{ "Tag": [], "Problem": "There are two people A and B. They can remove a coin at a time from a pile or split a pile into two non empty piles. The last person to remove the coin loses. The objective of the game is to lose. The game starts with A1 piles of single coins and A2 piles of two coins...How would the person lose based on the sequence (A1,A2,A3...)?\r\n\r\nI think this is the one about taking out a specific number that is bigger than the amount able to take out by one person but can be taken up if both people take the turn.\r\n\r\nI think it is similar to this:(A very basic game)\r\nA person can say 1,2 or 3numbers at a time. Two people keep saying the numbers until the person to say 21 loses. In this case, the second person always have to say up to the multiple of 4, which leads the person to victory.\r\n\r\nSo.. maybe there is a similar form of strategy to this game as well?", "Solution_1": "Amod lock this, the competition isnt over until 14th of october", "Solution_2": "Hey I read the downloads. I can't find which question is the same is this can you tell me?", "Solution_3": "lol its number 11 person A=michael, person B=robin. Your friend just changed a bit so that other people wouldn't notice. Its basically the same idea :ninja:\r\nIf someone to you the losing solution, then you friend could just write the exact opposite of the losing solution :roll: :wink:", "Solution_4": "Let:\r\n$n:$ The number of the piles\r\n$x:$ The total number of the coins\r\n$m:$ The number of the single coins\r\n\r\nIf $m$ is even and $x-n$ is also even then we will say that we have a [u]W-state[/u].\r\nElse we will say that we have an [u]L-state[/u].\r\n\r\n\r\nThe [b][u]objective[/u][/b] for each move is to reach a W-state.\r\n\r\n[color=brown]We will show that, every W-state can lead only to L-states.\nAlso, in every L-state, there is (at least one) move such that we can go to a W-state.[/color]\r\n\r\n\r\nFrom this point, when we say that $m$ changes, we will mean that its parity changes, that is, $m \\mod 2$ changes.\r\nWhen we say that $m$ keeps its value, we'll mean that $m \\mod 2$ keeps its value\r\n(for example, if the new value of $m$ is $m-2$, then we will say that $m$ doesn't change; it keeps its value)\r\n\r\n[hide=\"Let us first see the possible moves\"]\n$\\blacksquare$ Remove a coin from a pile with $k$ coins\n\nFor $x:$\nThe new value of $x$ is $x-1$. So $x$ [u]changes[/u]\n\nFor $n:$\nIf $k=1$ (if we remove from a pile with 1 coin) then, of course $n$ [u]changes[/u].\nElse it doesn't.\n\nIf $k=1$, then $m$ becomes $m-1$, so $m$ [u]changes[/u]\nIf $k=2$, then $m$ becomes $m+1$, so $m$ [u]changes[/u]\nIn any other case $m$ doesn't change.\n\nSummary:\n[quote]$x-n$ keeps its value if $k=1$\n$m$ keeps its value if $k>2$[/quote]\n\n\n$\\blacksquare$ Splitting\nWe split a pile of $k$ coins into two non-zero piles of $p$ and $q$ coins, such that $p+q=k$ (WLOG suppose that $p\\leq q$)\nOne should notice that $k$ must be at least 2.\n\nFor $n:$\nWe loose a pile but we obtain two new piles.\nSo $n$ becomes $n+1 \\Rightarrow n$ [u]changes[/u]\n\nFor $x:$\nOf course we have the same number of total coins, so $x$ [u]doesn't change[/u]\n\nFor $m:$\nWe saw that $k\\geq 2$. So we don't [u]loose[/u] any of the single coins by splitting. \nWe could only [u]obtain[/u] more single coins, depending on the values of $p,q$\n \nIf $11$, as we saw, $x-n$ changes, so we go to an L-state\nIf $k=1$, then $x-n$ keeps its value, but $m$ changes. That is an L-state.\n\nConclusion\nThis means that, if we \"give\" to the opponent a W-state, then he can't give us a W-state. \nHe'll give to us an L-state. \nOur goal is to convert an L-state to a W-state. Let's see how:\n------------------------------------------------------------------------[/hide]\n[hide=\"Now we will see that, if the state is L, then we can find a move, such that the result is a W-state\"]\nL-state means that at least one of $m,x-n$ is odd.\n\n[color=brown]Note\nIf $x-n$ is odd, then there is at least one pile with $k>1.$\nIndeed, if we have only single coins, then $x=n\\Rightarrow x-n$ is even[/color]\n\nWe'll check 3 cases:\n\n[hide=\"a) m odd and (x-n) even\"]\nSince $m$ is odd, there is at least one single coin.\nWe will remove this coin.\nRemember the rules of removing:\n[quote]$x-n$ keeps its value if $k=1$\n$m$ keeps its value if $k>2$[/quote]\nHere, $k=1$, so $x-n$ remains even and $m$ becomes even\nSo we have a W-state\n------------------------------------------------------------------------[/hide]\n[hide=\"b) m odd and (x-n) odd\"]\nWe have the sub-cases:\n\n$b.1)$ If there exists a pile with $k>2$ coins\nWe split this pile into two piles of $1$ and $k-1$ coins respectively.\nRemember the rules of splitting:\n[quote]$x-n$ changes\n$m$ changes if [u]just one[/u] of the new piles is a single coin[/quote] \n\nSo $x-n$ changes and $m$ changes $\\Rightarrow x-n \\equiv m \\equiv 0 \\mod 2$\n\n$b.2)$ If there is not a pile with $k>2$, then we can be sure that there exists at least one pile with $k=2$ coins (see note above)\nWe remove one coin from this pile\nRemember the rules of removing:\n[quote]$x-n$ keeps its value if $k=1$\n$m$ keeps its value if $k>2$[/quote]\n\nHere, it is $k=2$, so $x-n$ changes and $m$ changes $\\Rightarrow x-n \\equiv m \\equiv 0 \\mod 2$\n------------------------------------------------------------------------[/hide]\n[hide=\"c) m even and (x-n) odd\"]\nWe have the sub-cases:\n\n$c.1)$ If there is some pile with $k=2$ coins\nThen we split this pile\nRemember the rules of splitting:\n[quote]$x-n$ changes\n$m$ changes if [u]just one[/u] of the new piles is a single coin [/quote]\n\nSo $x-n$ changes and $m$ does not $\\Rightarrow x-n \\equiv m \\equiv 0 \\mod 2$\n\n\nc.2) If there is not a pile with $k=2$ coins\nThen we can be sure that there is some pile with $k>2$ coins (see note above)\nWe remove one coin from this pile\n\nRemember the rules of removing:\n[quote]$x-n$ keeps its value if $k=1$\n$m$ keeps its value if $k>2$[/quote]\n\nHere, it is $k>2$, so $x-n$ changes and $m$ doesn't $\\Rightarrow x-n \\equiv m \\equiv 0 \\mod 2$\n------------------------------------------------------------------------[/hide]\n----------------------------------------------------------------------------------------------------------\n[/hide]\r\n\r\n[size=134][color=blue]Conclusion[/color][/size]\r\nSo, we can always convert a W-state to an L-state\r\n\r\nIf the two players know this strategy, then the initial state will show who will be the winner.\r\n\r\nIf it is a W-state, then the first player can't convert it to a W-state, so he will loose\r\nIf it is an L-state, then the first player can convert it to a W-state, so he'll win", "Solution_5": "Nice proof pontios.\r\nI came up with a similar solution.\r\nI did\r\nLet A=pile of coins\r\n B=number of piles\r\n 0=even\r\nIf it is even(A)xodd(B)+odd(A)xeven(B)+even(A)xodd(B)+...\r\nor odd(A)xeven(B)+even(A)xodd(B)+odd(A)xeven(B)+...\r\nDuring the person B's turn; It is always a win.\r\nI have found this works up to any number of pile of coins up to 4. The only catch was that If person A had 1odd left he would lose. I was wondering would this strategy work? And up to what extent? Could someone contradict/point out/help me please?\r\nThanks.\r\n\r\nPS:I have concluded that pontios's solution is the most correct so far. However, is there any shorter solution?", "Solution_6": "Actually, when I posted my solution I couldn't simplify it more. And the proof was indeed hard. Thank you for revising this topic, because I had the chance to try it more. I wrote a script which calculates all the possible states starting from (1,2,3,4,5). \r\nFinally, I found a simpler solution to this problem.\r\n\r\nHowever, I can't understand some points in your post\r\n\r\n[quote=\"max1546\"]Let A=pile of coins\n B=number of piles\n 0=even\n[/quote]\nwhat is A?\n\n[quote=\"max1546\"]If it is even(A)xodd(B)+odd(A)xeven(B)+even(A)xodd(B)+...\nor odd(A)xeven(B)+even(A)xodd(B)+odd(A)xeven(B)+...\n[/quote]\r\n\r\nwhat does this sentence mean? :huh:", "Solution_7": "Sorry Pontios for being so unclear, I was rushing it.LOL\r\nhmm I'll try to explain some of the things you asked.\r\nRemember though, this MIGHT make you more confused.\r\nSo i'll do my best.\r\n\r\nIf there is 3coins stacked on top of each other, B=3, A=1.\r\nIf there are 6coins stacked so that it is 3coins per pile, it is B=3, A=2.\r\n\r\nThe other part was, I tried some trial and error and got the answer that if it is the person's turn and he starts off with;\r\n1(A)x1(B)=Certain chance of winning.\r\n1(A)x2(B)=Win\r\n2(A)x1(B)=Lose\r\n1(A)x3(B)=Win\r\n1(A)x4(B)=Lose\r\n1(A)x5(B)=Win\r\n2(A)x2(B)=Win\r\nAnd so on.I tried up to 4(A) and my sequence given worked. I was trying to take in to account every single move either could make and if the person who starts has a chance of winning no matter what the other person does, he wins. On the other hand, if the other person could win with at least one move no matter what the starting person does, I counted it as a loss.\r\n\r\nAhh, even I'm getting slightly confused. I hope this explained slightly more. Still I think your solution is the best so far. If I have time, I'll try and improve my solution and post it again. Hopefully it will be easier/clearer.\r\n\r\nThanks", "Solution_8": "Let's re-formulate the problem\r\n[quote]There are two people A and B. They can remove a coin at a time from a pile or split a pile into two non empty piles. \nThe last person to remove the coin [u][color=red]wins[/color][/u]. \n\nThe game starts with \nA1 piles of single coins \nA2 piles of two coins...\nA3 piles of three coins...\nHow would the person [u][color=red]win[/color][/u] based on the sequence (A1,A2,A3...)?\n[/quote]\r\n\r\n[hide=\"My new solution\"]\nEvery state can be represented by the sequence $A_{1}A_{2}A_{3}A_{4}...$\n\n\nLet $\\boxed{x = A_{1}\\mod 2}$ and $\\boxed{y=(\\sum{ A_{2k}})\\mod 2}$ \n\n(we don't need the numbers A3,A5,A7,...)\n\nEquivalently, we can say that we take the last binary digit of $A_{1}$ and $\\sum{ A_{2k}}$\n\n\n\nNext, we calculate the value $S=x \\vee y$ from the folowing table:\n\n$\\begin{array}{|c|c|c|}\\hline x & y & x \\vee y \\\\ \\hline 0 & 0 & 0 \\\\ \\hline 0 & 1 & 1 \\\\ \\hline 1 & 0 & 1 \\\\ \\hline 1 & 1 & 1 \\\\ \\hline \\end{array}$ (Boolean OR)\n\nAnd we are finished!\n\nThe state $(A_{1}A_{2}A_{3}A_{4}...)$is marked with the label $S$. \nWe will say that we have an [b]S[/b]-state\n\n\nIf you start playing from a [b]1[/b]-state, then you will [u]definitely[/u] win \nIf you start playing from a [b]0[/b]-state, then you will [u]definitely[/u] loose \n\nSo, after each move we should take care such that the result is a [b]0[/b]-state.\n\n[color=darkred]Note\n - The state $(A_{1}A_{2}A_{3}A_{4}...)=(1000...)$ is clearly a [b]1[/b]-state (if you start from this state, you'll win)\n - The state $(A_{1}A_{2}A_{3}A_{4}...)=(0000...)$ can be defined as a [b]0[/b]-state[/color][/hide]", "Solution_9": "[hide=\"Proof\"]\nAgain the proof is a litle complicated but no so much as the previous\n[hide=\"opponent s moves\"]\nThe opponent plays always starting from a [b]0[/b]-state, that is, $(x,y)=(0,0)$\n\nThat is, he has \n - an even number of single coins, and\n - an even number of even piles\n\n[i][u]Even pile[/u]: a pile consisting of 2 or 4 or 6, ... coins[/i]\n\nCan he make a move to go to a [b]0[/b]-state?\nThe answer is [b][u]no[/u][/b]\n\nLet's see the possible opponent's moves:\n\n$\\hline$\n[color=green][size=150]Removing[/size]\nRemove 1 single coin . Bad move! $x$ becomes $1$\nRemove 1 coin from a pile with $2$ coins. Bad move! $x,y$ become $1$\nRemove 1 coin from a pile with $m\\geq 3$ coins. Then we have:\n$A_{m}$ decreases by $1$\n$A_{m-1}$ increases by $1$\n\nSome of $m,m-1$ is even, so $y$ becomes $1$[/color]\n$\\hline$\n\n[color=brown][size=150]Splitting[/size]\nSplit the pile $A_{m+n}$ into two nonzero piles $A_{m},A_{n}$ (suppose $m\\geq n$)\n\nIf $m=n=1$ then he obtains two single coins ($x$ doesn't change) but he destroys one double-coin pile (so $y$ becomes 1)\n\nIf $m>n=1$ then he obtains 1 single coin ($x$ becomes 1)\n\nIf $n\\geq 2:$\n\nIf $m+n$ was odd, then [u]exactly one[/u] of $m,n$ is even (Bad move; $y$ becomes 1)\n\nIf $m+n$ was even:\n\nEither $m,n$ are both even\nOr they are both odd\n\nAt either case he destroys the y-balance[/color]\n\n\nSo the opponent can't turn the game to [b]0[/b]-state\n$\\hline$[/hide]\n[hide=\"Your moves\"]You start from a [b]1[/b]-state. Let's see if there is a way to turn the game to a [b]0[/b]-state\n\nIn the following, $m$ will represent a natural number $m \\in \\{0,1,2,...\\}$\nThere are three cases such that a state is a [b]1[/b]-state:\n\n$(x,y) = (1,0)$\n$(x,y) = (0,1)$\n$(x,y) = (1,1)$\n\n[hide=\"(1,0)\"]There is an odd number of single coins. You want to make it even.\n[color=red][b]Solution:[/b] Just remove a single coin[/color] ($x$ becomes 0)\n$\\hline$[/hide][hide=\"(0,1)\"]\nYou have an odd number of even piles.\n$\\blacktriangleright$ If there is a pile with $m+3$ coins \n[color=red][b]Solution: [/b] remove a coin from it.[/color]\nYou obtain a pile with $m+2$ coins, but you loose the initial pile with $m+3$ coins\nExactly one of $m+3$ and $m+2$ is even, so you adjust the y-balance ($y$ becomes 0)\n\n$\\blacktriangleright$ If there is not a pile with $m+3$ coins. Then you have only double coins (don't care about the single coins)\n[color=red][b]Solution: [/b]Split one double-coin pile into two single coins.[/color] So you adjust the y-balance\n$\\hline$[/hide][hide=\"(1,1)\"]\nYou have an odd number of single coins and an odd number of even piles\n\n$\\blacktriangleright$ If there is a pile with $m+3$ coins, then \n[color=red][b]Solution: [/b]split 1 coin[/color]\nYou obtain a single coin so $x$ becomes 0\nYou obtain a pile with $m+2$ coins, but you loose the initial pile with $m+3$ coins. For the same reasons as above, you adjust the value of $y$\n\n$\\blacktriangleright$ If there is not a pile with $m+3$ coins. Then you have only double coins (and some odd single coins)\n[color=red][b]Solution: [/b]Just remove one coin from a double-coin pile.[/color]\nThe double coin becomes single.\nSo you obtain one single and loose one double, as a result both the values of $x$ and $y$ are fixed\n$\\hline$[/hide]\n\nYou can see the diagram:\n\n$(x,y)=\\{\\begin{array}{{l}(1,0) \\Rightarrow \\text{Remove a single coin}} \\\\ \\\\ (0,1) \\Rightarrow \\exists m>0 : A_{m+3}>0 \\{\\begin{array}{l}\\exists \\Rightarrow \\text{Remove a coin from }A_{m+3}\\\\ \\not\\exists\\Rightarrow \\text{Split a coin from }A_{2}\\end{array} \\\\ \\\\ (1,1) \\Rightarrow \\exists m>0 : A_{m+3}>0 \\{\\begin{array}{l}\\exists \\Rightarrow \\text{Split a coin from }A_{m+3}\\\\ \\not\\exists\\Rightarrow \\text{Remove a coin from }A_{2}\\end{array} \\end{array}$\n\n\n\n\n\n[/hide][/hide]" } { "Tag": [ "AoPSwiki", "Asymptote", "vector", "LaTeX", "AMC 8" ], "Problem": "The AoPSWiki Words of the Week for Nov 8-14 are [[Asymptote]] (the graph concept, not the vector graphics language used with LaTeX) and [[AMC historical results]].", "Solution_1": "Shouldn't each diff be posted by the admins?", "Solution_2": "By \"AMC Historical Results\", do you mean perfect scores on AMC 8, or perfect scores on the AMC 12, or participating in IMO when you were 10 years old?", "Solution_3": "[quote=\"1=2\"]By \"AMC Historical Results\", do you mean perfect scores on AMC 8, or perfect scores on the AMC 12, or participating in IMO when you were 10 years old?[/quote]The community should decide these issues on the article's discussion page. I started off the discussion but the page should be whatever people think is most relevant." } { "Tag": [], "Problem": "I know I can solve all of them and I'm more than happy to explain (I write my explanation on computers) to you. I might be able to PM you...", "Solution_1": "i lived in North Carolina too!\r\n\r\nIn Charlotte, Mecklenburg district" } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "hi guys, i'm gr 12 this year, so everything for math is over for me, but i just want to know how u guys handle this so i can kick myself for this.\r\n\r\nok, i love math, i've done all of the waterloo books and more\r\nwhen i was preparing for comc this year, every contest i can get around 65, but actual thing i got 52 \r\nsame for euclid 2003, every practice i can get 80+ but i got 55 in 2003\r\nthis year, i think i can only get 7*\r\nboth from being too nervous\r\nhow do u guys get rid of nervousness so that u can do ur best?\r\nplus what materials do u guys work on?", "Solution_1": "i know i can do a lot better, because last year, i got 32 on hypatia, which is a 83rd in canada\r\ni know it is not very good compare to some of u guys, but it's good for me\r\nbtw, i got 111 on fermat because of nervous, and retarded mistakes because i was expecting 130+ because that's what i've been getting on my practice contests\r\n\r\nlong story short, i got my butt wooped because i was nervous, any ideas how to get past this? (other than, do more questions)", "Solution_2": "who cares... i was nervous too", "Solution_3": "um, comments like yours don't help, bugzpodder.\r\n\r\nhe wants someone to give advice. you shouldn't have said anything if you didn't have any.", "Solution_4": "sorry i was trying to say everyone is nervous... it didnt came out right", "Solution_5": "It gets better after you participate in more contests (I don't mean just practice, rather actually sitting in the test room).\r\n\r\nAt first, I can barely handle the stress of a 30 min contest. After a while, you just get used to the idea of math competitions. Now, sitting 4.5h for say USAMO isn't a problem for me.\r\n\r\nAs well, after you do more problems (especially the proving ones), you get a sense of what solutions are right and what are not.\r\n\r\nThen again, of the two COMCs that I've done, I never quite figured out where I lost the marks. The Olympiads are definitely better, at least for me. For both USAMO and APMO, I got marks divisible by 7 :).\r\n\r\nLastly, one of my philosophies for math contests is that, one's performance on a contest is no greater than 33% of his/her actual ability.", "Solution_6": "Remember it is just you and the problem. It is an icky little thing, but just slay it, take your time as you do with the practices. If you are getting panicky, do something else for a little while. I sometimes go for a walk to the toilet and back. Or eat something. Or take a few deep breaths. Feel your mind with thoughts of the problem, and not with thoughts like \"I must solve this to get a silver\".... concentrate, basically.", "Solution_7": ":D :D :D :D" } { "Tag": [ "probability", "expected value" ], "Problem": "A certain coin is weighted such that the chance of flipping heads is $ \\frac{1}{3}$ and the chance of flipping tails is $ \\frac{2}{3}$. Suppose that we win $ \\$3$ if we flip a heads on a coin toss, but lose $ \\$2$ if we flip tails. What is the expected value, in dollars, of our winnings after one flip? Express your answer as a common fraction.", "Solution_1": "By definition, this is $ 3\\cdot \\frac{1}{3}\\plus{}(\\minus{}2)\\cdot \\frac{2}{3} \\equal{}\\boxed{\\frac{\\minus{}1}{3}}$" } { "Tag": [ "analytic geometry", "inequalities", "calculus", "derivative", "function" ], "Problem": "If $a,b,c \\in [0,\\infty),bc\\ne 0 ,$ show that \\[1 \\le \\frac{a+b+c}{\\sqrt{a^{2}+b^{2}+c^{2}}}\\le \\sqrt{3}\\; .\\; \\; \\; \\; (\\mbox{corrected !})\\] [size=75][b]Try to generalize these inequalities.[/b][/size]", "Solution_1": "I think it generalized should be something like this:\r\n$1\\leq\\frac{a_{1}+a_{2}+...+a_{n}}{\\sqrt{a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}}}\\leq\\sqrt n$\r\nGiven that they are all >=0 and at least one is non-zero.\r\n\r\nTo prove this:\r\nFirst, note that scaling all the variables by some factor k leads to the same equation since the k cancels in the numerator and denominator. This means that we can restrict our interval to some finite interval, because anything outside that can be achieved by scaling. Take all the variables to be in the interval [0,1].\r\n\r\nNow I want to say that you only need to consider the endpoints, that is each variable=0 or 1, but I'm not sure why that would be allowed. If that works, then you can see that the minimum is 1 and the maximum is $\\sqrt n$.\r\n\r\nAnother way to prove the minimum:\r\nTake all the variables to be coordinates in n-dimensional space. Then the minimum path from the origin to that point is the denominator. The numerator is an alternative path. That means the numerator>=denominator, and equality gives when it =1.", "Solution_2": "For [b]Wumbate[/b]'s generalization, RMS-AM gives the RHS: \\[\\frac{a_{1}+\\ldots+a_{n}}{n}\\le\\sqrt{\\frac{a_{1}^{2}+\\ldots+a_{n}^{2}}{n}},\\] \\[\\frac{a_{1}+\\ldots+a_{n}}{\\sqrt{a_{1}^{2}+\\ldots+a_{n}^{2}}}\\le\\sqrt{n}.\\] Since all the $a_{i}$ are non-negative, then \\[a_{1}^{2}+\\ldots+a_{n}^{2}\\le (a_{1}+\\ldots+a_{n})^{2}.\\] Hence, \\[1\\le\\frac{a_{1}+\\ldots+a_{n}}{\\sqrt{a_{1}^{2}+\\ldots+a_{n}^{2}}},\\] giving the LHS.", "Solution_3": "Using Cauchy in Engel form,\r\n$\\frac{a_{1}^{2}}{1}+\\frac{a_{2}^{2}}{1}+\\cdots+\\frac{a_{n}^{2}}{1}\\geq \\frac{{(a_{1}+a_{2}+\\cdots+a_{n})}^{2}}{n}$\r\n$\\Rightarrow \\frac{a_{1}+a_{2}+\\cdots+a_{n}}{\\sqrt{a_{1}^{2}+a_{2}^{2}+\\cdots+a_{n}^{2}}}\\leq \\sqrt{n}$", "Solution_4": "Actually, the inequality cannot be strengthened past $1 \\le \\frac{a_{1}+a_{2}+...+a_{n}}{\\sqrt{a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}}}\\le \\sqrt{n}$, because setting $a_{1}=a_{2}=...=a_{n-1}=0$ yields a value of 1 for the expression. It is easy to prove the RHS using Cauchy, and the LHS follows as long as the terms are nonnegative.", "Solution_5": "[quote=\"K81o7\"]Actually, the inequality cannot be strengthened past $1 \\le \\frac{a_{1}+a_{2}+...+a_{n}}{\\sqrt{a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}}}\\le \\sqrt{n}$, because setting $a_{1}=a_{2}=...=a_{n-1}=0$ yields a value of 1 for the expression. ........[/quote]\r\nThanks for interest,", "Solution_6": "Even if $bc=0$, you can let $a=0$, and take the limit of the expression as $b \\to 0$ (with $c \\neq 0$). That value is the same: ie, by continuity, you can always make the expression as close to 1 as you want it to be.", "Solution_7": "[quote]... following problem arise: [b]prove or disprove that [/b]...... FALSE ![/quote]\r\nSorry, thanks for interest,Alex", "Solution_8": "Even if b=c>0, then you have this:\r\n$\\frac{a+2b}{\\sqrt{a^{2}+2b^{2}}}$\r\nNow take a->infinity and/or b->0 and this tends to 1. So it would still have 1 as a lower bound.\r\n\r\n\r\n\r\nFor the general inequality, I still want to know if there is a way to reduce the \"checking\" to values of 0 or 1.\r\n\r\nI think it would be nice if you could reduce all the variables to only 0 or 1 - it's easy, it happens to work, but I'm not sure with the reasoning.\r\n\r\nIf you take a partial derivative of the function with respect to one of the variables, and set that to 0, you will find the max and mins. Now if you assume all the other variables are 0 or 1, then the only interesting value of that variable is 0 or 1.\r\n\r\nI'm not too good with multivariable stuff, so if someone else could explain, that would be nice." } { "Tag": [], "Problem": "When are the student solutions for round 4 going to be posted? I'm curious to see some of the commended techniques for solving this round's problems.", "Solution_1": "Probably next week.", "Solution_2": "They are now posted." } { "Tag": [ "linear algebra", "matrix", "vector", "probability and stats" ], "Problem": "hi I 'm looking for informations about this kind of regression.\r\n\r\nWe try to estimate a symetric $ m$x$ m$ matrix, $ A$ for the following regression model : \r\n\r\n\\[ y_ty_t^T\\minus{}I_m\\equal{}A(x_{t}x_{t}^T\\minus{}I_m)A\\plus{}E_t\\] \r\n\r\nwith $ y_t,x_t$ two observations vectors of dimension $ m$, and $ E_t$ is a symetric matrix representing model errors.\r\n\r\nTo estimate this, $ A$ is calcutlated by minimising the the average Fr\u00f6benius norm of $ E_t$ matrices over the set of $ A$ in the space of symetric matrices. \r\n\r\nThis method is apparently known as a \"non linear least square\" method. I have tried to find some information about this kind of regression and estimation method in this particuliar context but i couldn't find anything. Basically I have found this strange regression in a statistical article where it is presented as \"classic\" but as I don't have a very large culture in statistics I didn't got much success on my googling research. \r\n\r\nIf anyone could lead me to some references or web links I would appreciate \r\n\r\nbye", "Solution_1": "If it was presented in a research article then your best bet is to look at the references. I don't mean to state the obvious but if it is listed as \"classic\" then several other papers in the references should have information pertaining to it." } { "Tag": [ "calculus", "derivative", "algebra", "polynomial", "induction", "real analysis", "real analysis solved" ], "Problem": "i dont know whether the problem is simple or not.It looks very harmless but I cant solve it. \r\nFind the nth derivative of e^(-x 2 ) \r\n\r\nLooks like I am getting stupid day by day. \r\nPlease hurry I want the solution desperately.\r\n\r\nI can feel that there will be some reduction or recursive type of formula but I want the result to involve only n,e,x.", "Solution_1": "Do you know [b]Hermite[/b] polynomials?", "Solution_2": "$(e^{-x^2})^{(n)}=(-1)^nH_n(x)e^{-x^2}$ with \r\n\r\n$H_0(x)=1$\r\n$H_1(x)=2x$\r\n$H_2(x)=4x^2-2$\r\n\r\nBy induction \r\n\r\n$H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x)$", "Solution_3": "f=e^(-x^2)\r\nf'=-2xe^(-x^2)\r\nf''=(-2x)^2 e^(-x^2)\r\nf'''=(-2x)^3 e^(-x^2)\r\n\r\nTherefore\r\n\r\nf^(n)=(-2x)^n e^(-x^2)", "Solution_4": "Product rule...", "Solution_5": "Just heard about them.I will surely be glad if u tell me more about that.\r\nI know how to solve recursions & generating functions.\r\nThats it.", "Solution_6": "Sorry but I didnt understand what Jayscholar has tried to say\r\nIs it really that simple!", "Solution_7": "[quote=\"the game\"]Sorry but I didnt understand what Jayscholar has tried to say\nIs it really that simple![/quote]\r\nNo, it isn't that simple. Jaysscholar is simply wrong, which is what jmerry was saying in #5.", "Solution_8": "[quote=\"jmerry\"]Product rule...[/quote]\r\nwhich states that $ (fg)' = f'g+fg'$", "Solution_9": "Certainly I know product rule.How will the knowledge of product rule help the solution of Hermite polynomials.I presume it will be solved by generating functions.", "Solution_10": "Thought it to be a lot easier than that.At least the recursion relation looked simple.Thanks any way." } { "Tag": [ "number theory", "greatest common divisor", "modular arithmetic", "relatively prime", "Divisibility Theory" ], "Problem": "If $n$ is a natural number, prove that the number $(n+1)(n+2)\\cdots(n+10)$ is not a perfect square.", "Solution_1": "The largest power of $ 2$ that divides $ \\frac{(n\\plus{}10)!}{n!}$ is $ 32$, so it cannot be a perfect square.", "Solution_2": "That doesn't quite work ($ n \\equal{} 54$, for example).\r\n\r\n\r\nSuppose the product is a perfect square. Among the ten terms in the product:\r\n\r\n- five are even\r\n- at most two are odd and divisible by 3\r\n- at most one is odd and divisible by 5\r\n- at most one is odd and divisible by 7\r\n\r\nSo at least one term is not divisible by any of 2,3,5,7. This term is relatively prime with the others since any common divisor of two terms is at most 9. So this term is a perfect square, and at least $ 11^{2}$.\r\n\r\nNow note that this is the only square in our product, otherwise we would have a square greater than or equal to $ 11^{2}$ differing from another square by at most 9.\r\n\r\nIn particular, there must be exactly one term not divisble by any of 2,3,5,7. So we must have equality in all of our counts, and no odd term can be divisible by 15, 21, or 35.\r\n\r\nConsider the odd terms divisible by 3. Neither are divisible by 2,5,7, and the gcd between either of these terms and any another one is at most 9, therefore a power of 3.\r\n\r\nSo we can factor out the largest power of 3 from each of these terms. Each quotient that results is relatively prime to all other terms and are both squares. So it follows that each odd term divisible by 3 is either a square or three times a square.\r\n\r\nWe know from above that neither can be squares, so both are three times a square. But this also produces two squares that are too close (in particular, they differ by 2). This is a contradiction, so the product cannot be a square.", "Solution_3": "We can prove it by using problem A9 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=150377]here[/url].\r\n\r\nFrom ten consecutive numbers we have one which is relatively prime to others. Remove this number from the list we have nine numbers which we can chose five from them which are consecutive. For five consecutive we will prove that there's one prime to others else. Indeed, there are at most two ones which is divisible by $ 3$, and if there're two ones then one is odd and one is even. In case we remove one which is even, so one left there're at most two even numbers from $ 4$ number left or $ 3$ for the left case. Remove them we have one odd left or two odd but in which there's one is not divisible by $ 3$. So there's one number which is odd, not divisible by $ 3$. This number from list $ 5$ consecutive numbers is relatively prime to others else. Remove it we get four numbers which always find two are consecutive, and then of cause relative each other. So we could find three numbers which is relatively to any $ 9$ numbers else. Now note that if $ a$ is square then $ a\\plus{}1,a\\plus{}2$ are not and $ a\\minus{}1,a\\minus{}2$ are not of $ a>2$. So in ten consecutive numbers it could not contains more than two squares. This proves the problem.", "Solution_4": "[quote=\"pluricomplex\"]\nSo we could find three numbers which is relatively to any $ 9$ numbers else. [/quote]\r\n\r\nthis isn't true. for example, take n=209, then only 211 is relatively prime to other nine numbers because 210 212 214 216 218 are even numbers;213 216 219 are multiple of 3; 210 and 215 are multiple of 5; 210 and 217 are multiple of 7\r\n\r\nthe mistake is that you may choose a number in the five consecutive numbers which is relatively prime to other $ 4$ numbers, but it maybe not relatively prime to other $ 9$ numbers.", "Solution_5": "[quote=\"Peter\"]If $ n$ is a natural number, prove that the number $ (n \\plus{} 1)(n \\plus{} 2)\\cdots(n \\plus{} 10)$ is not a perfect square.[/quote]\r\n\r\nThere is a much stronger result proven by Erdos, \"The product of consecutive integers is never a power\" you can check it at:\r\n[url]http://www2.renyi.hu/~p_erdos/1975-46.pdf[/url]", "Solution_6": "There was such a problem at Latvian Open Olympiad in Mathematics:\r\n\r\nIt is given that $ n$ is a natural even number. Let $ R \\equal{} n\\left(n\\plus{}1\\right)\\left(n\\plus{}2\\right)\\left(n\\plus{}3\\right)$.\r\n[list][i]a[/i]) Can $ R$ be a square of a natural number?[/list][list][i]b[/i]) Can $ R$ be a qube of a natural number?[/list]", "Solution_7": "I think all your solutions don't work, why?\nWe note that $n \\equiv 1 \\pmod{8},n \\equiv 1 \\pmod{9},n \\equiv 1 \\pmod{25}$\n$n \\equiv 43 \\pmod{49},n \\equiv 111 \\pmod{11^2}$, \nBy Chinese remainder theorem, there always exists $n$ satisfies all conditions above, then $11^2.2^8.3^4.7^2.5^2||(n+1)(n+2)...(n+10)$ so $mod(11,2,3,7,5)$ don't work!, In that cases, we only can consider the modular of prime number which is large enough but It's impossible", "Solution_8": "[quote=\"mszew\"][quote=\"Peter\"]If $ n$ is a natural number, prove that the number $ (n \\plus{} 1)(n \\plus{} 2)\\cdots(n \\plus{} 10)$ is not a perfect square.[/quote]\n\nThere is a much stronger result proven by Erdos, \"The product of consecutive integers is never a power\" you can check it at:\n[url]http://www2.renyi.hu/~p_erdos/1975-46.pdf[/url][/quote]\n\nThe URL cannot use anymore.\nI found a new one [url]http://www.renyi.hu/~p_erdos/1975-46.pdf[/url]", "Solution_9": "just consider divisibility by all primes less then $10$ and you'll face contradiction" } { "Tag": [], "Problem": "[color=purple]Trang chu\u0309 cu\u0309a ML co\u0301 ph\u00e2\u0300n Problem of day anh Tu\u00e2n co\u0301 th\u00ea\u0309 cho em bi\u00ea\u0301t \u0111i\u00ea\u0300u ki\u00ea\u0323n \u0111\u00ea\u0309 ba\u0300i toa\u0301n cu\u0309a mi\u0300nh \u0111\u01b0\u01a1\u0323c cho\u0323n ko a\u0323 :) \n Thanks s\u00ea\u0301p :D [/color]", "Solution_1": "T\u00f4i c\u0169ng mu\u1ed1n h\u1ecfi v\u1ea5n \u0111\u1ec1 n\u00e0y. Nh\u01b0ng t\u00f4i ngh\u0129 mu\u1ed1n \u0111\u01b0a b\u00e0i l\u00ean l\u00e0 do c\u00e1c master l\u00e0m." } { "Tag": [ "conics", "geometry proposed", "geometry" ], "Problem": "Is it possible to find three conics in the plane such that any straight line in the plane intersects at least two of the conics and through any point of the plane pass tangents to at least two of them?", "Solution_1": "Problem is too ambiguous..." } { "Tag": [ "topology", "function", "real analysis", "real analysis solved" ], "Problem": "could prove that $\\mathbb{R}^{\\mathbb{R}}$ is seprable or disprove it .it is not that hard that $\\mathbb{R}^{P(P(\\mathbb{R}))}$ is not seprable with product topology.", "Solution_1": "What is product topology?", "Solution_2": "Lemma: The cardinality of a separable space is at most that of the real numbers.\r\nProof: Take a countable dense subset. Every element of the space can be expressed as the limit of a sequence in that set. There are as many sequences as there are real numbers.\r\n\r\nNow, $\\mathbb{R}^{\\mathbb{R}}$ has larger cardinality than $\\mathbb{R}$, so it cannot be separable.", "Solution_3": "[quote=\"jmerry\"] Every element of the space can be expressed as the limit of a sequence in that set. [/quote]\r\nI would really love to see your proof of that, jmerry. It is not obvious at all if the base of topology at a point is not countable...", "Solution_4": "My previous post assumed a metric topology.", "Solution_5": "Yeah, I thought so too. The problem is that $\\mathbb R^{\\mathbb R}$ with the product topology is not metrizable...", "Solution_6": "Yes, I think it's not even first-countable. :?", "Solution_7": "yes not being first countable is almost obvious you can take these points :all with finite zero and the other index is 1 then the point that has all zero in all index is limit point of these sequence but you can not find the sequense such that its limit is all zero in all index so it is not first countable .", "Solution_8": "If the product topology is weak i.e the basis is of the form $\\Pi{ \\{U_{\\alpha}\\}}_{\\alpha \\in \\mathbb{R}}$ where all but finite $U_{\\alpha}$ are the whole $\\mathbb{R}$ and the others are open sets in $\\mathbb{R}$, then I think we can show that the space is seperable as follow :\r\n Let $I$ be the set of all horizontal intervals in $\\mathbb{R}^2$ whose endpoints have rational coordinates. This set is countable. Let $Q$ be the set of function $f$ ( and there fore $Q$ is a subset of $\\mathbb{R}^{\\mathbb{R}}$) whose graph deleted the points $(x,f(x)), f(x)=0$ is consisted of finite number of intervals in $I$. $Q$ is also countable. We will prove that $Q$ is densed in $\\mathbb{R}^{\\mathbb{R}}$.\r\n Now let $\\Pi {\\{{U_{\\alpha}\\}}_{\\alpha \\in \\mathbb{R}}}$ be an open set in the space as above.\r\nCall those open sets $U_{\\alpha}$ which are not the whole $\\mathbb{R}$, $U_1,\\ldots ,U_n$ and their indices are $\\alpha_1 < \\ldots <\\alpha_n$.\r\nLet $p_1,\\ldots ,p_n$ be rational points in $U_1,\\ldots ,U_n$ respectively. Choose rationals $s_i,t_i$ such that $s_1 < \\alpha_1 < t_1 < \\alpha_2 < \\ldots < \\alpha_n < t_n$\r\nConsider $f \\in Q$ the function that $f(s_i,t_i) = p_i$ and $0$ otherwise. It is easy to see that $f \\in \\Pi {\\{{U_{\\alpha}\\}}_{\\alpha \\in \\mathbb{R}}}$ because $f(\\alpha_i) = p _i\\in U_i$ and for other values $\\alpha$ we have $ f(\\alpha) \\in U_{\\alpha}$ because $U_{\\alpha} = \\mathbb{R}$", "Solution_9": "That looks like it works. This is still a monster.\r\n\r\nThe theorem proved here: if $A$ is separable and $B$ has a Hausdorff topology with a countable base, $A^B$ is separable in the product topology. This probably restricts the cardinality of $B$ to no more than that of the real numbers; I don't believe there are any separable (and Hausdorff) topological spaces larger in cardinality than $\\mathbb{R}^\\mathbb{R}.$", "Solution_10": "[quote=\"jmerry\"]I don't believe there are any separable (and Hausdorff) topological spaces larger in cardinality than $\\mathbb{R}^\\mathbb{R}.$[/quote]\r\nThat's right and easy to prove. Let $X$ be a Hausdorff space and let $D$ be a countable dense subset of $X$. For each point $x\\in X$ consider the set of all subsets of $D$ of the form $U\\cap D$ where $U$ is a neighborhood of $x$ in the topology of $X$. It is easy to check that the corresponding families of subsets are different for different $x$. Thus the cardinality of $X$ does not exceed that of $P(P(D))$, which is the cardinality of $\\mathbb R^{\\mathbb R}$." } { "Tag": [], "Problem": "is anyone else taking idea math? which class?\r\nI'm taking \"Math Circles\"", "Solution_1": "I might take Math circles at session 2." } { "Tag": [], "Problem": "hi all, pls help mi with this... :) \r\n\r\nwhy does the cis isomer has lower melting point but higher boiling point, solubility and dipole moment than the trans isomer?\r\n\r\n\r\nthx lots", "Solution_1": "Welcome to this forum.\r\n\r\nThe molecules of a cis isomer, as they are less simetric relative to the molecules of the trans isomer, don't pack so readily in an ordered crystaline structure, and so their melting points are, in general, higher.\r\n\r\nIn the cis isomer, the dipole moments of the two cis bonds add, whereas in trans isomer they cancel each other (this is specially true when the groups are the same). So, cis isomer has a greater dipole moment, i.e., is more polar. This higher polarity os cis isomer molecules explain why they have higher boiling points (stronger intermolecular forces) and better solubilities (stronger interactions with solvent)." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "find all [b]x,y[/b] and [b]z[/b] in positive integer:\r\n[b]1!+2!+...+x!=y z. [/b]", "Solution_1": "I think you mean $z \\geq 2$ otherwise there are many solutions :D \r\nAssume that $ z \\geq 2$ and $x >1$ (we clearly have that $(1,1,z)$ is a solution).\r\nLet $S(x)= 1! + \\cdots + x!$\r\n\r\n- If $z = 2$ : Then for $i \\geq 5$, we have $i! = 0 \\mod [10]$. Thus, if $x \\geq 4$, we have $S(x) = S(4) = 3 \\mod [10]$. But no square has unit digit $3$, so there is no solution in that case.\r\nDirect checking leads to the only solution $(3,3,2)$.\r\n\r\n- If $z \\geq 3$ : As above, we see that for $x \\geq 2$ we have $S(x) = 0 \\mod [3]$, so that $3$ has to divide $y$. But, since $ z \\geq 3$, it follows that $S(x) = 0 \\mod [27]$. It is easy to verify that this cannot happen when $x \\geq 8$.\r\nDirect checking shows that there is no solution for $x \\leq 7$, and we are done.\r\n\r\nTherefore, the solutions are $(x,y,z) = (1,1,z)$ and $(3,3,2)$.\r\n\r\nPierre." } { "Tag": [ "geometry", "inequalities", "trigonometry", "inequalities unsolved" ], "Problem": "Prove that for all acute triangle holds\r\n$\\sqrt3R\\geq\\frac{a^{2}+b^{2}+c^{2}}{a+b+c}.$", "Solution_1": "[quote=\"arqady\"]Prove that for all acute triangle holds\n$\\sqrt3R\\geq\\frac{a^{2}+b^{2}+c^{2}}{a+b+c}.$[/quote]\r\n\r\nI think it is not true.. Try $A \\rightarrow \\frac{\\pi}{2}$ and $B=C$.", "Solution_2": "[quote=\"Sung-yoon Kim\"][quote=\"arqady\"]Prove that for all acute triangle holds\n$\\sqrt3R\\geq\\frac{a^{2}+b^{2}+c^{2}}{a+b+c}.$[/quote]\n\nI think it is not true.. Try $A \\rightarrow \\frac{\\pi}{2}$ and $B=C$.[/quote]\r\nIf $a=\\sqrt2$ and $b=c=1$ then $R=\\frac{1}{\\sqrt2}$ and\r\n$\\sqrt3R\\geq\\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\\Leftrightarrow\\sqrt{\\frac{3}{2}}\\geq\\frac{4}{2+\\sqrt2},$ which true.", "Solution_3": "What about $A \\to 0$, $B=C$? Then $b=c$ and $a \\to 0$, but $R \\to \\frac{b}{2}$, right? (I suck at geometry so I hope I haven't misunderstood notation.) Then your expression approaches $\\frac{\\sqrt{3}}{2}> 1$, which false.", "Solution_4": "Thank you, Xevarion.", "Solution_5": "[color=darkblue][b]Remark.[/b] Suppose the isosceles triangle $ABC$ ($b=c$). If $a\\searrow 0$, then $b=c\\nearrow 2R$.\nThus, $\\frac{a^{2}+b^{2}+c^{2}}{a+b+c}=\\frac{a^{2}+2b^{2}}{a+2b}\\rightarrow b\\rightarrow 2R>R\\sqrt 3\\ !$ For example, the triangle $(1,4,4)\\ .$\n\n[b]Proposed problem.[/b] Let $ABC$ be a triangle for which there is $0<2r4\\ .$\nThe last inequality becomes :\n$(3t+2)^{2}(t-4)\\le 9t(t-1)^{2}\\Longleftrightarrow$ $6t^{2}+53t+16\\ge 0$, what is truly.\n[b]Remark.[/b] If $2r90^{\\circ}$ (obtuse $\\triangle$ : 2,3,4) ; if $x=4r$, then $C=90^{\\circ}$ (right $\\triangle$ : 3,4,5) ; if $x>4r$, then $C<90^{\\circ}$ (acute $\\triangle$ : 4,5,6 ).\n\n[b]Another counter example :[/b] $(1,3,\\sqrt{10})$ - right triangle.\n\n[b]Question.[/b] Exist $0<\\epsilon ,\\delta <20^{\\circ}$ so that the obtuse triangle $(\\cos \\epsilon\\ ,\\ \\cos (\\epsilon+\\delta )\\ ,\\ \\sin \\delta )$ with $R=1$ doesn't verify the relation $(1)$ ?? Indeed, for $0<\\epsilon ,\\delta <20^{\\circ}$ we have :\n$\\cos \\epsilon >\\cos (\\epsilon+\\delta)\\ ,\\ \\cos \\epsilon >\\sin\\delta\\ ;$ $\\cos\\epsilon <\\cos (\\epsilon+\\delta)+\\sin\\delta\\ ;$ $\\cos^{2}\\epsilon>\\cos^{2}(\\epsilon+\\delta)+\\sin^{2}\\delta\\ .$ [/color]", "Solution_6": "Thank you very much, Virgil!", "Solution_7": "ARQADY, see the my last message : I added a question with which I tried to find usually a counter exemple - obtuse triangle !" } { "Tag": [ "inequalities", "rearrangement inequality", "inequalities proposed" ], "Problem": "Let $a+b+c+d=4$ and $a,b,c,d\\geq0$.\r\nProve that $a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab\\leq4$", "Solution_1": "[quote=\"Parkdoosung\"]Let $a+b+c+d=4$ and $a,b,c,d\\geq0$.\nProve that $a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab\\leq4$[/quote]\r\nIt is necessary to prove, that $(a+b+c+d)^{4}\\geq64(a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab).$\r\nIf $a=\\min\\{a,b,c,d\\}$ and $b=a+x,$ $c=a+y,$ $d=a+z$ then it is killing. But it is very ugly.", "Solution_2": "[quote=\"Parkdoosung\"]Let $a+b+c+d=4$ and $a,b,c,d\\geq0$.\nProve that $a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab\\leq4$[/quote]\r\n\r\nLet ${p,q,r,s}={a,b,c,d}$ and $p \\geq q \\geq r \\geq s$. Then by rearrangement inequality,\r\n$a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab=a(abc)+b(bcd)+c(cda)+d(dab)$\r\n$\\leq p(pqr)+q(pqs)+r(prs)+s(qrs)=(pq+rs)(pr+qs)$\r\n$\\leq (\\frac{pq+rs+pr+qs}{2})^{2}=\\frac{1}{4}((p+s)(q+r))^{2}$\r\n$\\leq \\frac{1}{4}((\\frac{p+q+r+s}{2})^{2})^{2}$\r\n$=4$.\r\n\r\nEquality holds iff q=r=1 and p+s=2. So equality holds if two of them are equal to 1. Applying this, we can get the equality conditions $(a,b,c,d)=(1,1,1,1),(2,1,1,0)$ or any cyclic forms.\r\n\r\nNice problem!", "Solution_3": "And by this idea, we can solve that \r\n$a^{2}b+b^{2}c+c^{2}a \\leq \\frac{4}{27}$ if $a+b+c=1$\r\nwhich was from Canada.\r\n\r\nIf $\\{ p,q,r \\}= \\{ a,b,c \\}$, $p \\geq q \\geq r$, then as $pq \\geq pr \\geq qr$,\r\n$a^{2}b+b^{2}c+c^{2}a =a(ab)+b(bc)+c(ca) \\leq p(pq)+q(pr)+r(qr)$\r\n$=q(p^{2}+pr+r^{2}) \\leq q(p+r)^{2}= \\frac{1}{2}(2q)(p+r)(p+r)$\r\n$\\leq \\frac{1}{2}(\\frac{(2q)+(p+r)+(p+r)}{3})^{3}$\r\n$=\\frac{1}{2}(\\frac{2}{3})^{3}=\\frac{4}{27}$.", "Solution_4": "[quote=\"Sung-yoon Kim\"]And by this idea, we can solve that \n$a^{2}b+b^{2}c+c^{2}a \\leq \\frac{4}{27}$ if $a+b+c=1$\nwhich was from Canada.\n\nIf $\\{ p,q,r \\}= \\{ a,b,c \\}$, $p \\geq q \\geq r$, then as $pq \\geq pr \\geq qr$,\n$a^{2}b+b^{2}c+c^{2}a =a(ab)+b(bc)+c(ca) \\leq p(pq)+q(pr)+r(qr)$\n$=q(p^{2}+pr+r^{2}) \\leq q(p+r)^{2}= \\frac{1}{2}(2q)(p+r)(p+r)$\n$\\leq \\frac{1}{2}(\\frac{(2q)+(p+r)+(p+r)}{3})^{3}$\n$=\\frac{1}{2}(\\frac{2}{3})^{3}=\\frac{4}{27}$.[/quote]\r\n\r\nwo also have$a^{2}b+b^{2}c+c^{2}a+abc \\leq \\frac{4}{27}$ if$a+b+c=1$", "Solution_5": "Are there other inequalities where we can apply this :?:", "Solution_6": "Well, I found that $a^{2}bcd+b^{2}cde+c^{2}dea+d^{2}eab+e^{2}abc \\leq 5(\\frac{a+b+c+d+e}{5})^{5}$ can be solved by the same idea. I tried to generalize for the number of variables, but failed. But I got that if $2n$ satisfies, then $2n+1$ satisfies, so the remaining cases are even numbers bigger than or equal to 6.", "Solution_7": "[quote=\"Sung-yoon Kim\"] I found that $a^{2}bcd+b^{2}cde+c^{2}dea+d^{2}eab+e^{2}abc \\leq 5(\\frac{a+b+c+d+e}{5})^{5}$ can be solved by the same idea. [/quote]\r\nI see only that if $a+b+c+d+e=5$ and $a,b,c,d,e$ are non-negative then $a^{2}bcd+b^{2}cde+c^{2}dea+d^{2}eab+e^{2}abc+3abcde\\leq8.$", "Solution_8": "[quote=\"Sung-yoon Kim\"]I found that $a^{2}bcd+b^{2}cde+c^{2}dea+d^{2}eab+e^{2}abc \\leq 5(\\frac{a+b+c+d+e}{5})^{5}$ can be solved by the same idea.[/quote]\r\nNow I see it! :lol:", "Solution_9": "Also this one:\r\n\r\nIf:$x,y,z \\geq 0$, then: $3(x^{2}+y^{2}+z^{2}) \\geq 4(xy+yz)$", "Solution_10": "[quote=\"Sung-yoon Kim\"][quote=\"Parkdoosung\"]Let $a+b+c+d=4$ and $a,b,c,d\\geq0$.\nProve that $a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab\\leq4$[/quote]\n\nLet ${p,q,r,s}={a,b,c,d}$ and $p \\geq q \\geq r \\geq s$. Then by rearrangement inequality,\n$a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab=a(abc)+b(bcd)+c(cda)+d(dab)$\n$\\leq p(pqr)+q(pqs)+r(prs)+s(qrs)=(pq+rs)(pr+qs)$\n$\\leq (\\frac{pq+rs+pr+qs}{2})^{2}=\\frac{1}{4}((p+s)(q+r))^{2}$\n$\\leq \\frac{1}{4}((\\frac{p+q+r+s}{2})^{2})^{2}$\n$=4$.\n\nEquality holds iff q=r=1 and p+s=2. So equality holds if two of them are equal to 1. Applying this, we can get the equality conditions $(a,b,c,d)=(1,1,1,1),(2,1,1,0)$ or any cyclic forms.\n\nNice problem![/quote]\r\n\r\nI think this ineq is cyclic, it isn't symmetric sothat we can't assume that $a\\ge b\\ge c\\ge d$. :maybe:", "Solution_11": "We are not assuming that :wink: .", "Solution_12": "[quote=\"M4RI0\"]We are not assuming that :wink: .[/quote]\r\nOh, how stupid I am. I read too carelessly. Thank you very much", "Solution_13": "[quote=\"Sung-yoon Kim\"]And by this idea, we can solve that \n$a^{2}b+b^{2}c+c^{2}a \\leq \\frac{4}{27}$ if $a+b+c=1$\nwhich was from Canada.\n\nIf $\\{ p,q,r \\}= \\{ a,b,c \\}$, $p \\geq q \\geq r$, then as $pq \\geq pr \\geq qr$,\n$a^{2}b+b^{2}c+c^{2}a =a(ab)+b(bc)+c(ca) \\leq p(pq)+q(pr)+r(qr)$\n$=q(p^{2}+pr+r^{2}) \\leq q(p+r)^{2}= \\frac{1}{2}(2q)(p+r)(p+r)$\n$\\leq \\frac{1}{2}(\\frac{(2q)+(p+r)+(p+r)}{3})^{3}$\n$=\\frac{1}{2}(\\frac{2}{3})^{3}=\\frac{4}{27}$.[/quote]\r\n\r\nEdited: Sorry I don't really understand the solution but now I do.\r\n\r\nHmm.. that's different with assuming without loss of generality\r\n\r\nHahaha I think I am influenced by the post that tells it is cyclic and can't think clearly :D", "Solution_14": "[quote=\"BoesFX\"]\nBut, this is the incomplete solution.[/quote]\r\nWhy? Thank you.", "Solution_15": "[quote=\"Sung-yoon Kim\"]And by this idea, we can solve that \n$a^{2}b+b^{2}c+c^{2}a \\leq \\frac{4}{27}$ if $a+b+c=1$\nwhich was from Canada.\n\nIf $\\{ p,q,r \\}= \\{ a,b,c \\}$, $p \\geq q \\geq r$, then as $pq \\geq pr \\geq qr$,\n$a^{2}b+b^{2}c+c^{2}a =a(ab)+b(bc)+c(ca) \\leq p(pq)+q(pr)+r(qr)$\n$=q(p^{2}+pr+r^{2}) \\leq q(p+r)^{2}= \\frac{1}{2}(2q)(p+r)(p+r)$\n$\\leq \\frac{1}{2}(\\frac{(2q)+(p+r)+(p+r)}{3})^{3}$\n$=\\frac{1}{2}(\\frac{2}{3})^{3}=\\frac{4}{27}$.[/quote]\r\nWhy can we assume $p\\geq q\\geq r$? \r\n\r\nIs $p,q,r$ a permutation of $a,b,c$?", "Solution_16": "[quote=\"hitek\"]\nWhy can we assume $p\\geq q\\geq r$? \n\nIs $p,q,r$ a permutation of $a,b,c$?[/quote]\nYes!\n[quote=\"Sung-yoon Kim\"]\n\nIf $\\{ p,q,r \\}= \\{ a,b,c \\}$, $p \\geq q \\geq r$, then...[/quote] :wink:", "Solution_17": "Suppose that $ (x;y;z;t)$ is a permutation of $ a;b;c;d$ such that $ x\\geq y\\geq z\\geq t$\r\n=>$ xyz\\geq xyt\\geq xzt\\geq yzt$\r\nand $ x.xyz \\plus{} y.xyt \\plus{} z.xzt \\plus{} t.yzt\\geq a^{2}bc \\plus{} b^{2}cd \\plus{} c^{2}da \\plus{} d^{2}ab$\r\nThen \r\n$ x.xyz \\plus{} y.xyt \\plus{} z.xzt \\plus{} t.yzt \\equal{} (xy \\plus{} zt)(xz \\plus{} yt)\\leq\\frac {1}{4}(xz \\plus{} xy \\plus{} yt \\plus{} zt)^{2}\\leq 4$\r\nDone!", "Solution_18": "[quote=\"M4RI0\"]Also this one:\n\nIf:$ x,y,z\\geq 0$, then: $ 3(x^{2}\\plus{}y^{2}\\plus{}z^{2})\\geq 4(xy\\plus{}yz)$[/quote]\r\n$ (3x^{2}\\plus{}\\frac{3y^{2}}{2})\\plus{}(3z^{2}\\plus{}\\frac{3y^{2}}{2})\\geq\\frac{6}{\\sqrt{2}}(xy\\plus{}yz)\\geq 4(xy\\plus{}yz)$", "Solution_19": "I mean, there's an identical solution above, chien than.", "Solution_20": "And we prove by the same method:\r\n\r\n$ \\sum \\frac {a_k}{k^2} \\geq \\sum \\frac {1}{k}$\r\n\r\nFor $ a_k$ positive integers and distinct (ai#aj for all i,j)\r\nWich is from IMO 78\r\n :) :wink:", "Solution_21": "[quote=\"Sung-yoon Kim\"]And by this idea, we can solve that \n$ a^{2}b \\plus{} b^{2}c \\plus{} c^{2}a \\leq \\frac {4}{27}$ if $ a \\plus{} b \\plus{} c \\equal{} 1$\nwhich was from Canada.\n\nIf $ \\{ p,q,r \\} \\equal{} \\{ a,b,c \\}$, $ p \\geq q \\geq r$, then as $ pq \\geq pr \\geq qr$,\n$ a^{2}b \\plus{} b^{2}c \\plus{} c^{2}a \\equal{} a(ab) \\plus{} b(bc) \\plus{} c(ca) \\leq p(pq) \\plus{} q(pr) \\plus{} r(qr)$\n$ \\equal{} q(p^{2} \\plus{} pr \\plus{} r^{2}) \\leq q(p \\plus{} r)^{2} \\equal{} \\frac {1}{2}(2q)(p \\plus{} r)(p \\plus{} r)$\n$ \\leq \\frac {1}{2}(\\frac {(2q) \\plus{} (p \\plus{} r) \\plus{} (p \\plus{} r)}{3})^{3}$\n$ \\equal{} \\frac {1}{2}(\\frac {2}{3})^{3} \\equal{} \\frac {4}{27}$.[/quote]\r\n\r\n$ a,b,c$ must me positive, I think", "Solution_22": "[quote=\"Sung-yoon Kim\"][quote=\"Parkdoosung\"]Let $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 4$ and $ a,b,c,d\\geq0$.\nProve that $ a^{2}bc \\plus{} b^{2}cd \\plus{} c^{2}da \\plus{} d^{2}ab\\leq4$[/quote]\n\nLet $ {p,q,r,s} \\equal{} {a,b,c,d}$ and $ p \\geq q \\geq r \\geq s$. Then by rearrangement inequality,\n$ a^{2}bc \\plus{} b^{2}cd \\plus{} c^{2}da \\plus{} d^{2}ab \\equal{} a(abc) \\plus{} b(bcd) \\plus{} c(cda) \\plus{} d(dab)$\n$ \\leq p(pqr) \\plus{} q(pqs) \\plus{} r(prs) \\plus{} s(qrs) \\equal{} (pq \\plus{} rs)(pr \\plus{} qs)$\n$ \\leq (\\frac {pq \\plus{} rs \\plus{} pr \\plus{} qs}{2})^{2} \\equal{} \\frac {1}{4}((p \\plus{} s)(q \\plus{} r))^{2}$\n$ \\leq \\frac {1}{4}((\\frac {p \\plus{} q \\plus{} r \\plus{} s}{2})^{2})^{2}$\n$ \\equal{} 4$.\n\nEquality holds iff q=r=1 and p+s=2. So equality holds if two of them are equal to 1. Applying this, we can get the equality conditions $ (a,b,c,d) \\equal{} (1,1,1,1),(2,1,1,0)$ or any cyclic forms.\n\nNice problem![/quote]\r\n\r\nI don't see the point for introducing $ p, q, r, s$ to replace $ a, b, c, d$.\r\nAlso, how do one prove that \r\n\r\n$ a(abc) \\plus{} b(bcd) \\plus{} c(cda) \\plus{} d(dab)$\r\n$ \\leq p(pqr) \\plus{} q(pqs) \\plus{} r(prs) \\plus{} s(qrs)$?", "Solution_23": "[quote=\"Sung-yoon Kim\"][quote=\"Parkdoosung\"]Let $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} 4$ and $ a,b,c,d\\geq0$.\nProve that $ a^{2}bc \\plus{} b^{2}cd \\plus{} c^{2}da \\plus{} d^{2}ab\\leq4$[/quote]\n\n[b]Let $ {p,q,r,s} \\equal{} {a,b,c,d}$ and $ p \\geq q \\geq r \\geq s$.[/b] Then by rearrangement inequality,\n$ a^{2}bc \\plus{} b^{2}cd \\plus{} c^{2}da \\plus{} d^{2}ab \\equal{} a(abc) \\plus{} b(bcd) \\plus{} c(cda) \\plus{} d(dab)$\n$ \\leq p(pqr) \\plus{} q(pqs) \\plus{} r(prs) \\plus{} s(qrs) \\equal{} (pq \\plus{} rs)(pr \\plus{} qs)$\n$ \\leq (\\frac {pq \\plus{} rs \\plus{} pr \\plus{} qs}{2})^{2} \\equal{} \\frac {1}{4}((p \\plus{} s)(q \\plus{} r))^{2}$\n$ \\leq \\frac {1}{4}((\\frac {p \\plus{} q \\plus{} r \\plus{} s}{2})^{2})^{2}$\n$ \\equal{} 4$.\n\nEquality holds iff q=r=1 and p+s=2. So equality holds if two of them are equal to 1. Applying this, we can get the equality conditions $ (a,b,c,d) \\equal{} (1,1,1,1),(2,1,1,0)$ or any cyclic forms.\n\nNice problem![/quote]\r\nI don't know why we can assume this.", "Solution_24": "[quote=\"Sung-yoon Kim\"]Well, I found that $a^{2}bcd+b^{2}cde+c^{2}dea+d^{2}eab+e^{2}abc \\leq 5(\\frac{a+b+c+d+e}{5})^{5}$ can be solved by the same idea. I tried to generalize for the number of variables, but failed. But I got that if $2n$ satisfies, then $2n+1$ satisfies, so the remaining cases are even numbers bigger than or equal to 6.[/quote]\nLet $\\{a,b,c,d,e\\}=\\{p,q,r,s,t\\}$. WLOG $p\\le q\\le r\\le s\\le t$ and similarly we have $pqrs\\le prst\\le pqst\\le pqrt\\le qrst$. Therefore, by rearrangement:\n\n\n \\begin{eqnarray} &&\\text{LHS}=a(abcd)+b(bcde)+c(cdea)+d(deab) \\\\ \n&\\le& p(pqrs)+q(prst)+r(pqst)+s(pqrt)+t(qrst) \\\\ \n&=& qrs(p^2+t^2+3pt)=qrs[(p+t)^2+pt] \\\\ \n&\\le& \\frac{(2q)(2r)(2s)(p+t)(p+t)}{8}+pqrst \\\\\n &\\le& \\frac18(\\frac{2q+2r+2s+p+t+p+t}{5})^5+(\\frac{p+q+r+s+t}{5})^5 \\\\ \n&=& 5(\\frac{a+b+c+d+e}{5})^5 \\end{eqnarray}", "Solution_25": "[quote=\"Binomial-theorem\"]Let $\\{a,b,c,d,e\\}=\\{p,q,r,s,t\\}$. WLOG $p\\le q\\le r\\le s\\le t$ and similarly we have $pqrs\\le prst\\le pqst\\le pqrt\\le qrst$. [...][/quote]\nIt seems to be false :maybe:", "Solution_26": "[quote=\"shuchang\"][quote=\"Sung-yoon Kim\"]And by this idea, we can solve that \n$a^{2}b+b^{2}c+c^{2}a \\leq \\frac{4}{27}$ if $a+b+c=1$\nwhich was from Canada.\n\nIf $\\{ p,q,r \\}= \\{ a,b,c \\}$, $p \\geq q \\geq r$, then as $pq \\geq pr \\geq qr$,\n$a^{2}b+b^{2}c+c^{2}a =a(ab)+b(bc)+c(ca) \\leq p(pq)+q(pr)+r(qr)$\n$=q(p^{2}+pr+r^{2}) \\leq q(p+r)^{2}= \\frac{1}{2}(2q)(p+r)(p+r)$\n$\\leq \\frac{1}{2}(\\frac{(2q)+(p+r)+(p+r)}{3})^{3}$\n$=\\frac{1}{2}(\\frac{2}{3})^{3}=\\frac{4}{27}$.[/quote]\n\nwo also have$a^{2}b+b^{2}c+c^{2}a+abc \\leq \\frac{4}{27}$ if$a+b+c=1$[/quote]\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=534056&p=3541033#p3541033\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=768&sid=c3c5b1d6f4ddcaf0a1d3e963d9763e32#p768\nhttps://artofproblemsolving.com/community/c6h1136308p5308238", "Solution_27": "[quote=Sung-yoon Kim][quote=\"Parkdoosung\"]Let $a+b+c+d=4$ and $a,b,c,d\\geq0$.\nProve that $a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab\\leq4$[/quote]\n\nLet ${p,q,r,s}={a,b,c,d}$ and $p \\geq q \\geq r \\geq s$. Then by rearrangement inequality,\n$a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab=a(abc)+b(bcd)+c(cda)+d(dab)$\n$\\leq p(pqr)+q(pqs)+r(prs)+s(qrs)=(pq+rs)(pr+qs)$\n$\\leq (\\frac{pq+rs+pr+qs}{2})^{2}=\\frac{1}{4}((p+s)(q+r))^{2}$\n$\\leq \\frac{1}{4}((\\frac{p+q+r+s}{2})^{2})^{2}$\n$=4$.\n\nEquality holds iff q=r=1 and p+s=2. So equality holds if two of them are equal to 1. Applying this, we can get the equality conditions $(a,b,c,d)=(1,1,1,1),(2,1,1,0)$ or any cyclic forms.\n\nNice problem![/quote]\nclassical and nice\n\n", "Solution_28": "[quote=Binomial-theorem]\nLet $\\{a,b,c,d,e\\}=\\{p,q,r,s,t\\}$. WLOG $p\\le q\\le r\\le s\\le t$ and similarly we have $pqrs\\le prst\\le pqst\\le pqrt\\le qrst$. [/quote]\n\nThis is false. The inequality chain should be $pqrs\\le pqrt\\le pqst\\le prst\\le qrst$.", "Solution_29": "[quote=Parkdoosung]Let $a+b+c+d=4$ and $a,b,c,d\\geq0$.\nProve that $a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab\\leq4$[/quote]\n\na^2bc<((2a+b+c)/4)^4=((S+x)/4)^4=f(x)\nS=a+b+c+d\nX=a-d Y=b-a Z=c-b T=d-c\n(f(x)+f(y)+f(z)+f(t))/42$. Take $ A\\equal{}\\left(\\begin{array}{ccc} 0 & 1 & 1 \\\\ 1 & 0 & \\minus{}1 \\\\ 0 & 1 & 0\\end{array}\\right), B\\equal{}\\left(\\begin{array}{ccc} 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0\\end{array}\\right)$. Then $ A\\plus{}tB\\equal{}\\left(\\begin{array}{ccc} 0 & t\\plus{}1 & 1 \\\\ 1 & 0 & t\\minus{}1 \\\\ 0 & 1 & 0\\end{array}\\right)$, and you can check that the characteristic polynomial of this matrix is $ x^3\\minus{}1$ for any $ t$, which uniquely determines its similarity equivalence class. (since this polynomial has three unique roots, there is a basis of linearly independent eigenvectors)\r\n\r\nIn general, probably all we can show is that $ B$ is nilpotent, which I think we can do as follows:\r\n\r\nLet $ a_k(M)$ be the coefficient of $ x^k$ in the characteristic polynomial of $ M$. Then $ a_k: M_n \\minus{}> \\mathbb{C}$ is a continuous function satisfying $ a_k(\\alpha M) \\equal{} \\alpha ^{n\\minus{}k} a_k(M)$ for all $ \\alpha \\in \\mathbb{C}$. Suppose that we had $ a_k(B)\\neq 0$ for some $ k\\infty} {a_k(A\\plus{}tB)}\\equal{}\\lim_{t\\minus{}>\\infty} {t^{n\\minus{}k} a_k(t^{\\minus{}1}A\\plus{}B)}\\equal{}a_k(B) \\lim_{t\\minus{}>\\infty} {t^{n\\minus{}k}}\\equal{}\\pm \\infty$. So we must have $ a_k(B)\\equal{}0$ for all $ kNevermind, no it isn't. it should be just $ja_{n-j}$.", "Solution_5": "Hmm,\r\n\r\n\\[ s_0=r_1^0+r_2^0+\\cdots+r_n^0=\\underbrace{1+1+\\cdots+1}_{n \\ \\text{ones}}=n. \\]\r\n\r\nOr am I missing something?\r\n\r\n" } { "Tag": [ "geometry", "rectangle" ], "Problem": "How many $ 8''\\times8''$ square tiles are needed to tile a $ 6'\\times8'$ rectangular floor?", "Solution_1": "A 6' by 8' rectangle has an area of $ (6 \\times 12) \\times (8 \\times 12) \\equal{} 6912$ square inches. The tiles each have an area of $ 64$ square inches. Thus, there are $ 6912 \\div 64 \\equal{} \\boxed{108}$ square tiles needed.", "Solution_2": "If 6912/64=108 is too time consuming (it is, considering that this is countdown):\r\n\r\n6'=72\", 72/8=9\r\n8'=96\", 96/8=12\r\n9*12=108" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "I recently had this problem suggested by a person.It seems interesting and definitely may have been a problem given somewhere already.\r\nThe Fibinocci series is well known. On similar lines, can one characterize any recursive series for k sums i.e.,\r\nThe nth term equal the sum of the previous k terms. The nth term and the sum of n terms are to be found out. I know a good book on Recursive Series by Yasuhara. Such a problem is briefly discussed there.\r\nSathej", "Solution_1": "what was really your question?\r\ni think for $k$ terms it is surely like fibonacci numbers." } { "Tag": [ "trigonometry", "induction", "calculus", "calculus computations" ], "Problem": "$a\\otimes b=\\frac{a+b}{1+ab}\\\\ (..(1\\otimes 2)\\otimes 3)...)\\otimes n=?$\r\n\r\nedited ;)", "Solution_1": "[quote=\"PFloyd\"]$a\\otimes b=\\frac{a+b}{1+ab}\\\\ (..(a\\otimes b)\\otimes c)...)\\otimes n=?$[/quote]\r\nI don't understand. What do you want?", "Solution_2": "$a\\otimes b = \\tanh(\\tanh^{-1}a+\\tanh^{-1}b)$\r\n$a\\otimes b \\otimes c = \\tanh(\\tanh^{-1}a+\\tanh^{-1}b+\\tanh^{-1}c)$\r\n$a\\otimes b \\otimes c \\otimes d = \\tanh(\\tanh^{-1}a+\\tanh^{-1}b+\\tanh^{-1}c+\\tanh^{-1}d)$\r\n$\\cdots$", "Solution_3": "What is $\\tanh{\\alpha}$? I know $\\tan{\\alpha}$, then $\\tan{(a+b)}=\\frac{\\tan a+\\tan b}{1-\\tan{a}\\cdot\\tan{b}}$.", "Solution_4": "[quote=\"N.T.TUAN\"]What is $\\tanh{\\alpha}$? I know $\\tan{\\alpha}$, then $\\tan{(a+b)}=\\frac{\\tan a+\\tan b}{1-\\tan{a}\\cdot\\tan{b}}$.[/quote]\r\n\r\nIt's the hyperbolic tangent. I still don't understand the question. Are the ...s going from negative infinity to infinity?", "Solution_5": "No, I think PFloyd's question is from $1$ to $n$ :)", "Solution_6": "Let $a = 2 \\otimes \\cdots \\otimes n$. Then\r\n\r\n$1 \\otimes \\cdots \\otimes n = 1 \\otimes a = \\frac{1+a}{1+1\\cdot a}= 1$", "Solution_7": "brilliant :) thanks\r\n\r\nand what about $2\\otimes 3\\otimes...\\otimes n$?", "Solution_8": "$2\\otimes 3\\otimes\\cdots\\otimes n= \\frac{\\binom{n+1}{2}+(-1)^{n}}{\\binom{n+1}{2}-(-1)^{n}}$. Prove this by induction- I found it by calculating a few values and noticing the pattern." } { "Tag": [], "Problem": "Find all integers $ n > 1$ such that any prime divisor of $ n^6\\minus{}1$ is a\r\ndivisor of $ (n^3\\minus{}1)(n^2\\minus{}1).$\r\n\r\nBaltic way.", "Solution_1": "\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03c9 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c6\u03b9\u03b5\u03c1\u03c9\u03c3\u03b7 \u03c1\u03b5 \u0394\u03b7\u03bc\u03b7\u03c4\u03c1\u03b7.\r\n\u0392\u03b1\u03b6\u03c9 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c7\u03b5\u03b9\u03c1\u03b7 \u03bb\u03c5\u03c3\u03b7.(\u03c0\u03b5\u03c2 \u03bc\u03bf\u03c5 \u03b1\u03bd \u03c7\u03b1\u03bd\u03b5\u03b9 \u03ba\u03b1\u03c0\u03bf\u03c5 \u03ae \u03b1\u03bd \u03ba\u03b1\u03c4\u03b9 \u03b3\u03c1\u03b1\u03c6\u03c9 \u03b3\u03c1\u03b7\u03b3\u03bf\u03c1\u03b1-\u03c3\u03b7\u03bc\u03b5\u03c1\u03b1 \u03b5\u03c7\u03c9 \u03c4\u03b7\u03bd \u03b1\u03c0\u03bf\u03c6\u03bf\u03b9\u03c4\u03b7\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03c4\u03c1\u03b5\u03c7\u03c9 , soz :wink: )\r\n\u03b5\u03b9\u03bd\u03b1\u03b9 $ (n^6 \\minus{} 1) \\equal{} (n^3 \\minus{} 1)(n \\plus{} 1)(n^2 \\minus{} n \\plus{} 1)$\r\n\u0398\u03b5\u03c9\u03c1\u03c9 \u03c4\u03c5\u03c7\u03b1\u03b9\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03b4\u03b9\u03b1\u03c1\u03b5\u03c4\u03b7, $ p$, \u03c4\u03bf\u03c5 $ (n^2 \\minus{} n \\plus{} 1)$ \u03c4\u03bf\u03c4\u03b5 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03b9\u03b1\u03c1\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf $ (n \\minus{} 1)(n \\plus{} 1)(n^3 \\minus{} 1)$\r\n- \u0391\u03bd \u03b4\u03b9\u03b1\u03c1\u03b5\u03b9 \u03c4\u03bf $ n^3 \\minus{} 1$ \u03b5\u03c5\u03ba\u03bf\u03bb\u03b1 $ p \\equal{} 2$, \u03b1\u03c4\u03bf\u03c0\u03bf \u03b1\u03c6\u03bf\u03c5 $ n^2 \\minus{} n \\plus{} 1$\u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03bf \u03c0\u03b1\u03bd\u03c4\u03b1..\r\n-A\u03bd \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03b9 \u03c4\u03bf $ n \\minus{} 1$ \u03c4\u03bf\u03c4\u03b5 $ p|1$\u03ba\u03b1\u03b9 \u03b1\u03c4\u03bf\u03c0\u03bf\r\n- \u0391\u03bd \u03b4\u03b9\u03b1\u03c1\u03b5\u03b9 \u03c4\u03bf $ n \\plus{} 1$ \u03c4\u03bf\u03c4\u03b5 $ p|n \\minus{} 2$ \u03b1\u03c1\u03b1 $ p \\equal{} 3$\r\n\u03a3\u03c5\u03bd\u03b5\u03c0\u03c9\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 $ n^2 \\minus{} n \\plus{} 1 \\equal{} 3^x$\r\n\u03bf\u03bc\u03c9\u03c2 \u03b1\u03bd \u03c4\u03b7\u03bd \u03b8\u03b5\u03c9\u03c1\u03b7\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b1\u03bd \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf\u03b2\u03b1\u03b8\u03bc\u03b9\u03b1 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 n \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03b4\u03b9\u03b1\u03ba\u03c1\u03b9\u03bd\u03bf\u03c5\u03c3\u03b1 \u03c4\u03b5\u03bb\u03b5\u03b9\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03bf,\r\n\u0398\u03b5\u03c9\u03c1\u03c9 $ x > 1$ \u03c4\u03bf\u03c4\u03b5,\r\n$ 1 \\minus{} 4(1 \\minus{} 3^x) \\equal{} m^2 \\leftrightarrow 3 \\minus{} 4*3^x \\equal{} m^2 \\equal{} > 3|m \\equal{} > 1 \\minus{} 4*3^{x \\minus{} 1} \\equal{} 3z^2 \\equal{} > 3|1$ \u03b1\u03c4\u03bf\u03c0\u03bf..\r\n(\u03c4\u03bf \u03b1\u03c4\u03bf\u03c0\u03bf \u03c0\u03c1\u03bf\u03ba\u03c5\u03c0\u03c4\u03b5\u03b9 \u03b5\u03c0\u03b9\u03c3\u03b7\u03c2 \u03bc\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03c9\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 modulo9)\r\n\u03b1\u03c1\u03b1 $ x \\equal{} 1$, $ n(n \\minus{} 1) \\equal{} 2 \\equal{} > n \\equal{} 2$\r\n\u03bc\u03b5 \u03b5\u03c0\u03b1\u03bb\u03b7\u03b8\u03b5\u03c5\u03c3\u03b7 \u03bb\u03bf\u03b9\u03c0\u03bf\u03bd \u03bc\u03bf\u03bd\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03b7 $ n \\equal{} 2$. :)" } { "Tag": [ "geometry", "incenter", "trigonometry", "geometry solved" ], "Problem": "Does anyone has a prove for this one ?\r\n\r\nI is incenter , H is orthocenter of a triangle ABC . Then [tex] IH^2 = 2r^2 - 4R^2 \\cos A \\cos B \\cos C [/tex]\r\n\r\nThanks", "Solution_1": "See [url=http://groups.yahoo.com/group/Hyacinthos/messages/6147?viscount=41&expand=1]Hyacinthos messages #6147 and #6195[/url] for a trigonometric proof by Nikolaos Dergiades. This is the nicest proof I know for this identity.\r\n\r\n Darij" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "x,y,z\u22650 ; x\u00b2+y\u00b2+z\u00b2=1;\r\nprove that:\r\n\r\nb) $ \\frac {}$sqrt{xyz}${ } \\sqrt {3(x + y + z)}$ \u2265 $ \\frac {1}{x}$+$ \\frac {1}{y}$+$ \\frac {1}{z}$+xyz;\r\nc)1+$ \\sqrt {}$frac{y}{xz}${ }$\u2265xy+yz", "Solution_1": "[quote=\"kihe_freety5\"]x,y,z\u22650 ; x\u00b2+y\u00b2+z\u00b2=1;\nprove that:\n\nb) $ \\frac {}$sqrt{xyz}${ } \\sqrt {3(x + y + z)}$ \u2265 $ \\frac {1}{x}$+$ \\frac {1}{y}$+$ \\frac {1}{z}$+xyz;\nc)1+$ \\sqrt {}$frac{y}{xz}${ }$\u2265xy+yz[/quote]\r\n\r\n[b]retype the original[/b] :) \r\n\r\n$ x,y,z\\ge0$ ; $ x^2+y^2+z^2=1$\r\nprove that:\r\n\r\nb) $ \\frac{\\sqrt{xyz}}{\\sqrt {3(x + y + z)}} \\ge \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + xyz$\r\nc) $ 1+\\sqrt{\\frac{y}{xz}} \\ge xy+yz$" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "inequalities", "LaTeX" ], "Problem": "Suppose $ A$ is an n by n matrix.\r\n- The main diagonal elements of A are all -2, except positions (1,1) and (n,n), which are -3\r\n- The values on the diagonals directly above and directly below the main diagonal are all 1\r\n- The remaining values are 0\r\n\r\nFor example, for n=5 we have:\r\n[code]\n-3 1 0 0 0\n 1 -2 1 0 0\n 0 1 -2 1 0\n 0 0 1 -2 1\n 0 0 0 1 -3\n[/code]\r\n\r\nProve that the eigenvalues of $ A_n$ are all real and are all negative.", "Solution_1": "This is equivalent to showing that the eigenvalues of $ A_n \\plus{} 3I_n$ are real and less than $ 3$. Real follows by the spectral theorem. $ A_n \\plus{} 3I_n$ is the adjacency matrix of a graph $ G_n$ on $ n$ vertices constructed from a path graph to which a loop has been added at each vertex except the beginning and the end. Since each vertex has degree at most $ 3$, the eigenvalues are less than or equal to $ 3$. It's a general fact that if $ H$ is a proper subgraph of $ G$, the largest eigenvalue of $ H$ is strictly less than the largest eigenvalue of $ G$, and since $ G_n$ can be turned into a graph with largest eigenvalue exactly $ 3$ by adding self-loops at the beginning and end and an edge between them, the result follows.\r\n\r\nAlternately, you could show that $ \\det A_n \\neq 0$, which is probably straightforward by row-reduction. Incidentally, do you know the characteristic polynomial of $ A_n \\plus{} 2I_n$? It should be a straightforward perturbation of the corresponding Chebyshev polynomial.", "Solution_2": "The eigenvalues are real, simply because the matrix is real symmetric.\r\n\r\nAs for negative: you're asking for a proof that this matrix is negative definite. It's diagonally dominant, although not strictly diagonally dominant, and that should suffice as long as we can show that zero is not an eigenvalue.\r\n\r\nSuppose $ v$ is an eigenvector. Let $ v_j$ be the element of $ v$ such that $ |v_k|\\le |v_j|$ for all $ 1\\le k\\le 5.$ Multiply $ v$ by a scalar so that $ v_j = 1.$\r\n\r\nCase 1: suppose $ j$ is $ 2,3,$ or $ 4.$ Then the $ j$th entry of $ Av$ is \r\n\r\n$ v_{j - 1} - 2v_j + v_{j + 1} = v_{j - 1} - 2 + v_{j + 1}\\le 0.$\r\n\r\nThat last inequality comes from $ |v_{j - 1}|\\le 1$ and $ |v_{j + 1}|\\le 1.$\r\n\r\nBut if $ Av = \\lambda v,$ then the $ j$th entry of $ Av$ is $ \\lambda,$ so we have shown that $ \\lambda\\le 0.$\r\n\r\nCase 2: suppose $ j$ is $ 1$ or $ 5.$ WLOG, $ j = 1.$ Then\r\n\r\n$ \\lambda = (Av)_1 = - 3 + v_2\\le - 2.$\r\n\r\nSo in either case, we have $ \\lambda\\le0.$ Now all we have to do is rule out $ \\lambda = 0.$\r\n\r\nAs it turns out, it's quite easy to do row elimination in $ A.$\r\n\r\n(Note: TZF, you should learn the $ \\text{\\LaTeX}$ syntax for matrices - see this part of the post.)\r\n\r\n$ \\begin{bmatrix} - 3 & 1 & 0 & 0 & 0 \\\\\r\n1 & - 2 & 1 & 0 & 0 \\\\\r\n0 & 1 & - 2 & 1 & 0 \\\\\r\n0 & 0 & 1 & - 2 & 1 \\\\\r\n0 & 0 & 0 & 1 & - 3\\end{bmatrix}$ is row equivalent to $ \\begin{bmatrix} - 3 & 1 & 0 & 0 & 0 \\\\\r\n0 & - \\frac53 & 1 & 0 & 0 \\\\\r\n0 & 0 & - \\frac75 & 1 & 0 \\\\\r\n0 & 0 & 0 & - \\frac97 & 1 \\\\\r\n0 & 0 & 0 & 0 & - \\frac {20}9\\end{bmatrix}.$\r\n\r\nThat - row equivalence with no scalar multiplications and no row swaps - preserves determinant, so we may read off the telescoping product on the main diagonal that $ \\det A = - 20.$ Since that is not $ 0,$ $ A$ is invertible, and that means that $ 0$ is not an eigenvalue of $ A.$ Hence, all of the eigenvalues of $ A$ are negative, also known as $ A$ being negative definite.", "Solution_3": "[quote=\"Kent Merryfield\"]$ \\det A \\equal{} \\minus{} 20.$[/quote]\r\nIncidentally, $ A_n$ is the (negative of the) reduced Laplacian of the graph $ G'_{n \\plus{} 1}$ obtained by choosing a vertex in a cycle graph $ C_{n \\plus{} 1}$ and adding an extra edge from that vertex to each of its neighbors, so by [url=http://en.wikipedia.org/wiki/Kirchhoff%27s_theorem]Kirchhoff's theorem[/url] $ \\det A_n$ counts the number of spanning trees of this graph, which is obviously nonzero. In fact, a spanning tree can be obtained uniquely by deleting three edges in the following ways:\r\n\r\n- Delete three of the \"doubled\" edges. There are $ 4$ ways to do this.\r\n- Delete two \"doubled\" edges and any of the other non-doubled edges. There are $ 4(n \\minus{} 1)$ ways to do this.\r\n\r\nThis gives $ ( \\minus{} 1)^n \\det A_n \\equal{} 4n$.", "Solution_4": "[quote=\"Kent Merryfield\"] ... Let $ v_j$ be the element of $ v$ such that $ |v_k|\\le |v_j|$ for all $ 1\\le j\\le 5.$ ...[/quote]\r\nI think you mean :\r\nLet $ v_j$ be the element of $ v$ such that $ |v_k|\\le |v_j|$ for all $ 1\\le k\\le 5$\r\n\r\n[color=green][Yes. I fixed that. - K.M.][/color]", "Solution_5": "[b] Generral [/b]The fist we can see that if $ A,B$ are two Hermitian matrices such that $ A < 0$ and $ B \\leq 0$ then $ A \\plus{} B < 0$ Implies $ A \\plus{} B$ show that eigenvalues of $ A \\plus{} B$ are real and all negative.\r\n With $ a,b \\in \\mathbb{R}$ and $ C \\equal{} \\begin{bmatrix}a & b & 0 & 0 & 0 & ... & ... & 0 \\\\\r\nb & a & b & 0 & 0 & ... & ... & 0 \\\\\r\n0 & b & a & b & 0 & ... & ... & 0 \\\\\r\n0 & 0 & b & a & b & ... & ... & 0 \\\\\r\n... & ... & ... & ... & ... & ... & ... & ... \\\\\r\n0 & 0 & 0 & 0 & 0 & ... & b & a\\end{bmatrix}$ then $ C$ have eigenvalues are $ \\lambda_p \\equal{} a \\plus{} 2bcos(\\frac {p\\pi}{n \\plus{} 1}),1 \\leq p \\leq n , p\\in \\mathbb{N}$ ( you can read and proof it [b]homework of me [/b]:( ) \r\n Now \r\n $ \\begin{bmatrix} \\minus{} 3 & 1 & 0 & 0 & 0 & ... & ... & 0 \\\\\r\n1 & \\minus{} 2 & 1 & 0 & 0 & ... & ... & 0 \\\\\r\n0 & 1 & \\minus{} 2 & 1 & 0 & ... & ... & 0 \\\\\r\n0 & 0 & 1 & \\minus{} 2 & 1 & ... & ... & 0 \\\\\r\n... & ... & ... & ... & ... & ... & ... & ... \\\\\r\n0 & 0 & 0 & 0 & 0 & ... & 1 & \\minus{} 3 \\end{bmatrix} \\equal{} \\begin{bmatrix} \\minus{} 2 & 1 & 0 & 0 & 0 & ... & ... & 0 \\\\\r\n1 & \\minus{} 2 & 1 & 0 & 0 & ... & ... & 0 \\\\\r\n0 & 1 & \\minus{} 2 & 1 & 0 & ... & ... & 0 \\\\\r\n0 & 0 & 1 & \\minus{} 2 & 1 & ... & ... & 0 \\\\\r\n... & ... & ... & ... & ... & ... & ... & ... \\\\\r\n0 & 0 & 0 & 0 & 0 & ... & 1 & \\minus{} 2 \\end{bmatrix} \\plus{} \\begin{bmatrix} \\minus{} 1 & 0 & 0 & 0 & 0 & ... & ... & 0 \\\\\r\n0 & 0 & 0 & 0 & 0 & ... & ... & 0 \\\\\r\n0 & 0 & 0 & 0 & 0 & ... & ... & 0 \\\\\r\n0 & 0 & 0 & 0 & 0 & ... & ... & 0 \\\\\r\n... & ... & ... & ... & ... & ... & ... & ... \\\\\r\n0 & 0 & 0 & 0 & 0 & ... & 0 & \\minus{} 1 \\end{bmatrix} \\equal{} A \\plus{} B$ \r\nHere $ A$ have eigenvalues are $ \\lambda_p \\equal{} \\minus{} 2 \\plus{} 2cos(\\frac {p\\pi}{n \\plus{} 1})\\leq O,1 \\leq p \\leq n , p\\in \\mathbb{N}$ hence $ A \\leq O$\r\n $ B$ have eigenvalues are $ \\minus{} 1,0,0,...,0, \\minus{} 1$ Hence $ B < O$\r\nAs $ A \\plus{} B < O$ show that eigenvalues of $ A \\plus{} B$ are real and all negative.\r\n------------------------------------------------------", "Solution_6": "Thanks for all the interesting and varied replies! :) \r\n\r\nMy particular solution was just an algebraic one:\r\n$ \\lambda \\| \\mathbf{v} \\| ^2 \\equal{} (A\\mathbf{v})^T \\mathbf{v}$\r\n$ \\equal{} ( \\minus{} 3 v_1 \\plus{} v_2) v_1 \\plus{} ( \\minus{} v_1 \\plus{} 2 v_2 \\minus{} v_3) v_2 \\plus{}$ $ \\ldots \\plus{} ( \\minus{} v_{n \\minus{} 2} \\plus{} 2 v_{n \\minus{} 1} \\minus{} v_n) v_{n \\minus{} 1} \\plus{} (v_{n \\minus{} 1} \\minus{} 3 v_n) v_n$\r\n$ \\equal{} \\minus{} 2 v_1 ^2 \\minus{} (v_1 \\minus{} v_2)^2 \\minus{} \\ldots \\minus{} (v_{n \\minus{} 1} \\minus{} v_n)^2 \\minus{} 2 v_n ^2$\r\n$ \\leq 0$\r\n\r\nwith equality iff $ v_1 \\equal{} v_2 \\equal{} \\ldots \\equal{} v_n \\equal{} 0 \\iff \\mathbf{v} \\equal{} 0$\r\n\r\nBy the Spectral Theorem, there must be an orthonormal basis of eigenvectors, so $ \\mathbf{v} \\equal{} 0$ is not an option, so we must have $ \\lambda < 0$.\r\n\r\n[b]Note:[/b] This matrix originally came out of modeling a non-dissipative spring-mass system with multiple springs and masses. One of our homework assignments was to prove that the eigenvalues are negative, basically by saying that if they weren't, then energy conservation would be violated.", "Solution_7": "t0raji0u mentioned a graph-theoretic concept called the Laplacian. TZF was talking more about the real (one-dimensional) Laplacian, which is to say, $ \\frac {\\partial^2}{\\partial x^2}.$\r\n\r\nTZF's model is a discretized version of the wave equation, $ \\frac {\\partial^2u}{\\partial t^2} \\equal{} \\frac {\\partial^2u}{\\partial x^2},$ but with some extra wrinkle at the boundaries. It looks like you have a different, stronger spring at each end of the chain. If you used the same strength spring to attach this to the walls, you would have the matrix that QuyBac called $ A$ in his decomposition.\r\n\r\nAdded in edit:\r\n\r\nTZF, consider this variation. Take out the springs that connect you to the walls altogether. You should have the following matrix:\r\n\\[ \\begin{bmatrix}\\minus{}1&1&0&0&0\\\\1&\\minus{}2&1&0&0\\\\ 0&1&\\minus{}2&1&0\\\\ 0&0&1&\\minus{}2&1\\\\ 0&0&0&1&\\minus{}1\\end{bmatrix}\\]\r\nWhat can you say about the eigenvalues of this? What physical meaning might you attach to an eigenvalue of zero?", "Solution_8": "[quote=\"TZF\"]Thanks for all the interesting and varied replies! :) \n\nMy particular solution was just an algebraic one:\n$ \\lambda \\| \\mathbf{v} \\| ^2 \\equal{} (A\\mathbf{v})^T \\mathbf{v}$\n$ \\equal{} ( \\minus{} 3 v_1 \\plus{} v_2) v_1 \\plus{} ( \\minus{} v_1 \\plus{} 2 v_2 \\minus{} v_3) v_2 \\plus{}$ $ \\ldots \\plus{} ( \\minus{} v_{n \\minus{} 2} \\plus{} 2 v_{n \\minus{} 1} \\minus{} v_n) v_{n \\minus{} 1} \\plus{} (v_{n \\minus{} 1} \\minus{} 3 v_n) v_n$\n$ \\equal{} \\minus{} 2 v_1 ^2 \\minus{} (v_1 \\minus{} v_2)^2 \\minus{} \\ldots \\minus{} (v_{n \\minus{} 1} \\minus{} v_n)^2 \\minus{} 2 v_n ^2$\n$ \\leq 0$\n\nwith equality iff $ v_1 \\equal{} v_2 \\equal{} \\ldots \\equal{} v_n \\equal{} 0 \\iff \\mathbf{v} \\equal{} 0$\n\nBy the Spectral Theorem, there must be an orthonormal basis of eigenvectors, so $ \\mathbf{v} \\equal{} 0$ is not an option, so we must have $ \\lambda < 0$.\n.[/quote]\r\n I will add with real symmetric matrix .\r\nNow applied with Hermitian matrices . You can Interedted it example as:\r\n$ A \\equal{} \\begin{bmatrix} \\minus{} 5 & 2 \\minus{} i & 0 & 0 & 0 & ... & ... & ... & 0 \\\\\r\n2 \\plus{} i & \\minus{} 4 & 1 & 0 & 0 & ... & ... & ... & 0 \\\\\r\n0 & 1 & \\minus{} 2 & 1 & 0 & ... & ... & ... & 0 \\\\\r\n0 & 0 & 1 & \\minus{} 2 & 1 & ... & ... & ... & 0 \\\\\r\n... & ... & ... & ... & ... & ... & ... & ... & ... \\\\\r\n0 & 0 & 0 & 0 & 0 & ... & \\minus{} 2 & 1 & 0 \\\\\r\n0 & 0 & 0 & 0 & 0 & ... & 1 & \\minus{} 4 & 2 \\plus{} i \\\\\r\n0 & 0 & 0 & 0 & 0 & ... & 0 & 2 \\minus{} i & \\minus{} 5\\end{bmatrix}$\r\nProve that the eigenvalues of $ A$ are all real and are all negative." } { "Tag": [ "number theory", "prime numbers" ], "Problem": "Find the arithmetic mean of the prime numbers in this list: 21, 23, 25, 27, 29", "Solution_1": "first, we have to find the prime numbers\r\nin that list, the prime numbers are: 23 and 29\r\n\r\nthe mean of those two numbers are $ \\frac {23\\plus{}29} {2} \\equal{} \\frac {52} {2} \\equal{} 26$" } { "Tag": [ "complex analysis", "complex analysis solved" ], "Problem": "prove the followig form of weierstrass theorem : if f has an essential singularity at z=zo, then for given c belongs to C\r\nand epsilon > 0 for each delta > 0 there is a number a, |c-a|< epsilon, such that f(z)=a has infinitely many solutions in the disc\r\nB(zo,delta) centered at a and of radius delta.\r\n\r\nthx for ur help", "Solution_1": "[url=http://en.wikipedia.org/wiki/Casorati%E2%80%93Weierstrass_theorem]here[/url]" } { "Tag": [ "algebra", "polynomial", "linear algebra", "matrix", "ratio", "function" ], "Problem": "Solve system of equations in real numbers:\r\n$\\left\\{ \\begin{array}{l}y(x+y)^2=9 \\\\ y(x^3-y^3)=7 \\\\ \\end{array} \\right.$", "Solution_1": "(Division by y is allowed because if y were 0, we would have 0 = 9 and 0 = 7 so we know y is never 0)\r\n$\\{ \\begin{array}{l}y(x+y)^2=9 \\\\ y(x^3-y^3)=7 \\\\ \\end{array} \\Rightarrow \\{ \\begin{array}{l}(\\frac{x}{y}+1)^2=\\frac{9}{y^3} \\\\ (\\frac{x}{y}^3-1)= \\frac{7}{y^4} \\\\ \\end{array} $\r\n$\\rightarrow \\frac{x}{y} = M$\r\n$\\frac{(M+1)^8}{3^8}=\\frac{1}{y^{12}}$\r\n$\\frac{(M^3-1)^3}{7^3}=\\frac{1}{y^{12}}$\r\n$7^3(M+1)^8=3^8(M^3-1)^3$\r\n$(M-2)(6561M^8+12779M^7+22814M^6+16341M^5+13474M^4+2938M^3+6351M^2+3098M+3452) = 0$\r\n\r\nBy $M^3 - 1 = \\frac{7}{y^4}$ we know $M^3 - 1$ must be positive, so $M^3 > 1$ and $M > 1$ since it must be in the reals. With that we know the remaining 8th degree polynomial has no solution for $M$. So, $M = 2$ is the only solution. Plugging back in we get $y^3 = 1$ and $y^4 = 1$ which gives us $y = 1$. So, $2 = M = \\frac{x}{y} = \\frac{x}{1} = x$ and the only soln. is $(x,y) = (2,1)$.\r\nWhere did this system originate from and is there a nice non-ugly way to do it, i.e., factoring a 9th degree polynomial.", "Solution_2": "[quote=\"arqady\"]Solve system of equations in real numbers:\n$\\left\\{ \\begin{array}{l}y(x+y)^2=9 \\\\ y(x^3-y^3)=7 \\\\ \\end{array} \\right.$[/quote]\r\n\r\ni was thinking that $a=x+y$ and $b=xy$ would help\r\n\r\n$ya^2=9$\r\n\r\n$y(a-2y)( a^2+b) )=7$", "Solution_3": "[hide=\"Here are two solutions for this unbearable system.\"]\n[hide=\"The first method.\"] Observe that $01\\ }\\ .$\n \nFrom the ratio of the two equations we obtain the relation:\n\n$x-y=\\frac{7(x+y)^2}{9(x^2+xy+y^2)}=\\frac{7(t+1)^2}{9(t^2+t+1)}\\ ;$\n\n$\\frac xt=\\frac y1=\\frac{x+y}{t+1}=\\frac{x-y}{t-1}=\\frac{7(t+1)^2}{9(t^3-1)}\\ .$\n\nFrom the first equation of the system we obtain the relation\n\n$\\frac{7(t+1)^2}{9(t^3-1)}\\cdot \\left[\\frac{7(t+1)^3}{9(t^3-1)}\\right]^2=9\\Longleftrightarrow 7^3(t+1)^8=9^4(t^3-1)^3\\ ,\\ t>1$ a.s.o.\n[/hide][/hide]", "Solution_4": "[hide] $y(x^3-y^3)=7 \\Longleftrightarrow y(x-y)[(x+y)^2-xy]=7$ considering that : $y(x+y)^2=9$ we got:\n\n$7=(x-y)(9-xy^2)$ the results follow\n[/hide]", "Solution_5": "[quote=\"Virgil Nicula\"]\n$(y-1)(4y^8+4y^7+4y^6+31y^5+59y^4+59y^3+302y^2+680y+729)=0\\Longrightarrow$\n$\\boxed {\\ y=1\\ \\wedge\\ x=2\\ }\\ .$[/quote]\r\nWhy $4y^8+4y^7+4y^6+31y^5+59y^4+59y^3+302y^2+680y+729\\neq0?$ ;)\r\nThank you.", "Solution_6": "we want real solutions and that polynomial has no real solutions", "Solution_7": "[quote=\"Altheman\"]we want real solutions and that polynomial has no real solutions[/quote]\r\nWhy, Altheman? ;)", "Solution_8": "[hide=\"For Arqady\"] $y(x+y)^2=9\\ \\wedge \\ y(x^3-y^3)=7\\Longrightarrow 00, x\\neq y, a=\\frac{x}{y},b=\\frac{y}{x}, z\\in[x,y]$. prove that \r\n\\[\\frac{a}{1-b}+\\frac{b}{1-a}>\\frac{x+y+z}{\\sqrt[3]{xyz}}\\]", "Solution_1": "somebody..? :roll:", "Solution_2": "does anybody have a solution? or is it too easy? say something.. :mad:", "Solution_3": "i have a nice solution but let's see what you've got. :lol:", "Solution_4": "why isn't anyone interested??? :mad:", "Solution_5": "At first we notice that $\\frac{a}{1-b}+\\frac{b}{1-a}= 1+\\frac{x}{y}+\\frac{y}{x}$. We have to prove\r\n$1+\\frac{x}{y}+\\frac{y}{x}> \\frac{x+y+z}{\\sqrt[3]{xyz}}$. \r\nLet us put $f(x,y,z)= \\frac{x+y+z}{\\sqrt[3]{xyz}}$. Then\r\n$\\frac{\\partial f}{\\partial z}= \\frac{1}{3}\\cdot \\left(xyz\\right)^{-\\frac{1}{3}}\\cdot\\left((x+y)\\cdot z^{-1}+2\\right)$. \r\nThis expression is negative for $z<\\frac{x+y}{2}$ and positive for $z>\\frac{x+y}{2}$, so at the point $z=\\frac{x+y}{2}$ the function $f(x,y,z)$ reaches its minimum value. Thus the maximum value of $f(x,y,z)$ for fixed $x,y$ and $z\\in [x,y]$ can be reached only for $z\\in\\{x,y\\}$. We will prove the inequality for $z=x$, because the second case in analogous. \r\nThe inequality becomes\r\n$1+\\frac{x}{y}+\\frac{y}{x}>\\frac{2x+y}{\\sqrt[3]{x^{2}y}}$. \r\nTransforming it and returning to the notation with $a$, we get\r\n$1+a+a^{-1}> 2a^{\\frac{1}{3}}+a^{-\\frac{2}{3}}$. \r\nLet $g(t)=a^{t}$ for any $t\\in\\mathbb{R}$. Then $g$ is strictly convex for any $a\\in\\mathbb{R}^{+}\\setminus\\{1\\}$. We have $\\left(1,0,-1\\right)\\succ \\left(\\frac{1}{3}, \\frac{1}{3},-\\frac{2}{3}\\right)$, so by the Karamata inequality the desired inequality holds and is strict.", "Solution_6": "$1+\\frac{x}{y}+\\frac{y}{x}\\geq \\frac{x}{y}+\\frac{y}{z}+\\frac{z}{x}\\geq \\frac{x+y+z}{\\sqrt[3]{xyz}}$\r\n :)", "Solution_7": "my solution:\r\n$1+\\frac{x}{y}+\\frac{y}{x}=\\frac{x}{y}+\\frac{z}{z}+\\frac{y}{x}\\ge\\frac{x+y+z}{\\sqrt[3]{xyz}}$ :wink:" } { "Tag": [], "Problem": "What is the greatest number that cannot be obtained by adding only 3's, 4's, or a combination of 3's and 4's?", "Solution_1": "Use the Chicken Mcnugget Theorem, which yields the largest number which cannot be expressed as $ mx + ny$. In this case m = 3 and n = 4.\r\n\r\n$ 3 \\cdot 4 - 3 - 4 = \\boxed{5}$" } { "Tag": [ "calculus" ], "Problem": "Would anyone be willing to send me old Clemson Calculus Challenge problems? They will not send out old copies of exams. We have never gone, and it would be nice if we knew what to expect. Thank You", "Solution_1": "If I still had mine with me I would...\r\n\r\nIf I remember right, it was mainly AP stuff with more of the harder AP stuff and maybe a few that are a little harder than AP." } { "Tag": [ "geometry", "rectangle", "MATHCOUNTS", "number theory", "relatively prime" ], "Problem": "I need help with this problem. \r\n\r\n\r\nA 24 by 72 rectangle is split into 1 by 1 squares. A diaganol runs through a number of squares. How many of these 1 by 1 squares are passed through the diagonal?", "Solution_1": "I think the formula is $x+y-GCF(x,y)$.", "Solution_2": "This problem has been posted before.", "Solution_3": "So has that formula.", "Solution_4": "Yeah, I think it has posted atleast twice.", "Solution_5": "[quote=\"ckck\"]I need help with this problem. \n\n\nA 24 by 72 rectangle is split into 1 by 1 squares. A diaganol runs through a number of squares. How many of these 1 by 1 squares are passed through the diagonal?[/quote]\r\n\r\nIt has been posted many times actually, but for someone who's only been on here 2 weeks, they wouldn't know.\r\n\r\nI do suggest looking through old posts, but here you go :\r\n\r\nA 24 by 72 rectangle can be broken up into 1 by 3 rectangles as can be determined by dividing the dimensions by the GCF(24). In order to get from one corner to the other the line would pass through 24 of these 1 by 3 rectangles. In passing through each rectangle the line would pass through all 3 squares. So the line passes through 24x3 = 72 squares. \r\n\r\nIf you play around with a few examples(I suggest starting with relatively prime dimensions) you'll find the formula is x + y - gcf(x,y) as was posted. If you need proof of the formula, maybe you can get it from someone else in here.", "Solution_6": "See the attachment that I created for MathCounts tutoring sessions:", "Solution_7": "[quote=\"mysmartmouth\"]See the attachment that I created for MathCounts tutoring sessions:[/quote]\r\n\r\nDo you have any other ones? If so, can you post it?", "Solution_8": "Yes, I will find them and post them. Are they helpful? If so, I can make more.", "Solution_9": "[quote=\"mysmartmouth\"]Yes, I will find them and post them. Are they helpful? If so, I can make more.[/quote]\r\n\r\nYeah. There good practice. :)", "Solution_10": "I just remembered one of my former math team kids made a visual aide for this problem. I'll attach it. Not ground breaking or anything, but good for those visual learners:-)" } { "Tag": [ "geometry", "perimeter", "geometry unsolved" ], "Problem": "In the image, ABCD is a quadrilateral where $ \\overline{AB}\\, \\equal{} \\,\\overline{BC}$ , $ \\hat{DAB}\\, \\equal{} \\,80^o$ , and $ \\hat{CBA}\\, \\equal{} \\,40^o$. A point $ P$ exists so that the triangle $ DPC$ is equilateral. Calculate the perimeter of the triangle $ APB$.\r\n\r\n$ AB\\, \\equal{} \\,BC\\, \\equal{} \\,6$\r\n\r\n$ CD\\, \\equal{} \\,6$\r\n\r\nI can't solve this problem no matter what. I added the $ L$ for reference.\r\n\r\nThanks guys!", "Solution_1": "what is $ M$?", "Solution_2": "Sorry it was $ APB$. It's fixed." } { "Tag": [ "geometry proposed", "geometry", "Locus problems", "Geometric Inequalities", "Germany", "TST", "Team Selection Test" ], "Problem": "Let $d$ be a diameter of a circle $k$, and let $A$ be an arbitrary point on this diameter $d$ in the interior of $k$. Further, let $P$ be a point in the exterior of $k$. The circle with diameter $PA$ meets the circle $k$ at the points $M$ and $N$.\n\nFind all points $B$ on the diameter $d$ in the interior of $k$ such that\n\\[\\measuredangle MPA = \\measuredangle BPN \\quad \\text{and} \\quad PA \\leq PB.\\]\n(i. e. give an explicit description of these points without using the points $M$ and $N$).", "Solution_1": "If PM and PN intesect again the circle k at C and D respectively, it is well known by now that the center of k, the center of (CDP), the center of (MNP), that is, the midpoint of AP, and P form a parallelogram. It follows that B is the symmetrical of A about the center of k. I hope I am right...\r\n\r\nBest regards,\r\nsunken rock" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let a,b,c>0 Prove that\r\n$ (\\sum\\frac{a\\plus{}b}{c})^2 \\geq 18\\plus{}2(a^2\\plus{}b^2\\plus{}c^2)(\\frac{1}{a^2}\\plus{}\\frac{1}{b^2}\\plus{}\\frac{1}{c^2})$", "Solution_1": "Let a,b,c>0 ,then the inequality\r\n$ (\\sum\\frac{a\\plus{}b}{c})^2 \\geq (36\\minus{}9k)\\plus{}k(a^2\\plus{}b^2\\plus{}c^2)(\\frac{1}{a^2}\\plus{}\\frac{1}{b^2}\\plus{}\\frac{1}{c^2})$\r\nholds if and only if $ k \\leq1$.", "Solution_2": "Let a,b,c>0 Prove that\r\n$ 18 \\plus{} 2(a^2 \\plus{} b^2 \\plus{} c^2)(\\frac {1}{a^2} \\plus{} \\frac {1}{b^2} \\plus{} \\frac {1}{c^2}) \\geq3(\\sum\\frac{a\\plus{}b}{c}\\frac{b\\plus{}c}{a})$", "Solution_3": "[quote=\"tran300491\"]Let a,b,c>0 Prove that\n$ (\\sum\\frac {a + b}{c})^2 \\geq 18 + 2(a^2 + b^2 + c^2)(\\frac {1}{a^2} + \\frac {1}{b^2} + \\frac {1}{c^2})$[/quote]\r\n\r\n\r\nLet $ f(x)=\\sum ab(x-a)(x-b)$\r\n\r\n$ We have f(a)f(b)f(c)f(0) \\leq 0$\r\n\r\n$ --> \\Delta =B^2-4AC \\geq 0\n(A=\\sum ab;......)$\r\n\r\n$ -->(\\sum\\frac{a+b}{c})^2 \\geq 4(ab+bc+ca)(\\sum\\frac{1}{a^2})$\r\n\r\nand by IRAN 96 we have done :rotfl:", "Solution_4": "Try a=1,b=10,c=100,k=2!!", "Solution_5": "sorry \r\nthe inequality is only true if a,b,c is 3 sides of 1 triangle" } { "Tag": [ "\\/closed" ], "Problem": "Okay I look through the list of avatars and mine isnt there!!!\r\nI know this is not essential but it may end up as a bigger bug...\r\nI use internet explorer on a windows operating system.\r\nHelp would be appreciated. :maybe:", "Solution_1": "Did you just change your avatar? Because when I changed my avatar for a new one, it took a while (maybe a few hours or something) for it to appear in the classroom. Try during your next class...it should work.", "Solution_2": "No.\r\nActually, this problem has been bugging me for a couple weeks... :(", "Solution_3": "Also if you know, it would help if you (the person answering this) tell me what types of pic are accepeted in the classroom. (ie jpg,jpeg,gif...)", "Solution_4": "I think it has to do with the size.\r\nMine is the biggest it can be (80x100).\r\nMy previous one was smaller but it worked and was GIF. Now I have a big PNG and it doesn't load in the classroom... :huh:", "Solution_5": "Just put something like \r\n[code][img id=http://www.artofproblemsolving.com/Forum/images/avatars/776825435484df1668574f.png[/code]\r\n in the box when you click \"...\".", "Solution_6": "Okay that worked fine. Now if I resize the image as I have done, how do I get a new code for the pic?", "Solution_7": "Right Click / Control Click your avatar on \r\nhttp://www.mathlinks.ro/profile.php?mode=viewprofile&u=37836\r\nand choose Copy Image Location\r\nThen when you click ... type [img id= and then CTRL-P (Edit->Paste) and click OK/Enter whatever it says. That should work. :)", "Solution_8": "If you mean that it is missing in the classroom, this problem has (I hope) been fixed. The \"missing avatars\" should start appearing again tomorrow (Thursday).", "Solution_9": "Hurm mine still wasn't there..." } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Circle $l$ is tangent from inside to the circle $k$ , and touches the radius $PQ$ of the circle $k$ in the point $C$ . Let $A$ be the point on $k$ , and $B$ the point of the segment $CQ$ such that $AB$ is tangent to the circle $l$ which is perpendicular to $PQ$. Prove that $AC$ bisects $\\angle PAB$.", "Solution_1": "If I correctly understand you, then, in order to make the problem correct, PQ must be not just a radius, but a diameter of the circle k. The problem was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=23606 and http://www.mathlinks.ro/Forum/viewtopic.php?t=17446 .\r\n\r\n darij" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$c8(ad+bc)$", "Solution_1": "Counterexample: take $a = d = 2$, $b = c = 1$.", "Solution_2": "Right, I hope you changed it into the correct version now.", "Solution_3": "[quote=\"Arne\"]Counterexample: take $a = d = 2$, $b = c = 1$.[/quote]\r\n\r\nBut he said $c8(ad+bc)\\]", "Solution_5": "[quote=\"ehsan2004\"][quote=\"Arne\"]Counterexample: take $a = d = 2$, $b = c = 1$.[/quote]\n\nBut he said $c 005, I got scared and tried really hard other ways of doing this. now I know I was fine...thanks", "Solution_10": "why is there no loss of generality when we reduce the problem to just 1,2,3,4,5,6?", "Solution_11": "because the actual dimensions of the brick and the box are irrelevant, as long as each of the bricks dimensions has a corresponding dimension of the box that is larger.\r\n\r\nin other words, the 3 dimensions of the brick and the 3 dimensions of the box will have a specific order, like 1,2,3,4,5,6, but it doesn't matter what the actual dimensions are.", "Solution_12": "The first box fits into the second box, if, for some permutation $ \\sigma$, we have:\r\n$ a_{\\sigma(1)} < b_1$, $ a_{\\sigma(2)} < b_2$, $ a_{\\sigma(3)} < b_3$\r\n\r\nNow, when we pick six numbers from that set, it doesn't matter which exact [i]values[/i] we pick for the $ a$'s, but which rank they are within the list (largest, second largest ... smallest).\r\n\r\nSo, we can use any list of six distinct elements.\r\n\r\nBy the way, I think this is a really nice problem. :) \r\n\r\n\r\n[b]Edit:[/b][i] Same as above post[/i]", "Solution_13": "sooooooooooooo nice", "Solution_14": "[hide=What I did\nLet $a$ be the smaller box's dimension, and let $b$ be the bigger box's dimensions.\nWLOG, we can let the $6$ chosen numbers be $1,2,3,4,5,6.$\nWe look for which $3-3$ split for ${a,b}$ means the conditions described.\n\nLet's make a little chart:\n$a:$\n$b:$\n\nObserve that $1$ must be part of $a$ and $6$ must be part of $b.$\nFrom here, we just decide what could happen to each of $2,3,4,5.$\nThere are only $6$ arrangements, and trying every one of them, we realize that all but $1$ work. This one \"bad\" arrangement is when $a=1,4,5$ and $b=2,3,6.$\n\nTotal cases is $\\binom{6}{3}=20.$\n\nThus $\\dfrac{5}{20}=\\dfrac{1}{4}.$\nAdd the numerator and denominator, and we get $5.$\n\n[/hide]\n", "Solution_15": "[hide=What I did]\nLet $a$ be the smaller box's dimension, and let $b$ be the bigger box's dimensions.\nWLOG, we can let the $6$ chosen numbers be $1,2,3,4,5,6.$\nWe look for which $3-3$ split for ${a,b}$ means the conditions described.\n\nLet's make a little chart:\n$a:$\n$b:$\n\nObserve that $1$ must be part of $a$ and $6$ must be part of $b.$\nFrom here, we just decide what could happen to each of $2,3,4,5.$\nThere are only $6$ arrangements, and trying every one of them, we realize that all but $1$ work. This one \"bad\" arrangement is when $a=1,4,5$ and $b=2,3,6.$\n\nTotal cases is $\\binom{6}{3}=20.$\n\nThus $\\dfrac{5}{20}=\\dfrac{1}{4}.$\nAdd the numerator and denominator, and we get $5.$\n\n[/hide]\n\nAlso, can someone tell me the hook-length formula solution and how and why it works here? (I have no experience with this formula)", "Solution_16": "[hide = Solution]\nThe same as everyone above choose $1,2,3,4,5,6$....\n\nWe color the numbers with $a, b$ such that the amount of $b$ does not exceed the number of $a$ from left to right. This is just $$\\frac{C_3}{\\binom{6}{3}} = \\frac{\\frac{1}{4} \\binom{6}{3}}{\\binom{6}{3}} = \\frac{1}{4}$$where $C_n$ is the nth Catalan Number, so the answer is $5$." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $ n>1,k\\in\\mathbb{R}$ and $ 3^{n}\\minus{}1\\equal{}k\\times n$. Prove that $ n\\equal{}2p$", "Solution_1": "I don't understand your problem.\r\nIf $ n|3^{n}\\minus{}1$ then $ 2|n$\r\n$ k\\in R$?", "Solution_2": "yes, I want to prove that n is a odd ($ n\\equal{}2k$) if n such that above condition.\r\n\r\nPS TTsphn: \u00fd m\u00ecnh l\u00e0 ch\u1ee9ng minh $ n$ ch\u1ec9 c\u00f3 th\u1ec3 l\u00e0 s\u1ed1 ch\u1eb5n th\u00ec : $ 3^{n}\\minus{}1$ chia h\u1ebft cho $ n$", "Solution_3": "Let $ p$ is least prime divisible $ n$ , $ d\\equal{}ord_{3}(n)$ .\r\n\r\nWe have : $ 3^{n}\\minus{}1\\vdots p$,$ 3^{p\\minus{}1}\\vdots p$.\r\n\r\nThen :$ n\\vdots d,p\\minus{}1\\vdots d$.\r\n\r\n-Case 1 . $ p>2$ then$ p\\minus{}1\\vdots q$, $ q$ is a prime so $ n\\vdots q$ and $ p>q$ contracdiction .\r\n\r\n-Case $ p\\equal{}2$ we're done :)", "Solution_4": "WE can solve it very easy as following:\r\n$ p$ is smallest prime divisor of n\r\n$ p|\\gcd(3^{n}\\minus{}1,3^{p\\minus{}1}\\minus{}1)\\equal{}2$ so $ p\\equal{}2$\r\nI have some problems has same method example \r\n$ n|3^{n}\\plus{}2^{n}$ prove that $ 5|n$\r\n$ n|(\\frac{p\\plus{}1}{2})^{n}\\plus{}(\\frac{p\\minus{}1}{2})^{n}$ where p is a prime number .\r\nProve that $ p|n$\r\nMore..." } { "Tag": [ "conics", "ellipse", "geometry", "trigonometry", "vector", "calculus", "calculus computations" ], "Problem": "Once we know that the area bounded by a circle of radius $r$ is $\\pi r^{2}$, we can find the length of the circle as follows. When the radius is increased by amount $h$, the area grows by $\\pi((r+h)^{2}-r^{2})$. Dividing by $h$ and letting $h\\to 0$, we find the length of the circle $L(C)=2\\pi r$.\r\n\r\nNow try this with an ellipse: the area bounded by an ellipse with semi-axes $a$ and $b$ is $\\pi ab$. When both semi-axes are increased by $h$, the area grows by $\\pi((a+h)(b+h)-ab)$. Dividing by $h$ and letting $h\\to 0$, we find that $L(E)=\\pi (a+b)$. But this is wrong. Why?\r\n\r\n(In fact, $L(E)\\ge \\pi(a+b)$.)", "Solution_1": "Let $h$ be a small positive number, and consider the area between the ellipse with semi-axes $a$ and $b$ and that with semi-axes $a+h$ and $b+h.$ We have a narrow band of area. As mlok indicated, the area of this narrow band is $\\pi(a+b)h+o(h)$. Now, what is the width of the band? If the width were $h,$ then we would be perfectly justified in dividing by $h$ and letting $h\\to0.$ But is the width really $h?$\r\n\r\nConsider a parametric representation of the ellipses. Let $P(t)=(a\\cos t,b\\sin t)$ and $Q(t)=((a+h)\\cos t,(b+h)\\sin t).$ $P$ is a point on the inner rim, and $Q(t)$ is a point on the outer rim. The distance between $P(t)$ and $Q(t)$ is $h.$\r\n\r\nSo does that clinch it? Not so fast - yes, $P(t)$ and $Q(t)$ were \"corresponding\" points in some vague sense, but was the distance between then \"straight across\" the band? To answer that, we need to know what direction \"straight across\" should be. For that we'll take the normal direction to the inner ellipse. (That's not identical to the normal direction to the outer ellipse, but the difference in directions is small for $h$ small.) A unit vector in that normal direction is \r\n\r\n$\\vec{u}=\\frac{\\langle b\\cos t,a\\sin t\\rangle}{\\sqrt{b^{2}\\cos t+a^{2}\\sin t}}$\r\n\r\nWhereas a unit vector in the direction from $P(t)$ to $Q(t)$ is\r\n\r\n$\\vec{v}=\\langle \\cos t,\\sin t\\rangle.$\r\n\r\nThe two unit vectors are not identical. Their dot product, which is the cosine of the angle between them, is\r\n\r\n$\\vec{u}\\cdot\\vec{v}=\\frac{b\\cos^{2}t+a\\sin^{2}t}{\\sqrt{b^{2}\\cos^{2}t+a^{2}\\sin^{2}t}}.$\r\n\r\nIf $a\\ne b$ and $t$ is not a multiple of $\\frac{\\pi}2,$ that cosine is less than 1. The distance $h$ from $P(t)$ to $Q(t)$ is not the distance straight across the band, it is the length of an angled path. The distance straight across is $h$ times the dot product given above, which is less than $h.$ Since the width of the band averages to less than $h,$ then the arclength must be greater than $\\pi(a+b)$ to keep the area the same." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Problem. Give $ A \\equal{} {(a_1;...;a_n)}$ and $ B \\equal{} {(b_1;...;b_m)}$ with $ n \\leq m$:\r\na. Is injection from $ A$ to $ B$?\r\nb. Are injection and surjection from $ A$ to $ B$?", "Solution_1": "a. there's an injection as long as we can map any two elements in $ A$ to different elements in $ B$, and this is possible because $ n\\leq m$. \r\n\r\nb. for injection, see a. a map is a surjection if every element in $ B$ is the image of an element in $ A$. this is only possible if $ m\\equal{}n$." } { "Tag": [ "ceiling function" ], "Problem": "Tamantha wants to buy boxes in which to put her 77 compact discs. If each box holds 12 discs, how many does she need to buy?", "Solution_1": "She has 77 discs, and each box holds 12 discs. So we need to find $ \\lceil{\\frac{77}{12}}\\rceil$, which is $ \\boxed{7}$.\r\n\r\nNote that $ \\lceil{n}\\rceil$ means the smallest integer greaten than than $ n$." } { "Tag": [ "probability", "floor function", "trigonometry", "limit", "function", "vector", "modular arithmetic" ], "Problem": "For fixed integer m, let Xn have Binomial (n, 1/2) distribution. Let Yn = (Xn mod m) take values in Im = {0,1,2,..,m-1}\r\nSHow that the total variation distance between the distribution of Yn and uniform distribution on Im converges to 0 as n -> infinity.\r\n\r\nFirst of all, what is Yn???? Say Xn has probability (n choose k) (1/2)^n....does it mean Yn has probability from one of : (1 choose K)(1/2) ...... (m choose k) (1/2)^m???\r\n\r\nAnd what does it mean by Yn take values in Im???", "Solution_1": "What we actually have is this. For $ 0\\le k\\le m \\minus{} 1,$\r\n\r\n$ P(Y_n \\equal{} k) \\equal{} \\sum_{j \\equal{} 0}^{\\lfloor (n \\minus{} k)/m\\rfloor}P(X_n \\equal{} k \\plus{} jm) \\equal{} \\sum_{j \\equal{} 0}^{\\lfloor (n \\minus{} k)/m\\rfloor}\\binom{n}{k \\plus{} jm}\\frac1{2^n}.$\r\n\r\nMy gut instinct is to seek a transform method. I'll find the Fourier transform (on $ \\mathbb{Z}$) of the binomial distribution, then use Poisson summation ideas to find the discrete Fourier transform of the distribution of $ Y_n$ (which can be seen to be periodic).", "Solution_2": "Expanding on that idea:\r\n\r\nLet $ f_n(x)\\equal{}\\sum_{k\\equal{}0}^nP(X_n\\equal{}k)e^{\\minus{}2\\pi ikx}.$\r\n\r\nBy the binomial theorem, we can compute that $ f_n(x)\\equal{}\\frac1{2^n}(1\\plus{}e^{\\minus{}2\\pi ix})^n.$\r\n\r\nNow let $ g_n(\\ell)\\equal{}\\sum_{k\\equal{}0}^{m\\minus{}1}P(Y_n\\equal{}k)e^{\\minus{}2\\pi ik\\ell/m}.$\r\n\r\nThe theorem I'll quote (the proof is just a manipulation of sums) is that\r\n\r\n$ g_n(\\ell)\\equal{}f_n\\left(\\frac{\\ell}{m}\\right).$ Hence, $ g_n(\\ell)\\equal{}\\frac1{2^n}(1\\plus{}e^{\\minus{}2\\pi i\\ell/m})^n.$\r\n\r\nBut $ |1\\plus{}e^{2\\pi i\\ell/m}|\\equal{}2\\left|\\cos\\left(\\frac{\\pi \\ell}{m}\\right)\\right|.$ Since this is strictly less than $ 2$ for all $ \\ell\\ne0,$ we have this:\r\n\r\n$ \\lim_{n\\to\\infty}g_n(\\ell)\\equal{}\\begin{cases}1,&\\ell\\equal{}0\\\\0&\\ell\\ne 0\\end{cases}.$\r\n\r\nThis is exactly the discrete Fourier transform we would get from the \"discrete uniform\" distribution that has probability function $ \\frac1m$ at each value in $ \\{0,1,\\dots,m\\minus{}1\\}.$\r\n\r\nAs for the nature of the convergence: all norms are equivalent in a finite dimensional vector space.\r\n\r\n---\r\n\r\nNow I claim that there was nothing particularly special at all about the binomial distribution with $ p\\equal{}\\frac12.$\r\n\r\nClaim: Suppose $ X_j$ is an i.i.d. sequence of integer-valued random variables satisfying [condition to be named later]. Fix $ m.$ Define $ Y_n\\equiv\\sum_{j\\equal{}1}^nX_j\\pmod{m}.$ Then $ Y_n$ converges (in any reasonable sense) to the \"uniform discrete\" distribution on $ \\{0,1\\dots,m\\minus{}1\\}$ as $ n\\to\\infty.$\r\n\r\nCan someone fill in a reasonable [condition to be named later] and prove this? The techniques of this post should be sufficient.", "Solution_3": "Sorry for my stupidness,\r\nbut where does the summation come from?\r\nYn = Xn (mod m)\r\nSo how does the P(Yn=k) is a sum?\r\nDo you mind to explain this bit a little more?", "Solution_4": "Let's suppose $ m\\equal{}3$ and $ n\\equal{}8.$ Then $ X_8$ takes on the values $ 0,1,2,3,4,5,6,7,8$ with probabilities $ \\frac1{256},\\frac8{256},\\frac{28}{256},\\frac{56}{256},\\frac{70}{256},\r\n\\frac{56}{256},\\frac{28}{256},\\frac8{256},\\frac1{256}.$\r\n\r\nNow, $ Y_8\\equiv X_8\\pmod{3}.$ $ Y_8$ takes on only the value $ 0,1,2.$\r\n\r\nWe have $ Y_8\\equal{}0$ whenever $ X_8\\equiv 0\\pmod{3},$ which is whenever $ X_8\\equal{}0,3,$ or $ 6.$ Hence:\r\n\r\n$ P(Y_8\\equal{}0)\\equal{}P(X_8\\equal{}0)\\plus{}P(X_8\\equal{}3)\\plus{}P(X_8\\equal{}6)\\equal{}\\frac{1\\plus{}56\\plus{}28}{256}\r\n\\equal{}\\frac{85}{256}.$\r\n\r\nWe have $ Y_8\\equal{}1$ whenever $ X_8\\equiv 1\\pmod{3},$ which is whenever $ X_8\\equal{}1,4,$ or $ 7.$ Hence:\r\n\r\n$ P(Y_8\\equal{}1)\\equal{}P(X_8\\equal{}1)\\plus{}P(X_8\\equal{}4)\\plus{}P(X_8\\equal{}7)\\equal{}\\frac{8\\plus{}70\\plus{}8}{256}\r\n\\equal{}\\frac{86}{256}.$\r\n\r\nWe have $ Y_8\\equal{}2$ whenever $ X_8\\equiv 2\\pmod{3},$ which is whenever $ X_8\\equal{}2,5,$ or $ 8.$ Hence:\r\n\r\n$ P(Y_8\\equal{}2)\\equal{}P(X_8\\equal{}2)\\plus{}P(X_8\\equal{}5)\\plus{}P(X_8\\equal{}8)\\equal{}\\frac{28\\plus{}56\\plus{}1}{256}\r\n\\equal{}\\frac{85}{256}.$" } { "Tag": [ "function", "vector", "real analysis", "real analysis unsolved" ], "Problem": "let $ C$ be the algebra of functions defined on $ \\mathbb{R}$,continuous with complex values.\r\n\r\nlet $ p\\in\\mathbb{N}$ and $ E(p)$ the set of all functions $ f$ belonging to $ C$ for which there are two real $ (A,B)\\in\\mathbb{R}^{2}$, such that $ \\forall x\\in\\mathbb{R}$ we have $ |f(x)|\\leq\\ A\\plus{}B|x|^{p}$\r\n\r\nshow that: for every $ p$ we have $ E(p)\\subset E(p\\plus{}1).$", "Solution_1": "Use (weighted) AM-GM. $ |x|^p\\le \\frac{p}{p\\plus{}1}\\plus{}\\frac{1}{p\\plus{}1}|x|^{p\\plus{}1}$, so $ A\\plus{}B|x|^p\\le \\left(A\\plus{}\\frac{pB}{p\\plus{}1}\\right)\\plus{}\\frac{B}{p\\plus{}1}|x|^{p\\plus{}1}$.", "Solution_2": "thanks jmerry!\r\n\r\nI have another question:\r\n\r\nhow to prove that the vector space $ E(p)$ is infinite-dimensional?", "Solution_3": "$ E(p)$ certainly contains all bounded functions.\r\nLet $ f_i$ be a nonzero continuous function bounded by $ 1$ where $ f$ vanishes outside the interval $ [i,i\\plus{}1]$. Then the $ f_i$ are independent." } { "Tag": [], "Problem": "I am really stuck on this one, any input?\r\n\r\nLet S be the set of ordered pairs of integers that are defined in the following\r\nway:\r\n\u2022 (0, 0) 2 S.\r\n\u2022 If (a, b) 2 S then (a + 2, b + 3) 2 S.\r\n\u2022 If (a, b) 2 S then (a + 3, b + 2) 2 S.\r\nProve the following:\r\nFor every element (a, b) in S, we have that a + b is divisible by 5.\r\n\r\nThanks guys", "Solution_1": "[hide]Every ordered pair can be written as $ (2x \\plus{} 3y, 3x \\plus{}2y)$ for nonnegative integers $ x,y$. Also $ 5x \\plus{} 5y$ is divisible by $ 5$.[/hide]", "Solution_2": "How do you know that, say $ (1,1)$ in not in $ S$? I mean the following $ S$,\r\n\r\n$ S\\equal{}\\{(2x\\plus{}3y,3x\\plus{}2y),(1\\plus{}2s\\plus{}3t,1\\plus{}3s\\plus{}2t): x,y,s,t\\text{ non negative integers}\\}$\r\n\r\nalso satisfies the 3 conditions but makes the claim not true.\r\n\r\nMy point is that the definition of $ S$ doesn't say that $ S$ has $ (0,0)$ as its only \"initial point\", does it? Or perhaps it does?", "Solution_3": "Unfortunatly the text was not formatted properly. But you can assume that in the places where the number (2) are next to the round brackets are there to mean that the orderd pair belongs to S.\r\n\r\nIn other words:\r\n\r\n(0,0) belongs to S\r\n(a,b) belongs to S \r\n\r\nand so forth.", "Solution_4": "[hide=\"Solution\"]\nWe are given that the transformation are $ (x,y) \\rightarrow (x \\plus{} 2,y \\plus{} 3)$ and $ (x.y) \\rightarrow (x \\plus{} 3,y \\plus{} 2)$\n\nConsider $ S(x,y) \\equal{} x \\plus{} y$\n\nClearly for $ 5|S(0,0)$\n\nAssume that $ 5|S(x,y)$ for all points which have been generated so far.\n\nThen we can go to two more points. Consider one of them (the other one is obvious once the other is shown.\n\n$ S(x \\plus{} 2,y \\plus{} 3) \\equal{} x \\plus{} y \\plus{} 2 \\plus{} 3 \\equal{} x \\plus{} y \\plus{} 5$ But $ 5|x \\plus{} y$ therefore $ 5|x \\plus{} y \\plus{} 5 \\equal{} S(x,y)$\n[/hide]", "Solution_5": "When you write 5|S(0,0) what do you mean by 5bar?", "Solution_6": "Technically this problem is false as packham's point is legitimate, though it's clear what the intent is. There's two fixes I can see.\r\n1. Define a 2-dimensional sequence than a set.\r\n2. Let $ T$ be the intersection of all such sets $ S$ and prove stuff about $ T$ instead." } { "Tag": [ "geometry", "rectangle" ], "Problem": "Find the positive difference between the total number of rectangles and the total number of squares in a 5X3 grid. (please show how the number of recs and squares was acquired)", "Solution_1": "[hide=\"Solution\"]WLOG, assume $ n\\geq m$ in an $ m\\times n$ grid.\n\nA rectangle is defined by two horizontal and two vertical lines. There are $ \\binom{m + 1}{2}\\cdot\\binom{n + 1}{2}$ rectangles.\n\nIn the case of squares only, two horizontal lines and one vertical line define a square (WLOG, the second vertical line will always be below the first one). There are $ n\\cdot\\binom{m + 1}{2}$ squares.\n\\begin{align*}\\binom{m + 1}{2}\\cdot\\binom{n + 1}{2} - n\\cdot\\binom{m + 1}{2} & = \\binom{m + 1}{2}\\cdot\\left(\\binom{n + 1}{2} - n\\right) \\\\\n \\\\\n& = \\binom{m + 1}{2}\\cdot\\left(\\frac {(n + 1)!}{(n - 1)!2!} - n\\right) \\\\\n \\\\\n& = \\binom{m + 1}{2}\\cdot\\frac {n(n + 1) - 2n}{2} \\\\\n \\\\\n& = \\binom{m + 1}{2}\\cdot\\frac {n(n - 1)}{2} \\\\\n \\\\\n& = \\binom{m + 1}{2}\\cdot\\binom{n}{2}\\end{align*}\nAlthough I have no idea how to get this result with pure combinatorics, there are $ \\binom{m + 1}{2}\\cdot\\binom{n}{2}$ non-square rectangles in an $ m\\times n$ grid with $ n\\geq m$.\n\nFor this particular problem, $ \\binom{3 + 1}{2}\\cdot\\binom{5}{2} = 6\\cdot10 = \\boxed{60}$.[/hide]" } { "Tag": [ "geometry", "AMC", "AIME", "MATHCOUNTS", "probability", "function", "counting" ], "Problem": "I will be in Birmingham this Thursday and Friday. I will be guest teaching at several area schools including Vestavia Hills High School and the Alabama School of Fine Arts. I am trying to schedule visits with Pizitz Middle School and Hoover High School.\r\n\r\nIf there are any topics high school students would like to discuss before the AIME, put in your request now. If there is a preferred lesson for middle school students before state MATHCOUNTS, let me know and I will try to arrange material for these topics before my flight tomorrow.", "Solution_1": "Probability? I have a good amount of trouble with that, though I doubt anything I could be taught in an hour would help too much. \r\n\r\nGeneral AIME test-taking and problem-solving skills would help.", "Solution_2": "I have a great probability lecture virtually memorized.", "Solution_3": "Really...then I'm sure quite a few of us would appreciate it (Mrs. Tipton was absent the entire probability chapter, and Kevin and I are both fairly weak in the area).", "Solution_4": "I was there the last time you came to Birmingham... I thought the problems were too hard... at least for middle school/sixth grade... my dad thought I was a little stupid for not knowing these stuff and tried to explain. If you come to Pizitz again, you should try to teach something about geometry, or something from Mathcount, because mathcount district is this saturday", "Solution_5": "[quote=\"LuCy4EvA\"]I was there the last time you came to Birmingham... I thought the problems were too hard... at least for middle school/sixth grade... my dad thought I was a little stupid for not knowing these stuff and tried to explain. If you come to Pizitz again, you should try to teach something about geometry, or something from Mathcount, because mathcount district is this saturday[/quote]\r\n\r\nThat talk was very tough even for 8th graders. I wanted to display how much math there is that is out there to learn. If I come talk again, it will be on MATHCOUNTS level topics.\r\n\r\nHowever, Mrs. Chilton never got back to me, so it looks like I won't be visiting Pizitz this time.", "Solution_6": "Excellent talk you gave, thanks very much... I especially liked the generating functions idea...it really helps alot....\r\n\r\nJust to kind of go off on what you were saying at the end --\r\n\r\nGiven n variables, (ie. a + b + c), that are supposed to equal a number, m, and all the variables are non negative, then the number of possible n-tuples that equal m is [tex]{m+2\\choose n-1}[/tex] ?", "Solution_7": "[quote=\"Elemennop\"]Excellent talk you gave, thanks very much... I especially liked the generating functions idea...it really helps alot....\n\nJust to kind of go off on what you were saying at the end --\n\nGiven n variables, (ie. a + b + c), that are supposed to equal a number, m, and all the variables are non negative, then the number of possible n-tuples that equal m is [tex]{m+2\\choose n-1}[/tex] ?[/quote]\r\n\r\nWith $n$ variables, there are $n-1$ dividers around the $m$ unit objects. Of the $m+n-1$ objects and dividers, there are $\\binom{m+n-1}{n-1} = \\binom{m+n-1}{m}$ ways to order the $m$ identical unit objects and $n-1$ identical dividers. This is exactly like finding the number of ways to arrange $m$ A's and $n-1$ B's to make distinct \"words.\"\r\n\r\nI'm glad you enjoyed the talk. It's too bad I didn't have enough time to do generating functions justice.", "Solution_8": "Much thanks for the talk. It was very interesting and thought-provoking like always (like the class I should be at now...). \r\n\r\nWere you able to connect our online and real life identities? :) Maybe we should have mentioned it... \r\n\r\n[quote=\"MCrawford\"]With $n$ variables, there are $n-1$ dividers around the $m$ unit objects. Of the $m+n-1$ objects and dividers, there are $\\binom{m+n-1}{n-1} = \\binom{m+n-1}{m}$ ways to order the $m$ identical unit objects and $n-1$ identical dividers. This is exactly like finding the number of ways to arrange $m$ A's and $n-1$ B's to make distinct \"words.\"[/quote]\r\n\r\nAs a side note, this is equivalent to what I remember as the 'balls in urns' formula. That is, the number of ways that $b$ indistinguisable balls can be placed into $u$ distinguisable urns is $\\binom{b+u-1}{b}$. \r\n\r\nFor the [i]$n$ non-negative integer variables with sum $m$[/i] counting problem, each variable would be an urn and $m$ balls would make up the final sum. You could also think of it more like you did as $m$ objects lined up with $n-1$ dividers. The gaps between the objects and positions before and after are where the $n-1$ can be placed, so those locations would be the urns, and the dividers would be the balls. Gaps between ($m-1$) and positions before and after total to $m+1$ for urns. That and $n-1$ balls will give you the same thing when plugged into the formula. Whatever makes the most sense to you.\r\n\r\nBy the way, I learned about this in Zeitz's [i]Art and Craft of Problem Solving[/i].", "Solution_9": "Yes, binomial coefficients pop up everywhere. As long as you can make the analogies solidly, you can see how they tie an enormous amount of math together.\r\n\r\nI'm glad you're reading Zeitz. Great book.", "Solution_10": "What trip?", "Solution_11": "[hide]What trip??[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "How would I get $\\mathbb{N}$ in TeXnicCenter?\r\nIs there some package I need to use to view it? \r\n\r\nCurrently, when I type that, I'm gettng \"undefined control sequence\"\r\n\r\n\r\n\r\n\r\nNever mind, I added the 5 packages ( \\usepackage{amsmath}\r\n\\usepackage{amssymb}\r\n\\usepackage{amsthm}\r\n\\usepackage{array}\r\n\\usepackage{xy})\r\n\r\nto the beginning and it's fine now.\r\n\r\n\r\n\r\n\r\nHowever, something else happened, now it says:\r\n[quote]\nPackage amsmath warning: Foreign command \\atopwithdelims; \\frac or \\genfrac should be used instead on input line 392.[/quote]\n\nBut on line 392, all I have is [quote] Evaluate $\\sum_{k=0}^{n}2^{k}{9\\choose 3}$[/quote]\r\n\r\nwhat's wrong?", "Solution_1": "That's a LaTeX warning which won't stop the compilation (unlike an Error) and in this case is an informational message that in the amsmath package \\choose has been superseded by \\binom.\r\nIf you use\r\n[color=red]\\sum_{k=0}^{n}2^{k}\\binom{9}{3}[/color]\r\nyou will get $\\sum_{k=0}^{n}2^{k}\\binom{9}{3}$ with no warnings.\r\n\r\nTo explain part of the warning: \\genfrac allows one to define all sorts of fractions (different brackets, lines between top & bottom etc) and amsmath just defines \\frac and \\binom in terms of \\genfrac. You can do outrageous things with \\genfrac, for example: $\\genfrac{(}{]}{5pt}{2}{9}{3}$ :wink:", "Solution_2": "[quote=\"stevem\"]That's a LaTeX warning which won't stop the compilation (unlike an Error) and in this case is an informational message that in the amsmath package \\choose has been superseded by \\binom.\nIf you use\n[color=red]\\sum_{k=0}^{n}2^{k}\\binom{9}{3}[/color]\nyou will get $\\sum_{k=0}^{n}2^{k}\\binom{9}{3}$ with no warnings.\n\nTo explain part of the warning: \\genfrac allows one to define all sorts of fractions (different brackets, lines between top & bottom etc) and amsmath just defines \\frac and \\binom in terms of \\genfrac. You can do outrageous things with \\genfrac, for example: $\\genfrac{(}{]}{5pt}{2}{9}{3}$ :wink:[/quote]\r\n\r\nWow, cool, thanks. I decided to not use amsmath for the time being...", "Solution_3": "For the $\\mathbb{N}$ problem, you need the amsfonts package." } { "Tag": [], "Problem": "solve 1\\a + 1\\b + 1\\c + 1\\d + 1\\e = 7\\3.a,b,c,d,e are natural nos.", "Solution_1": "[hide=\"Solution\"]\n\nTake $ a\\ge b\\ge c\\ge d\\ge e$.\n\nNow $ \\frac73\\le \\frac5e$, so $ e\\le2$. Now try solving with the same procedure for a,b,c,d... [/hide]" } { "Tag": [], "Problem": "Let $n$ be a prime number. Find all $x\\in N$ such that\r\n$(1^n+2^n+...+x^n)+(1^n+2^n+...+(n-1)^n) = 1^n+2^n+...+(2n-1)^n$", "Solution_1": "Well i think is easy to see that (2n-1)^n is far much bigger than 1^n+2^n+3^n+...+n^n\r\n\r\nlets do it by induction...for 2 it happens let say for n-1 happens\r\n\r\nthe 1^(n-1)+2^(n-1)+...+(n-1)^(n-1) < (2n-3)^(n-1)lets multiply it by (n-1) and we get:\r\n\r\n(n-1)(2n-3)^(n-1) > (n-1) (1^(n-1)+2^(n-1)+...+(n-1)^(n-1))\r\n\r\n(n-1) (1^(n-1)+2^(n-1)+...+(n-1)^(n-1)) > 1^n+2^n+3^n+...+(n-1)^n\r\n\r\nthen (n-1) (2n-1)^(n-1) > (n-1)(2n-3)^(n-1) > 1^n+2^n+3^n+...+(n-1)^n\r\n\r\nand (n-1) (2n-1)^(n-1) + n^n > 1^n+2^n+3^n+...+n^n\r\n\r\nbut (n-1) (2n-1)^(n-1) < (2n-1)^n\r\n\r\nso the problem reduces that :\r\n\r\n1^n+2^n+...+(2n-2)^n + (2n-1)^n > 1^n+2^n+...+(2n-2)^n + 1^n+2^n+3^n+...+n^n\r\n\r\nso (2n-1) > k > (2n-2) but it can\u00b4t be so for n>1 there\u00b4s no solution.\r\n\r\nthe only solution is n=1 k=0" } { "Tag": [ "quadratics", "algebra", "inequalities proposed", "inequalities" ], "Problem": "Let x,y,z are real number such that x+y+z = 2004. Find the maximum of\r\nM = sinC*xy+sinA*yz+sinB*zx . Here ABC is a given triangle.", "Solution_1": "[quote=\"treegoner\"]Let $x,y,z$ be real number such that $x+y+z = 2004$. Find the maximum of\n$M = sinC\\cdot xy+sinA\\cdot yz+sinB\\cdot zx$ . Here $ABC$ is a given triangle![/quote]\r\n\r\n\r\n[color=darkred]Happy Christmas!![/color]\r\n\r\n :)", "Solution_2": "[quote=\"treegoner\"]Let x,y,z are real number such that x+y+z = 2004. Find the maximum of\n$M = xysinC+yzsinA+xzsinB$ . Here ABC is a given triangle.[/quote]\r\n\r\n The first: Merry Christmas( even i'm not a Christian :D )\r\nSecond: With this type of problem mean there is no restiction for variables x,y,z absolutely Lagrange undetermine multipliers will be the most effective.. \r\n If u don't want to use this method,the fact that : $M={2004}^2 \\frac{M}{{2004}^2}= \\frac{{2004}^2M}{(x+y+z)^2}$ will be useful by treat 2 variables( such as y,z as parametters ) and combine with the condition of positive delta for a quadratic equation of x only..Ok ( I hope my solution is not too abstract )" } { "Tag": [ "factorial" ], "Problem": "I've found that I can't do problems similar in this example...\r\n\r\nx^y, find the units digit in this number \r\n\r\nand other problems that are very similar to the example I gave.\r\n\r\nCan you guys help me approach to how to solve this type of problem?\r\n\r\nThanks", "Solution_1": "You basically need to find a pattern. For example, in 3^2004, I would see that 3*3=9, 9*3=27, 7*3=21, 1*3=3. So the pattern is 3, 9, 7, 1. Since 2004 has a remainder of 0 when dividing by 4, you know that the answer is 1.\r\n\r\nHope this helps.", "Solution_2": "Oh yeah...I understand now. Thanks.\r\n\r\nBut is there a more complex example maybe in the form of...\r\n\r\nx^y * z^n\r\n\r\nor something like that...the more complex the better...hehe", "Solution_3": "Its exactly the same. All you need to know that if you want the units digit of a*b, its the units digit of a * the units digit of b (this is what modulo 10 is). So you would just find x^y and z^n separately, multiply them, and take the units digit.", "Solution_4": "Yeah, eg:\r\n\r\nFind the units digit of:\r\n\r\n4^2004 * 1999^4\r\n\r\nUnits digit of 4^2004:\r\n\r\nwell look for the pattern (always fun):\r\n\r\n4*4 = 16, 4*16 = 64, 4*64 = 256\r\nSo it goes 4, 6, 4, 6, when we have 4^2, 4^3, 4^4. Even powers of 4 given units digits of 6.\r\n\r\nUnits digit of 1999^4, well since tens or any higher digit cannot influence the units digit, it is just the units digit of 9^4. which is 9*9*9*9, 81*81, or 1*1.\r\n\r\nMultiply these together, 1*6 = 6\r\n\r\nSo the units digit of 4^2004 * 1999^4 is 6! (that is just 6, not 6 factorial)\r\nNote the two examples of m^n, the former when m<>n.", "Solution_5": "How would you go about doing a question like this: ( 2^k)(5^300)=a 303 digit interger, what is the sum of the intergers?", "Solution_6": "I assume you meant the sum of the digits. There are several answers to this question, three to be precise.\r\n\r\nThe main point here is to note that [tex]2^{300}*5^{300}=10^{300}.[/tex] [tex]10^{300}[/tex] has 301 digits: a 1 followed by 300 0s. By increasing k by 1, you are multiplying the number by 2. So let's make a chart:\r\n[code]\nk=300 n= 100000...\nk=301 n= 200000...\nk=302 n= 400000...\nk=303 n= 800000...\nk=304 n= 1600000...\nk=305 n= 3200000...\nk=306 n= 6400000...\nk=307 n= 12800000...\nk=308 n= 25600000...\nk=309 n= 51200000...\nk=310 n=102400000...\n[/code]\r\n\r\nAs you can see n is a 303-digit number only for k=307, 308, or 309. Finding the sum of the digits is easy, because there are only 3 digits you have to add; the rest are 0s. So the possible answers are 11, 13, and 8.", "Solution_7": "Rep123, the question does not ask the sum of the digits of the smallest such integer, so I believe there are three valid answers.", "Solution_8": "Thanks both of you. 11 is the answer i needed, I missed the part of the question which said that k was the smallest positive interger.", "Solution_9": "Thanks everyone. You covered a lot of questions that I was about to ask.\r\n\r\nAre there anymore questions that are similar to this except the question is asking for something else and maybe not the units digit or the sum of the digits." } { "Tag": [ "number theory", "number theory proposed" ], "Problem": "Discuss the convergence of\r\n\\[ \\sum_{n\\equal{}1}^{\\infty}\\left(\\frac{p_n}{2p_{n\\plus{}1}\\minus{}p_{n\\plus{}2}}\\right)^n\\]\r\nwhere $ p_k$ is the k-th prime.", "Solution_1": "It is not converge, because exist h, suth that $ 2p_{n \\plus{} 1} < p_n \\plus{} p_{n \\plus{} 2} \\plus{} h$ for infinetely many n.", "Solution_2": "Of course (if it is true) it immediately implies that my series is divergent.\r\n\r\nI have checked in 3 books about number theory, but none of them contains the fact you wrote or even something similar.\r\n\r\n\r\nCould you explain it (give a sketch of proof) or say where it can be found?\r\n\r\n\r\ntkank you", "Solution_3": "It is easy to prove for h=0 (h>0 followed from h=0).\r\nLet exist only finite numbers n, suth that $ p_{n\\plus{}1}\\minus{}p_n0}\\to\\mathbb R_{>0},\\ x\\mapsto\\sqrt[n]x$ for $n=2,3,\\ldots$. Here, \"find\" means at least some algorithm which decides whether a given polynomial with rational coefficients has a root in $K$ or not.\r\n\r\n(I have no solution to this problem.)\r\n\r\nowk", "Solution_1": "Isn't this the maximal solvable extension of $\\mathbb{Q}$? (-> There are algorithms to compute Galois groups and deciding solvability)\r\n\r\n\r\noops, *real* -> so $\\mathbb{Q}^\\text{solv}\\cap \\mathbb{R}$?", "Solution_2": "This field is full of weird corner cases, but I can tell you this- it's as far from Galois as possible. The order is built into the algebra, and that means that it has no nontrivial automorphisms.\r\n\r\nIt's not the intersection of solvable numbers and real numbers; this field doesn't contain the roots of a typical irreducible (over $\\mathbb{Q}$) cubic with all three roots real.", "Solution_3": "uhm..\r\nwhat do you mean by \"the order is built into the algebra\"?\r\nand why can't it have nontrivial automorphisms? if you send $\\sqrt2 \\mapsto-\\sqrt2$, can't you extend this from an automorphism of $\\mathbb{Q}(\\sqrt2)$ to an automorphism of the whole field?", "Solution_4": "You can't send $\\sqrt{2}$ to $-\\sqrt{2}$, since one has a square root and the other doesn't. A number is positive in this field (a topological condition) if and only if it has a square root (an algebraic condition). Therefore, any automorphism must send positive numbers to positive numbers; this forces the automorphism to fix everything." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "[color=darkblue]Prove that: $ \\boxed{n^{n\\plus{}3}\\plus{}(n\\plus{}1)^{n\\plus{}3}<(n\\plus{}2)^{n\\plus{}3}, \\forall n \\in N^*}$[/color]", "Solution_1": "[quote=\"thanhnam2902\"][color=darkblue]Prove that: $ \\boxed{n^{n \\plus{} 3} \\plus{} (n \\plus{} 1)^{n \\plus{} 3} < (n \\plus{} 2)^{n \\plus{} 3}, \\forall n \\in N^*}$[/color][/quote]\r\nby binomial Newton's : \r\n $ (n \\plus{} 1 \\plus{} 1)^{n \\plus{} 3} \\geq (n \\plus{} 1)^{n \\plus{} 3} \\plus{} n(n \\plus{} 1)^{n \\plus{} 2} \\geq (n \\plus{} 1)^{n \\plus{} 3} \\plus{} n^{n \\plus{} 3}$ ( right because of $ (n \\plus{} 1)^{n \\plus{} 2}\\geq n^{n \\plus{} 2}$)", "Solution_2": "[color=darkblue]It's true and nice![/color]", "Solution_3": "it is a weak one." } { "Tag": [ "real analysis", "real analysis theorems" ], "Problem": "Is Picard's Theorem \r\n\r\nhttp://mathworld.wolfram.com/PicardsExistenceTheorem.html\r\n\r\nneccasary and sufficent for the differential equation\r\n\r\n$\\frac{dx}{dt} = f(x,t)$ with $x(t_0)=x_0$\r\n\r\nto have a unique solution.\r\n\r\nObviously it is sufficent, but is it neccasary?", "Solution_1": "Never mind, I have resolved my issues :)" } { "Tag": [ "quadratics", "number theory", "prime factorization", "algebra", "quadratic formula" ], "Problem": "Let $p\\ge 5$ be a prime number. Prove that $p^{a}+p^{b}$ is never a perfect square, not even a rational square, where $a,b$ are integers.", "Solution_1": "about perfect square , i have an idea\r\nassume that , there's exist x where $p^{a}+p^{b}=x^{2}$ , WLOG $a \\geq b$ , so that $x^{2}=p^{b}(p^{a-b}+1)$ , we can easily see that b is even and $p^{a-b}+1$ is also a perfect square\r\nlet $y^{2}=p^{a-b}+1$ , so that $(y-1)(y+1)=p^{a-b}$ , we obtain $p^{a-b}|2$ , but $p \\geq 5$ then it's a contracdition", "Solution_2": "your solution is almost right. but why should $p^{a-b}|2$?\r\n [hide=\"you can continue like this\"]since we have $y$ is even so $\\gcd(y+1,y-1)=1$ so $(y+1)(y-1)=p^{a-b}\\Longrightarrow y+1=p^{m},y-1=p^{n}$\nthus $p^{m}-p^{n}=2$,contradiction.[/hide]\r\nokey, now who can prove it for rational squares i.e. when $a$ or $b$ or both $<0$? :wink:", "Solution_3": "[quote=\"nayel\"]\nokey, now who can prove it for rational squares i.e. when $a$ or $b$ or both $<0$? :wink:[/quote]\r\n\r\nIt's the same argument. You'd multiply both sides by $p^{2}$ enough times to make both sides integral. The rationals have unique prime factorization over the integers with integer exponents, so you again get $p^{a-b}+1$. If we let one of $y \\pm 1$ then be $\\frac{p^{a-b}}{q}$ with q an integer, we have $p^{a-b}\\pm 2q =q^{2}$, quadratic formula,\r\n$q=\\pm q \\pm \\sqrt{q^{2}+p^{a-b}}$, that thing in the square root is clearly a multiple of $q^{2}$, so $q|p^{a-b}\\implies q|p$, so $q=1$, we're done..." } { "Tag": [ "\\/closed" ], "Problem": "enough said. IT'S UGLY", "Solution_1": "Use the ML skin if you want...", "Solution_2": "I was using Firefox when it was ugly. In Safari, it isn't ugly. Maybe it's Firefox?", "Solution_3": "Can you post a screen shot?", "Solution_4": "I don't know how to post a screenshot on a Mac. :(", "Solution_5": "Command (The apple logo) Shift 3.\r\nIf you want to select a certain region, use command shift 4.\r\nIt will then put a PDF on your desktop and will say \"Picture 1\".\r\n(That is what it does for the version of mac I have.)" } { "Tag": [ "logarithms", "algebra unsolved", "algebra" ], "Problem": "Find t:\r\n\r\n$ p^t\\plus{}\\frac{1}{p^t}\\equal{}\\frac{p\\plus{}1}{\\sqrt{p}}$", "Solution_1": "$ p^t \\plus{} \\frac {1}{p^t} \\equal{} \\frac {p \\plus{} 1}{\\sqrt {p}}$\r\n$ (p^t \\plus{} \\frac {1}{p^t}).\\sqrt {p} \\equal{} p \\plus{} 1$\r\n$ (p^t \\plus{} \\frac {1}{p^t}).p^{\\frac{1}{2}} \\equal{} p \\plus{} 1$\r\n$ p^{t\\plus{}\\frac{1}{2}} \\plus{} \\frac {1}{p^{t\\minus{}\\frac{1}{2}}} \\equal{} p \\plus{} 1$\r\n$ t\\equal{}\\frac{1}{2}$", "Solution_2": "Denote:\r\n\r\n$ p^t = x$\r\n\r\nthen:\r\n\r\n$ x + x^{ - 1} = \\frac {p + 1}{\\sqrt {p}}$\r\n\r\n$ \\sqrt {p} x^2 - x(p + 1) + \\sqrt {p} = 0$\r\n\r\n$ x_{1,2} = \\frac { p + 1 \\pm \\sqrt {\\left(p + 1\\right)^2 - 4 p}}{2 \\sqrt {p}}$\r\n\r\n$ x_{1,2} = \\frac { p + 1 \\pm \\sqrt {\\left(p - 1\\right)^2}}{2 \\sqrt {p}}$\r\n\r\n$ x_1 = \\sqrt {p}$\r\n\r\n$ x_2 = \\frac {1}{\\sqrt {p}}$\r\n\r\nSo:\r\n\r\n$ \\ln(x_1) = t \\ln p = \\frac {1}{2}\\ln p \\Rightarrow t = \\frac {1}{2}$\r\n\r\n$ \\ln(x_2) = t \\ln p = - \\frac {1}{2}\\ln p \\Rightarrow t = - \\frac {1}{2}$" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all positive integers $ n$ exceeding $ 1$ such that if $ 1 < k < n$ and $ (k, n) \\equal{} 1$ for all $ k$, then $ k$ is a prime.", "Solution_1": "[hide=\"Solution\"]\nAssume that if $ p_x$ is the lowest prime that does not divide $ n$, then $ p_x\\ge 8$. Let the primes below $ n$ be $ p_1, p_2, p_3, \\cdots, p_x, p_{x \\plus{} 1}, p_{x \\plus{} 2}, \\cdots, p_i$ in increasing order so that $ p_i$ is the largest prime below $ n$. Now, by Bertrand's Principle, we have that $ p_{x \\minus{} 1} > \\frac {p_x}{2}$ and $ p_{x \\minus{} 2} > \\frac {x_{x \\minus{} 1}}{2}$, meaning that $ p_{x \\minus{} 2}p_{x \\minus{} 1}p_x > \\frac {p_x^3}{8}\\ge p_x^2$. This means that $ n < p_{x \\minus{} 2}p_{x \\minus{} 1}p_x$ since or else, $ n > p_x^2$, but $ n$ is coprime with $ p_x$ as shown before, meaning that $ n$ is greater than $ p_x^2$, but coprime with it, yielding a contradiction. This means that $ p_1p_2p_3\\cdots p_{x \\minus{} 1}\\le n\\le p_{x \\minus{} 2}p_{x \\minus{} 1}p_x$. When $ p_x\\ge 5$, we get a contradiction, so $ p_x \\equal{} 2, 3$. Yet, $ p_x\\ge 8$, so we have a contradiction.\n\nNow, $ p_x\\le 7$. If $ p_x \\equal{} 7$, then we have that $ 2, 3, 5|n\\implies 30|n$, so $ n\\ge 30$. Yet, $ n\\le 49$ since $ n < p_x^2$ or else $ p_x^2$ is coprime with $ n$, but is not prime itself, yielding a contradiction. Thus, $ n \\equal{} 30$, which works. Now, if $ p_x \\equal{} 5$, then $ 2, 3|n$ and $ n\\le 25$. This gives that $ n$ could equal $ 24, 18, 12, 6$. All of these work. If $ p_x \\equal{} 3$, then $ n\\le 8$, and $ 2|x$. This gives that $ n \\equal{} 2, 4, 8$ work. Finally, if $ p_x \\equal{} 2$ then $ n \\equal{} 3$ works. We conclude that the solutions are $ 2, 4, 6, 8, 12, 18, 24, 30$. [/hide]" } { "Tag": [ "USAMTS", "geometry", "AMC", "AIME" ], "Problem": "In general USAMTS geometry, I had no idea on how to solve the problem. Last year, I managed to attempt 2 out of 4 geometry problems and got like 1/5 on each (surprisingly, I guessed on one of them and got the right answer :rotfl: ). So is there any good level problems or books that I can use to study for USAMTS level geometry? I want to at least get close to the answer but I can't and this is frustrating.\r\n\r\nI hope this is OK.. I don't mean to this year's round but all USAMTS geometry problems in general.", "Solution_1": "uh...well, I'd just do AIME geometry...\r\n\r\nI mean, I personally don't see USAMTS geometry problems being harder than #13-15 geometry problems on AIME\r\n\r\nOf course, a lot of the \"problems\" are actually proofs...but the basic knowledge of the math involved isn't much more than that needed for the AIME problems...\r\n\r\nBut maybe you should also look at[i] Challenging Problems in Geometry[/i]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $ x_1, x_2,..., x_n $ are nonegative real numbers then:\r\n$ (x{}_1x{}_2...x_n){}^n{}^-{}^1(x_1^2+x_2^2+...+x_n^2) \\le \\frac{2(x_1+x_2+...+x_n){}^2{}^c^(^n^)}{(2c(n))^c^(^n^)}$\r\nWhere $ c(n)= \\frac{n^2-n+2}{2}$ and $ n \\ge 2$", "Solution_1": "I'll post my solution soon :P", "Solution_2": "here is my solution:\r\n$(x_1+x_2+...+x_n)^2=\\displaystyle\\sum_{i=1}^{n}x_i^2 +\\displaystyle\\sum_{1\\le i \\prec j \\le n}{2x{}_ix{}_j}\r\n \\ge c(n)[(\\displaystyle\\sum_{i=1}^{n}x_i^2)(2^c^(^n^){}^-{}^1)(\\displaystyle\\prod_{i=1}^{n}x_i){}^n{}^-{}^1]^1^|^c^(^n^)$ By AGM and the result follows from it. I want to see other solutions. :?" } { "Tag": [ "algebra", "polynomial", "induction", "algebra unsolved" ], "Problem": "Find all the polynomials P(x) with the real coefficent satisfy\r\n$ P(x,y)\\equal{}P(x\\plus{}y,y\\minus{}x)$ for all the real number x,y", "Solution_1": "repeated application of the given relation produces $ p(x,y) \\equal{} p(x \\plus{} y,y \\minus{} x) \\equal{} p(2y, \\minus{} 2x) \\equal{} p(2y \\minus{} 2x, \\minus{} 2x \\minus{} 2y) \\equal{} p( \\minus{} 4x, \\minus{} 4y)$. so by induction, $ p(x,y) \\equal{} p(( \\minus{} 4)^nx,( \\minus{} 4)^ny)$ for all natural $ n$. this implies that $ p(x,y)$ must be constant." } { "Tag": [ "induction" ], "Problem": "How many strings with 6 numbers (formed using only 1 and 2) we can construct if we do not permit appearence of adjacent \"ones\". Example of working strings: 121212, 212121, 222221, 222122. We do not permit 221122, 112222, 112212 etc...\r\nThank you for your help", "Solution_1": "Try to write out a recursion, replacing $ 6$ with $ n$.", "Solution_2": "Very Thanks t0rajir0u. i don't believe it!!!\r\nFor $ n\\equal{}1$ we got two strings: (2) and (1)\r\nFor $ n\\equal{}2$ we have 3 strings: (21), (22), (12)\r\nFor $ n\\equal{}3$ we have 5 strings: (212), (221), (222), (122), (121) \r\nFor $ n\\equal{}4$ we have 8 strings: (2121), (2122), (2212), (2221), (2222), (1221),(1222), (1212) \r\n\r\nFibonnaci sequence!\r\nSo for $ n\\equal{}5$ we will have 5+8=13 strings and for $ n\\equal{}6$ we achieve 13+8=21 strings \r\n\r\nBut how can i make it formal? induction?", "Solution_3": "[hide=\"Hint\"]First let the first number in the string of length $ n$ be \"2\", then \"1\". How does the number of each of these relate to the number of strings with length $ n\\minus{}1$ and $ n\\minus{}2$?[/hide]" } { "Tag": [], "Problem": "A metal slug weighing 25.17g is added to a flask with a volume of 59.7mL. It is found that 43.7g of methanol (d = 0.791g/mL) must be added to the metal to fill the flask. What is the density of the metal?", "Solution_1": "[hide]The volume of methanol is $\\frac{43.7}{0.791}\\approx55.2465$. That means that 59.7-55.2465=4.4535 mL is taken up by the metal slug. The density of the slug is then $\\frac{25.17}{4.4535}\\approx5.65$ g/mL. [/hide]\r\n\r\nShouldn't this be physics or chemistry?", "Solution_2": "Physics? I think Chemistry would be fine..." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "I invented the following inequality, but I am not sure it is true :? :\r\nLet $a,b,c>0$ positive real numbers such that : $ab+bc+ca=1$\r\nProve that : $a^2bc+b^2ca+c^2ab \\leq \\frac{1}{3}$", "Solution_1": "And again... :(\r\nPlease, no more titles like \"Is it true ??\"! Titles should describe essences of problems!\r\n\r\nRead http://www.mathlinks.ro/viewtopic.php?p=140098 !!!", "Solution_2": "[quote=\"erdos\"]Let $a,b,c>0$ positive real numbers such that : $ab+bc+ca=1$\nProve that : $a^2bc+b^2ca+c^2ab \\leq \\frac{1}{3}$[/quote]\r\n\r\nOf course, this inequality is true: Since ab + bc + ca = 1, it rewrites as $a^2bc+b^2ca+c^2ab\\leq\\frac13\\left(ab+bc+ca\\right)^2$, or, equivalently, $ca\\cdot ab+ab\\cdot bc+bc\\cdot ca\\leq\\frac13\\left(ab+bc+ca\\right)^2$. But this is equivalent to $3\\left(ca\\cdot ab+ab\\cdot bc+bc\\cdot ca\\right)\\leq\\left(ab+bc+ca\\right)^2$, what is true since\r\n\r\n$\\left(ab+bc+ca\\right)^2-3\\left(ca\\cdot ab+ab\\cdot bc+bc\\cdot ca\\right)$\r\n$=\\frac12\\left(\\left(bc-ca\\right)^2+\\left(ca-ab\\right)^2+\\left(ab-bc\\right)^2\\right)\\geq 0$.\r\n\r\n Darij", "Solution_3": "[quote=\"Myth\"]And again... :(\nPlease, no more titles like \"Is it true ??\"! Titles should describe essences of problems!\n\nRead http://www.mathlinks.ro/viewtopic.php?p=140098 !!![/quote]\r\nI am really sorry :blush: :blush: :blush: I won't do that any more :lol: :lol: \r\nThanks [b]Darij[/b]" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "invariant", "function" ], "Problem": "Fie matricile $A,B \\in M_n(R),n \\in N^*.$\r\nDaca $A^2=B^3=I_n$ si $AB=BA$ demonstrati ca:$det(A+B+I_n)>0$.", "Solution_1": "Aceeasi tehnica se aplica foarte bine si aici.", "Solution_2": "I am sure there is a solution without diagonalisation\r\n\r\nLet's consider the matrix in $M_n(C)$ \r\n\r\n$A^2=I_n$ , A is a symetry it is diagonalisable the eiengenvalues are $\\pm 1$\r\n\r\n$B^3=I_n$, the polynomial $x^3-1=(x-1)(x-j)(x-j^2)$ split with simple roots\r\n\r\nB is diagonalisable\r\n\r\nBut $A,B$ commute they are diagonalisable in the same basis\r\n\r\n$A+B+I_n$ is similar to a diagonal matrix with eigenvalue $a+b+1$ \r\n\r\n$a=\\pm 1$, $b = 1,j,j^2$ so $a+b+1$ is not zero\r\n\r\n$det(A+B-xI_n)$ is real polynomial the root are conjugate \r\n\r\n$det(A+B+I_n)>0$ \r\n\r\nCan you post solution without diagonalisation ?", "Solution_3": "Presupunem prin absurd ca $\\det(A+B+I_n)=0$. Atunci $\\det(2AB+B^2)=0$. Cum detB<>0 rezulta det(2A+B)=0 si astfel prin inmultire cu detA (care e diferit de 0) rezulta $\\det(2I_n+AB)=0$.dar $(AB)^6=I_n$.Notam $D=2I_n+AB$\r\nsi obtinem prin aplicarea binomului lui Newton ca $D(D^5-12D^4+...-192I_n)=-63I_n$ de unde detD<>0 (contradictie)\r\n$\\det(A+B+I_n)=\\frac{1}{2^n}det(2A+2B+2I_n)=\\frac{1}{2^n}det[(A+I_n)^2+(\\sqrt2B^{-1})^2] >= 0$ \r\n$ B=(B^{-1})^2$", "Solution_4": "[quote=\"silviu\"]Fie matricile $A,B \\in M_n(R),n \\in N^*.$\nDaca $A^2=B^3=I_n$ si $AB=BA$ demonstrati ca:$det(A+B+I_n)>0$.[/quote]\r\n\r\nTo avoid diagonalization one can compute a resultant $\\Pi (\\alpha_i A + \\beta_j B + I)$ where $\\alpha_i^2 = \\beta_j^3 = 1$ are roots of unity. The result will be a invariant under $A \\to \\alpha A$ and $B \\to \\beta B$, therefore a function of $A^2$ and $B^3$, therefore a scalar $rI$. By computation one checks that it is nonzero, or note that $r$ is the resultant of the polynomials $(t-1)^2 - 1$ and $t^3 - 1$.", "Solution_5": "What is diagonalization ?" } { "Tag": [ "geometry", "angle bisector", "geometry proposed" ], "Problem": "I am posting some nice geometry problems I enjoyed during my holidays in Italy:\r\n\r\n1) The excircle of ABC opposite vertex C touches AB and AC at P and Q. The excircle opposite A touches AC and AB at U and X. Prove that the intersection of PQ and UX is equidistant from the lines AB and BC.", "Solution_1": "op02 project, Hungary problem 1. Again, I'm copying the wording from the book and my solution (pretty straightforward).\r\n\r\n[hide=\"Solution\"][color=blue][b]Problem.[/b] The excircle to the side AB of the triangle ABC touches the lines AB and AC at P and Q, respectively. The excircle to the side BC touches the lines AC and AB at U and X respectively. Prove that the intersection of the lines PQ and UX is equidistant from the lines AB and BC.[/color]\n\n[i]Solution.[/i] Let T be the orthogonal projection of the point C on the external angle bisector of the angle ABC. Of course, this external angle bisector passes through the centers I and J of the C-excircle and the A-excircle, respectively. We will prove that the point T lies on the lines PQ and UX.\n\nSince < CTJ = 90\u00b0, the triangle CTJ is right-angled, so that < TCJ = 90\u00b0 - < TJC. Since < CTJ = 90\u00b0 and < CUJ = 90\u00b0, the quadrilateral CTJU is cyclic, so that\n\n $\\measuredangle TUJ=\\measuredangle TCJ=90^{\\circ}-\\measuredangle TJC=90^{\\circ}-\\measuredangle BJC$\n $=90^{\\circ}-\\left(180^{\\circ}-\\measuredangle JBC-\\measuredangle JCB\\right)$ (since < BJC = 180\u00b0 - < JBC - < JCB by the sum of angles in triangle BJC)\n $=-90^{\\circ}+\\measuredangle JBC+\\measuredangle JCB$\n $=-90^{\\circ}+\\left(90^{\\circ}-\\frac{B}{2}\\right)+\\left(90^{\\circ}-\\frac{C}{2}\\right)$ (since the point J, being the A-excenter of triangle ABC, lies on the external angle bisectors of the angles ABC and BCA)\n $=90^{\\circ}-\\frac{B}{2}-\\frac{C}{2}=\\frac{180^{\\circ}-B-C}{2}=\\frac{A}{2}$ (since 180\u00b0 - B - C = A by the sum of angles in triangle ABC).\n\nOn the other hand, since < AXJ = 90\u00b0 and < AUJ = 90\u00b0, the quadrilateral AXJU is cyclic, and thus $\\measuredangle XUJ=\\measuredangle XAJ=\\frac{A}{2}$. Hence, < TUJ = < XUJ, and our point T lies on the line UX.\n\nSince < CTI = 90\u00b0, the triangle CTI is right-angled, and thus < TCI = 90\u00b0 - < TIC. Since < CTI = 90\u00b0 and < CQI = 90\u00b0, the quadrilateral CTIQ is cyclic, so that\n\n $\\measuredangle TQI=\\measuredangle TCI=90^{\\circ}-\\measuredangle TIC=90^{\\circ}-\\measuredangle BIC$\n $=90^{\\circ}-\\left(180^{\\circ}-\\measuredangle IBC-\\measuredangle ICB\\right)$ (since < BIC = 180\u00b0 - < IBC - < ICB by the sum of angles in triangle BIC)\n $=-90^{\\circ}+\\measuredangle IBC+\\measuredangle ICB$\n $=-90^{\\circ}+\\left(180^{\\circ}-\\left(90^{\\circ}-\\frac{B}{2}\\right)\\right)+\\frac{C}{2}$ (since the point I, being the C-excenter of triangle ABC, lies on the external angle bisector of the angle ABC and on the internal angle bisector of the angle BCA)\n $=\\frac{B}{2}+\\frac{C}{2}=\\frac{B+C}{2}=\\frac{180^{\\circ}-A}{2}$ (since B + C = 180\u00b0 - A by the sum of angles in triangle ABC)\n $=90^{\\circ}-\\frac{A}{2}$.\n\nMeanwhile, since < API = 90\u00b0 and < AQI = 90\u00b0, the quadrilateral APIQ is cyclic, and we obtain $\\measuredangle PQI=\\measuredangle PAI=90^{\\circ}-\\frac{A}{2}$. Hence, < TQI = < PQI, and our point T lies on the line PQ.\n\nHence, T is the intersection of the lines PQ and UX. Now, since the point T lies on the external bisector of the angle ABC, it is equidistant from the lines AB and BC. Qed..[/hide]\r\n\r\n[b]PS.[/b] The problem is actually equivalent to this one: http://www.mathlinks.ro/Forum/viewtopic.php?t=5438 . In fact, saying that the point T is equidistant from the lines AB and BC is equivalent to saying that it lies on the angle bisector of the angle CBX, i. e. that the line BT bisects the angle CBX.\r\n\r\n Darij", "Solution_2": "Nice game. Arne proposes an op02 problem and Darij carries out the \"copy&paste\" job on the forum. As there are quite a few problems the game might take some time... :D" } { "Tag": [ "rotation", "geometry", "geometric transformation", "reflection", "combinatorics proposed", "combinatorics" ], "Problem": "A (2n+1)x(2n+1) chessboard is covered by pieces of the shapes: (1), (2) and (3) (see below), which we can always rotate by 90 degrees and reflect across the horizontal and vertical axis. Prove that at least 4n+2 trominoes (3-square pieces) are used.\r\n\r\n[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=267[/img]\r\n\r\nI managed to prove that the number of trominoes(1) is at least 4n+3, but i'm not sure. Please verify me:\r\n\r\nI colored the board with 4 colors: A, B, C and D, this way(I'll draw here for n = 3)\r\n\r\nADADADA\r\nBCBCBCB\r\nADADADA\r\nBCBCBCB\r\nADADADA\r\nBCBCBCB\r\nADADADA\r\n\r\nSo, let:\r\na = number of trominoes that contain B, A, D (with A in corner)\r\nb = number of trominoes that contain A, B, C (with B in corner)\r\nc = number of trominoes that contain D, C, B (with C in corner)\r\nd = number of trominoes that contain C, D, A (with D in corner)\r\nThese are all the posibilities\r\nWe have ${(n+1)}^2$ 1x1 squares with A, ${n(n+1)}$ squares with B, ${n}^2$ squares with C, and ${n(n+1)}$ squares with D.\r\nAlso, each piece (2) or (3) always contains all 4 colors, A, B, C and D.\r\nLet m = number of squares(2) and t = number of pieces of type (3)\r\nSO:\r\n${a+b+d+m+t = {(n+1)}^2}$\r\n${a+b+c+m+t = {n(n+1)}}$\r\n${b+c+d+m+t = {n}^2}$\r\n${a+c+d+m+t = {n(n+1)}}$\r\nBy substracting second, third and forth equality from the first one, we obtain:\r\n${d-c = n+1}$\r\n${a-c = 2n+1}$\r\n${b-c = 2n+1}$\r\nAdding them, we get:\r\n${a+b+d-3c = 4n+3}$, which means ${a+b+c+d = 4n+3+4c \\geq 4n+3}$\r\nThis means the number of trominoes is at least 4n+3.\r\nPlease, tell me if I did something wrong.", "Solution_1": "Please help, i'm very curious", "Solution_2": "You use n in two different meanings which is not correct, but fortunately it doesn't affect the solution.", "Solution_3": "Indeed :) I'll modify it." } { "Tag": [ "algebra", "function", "domain" ], "Problem": "i know its an easy problem, but i can't remember how to solve it in algebric way.\r\n\r\n[img]http://img171.imageshack.us/img171/3563/1av9.gif[/img]\r\n\r\nany help will be appreciated\r\n\r\n\r\nthanks.", "Solution_1": "[hide] First find the intersection points by solving $\\sqrt{7-x}=5-|x|$. \n\n[b]Case 1[/b]: $x \\geq 0$.\nUpon solving, you get $x=3$ or $x=6$. The latter is extraneous.\n[b]Case 2[/b]: $x < 0$.\nUpon solving, you get $x=-2$ or $x=-9$. Again, the latter is extraneous.\n\nSo, now divide the number line into three regions: $x <-2$, $-2 < x < 3$, and $3 < x \\leq 7$ (since the domain of $\\sqrt{7-x}$ has to be taken into account).\n\nBy picking a test point from each region, you find that the first and the third satisfy the inequality.\n\nThus, the solution is $\\boxed{x\\leq-2\\,\\,\\,\\,\\,or\\,\\,\\,\\,\\,3\\leq x\\leq 7}$.\n\nI'm not 100% sure if this method works... :maybe: [/hide]" } { "Tag": [ "function", "real analysis", "real analysis solved" ], "Problem": "It is [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=155779#155779]easy[/url] to see that if $S$ is a countable set and $f : S \\times S \\to \\mathbb{R}^+$ is any function, then there exists a function $g : S \\to \\mathbb{R}^+$ such that $f(x,y) < g(x)g(y)$. It can [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=155779#155779]also[/url] be shown that this fails if $S$ has cardinality at least equal to the continuum. Can it be shown, without the Continuum Hypothesis, that this fails if $S$ is uncountable?", "Solution_1": "I don't know man.. This looks like it's suitable for the \"Open Questions\" section. I don't know your proof for that problem you mentioned, but the example I mentioned there relies more on the topological properties of the real line than on its cardinality. \r\n\r\nWe might want to investigate questions such as this one: is there a closed subset of $\\mathbb R$ having cardinality $S$? If the answer is positive, I think that property does indeed fail, but I'm not sure my argument is correct right now.\r\n\r\nEdit:\r\n\r\nActually, I don't think that condition I mentioned is enough.. :(", "Solution_2": "Are you both kidding? Or am I missing something?", "Solution_3": "What a pity!! Because I am pretty sure that closed subsets in $\\mathbb{R}$ cannot have cardinal between countable and continuum!!", "Solution_4": "Pretty much everything I said up there is stupid :). Sorry about that. The original problem does indeed seem to rely only on the uncountability.", "Solution_5": "Ok, grobber asked me to write my arguments.\r\nLet $S$ be an uncountable set, $f(x)=\\frac{1}{|x-y|}$ and $g:S \\to\\mathbb{R}$ be an arbitrary function. Consider the subsets $S_n=\\{x\\in S\\mid nc$ [i]that [/i]\r\n\r\n$1\\ge (a-1+\\frac{1}{b})(b-1+\\frac{1}{c})(c-1+\\frac{1}{a})$\r\n\r\ngoes as follows: \r\n\r\nUsing $abc=1$, assume $a\\ge 1\\ge b$\r\n\r\nThen manipulating, we see that $1-(a-1+\\frac{1}{b})(b-1+\\frac{1}{c})(c-1+\\frac{1}{a})=(c+\\frac{1}{c}-2)(a+\\frac{1}{b}-1)+\\frac{(a-1)(1-b)}{a}\\ge 0$, clearly, by assumption.\r\n\r\nWhen I see these algebraic manipulation insights to take advantage of order, I am amazed. I don't know how to think of them, to have that insight. Can someone help me out? Do you just manipulate classically (e.g. completing the square, factor, etc) and hope to get something, do you work backwards, are there specific things/expressions you look for and want to use, etc?\r\n\r\nThanks so much for your help, \r\n\r\nSunchips", "Solution_1": "Please, some help and comments from all who are experienced :help: :help: :help:", "Solution_2": "[quote=\"sunchips\"][b]IMO 2000, Problem 2[/b] Let $a$, $b$, $c$ be positive real numbers such that $abc=1$. Prove that\n\n$\\left(a-1+\\frac{1}{b}\\right)\\left(b-1+\\frac{1}{c}\\right)\\left(c-1+\\frac{1}{a}\\right)\\leq 1.$\n\nDo you just manipulate classically (e.g. completing the square, factor, etc) and hope to get something, do you work backwards, are there specific things/expressions you look for and want to use, etc? [/quote]If $abc=1$, then at least one of the ordered pairs $(b,c)$, $(c,a)$, or $(a,b)$ has its first element $\\leq1$ and the second $\\geq1$. One can, without loss of generality, take this ordered pair to be $(b,c)$.\n\n$1-\\left(a-1+\\frac{1}{b}\\right)\\left(b-1+\\frac{1}{c}\\right)\\left(c-1+\\frac{1}{a}\\right)$\n\n$\\equiv F(a,b,c)=F\\left(\\frac{d^2}{yz},\\frac{y}{d},\\frac{z}{d}\\right)\\equiv\\frac{G(y,z,d)}{yz^2d^3},$\n\nwhere $G(y,z,d)=G(y,y+s+t,y+t)$\n\n$=(s^2-st+t^2)y^4+3t(s^2+t^2)y^3+t^2(8s^2+3st+3t^2)y^2$\n\n$+t^2(2s^3+8s^2t+2st^2+t^3)y+s^2t^3(s+2t)\\geq0,$\n\nwhich is clearly true for $00$ [i]that [/i]\n\n$1\\ge (a-1+\\frac{1}{b})(b-1+\\frac{1}{c})(c-1+\\frac{1}{a})$\n\n[/quote]\nAfter substitution $a=\\frac{x}{y}$, $b=\\frac{y}{z}$ and $c=\\frac{z}{x}$ you can get something obvious. :wink:" } { "Tag": [ "Euler", "geometry", "circumcircle", "geometry proposed" ], "Problem": "Given a triangle $ ABC$ with altitudes $ AA_1$ and $ BB_1$. $ A_0$ and $ B_0$ are midpoints of $ BC$ and $ CA$ respectively. $ A_0B_0$ and $ A_1B_1$ meet at $ C'$. prove that $ CC'$ is perpendicular to the Euler Line of triangle $ ABC$", "Solution_1": "[b]REMARKS:[/b]\r\n[b]1) [/b]$ A_0B_0C_0 \\mbox{ pedal triangle of circumcircle O of triangle ABC, and } A_1B_1C_1 \\mbox{ pedal triangle of H;}$\r\n[b]2)[/b] $ \\mbox{H and O isogonal conjugate points in triangle ABC; and apply }$ \r\n[b]luisgeometria remark and Mashimaru solution for the PROBLEM of:[/b] http://www.mathlinks.ro/viewtopic.php?t=308889", "Solution_2": "Well, NSato has given an excellent solution using nine point circle poles and polars in the following topic:\r\n[url]http://www.mathlinks.ro/viewtopic.php?p=471718#471718[/url]", "Solution_3": "Dear Mathlinkers,\r\nyou can find a pure synthetic proof on\r\nhttp://perso.orange.fr/jl.ayme/ vol. 2 Les deux points de Schroeter p. 4-5\r\nSincerely\r\nJean-Louis" } { "Tag": [ "inequalities" ], "Problem": "What is the ordered 4-tuple of positive integers $(x,y,z,w)$, each as small as possible, that satisfies the simultaneous inequalities:\r\n$2x\\prec x+y\\prec x+z\\prec2y\\prec x+w\\prec y+z\\prec2z\\prec y+w\\prec z+w\\prec 2w$?", "Solution_1": "This was a CAML question, if I remember correctly. A number 6, also....\r\n\r\n\r\nI don't remember the answer though...I got it wrong on the test...", "Solution_2": "This was a problem which I brute forced during the competition. :)", "Solution_3": "Are $\\prec$'s $<$'s?\r\n\r\n[hide=\"If so...\"][hide=\"By Brute Force\"]The answer is (1,4,5,8)\n[/hide][/hide]", "Solution_4": "Yep", "Solution_5": "By brute force, I got the following\r\n\r\nI don't know if it is correct\r\n\r\n[hide=\"Here\"]\n\nw=1\nz=4\ny=6\nx=10\n\n[/hide]", "Solution_6": "how do u brute force this one? and how do u solve it w/o brute force? isnt the test only 30 mins too, so...", "Solution_7": "I plugged in random numbers, starting with w=1.\r\n\r\n\r\nThe CAML is only 30 minutes, but the first 4-5 questions only take ten minutes max.", "Solution_8": "This is a problem from the PAML, so it is probably a little different than the CAML. None of you have the correct answer. :(", "Solution_9": "[hide][tex](1,5,7,10)[/tex][/hide]\n\n\n\nDoesn't take that much time even with brute force.", "Solution_10": "[quote=\"ankur87\"]By brute force, I got the following\n\nI don't know if it is correct\n\n[hide=\"Here\"]\n\nw=1\nz=4\ny=6\nx=10\n\n[/hide][/quote]\r\nThat has it backwards =P .", "Solution_11": "That just goes to show\r\n\r\nNo matter how much math I try, English will always be my downfall.", "Solution_12": "What does: \r\nw=1 \r\nz=4 \r\ny=6 \r\nx=10\r\nhave to do with English? :) \r\nBy the way: Elemennop has the correct answer. Are you from PA?", "Solution_13": "I mixed up the letters\r\n\r\n\r\nIt's an attempt on my part to blame my being wrong on something other than math...to save grace." } { "Tag": [], "Problem": "The gasoline gauge on a van initially read $ \\frac 18$ full. When\n15 gallons of gasoline were added to the tank, the\ngauge then read $ \\frac 34$ full. How many more gallons\nwould be needed to fill the tank?", "Solution_1": "Let x be number of gallons when full:\r\n(3/4-1/8)x=15, so x=24, and the answer is 6" } { "Tag": [ "LaTeX", "binomial coefficients" ], "Problem": "I have this combinatorics book that does not contain solutions( :wallbash_red: ), I'm just wondering if anyone here can tell me if my solution is correct or wrong.\r\n\r\nProblem:\r\nProve that \r\n$ \\binom{2n}{n} \\equal{} \\binom{n}{0}^2 \\plus{} \\binom{n}{1}^2 \\plus{} \\binom{n}{2}^2 \\plus{} \\cdots \\plus{} \\binom{n}{n}^2$\r\n\r\nThe hint that the book gave,\r\n[hide=\"hint\"]\nConsider the coefficients of $ x^n$ in the following equation:\n$ (1 \\plus{} x)^{2n} \\equal{} (1 \\plus{} x)^n \\cdot (1 \\plus{} x)^n$\n[/hide]\n\nMy solution:\n[hide=\"solution\"]\nThe coefficient in the left side of the equation of $ x^n$ is $ \\binom{2n}{n}$, which means that the coefficient of $ x^n$ in the right hand of the equation is also $ \\binom{2n}{n}$.\n\nNow I start by expanding each $ (1 \\plus{} x)^n$:\n$ (1 \\plus{} x)^n \\cdot (1 \\plus{} x)^n \\equal{} \\left(\\binom{n}{0}1^n \\plus{} \\binom{n}{1}x \\plus{} \\cdots \\plus{} \\binom{n}{n}x^n\\right) \\cdot \\left(\\binom{n}{0}1^n \\plus{} \\binom{n}{1}x \\plus{} \\cdots \\plus{} \\binom{n}{n}x^n\\right)$\n\nSubstituting $ \\binom{n}{j}$ for $ \\binom{n}{n \\minus{} j}$, for $ 0\\le j\\le n$ in the second expansion:\n\n$ \\left(\\binom{n}{0}1^n \\plus{} \\binom{n}{1}x^1 \\plus{} \\cdots \\plus{} \\binom{n}{n}x^n\\right) \\cdot \\left(\\binom{n}{n}1^n \\plus{} \\binom{n}{n \\minus{} 1}x^1 \\plus{} \\cdots \\plus{} \\binom{n}{0}x^n\\right)$\n\nNow multiplying all the terms $ x^j \\cdot x^{n \\minus{} j}$ to equal $ x^n$ we find that the coefficients of $ x^j$ and $ x^{n \\minus{} j}$ are the same, $ \\binom{n}{j}$, adding all those up we get $ \\binom{2n}{n}x^n \\equal{} \\binom{n}{0}^2\\cdot x^n \\plus{} \\binom{n}{1}^2\\cdot x^n \\plus{} \\cdots \\plus{} \\binom{n}{n}^2\\cdot x^n$\n\nDividing by $ x^n$, we find that $ \\binom{2n}{n} \\equal{} \\binom{n}{0}^2 \\plus{} \\binom{n}{1}^2 \\plus{} \\cdots \\plus{} \\binom{n}{n \\minus{} 1}^2 \\plus{} \\binom{n}{n}^2$\n\nAny feedback on my solution will be great, thanks in advance.\n\nEdit: Fixed where JBL indicated for a 0, and changed the multiplication sign to a dot.\n[/hide]", "Solution_1": "That solution works.\r\n\r\nAlternatively, one can use [url=http://en.wikipedia.org/wiki/Vandermonde%27s_identity]Vandermonde's Identity[/url]: $ \\binom n0^2\\plus{}\\binom n1^2\\plus{}\\cdots\\equal{}\\binom n0\\binom nn\\plus{}\\binom n1\\binom n{n\\minus{}1}\\plus{}\\cdots\\equal{}\\binom{2n}n$, where $ \\binom nk\\equal{}\\binom n{n\\minus{}k}$ has been used.", "Solution_2": "[quote=\"tonypr\"]Substituting $ \\binom{n}{j}$ for $ \\binom{n}{n \\minus{} j}$, for $ 0\\le j\\le n$ in the second expansion:\n\n$ \\left(\\binom{n}{0}1^n \\plus{} \\binom{n}{1}x \\plus{} \\cdots \\plus{} \\binom{n}{n}x^n\\right) * \\left(\\binom{n}{n}1^n \\plus{} \\binom{n}{n \\minus{} 1}x \\plus{} \\cdots \\plus{} \\binom{n}{1}x^n\\right)$[/quote] The last binomial coefficient should be $ \\binom{n}{0}$, but otherwise it's okay. A Latex suggestion: use \\cdot instead of * for a nice multiplication dot.\r\n\r\nmath154, suggesting someone use Vandermonde to prove this is circularity at its worst ;)", "Solution_3": "[quote=\"JBL\"]math154, suggesting someone use Vandermonde to prove this is circularity at its worst ;)[/quote]\r\n\r\nEh, just apply the combinatorial idea behind it, then.", "Solution_4": "The combinatorial idea and the binomial coefficients proof that the book is hinting at are largely the same.\r\n\r\nThe combinatorial proof is to try to choose some $ n$ objects from a pile of $ 2n$ objects. We can obviously do this in $ \\binom{2n}{n}$ ways. Alternatively, we can break the $ 2n$ pile into two piles of $ n$ objects, and then choose, say, $ k$ objects from the first pile in $ \\binom{n}{k}$ ways. Then we need to choose $ n\\minus{}k$ more objects from the second pile, which can happen in $ \\binom{n}{n\\minus{}k}$ ways. Of course, $ \\binom{n}{k}\\equal{}\\binom{n}{n\\minus{}k}$, so this is equivalent to $ \\binom{n}{k}^2$. $ k$ can range from $ 0$ to $ n$, of course ($ 0$ objects from one pile, $ n$ from the other, and vice versa), so we have $ \\binom{2n}{n}\\equal{}\\sum_{k\\equal{}0}^{n}\\binom{n}{k}^2,$ QED." } { "Tag": [ "vector", "calculus", "integration", "function", "algebra", "domain", "real analysis" ], "Problem": "This problem is really getting at me. It's a very conceptual problem rather than a \"math-math\" problem. I don't even know where to start. \r\n\r\nIt's problem 129 in the pdf packet\r\nhttp://www.math.ufl.edu/%7Ehuang/calc3/fall2007.pdf", "Solution_1": "This is the most famous (I think) example of an irrotational vector field in $ \\mathbb{R}^2$ that is not conservative. You can easily show that the field is irrotational, by taking derivatives. Let the unit circle (traversed counterclockwise) be \r\n$ \\left \\{ \\begin{array}{l}\r\n x\\equal{}\\cos(t)\\\\\r\n y\\equal{}\\sin(t)\r\n \\end{array} \\right.$\r\nwith $ t \\in [0,2\\pi]$\r\nIf you evaluate your field along the unit circle, you find that there the field is $ \\minus{}\\sin(t)\\vec{i} \\plus{} \\cos(t)\\vec{j}$, that is exactly the tangent unit vector to the circle. So when you have to take the product between the tangent unit vector and the field along the circle you find 1, so your integral has to be >0. So this field is not conservative, cause its integral over the unit circle, that is a closed simple curve, is not equal to 0. The fact that the function $ \\arctan \\left( \\frac{y}{x} \\right)$ is a potential for the field in the region of the plane $ x>0$ is not a contraddition, cause a smooth irrotational field is conservative only on a simply connected domain. $ \\mathbb{R}^2 \\setminus \\{(0,0)\\}$ is not a simply connected domain, but the half-plane $ x>0$ is. So in this region a potential can exists." } { "Tag": [ "limit", "logarithms", "calculus", "calculus computations" ], "Problem": "consider $ f : \\mathbb{R^ \\plus{} } \\to \\mathbb{R^ \\plus{} }$, $ a > 1$ nondecreasing such that:\r\n$ \\lim_{x \\to \\plus{} \\infty} f(ax) \\minus{} f(x) \\equal{} 0$\r\nthen prove:\r\n$ \\lim_{x \\to \\plus{} \\infty} \\frac {f(x)}{ln(x)} \\equal{} 0$", "Solution_1": "Let $ f(e^u) \\equal{} g(u)$. We are given that $ \\lim_{x \\to \\infty} g(u \\plus{} \\ln a) \\minus{} g(u) \\equal{} 0$ and asked to prove that $ \\lim_{x \\to \\infty} \\frac{g(u)}{u} \\equal{} 0$. The result then follows by [url=http://en.wikipedia.org/wiki/Stolz-Ces%C3%A0ro_theorem]Stolz-Cesaro[/url].", "Solution_2": "how? could you please explain\r\n\r\nDo you mean using the fact that $ \\lim_{\\infty} \\frac {g(u)}{u} \\equal{} 0 \\Leftrightarrow \\forall u_n$ such that $ u_n$ diverges to $ \\infty$ we have $ \\frac {g(u_n)}{u_n}$ converges to $ 0$?" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Prove that $\\forall a,b,c>0$ we have :\r\n$(a^2+b^2+c^2)(\\frac{a^2}{(2a^2+bc)^2}+\\frac{b^2}{(2b^2+ca)^2}+\\frac{c^2}{(2c^2+ab)^2})\\ge 1$\r\n :) :)", "Solution_1": "[quote=\"nthd\"]Prove that $\\forall a,b,c>0$ we have :\n$(a^2+b^2+c^2)(\\frac{a^2}{(2a^2+bc)^2}+\\frac{b^2}{(2b^2+ca)^2}+\\frac{c^2}{(2c^2+ab)^2})\\ge 1$\n :) :)[/quote]\r\nI think this is wrong. Try $a=1,$ $b=c=2.$ ;)", "Solution_2": "[quote=\"arqady\"][quote=\"nthd\"]Prove that $\\forall a,b,c>0$ we have :\n$(a^2+b^2+c^2)(\\frac{a^2}{(2a^2+bc)^2}+\\frac{b^2}{(2b^2+ca)^2}+\\frac{c^2}{(2c^2+ab)^2})\\ge 1$\n :) :)[/quote]\nI think this is wrong. Try $a=1,$ $b=c=2.$ ;)[/quote]\r\nOh,soory.A month ago, I proved a similarly problem :) so I thought this was true ,too. :( .Thank you for pointing out my mistake." } { "Tag": [ "geometry", "ratio", "search", "vector", "geometry solved" ], "Problem": "Let $ABC$ be a triangle. $I$ is the midpoint of $BC$. An arbitrary line cut $AB, AC, AI$ at $P, Q, J$. Prove that:\r\n$\\frac{AB}{AP}+\\frac{AC}{AQ}=2\\frac{AI}{AJ}$.", "Solution_1": "An old problem. You can use Menelaus or Area Ratios, or search the site.", "Solution_2": "[quote=\"Lovasz\"]Let $ABC$ be a triangle. $I$ is the midpoint of $BC$. An arbitrary line cut $AB, AC, AI$ at $P, Q, J$. Prove that:\n$\\frac{AB}{AP}+\\frac{AC}{AQ}=2\\frac{AI}{AJ}$.[/quote]\r\n\r\nLet's use directed segments. The transversal theorem (Theorem 1 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=134343#p134343]http://www.mathlinks.ro/Forum/viewtopic.php?t=20289 post #1[/url]), applied to the points B, C, I on one line, to the point A not on this line, and to the collinear points P, Q, J on the lines BA, CA, IA, respectively, yields\r\n\r\n$CI\\cdot\\frac{BA}{PA}+IB\\cdot\\frac{CA}{QA}+BC\\cdot\\frac{IA}{JA}=0$.