",
"geometry",
"AMC",
"AIME",
"AMC 12",
"AMC 10"
],
"Problem": "Are they helpful? I am deciding on getting them and I want your thoughts on the books. What are their levels and what contests do they prepare you for? Are they worth the money? Thanks.",
"Solution_1": "They definitely are very helpful, at least in my opinion and what seems to be the majority of AoPSers. I haven't really started Volume 2 yet, but Volume 1 is extremely helpful for MathCounts and the AMCs, particularly the geometry section.",
"Solution_2": "They are very very very helpful! Vol 1 covers Mathcounts/AMC stuff, Vol 2 is more AIME/AMC/ARML stuff.\r\n\r\nVol 1 is recommended for middle school , Vol 2 is recommended for high school",
"Solution_3": "Just make sure that if you get them, you do a lot of the problems and actually learn math rather than being a passive reader.",
"Solution_4": "[quote=\"isabella2296\"]They definitely are very helpful, at least in my opinion and what seems to be the majority of AoPSers. I haven't really started Volume 2 yet, but Volume 1 is extremely helpful for MathCounts and the AMCs, particularly the geometry section.[/quote] did you make aime isabella",
"Solution_5": "If I was aiming for AIME via AMC12 (failed this year by a few mistakes, motivated to try AoPS now) should I just skip the first volume?",
"Solution_6": "Thanks to all. I will buy now.",
"Solution_7": "[quote=\"Ktk\"]If I was aiming for AIME via AMC12 (failed this year by a few mistakes, motivated to try AoPS now) should I just skip the first volume?[/quote]\r\nAoPS Vol.1 doesn't go into details about the stuff that's on the AMC 12 and not AMC 10, but it covers basics for the AMCs and just knowing all of that stuff really will help you more than knowing more advanced stuff from Vol. 2. So you should get Vol. 1."
}
{
"Tag": [
"\\/closed"
],
"Problem": "Recently, I've found that when I try to use the \"hide\" command, instead of applying the hide format, it will instead display the code. Was this an intentional modification, or a bug?",
"Solution_1": "Did you remember to close with (/hide)? (replace () with [])\r\n\r\n[hide=\"Hiding\"]...[/hide] works fine for me.",
"Solution_2": "Oops...\r\n\r\nAt least one time it was because I used the wrong slash. :oops: \r\n\r\n[hide=\"This is a test.\"] :maybe: [/hide]\r\n\r\nYep, that must have been it, because the above hide works just fine."
}
{
"Tag": [
"function",
"algebra proposed",
"algebra"
],
"Problem": "Prove that the elements of the set $ A\\equal{}\\{ (x,y)|x \\in \\mathbb{Z},y \\in \\mathbb{Z}, 2xy\\minus{}6x\\plus{}y\\equal{}0 \\}$ are on the graphic of the function $ f: \\mathbb{R} \\rightarrow \\mathbb{R}, f(2x\\minus{}3)\\equal{}8x\\minus{}7.$",
"Solution_1": "[quote=\"moldovan\"]Prove that the elements of the set $ A \\equal{} \\{ (x,y)|x \\in \\mathbb{Z},y \\in \\mathbb{Z}, 2xy \\minus{} 6x \\plus{} y \\equal{} 0 \\}$ are on the graphic of the function $ f: \\mathbb{R} \\rightarrow \\mathbb{R}, f(2x \\minus{} 3) \\equal{} 8x \\minus{} 7.$[/quote]\r\n\r\nUnfortunately, this is wrong : $ (\\minus{}1,6)\\in A$ but is not on the required graphic since $ f(\\minus{}1)\\equal{}1\\neq 6$ and $ f(6)\\equal{}29\\neq \\minus{}1$)"
}
{
"Tag": [
"AMC",
"AIME",
"USAMTS",
"USA(J)MO",
"USAMO"
],
"Problem": "Since this year seems to be easier than usual, what effect will that have on the cutoff? how far up will the cutoff move? Also, how is the cutoff determined? Is it the first score for which >x people got that score or above for some x?",
"Solution_1": "[quote=\"solafidefarms\"]Since this year seems to be easier than usual, what effect will that have on the cutoff? how far up will the cutoff move? Also, how is the cutoff determined? Is it the first score for which >x people got that score or above for some x?[/quote]\r\nI think that there are an x number of people who qualify for the AIME. Anyways, if the third round is uber hard than the qualifying score might be as low as 60-63.",
"Solution_2": "You start off with an index of 100 if you qualify through USAMTS, right?\r\n\r\nalso, people that qualify for the AIME through the USAMTS can qualify for the USAMO by both their index and the floor score, right?",
"Solution_3": "I think the index only matters in USAMO qualification is you are in 11th or 12th grade. Otherwise you go by floor which means if you score at least the score the lowest AIME score that the people that qualified by the index, you are in. So I don't think the USAMO index really does matter for us...\r\n\r\nBut I am interested in how qualifications to AIME is decided. It would feel good if I knew that even if I bombed the AMCs, I would still be in the AIME."
}
{
"Tag": [
"linear algebra",
"matrix",
"search",
"linear algebra unsolved"
],
"Problem": "DO any of u know what a \"[b][color=black][size=150]DIRAC MATRIX[/size][/color][/b]\" is?",
"Solution_1": "I have two words for you: [b]GOOGLE SEARCH[/b].",
"Solution_2": "THanks for that!!!!!\r\n\r\n\r\n@carcul- ia almost forgot that"
}
{
"Tag": [
"combinatorics unsolved",
"combinatorics"
],
"Problem": "There are n markers, each with one side white and the other side black,\r\naligned in a row so that their white sides are up. In each step, if possible, we\r\nchoose a marker with the white side up (but not one of the outermost markers),\r\nremove it and reverse the closest marker to the left and the closest marker to the\r\nright of it. Prove that one can achieve the state with only two markers remaining\r\nif and only if n - 1 is not divisible by 3.",
"Solution_1": "merged with http://www.mathlinks.ro/viewtopic.php?t=90046 .\r\n\r\n darij"
}
{
"Tag": [
"number theory proposed",
"number theory"
],
"Problem": "find all pairs of natural numbers $ (n,m)$ such that $ n!\\plus{}1\\equal{}m^2$\r\n\r\n\r\n$ (4,5),(5,11),(7,71)$ is this all ?",
"Solution_1": "This a very well known problem ( I think it is Brocard's problem) but unfortunately it hasn't been solved yet. I t seem that those are the only solutions but it is not known.\r\nIf someone can solve it I don't think he will post it here without publishing it :P :rotfl: \r\nBut I recommend you to try it, it's very nice\r\n\r\n\r\nDaniel"
}
{
"Tag": [
"calculus",
"derivative",
"real analysis",
"real analysis unsolved"
],
"Problem": "Show that $x^{n}+nx-1 =0 \\quad n\\geq 1$, has exactly one real root $a_{n}$. Does the series\r\n\\[\\sum_{n}a_{n}\\]\r\nconverge ?\r\n\r\n(I'm not sure about the problem. Think it should be \"one real [i]positive[/i] root\")",
"Solution_1": "It has at least one positive root by the Intermediate Value Theorem; its value at zero is negative, and it goes to $\\infty$ at $\\infty$. If $n$ is even, it has a negative root as well, so it should specify \"positive real root\".\r\nFor uniqueness, show that the derivative is positive for positive $x$; Rolle's theorem finishes the proof.\r\n\r\nThe series diverges: $\\frac1n