\r\n\r\nBut since the point I is the midpoint of the segment BC, we have $CI=IB=\\frac12\\cdot CB$ and BC = - CB, so this becomes\r\n\r\n$\\frac12\\cdot CB\\cdot\\frac{BA}{PA}+\\frac12\\cdot CB\\cdot\\frac{CA}{QA}+\\left(-CB\\right)\\cdot\\frac{IA}{JA}=0$.\r\n\r\nDivision by $\\frac12\\cdot CB$ transforms this into $\\frac{BA}{PA}+\\frac{CA}{QA}-2\\frac{IA}{JA}=0$, what is equivalent to $\\frac{BA}{PA}+\\frac{CA}{QA}=2\\frac{IA}{JA}$. But this is exactly your equation.\r\n\r\n Darij", "Solution_3": "Another: using vector or area." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Why is it that one has interchange rows in order to find the LU decomposition of the following matrix?\r\n\r\n\\[ \\left( {\\begin{array}{*{20}{c}}\r\n { \\minus{} 2} & 1 & 2 \\\\\r\n 4 & 1 & { \\minus{} 2} \\\\\r\n { \\minus{} 6} & { \\minus{} 3} & 4 \\\\\r\n\\end{array}} \\right)\\]\r\n\r\nI used a calculator to find LU but when I found its product I got \r\n\r\n\r\n\\[ \\left( {\\begin{array}{*{20}{c}}\r\n { \\minus{} 6} & { \\minus{} 3} & 4 \\\\\r\n { \\minus{} 2} & 1 & 2 \\\\\r\n 4 & 1 & { \\minus{} 2} \\\\\r\n\\end{array}} \\right)\\]\r\n\r\n\r\nThanks", "Solution_1": "We need a row interchange any time we get a zero in a would-be pivot position, starting from the upper left. We need a row interchange anytime an upper left square submatrix fails to be invertible - as happens here. Look at that $ 2\\times 2$ upper left submatrix. If you have to do a row interchange, then it isn't an LU decomposition any more; rather it is $ A\\equal{}PLU$ where $ P$ is a permutation, $ L$ is lower triangular, and $ U$ is upper triangular. That's exactly what happens here.\r\n\r\nWe determine the required permutation just by keeping track of the row swaps we perform.", "Solution_2": ":o cool thank you again" } { "Tag": [ "LaTeX", "blogs", "Support", "search", "\\/closed" ], "Problem": "How come the centering tag doesn't work in blogs?\r\n\r\n[hide=\"Example\"]\n\\[ \\frac {1}{2}\n\\]\n\n\\[ \\vdots\n\\]\n[/hide]\r\n\r\n[code]\n\\[\n\\frac {1}{2}\n\\]\n\n\\[\n\\vdots\n\\]\n[/code]", "Solution_1": "This question is best asked in Support so I am transferring it to that forum.", "Solution_2": "You can use mimetex. Search on google 'latex for blogger'.", "Solution_3": "Ugh, Mimetex is so ugly, and plus can't fix the centering issue. Why not just fix the built-in latexrender support for LaTeX?", "Solution_4": "Fixed." } { "Tag": [ "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "We define the sequence $ (a_n)$ by:\r\n$ a_0\\equal{}0, a_1\\equal{}1$ and:\r\n$ a_{n\\plus{}1}\\equal{}a_n\\plus{}sin$ $ a_{n\\minus{}1}$ if $ a_n>a_{n\\minus{}1}$\r\n$ a_{n\\plus{}1}\\equal{}a_n\\plus{}cos$ $ a_{n\\minus{}1}$ if $ a_n \\leq a_{n\\minus{}1}$\r\nFind $ lim$ $ a_n$.", "Solution_1": "Are you sure you typed everything correctly? The sequence, as it is defined now, seems to have no limit at all :?", "Solution_2": "Well, I tried it as written, experimentally, and this seems to converge rapidly to $ \\frac{\\pi}2.$ By tinkering around with the initial conditions, I can get it to converge to $ \\left(2k\\plus{}\\frac12\\right)\\pi$ for various integers $ k.$", "Solution_3": "Well, to converge to $ \\frac\\pi 2$, it would have to be eventually decreasing, so let $ b_n\\equal{}a_n\\minus{}\\frac \\pi 2$. We have $ b_{n\\plus{}1}\\equal{}b_n\\minus{}\\sin b_{n\\minus{}1}\\approx b_n\\minus{}b_{n\\minus{}1}$, when $ b_n$ are small and positive. But then you rather clearly overshoot $ \\frac\\pi 2$ in 3 steps. :maybe: Kent, are you sure that you are doing $ a_n\\plus{}\\cos a_{\\color{red}{n\\minus{}1}}$ in your spreadsheet and not $ a_n\\plus{}\\cos a_{\\color{red}{n}}$?", "Solution_4": "Ah ... fedja found my mistake. \r\n\r\nDoes not appear to converge.", "Solution_5": "$ \\liminf(a_n)\\approx{0.6035},\\limsup(a_n)\\approx{4.109}$." } { "Tag": [ "limit", "factorial", "function", "real analysis", "real analysis unsolved" ], "Problem": "Could someone explain me why \r\n\r\n $ \\lim_{n\\to \\infty} \\sum_{k \\equal{} 1}^n(\\frac {1}{k!} \\minus{} \\frac {1}{(k \\plus{} 1!)})$ equals $ \\lim_{n\\to \\infty} (1 \\minus{} \\frac {1}{(n \\plus{} 1!)})$ ?\r\n\r\nFound it in a book and I'm a little confused.:blush: \r\nThanks.", "Solution_1": "First of all, I believe it was meant to be $ n\\to\\infty$, since factorials (and Gamma function itself) are not defined for negative integers.\r\n\r\nSecond: notice that $ \\sum_{k \\equal{} 1}^{n}(\\frac {1}{k!} \\minus{} \\frac {1}{(k \\plus{} 1)!}) \\equal{} \\frac {1}{1!} \\minus{} \\frac {1}{2!} \\plus{} \\frac {1}{2!} \\minus{} \\frac {1}{3!} \\plus{} \\cdots \\plus{} \\frac {1}{(n \\minus{} 1)!} \\minus{} \\frac {1}{n!} \\plus{} \\frac {1}{n!} \\minus{} \\frac {1}{(n \\plus{} 1)!}$ so all members of the sum except for the first and the last cancel out, hence the result.", "Solution_2": "What book was it in? I don't know what to make of:\r\n\r\n$ \\lim_{n\\to\\minus{}\\infty}\\sum_{k \\equal{} 1}^{n}(\\frac{1}{k!}\\minus{}\\frac{1}{(k\\plus{}1!)})$\r\n\r\nBecause if we look at the partial sum... n=-3, for example:\r\n\r\n$ \\sum_{k \\equal{} 1}^{\\minus{}3}(\\frac{1}{k!}\\minus{}\\frac{1}{(k\\plus{}1!)})$ \r\n\r\nWe need to find -2!... and that's not defined, not even with a gamma function.\r\n\r\nSo, since this makes so little sense, give us some context-- what book is it?", "Solution_3": "$ \\infty$ indeed, sorry it's a typo. \r\nThanks for explaining, was really simple :D \\o/" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If $ a,b,c>0$ and $ abc\\equal{}1$, then\r\n$ \\sum(a\\plus{}\\sqrt{b}\\plus{}\\sqrt[3]{c})\\geq\\frac{9}{2}$", "Solution_1": "$ \\sum(a\\plus{}\\sqrt{b}\\plus{}\\sqrt[3]{c}) \\ge 9(abc)^{...}\\equal{}9 > \\frac{9}{2}$\r\n\r\n :oops:" } { "Tag": [ "limit", "integration", "calculus", "trigonometry", "calculus computations" ], "Problem": "calculate : $ \\lim_{n\\to\\infty}\\sum_{k\\equal{}0}^{n}\\frac{2^k}{C_n^{k}}$", "Solution_1": "[quote=\"mr.inequalities\"]calculate : $ \\lim_{n\\to\\infty}\\sum_{k \\equal{} 0}^{n}\\frac {2^k}{C_n^{k}}$[/quote]\r\nWhat is $ C_n^k$?", "Solution_2": ":D\r\n\r\n$ C_n^k\\equal{}\\dbinom{n}{k}$\r\n\r\nhttp://en.wikipedia.org/wiki/Binomial_coefficient", "Solution_3": "Well in that case the limit is certainly $ \\infty$. The contribution of just the last term alone is $ {2^n} \\to \\infty$. Surely the problem must be misstated?", "Solution_4": "sorry the problem is $ \\lim_{n\\to\\infty}\\sum_{k\\equal{}0}^n\\frac{2^k}{C_{2k}^k}$", "Solution_5": "The series does converge, as $ \\binom{2k}{k}$ is asymptotic to $ \\frac{2^{2k}}{\\sqrt{\\pi k}}.$ \r\n\r\nThe series is comparable to $ \\sum_k\\frac{\\sqrt{\\pi k}}{2^k}.$\r\n\r\nNow I'm looking for a probabilistic interpretation of the sum.", "Solution_6": "Numerical estimate: the sum appears likely to be exactly $ 2\\plus{}\\frac{\\pi}2.$", "Solution_7": "$ \\frac1{\\dbinom{2k}k}\\equal{}\\int_0^\\infty\\frac{kx^k}{(1\\plus{}x)^{2k\\plus{}1}}\\,dx,$ it's enough to transform the sum into an integral, remaining calculations are straightforward.", "Solution_8": "Note that $ k\\equal{}0$ must be treated separately in the formula given by |-.-|.\r\n\r\nIf we interchange sum and integral, and use the fact that $ \\sum_{k\\equal{}1}^{\\infty}kr^k\\equal{}\\frac{r}{(1\\minus{}r)^2},$ we get\r\n\r\n$ \\sum_{k\\equal{}0}^{\\infty}\\frac{2^k}{\\binom{2k}{k}}\\equal{} 1\\plus{}\\int_0^{\\infty} \\sum_{k\\equal{}1}^{\\infty}\\frac{k2^kx^k}{(1\\plus{}x)^{2k\\plus{}1}}\\,dx$\r\n\r\n$ \\equal{}1\\plus{}\\int_0^{\\infty}\\frac{2x(1\\plus{}x)}{(1\\plus{}x^2)^2}\\,dx$\r\n\r\n$ \\equal{}1\\plus{}\\int_0^{\\infty}\\frac{2x}{(1\\plus{}x^2)^2}\\,dx\\plus{} \\int_0^{\\infty}\\frac{2x^2}{(1\\plus{}x^2)^2}\\,dx\\equal{}1\\plus{}1\\plus{}\\frac{\\pi}2$\r\n\r\nThis verifies the numerical suggestion.", "Solution_9": "Maybe we can find the value by Telescoping Method.", "Solution_10": "Since the O.P. wants to know, how do we obtain the identity that [-.-] claims, namely\r\n\r\n$ \\frac1{\\dbinom{2k}k}\\equal{}\\int_0^\\infty\\frac{kx^k}{(1\\plus{}x)^{2k\\plus{}1}}\\,dx\\ ?$\r\n\r\nTry making the substitution $ u\\equal{}\\frac1{1\\plus{}x},$ so that $ x\\equal{}\\frac1u\\minus{}1.$ The integral becomes\r\n\r\n$ k\\int_0^1\\left(\\frac1u\\minus{}1\\right)^ku^{2k\\plus{}1}u^{\\minus{}2}\\,du$\r\n\r\n$ \\equal{}k\\int_0^1(1\\minus{}u)^ku^{k\\minus{}1}\\,du$\r\n\r\nThis is an instance of the well-known [b]Beta integral.[/b] Searching for that term should probably turn up plenty of hits. The general result is that for $ \\alpha,\\beta>0,$\r\n\r\n$ \\int_0^1(1\\minus{}u)^{\\alpha\\minus{}1}u^{\\beta\\minus{}1}\\,du\\equal{} \\frac{\\Gamma(\\alpha) \\Gamma(\\beta)} {\\Gamma(\\alpha\\plus{}\\beta)}.$\r\n\r\nIf we particularize this to $ \\alpha,\\beta$ integers, we get\r\n\r\n$ \\int_0^1(1\\minus{}u)^mu^n\\,du\\equal{}\\frac{m!n!}{(m\\plus{}n\\plus{}1)!}$\r\n\r\nSo $ k\\int_0^1(1\\minus{}u)^ku^{k\\minus{}1}\\,du\\equal{}\\frac{k\\cdot k!(k\\minus{}1)!}{(k\\plus{}k\\minus{}1\\plus{}1)!}\\equal{} \\frac{(k!)^2}{(2k)!}\\equal{}\\frac1{\\binom{2k}{k}}.$", "Solution_11": "thank's :)", "Solution_12": "A re-casting of the same computation, starting more directly from the Beta integral:\r\n\r\n$ \\frac{1}{\\binom{2k}{k}}\\equal{}k\\int_0^1x^k(1\\minus{}x)^{k\\minus{}1}\\,dx$\r\n\r\n$ \\sum_{k\\equal{}1}^{\\infty}\\frac{2^k}{\\binom{2k}{k}}\\equal{}\\sum_{k\\equal{}1}^{\\infty} k2^k\\int_0^1x^k(1\\minus{}x)^{k\\minus{}1}\\,dx$\r\n\r\n$ \\equal{}\\int_0^1\\frac1{1\\minus{}x}\\sum_{k\\equal{}1}^{\\infty}2^kx^k(1\\minus{}x)^k\\,dx$\r\n\r\n$ \\equal{}\\int_0^1\\frac1{1\\minus{}x}\\cdot\\frac{2x(1\\minus{}x)}{(1\\minus{}2x(1\\minus{}x))^2}\\,dx$\r\n\r\n$ \\equal{}\\int_0^1\\frac{2x}{(2x^2\\minus{}2x\\plus{}1)^2}\\,dx$\r\n\r\nFor this integral, let $ u\\equal{}2x\\minus{}1,$ so that $ x\\equal{}\\frac12(u\\plus{}1).$\r\n\r\n$ \\equal{}\\int_{\\minus{}1}^1\\frac{2(u\\plus{}1)}{(u^2\\plus{}1)^2}\\,du$\r\n\r\nSplit into even and odd parts; the odd part integrates to zero.\r\n\r\n$ \\equal{}4\\int_{0}^1\\frac1{(u^2\\plus{}1)^2}\\,du$\r\n\r\nNow make the trigonometric substitution $ u\\equal{}\\tan\\theta$ to obtain\r\n\r\n$ \\equal{}4\\int_0^{\\frac{\\pi}4}\\cos^2\\theta\\,d\\theta \\equal{} 2\\int_0^{\\frac{\\pi}4}1\\plus{}\\cos 2\\theta\\,d\\theta \\equal{}\\frac{\\pi}2\\plus{}1$\r\n\r\n(Note that I started the series at $ k\\equal{}1,$ so this agrees with the previous calculation.)\r\n\r\nThis is really the same argument that [-.-] gave; I just rearranged the details." } { "Tag": [], "Problem": "Jasper feels extreme gladness because today is the month of February and later on he'll become sad again for the rest of the months unless another February comes, Will he ever be happy again, $ 6^{333333333}$ months from now???", "Solution_1": "I feel like this question doesn't make much sense.\r\n\r\nFirst solution:\r\n\r\n[hide]Yes.\n\n$ 12 \\equal{} 2 \\times 6$. $ 6^{333333333} \\equal{} 6^{333333331} \\times 6^2 \\equal{} 6^{333333331} \\times 12 \\times 3$.\n\nTherefore, $ 6^{333333333}$ months from now, he will be at a February since $ 6^{333333333} \\rm{mod} 12 \\equiv 0$, which makes him happy.[/hide]\n\nSecond solution:\n\n[hide]$ 6^{333333333}$ is greater than $ 12$, so there must be a February some time between now and $ 6^{333333333}$ months from now, so he will be happy sometime in the middle.[/hide]", "Solution_2": "Of course $ \\boxed{YES}$, because he will have a lot of February(s) in the rest of his life and after he died by years as well :rotfl: .", "Solution_3": "waahahahaha... yuh right. :rotfl:", "Solution_4": "Honeslty, I made up the question cause I find it soooooo funny (It's still educational though)!! wahahahha!!\r\n\r\nBy the way... \"Jasper\" might not enjoy $ 6^{333333333}$ months from now because he might be already dead when that $ 6^{333333333}th$ will come...\r\n\r\n :rotfl:", "Solution_5": "[quote=\"maximos\"]Honeslty, I made up the question cause I find it soooooo funny (It's still educational though)!! wahahahha!!\n\nBy the way... \"Jasper\" might not enjoy $ 6^{333333333}$ months from now because he might be already dead when that $ 6^{333333333}th$ will come...\n\n :rotfl:[/quote]\r\nNo problem maybe he still can't be happy when is died until February come :rotfl:" } { "Tag": [], "Problem": "$a^2+b^2+c^2+d^2+4abcd=1.$ Prove that $a^2+b^2+c^2+d^2\\geq2\\sqrt2-2.$", "Solution_1": "[hide]\nFirst let's reduce it to the case $a,b,c,d > 0$. If two are negative, they cancel each other out so we may assume that at most one is negative. But in this case $4abcd$ is negative so the sum $a^2+b^2+c^2+d^2$ is greater than it is in the case in which $4abcd$ is positive. So proving it for positive $a,b,c,d$ is sufficient.\n\nLet $x = \\sqrt{abcd}$.\n\nWe have $4x \\le a^2+b^2+c^2+d^2$ by AM-GM, so $4x+4x^2 \\le 1$.\n\nThis simplifies to $(2x+1)^2 \\le 2 \\Rightarrow x \\le \\frac{\\sqrt{2}-1}{2} \\Rightarrow x^2 \\le \\frac{3-2\\sqrt{2}}{4}$.\n\nHence $a^2+b^2+c^2+d^2 = 1-4x^2 \\ge 1-4\\left(\\frac{3-2\\sqrt{2}}{4}\\right) = 2\\sqrt{2}-2$.[/hide]" } { "Tag": [ "probability" ], "Problem": "I have two stat question (don't know where to post it) :? \r\n\r\nAll bags entering a research facility are screened. 97% of the bags that contain forbidden material trigger an alarm. 15% of the bags that do not contain forbidden material also trigger the alarm. If 1 out of every 1000 bags entering the building contains forbidden material, what is the probability that a bag that triggers the alarm will actually contain forbidden material.\r\na) .00097\r\nb) .00640\r\nc) .03000\r\nd) .14550\r\ne) .97000\r\n\r\nA summer resort rents rowboats to customers but does not allow more than four people to a boat. Each boat is designed to hold no more than 800 pounds. Suppose the distribuiton of adult males who rent boats is normal with a mean of 190 pounds and standard deviation of 10 pounds. if the weights of individual passengers are independent, what is the probaility that a group of four adult male passengers will exceed the acceptable weight limit of 800 pounds?\r\nA .023\r\nb .046\r\nc .159\r\nd.317\r\ne.997\r\n\r\nThanks :)", "Solution_1": "Question 1:\r\n\r\nLet $F$ be a bag containing forbidden material, and $A$ be a bag that trips an alarm. Then \r\n\r\n$P(F | A) = \\frac{P(FA)}{P(A)} = \\frac{(0.97)(0.001)}{(0.97)(0.001) + (0.15)(0.999)} = 0.00643$,\r\n\r\nwhich is answer B.\r\n\r\nQuestion 2:\r\n\r\n$z = \\frac{x - \\mu}{\\sigma} = \\frac{800 - (4)(190)}{\\sqrt{(4)(10)^2}} = 2 \\Rightarrow P(z \\geq 2) \\approx 0.0228$,\r\n\r\nwhich is answer A. If my answers aren't clear to you, post more specific questions and I'll try to address them.", "Solution_2": "On this note, how is AP stat, the class? It's the only math i haven't taken at my (loser) school (with no ap calc bc), and i was wondering what kind of challenge it'd be. Taking Calc III concurrently with pre-cal based stat is... well, there's a gap.", "Solution_3": "[quote=\"quantum leap\"]On this note, how is AP stat, the class? It's the only math i haven't taken at my (loser) school (with no ap calc bc), and i was wondering what kind of challenge it'd be. Taking Calc III concurrently with pre-cal based stat is... well, there's a gap.[/quote]\r\n\r\nAP Stats is just easy. It's really not heavy on math; most of it in my opinion was just learning what things meant and what your calculations meant. I self-studied it about a week before the test and got a 5; admittedly, I didn't really learn much, but it tells you how hard the AP test is.", "Solution_4": "it's pretty easy, my teacher is a new one, his first time teaching ap stat, so i learned NOTHING in the class. i self studied everything. The only hard part is trying not to get confused on what formula to use. :huh:", "Solution_5": "Yeah and you can use your calculator on the test so just put all the formulas in and you're set :lol:\r\n\r\nI'm about to take it in like an hour.", "Solution_6": "I've heard it's really easy, but i'm wondering if it'd just be a waste of a class. Right now, I'm leaning towards taking it just because it's a least a math and and not an ap class which i don't like but am just taking for the credit.", "Solution_7": "[quote=\"paladin8\"]\nAP Stats is just easy.[/quote]\nOh, :oops: ouch, now I feel really dumb......\n\nand...\n\n[quote]Yeah and you can use your calculator on the test so just put all the formulas in and you're set[/quote]\r\n\r\n\"put the formulas in\" when there are tens of long formulas..... :oops: :roll:", "Solution_8": "[quote=\"chess64\"]Yeah and you can use your calculator on the test so just put all the formulas in and you're set :lol:\n\nI'm about to take it in like an hour.[/quote]\r\nExcept the AP booklet has all the formulae anyway, so you don't even need to do that :P", "Solution_9": "[quote=\"zanttrang\"][quote=\"chess64\"]Yeah and you can use your calculator on the test so just put all the formulas in and you're set :lol:\n\nI'm about to take it in like an hour.[/quote]\nExcept the AP booklet has all the formulae anyway, so you don't even need to do that :P[/quote]\r\n\r\nIt doesn't have all of them. I have a beastly program that gives you the assumptions and formula for every single test/interval that you learn. And I can't remember assumptions, so it's cool :P", "Solution_10": "[quote=\"chess64\"][quote=\"zanttrang\"][quote=\"chess64\"]Yeah and you can use your calculator on the test so just put all the formulas in and you're set :lol:\n\nI'm about to take it in like an hour.[/quote]\nExcept the AP booklet has all the formulae anyway, so you don't even need to do that :P[/quote]\n\nIt doesn't have all of them. I have a beastly program that gives you the assumptions and formula for every single test/interval that you learn. And I can't remember assumptions, so it's cool :P[/quote]\r\nPshh, that takes all the fun out of doing hypothesis tests and confidence intervals - assumptions are the most interesting part of one of those problems!", "Solution_11": "[quote=\"chess64\"]\nIt doesn't have all of them. I have a beastly program that gives you the assumptions and formula for every single test/interval that you learn. And I can't remember assumptions, so it's cool :P[/quote]\r\n\r\nThat's just cheap!!! :o :roll:", "Solution_12": "it's considered cheating to put notes on your calculator but no one really cares." } { "Tag": [ "function", "algebra", "polynomial", "calculus", "integration", "combinatorics unsolved", "combinatorics" ], "Problem": "Alice has two urns. Each urn contains four balls and on each ball a natural number is written. She draws one ball from each urn at random, notes the sum of the numbers written on them, and replaces the balls in the urns from which she took them. This she repeats a large number of times. Bill, on examining the numbers recorded, notices that the frequency with which each sum occurs is the same as if it were the sum of two natural numbers drawn at random from the range 1 to 4. What can he deduce about the numbers on the balls?", "Solution_1": "after some lengthy case by case split up we can say\nthe no's are following\n \nURN1--------- 1,2,3,4\nURN2----------1,2,3,4\n\nor\n\nURN1----------1,2,2,3\nURN2----------1,3,3,5\n\nor\n\nURN1----------1,3,3,5\nURN2----------1,2,2,3\n\nand these three are only possibilities.", "Solution_2": "Translating to generating functions, the question asks for factorizations of $(1 + x + x^2 + x^3)^2$ into two polynomials with nonnegative integral coefficients and coefficient sum equal to 4." } { "Tag": [], "Problem": "Ivan rents a car for $ \\$$25 a day and $ \\$$0.20 a mile. If he rents it for 4 days and drives it 400 miles, how many dollars does he pay?", "Solution_1": "$ 25 x4 days= 100 dollars.\n400 miles x$0.2 per mile= 80 dollars.\r\n\r\n100+80=180 dollars." } { "Tag": [], "Problem": "Prove if $n$ is a perfect pair number then:\r\n$\\sum\\limits_{\\left. d \\right|n} {\\frac{1}{d}} = 2$", "Solution_1": "[hide=\"Solution\"]$n$ is a number such that $f_1 + f_2 + ... + f_k = n$, where $f_k$ denotes the kth proper divisor of $n$. If we include all integer divisors of $n$ (which includes $f_{k+1} = n$), we get $f_1 + f_2 + ... + f_k + f_{k+1} = 2n$. Dividing this equation by $n$ gives $\\frac{f_1}{n} + \\frac{f_2}{n} + ... = 2$. Notice that since $f_i | n$ for all $1 \\le i \\le k$, we have $\\frac{f_i}{n} = \\frac{1}{f_q}$ for some integer $q$ (we also note that all integer divisors of a number $n$ must occur in pairs because perfect numbers cannot be square), so we have $\\sum_{i=1}^{k} \\frac{1}{f_i} = 2$, the desired result. Q.E.D.[/hide]" } { "Tag": [ "floor function", "number theory unsolved", "number theory" ], "Problem": "Let $ \\{x\\} \\equal{} x \\minus{} \\lfloor x \\rfloor$ \r\nProve that for every natural number $ n$ \r\n\r\n$ \\sum_{k \\equal{} 1}^{n^2} \\{ \\sqrt {k}\\} \\leq \\frac {n^2 \\minus{} 1}{2}$", "Solution_1": "$ \\sum_{k \\equal{} 1}^{n^{2}}\\{\\sqrt {k}\\} \\equal{} \\sum_{i \\equal{} 1}^{n^2}{(\\sqrt {k} \\minus{} \\lfloor \\sqrt {k}\\rfloor)}$.\r\n$ \\sum_{i \\equal{} 1}^{n^2}{\\lfloor \\sqrt {k}\\rfloor} \\equal{} \\sum_{i \\equal{} 1}^{n \\minus{} 1}{(2i \\plus{} 1)i} \\plus{} n$." } { "Tag": [], "Problem": "Fie $M$ mijlocul laturii $(AB)$ a triunghiului $ABC$ si $N$ punctul din plan pentru care $MN\\perp{AC}$ si $NB\\perp{BC}$. Fie $P$ punctul de pe latura $(BC)$ astfel incat lungimea segmentului $(NP)$ este egala cu raza cercului circumscris triunghiului $ABC$. Sa se arate ca $P$ este mijlocul laturii $(BC)$.", "Solution_1": "[b][color=blue]\nSa da triunghiul $ABC$ ce este inscris in cercul cu centrul in $O$ iar $M \\in \\left( {AB} \\right)$ este\nmijlocul laturii $AB$. Fie $R \\in AC$ astfel incat $MR \\bot AC$ si $\\left\\{ Q \\right\\} = MR \\cap BC$. \nFie $P \\in \\left( {BC} \\right)$ astfel incat $BP = PC$ si $N \\in MR$ astfel incat $NB \\bot BC$.\nSa se arate ca $NP$ este egal cu raza cercului circumscris triunghilui $ABC$\n[/color][/b]\r\n[img]http://img.photobucket.com/albums/v218/hudrea/mathlink50895.jpg[/img]\r\n[b][u]Solutie:[/u][/b] Evident $m\\left( {\\widehat{RQC}} \\right) = 90^0 - m\\left( {\\widehat{ACB}} \\right)$\r\nSe poate demonstra usor ca $B,P,O,M$ sunt patru puncte conciclice. (vezi $OM \\bot AB,OP \\bot BP$)\r\nDin $MP||AC \\Rightarrow m\\left( {\\widehat{MPB}} \\right) = m\\left( {\\widehat{ACB}} \\right)$ iar din $PM \\bot QR$ (vezi $MP||AC$) si din $NB \\bot BC$ \r\nrezulta ca $N,B,P,M$ sunt patru puncte conciclice si deci avem: \r\n$m\\left( {\\widehat{MNO}} \\right) = m\\left( {\\widehat{MPO}} \\right) = 90^0 - m\\left( {\\widehat{ACB}} \\right) \\Rightarrow $\r\n$ \\Rightarrow m\\left( {\\widehat{RQC}} \\right) = m\\left( {\\widehat{RNO}} \\right) \\Rightarrow NO||BC \\Rightarrow NBPO$ este dreptunghi $ \\Rightarrow NP = OB$", "Solution_2": "[hide=\"DEMONSTRATIE FARA CUVINTE\"][img]http://i19.photobucket.com/albums/b200/mariusdamian/2.png[/img][/hide]" } { "Tag": [], "Problem": "If 6 straight lines are drawn in a plane, what is the maximum number of points of inter section?\r\n\r\nis there a certain formula for this? if so, what is the solution please. :huh:", "Solution_1": "[quote=\"DaReaper\"]If 6 straight lines are drawn in a plane, what is the maximum number of points of inter section?\n\nis there a certain formula for this? if so, what is the solution please. :huh:[/quote]\r\n[hide]\nThe maximum number of points of intersection should be 6 choose 2, because you are looking for the number of ways for 2 lines out of 6 lines to be chosen.\n\nAlso, the formula for $ nCr$ is $ \\frac {n!}{r!(n \\minus{} r)!}$\n\nThe formula for maximum number of points with $ n$ lines would be $ \\frac {n!}{2(n \\minus{} 2)!}$\n[/hide]\r\nedit-oops forgot to hide", "Solution_2": "[quote=\"DaReaper\"]If 6 straight lines are drawn in a plane, what is the maximum number of points of inter section?\n\nis there a certain formula for this? if so, what is the solution please. :huh:[/quote]\r\nWhat do you mean by inner section?", "Solution_3": "not inner section, its intersection-crossing", "Solution_4": "[quote=\"i_like_pie\"][quote=\"DaReaper\"]If 6 straight lines are drawn in a plane, what is the maximum number of points of inter section?\n\nis there a certain formula for this? if so, what is the solution please. :huh:[/quote]\nWhat do you mean by inner section?[/quote]\r\nit's in[b]t[/b]er section (intersection) (in-ter-sec-tion by sound)", "Solution_5": "so does anyone know how to find the number of intersections?", "Solution_6": "Oh... I took \"inter section\" to be a typo for \"inner section\" while it should have been \"intersection\".\r\n\r\n[hide=\"Solution\"]For each pair of lines, there is one intersection. The maximum number is therefore $ \\binom{6}{2} \\equal{} \\boxed{15}$.\n\nTo guarantee that pairs always intersect in distinct points, the points can be placed in a circle. [/hide]", "Solution_7": "how does (6 2) = 15? what does (6 2) mean?", "Solution_8": "It is combinations. Look it up in the Wiki.\r\nEdit:[url]http://www.artofproblemsolving.com/Wiki/index.php/Combinations[/url]" } { "Tag": [], "Problem": "hello. kinda easy algebra problem i guess. \r\n\r\nFind the remainder when $ P(x)\\equal{}x^{2009}\\minus{}2x^{1004}\\plus{}3$ is divided by $ (x\\minus{}1)^2(x^4\\plus{}1)$", "Solution_1": "[quote=\"maths486\"]hello. kinda easy algebra problem i guess. \n\n[/quote]\r\nSTOP with these subjective comments of yours, AND\r\nthey're insincere. JUST post the problems straight.", "Solution_2": "@ Arrange your tan:\r\n\r\nwhat's your problem? people can always post their opinions in forums. There are tons and tons of topics here where people post their opinions on things. if you don't like my opinion, then just don't mind it. :mad:", "Solution_3": "hello, the remainder is $ 4x^5\\minus{}6x^4\\plus{}5x\\minus{}1$\r\nSonnhard.", "Solution_4": "maths 486 is right, everyone has an opinion.\r\nif u look at other posts in forums their opinionated as well\r\nbtw, how does it matter to u tan? :rotfl: :roll:", "Solution_5": "[quote=\"maths486\"]@ Arrange your tan:\n\nwhat's your problem? people can always post their opinions in forums. There are tons and tons of topics here where people post their opinions on things. if you don't like my opinion, then just don't mind it. :mad:[/quote]Although you can post your opinions, it is best that the opinion should stick with the topic in question. A remark like \"kinda easy algebra\" is, in my opinion (note that I'm referring to the subject), pointless. And yes, Arrange your tan has no right to claim whether they were insincere or not just because you have posted a pointless opinion. You can post a lot of pointless opinions but I doubt anyone would care except annoying AoPSers.", "Solution_6": "$ P(x) \\equal{} (x^{2009} \\minus{} 2009x \\plus{} 2008) \\minus{} 2(x^{1004} \\minus{} 1004x \\plus{} 1003) \\plus{} x \\plus{} 1$\r\n$ \\Longrightarrow (x \\minus{} 1)^2|P(x) \\minus{} (x \\plus{} 1).$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c,d$ real numbers, no all of them are zero. Prove that\r\n\r\n$ \\left.\\frac{a^2\\plus{}b^2\\plus{}c^2\\minus{}bc\\minus{}bd\\minus{}cd}{2a^2\\plus{}3b^2\\plus{}3c^2\\plus{}d^2}\\plus{}\\frac{a^2\\plus{}b^2\\plus{}d^2\\minus{}ac\\minus{}ad\\minus{}cd}{3a^2\\plus{}2b^2\\plus{}c^2\\plus{}3d^2}\\plus{}\\frac{a^2\\plus{}c^2\\plus{}d^2\\minus{}ab\\minus{}ad\\minus{}bd}{3a^2\\plus{}b^2\\plus{}2c^2\\plus{}3d^2}\\plus{}\\frac{b^2\\plus{}c^2\\plus{}d^2\\minus{}ab\\minus{}ac\\minus{}bc}{a^2\\plus{}3b^2\\plus{}3c^2\\plus{}2d^2} \\geq 0\\right.$\r\n\r\nWhen does equality hold?\r\n\r\nREMARK\r\nMaybe this inequality seems to be \"ugly\" and \"unusual\", but after solving it you'll see that the idea is really nice!", "Solution_1": "[quote=\"marin.bancos\"]Let $ a,b,c,d$ real numbers, no all of them are zero. Prove that\n\n$ \\left.\\frac {a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} bd \\minus{} cd}{2a^2 \\plus{} 3b^2 \\plus{} 3c^2 \\plus{} d^2} \\plus{} \\frac {a^2 \\plus{} b^2 \\plus{} d^2 \\minus{} ac \\minus{} ad \\minus{} cd}{3a^2 \\plus{} 2b^2 \\plus{} c^2 \\plus{} 3d^2} \\plus{} \\frac {a^2 \\plus{} c^2 \\plus{} d^2 \\minus{} ab \\minus{} ad \\minus{} bd}{3a^2 \\plus{} b^2 \\plus{} 2c^2 \\plus{} 3d^2} \\plus{} \\frac {b^2 \\plus{} c^2 \\plus{} d^2 \\minus{} ab \\minus{} ac \\minus{} bc}{a^2 \\plus{} 3b^2 \\plus{} 3c^2 \\plus{} 2d^2} \\geq 0\\right.$\n\nWhen does equality hold?\n[/quote] \r\nwlog $ a,b,c,d\\ge0$, as replacing $ a$ by $ |a|$ etc. can only decrease the LHS.\r\n\r\nMultiplying all by 2, we see it is equivalent to\r\n\r\n$ \\frac {(b\\plus{}c\\plus{}d)^2 }{2a^2 \\plus{} 3b^2 \\plus{} 3c^2 \\plus{} d^2} \\plus{} \\frac { (a\\plus{}c\\plus{}d)^2}{3a^2 \\plus{} 2b^2 \\plus{} c^2 \\plus{} 3d^2}$ $ \\plus{} \\frac {(a\\plus{}b\\plus{}d)^2 }{3a^2 \\plus{} b^2 \\plus{} 2c^2 \\plus{} 3d^2} \\plus{} \\frac { (a\\plus{}b\\plus{}c)^2}{a^2 \\plus{} 3b^2 \\plus{} 3c^2 \\plus{} 2d^2} \\le4$. \r\n\r\nNow it's just about cutting correctly. By C-S, \r\n\r\n$ LHS\\le \\frac {b ^2}{a ^2\\plus{}b ^2\\plus{}c^2}\\plus{} \\frac {c ^2}{a ^2\\plus{}b ^2\\plus{} c^2}\\plus{} \\frac { d^2}{b^2\\plus{} c^2\\plus{}d ^2}$ $ \\plus{} \\frac {a ^2}{a^2\\plus{}b^2\\plus{}d ^2}\\plus{} \\frac {c ^2}{ a^2\\plus{}c^2 \\plus{}d^2}\\plus{} \\frac {d ^2}{a^2\\plus{}b^2\\plus{}d ^2}$ \r\n\r\n$ \\plus{} \\frac {a ^2}{a^2\\plus{}c^2\\plus{}d ^2}\\plus{} \\frac {b ^2}{ a^2\\plus{}b^2 \\plus{}d^2}\\plus{} \\frac {d ^2}{a^2\\plus{}c^2\\plus{}d ^2}$ $ \\plus{} \\frac { a^2}{a ^2\\plus{}b ^2\\plus{}c^2}\\plus{}\\frac {b ^2}{b^2\\plus{} c^2\\plus{}d ^2}\\plus{} \\frac {c ^2}{b^2\\plus{} c^2\\plus{}d ^2}\\equal{}4.$ :lol: Did it.\r\n\r\nEquality holds iff $ a\\equal{}b\\equal{}c\\equal{}d\\neq0$.", "Solution_2": "Subtracting $ 1$ from each fraction and rearranging we have to prove\r\n$ S=\\frac{a^2+2b^2+2c^2+d^2+bc+cd+db}{2a^{2}+3b^{2}+3c^{2}+d^{2}}+\\frac{b^2+c^2+2d^2+2a^2+ac+cd+da}{3a^{2}+2b^{2}+c^{2}+3d^{2}}+\\frac{b^2+c^2+2d^2+2a^2+ab+bd+da}{3a^{2}+b^{2}+2c^{2}+3d^{2}}+\\frac{a^2+2b^2+2c^2+d^2+ab+bc+ca}{a^{2}+3b^{2}+3c^{2}+2d^{2}}\\leq 4$.\r\n\r\nNow $ S=\\frac{1}{2}\\frac{2a^{2}+3b^{2}+3c^{2}+d^{2}+b^2+c^2+d^2+2bc+2cd+2db}{2a^{2}+3b^{2}+3c^{2}+d^{2}}+\\frac{3a^{2}+2b^{2}+c^{2}+3d^{2}+a^2+c^2+d^2+2ac+2cd+2da}{3a^{2}+2b^{2}+c^{2}+3d^{2}}+\\frac{3a^{2}+b^{2}+2c^{2}+3d^{2}+a^2+b^2+d^2+2ab+2bd+2da}{3a^{2}+b^{2}+2c^{2}+3d^{2}}+\\frac{a^{2}+3b^{2}+3c^{2}+2d^{2}+a^2+b^2+c^2+2ab+2bc+2ca})$\r\n$ =2+\\frac{1}{2}(\\frac{(b+c+d)^2}{2a^{2}+3b^{2}+3c^{2}+d^{2}}+\\frac{(a+c+d)^2}{3a^{2}+2b^{2}+c^{2}+3d^{2}}+\\frac{(a+b+d)^2}{3a^{2}+b^{2}+2c^{2}+3d^{2}}+\\frac{(a+b+c)^2}{a^{2}+3b^{2}+3c^{2}+2d^{2}})$\r\n$ \\leq 2+\\frac{1}{2} (\\frac{b^2}{a^2+b^2+c^2}+\\frac{c^2}{a^2+b^2+c^2}+\\frac{d^2}{b^2+c^2+d^2}+\\frac{a^2}{a^2+b^2+d^2}+\\frac{c^2}{a^2+c^2+d^2}+\\frac{d^2}{a^2+b^2+d^2}+\\frac{a^2}{a^2+c^2+d^2}+\\frac{b^2}{a^2+b^2+d^2}+\\frac{d^2}{a^2+c^2+d^2}+\\frac{a^2}{a^2+b^2+c^2}+\\frac{b^2}{b^2+c^2+d^2}+\\frac{c^2}{b^2+c^2+d^2})$\r\n(using $ \\frac{a^2}{x}+\\frac{b^2}{y}+\\frac{c^2}{z}\\geq\\frac{(a+b+c)^2}{x+y+z}$, or better, [b]marin.bancos[/b]'s lemma.)\r\n$ =4$\r\n\r\nNow when the above lemma is applied to the first fraction, we get that when equality occurs, $ \\frac{b}{a^2+b^2+c^2}=\\frac{c}{a^2+b^2+c^2}%Error. \"iffb\" is a bad command.\n=c$. Similarly, the application of the lemma in the second fraction implies that when equality occurs, $ a=d$. Also, the application in the first fraction implies that when equality occurs, \r\n$ \\frac{c}{a^2+b^2+c^2}=\\frac{d}{b^2+c^2+d^2}\\iff \\frac{b}{a^2+2b^2}=\\frac{a}{a^2+2b^2}%Error. \"iffa\" is a bad command.\n=b$. Therefore, if equality occurs, then $ a=b=c=d$. Now if $ a=b=c=d$, all the fractions in the original inequality are zero, so that equality occurs. $ \\therefore$ equality occurs iff $ a=b=c=d$.", "Solution_3": "Thank you for your solutions to this \"ugly\" inequality. Both proofs follow the same idea of solving. I made this problem using this idea! \r\nDear [b]bakerbakura[/b], you are a little ironical! You know very well that the content of that lemma is something else. Of course, you've used Titu's lemma (or Cauchy-Engel, or Cauchy-Schwarz in a particular form), and this result could be considered as an application of it (see at http://www.mathlinks.ro/viewtopic.php?t=310550 ). But this doesn't mean that I claim you to admit that this result is my result.", "Solution_4": "So I guess I took a bit too long to type out my solution... :blush: [b]spanferkel[/b] got there before me...\r\n\r\nWhen I used the result above I actually took $ n \\equal{} 1$ in your lemma, so [b]technically[/b] it could be considered an application of your lemma. :P" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "a,b,c>=0\r\n\\[ \\left( {a \\plus{} b} \\right)\\left( {b \\plus{} c} \\right)\\left( {c \\plus{} a} \\right) \\ge 4\\sqrt [3]{{abc\\left( {a^2 \\plus{} bc} \\right)\\left( {b^2 \\plus{} ca} \\right)\\left( {c^2 \\plus{} ab} \\right)}}\r\n\\]", "Solution_1": "[quote=\"phatltv\"]a,b,c>=0\n\\[ \\left( {a \\plus{} b} \\right)\\left( {b \\plus{} c} \\right)\\left( {c \\plus{} a} \\right) \\ge 4\\sqrt [3]{{abc\\left( {a^2 \\plus{} bc} \\right)\\left( {b^2 \\plus{} ca} \\right)\\left( {c^2 \\plus{} ab} \\right)}}\n\\]\n[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=262971" } { "Tag": [], "Problem": "We at Art of Problem Solving are pleased to offer two books from the creator of MOEMS, Dr. George Lenchner. These books are [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=4]Creative Problem Solving in School Mathematics[/url] and [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=5]Math Olympiad Contest Problems for Elementary and Middle Schools[/url].", "Solution_1": "I have both of them! :)", "Solution_2": "Is AoPS going to also offer some previous tests?", "Solution_3": "I'm going to order them now!", "Solution_4": "I'm a teacher from Romania. How can I order these books?", "Solution_5": "Im ordering them once I finish my book!!", "Solution_6": "How does [i]Creative Problem Solving[/i] by Lenchner compare to AoPS's [i]Prealgebra[/i]? I am interested in the Lenchner book but if it is easier than [i]Prealgebra[/i] or covers the same material then maybe I don't need it since I'm already working through [i]Prealgebra[/i].", "Solution_7": "Thank you .. That must be great books", "Solution_8": "What are other books, can you please kindly mention" } { "Tag": [ "calculus", "function", "Putnam" ], "Problem": "This question is perhaps not so hard, but I'm putting it in this forum because it requires calculus. I'd like to see as many different solutions as possible.\r\n\r\nConsider any everywhere-differentiable function, $f: \\mathbb{R} \\to \\mathbb{R}$. Prove that there exists a number $a$ such that $-a = f(a)\\cdot f'(a)$.", "Solution_1": "Let g(x) = x 2 + f(x) 2.g is differentiable, continuous, so since g(x) --> infinity as x ---> +/- infinity (since g(x) >= x^2 for all x), we can find c, d distinct such that g(c) = g(d), then by Rolle's Theorem, g'(x) = 0 for some a between c and d, that is,\r\n\r\n2 f'(a)\u2022f(a) + 2a = 0 for some a. 2", "Solution_2": "I found the same solution. Does anyone have a substantially different one?", "Solution_3": "Well, I know of at least one other solution, but it's considerably less nice -- roughly, though, the idea is that f*f' cannot always be < -x when x > 0 and it cannot always be > -x when x < 0, so it must be equal to -x at some point. The solution you both found was essentially the way I came across the problem, but I'm curious if there are other possible solutions.\r\n\r\n\r\n*Random thought: if we assume that f has a convergent Taylor expansion, could that be used to prove it? (Not that it would be a [i]nice[/i] proof, but I'm just curious about other ways to solve the question -- not necessarily other nice ways.)", "Solution_4": "That solution resembles very much a Putnam problem from not so far back. I'll try to find it.\r\n\r\nEDIT: It was 1999 B4, and they're actually not that similar." } { "Tag": [ "LaTeX", "\\/closed" ], "Problem": "I noted that some special chars (like leq or x symbols) replaced with ? in old topics.", "Solution_1": "[quote=\"Myth\"]I noted that some special chars (like leq or x symbols) replaced with ? in old topics.[/quote]There should be no special chars to represent mathematic symbols. All should be translated into $\\LaTeX$ :)", "Solution_2": "Indeed, but I said about old topics (before April 2004).", "Solution_3": "[quote=\"Myth\"]Indeed, but I said about old topics (before April 2004).[/quote]Where possible and in their time limits moderators should edit those old topics to make them $\\LaTeX$ compatible. \r\n\r\nIf you are not sure what the old character was, just change the encoding of the page from UTF-8 to Western (ISO-8859-1) and you will see what they are.", "Solution_4": "I am solving April problems now, so I think I will meet \"broken\" characters not too often :)" } { "Tag": [ "integration" ], "Problem": "[color=blue]A solid, homogeneous ball with radius $R$. Before falling to the floor its center of mass is at rest, but the ball is spinning with angular velocity $\\omega_0$ about a horizontal axis through its centre. The lowest point of the ball is at a height $h$ above the floor.\nWhen released, the ball falls under gravity, and rebounds to a new height such that its lowest point is now $ah$ above the floor. The deformation of the ball and the floor on impact may be considered negligible. Ignore the presence of the air. The impact time, although small, is finite.\nThe mass of the ball is $m$, the acceleration due the gravity is $g$ the dynamic coefficient of friction between the ball and the floor is $\\mu$, and the moment of inertia of the ball about the axis which goes through its centre is $\\large I=\\frac{2}{5}mR^2$\nFind the minimum value of $\\omega_0$ for the situation that the ball slips during the entire impact time.[/color]", "Solution_1": "N = vertical force applied by the floor\r\nt = impact time\r\n\r\nLimit condition:\r\n$\\frac{2mR^2 \\omega_0}{5}=fRt$\r\n$\\omega_0=\\frac{5\\mu}{2R} \\cdot Nt$\r\n\r\nFrom momentum conservation (i think momentum is the right word; exscuse me if i'm wrong):\r\n$Nt=m(\\sqrt{2gh}+\\sqrt{2gha})$\r\n\r\nso i get $\\omega_0 \\geq \\frac{5\\mu}{2R} \\sqrt{2gh} (1+\\sqrt{a})$\r\n\r\nIs it correct?", "Solution_2": "bacco , sorry your solution is wrong\r\n\r\nanswer:\r\n\r\n$\\frac{7muy\\sqrt{2gh}(1+\\sqrt{a})}{2R}$\r\n\r\nhard and nice one\r\n\r\nphucnv87: have you had medal in IPHO? :D \r\n\r\nsincerely", "Solution_3": "Cuong,\r\ncould you write out how you derived that? :)", "Solution_4": "cuong, you went to IPhO? If you did, I guess I'll see you there this year since I'm gonna be on the US Team. ha ha, jk.\r\n\r\nyeah, bacco, you forgot to factor in that that no slip conditions occur if v > Wr. IE, if f/m(t) > WoR - (alpha)RT. In your equation, i think you assumed that W would be 0 after time t.", "Solution_5": "[color=blue]Here is the solution\nThe velocity of the ball's centre right before the collision\n$v_0=\\sqrt{2gh}$\nLet $v_{2x}$ and $v_{2y}$ are the horizontal and vertical components of the velocity of the ball's centre right after the collision\n$v_{2y}=-\\sqrt{2gah}$\nThe reduction of the momentum in vertical direction\n$mv_0-mv_{2y}=m(1+\\sqrt{a})\\sqrt{2gh}=\\int_{t_1}^{t_2}N(t)dt$\nThe augmentation of the mometum in horizontal direction\n$mv_{2x}=\\int_{t_1}^{t_2}f(t)dt$\nThe reduction of the momentum moment \n$I(\\omega_0-\\omega_2)=R\\int_{t_1}^{t_2}f(t)dt$\nwhere $\\omega_2$ is the angular velocity after the collision\nAnd we also know that\n$f=\\mu N$\nFrom the above equations, we can infer that\n$\\omega_2=\\omega_0-\\frac{1}{I}R\\mu m(1+\\sqrt{a})\\sqrt{2gh}$\nIn order to satisfy the condition of the problem, we must have \n$\\omega_2R\\geq v_{2x}$\n$v_{2x}=\\mu(1+\\sqrt{a})\\sqrt{2gh}$ \n$\\Longrightarrow \\omega_o\\geq\\frac{7\\mu\\sqrt{2gh}(1+\\sqrt{a})}{2R}$[/color]\r\n[color=green][b]Cuong[/b] I haven't taken part in any IPhO [/color]", "Solution_6": "Ok, I hadn't correctly understood the text. I thought that before and after the collision the horizontal speed should have been 0.", "Solution_7": "can we consider the normal force to be constant if there is no deformation of the surfaces?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $d(n)$ denote the number of positive divisors of a positive integer $n$. Prove that the sequence defined by $u_n = d(n^2 + 1)$ does not become strictly montonic from any point onwards.", "Solution_1": "oh! it reminds me the problem from vojtech jarnik\r\nhttp://prf.osu.cz/kma/dokumenty/vjaimc/j13problems1.html", "Solution_2": "It is Leningrad TST'98 (11 grade) for All-Russian MO.", "Solution_3": "Let it \\[ {f(n)=d(n^2-1)}. \\]\r\nSuppose that exists $n_0$ such that \r\n\\[{n_0 \\geq n \\Rightarrow f(n) < f(n+1)}\\]\r\nThere is a serie of lemmas which will be very useful in this proof (prove it!):\r\n\r\nLEMMA 1: \\[ f(n+k) \\geq f(n)+2k\\] for all $n_0 \\geq n$ and k.\r\nLEMMA 2: if n is even, then $f(n) \\leq n$.\r\n\r\nNow take k such that $k > n_0 - f(n_0)$ and $n_0+ k$ is even.\r\nThen \\[{ n_0+k \\geq f(n_0+k) \\geq f(n_0)+2k }\\]\r\nor $k \\leq n_0 - f(n_0)$, contradiction!\r\nAnd this is the end!" } { "Tag": [], "Problem": "A two-pan balance scale comes with a collection of weights.\r\nEach weight weighs a whole number of grams. Weights can be\r\nput in either or both pans during a weighing. To ensure any\r\nwhole number of grams up to 100 grams can be measured, what\r\nis the minimum number of weights needed in the collection?\r\n\r\nThank You!", "Solution_1": "I am not that sure if this is right, but powers of 2 accomplish everything. So I would say from 2^0 to 2^6 which 7 weights. Not sure if theirs anything else, cause powers of 3 do not work until 40.", "Solution_2": "Sorry, but the answer is 5. I kept getting 6; I just did weights of 50, 25, 12, 6, 3, 2\r\n\r\nThanks anyways! :lol:", "Solution_3": "[hide=\"Perhaps...\"]\nPowers of 3? Not sure...\n[/hide]", "Solution_4": "Yes. Powers of Three works. Answer is thus $ \\boxed{5}$ as FullMetalNaruto said.", "Solution_5": "So do you just guess and check, or is there some kind of formula you have to memorize? :huh: [/i]", "Solution_6": "Sorry to revive, buy you set up the equation $3^x>100$. Now we want the greatest possible value of $x$ so $x=\\boxed{5}$. Why we use powers of $3$, I do not know.", "Solution_7": "This is in another thread already that is pretty much right below your thread. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=363659[/url]\nIt's called \"Least Number of Weights Needed to Weigh up to 100 Grams\"." } { "Tag": [ "function", "induction", "logarithms", "algebra", "polynomial", "calculus", "calculus computations" ], "Problem": "Consider the function $ f(x) \\equal{} x\\ln(x)$, apply the Leibniz formula to the identity $ xf'(x) \\minus{} f(x) \\equal{} x$ on $ x\\in(0,\\infty)$ in order to compute the Taylor series of f(x) about the point $ 1$.\r\n\r\nWhat will happen to the $ x$ in the identity? :maybe: \r\n\r\nCan someone please help. Thank you very much. :P", "Solution_1": "how do you mean?", "Solution_2": "[quote=\"bookworm_vn\"]how do you mean?[/quote]\r\n\r\nI would want to know how to compute the Taylor series of $ f(x)$ about point $ 1$ with the Leibniz formula ([url]http://en.wikipedia.org/wiki/Leibniz_rule_%28generalized_product_rule%29[/url]). :) \r\n\r\nCan someone please help. Thank you. :P", "Solution_3": "You're assuming we can write $ f(x)\\equal{}\\sum_{n\\equal{}0}^{\\infty}a_n(x\\minus{}1)^n.$\r\n\r\nLet's rewrite $ xf'(x)\\minus{}f(x)\\equal{}x$ as $ [1\\plus{}(x\\minus{}1)]f'(x)\\minus{}f(x)\\equal{}1\\plus{}(x\\minus{}1)$\r\n\r\n$ f'(x)\\equal{}a_1\\plus{}2a_2(x\\minus{}1)\\plus{}3a_3(x\\minus{}1)^2\\plus{}4a_4(x\\minus{}1)^3\\plus{}\\cdots\\plus{}\r\n(n\\plus{}1)a_{n\\plus{}1}(x\\minus{}1)^n\\plus{}\\cdots$\r\n$ (x\\minus{}1)f'(x)\\equal{}0\\plus{}a_1(x\\minus{}1)\\plus{}2a_2(x\\minus{}1)^2\\plus{}3a_3(x\\minus{}1)^3\\plus{}\\cdots\\plus{}\r\nna_n(x\\minus{}1)^n\\plus{}\\cdots$\r\n$ \\minus{}f(x)\\equal{}\\minus{}a_0\\minus{}a_1(x\\minus{}1)\\minus{}a_2(x\\minus{}1)^2\\minus{}a_3(x\\minus{}1)^3\\minus{}\\cdots\\minus{}a_n(x\\minus{}1)^n\\minus{}\\cdots$\r\n\r\nAdding up the columns and equating this series to $ 1\\plus{}(x\\minus{}1),$ we get a chain of equations:\r\n\r\n$ a_1\\minus{}a_0\\equal{}1$\r\n$ 2a_2\\equal{}1$\r\n$ 3a_3\\plus{}a_2\\equal{}0$\r\n$ 4a_4\\plus{}2a_3\\equal{}0$\r\n$ \\dots$\r\n$ (n\\plus{}1)a_{n\\plus{}1}\\plus{}(n\\minus{}1)a_n\\equal{}0.$\r\n\r\nThis chain of equations does not determine $ a_0,$ but we know that from other sources. Since $ f(1)\\equal{}0,$ we have $ a_0\\equal{}0.$ Then:\r\n\r\n$ a_1\\equal{}1$\r\n$ a_2\\equal{}\\frac12$\r\n$ a_3\\equal{}\\minus{}\\frac13a_2\\equal{}\\minus{}\\frac16$\r\n$ a_4\\equal{}\\minus{}\\frac24a_3\\equal{}\\frac1{12}\\equal{}\\frac{2!}{4!}\\equal{}\\frac1{3\\cdot4}$\r\n$ a_5\\equal{}\\minus{}\\frac35a_4\\equal{}\\minus{}\\frac1{20}\\equal{}\\minus{}\\frac{3!}{5!}\\equal{}\\frac{(\\minus{}1)^5}{4\\cdot 5}$\r\n\r\nThe recursion $ a_{n\\plus{}1}\\equal{}\\minus{}\\frac{n\\minus{}1}{n\\plus{}1}\\,a_n$ and an induction can be used to find the general formula\r\n\r\n$ a_n\\equal{}\\frac{(\\minus{}1)^n}{(n\\minus{}1)n}$ for $ n\\ge2.$\r\n\r\n$ x\\ln x\\equal{}(x\\minus{}1)\\plus{}\\sum_{n\\equal{}2}^{\\infty}\\frac{(\\minus{}1)^n}{(n\\minus{}1)n}\\,(x\\minus{}1)^n.$\r\n\r\nThis series can be seen to converge on $ [0,2],$ and thus includes the continuous extension of $ f(x)$ to having a value at $ 0.$ Evaluation at the other end produces this formula for $ \\ln 2:$\r\n\r\n$ \\ln 2\\equal{}\\frac12\\left(1\\plus{}\\frac12\\minus{}\\frac1{2\\cdot3}\\plus{}\\frac1{3\\cdot4}\\minus{}\r\n\\frac1{4\\cdot5}\\plus{}\\cdots\\right).$\r\n\r\n[There were easier ways to get this particular series; this is really the general technique for finding series solutions to a differential equation (preferably a polynomial-coefficient linear differential equation) about an \"ordinary point.\"]" } { "Tag": [ "MATHCOUNTS", "search", "\\/closed" ], "Problem": "I've always wondered this. How do you become a mod? Is there a way to qualify, or did the admins just choose people who they thought were skilled at problem-solving?", "Solution_1": "You don't qualify. When a forum needs a moderator, the admins will decide who is a good, active user of that forum. Then they go into the Administrator Control Panel and add them to the Moderator group. For example, the Mathcounts forum has a lot of posts per day, so they must have someone watching it. Basically the only things you get as a mod are ability to lock, move, delete and split topics. They might be able to ban users, but I am not sure.", "Solution_2": "By using the search function.", "Solution_3": "[quote=\"minicon\"]You don't qualify. When a forum needs a moderator, the admins will decide who is a good, active user of that forum. Then they go into the Administrator Control Panel and add them to the Moderator group. For example, the Mathcounts forum has a lot of posts per day, so they must have someone watching it. Basically the only things you get as a mod are ability to lock, move, delete and split topics. They might be able to ban users, but I am not sure.[/quote]\r\n\r\nMost importantly they can post more than 3 smilies per post and have no 10 minute double posting restrictions. :rotfl: :rotfl:", "Solution_4": "Yes and some mods like 'x' can resemble hitler n can irritate you a lot :D :rotfl: and do whatever cyber shreyas said :D", "Solution_5": "[quote=\"ZetaX\"]By using the search function.[/quote]\r\n\r\n :rotfl: \r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186114]Here[/url] is where you can get more information.", "Solution_6": "[quote=\"madness\"]Yes and some mods like 'x' can resemble hitler n can irritate you a lot :D :rotfl: and do whatever cyber shreyas said :D[/quote]\r\nPlease explain...", "Solution_7": "that was a friendly jibe at my friend who's a mod :D \r\nSorry but i didnt want to post his name so that it doesnt affect him if people take that seriously :)" } { "Tag": [ "function", "Euler", "integration", "algebra", "partial fractions", "complex analysis", "calculus" ], "Problem": "How do you do sums like $ \\sum_{k\\equal{}0}^{\\infty}\\frac{1}{4k^2\\plus{}1}$ or $ \\sum_{k\\equal{}0}^{\\infty}\\frac{1}{4k^2\\plus{}4k\\plus{}1}$?", "Solution_1": "First: Does the denominator factor, so the sum telescopes (by partial fractions, even)?\r\nThat's a good sign that other trickery can be used.", "Solution_2": "Telelscopic sum will apply for summation $ \\sum_{k \\equal{} 0}^\\infty \\frac {4k}{4k^4 \\plus{} 1}$\r\n\r\n$ 4k^4 \\plus{} 1 \\equal{} 4k^4 \\plus{} 4k^2 \\plus{} 1 \\minus{} 4k^2 \\equal{} (2k^2 \\plus{} 1)^2 \\minus{} (2k)^2 \\equal{} (2k^2 \\plus{} 2k \\plus{} 1) (2k^2 \\minus{} 2k \\plus{} 1)$\r\n\r\nSo $ \\frac {4k}{4k^4 \\plus{} 1} \\equal{} \\left [ \\frac {1}{2k^2 \\minus{} 2k \\plus{} 1} \\minus{} \\frac {1}{2k^2 \\plus{} 2k \\plus{} 1} \\right ]$\r\n\r\nNote that $ 2k^2\\plus{}2k\\plus{}1 \\equal{} 2(k\\plus{}1)^2 \\minus{} 2(k\\plus{}1) \\plus{} 1$ and so a telescopic sum has been set up.", "Solution_3": "Sometimes it's also helpful to just look at the expansion and play with the numbers. Comparing it with other different series, even those better understood, may help.\r\n\r\n[hide=\"Answer for second sum\"] $ \\sum_{k \\equal{} 0}^{\\infty}\\frac {1}{4k^2 \\plus{} 4k \\plus{} 1} \\equal{} \\sum_{k \\equal{} 0}^{\\infty}\\frac {1}{(2k \\plus{} 1)^2} \\equal{} 1 \\plus{} \\frac {1}{3^2} \\plus{} \\frac {1}{5^2} \\plus{} \\frac {1}{7^2} \\plus{} \\frac {1}{9^2} \\plus{} ...$\n\nIt's clear the above is the sum of the reciprocals of the squares of the odd natural numbers.\n\nThis looks familiar to the Reimann Zeta Function: $ \\zeta(2) \\equal{} \\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{n^2} \\equal{} 1 \\plus{} \\frac {1}{2^2} \\plus{} \\frac {1}{3^2} \\plus{} ... \\equal{} \\prod_{p}\\left(\\frac {1}{1 \\minus{} \\frac {1}{p^2}}\\right) \\equal{} \\frac {\\pi^2}{6}$\n\nWe can also do this (inspired from [url=http://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function]a proof of Eulers Identity[/url]):\n\\[ \\frac {1}{2^2} \\plus{} \\frac {1}{4^2} \\plus{} \\frac {1}{6^2} \\plus{} ... \\equal{} \\frac {1}{2^2}\\zeta(2)\n\\]\n\n\\[ \\left(1 \\plus{} \\frac {1}{2^2} \\plus{} \\frac {1}{3^2} \\plus{} ... \\right) \\minus{} \\left(\\frac {1}{2^2} \\plus{} \\frac {1}{4^2} \\plus{} \\frac {1}{6^2} \\plus{} ...\\right) \\equal{} \\zeta(2) \\minus{} \\frac {1}{2^2}\\zeta(2) \\equal{} \\left(1 \\minus{} \\frac {1}{2^2}\\right)\\zeta(2)\n\\]\n\n\\[ 1 \\plus{} \\frac {1}{3^2} \\plus{} \\frac {1}{5^2} \\plus{} \\frac {1}{7^2} \\plus{} ... \\equal{} \\left(1 \\minus{} \\frac {1}{2^2}\\right)\\zeta(2)\n\\]\nTherefore\n\\[ \\sum_{k \\equal{} 0}^{\\infty}\\frac {1}{4k^2 \\plus{} 4k \\plus{} 1} \\equal{} 1 \\plus{} \\frac {1}{3^2} \\plus{} \\frac {1}{5^2} \\plus{} \\frac {1}{7^2} \\plus{}\\frac{1}{9^2}\\plus{} ... \\equal{} \\boxed{\\frac {\\pi^2}{8}}\n\\]\n[/hide]", "Solution_4": "[hide=\"First sum\"]$ \\sum_{k\\equal{}1}^{\\infty}\\frac{1}{4k^2\\plus{}1} \\equal{} \\frac{1}{4}\\sum_{k\\equal{}1}^{\\infty}\\frac{1}{k^2\\plus{}\\left(\\frac{1}{2}\\right)^2}$.\n\nNow, if I am in no mistake, $ \\sum_{k\\equal{}1}^{\\infty}\\frac{1}{k^2\\plus{}a^2} \\equal{} \\frac{\\pi a \\cosh(\\pi a)\\minus{}1}{2a^2}$\n\nwhich can be proved by using Complex Analysis or Fourier Series, for example. Now just consider a = 1/2.[/hide]", "Solution_5": "[quote=\"Patterns_34\"]Sometimes it's also helpful to just look at the expansion and play with the numbers. Comparing it with other different series, even those better understood, may help.\n\n[hide=\"Answer for second sum\"] $ \\sum_{k \\equal{} 0}^{\\infty}\\frac {1}{4k^2 \\plus{} 4k \\plus{} 1} \\equal{} \\sum_{k \\equal{} 0}^{\\infty}\\frac {1}{(2k \\plus{} 1)^2} \\equal{} 1 \\plus{} \\frac {1}{3^2} \\plus{} \\frac {1}{5^2} \\plus{} \\frac {1}{7^2} \\plus{} \\frac {1}{9^2} \\plus{} ...$\n\nIt's clear the above is the sum of the reciprocals of the squares of the odd natural numbers.\n\nThis looks familiar to the Reimann Zeta Function: $ \\zeta(2) \\equal{} \\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{n^2} \\equal{} 1 \\plus{} \\frac {1}{2^2} \\plus{} \\frac {1}{3^2} \\plus{} ... \\equal{} \\prod_{p}\\left(\\frac {1}{1 \\minus{} \\frac {1}{p^2}}\\right) \\equal{} \\frac {\\pi^2}{6}$\n\nWe can also do this (inspired from [url=http://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function]a proof of Eulers Identity[/url]):\n\\[ \\frac {1}{2^2} \\plus{} \\frac {1}{4^2} \\plus{} \\frac {1}{6^2} \\plus{} ... \\equal{} \\frac {1}{2^2}\\zeta(2)\n\\]\n\n\\[ \\left(1 \\plus{} \\frac {1}{2^2} \\plus{} \\frac {1}{3^2} \\plus{} ... \\right) \\minus{} \\left(\\frac {1}{2^2} \\plus{} \\frac {1}{4^2} \\plus{} \\frac {1}{6^2} \\plus{} ...\\right) \\equal{} \\zeta(2) \\minus{} \\frac {1}{2^2}\\zeta(2) \\equal{} \\left(1 \\minus{} \\frac {1}{2^2}\\right)\\zeta(2)\n\\]\n\n\\[ 1 \\plus{} \\frac {1}{3^2} \\plus{} \\frac {1}{5^2} \\plus{} \\frac {1}{7^2} \\plus{} ... \\equal{} \\left(1 \\minus{} \\frac {1}{2^2}\\right)\\zeta(2)\n\\]\nTherefore\n\\[ \\sum_{k \\equal{} 0}^{\\infty}\\frac {1}{4k^2 \\plus{} 4k \\plus{} 1} \\equal{} 1 \\plus{} \\frac {1}{3^2} \\plus{} \\frac {1}{5^2} \\plus{} \\frac {1}{7^2} \\plus{} \\frac {1}{9^2} \\plus{} ... \\equal{} \\boxed{\\frac {\\pi^2}{8}}\n\\]\n[/hide][/quote]\r\nI can't believe I missed that\r\n\r\nBy the way, is it true that the alternating version can't be done?", "Solution_6": "Try testing it with the alternating series test.. Does it converge or diverge?", "Solution_7": "Well obviously it converges. What I'm asking is whether it can be put in closed form or not.", "Solution_8": "[quote=\"0714446459923\"]Well obviously it converges. What I'm asking is whether it can be put in closed form or not.[/quote]\r\n\r\nit is called catalan's constant, no closed form is known as of yet. as a matter of fact, you should know that all of \r\n\r\n$ 1\\plus{}\\dfrac{1}{2^3} \\plus{}\\dfrac{1}{3^3} \\plus{}\\dfrac{1}{4^3} \\plus{}\\dfrac{1}{5^3} \\plus{} \\ldots$\r\n\r\n\r\n$ 1\\plus{}\\dfrac{1}{2^5} \\plus{}\\dfrac{1}{3^5} \\plus{}\\dfrac{1}{4^5} \\plus{}\\dfrac{1}{5^5} \\plus{} \\ldots$\r\n\r\n\r\n$ 1\\plus{}\\dfrac{1}{2^7} \\plus{}\\dfrac{1}{3^7} \\plus{}\\dfrac{1}{4^7} \\plus{}\\dfrac{1}{5^7} \\plus{} \\ldots$\r\netc.\r\n\r\n\r\nand\r\n\r\n\r\n$ 1\\minus{}\\dfrac{1}{3^2} \\plus{}\\dfrac{1}{5^2} \\minus{}\\dfrac{1}{7^2} \\plus{}\\dfrac{1}{9^2} \\minus{} \\ldots \\plus{}\\;\\minus{} \\ldots$\r\n\r\n\r\n$ 1\\minus{}\\dfrac{1}{3^4} \\plus{}\\dfrac{1}{5^4} \\minus{}\\dfrac{1}{7^4} \\plus{}\\dfrac{1}{9^4} \\minus{} \\ldots \\plus{}\\;\\minus{} \\ldots$\r\n\r\n\r\n$ 1\\minus{}\\dfrac{1}{3^6} \\plus{}\\dfrac{1}{5^6} \\minus{}\\dfrac{1}{7^6} \\plus{}\\dfrac{1}{9^6} \\minus{} \\ldots \\plus{}\\;\\minus{} \\ldots$\r\netc.\r\n\r\ndo not have any known closed forms.", "Solution_9": "[hide=\"First sum\"]Actually we have $ S \\equal{} \\sum_{k\\equal{}0}^{\\infty}\\frac{1}{4k^2\\plus{}1} \\equal{} 1 \\plus{} \\frac{1}{4} \\sum_{k\\equal{}1}^{\\infty}\\frac{1}{k^2 \\plus{} \\left(\\frac{1}{2}\\right)^2}$\n\nand $ \\sum_{k\\equal{}1}^{\\infty}\\frac{1}{k^2 \\plus{} a^2} \\equal{} \\frac{\\pi a \\coth(\\pi a) \\minus{} 1}{2a^2}$. Therefore, by considering a = 1/2 we get\n\n$ S \\equal{} \\frac{\\pi \\coth\\left(\\frac{\\pi}{2}\\right) \\plus{} 2}{4}$.[/hide]", "Solution_10": "Thank you very much. :maybe:", "Solution_11": "[quote=\"Carcul\"]$ \\sum_{k \\equal{} 1}^{\\infty}\\frac {1}{k^2 \\plus{} a^2} \\equal{} \\frac {\\pi a \\coth(\\pi a) \\minus{} 1}{2a^2}$[/quote]\r\n\r\nHow do you prove this?", "Solution_12": "Here is one wave to prove this identity\r\n\r\nRecall $ Sin(z) = z\\prod_{n=1}^{\\infty}\\left({1 - \\frac {z^2}{n^2\\pi^2}}\\right)$ . \r\nWe then take natural logs of both sides to find $ Log(Sin(z)) = Log(z) + \\sum_{n=1}^{\\infty}Log\\left({1 - \\frac {z^2}{n^2\\pi^2}}\\right)$ . \r\nDifferentiating both sides with respect to z gives us $ Cot(z) = 1/z + \\sum_{n=1}^{\\infty}\\frac{-\\frac{2z}{n^2\\pi^2}}{1 - \\frac {z^2}{n^2\\pi^2}}$ \r\nRearranging we find $ Cot(z) = 1/z - \\frac{2}{\\pi^2}\\sum_{n=1}^{\\infty}\\frac{z}{n^2 - \\frac {z^2}{\\pi^2}}$ Letting $ z = i\\pi\\phi$ \r\nwe find that ${ Cot(i\\pi\\phi)= -iCoth(\\pi\\phi) = \\frac{1}{i\\pi\\phi} - \\frac{2}{\\pi}\\sum_{n=1}^{\\infty}\\frac{i\\phi}{n^2 + \\phi^2}}$ \r\nSimplifying, one finds ${ \\frac{\\pi Coth(\\pi\\phi)}{2\\phi} = \\frac{1}{2\\phi^2} + \\sum_{n=1}^{\\infty}\\frac{1}{n^2 + \\phi^2}}$ or \r\n${ \\frac{\\pi\\phi Coth(\\pi\\phi)- 1} {2\\phi^2} = \\sum_{n=1}^{\\infty}\\frac{1}{n^2 + \\phi^2}}$ as desired", "Solution_13": "There are also Fourier series/Fourier transform methods based on the Poisson summation formula. Here's a possible starting point:\r\n\\[ \\frac{1}{k^2\\plus{}a^2}\\equal{}\\frac1{2a}\\int_{\\minus{}\\infty}^{\\infty}e^{\\minus{}a|x|}e^{ikx}\\,dx\\]", "Solution_14": "Or you can consider the Fourier series for the function f(x) = cosh(ax)." } { "Tag": [ "ARML", "geometry", "AMC", "AIME" ], "Problem": "After seeing [b]Altheman[/b]'s mock ARML individual round problems, I felt like making my own; but ARML is fast approaching, and I both do not have a full 8 problems and am not sure if these would be fair questions (and I'm not sure, as well, of the quality of these problems). :P So, I am just going to present them here for any who would like some practice (try to stay within the 10 or so minute time restriction, if you'd like :) ) \r\n\r\n[hide=\"Sheet 1\"]\n$(1)$ The equation $\\sqrt{ 2x^{2}+12-2 x \\sqrt{x^{2}+12}}=2x+\\sqrt{x^{2}+12}$ has at most two real roots. Find all of its real roots, or write \"$\\text{No Real Roots}$\" if there are none.\n\n$(2)$ Let $\\phi^{2}-\\phi-1=0$ and $\\phi > 0$. Evaluate $\\phi^{12}-\\frac{72}{\\phi^{3}}$.[/hide]\n\n[hide=\"Sheet 2\"]\n$(3)$ Find the sum of all $x$ such that $0 \\leq x < 100$ and $23$ divdes $P(x) = x^{4}-49x^{3}+72x^{2}-24x+299$.\n\n$(4)$ Evaluate $\\dbinom{90}{0}+\\dbinom{90}{6}+\\dbinom{90}{12}+\\ldots+\\dbinom{90}{90}$.[/hide]\n\n[hide=\"Sheet 3\"]\n$(5)$ $N$ is the largest integer less than $5000$ that has exactly $12$ positive divisors, including $1$ and itself. Find $N$.\n\n$(6)$ The region $\\mathcal{S}$ in the complex plane contains all $z$ such that:\n\\[\\frac{30}{ |z| }= \\frac{ 50}{ |z+1| }\\ \\text{ and }\\ |z| \\leq 3 \\]\nFind the area of $\\mathcal{S}$.\n[/hide]", "Solution_1": "Yay you have over a thousand posts...and as for the arml problems:\r\n\r\n[hide=\"Comments:\"]\n\nOverall, it is good\n\n1. denesting is your favorite...this is fine...i am not sure who would put no reals solutions as 0 is obviously a solution...\n\n2. is too computational...like i know you can just divide, but that takes a while with 12 degree...it is fine conceptually\n\n3. eww for that polynomial...i guess it is okay...this is a bit contrived....\n\nin mod 23...x^4-3x^3+3x^2-x=(x)(x-1)^3\n\n4. not reasonable. basically you have seen the problem or you have not. it is a nice problem and all...but it shows up way to often on aops\n\n5. ok...i am not sure how you do this quickly...p^2q^3 or pqr^2...i believe that 2^2*29*43=4988 works...but like who is going to see that...\n\n6. good ...it gets kind of messy\n\n2,6 seem familiar [from aime?]...which is fine for practice, but i doubt they would show up on an actual contest[/hide]\r\n\r\nalso this could get lost pretty easily...the topic title is not really eyecatching", "Solution_2": "[quote=\"Altheman\"]Yay you have over a thousand posts...and as for the arml problems:\n\n[hide=\"Comments:\"]\n\nOverall, it is good\n\n1. denesting is your favorite...this is fine...i am not sure who would put no reals solutions as 0 is obviously a solution...\n\n2. is too computational...like i know you can just divide, but that takes a while with 12 degree...it is fine conceptually\n\n3. eww for that polynomial...i guess it is okay...this is a bit contrived....\n\nin mod 23...x^4-3x^3+3x^2-x=(x)(x-1)^3\n\n4. not reasonable. basically you have seen the problem or you have not. it is a nice problem and all...but it shows up way to often on aops\n\n5. ok...i am not sure how you do this quickly...p^2q^3 or pqr^2...i believe that 2^2*29*43=4988 works...but like who is going to see that...\n\n6. good ...it gets kind of messy\n\n2,6 seem familiar [from aime?]...which is fine for practice, but i doubt they would show up on an actual contest[/hide]\n\nalso this could get lost pretty easily...the topic title is not really eyecatching[/quote]\n\nLol, yay, over a thousand. :D\n\n[hide=\"Stuffs (for Alex, mostly)\"]\nActually, for #1 i had meant to do the $f(x)= x \\implies f(f(x))=x$ thing... i think i might've copied down the problem wrong. i'll look... but w/e.\n\nAnd merr, i really like the idea behind #4. But yea, you're probably right, not that many ppl who haven't seen this before would be like, BAM, roots of unity filter.\n\nYea, I think #6 is similar to that one where its like 40/|z| = |z|/50 or something, where like there are two circles... idk. It's pretty similar. And #2, i think, is also similar to one that has the same idea in mind. :)[/hide]", "Solution_3": "[hide=\"I am actually hiding my posts for once\"]1 is $|x-\\sqrt{x^{2}+12}|=2x+\\sqrt{x^{2}+12}$\n\nso i don't know where that idea comes in...\n\nalso after solving it, the only solution is x=0...so you need a frustum...it is too guessable...ya kno...step your game up\nsam, i am curious, what was your solution to 5...[/hide]", "Solution_4": "Is the answer to the last one\r\n(225/256)pi?" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all $a,b \\in \\mathbb{N}$ such that $a^3=1+2+3+...+b-1+b$.", "Solution_1": "It is equavalent to solution elliptic equation:\r\n$2x^3=y(y+1).$ Solutions may be find elementary metods. Since $(y,y+1)=1, or \\ y=2z^3,y+1=t^3, or \\ y=z^3,y+1=2t^3,$ It is equavalent to equation Fermats vieve $t^3-2z^3=-1,$ with have only trivial solutions $z=1,t=1\\Longrightarrow x=1,y=1.$" } { "Tag": [], "Problem": "If to the numerator and denominator of the fraction $ \\frac{1}{3}$ you add its denominator, 3, the fraction will double.\r\nFind a fraction which will triple when its denominator is added to its numerator and to its denominator; find one that will quadruple.\r\n\r\n[i]from book \"The Moscow puzzles\" by Boris A. Kordemsky[/i]\r\n\r\n[u]Please hide Your solution . Write inside Hide ... /Hide[/u]", "Solution_1": "[hide=\"Solution\"]\n\n$ \\frac {1}{5}$ and $ \\frac {1}{7}$\n\n[/hide]", "Solution_2": "[hide=\"Generic Solution\"]\nSolving it generically, we start with a simple equation for fraction p/q and a number n for which it should multiply when the denominator is added to both the numerator and denominator:\n\n$ \\frac{p\\plus{}q}{2q} \\equal{} \\frac{np}{2}$\n\nThis immediately simplifies\n\n$ \\frac{p\\plus{}q}{2} \\equal{} np \\implies p\\plus{}q \\equal{} 2np$\n\nNow we have\n\n$ q \\equal{} 2np \\minus{} p \\equal{} p(n\\minus{}1)$\n\nThis allows us to choose any $ p$ and $ n$ to find the necessary denominator.\n\nExample for 3: $ q \\equal{} p(2n \\minus{} 1) \\equal{} 1(2 \\cdot 3 \\minus{} 1) \\equal{} 6 \\minus{} 1 \\equal{} 5$, so the fraction is $ \\frac{1}{5}$\n\nExample for 4: $ q \\equal{} p(2n \\minus{} 1) \\equal{} 1(2 \\cdot 4 \\minus{} 1) \\equal{} 8 \\minus{} 1 \\equal{} 7$, so the fraction is $ \\frac{1}{7}$\n\n\nNote that we can select any number we choose for $ p$, and using $ 1$ makes everything evaluate quickly. There are infinitely many examples for each $ n$, but we wish to use one that reduces the amount of calculations. If we assume $ p$ as 1, we get\n\n$ q \\equal{} 2n \\minus{} 1$\n\nWhich is easy to use.\n[/hide]", "Solution_3": "[hide]1/5 and 1/7.[/hide]", "Solution_4": "[quote=\"mathepic\"]\nSolving it generically, we start with a simple equation for fraction p/q and a number n for which it should multiply when the denominator is added to both the numerator and denominator:\n\n$ \\frac {p \\plus{} q}{2q} \\equal{} \\frac {np}{2}$\n\nThis immediately simplifies\n\n$ \\frac {p \\plus{} q}{2} \\equal{} np \\implies p \\plus{} q \\equal{} 2np$\n\nNow we have\n\n$ q \\equal{} 2np \\minus{} p \\equal{} p(n \\minus{} 1)$\n\n[/quote]\r\n\r\nwhat did you do here?\r\n\r\n$ \\frac {p \\plus{} q}{2q} \\equal{} \\frac {np}{2} \\implies \\frac {p \\plus{} q}{q} \\equal{} np\\implies p \\plus{} q \\equal{} npq$\r\n\r\n\r\ndo you mean $ \\frac{np}{q}$ for your second fraction?", "Solution_5": "Yes, it should be $ \\frac {np}{q}$.\r\n\r\nAny more of these questions,small_unicorn?", "Solution_6": "[hide=\"Solution\"]\nOK. Let x = the numerator and y equal the denominator.\n\n$ x/y$(3)=(x+y)/(2y)\n$ y(x\\plus{}y)$=2y(3x)\n$ x\\plus{}y\\equal{}6x$\ny=5x\n\nSo that would mean our answer is 1/5. [/hide]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "A circle is divided in $k$ sectors and a pawn is put in each sectors. \r\nWith each move, two pawns are moved, in two different direction, to their neighbour sectors. Is there a sequence of moves where, in the end, all pawns are in the same sectors?", "Solution_1": "For odd k it's obvious that if u chose a target sector and in each move put two pawns that are in the same distance from it 1 sector closer to it, in the end all pawns will end up in the target sector. \r\nFor even k it is not possible.", "Solution_2": "[quote=\"Kondr\"]\nFor even k it is not possible.[/quote]\r\n\r\nTo prove this, notice that both arcs which separate two pawns with two sectors between them are even, no matter what moves we perform, so the two pawns could never end up on the same sector." } { "Tag": [ "geometry", "circumcircle", "power of a point", "radical axis", "geometry solved" ], "Problem": "Given a quadrangle $ABCD$ with $AB|| CD$, $F$ is a point on $AB$ satisfying $FC=FD$, $AC \\cap BD = E$. Let $O_1,O_2$ are the circumcenters of $AFD$ and $BFC$ respectively. \r\nProve that $O_1O_2$ is perpendicular to $EF$.\r\n\r\nA minute before I thought that I solved this problem but I has just found a stupid mistake :(\r\n\r\n[i]Edited by Myth[/i]", "Solution_1": "Let circles $O_1$ and $O_2$ intersect AC and BD at P and Q respectively. We will show DCPQ is cyclic.\r\n$\\angle ABD+\\angle BAC=\\angle EDC+\\angle FDP=\\angle FDC+\\angle QDP$. Similarly, $\\angle ABD+\\angle BAC=\\angle FCD+\\angle PCQ$. But since FD=FC, $\\angle FDC=\\angle FCD$. It follows that $\\angle QDP=\\angle PCQ$. Therefore $D,C,P,Q$ are concyclic. Note that this implies $A,B,P,Q$ are also concylic. Consider the radical axis formed by the circumcircle of ABPQ, $O_1,O_2$. Since F lies on both AP and BQ, it also lies on the radical axis between $O_1,O_2$. The result follows since the radical axis is perpendicular to the line joining the centers.", "Solution_2": "Nice! :) \r\nI also tried to use radical axes idea, but I couldn't invent proper circles :(", "Solution_3": "My new solution is depends on yours, mecrazywong, but it\u2019s simplier than yours :D \r\nFirst we have $\\angle APD=\\angle AFD$\r\n\t $\\angle BQC=\\angle BFC$\r\nBut $\\angle AFD=\\angle BFC$ so $\\angle APD=\\angle BQC$ => $\\angle DQC=\\angle DPC$ => DCPQ is concyclic.\r\nAnd it\u2019s easy to get ABPQ is concyclic too.\r\nNow we have $EP.EA=EQ.EB$ and we get the result \r\n :)" } { "Tag": [ "algebra", "function", "domain", "calculus", "calculus computations" ], "Problem": "Let $ A$ be a set and $ f$ a continuous function. Find if the following statement is true or false: If $ f(A)$ is a closed interval and has more than one element, then $ A$ is a closed interval.", "Solution_1": "I think that the statement is false. For example $ f: (\\minus{}4\\pi,4\\pi)\\to{R},f(x)\\equal{}sinx$ has $ f(A)\\equal{}[\\minus{}1;1]$ which is a closed interval and the domain isn't", "Solution_2": "Then if $ f(A)$ is simply an interval that has at least one element, means that $ A$ is an interval?", "Solution_3": "Heh no. For $ f$ above we can adjoin any set to the interval.", "Solution_4": "Thanks. :)" } { "Tag": [ "calculus" ], "Problem": "Kindly note these rules and regulations before posting in this forum. These rules will be strictly implemented.\r\n\r\n1) The Advanced Physics forum is strictly for [color=red]High School - AP(Calculus based) , College, Olympiad or Research level [/color]Physics discussion. Esoteric , but popular topics like Relativity, String Theory can be discussed here. \r\n\r\n2) The Problems to be posted here MUST be accompanied by what the poster has done about it, how much headway he/she has made and what specific help he/she expects. Problem posted should be such that they require thought and not just application of a formula. [color=violet]Calculus[/color] based problems should be posted here. Kindly note that you should not expect your problem to be fully solved for you. [color=orange]NO Homework problems [/color]will be solved here. Those problems requring simple application of formulae should be posted in the [url=http://www.mathlinks.ro/Forum/index.php?f=405]Introductory Physics Forum[/url]. \r\n\r\n3) Discussion on [color=blue]Past Olympiad [/color]problems, details of Olympiad procedures etc. are welcome. But again, no Olympiad problem will be fully solved on this forum for anyone.\r\n\r\n4) If you need clarification of topics taught in higher College Physics classes or want to present your ideas on advanced topics, please do so here. [color=green]Theoretical discussions [/color]on advanced topics of research are welcome. But keep in mind that since few of the members on this forum are engaged in Physics research, you might not find your answers here. You are directed to [url=http://www.physicsforums.com]www.physicsforums.com[/url]for more advanced help. \r\n\r\n\r\nFor more information or clarification, kindly contact the Moderator. Further Suggestions are welcome. \r\n\r\nThank you.", "Solution_1": "IMPortant:\r\nDO hide all ur answers,Hints and suggestions that u post to a problem so that evryone gets equal chance :)", "Solution_2": "Can we post Quantum Physics problems here (e.x. Bohr Model)? \n\nThe reason I ask this is: in my textbook, this comes right after relativity so I think it should be OK for this forum but it is not mentioned here. \n\nThanks in advance!", "Solution_3": "[quote=\"Rushil\"]\n2) The Problems to be posted here MUST be accompanied by what the poster has done about it, how much headway he/she has made and what specific help he/she expects.[/quote]\n\nWhat if you made up the problem for others to do for fun? Surely you shouldn't post your own solution?\n\nAlso, for fun/recreational problems, should you post full solutions, or still give hints, partial solutions, etc., or should you post full solutions? If the answer is the second option, then how can you tell whether a problem is a homework problem or a recreational one?\n\nAll my questions also apply for the Introductory Physics Forum.", "Solution_4": "..\\8char" } { "Tag": [ "algebra", "polynomial", "Putnam", "algebra solved" ], "Problem": "Let P be a real polinomial with n distinct real roots that are all > 1.\r\nWe introduce Q like this:\r\nQ(x)=(x 2 +1)P(x)P'(x) + x[P(x) 2 + P'(x) 2 ] Prove that Q has at least 2n-1 real roots.", "Solution_1": "Maybe, it is not too difficult.", "Solution_2": "I 'll give you solution's problem.", "Solution_3": "Q(x)=(xP(x)+P'(x))(xP'(x)+P(x).", "Solution_4": "Q(x)=(xP(x)+P'(x))(xP'(x)+P(x)).\r\nT(x)=xP(x) you have :T(x) have n+1 roots\r\n=> T'(x) have n roots ,\r\nK(x)= P(X)e (-1/2)*x*x \r\n=> K'(X) have n-1 roots \r\nTHEN YOU HAVE RESULTS THAT WE CONSOLE", "Solution_5": "This is problem A5 on the 1981 Putnam: [url]http://www.kalva.demon.co.uk/putnam/putn81.html[/url]" } { "Tag": [], "Problem": "\u0392\u03c1\u03ae\u03ba\u03b1 \u03cc\u03c4\u03b9 \u03b7 \u03b4\u03b9\u03b1\u03ba\u03c1\u03af\u03bd\u03bf\u03c5\u03c3\u03b1 \u03c4\u03b7\u03c2 \u03c4\u03c1\u03b9\u03c4\u03bf\u03b2\u03ac\u03b8\u03bc\u03b9\u03b1\u03c2 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7\u03c2 $ az^3\\plus{}bz^2\\plus{}cz\\plus{}d\\equal{}0$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \r\n\r\n$ \\Delta\\equal{} 4b^3d\\minus{}b^2c^2\\plus{}4ac^3\\minus{}18abcd\\plus{}27a^2d^2$\r\n\r\n\u03a0\u03ce\u03c2 \u03cc\u03bc\u03c9\u03c2 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03b1\u03c5\u03c4\u03cc\u03c2 \u03bf \u03c4\u03cd\u03c0\u03bf\u03c2?", "Solution_1": "\u039f\u03c1\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $ f(x)\\equal{}ax^{3}\\plus{}bx^{2}\\plus{}cx\\plus{}d$ \u03bc\u03b5 $ a>0$ \u03ba\u03b1\u03b9 $ g(x)\\equal{}f^{'}(x)\\equal{}3ax^{2}\\plus{}2bx\\plus{}c$. \u0397 \u03b4\u03b9\u03b1\u03ba\u03c1\u03af\u03bd\u03bf\u03c5\u03c3\u03b1 \u03c4\u03b7\u03c2 $ g(x)$ \u03b5\u03af\u03bd\u03b1\u03b9 $ s_{1}\\equal{}b^{2}\\minus{}3ac$. \u0391\u03bd $ s_{1}>0$ \u03c4\u03cc\u03c4\u03b5 \u03b7 $ f^{'}(x)$ \u03ad\u03c7\u03b5\u03b9 2 \u03ac\u03bd\u03b9\u03c3\u03b5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ad\u03c2 \u03c1\u03af\u03b6\u03b5\u03c2 $ x_{1},x_{2}$ \u03bc\u03b5 $ x_{1}\\equal{}\\frac{\\minus{}b\\minus{}\\sqrt{s_{1}}}{3a}$ \u03ba\u03b1\u03b9 $ x_{2}\\equal{}\\frac{\\minus{}b\\plus{}\\sqrt{s_{1}}}{3a}$.\r\n\r\n\u0391\u03bd \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $ f(x_{1})\\equal{}\\frac{2b^{3}\\minus{}9abc\\plus{}(2b^{2}\\minus{}6ac)\\sqrt{s_{1}}\\plus{}27a^{2}d}{27a^{2}}$ \u03ba\u03b1\u03b9 $ f(x_{2})\\equal{}\\frac{2b^{3}\\minus{}9abc\\minus{}(2b^{2}\\minus{}6ac)\\sqrt{s_{1}}\\plus{}27a^{2}d}{27a^{2}}$, \u03c4\u03cc\u03c4\u03b5 \u03bc\u03b5 \u03c0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03b9\u03b1\u03c3\u03bc\u03cc \u03ba\u03b1\u03c4\u03b1 \u03bc\u03ad\u03bb\u03b7 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ f(x_{1})f(x_{2})\\equal{}\\frac{(2b^{3}\\minus{}9abc\\plus{}27a^{2}d^{2})^{2}\\minus{}(2b^{2}\\minus{}6ac)^{2}(b^{2}\\minus{}3ac)}{(27a^{2})^{2}}$ \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03c4\u03b9\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 $ f(x_{1})f(x_{2})\\equal{}\\frac{s_{2}}{27a^{2}}$ \u03cc\u03c0\u03bf\u03c5 $ s_{2}\\equal{}4ac^{3}\\plus{}4db^{3}\\plus{}27a^{2}d^{2}\\minus{}b^{2}c^{2}\\minus{}18abcd$.\r\n\r\n\u03a3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9, \u03b1\u03bd \u03b7 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 $ f(x)\\equal{}ax^{3}\\plus{}bx^{2}\\plus{}cx\\plus{}d\\Rightarrow f(x)\\equal{}0$ \u03ad\u03c7\u03b5\u03b9 3 \u03ac\u03bd\u03b9\u03c3\u03b5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ad\u03c2 \u03c1\u03af\u03b6\u03b5\u03c2 \u03c4\u03cc\u03c4\u03b5 $ s_{1}>0$ \u03ba\u03b1\u03b9 $ s_{2}<0$. \u0399\u03c3\u03c7\u03cd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03c1\u03bf\u03c6\u03bf, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03b1\u03bd \u03b9\u03c3\u03c7\u03cd\u03bf\u03c5\u03bd \u03cc\u03c4\u03b9 $ s_{1}>0$ \u03ba\u03b1\u03b9 $ s_{2}<0$ \u03c4\u03cc\u03c4\u03b5 \u03b7 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 $ f(x)\\equal{}0$ \u03ad\u03c7\u03b5\u03b9 3 \u03ac\u03bd\u03b9\u03c3\u03b5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ad\u03c2 \u03c1\u03af\u03b6\u03b5\u03c2. \u0395\u03ac\u03bd \u03c4\u03ce\u03c1\u03b1 \u03b9\u03c3\u03c7\u03cd\u03bf\u03c5\u03bd $ s_{1},s_{2}>0$ \u03c4\u03cc\u03c4\u03b5 \u03b7 $ f(x)$ \u03ad\u03c7\u03b5\u03b9 \u03bc\u03af\u03b1 \u03b1\u03c0\u03bb\u03ae \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae \u03c1\u03af\u03b6\u03b1 \u03ba\u03b1\u03b9 2 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ad\u03c2 \u03c3\u03c5\u03b6\u03cd\u03b3\u03b5\u03af\u03c2 \u03c1\u03af\u03b6\u03b5\u03c2. \u03a4\u03ad\u03bb\u03bf\u03c2 \u03b1\u03bd $ s_{2}\\equal{}0$ \u03c4\u03cc\u03c4\u03b5 \u03b7 $ f(x)$ \u03ad\u03c7\u03b5\u03b9 \u03bc\u03af\u03b1 \u03b1\u03c0\u03bb\u03ae \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae \u03c1\u03af\u03b6\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03af\u03b1 \u03b4\u03b9\u03c0\u03bb\u03ae \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae \u03c1\u03af\u03b6\u03b1. :wink:\r\n\r\n\u0395\u03ac\u03bd \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03cd\u03c0\u03b1\u03c1\u03be\u03b7 \u03c4\u03c9\u03bd \u03c1\u03b9\u03b6\u03ce\u03bd \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03c3\u03c4\u03b5\u03af\u03bb\u03b5\u03b9\u03c2 pm :)", "Solution_2": "\u0388\u03c3\u03c4\u03c9 $ r_{1},r_{2},r_{3}$ \u03bc\u03b5 $ r_{1} < r_{2} < r_{3}$ \u03bf\u03b9 3 \u03ac\u03bd\u03b9\u03c3\u03b5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ad\u03c2 \u03c1\u03af\u03b6\u03b5\u03c2 \u03c4\u03b7\u03c2 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7\u03c2 $ f(x) \\equal{} 0$. \u03a4\u03cc\u03c4\u03b5 \u03b1\u03c0\u03cc \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c4\u03bf\u03c5 Rolle \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \r\na)$ \\exists x_{1}\\in (r_{1},r_{2}): f^{'}(x_{1}) \\equal{} 0$\r\n\r\nb)$ \\exists x_{2}\\in (r_{2},r_{3}): f^{'}(x_{2}) \\equal{} 0$\r\n\r\n\u0386\u03c1\u03b1 \u03b7 $ f^{'}(x) \\equal{} 3ax^{2} \\plus{} 2bx \\plus{} c$ \u03ad\u03c7\u03b5\u03b9 2 \u03ac\u03bd\u03b9\u03c3\u03b5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ad\u03c2 \u03c1\u03af\u03b6\u03b5\u03c2 $ x_{1},x_{2}$.\r\n\r\n\u0391\u03c5\u03c4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b2\u03b1\u03c3\u03b9\u03ba\u03ae \u03c3\u03c7\u03ad\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03c3\u03c5\u03bd\u03b4\u03ad\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b1 $ x_{1},x_{2}$ \u03bc\u03b5 \u03c4\u03b7\u03bd $ f(x)$.\r\n\r\n\u03a4\u03af\u03c0\u03bf\u03c4\u03b1 \u03c4\u03bf \u03c3\u03c0\u03bf\u03c5\u03b4\u03b1\u03af\u03bf, \u03cc\u03c0\u03c9\u03c2 \u03c3\u03bf\u03c5 \u03b5\u03af\u03c0\u03b1 :P", "Solution_3": "\u03a0\u03b1\u03b9\u03b4\u03b9\u03ac \u03c0\u03bf\u03bb\u03cd \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03ac \u03c4\u03b1 \u03bb\u03ad\u03b5\u03b9 \u03b7 [url=http://en.wikipedia.org/wiki/Discriminant]Wikipedia[/url] \u03b3\u03b9\u03b1 \u03b4\u03b9\u03b1\u03ba\u03c1\u03af\u03bd\u03bf\u03c5\u03c3\u03b5\u03c2 \u03c0\u03bf\u03bb\u03c5\u03c9\u03bd\u03cd\u03bc\u03c9\u03bd \u03bf\u03c0\u03bf\u03b9\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03b2\u03b1\u03b8\u03bc\u03bf\u03cd! :o" } { "Tag": [ "videos" ], "Problem": "These are the rules in this game:\r\n\r\n1. Someone posts their hobby\r\n2.The next person then checks if his hobby is the same as one of the previous posts. If it is, then he posts the name of the username of the person who posted the hobby [b]first[/b] in spoiler tags.(if not, then do the same thing as in step 1)\r\n\r\neg: boy1: my hobby is video-gaming\r\n boy2: [hide=hobby]boy 1[hide]\r\nAt this point the hobby that boy1 posted gets 1 point.\r\n\r\n3. At the end of 2 days, I will check which hobby has the most points.(Math is NOT a hobby :P )\r\n\r\nMy hobby: Programming", "Solution_1": "My Hobby: tennis", "Solution_2": "Swimming\r\nAnd can you have more than one hobby? If so [hide=\"my other hobby is\"]batteredbutnotdefeated[/hide]", "Solution_3": "What do you mean by your second hobby?", "Solution_4": "@mathchampion, read the rules :wink:", "Solution_5": "My hobby is MUSIC.\r\n\r\nIsn't this going to be a bit complicated after there are 3 or 4 pages of hobbies to check?", "Solution_6": "[hide=\"Scores so far\"]\nbatteredbutnotdefeated\n[/hide]\r\nI think when you post your hobby you should also post your score\r\nAnd isn't it possible for people to lie about there hobby just to make sure somebody else doesn't get a point?", "Solution_7": "I think this thread is going to die soon...\r\n\r\n@supernerd: I'm depending on the good will of most AoPSers :rotfl: :rotfl: :rotfl:", "Solution_8": "My Hobby is playing video-games.", "Solution_9": "[b]doing math[/b] (it isn't just math...)\r\n\r\n[b]AoPS[/b]\r\n\r\nIf those don't count, nothing else." } { "Tag": [ "Inequality", "three variable inequality", "Muirhead", "IMO Shortlist", "IMO Longlist", "algebra", "inequalities proposed" ], "Problem": "Prove that for arbitrary positive numbers the following inequality holds \r\n\\[\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\leq \\frac{a^8 + b^8 + c^8}{a^3b^3c^3}.\\]", "Solution_1": "Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url] :)", "Solution_2": "[quote=\"orl\"]Prove that for arbitrary positive numbers the following inequality holds \n\\[\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\leq \\frac{a^8 + b^8 + c^8}{a^3b^3c^3}.\\][/quote]\r\n\r\nMultiplying this inequality with $a^3b^3c^3$, we get\r\n\r\n$a^2b^3c^3 + b^2c^3a^3 + c^2a^3b^3 \\leq a^8+b^8+c^8$.\r\n\r\nThis is trivial using the Muirhead inequality (in fact, $\\left(8;\\;0;\\;0\\right)\\succ\\left(3;\\;3;\\;2\\right)$). Without the Muirhead inequality, one can prove this by noting that AM-GM yields\r\n\r\n$a^2b^3c^3 = aabbbccc = \\sqrt[8]{a^8a^8b^8b^8b^8c^8c^8c^8} \\leq \\frac{a^8+a^8+b^8+b^8+b^8+c^8+c^8+c^8}{8}$\r\n$ = \\frac28 a^8 + \\frac38 b^8 + \\frac38 c^8$\r\n\r\nand similarly $b^2c^3a^3 \\leq \\frac28 b^8 + \\frac38 c^8 + \\frac38 a^8$ and $c^2a^3b^3 \\leq \\frac28 c^8 + \\frac38 a^8 + \\frac38 b^8$.\r\n\r\nSumming up, we get\r\n\r\n$a^2b^3c^3 + b^2c^3a^3 + c^2a^3b^3$\r\n$\\leq \\left(\\frac28 a^8 + \\frac38 b^8 + \\frac38 c^8\\right) + \\left(\\frac28 b^8 + \\frac38 c^8 + \\frac38 a^8\\right)$\r\n$ + \\left(\\frac28 c^8 + \\frac38 a^8 + \\frac38 b^8\\right)$\r\n$=a^8+b^8+c^8$,\r\n\r\nqed..\r\n\r\n Darij", "Solution_3": "I got the same solution as Darij. Most competitions don't allow 1 line proofs by Muirhead's inequality, but the fact that the inequality holds by Muirhead guarantees a solution by AM-GM or weighted AM-GM", "Solution_4": "and if we can use amgm,we can also use sos form", "Solution_5": "Actually, this is very easy problem: WLOG let $a \\geq b \\geq c$. Then by Chebyshev's inequality we have $3(a^8+b^8+c^8) \\geq (a^2+b^2+c^2)(a^6+b^6+c^6) \\geq (ab+bc+ca)(3a^2b^2c^2)=3(ab+bc+ca)(a^2b^2c^2)$ Now after dividing everything with $3a^3b^3c^3$ we have $\\frac{(a^8+b^8+c^8)}{a^3b^3c^3} \\geq \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$, and we're done.", "Solution_6": "Why isn't there some really troll solution yet? Ok, may I do that part...\n\nWe know that $a^2+b^2+c^2 \\ge ab+bc+ca$ for any reals.\n\nThus, $a^8+b^8+c^8 \\ge a^2b^4+b^4c^4+c^4a^4 \\ge a^2b^4c^2+b^2c^4a^2+c^2a^4b^2 \\ge a^2b^2c^2(a^2+b^2+c^2) \\ge (abc)^2(ab+bc+ca)$ which is the desired result.\n\nToo easy...", "Solution_7": "Very nice solution. ", "Solution_8": "$(5,-3,-3)\\succ(1,1,1)$", "Solution_9": "Multiply both sides by $a^3b^3c^3$, yielding $a^2b^3c^3 + b^2c^3a^3 + c^2a^3b^3 \\leq a^8+b^8+c^8$. Muirhead on $\\left(8,0,0\\right)\\succ\\left(3,3,2\\right)$ and dividing by $2$ yields the desired result.\n\n", "Solution_10": "Consider the sequences $(3,3,2),(8,0,0)$.Then: $ (3,3,2)\\prec (8,0,0)$\nBy Muirhead\u2019s theorem we obtain: $T [3,3,2] \\leq T [8,0,0]$\ni.e. $2(a^2b^3c^3+a^3b^2c^3+a^3b^3c^2) \\leq 2(a^8+b^8+c^8)$\n$\\implies a^2b^3c^3+a^3b^2c^3+a^3b^3c^2 \\leq a^8+b^8+c^8$\n$\\implies \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\leq \\frac{a^8 + b^8 + c^8}{a^3b^3c^3}$.$\\blacksquare$", "Solution_11": "Let $a,b,c>0.$ Prove that \n$$\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\leq\\frac{a^2 + b^2+ c^2}{abc} \\leq\\frac{a^5 + b^5 + c^5}{a^2b^2c^2}\\leq\\frac{a^8 + b^8 + c^8}{a^3b^3c^3}$$\n\n" } { "Tag": [ "LaTeX", "\\/closed" ], "Problem": "Sorry, but i want to know why cant i type latex? Where is [tex] tags? Please send me some help!", "Solution_1": "Hi, you can type LaTeX by inputing LaTeX code between [tex] and [/tex] tags, or between usual $ ... $ tags. \r\nMore info for general LaTeX is located here http://www.mathlinks.ro/LaTeX/AoPS_L_About.php or, if you want a quick guide, read this topic:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=5261\r\n\r\nYou can test LaTeX in the Test forum: http://www.mathlinks.ro/Forum/index.php?f=224 and you can ask further questions about LaTeX typing in the LaTeX forum: http://www.mathlinks.ro/Forum/index.php?f=123" } { "Tag": [], "Problem": "Find all positive integers $(x,y,z)$ such that:\r\n\r\n$xy+x-z=1$\r\n$xy+y+z=2006$", "Solution_1": "[quote=\"PenguinIntegral\"]Find all positive integers $(x,y,z)$ such that:\n\n$xy+x-z=1$\n$xy+y+z=2006$[/quote]\r\n\r\n[hide=\"Probably wrong\"]Add: $2xy+x+y=2007$\n\n$(2x+1)(2y+1)=4015=5\\cdot 11\\cdot 73$\n\nSo $(x,y)=(2,401)(10,182)(36,27)$ and permutations \n\n$(x,y,z)=\\boxed{(2,401,803)(401,2,1202)(10,182,1829)(182,10,2001)(36,27,1007)(27,36,998)}$[/hide]", "Solution_2": "[quote=\"ch1n353ch3s54a1l\"][quote=\"PenguinIntegral\"]Find all positive integers $(x,y,z)$ such that:\n\n$xy+x-z=1$\n$xy+y+z=2006$[/quote]\n\n[hide=\"Probably wrong\"]Add: $2xy+x+y=2007$\n\n$(2x+1)(2y+1)=4015=5\\cdot 11\\cdot 73$\n\nSo $(x,y)=(2,401)(10,182)(36,27)$ and permutations \n\n$(x,y,z)=\\boxed{(2,401,803)(401,2,1202)(10,182,1829)(182,10,2001)(36,27,1007)(27,36,998)}$[/hide][/quote]\r\n\r\nyour factorization is false", "Solution_3": "It looks correct to me, but then again this is Pre-Olympiad and I'm way out of my league.\r\n\r\n[hide]$2xy+x+y=2007$\n$4xy+2x+2y=4014$\n$4xy+2x+2y+1=4015$\n$(2x+1)(2y+1)=4015$[/hide]", "Solution_4": "oh haha...i did not read that carefully...ok" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "How many functions $f: \\{1,2,...,n\\}\\rightarrow \\{1,2,3,4,5\\}$ are there such that for any $k=1,2,...,n-1$ it holds that$|f(k+1)-f(k)|\\geq 3$?", "Solution_1": "Let $U_n$ be the desired number.\r\nDirect checking shows that $U_1=5$ and $U_2 = 6.$\r\n\r\nNow, assume that $n >2:$\r\nClearly, none of the numbers $f(k)$ can be equal to $3$. \r\nLet $a_n,b_n,c_n,d_n$ be the number of such functions such that $f(n)=1,2,4,5$ respectively. Thus $U_n = a_n+b_n+c_n+d_n.$\r\n\r\nAny 'good'function for $n+1$ is a good function when restricted to $\\{1, \\cdots, n \\}.$ Conversely, starting with a good function $f$ for $\\{1, \\cdots, n \\}$ we may $f(n+1)=4,5$ if $f(n)=1$, we may choose $f(n+1) = 5$ if $f(n)=2$, we may choose $f(n+1)=1$ if $f(n)=4$ and we may choose $f(n+1)=1,2$ if $f(n)=5.$\r\nIt follows that $a_{n+1} = c_n+d_n,b_{n+1} = d_n, c_{n+1} =a_n$ and $d_{n+1} = a_n + b_n.$\r\nThen $a_{n+1} + d_{n+1} = U_n$ so that $U_{n+1} = U_n + a_n + d_n = U_n + U_{n-1}.$\r\n\r\nSolving this linear recurrence, it easily follows (it may be convenient to define $U_0 = 1$) that $U_n = (\\frac 1 2 + \\frac 9 {10} \\sqrt 5)( \\frac {1 + \\sqrt 5} 2)^n + (\\frac 1 2 - \\frac 9 {10} \\sqrt 5)( \\frac {1 - \\sqrt 5} 2)^n .$\r\n\r\nPierre." } { "Tag": [ "algorithm", "USAMTS" ], "Problem": "In general, how do you grade computer programs that yield a solution? \r\n\r\nDo you want us to print out the entire program, or just explain in pseudocode?", "Solution_1": "[quote=\"1234567890\"]In general, how do you grade computer programs that yield a solution? \n\nDo you want us to print out the entire program, or just explain in pseudocode?[/quote]\r\nDon't give pseudocode.\r\n\r\nWe graders want a description in mathematical English why you expect the algorithm in your program to find the answer. We are interested in your mathematicial reasoning. If your program has some programming error in it that gave the wrong answer, the more fanatical graders might look through the code to point out the error (and maybe give you a little more partial credit if the error is minor), but that would be the only reason to give the actual code. Pseudocode would probably lack the bug, and thus would be useless.\r\n\r\nErin Schram\r\nUSAMTS grader" } { "Tag": [], "Problem": "The four-digit numeral 3AA1 is divisible by 9. What digit does A represent?", "Solution_1": "[hide=\" click to reveal hidden content\"]\n\nSince the sum of the digits must be divisible by $ 9$, $ A\\plus{}A\\equal{} 5$ or $ 14$; therefore $ A\\plus{}A\\equal{}14$ and so $ A\\equal{}7$.\n\n[$ \\frac{3771}{9} \\equal{}419$.]\n\n[/hide]", "Solution_2": "So you know that 3AA1 has to be divisible by 9..\r\nAnything divisible by 9= all the digits added up have to be divisible by 9,\r\nso then you know that 3+1=4..then if u want a multiple of 9, they have to be 9,18...\r\nBut, 9 won't work..and 18 does. You subtract 4 from 18=14..Then you divide that by 2, and get 7 as your answer.", "Solution_3": "ok. So\r\n[hide] 3AA1 has to be divisible by 9. So, it has to be divisible by 3....right? If you add up the digits for 3AA1, you have to get a multiple of nine. So, you already have a 4, when you add the 3 and the 1. So then A, A have to be the same. 9 would not work, because 9-4=5, and you can't split up nine into 4.5's. So our next choice, is eighteen. Eighteen works ,because 18-4=14, and 14 divided by 2=7 [/hide]" } { "Tag": [ "function" ], "Problem": "Find all functions $ f: \\mathbb{R}\\longrightarrow \\mathbb{R}$ such that\n\\[f(x\\plus{}y)\\plus{}f(y\\plus{}z)\\plus{}f(z\\plus{}x)\\ge 3f(x\\plus{}2y\\plus{}3z)\\]\nfor all $x, y, z \\in \\mathbb R$.", "Solution_1": "[hide=\"Hint 1\"]$ f$ is constant.[/hide]\n[hide=\"Hint 2\"]Prove that $ f(0)\\le f(t)$ and $ f(t)\\le f(0)$.[/hide]\n[hide=\"Big hint (do not open before you give up)\"]Plug in $ x \\equal{} t,y \\equal{} 0,z \\equal{} 0$; plug in $ x \\equal{} \\frac {t}{2},y \\equal{} \\frac {t}{2},z \\equal{} \\minus{} \\frac {t}{2}$.[/hide]", "Solution_2": "[hide=\"solution\"]\nSetting $ y = 0, z = 0 $ yields \\[ 2f(x) + f(0) \\ge 3 f(x) \\] or \\[ f(0) \\ge f(x) \\] for all $ x \\in \\mathbb{R} $.\n\nNow we let $ x = n, y = n $ and $ z = -n $ to obtain \\[ f(2n) + 2f(0) \\ge 3f(0) \\] or \\[ f(2n) \\ge f(0) \\] which is equivalent to \\[ f(x) \\ge f(0) \\] for all $ x \\in \\mathbb{R} $.\n\nThus, $ f(x) = f(0) = c $ for any real $ c $. Plugging into the equation gives $ 3c \\ge 3c $, which clearly always hold so the answer is $ \\boxed{f(x) \\equiv c} $ $ \\forall c \\in \\mathbb{R} $.\n\n[/hide]", "Solution_3": "$1 + 2 = 3$ again?!\n\nThe answer is $f(x)=c$ for all reals $c$; it's easy to see that these solutions work. Let $P(x,y,z)$ denote the given assertion.\n \nSetting $P(x,0,0)$ gives $3f(x) \\leq 2f(x) + f(0)$, so $f(x) \\leq f(0)$ for all reals $x$. Setting $P(\\frac{x}{2},\\frac{x}{2},-\\frac{x}{2})$ gives $3f(0) \\leq 2f(0) + f(x)$, so $f(0) \\leq f(x)$ for all reals $x$. Therefore $f(x) \\leq f(0) \\leq f(x)$, or $f$ is constant.\n\n[b]Remark.[/b] I really like this FE because both steps are motivated: you get the first from just doing stuff and you wonder, ``huh, how do I use the inequality?'' Then the solution is natural.", "Solution_4": "Really cute :)\nAs usual let $P(x,y,z)$ denote the given assertion.\n[list]\n[*]$P(x,0,0) \\implies f(x) \\le f(0)$ for all $x$.\n\n[*]$P\\left( \\dfrac{-3x}{2} , \\dfrac{3x}{2} , \\dfrac {-x}{2} \\right) \\implies 2f(0) \\le f(-2x) + f(x)$ which is also $\\le 2f(0) \\implies f(-2x) + f(x) = 2f(0)$.\n\n[*]$P(-x , 2x , -x) \\implies f(x) \\ge f(0)$.\n[/list]\nSo $f(x) = c$ for constant $c$.\n\nEdit: wait bruh how did I miss the more natural substitution shown above.", "Solution_5": "$P(x,0,0)\\Rightarrow f(x)\\le f(0)$\n$P(x,x,-x)\\Rightarrow f(2x)\\ge f(0)\\Rightarrow f(x)\\ge f(0)\\Rightarrow\\boxed{f(x)=c}$ for some $c\\in\\mathbb R$, which works.", "Solution_6": "$P(y,y,-y) \\implies f(2y)+f(0)+f(0) \\geq 3f(0) \\implies f(x) \\geq f(0) \\forall x$\n$P(-y,y,-y) \\implies f(0)+f(0)+f(-2y) \\geq 3f(-2y) \\implies f(0) \\geq f(-2y) \\implies f(0) \\geq f(x) \\forall x$\nTherefore, $f(x)=c$ where $c$ is a constant for all $x$", "Solution_7": "Only constant functions work.\n\nSumming this inequality cyclically, $$f(x+2y+3z) + f(y+2z+3x)+f(z+2x+3y) \\leq f(x+y) + f(y+z) + f(z+x).$$ On the other hand, substituting $P(2x+y, 2y+z, 2z+x)$, we obtain exactly the reverse inequality. As a result, equality must hold everywhere. Now, setting $y=z=0$, $$3f(x) = f(x) + 2f(0) \\implies f(x) = f(0)$$ is constant, as needed.", "Solution_8": "We claim the only solution is $f(x)=c$, where $c$ is a real constant. Note that this satisfies the given conditions. Let $P(x,y,z)$ denote the given assertion and $P(x,0,0)$ yields $f(x)\\le f(0)$. Then, $P(-z,-z,z)$ yields $f(0)\\le f(-2z)$ so $f(x)\\ge f(0)$. Hence, $f(x)=f(0)$ so we are done. $\\square$\n", "Solution_9": "Plugging in $(x,y,z)=(x,0,0)$ to get\n\n\\[3f(x) \\le 2f(x)+f(0)\\]\n\\[\\implies f(x) \\le f(0)\\]\n\nPlugging in $(x,y,z)=(x/2,x/2,-x/2)$ gives\n\n\\[3f(0) \\le f(x)+2f(0)\\]\n\\[\\implies f(x) \\ge f(0)\\]\n\nThus, $f(x)=f(0)$, so our answer is $\\boxed{f(x)=c, \\ c \\in \\mathbb{R}}$.", "Solution_10": "Substitute $(2k,0,0)$ and $(k,k-k)$ to find\n\\[f(2k) \\leq f(0), \\quad f(2k) \\ge f(0).\\]\n\nHence the constant $f(0)$ is the only value $f$ can take, so $\\boxed{f(x) = c, \\quad c \\in \\mathbb{R}}$, which evidently works. $\\blacksquare$", "Solution_11": "$f\\equiv c$ for some constant $c$ is the solution.\n\n$P(x,0,0) \\implies 3f(x) \\le f(x) + f(0) + f(x) \\implies f(x) \\le f(0)$.\n\n$P(-x,-x,x) \\implies 3f(-x + 2(-x) + 3x) \\le f(-2x) + f(0) + f(0) \\implies 3f(0) \\le f(-2x) + 2f(0) \\implies f(-2x) \\ge f(0)$.\nNow changing $-2x \\rightarrow x$, we get that $f(x) \\ge f(0)$.\n\nThus $f(x) = f(0)$. :yoda:", "Solution_12": "Letting $y = z = 0$ gives $f(x) \\leq f(0)$.Then $z = -x = -y$ gives $3f(0) \\leq f(2x) + 2f(0)$ so $f(0) \\leq f(2x)$, which is only possible if $f(x) = f(0) = c$.", "Solution_13": "The answer is $f \\equiv c$ for any real constant $c$, which works. Let $P(x, y, z)$ denote the given assertion. From $P(x, 0, 0)$ we have $3f(x) \\le 2f(x) + f(0)$, so $f(x) \\le f(0)$. From $P(\\tfrac{x}{2}, \\tfrac{x}{2}, -\\tfrac{x}{2})$ we have $3f(0) \\le f(x) + 2f(0)$, so $f(x) \\ge f(0)$. Hence $f(x) = f(0)$ for all $x$, which implies the solution set.", "Solution_14": "The answer is $f(x) = c$ only - clearly this works. Now\n\\[ P(x, 0, 0) \\implies 3f(x) \\le 2f(x) + f(0) \\implies f(x) \\le f(0) \\]\nso \n\\[ 3f(x+2y+3z) \\le f(x+y) + f(y+z) + f(z+x) \\le 3f(0). \\]\nNow $P(a, a, -a)$ implies that \n\\[ 3f(0) \\le f(2a) + 2f(0) \\le 3f(0) \\]\nso $f(0) \\le f(a)$ for all $a$, and we're done." } { "Tag": [ "probability" ], "Problem": "1. In a game in which there are 52 cards, and I draw 5, what are the chances (fraction) that I can make a straight flush, if As can be the low one or the high one?\r\n\r\n2. I roll 10 dices, in how many different ways can I row atleast one 1?\r\n\r\n3. I have fourteen blocks labled 1,2,3,4...14. I select one, then another, and then finally a third one, without replacement. What are the chances that the blocks I select are in increasing order in the way I chose them (so if I select 4,7,2 I have selected that, I cannot rearrange them).\r\n\r\nWhat are the chnaces that the blocks are in strictly increasing order ascending by 1 (so 1,2,3 or 2,3,4..)", "Solution_1": "Define flush :oops: \r\n#2[hide]Use complimentry counting. There are a total of $6^{10}$ possibilities. There are $5^{10}$ ways to not roll a one. Just subtract.[/hide]\n\nHint on #3: [hide] There are a total of $\\binom{14}{3}$ ways. Someone else can figure out the number of ways to have an increasing order.[/hide]", "Solution_2": "[quote=\"xpmath\"]Define flush :oops: \n#2[hide]Use complimentry counting. There are a total of $6^{10}$ possibilities. There are $5^{10}$ ways to not roll a one. Just subtract.[/hide]\n\nHint on #3: There are a total of $\\binom{14}{3}$ ways. Someone else can figure out the number of ways to have an increasing order.[/quote]\r\n\r\nactually there are a total of $\\binom{14}{3}$ ways for the numerator, as there are a total of $\\binom{14}{3}$ the numbers can be increasing :wink:", "Solution_3": "What would the denominator be then?", "Solution_4": "do you guys need answers to help out?", "Solution_5": "[quote=\"Art of Owna\"][quote=\"xpmath\"]Define flush :oops: \n#2[hide]Use complimentry counting. There are a total of $6^{10}$ possibilities. There are $5^{10}$ ways to not roll a one. Just subtract.[/hide]\n\nHint on #3: There are a total of $\\binom{14}{3}$ ways. Someone else can figure out the number of ways to have an increasing order.[/quote]\n\nactually there are a total of $\\binom{14}{3}$ ways for the numerator, as there are a total of $\\binom{14}{3}$ the numbers can be increasing :wink:[/quote]\r\nOh I see now. Every 3 numbers you choose have exactly 1 way to be increasing.", "Solution_6": "i need 25 posts." } { "Tag": [ "geometry", "3D geometry", "sphere" ], "Problem": "what is the radius of the sphere whose volume is 330 cubic meters?", "Solution_1": "[hide]\nThe volume of a sphere is given by $V = \\frac{4 \\pi r^{3}}{3}$.\nLet $V=330$ and solve for r. :maybe: \n[/hide]", "Solution_2": "Do I use 3.14 for pi when solving for r?", "Solution_3": "Depends on how precise you want the answer.", "Solution_4": "Usually you can just leave the answer in terms of pi.\r\nHowever, the directions on your homework or schoolwork might tell you to use 3.14 .", "Solution_5": "Why do you keep posting your homework problems anyway? AoPS isnt a place for people to do your work for you...", "Solution_6": "Yes that is right.\r\nHowever, solving for $r$ yields\r\n\r\n$r=\\sqrt[3]{\\frac{3V}{4\\pi }}$", "Solution_7": "Since no one has the answer yet...\r\n\r\n$r=\\sqrt[3]{\\frac{3V}{4\\pi }}$\r\n\r\n$r=\\sqrt[3]{\\frac{3(330)}{4\\pi}}$\r\n\r\n$r=\\sqrt[3]{\\frac{990}{12.56}}$\r\n\r\n$r \\approx \\sqrt[3]{78.82}$\r\n\r\n$r \\approx 8.878$ \r\n\r\n$\\approx 8.878 \\, \\text{meters}$\r\n\r\nLol not only is posting homework on AoPS wrong, but chances are we won't see the problem until it is too late. Try and to it yourself, won't you? :wink:", "Solution_8": "Can 8.878 meters be rounded off to the nearest tenths?", "Solution_9": "Depends on the requirements, but a good rule is to equal the significant numbers of the input\r\nwhich are 3. Thus an appropriate round off would be $8.88$", "Solution_10": "[quote=\"color\"]Depends on the requirements, but a good rule is to equal the significant numbers of the input\nwhich are 3. Thus an appropriate round off would be $8.88$[/quote]\r\n\r\nHe meant nearest tenths. Therefore the answer would be $8.9$.", "Solution_11": "uh, the answer is not 8.878, its 4.288. You accidently took the square root of 78.82 rather than the cube root. I realize this is an old thread, but just in case anyone starts looking :lol:", "Solution_12": "[quote=\"trivial34\"]uh, the answer is not 8.878, its 4.288. You accidently took the square root of 78.82 rather than the cube root. I realize this is an old thread, but just in case anyone starts looking :lol:[/quote]\r\n\r\nHuh. So it is. My mistake. I must have messed up on my calculator :blush: ." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "gcd(a,b)=1,a,b>0and there exists N\u2208Zsuch:aN+b is prime.\r\nprove that:there exists infinite n s.t:an+b is prime.\r\nI wanna a conformation.so please don't tell me that using Dirichlet Theorem,it's obvious...\r\nthx", "Solution_1": "[quote=\"bye\"]gcd(a,b)=1,a,b>0and there exists N\u2208Zsuch:aN+b is prime.\nprove that:there exists infinite n s.t:an+b is prime.\nI wanna a conformation.so please don't tell me that using Dirichlet Theorem,it's obvious...\nthx[/quote]\r\n\r\nI'm sorry, but I dont quite well understand your exercise. Are you just asking for a Dirichlet theorem proof ?\r\n\r\nIf so, Google is your friend : Just look for \"Dirichlet theorem proof\" and the first result seems ok (and too long to be copied here).\r\nIf not, maybe you could clarify your question.\r\n\r\nSorry if my request bother you.", "Solution_2": "i see a proof in a book.\r\nit says that if there exists n making the faction a prime.then dirichlet theorem is proved.i don't quite understand that.\r\ne.g if there exists a prime with the form 4n +1 .like 5,then there exists infinite primes with the form 4n+1.\r\ni think that there must be a conformation.you get a prime in the form,then you can get another prime in that form using the prime.i just don't know what the conformation is.", "Solution_3": "to make it more clear\r\nsuppose:an+b=p p is a prime\r\nthen there exists s satisfying:asp+b=q q is a prime\r\nwhat's s?", "Solution_4": "[quote=\"bye\"]e.g if there exists a prime with the form 4n +1 .like 5,then there exists infinite primes with the form 4n+1.[/quote]\r\nNo, that's not what the question is saying; $ a, b$ aren't fixed. You need the full strength of the given result. In this case, it would go as follows:\r\n\r\nThere exists a prime of the form $ 8n \\plus{} 1$, which must be greater than $ 5$. $ 17$ is such a prime.\r\n\r\nThere exists a prime of the form $ 20n \\plus{} 1$, which must be greater than $ 17$. $ 41$ is such a prime.\r\n\r\nThere exists a prime of the form $ 44n \\plus{} 1$, which must be greater than $ 41$...\r\n\r\nAnd so forth. Hence there exist infinitely many primes of the form $ 4n \\plus{} 1$.", "Solution_5": "well, i think i understand it...\r\nyeah...\r\nif an+b=p is a prime\r\nthen there exists l that:alp+b=q is a prime\r\nand so on...\r\nso there exists infinitive k s.t:\r\nak+b is a prime\r\nit seems easy now\r\nthanks t0rajir0u and pco...\r\n :lol:" } { "Tag": [], "Problem": "Given a number whose division by 3 gives remainder 1,division by 7 leaves remainder 3.What is the remainder if it is divided by 21?", "Solution_1": "is $10$ :D this number is 21k+10", "Solution_2": "I don't know if there's any way to find this.\r\nBut this is my way :blush: \r\nIf $n$ is divided by 7 gives 3 it could be $21k+3; 21k+10; 21k+17$\r\nMoreover, $n$ is divided by 3 gived 1 so $n$ must be $21k+10$" } { "Tag": [ "ratio" ], "Problem": "During the electrolysis of $ Zn(NO_{3})_{2}$ was released 3.36 liter of a gass. Find the mass of the metal emplaced in the cathode .", "Solution_1": "You could probably begin by writing the half reactions and determining what gas is evolving. Then you could use gas laws to go from volume to mass.", "Solution_2": "I can write the reacion :\r\n\r\n$ 2Zn(NO_{3})_{2}\\equal{}2Zn^{2\\plus{}}\\plus{}4(NO_{3})^{\\minus{}}$\r\n$ 8H_{2}O\\equal{}4H_{3}O^{\\plus{}}\\plus{}4OH^{\\minus{}}$\r\n\r\nCathode (-) | $ 2Zn^{\\plus{}2}\\plus{}4e^{\\minus{}}\\equal{}2Zn$\r\n_________________________________\r\n\r\nAnode (+) | $ 4OH^{\\minus{}}\\equal{}2H_{2}O\\plus{}O_{2}\\plus{}4e^{\\minus{}}$\r\n\r\n\r\nSo the reaction evolved is : $ 2Zn(NO_{3})_{2}\\plus{}2H_{2}O\\equal{}2Zn\\plus{}O_{2}\\plus{}4HNO_{3}$\r\n\r\nThan what shuld I do ? \r\n\r\n \r\n\r\n [quote=\"feynman_wannabe\"]Then you could use gas laws to go from volume to mass.\n[/quote]\r\n\r\n\r\nCan you explain it please", "Solution_3": "Culd somebody give me an answer ?", "Solution_4": "Since reaction conditions have not been mentioned, lets assume we carry out the process at STP. :P \r\n\r\nNow, as you yourself mentioned, Zn is deposited at the cathode. So we have to find its weight.\r\n\r\n3.36 litres of oxygen is equivalent to $ n\\equal{}\\frac {3.36}{22.4}$ moles of oxygen and hence $ 32n g$ of oxygen.\r\n\r\nAccording to Faraday's law, ratio of substances deposited/liberated at anode/cathode is in the ratio of their chemical equivalents. Note that one mole of $ Zn^{2\\plus{}}$ ions gains 2 moles of electrons while out of four moles of $ OH^\\minus{}$, 2 moles undergo change in oxidation but the change is equal to 2 (to form one mole of oxygen). Hence the equivalent weight of Zn is equal to $ \\frac {63.5}{2}$ and that of oxygen is equal to $ \\frac {32}{4}$.\r\n\r\nSo, $ \\frac {32n}{32/4}\\equal{}\\frac {x}{63.5/2}$, x=weight of zinc\r\n\r\n :)", "Solution_5": "Thank you hell_ever" } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra solved" ], "Problem": "Consider the polynomials : \r\n$F_{0}=1,F_{1}=X,...,F_{p}=X\\left( X-1\\right) \\left( X-2\\right) ...(X-p+1)$\r\none can easily show that $\\left( F_{0},F_{1},...,F_{n}\\right)$ is a basis of $\\mathbb R_{n}\\left[ X\\right]$\r\nBut how can we decompose an element $P$ of $\\mathbb R_{n}\\left[ X\\right]$ in that basis ? anyone can help ??", "Solution_1": "The nicest way to do this is with differences. Let $\\Delta f(n)=f(n+1)-f(n)$.\r\n\r\nThen $P(n)=P(0)F_0+\\Delta P(0)F_1+\\frac12\\Delta^2P(0)F_2+\\cdots+\\frac1{k!}\\Delta^k P(0)F_k+\\cdots$\r\nThis works because $\\Delta F_k=kF_{k-1}$\r\n\r\nIf we used $\\frac1{k!}F_k$ as the basis elements instead, this would be a very nice basis for the integer-valued (on the integers) polynomials." } { "Tag": [ "calculus", "integration", "probability and stats" ], "Problem": "suppose that $ v$ is a signed measure,$ f$ is $ L$-measurable.If the intergral $ f$ with respect to $ v$ exists,$ f.v$$ < 3$. Let $B = (b_{1}, \\ldots , b_{2006}), a_{k}\\leq n-1 \\Rightarrow b_{k}= a_{k}, a_{k}= n \\Rightarrow b_{k}= n-1$. We get $m_{B}> m_{A}$, hence $m_{A}$ can achieve the maximum value only if $n=3$.\r\n\r\nLet $f(i) = \\#\\{ a_{k}| a_{k}= i\\}$. Then $m_{A}= f(1) \\cdot f(2) \\cdot f(3) \\leq 669 \\cdot 669 \\cdot 668=298970748$\r\n\r\n$\\text{\\LaTeX}$ :wink:" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Suppose you add a polynomial of degree m to a polynomial of degree n, where m>n\r\n\r\nThen the result is a polynomial of degree_____\r\n\r\n\r\nSuppose you subtract a polynomial of degree m from a polynomial of degree n, where m>n\r\n\r\nThe result is a polynomial of degree______\r\n\r\n\r\n\r\nthanks for the help :)", "Solution_1": "Both the answers are m, subtracting or adding n from m just changes all of the terms from x^n down, but still keeps x^m" } { "Tag": [ "calculus", "integration", "limit", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "Prove that $ \\lim_{h\\minus{}>\\plus{}0}h\\sum_{n\\equal{}1}^{\\infty}\\frac{1}{1\\plus{}e^{nh}}\\equal{}\\int_{0}^{\\infty}\\frac{dx}{1\\plus{}e^x}$", "Solution_1": "Do you mean: $ \\lim_{h\\to 0^ \\plus{} }h\\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{1 \\plus{} e^{nh}} \\equal{} \\int\\limits_{0}^{\\infty}\\frac {dx}{1 \\plus{} e^{x}} \\; ?$", "Solution_2": "Let $ h>0$ and consider the integral $ I\\equal{}\\int_{1}^{\\infty}\\frac{dx}{1\\plus{}e^{xh}}$.\r\n\r\nWe have that \r\n\r\n$ I\\equal{}\\sum_{k\\equal{}1}^{\\infty}\\int_{k}^{k\\plus{}1}\\frac{dx}{1\\plus{}e^{xh}}$.\r\n\r\nand note that \r\n\r\n$ \\frac{1}{1\\plus{}e^{(k\\plus{}1)h}}<\\int_{k}^{k\\plus{}1}\\frac{dx}{1\\plus{}e^{xh}}<\\frac{1}{1\\plus{}e^{kh}}$.\r\n\r\nIt follows, by summation, that\r\n\r\n$ \\int_{1}^{\\infty}\\frac{dx}{1\\plus{}e^{hx}}<\\sum\\limits_{k\\equal{}1}^{\\infty}\\frac{1}{1\\plus{}e^{kh}}<\\frac{1}{1\\plus{}e^{h}}\\plus{}\\int_{1}^{\\infty}\\frac{dx}{1\\plus{}e^{hx}}$\r\n\r\nMultiply by h both sides of the preceding inequality and the problem reduces to calculating the limit\r\n\r\n$ \\lim\\limits_{h\\rightarrow 0}\\int_{1}^{\\infty}\\frac{hdx}{1\\plus{}e^{xh}}$.\r\n\r\nHowever, the substitution $ xh\\equal{}y$, solves the problem." } { "Tag": [ "limit", "topology", "calculus", "calculus computations" ], "Problem": "Can someone explain why the set of limit points $L\\left[\\mathbb{R}\\right]=\\varnothing$?", "Solution_1": "what is a limit point ?\r\nafter reading : http://mathworld.wolfram.com/LimitPoint.html\r\n\r\nEvery interval (a,b) from the reals (with a not equal to b) has an infinite number of real numbers in between them (1). Now suppose that L(R) is not the null set than tere exist such a point p that has the property that there exists an open set from R which contains the point p and only p but that contradicts (1).\r\nI hope i am right!", "Solution_2": "I understand a limit point to be a point $p$ in a set $X$ if there exists a sequence of points $p_{n}$ with $p\\neq p_{n}$ such that $\\lim_{n\\rightarrow\\infty}p_{n}=p$. I mean, if I had the interval $\\left(0,1\\right)$, then the set of limit points would be $L\\left[\\left(0,1\\right)\\right]=\\left[0,1\\right]$. I don't see why this doesn't extend to the real numbers and becomes $L\\left[\\mathbb{R}\\right]=\\left[-\\infty,\\infty\\right]$.", "Solution_3": "I think you understood the exact opposite... although.... are you sure about your first post?", "Solution_4": "The set of limit points of $\\mathbb{R}$ is $\\mathbb{R}$. Maybe you've got this confused with \"boundary\"?", "Solution_5": "[quote=\"jmerry\"]The set of limit points of $\\mathbb{R}$ is $\\mathbb{R}$. Maybe you've got this confused with \"boundary\"?[/quote]\r\n\r\nWhat do you mean!", "Solution_6": "I'll have to double-check with my instructor, but I clearly remembering him saying that the set of limit points of the real numbers is the null set.", "Solution_7": "Your instructor is wrong, then." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $ a$ and $ b$ be positive integers with $ (a,b) \\equal{} 1$. Show that there exist integers $ m$ and $ n$ for which $ a^m \\plus{} b^n\\equiv1\\mod{ab}$.", "Solution_1": "Since (a, b) =1 we need to find m and n such that $ a^{m}\\equal{}1(mod b)$ and $ b^{n}\\equal{}1(mod a)$. But clearly we can just take m = $ \\phi(b)$ and $ n\\equal{}\\phi(a)$.", "Solution_2": "oh god, I was about to answer the question as same way you have answered" } { "Tag": [ "linear algebra", "matrix", "calculus", "derivative", "linear algebra unsolved" ], "Problem": "Let $A=\\left(\\begin{array}{ccc}3 & 1 & 1 \\\\ 6 & 2 & 2 \\\\ 3a & b & b \\end{array}\\right)$ a matrix and $B$ the set of pairs $(a,b)\\in\\mathbb{R}^{2}$ such that $\\det(xA+yI_{3})$ is independent of $x$. Then $B$ is?", "Solution_1": "Let A be a (n,n) matrix and f:(x,y)->det(xA+yI); f is independent of x iff for all x,y D1f(x,y)=0 or trace(A.adj(xA+yI))=0 (where adj(u).u=det(u).I); y=0 implies det(A)=0. y<>0 implies trace(A.(I-x.A)^-1)=0 for x rather small or 0=trace(A.(I+xA+x^2.A^2+......)=trace(A)+xtrace(A^2)+.....; by differentiations in 0: 0=trace(A)=trace(A^2)=....=trace(A^n). One concludes from it that A is nilpotent.\r\n Conversely if A is nilpotent, we may suppose A upper triangular with diagonal 0; det(xA+yI)=y^n.\r\nConclusion: f is independent of x iff A is nilpotent.\r\n Here trace(A)=b=-5 and trace(A^2)=30+6a and a=-5." } { "Tag": [ "modular arithmetic" ], "Problem": "#1 We choose 2 numbers in between 1 and 100, inclusive, such that their difference is 7 and their product is a multiple of 5. How many ways can we do this?\r\n\r\n#2 How many 7-digit numbers are multiples of 388 and end in 388?\r\n\r\n#3 Bob chose four distinct digits from the set $ \\{1,2,3,4,5,6,7,8,9\\}$.With them he formed all the possible four digit number and added all the numbers formed. The result was $ 193,314$. Find the four digits Bob chose.", "Solution_1": "[hide=\"For #1:\"]\nThis should work...\n\nLet the two numbers be $ a$ and $ b$ such that $ a > b$.\n\nWe know that $ a \\minus{} b \\equal{} 7$ and $ ab\\equiv 0\\pmod{5}$, since only one of the two numbers can be a multiple of $ 5$ we can start with cases, one in which $ a$ is a multiple of $ 5$ and the other where $ b$ is a multiple of $ 5$.\n\nCase 1: ($ a$ is a multiple of $ 5$)\n\n The lowest value for $ a$ that satisfies the conditions is $ 10 \\equal{} 5(2)$ where $ b$ is $ 3$.\n\n The highest value for $ a$ that satisfies the conditions is $ 100 \\equal{} 5(20)$ where $ b$ is $ 93$.\n\nSo there are a total of $ 20 \\minus{} 2 \\plus{} 1 \\equal{} 19$ cases that work.\n\nCase 2: ($ b$ is a multiple of $ 5$)\n\n The lowest value for $ b$ that satisfies the conditions is $ 5$ where $ a$ is $ 12$.\n \n The highest value for $ b$ that satisfies the conditions is $ 90 \\equal{} 5(18)$ where $ a$ is $ 97$.\n\nSo there are a total of $ 18$ cases that work.\n\n\n------------------------------------------------\n\nIn total there are $ 19 \\plus{} 18 \\equal{} \\boxed{37}$ cases that work.\n[/hide]", "Solution_2": "Hey tony are you going to the Olimpiada de Mayo this saturday??", "Solution_3": "UNCOOL, U POSTED THAT JUST 1 MIN BEFORE I FINISHED TYPING :furious: \r\nFor #2:\r\n[hide]$ lcd(1000, 388) \\equal{} 97000$\n$ 10000/97 \\equal{} 103.0.......$\nthus there are 10 such numbers\nEDIT: the first digit of the number can be 0. so it should be 9000/97=92.78.....so 92 such numbers[/hide]", "Solution_4": "Could you explain what that is?", "Solution_5": "what [i]what [/i] is?\r\nanyways, i think tonypr's solution for #1 was wrong...\r\nFor #1:\r\n[hide]one of the numbers must be a multiple of 5\nfirst choose the number that must be a multiple of 5. there are 20 such numbers\nfor 5, 95, and 100, there is 1 choice for the 2nd number.\nfor the 17 numbers remaining, there are 2 choices each.\nthus there are $ 3 \\plus{} (2*17) \\equal{} 37$ ways total[/hide]", "Solution_6": "[hide=\"#3\"]$ 6666(a\\plus{}b\\plus{}c\\plus{}d)\\equal{}193314\\iff a\\plus{}b\\plus{}c\\plus{}d\\equal{}29$\n\nMaximum possible sum of distinct digits is $ 6\\plus{}7\\plus{}8\\plus{}9\\equal{}30$, hence the only possibility to obtain $ 29$ is $ (5,7,8,9)$[/hide]", "Solution_7": "[quote=\"modularmarc101\"]Could you explain what that is?[/quote]\r\n\r\nHe was looking for the lowest multiple of $ 388$ that ends in 3 zeros so that when added to a number that works like $ 3880388$ it will still be a multiple of $ 388$ and end in $ 388$\r\n\r\nSo any number in the form of $ 3880388 + 97000k$ ends in $ 388$ and is a multiple of $ 388$.\r\n\r\nWe want to find $ 1000000 < 3880388 + 97000k < 10000000$ \r\n\r\nSince we are only dealing with the thousands we don't need to worry about the $ 388$:\r\n\r\n\\begin{eqnarray*} 1000 & < & 3880 + 97k < 100000 \\\\\r\n1000 & < & 97(40 + k) < 10000 \\\\\r\n10 & < & 40 + k < 103 \\\\\r\n- 30 & < & k < 63 \\end{eqnarray*}\r\n\r\nFrom this we see that there are $ \\boxed{92}$ solutions.\r\n\r\nThe lowest of which is $ 3880388 - 29\\cdot 97000 = 3880388 - 2813000 = 1067388$\r\nThe highest is $ 3880388 + 62\\cdot 97 = 3880388 + 6014000 = 9894388$\r\n\r\nTo Modularmarc101, I'll try, it's that I have a competition that same day but it's in the univerisity in mayaguez so it shouldn't be that hard to be there.\r\n\r\nTo abcak, the solution may look different but our solutions are in a way the same, just that I counted the cases for the options separately while you did them together.", "Solution_8": "nice one for #3, Farenhajt.\r\n\r\nIs it kinda like adding it long if you know what i mean?", "Solution_9": "[quote=\"abcak\"]nice one for #3, Farenhajt.\n\nIs it kinda like adding it long if you know what i mean?[/quote]\r\n\r\nWell, if Bob could do it, so can we ;)\r\n\r\n(I'm just imagining various \"Bobs\" from problems like this, who set up on long addition of $ 24$ four-digit numbers just for the... fun of it :P)" } { "Tag": [ "function", "IMC", "college contests" ], "Problem": "Find all continuous functions $ f: \\mathbb{R}\\to\\mathbb{R}$ such that $ f(x)\\minus{}f(y)$ is rational for all real numbers such that $ x\\minus{}y$ is rational.", "Solution_1": "[hide=\"Solution\"]$ f(x) \\equal{} qx \\plus{} c$, where $ q\\in \\mathbb{Q}, \\ c\\in \\mathbb{R}$[/hide]\n\n[hide=\"Idea\"]Intermediate Value Theorem and density of set $ \\left\\{\\frac {k}{2^{m}}, \\ k\\in \\mathbb{Z}, m \\in \\mathbb{N}\\right\\}$ in $ \\mathbb{R}$.[/hide]\n\n[hide=\"Proof\"]For each $ n\\in \\mathbb{N}$ define $ g_{n}(x) \\equal{} f\\left(x \\plus{} \\tfrac{1}{2^{n}}\\right) \\minus{} f(x)$. Then $ g_{n}$ is continous and $ \\forall x\\in \\mathbb{R} \\quad g_{n}(x) \\in \\mathbb{Q}$, so from Intermediate Value Theorem, and from density of irrational numbers in $ \\mathbb{R}$, we obtain that $ g_{n}$ must be constant function. Let $ g_{n}(x) \\equal{} s_{n}$ and $ S \\equal{} \\left\\{\\frac {k}{2^{m}}, \\ k \\in \\mathbb{Z}, m \\in \\mathbb{N}\\right\\}$. Since $ f\\left(\\tfrac{k}{2^{m}}\\right) \\equal{} ks_{m} \\plus{} f(0)$ and $ s_{n \\plus{} m} \\equal{} \\frac {s_{m}}{2^{n}}$, we have $ \\forall t\\in S \\quad f(t) \\equal{} ts_{0} \\plus{} f(0)$. Now density of $ S$ in $ \\mathbb{R}$ implies that equality holds also for all $ t \\in \\mathbb{R}$, because $ f$ is continous, so the only solutions are $ f(x) \\equal{} qx \\plus{} c$, where $ q\\in \\mathbb{Q}, \\ c\\in \\mathbb{R}$. [/hide]" } { "Tag": [], "Problem": "What is the smallest possible real value of $ x^2 \\plus{} 8x$?", "Solution_1": "hello, the smallest possible real value is $ \\minus{}16$, it will be attained for $ x\\equal{}\\minus{}4$,because\r\n$ x^2\\plus{}8x\\equal{}(x\\minus{}4)^2\\minus{}16$.\r\nSonnhard.", "Solution_2": "[hide=\"Alternately\"]The minimum of the expression $ ax^2 \\plus{} bx \\plus{} c$, where $ a$, $ b$, and $ c$ are real numbers and $ a\\ne 0$, occurs when $ x \\equal{} \\minus{} \\frac {b}{2a}$. So the minimum of this expression is when $ x \\equal{} \\minus{} \\frac {8}{2(1)} \\equal{} \\minus{} 4$. Thus, the minimum value of $ x^2 \\plus{} 8x$ is $ ( \\minus{} 4)^2 \\plus{} 8( \\minus{} 4) \\equal{} 16 \\minus{} 32 \\equal{} \\boxed{ \\minus{} 16}$.[/hide]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "The order of the simple graph G is P. K is the max number such that G has K nonadjacent nodes (pairwise). B is the min number such that the nodes of G can be divded in B group(s) such that all nodes in each group is adjacent (pairwise). can we assert that K=B? prove your assertion. ( we can prove that K<= B )", "Solution_1": "It really does seem like the result should be true, though, doesn't it? But as the pentagon shows, we can have $ \\omega(G) \\cdot \\alpha(G) < |G|$. One fix is $ \\chi(G) \\cdot \\alpha(G) \\geq |G|$, and this fact is obvious. The last time I thought about this, I think that I decided that graph capacities could be a lot easier to understand if $ \\omega \\cdot \\alpha \\geq |G|$. Alas.", "Solution_2": "[quote=\"madadpoor\"]The order of the simple graph G is P. K is the max number such that G has K nonadjacent nodes (pairwise). B is the min number such that the nodes of G can be divded in B group(s) such that all nodes in each group is adjacent (pairwise). can we assert that K=B? prove your assertion. ( we can prove that K<= B )[/quote]\r\n\r\n[hide=\"here is my solution\"]The result is false and there is a simple counterexample with $ P\\equal{}5$. Let $ G$ have the $ 5$ vertices $ V_{1}$, $ V_{2}$, $ V_{3}$, $ V_{4}$, and $ V_{5}$. Let $ G$ have exactly $ 5$ edges: $ V_{1}V_{2}$, $ V_{2}V_{3}$, $ V_{3}V_{4}$, $ V_{4}V_{5}$, and $ V_{5}V_{1}$. It is fairly easy to see that for this graph $ G$ we have $ K\\equal{}2$ but $ B\\equal{}3$.[/hide]" } { "Tag": [ "algebra", "polynomial", "Vieta", "inequalities" ], "Problem": "The polynomial $p(x) = x^3 + ax^2 + bx + c$ has three real positive roots, whereas at least two of them are distinct. Prove that we have: \r\n\r\n\\[a^2 + b^2 + 18c > 0.\\]", "Solution_1": "$P(x)=(x-p)(x-q)(x-r)$ with $p\\not=q$, $p,q,r \\in \\mathbb{R}^+_0$\r\n\r\nc=-pqr\r\nb=pq+qr+rp\r\na=-p-q-r\r\n\r\nSo we need to prove $(p+q+r)^2 + (pq+qr+rp)^2 - 18pqr > 0$, which follows directly from AM-GM. (strict inequality since at least 2 are different)" } { "Tag": [], "Problem": "Like for example 5 to the power of 2 and 2 5ths.\r\nAnd how do I do, for example, 5 to the power of .65. How do I do those?\r\nThanks.", "Solution_1": "In the case of the first problem, you have $ 5^\\frac{12}{5}\\equal{}\\sqrt[5]{5^{12}}\\equal{}25\\sqrt[5]{25}$.\r\nExpressions like $ a^\\frac{b}{c}$ can be written as $ \\sqrt[c]{a^b}$.\r\n\r\nIn the second problem you could convert $ .65$ to a fraction ($ .65\\equal{}\\frac{65}{100}\\equal{}\\frac{13}{20}$) so that you have $ 5^\\frac{13}{20}\\equal{}\\sqrt[20]{5^{13}}$ which can't be simplified any further." } { "Tag": [], "Problem": "Three boys agree to divide a bag of marbles in the following manner. The first boy takes one more than half the marbles. The second takes a third of the number remaining. The third boy finds that he is left with twice as many marbles as the second boy. The original number of marbles:\r\n\r\n$ \\textbf{(A)}\\ \\text{is none of the following} \\qquad\r\n\\textbf{(B)}\\ \\text{cannot be determined from the given data}\\\\\r\n\\textbf{(C)}\\ \\text{is 20 or 26} \\qquad\r\n\\textbf{(D)}\\ \\text{is 14 or 32} \\qquad\r\n\\textbf{(E)}\\ \\text{is 8 or 38}$", "Solution_1": "Its $ D$ just by applying the numbers to the Problem.", "Solution_2": "[quote=\"Nandubetax\"]Its $ D$ just by applying the numbers to the Problem.[/quote]Why couldn't the original number of marbles be $ 8$?\r\nThe first boy would take $ 4 \\plus{} 1 \\equal{} 5$, leaving $ 3$ marbles, $ 1$ for the second boy and $ 2$ for the third one.\r\n\r\nWhy couldn't the original number of marbles be $ 20$?\r\nThe first boy would take $ 10 \\plus{} 1 \\equal{} 11$, leaving $ 9$ marbles, $ 3$ for the second boy and $ 6$ for the third one.\r\n\r\nCould it be that there were $ 68$ marbles? Or $ 602$?\r\n\r\nI'm sure [hide=\"that\"] the number remaining after the first boy took his share was a multiple of $ 3$.[/hide]\n\n[hide=\"and that \"]the original number was even.[/hide]", "Solution_3": "[quote=\"KMST\"][quote=\"Nandubetax\"]Its $ D$ just by applying the numbers to the Problem.[/quote]Why couldn't the original number of marbles be $ 8$?\nThe first boy would take $ 4 \\plus{} 1 \\equal{} 5$, leaving $ 3$ marbles, $ 1$ for the second boy and $ 2$ for the third one.\n\nWhy couldn't the original number of marbles be $ 20$?\nThe first boy would take $ 10 \\plus{} 1 \\equal{} 11$, leaving $ 9$ marbles, $ 3$ for the second boy and $ 6$ for the third one.\n\nCould it be that there were $ 68$ marbles? Or $ 602$?\n\nI'm sure [hide=\"that\"] the number remaining after the first boy took his share was a multiple of $ 3$.[/hide]\n\n[hide=\"and that \"]the original number was even.[/hide][/quote]This raises a good question: can we actually determine the exact number of marbles?\n[hide=\"Solution\"]Let $ M$ be the original number of marbles. Then, the first boy gets $ \\frac{M}2\\plus{}1\\equal{}\\frac{M\\plus{}2}2$ marbles and there are $ \\frac{M}2\\minus{}1\\equal{}\\frac{M\\minus{}2}2$ marbles left for the two other boys. The second boy gets $ \\frac13\\left(\\frac{M\\minus{}2}2\\right)\\equal{}\\frac{M\\minus{}2}6$ and the third boy gets the remaining marbles, $ \\frac23\\left(\\frac{M\\minus{}2}2\\right)$. So, the expression $ \\frac{M\\minus{}2}6$ must be an integer and thus $ M$ is of the form $ 6k\\plus{}2$. There are no more additional information from the problem that would help you determine how many marbles there were in the bag. Thus, the answer is $ \\bf (B)$.\n\nNote that $ \\bf (C), (D), (E)$ are all numbers of the form $ 6k\\plus{}2$.[/hide]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Hi,\r\n I want to draw a consecutive polygon of similar side inside a given polygon (of any number of sides) at a specified distance (varies).\r\nWhen the space inside the polygon is not enough to draw another polygon, I should stop drawing polygon.\r\n\r\nCan anyone tell me how to detect whether the a polygon at specified distance can be drawn inside the polygon or not.\r\n\r\nRefer attached figure for more idea.", "Solution_1": "you didn't say that your polygon must be regular. but if your polygon isn't regular, drawing lines parallel to its sides all at the same distance doesn't necessarily produce a polygon similar to it. so we'll assume your polygon is regular. then: \r\n\r\nfind the length of the apothem of the regular polygon and divide by your given distance, then take the floor - that's how many polygons you can have before you have to stop.", "Solution_2": "[quote=\"pleurestique\"]you didn't say that your polygon must be regular. but if your polygon isn't regular, drawing lines parallel to its sides all at the same distance doesn't necessarily produce a polygon similar to it. so we'll assume your polygon is regular. then: \n\nfind the length of the apothem of the regular polygon and divide by your given distance, then take the floor - that's how many polygons you can have before you have to stop.[/quote]\r\n\r\nSorry for not providing sufficient information\r\n\r\nMy Polygon may be either regular or irregular. Refer Fig Sample 1: In that polygon drawn in yellow should not be drawn. (it is different from the given shape,in that case i should stop draw polygon)\r\n\r\nRefer Sample 2: I am drawing parallel linse for the given polygon and find the intersection point and draw polygon with all the intersection point as vertex of polygon.\r\n\r\nKindly do help to solve the problem", "Solution_3": "if you want help then you'd better make sure you've expressed your problem clearly." } { "Tag": [ "calculus", "integration", "inequalities", "calculus computations" ], "Problem": "Let $ a,\\ b$ be constant numers such that $ a>b>0$. Find the volume of the set $ R$ in Eucledian space which satisfies the following system of inequalities:\r\n\r\n$ x\\geq 0,\\ y\\geq 0,\\ z\\geq 0,\\ max(x,\\ y,\\ z)\\leq a,\\ x\\plus{}y\\plus{}z\\minus{}min(x,\\ y,\\ z)\\leq a\\plus{}b$.", "Solution_1": "Can anyone solve the problem?" } { "Tag": [ "MATHCOUNTS" ], "Problem": "How can I get any previous Mathcounts competitions besides the 2003-2004 competitions?", "Solution_1": "try this- http://goodlymaths.0catch.com/#", "Solution_2": "You have to buy them from Mathcounts." } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "The characteristic function to an interval $ [a,b] \\subseteq [\\minus{} \\pi, \\pi]$ is an even or an odd function?", "Solution_1": "If the subset (sub-interval) which you choose $ [a,b]$ from $ [\\minus{}\\pi ,\\pi]$ is symmetric about $ 0$ then the the indicator function is an even function . :roll:" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $ a,b,c$ are positive real numbers and $ X,Y,Z \\in M_n{\\mathbb{(C)}}$ are Hermitian matrices and positive define such that :\r\n$ X^nY^n \\equal{} Y^nX^n$,\r\n$ X \\plus{} Y \\plus{} Z \\equal{} (a \\plus{} b \\plus{} c)I_n$,\r\n$ a^2X \\plus{} b^2Y \\plus{} c^2Z \\plus{} (abc)I_n \\equal{} 4XYZ$\r\nProve that : $ X \\equal{} \\frac {b \\plus{} c}{2}I_n$,$ Y \\equal{} \\frac {c \\plus{} a}{2}I_n$,$ Z \\equal{} \\frac {a \\plus{} b}{2}I_n$\r\n $ I_n$ is $ n$x$ n$ identity matrix . $ n \\in \\mathbb{N^{*}}$", "Solution_1": "[quote=\"QuyBac\"]$ X^nY^n \\equal{} Y^nX^n$,\n[/quote]Since $ X,Y>0$, the condition implies that $ X,Y$ commute.\n[quote]$ X \\plus{} Y \\plus{} Z \\equal{} (a \\plus{} b \\plus{} c)I_n$,[/quote]... such that $ Z$ commutes with $ X,Y$. So assume that everything is diagonal and solve the exercise simply over the reals." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "A={0<=a_{1}<=a_{2}<=...<=a_{k}<=k and a_1+...+a_k is divided by k.} and B={0<=a_{1}<=a_{2}<=...<=a_{k-1}<=k=a_{k} and a_1+...+a_k is divided by k.}\r\nProve that: |A|=2|B|", "Solution_1": "I don't understand the definitions of the sets :( . First I thought a_1,...,a_k are the elements of A but what are the elements of B then. Please explain it.", "Solution_2": "$A$ and $B$ are sets of sequences with those properties.", "Solution_3": "All right. Sorry for the idiot question but I'm not in best form today.\r\n\r\nMisha", "Solution_4": "Nice and easy:\r\nTake $k$ odd. By considering the substitutions $b_{i}=a_{i}+i$, since $k|1+2+...+k$ for odd $k$, it's clear there is a bijection between $A$ and\r\n$A'=\\left\\{ 1\\leq b_{1} < ... < b_{k} \\leq 2k : k | b_{1}+...+b_{k} \\right\\}$ and a bijection between $B$ and $B'=\\left\\{ 1\\leq b_{1} < ... < b_{k} = 2k : k | b_{1}+...+b_{k} \\right\\}$, so $|A|=|A'|$ and $|B|=|B'|$.\r\nIf $F$ is the family of all subsets of $[2k]$ with $k$ elements whose sum of elements is divisible with $k$, then $M \\in F \\Rightarrow [2k]-M \\in F$ so the number of $M \\in F$ s.t. $2k \\in M$ is equal to the number of $M \\in F$ s.t. $2k \\notin M$. In other words, $|B'|=|A'-B'| \\Leftrightarrow |A'|=2|B'|$ so $|A|=2|B|$. For even $k$ it's the same thing." } { "Tag": [ "factorial", "inequalities" ], "Problem": "Let k be a positive integer such that\r\n\r\n$((k!)!)!\\leq{((2006k^{2})!)!}$\r\n\r\nFind, with proof, the number of possible values of k.", "Solution_1": "Couldnt you just ignore the last two factorials, and then solve\r\n\r\n\r\n$k!\\leq{2006k^2}$ ?\r\n\r\n\r\n$(k-1)!\\leq{2006k}$\r\n\r\nI'm not even sure if this last line does anything.", "Solution_2": "Yeah, your on the right track. Now just explain how you know that it is valid to ignore the outside factorials.", "Solution_3": "So basically we just have to show that $a! \\leq b!$ if and only if $a \\leq b$\r\n\r\nso...\r\n\r\nSince $a! = a(a-1)!$ and we're given that $a \\geq 1$, we know that\r\n\r\n$a! \\geq (a-1)!$, so, (I don't know how to put this into a more rigorous proof), the result follows.\r\n\r\nDoes that cover everything or are there holes?", "Solution_4": "well your argument can constitute part of the \"if\" portion of the proof (where you assume $a\\leq{b}$ and prove the other assertion). You should use the same idea, but start by showing that $(a+1)!\\geq{a!}$. Then, say something about \"similarly\" and write $(a+2)!\\geq{(a+1)!}$. Then use this pattern to inductively write a string of inequalities that relate b to a $b!\\geq{(b-1)!}\\geq{\\ldots}\\geq{(a+1)!}\\geq{a!}$\r\n\r\nThink about the \"only if\" portion of the proof (where we assume $a!\\leq{b!}$ and try to prove the other assertion)" } { "Tag": [ "MATHCOUNTS", "AMC", "AIME", "linear algebra" ], "Problem": "Express the following infinetly repeating hexadecimal (base 16) into a decimal fraction: 0.1999DECDECDEC......\r\n\r\nI think that there is a special way to do this, but I don't know it. \r\nIs this too hard or too easy for Mathcounts?", "Solution_1": "There is a reasonable way to do this... but I'm leaving it for an MC student.", "Solution_2": "What grade are you in? Not 6, 7, or 8th is all I know.", "Solution_3": "[quote=\"236factorial\"]Express the following infinetly repeating hexadecimal (base 16) into a decimal fraction: 0.1999DECDECDEC......\n\nI think that there is a special way to do this, but I don't know it. \nIs this too hard or too easy for Mathcounts?[/quote]\r\n\r\nSimplify it into base 10... and you get 0.2000453453453.... the rest is self explanatory, I think...\r\n\r\n[hide] 0.2=1/5.... subtract 1/5 and you get 0.0000453453453...=$\\frac{151}{3330000}$, add 1/5 to that and you get: $\\frac{666151}{3330000}$...[/hide]\r\n\r\nEdit: woops, I messed up...", "Solution_4": "I think simplifying repeating decimals into base 10 is too hard for Mathcounts problems.", "Solution_5": "[quote=\"236factorial\"]Is this too hard or too easy for Mathcounts?[/quote]\r\nToo hard. This is also a messy, ugly problem that doesn't come out nicely. [What is the source of this problem?]\r\n\r\nHowever, the concepts are within the scope of MATHCOUNTS.\r\n\r\nLet's split this repeating hexadecimal into two parts, then follow the techniques you National competitors (should) already know for simplifying repeating decimals:\r\n\r\n$0.1999\\hbox{DECDEC}\\ldots_{16}=0.1999999\\ldots_{16}+0.0000453453\\ldots_{16}$\r\n\r\nNow, $0.1\\overline{9}_{16}=\\frac{1_{16}}{10_{16}}+\\frac{9_{16}}{\\hbox{F0}_{16}}=\\frac{24_{10}}{240_{10}}=\\frac{1}{10}$\r\n\r\n$0.0000\\overline{\\hbox{453}}_{16}=\\frac{\\hbox{453}_{16}}{\\hbox{FFF0000}_{16}}$\r\n\r\nConvert to decimal. (I did the first one for you.) Then add the fractions.", "Solution_6": "This problem comes from an ABACUS Math International Challenge for grades 5 and 6.\r\nI don't think that many 5th and 6th graders can understand base changes very well.", "Solution_7": "5th and 6th graders? That seems unlikely, despite being international, unless this was the hardest problem.", "Solution_8": "I feel sorry for those poor 5th and 6th graders... that's cruel and unusual punishment! *sigh* forced to solve harder-than-mathcounts math problems at such a young age *sigh* :D", "Solution_9": "You should see what the kids in Romania do ...", "Solution_10": "Yeah, their problems are harder than many AIME problems, from what Valentin Vornicu has posted.", "Solution_11": "Oh no - we are falling behind!", "Solution_12": "You just noticed?\r\n\r\nAs an additional note, you should see what the kids in Russia do!", "Solution_13": "There's a reason the Eastern European countries started the IMO! Their teams always make many of the top 5.", "Solution_14": "[quote=\"tarquin\"]You should see what the kids in Romania do ...[/quote]\r\nWould you believe a 6th grader from Romania is credited on the ABACUS Web Site as the author of this problem?", "Solution_15": "Pssh. In Korea, I'm only a little advanced. And I'm hopefully taking multivariable calc next year.", "Solution_16": "And how old are you? I know a guy who took Statistics in 8th grade, and now, in 11th grade, in our high school that has a highly gifted magnet which offers up through Discrete Math and Linear Algebra, he doesn't have any math to do.", "Solution_17": "I skipped a grade and I'm still 2 years ahead in math. Next year I might skip pre-cal (which I kinda did over the summer last year) and go to calc.", "Solution_18": "I'm in eighth grade and taking Math Analysis (precal)." } { "Tag": [ "number theory", "relatively prime", "number theory proposed" ], "Problem": "Prove that if $N$ is relatively prime with 10 then $N^{1}01$ ends with three digits, which are also the last three digits of N.\r\n\r\ndavron l.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=118937", "Solution_2": "It mean N is odd and $5\\not |N$, therefore $N^{4}=1(mod 16)$ and $N^{4}=1(mod 5)$. It give $N^{100}=1(mod 16)$ and $N^{100}=1(mod 5^{3})$. It mean $N^{100}=1(mod 2000)$, or $N^{101}=N(mod 2000)$." } { "Tag": [ "ratio", "geometry", "combinatorial geometry", "geometry proposed" ], "Problem": "Consider five points in the plane, no three collinear. The convex hull of this points has area $S$. Prove that there exist three points of them that form a triangle with area at most $\\frac{5-\\sqrt 5}{10}S$", "Solution_1": "can anybody post a solution for this one", "Solution_2": "This one was proposed by Zhen Wang and Ji Chen, see: [url=http://menet.math.ecnu.edu.cn/teaching/]Mathematics Teaching (Shanghai)[/url], 1988, No.2, p.40, Problem 155. The solution see 1988, No.3, p.41.\r\n\r\nIn March, 1989, Paul Erdos visited Colorado Springs. He offered $\\$10$ for the solution during his talk at the University of Colorado, and as a coauthor, Alexander Soifer immediately offered to match it. See:\r\n\r\nAlexander Soifer, [url=http://www.mathpropress.com/books/problemBooks/Triangle/]How Does One Cut a Triangle?[/url], Center for Excellence in Mathematics, Colorado Springs: 1990, p. 113." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello,\r\nTroubling over a series of rather simple problems.\r\n\r\n(i) Example of a group $G$ with infinite order such that every proper subgroup is of infinite index.\r\n\r\n(ii) Example of a group $G$ with infinite order such that every nontrvial subgroup is of finite index.\r\n\r\n(iii) Example of a group $G$ such that there exist nontrival subgroups of infinite index and proper subgroups of finite index.\r\n\r\nMy thoughts on the above are as follows:\r\n\r\n(i) $\\mathbb{Q}$ (Unsure as if this is correct as since $|\\mathbb Q| = |\\mathbb Z|$)\r\n\r\n(ii) $\\mathbb{Z}$ (I believe this is true since every group $2\\mathbb{Z}, 3\\mathbb{Z}, \\cdots$ is isomorphic to $\\mathbb Z_2,\\mathbb Z_3, \\cdots,$ respectively.)\r\n\r\n(iii) $\\mathbb{Z}_k\\times\\mathbb{Z}, k\\in\\mathbb N\\backslash\\left\\{0\\right\\}$ (Consider the subgroup $(0,\\mathbb Z)$ is infinite and $(\\mathbb Z_k, 0)$ is finite.) \r\n\r\nAm I on the right track? If not, could you please advise me on my poor logic!\r\n\r\nThanks,\r\nJNG", "Solution_1": "nZ is not isomorphic to Zn, but Z/nZ is isomorphic to Zn.\r\n(f: Z onto Zn defined by f(a) = \u00e2 is a surjective homomorphism of groups with kernel nZ)", "Solution_2": "Sorry, that is quite true. Poor notation on my part. \r\n\r\n-JNG", "Solution_3": "Consider $\\mathbb{Q} / \\mathbb{Z}$." } { "Tag": [ "geometry", "trigonometry" ], "Problem": "What is the basic difference between geometry and trigonometry at the high school level?\r\n\r\nThanks!", "Solution_1": "[quote=\"Interval\"]What is the basic difference between geometry and trigonometry at the high school level?\n\nThanks![/quote]\r\nWell, geometry includes triangles, circles and stuff. But mostly triangles. Trigonometry is the study of angles. You can learn more about trig from this http://mathworld.wolfram.com/Trigonometry.html", "Solution_2": "[quote=\"Interval\"]What is the basic difference between geometry and trigonometry at the high school level?\n\nThanks![/quote]\r\n\r\ngeometry is dealing with lines and circles and basic proofs. Geometry also includes [size=75]a little[/size] bit of trig like 30-60-90 triangles and such. you get into a lot of depth in trig in high school", "Solution_3": "I read that geometry means TO MEASURE THE EARTH.\r\n\r\nIs this true?\r\n\r\nAlso, trig is basically the study of angles and triangle sides in general, right?" } { "Tag": [ "modular arithmetic", "number theory" ], "Problem": "Prove that nineth power of a number is of the form 19m or 19m+-1 where m is a non negative integer", "Solution_1": "[hide]Clearly, any numbers of the form $ (19k)^9$ (which is $ 0 \\pmod {19}$) is equivalent to $ 0 \\pmod {19}$, so then we look at the factorings of $ x^9 \\minus{} 1$ and $ x^9 \\plus{} 1$:\n\n$ x^9 \\minus{} 1 \\equal{} (x \\minus{} 1)(x^2 \\plus{} x \\plus{} 1)(x^6 \\plus{} x^3 \\plus{} 1)$\n$ x^9 \\plus{} 1 \\equal{} (x \\plus{} 1)(x^2 \\minus{} x \\plus{} 1)(x^6 \\minus{} x^3 \\plus{} 1)$\n\nSo clearly, any number of the form $ 19k \\pm 1$ (which is $ 1$ or $ 18 \\pmod{19}$) follows this rule because then the $ (x \\pm 1)$ will be divisible by 19.\n\nAlso, anything in the form $ 19k\\pm \\{7,8,11,12\\}$ (which are $ 7,8,11,12 \\pmod{19}$) will follow this rule because then the $ x^2\\pm x \\plus{} 1$ term will be divisible by 19.\n\nAnd finally, those numbers for which $ x^3\\equiv 7,8,11,12 \\pmod{19}$ also follow the rule because the last factor is just equal to the second factor with $ x^3$ instead of $ x$. Those are $ \\{2,3,4,5,6,9,10,13,14,15,16,17\\}$, which complete all the remainders mod 19.\n\nSo in conclusion, the unavailability of Fermat's Little Theorem compels us to write a convoluted solution in place of what would've been a short one.[/hide]", "Solution_2": "Using Fermat's Little:\r\n\r\n\r\n[hide]If the number is a multiple of 19, then its nineth power is a multiple of 19. Otherwise, by Fermat's Little, we have x^18-1 = (x^9-1)(x^9+1) = 0, which implies that x^9 is plus or minus 1.[/hide]", "Solution_3": "Or we could brutally calculate all values of 1,2,3...,19 all to the power of 9 and then find the residue...\r\nIt wouldn't take really long" } { "Tag": [ "geometry", "rectangle", "probability" ], "Problem": "$Question 1$\r\nThere is a $100 \\times 10$ rectangular grid. All the $1000$ points are coloured either red, green or blue. \r\n\r\nProve that there exists a rectangle in the plane such that all four of its vert the same colour. Or, prove that it is not a necessary condition\r\n\r\n$Question 2$\r\nA frog performs a random walk on the integers. At each stage it jumps either one integer forwards or one integer back with equal probability. It starts its random walk from zero. Find the probability that it reaches $+10$ before it reaches $-20$.", "Solution_1": "N.2:\r\nsee here http://www.usamts.org/Forms/ExpectedValue.pdf\r\n\r\nactually i'm not sure that helps\r\n\r\ni'm not sure this is correct solution, here is my idea:\r\np(-20)=p(-10)^2 (easily seen)\r\nSo we find prob it reaches p(-10) resets, then reaches it again before p(10)\r\nwe can reduce problem to prob. it reaches -2 before 1\r\nP=1\\2 (1/2+1/2P) P=1/3" } { "Tag": [ "\\/closed" ], "Problem": "On the MathLinks skin, I noticed that the Resources button brings you to the same location as clicking on Links. (I was expecting the button to send me to the AoPS Resources page.) Is that the way it is supposed to be?", "Solution_1": "[quote=\"Ravi B\"]On the MathLinks skin, I noticed that the Resources button brings you to the same location as clicking on Links. (I was expecting the button to send me to the AoPS Resources page.) Is that the way it is supposed to be?[/quote]I didn't have time to complete everything on the skin, but that's where it's supposed to lead. The MathLinks skin is desgined to be more helpfull to one advanced user looking for advanced problems, so the links are adaptive.", "Solution_2": "Thanks." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ x,y,z$ be non-negative real numbers satisfy $ xy \\plus{} yz \\plus{} zx \\equal{} 1$.Prove that\r\n$ \\frac {1}{{1 \\plus{} yz \\plus{} {x^2}}} \\plus{} \\frac {1}{{1 \\plus{} zx \\plus{} {y^2}}} \\plus{} \\frac {1}{{1 \\plus{} xy \\plus{} {z^2}}} \\le \\frac {9}{5}$", "Solution_1": "\\[ \\Longleftrightarrow \\sum{\\left(\\frac {3}{5} \\minus{} \\frac {1}{1 \\plus{} yz \\plus{} x^2}\\right)}\\geq 0\\]\r\n\r\n\\[ \\Longleftrightarrow\\sum{\\frac {(x \\minus{} y)(x \\minus{} z) \\plus{} x(x \\minus{} y) \\plus{} x(x \\minus{} z)}{1 \\plus{} yz \\plus{} x^2}}\\geq 0\\]\r\n\r\n\\[ \\Longleftrightarrow\\sum{\\frac {1}{1 \\plus{} yz \\plus{} x^2}(x \\minus{} y)(x \\minus{} z)} \\plus{} \\sum{(x \\minus{} y)\\left(\\frac {x}{1 \\plus{} yz \\plus{} x^2} \\minus{} \\frac {y}{1 \\plus{} zx \\plus{} y^2}\\right)}\\geq 0\\]\r\n\r\n\\[ \\Longleftrightarrow\\sum{\\frac {1}{1 \\plus{} yz \\plus{} x^2}(x \\minus{} y)(x \\minus{} z)} \\plus{} \\sum{\\frac {2z(x \\plus{} y)}{(1 \\plus{} yz \\plus{} x^2)(1 \\plus{} zx \\plus{} y^2)}(x \\minus{} y)^2}\\geq 0\\]\r\nBut\r\n\\[ \\sum{\\frac {1}{1 \\plus{} yz \\plus{} x^2}(x \\minus{} y)(x \\minus{} z)}\\geq 0\\]\r\nit's true by $ V \\minus{} V Schur$,and the second part is obvious positive.\r\nI hope there isn't any mistakes.....\r\n\r\nI am sorry I made some typo,now I correct them. :blush:", "Solution_2": "[quote=\"Tourish\"]\n\\[ \\Longleftrightarrow \\sum{\\left(\\frac {3}{5} \\minus{} \\frac {1}{1 \\plus{} yz \\plus{} x^2}\\right)}\\geq 0\\]\n\n\\[ \\Longleftrightarrow\\sum{\\frac {(x \\minus{} y)(x \\minus{} z) \\plus{} x(x \\minus{} y) \\plus{} x(x \\minus{} z)}{1 \\plus{} yz \\plus{} x^2}}\\geq 0\\]\n\n\\[ \\Longleftrightarrow\\sum{\\frac {1}{1 \\plus{} yz \\plus{} x^2}(x \\minus{} y)(y \\minus{} z)} \\plus{} \\sum{(x \\minus{} y)\\left(\\frac {x}{1 \\plus{} yz \\plus{} x^2} \\minus{} \\frac {y}{1 \\plus{} zx \\plus{} y^2}\\right)}\\geq 0\\]\n\n\\[ \\Longleftrightarrow\\sum{\\frac {1}{1 \\plus{} yz \\plus{} x^2}(x \\minus{} y)(y \\minus{} z)} \\plus{} \\sum{\\frac {2z(x \\plus{} y)}{(1 \\plus{} yz \\plus{} x^2)(1 \\plus{} zx \\plus{} y^2)}(x \\minus{} y)^2}\\geq 0\\]\nBut\n\\[ \\sum{\\frac {1}{1 \\plus{} yz \\plus{} x^2}(x \\minus{} y)(y \\minus{} z)}\\geq 0\\]\nit's true by $ V \\minus{} V Schur$,and the second part is obvious positive.\nI hope there isn't any mistakes.....[/quote]\r\nYour proof is very nice,Tourish. I think this is very natural way to prove it.\r\n[hide=\"P/s\"]I will wait for a proof by Cauchy-Schwarz inequality.[/hide]", "Solution_3": "[quote=\"Tourish\"]\n\\[ \\Longleftrightarrow\\sum{\\frac {(x \\minus{} y)(x \\minus{} z) \\plus{} x(x \\minus{} y) \\plus{} x(x \\minus{} z)}{1 \\plus{} yz \\plus{} x^2}}\\geq 0\\]\n[/quote]\r\n\r\nnice step!", "Solution_4": "where did you use the caushy shwarz ineq.?", "Solution_5": "Can someone give me another proof by Cauchy-Schwarz?\r\nAnd the double inequality which I've posted in this link:\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=317946" } { "Tag": [], "Problem": "Hi please could someone help me into solving this equations , please I will be very happy to know the process of solving ,, :( :( :( \r\n\r\nThank you very much!!!!\r\n\r\n\r\n\r\n6x + 7 - x - 11 = 0\r\n 2 2\r\n\r\nand the second one:\r\n\r\nx + 3 = 1 + x + 1\r\n x -2 2(x-3)", "Solution_1": "For the first one you add all like terms. So 6x+x=7x. And 7-11=-4.So we have 7x-4=0.Add the four on the other side so we have 7x=4. Then divide both sides by seven to isolate the variable and x=4/7.\r\n\r\nThe second one we do the same thing. So x+3=x+2. Subtract the x from both sides and we get 3=2, which can't be correct so no solutions or empty set. I may have done this wrong though.\r\nWhats about the x-2 2(x-3)?", "Solution_2": "Hmm, I believe the two questions are two system of equations. Perhaps you could fix any typos?", "Solution_3": "Thank you very much my friends,,, :blush: Iam sorry but I cant find a way to copy paste from m.word please find the attached pdf file with the problematic equations on this link http://www.mathlinks.ro/Forum/viewtopic.php?t=121429 \r\n\r\n\r\nthank you very much!!!!", "Solution_4": "I'm gonna go out on a limb and assume that your last expression \"x -2 2(x-3)\" was supposed to be the equation $x-2=2(x-3)$ even if my assumption is incorrect, this will be a sample solution to a two-step algebraic equation from me to you. To solve for $x$ you would first distribute the $2$ on the right side of the equation which would give $x-2=2x-3*2$ then isolate your variable (x in this case) to one side of the equation and have your constant(s) on the other side. $x-2-x=2x-6-x \\rightarrow-2=x-6 \\rightarrow x-6+6=-2+6 \\rightarrow x=4$" } { "Tag": [ "email", "search", "articles", "MATHCOUNTS" ], "Problem": "[url=http://mathworld.wolfram.com]MathWorld[/url] has recieved an email from [url=http://www.mersenne.org]GIMPS[/url] (Great Internet Mersenne Prime Search) reporting that a 42nd Mersenne Prime has been found! Click [url=http://mathworld.wolfram.com/news/2005-02-18/mersenne/]here[/url] for the article and more info on Mersenne Primes.\r\n\r\n(Note: All 3 links will take you outside of AoPS)", "Solution_1": "Cool! Thanks for sharing. By the way, it was great to meet you and Sly Si in person at the MATHCOUNTS chapter meet.", "Solution_2": "An update:\r\n\r\nThe GIMPS website has now officially reported that they have found a potential new mersenne prime. They are in the process of verifying that it is indeed prime, and they will hopefully have announced the actual number and the name of the person that found it by the weekend.\r\n\r\nA few notes about the candidate (assuming it is indeed prime): it will be the largest known prime number, but it is not long enough to win the US$100,000 prize for finding the first 10,000,000-digit prime number. Also, this will be the 8th mersenne prime found by GIMPS, all of which were the largest known primes in the world upon their discovery.", "Solution_3": "Another update:\r\n\r\nGIMPS has announced that its new mersenne prime has been independently verified to be prime; the number is [size=150]2^25,964,951-1[/size].", "Solution_4": "yeah... I was at the Rice Math contest and the guest lecutrer mentioned that 42nd mersenne prime was found on 18th. did they say how many digits?? How many more until they can give out the 100grand for the 100 billion or 100 million digit mersenne prime?", "Solution_5": "Seems I have to put my GIMPS program back in the background, just in case :D :D", "Solution_6": "and now there seems to be another one \r\n\r\non december 16 (or 15?), 2005, a new mersenne prime was found.\r\n\r\nit's still being verified\r\n\r\nsee http://www.mersenne.org", "Solution_7": "Nowadays, which is the bigger prime found????" } { "Tag": [ "Duke", "college" ], "Problem": "Hey guys, I plan on double-majoring in chemistry and economics. Which colleges would you suggest I apply to, not considering the Ivy leagues? From what I've heard, Duke, U of Chicago, Northwestern are good choices? Any others?", "Solution_1": "U Chicago is definitely a good choice. I go to Hopkins and I've heard that the econ department is one of the best in the nation, as well as the chemistry department.", "Solution_2": "I am wondering why don't you want to apply to Ivy League? I have heard they are very good :) :maybe:", "Solution_3": "The Ivy League is nothing more than an athletic conference. Yes, the schools are quite good but there are plenty of non-Ivy schools that are just as good if not better.", "Solution_4": "yeah i know that... im wondering with which sport a person will be accepted easily in Ivy league universities? soccer, basketball, tennis,swimming,etc", "Solution_5": "In most cases, you have to be recruited by the coach if you want to actually get on a top team. Some teams are walk-ons, others are not. And if you are not being recruited, it doesn't really matter which sport you play-rather what you get out of it. And no one gets accepted easily." } { "Tag": [ "algebra", "polynomial", "limit", "algebra unsolved" ], "Problem": "Let $1,\\dots,999$ be the roots of $P(x)$, $deg P(x)=999$. Does there exist two polynomials $S(x), \\deg S(x)=37$, $Q(x), \\deg Q(x)=27$ such that $P(x)=Q(S(x))$?", "Solution_1": "Let $1<\\alpha_{1}<\\cdots<\\alpha_{r}<999$ be the real roots of $S'$ between $1$ and $999$ ($r\\leq \\deg S'=36$), such that $S$ is strictly increasing or decreasing on each of the intervals $[1,\\alpha_{1}],(\\alpha_{1},\\alpha_{2}],\\ldots,(\\alpha_{r-1},\\alpha_{r}],(\\alpha_{r},999]$. This gives $r+1\\leq 37$ strictly increasing or decreasing sequences (possibly empty):\r\n\r\n$S(1),S(2),\\ldots,S([\\alpha_{1}])$\r\n$S([\\alpha_{1}]+1),S([\\alpha_{1}]+2),\\ldots,S([\\alpha_{2}])$\r\n$\\vdots$\r\n$S([\\alpha_{r}]+1),S([\\alpha_{r}]+2),\\ldots,S(999)$\r\n\r\nNow $Q(S(k))=P(k)=0$ for all $k=1,\\ldots,999$, and there are at most $\\deg Q=27$ different roots of $Q$. Therefore $S$ can take at most $27$ different values on $\\{1,2,\\ldots,999\\}$. And so each of the $r+1\\leq 37$ sequences above must have length $\\leq 27$. But these sequences cover all $999=27\\cdot 37$ numbers $1,2,\\ldots,999$. So we must have $r+1=37$ sequences, each of length exactly $27$, and which all run through all $27$ different roots of $Q$. In particular $[\\alpha_{1}]=27$ and $[\\alpha_{2}]=54$. Now if $S$ were both increasing or decreasing on $[1,\\alpha_{1}]$ and $(\\alpha_{1},\\alpha_{2}]$, then we would have $54$ different values $S(1),S(2),\\ldots,S(54)$. Contradiction! \r\nTherefore the sequences $S(1),S(2),\\ldots,S(27)$ and $S(28),S(29),\\ldots,S(54)$ run through the roots of $Q$ in converse order. And this gives $S(k)=S(55-k)$ for all $k=1,\\ldots,27$. \r\nBut now the polynomial $S(X)-S(55-X)$ has degree $\\leq 37$ and $54$ different roots $1,2,\\ldots,54$. And then $S(X)=S(55-X)$ and $\\lim_{X\\rightarrow+\\infty}S(X)=\\lim_{X\\rightarrow+\\infty}S(55-X)=\\lim_{X\\rightarrow+\\infty}S(-X)$, which contradicts $\\deg S=37$ odd." } { "Tag": [ "blogs", "ARML" ], "Problem": "This might have been posted somewhere more general, since this is his site, but Mr. Rusczyk (the Las Vegas site coordinator) passed out from the heat in the testing room. He was probably ok after getting water, but if not, get well soon!", "Solution_1": "They said it was also from food poisoning, wasn't it?", "Solution_2": "I thought he said something about food--heat and food poisoning, then. That sounds more serious, so again, get well soon.\r\n\r\nDoes anyone know whether the bad food was the cafeteria food or something he'd eaten before?", "Solution_3": "the cafetira food was defenatly not the best; but considering how much i ate i think we can well assume it was probably not from there :D :D \r\n\r\ni was excited when i heard Mr. Rusczyk was the chair for the LV site.\r\nanyways...i hope Mr. Rusczyk gets well soon.", "Solution_4": "Ouch, heat plus food poisoning (or so rumor has it). Either way, best wishes for a quick and pleasant recovery.", "Solution_5": "Yeah, I hope he gets better. Heat exhaustion is awful.", "Solution_6": "I am glad that GA went to Penn state. Penguins like me do worse in heat. \r\n\r\nMr. Rusczyk got food poisoning at LV? I feel lucky for just throwing up my breakfast at Penn state. The food there was edible, just not digestable. \r\n\r\nHopefully he gets better.", "Solution_7": "I hope you feel better soon Mr. Rusczyk! The food at PennState was OK, I didn't really like it. But I stocked up on donuts!", "Solution_8": "according to what i read in Mr. Crawford's and Mr. Rusczyk's blog; he (Mr. Rusczyk) is fine :)", "Solution_9": "Mr. Rusczyk is OK, he is back!!! He just made a post here\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=527965#p527965]http://www.mathlinks.ro/Forum/viewtopic.php?p=527965#p527965[/url]", "Solution_10": "Yay! He is better!\r\n\r\nI didn't know that he passed out.\r\n\r\nAt Penn State, the food was O.K. (not good, but O.K.), and there was PLENTY of water (it rained all day Friday and part of Saturday). I'm glad at Penn State, because nothing MAJOR happened there (apart from a couple minor elevator incedents...)", "Solution_11": "Excellent! That's a relief. :-)\r\n\r\n[quote=\"10000th User\"]Mr. Rusczyk is OK, he is back!!! He just made a post here\n\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=527965#p527965]http://www.mathlinks.ro/Forum/viewtopic.php?p=527965#p527965[/url][/quote]", "Solution_12": "i went to penn state with the GA arml team. we ate at some cracker-barrel style restaurant near penn state that gave me food poisoning. the food tasted great when i was eating it, but later i found myself so sick of the taste. even when i think of it now i get this urge to throw up. i never want to go to cracker-barrel again in my life.", "Solution_13": "Thanks for the sentiments. Just to clarify - I didn't quite pass out, though I came close at times. It was food poisoning, and I can't blame UNLV, as I never had time to hit the cafeteria there - I ate at various restaurants or random snacks we picked up for coaches' meetings or the grading room. Fortunately, there were many experienced ARML people there and an army of volunteers, so I wasn't needed Saturday." } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "induction", "complex numbers", "linear algebra unsolved" ], "Problem": "Let $ A \\in Mn(\\mathbb{R})$and for all $ X\\in Mn(\\mathbb{R}): \\det(A\\plus{}X)\\equal{}\\det(X)$ .Prove that $ A$ is zero.", "Solution_1": "let $ p(x) \\equal{} \\det (xI \\minus{} A)$\r\nwe have for any real $ x$,\r\n$ p(x) \\equal{} \\det (xI \\minus{} A) \\equal{} \\det (xI \\minus{} A \\plus{} A) \\equal{} x^n$\r\nThus $ p(x) \\minus{} x^n$ has infinite roots.\r\nWhich means $ p(x) \\equal{} x^n$ for $ x \\in \\mathbb{C}$\r\nThus the eigenvalue of $ A$ is only $ 0$\r\nWe can just consider $ A$ is the jordan block case.\r\nAnd if $ A \\ne 0$ to construct $ X$ to find a contradiction :) .\r\n\r\n\r\nBut I still have problem,if we change the condition that $ A$ is a complex matrix how to slove it(In this case,you can't assume that $ A$ is just the Jordan block):?: ,which is:\r\nLet $ A \\in Mn(\\mathbb{C})$and for all $ X\\in Mn(\\mathbb{R}): \\det(A \\plus{} X) \\equal{} \\det(X)$ .Prove that $ A$ is zero.\r\n\r\nEdit:I have an idea to slove this:\r\nIf we consider $ \\det (A \\plus{} X) \\minus{} \\det X$ is an polynomial of $ X$ which have $ n^2$ variables.\r\nThen the polynomial is zero for all $ X \\in Mn(\\mathbb{R})$\r\nWhich means $ \\det (A \\plus{} X) \\minus{} \\det X \\equal{} 0$ is an identity.\r\n(Because:\r\nWe can use induction(on $ n$) to prove that an polynomial $ p(x)$ with $ n$ variables have all $ p(x) \\equal{} 0,x\\in \\mathbb{R}^n$ then $ p(x) \\equal{} 0$)\r\nThen we have, $ \\det(A \\plus{} X) \\equal{} \\det(X)$,for all $ X \\in Mn(\\mathbb{C})$\r\nThe rest is similar to above.\r\n\r\n\r\nIs my approach feasible?", "Solution_2": "When I get back to the dormitory,I find the following nice theorem will still be true if we use something named Hilbert Nullstellenatz.(Which can be proved by induction)\r\n\r\nLet $ A \\in M_n(\\mathbb{C})$ and for all $ X$ with entities $ x_{ij} \\in \\{a,b\\}$($ a,b$ are any fixed distinct complex numbers),we have $ \\det (A \\plus{} X) \\equal{} \\det (X)$,then$ A \\equal{} 0$.\r\n\r\nIsn't it nicer :lol: ?" } { "Tag": [ "Putnam", "integration", "trigonometry", "function", "calculus", "induction", "college contests" ], "Problem": "Putnam 1985\r\n\r\nA5) Let $I_{m}=\\int_{0}^{2\\pi}\\cos (x)\\cos (2x)\\cdots\\cos (mx) dx$. For which integers $m$, $1\\leq m\\leq 10$ is $I_{m}\\neq 0$?\r\n\r\nI made a plot with maple to help me figure it out, the plot is of the functions $y=\\prod_{k=1}^{m}\\cos (kx),$ for $1\\leq m\\leq 10,$ the blue curves are the ones whose integral over $\\left[ 0,2\\pi\\right]$ is non-zero, and the red ones have zero as the value of its integral.\r\n\r\n[url=http://123pichosting.com/viewer.php?id=2117Putnam_1985_A5.jpg][img]http://123pichosting.com/thumbs/2117Putnam_1985_A5.jpg[/img][/url]\r\n\r\n[hide=\"my solution\"]Set $f_{m}(x)=\\prod_{k=1}^{m}\\cos (kx),$ for $1\\leq m\\leq 10.$ Recall that $\\cos (x)=\\frac{1}{2}(e^{ix}+e^{-ix})$ so that we may write $f_{m}(x)=\\prod_{k=1}^{m}\\frac{1}{2}(e^{ikx}+e^{-ikx})$. Notice that \\[f_{m}(x+\\pi )=\\prod_{k=1}^{m}\\frac{1}{2}(e^{ik(x+\\pi)}+e^{-ik(x+\\pi)}) = \\prod_{k=1}^{m}\\frac{1}{2}(e^{ik\\pi}e^{ikx}+e^{-ik\\pi}e^{-ikx})\\] here I note that $e^{ik\\pi}= (e^{i\\pi})^{k}=(-1)^{k}$ and that $e^{-ik\\pi}= (e^{i\\pi})^{-k}=(-1)^{-k}=(-1)^{k}$ so we have \\[f_{m}(x+\\pi )=\\prod_{k=1}^{m}\\frac{(-1)^{k}}{2}(e^{ikx}+e^{-ikx}) =(-1)^{\\sum_{k=1}^{m}k}\\prod_{k=1}^{m}\\frac{1}{2}(e^{ikx}+e^{-ikx})\\] \\[=(-1)^{\\frac{m(m+1)}{2}}\\prod_{k=1}^{m}\\cos (kx)=(-1)^{\\frac{m(m+1)}{2}}f_{m}(x)\\] ok, so here is the punchline, \\[I_{m}=\\int_{0}^{2\\pi}f_{m}(x)dx= \\int_{0}^{\\pi}f_{m}(x)dx+\\int_{\\pi}^{2\\pi}f_{m}(x)dx\\] \\[= \\int_{0}^{\\pi}f_{m}(x)dx+\\int_{0}^{\\pi}f_{m}(x+\\pi )dx= \\int_{0}^{\\pi}\\left[ f_{m}(x)+f_{m}(x+\\pi )\\right] dx\\] \\[= \\int_{0}^{\\pi}\\left[ f_{m}(x)+(-1)^{\\frac{m(m+1)}{2}}f_{m}(x)\\right] dx= \\int_{0}^{\\pi}f_{m}(x)\\left[ 1+(-1)^{\\frac{m(m+1)}{2}}\\right] dx\\] which is non-zero for $m=3,4,7,8$ since $(-1)^{\\frac{m(m+1)}{2}}=-1,-1,1,1,-1,-1,1,1,\\ldots$ for $m=1,2,3,\\ldots$.[/hide]", "Solution_1": "Can you set up a recurrence from $I_{m+1}$ to $I_{m}$?\r\nI will look for a way :wink:", "Solution_2": "The [hide=\"answer\"]$I_{n}$ is nonzero if and only if $1+2+\\cdots+m=\\frac{m(m+1)}2$ is odd, which occurs whenever $n$ is equivalent to 1 or 2 mod 4.[/hide]\r\n\r\nFor the idea behind that answer, see this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103871]closely related problem[/url]\r\n\r\n(All right, it's basically equivalent to your hidden solution)", "Solution_3": "[quote=\"benorin\"]Putnam 1985\n\nA5) Let $I_{m}=\\int_{0}^{2\\pi}\\cos (x)\\cos (2x)\\cdots\\cos (mx) dx$. For which integers $m$, $1\\leq m\\leq 10$ is $I_{m}\\neq 0$?\n[hide=\"my solution\"]Set $f_{m}(x)=\\prod_{k=1}^{m}\\cos (kx),$ for $1\\leq m\\leq 10.$ Recall that $\\cos (x)=\\frac{1}{2}(e^{ix}+e^{-ix})$ so that we may write $f_{m}(x)=\\prod_{k=1}^{m}\\frac{1}{2}(e^{ikx}+e^{-ikx})$. Notice that\n\\[f_{m}(x+\\pi )=\\prod_{k=1}^{m}\\frac{1}{2}(e^{ik(x+\\pi)}+e^{-ik(x+\\pi)}) = \\prod_{k=1}^{m}\\frac{1}{2}(e^{ik\\pi}e^{ikx}+e^{-ik\\pi}e^{-ikx}) \\]\nhere I note that $e^{ik\\pi}= (e^{i\\pi})^{k}=(-1)^{k}$ and that $e^{-ik\\pi}= (e^{i\\pi})^{-k}=(-1)^{-k}=(-1)^{k}$ so we have\n\\[f_{m}(x+\\pi )=\\prod_{k=1}^{m}\\frac{(-1)^{k}}{2}(e^{ikx}+e^{-ikx}) =(-1)^{\\sum_{k=1}^{m}k}\\prod_{k=1}^{m}\\frac{1}{2}(e^{ikx}+e^{-ikx}) \\]\n\n\\[=(-1)^{\\frac{m(m+1)}{2}}\\prod_{k=1}^{m}\\cos (kx)=(-1)^{\\frac{m(m+1)}{2}}f_{m}(x) \\]\nok, so here is the punchline,\n\\[I_{m}=\\int_{0}^{2\\pi}f_{m}(x)dx= \\int_{0}^{\\pi}f_{m}(x)dx+\\int_{\\pi}^{2\\pi}f_{m}(x)dx \\]\n\n\\[= \\int_{0}^{\\pi}f_{m}(x)dx+\\int_{0}^{\\pi}f_{m}(x+\\pi )dx= \\int_{0}^{\\pi}\\left[ f_{m}(x)+f_{m}(x+\\pi )\\right] dx \\]\n\n\\[= \\int_{0}^{\\pi}\\left[ f_{m}(x)+(-1)^{\\frac{m(m+1)}{2}}f_{m}(x)\\right] dx= \\int_{0}^{\\pi}f_{m}(x)\\left[ 1+(-1)^{\\frac{m(m+1)}{2}}\\right] dx \\]\nwhich is non-zero for $m=3,4,7,8$ since $(-1)^{\\frac{m(m+1)}{2}}=-1,-1,1,1,-1,-1,1,1,\\ldots$ for $m=1,2,3,\\ldots$.[/hide][/quote]\r\n\r\nI think this is the best solution that i have seen to this problem. (mainly because it is the only one so far that i have been able to follow)", "Solution_4": "question, how do you know that $2\\int_{0}^{\\pi}f_{m}(x)dx\\neq 0$?", "Solution_5": "I have the following solution, which does not use $\\cos x=\\frac{e^{ix}+e^{-ix}}{2}$.\nLet $c_k(x)=cos(kx)$ for each integer $k$. By standard trigonometric identities, $c_kc_l=\\frac{c_{k+l}+c_{k-l}}{2}$. Multiplying by $c_m$ yields $c_kc_lc_m=\\frac{c_{k+l}c_m+c_{k-l}c_m}{2}=\\frac{c_{k+l+m}+c_{k+l-m}+c_{k-l+m}+c_{k-l-m}}{4}$.\n\nContinuing this way, and using the fact that $c_k=c_{-k}$ we can prove by (a very neat) induction that \n\n$2\\prod_{i=1}^m{c_{k_i}}=\\frac{\\sum_b{c_{b}}}{2^{m-1}}$\n\nWhere the sum on the right hand side is taken over all possible $b=\\pm k_1\\pm k_2\\pm \\ldots \\pm k_m$. In our problem, $k_i=i$.\n\nSince $\\int_0^{2\\pi} c_k(x) \\,dx= 2\\pi$ iff $k=0$ and is $0$ otherwise, we have $I_m=\\frac{\\sum_b{\\int_0^{2\\pi} c_b(x)\\,dx}}{2^m}\\neq 0 $ iff for at least one $b$ we have $\\int_0^{2\\pi} c_b(x)\\,dx \\neq 0$, which happens iff we can choose signs in $\\pm 1 \\pm 2 \\pm \\ldots \\pm m$ that yield $0$. \n\nBy parity, we may discard $m=1,2,5,6,9,10$ as in the other solutions. For the other numbers we use $1-2+3$, $1-2-3+4$, $-1+2-3-4+5-6+7$ and $1-2-3+4+5-6-7+8$." } { "Tag": [], "Problem": "1) Solve for $x,y \\in \\mathbb{Z}$, $y \\neq 0$:\r\n\r\n$x^2 + y^3 = (x+y)^2$\r\n\r\n\r\n2) Again, solve for $x,y \\in \\mathbb{Z}$, $y \\neq 0$:\r\n\r\n$x^3 + y^3= (x-y)^4$", "Solution_1": "[hide=\"Problem 1\"]$x^{2}+y^{3}=\\left(x+y\\right)^{2}\\implies y^{2}+y-2x=0$\n\n$\\implies y=-\\frac{1}{2}\\pm\\frac{\\sqrt{1+8x}}{2}$\n\nThen $8x=\\left(k-1\\right)\\left(k+1\\right)$, so there are an infinite number of solutions.[/hide]" } { "Tag": [ "calculus", "integration", "AMC", "USA(J)MO", "USAMO" ], "Problem": "Some chessmen are placed on an $ n$ by $ n$ chessboard, with positive integral $ n$, such that:\r\n\r\n1. Every square without a chess piece shares a side with one that has a chess piece.\r\n2.Given any two squares that contain chessmen, there is a sequence of squares containing chessmen, starting and ending with the given squares, such that every two consecutive squares share a side.\r\n\r\nProve that at least $ \\frac{n^2\\minus{}2}{3}$ chessmen are on the board.\r\n\r\n[hide=\"my solution\"]\nLet us set $ x$ as the number of chessmen. We know that there are $ n^2$ squares that must either be covered by a chess piece or adjacent to a square that is covered by a chess piece because of rule 1.\n\nWe can \"tile\" the board with chessmen by first laying down one, and then laying another adjacent to that one, and another adjacent to that one, et cetera, thus adhering to rule 2.\n\nThe first chess piece \"tiles\" the square, and the four adjacent squares around it. All of the chessmen that are added after that one \"tile\" a maximum of three squares after the first. So, the maximum number of squares that are covered can be represented in terms of $ x$; the initial five \"tiled\" by the original chess piece, plus the three \"tiled\" by every next chess piece:\n\n$ 5 \\plus{} 3(x \\minus{} 1)$\n\nThen, for sufficient $ x$,\n\n$ 5 \\plus{} 3(x\\minus{}1) \\geq n^2$\n\nSolving for $ x$, we have\n\n$ x \\geq \\frac{n^2\\minus{}2}{3}$,\n\nproving our original theorem.\n[/hide]", "Solution_1": "Source: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=340035#340035]USAMO 1999, Problem 1[/url]" } { "Tag": [ "videos", "Support" ], "Problem": "Is there a certain amount of questions you need to finish, before getting videos? I already did 62+ problems. I have done 100+ now, and I have only gotten 2 more videos. Is there a \"Complete 110+ problems, get 2 xtra videos?\" I like the videos, and want to get more. Just a question :)", "Solution_1": "You need to pass subjects in order to access more videos. So, if you just answer \"1\" for 60 problems to get to the videos, you'll only have a few videos you can see before you actually start doing the problems, getting them right, passing some subjects, and earning access to more videos.", "Solution_2": "Unless, of course, the answer to all 60 problems IS 1.", "Solution_3": "I don't think that's a good idea. You should click on [list]Video lessons[/list] after every problem.", "Solution_4": "Since AoPS is nonprofit, they don't want random people who aren't even trying to learn taking advantage of their videos, so you have to answer a certain amount of correct answers to watch.", "Solution_5": "This thread is like 4 years old. Why was it bumped?", "Solution_6": "I don't know! But the videos are helpful. I think that this type of security is great, anyway." } { "Tag": [ "conics", "geometry", "3D geometry", "ellipse", "parabola" ], "Problem": "can you guys help me the question is attached! :huh: \r\nthanks in advance!", "Solution_1": "In general, the type of conic section is determined by the angle of the plane, and whether or not it passes through the vertex. Let's denote the angle the plane forms with the base of the cone (in degrees) $\\theta$, and the angle that the base of the cone forms with its side (also in degrees) $\\alpha$.\r\n\r\nFirst, the conic sections when the plane does not pass through the vertex: if $\\theta = 0$, this conic section will be a circle; if $0<\\theta<\\alpha$, the conic will be an ellipse; If $\\theta=\\alpha$, then the conic will be a parabola; If $\\alpha<\\theta\\le90$, then the conic will be a hyperbola.\r\n\r\nNow I will cover the degenerate conic sections, which happen when the plane passes through the origin: if $0\\le\\theta<\\alpha$, then the conic section will be a single point; if $\\theta=\\alpha$, then the conic section will be a single line; if $\\alpha<\\theta\\le90$, then the conic section will be a pair of intersecting lines.\r\n\r\nIf you have further questions, feel free to ask." } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "$ a,b$ are odd integers\r\n\\[ \\begin{array}{l} a_1 \\equal{} a \\\\\r\na_2 \\equal{} b \\\\\r\n\\vdots \\\\\r\n\\vdots \\\\\r\na_n \\equal{} largest ..odd..divisor..of(a_{n \\minus{} 1} \\plus{} a_{n \\minus{} 2} ) \\\\\r\n\\end{array}\r\n\\]\r\nProve that there exists a natural number $ N$ such that\r\n\\[ \\forall k \\ge N,a_k \\equal{} \\gcd (a,b)\r\n\\]", "Solution_1": "Posted before. See here:\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?search_id=1870291191&t=58207[/url]" } { "Tag": [ "geometry", "3D geometry", "sphere" ], "Problem": "My next forum game is coming up. Sign up for it if you know you will be on AoPS at least once every two days! You must have joined AoPS on or before November 19, 2009 to join this game, and you must have been on AoPS for at least a year to take spots 11 through 15. Provide your gender, and pick a number between 10 and 99 inclusive that no user who has previously signed up has picked. \r\n\r\n1. monkeygirl13 [17 - 85,52,42,57] - female\r\n2. westiepaw [37 - 77,55,71,82] - female\r\n3. wrightboy10 [52 - 10,59,58,10] - male\r\n4. Richard_Min [99 - 33,18,42,93] - male\r\n5. Butler [50 - 79,83,38,62] - male\r\n6. randomguy64 [42 - 46,86,82,66] - male\r\n7. tennis123 [21 - 44,38,61,46] - female\r\n8. jjfun1 [33 - 78,63,25,98] - female\r\n9. andrewjjiang97 [13 - 70,11,65,95] - male \r\n10. vahalla [59 - 52,11,62,28] - male\r\n11. ksun48 [98 - 93,27,42,87] - male\r\n12. AIME15 [62 - 21,83,38,18] - male\r\n13. aleph0 [61 - 77,56,24,80] - male\r\n14. Yoshi [97 - 77,46,20,88] - male\r\n15. bluecarneal [15 - 46,43,79,48] - male", "Solution_1": "/join\r\nspot 11\r\n98\r\nmale", "Solution_2": "Spot 10.\r\n\r\nMy number is 59.", "Solution_3": "What is ATTOS going to be about?", "Solution_4": "i join \r\n50", "Solution_5": "1. \r\n2. \r\n3. \r\n4. \r\n5. \r\n6. \r\n7. \r\n8. \r\n9. \r\n10. vahalla, male? (59)\r\n11. ksun48, ????? (98)\r\n12. \r\n13. \r\n14. \r\n15. bluecarneal, male (15)\r\nbutler, pick spot, male, (50)", "Solution_6": "1.\r\n2.\r\n3.\r\n4.\r\n5.\r\n6.\r\n7.\r\n8.\r\n9. andrewjjiang97, male, (13)\r\n10. vahalla, male? (59)\r\n11. ksun48, ????? (98)\r\n12.\r\n13.\r\n14.\r\n15. bluecarneal, male (15)\r\nbutler, pick spot, male, (50)", "Solution_7": "1. \r\n2. \r\n3. \r\n4. \r\n5. \r\n6. \r\n7. \r\n8. \r\n9. andrewjjiang97, male, (13) \r\n10. vahalla, male? (59) \r\n11. ksun48, ????? (98) \r\n12. \r\n13. aleph0, male, (61)\r\n14. \r\n15. bluecarneal, male (15) \r\nbutler, pick spot, male, (50)", "Solution_8": "Wow, I am indeed surprised that within half an hour so many people replied, it wasn't like this for LTM, you know. Also, no need to keep track of who's in; I'll update the first post for you. As for the question on what ATTOS is about, the reason I have a separate signup page is for the uncertainty in the air...NEVR FEER TZE UNNOEN.", "Solution_9": "1.\r\n2.\r\n3.\r\n4.\r\n5.\r\n6.\r\n7.\r\n8.\r\n9. andrewjjiang97, male, (13)\r\n10. vahalla, male? (59)\r\n11. ksun48, ????? (98)\r\n12. AIME15, male, (62)\r\n13. aleph0, male, (61)\r\n14. \r\n15. bluecarneal, male (15)\r\nbutler, pick spot, male, (50)\r\n\r\n@phire: sorry i had to do that", "Solution_10": "By the way, because I know someone will ask, those four numbers in the first post after your number are randomly generated.", "Solution_11": "[quote=\"PhireKaLk6781\"]By the way, because I know someone will ask, those four numbers in the first post after your number are randomly generated.[/quote]\r\n\r\nwhat are they for?", "Solution_12": "Spot 7, 3, female", "Solution_13": "spot 5, male, 3", "Solution_14": "1. monkeygirl13 (17) female\r\n2. \r\n3. \r\n4. \r\n5. piegragh, male\r\n6. \r\n7. tennis123 (3) female\r\n8. \r\n9. andrewjjiang97, male, (13) \r\n10. vahalla, male? (59) \r\n11. ksun48, ????? (98) \r\n12. AIME15, male, (62) \r\n13. aleph0, male, (61) \r\n14. \r\n15. bluecarneal, male (15) \r\nbutler, pick spot, male, (50)", "Solution_15": "Current replacements list: \r\npiegraph\r\nspaceguy524", "Solution_16": "At\r\nThe\r\nT_ _ _ _ _ _ _\r\nOf\r\nS_ _ _ _ _ _\r\n\r\nThis makes sense, but what are the other two words?", "Solution_17": "I want to say \"at the threshold of silence\"", "Solution_18": "He told me that the second letter of the word starting with s was p, but \"threshold\" sounds pretty good.", "Solution_19": "spartan?\r\nI can't think of anymore sp 7 letter words.\r\n\r\nEDIT: forgot I had a crossword puzzle solver.\r\n\r\nthere are a lot more, but none would be relevant.", "Solution_20": "I used a crossword puzzle to read off possibilities. They were in alphabetical order. He said I was close when I typed \"spheres\" but farther when I typed \"spirits\".", "Solution_21": "At The Treasure of Something?", "Solution_22": "At the Threshold of Spiders? Idk, that might work. :P", "Solution_23": "At the Treasure of Silence?", "Solution_24": "maybe its At The Threshold of Silence.", "Solution_25": "[quote=\"aleph0\"]He (PhireKalk) told me that the second letter of the word starting with s was p, but \"threshold\" sounds pretty good.[/quote]\r\n\r\nSo the last word probably isn't 'silence'.", "Solution_26": "At The Threshold Of Spleens? lol not\r\n\r\nAt The Threshold of Sp_ _ _ _ _", "Solution_27": "Threshold doesn't have the correct amount of letters, so it can't be it.", "Solution_28": "Wow, I could probably even make a game just for guessing the name of the next game at this rate.", "Solution_29": "At the Threshold of Silence, does sound like the name of a game, or a book.\r\n\r\nI thought I was right...." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Hi people... :) \r\nI just found an inequality in a paper of the Peruvian training ....\r\n\r\nI'd LIKE TO know if this is an original constraction!! :?\r\nIt has an introduction ...\r\n\r\nSummarizing....\r\n\r\nLet $x,y,z > 0$ real numbers,\r\nwe know that: \r\n\r\n$\\frac{a^3 + b^3 + c^3 }{abc} \\geq 3$, (directly MA-MG)\r\n\r\nbut the left part could be re-writed like that:\r\n\r\n\\[ \\frac{a^3 + b^3 + c^3 }{abc} \\geq (\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a}) + (\\frac{a}{b} - \\frac{b}{a}) + (\\frac{b}{c} - \\frac{c}{b}) + (\\frac{c}{a} - \\frac{a}{c}). \\]\r\n\r\nProve this...\r\n\r\nCarlos Bravo :)\r\nLIMA-PERU", "Solution_1": "sincerely i have not seen this inequality before, but i can't guarantee that this is original. it's not so difficult, indeed is very nice and simple, here's my solution\r\n\r\nmultiply by $abc$ to obtain that your inequality is equivalent to\r\n$a^3+b^3+c^3+a^2b+ac^2+b^2c\\geq2(ab^2+a^2c+bc^2)$, \r\nbut by am-gm we have that\r\n$a^3+ac^2\\geq2a^2c$, $b^3+a^2b%Error. \"geqab\" is a bad command.\n^2$, and $c^3+b^2c\\geq2bc^2$, \r\nso we're done! :D" } { "Tag": [], "Problem": "\u039d\u0394\u039f--->\u0391\u03bd $ 24/xy\\plus{}1$ \u03c4\u03bf\u03c4\u03b5 $ 24/x\\plus{}y$ $ x,y$ \u03b1\u03bd\u03b7\u03ba\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf\u03c5\u03c2 \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03bf\u03c5\u03c2.\r\n(\u03bc\u03b5 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03b5\u03c1\u03b5\u03b9 \u03c0\u03bf\u03bb\u03c5 \u03bd\u03b1 \u03b4\u03c9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03b5\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2)", "Solution_1": "\u03a3\u03b1\u03bd \u03bc\u03b9\u03b1 \u03c0\u03c1\u03c9\u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03c9\u03c2 $ gcd(x,24) \\equal{} 1$ \u03ba\u03b1\u03b9 $ x(y \\plus{} x) \\equal{} xy \\plus{} x^2 \\equiv xy \\plus{} 1 \\equiv 0 \\mod 24$ \u03b1\u03c0\u03bf \u03bf\u03c0\u03bf\u03c5 $ 24|x \\plus{} y$\r\n\r\n\u03a3\u03b1\u03bd \u03bc\u03b9\u03b1 \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03b7 \u03c0\u03c1\u03bf\u03c3\u03b5\u03b3\u03b3\u03b9\u03c3\u03b7, \u03b1\u03bb\u03bb\u03b1 \u03bf\u03c7\u03b9 \u03bc\u03b5 \u03ba\u03b1\u03bc\u03b9\u03b1 \u03bc\u03b5\u03b3\u03b1\u03bb\u03b7 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b1, \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b8\u03b5\u03c9\u03c1\u03b7\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf \u03c4\u03c1\u03b9\u03c9\u03bd\u03c5\u03bc\u03bf \u03bc\u03b5 \u03c1\u03b9\u03b6\u03b5\u03c2 \u03c4\u03b1 $ x,y$ \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03c4\u03bf \u03c7\u03b5\u03b9\u03c1\u03b9\u03c3\u03c4\u03b5\u03b9\u03c2 \u03c9\u03c2 \u03b5\u03be\u03b7\u03c2.\r\nE\u03c3\u03c4\u03c9 \u03bf\u03c4\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03bf $ P(x) \\equal{} x^2 \\plus{} bx \\plus{} c$ \u03bc\u03b5 $ 24|c \\plus{} 1$ .To\u03c4\u03b5 \u039codulo 24 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 $ P(x) \\equal{} bx \\plus{} c \\plus{} 1\\equal{}bx$ \u03b1\u03c0\u03bf \u03bf\u03c0\u03bf\u03c5 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03c5\u03bc\u03b5\u03bd\u03bf.", "Solution_2": "\u0395\u03c3\u03c4\u03c9 \u03bf\u03c4\u03b9 $ x \\plus{} y \\equal{} a$ , $ x \\minus{} y \\equal{} b$ \u03c4\u03bf\u03c4\u03b5 \u03b7 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03bc\u03b1\u03c2 \u03bc\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03b9\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c9\u03c2 \u03b5\u03be\u03b7\u03c2:\r\nN\u0394\u039f-->\u0391\u039d $ 24/(a^2 \\minus{} b^2 \\plus{} 4)/4$ \u03c4\u03bf\u03c4\u03b5 \u03ba\u03b1\u03b9 $ 24/a$\r\n\u03b1\u03bd \u03bf $ 3$ \u03b4\u03b5\u03bd \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03b9 \u03c4\u03bf\u03bd $ a$ \u03c4\u03bf\u03c4\u03b5 $ a^2 \\equal{} 1mod4 \\equal{} > b^2 \\equal{} 2mod3$ ATO\u03a0\u039f \u0391\u03c1\u03b1 $ a \\equal{} 0mod3$\r\n\u039c\u03b5\u03bd\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03c1\u03b5\u03c5\u03bd\u03b7\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b1\u03bd \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03b9\u03c4\u03b5 \u03bc\u03b5 \u03c4\u03bf $ 32$\r\n\u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 $ 32/a^2 \\minus{} b^2 \\plus{} 4$ 8a \u03b4\u03b5\u03b9\u03be\u03c9 \u03bf\u03c4\u03b9 \u03ba\u03b1\u03b9 $ 32/a$\r\n\u03b5\u03c0\u03b5\u03b9\u03b4\u03b7 $ 32/a^2 \\minus{} b^2 \\plus{} 4 \\equal{} > a \\equal{} bmod2$\r\n\u03b5\u03c3\u03c4\u03c9 \u03bf\u03c4\u03b9 $ a \\equal{} 8k \\plus{} z$ , $ b \\equal{} 8k \\plus{} m$ \u03c4\u03bf\u03c4\u03b5 $ a^2 \\minus{} b^2 \\plus{} 4 \\equal{} z^2 \\minus{} m^2 \\plus{} 4mod32$ \u03bc\u03b5 $ z,m \\equal{} {0,1,2,3,4,5,6,7,8}$\r\n\u03c4\u03b5\u03bb\u03bf\u03c2 \u03b5\u03c0\u03b5\u03b9\u03b4\u03b7 $ z \\equal{} mmod2$ \u03b5\u03c5\u03ba\u03bf\u03bb\u03b1 \u03b2\u03bb\u03b5\u03c0\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 $ z \\equal{} 0$ \u03b1\u03c1\u03b1 $ 8/a$ ." } { "Tag": [ "continued fraction" ], "Problem": "[b] Let $M$ be a positive integer , $M\\le 100$ . Using only $M$ perfect one ohm resistors, construct a resistance of [/b] [color=red] [b] E [/b] [/color] [b] ohms ,\n were [/b] [color=red] [b] E [/b] [/color] [b] approximate ,, $e$ \" (Napier's constant) with at least five decimals (only series-parallel circuits are allowed). [/b]\r\nSee also :\r\n [1] Problem E2459 proposed by A.A.Mullin in Amer.Math.Monthly (1974)\r\n [2] Problem 393 from Matematicki Vesnik 13(28)(1976),solution in 15(1978).", "Solution_1": "I can do six decimal places with M=18, but I don't use only paralell-series circuits. R is approximately 2.71828103... if you label vertices 0 to 13 and join the pairs\r\n\r\n0 7\r\n0 4\r\n1 3\r\n1 5\r\n2 8\r\n2 10\r\n2 4\r\n3 7\r\n3 11\r\n3 4\r\n5 8\r\n5 9\r\n5 10\r\n9 11\r\n10 4\r\n11 4\r\n4 6\r\n6 12\r\n\r\nThe requested resistance is measured between nodes 0 and 12.", "Solution_2": "Bus, \r\nhow did you come about doing that? I'm afraid I'm not very good with circuits, is there a method for figuring out those resistances?", "Solution_3": "(I'm kubus, not bus:)\r\n\r\nTo calculate resistance of a non-parallel-series circuit, you can either do some clever tricks (triangle-star exchange, cutting and pasting parts, ...) or simply write the potentials (voltages) in the nodes as N unknows and use the Ohm's and Kirchhoff's laws to write N linear equations. Then just solve. I wrote a program some time ago to do this for me, I've started a thread about finding a scheme with resistance equal to $\\pi \\approx 3.14159...$ a couple of months ago.\r\n\r\nIf you only want parallel-series circuits, you can make use of the continued fraction of e:\r\n$e \\approx 2+\\frac{1}{1+\\frac{1}{2+\\frac{1}{1+\\frac{1}{1+\\frac{1}{4+\\cdots}}}}}$\r\nThis way, with 21 resistors, you can get right 7 decimal digits.", "Solution_4": "Whoops, I just saw the flag, and he's been posting more frequently lately... sorry :blush: \r\n\r\nAnd, thanks for the reply =)", "Solution_5": "[quote=\"kubus\"]This way, with 21 resistors, you can get right 7 decimal digits.[/quote]\r\nHi kubus, can you prove that [color=red][b] 21 [/b] [/color] is [b] the minimal number of resistors [/b] in order to obtain an approximation for $e$ with at least [color=red][b] 7 [/b] [/color] decimals ?", "Solution_6": "I certainly can not prove this, moreover, I doubt it is the minimal number of resistors.\r\n\r\nA kind of \"proof\" would be to write a program to generate ALL the circuits with 1,2,3,4, ... resistors and see what is the minimum number. This could be really time-consuming, but I'll think about it.\r\n\r\nBoth of my examples were just first that came to my mind (or first spat out by the computer). I am sure you can do better in both cases." } { "Tag": [ "inequalities" ], "Problem": "Prove that if a, b, c are the lengths of the sides of a triangle then\r\n\r\n3/2 <= a/(b + c) + b/(c + a) + c/(a + b) < 2.\r\n\r\nThis one is easy 8-)", "Solution_1": "Still working hard on those ineg\r\n\r\n\r\n* 3/2 <= a/(b + c) + b/(c + a) + c/(a + b) \r\n\r\nI think it works for every a,b,c No :?: \r\n\r\n\r\nWolog a >= b >= c\r\nthen 1/a+b <= 1/a+c <= 1/b+c\r\n\r\nBy Chebyshev : \r\n\r\na/(b + c) + b/(c + a) + c/(a + b) >= 1/3 (a+b+c)(1/a+b + 1/a+c + 1/b+c))\r\n\r\nBy Cauchy : (:Sigma: ai)(:Sigma: 1/ai) >= n:^2: (if n ai)\r\n\r\nSo (1/a+b + 1/a+c + 1/b+c))(a+b + a+c + b+c) >= 9\r\n\r\na/(b + c) + b/(c + a) + c/(a + b) >= 1/3 (a+b+c)*9/2/(a+b+c) = 3/2\r\n\r\n\r\n* a/(b + c) + b/(c + a) + c/(a + b) < 2\r\n\r\ns = (a+b+c)/2, then b+c > s, c+a > s, b+a > s for a triangle.\r\nso a/(b + c) + b/(c + a) + c/(a + b) < (a+b+c)/s = 2\r\n\r\n\r\nThis one was not the most difficult I think :)", "Solution_2": "Indeed. The first inequality is true for any a, b, c > 0.\r\nIt is called \"Nesbitt inequality\".\r\nRemember this one since it is very often useful !\r\n\r\nNumerous proofs are available :\r\nCauchy ; \r\nRearrangement ;\r\nChebyshef ...\r\n\r\nThe right inequality is also quite easy 8-)" } { "Tag": [ "calculus", "geometry", "trigonometry", "function", "conics", "vector", "logarithms" ], "Problem": "I'm just curious: How many of you took Pre-Calculus in 9th grade? Just wondering, 'cause I want to know how you can deal with taking such a hard subject!!!", "Solution_1": "I took precal in grade 9. I found it to be mostly a repeat of material I had learned in the past few years (algebra, algebra 2, and geometry), although that may be because I had a really good geometry teacher and I took algebra 2 twice because public schools are stupid, so I had a good foundation in those subjects. \r\n\r\nOur classes were 50 min, 5 days/week. I probably played about 3 hours of calculator tetris every week in that class.", "Solution_2": "I am actually taking Precalculus right now in the 7th grade. It really is review if you have taken Algebra, Algebra II, and Geometry like Xevarion said.", "Solution_3": "It's possible that at his school, the precal curriculum does include a lot of new material. \r\n\r\nSome important things that were in my precal curriculum were trig (memorize those trig identities!) and getting familiar with functions like exponentials and log and rational functions. \r\n\r\nIf you haven't seen much about those functions before in geometry and algebra, learning that material could be hard. Make sure you study that stuff and understand how to manipulate things abstractly and also the concrete meaning of what you're doing. \r\n\r\nAnother tip: always do your homework. Pay attention in class (unless you're totally sure you already understand all the material from the lesson).", "Solution_4": "I took it... never. I did do calculus in tenth grade, but I skipped precalculus before that. The key is that I taught myself a lot of algebra and trig while I was bored in geometry.", "Solution_5": "I took it in 9th grade after having skipped Algebra II. I learned a great deal in that class, in particular about trig identities, law of cosines, law of sines, conic sections, vectors, complex numbers, and e. I think I worked harder (and learned more) during that year than in any other year I was in high school. Doing a LOT of AHSMEs (now called AMCs) helped reinforce the concepts that I was learning in precal at the time.\r\n\r\nThe book that we used is still popular today - Brown and Robbins' [u]Advanced Mathematics[/u], although I think the book has gone through a lot of changes since then.", "Solution_6": "9th grade as well. I had already taken the EPGY Precalculus course which I found extremely difficult, so I decided a review would be worthwhile. Boy, was I wrong. Never trust public school math education. I did manage to almost never have HW from that class though, since I always did it during class.", "Solution_7": "I never actually took precalc - I took Honors Geometry in 9th grade (they laughed it off when I asked to skip; although the next year I managed to yell my way into calculus). Honestly, a lot of precalc is a direct extension of algebra II. The main things that may be new are dealing with exponentials, dealing with logarithms, and dealing with trig.\r\n\r\nExponentials and logarithms are really the same thing; it just takes some seeing to get used to. As for trig, I was having a bit of trouble at first. Then, my dad got out his old math book, which had a page full of trig identities and said, derive all of these out yourself, and if you make it to the end, you must have figured trig out. And it really helped." } { "Tag": [ "function", "floor function", "induction" ], "Problem": "Let $ f: \\mathbb{N} \\rightarrow \\mathbb{N}$ be a function such that $ f(n\\plus{}1) > f(n)$ and $ f(f(n)) \\equal{} 3n$ for all $ n$. Evaluate $ f(2001)$.", "Solution_1": "[hide]\nConsider $ f(n) \\equal{} \\lfloor \\sqrt {3}n \\rfloor$ The result follows\n[/hide]", "Solution_2": "Well how did you get to that function? :huh:", "Solution_3": "[quote=\"jimhu\"][hide]\nConsider $ f(n) \\equal{} \\lfloor \\sqrt {3}n \\rfloor$ The result follows\n[/hide][/quote]\r\nSorry, how does it follow? $ \\left\\lfloor\\sqrt3\\lfloor\\sqrt3\\cdot1\\rfloor\\right\\rfloor\\equal{}1<3\\equal{}f(f(1))$.", "Solution_4": "[hide=\"WRONG Solution - DO NOT DISILLUSION YOURSELF!\"]\nNote that $ f(f(1)) \\equal{} 3$\nNow, if $ f(1) \\equal{} x$ and $ f(x) \\equal{} 3$, clearly $ x$ must be $ 2$. Then $ f(3) \\equal{} 3(2) \\equal{} 6$, and for every $ a_n$, $ a_n \\equal{} 3a_{n \\minus{} 2}$. Thus if $ a_1 \\equal{} 1$, $ \\{a_n\\} \\equal{} \\{1, 2, 3, 6, 9, 12, \\dots\\}$. From this, it follows that $ f(x) \\equal{} \\begin{cases} \\frac 32 x & \\text{if } 2\\ |\\ x \\\\\n2x & \\text{if } 2\\not|\\ x \\end{cases}$\n\nThen it becomes apparent that $ f(2001) \\equal{} \\boxed {4002}$.\n[/hide]", "Solution_5": "Tisk Tisk yongie, it nshould be 9,12 not 9,18. This is the most natural approach to solving the problem. Namely, solving enough numbers to find a pattern.\r\n\r\nf(1) cannot equal 1, since then f(f(1))=1\r\nIf $ f(1)\\equal{}k\\ge3$ then $ 3\\equal{}f(f(1))\\equal{}f(k)\\ge f(3)>f(1)\\equal{}k$ contradiction\r\nSo f(1)=2\r\nAnd therefore f(2)=f(f(1))=3\r\nf(3)=f(f(2))=6\r\nf(6)=f(f(3))=9\r\nWhich means that f(4)=7 and f(5)=8 since they are squeezed in between f(3) and f(6).\r\nf(7)=12\r\nf(8)=15\r\nf(9)=18\r\nf(12)=21\r\nSo f(10)=19 f(11)=20\r\nf(15)=24\r\nSo f(13)=22 f(14)=23\r\nf(18)=27\r\nSo f(16)=25 f(17)=26\r\nf(19)=30\r\nf(20)=33\r\n\r\n1 2\r\n2 3\r\n3 6\r\n4 7\r\n5 8\r\n6 9\r\n7 12\r\n8 15\r\n9 18\r\n10 19\r\n11 20\r\n12 21\r\n13 22\r\n14 23\r\n15 24\r\n16 25\r\n17 26\r\n18 27\r\n19 30\r\n20 33\r\n\r\nProve by induction that the differences between consecutive numbers goes 1 3 1 1 1 3 3 3 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 Where there are 3 times as many as previous (use induction on the powers of 3 should be useful). Then bash out 2001. This should work.", "Solution_6": "Oh... :blush: Never bothered to look at $ f(4)$... *whacks himself for being stupid again* :oops: \r\n\r\n(just kidding)\r\n\r\nContinuing his list:\r\n\r\n1 2 \r\n2 3 \r\n3 6 \r\n4 7 \r\n5 8 \r\n6 9 \r\n7 12 \r\n8 15 \r\n9 18 \r\n10 19 \r\n11 20 \r\n12 21 \r\n13 22 \r\n14 23 \r\n15 24 \r\n16 25 \r\n17 26 \r\n18 27 \r\n19 30 \r\n20 33 \r\n21 36\r\n22 39\r\n23 42\r\n24 45\r\n25 48\r\n26 51\r\n27 54\r\n28 55\r\n29 56\r\n30 57\r\n31 58\r\n32 59\r\n\r\nUsing induction, we can see that if $ 2 \\cdot 3^k < n < 3^{k \\plus{} 1}$, then $ f(n) \\equal{} 3^{k \\plus{} 1} \\plus{} 3(n \\minus{} 2 \\cdot 3^k)$, or $ 3n \\minus{} 3^{k \\plus{} 1}$. (Where $ k$ is a natural number.) We note that $ 2 \\cdot 3^6 < 2001 < 3^7$, so it follows that $ f(2001) \\equal{} 6003 \\minus{} 2187 \\equal{} \\boxed {3816}$.\r\n\r\nP.S. My name is Yongyi, not yongie... :|" } { "Tag": [ "ratio" ], "Problem": "17.In a college 240 fruits are distributed among the students. Find the number of students who take all the three fruits. Followng are the observations:\r\n\r\ni.Every student takes at least one fruit \u2013 mango, banana or apple and at most one fruit of each type.\r\n\r\nii.Ratio of number of students who take only a mango to those who take only an apple to those who take only a banana is 9:4:3\r\niii.Ratio of number of students taking banana and an apple but not a mango to those taking an apple and a mango but not a banana to those taking a mango and a banana but not an apple is 4:6:3\r\n\r\niv.Ratio of Number of students taking at least mango, at least banana, at least apple is 5:3:4", "Solution_1": "[hide=\"Good hint\"]Use a Venn-Diagram. That really helps on problems like these. [/hide]", "Solution_2": "Can someone solve the problem? When I tried it with my method, it won't work. :(", "Solution_3": "even better than a venn diagram is a generalized venn diagram- without having to draw circles!! the mathematical name for this is [b]The Principle of Inclusion-Exclusion[/b] of PIE (try searching the forums for it...)\r\n\r\nEDIT: better yet, there is an entry in the AoPS wiki!!", "Solution_4": "I don't think that will work because we only have ratios now.", "Solution_5": "It works fine. We just interpret, for example, $9: 4: 3$ to mean the quantities $9x, 4x, 3x$ and we continue assigning variables until we get a (rather large but solvable) system.", "Solution_6": "Yay to rather large but solvable systems!!\r\nand computer lol...", "Solution_7": "can you solve it??? i think the question is wrong....not sure", "Solution_8": "$M: B: A: M\\cap B: B\\cap A: M\\cap A: M\\cap B\\cap A=9: 3: 4: 3: 4: 6: 2$\r\n\r\nSo, 240 fruit is impossible.\r\n\r\n[hide=\"Outline\"] Draw the Venn Diagram. Fill in $9,3,4$. From iii, fill in $3x, 4x, 6x$. Call the center $y$.\nFrom iv, we get the easily solved system:\n$3+7x+y=3z$\n$9+9x+y=5z$\n$4+10x+y=4z$ [/hide]" } { "Tag": [ "puzzles" ], "Problem": "There was a death and the victim was identified as A, who recently was married. The police went to the crime scene and they reported the death as a suicide. \r\n\r\nLater that day, after the police left, detective B, hired by A's friend who thought it was a murder, searched the crime scene and found a note the police missed.\r\n\r\nIt read,\r\n\r\n\"4,3: 8,1:_: 9,1: 2,1: 7,4:_: 6,1:9,3:_: 9,1: 4,3: 3,3: 3,2: !\"\r\n\r\nDetective B took out his cell phone and started dialing the police to tell them about his findings. Once B opened the phone to dial, he immediately screamed out, \"I Solved It!\"\r\n\r\nWho was the murderer and how did B find out?", "Solution_1": "[hide=\"Hint\"]\n[code]\n 1 2 3\n ABC DEF\n\n 4 5 6\nGHI JKL MNO\n\n 7 8 9\nPQRS TUV WXYZ\n\n 0\n[/code][/hide]\n[hide=\"Answer\"]\nIT_WAS_MY_WIFE!\n[/hide]" } { "Tag": [ "\\/closed" ], "Problem": "What are bots and what role do they serve on the board?", "Solution_1": "[[AoPS_forums_tutorial#Bot]]\r\n\r\nThat explains a lot.", "Solution_2": "http://en.wikipedia.org/wiki/Internet_bot\r\n\r\nAlso, a user called Yahoo Slurp! was mistaken for a bot once.", "Solution_3": "[quote=\"Temperal\"]http://en.wikipedia.org/wiki/Internet_bot\n\nAlso, a user called Yahoo Slurp! was mistaken for a bot once.[/quote]\r\nWas that because he made some posts?" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "Compute the integral: $\\displaystyle \\int \\frac{x^{3}+x+1}{x^{4}+x^{2}+1} dx.$", "Solution_1": "$\\frac{x^3+x+1}{x^4+x^2+1} = \\frac{1}{2(x^2-x+1)} + \\frac{2x+1}{2(x^2+x+1)}$.\r\n\r\nFor the first integral on the right,\r\n\r\n$\\int \\frac{dx}{x^2-x+1} = \\int \\frac{d(x-\\frac{1}{2})}{(x-\\frac{1}{2})^2+\\frac{3}{4}} = \\frac{1}{\\sqrt{\\frac34}}\\arctan \\frac{x-\\frac{1}{2}}{\\sqrt{\\frac34}} + c$\r\n\r\nand for the second one, notice that $(x^2+x+1)' = 2x + 1$ hence it is $\\frac{1}{2}\\ln(x^2+x+1) + c$.\r\n\r\nTherefore the result is\r\n\r\n$\\frac{1}{2}\\ln(x^2+x+1) + \\frac{1}{\\sqrt3}\\arctan \\frac{2x-1}{\\sqrt3} + c$", "Solution_2": "$\\text{I saw a different }(but\\;ridiculously\\;longer)\\text{ method . . .}$\r\n\r\n$\\displaystyle{\\text{We have: }\\int \\frac{x^3 + x + 1}{x^4 + x^2 + 1}\\;dx }$\r\n\r\n\r\n$\\text{Multiply top and bottom by }x^2 - 1:$\r\n\r\n$\\displaystyle{\\frac{x^2-1}{x^2-1}\\cdot\\frac{x^3+x+1}{x^4+x^2+1} \\;= \\;\\frac{x^5-2x^4+x^2-x-1}{x^6-1} \\;= \\;\\frac{x^5 - 2x^3 + x^2 - x - 1}{(x-1)(x+1)(x^2-x+1)(x^2+x+1)} }$\r\n\r\n$\\text{ . . . and apply Partial Fractions.}$", "Solution_3": "Well the first method applied partial fractions from the beginning...." } { "Tag": [ "logarithms", "floor function", "limit", "calculus", "integration", "calculus computations" ], "Problem": "1. Prove that the sequence $ x_n\\equal{} \\left\\{1\\plus{} \\frac{1}{2}\\plus{} \\frac{1}{3}\\plus{} \\dots \\plus{} \\frac{1}{n} \\right\\}\\minus{} \\left\\{ \\ln{n} \\right\\}$ is divergent where $ \\{ x \\}\\equal{} x \\minus{} \\lfloor x \\rfloor$ is the fractional part of $ x \\in \\mathbb{R}$.", "Solution_1": "First show(or assume that we know) that $ \\lim \\sum\\frac{1}{n}\\minus{}\\ln n$ exists. Then your statement follows from the fact that $ [\\sum \\frac 1n] \\minus{}[\\ln n]$ takes values 1 and 0 infinitely often: indeed, if the limit of the difference of fractional parts exists, then the same would be true for integer parts.", "Solution_2": "We don't quite need convergence- just that $ \\sum_{k\\equal{}1}^n \\frac1k\\minus{}\\ln n$ is bounded away from any integers for large $ n$. This is easy with standard integral estimates. Then the two parts increment at different times, and we get infinitely many jumps up and down.", "Solution_3": "[quote=\"Yustas\"] $ [\\sum \\frac 1n] \\minus{} [\\ln n]$ takes values 1 and 0 infinitely often.[/quote]\r\n\r\nI`ll apreciate very much if you can show a rigurous proof for this part. :D\r\n\r\nThanks" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "AIME I" ], "Problem": "I found several problems with the AMC and AIME state statistics\r\n\r\n\r\nFor example, on the amc 12b, it seems like Wendy Hou, who got the 133.5, was classified as a male.\r\n\r\n\r\n\r\nIn addition, what's really weird are the AIME I statistics.\r\n\r\nApparently, it says that there was 2 sophomores who got 10s, no sophomores with 9's, and 1 sophomore with an 8. So that means at most 3 people made the USAMO who are sophomores.\r\n\r\nFlorida Results for sophomores in AIME I are:\r\n\r\n10's: Rui Yang, Gaku, Liu, Wendy Hou\r\n9's: Chenyu Lin, Saiem Gilani\r\n8's: Marcel Nunez\r\n\r\nSo the 8 is right, and Wendy was included as a female 10. However, only one male, either gaku or rui, was reported as a 10, and neither myself nor saiem were accounted for. Marcel was reported, however.\r\n\r\nSince all 6 of us made the USAMO, there must be some problem with the statistics.", "Solution_1": "The statistics, or the bubbling skill of the participants? I forgot to bubble my gender on the AMC10B.", "Solution_2": "yea I remember I was specifically instructed at some point not to bubble in grade, gender, age as there wasn't any point", "Solution_3": "for the AIME, I'm positive I bubbled in right.\r\n\r\nBy looking at the USAMO qualifiers list and the AIME statistics, the two conflicts.", "Solution_4": "Aw, cut them some slack...", "Solution_5": "Seriously, I think people have taken it a bit to the extreme. The AMC is trying their best, I'm sure they will realize their mistakes and fix them as soon as possible. It's not easy running a competition of this size! I for one am thankful for all the work they put it to keep this set of competitions the best and most prestigious of all :)" } { "Tag": [ "algebra", "polynomial", "Vieta", "number theory proposed", "number theory" ], "Problem": "Show that there exist integers $ j,k,l,m,n$ greater than $ 100$ such that:\r\n\r\n$ j^2\\plus{}k^2\\plus{}l^2\\plus{}m^2\\plus{}n^2\\equal{}jklmn\\minus{}12.$", "Solution_1": "It has a solution (2,2,2,2,2) and we can see the equation about j.\r\n\r\nAnd we can use vieta jumping\r\nThen, (2,2,2,2,2^4-2) also can be a solution\r\n\r\nrepeat it.", "Solution_2": "we can take j=k=l=m=n & find j = 2\n\nSo initial seed of solution is (2,2,2,2,2)\n\nWLOG 2 <= j <= k <= l <= m <= n\n\nUsing vieta we can see (i,j) is solution for quadratic equation and i = klmn - j >= 8n -j >= 7n so new root is bigger than \"n\"\n\nSo (klmn-j,k,l,m,n) is solution , because of symmetry and root flipping (k,l,m,n, klmn-j) is the solution.. Applying on this seed \n\n(2,2,2,2,2) ==> ( 2,2,2,2,14) ==> (2,2,2,14,110) == > (2,2,14,110, 6158) ...... we can easily see infinitely many solution beyond 100 can be constructed using this path\n" } { "Tag": [ "MATHCOUNTS", "Ring Theory" ], "Problem": "Does anyone know(if there is a place) where I can find a timeline for mathcounts? i.e. chapter, state, and national competition dates? Because there's stuff I'm thinking about signing up for a field trip and history day and I want to make sure it wouldn't clash with anything.", "Solution_1": "Well, it depends on your chapter and state, so you have to contact them. And since the National Location is TBA, I think it's date is too.", "Solution_2": "[quote=\"maybach\"] And since the National Location is TBA, I think it's date is too.[/quote]\r\n\r\nIt's in Orlando.", "Solution_3": "oh. I thought that was unofficial and could change........\r\n\r\nBut check with your state tennis123", "Solution_4": "Look [url=https://mathcounts.org/Page.aspx?pid=294]here[/url] for important dates (yes, nationals are in Orlando May 6-9). You can check [url=http://mathcounts.org/Document.Doc?id=411]here[/url] to see if your school has registered. When registration closes and it gets close to the State Competition, they will post its time, date, and place on the MC website.\r\n\r\nEDIT: Sorry, fixed the link.", "Solution_5": "Wait, Orlando again?", "Solution_6": "Yes. I think it might be for 2011 too. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=294840]this[/url] thread." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "parallelogram", "vector", "arithmetic sequence", "Pythagorean Theorem" ], "Problem": "A trirectangular tetrahedron has the property that the areas of its four faces are in arithmetic progression and the smallest has area 4 + 3:rt2:. Compute the volume of the tetrahedron.\r\n\r\n(A trirectangular tetrahedron is one in which the three edges emanating from one of the vertices are all perpendicular to each other.)\r\n\r\n(Thanks for the catch, Fiery. I copied this from Intermediate since no one there was doing it.)", "Solution_1": "\"area 4+3\"? typo there?", "Solution_2": "Just a first observation: the area of the face which isn't a right triangle is the largest. I hope I'll be back for more :D", "Solution_3": "Really, this isn't a \"nice\" problem. You have to know something about trirectangular tetrahedra in order to make it not a dreadful mess to do.", "Solution_4": "I think the 4+3 :rt2: side length scared people away. :)\r\n\r\nBy de Gua's Theorem, we know that the sum of the squares of the three faces adjacent to the trihedral angle equals the square of the other face, so that\r\n\r\n(4 + 3*Sqrt[2])^2 + (4 + 3*Sqrt[2] + d)^2 + (4 + 3*Sqrt[2] + 2d)^2 = (4 + 3*Sqrt[2] + 3d)^2\r\n\r\nSolving, we find d=3+2 :rt2: \r\n\r\nThe sides are thus 4+3 :rt2: , 7+5 :rt2: , 10+7 :rt2: , so\r\n\r\nV = (1/6)(4+3 :rt2: )( 7+5 :rt2: )( 10+7 :rt2:)\r\nV = 577/3 + 136*Sqrt(2)\r\n\r\nBy the way, what is the source of the problem?", "Solution_5": "How had you heard of de Gua's Theorem? I hadn't heard of it until you mentioned it.", "Solution_6": "The source of the problem is my own twisted mind. I hadn't ever heard of de Gua's theorem by name, but it was given to me as a fun excercize at one point, and I think I may have posted a question about it on the board. (Is there any deeper reason relating to Pythagorean Theorem that this should be true? And consensus seemed to be not.)", "Solution_7": "I was wondering about a deeper reason relating to the Pythagorean Theorem as well. I imagine that there should be, but I don't see it immediately. Heron's Theorem just seems too messy.", "Solution_8": "Well, the proof of de Gua's is pretty straightforward (directly from Heron's, though some algebra required). I actually looked up the theorem first before proving it though, since Joel mentioned:\r\n\r\n[quote]You have to know something about trirectangular tetrahedra in order to make it not a dreadful mess to do.[/quote]", "Solution_9": "A faster way to prove it is to use the fact that the area of a parallelogram spanned by vectors v1 and v2 is ||v1 x v2||. So the area of a triangle with those vectors as sides is 1/2*||v1 x v2||. The theorem is not hard to prove from there.\r\n\r\nI do, however, believe that there's an intuitive arguement for why it is true. I seem to recall discovering it myself when I was in high school - before I knew about Heron's Formula or cross products. But I can't for the life of me remember how.\r\n\r\nNice problem by the way Joel!", "Solution_10": "By the way, why did you choose an area of 4+3 :rt2: ? Why not just like an area of 2 or something? :)", "Solution_11": "Because I checked it first with an area of 1, and you get a whole bunch of things like :sqrt:(4 + 3 :rt2:) or some such thing, and by multiplying all the edge lengths through I could make it go away, with the downside of multiplying the area by something not too pretty to get that.\r\n\r\nI just proved it using Hero's and areas -- it's pretty straighforward if a bit messy. I don't know how else one would prove it.", "Solution_12": "Try it via the way I suggested above JBL - using cross products. It turns out very nicely.\r\n\r\nI think I remember the way that I found it in high school. I think I expressed the volume of the tetrahedron in different ways, using different triangles as bases, and set them equal or manipulated them in some way. The key is to find the altitude from that face to the origin." } { "Tag": [ "inequalities" ], "Problem": "Shown as the following,", "Solution_1": "As an expression for $ T_n$, I have worked out:\r\n[hide]$ T_n \\equal{} \\frac {[(n \\plus{} 1)!]^4(2^{4n \\plus{} 2})}{[(2n \\plus{} 2)!]^2(n \\plus{} 1)}$\n\nI don't know how to continue if I try to use this.\n\nThe algebra for getting this formula could be shown at a later time. (EDIT)\n\n[/hide]", "Solution_2": "Denoting \r\n\\[ T{'_n} \\equal{} \\left( {1 \\minus{} \\frac{1}{{{2^2}}}} \\right)\\left( {1 \\minus{} \\frac{1}{{{4^2}}}} \\right) \\ldots \\left( {1 \\minus{} \\frac{1}{{{{\\left( {2n} \\right)}^2}}}} \\right)\\]\r\n\\[ T'{'_n} \\equal{} \\left( {1 \\minus{} \\frac{1}{{{4^2}}}} \\right)\\left( {1 \\minus{} \\frac{1}{{{6^2}}}} \\right) \\ldots \\left( {1 \\minus{} \\frac{1}{{{{\\left( {2n \\plus{} 2} \\right)}^2}}}} \\right)\\]\r\nWe have $ T{'_n}=2(ab+bc+ca)", "Solution_2": "any idea I am waiting for its solution Thanks", "Solution_3": "any one helps me ?thanks :?:", "Solution_4": "[quote=\"stevenwang1992\"]any one helps me ?thanks :?:[/quote]\r\ni think you should learn LATEX before you post problem ! :blush:", "Solution_5": "I guess so too.But I really donnot have time. :blush:", "Solution_6": "[quote=\"stevenwang1992\"]How to prove $ \\sum\\frac {a}{a \\plus{} \\sqrt {a^2 \\plus{} 3bc}}\\leq1$ where $ a, b, c$ are positive[/quote]$ 1\\geq\\sum\\frac {a}{\\sqrt {a^2 \\plus{} 3bc} \\plus{} a} \\equal{} \\sum\\frac {a(\\sqrt {a^2 \\plus{} 3bc} \\minus{} a)}{3bc}\\Longleftrightarrow$\r\n\r\n$ \\sum{a^2\\sqrt {a^2 \\plus{} 3bc}}\\leq a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3abc.$\r\n\r\nUsing Cauchy inequlity, we have\r\n\r\n$ \\left(\\sum{a^2\\sqrt {a^2 \\plus{} 3bc}}\\right)^2\\leq \\sum{\\frac {a^2}{b^2 \\plus{} c^2 \\minus{} bc \\plus{} ca \\plus{} ab}}\\sum{a^2(a^2 \\plus{} 3bc)(b^2 \\plus{} c^2 \\minus{} bc \\plus{} ca \\plus{} ab)}$\r\n\r\n$ \\equal{} (a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3abc)^2 \\minus{} \\frac {a^2b^2c^2(a \\plus{} b \\plus{} c)^2(a^2 \\plus{} b^2 \\plus{} c^2)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} ca \\minus{} ab)}{\\prod{(b^2 \\plus{} c^2 \\minus{} bc \\plus{} ca \\plus{} ab)}}$\r\n\r\n$ \\leq (a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3abc)^2.$", "Solution_7": "It's my old problem. \r\n[quote=\"Ji Chen\"]\nUsing Cauchy inequlity, we have\n\n$ \\left(\\sum{a^2\\sqrt {a^2 \\plus{} 3bc}}\\right)^2\\leq \\sum{\\frac {a^2}{b^2 \\plus{} c^2 \\minus{} bc \\plus{} ca \\plus{} ab}}\\sum{a^2(a^2 \\plus{} 3bc)(b^2 \\plus{} c^2 \\minus{} bc \\plus{} ca \\plus{} ab)}$\n\n$ \\equal{} (a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3abc)^2 \\minus{} \\frac {a^2b^2c^2(a \\plus{} b \\plus{} c)^2(a^2 \\plus{} b^2 \\plus{} c^2)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} ca \\minus{} ab)}{\\prod{(b^2 \\plus{} c^2 \\minus{} bc \\plus{} ca \\plus{} ab)}}$\n[/quote]\r\nVeru nice identity, Jichen. Thank you very much for your nice proof :lol:", "Solution_8": "[quote=\"stevenwang1992\"]How to prove $ \\sum_{cyc}\\frac {a}{a \\plus{} \\sqrt {a^2 \\plus{} 3bc}}\\leq1$ where a b c are positive[/quote]\r\nHere is my proof:\r\n$ \\sum_{cyc}\\frac {a}{a \\plus{} \\sqrt {a^2 \\plus{} 3bc}}\\leq1\\Leftrightarrow\\sum_{cyc}a^2\\sqrt {a^2 \\plus{} 3bc}\\leq a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3abc\\Leftrightarrow$( after squaring ) \r\n$ \\Leftrightarrow2\\sum_{cyc}a^2b^2\\sqrt {(a^2 \\plus{} 3bc)(b^2 \\plus{} 3ac)}\\leq\\sum_{cyc}(2a^3b^3 \\plus{} 3a^4bc \\plus{} 3a^2b^2c^2)\\Leftrightarrow$( after squaring )\r\n$ \\Leftrightarrow\\sum_{cyc}(9a^6 \\plus{} 62a^4bc \\plus{} 18a^3b^3 \\plus{} 39a^2b^2c^2)\\geq$\r\n$ \\geq8\\sum_{cyc}a^2(a^2 \\plus{} 3bc)\\sqrt {(b^2 \\plus{} 3ac)(c^2 \\plus{} 3ab)}.$\r\nBut $ 8\\sum_{cyc}a^2(a^2 \\plus{} 3bc)\\sqrt {(b^2 \\plus{} 3ac)(c^2 \\plus{} 3ab)}\\leq4\\sum_{cyc}a^2(a^2 \\plus{} 3bc)(b^2 \\plus{} c^2 \\plus{} 3ab \\plus{} 3ac).$\r\nHence, it remains to prove that\r\n$ \\sum_{cyc}(9a^6 \\plus{} 62a^4bc \\plus{} 18a^3b^3 \\plus{} 39a^2b^2c^2)\\geq$\r\n$ \\geq4\\sum_{cyc}a^2(a^2 \\plus{} 3bc)(b^2 \\plus{} c^2 \\plus{} 3ab \\plus{} 3ac).$\r\nBut $ \\sum_{cyc}(9a^6 \\plus{} 62a^4bc \\plus{} 18a^3b^3 \\plus{} 39a^2b^2c^2)\\geq$\r\n$ \\geq4\\sum_{cyc}a^2(a^2 \\plus{} 3bc)(b^2 \\plus{} c^2 \\plus{} 3ab \\plus{} 3ac)\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{sym}(4.5a^6 \\minus{} 12a^5b \\minus{} 4a^4b^2 \\plus{} 9a^3b^3 \\plus{} 31a^4bc \\minus{} 48a^3b^2c \\plus{} 19.5a^2b^2c^2)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow6\\left(\\sum_{cyc}(a^3 \\minus{} a^2b \\minus{} a^2c \\plus{} abc)\\right)^2 \\plus{}$\r\n$ \\plus{} \\sum_{sym}(1.5a^6 \\plus{} 2a^4b^2 \\minus{} 3a^3b^3 \\plus{} 7a^4bc \\minus{} 12a^3b^2c \\plus{} 4.5a^2b^2c^2)\\geq0,$ which is obvious.", "Solution_9": "Dear Ji Chen,\r\n\r\nI would like to know, if possible, how you found this wonderful formula.\r\n\r\nDid you use a special algorithm by computer?\r\n\r\n\r\nThank you very much. :)\r\n\r\n[quote=\"Ji Chen\"]$ \\sum{\\frac {a^2}{b^2 \\plus{} c^2 \\minus{} bc \\plus{} ca \\plus{} ab}}\\sum{a^2(a^2 \\plus{} 3bc)(b^2 \\plus{} c^2 \\minus{} bc \\plus{} ca \\plus{} ab)}$\n\n$ \\equal{} (a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3abc)^2 \\minus{} \\frac {a^2b^2c^2(a \\plus{} b \\plus{} c)^2(a^2 \\plus{} b^2 \\plus{} c^2)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} bc \\minus{} ca \\minus{} ab)}{\\prod{(b^2 \\plus{} c^2 \\minus{} bc \\plus{} ca \\plus{} ab)}}$\n\n[/quote]", "Solution_10": "Is there a simple way to do it like sos?thanks :blush:", "Solution_11": "Dear arqady,\r\n\r\nI checked your proof and it is correct but also NEAT, CLEAR and ELEGANT :wink: \r\n\r\nCongratulations for this proof of a very hard inequality :) \r\n\r\n[quote=\"arqady\"][quote=\"stevenwang1992\"]How to prove $ \\sum_{cyc}\\frac {a}{a \\plus{} \\sqrt {a^2 \\plus{} 3bc}}\\leq1$ where a b c are positive[/quote]\nHere is my proof:\n$ \\sum_{cyc}\\frac {a}{a \\plus{} \\sqrt {a^2 \\plus{} 3bc}}\\leq1\\Leftrightarrow\\sum_{cyc}a^2\\sqrt {a^2 \\plus{} 3bc}\\leq a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3abc\\Leftrightarrow$( after squaring ) \n$ \\Leftrightarrow2\\sum_{cyc}a^2b^2\\sqrt {(a^2 \\plus{} 3bc)(b^2 \\plus{} 3ac)}\\leq\\sum_{cyc}(2a^3b^3 \\plus{} 3a^4bc \\plus{} 3a^2b^2c^2)\\Leftrightarrow$( after squaring )\n$ \\Leftrightarrow\\sum_{cyc}(9a^6 \\plus{} 62a^4bc \\plus{} 18a^3b^3 \\plus{} 39a^2b^2c^2)\\geq$\n$ \\geq8\\sum_{cyc}a^2(a^2 \\plus{} 3bc)\\sqrt {(b^2 \\plus{} 3ac)(c^2 \\plus{} 3ab)}.$\nBut $ 8\\sum_{cyc}a^2(a^2 \\plus{} 3bc)\\sqrt {(b^2 \\plus{} 3ac)(c^2 \\plus{} 3ab)}\\leq4\\sum_{cyc}a^2(a^2 \\plus{} 3bc)(b^2 \\plus{} c^2 \\plus{} 3ab \\plus{} 3ac).$\nHence, it remains to prove that\n$ \\sum_{cyc}(9a^6 \\plus{} 62a^4bc \\plus{} 18a^3b^3 \\plus{} 39a^2b^2c^2)\\geq$\n$ \\geq4\\sum_{cyc}a^2(a^2 \\plus{} 3bc)(b^2 \\plus{} c^2 \\plus{} 3ab \\plus{} 3ac).$\nBut $ \\sum_{cyc}(9a^6 \\plus{} 62a^4bc \\plus{} 18a^3b^3 \\plus{} 39a^2b^2c^2)\\geq$\n$ \\geq4\\sum_{cyc}a^2(a^2 \\plus{} 3bc)(b^2 \\plus{} c^2 \\plus{} 3ab \\plus{} 3ac)\\Leftrightarrow$\n$ \\Leftrightarrow\\sum_{sym}(4.5a^6 \\minus{} 12a^5b \\minus{} 4a^4b^2 \\plus{} 9a^3b^3 \\plus{} 31a^4bc \\minus{} 48a^3b^2c \\plus{} 19.5a^2b^2c^2)\\geq0\\Leftrightarrow$\n$ \\Leftrightarrow6\\left(\\sum_{cyc}(a^3 \\minus{} a^2b \\minus{} a^2c \\plus{} abc)\\right)^2 \\plus{}$\n$ \\plus{} \\sum_{sym}(1.5a^6 \\plus{} 2a^4b^2 \\minus{} 3a^3b^3 \\plus{} 7a^4bc \\minus{} 12a^3b^2c \\plus{} 4.5a^2b^2c^2)\\geq0,$ which is obvious.[/quote]", "Solution_12": "[quote=\"Ji Chen\"][quote=\"stevenwang1992\"]How to prove $ \\sum\\frac {a}{a \\plus{} \\sqrt {a^2 \\plus{} 3bc}}\\leq1$ where $ a, b, c$ are positive[/quote][/quote]\r\nThe following is my proof:\r\n\r\nLet $ x\\equal{}\\frac{a}{a\\plus{}\\sqrt{a^2\\plus{}3bc}},y\\equal{}\\frac{b}{b\\plus{}\\sqrt{b^2\\plus{}3ca}},z\\equal{}\\frac{c}{c\\plus{}\\sqrt{c^2\\plus{}3ab}}.$ Then $ x,y,z <\\frac{1}{2},$ and we find that $ \\frac{bc}{a^2}\\equal{}\\frac{1\\minus{}2x}{3x^2},\\frac{ca}{b^2}\\equal{}\\frac{1\\minus{}2y}{3y^2},\\frac{ab}{c^2}\\equal{}\\frac{1\\minus{}2z}{3z^2},$ from which it follows that\r\n$ (1\\minus{}2x)(1\\minus{}2y)(1\\minus{}2z) \\equal{}27x^2y^2z^2.$\r\nWe are required to prove $ x\\plus{}y\\plus{}z \\le 1.$ The method we choose to do this work is contradiction. Let us suppose that there exists some positive numbers $ x,y,z <\\frac{1}{2}$ so that $ x\\plus{}y\\plus{}z>1.$ By this assumption, we have\r\n$ (1\\minus{}2x)(1\\minus{}2y)(1\\minus{}2z) <[(x\\plus{}y\\plus{}z)\\minus{}2x][(x\\plus{}y\\plus{}z)\\minus{}2y][(x\\plus{}y\\plus{}z)\\minus{}2z]\\equal{}(x\\plus{}y\\minus{}z)(y\\plus{}z\\minus{}x)(z\\plus{}x\\minus{}y)<(x\\plus{}y\\minus{}z)(y\\plus{}z\\minus{}x)(z\\plus{}x\\minus{}y)(x\\plus{}y\\plus{}z)^3 \\equal{}[2(x^2y^2\\plus{}y^2z^2\\plus{}z^2x^2)\\minus{}(x^4\\plus{}y^4\\plus{}z^4)](x\\plus{}y\\plus{}z)^2 \\le 3[2(x^2y^2\\plus{}y^2z^2\\plus{}z^2x^2)\\minus{}x^4\\minus{}y^4\\minus{}z^4](x^2\\plus{}y^2\\plus{}z^2).$\r\nOn the other hand, from the Schur's Inequality (the special case third degree), we have\r\n$ [2(x^2y^2\\plus{}y^2z^2\\plus{}z^2x^2)\\minus{}(x^4\\plus{}y^4\\plus{}z^4)](x^2\\plus{}y^2\\plus{}z^2) \\le 9x^2y^2z^2,$\r\nso from the above estimation, we get\r\n$ (1\\minus{}2x)(1\\minus{}2y)(1\\minus{}2z)<27x^2y^2z^2,$\r\nwhich is contradiction, and hence our proof is completed.", "Solution_13": "[quote=\"manlio\"]Dear arqady,\n\nI checked your proof and it is correct but also NEAT, CLEAR and ELEGANT :wink: \n\nCongratulations for this proof of a very hard inequality :) \n\n[/quote]\r\nThank you!", "Solution_14": "[i]A very good solution , Can , Congratulation . :lol: [/i]", "Solution_15": "[img]http://s13.sinaimg.cn/middle/006ptkjAzy74X63CkTqbc&690[/img]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Alice and Bob play a game on a 6 by 6 grid. On his or her turn, a player chooses a rational number not yet appearing in the grid and writes it in an empty square of the grid. Alice goes first and then the players alternate. When all squares have numbers written in them, in each row, the square with the greatest number in that row is colored black. Alice wins if she can then draw a line from the top of the grid to the bottom of the grid that stays in black squares, and Bob wins if she can't. (If two squares share a vertex, Alice can draw a line from one to the other that stays in those two squares.) Find, with proof, a winning strategy for one of the players.", "Solution_1": "I am not sure, but I think I solved it :\r\nWe will not allow the first player to pass the second row.\r\nWe take the numbering of collumns \r\n0 1 2 3 4 5\r\n\r\nNow put the numbers \r\n0 1 2 3 4 5\r\n4 5 0 1 2 3\r\nIn the first 2 rows of another table.\r\nAny time the first player puts a number in one of x, we put a number in the other x so that the relationship (bigger, smaller ) between the same numbers in both rows is same. Because nr are rational, we can do it.\r\nNow, obviously, biggest number will not be neighbours :P", "Solution_2": "[quote=\"DusT\"]\nNow, obviously, biggest number will not be neighbours :P[/quote]\r\n\r\nObviously?\r\n\r\n$\\begin{array}{cccccc}\r\n0&2&4&6&8&\\mathbf{10}\\\\\r\n9&11&1&3&5&\\mathbf{15}\r\n\\end{array}&$", "Solution_3": "Here's a way:\r\n\r\nDenote columns as c1, c2, c3, c4, c5, c6.\r\n\r\nShow that Bob has the ability to do one of the following in any row:\r\n(the basic strategy for Bob is to always place numbers in the row Alice put hers)\r\n#1 Force the black square to be in one of c1, c2, or c3.\r\n#2 Force the black square to be in one of c4, c5, or c6.\r\n#3 Force the black square to be in one of c1, c2, c5, c6.\r\n\r\nNow let Bob execute #1 on row 1, #3 on row 2, and #2 on row 3.\r\n\r\nIf, in the second row, black square is in one of c1, c2, then it cannot be connected to row 3. (case 1)\r\n\r\nIf, in the second row, black square is in one of c5, c6, then it cannot be connected to row 1. (case 2)\r\n\r\nIn either case, Bob wins.\r\n\r\n_: Grid boxes\r\nX: Possible locations of black square\r\n\r\nCase 1:\r\nX X X_ _ _\r\nX X _ _ _ _\r\n_ _ _ X X X\r\n\r\nCase 2:\r\nX X X_ _ _\r\n_ _ _ _ X X\r\n_ _ _ X X X\r\n\r\nSo in fact, Bob wins in a 6x3 grid.", "Solution_4": "Your solution is very similar to my one. Only my one is based on more explicit results:\r\n1. Bob can force the maximum in a row to lie in a group of 3 squares he chooses.\r\n2 Bob reaches the maximum in row ! to be on c1,c2,c3, on row 2 on c4,c5,c6, on row 3 on c1,c2,c6.\r\nthis is his winning strategy", "Solution_5": "Yeah, obiously!\r\nReply tu lugash\r\n0 2 4 6 8 10\r\n9 11 1 3 5 15\r\nThe second array is not sorted in the same way.\r\nIt should be\r\n0 2 4 6 8 10\r\n11 15 1 3 5 9", "Solution_6": "Hmm... DusT\r\n\r\nHow do you take care of cases in which Alice places her first number in the first row but the second number the second row, so that Bob cannot always place his number according to \r\n\r\n012345\r\n450123\r\n\r\n?\r\n\r\nFor example, let's say the first two moves are as follows:\r\n\r\n_ _ 2 _ _ _\r\n_ _ _ _ 3 _\r\n_ _ _ _ _ _\r\n\r\n(Alice places 2, Bob places 3)\r\n\r\n_ _ 2 _ _ _\r\n1 _ _ _ 3 _\r\n_ _ 4 _ _ _\r\n\r\n(Alices places 1, Bob places 4)\r\n\r\n_ _ 2 _ 5 _\r\n1 _ _ _ 3 _\r\n_ _ 4 _ _ _\r\n\r\nNow Alice places 5, but Bob's counterspot is already taken.", "Solution_7": "Sorry if I explain wrong , but I mean whean Alice puts a number in square x, Bob puts the number in the other x, so that the arrangement of the numbers remains the same :P\r\nHope no it's clear", "Solution_8": "Yes, but my point was that what if the other \"x\" is already taken?", "Solution_9": "Isn't it obvious that there are only 2 letters of the same type, so when the 1st player takes one, we take the other, so either both are taken, or none is taken.\r\nNow, it is solved I think :)", "Solution_10": "Isn't this sorta obvious? Bob always makes the last move. He knows which square will be black in every row except the one in which the last empty square is. If he sees that this square will complete the line, he can always choose a sufficiently small number. If it won't, then we can always choose a sufficiently large number.", "Solution_11": "No, that doesn't work -- given the black squares in the first five rows, there might be more than one possible position for a black square in the sixth row that Alice can use to draw her line.", "Solution_12": "Er, how? Can you give an example?", "Solution_13": "I think that the confusion here is that a \"line\" doesn't mean a straight line. It means a path from the top to the bottom where adjacent squares share a vertex.", "Solution_14": "If the first square in rows 1 through 5 and the second square in row 6 is colored black, there exists a line (in the Euclidean sense) that connects the top edge to the bottom edge and passes only though black squares.", "Solution_15": "Is it necessary for each player to choose a rational number??????", "Solution_16": "[hide=\"Slightly different\"]\nWe claim in each row, Bob can choose, prior to the game's beginning, three \"desired\" squares such that one of those will turn out black. To do this, if he plays in the same row as Alice every turn. If Alice plays in one of the desired black squares, he plays a smaller number in an undesired square. If she plays in an un-desired square, he plays in a desired one.\n\nThen, using chessboard notation, have Bob desire a1, a2, a3, b1, b2, b6, c4, c5, c6. Clearly, Bob wins.[/hide]", "Solution_17": "I think I am interpreting this problem wrong or this is trivial. Kindly clarify:\n\nThe grid has 36 (even) number of squares so Bob will make the last move. Clearly he can fill the number such that if the greatest numbers of the other rows lie in that column, he can choose a sufficiently small number. If not, a sufficiently large number suffices. ", "Solution_18": "In the 6 by 6 grid, shade in the first 3 squares in the first and fourth rows, the first, 5th, and 6th squares in the 2nd and 5th rows, and the 3rd, 4th, and 5th squares in the 3rd and 6th rows. If the black squares were located in any of these shaded squares, a path wouldn't exist. Hence, Bob can win if he can force the black squares to be in these shaded squares. To do this, if Alice writes a number in a shaded square, Bob writes a smaller number in an unshaded square in the same row. If Alice writes a number in an unshaded square, but writes a larger number in a shaded square in the same row. ", "Solution_19": "[quote=billzhao]Alice and Bob play a game on a 6 by 6 grid. On his or her turn, a player chooses a rational number not yet appearing in the grid and writes it in an empty square of the grid. Alice goes first and then the players alternate. When all squares have numbers written in them, in each row, the square with the greatest number in that row is colored black. Alice wins if she can then draw a line from the top of the grid to the bottom of the grid that stays in black squares, and Bob wins if she can't. (If two squares share a vertex, Alice can draw a line from one to the other that stays in those two squares.) Find, with proof, a winning strategy for one of the players.[/quote]\n\n\nBob wins no matter what.\n(Refer to the figure)Bob's strategy is to make sure that no two black squares share a vertice.He puts numbers in any row in an increasing order.\n\n$\\;\\;\\;\\;\\; \\bullet $ If Alice decides to follow along,by trying to disrupt Bob's strategy,then Bob starts going for $3 \\cdot 3$ squares,there are $14$ of these.He changes the Corner's rather than the middle if Alice has fixed the middle and if Alice hasn't fixed the middle he gors for the middle.\nAlice can't disrupt it;there are an even number of squares i.e Bob finishes it.\n\n$\\;\\;\\;\\;\\; \\bullet $ If Alice decides not to play along then Bob goes for squares of the same parity;this is a clever strategy i.e Alice can't give Bob a last minute surprise;If after some point she decides to play along then Bob is safe;All the odd parity/even parity squares are reeserved so in order to move to the bottom of the board Alice would need to use an square of the parity Bob choose but this is impossible since all of these are white squares.", "Solution_20": "not sure if this works but any line going through the top and to the bottom has to pass through only one square in at least one row\n\nalso bob can make sure that his turns are the turns that color squares black\n\nso on each of these turns he has two choices for what the black square will be\n\nand im pretty sure this implies bob will always win", "Solution_21": "[hide = Solution] This took me far longer than it should have.\n\nI claim that Bob can always win.\nConsider the $2 \\times 6$ board below: \n\\[\n\\begin{bmatrix}\n1 & 2 & 3 & 4 & 5 & 6 \\\\\n4 & 5 & 6 & 1 & 2 & 3\n\\end{bmatrix}.\n\\]\nThe strategy is that if Alice colors $x$ Bob colors the other $x$. We then have that with this coloring Alice may never obtain a path where two black squares share a vertex. We then have for the remaining $4 \\times 6$ portion of the board, Bob may play in the same way. $\\blacksquare$ [/hide]", "Solution_22": "We claim that Bob always wins. Bob's winning strategy is to always place in the same row as Alice. Consider any row, and we claim that Bob can limit the black square's position to three of these squares. If Alice places a number in one of the squares Bob wants black, then Bob can place a smaller number in a square he wants white, and vice versa. Thus, our claim is true.\n\nBob can limit row 1's to the rightmost three columns, row 2's to the leftmost two columns and rightmost column, and row 3's to the leftmost three columns. Starting from the top row, there is no way to draw a line that gets past row $3,$ so we're done.", "Solution_23": "Divide the board into 12 \"zones\", each of which is a set of 3 (not necessarily connected) cells, which we call $A1,A2,B1,B2,C1,C2,D1,D2,E1,E2,F1,F2$, in the following manner:\n\n$$\\begin{matrix}\nA1 & A1 & A1 & A2 & A2 & A2\\\\\nB2 & B2 & B2 & B1 & B1 & B1\\\\\nC1 & C1 & C2 & C2 & C2 & C1\\\\\nD1 & D1 & D1 & D2 & D2 & D2\\\\\nE2 & E2 & E2 & E1 & E1 & E1\\\\\nF1 & F1 & F2 & F2 & F2 & F1\\\\\n\n\\end{matrix}$$\n\nBob can then use the following strategy:\\\\\n\nIf Alice writes a number in a \"1\" zone, Bob comes up with a number smaller than all numbers on the board (which is always possible) and writes it in a \"2\" zone of the same letter as Alice's zone.\\\\\nIf Alice writes a number in a \"2\" zone, Bob comes up with a number larger than all numbers on the board (which is again possible) and writes it in a \"1\" zone of the same letter as Alice's zone.\\\\\n\nNote that the only time when the largest number in a row can be in a \"2\" zone is right after Alice writes a number in that zone, but using this strategy Bob will immediately overturn that by placing a larger number in the corresponding \"1\" zone. Thus, by the end of the process, all black cells are in \"1\" zones. This clearly prevents Alice from winning, so Bob wins are we are done.\n\n", "Solution_24": "We claim that $\\boxed{\\text{Bob}}$ is the winner. \n\nThe key in this problem is that Bob can always prevent Alice from taking any $3$ squares he chooses. Indeed say on one of Alice's turns she plays in a square he wishes to be white. Then he can simply play a higher number in a square that he wishes to be black and so on. Now the finish is obvious. Restrict Alice to the square $(1, 2, 3)$ in row $1$. On row $2$ restrict Alice to the squares $(1, 5, 6)$. Then on the third row restrict her to the square $(4, 5, 6)$. Clearly Alice cannot win." } { "Tag": [], "Problem": "Let a,b is the positive integer numbers, (a,b)=1. Prove that for all number greater than ab-a-b can be expressed in the form ax+by(x,y are positive integer numbers) :D", "Solution_1": "Hi TRAN THAI HUNG. See here: http://en.wikipedia.org/wiki/Coin_problem\r\n\r\nand here: http://www.math.ups.edu/~mspivey/FrobFinal.pdf\r\n\r\nHope this is helpful.", "Solution_2": "Thanks a lot for your problem, dear friend.\r\n\r\nIn fact, a friend of mine gave me another similar problem about a month ago:\r\n$ a,b\\in\\mathbb N; a > b$. Define $ S \\equal{} \\{ax \\plus{} by \\ | \\ x,y\\in\\mathbb N_0\\}$\r\nIf $ S$ misses exactly $ 35$ integers [b]including[/b] $ 58$; what are the values of $ a$ and $ b$?" } { "Tag": [ "number theory" ], "Problem": "what is the remainder when 3^25586 is divided by 7?", "Solution_1": "powers of 3 (mod 7)\r\n\r\n$3^1 = 3$ (mod 7)\r\n$3^2 = 2$ (mod 7)\r\n$3^3 = 6$ (mod 7)\r\n$3^4 = 4$ (mod 7)\r\n$3^5 = 5$ (mod 7)\r\n$3^6 = 1$ (mod 7)\r\n$3^7 = 3$ (mod 7)\r\n\r\nand the pattern repeats.\r\n\r\n25586= 2 (mod 6)\r\n\r\nand so\r\n\r\n$3^{25586} = 3^2 = 2$ (mod 7)\r\n\r\nso the answer is 2", "Solution_2": "What about solving it without using mod 7?", "Solution_3": "you could always do this:\r\n\r\n(3^0)/7= remainder 1\r\n(3^1)/7= remainder 3\r\n(3^2)/7= remainder 2\r\n(3^3)/7= remainder 6\r\n(3^4)/7= remainder 4\r\n(3^5)/7= remainder 5\r\n\r\n\r\nand just keep going and find a pattern.\r\nit's pretty much the same thing as using mods, but it doesn't look as scary.", "Solution_4": "Thanks. It almost the same, but is better for people that don't know mods... that's what I wanted.... thanks so much!", "Solution_5": "Using Fermat's Little Theorem, your life becomes a lot easier :lol: :\r\n3^6 congruent to 1 (mod 7)\r\nRaising both sides to the 4264th power,\r\n3^25584 congruent to 1 mod 7\r\nMultiplying both side by 3^2,\r\n3^25586 congruent to 3^2 (mod 7) congruent to 2 (mod 7)\r\nThe answer is 2.", "Solution_6": "I would agree with tim1234133 to get 2 mod 7 so remainder is 2" } { "Tag": [ "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "59. A cyclic group of order 15 has an element $ x$ such that the set $ \\{x^3, x^5, x^9\\}$ has exactly two elements. How many elements are in the set $ \\{x^{13n}: n\\mbox{ is a positive integer} \\}$?\r\n\r\n\r\n60. If $ S$ is a ring with the property that $ s\\equal{}s^2$ for each $ s$ in $ S$, which of the following must be true?\r\n\r\nI. $ s\\plus{}s\\equal{}0$ for each $ s$ in $ S$.\r\n\r\nII. $ (s\\plus{}t)^2 \\equal{} s^2 \\plus{} t^2$ for each $ s,t$ in $ S$.\r\n\r\nIII. $ S$ is commutative.", "Solution_1": "59. Obviously x is not e, so it cannot be $ x^3\\equal{}x^5$ or $ x^5\\equal{}x^9$. Then it must be $ x^3\\equal{}x^9$, ie\r\n$ x^6\\equal{}e$. Thus x must be of order 3 and $ x^{13n}$ can be x, $ x^2$ or e, which are 3 diff. elements.\r\n\r\n60. I. $ s\\plus{}s\\equal{}(s\\plus{}s)^2\\equal{}s^2\\plus{}s^2\\plus{}s^2\\plus{}s^2\\equal{}s\\plus{}s\\plus{}s\\plus{}s$, so it must be $ s\\plus{}s\\equal{}0$\r\nII. $ (s\\plus{}t)^2\\equal{}s\\plus{}t\\equal{}s^2\\plus{}t^2$\r\nIII. $ (s\\plus{}t)^2\\equal{}s^2\\plus{}st\\plus{}ts\\plus{}t^2\\equal{}s^2\\plus{}t^2$. thus we have $ st\\plus{}ts\\equal{}0$, but $ st\\plus{}st\\equal{}0$, so\r\n it follows that $ st\\equal{}ts$\\,.\r\nAll claims are true." } { "Tag": [ "Support" ], "Problem": "State your thoughts. You start the discusssion", "Solution_1": "Global politics so what do we believe? Well I can tell u I firmly believe in democracy and I see the why there is a monarchy but I dont understand why there is a ditartorship. I mean thats like liveing in a war based country that all u do in ur life is train to fight and for what. Its like they are asking to start or get involved in a war. I dont agree with it but that isnt going to help therfore I have to do is accept it. A monarchy I dont totally and completely agree with but its better than a dictatorship. I believe it should be the peoples vote that counts thats why I like a democracy. Now dont ask me to choose between being a democrat, republican, or whig because I refuse to. Im an optimist when it comes to national politics. I will support the president no matter what even though I may not like what hes doing or who he is but I will tell u I tend to favor the republican point of view. The fact is when it comes to politics Im not much of a talker because almost no one shares my point of view but like I say u cant judge a person when u dont know all the facts the same goes for any president or the person precideing over the country, state, ect. u arent in there shoes so u dont know everything they do. I also dont think u should gripe about the president or leader unless u voted then u have that right.", "Solution_2": "aalindsey -- what do you think the difference between a dictatorship and monarchy is? The only thing I can think of is that the kings claim to be inherently higher in social status than the other members of society, and that monarchies tend to come along with established aristocracies.\r\n\r\nAlso, could you please try to work on your spelling and grammar? I have difficulty reading what you write.", "Solution_3": "Have most of you adopted your own philosophies about politics and government, or have your parents corrupted you since Day 1? :)", "Solution_4": "It's interesting.. I find that most people living under democracy tend to get taught that monarchies, dictatorships, communism, etc. are cruel to the people somehow. Yet my parents (who lived in Communist China for 30+ years) worshipped Mao and were satisfied with what they had under Communist rule. \r\n\r\n(Of course, once we immigrated here, their attitudes towards the Communist Party went from reverent to spiteful, but, eh, that's another story.)\r\n\r\nI guess you could always say that the people living under \"oppression\" are unaware that they are indeed being oppressed.. but I hesitate to say that all non-democratic societies are necessarily unfair to their people. I personally am very grateful to be here instead of there, but if I had grown up in China maybe I wouldn't have minded so much.", "Solution_5": "wyra -- what was the social status of your parents in China? That makes a very big difference in how they were treated and what their opinions would have been about the society.\r\n\r\naalindsey -- \"I will support the president no matter what even though I may not like what hes doing ...\"\r\nSo you think that it's better to step aside while your country does something that you find morally repugnant than it is to speak out about it, because it's your country? No thanks -- I'll be at the peace rally.\r\n\r\nwyra again -- \"but I hesitate to say that all non-democratic societies are necessarily unfair to their people.\"\r\nIn addition, I would hesitate to say that democratic societies are ever fair to their people. In order to get fairness, you have to trade away a certain amount of democracy, at least on larger scales. (In small, homogenous situations this is not necessarily the case, but in anything resembling a modern nation I think this is fundamentally true.) \r\n\r\nWhat was it Churchill said? \"Democracy is the worst form of government except for all those others that have been tried.\"\r\n\r\n\r\nBy the way, you might all want to try out http://www.nationstates.net -- it's a very interesting game that essentially lets you guide the policy for a nation. It's a little simplistic sometimes, but it's fun.", "Solution_6": "What you said about social status.. that's very true. \r\n\r\nMy grandpa (mom's dad) actually invented some things for the government. Instead of paying him, of course, they gave him a tiny apartment to live in. My mom got off easy during the Cultural Revolution as well; she tended pigs for a few years. As a high school student (before the CR) she was an active young Communist.. though she shudders now to talk about it.\r\n\r\nMy grandpa (dad's dad) was a \"capitalist,\" so once the Communists came to power my grandpa basically lived in fear of his life. My dad wasn't so much in support of the gov't as he was indifferent I guess. He walked across the eastern coast of China spreading the word of Mao during the CR, then went off to work on the mountainside (coal mines) for a good 6 years. \r\n\r\nAfter that I guess you could say he was a little annoyed with the gov't. \r\n\r\nI meant \"satisfied\" in the sense that they were happy with what they did have -- some food, some water, their lives. Well, alright, I am painting a very bleak picture after all. :| \r\n\r\nAnd yes, not all democracies are kind to their people.. or even have true democratic ideals. I am very tempted to talk about this country, for instance.. but I will control myself. Ahem.\r\n\r\n\r\nBy the way, Joel, this is Linda. (Hi!)", "Solution_7": "Hey! Go figure. Welcome to AOPS :) Now you're following me around online, after showing up in all the classes I visited at Stuy ...\r\n\r\nOkay, so humour has never been my strong suit. What inspired you to join the forum?", "Solution_8": "Hahaha.\r\n\r\nWell Long kept telling me about it in Sept., so I figured I should take a look. Yeah, it took me a few months to finally stop being lazy about it but I like the AoPS forum. A lot of nice problems and people with very nice solutions. \r\n\r\nAnd now that I am dutifully slacking off in my classes, I have plenty of time to spare. :D" } { "Tag": [ "inequalities" ], "Problem": "If $ x < a < 0$ means that $ x$ and $ a$ are numbers such that $ x$ is less than $ a$ and $ a$ is less than zero, then:\r\n\r\n$ \\textbf{(A)}\\ x^2 < ax < 0 \\qquad\\textbf{(B)}\\ x^2 > ax > a^2 \\qquad\\textbf{(C)}\\ x^2 < a^2 < 0$\r\n$ \\textbf{(D)}\\ x^2 > ax\\text{ but }ax < 0 \\qquad\\textbf{(E)}\\ x^2 > a^2\\text{ but }a^2 < 0$", "Solution_1": "clearly B. Since $ a$ and $ x$ are both negative, $ ax$ is positive. We also know $ a^2$ and $ x^2$ are positive, by trivial inequality, and $ x^2 > a^2$ since $ x < a < 0$ clearly gives $ x$ the larger absolute value. Since these two squares are not equal and positive, we then know their geometric mean, $ ax$, is between them. Thus $ B$ is the answer." } { "Tag": [ "inequalities", "geometry", "incenter", "trigonometry", "geometry proposed" ], "Problem": "Given triangle $ABC$ with incenter $I,A'B'C'$ is cevian triangle of $I$ prove that\r\n$IA+IB+IC\\ge 2(IA'+IB'+IC')$", "Solution_1": "after a van aubel and some computations, if i am not mistaken, the inequality is equivalent with : \r\n\r\n$\\sum{\\frac{p-a}{a}\\cdot \\sin{\\frac{A}{2}}}\\geq \\sum{\\sin{\\frac{A}{2}}}$", "Solution_2": "Are you sure? can you show clearly ? and moreover how do you prove your posted inequality?", "Solution_3": "Can you show a solution pohoatza or everybody? actually I made a mistake when I creat this problem, I can prove the weaker inequality,only. Could anybody show a solution? :)", "Solution_4": "I took a look over it, and i realised I did a mistake, here is a complete proof:\r\n\r\nFrom the bisector theorem in $\\triangle{ABA'}$, we have that $\\frac{AI}{IA'}=\\frac{b+c}{a}$, so $IA'=\\frac{a\\cdot IA}{b+c}$.\r\n\r\nBut $AI = \\frac{r}{\\sin{\\frac{A}{2}}}$, so we can rewrite the inequality as:\r\n\r\n$\\sum{\\frac{1}{\\sin{\\frac{A}{2}}}}\\geq \\sum{\\frac{2a}{(b+c)\\sin{\\frac{A}{2}}}}$.\r\n\r\n$\\iff \\sum{\\frac{1}{\\sin{\\frac{A}{2}}}(1-\\frac{2a}{b+c})}\\geq 0$\r\n\r\n$\\iff \\sum{\\frac{1}{\\sin{\\frac{A}{2}}}(\\frac{b+c-2a}{b+c})}\\geq 0$\r\n\r\nThus it is enough to prove that $\\sum{\\frac{b+c-2a}{\\sin{\\frac{A}{2}}}}\\geq 0$, which is [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=533[/url] :)", "Solution_5": "[color=purple] Hi gemath ! This is a long time i don't meet you in Mathlinks and i miss you :P .How are you ? I hope you are sick :D .The above porblem is not hard but it's really a beautiful problem because the proff of mine using geometry and claasic inequality.\n Here is my proff :\n Letting $l_{a}, l_{b},l_{c}$ are three angle bisectors of triangle ABC. In triangle it's easy to check that $IA+ID=l_{a}$ thus your inequality is equipvalent to \n $IA+IB+IC\\ge 2(l_{a}+l_{b}+l_{c}-IA-IB-IC) \\longleftrightarrow IA+IB+IC\\ge\\frac{2}{3}(l_{a}+l_{b}+l_{c})(H)$\n It's not hard to show that $IA=\\sqrt{\\frac{bc(p-a)}{p}}and l_{a}=\\frac{2\\sqrt{bcp (p-a)}}{b+c}$ .Construct two similar relations you can see we need to prove \n $\\frac{\\sqrt{bc(p-a)}+\\sqrt{ca(p-b)}+\\sqrt{ab(p-c)}}{\\sqrt{p}}\\ge \\frac{4\\sqrt{p}}{3}[\\frac{\\sqrt{bc(p-a)}}{b+c}+\\frac{\\sqrt{ca(p-b)}}{c+a}+\\frac{\\sqrt{ba(p-a)}}{b+a}] \\longleftrightarrow \\sqrt{bc(p-a)}+\\sqrt{ca(p-b)}+\\sqrt{ab(p-c)}\\ge\\frac{4p}{3}(\\frac{\\sqrt{bc(p-a)}}{b+c}+\\frac{\\sqrt{ca(p-b)}}{c+a}+\\frac{\\sqrt{ba(p-a)}}{b+a})(2)$\nWithuot loss of generality we assume that $a\\ge b\\ge c$ .According to Steiner-Lehmus theorem we have $l_{a}\\le l_{b}\\le l_{c}$ thus $\\frac{\\sqrt{bc(p-a)}}{b+c}\\le \\frac{\\sqrt{ca(p-b)}}{c+a}\\le \\frac{\\sqrt{ba(p-a)}}{b+a}$ and $b+c \\le c+a \\le a+b$ \n hence applying Trebyshev inequality we get\n$(b+c)\\frac{\\sqrt{bc(p-a)}}{b+c}+(c+a)\\frac{\\sqrt{ca(p-b)}}{c+a}+(a+b)\\frac{\\sqrt{ba(p-a)}}{b+a}\\ge \\frac{1}{3}(a+b+b+c+c+a)(\\frac{\\sqrt{bc(p-a)}}{b+c}+\\frac{\\sqrt{ca(p-b)}}{c+a}+\\frac{\\sqrt{ba(p-a)}}{b+a})$ which is $(2)$. And the proff is finished.\n By similar method we can prove\n $\\frac{1}{l_{a}}+\\frac{1}{l_{b}}+\\frac{1}{l_{c}}\\le \\frac{2}{3r}(\\sin\\frac{A}{2}+\\sin\\frac{B}{2}+\\sin\\frac{C}{2})$ \nwhich is equipvalent to\n$\\frac{1}{l_{a}}+\\frac{1}{l_{b}}+\\frac{1}{l_{c}}\\le\\frac{2}{3}(\\frac{1}{IA}+\\frac{1}{IB}+\\frac{1}{IC}) (K)$\n Check $(H)and (K)$ please ! :D\n I hope you(gemath, pohoatza and other mathlinker..) can help me for this problem\n$\\frac{1}{m_{a}}+\\frac{1}{m_{b}}+\\frac{1}{m_{c}}\\le \\frac{2}{3r}(cosA+cosB+cosC)$\n:D [/color]", "Solution_6": "I think there are some typos in your post, I think that the original inequality with $l_{a}$, $l_{b}$, $l_{c}$, is $\\sum{\\frac{1}{l_{a}}}\\leq \\frac{2}{3r \\cdot \\sum{\\sin{\\frac{A}{2}}}}$, posted by gemath here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=816658#816658[/url].\r\n\r\nNow if in your proposed inequality with $\\frac{1}{m_{a}}$, it is the same typo, and the $\\sum{\\cos{A}}$, is at the denominator, the same method works:\r\n\r\n$\\sum{\\frac{1}{m_{a}}}\\leq \\sum{\\frac{1}{h_{a}}}= \\frac{1}{r}\\leq \\frac{2}{3r \\cdot \\sum{\\cos{A}}}$.\r\n\r\nIf not, the inequality is really trivial, because\r\n$\\sum{\\frac{1}{m_{a}}}\\leq \\sum{\\frac{1}{h_{a}}}= \\frac{1}{r}\\leq \\frac{2}{3r \\cdot \\sum{\\cos{A}}}\\leq \\frac{2}{3r}\\cdot (\\sum{\\cos{A}})$, the last one beeing equivalent with $(\\sum{\\cos{A}})^{2}\\geq 1$, which is obvious, because $\\sum{\\cos{A}}= 1+\\frac{r}{R}> 1$.", "Solution_7": "[color=purple] Sorry pohoatza the inequality you 've proved is fail. \n It must be $\\sum\\frac{1}{m_{a}}\\le \\frac{3}{2r}(\\sum\\cosh)$\n Sorry because i made you loss time. [/color]" } { "Tag": [ "symmetry", "quadratics", "LaTeX", "algebra", "algebra proposed" ], "Problem": "The problem is to find X,Y,Z where\r\n\r\nX^2 = Y+a\r\nY^2 = Z+a\r\nZ^2 = X+a\r\n\r\nI found this to be an interesting problem. \r\nIt looks very easy but is a bit tricky. \r\nWhat do you think? \r\n\r\nRay", "Solution_1": "What is the solution space $ S$ for $ x,y,z$ and what is the definition space $ D$ for $ a$ ? (is this the tricky part :lol: ?)\r\n\r\nE.g. for $ D\\equal{}S\\equal{}\\mathbb{C}$ any solution of $ 0\\equal{}((x^2\\minus{}a)^2\\minus{}a)^2\\minus{}(x\\plus{}a)$ and then $ y\\equal{}x^2\\minus{}a,z\\equal{}y^2\\minus{}a$ will do it.\r\n\r\nIf you meant for instance $ D\\equal{}S\\equal{}\\mathbb{Z}$, then you could proceed as follows:\r\n\r\nCase 1: Not all of the variables $ x,y,z$ are equal. Using symmetry wlog we assume $ y\\not\\equal{}x$.\r\n\r\nSubtracting the original equations from each other we get $ x^2\\minus{}y^2\\equal{}y\\minus{}z\\,,\\,y^2\\minus{}z^2\\equal{}z\\minus{}x\\,,\\,x^2\\minus{}z^2\\equal{}y\\minus{}x$. This yields after successive substitution:\r\n$ y\\minus{}x\\equal{}(x\\minus{}z)(x\\plus{}z)\\equal{}(x\\plus{}z)(z^2\\minus{}y^2)\\equal{}\\ldots \\equal{}(x\\plus{}z)(z\\plus{}y)(y\\plus{}x)(y\\minus{}x)$ .i.e. $ 1\\equal{}(x\\plus{}z)(z\\plus{}y)(y\\plus{}x)\\Longrightarrow x\\plus{}z,z\\plus{}y,y\\plus{}x\\in\\{\\pm 1\\}$ and there can only be none or two expressions be equal to $ \\minus{}1$, ie. we have at least 2 expressions of the same value. Subtraction of two expressions of the same value yields then the identity of two elements and the remaining expression then that the sum of these 2 identical elements is $ \\pm 1$, hence they are not integers, a contradiction.\r\n\r\nCase 2: $ x\\equal{}y\\equal{}z$\r\n\r\nChecking the resulting quadratic equation $ (2x\\minus{}1)^2\\equal{}4a\\plus{}1$ this permits an integer solution, iff $ 4a\\plus{}1$ is a perfect square", "Solution_2": "The tricky part comes after the 8th degree equation. This is a problem posed and solved by the late Ramanujan. He reduces the 8th degree equation to a 6th degree. After that (the tricky part) is to decompose the 6th degree equation into two equations of the third degree which can be solved by the usual methods. Perhaps this is a good high school algebra problem? Then based on the solution he comes up with various identities such as (sorry I didn't get a chance to get up to speed with LATEX yet): 1+4sin(10) = sqrt(11-2sqrt(11+2sqrt(11-2sqrt(11...etc) Definitely he was a genius with algebraic manipulation and coming up with all kinds of formulas and series. This particular problem comes from: \"Collected Papers of Ramanujan\" Chelsea Publishing Company, New York, reprinted 1962, Library of Congress Catalog Card No. 62-8326, page 327, Question 507 (v,240). These problems were published in the Journal of the Indian Mathematical Society. Perhaps this particular problem was one of the simpler ones but still is useful for a good high school level algebraic workout requiring a few insights here and there to produce the final solution. The mathematical package MAPLE only seems to produce a partial solution if one wants to use that for checking purposes. \r\n\r\nThe 8th degree equation\r\n$ x^8 \\minus{} 4x^6a \\plus{} 2x^4(3a^2 \\minus{} a) \\minus{} 4x^2(a^3 \\minus{} a^2) \\minus{} x \\plus{} a^4 \\minus{} 2a^3 \\plus{} a^2 \\minus{} a \\equal{} 0$\r\n\r\nis divided by \r\n$ x^2 \\minus{} x \\minus{} a$\r\n\r\nto produce the 6th degree equation:\r\n$ x^6 \\plus{} x^5 \\plus{} x^4(1 \\minus{} 3a) \\plus{} x^3(1 \\minus{} 2a) \\plus{} x^2(1 \\minus{} 3a \\plus{} 3a^2) \\plus{} x(1 \\minus{} 2a \\plus{} a^2) \\plus{} (1 \\minus{} a \\plus{} 2a^2 \\minus{} a^3) \\equal{} 0$\r\n\r\nThis is then reduced to two cubic equations and solved (I didn't find this part obvious).\r\nLet $ \\alpha, \\beta, \\delta, \\alpha', \\beta', \\gamma'$ be roots of this equation. Then since $ y$ and $ z$ are also roots, we may suppose that\r\n\r\n$ \\alpha^2 \\equal{} a\\plus{}\\beta, \\alpha'^2 \\equal{} a\\plus{}\\beta',$\r\n$ \\beta^2 \\equal{} a\\plus{}\\gamma, \\beta'^2 \\equal{} a\\plus{} \\gamma',$\r\n$ \\gamma^2 \\equal{} a\\plus{}\\alpha, \\gamma'^2 \\equal{} a\\plus{}\\alpha',$\r\n\r\nLet $ \\alpha\\plus{}\\beta\\plus{}\\gamma \\equal{} u, \\alpha'\\plus{}\\beta'\\plus{}\\gamma' \\equal{}v;$\r\nthen we see that $ \\alpha^2\\plus{}\\beta^2\\plus{}\\gamma^2 \\equal{} 3a\\plus{}u$\r\nand\r\n$ \\alpha\\beta\\plus{}\\beta\\gamma\\plus{}\\gamma\\alpha \\equal{} \\frac{1}{2}(u^2\\minus{}u\\minus{}3a)$\r\n\r\nThen a few more steps and we get:\r\n\r\nThen the sextic is identical with:\r\n $ (x^3\\minus{}x^2u\\plus{}\\frac{1}{2}x(u^2\\minus{}u\\minus{}3a)\\minus{}\\frac{1}{6}(u^3\\minus{}2u^2\\minus{}7au\\minus{}u\\minus{}3a))(x^3\\minus{}x^2v\\plus{}\\frac{1}{2}x(v^2\\minus{}v\\minus{}3a)\\minus{}\\frac{1}{6}(v^3\\minus{}2v^2\\minus{}7av\\minus{}v\\minus{}3a))$\r\n\r\nThen eventually Ramanujan comes up with:\r\n$ u\\plus{}v\\equal{}\\minus{}1$ and $ uv \\equal{} 2\\minus{}a$\r\n\r\nTherefore\r\nu = $ \\minus{}\\frac{1}{2}(1\\plus{}\\sqrt(4a\\minus{}7))$ , $ v \\equal{} \\minus{}\\frac{1}{2}(1\\minus{}\\sqrt(4a\\minus{}7))$\r\n\r\nThis is substituted into the equation above involving $ x,u,v$ and $ a$ to produce two cubic equations.\r\n\r\n$ x^3\\plus{}\\frac{1}{2}x^2(1\\plus{}\\sqrt(4a\\minus{}7))\\minus{}\\frac{1}{2}x(2a\\plus{}1\\plus{}\\sqrt(4a\\minus{}7))\\plus{}\\frac{1}{2}(a\\minus{}2\\plus{}a\\sqrt(4a\\minus{}7))\\equal{}0$\r\n$ x^3\\plus{}\\frac{1}{2}x^2(1\\minus{}\\sqrt(4a\\minus{}7))\\minus{}\\frac{1}{2}x(2a\\plus{}1\\minus{}\\sqrt(4a\\minus{}7))\\plus{}\\frac{1}{2}(a\\minus{}2\\minus{}a\\sqrt(4a\\minus{}7))\\equal{}0$\r\n\r\nRay", "Solution_3": "I think Ramanujan made an error in his solution.\r\n\r\nRay", "Solution_4": "Never mind he didn't make any error. Easy to make a mistake with a lot of tedious algebra. Definitely a good homework problem for highschool.\r\n\r\nRay" } { "Tag": [ "inequalities", "function", "inequalities theorems" ], "Problem": "I come the idea inspired by the method $siuhochung$ often used:\r\n$theorem$,let, $f$ is a function increasing in $[0,+\\infty)$\r\nand $f(x) \\geq 0$\r\nlet $a_1 \\geq a_2 \\geq a_3 \\geq \\cdots \\geq a_n$\r\n$g_1,g_2, \\cdots,g_n$ are functions of $a_1 , a_2,a_3 , \\cdots , a_n$\r\nif we have $\\sum_{i=1}^kg_i \\geq 0$ and $g_1+g_2+\\cdots+g_n \\geq 0$\r\nthen we have:$f(a_1)g_1+f(a_2)g_2+\\cdots+f(a_n)g_n \\geq 0$\r\nanybody who used this $theorem$ solved a inequality,please post here,\r\nthank you :)", "Solution_1": "let $g_1=(a-b)(a-c),g_2=(b-a)(b-c),g_3=(c-a)(c-b)$,$n=3$\r\nit is schur.", "Solution_2": "sorry,I find my theorem is easy and useless :blush:" } { "Tag": [ "linear algebra", "matrix", "function", "vector", "linear algebra unsolved" ], "Problem": "Let $A,B\\in\\mathbb{C}^{n\\times n}$ with $\\rho(AB - BA) = 1$. Show that $(AB - BA)^2 = 0$.", "Solution_1": "we wil work with endomorphism.\r\nSuppose that $Rank(uv-vu)=1$. One can find a basis where the matrix $M$ of $uv-vu$ is null\r\neverywhere, except on the first line. (because $rank(M)=1$) But $Tr(M)=0$ so $M_{1,1}=0$.\r\nConclusion follows ...", "Solution_2": "[quote=\"alekk\"]we wil work with endomorphism.\nSuppose that $Rank(uv-vu)=1$. One can find a basis where the matrix $M$ of $uv-vu$ is null\neverywhere, except on the first line. (because $rank(M)=1$) But $Tr(M)=0$ so $M_{1,1}=0$.\nConclusion follows ...[/quote]\r\n\r\nwhy trace $M$ equals $0$ ?", "Solution_3": "$Tr(AB)=TR(BA$ ..", "Solution_4": "Or, in matrix language:\r\n\r\nSince $\\rho(AB-BA)=1$, we have $n-1$ eigenvalues equal to zero. Since $\\Tr(AB)=\\Tr(BA)$ we also have the sum of the eigenvalues zero, so all eigenvalues are zero.\r\nNow consider the Jordan Form of $C=AB-BA$. This matrix has rank 1 with all diagonal elements zero, thus is of the form $C_{i,j}=\\delta_{12}$, and thus is $C^2$ trivially the nullmatrix.", "Solution_5": "I'm sorry, but I don't understand which function is $\\rho$...", "Solution_6": "Note that the result in this problem also hold for any field $ F $ , not only complex field.\n\nDefine $C = AB - BA $ .Since rank $C $ is 1 , then there exist a non zero column vectors $u $ and $v $ such that $uv^{t} $ = $C$\n\nSince tr $ AB$ = tr$BA$ , we get that tr$C$ = $0$ , or we can say that $v^{t}u$ = $ 0$.\n\nHence we get that $C^2$ = $(uv^{t})^2$ = $uv^{t}uv{t}$ = $u0v^{t}$ = $0$\n\n(assuming that $ \\rho $ is a rank function :lol: )", "Solution_7": "[quote=\"alekk\"]One can find a basis where the matrix $M$ of $uv-vu$ is null\neverywhere, except on the first line. (because $rank(M)=1$)[/quote] I am quite sure this is not true. In this particular problem, this is indeed the case but only due to the fact that all eigenvalues are zero (you can consider Jordan form as Peter has). However, you need to use that $tr M =0$. By the way, my solution is the same as removablesingularity's.", "Solution_8": "I understand upto the point where we know that the eigenvalue of $AB-BA$ are all 0. I didn't understand this.This matrix has rank 1 with all diagonal elements zero, thus is of the form $C_{i,j}=\\delta_{12}$ (sorry I don't know this notation), and thus is $C^2$ trivially the nullmatrix.\nalso in solution, this is also written, In the Jordan canonical form there can only\nbe one 2 \u00d7 2 cage. I also didn't understand this. Someone please explain. Thank you.", "Solution_9": "[quote=Adamsawyer]I understand upto the point where we know that the eigenvalue of $AB-BA$ are all 0. I didn't understand this.This matrix has rank 1 with all diagonal elements zero, thus is of the form $C_{i,j}=\\delta_{12}$ (sorry I don't know this notation), and thus is $C^2$ trivially the nullmatrix.[/quote]\n\nI agree that the notation $C_{i,j}=\\delta_{12}$ means nothing.\nWhat the poster meant is that the Jordan form of a rank 1 nilpotent matrix must have only Jordan blocks (relative to the eigenvalue 0) of order 1 or 2 (in fact, only one Jordan block of order 2).", "Solution_10": "[url= https://www.imc-math.org.uk/imc2000/prob_sol1.pdf] problem 3 first day[/url]\nhttps://www.imc-math.org.uk/imc2000/prob_sol1.pdf" } { "Tag": [], "Problem": "For an unknown substance Cux(CO3)y, is it possible to determine the empirical formula if I only know the mass of CO2?", "Solution_1": "i'd say no b/c you'll need to convert them to mol and you can't do that with only mass of $CO_2$", "Solution_2": "Yeah, you gotta know the amount of the substance or the copper." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "It's too interesting:\r\nFind all triples $(f,g,h)$ satisfying\r\n\\[\\frac{f(x)-g(y)}{x-y}=h\\left(\\frac{x+y}{2}\\right)\\] for all $x\\neq y$.\r\n\r\n[i]Edited by Myth[/i]", "Solution_1": "I mean find f,g,h.\r\nI predict that f,g,h are poliminals; degf,degg<=2", "Solution_2": "Something strange is happening here....\r\nIf they would be continuous functions, making $x\\rightarrow y$ we would get that $f(x)=g(x)$.\r\nAlso, it is very easy to get that $f(x)=f(-x)$ by putting $y=-x$.\r\nA little work could be done to find the solutions :) :D\r\nSo I think $f(x)=kx^2$ would be a solution....\r\nIf the functions wouldn't be continuous, i don't know what to say, but it would be a very bad function.", "Solution_3": "Oh, the solution I find is more interesting: f(x)=g(x) for all x, and h=f'(x), (like Lagrange :) )\r\n So, we put x=-y, y=-x, x=0 ,y=0 we get f(x)-g(x)=const!\r\n Using this in the original equation (note that g(x)=f(x)-const), put x=-y, y=-x, we also get f(x)=g(x) (const=0).\r\n After that we can see the origin equation as the equation with h by getting \r\n f(x)-f(0)=xh(x/2) (put y=0),\r\n we have : (xh(x/2)-yh(y/2))/(x-y)=h((x+y)/2)\r\n Put H(x)=h(x)-h(0), we get\r\n (xH(x/2)-yH(y/2))/(x-y)=H((x+y)/2) and H(-x)=-H(x)\r\n Put x=-y, y=-x we get H((x+y)/2)=A(x+y)/2 or H(x)=Ax\r\n So h(x)= 2ax+b and f(x)=g(x)=ax^2 +bx+c" } { "Tag": [ "analytic geometry", "graphing lines", "slope", "geometry", "3D geometry", "sphere", "quadratics" ], "Problem": "Let there be the equation : $ x^{n}+y^{n}+z^{n}=2$ . Prove that for each $ \\vee n \\in \\{1,2,3,4\\}$ , these equation has an infinity of rational roots .\r\n I've posted this [url=http://www.mathlinks.ro/viewtopic.php?t=156176]here[/url] before in romanian , but I've decided to translate it in English , too .", "Solution_1": "$ n = 1$ is trivial.\r\n\r\n[hide=\"n=2\"] First we must exhibit a rational solution. Let $ z = 1$. Then we are trying to solve $ x^{2}+y^{2}= 1$, which has its first nontrivial rational solution at $ x = \\frac{3}{5}, y = \\frac{4}{5}$. \n\nThe graph $ x^{2}+y^{2}+z^{2}= 2$ describes a sphere.\n\n[b]Claim:[/b] A line of rational slope passing through a rational point on the sphere intersects it at a second rational point.\n\nProof: By \"a line of rational slope\" I mean a line that can be represented as the intersection of two planes\n\n$ ax+by+cz = d$\n$ ex+fy+gz = h$\n\nWith rational coefficients. Substituting those equations into $ x^{2}+y^{2}+z^{2}= 2$ we can produce a quadratic in any of the variables (say $ x$) with rational coefficients. Since it has one rational root (the original rational point) it must have another (by Vieta's, etc.). For rational $ x$ we produce rational $ y, z$ from the first two equations.\n\nSo starting with our rational solution we can draw any line with rational slope to another rational solution. QED. [/hide]", "Solution_2": "[hide=\"$ n=1$\"]Choose $ (x,y,z) = (2,k,-k)$ with $ k$ rational.[/hide]\n\n[hide=\"$ n=2$\"]Choose $ (x,y,z) = \\left(1,\\frac{m^{2}-n^{2}}{m^{2}+n^{2}},\\frac{2mn}{m^{2}+n^{2}}\\right)$ with $ m$ and $ n$ rational.[/hide]\n\n[hide=\"$ n=3$\"]Choose $ (x,y,z) = (6k^{3}+1,1-6k^{3},-6k^{2})$ with $ k$ rational.[/hide]\r\n\r\n(These are hide tags, btw. :P )", "Solution_3": "I'll appreciate any feedback as to how this solution would constitute as an acceptable proof:\r\n[hide]\nClaim: There is an infinite rational number between $ 0$ and $ 1$.\nProof: Let $ A_{n}= (\\frac{1}{2})^{n}$. We know that $ \\frac{1}{2}$ is a rational number, and the product of two rational number results in another rational number. Therefore the function $ A_{n}$ would always result in a rational number for $ n \\in (0, infinity)$. For $ n \\in (0, infinity)$, $ 0 \\le A_{n}\\le 1$ which concludes our claim.\n------\nArbitrarily, let $ z=0$. Then the original equation becomes:\n$ x^{n}+y^{n}= 1$\n$ y^{n}= 1-x^{n}$\nThis relation shows that as x increases, y decreases. Similarly, as x decrease, y increase.\n\nRegardless of any positive integer $ n$, two solutions to $ y^{n}= 1-x^{n}$ is $ (0,1) and (1,0)$. This shows that as the variable $ x$ goes from value $ 1$ to value $ 0$, the variable $ y$ goes from value $ 0$ to value $ 1$. What this means is that any number, ration or irrational, between $ 0$ and $ 1$ would be a solution for $ x$, and a corresponding number between $ 0$ and $ 1$ would likewise be a solution for $ y$.\n\nThe equation $ y^{n}= 1-x^{n}$ shows that if any one of the variable is rational, then so is the other.\n\nBecause there is an infinite rational number between $ 0$ and $ 1$, there is an infinite rational solution for $ (x, y)$ in the equation $ y^{n}= 1-x^{n}$ , thus there is an infinte ration solution to $ x^{n}+y^{n}+z^{n}= 2$.\n[/hide]", "Solution_4": "[quote=\"p4fn2w\"]The equation $ y^{n}= 1-x^{n}$ shows that if any one of the variable is rational, then so is the other.[/quote]\n\nTrivially false; take $ n = 3, x =-1$ (for example.)\n\n[quote=\"p4n2w\"]Because there is an infinite rational number between $ 0$ and $ 1$, there is an infinite rational solution for $ (x, y)$ in the equation $ y^{n}= 1-x^{n}$[/quote]\r\n\r\nEr, no. This would imply the falsehood of Fermat's Last Theorem.", "Solution_5": "Other solutions for $ n=3$ and $ n=4$ ?" } { "Tag": [], "Problem": "How many times does the graph of the equation f(x) = x^5 - 3x^4 + 9x^3 -19x^2 +18x - 6 cross the x axis? \r\n\r\nI solved the problem and got the correct answer by using descartes rule of signs and the rational root theorem. I was wondering if there is a faster way to do it...", "Solution_1": "[quote=\"mr. math\"]How many times does the graph of the equation f(x) = x^5 - 3x^4 + 9x^3 -19x^2 +18x - 6 cross the x axis? \n\nI solved the problem and got the correct answer by using descartes rule of signs and the rational root theorem. I was wondering if there is a faster way to do it...[/quote]\r\nYes, there's a very simple way. Find the degree of polynomial. In this case it is 5. So, it will have 5 roots. So, it will cross the x-axis at most 5 times when there are unique solutions. But if some of the roots happen to be equal then it will cross x-axis less than 5 times.", "Solution_2": "[quote=\"Lakshya\"]Yes, there's a very simple way. Find the degree of polynomial. In this case it is 5. So, it will have 5 roots. So, it will cross the x-axis at most 5 times when there are unique solutions. But if some of the roots happen to be equal then it will cross x-axis less than 5 times.[/quote]\r\n... or if there are imaginary roots.\r\n\r\n[hide]Note that the coefficients add to $ 0$. That means $ x\\equal{}1$ is a solution. Factor out $ (x\\minus{}1)$:\n$ (x\\minus{}1)(x^4\\minus{}2x^3\\plus{}7x^2\\minus{}12x\\plus{}6)$\n\nHey, the coefficients of the quartic add to $ 0$. That means we have another factor of $ (x\\minus{}1)$:\n$ (x\\minus{}1)^2 (x^3\\minus{}x^2\\plus{}6x\\minus{}6)$\n\nAnd again, the coefficients of the cubic add to $ 0$. So factor out another $ (x\\minus{}1)$:\n$ (x\\minus{}1)^3 (x^2\\plus{}6)$\n\nNow we want the number of distinct real solutions to $ (x\\minus{}1)^3 (x^2\\plus{}6) \\equal{} 0$. $ x^2\\plus{}6 \\geq 0$, so divide it out to get $ (x\\minus{}1)^3 \\equal{} 0$. The only solution is $ x\\equal{}1$, thus the graph crosses the x-axis once.[/hide]" } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "is it true (and if it is the prove it) that for all positive $ a,b,c$\r\n$ \\frac {3}{a \\plus{} b \\plus{} c}\\leq {\\frac {1}{\\sqrt {ab}}} \\plus{} {\\frac {1}{ \\sqrt {bc}}} \\plus{} {\\frac {1}{\\sqrt {ca}}}$ ?\r\nThanks", "Solution_1": "Use Cauchy and then AM-GM for RHS", "Solution_2": "it is true...\r\n$ (a+b+c)(\\frac {1}{\\sqrt {ab}} + {\\frac {1}{\\sqrt {bc}} + {\\frac {1}{\\sqrt {ca}})\\geq3\\sqrt[3]{abc}\\times3\\sqrt[3]{\\frac {1}{\\sqrt {ab}}\\times\\frac {1}{\\sqrt {bc}}\\times\\frac {1}{\\sqrt {ca}}}=9>3}}$", "Solution_3": "Ok, it's very easy :blush: \r\nthanks", "Solution_4": "[quote=\"ali666\"]it is true...\n$ (a + b + c)(\\frac {1}{\\sqrt {ab}} + {\\frac {1}{\\sqrt {bc}} + {\\frac {1}{\\sqrt {ca}})\\geq3\\sqrt [3]{abc}\\times3\\sqrt [3]{\\frac {1}{\\sqrt {ab}}\\times\\frac {1}{\\sqrt {bc}}\\times\\frac {1}{\\sqrt {ca}}} = 9 > 3}}$[/quote] IT IS TRY" } { "Tag": [ "algebra", "system of equations" ], "Problem": "Solve in $ \\mathbb{ R}$:\r\n$ \\frac{x^2}{y}\\minus{}\\frac{4y^2}{x}\\equal{}1$\r\n$ 3x\\minus{}10y\\equal{}\\frac{1}{2}$", "Solution_1": "After I equate 2 equations, the outcome is\r\n\r\n$ x^3\\minus{}6x^2y\\plus{}20xy^2\\minus{}4y^3\\equal{}0$", "Solution_2": "One more:\r\nSolve in $ \\mathbb{R}:$\r\n$ \\frac {2x^3}{y} \\plus{} \\frac {3y^3}{x} \\equal{} 1\\minus{} 8xy$\r\n$ 8x^2\\plus{}9y^2\\equal{} \\sqrt2$", "Solution_3": "We change the system of equations into a homogeneous form as follows\r\n\r\n$ 2\\sqrt{2}x^4\\minus{}8x^3y\\plus{}8\\sqrt{2}(xy)^2\\minus{}9xy^3\\plus{}3\\sqrt{2}y^4\\equal{}0$", "Solution_4": "[quote=\"great math\"]We change the system of equations into a homogeneous form as follows\n\n$ 2\\sqrt {2}x^4 \\minus{} 8x^3y \\plus{} 8\\sqrt {2}(xy)^2 \\minus{} 9xy^3 \\plus{} 3\\sqrt {2}y^4 \\equal{} 0$[/quote]\r\nAnd how do you solve it?", "Solution_5": "Once it's homogeneous let $ y \\equal{} mx$ and solve for possible values of $ m$. :)", "Solution_6": "[quote=\"JavaMan\"]Once it's homogeneous let $ y \\equal{} mx$ and solve for possible values of $ m$. :)[/quote]\r\nYeah! Of couse it is!\r\nBut what I mean is to solve the equation of the fourth degree.", "Solution_7": "hello, to the first system, solving the second equation for $ y$ we get\r\n$ y\\equal{}\\frac{3}{10}x\\minus{}\\frac{1}{20}$ inserting this in the first equation, simplifying and factoring we get\r\n$ \\frac{1}{100}\\left(\\frac{178x^3\\minus{}492x^2\\plus{}82x\\plus{}1}{(6x\\minus{}1)x}\\right)\\equal{}0$\r\nso we have to solve\r\n$ 178x^3\\minus{}492x^2\\plus{}82x\\plus{}1\\equal{}0$, with the following real solution\r\n$ x\\equal{}\\minus{}{\\frac {5}{66537402}}\\,\\sqrt [3]{3808521050288486841\\plus{}\r\n82272613989570141\\,\\sqrt {2409}}\\plus{}91430\\,{\\frac {1}{\\sqrt [3]{\r\n3808521050288486841\\plus{}82272613989570141\\,\\sqrt {2409}}}}\\plus{}{\\frac {41}{446\r\n}}$\r\nSonnhard.", "Solution_8": "Hello, the second system has no real solutions.\r\nSonnhard." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b,c > 0$ such that $ a^{2} \\plus{} b^{2} \\plus{} c^{2} \\equal{} 1$, and let\r\n$ A \\equal{} \\displaystyle \\frac {1}{1 \\minus{} ab} \\plus{} \\frac {1}{1 \\minus{} bc} \\plus{} \\frac {1}{1 \\minus{} ca}, B \\equal{} \\displaystyle \\frac {1}{1 \\minus{} \\left(\\frac {a \\plus{} b}{2}\\right)^{2}} \\plus{} \\frac {1}{1 \\minus{} \\left(\\frac {b \\plus{} c}{2}\\right)^{2}} \\plus{} \\frac {1}{1 \\minus{} \\left(\\frac {c \\plus{} a}{2}\\right)^{2}},$ and\r\n$ C \\equal{} \\displaystyle \\frac {1}{1 \\minus{} a^{2}} \\plus{} \\frac {1}{1 \\minus{} b^{2}} \\plus{} \\frac {1}{1 \\minus{} c^{2}}.$\r\n\r\nProve that (1) $ A \\plus{} C\\geq 9$, (2) $ 7B \\plus{} 3C\\geq 45$.", "Solution_1": "Yes, (2) has been unsolved for a long time. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=81709]here[/url]. Have you found a solution for it ? :roll:", "Solution_2": "(1) is easy (it can be proved by Cauchy-Schwarz), (2) is harder. My solution does not use any big calculations but it is rather long. See it here: http://can-hang2007.blogspot.com/2010/01/inequality-64-w-berndt.html", "Solution_3": "Will you please post the solution of (1), can_hang2007 ?", "Solution_4": "[quote=\"Shanku\"]Will you please post the solution of (1), can_hang2007 ?[/quote]\r\nVery please, of course. :-) See it here: http://can-hang2007.blogspot.com/2010/01/inequality-68-w-berndt.html" } { "Tag": [], "Problem": "What is a formula for the pattern shown below.\r\n\r\n0,1,3,6,10,15", "Solution_1": "[quote=\"shyce\"]What is a formula for the pattern shown below.\n\n0,1,3,6,10,15[/quote]\r\n\r\n[hide]\nIf you double each term in the sequence, you get 0, 2, 6, 12, 20, 30...\n\nIt's clear that each number in the sequence is $0\\cdot 1, 1\\cdot 2, 2\\cdot 3, 3\\cdot 4, ...$\n\nSo each term in the new sequence is $x(x-1)$\n\nDivide by 2 to get $\\frac{x(x-1)}{2}$\n[/hide]", "Solution_2": "Basically, it's just\r\n\r\n$\\sum\\limits_{i = 0}^n {i = \\frac{{n(n - 1)}}{2}}$", "Solution_3": "EDIT: \r\n\r\nBTW, it's the formula for the $n+1$th triangular number" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "[i]Let $ a_{1}$, $ a_{2}$ ,....,$ a_{n }$ , $ b_{1}$, $ b_{2}$,..., $ b_{n}$ be real numbers\n Prove that \n $ \\sum_{i \\equal{} 1}^n a_i.b_i \\plus{} \\sqrt { ( \\sum_{i \\equal{} 1}^n a_i ^2 )(\\sum_{i \\equal{} 1}^n b_i ^2) }$ $ \\geq$ $ \\frac {2}{n}.( \\sum_{i \\equal{} 1}^n a_i )(\\sum_{i \\equal{} 1}^n b_i)$[/i]\r\n Argady , Nayel , Jichen ... can you help me? :blush:\r\n\r\n[i] [u]Problem 2[/u]\n Let a,b,c be nonnegative real numbers such that a+b+c =3[/i] Prove that\r\n $ \\frac{5 \\minus{}3bc}{ 1\\plus{}a} \\plus{} \\frac{5 \\minus{} 3ca}{1\\plus{}b} \\plus{}\\frac{ 5\\minus{}3ab}{1\\plus{}c} \\geq ab \\plus{}ba \\plus{}ca$", "Solution_1": "[quote=\"nkht-tk14\"][i]Let $ a_{1}$, $ a_{2}$ ,....,$ a_{n }$ , $ b_{1}$, $ b_{2}$,..., $ b_{n}$ be real numbers\n Prove that \n $ \\sum_{i \\equal{} 1}^n a_i.b_i \\plus{} \\sqrt { ( \\sum_{i \\equal{} 1}^n a_i ^2 )(\\sum_{i \\equal{} 1}^n b_i ^2) }$ $ \\geq$ $ \\frac {2}{n}.( \\sum_{i \\equal{} 1}^n a_i )(\\sum_{i \\equal{} 1}^n b_i)$[/i]\n Argady , Nayel , Jichen ... can you help me? :blush:\n\n[i] [u]Problem 2[/u]\n Let a,b,c be nonnegative real numbers such that a+b+c =3[/i] Prove that\n $ \\frac {5 \\minus{} 3bc}{ 1 \\plus{} a} \\plus{} \\frac {5 \\minus{} 3ca}{1 \\plus{} b} \\plus{} \\frac { 5 \\minus{} 3ab}{1 \\plus{} c} \\geq ab \\plus{} ba \\plus{} ca$[/quote]\r\n\r\n\r\n$ b\\equal{}\\frac{x_{1}\\plus{}...\\plus{}x_{n}}{n}$ ->\r\n$ \\sum_{i \\equal{} 1}^n a_i.b_i \\plus{} \\sqrt { ( \\sum_{i \\equal{} 1}^n a_i ^2 )(\\sum_{i \\equal{} 1}^n b_i ^2) }$ $ \\geq$ $ \\frac {2}{n}.( \\sum_{i \\equal{} 1}^n a_i )(\\sum_{i \\equal{} 1}^n b_i)$--> \r\n$ \\sqrt{(a_{1}^{2}\\plus{}...\\plus{}a_{n}^{2})(x_{1}^{2}\\plus{}...\\plus{}x_{n}^{2})}\\geq(a_{1}(B\\minus{}x_{1})\\plus{}...\\plus{}a_{n}(B\\minus{}x_{n}))$\r\n\r\n\r\n$ (2B\\minus{}x_{1})^{2}\\plus{}...\\plus{}(2B\\minus{}x_{n})^{2}\\equal{}4B^{2}\\minus{}4Bx_{1}\\plus{}x_{1}^{2}\\plus{}...\\plus{}4B^{2}\\minus{}4Bx_{n}\\plus{}x_{n}^{2}\\equal{}4nB^{2}\\minus{}4B(x_{1}\\plus{}...\\plus{}x_{n})\\plus{}x_{1}^{2}\\plus{}...\\plus{}x_{n}^{2}\\equal{}x_{1}^{2}\\plus{}...\\plus{}x_{n}^{2}$ ---->\r\n\r\n$ \\sqrt{(a_{1}^{2}\\plus{}...\\plus{}a_{n}^{2})(x_{1}^{2}\\plus{}...\\plus{}x_{n}^{2})}\\equal{}\\sqrt{(a_{1}^{2}\\plus{}...\\plus{}a_{n}^{2})((2B\\minus{}x_{1})^{2}\\plus{}...\\plus{}(2B\\minus{}x_{n})^{2})}\\geq(a_{1}(B\\minus{}x_{1})\\plus{}...\\plus{}a_{n}(B\\minus{}x_{n}))$ proved.", "Solution_2": "@yaya_yc: are you YAYA TOURE??", "Solution_3": "[quote=\"nkht-tk14\"]\n[i] [u]Problem 2[/u]\n Let a,b,c be nonnegative real numbers such that a+b+c =3[/i] Prove that\n $ \\frac {5 \\minus{} 3bc}{ 1 \\plus{} a} \\plus{} \\frac {5 \\minus{} 3ca}{1 \\plus{} b} \\plus{} \\frac { 5 \\minus{} 3ab}{1 \\plus{} c} \\geq ab \\plus{} ba \\plus{} ca$[/quote]\r\nAfter expanding and homogenization we obtain:\r\n$ \\sum_{cyc}\\left(\\frac {5 \\minus{} 3bc}{ 1 \\plus{} a}\\minus{}ab\\right)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(5a^5\\plus{}14a^4b\\plus{}14a^4c\\minus{}19a^3b^2\\minus{}19a^3c^2\\plus{}32a^3bc\\minus{}27a^2b^2c)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow5\\sum_{cyc}(a^5\\minus{}a^4b\\minus{}a^4c\\plus{}a^3bc)\\plus{}19\\sum_{cyc}(a^4b\\plus{}a^4c\\minus{}a^3b^2\\minus{}a^3c^2)\\plus{}$\r\n$ \\plus{}27\\sum_{cyc}(a^2\\minus{}ab)abc\\geq0,$ which true by Schur and Muirhead.", "Solution_4": "[quote=\"nkht-tk14\"][i]Let $ a_{1}, a_{2},\\cdots,a_{n }, b_{1}, b_{2},\\cdots, b_{n}$ be real numbers. Prove that\n\n$ \\sum_{i = 1}^n a_{i}b_{i} + \\sqrt {( \\sum_{i = 1}^n a_{i}^2)(\\sum_{i = 1}^n b_{i}^2)}\\geq\\frac {2}{n}(\\sum_{i = 1}^n a_{i})(\\sum_{i = 1}^n b_{i})$.[/i][/quote]$ \\sqrt {( \\sum_{i = 1}^n a_{i}^2)(\\sum_{i = 1}^n b_{i}^2)} = \\sqrt {( \\sum_{i = 1}^n a_{i}^2)[\\sum_{i = 1}^n(\\frac {2}{n}\\sum_{j = 1}^n b_{j} - b_{i})^2]}$\r\n\r\n$ \\geq\\sum_{i = 1}^n a_{i}(\\frac {2}{n}\\sum_{j = 1}^n b_{j} - b_{i}) = \\frac {2}{n}(\\sum_{i = 1}^n a_{i})(\\sum_{j = 1}^n b_{j}) - \\sum_{i = 1}^n a_{i}b_{i}$.\r\n\r\nBy the way, if $ a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}$ be real numbers, then\r\n\r\n$ 2(a_{1}b_{1}c_{1} + a_{2}b_{2}c_{2}) + \\sqrt {2(a_{1}^2 + a_{2}^2)(b_{1}^2 + b_{2}^2)(c_{1}^2 + c_{2}^2)}$\r\n\r\n$ \\geq(a_{1} + a_{2})(b_{1} + b_{2})(c_{1} + c_{2})$.\r\n\r\n[hide=\"Proof\"] $ 2(a_{1}^2+a_{2}^2)(b_{1}^2+b_{2}^2)(c_{1}^2+c_{2}^2)-[(a_{1}+a_{2})(b_{1}+b_{2})(c_{1}+c_{2})-2(a_{1}b_{1}c_{1}+a_{2}b_{2}c_{2})]^2$\n\n$ =(a_{1}b_{1}c_{1}+a_{2}b_{1}c_{1}+a_{1}b_{2}c_{1}+a_{1}b_{1}c_{2}-a_{1}b_{2}c_{2}-a_{2}b_{1}c_{2}-a_{2}b_{2}c_{1}-a_{2}b_{2}c_{2})^2$\n\n$ \\geq 0\\Longrightarrow$\n\n$ \\sqrt{2(a_{1}^2+a_{2}^2)(b_{1}^2+b_{2}^2)(c_{1}^2+c_{2}^2)}$\n\n$ \\geq(a_{1}+a_{2})(b_{1}+b_{2})(c_{1}+c_{2})-2(a_{1}b_{1}c_{1}+a_{2}b_{2}c_{2})$.[/hide]" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "sphere", "geometry theorems" ], "Problem": "hi ! i have question ,where is centre of ball in tetrahedron? :)", "Solution_1": "suppose the tetrahedron is $ABCD$ draw $6$ sphere with the diameters $AB,AD,BD ,BC,AC,CD$.\r\n\r\nthe intersection of these spheres will be a point which is the center of the tetrahedron..." } { "Tag": [ "MATHCOUNTS", "number theory", "prime factorization" ], "Problem": "The prime factorization of a number X is $a^a$ $b^b$ $c^c$ where a, b, and c are unique prime numbers. If X has 72 factors, what is the value of X?", "Solution_1": "[hide=\"solution\"]\nNumber of factors = 72 = (a + 1)(b + 1)(c + 1)\na, b, and c must be prime, so a = 2, b = 3, and c = 5.\n(2^2)(3^3)(5^5)\n(4)(27)(3125)\n(108)(3125)\n312500 + 3125(8)\n312500 + 25000\n[b]337500[/b]\n[/hide]\r\n\r\nAll without a calculator. SPESH-UAL!!!", "Solution_2": "[hide](a+1)(b+1)(c+1)=72\n(a+1)(b+1)(c+1)=2*2*2*3*3\nSo a can be 3(2*2-1), b can be 2(3-1), and c can be (6-1)=5, for a total of $3^3 2^2 5^5=337500$[/hide]\r\n\r\nBilly", "Solution_3": "[quote=\"frost13\"]The prime factorization of a number X is $a^a$ $b^b$ $c^c$ where a, b, and c are unique prime numbers. If X has 72 factors, what is the value of X?[/quote]\r\n\r\n[hide=\"solution\"]$(a+1)(b+1)(c+1)=72$\n\n$72=2^33^2$\n\n$a+1=6$\n$b+1=4$\n$c+1=3$\n\n$a=5$\n$b=3$\n$c=2$\n\n$x=2^23^35^5$[/hide]", "Solution_4": "So is there a formula, based on what you all wrote, for the number of factors a number has? Sorry I'm jw because I'm preparing for the States and want to know every formula that I can.", "Solution_5": "hockey, \r\n\r\nthere is a formula, and it can be manipulated for various problems. I'm attaching a document of various formulas, shortcuts to common mathcounts questions for you. Good Luck, \r\n\r\nfrost", "Solution_6": "hockey, \r\n\r\nthere is a formula, and it can be manipulated for various problems. I'm attaching a document of various formulas, shortcuts to common mathcounts questions for you. Good Luck, \r\n\r\nfrost", "Solution_7": "[quote=\"Treething\"][hide=\"solution\"]\nNumber of factors = 72 = (a + 1)(b + 1)(c + 1)\na, b, and c must be prime, so a = 2, b = 3, and c = 5.\n(2^2)(3^3)(5^5)\n(4)(27)(3125)\n(108)(3125)\n312500 + 3125(8)\n312500 + 25000\n[b]337500[/b]\n[/hide]\n\nAll without a calculator. SPESH-UAL!!![/quote]\r\n\r\nWhat's so Spesh-ual? :huh:" } { "Tag": [ "USAMTS", "search", "LaTeX", "Asymptote", "geometry", "AMC", "AIME" ], "Problem": "Round 1 problems of Year 19 of the USA Mathematical Talent Search are now available at [url]http://www.usamts.org[/url]. The postmark deadline for solutions is October 9, 2007. A copy of the rules is available with the problems; please read them carefully.\r\n\r\nIMPORTANT REGISTRATION INFORMATION: If you have not already registered online for Year 19, please do so at http://www.usamts.org. If you participated in Year 18, all you have to do to register for Year 19 is log in to [url]http://www.usamts.org[/url], then set your grade in the Profile section of the My USAMTS pages.\r\n\r\nWe hope you enjoy the problems, and we look forward to working with you on the USAMTS this year.", "Solution_1": "I noticed a few bugs in the LaTeX template for the solutions form.\r\n\r\n1) The include file is labeled usamts, but the line in the solutions form reads \"\\include{usamts[b]_blank[/b]}\".\r\n2) Solution 1 has a \"\\ \" on the line above the comment.", "Solution_2": "Thanks! I think these are fixed now.", "Solution_3": "Are the templates the same as last year?", "Solution_4": "Great set of problems! I like having the sets during vacations because now I can write up solutions before school starts.", "Solution_5": "Um I faxed my entry form, under 13 permission form, and regular permission form to the designated location, but it still says that the people didn't receive it on my Profile page. I faxed it like a week ago, so am I being hasty or is there some problem. My ID is 9524, if that helps.\r\n\r\nP.S. Looooooooove the questions. :lol:", "Solution_6": "In the .tex file, it says to enter your AoPS username. Is it supposed to be your USAMTS username?", "Solution_7": "[quote=\"LeFromage\"]In the .tex file, it says to enter your AoPS username. Is it supposed to be your USAMTS username?[/quote]\r\n\r\ni made them the same, just in case... but maybe its supposed to be your USAMTS username.\r\n\r\nthe problems are very nice. :)", "Solution_8": "Templates the same (but make sure you update the year).\r\n\r\nThat should be USAMTS username (I just fixed it on the site.)\r\n\r\nWe won't start filing the forms we receive until next month, so don't worry yet about it saying your forms are not received.", "Solution_9": "One more thing I noticed: on the fourth page of the USAMTS entry form (http://www.usamts.org/Forms/USAMTSEntryForm.pdf), it still says \"Year 18\" and \"2006-7\".", "Solution_10": "Would it be possible to find the tex file of the problems, specifically the 18-gon diagram?", "Solution_11": "I think you should be able to build that diagram yourself in asymptote, with a little elbow grease. Obviously we shouldn't post it on the forum, though.\r\n\r\nBy the way, do you prefer that we save asymptote images as separate files and send just their pdf, that we also send the .asy file, or that we keep the asymptote code within the .tex file?", "Solution_12": "Are the grader's going to print out the file and grade it? If I create a diagrma in geometry sketchpad, it becomes blurry in tex when on a 24\" screen (don't know what screen resolution and all the grader's might use), so are they going to print it?", "Solution_13": "I seriously doubt that graders will take off for a diagram being blurry, unless there is text that is so small as to be illegible. But, if you are really worried about it, create your pictures in Asymptote. That way you can just write the source code in your latex document, without any fuss.", "Solution_14": "The reason is that I don't want to do it in asymptote, because it takes too much time", "Solution_15": "[quote=\"roadnottaken\"]I seriously doubt that graders will take off for a diagram being blurry, unless there is text that is so small as to be illegible. But, if you are really worried about it, create your pictures in Asymptote. That way you can just write the source code in your latex document, without any fuss.[/quote]\r\nDude, you're answering legitimate questions directed to the usamts people with things like \"I think\" or \"I believe\"\r\n\r\nLet the question be answered, your reply didn't actually help anyone..", "Solution_16": "[quote=\"#H34N1\"]Are the grader's going to print out the file and grade it? If I create a diagrma in geometry sketchpad, it becomes blurry in tex when on a 24\" screen (don't know what screen resolution and all the grader's might use), so are they going to print it?[/quote]\r\nYes, they print it out.\r\nI asked that once in a mathjam.", "Solution_17": "hey, just wondering. we're allowed to write everything by hand (in pen) and fax the answers to them right? \r\n\r\nJust want to confirm.\r\n\r\nthanks", "Solution_18": "[quote=\"Geeimatree\"]hey, just wondering. we're allowed to write everything by hand (in pen) and fax the answers to them right?[/quote]Yes. Alternatively, you can mail them.", "Solution_19": "pretty interesting. I got one in the mail :)", "Solution_20": "[quote=\"moogra\"]pretty interesting. I got one in the mail :)[/quote]\r\n\r\nThey send them out to people who took the AIME, or something like that. Or maybe people who bought some Aops books. Something like that...", "Solution_21": "I know for a fact that not all of their criteria are related to AoPS; the way I originally discovered this website was by being mailed a USAMTS form.", "Solution_22": "[quote=\"mustafa\"][quote=\"moogra\"]pretty interesting. I got one in the mail :)[/quote]\n\nThey send them out to people who took the AIME, or something like that. Or maybe people who bought some Aops books. Something like that...[/quote]\r\n\r\nWell I also read on the USAMTS page that the don't send out the problems. I guess I was just lucky.", "Solution_23": "[quote=\"moogra\"][quote=\"mustafa\"][quote=\"moogra\"]pretty interesting. I got one in the mail :)[/quote]\n\nThey send them out to people who took the AIME, or something like that. Or maybe people who bought some Aops books. Something like that...[/quote]\n\nWell I also read on the USAMTS page that the don't send out the problems. I guess I was just lucky.[/quote]\r\n\r\nNah, I got the problems too", "Solution_24": "That stuff looks hard- very hard", "Solution_25": "I've just sent my solutions by fax; however, for some ridiculous reason, the secretary of my mother's office, where she faxed for me, did it thrice... And I'm also sorry for forgetting copying the USAMTS# on the coversheet (all other things are copied)... Do these things matter?\r\n\r\nB.T.W.: I've registered on the USAMTS website and have never received its e-mail (I've verified it's correct) Does anybody know why?\r\n\r\nThanks.", "Solution_26": "[quote=\"timwu\"]I've just sent my solutions by fax; however, for some ridiculous reason, the secretary of my mother's office, where she faxed for me, did it thrice... And I'm also sorry for forgetting copying the USAMTS# on the coversheet (all other things are copied)... Do these things matter?\n\nB.T.W.: I've registered on the USAMTS website and have never received its e-mail (I've verified it's correct) Does anybody know why?\n\nThanks.[/quote]\r\n\r\nDon't worry about faxing too many times. I'm assuming you have your USAMTS number on your solutions. As for the emails, maybe your spam blocker nailed the.?", "Solution_27": "Thank you for reply;I did copy the number on every page, and I'm going to check if there's some spam blocker in my g-mail.\r\nB.T.W.: The questions are very nice, :)", "Solution_28": "What e-mail are you talking about?", "Solution_29": "Another question: when will the comfirmation check mark appear on my \"My Scores\" of the USAMTS website? Because it seems that my e-mail doesn't work (I've checked my e-mail, there's no spasm or trash e-mail at all.)\r\n\r\nTo 123456789: I mean the system e-mail from USAMTS website after a user registers.", "Solution_30": "We'll probably start filing the papers in a week or so." } { "Tag": [], "Problem": "1)Show that $a^{12}-b^{12}$ is divisible by 91 if a and b are both prime to 91.\r\n\r\n2)Show that $n^{36}-1$ is divisible by 3374 if n is prime to 2,3,19 and 37.\r\n\r\n3)If m is a prime number and a,b are less then m,prove that\r\n$a^{m-2}+a^{m-3}b+a^{m-4}b^2+......b^{m-2}$\r\n is a multiple of m.", "Solution_1": "Just correcting a few typos :) :\r\n[quote=\"shadysaysurspammed\"]1)Show that $a^{12}-b^{12}$ is divisible by 91 if a and b are both prime to 91.\n\n2)Show that $n^{36}-1$ is divisible by 3374 if n is prime to 2,3,19 and 37.\n[/quote]", "Solution_2": "yes,sorry for the typos,edited now." } { "Tag": [ "Alcumus", "Support" ], "Problem": "Something is wrong. No matter what I do my Karma stays at 1000!", "Solution_1": "Well it finally started working right. It really wouldnt change for a while, though.", "Solution_2": "You were probably in diagnostic mode; you hadn't done enough questions. Are you a new user for Alcumus? Look in the instructions.", "Solution_3": "Yeah I just started going on Alcumus. Its working now.", "Solution_4": "Please read the [url=http://www.artofproblemsolving.com/Alcumus/Instructions.php]instructions[/url] as it contains information about such things." } { "Tag": [ "function" ], "Problem": "Let $P_{n}$ be the number of ways to make $\\$n$ by only using nickels, dimes, and quarters. Compute the remainder when $P_{2007}$ is divided by $1000$.", "Solution_1": "Since I don't know all of the tricks with this method, i used an altered [hide] generating function argument. [hide=\"And the answer I got is:\"] \\[\\boxed{115}\\] [/hide] I'll write my solution up later, whether or not it's correct. I just don't have the time now. :) [/hide]" } { "Tag": [ "algebra", "polynomial", "logarithms", "Functional Analysis", "real analysis", "real analysis unsolved" ], "Problem": "Here's a very beautiful problem: suppose than $ k_n$ is a sequence of positive integers with the following property: any series $ \\sum_{n}{a_n e^{ixk_n}}$ that converges uniformly on $ \\mathbb {R}$ converges actually normally, that is $ (a_n)_n$ is in $ l^1$. Then there are constants $ c>1$, $ d>0$ such that $ k_n\\geq d c^n$ for all $ n$. Can you exhibit a series not having the above property if the sequence $ k_n$ is polynomial?", "Solution_1": "Is that result by Kolmogorov?", "Solution_2": "If I were to guess, I would rather attribute this result to Khintchin (though historically I may be completely wrong here, it is closer to his work in spirit). If you have a sum $ \\sum_{j\\equal{}1}^n\\pm e^{ik_jx}$ with $ 0\\le k_j\\le N$, its $ L^\\infty$-norm does not exceed $ C\\sqrt {n\\log N}$ with high probability. This (together with some standard tricks from functional analysis) allows to conclude that under our condition $ n\\le C\\log N$, which is equivalent to what harazi asked. To exhibit an explicit series for a given sequence $ k_j$ is a much harder problem.", "Solution_3": "[hide=\"2Fedor\"]\u042f \u043d\u0435 \u043f\u043e\u043d\u044f\u043b \u0441\u0432\u044f\u0437\u0438 :blush: \u042f \u0442\u0430\u043a \u0434\u0443\u043c\u0430\u044e, \u043e\u043d \u0431\u044b \u043d\u0435 \u0434\u0430\u043b \u0442\u0443\u0442 \u044d\u0442\u0443 \u0437\u0430\u0434\u0430\u0447\u0443, \u0435\u0441\u043b\u0438 \u0431\u044b \u0443 \u043d\u0435\u0435 \u043d\u0435 \u0431\u044b\u043b\u043e \u0434\u043e\u0441\u0442\u0430\u0442\u043e\u0447\u043d\u043e \u043a\u043e\u0440\u043e\u0442\u043a\u043e\u0433\u043e \u0440\u0435\u0448\u0435\u043d\u0438\u044f, \u043a\u043e\u0442\u043e\u0440\u043e\u0435 \u043f\u043e\u043b\u043d\u043e\u0441\u0442\u044c\u044e \u043c\u043e\u0436\u043d\u043e \u0442\u0443\u0442 \u0438\u0437\u043b\u043e\u0436\u0438\u0442\u044c.[/hide]", "Solution_4": "Well, that book does not say who is the author of this theorem (which is a bit strange, because it seems to be rather accurate with all the other results presented...), but the solution it gives and which I have is exactly the one presented by Fedja. And I have to confess that I don't have explicit examples for practically no nontrivial situation." } { "Tag": [], "Problem": "Subiectele si solut de la etapa judeteana din 01 martie 2008 sunt postate pe site-ul http://www.mategl.com", "Solution_1": "Please , \r\n can you send some other links, where I can find problems and solutions of Romanian competitions , like this nice Web site above ?\r\n Thanks - babis", "Solution_2": "Cum vi s-au parut subiectele la a unspea si a doispea? La a doispea mi s-au parut mai grele ca in anii trecuti.", "Solution_3": "Le puteti gasi si la http://mateforum.ro/viewtopic.php?t=1184" } { "Tag": [], "Problem": "http://www.cnn.com/video/#/video/us/2008/08/08/legrand.haunted.school.wspa", "Solution_1": "The related story can be found here: http://www.abc2news.com/news/local/story.aspx?content_id=caf271c6-f450-4499-84f8-0fa4fdd34d92&rss=702" } { "Tag": [ "algebra", "function", "domain", "topology" ], "Problem": "$ S^1\\equal{}\\{(x,y)\\in \\mathbb{R}^2 | x^2\\plus{}y^2\\equal{}1\\}$.\r\nLet $ x\\equal{}(x_1,x_2)\\in S^1$,$ \\minus{}x\\equal{}(\\minus{}x_1,\\minus{}x_2)\\in S^1$.\r\nDefine map $ r: S^1\\to S^1$ as $ \\forall \\, x\\in S^1, r(x)\\equal{}\\minus{}x$.\r\nHow to prove the continuity of $ r$?\r\n\r\nthanks", "Solution_1": "Consider that map as a map between $ \\mathbb{R}^2 \\rightarrow \\mathbb{R}^2$. Clearly its a finite dimensional linear map. These maps are continuous. When restricted to $ S^1$ on the domain, then the map is also continuous. Then its trivial to check that the codomain is $ S^1$ because multiplication by $ \\minus{}1$ doesn't change the absolute value.", "Solution_2": "More directly, $ \\parallel{}f(x) \\minus{} f(y)\\parallel{} \\equal{} \\parallel{}\\minus{}x \\minus{} (\\minus{}y)\\parallel{} \\equal{} \\parallel{}x\\minus{}y\\parallel{}$. Any function for which there exists a positive constant $ \\lambda$ (in our case $ \\lambda \\equal{} 1$) such that $ \\parallel{}f(x) \\minus{} f(y)\\parallel{} \\leq \\lambda \\parallel{}x\\minus{}y\\parallel{}$ can be easily checked by definition to be continuous (such functions are called lipschitzian; if $ \\lambda < 1$ they are called contractions).", "Solution_3": "[quote=\"LkNsngth\"]When restricted to $ S^1$ on the domain, then the map is also continuous.[/quote]This is not sufficient to be precise. Note that you have proven that the composition $ S^1\\to S^1\\to\\mathbb R^2$ is continuous. But why is $ S^1\\to S^1$ is continuous? There is one more argument needed.\r\n\r\n@ water2011: Are all your postings homework? It smells like ... :mad: :wink:" } { "Tag": [ "geometry", "3D geometry", "sphere", "calculus", "calculus computations" ], "Problem": "This problem can be solved by using Calculus, but apparently there's an easy intuitive method as well.\r\n\r\nIf you cut a sphere through the centre with a cylindrical hole of diameter l, what is the remaining volume of the sphere (in terms of l)?\r\n\r\nA hint is that the answer is independent of the radius of the sphere - in fact, the intuitive method uses this fact in its argument.", "Solution_1": "Oh, come on, isn't anybody going to help me??? :(", "Solution_2": "To get a volume independent of the radius of the sphere, its not the diameter of the hole that should be fixed, but rather the height of the cylinder. (Or if you think of the remaining object as something like a napkin ring, its standing height.)", "Solution_3": "So, if you ignore the hint and the volume is dependent on the radius of the sphere, what would the answer be???", "Solution_4": "Kent Merryfield is right. The volume of the remaining solid should be $\\frac{4}{3}(R^2-I^2)^{\\frac{3}{2}}\\pi$, where $R$ is the radius of the sphere. I think the trick Melissa has in her mind is the use of the following theorem(by the way, this theorem was first discovered by Zu Heng, the son of the famous Chinese mathematician Zu Chongzi who obtained the numerical value of $\\pi$ accurate to 7 decimal places):\r\n\r\nLet $A$ and $B$ be two solids which stand on their bases on a plane $V$, with equal height. If the areas of the intersections of $A$, $B$ with any plane $V'$ parallel to $V$ are always equal, then $A$ and $B$ have the same volume.\r\n\r\nIn fact, the remaining solid in the problem has the same volume as a sphere of radius $\\sqrt{R^2-I^2}$, by the above theorem." } { "Tag": [ "probability", "function", "combinatorics unsolved", "combinatorics" ], "Problem": "What is the probability that 10 coins selected randomly from 100 coins add up to $ 1?\r\nAssume the 100 coins consist randomly of pennies, nickels, dimes, and quarters. Each coin has an equal probability of being selected.\r\n\r\nI guess generating functions are required to find the number of ways.", "Solution_1": "By simple inspection we find the possible decomposition of 100 as the sum of 10 numbers out of 1,5,10,25: obviously, the task is impossible with at least 4 quarters, with 3 quarters you need less than 5 between nickels and dimes, so you need to have pennies, and you cannot have as many as 10, so you have exactly 5, leaving 20 cents in 2 coins, and the only possibility is $ 100 \\equal{} 3\\cdot25 \\plus{} 2\\cdot10 \\plus{} 5\\cdot1$; if you have exactly 2 quarters, and you have pennies, you need 45 cents in 3 coins which are nickels and dimes, absurd, so you have only nickels and dimes left such that $ 10d \\plus{} 5n \\equal{} 50$, $ d \\plus{} n \\equal{} 8$, and $ 100 \\equal{} 2\\cdot25 \\plus{} 2\\cdot10 \\plus{} 6\\cdot5$ is the only possibility; with exactly 1 quarter, you need at least 6 dimes or you don't reach 100, but you cannot have more than 7 or you have too much, so either 6 or 7 dimes, leaving 3 or 2 coins for 15 or 5 cents, the second case is impossible and the first case yields $ 100 \\equal{} 1\\cdot25 \\plus{} 6\\cdot10 \\plus{} 3\\cdot5$; with no quarters, only $ 100 \\equal{} 10\\cdot10$.\r\n\r\nThe total number of favorable cases is then\r\n\\[ \\binom{25}{3}\\binom{25}{2}\\binom{25}{5} \\plus{} \\binom{25}{2}\\binom{25}{2}\\binom{25}{6} \\plus{} \\binom{25}{1}\\binom{25}{6}\\binom{25}{3} \\plus{} \\binom{25}{10},\r\n\\]\r\nwhile the total number of possible cases is $ \\binom{100}{10}$, since you need to have 25 coins of each one of the four types. All cases are equally probable since all coins have equal probability of being selected. Divide favorable into possible and you get the result..." } { "Tag": [ "percent" ], "Problem": "A stock loses $ 10\\%$ of its value on Monday. On Tuesday it loses \n$ 20\\%$ of the value it had at the end of the day on Monday. What is \nthe overall percent loss in value from the beginning of Monday to \nthe end of Tuesday?", "Solution_1": "On Monday, it goes from $ 100\\%\n$, to $ 100\\% - 10\\%(100\\%) = 90%\n$. Then, on Tuesday, it goes from $ 90\\%%\n$ to $ 90\\% - 20\\%(90\\%) = 72\\%\n$. Therefore, our total loss is $ 100\\% - 72\\% = \\boxed{28}\\%\n$" } { "Tag": [ "number theory", "least common multiple", "superior algebra", "superior algebra solved" ], "Problem": "Find the smallest n\\in N* s.t. for any permutation x in S9 we have xn=e (the identity permutation). The same question for permutations in A9 (even permutations in S9).", "Solution_1": "The greatest n ?", "Solution_2": "If you need the lowest, I got the first part. It is known that the order of a permutation is the lcm of the lengths of its cycles. n is the lcm of all this orders. The number does not contain any prime greater than 9.\r\n\r\nThe power of 2 is at most 3 as we may have a cycle of length 8.\r\nThe power of 3 is at most 2 as we may have a cycle of length 9\r\nThe power of 5 is at most 1 as we may have a cycle of length 5\r\nThe power of 7 is at most 1 as we may have a cycle of length 7\r\n\r\nSo the number is 8*9*5*7 :)", "Solution_3": "Right! You need th smallest n, of course (my mistake). I edited the post. Sorry :).", "Solution_4": "For the case of An I will use the following result:\r\n\r\nFor any cycle we have (a1,...an)=(a1,an-1)(a1,an-2)...(a1,a2) \r\n\r\nso one cycle is even if its length is odd and odd if its lengfth is even. A permutation is even only if it contains an even number of cycles of even length.\r\n\r\nIn the argument for Sn we should change only the power of 2 because it involves the even length cycles. The power of 2 is at most 2 when we have 2 cycles of length 4. So the number is 4*9*5*7 :)" } { "Tag": [ "geometry", "circumcircle", "congruent triangles", "geometry proposed" ], "Problem": "Given a circle and its center $O$, a point $A$ inside the circle and a distance $h$, construct a triangle $BAC$ with $\\angle BAC = 90^\\circ$, $B$ and $C$ on the circle and the altitude from $A$ length $h$.", "Solution_1": "The circle is known, so let $R$ be its radius and $AD=h$ the altitude of $\\triangle ABC$.\r\n\r\nThe triangle $ABC$ is right so we have $AD^2 = BD \\cdot DC$ \r\n\r\nBut $-BD \\cdot DC$ is also the power of the point $D$ with respect to the circle $(O,R)$\r\n\r\n$-AD^2 = OD^2-R^2 \\Rightarrow OD^2 = R^2-h^2$\r\n\r\n\r\n$\\hline$\r\n\r\nSo, we construct the circles $w_1 = (O, \\sqrt{R^2-h^2})$ and $w_2 = (A,h)$.\r\n\r\nIf the two circles have common point(s), then take $D\\in w_1 \\cap w_2$.\r\n\r\nThe perpendicular line to $AD$ intersects the circle $(O,R)$ at the points $BC$.\r\n\r\n$\\hline$\r\n\r\nNow we should check if the constructed figure has the desired properties\r\n\r\nFrom it's construction, $AD$ is the altitude of the triangle $ABC$\r\n\r\n$OD=\\sqrt{R^2-AD^2} \\Rightarrow OD^2 = R^2-AD^2 \\Rightarrow -AD^2 = OD^2-R^2 \\Rightarrow$ \r\nthe power of $D$ w.r.t. $(O,R)$ is $-AD^2,$ but it is also $-BD \\cdot DC \\Rightarrow$\r\n\r\n$AD^2=BD \\cdot DC \\Rightarrow \\frac{BD}{AD} = \\frac{AD}{DC} \\Rightarrow$\r\n\r\nThe right triangles $BDA,ADC$ are similar $\\Rightarrow \\angle DBA = \\angle DAC$\r\n\r\n$\\angle BAC = \\angle BAD + \\angle DAC \\Rightarrow$\r\n$\\angle BAC = \\angle BAD + \\angle DBA \\Rightarrow$\r\n$\\angle BAC = 90^o$\r\n\r\n\r\n$\\hline$\r\n\r\nAt first we should notice that $h$ must be less than (or equal to) $R$ because $OD^2 = R^2-h^2$\r\n(if $h=R$ then $BC$ is a diameter)\r\n\r\nAlso the circles $w_1,w_2$ must have common point(s).\r\n\r\nSo we must have $|OA-h|\\leq OD\\leq OA+h \\Leftrightarrow$\r\n\r\n$|OA-h|\\leq \\sqrt{R^2-h^2}\\leq OA+h$\r\n\r\n\r\nIf one of the equalities holds, then the circles $w_1,w_2$ are touching, so we have one solution for $D$.\r\n\r\nIf $|OA-h|< \\sqrt{R^2-h^2} < OA+h$, then the circles have two common points $D,D'$ and we get two congruent triangles $ABC$ and $AB'C'$", "Solution_2": "Draw the circle with center $A$ and radii $h$. We denote it by $(A)$.\n\nLet $X$ and $Y$ be the intersections of $(A)$ and $(O)$. Let $D \\equiv AO \\cap XY$. Clearly $AO \\perp XY$.\n\nNow construct a circle $T$ with center $A$ and radii $\\frac{AD\\cdot AO}{h}$. I think we can do this, otherwise, we can just construct a segment with this length and proceed.\n\nNotice that the radii of the circle $T$ is greater than $AD$ since we obviously have $AO \\ge h$.\n\nHence circle $T$ has at least one intersection with $XY$, let it be $E$. Let $E'$ be the intersecion of $AE$ with the circumcircle of $EDO$. We have, $\\angle AE'O = \\angle EDO = 90$. Also, $AE \\cdot AE' = AD \\cdot AO \\implies AE' = \\frac{AD \\cdot AO}{AE} = h$. Let $B$ and $C$ be the intersections of $E'O$ with $(O)$. Triangle $ABC$ satisfies the conditions.\n\nIf we had $AO = h$ then just draw the perpendicular to $AO$ at $O$.", "Solution_3": "Construct the circle $(A,h)$, draw the tangents from $O$ to it, they are the two diameters $BC$.\n\nBest regards,\nsunken rock", "Solution_4": "[quote=\"sunken rock\"]Construct the circle $(A,h)$, draw the tangents from $O$ to it, they are the two diameters $BC$.\n\nBest regards,\nsunken rock[/quote]\nBut we can't draw tangents when $O$ is in the circle $(A,h)$." } { "Tag": [], "Problem": "If $ a,b,c \\in \\mathbb{Z}$ so that $ 8a\\minus{}10b\\plus{}27c\\equal{}0$ then prove that:\r\n\r\n$ \\frac{(2a\\minus{}b)(2b\\minus{}3c)(3c\\plus{}2a)}{60} \\in \\mathbb{Z}$.", "Solution_1": "[hide=\"Solution\"]\nFrom the given equation we get that:\n\n$ 2a-b=3(3b-9c-2a)$\n\n$ 2b-3c=8(a-b+3c)$\n\n$ 3c+2a=10(3c+a-b)$\n\nMultiplying these three together we get and dividing by $ 60$:\n\n$ \\dfrac{(2a-b)(2b-3c)(3c+2a)}{60}=4(3b-9c-2a)(a-b+3c)(3c+a-b)$\n\nAnd since $ a,b,c\\in \\mathbb{Z}$ then so is $ 4(3b-9c-2a)(a-b+3c)(3c+a-b)$.\n[/hide]" } { "Tag": [ "limit", "calculus", "integration", "function", "calculus computations" ], "Problem": "These were from a math test at school, and I'm not completely sure I got the right answers:\r\n\r\nEvaluate the following limits:\r\n\r\n$ \\lim_{n\\to{\\infty}} (\\minus{}1)^n * cos((\\frac{2n\\plus{}1}{2})\\pi)$\r\n\r\n$ \\lim_{n\\to{\\infty}} (\\minus{}1)^n * sin((\\frac{2n\\plus{}1}{2})\\pi)$", "Solution_1": "Both sequences are periodic, so just evaluate the first few terms. \r\n\r\nLimits do not belong in the High School forums.", "Solution_2": "I guess that depends where are you from... in Croatia we learn it in high school :)", "Solution_3": "[quote=\"t0rajir0u\"]Both sequences are periodic, so just evaluate the first few terms. \n\nLimits do not belong in the High School forums.[/quote]\r\nI'm sorry, we learn limits at our high school so I assumed everyone did =/.\r\n\r\nI'm still a bit confused as to how evaluating the first few terms gets the limits for those, though.", "Solution_4": "You'll understand once you evaluate the first few terms.", "Solution_5": "[hide=\"solutions\"]1. 0\n2. 1[/hide]", "Solution_6": "Alright, I see how for all positive integer values of n, the first sequence always evaluates as 0, and the second sequence always evaluates as 1. But in between integer values of n, the evaluation jumps around a lot, so doesn't that mean that there's no limit?", "Solution_7": "Hmm. On second thought, the problem is somewhat ambiguous. Usually, though, when we use the letter $ n$ we are talking about sequences. $ (\\minus{}1)^n$, in particular, is very poorly defined at non-integers.", "Solution_8": "Forgive me; I am but an 8th grader, but I know some stuff about limits.\r\nBut perhaps my knowledge is inadequate.\r\n\r\nI thought that ${ \\lim_{n\\to\\{\\infty}} ( - 1)^n$ does not exist because any integral value of n will result in -1 or 1, so there is no definite limit. How can you even evaluate the problem if the problem involves a limit that does not exist?\r\n\r\nUnless you mean:\r\n$ \\lim_{n\\to{\\infty}} [( - 1)^n * cos((\\frac {2n + 1}{2})\\pi)]$\r\nand\r\n$ \\lim_{n\\to{\\infty}} [( - 1)^n * sin((\\frac {2n + 1}{2})\\pi)]$", "Solution_9": "[quote=\"MathAndKnowledge\"]\nUnless you mean:\n$ \\lim_{n\\to{\\infty}} [( \\minus{} 1)^n * cos((\\frac {2n \\plus{} 1}{2})\\pi)]$\nand\n$ \\lim_{n\\to{\\infty}} [( \\minus{} 1)^n * sin((\\frac {2n \\plus{} 1}{2})\\pi)]$[/quote]\n\nI never thought it could be something else.. It's quite unusual for a part of an expression to be evaluated using the $ \\lim$ operator, if actually not incorrect to assume it does. It's like setting a variable of a function to a value but only for certain places in the same expression. If $ n\\rightarrow \\infty$ then it's commonly (and I repeat, if not actually definitely) for all $ n$ in the expression to be evaluated in such manor.\n\n[quote=\"Flame\"]\nI guess that depends where are you from... in Croatia we learn it in high school Smile\n[/quote]\r\n\r\nBome da..", "Solution_10": "[quote=\"Lulze\"]These were from a math test at school, and I'm not completely sure I got the right answers:\n\nEvaluate the following limits:\n\n$ \\lim_{n\\to{\\infty}} ( \\minus{} 1)^n * cos((\\frac {2n \\plus{} 1}{2})\\pi)$\n\n$ \\lim_{n\\to{\\infty}} ( \\minus{} 1)^n * sin((\\frac {2n \\plus{} 1}{2})\\pi)$[/quote]\r\n\r\n$ \\lim_{n\\to{\\infty}} ( \\minus{} 1)^n * cos((\\frac {2n \\plus{} 1}{2})\\pi) \\equal{} \\lim_{n\\to{\\infty}} ( \\minus{} 1)^n.cos( \\frac {\\pi}{2} \\plus{} n \\pi) \\equal{} 0$\r\n\r\n$ \\lim_{n\\to{\\infty}} ( \\minus{} 1)^n * sin((\\frac {2n \\plus{} 1}{2})\\pi) \\equal{} \\lim_{n\\to{\\infty}} ( \\minus{} 1)^n. sin( \\frac {\\pi}{2} \\plus{} n \\pi) \\equal{} 1$" } { "Tag": [ "geometry", "3D geometry", "sphere", "Support", "MIT", "college", "Harvard" ], "Problem": "Goin by the amt of money on stake for research ppl, looks like ill make more money marketing for chinna vayasu sema talent than doing research in mathematics...", "Solution_1": "@chappli ...\r\npeople don't do research for money ofc whatever comes in the way won't be or shudn't be denied :D \r\nthey do that for fun or atleast that's what i think it's os good to be pais for what u like doing the most :D", "Solution_2": "[quote=\"pardesi\"]@chappli ...\nthey do that for fun or atleast that's what i think it's os good to be pais for what u like doing the most :D[/quote]\r\n\r\nPerfect point! Sadly researchers dont get paid too much today. Neither do teachers.... Thats why we are going to the dogs and progressing narrowly. :(", "Solution_3": "but things have certainly improved what they were 2-3 years ago just look at the number of peopl applied for NEST :D ...\r\nyes money is imp.....but u shud do something for ur nation(ofc - paying taxes :P )", "Solution_4": "[quote=\"pardesi\"]but things have certainly improved what they were 2-3 years ago just look at the number of peopl applied for NEST :D ...\nyes money is imp.....but u shud do something for ur nation(ofc - paying taxes :P )[/quote]\r\n\r\nNo, If our teachers are paid more? Then surely they will be encouraged to give their heart and soul to take class. I personally know that in sum schools middle school teachers are being paid paltry amounts and in some even high school teachers.\r\n\r\nMoney is imp. Only if you have some stability by the side, you will do research happily giving your heart and soul knowing that you are earning bread for your family...And majority go for job after 4 year courses except the focussed few.\r\n\r\nDoes India not have the quality to produce scientists with superior technology? Does India not have the tech skills to produce a desi version of Google, which can go one up? Or can we not produce people who can solve the Yang Mills Mass-Gap problem or Birch & ... Dyer??? Surely we can.\r\n\r\nPardesi, start a new thread, this will be an interesting topic.", "Solution_5": "Guys cool down.. Plz understand wat I meant.. I jus meant researchers ought to be paid more than wat they are given.. Its such a sad sight in India.. As a matter of fact, I will only do research in my career as I personally feel job satisfaction is much more important for me as a human being..I want to work for myself, and not for another guy and his country..\r\nAnd I think shre very well knows what my plans in life are, so I thot u wud atleast understand what I meant..Im disappointed by ur interpretations!!", "Solution_6": "I think this problem would change as more and more people are getting to understand the valid point which Chappli put forward......\r\nI think money for resarch would not be a problem from now on, keeping in mind so many are ready to sponsor researchers", "Solution_7": "yes money is not a problem..problem is mindset...even a millionare can be 'poor' :D", "Solution_8": "I dont think its a question of getting paid at all. \r\n\r\nBut as others have pointed out, the research potential at India is very very minimal, and to top that, the recognition and FUNDING for research is barely sufficient, so it productively prevents more people from going into research. If you work for a top notch robotics or bio tech corp, YOU get paid well enough, its that your project is put on hold until and unless sponsors deem it fit to go into production\r\n\r\nOn the other hand, abroad, wider ranging research opportunities are infinitely more available, and in all spheres, but then again, that could be because they are more developed and the govt can afford to fund such research.\r\n\r\nOur institutions are doing pretty well at the outset, but then if only they were to accept colloborations with the institutes of the likes of those abroad, and by that I dont only mean US, even places like Singapore, where resources are technically not abounding, we could do much better\r\n\r\nIIT is definitely on par with the top technical universities in the world, just that, the exposure to students is very minimal, and placements are becoming an issue day by day, with more people preferring to do an MBA to settle in the corporate grind rather than go into research.....", "Solution_9": "And as someone who lectured in our school mentioned, iit has lesser of practical training than many other foreign univs which are @ lower standards than iit :(", "Solution_10": "i knwo the research oppurtunities outside are far far far better than in india...but i wud prefer anyday doing it here because doing research outside is as good as selling pancakes(Corporate) in India :) \r\nit's a duty we must do something for our nation....\r\nif it so happens that say 'x' loves do research on 'y' then he/she shud do it here otherwise that find will go 'waste'..\r\nour society has given us many things and it deserves much more respect....\r\none C.V.Raman in a century is better than 20 Haragobind Khoronas :rotfl:", "Solution_11": "@Rohit: yes, that is one of the predominant cause of concern here\r\n\r\n@pardesi: You completely misinterpreted what I said... I never talked about opportunities at all, nor do I undermime the intellect of all those who want to pursue research. But fact is, funding, very clearly monetary support for taking up research isnt available. Ultimately you may argue that we dont need support if we've got the will, but I say that complete rubbish.... You need equipment, you need samplings, and you need testing outlets(for all branches of science perhaps, maybe not math) and for that you need funding, and for that your govt should support research or you should have ties with other sponsors who are willing to fund such research.\r\n\r\nI said India lacks INITIATIVE... Like what baby said, IIT's are far superior to colleges like say, Michigan School of Engineering, Georgia Tech or Viriginia Tech, but they give hands on exp.. No guesses as to which one potential research scientists would prefer dont you think?\r\n\r\nAnd I completely resent that statement about Raman and Khurana. No scientist is less than the other, quite literally. It doesnt do good to downsize one. \r\n\r\nAs a biologist, I think that Khurana's contribution is equally important, for without that, stem cell research of today would be damn near impossible. So never undermine anyone..", "Solution_12": "in india in every decade research hits new lows :| \r\nnot a single indian research paper published in \"nature\" or \"science\" since 2 or 3 decades\r\nindias best journal has an impact factor of 0.6 against the world average of 8.45\r\nin india one research paper of citation above 1000 is something like 1 in 100000 :roll: \r\nfor the matter of fact not a single indian research paper has had citation above 500 in this decade :(", "Solution_13": "I think for certain people, Parental Pressure would be a problem, They might just want them to be like the rest", "Solution_14": "[quote=\"anitsahu\"]in india in every decade research hits new lows :| \nnot a single indian research paper published in \"nature\" or \"science\" since 2 or 3 decades\nindias best journal has an impact factor of 0.6 against the world average of 8.45\nin india one research paper of citation above 1000 is something like 1 in 100000 :roll: \nfor the matter of fact not a single indian research paper has had citation above 500 in this decade :([/quote]\r\n\r\nYes that's what govi would say, that any indian scientist would rather put his work in a foreign journal rather than put his work in an indian journal and not get recognition :(\r\n\r\nEDIT: For once anit we agree on a topic :P :rotfl:", "Solution_15": "hi i am rt.\r\nwanna study nuclear physics or molecular exobiology from MIT or Harvard or Princeton. :rotfl: :rotfl: \r\n\r\nbtw shruti is at Penn State and Ashwath at Princeton......keep it up people....u all r geniuses........", "Solution_16": "dont worry,u'll be among them by next year :) \r\n\r\nAnd my views abt Hitlerism (:P),really it can do good\r\ni endorse it fully,but dont u think some of us may land up in concentration camps then :rotfl: \r\n\r\nIts time for India to attack US:those bloody hells attract Indians like *******:", "Solution_17": "no one discusses abt lord hitler :omighty: without me pokin my nose\r\n\r\nwe wudn't land up in conc camps but \r\n[hide]\nlok sabha, vidhan sabha etc wud have to be disinfected using zyklon b :rotfl: \n[/hide]\r\n\r\nattacking america is not the solution, its suicide \r\n\r\ni think dis discussion shud be moved to metathread", "Solution_18": "[quote=\"rt_08\"]Ashwath at Princeton[/quote]\r\n\r\nNot yet, sadly. I have four more months of waiting to survive before I leave for Princeton.", "Solution_19": "what rank is the cutoff for the Integrated MSc in chemistry ????? can I get it in IIT-B ????? heard it's got best faculty for chem. \r\n\r\nsome advice please\r\n\r\nmy marks are[hide]220 around that :blush: :blush: [/hide]", "Solution_20": "[quote=\"ashwath.rabindranath\"]Not yet, sadly. I have four more months of waiting to survive before I leave for Princeton.[/quote]\r\nofc u'll survive:rofl:\r\nbest of luck for that!!!\r\n\r\n\r\nand plz plz do come back to INdia :o", "Solution_21": "yes ashwath do come back to india.......v need u all........[b]JAI HO[/b]", "Solution_22": "A tradition doesn't bind you with a compulsion that you can't be different from others.\r\nWhen I say I WILL come back, I mean it. Ofc, the time period may vary in some highly improbable scenario like me getting the opportunity to do some pathbraking research that benefits ALL mankind; Not a particular country.\r\n\r\nAshwanath? Can you post your full profile? Besides INMO TC.", "Solution_23": "there is no use coming back after say 50yrs :rotfl: coz ppl will then call u NRA (Non reliable americans)", "Solution_24": "[quote=\"rt_08\"]yes ashwath do come back to india.......v need u all........[b]JAI HO[/b][/quote]\r\n\r\nWhat for? \r\n\r\nAnd for the record, my name is Ashwath. Not Asswat. Not Asvanath. Not Arshavd (yes, people have been very cruel to my name before). Not Aswath. ASHWATH.\r\n\r\nLet's talk about the future later. Right now, I don't see myself returning to India for the next 10-15 years once I leave, except for holidays and stuff.", "Solution_25": "[quote=\"rituraj007\"]there is no use coming back after say 50yrs :rotfl: coz ppl will then call u NRA (Non reliable americans)[/quote]\r\nI'll be glad if I even live that long :P\r\nSo the stipulated time period is much lesser than that :P\r\nAnd don't worry, I'm a poetic idiot! Never manged to take life;'s practicality as it should be. :blush:\r\nReminds me of a long discussion with rt once. Some interview ques he got. He'll probably elaborate with more clarity :)", "Solution_26": "well interview s***** big time and i am in no mood of elaborating that acriminous session....... :mad:", "Solution_27": "GRRRRRRRRRR!!!!!!!!!!!\r\nNow I've been rejected from both Summer Science Program and International Summer School for Young Physicists. Just got my rejection letter for the latter. Some South Indian Champu got in instead. http://www.perimeterinstitute.ca/Outreach/Students/ISSYP_2009_Participant_List/ (Note that 50% of the international participants have surnames starting with k/l/m) From what google shows, he doesn't have any big physics olympiad/kvpy/ntse etc. I thought my application had enough passion in it to get through :(\r\nIf I can't even get into a good summer program, what hope do I have left for top US Univs..... :(", "Solution_28": "Dont worry yaar,US admission procedures are far more different and in depth in nature.Rejection from a Summer Program nvr means that ur chances are low at an ivy or a good US university :)", "Solution_29": "WOW! Worth reading from start! :) " } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all the continuous functions $ f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that:\r\n\r\n$ f(x\\plus{}y\\minus{}xy)\\equal{}f(x\\plus{}y)\\minus{}f(x)f(y)$, for all $ x,y \\in \\mathbb{R}$", "Solution_1": "$ P(x,0): f(x)f(0)\\equal{}0$. $ f(x)\\equal{}0$ is one solution.\r\nNow assume $ f(0)\\equal{}0$ and let $ a\\equal{}f(1)$.\r\n$ P(x,1): f(x\\plus{}1)\\equal{}a(f(x)\\plus{}1)$ (*).\r\n$ P(1,\\minus{}1): a(1\\plus{}f(\\minus{}1))\\equal{}0$\r\nIf $ a\\equal{}0$, (*) implies $ f(x)\\equal{}0$. So $ f(\\minus{}1)\\equal{}\\minus{}1$.\r\n$ P(x,\\minus{}1): f(2x\\minus{}1)\\equal{}f(x\\minus{}1)\\plus{}f(x)$ (**)\r\nSo $ f(3)\\equal{}f(2)\\plus{}a$, \r\n(*) gives $ f(2)\\equal{}a^2\\plus{}a$ and $ f(3)\\equal{}a^3\\plus{}a^2\\plus{}a$\r\nHence, $ a^3\\plus{}a^2\\plus{}a\\equal{}a^2\\plus{}2a\\Rightarrow a\\equal{}0,\\pm1$. We already exclude $ a\\equal{}0$.\r\nIf $ a\\equal{}\\minus{}1$, (*) gives $ f(x\\plus{}1)\\plus{}f(x)\\equal{}\\minus{}1$ and then (**) implies $ f(2x\\minus{}1)\\equal{}\\minus{}1$, contradiction with $ f(0)\\equal{}0$.\r\n\r\nHence, $ a\\equal{}1$ and $ f(x\\plus{}1)\\equal{}f(x)\\plus{}1$.\r\nWe have $ f(n)\\equal{}n,\\forall n\\in\\mathbb Z$ and $ f(x\\plus{}n)\\equal{}f(x)\\plus{}n$.\r\n$ P(x,\\minus{}n\\plus{}1): f(nx)\\equal{}nf(x)$. So $ f(r)\\equal{}r,r\\in\\mathbb Q$ and $ f(rx)\\equal{}rf(x)$\r\n$ f(x\\plus{}r)\\equal{}rf(\\frac xr\\plus{}1)\\equal{}rf(\\frac xr)\\plus{}r\\equal{}f(x)\\plus{}r$ (F1)\r\n$ P(x,\\minus{}x): f(x^2)\\equal{}\\minus{}f(x)f(\\minus{}x)\\equal{}f(x)^2$ (F2)\r\nSo $ f(x)\\ge0,\\forall x\\ge0$.\r\nFor any $ x$, we can find rational sequences $ r_n\\uparrow x,R_n\\downarrow x$\r\n$ f(x)\\equal{}f(r_n)\\plus{}f(x\\minus{}r_n)\\ge f(r_n)\\equal{}r_n$ and similarly $ f(x)\\le f(R_n)\\equal{}R_n$.\r\nHence, $ f(x)\\equal{}x$.\r\n\r\nEither $ f(x)\\equal{}0$ or $ f(x)\\equal{}x$. We do not need continuity." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "prove that for all positive integers $n,m$ so that $m$ is odd we have\r\n $ 3^{m} n \\mid\\sum_{k=0}^{m}\\binom {3m}{3k} \\ (3n-1)^k. $", "Solution_1": "it's Romanian ROM TST 2004: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=3%5Em%2An&t=5487" } { "Tag": [ "limit", "trigonometry", "function", "calculus", "calculus computations" ], "Problem": "There is a classic example to show that if the limit produced by application of L'Hopital does not exist, then the original limit may or may not exist: $ \\lim_{x\\to\\infty} \\frac {x \\plus{} \\sin x}{x}.$\r\n\r\nI wish to disprove: Suppose $ f$ and $ g$ are differentiable functions on an open interval containing $ x \\equal{} a$, except possibly at $ x \\equal{} a$, and that $ \\lim_{x\\to a} f(x) \\equal{} \\lim_{x\\to a} g(x) \\equal{} 0$. If $ \\lim_{x\\to a} \\frac {f'(x)}{g'(x)}$ does not exist, then $ \\lim_{x\\to a} \\frac {f(x)}{g(x)}$ does not exist.", "Solution_1": "Can anyone provide a counter-example?", "Solution_2": "Try $ \\frac{x^2 \\sin (1/x)}{x}$ as $ x \\rightarrow 0$.", "Solution_3": "Thanks. Should have thought of that, using $ \\minus{}1 \\leq \\sin(1/x) \\leq 1.$", "Solution_4": "Or just turn $ x$ upside-down in the first example and use ${ \\frac{\\frac{1}{x} + \\sin \\frac{1}{x}}{\\frac{1}{x}}}$. (Of course, these all work because $ \\sin x$ is like $ 1$ but squiggly, while $ \\frac{d}{dx} \\sin x$ is still like $ 1$ but squiggly instead of being like $ 0$.)\r\n\r\n(The terminology \"like 1 but squiggly\" is a fabulous way of thinking about things that I'm either paraphrasing or directly quoting from Kent Merryfield.)", "Solution_5": "[quote=\"JBL\"](The terminology \"like 1 but squiggly\" is a fabulous way of thinking about things that I'm either paraphrasing or directly quoting from Kent Merryfield.)[/quote]\r\nI think I usually say \"wiggly\" or \"wigglier\" rather than than \"squiggly\" - but it works with either word." } { "Tag": [ "AMC", "AMC 12", "USA(J)MO", "USAMO", "AIME", "geometry" ], "Problem": "I love doing math but I have to take AP classes to get into a good college. But I thought maybe colleges would be impressed by having a high score on the AMC/AIME/USAMO so I could possibly drop my AP and honors classes and have more time to train in mathematics for these competitions. I've heard that it is more rare to make a perfect score on the AMC12 and only 10,000 people are invited to the AIME. I'm sure the people who qualify for the USAMO or IMO must sacrifice their grades in order to be that good. Would it be wise to concentrate on something I love and will major in college or take honors and AP classes and have less time to do math.", "Solution_1": "[quote=\"jobro_gold\"]I'm sure the people who qualify for the USAMO or IMO must sacrifice their grades in order to be that good.[/quote]\r\n\r\nThat's the part I am not at all sure about. I rather suspect that there is a sufficient number of USAMO qualifiers who are really good on AP tests (including AP tests in non-math subjects) that it would be a dangerous strategy to concentrate only on USAMO to the exclusion of taking a challenging general curriculum in high school. \r\n\r\nIn general, I advise parents and young people not to view college admission requirements as a reason NOT to do something, but as a reason to DO something. So the thinking I would apply to the issue you are concerned about is, \"Hey, lots of kids in my high school are taking AP classes, and I can keep up with them, but I'm also going to prepare for the AMC series of contests and try my best to make MOP.\" Note that it's at least theoretically possible to STILL get dinged in college admission to your favorite school if you are an AP scholar with distinction and an IMO team member and someone who scored a pair of 800s on the SAT I (although that wouldn't be likely :D ). Colleges aren't required to admit anybody--not in the United States they aren't, if they are in the top tier of privately operated colleges. \r\n\r\nEspecially because many of the mistakes I read about here on AoPS that cause people to crash on the AIME are essentially reading mistakes, I think you are best advised doing whatever you can, throughout your high school years, to build up your reading ability. One can never read too well to be good college material. If an AP English course helps you read better than you have ever read before, that would be a good reason to take the course, whether or not you think you need the score on the AP exam. \r\n\r\nPick the path that fits you, then work hard to move forward along it. \r\n\r\nGood luck!", "Solution_2": "I honestly think that no English class is going to help you read the questions on the AIME more accurately. The only thing that can do that is not being in such a hurry. You get twelve minutes per question if you want to get a perfect score, so if you're trying to solve each problem in two minutes, that's not so good. (Wow, I really should listen to my own advice.)", "Solution_3": "[quote=\"jobro_gold\"]I have to take AP classes to get into a good college.[/quote]\r\n\r\nI think the concept, which I have complained repeatedly about before, that anyone \"has\" to do anything to get into a \"good\" college is just wrong. There are a wide variety of things which will help a student get into selective universities, and many different students have many different combinations of these studies, attributes, activities, and whatnot. Perhaps more annoying, there are many different students, who had very similar combinations of activities, studies, attributes, and whatnot, who did [i]not[/i] get into the same selective universities. It is highly unlikely that by picking between various activities, you will increase your chances of getting into college; more annoyingly, there is no one, anywhere, who would be able to tell you definatively which combinations would be best. This, I guess, is just a corollary to tokenadult's mantra that \"college admission requirements should not be a reason NOT to do something, but should be a reason to DO something.\" Frankly, you're probably best off doing whatever you prefer to do, because doing what you like reverberates in other areas: it makes for something to talk about at interviews, it makes for something to write about in essays, and it makes for something that you are more likely to do in your free time and thus more likely to do well in. \r\n\r\nAs for my personal interaction with the things, I took 5 AP classes/tests, counting 2 half-year classes and both Phsyics C exams, during high school.", "Solution_4": "What JBL said is exactly what I was thinking. Not everyone has to take the same course of action (CoA :D ) to be accepted into a decent college. Anyway, I enjoy doing math and I'm extremely stressed out by the AP class I have now, AP World History. So I say, the only way I'm going do work ( I'm a moderate slacker and procrastinator) is if I'm having fun. Then, it doesn't seem like work anymore. But most people don't even understand why I do math for fun so how can they understand my ideology. But, if I could get some opinions from college staff members - I know their must be some who go to AoPS - that would be EXTREMELY :shock: helpful." } { "Tag": [], "Problem": "It is impossible to find two positive integers such that the square of one is exactly twice the square of the other. However, there are integers\nsuch that the square of one is one less than twice the square of the other.\nFind the product of the smallest two positive integers where this is true.", "Solution_1": "Testing leads to (5,7), so the product is 35.", "Solution_2": "quick question here\r\nwhy can't it be (1,1)? \r\nthe question doesn't state that they have to be different" } { "Tag": [], "Problem": "hey i saw this qsn in a test i wrote. i had no idea how to do it. :(\r\n\r\nIf 2 moles of A and 3 moles of B are mixed to form an ideal soln, vap pressure of A and B are 120 and 180 mm of Hg respectively. \r\n\r\n1. What are the compositions of A and B in the vapour phase when the first traces of vapours are formed??\r\n\r\n2. At what pressure will the last traces of liquid disappear?\r\n\r\nAny ideas on how to proceed?", "Solution_1": "have u given the vapour presuure of the pure component or the pressures at equilibrium", "Solution_2": "Its very easy\r\n[hide=\"idea\"]\n1.separately find xa and xb and use raoult's law to find Pressure in vapour phase\n2.Then use Dalton's law of partial pressures then that's all\n[/hide]", "Solution_3": "[quote=\"valeriummaximum\"]have u given the vapour presuure of the pure component or the pressures at equilibrium[/quote]\r\n\r\npure component.", "Solution_4": "chemrock-- you will only get the conc of the two components at equilibrium i.e the equlibrium vapour pressure", "Solution_5": "[quote=\"chemrock\"]\n2.Then use Dalton's law of partial pressures then that's all\n[/hide][/quote]\r\n\r\ncan u explain a bit more? how do i connect that to the composition when the first traces of vapours are formed??", "Solution_6": "It's just a twist that's all:\r\nLet Pa0 & Pb0 be the pressures of pure solvent.\r\nP=Pa0.xa + Pb0.xb\r\nP.x'a=Pa0.xa\r\nand then x'a is the compostion in vapour phase of A.", "Solution_7": "Vp of the pure component is given \r\nAlso you are not given the Vp at equilibrium..\r\nSo,I think some data is missing???", "Solution_8": "@chemrock: your formulae are right no doubt. :) But then how do the answers to the 2 qsns i asked differ??", "Solution_9": "I think the second is related to boiling point that is vapour pressure =atmospheric pressure :maybe: :maybe: :maybe:" } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Assume that the series $ \\displaystyle\\sum_{n\\equal{}1}^{\\infty}a_n$ is convergent and $ a_n>0$. Show that there exist a sequence $ b_1\\leq b_2\\leq\\ldots\\leq b_n\\leq\\ldots$ such that $ \\displaystyle\\lim_{n\\to\\infty}b_n\\equal{}\\infty$ and the series $ \\displaystyle\\sum_{n\\equal{}1}^{\\infty}a_nb_n$ is convergent.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=202699", "Solution_2": "Thanks, beautiful solution..." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let $ x,y\\in\\mathbb{R}_\\plus{}$ such that $ x^n\\plus{}y^n\\equal{}1(n>1,n\\in\\mathbb{N}_\\plus{})$,\r\n\r\nThen find the maximum of $ x\\plus{}4y$", "Solution_1": "write the n-th power of x+4y on the left side, and the things only left is to fill in the blank\r\n\r\nWhat is the generalized form of Cauchy-Schwarz inequation?\r\n\r\nWell, not every inequality member know it. I'll show it here.\r\n\r\nFor positive integers m,n and positive reals from x11 to xmn(Failed to use subscripts here, you know what I say, don't you?)\r\n(x11^m+x12^m+\u2026\u2026+x1n^m)(x21^m+x22^m+\u2026\u2026+x2n^m)\u2026\u2026(xm1^m+xm2^m+\u2026\u2026xmn^m)\r\nis not less than\r\n(x11x21\u2026\u2026xm1+x12x22\u2026\u2026xm2+\u2026\u2026+x1nx2n\u2026\u2026xmn)^m", "Solution_2": "let $ a\\equal{}2^{\\frac {2n}{n\\minus{}1}}$.\r\nwith AM-GM:\r\n$ \\frac {(n\\minus{}1)a}{a\\plus{}1} \\plus{} y^n \\geq 4ny.(a\\plus{}1)^{\\frac {\\minus{}n\\plus{}1}{n}}$\r\nand\r\n$ \\frac {n\\minus{}1}{a\\plus{}1} \\plus{} x^n \\geq nx.(a\\plus{}1)^{\\frac {\\minus{}n\\plus{}1}{n}}$\r\nso\r\n$ (n\\minus{}1)\\plus{}x^n\\plus{}y^n \\geq n.(x\\plus{}4y).(a\\plus{}1)^{\\frac {\\minus{}n\\plus{}1}{n}}$\r\nso\r\n$ x\\plus{}4y \\leq (a\\plus{}1)^{\\frac {n\\minus{}1}{n}}$." } { "Tag": [ "inequalities", "trigonometry", "function" ], "Problem": "Triangle ABC is acute-angled\r\n\r\n\r\nProve that SinA +SinB+SinC > CosA+ CosB +CosC\r\n\r\nThere are many possible proofs of this result", "Solution_1": "Here is one of the proofs.\r\nTriangle $ABC$ is acute-angled so the both sides of the given inequality are positive.Using this we can square this inequality.So we have to prove that\r\n$\\sum_{cyclic}{(\\cos^{2}{A}-\\sin^{2}{A})}$$+\\sum_{cyclic}(\\cos{A}\\cdot\\cos{B}-\\sin{A}\\cdot\\sin{B})<0$\r\nUsing that $\\sin^{2}{A}+\\cos^{2}{A}=1$ we get\r\n$\\sum_{cyclic}{(2\\cdot\\cos^{2}{A}-1)}+2\\cdot\\sum_{cyclic}{\\cos(A+B)} <0$ or\r\n $t=2\\cdot\\sum_{cyclic}{\\cos{A}\\cdot(\\cos{A}-1)} <3$.\r\nBut $0<\\cos{x}<1$ there $0 \\cos(3\\pi/4 - A - B)$\r\n\r\nThe RHS is $\\cos(3\\pi/4)\\cos(A+B) + \\sin(3\\pi/4)\\sin(A+B)$\r\n\r\nThe first term is clearly negative, so we need to prove that\r\n\r\n$\\sin A + \\sin B > \\sin(3\\pi/4)\\sin(A+B) $\r\n\r\nBut \r\n\r\n$\\sin(3\\pi/4)\\sin(A+B) = \\sqrt{2}/2 \\sin A \\cos A + \\sqrt{2}/2 \\sin B \\cos B$\r\n\r\nSince $\\cos A, \\cos B \\leq 1$, we are done.\r\n\r\nI really wanted a proof by convexity/concavity, though :(", "Solution_3": "One more note. The LHS is > 2, the RHS is < 2. Anyone have a simple way of proving this, which trivializes the result?", "Solution_4": "looks like jensen's to me", "Solution_5": "Jensen's was (of course) the first thing I tried.\r\n\r\n$f(x) = \\sin x - \\cos x$ is not convex over $(0,\\pi]$\r\n\r\nHowever, there's a cool trick used when finding the minimum of $\\cos x +\\cos y + \\cos z$ where $x,y,z$ are angles of a triangle, I wonder if it will work here.", "Solution_6": "Blahblahblah, You wanted to apply convexity/concavity.In this problem you can only apply Jensens inequality to suitable bounding functions as follows:\r\n\r\nSinA =2SinA/2.CosA/2 >2(SinA/2)^2 , as angles are acute.\r\n\r\nTherefore SinA +SinB +SinC>2[ (SinA/2)^2 +(SinB/2)^2 +(SinC/2)^2 ]\r\n\r\nThe function (SinX/2)^2 is convex on the interval 03/2\r\n\r\nNow we find an upper bound for the sum of the cosines \r\n\r\nCosA +Cos B =2cosA+B/2.CosA-B/2\r\n <=2 sinC/2\r\n\r\nTHerefore CosA +COsB +CosC <=SInA/2 +SinB/2 +SinC/2\r\n\r\nThe function SinX/2 is concave on the intereval 02 in an acute angled triangle\r\nSinA +SinB+SinC >SinA^2 +SinB^2 +SinC^2 = 2 +2CosACosBCosC >2 since the cosines of acute angles are positive", "Solution_7": "At first, I just wanted to note that in Armo's nice solution\r\n\r\n[quote=\"Armo\"]Here is one of the proofs.\nTriangle $ABC$ is acute-angled so the both sides of the given inequality are positive.Using this we can square this inequality.So we have to prove that\n$\\sum_{cyclic}{(\\cos^{2}{A}-\\sin^{2}{A})}+\\sum_{cyclic}(\\cos{A}\\cdot\\cos{B}-\\sin{A}\\cdot\\sin{B})<0$\nUsing that $\\sin^{2}{A}+\\cos^{2}{A}=1$ we get\n$\\sum_{cyclic}{(2\\cdot\\cos^{2}{A}-1)}+2\\cdot\\sum_{cyclic}{\\cos(A+B)} <0$ or\n $t=2\\cdot\\sum_{cyclic}{\\cos{A}\\cdot(\\cos{A}-1)} <3$.\nBut $0<\\cos{x}<1$ there $0 2$ and $\\cos A + \\cos B + \\cos C \\leq \\frac32$. The first of these two inequalities was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=5379 ; the second one is well-known and holds not only for acute triangles.\r\n\r\n Darij", "Solution_8": "This is easy! :D \r\n$\\cos A+ \\cos B=2\\cos \\frac{A+B}{2}\\cos \\frac{A-B}{2}=2\\sin \\frac{C}{2}\\cos \\frac{A-B}{2}\\leqq2\\sin \\frac{C}{2}$, similarly $\\cos B+\\cos C\\leqq2\\sin \\frac{A}{2}$, $\\cos C+\\cos A\\leqq2\\sin \\frac{B}{2}$, \r\nadding up these inequality and dividing by 2 both of sides,we have $\\cos A+\\cos B+\\cos C\\leqq\\sin \\frac{A}{2}+\\sin \\frac{B}{2}+\\sin \\frac{C}{2}<\\sin A+\\sin B+\\sin C$,because when $0<\\frac{A}{2}<\\frac{\\pi}{4}$, we get $\\frac{1}{\\sqrt{2}}<\\cos \\frac{A}{2}<1$.Thus we obtain $\\sin\\frac{A}{2}<\\sin A$ etc.\r\n\r\nkunny" } { "Tag": [ "inequalities", "number theory proposed", "number theory" ], "Problem": "prove that :$ [1,2,3,...,n] \\geq 2^{n\\minus{}1}$", "Solution_1": "[quote=\"stvs_f\"]prove that :$ [1,2,3,...,n] \\geq 2^{n \\minus{} 1}$[/quote] \r\n\r\nHemmm, what's the meaning of this inequality ? interval greater than integer ? :huh:", "Solution_2": "($ n \\in N$)what is the problem?", "Solution_3": "[quote=\"stvs_f\"]($ n \\in N$)what is the problem?[/quote]\r\n\r\nI'm sorry, but I dont understand the notation $ [1,2,...,n]$. What is the meaning of this expression ?", "Solution_4": "[quote=\"stvs_f\"]prove that :$ [1,2,3,...,n] \\geq 2^{n \\minus{} 1}$[/quote]\r\nthat mean: prove that: $ lcm(1,2,...,n)\\geq 2^n$", "Solution_5": "then prove that lcm(1,...,n)<4^(n-1) and deduce that 1/(a) + 1/(a+b) + ... + 1/(a+nb) isn't an integer(a and b are positive coprime numbers)", "Solution_6": "[quote=\"math10\"][quote=\"stvs_f\"]prove that :$ [1,2,3,...,n] \\geq 2^{n \\minus{} 1}$[/quote]\nthat mean: prove that: $ lcm(1,2,...,n)\\geq 2^n$[/quote]\r\n $ lcm(1,2,...,n)\\geq 2^{n\\minus{}1}$", "Solution_7": "any solution? :(" } { "Tag": [ "integration", "logarithms", "MIT", "college", "real analysis", "real analysis unsolved" ], "Problem": "Here are three series including binom.\r\n(1)$ \\sum_{n\\equal{}1}^{\\infty }\\frac{(\\minus{}1)^{n}}{n\\binom{2n}{n}}$\r\n(2)$ \\sum_{n\\equal{}1}^{\\infty }\\frac{(\\minus{}1)^{n}}{n^{2}\\binom{2n}{n}}$\r\n(3)$ \\sum_{n\\equal{}1}^{\\infty }\\frac{(\\minus{}1)^{n}}{n^{3}\\binom{2n}{n}}$", "Solution_1": "hello, do you want to know the values of your sums?\r\nSonnhard.", "Solution_2": "Use the identity $ \\int_{0}^{1} t^n (1 \\minus{} t)^{n\\minus{}1} \\, dt \\equal{} \\frac{n! (n\\minus{}1)!}{(2n)!} \\equal{} \\frac{1}{n {2n \\choose n}}$. This takes care of the first two sums but the third I'm not completely sure about.", "Solution_3": "hello, for the first sum i have got\r\n$ \\sum_{n \\equal{} 1}^{\\infty}\\frac {( \\minus{} 1)^n}{n\\binom{2n}{n}} \\equal{} \\minus{} \\frac {2}{5}\\sqrt{5}arsinh(\\frac {1}{2})$.\r\nSonnhard.", "Solution_4": "Thanks for your replies.I solved (2),and it is $ \\minus{}\\ln^{2} \\frac{\\sqrt{2}\\minus{}1}{2}$.\r\nBut (3) is still difficult.", "Solution_5": "hello, for 2) i have got\r\n$ \\sum_{n\\equal{}1}^{\\infty}\\frac{(\\minus{}1)^n}{n^2\\binom{2n}{n}}\\equal{}\\minus{}2arsinh(\\frac{1}{2})$\r\nand for 3)\r\n$ \\sum_{n\\equal{}1}^{\\infty}\\frac{(\\minus{}1)^n}{n^3\\binom{2n}{n}}\\equal{}\\minus{}\\frac{2}{5}\\zeta(3)$.\r\nSonnhard.", "Solution_6": "Allow me to change the sign operator, after all it makes no difference.\r\n\r\nI can only confirm Sonnhard's statement, that the first series $ \\sum_{n \\equal{} 1}^\\infty \\frac {( \\minus{} 1)^{n \\minus{} 1}}{n\\, {2n\\choose n}}$ \r\n\r\nconverges to $ \\frac25 \\, \\text{arsinh}\\left(\\frac12\\right)\\cdot \\sqrt {5}$, or equivalently to $ \\frac2{\\sqrt {5}} \\, \\text{arsinh}\\left(\\frac12\\right)$ .\r\n\r\nYou also get this answer from Maple and Mathematica.\r\n\r\nThe second series $ \\sum_{n \\equal{} 1}^\\infty \\frac {( \\minus{} 1)^{n \\minus{} 1}}{n^2\\, {2n\\choose n}}$ is $ 2\\, \\text{arsinh}^2\\left(\\frac12\\right) \\equal{} 2\\log^2\\phi$ ,\r\n\r\nwhere $ \\phi \\equal{} \\frac {1 \\plus{} \\sqrt {5}}{2}$ stands for the golden ratio. \r\n\r\nThis value is also stored in both computer algebra systems.\r\n\r\nThe third series $ \\sum_{n \\equal{} 1}^\\infty \\frac {( \\minus{} 1)^{n \\minus{} 1}}{n^3\\, {2n\\choose n}}$ seems to be the most difficult of them.\r\n\r\nHere the value $ \\frac25 \\zeta(3)$ is only available in Mathematica.\r\n\r\nIf you want to know, how to prove, that these series match with their corresponding values,\r\n\r\nyou can try to read this [url=http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Unendliche_Reihen#Reihen_mit_zentrierten_Binomialkoeffizienten]part[/url] of the German wikibooks formula collection.", "Solution_7": "For one method of proving (3), check equations 36, 37 of this\r\nhttp://mathworld.wolfram.com/AperysConstant.html" } { "Tag": [ "function", "integration", "calculus", "real analysis", "complex analysis", "complex analysis unsolved" ], "Problem": "Fix $a > 0$, $p > 1$, and consider any function $f$ s.t. the Fourier inversion formula holds. Also define the Fourier transform $\\hat{f}: t \\mapsto \\int_{-\\infty}^{\\infty}f(x) e^{-2\\pi i t x}\\, dx$, and assume $\\hat{f}(t) = O\\left(e^{-\\frac{2\\pi a^{p}}{p}|t|^{p}}\\right)$ as $|t| \\to \\infty$. Now let $q = \\frac{p}{p-1}$ so that $\\frac{1}{p}+\\frac{1}{q}= 1$. Prove that $f$ can be extended to a entire fct. (easy) and for $b > \\frac{2\\pi}{qa^{q}}$, there exists $A_{b}> 0$ s.t. $|f(z)| \\le A_{b}e^{b|z|^{q}}$ for all $z \\in \\mathbb{C}$ (hard).\r\n\r\nHint: $|ty| \\le |t|^{p}\\frac{a^{p}}{p}+|y|^{q}\\frac{1}{qa^{q}}$ by AM-GM.", "Solution_1": "Aww... no takers? :) I guess this forum doesn't like Fourier analysis very much.", "Solution_2": "[quote=\"fairyofwind\"]\n\nHint: $|ty| \\le |t|^{p}\\frac{a^{p}}{p}+|y|^{q}\\frac{1}{qa^{q}}$ by AM-GM.[/quote]\r\n\r\nDont you mean \"Holder inequality\".\r\nHa ha!", "Solution_3": "Well... weighted AM-GM. Blah.", "Solution_4": "Ignore sylow_theory, he's just trolling. You'll get more responses on Fourier analysis problems in the Calculus - Real Analysis section though; it gets a lot more traffic than does this forum." } { "Tag": [ "quadratics", "geometry theorems", "geometry" ], "Problem": "I want write a theorem with this i discovered in the high school when i was 16 years old\r\n\\[ \\begin{array}{l} {\\rm x}^{\\rm 2} \\plus{} (y \\minus{} 1)^2 \\equal{} 4{\\rm } \\\\\r\n{\\rm (0}{\\rm ,1)} \\\\\r\n\\left\\{ {_{\\frac {{\\partial (y \\minus{} 1)^2 }}{{\\partial y}} \\equal{} \\frac {{\\partial (y^2 \\minus{} 2y \\plus{} 1)}}{{\\partial y}} \\equal{} 2y \\minus{} 2 \\equal{} 0 \\Leftrightarrow y \\equal{} 1}^{\\frac {{\\partial x^2 }}{{\\partial x}} \\equal{} 2x \\equal{} 0 \\Leftrightarrow x \\equal{} 0} } \\right. \\\\\r\n\\end{array}\r\n\\]\r\n:(", "Solution_1": "[b]Claim:[/b] Let $ (a, b)$ be a point on a circle $ (x \\minus{} h)^2 \\plus{} (y \\minus{} k)^2 \\equal{} r^2$. The equation of the tangent line at $ (a, b)$ is $ (a \\minus{} h)(x \\minus{} h) \\plus{} (b \\minus{} k)(x \\minus{} k) \\equal{} r^2$.\r\n\r\n[b]Generalization:[/b] Let $ \\mathbf{P}$ be a point in the zero set of a [url=http://en.wikipedia.org/wiki/Quadratic_form]quadratic form[/url] $ \\mathbf{XAX}^T \\equal{} 0$. The equation of the tangent space at $ \\mathbf{P}$ is $ \\mathbf{PAX}^T \\equal{} 0$.\r\n\r\nEdit: I seem to have misread your post. I'm not sure what you're trying to say. It's true that $ \\frac {d(x \\minus{} h)^2}{dx} \\equal{} 0 \\Leftrightarrow h \\equal{} 0$ because the rate of change of a quadratic is zero only at its vertex. There's nothing else going on here." } { "Tag": [ "AMC", "AMC 8", "AMC 10", "AIME", "MATHCOUNTS", "AMC 10 A" ], "Problem": "how come you guys never talk about AMC8 in this forum ????\r\n\r\nand never make mock AMC8????\r\n\r\nand another problem....\r\n\r\nmy brother got the first place in school last year in AMC8 but never got invitation for AMC10....", "Solution_1": "[quote=\"shinwoo\"]\nand another problem....\n\nmy brother got the first place in school last year in AMC8 but never got invitation for AMC10....[/quote]\r\n\r\nFirst place in school or a perfect score - you don't need an invitation to do the AMC10. Someone in the school may suggest/recommend.\r\n\r\nIn our school, we recommend AMC10 to students who we think have some ability to do it - especially if they are in middle school. Based on the response we inform the high school about the number of test takers so they get included in the count for ordering tests. We do not want students to feel overwhelmed or have a bad experience. \r\n\r\nIn high school they sign up if interested. \r\n\r\nNaga", "Solution_2": "[quote=\"shinwoo\"]how come you guys never talk about AMC8 in this forum ????\n\nand never make mock AMC8????\n\n[/quote]\r\n\r\nHere's a link to a mock AMC8. Those discussions were hot before November. \r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=14666\r\n\r\nNow you will hear more about AIME and then USAMO. Simply because the AIME is coming up and that will be followed by the USAMO.\r\n\r\nNaga", "Solution_3": "huh??\r\n\r\nthey are AMCB\r\n\r\nnot AMC 8\r\n\r\nB :!=: 8", "Solution_4": "The mock AMC tests here have letters, to indicate that they are mock tests. \r\n\r\nThe interest in each AMC test level is seasonal. Now we have just been in the AMC 10/12 season, so everyone was talking about those tests, and people are beginning now to talk about the AIME, which comes next. Stay here for a year and you will see each test mentioned in turn.", "Solution_5": "but\r\n\r\nlast year, we never talked about AMC8 and I screwed up the test pretty good... :(", "Solution_6": "Try [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15878[/url] for the Mock AMC 8. And I think part of the reason that people aren't talking about it is because not that many people are trying to do well on it. If you want practice, just keep doing the problems on the website, and look at some of the MathCounts resources too.", "Solution_7": "AMC8 is not big at all. The people who score high on the AMC8 are usually more focused on the AMC10. That's why it isn't very popular.", "Solution_8": "[quote]The people who score high on the AMC8 are usually more focused on the AMC10.[/quote]\r\n\r\nNot only focused on AMC10. The top MATHCOUNTS students can usually score well in AMC10. I am sure that you will see a few perfect scorers from middle school in this year's AMC10A because it was too easy.\r\n\r\nIMO, if you participate MATHCOUNTS in middle school, AMC8 and AMC10 will come along like a piece of cake... \r\n\r\n-My 22" } { "Tag": [ "trigonometry" ], "Problem": "Let $a_1=\\sqrt2$. For $n\\ge2$, define $a_n=\\sqrt{2+a_{n-1}}$. Then $a_4=2\\cos\\theta$ for which of the following values of $\\theta$? \r\n\r\n(a)$\\frac{\\pi}{12}$\r\n\r\n(b)$\\frac{\\pi}{16}$\r\n\r\n(c)$\\frac{\\pi}{18}$\r\n\r\n(d)$\\frac{\\pi}{24}$\r\n\r\n(e)$\\frac{\\pi}{32}$", "Solution_1": "[hide]$a_n=2\\cos \\frac{\\pi}{2^{n+1}}$\n\nBase case: $a_1=\\sqrt{2}=2*\\frac{\\sqrt{2}}{2}=2\\cos \\frac{\\pi}{4}$.\n\nInductive step: Let $\\theta=\\frac{\\pi}{2^{n+1}}$. Then $a_{n+1}=\\sqrt{2+2\\cos \\theta}=2\\sqrt{\\frac{1+\\cos \\theta}{2}}=2\\sqrt{\\cos^2 \\frac{\\theta}{2}}=2\\cos \\frac{\\theta}{2}$.\n\nSetting $n=4$, $a_4=2\\cos \\frac{\\pi}{32}\\Rightarrow \\boxed{E}$.[/hide]", "Solution_2": "[hide=\"Answer\"]Well, I have it memorized that $2\\cos \\frac{\\pi}{2^n}$ is the sum of $n-1$ $\\sqrt{2}$'s successively inside the previous one. So, $\\sqrt{2+\\sqrt{2+\\sqrt{2+\\sqrt{2}}}}=2\\cos \\frac{\\pi}{32}\\Rightarrow \\boxed{E}$.\n\nFor a more in-depth solution, read scorpius'. He probably explained it better than I can. :P[/hide]" } { "Tag": [ "inequalities", "AMC", "USA(J)MO", "USAMO", "algebra proposed", "algebra" ], "Problem": "Suppose $x_1,...,x_n>0$ and $\\sum_{i,j}{|x_i-x_j|}=\\sum_{i,j}{|1-x_i x_j|}$. Prove that $x_1+x_2+...+x_n=n$.", "Solution_1": "strange problem ;) well,what does[color=red] \"adapted\" by harazi [/color]mean?", "Solution_2": "I know it is strange and especially I think difficult. \"Adapted\" is a word I like to put in front of those \"creations\" of mine which are easily derived from some famous problems, but which, strangely, cannot be seen so easy even by experienced people. This one is an example.", "Solution_3": "Too hard (at least for me!)\r\nDear harazi,could you give me some hints?", "Solution_4": "I really want to know the solution. :P", "Solution_5": "I believe you made a mistake. Check the case where $a_1=1$ and $a_2=0.5$.", "Solution_6": "I've checked and you're wrong.", "Solution_7": "I didn't notice that $i,j$ need not be distinct; sorry.", "Solution_8": "Harazi, did you adapt the result from the inequality in USAMO 2000 Problem 6?", "Solution_9": "Of course! ;) And you know, I really hoped someone who does not see that will come with an easy solution. But I guess this is impossible. :D", "Solution_10": "Your clever plan may have worked. Your question got me thinking about the USAMO 2000 inequality. I just posted a proposed simple solution to that inequality in the topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15292]The hardest USAMO inequality problem[/url].", "Solution_11": "My plan always works but we must have some patience and the results quickly follow. I said it there and I say it here: this is the most beautiful solution to an inequality that I have ever seen.", "Solution_12": "Coming from you, a master of inequalities, that is high praise. Thanks!", "Solution_13": "I haven't finished my solution yet but here is my idea : \r\n The hypothesis is equivalent to :\r\n $ \\sum_{1\\ leq i \\leq n}|1-x_i^2| = 2.(\\sum_{1 \\leq i < j \\leq n}|x_i-x_j| - \\sum_{1 \\leq i < j \\leq n}|1-x_ix_j|) $ \r\n[b] Case 1 : [/b] $ x_i \\geq x_j \\geq 1 $ \r\n [b] Case 2:[/b] $ 1 \\geq x_i \\geq x_j $ \r\n In 2 cases , we will prove : \r\n $ |1-x_i^2| + |1-x_j^2| \\geq 4.(|x_i-x_j| - |1-x_ix_j|) $ \r\n It's easy ;) \r\n [b] Case 3 :[/b] $ x_i \\geq 4 , x_j \\leq 1 $ , we can easily show : $ x_i^2-x_j^2 \\geq 4(x_i-x_j) $ \r\n [b]Case 4 :[/b] $ 4 \\geq x_i \\geq 1 \\geq x_j >0 $ : not yet :) \r\n[/b]", "Solution_14": "in fact ,I still do not know how to solve it,help!", "Solution_15": "How about your solution , Harazi ?", "Solution_16": "[hide=\"Hint\"]Use the identity\n\\[\n \\left| p - q \\right| = p + q - 2 \\min(p, q)\n\\]\nto get rid of all the absolute values. You can then use the USAMO 2000 inequality.\n[/hide]", "Solution_17": "Moubi asked me to write a more complete solution, so here goes. By replacing the absolute values with minimums as my hint recommended, we get the equation\r\n\\[\r\n \\sum_{i,j} [x_i + x_j - 2 \\min(x_i, x_j)] \r\n = \\sum_{i, j} [1 + x_i x_j - 2 \\min(x_ix_j, 1)].\r\n\\]\r\nBy grouping the terms not involving minimum together, we get the equation\r\n\\[\r\n -2 \\sum_{i, j} \\min(x_i, x_j) \r\n = \\sum_{i,j} (x_i-1)(x_j-1) - 2 \\sum_{i, j} \\min(x_i x_j, 1).\r\n\\]\r\nBut the first sum on the right-hand side is just the square of $\\sum_i (x_i - 1)$. Thus we get\r\n\\[\r\n -2 \\sum_{i,j} \\min(x_i, x_j)\r\n = \\left(\\sum_i x_i - n\\right)^2 - 2 \\sum_{i, j} \\min(x_i x_j, 1).\r\n\\]\r\nNow let's apply the USAMO 2000 inequality with $a_i = x_i$ and $b_i = 1$. We get the inequality\r\n\\[\r\n \\sum_{i, j} \\min(x_i, x_j) \\ge \\sum_{i, j} \\min(x_i x_j, 1). \r\n\\]\r\nDoubling the last equation and adding it to the second-to-last equation, all the minimum terms cancel out, and we are left with the inequality\r\n\\[\r\n 0 \\ge \\left(\\sum_i x_i - n \\right)^2.\r\n\\]\r\nThe only nonpositive square that I know of is zero. Thus we get\r\n\\[\r\n \\sum_i x_i = n.\r\n\\]\r\nThat's what we wanted.", "Solution_18": "Harazi this is the greatest problem I have ever seen :lol: But the solution of Ravi B is unpredictable. It is all the money.WOW :first:", "Solution_19": "Thanks.\r\n\r\nYes, harazi did come up with a cool problem. Without the hint that his inequality was derived from another famous problem, it would have been a lot harder.", "Solution_20": "Bump. This one is a hard problem!" } { "Tag": [], "Problem": "source: Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu\r\n\r\n\r\nProve that if $a_{1}0$\r\n\\[ f(x+y)\\geq f(x)+yf(f(x))\\]", "Solution_1": "Here's the outline:\r\n\r\nFirst of all, it's obvious that $f$ should be strictly increasing. Then, it's also obvious that $\\displaystyle \\lim_{x\\to \\infty}f(x)=\\infty$. Let $g(x)=f(x)-x$. Then $g(x+y)-g(x)\\ge y(f(f(x))-1)\\to \\infty$ when $x\\to \\infty$ for fixed $y$. This shows that $g(x)\\to \\infty$ when $x\\to \\infty$ (we also use the fact that $g$ becomes strictly increasing from a certain point on). \r\n\r\nNow let $y=1$. We have $1>\\frac {f(f(x))}{f(x+1)}\\ (*)$, but for large enough $x$, $f(x)>x+1\\Rightarrow f(f(x))>f(x+1)\\Rightarrow$ we have a contradiction with $(*)$.", "Solution_2": "it's also from IMC2001 in Czech Republic." } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "all odd prime factors of numbers which are 1+ an odd power of 3\r\nare -1 mod 3", "Solution_1": "[quote=\"shirshendu\"]all odd prime factors of numbers which are 1+ an odd power of 3\nare -1 mod 3[/quote]\nDid you mean:\n[quote=\"shirshendu\"]Let $n=1+3^{2k+1}$ for some $k\\in\\mathbb{N}$. Then for any odd prime $p\\mid n$ we have $p\\equiv-1\\pmod{3}$. [/quote]\r\n :?: \r\nIf you meant this, something is wrong here. Take $k=1$. Then $n=1+3^{3}= 28$ and $7\\mid n$, but $7\\equiv 1\\pmod{3}$. :maybe:", "Solution_2": "If odd prime $p|1+3^{2k+1}$, then $(\\frac{-3}{p})=1\\Longrightarrow p=1(mod \\ 6)$." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Show that there exists an infinite sequence $ (a_n)_{n\\equal{}0,1,2,\\dots}$ of positive integers such that for all non-negative integers $ N$ the number $ \\sum_{k\\equal{}0}^{N}a_k^2$ is the square of an integer. Find a reccurence formula for such a sequence.", "Solution_1": "Take $ a_0\\equal{}3$ and suppose that $ a_0,a_1,...,a_{n\\minus{}1}$ are integers with the property stated. We want to construct $ a_n$ so that $ a^2\\plus{}a_n^2\\equal{}c^2$ where $ a$ is an integer such that $ a^2\\equal{}\\sum_{i\\equal{}0}^{n\\minus{}1}a_i^2$ and $ c$ is some integer. If $ a$ is even, take $ a_n\\equal{}\\frac{a^2}{4}\\minus{}1$. If $ a$ is odd, take $ a_n\\equal{}\\frac{a^2\\minus{}1}{2}.$" } { "Tag": [ "counting", "distinguishability", "email", "AMC", "AMC 8", "MATHCOUNTS", "search" ], "Problem": "Ann has 3 roses and 3 carnations. She will put two flowers in each of three pots. How many different arrangements of potted flowers are there?", "Solution_1": "[quote=\"mliang\"]Ann has 3 roses and 3 carnations. She will put two flowers in each of three pots. How many different arrangements of potted flowers are there?[/quote]\r\n[hide]$\\frac{6!}{3!3!2!}=10$ I think[/hide]", "Solution_2": "I can't explain my method really well, so I'll try the best I can do:\r\nI agree with bpms:\r\n[hide=\"I think I have it\"] Without any restriction (type of flowers, how many flowers per pot), we have $6!$ ways. We divide this by $3!$ and by another $3!$ to account for the types of flowers. We must also divide by $2!$ to account for the number of flowers per pot. Doing this gets $10$ as the answer. [/hide]", "Solution_3": "[hide]\nfirst solution: if the flowers are being arranged only by the type.\n\nlet C = carnation and R = rose\n\nWe have three flower pots X, Y and Z. \n\nThe possible arrangements are:\n \n X______Y______Z\n\nCR_____RR_____CC\n\nCR_____CC_____RR\n\nCC_____CR_____RR\n\nCC_____RR_____CR\n\nRR_____CR_____CC\n\nRR_____CC_____CR\n\nCR_____CR_____CR\n\nThat would be 7 different arrangements.\n\nsecond solution: if the carnations were all different and if all the roses were all different. (for example, their color; something that gives them individuallity)\n\n\nfor pot X we would have 2 flowers to pick out of 6. The number of possible combinations for pot X would be:\n\n$_{6}C_{2}= \\frac{_{6}P_{2}}{2!}= 15$\n\nfor pot Y we would have 2 flowers to pick out of 4. The number of possible combinations for pot Y would be:\n\n$_{4}C_{2}= \\frac{_{4}P_{2}}{2!}= 6$\n\nfor pot Z we would have only 2 flowers left, which would leave only one possible combination.\n\nso that would be $15 \\times 6 \\times 1$ or $90$ different combinations\n\nfor pot Y if we would have 2 flowers to pick out of 6, the number of ADDITIONAL possible combinations for pot Y would be $9$\n\nfor pot Z if we would have 2 flowers to pick out of 4, the number of ADDITIONAL possible combinations for pot Z would be $5$\n\nso that would be $9 \\times 5 \\times 1$ or $45$ ADDITIONAL combinations.\n\nfor pot Z if we would have 2 flowers to pick out of 6, the number of ADDITIONAL possible combinations for pot Z would be $9$\n\nso that would be $9$ ADDITIONAL combinations.\n\nthe total number of arrangements would be $90+45+9$ or $144$\n\n[/hide]", "Solution_4": "Are the pots distinguishable?", "Solution_5": "mliang, do you have the answer? I don't think 10 is correct...", "Solution_6": "[quote=\"frost13\"]mliang, do you have the answer? I don't think 10 is correct...[/quote]\r\nCan you please post your reasoning, sir? I did this the same way as vishalarul, but I may be wrong...", "Solution_7": "[quote=\"mliang\"]Ann has 3 roses and 3 carnations. She will put two flowers in each of three pots. How many different arrangements of potted flowers are there?[/quote]\r\n[hide]If the pots and flowers are both indistinguishable, there are $2$ arrangements.\n\nIf the pots and flowers are distinguishable, then there are $252$ arrangments.\n\nIf the pots are indistiguishable but the flowers are, there are $72$ arrangements.\n\nIf the pots are distiguishable but the flowers aren't, there are $7$ arrangements.[/hide]", "Solution_8": "no need for the formality bmps :lol: \r\n\r\nHere's how I see it.\r\n[hide]\nrc rc rc\ncc cr rr\n\nAny other arrangement is the same as one of these two the way I see it. I admit the question is vague and 13375P34K43V312 is right to ask if the pots are distinguishable. I believe that since we are not told if the pots are labeled or fixed in position, we have to assume we can not tell the difference from one pot to the next, just as I would assume I can't tell the difference from one carnation to another.\n\nDo you see what I mean. If the source was posted I could check the answer, but I am in no mood to sort through every question ever written:-)\n[/hide]\r\n\r\nThat only helps if you believe me....and of course if I am right. I guess it would be embarrasing if I am not :lol:", "Solution_9": "[quote=\"frost13\"]no need for the formality bmps :lol: [/quote]\r\nIt's a habit, I grew up learning to be respectful to elders.\r\nAnyways, I thought that the flowers were distinct, but the pots aren't. It definitely matters...\r\nIf pots and flowers are indistinguishable than this is similar to this old nationals sprint:\r\nFour indistinguishable red chips and two indistinguishable white chips are arranged in a circle. How many different arrangements are there?", "Solution_10": "[quote=\"bpms\"][quote=\"frost13\"]no need for the formality bmps :lol: [/quote]\nIt's a habit, I grew up learning to be respectful to elders.\n[/quote]\r\n\r\nYeah, like I said the question is pretty vague.\r\n\r\nBeing respectful is great and your parents taught you well, but at 33 I'm not ready to be considered an \"elder\" just yet :lol:", "Solution_11": "The pots could be distinguishable just by how they are situated in relation to each other.", "Solution_12": "I have already posted the answer for each case, so all we need is a clarification from mliang.", "Solution_13": "[quote=\"frost13\"][quote=\"bpms\"][quote=\"frost13\"]no need for the formality bmps :lol: [/quote]\nIt's a habit, I grew up learning to be respectful to elders.\nquote]\n\nYeah, like I said the question is pretty vague.\n\nBeing respectful is great and your parents taught you well, but at 33 I'm not ready to be considered an \"elder\" just yet :lol:[/quote][/quote]\r\nBy elder I mean any adult that is out of college pretty much...", "Solution_14": "hey bmps, didn't want to keep going back and forth on the forum. You didn't offend me or anything with the \"elder\" comment, I was just kidding with you. \r\n\r\nDid you get my email with the AMC8 solution scan?\r\n\r\nEdited : Seriously I am losing my mind, I meant that to be a PM :lol:", "Solution_15": "[quote=\"frost13\"]hey bmps, didn't want to keep going back and forth on the forum. You didn't offend me or anything with the \"elder\" comment, I was just kidding with you. \n\nDid you get my email with the AMC8 solution scan?[/quote]\r\nYeah, thanks a lot for that.\r\nAnyways, does anyone get I like pie's solution, some of it doesn't seem to be right...", "Solution_16": "[hide=\"Solutions\"]If the pots and flowers are both indistinguishable, there are $\\boxed{2}$ arrangements: $\\text{cc rr cr}$ and $\\text{cr cr cr.}$\n\nIf the pots are distiguishable but the flowers aren't, there are $\\boxed{7}$ arrangements. The pots in the first arrangement (from above) can be switched to make six arrangements, plus the other one.\n\nIf the pots are indistiguishable but the flowers are, there are $\\boxed{72}$ arrangements. Take the original two arrangements. In each arrangement, all roses can be switched for six different arrangements, and the same can be done for the carnations. Therefore, there are $2\\cdot6^{2}=72$ arrangements.\n\nIf the pots and flowers are distinguishable, there are $\\boxed{252}$ arrangments. Here, we take the number of arrangements if only the pots are distinguishable, $7,$ and again multiply this by $6^{2}.$[/hide]\r\nStill waiting for a clarification...", "Solution_17": "[quote=\"i_like_pie\"][hide=\"Solutions\"]If the pots and flowers are both indistinguishable, there are $\\boxed{2}$ arrangements: $\\text{cc rr cr}$ and $\\text{cr cr cr.}$\n\nIf the pots are distiguishable but the flowers aren't, there are $\\boxed{7}$ arrangements. The pots in the first arrangement (from above) can be switched to make six arrangements, plus the other one.\n\nIf the pots are indistiguishable but the flowers are, there are $\\boxed{72}$ arrangements. Take the original two arrangements. In each arrangement, all roses can be switched for six different arrangements, and the same can be done for the carnations. Therefore, there are $2\\cdot6^{2}=72$ arrangements.\n\nIf the pots and flowers are distinguishable, there are $\\boxed{252}$ arrangments. Here, we take the number of arrangements if only the pots are distinguishable, $7,$ and again multiply this by $6^{2}.$[/hide]\nStill waiting for a clarification...[/quote]\r\nOkay, never mind your method makes more sense then mine.", "Solution_18": "I agree with Ilikepie\r\n\r\n[hide] 6! divided by 3! divided by 3! divided by 2!. 6!=720, divided by 3! equals 6= 120. Divided by 3! = 20. Divided by 2! = 10. :P [/hide]", "Solution_19": "I found the actual mathcounts competition problem stated as :\r\n\r\n\"Amy had three Powerful Pink geraniums and three Roaring Red geraniums. If she plants two flowers in each of three pots, how many distinct 3-pot arrangements could she make\"\r\n\r\nIdon't know if mliang just rephrased it for typing ease, but this was in an actual competition.\r\n\r\n[hide=\" answer\"]\n2 arrangements\n[/hide]", "Solution_20": "Sorry for the clarification delay people. I just got the answer yesterday. It was one of the practice problems from my coach. Frost has the exact problem posted. I was confused about the distinguishment as well. But yes, it is only 2 ways. I guess pots and flowers are both undistinguishable.\r\n@frost, how did you find the problem? it was from a really old sprint.", "Solution_21": "[quote=\"mliang\"]Sorry for the clarification delay people. I just got the answer yesterday. It was one of the practice problems from my coach. Frost has the exact problem posted. I was confused about the distinguishment as well. But yes, it is only 2 ways. I guess pots and flowers are both undistinguishable.\n@frost, how did you find the problem? it was from a really old sprint.[/quote]\r\nWhich one was it from?", "Solution_22": "From 1994 I think.", "Solution_23": "well frost may/may not (don't want to say anything.....) have a lot of tests\r\n\r\n-jorian", "Solution_24": "i didn't find the test actually. I have all problems from like 1983 - 1995 sorted by topic. I just opened the document and did a search for \"flower\"\r\n\r\nAnd since I keep getting requests for a document I made my 2nd or 3rd coaching called \"Tips and Tricks\" that somehow ended up posted on Goodlymaths and is now not working, here it is. I haven't had a chance to update it with a lot of the \"tips or tricks\" I have learned since, but here's how it stands.", "Solution_25": "[quote=\"frost13\"]i didn't find the test actually. I have all problems from like 1983 - 1995 sorted by topic. I just opened the document and did a search for \"flower\"[/quote]\r\nEducaide?", "Solution_26": "[quote=\"bpms\"][quote=\"frost13\"]i didn't find the test actually. I have all problems from like 1983 - 1995 sorted by topic. I just opened the document and did a search for \"flower\"[/quote]\nEducaide?[/quote]\r\n\r\nyup", "Solution_27": "thanks for the document\r\n\r\ni will have time to look at it later\r\n\r\n-jorian", "Solution_28": "Yeah Mr. Frost, thanks. :D" } { "Tag": [ "geometry", "circumcircle", "algebra", "function", "domain", "combinatorial geometry", "combinatorics unsolved" ], "Problem": "Does there exist a set of $n > 2$ points in the plane such that no three are collinear and the circumcenter of any three points of the set is also in the set? \r\n\r\n[i]Brasil 1996[/i]", "Solution_1": "No, there is no such set.\r\n\r\nSuppose, for a contradiction, that there exists such a finite set, say $S$.\r\nLet $P,Q$ be two consecutive vertices on the convex hull of $S$. Then, all the other points from $S$ are on the same side of the line $(PQ)$, and none is on this line.\r\nLet $E$ be the domain delimited by the line $(PQ)$ and the two lines perpendicular to $(PQ)$ from $P$ and $Q$, and on the same side of $(PQ)$ than the points of $S$.\r\n\r\nSince $|S| \\geq 3$ there is at least one point in $S$ which is not $P$ or $Q$, and then the center of the circumcircle of $PQM$, say $\\Omega$, belongs to $E$ (since it is on the perpendicular bissector of the segment $PQ$).\r\nIt follows that the set $S'$ of the points from $S$ which belongs to $E$ is non-empty, and finite.\r\nLet $M \\in S'$ such that the distance of $M$ to the line $(PQ)$ is minimal.\r\nSince $M \\in S'$ we deduce that both angles $MPQ$ and $MQP$ are not greater than $\\frac{\\pi} 2.$\r\n- If the angle $PMQ$ is not greater than $\\frac{\\pi} 2$ then the circumcenter of $MPQ$, say $\\Omega$ again, is an interior point of $PMQ$, which is impossible from the minimality of $M$, or is the midpoint of the segment $PQ$ which is impossible from above.\r\n- If the angle $PMQ$ is greater than $\\frac{\\pi} 2$ then $\\Omega$ is on the other side of $(PQ)$ than $M$ which is a contradiction too.\r\n\r\nPierre.", "Solution_2": "is this combinatorics -_-?", "Solution_3": "Yes.\r\nIt is combinatorial geometry so it is combinatorics.\r\n\r\nPierre.", "Solution_4": "Here is my solution.\r\nThere is no such set.\r\n\r\nSuppose, for a contradiction, that there exists such a finite set, say S. \r\n\r\nLet $d$ be a smallest distance between any two points of $S$. Suppose $AB=d$, where $A, B \\in S$.\r\n\r\nCall $l$ is the perpencular bisector of the segment $AB$.\r\n\r\nBecause $n\\ge 3$, there is at least one point $C$ in $S$ which is not $A$ and $B$\r\nand then the center of the circumcircle of $ABC$ , say $D$ , belongs to $l$.\r\n.Furthere more, the center of the circumcircle of $ABD$, say $E$, belongs to $l$,too.\r\nBesides, no three point of S are collinear, hence the center of there circumcircle of $ABE$ is $D$.\r\nTherefore, we have $DA=DB=DC=CA=CB$, it follows that $\\hat{ADB}=\\hat{ACB}=120^{o}$.\r\nthus, $CD=\\frac{d}{\\sqrt{3}}2 such that there exist n points in plane such that the circumcenter of any three points of the set is also in the set.", "Solution_6": "Choose any $n$ collinear points ;) \r\n\r\nPierre." } { "Tag": [ "trigonometry", "geometry unsolved", "geometry" ], "Problem": "Let ABC be an acute triangle. Circle k with diameter AB intersects AC and BC again at M and N respective ly. The tangents to k at M and N meet at point P. Given that CP=MN, determine angle ACB.", "Solution_1": "[quote=\"Beat\"]Let ABC be an acute triangle. Circle k with diameter AB intersects AC and BC again at M and N respective ly. The tangents to k at M and N meet at point P. Given that CP=MN, determine angle ACB.[/quote]\r\nTry to prove that,if $H$ is orthocentre of $ABC$,then $P$ is midelpoint of $CH$,and we have that $CP=\\frac{CH}{2}=Rcos\\angle{C}$,also $MN=ABcos\\angle{C}$=>\r\n$CP=MN$<=>$cos\\angle{C}=1/2$=>$\\angle{C}=60^0$", "Solution_2": "For the middlepoint $D$ of the side $AB$ we have $PD=\\frac{c}{2\\cos (90^{\\circ}-C)}=$ $\\frac{c}{2\\sin C}=$ $R=2\\cdot \\frac R2$, i.e. the point $P$ is the middlepoint of the segment $[CH].$\r\n\r\n[b]Remark.[/b] The middlepoint $U$ of the segment $[CH]$ and the middlepoint $D$ of the side $[AB]$ belong to the Euler's circle $C\\left(E,\\frac R2\\right)$ and the segment $[DU]$ is its diameter, i.e. $DU=R.$", "Solution_3": "I`m sick of not clear explanations. Can you write from where did you get those equalities!\r\n\r\nDoes only darij can explain in this forum?", "Solution_4": "[quote=\"Beat\"]I`m sick of not clear explanations. Can you write from where did you get those equalities!\n\nDoes only darij can explain in this forum?[/quote]\r\nWhat do you can't understand? :?", "Solution_5": "I cannot understand why when i`m asking for help you are answering me:\r\nTry to prove that,..\r\nDo you think that i did`n tried?", "Solution_6": "Ok,ok,ok!!!!!!! :roll: \r\nHer is explaining to my solution.\r\nWe have that $H\\in{AN}\\cap{BM}$,and $\\angle{CMH}=\\angle{CNH}=90^0$=>\r\n$CNHM$ inscribed in circle with deometre $CH$.=>If$O$ is midelpoint of $CH$=>\r\n$OM=ON$ and $\\angle{OMN}=\\angle{ONM}=90^0-\\angle{C}$,but also$\\angle{MAN}= \\angle{MBN}=90^0-\\angle{C}$=>$P=O$\r\nNow I hope you can understand evrithing in my solution :) .", "Solution_7": "Thanks Tiks :yup:", "Solution_8": "[quote=\"Beat\"]Thanks Tiks :yup:[/quote]\r\nYou are welcome [b]Beat [/b] ;)" } { "Tag": [ "vector", "linear algebra", "matrix", "LaTeX" ], "Problem": "if v=(0 1 0) and w=(1 0 0) what's v*w?\r\nthe 3d vectors are so hard to grasp so does the right hand rule, anyone can explain it to me?", "Solution_1": "Try representing the vectors as their components i+j+k. Then, the cross product is essentially the determinant of the matrix below.\r\n$ 3 \\times 3$ matrix\r\n\\[ \\begin{pmatrix}\r\ni & j & k \\\\\r\na & b & c \\\\\r\nd & e & f\r\n\\end{pmatrix} \\]\r\n\r\nWhere the first component vector is $ (a,b,c)$ and the second component vector is $ (d,e,f)$.", "Solution_2": "[quote=\"cdymdcool\"]if v=(0 1 0) and w=(1 0 0) what's v*w?\nthe 3d vectors are so hard to grasp so does the right hand rule, anyone can explain it to me?[/quote]\r\nUsing that rule, the cross product is (0, 0, -1) (sorry no Latex) :(", "Solution_3": "Wow I just studied this yesterday in review of physics =o\r\n\r\nRegarding hte right hand rule, hold your hand in a fist like manner but like keep your fingers straight after it curls back. The quickest path is then the way your finger points\r\n\r\nIts kind of hard to describe in words; does your book have a picture?", "Solution_4": "the way i learned to picture the right hand rule was a guy lying on his back just after waking up with his hands perpendicular to his body.\r\n\r\nit's a stretch, but i assure you that you will not forget the concept", "Solution_5": "Oh...that rule.\r\n\r\nPut your right hand/palm on a wheel. When you push your hand out to make the wheel spin, if your thumb is up, it's positive.\r\n\r\nWhy can't you just remember that counterclockwise is positive? :mad: It's the same for vector bearings in physics." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "algebra proposed", "algebra" ], "Problem": "Let $ a_{i} \\in N, i \\equal{} 1,2,3...k, k \\in N$. Find the values of $ k$ for which the equation\r\n\r\n$ a_{1} \\plus{} a_{2} \\plus{} a_{3} ... \\plus{} a_{k} \\equal{} a_{1}a_{2}a_{3} ... a_{k}$\r\n\r\nhas integer solutions.", "Solution_1": "see USAMO 2006 problem 4.", "Solution_2": "[quote=\"CDP100\"]Let $ a_{i} \\in N, i \\equal{} 1,2,3...k, k \\in N$. Find the values of $ k$ for which the equation\n\n$ a_{1} \\plus{} a_{2} \\plus{} a_{3} ... \\plus{} a_{k} \\equal{} a_{1}a_{2}a_{3} ... a_{k}$\n\nhas integer solutions.[/quote]\r\n\r\nFor any k.\r\n\r\n$ k\\equal{}1$ : $ 1\\equal{}1$\r\n$ k\\equal{}2$ : $ 2\\plus{}2\\equal{}2\\times 2$\r\n$ k>2$ : $ 1\\plus{}1\\plus{}1\\plus{}\\ldots\\plus{}1\\plus{}2\\plus{}k\\equal{}2k$" } { "Tag": [], "Problem": "If $ a^x \\equal{} c^q \\equal{} b$ and $ c^y \\equal{} a^z \\equal{} d$, then\r\n\r\n$ \\textbf{(A)}\\ xy \\equal{} qz \\qquad\\textbf{(B)}\\ \\frac {x}{y} \\equal{} \\frac {q}{z} \\qquad\\textbf{(C)}\\ x \\plus{} y \\equal{} q \\plus{} z \\qquad\\textbf{(D)}\\ x \\minus{} y \\equal{} q \\minus{} z$\r\n$ \\textbf{(E)}\\ x^y \\equal{} q^z$", "Solution_1": "[hide=\"Solution\"]\nSuppose we solve for $ a$ and $ c$ in terms of $ b$, we have:\n\n$ a^x \\equal{} b \\implies a \\equal{} b^{\\frac{1}{x}}\\\\\nc^q \\equal{} b \\implies c \\equal{} b^{\\frac{1}{q}}$\n\nNow we plug that information into the equation $ c^y \\equal{} a^z$.\n\n$ (b^{\\frac{1}{q}})^y \\equal{} (b^{\\frac{1}{x}})^z\\\\\nb^{\\frac{y}{q}} \\equal{} b^{\\frac{z}{x}}\\\\\n\\frac{y}{q} \\equal{} \\frac{z}{x}\\\\\n\\boxed{(A) xy \\equal{} qz}$ \n[/hide]", "Solution_2": "I dont think your solution is correct. What if $q=x=0$ so $b=1$? You cannot divide by 0.", "Solution_3": "I think if $q = x = 1$, $xy = qz = 0$ (right)\nThen we do as TachyonPulse", "Solution_4": "You could also do this problem using logarithms.\n$a^x=b$\n$\\therefore x=log_ab$\n$c^q=b$\n$\\therefore q=log_cb$\n$c^y=d$\n$\\therefore y=log_cd$\n$a^z=d$\n$\\therefore z=log_ad$\n\n$x \\cdot y = log_ab \\cdot log_cd = log_ad\\cdot log_cb = z \\cdot q = q \\cdot z$" } { "Tag": [], "Problem": "ABC is an equalirateral triangle .and (O)is its circumcircle.\r\n\r\nfor each point like P on (O) inwich PB>PAandPC,prove:\r\n\r\nPA+PC=PB", "Solution_1": "The point is rotating $60^\\circ$. :lol:", "Solution_2": "excuse me kunny,what do u mean by :the point is rotating 60.?", "Solution_3": "Hint\r\n\r\n[hide]This is a special case of the Ptolemy theorem.[/hide]", "Solution_4": "thanks arne,\r\n ;) \r\nim waiting for theother solutions.", "Solution_5": "Special case can have some special solutions.\r\n\r\nLet $AP \\cap BC = X$. Then $\\angle{BXP}=\\angle{BAP}+60^o=\\angle{BCP}+60^o=\\angle{ACP}$. So:\r\n\r\n$\\frac{\\triangle{BXP}}{\\triangle{ACP}}=\\frac{BX\\cdot XP}{AC \\cdot CP}$ and $\\frac{\\triangle{PXC}}{\\triangle{ACP}}=\\frac{PX \\cdot XC}{AC \\cdot CP}$\r\n\r\nHence: $\\frac{\\triangle{BXP}+\\triangle{CXP}}{\\triangle{ACP}}=\\frac{\\triangle{BPC}}{\\triangle{ACP}} =\\frac{XP \\cdot BC}{AC \\cdot CP}$\r\n\r\nSince $\\angle{BXA}=180^o-\\angle{ACP}$, so we can use the same trick on $\\triangle{BXA}, \\triangle{AXB}$ and $\\triangle{ACP}$. We will get: \r\n\\[\\frac{\\triangle{ABC}}{\\triangle{ACP}}=\\frac{AX \\cdot BC}{AC \\cdot CP}\\]\r\nAdd it to the previous equation we got: \r\n\\[\\frac{S_{ABPC}}{\\triangle{ACP}}=\\frac{BC \\cdot AP}{AC \\cdot CP}\\]\r\nSimilarly, we have $\\frac{S_{ABPC}}{\\triangle{ABP}}=\\frac{AP \\cdot BC}{BP \\cdot AB}$\r\n\r\nSince $\\triangle{ABP}+\\triangle{ACP}=S_{ABPC}$, so we will end with $\\frac{AC \\cdot CP+ BA \\cdot BP}{AP \\cdot BC}=1$. \r\nBecause $AB=AC=BC$, so we finally get $PA=BP+CP$.", "Solution_6": "[quote=\"kunny\"]The point is rotating $60^\\circ$. :lol:[/quote]\r\nCan you please show your proof, kunny?", "Solution_7": "I suppose that the line BC separes the points A,P. Let $R\\in PC$ be the point for which the line AC separes the points P,R and $m(\\angle BAR)=m(\\angle BAP)$. Easily shows that the triangle APR is equilateral and CR=PB, i.e. PA=PB+PC.", "Solution_8": "thanks to levi and shobber ;)" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Is there a number K such that:\r\n \r\n (k)(k+1)(k+2)\u00b7\u00b7\u00b7(k+2006)/2007!\r\n\r\nends in exactly 2007 zeros?", "Solution_1": "Let $ ord_2(k\\plus{}2006)\\equal{}n\\ge 11, ord_5(k\\plus{}2006)\\equal{}m\\ge 5$, then $ ord_2(C_{k\\plus{}2006}^{2007})\\equal{}n,ord_5(C_{k\\plus{}2006})\\equal{}m$.\r\nWe can choose $ k\\equal{}10^{2007}\\minus{}2006$.", "Solution_2": "A condition was missing: all of the numbers k,k+1,k+2,k+3.....k+2006 end in, at most three zeros.\r\n :)" } { "Tag": [], "Problem": "Aflati inversa functiei f:(0, infinit) cu valori in (0, infinit), f(x) = radical din x + radical de ordinul 3 din x. \r\n\r\nScue, nu stiu cum sa bat aici radicalul. \r\n\r\nMultumesc", "Solution_1": "Tu trebuie sa rezolvi ecuatia $y=\\sqrt{x} + \\sqrt[3]{x}$. Fie $x=z^6$. Atunci $y=z^3+z^2$. Apoi vezi http://mathworld.wolfram.com/CubicFormula.html ...\r\n\r\nProblema e foarte urata :D", "Solution_2": "Intr-adev\u0103r, problema este ur\u00e2t\u0103. Ideea de rezolvare se mai poate g\u0103si \u015fi \u00een \"Tabele \u015fi formule matematice\" de E. Rogai.", "Solution_3": "Va multumesc foarte mult. :) \r\n\r\nIntrebarea este ce cauta intr-un manual de clasa a zecea? Teoria necesara rezolvarii tine de anul intai facultate. \r\nsi mai este o intrebare. Autorii de manuale ar trebui obligati prin lege sa anexeze si solutiile complet rezolvate la probleme. Profesorii oricum nu-ti spun solutiile la problemele nerezolvate in clasa. \r\n\r\nNumai bine", "Solution_4": "[quote=\"jeanvaljean\"]Va multumesc foarte mult. :) \n\nIntrebarea este ce cauta intr-un manual de clasa a zecea? Teoria necesara rezolvarii tine de anul intai facultate. \nsi mai este o intrebare. Autorii de manuale ar trebui obligati prin lege sa anexeze si solutiile complet rezolvate la probleme. Profesorii oricum nu-ti spun solutiile la problemele nerezolvate in clasa. [/quote]\r\nM-am uitat si eu in manual si nu ai dreptate. Problema nu cere determinarea inversei, ci doar studiul injectivitatii si surjectivitatii. E adevarat ca surjectivitatea e mai greu de argumentat, dar in nici un caz nu ai nevoie de cunostinte de facultate. Toate cele bune!", "Solution_5": "Ai dreptate draga Bogdan. Uitasem enuntul initial deoarece aveam problema in cap de mult timp. \r\ntotusi, cum se poate justifica surjectivitatea utilizand numai cunostinte de clasele noua si zece? Fara analiza sau ecuatii de gradul trei. Eu nu vad momentan decat metoda grafica; schitand graficul si paralela la Ox etc. Daca ai timp sugereaza-mi ideea. \r\n\r\nMultumesc" } { "Tag": [], "Problem": "Solve the equation\r\n\r\n$ [x] \\times (x) = x $ where $ [x] $ is the greatest integer fn and $(x) $ is the fractional part of x.\r\n\r\nIs there any general method for such problems....", "Solution_1": "It looks challenging and smart!!!\r\nI tried it , unfortunately not finding the complete solution. :( \r\nI thought that since $0\\leq(x)<1$ we get $[x]<0$ but then.......?\r\nDo you have any hint? \r\n\r\nI am not so acquainted with such problems\r\nalthough I like them! :)", "Solution_2": "[quote=\"Rushil\"]Solve the equation\n\n$ [x] \\times (x) = x $ where $ [x] $ is the greatest integer fn and $(x) $ is the fractional part of x.\n\nIs there any general method for such problems....[/quote]\r\n\r\nThe soln. set should be { b2/(b-1) where b is a [b]-ve[/b] integer or 0 }\r\n\r\nPlease verify the ans.", "Solution_3": "How did you reach that?? :)", "Solution_4": "try considering different groups of values for $x$:\r\n\r\n1) $x=0$ is a trivial solution\r\n\r\n2) $x$ cannot be positive because either:\r\n$02/(b-1) where b is some -ve integer or zero.\r\n\r\nSince , zero is the trivial solution, let's test for say b= -3\r\n\r\nwe have x= (-3)2/(-3-1) = -9/4\r\n\r\n Also [x] = [-9/4]=[-2.25] = -3\r\n\r\n and (x) = (-2.25) = (-3 + 0.75 )= 0.75\r\n\r\nSo [x](x)= -3 * 0.75 = -3*(3/4) = -9/4 = x\r\n\r\nSimilarly we can prove that any number of the form b2/(b-1) where b is some -ve integer or zero, is a solution to the given equation.\r\n\r\nAlso we note that x=-1/2 is a solution .", "Solution_6": "[quote=\"socrates\"]How did you reach that?? :)[/quote]\r\n\r\nA small correction in my solution..note the square on b..\r\n\r\nshall xplain the solution later to u ...'m succumbing to my workload (at office :D ) now", "Solution_7": "oh ok yours is right. i misundertood (x) for negative numbers to be just what's after the decimal point, i.e. i made (-2.25) to be .25 instead of .75", "Solution_8": "[quote=\"L_Li\"]oh ok yours is right. i misundertood (x) for negative numbers to be just what's after the decimal point, i.e. i made (-2.25) to be .25 instead of .75[/quote]\r\n\r\nActually, by definition (x) or {x},i.e. the fractional part of any real no x,is defined as \r\n\r\n(x) = x- [x].\r\n\r\nSo it turns out to be something which always lies in the range [ 0, 1) for any real no. x.", "Solution_9": "Well you could let the fractional part be a and the integer part b, then you know\r\n$ab=a+b$ where $a\\in (0, 1)$ and b is a negative integer...\r\nThen for each b, there is an a which works. You find it's something like\r\n$-\\frac{b^2}{b+1}$ which gives us $a=\\frac{b}{b+1}$\r\nThis gives us 0, -1/2, -4/3, -9/4, -16/5, ....", "Solution_10": "If $x$ is negative, say $-2.25$, is the fractional part $\\frac14$ or $-\\frac14$?", "Solution_11": "[quote=\"nat mc\"]If $x$ is negative, say $-2.25$, is the fractional part $\\frac14$ or $-\\frac14$?[/quote]\r\nThe fractional part is $\\frac{3}{4}$, because $-2.25$ rounded down gives $-3$.", "Solution_12": "So did we agree the answer is $x=-\\frac{b^2}{b+1}$ with $b\\in\\mathbb{N}\\cup\\{0\\}$?", "Solution_13": "[quote=\"nat mc\"]So did we agree the answer is $x=-\\frac{b^2}{b+1}$ with $b\\in\\mathbb{N}\\cup\\{0\\}$?[/quote]\r\n\r\nYes,\r\n\r\nactually the solution -b2/(b+1) where b is a natural no. or zero\r\n\r\nis equivalent to b2/(b-1)where b is a NEGATIVE integer or zero\r\n\r\nwe can prove that easily", "Solution_14": "indeed the answer IS $ x = \\frac{b^2}{b-1} $ where $ b \\leq 0 $\r\n\r\n[color=green]EDIT : Sorry there was a typo!!![/color]", "Solution_15": "[quote=\"Rushil\"]indeed the answer IS $ x = \\frac{b^2}{b+1} $ where $ b \\leq 0 $[/quote]\r\n\r\nShouldn't it be b2/(b-1) for b <= 0\r\n\r\nand NOT b2/(b+1) for b <= 0 ,for if b=-1, the solution b2/(b+1) is NOT defined.\r\n\r\nActually there are two equivalent solutions\r\n\r\n x = -b2/(b+1) where b is some Natural no. or zero\r\n\r\nand x = b2/(b-1) where b is some Negative integer or zero" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Solve the equation with $ x,y,z\\in \\mathbb Z^{\\plus{}}$: $ x^x\\plus{}2y^y\\equal{}3z^z$", "Solution_1": "[hide=\"hint\"]\n$ x\\equal{}y\\equal{}z$\nuse the fact that $ 2 \\le (1\\plus{}\\frac{1}{z})^z<3$.\n[/hide]", "Solution_2": "$ x^{x}\\plus{}2y^{y}\\equal{}3z^{z}$ \r\nIf $ x\\equal{}y\\implies x\\equal{}y\\equal{}z$ \r\nSupposex,y,z different .Hence $ max (x,y,z)\\ge 3$ .If $ z\\equal{}max(x,y,z)$ \r\n$ 3z^{z}\\equal{}2z^{z}\\plus{}z^{z}>2y^{y}\\plus{}x^{x}$ \r\nIf $ x\\equal{}max(x,y,z)$ we have \r\n$ x^{x}\\equal{}x\\cdot x^{x\\minus{}1}>3z^{z}$ ,$ (x\\ge 3,x\\minus{}1\\ge z)$ \r\nSimilary if $ x\\equal{}max(x,y,z)$ we have $ x^{x}>3z^{z}$\r\nAnswers :(n,n,n) :P" } { "Tag": [], "Problem": "Teepees are made such that the length of a supporting pole is at least four feet longer than the diameter of the circular base. If a teepee has a base circumference of 40 feet, at least how long must a supporting pole be? Express your answer to the nearest whole number.", "Solution_1": "The diameter is 40/pi. So the supporting pole must be at least 40/pi + 4, which is approximately 17", "Solution_2": "You got that right", "Solution_3": "Welcome to AoPS chatting system!\n" } { "Tag": [], "Problem": "anyone who knows how to solve for \"n\" in permutations \r\ngiven:\r\n\r\nP(n,5)=2520", "Solution_1": "You are to find $n\\geq 5$ such that $n(n-1)(n-2)(n-3)(n-4)=2520.$\r\nThat is to say, express $2520$ as the product of 5 consecutive numbers.", "Solution_2": "Yeah, unless you just want to, you shouldn't actually solve this. I would recomment just plugging in values of n>5 to check, it shouldn't take you long." } { "Tag": [ "integration", "trigonometry", "logarithms", "calculus", "calculus computations" ], "Problem": "If $ \\int_{1}^{\\infty}\\frac{ (x^{3} \\plus{} 3 ) dx}{x^{6}( {x^{2} \\plus{} 1})}$ = $ \\frac{a \\plus{} b{\\pi}}{c}$\r\n\r\nthen find a,b,c .", "Solution_1": "try substituting $ x \\equal{} \\tan \\theta$", "Solution_2": "or $ x\\equal{}\\frac{1}{t}.$", "Solution_3": "@ NORA and KUNNY \r\n\r\nactually i have got the answer but i don know if its correct.\r\nplese give ur answers so that i can check with mine", "Solution_4": "[quote=\"vijaymenon\"]If $ \\int_{1}^{\\infty}\\frac { (x^{3} \\plus{} 3 ) dx}{x^{6}( {x^{2} \\plus{} 1})}$ = $ \\frac {a \\plus{} b{\\pi}}{c}$\n\nthen find a,b,c .[/quote]\r\n\r\nLet $ x\\equal{}\\frac{1}{t}$, we have $ \\int_{1}^{\\infty}\\frac { (x^{3} \\plus{} 3 ) dx}{x^{6}( {x^{2} \\plus{} 1})}\\ dx\\equal{}\\int_0^1 \\frac{t^3\\plus{}3t^6}{t^2\\plus{}1}\\ dt$\r\n\r\n$ \\equal{}\\int_0^1 \\left(t\\minus{}\\frac{t}{t^2\\plus{}1}\\right)\\ dt\\plus{}3\\int_0^1 \\left(t^4\\minus{}t^2\\plus{}1\\minus{}\\frac{1}{t^2\\plus{}1}\\right)\\ dt$\r\n\r\n$ \\equal{}\\left[\\frac{t^2}{2}\\minus{}\\frac{1}{2}\\ln (t^2\\plus{}1)\\right]_0^1\\plus{}3\\left[\\frac{t^5}{5}\\minus{}\\frac{t^3}{3}\\plus{}t\\right]_0^1\\minus{}3\\cdot \\frac{\\pi}{4}$\r\n\r\n$ \\equal{}\\frac{69}{15}\\minus{}\\frac{1}{2}\\ln 2\\minus{}\\frac{3}{4}\\pi$", "Solution_5": "thank u Kunny :) :)" } { "Tag": [], "Problem": "Demonstrati ca exista o infinitate de patrulatere convexe pentru care daca stim lungimile a trei laturi, putem afla si lungimea ultimei laturi. :)", "Solution_1": "Exista o infinitate de patrulatere convexe circumscriptibile.", "Solution_2": "Da, asa e, folosim Lema lui Pitot cu patrulaterele circumscriptibile." } { "Tag": [], "Problem": "Is nth a word?\r\n\r\nApparently it doesn't have any vowels, but it is defined in many dictionaries, used as an adjective. Vote, post, support.", "Solution_1": "nth as in the nth side of a figure.\r\n\r\nI guess we math people use it a lot. However, I don't think it is a real word.... :|", "Solution_2": "I say yes, but only in mathematical terminology (whatever that is). It is similar to \"an n-gon\".", "Solution_3": "Its in the dictionary - its a word.", "Solution_4": "It's used in real life when you want to say a number is large, as in \"for the nth time, could you please stop?\".\r\n\r\nAlso, a word doesn't need to have vowels, take [url=http://en.wiktionary.org/wiki/cwm]cwm[/url], for instance (although you could say that w is being used as a vowel, I guess).", "Solution_5": "The letters A, B, C, D, E, F, G, H, I, ... are in the dictionary as \"The [i]nth[/i]( :rotfl: ) word in the English alphabet.\" So there are entries in the dictionary without vowels." } { "Tag": [ "LaTeX" ], "Problem": "Can I format nicely an illustration of the long division involved with calculating the first few terms of a Laurent series, say for $ \\frac {z}{Sin^2(z)}$. I tried using the \\polynom package but that seems to be inapplicable (didn't work).\r\n\r\nConsider for example, the first three sets of calculations for\r\n\\[ \\frac {z}{z^2 \\minus{} \\frac {z^4}{3} \\plus{} \\frac {2z^6}{45} \\plus{} \\cdots}\r\n\\]\r\nI saw the tabular example given in another post here and I can start the code:\r\n\\[ \\setlength{\\tabcolsep}{30pt} z^2 \\minus{} \\frac {z^4}{3} \\plus{} \\frac {2z^6}{45} \\plus{} \\cdots\\,\\begin{tabular}{|lllll} \\hline z \\\\\r\n\\end{tabular}\r\n\\]\r\nSo the first line under the z (from multiplication by $ 1/z$ would be $ \\minus{} (z \\minus{} \\frac {z^3}{3} \\plus{} \\frac {2z^5}{45})$ that would give the first term of $ 1/z$. Next would be the multiplication by $ z/3$, and so on. Guess I'd be interested in perfectly formatting it (everything aligned nicely) but maybe that's a tough request to ask LaTex. How do I place the result over the division symbol: $ 1/z \\plus{} z/3 \\plus{} z^3/15 \\plus{} \\cdots$?\r\n\r\nThe code here does the first line under z but does not seem to work in the forum. Also when I attempt to use \\frac it does not compile in TeXnicCenter also:\r\n\r\n[code]\n\\setlength{\\tabcolsep}{30pt}\nz^3-\\frac{z^4}{3}+\\frac{2z^6}{45}+\\cdots \\,\n\\begin{tabular}{|lllll}\n\\hline\n\\begin{align*}\n&z\\\\\n-&\\!\\!\\!(z-z^3/3+2z^5/45)\\\\\n\\hline\n\\end{align*}\n\\end{tabular}\n[/code]\r\n\r\nAny ideas?\r\nShaw", "Solution_1": "The problem is not \\frac but the fact that you are trying to use the align environment with its associated new lines inside a cell. A cell cannot contain more than one line so solve the problem by putting align inside a parbox.\r\n[code]\\setlength{\\tabcolsep}{30pt} \n$z^3-\\frac{z^4}{3}+\\frac{2z^6}{45}+\\cdots \\,$ \n\\begin{tabular}{|lllll} \n\\hline \n\\parbox{1cm}{\\begin{align*} \n-&(z-z^3/3+2z^5/45)\\\\ \n\\hline \n\\end{align*}} \n\\end{tabular}[/code]", "Solution_2": "Also have a look at the code at http://www.physicsforums.com/showthread.php?p=373931 which you may want to adapt.", "Solution_3": "Thanks Steve! I modified the PF code for my problem. It's just about perfect. May need to tweak it a bit. I'm unable to post it here though: \r\n\r\n[code]\\begin{equation*}\n\\begin{array}{rc@{}c}\n& \\multicolumn{2}{l}{\\, \\, \\, \\frac{1}{z}+\\frac{z}{3}+\\frac{z^3}{15}-\\dotsb} \\vspace*{0.12cm} \\\\ \\cline{2-3}\n\\multicolumn{1}{r}{z^2-\\frac{z^4}{3}+\\frac{2z^6}{45}-\\dotsb \\hspace*{-4.8pt}} & \\multicolumn{1}{l}{ \\hspace*{-5.6pt} \\Big) \\hspace*{4.6pt} z} \\\\\n& \\multicolumn{2}{l}{\\, \\, \\, z-\\frac{z^3}{3}+\\frac{2z^5}{45}-\\dotsb} \\vspace*{0.12cm} \\\\ \\cline{2-3}\n\n& \\multicolumn{2}{l}{\\, \\, \\, \\phantom{1{}-{}} \\frac{z^3}{3}-\\frac{2z^5}{45}+\\dotsb} \\\\\n& \\multicolumn{2}{l}{\\, \\, \\, \\phantom{1{}-{}} \\frac{z^3}{3}-\\frac{z^5}{9}+\\frac{2z^7}{135}+\\dotsb} \\vspace*{0.12cm} \\\\ \\cline{2-3}\n& \\multicolumn{2}{l}{\\, \\, \\, \\phantom{z{}-{}\\frac{z^3}{3}{}-{}} \\frac{z^5}{15}-\\dotsb}\n\\end{array}\n\\end{equation*}\n[/code]", "Solution_4": "Works fine on AoPS as well if you use dollars instead of \\begin/\\end{equation}: I would spread the lines a bit though...\r\n\\[ \\begin{array}{rc@{}c}\r\n& \\multicolumn{2}{l}{\\, \\, \\, \\frac{1}{z}\\plus{}\\frac{z}{3}\\plus{}\\frac{z^3}{15}\\minus{}\\dotsb} \\vspace*{0.12cm} \\\\ \\cline{2\\minus{}3}\r\n\\multicolumn{1}{r}{z^2\\minus{}\\frac{z^4}{3}\\plus{}\\frac{2z^6}{45}\\minus{}\\dotsb \\hspace*{\\minus{}4.8pt}} & \\multicolumn{1}{l}{ \\hspace*{\\minus{}5.6pt} \\Big) \\hspace*{4.6pt} z} \\\\\r\n& \\multicolumn{2}{l}{\\, \\, \\, z\\minus{}\\frac{z^3}{3}\\plus{}\\frac{2z^5}{45}\\minus{}\\dotsb} \\vspace*{0.12cm} \\\\ \\cline{2\\minus{}3}\r\n\r\n& \\multicolumn{2}{l}{\\, \\, \\, \\phantom{1{}\\minus{}{}} \\frac{z^3}{3}\\minus{}\\frac{2z^5}{45}\\plus{}\\dotsb} \\\\\r\n& \\multicolumn{2}{l}{\\, \\, \\, \\phantom{1{}\\minus{}{}} \\frac{z^3}{3}\\minus{}\\frac{z^5}{9}\\plus{}\\frac{2z^7}{135}\\plus{}\\dotsb} \\vspace*{0.12cm} \\\\ \\cline{2\\minus{}3}\r\n& \\multicolumn{2}{l}{\\, \\, \\, \\phantom{z{}\\minus{}{}\\frac{z^3}{3}{}\\minus{}{}} \\frac{z^5}{15}\\minus{}\\dotsb}\r\n\\end{array}\\]", "Solution_5": "Can't seem to get the spacing between the horizontal lines just right but that's ok for now. Thanks guys!\r\n\r\n\\[ \\begin{array}{rc@{}c}\r\n& \\multicolumn{2}{l}{\\, \\, \\, \\frac{1}{z}+\\frac{z}{3}+\\frac{z^3}{15}-\\dotsb} \\vspace*{0.12cm} \\\\ \\cline{2-3}\r\n\\multicolumn{1}{r}{z^2-\\frac{z^4}{3}+\\frac{2z^6}{45}-\\dotsb \\hspace*{-4.8pt}} & \\multicolumn{1}{l}{ \\hspace*{-5.6pt} \\Big) \\hspace*{4.6pt} z} \\\\\r\n& \\multicolumn{2}{l}{\\, \\, \\, z-\\frac{z^3}{3}+\\frac{2z^5}{45}-\\dotsb} \r\n\\\\\r\n\\vspace{1mm}\r\n\\cline{2-3}\r\n\\vspace{1mm}\r\n& \\multicolumn{2}{l}{\\, \\, \\, \\phantom{1{}-{}}\r\n \\frac{z^3}{3}-\\frac{2z^5}{45}+\\dotsb} \\\\\r\n& \\multicolumn{2}{l}{\\, \\, \\, \\phantom{1{}-{}} \\frac{z^3}{3}-\\frac{z^5}{9}+\\frac{2z^7}{135}+\\dotsb} \\vspace*{0.12cm}\r\n\\\\\r\n\\vspace{1mm}\r\n\\cline{2-3}\r\n\\vspace{1mm}\r\n& \\multicolumn{2}{l}{\\, \\, \\, \\phantom{z{}-{}\\frac{z^3}{3}{}-{}} \\frac{z^5}{15}-\\dotsb}\r\n\\end{array}\\]", "Solution_6": "I'm working on another residue problem and created another version of long division which works well. Some notes:\r\n\r\n(1) I eliminated the multicolumn syntax above. The code below is simpler. Each term has a separate column. So the figure below has six columns.\r\n\r\n(2) The six columns give better control over the length of the horizontal lines.\r\n\r\n(3) In order to create a nicer spacing above and below the horizontal lines, I used the phantom operator with a \\Big) command. This forces the succeeding line to be larger thus giving a nicer spacing.\r\n\r\n(4) In the particular case below, I needed to enlarge the column containing the \\cdots so that the other elements in that column would fit.\r\n\r\n(5) The case below is for even $ n$ just for the record.\r\n\\[ \\begin{array}{llllll} & w^{n - 2} + & \\pi w^{n - 3} + & \\phantom{tk}\\cdots\\phantom{tk} + & \\pi^{n - 2} + &{ \\frac {\\pi^{n - 1}}{w - \\pi}} \\\\\r\n\\cline{2 - 6} w - \\pi & \\hspace{ - 6pt}\\Big)w^{n - 1} \\\\\r\n& w^{n - 1} - & \\pi w^{n - 2} \\\\\r\n\\cline{2 - 6} \\phantom{t\\Big)} & & \\pi w^{n - 2} \\\\\r\n& & \\pi w^{n - 2} + & \\pi^2w^{n - 3} \\\\\r\n\\cline{3 - 6} \\phantom{t\\Big)} & & & \\pi^2w^{n - 3} \\\\\r\n\\phantom{t\\Big)} & & & \\vdots \\\\\r\n\\cline{4 - 6} \\phantom{t\\Big)} & & & \\cdots & & \\pi^{n - 1} \\end{array}\r\n\\]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Positive numbers $ x,y,z$ are such that $ xy + yz + zx = 3$ . Prove the inequality\r\n\r\n$ A.\\frac {x + y + z}{3}\\ge{\\sqrt [21]{\\frac {x^3 + y^3 + z^3}{3}} \\ge{\\sqrt [14]{\\frac {x^2 + y^2 + z^2}{3}}.}}$\r\n\r\n$ B.\\frac {x + y + z}{3}\\ge{\\sqrt [6]{\\frac {x^2 + y^2 + z^2}{3}} \\ge{\\sqrt [18]{\\frac {x^3 + y^3 + z^3}{3}}.}}$", "Solution_1": "[quote=\"fjwxcsl\"]Positive numbers $ x,y,z$ are such that $ xy + yz + zx = 3$ . Prove the inequality\n\n$ A.\\frac {x + y + z}{3}\\ge{\\sqrt [21]{\\frac {x^3 + y^3 + z^3}{3}} \\ge{\\sqrt [14]{\\frac {x^2 + y^2 + z^2}{3}}.}}$\n\n$ B.\\frac {x + y + z}{3}\\ge{\\sqrt [6]{\\frac {x^2 + y^2 + z^2}{3}} \\ge{\\sqrt [18]{\\frac {x^3 + y^3 + z^3}{3}}.}}$[/quote]\r\n$ \\frac {x + y + z}{3}\\ge{\\sqrt [16]{\\frac {x^3 + y^3 + z^3}{3}}}$ stronger. :wink: \r\nSee also here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=78337", "Solution_2": "[quote=\"arqady\"][quote=\"fjwxcsl\"]Positive numbers $ x,y,z$ are such that $ xy + yz + zx = 3$ . Prove the inequality\n\n$ A.\\frac {x + y + z}{3}\\ge{\\sqrt [21]{\\frac {x^3 + y^3 + z^3}{3}} \\ge{\\sqrt [14]{\\frac {x^2 + y^2 + z^2}{3}}.}}$\n\n$ B.\\frac {x + y + z}{3}\\ge{\\sqrt [6]{\\frac {x^2 + y^2 + z^2}{3}} \\ge{\\sqrt [18]{\\frac {x^3 + y^3 + z^3}{3}}.}}$[/quote]\n$ \\frac {x + y + z}{3}\\ge{\\sqrt [16]{\\frac {x^3 + y^3 + z^3}{3}}}$ stronger. :wink: \nSee also here:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=78337[/quote]\r\n\r\nThank you,arqady!\r\n\r\n We have the strongest and best\r\n\r\n\r\n$ AB.\\frac {x + y + z}{3}\\ge{\\sqrt [6]{\\frac {x^2 + y^2 + z^2}{3}} \\ge{\\sqrt [16]{\\frac {x^3 + y^3 + z^3}{3}}\\ge{\\sqrt [11]{\\frac {x^2 + y^2 + z^2}{3}}.}}}$\r\n\r\nThis is the most perfect!", "Solution_3": "[quote=\"fjwxcsl\"]\n We have the strongest and best\n\n\n$ AB.\\frac {x + y + z}{3}\\ge{\\sqrt [6]{\\frac {x^2 + y^2 + z^2}{3}} \\ge{\\sqrt [16]{\\frac {x^3 + y^3 + z^3}{3}}\\ge{\\sqrt [11]{\\frac {x^2 + y^2 + z^2}{3}}.}}}$\n\nThis is the most perfect![/quote]\r\nThe right inequality follows from $ \\sqrt[3]{\\frac{x^3+y^3+z^3}{3}}\\geq\\sqrt{\\frac{x^2+y^2+z^2}{3}}$ and $ x^3+y^3+z^3\\geq3.$", "Solution_4": "[quote=\"arqady\"][quote=\"fjwxcsl\"]\n We have the strongest and best\n\n\n$ AB.\\frac {x + y + z}{3}\\ge{\\sqrt [6]{\\frac {x^2 + y^2 + z^2}{3}} \\ge{\\sqrt [16]{\\frac {x^3 + y^3 + z^3}{3}}\\ge{\\sqrt [11]{\\frac {x^2 + y^2 + z^2}{3}}.}}}$\n\nThis is the most perfect![/quote]\nThe right inequality follows from $ \\sqrt [3]{\\frac {x^3 + y^3 + z^3}{3}}\\geq\\sqrt {\\frac {x^2 + y^2 + z^2}{3}}$ and $ x^3 + y^3 + z^3\\geq3.$[/quote]\r\n\r\nYou are right,arqady!\r\n\r\n$ \\sqrt [3]{\\frac {x^3 + y^3 + z^3}{3}}\\geq\\sqrt {\\frac {x^2 + y^2 + z^2}{3}}$ \r\n\r\nis stronger than \r\n\r\n\\[ {\\sqrt [16]{\\frac {x^3 + y^3 + z^3}{3}}\\ge{\\sqrt [11]{\\frac {x^2 + y^2 + z^2}{3}}.}}\\]", "Solution_5": "[quote=\"arqady\"][quote=\"fjwxcsl\"]Positive numbers $ x,y,z$ are such that $ xy + yz + zx = 3$ . Prove the inequality\n\n$ A.\\frac {x + y + z}{3}\\ge{\\sqrt [21]{\\frac {x^3 + y^3 + z^3}{3}} \\ge{\\sqrt [14]{\\frac {x^2 + y^2 + z^2}{3}}.}}$\n\n$ B.\\frac {x + y + z}{3}\\ge{\\sqrt [6]{\\frac {x^2 + y^2 + z^2}{3}} \\ge{\\sqrt [18]{\\frac {x^3 + y^3 + z^3}{3}}.}}$[/quote]\n$ \\frac {x + y + z}{3}\\ge{\\sqrt [16]{\\frac {x^3 + y^3 + z^3}{3}}}$ stronger. :wink: \nSee also here:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=78337[/quote]\n\n${ \\frac {x + y + z}{3}\\ge (\\frac {x^3 + y^3 + z^3}{3}})^{\\frac{1}{k}}$ stronger. :wink:", "Solution_6": "[quote=\"fjwxcsl\"]\n\n We have the strongest and best\n\n\n$ AB.\\frac {x + y + z}{3}\\ge{\\sqrt [6]{\\frac {x^2 + y^2 + z^2}{3}} \\ge{\\sqrt [16]{\\frac {x^3 + y^3 + z^3}{3}}\\ge{\\sqrt [11]{\\frac {x^2 + y^2 + z^2}{3}}.}}}$\n\nThis is the most perfect![/quote]\r\nThe minimal $ k$ for which the inequality\r\n\\[ \\left(\\frac {x^2 + y^2 + z^2}{3}\\right)^k \\geq\\frac {x^3 + y^3 + z^3}{3}\\]\r\nis true for all non-negative numbers $ x,$ $ y$ and $ z$ such that $ xy+xz+yz=3$\r\nis $ k=\\max_{t\\geq0}\\frac{3(t^2+2t+2)(t^2+2)}{2(t+2)(t^3+2)}=2.526828496...$" } { "Tag": [ "Putnam", "conics", "ellipse", "geometry", "perimeter", "college contests" ], "Problem": "Putnam 1950/B2) An ellipse with semi-axes a and b has perimeter length $p(a, b)$. For $\\frac{b}{a}$ near 1, is $\\pi (a+b)$ or $2\\pi \\sqrt{ab}$ the better approximation to $p(a, b)$?\r\n\r\n[hide=\"My weak (without any hint of rigor) physics like solution to this using differentials\"]For clarity, I shall do [color=blue]the calculation for a circle[/color] of radius $r$ first: \n\n[color=blue]The area of such an circle is given by $A(r)=\\pi r^{2}$. Now one might approximate the circumference $C$ of said circle by the method of differentials according to the following geometric observation: that the $C$ is approximately given by the difference of areas of concentric circles with radii $r$ and $r+\\Delta r$ divided by the \"width\" of the circular annulus, namely $\\Delta r$, that is put\n\\[C\\approx \\frac{A(r+\\Delta r)-A(r)}{\\Delta r}= \\frac{\\Delta A}{\\Delta r}\\]\nat this point we may use differentials (wherein we put $\\Delta r = dr$) to approximate the difference of areas to get\n\\[C\\approx \\frac{dA}{dr}= \\frac{d}{dr}\\left(\\pi r^{2}\\right) = 2\\pi r \\]\n[/color]\n\nNow, as for that same [color=red]calculation for said ellipse[/color]:\n[color=red]\nRecall that the area of such an ellipse is given by $A(a,b)=\\pi ab$ so that reasoning similar to the method of approximating $C$ by geometric observation used the above, we boldly state that the perimeter $p(a,b)$ is approximately given by\n\\[p(a,b)\\approx \\frac{A(a+\\Delta a, b+\\Delta b)-A(a,b)}{\\Delta r}= \\frac{\\Delta A}{\\Delta r}\\]\nwhere the quantity $\\Delta r$ is something like, say the mean width of the elliptical annulus, but since we know that $\\frac{b}{a}$ is near 1, we have\n\\[a\\approx b\\Rightarrow \\Delta a\\approx \\Delta b\\Leftrightarrow da\\approx db\\Rightarrow \\frac{da}{db}\\approx 1 \\]\nso set $\\Delta r= dr \\approx da\\approx db$ and hence by the method of differentials we have\n\\[p(a,b)\\approx \\frac{dA}{\\Delta r}= \\frac{\\pi b\\da+\\pi a\\db}{dr}\\approx \\frac{\\pi b\\dr+\\pi a\\dr}{dr}= \\pi (a+b) \\]\nand for $a=b: =r$ this is the same as the above approximation for $C$.[/color] [/hide]\r\n\r\nEdit: fixed dimensionality mix-up (that is I had volumes in there, but now I have areas).", "Solution_1": "Look at what I did [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=55166]here[/url] and tell me whether it helps.", "Solution_2": "[quote=\"Kent Merryfield\"]Look at what I did [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=55166]here[/url] and tell me whether it helps.[/quote]\r\n\r\nMaybe [url=http://www.kalva.demon.co.uk/putnam/psoln/psol508.html]this[/url] would be interesting to you.\r\n\r\nBTW, what do you think my solution to this would have been scored as?" } { "Tag": [ "inequalities", "function", "induction", "inequalities unsolved" ], "Problem": "$a_1 ,a_2,. . . , a_n$ are real numbers such that \r\n$a_1 \\le a_2 \\le . . . \\le a_n$ prove that \r\n\r\n$\\sum _{i=1}^{n} a_ia_{i+1}^4 \\ge \\sum _{i=1}^{n} a_i^4 a_{i+1} $\r\n\r\n$a_{n+1}=a_1$", "Solution_1": "In http://www.mathlinks.ro/Forum/viewtopic.php?t=22923 , a more general result was shown:\r\n\r\n[color=blue]If $f:\\mathbb{R}\\to\\mathbb{R}$ is a convex function, and $a_1$, $a_2$, ..., $a_n$ are real numbers such that $a_1\\leq a_2\\leq ...\\leq a_n$, then\n\n$\\sum_{i=1}^n a_{i+1}f\\left(a_i\\right)\\leq\\sum_{i=1}^n a_if\\left(a_{i+1}\\right)$;\n\nhereby, we set $a_{n+1}=a_1$.[/color]\r\n\r\nYour problem follows from this general result if we take the convex function $f\\left(x\\right)=x^4$.\r\n\r\n Darij", "Solution_2": "the proof is by using induction on $n$ .we can simply show that it's enough to prove the problem \r\nfor $n=3$ .so we have \r\n$a_1=x ,a_2=y,a_3=z$ and put $S=xy^4+yz^4+zx^4$ and $T=yx^4+zy^4+x^4$ so we have \r\n$S-T=(x-y)y^4 +y(z^4-x^4)+xz(x^3-z^3)=(x-y)y^4-y(z^2+x^2)(z+x)+xz(x^2+xz+z^2))\\\\\r\n=(x-z)(y^4-yz^3-yz^2x-yzx^2-yx^3+x^3z+x^2z^2+xz^3)\\\\\r\n=(x-z)(x-y)(-y(x^2+xy+y^2)+z^3+x^2z+xz^2)\\\\\r\n=(x-z)(x-y(-yx^2-xy^2-y^3+z^3+x^2z+xz^2)\\\\\r\n=(x-z)(x-y)(z-y)(z^2+zy+y^2+x^2+xy+yz)\\\\\r\n=\\frac{1}{2}(x-z)(x-y)(z-y)((x+y)^2+(x+z)^2+(y+z)^2)\\ge0$\r\n :lol:" } { "Tag": [ "quadratics", "ARML", "number theory proposed", "number theory" ], "Problem": "Let a,b ,n be positive integers and p be prime number.Determine all pair of (a,b) which satisfies the following equation.\r\n\r\n\\[a^3+b^3=p^n\\]", "Solution_1": "Let $d = \\gcd(a,b)$. Then $d = p^t$ for some $t$, and we may divide each term of the equation by $p^{3t}$.\r\nThus, we assume that $\\gcd(a,b) = 1$. (1)\r\nThus $p^n = (a+b)(a^2 - ab + b^2) = (a+b)((a+b)^2-3ab)$\r\nSince $\\gcd(a,b) = 1$ we have $\\gcd(a+b,(a+b)^2-3ab)=1$ or $3$.\r\n\r\n- If $\\gcd(a+b,(a+b)^2-3ab)=1$ then since $a+b > 1$, we have $a+b = p^n$ and $a^2+b^2-ab = 1$.\r\nWlog, we may assume that $a \\geq b$.\r\nIf $a > 1$ then $a> b$ (they are coprime thus not equal) so that $a^2 + b^2 = 1 +ab \\leq 1 + a(a-1)$ and then $a+b^2 \\leq 1$ which is impossible. Thus $a=b=1$ and $p=2$ and $n=1$.\r\nReturning to (1) it leads to the solutions $a=2^t=b$ and $p=2$, $n = 3t+1$.\r\n\r\n- If $\\gcd(a+b,(a+b)^2-3ab)=3$ then $p=3$ and $a+b = 3^{\\alpha}$ and $(a+b)^2 -3ab = a^2 + b^2 - ab = 3^{\\beta}$, for some integers $\\alpha \\geq 1$ and $\\beta \\geq 0$.\r\nMoreover neither $a$ nor $b$ is divisible by $3$ otherwise both are divisible by $3$ which contradicts that they are coprime.\r\nBut $3^{2 \\alpha} - 3ab = 3^{\\beta}$ so that $\\beta = 1$.\r\nIt follows that $ab = 3^{2 \\alpha -1} - 1$ and $a+b = 3^{\\alpha}$.\r\nThus $a,b$ ae the solutions of the quadratic $X^2 - 3^{\\alpha} X + 3^{2\\alpha-1} - 1 = 0$.\r\nIts discriminant is $\\Delta = 4 - 3^{2\\alpha -1}$ which has to be a non-negative integer. This forces $\\alpha = 1$ thus $n=2$ and $a=1,b=2$ (or $a=2,b=1$).\r\nBack to (1) this leads to the solutions $a=3^t,b=2\\cdot 3^t,p=3,n=3t+2$ and $b=3^t,a=2\\cdot 3^t,p=3,n=3t+2$.\r\n\r\nPierre.", "Solution_2": "Bonjour,Pierre!\r\n\r\nMerci beaucoup de votre reponse.\r\n\r\nThere is no doubt about your solution.You are solving rapidly.\r\nMay be this problem is EASY for you.Is this problem famous?\r\n\r\nkunny", "Solution_3": "It also appeared in Hungary 2000.", "Solution_4": "I do not know if it is famous but it is quite straightforward (even if I did some usless things :D ).\r\n\r\nPierre.", "Solution_5": "Hmm...this problem looks very familiar, actually, a problem from ARML Power Question during 1990's. I forgot the year though :(" } { "Tag": [ "calculus", "integration", "rotation", "algebra unsolved", "algebra" ], "Problem": "On the table before you is a cylindrical ice cream cake with chocolate icing on top. From it you cut successive wedges of angle $ x$, where $ x$ is [b]arbitrary[/b]. Each time a wedge is cut, it is turned upside-down and reinserted into the cake.\r\n :arrow: Prove that after a [b]finite[/b] number of such operations, all the icing is back on top of the cake.\r\n\r\nIt's really a surprising conclusion... But can you give a proof? $ x$ is really arbitrary,and it can be irregular.", "Solution_1": "[quote=\"ZHANGWENZHONGKK\"]On the table before you is a cylindrical ice cream cake with chocolate icing on top. From it you cut successive wedges of angle $ x$, where $ x$ is [b]arbitrary[/b]. Each time a wedge is cut, it is turned upside-down and reinserted into the cake.\n :arrow: Prove that after a [b]finite[/b] number of such operations, all the icing is back on top of the cake.\n\nIt's really a surprising conclusion... But can you give a proof? $ x$ is really arbitrary,and it can be irregular.[/quote]\r\nIt's wrong, IMHO.\r\n\r\nLet the sequence $ x_1 \\equal{} x\\in(0,2\\pi)$ and $ x_{k \\plus{} 1} \\equal{} 2\\pi\\{\\frac {x_k \\plus{} x}{2\\pi}\\}$ (where $ \\{u\\}$ is the fractional part of $ u$).\r\n$ x_k$ is the part of the top which is either iced either non-iced (depending on the number of full circles we did) after $ k$ operations.\r\n\r\nIf we want the top full iced or full non_iced after $ n$ operations, we need $ x_n \\equal{} 0$ and, since we clearly have $ x_k \\equal{} 2\\pi\\{\\frac {kx}{2\\pi}\\}$, we need $ \\frac {x}{2\\pi}\\in\\mathbb Q$", "Solution_2": "The scary thing is, the problem is correct as stated (think about what happens to the icing when it is flipped, it is also reversed!!!!)\r\n\r\nBut seriously, it is correct.", "Solution_3": "[quote=\"rofler\"] But seriously, it is correct.[/quote]\r\n\r\nIf the problem is correct, I'm seriously interested in the proof, or in any flaw in my demo above.\r\n\r\n :ewpu:", "Solution_4": "http://demonstrations.wolfram.com/TheCakeIcingPuzzle/\r\n\r\n:)", "Solution_5": "[quote=\"rofler\"]http://demonstrations.wolfram.com/TheCakeIcingPuzzle/\n\n:)[/quote]\r\n\r\nHumm, there is no demonstration in this link. Just an assertion and an animation which demonstrate nothing.\r\nCould you just give us a demo or point the flaw in my own one ?\r\n\r\nThanks.", "Solution_6": "The flaw I believe is that if a wedge that is partially iced and partially not iced is flipped, the positions of the iced and non-iced sides reverse. So\r\n\r\nPlain\r\nIced\r\nIced\r\n\r\nflips over to become\r\n\r\nPlain\r\nPlain\r\nIced\r\n\r\nas opposed to\r\n\r\nIced\r\nPlain\r\nPlain", "Solution_7": "[quote=\"MellowMelon\"]The flaw I believe is that if a wedge that is partially iced and partially not iced is flipped, the positions of the iced and non-iced sides reverse.[/quote]\r\n\r\nWow ! you're right! I dont know if the problem is true, but my own counter-demo is surely false ! \r\nThanks a lot. I used a bad model. :blush:", "Solution_8": ":) Mr Zuming Feng picked it into his lecture at classes in China TST Activity. If you believe him, then, please believe me. But he forgot at last to give the solution.\r\n:!: I still remember that his last hint was that, never prove that at a certain moment all the chocolate icing is on the bottom. Although this makes the problem more and more peculiar, I wish my hint would save your time.\r\n\r\nSo, dear MellowMelon, you'd better change your idea.", "Solution_9": "I have solved it - the result is true. I cannot fathom how Peter Winkler managed to come up with this problem.\r\n\r\n[hide=\"solution\"]If $ x$ is an integral divisor of $ 2\\pi$ the result is obvious. Assume $ x$ is not such a divisor.\n\nWe will define the operation $ cut$ on a cake to mean cutting and flipping slices $ k$ times, going counterclockwise from the positive x-axis such that $ kx < 2\\pi$ and $ (k \\plus{} 1)x > 2\\pi$, and then rotating the cake by $ p \\equal{} 2\\pi \\minus{} kx$ degrees counterclockwise. This is equivalent to cutting and flipping $ k$ slices and rotating the cake to be ready for the next $ cut$.\n\nWe will define -1 as white and 1 as chocolate. So if we have a wedge of color 1, it's a chocolate wedge.\n\nWe will define an $ awesome_n$ cake to consist of $ n$ alternating $ p$ wedges of color $ ( \\minus{} 1)^{n \\plus{} 1}$ and $ x \\minus{} p$ wedges of color $ ( \\minus{} 1)^n$ and ending in a $ p$ wedge, going counterclockwise.\n\nThe complete chocolate cake, which we start out with, is a $ awesome_0$ cake. That is our base case.\n\nTaking any $ awesome_n$ cake, where $ n \\leq k$, and performing $ cut$ on it will flip each $ p$, $ x \\minus{} p$ group to become a $ x \\minus{} p$ of color $ ( \\minus{} 1)^{n \\plus{} 1}$, $ p$ of color $ ( \\minus{} 1)^n$ group. It will leave a $ p$ slice of the same color as the other $ p$ slices below the positive x-axis. The cake will then be rotated $ p$ radians to become an $ awesome_{n \\plus{} 1}$ cake.\n\nPerforming this operation a finite number of times will finally lead us to the $ awesome_{k \\plus{} 1}$ cake.\n\nWe then define a $ spectacular_n$ cake to be an $ awesome_n$ cake, except with the wedges going clockwise instead of counterclockwise.\n\nThe $ awesome_{k \\plus{} 1}$ cake is also the $ spectacular_{k \\plus{} 1}$ cake. That is our base case.\n\nTaking any $ spectacular_n$ cake, where $ n \\geq 1$, and performing $ cut$ on it will flip each $ p$, $ x \\minus{} p$ group to become a $ x \\minus{} p$ of color $ ( \\minus{} 1)^{n \\plus{} 1}$, $ p$ of color $ ( \\minus{} 1)^n$ group. It will leave a $ x \\minus{} p$ slice of color $ ( \\minus{} 1)^{n \\plus{} 1}$ in the group reached by going counterclockwise from the positive x-axis, and will leave the $ p$ slice of color $ ( \\minus{} 1)^{n \\plus{} 1}$ below the positive x-axis. It will then rotate the figure to become the $ spectacular_{n \\minus{} 1}$ cake.\n\nPerforming this operation a finite number of times will lead us to the $ specactular_0$ cake, which is the chocolate cake.\n\nNotes: keeping track of how many total cuts have occurred will lead to the formula $ 2k(k \\plus{} 1)$ total $ x$ cuts. I left it out of the proof because it's not really related.\nNotesnotes: again, that's assuming non-integral divisors of $ 2\\pi$.\n[/hide]", "Solution_10": "[quote=\"Differ\"]I have solved it - the result is true. [/quote]\r\n\r\nWow, great congrats ! :clap: \r\n\r\nAnd very nice demo, indeed. I had some difficulties to figure out each case but finally it's OK.\r\n\r\nGreat!", "Solution_11": ":clap: :clap: \r\nAnd the demonstration is also excellent. Sometimes I forget to find ideas there." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Find $ a,b$ such that $ Max|\\sqrt {x} \\plus{} ax \\plus{} b|$ is smallest possible, where $ a,b$ are 2 real numbers and $ x \\in [0;16]$", "Solution_1": "Come on! Yeah!", "Solution_2": "Another ricardokaka or lalala clone... Lock this up mods, please.\r\n\r\n(M&Y, already posted by some of above mentioned:))" } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "prove that \r\n$ detA \\in Z$", "Solution_1": "Do you assume that each $ a_i$ is an integer? If so, the conclusion is clear. Otherwise it is not right. For example, let $ n\\equal{}2$ and $ \\det A\\equal{}a_2\\minus{}a_1$.", "Solution_2": "oh i am sorry,$ a_{i}\\in Z$", "Solution_3": "The exact proof depends on your definition of determinant, but in any case $ \\det A$ will be equal to some sum of products of integers, hence an integer itself.", "Solution_4": "[quote=\"tuandokim\"]prove that \n$ detA \\in Z$[/quote]\r\n Here : $ A \\equal{} \\left[\\begin{array}{ccccc}1 & 1 & ... & 1 \\\\\r\nC_{a_1}^{1} & C_{a_2}^{1} & ... & C_{a_n}^{1} \\\\\r\nC_{a_1}^{2} & C_{a_2}^{2} & ... & C_{a_n}^{2} \\\\\r\n... & ... & ... & ... \\\\\r\nC_{a_1}^{n \\minus{} 1} & C_{a_2}^{n \\minus{} 1} & ... & C_{a_n}^{n \\minus{} 1}\\end{array}\\right]$\r\n You can find $ det(A)$ \r\n1. It is easy to give $ \\det(A) \\in \\mathbb{Z}$ by formular $ det(A) \\equal{} \\sum_{\\sigma \\in S_{n}}\\epsilon (\\sigma) a_{\\sigma_{1}1}a_{\\sigma_{2}2}...a_{\\sigma_{n}n}$ here $ a_{\\sigma_{k}k} \\in \\mathbb{Z} ,k \\in {1,2,...,n}$\r\n2. By $ C_{x}^{k\\minus{}1} \\equal{} \\frac {x(x \\minus{} 1)....(x \\minus{} k \\plus{} 2)}{(k \\minus{} 1)!} \\equal{} \\frac {x^{k \\minus{} 1}}{(k \\minus{} 1)!} \\plus{} p_{k \\minus{} 2}(x) ,k \\in \\{1,2,3,...,n\\}$ \r\n easy to calculate :\r\n $ \\det(A) \\equal{} \\frac {1}{1!2!3!...(n \\minus{} 1)!}\\det\\left[\\begin{array}{ccccc}1 & 1 & ... & 1 \\\\\r\na_1 & a_2 & ... & a_n \\\\\r\na_1^{2} & a_2^{2} & ... & a_n^{2} \\\\\r\n... & ... & ... & ... \\\\\r\na_1^{n \\minus{} 1} & a_2^{n \\minus{} 1} & ... & a_n^{n \\minus{} 1}\\end{array}\\right] \\equal{} \\frac {1}{1!2!3!...(n \\minus{} 1)!}\\prod_{1 \\leq i < j \\leq n}{(a_j \\minus{} a_i)}$" } { "Tag": [ "calculus", "integration", "Gauss", "inequalities", "derivative", "LaTeX", "geometry" ], "Problem": "let f(n)=n^(1/n)-1. for example, if n=2, then its sqrt2-1. if n=3, then its cbrt3-1.\r\nprove that f(1)+f(2)+f(3)+f(4)+... is infinite\r\n[hide=\"hint\"] prove that this series never goes below one and f(n) where n is sufficiently large is very close to one[/hide]\n[hide=\"big hint\"] show that f(k) where k>3 is more than 1/k[/hide]\r\nlet g(a)=f(a)+1\r\nprove that g(1)*g(2)*g(3)*g(4)*... is divergent also", "Solution_1": "Very nice problem Scrambled. It could perhaps more easily be stated by saying:\r\n\r\nProve that $\\sum_{n=1}^{\\infty} (\\sqrt[n]{n} - 1)$ diverges.\r\n\r\nThis may properly belong to the Calculus forum (though there are non-calculus ways of attacking it).", "Solution_2": "Why would you want to use calculus?\r\n\r\nApplying the integral test gets ugly, unless I'm missing something.\r\n\r\nBut for $n\\geq 3$, we have $\\left(1+\\frac {1}{n}\\right)^n(n+1)^n$ for $n>2$", "Solution_3": "true, but the fact you mentioned which makes it diverge (if you didnt look at the hint) was supposedly hard to see (seeing that gauss didnt catch that fact). \r\nthe next result you stated (n+1)^n0$ and $ a^{2}+b^{2}+c^{2}=3$.Pro that $ \\frac{1}{a+b}$+$ \\frac{1}{b+c}$+$ \\frac{1}{c+a}$ <=$ \\frac{3}{2}$[quote]:lol:[/quote]", "Solution_1": "It's not true , try $ a=\\frac{\\sqrt{14}}{\\sqrt{5}}, b=c=\\frac{1}{\\sqrt{10}}$", "Solution_2": "The inequality inversely holds:\r\n\r\n(a^2+b^2+c^2)/3*(1/(a+b)+1/(b+c)+1/(c+a))^2-(3/2)^2>=0\r\n\r\n<=>4*F[6][0,1]+28*F[6][0,2]+49*F[6][0,3]+192*F[6][1,1]+92*F[6][0,4]+116*F[6][1,2]>=0\r\n\r\n(F[6][0,1]=sum(a^4*(a-b)*(a-c))>=0;\r\nF[6][0,2]=sum(a^3*(b+c)*(a-b)*(a-c))>=0;\r\nF[6][0,3]=sum(a^2*(b-c)^2*(a-b)*(a-c))>=0;\r\nF[6][1,1]=sum(a^2*b*c*(a-b)*(a-c))>=0;\r\nF[6][0,4]=sum(b^2*c^2*(a-b)*(a-c))>=0;\r\nF[6][1,2]=sum(a*b*c*(b+c)*(a-b)*(a-c))>=0).", "Solution_3": "[quote=\"nguoigo\"]For$ a,b,c>0$ and $ a^{2}+b^{2}+c^{2}=3$.Pro that $ \\frac{1}{a+b}$+$ \\frac{1}{b+c}$+$ \\frac{1}{c+a}$ <=$ \\frac{3}{2}$[/quote]\r\n$ 2(a+b+c)\\left(\\sum_{cyc}\\frac{1}{a+b}-\\frac{3}{2}\\right)=\\sum_{cyc}\\left(\\frac{a+c}{b+c}+\\frac{b+c}{a+c}-2\\right)+3(3-a-b-c)\\geq0.$", "Solution_4": "For$ a,b,c>0$ and $ \\sqrt{a}+\\sqrt{b}+\\sqrt{c}=3$.Pro that $ \\frac{1}{a+b}$+$ \\frac{1}{b+c}$+$ \\frac{1}{c+a}$ <=$ \\frac{3}{2}$ :blush:\r\n I'm sorry!!!", "Solution_5": "[quote=\"nguoigo\"]For$ a,b,c>0$ and $ \\sqrt{a}+\\sqrt{b}+\\sqrt{c}=3$.Pro that $ \\frac{1}{a+b}$+$ \\frac{1}{b+c}$+$ \\frac{1}{c+a}$ <=$ \\frac{3}{2}$ :blush:\n I'm sorry!!![/quote]\r\nTry $ a=7.84$ and $ b=c=0.01.$ :wink:", "Solution_6": "It is still wrong.Take $ a\\rightarrow 0,b\\rightarrow 0$ :D" } { "Tag": [ "linear algebra", "matrix", "absolute value", "combinatorics proposed", "combinatorics" ], "Problem": "In a chess tournament , each of 2n players plays every other player once in each of two rounds. A win is worth 1 point ,a draw is worth 0.5 point and a loss is worth nothing. Prove that if for every player , the total score in the first found differs from that in the second round by at least n points ,then this difference is exactly n points for every player.", "Solution_1": ".....Any ideas....", "Solution_2": "We show that the sum of differences does not exceed 2n^2 if the difference for each player is over n, then the difference has to be n. \r\n\r\nConstruct a matrix that describes round 1. Make the entry for (i,j) the points i obtained for playing j, with possible entries being 0, 0.5, 1. Ignore all (i,i) entries. Do the same for round 2. \r\n\r\nSubtract the matrix for round 1 to the matrix for round 2. notice that in the result matrix the possible entries are 0, -0.5, 0.5, 1, - 1 with (i,j) = -(j,i) given i != j. \r\n\r\nThe absolute value for the sum of each column must be greater than or equal to n. However, the sign may be either negative or positive. Suppose there are m players for whom the difference is positve. Rearrage the matrix such that these m players have columns 1 to m. So sum for the positve differences is |(1,2) + (1,3) + (1,4)... (1,n)| + |(2,1) + (2,3)...| +...+ |(m,1) + (m,2)...+ (m,n)|. Since they are all of the same sign (i.e. positive) we may remove the absolute value sign. Now notice that for any (i,j), j < m, there exist (j,i) with (j,i) = -(i,j), so they cancel out. Exactly m^2 - m numbers cancel out that way and there are (2n-1)m components to the sum. So with the numbers that cancel out there are exactly 2nm-m^2 components. Since the maximal entry is 1, the maximal sum is 2nm-m^2. The same can be done for all negative differrences. \r\n\r\nThus the maximal for m positve differences and k negative differences with m + k = 2n is 2n(m+k) - m^2 - k^2 = 2n^2 - m^2 - k^2. But m^2 - k^2 is minimal if m = k = n. So the maximal is 2n(2n) - n^2 - n^2 = 2n^2, as discribed in the initial statement. So the only possible differences are n for each player." } { "Tag": [ "calculus", "integration", "number theory unsolved", "number theory" ], "Problem": "Can anyone please send a solution for this:-\r\n\r\nProve that on the average the number of ways of expressing a positive integer n as a sum of two integral squares \r\n $ x^{2}$+$ y^{2}$=n, is $ \\pi$", "Solution_1": "hi anup welcome to the forum i am moving this to numnber theory section", "Solution_2": "http://www.mathlinks.ro/viewtopic.php?t=150965", "Solution_3": "I belive it! :lol:" } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "Let $k$ be a given positive integer. De\ufb01ne $x_{1}= 1$ and, for each $n > 1$, set $x_{n+1}$ to be the smallest positive integer not belonging to the set $\\{x_{i}, x_{i}+ik | i = 1, . . . , n\\}$. Prove that there is a real number $a$ such that $x_{n}= [an]$ for all $n \\in\\mathbb{ N}$.", "Solution_1": "lemma: two irretional numbers a,b such that $\\frac{1}{a}+\\frac{1}{b}=1$ then $[an]$ and $[bn]$ is partition of N\r\nproof:exercise for you\r\n consider a the positive root of equation $\\frac{1}{a}+\\frac{1}{a+k}=1$\r\ntry to prove it's the required number.", "Solution_2": "Solution above same as solution I know :P \r\nlemma is Beatty Theorem , here is a nice apply http://www.mathlinks.ro/Forum/viewtopic.php?t=136049", "Solution_3": "Sorry, but I haven't had sufficient understanding about the Beatty Theorem and the lemma written by dhthstn. Can you explain more clearly and post the proof ? I really appreciate that, thank you.. :)", "Solution_4": "If you search on forum for keywords : beatty theorem then you will have http://www.mathlinks.ro/Forum/viewtopic.php?highlight=theorem+beatty&t=26352" } { "Tag": [ "calculus", "integration", "geometry", "trigonometry", "calculus computations" ], "Problem": "Find the area of the surface \r\n\r\nS=Int(Int(sqrt(1+(dz/dx)^2+(dz/dy)^2))\r\n\r\non the plane x+3y-z=2 inside the cylinder x^2+y^2=9\r\n\r\n[Note that it does not tell you the limits nor what you are supposed to be integrating with respect to]\r\n\r\nHow are you suppose to do this? Find the partials and sub them into the S equation? What does S stand for? Isn't S the double integral of the magnitude of velocity? Argh!!!", "Solution_1": "$S$ is the surface area. You are given a formula for it: $S=\\iint_{D}\\sqrt{1+\\left(\\frac{\\partial z}{\\partial x}\\right)^{2}+\\left(\\frac{\\partial z}{\\partial y}\\right)^{2}}\\,dA$. This formula is valid for the part of the graph $z=z(x,y)$ above (or below -- which is irrelevant) a region $D$ in the $xy$-plane. Another way to put it: $D$ is the projection of the surface onto the $xy$-plane. In your case $z=x+3y-2$ and the region $D$ is a disk of radius $3$.", "Solution_2": "this can be done in an elementary way...consider the cosine of the angle formed with the given plane and a plane perpendicular to the cylinder", "Solution_3": "Working off mlok's post:\r\n\r\n\\[S=\\iint_{D}\\sqrt{1+\\left(\\frac{\\partial z}{\\partial x}\\right)^{2}+\\left(\\frac{\\partial z}{\\partial y}\\right)^{2}}\\,dA=\\iint_{D}\\sqrt{1+1^{2}+3^{2}}\\,dA \\]\r\n\\[= \\sqrt{11}\\iint_{D}\\,dA = \\sqrt{11}\\mbox{Area}(D) = \\sqrt{11}\\pi 3^{2}=9\\sqrt{11}\\pi\\]" } { "Tag": [ "IMO", "IMO 2008" ], "Problem": "Lisa Sauermann (Germany):\t 7\t2\t7\t7\t5\t7\t35\t12\tGold medal\r\nZhuo Chen (People's Republic of China):\t7\t7\t7\t7\t7\t0\t35\t12\tGold medal\r\n\r\nCheck: http://www.imo-official.org/year_individual_r.aspx?year=2008&column=total&order=desc&gender=filter&nameform=western&language=en\r\n\r\nBut actually Germany wins as the German contestant solved problem 3 and 6! :D :first:", "Solution_1": "Why Farhod Haydarov \u2640 from Uzbekistan is in that list? :D", "Solution_2": "[quote=\"Yuriy Solovyov\"]Why Farhod Haydarov \u2640 from Uzbekistan is in that list? :D[/quote]\r\n\r\nI guess Farhod likes girls. But I must admit somebody must have scanned the linked list thoroughly to find out the odd one! :D", "Solution_3": "[quote=\"orl\"][quote=\"Yuriy Solovyov\"]Why Farhod Haydarov \u2640 from Uzbekistan is in that list? :D[/quote]\n\nI guess Farhod likes girls. But I must admit somebody must have scanned the linked list thoroughly to find out the odd one! :D[/quote]\r\n\r\nthere is another typo : Redi Haderi \u2640 from Albania :D", "Solution_4": "Also\r\nJuan Andr\u00e9s Dodyk \u2640. So there are probably also some girls listed as boys.", "Solution_5": "Actually it's the other way round: Juan is a boy who is listed as a girl at the IMO webpage.\r\n\r\nAnyway, isn't it a bit funny? :P" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Give a,b,c are positive real and min{a,b,c} \\geq 1/4max{a,b,c}.\r\nProve that:\r\na 2 b(a-b)+b 2 c(b-c)+c 2 a(c-a) \\geq 0", "Solution_1": "The Kalva website gives the intial condition as a, b, and c are the sides of a triangle, not min{a,b,c}>=1/4max{a,b,c}. I'll assume that you intend the former.\r\n\r\nLet a=y+z, b=z+x, c=x+y. The triangle conditions implies that x,y,z>0. Then the inequality to be proven becomes, after some trivial transformations, \r\n\r\ny^3 x + z^3 y + x^3 z >= xyz(x+y+z). \r\n\r\nObserve that 2/7 y^3 x + 1/7 z^3 y + 4/7 x^3 z >= x^2yz, by the weighted AM-GM inequality. Adding this inequality to its cyclic counterparts gives the required result.", "Solution_2": "No, I think it is a new inequality made with an inspiration from the IMO one.\r\n\r\nPierre.", "Solution_3": "You are right ,pbornsztein?. :) I know the IMO problem Michael Lipnowski tell us have a very short solution.May be my inequality is created from that proof.", "Solution_4": "No idea about this one ????? :(", "Solution_5": "can you tell me why we put min{a,b,c}>=1/4max{a,b,c}", "Solution_6": "I still don't have solution yet,but I think min{a,b,c}>=1/4max{a,b,c} to let we have \\sqrt a, \\sqrt b, \\sqrt c are three sides of a triangles.So we can let a=(x+y) 2 ,b=(y+z) 2 ,c=(x+z) 2 .....", "Solution_7": "Finally,I found the greater condition for this one.That is \\sqrt ab, \\sqrt ac, \\sqrt bc are three sides of a triangle which can be deduced from min{a,b,c} 1/4max{a,b,c}. \r\nSo do the new or old problem,your choice...{However, i still wait the answer for both}. :)", "Solution_8": "i have the solution for your inequality, anhcuong\r\nwe have\r\na 2 b(a-b)+b 2 c(b-c)+c 2 a(c-a) \\geq 0\r\n<=> (a-c) 2 (ac+ab-bc)+(b-c) 2 (bc+ac-ab)+(b-a) 2 (ab+bc-ac) \\geq 0\r\nlet a be the max{a,b,c}\r\ncase 1: a \\geq b \\geq c \\geq a/4 (it's easy to check)\r\ncase 2: a \\geq c \\geq b \\geq a/4 \r\nfrom b \\geq a/4 => (b-a) 2 (2bc-ac) \\geq -1/2(b-a) 2 ac (1)\r\nfrom cauchy: [(a-c) 2 +(c-b) 2] \\geq 1/2(b-a) 2 ac (2)\r\nb(a-c) 2 [(b-a) 2 -(c-b) 2] \\geq 0 (3)\r\nb(c-a) 2 (a-c) \\geq 0 (4)\r\n(1)+(2)+(3)+(4) and we have what we want :D \r\nfor this problem i think you can make it tigher, right? anhcuong\r\nplease show me your result", "Solution_9": "So,the first problem is proved by a nice solution:) .Please tell me how can you change the first one to (a-c) 2 (ac+ab-bc)+(b-c) 2 (bc+ac-ab)+(b-a) 2 (ab+bc-ac) \\geq 0 \r\nAnd what about the second condition, \\sqrtab, \\sqrtac, \\sqrtbc are three sides of a triangle.I wait for your ideas...", "Solution_10": "Anh Cuong\r\ni have another solution\r\nwe have the solution with teacher said\r\ni don't remember that carefully\r\n c>b>a\r\nso if we have a>b/2 this is end\r\nif a1/4c and 3a>x\r\nput b=y+a so a0\r\nsorry but i only prove that with min{a,b,c} \\geq 1/4max{a,b,c}\r\ncan you help me with this way", "Solution_11": "Sorry,but I don't think this one helps .Let try it and replyed to me when you find something.But what about the second condition of mine.Think about it too,ok??? :)", "Solution_12": "hi anhcuong. I have the solution for your ineq with the second condition\r\nand here's my solution\r\na 2 b(a-b)+b 2 c(b-c)+c 2 a(c-a) \\geq 0 \r\n<=> (a-c) 2 (ac+ab-bc)+(b-c) 2 (bc+ac-ab)+(b-a) 2 (ab+bc-ac) \\geq 0 (*)\r\nwe can assume a be the max of {a,c,b}\r\ncase1: a \\geq b \\geq c (easy)\r\ncase2: a \\geq c \\geq b\r\nbecause \\sqrt ab, \\sqrt bc, \\sqrt ca are the sides of a triangle so we can assume that\r\nab=(x+y) 2 \r\nbc=(y+z) 2 \r\nca=(z+x) 2 \r\nwith x,y,z > 0 \r\nthen we have\r\n(*) <=> (a-c) 2 (x 2 +xy+xz-yz) + (b-c) 2 (z 2 +zy+zx-xy) + (a-b) 2 (y 2 +yx+yz-xz) \\geq 0\r\nfrom a \\geq c \\geq b we have x \\geq z \\geq y\r\nfrom cauchy: (a-c) 2 x 2 + (b-c) 2 z 2 \\geq 2abs((a-c)(b-c))xz (1)\r\nxz[(a-c) 2 +(b-c) 2 ]=xz[(a-b) 2 - 2abs((a-c)(b-c))xz] (2)\r\n(a-b) 2 (y 2 +yx+yz) +(a-c) 2 (xy-yz) + (b-c) 2 (zy-xy) \\geq 0 (3)\r\n(1)+(2)+(3) we have your inequality :D \r\nis this the same as your solution, anhcuong\r\nif not, can you show me how you prove it, ok?", "Solution_13": "hi anhcuong's problem is very beautyful but arosis i's prove", "Solution_14": "Very nice solution for it,arosisi. :) I have two another solution and it don't need to examine two case.However it can be proved by BCS.... :)" } { "Tag": [], "Problem": "Could someone provide me with some information regarding the quality of Harvard's physics department? I am mostly interested in quantum physics and its other sub divisions.", "Solution_1": "It's one of the best in the country. Was there something more specific you wanted to know?" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Find the constants x,y,z such that in any acute triangle we have \r\n a^2+b^2+c^2<=xR^2+yRr+zr^2 and such that this is the strongest inequality of this type. I don't have a solution.", "Solution_1": "In any triangle we have a^2+b^2+c^2=2(s^2-r^2-4Rr)\r\nand 4R+r>=s*sqrt3. s=semiperimeter\r\nMaybe this will help you...", "Solution_2": "So you must find the best inequality of the form s<=f(R,r).\r\nI will take a look at my notes..and i will keep in touch.", "Solution_3": "a^2+b^2+c^2<=xR^2+yRr+zr^2\r\nx=9; y=-6; z=12;\r\n\r\ncould this be it?", "Solution_4": "I would like to see your solution Langragia....", "Solution_5": "well... the solution is actually based on the following fact:\r\n\r\na^2+b^2+c^2<=4(R^2+3Rr-r^2-sqrt(R)(R-2r)^(2/3)); an inequality which is well-known; this is the smoothest inequality!! (which involves R,r); then it's easy to show that a^2+b^2+c^2<=xR^2+yRr+zr^2 for\r\nx=9; y=-6; z=12;\r\n\r\n\r\ncheers! :D :D" } { "Tag": [], "Problem": "[img]http://img106.imageshack.us/img106/6989/mathsvw0.jpg[/img]", "Solution_1": "[hide]$ \\sqrt{11...1\\minus{}22...2}\\equal{}\\sqrt{\\frac{10^{2000}\\minus{}1}{9}\\minus{}2\\frac{10^{1000}\\minus{}1}{9}}$\n\n$ \\equal{}\\sqrt{\\frac{10^{2000}\\minus{}2\\cdot10^{1000}\\plus{}1}{9}}$\n\n$ \\equal{}\\sqrt{\\frac{(10^{1000}\\minus{}1)^2}{9}}$\n\n$ \\equal{}\\frac{10^{1000}\\minus{}1}{3}$\n\nam I a genius now? :P [/hide]", "Solution_2": "You would be a small genius if you wouldn't now the \"trick\" for 111...11. :p", "Solution_3": "[quote=\"Amazigh\"]You would be a small genius if you wouldn't now the \"trick\" for 111...11. :p[/quote]\r\n :D \r\nlol dik chi li kisla7 lhad 97ab d lihoud :rotfl: khso y kon genie zamel lakhor ma3arfchi bi ana aklakh nass homa sohyoun biba lilk a khatabi" } { "Tag": [ "abstract algebra", "topology", "superior algebra", "superior algebra theorems" ], "Problem": "Let $\\mathcal F,\\ \\mathcal F',\\ \\mathcal G$ be sheaves of abelian groups on a topological space $X$. \r\nAssume that \\[0\\to\\mathcal F'\\to\\mathcal F\\to\\mathcal G\\to 0\\] is an exact sequence of sheaves and that $\\mathcal F'$ is flasque i.e. for every pair of open subsets $U\\subseteq V$, the restriction map $\\mathcal F'(V)\\to\\mathcal F'(U)$ is surjective.\r\n\r\nProve that there is an exact sequence of abelian groups:\r\n$0\\to\\mathcal F'(X)\\to\\mathcal F(X)\\to\\mathcal G(X)\\to 0$", "Solution_1": "It is well-known that an exact sequence of sheaves $0 \\rightarrow \\mathcal{F'}\\xrightarrow{f}\\mathcal{F}\\xrightarrow{g}\\mathcal{F''}\\rightarrow 0$ induces an exact sequence of abelian groups $0 \\rightarrow \\mathcal{F'}(U) \\xrightarrow{f(U)}\\mathcal{F}(U) \\xrightarrow{g(U)}\\mathcal{F''}(U)$, for any open set $U\\subset X$. So we only need to prove that the map $\\mathcal{F}(U) \\xrightarrow{g(U)}\\mathcal{F''}(U)$ is surjective.\r\n\r\nLet $P\\in U$. Since the map of sheaves $\\mathcal{F}\\xrightarrow{g}\\mathcal{F''}$ is surjective, it follows that the induced map on stalks $\\mathcal{F}_{P}\\xrightarrow{g_{P}}\\mathcal{F''}_{P}$ is also surjective. Pick $s\\in \\mathcal{F''}(U)$. There is a neighborhood $W$ of $P$, $W\\subset U$ and a section $r\\in \\mathcal{F}(W)$ such that $g(W)(r)=s$. Pick a maximum such preimage, as shown below.\r\n\r\n Let $M=\\{(V,t) \\mid P\\in V\\subset U \\ \\mbox{and}\\ s\\restriction_{V}=g(V)(t)\\}.$ By the surjectivity of $g_{P}$, $M$ is nonempty. An easy argument shows that $M$, equipped with the order relation $(V_{1},t_{1})\\leq(V_{2},t_{2})$ iff $V_{1}\\subseteq V_{2}\\ \\mbox{and}\\ t_{1}=t_{2}\\restriction_{V_{1}}$, is a totally ordered set. Hence, by Zorn's Lemma there is a maximal element in $M$. Suppose the maximal element is $(W,r)$ and suppose $W\\neq U$. Let $Q\\in U\\setminus W$. By the same argument as before, there is an open set $W'\\subset U$, $W' \\not\\subset W$ and a section $r' \\in \\mathcal{F}(W')$ such that $r'=s\\restriction_{W'}$.\r\n\r\nWe would like to glue $r$ and $r'$ together, but this may not be possible. However, $(r-r')\\restriction_{W\\cap W'}=0$, so $(r-r')\\restriction_{W\\cap W'}=f(W\\cap W')(\\phi)$ for some $\\phi \\in \\mathcal{F'}(W\\cap W')$. Since $\\mathcal{F}$ is flasque, $\\phi$ extends to a section in $\\mathcal{F'}(U)$. Now glue together $r$ and $r''\\in \\mathcal{F}(W')$, where $r''=f(W')(\\phi)-r\\restriction_{W\\cap W'}$ (this is possible since they restrict to the same map on $W\\cap W'$). We obtain a section on $W\\cup W'$ which is taken by $f(W\\cap W')$ to $s$. This contradicts the maximality assumption for $(W,r)$." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "10.Through the point $P$ inside of the given circle are driven two perpendicular lines which meets the circle at $A$ and $B$. Find the locus of the projections of $P$ onto $AB$. $\\clubsuit$", "Solution_1": "[color=darkred][b][u]Lermma. The properties of the cyclic and orthodiagonal quadrilateral.[/u][/b]\n\nLet $ABCD$ be a cyclic, orthodiagonal $(AC\\perp BD)$ and convex quadrilateral inscribed in the circle $w=C(O,R)\\ .$ We will use the following standard notation : $AB=a$, $BC=b$, $CD=c$, $DA=d$,$AC=e$, $BD=f\\ .$ Denote the intersection $E\\in AC\\cap BD$ and the middlepoints $M$, $N$, $P$, $Q$ of the sides $[AB]$, $[BC]$, $[CD]$, $[DA]$ respectively. Then there are the following properties :\n\n$1.\\blacktriangleright ME\\perp CD\\ ;\\ \\delta_{AB}(O)=\\frac 12\\cdot CD\\ ,$ where $\\delta_{AB}(O)$ is the distance of the point $O$ to the line $AB\\ .$\n\n$2.\\blacktriangleright \\delta_{AB}(O)+\\delta_{BC}(O)+\\delta_{CD}(O)+\\delta_{DA}(O)=p\\ .$\n\n$3.\\blacktriangleright \\sigma [ABCD]=\\frac 12(ac+bd)=\\frac 12ef\\ .$\n\n$4.\\blacktriangleright a^2+c^2=b^2+d^2=4R^2\\ ;\\ MP=NQ=\\frac 12\\sqrt{e^2+f^2}\\ .$\n\n$5\\blacktriangleright e^2+f^2=4\\left[R^2+p_w (E)\\right]\\ ,$ where $p_w(E)=R^2-OE^2$ is the power of the point $E$ w.r.t. the circle $w\\ .$\n\n$6.\\blacktriangleright$ The points $M,N,P,Q$ and the projections of the point $P$ on the sides of the quadrilateral $ABCD$ belong to the circle with the centre in the middlepoint of the segment $OE$ and the length of the diameter equal to $R^2+p_w(E)=2R^2-OE^2$ ([u]the circle of the eight points[/u]) .[/color]\r\n\r\n[color=blue][b]Remark.[/b] The proposed problem is a immediate consequence of the above mentioned $6^{th}$- property of a cyclic and orthodiagonal quadrilateral.[/color]" } { "Tag": [ "rate problems" ], "Problem": "A man travels $ m$ feet due north at $ 2$ minutes per mile. He returns due south to his starting point at $ 2$ miles per minute. The average rate in miles per hour for the entire trip is:\r\n\r\n$ \\textbf{(A)}\\ 75\\qquad \r\n\\textbf{(B)}\\ 48\\qquad \r\n\\textbf{(C)}\\ 45\\qquad \r\n\\textbf{(D)}\\ 24\\qquad\\\\ \r\n\\textbf{(E)}\\ \\text{impossible to determine without knowing the value of }{m}$", "Solution_1": "The rate for going north is $ .5mpm$ (miles per minute). The rate for going south is $ 2mpm$. \r\n\r\nThe total minutes traveled north is $ 2m$. The total minutes traveled south is $ .5m$. That is a total of $ 2.5m$ minutes.\r\n\r\nThe total miles traveled north is m. The total miles traveled south is m.\r\n\r\nTo get the average, we divide the total miles by the total time (in minutes). $ \\frac{miles}{minutes}\\equal{}\\frac{m\\plus{}m}{2.5m}\\equal{}\\frac{2m}{2.5m}\\equal{}\\frac{4}{5}$\r\n\r\nWe have a rate of $ \\frac{4}{5}mpm$. We multiply that by 60 (because there is 60 minutes in an hour). Our final answer is $ \\frac{4}{5}\\times60\\equal{}48$. That is answer choice $ \\mathbf{(B)}$." } { "Tag": [ "ratio", "geometry", "algebra proposed", "algebra" ], "Problem": "Triangle $ ABC$ is inscript in a circumference $ \\Gamma$. A chord $ MN\\equal{}1$ of $ \\Gamma$ intersects the sides $ AB$ and $ AC$ at $ X$ and $ Y$ respectively, with $ M$, $ X$, $ Y$, $ N$ in that order in $ MN$. Let $ UV$ be the diameter of $ \\Gamma$ perpendicular to $ MN$ with $ U$ and $ A$ in the same semiplane respect to $ MN$. Lines $ AV$, $ BU$ and $ CU$ cut $ MN$ in the ratios $ \\frac{3}{2}$, $ \\frac{4}{5}$ and $ \\frac{7}{6}$ respectively (starting counting from $ M$). Find $ XY$", "Solution_1": "With analitycal geometry the value is 1/11", "Solution_2": "[quote=\"castigioni\"]With analitycal geometry the value is 1/11[/quote]\nyou did get the answer,but it can be solved by just some ratio substitutions,not analytical.", "Solution_3": "shall I give some hint here?\nthe key step(at least it seems to me) is to notice if chord $AB,MN$ intersect at $X$,then\n$\\frac{MX}{XN}=\\frac{MA*MB}{AN*BN}$.", "Solution_4": "1/11 is the correct answer. Sorry I plugged in a wrong number and came up with a different numerical result. Please delete my post above. Thanks.", "Solution_5": "http://voducdien.wordpress.com/2011/12/02/problem-2-of-argentine-mathematical-olympiad-2008/" } { "Tag": [ "logarithms", "limit", "function", "Euler", "calculus", "calculus computations" ], "Problem": "Having: $f:R \\to (0, + \\infty )$ satisfating: \r\n\\[\r\nf(x) \\cdot \\ln f(x) = e^x ,\\forall x \\in R\r\n\\]\\[\r\n\\mathop {\\lim }\\limits_{x \\to \\infty } (1 + \\frac{{\\ln x}}{{f(x)}})^{\\frac{{f(x)}}{x}} = ?,\\forall x \\in R.\r\n\\]", "Solution_1": "We have $f(x)\\ln{f(x)}=e^{x}$ thus $f(x)>e^{\\frac{x}{2}}$ $\\forall x\\in{\\mathbb{R}}$.It is clear that we have to count $\\lim({\\frac{f(x)}{x}}\\cdot\\ln({1+\\frac{\\ln{x}}{f(x)}}))$.Since $f(x)>e^{\\frac{x}{2}}$ then $\\frac{\\ln{x}}{f(x)}{\\rightarrow}{0}$ so ${\\frac{f(x)}{x}}\\cdot\\ln({1+\\frac{\\ln{x}}{f(x)}){\\sim}\\frac{f(x)}{x}}{\\cdot}\\frac{\\ln{x}}{f(x)}=\\frac{\\ln{x}}{x} {\\rightarrow}{0}$ so the initial limit is $1$.", "Solution_2": "[quote=\"silviu\"]\n$f(x) \\cdot \\ln f(x) = e^x ,\\forall x \\in R$\n[/quote]\r\n\r\n$f(x)$ is related to $W$ Lambert's function proposed by Lambert 1758 studied by Euler 1779\r\n\r\n$f(x)=exp(W(e^x))$", "Solution_3": "What is asymptotic expansion of $W$ ?", "Solution_4": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23692" } { "Tag": [], "Problem": "Just wondering, who here goes to the Waterloo Math Circles? I'm the Indian guy in the senior class looking lost the whole time :blush: .", "Solution_1": "me! :D\r\n\r\naww.. it says my message is too small.. hopefully it's long enough now..", "Solution_2": "I *may* be making an appearance as an instructor.", "Solution_3": "yeah I used to go regulary, didn't go for a year, now going again", "Solution_4": "It's a good place to waste time in :D \r\nNo, actually one can learn some formulas :)" } { "Tag": [], "Problem": "What is the measure of angle 4 if $ m\\angle 1 \\equal{} 76^{\\circ}, m\\angle 2 \\equal{} 27^{\\circ}$ and $ m\\angle 3 \\equal{} 17^{\\circ}$?\n\n[asy]draw((0,0)--(4,8)--(10,0)--cycle,linewidth(1));\ndraw((0,0)--(5,3)--(10,0),linewidth(1));\nlabel(\"2\",(1,1.2));\nlabel(\"1\",(4,7.75),S);\nlabel(\"4\",(5,3),S);\nlabel(\"3\",(8,1.8));[/asy]", "Solution_1": "A1 + A2 + A3 + (360-A4) = 360\r\nA4 = A1 + A2 + A3 = 76+27+17 = 120 degrees", "Solution_2": "[asy]size(300);\ndraw((0,0)--(4,8)--(10,0)--cycle,linewidth(1));\ndraw((0,0)--(5,3)--(10,0),linewidth(1));\ndraw((5,3)--(200/29,120/29),dashed+linewidth(1));\nlabel(\"2\",(0+1/2,0+1/2),NE);\nlabel(\"1\",(4,7.75),S);\nlabel(\"4=1+2+3\",(5,3-1/2),S);\nlabel(\"3\",(10-3/4,0+3/4),WNW);\nlabel(\"1+2\",(200/29,120/29-1/2),SSW);[/asy]\r\n\r\nThus, $ \\angle 4\\equal{}\\angle 1\\plus{}\\angle 2\\plus{}\\angle 3\\equal{}76^\\circ\\plus{}27^\\circ\\plus{}17^\\circ\\equal{}\\boxed{120^\\circ}$." } { "Tag": [ "group theory", "abstract algebra", "Ring Theory", "superior algebra", "regular element" ], "Problem": "Prove that there does not exist a ring with exactly 5 regular elements.\r\n($ a$ is called a regular element if $ ax \\equal{} 0$ or $ xa \\equal{} 0$ implies $ x \\equal{} 0$.)\r\n\r\nA ring is not necessarily commutative, does not necessarily contain unity element, or is not necessarily finite.", "Solution_1": "You should however specify what you understand by ring: does it have a unity, for instance? A very similar problem was discussed very long time ago: if $ A$ is a unitary ring, then it cannot have $ 5$ invertible elements. If I remember correctly, the solution used actually that they were regular, so maybe in this case one can adapt the proof. When the ring has no unity, I guess it's more tricky.", "Solution_2": "OK, I will.\r\n\r\nP.S. [i]Here the ring does not necessarily contains unity element, but one part of my proof was that it has the unity element.[/i]", "Solution_3": "hm, if the ring has a unit, then it's easy, right? it seems like the regular elements form a cyclic group of order 5 and then nobody can be $ \\minus{}1$ (since it would then have even order)", "Solution_4": "The ring can be of characteristic 2.", "Solution_5": "The set $ G$ of regular elements of the ring $ R$ is obviously closed under multiplication.\r\nThe left-multiplication in $ G$ with any $ a\\in G$ is injective, and since $ G$ is finite, it is also surjective.\r\nAnalog for the right-multiplication, so the equations $ ax \\equal{} b$ and $ ya \\equal{} b$ have solutions $ x,y\\in G$ for any $ a,b\\in G$.\r\nThis already gives $ G$ a group, so $ G\\cong C_5$ cyclic of order $ 5$ and has a neutral element $ 1$.\r\n($ 1$ is not necessarily neutral in $ R$)\r\n\r\nBut now as MysticTerminator pointed out we get $ \\{\\pm 1\\}$ a subgroup of $ G$, and by Lagrange we get $ 1 \\equal{} \\minus{} 1$.\r\nThen for $ 1\\not \\equal{} c\\in G$ we get the canonical ring homo $ \\Phi: \\mathbb{F}_2[X]\\to R, 1\\mapsto 1,X\\mapsto c$.\r\n\r\nThe image $ S$ of $ \\Phi$ is a subring of $ R$ with neutral element $ 1$, and since $ G \\equal{} \\langle c\\rangle \\subseteq S$, \r\nwe get in fact every regular element of $ R$ a unit in $ S$.\r\n\r\nConversely any unit $ a$ of $ S$ must be a regular element of $ R$,\r\nbecause otherwise assume WLOG $ xa \\equal{} 0$ with some $ 0\\not \\equal{} x\\in R$.\r\nBut then for $ b\\in S$ with $ ab \\equal{} 1$ we get $ x1 \\equal{} xab \\equal{} 0$, which contradicts $ 1\\in G$ regular in $ R$.\r\nSo $ S$ has $ 5$ units.\r\n\r\n$ c^5 \\equal{} 1$ by Lagrange, so the kernel of $ \\Phi$ contains $ X^5 \\minus{} 1 \\equal{} (X \\minus{} 1)(X^4 \\plus{} X^3 \\plus{} X^2 \\plus{} X \\plus{} 1)$.\r\nLet $ \\zeta$ be root of $ X^4 \\plus{} X^3 \\plus{} X^2 \\plus{} X \\plus{} 1$ over $ \\mathbb{F}_2$.\r\nThen $ \\mathbb{F}_2(\\zeta) \\equal{} \\mathbb{F}_{2^n}$ for some $ n\\le 4$, \r\nand by Lagrange $ 5 \\equal{} |\\langle \\zeta\\rangle|$ divides $ |\\mathbb{F}_{2^n}^\\times| \\equal{} 2^n \\minus{} 1$, which gives $ n \\equal{} 4$ \r\nand $ X^4 \\plus{} X^3 \\plus{} X^2 \\plus{} X \\plus{} 1$ the irreducible minimalpolynomial of $ \\zeta$ over $ \\mathbb{F}_2$.\r\n\r\nAnd now the [url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Chinese remainder theorem[/url] gives $ S$ isomorphic to a factor-ring of\r\n$ \\mathbb{F}_2[X]/(X^5 \\minus{} 1)\\cong \\mathbb{F}_2[X]/(X \\minus{} 1)\\times \\mathbb{F}_2[X]/(X^4 \\plus{} X^3 \\plus{} X^2 \\plus{} X \\plus{} 1)\\cong \\mathbb{F}_2\\times \\mathbb{F}_{16}$,\r\nso $ S$ is isomorphic to $ 0$ or $ \\mathbb{F}_2$ or $ \\mathbb{F}_{16}$ or $ \\mathbb{F}_2\\times \\mathbb{F}_{16}$, and has either $ 1$ or $ 15$ units.\r\nContradiction!", "Solution_6": "Generalization:\r\n\r\nCharacterize those finite semigroup $ G$ with the property that there is a ring whose semigroup of regular elements is isomorphic to $ G$.\r\n\r\n\r\n(Just as in the initial problem, the ring need not have a unit, be commutative, finite, etc.)", "Solution_7": "I pretty much proved above, that $ G$ must be the unit-group of some finite unitary ring.\r\n(take $ S \\equal{} \\sum_{g\\in G}\\mathbb{Z}g$ the finite unitary subring with $ G \\equal{} S^\\times$)\r\nBut beyond that? Take the unit group $ G$ of any group algebra $ RH$ for any finite unitary ring $ R\\not \\equal{} 0$ \r\nand any finite group $ H$. Then always $ H\\le G$.\r\n\r\nEDIT\r\nOk, here is something better using [url=http://en.wikipedia.org/wiki/Artin-Wedderburn_theorem]Artin\u2013Wedderburn[/url] and [url=http://en.wikipedia.org/wiki/Wedderburn_theorem]Wedderburn[/url].\r\nIn fact every finite unitary ring $ R$ has a nilpotent ideal $ J(R)$,\r\nwith $ R/J(R)\\cong \\mathcal{M}_{n_1}(\\mathbb{F}_{q_1})\\times \\cdots\\times\\mathcal{M}_{n_r}(\\mathbb{F}_{q_r})$ for some finite fields $ \\mathbb{F}_{q_1},\\ldots,\\mathbb{F}_{q_r}$ and some $ n_1,\\ldots,n_r\\in\\mathbb{N}$.\r\n\r\nThen the unit-group $ G \\equal{} R^\\times$ has the nilpotent normal subgroup $ N \\equal{} 1 \\plus{} J(R)\\unlhd G$ with\r\n$ G/N\\cong (R/J(R))^\\times\\cong GL_{n_1}(\\mathbb{F}_{q_1})\\times \\cdots\\times GL_{n_r}(\\mathbb{F}_{q_r})$.\r\n\r\nFor that show, that the group homo $ R^\\times\\to (R/J(R))^\\times$ is an epi with kernel $ N \\equal{} 1 \\plus{} J(R)$.\r\nAnd for $ 1 \\plus{} J(R)$ nilpotent we need $ J(R)$ nilpotent and $ [1 \\plus{} J(R),1 \\plus{} J(R)^n]\\le 1 \\plus{} J(R)^{n \\plus{} 1}$ for all $ n\\in\\mathbb{N}$.\r\nFor that take $ x\\in J(R),y\\in J(R)^n$, then\r\n$ [1 \\plus{} x,1 \\plus{} y] \\equal{} (1 \\plus{} x)^{ \\minus{} 1}(1 \\plus{} y)^{ \\minus{} 1}(1 \\plus{} x)(1 \\plus{} y) \\\\\r\n\\hspace*{0.77in} \\equal{} (1 \\plus{} x)^{ \\minus{} 1}(1 \\plus{} y)^{ \\minus{} 1}((1 \\plus{} y)(1 \\plus{} x) \\plus{} xy \\minus{} yx) \\\\\r\n\\hspace*{0.77in} \\equal{} 1 \\plus{} (1 \\plus{} x)^{ \\minus{} 1}(1 \\plus{} y)^{ \\minus{} 1}(\\underbrace{xy \\minus{} yx}_{\\in J(R)^{n \\plus{} 1}})\\in 1 \\plus{} J(R)^{n \\plus{} 1}$" } { "Tag": [ "calculus", "integration", "real analysis", "real analysis solved" ], "Problem": "Compute lim(n->00) \\int (n to n+1) {x 2 *sin(1/x)}dx\r\n{x}=decimal part of x", "Solution_1": "Is it 1/2 ?", "Solution_2": "I think it is, indeed, 1/2. We write it as \\int{n->n+1}f(x)dx-\\int{n->n+1}[f(x)]dx, where f(x)=x2sin(1/x). It's easy to see that \\int{n->n+1}[f(x)]dx-\\int{n->n+1}ndx->0, so \\int{n->n+1}[f(x)]dx-n->0, so we have to compute the limit of \\int{n->n+1}f(x)dx-n=\\int{n->n+1}f(x)dx-\\int{n->n+1}xdx+1/2=\\int{n->n+1}(f(x)-x)dx+1/2, and since f(x)-x->0 as x->00, it means that the limit is 1/2." } { "Tag": [ "calculus", "integration", "algebra unsolved", "algebra" ], "Problem": "Hello,\r\n Find all integral pairs ($a,c$) such that $a^{4}$ + $6a^{3}$ + $11a^{2}$ + $6a+1$ = $q$($a^{2}$ - $1$)($c^{2}$ - $1$)$/$$(a^{2}$ + $c^{2})$, where $q$ = $2^{x}$.$5^{y}$.$7^{z}$..... i.e.,product of some arbitrary powers of alternate primes.", "Solution_1": "Sorry this problem should be in Algebra Unsolved Problems Forum; not in the proposed and own problems!", "Solution_2": "So, how do you know this is an easy one? :wink: \r\n\r\nPierre.", "Solution_3": "I've solved it :lol: and anyway it is a regional olympiad problem for 9th and 10th graders; so it should be easy.", "Solution_4": "Still no takers for this problem?...\r\n[hide=\"Hint\"]Try using Number Theory and modulo properties[/hide]", "Solution_5": "Is this problem that tough :?: :wink:", "Solution_6": "i actually don't get what you mean by product of powers of [b]alternate[/b] primes", "Solution_7": "Oh, I am sorry for the typo. q is the product of arbitrary powers of alternate primes i.e., q = $2^{x}.5^{y}.11^{z}...$.Here alternate primes mean 2,5,11,17,23,..." } { "Tag": [ "Support" ], "Problem": "What does the rank mean? Is a higher or lower rank better?", "Solution_1": "It's just your rank compared to other users. Higher is better, generally.", "Solution_2": "[quote=\"levans\"]It's just your rank compared to other users. Higher is better, generally.[/quote]\r\n\r\nso a rank of 9 is better than a rank of 10?", "Solution_3": "Isn't 1 of 32 better than 15 of 32...\r\n\r\nLower is better, I think... :o", "Solution_4": "So like as of now am 29th in Karma out of 1433. That means there are 28 people better than me, and 1404 below me.", "Solution_5": "#1 is first place, #2 is second, etc." } { "Tag": [ "trigonometry", "parameterization", "algebra", "polynomial", "function", "inequalities proposed", "inequalities" ], "Problem": "Let $\\varphi = \\frac{1+\\sqrt{5}}{2}$ and $f(x)= 1+\\varphi \\cos{x}+A\\cos{2x}+B\\cos{3x}\\;$ with real parameters $A$ and $B$. \r\nIt is known that $f(x) \\ge 0$ for all $x \\in [0,\\pi]$ . Find $A$ and $B.$", "Solution_1": "Here what I think for this one \r\n\r\n$f(x)$ is even trigonometric polynomial so for any $x \\in [-\\pi;\\pi]$ $f(x) \\geq 0$\r\n\r\n$f$ is $2\\pi$ periodic function so $f(x)$ is positive for any $x\\in \\mathbb R$\r\n\r\n\r\n\r\nNow a problem posted by harazi said $f$ a trigonometric polynomial is postive on $\\mathbb R$ then there exist $g$ such that $f(x)=|g(e^{ix})|^2$", "Solution_2": "Let $ f(x)=1+\\varphi\\cos{x}+A\\cos{2x}+B\\cos{3x} .$ Observe that $\\varphi=2\\cos{\\frac{\\pi}{5}} $. If we denote $ \\Omega=(A-B)-(\\varphi-1) , $ then we obtain\r\n\\[ f(\\pi)= \\Omega \\; \\; ,\\; \\;\r\nf\\left(\\frac{3\\pi}{5}\\right) = -\\; \\frac{\\varphi}{2}\\Omega \\; \\; .\\]\r\nBecause $ f(\\pi)\\ge 0 $ and $\\; f\\left(\\frac{3\\pi}{5}\\right)\\ge 0\r\n$ we must have $\\Omega=0 $, or $\\; (**)\\; \\; A-B= \\varphi-1\\; . $ Let $\\; x:=\\arccos{t}\\; ,\\; t\\in [-1,1] $ and\r\n\\[ G(t):=f(\\arccos{t})= 1+\\varphi t+A(2t^2-1)+B(4t^3-3t)\\]\r\nor using (**), one finds $ G(t)=1+\\varphi t +A(2t^2-1)+(A-\\varphi+1)(4t^3-3t)\\; .$\r\nIn order to have $\\; G(t)\\ge 0\\; ,\\; \\forall t\\in [-1,1], $ we must have the representation\r\n\\[ G(t)=(1+x)(ax+b)^2+(1-x)(cx+d)^2\\; \\; \\; ,\\; \\; a,b,c,d\\in {\\mathbb R}.\\]\r\nIn our case $ G(-1)=G\\left(\\cos\\frac{3\\pi}{5} \\right)=0\\; ,\\;\r\nG(0)=1-A. $ Therefore $ G(t)=\\lambda(1+x)\\left(x-\\cos{\\frac{3\\pi}{5}}\\right)^2\\; \\; ,\\; \\; \\lambda>0 \\; . $\r\nAbove conditions imply $ A=\\frac{2\\sqrt{5}}{5}=0.8944719...\\; ,\\;\r\nB=\\frac{5-\\sqrt{5}}{10}=0.2763932...\\; .$" } { "Tag": [ "geometry", "parallelogram", "geometry unsolved" ], "Problem": "Let $ABCD$ be a convex cuadrilateral, such that $AB=CD$, but it hasn't got any opposite parallel sides. Let $M$ and $N$ be the midpoints of $AD$ and $BC$. Prove that the angle between $MN$ and $AB$ and the angle between $MN$ and $CD$ are equal.", "Solution_1": "draw the parallelograms $AMB'B$ and $DMC'C$. then since $BB'$ is parallel and equal in length to $CC'$ (because they're both parallel and equal to $AM$ and $MD$), and since $BN = NC$, it follows that $B', N, C'$ are collinear and $B'N = NC'$. finally, since $MB' = MC'$, triangles $MB'N$ and $MC'N$ are then congruent, so $\\angle B'MN = \\angle C'MN$, and the result follows." } { "Tag": [ "trigonometry" ], "Problem": "Find $ n\\in Z$ for $ z\\equal{}(1\\plus{}i)^n\\minus{}(1\\minus{}i)^n$ is a complex number. \r\n\r\nA) $ n\\equal{}2k, k \\in Z$\r\n\r\nB) $ n\\equal{}4k, k \\in Z$\r\n\r\nC) $ n \\neq 4k, k \\in Z$\r\n\r\nD) $ n \\in Z$", "Solution_1": "$ \\begin{eqnarray*}\r\n(1+i)^n-(1-i)^n\r\n& = & (\\sqrt{2} \\text{cic}\\frac{\\pi}{4})^n - (\\sqrt{2} \\text{cic}\\frac{-\\pi}{4})^n \\\\\r\n& = & 2^{n/2} \\text{cic}\\frac{n\\pi}{4} - (2^{n/2} \\text{cic}\\frac{-n\\pi}{4}) \\\\\r\n& = & 2^{n/2} \\left(\\cos \\frac{n\\pi}{4} + i \\sin \\frac{n\\pi}{4}\\right) + (2^{n/2} \\left(\\cos \\frac{n\\pi}{4} - i \\sin \\frac{n\\pi}{4}\\right)) \\\\\r\n& = & 2^{(n+2)/2} i \\sin \\frac{n\\pi}{4}\\right)\r\n\\end{eqnarray+}$", "Solution_2": "I'm pretty sure there is a missing keyword in the problem for otherwise the problem would be extremely trivial, according to definitions... :ninja:", "Solution_3": "[quote=\"thanhnam2902\"]Find $ n\\in Z$ for $ z \\equal{} (1 \\plus{} i)^n \\minus{} (1 \\minus{} i)^n$ is a complex number. \n\nA) $ n \\equal{} 2k, k \\in Z$\n\nB) $ n \\equal{} 4k, k \\in Z$\n\nC) $ n \\neq 4k, k \\in Z$\n\nD) $ n \\in Z$[/quote]\r\n\r\nif $ n\\equal{}4k , k \\in Z$ then $ (1\\minus{}i)^{n} \\minus{}(1\\minus{}i)^{n}$ is not a cmplex number. so $ n \\neq 4k ,k\\in Z$ or $ \\boxed C$", "Solution_4": "[quote=\"enndb0x\"][quote=\"thanhnam2902\"]Find $ n\\in Z$ for $ z \\equal{} (1 \\plus{} i)^n \\minus{} (1 \\minus{} i)^n$ is a complex number. \n\nA) $ n \\equal{} 2k, k \\in Z$\n\nB) $ n \\equal{} 4k, k \\in Z$\n\nC) $ n \\neq 4k, k \\in Z$\n\nD) $ n \\in Z$[/quote]\n\nif $ n \\equal{} 4k , k \\in Z$ then $ (1 \\minus{} i)^{n} \\minus{} (1 \\minus{} i)^{n}$ is not a cmplex number. so $ n \\neq 4k ,k\\in Z$ or $ \\boxed C$[/quote]How is that not a complex number? If it is according to this 'definition', the fundamental theorem of algebra would be false...", "Solution_5": "Sounds like there may be some language barriers here.\r\n\r\nIn the American(/English?) math system:\r\nA real number is your standard number line, with $ 0$, $ \\minus{}1$, $ \\pi$, and everything else.\r\nA complex number is any number of the form $ a\\plus{}bi$ for $ a,b$ real numbers. This [i]includes[/i] real and imaginary numbers.\r\n\r\n10000th user's remark is that $ (1\\plus{}i)^n \\minus{} (1\\minus{}i)^n$ is always complex, which is true. It might be real or it might not be, but that doesn't change whether it is complex or not.\r\n\r\nMaybe the problem meant \"not real\", but that doesn't make a lot of sense.", "Solution_6": "Question can be fixed: Find $ n\\in Z$ when $ z$ is a non-real complex number.", "Solution_7": "Can't we just substitute 1 for n and 4 for n? This is clearly enough to weed out the other answers, leaving C. :maybe:", "Solution_8": "Thank you for your post. :P" } { "Tag": [ "probability" ], "Problem": "could use some help with these... :oops: \r\n\r\n1)\r\n\r\nset up the letters A,F,K,N,O,T alphabetically and for every combination set a nummber it represents\r\n\r\na) what nummber is ANKFOT ?\r\nb) what does nr 711 spell?\r\n\r\nafterwards ill come with more", "Solution_1": "I'm assuming that you mean the first permutation alphabetically is 1, the second is 2, and so on:\r\n\r\n[hide] \n\na) The first permutation is AFKNOT. There are 24 words where F is the second letter and 24 where is is the third letter (ahead of ANKFOT, i.e. starting with A). ANKFOT is the 7th with A first and F fourth. So it is number 55. \n\nb) There are 720 total permutations. So 711 very near the end. Clearly it must start with TO. There are 6 permutations of the last four with N first and 6 with K, so K is next. It is the third of these 6, so F is next (the middle letter). It is the first of these , so A is next and then N. The word is TOKFAN.\n\n \n\n[/hide]" } { "Tag": [ "logarithms", "calculus", "calculus computations" ], "Problem": "compute this limit:\r\n$ u_n\\equal{} (n\\plus{}1)^{1\\plus{}\\frac{1}{n\\plus{}1}}\\minus{} n^{1\\plus{}\\frac{1}{n}}$\r\nHere's what i did:\r\n$ n^{1\\plus{}\\frac{1}{n}}\\equal{}n.e^{\\frac{ln(n)}{n}}$\r\n$ \\equal{}n.(1\\plus{}\\frac{ln(n)}{n}\\plus{}o(\\frac{ln(n)}{n}))$\r\n$ \\equal{}n\\plus{}ln(n)\\plus{}no(\\frac{ln(n)}{n}))$\r\nde meme\r\n$ (n\\plus{}1)^{1\\plus{}\\frac{1}{n\\plus{}1}}\\equal{}n\\plus{}1 \\plus{} ln(n\\plus{}1) \\plus{}(n\\plus{}1)o(\\frac{ln(n\\plus{}1)}{n\\plus{}1}))$\r\nThe difference gives:\r\n$ u_n\\equal{}1 \\plus{} (n\\plus{}1).o(\\frac{ln(n\\plus{}1)}{n\\plus{}1})) \\minus{} n.o(\\frac{ln(n)}{n}))\r\n \\equal{}1 \\plus{} no(1) \\plus{} o(1)$\r\n\r\nwich dont give the limit.\r\nThanks.\r\nNB:The limit is 1", "Solution_1": "We can use that $ e^x\\equal{}1\\plus{}x\\plus{}O(x^2)$ as $ x\\to 0.$ We don't need to limit ourselves to saying $ o(x)$ when we have $ O(x^2).$\r\n\r\nNote also that $ \\frac{\\ln(n\\plus{}1)}{n\\plus{}1}$ is asymptotically the same as $ \\frac{\\ln n}{n}.$ Lump those together in your $ O$ calculations.\r\n\r\n$ (n\\plus{}1)\\left(1\\plus{}\\frac{\\ln(n\\plus{}1)}{n\\plus{}1}\\plus{}O\\left(\\frac{\\ln^2n}{n^2}\\right)\\right)\\minus{}n\\left(1\\plus{}\\frac{\\ln n}{n}\\plus{}O\\left(\\frac{\\ln^2n}{n^2}\\right)\\right)$\r\n\r\n$ \\equal{}1\\plus{}\\ln(n\\plus{}1)\\minus{}\\ln n\\plus{}O\\left(\\frac{\\ln^2n}{n}\\right)$\r\n\r\nSince $ \\ln(n\\plus{}1)\\minus{}\\ln n\\equal{}\\ln\\left(1\\plus{}\\frac1n\\right)\\equal{}O\\left(\\frac1n\\right),$ we have everything except the $ 1$ tending to zero and the limit is $ 1.$" } { "Tag": [ "number theory solved", "number theory" ], "Problem": "The natural numbers x,y satisfy 2x^2 + x = 3y^2 + y.\r\nProve that x-y, 2x+2y+1 and 3x+3y+1 are squares !", "Solution_1": "2xx+x=3yy+y\r\nx-y+2xx-2yy=yy\r\n(x-y)(2x+2y+1)=yy\r\n(x-y, 2x+2y+1) =1\r\nTherefore x-y, 2x+2y+1 are perfect squares.\r\nx-y+3xx-3yy=xx\r\n(x-y)(3x+3y+1)=xx\r\nTherefore 3x+3y+1 is a perfect square too." } { "Tag": [ "probability" ], "Problem": "Two players [b]A [/b]and [b]B[/b] are playing the final of a championship which consists of a series of matches. Probability of [b]A[/b] winning the match is [b]2/3 [/b]and that of [b]B[/b] is [b]1/3[/b]. The winner will be the one who is ahead by [b]2 [/b]games as compared to the other player and wins at least [b]6[/b] games. It is given that [b]B[/b] has won the first four games, find the probability that [b]A [/b]wins the championship.", "Solution_1": "This is a little messy and involved. Also, most of the arithmetic is omitted.\r\n\r\nLet us denote by $p_{1}$ the probability that $A$ wins if $A$ is ahead by 1 and can win on her next turn (so, has at least 5 wins); by $p_{0}$ the probability that $A$ wins if $A$ and $B$ are tied (and each has at least 4 wins, so can win with 2 more wins), and by $p_{-1}$ the probability that $A$ wins if $B$ is ahead by 1 and $B$ can win on his next turn (so, has at least 5 wins). Then we have $p_{1}= \\frac{2}{3}+\\frac{1}{3}p_{0}$, $p_{0}= \\frac{2}{3}p_{-1}+\\frac{1}{3}p_{1}$ and $p_{-1}= \\frac{2}{3}p_{0}$. We can solve these equations to get $p_{0}= \\frac{4}{5}$, $p_{-1}= \\frac{8}{15}$ and $p_{1}= \\frac{14}{15}$.\r\n\r\nNow, let $p(m, n)$ be the probability that $A$ wins given that the current state of the game is that $A$ has $m$ wins and $B$ has $n$. We wish to compute $p(0, 4)$. Note that $p(m, n) = \\frac{2}{3}p(m+1, n)+\\frac{1}{3}p(m, n+1)$ and that $p(k, k) = p_{0}$, $p(k+1, k) = p_{1}$ and $p(k, k+1) = p_{-1}$ for $k \\geq 4$. This allows us to work backwards to find that the answer is $p(0, 4) = \\frac{1088}{3645}$.\r\n\r\nIncidentally, starting the game by giving $B$ a head-start of 3 makes it very close to even (slightly to $A$'s advantage)." } { "Tag": [ "limit", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "[color=darkred]Prove that for any $ n\\in N^{*}$ the equation $ x^{n}\\plus{}nx\\equal{}1$ has only one positive root $ x_{n}$ and $ x_{n}\\searrow 0$, $ \\lim_{n\\to\\infty}\\left(1\\plus{}x_{n}\\right)^{n}\\equal{}e$.[/color]", "Solution_1": "[quote=\"Virgil Nicula\"][color=darkred]Prove that for any $ n\\in N^{*}$ the equation $ x^{n}\\plus{}nx \\equal{} 1$ has only one positive root $ x_{n}$ and $ x_{n}\\searrow 0$, $ \\lim_{n\\to\\infty}\\left(1\\plus{}x_{n}\\right)^{n}\\equal{} e$.[/color][/quote]\r\nlet $ n\\to\\plus{}oo$ therefore\r\nwe have $ x^{n}\\plus{}nx \\equal{} 1$ therefore we can easly show that $ x\\to 0$.\r\nok then $ (1\\minus{}nx)^{\\frac{1}{n}}\\equal{} x\\to 0$ therefore we can say that \r\n$ x$=$ \\frac{1}{n}\\plus{}\\alpha_{n}$ where $ (\\alpha_{n})^{\\frac{1}{n}}\\equal{}o(\\frac{1}{n})$\r\ntherefore\r\n$ (1\\plus{}x)^{n}\\equal{} (1\\plus{}\\frac{1}{n}\\plus{}o(\\frac{1}{n}))^{n}\\equal{} e^{n(\\ln (1\\plus{}\\frac{1}{n}\\plus{}o(\\frac{1}{n})))}\\equal{} e^{n(\\frac{1}{n}\\plus{}o(\\frac{1}{n}))}\\equal{} e*e^{o(1)}\\to e$ when $ n\\to\\plus{}oo$", "Solution_2": "[quote=\"Virgil Nicula\"][color=darkred]Prove that for any $ n\\in N^{*}$ the equation $ x^{n}\\plus{}nx \\equal{} 1$ has only one positive root $ x_{n}$ and $ x_{n}\\searrow 0$, $ \\lim_{n\\to\\infty}\\left(1\\plus{}x_{n}\\right)^{n}\\equal{} e$.[/color][/quote]\r\n\r\nYou also posted here (pb 7) \r\nhttp://www.mathlinks.ro/viewtopic.php?t=70659" } { "Tag": [], "Problem": "How many distinct values are possible for the expression $ 8 \\plus{} 4 \\times 6 \\div 2$ if one set of parentheses may be placed in the expression?", "Solution_1": "Well first, let's see how many ways we can place the parentheses and THEN solve each one.\r\n$ (8 \\plus{} 4)x6/div2$\r\n$ 8 \\plus{} (4x6)/div2$\r\n$ 8 \\plus{} 4x(6/div2)$\r\n$ (8 \\plus{} 4x6)/div2$\r\n$ 8 \\plus{} (4x6/div2)$\r\n$ (8 \\plus{} 4x6/div2)$\r\n\r\nSo now we see that there are 6 ways that the parentheses can be put. Now let's solve them to find each distinct answer.\r\n\r\n$ (8 \\plus{} 4)x6/div2$ Solve in parentheses first\r\n$ 12x6/div2$ Now solve from left to right\r\n$ 72/div2$ Divide\r\n$ 36$\r\n\r\nNow we have ONE distinct answer, 36. Now let's solve the second one.\r\n\r\n$ 8 \\plus{} (4x6)/div2$ Solve in parentheses first\r\n$ 8 \\plus{} 24/div2$ Divide according to order of operations\r\n$ 8 \\plus{} 12$ Add\r\n$ 20$\r\n\r\nNow we have a second distinct answer, 20. Currently we have two different answers. Now to solve the third one.\r\n\r\n$ 8 \\plus{} 4x(6/div2)$ Solve in parentheses first\r\n$ 8 \\plus{} 4x3$ Multiply\r\n$ 8 \\plus{} 12$ Add\r\n$ 20$\r\n\r\nNow we have a third answer... which apparently is NOT distinct, it's a repeat from the last equation; so we still have two different answers. \r\n\r\n$ (8 \\plus{} 4x6)/div2$ Solve in parentheses first\r\n$ (8 \\plus{} 24)/div2$ Add\r\n$ 32/div2$ Divide\r\n$ 16$\r\n\r\nNow we have third distinct answer, 16. Now for the fourth one (currently we have 3 different answers).\r\n\r\n$ 8 \\plus{} (4x6/div2)$ Solve in parentheses first\r\n$ 8 \\plus{} (24/div2)$ Divide\r\n$ 8 \\plus{} (12)$ Add\r\n$ 20$\r\n\r\nNow we see ANOTHER repeat, 20.\r\n\r\n$ (8 \\plus{} 4x6/div2)$ Solve in parentheses first; multiply\r\n$ (8 \\plus{} 24/div2)$ Divide\r\n$ (8 \\plus{} 12)/div$ Add\r\n$ 20$\r\n\r\nWe see 20 again. We have four distinct solutions, therefore the answer is $ 4$", "Solution_2": "i think u mean 3...", "Solution_3": "There are $6$ ways to place parenthesis around $2$ or more terms in the expression.\n\n$(8+4\\times 6\\div 2)=20$\n\n$ (8+4\\times 6)\\div 2=10$\n\n$ 8+(4\\times 6\\div 2)=20$\n\n$(8+4)\\times 6\\div 2=36$\n\n$ 8+(4\\times 6)\\div 2=20$\n\n$ 8+4\\times (6\\div 2)=20$\n\nThus there are only $\\boxed {3}$ distinct solutions." } { "Tag": [ "integration", "limit", "calculus", "derivative", "calculus computations" ], "Problem": "$f$ continous on $\\mathbb{R}$ and $k \\in \\mathbb{R}$ such as\r\n$\\int_{0}^{x}f(t) dt = \\frac{x}{2} ( f(x)+k ) $ for every $x \\in \\mathbb{R}$\r\nhow much is $f(0)$ ?", "Solution_1": "Divide both sides by $x$ and let $x\\to 0$: you'll get $f(0)=\\frac12(f(0)+k)$.", "Solution_2": "How do you show\r\nthat $\\int_{0}^{x}\\frac{1}{x}f(t) dt=f(0)$?", "Solution_3": "He just skipped a couple steps. Use L'hopital's rule:\r\n\r\n[hide]$\\int_{0}^{x}f(t) dt = \\frac{x}{2}[f(x)+k )]$\n$\\frac{\\int_{0}^{x}f(t) dt}{x}= \\frac{1}{2}[f(x)+k]$; Divide by $x$\n$\\lim_{x \\rightarrow 0}\\frac{\\int_{0}^{x}f(t) dt}{x}=\\lim_{x \\rightarrow 0}\\frac{1}{2}[f(x)+k]$; Take the limit of both sides as $x \\rightarrow 0$. On the left we have $\\frac{0}{0}$ so we may proceed to take a derivative of both the top and the bottom by L'hopital's rule.\n$\\lim_{x \\rightarrow 0}\\frac{f(x)}{1}=\\frac{1}{2}[f(0)+k]$\n$f(0)=\\frac{1}{2}[f(0)+k]$\n$2f(0)=f(0)+k$\n$\\boxed{f(0)=k}$[/hide]", "Solution_4": "skimnc - quoting L'H\u00f4pital is circular reasoning. To use the rule requires one to know the derivative - but notice that if $F(0)=0,$ then $\\lim_{x\\to 0}\\frac{F(x)}x$ is the very definition of $F'(0).$\r\n\r\nIn this case $F(x)=\\int_{0}^{x}f(t)\\,dt.$ The reason we know its derivative at zero is the Fundamental Theorem of Calculus. The hypothesis \"$f$ continous on $\\mathbb{R}$\" - excuse me, fixing the spelling, [b]continuous[/b] on $\\mathbb{R}$ - is there because it's needed for the FTC. But that's also a little stronger than you need - you only need assume that $f$ is continuous at $0$ and integrable on an interval containing zero.", "Solution_5": "Well yes the problem stated it was continuous at zero so I assumed he knew that. If it weren't then the right side would not be valid either. I was just branching off what mlok had done." } { "Tag": [], "Problem": "In the link CONTEST we can find only the current round of the contest 7. Where I can find the others contest (the problems).", "Solution_1": "I'm slowly rebuilding the contest page. However it will take some time for that to happen." } { "Tag": [ "Putnam", "function", "number theory", "greatest common divisor", "search", "calculus", "calculus computations" ], "Problem": "Find all continuously differentiable real functions f such that for every rational number q, the number\r\nf(q) is rational and has the same denominator as q.\r\n\r\n(The denominator of a rational number q is the unique\r\npositive integer b such that q = a/b for some integer a\r\nwith gcd(a, b) = 1.) (Note: gcd means greatest common\r\ndivisor.)", "Solution_1": "This was B5, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244054]posted here[/url].", "Solution_2": "oh! thanks , din knew the location before.\r\n\r\n\r\ncan u give me the links of other 2008 putnam q too ?", "Solution_3": "Just search for Putnam 2008. All 12 problems were posted by Dr. Merryfield, with proper naming \"Putnam 2008 problem XY\".\r\n\r\nhttp://www.mathlinks.ro/search.php is your friend. :lol:" } { "Tag": [ "vector", "invariant" ], "Problem": "In short: Positron KE=$2.000$ MeV collides with electron at rest. The two are annihilated. Two photons produced. One in direction of positron, other in opposite direction. Find energies of photons.\r\n\r\nMy solution: Conservation of Energy yields $E_{0e^{+}}+KE_{e^{+}}+E_{0e^{-}}=E_{p_{1}}+E_{p_{2}}$ The entire left side is known and the energies of the photons on the right are solely kinetic (since they are photons). \r\n\r\nNow conservation of momentum yields $p_{e^{+}}=p_{p_{1}}+p_{p_{2}}$. Here is where it is shaky. I said that because we knew KE of all three of those, we can say momentum is $\\frac{KE}{c}$ (where we have our KE in MeV), so $\\frac{KE_{e^{+}}}{c}=\\frac{KE_{p_{1}}}{c}-\\frac{KE_{p_{2}}}{c}$. Should I use a relativistic conversion or something? My final answers are $0.511 ~MeV$ for the one travelling opposite direction and $2.511 ~MeV$ for the one travelling in the same direction.", "Solution_1": "well, here's what i would do to get rid of your admittedly wishy-washy step:\r\n\r\nyou know the momentum of the positron because you can calculate velocity from the relativistic expression for kinetic energy ($K = mc^{2}(\\gamma-1)$) and plug into the relativistic expression for momentum ($p = \\gamma mv$). then, photons have no rest mass, so their momentum is just $\\frac{E}{c}$. so we have 2 equations, 2 unknowns like before, so you can solve.\r\n\r\nidk if that changes anything, compute it yourself.", "Solution_2": "Thanks, that is actually how I did the problem the first time. It does give a different answer. So this raises a couple of questions. Why does the first way not work? Guess: It doesn't take into consideration rest energy/mass. (Meaning for a particle with rest mass $p\\neq \\frac{KE}{c}$).\r\n\r\nQuestion 2: So let's take into consideration all the energy. We have $E^{2}=(mc^{2})^{2}+p^{2}c^{2}$ and so momentum is $p=\\sqrt{\\frac{E^{2}}{c^{2}}-m^{2}c^{2}}$ of which all the values are known. This yields yet another answer. Of course, I am notoriously bad at typing things into a calculator :oops: so that could be the problem.", "Solution_3": "your guess as to why your first way doesn't work is correct. that can be easily verified by comparing your faulty expression $p = \\frac{E}{c}$ to the general equation $E^{2}= (pc)^{2}+(mc^{2})^{2}$.\r\n\r\nyour second way ought to work, but it looks like it would just generate irritating algebra, since you'd have to substitute $E = mc^{2}+K$ and expand it, simplify, whatever.", "Solution_4": "Problems of this kind often lead to irritating equations... That's why 4-vectors are the way to go! :)", "Solution_5": "What do these 4- vectors mean", "Solution_6": "4-vectors are four dimensional vectors -- for example, $(ct, x, y, z)$, where $t$ is the time, $c$ the speed of light and $x, y, z$ are the coordinates. They make certain relativity problems much easier because they have some neat properties (their scalar product is invariant under Lorentz transformations). Try googling them for more information, I don't have the time to elaborate right now :maybe:", "Solution_7": "I've used four-vectors in relativity. Is it possible to apply this in Compton Effect/Pair-production type problems? I'd be really interested to see considering how much easier relativity becomes with them.", "Solution_8": "Of course you can. For example, the formula for the Compton effect can be derived in a few lines using 4-vectors :)" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ A$ be a ring such that $ 1 \\plus{} 1 \\neq 0$ and $ |U(A)| \\equal{} n$. If the product of the elements of $ U\\left(A\\right)$ in some specific order is $ 1$ (e. g. if there exists an order in which the product of the elements of $ U\\left(A\\right)$ equals $ 1$), then prove that $ 4\\mid n$.", "Solution_1": "Sorry, what does \"in a specific order\" mean?", "Solution_2": "The ring isn't supposed to be commutative, so \"the product of the elements in U(A)\" isn't welldefined without defining in which order the elements shall be multiplied.", "Solution_3": "A friend of mine got this one.\r\n\r\nWe prove the more general statement that if $ G$ is a finite group containing an element of order 2 such that the product of the elements of $ G$ in some order is $ 1$, then $ 4$ divides $ |G|$. The result follows, since the conditions imply that $ \\minus{}1$ is an element of order 2.\r\n\r\nWe assume for contradiction that $ |G|\\equiv 2\\ (mod\\ 4)$. (clearly $ |G|$ is even)\r\n\r\nLet $ G$ act on itself by left multiplication, then each $ g\\in G$ can be written naturally as a permutation in $ S_n$. Let $ \\phi(g)$ be the sign of this permutation. Let $ x$ be an element of order 2, and note that the permutation assigned to $ x$ is the product of $ |G|/2$ transpositions, so it must be odd, hence $ \\phi(x)\\equal{}\\minus{}1$. It follows that $ \\phi(xy)\\equal{}\\minus{}\\phi(y)$ for all $ y$, so that $ |\\phi^{\\minus{}1}(1)| \\equal{} |\\phi^{\\minus{}1}(\\minus{}1)|$.\r\n\r\nApplying $ \\phi$ to the equation $ x_1 x_2 ... x_n \\equal{} 1$ gives $ (\\minus{}1)^k \\equal{} 1$, where $ k\\equal{}|\\phi^{\\minus{}1}(\\minus{}1)|\\equal{}|G|/2$, hence $ |G|/2$ is even, so $ 4$ divides $ |G|$, contradiction. We conclude that $ 4$ divides $ |G|$." } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "Got a question, just can't figure it. Fairly easy I'm sure but I'm just not understanding it.\r\n\r\nA right Triangle the hypotenuse is 10 meters. One leg(base) is 2 meters longer than the other. What are the lengths of the legs? The drawing shows the hypotenuse as \"10m\", the base as \"x + 2m\" and the other leg as \"x\". I'm just not seeing the way of Factoring to figure out what the legs will be. Thanks for any help in advance!!!", "Solution_1": "huh? isnt this trivial? anyway, solving \\[ x^2+(x+2)^2=100 \\] gives (x+8)(x-6)=0 from which u get that one of the sides has length 6 and hence the other has 8. from now on, please post your questions in the appropriate forum. thanks" } { "Tag": [ "integration", "logarithms", "real analysis", "real analysis solved" ], "Problem": "Evaluate the following definite integrals.\r\n\r\n(1)$\\int^e_1 5^{\\ln x}dx$\r\n\r\n(2)$\\int^1_0 \\frac{x+1}{(x^2+1)^2}dx$", "Solution_1": "To 1): $\\frac{5e-1}{\\ln5+1}$", "Solution_2": "for the first one it is easy to see that $5^{\\ln x}=x^{\\ln 5}$...\r\n\r\nfor the second one we have \r\n$\\displaystyle\\int\\frac{x+1}{(x^2+1)^2}dx=-\\frac12\\int\\left(\\frac1{x^2+1}\\right)'dx+\\text{arctan}x+\\frac12\\int x\\left(\\frac1{x^2+1}\\right)'dx=-\\frac12\\frac1{1+x^2}+\\text{arctan}x+\\frac12\\frac x{x^2+1}-\\frac12\\text{arctan}x+C=\\frac{x-1}{2(1+x^2)}+\\frac12\\text{arctan}x+C$.\r\nNow evaluate it in $0$ and $1$." } { "Tag": [ "inequalities", "induction", "inequalities open" ], "Problem": "[quote=\"tranvinhphuctk14\"]Disprove or Prove that:\nIf $a,b,c>0$ and $k \\in N$ then:\n\n$\\large \\sqrt[2k]{\\frac{a}{b+c}}+\\sqrt[2k]{\\frac{b}{a+c}}+\\sqrt[2k]{\\frac{c}{a+b}}> 2.$[/quote]", "Solution_1": "More generally, for $k\\in\\mathbb{R},k\\geq 2$ we have \\[\\sqrt[k]{\\frac{a}{b+c}}+\\sqrt[k]{\\frac{b}{c+a}}+\\sqrt[k]{\\frac{c}{a+b}}>2.\\]", "Solution_2": "do you mean $k\\ge 2$ because the right side isn't satisfied when $k=1$\r\n\r\nif you do, then the right side can be easily proven using induction", "Solution_3": "Right side easily proved with induction? The RHS is 2... do you mean the greater than side? Because this doesn't follow from induction either... we prove the opposite:\r\nWe want $\\sqrt[k+1]{\\frac{a}{b+c}}+\\sqrt[k+1]{\\frac{b}{c+a}}+\\sqrt[k+1]{\\frac{c}{a+b}}\\geq \\sqrt[k]{\\frac{a}{b+c}}+\\sqrt[k]{\\frac{b}{c+a}}+\\sqrt[k]{\\frac{c}{a+b}}$\r\nBut we have the exact opposite is true because $\\sqrt[k+1]a\\leq \\sqrt[k]a$", "Solution_4": "nvm. there used to be a two part inequality but he editted it out", "Solution_5": "oh what did it say?\r\n(I notice Bodan has a knack for editting, look at his sig. then look at the 4 edits on a rather small post...)", "Solution_6": "you can have a look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76277]here[/url]" } { "Tag": [ "function" ], "Problem": "hi,\r\n\r\nwe can prove using the following function f that the number of N is the same of the number of pair numbers :D .\r\n\r\nf:$ N\\to P$\r\n $ n\\to 2n$\r\n\r\nP = pair numbers.\r\n\r\n\r\nf is a bijetion .\r\n\r\ndon't you find that a little bit illogical :D?", "Solution_1": "Possibly. That's why we have the notion of [url=http://en.wikipedia.org/wiki/Schnirelmann_density]Schnirelmann density[/url]." } { "Tag": [ "inequalities", "calculus", "inequalities solved" ], "Problem": "Show that any real values satisfying x+y+z=5 and xy+yz+zx=3\r\nlies between -1 and 13/3.", "Solution_1": "First of all, x^2+y^2+z^2=19.\r\n\r\nFor the >=-1 part:\r\n\r\nLet's assume x<-1 => y+z>6 => xy+xz=x(y+z)<-6 => yz>9 => y^2+z^2>18 => x^2<1 => false since x<-1, so |x|>1, so we have a contradiction.\r\n\r\nFor the <=13/3 part:\r\n\r\nLet's assume x>13/3 => 0 yz<1/9 => x(y+z)>26/9 => y+z>26/9x => 5>x+26/9x => 9x^2-45x+26<0 => 9(x-13/3)(x-2/3)<0 => x is in (2/3,13/3), contradiction because x>13/3.\r\n\r\nWe can find examples of values -1 and 13/3:\r\n\r\nthe triplets (x,y,z) = (-1,3,3) and (13/3,1/3,1/3) are both solutions.\r\n\r\nI hope it's correct (I might have made some mistakes on the way).", "Solution_2": "It is right, Grobber!", "Solution_3": "Here is my solution\r\nWe know that (x+y) 2 \\geq 4xy\r\nBut x+y=5-z and 4xy=4z(5-z) and after some more calculs we have -3z 2 +10z+13 \\geq 0\r\n That means that -1 \\leq z \\leq 13/3\r\n Idem for x,y\r\n\r\n\r\n\r\n\r\n\r\n Hope my solution is corect! :) :) :)" } { "Tag": [], "Problem": "Square $ ABCD$ has side length 2. A semicircle with diameter $ AB$ is constructed inside the square, and the tangent to semicircle from $ C$ intersects side $ AD$ at $ E$. What is the length of $ CE$?\r\n\r\n[img]http://mathprob.com/_private/get.php/c6c5992d7d4c0d51827a56e3dff9f310/da_square.PNG[/img]", "Solution_1": "[hide]From graphing I got $ CE \\equal{} \\frac{5}{2}$[/hide]", "Solution_2": "[hide]\nLet DE = x. Then $ EA \\equal{} 2\\minus{}x$. Next, let T be the point of tangency between the semicircle and segment CE. Then $ CE^2 \\equal{} DE^2 \\plus{} DC^2 \\equal{} x^2 \\plus{} 4 \\equal{} (CT \\plus{} TE)^2$. But CT = CB = 2, and TE = EA = 2-x, so\n\n$ (2\\plus{}2\\minus{}x)^2 \\equal{} x^2 \\plus{} 4$,\n\nfrom which it easily follows that x = 3/2, and CE = 5/2.\n[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $ x,y,z,$ be the natural numbers satisfying equation $ x^2 \\plus{} y^2\\equal{}z^2$ \r\n\r\nShow that for every x,y,z numbers, $ (z\\minus{}x)(z\\minus{}y)\\equal{}2n^2$ where $ n$ is a natural number too.", "Solution_1": "[quote=\"cenx\"]Let $ x,y,z,$ be the natural numbers satisfying equation $ x^2 \\plus{} y^2 \\equal{} z^2$ \n\nShow that for every x,y,z numbers, $ (z \\minus{} x)(z \\minus{} y) \\equal{} 2n^2$ where $ n$ is a natural number too.[/quote]\r\n\r\n$ (z \\minus{} x)(z \\minus{} y)\\equal{}z^2\\minus{}zx\\minus{}zy\\plus{}xy\\equal{}\\frac{x^2\\plus{}y^2\\plus{}z^2}{2}\\minus{}zx\\minus{}zy\\plus{}xy\\equal{}2 \\left(\\frac{x\\plus{}y\\minus{}z}{2}\\right)^2$\r\n\r\nIt's easy to check by modulo 4 that if $ 2|z$ then $ 2|x$ and $ 2|y$ and if $ 2$ doesn't divide $ z$ then $ 2$ only divides one of $ x,y$.", "Solution_2": "[quote=\"cenx\"]Let $ x,y,z,$ be the natural numbers satisfying equation $ x^2 \\plus{} y^2 \\equal{} z^2$ \n\nShow that for every x,y,z numbers, $ (z \\minus{} x)(z \\minus{} y) \\equal{} 2n^2$ where $ n$ is a natural number too.[/quote]\r\n\r\nSince general solution of $ x^2\\plus{}y^2\\equal{}z^2$ are $ (x,y,z)\\equal{}(2abc,a^2c\\minus{}b^2c,a^2c\\plus{}b^2c)$ and $ (x,y,z)\\equal{}(a^2c\\minus{}b^2c,2abc,a^2c\\plus{}b^2c)$, we get :\r\n\r\n$ (z \\minus{} x)(z \\minus{} y) \\equal{}(a^2c\\plus{}b^2c\\minus{}2abc)(a^2c\\plus{}b^2c\\minus{}a^2c\\plus{}b^2c)$ $ \\equal{}2b^2c^2(a\\minus{}b)^2$\r\n\r\nQ.E.D." } { "Tag": [ "LaTeX" ], "Problem": "How would I write \r\n\r\n{y|y is a former president}\r\n\r\nin $\\LaTeX$? Or on this forum?", "Solution_1": "Is this what you want?\r\n\r\n$\\{y|y \\text{ is a former president}\\}$\r\n\r\n[code]\\{y|y \\text{ is a former president}\\}[/code]", "Solution_2": "I think, it is better way to use \"\\mid\" command for \"|\".\r\n\\[\\{y\\mid y \\text{ is a former president}\\}.\\]", "Solution_3": "I did not know about \\mid. Thanks for sharing :)", "Solution_4": "Thanks!" } { "Tag": [ "modular arithmetic", "Congruences" ], "Problem": "Prove that for $n\\geq 2$, \\[\\underbrace{2^{2^{\\cdots^{2}}}}_{n\\text{ terms}}\\equiv \\underbrace{2^{2^{\\cdots^{2}}}}_{n-1\\text{ terms}}\\; \\pmod{n}.\\]", "Solution_1": "Taking a closer look at my proof in http://www.mathlinks.ro/viewtopic.php?p=849418.", "Solution_2": "Induction", "Solution_3": "[quote=lazizbek42]Induction[/quote]\n\nYes, I solved it that way but the induction isn't easy. ", "Solution_4": "For completeness' sake, I will simply copy my recent reply on [url]https://artofproblemsolving.com/community/c146h150506p849417[/url].\n\nWe will show that with $a_1 = 2, a_2 = 2^2, a_3 = 2^{2^{2}}, \\ldots$, we have that $a_i \\equiv a_j \\pmod{n}$ for all $i,j > \\log_2(n)$. First note that if $a_k$ is congruent to $a_{k+1} = 2^{a_k}$ modulo $n$ for some $k$, then $a_k \\equiv {a_j} \\pmod{n}$, for all $j > k$ by induction. This means that we are done as soon as we can find two consecutive terms which are congruent modulo $n$. Now assume by induction that $a_k \\equiv a_{k+1} \\pmod{n}$ for all $n < 2^k$, which is trivially true for $k = 1$. We will prove that $a_{k+1} \\equiv a_{k+2} \\pmod{n}$ holds for all $n < 2^{k+1}$. So assume $2^k \\le n < 2^{k+1}$ and write $n = l2^m$ with $l$ odd. \\\\\n\nFirst we will deal with the even case; assume that $m \\ge 1$. We actually intend to show that already $a_k \\equiv a_{k+1} \\pmod{n}$ in this case. First it is clear that $a_i \\equiv a_j \\equiv 0 \\pmod{2^m}$ for all $i,j \\ge k \\ge m$, so it is sufficient to show $a_k \\equiv a_{k+1} \\pmod{l}$. But since $l < 2^k$, this already follows from the induction hypothesis. \n\nSo now let's move on to the odd case and assume $m = 0$. Since $\\varphi(n)$ is even, we obtain (either by what we just proved or the induction hypothesis) that $a_{k+1} \\equiv a_k \\pmod{\\varphi(n)}$. But by Euler's Theorem this implies $a_{k+1} = 2^{a_{k}} \\equiv 2^{a_{k+1}} = a_{k+2} \\pmod{n}$. " } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "linear algebra unsolved" ], "Problem": "Let $a_{ij}$ denote a real symetric matrix with coefficients in $[-1,1]$. Define the huge polynomial \r\n $p(x)=\\sum_{A\\in \\{1,2,...,n\\}}{X^{|A|}\\cdot\\prod_{i\\in A}{\\prod_{j\\in A^{c}}{a_{ij}}}}$. Here $A^c$ is the complementary set of $A$ (the numbers that are not in $A$). Prove that this polynomial has all its roots on the unit circle.", "Solution_1": "No activity in linear algebra forum? Any idea on this one? The hint given was to consider the polynomial in $n$ variables $X_1,...,X_n$ obtained replacing $X^{|J|}$ with $\\prod_{j\\in J}{X_j}$, but even with this \"hint\" I have no idea." } { "Tag": [ "calculus", "integration", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Okay, so I am modeling a rocket, and for simplicity sake, let's say it is governed by\r\n\r\n$ \\frac {dv}{dt} \\equal{} v$\r\n\r\nSo seperation of variables -> $ \\frac {dv}{v} \\equal{} dt.$\r\n\r\nTake the definite integral of both sides, \r\n\r\n$ \\int^{v_f}_{v_i} \\frac {dv}{v} \\equal{} \\int^{t}_{0} dt$\r\n\r\n$ \\ln|v_f| \\minus{} \\ln|v_i| \\equal{} t$\r\n$ \\ln|\\frac {v_f}{v_i}| \\equal{} t$\r\n$ |\\frac {v_f}{v_i}| \\equal{} e^t$\r\n\r\nOkay, this is where I am getting confused.\r\n\r\nSo is it $ |v_f| \\equal{} |v_i|e^t$ ?\r\n\r\nBut that doesn't really make sense since that would essentially be two equations:\r\n\r\n$ v_f \\equal{} v_i(e^t)$ and $ v_f \\equal{} \\minus{} v_i(e^t)$\r\n\r\nObviously it can't be both. Do I need some type of extra information to see which solution is to the original differential equation is the one that actually models my rocket? That doesn't make since to me since this is just a first order system, it should only require one initial condition. I graphed both, and it was clear that $ v_f \\equal{} v_i(e^t)$ was the solution that made the most sense, but then that was like we never used absolute values. So when is it necessary to use the absolute values when integrating $ \\frac {1}{x}$.\r\n\r\nThanks for the help!", "Solution_1": "I've never particularly liked the $ \\ln |x|$ formula- it gives the false impression that you can cross the origin. How would I put it? The most general antiderivative of $ \\frac1x$ is $ \\ln(cx)$ for some nonzero constant $ c$. On the real line, we use a negative $ c$ if $ x$ is negative, and a positive $ c$ if $ x$ is positive; we can't have both, because $ \\frac1x$ is not integrable on any interval containing the origin.\r\nIf $ x$ has units, we put the reciprocal of those units into $ c$; we can only take logarithms of unitless numbers.\r\n\r\nHow does that apply here? We want $ t\\equal{}0$ when $ v\\equal{}v_i$, so we use $ \\ln\\left(\\frac{v}{v_i}\\right)$ as the antiderivative. Solving $ \\ln\\left(\\frac{v_f}{v_i}\\right)\\equal{}t$, we get $ v_f\\equal{}v_i\\cdot e^t$.\r\n\r\nA short version of what I think:\r\n$ \\ln(x)$ is an antiderivative of $ \\frac1x$ on $ (0,\\infty)$.\r\n$ \\ln(\\minus{}x)$ is an antiderivative of $ \\frac1x$ on $ (\\minus{}\\infty,0)$.\r\n$ \\ln|x|$ is not an antiderivative of $ \\frac1x$ on $ (\\minus{}\\infty,\\infty)$.", "Solution_2": "That makes a lot of sense, thanks.\r\n\r\nWould there ever be a case in some type of other problem that the final result would be $ y \\equal{} \\minus{} a(e^t)$ instead of $ y \\equal{} a(e^t)$ ?", "Solution_3": "Yes- that goes with the initial condition $ y(0)\\equal{}\\minus{}a$." } { "Tag": [ "number theory", "least common multiple" ], "Problem": "What is the least common multiple of the first ten positive integers?", "Solution_1": "$ lcm (1,2,3,4,5,6,7,8,9,10) \\equal{} 1 * 2 * 3 * 2 * 5 * 7 * 2 * 3 \\equal{} \\boxed{2520}$", "Solution_2": "You can just do 1*5*7*8*9 because you need all prime numbers so 5 and 7, but 8 covers for 2 and 9 covers for 3. This gives you 2520." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ \\forall a,b,c>0$,distinct from $ 1$, such that $ a\\plus{}b\\plus{}c\\equal{}1$,prove that:\r\n\r\n$ \\frac{8a^3}{b\\plus{}c}\\plus{} \\frac{8b^3}{c\\plus{}a}\\plus{} \\frac{8c^3}{a\\plus{}b} \\ge (1\\minus{}a)^2\\plus{}(1\\minus{}b)^2\\plus{}(1\\minus{}c)^2$", "Solution_1": "We have\r\n\r\n$ LHS \\ge 4(a^2\\plus{}b^2\\plus{}c^2) \\ge RHS$\r\n\r\nIt is done here.", "Solution_2": "[hide=\"@Obel1x\"]\nTe kom tregu qe esht veshtir me kriju mosbarazime t'zorshme :P \n\n[/hide]" } { "Tag": [ "geometry", "geometric transformation", "dilation" ], "Problem": "The hydrogenation of an alkyne $ R\\minus{}C \\equiv C\\minus{}R'$ using lithium ($ Li$) and liquid ammonia ($ NH_3$) results in an alkene and the addition is anti-stereochemistry.\r\n\r\nHow can we explain this phenomenon?", "Solution_1": "That phenomenon is explained by the mechanism of the reaction. First, note that when an alkali metal is dissolved in anhydrous liquid ammonia, one obtains a deep blue solution (when dilute, but with a bronze appearence when we have concentrated solutions) that behaves much like a solution of sodium ions and solvated electrons - in fact, it is experimentally found that the electrons are delocalised all over the solution, as they would be in a metal. These solutions are very usefull reducing agents, and the reactions that use them are called [i]dissolving metal reductions[/i] - another very important reaction of this type is the Birch reduction. \r\nIn the mechanism that follows, I will represent the single electron transfers as if they occur from lithium atoms. In the first step, the lithium atom transfers one electron to one of the carbons of the triple bond, producing the radical-anion I. Next, a proton transfer from ammonia gives the vinylic radical II. \r\nNote that, although not shown, the species I and II exist as an equilibrium between both possible conformations: the one shown and another with both R groups in the same side - this is similar to the cis-trans interconversion of oximes (RR'C=N-OH). \r\nIn the next step, a second electron transfer from a lithium atom gives the trans-vinylic anion III: now, it is this step that determines the anti stereochemistry of the reduction, as the trans anion is preferably formed in order to minimize steric repulsions between the R groups. \r\nFinally, a second proton transfer from ammonia completes the mechanism to produce the final product, the trans-alkene IV.", "Solution_2": "Ah, that denotes the mechanism in a much clearer way than in the book thanks!" } { "Tag": [], "Problem": "Three circles are arranged in a row so that each is tangent to the circles next to it. The radii of the two cirlces at the two ends are 5 and 3. What is the length of the line segment AB that passes through the center of each circle? (hint: endpoint A is on the larger circle and endpoint B is on smaller circle)\r\n\r\n[hide] my guess was that it had to be 16 + something (16 is the sum of the diameters of the two outer circles) , but i can figure out what[/hide]", "Solution_1": "Your problem is ambiguous; there are no restrictions on the size of the middle circle. All you can say is that $ AB\\ge16$." } { "Tag": [ "search" ], "Problem": "Guess the game from the screenshot.\r\n\r\nTry to guess the game from the screenshot (name & platform). The correct guesser posts a new screenshot afterwards.\r\n\r\nRules:\r\n-No unreleased games\r\n-If you get the image online, try to reupload the image onto something like http://www.imageshack.us (or just attach it here) with [b]a different filename[/b] so it isn't immediately obvious what the game is.\r\n-If no one figures it out after three days, a new screenshot may be posted.\r\n-Only one guess per post, and no double posting.\r\n\r\nFirst one:\r\n[img]http://img292.imageshack.us/img292/3677/snshotop1.png[/img]", "Solution_1": "wario ware of ds?", "Solution_2": "actually it's from the gba but close enough :lol: \r\n\r\nyour turn.", "Solution_3": "[img]http://me.yogurttree.com/junggi/asdgasdg.jpg[/img]\r\npossibly the best game ever created for .", "Solution_4": "[quote=\"junggi\"][img]http://me.yogurttree.com/junggi/asdgasdg.jpg[/img]\npossibly the best game ever created for .[/quote]\r\nSSB:M\r\n\r\nSomeone else can post the pictar....", "Solution_5": "[quote=\"junggi\"][img]http://me.yogurttree.com/junggi/asdgasdg.jpg[/img]\n[/quote]\r\n :omighty: :omighty: :omighty: ", "Solution_6": "[quote=\"ragnarok23\"][quote=\"junggi\"][img]http://me.yogurttree.com/junggi/asdgasdg.jpg[/img]\n[/quote]\n :omighty: :omighty: :omighty: [/quote]\r\nI second that.", "Solution_7": "To keep it going.\r\n[img]http://img53.imageshack.us/img53/2601/screenshotofyu8.png[/img]", "Solution_8": "looks like pokemon", "Solution_9": "[img]http://www.lunchlady.org/uploaded/images/emeraldsword.jpg[/img]", "Solution_10": "[quote]looks like pokemon[/quote]\r\nYeah, that's right.\r\n\r\nStepmania, PC (you only use C400?)\r\n\r\n\r\n[img]http://i9.photobucket.com/albums/a72/randomdragoon/screenshotofavg.png[/img]", "Solution_11": "PRESS R OR Z TWICE TO DO A BARREL ROLL\r\n\r\nand yeah, i only use c400\r\n\r\ni should move up, but i'm too lazy to figure out how to use c mods other than 400 (yeah, that's right)", "Solution_12": "[quote=\"randomdragoon\"][/quote]\r\nor something like that\r\n\r\n\r\n[img]http://me.yogurttree.com/junggi/asdgdsahah.jpg[/img]", "Solution_13": "[quote=\"junggi\"][quote=\"randomdragoon\"][/quote]\nor something like that\n\n\n[img]http://me.yogurttree.com/junggi/asdgdsahah.jpg[/img][/quote]\r\nit's Playstation! (my best guess lol)\r\nokay it's DBZ rite?\r\n\r\nhere's a nice text-based game:\r\n[img]http://img235.imageshack.us/img235/3488/nznnskldjflkdjlksfjcp4.png[/img]", "Solution_14": "It's not Star Fox 64. It's Star Fox for the SNES.\r\n\r\nAlso, don't post another screenshot if your answer is a total guess or obviously wrong.", "Solution_15": "[quote=\"randomdragoon\"]It's not Star Fox 64. It's Star Fox for the SNES.\n[/quote]\r\ndoh how could I be so stupid, i mean just look at the graphics, ewwww", "Solution_16": "[quote=\"junggi\"][quote=\"randomdragoon\"]It's not Star Fox 64. It's Star Fox for the SNES.\n[/quote]\ndoh how could I be so stupid, i mean just look at the graphics, ewwww[/quote]\r\njunggi is my guess close?", "Solution_17": "[quote=\"ProtestanT\"][quote=\"junggi\"][quote=\"randomdragoon\"]It's not Star Fox 64. It's Star Fox for the SNES.\n[/quote]\ndoh how could I be so stupid, i mean just look at the graphics, ewwww[/quote]\njunggi is my guess close?[/quote]\r\nnot at all, my friend, not at all.\r\nIf it was DBZ, they would be flying, firing a bunch of weird fireball thingies and have huge yellow hair.", "Solution_18": "I'm guessing it's some version of Tekken.", "Solution_19": "[quote=\"randomdragoon\"]I'm guessing it's some version of Tekken.[/quote]\r\nyup, tekken 5.", "Solution_20": "[quote=\"ProtestanT\"]\nit's Playstation! (my best guess lol)\nokay it's DBZ rite?\n\nhere's a nice text-based game:\n[img]http://img235.imageshack.us/img235/3488/nznnskldjflkdjlksfjcp4.png[/img][/quote]\r\nI believe this game is Nuke Zone, a browser-based game for the PC.\r\nEDIT: Yeah, visited their website and wikipedia entry, definately is.\r\n\r\nNext:\r\n[img]http://i9.photobucket.com/albums/a72/randomdragoon/screenshotofavg2.jpg[/img]", "Solution_21": "Bust a Move # whatever", "Solution_22": "Correct! Your turn.", "Solution_23": "i was wonderin how u figured it out...then i googled \"command centre missile silo research lab shipyard airfield war factory barracks power plant advanced turret headquarter torpedo launcher sam site machinegun\", and most of the top links were about nukezone... i think the SAM site gave it away", "Solution_24": "[quote=\"ProtestanT\"]i was wonderin how u figured it out...then i googled \"command centre missile silo research lab shipyard airfield war factory barracks power plant advanced turret headquarter torpedo launcher sam site machinegun\", and most of the top links were about nukezone... i think the SAM site gave it away[/quote]\r\nActually, I just went on wikipedia for a list of browser-based MMORPGs. It was sorted by category, so I could immediately skip all of the medieval/fantasy/etc. categories, and go straight to the proper category, which had about eight games in it. It didn't take much longer to realize that the wikipedia screenshot for Nuke Zone looked extremely similar.", "Solution_25": "Here's one while you wait:\r\n\r\n\r\nIt isn't actually me playing - its from a friend.", "Solution_26": "Age of Empires #something.", "Solution_27": "AoE3 :D :D\r\nand hes Ottoman.", "Solution_28": "[img]http://www.garia.net/g6/anewdawn.jpg[/img]\r\n\r\nsweet game really.", "Solution_29": "Eve Online, game for PC.\r\n\r\nThis one is a really fun game:\r\n[img]http://img329.imageshack.us/img329/9396/screenshotoa3.jpg[/img]", "Solution_30": "kyy's picture looked like a tree trunk, and random's looked like crabs", "Solution_31": "Haha, search up some more ships from Eve. That one isn't even the ugliest...Try \"Eve Ships Mimitar,\" they have trash cans...fly swatters...whatnot.", "Solution_32": "[img]http://games.kikizo.com/media/halo2_june04/06c.jpg[/img]\r\n\r\nthe best game =)", "Solution_33": "Pokemon Colloseum", "Solution_34": "HALO 2\r\nhttp://img120.imageshack.us/my.php?image=picture52tj3.png\r\n\r\nIts a great game, either this or original would be OK.[/code]" } { "Tag": [ "logarithms" ], "Problem": "Could anyone give me any hints for this AOPS Vol 2 Logarithm Problem?\r\n\r\nIf log 36 = a and log 125 = b, express log(1/12) in terms of a and b.", "Solution_1": "[hide=\"hint\"]\n$ \\log{5} \\equal{} \\log{\\frac{10}{2}}$[/hide]", "Solution_2": "hello, what base has your logarithm?\r\nSonnhard.", "Solution_3": "Thanks for the hints. I solved it.", "Solution_4": "[quote=\"Dr Sonnhard Graubner\"]hello, what base has your logarithm?\nSonnhard.[/quote]\r\nWhen there isn't a base written it is assumed to be in base 10 :wink:" } { "Tag": [ "ratio", "real analysis", "real analysis unsolved" ], "Problem": "Show that there is real a s.t \r\n 1/3\u2264{a^n }\u22642/3\r\nholds for all positive integers n. (where {x} is fractional part of x) \r\nto view all the Some problems from the Uzbekistan universities olympiads visit \r\nhttp://www.mathlinks.ro/Forum/topic-81099.html \r\n(alekk if you have solution , please send)", "Solution_1": "Use the golden ratio, or some number like that. What are they called again? [url=http://mathworld.wolfram.com/PisotNumber.html]Pisot Numbers.[/url]" } { "Tag": [ "abstract algebra", "Ring Theory", "superior algebra", "superior algebra solved" ], "Problem": "Let $R$ be a P.I.D. and $F$ its field of fractions. Suppose $S$ is a ring with $R \\subset S \\subset F$.\r\n\r\n(a) Show that all elements $\\alpha \\in S$ can be written as $a/b$, where $a,b \\in R$ and $1/b \\in S$.\r\n\r\n(b) Show that $S$ is a P.I.D.\r\n\r\n(c) Show that if $S$ is finitely generated as an $R$-module, then $S = R$.\r\n\r\nThanks!", "Solution_1": "Are you sure about (c)? Consider $R = Z[\\sqrt{-3}]$ and $S = Z\\left[\\frac{1+\\sqrt{-3}}{2}\\right]$.", "Solution_2": "Let $T=\\{x\\in R|\\frac{1}{x}\\in S\\}$. Then merely by the ring axioms for $S$, $T$ is multiplicatively closed. Obivously, $T^{-1}R$ naturally injects in $S$.\r\n\r\n(a) Let $\\alpha\\in S$. Then $\\alpha=\\frac{a}{b}$ for some coprime $a,b\\in R$.\r\nThere exist $c,d\\in R$ such that $ac+bd=1$. We have $\\frac{ac+bd}b=c\\alpha+d\\in S$, so $\\frac{1}{b}\\in S$.\r\nIt is now clear that $S\\simeq T^{-1}R$.\r\nFor this reason, (b) should now be purely theoretic: a localisation of a PID is PID.\r\n\r\n(c) $R$ is noetherian. $S$ is finitely generated as an $R$ module, so it is a noetherian $R$-module. Assume it is different from $R$. Then it contains some $\\frac{1}{b}$ for some $b\\in R$ that is not a unit in this ring. Consider the sub $R$-modules $R/b^{n}$ for $n\\in\\mathbb N^{*}$. These form an increasing sequence of sub-modules that never becomes stationary.\r\n----------------------------------------------------------------------------\r\nrgep's example is not appropriate for this problem's context because $\\mathbb Z[\\sqrt{-3}]$ is not PID. I think it is not even factorial." } { "Tag": [ "MATHCOUNTS", "geometry" ], "Problem": "the vertices of a triangle are thepoints of intersection of the lines y=-x-1, x=2, and y=1/5x+13/5. find an equation of the circle passing through all three vertices", "Solution_1": "when they give you an equation and they want you to find the area how DO you go about doing it?\r\ni have no clue at all. one example is the contest #2 in the getting started section. how do you solve a problem like that?", "Solution_2": "Hint: Since the AMC contest is done you can go look at the solution for problem number 21 and compare it to this one...", "Solution_3": "[quote=\"Leeyuhang\"]when they give you an equation and they want you to find the area how DO you go about doing it?\ni have no clue at all. one example is the contest #2 in the getting started section. how do you solve a problem like that?[/quote]\r\n\r\nI presume you are talking about finding the area of a circle given its equation.. well, to find the area, you have to find the radius. You can rearrange the formula for a circle in such a way that tells you the radius.", "Solution_4": "have you taken algebra II leeyuhang? its in there about equations of circles", "Solution_5": "no, i don't but i'll look in my dad's collection of textbooks.\r\nisn't algebra 2 also called intermediate algebra or something like that?", "Solution_6": "well basically the circle equation is of the form (x-h)^2+(y-i)^2=r^2\r\n\r\nwhere (x,y) is a point on the circle, and (h,i) is the origin, and r is the radius. dont get mad if i messed that up, i think...im floating", "Solution_7": "origin as in 0,0 or the circle? does h and i stand for something?\r\nfloating?", "Solution_8": "(h,i) is the centre of the circle.", "Solution_9": "yeah, as i said...im floating aka delirious" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let A and B be two subgroups of a group G. If |A|=p, a prime integer, then show that either A\u22c2B={e} or A\u2286B.\r\n\r\nthank you.... :)", "Solution_1": "just guess the amount of subgroups of A ;)", "Solution_2": "[hide=\"Umm..\"]Let A and B be subsets of a group G, and let |A|=p where p is prime.\nSuppose $ A \\cap B \\not= \\{ e \\}$. Since we know that $ A \\cap B \\not= \\emptyset$ ,there is $ g \\in {A \\cap B}$ such that $ g \\not= e$. Since $ o(g)=p$, so $ ||=p$. That is $ < g > = A$. But we know that ${ \\subseteq B}$. Thus $ A \\subseteq B$.[/hide]" } { "Tag": [ "inequalities", "trigonometry", "geometry", "trig identities", "Law of Sines", "geometry proposed" ], "Problem": "$ \\triangle\\ ABC$ $ \\implies$ $ \\max\\ \\{b,c\\} < \\sqrt {R(a \\plus{} 2m_a)}\\ (*)\\ ,$ where $ m_a$ is the length of the $ A$- median.\r\n[b]Remark.[/b] For $ a \\equal{} b$ and $ c\\rightarrow 0$ , we\"ll have $ m_a\\rightarrow \\frac b2$ , $ R\\rightarrow\\frac b2$ and $ (*)$ becomes \"[b]equality[/b]\" !", "Solution_1": "[quote=\"Virgil Nicula\"]$ \\triangle\\ ABC$ $ \\implies$ $ \\max\\ \\{b,c\\} < \\sqrt {R(a \\plus{} 2m_a)}\\ (*)\\ ,$ where $ m_a$ is the length of the $ A$- median.\n[b]Remark.[/b] For $ a \\equal{} b$ and $ c\\rightarrow 0$ , we\"ll have $ m_a\\rightarrow \\frac b2$ , $ R\\rightarrow\\frac b2$ and $ (*)$ becomes \"[b]equality[/b]\" ![/quote]\r\n[hide=\"Solution\"]\nWithout loss of generality, allow us to say that $ b\\ge c$. Now, we see that $ m_a\\plus{}\\frac{a}{2}> b\\implies 2m_a\\plus{}a> 2b\\implies \\sqrt{2m_a\\plus{}a}>\\sqrt{2b}$. The Law of Sines gives us that \\[ R\\equal{}\\frac{b}{2\\sin \\angle ABC}\\implies \\sqrt{R(2m_a\\plus{}a)}>\\sqrt{2bR}\\equal{}\\sqrt{2b(\\frac{b}{2\\sin \\angle ABC})}\\equal{}\\frac{b}{\\sqrt{\\sin \\angle ABC}}>b\\] and we are done. [/hide]" } { "Tag": [ "videos", "function", "algorithm", "AMC", "AIME", "probability", "expected value" ], "Problem": "I used to be a beast at minesweeper, but I\"m out of practice now. What are your best times?\r\n\r\nbeginner:4, average around 9\r\nintermediate: 36, average around 45\r\nexpert: 114, average around 140", "Solution_1": "not as good as you... but here are my best and average times:\r\n\r\nBeginner: Best-6, Average-10\r\nIntermediate: Best-40, Average-51\r\nExpert: Best:152, Average-170", "Solution_2": "Beginner: 6\r\nIntermediate: 42\r\nExpert: 999 :P I'm too lazy to beat it\r\n\r\nThose are my best times.", "Solution_3": "I never even beat expert once.", "Solution_4": "(Best times:)\r\n\r\nBeginner: 7\r\nIntermediate: ... 42?\r\nExpert: 128\r\n\r\nI stopped playing when I cramped my hand too much clicking. :P", "Solution_5": "[quote=\"t0rajir0u\"]\n\nI stopped playing when I cramped my hand too much clicking. :P[/quote]\r\n\r\nSame. I used to be able to play for 2 hours straight, then I realized I had played for 2 hours, then I immediately began to feel the fatigue.", "Solution_6": "YES! After about two months, I finally got a new Intermediate best time: 40 seconds! :D", "Solution_7": "You know there's a lot of luck involved in minesweeper as well.", "Solution_8": "[quote=\"SplashD\"]You know there's a lot of luck involved in minesweeper as well.[/quote]\r\nmy friend just clicked randomly and got 1 second twice in a row. :o", "Solution_9": "Expert always comes down to luck and I always get unlucky.", "Solution_10": "I play a lot of \"customized\" minesweeper games (30*24 with 200 mines). I did beat it once, with 432 seconds.", "Solution_11": "[quote=\"sq53e\"]I play a lot of \"customized\" minesweeper games (30*24 with 200 mines). I did beat it once, with 432 seconds.[/quote]\r\n\r\nHas anyone every beat customized with a 24*30 board and 667 mines?", "Solution_12": "[quote=\"SplashD\"]Has anyone every beat customized with a 24*30 board and 667 mines?[/quote]\r\n\r\nThat's like impossible... unless you're really, really lucky", "Solution_13": "It's actually possible to beat it if you know the cheat codes.", "Solution_14": "[quote=\"sq53e\"]It's actually possible to beat it if you know the cheat codes.[/quote]\r\n\r\nCheat codes? What cheat codes?", "Solution_15": "For me: \r\nBeginner: 2 seconds\r\nIntermediate: 39 seconds\r\nExpert: 89 seconds", "Solution_16": "[quote=\"louiethechewy123\"][quote=\"moogra\"]It's not really reviving when it's not even on the bottom half of the first page. Reduced my expert time to 229[/quote]\n\nWhat? This is the 7th page...[/quote]\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/index.php?f=360]This page?[/url]\r\n\r\nI wouldn't ever revive stuff on the 7th page, but as it was in the upper middle of the first page, it's okay.\r\n\r\nI got 54 in intermediate today, 6 in beginner (average 15ish). I need to get better beginner average-wise.", "Solution_17": "Ok... Got 113 in beginner :)", "Solution_18": "1 second custom (snapshot is attached :rotfl: )\r\n78 seconds beginner", "Solution_19": "HOW can you do 10 clicks in one second?\r\n :o", "Solution_20": "This is what I did:\r\n\r\nCustom --> Height 24 --> Width 30 --> Mines 10 --> OK\r\n\r\nClick a random square. A long as you don't click a mine or number, pretty much the whole screen goes out. Repeat this process and eventually you will get a one click win!", "Solution_21": "I used to play all three levels on my old Windows XP computer but ever since I switched to my current computer two years ago, I\u2019ve been playing only the Advanced level on the Windows Vista version.\r\n\r\n[list][img]http://i638.photobucket.com/albums/uu109/susannah_dingley/Minesweeper.jpg[/img][/list]\r\nSo my best time at Advanced level on my new computer is 198 seconds \u2013 and I win between two and three out of every ten games. I\u2019ve forgotten my best times on my old computer, but I most probably have solved Advanced Minesweeper in much less than 198 seconds before. I was younger and my reflexes were quicker then. :P", "Solution_22": "I beat the 26*30, 668 mines game.\r\nThe screen print is included.", "Solution_23": "What I find strange about the new WINDOWS XP's is that they always allow you to repeat the one you just did. :huh:", "Solution_24": "I stinK at it", "Solution_25": "I beat a 500-mine game once, but I tried ... let's see ... 109 times before completing it.\nEDIT: It took me 2 hours. xD", "Solution_26": "I just got a new computer, so I don't remember my best times..... but since this computer is a touchpad and I haven't gotten around to getting a mouse, I probably won't be getting good times anytime soon. :P", "Solution_27": "I've played a bit more and have gotten some better times:\nFlagging:\nBeginner 1 (3bv = 1 lol)\nIntermediate 25? was on old computer\nExpert 88\n\nNon-flagging\nBeginner: 4.71 (3bv=14) (minesweeper clone)\nIntermediate: 30.20 (3bv=65) (hi-games.net)\nExpert: 126 (3bv = 174) (hi-games.net)\n\n3bv/s:\nBeginner: 3.28 (arbiter)\nIntermediate: 2.15\nExpert: something crappy\n\nIOE:\nBeginner: 1.00 (clone)\nIntermediate: 0.905 (clone)\nExpert: something crappy", "Solution_28": "My best times:\nBeginner: 10\nIntermediate: 34\nExpert: 88\n\nMy average times:\nBeginner and Intermediate: I haven't played that again recently...\nExpert: 120", "Solution_29": "Expert: -52 minutes (end_time - start_time)\n[negative time run]" } { "Tag": [ "algebra", "linear equation" ], "Problem": "How many ways can you take 4 numbers of the set {$ 1,2,3,4,5,6,7,8$} to make a sum of $ 18$?", "Solution_1": "Must they be distinct? Let's assume yes:\r\n$ 1,2,7,8$\r\n$ 1,3,6,8$\r\n$ 1,4,5,8$\r\n$ 1,4,6,7$\r\n$ 2,3,5,8$\r\n$ 2,3,6,7$\r\n$ 2,4,5,7$\r\n$ 3,4,5,6$", "Solution_2": "This can also maybe be done by generating a fn :maybe: :)", "Solution_3": "is there a way to do this using counting principles? i.e nCr? Sorry if this doesn't make sense, i'm new to competitive, harder, math and am trying to understand a faster way of doing it rather the slow way of trial and error.", "Solution_4": "[hide=\"Well...\"] If the numbers [i]aren't[/i] distinct (and if order matters), we want the coefficient of $ x^{18}$ in $ (x \\plus{} x^2 \\plus{} ... \\plus{} x^{8})^4$, to which we can apply the [url=http://en.wikipedia.org/wiki/Multinomial_theorem]Multinomial theorem[/url], although for this problem size it doesn't help much. This approach becomes much more helpful if $ 4$ is replaced with a much larger number. [/hide]", "Solution_5": "[quote=\"LightEntropy\"]is there a way to do this using counting principles? i.e nCr? Sorry if this doesn't make sense, i'm new to competitive, harder, math and am trying to understand a faster way of doing it rather the slow way of trial and error.[/quote]\r\n\r\n[hide=\"Yes.\"](You may want to read chapters 13 and 15 of [url=http://www.maths.mq.edu.au/~wchen/lndmfolder/lndm.html]this[/url] to understand what I am doing, although I try to explain the method decently.)\n\nIf the numbers don't need to be distinct and if order matters, this approach works:\n\nSet it up like a linear equation:\n\n$ a_1 \\plus{} a_2 \\plus{} a_3 \\plus{} a_4 \\equal{} 18$\n\nWe're trying to find the number of solutions for that equation when $ a_1,a_2,a_3,a_4 \\in \\{1,2,3,4,5,6,7,8\\}$.\n\nThe number of solutions of that equation when $ a_1,a_2,a_3,a_4 \\in \\{1,2,3,4,5...\\}$ is the same as the number of solutions of $ b_1 \\plus{} b_2 \\plus{} b_3 \\plus{} b_4 \\equal{} 14$ when $ b_1,b_2,b_3,b_4 \\in \\{0,1,2,3,4,5...\\}$ because $ (a_1 \\minus{} 1) \\plus{} (a_2 \\minus{} 1) \\plus{} (a_3 \\minus{} 1) \\plus{} (a_4 \\minus{} 1) \\equal{} 14$. Basically, we dropped them all down one and it still works the same way.\n\nThe number of solutions to $ a_1 \\plus{} a_2 \\plus{} a_3 \\plus{} \\cdots \\plus{} a_k \\equal{} n$ without any restrictions is $ {n \\plus{} k \\minus{} 1 \\choose n}$. (That's $ _{n \\plus{} k \\minus{} 1}C_n$, but the other way, which is called a binomial coefficient, is more convenient).\n\nSo without restrictions, we have:\n\n$ {14 \\plus{} 4 \\minus{} 1 \\choose 14} \\equal{} {17 \\choose 14} \\equal{} 680$.\n\nNow we want to get rid of all of the cases where any $ b_i > 7$ (remember, we dropped everything down one in order to make the first number in those sets be zero, so we're now using 7 instead of 8). So for every $ i \\equal{} 1,2,3,4$:\n\n$ S_i \\in \\{(b_1,...b_4): b_i > 7\\}$\n\nAnd from the 680 solutions from before, we'll subtract $ |S_1 \\cup S_2 \\cup S_3 \\cup S_ 4|$; that is, we're subtracting every case where $ b_i > 7$. \n\n$ |S_1| \\equal{} |S_2| \\equal{} |S_3| \\equal{} |S_4| \\equal{} {14 \\plus{} 4 \\minus{} 8 \\minus{} 1 \\choose 14 \\minus{} 8} \\equal{} {9 \\choose 6}$.\n\nNote that this is equal to the number of solutions in $ (b_1 \\plus{} 8) \\plus{} b_2 \\plus{} b_3 \\plus{} b_4 \\equal{} 14$ We got the 8 from the fact that the limit on the other set is 7, and then we add one.\n\nThere are also 4 sets that we counted equal to the binomial coefficient up above, so later, we'll multiply that by 4.\n\nFor every $ |S_n \\cap S_m|$ when $ s,n \\equal{} 1,2,3,4$, the number of solutions to\n\n$ (b_1 \\plus{} 8) \\plus{} (b_2 \\plus{} 8) \\plus{} b_3 \\plus{} b_4 \\equal{} 14$\n$ b_1 \\plus{} b_2 \\plus{} b_3 \\plus{} b_4 \\equal{} \\minus{} 2$\n\nis zero. Note that its corresponding binomial coefficient\n\n{14+4-16-1 \\choose 14-16} = {1 \\choose -2}\n\nis meaningless. So that leaves us with\n\n$ |S_1 \\cup S_2 \\cup S_3 \\cup S_ 4| \\equal{} 4{9 \\choose 6} \\equal{} 336$\n\nThe solution is then\n\n$ {17 \\choose 14} \\minus{} 4{9 \\choose 6} \\equal{} 680 \\minus{} 336 \\equal{} \\boxed{344}$.\n\nWhen you get used to it, it's a lot quicker than it seems at first. When you keep using this method, you'll soon see a pattern that'll be easily identifiable for you personally.[/hide]", "Solution_6": "[b]@ LordCarbo:[/b] [i]I think order doesn't matter ;) [/i]\r\n\r\n[hide=\"Solution\"]Let's subtract 1 from each of the numbers, so that we want to pick numbers from 0, 1, ... 7 that add to 14.\nNote that these numbers sum to 7*8/2 = 28, so the other four numbers will add to 28-14 = 14 as well.\n\nThus, we want to place the numbers into the format _+_+_+_=_+_+_+_ where numbers on the LHS were chosen and numbers on the RHS were not chosen\n\nWe know that sets such that if a is in the subset then 7-a is also in the subset will automatically satisfy this. This way, if we mark the numbers from 0 to 7 which are chosen, they will be symmetric, and it suffices to chose two of the four numbers 0 1 2 3. There are 4C2 = 6 ways of doing this.\n\nThe only other possibility is if on one side we chose _XX_ and on the other side we chose X__X. There are two ways of picking the sides, so there are 2 ways here.\n\nThis gives a total of 8 ways.[/hide]", "Solution_7": "Sorry for the confusion guys, order does not matter and the terms must be distinct." } { "Tag": [ "Sequence", "algebra", "Inequality", "IMO Shortlist" ], "Problem": "Let $c_1, \\ldots, c_n \\in \\mathbb{R}$ with $n \\geq 2$ such that \\[ 0 \\leq \\sum^n_{i=1} c_i \\leq n. \\] Show that we can find integers $k_1, \\ldots, k_n$ such that \\[ \\sum^n_{i=1} k_i = 0 \\] and \\[ 1-n \\leq c_i + n \\cdot k_i \\leq n \\] for every $i = 1, \\ldots, n.$\r\n\r\n[hide=\"Another formulation:\"]\nLet $x_1, \\ldots, x_n,$ with $n \\geq 2$ be real numbers such that \\[ |x_1 + \\ldots + x_n| \\leq n. \\] Show that there exist integers $k_1, \\ldots, k_n$ such that \\[ |k_1 + \\ldots + k_n| = 0. \\] and \\[ |x_i + 2 \\cdot n \\cdot k_i| \\leq 2 \\cdot n -1 \\] for every $i = 1, \\ldots, n.$ In order to prove this, denote $c_i = \\frac{1+x_i}{2}$ for $i = 1, \\ldots, n,$ etc.\n[/hide]", "Solution_1": "Let $S=\\sum_{j=1}^{n} c_j$.\r\n$k_i=[\\frac{-c_i}{n}]+[\\frac{Si}{n}]-[\\frac{S(i-1)}{n}]$.", "Solution_2": "Please be so kind and write a full solution. :)", "Solution_3": "Exactly, let $\\frac{-c_i}{n}=m_i+x_i,m_i=[\\frac{-c_i}{n}],l_0=0;l_i=[1+\\sum_{j=1}^i x_j],k_i=m_i+l_i-l_{i-1}$.\r\nWe have $\\sum_j k_j=\\sum_j m_i +1+[\\sum_j x_j] \\Longrightarrow \\sum_j \\frac{-c_j}{n}<\\sum_i k_i \\le 1+\\sum_j \\frac{-c_j}{n}.$\r\nTherefore if $0<\\sum_j c_j \\le n$, then $\\sum_j k_j=0.$ \r\nWe have $\\frac{-c_i}{n}-1=3 points are given in plane s.t. they all dont lie on a single line,then they determine atleast n-1 different slopes ,where equality is possible only if n is odd and n>=5 :welcome:", "Solution_1": "You can find this in the Chapter 10 of [u]Proofs from the book[/u] written by Martin Aigner and Gunter M Ziegler : [url=http://books.google.com/books?vid=ISBN3540404600&id=KvQr9l0wgf8C&pg=PA59&lpg=PA59&dq=the+slope+problem&sig=QGXdWe9ogcbJG294Zi19elJHKKU]The slope problem.[/url]" } { "Tag": [ "search", "AMC", "AIME", "\\/closed" ], "Problem": "Does anyone know if I can get a few (3-5) sample WOOT problems? My dad is reluctant to spend $ \\$$695 and wants to see if I am ready.", "Solution_1": "Take a look at the PDF on [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOTSchedule.php]the schedule page[/url]. You'll find some sample problems there.", "Solution_2": "Is there any place to find the answers?", "Solution_3": "We haven't published answers -- you can ask the problems on the message board (or search for them, some have probably been discussed before).", "Solution_4": "Also, I think one of the wonderful things about taking an AoPS course is that you can register for the class, and if its too difficult, you can still quit before 2 weeks (I think) after the course's start.", "Solution_5": "So how much should I be able to score on AIME before WOOT? I want to make sure I take it at the right time.", "Solution_6": "Look here:\r\nhttp://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php\r\nSpecifically, this is the question to your answer:\r\n\"Students who are not capable of consistently scoring 5 or higher on the American Invitational Mathematics Exam should strongly consider waiting until they have more experience before joining WOOT.\"", "Solution_7": "Also see [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=257]here[/url]. You might also want to attend the math jam on September 8.", "Solution_8": "[quote=\"rrusczyk\"]Take a look at the PDF on [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOTSchedule.php]the schedule page[/url]. You'll find some sample problems there.[/quote]\r\n\r\nJust one question, those sample problems, are they easier, aligned or harder than the program itself?", "Solution_9": "They're fairly representative of the spread. You can focus on one end or the other in the course of WOOT, depending on your interest and ability.", "Solution_10": "No, no, no I never considered taking WOOT :P \r\n\r\nJust wanted to know at what difficulty is it worked there, gathering info on everything :) :roll:" } { "Tag": [ "absolute value" ], "Problem": "Um... lets say that there is a clock that when the hour hand makes one revolution around the center, the minute hand makes 16, and the second hand makes 36. Yah, wierd... so the question is not counting when they meet at the beginning and end, how many times do all 3 hands meet during one revolution of the hour hand?", "Solution_1": "Easier problems for people who couldn't solve the original question (these feature the same concept):\n\nQuestion 1 wrote:It is 4'o clock now. When is the next time the hour hand and the minute hand meet?\nHint:[hide]Think about how many degrees/radian the hands move/hour[/hide]\nIf you apply it to different angle differences between the hands...\nQuestion 2 wrote:The hour hand and the minute hand is together now. (The answer to Question 1) When will the hands make a right angle?\n\nRemember that [hide]Question 2's answer must be later than Question 1's answer[/hide].\n\n\n\nI think you'll get how to solve Chinaboy's question now.", "Solution_2": "Ok. \n\n\n\n[hide]Hour hand - (moves 6 degrees every 12 minutes),(moves .5 degrees every 1 minute).\n\n\n\nMinute hand - (moves 6 degrees every minute)\n\n\n\nBetween the minute hand and the hour hand, the amount of degrees is 120. We want to know the time when the difference of degrees between the hour hand and the minute hand is zero.\n\n\n\nHere is the equation...\n\n120+.5m-6m=0.\n\n\n\nSolving for m, we get m=21.8[/hide]\n\n\n\nAre there any better ways to solve this?", "Solution_3": "I do it that way, too. The only difference is 120 + .5x = 6x which is the same thing as yours... but then there always might be easier shortcuts...", "Solution_4": "how?!@?! help requested! :!:", "Solution_5": "Barnacle, to help with Tare's problem here is a little more detail\r\n\r\n\r\n[hide]\nSince the hour hand moves 360 degrees in 12 hours, the hour hand moves 30 degrees per hour (30h) So for example at 1:00 the hands are 30(1) degrees apart as the minute hand is back in it's original place.\n\nNow, since the minute hand of a normal clock moves 360 in 60 minutes, it moves 6 degrees per minute (6m). But since after 1:00 the minute hand is actuallu catching up to the hour hand(until it passes it), we can call that minus 6m.\n\nFinally, each minuts the hour hand moves 30/60 = 1/2 a degree (further away from the minute hand...again until they have crossed) so call that plus .5m\n\nWe have the distance between the hands then as 30h - 6m + .5m = 30h - 5.5m\n\nIf you want to know when the hands are in the same place, set that equal to zero\n\n30h - 5.5m = 0\n\nSince this problem asked for the first time they cross after 4:00, substitute 4 for h giving you\n\n30(4) - 5.5m = 0 or 120 - 5.5m = 0 which is what everyone else gave as the formula.\n\nm = 21.8181...\n\nBe careful when using the 30h - 5.5m formula as once the minute hand passes the hour hand you can get negative answers, so the answer is actually the absolute value of the formula. Also for times like 11:02, you will get values well over 180 degrees, in these cases subtract your value from 360 to get the actual angel(non-reflex)\n\nHope that helps. Try using this logic to do the original problem with the crazy clock :) \n[/hide]" } { "Tag": [ "ARML", "AMC", "AIME", "USA(J)MO", "USAMO", "geometry" ], "Problem": "Can somebody give me any information about ARML 2009 for WA states, I'm in Camas.", "Solution_1": "Just saw this ...\r\n\r\nYes WA will have teams for 2009 ARML. We are not one of those better organized states. Typically this kicks off in Spring (March / April). There seems to be higher interests now and hopefully there will be more teams from WA this year. Note that new teams would have to start with Division B (currently our Gold team competes in Division A, and Red team competes in Division B).\r\n\r\nI am also running the 2008-2009 ARML Power Contest, this power team is made up by a selected group of students from local schools (Bellevue, Redmond, Garfield, Lakeside, TJ, Interlake, Nathan Hale, etc.). Due to the collaborative problem solving nature it is not feasible to includes students far away from puget sound area. With November Round 1, we are currently tied for 4th place. Round 2 is scheduled Feb 7.\r\n\r\nMore to come after Mu Alpha Theta State competition.", "Solution_2": "In order to get selected to the 2009 WA ARML team does one have to participate in the 2008-2009 power contest first?\r\n\r\nI was also a little curious on the criteria for getting selected.. what counts?", "Solution_3": "No, but the opposite might hold some truth, let me explain.\r\n\r\nMr. Norris (of TJHS) runs WA ARML team, I organize the practices (including mini-ARML / mockup with Oregon team last year) and recommend team members. I don't see any particular reason to restrict the number of teams, but each team should have 15 students (the relay format requires students break into 5 sub-teams of 3 each). If we are considering more than two teams which is what WA has in the last couple of years, we need to get multiple of 15 committed students. The cost is about $400 or so per student ($75 fee, plus air fair to/from Las Vegas, some meals, and misc expenses.). Math enthusiastic students who are up to ARML level challenge are all encouraged to join, in my view. In 2008 MIC and Fall Classic, some parents and coaches talked to me about ARML, I believe with the interest level we are likely to get more than 2 teams this year, but let me check with Mr. Norris first. Please note that currently Gold team competes in Division A, Red team competes in Division B. New team will start in Division B.\r\n\r\nARML Power Contest is a write-in 2 round contest (Nov and Feb). Each round is 45min, questions are similar to ARML Contest Power Round. I am the sponsor and organizer for WA ARML Power Contest (2007-2008, and 2008-2009). Team size is smaller (around 10 or so), team members are typically among the top math students and contest winners in WA. AMC-12, AIME scores are key indicators (e.g. if you are USAMO qualifier and you are local to Puget Sound area, you will be invited). Local competitions are looked into as well (e.g. MIC, Mu Alpha Theta). Typically Power team becomes the core of our Gold team. Puget Sound is the primary target area due to practical consideration of team practices and taking the tests together and this area has the highest concentration of talents in WA. If you are interested for next year's Power Contest, drop me a PM.\r\n\r\nTo answer your question: you don't have to be in ARML Power team to participate ARML. If you are contributing member of WA ARML Power Team, chances are you will be placed in WA Gold team.\r\n\r\nPlease check ARML site (http://www.arml.com) to learn more.", "Solution_4": "Thanks for the clarification and your insights, minghu.\r\n\r\nI will send you a PM; check your inbox" } { "Tag": [ "geometry", "ratio" ], "Problem": "In triangle $ABC$, point $D$ divides $BC$ in the ratio 1:2, and $F$ divides $AD$ in the ratio 1:2. Line $BF$ intersects line $AC$ in point $E$. What is the ratio of the area of figure $EFDC$ to the area of triangle $ABF$?", "Solution_1": "[hide]$\\frac{40}{7}$[/hide]", "Solution_2": "Is it $2AF=FD$ and $2BD=DC$? Or the other way around? :?", "Solution_3": "[quote=\"lotrgreengrapes7926\"]Is it $2AF=FD$ and $2BD=DC$? Or the other way around? :?[/quote]\r\nYes, it is $2AF = FD$ and $2BD = DC$. I am not sure how to include a diagram in PDF inline in a message. Can anybody say how to include a PDF diagram inline in a message? Is the only way to include a diagram is including it as an attachment?", "Solution_4": "[hide] :idea: ABC=BFD; AEF=EFD\nABF+AEF= BFD+EFD => ABE=BED and BED=EDC \n=> ABE=BED=EDC=1/3ABC\n Therefore : ADC-AEB=FECD-ABF\n=> 1/2ABC-1/3ABC=FED+BED-BFD\n=> 1/6ABC=2FED\n=> 1/12ABC=FED\n BFD=BED-EFD=1/3ABC-1/12ABC=1/4ABC\n EFDC=DEC+FED=1/3ABC+1/12ABC=5/12ABC\nSo that: BFD : EFDC=(1/4ABC):(5/12ABC)=3/5[/hide]", "Solution_5": "[quote=\"TRAN THAI HUNG\"] ABC=BFD; AEF=EFD[/quote]\r\nI think you mean $ABF$ for the first one. And I don't think this is right. I got $2ABF=BFD$ and $2AEF=EFD$. :?" } { "Tag": [ "function", "number theory unsolved", "number theory" ], "Problem": "Find all functions $ f : \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that \\[ f(x\\plus{}y) \\plus{} f(xy) \\equal{} f(x) \\minus{} f(y) \\plus{}1,\\] for all $x,y \\in \\mathbb{Z}$.", "Solution_1": "there have to be some mistake here... I mean, plug in it $ x \\equal{} y \\equal{} 0$ and you get $ f(0\\plus{}0)\\plus{}f(0.0) \\equal{} f(0)\\minus{}f(0)\\plus{}1 \\equal{} 1$, that is $ 2f(0) \\equal{} 1$ and $ f(0) \\equal{} \\dfrac{1}{2}$ which is not in $ \\mathbb{Z}$, contradiction!.", "Solution_2": "[quote=\"Lousin Garckz\"]there have to be some mistake here... I mean, plug in it $ x \\equal{} y \\equal{} 0$ and you get $ f(0 \\plus{} 0) \\plus{} f(0.0) \\equal{} f(0) \\minus{} f(0) \\plus{} 1 \\equal{} 1$, that is $ 2f(0) \\equal{} 1$ and $ f(0) \\equal{} \\dfrac{1}{2}$ which is not in $ \\mathbb{Z}$, contradiction!.[/quote]\r\n\r\nSorry, this the correct one \r\n$ f(x\\plus{}y) \\plus{} f(xy) \\equal{} f(x)f(y) \\plus{}1$, $ f : \\mathbb{Z} \\rightarrow \\mathbb{Z}$", "Solution_3": "[quote=\"rajelaz\"] Sorry, this the correct one \n$ f(x \\plus{} y) \\plus{} f(xy) \\equal{} f(x)f(y) \\plus{} 1$, $ f : \\mathbb{Z} \\rightarrow \\mathbb{Z}$[/quote]\r\n\r\nLet $ P(x,y)$ be the assertion $ f(x \\plus{} y) \\plus{} f(xy) \\equal{} f(x)f(y) \\plus{} 1$\r\nLet $ f(1)\\equal{}a$\r\n\r\nIf $ a\\equal{}1$, $ P(x\\minus{}1,1)$ $ \\implies$ $ f(x)\\plus{}f(x\\minus{}1)\\equal{}f(x\\minus{}1)\\plus{}1$ and so $ f(x)\\equal{}1$ $ \\forall x$\r\n\r\nConsider now $ a\\neq 1$\r\n$ P(0,0)$ $ \\implies$ $ (f(0)\\minus{}1)^2\\equal{}0$ and so $ f(0)\\equal{}1$\r\n$ P(\\minus{}1,1)$ $ \\implies$ $ (a\\minus{}1)f(\\minus{}1)\\equal{}0$ and so $ f(\\minus{}1)\\equal{}0$\r\n$ P(\\minus{}2,1)$ $ \\implies$ $ 1 \\equal{} (1\\minus{}a)f(\\minus{}2)$ and so $ 1\\minus{}a|1$ and so $ a\\in\\{0,2\\}$\r\n\r\nIf $ a\\equal{}2$ $ P(x,1)$ $ \\implies$ $ f(x\\plus{}1)\\equal{}f(x)\\plus{}1$ and we got the solution $ f(x)\\equal{}x\\plus{}1$ $ \\forall x$\r\n\r\nIf $ a\\equal{}0$ $ P(x,1)$ $ \\implies$ $ f(x\\plus{}1)\\equal{}1\\minus{}f(x)$ and so $ f(x\\plus{}2)\\equal{}f(x)$ and we got the solution $ f(x)\\equal{}0$ if $ x$ odd and $ f(x)\\equal{}1$ if $ x$ even.\r\n\r\nSo three solutions :\r\n\r\n$ f(x)\\equal{}x\\plus{}1$ $ \\forall x$\r\n$ f(x)\\equal{}1$ $ \\forall x$\r\n$ f(x)\\equal{}0$ if $ x$ odd and $ f(x)\\equal{}1$ if $ x$ even" } { "Tag": [ "factorial", "real analysis", "real analysis solved" ], "Problem": "i've been doin crap with factorials and the harmonic series and i've made a mistake that i cant figure out. the end result is 1 = ( n! ) ^ 2, which is reminiscent of the 3=1 or 2=1 puzzles, so its like one of those puzzles, but i dont think it breaks a basic rule\r\n\r\n\r\n\r\nln(a/b) = ln(a) - ln(b)\r\n\r\nalso:\r\n\r\nln(a) + ln(b) = ab\r\n\r\n(all rheiman sums from 1 to n)\r\n\r\n[SUM] ln(1/ i) = ln(1/1) + ln(1/2) + ln(1/3) ... ln(1/n)\r\n\r\nor\r\n\r\n[SUM] ln( 1 / i ) = ln ( 1/1 * 1/2 * 1/3 ... 1/n )\r\n\r\n[SUM] ln( 1 / i ) = ln ( 1/n! )\r\n\r\n\r\n[SUM] ln( i ) = ln ( 1 * 2 * 3 ... n )\r\n = ln( n! )\r\n\r\nnow the application (and my problem) :\r\n\r\nln( 1 / i ) = ln(1) - ln(i)\r\n\r\nthese are identical values just different representations, so:\r\n\r\n[SUM] ln( 1 / i ) = [SUM] [ ln(1) - ln( i ) ]\r\n\r\n[SUM] ln( 1 / i ) = [SUM] ln( i ) // ln(1) = 0\r\n\r\nln( 1 / i! ) = ln( i! ) // each raised to a power of e\r\n\r\n1 / i! = i! //which doesnt make any sense or even more nonsense\r\n\r\n1 = ( i! )^2\r\n\r\ni cant see what i did wrong, help", "Solution_1": "[quote=\"oooooh ya\"][SUM] ln( 1 / i ) = [SUM] [ ln(1) - ln( i ) ]\n\n[SUM] ln( 1 / i ) = [SUM] ln( i ) // ln(1) = 0\n[/quote]\r\nThat transition is pure nonsense.\r\nI don't even know what you mean by //, but this has to be nonsense.", "Solution_2": "[quote=\"oooooh ya\"][SUM] ln( 1 / i ) = [SUM] [ ln(1) - ln( i ) ]\n[/quote]\nCorrect\n[quote]\n[SUM] ln( 1 / i ) = [SUM] ln( i ) // ln(1) = 0\n[/quote]\r\nsign error, should be \r\n\r\n[SUM] ln( 1 / i ) = [b]-[/b] [SUM] ln( i )" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": ":?: find all polynomials with integer coeff.P,s.t.P(x)=2^n,has integer sol.for all naturals n.\r\nany :idea:", "Solution_1": "does 0 belong to natural interger?if so,I guess I've got a solution,if not\r\nI can just solve the situation when the constant A0 is odd\r\ndefine S(a,b)={x|P(x)=2^a(mod2^(a+1)),x belong to{1,2.....2^b}} (a=t,P(x) is strickly increase,\r\nconsider an interger \"r\" s.t. the solution of P(x)=2^r is larger than t,\r\nso when m>r,there is only one solution s.t.P(x)=2^s,\r\ndefine the solution with Gm,then {Gm(m>r)}is strickly increase,\r\nand Gm=Xm(mod2^(m+1))\r\nfrom the fact that {Ya,Ya+2^(a+1)}={X(a+1),Y(a+1)}\r\nit's easy to prove that there exsist infinite \"c\">m \r\ns.t.X(c+1)>=X(c)+2^c :huh: \r\n\r\nfor each of the c,P(Gc)=2^c,P(Gc+2^c)==2,we can prove when c is large enough,\r\nP(Gc+2^c)>=2^(c+1), :huh: which is a contrariety\r\n\r\nas for deg(P(X))=1 or 0,it's easy :P \r\n that I didn't provethe sentenses end with :huh:. but I think to prove them is not hard", "Solution_2": ":lol: ,nice approach\r\n 0 isnt natural no.,but i think yours approach may work :D \r\n will tell if get :roll:" } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "LaTeX", "calculus computations" ], "Problem": "I know the solution from the back of the book but i cant seem to work it out right.\r\n\r\nInt from pi to 0 : tan[x/2] \r\n\r\nusing the natural log to int. The solution the book gives is ln2. How do I work this out?\r\nThank you!!![/b]", "Solution_1": "$\\int_0^{\\pi}\\tan\\left(\\frac x2\\right)\\,dx=\\infty.$\r\n\r\nWhat, exactly, is the problem you're looking at?", "Solution_2": "yeah thats pretty much it only its supposed to be\r\n\r\n$ \\int_0^{\\frac{\\pi}2}\\tan\\left(\\frac x2\\right)\\,dx.$\r\n\r\n[moderator edit]\r\n\r\nfrom 0 to pi/2\r\n\r\nI was working it out using \r\n\r\ntan(u)du \r\nand I wind up with \r\n2ln sec|u| evaluated from 0 to pi \r\n\r\nor as an alternative i get \r\n\r\n-ln cos|u| evauated from 0 to pi \r\n\r\nI am sorry I'm new and I am not familiar with using the displaystyle", "Solution_3": "Let's just push the substitution through. You're clearly making the substitution\r\n\r\n$u=\\frac x2,\\,x=2u,\\,dx=2du.$\r\n\r\nJust remember now that although $x$ goes from $0$ to $\\frac{\\pi}2,$ $u$ goes from $0$ to $\\frac{\\pi}4.$ That's $\\frac{\\pi}4,$ not $\\pi.$ So we now have\r\n\r\n$2\\int_0^{\\frac{\\pi}4}\\tan u\\,du.$\r\n\r\nFinish it from there.", "Solution_4": "That helps a lot. Thank You\r\n\r\nI plug in -2ln(cos(pi/4)) - -2ln(cos(0))\r\nand I get .693147... which is equal to ln2\r\n\r\nBut what I dont understand is why the book writes the solution as ln2\r\nis there something that I'm missing. Can that evaluation be simplified to avoid getting the decimal?\r\n\r\nThanks again!!", "Solution_5": "Tell me: what is $\\cos\\left(\\frac{\\pi}4\\right)$ ? The exact value, not the decimal. This is one of the values you must know.\r\n\r\nSecond issue: is there a way of rewriting $\\ln\\left(\\sqrt{u}\\right)$ so that it doesn't have any square root symbols? And can you simplify $\\ln\\left(\\frac1u\\right)$?", "Solution_6": "We just started going over natural logs in class so I'm a little shaky with some of the specifics of simpilfying and finding dy/dx and ints, but this is how I approached it.\r\n\r\ncos(pi/4) = :rt2: / 2\r\n\r\nthen: -2ln( :rt2: /2) \r\n= -2ln( :rt2: ) - -ln(2)\r\n= -2ln(2^(1/2)) - -ln2\r\n= -ln2 - -ln2 ?\r\nI must be doing something wrong.\r\nAs for the ln(1/u) cant that be written as\r\nln1-ln(u)\r\nIs that what you meant?", "Solution_7": "[quote=\"illcrowflu\"]That helps a lot. Thank You\n\nI plug in -2ln(cos(pi/4)) - -2ln(cos(0))\nand I get .693147... which is equal to ln2\n\nBut what I dont understand is why the book writes the solution as ln2\nis there something that I'm missing. Can that evaluation be simplified to avoid getting the decimal?\n\nThanks again!![/quote]Just enjoying this--\r\nThe right hand gives you zero--from the cos $\\rightarrow$ 1; ln 0[b].[/b]\r\nThe left, a multiple of ln(cos(:pi:/4)) --by -2 -- can be rewritten with that [b]multiplier in the exponent[/b] ([b]inside the log[/b]). ~Square 1/ (1/:sqrt:2)... Tada. But I'm not familiar to this forum, so I might well be clueless here.\r\n :downto:\r\n\r\nP.S. Yeah, I hope I wasn't giving anything too much away.", "Solution_8": "[quote]As for the ln(1/u) cant that be written as\nln1-ln(u) [/quote]\r\nI'd never let a calculus student of mine get away with leaving $\\ln1$ in an answer. Come on, now, $\\ln1=0.$ What I wanted you to say was that $\\ln\\left(\\frac1u\\right)=\\ln(u^{-1})=-\\ln(u).$\r\n\r\n$-2\\ln\\left(\\frac{\\sqrt{2}}2\\right)=-2\\ln\\left(\\frac1{\\sqrt{2}}\\right) = -2\\ln(2^{-1/2}) = -2\\left(-\\frac12\\right)\\ln2=\\ln2.$\r\n\r\nfedwini, I'd edit your bad LaTeX code to make it readable if I could figure out what you were trying to say, but I can't. It looks like you're trying to use LaTeX interchangably with the emoticons just as a collection of symbols. That's not a good way of doing it. Try writing expressions entirely in LaTex. For instance, (dollar)\\frac{\\sqrt{2}}{2}(dollar) becomes $\\frac{\\sqrt{2}}{2}$.", "Solution_9": "Thanks a lot Kent you were a big help - I appreciate your time.\r\nBTW I am new to the forum but I plan to keep posting as it has been so helpful.\r\n\r\nFor everyone else's sake how do I write the equations out like you were rather than using regular ascii characters?", "Solution_10": "[quote]For everyone else's sake how do I write the equations out like you were rather than using regular ascii characters?[/quote]\r\nIt's called LaTeX. There's some information posted over on the side, as one of the site guides, and there's also a LaTeX forum for questions and problem.\r\n\r\nGo look at your own #3 post on this topic, which I edited, to get an idea.\r\n\r\nA dollar sign puts you into LaTeX mode, and another dollar sign gets you out. Within that, a backslash introduces a named object, with the names mostly being mnomonic: \\int for integral, \\infty for infinity, \\frac for fraction, and so on. ^ is for superscript, _ for subscript. Curly braces {} delineate and enclose the scope of each object. Thus \\frac{x^2+1}{2x+e^{x-1}} becomes $\\frac{x^2+1}{2x+e^{x-1}}.$", "Solution_11": "[quote=\"Kent Merryfield\"][quote]$\\ldots$\nfedwini, I'd edit your bad LaTeX code to make it readable if I could figure out what you were trying to say, but I can't. It looks like you're trying to use LaTeX interchangably with the emoticons just as a collection of symbols. That's not a good way of doing it. Try writing expressions entirely in LaTex. For instance, (dollar)\\frac{\\sqrt{2}}{2}(dollar) becomes $\\frac{\\sqrt{2}}{2}$.[/quote][/quote]I've got that fixed now--and I do need to use better LaTeX. There is one thing that I wanted to mention, however: (illcrowflu,) Hovering over any LaTeX picture shows the coding, but ~ If you cannot get that to work for you with long strings, quoting someone else's message will always show you what they wrote--though you'd intend to cancel out of the Reply.", "Solution_12": "Actually, selecting and copying that part of the post should work. If you paste that in, you'll get the code, without the dollar signs.", "Solution_13": "[quote=\"jmerry\"]Actually, selecting and copying that part of the post should work. If you paste that in, you'll get the code, without the dollar signs.[/quote]Good deal--I mean good to know. :lol:" } { "Tag": [ "function", "floor function", "algebra proposed", "algebra" ], "Problem": "Let $ [x]$ denote the integer part of $ x$, i.e., the greatest integer not exceeding $ x$. If $ n$ is a positive integer, express as a simple function of $ n$ the sum \r\n\r\n$ \\left[\\frac{n\\plus{}1}{2}\\right] \\plus{} \\left[\\frac{n\\plus{}2}{4}\\right]\\plus{}...\\plus{}\\left[\\frac{n\\plus{}2^k}{2^{k\\plus{}1}}\\right]\\plus{}...$\r\n\r\n\r\nBig hint (try to solve it w/o):\r\n\r\n[hide]$ \\lfloor{x\\plus{}\\frac{1}{2}}\\rfloor \\equal{} \\lfloor{2x}\\rfloor\\minus{}\\lfloor{x}\\rfloor$[/hide]", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125439\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=34668\r\n\r\nThere's also at least one other thread where this problem was posted in which I gave a slightly different solution. \r\n\r\nI also think it would be very instructive, although overkill, to give a proof by [b]generating functions.[/b]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "$ \\begin {cases} x^2\\plus{}y^2\\equal{}z^2\\\\xz\\equal{}y^2\\\\xy\\equal{}10\\end {cases}$", "Solution_1": "[quote=\"thanhnam2902\"]$ \\begin {cases} x^2 \\plus{} y^2 \\equal{} z^2 \\\\\nxz \\equal{} y^2 \\\\\nxy \\equal{} 10\\end {cases}$[/quote]\r\n\r\nSee that $ x,y,z$ are not equal to zero. Then \r\n\\[ \\Rightarrow x^2 \\plus{} xz \\equal{} z^2 \r\n\\]\r\n\\[ \\Leftrightarrow t^2 \\plus{} t \\minus{} 1 \\equal{} 0,t \\equal{} \\frac{x}{z}\r\n\\]\r\n\\[ \\Rightarrow \\left[ \\begin{array}{l}\r\n t \\equal{} \\frac{{ \\minus{} 1 \\plus{} \\sqrt 5 }}{2} \\\\ \r\n t \\equal{} \\frac{{ \\minus{} 1 \\minus{} \\sqrt 5 }}{2} \\\\ \r\n \\end{array} \\right.\r\n\\]", "Solution_2": "[quote=\"Honey_S\"][quote=\"thanhnam2902\"]$ \\begin {cases} x^2 \\plus{} y^2 \\equal{} z^2 \\\\\nxz \\equal{} y^2 \\\\\nxy \\equal{} 10\\end {cases}$[/quote]\n\nSee that $ x,y,z$ are not equal to zero. Then\n\\[ \\Rightarrow x^2 \\plus{} xz \\equal{} z^2\n\\]\n\n\\[ \\Leftrightarrow t^2 \\plus{} t \\minus{} 1 \\equal{} 0,t \\equal{} \\frac {x}{z}\n\\]\n\n\\[ \\Rightarrow \\left[ \\begin{array}{l} t \\equal{} \\frac {{ \\minus{} 1 \\plus{} \\sqrt 5 }}{2} \\\\\nt \\equal{} \\frac {{ \\minus{} 1 \\minus{} \\sqrt 5 }}{2} \\\\\n\\end{array} \\right.\n\\]\n[/quote]\r\nThank you. Oh! This system is easy." } { "Tag": [ "probability", "function", "expected value", "probability and stats" ], "Problem": "Brief interruptions to my work occur on average every ten minutes and the number of interruptions in any given time period has a Poisson distribution. Given that an interruption has just occurred, find the probability that I will have less than t minutes to work before the next interruption. If the random variable T is the time I have to work before the next interruption, find the probability density function of T.\r\n\r\nI need an uninterrupted half hour to finish an important paper . Show that the expected number of interruption before my first uninterrupted period of half an hour or more is $ e^3 \\minus{}1$. Find also the expected length of time between interruptions that are less than half an hour apart. Hence write down the expected wait before my first uninterrupted period of half an hour or more.", "Solution_1": "if you have already heard about Poisson processes, you immediately know that the time length between two consecutive interruptions is exponentially distributed with mean $ 10$ minutes. From here, everything is straightforward.\r\n1: $ T$ has density $ \\lambda \\exp(\\minus{}\\lambda t) \\, dt$ on $ (0,\\infty)$ with $ \\lambda \\equal{} \\frac{1}{10 \\text{ min}}$\r\n2: properties of the geometric distribution gives you the expected number of interruption before the first uninterrupted period of half an hour or more, etc ...\r\n\r\nIs this OK ?", "Solution_2": "I'm not familiar with the geometric distribution, but I solved it like this:\r\n\r\n$ p \\equal{} P(T > 30) \\equal{} e^{ \\minus{} 3}$\r\n\r\nExpected number of interruptions before the first uninterrupted half hour:\r\n\r\n$ 0\\cdot p \\plus{} 1\\cdot (1 \\minus{} p) p \\plus{} 2 \\cdot (1 \\minus{} p)^2 \\cdot p \\plus{} 3 \\cdot (1 \\minus{} p)^3 \\cdot p \\plus{} \\cdots \\equal{} p (\\frac {1}{p^2} \\minus{} \\frac {1}{p}) \\equal{} e^3 \\minus{} 1$\r\n\r\nFor the expected time to pass before the first uninterrupted half hour I calculated about 2h 40min. What's your solution?\r\nAre you studying in Cambridge?", "Solution_3": "1: yes, your solution is essentially what I had in mind (http://en.wikipedia.org/wiki/Geometric_distribution)\r\n2: For the expected time to pass before the first uninterrupted half hour it is the same idea:\r\n\\[ \\equal{} E[T|T < 30] \\cdot p\\plus{}2\\cdot E[T|T < 30] \\cdot (1\\minus{}p) p\\plus{}3 \\cdot E[T|T < 30] \\cdot (1\\minus{}p)^{2}\\cdot p\\plus{}... \\]" } { "Tag": [], "Problem": "I'm not sure if this is the right forum.Anyway here's the question:-\r\nIn a convex quadrileteral ABCD the angle bisectors of angle ABC and angle CAB meet at point [b]P[/b].The angle bisectors of angle BAD and angle DBA meet at [b]Q[/b].Prove that if A,B,C,D are concyclic then so is A,B,P,Q.", "Solution_1": "Since $ ABCD$ is cyclic, $ \\angle DAC \\equal{} \\angle DBC \\equal{} x$. Then, letting $ \\angle CAB \\equal{} y$, $ \\angle DBA \\equal{} z$, we see through simple angle chasing that \r\n\r\n$ \\angle BPA \\equal{} 180 \\minus{} \\frac{\\alpha\\plus{}\\beta\\plus{}\\gamma}{2} \\equal{} \\angle BQA$. So $ ABPQ$ is cyclic." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all functions $f(n)$ such that $f(2n) = f(n+1)^2 - f(n-1)^2$\r\n\r\n1) $f : \\mathbb{N} \\rightarrow \\mathbb{N}$\r\n2) $f : \\mathbb{R} \\rightarrow \\mathbb{R}$", "Solution_1": "Is it really an olympiad problem?? :maybe: \r\n\r\nI'm getting a lot of weird stuff, yet I still can't manage to conclude.\r\n\r\nFor the 1st one, I think there are none, but I can't prove it. (I can tell you what I have so far)\r\n\r\nFor the 2nd one, I guess there are infinitely many, but I don't think there's an easy characterization. (I can show you a few examples)" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Lets talk about this game of extremes.\r\nEither you know about it, or dont, and love it, or hate it.\r\nPost igns, character info, and try to convince me to quit.\r\n ;)", "Solution_1": "LOL, you could do homework with all that time. :)", "Solution_2": "well, i only play on weekends, so if you're implying me doing extra work upon extra work... :P \r\n\r\nany other arguements?", "Solution_3": "what is maplestory? :?", "Solution_4": "Its an addicting free online mmorpg much like runescape, though i've never played runescape. Its 2D sidescroll, and may sound shabby at first, but most definitely isn't. I would emphasize the addicting part.", "Solution_5": "So I'm guessing that your avatar is your Maplestory character?", "Solution_6": "not my character. just a random avatar i found. i find that ms character make good avatars... :D", "Solution_7": "can you give me a link to maplestory?", "Solution_8": "[quote=\"Stratigizer\"]Lets talk about this game of extremes.\nEither you know about it, or dont, and love it, or hate it.\nPost igns, character info, and try to convince me to quit.\n ;)[/quote]\r\n\r\nOh this game... I tried to start it but it wont get on my computer... I think that its pretty good... a little like runescape... I did make an account though i think. mabey on someone else's com.\r\n\r\nits on http://www.maplestory.com", "Solution_9": "actually, that site is just a portal.\r\n\r\ngoto http://www.mapleglobal.com for the site, download, and a bit of info.\r\n\r\nits completely free, which adds to the awesomeness.", "Solution_10": "is there any way to play without downloading it? it would take nearly 5 hours on just the .exe file alone on my computer.", "Solution_11": "[quote=\"MathFiend\"]is there any way to play without downloading it? it would take nearly 5 hours on just the .exe file alone on my computer.[/quote]\r\n\r\nyea. it says 56 hours for my com.", "Solution_12": "[quote=\"Spencer\"][quote=\"MathFiend\"]is there any way to play without downloading it? it would take nearly 5 hours on just the .exe file alone on my computer.[/quote]\n\nyea. it says 56 hours for my com.[/quote]\r\n\r\nIm trying to download right now...\r\n\r\nHey, if I download the thing on the bottom, then the main game loads for only likke 1 hour.", "Solution_13": "I used to play this...\r\nGot to be pretty boring though\r\nI mean killing the same things over and over....\r\nWhat the point??\r\nMy opinion\r\n[hide]Quit before it takes over ur life!!![/hide]\nI got some people at my school who talk about it for hours\nIt like the only thing they do \nThis is better than Maplestory\n[hide] :ddr: [/hide]", "Solution_14": "If trying to download the game,\r\nDo the parts\r\nIt is way faster\r\nIt takes 4 hours on my comp to download all of it\r\nbut only 2 hours with parts", "Solution_15": "Well, you only have pretty much 1 choice for armor.\r\nuse the two-handed swords, scimitar and lionheart.\r\ntho they are more expensive, they look cooler and are the most balanced.\r\n\r\nanyways, i always liked magic claw better, as the terrain is no obstacle for attacking. and regular [i]claws[/i] are for theives, yes, the throwing star casters...\r\nanyways, im still 46, doing all these awesome new quests!\r\n[/i]", "Solution_16": "Well, you only have pretty much 1 choice for armor.\r\nuse the two-handed swords, scimitar and lionheart.\r\ntho they are more expensive, they look cooler and are the most balanced.\r\n\r\nanyways, i always liked magic claw better, as the terrain is no obstacle for attacking. and regular [i]claws[/i] are for theives, yes, the throwing star casters...\r\nanyways, im still 46, doing all these awesome new quests!", "Solution_17": "Well, you only have pretty much 1 choice for armor.\r\nuse the two-handed swords, scimitar and lionheart.\r\ntho they are more expensive, they look cooler and are the most balanced.\r\n\r\nanyways, i always liked magic claw better, as the terrain is no obstacle for attacking. and regular [i]claws[/i] are for theives, yes, the throwing star casters...\r\nanyways, im still 46, doing all these awesome new quests!", "Solution_18": "Well, you only have pretty much 1 choice for armor.\r\nuse the two-handed swords, scimitar and lionheart.\r\ntho they are more expensive, they look cooler and are the most balanced.\r\n\r\nanyways, i always liked magic claw better, as the terrain is no obstacle for attacking. and regular [i]claws[/i] are for theives, yes, the throwing star casters...\r\nanyways, im still 46, doing all these awesome new quests!", "Solution_19": "umm... spam???\r\n\r\nwhat is wrong with your posts?", "Solution_20": "WOAA!! I've seen plenty of double posts, but never a quadruple post :oops_sign: :w00tb: \r\n\r\n\r\nHehe I'm a hardcore RS fan.\r\n\r\nMaplestory has only one skill: BEATING CUTE FUZZY THINGS TILL THEY YELL :help: \r\n\r\nSeems a bit too simple for me. On the other hand RS has 21 skills.\r\n\r\nFarmers for life.", "Solution_21": "LVL 57 finally ice-lightning if u guys forgot\r\nI got 3 anakamoon\r\nIf anyone is on scania, I may give u guys one.\r\nmy name is [i]yomama789[/i]\r\nDo any of u guys have BC ores, I need 50 of them and I onli get 20 right now", "Solution_22": "[quote=\"Stratigizer\"]Well, you only have pretty much 1 choice for armor.\nuse the two-handed swords, scimitar and lionheart.\ntho they are more expensive, they look cooler and are the most balanced.\n[/quote]\r\n\r\nI'm 31, blunt weapon page.\r\nAnd you say I have only 1 choice, but I'm still not too knowing of everything MapleStory, could you tell me which armor I should get? Lvl requirement 30 or 35, stuff I'd keep until 40.\r\nAnd I have lvl 25 armor, if there's any you think would be worth keeping until 40, please advise!", "Solution_23": "um, yeah, i have no clue what happened there...sorry about that!\r\n\r\nim not sure about the stat requirements, but at lv 30, your top is the green hwarang shirt or w/e you call them, black martial arts pants, the shoe you get from nella's last quest, and it doesn't really matter the glove.\r\n\r\nlv 35 you should get brown/blue helm, forgot the name, and white/black chainmail, which is overall. keep your lv 30 shoes, and you should have lv 35 gloves from john's sleepywood jump quest.\r\n\r\nIf you have an orihalcon burgernet helm, keep it til 35, and if you have venon, or perhaps the lv 25 glove, keep until 35. however, i dont think you should keep anything til lv 40. stat-wise, its not beneficial compared to cost, and it just doesn't look as cool!\r\n\r\ngoto http://www.hidden-street.net to look at the stats and requirements, but also, to everyone who doesn't know, goto http://www.happymapling.net for a look at equips up to lv 100!\r\n\r\nok, enough of that...", "Solution_24": "Ludi's coming out soon\r\nI'm so prepared for it\r\nif no one believes me go to mapleglobal to look", "Solution_25": "so...what do you think of ludi?\r\n\r\n48 now", "Solution_26": "It's okay, I want to try the pq though.\r\nI uninstalled MapleStory because it was distracting me from mathcounts -_-", "Solution_27": "pq is what, 35-50?\r\nbut it seems like the other pq, it just takes too long to get in ><\r\n\r\nI guess state hasn't happened yet? Anyways, good luck! and I'm from Texas too, so go Texas ;)", "Solution_28": "i love maple lol\r\n\r\nthe best part of maple is calculating profits regarding scrolling and junk, i'm soo rich from just doing nothing but scrolling. \r\n15dex 6 slots left sauna robe \r\n11 att gloves \r\netc XD", "Solution_29": "yeah, i too just discovered my lucky hand in scrolling. my overall robe is +6, with 3 10% def scrolls. i also have +12 jump +4speed shoes, with like 2 10% jump scrolls...\r\nalso, my friend borrowed a fish spear and tried 3 10% for spear att, and they all worked.\r\n\r\nyeah, seems like its gonna be all 10% for me =)" } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "Prove that if [b]K[/b] is a field and [i]x_1, x_2, ... x_n[/i] are elements of algebric closer of [b]K[/b], then [b]K[/b]([i]x_1, x_2, ... x_n[/i]) is a collection of all the expressions of the form [b]f[/b]([i]x_1, x_2, ... x_n[/i]) in wich [b]f[/b] beloning to [b]K[X_1,...X_n][/b] is a polynomial such that [i]deg[/i](X_i) [b]f[/b] < [b][K([i]x_i[/i]):K][/b]", "Solution_1": "Induction." } { "Tag": [], "Problem": "The rules of this game are as follows:\r\n\r\n1. Each player gets to pick 5 things they would have with them if they were stranded on a barren wasteland. These items cannot be larger than a house.(e.g. a tent would be a good idea)\r\n2. players can trade items by typing the username, the item that they want, and the item they are willing to trade from the person they want to trade with. The person can then decide whether to trade or not.\r\n3. every week or so I will post a problem. Each player will PM me their solutions to me, and the players who get the problem wrong must give up an item.\r\n4. The players with insufficient items for survival will be eliminated.(I will be the one to decide who has insufficient items)\r\n5. There will be a max of 5 players per game.\r\n6. The winner of this game may host the next one.\r\n\r\n[hide=sign ups][/hide]", "Solution_1": "Sign up...", "Solution_2": "sign up\r\n\r\n[hide=\"signups\"]\nBugi\nwestiepaw\n[/hide]", "Solution_3": "sign up\r\n\r\n[hide=\"signups\"]\nBugi\nwestiepaw\nspaceguy524\n[/hide]", "Solution_4": "Sign up.\r\n\r\n[hide=\"signups\"]\nBugi\nwestiepaw\nspaceguy524\ncyberspace\n[/hide]\r\n\r\nOne more...", "Solution_5": "[b]sign up[/b]", "Solution_6": "WAIT! Never mind. There's already 5 people. :P", "Solution_7": "o.k.... The competition begins.\r\n\r\nYou 5... Pick your 5 items, and be smart about them. If you're not smart about them, you might just die when the game starts. :)", "Solution_8": "What does the barren island have on it? Is there any hope of escape, or should I choose items for long-term survival, rather than escape? Thanks!", "Solution_9": "Drop out, sorry.", "Solution_10": "Ok, i will sign up since Bugi dropped out." } { "Tag": [ "geometry", "trapezoid" ], "Problem": "Let a circle be inscribed in an isosceles trapezoid with base lengths $ m$ and $ n$. Express the diameter $ d$ of the circle in terms of $ m$ and $ n$.", "Solution_1": "[quote=\"123456789\"]Let a circle be inscribed in an isosceles trapezoid with base lengths $ m$ and $ n$. Express the diameter $ d$ of the circle in terms of $ m$ and $ n$.[/quote]\r\n\r\n[hide]\nDraw the segments from the center of the circle to each vertex of the trapezoid, and the perpendiculars to each edge.\nIt's easy to see that the lateral sides must be $ \\frac {m \\plus{} n}{2}$ by congruencies.\nDraw the height from the upper left vertex. Use the pythagorean theorem.\n$ \\frac {m \\minus{} n}{2}^2 \\plus{} h^2 \\equal{} \\frac {m \\plus{} n}{2}^2$\n$ \\frac {m^2 \\plus{} n^2}{4} \\minus{} \\frac {mn}{2} \\plus{} h^2 \\equal{} \\frac {m^2 \\plus{} n^2}{4} \\plus{} \\frac {mn}{2}$\ncancel, combine like terms\n$ h^2 \\equal{} mn$\nor in other words\n$ (2r)^2 \\equal{} mn$\n$ r \\equal{} \\frac {\\sqrt {mn}}{2}$\n[/hide]\r\n\r\ntoo simple..someone tell me what i did wrong\r\n\r\nedit: yes alan i meant circle", "Solution_2": "uh, triangle???", "Solution_3": "[quote=\"alanchou\"]uh, triangle???[/quote]\r\n\r\noh heh that was a typo, a freudian slip to be exact :rotfl:", "Solution_4": "[quote=\"CircleSquared\"][quote=\"123456789\"]Let a circle be inscribed in an isosceles trapezoid with base lengths $ m$ and $ n$. Express the diameter $ d$ of the circle in terms of $ m$ and $ n$.[/quote]\n\n[hide]\nDraw the segments from the center of the circle to each vertex of the trapezoid, and the perpendiculars to each edge.\nIt's easy to see that the lateral sides must be $ \\frac {m \\plus{} n}{2}$ by congruencies.\nDraw the height from the upper left vertex. Use the pythagorean theorem.\n$ \\frac {m \\minus{} n}{2}^2 \\plus{} h^2 \\equal{} \\frac {m \\plus{} n}{2}^2$\n$ \\frac {m^2 \\plus{} n^2}{4} \\minus{} \\frac {mn}{2} \\plus{} h^2 \\equal{} \\frac {m^2 \\plus{} n^2}{4} \\plus{} \\frac {mn}{2}$\ncancel, combine like terms\n$ h^2 \\equal{} mn$\nor in other words\n$ (2r)^2 \\equal{} mn$\n$ r \\equal{} \\frac {\\sqrt {mn}}{2}$\n[/hide]\n\ntoo simple..someone tell me what i did wrong\n\nedit: yes alan i meant circle[/quote]It's meant to be simple. It's High School Basics. And I was asking for $ d$, but $ r$ is just as good." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$ 1)$ Given $ a, b, c \\geq\\ 0$ and $ a\\plus{}b\\plus{}c\\equal{}3$. \r\nProve that: $ \\frac {7a^2 \\plus{} 9}{b^2 \\plus{} c^2} \\plus{} \\frac {7b^2 \\plus{} 9}{c^2 \\plus{} a^2} \\plus{} \\frac {7c^2 \\plus{} 9}{a^2 \\plus{} b^2}\\ge 24$\r\n$ 1)$ Given $ a, b, c \\geq\\ 0$ and $ ab\\plus{}bc\\plus{}ca\\equal{}3$. \r\nProve that: $ \\frac {4a^2 \\plus{} 3}{b^2 \\plus{} c^2} \\plus{} \\frac {4b^2 \\plus{} 3}{c^2 \\plus{} a^2} \\plus{} \\frac {4c^2 \\plus{} 3}{a^2 \\plus{} b^2}\\ge \\frac {21}{2}$\r\n\r\nI see some long proofs in here: http://diendantoanhoc.net/forum/index.php?showtopic=45718\r\nIndeed, both of them can be solve easily by CS and Schur ineq :)", "Solution_1": "I'm also a fan of Vasc's problems , note that the first and the second one are equivalents so I only write the solution of the frst one:\r\n\r\nit's equivalent to:\r\n\r\n$ 7\\sum{ \\frac{a^2}{b^2+c^2}} +( \\sum{a})^2 \\sum{\\frac{1}{b^2+c^2}} \\geq 24$\r\n\r\nusing Cauchy it follows that :\r\n\r\n$ 7\\sum{ \\frac{a^2}{b^2+c^2}} \\geq 7\\frac{(\\sum{a^2})^2}{2\\sum{(ab)^2}}$\r\n\r\nand $ \\sum{ \\frac{1}{b^2+c^2} } \\geq \\frac{ 3+ \\frac{(\\sum{a^2})^2}{2\\sum{(ab)^2}}{\\sum{a^2}}$\r\n\r\nWLOG $ a+b+c=1$ it's equivalent to :\r\n$ f(r)= 4-23q+23q^2+20q^3+42r-96rq \\geq 0$ and thus f is increasing\r\nand it's easy to check that : if $ q\\leq \\frac{1}{4}$ $ f(0 )\\geq 0$ and if $ q\\geq \\frac{1}{4}$ $ f(\\frac{4q-1}{9}) \\geq 0$" } { "Tag": [], "Problem": "Are given 25 numbers . The sum of each 4 of them is positive prove that the sum of all those numbers is positive .", "Solution_1": "can't we just add all the \"4-possibilities\" up? :maybe: \r\nbecause then each number would appear $ \\binom{24}{3}$ times,I mean let $ x_1,x_2,\\ldots ,x_{25}$ be our 25 numbers,and we had:\r\n\r\n$ x_1 \\plus{} x_2 \\plus{} x_3 \\plus{} x_4 \\equal{} a_1$\r\n.\r\n.\r\n.\r\n$ x_{22} \\plus{} x_{23} \\plus{} x_{24} \\plus{} x_{25} \\equal{} a_{\\binom{25}{4}}$\r\n\r\nwhere $ a_i$ are positive numbers for $ i \\equal{} 1,2,\\ldots ,\\binom{25}{4}$\r\n\r\nnow add all the equalities we had above,we would end up with:\r\n\r\n$ \\binom{25}{3}\\sum_{i \\equal{} 1}^{25} x_i \\equal{} \\sum_{i \\equal{} 1}^{\\binom{25}{4}} a_i > 0$\r\n\r\n$ \\Rightarrow \\sum_{i \\equal{} 1}^{25} x_i > 0$\r\n :maybe:" } { "Tag": [ "inequalities", "geometry", "perimeter", "rectangle", "triangle inequality", "Pythagorean Theorem" ], "Problem": "A convex polygon does not extend outside a square side 1. Prove that the sum of the squares of its sides is at most 4.", "Solution_1": "[hide=\"Hint\"]\nDivide into right triangles. Then use the fact that $ x^{2}\\leq x \\ \\forall x \\leq 1$.[/hide]", "Solution_2": "[hide=\"Interesting\"]Consider the orthagonal components of each side respective to the square. By the triangle inequality, $ a+b \\geq \\sqrt{a^{2}+b^{2}}$ for component lengths $ a$ and $ b$. Note that the sum of the lengths of these components equals the perimeter of a binding rectangle of that polygon.\n\nLet us orient our components such that the achieved binding rectangle has edges parallel to those of the square. This rectangle must be fully contained within the square, and hence no component can have length greater than 1. It follows that the square of the length of a component cannot exceed the length of that component.\n\nNow, note that the sum of the squares of the side lengths equals the sum of the squares of the lengths of the components (Pythagorean Theorem), which does not exceed the sum of the lengths of the components, which is equivalent to the perimeter of the binding rectangle.\n\nSince neither the length nor width of this binding rectangle can exceed 1, the perimeter of this rectangle cannot exceed 4. Transitively, the sum of the squares of the side-lengths of the polygon cannot exceed 4.\n\nTesting the unit square case gives that this bound is indeed achievable, so 4 is the best upper bound. [/hide]\r\n\r\n...\r\nBah my computer stalled. Anyway, it turns out that this is basically what cinco suggested." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "f(x+y)=f(x)f(y) and f(a)=1990. Prove that f is injectiv.", "Solution_1": "I can't see why you can't take $ f(x) \\equal{} e^{g(x)}$, where $ g(x)$ is a pathological solution of the Cauchy equation with some of the Hamel basis elements (that don't intersect with $ a$) set to $ 0$ by $ g$.", "Solution_2": "You can set that only if this funciton is monotone or injectiv.", "Solution_3": "Why? The function $ f: \\mathbb R\\to\\mathbb R$ can be shown to have values in $ \\mathbb R^\\plus{}$ only such that MellowMelon's reasoning does work. Can you provide a proof?" } { "Tag": [ "geometry", "3D geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "A cube is divided into n^3 little cubes, prove that we can assign a DIFFERENT integer to each little cube so that the sum of each of the 3n^2 rows parallel to an edge is zero.", "Solution_1": "edit: it means the sum of numbers in the n little cubes in all 3n^2 rows is zero", "Solution_2": "for $ n\\equal{}2$ it's obviously umtrue :wink: since $ a_{1,1,1} \\plus{} a_{1,1,2} \\equal{} 0$ and also $ a_{1,1,1} \\plus{} a_{1,2,1} \\equal{}0$ so it must be\r\n$ a_{1,2,1} \\equal{} a_{1,1,2}$ which is a contraddiction since the numbers were supposed to be different.", "Solution_3": "Sorry, the problem states that n >=3" } { "Tag": [ "USAMTS", "LaTeX", "modular arithmetic" ], "Problem": "In general does $ \\sum a\\plus{}b$ mean $ (\\sum a)\\plus{}b$ or $ \\sum (a\\plus{}b)$?", "Solution_1": "Umm well is this a question that's attempting to get an answer for one of the USAMTS problems?\r\n\r\nIf so, you just have to look it up on your own.", "Solution_2": "In context, this depends on whether $ b$ has a term that contains the index of the summation. This notation is ambiguous, though; to be precise, you should add parentheses appropriately.", "Solution_3": "Typically, $ \\sum a \\plus{} b \\equal{} (\\sum a) \\plus{} b$. Heh, I see what problem you're looking at... I don't think this counts as cheating though. :)", "Solution_4": "[quote=\"Brut3Forc3\"]Typically, $ \\sum a + b = (\\sum a) + b$. Heh, I see what problem you're looking at... I don't think this counts as cheating though. :)[/quote]\r\nIf it's not considered cheating for telling this, I believe it's $ \\sum{(a + b)}$ cuz in $ \\text{\\LaTeX}$, when you just type a+b in the brackets for sum, your exact original reading shows up.", "Solution_5": "Brackets or parenthesis? Brackets usually disappear on everything...", "Solution_6": "I meant these brackets: { }", "Solution_7": "[quote=\"gauss1181\"]I meant these brackets: { }[/quote]\r\n\r\nYeah, if you type a latex expression and use brackets as parenthesis, they disappear. The summation notation in latex does not require brackets...\r\n\r\nusually I type \\displaystyle\\sum_{i=1}^5 x^2 rather than usually I type \\displaystyle\\sum_{i=1}^5{x^2}\r\n\r\nand it does the exact same thing.", "Solution_8": "[quote=\"pythag011\"][quote=\"gauss1181\"]I meant these brackets: { }[/quote]\n\nYeah, if you type a latex expression and use brackets as parenthesis, they disappear. The summation notation in latex does not require brackets...\n\nusually I type \\displaystyle\\sum_{i=1}^5 x^2 rather than usually I type \\displaystyle\\sum_{i=1}^5{x^2}\n\nand it does the exact same thing.[/quote]\r\nHowever, when one of the \\mod things enclose something in brackets, it's also in parenthesis. I think it's \\pmod. 4\\pmod{3 3k4} shows up like this: $ 4\\pmod{3 3k4}$", "Solution_9": "[quote=\"ProtestanT\"][quote=\"pythag011\"][quote=\"gauss1181\"]I meant these brackets: { }[/quote]\n\nYeah, if you type a latex expression and use brackets as parenthesis, they disappear. The summation notation in latex does not require brackets...\n\nusually I type \\displaystyle\\sum_{i=1}^5 x^2 rather than usually I type \\displaystyle\\sum_{i=1}^5{x^2}\n\nand it does the exact same thing.[/quote]\nHowever, when one of the \\mod things enclose something in brackets, it's also in parenthesis. I think it's \\pmod. 4\\pmod{3 3k4} shows up like this: $ 4\\pmod{3 3k4}$[/quote]\r\n\r\n\\pmod is not a symbol. \\sum is.", "Solution_10": "Just assume one and solve the problem with it.", "Solution_11": "[quote=\"t0rajir0u\"]In context, this depends on whether $ b$ has a term that contains the index of the summation. This notation is ambiguous, though; to be precise, you should add parentheses appropriately.[/quote]\r\n\r\nI totally agree. Just use your common sense and knowledge of summation notation. :wink:" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "If I have a contour diagram of:\r\n\r\n[url=http://imageshack.us][img]http://img118.imageshack.us/img118/9608/graphdu7.jpg[/img][/url]\r\n\r\n\r\nthen what would the graph of:\r\n\r\nf(-x,y) = f(x,y)\r\n\r\nand \r\n\r\nf(-x,y) = -f(x,y)\r\n\r\nlook like??\r\n\r\nmy guess is that the first one is the same as the original one translated with the y-axis and the second one translated with the x-axis. Is this right?", "Solution_1": "[quote=\"EquinoX\"]f(-x,y) = f(x,y) \n...\ntranslated with the y-axis[/quote] Hmm?", "Solution_2": "[quote=\"JBL\"][quote=\"EquinoX\"]f(-x,y) = f(x,y) \n...\ntranslated with the y-axis[/quote] Hmm?[/quote]\r\n\r\nWhy a Hmmm??", "Solution_3": "Because neither of the lines I quoted make sense ;)", "Solution_4": "What I meant translated with the y-axis is pretend that the y-axis is a mirror, so therefore everything to the right of the y-axis will be reflected to the left of the y-axis. That's what I assume \r\n f(-x,y) = f(x,y) looks like and I just want to confirm whether this is correct or not?", "Solution_5": "Your notation is wrong. You are talking about f(-x,y)=0, not f(-x, y)=f(x,y).. There is a big difference.. Suppose you have f(x,y) = x+y. Then f(-x,y)=-x+y. So, if you plot x+y=0 you get y = -x, and plotting -x+y=0 you get y=x. This goes by your assumption. But f(x,y) = f(-x,y) is x+y = -x+y => x=0, and this has nothing to do with what you are talking about.", "Solution_6": "Yes what I am trying to say is to find what the function f(-x,y) looks like if f(x,y) is on the picture above..." } { "Tag": [ "ratio", "geometry", "geometric transformation", "rotation", "power of a point", "radical axis", "geometry proposed" ], "Problem": "Given the triangle $ABC$ we consider the points $X,Y,Z$ such that the triangles $ABZ,BCX,CAZ$ are equilateral, and they don't have intersection with $ABC$. Let $B'$ be the midpoint of $BC$, $N'$ the midpoint of $CY$, and $M,N$ the midpoints of $AZ,CX$, respectively. Prove that $B'N' \\bot MN$.", "Solution_1": "Cute one :).\r\n\r\nConsider the transformation we get by composing the spiral similarity cetnered at $B$ of angle $\\frac\\pi 6$ and ratio $\\frac{\\sqrt 3}2$ with the rotation of angle $\\frac \\pi 3$ centered at $C$ (the rotation comes first, then the spiral similarity). This transformation is again a spiral similarity of angle $\\frac \\pi 2$. The image of $Y$ is $M$, and the image of $B$ is $N$, which means that the image of the line $BY$ is $MN$, which shows that $BY\\perp MN$. Now, since $BY\\|B'N'$, we get the conclusion.\r\n\r\nP.S. the above even gives the ratio of the lengths of the two segments.", "Solution_2": "wow! your solutions is Great! Another nice idea is to use the fact that $MN$ is parallel to the line between the centers of $ABZ,BCX$. Hence by radical axis, and since the Fermat point of $ABC$ belongs to $BY$ and to the intersection of the circles $ABZ,BCX$, we get $MN\\bot BY$ and the conclusion follows.", "Solution_3": "jojo, excelent, es una muy interesante soluci\u00f3n la de grobber.\r\n\r\nUna m\u00ednima generalizaci\u00f3n del problema es:\r\n\r\nGiven the triangle ABC we consider the points A', B', C' such that the triangles BCA', CAB', ABC' are equilateral an externals to ABC. Let Ca, Ba, Ab, Cb, Bc y Ac be the midpoints of BA', A'C, CB', B'A, AC' y C'B. The intersection of BaBc and CaCb is X, the intersection of AbAc and CaCb is Y, and the intersection of BaBc and AbAc is Z. Prove that XYZ is homotetic with the triangle NaNbNc (centers of BCA', CAB', ABC'), with center F and reason 1/2. (perd\u00f3n por mi ingl\u00e9s, is terrible).", "Solution_4": "pretty easy I think, the rotation (CCA :pi:/3 2) of BY from C is AX, and the rotation (CA :pi:/6 2/ :sqrt: 3) of MN from B is AX, too. So we get it.", "Solution_5": "3500th post :thumbup: :coolspeak: :clap2: \n\nComplex number bash ftw.\n\n[hide=\"Solution\"]\n[hide=\"Hints from AoPS\"]\n127) Let $A$ be the origin. What must be true about $(m-n)/(n'-b')$ if and only if $\\overline{MN} \\perp \\overline{B'N'}$?\n85) If you're stuck with a ratio that you'd like to show is inamginary, here's a hint. If $\\frac{z_1}{z_2}=\\frac{z_3}{z_4}=k$, then what does $\\frac{z_1b+z_3c}{z_2b+z_4c}$ equal? [/hide]\nLet $A$ be the origin\n\nWe desire: $\\frac{m-n}{b'-n'}$ being imaginary.\n\nWe know: $b-a=\\zeta(z-a)\\implies b=\\zeta*z$\n\nAlso, $x-c=\\zeta(b-c)$ and $y=\\zeta(c)$. \n\nNow we have a HIGHLY computationally intense problem so let's try it. \n\n$m=\\frac{z+a}{2}=\\frac{\\frac{b}{\\zeta}+a}{2}$. Next, $n=\\frac{x+c}{2}=\\frac{2c+\\zeta(b-c)}{2}$. Also, $b'=\\frac{b+c}{2}$ and $n'=\\frac{c+y}{2}=\\frac{c+\\zeta(c)}{2}$.\n\n$\\frac{m-n}{b'-n'}=\\frac{\\frac{b}{\\zeta}-2c-\\zeta(b-c)}{b-\\zeta*c}$ which we need to be imaginary.\n\nThis is the same thing as $\\frac{\\frac{b}{\\zeta}-2c-\\zeta(b)+\\zeta*c}{b-\\zeta*c}$\n\nThis is also the same as $\\frac{b-\\zeta^2(b)-2\\zeta*c+\\zeta^2*c}{b\\zeta-\\zeta^2*c}$\n\nTherefore, we get $\\frac{b(1-\\zeta^2)+c(\\zeta^2-2\\zeta)}{b\\zeta+c(-\\zeta^2)}$\n\nWe now prove $\\frac{1-\\zeta^2}{\\zeta}=\\frac{\\zeta^2-2\\zeta}{-\\zeta^2}$.\n\nThis is crucial in the next part of the problem.\n\n$\\frac{1-\\zeta^2}{\\zeta}=\\frac{\\zeta-2}{-\\zeta}$ or $1-\\zeta^2=-\\zeta+2$ Therefore, we get $\\zeta^2-\\zeta+1=0$ which is true by the definition of $\\zeta$.\n\nYay, lemma proven.\n\nFind $\\frac{bz_1+cz_3}{bz_2+cz_4}$ where $\\frac{z_1}{z_2}=\\frac{z_3}{z_4}$.\n\nWe note that letting this value be $k$, we get $z_1=kz_2$.\n\nTherefore, we have $\\frac{bkz_2+ckz_4}{bz_2+cz_4}=k$.\n\nTherefore, we must now prove that $\\frac{1-\\zeta^2}{\\zeta}$ is imaginary.\n\nSince $z$ is a primitive 6th root of unity, we get $z=e^{\\frac{\\pi}{3}i}$ or $z=e^{\\frac{5\\pi}{3}i}$. \n\nTherefore, $z=\\frac{\\sqrt3}{2}+\\frac{i}{2}$ or $z=\\frac{1}{2}-\\frac{i\\sqrt3}{2}$.\n\nIn the first case, we get $z^2=e^{\\frac{2\\pi}{3}i}$ or therefore it is the same as $\\frac{-1}{2}+\\frac{\\sqrt3}{2}i$\n\nWe then get $\\frac{\\frac32-\\frac{\\sqrt3}{2}i}{\\frac{\\sqrt3}{2}+\\frac{i}{2}}$\n$= \\frac{3-\\sqrt3*i}{\\sqrt3+i}=\\frac{(3-\\sqrt3*i)(\\sqrt3+i)}{4}=\\frac{3-4\\sqrt{3}i-3}{4}=-\\sqrt{3}i$.\n\nIn the second case, we get $z^2=e^{\\frac{4\\pi}{3}i}=-\\frac{1}{2}-\\frac{\\sqrt3}{2}i$.\nAlso, we get $\\frac{1-z^2}{z}=\\frac{3+\\sqrt3i}{1-\\sqrt3i}=\\frac{(3+\\sqrt3i)(1+\\sqrt3i)}{4}=\\frac{4\\sqrt3i}{4}=\\sqrt3i$.\n\nWe are finally done. $\\Box$. [/hide]", "Solution_6": "We denote the affix of each uppercase letter by a lowercase letter.Then by rotation formula we have,\n\n$x=c+(b-c)\\epsilon$\n$y=a+(c-a)\\epsilon$\n$z=b+(a-b)\\epsilon$\n\nThus $n=c+\\frac{(b-c)\\epsilon}{2}$\n $n'=\\frac{(c+a)+(c-a)\\epsilon}{2}$\n $m=\\frac{(a+b)+(a-b)\\epsilon}{2}$\n $b'=\\frac{b+c}{2}$ \nIt is just calculation to see that $\\frac{m-n}{b'-n'}=i\\mathbb{R}$ for some nonzero real $\\mathbb{R}$.\n\n[u][b]Note[/b][/u]:Here $\\epsilon$ represents $cos60^{\\circ}+isin60^{\\circ}$.\n\nBye.\n\nMaths is the doctor of science...\n\nSayantan..", "Solution_7": "Dear Mathlinkers,\nfor a more general proof, you can see\nhttp://perso.orange.fr/jl.ayme vol. 16 p. 43-44\n\nSincerely\nJean-Louis" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all prime divisors $ p$(p<1500) of $ 2^{37}\\minus{}1$\r\nI tried to find greater prime divisors,but failed", "Solution_1": "You can chek only $ p\\equal{}1\\mod 37$ and $ p\\equal{}\\pm 1\\mod 8$.\r\nIt mean $ p\\equal{}1\\mod 296$ or $ p\\equal{}223\\mod 296$.", "Solution_2": "hello, i have $ 2^{37}\\minus{}1\\equal{}223\\cdot616318177$.\r\nSonnhard." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "As the title says :lol:", "Solution_1": "This must have been posted before.\r\n\r\n[hide=\"Hint\"]Show that the product of four consecutive integers is one less than a perfect square.[/hide]", "Solution_2": "[hide]$ x(x\\plus{}1)(x\\plus{}2)(x\\plus{}3) \\plus{} 1 \\equal{} x^4 \\plus{} 6x^3 \\plus{} 11x^2 \\plus{} 6x \\plus{} 1 \\equal{} (x^2 \\plus{} 3x \\plus{} 1)^2$[/hide]\r\n\r\nWe use the fact that if $ m$ is a perfect square then $ m\\minus{}1$ can not be a perfect square. :)", "Solution_3": "[quote=\"B_Gasperov\"]We use the fact that if $ m$ is a perfect square then $ m \\minus{} 1$ can not be a perfect square. :)[/quote]\r\n\r\nWhat about $ m \\equal{} 1$?\r\n\r\nThe actual property you're using is that between two consecutive perfect squares there isn't another perfect square. This is true of any monotonic sequence of integers and can be used to solve several other related problems.", "Solution_4": "everybody found my solution.", "Solution_5": "This is part a) of a problem from the Spanish olympiad. It can be a perfect square only if one of the four terms equals zero.\r\n\r\nnow part b):\r\n\r\nProve that the product of 4 consecutive positive integers is never a perfect cube.", "Solution_6": "[quote=\"Pul de Algodoncito\"]Prove that the product of 4 consecutive positive integers is never a perfect cube.[/quote]\r\n\r\nLet $ n(n\\plus{}3)\\equal{}2k$. Then $ (n\\plus{}1)(n\\plus{}2)\\equal{}2k\\plus{}2$. Thus \\[ P\\equal{}n(n\\plus{}1)(n\\plus{}2)(n\\plus{}3)\\equal{}4k(k\\plus{}1)\\]\r\nLet $ P\\equal{}(2t)^3$. Then $ 4k(k\\plus{}1)\\equal{}8t^3\\implies k(k\\plus{}1)\\equal{}2t^3$. So the only possible solution to this equation is $ k\\equal{}1,t\\equal{}1$, as $ (k,k\\plus{}1)\\equal{}1$. But $ k\\equal{}1$ implies $ n(n\\plus{}3)\\equal{}2$, which is impossible. Hence the conclusion.", "Solution_7": "Let the 4 consecutive positive integers be x,x+1,x+2 and x+3\r\nTheir product = {x*(x+3})*{(x+1)*(x+2)} = (x^2+3x+1)^2 -1.\r\nSo, the product of them can't be a perfect square." } { "Tag": [ "trigonometry", "geometry", "area of a triangle", "Heron\\u0027s formula" ], "Problem": "I can't do this problem... :( \r\n\r\nTwo circles have radii a,b. The distance between their centers is d. They intersect at P and Q. The distance from P to Q is?\r\n\r\nExpress your answering using a,b,and/or d.", "Solution_1": "this seems an easy problem. let O1, O2 be the centers of the circles. draw the O1a, where it's tangent to the second circle (that would be point P), and draw O2b were it's tangent to the first one(point P). then draw the line, d. it is obvious that the angle O1PO2 is 90 :^o: and all you need to do now is to figure out the altitue of the line PH, that is orthagonal to D, and then multiply it by two. and that would be the length of PQ.", "Solution_2": "[hide]Oops messed up the first time.\n$2a \\sin \\left( \\cos^{-1} \\left( \\frac {b^2 - a^2 - d^2}{-2ad} \\right ) \\right)$[/hide]", "Solution_3": "Here were the choices.. Since no answer doesn't seem to give me \"variable\" answer..\r\n\r\nA) $\\frac{[d^2 - (a-b)^2]^{\\frac{1}{2}} [(a+b)^2 - d^2]^{\\frac{1}{2}}}{d}$\r\n\r\nB) $(a^2 + b^2 - d^2)^{\\frac{1}{2}}$\r\n\r\nC) $[(a-b)(a+b)]^{\\frac{1}{2}}$\r\n\r\nD) $\\frac{a+b+d}{3}$\r\n\r\nE) None of the above", "Solution_4": "[hide]Answer: [b]A [/b]\n\n Using Heron's formula, $\\triangle PO_1O_2 =\\frac{1}{2}dh=\\frac{1}{2}\\sqrt{\\frac{a+b+d}{2}\\cdot \\frac{a+b-d}{2}\\cdot \\frac{b+d-a}{2}\\cdot \\frac{d+a-b}{2}}$[/hide]", "Solution_5": "think of the triangle with vertices of the centers and $P$. The sides are $a,b,d$. The distance from $P$ to the base is $\\displaystyle \\frac{1}{2}$ of $PQ$.\r\nTo find the height, use Heron's formula.\r\n$\\displaystyle \\frac{1}{2}dh=\\sqrt{\\frac{a+b+d}{2}\\cdot \\frac{a+b-d}{2}\\cdot \\frac{b+d-a}{2}\\cdot \\frac{d+a-b}{2}}$\r\n\r\n$h={2}\\displaystyle \\frac{\\sqrt{\\frac{a+b+d}{2}\\cdot \\frac{a+b-d}{2}\\cdot \\frac{b+d-a}{2}\\cdot \\frac{d+a-b}{2}}}{d}$\r\n\r\n$PQ={4}\\displaystyle \\frac{\\sqrt{\\frac{a+b+d}{2}\\cdot \\frac{a+b-d}{2}\\cdot \\frac{b+d-a}{2}\\cdot \\frac{d+a-b}{2}}}{d}$\r\n\r\n$PQ=\\frac{\\sqrt{d^2 - (a-b)^2} \\sqrt{(a+b)^2 - d^2}}{d}$\r\n\r\nAnswer is $\\text {A}$" } { "Tag": [ "function", "\\/closed" ], "Problem": "This is the second time I'm posting this, but its back again!\r\n\r\nEvery time I post in the MC forum, I get this:\r\n\r\n[code]\nSQL requests not achieved\n\nDEBUG MODE\n\nSQL Error: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' 19209) GROUP BY ug.user_id' at line 7\n\nSQL Request: SELECT MAX(a.auth_value) as allowed, ug.user_id FROM phpbb_auths a, phpbb_user_group ug WHERE a.obj_type = 'f' AND a.obj_id = 132 AND a.auth_name = 'auth_watch_forum' AND a.group_id = ug.group_id AND ug.user_id IN (, 19209) GROUP BY ug.user_id\nLine : 1189\nFile : functions_post.php\n[/code]\r\n\r\nCan someone please see what's wrong?", "Solution_1": "[quote=\"#H34N1\"]This is the second time I'm posting this, but its back again!\n\nEvery time I post in the MC forum, I get this:\n\n[code]\nSQL requests not achieved\n\nDEBUG MODE\n\nSQL Error: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' 19209) GROUP BY ug.user_id' at line 7\n\nSQL Request: SELECT MAX(a.auth_value) as allowed, ug.user_id FROM phpbb_auths a, phpbb_user_group ug WHERE a.obj_type = 'f' AND a.obj_id = 132 AND a.auth_name = 'auth_watch_forum' AND a.group_id = ug.group_id AND ug.user_id IN (, 19209) GROUP BY ug.user_id\nLine : 1189\nFile : functions_post.php\n[/code]\n\nCan someone please see what's wrong?[/quote]\r\nThe post still gets made, right?", "Solution_2": "Yes, but the messages is quite annoying.", "Solution_3": "[quote=\"#H34N1\"]Yes, but the messages is quite annoying.[/quote]\r\n\r\nyes that has happened to me many times and i recall someone else posting this bug somewhere...", "Solution_4": "That was me :roll: \r\n\r\nI couldn't find it so I posted it again.", "Solution_5": "Bug reproduced, and finally fixed, thanks to Darij :)" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all triples of positive integers $ (p,q,n)$, with $ p$ and $ q$ primes, satisfying:\r\n\r\n$ p(p\\plus{}3)\\plus{}q(q\\plus{}3)\\equal{}n(n\\plus{}3)$.", "Solution_1": "Here is my solution.\r\nBy $ mod \\ 3$, we have $ p^2 \\plus{} q^2 \\equiv \\ n^2$. Hence, $ p$ or $ q$ is equal to $ 3$, otherwise $ n^2 \\equiv \\ 2$. Now, w.l.o.g. suppose that $ p \\equal{} 3$, so the equation becomes $ 18 \\plus{} q(q \\plus{} 3) \\equal{} n(n \\plus{} 3)$. Solving like a 2-nd degree equation in $ n$, we have the discriminant equals to $ \\Delta \\equal{} \\sqrt {4q(q \\plus{} 3) \\plus{} 81}$. Since $ n$ is a integer and $ \\Delta$ is odd, $ 4q(q \\plus{} 3) \\plus{} 81 \\equal{} (2k \\plus{} 1)^2$, where $ k$ is a nonnegative integer. Hence $ q(q \\plus{} 3) \\equal{} (k \\minus{} 4)(k \\plus{} 5)$. \r\nNow suppose that $ k \\equal{} qt \\minus{} 5$, where $ t$ is a nonnegative integer. So $ q \\plus{} 3 \\equal{} (qt \\minus{} 9)t$. So $ q \\plus{} 3\\geq qt \\minus{} 9$, so $ t\\leq \\frac{q\\plus{}12}{q}$, and it's easy to see that $ t\\leq 7$. By brute force, we can observe that $ t \\equal{} 5$ and $ p \\equal{} 2$. Substituting in the equation, the only positive integer $ n$ is $ 4$.\r\nAnalogously, the case $ k \\equal{} qt \\plus{} 4$ don't generate solutions.\r\nThe unique triple $ (p,q,n)$ is $ (2,3,4)$.", "Solution_2": "The above solution is incorrect, as $3(3 + 3) + 7(7 + 3) = 88 = 8(8 + 3)$. Here's my attempt.\n\n\nConsidering the expression modulo 3, we note that the only quadratic residues modulo 3 are 0 and 1, thus \n$$\np(p + 3) + q(q + 3) = n(n + 3) \\iff p^2 + q^2 \\equiv n^2 \\pmod{3},\n$$\nwill have solutions only if $p^2 + q^2 \\equiv 0$ or $1 \\pmod{3}$.\n\nIf $p^2 + q^2 \\equiv 0$, then $p = q = 3$, but then there is no solutions for $n$.\n\nThis implies that one of $p$ or $q$ must be 3. WLOG let $p = 3$, then\n$$\n18 + q(q + 3) = n(n + 3) \\implies 18 = (n-q)(n + q + 3).\n$$\nChecking factor pairs of 18, we find that the only solutions are $(p, q, n) = (3, 2,4)$ and $(3, 7, 8)$,\nso all of the solutions to the equation are\n$$\n(p, q, n) \\in \\{(2, 3, 4), \\; (3, 2, 4),\\; (3, 7, 8), \\; (7, 3, 8)\\}.\n$$", "Solution_3": "As people above me have proven, one of the primes p or q equals 3. WLOG q=3\nThen we expand and get 18+q\u00b2-n\u00b2=3(n-q) using diffrence of squares and changing sides we get 18=(n-q)(3+q+n) so we have n>q and because q\u22652 we have 3+q+n\u22658 and 3+q+n divides 18 so either 3+q+n=9 and n-q=2 or 3+q+n=18 and n-q=1" } { "Tag": [], "Problem": "I had a simple conceptual doubt once and forgot all about that. Now that I recall it could someone help me out...\r\n\r\nThere is a block placed on a rough horizontal surface such that it just touches a wall perpendicular to the floor. Now if a slight increasing external force is applied pushing it against the wall then what are the normal forces (from the wall) and the frictional forces.\r\n\r\nI mean suppose the limiting friction is 10N. And a force of 5N is applied then is the entire 5N opposed by friction or the normal force of the wall. If not then how does the 5N divide among the two opposing forces ?\r\n\r\nI am sorry if my doubt is really stupid :-) .But please help..", "Solution_1": "The 5N countering the force pushing against the wall is the normal force, but I don't see where you're getting the frictional force from.\r\n\r\nConstructing the free body diagram for this system, if we suppose the walls are below and to the right of the block, you should get four forces as follows:\r\n-Pushing force to the right\r\n-Weight acting downwards\r\n-Normal force acting upwards\r\n-Normal force acting left\r\n\r\nThe left normal force counters the right force and the upwards normal force counters the downwards weight.", "Solution_2": "Thanks its clear now...\r\nThere is no friction when there is no tendancy to move" } { "Tag": [ "LaTeX", "irrational number" ], "Problem": "How can I solve $\\sqrt{164}$ without using a calculator?", "Solution_1": "[quote=\"blueshark\"]How can I solve sqrt{164} without using a calculator?[/quote]To make that appear as $\\sqrt{164}$, just type \\sqrt{164} and type dollar signs before and after it. If you're going to be around here for a while you should learn $\\LaTeX$. It's very easy.\r\n\r\nAnyway, I'm not sure what you mean when you say you want to \"solve $\\sqrt{164}$.\" If you want to find it's decimal expansion, look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48990]here[/url]. It's an irrational number, so it will have an infinite nonrepeating decimal expansion, so you won't be able to actually write $\\sqrt{164}$ completely as a decimal.", "Solution_2": "By solve,do u mean like till infinite digits??cause we can write sqroot164 as 2*sqroot41 and then find sqroot 41 by the normal method(as taught in school in around 8th grade).We can find the appprox value by this method.", "Solution_3": "Thanks for your reply. What I meant to ask is: What is\r\n$\\sqrt{164}$ correct to the nearest integer? How can I find the answer without using a calculator?\r\n\r\nFor example, I know how to solve problems like $\\sqrt{169}$ which equals 13 because 13 times 13 = 169, right? The same for applies for the sqaure root of 36, 25, 81, 9, 4, 16, 100, etc. I recall my teacher saying that numbers like 4, 9, 16, 25, 36, 81, and 100 are called perfect squares, right? \r\n\r\nHow do I answer the question above without using a calculator?", "Solution_4": "something like this ?\r\n\r\n$\\sqrt{144}<\\sqrt{164}<\\sqrt{169}$ so $12<\\sqrt{164}<13$ :huh: \r\n\r\nThis is basically through guessing the approximate like what is the nearest integer for\r\n\r\n$\\sqrt{3724}$ ? So you know $\\sqrt{3600}=60$ so the integer should be starting with $6$ . Then just work out some number $61^2=3721$ , $62^2=3844$ ,$63^2=3969$ and you will get the nearest integer to be $61$ :)", "Solution_5": "i'm sorry for the horrifying diagram,but thats the method i follow for findin square roots.if u multiply 6.4 by 2,u get 12.8 and sqroot 164 is 12.806.....", "Solution_6": "[hide]\nsince 164 is between $12^2$ and $13^2$ the answer is betwen 12 and 13. 164 is closer to 169 than 144 so the answer is definetly bigger than 12.5. Seeing that 164 is only 5 off from 169 and 20 off from 144 we can estimate that the answer is 12.8 or 12.9. Trying 12.8 i get 163.84 so 12.8 is close. Trying 12.9 i get 166.41 so the answer is approximatley 12.8.[/hide]", "Solution_7": "If you want to find the value of $\\sqrt{164}$..\r\n\r\n[hide]Find the nearest perfect square, 169, and take the square root of that, which is 13. Divide 164 by 13, to get (I like to cheat with calculators) 12.62. Take the average of that and 13 to get 12.81. Now divide 164 by 12.81 to get 12.805, take the average of 12.81 and 12.805 to get about 12.806. Continue until you get to the desired place. [/hide]", "Solution_8": "[hide]\n164 is in between 12^2=144 and 13^2=169,\n164 is closer to 169 then 144, so it will be more than 12.5.\nTry 12.8, you get 163.84\nTry 12.9, you get 166.41\n163.84 is closer to 164 than 166.41\n$\\sqrt{164} \\approx \\framebox{12.8}$[/hide]", "Solution_9": "I hope you mean 12.8 in boxes :) \r\n\r\nThat's a method too. There are many.", "Solution_10": "http://www.mathpath.org/Algor/algor.square.root.htm", "Solution_11": "I want to thank all the tutors who took time to explain my question.", "Solution_12": "$\\sqrt{144}$<$\\sqrt{164}$<$\\sqrt{169}$\r\n\r\nSo, 12<$\\sqrt{164}$<13", "Solution_13": "well it is my first post in this sub-forum \r\n$\\sqrt{164} \\approx \\frac{333}{26}$ :D :D \r\nI use this formula \\[ \\sqrt{x} \\approx \\frac{2x + p}{2 \\sqrt{x+p}} \\]\r\ni choose $p=5$.\r\nwant a proof??? :)", "Solution_14": "I don't know if this was posted earlier, but you could always use the binomial theorem. ;)" } { "Tag": [ "algebra unsolved", "algebra", "combinatorics", "Combinatorial sum" ], "Problem": "Find The Closed form of:\r\n\r\n$\\sum_{k=0,n}(-1)^k(n-k)!(n+k)!$", "Solution_1": "It is: $(2n)!(1-\\frac{1}{2(n+1)})$ if I didn't miscalculated.\r\n\r\nI used a little changed Australian problem: $\\sum _0 ^{2n} \\frac{(-1)^k}{C_{2n} ^k}=2-\\frac{1}{n+1}$", "Solution_2": "which year's TST was this from?", "Solution_3": "[quote=\"Megus\"]It is: $(2n)!(1-\\frac{1}{2(n+1)})$ if I didn't miscalculated.\n\nI used a little changed Australian problem: $\\sum _0 ^{2n} \\frac{(-1)^k}{C_{2n} ^k}=2-\\frac{1}{n+1}$[/quote]\r\nhow to prove the australian problem?", "Solution_4": "you can use induction.\r\ni'll try to write a detailed solution in a moment", "Solution_5": "seems that i'm wrong :(", "Solution_6": "[quote=\"perfect_radio\"]\nFor $n=1$ we have $\\frac{1}{C_{2}^0-C_{2}^1+C_{2}^2}=2-\\frac{1}{2}$\n$\\sum _{k=0} ^{2n} \\frac{(-1)^k}{C_{2n} ^k}=2-\\frac{1}{n+1}$[/quote]\r\n\r\nThere, I think the [ was supposed to be a { ;)", "Solution_7": "Where can I find a complete solution?", "Solution_8": "To prove australian problem first prove that:\r\n\r\n$\\frac{1}{C _{2n}^k}=\\frac{2n+1}{2n+2}(\\frac{1}{C_{2n+1}^{k}}+\\frac{1}{C_{2n+1}^{k+1}})$", "Solution_9": "i just checked.. it's from the 2001 TST. you can find them at http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html, though there aren't any solutions posted." } { "Tag": [ "search", "function" ], "Problem": "Heloo everyone.\r\nI will be very pleased if s.o. of you can write me links or problems,which have proofed with Stewart's theorem.\r\nThank you.", "Solution_1": "Use the Search function to get some problems involving Stewart's theorem. [url]http://www.mathlinks.ro/Forum/search.php[/url]\r\nAnd also it not regards Middle School, of course. :!:", "Solution_2": "[quote=\"Andreas\"]Use the Search function to get some problems involving Stewart's theorem. [url]http://www.mathlinks.ro/Forum/search.php[/url]\nAnd also it not regards Middle School, of course. :!:[/quote]\r\nThank you" } { "Tag": [ "limit" ], "Problem": "\u039b\u03bf\u03b9\u03c0\u03bf\u03bd \u03bd\u03b1 \u03bc\u03b9\u03b1 \u03bb\u03b9\u03b3\u03bf \u03c0\u03b5\u03c1\u03b9\u03b5\u03c1\u03b3\u03b7 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7.\u039a\u03b1\u03bb\u03bf\u03ba\u03b1\u03b9\u03c1\u03b9\u03bd\u03b7 \u03c0\u03b5\u03b9\u03c4\u03b5....\r\n\u03a4\u03b7\u03bd \u03b1\u03c6\u03b9\u03b5\u03c1\u03c9\u03bd\u03c9 \u03c3\u03b5 \u03bf\u03bb\u03bf\u03c5\u03c2 .\r\n\r\n\u0391\u03bd $a,b,c,d \\in \\mathbb{R}$ \u03bc\u03b5 \r\n\r\n$a+b+c+d<-1$\r\n$a-b+c-d<-1$\r\n$\\mid 4a+2c \\mid=\\mid 3b+d \\mid$\r\n$\\mid b \\mid \\leq 1$\r\n\r\n\u03bd\u03b4\u03bf $ac\\leq \\frac{3}{8}$\r\n\r\n :)", "Solution_1": "\u03a9\u03c1\u03b1\u03af\u03b1 \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 Socrates :) . \u03a0\u03bf\u03cd \u03c4\u03b7\u03bd \u03b2\u03c1\u03ae\u03ba\u03b5\u03c2 ??\r\n\u039d\u03b1 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b5\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03b7 \u03bb\u03cd\u03c3\u03b7 : \r\n\r\n[hide=\"\u039b\u03cd\u03c3\u03b7\"]\u0398\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03bf \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf $p(x) = a x^{4} +b x^{3} + c x^{2} +d x +1$.\n\n\u0388\u03c7\u03bf\u03c5\u03bc\u03b5 $p(1)=a+b+c+d +1<0$ \u03ba\u03b1\u03b9 $p(-1)=a-b+c-d +1<0$ .\n\n\u0391\u03c6\u03bf\u03cd $p(0)=1>0$ \u03c4\u03bf \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf \u03ad\u03c7\u03b5\u03b9 \u03b4\u03cd\u03bf \u03c1\u03af\u03b6\u03b5\u03c2 $\\rho_{1} \\in (-1,0)$ \u03ba\u03b1\u03b9 $\\rho_{2} \\in (0,1)$.\n\n[u][b]\u03a0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 1[/b][/u]\n\n\u0386\u03bd $\\boxed{a>0}$ \u03b1\u03c6\u03bf\u03cd $\\lim_{x \\to + \\infty}p(x)=\\lim_{x \\to -\\infty}p(x)=+ \\infty$ \u03ba\u03b1\u03b9 $p(1), \\, p(-1) \\, <0$ \n\n\u03c4\u03bf \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf \u03ad\u03c7\u03b5\u03b9 \u03ac\u03bb\u03bb\u03b5\u03c2 \u03b4\u03cd\u03bf \u03c1\u03af\u03b6\u03b5\u03c2 $\\rho_{3} \\in (-\\infty,-1)$ \u03ba\u03b1\u03b9 $\\rho_{4} \\in (1, \\infty)$ .\n\n[u][b]\u03a0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 2[/b][/u]\n\n$\\boxed{a<0}$\n\n\u0388\u03c0\u03af\u03c3\u03b7\u03c2 $p'(x)=4ax^{3}+3bx^{2}+2cx+d$ \u03ac\u03c1\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7\n\n$\\\\ \\mid 4a+2c \\mid=\\mid 3b+d \\mid$ \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 $p'(1)=0$ \u03ae $p'(-1)=0$ .\n\n\u03a5\u03c0\u03bf\u03b8\u03ad\u03c4\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $p'(1)=4a+3b+2c+d=0$. \n\n\u0391\u03c5\u03c4\u03cc \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c3\u03c5\u03bc\u03b2\u03b5\u03af \u03cc\u03c4\u03b1\u03bd $(4a+2c)(3b+d) \\leq 0$ .\n\n$\\bullet$ \u0386\u03bd $(4a+2c)(3b+d) =0$ \u03c4\u03cc\u03c4\u03b5 $p'(-1)$ \u03c3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 \u03c4\u03bf $p'(x)$ \u03ad\u03c7\u03b5\u03b9 \u03c4\u03c1\u03b5\u03b9\u03c2 \u03c1\u03af\u03b6\u03b5\u03c2.\n\n\n$\\bullet$ \u0391\u03bd $4a+2c <0$ \u03ba\u03b1\u03b9 $3b+d>0$ \u03c4\u03cc\u03c4\u03b5 $p'(-1)=(3b+d)-(4a+2c) >0$.\n\u03a3\u03b5 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b1\u03bd \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $p'(x)>0$ \u03b3\u03b9\u03b1 $x \\in (- \\infty ,1)$ \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 \n\u03ac\u03c4\u03bf\u03c0\u03bf \u03b3\u03b9\u03b1\u03c4\u03af \u03c4\u03bf \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf \u03ad\u03c7\u03b5\u03b9 \u03c1\u03af\u03b6\u03b1 \u03c3\u03c4\u03bf $(0,1)$ .\n\u0386\u03c1\u03b1 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03b1\u03bb\u03bb\u03ac\u03b6\u03b5\u03b9 \u03c0\u03c1\u03cc\u03c3\u03b7\u03bc\u03bf \u03ba\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c0\u03ac\u03bb\u03b9 3 \u03c1\u03af\u03b6\u03b5\u03c2 .\n\n\n$\\bullet$ \u0391\u03bd $4a+2c >0$ \u03ba\u03b1\u03b9 $3b+d<0$ \u03c4\u03cc\u03c4\u03b5 $p'(-1)<0$ .\n\u03a4\u03ce\u03c1\u03b1 \u03b1\u03c6\u03bf\u03cd $\\lim_{x \\to - \\infty}p'(x)=+ \\infty$ \u03c4\u03bf $p'(x)$ \u03ad\u03c7\u03b5\u03b9 \u03c1\u03af\u03b6\u03b1 $\\xi \\in (- \\infty ,-1)$ \u03ac\u03c1\u03b1 \u03ad\u03c7\u03b5\u03b9 3 \u03c1\u03af\u03b6\u03b5\u03c2 .\n[u][b]\n\u03a0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 3[/b][/u]\n\n$\\\\ a=0$ :D \n\n\u0391\u03c0\u03cc \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03c4\u03bf $p''(x)=12ax^{2}+6bx+2c$ \u03ad\u03c7\u03b5\u03b9 2 \u03c1\u03af\u03b6\u03b5\u03c2\n\u03ac\u03c1\u03b1 $D = 36b^{2}-96ac \\geq 0 \\Rightarrow \\frac{3}{8} \\cdot b^{2} \\geq ac$\n\n\u03ba\u03b1\u03b9 \u03b1\u03c6\u03bf\u03cd $|b| \\leq 1$ \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 $\\boxed{ac \\leq \\frac{3}{8}}$ .[/hide]\r\n\r\n\u03a0\u03ae\u03c1\u03b1 \u03c0\u03bf\u03bb\u03bb\u03ad\u03c2 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03bc\u03bf\u03c5 \u03b1\u03c1\u03ad\u03c3\u03b5\u03b9. \r\n\u0398\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ae\u03c3\u03c9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03c4\u03b9 \u03c0\u03b9\u03bf \u03ba\u03bf\u03bc\u03c8\u03cc \u03b1\u03cd\u03c1\u03b9\u03bf . :)" } { "Tag": [ "algebra", "polynomial", "vector", "complex numbers", "superior algebra", "superior algebra unsolved" ], "Problem": "1. Let $\\alpha, \\beta$ be complex numbers of degree 3 over $\\mathbb{Q}$. Determine the possibilities for $[\\mathbb{Q}(\\alpha, \\beta) : \\mathbb{Q}]$.\r\n\r\n2. Let $\\alpha, \\beta$ be complex roots of irreducible polynomials $f(x), g(x) \\in \\mathbb{Q}[x]$. Prove that $f(x)$ is irreducible in $\\mathbb{Q}[\\alpha]$ iff $g(x)$ is irreducible in $\\mathbb{Q}[\\beta]$.", "Solution_1": "1. Suppose $[\\mathbb{Q}(\\alpha): \\mathbb{Q}]=3$ and $[\\mathbb{Q}(\\beta): \\mathbb{Q}]=3$. Then $3~\\leq[\\mathbb{Q}(\\alpha,\\beta): \\mathbb{Q}]\\leq~9$. \r\nIf $\\mathbb{Q}(\\alpha)=\\mathbb{Q}(\\beta)$ then $[\\mathbb{Q}(\\alpha,\\beta): \\mathbb{Q}]=3$. If no common basis elements of $\\mathbb{Q}(\\alpha)$ and $\\mathbb{Q}(\\beta)$ as vector spaces over $\\mathbb{Q}$, then $[\\mathbb{Q}(\\alpha,\\beta): \\mathbb{Q}]=9$.", "Solution_2": "[quote=\"abdul\"]If no common basis elements of $\\mathbb{Q}(\\alpha)$ and $\\mathbb{Q}(\\beta)$ as vector spaces over $\\mathbb{Q}$, then $[\\mathbb{Q}(\\alpha,\\beta): \\mathbb{Q}]=9$.[/quote]\r\nThis is not true, you can still get degree $6$.\r\n\r\n2. is a matter of degree calculation, too.", "Solution_3": "When does this equality $[\\mathbb{Q}(\\alpha,\\beta): \\mathbb{Q}]=9$ hold?", "Solution_4": "$\\alpha = \\sqrt[3]{2}, \\beta = \\sqrt[3]{3}$.", "Solution_5": "This what i think too when i made $9$ as upper bound.\r\nBut how can still get degree $6$ if no common basis elements of $\\mathbb{Q}(\\alpha)$ and $\\mathbb{Q}(\\beta)$?\r\n\r\nI just read about field extensions for first time some hours ago..", "Solution_6": "You can even have $\\mathbb{Q}(\\alpha)=\\mathbb{Q}(\\beta)$ but disjoint bases, simply as bases can be choosen quite random (you mean something else than bases I think).\r\nOne example for degree $6$: $\\alpha = \\sqrt[3]{2}$ and $\\beta = \\zeta_{3}\\sqrt[3]{2}$ with $\\zeta_{3}^{3}=1 , \\zeta_{3}\\neq 1$.", "Solution_7": "Thank you, ZetaX." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "This is MIC round 3 and its solution... We are looking for your ideas. :)\r\n\r\nRegards\r\nVo Quoc Ba Can", "Solution_1": "[quote=\"toanhocmuonmau\"]This is MIC round 3 and its solution... We are looking for your ideas. :)\n\nRegards\nVo Quoc Ba Can[/quote]\r\nOne of it was from here:http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=48&Id=14604&page=2 :)" } { "Tag": [], "Problem": "A recursive sequence has terms in the form $ a_n/b_n$ and is defined such that all even numbered terms are the reciprocal of the immediately preceding term and in all odd numbered terms, $ a_n$ and $ b_n$ are equal to $ a_{n \\minus{} 2} \\plus{} 1$ and $ b_{n \\minus{} 2} \\plus{} 1$ respectively. If $ t_1 \\equal{} 2/3$ what is: \r\n\r\n$ (a_{2005} \\plus{} a_{2006} \\plus{} a_{2007} \\plus{} a_{2008})/(b_{2005} \\plus{} b_{2006} \\plus{} b_{2007} \\plus{} b_{2008})$", "Solution_1": "[hide=\"Solution\"]\n\nNotice that\n\n$ a_{2k} \\equal{} a_{2k \\minus{} 1} \\plus{} 1$ and\n$ a_{2k \\plus{} 1} \\equal{} a_{2k}$\nfor any integer k.\n\nAlso notice that\n\n$ b_{2j} \\equal{} b_{2j \\minus{} 1} \\minus{} 1$ and\n$ b_{2j \\plus{} 1} \\equal{} b_{2j} \\plus{} 2$\n\nWe can turn the recursive definition of sequence $ \\{t\\}$ into a... a... (the word is slipping from my mind) definition where you can find the term with the index.\n\n$ a_x \\equal{} (x \\plus{} 3)/2$ if x is odd\n$ a_x \\equal{} (x \\plus{} 4)/2$ if x is even\n$ b_x \\equal{} (x \\plus{} 5)/2$ if x is odd\n$ b_x \\equal{} (x \\plus{} 2)/2$ if x is even\n\nTherefore,\n\n$ (a_{2005} \\plus{} a_{2006} \\plus{} a_{2007} \\plus{} a_{2008})/(b_{2005} \\plus{} b_{2006} \\plus{} b_{2007} \\plus{} b_{2008})$\n\n$ \\equal{} (1004 \\plus{} 1005 \\plus{} 1005 \\plus{} 1006)/(1005 \\plus{} 1004 \\plus{} 1006 \\plus{} 1005)$\n\n$ \\equal{} 1$\n\nWe can generalize this. For indices $ 4n \\plus{} 1, 4n \\plus{} 2, 4n \\plus{} 3, 4n \\plus{} 4$ (a is an integer),\n\n$ (a_{4n \\plus{} 1} \\plus{} a_{4n \\plus{} 2} \\plus{} a_{4n \\plus{} 3} \\plus{} a_{4n \\plus{} 4})/(b_{4n \\plus{} 1} \\plus{} b_{4n \\plus{} 2} \\plus{} b_{4n \\plus{} 3} \\plus{} b_{4n \\plus{} 4})$\n\n$ \\equal{} [(2n \\plus{} 2) \\plus{} (2n \\plus{} 3) \\plus{} (2n \\plus{} 3) \\plus{} (2n \\plus{} 4)]/[(2n \\plus{} 3) \\plus{} (2n \\plus{} 2) \\plus{} (2n \\plus{} 4) \\plus{} (2n \\plus{} 3)]$\n\n$ \\equal{} 1$\n\n[/hide]", "Solution_2": "[quote=\"Aryth\"][...] (the word is slipping from my mind) [...][/quote]\r\n\r\nExplicit? :wink:", "Solution_3": "Good call :P" } { "Tag": [ "function", "inequalities proposed", "inequalities" ], "Problem": "Let $ n \\geq 3$ be an odd integer. Determine the maximum value of\r\n\\[ \\sqrt{|x_{1}\\minus{}x_{2}|}\\plus{}\\sqrt{|x_{2}\\minus{}x_{3}|}\\plus{}\\ldots\\plus{}\\sqrt{|x_{n\\minus{}1}\\minus{}x_{n}|}\\plus{}\\sqrt{|x_{n}\\minus{}x_{1}|},\\]\r\nwhere $ x_{i}$ are positive real numbers from the interval $ [0,1]$.", "Solution_1": "[hide=\"no proof\"]\n$ n \\minus{} 2 \\plus{} \\sqrt {2}$. My proof is inelegant.\n[/hide]", "Solution_2": "[quote=\"JoeBlow\"][hide=\"no proof\"]\n$ n \\minus{} 2 \\plus{} \\sqrt {2}$. My proof is inelegant.\n[/hide][/quote]\r\nAfter solving it I understand what you mean! I solved it making extensive use of mixing variables, to reduce it to maximizing a function defined for each partition of n.", "Solution_3": "We have a continuos function on a compact set $ [0,1]^n$, hence there is an optimal point $ (x_1,...,x_n)$.\r\nNote now that\r\n0) impossible to have $ x_{i \\minus{} 1} \\equal{} x_{i} \\equal{} x_{i \\plus{} 1}$;\r\n1) if $ x_i\\leq x_{i \\minus{} 1}$ and $ x_i\\leq x_{i \\plus{} 1}$, then $ x_i \\equal{} 0$;\r\n2) if $ x_i\\geq x_{i \\minus{} 1}$ and $ x_i\\geq x_{i \\plus{} 1}$, then $ x_i \\equal{} 1$;\r\n3) if $ x_{i \\plus{} 1}\\leq x_i\\leq x_{i \\minus{} 1}$ or $ x_{i \\minus{} 1}\\leq x_i\\leq x_{i \\plus{} 1}$, then $ x_i \\equal{} \\frac {x_{i \\minus{} 1} \\plus{} x_{i \\plus{} 1}}{2}$.\r\nIt follows that $ (x_1,...,x_n)$ looks like\r\n\\[ (0,\\frac {1}{k_1},\\frac2{k_1},...,1,\\frac {k_2 \\minus{} 1}{k_2},...,\\frac2{k_2},\\frac1{k_2},0,\\frac1{k_3},...,\\frac1{k_l}),\r\n\\]\r\nwhere $ k_1$, $ k_2$, ..., $ k_l$ are natural numbers, $ k_1 \\plus{} k_2 \\plus{} ... \\plus{} k_l \\equal{} n$, $ l$ is even clearly. Then the function is this point equals\r\n\\[ S \\equal{} \\sqrt {k_1} \\plus{} \\sqrt {k_2} \\plus{} ...\\sqrt {k_l}.\r\n\\]\r\nUsing the fact that $ l$ is even and $ \\sqrt {k} < \\sqrt {k \\minus{} 1} \\plus{} 1$ we conclude that maximal possible value of $ S$ is $ n \\minus{} 2 \\plus{} \\sqrt 2$ ($ l \\equal{} n \\minus{} 1$, $ k_1 \\equal{} k_2 \\equal{} ... \\equal{} k_{l \\minus{} 1} \\equal{} 1$, $ k_l \\equal{} 2$ in this case).\r\n\r\nIt is not a real olympiad problem in sense of inner beauty.", "Solution_4": "since $ n$ is odd, there must be an $ i$ such that both $ x_i$ and $ x_{i \\plus{} 1}$ are both belong to $ [0,\\frac {1}{2}]$ or $ [\\frac {1}{2},1]$. without loss of generality let $ x_1\\leq x_2$ and $ x_1$, $ x_2$ belong to $ [0,\\frac {1}{2}]$. We can prove that $ \\sqrt {x_2 \\minus{} x_1} \\plus{} \\sqrt {Ix_3 \\minus{} x_2I}\\leq \\sqrt {2}$. If $ x_3 > x_2$, $ \\sqrt {x_2 \\minus{} x_1} \\plus{} \\sqrt {x_3 \\minus{} x_2}\\leq 2\\cdot \\sqrt {\\frac {x_3 \\minus{} x_1}{2}}\\leq \\sqrt {2}$; else $ x_1,x_2,x_3$ are all belong to $ [0,\\frac {1}{2}]$. Hence, $ \\sqrt {x_2 \\minus{} x_1} \\plus{} \\sqrt {Ix_3 \\minus{} x_2I}\\leq \\sqrt \\frac {1}{2} \\plus{} \\sqrt \\frac {1}{2}$. Also all of the other terms of the sum are less then or equal to $ 1$. summing them gives the desired result.\r\nexample is $ (0,\\frac {1}{2},1,0,1,\\ldots ,1)$\r\nnote: all the indices are considered in modulo $ n$", "Solution_5": "We can assume there exists some $ i$ for which $ x_i \\leq x_{i\\plus{}1} \\leq x_{i\\plus{}2}$ because if it didnt exist: we can WLOG suppose $ x_1\\equal{}max{x_i}$. Now $ x_1 \\geq x_2$. It follows that $ x_2 \\leq x_3$ because if not our assumption would be faulty. Now we continue, it would be: $ x_1 \\geq x_2 \\leq x_3 \\geq x_4 .... \\leq x_n \\geq x_1$, because n is odd. It's a contradiction that $ x_1\\equal{}max{x_i}$. So there exists $ i$ such that $ x_i \\leq x_{i\\plus{}1} \\leq x_{i\\plus{}2}$. Now we can use AM-QM: $ \\sqrt{_x{i\\plus{}2}\\minus{}x_{i\\plus{}1}}\\plus{}\\sqrt{x_{i\\plus{}1}\\minus{}x_i} \\leq \\sqrt{2}\\sqrt{x_{i\\plus{}2}\\minus{}x_i}$. Now, maximum is attained if all roots are equal to $ 1$. Now it easy to check this is attainable, if $ i$ is even, then $ (x_1,x_2,...,x_i,x_{i\\plus{}1},x_{i\\plus{}2},...,x_n)\\equal{}(0,1,...,1,\\frac{1}{2},0,...,1)$. Similarly if $ i$ is odd. The maximum is $ n\\minus{}2\\plus{}\\sqrt{2}$.", "Solution_6": "I think that this is pretty similar to IMO 2021/2. Here we can use shifting to let the biggest number be $1$ or the smallest to be $0$, and then induction will probably finish. " } { "Tag": [ "linear algebra", "matrix" ], "Problem": "$ A*\\begin{pmatrix}\r\n 2&4\\\\\r\n 1&3\r\n \\end{pmatrix}\\equal{}\\begin{pmatrix}\r\n 2&4\\\\\r\n 1&3\r\n \\end{pmatrix}*\\begin{pmatrix}\r\n 1&1\\\\\r\n 11&5\r\n \\end{pmatrix}$\r\n$ A\\equal{}\\begin{pmatrix}\r\n 1&1\\\\\r\n 11&5\r\n \\end{pmatrix}$where A is 2x2 matrix", "Solution_1": "No. it's not true" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find all functions $ f : \\mathbb{R}\\to\\mathbb{R}$ such that for any $ x,y\\in \\mathbb R$,\n\\[ f(x^2 - y^2) = (x - y)\\cdot [f(x) + f(y)].\\]", "Solution_1": "[quote=\"cnyd\"]$ f: R\\rightarrow R$ $ ,$ $ \\forall x,y\\in\\ R$ $ f(x^2 \\minus{} y^2) \\equal{} (x \\minus{} y).[f(x) \\plus{} f(y)]$[/quote]\r\n\r\n[hide=\"My solution\"]\nLet $ P(x,y)$ the assertion $ f(x^2\\minus{}y^2)\\equal{}(x\\minus{}y)(f(x)\\plus{}f(y))$\n\n$ P(x,x)$ $ \\implies$ $ f(0)\\equal{}0$\n$ P(x,0)$ $ \\implies$ $ f(x^2)\\equal{}xf(x)$\n$ P(\\minus{}x,0)$ $ \\implies$ $ f(x^2)\\equal{}\\minus{}xf(\\minus{}x)$ $ \\implies$ $ f(\\minus{}x)\\equal{}\\minus{}f(x)$\n$ P(x,\\minus{}y)$ $ \\implies$ $ f(x^2\\minus{}y^2)\\equal{}(x\\plus{}y)(f(x)\\minus{}f(y))$\n\nSo $ (x\\minus{}y)(f(x)\\plus{}f(y))\\equal{}(x\\plus{}y)(f(x)\\minus{}f(y))$ and so $ xf(y)\\equal{}yf(x)$ and so $ \\frac{f(x)}{x}\\equal{}\\frac{f(y)}{y}$ $ \\forall x,y\\neq 0$\n\nSo $ f(x)\\equal{}ax$ $ \\forall x$ and it is easy to check back that this solution fits the initial equation.\n[/hide]", "Solution_2": "$ \\frac{f(x)}{x}\\equal{}\\frac{f(y)}{y}$ $ \\implies$ $ f(x) \\equal{} ax$ Could you explain better?", "Solution_3": "[quote=\"cnyd\"]$ \\frac {f(x)}{x} \\equal{} \\frac {f(y)}{y}$ $ \\implies$ $ f(x) \\equal{} ax$ Could you explain better?[/quote]\r\n\r\n$ \\forall x\\neq 0$ $ \\frac{f(x)}{x}\\equal{}\\frac{f(1)}{1}$ and so $ f(x)\\equal{}f(1)x$ $ \\forall x$", "Solution_4": "aah thanks :lol: :D" } { "Tag": [ "geometry", "geometric transformation" ], "Problem": "A particle moves through the first quadrant of the figure as follows: during the first minute it moves from the origin to (1,0). Then, it continues to follow the pattern indicated in the figure, going back and forth between the positive x and y axes, moving one unit of distance parallel to an axis in one minute. Which point will the particle reach after exactly 2 hours? \r\n\r\n\r\n[img]http://www.kangurusa.com/clark/pdb/kangur_2005_0708/output/kangur_2005_0708_30.png[/img]", "Solution_1": "Um I'm sorry I can't really understand how the pattern goes, can you clarify that?", "Solution_2": "I think I'm misreading the diagram or the question somehow, but I'll try anyway.\r\n\r\n[hide=\"Solution\"]2 hours is 120 minutes, so 120 moves are made. The pattern tells us that the particle essentially moves from $ (a,b)$ to $ (a\\plus{}4,b\\plus{}2)$ every 22 minutes (if I read your diagram correctly...)\n\n$ 120\\equal{}5\\cdot 22\\plus{}10$, so we will make 5 of these translations from $ (0,0)$. After 110 minutes, the particle is at $ (4\\cdot 5,2\\cdot 5)\\equal{}(20,10)$.\n\nNow we have 10 moves left. Checking the diagram, if we start at $ (a,b)$, we move to $ (a\\plus{}3,b\\plus{}1)$ in 10 steps, so we end up at $ \\boxed{(23,11)}$.[/hide]" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "Prove that a continuous function $ f: M \\rightarrow \\mathbb{R}$, all of whose values are integers, is constant provided that $ M$ is connected.\r\n\r\nIn trying to start this problem as I started looking at $ f^{ \\minus{} 1}(i)$ where $ i \\in \\mathbb{Z}$, is $ f^{ \\minus{} 1}(i)$ open and why?", "Solution_1": "$ f^{\\minus{}1}(\\{i\\})\\equal{}f^{\\minus{}1}(i\\minus{}\\tfrac12,i\\plus{}\\tfrac12).$\r\n\r\nThat's why it's open. And it's also closed, since $ \\{i\\}$ is a closed set.", "Solution_2": "So basically you're saying that for whatever size ball we try to set up around $ i$, that ball still contains $ i$ and nothing else right?", "Solution_3": "There's a theorem in my book that states:\r\n\r\nIf $ M$ is connected, $ f: M \\rightarrow N$ is continuous, and $ f$ is onto, then $ N$ is connected. The continuous image of a connected is a connected.\r\n\r\nIn writing my proof, can I invoke the theorem above. Then say that because the image of $ f$ has to be connected, and that $ \\mathbb{Z}$ is a disjoint union of clopen proper singleton sets (defn. of disconnected), then $ f$ has to have a singleton set as an image, and in our set $ \\mathbb{Z}$ - a constant.\r\n\r\nDo I have to be more specific, or can this suffice?" } { "Tag": [ "function", "Asymptote", "calculus", "calculus computations" ], "Problem": "For any real $a$ find the number of real roots for the equation $x=\\tanh ax$", "Solution_1": "Write the equation as $\\frac{t}{a}=\\tanh t$. Since $t=0$ is always a root, it suffices to look at $t>0$. Two important oservations are: (1) $\\tanh$ is a concave function with a horizontal asymptote; and (2) $(\\tanh t)'=1$ at $t=0$." } { "Tag": [], "Problem": "Ben left at 10:30 from Toronto and arrived in Sudbury at 2:04. He drove from Toronto to Parry Sound on a 4 lane highway at 100km/hr. Then from Parry Sound to Sudbury he drove on a 2 lane highway at 90km/hr. If the distance from Toronto to Parry Sound is 335 km, at what time did he switch from the 4 lane to 2 lane highway?", "Solution_1": "Total time is 3 hours, 34 minutes. $ t\\equal{}\\frac{d}{v}\\Rightarrow 90D\\plus{}100d\\equal{}3.5\\bar{6}$ and $ D\\plus{}d\\equal{}335$ Now we have two equations and two unknowns. Once we solve for $ d$, we have $ 100d\\equal{}t$ so we can get the amount of time to get to the 2-lane highway and just add it to 10:30." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "is this a planar graph? If yes, how does its dual graph look like?\r\n[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=266[/img]", "Solution_1": "It is planar and the dual graph has vertices A, B, C and edges AB, AB, BC, BC, AC, AC. If you want to know how to construct it, look at http://mathworld.wolfram.com/DualGraph.html" } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "can you integrate the following ? :D \r\n\r\n$ \\textbf{(i)}\\;\\;\\;\\int_0^1\\;\\;\\left(\\dfrac{\\sin^{\\minus{}1}\\,x}{x}\\right)^2\\;\\;dx$\r\n\r\n$ \\textbf{(ii)}\\;\\;\\;\\int_0^1\\;\\;\\dfrac{\\sin^{\\minus{}1}\\,x}{\\sqrt{\\;1\\plus{}x^2\\;}}\\;\\;dx$", "Solution_1": "hello, for $ (i)$ it is 4Catalan's constant -$ \\frac{1}{4}\\pi^2$.\r\nSonnhard.", "Solution_2": "After substitution $ x\\equal{}\\sin t$ the integral becomes $ \\int_0^{\\frac{\\pi}{2}} \\left(\\frac{t}{\\sin t}\\right)^2 \\, \\cos t \\; dt$\r\n\r\nAnd this is $ \\int_0^{\\frac{\\pi}{2}} t^2 \\; \\left(\\frac{\\minus{}1}{\\sin t}\\right)' \\, dt\\equal{}\\left[\\frac{\\minus{}t^2}{\\sin t}\\right]_0^{\\frac{\\pi}{2}}\\plus{}\\int_0^{\\frac{\\pi}{2}} \\frac{2t}{\\sin t} \\, dt\\equal{}\\minus{}\\frac{\\pi^2}{4}\\plus{}4\\, \\underbrace{\\int_0^{\\frac{\\pi}{4}} \\frac{2t}{\\sin 2t} \\, dt}_{\\equal{}J}$\r\n\r\nWhereas $ J\\equal{}\\int_0^{\\frac{\\pi}{4}} \\frac{t}{\\sin t \\; \\cos t} \\, dt\\equal{}\\minus{}\\int_0^{\\frac{\\pi}{4}} t \\, \\frac{1}{\\cot t} \\, \\frac{\\minus{}1}{\\sin^2 t} \\, dt\r\n\\equal{}\\underbrace{\\Big[\\minus{}t \\, \\ln\\cot t \\Big]_0^{\\frac{\\pi}{4}}}_{\\equal{}0}\\plus{}\\int_0^{\\frac{\\pi}{4}} \\ln\\cot t \\; dt$\r\n\r\n$ \\equal{}\\int_0^1 \\frac{\\minus{}\\ln u}{1\\plus{}u^2} \\, du\\equal{}\\underbrace{\\Big[\\minus{}\\ln u \\; \\arctan u \\Big]_0^1}_{\\equal{}0}\\plus{}\\int_0^1 \\frac{\\arctan u}{u} \\, du\\equal{}G$ , hence $ I\\equal{}4\\, G\\minus{}\\frac{\\pi^2}{4}$ .", "Solution_3": "After substitution $ x\\equal{}\\sin t$ the second integral $ I$ becomes $ \\int_0^{\\frac{\\pi}{2}} \\frac{t}{\\sqrt{1\\plus{}\\sin^2 t}}\\, \\cos t \\, dt$ .\r\n\r\nAnd this is $ \\int_0^{\\frac{\\pi}{2}} t \\, \\left[\\text{arsinh}(\\sin t)\\right]' \\, dt\\equal{}\\underbrace{\\left[t\\, \\text{arsinh}(\\sin t)\\right]_0^{\\frac{\\pi}{2}}}_{\\equal{}\\frac{\\pi}{2} \\, \\text{arsinh} 1}\r\n\\minus{}\\underbrace{\\int_0^{\\frac{\\pi}{2}} \\text{arsinh}(\\sin t)\\, dt}_{\\equal{}J}$ .\r\n\r\nBecause of $ \\text{arsinh} z\\equal{}\\sum_{n\\equal{}0}^\\infty {\\minus{}\\frac12 \\choose n} \\frac{z^{2n\\plus{}1}}{2n\\plus{}1}\\equal{}\\sum_{n\\equal{}0}^\\infty (\\minus{}1)^n \\, \\frac{1}{2^{2n}} \\, {2n\\choose n} \\, \\frac{z^{2n\\plus{}1}}{2n\\plus{}1}$\r\n\r\n$ J\\equal{}\\sum_{n\\equal{}0}^\\infty (\\minus{}1)^n \\frac{1}{2^{2n}} {2n\\choose n} \\frac{1}{2n\\plus{}1} \\int_0^{\\frac{\\pi}{2}} \\sin^{2n\\plus{}1} t \\; dt$, whereas $ \\int_0^{\\frac{\\pi}{2}} \\sin^{2n\\plus{}1} t\\; dt\\equal{}\\frac{1}{2n\\plus{}1} \\left[\\frac{1}{2^{2n}} {2n\\choose n}\\right]^{\\minus{}1}$ .\r\n\r\nTherefore $ J\\equal{}\\sum_{n\\equal{}0}^\\infty (\\minus{}1)^n \\frac{1}{(2n\\plus{}1)^2}\\equal{}G$ , and hence $ I\\equal{}\\frac{\\pi}{2} \\, \\text{arsinh}\\, 1\\minus{}G$." } { "Tag": [ "inequalities", "function", "logarithms", "trigonometry", "geometry solved", "geometry" ], "Problem": "How to prove that: In any triangle $\\triangle ABC$, \\[8 \\omega ^3 \\le ABC\\], where $\\omega$ is the Brocard angle?", "Solution_1": "Nice problem, in $Prasolov's$ book you can find a nice solution using Jensen(and a couple of other inequalities with Brocard's angle)!", "Solution_2": "Do you mind to post the solution here? I don't have this book... Thanks so much. :)", "Solution_3": "[quote=\"yptsoi\"]How to prove that: In any triangle $\\triangle ABC$, \\[8 \\omega ^3 < ABC\\], where $\\omega$ is the Brocard angle?[/quote]\r\n\r\nYptsoi, I guess you meant the inequality to be $8\\omega^3\\leq ABC$ (the sign is $\\leq$, and equality is obtained for equilateral triangles ABC).\r\n\r\nIn fact, this is the famous [url=http://mathworld.wolfram.com/YffConjecture.html][i]Yff inequality[/i][/url]. Here is the proof as given in Prasolov's book:\r\n\r\nConsider the function $f\\left(x\\right)=\\ln\\frac{x}{\\sin x}=\\ln x-\\ln\\sin x$. The functions\r\n\r\n$f^{\\prime}\\left(x\\right)=\\frac{1}{x}-\\cot x$ and $f^{\\prime\\prime}\\left(x\\right)=\\frac{1}{\\sin^2 x}-\\frac{1}{x^2}$\r\n\r\nare both easily seen to be positive for $02 \\Rightarrow (\\frac{3}{5})^x\\plus{}(\\frac{4}{5})^x <(\\frac{3}{5})^2\\plus{}(\\frac{4}{5})^2\\equal{}1$\r\n +If $ x<2 \\Rightarrow (\\frac{3}{5})^x\\plus{}(\\frac{4}{5})^x >(\\frac{3}{5})^2\\plus{}(\\frac{4}{5})^2\\equal{}1$\r\nSo $ x\\equal{}2$ is the only one solution of the equation.", "Solution_3": "fermat's last theorem says that for any diophantine equation a^x + b^x = c^x, no x for which x>2 produces integer solutions for a, b, and c. the proof is very interesting, but it delves into very deep number theory. check it out.", "Solution_4": "[quote=\"SimonM\"][quote=\"sagar_astakala\"]Can we prove that x = 2 is the only one solution of the equation \n\n3^x + 4^x = 5^x ?[/quote]\n\nYou could claim this is due to Fermat's Last Theorem. [/quote]\n\nThe question is for real $ x$. NguyenDungTN has the correct solution. \n\n[quote=\"Riders on the Storm\"]fermat's last theorem says that for any diophantine equation a^x + b^x = c^x, no x for which x>2 produces integer solutions for a, b, and c. the proof is very interesting, but it delves into very deep number theory. check it out.[/quote]\r\n\r\nInteger $ x$, and solutions such that $ abc \\neq 0$. Wiles' proof is beyond undergraduate study.", "Solution_5": "I think that this trivial solution is also the only one to $ 3^x\\plus{}4^y\\equal{}5^z$ in integers $ x,y,z$.", "Solution_6": "[quote=\"t0rajir0u\"]Integer $ x$, and solutions such that $ abc \\neq 0$. Wiles' proof is beyond undergraduate study.[/quote] \r\n\r\nAnd some postgraduates or lecturers too!", "Solution_7": "[quote=\"t0rajir0u\"][quote=\"SimonM\"][quote=\"sagar_astakala\"]Can we prove that x = 2 is the only one solution of the equation \n\n3^x + 4^x = 5^x ?[/quote]\n\nYou could claim this is due to Fermat's Last Theorem. [/quote]\n\nThe question is for real $ x$. NguyenDungTN has the correct solution. \n\n[quote=\"Riders on the Storm\"]fermat's last theorem says that for any diophantine equation a^x + b^x = c^x, no x for which x>2 produces integer solutions for a, b, and c. the proof is very interesting, but it delves into very deep number theory. check it out.[/quote]\n\nInteger $ x$, and solutions such that $ abc \\neq 0$. Wiles' proof is beyond undergraduate study.[/quote]\r\n\r\nsorry i was confused. i thought you just wanted to prove that it doesnt work for x=3 in general. i think euler had an excellent proof of this." } { "Tag": [ "function", "calculus", "derivative", "calculus computations" ], "Problem": "My question is how to find the criticals numbers for this function (solve for x for this particular function) and find the intervals of increasing and decreasing? \r\n\r\nx^2 - 1/x", "Solution_1": "1st step: find the derivative ($2x+1/x^{2}$)\r\n2nd step: identify the values of $x$ for which the derivative does not exist\r\n3rd step: identify the values of $x$ for which the derivative is equal to $0$. \r\n\r\nThe numbers found in steps 2 and 3 are critical numbers. When you plot then on the real line, you will get a partition of the line into some collection of intervals. The derivative has constant sign within each interval. You can determine the sign by testing some point of each interval. (+) indicates increasing function, (-) decreasing.", "Solution_2": "So in this case how do i solve for x when 2x + 1/x^2. Im not sure what method to use after i recuced it down to 2x^3 + 1 then 2x^3 = -1, im just wondering how do you solve it now?", "Solution_3": "$x^{3}=-\\frac12$, hence $x=-\\frac{1}{2^{1/3}}$", "Solution_4": "Just one more question. When you said plug in values im just wondering which values and which equation, the derivative of the function or the original equation?", "Solution_5": "You want to know the sign of the derivative in each interval such as $(0,\\infty)$, so you pick a point, e.g. $x=1$ and plug it into the derivative. Got a positive number? The derivative is positive, and the original function is increasing." } { "Tag": [ "inequalities", "geometry", "circumcircle", "inradius", "geometry open" ], "Problem": "Let $ \\triangle ABC$. Find the maximal constant $ k$ such that the following inequality holds :\r\n\\[ \\frac {r}{R} \\plus{} k\\frac {(a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2}{R^2}\\le \\frac {1}{2}\\]\r\n[hide] I knew $ k \\ge \\frac {1}{16}$ :( [/hide]", "Solution_1": "1993,I got\nIn $ \\triangle ABC$. the following inequality holds :\n\\[ R\\ge \\22r \\plus{} \\frac {(a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2}{8R}.\\]", "Solution_2": "In $ \\triangle ABC$. the following inequality holds :\\[2r+\\frac {1}{8R}\\sum(b-c)^2\\le R\\le 2r+\\frac {3}{16r}\\sum(b-c)^2.\\]\n\\[2r+\\frac {1}{16R}\\left(\\sum|b-c|\\right)^2\\le R\\le 2r+\\frac {1}{16r}\\left(\\sum|b-c|\\right)^2.\\]", "Solution_3": "Thanks a lot for keeping this problem tuned up over three years.\n\nI have had a solution for $k = 20$ and any constant backwards to 16. \n\nThat still remains hidden to me of how this solution performs after that?\n\nInteresting question then I think? Hopefully some fellows could raise a certain for this and I could close up this 3 year old post.\n\nThanks a lot sqing and have a nice 12/12/12 date.", "Solution_4": "[quote=\"Allnames\"]Thanks a lot for keeping this problem tuned up over three years.\n\nI have had a solution for $k = 20$ and any constant backwards to 16. \n\nThat still remains hidden to me of how this solution performs after that?\n\nInteresting question then I think? Hopefully some fellows could raise a certain for this and I could close up this 3 year old post.\n\nThanks a lot sqing and have a nice 12/12/12 date.[/quote]\na nice 12/12/12 date.\nGood luck.\nThank Allnames.\n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2347259#p2347259:\n that in any $\\triangle\\, ABC$ with side-lengths $a,b,c$ , circumradius $R$ and inradius $r$ the following inequality holds: \n$\\boxed{\\ \\frac 1{16R}\\cdot\\left(\\sum\\, |b-c|\\right)^2+2r\\, \\le\\, R\\, \\le\\, 2r+\\frac 1{16r}\\cdot\\left(\\sum\\, |b-c|\\right)^2\\ }$ ." } { "Tag": [ "function", "logarithms", "complex analysis", "complex analysis unsolved" ], "Problem": "It is know that $ e^{f(z)}$ is an entire function provided $ f(z)$ is also an entire function. My question is as follows: Is $ e^{e^z}$ an entire function? Since $ e^{e^z}$ never equals to $ 0$ and $ 1$ then it contradicts to the Little Picard Theorem, am I wrong? :(", "Solution_1": "It is true that $ e^{e^z}\\ne 0$ but you have to reconsider your idea about $ 1$ (Hint: $ 0$ is not the only possible value of $ w$ for which $ e^w\\equal{}1$ ;) ).", "Solution_2": "Oh thanks fedja :oops: , I did figure out the problem, indeed, $ e^{\\minus{}i\\pi} \\equal{} 1$ and $ e^{\\ln(\\minus{}i\\pi)} \\equal{} \\minus{}i \\pi$." } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "O is a point inside \u25b3ABC,OA=OB+OC.R,S are inside \u25b3ABO,ACO. A,R,O,C are concyclic and AR=RC A,B,O,S are concyclic and AS=BS \r\nProve: the circumcircles of \u25b3ROB \u25b3SOC have only one common inner tangent", "Solution_1": "here is my solution:\r\n[url]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=15715[/url]", "Solution_2": "Thank you,dear mr.danh :lol: \r\nI know nothing about invertion,but I will try to understand" } { "Tag": [ "factorial", "induction" ], "Problem": "Suppose that any $ n\\in \\mathbb{N}$ and $ n\\ge 2$ ,\r\n\r\nprove that $ 2!4!6!\\cdots(2n)! >\\left((n+1)!\\right)^n$", "Solution_1": "Can this problem be done with induction??\r\n\r\nFor $ n\\equal{}2$, $ 2!.4!\\equal{}48>(3!)^2\\equal{}36$\r\nNow assume the statement is true for $ n\\equal{}k$\r\n$ 2!.4!.\\dots.(2k)!>((k\\plus{}1)!)^k$\r\n\r\nThen it follows that \r\n$ 2!.4!.\\dots (2k)!.(2k\\plus{}2)!>(2k\\plus{}2)!((k\\plus{}1)!)^k$ (by assumption)\r\n$ RHS>((k\\plus{}1)!)^{k\\plus{}1}$\r\nThis follows from $ (2k\\plus{}2)!>(k\\plus{}1)!$, so $ (2k\\plus{}2)!((k\\plus{}1)!)^k>((k\\plus{}1)!)^k.(k\\plus{}1)!\\equal{}((k\\plus{}1)!)^{k\\plus{}1}$\r\nWhich means the statement is true for $ n\\equal{}k\\plus{}1$\r\nThen, it's proved by induction\r\n\r\nI'm not sure that my solution is correct. I just learnt how to use induction :P", "Solution_2": "[quote=\"IW@IT\"]Which means the statement is true for $ n \\equal{} k \\plus{} 1$[/quote] Nope -- you need the RHS of the final expression to be $ ((k \\plus{} {\\bf 2})!)^{k \\plus{} 1}$, so the comparison you did between $ (2k \\plus{} 2)!$ and $ (k \\plus{} 1)!$ needs to be between $ (2k \\plus{} 2)!$ and $ (k \\plus{} 2)^{k \\plus{} 1} (k \\plus{} 1)!$. This is a bit trickier to argue than what you did, but not terribly much :)" } { "Tag": [ "calculus", "induction", "geometric series", "real analysis", "real analysis unsolved" ], "Problem": "I need to show the collection of positive integers without 0 in their decimal expansion sum to a number less than 90. Is this a variation of the Cantor set and how do I start this problem?\r\n\r\nThanks.", "Solution_1": "$91$ has no $0$ in its decimal expansion and $91>90$ ? I think I misunderstood the problem ...", "Solution_2": "No. The problem is this classic (which I heard in my calculus class 35 years ago):\r\n\r\nLet $S=\\{n\\in \\mathbb{N}:$ the decimal repesentation of $n$ does not contain the digit $0\\}.$\r\n\r\nProve that $\\sum_{n\\in S}\\frac1n\\le 90.$\r\n\r\nThis is \r\n\r\n$1+\\frac12+\\cdots \\frac19+\\frac1{11}+\\cdots+\\frac1{99}+\\frac1{111}+\\cdots$\r\n\r\n[toadsmack7 left out the word \"reciprocal.\" Without that word, the problem is nonsense.]\r\n\r\nI suppose you could find a relationship to a Cantor set if you tried, but you don't have to and that may have little to do with solving the problem.", "Solution_3": "Yes, sorry about that. Kent has the formula right. Any ideas on how to solve this??", "Solution_4": "It seemed like that the most reasonable approach would be to consider numbers with the same number of digits at a time. \r\n\r\nSo for one digit we get: \\[\\sum_{k=1}^{9}\\frac{1}{k}\\] .\r\n\r\nThe reason we think about number of digits at a time is if we think about the \"chance\" that a number has no zeroes, it is something like $.9^{n-1}$ where $n$ is the number of digits (the first digit cannot be zero). So it seems like we might be able to get a nice geometric series that we can actually compute for. The only question is which $.9^{n-1}$ of the total sum do we keep and which do we throw out? Well out of the numbers $\\frac{1}{10}, \\dots, \\frac{1}{19}$ we throw out $\\frac{1}{10}$. This is the biggest one of these $10$. This is promising, we throw out $20^{-1}$ out of the next ten, etc. etc. In other words, we throw the \"biggest\" tenth in some sense. So we expect that we will be able to bound the entire sum by: \\[\\sum_{n=0}^{\\infty}(\\frac{9}{10})^{n}\\sum_{k=10^{n}}^{10^{n+1}-1}\\frac{1}{k}\\] So now we actually establish this inductively. \r\n\r\nActually, we don't since I have a headache. But the induction is not very tricky. Notice also that the bound should be not bad since the approximation gets pretty much arbitrarily good as we go along. So now to bound this sum, we pull out the handy dandy bound of $\\log(x) < \\sum_{k=1}^{x}\\frac{1}{k}< \\log(x)+1$ as usual (just look at what happens when we integrate). So our bound is going to be about $(\\log(10)+1) \\sum_{n=0}^{\\infty}(\\frac{9}{10})^{n}= 30,40$ ish. Hmm, why the 90 then? Did I make some mistake?", "Solution_5": "The problem has been posted and solved before. Here is one link:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=104330[/url].", "Solution_6": "Haha, there are so many easier ways. I guess I should have realized that $90 = 9+9\\frac{9}{10}+9(\\frac{9}{10})^{2}+\\cdots$. Yay for cryptic comment." } { "Tag": [ "videos" ], "Problem": "I found a fun video forum. Videos are update regularly. See http://www.funnywebvideos.com/", "Solution_1": "The page looks pretty good.....only thing, it seems to take quite some time to load...", "Solution_2": "I was quite surprised to see the man who changed the history of skateboarding can stiil laugh after that accident......", "Solution_3": "the dude scaling the wall was amazing" } { "Tag": [], "Problem": "Fie un cerc de centru $O$ si raza $r$, $AB$ diametru. $P$ un punct fixat pe $AB$.\r\nFie $M$ si $N$ doua puncte pe cerc, $P$ apartine lui $MN$.\r\nSa se afle pozitia punctelor $M$ si $N$ pe cerc astfel incat aria lui $AMBN$ sa fie maxima.", "Solution_1": "[quote=\"dina013\"][color=darkred][b]Fie un punct fix $P$ care apartine unui diametru fix $AB$ al cercului $w$. Notam doua puncte mobile $\\{M,N\\}\\subset w$ pentru care $P\\in MN$.\nSa se stabileasca pozitiile punctelor $M$ si $N$ pentru care aria patrulaterului $AMBN$ este maxima.[/b][/color][/quote]\n[color=darkblue][b][u]Demonstratie.[/u][/b] Se observa ca $[AMBN]=[ABM]+[ABN]=\\frac{AB}{OP}\\cdot [OPM]+\\frac{AB}{OP}\\cdot [OPN]=$ $\\frac{AB}{OP}\\cdot ([OPM]+[OPN])=$ $\\frac{AB}{OP}\\cdot [OMN]$.\nAsadar, aceasta problema se reduce la urmatoarea problema cunoscuta (vezi o carte mai veche, [b]\"Maxime si minime\"[/b] de [u]Ion Ionescu[/u]) :[/color] \n\n[quote][color=darkred][b]Fie un punct fix $P$ din interiorul cercului dat $w=C(O)$. Notam doua puncte mobile $\\{M,N\\}\\subset w$ pentru care $P\\in MN$.\nSa se stabileasca pozitiile punctelor $M$ si $N$ pentru care aria triunghiului $MON$ este maxima.[/b][/color][/quote]\r\n\r\n[color=darkblue][b]P.S.[/b] Voi reveni curand tot in acest mesaj (sunt chemat de sus la masa de pranz !).[/color]" } { "Tag": [], "Problem": "The problem marathon posted by Silverfalcon has not had enough visitors recently. If you have an interesting problem that you think would help others, please answer the previous question and post it. Thanks. :)", "Solution_1": "hm... not sure what that is... can you post a link?", "Solution_2": "mabye if some of us link the basics marathon to our signature, we'd get some more hits...... I'll consider doing it......", "Solution_3": "i might do it", "Solution_4": "I love marathons!", "Solution_5": "math92, could you do that?" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Serre says that for a finite number of elements $ q_1,...,q_n \\in \\bf{Q}$ and $ \\epsilon_1,...,\\epsilon_n \\in \\{ \\minus{} 1, \\plus{} 1\\}$ one can find an element $ x \\in \\bf{Q}$ such that the Hilbert symbol satisfies $ (x,q_i) \\equal{} \\epsilon_i$. What goes wrong if we want to do this for infinitely many $ q_i$?", "Solution_1": "Principielly it's the same as for the Chinese remainder theorem. You construct it in some way, and this is a purely algebraic construction, so no limit-like things can be done (in both theorems)." } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "Including the endpoints, how many ordered pairs on the line segment with endpoints $ (0,0)$ and $ (16,8)$ have both integer coordinates?", "Solution_1": "the first thing we should see is that this line has slope $ \\frac{1}{2}$\r\n\r\nfrom there, we can see that for every $ y$, there will be an $ x$ that has an integer value\r\n\r\nso, we just find how many $ y$'s there are, and we will have our answer\r\n\r\nwe have the integers $ 0\\minus{}8$ for $ y$ values, so there are $ 9$ ordered pairs that have integer coordinates" } { "Tag": [ "Gauss", "quadratics", "number theory unsolved", "number theory" ], "Problem": "Let$ n $ be natural number and $n=p_1p_2...p_k$ and $p_i$ are odd prime numbers.prove that:\r\n\r\n1- $\\frac{p_1^2-1}{8}+...+\\frac{p_k^2-1}{8}\\equiv \\frac{(p_1p_2...P_k)^2-1}{8}$ $(\\mod 2)$\r\n\r\n\r\n2- $\\left(\\frac{2}{n}\\right)=(-1)^{\\frac{n^2-1}{8}}$", "Solution_1": "second comes straight from the Gauss lemma for the Jacobi symbol", "Solution_2": "[quote=\"Mathx\"]Let$ n $ be natural number and $n=p_1p_2...p_k$ and $p_i$ are odd prime numbers.prove that:\n\n1- $\\frac{p_1^2-1}{8}+...+\\frac{p_k^2-1}{8}\\equiv \\frac{(p_1p_2...P_k)^2-1}{8}$ $(\\mod 2)$\n\n\n2- $\\left(\\frac{2}{n}\\right)=(-1)^{\\frac{n^2-1}{8}}$[/quote]\r\n\r\nFrom the Gauss lemma we now that $\\left(\\frac{2}{p}\\right)=(-1)^{\\frac{p^2-1}{2}}$ for odd $p$ prime.\r\n\r\nNow let's divide primes on these of the form $8k+1,8k+3,8k+5,8k+7$.\r\nThus for $p=8k+1$ : $\\frac{p^2-1}{8}= 0$ $(\\mod 2)$\r\n $p=8k+3$ : $\\frac{p^2-1}{8}= 1$ $ (\\mod 2)$\r\n $p=8k+5$ : $\\frac{p^2-1}{8}= 1$ $(\\mod 2)$\r\n $p=8k+7$ : $\\frac{p^2-1}{8}= 0 $ $(\\mod 2)$\r\nOn the right side we have $\\frac{(p_1p_2...p_k)^2-1}{8} = \\frac{7^{2l_1}5^{2l_2}3^{2l_3} -1}{8} = \\frac{9^{l_2+l_3}-1}{8}$ $(\\mod 2)$, where\r\n$l_1,l_2,l_3$ are exponents of respectively - $7,5,3$ and where $l_2+l_3$ is the number of prime divisors of the form $8k+3$ and $8k+5$.\r\n\r\nIf the sum of numbers of primes of the form $p=8k+3$,$p=8k+5$ is even then the left side is $0$ $(\\mod 2)$.\r\nThe right side is $\\frac{81^n-1}{8} = 0$ $(\\mod 8)$ (cause $\\frac{80}{8}= 0$ $(\\mod 2)$, where $n$ is a positive integer.\r\nIf the sum of primes of the form $p=8k+3,p=8k+5$ is odd then the left side is $1$ $(\\mod 2)$.\r\nThe right side is $\\frac{9*81^n-1}{8}=\\frac{9(81^n-1)}{8}+1=1$ $(\\mod 2)$.\r\nThus we always have\r\n$\\frac{p_1^2-1}{8}+...+\\frac{p_k^2-1}{8}\\equiv \\frac{(p_1p_2...P_k)^2-1}{8}$ $(\\mod 2)$, so\r\n$\\left(\\frac{2}{n}\\right)=\\prod_{1 \\leq i \\leq k}\\left(\\frac{2}{p_i}\\right)=(-1)^{\\frac{p_1^2-1}{8}+...+\\frac{p_k^2-1}{8}}=(-1)^{\\frac{(p_1p_2...P_k)^2-1}{8}}$ (applying the quadratic reciprocity theorem - $\\left(\\frac{p_i}{2}\\right)=1$, q.e.d." } { "Tag": [ "Asymptote" ], "Problem": "How can I use (x,y,z) i.e. 3-dimensional points in asymptote?\r\n\r\n[asy]draw((0,0,0)--(1,1,1));[/asy]\r\n\r\nThat doesn't work but that is the sort of hting I want to do. How would I do it?", "Solution_1": "I believe this would work:\r\n\r\n[code]\n[asy]\nimport three;\ntriple stuffgoeshere=(0,0,0);\ndot(stuffgoeshere, red);\ndraw(O--(1,1,1), blue);\n[/asy]\n[/code]\r\n\r\n[asy]import three;\ntriple stuffgoeshere=(0,0,0);\ndot(stuffgoeshere, red);\ndraw(O--(1,1,1), blue);[/asy]\r\n\r\nOf course, you could make much more complicated stuff than that, but its just to get you started. I hope you understand the code.", "Solution_2": "[asy]import three;\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);[/asy]\r\nHm thank you very much." } { "Tag": [ "percent" ], "Problem": "The BigTown Outdoors Club recently held an election for new officers. The turnout for the special election meeting was good. Of the 700 members, 541 showed up to vote. To the nearest tenth of a percent, what percent of the members voted in the special election?", "Solution_1": "can we use a calculator?\r\n[hide]77.3%[/hide]", "Solution_2": "I guess you can use a caculator.", "Solution_3": "that's what i did", "Solution_4": "[hide]541/700*100% ~ 77.3%[/hide][/hide]", "Solution_5": "[hide]541/700*100=77.3% (rounded to the nearest tenth)[/hide]", "Solution_6": "[quote=\"colts18\"]The BigTown Outdoors Club recently held an election for new officers. The turnout for the special election meeting was good. Of the 700 members, 541 showed up to vote. To the nearest tenth of a percent, what percent of the members voted in the special election?[/quote]\r\n\r\n[hide]541/700=.7728=77.28%[/hide]", "Solution_7": "[hide=\"to jli\"]it says to the nearest tenth percent. your answer is to the hundreth percent ;) [/hide]", "Solution_8": "[hide]77.3%, 541/700 = 77.3%[/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a, b, c$ be positive numbers.Prove:\r\n$ \\frac{a^3}{a^2\\plus{}2bc}\\plus{}\\frac{b^3}{b^2\\plus{}2ac}\\plus{}\\frac{c^3}{c^2\\plus{}2ab}\\ge\\frac{a^3\\plus{}b^3\\plus{}c^3}{a^2\\plus{}b^2\\plus{}c^2}$", "Solution_1": "$ (b\\minus{}c)^2\\ge 0 \\Leftrightarrow a^2\\plus{}b^2\\plus{}c^2\\ge a^2\\plus{}2bc\\Leftrightarrow \\dfrac{a^3}{a^2\\plus{}b^2\\plus{}c^2}\\le \\dfrac{a^3}{a^2\\plus{}2bc}$.\r\nAplying cyclic sum, we get the result", "Solution_2": "It's very easy :(.\r\nto be continue\r\n$ Let: a,b,c \\ge 0$\r\nch\u1ee9ng minh:\r\n$ \\sum\\frac{a}{b\\plus{}c} \\plus{}\\frac{3abc}{2(a^2b\\plus{}b^2c\\plus{}c^2a)} \\ge 2$\r\n\r\nP/s:From maths.vn", "Solution_3": "How about Titu's Lemma?" } { "Tag": [ "ratio", "geometry", "vector", "geometry solved" ], "Problem": "Given two triangles and such that the lengths of the sides of the first triangle are the lengths of the medians of the second triangle. Determine the ratio of the areas of these triangles.", "Solution_1": "The ratio is 3/4; we just construct the triangle having the medians as sides by formung some parallelograms. This is possible because the sum of the vectors AM, BN, CP (where M, N, P are midpts of BC, CA, AB) is 0, so there is a triangle having sides equal to AM, BN, CP and parallel to these.", "Solution_2": "Indeed Grobber !" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Prove that among sixteen consecutive integers it is always possible to find one that is relatively prime to all the rest.", "Solution_1": "I don't think this belongs in the MATHCOUNTS forum", "Solution_2": "Hmm....this looks familiar. Exactly the same as number 21b from the midterm. \r\n\r\nI have no clue how to do it." } { "Tag": [ "geometry", "perimeter" ], "Problem": "Each side of hexagon ABCDEF has a length of at least 5 cm and AB = 7 cm. How many centimeters are in the least possible perimeter of hexagon ABCDEF?", "Solution_1": "[hide]We set $ \\overline{AB}\\equal{}7$ and let all the other sides have length $ 5$. Then, \\[ 5\\cdot5\\plus{}7\\equal{}\\boxed{32}\\]\n$ \\Box$[/hide]" } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Hi community ,\r\n i'm paining with this little problem.\r\n\r\n[quote]let ABC an isocele triangle of apex(vertex) B and M be an arbitrary point of the circumcircle $ \\Gamma$ of ABC .\nlet H be the orthogonal projection of C on the line (BM) et I the point of intersection of the two lines (CH) and (AM).\nWhat is the locus (geometric place) of I when M describes $ \\Gamma$\t?[/quote]\r\n\r\ni can see that this is gonna be :\r\n[hide]the circle of center B and whose radius is the common measure of BA and BC [/hide]\r\n\r\ni've drawn this figure with GeoGebra ; i hope it helps :\r\n\r\n[img]http://safaelmarwa.googlepages.com/olymp5fig2.PNG[/img]\r\n\r\nbut i didn't succeed in proving it .\r\n\r\ni need your help.\r\n\r\nthanks.", "Solution_1": "[hide=\"note\"]By angle chasing, it's easy to see that $ \\angle AIC \\equal{} 1/2\\angle ABC$[/hide]", "Solution_2": "Indeed man , on this arch it's true but i don't see this holds when M lies on one of the two other archs ? \r\nBesides :maybe: i'm still unsure about the case of M = B , is the line (BM) then reduced to the point B ? or is it the tangent to the circle at B ? \r\nWith Geogebra software , dragging M will drag I and show you a bit what 'll happen , but the result is a bit weird ..\r\nAny more suggestions to completely solve this problem ?", "Solution_3": "Denote N is midpoint of AC then we have NH is the mid-line of AI which means NH=1/2AI and MH//AI", "Solution_4": "[quote=\"plane geometry\"]Denote N is midpoint of AC then we have NH is the mid-line of AI which means NH=1/2AI and MH//AI[/quote]\r\n\r\nI can't see what you're talking about :oops: \r\n\r\nHey community , F1 is still pressed here :huh: ?" } { "Tag": [ "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "probably it's standard, but i found it just nice (and i could be interested in \"more elementary\" solutions)..\r\n\r\nprove that $ \\prod_{k\\equal{}0}^\\infty \\mathbb{Z}$ is not projective (as a $ \\mathbb{Z}$-module).", "Solution_1": "$ \\mathbb{Z}$ PID => (free = projective). and it is not free.", "Solution_2": "i didn't find it obvious to prove that it's not.\r\nat least, i don't actually see it, i just needed to show it.\r\n\r\nby the way, isn't there way to prove it straight by definition, without involving the theorem? i used that too, but professor gave us this exercise before stating the theorem." } { "Tag": [ "probability", "algorithm" ], "Problem": "I am a visual basic fanatic. Here we discuss Visual Basic 5.0/6.0 Programming.", "Solution_1": "Even though I dont like visual basic I took it last term in college. With hard work, I managed to get an A. What do you find so interesting about it?", "Solution_2": "Its interesting because think about some probability problems. You can program to simulate the condition and do lets say 100000 trials and you can find the probability by brute forcing.", "Solution_3": "Heh, You know what Dijkstra said about BASIC, don't you?\r\n[hide]It is practically impossible to teach good programming to students that have had a prior exposure to BASIC: as potential programmers they are mentally mutilated beyond hope of regeneration\n[/hide]", "Solution_4": "Who's Dijkstra?", "Solution_5": "I think Visual C++.NET is better. :P", "Solution_6": "I agree...\r\n\r\nBut visual basic is a programming language that really doesnt have a finite structure to it. Kind of like HTML! You can get away with doing almost anything and the code will run fine.\r\n\r\nYou should get into programming a bit more advanced that VB, C++.NET is a lot more capable and cleaner that VB.\r\n\r\nThe only problem is that you develop a crutch after you program in it...just cuz of the nature of the language.", "Solution_7": "[quote=\"henryyangrui\"]Who's Dijkstra?[/quote]\r\nDijkstra was a famous Computer Scientist (He died in 2002, I believe). His single source shortest path finding algorithm (Known as Dijkstra's algorithm) is why he is so famous (Though he may have also probably found many other famous algorithms which I may not be knowing of). See his wikipedia page here: [url]http://en.wikipedia.org/wiki/Edsger_Dijkstra[/url]" } { "Tag": [ "search", "trigonometry", "rotation", "algebra", "function", "domain", "analytic geometry" ], "Problem": "I'm not realy a math type person, But if i know the steps in solving a problem generaly it isn't a problem. Now the reason im posting is because i dont know the steps in solving this one, Nor how to search for or describe the problem. I can however. represent it visualy as follows. I know it's not the best drawn but hopefully someone with a kind heart can lend me a hand. I feel kind if like i should know this, and feel a bit silly needing to get help :oops: . I figured this was the appropriate section based on my own expirence with this type of problem.\r\n\r\n[img]http://img132.imageshack.us/img132/746/untitledxw5.png[/img]", "Solution_1": "The green angle looks like 90 degrees.\r\n\r\nIf you were talking about the triangles angle it's simple,\r\n\r\n$\\tan \\theta = \\frac{5}{4}$\r\n$\\theta = \\tan^{-1}(\\frac{5}{4})$", "Solution_2": "Also, you can use the law of cosines.\r\n\r\n$c^{2}=a^{2}+b^{2}-2ab\\cos{C}$", "Solution_3": "The blue object rotates, And im looking for the angle between its front and the bottom of the triangle. It looks near 90 degrees in this example but it changes as the object is rotated... \r\n\r\nAlso the green was drawn incorrectly, sorry, I need the equation for finding this as the blue rotates around its point.\r\n\r\n[img]http://img140.imageshack.us/img140/8656/untitled2rn0.png[/img]", "Solution_4": "How fast is the blue object rotating? (I'm guessing it's rotating in the clockwise direction and you want the angle between the horizontal [from the triangles leg] and the blue line).\r\n\r\nIs this for a game by any chance?", "Solution_5": "It isnt rotating at any particular speed, and i don't think we need to incoperate speed, i need only calculate the angle at the time of calculation. I't may be rotating in either direction, but i don't think it's relevent if im only calculating it as a current position.\r\n\r\nWhy do you ask if it is for a game?", "Solution_6": "Let $\\angle G$ be that green angle you drew. We know that complementary angle to the left of it is $tan\\frac{5}{4}$. If that blue cross has rotated $x^\\circ$ clockwise then we know that $x$ has the domain $[0,tan\\frac{5}{4}]$. So we have\r\n\\[\\angle G=90-(tan\\frac{5}{4}-x) \\]\r\nas your formula.", "Solution_7": "Can you provide the information you know (such as coordinate of the top of the blue line with respect to the center), because I'm unsure as to what I can use to base calculations upon. (if that makes sense..)\r\n\r\nI was just curious :D", "Solution_8": "Say the center of the object is 0 (likley wont be but does this matter?)\r\n\r\nAnd the point at the top of the triangle is at -4,5 or so, These are just random... Mabey i should attempt to find away to express it more clearly.\r\n\r\nAs for it having to do with a game, Not exactly..\r\n\r\n\r\n\r\nEdit: To be more broad, I need to make the object face a givin point, They can both be anywhere nearby one another on the plane (within 20 or so units)" } { "Tag": [ "floor function", "integration" ], "Problem": "Evaluate\r\n\\[ \\left\\lfloor \\sum_{n = 1}^{10^9} n^{- 2 / 3} \\right\\rfloor \\]", "Solution_1": "Note that \r\n\r\n$\\int^N_1 n^{-2/3} \\ dn < \\sum^N_{n=1} n^{-2/3} < 1 + \\int^N_1 n^{-2/3} \\ dn$\r\n\r\n$\\Rightarrow 2997 < \\sum^{10^9}_{n=1} n^{-2/3} < 2998$\r\n\r\nso $\\left\\lfloor \\sum_{n = 1}^{10^9} n^{- 2 / 3} \\right\\rfloor = 2997$." } { "Tag": [], "Problem": "convert cyclo heptane---------> methyl benzene", "Solution_1": "Well, plz tell me if dis is fine:\r\n\r\ncycloheptane ---------> Br2 (free radical path) bromocyclo heptane \r\nthen Dehydrohalogenation using sm alc KOH.\r\nThen Ozonolyse it.\r\nThen Wolff Kishner Reduction.\r\nLastly, Aromatization with Mo2O3 or V2O5 supported by Al2O3 at 10-20 atm and 773K.", "Solution_2": "wont the last step form a seven membered aromatic system.....\r\n\r\n\r\n\r\nalso...\r\ni think this conversion works out in just 1 step..", "Solution_3": "This conversion is kind of ridiculous... \r\n\r\nAnyway, just use a suitable metal catalyst, or a transition metal salt like $ \\ce{RhCl3}$.", "Solution_4": "Why ridiculous???", "Solution_5": "1. No one would ever think of making toluene from cycloheptane, since toluene is readily available in every laboratory.\r\n\r\n2. Conversion of saturated carbon rings to benzene rings is seldom used in synthesis. Perhaps the rare examples start with some ring system that have already some insaturation.", "Solution_6": "I dont think it gives a seven member aromatic ring. I think toluene is formed. i ve seen it in sm book. ", "Solution_7": "What seven membered aromatic ring could form?", "Solution_8": "u are right Euclidean Geometer....\r\n\r\nIt forms the required compound only...\r\n\r\nBut this reaction can be directly done by aromatization i.e treating cycloheptane in a red hot iron tune....(not actually mine--chemrock's method)....\r\n\r\nMy method is trating cycloheptane with Se and we get toluene as the product", "Solution_9": "exactly the topic i was gonna start :D \r\nCarcul u once asked me for a referrence to convert cyclohexane into benzene using Pd/Se/C but i cannot find it. But that was what my prof told me. Do u know the mechanism for that? Is it by adsorption or sth related to surface chemistry? :maybe: . Also are there other reagents (other than using allylic chlorination?) :)", "Solution_10": "Advice: don't waste your time with that. Such processes have at best only an academic interest.", "Solution_11": "i know this was to come ....i didn't want to disturb the discussion...but having been here for more than 1.5 year this was tentative :D" } { "Tag": [ "algebra", "polynomial", "quadratics", "quadratic formula", "system of equations" ], "Problem": "1,1,2,3,5,8,13,21,34,55,89,144,233,....\r\n\r\nIs there a formula for this so as called the \"Fibonnaci Sequence\"\r\n\r\n :rotfl: :blush:", "Solution_1": "Yes...Binet's Formula: $F_{n}=\\frac{1}{\\sqrt{5}}[(\\frac{1+\\sqrt{5}}{2})^{n+1}-(\\frac{1-\\sqrt{5}}{2})^{n+1}]$... :|", "Solution_2": "The question's been discussed (and locked, too...)\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130146[/url]", "Solution_3": "There's a few formulas.\r\n\r\nBinet's is the well known one.\r\nAnother is the general recursive relationship:\r\nLet $F_{n}$ be defined as the Nth Fibonacci number.\r\nDefine $F_{0}= 1$ and $F_{1}= 1$\r\n\r\nThen the general recursive relation is $F_{n}= F_{n-2}+F_{n-2}$\r\nOr, $F_{n+2}=F_{n+1}+F_{n}$\r\n\r\nHere's how we'd go about solving this for Binet's Formula (beware that this is well above MC Level, but I'll put it in for anyone who is interested:\r\n\r\nFor all linear recursions, we can generate a characteristic polynomial that defines the recursion. For example, the polynomial for the Fibonacci numbers is \r\n$x^{2}= x+1$\r\n(This comes from the fact that the Nth term becomes $x^{0}$, and the N+1th term becomes $x^{1}$)\r\n\r\nRearranging this equation, we get\r\n$x^{2}-x-1 = 0$\r\n\r\nNow we can set up a general recurrence formula, in the form \r\n$F_{n}= A\\cdot(R_{1})^{n}+B\\cdot(R_{2})^{n}$, where A and B are constants, and $R_{1}$ and $R_{2}$ are the roots of the characteristic polynomial.\r\n\r\nThe roots of the above eqn, by the Quadratic Formula, are\r\n$\\frac{1 \\pm \\sqrt{5}}{2}$ Call this $\\phi$\r\n\r\nThus, we can sub these in, and set up a system of equations:\r\n$F_{0}= A(\\phi)^{0}+B(\\phi)^{0}$\r\n$F_{1}= A(\\phi)^{1}+B(\\phi)^{1}$\r\n\r\n$1 = A+B$\r\n$1 = A\\phi+B\\phi$\r\n\r\nSolving this yields values for A and B, which leads us to the explicit formula for the Fibonacci Series (Binet's Formula).\r\n\r\nHere's a practice problem to try.\r\n\r\nThere is a recurrence relation defined as $F_{n}= 3F_{n-1}-2F_{n}$ with initial values $F_{0}= 2$, $F_{1}= 3$\r\n\r\n[hide=\"Step One: Define the characteristic polynomial\"]\n$F_{n+2}= 3F_{n+1}-2F_{n}$\n$x^{2}= 3x-2$\n$x^{2}-3x+2 = 0$[/hide]\n\n[hide=\"Step Two: Find the Roots\"]\n$(x-2)(x-1) = 0$\n$x = 2; 1$[/hide]\n\n[hide=\"Step Three: Set up the recurrence relation\"]\n$F_{n}= A(r_{1})^{n}+B(r_{2})^{n}$\n$F_{n}= A(2)^{n}+B(1)^{n}$\n$F_{n}= A(2)^{n}+B$[/hide]\n\n[hide=\"Step Four: Set up system of equation to find A and B\"]\n$2 = A(2)^{0}+B$\n$3 = A(2)^{1}+B$[/hide]\n\n[hide=\"Step Five: Solve system, getting values of constants A and B\"]\n$2 = A+B$\n$3 = 2A+B$\n$A = 1; B = 1$[/hide]\n\n[hide=\"Step Six: Finish the explicit formula\"]\n$F_{N}= 2^{n}+1$[/hide]", "Solution_4": "okay guys i was using the above formulas to make a program to find the nth term in the fibonnacci sequence. they didn't work and they were off in terms of the signs. so i went to wolframs math world and i got binets formula and it really does work\r\n\r\nhere is the binets formula that i obtained from math world\r\n\r\n{(1+sqrt50^N)-(1-sqrt5)^N} divided by(2^N times sqrt 5)\r\n\r\nN represents the term in the sequence, e.g. first term, third term", "Solution_5": "[quote=\"learner1729\"]okay guys i was using the above formulas to make a program to find the nth term in the fibonnacci sequence. they didn't work and they were off in terms of the signs. so i went to wolframs math world and i got binets formula and it really does work\n\nhere is the binets formula that i obtained from math world\n\n{(1+sqrt50^N)-(1-sqrt5)^N} divided by(2^N times sqrt 5)\n\nN represents the term in the sequence, e.g. first term, third term[/quote]\r\n\r\nSorry to dissappoint you but that was [u]just[/u] posted. Please do not post things that have already been posted.", "Solution_6": "all right but mine is a whole lot easier to understand, just one variable to plug in", "Solution_7": "and how many are in the first 1?...", "Solution_8": "i don't understand what u mean", "Solution_9": "who r u????????venky??????", "Solution_10": "[quote=\"DiscreetFourierTransform\"]There's a few formulas.\n\nBinet's is the well known one.\nAnother is the general recursive relationship:\nLet $F_{n}$ be defined as the Nth Fibonacci number.\nDefine $F_{0}= 1$ and $F_{1}= 1$\n\nThen the general recursive relation is $F_{n}= F_{n-2}+F_{n-2}$\nOr, $F_{n+2}=F_{n+1}+F_{n}$\n\nHere's how we'd go about solving this for Binet's Formula (beware that this is well above MC Level, but I'll put it in for anyone who is interested:\n\nFor all linear recursions, we can generate a characteristic polynomial that defines the recursion. For example, the polynomial for the Fibonacci numbers is \n$x^{2}= x+1$\n(This comes from the fact that the Nth term becomes $x^{0}$, and the N+1th term becomes $x^{1}$)\n\nRearranging this equation, we get\n$x^{2}-x-1 = 0$\n\nNow we can set up a general recurrence formula, in the form \n$F_{n}= A\\cdot(R_{1})^{n}+B\\cdot(R_{2})^{n}$, where A and B are constants, and $R_{1}$ and $R_{2}$ are the roots of the characteristic polynomial.\n\nThe roots of the above eqn, by the Quadratic Formula, are\n$\\frac{1 \\pm \\sqrt{5}}{2}$ Call this $\\phi$\n\nThus, we can sub these in, and set up a system of equations:\n$F_{0}= A(\\phi)^{0}+B(\\phi)^{0}$\n$F_{1}= A(\\phi)^{1}+B(\\phi)^{1}$\n\n$1 = A+B$\n$1 = A\\phi+B\\phi$\n\nSolving this yields values for A and B, which leads us to the explicit formula for the Fibonacci Series (Binet's Formula).\n\nHere's a practice problem to try.\n\nThere is a recurrence relation defined as $F_{n}= 3F_{n-1}-2F_{n}$ with initial values $F_{0}= 2$, $F_{1}= 3$\n\n[hide=\"Step One: Define the characteristic polynomial\"]\n$F_{n+2}= 3F_{n+1}-2F_{n}$\n$x^{2}= 3x-2$\n$x^{2}-3x+2 = 0$[/hide]\n\n[hide=\"Step Two: Find the Roots\"]\n$(x-2)(x-1) = 0$\n$x = 2; 1$[/hide]\n\n[hide=\"Step Three: Set up the recurrence relation\"]\n$F_{n}= A(r_{1})^{n}+B(r_{2})^{n}$\n$F_{n}= A(2)^{n}+B(1)^{n}$\n$F_{n}= A(2)^{n}+B$[/hide]\n\n[hide=\"Step Four: Set up system of equation to find A and B\"]\n$2 = A(2)^{0}+B$\n$3 = A(2)^{1}+B$[/hide]\n\n[hide=\"Step Five: Solve system, getting values of constants A and B\"]\n$2 = A+B$\n$3 = 2A+B$\n$A = 1; B = 1$[/hide]\n\n[hide=\"Step Six: Finish the explicit formula\"]\n$F_{N}= 2^{n}+1$[/hide][/quote]\r\n\r\ni'm confused\r\n\r\nwhat is the $F_{N}= 2^{n}+1$ for?\r\n\r\nnot for fibonacci i think. i didn't work.\r\n\r\njorian", "Solution_11": "it's for the recursion he defined.", "Solution_12": "it was too complex so i had a hard time trying to understand it\r\n\r\n-jorian" } { "Tag": [ "trigonometry", "analytic geometry", "calculus", "conics", "ellipse", "derivative", "geometry" ], "Problem": "Is there anything wrong with the formula $F = G\\frac{m_{1}m_{2}}{r^{2}}$ ?\r\n\r\nI used it to develop a long equation about the speed of the planet orbiting the object at any given moment and tested it with Earth, Neptune and Jupiter and in all cases I got about 20% error.\r\n\r\nThe formula I developed, by the way, is\r\n\r\n$v = \\sqrt{GM\\frac{(a-c\\cos \\theta)^{2}}{b^{4}}\\sqrt{\\frac{b^{8}\\cos^{2}\\theta}{a^{4}(a-c\\cos \\theta)^{2}}+b^{2}\\left( 1-\\frac{b^{2}\\cos \\theta}{a^{2}(a-c\\cos \\theta)}\\right)}}$,\r\n\r\nwhere v is the speed of the planet, G is the gravitational constant, M is the mass of the object being orbited, a is the semimajor axis of the orbit, b is the semi-minor axis, c is $\\sqrt{a^{2}-b^{2}}$ and $\\theta$ is the angle made between the orbiting object, the object being orbited and the point where the orbiting object is the farthest away.", "Solution_1": "$F = G\\frac{m_{1}m_{2}}{r^{2}}$\r\n\r\nwe cannot proove that this formula is true or false. but for pratical purpose this formula gives the rigth solution.\r\n\r\nyour problem can be:\r\n\r\nin the set up of your problem you did not considerate that in the universe have more than 3 objects :)\r\n\r\nor a mistake in simplification\r\n\r\nif you had said how you derive this formula, because there is best formulas for the velocity of planets. try use polar coordinates\r\n\r\nit's impossible to help you. you just said \"this formula is rigth\" but it's impossible for we deduce what problem you made just with the answer\r\n\r\nwhat's the question for 42? dump", "Solution_2": "What I did was really, really long, and I did it three times and I got the same equation, so it can't be the simplifying. I did use polar coordinates, hence the theta. I'm not sure what you mean by universe having more than 3 objects. My derivation of the formula was completely mathematical. I only applied it to the planets to see if I was right.\r\n\r\nWhat I was asking was, is there a more accurate form of the formula $F = G\\frac{m_{1}m_{2}}{r^{2}}$?", "Solution_3": "Yes, but it involves general relativity.", "Solution_4": "you don't need to use Einstein on the problem dude.\r\n\r\nKepler did a lot of things without calculus and another usefull things.\r\n\r\nwhat I mean for more than 3 planets is\r\n\r\nIn the movement of the earth on the sun, for our calculation we need to think that. MArs, pluto, etc. all those planets have a interference on the movement of the earth. it's not a perfect ellipse, but the equation says it is a ellipse\r\n\r\nand what i mean for using Polar coordinates is the following.\r\n\r\nuse the equation of an ellipse, in pollar coordinates, with some algebra an calculus you can have the \"velocity\" that is equal to the derivative\r\n\r\nyou can see sites about derivative in polar coordinates\r\nand the First Law of Kepler, the deduction and proof.\r\n\r\nthe solution is in Kepler", "Solution_5": "Of course, for everyday purposes you don't need GR. I just pointed out that there [i]is[/i] a more precise law...", "Solution_6": "Yes! That is exactly what I did! The polar coordinates for an ellipse!\r\n\r\nKepler doesn't tell me anything. He just says T^3 = a^2 for all planets and that the areas are the same when the planet moves the same amount of time. How am I supposed to get the speed of the object using Kepler's laws? I doubt the planets have that much amount of an effect on earth that it decreases its velocity by 20%. There has to be a better explanation.", "Solution_7": "I have not been able to make sense out of that formula - it has an awful lot of moving parts, with roots inside roots and exponents up to 8.\r\n\r\nThere should be a much, much simpler formula for velocity based on simple conservation of energy:\r\n\r\n$-\\frac{GMm}r+\\frac12mv^{2}=C,$ where $r$ is the (variable) distance and $C$ is a constant.\r\n\r\nThere is one slight (classical) improvement on the accuracy. Instead of the smaller object revolving around the larger, they both revolve around their common center of mass. Take that into account and you wind up multiplying or dividing various things by a factor of $\\frac{M}{M+m}.$ Of course, even for Jupiter and the sun, that's pretty close to 1 - no 20% lurking in that correction.", "Solution_8": "err emilgouliev. what you said is the third law of kepler.\r\n\r\nthe first is what we want.\r\nWith it we have the equation of the movement of a planet using the mass of the sum, the constant G, the actual position of the plannet and the velocity\r\n\r\nsee http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion\r\n\r\nI think a better solution is\r\n\r\n${x=r \\cos (\\theta )}$ (Parametric equations of the movement. r is not constant)\r\n${y=r \\sin (\\theta )}$\r\n\r\n${r=\\frac{p}{\\epsilon \\cos (\\theta )+1}}$ (In PolarCoordinates)\r\n\r\n${v=\\frac{dy}{dx}=\\frac{\\frac{dy}{d\\theta }}{\\frac{dx}{d\\theta }}=\\frac{r \\cos (\\theta )+\\frac{dr}{d\\theta }\\sin (\\theta )}{\\cos (\\theta ) \\frac{dr}{d\\theta }-r \\sin (\\theta )}}$\r\n\r\nAfter a few calculations\r\n${\\frac{dy}{dx}=-(\\epsilon+\\cos (\\theta )) \\csc (\\theta )}$\r\n\r\nthe next problem is just to express $P, \\epsilon$ in terms of the others constants.", "Solution_9": "yagaron: $v$ does not equal $\\frac{dy}{dx}.$ As a scalar, $v$ equals $\\frac{ds}{dt},$ where $s$ measures arclength.\r\n\r\n$v=\\sqrt{\\left(\\frac{dr}{dt}\\right)^{2}+\\left(r\\frac{d\\theta}{dt}\\right)^{2}}.$\r\n\r\nTo which we can add two more conditions.\r\n\r\nThe conservation of momentum condition:\r\n\r\n$r\\frac{d\\theta}{dt}=\\text{constant}$\r\n\r\nand the conservation of energy condition, which I gave above.", "Solution_10": "What I did was to use $a_{c}= \\frac{v^{2}}{r_{c}}$. I substituted in $\\frac{GM}{r^{2}}$ for $a_{c}$, and $r\\cos \\theta-\\frac{b^{2}r\\cos \\theta}{a}$ for $r_{c}$, and then substitute in $\\frac{b^{2}}{a-c\\cos \\theta}$ for $r$, which gave me that whole thing.\r\n\r\nIs there a way to find $\\frac{d\\theta}{dt}$ using Kepler's laws?", "Solution_11": "[quote=\"emilgouliev\"]What I did was to use $a_{c}= \\frac{v^{2}}{r_{c}}$. I substituted in $\\frac{GM}{r^{2}}$ for $a_{c}$.[/quote]\n\nThis is not accurate. You're using a formula for centripetal acceleration that is only accurate for objects moving in a circle.\n\n[quote=\"emilgouliev\"]\nIs there a way to find $\\frac{d\\theta}{dt}$ using Kepler's laws?[/quote]\r\n\r\nAngular Momentum, $\\omega = \\frac{d\\theta}{dt}$.", "Solution_12": "ohh ya...\r\n\r\nthanks Kent. i forgot!!! we are using vectors.\r\n\r\nsorry for the mistake", "Solution_13": "[quote=\"gauss202\"][quote=\"emilgouliev\"]What I did was to use $a_{c}= \\frac{v^{2}}{r_{c}}$. I substituted in $\\frac{GM}{r^{2}}$ for $a_{c}$.[/quote]\n\nThis is not accurate. You're using a formula for centripetal acceleration that is only accurate for objects moving in a circle.\n\n[quote=\"emilgouliev\"]\nIs there a way to find $\\frac{d\\theta}{dt}$ using Kepler's laws?[/quote]\n\nAngular Momentum, $\\omega = \\frac{d\\theta}{dt}$.[/quote]\r\n\r\nNo, it is. You were probably taught that $a = \\frac{v^{2}}{r}$ works in a circle, but in fact it works for any curved path. The $r_{c}$ is the radius of the circle that's tangent to the path at that point.\r\n\r\nI found my mistake, by the way. I forgot to subtract the $c$ when doing the radius. The new equation is:\r\n\r\n$v = \\sqrt{GM \\frac{(a-c\\cos \\theta)^{2}}{b^{3}}\\sqrt{1-\\left(\\frac{b^{2}\\cos \\theta}{a(a-c\\cos \\theta)}-\\frac{c}{a}\\right)^{2}+\\left( \\frac{b^{2}}{a^{2}}\\right) \\left( \\frac{b^{2}\\cos \\theta}{a-c\\cos \\theta}-c \\right)^{2}}}$, if $0 < \\theta < \\frac{\\pi}{2}$ or $\\frac{3\\pi}{2}< \\theta < 2\\pi$\r\n\r\n$v = \\sqrt{GM \\frac{(a-c\\cos \\theta)^{2}}{b^{3}}\\sqrt{1-\\left( \\frac{c}{a}-\\frac{b^{2}\\cos \\theta}{a(a-c\\cos \\theta)}\\right)^{2}+\\left( \\frac{b^{2}}{a^{2}}\\right) \\left( c-\\frac{b^{2}\\cos \\theta}{a-c\\cos \\theta}\\right)^{2}}}$, if $\\frac{\\pi}{2}< \\theta < \\frac{3\\pi}{2}$\r\n\r\nThis new equation gave an amazing .0004% error for Earth.", "Solution_14": "[quote=\"emilgouliev\"]\nNo, it is. You were probably taught that $a = \\frac{v^{2}}{r}$ works in a circle, but in fact it works for any curved path. The $r_{c}$ is the radius of the circle that's tangent to the path at that point.[/quote]\r\n\r\nMy objection isn't to the \"r\" it is to the \"v\". The v in the formula for centripetal acceleration is only the tangential component of the velocity. In a circle, of course, the tangential velocity is equal to the total velocity - but in an elliptical orbit they are not [i]exactly[/i] the same.\r\n\r\nIn an orbit with relatively low eccentricity this error should be very small. Or it may also be possible that this error has a small net effect over the course of an entire orbit.", "Solution_15": "[quote=\"emilgouliev\"]\nNo, it is. You were probably taught that $a = \\frac{v^{2}}{r}$ works in a circle, but in fact it works for any curved path. The $r_{c}$ is the radius of the circle that's tangent to the path at that point.[/quote]\r\n\r\nWell, there are several problems with that.\r\n\r\nThat is the formula you get if you decompose the acceleration into two components, one normal to the path and one tangent to the path.\r\n\r\nThe acceleration tangent to the path is $\\frac{d}{dt}|v|.$ That is, it is the rate of change of the speed (and hence dependent on the rate of change of the kinetic energy). Note that for the body in elliptical orbit, the speed is not constant - it is greatest when the body is closest to the sun and least when the body is furthest from the sun.\r\n\r\nThe acceleration normal to the path is indeed equal to $\\frac{v^{2}}{r_{c}}$ where $r_{c}$ is the radius of curvature (the distance to the center of the best-approximating circle). However, please note two issues. The first is that $r_{c}\\ne r.$ At both aphelion and perihelion, $r_{c}r.$ Note also that the center of that approximating circle is not where the sun is - that is, the normal to the curve does not generally point toward the sum. The total acceleration - the vector sum of the two components - does point toward the sun, but in general, neither the tangent nor the normal components of the acceleration point in that direction.\r\n\r\nIn retrospect, that decomposition into normal and tangential accelerations is [b]not[/b] useful in this instance, and I would not recommend pursuing this line any further.\r\n\r\nIt is clear to me now that emilgouliev's formula has to be incorrect, because of the concerns expressed above. I would strongly suggest scrapping this approach and starting over. Don't directly use the acceleration at all, rather derive the velocity from conservation of energy.", "Solution_16": "[quote=\"gauss202\"]\nIn a circle, of course, the tangential velocity is equal to the total velocity - but in an elliptical orbit they are not [i]exactly[/i] the same.[/quote]\n\nOK.\n\n[quote=\"Kent Merryfield\"]Note that for the body in elliptical orbit, the speed is not constant - it is greatest when the body is closest to the sun and least when the body is furthest from the sun. [/quote]\n\nWhich is why I have the theta in the equation. When you put in $\\pi$ for $\\theta$ you get the maximum speed, and if you put $0$ for $\\theta$, you get the minimum speed.\n\n[quote=\"Kent Merryfield\"]The acceleration normal to the path is indeed equal to \\frac{v^{2}}{r_{c}} where r_{c} is the radius of curvature (the distance to the center of the best-approximating circle). However, please note two issues. The first is that r_{c}\\ne r. At both aphelion and perihelion, r_{c}r. Note also that the center of that approximating circle is not where the sun is - that is, the normal to the curve does not generally point toward the sum. The total acceleration - the vector sum of the two components - does point toward the sun, but in general, neither the tangent nor the normal components of the acceleration point in that direction. [/quote]\n\nI already took those into consideration when creating my equation.\n\n[quote=\"Kent Merryfield\"]It is clear to me now that emilgouliev's formula has to be incorrect, because of the concerns expressed above.[/quote]\r\n\r\nThe only reason it is incorrect is because of what gauss202 pointed out, so besides that you should not have any problems with it.", "Solution_17": "By the way, where did you find data about the velocities of the planets in terms of $\\theta$?", "Solution_18": "I didn't, actually. I only used the maximum and the minimum velocties, although I should the othe ones, too. My physics teacher found another mistake in it, which I have to fix.", "Solution_19": "[quote]\n\n${r=\\left\\{\\frac{p \\cos (\\theta )}{\\epsilon \\cos (\\theta )+1},\\frac{p \\sin (\\theta )}{\\epsilon \\cos (\\theta )+1},0\\right\\}}$\n${v=\\frac{dr}{dt}}$ , ${a=\\frac{dv}{dt}}$\n\n${V^{2}=v.v=\\frac{p^{2}\\left(\\epsilon^{2}+2 \\cos (\\theta ) \\epsilon+1\\right) \\left(\\frac{d\\theta }{dt}\\right)^{2}}{(\\epsilon \\cos (\\theta )+1)^{4}}}$\n\n${\\text{An}=a.\\frac{(v\\times a)\\times v}{\\sqrt{(v\\times a).(v\\times a) (v.v)}}=\\frac{G M}{r.r}}$\n\n${\\left(\\frac{d\\theta }{dt}\\right)^{2}=\\frac{G M (\\epsilon \\cos (\\theta )+1)^{3}\\sqrt{\\epsilon^{2}+2 \\cos (\\theta ) \\epsilon+1}}{p^{3}}}$\n\n${V^{2}=\\frac{G M \\left(\\epsilon^{2}+2 \\cos (\\theta ) \\epsilon+1\\right)^{3/2}}{p (\\epsilon \\cos (\\theta )+1)}}$\n[/quote]\r\n\r\nI think this is rigth, or not?", "Solution_20": "err....\r\n\r\nit should be an elipse, not a circle\r\nand, since u used \"r\", it's a considerable error\r\n\r\nthat's the error", "Solution_21": "[quote=\"Fabiomr1\"]err....\n\nit should be an elipse, not a circle\nand, since u used \"r\", it's a considerable error\n\nthat's the error[/quote]\r\n\r\nWhat are you talking about?", "Solution_22": "emilog, didnt you saw my formulae?", "Solution_23": "Is the planet that is orbited fixed, or are the planets orbiting each other around the centre of mass?", "Solution_24": "yagaron, I'm actually confused on what stands for what.\r\n\r\nThaakisfox, the planet being orbited is fixed.", "Solution_25": "Then why all the fuss with that long formula?\r\nHere is a much simpler one:\r\nThe orbiting component of the velicity of the orbiting planet:\r\n$v_{\\phi}=\\frac{\\Psi}{R}= \\frac{\\Psi}{p}(1+e\\cos \\phi)$\r\nHere:\r\n$\\Psi =\\frac{2\\pi}{T}\\sqrt{a^{3}p}=\\sqrt{GMp}$ where $p$ is the parameter of the ellipse, $e$ is the excentricity, $a$ is the semi-major axis, $T$ is the period of orbit of the planet.\r\nThe radial component of the velocity:\r\n$v_{R}= \\dot R =\\frac{dR}{d\\phi}\\frac{d\\phi}{dt}=\\frac{pe \\dot \\phi\\sin\\phi}{(1+e\\cos\\phi)^{2}}=\\frac{R^{2}\\dot\\phi}{p}e \\sin\\phi$\r\nSo\r\n$v_{R}=\\frac{\\Psi}{p}e\\sin\\phi$\r\nSo the square of the velocity:\r\n$v^{2}=v_{R}^{2}+v_{\\phi}^{2}=\\frac{{\\Psi}^{2}}{p^{2}}(e^{2}\\sin^{2}\\phi+(1+e\\cos\\phi)^{2})$\r\nSo from here:\r\n$v^{2}=\\frac{{\\Psi}^{2}}{p}\\left ( \\frac{e^{2}-1}{p}+\\frac{2}{R}\\right )$\r\nBut we also know that:\r\n$\\frac{1-e^{2}}{p}=\\frac{1}{a}$\r\nSo the velocity of the orbiting planet:\r\n$v=\\Psi \\sqrt{\\frac{1}{p}\\left (\\frac{2}{R}-\\frac{1}{a}\\right )}=\\sqrt{GM\\left (\\frac{2}{R}-\\frac{1}{a}\\right )}$\r\n\r\nNow this formula looks much simpler to me :D" } { "Tag": [ "inequalities", "inequalities theorems" ], "Problem": "Can you help me prove this inequalities by using Jensen inequality?\r\nThanks.", "Solution_1": "look here.\r\nhttp://www.mathlinks.ro/viewtopic.php?p=1346817#1346817" } { "Tag": [ "number theory", "prime numbers" ], "Problem": "The question is this:\r\n\r\nthe number 24 can be made by multiplying 4 prime numbers, 2, 3, 2, 2.\r\n\r\nhow many primes must be multipled to make 2400? is there a formula to help me determine that?", "Solution_1": "well the thing is to take the square root\r\n\r\n$\\sqrt{2400}\\approx 48$\r\n\r\nthat means u divide 2400 by all primes less than 48...\r\n\r\nwhen you have a quotient you divide until you reach a prime\r\n\r\nI know it might be confusing :lol:", "Solution_2": "nonononono\r\nJust start dividing.\r\n2400/2=1200 2\r\n1200/2=600 2\r\n600/2=300 2\r\n300/2=150 2\r\n150/2=75 2\r\n75/3=25 3\r\n25/5=5 5\r\n5/5=1 5\r\n\r\nLooking at the crooked right column :D we see that we have 5 2s, a 3, and 2 5s, which is a total of $8$", "Solution_3": "yeah kstan013's way is much much simpler in this case as the number 2400 is convieniant but my way works for any number", "Solution_4": "[quote=\"epatjn\"]The question is this:\n\nthe number 24 can be made by multiplying 4 prime numbers, 2, 3, 2, 2.\n\nhow many primes must be multipled to make 2400? is there a formula to help me determine that?[/quote]\r\n2400/2=1200/2=600/2=300/2=150/2=75\r\nSO just 2, and the prime factors of 75 which are 3 and 5", "Solution_5": "[quote=\"epatjn\"]The question is this:\n\nthe number 24 can be made by multiplying 4 prime numbers, 2, 3, 2, 2.\n\nhow many primes must be multipled to make 2400? is there a formula to help me determine that?[/quote]\r\n[hide]Knowing that $24=2^{3}\\times3$, and $100=2^{2}\\times5^{2}$, you multiply them together to get $2400=2^{5}\\times3\\times5^{2}$[/hide]", "Solution_6": "Yes, all you have to do is make a factor tree, simple as that.\r\n[hide]You and up with 2 to the power of 5, 5 squared, and 3, so 8 in all. [/hide]Right?", "Solution_7": "[quote=\"lifesapain\"]Yes, all you have to do is make a factor tree, simple as that.\n[hide]You and up with 2 to the power of 5, 5 squared, and 3, so 8 in all. [/hide]Right?[/quote]\r\n\r\n...as a future warning please do not revive topics more than 5 days old" } { "Tag": [ "group theory", "abstract algebra", "Galois Theory", "superior algebra", "superior algebra solved" ], "Problem": "How can I find the splitting field of $x^{p}-x-a \\in \\mathbb{F}_{p}[x]$ where $a \\neq 0$, and why should the Galois group of this splitting field be cyclic? \r\n\r\nElementary solutions are preferred since google gave me only advanced answers :)", "Solution_1": "first obverse that $x^{p}-x-a$ is separable and has no root in $\\mathbb{F}_{p}$, and with every root $\\alpha$ (we fix one in a splitting field), $\\alpha+i$ is also a root. so $\\alpha+i$, where $i$ run's through $\\mathbb{F}_{p}$, are all roots. thus, the splitting field is $\\mathbb{F}_{p}(\\alpha)$, and we have an injective homomorphism $G \\to \\mathbb{F}_{p}, f \\mapsto f(\\alpha)-\\alpha$, where $G$ denotes the galois group. but $\\mathbb{F}_{p}$ has only the two obvious subgroups, so that $G \\cong \\mathbb{F}_{p}$.", "Solution_2": "Actually, every extension of finite fields is Galois with cyclic Galois group, canonically generated by the Frobenius automorphism. owk", "Solution_3": "@owk: indeed :-). but I think the concrete situation here is also interesting ;-)." } { "Tag": [ "function", "superior algebra", "superior algebra unsolved" ], "Problem": "Wasn't sure if I should post this here, but...\r\n\r\nIf f: A -> B let us define f_image: P(A) -> P(B) by\r\nf_image(E) = {b in B | b = f(x) for some x in E}.\r\nProve that the function f_image is onto if and only if f is onto.\r\n\r\nP(A) means the power set of A.\r\n\r\nSo far, all I have is the following:\r\nProve that: If f is onto, then f_image is onto.\r\nAssume f is onto. We want f_image is onto, or, for all Y in P(B), there exists an X in P(A) such that f_image(X) = Y. \r\n\r\nI know it isn't much, but I really have no idea what to do now. Can anyone help? Tell me if you need something clarified.\r\n\r\n[Edit] I forgot - please reply ASAP if you can. I need this by Friday.", "Solution_1": "I hope you are not asking anybody here to do your homework. Just use the definitions, and common sense. You'll be able to do it.", "Solution_2": "No, I'm not trying to get people to do my homework. I'm just asking someone if they could point me in the right direction...I really have no idea how to continue." } { "Tag": [ "algebra", "polynomial" ], "Problem": "it should be [hide](2x:^2:+5)(2x-3)[/hide]", "Solution_1": "Yes, but I think that it would be nice to know how you managed to conjure the right expressions to multiply.", "Solution_2": "Q: 4x:^3:-6x:^2:+10x-15\n\nA: [hide](2x:^2:+5)(2x-3)[/hide]\n\n\n\nSolution:[hide]well, hmmmmmmm...whered my pencil go...aha...*pats himself on the back*...well, the constants equal -15. and on a hunch i would use 3 and 5. nevermind the signs for now. i knew that a polynomial times :pm: 3 or :pm: 5 equaled -6x:^2:. 3 would divide that. so then we know that 2x:^2: and -3 are in the answer. they have to be on different factors though, if you want to multiply them. also, -3 times a number equals -15. so the other constant is 5. then its easy from there. just plug in what you know. and you get: [hide](2x:^2:+5)(2x-3)[/hide][/hide]\n\n\n\nif this didnt make sense to you, cause i didnt really explain it nicely, just ask.", "Solution_3": "Well, a nice little trick that works with a couple (not many, but some) factorising questions:\n\n[hide]\n\n4x:^3:-6x:^2:+10x-15 = x:^2:(4x-6) + 5(2x-3) = 2x:^2(2x-3)+5(2x-3) = (2x:^2:+5)(2x-3).\n\n[/hide]", "Solution_4": "thats a cool trick sphink. my solution may have been longer, but it only took about 10 seconds for me, but that may not occur all the time. your way would be better for cases one is unsure of", "Solution_5": "Rep,\r\n\r\nDo you know the rational roots theorem?\r\n\r\nAny rational root of a polynomial has a numerator that is a divisor of the constant of the polynomial and a denominator that is a divisor of the lead coefficient of the polynomial. Can you see why this is true?", "Solution_6": "thats good to know\r\n\r\n\r\n\r\nP.S. \r\nwow, i didnt realize that i passed the 300 mark. yeha" } { "Tag": [], "Problem": "What is the smallest $ n$ for which $ 1\\plus{}\\frac{1}{2}\\plus{}\\frac{1}{3}\\plus{}\\frac{1}{4}\\plus{}\\cdots\\plus{}\\frac{1}{n}>3?$", "Solution_1": "I think we have to bruteforce this. But it's not too bad, especially since there are no nasty repeating decimals.\r\n[code]\n1/1 = 1.000000 (1.000)\n1/2 = 0.500000 (1.500)\n1/3 = 0.333333 (1.833)\n1/4 = 0.250000 (2.083)\n1/5 = 0.200000 (2.283)\n1/6 = 0.166667 (2.450)\n1/7 = 0.142857 (2.593)\n1/8 = 0.125000 (2.718)\n1/9 = 0.111111 (2.829)\n1/10= 0.100000 (2.929)\n1/11= 0.090909 (3.020)\n[/code]\r\n\r\nSo n is 11.", "Solution_2": "Hi Everybody,\r\n\r\nI don't quite understand how you did the graph. Could you elaborate? Thanks!\r\n\r\n--mathdriven", "Solution_3": "sorry to revive this, but can anyone tell me a way to do it without brute forcing?\nCan you pm me if you know?", "Solution_4": "Can you PM me too?", "Solution_5": "Any easy Way for doing this" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Does $ a^4 \\plus{} b^4 \\plus{} c^4 \\leqslant 2(a^2 b^2 \\plus{} b^2 c^2 \\plus{} c^2 a^2 )$ hold for every $ a,b,c$ if it is known that $ a^2 \\plus{} b^2 \\plus{} c^2 \\leqslant 2(ab \\plus{} bc \\plus{} ca)$.", "Solution_1": "why do you posted this problem two time,\r\nsee her, http://www.mathlinks.ro/viewtopic.php?t=271098" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $a,b,c$ are positive reals and $ab+bc+ca=3$ prove that\r\n\r\n$a+b+c\\geq abc+2$", "Solution_1": "Firstly we prove that $a+b+c\\geq 3.$\r\nIndeed, $a+b+c\\geq 3$ $\\Leftrightarrow$ $(a+b+c)^2\\geq 3(ab+bc+ca)$, obvious.\r\nFrom $ab+bc+ca=3$, we obtain \\[ a^2 b^2+b^2 c^2 +c^2 a^2+2abc(a+b+c)=9\\geq 3abc(a+b+c)\\Longrightarrow \\] \r\n\\[ \\Longrightarrow abc(a+b+c)\\leq 3 \\Longrightarrow abc \\geq\\dfrac{3}{a+b+c}. \\]\r\nIt follows \\[ a+b+c\\geq abc+2 \\Leftrightarrow abc\\geq a+b+c-2. \\]\r\nThen its enough to prove that $\\dfrac{3}{a+b+c}\\leq a+b+c-2.$\r\nWe have $\\dfrac{3}{a+b+c}\\leq a+b+c-2\\Leftrightarrow (a+b+c+1)(a+b+c-3)\\geq 0$ and hence the conclusion.", "Solution_2": "Actually, this problem appeared on JBMO shortlist in 2003. I gave for this inequality two solutions. One of them is exactly like Marius, but the other one is much more easier. Here it goes: \r\n\r\nBy AM-GM we get $3=ab+bc+ca\\geq3\\sqrt [3] {abc}$.From here it results that $abc\\leq 1.$\r\nSo if we prove that $a+b+c\\geq 3$ we are done here. But this follows from this well-known inequality $(a+b+c)^{2}\\geq 3(ab+bc+ca).$ :D ;)", "Solution_3": "If$a,b,c\\in R: a^2+b^2+c^2=2$ we have:\r\n $\\mid a+b+c-abc\\mid\\le 2$" } { "Tag": [ "induction", "modular arithmetic" ], "Problem": "1.given tha d(n) is the highest odd factor n.eg dn(90)=15.\r\nfind.d(1)+...+d(2^99)\r\nif you solve this problem explain.\r\n2.prove that (n)^5 -(n) is divisible by 15.\r\nmy soluting involved using fermat's and the chinese rem's.\r\n(n)^5=n(mod5).\r\nmaybe you may use induction :D [quote]all laws of motion are the same in all uniformly moving frames of refrence[/quote]", "Solution_1": "1. Isn't $ d(90)\\equal{}45$?\r\n2. It's easy by Little Fermat's theorem, but it can also be done like this:\r\n$ n^5\\minus{}n\\equal{}n(n^4\\minus{}1)\\equal{}(n\\minus{}1)n(n\\plus{}1)(n^2\\plus{}1)$ Now try $ n\\equal{}5a\\plus{}k$ for $ k\\in \\{0,1,2,3,4\\}$.", "Solution_2": "[hide=\"Three Solutions for the 2nd Part\"]\n1st solution: We know that $ 5 \\mid n^5 \\minus{} n$ by Fermat's Little Theorem. Now $ n^5 \\minus{} n \\equal{} n(n^4 \\minus{} 1) \\equal{} n(n \\minus{} 1)(n \\plus{} 1)(n^2 \\plus{} 1)$, which clearly contains a multiple of 3.\n2nd solution: 5 and 3 are small mods, we can just try all numbers $ \\mod 5$ and $ \\mod 3$.\n3rd solution: $ n^5 \\minus{} n \\equiv n^5 \\plus{} 5n^4 \\plus{} 10n^3 \\plus{} 10n^2 \\plus{} 5n \\plus{} 1 \\minus{} 1 \\minus{} n \\equiv (n \\plus{} 1)^5 \\minus{} (n \\plus{} 1) \\pmod{5}$. \nSince this is equivalent to $ 0$ for $ n \\equal{} 1$, it is equivalent to $ 0$ for all integers.\n\nAlso, we can actually say that $ 30 \\mid n^5 \\minus{} n$, and for odd $ n$, $ 240 \\mid n^5 \\minus{} n$ (by considering this in $ \\mod 16$)[/hide]", "Solution_3": "i'll take #1\r\n\r\nso i'm assuming that $ d(1) \\plus{} \\cdots \\plus{} d(2^{99})$ means, like, $ d(1) \\plus{} d(2) \\plus{} d(3) \\plus{} ... \\plus{} d(2^{99})$. if we're only summing powers of 2, the problem is pretty trivial\r\n\r\nso between $ 1$ and $ 2^{99}$, there are $ 2^{98}$ numbers $ k_0$ such that $ 2^0$ is the greatest power of two going into them. so we sum the first $ 2^{98}$ odd numbers to get the sum of all $ d(k)$. that's $ (2^{98})^2 \\equal{} 2^{196}$\r\n\r\nsimilarly, there are $ 2^{97}$ numbers $ k_1$ such that $ 2^1$ is the greatest power of two going into them. so $ d(k_1)\\equal{}\\frac{k_1}{2}$. we sum the first $ 2^{97}$ odd numbers and we get $ 2^{194}$.\r\n\r\nwe continue this trend and our final answer is $ 2^0 \\plus{} 2^2 \\plus{} 2^4 \\plus{} \\cdots \\plus{} 2^{196}$." } { "Tag": [ "vector", "inequalities", "geometry", "geometry solved" ], "Problem": "Let ABCD is a a quadrilateral. M,N is the midpoint of AB and CD. P is the point of intersection of AC and BD. Prove that:\r\n area(MNP) \\leq 8/27 area(ABCD).\r\nIt is maybe not too difficult but it is so strange when I use vector and find a fantasy result. By this result, I can't say that inequality is right . However, help me prove it.", "Solution_1": "In fact,that inequality must be:\r\narea(MNP) \\leq 4/27 area(ABCD).", "Solution_2": "[quote=\"Anh Cuong\"]In fact,that inequality must be:\narea(MNP) \\leq 4/27 area(ABCD).[/quote]\r\nwith x,y be unit vectors\r\ncan we have (x-y)/((1+x)(1+y)) \\leq 8/27\r\ncan you show me?", "Solution_3": "If you do well, master, my inequality seem wrong. I am not sure whether my problem is right , you can prove or not. :(", "Solution_4": "This thread is a mess. Let's put things at the right places.\r\n\r\nIn the following, I will denote by $\\left[P_1P_2...P_n\\right]$ the directed area and by $\\left|P_1P_2...P_n\\right|$ the non-directed area of a polygon $P_1P_2...P_n$.\r\n\r\nAt first, for the inequalities we are discussing here, the quadrilateral ABCD must be assumed to be convex; in fact, if the quadrilateral ABCD can be concave, there is no real number k such that $\\left|MNP\\right| \\leq k \\cdot \\left|ABCD\\right|$ holds for every quadrilateral ABCD. But still, the inequality $\\left|MNP\\right| \\leq \\frac{4}{27} \\cdot \\left|ABCD\\right|$ holds not for every convex quadrilateral ABCD. On the other hand, the inequality $\\left|MNP\\right| \\leq \\frac{8}{27} \\cdot \\left|ABCD\\right|$ does hold for every convex quadrilateral ABCD, and actually it can be replaced by a stronger inequality: $\\left|MNP\\right| < \\frac14 \\cdot \\left|ABCD\\right|$. In the following, I will prove this stronger inequality. Note that the constant $\\frac14$ in this inequality cannot be replaced by a better one; in fact, when the point B approaches the point C, the values |MNP| and $\\frac14 \\cdot \\left|ABCD\\right|$ get arbitrarily close to each other.\r\n\r\nSo it remains to prove the inequality $\\left|MNP\\right| < \\frac14 \\cdot \\left|ABCD\\right|$. First I will prove a lemma:\r\n\r\n[b]Lemma 1.[/b] If ABCD is a quadrilateral, if M and N are the midpoints of the segments AB and CD, respectively, and if P is the point of intersection of the diagonals AC and BD, then the directed areas of triangles ADP, BCP and MNP satisfy the following equation:\r\n\r\n$\\left[MNP\\right] = \\frac14 \\cdot \\left(\\left[ADP\\right] + \\left[BCP\\right]\\right)$.\r\n\r\n[i]Proof.[/i] There are many different ways to prove Lemma 1. For instance, you can note that Lemma 1 follows from applying Theorem 3.14 of H. S. M. Coxeter, S. L. Greitzer, [i]Geometry Revisited[/i], Toronto - New York 1967, to the (self-intersecting) quadrilateral CBDA. But one can also easily prove Lemma 1 using the cross product of vectors:\r\n\r\nIt is well-known that the directed area $\\left[P_1P_2P_3\\right]$ of a triangle $P_1P_2P_3$ equals $\\frac12 \\cdot \\overrightarrow{P_3P_1}\\times\\overrightarrow{P_3P_2}$. Thus,\r\n\r\n$\\left[ ADP\\right] =\\frac{1}{2}\\cdot \\overrightarrow{PA}\\times\\overrightarrow{PD}$;\r\n$\\left[ BCP\\right] =\\frac{1}{2}\\cdot \\overrightarrow{PB}\\times\\overrightarrow{PC}$;\r\n$\\left[ MNP\\right] =\\frac{1}{2}\\cdot \\overrightarrow{PM}\\times\\overrightarrow{PN}$.\r\n\r\nOn the other hand,\r\n\r\n$\\left( \\overrightarrow{PA}+\\overrightarrow{PB}\\right) \\times \\left( \\overrightarrow{PC}+\\overrightarrow{PD}\\right)$\r\n$=\\overrightarrow{PA}\\times \\overrightarrow{PC}+\\overrightarrow{PA}\\times \\overrightarrow{PD}+\\overrightarrow{PB}\\times \\overrightarrow{PC}+\\overrightarrow{PB}\\times \\overrightarrow{PD}$.\r\n\r\nNow remember that the cross product of two collinear vectors is 0. Since the vectors $\\overrightarrow{PA}$ and $\\overrightarrow{PC}$ are collinear, we thus have $\\overrightarrow{PA}\\times \\overrightarrow{PC} = 0$. Similarly, $\\overrightarrow{PB}\\times \\overrightarrow{PD} = 0$. Thus,\r\n\r\n$\\left( \\overrightarrow{PA}+\\overrightarrow{PB}\\right) \\times \\left( \\overrightarrow{PC}+\\overrightarrow{PD}\\right) =\\overrightarrow{PA}\\times\\overrightarrow{PD}+\\overrightarrow{PB}\\times \\overrightarrow{PC}$.\r\n\r\nBut since the points M and N are the midpoints of the segments AB and CD, respectively, we have $\\overrightarrow{PM}=\\frac{\\overrightarrow{PA}+\\overrightarrow{PB}}{2}$ and $\\overrightarrow{PN}=\\frac{\\overrightarrow{PC}+\\overrightarrow{PD}}{2}$. Thus,\r\n\r\n$\\left[ MNP\\right] =\\frac{1}{2}\\cdot \\overrightarrow{PM}\\times\\overrightarrow{PN}=\\frac{1}{2}\\cdot \\frac{\\overrightarrow{PA}+\\overrightarrow{PB}}{2}\\times \\frac{\\overrightarrow{PC}+\\overrightarrow{PD}}{2}$\r\n$=\\frac{1}{8}\\cdot \\left( \\overrightarrow{PA}+\\overrightarrow{PB}\\right)\\times \\left( \\overrightarrow{PC}+\\overrightarrow{PD}\\right) =\\frac{1}{8}\\cdot \\left( \\overrightarrow{PA}\\times \\overrightarrow{PD}+\\overrightarrow{PB}\\times \\overrightarrow{PC}\\right)$\r\n$=\\frac{1}{4}\\cdot \\left( \\frac{1}{2}\\cdot \\overrightarrow{PA}\\times\\overrightarrow{PD}+\\frac{1}{2}\\cdot \\overrightarrow{PB}\\times\\overrightarrow{PC}\\right) =\\frac{1}{4}\\cdot \\left( \\left[ ADP\\right] +\\left[BCP\\right] \\right)$,\r\n\r\nwhat proves Lemma 1.\r\n\r\nNow, the equation\r\n\r\n$\\left[MNP\\right] = \\frac14 \\cdot \\left(\\left[ADP\\right] + \\left[BCP\\right]\\right)$,\r\n\r\nwhen converted to non-directed areas, yields\r\n\r\n$\\left|MNP\\right| = \\frac14 \\cdot \\left|\\left|ADP\\right| - \\left|BCP\\right|\\right|$\r\n\r\n(in fact, we get the - sign since, because of the convexity of the quadrilateral ABCD, the triangles ADP and BCP have different orientations and thus their directed areas differ in sign). But clearly\r\n\r\n$\\left|\\left|ADP\\right| - \\left|BCP\\right|\\right| < \\max\\left\\{\\left|ADP\\right|;\\;\\left|BCP\\right|\\right\\}$.\r\n\r\nBut since the quadrilateral ABCD is convex, we have |ADP| < |ABCD| and |BCP| < |ABCD|, thus\r\n\r\n$\\max\\left\\{\\left|ADP\\right|;\\;\\left|BCP\\right|\\right\\} < \\left|ABCD\\right|$.\r\n\r\nAs a result, $\\left|\\left|ADP\\right| - \\left|BCP\\right|\\right| < \\left|ABCD\\right|$, and thus\r\n\r\n$\\left|MNP\\right| = \\frac14 \\cdot \\left|\\left|ADP\\right| - \\left|BCP\\right|\\right| < \\frac14 \\cdot \\left|ABCD\\right|$,\r\n\r\nproving our initial inequality.\r\n\r\n Darij" } { "Tag": [ "USAMTS" ], "Problem": "Look at the title for the first sentence\r\n\r\nBut you have to sign up with a special form, and I turn 13 september 30th, so would it just be easier to sign up when I turn 13 or sign up now?", "Solution_1": "It's not going to matter whether you sign up now, or you sign up on Oct. 14th, so I guess you should do whatever is easier.\r\nHowever, there is nothing stopping you from working on the problems now.", "Solution_2": "If you haven't yet registered on the USAMTS web site, then the simplest thing is for you to wait and register after you've turned 13. However, if you register while you're still under 13, then we'll need the parental permission form." } { "Tag": [ "search", "y is this so ald", "screensavers" ], "Problem": "Does anyone know where I can find some good math screensavers? I tried google but I couldn't find anything good.", "Solution_1": "Hi joml88 and Everyone,\r\n\r\n[quote=\"joml88\"]Does anyone know where I can find some good math screensavers? I tried google but I couldn't find anything good.[/quote]\r\n\r\nThere are many screensavers that use fractal art that are pretty neat. Try the search term \"fractal screen savers\" using < [url]http://www.yahoo.com[/url] >. I got several hits.\r\n\r\nI personally run a program called ZetaGrid as my screensaver. ZetaGrid is a grid computing project to calculate the zeroes of the Reimann zeta function. The aim is to either find statistics about the zeroes that might help generate ideas about how to prove the Reimann hypothesis, or to actaully find a zero of the Reimann zeta function that does not lie on the line [tex]1/2+bi[/tex] in [tex]\\mathbb{C}. [/tex] The program can be run as a screensaver. So while you're not using your compuer, it can be doing some fancy mathematics! :lol: My computer has computed over 170 million zeroes of the Reimann zeta function to date. The URL for the ZetaGrid project is < [url]http://www.zetagrid.net/zeta/index.html[/url] >.\r\n\r\nSince I brought up the topic of grid computing, one can find a list of active distributed computing projects at < [url]http://www.aspenleaf.com/distributed/ap-math.html[/url] >.\r\n\r\nAll the best,\r\n\r\nmathfreak", "Solution_2": "how does the screensaver look like? (be4 I install it)", "Solution_3": "Hi Peter VDD and Everyone,\r\n\r\n[quote=\"Peter VDD\"]how does the screensaver look like? (be4 I install it)[/quote]\r\n\r\nThe screensaver for ZetaGrid is nothing fancy. It's a box that changes position on one's computer screen every few minutes. It gives the current time and also shows your computer's progress on its current work unit. It also tells you how many zeroes your computer is calculating per second (?). \r\n\r\nYours truly,\r\n\r\nmathfreak", "Solution_4": "14 year bump", "Solution_5": "# why?!?!?!?!?", "Solution_6": "5 year bump" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Given $ a, b, c$ are real numbers. Prove that: $ (a \\minus{} b)^4 \\plus{} (b \\minus{} c)^4 \\plus{} (c \\minus{} a)^4 \\geq\\ 8(a \\minus{} b)^2(c \\minus{} a)(c \\minus{} b)$", "Solution_1": "We only did (c-a)(c-b)>0.And we also can assume $ a \\minus{} b\\ge0$\r\nThen we identify $ a \\minus{} b \\equal{} x,c \\minus{} b \\equal{} y,c \\minus{} a \\equal{} z$.Now $ x,y,z\\ge0$\r\nAnd x+z=y We need to prove $ x^4 \\plus{} (x \\plus{} z)^4 \\plus{} y^4 \\equal{} x^4 \\plus{} y^4 \\plus{} z^4\\ge8x^2yz \\equal{} 8x^2(x \\plus{} z)z$\r\nThen we need to prove:$ x^4 \\plus{} z^4 \\plus{} 2xz^3\\ge 2x^3z \\plus{} x^2z^2$.\r\nWe assume x/z=p.Then we need to prove $ p^4 \\minus{} 2p^3 \\minus{} p^2 \\plus{} 2p \\plus{} 1\\ge0$\r\nAnd I think this one maybe need computer to calculate the min.I guess it's not a simple digit.", "Solution_2": "I think U have a mistakes in calculus. Notice that equality occurs when $ a \\equal{} b \\equal{} c.$ In your poof, I don't see this case. But indeed, it's very simple and I said that \"It's a joke\". My solution is very short :) \r\nBesides, equality occurs when $ a \\equal{} 0; b \\equal{} 1; c \\equal{} \\frac{1\\plus{}\\sqrt[]{5}}{2}$...\r\nDo U think can_hang's lenma works here ?", "Solution_3": "[quote=\"nguoivn\"]I think U have a mistakes in calculus. Notice that equality occurs when $ a \\equal{} b \\equal{} c.$ In your poof, I don't see this case. But indeed, it's very simple and I said that \"It's a joke\". My solution is very short :) \nBesides, equality occurs when $ a \\equal{} 0; b \\equal{} 1; c \\equal{} \\frac {1 \\plus{} \\sqrt []{5}}{2}$...\nDo U think can_hang's lenma works here ?[/quote]\r\nI'm not wrong.But I didn't find it.\r\n$ \\equal{} (p^2 \\minus{} p \\minus{} 1)^2\\ge0$", "Solution_4": "Sorry, your proof is trues. And my solution is here (it's also the way to creat it) :) \r\nhttp://www.maths.vn/forums/showthread.php?t=9244&page=15" } { "Tag": [ "ratio" ], "Problem": "The fractional part of a number is the difference between the number and the largest integer not exceeding the number. For example, the fractional part of $ 2\\frac{3}{4}$ is $ \\frac{3}{4}$, and the fractional part of 4 is 0.\n\nWhat is the ratio of the fractional part of the sum of $ 2\\frac{2}{3}$, $ 3\\frac{3}{4}$, and $ 4\\frac{4}{5}$ to the sum of their fractional parts?", "Solution_1": "The sum of these three mixed numbers is $ 11\\frac{13}{60}$. The sum of their fractional parts is $ \\frac{2}{3} \\plus{} \\frac{3}{4} \\plus{} \\frac{4}{5} \\equal{} 2\\frac{13}{60}$. Thus, the ratio is $ \\frac{133}{673}$.\r\n\r\nEDIT: Stupidity fail. It's easier to leave everything as a common fraction, which lets you do the ratio more easily.", "Solution_2": "Actually, it asks for the ratio of the fractional part of the sum (which would be $ \\frac{13}{60}$) to the sum of their fractional parts (which is $ 2\\frac{13}{60}\\equal{}\\frac{133}{60}$). So our final answer would be $ \\frac{\\frac{13}{60}}{\\frac{133}{60}}\\equal{}\\boxed{\\frac{13}{133}}$" } { "Tag": [], "Problem": "We have 2001 pieces of cheese and we know the weight of every piece. Is it possible to chose one piece, to cut it in two parts (not necessarily equal) and then to arrange the pieces of cheese obtained in two sack of the same weight everyone with 1001 pieces?", "Solution_1": "Yes.\r\n[hide]Proof:\nFirst, label the initial 2001 pieces $m_1, m_2, m_3, ..., m_{2001}$ in increasing order from smallest to largest. Then make two piles, $P_1$ and $P_2$. Put $m_1$ into $P_1$ and note that $P_1-P_2=m_1$. Then put $m_2$ into $P_2$ and the difference changes to $m_2-m_1$, which is less than $m_2$. Then put m_3 into $P_1$, and the difference changes to $m_3-m_2+m+1=1 won't get any plausible results, trust me, I have been there already... :(\r\n\r\nI guess the key is what to do with the a^2-ab+b^2...(I think everyone's noticed that it's equal to (a^3+b^3)/(a+b), but it doesn't seem to yield anything... :? )", "Solution_9": "[quote=\"billzhao\"]First of all, it's not symmetric (it's only cyclic).\n\nSecondly, assume an order won't help Maverick's argument.[/quote]\r\n\r\nYou're right! I mixed again everything up :(", "Solution_10": "Don't worry DanielKirchner about your errors and remember that our experience is the amount of our errors, but probably to have too much experience, in live, is not too much pleasant ;) \r\n\r\nHowever the original inequality I found on my notes was like that\r\n\r\n$ \\frac{a^3(a+b)}{a^3+b^3}+ \\frac{b^3(b+c)}{b^3+c^3}+ \\frac{c^3(c+a)}{c^3+a^3} \\geq 3\\frac{(ab+bc+ca)}{a+b+c}$\r\n\r\nand this is a well known old conjecture never solved.\r\n\r\nProbably my semplification is not a good thingh to perform :( \r\n\r\nProbably Cauchy can help us to solve it, but perhaps it is very hard.", "Solution_11": "[quote=\"manlio\"]For $a,b,c$ positive real numbers such that $a+b+c=1$ prove the ineq\n\n$\\frac{a^3}{a^2-ab+b^2}+\\frac{b^3}{b^2-bc+c^2}+\\frac{c^3}{c^2-ca+a^2} \\geq 3(ab+bc+ca) $[/quote]\r\n\r\nWhat about $a=\\frac16$, $b=\\frac13$, $c=\\frac12$ ?\r\n\r\n Darij", "Solution_12": "[quote=\"darij grinberg\"]\n\nWhat about $a=\\frac16$, $b=\\frac13$, $c=\\frac12$ ?\n\n Darij[/quote]\r\n\r\nIndeed, that's a counterexample for both inequalities (the simplification by manlio and the other one)\r\n\r\n$LHS= \\frac {8}{9}$\r\n$RHS= \\frac {11}{12}$\r\n\r\nIts obvious that $RHS > LHS$", "Solution_13": "[quote=\"manlio\"]$ \\frac{a^3(a+b)}{a^3+b^3}+ \\frac{b^3(b+c)}{b^3+c^3}+ \\frac{c^3(c+a)}{c^3+a^3} \\geq 3\\frac{(ab+bc+ca)}{a+b+c}$[/quote]\r\n\r\nAs I've said, this one is not correct, but it seems that the minimum of\r\n\r\n$\\frac{\\frac{a^3(a+b)}{a^3+b^3}+ \\frac{b^3(b+c)}{b^3+c^3}+ \\frac{c^3(c+a)}{c^3+a^3}}{\\left(\\frac{(ab+bc+ca)}{a+b+c}\\right)}$\r\n\r\nis somewhere around 2.902. Exact value?\r\n\r\n Darij" } { "Tag": [], "Problem": "Sprint Round\r\n27. What is the greatest integer value of n such that (n 2 + 2n + 5)/(n - 3) is an integer?\r\n\r\nTarget Round\r\n8. The sum of a list of positive integers is 20. What is the maximum possible product of the integers in that list?\r\n\r\nI need help finding an easy, effective method for solving these kinds of problems.\r\n\r\n[b]PLEASE INCLUDE SOLUTIONS, NOT JUST ANSWERS!! I ALREADY HAVE THE ANSWERS![/b]\r\n\r\nSolutions that involve guess & check are fine - as long as they are educated and effective under time pressure.", "Solution_1": "27\r\n[hide]Do the division: $\\frac {n^2 + 2n + 5}{n - 3} = n + 5 + \\frac {20}{n-3}$. The largest number that the denominator can be and the fraction still remain an integer is 20, so the largest value of $n$ is $23$.[/hide]", "Solution_2": "8\r\n[hide]Know that in order to achieve the maximum product you must have as many 3's as possible. The answer is then $3^6 \\cdot 2 = 1458$.[/hide]", "Solution_3": "[quote=\"tarquin\"]8\n[hide]Know that in order to achieve the maximum product you must have as many 3's as possible. The answer is then $3^6 \\cdot 2 = 1458$.[/hide][/quote]\r\n\r\nHow would you discover that three is the best given a short amount of time?", "Solution_4": "Hey, who wants to prove that $3^a > b^c$ where $3a = bc$ and $b$ is a prime number ?", "Solution_5": "Yeah really, it's just something you have to know. \r\nBut there are some times where as many threes as possible is not the best. That is when the remainder is 1 when the sum is divided by three. There you should instead use 4 instead of 3x1", "Solution_6": "[hide]Do the long division. We find that $\\frac{n^{2}+2n+5}{n-3}=n+5+\\frac{20}{n-3}$. So whenever we plug an integer into that, we will receive an integer, as long as $20$ evenly divides $n-3$. The largest integer $n$ for that to work is $n-3=20$, so $n=23$.[/hide]", "Solution_7": "Actually you don't have to do long division, you can do synthetic, it's much easier", "Solution_8": "[quote=\"nat mc\"]Actually you don't have to do long division, you can do synthetic, it's much easier[/quote]Whats synthetic division???", "Solution_9": "So if you have n 2 +2n+5 and you divide it by n-3, then you put this\r\n\r\n3 l 1 2 5\r\n l\r\n3 is the number you are subtracting and 1,2,and 5 are coefficients\r\nBring the 1 down\r\n\r\n3 l 1 2 5\r\n l\r\n 1\r\n\r\nNow multiply 1 by 3 and put it under the 2\r\n\r\n3 l 1 2 5\r\n l 3 \r\n 1\r\n\r\nNow do 2+3 and put it under the last 3\r\n\r\n3 l 1 2 5\r\n l 3 \r\n 1 5\r\n\r\nNow multiple 5 by 3 and put it under the other 5\r\n\r\n3 l 1 2 5\r\n l 3 15 \r\n 1 5\r\nNow do 5+15 and put it under the 15\r\n\r\n3 l 1 2 5\r\n l 3 15 \r\n 1 5 20\r\n1, 5, and 20 are the coefficients of the answer 1 degree less than the first equation, so it is n+5+20/(n-3)\r\nAnswer to question=23", "Solution_10": "The reason that 3 gives you the greatest answer is because its closest to e.", "Solution_11": "[quote=\"Phelpedo\"]How would you discover that three is the best given a short amount of time?[/quote]\r\n[hide]We have already been told the answer is $2\\times 3\\times 3\\times 3\\times 3\\times 3\\times 3$. Let's try to forget that, and pretend we don't know anything about the answer.\n\nSuppose one member of the final set is $5$. We could replace the $5$ with $2$ and $3$. (I'll write it as $5\\rightarrow 2+3.$) The sum would remain the same, but the product would be larger. Thus, there cannot be a $5$ in the final set. Similarly, there cannot be any number larger than $5$.\n\nSuppose one member of the final set is $1$. We could combine the $1$ with any other member of the set. The sum would remain the same, but the product would be larger. For example, if the set contained a $1$ and a $3$, you could combine these elements and replace them with a $4$. $(1+3\\rightarrow 4).$ The sum is the same, but the product is larger.\n\nAll we have left to consider are $2$, $3$, and $4$. Suppose the final set contains three 2's. You could replace them with $2+2+2\\rightarrow3+3$. The sum is the same, but the product is larger. Therefore, the final set cannot contain 3 or more 2's.\n\nSuppose the final set contains a $2$ and a $4$. You could replace them with $2+4\\rightarrow3+3$. The sum is the same, but product is larger. Therefore, the final set cannot contain both 2's and 4's.\n\nSuppose the final set contains two 4's. You could replace them with $4+4\\rightarrow3+3+2$. Again, the sum is the same, but the product is larger. Therefore, the final set cannot contain two or more 4's.\n\nWhat we are left with is a set that can contain nothing but 2's, 3's, and 4's; a maximum of two 2's; a maximum of one 4 (but not both 2's and 4's).\n\nNow, there is a unique solution to the problem, given the above observations.[/hide]", "Solution_12": "Does this mean that for any positive a, b, and c, $e^a > b^c$, where $ae = bc$" } { "Tag": [ "induction", "floor function", "ceiling function", "inequalities", "combinatorics unsolved", "combinatorics" ], "Problem": "Suppose we have $n$ one's in a row .In each move we can delete two numbers like $a,b$ and replace them with $(a+b)/4$\r\n.We continue these moves till just one number remains in the row .Prove that the remaining number is not smaller than $\\frac{1}{n}$.", "Solution_1": "Let $m_{n}$ be the smallest possible last resulting number. It is clear that \\[m_{n}= \\min_{s+t=n}\\frac{m_{s}+m_{t}}{4}.\\] Let's prove by induction that $m_{n}\\geq \\frac{1}{n}$.\r\nThe base case $n=1$ is trivial. \r\nSuppose that for all $k1+\\frac{1}{2}+...+\\frac{1}{m}$", "Solution_1": "Each positive integer not greater than $m$ has prime factorization which only uses primes not greater than $m$. Moreover, using the unicity of the prime factorization it is clear that for each $k \\leq m$ the term $\\frac 1 k$ appears exactly once in the expansion of $P = (1 + \\frac 1 {p_1} + \\frac 1 {p_1^2} + \\cdots) (1 + \\frac 1 {p_2} + \\frac 1 {p_2^2} + \\cdots) \\cdots (1 + \\frac 1 {p_n} + \\frac 1 {p_n^2} + \\cdots).$\r\nMoreover, it is also clear that some terms of that expansion are of the form $\\frac 1 k$ where $k > m.$\r\nIt follows that $S > 1 + \\frac 1 2 + \\cdots + \\frac 1 m.$\r\nSince $ (1 + \\frac 1 {p_i} + \\frac 1 {p_i^2} + \\cdots) = \\frac {p_i} {p_i - 1}$ the result follows.\r\n\r\nPierre." } { "Tag": [ "inequalities", "calculus", "derivative", "function", "inequalities proposed" ], "Problem": "If $ a$, $ b$, $ c$ are positive reals, prove that\r\n\r\n$ \\left(a^{2}\\plus{}b^{2}\\plus{}c^{2}\\right)\\left(\\frac{a^{2}}{\\left(a^{2}\\plus{}bc\\right)^{2}}\\plus{}\\frac{b^{2}}{\\left(b^{2}\\plus{}ca\\right)^{2}}\\plus{}\\frac{c^{2}}{\\left(c^{2}\\plus{}ab\\right)^{2}}\\right)\\geq\\frac{9}{4}$.", "Solution_1": "I think it's weaker than Iran1996's . \r\n Put $ a^2=yz , b^2=xz , c^2=xy $ \r\n The ineq is equivalent to $ A= (xy+yz+xz) . \\sum(\\frac{1}{(x+\\sqrt{yz})^2}) \\geq 9/4 $ \r\n By AM-GM : $ \\sqrt{yz} \\leq (y+z)/2 $ \r\n So $ A \\geq (xy+yz+xz). \\sum(\\frac{4}{(2x+y+z)^2}) \\geq 9/4 $ : It's just a different formation of I96", "Solution_2": "The last inequality is not true (see x=y=0). :P", "Solution_3": "vasc x,y,z cannot be 0 because a,b,c>0.", "Solution_4": "just :P :D", "Solution_5": "The inequality is very hard. I think the following is less hard (but also very hard):\r\n$\\sum \\frac{1}{6a^2+bc} \\ge \\frac{9}{7(ab+bc+ca)}$.\r\nHas somebody a solution ?", "Solution_6": "[quote=\"Vasc\"]The inequality is very hard. I think the following is less hard (but also very hard):\n$\\sum \\frac{1}{6a^2+bc} \\ge \\frac{9}{7(ab+bc+ca)}$.\nHas somebody a solution ?[/quote] \r\n$ \\sum \\frac{1}{(k-1)a^2+bc} \\ge \\frac{9}{k(ab+bc+ca)} $.\r\n Is it still true for k =6 ? I \"proved\" the ineq is wrong for k= 1,2,3,4,5 \r\n Do you have a solution , Vasc ?", "Solution_7": "I think with a,b,c >0 andc=min(a,b,c)\r\n(ab+bc+ca)(2/(c+(ab)^(1/2))^2+1/(a+b)^2)>=9/4\r\nIs it right?", "Solution_8": "[quote=\"nthd\"]I think with a,b,c >0 andc=min(a,b,c)\n(ab+bc+ca)(2/(c+(ab)^(1/2))^2+1/(a+b)^2)>=9/4\nIs it right?[/quote]\r\nIt is not right (see $c=0$).\r\n\r\nNow, related to\r\n$\\displaystyle \\sum \\frac{1}{pa^2+bc} \\ge \\frac{9}{(p+1)(ab+bc+ca)}$,\r\nI think that it is true for $p\\ge 3+\\sqrt{7}$, but I have no solution.", "Solution_9": "I think you are right :) ,Vasc.", "Solution_10": "mda.........vasc $c>0$. you can not take $c=0$ :?", "Solution_11": "he can take $c\\rightarrow 0$. But could anyone solve this very hard inequality ?", "Solution_12": "view [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=55584&highlight=#[/url]", "Solution_13": "I found...\r\n\r\n $(a^2+b^2+c^2)[\\frac{a^2}{(a^2+bc)^2}+\\frac{b^2}{(b^2+ac)^2}+\\frac{c^2}{(c^2+ab)^2}] \\ge \\frac{9}{4}$ <->\r\n $(a^2+b^2+c^2)[(\\frac{a}{a^2+bc})^2+(\\frac{b}{b^2+ac})^2+(\\frac{c}{c^2+ab})^2] \\ge \\frac{9}{4}$ <->\r\n $(a^2+b^2+c^2)[(\\frac{a^2}{a^3+abc})^2+(\\frac{b^2}{b^3+abc})^2+(\\frac{c^2}{c^3+abc})^2] \\ge \\frac{9}{4}$.\r\n\r\n Consider now $f(x)=(\\frac{x^2}{x^3+P})^2$, where we can assume that $P=abc$ is a constant. The problem it\u00b4s demonstrate that the second derivative of this functions is always positive for all x positive real. But it\u00b4s possible and you must believe me...\r\n Now we make:\r\n\r\n $(\\frac{a^2}{a^3+P})^2+(\\frac{b^2}{b^3+P})^2+(\\frac{c^2}{c^3+P})^2 \\ge 3(\\frac{S^2}{S^3+P})^2$ where we have $S=\\frac{a+b+c}{3}$ and we must note that $S \\ge \\sqrt[3]{P}$ -> $S^3 \\ge P$ and now we have $2S^3 \\ge S^3+P$ -> $\\frac{1}{S^3+P} \\ge \\frac{1}{2S^3}$ -> $\\frac{S^2}{S^3+P} \\ge \\frac{1}{2S}$ -> $(\\frac{S^2}{S^3+P})^2 \\ge \\frac{1}{4S^2}$ -> $(a^2+b^2+c^2)3(\\frac{S^2}{S^3+P})^2 \\ge 3(a^2+b^2+c^2)\\frac{1}{4S^2}$ and so it\u00b4s proved because by Jensen\u00b4s Inequalitie again we have $(a^2+b^2+c^2) \\ge 3S^2$", "Solution_14": "[quote=\"mffmartinelli\"]Consider now $f(x)=(\\frac{x^2}{x^3+P})^2$, where we can assume that $P=abc$ is a constant. The problem it\u00b4s demonstrate that the second derivative of this functions is always positive for all x positive real. But it\u00b4s possible and you must believe me...[/quote]\r\n\r\nI don't believe you, because calc101.com does not believe you. I don't understand why so many people claim a function is convex.", "Solution_15": "Yes. You are write, I had a mistake when I calculate the second derivate of this function. I\u00b4m sorry. In some interval this function changes concave. But is a \"small\" interval... :lol:", "Solution_16": "[quote=\"nthd\"]Let $ a$, $ b$, $ c$ are positive reals, prove that\n\n$ \\frac {4a^{2}}{\\left(a^{2} \\plus{} bc\\right)^{2}} \\plus{} \\frac {4b^{2}}{\\left(b^{2} \\plus{} ca\\right)^{2}} \\plus{} \\frac {4c^{2}}{\\left(c^{2} \\plus{} ab\\right)^{2}}\\geq\\frac {9}{a^{2} \\plus{} b^{2} \\plus{} c^{2}}.$[/quote]$ \\sum{\\frac {4a^{2}}{\\left(a^{2} \\plus{} bc\\right)^{2}}} \\minus{} \\frac {9}{a^{2} \\plus{} b^{2} \\plus{} c^{2}}\\equiv\\frac {F(a,b,c)}{\\left(a^{2} \\plus{} bc\\right)^{2}\\left(b^{2} \\plus{} ca\\right)^{2}\\left(c^{2} \\plus{} ab\\right)^{2}\\left(a^{2} \\plus{} b^{2} \\plus{} c^{2}\\right)}.$\n\n$ F(a,b,c) \\equal{} F(a,a \\plus{} s,a \\plus{} t)$\n\n$ \\equal{} 32\\left(s^2 \\minus{} st \\plus{} t^2\\right)a^{10} \\plus{} 64(s \\plus{} t)\\left(s^2 \\plus{} t^2\\right)a^9$\n\n$ \\plus{} 8\\left(11s^4 \\plus{} 14s^3t \\plus{} 69s^2t^2 \\plus{} 14st^3 \\plus{} 11t^4\\right)a^8$\n\n$ \\plus{} 4(s \\plus{} t)\\left(22s^4 \\plus{} 11s^3t \\plus{} 203s^2t^2 \\plus{} 11st^3 \\plus{} 22t^4\\right)a^7$\n\n$ \\plus{} \\left(56s^6 \\plus{} 140s^5t \\plus{} 867s^4t^2 \\plus{} 1122s^3t^3 \\plus{} 867s^2t^4 \\plus{} 140st^5 \\plus{} 56t^6\\right)a^6$\n\n$ \\plus{} 2(s \\plus{} t)\\left(10s^6 \\plus{} 39s^5t \\plus{} 299s^4t^2 \\plus{} 153s^3t^3 \\plus{} 299s^2t^4 \\plus{} 39st^5 \\plus{} 10t^6\\right)a^5$\n\n$ \\plus{} \\left(3s^8 \\plus{} 38s^7t \\plus{} 371s^6t^2 \\plus{} 584s^5t^3 \\plus{} 487s^4t^4 \\plus{} 584s^3t^5 \\plus{} 371s^2t^6 \\plus{} 38st^7 \\plus{} 3t^8\\right)a^4$\n\n$ \\plus{} 2st(s \\plus{} t)\\left(3s^6 \\plus{} 59s^5t \\plus{} 88s^4t^2 \\plus{} 11s^3t^3 \\plus{} 88s^2t^4 \\plus{} 59st^5 \\plus{} 3t^6\\right)a^3$\n\n$ \\plus{} s^2t^2\\left(19s^6 \\plus{} 96s^5t \\plus{} 79s^4t^2 \\plus{} 22s^3t^3 \\plus{} 79s^2t^4 \\plus{} 96st^5 \\plus{} 19t^6\\right)a^2$\n\n$ \\plus{} 2s^3t^3(s \\plus{} t)\\left(8s^4 \\plus{} 3s^3t \\minus{} 2s^2t^2 \\plus{} 3st^3 \\plus{} 8t^4\\right)a$\n\n$ \\plus{} s^4t^4\\left(2s^2 \\minus{} 3st \\plus{} 2t^2\\right)\\left(2s^2 \\plus{} 3st \\plus{} 2t^2\\right)\\geq 0,$\n\nwhich is clearly true for $ a \\equal{} \\min\\{a,b,c\\}.$\n\n[b]Remark[/b] nthd's inequality holds for real $ a,b,c$ with $ bc \\plus{} ca \\plus{} ab \\ne 0.$\n[quote=\"Vasc\"]If $ a,b,c$ are nonnegative numbers such that $ bc \\plus{} ca \\plus{} ab > 0$, then\n\n$ \\frac {7}{6a^2 \\plus{} bc} \\plus{} \\frac {7}{6b^2 \\plus{} ca} \\plus{} \\frac {7}{6c^2 \\plus{} ab} \\ge \\frac {9}{ bc \\plus{} ca \\plus{} ab}.$[/quote]$ \\sum {\\frac {7}{6a^2 \\plus{} bc}} \\minus{} \\frac {9}{bc \\plus{} ca \\plus{} ab}$\r\n\r\n$ \\equal{} \\sum{\\frac {(b \\minus{} c)^2\\left[6a^4 \\plus{} 322a^2bc \\plus{} 21abc(b \\plus{} c) \\plus{} 36b^2c^2\\right]}{\\left(6a^2 \\plus{} bc\\right)\\left(6b^2 \\plus{} ca\\right)\\left(6c^2 \\plus{} ab\\right)(bc \\plus{} ca \\plus{} ab)}}\\geq0.$\r\n\r\nBy the way, Vasc inequality holds for real $ a,b,c$ with $ a^2 \\plus{} 57bc>0,b^2 \\plus{} 57ca>0,c^2 \\plus{} 57ab>0.$" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "The positive integrer number $n$ has $1994$ digits. $14$ of its digits are $0$'s and the number of times that the other digits: $1, 2, 3, 4, 5, 6, 7, 8, 9$ appear are in proportion $1: 2: 3: 4: 5: 6: 7: 8: 9$, respectively. Prove that $n$ is not a perfect square.", "Solution_1": "0 -> 14\r\n1 -> 44\r\n2 -> 88\r\n3 -> 132\r\n4 -> 176\r\n5 -> 220\r\n6 -> 264\r\n7 -> 308\r\n8 -> 352\r\n9 -> 396\r\n\r\nThe number of times each digit apears in $n$ is listed above.\r\nTake $n$ mod.$9$ . We get $n\\equiv 1*44+2*88+3*132+...+9*396\\equiv 3$, Which is not a square." } { "Tag": [], "Problem": "Can anyone share a procedure or a reference to a procedure to prepare and observe sperm cells under a microscope?\r\n\r\nWhat magnification would be appropriate?\r\n\r\nThanks.", "Solution_1": "Hi,\r\n\r\nThis site has a link to microscope procedures.\r\n\r\nhttp://www.freewebs.com/ads-online/\r\n\r\nand this site has information on magnification\r\n\r\nhttp://ask.metafilter.com/mefi/20996\r\n\r\nHope that helps." } { "Tag": [ "absolute value", "algebra proposed", "algebra" ], "Problem": "Let $z_{1},z_{2},...,z_{n}\\in\\mathbb{C}$ such that $|z_{1}|=|z_{2}|=...=|z_{n}|>0$. Prove that $\\frac{(z_{1}+z_{2})(z_{2}+z_{3})...(z_{n}+z_{1})}{z_{1}z_{2}...z_{n}}\\in\\mathbb{R}.$", "Solution_1": "i think it doesn't hold for $n=3$ allready. try $1,1+i,1+2i$ and you get\r\n$\\frac{(1+(1+i))((1+i)+(1+2i))((1+2i)+1)}{1(1+i)(1+2i)}=$\r\n$2\\frac{(2+i)(2+3i)}{1+2i}=2\\frac{1+8i}{1+2i}=\\frac{2}{5}(1+8i)(1-2i)=\\frac{2}{5}(17+6i)$\r\ndoesn't work for $n\\geq 4$ because you put these three numbers for $z_{1},z_{2},z_{3}$ and $1$ for all the rest...\r\nbut it holds for $n=2$ :wink:", "Solution_2": "[quote=\"goc\"]i think it doesn't hold for $n=3$ allready. try $1,1+i,1+2i$ and you get\n$\\frac{(1+(1+i))((1+i)+(1+2i))((1+2i)+1)}{1(1+i)(1+2i)}=$\n$2\\frac{(2+i)(2+3i)}{1+2i}=2\\frac{1+8i}{1+2i}=\\frac{2}{5}(1+8i)(1-2i)=\\frac{2}{5}(17+6i)$\ndoesn't work for $n\\geq 4$ because you put these three numbers for $z_{1},z_{2},z_{3}$ and $1$ for all the rest...\nbut it holds for $n=2$ :wink:[/quote]\r\nBut $|z_{1}|=|z_{2}|=...=|z_{n}|$.", "Solution_3": "ok,i'm an idiot. :blush: \r\nbut if it holds for all $n$ then we would have that $\\frac{(z_{n}+z_{1})(z_{n}+z_{n-1})}{z_{n}(z_{n-1}+z_{1})}$ must be real for those kind of $z$ or $\\frac{(1+a)(1+b)}{a+b}$ is real where $a=\\frac{z_{1}}{z_{n}}$ and $b=\\frac{z_{n-1}}{z_{n}}$ and $|a|=|b|=1$ and then \r\n$\\frac{1+ab}{a+b}$ real \r\n$(1+ab)(a^{*}+b^{*})$ real ($aa^{*}=1$)\r\n$2(a^{*}+b^{*})$ real which is not true...\r\ncould you find my mistake?", "Solution_4": "Let $U = \\prod_{i=1}^{n}\\frac{z_{i}+z_{i+1}}{z_{i}}= \\prod_{i=1}^{n}\\left( 1+\\frac{z_{i+1}}{z_{i}}\\right)$.\r\nThen, we have that $\\overline U = \\prod_{i=1}^{n}\\left( 1+\\frac{z_{i}}{z_{i+1}}\\right)$, since $\\frac{z_{i+1}}{z_{i}}$ has absolute value $1$, i.e. its conjugate is its inverse.\r\nThat product is nothing but $U$, so we're done." } { "Tag": [], "Problem": "[color=darkred][size=134][u]An easy extension of the [b]Eyeball's theorem.[/b][/u][/size] Let $ \\{\\begin{array}{c}w_{1}= C(O_{1},R_{1})\\\\ \\\\ w_{2}= C(O_{2},R_{2})\\end{array}$ be two circles, $ O_{1}\\not\\equiv O_{2}$.\n\nDenote the line $ d = O_{1}O_{2}$. Let $ M\\not\\in O_{1}O_{2}$ be a point so that $ \\frac{MO_{1}}{MO_{2}}=\\frac{R_{1}}{R_{2}}$.\n\nChoose the points $ \\{\\begin{array}{c}X\\in MO_{1}\\cap w_{1}\\\\ \\\\ Y\\in MO_{2}\\cap w_{2}\\end{array}$ for which denote $ \\{\\begin{array}{c}P\\in d\\ ,\\ XP\\perp d\\\\ \\\\ Q\\in d\\ ,\\ YQ\\perp d\\end{array}$. Prove that $ XP = YQ$.\n\n[b]Remark.[/b] You can find this theorem at the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=127056 (the fifth question - user [b]Ilthigore[/b])[/color]", "Solution_1": "[hide]Draw perpendicular from M to d and label it A.\n$ \\triangle{XO_{1}P}\\sim\\triangle{MAO_{1}}$ and $ \\triangle{MAO_{2}}\\sim\\triangle{YQO_{2}}$ since MA is parallel to XP and YQ.\n$ MO_{1}/MO_{2}\\equal{}R_{1}/R_{2}\\equal{}XO_{1}/YO_{2}$\nBy similarity and the above, MA/YQ=MA/XP so YQ=XP[/hide]" } { "Tag": [ "MATHCOUNTS", "geometry", "rhombus", "perimeter" ], "Problem": "Let's start an Indiana Mathcounts marathon.\r\n\r\nNew Problem:\r\n\r\n$ ab\\equal{}3,bc\\equal{}9,ac\\equal{}12$ What is $ a$? If $ a,b,c>0$", "Solution_1": "2.\r\n\r\nfine i won't be a jerk.\r\n\r\n[hide](abc)^2 = 3 * 9 * 12 so abc = 18 so a = 18/9 = 2[/hide]", "Solution_2": "Let's leave this to middle school students cough cough", "Solution_3": "[hide=\"NP\"]\nIf 1 diagonal of a rhombus is 4 meters greater than the other, and the perimeter of the rhombus is 40, what is the area of the rhombus? [/hide]" } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Find all primes $ p$ such that $ p\\equal{}a^2\\plus{}b^2\\plus{}c^2$ and $ p|a^4\\plus{}b^4\\plus{}c^4$ for some $ a,b,c \\in N$.", "Solution_1": "[quote=\"KDS\"]Find all primes $ p$ such that $ p \\equal{} a^2 \\plus{} b^2 \\plus{} c^2$ and $ p|a^4 \\plus{} b^4 \\plus{} c^4$ for some $ a,b,c \\in N$.[/quote]\r\n\r\nLet $ a\\ge b\\ge c\\ge 1$ and so $ p\\ge 3$\r\n\r\nWe get $ p|(a^2\\plus{}b^2\\plus{}c^2)^2\\minus{}(a^4\\plus{}b^4\\plus{}c^4)$ and so $ p|2(a^2b^2\\plus{}a^2c^2\\plus{}b^2c^2)$ and so $ p|a^2b^2\\plus{}a^2c^2\\plus{}b^2c^2$ (since $ p$ is prime $ \\ge 3$)\r\n\r\n$ \\implies$ $ p|a^2(a^2\\plus{}b^2\\plus{}c^2)\\minus{}(a^2b^2\\plus{}a^2c^2\\plus{}b^2c^2)$\r\n\r\n$ \\implies$ $ p|a^4\\minus{}b^2c^2\\equal{}(a^2\\minus{}bc)(a^2\\plus{}bc)$. So :\r\n\r\nIf $ a^2\\ne bc$ then $ 0a^2\\plus{}bc$\r\nSo $ a^2\\equal{}bc$ and so $ a\\equal{}b\\equal{}c$ and then $ p\\equal{}3a$ and the only solution is $ p\\equal{}3$ (else $ p$ would not be prime).\r\n\r\nHence the unique solution $ (a,b,c)\\equal{}(1,1,1)$ and $ p\\equal{}3$ which, indeed, is a solution.", "Solution_2": "Equivalently, $ \\minus{}c^4 \\equiv a^4 \\plus{} b^4 \\equal{} (a^2 \\plus{} b^2)^2 \\minus{} 2a^2b^2 \\equiv c^4 \\minus{} 2a^2b^2 \\pmod{p}$, hence $ p \\mid 2(ab\\minus{}c^2)(ab\\plus{}c^2)$, etc." } { "Tag": [ "college", "search", "real analysis", "real analysis theorems" ], "Problem": "I'm sure you guys know the book \"Berkeley Problems in Mathematics.\"\r\n\r\nDo you guys know about more books like this one?", "Solution_1": "I just know there exists someting called the Berkely Maths circle. What is this book?", "Solution_2": "u can go for IA Maron 's book", "Solution_3": "Go to Spriger's web , and try to search to find the book of your. And I want to know about this book on your opion , give me please. I had to find list of all book on this series , but It's vary on country to country, China Edition and western.\r\n\r\n :mad: I like the problem in analytical form , not set form. \r\n\r\n\r\nKong\r\n\r\nThailand", "Solution_4": "Problems in Mathemaical analysis I,II,III, W.J.Kaczor-M.T.Nowak\r\nPolya-Szego, Problems in analysis, Springer" } { "Tag": [ "geometry", "perimeter", "geometric transformation", "reflection", "rotation", "function", "analytic geometry" ], "Problem": "Are you counting triangles as equivalent under reflection? For example is a triangle with sides (3, 4, 5) the same as a triangle with sides (4, 3, 5), with both read clockwise?", "Solution_1": "Not counting reflections or rotations at all:\r\nP=3, (1,1,1)\r\nP=4 { }\r\nP=5 (1,2,2)\r\nP=6 (2,2,2)\r\nP=7 (2,3,2), (1,3,3)\r\nP=8 (2,3,3)\r\nP=9 (3,3,3), (2,3,4), (1,4,4)\r\n\r\nI can't think of any pretty functions that go f(x) = {1,0,1,1,2,1,3,...} for x = {3,4,5,6,7,8,9}.", "Solution_2": "[color=cyan]I think the problem is better considered when we look at odd perimeters and even perimeters seperately. For odds, the first few values (starting with 3) are 1, 1, 2, 3, 4, 5, 7, 8, . . . and for evens (starting with 4), 0, 1, 2, 3, 4, 5, 7, . . . which seems to me to have a definite posibility for a nice pattern, if you make your lists twice or thrice (a great word!) as long as mine are. Although, also maybe not.[/color]", "Solution_3": "[color=red]\n\nin fact, there does exist such a formula. however, it is not particularly nice. \n\nhere it is:\n\nlet a+b+c=n. then, let 8N = n :^2: -6n+8.\n\nif the number of incongruent triangles is T, then\n\n(if n even) T=N(n :^2: /48)\n(if n odd) T=N((n :^2: +6n)/48)\n\nto prove this, one would first count the N=number of (ordered) triples (a,b,c) such that a+b+c=n and a+b>c, b+c>a, a+c>b (which can be done very nicely by considering lattice points inside triangles in the coordinate plane), and then subtract off the number of (triples, or septuples) of (a,b,c). \n\n\n[/color]" } { "Tag": [ "algebra", "linear equation" ], "Problem": "a woman invested 10,000 in two accounts. One of the accounts pays 8% interestand the other 15%. how much is invested in each account if the total interest equals 1,080?\r\n\r\nI need some help coming up with the formula to solve.. I thought I had it figured out but now I am clueless... Any help would be great!! Thanks!", "Solution_1": "I think I know how to do this. Let's see...\r\n\r\nSet the amount of interest money in each account as variables...\r\n\r\nClues\r\n[hide]If the amount of interest in one account is 0.15 x... What will be the amount of interest in the other account in terms of x?[/hide]\n\n[hide]If there's 10 fruits consisting of apples and bananbas, and there's x apples, how many will be bananbas?...[/hide]\n\n[hide]0.08(10,000-x)[/hide]\n\n[hide]So the amount of interest in one account is [b]0.15x [/b]and the amount of interest in the other account is [b]0.08(10,000-x). [/b]Agree So Far?[/hide]\n\nWhat does the problem say about the interests of the accounts\n\n[hide]Total Interest = 1,080 [/hide]\n\nSet it up as an equation\n\n[hide]0.15x+0.08(10,000-x)=1080[/hide]\n\nGot it?\n\nThen solve for x AND ________\n\n[hide] I got x=4000, so 10,000-x=6000[/hide]\n\n[hide] so 4000 is invested in the account that pays 15% interest and the 6000 is invested in the account that pays 8%[/hide]\r\nCHECK YOUR WORK\r\n\r\nAny question?\r\n\r\nsorry, I think you know where the dollars signs should be placed... so make sure to include them", "Solution_2": "I didn't read the previous post because it took too long. ;) No offense of course. ;) \r\n\r\nLet x be the money invested in the account with 8% interest. Let y be for 15%.\r\n$x+y=10000$ and $.8x+.15y=1080$\r\n\r\nNow, just solve it as a linear equation to get that y=4000 and x=6000. \r\n\r\nDone! :)", "Solution_3": "Awesome!! Thank you both for your help!! :)", "Solution_4": "[hide=\"equations\"]x+y=10000\n.08x+.15y=1080. \n\nSolve to get 4000 and 6000. \n\nOh well, someone else posted an answer so don't mind this too much. [/hide]" } { "Tag": [ "AMC", "AIME" ], "Problem": "Let $P_1$ be a regular r-gon and $P_2$ be a regular $s-gon$ $(r\\ge s\\ge3)$ such that each interior angel of $P_1$ is $59/58$ as larger as each interior angle of $P_2.$ What is the largest possible value of $s$?", "Solution_1": "[hide]117? solution later, if its right.[/hide]" } { "Tag": [], "Problem": "Find all of the real and complex solutions (if any exist) of the equation $x^{7}+7^{7}=(x+7)^{7}$.\r\n\r\nThanks", "Solution_1": "See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=116936[/url] with $a=7$.", "Solution_2": "Thanks So Much...", "Solution_3": "this problem is the same as \r\n\r\n$\\begin{eqnarray*}(x+1)^{7}-x^{7}-1 &=& x^{6}+3x^{5}+5x^{4}+5x^{3}+3x^{2}+x \\\\ &=&(x)(x+1)(x^{4}+2x^{3}+3x^{2}+2x+1)\\\\ &=&(x)(x+1)(x^{2}+x+1)^{2}$\r\n\r\nwhere we get the second line because $0$,and $-1$ are clearly roots, then we get the third line from a common factorization\r\n\r\nthe solutions to the proposed problem are just the solutions to the above equation, just they are multiplied by $a$", "Solution_4": "Wait, so it doesn't make a difference whether or not it is 7^7 instead of 1^7...", "Solution_5": "[quote=\"mathgeniuse^ln(x)\"]Wait, so it doesn't make a difference whether or not it is 7^7 instead of 1^7...[/quote]\r\n\r\n$x^{7}+a^{7}=(x+a)^{7}$\r\n$\\left(\\frac{x}{a}\\right)^{7}+1=\\left(\\frac{x}{a}+1\\right)^{7}$\r\n$y^{7}+1=(y+1)^{7}$\r\n\r\nso they are equivalent problems (if $a=0$, then any $x$ is a solution)", "Solution_6": "If you have $a$ instead of $1$, solutions for $1$ must be multiplied by $a$ at the end." } { "Tag": [ "ratio", "exterior angle" ], "Problem": "In a given regular polygon, the ratio of the number of degrees in an interior angle to the number of degrees in an exterior angle is 7:2. How many sides does the polygon have?", "Solution_1": "[hide]The number of degrees of each interior angle is $\\frac{180(k-2)}{k}$. Each exterior angle is $\\frac{360}{k}$. Given the ratio of 7:2, we have \n\n$2(\\frac{180(k-2)}{k})=7(\\frac{360}{k})$\n\nThe k's cancel. \n\n$2(180k-360)=7(360)$\n\n$360(k-2)=360(7)$\n\n$k=9$[/hide]", "Solution_2": "Thanks. I knew that I needed to use 180(n-2)/n and 36/n but just did not know how to set up the proper equations.\r\nI got it now.", "Solution_3": "[hide]\n7+2=9(duh!)\n180/9=20.\n\nexterior angle=40.\n\n360/n=40\n40n=360\nn=9.\n\nTHIS IS A NONAGON.\n[/hide]", "Solution_4": "Good stuff here.\r\n\r\nThanks." } { "Tag": [ "geometry", "circumcircle", "power of a point", "radical axis", "geometry unsolved" ], "Problem": "Let B and C be arbitrary points on sides AP and PD respectively of an acute triangle APD. The diagonals of the quadrilateral ABCD meet at Q, and $ H_1,H_2$ are orthocenters of triangle APD and BPC respectively. Prove that if $ H_1H_2$ pass through point X which is the intersection of the circumcircles of triangles ABQ and CDQ other than Q, then it must also pass through Y which is the intersection of the circumcircles of triangles BCQ and ADQ other than Q.", "Solution_1": "$\\angle APD \\equiv \\angle BPC$ is not right $\\Longrightarrow$ $H_1 \\not\\equiv H_2$ are different from each other. $\\mathcal U, \\mathcal V$ are circles on diameters $AC, BD.$ Powers of $H_1$ to $\\mathcal U, \\mathcal V$ are both equal to half the power of $H_1$ to circumcircle $\\odot(APD),$ while powers of $H_2$ to $\\mathcal U, \\mathcal V$ are both equal to half the power of $H_2$ to cirumcircle $\\odot(BPC)$ $\\Longrightarrow$ $H_1H_2$ is radical axis of $\\mathcal U, \\mathcal V.$ Inversion $\\mathbb Q$ with center $Q$ takes quadrilateral $ABCD$ into quadrilateral $A'B'C'D',$ the opposite sides of which intersect at inversion images $X' \\equiv A'B' \\cap C'D'$ of $X$ and $Y' \\equiv B'C' \\cap D'A'$ of $Y.$ Circles $\\mathcal U, \\mathcal V$ on diameters $AC, BD$ go to circles $\\mathcal U', \\mathcal V'$ on diameters $A'C', B'D'.$ $\\mathcal Z'$ is a circle on diameter $X'Y'.$ Powers of the orthocenter of $\\triangle A'X'D'$ to $\\mathcal U', \\mathcal V', \\mathcal Z'$ are all equal to half its power to the circumcircle $\\odot A'X'D.'$ Similar statements hold for orthocenters of $\\triangle B'X'C', \\triangle A'Y'B', \\triangle C'Y'D'$ $\\Longrightarrow$ these circles are coaxal (and their collinear centers $U', V', Z'$ make Newton line of the quadrilateral $A'B'C'D'$). Let $\\mathcal Z$ be inversion image of $\\mathcal Z'$ under $\\mathbb Q.$ Then $\\mathcal Z$ is a circle through $X, Y$ coaxal with $\\mathcal U, \\mathcal V$ and $H_1H_2$ is their common radical axis. \n\nAssume $X \\in \\mathcal Z$ is on the radical axis $H_1H_2$ of the pencil $\\{\\mathcal U, \\mathcal V, \\mathcal Z\\}$ $\\Longrightarrow$\n\n1. Either $\\mathcal Z$ is a line, identical with the radical axis $H_1H_2$ of this pencil (regardless of whether the pencil is intersecting or not) $\\Longrightarrow$ $Y \\in \\mathcal Z$ is also on $H_1H_2.$\n\n2. Or the pencil is intersecting and $X \\in \\mathcal Z$ is identical with one intersection of this pencil. Feet of perpendiculars from $X$ to $AQC, BQD, ABP, PCD$ are collinear on a common Simson line $x$ of $\\triangle ABQ, \\triangle CDQ$ with the pole $X$ $\\Longrightarrow$ $x$ is also common Simson line of $\\triangle APC, \\triangle BPD$ with the pole $X$ $\\Longrightarrow$ circles $\\mathcal U, \\mathcal V$ are circumcircles of the last two triangles $\\Longrightarrow$ $P$ is the other intersection of the circles $\\mathcal U, \\mathcal V$ on diameters $AC, BD$ (Clifford circle theorem) $\\Longrightarrow$ $AP \\perp PD$, which contradicts the problem condition that the $\\triangle APD$ is acute." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "geometry" ], "Problem": "How's everyone's summer AMC Preparation going? I haven't been doing too much lately, just a weekend problem solving seminar but I need to do more work to prep, that's the only way to get good IMO, solve problems... Gah, once summer assignments are in I'm gonna go straight up on math... \r\n\r\n[b]Vaya con Dios![/b]\r\n\r\nIW.OAV", "Solution_1": "I've been doing problems and reading out of the AoPS everyday, and once I am done with that, Volume 2 and once I am done with that ACoPS. My goal for this year is to just make USAMO which I think is a very realistic goal plus I plan to participate in every single math oppurtunity there is (in the bay area).", "Solution_2": "[quote]Vaya con Dios![/quote]\r\n\r\nI like that comment. Hablaste espanol?", "Solution_3": "The Bay Area is an excellent place to live in terms of mathematical opportunities. There are at least nine Saturday competitions for you to go to.", "Solution_4": "Is there a list of Bay Area contests and math oppurtunities?", "Solution_5": "There's a list of competitions with sample problems on my website at [url]http://www.albanyconsort.com/simon[/url]. Then click on the LOCAL MATH COMPETITIONS link. There's a bunch of other fun stuff there. Enjoy.", "Solution_6": "Math contests by state:\r\nhttp://www.mualphatheta.org/Links/MAO_Links_Contests_Regional.htm \r\n\r\nGreat resource for pple in and around SFBA:\r\nhttp://www.albanyconsort.com/contests/contests.html\r\n\r\nHarker (private school in Bay Area) goes to almost every math contest. See their site for some contests - http://www.harker.org/announcements/2001_2002/math0102.htm will help, I hope.", "Solution_7": "Yay. I'm on Hann's math contest reference list.", "Solution_8": "Haha - took me a long time to realize that was your site. I found it through google last year... Nice site, btw. Only problem I have with it is the home page...too colorful for me. =P", "Solution_9": "Yes, it is a bit colorful, but I wanted a nice math background. Any other suggestions?" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "I like this one :\r\n\r\nLet $n$ be an integer.\r\nProve that there exists a finite graph with no triangle and with chromatic number greater than $n$.\r\n\r\nPierre.", "Solution_1": "From a k-chromatic triangle free graph G lets construct a k+1 chromatic triangle free graph G'. Let [i]X[/i](G) be the chromatic number of G. Let V(G)= {v1,...,vn}. Beginning with G add vertices X={x1,...,xn} and one more vertex y. Add edges to make x_i adjacent to all of N(v_i), and let N(y)= X. Clearly X is an idependet set in G'. Hence the other vertices of any triangle containing x_i belong to V(G) and are neighbors of v_i. This would complete a triangle in G, which can't exist. Therefore G' is triangle free.\r\nA proper k-coloring f of G extends to a proper k+1 coloring of G' by setting f(x_i)=f(v_i) and f(y)=k+1. Therefore [i]X[/i](G')<=[i]X[/i](G) + 1. So to obtain equality we show that [i]X[/i](G) < [i]X[/i](G'). So we consider any proper coloring of G' and obain from it a proper coloring of G using fewer colors. \r\nLet g be a proper k-coloring of G'. By changing the names of colors we may assume that g(y)=k. This restricts g to {1,...,k-1} on X. In V(G), it may use all k colors. Let A be the set of vertices in G on which g uses color k; we change the colors used on A to obain a proper k-1 coloring of G. For each v_1 in A we change the color of v_1 to g(x_i). Because all the vertices in A have color k under g, no two vertices of A are adjacent. Thus we need only check edges of the form v_i,v' with v_i in A and v' in V(G) - A. If v' is adjacent to v_i then by construction also v' is adjacent to x_i, which yields that g(v') and g(x_i) are different. Since we change the color on v_i to g(x_i) our change does not violate the edge v_i,v'. So the modified coloring of V(G) is a proper k-1 coloring of G.", "Solution_2": "Very nice!!!", "Solution_3": "My solution is different :\r\nLet $k=2^n$, let's consider the grid of the points of the plane with integer and positive coordinates. Let $\\Delta_i$ be the line with equation $x=i$, where $i=1,...,k$.\r\nNow consider the graph whose vertices are the grid points where $1 \\leq a,b \\leq k$. And join $(a,b)$ and $(c,d)$ by an edge iff $a+b=c$ or $c+d = a$, that is join $(a,b)$ to all the points of the line $\\Delta_{a+b}$ and only to them.\r\nAssume that $(a,b),(c,d),(e,f)$ form a triangle. Wlog, we may assume that $a \\leq c \\leq e$. In that case $a+b=c=e$ and $c+d=e$, which forces $d=0$, a contradiction. Thus, there is no triangle.\r\n\r\nNow, let $a,b$ be two positive integers both not greater than $k$, and assume that $a 2009.$ Then, $ a_3 < a_1$ is equivalent to $ \\frac{a_1\\plus{}2009}{a_2\\plus{}1} < a_1,$ or multiplying out, $ a_1\\plus{}2009 < a_1a_2 \\plus{} a_1,$ which is true, so $ a_3 < a_1.$ However, this means that the sequence $ b_i$ with $ b_1 \\equal{} a_2$ and $ b_2 \\equal{} a_3$ will also work but have smaller sum of first two numbers.\n\nNow, we prove that $ a_1a_2 \\ge 2009.$ Assume that $ a_1a_2 < 2009.$ Then similarly as last time, we can prove that $ a_1 < a_3.$ Also, we can prove that $ a_2a_3 < 2009,$ or $ \\frac{(a_1\\plus{}2009)a_2}{a_2\\plus{}1} < 2009,$ or $ a_1a_2 \\plus{} 2009a_2 < 2009a_2\\plus{}2009$ which is true, so $ a_2a_3 < 2009.$ Thus, induction gives us that $ a_ia_{i\\plus{}1} < 2009$ and $ a_i < a_{i\\plus{}2}.$ This is implies that the $ a_i$ will get arbitrarily large because they are integers, so $ a_ia_{i\\plus{}1} < 2009$ cannot be satisfied.\n\nThus, $ a_1a_2 \\equal{} 2009,$ and the minimum sum for this is $ a_1 \\equal{} 41, a_2 \\equal{} 49$ which works because the $ a_i$ just goes 41,49,41,49...[/hide]", "Solution_2": "Is this the easiest solution. It seems to be the hardest problem to actually prove on this AIME.", "Solution_3": "Simple proof by showing that $ a_n \\equal{} a_{n \\plus{} 2}$\r\n[hide=\"Another Approach\"]\nRearranging the given relationship, we have\n\\[ a_{n \\plus{} 2} \\plus{} a_{n \\plus{} 2}a_{n \\plus{} 1} \\equal{} a_n \\plus{} 2009 (E1)\n\\]\nChanging $ a_{n \\plus{} 2}$ to $ a_{n \\plus{} 3}$, we now have\n\\[ a_{n \\plus{} 3} \\plus{} a_{n \\plus{} 3}a_{n \\plus{} 2} \\equal{} a_{n \\plus{} 1} \\plus{} 2009 (E2)\n\\]\nE2-E1 gives us,\n\\[ a_{n \\plus{} 3} \\minus{} a_{n \\plus{} 1} \\equal{} \\frac {a_{n \\plus{} 2} \\minus{} a_n}{a_{n \\plus{} 2} \\plus{} 1}\n\\]\nAssuming all terms are positive integers,\n\\[ a_{n \\plus{} 2} \\plus{} 1 > 1\n\\]\nLet $ D_n$ be $ a_{n \\plus{} 2} \\minus{} a_n$, if $ D_1$ is not 0, then $ D_1 > D_2 > D_3....$ forms a contradiction since all Ds must always be positive integers. (Fermat's method of infinite descent)\nThen $ D_1 \\equal{} D_2 \\equal{} D_3 \\equal{} ... \\equal{} D_n \\equal{} 0$\nThis question hence becomes $ a_n \\equal{} a_{n \\plus{} 2}$[/hide]", "Solution_4": "Could you show the general idea?\r\n\r\nEdited:very beautiful", "Solution_5": "Copied the typing regulation line by line... DONE", "Solution_6": "Wow that's pretty nice gf4848.", "Solution_7": "[hide]Let $ a_1 \\equal{} a$ and $ a_2 \\equal{} b$. Then compute $ a_3 \\equal{} \\frac {a \\plus{} 2009}{b \\plus{} 1}$ and $ a_4 \\equal{} \\frac {(b \\plus{} 2009)(b \\plus{} 1)}{a \\plus{} b \\plus{} 2010}$. $ a_4$ is an integer, so $ (b \\plus{} 2009)(b \\plus{} 1) \\equal{} b^2 \\plus{} 2010b \\plus{} 2009\\equiv 0\\pmod{a \\plus{} b \\plus{} 2010}\\Rightarrow b^2 \\plus{} 2010b \\plus{} 2009 \\minus{} b(a \\plus{} b \\plus{} 2010) \\equal{} 2009 \\minus{} ab\\equiv 0 \\pmod{a \\plus{} b \\plus{} 2010}\\Rightarrow ab \\equiv 2009 \\pmod{a \\plus{} b \\plus{} 2010}$. \n\nIf $ ab \\equal{} 2009$, the minimum of $ a \\plus{} b$, $ a$ and $ b$ are integers is $ 41$ and $ 49$, and we can verify that this yields $ a_1 \\equal{} a_3 \\equal{} a_5 \\equal{} \\ldots \\equal{} 41$ and $ a_2 \\equal{} a_4 \\equal{} a_6 \\equal{} \\ldots \\equal{} 49$. \n\nIf $ ab\\neq 2009$, then $ ab\\ge 2009 \\plus{} (a \\plus{} b \\plus{} 2010) \\equal{} a \\plus{} b \\plus{} 4019$. Then $ \\sqrt {ab} > \\sqrt {4019} > 63$. By AM-GM, $ a \\plus{} b\\ge 2\\sqrt {ab} > 126$. This doesn't give a minimum, so the answer is when $ a\\plus{}b \\equal{} 41\\plus{}49\\equal{}\\boxed{090}$.\n[/hide]", "Solution_8": "this is what i did:\r\n\r\nyou can see that some sequence of integers has to repeat\r\nif a_1 = a_2,\r\nthen a_i = sqrt(2009), not integer\r\n\r\ntest a_1 = a_3\r\n\r\nyou get a nice expression of a_1 * a_2 = 2009\r\na_1 + a_2 is minimum when they are as close to each other as possible\r\nso you get min (a_1 + a_2) = 41 + 49 = 90\r\n\r\na_1 = a_4 gets too complicated..so i ignored\r\n\r\nnot a full solution, but still worked :D", "Solution_9": ":( ....\r\nDPatrick has STOLEN my solution.", "Solution_10": "[quote=\"gf4848\"]:( ....\nDPatrick has STOLEN my solution.[/quote]\r\n\r\nAlas, the horrors of AoPS admins. :o :( \r\n\r\nBut really, you should be proud of having such an elegant solution.", "Solution_11": "This one seems rather different from the ones already posted.\r\n[hide]\nLet $ b_k: \\equal{} a_k \\plus{} 1$, to simplify notation. The text now reads\n\\[ b_{n \\plus{} 1}b_{n \\plus{} 2} \\equal{} 2008 \\plus{} b_n \\plus{} b_{n \\plus{} 1},\n\\]\nfrom which we can infer\n\\[ b_{n \\plus{} 1} \\mid 2008 \\plus{} b_n \\plus{} b_{n \\plus{} 1},\n\\]\n\n\\[ b_{n \\plus{} 2} \\mid 2008 \\plus{} b_n \\plus{} b_{n \\plus{} 1}.\n\\]\nBoth are valid for all $ n$. Let us shift the indices down by one in the second formula, and subtract it from the first, to get $ b_{n \\plus{} 1} \\mid b_{n \\plus{} 1} \\minus{} b_{n \\minus{} 1}$. This implies that $ b_{n \\plus{} 1} \\mid b_{n \\minus{} 1}$, and thus from a certain point on we have $ b_n \\equal{} b_{n \\plus{} 2} \\equal{} b_{n \\plus{} 4} \\equal{} \\dots$, that is, the succession $ b_n$ is made with two alternating values, $ a,b,a,b,a,b,\\dots$. Since the relation can be inverted to yield $ b_n$ in terms of $ b_{n \\plus{} 2}$ and $ b_{n \\plus{} 1}$, this implies that the succession is in that form starting from the $ b_1$. Now it is clear that the only two conditions needed are $ a \\equal{} \\frac {a \\plus{} 2009}{b \\plus{} 1}, b \\equal{} \\frac {b \\plus{} 2009}{a \\plus{} 1}$, from which we get $ ab \\equal{} 2009$.\n(look, ma, no inequalities! :))\n[/hide]", "Solution_12": "Can you please explain the logic behind Fermat's method of infinite descent (used in [b]gf4848[/b]'s solution)? I read a wikipedia article on it with an example about the rationality of $\\sqrt{2}$, but I still don't completely understand it. Thank you :)", "Solution_13": "What you call it doesn't really matter. I'd call it the Pigeonhole Principle.\n\n[hide]Define the integer sequence $D_n = a_{n+2} - a_n$, and assume for the sake of contradiction that $D_1 \\neq 0$.\n\nWe know the sequence $D$ satisfies the equality $ D_{n+1} = \\frac{D_n}{a_{n+2}}$.\n\nTaking the magnitude and using $a_{n+2} > 0$, we get: $0 < | D_{n+1}| < |D_{n}|$\n\n$\\blacktriangleright$ Transitively: $0 < |D_i | < |D_j|$ for all $i > j \\; $.\n\nWe know $|D_i | \\in \\mathbb{N}$, so $|D_i| \\in \\{ 1 \\ldots |D_1| -1 \\}$ for all $i > 1$.\n\nConsider that set of values as $|D_1| - 1$ \"holes\", with each of $+\\infty$ sequence term $|D_i|$ as a \"pigeons\". By the Pigeonhole Principle, there must exist two distinct equal terms $|D_i | = |D_j|$ for some $i > j$, which contradicts the statement at $\\blacktriangleright$.\n\nIt follows that our assumption of $|D_1| \\neq 0$ must be false, and hence, $|D_1| = 0 \\implies D_1 = 0 \\implies \\boxed{\\forall i \\in \\mathbb{N}, \\; : \\; D_i = 0}$.[/hide]\n\n\n[b]@ gf4848:[/b][i] It's not immediately clear that $D_i$ must be positive.[/i]\n[b]Edit:[/b] You're right, I've fixed my post.", "Solution_14": "I think I understand it :). (I think $i > j$ though for fifth line? o.o unless I misunderstood your detailed response XD)\n\nThank you very much! :) :) I think I understand it much better now :)", "Solution_15": "Sure. By the way, a proof by \"infinite descent\" is basically a particular type of proof by contradiction, saying that you cannot have an infinite subset of a finite set. In that sense, it can always be expressed in terms of the Pigeonhole Principle. In the example here, we wanted an infinite set of unique $| D_i |$'s that had to be a subset of $ \\{ 1, \\; \\ldots \\| D_1 \\| -1 \\}$, which is impossible.\n\nIn the $\\sqrt{2}$ irrationality case, the argument was that if $(p_0 , q_0)$ is a solution, then $(p_1, q_1) = \\left( \\frac{p_0}{2} , \\; \\frac{q_0}{2} \\right)$ is another, strictly smaller solution. Thus, we can generate an infinite set of correct solutions no larger than $(p_0, q_0)$.\n\nHowever, the set of solutions no larger than $(p_0, q_0)$ is clearly a subset of the set of ordered pairs no larger than $(p_0, q_0)$, and the latter is a finite set. Thus, we've arrived at a contradiction.\n\nNote that I haven't defined \"smaller\" -- I could have used $(w, x) < (y, z) \\iff w1 \\iff b_2>b_4$. Since $b_2=b_4(b_3-b_5+1)$, and $b_2>b_4$, we have $b_3-b_5+1>1 \\iff b_3>b_5$. Continuing this reasoning gives that if for there exists some $k$ such that $b_k>b_{k+2}$, then the sequence of odd indices and even indices are strictly decreasing, and for large enough $k$ we can get a contradiction. Thus, $b_k=b_{k+2}$ for all $k$.\n\nAnswer extraction is easy from here: we now have \\[b_1+b_2+2008=b_1b_2\\iff (b_1-1)(b_2-1)=a_1a_2=2009\\] so the minimum of $a_1+a_2$ is $41+49=\\boxed{090}$.\n[/hide]", "Solution_20": "Can I have a hint on how to start this problem? It seems rather difficult! :read: ", "Solution_21": "Try to show that the sequence has to have period of length 2.", "Solution_22": "[quote=channing421]Try to show that the sequence has to have period of length 2.[/quote]\n\nI'm just revisiting this now for some reason, but I found several terms and didn't find a period, I don't really know how to show $a_n = a_{n+2}$..", "Solution_23": "It doesn't hold for all sequences; you must use the condition that all terms in the sequence are positive integers (it is intuitively true that the sequence must be periodic; otherwise, it'd be hard to make all of the terms positive integers!)." } { "Tag": [], "Problem": "\u0393\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $a,b,c\\geq 0$ \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 \r\n$\\frac{a(b^2+1)(c^2+1)}{[1+b^2c^2+(b+c)^2](a^2+1)}+\\frac{b(c^2+1)(a^2+1)}{[1+a^2c^2+(a+c)^2](b^2+1)}+\\frac{c(a^2+1)(b^2+1)}{[1+b^2a^2+(a+b)^2](c^2+1)}\\leq 1$", "Solution_1": "Giati kaneis den edose prosoxi se aftin ?\r\nLoipon, iparxei mia lisi se 2 seires :wink:", "Solution_2": "\u039d\u03b1\u03b9 \u03a3\u03c4\u03ad\u03c1\u03b3\u03b9\u03bf \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 :D .\u039a\u03b1\u03bb\u03ce\u03c2 \u03ae\u03c1\u03b8\u03b5\u03c2 \u03c0\u03ac\u03bb\u03b9 :lol:", "Solution_3": "Ena hint?\r\n\r\neinai kleisth, isn't it?\r\n\r\n\r\n :maybe:", "Solution_4": "\u0388\u03bd\u03b1 hint : \u039c\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03b9\u03c3\u03bc\u03cc\u03c2 :D", "Solution_5": "\u0391\u03bd \u03ba\u03b9 \u03b8\u03b1 \u03ae\u03b8\u03b5\u03bb\u03b1 \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03c3\u03b5 \u03b4\u03c5\u03bf \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03c4\u03bf\u03c5 \u03a3\u03c4\u03ad\u03c1\u03b3\u03b9\u03bf\u03c5 \u03b8\u03b1 \u03ae\u03b8\u03b5\u03bb\u03b1 \u03bd\u03b1 \u03c1\u03ce\u03c4\u03b7\u03c3\u03c9 \u03bc\u03ae\u03c0\u03c9\u03c2 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03b1\u03c0\u03b1\u03b3\u03c9\u03b3\u03ae \u03c3\u03b5 \u03ac\u03c4\u03bf\u03c0\u03bf \u03b4\u03b7\u03bb\u03b1\u03b4\u03b7 \u03b1\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u0391 \u03bc\u03ad\u03bb\u03bf\u03c2 >1 ? \u03b8\u03b1 \u03ae\u03b8\u03b5\u03bb\u03b1 \u03bd\u03b1 \u03bc\u03b1\u03c2 \u03c0\u03b5\u03af \u03bf \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03cc\u03c2 \u03b1\u03bd \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03bf\u03c4\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b2\u03b3\u03b5\u03af", "Solution_6": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd\u03c2 \u03ba\u03b1\u03b9\u03c1\u03cc\u03c2 \u03c0\u03ad\u03c1\u03b1\u03c3\u03b5......\r\n\u0398\u03ad\u03c4\u03bf\u03c5\u03bc\u03b5 $x=\\frac{2a}{a^{2}+1}$ \u03ac\u03c1\u03b1 $0\\leq x\\leq 1$ .\r\n\u038c\u03bc\u03bf\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03ac\u03bb\u03bb\u03b1 .\u0388\u03c4\u03c3\u03b9 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03bc\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd :\r\n$\\frac{x}{1+yz}+\\frac{y}{1+zx}+\\frac{z}{1+xy}\\leq 2$ \r\n\r\n\u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03b9\u03b1 \u03b8\u03ad\u03bb\u03c9 \u03bd\u03b1 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b5\u03c4\u03b5 \u03cc\u03c0\u03c9\u03c2 \u03b4\u03b5\u03bd \u03c4\u03b7\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 :wink: \r\n\u03a4\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03bb\u03ad\u03bd\u03b5 \u03c4\u03bf \u03b5\u03be\u03ae\u03c2 :(\u03c4\u03bf\u03c5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03b1\u03c5\u03c4\u03ac \u03c0\u03bf\u03c5 \u03ad\u03c7\u03c9 \u03b5\u03b3\u03ce.....)\r\n\u03a5\u03c0\u03bf\u03b8 \u03a7\u0392\u03a4\u0393 \u03cc\u03c4\u03b9 $0\\leq x\\leq y\\leq z\\leq 1$ \r\n\u0386\u03c1\u03b1 \r\n$\\frac{x}{1+yz}+\\frac{y}{1+zx}+\\frac{z}{1+xy}\\leq\\frac{x+y+z}{1+xy}\\leq 2$ \r\n\r\n\u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03c9\u03c1\u03b1\u03af\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03b1\u03bb\u03bb\u03ac \u03b3\u03b9\u03b1\u03c4\u03af \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c8\u03ac\u03be\u03bf\u03c5\u03bc\u03b5 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03c4\u03b9 \u03c4\u03bf \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03cc...... :)", "Solution_7": "\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad \u03bf\u03c4\u03b9 \u03b5\u03bd\u03bd\u03bf\u03b5\u03af\u03c2 \r\n\u03c7= 2*\u03b1/\u03b1^2+1 \u03ba\u03b9 0<\u03c7=<1\r\n\u03b4\u03b9\u03bf\u03c4\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03cc\u03c0\u03c9\u03c2 \u03c4\u03bf \u03ad\u03b3\u03c1\u03b1\u03c8\u03b5\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c3\u03b1\u03bd \u03bc\u03b9\u03b1 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 \u03bc\u03b5 \u03b2\u03ac\u03c3\u03b7 \u03c4\u03bf \u03c7 \r\n\u03b4\u03b5\u03bd \u03c4\u03bf \u03ba\u03ac\u03bd\u03c9 \u03ba\u03b1\u03ba\u03bf\u03c0\u03c1\u03bf\u03b5\u03c1\u03b1\u03b9\u03c4\u03b1 \u03b1\u03bb\u03bb\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c0\u03bf\u03c6\u03c5\u03b3\u03b7 \u03c0\u03b1\u03c1\u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b5\u03c9\u03bd \r\n\u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad \u03b1\u03bd \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03b4\u03b9\u03bf\u03c1\u03b8\u03c9\u03c3\u03ad \u03c4\u03bf", "Solution_8": "\u03b5\u03be\u03c5\u03c0\u03bd\u03bf \u03c3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03b5.\u03b1\u03bc\u03b2\u03c1\u03bf\u03c3\u03b9\u03b5 \u03b1\u03c5\u03c4\u03bf \u03b5\u03bd\u03bd\u03bf\u03b5\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03c9\u03c2.\u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03b1\u03bb\u03bb\u03b7 \u03bb\u03c5\u03c3\u03b7 ?\u03b3\u03b9\u03b1\u03c4\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03b1 \u03c4\u03bf \u03b8\u03b5\u03bc\u03b1 \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03b5\u03c5\u03b5\u03b9.\u03b1\u03bb\u03b7\u03b8\u03b5\u03b9\u03b1 \u03c0\u03bf\u03b9\u03bf \u03b7\u03c4\u03b1\u03bd \u03c0\u03bf\u03b9\u03bf \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03bf \u03b7 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1 \u03c4\u03bf\u03c5 \u03c3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd \u03ae \u03c4\u03bf \u03c6\u03b5\u03c4\u03b9\u03bd\u03bf \u03c4\u03b5\u03c4\u03b1\u03c1\u03c4\u03bf \u03b8\u03b5\u03bc\u03b1 \u03c4\u03bf\u03c5 \u03bc\u03b9\u03ba\u03c1\u03bf\u03c5 \u03b1\u03c1\u03c7\u03b9\u03bc\u03b7\u03b4\u03b7?", "Solution_9": "\u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf", "Solution_10": "\u03a4\u03bf \u03b4\u03b9\u03cc\u03c1\u03b8\u03c9\u03c3\u03b1 \u03c4\u03bf \u03c4\u03c5\u03c0\u03bf\u03b3\u03c1\u03b1\u03c6\u03b9\u03ba\u03cc..........\r\n\u03a3\u03c4\u03ad\u03c1\u03b3\u03b9\u03bf \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ae???? :)", "Solution_11": "\u03c3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd \u03b8\u03b5\u03c2 \u03bc\u03b9\u03b1 \u03bb\u03c5\u03c3\u03b7 \u03c7\u03c9\u03c1\u03b9\u03c2 \u03bd\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03b4\u03b9\u03b1\u03c4\u03b1\u03be\u03b5\u03b9\u03c2 \u03c3\u03b5 \u03b1\u03c5\u03be.\u03c3\u03b5\u03b9\u03c1\u03b1? \u03b3\u03b9\u03b1\u03c4\u03b9 \u03c7/zy+1<=x/xyz+1 to idio isxyei kai gia ta alla 2 klasmata \u03b1\u03c1\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b8\u03b5\u03c4\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b9\u03c2 3 \u03c3\u03c7\u03b5\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03bb\u03c5\u03bd\u03b5\u03c4\u03b1\u03b9", "Solution_12": "\u0393\u03b9'\u03b1\u03c5\u03c4\u03bf \u03c0\u03bf\u03c5 \u03c1\u03c9\u03c4\u03b7\u03c3\u03b5 \u03bf \u039a\u03c9\u03c3\u03c4\u03b1\u03c2 \u03b4\u03b7\u03bb\u03b1\u03b4\u03b7 \u03c0\u03b9\u03b1 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b1\u03c0\u03bf \u03c4\u03b9\u03c2 2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03b7 \u03c0\u03b9\u03c3\u03c4\u03b5\u03c5\u03c9 \u03bf\u03c4\u03b9 \u03b1\u03c0\u03b1\u03bd\u03c4\u03b7\u03c3\u03b7 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b4\u03c9\u03c3\u03bf\u03c5\u03bc\u03b5!\u0397\u03bb\u03c5\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd \u03bc\u03b5 \u03bc\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03b9\u03c3\u03bc\u03bf \u03b7\u03c4\u03b1\u03bd \u03c0\u03bf\u03bb\u03c5 \u03ba\u03b1\u03bb\u03b7!\u0392\u03b5\u03b2\u03b1\u03b9\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03c6\u03b5\u03c4\u03b9\u03bd\u03b7 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03c0\u03b5\u03c3\u03b5 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03bc\u03b9\u03ba\u03c1\u03bf\u03c5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03c9 \u03bd\u03b1 \u03c0\u03c9 \u03bf\u03c4\u03b9 \u03b7\u03c4\u03b1\u03bd \u03c0\u03bf\u03bb\u03c5 \u03ba\u03b1\u03bb\u03b7 \u03ba\u03b1\u03b9 \u03b7 \u03bb\u03c5\u03c3\u03b7 \u03c4\u03b7\u03c2 \u0395\u039c\u0395 \u03b1\u03ba\u03bf\u03bc\u03b1 \u03ba\u03b1\u03bb\u03c5\u03c4\u03b5\u03c1\u03b7!\u0391\u03bd \u03ba\u03b1\u03b9 \u03c5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd \u03ba\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03b5\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2... (\u03c7\u03c9\u03c1\u03b9\u03c2 \u03c7\u03c1\u03b7\u03c3\u03b7 \u03c4\u03c1\u03b9\u03b3\u03bf\u03bd\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03c9\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd)... :)", "Solution_13": "\u03a6\u03af\u03bb\u03b5 \u0394\u03b7\u03bc\u03ae\u03c4\u03c1\u03b7 \u03b5\u03b3\u03c9 \u03c4\u03b7\u03bd \u03ad\u03c7\u03c9 \u03bb\u03cd\u03c3\u03b5\u03b9 \u03c7\u03c9\u03c1\u03b9\u03c2 \u03c4\u03c1\u03b3\u03c9\u03bd\u03bf\u03bc\u03b5\u03c4\u03c1\u03af\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03c5\u03b5\u03c4\u03b1\u03b9 \u03bf\u03c4\u03b9 1/3>=\u03c7*\u03c5> -1 \u03ac\u03c1\u03b1 2/9>=\u039a>=-2 \u03b7 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c5\u03ba\u03bf\u03bb\u03ae \u03bc\u03b5 \u03c4\u03c1\u03b9\u03ce\u03bd\u03c5\u03bc\u03bf", "Solution_14": "\u039a\u03b1\u03bb\u03b7 \u03c3\u03ba\u03b5\u03c8\u03b7 \u03c6\u03b9\u03bb\u03b5 \u03bc\u03bf\u03c5 \u0391\u03bc\u03b2\u03c1\u03bf\u03c3\u03b9\u03b5.\u0395\u03b3\u03c9 \u03b5\u03c7\u03c9 \u03b2\u03c1\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b5\u03be\u03b7\u03c2 \u03b1\u03c0\u03bb\u03b7 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7:\u03b8\u03b5\u03c4\u03c9 x+y=a kai x-y=b \u03bf\u03c0\u03bf\u03c4\u03b5 \u03b7 \u03c0\u03c1\u03bf\u03c2 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1 \u03bc\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03b9\u03b6\u03b5\u03c4\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03bf\u03b4.\u03b5\u03c5\u03ba\u03bf\u03bb\u03b1. :)", "Solution_15": "\u03b1\u03bd \u03bb\u03c5\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b5\u03c4\u03c3\u03b9 \u03c4\u03bf 4\u03bf \u03b8\u03b5\u03bc\u03b1 \u03c4\u03bf\u03c5 \u03b1\u03c1\u03c7\u03b9\u03bc\u03b7\u03b4\u03b7 \u03c0\u03b9\u03c3\u03c4\u03b5\u03c5\u03c9 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03c9 \u03bf\u03c4\u03b9 \u03c4\u03bf\u03c5 \u03c3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03bf\u03c4\u03b5\u03c1\u03bf.\u03b3\u03b9\u03b1\u03c4\u03b9 \u03b8\u03b5\u03bb\u03b5\u03b9 \u03bc\u03b5\u03b3\u03b1\u03bb\u03b7 \u03c0\u03c1\u03bf\u03c3\u03bf\u03c7\u03b7 \u03c3\u03c4\u03bf \u03bc\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03b9\u03c3\u03bc\u03bf. :o", "Solution_16": "\u039d\u03b1\u03b9 \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03bf\u03c4\u03b9 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b4\u03af\u03ba\u03b9\u03bf \u03c6\u03af\u03bb\u03b5 \u039a\u03ce\u03c3\u03c4\u03b1 \u03ba\u03b1\u03b8\u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03ae \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03b5\u03c1\u03ac\u03c3\u03c4\u03b9\u03b1 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03bc\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03af\u03c3\u03b5\u03b9\u03c2 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1" } { "Tag": [ "geometry", "\\/closed" ], "Problem": "In the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=3184&view=next#,\r\n[quote=\"manlio\"]\nLet $d_1 \\leq d_2 \\leq d_3$ where $d_1,d_2,d_3$ are the distances of the circumcenter of a triangle to its [b][u]excentres[/u][/b]. Prove that:\n\\[ 1)\\ R {k+3 \\choose k}$ for $k \\geq 3$. Thus, we must have $n > k + 3$ when $k \\geq 3$. Also note that\n${k + 4 \\choose k} - (k^3 + 1) = \\frac1{24}(k+1)((k+4)(k+3)(k+2)-24(k^2-k+1)) = \\frac1{24}k(k+1)(k-5)(k-10)$ \nThis final expression is positive when $k > 10$, so in those cases we must have $n < k + 4$. Combining this result with that of the previous line, there are no solutions for $k > 10$. We also note from this factorization that $k = 5, 10$ give us solutions. The only cases left to check are $5 < k < 10$, and in these it suffices to note that ${11\\choose6} > 6^3+ 1, {12\\choose 7} > 7^3 + 1, {13\\choose 8} > 8^3 + 1$ and ${14\\choose 9} > 9^3+1$, so there are no solutions in these cases, either.[/hide]", "Solution_2": "thanks for help :)" } { "Tag": [ "limit", "logarithms", "calculus", "calculus computations" ], "Problem": "Prove that $ \\lim_{x \\to 0} x^{x} \\equal{} 1$.", "Solution_1": "hello, rewriting $ x^x$ in the following form\r\n$ x^x = e^{\\frac {\\ln(x)}{\\frac {1}{x}}}$ and applying L'Hospitals rule we get\r\n${ e^{\\lim_{x \\to 0 + }\\frac {\\frac {1}{x}}{{\\frac { - 1}{x^2}}}} = e^{\\lim_{x \\to 0 + } - x}} = e^{0} = 1$\r\nSonnhard.", "Solution_2": "Thanks Sonnhard. (You've written $ x$ instead of $ e$ just after the $ \\equal{}$ sign in your second line.)\r\n\r\nI was hoping for something without l'Hospital's rule (spelling your way, as the circumflex, which I can't do, indicates that \"s\" was once there).", "Solution_3": "hello, thank you for your hint, i hope it is correct now.\r\nSonnhard.", "Solution_4": "Does this means proving the limit exist? I hesitated as I was not sure if the limit even exists or not.", "Solution_5": "Hi KenHungKK. I guess so. Do you have a proof (assuming that the limit exists)?\r\n\r\n\r\n\r\nHi Sonnhard. I can't see what you've edited but your typo is still there.\r\n\r\n$ x^{x} \\equal{} e^{\\ln x^{x}} \\equal{} e^{x \\ln x} \\equal{} e^{\\frac{\\ln x}{\\frac{1}{x}}}$.\r\n\r\nSo, $ x^{x} \\equal{} e^{\\frac{\\ln (x)}{\\frac{1}{x}}}$; on your second line you've written $ x^{x} \\equal{} x^{\\frac{\\ln (x)}{\\frac{1}{x}}}$.", "Solution_6": "[u]Another Way:[/u] $ \\lim_{x \\to 0^ \\plus{} } x^x \\equal{} \\lim_{x \\to 0^ \\plus{} } e^{x \\ln x} \\equal{} e^0 \\equal{} 1$", "Solution_7": "hello, but here one must know that $ \\lim_{x \\to 0\\plus{}}x\\ln(x)\\equal{}0$.\r\nSonnhard.", "Solution_8": "And how do we show that $ \\lim_{x \\to 0^{\\plus{}}} x \\ln x \\equal{} 0$?", "Solution_9": "How about first showing that $ \\lim_{x \\to \\infty} \\frac{1}{x} \\ln x \\equal{} 0$?\r\n\r\nLet $ x\\equal{} e^{t}$. Then $ \\frac{1}{x} \\ln x \\equal{} e^{\\minus{}t}\\cdot t \\equal{} te^{\\minus{}t} \\to 0\\ (t \\to \\infty)$\r\n\r\nNow let $ x \\equal{} \\frac{1}{u}$. Then $ u \\ln u^{\\minus{}1} \\equal{} \\minus{}u\\ ln u \\to 0 \\ (u \\to 0^{\\plus{}})$.\r\n\r\nTherefore $ x \\ln x \\to 0 \\ (x \\to 0^{\\plus{}})$." } { "Tag": [ "quadratics", "algebra", "difference of squares", "special factorizations" ], "Problem": "1. If $ 0n\\plus{}1$, we have $ f(n)\\geq n\\plus{}2$ and so $ f(f(n))\\leq n^2\\plus{}3n\\plus{}3\\minus{}(n\\plus{}2)^2\\equal{}\\minus{}n\\minus{}1<0$, which is impossible. So $ f(n)\\leq n\\plus{}1$ $ \\forall n$\r\n\r\nIf $ 0n^2\\plus{}3n\\plus{}3\\minus{}(n\\plus{}1)^2\\equal{}n\\plus{}2>f(n)\\plus{}1$ and so $ f(m)>m\\plus{}1$ with $ m\\equal{}f(n)$ and this is impossible (see line above).\r\n\r\nSo we must have $ f(n)\\equal{}n\\plus{}1$ $ \\forall n$ and it is easy to check that this necessary condition is sufficient." } { "Tag": [ "probability", "puzzles" ], "Problem": "There are 7 hat colors. 7 people get 1 hat each which can be any colors e.g. they can be all the same. Each person is able to see the colors of all the rest. When everyone is ready, they have to guess their color by saying out SIMULTANEOUSLY. They will win if at least 1 person guesses correctly. They can make a plan in advance.\r\n\r\nHow do they win?", "Solution_1": "What types of colors are the hats? I mean, what if one of them is like \"my hat is orange\" and the guy who checks if they said it right says \"no, it's macaroni and cheese orange\" :maybe:", "Solution_2": "In the original problem, it is rainbow color. It's easier to say color 1-7.", "Solution_3": "oh, oh, oh, i totally remember doing a problem like this in sixth grade, if only i could remember the answer. \r\n\r\nI'll get back to you on this.", "Solution_4": "What are they allowed to do before everyone is ready? I'm assuming they aren't allowed to talk... I have an idea though.[hide]Suppose they are standing in a circle. Before hand, they number the colors 1-7. Before everyone is ready, they hold up the number of the color of the hat of the person to the right of them, and look to see what the person to the left of them held up for their own color. This way everybody gets it right.\n\nI think something is wrong with my answer, but not sure what. :( [/hide]", "Solution_5": "[quote=\"laughinghead505\"]Suppose they are standing in a circle. Before hand, they number the colors 1-7. Before everyone is ready, they hold up the number of the color of the hat of the person to the right of them, and look to see what the person to the left of them held up for their own color. This way everybody gets it right.\n\nI think something is wrong with my answer, but not sure what. [/quote]\r\n\r\nI don't think they are allowed to communicate to each other riight before the guess, but they can only plan before the hats are handed out.", "Solution_6": "OH! OH! I think I know!!! :lol: [hide]They all stand in a circle. Before hand, they chose one person to save. When they are all gathered, the other six will slouch or do something like that (decided before hand) if the color of the hat of the person who was chosen is the same as the color of the person next to them. That way the one person will know his color.\n\nIs it right??? :) [/hide]", "Solution_7": "and if there isn't a person with the same color they will do nothing\r\n\r\nbut I do not think that this solution is allowed :D", "Solution_8": "Why not? :(", "Solution_9": "it is not... maths :D", "Solution_10": "They are not allowed to send any signal. Only thing they can do is saying out a color.\r\n\r\n[hide=\"Hint\"]Try 2 people with 2 colors first.[/hide]", "Solution_11": "[hide=\"2 people\"]They decide before hand that one of them will say the color of the other person and the other person will say the color that is not the other person's color. For example: A and B are the two people and the colors of the hats are 1 and 2. They decided that A will say the color of B and B will say the color that is not of A. If A's hat is 1 and B's hat is 2, then A will say 2 and B will say 2.[/hide]", "Solution_12": "That's right! :lol:", "Solution_13": "What's about seven people?", "Solution_14": "there's several strategies that give them a good probability of winning, but can they always win? \r\n[hide=\"for example\"]they could all say the color of the person to the right of them. the probability this would work is the probability that two adjacent people have the same color. \n[/hide]", "Solution_15": "There's a way to be sure that at least one person answers it correctly.", "Solution_16": "I'm stuck. :(", "Solution_17": "I think that it is impossible because no matter what color everyone says, they could just plain be wrong. Is this a trick question?", "Solution_18": "Well, with two people, no matter what they say they could just be plain wrong, but there's a way to do it... :maybe:", "Solution_19": "Oh...I see how to do it now.\r\n[hide]\nThey together, assign each color a number, ie 1 2 3 4 5 6 0. When they go out there, the first person says the color with number congruent to one plus the additive inverse of the sum of all the other hats (modulo 7). The second person says 2 plus the sum of the additive inverse of the sum he sees. And so on. If the sum of all of the hats there is $k\\mod{7}$, then the person with number congruent to $k\\mod{7}$ will answer correctly (and only that person)\n\nThis is true because the first person is saying what color his hat will be if the sum is 1mod7, the second person is saying what color his hat will be if the sum is 2mod7, and so on. One of them is right.\n[/hide]", "Solution_20": "Finally got it.\r\n\r\nNumber the people 0 to 6. Number the colour 0 to 6. Let $a_{n}$ be the colour that #n person have.\r\n\r\nLet N be the sum of products of the number of colour and the number of person :roll: \r\nie) $\\Sigma n a_{n}$\r\nNote that $a_{0}$ does not affect N.\r\n\r\nThis is how they guess:\r\nperson 1 will guess so that the N according to the guess and the others colour is $1+a_{0}$ mod 7. Likewise, person n will guess so that N is congruent to $n+a_{0}$ mod 7, where n>0. This is possible because (n,7)=1. person 0 will guess N mod 7. This is also possible, because this person knows N.\r\n\r\nThis is why it works:\r\nThere are 7 possible value of N mod 7, 0 to 6. Person 1 to 6 are covering every possible case but $a_{0}$ case. Yet, person 0 is guessing N mod 7 as his/her colour. Therefore, we have covered all cases.\r\n\r\nwarning: This solution may not be clear, since I'm writing this near midnight :sleep2: (11:42 to be exact). Ask me to clear things up.", "Solution_21": "[hide=\"My solution\"]Let the colors be 0,1,2,3,4,5,6. Each person finds the sum of the others' hat colors.\nPerson 1 says color that makes the sum congruent 1 mod 7.\nPerson 2 says color that makes the sum congruent 2 mod 7.\nand so on.\n\nThis works because their hat colors sum has to congruent either 0,1,2,3,4,5 or 6.\n\nFor my solution, exactly one person answers correctly.[/hide]", "Solution_22": "Wow! That's a nice solution! :lol:", "Solution_23": "[hide=\"stupid solution\"]For the first person, one of the others gives him a secret and almost invisible sign, telling him what color his hat is. Then the others do that. Therefore, everyone is saved.[/hide]" } { "Tag": [ "function", "geometry", "geometric transformation", "reflection", "vector", "MATHCOUNTS", "number theory" ], "Problem": "How would you go about proving:\r\n\r\n1) Additive and commutative properties of addition and multiplication?\r\n\r\n2) Theorems related to the above (example, subtraction and division properties)?\r\n\r\n3) $ a^{x}\\times a^{y}= a^{x+y}$ ?\r\n\r\n4) That the inverse of a function such as $ y = x^{2}+4$ can be found by interchanging the x and y variables?\r\n\r\nI know it's quite possible that these are all accepted as definitions, but if there's the slightest chance that any of them can be derived from something, I want to know how. Hopefully no-one found that confusing :oops:", "Solution_1": "Well, for 3), that would equal $ (a_{1}* a_{2}*.... a_{x})(a_{1}* a_{2}*...... a_{y}) = (a_{1}* a_{2}........a_{x+y})$, so the exponent for $ a$ would be $ x+y$.", "Solution_2": "Definition of an inverse function is that given $ y=f(x)$ and $ y=f^{-1}(x)$ is its inverse:\r\n\r\n$ f(f^{-1}(x)) = f^{-1}(f(x)) = x$. So essentially $ y= f^{-1}(x)$ is the reflection of $ y=f(x)$ about the line $ y=x$. Which is why all you do is inter-change the variables.", "Solution_3": "For no.2)Subtraction and Division is defined as the inverse of addition and multiplication respectively, so theorems about subtraction and division comes from no.1)", "Solution_4": "[quote=\"Prime factorization\"]How would you go about proving:\n\n1) Additive and commutative properties of addition and multiplication?\n\n2) Theorems related to the above (example, subtraction and division properties)?\n\n3) $ a^{x}\\times a^{y}= a^{x+y}$ ?\n\n4) That the inverse of a function such as $ y = x^{2}+4$ can be found by interchanging the x and y variables?\n[/quote]\r\n\r\nThe first 3 can be proved using a formal system with the Peano Axioms (check out the book Godel Escher Bach for a very lovely explanation of formals systems and the foundations of mathematics and lots of other cool math-philosophical subjects). The difficulty discovered in 1931 by Godel was that while you can prove arithmetic expressions are true using any suitable Formal System, there is no way to prove that any given formal system is both contradiction-free and complete.\r\n\r\nThe last does not require a proof. It is a definition. And as my theory of computation teach always told us, a definition is never wrong. It could be useless, but it's never wrong. The way you know it is a definition is that it simply gives a name to a statement. In the case of inverse functions, when you say:\r\n\r\n$ y = f^{-1}(x)$, what is actually meant is the function which satisfies $ x = f^{-1}(y)$ (if one even exists). It is very important to be able to recognize the difference between definitions and theorems. Anyone can come up with definitions for any field. What matters is how useful those definitions are when it comes time to start proving theorems.", "Solution_5": "According to [url]http://en.wikipedia.org/wiki/Peano_axioms[/url], these axioms seem to only hold for natural numbers :( I don't see why 4.7 + 3.8 wouldn't equal 3.8+4.7, which is the commutative property of addition if I'm not mistaken.\r\n\r\n :huh:\r\n\r\nAlso, from what I read, the concept seems a little beyond elementary as it involves number theory and set theory. I'm trying to avoid such topics until I'm older and prepared for such material :blush:\r\n\r\nYou sure there isn't a simpler way to solve it? Perhaps you could show me a proof of just one axiom so I could pick up on the concept.", "Solution_6": "[quote=\"Prime factorization\"]According to [url]http://en.wikipedia.org/wiki/Peano_axioms[/url], these axioms seem to only hold for natural numbers :( I don't see why 4.7 + 3.8 wouldn't equal 3.8+4.7, which is the commutative property of addition if I'm not mistaken.[/quote]\n\nRational and Real numbers are just illusions of integers. A rational number is merely an ordered pair of Integers. A real number is just a function from Integers into the Rationals whose images tend to closer together as the argument of the function increases (Cauchy Sequences). \n\nThe math is no fun to work out. Proofs at this level are extremely mechanical and don't have the elegance of \"higher level\" proofs found in the rest of mathematics.\n\n\n\n[quote]Also, from what I read, the concept seems a little beyond elementary as it involves number theory and set theory. I'm trying to avoid such topics until I'm older and prepared for such material :blush:[/quote]\n\nLearn as much as you can as soon as you have the time, and you'll surprise yourself someday with how much of it you actually understood.\n\n\n\n[quote]You sure there isn't a simpler way to solve it? Perhaps you could show me a proof of just one axiom so I could pick up on the concept.[/quote] \r\n\r\nThere is a good example of the transitive property of integer addition in Godel Escher Bach, the book I spoke up. However, it takes two solid pages. These proofs aren't elegant. If you know anything about programming, Euclid's proof of the infinitude of primes would be analogous to a program written in Python, Java, or C, while elementary proofs like this would be like programming in assembler or raw machine code. \r\n\r\nThe proofs simply demonstrate the correctness, they do not paint a satisfactory picture of why those statements in arithmetic are true. The best proof for human consumption about the nature of arithmetic are probably given in terms of wooden blocks or small stones. Rigorous proofs are extremely useful when you can't visualize what you're talking about (for example, infinite dimension vector spaces or wave functions in Physics). They aren't really interesting when it comes to the fundamentals.", "Solution_7": "[quote=\"Prime factorization\"]these axioms seem to only hold for natural numbers :( I don't see why 4.7 + 3.8 wouldn't equal 3.8+4.7[/quote]\r\n\r\nAs I said in a previous MathCounts thread, definitions are only kept because they are useful. The structure of the rationals and the reals is defined so that they continue to satisfy properties like associativity, commutativity, and distributivity.\r\n\r\nThis is distinct from the topic of the rigorous construction of the reals, which is fairly complicated and beyond the scope of this board." } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "linear algebra", "matrix" ], "Problem": "Let R be an integral domain. and $ S \\equal{} \\{1,a,a^2,...\\}, a\\not \\equal{} 0$\r\nIs $ R_S$, the localization of $ R$ at $ S$ isomorphic to $ R[x] / ( \\minus{} 1 \\plus{} ax)$ ?\r\n\r\nI am having trouble proving this. I got that $ R[X] /( \\minus{} 1 \\plus{} ax)$ is equivalent to $ F \\equal{} \\{rf \\text{\\big{|}} \\minus{} 1 \\plus{} af \\equal{} 0, r\\in R\\}$ so $ f \\equal{} a^{ \\minus{} 1}$. So $ F \\equal{} \\{\\frac ra\\ | r\\in R\\}$. Is this $ R_S$?? I can't prove the isomophism part :(\r\n $ T[x] / (x^2 \\plus{} 1)$ is equivalent to all elements of the form $ t_1 \\plus{} t_2 \\alpha$ where $ \\alpha^2 \\plus{} 1 \\equal{} 0$, that is $ \\alpha$ is a root of the polynomial we're modding with and is easily extended to problems like $ \\mathbb{Z}_2[X] / x^2 \\plus{} x \\plus{} 1 \\equal{} \\{ a \\plus{} b\\beta \\}$ where $ \\beta$ is a root of $ x^2 \\plus{} x \\plus{} 1$ (Hence it has 4 elements =) ).\r\n\r\nThanks a bunch", "Solution_1": "For the picture to make sense, you have to include the higher-degree elements. I'd go with constructing the appropriate homomorphisms in both directions, and showing they're mutual inverses.\r\n\r\nFrom $ R_S$, we set $ f\\left(\\frac{b}{a^n}\\right)\\equal{}[bx^n]$.\r\nFrom $ R[x]$, we set $ g(x)\\equal{}\\frac1a$ and generate the rest from there; since $ g(ax\\minus{}1)\\equal{}a\\cdot \\frac1a\\minus{}1\\equal{}1\\minus{}1\\equal{}0$, $ g$ factors through the quotient $ R[x]/(ax\\minus{}1)$; call the map from the quotient $ g'$\r\n\r\nWe have $ g'\\circ f\\left(\\frac{b}{a^n}\\right)\\equal{}g'([bx^n])\\equal{}b\\cdot\\left(\\frac1a\\right)^n\\equal{}\\frac{b}{a^n}$. That direction is the identity.\r\nWe have $ f\\circ g'([b_0\\plus{}b_1x\\plus{}\\cdots \\plus{}b_nx^n])\\equal{}f\\circ g'(b_0)\\plus{}f\\circ g'(b_1x)\\plus{}\\cdots \\plus{}f\\circ g'(b_nx^n)$\r\n$ \\equal{}f(b_0)\\plus{}f\\left(\\frac{b_1}{a}\\right)\\plus{}\\cdots\\plus{}f\\left(\\frac{b_n}{a^n}\\right)\\equal{}[b_0]\\plus{}[b_1x]\\plus{}\\cdots\\plus{}[b_nx^n]$\r\nThat's the identity, too.\r\n\r\nYou my want to check that these are homomorphisms as well. As a side bit, this shows that every class in the polynomial quotient has monomial representatives at high enough degree; those are the common denominator.", "Solution_2": "Here another more conceptual approach which does not boils down to perform calculations with elements over and over again. In the sequel every ring is meant to be unital and commutative.\r\n\r\nNote that the localization $ R_S$ w.r.t. a multiplicative subset $ S$ fulfils a [b]universal property[/b] by which it is [u]uniquely[/u] (up to [u]unique[/u] isomorphism) determined: \r\n\r\nGiven any ring homomorphism $ \\varphi: R\\to T$ with $ \\varphi(S)\\subset T^*$ (the group of units of $ T$), then there is a [u]unique[/u] ring homomorphism $ \\varphi_S: R_S\\to T$ with $ \\varphi_S\\circ i_S\\equal{}\\varphi$ where $ i_S: R\\to R_S$ denotes the (natural) ring homomorphism defined by $ i_S(r)\\equal{}\\frac{r}{1}$:\r\n\\[ \\begin{matrix}\r\nR& &\\\\\r\n\\downarrow_{i_S}& \\searrow^\\varphi &\\\\\r\nR_S&\\stackrel{\\exists !\\varphi_S}{\\to}&T\r\n\\end{matrix}\\]\r\n\r\n\r\n\r\nAfter this general remark the claim boils down to show that $ R[X]/(aX\\minus{}1)$ fulfils the universal property of $ R_S$. For this use the universal properties of (i.) the polynomial ring and (ii.) quotient rings. After some training this can be seen just by \"ocular inspection\". ;)\r\n\r\n[color=blue]BTW: How can commutative diagrams be coded in this forum's postings? This above does not look too nice to me. [/color]:(", "Solution_3": "In fact, in general, it is simpler to construct [b]commutative[/b] localizations as quotients of polynomial rings since one can employ prior-known universal properties of polynomial rings and factor rings. For examples of this approach see e.g. Rotman, Advanced Modern Algebra or [1], though neither presentation is as simple as it could be. Note: The reason that the pair construction of fractions is ubiquitous is because it is needed in the noncommutative case -- not because the pair method is simpler or more natural. For more on the noncommutative case, see [2], esp. the survey by the renowned expert Paul M. Cohn: Localization in general rings, a historical survey, p.15. See also the many other interesting articles in that book, which is freely available online via the link [2] below.\r\n\r\n[1] Voloch. Rings of fractions the hard way.\r\nhttp://www.ma.utexas.edu/users/voloch/Homework/fractions.pdf\r\n\r\n[2] Noncommutative localization in algebra and topology\r\nICMS Edinburgh 2002, Edited by Andrew Ranicki\r\nhttp://www.maths.ed.ac.uk/~aar/books/nlat.pdf#page=15", "Solution_4": "Third Edition: There are packages for commutative diagrams, but they're not installed on the forum. Your best bet is to make a picture elsewhere and embed it.\r\nSee also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=250471]this thread[/url]." } { "Tag": [ "trigonometry", "inequalities" ], "Problem": "If $0\\le\\theta\\le2\\pi$ find all the values of $\\theta$ such that $\\sin2\\theta\\le\\sin\\theta$\r\n[i]Descartes 1982 3(b)[/i]\r\n\r\nOh, this is easy I thought to myself as I make a mistake I seem to do every time I'm confronted with a similar situation:\r\n\r\n[hide=\"Wrong solution\"]\n$\\sin2\\theta\\le\\line\\theta$\n$2\\sin\\theta\\cos\\theta\\le\\sin\\theta$\n$2\\cos\\theta\\le1$ <---- That's the kicker\n$\\cos\\theta\\le\\frac{1}{2}$\nTherefore $\\frac{2}{3}\\pi\\le\\theta\\le\\frac{4}{3}\\pi$\n[/hide]\n\nThis is the answer at the back of the book:\n[hide]\n$\\sin2\\theta=\\sin\\theta$\n$2\\sin\\theta\\cos\\theta=\\sin\\theta$\n$\\sin\\theta(2\\cos\\theta-1)=0$ <---- Should have done this\n$\\sin\\theta=0$ or $\\cos\\theta=\\frac{1}{2}$\nThe graphs intersect at $\\theta=0,\\frac{\\pi}{3},\\pi,\\frac{5\\pi}{3},2\\pi$\nThus, $\\sin2\\theta\\le\\sin\\theta$ if $\\theta=0$ or $\\frac{\\pi}{3}\\le\\theta\\le\\pi$ or $\\frac{5\\pi}{3}\\le\\theta\\le2\\pi$\n[/hide]\r\n\r\nSo what is conceptually wrong with the manipulation I made?", "Solution_1": "$2cos(\\theta)$ is not $\\le 1$. Try $\\theta = 0$.", "Solution_2": "Hmmm very true, I didn't actually look at the inequality while I was solving I just divided $\\sin\\theta$ throughout and kept on truckin'... So my question is, why doesn't that division work out to a correct solution? (Other than the fact that it does not in this specific case, or is it just this specific case?)\r\n\r\nEDIT: \r\nbasically I'm asking why you can't divide by $\\sin\\theta$ throughout in $2\\sin\\theta\\cos\\theta=\\sin\\theta$ so that it results in $2\\cos\\theta=1$. Logically in my head there's no reason why this shouldn't work. Obviously I'm missing something.", "Solution_3": "[hide]What you did was that you should have brought the $\\sin\\theta$ over.\n\nSo, it'd be:\n\n$2\\sin\\theta\\cos\\theta=\\sin\\theta$\n$2\\sin\\theta\\cos\\theta-\\sin\\theta=0$\n$\\sin\\theta(2\\cos\\theta-1)=0$ by factoring.\n\nAlso, you made a mistake since $2\\cos\\theta\\not\\leq 1$. It is only true if $\\theta\\neq 0$.[/hide]\r\n\r\nedit: 2 responses by the time I post :roll:", "Solution_4": "UPDATE:\r\nSo I had an epiphany while eating dinner (if you can call something so simple epiphanic... eh):\r\n\r\nAs an example:\r\n\r\n$\\frac{x^{2}-4}{x+2}$ can be simplified to $x-2$, but the fact remains that $x\\not=-2$ although it's not apparent in the simplification.\r\n\r\n$2\\cos x\\sin x=\\sin x$ can be simplified to $2\\cos x=1$ but we are omitting the trivial case when $\\sin x=0$ which is similarly unapparent after simplification like in the above example.\r\n\r\nEDIT: I'm pretty sure this reasoning is still faulty... anyone care to help me over this ridiculous mental block :P?\r\n\r\nEDIT 2: After further consideration, I don't think this reasoning is faulty because if you plot $2\\cos x\\sin x-\\sin x$ versus $2\\cos x-1$ the intercepts are the same.\r\n\r\nCould someone please just put me out of my misery and settle this? I hate being stumped on such trivial things.", "Solution_5": "I don't know what you are asking for, but I think you shouldn't have divided the inequality by $\\sin x$ in the first place because $\\sin x$ can be $0$ or negative. If it is negative, then the sign should face the opposite way.\r\n\r\n$-3>-5$\r\ndivide by $-1$(which is also a possible sine value) and you get\r\n$3>5$\r\nfirst but you change signs to make it correct\r\n$3<5$\r\n\r\nThe answer key solution you gave was very clear though. That's how my math teacher would do it, since she doesn't perform operations with inequalities at all :ninja:", "Solution_6": "[quote=\"#H34N1\"][hide]What you did was that you should have brought the $\\sin\\theta$ over.\n\nSo, it'd be:\n\n$2\\sin\\theta\\cos\\theta=\\sin\\theta$\n$2\\sin\\theta\\cos\\theta-\\sin\\theta=0$\n$\\sin\\theta(2\\cos\\theta-1)=0$ by factoring.\n\nAlso, you made a mistake since $2\\cos\\theta\\not\\leq 1$. It is only true if $\\theta\\neq 0$.[/hide]\n\nedit: 2 responses by the time I post :roll:[/quote]\r\nUhhh, dead, consider $\\theta=60$\r\n$\\sqrt{3}>1$", "Solution_7": "$60\\neq 0$." } { "Tag": [ "conics", "hyperbola" ], "Problem": "Find the number of real solutions of the equation $ (x\\plus{}1)(3^x\\minus{}2)\\equal{}1$. This was asked in an objective test but I am not giving the options though. I am doubtful on this.", "Solution_1": "It just so happens that there are two real options. One is slightly less than 1, and the other is about -1.5, but neither is rational. Unfortunately, I can't tell you how to get those answers, as I used a calculator=)", "Solution_2": "You don't need to solve the equation to find the number of solutions; consider graphs. Rearranging the equation, we obtain:\r\n\r\n$ 3^x \\equal{} \\frac {1}{x \\plus{} 1} \\plus{} 2.$\r\n\r\nClearly, $ 3^x$ will intersect once with the upper part of the hyperbola. The lower part of the hyperbola will also intersect the $ 3^x$ once because it is shifted up by $ 2$ and therefore, is above the $ x$-axis and must intersect with the exponent function. Thus, there are $ \\boxed{2}$ real solutions to the equation.\r\n\r\nOf course, this not very rigorous. I guess you could consider monotonicity.", "Solution_3": "Thanks for your replies and I too got 2 real solutions and using many applications of logarithms.Dont remember my solution and neither can I try now due to bad health.\r\n\r\nThanks again for your replies. Infact, I remember getting answers near $ \\plus{}$ or $ \\minus{}$ $ \\sqrt {2}$ goldenboy1.618 pointed it out too though he said one solution is slightly less than $ 1$ , but definitely my approach yielded two real solutions." } { "Tag": [ "pigeonhole principle" ], "Problem": "there are 15 subjects, 18 students. Each student pick 4 subjects. \r\nprove that there exist 2 student with 2 common subjects.", "Solution_1": "As easy as this problem may be, I'm not sure if proof questions belong in the High School Basics forum.", "Solution_2": "yes, it's a basic proof :)\r\nbut i dont know how to write it!\r\n(don't think it belongs to the HSB but also the intermediate (too easy for that)", "Solution_3": "thought this would be easy but new tothe subject, I don't know how to write it out.\r\n\r\ncan someone show me how :( :oops:", "Solution_4": "This is more of a conbinatorics problem, then the pigeonhole step is pretty trivial.", "Solution_5": "[hide]\nThere's 18 students, each have 4 subjects. There are $\\binom{4}{2}= 6$ different pairs of subjects each person does.\n\nIf we could fill all the 18 * 6 = 108 pairs of subjects so that no two are equal, then there wouldn't exist 2 students with 2 common subjects.\n\nThere are $\\binom{15}{2}= 105$ different pairs of subjects, which is less than 108 pairs required by the 18 students so that no 2 have 2 common subjects.\n[/hide]" } { "Tag": [ "inequalities", "logarithms", "geometry", "rectangle", "limit", "calculus", "calculus computations" ], "Problem": "\\[ 1\\plus{}\\frac{1}{2}\\plus{}\\frac{1}{3}\\plus{}\\cdots \\plus{}\\frac{1}{m} \\leq 1 \\plus{} \\log m\\]\r\n\r\nthanks", "Solution_1": "$ lg(n\\plus{}1)\\minus{}lg(n) \\geq \\frac{1}{n}$", "Solution_2": "[quote=\"andreica\"]$ lg(n \\plus{} 1) \\minus{} lg(n) \\geq \\frac {1}{n}$[/quote]\r\n\r\nit should be $ lg(n \\plus{} 1) \\minus{} lg(n) \\geq \\frac {1}{n\\plus{}1}$.", "Solution_3": "Draw a picture. This picture should include the following:\r\n\r\nThe curve $ y \\equal{} \\frac1x$ for $ 1\\le x\\le m.$\r\n\r\nBoxes of width $ 1$ that lie under the curve. That is, draw the rectangles with sides parallel to the axes whose lower left and upper right vertices are, respectively:\r\n$ (1,0)$ and $ \\left(2,\\frac12\\right).$\r\n$ (2,0)$ and $ \\left(3,\\frac12\\right).$\r\n$ \\dots$\r\n$ (m \\minus{} 1,0)$ and $ \\left(m,\\frac1m\\right).$\r\n\r\nAnd, for good measure, the square with lower left and upper right vertices $ (0,0)$ and $ (1,1).$\r\n\r\nCan you see the inequality in that picture?\r\n\r\nA slightly different picture should get you \r\n$ \\ln m\\le 1 \\plus{} \\frac12 \\plus{} \\frac13 \\plus{} \\cdots \\plus{} \\frac1{m \\minus{} 1}.$", "Solution_4": "How to compute $ \\lim_{m\\to \\infty} 1\\plus{} \\log m \\minus{} 1\\minus{}\\frac{1}{2}\\minus{}\\cdots\\minus{}\\frac{1}{m}$ ?", "Solution_5": "It's $ 1\\minus{}\\gamma$, where $ \\gamma$ is a famous constant defined by a very similar limit. Although it's not too hard to approximate $ \\gamma$, very little is known about the properties of the exact number.\r\n\r\nDon't look for an exact value here. Just estimate." } { "Tag": [ "geometry", "Support", "\\/closed" ], "Problem": "When a tag is added by clicking in the line of tag inputs above the message input area, and then manually deleted (meaning with backspace or delete), a closing tag is automatically inserted at the end of the post when submitted even though the original starting tag was already deleted. This causes some posts to unintentionally have a tag at the end of it dangling there in the middle of nowhere. The presence of these \"dangling tags\" is not that much of a problem, but is there a way to make the system check to make sure the start tag still exists before automatically implementing the end tag, so that users would not have to check and re-edit the extra tag out? Thanks for the thought input.", "Solution_1": "Or, you could spend an extra 15 seconds to preview your post, checking for such errors as well as other problems, which is something that everyone should probably do anyhow. Not every minor inconvenience deserves a post in the support forum. :roll:", "Solution_2": "As JBL said, you can preview to make sure. Also, you can look at the top and if there is a * beside one of them, that means that it is going to show up at the end. \r\n\r\nMy solution: If you want to bold something, hide something, ... , you can actually type the code, or you could select the text then click the button. This adds both the start and end at the same time." } { "Tag": [], "Problem": "Prove that $ 10|a$ where $ a\\equal{}33^{77}\\plus{}77^{33}.$", "Solution_1": "[hide=\"hint\"]In order for $ 10|a$, the last digit of $ 33^{77} \\plus{} 77^{33}$ must be a 0. So we only need to consider the last digits of $ 33$ and $ 77$, namely $ 3$ and $ 7$. Look for a pattern of the last digits of $ 3^k$ and $ 7^k$.[/hide]", "Solution_2": "hello, we get\r\n$ 33^{77} \\equiv 3 \\mod 10$\r\nand\r\n$ 77^{33} \\equiv 7 \\mod 10$\r\nso our sum $ 33^{77}\\plus{}77^{33} \\equiv 3\\plus{}7 \\equiv 0\\mod 10$.\r\nSonnhard.", "Solution_3": "[quote=\"Dr Sonnhard Graubner\"]hello, we get\n$ 33^{77} \\equiv 3 \\mod 10$\nand\n$ 77^{33} \\equiv 7 \\mod 10$\nso our sum $ 33^{77} \\plus{} 77^{33} \\equiv 3 \\plus{} 7 \\equiv 0\\mod 10$.\nSonnhard.[/quote]\r\n\r\nNote that 33 and 77 are both congruent to 1 in mod 4, which is why they have their corresponding units digits.\r\n\r\npowers of units digit 3 go in cycle: 3, 9, 7, 1 and repeats in cycles of 4\r\npowers of units digit 7 go in cycle: 7, 9, 3, 1 and repeats in cycles of 4", "Solution_4": "$ 33^{77}\\plus{}77^{33}\\equiv 3^{77}\\plus{}(\\minus{}3)^{33}\\equiv 3^{77}\\minus{}3^{33}\\equiv 3^{33}(3^{44}\\minus{}1)\\bmod 10$\r\n\r\nAnd $ 3^{44}\\equiv (\\minus{}1)^{22}\\equiv 1$, so this is $ 1\\minus{}1\\equal{}0$ $ \\Box$" } { "Tag": [], "Problem": "In how many ways can two dice be rolled to yield a sum divisible by 3?", "Solution_1": "The sum needs to be a 3, 6, 9, or 12\r\nFor 3:\r\n\r\n1,2\r\n2,1\r\n\r\nFor 6:\r\n\r\n1,5\r\n2,4\r\n3,3\r\n4,2\r\n5,1\r\n\r\nFor 9:\r\n\r\n3,6\r\n4,5\r\n5,4\r\n6,3\r\n\r\nFor 12:\r\n\r\n6,6\r\n\r\nSo there are $ 2 \\plus{} 5 \\plus{} 4 \\plus{} 1 \\equal{} \\boxed{12}$ ways." } { "Tag": [], "Problem": "A packer sorts eggs by size. She finds $ 5\\frac13$ dozen large eggs in the first 24 dozen eggs that she sorts. At this rate, how many large eggs can she expect to find in 60 dozen eggs?", "Solution_1": "24 is 2/5 of 60, so multiply $ 16/3\\times 5/2\\equal{}80/6\\equal{}40/3$" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find all fundtionsf such that f(2008)=$ a^3$;f(2010)=$ (a\\plus{}1)^3$ (with a is a integer)", "Solution_1": "I don't understand. Without any restrictions on $ f$, the other values of $ f(n)$ are arbitrary." } { "Tag": [ "linear algebra", "matrix", "invariant", "geometry", "geometric transformation", "rotation", "linear algebra unsolved" ], "Problem": "Let $ A\\in\\mathbb{R}^{n\\times n}$ be an irreducible doubly stochastic matrix, how can I show that its cyclic index divides $ n$?", "Solution_1": "We use Perron-Frobenius theorem.\r\nLet $ A$ be an irreducible and nonnegative $ n\\times{n}$ matrix and let $ r$ its spectral radius. Then $ r$ is an eigenvalue of $ A$ and if $ \\lambda$ is an other eigenvalue then $ |\\lambda|\\leq{r}$. \r\nMoreover if $ A$ has exactly $ p$ eigenvalues of modulus $ r$ then the set of all its eigenvalues is invariant under rotations by $ 2\\pi/p$ about the origin ($ p$ is the cyclic index). Thus the $ A$-eigenvalues of modulus $ r$ are vertex of a regular $ p$-polygon and are single eigenvalues.\r\n Moreover, on each circle of center the origin and of radius $ s.<", "Solution_2": "oops...\r\ni forgot to put this in advanced section.\r\ncan i move it?", "Solution_3": "[quote]oops...\ni forgot to put this in advanced section.\ncan i move it?[/quote]\r\n\r\n:lol: That explains my utter confusion when I read this. What subject do you learn about that in?" } { "Tag": [ "LaTeX" ], "Problem": "Is there a way I can convert a whole LaTeX document into multiple jpeg's, png's, etc.?", "Solution_1": "You can do this online at [url=http://sciencesoft.at/latex/?lang=en]ScienceSoft.at[/url] or do it yourself by the method that they use:\r\n[quote=\"ScienceSoft.at\"]LaTeX translates the input source text into the DVI-format (DeVice-Independent) - which is converted into the EPS-format Encapsulated PostScript) by means of the tool dvips (DVI-to-PostScript). Then, Ghostscript is used to convert the corresponding EPS-file into the PNG-format (Portable Network Graphics) which is fit for the world wide web.[/quote]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "In how many ways can we $N$ [b]equal[/b] books put on $N+1$ different shelves?\r\n\r\nGeneraly, in how many ways can we $N$ [b]equal[/b] books put on $M$ different shelves?", "Solution_1": "The all you need is to avaluate the number of sequances $N$ letters $b$ and $M-1$ letters $|$, which separates books to stand them on shelves.\r\nIf there is no $b$ between two $|'s$ it means there is nothing to stand on corresponding shelve. As a result you need to choose $N$ places in the sequance for $b's$ or equivalently $M-1$ places for $|'s$.\r\nIf it is forbidden (?) to have empty shelves, just stand one book on the each shelve and reduce problem to above. Of course in this case $N$ must be no less than $M$.", "Solution_2": "In general, if you have $n$ books numbered $1$ to $n$ and you want to place them on $m$ shelves such that book $i$ comes at least $x$ shelves before book $(i+1)$ for all $1 \\leq i < n$, then the number of ways is $\\binom{m-(n-1)(x-1)}{n}$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $ a_1,a_2,...,a_n$ be $ n\\geq 4$ positive integers whose sum is $ 2n \\minus{} 1$. Prove that there are some consecutive terms $ a_i,a_{i \\plus{} 1},...,a_j$ $ (j > i)$, whose average is an integer.", "Solution_1": "It is not true. For example $ n\\equal{}2,a_1\\equal{}1,a_2\\equal{}2,a_1\\plus{}a_2\\equal{}3\\equal{}2n\\minus{}1$.", "Solution_2": "Forgot an important condition. Edited!", "Solution_3": "Anybody has an idea?" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Using that $ (1\\plus{}\\frac{1}{t} )^t \\rightarrow e$ as $ t\\rightarrow \\infty$, $ \\alpha, \\beta \\in R$. Find the folowing limits\r\n\r\ni) $ (\\frac{n\\plus{}\\alpha}{n\\plus{}\\beta})^n$ as $ n\\rightarrow \\infty$\r\nii) $ (1\\plus{}\\frac{\\alpha}{n})^{\\beta n}$ $ n\\rightarrow \\infty$", "Solution_1": "hello, for i) write your term in the form\r\n$ \\left((1+\\frac{\\alpha-\\beta}{n+\\beta})^{\\frac{n+\\beta}{\\alpha-\\beta}\\right)^{\\frac{(\\alpha-\\beta)n}{n+\\beta}}}$\r\nthe searched limit is $ e^{\\alpha-\\beta}$.\r\nSonnhard." } { "Tag": [], "Problem": "Is anybody here taking biochemistry classes? If so could i post some biochemistry topics?", "Solution_1": "I know Rushil is studying/taking organic, and I'm studying organic on my own, but I don't know if anybody is doing biochemistry.\r\n\r\nNothing/No one is saying you can't post them, though. At least, I'm not. :lol:", "Solution_2": "As part of our general organic chem course, we cover Biomolecules - Carbohydrates and Amino acids. I am presently studying these ... I'd love to see questions on these. Which other topics are there in biochem??\r\n\r\nAlso, inteluser, I live in India where EVERYONE is supposed to take a standard chem course having a bulk of organic chem in Class XI and XII ... just for your info. Also, my favourite topics in Chem are (in order) m- Physical Chem.. (full advanced quantum etc. (not taught - just my super interest!) ), Organic and then (obviously) Inorganic!!!", "Solution_3": "i know biochem, so yes post some.", "Solution_4": "Still waiting for your questions, sefu!" } { "Tag": [ "algebra", "function", "domain", "linear algebra", "matrix", "logarithms", "calculus" ], "Problem": "Find the continous morphisms from the set of complex numbers to the set of nxn invertible complex matrices.", "Solution_1": "What operation or operations are you preserving with this morphism? Are you regarding the complex numbers as a group under addition?", "Solution_2": "Oh, sorry, in fact the domain of the morphism is the unit circle in the complex plane. Sorry again.", "Solution_3": "What's your opinion, Harazi? I think they're of the form $f(z)=A^{-1}diag(z^{n_1},\\ldots,z^{n_p})A$, where $A$ is a fixed invertible matrix, and $n_k$ are fixed integers. \r\n\r\nI really don't have the necessary time right now, so I might be wrong because I'm in a hurry, but we can first look at those complex numbers which are roots of unity. Their images are diagonalizable matrices which commute, so they are simultaneously diagonalizable by some matrix $A$. After this, by continuity, we can show that all the matrices in the image are diagonalizable by $A$ and have complex numbers of modulus $1$ on the diagonal. After that, we break the homomorphism into $n$ distinct homomorphisms from the unit circle to itself, the $k$'th one taking $z$ to the $k$'th entry on the diagonal of $f(z)$. Since these are continuous, I believe they're of the form $f_k(z)=z^{n_k}$.\r\n\r\nIt looks Ok to me right now, but I bet it's not... :?", "Solution_4": "I think it is ok. :D", "Solution_5": "Find the continous morphisms from the set of REAL numbers to the set of nxn invertible REAL matrices", "Solution_6": "Again, here, by reals, you mean the multiplicative group $(\\mathbb R^*,\\cdot)$, right?", "Solution_7": "I mean $(R,+)$", "Solution_8": "i mean $f(x+y)=f(x).f(y)$ for every real $x$ and $y$, $f(x)$ in $Gln(R)$, $f$ continuous.", "Solution_9": "Shame on me for not seeing what the answer was earlier :). All continuous $f$ group homomorphisms from $(\\mathbb R,+)$ to $(Gl_n(\\mathbb R),\\cdot)$ are of the form $f(t)=\\exp(tX),\\ \\forall t\\in\\mathbb R$ for a (unique) real $n\\times n$ matrix $X$. \r\n\r\nLet's prove the corresponding assertion for $f: \\mathbb R\\to Gl_n(\\mathbb C)$ first, namely that all such continuous homomorphisms are of the form $f(t)=\\exp(tX)$ for a unique complex $n\\times n$ matrix $X$. \r\n\r\nWe have $f(0)=I$, so for small enough $t$, we'll have $\\|f(t)-I\\|<1$, where $\\|\\cdot\\|$ is the [url=http://planetmath.org/encyclopedia/MatrixFNorm.html]Frobenius norm[/url]. This means that for small $t,t'$, the [url=http://planetmath.org/encyclopedia/MatrixLogarithm.html]principal logarithms[/url] $\\ln f(t),\\ln f(t'),\\ln f(t+t')=\\ln(f(t)\\cdot f(t'))$ are defined, and they satisfy $\\ln f(t+t')=\\ln f(t)+\\ln f(t')$ (in other words, $t\\mapsto \\ln f(t)$ is a continuous local homomorphism around $0$ from $(\\mathbb R,+)$ to $(\\mathcal M_n(\\mathbb C),+)$). It's easy to conclude from here that there is a complex matrix $X$ such that $\\ln f(t)=tX$ for all small enough $t$, which by exponentiation gives $f(t)=\\exp(tX)$ (again, for small $t$). This relation is extended to the entire real line by continuity, and the existence of $X$ is proven. The uniqueness follows by observing that $t\\mapsto \\exp(tX)$ is differentiable, and $X$ is the derivative of this map at $t=0$, so $X$ is actually uniquely defined by $f$.\r\n\r\nFor our problem, $X$ has to be real, since $\\exp(tX)$ is a real matrix for all $t$, so when we take the derivative of $t\\mapsto \\exp(tX)$ at $t=0$ we'll get a real matrix, which is $X$.", "Solution_10": "[quote=\"grobber\"]What's your opinion, Harazi? I think they're of the form $f(z)=A^{-1}diag(z^{n_1},\\ldots,z^{n_p})A$, where $A$ is a fixed invertible matrix, and $n_k$ are fixed integers. \n\nI really don't have the necessary time right now, so I might be wrong because I'm in a hurry, but we can first look at those complex numbers which are roots of unity. Their images are diagonalizable matrices which commute, so they are simultaneously diagonalizable by some matrix $A$. After this, by continuity, we can show that all the matrices in the image are diagonalizable by $A$ and have complex numbers of modulus $1$ on the diagonal. After that, we break the homomorphism into $n$ distinct homomorphisms from the unit circle to itself, the $k$'th one taking $z$ to the $k$'th entry on the diagonal of $f(z)$. Since these are continuous, I believe they're of the form $f_k(z)=z^{n_k}$.\n\nIt looks Ok to me right now, but I bet it's not... :?[/quote]\r\n\r\nI dont see why the continuous homomorphismos from the unit circle to itself have the form $f(z)=z^k$??", "Solution_11": "Let's identify the circle group with $\\mathbb R/\\mathbb Z$, in order to use additive notation. The statement then becomes: all continuous endomorphisms of $\\mathbb R/\\mathbb Z$ are of the form $x\\mapsto kx$, for some integer $k$. \r\n\r\nChoose a small neighborhood $U$ of $0$ (say, $U\\subset\\left(-\\frac 12,\\frac 12\\right)$), and let $V\\subset\\left(-\\frac 12,\\frac 12\\right)$ be a neighborhood of $0$ so small that whenever $x\\in V,\\ x$ is mapped into $U$. Now take $n$ so large that $\\frac 1n\\in V$. Since $n\\cdot\\frac 1n=0,\\ \\frac 1n$ must be mapped onto $\\frac kn\\in U$ for some integer $k$. Now, for positive integers $t,\\ \\frac 1^{2^tn}$ must be mapped onto $\\frac k{2^tn}$ plus some integer, but since this image must also lie inside $\\left(-\\frac 12,\\frac 12\\right)$, that extra integer must be zero. Our homomorphism thus sends every element of the group of the form $\\frac m{2^tn}\\ (*)$ onto $\\frac{km}{2^tn}$, and since the elements of the form $(*)$ form a dense set in our group, we're done: every element $x$ is sent to $kx$." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "This is cute:\r\n\r\nFind all primes $p, q$ such that $pq$ divides $2^p + 2^q$.", "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=20918", "Solution_2": "It is a vicious circle since I took it from here to propose it in our program... :D \r\n\r\nPierre.", "Solution_3": "[quote=\"pbornsztein\"]It is a vicious circle since I took it from here to propose it in our program... :D \n\nPierre.[/quote] :D :D :D" } { "Tag": [ "limit", "logarithms", "inequalities", "Euler", "number theory unsolved", "number theory" ], "Problem": "Stimate the following product ($p_j$ represents the $j$-th prime number)\r\n\\[\r\n\\prod_{j=1}^m \\frac{p_j}{p_j-1},\r\n\\]\r\nan use the stimation to calculate $\\displaystyle \\lim_{m\\to \\infty} \\prod_{j=1}^m \\frac{p_j}{p_j-1}.$.\r\n\r\nRX", "Solution_1": "It is known that for every $x \\geq 2$, we have $\\prod_{p \\leq x} \\frac p {p-1} \\geq \\log (x).$\r\n\r\nIt follows that the desired limit is $+ \\infty .$\r\n\r\nPierre.", "Solution_2": "[quote=\"pbornsztein\"]It is known that for every $x \\geq 2$, we have $\\prod_{p \\leq x} \\frac p {p-1} \\geq \\log (x).$\n\nIt follows that the desired limit is $+ \\infty .$\n\nPierre.[/quote]\r\nThx", "Solution_3": "pbornsztein, could you please give a proof of the inequality you used ?", "Solution_4": "Write $\\prod_{p\\le x}\\frac p{p-1}$ as $\\prod_{p\\le x}\\frac 1{1-\\frac 1p}$, which is equal to $\\prod_{p\\le x}(1+\\frac 1p+\\frac 1{p^2}+\\ldots)$, and this is the sum of the recirpocals of all the numbers having in their decomposition only primes less than $x$, which is larger than $\\sum_{n\\le x}\\frac 1n$. Now, $\\sum_{i=1}^n\\frac 1i-\\ln (n+1)$ tends to a positive limit (the Euler constant $\\gamma$), so that suffices for the problem.", "Solution_5": "[quote=\"RobertuX\"]Stimate the following product ($p_j$ represents the $j$-th prime number)\n\\[\n\\prod_{j=1}^m \\frac{p_j}{p_j-1},\n\\]\nan use the stimation to calculate $\\displaystyle \\lim_{m\\to \\infty} \\prod_{j=1}^m \\frac{p_j}{p_j-1}.$.\n\nRX[/quote]\r\nMore easy saying that this product tends to $\\zeta(1)=+\\infty$. But thx a lot of ur comments.", "Solution_6": "[quote=\"grobber\"]and this is the sum of the recirpocals of all the numbers having in their decomposition only primes less than $x$, which is smaller than $\\sum_{n\\le x}\\frac 1n$. .[/quote]\r\n\r\nCould you explain it for me ? I can't understand why :?", "Solution_7": "I meant to say larger instead of smaller. I have edited my previous message." } { "Tag": [ "function", "algebra", "polynomial", "real analysis", "real analysis unsolved" ], "Problem": "Prove that for every real continuous function $f$ on $[a, b]$, one can find polynomial with integer coeficients, such that for every $a > 0$, there exists $n \\in N$ and $P_n$ with $\\mbox{deg} P_n \\leqslant n$, and \r\n\\[ |f (x) - P_n (x)| \\leqslant a, \\quad x \\in [a, b]. \\]", "Solution_1": "[quote=\"alexilic\"]Prove that for every real continuous function $f$ on $[a, b]$, one can find polynomial with integer coeficients, such that for every $a > 0$, there exists $n \\in N$ and $P_n$ with $\\mbox{deg} P_n \\leqslant n$, and \n\\[ |f (x) - P_n (x)| \\leqslant a, \\quad x \\in [a, b]. \\][/quote]\nI think it is something wrong here with the double using of $a$, i think in [quote=\"alexilic\"] such that for every $a > 0$, there exists [/quote] you meant some another $\\epsilon$, i.e. you rpobvably meant that $f$ can be uniformly approximated by polynomials from $\\mathbb{Z}[X]$.", "Solution_2": "For every $f \\in C [a, b]$, prove that there exist array of polynomials $P_n (x) \\in Z [X]$, such that\r\n\\[ P_n \\rightarrow f, n \\rightarrow \\infty \\] for every $x \\in [a, b]$.", "Solution_3": "There are several ways of showing this. I'll use this:\r\nWLOG, you may assume that $[a,b] \\subset (0,1)$. Since the set $\\{\\frac{k}{2^n}, k,n \\in \\mathbb{N}_0 \\}$ is dense in $(0,1)$ we only need to prove that $\\frac{1}{2}$ can be approximated by polynomials with integers coefficients with arbitrarily small accuracy i.e. for any $\\epsilon >0$ there exists $p \\in \\mathbb{Z}[X]$, such that $||\\frac{1}{2}-p||_{[a,b]} < \\epsilon$ (*). If we get this we can coclude that any real from $(0,1)$ can be approximated by polynomiasl from $\\mathbb{Z}[X]$ and such all the polynomials from $\\mathbb{R}_n[X]$ as well and know we can just use Weirstrass approximation theorem that states that our function $f$ can be approximated by some polynomial from $\\mathbb{R}_n[X]$. So we have to prove (*):\r\n$\\frac{1}{2}=\\frac{1-x}{1-(1-2(1-x))}= \\sum_{k=0}^{\\infty} (1-x)(1-2(1-x))^k$, and as our approximation polynomials we can choose tha partial sum of this infinite series which as you may prove converges uniformly on $[a,b] \\subset (0,1)$.", "Solution_4": "(beware : for $[a,b]=[0;10]$ it does not work)", "Solution_5": "Yes, for the case, when the endpoints of our interval are integer as in this case of $[0,10]$ we also need the condition $f(a),f(b) \\in \\mathbb{Z}$.", "Solution_6": "So, can someone explain general solution?\r\nThx", "Solution_7": "[quote=\"alexilic\"]So, can someone explain general solution?\nThx[/quote]\r\nWell, as for me i don't know the answer in general. What is also that interests me, you didn't say anything about you $[a,b]$ and function $f$ except that it is continious. But not every function can be approximated with integer polynomials on every interval. I'll try to sum up the cases which i know about:\r\n\r\n1). If $[a,b]$ doesn't contain integers then i showed a proof in post #4.\r\n\r\n2). If $[a,b]$ does contain integer, then you must have at least these resctrictions on our fucntion $f$: $f(\\mathbb{Z}\\bigcap [a,b]) \\in \\mathbb{Z}$, i don't know whether these resctrictions are SUFFICIENT. For example if $[a,b]=[0,1]$ i know that they are sufficient(use Bernstein polynomials, to be accurate kinda integer part of them). \r\nIf $[a,b]=[-1,1]$ i know that they are also sufficient. But the general case...i really don't know.\r\n\r\n\r\n\r\nQuestion to others: Who knows the result in general.", "Solution_8": "[quote=\"alekk\"](beware : for $[a,b]=[0;10]$ it does not work)[/quote]\r\n\r\nAh, i can also prove that we the result of the problem is not true if $b-a \\geq 4$. So, yopu must have at least add this condition to the problem statement. It is probably what Alekk was talking about.\r\n\r\nThis neccessary condition $b-a <4$ comes from the Chebysheff theoprem that $||p||_{[a,b]} \\geq 2 (\\frac{b-a}{4})^n$ and in the case of $b-a \\geq 4$ we would have that $||p_n -p_m || \\geq 2$ and this implies that $p_n \\to f$ uniformly is surely impossible due to Cauchy criterion for uniform convergence." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Assume $ a_{n} \\geq a_{n+1}$, for $ \\forall$ $ n \\geq 1$\r\n\r\nand $ \\lim_{n \\rightarrow \\infty}a_{n}=0$\r\n\r\ndefine: $ b_{n}=a_{n}-2a_{n+1}+a_{n+2}$, for $ \\forall$ $ n \\geq 1$\r\n\r\nassume $ b_{n}\\geq0$, for $ \\forall$ $ n\\geq1$\r\n\r\nshow that: $ \\sum_{n=1}^{\\infty}nb_{n}$ converge\r\n\r\n[color=red]My Idea[/color]\r\n\r\nI want to use Cauchy convergence principle....\r\n\r\nso need to compute: $ |mb_{m}+(m+1)b_{m+1}+...+nb_{n}|$\r\n\r\nthe above equal: \r\n$ |m[(a_{m}-a_{m+1})-(a_{m+1}-a_{m+2})]+(m+1)[(a_{m+1}-a_{m+2})-(a_{m+2}-a_{m+3})]+\n...+n[(a_{n}-a_{n+1})- (a_{n+1}a_{n+2})]|$\r\n\r\n$ =|m(a_{m}-a_{m+1})+a_{m+1}-a_{n+1}-n(a_{n+1}-a_{n+2})|$\r\n\r\nBut I can't go any further......", "Solution_1": "isn't $ 1/\\sqrt n$ a counterexample?\r\n\r\n[quote=\"Xingyuan\"]Assume $ a_{n} \\geq a_{n \\plus{} 1}$, for $ \\forall$ $ n \\geq 1$\n\nand $ \\lim_{n \\rightarrow \\infty}a_{n} \\equal{} 0$\n\ndefine: $ b_{n} \\equal{} a_{n} \\minus{} 2a_{n \\plus{} 1} \\plus{} a_{n \\plus{} 2}$, for $ \\forall$ $ n \\geq 1$\n\nassume $ b_{n}\\geq0$, for $ \\forall$ $ n\\geq1$\n\nshow that: $ \\sum_{n \\equal{} 1}^{\\infty}nb_{n}$ converge\n\n[color=red]My Idea[/color]\n\nI want to use Cauchy convergence principle....\n\nso need to compute: $ |mb_{m} \\plus{} (m \\plus{} 1)b_{m \\plus{} 1} \\plus{} ... \\plus{} nb_{n}|$\n\nthe above equal: \n$ |m[(a_{m} \\minus{} a_{m \\plus{} 1}) \\minus{} (a_{m \\plus{} 1} \\minus{} a_{m \\plus{} 2})] \\plus{} (m \\plus{} 1)[(a_{m \\plus{} 1} \\minus{} a_{m \\plus{} 2}) \\minus{} (a_{m \\plus{} 2} \\minus{} a_{m \\plus{} 3})] \\plus{} ... \\plus{} n[(a_{n} \\minus{} a_{n \\plus{} 1}) \\minus{} (a_{n \\plus{} 1}a_{n \\plus{} 2})]|$\n\n$ \\equal{} |m(a_{m} \\minus{} a_{m \\plus{} 1}) \\plus{} a_{m \\plus{} 1} \\minus{} a_{n \\plus{} 1} \\minus{} n(a_{n \\plus{} 1} \\minus{} a_{n \\plus{} 2})|$\n\nBut I can't go any further......[/quote]", "Solution_2": "[quote=\"Arrigo-Sacchi\"]isn't $ 1/\\sqrt n$ a counterexample?\n[/quote]\r\n\r\nI think $ \\frac{1}{\\sqrt{n}}$ is not a counterexample.......\r\n\r\nbecause:\r\n\r\nIf $ a_{n}\\equal{}\\frac{1}{\\sqrt{n}}$\r\n\r\nthen $ b_{n}\\equal{}\\frac{1}{\\sqrt{n}}\\minus{}\\frac{2}{\\sqrt{n\\plus{}1}}\\plus{}\\frac{1}{\\sqrt{n\\plus{}2}}$\r\n\r\nobviously, $ b_{n}\\geq0$\r\n\r\nand $ \\sum_{n\\equal{}1}^{\\infty}nb_{n}\\equal{}\\sum_{n\\equal{}1}^{\\infty}n\\left(\\frac{1}{\\sqrt{n}}\\minus{}\\frac{2}{\\sqrt{n\\plus{}1}}\\plus{}\\frac{1}{\\sqrt{n\\plus{}2}}\\right)$\r\n\r\nI guess the above series converges if and only if:\r\n\r\n$ \\sum_{n\\equal{}1}^{\\infty}\\left(\\sqrt{n}\\minus{}2\\sqrt{n\\plus{}1}\\plus{}\\sqrt{n\\plus{}2}\\right)$ converges $ (*)$\r\n\r\n[color=red](I can't give a proof)[/color]\r\n\r\nobviously, $ \\sum_{n\\equal{}1}^{\\infty}\\left(\\sqrt{n}\\minus{}2\\sqrt{n\\plus{}1}\\plus{}\\sqrt{n\\plus{}2}\\right)$ is converges\r\n\r\nso if we work out $ (*)$, we can get the conclusion that its not a counterexample.....", "Solution_3": "[quote=\"Xingyuan\"]\n\nI think $ \\frac {1}{\\sqrt {n}}$ is not a counterexample.......\n[/quote] \r\n\r\nYes, I made the error of considering $ \\sum na_n$ instead of $ \\sum nb_n.$", "Solution_4": "[quote=\"Arrigo-Sacchi\"][quote=\"Xingyuan\"]\n\nI think $ \\frac {1}{\\sqrt {n}}$ is not a counterexample.......\n[/quote] \n\nYes, I made the error of considering $ \\sum na_n$ instead of $ \\sum nb_n.$[/quote]\r\n\r\nBut it is a good example :D", "Solution_5": "Hint: Let $ c_n \\equal{} a_n \\minus{} a_{n\\plus{}1}$. Then $ b_n \\equal{} c_n \\minus{} c_{n\\plus{}1}$. Use summation by parts on $ \\sum_{n\\equal{}1}^{N} n(c_n \\minus{} c_{n\\plus{}1})$." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "(7-2z)^(1/3)=(2z-5)^(1/2) where z=x+iy find x,y", "Solution_1": "$ (7\\minus{}2z)^2\\equal{}(2z\\minus{}5)^3$\r\n\r\nlet $ w\\equal{}2z\\minus{}5$, then this looks like $ (2\\minus{}w)^2\\equal{}w^3$\r\n\r\n$ 0\\equal{}w^3\\minus{}w^2\\plus{}4w\\minus{}4\\equal{}w^2(w\\minus{}1)\\plus{}4(w\\minus{}1)\\equal{}(w\\minus{}2i)(w\\plus{}2i)(w\\minus{}1)$\r\n\r\nso.. $ z\\equal{}\\frac{w\\plus{}5}{2}$, which are $ \\frac{5}{2} \\pm i, 3$", "Solution_2": "thanks thanks thanks thanks thanks" } { "Tag": [ "linear algebra", "matrix", "LaTeX" ], "Problem": "Which package is the bmatrix, pmatrix, vmatrix etc. commands in? I searched around and couldn't find it.", "Solution_1": "amsmath. See [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math Mode[/url]." } { "Tag": [], "Problem": "Any1 got any tips on how to get hold of hacks or cheats for BattleOn AQ.", "Solution_1": "...Cheats? I think very few people cheat on this forum and I'm thinking very few even play that game/heard of it. I don't think I've ever heard of this game but cheating and hacking is only for noobs.", "Solution_2": "i have never heard of this game either and what is noob?", "Solution_3": "[quote=\"math92\"]i have never heard of this game either and what is noob?[/quote]\r\n\r\nA noob is a player that either sucks, cheats/hacks, whines continuously, or any combination of the three.", "Solution_4": "oh, well i am not a noob. :)" } { "Tag": [ "trigonometry" ], "Problem": "In any triangle ABC, if $ BC\\plus{}AC\\equal{}2AB$, show that $ \\cot(A/2)\\plus{}\\cot(B/2)\\equal{}2\\cot(C/2)$", "Solution_1": "$ a\\plus{}b\\equal{}2c \\Longleftrightarrow 2s\\minus{}(a\\plus{}b)\\equal{}2s\\minus{}2c \\Longleftrightarrow s\\minus{}a\\plus{}s\\minus{}b\\equal{}2(s\\minus{}c)$\r\n\r\n$ \\Longleftrightarrow \\frac{s(s\\minus{}a)}{[ABC]}\\plus{}\\frac{s(s\\minus{}b)}{[ABC]}\\equal{}\\frac{2s(s\\minus{}c)}{[ABC]}$\r\n\r\nNow using [u][b]Heron[/b]'s formula[/u] $ \\implies \\sqrt{\\frac{s^2(s\\minus{}a)^2}{s(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}}\\plus{}\\sqrt{\\frac{s^2(s\\minus{}b)^2}{s(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}}\\equal{}$ $ 2\\sqrt{\\frac{s^2(s\\minus{}c)^2}{s(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}}$\r\n\r\n$ \\Longleftrightarrow \\sqrt{\\frac{s(s\\minus{}a)}{(s\\minus{}b)(s\\minus{}c)}}\\plus{}\\sqrt{\\frac{s(s\\minus{}b)}{(s\\minus{}a)(s\\minus{}c)}}\\equal{}2\\sqrt{\\frac{s(s\\minus{}c)}{(s\\minus{}a)(s\\minus{}b)}}$ $ \\Longleftrightarrow \\cot\\frac A2\\plus{}\\cot\\frac B2\\equal{}2\\cot\\frac C2$ .", "Solution_2": "[color=darkblue]Other relations in triangles having [b]sides in aritmethic progression[/b]: $ \\left\\|\\begin{array}{ccc}\\frac{2}{h_{a}} \\equal{} \\frac{1}{h_{b}}\\plus{}\\frac{1}{h_{c}}\\\\\\\\\n4\\cos A\\plus{}\\cos\\left(B\\minus{}C\\right)\\equal{}3\\\\\\\\\n\\sin\\frac{A}{2}\\equal{}\\frac{1}{2}\\cos\\frac{B\\minus{}C}{2}\\end{array}\\right\\|$ and $ \\left\\|\\begin{array}{ccc}3\\tan\\frac{B}{2}\\tan\\frac{C}{2}\\equal{}1\\\\\\\\\n3\\sin^{2}\\frac{A}{2}\\equal{}\\sin B\\sin C \\\\\\\\ \n2\\cos A\\plus{}\\cos B\\plus{}\\cos C\\equal{}2 \\end{array}\\right\\|$ .[/color]", "Solution_3": "[quote=\"Mateescu Constantin\"]$ a \\plus{} b \\equal{} 2c \\Longleftrightarrow 2s \\minus{} (a \\plus{} b) \\equal{} 2s \\minus{} 2c \\Longleftrightarrow s \\minus{} a \\plus{} s \\minus{} b \\equal{} 2(s \\minus{} c)$\n\n$ \\Longleftrightarrow \\frac {s(s \\minus{} a)}{[ABC]} \\plus{} \\frac {s(s \\minus{} b)}{[ABC]} \\equal{} \\frac {2s(s \\minus{} c)}{[ABC]}$\n\nNow using [u][b]Heron[/b]'s formula[/u] $ \\implies \\sqrt {\\frac {s^2(s \\minus{} a)^2}{s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}} \\plus{} \\sqrt {\\frac {s^2(s \\minus{} b)^2}{s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}} \\equal{}$ $ 2\\sqrt {\\frac {s^2(s \\minus{} c)^2}{s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}}$\n\n$ \\Longleftrightarrow \\sqrt {\\frac {s(s \\minus{} a)}{(s \\minus{} b)(s \\minus{} c)}} \\plus{} \\sqrt {\\frac {s(s \\minus{} b)}{(s \\minus{} a)(s \\minus{} c)}} \\equal{} 2\\sqrt {\\frac {s(s \\minus{} c)}{(s \\minus{} a)(s \\minus{} b)}}$ $ \\Longleftrightarrow \\cot\\frac A2 \\plus{} \\cot\\frac B2 \\equal{} 2\\cot\\frac C2$ .[/quote]\r\n\r\n\r\nCan you show how do you get that\r\n$ \\sqrt{\\frac{s(s\\minus{}a)}{(s\\minus{}b)(s\\minus{}c)}}\\equal{}\\cot\\frac{A}{2}$\r\nP.S:Sorry for the ignorance", "Solution_4": "[quote=\"ridgers\"][quote=\"Mateescu Constantin\"]$ a \\plus{} b \\equal{} 2c \\Longleftrightarrow 2s \\minus{} (a \\plus{} b) \\equal{} 2s \\minus{} 2c \\Longleftrightarrow s \\minus{} a \\plus{} s \\minus{} b \\equal{} 2(s \\minus{} c)$\n\n$ \\Longleftrightarrow \\frac {s(s \\minus{} a)}{[ABC]} \\plus{} \\frac {s(s \\minus{} b)}{[ABC]} \\equal{} \\frac {2s(s \\minus{} c)}{[ABC]}$\n\nNow using [u][b]Heron[/b]'s formula[/u] $ \\implies \\sqrt {\\frac {s^2(s \\minus{} a)^2}{s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}} \\plus{} \\sqrt {\\frac {s^2(s \\minus{} b)^2}{s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}} \\equal{}$ $ 2\\sqrt {\\frac {s^2(s \\minus{} c)^2}{s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}}$\n\n$ \\Longleftrightarrow \\sqrt {\\frac {s(s \\minus{} a)}{(s \\minus{} b)(s \\minus{} c)}} \\plus{} \\sqrt {\\frac {s(s \\minus{} b)}{(s \\minus{} a)(s \\minus{} c)}} \\equal{} 2\\sqrt {\\frac {s(s \\minus{} c)}{(s \\minus{} a)(s \\minus{} b)}}$ $ \\Longleftrightarrow \\cot\\frac A2 \\plus{} \\cot\\frac B2 \\equal{} 2\\cot\\frac C2$ .[/quote]\n\n\nCan you show how do you get that\n$ \\sqrt {\\frac {s(s \\minus{} a)}{(s \\minus{} b)(s \\minus{} c)}} \\equal{} \\cot\\frac {A}{2}$\nP.S:Sorry for the ignorance[/quote]\r\n\r\nsee here http://www.mathlinks.ro/viewtopic.php?t=322695" } { "Tag": [], "Problem": "[img]http://img171.imageshack.us/img171/1598/55873420xp0.jpg[/img]", "Solution_1": "[quote=\"Aafour\"][img]http://img171.imageshack.us/img171/1598/55873420xp0.jpg[/img][/quote]\r\n\r\n\r\n$ y\\equal{}2x\\plus{}2$ --> parallel --> $ y\\equal{}2x\\plus{}b$ ---> $ (\\minus{}2.4)$ --> $ 4\\equal{}2*(\\minus{}2)\\plus{}b$ --> $ b\\equal{}8$\r\n\r\n$ y\\equal{}2x\\plus{}8$ ---> $ y\\equal{}0$ --> $ 0\\equal{}2x\\plus{}8$ --> $ x\\equal{}\\minus{}4$ --> $ (\\minus{}4;0)$" } { "Tag": [ "polynomial", "algebra", "Summation", "equation", "Diophantine equation", "IMO Shortlist", "IMO Longlist" ], "Problem": "Suppose that $p$ and $q$ are two different positive integers and $x$ is a real number. Form the product $(x+p)(x+q).$ Find the sum $S(x,n) = \\sum (x+p)(x+q),$ where $p$ and $q$ take values from 1 to $n.$ Does there exist integer values of $x$ for which $S(x,n) = 0.$", "Solution_1": "Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url] :)" } { "Tag": [ "geometry", "AMC" ], "Problem": "Hi everyone...\r\n\r\nDoes anyone know where around or in MD (or which county/schools) offer the AMC B test. Please reply/PM me as soon as possible. I really want to take this test. \r\n\r\nThanks", "Solution_1": "http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml\r\n\r\nI heard you have to register by the 14th or something.", "Solution_2": "If you live in Maryland, you could probably take it with my math club, the Howard Area Homeschoolers. PM CatherineAsaro for more information.", "Solution_3": "[quote=\"E^(pi*i)=-1\"]If you live in Maryland, you could probably take it with my math club, the Howard Area Homeschoolers. PM CatherineAsaro for more information.[/quote]\r\n\r\nthank you for your offer. I have emailed Catherine Asaro already, and I was wondering if I need to complete further registration or has everyone already been registered?", "Solution_4": "I'm not sure, but I would assume you wouldn't need to register." } { "Tag": [], "Problem": "Determine all solutions k, n1, n2, ..., nk to\r\n\r\nn1|2n2 - 1\r\nn2|2n3 - 1\r\n...\r\nnk|2n1 - 1.\r\n\r\nI like this problem.", "Solution_1": "I think this lemma might help:\n\n\n\nLemma: [hide] if a is divisible by a smallest prime p and b is divisible by a smallest prime q, and p <= q, then a cannot divide 2^b - 1.\n\n\n\nProof:\n\nfirst, clearly p!=2.\n\nif a | 2^b - 1, then 2^b == 1 (mod a ) => 2^b == 1 (mod p).\n\n\n\nLet k be the order of 2 (mod p).\n\nThen k <= p-1, but k|b. But since the smallest prime factor of b is q > p, k cannot divide b unless k = 1. But k cannot equal 1 since otherwise 2==1(mod p) => p | 1.\n\n\n\nThis completes the proof.\n\n\n\n\n\n\n\n[/hide]", "Solution_2": "From the lemma: \n\n\n\n[hide]\n\nThe only solution is when all are equal to 1 (but k can be anything in this case).\n\n(Assuming all are natural... otherwise we can also have all of them equal to 0)\n\n\n\nNote we cannot have one value equal to one and another equal to something greater than one, since one equalling one implies all others are equal to one. (Just work backwards: 2^1 -1 = 1)\n\n\n\nWe assume that none are equal to one. Then each has a unique nonempty prime factorization.\n\nsuppsose all these conditions are true. Then\n\n\n\nif p[i] is the smallest prime divisor of n[i]:\n\n\n\np[1] > p[2] > p[3] > ... > p[k] > p[1].\n\n\n\nA contradiction.\n\n\n\n[/hide]\n\n\n\nSorry for the sloppiness . Hope it works.", "Solution_3": "It does :)", "Solution_4": "Is this the way you approached the problem? Or, did you have a different solution, because I would be interested to see it.", "Solution_5": "My solution is a little bit different, but it comes down to the same.\r\n\r\nI defined \r\np' = min{p[b]|[/b]p is a prime such that p|ni for a certain 1 :le: i :le: k}.\r\n\r\nThen we can show that one of the ni must have a prime divisor p\" < p'.\r\n(In fact, if that p'|nk, then p\" < p' must divide nk + 1.\r\nActually it is the same method as in your lemma ...)\r\n\r\nHence contradiction, and the result follows if we assume k :ge: 2.\r\n\r\nIf k = 1 then the problem reduces to finding all n st n|2n - 1.\r\nThis is not so difficult.", "Solution_6": "actually u dont need the special case. if k = 1 then nk IS nk+1", "Solution_7": "Isn't that what said? Or am I missing something ...", "Solution_8": "[quote=\"Arne\"]Hence contradiction, and the result follows if we assume k :ge: 2.\n\nIf k = 1 then the problem reduces to finding all n st n|2n - 1.\nThis is not so difficult.[/quote]\r\n\r\nFirstly its just me being picky.\r\nMy only \"point\" was that you dont have to assume k>=2.\r\n" } { "Tag": [ "ratio" ], "Problem": "A shoemaker died, leaving behind 17 pair's of men's shoes and boots. In his last will and testament he disposed of them as follows:\r\n\r\n$ \\frac{1}{2}$ of the collection to his youngest, hardworking, son; $ \\frac{1}{3}$ to the middle son, and $ \\frac{1}{9}$ to \"Lazybones\", the eldest son.\r\n\r\nUnderstanding that it was not possible to divide 17 pairs into halves, thirds, and ninths, the exasperated executor of the will tossed in his own footwear, making it a total of eighteen pairs, whihc he divided as follows;\r\n\r\n$ \\frac{1}{2}$, or 9 pairs, to the youngest son; $ \\frac{1}{3}$, or 6 pairs, to the middle son; and $ \\frac{1}{9}$, to the eldest son.\r\n\r\nUpon realising that adding up 9,6 and 2 made 17, the execytor elatedly retrieved his own pairs of shoes and closed the estate.", "Solution_1": "That\u2019s because $ \\frac {1}{2}$, $ \\frac {1}{3}$ and $ \\frac {1}{9}$ do not add up to 1. The agent was only splitting the 17 pairs of footwear items in the ratios $ \\frac {9}{17}$, $ \\frac {6}{17}$ and $ \\frac {2}{17}$, so his own pair didn\u2019t enter the calculation.", "Solution_2": "This property can be generalised by finding all values of n and a and b and c such that\r\n(n+1)/a + (n+1)/b + (n+1)/c = 1 and 1/a + 1/b + 1/c not equal to 1" } { "Tag": [ "function", "analytic geometry", "trigonometry", "quadratics", "trig identities" ], "Problem": "Here are eight tough questions which I am unable to solve.\r\nPlease help me by explaining as well as showing the step\\solutions.\r\nThank you in advance.\r\n\r\nQuestion 1:\r\n[hide]Given that $ Cosx = Cosy$ and $ Sinx = - Siny$, Show that $ Sin^2 \\frac {x + y}{2} = 0$\n[/hide]\nQuestion 2:\n[hide]The number of hours of day light in a certain town is given by $ y = 3sin\\frac {2\\pi}{365}(x - 79) + 12$,\nwhere $ x$ is the number of days after January 1.\na) What is the amplitude of the function?\nb) What is the period of the function?\nc) How many hours of daylight are there on the longest day of the year?\nd) How many hours of daylight are there on the shortest day of the year?\n[/hide]\nQuestion 3:\n[hide]Let AC and BD are tow chords of a circle with centre O such that they intersect at right angles inside the circle at a point M. Suppose K and L are mid points of the arc AB and chords Cd respectively. Prove that OKML is a parallelogram.\n[img]http://www.fileden.com/files/2006/11/16/384824/Question%203%20image.bmp[/img]\n[/hide]\nQuestion 4:\n[hide]Determine the values of $ a$ and $ c$ for which\n[img]http://www.fileden.com/files/2006/11/16/384824/question%204%20image.bmp[/img]\n[/hide]\nQuestion 5:\n[hide]Two vertices of a triangle are (5,-1) and (-2,3). If the orthocentre of the triangle is the origin, find the coordinates of the third vertex.\n[/hide]\nQuestion 6:\n[hide]If $ sec (\\theta - \\alpha), sec\\theta$ and $ sec (\\theta + \\alpha)$ are in A.P., prove that $ cos\\theta = \\sqrt2 cos \\frac {\\alpha}{2}$\n[/hide]\nQuestion 7:\n[hide]Show that $ 2tan^ - ^1 \\frac {1}{5} + sec^ - ^1 \\frac {5\\sqrt2}{7} + 2tan^ - ^1 \\frac {1}{8} = \\frac {\\pi}{4}$\n[/hide]\nQuestion 8:\n[hide]Balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row of two balls, three in third row and so on. If 669 more balls are added then all the balls can be arranged in the shape of a square and each of sides then contains 8 balls less than the side of the triangle had. Determine the initial number of balls.[/hide]", "Solution_1": "[quote=\"CyanSnow\"]Here are eight tough questions which I am unable to solve.\nPlease help me by explaining as well as showing the step\\solutions.\nThank you in advance.\n\nQuestion 1:\n[hide]Given that $ Cosx \\equal{} Cosy$ and $ Sinx \\equal{} \\minus{} Siny$, Show that $ Sin^2 \\frac {x \\plus{} y}{2} \\equal{} 0$\n[/hide]\n[/quote]\n\n[hide=\"Hint for 1\"]\nConsider the quadrants\nWhat does that tell you about the bisector of x and y?\n[/hide]", "Solution_2": "$ 1)$ We know that $ \\cos x \\equal{} \\cos y$ and $ \\sin x \\equal{} \\minus{} \\sin y$.\r\n\r\n[hide]So that must hold true for all $ x$ and $ y$. Thus $ \\cos \\left( \\frac {x}{2} \\right) \\equal{} \\cos \\left( \\frac {y}{2} \\right)$ and $ \\sin \\left( \\frac {y}{2} \\right) \\equal{} \\minus{} \\sin \\left( \\frac {x}{2} \\right)$.\n\nThus:\n\n$ \\cos \\left( \\frac {x}{2} \\right) \\cdot \\sin \\left( \\frac {x}{2} \\right) \\equal{} \\cos \\left( \\frac {y}{2} \\right) \\cdot \\left[ \\minus{} \\sin \\left( \\frac {y}{2} \\right)\\right]$; substitute what we know is true\n$ \\iff \\cos \\left( \\frac {x}{2} \\right) \\cdot \\sin \\left( \\frac {x}{2} \\right) \\plus{} \\cos \\left( \\frac {y}{2} \\right) \\cdot \\sin \\left( \\frac {y}{2} \\right) \\equal{} 0$; bring over to other side\n$ \\iff \\sin \\left( \\frac {x}{2} \\plus{} \\frac {y}{2} \\right) \\equal{} 0$; use trig identities\n$ \\iff \\sin \\left( \\frac {x \\plus{} y}{2} \\right) \\equal{} 0$\n$ \\iff \\boxed{\\sin^2 \\left( \\frac {x \\plus{} y}{2} \\right) \\equal{} 0}$; just square it on this step[/hide]\n\n$ 2)$ Given $ y \\equal{} 3 \\sin \\left[ \\frac {2\\pi}{365} \\cdot (x \\minus{} 79) \\right] \\plus{} 12$ where $ x$ is the number of days after January $ 1^{\\text{st}}$ and $ y$ is the number of hours of daylight,\n\n[hide]$ a)$ The amplitude of the function is $ 3$. This is just the number that $ \\sin$ is multiplied by.\n\n$ b)$ When $ y\\equal{}\\sin (bx\\plus{}c)$, the period is given by $ P\\equal{}\\frac{2\\pi}{b}$.\n\nThus:\n\n$ P \\equal{} \\frac{2\\pi}{\\frac{2\\pi}{365}}$\n$ \\implies \\boxed{P\\equal{}365}$ which is obvious since there are 365 days in the year.\n\n$ c)$ The maximum of the $ \\sin$ function is $ 1$. Thus the maximum of $ y$ is given by $ y \\equal{} 3 \\cdot 1 \\plus{} 12 \\equal{} \\boxed{15}$\n\n$ d)$ The mininum of the $ \\sin$ function is $ \\minus{}1$. Thus the mininum of $ y$ is given by $ y\\equal{}3 \\cdot (\\minus{}1) \\plus{}12 \\equal{} \\boxed{9}$.[/hide]", "Solution_3": "Ok...Thank you for the help!\r\n\r\nNow would you please guide me to question 5 and 6?", "Solution_4": "[quote=\"CyanSnow\"]If $ \\sec (\\theta - \\alpha), \\sec\\theta$ and $ \\sec (\\theta + \\alpha)$ are in A.P., prove that $ \\cos\\theta = \\sqrt2 \\cos \\frac {\\alpha}{2}$[/quote]Some trig identities :| \r\n$ \\frac{1}{\\cos{\\theta}\\cos{\\alpha}+\\sin{\\theta}\\sin{\\alpha}}$, $ \\frac{1}{\\cos{\\theta}}$, $ \\frac{1}{\\cos{\\theta}\\cos{\\alpha}-\\sin{\\theta}\\sin{\\alpha}}$ are in arithmetic progression.\r\n\\begin{eqnarray*}(\\cos{\\theta}\\cos{\\alpha})^2-(\\sin{\\theta}\\sin{\\alpha})^2&=&\\frac{\\cos{\\theta}(\\cos{\\theta}\\cos{\\alpha}+\\sin{\\theta}\\sin{\\alpha})+\\cos{\\theta}(\\cos{\\theta}\\cos{\\alpha}-\\sin{\\theta}\\sin{\\alpha})}{2}\\\\\r\n&=&\\cos^2{\\theta}\\cos{\\alpha}\\end{eqnarray*}\r\n$ (\\cos{\\theta}\\cos{\\alpha})^2-(1-\\cos^2{\\theta})\\sin^2{\\alpha}=\\cos^2{\\theta}\\cos{\\alpha}$\r\n$ \\cos^2{\\theta}=\\frac{\\sin^2{\\alpha}}{\\cos^2{\\alpha}+\\sin^2{\\alpha}-\\cos{\\alpha}}=\\frac{\\sin^2{\\alpha}}{1-\\cos{\\alpha}}\\implies \\cos{\\theta}=\\frac{\\sin{\\alpha}}{\\sqrt{1-(1-2\\sin^2{\\frac{\\alpha}{2}})}}$\r\n$ \\implies\\cos{\\theta}=\\frac{2\\sin{\\frac{\\alpha}{2}}\\cos{\\frac{\\alpha}{2}}}{\\sqrt{1-(1-2\\sin^2{\\frac{\\alpha}{2}})}}=\\sqrt{2}\\cos{\\frac{\\alpha}{2}}$", "Solution_5": "8. [hide]Let $ N\\equal{}\\frac{n(n\\plus{}1)}{2}$, (where n is the no. of rows in the equilateral triangle) be the original no. of balls.so,we have\n$ N\\plus{}669\\equal{}(n\\minus{}8)^{2}$\nsolving the quadratic, we get $ n\\equal{}55 \\Rightarrow N\\equal{}770$[/hide]", "Solution_6": "is the ans of ques 4 =\r\na=-3/2\r\nc=1/2\r\nis the ans of ques 5 =(-4,-7)" } { "Tag": [ "trigonometry" ], "Problem": "1. 64sin10sin30sin50sin70sin90 = ? \r\n 2. Let cos(A+B) = 3/root10 and cosA = 2/root5 determine cosB/cos(2A+B)", "Solution_1": "1. [hide] Apply the double angle identity many, many times to $ sin10$. \\[ sin10 \\equal{} \\frac {sin20}{2cos10} \\equal{} \\frac {sin40}{4(cos10)(cos20)} \\equal{} \\frac {sin80}{8(cos10)(cos20)(cos40)}.\\] Now use $ sin(x) \\equal{} cos ( 90 \\minus{} x)$. We get \\[ \\frac {cos10}{8(cos10)(cos20)(cos40)} \\equal{} \\frac {1}{8(cos20)(cos40)}.\\]\n\nIf we apply this previous identity to the $ sin50$ and $ sin70$. You will see that all the bad angles cancel and we are left with $ 64/16 \\equal{} 4$ .[/hide]", "Solution_2": "[hide=\"Hint for (1)\"]\nuse $ \\sin A. sin (60\\minus{}A). \\ sin(60\\plus{}A) \\equal{} \\frac{1}{4} \\sin 3A$\n\nwhere the angles are in degrees.\n\n[/hide]\n\n[hide=\"hint (2)\"]\nusing expansion of $ \\cos (A\\plus{}B)$ find $ \\cos B$\n\nnow it should be simple!\n[/hide]", "Solution_3": "[quote=\"Nora.91\"][hide=\"Hint for (1)\"]\nuse $ \\sin A. sin (60 \\minus{} A). \\ sin(60 \\plus{} A) \\equal{} \\frac {1}{4} \\sin 3A$\n\nwhere the angles are in degrees.\n\n[/hide]\n\n[hide=\"hint (2)\"]\nusing expansion of $ \\cos (A \\plus{} B)$ find $ \\cos B$\n\nnow it should be simple!\n[/hide][/quote]\r\n\r\nJust out of curiosity, did you know the formula $ \\sin A. sin (60 \\minus{} A). \\ sin(60 \\plus{} A) \\equal{} \\frac {1}{4} \\sin 3A$ or did you just figure it out when you did the problem.", "Solution_4": "[quote=\"kurt.math\"]\nJust out of curiosity, did you know the formula $ \\sin A. sin (60 \\minus{} A). \\ sin(60 \\plus{} A) \\equal{} \\frac {1}{4} \\sin 3A$ or did you just figure it out when you did the problem.[/quote]\r\n\r\ni knew it.\r\n\r\n\r\n$ \\sin 3A \\equal{} 3 \\sin A \\minus{} 4 \\sin ^3 A \\\\ \\equal{} 4 \\sin A ( (\\frac{\\sqrt{3}}{2})^2 \\minus{} \\sin ^2 A ) \\equal{} 4 \\sin A (\\sin (60 \\minus{} A). \\sin(60 \\plus{} A) )$", "Solution_5": "64sin10sin30sin50sin70sin90 = 32cos80cos40cos20 = (16sin40cos40cos80)/sin20 \r\n\r\n = (8sin80cos80)/sin20 = (4sin160)/sin20 = 4 \r\n\r\n\r\n - - 5555 thank you everybody" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "If $ a$, $ b$, $ c$ are nonnegative reals satisfying $ a^2\\plus{}b^2\\plus{}c^2\\equal{}1$, then prove that\r\n\r\n$ \\frac {1}{1 \\minus{} a^2} \\plus{} \\frac {1}{1 \\minus{} b^2} \\plus{} \\frac {1}{1 \\minus{} c^2} \\plus{} \\frac {1}{1 \\minus{} ab} \\plus{} \\frac {1}{1 \\minus{} bc} \\plus{} \\frac {1}{1 \\minus{} ca} \\geq 9$.", "Solution_1": "[quote=\"vedran6\"]If $ a$, $ b$, $ c$ are nonnegative reals satisfying $ a^2\\plus{}b^2\\plus{}c^2\\equal{}1$, then prove that\n\n$ \\frac {1}{1 \\minus{} a^2} \\plus{} \\frac {1}{1 \\minus{} b^2} \\plus{} \\frac {1}{1 \\minus{} c^2} \\plus{} \\frac {1}{1 \\minus{} ab} \\plus{} \\frac {1}{1 \\minus{} bc} \\plus{} \\frac {1}{1 \\minus{} ca} \\geq 9$.[/quote]\nMy ugly proof:\n$\\sum_{cyc} \\frac {1}{1 \\minus{} a^2}+\\sum_{cyc} \\frac {1}{1 \\minus{} ab}=\\sum_{cyc} \\frac {(3a+b+c)^2}{(3a+b+c)^2(1 \\minus{} a^2)}+\\sum_{cyc} \\frac {(a+b)^2}{(a+b)^2(1 \\minus{} ab)}\\geq$\n$\\geq\\frac{25(a+b+c)^2}{\\sum\\limits_{cyc}(3a+b+c)^2(1 \\minus{} a^2)}+\\frac{4(a+b+c)^2}{\\sum\\limits_{cyc}(a+b)^2(1 \\minus{} ab)}\\geq9$,\nwhere the last inequality we can prove by $uvw$'s technique." } { "Tag": [ "algebra", "polynomial", "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "supposedly a very easy problem\r\n is a semi group.prove that there exists an element 'a' in the semigroup A s.t a*a=a.\r\nbest regards\r\nvasu", "Solution_1": "Maybe it should be finite or something? Otherwise, it's clearly not true. Take the semigroup generated by $X$ in a polynomial ring $K[X]$, for instance.\r\n\r\nEdit:\r\n\r\nAnd if $A$ is indeed finite, then the problem is easy:\r\n\r\nTake some $x\\in A$ and two positive integers $m(u+2)m$, and then take $a=x^{(u+1)(n-m)}$." } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "real analysis", "real analysis solved" ], "Problem": "Evaluate the following definite integrals.\r\n\r\n(1)$\\int^{16}_1\\frac{dx}{\\sqrt{x}+\\sqrt[4]{x}}$\r\n\r\n(2)$\\int^{\\frac{\\pi}{3}}_0 \\frac{dx}{\\sin x+\\sqrt{3}\\cos x} dx$", "Solution_1": "For the first one put $x=t^4$ so $t\\in[1,2]$.\r\nresults our integral is equal to \r\n$\\displaystyle\\int_1^2\\frac{4t^3}{t+t^2}dt=4\\int_1^2\\frac{t^2}{t+1}dt=4\\int_1^2\\left(t-1+\\frac1{t+1}\\right)=8-1-(8-4)+4\\ln\\frac32=3+4\\ln\\frac32$\r\n\r\nfor the second one... rewrite $ \\sin x+\\sqrt3\\cos x=2(\\frac12\\sin x+\\frac{\\sqrt3}2\\cos x)=2\\cos(x-\\pi/6)$....", "Solution_2": "Thank you for your reply, Kuba.\r\n\r\nFor the first one, the answer is $2+4\\ln \\frac{3}{2}$.\r\n\r\nkunny" } { "Tag": [], "Problem": "Prove that the expression \r\n\\[ m^5 + 3m^4n - 5m^3n^2-15m^2n^3+4mn^4+12n^5 \\] cannot have the value $33$, regardless of what integers are substituted for $m$ and $n$.", "Solution_1": "[hide]\nNote that we have two sets of coefficients with $(1, -5, 4)$, thus we can factor to\n$3n(m^4-5m^2n^2+n^4) + m(m^4-5m^2n^2 + n^4)$\n$(m+3n)(m^2-n^2)(m^2-4n^2)$\nFor this expression to equal $33$, we either have to have 1 factor equal $33$, and the other two equal $1$, or two factors of $3$ and $11$, and another of $1$. Regardless, one factor must be equal to $1$.\nThe 2nd and 3rd factors cannot equal 1, by contradiction. The two possibilities are $m^2 = 1 + n^2$ and $m^2 = 1 + 4n^2 = 1 + (2n)^2$, both cases only have solutions $m = \\pm 1, n = 0$, which never results in $33$.\nAssume the first factor equals 1. Then $m = 1-3n$, and plugging back into the expression, the remaining part is $(8n^2 - 6n + 1)(5n^2 - 6n + 1)$. Setting both factors equal to 1 yields one solution, $n=0$, which does not work. Thus the only possibility is that one factor is 3 and the other is 11. Setting both equal to 3 yields one solution, $n=1$, for the first factor. However plugging it into the second factor gives 0, which is not 11.\nTherefore there are no integer solutions.\n[/hide]" } { "Tag": [], "Problem": "What sort of chemistry competitions are out there besides the USNCO and where can I find practice tests/previous tests for them and the USNCO?", "Solution_1": "International Chemistry Olympiads." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "If $a,b,c,d \\in \\mathbb{R}_{+}$ and $a+b +c +d =1$, show that \\[ ab +bc +cd \\leq \\dfrac{1}{4}. \\]", "Solution_1": "Look at this:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=30787[/url]" } { "Tag": [ "geometry", "3D geometry", "combinatorics proposed", "combinatorics" ], "Problem": "A cube of edge $ n$ is made of $ n^{3}$ unit cubes, and there are $ 6n^{2}$ unit squares on the surface of the cube. At most how many of these squares can be chosen so that no two have a common edge?", "Solution_1": "Anybody? :)", "Solution_2": "Here is the idea for the case $ n = 2k$ is even. The surface of the cube can be broken down into $ 4n$ disjoint \"ribbons,\" each of which is an odd cycle encircling one of the vertices of the cube. For example, eight of these ribbons consist of the three squares containing a given vertex of the cube. The next eight (if $ k \\geq 2$) consist of the six squares which border the first three, and the additional three squares needed to make a circuit out of them. And so on. (I'm sorry if this isn't clear.) Thus, we have eight cycles of length 3, eight cycles of length 9, ..., and eight cycles of length $ 3(n-1)$. The cycles of length 3 may have only one colored square; the cycles of length 9 may have only 4 colored squares; ...; and the cycles of length $ 3(n-1)$ may have only $ \\frac{3(n-1)-1}{2}= 3k-2$ colored squares. Thus, there are at most $ 8 \\sum_{i = 1}^{k}3i-2 =12k^{2}-4k = 3n^{2}-2n$ colored squares.\r\n\r\nIt is still necessary to find a coloring for which this maximum is achieved (it's not too hard to do, but it's unpleasant to describe without a diagram) and to do the odd case. I believe for $ n = 2k-1$ the situation is essentially the same; there is a little trick to be done at the end, since it is not possible to decompose the cube entirely into odd cycles in the same way, but the main idea is still good." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Evaluate the minimum value of $\\frac{sen^{3}x}{cosx}+\\frac{cos^{3}x}{senx}=-\\frac{3}{4}$", "Solution_1": "it's easy to show that the result is periodic\r\n\r\nSo the minimun will be ${-\\infty }$\r\n\r\nif you rewrite your question will be possible to solve" } { "Tag": [], "Problem": "Express $1+\\frac{1}{2+\\frac{1}{3}}$ as a common fraction.", "Solution_1": "[quote=\"rcv\"]Express $1+\\frac{1}{2+\\frac{1}{3}}$ as a common fraction.[/quote]\r\n\r\n[hide]\nWork your way up.\n\n2 + 1/3 = 7/3\n\n1 / (7/3) = 3/7\n\n1 + 3/7 = 10/7\n[/hide]\r\n\r\n :)", "Solution_2": "[quote=\"rcv\"]Express $1+\\frac{1}{2+\\frac{1}{3}}$ as a common fraction.[/quote]\r\n[hide=\"solution\"]\n1 + (1/(2 + 1/3))\n1 + (1/(7/3))\n1 + (3/7)\n10/7\n[/hide]", "Solution_3": "[quote=\"rcv\"]Express $1+\\frac{1}{2+\\frac{1}{3}}$ as a common fraction.[/quote]\r\n\r\n[hide]$1+\\frac{1}{\\frac{7}{3}}=1+\\frac{3}{7}=\\frac{10}{7}$[/hide]", "Solution_4": "[quote=\"rcv\"]Express $1+\\frac{1}{2+\\frac{1}{3}}$ as a common fraction.[/quote]\r\n\r\n[hide]2 + 1/3 = 7/3\n\n1/(7/3) = 3/7\n\n1+3/7 = 10/7[/hide]" } { "Tag": [], "Problem": "Any prime number $ p \\equiv 1\\ mod 4$ can be expressed as a sum of two squares.\r\nThat is the sequence $ 5 ,13,17,29,37.....................$ each term can be expressed as sum of two squares.\r\nIs this true or not ?", "Solution_1": "Its [url=http://en.wikipedia.org/wiki/Proofs_of_Fermat's_theorem_on_sums_of_two_squares]true.[/url]" } { "Tag": [ "puzzles" ], "Problem": "Sheikh's Heritage\r\nAn Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower will win. The brothers, after wandering aimlessly for days, ask a wise man for advise. After hearing the advice they jump on the camels and race as fast as they can to the city. What does the wise man say?", "Solution_1": "[hide=\"i knew this\"]they switched camels[/hide]", "Solution_2": "I heard this too, even though I couldn't remember the answer :D ." } { "Tag": [ "inequalities", "number theory proposed", "number theory" ], "Problem": "Find all the nonegative integers satisfying m,n such that:\r\n\r\n 1!3!5!.........(2n-1)!=m!", "Solution_1": "its hard !", "Solution_2": "No,it isn't.It's very easy. :P It had been given at Russian Olympiad.I'll post a solution tomorrow. ;) Now I'm tired...", "Solution_3": "Suppose $n > 2$, whence $m > 3$\r\nLet $p_n$ denote the $n_{%Error. \"tmop\" is a bad command.\n{th}}$ prime. Let $p_k$ be the greatest prime divisor of $m$.\r\nThen $p_k \\leq m < p_{k + 1}$ (by definition) and $p_k \\leq 2 n - 1 < p_{k + 1}$.\r\n(If $2 n - 1 < p_k$, then $p_k \\nmid \\prod_{j = 1}^n \\left( 2 j - 1 \\right) !$ but $p_k |m!$. If $2 n - 1 \\geq p_{k + 1}$, then $p_{k + 1} | \\prod_{j = 1}^n \\left( 2 j - 1 \\right) !$ but $p_{k + 1} \\nmid m!$. A contradiction in both cases). Moreover\r\n\\begin{eqnarray*} \\prod_{j = 1}^{n - 1} \\left( 2 j - 1 \\right) ! & = & \\frac{m!}{( 2 n - 1 ) !}\\\\ & = & \\prod^m_{j = p_k + 1} j \\end{eqnarray*}\r\nis not divisble by $p_k$ (since $2 p_k > p_{k + 1} - 1 \\geq m$). Therefore $2 ( n - 1 ) - 1 = 2 n - 3 < p_k$.\r\nSince $2 n - 1 \\geq p_k$ and $2 n - 3 < p_k$ we have $2 n - 1 = p_k$ or $2 n - 1 = p_k + 1$.\r\nBy asumption $n > 2$ hence $2 n - 2 = p_k$ is impossible. Therefore $2 n - 1 = p_k$.\r\nWhence\r\n\\begin{eqnarray*} ( 2 n - 2 ) ! \\prod^{n - 1}_{j = 1} \\left( 2 j - 1 \\right) ! & = & \\frac{m!}{2 n - 1}\\\\ & = & \\frac{m!}{p_k} \\end{eqnarray*}\r\n\r\nEDIT: Mistake, to be corrected here.", "Solution_4": "That is the proof of V.P.Beskarev, the author of the problem.I didn't read it very well, but I think it's correct, {x}.\r\nNow, please solve this equation, composed by the same V.P.Beskarev:1!3!5!...(2m-1)!=(m(m+1)/2)!. ;)", "Solution_5": "Isn't this a particular case of the preceding problem ?\r\nThat is, it suffice to solve $( m, 1 + 2 + \\ldots + m ) = ( 1, 1 ) \\mbox{ or } (2, 3 )$. Hence $m = 1$ or $m = 2$.\r\nOr is there some shorter, nicer solution we are supposed to find ? :P (I guess yes, so I'll try to find one)\r\nAlso, did V.P Beskarev publish his results in some journal ? Do you have any reference ?", "Solution_6": "The solutions of this equation are m=1,2,3,4. ;) Do you want to know a different way to solve it?", "Solution_7": "Here it is.We'll use the Theoerem of Bertrand.Between a and 2a there are at least 1 prime number.We notice p the prime number which is between 2x-1 and 2(2x-1).If we had the inequality 2(2x-1)x(x+1)/2=>x=<6.We only obtain the previous solutions.This is the solution of V.P.Beskarev in Kvant Magazine. ;)", "Solution_8": "Sorry, I've made a mistake, beginning with \"If $m>p_k$\".\r\nI will correct it." } { "Tag": [ "symmetry" ], "Problem": "Given two points A and B, and some point C on segment AB which is not the midpoint of AB, construct some point C' on line AB such that \r\n\r\n[tex]\\displaystyle\\frac{AC'}{C'B} = \\frac{AC}{CB}[/tex]\r\n\r\nusing only a straightedge.\r\n\r\n(NOTE: These are undirected edge lengths).", "Solution_1": "[hide]For this to be true, C' must be the same point as C. \n\n\n\nGiven a point C and another point C', if AC>AC', then CBC'B. Due to this inverse nature, the only way for the equation to hold true would be if C is the same point as C'.\n\n[/hide]", "Solution_2": "The problem says C' lies on \"line AB\" not \"segment AB.\"\r\n\r\nSuch a point C' does exist, as long as C is not the midpoint of segment AB.\r\n\r\nSo... construct it :).", "Solution_3": "Are we allowed to know any definite values during the construction?\n\n\n\n[hide]Point C' can be defined as being AC/(CB-AC) away from point A if AC>CB (C' would be on the opposite side of point A from point B. Point C' would have to be the same as point C if C' was in segment AB, and if it were on the other side of point B, the problem would become impossible), or it can be defined as CB/(AC-CB) away from point B in the same manner if ACCB for simplicity's sake. The same thing applies if AC CB. Let's have a = AC and b = CB. Construct a triangle with sides (a + b) and (a - b) (the angle between them doesn't matter - for convenience, let's suppose that these two sides are perpendicular). Construct a segment of length b that is parallel to the side (a - b), and touches both other sides of the triangle (or their extensions). Then, the segment between the vertice opposite length (a - b) and the intersection of the segments of length (a + b) (or its extension) and (b) (call it c) has the required length. \n\n\n\nProof: We need to construct length c such that\n\n\n\na : b = (a + b + c) : c\n\n\n\nRearrange to get\n\n\n\nac = ab + b2 + bc\n\nac - bc = ab + b2\n\n(a - b)c = b(a + b)\n\nc : b = (a + b) : (a - b)\n\n\n\nThis is the similar triangle construction in the diagram.\n\n[/hide]", "Solution_5": "Coder wrote:Are we allowed to know any definite values during the construction?\n\n[hide]Point C' can be defined as being AC/(CB-AC) away from point A if AC>CB (C' would be on the opposite side of point A from point B. Point C' would have to be the same as point C if C' was in segment AB, and if it were on the other side of point B, the problem would become impossible), or it can be defined as CB/(AC-CB) away from point B in the same manner if ACCB for simplicity's sake. The same thing applies if AC0,$ there exists $ n$ such that \r\n\r\n$ \\sum_{i\\equal{}n}^{\\infty}\\mu(A_i)<\\epsilon.$\r\n\r\nHence, $ \\mu(Z)<\\epsilon$ for all $ \\epsilon>0,$ from which we deduce that $ \\mu(Z)\\equal{}0.$\r\n\r\n==================\r\n\r\nNote in both cases the centrality of the hypothesis that $ \\sum_{i\\equal{}1}^{\\infty}\\mu(A_i)<\\infty.$ Would it be enough just to have $ \\lim_{i\\to\\infty}\\mu(A_i)\\equal{}0?$\r\n\r\n(Yes, I already know the answer to that; I'm looking for a response from a student, possibly one taking this course.)" } { "Tag": [ "AwesomeMath", "summer program", "trigonometry" ], "Problem": "I vote Bubala as funniest of the camp :rotfl: \r\n\r\nTeam Contest 1: Amazing 11-variable bogus proof\r\n\r\nTeam Contest 2: My wrist!!!!!!!\r\n\r\n:D\r\n\r\nVote in the poll...", "Solution_1": "I was there, LOL :D \r\n\r\nDELTA team will rise again!!!!!!!", "Solution_2": "Gabriel wasnt that bad... :maybe: \r\nnot_trig, are you damien?", "Solution_3": "Gabriel and bubula rock my socks.", "Solution_4": "Yes, asdf4. LOOK AT THE YEARBOOK :D \r\n\r\nTeam $\\gamma$ [b]PWNED[/b] team $\\delta$. Don't try it, Ben.\r\n\r\nBTW, who voted 10!?!?!?!? Gabriel is a genius, but he's not that great a teacher.", "Solution_5": "dude gabriel and i love each other so much it's not even funny\r\n\r\n\"What the s***? I don't get it at all...\"\r\n\"Who said that? Oh, you. You should just go back to sleep.\"", "Solution_6": "Treething:\r\n\r\n\"You. Out.\"", "Solution_7": "I gave Gabriel a 10 because he's hilarious and he laughed at the Dorin-speedo joke.\r\n\r\nand because he curses all the time and yells at damien.\r\n\r\nPS~ bubala SO should've won that bs-proof award...", "Solution_8": "NO, Vitek should have won the BS proof award, :rotfl: . Remember the team contests?", "Solution_9": "i think a certain BS proof that actually got credit beats them all?", "Solution_10": "Bubala should have won the BS proof; I agree. He didn't get any credit (1 -1)\r\n\r\nYou love Gabriel because he yelled at me? \r\n@#$%^{&}amp;*)!@#\n$%^&*(!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!", "Solution_11": "[quote=\"mathclass\"]Gabriel and bubula rock my socks.[/quote]\r\n\r\nOMG bubs you have an admirer! I wonder who this is hmmmm hyperness suggests Patricia Li but eh", "Solution_12": "whoa patricia and bubs would be interesting", "Solution_13": "nope, not patricia\r\n\r\ni'm patricia's husband. or wife. either one works.", "Solution_14": "I heard Bubala was genuinely upset he didn't get any points on the giant bogus proof that he pulled an all-nighter to concoct (\"z looks like 2, so let's let them be equal...\") :D The wrist thing was a classic :rotfl:", "Solution_15": "Actually, it was \"h looks like k, so I used them interchangeably\". \r\n\r\nz=2 for all z was attributed to him, though I stole it. :ninja:", "Solution_16": "And also in all cases y=z since they look really similar in cursive (at least that's the way I see it).\r\n\r\nStop making fun of Gabriel by the way.", "Solution_17": "Gabriel's the man. He's got 11 votes for 10." } { "Tag": [ "geometry", "3D geometry", "sphere" ], "Problem": "10 planes cut through a sphere so that the sphere is divided into $n$ pieces. What is the maximum possible value of $n$?", "Solution_1": "[quote=\"frt\"]10 planes cut through a sphere so that the sphere is divided into $n$ pieces. What is the maximum possible value of $n$?[/quote]\r\n\r\n[hide] $28?$[/hide]", "Solution_2": "[quote=\"Jos\u00e9\"][quote=\"frt\"]10 planes cut through a sphere so that the sphere is divided into $n$ pieces. What is the maximum possible value of $n$?[/quote]\n\n[hide] $28?$[/hide][/quote]\r\nGot proof?", "Solution_3": "[quote=\"Jos\u00e9\"][hide] $28?$[/hide][/quote]\r\nSorry, that's not correct. The answer is more than that.", "Solution_4": "[quote=\"frt\"][quote=\"Jos\u00e9\"][hide] $28?$[/hide][/quote]\nSorry, that's not correct. The answer is more than that.[/quote]\r\n\r\nThanks, anyway.", "Solution_5": "i believe the answer is [hide]176[/hide]\r\n\r\ngeneralization: try and find the number of sections using $k$ cuts" } { "Tag": [ "Recursive Sequences" ], "Problem": "Let $\\{u_{n}\\}_{n \\ge 0}$ be a sequence of integers satisfying the recurrence relation $u_{n+2}=u_{n+1}^2 -u_{n}$ $(n \\in \\mathbb{N})$. Suppose that $u_{0}=39$ and $u_{1}=45$. Prove that $1986$ divides infinitely many terms of this sequence.", "Solution_1": "$ r_n\\equiv u_n (\\mod 1986)$ which $ 0\\leq r_n\\leq 1985$\r\nBecause $ \\{r_n\\}$ take finite value so it has $ k,l\\in N$ such that : \r\n$ k > l$\r\n$ r_k \\equal{} r_l,r_{k \\plus{} 1} \\equal{} r_{l \\plus{} 1}$ (1)\r\n We prove that $ r_{n} \\equal{} r_{n \\plus{} k \\minus{} l},\\forall n$\r\nBecause (1) we has :$ r_{k \\minus{} 1}\\equiv r_{l \\minus{} 1} (\\mod 1986)$ so $ r_{l \\minus{} 1} \\equal{} r_{k \\minus{} 1}$\r\n $ r_{k \\plus{} 2} \\equal{} \\equiv r_{l \\plus{} 2}(\\mod 1986)$ so $ r_{k \\plus{} 2} \\equal{} r_{l \\plus{} 2}$\r\nContine this progress we has $ r_{n} \\equal{} r_{n \\plus{} k \\minus{} l},\\forall n\\in N$\r\nBecause $ r_2\\equiv 0(\\mod 1986)$ so $ r_{h(k \\minus{} l) \\plus{} 2}\\equiv 0(\\mod 1986)$\r\nSo has infinite $ u_n$ such that $ 1986|u_n$", "Solution_2": "this is not a problem in China 1991.\r\nIn fact, it is in Canada 1986.\r\n\r\nmaybe that is the reason why the sourse of M27 was shown as \"Canada 1986\":http://www.mathlinks.ro/viewtopic.php?p=849911#849911", "Solution_3": "Prove TTsphn is not complete. May be period $ r_{k\\plus{}T}\\equiv r_k \\ \\forall k>C$. In these case we can continue sequense $ n\\to n\\minus{}1$: $ u_n\\equal{}u_{n\\plus{}1}^2\\minus{}u_{n\\plus{}2}$, therefore $ C\\equal{}0$." } { "Tag": [ "geometry" ], "Problem": "If $s$ is the semiperimeter prove or disrove the following\r\n$8s^3<9^s$", "Solution_1": "Er, does the fact its the semiperimeter have anything to do with anything???", "Solution_2": "Strange indeed. The question seems to be rubbish.", "Solution_3": "I think he means ... of a polygon with area 1. Or something. \r\n\r\nAs it's true from 3/2 onwards... (and the optimal is a circle, where the semiperimeter would only be pi/2)\r\n\r\nSeems quite nice thus :)", "Solution_4": "It is a part of a problem which I tried to solve. I reach at this and I couldn't find an answer. I make some computations and this ineq is true I think but how could someone solve it? Help me.", "Solution_5": "It is not true for all s. For >3/2, it's sufficient to state that \r\n$9^s-8s^3$ is increasing.", "Solution_6": "Give me a counterexample please", "Solution_7": "everything between 1.3 and 1.5 should be a counterexample?", "Solution_8": "[quote=\"Peter VDD\"]everything between 1.3 and 1.5 should be a counterexample?[/quote]\r\n\r\nOk thanks" } { "Tag": [ "AIME", "modular arithmetic", "function", "algebra", "binomial theorem", "number theory" ], "Problem": "Let $a_n = 6^n + 8^n$. Determine the remainder on dividing $a_{83}$ by 49.", "Solution_1": "[hide=\"Solution\"]First, we try to find a relationship between the numbers we're provided with and 49. We realize that $49=7^2$ and both $6$ and $8$ greater or less than 7 by $1$. \n\nExpressing the numbers in terms of $7$, we get $(7-1)^{83}+(7+1)^{83}$.\n\nApplying the Binomial Theorem, half of our terms cancel out and we are left with $2(7^{83}+3403*7^{81}+\\cdots + 83*7)$. We realize that all of the terms in this big jumble of numbers are divisible by 49 except the final term.\n\nAfter some quick division, our answer is $35$.[/hide]", "Solution_2": "I know this is a really old bump and all, but I want to do the AIME problems systematically. Can someone explain what one has to divide to get 35? :huh:", "Solution_3": "[hide]\n\nThe divisor can be rewritten as $7^2$.\n\nRewrite the dividend as: $(7-1)^{83}+(7+1)^{83}$\n\nWhen we expand this out using the Binomial Theorem, we'll get the sum of a bunch of different powers of $7$ multiplied by the corresponding coefficient. We are dividing by $7^2$, so any term with $7$ raised to a power greater than or equal to $2$ is going to leave no remainder, so we're only concerned with the terms where $7$ is raised to either the first or zero power.\n\nThose terms are: $\\displaystyle{83\\choose 82}7^1(-1)^{82}+\\displaystyle{83\\choose 83}7^0(-1)^{83}+\\displaystyle{83\\choose 82}7^1(1)^{82}+\\displaystyle{83\\choose 83}7^0(1)^{83}$\n\nTheir sum equals: $2\\times83\\times7=1162$\n$1162\\equiv \\boxed{35} \\bmod{49}$\n\n[/hide]", "Solution_4": "Oh, I get it.\r\n\r\nThe $-1$ and $1$ terms will cancel out. The whole thing is multiplied by $2$ because there are two seperate \"binomials\".\r\n\r\nThanks.", "Solution_5": "Old topic, but I thought this was a particularly clever solution, and needed to be added. It's probably the fastest way I've seen to do this problem.\r\n\r\n[hide]\nWe are looking for $ 6^{83} \\plus{} 8^{83} \\pmod{49}$\nWe know $ \\phi(49)\\equal{}42$ so we can subtract out 42 from the exponents:\n$ 6^{41} \\plus{} 8^{41} \\pmod{49}$\nAnd again:\n$ 6^{\\minus{}1} \\plus{} 8^{\\minus{}1} \\pmod{49}$\n\nSo now this becomes a question of finding inverses mod 49.\nTo find the inverse of $ a \\bmod b$, we can solve the Diophantine equation $ ax \\plus{} by \\equal{} 1$. The value we get for $ x$ is the inverse we are looking for.\n\nSo looking at $ 6x \\plus{} 49y \\equal{} 1$, we quickly see $ x\\equal{}\\minus{}8$ and $ y\\equal{}1$.\nLooking at $ 8x\\plus{}49y\\equal{}1$, we see $ x\\equal{}\\minus{}6$ and $ y\\equal{}1$.\n\nSo we can simplify\n\n$ 6^{\\minus{}1} \\plus{} 8^{\\minus{}1} \\equiv \\minus{}8 \\plus{} \\minus{}6 \\equiv \\boxed{35} \\pmod{49}$\n[/hide]", "Solution_6": "[hide]That's really nice. You can also finish it by\n\\[ 6^{ \\minus{} 1} \\plus{} 8^{ \\minus{} 1} \\equal{} \\frac {8 \\plus{} 6}{48}\\equiv\\frac {14}{ \\minus{} 1}\\pmod{49} \\equal{} \\minus{} 14\\equiv35\\pmod{49},\n\\] instead of finding inverses.\n[/hide]", "Solution_7": "[quote=\"towersfreak2006\"][hide]That's really nice. You can also finish it by\n\\[ 6^{ \\minus{} 1} \\plus{} 8^{ \\minus{} 1} \\equal{} \\frac {8 \\plus{} 6}{48}\\equiv\\frac {14}{ \\minus{} 1}\\pmod{49} \\equal{} \\minus{} 14\\equiv35\\pmod{49},\n\\]\ninstead of finding inverses.\n[/hide][/quote]Very clever!", "Solution_8": "[quote=\"mysmartmouth\"]Old topic, but I thought this was a particularly clever solution, and needed to be added. It's probably the fastest way I've seen to do this problem.\n\n[hide]\nWe are looking for $ 6^{83} \\plus{} 8^{83} \\pmod{49}$\nWe know $ \\phi(49) \\equal{} 42$ so we can subtract out 42 from the exponents:\n$ 6^{41} \\plus{} 8^{41} \\pmod{49}$\nAnd again:\n$ 6^{ \\minus{} 1} \\plus{} 8^{ \\minus{} 1} \\pmod{49}$\n\nSo now this becomes a question of finding inverses mod 49.\nTo find the inverse of $ a \\bmod b$, we can solve the Diophantine equation $ ax \\plus{} by \\equal{} 1$. The value we get for $ x$ is the inverse we are looking for.\n\nSo looking at $ 6x \\plus{} 49y \\equal{} 1$, we quickly see $ x \\equal{} \\minus{} 8$ and $ y \\equal{} 1$.\nLooking at $ 8x \\plus{} 49y \\equal{} 1$, we see $ x \\equal{} \\minus{} 6$ and $ y \\equal{} 1$.\n\nSo we can simplify\n\n$ 6^{ \\minus{} 1} \\plus{} 8^{ \\minus{} 1} \\equiv \\minus{} 8 \\plus{} \\minus{} 6 \\equiv \\boxed{35} \\pmod{49}$\n[/hide][/quote]\r\n\r\nWhat do you mean by $ \\phi(49) \\equal{} 42$", "Solution_9": "[url=http://www.artofproblemsolving.com/Wiki/index.php/Totient_function]Euler's Totient Function[/url]\r\n\r\n[url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_theorem]Euler's Totient Theorem[/url]", "Solution_10": "[quote=towersfreak2006][hide]That's really nice. You can also finish it by\n\\[ 6^{ \\minus{} 1} \\plus{} 8^{ \\minus{} 1} \\equal{} \\frac {8 \\plus{} 6}{48}\\equiv\\frac {14}{ \\minus{} 1}\\pmod{49} \\equal{} \\minus{} 14\\equiv35\\pmod{49},\n\\] instead of finding inverses.\n[/hide][/quote]\n\nMassive bump, but how is 14/48 congruent to -14 in mod 49?", "Solution_11": "Because $48$ is $-1 \\mod 49$. So, $\\dfrac {14}{-1} = -14$", "Solution_12": "but is it a correct way? 48 is below 14 , so how can we say this? 14/48 is congruent to -14 mod 49 ?", "Solution_13": "Saying $\\frac{}{48}$ doesn't really mean dividing by $48$; it's just a shorthand way of writing the inverse of $48 \\bmod 49$. Which turns out to be $-1$.", "Solution_14": "[quote=brainiac1]Saying $\\frac{}{48}$ doesn't really mean dividing by $48$; it's just a shorthand way of writing the inverse of $48 \\bmod 49$. Which turns out to be $-1$.[/quote]\n\nOk got it .. so this means 48 ^ (-1) mod 49 ..and which is equal to 49 ..", "Solution_15": "$\\phi(49)=49(1-\\frac{1}{7})=42$\n$6^{83} \\mod {49}=6^{83 \\mod {42}} \\mod {49}=6^{-1} \\mod {49}$\nSimilarly: $8^{83} \\mod {49}=8^{-1} \\mod {49}$.\n\nAdding and applying Euclid's algorithm and common sense gives: $35 \\mod {49}$", "Solution_16": "$\\phi (49)=42$\n$6^{\u20131}+8^{\u20131}\\equiv 35 (mod49)$", "Solution_17": "hello,,, i saw one person on blog clowning this problem with very hard method,,, so this must be fixed,,, we will fix it now\n\n[hide=Solution]From Euler's Theorem we have $6^{42} \\equiv 1 \\pmod {49}$ and $8^{42} \\equiv 1 \\pmod {49}$. This means our expression is equal to $6^{-1}+8^{-1} \\equiv 7 \\cdot 24^{-1} \\pmod {49}$. Note that $24 \\cdot 2 \\cdot -1 \\equiv 1 \\pmod {49}$, so $24 \\cdot 47 \\equiv 1 \\pmod {47}$. We now just need $7 \\cdot 47 \\pmod {49} = \\boxed{35} \\pmod {49}$.[/hide]", "Solution_18": "[quote=surge]I know this is a really old bump and all, but I want to do the AIME problems systematically. Can someone explain what one has to divide to get 35? :huh:[/quote]\n\n[quote=mysmartmouth]Old topic, but I thought this was a particularly clever solution, and needed to be added. It's probably the fastest way I've seen to do this problem.\n\n[hide]\nWe are looking for $ 6^{83} \\plus{} 8^{83} \\pmod{49}$\nWe know $ \\phi(49)\\equal{}42$ so we can subtract out 42 from the exponents:\n$ 6^{41} \\plus{} 8^{41} \\pmod{49}$\nAnd again:\n$ 6^{\\minus{}1} \\plus{} 8^{\\minus{}1} \\pmod{49}$\n\nSo now this becomes a question of finding inverses mod 49.\nTo find the inverse of $ a \\bmod b$, we can solve the Diophantine equation $ ax \\plus{} by \\equal{} 1$. The value we get for $ x$ is the inverse we are looking for.\n\nSo looking at $ 6x \\plus{} 49y \\equal{} 1$, we quickly see $ x\\equal{}\\minus{}8$ and $ y\\equal{}1$.\nLooking at $ 8x\\plus{}49y\\equal{}1$, we see $ x\\equal{}\\minus{}6$ and $ y\\equal{}1$.\n\nSo we can simplify\n\n$ 6^{\\minus{}1} \\plus{} 8^{\\minus{}1} \\equiv \\minus{}8 \\plus{} \\minus{}6 \\equiv \\boxed{35} \\pmod{49}$\n[/hide][/quote]\n\n[quote=cdymdcool][quote=\"mysmartmouth\"]Old topic, but I thought this was a particularly clever solution, and needed to be added. It's probably the fastest way I've seen to do this problem.\n\n[hide]\nWe are looking for $ 6^{83} \\plus{} 8^{83} \\pmod{49}$\nWe know $ \\phi(49) \\equal{} 42$ so we can subtract out 42 from the exponents:\n$ 6^{41} \\plus{} 8^{41} \\pmod{49}$\nAnd again:\n$ 6^{ \\minus{} 1} \\plus{} 8^{ \\minus{} 1} \\pmod{49}$\n\nSo now this becomes a question of finding inverses mod 49.\nTo find the inverse of $ a \\bmod b$, we can solve the Diophantine equation $ ax \\plus{} by \\equal{} 1$. The value we get for $ x$ is the inverse we are looking for.\n\nSo looking at $ 6x \\plus{} 49y \\equal{} 1$, we quickly see $ x \\equal{} \\minus{} 8$ and $ y \\equal{} 1$.\nLooking at $ 8x \\plus{} 49y \\equal{} 1$, we see $ x \\equal{} \\minus{} 6$ and $ y \\equal{} 1$.\n\nSo we can simplify\n\n$ 6^{ \\minus{} 1} \\plus{} 8^{ \\minus{} 1} \\equiv \\minus{} 8 \\plus{} \\minus{} 6 \\equiv \\boxed{35} \\pmod{49}$\n[/hide][/quote]\n\nWhat do you mean by $ \\phi(49) \\equal{} 42$[/quote]\n\n[quote=RamanConjecture][quote=towersfreak2006][hide]That's really nice. You can also finish it by\n\\[ 6^{ \\minus{} 1} \\plus{} 8^{ \\minus{} 1} \\equal{} \\frac {8 \\plus{} 6}{48}\\equiv\\frac {14}{ \\minus{} 1}\\pmod{49} \\equal{} \\minus{} 14\\equiv35\\pmod{49},\n\\] instead of finding inverses.\n[/hide][/quote]\n\nMassive bump, but how is 14/48 congruent to -14 in mod 49?[/quote]\n\n[quote=targo___]but is it a correct way? 48 is below 14 , so how can we say this? 14/48 is congruent to -14 mod 49 ?[/quote]\n\n[quote=Paragdey12]$\\phi (49)=42$\n$6^{\u20131}+8^{\u20131}\\equiv 35 (mod49)$[/quote]\n\n[quote=OlympusHero]hello,,, i saw one person on blog clowning this problem with very hard method,,, so this must be fixed,,, we will fix it now\n\n[hide=Solution]From Euler's Theorem we have $6^{42} \\equiv 1 \\pmod {49}$ and $8^{42} \\equiv 1 \\pmod {49}$. This means our expression is equal to $6^{-1}+8^{-1} \\equiv 7 \\cdot 24^{-1} \\pmod {49}$. Note that $24 \\cdot 2 \\cdot -1 \\equiv 1 \\pmod {49}$, so $24 \\cdot 47 \\equiv 1 \\pmod {47}$. We now just need $7 \\cdot 47 \\pmod {49} = \\boxed{35} \\pmod {49}$.[/hide][/quote]\n\nEvery single post is a bump lol", "Solution_19": "We apply Euler\u2019s Theorem: $\\phi(49) = 42.$ $$6^{83} + 8^{83} \\equiv 6^{-1} + 8^{-1} \\equiv 41 + 43 \\equiv \\boxed{035} \\pmod{49}.$$", "Solution_20": "Nice, problem, this was in my AMSP class. I took session 1.\n\nWe have $\\phi(49)=42$, so by Euler's, $6^{83} + 8^{83} \\equiv 6^{-1} + 8^{-1} \\equiv 41 + 43 \\equiv 35 \\pmod{49}.$", "Solution_21": "this is an okayish problem having the euler totient function\n and at last we have 35\n\n ", "Solution_22": "We have\n\n\\[6^{83}+8^{83} \\equiv \\frac{1}{6}+\\frac{1}{8} = \\frac{14}{48} \\equiv \\frac{14}{-1} \\equiv \\boxed{35} \\pmod{49}.\\]", "Solution_23": "[hide]Substituting $6=7-1$ and $8=7+1$ and using the binomial theorem will do the trick but Euler's totient offers a quicker solution. We know that $6^{84}=(6^{42})^2= 1 \\mod 49$ and $8^{84}=(8^{42})^2= 1 \\mod 49$. Therefore, $6^{83} = 6^{-1} \\mod 49$ and $8^{83} = 8^{-1} \\mod 49$. Finally, we can find that $6^{-1}+ 8^{-1}=41 + 43 \\mod 49 = 35 \\mod 49$.[/hide]", "Solution_24": "Actually everyone above is wrong \n\nIt's $\\varphi$ not $\\phi=\\frac{1+\\sqrt{5}}{2}$\n:nerd:", "Solution_25": "Nope... both work... you're the one who's wrong...\n" } { "Tag": [ "geometry", "3D geometry", "calculus", "calculus computations" ], "Problem": "My teacher gave me a related rates worksheets where we apply derivatives. I have solved all of them except this one:\r\n\r\nSuppose you are drinking root beer from a conical paper cup. The cup has a diameter of 8 cm and depth of 10cm. As you suck on the straw, root beer leaves the cup at the rate 7 cm^3/sec. At what rate is the level of the liquid in the cup changing when the liquid is 6cm deep?", "Solution_1": "Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of $ 8 ~\\text{cm}$ and depth of $ 10 ~\\text{cm}$. As you suck on the straw, root beer leaves the cup at the rate $ 7 ~\\frac {\\text{cm}^3}{\\text{sec}}$. At what rate is the level of the liquid in the cup changing when the liquid is $ 6 ~\\text{cm}$ deep?\r\n\r\nIf my final answer is wrong I apologize, its been a while since I've had to do these. I'm fairly positive it is correct though.\r\n\r\n[b]Hint:[/b]\r\n[hide]If you were able to solve others you should note that the radius is directly proportional to the height in a cone. That is, $ r \\equal{} k\\cdot h \\implies 4 \\equal{} 10k \\implies k \\equal{} \\frac {2}{5}$. So $ \\boxed{\\frac {r}{h} \\equal{} \\frac {2}{5}}$ and use this in solving.[/hide]\n\n[b]Solution:[/b]\n[hide]Note that $ \\frac {dV}{dt} \\equal{} \\minus{} 7$\n\n$ V \\equal{} \\frac {1}{3} \\pi r^2 h$\n$ V \\equal{} \\frac {1}{3} \\pi \\left( \\frac {2h}{5} \\right)^2 h$\n$ V \\equal{} \\frac {4 \\pi h^3}{75}$\n\nThus:\n$ \\frac {dV}{dt} \\equal{} \\frac {4 \\pi}{75} \\cdot 3h^2 \\cdot \\frac {dh}{dt}$\n$ \\minus{} 7 \\equal{} \\frac {4 \\pi}{25} \\cdot 6^2 \\cdot \\frac {dh}{dt}$\n$ \\boxed{ \\minus{} \\frac {175}{144 \\pi} \\equal{} \\frac {dh}{dt}}$[/hide]", "Solution_2": "[quote=\"skimnc\"]Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of $ 8 ~\\text{cm}$ and depth of $ 10 ~\\text{cm}$. As you suck on the straw, root beer leaves the cup at the rate $ 7 ~\\frac {\\text{cm}^3}{\\text{sec}}$. At what rate is the level of the liquid in the cup changing when the liquid is $ 6 ~\\text{cm}$ deep?\n\nIf my final answer is wrong I apologize, its been a while since I've had to do these. I'm fairly positive it is correct though.\n\n[b]Hint:[/b]\n[hide]If you were able to solve others you should note that the radius is directly proportional to the height in a cone. That is, $ r \\equal{} k\\cdot h \\implies 4 \\equal{} 10k \\implies k \\equal{} \\frac {2}{5}$. So $ \\boxed{\\frac {r}{h} \\equal{} \\frac {2}{5}}$ and use this in solving.[/hide]\n\n[b]Solution:[/b]\n[hide]Note that $ \\frac {dV}{dt} \\equal{} \\minus{} 7$\n\n$ V \\equal{} \\frac {1}{3} \\pi r^2 h$\n$ V \\equal{} \\frac {1}{3} \\pi \\left( \\frac {2h}{5} \\right)^2 h$\n$ V \\equal{} \\frac {4 \\pi h^3}{75}$\n\nThus:\n$ \\frac {dV}{dt} \\equal{} \\frac {4 \\pi}{75} \\cdot 3h^2 \\cdot \\frac {dh}{dt}$\n$ \\minus{} 7 \\equal{} \\frac {4 \\pi}{25} \\cdot 6^2 \\cdot \\frac {dh}{dt}$\n$ \\boxed{ \\minus{} \\frac {175}{144 \\pi} \\equal{} \\frac {dh}{dt}}$[/hide][/quote]\r\n\r\nI just checked over your work. Nice job!", "Solution_3": "Okay, thank you very much. Makes sense now. I just wasn't sure where to begin, and I'm still a little shaky on implicit differentiation." } { "Tag": [], "Problem": "I know nothing about c++ programming.Yet I need to program unknown math into a computer program.Can anyome help me\r\nThanks\r\nunknowm math", "Solution_1": "You might want to ask some more specific questions to make it easier for others to help you. :wink:\r\nThanks.", "Solution_2": "Hi\r\nWhat I need is to program numbers in the computer .As of now I have done this in excel.I had one of the best programmer in the world from a small island .He was over Boland and Microsoft as I said one of the best in the world.\r\n\r\nHe knew what I had really worked But the trust was only one way .I trusted him to do the right thing but he had stops in the program and when I found out I told him he was out. I know I would lose one of the best programmer in the world but I needed to have somebody that I could trust and he didn't fit that bill. \r\n\r\nWhat the program can do is to pick an event before it will happen.This program would be worth many millions of dollars when finish. It's a self proven system that must work because it only uses only the Math Laws.The concept I invented myself.\r\n\r\nI have the lay out still, so it wouldn't be that hard to finish.Who ever can finish it may use it at the end to keep.\r\nThanks" } { "Tag": [ "IMO", "IMO Shortlist" ], "Problem": "IMO 1967 partial IMO LongList:", "Solution_1": "IMO LongList 1967 to 1977 problems and solutions." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let a,b are natural numbers for which a|(b+1) and b|(a^2 -2). Prove that (b+1)/2 is a square of natural number.\r\n\r\n :flex:", "Solution_1": "The first condition implies the existence of integer $ c$ such that $ b\\equal{}ac\\minus{}1,\\frac{b\\plus{}1}{2}\\equal{}\\frac{ac}{2}$. So the second condition implies that $ ac\\minus{}1$ divides $ a^2\\minus{}2$. Therefore, $ ac\\minus{}1$ also divides\r\n\\[ a(ac\\minus{}1)\\minus{}c(a^2\\minus{}2)\\equal{}2c\\minus{}a\\]\r\nso that $ 2c\\minus{}a\\equal{}d(ac\\minus{}1)$ for some integer $ d$. As $ a$ and $ b$ are positive, $ c$ is positive, and $ a\\equal{}c\\equal{}1$ does not satisfy the equation for any $ d$. So $ ac\\minus{}1$ positive. We then have\r\n\\[ d\\plus{}1\\equal{}\\frac{2c\\minus{}a}{ac\\minus{}1}\\plus{}1\\equal{}\\frac{ac\\minus{}a\\plus{}2c\\minus{}1}{ac\\minus{}1}\\equal{}\\frac{(a\\plus{}2)(c\\minus{}1)\\plus{}1}{ac\\minus{}1}>0\\]\r\nand (unless $ a\\equal{}1$)\r\n\\[ 1\\minus{}d\\equal{}1\\minus{}\\frac{2c\\minus{}a}{ac\\minus{}1}\\equal{}\\frac{ac\\plus{}a\\minus{}2c\\minus{}1}{ac\\minus{}1}\\equal{}\\frac{(a\\minus{}2)(c\\plus{}1)\\plus{}1}{ac\\minus{}1}>0\\]\r\nSo there are two cases: $ a\\equal{}1$ or $ d\\equal{}0$. In the first case, $ c\\minus{}1$ divides both $ 2c\\minus{}1$ and $ 2c\\minus{}2$, therefore 1, so that $ c\\equal{}2$ giving $ \\frac{ac}{2}\\equal{}1$, a square. In the second, $ a\\equal{}2c$ so $ \\frac{ac}{2}\\equal{}c^2$, as desired." } { "Tag": [], "Problem": "A guidance counselor is planning schedules for 30 students. Sixteen students say they want to take French, 16 want to take Spanish, and 11 want to take Latin. Five say they want to take both French and Latin, and of these, 3 wanted to take Spanish as well. Five want only Latin, and 8 want only Spanish. How many students want French only?\r\n1 day ago - 2 days left to answer.", "Solution_1": "Are you going to get in trouble for cheating? It appears as though you've copy-pasted this from some sort of online math course, and have forgotten to delete the thing saying how long you have.", "Solution_2": "yeah, I had the same thoughts from the \"1 day ago - 2 days left to answer\" How often are there deadlines for problems posted on AoPS? Some problems probably have been discussed for a very long time before the answer was final, but all of a sudden there's this", "Solution_3": "This isn't even a hard problem. Just use deductive reasoning.", "Solution_4": "You might find [url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]PIE[/url] useful in solving this type of problem." } { "Tag": [ "blogs", "ARML", "AMC", "USA(J)MO", "USAMO", "HMMT", "MATHCOUNTS" ], "Problem": "Does any AoPSers attend your school?", "Solution_1": "currently, I've gotten 2 people in my school district to join AoPS. 4 more are joining very soon. And once my math club gets rolling, my blog will be used to detail what goes on at each meeting and what to expect, so everyone will have to join.\r\n\r\nOf course, none of these people are active.... :sad:", "Solution_2": "Nope. Not a single one. Not even me.\r\nlol only one answer to the poll even makes sense for those who can answer it.", "Solution_3": "Come on, you know what he means :P. Or maybe some people would have to answer \"What's AoPS?\" :D", "Solution_4": "um...... me???? duh.... but no one else, and I am not that interested in bringing others either, since they dont want to do, or be interested in the sort of competitions that I like. I dont like my school's policy of prioritizing Mu Alpha Theta competition, rather than something like ARML or USAMO, or even some other ones like HMMT or BAMO (although we cant participate). I am beginning to restrain from letting my classmates know about the math i do out of school, because I know that they will soon be disinterested in the concept of Olympiad and hard math, and just be focused on Class subjects and et cetera, which I think in my school has a slow curriculum anyways. \r\n\r\nSo, no other kids from our school are AoPSer.", "Solution_5": "well, no one around at my school has an ounce of math talent (includes me) as only 1 person (or 2 or 3 max) got over 100 on the AMC's :( I don't tell anyone else about AoPS as they don't really care, but oh well... our math club isn't very constructive; we're more concerned with rearranging the keys on the keyboard than learning math :D", "Solution_6": "joml88 introduced me and some other members to AoPS. actually he tortured me until i joined. now our entire team is registered although most ppl only post in the blog, if ever", "Solution_7": "A couple people from my school are on AoPS, though it's not because I told them. I really don't do much to promote AoPS because I get the feeling that not very many people would appreciate what this community has to offer. However, our math team is pretty strong, so at least I'm not too alone.", "Solution_8": "I think I'm the only one from my school here, *shrugs* our math team is pretty good, but there aren't too many people who really do it because they like to do it, mostly it's because they have to or because they think colleges will like it *shrugs*", "Solution_9": "My sister \"colts18\" goes to my school.", "Solution_10": "I know some people that are on AoPS as users, but they hardly ever post on the forum. They only have an AoPS account for USAMTS.", "Solution_11": "sheepwarrior made me get an account -_-;;", "Solution_12": "I am a \"wanna be\". I just talk on the forum. I have never read the AoPs books, I don't kinow why. Another person at my school bought both AoPS books, and read both. We are #1 and #2 in our school in math.", "Solution_13": "None that I know of. It is a little too bad because there are alot of strong math kids in my school. Although I do know of a couple kids who bought the Aops books and seem to like them.", "Solution_14": "I'm a newbie here and our school has no math team or math club or anything math related.. except for math classes. Sucks.", "Solution_15": "You live in Calif and your school has no math team? Where in California do you live???? Certainly not the SFBA, right? Maybe LA or something...", "Solution_16": "I used to live in SFBA! When I was like 2! But now I live in southern California.. in Santa Ana.. in Orange County, yeah, around LA.", "Solution_17": "heh, i live in a town...so I'm the only one in the town somewhat interested in math. sad eh", "Solution_18": "yeah, well how big is your town? You live in Alberta...", "Solution_19": "[quote=\"bubala\"]yeah, well how big is your town? You live in Alberta...[/quote]\r\n\r\nMiddle of no where alberta actually. 15 000 people in my town, heh. I'm lucky though, I live incredibly close to the city that its not a problem living in Okotoks", "Solution_20": "aznness/uberl33tmage and jimli go to the same school as I do. There's also someone who lurks occasionally but doesn't have an account. There may be others, I wouldn't know.", "Solution_21": "[quote=\"ambierona\"]I used to live in SFBA! When I was like 2! But now I live in southern California.. in Santa Ana.. in Orange County, yeah, around LA.[/quote] Arent there still good math schools in SoCal like Mark Keppel?", "Solution_22": "zero", "Solution_23": "[quote=\"math92\"]My sister \"colts18\" goes to my school.[/quote]\r\nYes that is true.", "Solution_24": "Even though Edmonton has population of 1 million, I am the only AoPSer(I'm quite sure about this) :( . So, I assume no one at my school would be AoPSer....", "Solution_25": "I know two people in Aops in my school uberl33tmage and lynelle ye. We all were on a Mathcounts team. I have both Aops books but no independent study. ccy is also a person in my school but he's not really on our team or anything much.", "Solution_26": "[quote=\"furious\"]joml88 introduced me and some other members to AoPS. actually he tortured me until i joined. now our entire team is registered although most ppl only post in the blog, if ever[/quote]\r\n\r\nHehe, that just about says it for me....except I dont' remember \"torturing\" furious (to join AoPS that is... ;) )", "Solution_27": "i tortured colts18!!! :D", "Solution_28": "i think math92 should of put i don't know in there,too because i don't know, but i'm preety sure that there isn't anybody in my school", "Solution_29": "ast92391 goes to my school, but I think he only joined cause our mathcounts coach said we should go to AoPS. ineedmathhelp used to go to my school, and will go to my school next year.\r\n\r\n[quote=\"jb05\"]lol only one answer to the poll even makes sense for those who can answer it.[/quote]\r\n\r\nTechnically, a MathLinkser who never looks at the top right corner could answer option 3.", "Solution_30": "I have a strange feeling that a decent amount of kids know about AoPS in miami...", "Solution_31": "I'm the only one at my school who's any good at competitive math. :( There are many people who are good at \"school math\" as in Algebra II, or whatever they're in, but they do not know... [hide]The Art of Problem Solving![/hide] So I'm the only AoPSer at my school. There's another kid I'm trying to convince, but...", "Solution_32": "I'm the only AoPSer who goes to my school. I've told a couple people, but I don't think any of them signed up.", "Solution_33": "There's one other at my school. Of course, there aren't that many because we're mainly a humanities school...", "Solution_34": "[quote=\"mathbomberII\"]i think math92 should of put i don't know in there,too because i don't know, but i'm preety sure that there isn't anybody in my school[/quote]\r\ncolts18???she goes to my school", "Solution_35": "[quote=\"yif man12\"]There's one other at my school. Of course, there aren't that many because we're mainly a humanities school...[/quote]\r\nthat would be jsrosen right?\r\n\r\nIn our school, everyone in our math enrichment class has to sign up. But there are only a handful of people who actually want to be here", "Solution_36": "I'm probably the only one here. There is another kid that I think is really good at problem solving type of stuff (he got a 97 on the AMC 10 without even knowing what it was 15 minutes before he took it).\\, but I don't think he would want to join.", "Solution_37": "[quote=\"Scrambled\"][quote=\"yif man12\"]There's one other at my school. Of course, there aren't that many because we're mainly a humanities school...[/quote]\nthat would be jsrosen right?\n\nIn our school, everyone in our math enrichment class has to sign up. But there are only a handful of people who actually want to be here[/quote]\r\n\r\n :) Yup. Though he doesn't really post anymore. He comes and looks around once every month or so.", "Solution_38": "We only have two people in Shanghai American School. Me and sheepwarrior." } { "Tag": [ "symmetry", "ratio", "geometry", "3D geometry", "icosahedron", "group theory", "superior algebra" ], "Problem": "The symmetry group of a collection of points $ C$ in the plane is the set of orthogonal transformations in $ \\mathbb{R}^2$ that restrict to a bijection on $ C$. For example, the symmetry group of the vertices of a regular $ n$-gon is the dihedral group $ D_n$.\r\n\r\nWhat is the maximal size of the symmetry group of a collection of $ n$ points? (Can one do better than the dihedral group?)\r\n\r\nHow does the answer change in three dimensions? $ d$ dimensions?", "Solution_1": "In two dimensions, we can embed the symmetry group in the dihedral group on the convex hull.\r\n\r\nIn three dimensions, the largest ratio of symmetry group size to set size comes for the icosahedron, with 12 vertices and 120 elements in the group. We're limited by the low number of regular polyhedra, and for large $ n$ the best possible is about 4n: the dihedral group plus flips, achieved for a double pyramid.\r\n\r\nWe should require that the collection encloses a $ d$-dimensional volume; otherwise, we get large groups acting on the empty dimensions.", "Solution_2": "You might want to read [url=http://en.wikipedia.org/wiki/Coxeter_group]here[/url] for more information on finite symmetry groups." } { "Tag": [ "geometry", "3D geometry", "pyramid", "ratio" ], "Problem": "A pyramid with a square base is cut by a plane that is parallel to its base and is $ 2$ units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid?\r\n\r\n$ \\textbf{(A)}\\ 2\\qquad \r\n\\textbf{(B)}\\ 2 \\plus{} \\sqrt{2}\\qquad \r\n\\textbf{(C)}\\ 1 \\plus{} 2\\sqrt{2}\\qquad \r\n\\textbf{(D)}\\ 4\\qquad \r\n\\textbf{(E)}\\ 4 \\plus{} 2\\sqrt{2}$", "Solution_1": "[hide]Let $ h$ be the altitude of the original, and $ h \\minus{} 2$ altitude of the smaller pyramid. The smaller pyramid and the original one are similar with coefficient $ k \\equal{} \\frac {h \\minus{} 2}{h}$. Let surface area of the original be $ S_1$ and of the small one $ S_2$. Since all the areas are proportional with coefficient $ k^2 \\implies S_2 \\equal{} k^2S_1$. Let's substitute $ S_1$ using $ S_1 \\equal{} 2S_2$ identity. We get $ S_2 \\equal{} 2k^2S_2 \\implies k^2 \\equal{} \\frac {1}{2}$. Using $ k \\equal{} \\frac {h \\minus{} 2}{h}$, we get $ h \\equal{} 4 \\plus{} 2\\sqrt {2}$.\n\nThe answer is $ \\textbf{(E)}$.[/hide]", "Solution_2": "[hide=\"alternate solution\"]\nThe two pyramids are similar therefore if their suface area has a raito $ \\frac{1}{2}$, there legths have a ratio of $ \\frac{1}{\\sqrt{2}}$\n\n$ \\Rightarrow k\\minus{}\\frac{1}{\\sqrt{2}}k \\equal{} 2 \\Leftrightarrow k \\equal{} \\frac{2\\sqrt{2}}{\\sqrt{2}\\minus{}1}\\cdot \\frac{\\sqrt{2}\\plus{}1}{\\sqrt{2}\\plus{}1} \\equal{} 4\\plus{}2\\sqrt{2} \\longrightarrow \\boxed{E}$[/hide]" } { "Tag": [], "Problem": "I have an arithmetic sequence. The first term of the first sequence is $ 0$. The second term of the first sequence is the first term of the first sequence plus the first term of the second sequence. Similarly, the third term of the first sequence is the second term of the first sequence plus the second term of the second sequence. If the fifth term of the second sequence is $ 3$, what is the fifth term of the first sequence?", "Solution_1": "Suppose the first sequence went up by $ n$ - that is, the first term is 0, the second is $ n$, the third $ 2n$, and so on. Similarly, let the first term of the second sequence be $ k$, and let it go up by intervals of $ d$, so that the second term is $ k \\plus{} d$ and the third $ k \\plus{} 2d$.\r\n\r\nWe can write some equations: $ n \\equal{} 0 \\plus{} k$, from the second sentence, which implies that $ n \\equal{} k$.\r\n\r\nNow, the third sentence gives us $ 2n \\equal{} n \\plus{} k \\plus{} d$, or $ n \\equal{} k \\plus{} d$, meaning $ 0 \\equal{} d$. From the last sentence, we see that $ k \\equal{} 3$, so the second sequence looks like 3, 3, 3, ... \r\n\r\nSince $ n \\equal{} k$, that means $ n \\equal{} 3$, so the fifth term of the first sequence is $ 4 \\times 3 \\equal{} \\boxed{12}$." } { "Tag": [ "trigonometry" ], "Problem": "Here is an interesting but easy challege. \r\n\r\nPlease do without using calculator.\r\n\r\n$\\sin (\\cos^{-1} \\frac {\\sqrt 5}{5}) = \\frac {a \\cdot \\sqrt b}{c}$\r\n\r\nWhere a,b, and c don't necessarily represent the different numbers.\r\n\r\nFind $a+b+c$.", "Solution_1": "[hide]I did it by drawing a triangle with side lengths $\\sqrt 5$, $2\\sqrt5$, and 5...\n$ \\sin (\\cos^{-1} \\frac {\\sqrt 5}{5}) = \\frac {2 \\cdot \\sqrt 5}{5}\\\\\na+b+c = 2 + 5 + 5 = 12$[/hide]", "Solution_2": "The problem should have imposed some restrictions on a, b, and c, such as they are all integers, b is square-free, and a and c are relatively prime.", "Solution_3": "[quote=\"Ravi B\"]The problem should have imposed some restrictions on a, b, and c, such as they are all integers, b is square-free, and a and c are relatively prime.[/quote]\r\n\r\nNot to mention the sign .. as there can be two answers.", "Solution_4": "it is implied that $\\cos^{-1} x$ is the principal value...\r\non another note, verify $\\tan^2 x+1=\\sec^2 x$.", "Solution_5": "[quote=\"chenwb\"]on another note, verify $\\tan^2 x+1=\\sec^2 x$.[/quote]\r\n\r\nMultiplying through by $\\cos^2{x}$ gives the well-known trig identity $\\sin^2{x}+\\cos^2{x}=1$.", "Solution_6": "[hide][quote=\"Silverfalcon\"]Here is an interesting but easy challege. \n\nPlease do without using calculator.\n\n$\\sin (\\cos^{-1} \\frac {\\sqrt 5}{5}) = \\frac {a \\cdot \\sqrt b}{c}$\n\nWhere a,b, and c don't necessarily represent the different numbers.\n\nFind $a+b+c$.[/quote]\n\n$\\frac{\\sqrt 5}{5}=\\frac{1}{\\sqrt{5}}$\n\nthe hypotenus is $\\sqrt 5$\n\nthe other side is $\\sqrt{5-1}=2$\n\nso $\\sin \\theta=\\frac{2}{\\sqrt 5}=\\frac{2\\cdot \\sqrt 5}{5}$\n\n$2+5+5=12$[/hide]", "Solution_7": "$\\tan^2 x+1=\\sec^2 x$\r\n\r\n$\\frac{opposite^2}{adjacent^2}+\\frac{adjacent^2}{adjacent^2}=\\frac{hypotenus^2}{adjacent^2}$\r\n\r\n$\\frac{opposite^2+adjacent^2}{adjacent^2}=\\frac{hypotenus^2}{adjacent^2}$\r\n\r\n$\\frac{hypotenus^2}{adjacent^2}=\\frac{hypotenus^2}{adjacent^2}$\r\n\r\nTADA!" } { "Tag": [], "Problem": "Find all $ n \\in \\mathbb{N}$ such that :\r\n$ n2^{n\\minus{}1} \\plus{} 1$ is a perfect square.", "Solution_1": "[hide]Let the square be $ x^2$. Then $ x^2\\minus{}1\\equal{}(x\\plus{}1)(x\\minus{}1)\\equal{}n(2^{n\\minus{}1})$. But $ x\\plus{}1$ and $ x\\minus{}1$ are different mod 4, so one of them must not be divisible by 4, so one of them is divisible by $ 2^{n\\minus{}2}$. Thus x is at least $ 2^{n\\minus{}2}\\minus{}1$, so $ x^2\\minus{}1\\geq(x\\plus{}1)(x\\minus{}1)\\equal{}2^{2n\\minus{}4}\\minus{}2^{n\\minus{}1}$, so $ 2^{n\\minus{}3}\\minus{}1\\leq{n}$. Thus n<6, so we just have to try n=1, 2, 3, 4, 5. n=1 yields 2, not a square. n=2 yields 5, also not a square. n=3 yields 13, also a nonsquare. n=4 yields 33, a nonsquare, and n=5 yields 81, which is a square. Thus the only solution is n=5.[/hide]" } { "Tag": [ "function", "integration", "calculus", "real analysis", "real analysis unsolved" ], "Problem": "I got this problem on an exam today: let $f : (a,b) \\to \\mathbb{R}$ be a function, with $a,b \\in \\mathbb{R} \\cup \\{\\pm \\infty\\}$, which is derivable and convex on $(a,b)$. Prove that it has limits in $a$ and $b$. \r\n\r\n[hide=\"Hint\"]Use the fact that $f'$ exists and is non-decreasing, thus now allowing random 0-s to appear.[/hide]", "Solution_1": "Really quite straightforward. Pick $a_n\\rightarrow a$. Then we have $f(a_n)=f(x)-\\int_{a_n}^xf'(t)dt$. Since the integrals converge ($f'$ is non-decreasing) the left-hand side converges too. The same argument works for $b$.\r\n\r\nMisha" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $ A \\equal{}[a_{i,j}]_{nxn}$ satisfying the following condition:\r\n \r\n $ a_{i,j}\\equal{}0$ if $ i\\equal{}j$ \r\n and \r\n $ a_{i,j}\\equal{}1$ or $ p$ if $ i$#$ j$. where $ p$ is a prime number $ >2$.\r\n Prove that : $ rank(A) \\geq {n\\minus{}1}$", "Solution_1": "You don't need $ p$ to be prime.\r\n$ A\\plus{}I_n$ or $ A\\plus{}pI_n$ has rank 1, hence $ \\minus{}1$ or $ \\minus{}p$ is an eignevalue with geometric multiplicity $ n\\minus{}1$, as a result $ 0$ is at most an eigenvalue with geometric multiplicity $ 1$, hence $ \\mathrm{rank} A\\geqslant n\\minus{}1$.", "Solution_2": "JC_math, why \"$ A \\plus{} I_n$ or $ A \\plus{} pI_n$ has rank 1\"? The matrix\r\n\r\n$ A \\equal{} \\begin{bmatrix} 0 & 3 & 1 \\\\\r\n1 & 0 & 3 \\\\\r\n3 & 3 & 0 \\\\\r\n\\end{bmatrix}$\r\n\r\nis in the conditions of the problem and $ rank(A \\plus{} I) \\equal{} rank(A \\plus{} 3I) \\equal{} 3 \\equal{} rank(A)$.", "Solution_3": "Sorry but I read this\r\n[quote=\"minhthutt\"] $ a_{i,j} \\equal{} 1$ or $ p$ if $ i$#$ j$. where $ p$ is a prime number $ > 2$.[/quote]\r\ndiiferently, i.e. all the $ a_{i,j}$ for $ i\\neq j$ are either equal to 1 or $ p$.\r\n\r\nHere is another approach.\r\nConsider $ A$ as an element of $ \\mathcal{M}_n(\\mathbb{Z})$, then reduce in $ \\mathbb{Z}/(p\\plus{}1)\\mathbb{Z}$, you have $ \\mathrm{rank}_{\\mathbb{Z}/(p\\plus{}1)\\mathbb{Z}} (A) \\equal{} n\\minus{}1$, hence $ \\mathrm{rank}_{\\mathbb{Z}} (A) \\geqslant n\\minus{}1$,...", "Solution_4": "I have different version of solution, please check whether my answer is true (or not)\r\n[hide=\"answer\"]\nLook the matrix over $ Z_2$ (So that if the number is even we give value $ 0$ and if the number is odd we take the value $ 1$), then the matrix will be : $ A\\equal{}{a}_{ij}$ where $ {a}_{ij}\\equal{}0$ for all $ i\\equal{}j$ and $ {a}_{ij}\\equal{}1$ if $ i$ isn't equal to $ j$ (By noting that all prime that is greater than $ 2$ is an odd number.\nBy some elementary row operations we can obtain that for the first $ n\\minus{}1$ columns in that matrix we have that $ {a}_{ij}\\equal{}1$ if $ i\\equal{}j$ or $ i\\equal{}j\\plus{}1$ and $ 0$ in the other place (in the first $ n\\minus{}1$ columns in $ A$).Check that all of the $ n\\minus{}1$ rows is linearry independent so $ rank({A}_{{Z}_{2}}) \\geq n\\minus{}1$ and $ rank(A) \\geq rank({A}_{{Z}_{2}})$ so that $ rank(A) \\geq n\\minus{}1$. So you don't need $ p$ to be prime number greater than $ 2$, you just need $ p$ to be an odd number.\nI am sorry if my answer is wrong.\n[/hide]" } { "Tag": [ "ratio", "geometry", "3D geometry", "prism" ], "Problem": "Television screen sizes are the diagonal length of the rectangular screen. Hector recently changed from watching television witha 13-inch screen to a television with a similar 19-inch screen. If a boxcar appeared 8 inches long on the 13-inch screen how long, to the nearest inch, will it appear on the 19-inch screen?\r\n\r\nsource: 2005-06 Preparing for the ACT booklet.", "Solution_1": "[quote=\"nonie\"]Television screen sizes are the diagonal length of the rectangular screen. Hector recently changed from watching television witha 13-inch screen to a television with a similar 19-inch screen. If a boxcar appeared 8 inches long on the 13-inch screen how long, to the nearest inch, will it appear on the 19-inch screen?\n\nsource: 2005-06 Preparing for the ACT booklet.[/quote]\r\n\r\n[hide]Uh...\nNOt sure if i misunderstood this problem....\n$\\frac{13}{8}=\\frac{19}{x}$\n$x\\approx 12$[/hide]", "Solution_2": "[quote=\"ch1n353ch3s54a1l\"][quote=\"nonie\"]Television screen sizes are the diagonal length of the rectangular screen. Hector recently changed from watching television witha 13-inch screen to a television with a similar 19-inch screen. If a boxcar appeared 8 inches long on the 13-inch screen how long, to the nearest inch, will it appear on the 19-inch screen?\n\nsource: 2005-06 Preparing for the ACT booklet.[/quote]\n\n[hide]Uh...\nNOt sure if i misunderstood this problem....\n$\\frac{13}{8}=\\frac{19}{x}$\n$x\\approx 12$[/hide][/quote]\r\n\r\nNo , you're right. :)", "Solution_3": "it is 12 inches\r\n\r\nyou need a proportion . 8/13= x/19\r\ncross multiply\r\n19x8= 152\r\nthen divided 152/13= 11.69 round to 12 :D", "Solution_4": "Yep, they've covered it above. Also, some notes on similar figures:\r\n-You can use any common linear measure to set up a proportion (like we did in this problem)\r\n-You can use the square of any common linear measure to relate areas. So if you were asked the ratio of the areas of the two screens, it would be (19/13)^2. \r\n-You can use the cube of any common linear measure to relate volume (if you have similar prisms, for instance).\r\n\r\nThose can be useful and can save a lot of work." } { "Tag": [ "college", "calculus", "calculus computations" ], "Problem": "Hi I was wondering if there are any good books out there with a collection of differential equations problems which include ODEs and PDEs. Currently, the only problem book that I have which have some differential equations problems is the \"Berkeley Problems in Mathematics\". I was wondering if there are any other books that people here can recommend. Thanks.", "Solution_1": "Do you want exclusively a problem book devoted to differential equations? I'm not sure one of those exists. If you pick up a differential equations textbook, chances are, you'll get far more problems that just a problem book.", "Solution_2": "my class is currently using Partial Differential Equations by Evans and its a pretty reader friendly book. although i cant compare it to other books of its kind yet, i like it." } { "Tag": [ "algebra", "polynomial", "linear algebra", "matrix" ], "Problem": "Have anyone know if there is in Linear Algebra a theorem like this:\r\n\"Two matrices are similar [u]if and only if[/u] have the same minimal & charecteristic polynomials\"\r\nI need this theorem (if exists) in order to solve an exercise.\r\nIf exists a such theorem could someone post a link to its proof??\r\n\r\nThanks in advance", "Solution_1": "This doesn't hold. What I posted in http://www.mathlinks.ro/viewtopic.php?t=175802 may be the best result of such kind which holds over arbitrary fields. I think some other topic gives a detailed proof for the algebraically closed field case, but I can't find it.\r\n\r\n darij", "Solution_2": "It does hold for $ 3\\times 3$ matrices, if it can help you, but for $ n\\times n$, $ n\\geq 4$, no. (Just like Darij said, and shown in the linked thread)", "Solution_3": "[quote=\"hsiljak\"]It does hold for $ 3\\times 3$ matrices, if it can help you, but for $ n\\times n$, $ n\\geq 4$, no. (Just like Darij said, and shown in the linked thread)[/quote]\r\n\r\nMy case is 2x2 matrices. does it hold?? and Why??\r\n\r\nThank you", "Solution_4": "Yes, it does hold. I was always bad at algebra (mental note: study algebra this summer), but I think you could prove it on Jordan blocks... Correct me if I'm wrong :)", "Solution_5": "The $ 2 \\times 2$ case is degenerate because there are only two possible Jordan block decompositions: one block or two blocks. In the one-block case, the characteristic and minimal polynomials are both $ (x \\minus{} \\lambda)^2$. In the two-block case, the characteristic and minimal polynomials coincide if eigenvalues are distinct and the minimal polynomial is of degree $ 1$ if they aren't. That is, the map from Jordan blocks to characteristic and minimal polynomials is injective, which isn't true in general.\r\n\r\nFor what it's worth, it's pretty easy to show that a statement of this form can't hold for sufficiently large $ n$: the number of Jordan block decompositions for an $ n \\times n$ matrix with a single eigenvalue of multiplicity $ n$ is $ B_n$, which grows faster than exponentially in $ n$, whereas knowing characteristic and minimal polynomials only gets you polynomial information in $ n$." } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "Why is a compact operator automatically bounded?", "Solution_1": "Because the image of the unit ball is contained in a compact set and compact sets are always bounded.", "Solution_2": "and you should also ask yourself why it is not invertible when working with infinite dimensional spaces .." } { "Tag": [ "trigonometry", "algebra proposed", "algebra" ], "Problem": "Prove that :\r\n\r\n$ \\cos^7\\ x \\plus{} \\cos^7\\ \\left(x \\plus{} \\frac {2\\pi}{3}\\right) \\plus{} \\cos^7 \\ \\left(x \\plus{} \\frac {4\\pi}{3}\\right) \\equal{} \\frac {63}{64}\\cos \\ 3x\\ , \\ (\\forall)x\\in \\mathbb{R}$", "Solution_1": "hello, i think it must be\r\n$ \\cos^7(x) \\plus{} \\cos^7(x \\plus{} \\frac {2\\pi}{3}) \\plus{} \\cos^7(x \\plus{} \\frac {4\\pi}{3}) \\equal{} \\frac {63}{64}\\cos(3x)$\r\nSonnhard.", "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]hello, i think it must be\n$ \\cos^7(x) \\plus{} \\cos^7(x \\plus{} \\frac {2\\pi}{3}) \\plus{} \\cos^7(x \\plus{} \\frac {4\\pi}{3}) \\equal{} \\frac {63}{64}\\cos(3x)$\nSonnhard.[/quote]\r\n\r\nYou're right . It was an idiotic typo of mine .", "Solution_3": "$ a\\equal{}cos(x),b\\equal{}cos(x\\plus{}\\frac{2\\pi}{3}),c\\equal{}cos(x\\plus{}\\frac{4\\pi}{3})$ are roots of $ f(y)\\equal{}4y^3\\minus{}3y\\minus{}cos(3x)\\equal{}0$.\r\nTherefore $ P_m\\equal{}a^m\\plus{}b^m\\plus{}c^m\\equal{}cos^mx\\plus{}cos^m(x\\plus{}\\frac{2\\pi}{3})\\plus{}cos^m(x\\plus{}\\frac{4\\pi}{3})$ satisfyed \r\n\\[ P_{m\\plus{}3}\\minus{}3P_{m\\plus{}1}\\minus{}cos(3x)P_m\\equiv 0.\\]\r\nWe had $ P_0\\equal{}3,P_1\\equal{}0,P_2\\equal{}(a\\plus{}b\\plus{}c)^2\\minus{}2(ab\\plus{}bc\\plus{}ca)\\equal{}0\\minus{}2(\\minus{}3/4)\\equal{}3/2.$\r\nTherefore $ P_7\\equal{}\\frac 14(3 P_5\\plus{}cos(3x)P_4)\\equal{}\\frac{1}{16}(9P_3\\plus{}6cos(3x)P_2\\plus{}cos^2(3x)P_1)$ or\r\n$ P_7\\equal{}\\frac{9}{16}(P_3\\plus{}cos(3x))\\equal{}\\frac{9}{16}(\\frac 34 \\plus{}1)cos(3x)\\equal{}\\frac{63}{64}cos(3x)$." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let x, y, and z be positive real numbers with $ x \\plus{} y \\plus{} z \\equal{} 1$. Prove that\r\n\\[ \\frac{xy}{\\sqrt{xy\\plus{}yz}} \\plus{} \\frac{yz}{\\sqrt{yz\\plus{}zx}} \\plus{} \\frac{zx}{\\sqrt{zx\\plus{}xy}} \\leq \\frac{1}{\\sqrt{2}}.\\]", "Solution_1": "$ \\sum \\frac{xy}{\\sqrt{xy+yz}} = \\sum x \\sqrt{ \\frac{y}{1-y}} \\le \\sqrt{\\frac{\\sum xy}{1-\\sum xy}}$ (Jensen's on $ f(x) = \\sqrt{\\frac{y}{1-y}}$ with x,y,z as weights)\r\n\r\n$ (\\sum x)^2 \\ge 3 \\sum xy \\Rightarrow \\sum \\xy \\le \\frac{1}{3}$\r\n\r\nHence $ \\sqrt{\\frac{\\sum xy}{1-\\sum xy}} = \\sqrt{\\frac{1}{1-\\sum xy}-1} \\le \\frac{1}{\\sqrt 2}$", "Solution_2": "For $ f(x) \\equal{} \\sqrt{\\frac{x}{1\\minus{}x}}$\r\n$ f''(x) \\equal{} \\frac{4x\\minus{}1}{4x^{3/2}(1\\minus{}x)^{5/2}}$ is not concave on $ (0,1)$. Therefore you cannot apply Jensen's inequality.", "Solution_3": "[quote=\"MayankM\"]Let x, y, and z be positive real numbers with $ x \\plus{} y \\plus{} z \\equal{} 1$. Prove that\n\\[ \\frac {xy}{\\sqrt {xy \\plus{} yz}} \\plus{} \\frac {yz}{\\sqrt {yz \\plus{} zx}} \\plus{} \\frac {zx}{\\sqrt {zx \\plus{} xy}} \\leq \\frac {1}{\\sqrt {2}}.\\]\n[/quote]\r\n\r\n$ \\frac {xy}{\\sqrt {xy \\plus{} yz}} \\plus{} \\frac {yz}{\\sqrt {yz \\plus{} zx}} \\plus{} \\frac {zx}{\\sqrt {zx \\plus{} xy}}$\r\n\r\n$ \\le \\frac {\\sqrt{2}xy}{\\sqrt {xy} \\plus{} \\sqrt{yz}} \\plus{} \\frac {\\sqrt{2}yz}{\\sqrt {yz} \\plus{} \\sqrt{zx}} \\plus{} \\frac {\\sqrt{2}zx}{\\sqrt {zx} \\plus{} \\sqrt{xy}}$\r\n\r\n$ \\equal{} \\frac {xy\\plus{}yz}{\\sqrt{2}(\\sqrt {xy} \\plus{} \\sqrt{yz})} \\plus{} \\frac {yz\\plus{}zx}{\\sqrt{2}(\\sqrt {yz} \\plus{} \\sqrt{zx})} \\plus{} \\frac {zx\\plus{}xy}{\\sqrt{2}(\\sqrt {zx} \\plus{} \\sqrt{xy})},$\r\n\r\n\\[ \\sum {\\frac{{xy \\plus{} yz}}{{\\sqrt {xy} \\plus{} \\sqrt {yz} }}} \\le x \\plus{} y \\plus{} z\\]\r\n\\[ \\Leftrightarrow \\sum {\\left( {\\sqrt x \\minus{} 2\\sqrt y \\plus{} \\sqrt z } \\right)\\left( {\\frac{{z \\plus{} x}}{{\\sqrt z \\plus{} \\sqrt x }}} \\right)} \\ge 0\\]\r\n\\[ \\Leftrightarrow \\sum {\\left( {\\sqrt x \\minus{} \\sqrt y } \\right)\\left( {\\frac{{z \\plus{} x}}{{\\sqrt z \\plus{} \\sqrt x }} \\minus{} \\frac{{y \\plus{} z}}{{\\sqrt y \\plus{} \\sqrt z }}} \\right)} \\ge 0\\]\r\n\\[ \\Leftrightarrow \\sum {\\frac{{\\left( {\\sqrt x \\minus{} \\sqrt y } \\right)^2 \\sqrt {xy} }}{{\\left( {\\sqrt z \\plus{} \\sqrt x } \\right)\\left( {\\sqrt y \\plus{} \\sqrt z } \\right)}}} \\ge 0.\\]", "Solution_4": "i was kinda betting on f being concave :lol:", "Solution_5": "Kuing I'm unable to follow this step.\r\n\\[ \\frac{\\sqrt{2}xy}{\\sqrt{xy}\\plus{}\\sqrt{yz}}\\plus{}\\frac{\\sqrt{2}yz}{\\sqrt{yz}\\plus{}\\sqrt{zx}}\\plus{}\\frac{\\sqrt{2}zx}{\\sqrt{zx}\\plus{}\\sqrt{xy}} \r\n\\\\\r\n \\equal{}\\frac{xy\\plus{}yz}{\\sqrt{2}(\\sqrt{xy}\\plus{}\\sqrt{yz})}\\plus{}\\frac{yz\\plus{}zx}{\\sqrt{2}(\\sqrt{yz}\\plus{}\\sqrt{zx})}\\plus{}\\frac{zx\\plus{}xy}{\\sqrt{2}(\\sqrt{zx}\\plus{}\\sqrt{xy})},\\]", "Solution_6": "[quote=\"MayankM\"]Kuing I'm unable to follow this step.\n\\[ \\frac {\\sqrt {2}xy}{\\sqrt {xy} \\plus{} \\sqrt {yz}} \\plus{} \\frac {\\sqrt {2}yz}{\\sqrt {yz} \\plus{} \\sqrt {zx}} \\plus{} \\frac {\\sqrt {2}zx}{\\sqrt {zx} \\plus{} \\sqrt {xy}} \\\\\n\\equal{} \\frac {xy \\plus{} yz}{\\sqrt {2}(\\sqrt {xy} \\plus{} \\sqrt {yz})} \\plus{} \\frac {yz \\plus{} zx}{\\sqrt {2}(\\sqrt {yz} \\plus{} \\sqrt {zx})} \\plus{} \\frac {zx \\plus{} xy}{\\sqrt {2}(\\sqrt {zx} \\plus{} \\sqrt {xy})},\\]\n[/quote]\r\n\r\n\\[ A \\equal{} \\sum {\\frac{{xy}}{{\\sqrt {xy} \\plus{} \\sqrt {yz} }}} ,B \\equal{} \\sum {\\frac{{yz}}{{\\sqrt {xy} \\plus{} \\sqrt {yz} }}}\\]\r\n\\[ A \\minus{} B \\equal{} \\sum {\\frac{{xy \\minus{} yz}}{{\\sqrt {xy} \\plus{} \\sqrt {yz} }}} \\equal{} \\sum {\\frac{{\\left( {\\sqrt {xy} \\plus{} \\sqrt {yz} } \\right)\\left( {\\sqrt {xy} \\minus{} \\sqrt {yz} } \\right)}}{{\\sqrt {xy} \\plus{} \\sqrt {yz} }}} \\equal{} \\sum {\\left( {\\sqrt {xy} \\minus{} \\sqrt {yz} } \\right)} \\equal{} 0\\]\r\n\\[ A \\equal{} B\\]" } { "Tag": [ "inequalities", "analytic geometry", "graphing lines", "slope", "absolute value" ], "Problem": "Alright, there's another problem that just stumped me this time.\r\nI'm solving for $X$.\r\n\r\n$|x + 1/3| = |x - 3|$\r\n$-x...............-x$\r\n\r\n$1/3 = -3$\r\n\r\n:| That's not right....\r\n\r\nThe answer to the problem is $X=4/3$ but I don't see how that can be accomplished.", "Solution_1": "I got it solved, nevermind. I'd delete the thread if I could.", "Solution_2": "Are you sure? Why not post your solution?", "Solution_3": "Hehe, your subtracting x from both sides is pretty much just ignoring the absolute values all together. Why's the title have \"inequalities\" tho?", "Solution_4": "If one part of the absolution process does not work, that means there might be only one possible answer left for X.\r\nThere isn't always two answers you can get.\r\nEither they can share the same answer, or there is only one answer for both and [b]not[/b] two answers by doing the problem [b]two different ways. [/b]\r\n\r\n$|x + \\frac{1}{3}| = |x - 3|$\r\n\r\nSO:\r\n\r\n$-x - \\frac{1}{3} = x - 3$\r\n\r\n$+x$\r\n\r\n$-\\frac{1}{3} = 2x - 3$\r\n\r\n$+3$\r\n\r\n$\\frac{8}{3} = 2x$\r\n\r\n$8/3/2 = \\frac{2x}{2} = x = \\frac{4}{3}$\r\n\r\n$x = \\frac{4}{3}$", "Solution_5": "Another way is by squaring both side ,\r\n\r\n$x^2+\\frac{2}{3}x+\\frac 19 =x^2-6x+9$ \r\n\r\nSimplifying and get $x=\\frac 43$ :D", "Solution_6": "Also, you can think of it graphically. Absolute value graphs are just where the line is \"reflected\" at the x-axis, cuz the y-value can't go below 0. They still keep their slope, so theres only one intersection of your two." } { "Tag": [ "geometry" ], "Problem": "So much have been mentioned about the application of fractal geometry in physics. Anyone know if there are any applications of fractal geometry in economics? :oops:", "Solution_1": "[url=http://classes.yale.edu/fractals/RandFrac/welcome.html]Random Fractals and the Stock Market[/url]\r\nYou can find more stuff like that with google. I'm not a big fan of it.", "Solution_2": "Application of fractal geometry in finance is well known.\r\nfor example, paths of Brownian Motion are fractals.\r\nHow about the other parts of economics apart from finance?\r\nDo we have applications of fractal geometry in microeconomics or industrial economics? :roll:" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find all the real solutions $(x,y,z)$ of the system:\r\n\r\n$xyz = 8$\r\n\r\n$x^2y+y^2z+z^2x = 73$\r\n\r\n$x(y-z)^2+y(z-x)^2+z(x-y)^2 = 98$", "Solution_1": "from the third equation , we have $xy^2+yz^2+zx^2 =73$\r\nthat implies $x^2y+y^2z+z^2x=xy^2+yz^2+zx^2$\r\nthus $(x-y)(y-z)(z-x)=0$\r\nand then it is easy to find roots" } { "Tag": [ "trigonometry", "calculus", "calculus computations" ], "Problem": "how would you sketch a phase portrait of\r\n\r\n$ x''\\equal{}\\cos x$?\r\n\r\nThank you.", "Solution_1": "In Maple\r\n\r\n[img]http://img508.imageshack.us/img508/9675/clipboard01hg0.jpg[/img]\r\n\r\n :roll:" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that for every $n \\geq 2$, there are $n$ positive integers $a_1,a_2,...,a_n$ such that\r\n\\[ \\frac{1}{a_{1}}+\\frac{1}{a_{2}}+...+\\frac{1}{a_{n}} = 1 + \\frac{1}{a_{1}a_{2}...a_{n}}. \\]", "Solution_1": "It is obviosly $a_1=2;a_n=\\prod_{i=1}^{n-1} a_i +1$.", "Solution_2": "Obviously it isn't :D", "Solution_3": "[quote=\"Rust\"]It is obviosly $a_1=2;a_n=\\prod_{i=1}^{n-1} a_i +1$.[/quote]\r\nYou are wrong:\r\n$\\frac{1}{2}+\\frac{1}{3}=\\frac{5}{6}\\not=\\frac{7}{6}=1+\\frac{1}{6}$.", "Solution_4": "[hide]\nDefine $x_k$ by $x_1=2$ and $x_{k+1}=x_k^2-x_k+1$.\n\nLemma 1: $\\sum_{j=1}^{n-1} \\frac{1}{x_j}=1-\\frac{1}{x_n-1}$ for $n\\geq 2$.\n\nProof: The base case checks because $\\frac 12 =1-\\frac 12 =1-\\frac{1}{3-1}$ and $x_2=3$.\n\nFor the inductive step, $\\sum_{j=1}^n \\frac{1}{x_j}=1-\\frac{1}{x_n-1}+\\frac{1}{x_n}=1-\\frac{1}{x_n(x_n-1)}=\\frac{1}{x_{n+1}-1}$, so the identity holds.\n\nLemma 2: $x_{k+1}=1+\\prod_{j=1}^k x_j$.\n\nProof: Again, clearly visible for $k=2$. The inductive step works because $x_{k+2}=1+x_{k+1}(x_{k+1}-1)=1+x_{k+1}\\prod_{j=1}^k x_j=1+\\prod_{j=1}^{k+1} x_j$.\n\nLet $a_j=x_j$ for $j\\leq n-1$. Applying Lemma 1, $\\sum_{j=1}^n \\frac{1}{a_j}=1-\\frac{1}{x_n-1}+\\frac{1}{a_n}$. The right side of the original equation equals $1+\\frac{1}{(x_n-1)a_n}$ from Lemma 2.\n\nSubtracting off 1 and multiplying to clear fractions, $x_n-1-a_n=1\\Rightarrow a_n=x_n-2$.\n\nTherefore the n-tuple $(x_1,x_2,\\cdots ,x_{n-1},x_n-2)$ is a solution for any $n\\geq 2$.\n[/hide]", "Solution_5": "Thank you, nice solution. I think that Rust's construction almost works after all since you can take $(a_1,a_2,...a_{n-1},a_n-2)$, which have the desired property.", "Solution_6": "There are only one solution (obviosly):\r\n$a_1=2,a_k=\\prod_{i=1}^{k-1}a_i +1,1=0}\r\n\r\nThe limits of the integral of dy is already clear (0=3$ (I'm using the well known and easy to prove fact that the only positive integer solution $(a,b),a>1$ to $3^a=2^b+1$ is $(a,b)=(2,3)$). Then take $n_3=n_2p_3$. Just like above, we deduce that $2^{p_3}+1$ has a prime factor $p_4$ which is coprime to $n_3$, and take $n_4=n_3p_4$, and so on. $n_k$ will have exactly $k$ prime factors and will satisfy $n_k|2^{n_k}+1$.", "Solution_2": "[quote=\"Valentin Vornicu\"]Does there exist a positive integer $n$ such that $n$ has exactly 2000 prime divisors and $n$ divides $2^n + 1$?[/quote]\r\n\r\nYou can find my paper for a general problem of this problem in [url=http://www.math.ust.hk/excalibur/v9_n3.pdf]This file[/url]", "Solution_3": "Let $N=2^n+1$. We will assume for the sake of contradiction that $n|N$.\r\n\r\n$2^n+1 \\equiv 0$ (mod $n$) $\\Rightarrow 2^n \\equiv -1$ (mod $n$). So 2 does not divide $n$, and so $n$ is odd.\r\n\r\nSelect an arbitrary prime factor of $n$ and call it $p$. Let's represent $n$ in the form $p^am$, where $m$ is not divisible by $p$.\r\n\r\nNote that $p$ and $m$ are both odd since $n$ is odd. By repeated applications of Fermat's Little Theorem:\r\n\r\n$N = 2^n+1 = 2^{p^am} + 1 = (2^{p^{a-1}m})^p + 1 \\equiv 2^{p^{a-1}m} + 1$ (mod $p$)\r\n\r\nContinuing in this manner, and inducting on k from 1 to $a$,\r\n\r\n$2^{p^{a-k}m}+1 \\equiv (2^{p^{a-k-1}m})^p + 1$ (mod $p$) $\\equiv 2^{p^{a-k-1}m} + 1$ (mod $p$)\r\n\r\nSo we have $N \\equiv 2^m+1$ (mod $p$)\r\n\r\nSince $p$ is relatively prime to $m$, $N \\equiv 1+1$ (mod $p$) $\\equiv 2$ (mod $p$)\r\n\r\nSince $p$ is odd, $N$ is not divisible by $p$. Hence $N$ is not divisible by $n$. So we have a contradiction, and our original assumption was false, and therefore $N$ is still not divisible by $n$.", "Solution_4": "Hmmm... I made a mistake here somewhere but I do not see it.", "Solution_5": "Yes, since there are lots of integers $n$ such that $n$ divides $2^n + 1$, and the statement is true!", "Solution_6": "There exists a pretty beautiful generalization:\r\n\r\n\"\r\nLet $s, a, b$ positive integers, such that $GCD(a,b) = 1$ and $a+b$ is not a 2-power.\r\nShow that there exists infinitely many $n \\in N$ such that\r\n\r\n--- $n=p_1^{e_1} \\cdot p_2^{e_2} \\cdot p_3^{e_3} \\cdots p_s^{e_s} \\cdot$ is the canonical factoring of $n$.\r\n \r\n--- $n|(a^n+b^n)$\r\n\"", "Solution_7": "[quote=\"Valentin Vornicu\"]Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n \\plus{} 1$?[/quote]\r\nHere is a solution that I don't think has been mentioned yet:\r\n[hide=\"Solution\"]\nWe proceed with induction. Clearly, $ n\\equal{}9$ works for the base case. Now, we will show that we can repeat an algorithm so that we can find a prime factor, $ p$, that is a factor of $ 2^n\\plus{}1$, but not of $ n$. Once this is done, we simply have that $ np|(2^n\\plus{}1)|(2^{np}\\plus{}1)$ and $ np$ has one more prime divisor than before. We use induction to find this algorithm. For the base case ($ n\\equal{}9$), we can find $ 19$ is a factor of $ 2^9\\plus{}1$, so the base case is proven. Now, let us consider $ 2^{np}\\plus{}1$, where $ p$ is a prime divisor if $ 2^n\\plus{}1$ and not of $ n$ (which is the inductive hypothesis.) $ 2^{np}\\plus{}1\\equal{}(2^n\\plus{}1)(2^{(p\\minus{}1)n}\\minus{}2^{(p\\minus{}2)n}\\plus{}\\cdots \\plus{}1)$. Set $ 2^n\\equal{}x$. Then, this is $ 2^{np}\\plus{}1\\equal{}(x\\plus{}1)(x^{p\\minus{}1}\\minus{}x^{p\\minus{}2}\\plus{}\\cdots \\plus{}1)$. We compute the $ \\gcd$ of both factored terms by computing the remainder of the second when divided by the first. Plug $ x\\equal{}\\minus{}1$ in, so we get that the remainder is $ p$, meaning that $ p$ divides the $ \\gcd$ of $ x\\plus{}1$ and $ x^{p\\minus{}1}\\minus{}x^{p\\minus{}2}\\plus{}\\cdots \\plus{}1$. In fact, this $ \\gcd$ is $ p$ since $ p|(x\\plus{}1)$. Then, $ np^2|(x\\plus{}1)(x^{p\\minus{}1}\\minus{}x^{p\\minus{}2}\\plus{}\\cdots \\plus{}1)\\equal{}x^p\\plus{}1$. This means that $ np^2$ divides $ 2^{np}\\plus{}1$, so $ np^2$ divides $ 2^{np^2}\\plus{}1$. We do this again, but now we set $ y\\equal{}2^{np}$, so $ 2^{np^2}\\plus{}1\\equal{}(y\\plus{}1)(y^{p\\minus{}1}\\minus{}y^{p\\minus{}2}\\plus{}\\cdots \\plus{}1)$. Again, the $ \\gcd$ between the factored terms is $ p$. Since $ p^2|(y\\plus{}1)$, we have that $ p$ can only divide $ y^{p\\minus{}1}\\minus{}y^{p\\minus{}2}\\plus{}\\cdots \\plus{}1$ once. Yet, $ y^{p\\minus{}1}\\minus{}y^{p\\minus{}2}\\plus{}\\cdots \\plus{}1>p$ for $ p>3$, so another prime factor that is not a factor of $ np^2$ divides $ y^{p\\minus{}1}\\minus{}y^{p\\minus{}2}\\plus{}\\cdots \\plus{}1$, so we are done. [/hide]", "Solution_8": "For any i, [code]2^3^i+1[/code] is divisible by [code]3^i[/code] (the proof is easy with euler`s theorem+induction and maybe with primitive roots (2 is primitive root modulo 3^i for any i)). Hence, for i=1999, [code]3^1999[/code] has 2000 divisors and it satisfies the asked in the problem.\r\n\r\nit is correct??\r\n\r\nplease , answer me.\r\n\r\nbye\r\n\r\nsorry for my english.", "Solution_9": "Well, the point is to find a number which has exactly $ 2000$ [b]prime[/b] divisiors, whereas $ 3^{1999}$ has only one ($ 3$).\r\n\r\nBut it is a very nice thought to look at powers of $ 3$, when the problem considers powers of $ 2$ (vide grobber's solution, which I do not completely understand yet, but it seems very nice and simple)", "Solution_10": "[hide=\"Solution overkilling with Zsigmondy's\"]\n\nBy Zsigmondy's we know that there exists a prime $p$ such that $p\\mid 2^i+1$ but $p\\nmid 2^j+1$ for $j 3$ is Korean, \tthere exists a prime $p$ not dividing $n$ \tsuch that $np$ is Korean too.\n\n[i]Proof.[/i] \tI claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$, \twhich exists by Zsigmondy theorem. \tObviously $p \\neq 2$. \tThen: \t[list] \t\t[*]Since $p \\nmid 2^{\\varphi(n)}-1$ it follows then that $p \\nmid n$. \t\t[*]Moreover, $p \\mid 2^n+1$ since $p \\nmid 2^n-1$; \t[/list] \tHence $np \\mid 2^{np} + 1$ by Chinese Theorem, \tsince $\\gcd(n,p) = 1$. $\\blacksquare$\n\nEDIT: The version of the proof I posted four years ago was incorrect. This one should work.", "Solution_12": "I feel like it isn't interesting to make any new remark on this problem, but anyway I'm posting my approach too.\n\nIf we show that for any positive integer k, there exists a positive integer n with exactly k distinct prime divisors such that n | 2^n + 1, then we are done, since the problem asks us to examine a special case, more exactly k = 2000. Furthermore, we can even show that we can find these n's divisible by a power of 3, which will help us on our proof.\n\nWe use induction on k. k = 1, we can choose n(1) = 3, which clearly satisfies the conditions. Assume that k >= 1, and there exists n(k) = 3^t * m, where gcd(3, m) = 1, and m has exactly (k - 1) distinct prime divisors. So, we have n(k) | 2^(n(k)) + 1.\nBefore generating n(k+1) from n(k), let us look at the number 3n(k), which clearly has k distinct prime divisors. 2^(3n(k)) + 1 = (2^(n(k)) + 1)(2^(2n(k)) - 2^(n(k)) + 1). Since we must have n(k) always odd because of the fact that n(k) = 1 (mod 2), we deduce that 3 | (2^(2n(k)) - 2^(n(k)) + 1), so we have that 3n(k) | 2^(3n(k)) + 1. It is enough to find a prime p, such that p | 2^(3n(k)) + 1 and p doesn't divide (2^(n(k)) + 1), which could guarantee us that p doesn't divide n(k) and consequently, we could generate n(k + 1) = 3n(k)*p, which could clearly work by observing that 2^(3n(k)*p) + 1 = (2^(n(k)) + 1)(2^(2n(k)) - 2^(n(k)) + 1)*A. But, since we can pick up this prime p by Zsigmondy, we are done.\n\nNote that in case of not using Zsigmondy, we can observe that gcd(a^2 - a + 1, a + 1) = gcd(3, a + 1) = 1 if a is not 2 mod 3 and 3 if a = 3k + 2. But if a = 3k + 2, then a^2 - a + 1 is divisible by 3 but not by 9, so we could pick up any prime p that divides (a^2 - a + 1) / 3.", "Solution_13": "Let $n=\\prod_{i=1}^{2000}{p_i}$ and wlog let $p_12$,and this is true because $GCD(a+1,a^2-a+1)$ is at most $3$ and $a^2-a+1>3$ for $a>2$ so we are done.", "Solution_15": "[quote=Msn05][quote=v_Enhance]Answer: Yes.\n\nWe say that $n$ is [i]Korean[/i] if $n \\mid 2^n+1$. First, observe that $n=9$ is Korean. Now, the problem is solved upon the following claim:\n\n[b]Claim[/b]: \tIf $n > 3$ is Korean, \tthere exists a prime $p$ not dividing $n$ \tsuch that $np$ is Korean too.\n\n[i]Proof.[/i] \tI claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$, \twhich exists by Zsigmondy theorem. \tObviously $p \\neq 2$. \tThen: \t[list] \t\t[*]Since $p \\nmid 2^{\\varphi(n)}-1$ it follows then that $p \\nmid n$. \t\t[*]Moreover, $p \\mid 2^n+1$ since $p \\nmid 2^n-1$; \t[/list] \tHence $np \\mid 2^{np} + 1$ by Chinese Theorem, \tsince $\\gcd(n,p) = 1$. $\\blacksquare$\n\nEDIT: The version of the proof I posted four years ago was incorrect. This one should work.[/quote]\n\nWhy $p \\nmid 2^{\\varphi(n)}-1$?[/quote]\n\nBecause $\\varphi(n) 3$ is Korean, \tthere exists a prime $p$ not dividing $n$ \tsuch that $np$ is Korean too.\n\n[i]Proof.[/i] \tI claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$, \twhich exists by Zsigmondy theorem. \tObviously $p \\neq 2$. \tThen: \t[list] \t\t[*]Since $p \\nmid 2^{\\varphi(n)}-1$ it follows then that $p \\nmid n$. \t\t[*]Moreover, $p \\mid 2^n+1$ since $p \\nmid 2^n-1$; \t[/list] \tHence $np \\mid 2^{np} + 1$ by Chinese Theorem, \tsince $\\gcd(n,p) = 1$. $\\blacksquare$\n\nEDIT: The version of the proof I posted four years ago was incorrect. This one should work.[/quote]\n\nWhy $p \\nmid 2^{\\varphi(n)}-1$?[/quote]\n\nBecause $\\varphi(n)p_k$ such that for all positive integers $e_{k+1}$, it is true that $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the problem condition, where we can make $m$ any positive integer we want. \\\\\\\\\n[color=#9a00ff][b]C1:[/b][/color] First, before we begin, I claim that if any prime $q\\mid n$, then $nq$ also satisfies problem conditions. This will then prove the part where we can make $m$ whatever we want. This is because \n\\[n\\mid 2^n+1 \\iff \\nu_q(n)\\leq \\nu_q(2^n+1),\\]\nand by LTE (which we can use, since $n\\mid 2^n+1$ implies that $q\\mid 2^n+1$), this gives us that \n\\[\\nu_q(nq)=1+\\nu_q(n)\\leq 1+\\nu_q(2^n+1)=\\nu_q(2^{nq}+1),\\]\nwhich means that $nq$ also satisfies the problem conditions, as desired. Therefore, by induction, we also get that $nq^m$ also satisfies problem conditions. \\\\\\\\\n[color=#9a00ff][b]C2:[/b][/color] Now, I claim that there exists a prime $p_{k+1}$ such that for all positive integers $e_{k+1}$, $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the conditions. For simplicity, from here on out, I will refer to $p_{k+1}$ as $r$ and $p_k$ as $q$. Note that this implies that \n\\[r\\mid 2^n+1 \\iff ord_r(2)\\mid 2n,\\]\nand since $ord_r(2)\\mid \\phi(r)=r-1$, we get that $ord_r(2)\\mid \\gcd(2n,r-1)$. First, I claim that there always exists an $r$ that divides $2^{q^c}+1$ for some $c$ we can choose such that $r>q$. This is clearly true by Zsigmondy's theorem. I now claim that if we take this $r$, it is true that $rn$ also satisfies problem conditions. This, combined with [color=#9a00ff][b](C1)[/b][/color], will prove our master claim, which states that $nr^e_{k+1}$ satisfies problem conditions for any $e_{k+1}$. \\\\\\\\\nTo prove this, first make sure that our previous $n$ is divisible by the $q^c$ we chose. We can do this by making another $n$ that satisfies problem conditions by multiplying it by a power of $q$, which we can do by [color=#9a00ff][b](C1)[/b][/color]. Next, by LTE, we have,\n\\[\\nu_r(rn)=1\\leq \\nu_r(2^{rn}+1)=\\nu_r(2^n+1)+1,\\]\nwhich we can do since we know that $q^c \\mid n$ and $r\\mid 2^{q^c}+1$, which gives that $r\\mid 2^n+1$. This means that $rn$ satisfies problem conditions, since $rn\\mid 2^{rn}+1$, which combined with [color=#9a00ff][b](C1)[/b][/color], proves our inductive step claim. \\\\\\\\\nFinally, to complete our induction, note that $3\\mid 2^3+1$. Therefore, there exists a positive integer $n$ such that $n\\mid 2^n+1$ with exactly $k$ distinct prime divisors for all positive integers $k$, finishing the problem. " } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "What is the inverse of the function $f: (2, +\\infty)\\to R$, $f(x)= \\frac{2x^2-7x-9}{x-2}$. How can i use the fact that $(f^{-1}(y))' = \\frac{1}{f'(x)}$ ?", "Solution_1": "does anybody have a clue ?" } { "Tag": [ "research" ], "Problem": "Suppose I have an idea in mind. How will I know if its worth writing a paper on and not plain trash?\r\n\r\nSuppose I ask for the opinion of a prof and he says its plain trash but uses it for his own interests. \r\nSo whose opinion should I consider relevant in evaluating my idea?\r\n\r\nWhat is the formal procedure for sending papers/articles to a journal?\r\n\r\nDoes \"research\" mean doing \"genius\" stuff and out-of-the-blue stunning results only??\r\n\r\n\r\n\r\nThanks.", "Solution_1": "You could send it to a professor with a sufficiently well-worded non-disclosure agreement. Though that's probably not necessary. First of all, unless your idea is something really monumental (like a proof of the Riemann Hypothesis or something), most professors will not covet your result enough to even contemplate stealing it. Furthermore, most professors would not risk their integrity by stealing the work of someone who could easily prove that the idea was theirs (which you probably could if required).\r\n\r\nSo basically, if you genuinely believe that you have a noteworthy result, find a professor who will take you seriously. Then politely show them your work and ask for feedback. If it will make you feel more comfortable, put a simple statement of proprietary rights somewhere at the beginning of your paper.\r\n\r\nAfter you've had someone look at it and give you feedback, then if they think it is journal worthy you can find an appropriate journal to submit it to. You'd contact the journal directly for the correct method for submitting it. Each journal will have different requirements for formatting and methods of submission.\r\n\r\nResearch doesn't have to be mind-blowingly out of the blue to be worthy of publication. It just has to be something with enough originality or new work to be worth other people to read.", "Solution_2": "[quote=\"romanovodka\"]Suppose I have an idea in mind. How will I know if its worth writing a paper on and not plain trash?\n...\nSo whose opinion should I consider relevant in evaluating my idea?[/quote] The only way to do this is to talk to other people. Professors are a good bet, because they're most likely to be familiar with the state of knowledge. But any person with such familiarity would suffice. \n\n[quote=\"romanovodka\"]Suppose I ask for the opinion of a prof and he says its plain trash but uses it for his own interests.[/quote] This is unlikely to happen. But, there are several ways to protect yourself against it. The most obvious is to stick a draft of a paper in some public place that demonstrates your priority. An appropriate part of this forum, for example, but more usually the [url=http://arxiv.org]arXiv[/url].\n\n[quote=\"romanovodka\"]What is the formal procedure for sending papers/articles to a journal?[/quote] This depends on the journal -- any journal website should have a link to \"Instructions for Authors\" or \"Guide to submission\" or something on the front page.\n\n[quote=\"romanovodka\"]Does \"research\" mean doing \"genius\" stuff and out-of-the-blue stunning results only??[/quote] No. But it does generally mean producing results that have not been proved before, or new proofs of known results that show something new and interesting.", "Solution_3": "Another question: \r\nSuggestions for journals for someone interested in Algebra generally (more specifically Representation Theory?) What low level journals would be good for someone to publish a paper in?", "Solution_4": "[quote=\"Vinoth\"]Another question: \nSuggestions for journals for someone interested in Algebra generally (more specifically Representation Theory?) What low level journals would be good for someone to publish a paper in?[/quote]\r\nOne method to find a good journal to submit to: Look at the bibliography of your paper, and see what journals recently published the related results you used." } { "Tag": [ "LaTeX" ], "Problem": "How does one make amazing PDFs like [url=http://www.ctan.org/tex-archive/macros/latex/contrib/pdftricks/manual.pdf]this[/url] one?", "Solution_1": "Use [url=http://latex-beamer.sourceforge.net/]Beamer[/url]. Here's [url=http://www.sixthform.info/steve/Slideb.pdf]one[/url] that I did a couple of years ago using beamer - save it then run it (so it doesn't run inside your Browser). Press CTRL-L to start and then the arrow buttons (or mouse) to go through it. ESC goes back to the normal screen. \r\nBeamer has lots of ready made templates but you can take a little while to get used to. There are lots of tutorials to be found on the net, for example, [url=http://www.math.umbc.edu/~rouben/beamer/]A Beamer Quickstart[/url]. \r\nOnce you get used to it you'll love it and it can produce stunning presentations.", "Solution_2": "Alternatively, you can use prosper, HA-prosper, or the latest replacement of these, namely [url=http://www.ctan.org/tex-archive/macros/latex/contrib/powerdot/]powerdot[/url]. A nice introduction to powerdot can be found [url=http://csgsa.cs.clemson.edu/seminar/tools06/docbook/html/ch02s04.html]here[/url].\r\n\r\nI personally use [url=http://amath.colorado.edu/documentation/LaTeX/prosper/]prosper[/url], because I found that there is currently more help available online for this package, as it is older and more widely used. I am not sure what exactly has been fixed or upgraded in HA-prosper and powerdot; prosper is easy to use and works just fine for all practical purposes." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $x;y;z >0$ and satisfying $x+y+z=xyz$\r\nFind max of $S=(1-x)(1-y)(1-z)$", "Solution_1": "See $x\\rightarrow 0$ and $y,z \\rightarrow \\infty$.\r\nBut\r\n$S \\ge (1-\\sqrt 3)^{3}$.", "Solution_2": "Other problem:\r\nLet $x;y;z>0$ and satisfying $x+y+z=xyz$.Find max:\r\n$A=(x-1)(y-1)(z-1)$" } { "Tag": [ "geometry", "trapezoid", "trigonometry", "geometry unsolved" ], "Problem": "PQR is a triangle in which there is an inside trapezoid ABCD. AB parralel to QR, BC parralel PQ and AD parallel PR. Also, area of PQR is 1 unit. Find out maximum possible area of trapezoid ABCD \r\n\r\n(from Indian Embassy practice papers)", "Solution_1": "[hide]Let $E$ be the area of $ABCD$ and $F=1-E$ the rest area of $\\triangle PQR$\n\nLet $m=\\frac{PA}{PQ}=\\frac{PB}{PR}$. \nThen\n\n$QA = (1-m)PQ$\n$RB = (1-m)PR$\n$QD = RC = (1-m)QR$\n\nAlso, it should be $m>1-m \\Leftrightarrow \\boxed{m>\\frac{1}{2}}$ or else $ABCD$ will be not a trapezoid. \n\nSo $(PAB) = m^{2}, \\ \\ (QAD) = (RBC) = (1-m)^{2}\\Rightarrow $\n \n\n$F=m^{2}+2(1-m)^{2}$\n$= 3m^{2}-4m+2$\n\nThen $E=1-F =-3m^{2}+4m-1$\n\n$E$ has a maximum value for $m=-\\frac{b}{2a}= \\frac{2}{3}$\n\nSo we find that the maximum value of the area is $E_{max}=\\frac{1}{3}$[/hide]", "Solution_2": "how does (PAB) = m^2?", "Solution_3": "It is obvious that the base of trapezoid has to pass centroid for max area.\r\n\r\n M.T.", "Solution_4": "[quote=\"shyamsundar23\"]how does (PAB) = m^2?[/quote]\r\n\r\n$(PQR) = \\frac{1}{2}PQ\\cdot PR\\cdot \\sin P$\r\n\r\n$(PAB) = \\frac{1}{2}PA\\cdot PB\\cdot \\sin P = \\frac{1}{2}(m\\cdot PQ)\\cdot (m\\cdot PR)\\cdot \\sin P= m^{2}\\cdot \\frac{1}{2}PQ\\cdot PR\\cdot \\sin P$\r\n\r\n$= m^{2}\\cdot (PQR)$" } { "Tag": [ "induction", "number theory", "prime factorization" ], "Problem": "Prove that a list can be made of all the subsets of a finite set in such a way that\r\n(i) the empty set is frst in the list;\r\n(ii) each subset occurs exactly once;\r\n(iii) each subset in the list is obtained either by adding one element to the preceding subset or by deleting one element of the previous subset.", "Solution_1": "[hide=\"Solution\"]\nWe'll give the construction for a countably infinite set such that truncating the list at $ 2^{n}$ gives the desired list for an $ n$ element set.\n\nFirst of all, it's equivalent to solve the problem for the set $ \\{0,1,2,3,\\ldots,n\\minus{}1\\}$, so we'll only do it for that.\n\nThe first term in the list, $ S_{0}$ is the empty set, of course.\n\nFor $ i\\geq 1$, let $ S_{i}\\equal{} S_{i\\minus{}1}\\Delta\\{a_{i}\\}$, where $ \\Delta$ is the [url=http://en.wikipedia.org/wiki/Symmetric_difference]symmetric difference[/url], and $ a_{i}$ is the exponent of the 2 in the prime factorization of $ i$ (so $ a_{2}\\equal{} 1$, $ a_{24}\\equal{} 3$, etc.). Then the desired list for an $ n$ element set is $ S_{0}, S_{1}, S_{2}\\ldots S_{2^{n}\\minus{}1}$.\n\nObviously (i) is satisfied, and (iii) is satisfied by the properties of the symmetric difference. We have $ 2^{n}$ elements in the list, the number of subsets of the set, so if we prove that no two $ S_{i}$ are the same, we have proven (ii).\n\nAgain by the properties of the symmetric difference, it suffices to show that no nonempty subsequence of $ \\{a_{i}\\}$ containing consecutive terms contains every term an even number of times. I will state without proof that if a consecutive subsequence of $ \\{a_{i}\\}$ contains at least two equal elements $ x$, it also contains a term that is at least $ x\\plus{}1$. I might come back and prove this is true later. So suppose such a consecutive subsequence exists, and take the highest term. It must occur at least two times. But then there is another term that is higher. Contradiction. And we're done.\n\nFor those who want a more concrete look at the answer, here are the first few $ S_{i}$:\n\n[hide=\"First 16 terms, showing how to do it for up to a 4 element set\"]\n$ S_{0}\\equal{}\\{\\}$\n$ S_{1}\\equal{}\\{0\\}$\n$ S_{2}\\equal{}\\{0,1\\}$\n$ S_{3}\\equal{}\\{1\\}$\n$ S_{4}\\equal{}\\{1,2\\}$\n$ S_{5}\\equal{}\\{0,1,2\\}$\n$ S_{6}\\equal{}\\{0,2\\}$\n$ S_{7}\\equal{}\\{2\\}$\n$ S_{8}\\equal{}\\{2,3\\}$\n$ S_{9}\\equal{}\\{0,2,3\\}$\n$ S_{10}\\equal{}\\{0,1,2,3\\}$\n$ S_{11}\\equal{}\\{1,2,3\\}$\n$ S_{12}\\equal{}\\{1,3\\}$\n$ S_{13}\\equal{}\\{0,1,3\\}$\n$ S_{14}\\equal{}\\{0,3\\}$\n$ S_{15}\\equal{}\\{3\\}$\n[/hide]\n[/hide]", "Solution_2": "I prefer a nice\r\n[hide=\"induction\"]\nBase case: $ |S|\\equal{}0$\n$ \\{\\}$ Done.\n\nInductive step. Take away one element, $ a$ from $ S$ to get $ S'$. Then, create the sequence of subsets of $ S'$, $ S_{0}, S_{1}, S_{2},\\ldots, S_{m}$. Then you append $ S_{m}\\cup\\{a\\}, S_{m\\minus{}1}\\cup\\{a\\},\\ldots, S_{0}\\cup\\{a\\}$. Done.\n\n[hide=\"The first few of mine\"]\nNotice that they all begin the same way, so I'll just note where to stop for each size of $ S$.\n{}\n---- 0\n{0}\n---- 1\n{0,1}\n{1}\n---- 2\n{1,2}\n{0,1,2}\n{0,2}\n{2}\n---- 3\n{2,3}\n{0,2,3}\n{0,1,2,3}\n{1,2,3}\n{1,3}\n{0,1,3}\n{0,3}\n{3}\n---- 4\n[/hide]\nThese are actually the same sets as given above, but generated inductively instead of explicitly.\n[/hide]", "Solution_3": "I would hardly say mine were generated explicitly; it was recursive, at best. :P\r\n\r\nThat's a much cleaner way than mine to do it; don't have to do any work to prove all of the terms in the sequence are different." } { "Tag": [ "algebra", "polynomial", "modular arithmetic", "algebra unsolved" ], "Problem": "Let be given an odd prime p and the polynomial :\r\n $ T(x) = (p-1)x^p-x -1 $ . \r\n Prove that : there exists an infinite number of possitive integers a such that $ T(a) $ is divisible by $ p^p $", "Solution_1": "Suppose you've found $a_n$ s.t. $p^n|T(a_n)$. We now look for numbers of the form $a_{n+1}=a_n+tp^n,\\ t\\in\\overline{0,p-1}$ s.t. $p^{n+1}|T(a_{n+1})$. \r\n\r\nWe have $T(a_n+tp^n)=T(a_n)+tp^nT'(a_n)+p^{n+1}M$. We divide this by $p^n$ to get $\\frac{T(a_n+tp^n)}{p^n}=\\frac{T(a_n)}{p^n}+tT'(a_n)+pM$, and we want $t\\in\\overline{0,p-1}$ s.t. the RHS is divisible by $p$. We can clearly choose such a $t$, since $T'(a_n)\\equiv -1\\pmod p$, so as $t$ goes through $0,1,\\ldots,p-1\\pmod p$, so does the RHS. \r\n\r\nWe have shown that for all $n$, there is an integer $a_n$ s.t. $p^n|T(a_n)$, and if there is one such integer, there are infinitely many. In particular, this holds for $n=p$, as desired." } { "Tag": [ "geometry", "geometric transformation", "vector", "analytic geometry", "algebra solved", "algebra" ], "Problem": "A region in the plane has area \\geq 1. Prove that it contains some points (a, b) and (c, d) such that both a - c and b - d are integers.", "Solution_1": "I think it is false if you only suppose that the area is \\geq 1. A counter-example is the interior (not the sides) of a square with side 1.\r\n\r\nIn another hand, the result is true if it is assumed that the area is > 1 (it is known as Blichfeldt's lemma, and uses as a starting point the prove Minkowski's theorem in geometry of numbers) :\r\n\r\nLet E be a set with area > 1.\r\nConsider the usual square grid with side 1.\r\n\r\nDecompose E as E = \\cup ( E \\cap S) where the union is over the squares S of the grid. Let T(S) = t( (E \\cap S)' ) where t is the translation such that t(S) = S_0 the square with vertices (0,0),(0,1),(1,1) and (1,0), and (E \\cap S)' denotes the set (E \\cap S) without the sides of S.\r\nNote that each of this translation has vector with integer coordinates.\r\n\r\nWe have area[E] = sum area[(E \\cap S)'] = sum [T(S)] > 1 = area[S_0] It follows that two of the sets T(S) must have a non empty intersection. Then there exists M(x,y) in the interior of S_0 which belongs to some T(S) and T(S') for some distinct S,S' of the grid. Using the reciprocal translations we deduce that there are two distinct points of E which have coordinates of the form (x+a,y+b) and (x+c,y+d) respectively, where a,b,c,d are integers. And we are done\r\n\r\nPierre." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "function", "geometry", "3D geometry" ], "Problem": "I saw this problem in Engel, USAMO 1980; \r\n\r\n$If a,b,c$ are 3 reals that can take any value from 0 to 1 (both inclusive); prove that $(a/(b+c+1))+(b/(c+a+1))+(c/(a+b+1))+(1-a)(1-b)(1-c)$ can take a maximum value of 1.\r\nHe has proved it by the following method:\r\n\r\n\" LHS = f(a,b,c) ; f is strictly convex in each variable, and so is defined on a closed convex cube. \"\r\nI understand this. But then he says, \"thus, the function assumes extremal values at extremal points\". How is this ? \r\n\r\nHe cites some theorem of Weierstrass. Please help. I need to get this cleared quick. Thanks, in advance.", "Solution_1": "The values of $a,b,c$ are defined for $[0,1]$\r\n\r\nThis means that the function takes its values from the unit cube in $\\mathbb{R}^{3}$ space.\r\n\r\nNow, look at each variable and hold the other two constant. Then the function in Convex in that variable, which you can check by taking second derivities. Since its convex, that means it assumes the maximum on one or both of its endpoints. Since the same is true for the other two variables, you know the whole function takes on its maximum at one of the points\r\n\r\n$(0,0,0),(0,0,1),(0,1,1),(1,1,1)$\r\n\r\n(We dont have to check all the values since ots symmetric in its variables)\r\nHope this helps." } { "Tag": [], "Problem": "If $ x^2\\minus{}y^2\\equal{}27$ and $ x^3\\plus{}y^3\\equal{}4941$, find $ 2xy$. \r\n\r\nAgain, as in the previous thread, is there an \"elegant\" solution?", "Solution_1": "If you factor things out, you will get\r\n\r\n$ (x\\plus{}y)(x\\minus{}y)\\equal{}27$\r\n$ (x\\plus{}y)(x^2\\minus{}xy\\plus{}y^2)\\equal{}(81)(61)$\r\n\r\nIf we have $ x\\plus{}y\\equal{}27$ and $ x\\minus{}y\\equal{}1$ (Sorry, I couldn't really explain anything legit other than guess and check), then solving the system gives $ x\\equal{}14$ and $ y\\equal{}13$. Now to check this:\r\n\r\n$ (27)(14^2\\minus{}14*13\\plus{}13^2)\\equal{}(27)(196\\plus{}169\\minus{}182)\\equal{}(27)(183)\\equal{}(81)(61)$\r\n\r\nwhich works.\r\n\r\nSo $ 2xy\\equal{}2(13)(14)\\equal{}\\boxed{364}$.", "Solution_2": "we can use Cardano methode to solve a qubic equation we will get y= 13 \r\nand x= 14" } { "Tag": [ "geometry", "rectangle", "ratio", "perimeter", "ceiling function", "floor function", "number theory" ], "Problem": "Let $ ABCD$ be a rectangle with $ AB \\equal{} m,BC \\equal{} n$ and $ m,n\\in\\mathbb{N},(m,n)\\equal{}1$.Rectangle $ ABCD$ is divided into squares of unit length and colored as chess board.\r\nLet $ X$ be the sum of all white pieces,that are obtained by intersecting unit squares by the line $ AC$,similarly $ Y$ is the sum of all black pieces,that are obtained by intersecting unit squares by the line $ AC$.\r\nFind the value $ \\frac {X}{Y}$ in terms of $ m,n$.", "Solution_1": "It's important to specify whether the unit square touching point $ A$ is white or black, right? I'll assume it's white here. Here's a proof with a few sizable holes I don't care to plug.\r\n\r\nFirst, note that if $ m \\equal{} da, n \\equal{} db$ with $ (a,b) \\equal{} 1$ (so $ (m,n) \\equal{} d$) then we can subdivide the $ m \\times n$ rectangle into $ d$ $ a \\times b$ rectangles so that they don't overlap and $ AC$ passes through all of them. $ AC$ passes through the same color squares in each rectangle, so the ratio $ X/Y$ for the $ m \\times n$ rectangle is the same as the ratio for the $ a \\times b$ rectangle. Therefore it suffices to consider $ (m,n) \\equal{} 1$.\r\n\r\nObserve now that when $ (m,n) \\equal{} 1$ that we can not pass through the vertex of any unit square. So line $ AC$ only passes through edges of unit squares. This means that if we mark the squares line $ AC$ touches we can walk from the square touching $ A$ to the square touching $ C$ through only marked squares. In any such walk (marked squares or not) we always pass through the same number of squares, so if we consider the walk around the perimeter we get $ m \\plus{} n \\minus{} 1$ squares. This means that line $ AC$ passes through $ m \\plus{} n \\minus{} 1$ squares.\r\n\r\nSince we only pass through the boundaries of the unit squares, $ AC$ always passes through a white, then a black, then a white, and so on. By the assumption in the first paragraph, we touch a white first. If $ m \\plus{} n \\minus{} 1$ is even, we pass through $ \\frac{m \\plus{} n \\minus{} 1}{2}$ white and black squares both, and the desired ratio is $ 1$. If $ m \\plus{} n \\minus{} 1$ is odd, we pass through $ \\frac{m \\plus{} n}{2}$ white squares and $ \\frac{m \\plus{} n \\minus{} 2}{2}$ black squares, and the desired ratio is $ \\frac{m\\plus{}n}{m\\plus{}n\\minus{}2}$.", "Solution_2": "You do not understand the statement,$ X$ is not the number of black unit squares,which are intersected by $ AC$,it is the sum of length of the pieces,which lies on $ AC$ and inside black squares.\r\nFor example:\r\nIf $ m\\equal{}n\\equal{}2$,then $ X\\equal{}\\sqrt{2}$", "Solution_3": "If $ m \\equal{} n \\equal{} 2$ then according to the clarification either $ X \\equal{} 0$ or $ X$ is undefined :maybe:\r\n\r\nThe reduction to relatively prime $ m, n$ is still valid. If one of $ m, n$ is even then the answer is $ X \\equal{} 1$, so the only interesting part is for $ m, n$ odd. A little experimenting suggests that the answer will always be $ \\frac{\\left\\lceil \\frac{m\\plus{}n}{2} \\right\\rceil}{\\left\\lfloor \\frac{m\\plus{}n}{2} \\right\\rfloor}$. This sounds entirely reasonable to me, but I haven't tried to prove it.", "Solution_4": "$ X$ doesn't represent the ratio he's asking for, but you are right that $ X/Y \\equal{} 0$ or $ X/Y$ is undefined." } { "Tag": [], "Problem": "I'm working on chapter 7, and was wondering if anyone is kind and could tell me the answers to the problems in the back. :lol:", "Solution_1": "which volume?", "Solution_2": "oh oops I forgot to say that. volume 1. :D" } { "Tag": [], "Problem": "How many positive integers $ n$ satisfy \\[ (n \\plus{} 8)(n \\minus{} 3)(n\\minus{}12)<0\\]", "Solution_1": "This could only be negative if all three are negative or 1 is negative while the the other two are positive.\r\n\r\nCase 1: All 3 terms are negative -\r\n\r\nThis occurs when n < -8, but clearly, there are no positive solutions for this.\r\n\r\nCase 2: One term is negative while the otehr two are positive -\r\n\r\n(n+8) and (n-3) are the two biggest terms, so in order for only one of them to be negative, then (n-12) must be it. (n-3) is the second smallest, and this must be positive, so n > 3. However, (n-12) is less than 0, so n < 12. The integer solutions are\r\n\r\n$ 4\\le n\\le 11$\r\n\r\n\r\n\r\nTherefore, there are a total of (11-4+1) = 8 positive integer solutions." } { "Tag": [ "calculus", "inequalities", "algebra" ], "Problem": "Show that for all n > 1 , \r\n$\\sqrt{n}+\\sqrt{ n+1}+. . .+\\sqrt{n^{2}-1}<\\frac{2}{3}(n^{3}-n^{\\frac{3}{2}})$", "Solution_1": "This question looks to me like it more properly belongs in a calculus forum, although perhaps there are elementary ways of solving it.", "Solution_2": "[hide]\nIt would follow immediately from the inequality for any positive integer $m$:\n\\[\\sqrt{m}<\\frac{2}{3}((m+1)^{3/2}-m^{3/2})\\]\nBut we can rationalize the numerator on the right hand side:\n\\begin{eqnarray*}\\frac{2}{3}((m+1)^{3/2}-m^{3/2})&=&\\frac{2}{3}\\cdot{(m+1)^{3}-m^{3}}{(m+1)^{3/2}+m^{3/2}\\\\ &=&\\frac{2}{3}\\cdot \\frac{m^{2}+m(m+1)+(m+1)^{2}}{(m+1)^{3/2}+m^{3/2}}\\\\ &=&\\sqrt{m}\\cdot \\frac{2}{3}\\cdot \\frac{1+r^{2}+r^{4}}{1+r^{3}}\\\\ }\\end{eqnarray*}\nwhere $r=\\sqrt{\\frac{m+1}{m}}>1$. It only remains to show that\n\\[\\frac{1+r^{2}+r^{4}}{1+r^{3}}>\\frac{3}{2}\\]\nBut the numerator factors into $(1+r+r^{2})(1-r+r^{2})$ and the denominator is $(1+r)(1-r+r^{2})$ so the left hand side of the above is actually $\\frac{1+r+r^{2}}{1+r}$. The fact that this is greater than $\\frac{3}{2}$ is a result of\n\\[2(1+r+r^{2})-3(1+r)=2r^{2}-r-1=(r-1)(2r+1)>0\\]\n[/hide]" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Prove that:\r\n$ \\sum_{n\\equal{}1}^{\\infty}\\frac{F_n}{10^n}\\equal{}\\frac{10}{89},\\sum_{n\\equal{}1}^{\\infty}\\frac{(\\minus{}1)^{n\\plus{}1}F_n}{10^n}\\equal{}\\frac{10}{109}$", "Solution_1": "Let,\r\n\\[ S\\equal{}\\sum_{i\\equal{}0}^\\infty \\frac{F_i}{k^{i\\plus{}1}}\\]\r\nwhere $ k$ is a nonzero real number.\r\nSuppose the series converges. Then, by Binet's formula,\r\n$ \\sum_{i\\equal{}0}^\\infty \\frac{F_i}{k^{i\\plus{}1}} \\equal{} \\frac{1}{\\sqrt{5}k}\\left[ \\sum_{i\\equal{}0}^\\infty \\left( \\frac{1\\plus{}\\sqrt{5}}{2k} \\right)^i \\minus{} \\left( \\frac{1\\minus{}\\sqrt{5}}{2k} \\right)^i \\right]$\r\n$ \\equal{} \\frac{1}{\\sqrt{5}k}\\left[ \\frac{1}{1\\minus{}(1\\plus{}\\sqrt{5})/2k} \\minus{} \\frac{1}{1\\minus{}(1\\minus{}\\sqrt{5})/2k} \\right]$\r\n$ \\equal{} \\frac{1}{\\sqrt{5}k}\\left( \\frac{1}{2k\\minus{}1\\minus{}\\sqrt{5}} \\minus{} \\frac{1}{2k\\minus{}1\\plus{}\\sqrt{5}} \\right)$\r\nThus,\r\n\\[ S\\equal{}\\frac{1}{k^2\\minus{}k\\minus{}1}\\]\r\n\r\nwhen $ k\\equal{}10$, $ \\sum_{i\\equal{}0}^\\infty \\frac{F_i}{10^{i\\plus{}1}} \\equal{} \\frac{1}{89}$\r\nThus,\r\n\\[ \\sum_{i\\equal{}0}^\\infty \\frac{F_i}{10^i}\\equal{}\\frac{10}{89}\\]\r\n\r\nwhen $ k\\equal{}\\minus{}10$, $ \\sum_{i\\equal{}0}^\\infty \\frac{F_i}{(\\minus{}10)^{i\\plus{}1}} \\equal{} \\frac{1}{109}$\r\nThus,\r\n\\[ \\sum_{i\\equal{}0}^\\infty \\frac{F_i}{10^i}\\equal{}\\frac{10}{109}\\]" } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "We have divided a 1999-sided regular polygon into triangles by drawing its diagonals . Prove that there exists exactly one acute angled triangle .", "Solution_1": "[quote=\"Toxic_Frog\"]We have divided a 1999-sided regular polygon into triangles by drawing its diagonals . Prove that there exists exactly one acute angled triangle .[/quote]\r\n\r\nSuspiciously homework-looking again, so a hint only:\r\n\r\nHow does one infer whether a triangle is acute-angled from the position of its circumcenter?\r\n\r\n darij", "Solution_2": "thx a lot \r\n[quote]\n Suspiciously homework-looking again\n[/quote]\r\n\r\nNo proposed in a mock olympiad.", "Solution_3": "Seems to be a VERY simple olympiad problem then.\r\n\r\n Darij" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all continuos function $f: R\\to R$ such that:\r\n$f(x+2f(y))=f(x)+f(y)+y$\r\nI need a nice solution!!", "Solution_1": "Now, let $f(0)=a$ \r\n\r\nPut $x=0$ we have: $f(2f(y))=y+f(y)+a$ $\\forall y\\in R$. $(1)$.\r\n\r\nPut $y=0$ in $(1)$ we have $f(2a)=2a$ $(2)$.\r\n\r\nPut $y=2f(z)$ we have :\r\n\r\n$f(x+2f(2f(z)))=f(x)+2f(z)+f(2f(z))=f(x)+3f(z)+z+a.$ $(3)$\r\n\r\nBy other hand, $f(x+2f(2f(z)))=f(x+2z+2a+2f(z))=f(x+2z+2a)+z+f(z)$ $(4)$.\r\n\r\n$(3)$ and $(4)$ follow that $f(x+2z+2a)=f(x)+2f(z)+a$ $\\forall x;z\\in R$. $(*)$.\r\n\r\nPut $x=z=0$ in $(*)$, then $f(2a)=4a$ $(5)$.\r\n\r\nFrom $(2)$ and $(5)$: $a=0$.Thus, (*) become : $f(x+2z)=f(x)+2f(z)$ $\\forall x;z\\in R$. $(**)$\r\n\r\nPut $x=0$ in $(**)$ we have $f(2z)=2f(z)$.hence $f(x+y)=f(x+2\\frac{y}{2})=f(x)+2f(\\frac{y}{2})=f(x)+f(y)$\r\n\r\nTherefore: $f(x+y)=f(x)+f(y) \\forall x;y\\in R$ $(***)$.\r\n\r\nBecause $f$ is continuos, thus $f(x)=\\alpha x$ \r\n\r\n.But, $f(2f(y))=y+f(y)+a$ $\\forall y\\in R$,$\\Rightarrow$ $2\\alpha^2=1+\\alpha$, thus $\\alpha =1$ or $-\\frac{1}{2}$.\r\n\r\nTherefore, $f(x)=x$ and $f(x)=-\\frac{x}{2}$ are all solutions.", "Solution_2": "[quote=\"Hong Quy\"][b]lemma[/b]: if $f: R\\to R$ satisfies $2f(f(y))=y+f(y) \\forall y\\in R$ then $f$ must be continous at $x_0$ (for some $x_0$).\n\n The lemma is very easy, you should prove it by yourself. [/quote]\r\nReally Hong Quy, I'm verry interested to see this very easy proof ! :D Actually that's impossible !\r\nHave a look ..\r\nLet $B$ be a base of the $R$ (real number) vectorial space on the $Q$(rational number) field.\r\nOn this vectorial space I define the linear functions $f$ such that $\\forall b\\in B\\ f(b)=\\lambda_b*b$, where $\\lambda_b\\in \\{1,-\\frac 12\\}$\r\nThere are $2 ^{card(R)}$ of such functions and they satisfy the relation ship but only two are continous !!\r\n\r\n :cool:", "Solution_3": "[quote]Really Hong Quy, I'm verry interested to see this very easy proof ! :D Actually that's impossible ![/quote]\r\n\r\n\r\n\r\n You are right. :blush:\r\n\r\nI've edit my solution without lemma.In fact,I hadn't seen the condition \"$f$ is continuos\" before I solved the problem.", "Solution_4": "Yes,Hong Quy.I got a similar one,after posting.May this problem has an unique solution,as above!!" } { "Tag": [ "number theory", "prime factorization" ], "Problem": "The product of a set of positive integers is 144. What is the least possible sum of this set of positive integers?", "Solution_1": "The prime factorization of 144 is 2^4 * 3^2. SInce these are its smallest factors, the answer is (2*4) + (3*2) = 8 + 6 = [u]14[/u].\r\nQED :)", "Solution_2": "can someone tell me what makes this true?\n\nand will this be true for everything", "Solution_3": "The closer the 2 factors are, the smaller the sum, so a perfect square would satisfy the need.", "Solution_4": "Wait, the problem says \" a [b]SET[/b] of positive integers\", and a set shouldn't have repeated elements. Shouldn't the answer be $2+3+4+6=15$?" } { "Tag": [], "Problem": "[b]Hello[/b]\r\n\r\nI would like some help understanding the Chi-Square Test...\r\n\r\nThis illustration is about rocks - yeah interesting!\r\n\r\nI can use the formula to get the results but I would like to know how to interpret\r\n\r\nX2 (4) = 9.4 < .05\r\n\r\n(I think I have written this correctly.)\r\nThe table of critical values in my book for chi square shows that with four degrees of freedom the critical value of chi square is 9.49 at the 0.05 level of significance. \r\nThe example says to reject the null hypothesis and conclude that \"on the evidence we have collected the reliance on cereal production as a main source of income is affected by rock type. \"\r\n\r\nI'm not sure what is significant about the result of this test\r\n :?:\r\n\r\np.s.\r\nsorry i put the tables up in the wrong order...", "Solution_1": "Since the calculated p value is greater than the critical value, you reject the null hypothesis, which originally states that there is no statistically significant difference between the results you observed and those expected. So basically it's saying that your data does not really match your original hypothesis. Judging from glancing briefly at your tables, your original hypothesis must have been something like:\r\n\r\nThe rock type does not affect the number of cereal farms.\r\n\r\nAnd since the total was 26 cereal farms, and there were 5 rock types, the results that would have supported that hypothesis would be if each rock type grew approximately 5.2 farms. However, according to the chi-square test, your results deviate too much to fit that model and therefore rock type does affect the number of cereal farms.\r\n\r\nI hope that answers your questions. :)" } { "Tag": [ "articles", "LaTeX" ], "Problem": "I have written the definitions in my works as\r\n\\begin{defn}\r\n\\label{finiteset}\r\nA set $ A$ is (dollar)\\operatorname{finite}(dollar) if there exists $ n\\in \\mathbb{Z}$ s.t. $ |A| \\equal{} n$.\r\n\\end{defn}\r\n\r\nWell, that was only an example of the form I have written my definitions, of course I won't define such a trivial things. Now, the problem is that I have to write one paper in Finnish and I have to define a term containing the letter \u00e4. How to do that because\r\n\r\n\\begin{defn}\r\n\\label{aarellinen joukko}\r\nJoukko $ A$ on (dollar)\\operatorname{\u00e4\u00e4rellinen}(dollar), jos on olemassa $ n\\in \\mathbb{Z}$ jolle $ |A| \\equal{} n$\r\n\\end{defn}\r\n\r\noutputs only\r\nJoukko $ A$ on rellinen, jos on olemassa $ n\\in \\mathbb{Z}$ jolle $ |A| \\equal{} n$?\r\n\r\nNeither\r\n\\begin{defn}\r\n\\label{aarellinen joukko}\r\nJoukko $ A$ on (dollar)\\operatorname{\\\"a\\\"arellinen}(dollar), jos on olemassa $ n\\in \\mathbb{Z}$ jolle $ |A| \\equal{} n$\r\n\\end{defn}\r\nwon't work.\r\n\r\nHere is the beginning of the tex-file:\r\n\r\n\\documentclass[a4paper]{article}\r\n\\usepackage[T1]{fontenc}\r\n\\usepackage[latin1]{inputenc}\r\n\\usepackage[finnish]{babel}\r\n\\usepackage{amssymb}\r\n\\usepackage{amsmath}\r\n\\usepackage{stmaryrd}\r\n\\usepackage{euscript}\r\n\\begin{document}\r\n\\newtheorem{lau}{Lause}[section]\r\n\\newtheorem{defn}[lau]{M\\\"a\\\"aritelm\\\"a}\r\n\\newtheorem{merk}[lau]{Merkint\\\"a}\r\n\\newtheorem{lem}{Lemma}[section]\r\n\\newtheorem{cor}[lau]{Korollaari}\r\n\\newtheorem{prop}{Propositio}[section]\r\n\\newtheorem{teor}{Teoreema}[section]\r\n\\begin{titlepage}", "Solution_1": "The problem is because you want the accent in math-mode. So instead of \\\"{a} you use \\ddot{a} as in\r\n$ \\operatorname{\\ddot{a}\\ddot{a}rellinen}$.\r\nYou'll find a list of math-mode symbols on page 56 of [url=http://tug.ctan.org/tex-archive/info/symbols/comprehensive/symbols-a4.pdf]The Comprehensive LaTeX Symbol List[/url]" } { "Tag": [ "inequalities", "inequalities proposed", "109" ], "Problem": "Given $ a, b, c > 0$ and $ abc\\equal{}1$. Prove that\r\n\\[ \\frac{1}{(1\\plus{}a)^2(b\\plus{}c)}\\plus{}\\frac{1}{(1\\plus{}b)^2(c\\plus{}a)}\\plus{} \\frac{1}{(1\\plus{}c)^2(a\\plus{}b)} \\le \\frac{3}{8}\\]\r\n\r\nIt exist a nice proof by Am-Gm :)", "Solution_1": "$ \\frac {3}{8} \\minus{} \\sum{\\frac {1}{(a \\plus{} 1)^2(b \\plus{} c)}} \\equal{} \\frac {3}{8} \\minus{} \\sum{\\frac {bc\\sqrt [3]{a}}{(\\sqrt [3]{a^2} \\plus{} \\sqrt [3]{bc})^2(b \\plus{} c)}}$\r\n\r\n$ \\equiv\\frac {3}{8} \\minus{} \\sum{\\frac {xy^3z^3}{(x^2 \\plus{} yz)^2(y^3 \\plus{} z^3)}}$\r\n\r\n$ \\equal{} \\sum\\frac {x(y \\minus{} z)^2}{16(x^2 \\plus{} yz)(y^2 \\plus{} zx)^2(z^2 \\plus{} xy)^2}\\bigg[x^3\\left(3y^4 \\plus{} 2y^3z \\plus{} 8y^2z^2 \\plus{} 2yz^3 \\plus{} 3z^4\\right)$\r\n\r\n$ \\plus{} \\frac {2x^2yz}{y^3 \\plus{} z^3}\\left(3y^6 \\plus{} 7y^5z \\plus{} 9y^4z^2 \\plus{} 12y^3z^3 \\plus{} 9y^2z^4 \\plus{} 7yz^5 \\plus{} 3z^6\\right) \\plus{} 3xyz\\left(y^4 \\plus{} z^4\\right)$\r\n\r\n$ \\plus{} \\frac {2y^3z^3}{y^3 \\plus{} z^3}\\left(y^2 \\plus{} z^2\\right)\\left(3y^2 \\plus{} 5yz \\plus{} 3z^2\\right)\\bigg]\\geq0.$\r\n\r\n\r\nBy the way, if $ a, b, c$ are nonnegative numbers such that $ bc \\plus{} ca \\plus{} ab \\plus{} abc\\geq 4,$ then\r\n\r\n$ \\sum{\\frac {1}{(b \\plus{} c)[4a^2 \\plus{} (\\sqrt {97} \\minus{} 1)a \\plus{} 4]}}\\leq \\frac {\\sqrt {97} \\minus{} 7}{32},$\r\n\r\nwith equality if $ a \\equal{} 0,b \\equal{} c \\equal{} 2.$\r\n\r\nSee also : http://www.mathlinks.ro/viewtopic.php?t=314506", "Solution_2": "[quote=\"Ji Chen\"]$ \\frac {3}{8} \\minus{} \\sum{\\frac {1}{(a \\plus{} 1)^2(b \\plus{} c)}} \\equal{} \\frac {3}{8} \\minus{} \\sum{\\frac {bc\\sqrt [3]{a}}{(\\sqrt [3]{a^2} \\plus{} \\sqrt [3]{bc})^2(b \\plus{} c)}}$\n\n$ \\equiv\\frac {3}{8} \\minus{} \\sum{\\frac {xy^3z^3}{(x^2 \\plus{} yz)^2(y^3 \\plus{} z^3)}}$\n\n$ \\equal{} \\sum\\frac {x(y \\minus{} z)^2}{16(x^2 \\plus{} yz)(y^2 \\plus{} zx)^2(z^2 \\plus{} xy)^2}\\bigg[x^3\\left(3y^4 \\plus{} 2y^3z \\plus{} 8y^2z^2 \\plus{} 2yz^3 \\plus{} 3z^4\\right)$\n\n$ \\plus{} \\frac {2x^2yz}{y^3 \\plus{} z^3}\\left(3y^6 \\plus{} 7y^5z \\plus{} 9y^4z^2 \\plus{} 12y^3z^3 \\plus{} 9y^2z^4 \\plus{} 7yz^5 \\plus{} 3z^6\\right) \\plus{} 3xyz\\left(y^4 \\plus{} z^4\\right)$\n\n$ \\plus{} \\frac {2y^3z^3}{y^3 \\plus{} z^3}\\left(y^2 \\plus{} z^2\\right)\\left(3y^2 \\plus{} 5yz \\plus{} 3z^2\\right)\\bigg]\\geq0.$\n\n\nBy the way, if $ a, b, c$ are nonnegative numbers such that $ bc \\plus{} ca \\plus{} ab \\plus{} abc\\geq 4,$ then\n\n$ \\sum{\\frac {1}{(b \\plus{} c)[4a^2 \\plus{} (\\sqrt {97} \\minus{} 1)a \\plus{} 4]}}\\leq \\frac {\\sqrt {97} \\minus{} 7}{32},$\n\nwith equality if $ a \\equal{} 0,b \\equal{} c \\equal{} 2.$\n\nSee also : http://www.mathlinks.ro/viewtopic.php?t=314506[/quote]\r\nthere is no way that this was hand done am i right??\r\n[hide=\"if it was due to a genius computer program otherwise omit this\"]\nwat is the fun in inequalities if a computer is gonna solve it instead of you?? \ndo we do inequalities because we like it or because we [b]HAVE[/b] to get solutions with so much desperation as to using these programs??? \ni am extremely sorry about the above if you have not used a computer programme or any other device other than your brain for this..\n[/hide]", "Solution_3": "$ \\sqrt {97}$ is ugly.But the following inequality is nice:\r\n\r\nIf $ a,b,c > 0$ and $ bc \\plus{} ca \\plus{} ab \\plus{} abc \\equal{} 4,$ then \r\n\r\n$ \\frac {1}{(b \\plus{} c)(1 \\plus{} a)^{2}} \\plus{} \\frac {1}{(a \\plus{} c)(1 \\plus{} b)^{2}} \\plus{} \\frac {1}{(a \\plus{} b)(1 \\plus{} c)^{2}}\\leq \\frac {3}{8}.$\r\n\r\n[b]Proof[/b] $ a \\equal{} \\frac {2u}{w \\plus{} v},b \\equal{} \\frac {2v}{u \\plus{} w},c \\equal{} \\frac {2w}{u \\plus{} v}$,\r\n\r\n$ \\frac {1}{(b \\plus{} c)(1 \\plus{} a)^{2}} \\plus{} \\frac {1}{(a \\plus{} c)(1 \\plus{} b)^{2}} \\plus{} \\frac {1}{(a \\plus{} b)(1 \\plus{} c)^{2}}\\leq \\frac {3}{8}$\r\n\r\n=> $ \\frac {3}{8} \\minus{} \\frac {1}{(\\frac {2v}{u \\plus{} w} \\plus{} \\frac {2w}{u \\plus{} v})(1 \\plus{} \\frac {2u}{w \\plus{} v})^{2}}$$ \\minus{} \\frac {1}{(\\frac {2u}{w \\plus{} v} \\plus{} \\frac {2w}{u \\plus{} v})(1 \\plus{} \\frac {2v}{u \\plus{} w})^{2}}$ $ \\minus{} \\frac {1}{(\\frac {2v}{u \\plus{} w} \\plus{} \\frac {2u}{w \\plus{} v})(1 \\plus{} \\frac {2w}{u \\plus{} v})^{2}}$\r\n\r\nso it's equivalent to prove\r\n\r\n$ \\equal{} > 107(\\sum_{\\text{sym}}v^{9}u^{3} \\minus{} \\sum_{\\text{sym}}v^{7}u^{5}) \\plus{} 136(\\sum_{\\text{sym}}v^{8}u^{4} \\minus{} \\sum_{\\text{sym}}v^{6}u^{6})$ \r\n\r\n$ \\plus{} 59(\\sum_{\\text{sym}}v^{9}u^{3} \\minus{} \\sum_{\\text{sym}}v^{6}u^{3}w^{3}) \\plus{} 259(\\sum_{\\text{sym}}v^{7}u^{4}w \\minus{} \\sum_{\\text{sym}}v^{6}u^{5}w)$ \r\n$ \\plus{} 310(\\sum_{\\text{sym}}v^{9}u^{2}w \\minus{} \\sum_{\\text{sym}}v^{5}u^{5}v^{2})$ $ \\plus{} 10(\\sum_{\\text{sym}}v^{7}u^{4}w \\minus{} \\sum_{\\text{sym}}v^{6}u^{4}w^{2})$\r\n\r\n$ \\plus{} (\\sum_{\\text{sym}}v^{9}u^{3} \\plus{} 12\\sum_{\\text{sym}}v^{11}u \\plus{} 37\\sum_{\\text{sym}}v^{9}u^{2}v^{1} \\plus{} 60\\sum_{\\text{sym}}v^{10}uv \\plus{} 63\\sum_{\\text{sym}}v^{7}u^{4}v$\r\n\r\n$ \\plus{} 72\\sum_{\\text{sym}}v^{10}u^{2} \\plus{} 634\\sum_{\\text{sym}}v^{8}u^{3}v \\plus{} 887\\sum_{\\text{sym}}v^{7}u^{3}v^ {2} \\minus{} 1766\\sum_{\\text{sym}}v^{5}u^{4}v^{3})$ \r\n\r\n$ \\plus{} (62\\sum_{\\text{sym}}v^{9}u^{2}v \\plus{} 462\\sum_{\\text{sym}}v^{8}u^{2}v^{2} \\minus{} 3144u^{4}v^{4}w^{4}) \\geq 0 .$\r\n$ \\equal{} > 107(\\sum_{\\text{sym}}v^{9}u^{3} \\minus{} \\sum_{\\text{sym}}v^{7}u^{5}) \\plus{} 136(\\sum_{\\text{sym}}v^{8}u^{4} \\minus{} \\sum_{\\text{sym}}v^{6}u^{6})$ \r\n\r\n$ \\plus{} 59(\\sum_{\\text{sym}}v^{9}u^{3} \\minus{} \\sum_{\\text{sym}}v^{6}u^{3}w^{3}) \\plus{} 259(\\sum_{\\text{sym}}v^{7}u^{4}w \\minus{} \\sum_{\\text{sym}}v^{6}u^{5}w)$ \r\n$ \\plus{} 310(\\sum_{\\text{sym}}v^{9}u^{2}w \\minus{} \\sum_{\\text{sym}}v^{5}u^{5}v^{2})$ $ \\plus{} 10(\\sum_{\\text{sym}}v^{7}u^{4}w \\minus{} \\sum_{\\text{sym}}v^{6}u^{4}w^{2})$\r\n\r\n$ \\plus{} (\\sum_{\\text{sym}}v^{9}u^{3} \\plus{} 12\\sum_{\\text{sym}}v^{11}u \\plus{} 37\\sum_{\\text{sym}}v^{9}u^{2}v^{1} \\plus{} 60\\sum_{\\text{sym}}v^{10}uv \\plus{} 63\\sum_{\\text{sym}}v^{7}u^{4}v$\r\n\r\n$ \\plus{} 72\\sum_{\\text{sym}}v^{10}u^{2} \\plus{} 634\\sum_{\\text{sym}}v^{8}u^{3}v \\plus{} 887\\sum_{\\text{sym}}v^{7}u^{3}v^ {2} \\minus{} 1766\\sum_{\\text{sym}}v^{5}u^{4}v^{3})$ \r\n\r\n$ \\plus{} (62\\sum_{\\text{sym}}v^{9}u^{2}v \\plus{} 462\\sum_{\\text{sym}}v^{8}u^{2}v^{2} \\minus{} 3144u^{4}v^{4}w^{4}) \\geq 0.$\r\n \r\nwhich is clearly true by Muirhead theorem.", "Solution_4": "[quote=\"nguoivn\"]Given $ a, b, c > 0$ and $ abc \\equal{} 1$. Prove that\n\\[ \\frac {1}{(1 \\plus{} a)^2(b \\plus{} c)} \\plus{} \\frac {1}{(1 \\plus{} b)^2(c \\plus{} a)} \\plus{} \\frac {1}{(1 \\plus{} c)^2(a \\plus{} b)} \\le \\frac {3}{8}\\]\nIt exist a nice proof by Am-Gm :)[/quote]\r\nsee any solution here http://ddbdt.co.cc/forum/ba-t-ae-ng-tha-c-sa-ng-ta-o/347-la-m-ma-nh-1-ba-i-ca-arqady.html#post1511", "Solution_5": "[quote=\"jingjun\"]If $ a,b,c > 0$ and $ bc \\plus{} ca \\plus{} ab \\plus{} abc \\equal{} 4,$ then \n\n$ \\frac {1}{(1 \\plus{} a)^2(b \\plus{} c)} \\plus{} \\frac {1}{(1 \\plus{} b)^2(c \\plus{} a)} \\plus{} \\frac {1}{(1 \\plus{} c)^2(a \\plus{} b)}\\leq \\frac {3}{8}.$\n\n[b]Proof[/b] Substitute $ a \\equal{} \\frac {2u}{v \\plus{} w},b \\equal{} \\frac {2v}{w \\plus{} u},c \\equal{} \\frac {2w}{u \\plus{} v},$ where $ u,v,w > 0.$ Then[/quote]$ (2u \\plus{} w \\plus{} v)^2(2v \\plus{} u \\plus{} w)^2(2w \\plus{} v \\plus{} u)^2\\left[3\\minus{} \\sum{\\frac {8}{(1 \\plus{} a)^2(b \\plus{} c)}}\\right]$\r\n\r\n$ \\equal{} \\sum\\frac { (v \\plus{} w)(v \\minus{} w)^2}{u(v \\plus{} w) \\plus{} v^2 \\plus{} w^2}\\Big[2u^5 \\plus{} 3u^4(v \\plus{} w) \\plus{} 2u^3(v^2 \\plus{} 15vw \\plus{} w^2)$\r\n\r\n$ \\plus{} u^2(v \\plus{} w)(9v^2 \\plus{} 89vw \\plus{} 9w^2) \\plus{} u(14v^4 \\plus{} 117v^3w \\plus{} 198v^2w^2 \\plus{} 117vw^3 \\plus{} 14w^4)$\r\n\r\n$ \\plus{} 2(v \\plus{} w)(v^2 \\plus{} 5vw \\plus{} w^2)(3v^2 \\plus{} 5vw \\plus{} 3w^2)\\Big]\\geq0.$\r\n\r\n\r\n[b]Remark[/b] This cool substitution see also : http://www.mathlinks.ro/viewtopic.php?t=169195\r\n\r\n\r\nBy the way, let $ a,b,c$ are nonnegative numbers such that $ bc \\plus{} ca \\plus{} ab \\plus{} kabc \\geq 3 \\plus{} k,$ then \r\n\r\n$ \\frac {1}{(1 \\plus{} a)^2(b \\plus{} c)} \\plus{} \\frac {1}{(1 \\plus{} b)^2(c \\plus{} a)} \\plus{} \\frac {1}{(1 \\plus{} c)^2(a \\plus{} b)}\\leq \\frac {3}{8}$\r\n\r\nholds if and only if $ k\\geq \\frac {\\sqrt [3]{332119 \\plus{} 5400\\sqrt {3945}}}{27} \\minus{} \\frac {1679}{27\\sqrt [3]{332119 \\plus{} 5400\\sqrt {3945}}} \\minus{} \\frac {47}{27} \\equal{} 0.79199\\cdots.$", "Solution_6": "[quote=\"nguoivn\"]Given $ a, b, c > 0$ and $ abc \\equal{} 1$. Prove that\n\\[ \\frac {1}{(1 \\plus{} a)^2(b \\plus{} c)} \\plus{} \\frac {1}{(1 \\plus{} b)^2(c \\plus{} a)} \\plus{} \\frac {1}{(1 \\plus{} c)^2(a \\plus{} b)} \\le \\frac {3}{8}\\]\nIt exist a nice proof by Am-Gm :)[/quote]\nSee here: http://canhang2007.wordpress.com/2009/12/03/inequality-86-t-q-anh/\n[quote=\"jingjun\"]$ \\sqrt {97}$ is ugly.But the following inequality is nice:\n\nIf $ a,b,c > 0$ and $ bc \\plus{} ca \\plus{} ab \\plus{} abc \\equal{} 4,$ then \n\n$ \\frac {1}{(b \\plus{} c)(1 \\plus{} a)^{2}} \\plus{} \\frac {1}{(a \\plus{} c)(1 \\plus{} b)^{2}} \\plus{} \\frac {1}{(a \\plus{} b)(1 \\plus{} c)^{2}}\\leq \\frac {3}{8}.$\n[/quote]\r\nSee here: http://canhang2007.wordpress.com/2009/12/03/inequality-87-unknown-author/" } { "Tag": [], "Problem": "Hi,\r\n\r\nI realized that the second Russian team participates in Bulgarian NMO, same for Romania (but 12 grades are excluded). China participates in the Russian NMO. And the fourth German IMO team participates in the China TSTs. :D Just kidding. But what else do you know about such traditions to take part in other national competitions ? :)", "Solution_1": "Moldavian team participates in Romanian NMO. This year second Bulgarian team participated there too.", "Solution_2": "This is the first year in a long time when Canadians were allowed to officially write the USAMO.", "Solution_3": "Yes, the Moldovian team always participates in the Romanian NMO since 1994 (or even earlier). We (the Romanians) used to send a team to Moldovian NMO (that was untill 2000): the first students in Bucharest, that didn't qualify for the National Olympiad would go to the Moldovian NMO. \r\n\r\nI belive that in the next years this kind of \"exchange\" programs between NMOs will become something normal, between most of the leading IMO countries (and not only).", "Solution_4": "It's indeed a tradition for China and Russia participating at each other's MO.\r\nIn CMO2004 the Russian team got 1 gold and 5 silver.", "Solution_5": "How many golds are usually awarded at CMO2004?\r\nAnd what was the gold cut-off this year?" } { "Tag": [], "Problem": "I looked @ collegeboard's practice sat 2 math and it seemed pretty easy, as long as you covered the material. \r\n\r\nFor those people who took the SAT 2s math, on scale of 1-10 and the sat regular math is a 3, how difficult is the sat 2 math?\r\n\r\nAre there any other places i can get practice problems? I searched google for a while and no luck", "Solution_1": "hmm... i would say a 5 (using your relative scale and guesstimating for you. for me it was like a 1) then again, don't trust me since i walked in blind and walked out with an 800.\r\n\r\nbut seriously don't base everything on my opinion as mine may not correspond to yours.", "Solution_2": "anyone else? tt", "Solution_3": "this may be in a math forum, but you're in the wrong place for opinions on this.\r\n\r\ntry Yahoo! Answers or something.", "Solution_4": "If you remember concepts well, then math SAT Subject Test should be a breeze\r\n\r\nThe trickiest stuff on there are like... sums of geometric series.\r\nTo the average math class student, that type of formula is forgotten quickly, but if you remember things like that, then you'll do fine.\r\n\r\njust make sure you watch the clock", "Solution_5": "If you look at the average range of SAT scores the average (well reputed) college accepts, the SAT II math score average is generally significantly higher than the regular SAT math score average. I know many people who took the SAT II and got an 800, but got less than that on the regular SAT math. With this in mind, I would say that the SAT II is much easier. Note that the problems might not be that much easier, but scoring is more lenient." } { "Tag": [ "vector", "linear algebra", "matrix", "algebra", "system of equations", "linear algebra unsolved" ], "Problem": "1. Can any of you tell me what an [i]affine space[/i] is? A formal definition is nowhere to be found on the Internet...\r\n\r\n2. Is the solution set of an $ m \\times n$ non-homogeneous system of equations an examples of an affine space? What would its dimension be?\r\n\r\nI look forward to hearing from you soon.\r\n\r\nBest Regards.", "Solution_1": "See the [url=http://en.wikipedia.org/wiki/Affine_space]Wikipedia article[/url].", "Solution_2": "Thanks for taking the time to reply, comrades.\r\n\r\nFrom [b]LkNsngth[/b]'s post I deduce that the first question in 2 has a negative answer. The [b]wikipedia[/b] article makes me think that the answer must be afirmative, though... What's going on here?\r\n\r\nAlso, I'm still doubtful about the dimension-[i]thingy[/i]. How could I go about calculating it?\r\n\r\nSalutations.", "Solution_3": "i accidentally posted my answer to this in a new topic, here is the link to it http://www.mathlinks.ro/viewtopic.php?t=233136 (sorry).", "Solution_4": "The solution to a set of linear equations is a translate of a vector space (the nullspace of the corresponding matrix) and so can be given the structure of an affine space. Its dimension is the dimension of the corresponding nullspace, or the nullity of the corresponding matrix.", "Solution_5": "[quote=\"t0rajir0u\"] Its dimension is the dimension of the corresponding nullspace, or the nullity of the corresponding matrix.[/quote]\r\n\r\nEven if the system is non-homogeneous? \r\n\r\nThanks again for taking the time to reply.", "Solution_6": "Yes. If the system $ \\mathbf{Ax} \\equal{} \\mathbf{b}$ has a solution $ \\mathbf{x} \\equal{} \\mathbf{x}_0$, then the set of all solutions takes the form $ \\mathbf{x} \\equal{} \\mathbf{x}_0 \\plus{} \\mathbf{v}$ where $ \\mathbf{v}$ is in the nullspace of $ \\mathbf{A}$ - like I said, the solution set is a translate of the nullspace. There are two senses in which a translate of a vector space can be given a meaningful dimension:\r\n\r\n- as an affine space (it has the same dimension as the vector space, which I believe is by definition), or\r\n\r\n- as an [url=http://en.wikipedia.org/wiki/Algebraic_variety]algebraic variety[/url]. The notion of dimension here is much more subtle, but the two concepts agree in this case.", "Solution_7": "Things are much clearer now, esteemed t0rajir0u.\r\n\r\nThanks for caring, man! :lol:" } { "Tag": [ "logarithms", "real analysis", "real analysis solved" ], "Problem": "Let $x_n$ be a sequence that converges to 0 and such that $| n(n+1)x_{n+1}-n^2 x_n|$ is bounded. Does it follow that $nx_n$ converges?", "Solution_1": "No, $n x_n$ can diverge. For example, define $x_n = \\frac{\\ln n}{n}$.", "Solution_2": "[quote=\"harazi\"]Let $x_n$ be a sequence that converges to 0 and such that $| n(n+1)x_{n+1}-n^2 x_n|$ is bounded. Does it follow that $nx_n$ converges?[/quote]\r\n\r\nIt says $|u_{n+1} - u_n| = O(1/n)$ where $u_n = nx_n$, so that $u_n = O( \\log n )$ and any asymptotic behavior of $u_n$ consistent with this estimate can be realized." } { "Tag": [ "limit", "LaTeX", "algebra unsolved", "algebra" ], "Problem": "Given 2 real numbers $x\\ge 0$ and $\\alpha\\ge 1$\r\nDefine the sequence $\\{a_{n}\\}$ as follow : $a_{1}=x \\ ; \\ a_{n+1}=1+\\frac{a_{n}^{2}}{n^{\\alpha}}\\ \\forall n\\ge 1$\r\n1/ Prove that $\\exists \\delta=\\delta(\\alpha)$ such that \r\n if $x<\\delta$ then $\\lim_{n\\rightarrow+\\infty}a_{n}=1$ \r\n if $x\\ge \\delta$ then $\\lim_{n\\rightarrow+\\infty}a_{n}=+\\infty$\r\n2/ Prove that $\\delta(\\alpha)$ is increasing", "Solution_1": "[quote=\"Stun\"]Given 2 real numbers $x\\ge 0$ and $\\alpha\\ge 1$\nDefine the sequence $\\{a_{n}\\}$ as follow : $a_{1}=x \\ ; \\ a_{n+1}=1+\\frac{a_{n}^{2}}{n^{\\alpha}}\\ \\forall n\\ge 1$\n1/ Prove that $\\exists \\delta=\\delta(\\alpha)$ such that \n if $x<\\delta$ then $\\lim_{n\\rightarrow+\\infty}a_{n}=1$ \n if $x\\ge \\delta$ then $\\lim_{n\\rightarrow+\\infty}a_{n}=+\\infty$\n2/ Prove that $\\delta(\\alpha)$ is increasing[/quote]\r\n\r\nWhat does the backwards E notation mean?", "Solution_2": "Look at the LaTEx code, it meas exists. :wink:", "Solution_3": "can anyone give a solution ? :maybe:" } { "Tag": [ "trigonometry", "geometry", "AMC", "AIME", "USA(J)MO", "USAMO", "MATHCOUNTS" ], "Problem": "Is trig helpful for AMC [b]10[/b]? \r\n\r\nthank you", "Solution_1": "AMC 10 typically does not cover trigonometry, but AMC 12 does.", "Solution_2": "It IS helpful tho", "Solution_3": "I asked this a bit ago. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=234122", "Solution_4": "A Question: \r\n\r\nIs trig needed for AMC 12?\r\n\r\nIf so then what specific areas should you be familiar with?", "Solution_5": "Yes, it's asked in the AMC 12.\r\nI don't quite get what you mean by which areas. Just be familiar with them all I guess, although I personally don't know too much the part that deals with graphs.\r\n\r\nPerhaps you should look at the amc faq?", "Solution_6": "Trig could be helpful for some of the harder geo problems in AMC 10, I believe.\r\n(Such as the area of a triangle formula that uses sin)", "Solution_7": "So would a thorough read of the trig in Volume 2 be sufficient?", "Solution_8": "Probably...but I think the basic trig in Intro to Geo and AoPS Volume 1 would help just enough, but if you want more clarification on trig, or better methods, AoPS Volume 2 surely could help.", "Solution_9": "i would suggest getting Plane Trigonometry by S.L. Loney. Its a classic and explains concepts very well.", "Solution_10": "Trigonometry in AMC-12 is such a broad topic. It can mean simple computation of sine, cosine, and tangent of triangles to complex numbers and use of binomial theorem with DeMoivre's. So technically, anything that is beyond algebra II on AoPS volume II and is not part of NT can be considered as \"area\" of trigonometry.\r\n\r\nFor AMc-10 and trigonometry, I think it helps depending on the person. When I was taking AMCs, I hated geometry and solved almost all geometry with trigonometry (with calculator too but that's not for you folks anymore! :) ). General idea can even help to give you what kind of construction is needed so definitely, keep in mind but you don't have to use it.\r\n\r\nWhen my academics settle down bit, I need to combine the advices and tips for studying AMCs because this type of question arises every year about the same time.", "Solution_11": "no problem in AMC 10 will REQUIRE trig. but trig. can be helpful on some problems.(2008 AMC 10 #25, which is also 2008 AMC 12 #22).", "Solution_12": "[quote=\"cognos599\"]no problem in AMC 10 will REQUIRE trig. but trig. can be helpful on some problems.(2008 AMC 10 #25, which is also 2008 AMC 12 #22).[/quote]\r\n\r\nYeah, the trig solution was easier, but the other didn't require it.", "Solution_13": "Knowing anything is helpful in doing anything.", "Solution_14": "[quote=\"diophantient\"]Knowing anything is helpful in doing anything.[/quote]\r\nThis is not true. I suggest to you, diophantient, not to make such claims before carefully evaluating them.\r\nUsing the word \"anything\" twice, namely, is the mistake in the statement.\r\nCounterexample: If you are asked to evaluate $ 1 \\plus{} 2 \\plus{} 3 \\plus{} \\cdots \\plus{} 10$, then will knowing the fact that $ \\frac {a \\plus{} b}{2}\\ge \\sqrt {ab}$ for positive integers $ a,b$ be helpful? I'm sure that some crazy solution can be found using that fact, but it is certainly not \"helpful\" to use it.\r\n\r\nI think what you are trying to say is that one should be familiar with any topic in mathematics in order to get an extra edge on contests such as the AMCs. That is wrong, too. I'm sure that knowing something like Fermat's Last Theorem will not be helpful to you on the AMCs (I'm not talking about AIME, USAMO, and beyond, just AMC).\r\n\r\nWhat is correct is to say that it can be helpful to know certain topics in order to solve contest problems such as those on the AMCs, even if the topics aren't necessary to know in order to solve the problems. This is certainly true for trigonometry: trigonometry is extremely usable on AMC and even MATHCOUNTS problems (and helpful, in that it provides a faster solution than non-trigonometry methods).\r\nExample: Find the area of a triangle with two side lengths of $ 12$ and $ 15$ and the included angle of those two sides having an angle measure of $ 120^\\circ$.\r\n\r\nThis type of problem actually does show up on MATHCOUNTS (and maybe AMC, although I'm not quite sure) problems, and it is extremely easy to solve with the formula $ A \\equal{} \\frac {1}{2}ab\\sin C$, where $ A$ is the area of a triangle with two side lengths of $ a$ and $ b$ and the included angle of those two sides having an angle measure of $ C$ radians (or $ C\\cdot\\frac {180}{\\pi}$ degrees).\r\nOther problems (for example, the one mentioned by cognos599) on the AMC 10 can be solved quicker with trigonometry.\r\n\r\nSo, to make a long answer short, yes, trigonometry is helpful for the AMC 10." } { "Tag": [ "complex analysis", "function", "limit", "complex numbers", "topology", "real analysis", "algebra" ], "Problem": "What are the reasons why normal proving techniques have failed when it comes to the riemann hypothesis? Please post specific attempts/failed techniques if you can.", "Solution_1": "I have no clue what a \"normal\" proving technique is.\r\nThe problem is difficult, that's all.", "Solution_2": "[quote=\"ZetaX\"]I have no clue what a \"normal\" proving technique is.\nThe problem is difficult, that's all.[/quote]\r\n\r\nI would consider any technique that has been used in the past to prove something, normal.", "Solution_3": "The most obvious approach is to prove that the reciprocal $ 1/\\zeta(s)$ is holomorphic in the half-plane $ \\re s > \\frac12$. There is a [i]formal [/i]Dirichlet series for it; all that's left to prove is that the series converges. :P \r\n\r\nThe book \"Scattering theory for automorphic functions\" by Lax and Phillips has a section \"How not to prove the Riemann Hypothesis\".", "Solution_4": "This will be a silly question. However, could anybody tell me what exactly is Riemann's Hypothesis asking? I checked Wiki and only find unexplained nouns as zeta-function.", "Solution_5": "For some $ s$ (namely those complex $ s$ with real part $ >1$) the sum $ \\zeta(s) : \\equal{} \\sum_{n\\equal{}1}^{\\infty} \\frac{1}{n^s}$ converges. A bit more checking gives, that for those $ s$ it is complex differentiable (so $ \\lim_{h \\to 0} \\frac{\\zeta(s\\plus{}h)\\minus{}\\zeta(s)}{h}$ exists, where $ h$ can be complex).\r\nNow there is a fundamental theorem from complex analysis: Let $ U \\subset \\mathbb C$ have an accumalation point (a point where infinitely many elements of $ U$ are; e.g. if $ U$ is a nonempty open set); If two complex differentiable functions $ f,g$ are defined on connected sets $ X,Y \\subset \\mathbb C$ and agree on $ U \\subset X,Y$, then $ f$ and $ g$ are the same on all of $ X\\cap Y$. As a result, there is at most one complex differentiable function that continues a given such function to a connected set $ X$ (note that this is essentially different from real analysis where e.g. a function on an interval can be extended to the reals in many ways).\r\nNow there exists a complex differentiable function $ \\zeta$ that is defined on $ \\mathbb C \\backslash \\{1\\}$ and agrees with that function mentioned above. And by that theorem, this function is the only such. We call it the Zeta-function.\r\nIt has some properties (e.g. a functional equation relating $ \\zeta(s)$ to $ \\zeta(1\\minus{}s)$) and has some \"trivial\" zeros at negative even integers (and no other real zeros). The Riemann hypothesis says: all other zeros have real part $ \\frac 12$.", "Solution_6": "One can avoid the discussion of analytic continuation by considering $ (1\\minus{}2^{1\\minus{}s})\\zeta(s)\\equal{}\\sum_{n\\equal{}1}^{\\infty}\\frac{(\\minus{}1)^{n\\minus{}1}}{n^s}$. This series converges in the half-plane $ \\mathrm{Re}\\,s>0$, and the hypothesis is that its sum is not equal to zero unless $ \\mathrm{Re}\\,s\\equal{}\\frac12$." } { "Tag": [ "floor function" ], "Problem": "(1) If $ \\frac {a!}{b!}$ is a multiple of $ 4$ but not a multiple of $ 8$, then what is the maximum value of $ a \\minus{} b$?\r\n\r\n(2) Less than $ 50$ people are at a party. Each person shakes everyone else's hand. If there is an odd number of total handshakes at the party, what is the largest number of people that could be at the party?\r\n\r\nThanks a lot!", "Solution_1": "[hide=\"1\"] We assume $ a\\ge b$ (or else $ \\frac{a!}{b!}\\not\\in\\mathbb{Z}$). So, $ \\frac{a!}{b!}\\equal{}a(a\\minus{}1)(a\\minus{}2)\\cdots(b\\plus{}1)$. Notice that there are $ a\\minus{}b$ terms in this product, so what we are maximizing in the number of terms. We claim that the maximum number of terms is $ 3$, which is achieved when $ a\\equal{}4t\\plus{}1,b\\equal{}4t\\minus{}2$. Suppose there was a greater number of terms that satisfies the conditions. Notice that $ b\\plus{}1,b\\plus{}2,...,a\\minus{}1,a$ are a sequence of $ a\\minus{}b$ terms where $ a\\minus{}b\\ge4$. Thus, there is a multiple of $ 4$ among them. This multiple of $ 4$ lies in one of the halves of the sequence, each of which has at least $ \\lfloor\\frac{a\\minus{}b}{2}\\rfloor\\ge2$. Thus, in the other half there are still left at least 2 consecutive integers. Thus one of them must be a multiple of $ 2$ and so the total product is divisible by $ 8$.\n[/hide]\n\n[hide=\"2\"] The total number of handshakes is $ \\binom{n}{2}$ where $ n$ is the number of people. This is equal to $ \\frac{n(n\\minus{}1)}{2}$. For this to be odd, we must have $ n$ or $ n\\minus{}1$ leave a remainder of $ 2$ when divided by $ 4$. Thus, the greatest amount of people is $ 47$.[/hide]", "Solution_2": "I don't get how this first statement is made:\r\n\r\n[quote]We assume $ a\\ge b$ (or else $ \\frac {a!}{b!}\\not\\in\\mathbb{Z}$). So, $ \\frac {a!}{b!} \\equal{} a(a \\minus{} 1)(a \\minus{} 2)\\cdots(b \\plus{} 1)$.[/quote]", "Solution_3": "$ \\frac{a!}{b!}\\equal{}\\frac{a(a\\minus{}1)(a\\minus{}2)\\cdots(b\\plus{}1)b(b\\minus{}1)(b\\minus{}2)\\cdots3\\cdot2\\cdot1}{b(b\\minus{}1)(b\\minus{}2)\\cdots3\\cdot2\\cdot1}\\equal{}a(a\\minus{}1)\\cdots(b\\plus{}1)$" } { "Tag": [ "AoPS Books", "LaTeX" ], "Problem": "What is (are) they? They look nice. I want to use them :) .", "Solution_1": "I know this is an old thread, but I'd like to answer my own question in case anyone else wants to know. The AoPS books use Palatino (at least Volume 1 & 2 do, I don't have the other books), as well as the matching math font.\r\n\r\n[code]\\usepackage{mathpazo}[/code]\r\n\r\nIn my opinion, Palatino looks better than the default Computer Modern font. (The latter places heavy emphasis on the vertical strokes (imitates a horizontal calligraphic pen); the former has the thickest lines on a slightly offset angle (imitates a tilted calligraphic pen).)" } { "Tag": [ "trigonometry" ], "Problem": "$\\alpha=\\frac{\\pi}{7}$\u3068\u3059\u308b\u3002\u3053\u306e\u3068\u304d\u3001\u6b21\u306e\u7b49\u5f0f\u3092\u8a3c\u660e\u305b\u3088\u3002\r\n\\[ \\frac{1}{\\sin{\\alpha}}= \\frac{1}{\\sin{2\\alpha}} +\\frac{1}{\\sin{3\\alpha}} \\]", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=54521" } { "Tag": [ "ratio", "geometry proposed", "geometry" ], "Problem": "Let $ABC$ be an isosceles triangle at angle $A$ and $\\hat{A}= 36^{o}$.Prove that the ratio $\\frac{BA}{BC}$ is irrational.", "Solution_1": "The triangle can be seen in a regular pentagon.\r\n\r\nRemark the golden ratio, we have ${\\frac{BA}{BC}=\\frac{\\sqrt{5}+1}{2}}.$" } { "Tag": [], "Problem": "In a party of $n$ persons, every set of $k$ persons have exaclty one common friend. What is the number of friends of the person with most friends.", "Solution_1": "PLEASE A MODERATOR DELETE THIS POST. THIS PROBLEM IS ALREADY POSTED IN THE COMBINATORICS SECTION.. THANKS" } { "Tag": [ "linear algebra", "matrix", "vector", "linear algebra unsolved" ], "Problem": "Diagonalize the matrix B = \r\n\r\n$ \\left(\\begin{array}{cc}{acc}\r\n1 & 2 \\\r\n3 & 4 \\\r\n\\end{array}\\right)$\r\n\r\nby a change of basis.\r\n~~~\r\n\r\nI don't know how to diagonalize \"by a change of basis.\" How do you do that?\r\n\r\nI tried to find the determinant of $ \\lambda I - B$. I got real, but ugly, eigenvalues of $ \\frac{5 \\pm \\sqrt{33}}{2}$. I tried to find the eigenvector for $ \\frac{5 + \\sqrt{33}}{2}$, but I wasn't successful. (I got the zero vector.)", "Solution_1": "Let's try the eigenvalue $ \\lambda=\\frac{5+\\sqrt{33}}2.$ We want to find the null space of \r\n\r\n$ B-\\lambda I=\\begin{pmatrix}\\frac{-3-\\sqrt{33}}2&2\\\\\r\n3&\\frac{3-\\sqrt{33}}2\\end{pmatrix}.$\r\n\r\nNow, I'm not going to spend any time row reducing this. If we've done things right, it is singular, and after subtracting the appropriate multiple of the the top row, the second row should be zero. Now find some vector such that the top row times it is zero. We can just rearrange what we see to get\r\n\r\n$ x=\\begin{pmatrix}2\\\\ \\frac{3+\\sqrt{33}}2\\end{pmatrix}.$\r\n\r\nAt this point, I will do one check:\r\n\r\n$ \\begin{pmatrix}3&\\frac{3-\\sqrt{33}}2\\end{pmatrix}\r\n\\begin{pmatrix}2\\\\ \\frac{3+\\sqrt{33}}2\\end{pmatrix}=\r\n6+\\frac{9-33}4=0.$\r\n\r\nIt checks. This is our eigenvector.\r\n\r\nFor the other eigenvector, we should just be able to just reverse the sign of the square root.\r\n\r\nNow, let $ P=\\begin{pmatrix}2&2\\\\ \\frac{3+\\sqrt{33}}2&\\frac{3-\\sqrt{33}}2\\end{pmatrix}.$\r\n\r\nWe explicitly compute that $ P^{-1}=-\\frac1{2\\sqrt{33}}{\\begin{pmatrix}\\frac{3-\\sqrt{33}}2&-2\\\\ -\\frac{3+\\sqrt{33}}2&2\\end{pmatrix}}$\r\n\r\nThen $ P^{-1}BP=\\begin{pmatrix}\\frac{5+\\sqrt{33}}2&0\\\\ 0&\\frac{5-\\sqrt{33}}2\\end{pmatrix}.$\r\n\r\nThere were a couple of arbitrary choices in that process; it would be possible for a correct answer to look a little different.", "Solution_2": "Thank you!" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f: [a,b]\\minus{}> R$ a function continuos on $ [a,b]$ and derivable on $ (a,b)$ with $ f(a)\\equal{}f(b)$ .\r\nProve that exist $ c_{1} , c_{2} \\in (a,b)$ s . t . \r\n$ af^{\\prime} (c_{1}) \\plus{}bf^{\\prime} (c_{2}) \\equal{}0$", "Solution_1": "What's wrong with taking $ c_1\\equal{}c_2\\equal{}c$ where $ f'(c)\\equal{}0$ (such $ c$ exists according to the standard Rolle's theorem). I'm pretty sure you meant something else, but what? :?" } { "Tag": [ "Support", "function", "geometry" ], "Problem": "Gabriel A. Moens, from Brigham Young University Law Review once stated \u201cthe idea that church and state have different roles in society and that respectful separation between them should be maintained is essentially a good idea.? Because I agree with those words, I must negate the resolution: Democracy is best served by a strict separation of church and state. \r\n\r\n\r\n\r\nPoint of Clarification: This resolution must first be established in the United States before we go beyond its borders. If we try to worry about every democracy, we interfere with laws set up by separate countries.\r\n\r\nV: Prosperity, in that people are more prosperous when religion is able to be valued by a governments citizens.\t\r\nVC: Democracy, because when we value support religious beliefs, we help the success of our democratic governments. This makes them more prosperous.\r\n\r\nIf a man can survive and prosper without something, then that object isn\u2019t that important. This leads me to my first contention.\r\n\r\nContention I: Democracies can function and flourish without a strict separation of church and state. Great Britain, a democratic country has civil rights, an effective economic system and working parliament. But it also has a state church, which is the Anglican Church. Yet England is a great example of a supreme democratic nation. France, Spain, and England were all religious nations when they abolished slavery, which was before the United States did. India, another example, is a well functioning democracy. It has a majority of Hindus. India\u2019s state-church interaction benefits the lower caste population that\u2019s being discriminated by the Hindu religion by granting them money. Indonesia, another successful democratic country with a predominant religion of Muslim had a female prime minister. Even though under Islamic Laws, women are placed inferior to men, a woman led an Islamic nation before. Social Welfare, greatly valued in democracies can be hindered by a strict separation of church and state. Religious organizations best supports social welfare, especially in the area o f addiction treatment. Faiths-based organizations often have a higher success rate then secular treatments. If a strict separation were in place, it would have denied public access to such organizations. \r\n\r\nContention II: A strict separation of church and state doesn\u2019t guarantee and can restrict freedoms. In some cases, Native Americans were allowed to continue to use a hallucinogenic drug called peyote. Had a strict separation been in place, the Native Americans would not have been granted this right. The same pertains to the nuns working for free in hospitals instead of being paid minimal wage. Exceptions are often made for religious reasons. A strict separation also hinders an individual\u2019s rights to expression. In France, public schools banned the wearing of any religious symbols. Headscarves are a predominant symbol in Muslim faith, but the schools thought by banning them; they were giving the Muslim girls more freedom from their religion. However, the Muslim girls protested, declaring that by banning a harmless religious symbol, they were being discriminated. Once again, we must realize that by a strict separation of church and state, anything somewhat religious in any public school and government official building would be banned. This brings us to my third contention.\r\n\r\nContention III: This resolution is severely detrimental to the education process. \r\nAs stated in the definition provided, strict separation is having no correlation. Schools are one form of a governmental institution, hence the federal funding, state funding, et cetera. By such a regulation as the resolution proposes, classes in public schools would be harmed. \r\nSub point A: Classes such as World History would be harmed, because much of what is taught comes from religious based events, such as Egyptian study, Roman Empire, the Greek civilization, and England's empire. \r\nSub point B: Many other classes would be harmed as well, including Speech and Debate. Under this resolution, we could not even discuss this type of topic, because it brings religious views into a governmental institution, and would be violating the resolution by affirming it.\r\n\r\nA strict separation of church and state not only hinders individual rights, it is also not necessary to best serve democracy. When a strict separation is in place, the government is unable to support any religious groups, thus denying charities from helping the less fortunate. For these reasons and more, I urge you to vote negative.", "Solution_1": "Good Case. Comments:\r\n\r\nI don't agree with the observation. The resolution is not limited to the United States, and without proving why it's important to establish it here, it's not an observation but rather a contention - not to mention a contention which shouldn't belong.\r\n\r\nI don't think you should run democracy as a criterion to judge prosperity as that's contendable. Democracy has not always been the most prosperous type of government in history. Use it as a value if anything, and you must specify what aspects of democracy you are refering to.\r\n\r\nI'd rephrase your first contention: Democracies flourish under non-strict separation of church and state. \r\n\r\nYou must be aggressive and use attacking arguments. Don't go with the neutral statement.\r\n\r\n ;) Where did you get your second contention from? Honestly.\r\n\r\nAs an AFF, my definition would be a separation of church and state as a separation of religious and state institutions, as defined by Wikepedia and many other dictionaries. The resolution does not specifically refer to religion itself, only that religion and state and made separate entities which do not directly impact each other. \r\n\r\nYou prove that strict separation is bad, but what if the AFF goes on the offensive? You haven't clearly stated why non-strict is necessarily better.\r\n\r\nOtherwise, very good case.", "Solution_2": "I believe it's a fairly good case. :P \r\n\r\nBut even so, there's a few problems I see. The greatest is probably the fact that most of what you're doing is presenting examples instead of offering arguments. Debate is about your ability to argue, not find facts. I must say though, it is a very interesting approach. I generally see people with not enough \"tangible proof\" in LD, instead of too much. It's also kind of short. I'm not sure what time limits you compete under, but you should always try to hit within 10 seconds of your full time on the first speech. In my non-expert opinion, that just shows that you actually do have a lot to say and back up your case with. Having 8 minutes and making a 2 minute speech would definitely not be something I want to do too often, unless of course I'm on NEG and the AFF case that just got presented was absolutely horrendous.\r\n\r\nOh yeah...contention III doesn't need to be there. I see absolutely no way in which it actually benefits your case, and heck...I'd even argue it's kind of off-topic.", "Solution_3": "You could argue your contention three, but not on the grounds your presenting.\r\n\r\nAlso, it's a NEG MithsApprentice, so I think it is of decent length.\r\n\r\nHe must reserve at least 3 minutes for rebuttal.", "Solution_4": ";) You post your case up here for us to critique it so I'm going to be honest.\r\n\r\n With this case, you will not qualify into whatever you are trying for. Is this for states? I know it's not for nationals because I can tell you are a fairly new LD debater just from your case. I can tell you now, you have huge wholes in your case that need to be patched up.\r\n Here in Hawaii, we just had N.I.T on another island, Kauai, and the day I left I heard a fellow LDer going to the same debate as I was. Her case was very similar to yours and like I predicted, she didn't even make it through the 5th round. (If you haven't figured, I am a high school LD Debater from Kahuku High School in Hawaii on the island of Oahu. I am currently a senior.)\r\n Lets help you patch up you case. First of all, you do know that the U.S is not a Democracy...don't you? We are a Republic. Why? We have representatives. \"...to the Republic for which it stands\" an example for you from our own Pledge of Allegiance.\r\n Secondly, your case is not really leading to the point of you value and value criterion. Besides, Democracy as a VC? Democracy is a very weak bases and usually used as a value. How about the Value \"Societal Welfare\" and VC \"Status Quo\"? It is easily linked to EVERYTHING in Neg and also you can make arguments (which you really need because I see no arguments whatsoever).\r\n Moving on to your contentions. Awesome, Awesome Contention 3, but know EVERYTHING with each country you mention or else your opponent can severely rip your case apart. Contention 2 gives your opponent the chance to take the win. I say scratch the whole C2. No matter what religion or culture says peyote is okay, it is still an illegal drug and NO ONE can get away with it. Native Americans should be able to smoke peyote cause its apart of their religion but other people who smoke it for personal needs cannot? That is the biggest whole in your case. No matter what, whether there is a strict separation or not, what is illegal is illegal! No bends, no turns, no leaning for ANYONE. I must admit though, the muslim girls and headscarves, very good arguements. Maybe even faith-based programs would be good. For example, Alcoholics Anonymous or Noacrotics Anonymous. Maybe for the criterion I gave you, a good contention would something like \"Religious organizations can assist the government's achievement of societal welfare\". Just a suggestion. Okay now your Contention 3. WoW! You either need to drop it or you need to work on it and go deeper into it. Tell WHY it is important and HOW it would be dtrimental in the EXACT way.Education process, not a very good agrument topic on you part. Very good for your opponent to poke holes in. How about something like \"The resolution is severely detrimental to society\". You can say that 'by strictly separating Church and state,your opponent is advocating no intervention by government. By allowing acts such as; Allowing people to torturine themselves or others at their consent as a religious rite to show dedication to their God, every criminal will be commiting crimes in the name of their God. The federal government will not have the power to interfere. Thus putting society at risk'. Just another suggestion.\r\n You do have a really awesome case. It just that you have many big topics that have a strong Affirmative side to it that could kill your case IF your opponent was smart and knew what they were doing.\r\n You have an awesome quote. I'm Mormon also, in fact I am on a B.Y.U-Hawaii computer right now. (My Grandfather is a professor here). Keep up the good work and stick with LD. Just be confident and stay cool, calm, and collected. Good luck! If you need any help or there is anything I can do for you, just e-mail me. Waugh@hawaii.edu", "Solution_5": "If you want to take a look at my case, you can e-mail me and i'll e-mail it back to you. You can take ideas from it but I would advise you not to copy it straight because their might be things that you dont really know about and will cause you to lose. But you can always use it as just an outline. :lol: Okay. Aloha.[/quote]" } { "Tag": [ "algebra solved", "algebra" ], "Problem": "Find all $: R \\to R : \\ \\ f(x^2+f(y))=y+f^2(x)$", "Solution_1": "Firstly if $f(x)=f(y)$ then $f(a^2+f(x))=f(a^2+f(y))$\r\nFrom here $x+f(a)^2=y+f(a)^2 \\Longrightarrow x=y$\r\nWe get that $f(x)$ is injective.\r\nTaking $x=0$ we get $f(f(y))=y+f(0)^2$ $(1)$\r\nSo there exists $t\\in R$ such that $f(t)=0$\r\nTaking $x=y=t$ we get $f(t^2)=t \\Longrightarrow f(f(t^2))=f(t) \\Longrightarrow t^2+f(0)^2=0$\r\nSo $t=f(0)=0$\r\nTaking $y=0$ we get $f(x^2)=f(x)^2=f(-x)^2$\r\nFrom injectivity of $f$ we get $f(-x)=-f(x)$\r\nTaking $y=0$ we get $f(x^2)=f(x)^2$.So $f(x)\\ge 0$ if and only if $x\\ge 0$ $(2)$\r\nTaking $y=-x^2$ we have $f(x^2+f(-x^2))=-x^2+f(x)^2$\r\n$f(x^2-f(x^2))=f(x)^2-x^2$\r\nFrom $(2)$ we get that $f(x)=x$\r\nThat's all ;)" } { "Tag": [ "LaTeX" ], "Problem": "If [i]a [/i]and [i]b [/i]are distince, odd primes, then how many distinct positive factors does [i]4a(squared)b(cubed)[/i] have.\r\n\r\nThanks", "Solution_1": "[hide]Ok since 4=2^2. Then you have 2^2*a^2*b^2. Using this you add one to each exponent and you can find 3,3,3. You multiply the exponents so it would be 3*3*3. The answer is 27.[/hide]", "Solution_2": "[hide]The number is $2^2a^2b^3$. Each factor can be divided by $2^0,\\ 2^1,\\ 2^2$, it can be divided by $a^0,\\ a^1,\\ a^2$ and so on. Thus there are $3\\times 3\\times 4=36$ factors because every factor is equal to $2^xa^yb^z$ where $x$ can be 0, 1, or 2, and so on.[/hide]\r\n\r\nI would recommend learning and using LaTex (like above)." } { "Tag": [], "Problem": "Determine if $ x^2\\plus{}xy\\plus{}y^2\\equal{}2$ has a solution in $ \\mathbb{Q}$.", "Solution_1": "Solution.\r\n[hide]\nWe want to find solutions of $ m^2 \\plus{} mn \\plus{} n^2 \\equal{} 2k^2$ in integers (and $ k\\neq 0$).\n\nLet $ (m,n,k)$ - solution with minimum value of $ |k|$ ($ k\\neq 0$).\nIf $ m,n$ both odd, then left side is odd.\n\nIf one of $ m,n$ is even (for example, $ m$), then $ m^2 \\plus{} mn$ is divisible by 2, so $ n^2$ is divisible by 2. So, we have $ m,n$ - even. Thus, $ m^2 \\plus{} mn \\plus{} n^2$ is divisible by 4. It means that $ 2k^2$ is divisible by 4. So, $ k$ is even.\n\n$ m,n,k$ - even. So, $ (\\frac {m}{2},\\frac {n}{2},\\frac {k}{2})$ - solution too. But we chose solution $ (m,n,k)$ with a minimum value $ |k|$. Contradiction.\n[/hide]" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $S_{XYZ}(l)$ be the set of all $T$-s that $TX^2+TY^2+TZ^2=l$, where $l$ is a constant (of course it is big enough). \r\nNow imagine a triangle $ABC$ and points $A'$ and $C'$ inside the $\\angle ABC$, which satisfy to following condidtions: $AA'=CC'$ and $AA'||CC'$. Prove that for every (big enough) $m$, we can find out a $n$(this is big enough too), which satisfies this: $S_{ABC}(m)=S_{A'BC'}(n)$", "Solution_1": "[quote=\"Limp Bizkit\"]Let $S_{XYZ}(l)$ be the set of all $T$-s that $TX^2+TY^2+TZ^2=l$, where $l$ is a constant (of course it is big enough). \nNow imagine a triangle $ABC$ and points $A'$ and $C'$ inside the $\\angle ABC$, which satisfy to following condidtions: $AA'=CC'$ and $AA'||CC'$. Prove that for every (big enough) $m$, we can find out a $n$(this is big enough too), which satisfies this: $S_{ABC}(m)=S_{A'BC'}(n)$[/quote]\r\nIf $M$ is the medians intersection point of triangle $XYZ$\r\nWe know the following formula \r\n$PM^2=\\frac{1}{9}(3(PA^2+PB^2+PC^2)-(AB^2+AC^2+BC^2))$\r\nSo the set $S_{XYZ}(l)$ is a circle with centre in $M$.\r\nIf $S_{ABC}(m)=S_{A'BC'}(n)$=>the medians intersections points of triangles $ABC$ and $A'BC'$ are same, BUT this condition is wrong, therefore the problem is wrong too :rotfl: .\r\nI think in the correct version of this problem asking to prove that for every (big enough) $m$, we can find out a $n$( big enough too), which satisfies this: $S_{ABC'}(m)=S_{A'BC}(n)$,now anything is true,becaus how we can see the triangles $ABC'$ and $A'BC$ have same medians intersection point ;),we can choose $n$ such that the radius of circles are equal,and the problem is done. \r\n [hide=\"Only for Limp Bizkit\"]Ara Neso xntirt sxalaaaaaaaaaaaaa :rotfl: :rotfl: [/hide]" } { "Tag": [ "Princeton", "college", "MIT", "inequalities", "Putnam", "real analysis", "Gauss" ], "Problem": "Hello. Please post here names of youngest perfect scorers or gold medallist at IMO. Comments would be welcomed . THANKS", "Solution_1": ":? \r\n\r\nThe youngest perfect scorer was the former GDR participant Pavel Kroeger. \r\nWhen he attended 7th grade, he managed this performance.\r\nA student from Australia was the youngest participant. \r\nIf I am not mistaken he was 11 years during his first IMO. \r\nIn the next days I could provide more detailed records.", "Solution_2": ":D \r\nMay be", "Solution_3": "[quote=\"new\"]:? \n\nThe youngest perfect scorer was the former GDR participant Pavel Kroeger. \nWhen he attended 7th grade, he managed this performance.\nA student from Australia was the youngest participant. \nIf I am not mistaken he was 11 years during his first IMO. \nIn the next days I could provide more detailed records.[/quote]\r\n\r\nterry tao (australia) was that australian student. he got gold when he was 13, after which he went to uni i think. if he had continued, he would have easily got about 5 Gold ...", "Solution_4": "now did that tao guy had a \"gold childhood\" or what? :P he didn't score max. points though, so I guess he gaved up pretty quickly :)", "Solution_5": "um, tao didnt do the IMO after he was 13 for some reason :? . (anyway, he was the one that proved we can find an arbitrarily long sequence of primes in a.p.).", "Solution_6": "Terry went off to uni afterwards, making himself ineligible.. (might have been 1 year in between or something though..) \r\nOh yeah. Vinoth, get back to doing problems :-P", "Solution_7": "Terry Tao is the same one involved in the recent proof that there are primes in arithmetic progressions of length k for any natural number k.\r\n\r\nI believe he was the youngest to receive a Gold Medal at IMO. He did so in 1988 at (just barely) the age of 13 - (his 13th birthday was during the week of the competition). He also recieved a Bronze Medal in 1986 and a Silver Medal in 1987.\r\n\r\nRight after his third time at IMO he went to Flinders University of South Australia where he got his Bachelors and Masters degrees at the age of 16. He then went to Princeton University where he got his PhD in 1996 at the age of 20. He now teaches and works at the University of California, Los Angeles.\r\n\r\nIf he had competed in IMO until he was 19 years old it's safe to say that he could have won as many as 7 Gold Medals and 9 Medals in total.", "Solution_8": "What I find funny about [url=http://www.math.ucla.edu/~tao/]Terrence Tao[/url] is that he gives the [url=http://www.math.ucla.edu/~tao/putnam/]Canadian Mathematical Olympiad 1999[/url] as regular course work. :lol:\r\n\r\nThis course was designed as preparation for the Putnam contest.", "Solution_9": "Well, \r\n\r\nBombieri could prove the Prime Number Theorem at age 11.\r\nSelberg was reading Ramanujan's collected works in his early/mid teen years.\r\nDeligne read Bourbaki commutative algebra in high school.\r\nDon Zagier started MIT at age 12 or so, and won the Putnam competition\r\nduring one of his years there. He went to do his PhD with Atiyah in his\r\nlate teens. \r\nKazhdan started working with Gelfand around age 13.\r\nBeilinson was learning Grothendieck-Riemann-Roch in his mid teens.\r\nGrothendieck independently rediscovered Lebesgue measure theory\r\nat age 15. \r\n\r\nOne of the basic nonlinear binary codes (Kerdock? Nordstrom? I forgot)\r\nis named after a 14-year-old British student who attended a lecture about \r\ncoding theory and worked out the example.\r\n\r\nJohn Conway made some discoveries in knot theory during high school.\r\nSimon Norton got 3 gold medals at IMO and as a child-prodigy type,\r\nmay have been fairly young for the first one. \r\n\r\nBau Chau Ngo who works on geometric Langlands' program and recently\r\nproved the Fundamental Lemma (25 yrs notorious open problem) with\r\nhis advisor Laumon, got a perfect score at IMO at a fairly young age,\r\n14 or 15. \r\n\r\nNoam Elkies was also 14-15 for his first IMO gold, a few years before\r\ndisproving Euler's conjecture on sums of fourth powers. \r\n\r\nTerry Tao was young but not particularly spectacular at the IMO.\r\nHe became spectacular later, during and after graduate school.", "Solution_10": "[quote=\"fleeting_guest\"]Don Zagier started MIT at age 12 or so, and won the Putnam competition\nduring one of his years there. He went to do his PhD with Atiyah in his\nlate teens. [/quote]\r\n\r\nFunny story about Don Zagier, actually. He graduated high school speaking something like 7 languages when he was 12, I think. Maybe 13. He applied to Oxford, but they told him that they don't accept anyone under age 16 as an undergraduate. So he went to MIT for 2 years, graduated, and at 15 was accepted as an Oxford Ph.D. student since they didn't have age restrictions on who could seek higher degrees.", "Solution_11": "[quote=\"orl\"]What I find funny about [url=http://www.math.ucla.edu/~tao/]Terrence Tao[/url] is that he gives the [url=http://www.math.ucla.edu/~tao/putnam/]Canadian Mathematical Olympiad 1999[/url] as regular course work. :lol:\n[/quote]\r\n\r\nQ5 is actually somewhat difficult, IMHO. I have not yet seen a solution that streamline with typical inequality solutions", "Solution_12": "I'd forgotten the Zagier story about Oxford. I think Gross tells it\r\noccasionally in connection with the Gross-Zagier formula. \r\n\r\nAlso, I think Drinfeld's IMO 24k was at a young age, 14-15.\r\n\r\nKontsevich would have been a natural for the IMO, and posed problems in \r\nKvant, but as far as I know was not involved in the competition. \r\nHe started university in Moscow several years early (15 or so) and\r\nI think also Gelfand's seminars, so he may have had better things to do\r\nat the relevant age.", "Solution_13": "[quote=\"fleeting_guest\"]I'd forgotten the Zagier story about Oxford. I think Gross tells it\noccasionally in connection with the Gross-Zagier formula. [/quote]\r\n\r\nThat's where I heard it :)", "Solution_14": "r these guys geniuses or just their efforts?", "Solution_15": "[quote=\"Tokyo\"]r these guys geniuses or just their efforts?[/quote]I think they are more of a genius than a hard worker (but they are certainly both). :roll:", "Solution_16": "And I disagree with Valentine. I think that they are actually more a result of having trained themselves to be [i]efficient[/i] thinkers rather than being \"natural\" geniuses or concientious hard workers.\r\n\r\nAs one of my math teachers once said -- its better to work efficiently than to work hard.\r\n\r\nEffective thinkers are people who've somehow, either consciously or subconsciously, learned to synthesize information in an unusual way. How they learn to do this is probably a combination of random circumstances and deliberate conditioning.\r\n\r\nOnce they've begun to think efficiently, and have had positive experiences from doing so, it then becomes a positive feedback system. They can continually use that knowledge to learn and sysnthesize even more knowledge and information! So its sort of an exponential learning curve based on their random initial conditions and the initial conditions of their childhood experiences. I think the element of random circumstances is what makes it impossible to recreate such individuals at will.\r\n\r\nEither way, I think it's an oversimplification to attribute such achievements to either unexplainable \"genius\" or simply to self-willed \"hard work.\"\r\n\r\nThere's probably nothing particularly different about them genetically than from any other normal bright person, and they probably don't work \"harder\" than even some students who fight to make good scores on even much simplier math exams.\r\n\r\nBy looking at each of these people individually, I bet you could find things in each of their personalities or childhoods which created an initial condition that fertilized their ability to think very efficiently.", "Solution_17": "[quote=\"gauss202\"]And I disagree. I think that they're actually more a result of having trained themselves to be [i]efficient[/i] thinkers rather than them being natural geniuses or necessarily hard workers.\n[/quote]I was not talking only about their high-school acomplishments, but rather more about their college ones ;)", "Solution_18": "[quote=\"fleeting_guest\"]Terry Tao was young but not particularly spectacular at the IMO.\nHe became spectacular later, during and after graduate school.[/quote]\r\n\r\nI wouldn't exactly say that. Scoring Bronze, Silver, and Gold at IMO at the ages of 11-13 is pretty spectacular in itself. It is really the age factor that is the most impressive thing about all of the IMO participants, because the 19 year age limit is rather arbitrary.\r\n\r\nIf you allowed people to participate until they were 25 or 30 or older, it would be a completely different competition. It is the fact that such young people can answer such difficult questions that is impressive. So the younger, the even more impressive to me. It is clear that if some of those who went to college at an early age had stayed to compete for longer they would have won many more Gold Medals and probably come close to some perfect scores.", "Solution_19": "Hmmm, This is an nice intellectual disquisition in here; but can anyone elaborate/elucidate the explanation presented by gauss and valentine?", "Solution_20": "[quote]Hello. Please post here names of youngest perfect scorers or gold medallist at IMO. Comments would be welcomed . THANKS[/quote]\r\nTerence Tao must surely have been the youngest gold medallist. But he didn't get a perfect score, and nor did Pawel Kroger.\r\n[url=http://tinyurl.com/ye5nvzb]Kroger[/url] got 40 and 27 in 1972-73.\r\n[url=http://tinyurl.com/y8ueg5t]Tao[/url] got 19, 40, 34 in 1986-88.\r\n\r\nSo who [b]does[/b] hold the record as the youngest ever perfect scorer?\r\n\r\nDavid", "Solution_21": "There was a legend in my high school, now a well known mathematician, who graduated a few years before I attended who just might fit the bill, Noam Elkies.\n\nFrom Wiki: born August 25, 1966 and \n\nhttp://www.imo-official.org/participant_r.aspx?id=10464 shows a perfect score in 1981 which would have made him 14 at the time.", "Solution_22": "[quote=\"davidstone\"][quote]Hello. Please post here names of youngest perfect scorers or gold medallist at IMO. Comments would be welcomed . THANKS[/quote]\nTerence Tao must surely have been the youngest gold medallist. But he didn't get a perfect score, and nor did Pawel Kroger.\n[url=http://tinyurl.com/ye5nvzb]Kroger[/url] got 40 and 27 in 1972-73.\n[url=http://tinyurl.com/y8ueg5t]Tao[/url] got 19, 40, 34 in 1986-88.\n\nSo who [b]does[/b] hold the record as the youngest ever perfect scorer?\n\nDavid[/quote]\n\nGreetings to Ireland. [url=http://sites.google.com/site/pawelkroeger/]Pawel Kr\u00f6ger[/url] got indeed a full score of 40/40 at 14th IMO 1972 in Poland. Please check page 32/96 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=87&t=295730]here.[/url] It is written that he has been in grade 7 at this time which means he has been around age 13 at this time." } { "Tag": [ "vector", "invariant" ], "Problem": "A particle of mass $M$ and energy $E$ collides elastically with a particle of mass $m$ which was resting before the collision. Find the energy of the first particle after the collision. Assume that both particles move only in one dimension.", "Solution_1": "well djole is $M$ the inertial mass or the recorded mass of the moving body", "Solution_2": "There is a convention that when you say that a relativistic particle has a mass of $m$, it is understood that this is the rest mass. I'm a follower of this rule, so...", "Solution_3": "well thanks . i just wanted to clear my doubt because i follow a convention that says $M_{0}$ as the particles inertial mass and $M$ as it's relativistic mass", "Solution_4": "this problem can be dun by normal method + [i]patience[/i]", "Solution_5": "Can you do it? :P It's not enough to say that it can be done ;)", "Solution_6": "well u r right the thing is it can be done and i know how to do it .the only thing is that i wrote it out of loss of patience .ok let me give the sol^n soon", "Solution_7": "I'm waiting :) Take your time, it's quite tricky...", "Solution_8": "hmmm... let me post my approach and see\r\n let's solve the problem by using $four-vectors$\r\n\r\nfor simplicity i take the inertial mass of 1st body as $M_{0}$ ana inertial mass of 2nd body as $m_{0}$\r\n\r\nso before collision \r\n$\\vec P_{1}$ $=$ $M_{0}$ $\\gamma_{1}$ $(c,v_{1},0,0)$\r\n$\\vec P_{2}$ $=$ $m_{0}$ $\\gamma_{2}$ $(c,0,0,0,)$\r\n\r\nafter collision,\r\n\r\n$\\vec P_{1}^{'}$ $=$ $M_{0}$ $\\gamma_{1}^{'}(c,v_{1}^{'},0,0)$\r\n\r\n$\\vec P_{2}^{'}$ $=$ $m_{0}$ $\\gamma_{2}^{'}(c,v_{2}^{'},0,0,)$\r\n\r\nbut by the conservation laws\r\n$\\vec P_{1}$ $+$ $\\vec P_{2}$ $=$ $\\vec P_{1}^{'}$ $+$ $\\vec P_{2}^{'}$\r\n\r\nthis implies, $|\\vec P_{1}^{'}|^{2}$ $=$ $( \\vec P_{1}^{'}$ $+$ $\\vec P_{2}^{'}-\\vec P_{2})^{2}$\r\n\r\nalso since this is an elastic collision\r\n\r\nwe have ,\r\n$M_{0}$ $( \\gamma_{1}-1 )$ $+$ $m_{0}$ $( \\gamma_{2}-1 )$ $=$ $M_{0}$ $( \\gamma_{1}^{'}-1 )$ $+$ $m_{0}$ $( \\gamma_{2}^{'}-1 )$\r\n\r\nalso we can equate the corresponding parts of the momentom conservation equation using \r\n\r\n$|\\vec P_{1}|^{2}$ $=$ $-M_{0}^{2}c^{2}$\r\nsimilarly we get \r\n$|\\vec P_{2}|^{2}$ $=$ $-m_{0}^{2}c^{2}$\r\n$|\\vec P_{1}^{'}|^{2}$ $=$ $-M_{0}^{2}c^{2}$\r\n$|\\vec P_{2}^{'}|^{2}$ $=$ $-m_{0}^{2}c^{2}$\r\n\r\nalso, $\\vec P_{1}.\\vec P_{2}$ $=$ $-M_{0}\\gamma_{1}m_{0}\\gamma_{2}c^{2}$ \r\nsimilarly get the other dot products..\r\n\r\nalso $\\gamma_{2}$ $=$ $1$ \r\nfurther \r\n$\\gamma_{1}$ $=$ $\\frac{E}{M_{0}c^{2}}$\r\n\r\nusing this we tend to get \r\n$\\gamma_{2}^{'}$ ehich gives us \r\n$E^{'}$ $=$ $M_{0}$ $\\gamma_{2}^{'}$ $c^{2}$\r\n\r\nthis is one approach the other is to use the centre of mass frame of refrence for analysis after collision which is an inertial frame of refrence (since our initial frame of refrence was inertial though there doesn't exist something called inertial) \r\n\r\nwell abnyways we use this fundamental invariant property of he centre of mass frame of refrence which states that \r\n\r\nin the centre of mass frame $\\sum \\vec P_{i}$ $=$ $(\\frac{E_{tot}}{c},0,0,0)$\r\n hence equate both sides by the previous obtained equations.....", "Solution_9": "The general procedure is correct. I hope you'll at least find the time to write down the expression for the final result. I have two questions:\r\n\r\n- why is there a minus in $|\\vec P|^{2}=-M^{2}c^{2}$?\r\n\r\n- I think you need to find $\\gamma'_{1}$, not $\\gamma'_{2}$, but that's probably a typo. You're supposed to find the energy of $M$ after the collision.", "Solution_10": "surely i will write down the final expression for i will be a stisfied man then\r\n\r\nthe second one is surely a typo. :) \r\n\r\nbut i have no clue as to why i did the first one as there was modulus on the vector... :o" } { "Tag": [], "Problem": "Express $ (0.5 \\plus{} \\frac23)^2$ as a common fraction.", "Solution_1": "$ \\frac{1}{2} \\plus{} \\frac{2}{3} \\equal{} \\frac{7}{6}$, and that squared is $ \\boxed{\\frac{49}{36}}$." } { "Tag": [ "geometry", "circumcircle", "blogs", "geometric transformation", "homothety", "cyclic quadrilateral", "power of a point" ], "Problem": "There is cyclic quadrilateral ABCD(center O) \r\n\r\nAnd intersection of AC,BD is E.\r\n\r\nThere is a point P is ABCD and X,Y,Z,W is circumcenter of ABP,BCP,CDP,DAP\r\n\r\nThen,\r\n\r\nXZ, YW, OE are concurrent!!", "Solution_1": "This is the Generalization of the China TST 2006 geometry problem,though it is not hard,maybe even easy.But anyway this is nice application of the pole-polar theory.\r\nProof:\r\nLet $ \\omega(O),\\omega_1(O_1),\\omega_2(O_2),\\omega_3(O_3),\\omega_4(O_4)$ be the circumcircles of the $ ABCD,\\triangle APB,\\triangle BPC,\\triangle CPD,\\triangle DPA$,respectively.Denote $ \\omega_1\\cap\\omega_3 \\equal{}P,N$ and $ \\omega_2\\cap \\omega_4 \\equal{}P,M$.Then\r\nThe point of intersection of $ O_1O_3$ and $ O_2O_4$ is the circumcenter of $ \\triangle PNM$.\r\nConsider circumcircle of $ \\triangle PNM$,denote it as $ \\zeta(I)$.Now we must prove that $ I,O,E$ are collinear.\r\nDenote $ AD\\cap BC\\equal{} F$,$ AB\\cap CD\\equal{}G$.Then it is well known that $ OE\\perp FG$($ FG$ is the polar of $ E$ w.r.t $ \\omega$).Hence \r\nit is sufficiently to prove that $ OI\\perp EG$.\r\nIt is well known that radical axes of the 3 circles passes through one point,hence\r\n$ PM,AD,BC$ are concurrent.\r\n$ PN,AB,CD$ are concurrent.\r\nSince $ PM$ is the radical axe of $ \\zeta$ and $ \\omega_2$,and $ AD$ is radical axe of $ \\omega$ and $ \\omega_2$,thus radical axe\r\nof $ \\zeta$ and $ \\omega$ passes through $ F\\equal{}PM\\cap AD$.Similarly the radical axe of $ \\zeta$ and $ \\omega$ passes through $ G$,hence\r\n$ FG$ is radical axe of $ \\zeta$ and $ \\omega$,and we are done.", "Solution_2": "[quote=\"Erken\"]This is the Generalization of the China TST 2006 geometry problem,though it is not hard,maybe even easy.But anyway this is nice application of the pole-polar theory.\nProof:\nLet $ \\omega(O),\\omega_1(O_1),\\omega_2(O_2),\\omega_3(O_3),\\omega_4(O_4)$ be the circumcircles of the $ ABCD,\\triangle APB,\\triangle BPC,\\triangle CPD,\\triangle DPA$,respectively.Denote $ \\omega_1\\cap\\omega_3 \\equal{} P,N$ and $ \\omega_2\\cap \\omega_4 \\equal{} P,M$.Then\nThe point of intersection of $ O_1O_3$ and $ O_2O_4$ is the circumcenter of $ \\triangle PNM$.\nConsider circumcircle of $ \\triangle PNM$,denote it as $ \\zeta(I)$.Now we must prove that $ I,O,E$ are collinear.\nDenote $ AD\\cap BC \\equal{} F$,$ AB\\cap CD \\equal{} G$.Then it is well known that $ OE\\perp FG$($ FG$ is the polar of $ E$ w.r.t $ \\omega$).Hence \nit is sufficiently to prove that $ OI\\perp EG$.\nIt is well known that radical axes of the 3 circles passes through one point,hence\n$ PM,AD,BC$ are concurrent.\n$ PN,AB,CD$ are concurrent.\nSince $ PM$ is the radical axe of $ \\zeta$ and $ \\omega_2$,and $ AD$ is radical axe of $ \\omega$ and $ \\omega_2$,thus radical axe\nof $ \\zeta$ and $ \\omega$ passes through $ F \\equal{} PM\\cap AD$.Similarly the radical axe of $ \\zeta$ and $ \\omega$ passes through $ G$,hence\n$ FG$ is radical axe of $ \\zeta$ and $ \\omega$,and we are done.[/quote]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nOh, thank you. \r\n\r\nBut I don't know why FG is polar of E.....", "Solution_3": "You need to know some properties of harmonic fours and polar theory.If you know them,see here:\r\nhttp://www.mathlinks.ro/viewtopic.php?t=168866", "Solution_4": "See my blog for the proof of the fact above :)", "Solution_5": "Very nice solution, Erken :) I have an another solution base on polar theory, but it is worse than yours :) \r\nLet $ S \\in XZ \\cap YW$\r\nDenote $ u$ is the polar of the point $ U$ wrt $ (O)$, and $ (mn)$ is the pole of the line passes through $ M,N$ wrt $ (O)$\r\nLet $ A_1 \\in WX \\cap AA$, and $ B_1, C_1, D_1$ are defined similarly.\r\nWe have $ A_1,B_1, C_1, D_1$ are collinear, because all of them lie on the radical axis of the circle $ (O)$ and the point-circle $ P$\r\nThen $ A(wx) = a_1, B(xy) = b_1, C(yz) = c_1, D(zw) = d_1$ are concurrent at a point, called $ Q$\r\nFrom this, and not that $ (wx)(xy) = x \\parallel AB, (xy)(yz) = y %Error. \"parralel\" is a bad command.\nBC, (yz)(zw) = z \\parallel CD, (zw)(wx) = w \\parallel DA$, we have there exists a homothety $ f$, centered at $ Q$ such that: $ f(A) = (wx), f(B) = (xy) , f(C) = (yz), f(D) = zw$\r\nNow let $ AB \\cap CD = F, AD \\cap BC = G$, we have $ f(F) = x \\cap z = (xz), f(G) = (yw)$\r\nSo $ FG \\parallel (xz)(yw) = s$, then $ E$ - the pole of $ FG$, $ S$ - the pole of $ s$ and $ O$ are collinear." } { "Tag": [ "calculus", "integration", "floor function", "calculus computations" ], "Problem": "{x}\r\n\u222b 14 + [x] dx where {.} denotes fractional part and [.] denotes integral part\r\n0", "Solution_1": "The expression you've written is meaningless: $ x$ gets either to be the variable of integration or involved in the bounds, not both. The typical way to make something like this sensible is to guess that you meant $ \\int_0^{\\{t\\}} 14 \\plus{} \\lfloor x \\rfloor \\, dx$, a not very interesting integral (since the integrand is constant on the interval over which we're integrating).", "Solution_2": "How about \r\n[x]\t\r\n\u222b 2x \u2013 [x] dx where [.] denotes greatest integer\r\n0", "Solution_3": "Please re-read the first sentence of my previous post and think about its content. If you come to understand its content, edit your post appropriately. If you don't understand its content, try to form a precise statement of what you don't understand and I or someone else will help explain it to you.", "Solution_4": "why can't one variable used for both purposes?", "Solution_5": "How the hell can you integrate something over a variable interval ?", "Solution_6": "Right: the notation $ \\int_a^b f(x)\\, dx$ means that you fix an interval $ [a, b]$ and then do the process known as definite integration over this interval. The $ x$ in $ f(x)dx$ is a placeholder -- it doesn't exist outside the integral. This is very similar to the fact that $ \\sum_{i \\equal{} 1}^{5} i \\equal{} 1 \\plus{} 2 \\plus{} 3 \\plus{} 4 \\plus{} 5$ -- the index of summation doesn't have any meaning outside the sum.", "Solution_7": "But your answer seems to be wrong\r\n\r\n0<{x}<1, then in (0,1), [x] = 0\r\n\r\n\r\nthe answer thus should be 14{x}", "Solution_8": "I didn't give an answer, so I'm not sure how you think that it could be wrong. :wink:" } { "Tag": [], "Problem": "The Wildcats scored an average of 60 points per game during the 1998 season. They averaged 72 points for each of their 13 home games. How many points did they average for each of their 12 away games?", "Solution_1": "Fromt he 72 point games, the team scored 13*12 points over 60.\r\nThus, to make up for it, the team must score 13*12 points below 60 in the 12 days, or an average of 13 points under 60, which is 47.", "Solution_2": "Or: \r\n\r\nThey score a total of 60*25=1500 pointz\r\n72*13 in home games.which is 936\r\nThus in away games they scored 1500-936=564\r\ndivide by 12\r\n\r\n47", "Solution_3": "Or we could use weighted averages. Letting x be the number of points below 60, 12x=13x12 so x=13, so 60-13=47. Pwnt.\r\n\r\n\r\nEDIT: 1200th post! w00t! :D", "Solution_4": "Or:\r\n\r\n13 home games: 13*12 pts up\r\n12 away games: 12*13 pts down => 13 pts below average => 47 pts", "Solution_5": "@mewto and james: thats the same solution I put." } { "Tag": [ "geometry", "incenter", "trigonometry", "angle", "midpoint", "IMO Shortlist" ], "Problem": "The incenter of the triangle $ ABC$ is $ K.$ The midpoint of $ AB$ is $ C_1$ and that of $ AC$ is $ B_1.$ The lines $ C_1K$ and $ AC$ meet at $ B_2,$ the lines $ B_1K$ and $ AB$ at $ C_2.$ If the areas of the triangles $ AB_2C_2$ and $ ABC$ are equal, what is the measure of angle $ \\angle CAB?$", "Solution_1": "[quote=\"orl\"]The incenter of the triangle $ ABC$ is $ K.$ The midpoint of $ AB$ is $ C_1$ and that of $ AC$ is $ B_1.$ The lines $ C_1K$ and $ AC$ meet at $ B_2,$ the lines $ B_1K$ and $ AB$ at $ C_2.$ If the areas of the triangles $ AB_2C_2$ and $ ABC$ are equal, what is the measure of angle $ \\angle CAB?$[/quote]\r\n[hide=\"Solution\"]\nLet the tangency points between the incircle with $ AB$ and $ AC$ respectively be $ D$ and $ E$. Let the midpoints of $ CD$ and $ BE$ be $ Y$ and $ X$ respectively. Furthermore, let the semiperimeter of $ \\triangle ABC$ be $ s$, $ AB \\equal{} c$, $ BC \\equal{} a$, and $ AC \\equal{} b$. It is well-known that $ B_1$, $ X$, and $ K$ are collinear as are $ C_1$, $ Y$, and $ K$. Notice that\\[ 1\\equal{}\\frac {[AB_2C_2]}{[ABC]} \\equal{} \\frac {\\frac {AC_2\\cdot AB_2\\sin \\angle BAC}{2}}{\\frac {AB\\cdot AC\\sin \\angle BAC}{2}} \\equal{} \\frac {AC_2}{AB}\\cdot \\frac {AB_2}{AC}\\]Apply Menelaus to $ \\triangle AEB$ with transversal $ C_2XB_1$, we have that $ \\frac {AC_2}{C_2B}\\cdot \\frac {BX}{XE}\\cdot \\frac {EB_1}{AB_1} \\equal{} 1$. We have that $ BX \\equal{} XE$, so\\[ \\frac {AC_2}{C_2B} \\equal{} \\frac {AB_1}{EB_1} \\equal{} \\frac {\\frac {b}{2}}{\\frac {b}{2} \\minus{} (s \\minus{} a)} \\equal{} \\frac {b}{b \\minus{} (a \\plus{} b \\plus{} c \\minus{} 2a)} \\equal{} \\frac {b}{a \\minus{} c}\\]which means that $ \\frac {C_2B}{AC_2} \\equal{} \\frac {a \\minus{} c}{b}$, so $ \\frac {AB}{AC_2} \\equal{} 1 \\plus{} \\frac {C_2B}{AC_2} \\equal{} \\frac {a \\plus{} b \\minus{} c}{b}$. Similarly, $ \\frac {AC}{AB_2} \\equal{} \\frac {a \\plus{} c \\minus{} b}{c}$, so\\[ 1 \\equal{} \\frac {AC}{AB_2}\\cdot \\frac {AB}{AC_2} \\equal{} \\frac {a^2 \\minus{} (b \\minus{} c)^2}{bc} \\equal{} \\frac {a^2 \\minus{} b^2 \\minus{} c^2 \\plus{} 2bc}{bc}\\]\\[ bc \\equal{} a^2 \\minus{} b^2 \\minus{} c^2 \\plus{} 2bc\\]\\[ a^2 \\equal{} b^2 \\plus{} c^2 \\minus{} bc\\]\\[ \\cos \\angle BAC \\equal{} \\frac {b^2 \\plus{} c^2 \\minus{} a^2}{2bc} \\equal{} \\frac {1}{2} \\equal{} \\cos 60\\]from which it follows that $ \\angle BAC \\equal{} 60$. [/hide]", "Solution_2": "One can use [b]Cristea's theorem ( the Theorem of Transversal )[/b] to find $ AB_2$ and $ AC_2$, getting the same value for the ratio $ \\frac{AB \\cdot AC}{AB_2 \\cdot AC_2}$ as to [i]The QuattoMaster 6000[/i].\r\n\r\nBest regards,\r\nsunken rock", "Solution_3": "My Method\uff1a\n\nLet $BK, CK$ intersect $AC, AB$ at $D, E$ respectively\uff0c$CC_3 \\parallel B_1K, BB_3 \\parallel C_1K,CC_3 \\cap BB_3=F,FJ \\parallel AB, FL \\parallel AC$\nThen $C_2, B_2$ are the midpoints of $AC_3, AB_3$ respectively\uff0c$K$ is the common midpoint of $AF, EJ, DL$ . $\\triangle BEK \\cong \\triangle BGK$\n$\\frac{AC}{AC_3}=\\frac{FC}{FC_3}=\\frac{CJ}{2KJ}$\uff0csimilarly\uff0c$\\frac{AB}{AB_3}=\\frac{BL}{2KL}$\uff0c$S(AB_3C_3)=4S(AB_2C_2), BK$ bisector $EG,GL \\parallel BK$\n\nSo on the basis of this, we can get\uff1a\n\n$S(ABC)=S(AB_2C_2) \\Longleftrightarrow 4S(ABC)=S(AB_3C_3) \\Longleftrightarrow \\frac{CJ}{KJ}=\\frac{KL}{BL}$\n\n$\\Longleftrightarrow \\frac{CG}{BG}=\\frac{KL}{BL} \\Longleftrightarrow GL \\parallel CK \\Longleftrightarrow CK$ bisector $DG \\Longleftrightarrow \\triangle CDK \\cong \\triangle CGK$\n\n$\\Longleftrightarrow \\angle KEB+\\angle KDC=\\angle KGB+\\angle KGC=180^{\\circ} \\Longleftrightarrow A, D, K, E$ are on a circle \n\n$\\Longleftrightarrow \\angle BAC+\\angle BKC=180^{\\circ} \\Longleftrightarrow \\angle BAC=60^{\\circ}$ . [b]Q. E. D.[/b]" } { "Tag": [ "function", "algebra", "polynomial", "integration", "complex analysis", "Functional Analysis" ], "Problem": "Consider the Banach space of all continuous functions on the closed unit disc with the maximum norm.\r\n\r\nShow that the set of polynomial functions is not dense in this Banach space.", "Solution_1": "More interesting question: what is the closure of the polynomials?", "Solution_2": "A proof that $ f(z)\\equal{}\\overline{z}$ is not in the closure of the polynomials:\r\n\r\nLet $ P(z)$ be any polynomial. Let $ C$ be the unit circle, with its standard parameterization.\r\n\r\n$ \\int_C(f(z)\\minus{}P(z))\\,dz\\equal{}\\int_Cf(z)\\,dz$ by Cauchy's Theorem.\r\n\r\n$ \\int_Cf(z)\\,dz\\equal{}\\int_C\\overline{z}\\,dz\\equal{} i\\int_0^{2\\pi}e^{\\minus{}i\\theta}e^{i\\theta}\\,d\\theta\\equal{}2\\pi i.$\r\n\r\nHence,\r\n\r\n$ 2\\pi\\equal{}\\left|\\int_C(f(z)\\minus{}P(z))\\,dz\\right|\\le\\int_C|f(z)\\minus{}P(z)|\\,|dz| \\le2\\pi\\|f\\minus{}P\\|_{\\infty}$\r\n\r\nwhere $ \\|f\\minus{}P\\|_{\\infty}\\equal{}\\sup_{z\\in\\overline{D}}|f(z)\\minus{}P(z)|$ is the uniform norm.\r\n\r\nThus $ \\|f\\minus{}P\\|_{\\infty}\\ge 1$ and $ f$ cannot be approximated by polynomials.\r\n\r\n-----\r\n\r\nThe closure of the polynomials in $ C(\\overline{D})$ is usually called $ A.$ It is the set of functions continuous on the closed unit disk and holomorphic on the interior of the disk. For such a function, the Cesaro means of the partial sums of the power series centered at zero should give a sequence of polynomials converging uniformly to the function.\r\n\r\nOn the other hand, the set of polynomials in both $ z$ and $ \\overline{z}$ is dense in $ C(\\overline{D}),$ by a version of the Weierstrass approximation theorem." } { "Tag": [], "Problem": "OK, it has become clear that this forum has been overrun by spam. This is going to stop now.\r\n\r\n[b]The Games & Fun Forum will be locked for 24 hours[/b], so that we can clean out all the spam, and so those of you who have been abusing this forum can cool off. When the forum is reopened tomorrow, users who spam will be subject to banning without warning.\r\n\r\nSpam is unacceptable. All posts should have some [b]meaningful[/b] content, even here in Games & Fun. Spam not only interferes with users who wish to have a coherent discussion, it also unnecessarily taxes our database and our web server. \r\n\r\nIn the future, there will be a zero-tolerance policy on \"3\"'ing, pointless ziggurats, and other egregious spam.\r\n\r\nLet me remind you of our user conduct policy (which is in full at http://www.artofproblemsolving.com/Legal/AoPS_L_Comm_Terms.php )\r\n\r\n[quote]a. not to harm minors or cause minors to be harmed in any way;\n\nb. not to forge headers of postings, e-mails, or otherwise manipulate identifiers in order to disguise the origin of any Content transmitted through the Services;\n\nc. not to impersonate any person or entity, including, but not limited to, an AoPS employee, forum moderator, guide or guest, or falsely state or otherwise misrepresent your affiliation with any person or entity;\n\nd. not to upload, post, e-mail, transmit or otherwise make available any Content that is unlawful, harmful, threatening, abusive, harassing, tortuous, defamatory, vulgar, obscene, libelous, invasive of another's privacy, hateful, or racially, ethnically or otherwise objectionable;\n\ne. not to upload, post, e-mail, transmit or otherwise make available any Content that you do not have a right to make available under any law or under contractual or fiduciary relationships (such as inside information, proprietary and confidential information learned or disclosed as part of employment relationships or under nondisclosure or confidential disclosure agreements);\n\nf. not to upload, post, e-mail, transmit or otherwise make available any information that you consider confidential in nature, which posting may cause you or any third party to lose rights to the intellectual property associated with the information;\n\ng. not to upload, post, e-mail, transmit or otherwise make available any Content that infringes any patent, trademark, trade secret, copyright or other proprietary and lawful rights of any party;\n\nh. not to upload, post, e-mail, transmit or otherwise make available any unsolicited or unauthorized advertising, promotional materials, \"junk mail,\" \"spam,\" \"chain letters,\" \"pyramid schemes,\" or any other form of solicitation. You understand that such postings may violate laws of various jurisdictions, and that you may be subject to prosecution or to civil suit;\n\ni. not to intentionally or unintentionally violate any local, state, national or international law;\n\nj. not to \"stalk\" or otherwise harass another; or collect or store personal data about other users;\n\nk. to upload, post, e-mail, transmit or otherwise make available only material which the User has created and to which the User owns the copyrights; and\n\nl. to report any violations of the above immediately to AoPS.\n[/quote]Parts (d) and (h) of the above are particularly significant, and are what have resulted in this 24-hour \"cooling-off\" period.\r\n\r\nWe provide this forum -- for free -- for your enjoyment. Don't abuse the privilege.", "Solution_1": "I do not see how (d) or (h) is violated in G&FF.", "Solution_2": "some forums (like slot machine) are locked\r\n\r\ncan you unlock them?\r\n\r\nand if you can't, is it ok if we start a new thread?\r\n\r\nthanks", "Solution_3": "[quote=\"akalra1\"]some forums (like slot machine) are locked\n\ncan you unlock them?\n\nand if you can't, is it ok if we start a new thread?\n\nthanks[/quote]\r\n\r\nObviously not, otherwise they wouldn't have been locked.", "Solution_4": "I still do not see how (d) or (h) is violated in G&FF.", "Solution_5": "[quote=\"Klebian\"][quote=\"akalra1\"]some forums (like slot machine) are locked\n\ncan you unlock them?\n\nand if you can't, is it ok if we start a new thread?\n\nthanks[/quote]\n\nObviously not, otherwise they wouldn't have been locked.[/quote]\r\n\r\nwhat was wrong with slot machine\r\n\r\nwas that too spammy?", "Solution_6": "can we start a new \"clean\" slot machine thread?\r\n\r\ni liked it", "Solution_7": "For now, do not introduce a new slot machine thread.", "Solution_8": "yeah, it was pretty stupid", "Solution_9": "WHAT is the MAX KB for an upload of an unharmful not spamming picture?", "Solution_10": "What was wrong with that game Duck, Duck, Goose?", "Solution_11": "jesus guys.\r\n\r\nstop asking questions and use common sense, if you type up a post and it looks like something i'd say, it's spam. VIEWER DISCRETION IS ADVISED." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Determine, with proof, all integers $ x$ for which $ x(x\\plus{}1)(x\\plus{}7)(x\\plus{}8)$ is a perfect square.", "Solution_1": "Let $ u\\equal{}x\\plus{}4$. This transforms the product into $ (u\\plus{}3)(u\\minus{}3)(u\\plus{}4)(u\\minus{}4)\\equal{}(u^2\\minus{}9)(u^2\\minus{}16)$.\r\nNow let $ v\\equal{}u^2\\minus{}12$. Then we have $ (u^2\\minus{}9)(u^2\\minus{}16)\\equal{}(v\\minus{}4)(v\\plus{}4\\minus{}1)\\equal{}(v^2\\minus{}16)\\minus{}(v\\minus{}4)\\equal{}v^2\\minus{}v\\minus{}12$. This will be between $ (v\\minus{}1)^2$ and $ v^2$ if $ v> 13$. We have $ v\\equal{}u^2\\minus{}12\\equal{}(x\\plus{}4)^2\\minus{}12$ so $ v>13$ if $ x>1$ or $ x<\\minus{}9$. The cases of x in between are dealt with by brute force so $ x\\equal{}1,0,\\minus{}1,\\minus{}4,\\minus{}7,\\minus{}8,\\minus{}9$" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "A square has all of its vertices on lattice points. The length of the square is $\\sqrt{p}$ where $p$ is prime. Find the number of lattice points that are contained in the square.", "Solution_1": "i) To prove don't have other lattice point in each segment trought 2 lattice points with size $\\sqrt{p}$, for any prime $p$, is easy. Suppose wlog that one of the points is $A=(0,0)$, and the other $B=(a,b)$. If one point $C=(c,d)$ are in AB, we will have that $c = ka$ and $d = kb$, for any positive rational $k$. Suppose that $k = \\frac{x}{y}$, where $gcd(x,y)=1$ and $y=0 there is no real solution to the equation\r\nx^2000/2001^n + x^1999/2000^n + ... +x^2/3^n + x/2^n + 1 = 0.\r\n\r\nGillis 1984:\r\nif x + y + z = 2pi and cos x cos y cos z = 1/4\r\nprove 4cos 3x cos 3y cos 3z - 12cos 2x cos 2y cos 2z = 1.\r\n\r\nactually I have two nice solutions to the first one but only brute force solution to the second.", "Solution_1": "by reccurence for n=0 the equation :\r\n$ 1\\plus{}x\\plus{}x^2\\plus{}...\\plus{}x^{2000}\\equal{}0$ has no real solution \r\nwe suppose that the equation $ x^{2000}/2001^n \\plus{} x^{1999}/2000^n \\plus{} ... \\plus{}x^2/3^n \\plus{} x/2^n \\plus{} 1 \\equal{} 0$ have no real solution ,\r\nsuppose for exemple that $ x^{2000}/2001^n \\plus{} x^{1999}/2000^n \\plus{} ... \\plus{}x^2/3^n \\plus{} x/2^n \\plus{} 1>0$ for all real number.\r\nwe considere the polynomial $ P(x)\\equal{}x\\plus{}x^2/2^{n\\plus{}1}\\plus{}...\\plus{}x^{2001}/2001^{n\\plus{}1}$\r\nwe have $ P'(x)\\equal{}x^{2000}/2001^n \\plus{} x^{1999}/2000^n \\plus{} ... \\plus{}x^2/3^n \\plus{} x/2^n \\plus{} 1>0$ so P(x) is bijective ,then the equation $ P(x)\\equal{}0$ have one solution which is x=0 ,so conclusion the equation $ 1\\plus{}x/2^{n\\plus{}1}\\plus{}...x^{2000}/2^{n\\plus{}1}\\equal{}0$ have no real solution, we are done :)" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $\\mathcal S$ denote the set of points inside or on the border of a triangle $ABC$, without a fixed point $T$ inside the triangle. Show that $\\mathcal S$ can be partitioned into disjoint closed segemnts. \r\n\r\n[i]Yugoslavia[/i]", "Solution_1": "Suppose the point $T$ is strictly inside the triangle, and not on the borders (the other cases will follow from this construction). Let $\\ell_a,\\ell_b,\\ell_c$ be the lines parallel to $BC,CA,AB$ respectively which pass through $T$. Our set of segments will consist of:\r\n\r\n(a) those segments parallel to $BC$, with endpoints on $AB,AC$, lying between $\\ell_a$ and $BC$ (but not on $\\ell_a$);\r\n\r\n(b) the segments parallel to $CA$, with endpoints on $AB,\\ell_a$, lying between $\\ell_b$ and $CA$ (not on $\\ell_b$);\r\n\r\n(c) the segments parallel to $AB$, with endpoints on $\\ell_a,\\ell_b$, lying between $\\ell_c$ and $AB$." } { "Tag": [ "calculus", "search", "Support", "real analysis" ], "Problem": "Hi everyone!\r\n\r\nI've known that Mathlinks is a good forum for mathematicians. They share excellent problems, that were posted at various contests\\Olympiads, or that they simply find really worth solving and they help each other.\r\n\r\nI hoped that this Physics forum would act in the same way. I hoped to meet many people who share a passion for Physics and who are willing to post/try to solve good (read as difficult) problems. I was partially right, but only partially.\r\nI've seen topics like \"Homework problems\" which obviously speak that these problems are not quite... Olympiad problems, or Very difficult problems.\r\nI am not the owner or moderator of the forum, and that is why I am not to decide what should be permitted and what should not, but to me such topics are disappointing, and I am against them. Why should this forum turn into one where people post homework problems, and others solve them. Not only is this unfair, but also irrational. It's irrational to solve lots of low difficulty problems, because they worsen your skills and waste your time.\r\n\r\nThat is why, I ask you all to post only trully challenging problems.\r\n\r\nThank you.\r\n\r\n\r\nPS - Please express your opinions on the matter.", "Solution_1": "I quite agree with this complaint, and although I haven't been a member for long, I can see that a large portion of these problems are of easy or (barely) medium difficulty. Now, I appreciate that there are people who need help with their homework -- I don't know whether this forum is intended for them. If it is, then there's nothing to be done about it; everything should remain untouched. If it isn't, then this forum clearly needs more rigorous rules and moderators. \r\n\r\nWell, it seems I've been the first one to vote -- I voted that this should be a forum for posting challenging problems. I'd really like to hear the opinions of the other frequent visitors of this forum. I'd also like to see the response of the authorities. \r\n\r\nI hope that people will take this matter seriously and decide what should be done with the Physics forum before it becomes a disappointment to those who want challenging instead of school problems (and I hope that the number of these people isn't small!)", "Solution_2": "We need a subforum \"Homework Problems.\" The Calculus and Real Analysis forum did something similar by creating a \"Calculus Computation and Tutorials\" forum. AoPS/mathlinks isn't an ideal place for basic-level homework help in physics--physicsforums.com is much better for this. However, we still need to accomodate those who come here for homework help.\r\nWe also need a new moderator. xxreddevilzxx isn't active on this forum, so we still have outdated announcements, and the forum needs FAQ to answer the recurring questions about the US screening exams.", "Solution_3": "2 schulmannerism: Good ideas! How about writing an e-mail to the owner of forum, or the main moderator, in order to draw his/her attention to this matter. The only problem would probably be the realtively small number of people who post on this forum. Though improving the forum will help it become more popular.", "Solution_4": "I've been away for a long time from this forum coz I was (and am still veryyyy busy in IIT-JEE prep - I hope you know about it!!)...\r\n\r\nWhile I was here, I found that people like kubus, yetti , me , daveed etc. (sorry if I missed someone out!) discussed great theoretical problems and some interesting and challenging problems requiring great though as well ( search the forum to find these gems!!) - I even started a weekly contest to encourage and further discuss the great problems of Physics --- but the contest went null as almost no one participated --- now I see that after a break of a month when I return , the forum is littered(no offence please ~!!) with simple \"formula based problem\"\" I do not wish to offend anyone --- it is true that when people who are also studying physics visit mathlinks , they will obviously post all problems -easy or challenging discussions - here !! I strongly second the motion of creation of an appropriately named subforum for simple forums (the title should be such that first time posters should post easy problems in that subforum!)..... We should also suggest when the Physics forum page opens that greater detailed and more specific discussions 'might' be better posted at physicsforums.com (of course along with that - you should post it here to share it here!!)\r\n\r\nRegarding the moderator --- the present one is really helpful when you approach him and always there - he's probably more engaged in his Maths forums and probably gives a little lower priority to Physics(no offence please)... I was very active in this forum and will be 2 months hence once my all important IIT exam(Indian Institute of Technology - the toughest Physics, MAth, Chem combined exam in the WORLD!) is over..... I have always wanted to be a part of moderating this forum.... Valentin told me to wait --- I feel the time has come--- I would like to nominate myself!! :D (Please note that I will be active only after 2 months -- however, since the issue hs come up - I post my candidacy right now!! :lol: )", "Solution_5": "Rushil, I will personally support you in case there will be a voting!", "Solution_6": "I would prefer it to be a hard problems forum, but then I feel it really doesn't matter if some homework problems are posted once in a while.......Homework problems can, at times, be a key to understanding concepts pretty well.........The problem with problems like those from the physics olympiads is that the minute someone from somewhere near the eigth or ninth grade sees them, they might really feel that the stuff is going above their head and leave....On the contrary, when you have a mixture of both simple and difficult problems, it should lead to a higher number of active participants in the forum.... And well, if some member posts homework problems for say, a couple of months, he's bound to look at the other problems as well....So basically, it's like letting him/her slowly mature and thereby understand things better.......When you have only the tough Olympiad problems, it's somewhat like discouraging young talent......All the same, the more experienced members do need to go on with the olympiad style of problems. I quite look forward to seeing a mixture of both, though I'll prefer a majority of the olympiad problems while at the same time not ignoring the homework problems.", "Solution_7": "Well, it's fairly clear that the best solution would simply be to have a subforum for \"computations and tutorials\", as previously suggested, and hopefully the small number of people won't be a problem, as the forum seems to be growing in number (as opposed to the state forums, which have definitely been declining since near the start). Seeing as I don't think anybody has written a PM to Mr. Rusczyk or another administrator yet, as implied from the posts above, I'll do so now... this should probably solve all problems.", "Solution_8": "Please do so... Valentin i s usually more active on the forum --- maybe you should PM him too!!!", "Solution_9": "Hey... not a seperate forum --- it should be a SUBFORUM !!! Otherwise everyone will be confused!!! \r\n\r\n\r\nAlso, we should start moving topics to the easy forum if we consider it to be so ---- the moderator can do it only!! :D", "Solution_10": "[quote=\"Rushil\"]Please do so... Valentin i s usually more active on the forum --- maybe you should PM him too!!![/quote]No need to, my eyes see everything :police: :yoda: :P\r\n\r\nRegarding the issues presented in this topic, the following measures will be taken: \r\n\r\n- a second moderator will be assigned (by the way, are there any other nominations besides Rushil?) \r\n- a homeworks help subforum will be created and all the homework stuff should be dealt with there (remember still that some people might just be at the \"getting started\" level in physics and not really want to do homework - don't push people just because they are asking easy questions ;) )\r\n\r\nYou can further discuss nominations for moderators (though keep in mind that the final decision will [b]not[/b] be based (entirely) on the \"democratic\" vote :P) and naming of the homework help subforum (for now I'll just name it \"Homework Problems & Tutorials\").", "Solution_11": "That's great Valentin!!\r\n\r\nEagerly waiting your decision and the reformating of the Physics forum!!!", "Solution_12": "Although the creation of the new forum (Introductory Physics) was much needed, the entire Physics section still needs moderating. Seeing that nobody except [b]Rushil[/b] has publicly nominated himself for the position of the new moderator -- together with the fact that even [b]Rushil[/b] will be unable to devote his time to the forum for a few month -- has made me decide to offer myself as the moderator before [b]Rushil[/b] is back. \r\n\r\nHowever, since I am rather new on this forum, I will gladly step back in case that a senior of mine (the number of such people on this forum is easily very large) wishes to take these chores onto himself -- I have no desire to hinder someone's years-old plans to become a moderator :) -- in other words, I am willing even to withdraw when [b]Rushil[/b] (or almost anybody else, for that matter) feels eager to take on moderating. I just want to have some authority active here (and this authority doesn't need to be me) who'd sort topics into the right forums and so on, just so the chaos in this section does not increase further. A couple of months is not a very long time, but I think that the Physics section has no need to wait that long for attention.\r\n\r\nSeeing the lack of response to the Administrator's call, I can only hope that other people who want to moderate the two Physics forums will announce themselves here. In the meantime, I hope that my offer will be considered in due time. Again, I have no pretensions to gain a position on this forum :) -- it's just that there doesn't seem to be anybody else with time and will to do this and I'd like to see these forums in good shape.\r\n\r\nThanks!", "Solution_13": "That's great , Djole!!\r\n\r\nSince youve started the idea of reform of the forum, i'm sure you'll make a great moderator! I support you(as you do me!! :D )... Anyways, I hope Valentin can make both of us moderators ---- there's no reason for you to step down later --- it'll be great!! \r\n\r\nThere's no 'year-old' plan to be a moderaot - so dont think you are obstructing anyone! \r\nIt'll be great if more people can partcipate in the organisations!!\r\n\r\nStill waiting Valentin....\r\n\r\n\r\n[size=150][color=red]Valentin, Where are you??!!!!!![/color][/size]" } { "Tag": [ "algebra", "polynomial", "inequalities", "function", "algebra unsolved" ], "Problem": "Let real coefficient polynomial $f(x) = x^n + a_1 \\cdot x^{n-1} + \\ldots + a_n$ has real roots $b_1, b_2, \\ldots, b_n$, $n \\geq 2,$ prove that $\\forall x \\geq max\\{b_1, b_2, \\ldots, b_n\\}$, we have\r\n\\[f(x+1) \\geq \\frac{2 \\cdot n^2}{\\frac{1}{x-b_1} + \\frac{1}{x-b_2} + \\ldots + \\frac{1}{x-b_n}}.\\]", "Solution_1": "Put $x-b_i=a_i$. We must prove that for positive numbers $a_i$ we have $ (1+a_1)...(1+a_n)(1/a_1+...+1/a_n)\\geq 2n^2$. USe Huygens inequality to find that the LHS is greater than or equal to $n(1+G)^n/G$ where $G$ is the geometric mean of $a_i$. A study of the function $\\frac{(1+x)^n}{x}$ shows that it has minimal value in $\\frac{1}{n-1}$ and this one is $ (n-1)(n/(n-1))^n$ obviously greater than $2n$.", "Solution_2": "[quote=\"harazi\"]Put $ x \\minus{} b_i \\equal{} a_i$. We must prove that for positive numbers $ a_i$ we have $ (1 \\plus{} a_1)...(1 \\plus{} a_n)(1/a_1 \\plus{} ... \\plus{} 1/a_n)\\geq 2n^2$. USe Huygens inequality to find that the LHS is greater than or equal to $ n(1 \\plus{} G)^n/G$ where $ G$ is the geometric mean of $ a_i$. A study of the function $ \\frac {(1 \\plus{} x)^n}{x}$ shows that it has minimal value in $ \\frac {1}{n \\minus{} 1}$ and this one is $ (n \\minus{} 1)(n/(n \\minus{} 1))^n$ obviously greater than $ 2n$.[/quote]\r\nCan anyone post other solution?(This problem is not difficult but I can not solve this problem by other)", "Solution_3": "Well , you can prove the last ineq : $ \\frac{(1\\plus{}x)^n}{x} \\ge 2n$ like this\r\n$ \\frac{(1\\plus{}x)^n}{x} \\ge \\frac{1}{x}\\plus{}n\\plus{}\\frac{ n(n\\minus{}1)}{2x} \\ge n\\plus{}\\sqrt{ 2n(n\\minus{}1)} \\ge 2n$", "Solution_4": "[quote=\"math10\"][quote=\"harazi\"]Put $ x \\minus{} b_i \\equal{} a_i$. We must prove that for positive numbers $ a_i$ we have $ (1 \\plus{} a_1)...(1 \\plus{} a_n)(1/a_1 \\plus{} ... \\plus{} 1/a_n)\\geq 2n^2$. USe Huygens inequality to find that the LHS is greater than or equal to $ n(1 \\plus{} G)^n/G$ where $ G$ is the geometric mean of $ a_i$. A study of the function $ \\frac {(1 \\plus{} x)^n}{x}$ shows that it has minimal value in $ \\frac {1}{n \\minus{} 1}$ and this one is $ (n \\minus{} 1)(n/(n \\minus{} 1))^n$ obviously greater than $ 2n$.[/quote]\nCan anyone post other solution?(This problem is not difficult but I can not solve this problem by other)[/quote]\r\n$ x\\minus{}b_i \\equal{} c_i \\Rightarrow (c_1\\plus{}1)...(c_n\\plus{}1)\\left ( \\frac{1}{c_1} \\plus{} ... \\plus{} \\frac{1}{c_n} \\right ) \\ge 2n^2$\r\n$ \\frac{1}{c_1} \\plus{} ... \\plus{} \\frac{1}{c_n} \\ge n\\sqrt[n]{\\frac{1}{c_1c_2...c_n}}$\r\n$ c_i\\plus{}1 \\equal{} c_i\\plus{}(n\\minus{}1)\\frac{1}{n\\minus{}1} \\ge n\\sqrt[n]{c_i\\frac{1}{(n\\minus{}1)^{n\\minus{}1}}}$\r\nSo $ LHS \\ge n^{n\\plus{}1}\\frac{1}{(n\\minus{}1)^{n\\minus{}1}} \\ge 2n^2 \\iff$\r\n$ \\left ( \\frac{n}{n\\minus{}1} \\right )^{n\\minus{}1} \\ge 2 \\iff$\r\n$ \\left ( 1 \\plus{} \\frac{1}{m} \\right )^m \\ge 2, m \\equal{} n\\minus{}1 \\ge 1$\r\nBut $ \\left ( 1 \\plus{} \\frac{1}{m} \\right )^m \\equal{} \\sum_{i\\equal{}0}^m \\binom{m}{i} m^{\\minus{}i} \\ge \\binom{m}{0}\\plus{}\\binom{m}{1}m^{\\minus{}1} \\equal{} 2$, and we are done." } { "Tag": [ "AMC" ], "Problem": "Are these worth buying... Are they helpful and what do they cover?", "Solution_1": "There have been a few discussions about this already:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=arbelos&t=32200\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=arbelos&t=27684\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=arbelos&t=12785\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=arbelos&t=8937" } { "Tag": [ "trigonometry", "algebra", "polynomial", "quadratics" ], "Problem": "Find the exact value of $\\sin1^{\\circ}$ (or at least outline a process)", "Solution_1": "$\\sin 15^{\\circ} = \\frac{\\sqrt{6} - \\sqrt{2}}{4}$, then solve a polynomial of degree 15 that you obtain through Euler. :)\r\n\r\nAlternately, if you feel like evaluating the Taylor Series of $\\sin \\frac{\\pi}{180}$...", "Solution_2": "This must be impossible. If you could find the exact value of the sin of one degree, you could use addition formulae to find the exact value of sin of every other degree angle, and I know that this is not possible. Do you mean a process for approximating the number? In that case, you could either simply use a McLaurin/Taylor series, or figure out a high-degree polynomial, one of whose (non-analytic) roots would be the proper number.", "Solution_3": "[hide=\"an outline\"]$\\sin 45^\\circ=\\cos 45^\\circ=\\frac{\\sqrt{2}}{2}$.\n\n$\\cos 36^\\circ=\\frac{\\sqrt{5}+1}{4}$ (and from this $\\sin 36^\\circ$ can be found). This can be shown by using $\\cos 36^\\circ \\cos 72^\\circ=\\frac 14$ (since multiplying by $4\\sin 36^\\circ$ gives $\\sin 36^\\circ$), expanding in terms of $\\cos 36^\\circ$, and getting rid of the extraneous root -1/2.\n\n$\\cos 20^\\circ$ (and therefore $\\sin 20^\\circ$) can be found since $4\\cos^3 20^\\circ -3\\cos 20^\\circ-\\frac 12=0$, and then the cubic can be solved.\n\nSince $1=9(45-36)-4(20)$, the addition formula can be used enough times to find $\\sin 1^\\circ$.[/hide]", "Solution_4": "[quote=\"scorpius119\"]\n$\\cos 20^\\circ$ (and therefore $\\sin 20^\\circ$) can be found since $4\\cos^3 20^\\circ -3\\cos 20^\\circ-\\frac 12=0$, and then the cubic can be solved.\n[/quote]\r\n\r\nNot by the cubic formula - that method will force you to calculate a root that evaluates to an expression in terms of $\\cos 20^{\\circ}$, which takes you right back where you started. I also don't think a simple linear term can be factored out...\r\n\r\nIn any case, the polynomial of degree 15 was kind of a joke :) If you wanted an approximation, the Taylor series converges rather quickly - the first two terms gives $\\sin \\frac{\\pi}{180} \\approx 0.017452406424$ while calculator gives $\\sin \\frac{\\pi}{180} \\approx 0.017452406437$ (but of course, the calculator doesn't do more than evaluate the Taylor series to sufficient digits...)\r\n\r\nThe question, then, is what you mean by \"exact value\" - a closed radical expression? This is clearly impossible - at best, finding $\\sin \\frac{\\pi}{180}$ in terms of $\\sin \\frac{\\pi}{5}$ requires dividing by $36 = 2^2 \\times 3^2$, a process equivalent to solving two quadratics and two cubics - and both cubics cannot be solved except in terms of the $\\sin$ values they are trying to find!\r\n\r\n(Edit: However, by this logic it would be entirely feasible to find a closed form for $\\sin 9^{\\circ}$. It would be very ugly :( )", "Solution_5": "Edit: My mistake $90 - 1 = 99$ between the hours of 11 p.m. and 1 a.m. Don't mock me, it makes me sad.\r\n \r\n$\\sin 1^{\\circ} = \\cos 89^{\\circ} = \\cos \\frac{89\\pi}{180}$\r\n$2\\cos a = \\sqrt{2+2\\cos 2a}$ when $\\cos a$ is positive and\r\n$2\\cos a = -\\sqrt{2+2\\cos 2a}$ when $\\cos a$ is negative \r\n$\\Rightarrow 2\\cos\\frac{89\\pi}{180} = \\sqrt{2+2\\cos\\frac{89\\pi}{90}} = \\sqrt{2-\\sqrt{2+2\\cos\\frac{89\\pi}{45}}} = \\sqrt{2-\\sqrt{2+2\\cos\\frac{\\pi}{45}}}$\r\nIt can be proven that $2\\cos\\frac{\\pi}{45}$ is ($\\frac{\\pi}{45} = \\frac{91\\pi}{2^{12} - 1}$ is how we find our string of signs) equal to the the infinite radical of the form $\\sqrt{2 + \\sqrt{2 + \\sqrt{2 \\cdots}}}$, but it has a string of 12 signs which is too long to actually put under a radical, so we will just say, the repeating string $(\\overline {++++---+--+-})$. [hide=\"(Extra)\"]See if you can prove it, it is actually proven generally by me in other [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59094&start=20]post,[/url] sort of.[/hide]\r\nSubbing back in, we get:\r\n$2\\sin 1^{\\circ} = 2\\cos\\frac{89\\pi}{180} = \\sqrt{2-\\sqrt{2+2\\cos\\frac{\\pi}{45}}}$\r\n$= \\sqrt{2 - \\sqrt{2 + \\sqrt{2 + \\sqrt{2 + \\sqrt{2 + \\sqrt{2 + \\cdots}}}}}}$\r\n$= [-+]\\Vee (++++---+--+-) = (-+ \\overline {++++---+--+-})$", "Solution_6": "[quote=\"Pork_Chop8\"]$\\sin 1^{\\circ} = \\cos 99^{\\circ}$...[/quote]\r\n\r\nAm I missing something here? Some recent groundbreaking research in trigonometry? Last time I checked, $\\sin 1^{\\circ} = \\cos 89^{\\circ}=-\\cos 91^{\\circ}$", "Solution_7": "$\\sin x^{\\circ} = \\cos (90^{\\circ} - x^{\\circ})$\r\n\r\nso yes \r\n\r\n$\\sin 1^{\\circ} = \\cos 89^ {\\circ}$", "Solution_8": "well, i was thinking that we have sin 36 and sin 30, so use subtraction to get sin 6, then half to sin 3, then triple angle general cubic (solve with cardano) to get sin 1", "Solution_9": "Ah, I realized that trying to find $\\cos 20^\\circ$ in terms of radicals is useless - the result is $\\frac{\\sqrt[3]{1+i\\sqrt{3}}+\\sqrt[3]{1-i\\sqrt{3}}}{2\\sqrt[3]{2}}$, which is just DeMoivre's Theorem (this would be like saying that $\\cos 72^\\circ$ is half the sum of two fifth roots of 1, not very useful).\r\n\r\nAt least it would be possible to find (and, in fact, construct with compass/straightedge) the value of $\\sin 3^\\circ$.", "Solution_10": "[quote=\"Altheman\"]well, i was thinking that we have sin 36 and sin 30, so use subtraction to get sin 6, then half to sin 3, then triple angle general cubic (solve with cardano) to get sin 1[/quote]\r\n\r\nCardano would give your answer in terms of $\\sin 1^{\\circ}$, I have already tried trisecting a similar angle with it :|", "Solution_11": "we have sin 18 which is phi/2 and sin 15,\r\nfind sin 3 by sin(18-15). Then use triple angle formula for sin (sin 3x = 3sinx - 4sin^3(t))and solve for sin 1 using the cubic formula", "Solution_12": "But unfortunately, it doesn't work since at some point you have to take a cubic root from a complex number, and using the trigonometric form (or the exponential, which is basically the same) takes you right back at the start. (You end up with $\\sqrt[3]{\\cos 3^\\circ+i\\sin 3^\\circ}=\\cos 1^\\circ+i\\sin 1^\\circ$, and there's no other way to take the root)." } { "Tag": [ "inequalities" ], "Problem": "The positive integers a, b, c, d are such that\r\na^2 + b/(a+c) = d\r\nProve that d\u2264b+(c-1)^2", "Solution_1": "Let a = 5, b = 16, c = 3 then d = 27.\r\nOur inequality 27 =< 16 + 4 = 20 so counterexample.", "Solution_2": "Ooops typo!!\r\nIn fact it's (a 2 +b)/(a+c)=d" } { "Tag": [ "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "I was following a topic at the following url: http://mathforum.org/kb/thread.jspa?threadID=1545458&tstart=60\r\n\r\nThe question was the following: Define a subspace of the real line that can be represented as the union of two locally compact subspaces, one of which is closed and the other open, and that this is not a locally compact space.\r\n\r\nThe solution was given as: Take A = (0,1/2) U (3/4,5/6) U (7/8,9/10) U ... and B = {1}. Then A is open, B is closed, A and B are locally compact, but their union isn't.\r\n\r\nI still don't understand why A U B would NOT be locally compact.\r\n\r\nI was given the answer that \"Because no neighborhood of 1 in A U B is compact.\" I don't understand why it wouldn't. \r\n\r\nForgive my ignorance here but could anyone actually provide a proof of the above problem? \r\n\r\nR.D.", "Solution_1": "A neighborhood of $\\{1\\}$ is a set $N$ such that $1$ is contained in the interior of $N$. Since we are dealing with the topology induced from $\\mathbb R$, this means: $\\exists\\ r>0$ such that $(A\\cup B)\\cap (1-r,1+r)\\subseteq N$. It follows that $N$ contains one (in fact infinitely many) of the open intervals that comprise $A$. Consequently, $N$ is not closed in $\\mathbb R$. Finally, recall that a subset of $\\mathbb R$ is compact iff it is closed and bounded." } { "Tag": [], "Problem": "How do you do problems in the form:\r\n\r\nWhat is the smallest $ n$ so that $ 71$ divides $ 7n \\plus{} 1$?", "Solution_1": "[b]7n+1= 0 (mod 71)\n7n = -1 (mod 71)\n7n = 70 (mod 71)\nn = 10.[/b]", "Solution_2": "First note that $ 7n\\plus{}1\\ge 71$ obvsiouly for $ n\\equal{}10$ we get $ \\frac{71}{71}\\equal{}1$ which is the smallest $ n$ for which $ 71|7n\\plus{}1$." } { "Tag": [], "Problem": "Drexel University\u2019s Goodwin College of Professional Studies offers tuition remission for prospective middle- and high-school math and science teachers who enroll in a federally funded program to teach in high-need urban areas. Successful candidates will earn state teacher certification and can choose to teach in the School District of Philadelphia (SDP), Chester/Upland, or Mobile County, Alabama. All course work is online and begins this summer. Candidates will work as classroom teachers while they complete course work towards certification. Salary bonuses and other financial incentives are available for SDP teachers. \r\n\r\nA bachelor\u2019s degree with a minimum GPA of 3.0 from an accredited college or university (with major course work in math or science) and successful completion of the Praxis I (PPST Reading, Writing, Math) exam are required for admission. Official college transcripts and regular access to the Internet are required. A three-year commitment to teaching within designated school district is obligatory. \r\n\r\nFor more information and other program requirements, please contact Tom Yates, Drexel University: tty22@drexel.edu", "Solution_1": "I am also in Drexel University!! They are the ones who host the mathforum.", "Solution_2": "Well, is there anybody who would like to do what Tom Yates proposed?" } { "Tag": [], "Problem": "If you deposit 150 per month on an account that has an annual compound interest of 3percent how much you will have after 10 years? When will kyou exceed $100,000? How much do you need to put per month if you want to have 100,000 in ten years?", "Solution_1": "Doesn't it depend on how often it is compounded, and is the interest calculated after the money is deposited or before?", "Solution_2": "is it compounded monthly, quaterly, or annually??? whatever the compounding period, the best way to do this is by drawing a time line. we did this in math class last year...it's a piece of cake that way...u don't need any formulas or any special equations. \r\n\r\ne.g. if u have $X at the beginning and the interest rate is x% per annum and is compounded monthly, first divide x by 12. after the first month, you'll have (1+x/12)X, after second month, you'll have (1+x/12)^2*X, etc. ends up being a geometric sequence...sum it and u'll get the desired number. \r\n\r\ncheers!" } { "Tag": [ "inequalities", "geometry proposed", "geometry" ], "Problem": "Let $\\triangle ABC$ and the points $M$, $N$, $P$ which belong to the lines $BC$, $CA$, $AB$ respectively. Prove that:\r\n\r\n$\\blacksquare\\ 1^{\\circ}.\\ \\boxed {AM^2+BN^2+CP^2\\ge MN^2+NP^2+PM^2+2\\left(MB\\cdot MC+NC\\cdot NA+PA\\cdot PB\\right)}.$\r\n\r\n$\\blacksquare\\ 2^{\\circ}.$ The triangles $ABC$ and $MNP$ have same centroid if and only if exists the following relation: \\[ \\boxed {\\ AM^2+BN^2+CP^2=MN^2+NP^2+PM^2+2\\left( MB\\cdot MC+NC\\cdot NA+PA\\cdot PB\\right)\\ }\\ . \\] [hide=\"Here are some remarks.\"] See Lemma 3 from the my last reply \nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=67453\nand the my proposed inequality from \nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=68437 [/hide]", "Solution_1": "Let G be the centroid of triangle ABC.From Leibniz we have $\\displaystyle MG^2=\\frac{MA^2+MB^2+MC^2}{3}-\\frac{AB^2+BC^2+CA^2}{9}$,and the analogs for N and P we summ them up and we get $\\displaystyle \\sum MG^2=\\frac{\\sum MA^2-2\\sum MA\\cdot MB}{3}\\geq {\\frac\\sum MN^2}{3}$ so we got the desired ineq.", "Solution_2": "[quote=\"mateivld\"]Let G be the centroid of triangle ABC.From Leibniz we have $\\displaystyle MG^2=\\frac{MA^2+MB^2+MC^2}{3}-\\frac{AB^2+BC^2+CA^2}{9}$,and the analogs for N and P we summ them up and we get $\\displaystyle \\sum MG^2=\\frac{\\sum MA^2-2\\sum MA\\cdot MB}{3}\\geq \\frac{\\sum MN^2}{3}$,from Leibniz ineq, so we got the desired ineq." } { "Tag": [], "Problem": "I found this image online;\r\nRead the Poll\r\n[img]http://img002.picture2life.net/5663304/kj_web-large_medium.png[/img]", "Solution_1": "The image does not move , as u said is an optical effect . :wink:\r\n\r\nSee the figure closely and then you'll realize that it does not move , It seems with movement when u see it with a certain distance to the scream", "Solution_2": "How does that optical illusion work?", "Solution_3": "I think it's because of your tiny eye movements or something. :huh:", "Solution_4": "Check out this one!! :D :D :D", "Solution_5": "whoa.... that's cool... and kinda creepy", "Solution_6": "lol... i know but i chose it because i thought that wuz super cool!!!", "Solution_7": "well it IS super cool so good choice", "Solution_8": "hmm... should i post another one?", "Solution_9": "sure! go for it!!", "Solution_10": "http://www.michaelbach.de/ot/\r\n\r\nThere were so many cool ones.\r\n\r\nCheck it out!!", "Solution_11": "YAY! Illusions.... I grew up with them but my eyes made the dang images look like magic. :rotfl: Yeah, I was a strange child.", "Solution_12": "hmm, should we post more optical illusions?\r\n :D :D\r\n\r\nCOUNT THE NUMBER OF BLACK DOTS!!!!!!! :wink:", "Solution_13": "i'm up for the task.\r\n\r\nAre these rows straight? Or not?\r\n\r\n[img]http://ccl.northwestern.edu/netlogo/models/models/Sample%20Models/Art/Optical%20Illusions.png[/img]", "Solution_14": "haha, nice one! :wink: \r\n\r\nLook at this one!!\r\n\r\nSAY THE COLOR NAME OF THE WORD OUT LOUD!!!! (NOT WHAT THE WORD SAYS) GL!", "Solution_15": "Wow, you're good at those kind of stuff.", "Solution_16": "I could only find 5. Hints?????", "Solution_17": "Baby counts, and look in the sky.", "Solution_18": "whoa...how'd you get nine? i can only see 4, including the baby", "Solution_19": "That's what I once said.\r\nI can faintly see one in the lower left hand corner in the sky...\r\n\r\nI don't think lots of people can see what I can see...", "Solution_20": "Oh yes they can. :)", "Solution_21": "[hide=\"I see 8 people\"]\nthe obvious huge one\nthe woman in the big man's hair\nher baby\nthe man where his face is the big man's eye\nway to the right, at the bottom of the sky, there is a sideways face facing right\nsame thing for the left side of the arch, between the bird and the arch, facing left\nto the right of the black bird is a faint other bird, which is the eye of a person\nto the left of the black bird is a faint face facing forward\n\nnine if the thing at the bottom left counts[/hide]", "Solution_22": "On the right track :).\r\nAnyways,\r\nhere's one.\r\n\r\n[img]http://www.guy-sports.com/fun_pictures/jf_dots1a.jpg[/img]\r\nCan you control the spinning by relaxing your eyes?", "Solution_23": "If you look straight at them and don't allow your eyes to twitch or move at all, they stop moving. :D", "Solution_24": "How do you post images????", "Solution_25": "[code][img]url[/img]\n[/code]\r\nFor example:\r\n[img]http://whywedoit.files.wordpress.com/2009/04/smile.jpg[/img]", "Solution_26": "Also, you can use the File, Browse thing at the bottom as well.\r\n\r\nLike this:", "Solution_27": "oh!!\r\nI will post one!\r\nAlright, stare at the center!!What happens to the circle?\r\n[img]http://www.freebiehome.co.uk/images/stare-optical-illusion.gif[/img]", "Solution_28": "It trembles.", "Solution_29": "lol .. heh..it does tremble" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "In how many ways can 6 faces of a cube be painted with 6 different colours????\r\n\r\n \r\n\r\nPlease explain the answer also..", "Solution_1": "Are alowed to paint two different 'faces' with the same colour ? \r\nIf yes :there are $ 6^{6}$ ways of painting them \r\nIf no : there are $ 6! \\equal{} 6\\cdot 5\\cdot 4\\cdot 3 \\cdot 2\\cdot 1$ ways of painting .", "Solution_2": "The solution is more complex than that if two \"cubes\" are said to be equivalent if one can be achieved by rotating the other in three-dimensional space. Then, for instance, there are only $ 5\\cdot 3\\cdot 2\\equal{}30$ possibilities for the \"non-repetition\" case, and other cases can be found using Polya enumeration (or Burnside's lemma).", "Solution_3": "The answer given is 12", "Solution_4": "It would help hugely if you would write the problem exactly as it is given. They gave the correct answers for the problem you posted, but apparently we were wrong and the answer is 12. I can't see how you could possibly get 12, but maybe I'm mistaken.", "Solution_5": "I got the answer 15, the number of ways to group 6 colors into groups of 2." } { "Tag": [ "search", "AoPSwiki", "algorithm" ], "Problem": "I need some help solving problems that say things like:\r\n\r\nIf in a game you can score 3 points, 4 points, or 7 points, what is the highest score that can't be obtained?\r\n\r\nI don't need to solve the problem, but I need some help with the reasoning. I don't want to do guess and check. I think I have seen a formula for these types of problems, but I can't remember.", "Solution_1": "I'm not sure if there's a formula, but I think you can generate functions.", "Solution_2": "a proof would be nice. :)", "Solution_3": "Thanks. I did a search for it and found [url=http://www.artofproblemsolving.com/Wiki/index.php/Chicken_McNugget_Theorem]this[/url] form good old AoPSWiki, and it has a proof. Just wondering, what about three numbers?", "Solution_4": "eh, for two numbers it's\r\n$ \\ x*y\\minus{}(x\\plus{}y)$\r\nwell i think :|", "Solution_5": "That is the expanded version of athunder's.", "Solution_6": "Assume that $ n$ is the largest number that cannot be achieved only using 3, 4, or 7.\r\n\r\nThen, $ n$ must be smaller or equal to $ 3\\cdot4 \\minus{} 3 \\minus{} 4 \\equal{} 5$ since anything that can be achieved using 3 and 4 can be achieved using 3, 4, and 7.\r\n\r\nRepeating this process, we have $ n\\le 5$, $ n\\le 11$, and $ n\\le 17$.\r\n\r\nHence, we conclude that 5 is the largest number that cannot be achieved using 3, 4, and 7.", "Solution_7": "[quote=\"athunder\"]Well...its a theorem if you want a proof look it up.\n\nI do not think there is one for any over two.[/quote]\r\nI tried to figure out one for three numbers (and I epic failed).\r\nIs there one for 4 numbers?", "Solution_8": "There is no known formula for 3 or more - search Frobenius Coin Problem on Wikipedia.", "Solution_9": "Its also called the dartboard formula or dartboard trick. \r\n\r\nTo make this not all pointless, i'll make a problem and solve it. In a dart, you can either attain 5 or 3 points. Whats the largest number that cannot be attained. The answer would be $ (5*3)\\minus{}5\\minus{}3$ or $ 7$.", "Solution_10": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43173]this[/url] for a bash of mine.\r\n\r\nBasically, you attempt to create every number from 0 to n-1 (mod n) where n is the smallest of the three numbers. Then take the largest number you've created and subtract n from it.\r\n\r\nIn this case:\r\n\r\n0 (mod 3) is 3\r\n1 (mod 3) is 4\r\n2 (mod 3) is 8\r\n\r\nThe largest of those three numbers is 8. Subtracting 3 from it, our answer is 5.\r\n\r\nIt's not a formula, more of an algorithm. If you try to create a proof, I bet it's pretty similar to the proof for two variables in Volume 2. Creating the n numbers should be pretty intuitive. Do you see why we subtract n at the end?\r\n\r\nEDIT: Yeah, uh, (n-1)+1 is not n-1. I can see why I didn't win Mathcounts.", "Solution_11": "[quote=\"xpmath\"]\nBasically, you attempt to create every number from 0 to n-1 (mod n) where n is the smallest of the three numbers. Then take the largest number you've created and subtract n from it.\n\nIn this case:\n\n0 (mod 3) is 3\n1 (mod 3) is 4\n2 (mod 3) is 8\n\nThe largest of those three numbers is 8. Subtracting 3 from it, our answer is 5.\n[/quote]\r\n\r\nHold on. For the 2 mod 3 part, how did you get 8 specifically? You could have gotten 5, or 8, or 11 etc. Same with the 0 mod 3. you could have gotten, say, 12, and since that is the largest number you created, subtracted 3 to get 9 (which is wrong)\r\n\r\nAnd would fantasylovers way work for more than 3 numbers?", "Solution_12": "Sorry, I forgot to mention; you can only use the numbers mentioned in the problem. So, you can only use 3, 4, and 7 to construct the numbers (mod 3). You're trying to make the smallest number possible for each residue. And it'll be pretty much a waste of time if you use anymore 3s after the 0 (mod 3) as it doesn't change the residue.\r\n\r\nFantasy's method works for this since there are small numbers. Try 4, 9, 11 from my link. You'd have to check numbers less than or equal to 23.\r\n\r\nI don't think I'm doing a good job of explaining at 10:30, so ask any questions you have.", "Solution_13": "[quote=\"xpmath\"]For example, consider 4, 9, and 11. We have to find the smallest sum of 4s, 9s, and 11s so we end up with numbers congruent to $ 0, 1, 2, 3 \\pmod{4}$.\n\nSmallest 0 (mod 4): 4\nSmallest 1 (mod 4): 9\nSmallest 2 (mod 4):18\nSmallest 3 (mod 4): 11\n\nOur answer is 18-4=14. [/quote]\r\n\r\nOk, Im extremely dumb and have no idea what residue is. Can you try to explain to me how you got the numbers 4,9,18,11, instead of, say, 4,9,14,11 or 4,9,10,7", "Solution_14": "Err...I'm not sure.\r\n\r\nHere, I'll try explaining again.\r\n\r\nSo, in this case, we have the find the largest number that cannot be created by adding any number of 3s, 4s, or 7s.\r\n\r\nHere's an algorithm, that (I think) will work:\r\n\r\n[hide=\"algorithm, on this example\"]First, take the smallest of the three numbers you're allowed to use, which is 3. \n\nNext, list the possible remainders a number can have when divided by 3, which are 0, 1, and 2.\nNow, find the smallest number that can be created by adding any number of 3s, 4s, and 7s, for each remainder:\n\nThe smallest number that we can make with 3s, 4s, and 7s, which leaves a remainder of 0 when divided by 3, is just 3 itself.\n\nThe smallest number that we can make with 3s, 4s, and 7s, which leaves a remainder of 1 when divided by 3, is 4.\n\nThe smallest number that we can make with 3s, 4s, and 7s, which leaves a remainder of 2 when divided by 3, is 4+4=8.\n\nNow take the largest of the numbers you made; in this case, it's 8. From this, subtract 3 (the smallest of the three numbers you could use), to get your answer, 5.[/hide]\n\n[hide=\"algorithm with 4, 9, 11\"] The smallest of the three numbers is 4.\n\nWe now have to make the smallest numbers, $ a, b, c, d$, that leave remainders of 0, 1, 2, and 3 when divided by 4, respectively. Note that we can [b]only[/b] use the numbers 4, 9, and 11 to make these numbers.\n\nWe can see that the smallest number that leaves a remainder of 0 when divided by 4 that we can make is 4 itself.\n\nThe smallest that leaves a remainder of 1 when divided by 4 is 9.\n\nThe smallest that leaves a remainder of 2 when divided by 4 is 9+9=18.\n\nThe smallest that leaves a remainder of 3 when divided by 4 is 11.\n\nThe greatest of those four numbers that we just got is 18. From here, we subtract 4 to get the answer, 14.[/hide]\r\n\r\nEh you edited. Basically, we can only use the numbers in the problem. You can't make 14 using 4s, 9s, or 11s.\r\n\r\nAlso a residue (mod m) is just any remainder that's possible when dividing a number by m. [url=http://www.artofproblemsolving.com/Wiki/index.php/Residue_class]Residue class[/url], as said on the Wiki.", "Solution_15": "OK now I get it clearly. Thanks! :lol:", "Solution_16": "Sry one more question. Wouldn't 0 mod n be sorta trivial since it would obviously be n, and n-n is 0? That would only be the right answer If 1 were one of the numbers,right?", "Solution_17": "Yeah because if 1 is one of the numbers, you can pretty much make every natural number...\r\n\r\nAlso, yes, for 0 (mod n), you obviously use n :P" } { "Tag": [ "inequalities" ], "Problem": "For a>b>0 and n positive integer > 1 show that for k > = 0\r\n$ (a^n \\plus{} k^n)^{1/n} \\minus{} (b^n \\plus{} k^n)^{1/n} < \\equal{} a \\minus{} b$\r\n :P", "Solution_1": "this has LMV written all over it\r\ndefine $ f(x)\\equal{}(x^{n}\\plus{}y)^{\\frac{1}{n}}$ where $ x,y \\in \\mathbb{R}^{\\plus{}}\\plus{}0$ 0 only for y :wink: \r\nnow $ f'(x)\\equal{}\\frac{x^{n\\minus{}1}}{(x^{n}\\plus{}y)^{\\frac{n\\minus{}1}{n}}} \\leq1$\r\nso \r\n$ \\frac{f(a)\\minus{}f(b)}{a\\minus{}b} \\equal{}f'(c) \\leq 1$\r\nequality only holding when $ k\\equal{}0$\r\nresult follows", "Solution_2": "[quote=\"pardesi\"]this has LMV written all over it\ndefine $ f(x) \\equal{} (x^{n} \\plus{} y)^{\\frac {1}{n}}$ where $ x,y \\in \\mathbb{R}^{ \\plus{} } \\plus{} 0$ 0 only for y :wink: \nnow $ f'(x) \\equal{} \\frac {x^{n \\minus{} 1}}{(x^{n} \\plus{} y)^{\\frac {n \\minus{} 1}{n}}} \\leq1$\nso \nresult follows[/quote]\r\n\r\nHow did ya get f'(x) as that value isn't it 1/$ n^2$ times that value? :maybe:\r\n\r\nNo sorry i figured it out :blush:", "Solution_3": "More solns in a different approach would be welcome :) :wink:\r\nWhat the hell is the 10 mins rule for??? :mad:", "Solution_4": "may i know what is lmv??", "Solution_5": "Lagrange's Mean Value Theorem", "Solution_6": "Actually this sum is a sum from an IIT book for inequalities you know so try using inequalities :P\r\n\r\n@ Shreyas : Please see the private msg." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "Say we have a number like: 541685414. Can it be written as a sum of three perfect cubes? Can someone show me the steps required to tackle a problem like this, because I'm not really interested in the final answer.\r\n\r\nThanks.", "Solution_1": "You can try modulo 7 and 9 in general case. But for your particular example it doesn't work :(", "Solution_2": "Why modulo 7 or 9? What happens then?", "Solution_3": "Cubes gives remainders -1, 0, +1 modulo 7 or 9. Threfore, sum of two cubes can't be equal to 7k+3, or sum of three cubes can't be equal to 9k+4 and 9k+5." } { "Tag": [ "AMC 10", "AMC" ], "Problem": "Just wondering for people who have used the book: First Steps for Math Olympians:\r\n\r\nDo you find it useful in preparing for the AMC 10?\r\nIf I did AoPS V1 and did all of the problems, would I need it or would it be overkill?\r\n\r\nAlso, will doing the AMC 10 online class boost my score from a current 110-111 average to 120+?\r\n\r\nThanks guys", "Solution_1": "Definitely FSMO will help, however i recommend AoPs v1 b/c it boosted my score alot. the class is good to.", "Solution_2": "You can also find a lot on the AMC website:\r\nhttp://www.unl.edu/amc." } { "Tag": [], "Problem": "Find all ordered pairs $ a,\\ b,\\ c$ of positive integers such that $ \\left\\{\r\n\\begin{array}{ll}\r\nab\\plus{}c\\equal{}13 &\\quad \\\\\r\n a\\plus{}bc\\equal{}23 &\\quad\r\n\\end{array}\r\n\\right.$", "Solution_1": "[hide=\" this is too easy\"]\n$ (a\\plus{}bc)\\minus{}(ab\\plus{}c)\\equal{}10$\n$ a\\minus{}ab\\minus{}c\\plus{}bc\\equal{}10$\n$ a(1\\minus{}b)\\minus{}c(1\\minus{}b)\\equal{}10$\n$ (c\\minus{}a)(b\\minus{}1)\\equal{}10 \\implies b\\minus{}1\\in\\{1,2,5,10\\}$\n\nchecking all 4 cases we get 2 solutions: $ (1,2,11)$ and $ (1,11,2)$[/hide]", "Solution_2": "Not perfect! :(", "Solution_3": "Not perfect in a way not entirely correct or in a way this could have been solved in a nicer way?", "Solution_4": "You have almost solved the problem, but you forgot one more pair.", "Solution_5": "The only thing that could work is $ (23, 0, 13)$ but $ 0$ is not a positive integer.", "Solution_6": "Right, but you can find another pair.", "Solution_7": "[hide=\"The third\"]You neglected to record $ (2,3,7)$, which results from $ b\\minus{}1 \\equal{} 2 \\implies b\\equal{}3$[/hide]", "Solution_8": "Yeah, you are right. I had a typo in a notebook and was too lazy to check XD", "Solution_9": "Now the problem has been finished. :lol: \r\n\r\n[hide=\"Answer\"]$ \\{a,\\ b,\\ c\\}\\in{\\{1,\\ 2,\\ 11\\}, \\{1, \\ 11,\\ 2\\},\\ \\{2,\\ 3,\\ 7\\}}$.[/hide]" } { "Tag": [], "Problem": "A five-legged Martian has a drawer full of socks, each of which is red,\nwhite or blue, and there are at least five socks of each color. The Martian\npulls out one sock at a time without looking. How many socks must the\nMartian remove from the drawer to be certain there will be 5 socks of the\nsame color?", "Solution_1": "The worst case scenario is if we get $ 4$ of each of the $ 3$ colors.\r\n\r\nThis gives $ 12$ socks. The next sock must be a matching sock\r\n\r\nto one of the colors, giving $ \\boxed{13}$ as the answer." } { "Tag": [], "Problem": "If $ y \\minus{} x \\equal{} 7$, what is the value of $ x \\minus{} y$ ?", "Solution_1": "There is a very useful identity $ (a\\minus{}b)\\equal{}\\minus{}(b\\minus{}a)$.\r\n\r\nThis can be proven by multiplying $ (a\\minus{}b)$ by $ \\minus{}1$ to get $ (\\minus{}a\\plus{}b)$, and then rearranging terms to get $ (b\\minus{}a)$.\r\n\r\nSo, we substitute $ y$ for $ a$ and $ x$ for $ b$; if $ y\\minus{}x\\equal{}7$, then $ x\\minus{}y\\equal{}\\boxed{\\minus{}7}$.", "Solution_2": "Can you edit it so that the solution is hidden?\n[hide=\"Psst!\"]\nTo hide your message, write it in this format: [quote] [hi de=\"Title of the message\"] \"Message\" [/hi de][/quote]\nignoring the spaces in between hi and de. [/hide]", "Solution_3": "$[hide=\"Why??????????\"]\"Why?\"[/hide]$", "Solution_4": "[code][hide=\"Why??????????\"]Why?[/hide][/code]\n^That's the code for it", "Solution_5": "Who cares about the spaces? LOL", "Solution_6": "I like HTML, no spacing errors", "Solution_7": "[hide=\"Why??????????\"]Why?[/hide]\nI did [hide = \" Why??????????\"]Why?[/hide] or something" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "find all possible value of $ \\LARGE F_{i}(t) (0 \\leq t \\leq 1,1 \\leq i \\leq n)$\r\ndefine:$ \\LARGE F \\equal{} max_{(0 \\leq x_{i} \\leq 1)} \\left| F_{1}(x_{1}) \\plus{} F_{2}(x_{2}) \\plus{} ...... \\plus{} F_{n}(x_{n}) \\minus{} x_{1}x_{2}......x_{n} \\right|$\r\ndetermine $ F_{min}$\r\n\r\n\r\n\r\n\r\n\r\nnot typical problem.Our Question for thinking\r\nhard to me :|", "Solution_1": "[b] anybody have thoughts on this problem?[/b] :maybe:" } { "Tag": [ "linear algebra", "matrix", "vector" ], "Problem": "Let $ A\\equal{}span\\{(1,2,3,4), (4,3,2,1), (2,3,4,5), (1,1,1,1)\\}\\subset R^4$ and $ B\\equal{}span\\{(\\minus{}1,\\minus{}2, 1, 0), (1,\\minus{}3,0,1)\\}$\r\nDo there exist a subspace $ C\\subset B$ such that $ A\\oplus C\\equal{}R^4$?\r\n\r\n$ A\\oplus C\\equal{}R^4$ means that: $ R^4\\equal{}A\\plus{}C$ and $ A\\cap C\\equal{}{0}$", "Solution_1": "Such a subspace $ C$ exists if and only if $ A\\plus{}B\\equal{}\\mathbb R^4$. The necessity should be clear. For sufficiency, suppose that $ A\\plus{}B\\equal{}\\mathbb R^4$. Choose a basis of $ A$ and a basis of $ B$. Add the elements of the basis of $ B$ to the basis of $ A$ one by one, while making sure that the union is linearly independent. When you can't do this any more, you have a basis of $ \\mathbb R^4$. The added basis elements from $ B$ will be the basis of $ C$. \r\n\r\nBecause of the above, the question can be answered by forming a 6x4 matrix of all given vectors and computing its rank.", "Solution_2": "I forgot the second point of this problem. Here it is:\r\nFind or prove that it's impossible to find such subspaces of $ R^4$- $ U, V$ ,that:\r\n\\[ R^4\\equal{}A\\oplus U\\equal{}U\\oplus V\\equal{}V\\oplus A\\]\r\n(A is the same as above)", "Solution_3": "That is possible if and only if $ \\mathrm{dim}(A)\\equal{}2$. Informally, $ U$ and $ V$ can be two \"random\" two-dimensional subspaces. Formally, you choose $ U$ to be some complement of $ A$ and then think of what $ V$ should be." } { "Tag": [], "Problem": "Which of these three numbers is between the other two on the number line? Answer with the letter only.\n\nA) $ 1/\\sqrt[3]5$\n\nB) $ 2\\minus{}\\sqrt2$\n\nC) $ 8\\minus{}\\sqrt{55}$", "Solution_1": "[quote=\"GameBot\"]Which of these three numbers is between the other two on the number line? Answer with the letter only.\n\nA) $ 1/\\sqrt [3]5$\n\nB) $ 2 - \\sqrt2$\n\nC) $ 8 - \\sqrt {55}$[/quote]\r\n\r\nLets start by comparing B and C.\r\n\r\n$ 2-\\sqrt2 \\? 8-\\sqrt{55}$\r\n$ \\sqrt{55}-\\sqrt{2} \\? 6$\r\n$ 57-2\\sqrt{110} \\? 36$\r\n$ 21 \\? 2\\sqrt{110}$\r\n$ \\frac{441}{4} \\? 110$\r\nSo B is larger then C.\r\n\r\n$ \\frac{1}{\\sqrt{3}} \\? 2-\\sqrt{2}$\r\n$ 1 \\? 2\\sqrt{3}-\\sqrt{6}$\r\n$ 7+2\\sqrt{6} \\? 12$\r\n$ 2 \\sqrt{6} \\? 5$\r\n$ 24 \\? 25$\r\n\r\nSo B is larger then A.\r\n\r\n$ \\frac{1}{\\sqrt{3}} \\? 8-\\sqrt{55}$\r\n$ 1 \\? 8\\sqrt{3}-\\sqrt{165}$\r\n$ 166+2\\sqrt{165} \\? 192$\r\n$ 2\\sqrt{165} \\? 26$\r\n$ 660 \\? 676$\r\nSo C is larger then A\r\n\r\nSo B>C>A, and our answer is C.", "Solution_2": "You can just put the numbers into your calculator and then compare them if you wanted the easy way to the answer or if you were lazy.... :wink:" } { "Tag": [], "Problem": "Each valve A, B, and C, when open, releases water into a pool at its own constant rate. With all three valves open, the tank fills in 1 hour, with only valves A and C open it takes 1.5 hours, and with only valves B and C open it takes 2 hours. How long will it take to fill the tank with only valves A and B open?", "Solution_1": "[hide=\"Answer\"]Valves $A$ and $C$ will fill $\\frac{2}{3}$ of the tank in 1 hour, and valves $B$ and $C$ will fill $\\frac{1}{2}$ of the tank in one hour. $\\frac{1}{A}+\\frac{1}{B}+\\frac{1}{C}=1$ and $\\frac{1}{A}+\\frac{1}{C}=\\frac{2}{3}$, so $1-\\frac{2}{3}=\\frac{1}{3}=\\frac{1}{B}$. $\\frac{1}{B}+\\frac{1}{C}=\\frac{1}{2}$, so $\\frac{1}{C}=\\frac{1}{6} \\Rightarrow \\frac{1}{A}+\\frac{1}{6}=\\frac{2}{3}$, so $\\frac{1}{A}=\\frac{1}{2}$, and $\\frac{1}{A}+\\frac{1}{B}=\\frac{5}{6}$ of the tank filled in one hour, so the entire tank is filled in $72$ minutes.[/hide]\r\n\r\n[b]EDIT:[/b] Modified solution to make a little more sense.", "Solution_2": "[quote=\"4everwise\"]Each valve A, B, and C, when open, releases water into a pool at its own constant rate. With all three valves open, the tank fills in 1 hour, with only valves A and C open it takes 1.5 hours, and with only valves B and C open it takes 2 hours. How long will it take to fill the tank with only valves A and B open?[/quote]\r\n\r\n\r\n[hide]\nTo solve this problem, it is essential to know that the time needed to finish filling pool divided by all the members equals 1:\nI do realize that this solution is worded well, but you will understand by looking at the solution :o \n$\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=1$\n$\\frac{3}{2a}+\\frac{3}{2b}=1$ \n$\\frac{2}{b}+\\frac{2}{c}=1$ \n\nSolve for a and b in terms of c, and then substitute c back in to equation 2 and 3\n\nAns. $\\frac{3}{2}$ hours :D [/hide]", "Solution_3": "[hide]Speeds a,b,c\na+b+c=1\na+c=1/1.5\nb+c=1/2\na+b=1-1/1.5+1-1/2=5/6\ntime=1/(5/6)=6/5 hours[/hide]" } { "Tag": [ "search", "real analysis", "real analysis unsolved" ], "Problem": "Here is an interesting summation that I came up with. I am hoping to check to make sure that I evaluated it correctly. How would you evaluate it? What did you come up with?\r\n\r\n$\\sum_{n=2}^{\\infty}\\frac{(-1)^{n+1}4^{n}(6n^{2}+9n+3){\\pi}^{2n+1}}{(2n)!}$\r\n\r\n[hide=\"Here is my answer.\"]$-30{\\pi}^{3}$[/hide]", "Solution_1": "This is a Ramanujan-like series, similar sums have been obtained by Guillera and others by using integers relation method, you might want to search the Ramanujan journal for papers on such topics.", "Solution_2": "$6n^{2}+9n+3 = \\frac{3}{2}(2n)(2n-1)+6\\cdot 2n+3$, so we can re-write the sum as\r\n\\[\\sum_{n = 2}^{\\inf}-\\frac{3\\pi}{2}\\left(\\frac{(2\\pi i)^{2n}}{(2n-2)!}\\right)-6\\pi \\frac{(2\\pi i)^{2n}}{(2n-1)!}-3\\pi \\frac{(2\\pi i)^{2n}}{(2n)!}\\]\r\nand then you can calculate it from the power series for $\\cosh$ and $\\sinh$." } { "Tag": [], "Problem": "Does anyone here play on chess live? If you do, please list your handle here. Mine is johnnyfox :) if you need with chess live contact me there im an unofficial SR(too young to be a real one)but i know the site well", "Solution_1": "smallpawn but hopefully will change it to \"whatev\" someday", "Solution_2": "Nah, you'll be too lazy lazy lazy. Plus, you should change it to ORZNORZ!!!\r\n\r\nBy the way, ORZNORZ is the name of our iTest team. Do you or Jeremy want to be on it? PM me.", "Solution_3": "chess live is a cool place, its a newer server though, so it has less people than ICC and other places like that :(", "Solution_4": "Anyone know how we can get a computer account on USCL?", "Solution_5": "hope this helps... some info on computer applying and computer usage :) hope to play you there!\r\n\r\ncomputer_app \r\nThis topic is generated from a server help file and is directed at the command line interface that underlies Chess Live. While it requires more knowledge to use, this text interface allows great flexibility in using the server. See the help on Command Line Interface for more details.\r\n\r\nIf you want to have a computer account, please send a message to Computers with ALL of the following info (please use this as a form): \r\n\r\nFull name: Handle: Requested Handle: (must be available) Engine name: Programmer: Hardware: Software: Variant suported: Manual or Automatic operation: Login times: Estimated rating: Main time control and variant: Reason for having a computer account: \r\n\r\nBe sure to read the computers help file for the guidelines of a computer account. Be advised that the decision taken by the computer commitee is final and without appeal. \r\n\r\nHere is the basis upon which computer accounts are granted: \r\n\r\n- Priority is given to those who can play variants. - Priority is given to automatic computers (not need of an operator.) - Priority is given to those who have had an account here for some time. - Priority is given to those which are built from home grown code and operated by the programmer. \r\n\r\nOf course, the main criteria in deciding on which computer accounts are allowed is the current demand. \r\n\r\n------------------------------------------------------------------------------\r\n\r\nGeneral rules regarding computer accounts \r\n\r\nComputers may register and play rated chess matches. However, computers playing rated games are required to be registered so that users can know whether they are playing against humans or computers. Using an unregistered computer to play rated chess matches on this server is considered to be abusive behavior and will result in sanctions; see etiquette. \r\n\r\nTo request a computer account, visit the Computers section of the Chess Live web-site. At this time, computer accounts will only be granted for computers of U2000 strength! You will find an application form and other information there. It takes some technical skill to get a Computer account running, as you must use WinBoard rather than the Chess Live client to control your chess engine, and you must spend some effort configuring it. There is a How to Get Crafty working on USCL page which can help you out with this. Note that Chess Live staff cannot for the most part help you configure WinBoard and the engine in question.\r\n\r\nSpecial Notes: \r\n\r\nServer admins reserve the right to limit the number of computer accounts registered on the server and may seek to ensure a wide range of playing strength of these accounts. \r\nComputer accounts that want to be open for challenges for suicide chess games need to do the following: have the 'f1' formula variable set to \"suicide\" (with no extra spaces of any kind); you need not use 'f1' as part of the 'formula' setting. Computer accounts without 'f1' set to \"suicide\" will have challenges for suicide chess declined automatically. In the same fashion as for suicide, set 'f2' to \"crazyhouse\" if you wish for your computer account to be open for crazyhouse challenges. \r\nComputer accounts may only play in tournaments that are explicitly announced to be open to computers when the tournament is opened. Joining a tourney with a computer account when it was not authorized may result in an immediate ban from tourneys. \r\nIf computer account is not used for more than 3 months it may be removed WITHOUT any warning. We want the computers on the server to be active. \r\nComputers should always keep information about hardware/software in finger notes. This must include (at a minimum): the owner of the computer account, the type of machine it is being run on, and the engine that is being used by the account. Providing this information is mandatory, and not having it in your computer's finger notes may result in a loss of the account. \r\nRated games between a computer and account and its operator are NOT allowed. Play your computer unrated only if you want to test it. \r\nAbuse of any kind by either the owner or the computer account itself may result in the removal of the account in addition to other sanctions. \r\nHow to start a game against a computer \r\nComputer accounts on Chess Live consist of a variety of chess programs. There are computers for players of all skill levels. Some computers are intentionally configured to play below their potential strength in order to assist beginning and intermediate players. The majority of computers, however, are very strong. \r\n\r\n\r\nHow to see what computer accounts are currently on-line\r\nIf you decided to play a game or few against a computer and want to see which computers are currently available for a game, To do this, check for any players in the Players Tab of the Info Window for any players whose name is followed by a (C) badge. \r\n\r\n\r\nHow to find computer's strength and other qualifications\r\nAfter you found a computer you might like to play, you would want to check its qualifications to make sure that it suits your playing interests. To do that, right click over their name in the players window and select Profile, \r\nAfter you did that, on your screen you will see this computer's Profile which will contain it's rating table and any notes that the operator has set for that particular account. \r\n\r\nHow to start a game against a computer \r\nThere are two ways: \r\n\r\nCheck in the Seek Tab of the Info Window for any square dots, these represent computer seeks. If you hold the mouse over any of these, you will see the name of the computer, its rating, and the type of game being offered. Double clicking on the seek will begin the game. \r\nChallenge the computer directly. However, before challenging the selected computer to a game, make sure it plays the variant that you want. Some computers are created specifically to play wild, suicide, crazyhouse, etc. variants. When you are ready to play, the easiest way to issue a challenge is to go to \"Actions\" menu, and selecting \"Challenge Opponent\" option. Follow the same procedure as you would against a regular opponent. Remember that computer accounts have the same privileges as regular user accounts (expect for some technical commands). \r\n\r\nIf at any time during the game against a computer you find that something goes wrong, please message its operator! The name of the operator can be found in computer's finger notes. It is very important that you let the operator know about anything that goes wrong with the computer. Most of the time, automatic computer accounts are unattended and therefore the operator has no idea if something goes wrong. If the computer seriously misbehaves, contact an on-line administrator. \r\n-----------------------------------------------------------------------------------", "Solution_6": "I'm going to assume that was copy+pasted.....\r\n\r\nI play in FICS, USCL was too empty and I could never find human ppl to play", "Solution_7": "Yes, please CITE everything you quote off another source. It's the legal and polite thing to do.\r\n\r\nOtherwise, I was kinda wondering if anyone has one. I play on USCL sometimes and I would kinda like to have a computer account as well.", "Solution_8": "i haven't been on in such a long time but i hear my account champchess still has its TM abilities, maybe something to do a little this summer.", "Solution_9": "How do you get to it? Do you have to download anything to use it? Is it welcome to people under the age of 18? Please give me your answ3ers!", "Solution_10": "My username is mathprodigy. (Not exactly true, but.... :D)\r\n\r\nIf you challenge me and tell me you're from AoPS, I always accept.", "Solution_11": "I'm knightmair. I am an ex-SR and ex-TM (both removed due to inactivity) but I have a good 300 tournaments run and 30 online days logged.", "Solution_12": "I unfortunately don't play on US chess live.\r\n\r\nDoes anyone play on FICS? If so, add me! My user name is JavaManIssa", "Solution_13": "Just downloaded Chess Live. Handle=fireemblem", "Solution_14": "does anybody know a free bughouse site (preferably no downloads) where 2 people can practice?", "Solution_15": "Yeah, that will be awesome. Also, do anybody know a download for four player chess? Heard it's really fun.", "Solution_16": "Anyone play on FICS? I'm totient.", "Solution_17": "I play on FICS, I added you to my GNotify.. it says you've never connected or played any rated games :ninja:", "Solution_18": "[quote=\"JavaMan\"]I play on FICS, I added you to my GNotify.. it says you've never connected or played any rated games :ninja:[/quote]\r\n\r\nLol, that's because I just made my FICS account right before I posted. :lol: I'll try to get on sometime soon. What's your handle?", "Solution_19": "mine is brick. i couldn't thnk of anything else. i'm not on much though. i might be on more in the future, when i get a faster computer.", "Solution_20": "My handle's JavaManIssa\r\n\r\nJeesh Brick, you last disconnected 1998 :lol:" } { "Tag": [], "Problem": "A rectangular box is 4 cm thick, and its square bases measure 16 cm by 16 cm. What is the distance, in centimeters, from the center point $ P$ of one square base to corner $ Q$ of the opposite base? Express your answer in simplest terms.\n\n[asy]import three;\ndraw((0,0,1/4)--(1,0,1/4)--(1,1,1/4)--(0,1,1/4)--(0,0,1/4)--cycle3,linewidth(2));\n\ndraw((0,1,0)--(1,1,0),linewidth(2));\ndraw((1,1,0)--(1,0,0),linewidth(2));\ndraw((0,1,0)--(0,1,1/4),linewidth(2));\ndraw((1,1,0)--(1,1,1/4),linewidth(2));\ndraw((1,0,0)--(1,0,1/4),linewidth(2));\n\ndot((1/2,1/2,1/4));\ndot((0,1,0));\n\nlabel(\"P\",(1/2,1/2,1/4),W);\nlabel(\"Q\",(0,1,0),E);[/asy]", "Solution_1": "[hide]12 if I am correct.[/hide]", "Solution_2": "[quote=\"thdanh90\"][hide]12 if I am correct.[/hide][/quote]\r\n\r\nyou know, people would appreciate it if you included HOW you got your answer", "Solution_3": "[hide][asy]import three;\ndraw((0,0,1/4)--(1,0,1/4)--(1,1,1/4)--(0,1,1/4)--(0,0,1/4)--cycle3,linewidth(2));\n\ndraw((0,1,0)--(1,1,0),linewidth(2));\ndraw((1,1,0)--(1,0,0),linewidth(2));\ndraw((0,1,0)--(0,1,1/4),linewidth(2));\ndraw((1,1,0)--(1,1,1/4),linewidth(2));\ndraw((1,0,0)--(1,0,1/4),linewidth(2));\n\ndot((1/2,1/2,1/4));\ndot((0,1,0));\n\nlabel(\"P\",(1/2,1/2,1/4),W);\nlabel(\"Q\",(0,1,0),E);\nlabel(\"R\",(0,1,1/4),NE);[/asy]\n$ PR=8\\sqrt{2}, QR=4$\nSince $ \\angle PRQ = 90^{\\circ}$, we have $ PQ^2=(8\\sqrt{2})^2 + 4^2=128+16=144 \\Rightarrow PQ=12$.[/hide]" } { "Tag": [ "trigonometry", "function", "calculus", "calculus computations" ], "Problem": "Investigate the stability of the constant solutions $ (u_{n\\plus{}1} \\equal{} u_n)$ of the discrete equation\r\n\\[ u_{n\\plus{}1} \\equal{} 4u_n (1 \\minus{} u_n)\\]In the case $ 0 \\le u_0 \\le 1$, use the substitution $ u_0 \\equal{} \\sin^2 \\theta$ to find the general solution and verify your stability results. Can you find an explicit form of the general solution in the case $ u_0 > 1$?", "Solution_1": "I want to type for you but I don't know how...(I'm a computer illiteracy, sorry.) It's about the logistic map. See the book [Mathematical Methods for Physicists] by Arfken. \r\n\r\n Thinking graph of the function $ y \\equal{} 4x(1 \\minus{} x)$and $ y\\equal{}x$ will help.", "Solution_2": "The interation $ u_{n+1}=f(u_n)$ can only possibly converge to a fixed point of the function, namely an $ x$ such that $ x=f(x).$\r\n\r\nSuch a fixed point $ x$ is stable (all sequences starting sufficiently close to $ x$ converge to $ x$) if an only if $ |f'(x)|<1.$ If $ |f'(x)|>1,$ then the fixed point $ x$ is unstable, and sequences that start very close to (but not exactly at) $ x$ do not remain very close to $ x.$\r\n\r\n(Point for which $ |f'(x)|=1$ require more delicate analysis; I will not attempt such analysis below.)\r\n\r\nThe family of logistic mappings $ f(x)=rx(1-x)$ maps $ [0,1]$ into $ [0,1]$ as long as $ 0\\le r\\le 4.$\r\n\r\nSolving $ x=rx(1-x)$ gives us two roots: $ x=0$ and $ x=1-\\frac1r.$\r\n\r\n$ f'(x)=r$ for $ x=0$ and $ f'(x)=2-r$ for $ x=1-\\frac1r.$\r\n\r\nFor $ 0\\le x<1,$ the point $ x=0$ is a stable fixed point and the point $ x=1-\\frac1r$ does not lie in $ [0,1]$ and cannot be accessed by a sequence starting inside $ [0,1]$. Hence, $ x=0$ is the unique stable solution, and all sequences starting in $ [0,1]$ tend to zero.\r\n\r\nFor $ 1 \\frac{\\sin^2 x}x = \\frac{1}{2}(\\frac{1}x - \\frac{\\cos 2x}x) \\]" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "To side AC of right triangle (0$ and $ a\\in K$, if $ f(x)\\equal{}x^p\\minus{}x\\minus{}a$ has no root in $ K$, then $ f(x)$ is irreducible in $ K[x]$.", "Solution_1": "in fact, $ f$ splits into linear factors or it is irreducible. there are two well-known proofs for this:\r\n\r\n1) if $ c$ is a root in a splitting field (of $ f$), then obviously $ c \\plus{} i$ are the other ones, where $ i$ runs through $ \\mathbb{F}_p$. in particular, $ K(c)$ is a splitting field. with $ f' \\equal{} \\minus{} 1$ we see that $ f$ is separable. hence $ K(c)$ is a galois extension of $ K$. our aim is to prove that the galois group is isomorphic to $ \\mathbb{Z}/p$, for then $ p \\equal{} |K(c) : K|$ is the degree of an irreducible factor of $ f$, thus $ f$. observe that $ Gal(K(c) : K) \\to \\mathbb{Z}/p , \\sigma \\mapsto \\sigma(c) \\minus{} c$ is a well-defined injective group homomorphism, with nontrivial image (else $ c \\in K$), thus an isomorphism since $ p$ is prime.\r\n\r\n2) $ f$ is invariant under the automorphism $ x \\mapsto x \\plus{} 1$ of $ K[x]$. thus $ \\mathbb{Z}/p$ acts on the monic irreducible factors $ g$ of $ f$. if some $ g$ is a fixed point, $ g(x) \\minus{} g(0)$ has every $ x \\in \\mathbb{F}_p$ as a root, thus has degree $ \\geq p$ and $ f \\equal{} g$ is irreducible. otherwise all orbits have $ p$ elements, which yield $ p$ distinct monic irreducible factors and thus a splitting of $ f$.", "Solution_2": "Suppose $ f\\equal{}gh$ is a nontrivial factorization of $ f$ in $ K[x]$ with $ g,h$ monic. Let $ u$ be a root of $ f$ in some splitting field. The remaining roots (in that splitting field) are $ u\\plus{}1,u\\plus{}2,\\ldots,u\\plus{}p\\minus{}1$. Write $ g\\equal{}x^k\\plus{}c_{k\\minus{}1}x^{k\\minus{}1}\\plus{}\\cdots\\plus{}c_1x\\plus{}c_0$ where $ c_i\\in K$ and $ 0b$, in other words it uses that $ Y$ is totally ordered.\r\nI also used that every finite set has a minimum. This is an equivalent formulation of being totally ordered.", "Solution_4": "We must be using different definitions because in Munkres an ordered set is comparable which is what we are assuming." } { "Tag": [ "inequalities", "ratio" ], "Problem": "Prove or disprove that for all numbers $ 0 \\le p \\le 1$ and $ 0 \\le q \\le 1$, $ 2pq > p^2$ and $ 2pq > q^2$.", "Solution_1": "I think you made a typo: If $ p,q\\equal{}0$, then $ p^2\\equal{}0\\equal{}2pq$ and $ q^2\\equal{}2pq$. I believe it should read greater than or equal to, not just greater than.", "Solution_2": "[quote=\"7h3.D3m0n.117\"]Prove or disprove that for all numbers $ 0 \\le p \\le 1$ and $ 0 \\le q \\le 1$, $ 2pq > p^2$ and $ 2pq > q^2$.[/quote]\r\nFor p=0.3 and q=0.1\r\n$ p^2\\equal{}0.09$ and $ q^2\\equal{}0.01$\r\n$ 2pq\\equal{}0.06 < p^2$", "Solution_3": "[quote=\"7h3.D3m0n.117\"]Prove or disprove that for all numbers $ 0 \\le p \\le 1$ and $ 0 \\le q \\le 1$, $ 2pq > p^2$ and $ 2pq > q^2$.[/quote]\r\n\r\nSince this is Hardy-Weinberg I think 7h3.D3m0n.117 meant that $ p \\plus{} q \\equal{} 1,\\ 0 \\le p,q \\le 1$. But the inequalities don't make sense no matter how you try to think about them .. (unless its supposed to be really easy to disprove)", "Solution_4": "[quote=\"azjps\"]Since this is Hardy-Weinberg I think 7h3.D3m0n.117 meant that $ p \\plus{} q \\equal{} 1,\\ 0 \\le p,q \\le 1$. But the inequalities don't make sense no matter how you try to think about them .. (unless its supposed to be really easy to disprove)[/quote]That is exactly a missing condition ($ p\\plus{}q\\equal{}1$) in Hardy-Weinberg. For more info, if anyone is interested in Biology, here is a link to it\r\n\r\n[url]http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/H/Hardy_Weinberg.html[/url]", "Solution_5": "Since I originally placed this in HSB, I meant to prove that $ 2pq$ (representing the frequency of heterozygous genotypes) would always be greater than $ p^2$ or $ q^2$, since by viewing a monohybrid Punnet Square, the ratio of homozygous dominant:homozygous reccessive:heterozygous would be 1:1:2.", "Solution_6": "[quote=\"7h3.D3m0n.117\"]Prove or disprove that for all numbers $ 0 \\le p \\le 1$ and $ 0 \\le q \\le 1$, $ 2pq > p^2$ and $ 2pq > q^2$.[/quote]\n\n[quote=\"7h3.D3m0n.117\"]Since I originally placed this in HSB, I meant to prove that $ 2pq$ (representing the frequency of heterozygous genotypes) would always be greater than $ p^2$ or $ q^2$, since by viewing a monohybrid Punnet Square, the ratio of homozygous dominant:homozygous reccessive:heterozygous would be 1:1:2.[/quote]\r\n\r\nCounterexample for quote #1: let $ p\\equal{}0.1$. It's an 'and' statement, so one of the inequalities is false implies the claim is false.\r\n\r\nQuote #2 is too weak and $ 2pq$ is at least greater than $ p^2$ or $ q^2$\r\n\r\nQuote #1 can be true for certain values of $ p$. Find that interval of values for $ p$ for which quote #1 is true, that is, for which $ 2pq > p^2$ [u]and[/u] $ 2pq > q^2$." } { "Tag": [ "CEMC" ], "Problem": "Does anyone know when the results for FGH will be released?", "Solution_1": "i am usually the first to know about these things (my teacher is in the loop and he tells it to us as soon as he hear something)... I'll let you guys know if i hear these things", "Solution_2": "Ok thx.\r\n\r\nFor once, I actually think I did pretty good on Fryer that's why I'm getting more and more anxious.", "Solution_3": "I just got my result from Fryer today.\r\n\r\nI scored a 32...in my opinion, pretty bad.\r\n\r\nSecond in my school.", "Solution_4": "[quote=\"kool_dudy\"]I just got my result from Fryer today.\n\nI scored a 32...in my opinion, pretty bad.\n\nSecond in my school.[/quote]\r\n\r\nAverage for Fryer was nearly 27.", "Solution_5": "I'm really disappointed now. I just went to CEMC and found out that gold standard is at 36 and I'm only eligible for bronze...:(" } { "Tag": [ "geometry", "search", "combinatorics proposed", "combinatorics" ], "Problem": "Given 18 points in the plane, no three collinear, so that they form 816 triangles. The sum of the area of these triangles is A. Six are colored red, six green and six blue. Show that the sum of the areas of the triangles whose vertices are the same color does not exceed A/4.", "Solution_1": "Have a look here :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?search_id=1657994991&t=3300\r\n\r\nPierre." } { "Tag": [], "Problem": "http://www.zeropages.co.uk/norpton\r\n\r\nhope you like it and good luck.", "Solution_1": "i can't find the game.", "Solution_2": "i'm stuck on level 3", "Solution_3": "[quote=\"math92\"]i can't find the game.[/quote]\r\n\r\n[hide=\"It is there.\"]The way to the first level isn't as simple as clicking a link. Unless you're not doing it on your own.[/hide]", "Solution_4": "i still can't get there :(", "Solution_5": "[hide=\"Hint\"]Try using the things under \"The following things will help you a lot\" on the hints page.[/hide]", "Solution_6": "i am stuck on level 4 cant find the password or username :blush: can anyone help?", "Solution_7": "This is what I figured out for level 3:\r\n[hide]The stuff in the corner is assembly code, and according to the assembly code tutorial on the main site, it does nothing. Also, there's the title, \"dont leave me!\" and the picture, in which the word LEAVE can be seen in red. I wonder if this refers to going \"up\" a page in order to find the tutorial? Or does it refer to some assembly command? I also don't get \"the slash goes inbetween\" or \"the power to create everything.\"[/hide]", "Solution_8": "I made it to level 5 of notpron, and am currently at level 2 of norpton. Keep adding more stuff!", "Solution_9": "Hint for level 3: your on the right track but you need to know what it does", "Solution_10": "[quote=\"math92\"]i still can't get there :([/quote]\r\n\r\nMe too. :( I've tryied to change the URL ... but nothing(level.htm, start.htm, game.htm). They don't work.", "Solution_11": "u need to decode the numbers/letters. its one word look at the other hints", "Solution_12": "Right now my browser is having trouble opening your website..." } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "If into a finite unitar ring A, 2 is nilpotent, then card(A)=2^n ?", "Solution_1": "If $ 2$ is nilpotent, the additive order of any element is a power of $ 2$. Apply Cauchy's theorem.", "Solution_2": "Why the additive order of any element is a power of 2 ?", "Solution_3": "Because if $ 2^n\\equal{}0$, $ 2^n x \\equal{} 0$." } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "is $X^{6}+5$ reductible in $Z_{7}$ ?", "Solution_1": "By Fermats little theorem, $X^{6}\\equiv1(\\mod7)$ so that\r\n$X^{6}+5\\equiv6(\\mod7)$. Hence the polynomial is irreducible.", "Solution_2": "I wonder why there are so many people thinking that irreducible=no_roots ...", "Solution_3": "Here is the long way to do it.\r\n\r\nabdul show it has no linear factors (but it can still be reduccible).\r\n\r\nThen there are only two possible cases:\r\n\r\n$x^{6}+5 = (x^{2}+ax+b)(x^{4}+cx^{3}+dx^{2}+ex+f)$\r\n\r\nOr,\r\n\r\n$x^{6}+5=(x^{3}+ax^{2}+bx+c)(x^{3}+dx^{2}+ex+f)$\r\n\r\nAnd compare coffeicients after expanding.", "Solution_4": "If a is a root of $x^{6}+5$, then the other roots are $2a,3a,4a,5a,6a$.\r\n$\\phi_{i}: \\mathbb{F}_{p}(a) \\Rightarrow \\mathbb{F}_{p}(a)$ generated by fixing $\\mathbb{F}_{p}$ and $a \\Rightarrow ia$ defines a $F_{p}$ automorphism. Therefore all these 6 roots belong to the same irreducible polynomial in F_p[x]. So it is irreducible.", "Solution_5": "nici one Soarer and thanks abdul, that's what i thought too . the whole problem was something like : find all a in $Z_{7}$ so as $X^{6}+aX+5$ is irreducible . and i said that the only solution was for $a=0$ , because $X^{6}=1$ so the polynom is actually $aX+6$ when x is not 0 . whatever $a$ different from 0 we take , this has a root. so the initial polynom is reducible. am i right ?", "Solution_6": "[quote=\"Soarer\"]If a is a root of $x^{6}+5$, then the other roots are $2a,3a,4a,5a,6a$.\n$\\phi_{i}: \\mathbb{F}_{p}(a) \\Rightarrow \\mathbb{F}_{p}(a)$ generated by fixing $\\mathbb{F}_{p}$ and $a \\Rightarrow ia$ defines a $F_{p}$ automorphism.[/quote]\r\nCan somebody show me that \"$a \\Rightarrow ia$ defines a $F_{p}$ automorphism\" explicitly?", "Solution_7": "you're just permuting roots of an irreducible polynomial: in this case, it suffice adjoining one single root to have the splitting field, so fixing the image of a root fixes the automorphism of the field..\r\nis this what you wanted to ask, or should i go further into details?", "Solution_8": "Thank you,mago.\r\nIn fact,I want to ask how can I see such automorphism exist?\r\n\r\nWe have for any field $K$,$u,v$ is algebraic over $K$,there exist a isomorphism $f: K(u) \\to K(v)$,such that fixed $K$ and $u \\to v$ if and only if $u,v$ are the roots of the same irreducible polynomial.\r\n\r\nSo I am sorry to say that $Soarer$ 's proof is nonsense.\r\nIf his proof works,then consider the problem:\r\ndoes $x^{12}+4$ reducible over $Z_{13}$\r\n\r\nWe can use his way to get $x^{12}+4$ is irreducible over $Z_{13}$.But this is obviously wrong,because:$x^{12}+4=(x^{6})^{2}+4x^{6}+4-4x^{6}=(x^{6}+2)^{2}-4x^{6}=(x^{6}+2x^{3}+2)(x^{6}-2x^{3}+2)$\r\n\r\nI am sorry if what I wrote is wrong.\r\n\r\n[Edited]", "Solution_9": "$x^{6}+5=x^{6}-9=(x^{3}+3)(x^{3}-3)$. It factors.", "Solution_10": "Thank you,jmerry." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "A triangle whose all sides have length not smaller than $1$ is inscribed in a square of side length $1$. Prove that the center of the square lies inside the triangle or on its boundary.", "Solution_1": "If a triangle XYZ doesn't contain the center of the square, there is a line through the center of the square that doesn't intersect XYZ.\r\nProof: Suppose that every line through the center intersects XYZ. If the four vertices of a convex quadrilateral are part of a convex figure, then all points that line inside that quadrilateral are part of the convex figure. By taking two lines through the origin and finding their intersections with the triangle, we can find a convex quadrilateral containing the origin, inside of XYZ.\r\n\r\nDraw a line perpendicular to the line through the center of the square, as shown in the diagram.\r\n\r\nAll sides of the triangle must lie in the union of the blue and yellow regions. \r\n\r\nThe quadrilaterals surrounding the blue and yellow regions are concyclic, inscribable in circles of diameter $\\sqrt{2}/2$. Any line segment of length 1 that lies in the union of the colored regions must lie in both, since $1>\\sqrt{2}/2$\r\n\r\nIf X lies in the blue region, then Y lies in the yellow region, then Z lies in the blue region, then X lies in the yellow region, a contradiction.", "Solution_2": "[quote=\"schulmannerism\"]\n By taking two lines through the origin and finding their intersections with the triangle, we can find a convex quadrilateral containing the origin, inside of XYZ.[/quote]\r\n\r\nIs it always that there must be a quadrilateral containing the origin?" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "For $x;y;z$ are positive real numbers satisfying $3x^{3}+4xy+3y^{2}=14$\r\nFind the least value of the expression:$A=x^{2}+y^{2}$", "Solution_1": "chien than, Are you sure that $3x^{3}+4xy+3y^{2}=14?$\r\nI think that $3x^{2}+4xy+3y^{2}=14.$ :maybe:", "Solution_2": "If $3x^{2}+4xy+3y^{2}=14$ then it's very easy because $5(x^{2}+y^{2})\\geq3x^{2}+4xy+3y^{2}=14$ $\\rightarrow$ $A=\\frac{14}{5}$" } { "Tag": [ "logarithms", "trigonometry", "calculus", "calculus computations" ], "Problem": "can you show this ? :D [color=white]............ [/color](i know i can't) ! LOL ! :rotfl: \r\n\r\n\r\n\r\n$ \\sum_{k=1}^\\infty\\;\\;\\frac{(-1)^{k+1}\\cdot k^{2}}{1+k^{3}}\\;\\;=\\;\\;\\boxed{\\frac{1}{3}\\left( \\ln \\left( \\frac{e}{2}\\right) \\;+\\;\\frac{\\pi}{\\cosh \\left( \\frac{\\pi\\;\\sqrt 3}{2}\\right) }\\right) }$", "Solution_1": "$ \\sum_{k=1}^{\\infty }\\frac{(-1)^{k+1}k^{2}}{k^{3}+1}=\\sum_{k=1}^{\\infty }\\left(\\frac{(2 k-1)^{2}}{(2 k-1)^{3}+1}-\\frac{(2 k)^{2}}{(2 k)^{3}+1}\\right)=\\sum_{k=1}^{\\infty }\\left(\\frac{1}{6 k}-\\frac{1}{3 (2 k+1)}\\right)+\\sum_{k=1}^{\\infty }\\left(\\frac{1-4 k}{3 \\left(4 k^{2}-2 k+1\\right)}-\\frac{3-4 k}{3 \\left(4 k^{2}-6 k+3\\right)}\\right)$\r\n\r\ni don know how to calculate the last sum. the first is simple.", "Solution_2": "$ \\frac{(-1)^{k+1}3k^{2}}{1+k^{3}}=\\frac{(-1)^{k+1}}{k+1}+\\frac{(-1)^{k+1}}{k-\\frac12+i\\frac{\\sqrt{3}}{2}}+\\frac{(-1)^{k+1}}{k-\\frac12-i\\frac{\\sqrt{3}}{2}}$\r\n\r\n$ \\sum_{k=1}^{n}\\frac{(-1)^{k+1}}{k-\\frac12+z}=\\sum_{k=0}^{n-1}\\frac{(-1)^{k}}{k+\\frac12+z}=\\sum_{k=0}^{n-1}\\frac{(-1)^{k+1}}{-k-\\frac12-z}=\\sum_{k=-n+1}^{0}\\frac{(-1)^{k+1}}{k-\\frac12-z}$\r\n\r\n$ \\Rightarrow \\sum_{k=1}^{n}\\left(\\frac{(-1)^{k+1}}{k-\\frac12+z}+\\frac{(-1)^{k+1}}{k-\\frac12-z}\\right) =\\sum_{k=-n+1}^{n}\\frac{(-1)^{k+1}}{k-\\frac12-z}\\to \\frac{\\pi}{\\cos \\pi z}$\r\n\r\nThat is $ \\frac{\\pi}{\\cosh \\frac{\\pi \\sqrt{3}}{2}}$ for $ z=i\\frac{\\sqrt{3}}{2}$ and $ \\sum_{k=1}^{n}\\frac{(-1)^{k+1}}{k+1}\\to \\ln \\frac{e}{2}$\r\n\r\nSee also: [url]http://mathworld.wolfram.com/FoxTrotSeries.html[/url]" } { "Tag": [ "geometry", "3D geometry", "calculus", "calculus computations" ], "Problem": "The base of a cone-shaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 below the base. The tank is filled to a depth of 7 feet, and water is flowing out of the tank at a rate of 3 cubic feet per minute.\r\n\r\nFind the rate of change of the depth of the water in the tank.", "Solution_1": "i really need someone to show this one to me", "Solution_2": "Is water flowing out of the tank from the bottom (i.e., from the vertex)?\r\n\r\nHave you tried to solve this by your own?" } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "We have two distinct positive integers $a,b$ with $a|b$. Each of $a,b$ consists of $2n$ decimal digits. The first $n$ digits of $a$ are identical to the last $n$ digits of $b$, and vice versa. Determine $a,b$.", "Solution_1": "[hide=\"solution\"]\nLet $ a\\equal{}x10^n \\plus{}y$ and $ b\\equal{}y10^n \\plus{}x$. We start by using polynomial long division (with the \"variable\" being $ 10^n$), which gives $ \\frac{y}{x}$ as the quotient and $ x\\minus{}\\frac{y}{x}$ as the remainder. In order for $ a$ to divide $ b$, the remainder ($ x\\minus{}\\frac{y}{x}$) must equal zero, so $ x^2\\equal{}y$. Remember, however, the requirement that $ x$ and $ y$ have the same number of digits. Since squaring any positive integer greater than 3 results in the number of digits increasing, the only possible values for $ a$ and $ b$ are $ (24,42)$ and $ (39,93)$.\n[/hide]", "Solution_2": "[quote=\"foxjwill\"][hide=\"solution\"]\nLet $ a \\equal{} x10^n \\plus{} y$ and $ b \\equal{} y10^n \\plus{} x$. We start by using polynomial long division (with the \"variable\" being $ 10^n$), which gives $ \\frac {y}{x}$ as the quotient and $ x \\minus{} \\frac {y}{x}$ as the remainder. In order for $ a$ to divide $ b$, the remainder ($ x \\minus{} \\frac {y}{x}$) must equal zero, so $ x^2 \\equal{} y$. Remember, however, the requirement that $ x$ and $ y$ have the same number of digits. Since squaring any positive integer greater than 3 results in the number of digits increasing, the only possible values for $ a$ and $ b$ are $ (24,42)$ and $ (39,93)$.\n[/hide][/quote]Your answer is not correct, 24 does not divide 42, nor does 39 divide 93.", "Solution_3": "[quote=\"Valentin Vornicu\"][quote=\"foxjwill\"][hide=\"solution\"]\nLet $ a \\equal{} x10^n \\plus{} y$ and $ b \\equal{} y10^n \\plus{} x$. We start by using polynomial long division (with the \"variable\" being $ 10^n$), which gives $ \\frac {y}{x}$ as the quotient and $ x \\minus{} \\frac {y}{x}$ as the remainder. In order for $ a$ to divide $ b$, the remainder ($ x \\minus{} \\frac {y}{x}$) must equal zero, so $ x^2 \\equal{} y$. Remember, however, the requirement that $ x$ and $ y$ have the same number of digits. Since squaring any positive integer greater than 3 results in the number of digits increasing, the only possible values for $ a$ and $ b$ are $ (24,42)$ and $ (39,93)$.\n[/hide][/quote]Your answer is not correct, 24 does not divide 42, nor does 39 divide 93.[/quote]\r\n\r\nOy! I was so excited that I had found a solution that I forgot to check to see if it was correct! >_<", "Solution_4": "Using the well-known properties of the decimal expansion of $ \\frac{1}{7}$, it's easy to find:\r\n\r\n$ 142857 \\times 6 \\equal{} 857142$.\r\n\r\nI'm quite sure there are no other solutions, but I need some more time to find a clean proof." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "let $ \\\\a_n\\equal{}\\frac{(2\\plus{}\\sqrt{3})^{2^n}\\plus{}(2\\minus{}\\sqrt{3})^{2^n}}{2}, n\\geq1.$\r\n\r\n$ \\\\p|a_n$ and $ 12|p\\minus{}1,p\\minus{}prime$ then prove that $ 2^{n\\plus{}2}|p\\minus{}1$", "Solution_1": "It is an good problem . \r\nIf $ p\\equiv 1 (\\mod 12 )$ then we have : \r\n$ (\\frac {3}{p}) \\equal{} 1$ so exist an integer m such that $ m^2\\equiv 3 (\\mod p)$\r\nSo $ a_n\\equiv (m \\plus{} 2)^{2^m} \\plus{} (2 \\minus{} m)^{2^n} (\\mod p)$\r\n$ \\Rightarrow p|(2 \\plus{} m)^{2^n} \\plus{} (2 \\minus{} m)^{2^n}$\r\n$ \\Rightarrow p|a^{2^n} \\plus{} 1$\r\nWhere $ a \\equal{} (\\frac {m \\plus{} 2}{2 \\minus{} m}) (\\mod p )$\r\nBecause $ (m \\plus{} 2)(2 \\minus{} m)\\equiv 1 (\\mod p)$ so $ a$ is a quadric reside mod p . \r\nIt mean that there exist an integer $ x$ such that $ x^2\\equiv a (\\mod p )$\r\nSo $ p|x^{2^{n \\plus{} 1}} \\plus{} 1$\r\nBecause $ ord_x(\\mod p) \\equal{} 2^{n \\plus{} 2}$ so $ 2^{n \\plus{} 2}|p \\minus{} 1$ \r\n Problem claim .", "Solution_2": "[quote=\"TTsphn\"]It is an good problem .[/quote]\r\n\r\nMaybe, but not sufficientely to be posted three times in a row :wink: \r\nTwo times deleted...\r\n\r\nPierre.", "Solution_3": "Thank you very much help me. :)" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "Given $ f: (a,b) \\to \\mathbb{R}$ is both continuous and differentiable, find the following in terms of $ f(x)$ and $ f'(x)$:\r\n\r\n[list]$ \\lim_{h \\to 0} \\frac {f(x \\plus{} 2h) \\minus{} f(x)}{h}$[/list]\r\nObviously, $ f'(x) \\equal{} \\lim_{h \\to 0} \\frac {f(x \\plus{} h) \\minus{} f(x)}{h}$, but I don't know how to deal with the $ 2h$. Thanks.", "Solution_1": "What can you do with $ \\lim_{h\\to0}2\\cdot\\frac{f(x\\plus{}2h)\\minus{}f(x)}{2h}\\ ?$", "Solution_2": "[quote=\"Kent Merryfield\"]What can you do with $ \\lim_{h\\to0}2\\cdot\\frac {f(x \\plus{} 2h) \\minus{} f(x)}{2h}\\ ?$[/quote]\r\nGot it. Doh! Thanks." } { "Tag": [ "probability" ], "Problem": "Two six-sided dice are rolled. Given that at least one die shows a one, what is the probability that both of them are ones? Express your answer as a common fraction.", "Solution_1": "There are eleven ways we get atleast a single one, all of the same probability.\r\n1-1\r\n1-2\r\n1-3\r\n1-4\r\n1-5\r\n1-6\r\n2-1\r\n3-1\r\n4-1\r\n5-1\r\n6-1\r\nOut of these, only the first one, or one choice out of 11, has both dice one's.\r\nAns: $ \\boxed{1/11}$" } { "Tag": [ "summer program", "MathPath", "email" ], "Problem": "Hello,\r\n\r\nAre all of the seats in MathPath taken up already? I want to know if I can apply if possible. I know this is late but I didn't find out about these camps until this summer, and now I'm regretting it. :(\r\n\r\nThank you\r\nrts2007", "Solution_1": "Hello,\r\n\r\nYou can go to http://www.mathpath.org and click on the link to email Prof Maurer. Ask him if seats are still open and application procedures. Since the MathPath website says seats are still open you can probably still apply.", "Solution_2": "It starts in exactly 14 days (June 28). The deadline was the end of April, but it was extended to the middle of May. You should contact them. I found out in the middle of May, and I only had a few days to decide. I didn't apply, since I wasn't really ready. \r\n\r\nMy advice is to plan for next year. Find the dates, as other camps my have conflicting dates. Plan so that there is no overlap. Find all the information you will need (recommendations, materials, $ \\$\\$\\$$). Check regularly for the admission test. Apply early, since some camps have an early bird fee. \r\n\r\nIf I am wrong, please correct me. I have never been to a summer camp :oops: .", "Solution_3": "Yes, best of luck, but I would think, with camp just 2 weeks away, they would be set. \r\n\r\nAlso, levans, are you going to be there?", "Solution_4": "Dear rts2007,\r\n\r\n levans is right as to what you should do.\r\n\r\n We can still make room for you at camp, but you have to move fast. You have to do the same application as everyone else, and that usually takes a week to put together. Even if you see how to solve all the Qualifying Test problems right away, it still takes time to write them up clearly, which counts a lot with us. And you have to line up your recommenders and get them to submit their recommendations (getting [i]them[/i] to hurry up is often difficult). And if we do admit you, then unless you live within driving distance of Colorado Springs, you have to book airfare [i]after[/i] we admit you. Maybe you'll get a great last minute fare, but more likely it will be expensive.\r\n\r\n We are game to consider your application if you and your parents are willing to do it. But I would start right now, this weekend, working on the Qualifying Test. And you still have to email me through the MathPath website so I can give you further speed-up suggestions.\r\n\r\n Oh, pinkmuskrat (who will be attending MathPath) asks if levans will be there. Yes, for the first two weeks.\r\n\r\n Prof Maurer (= Mr M at camp)" } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ R$ be an injective $ R$ module ,prove that:\r\n$ Ann(I \\cap J ) \\equal{} Ann(I) \\plus{} Ann(J)$. $ I$ and $ J$ are ideals of $ R$.", "Solution_1": "What's $ I$ and $ J$?", "Solution_2": "I edited it.", "Solution_3": "any solution?", "Solution_4": "[quote=\"artin\"]Let $ R$ be an injective $ R$ module ,prove that:\n\n$ Ann(I \\cap J ) \\equal{} Ann(I) \\plus{} Ann(J)$. $ I$ and $ J$ are ideals of $ R$.[/quote]\r\n\r\n$ Ann(I) \\plus{} Ann(J) \\subseteq Ann(I \\cap J)$ is trivial. Now let $ x \\in Ann(I \\cap J).$ Then the map $ f: I \\plus{} J \\longrightarrow R$ defined by $ f(a\\plus{}b)\\equal{}ax, \\ \\ a \\in I, b \\in J,$ is a \r\n\r\nwell-defined $ R$ homomorphism and thus, since $ R$ is self-injective, there exists $ r \\in R$ such that $ ax\\equal{}f(a\\plus{}b)\\equal{}(a\\plus{}b)r,$ for all $ a \\in I, b \\in J.$ Particularly \r\n\r\n$ ax\\equal{}ar$ and $ br\\equal{}0$ for all $ a \\in I, b \\in J.$ hence $ x\\minus{}r \\in Ann(I)$ and $ r \\in Ann(J).$ Thus $ x \\in Ann(I) \\plus{} Ann(J)$ and the proof is complete." } { "Tag": [ "algebra", "polynomial", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $\\displaystyle \\mathcal K$ be a field with $\\displaystyle 27$ elements. Prove that there is $\\displaystyle a \\in \\mathcal K$ such that $\\displaystyle a^3 = a + 2$.\r\n\r\n\r\n[i]Mircea Becheanu[/i]", "Solution_1": "hi,\r\nconsider $\\mathbb{F}_3$,the polynomial $x^3 -x -2$ is irreducible over it,so the field obtained by adjoining a root of it to $\\mathbb{F}_3$ is a field of order 27,now since all fields of order 27 arre isomorphic,every such field has an element $\\alpha$ s.t $\\alpha ^3 =\\alpha +2$.", "Solution_2": "I think everybody who knows a little bit about finite fields has thought of rohith's solution ;-)\r\n \r\nIs there another \"elementary\" proof? :-)", "Solution_3": "Hi.\r\n\r\nThere is something like an \"elementary\" proof:\r\n\r\nConsider the map $\\phi: \\mathbb{F}_{27}\\rightarrow\\mathbb{F}_{27}: x\\mapsto x^3-x$ which is $\\mathbb{F}_3$-linear.\r\nSince every element a of $\\mathbb{F}_{27}$ is a solution of $x^{27}-x=0$, the third power of this map is zero: it is nilpotent.\r\nThe kernel of $\\phi$ is the subfield $\\mathbb{F}_3$. Since $\\phi$ ist nilpotent, $ker(\\phi)$ is a subset of $im(\\phi)$. So there must be an element $y\\in\\mathbb{F}_{27}$ with $\\phi(y)=2$, this is the element we've been looking for.\r\n\r\nIs this elementary enough?\r\n\r\nmfg Gockel." } { "Tag": [ "vector", "algebra", "polynomial", "linear algebra", "matrix", "inequalities", "induction" ], "Problem": "Let $ V$ be a finite dimensional vector space and $ T: V\\to V$ be a linear transformation such that $ \\exists n\\in \\mathbb{Z}^ \\plus{}$ with $ T^n \\equal{} 0$. Let $ r(k) \\equal{} \\text{rk }(T^k)$. Show that $ r(k)$ is strictly decreasing when $ r(k) > 0$.\r\n\r\nI can show that $ r(k)$ is non-increasing by showing that $ \\text{im }(T^{k \\plus{} 1})$ is a subspace of $ \\text{im }(T^k)$, but I'm not sure where to go from here. I also considered trying to show that $ \\text{dim(ker}(T^k)) < \\text{dim(ker}(T^{k \\plus{} 1}))$, but I can't find a way to show that $ \\exists \\vec{v}\\in V$ such that $ T^{k \\plus{} 1}\\vec{v} \\equal{} \\vec{0}$ and $ T^{k}\\vec{v}\\neq \\vec{0}$.", "Solution_1": "As a matter of notation and thought pattern, I always prefer working with matrices to working with linear transformations and subspaces.\r\n\r\nBy Schur's lemma, if all the characteristic polynomial can be completely factored, than any matrix is similar to an upper triangular matrix with the eigenvalues on the main diagonal.\r\n\r\nAs a corollary of that, any nilpotent matrix is similar to a strictly upper triangular matrix.\r\n\r\nTry proving this as a theorem about strictly upper triangular matrices.", "Solution_2": "Of course there is a minimal n for which $ T^n\\equal{}0$. Then we can say there is a $ v$ for which $ T^{n\\minus{}1}(v)\\ne 0$.\r\n\r\nIf V has dimension $ N$, in the case $ N\\le n$, we have that\r\n\r\n$ v$, $ Tv$, ... $ T^Nv$ is a basis for $ V$ from which your claim follows.\r\n\r\n\r\n\r\nNow it suffices to consider the case $ N>n$... \r\n\r\nPerhaps in this case we have to be more careful with our choice of v... Say $ \\{v_1,...,v_n\\}$ is a basis for $ V$. Then we can choose a $ v_i$ such that $ T^{n\\minus{}1}(v)\\ne 0$. Then I think that the same proof still holds but we note that the dimension can go down by more than 1.\r\n\r\n(extent v, Tv,..., T^{n-1}v to a basis of V...)", "Solution_3": "[url=http://planetmath.org/encyclopedia/FrobeniousInequality.html]The Frobenius inequality[/url] yields $ \\mathrm{rank}A^{a\\plus{}b}\\plus{}\\mathrm{rank}A^{b\\plus{}c}\\leq\\mathrm{rank}A^b\\plus{}\\mathrm{rank}A^{a\\plus{}b\\plus{}c}$. The rest is combinatorics ;)\r\n\r\n darij", "Solution_4": "I think it's a rare case of endemic blindness and beating around the bush. The result is trivial. :) \r\n\r\nWe have $ r(k) : \\equal{} \\textrm{rk}(T^k) \\equal{} \\dim \\textrm{Im}(T^k)$, and obviously $ \\textrm{Im}(T^{k\\plus{}1}) \\subseteq \\textrm{Im}(T^k)$. Assume $ r(k\\plus{}1) \\equal{} r(k) > 0$; this means $ \\textrm{Im}(T^{k\\plus{}1}) \\equal{} \\textrm{Im}(T^k)$, i.e. $ T^{k\\plus{}1}(V) \\equal{} T^k(V)$. But then, by trivial induction, $ T^{m}(V) \\equal{} T^k(V)$ for all $ m>k$, absurd, since $ T^n(V) \\equal{} \\{0\\}$.\r\n\r\n[color=red]In one line, once the sequence of nested subspaces $ T^k(V)$ contains two consecutive equal terms, it becomes stationary, which according to the givens can only occur for $ \\{0\\}$.[/color]" } { "Tag": [ "modular arithmetic", "algebra", "binomial theorem" ], "Problem": "Find the remainder when $ 3^{2003}$ is divided by $ 28$.", "Solution_1": "$ 3^3\\equiv \\minus{}1\\pmod{28}$\r\n\r\n$ 3^6\\equiv (\\minus{}1)^2\\equiv 1\\pmod{28}$\r\n\r\n$ 3^{2003}\\equal{}3^{1998}\\cdot3^5\\equiv 1^{333}\\cdot3^5\\equiv 3^5\\equiv 19\\pmod{28}$\r\n\r\n[hide=\" \"] \nI bet someone will try to explain this better, except if electron proton posts that he didn't understand the solution right away [/hide]", "Solution_2": "hello, note that $ 3^3\\equiv \\minus{}1 \\mod 28$ so we get $ 3^{2001}\\equal{}3^{3\\cdot667}\\equiv\\minus{}1\\mod 28$ and from here we get $ 3^{2003}\\equiv\\minus{}9\\mod 28$\r\nSonnhard.", "Solution_3": "[hide=\"Using binomial theorem\"]\nWrite it as $ 3.(28\\minus{}1)^{667}$ and apply binomial theorem.\n19 will be the answer.\n\n[/hide]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Given that $ 2^{2004}$ is a 604-digit number with leafing digit 1. Determine the number of elements in the set $ \\{2^0,2^1,\\cdots,2^{2003}\\}$ with leafing digit 4. (The leafing digit is the first digit on the left hand)", "Solution_1": "Wasn't this problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1368329#1368329]just posted[/url]?" } { "Tag": [], "Problem": ":arrow: \u0394\u03cd\u03bf \u03ba\u03cd\u03ba\u03bb\u03bf\u03b9 \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf \u03c4\u03ad\u03bc\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9. \u0388\u03c3\u03c4\u03c9 $ A$ \u03ad\u03bd\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c4\u03bf\u03bc\u03ae\u03c2. \u039e\u03b5\u03ba\u03b9\u03bd\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b1\u03c5\u03c4\u03cc\u03c7\u03c1\u03bf\u03bd\u03b1 \u03b1\u03c0\u03cc \u03c4\u03bf $ A$ \u03b4\u03cd\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03ba\u03b9\u03bd\u03bf\u03cd\u03bd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03ad\u03c2 \u03c4\u03b1\u03c7\u03cd\u03c4\u03b7\u03c4\u03b5\u03c2, \u03ba\u03b1\u03b9 \u03c4\u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1 \u03c4\u03b1\u03be\u03b9\u03b4\u03b5\u03cd\u03b5\u03b9 \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03bf \u03b4\u03b9\u03ba\u03cc \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf \u03c3\u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03ba\u03b1\u03c4\u03b5\u03cd\u03b8\u03c5\u03bd\u03c3\u03b7. \u039c\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03bc\u03b9\u03b1 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c4\u03c1\u03bf\u03c6\u03ae \u03c4\u03b1 \u03b4\u03cd\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03b5\u03c0\u03b9\u03c3\u03c4\u03c1\u03ad\u03c6\u03bf\u03c5\u03bd \u03c4\u03b1\u03c5\u03c4\u03cc\u03c7\u03c1\u03bf\u03bd\u03b1 \u03c3\u03c4\u03bf $ A$ . \u0394\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ P$ \u03c3\u03c4\u03bf \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03ce\u03c3\u03c4\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03c3\u03c4\u03b9\u03b3\u03bc\u03ae \u03b7 \u03b1\u03c0\u03cc\u03c3\u03c4\u03b1\u03c3\u03b7 \u03c4\u03c9\u03bd \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03cc \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03ae.", "Solution_1": "\u039c\u03ae\u03c0\u03c9\u03c2 \u03b5\u03bd\u03bd\u03bf\u03b5\u03af\u03c2 \u03c4\u03bf \u03b1\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd \u03b1\u03c0\u03bf\u03c3\u03c4\u03ac\u03c3\u03b5\u03c9\u03bd???", "Solution_2": "\u039d\u03b1\u03b9 \u03c1\u03b5 \u03c3\u03c5" } { "Tag": [ "\\/closed" ], "Problem": "Hello, \r\n\r\nI am new to this forum and would like to know if teachers and/or parents are also taking the AoPS courses. Although I have been looking through this site for my son, I feel that I could benefit from the courses as well. Have other adults tried the courses, and if so, what were their thoughts?\r\n\r\nThanks", "Solution_1": "This should probably go in the classes forum. \r\n\r\nI am not sure if any parents take the classes. There aren't any that I know of. I don't really see any problem with teachers/parents/etc taking the classes. I can definitely see the benefit of teachers and parents taking it. Then they can teach their children what they have learned. All in all the classes are a ton of fun and you learn so much.", "Solution_2": "Moved it.\r\n\r\nI think that there should be no problem b/c:\r\n(a) the internet is anonymous, so nobody knows your age\r\n(b) the classes are not geared towards a specific age group, I don't think (just specific tests that HS kids take...)\r\n\r\nThere are a lot of adults on this forum, but they are mostly the owners of the website or math professors or something like that.", "Solution_3": "[quote=\"Thomas\"]Hello, \n\nI am new to this forum and would like to know if teachers and/or parents are also taking the AoPS courses. Although I have been looking through this site for my son, I feel that I could benefit from the courses as well. Have other adults tried the courses, and if so, what were their thoughts?\n\nThanks[/quote]\r\n\r\nWe do have adults take our classes. We have had at least one teacher take several of our courses. We have no age limitations. If you feel our courses would be beneficial to you then we would be happy to have you in class." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let S(n,p) denote the number of surjections from a set with a cardinal equal to n to a set with a cardinal equal to p.\r\n\r\n1/ Calculate S(n,n) and S(p+1,p)\r\n2/ Show that :\r\n$S(n,p) = \\sum_{k=1}^{n} \\binom{n}{k}S(n-k,p-1)$\r\n3/ Show that :\r\n$\\sum_{k=0}^{p} \\binom{p}{k}S(n,k) = p^{n}$\r\n\r\n\r\nEnjoy :D", "Solution_1": "It's all about Stirling numbers of the second kind.", "Solution_2": "Hmm... could you explain what you mean by this?\r\n\r\nNota : I know what those numbers are, but I can't see how they help with this problem.. :blush:", "Solution_3": "S(n,p) = p!*S2(n,p), where S2(n,p) is Striling number of the second kind.\r\n\r\nIn particular, formula 3) is exacly formula (10) at [url]http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html[/url] for x=m=p.", "Solution_4": "Ok, thanks a lot Maxal! :D" } { "Tag": [ "geometry", "inradius", "trigonometry", "geometry unsolved" ], "Problem": "Construct a triangle given its three exradii.", "Solution_1": "Denote a = BC, b = CA, c = AB the sides, $\\alpha = \\angle A,\\ \\beta = \\angle B,\\ \\gamma = \\angle C$, the angles $r, r_a, r_b, r_c$ the inradius and exradii against the vertices A, B, C, s the semiperimeter and $\\triangle$ the area of the triangle $\\triangle ABC$.\r\n\r\n$r = (s - a) \\tan \\frac \\alpha 2 = (s - b) \\tan \\frac \\beta 2 = (s - c) \\tan \\frac \\gamma 2$\r\n\r\n$r_a = s \\tan \\frac \\alpha 2,\\ r_b = s \\tan \\frac \\beta 2,\\ r_c = s \\tan \\frac \\gamma 2$\r\n\r\n$r_ar_b + r_br_c + r_cr_a = s^2 \\left( \\tan \\frac \\alpha 2 \\tan \\frac \\beta 2 + \\tan \\frac \\beta 2 \\tan \\frac \\gamma 2 + \\tan \\frac \\gamma 2 \\tan \\frac \\alpha 2 \\right) =$\r\n\r\n$= s^2 \\left( \\frac{r^2}{(s - a)(s - b)} + \\frac{r^2}{(s - b)(s - c)} + \\frac{r^2}{(s - c)(s - a)} \\right) =$\r\n\r\n$= \\triangle^2 \\frac{3s - (a + b + c)}{(s - a)(s - b)(s - c)} = s^2$\r\n\r\nLet F be the 1st Fermat point of a triangle $\\triangle XYZ$ such that $FX = r_a,\\ FY = r_b,\\ FZ = r_c$. Since $\\angle XFY = \\angle YFZ = \\angle ZFX = 120^\\circ$, the triangle $\\triangle XYZ$ is trivial to construct. The area of this triangle is\r\n\r\n$|\\triangle XYZ| = \\frac 1 2 \\sin 120^\\circ \\left(r_ar_b + r_br_c + r_cr_a\\right) = \\frac{s^2 \\sqrt 3}{4}$\r\n\r\nConstruct an equilateral triangle with the same area - its side will be s, the semiperimeter of the triangle $\\triangle ABC$. For example, denote x = YZ and let $h_x$ be the X-altitude of this triangle. Then\r\n\r\n$|\\triangle XYZ| = \\frac 1 2 h_x x = \\frac{s^2 \\sqrt 3}{4},\\ \\ h_x = \\frac{s^2 \\sqrt 3}{2x}$\r\n\r\nDraw an equilateral triangle with the altitude x = YZ. Its side is $p = \\frac{2x}{\\sqrt 3}$. Construct the square root \r\n\r\n$\\sqrt{h_x p} = \\sqrt{\\frac{s^2 \\sqrt 3}{2x} \\cdot \\frac{2x}{\\sqrt 3}} = s$\r\n\r\nOnce the semiperimeter s of the triangle $\\triangle ABC$ is constructed, all its angles can be constructed and then the triangle itself. For example, draw the circle $C_a(E_a, r_a)$, pick an arbitrary point P on this circle and draw a tangent to $C_a$ at P. Let A be a point on this tangent, such that AP = s. Construct the other tangent AQ from A to $C_a$ by reflecting P in $AE_a$. The line $AE_a$ bisects the angle $\\angle PAQ$ and a normal to $AE_a$ at A bisects the complementary angle. Draw a circle $C_b(E_b, r_b)$ centered on this normal and tangent to the lines AP, AQ. One of the lines AP, AQ is the common internal tangent of the circles $C_a, C_b$, say the line AQ. Draw the other common internal tangent of these 2 circles by reflecting the common internal tangent AQ in the center line $E_aE_b$. The other common internal tangent meets the lines AP, AQ at points B, C, respectively, completing the construction of the triangle $\\triangle ABC$." } { "Tag": [], "Problem": "If $ \\pi\\equal{}3.1415926...$, what is the exact value of $ |\\pi\\minus{}3.14|\\plus{}|\\pi\\minus{}\\frac{22}{7}|$? Express your answer as a common fraction.", "Solution_1": "It's $ \\frac{22}{7}\\minus{}3.14\\equal{}\\frac{2200}{700}\\minus{}\\frac{2198}{700}\\equal{}\\frac{1}{350}$ as $ 3.14<\\pi<\\frac{22}{7}$", "Solution_2": "How do you get that?", "Solution_3": "$ |a\\minus{}b|\\equal{}a\\minus{}b$ iff $ a\\geq b$, and $ |a\\minus{}b|\\equal{}b\\minus{}a$ iff $ a0$\r\n\r\n$ |\\pi\\minus{}\\frac{22}{7}|\\equal{}|3.14159\\minus{}3.14285|\\equal{}3.14285\\minus{}3.14159\\equal{}\\frac{22}{7}\\minus{}\\pi$\r\n\r\nThus, $ |\\pi\\minus{}3.14|\\minus{}|\\pi\\minus{}\\frac{22}{7}|\\equal{}\\pi\\minus{}3.14\\plus{}\\frac{22}{7}\\minus{}\\pi\\equal{}\\frac{1}{350}$ \r\n\r\nI remember one guy on my team said he left it as 2/700 :rotfl:", "Solution_5": "[quote=\"Mewto55555\"]I remember one guy on my team said he left it as 2/700 :rotfl:[/quote]\r\n\r\nThat guy happens to be me... :P", "Solution_6": "How do you get $3.14$ + $22/7$ quickly?", "Solution_7": "There isn't really a fast way to find $\\dfrac{22}{7}-3.14$ other than noting their common denominator is $\\gcd(100,7)=700.$", "Solution_8": "Got it, thanks", "Solution_9": "[quote=hamon][quote=priceyfalcon]How do you get $3.14$ + $22/7$ quickly?[/quote]\n\nI don't think we need a 12 year bump here[/quote]\n\nAccording to the sheriff, who gave me a report ban for reporting this, you're supposed to bump alcumus (and I guess FTW), as it's open forever." } { "Tag": [], "Problem": "According to their website ([url]http://www.usaco.org[/url]), they're starting the contest in October this year, instead of November. That's a total of 7 contests this year.", "Solution_1": "jeese, i dunno if i'll be able to do all of those. Rarely do I have 3-4 open hours on a weekend.", "Solution_2": "you don't have to do all of the rounds in order to participate. Just do what you can.\r\n\r\nbtw - People who are interested in USACO might want to take advantage of the new discussion forums there.", "Solution_3": "Just wanted to bump this, since the compeition is this weekend.", "Solution_4": "Are you talking about the online 'practice' contests or the actual USACO contests part of the selection for IOI???", "Solution_5": "I'm pretty sure that this is part of the selection for the US IOI team.", "Solution_6": "Nope, that's just assignment for the appopriate division. It's written in rules that it doesn't count to IOI team selection.", "Solution_7": "So how did everyone do?" } { "Tag": [], "Problem": "The number of real values of $ x$ satisfying the equation $ 2^{2x^2 \\minus{} 7x \\plus{} 5} \\equal{} 1$ is:\r\n\r\n$ \\textbf{(A)}\\ 0 \\qquad \\textbf{(B)}\\ 1 \\qquad \\textbf{(C)}\\ 2 \\qquad \\textbf{(D)}\\ 3 \\qquad \\textbf{(E)}\\ \\text{more than 4}$", "Solution_1": "All zeroes of $ 2x^2\\minus{}7x\\plus{}5\\equal{}0$ satisfy this (and they are the only solutions). Since the discriminant is $ 9>0$, there are $ \\boxed{\\textbf{(C)}\\ 2}$ solutions.", "Solution_2": "Solvent\n$2^\\gamma=1\\implies\\gamma=0$\n$2x^2-7x+5=\\gamma$\n$\\gamma=0$\n$\\implies $2$x$^2$-7$x$+5$=0 (Transit law)\nx identical to $1$, x idempotent to $-\\frac52$ solutes\n--> Double (C)\n\ngod is greta\n-tony BARKER", "Solution_3": "Note that the only solutions to this equality occur when you choose the roots of the polynomial $f(x)=2x^2 - 7x + 5$. From here, there are a couple of approaches (some displayed above), but the two I will use are Descartes Rule of Signs and the quadratic formula. \n\n[b]Solution 1[/b]: Using DRoS, we see that $f(x)=2x^2 - 7x + 5$ has two changes in sign which means that it has two positive real roots. If we take $f(-x)=2x^2 +7x + 5$, we see that there are $0$ changes in sign so it has no negative roots. Hence, the answer is $ \\boxed{\\textbf{(C)}\\ 2}$.\n\n[b]Solution 2[/b]: Using the quadratic formula, we find that $f(x)=2x^2 - 7x + 5$ has two real roots, $\\frac{5}{2}$ and $1$. This not only confirms our previous solution, but also gives us the answer again $ \\boxed{\\textbf{(C)}\\ 2}$.\n\n[b]Disclaimer[/b]: I know Descartes only gives the maximum amount of roots a polynomial can have and that solution 1 is essentially cheese, but it typically works for second and third degree polynomials (there are exceptions, as always) in my experience and it is basically confirmed to be true by solution 2. Just wanted to put that out there because you should [i]never[/i] use Descartes as viable proof for the roots of a polynomial." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Calculate: $ \\int\\frac{\\cos (x)}{\\sqrt{\\cos (2x)}}\\, dx$.", "Solution_1": "$ \\int \\frac {\\cos x}{\\sqrt{\\cos 2x}}\\ dx \\equal{} \\int \\frac {1}{\\sqrt {1 \\minus{} \\sin ^ 2 x}} (\\sin x)'\\ dx \\equal{} \\sin ^{ \\minus{} 1}(\\sin x) \\plus{} C$.", "Solution_2": "hello kunny, i think it's $ \\cos(2x)\\equal{}1\\minus{}2\\sin^2(x)$.\r\nSonnhard.", "Solution_3": "Oops!\r\n\r\n$ \\int \\frac{\\cos x}{\\sqrt{\\cos 2x}}\\ dx\\equal{}\\frac{1}{\\sqrt{2}}\\int \\frac{1}{\\sqrt{1\\minus{}(\\sqrt{2}\\sin x)^2}}(\\sqrt{2}\\sin x)'\\ dx$\r\n\r\n$ \\equal{}\\frac{1}{\\sqrt{2}}\\sin ^ {\\minus{}1} (\\sqrt{2}\\sin x)\\plus{}C$.", "Solution_4": "hello, here is my solution\r\nby substituting $ t \\equal{} \\sqrt {2\\cos^2(x) \\minus{} 1}$ we get $ dt \\equal{} \\frac { \\minus{} 2\\cos(x)\\sin(x)}{\\sqrt {2\\cos^2(x) \\minus{} 1}}dx$ and from here $ dx \\equal{} \\frac {\\sqrt {2\\cos^2(x) \\minus{} 1}}{ \\minus{} 2\\cos(x)\\sin(x)}dt$ and $ \\sin(x) \\equal{} \\pm\\sqrt {\\frac {1 \\minus{} t^2}{2}}$ and our integral is\r\n$ \\pm\\frac{\\sqrt {2}}{2}\\int \\frac {dt}{\\sqrt {1 \\minus{} t^2}} \\equal{} \\pm\\frac{\\sqrt {2}}{2}\\arcsin(t) \\plus{} C$.\r\nSonnhard." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Prove that:\r\n$ \\sum_{n\\equal{}1}^{\\infty}\\frac{1}{F_n}>\\frac{803}{240}$", "Solution_1": "In this problem we have: $ F_0\\equal{}1, F_1\\equal{}1$ or $ F_1\\equal{}1, F_2\\equal{}1$?", "Solution_2": "$ F_1\\equal{}F_2\\equal{}1$" } { "Tag": [], "Problem": "Does anyone have any recommendations for undergraduate research opportunities during the summer? Also, could people share their experiences?", "Solution_1": "Hello,\nI think you can try REUs, but you must be a US Citizen or a Permanent Citizen to participate in them. They are also quite competitive, but worth applying to. Of course, you can get in touch with a mathematics professor, ask him or her about some interesting mathematics problems to look into or books to read. When school starts again, you can talk to him or her about some new thoughts you've had, etc. You can also intern at various places as a trader, etc. I know JP Morgan and DE Shaw are always looking for interns, especially those who are more mathematically inclined. Best of luck in your endeavors!" } { "Tag": [ "geometry", "geometric transformation", "reflection", "trapezoid", "geometry unsolved" ], "Problem": "Does anyone know how to do this:\r\n\r\n[color=green]Prove or disprove by giving a counter example.\n\n\"If two medians of a triangle are of the same length, thenthe triangle is isosceles\"[/color]\r\n\r\nThanks very much!!! :lol:", "Solution_1": "Denote the reflections $A'$, $B'$ of the points $A$, $B$ w.r.t. the midpoints of the sides $[BC]$, $[CA]$ respectively. Then $m_a=m_b$ $\\Longleftrightarrow$ $AA'=BB'$ $\\Longleftrightarrow$ the quadrilateral $ABA'B'$ is a isosceles trapezoid $\\Longleftrightarrow$ $BA'=AB'$ $\\Longleftrightarrow$ $AC=BC$ $\\Longleftrightarrow$ $a=b\\ .$" } { "Tag": [ "analytic geometry", "vector", "symmetry" ], "Problem": "Prove that the coordinates of the centroid of a coordinatized triangle are $ (a,b)$ where $ a$ is the arithmetic mean of the x-coordinates of the vertices of the triangle and $ b$ is the arithmetic mean of the y-coordinates of the triangle.", "Solution_1": "[hide=\"Solution\"]Let $ \\vec{P}$ be the centroid. This is equivalent to proving that \\[ \\vec{P}=\\frac{\\vec{A}+\\vec{B}+\\vec{C}}{3}\\]$ \\vec{P}$ is $ 2/3$ of the way from $ A$ to the midpoint of $ BC$, so \n\\begin{eqnarray*}\n\\vec{P}&=&\\vec{A}+\\frac{2}{3}\\left(\\frac{\\vec{B}+\\vec{C}}{2}-\\vec{A}\\right)\\\\\n&=&\\frac{\\vec{A}+\\vec{B}+\\vec{C}}{3}\\end{eqnarray*}And we're done.[/hide]", "Solution_2": "How would we have proven it without the knowledge that the centroid divides its distance to the vertex as 2/3 of the distance from the vertex to the midpoint of the opposite side?\r\n\r\n(BTW, is there an elegant way to prove it? I know a way using vectors.)", "Solution_3": "without vectors:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229477[/url]", "Solution_4": "Of course these proofs are wrong because they assume the existence of the centroid.\r\n\r\nDefn: The centroid is the intersection of the medians of a triangle.\r\n\r\nFirst, we have to check that our definition makes sense. That is to say, we need to show that the three medians are concurrent.\r\n\r\nWe can easily verify through vectors that $ G\\equal{}\\frac{A\\plus{}B\\plus{}C}{3}$ lies on the line that connects $ A$ and $ \\frac{B\\plus{}C}{2}$. \r\n\r\nBy symmetry of argument, $ G$ lies on all of the medians, hence $ G$ is the centroid (and our definition makes sense).", "Solution_5": "Can we prove it without using coordinates at all? Also the existence and uniqueness of the centroid.", "Solution_6": "Of course.\r\n\r\nExistence: use Ceva's theorem.\r\n\r\nI don't think that uniqueness is really something to focus on here. The intersection of two lines is usually a point..." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all $ x,y,z\\in N$ such that\r\n$ 4^x\\plus{}4^y\\plus{}4^z$ is a perfect square.", "Solution_1": "Posted before at least 3 times.", "Solution_2": "More \r\nFind all $ (x,y,z)\\in N$ \r\n$ a^{2x}\\plus{}a^{2y}\\plus{}a^{2z}$ is a perfect square." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Give a,b,c,d are positive numbers, and a+b+c+d=4, prove that\r\n\\[ 5(ab\\plus{}bc\\plus{}cd\\plus{}da\\plus{}ac\\plus{}bd)\\leq abc\\plus{}bcd\\plus{}cda\\plus{}dab\\plus{}16.\\]\r\n\\[ 3(ab\\plus{}bc\\plus{}cd\\plus{}da\\plus{}ac\\plus{}bd)\\leq abc\\plus{}bcd\\plus{}cda\\plus{}dab\\plus{}14.\\]", "Solution_1": "[quote=\"shijin93\"]Give a,b,c,d are positive numbers, and a+b+c+d=4, prove that\n\n\\[ 3(ab \\plus{} bc \\plus{} cd \\plus{} da \\plus{} ac \\plus{} bd)\\leq abc \\plus{} bcd \\plus{} cda \\plus{} dab \\plus{} 14.\n\\]\n[/quote]\r\n$ \\frac{22}{9}(ab \\plus{} bc \\plus{} cd \\plus{} da \\plus{} ac \\plus{} bd)\\leq abc \\plus{} bcd \\plus{} cda \\plus{} dab \\plus{} \\frac{32}{3}$ is true too.", "Solution_2": "[quote=\"shijin93\"]Give a,b,c,d are positive numbers, and a+b+c+d=4, prove that\n\\[ 5(ab \\plus{} bc \\plus{} cd \\plus{} da \\plus{} ac \\plus{} bd)\\leq abc \\plus{} bcd \\plus{} cda \\plus{} dab \\plus{} 16.\n\\]\n\n\\[ 3(ab \\plus{} bc \\plus{} cd \\plus{} da \\plus{} ac \\plus{} bd)\\leq abc \\plus{} bcd \\plus{} cda \\plus{} dab \\plus{} 14.\n\\]\n[/quote]\r\n1)$ a\\equal{}b\\equal{}c\\equal{}d\\equal{}1, 30<20$ :wink: \r\n2) My solution very ugly(Murhed, Schur)", "Solution_3": "Sorry I mistyped it. It must be this\r\n\\[ 5(ab\\plus{}bc\\plus{}ca\\plus{}da\\plus{}db\\plus{}cd)\\leq abc\\plus{}bcd\\plus{}cda\\plus{}dab\\plus{}26.\\]" } { "Tag": [ "quadratics", "calculus", "calculus computations" ], "Problem": "How do you check in general whether a differential is exact?\r\nMy teacher said something about \"Euler's Criterion\" but that's about Quadratic Reciprocity and has nothing to do with exact differentials (I think).", "Solution_1": "In two dimensions, $ dF \\equal{} M(x,y)dx \\plus{} N(x,y)dy$ is an exact differential iff $ \\frac {\\partial M}{\\partial y} \\equal{} \\frac {\\partial N}{\\partial x}$. \r\n\r\nThe corresponding criterion for $ n$ dimensions involves $ \\frac {n(n \\minus{} 1)}{2}$ such relations, and can be similarly derived from the fact that $ \\frac {\\partial^2 F}{\\partial x_i \\partial x_j} \\equal{} \\frac {\\partial ^2 F}{\\partial x_j \\partial x_i}$ for all $ 1\\leq i,j\\leq n$. See [url=http://mathworld.wolfram.com/ExactDifferential.html]MathWorld[/url] for the two-dimensional case, or [url=http://en.wikipedia.org/wiki/Exact_differential]Wikipedia[/url] for a slightly more general discussion." } { "Tag": [ "algebra", "polynomial", "least common multiple", "linear algebra", "linear algebra unsolved" ], "Problem": "Let $P, Q$ be polynomials of degree $m, n$ respectively. Prove that the necessary and sufficient condition for$P$ and $Q$have common root is that the following system is linear independent:\r\n$(P, XP, X^{2}P,..., X^{n-1}P, Q, XQ, X^{2}Q, ..., X^{m-1}Q)$", "Solution_1": "This needs context; what is $X$?", "Solution_2": "$X \\in R$ is the variable of the polynomials.", "Solution_3": "It's still trivially wrong; the condition should be linear dependence rather than linear independence. With that fix, it's routine (over an algebraically closed field). Any common factor gives a linear dependence relation, and any linear dependence relation gives that the least common multiple has degree less than $m+n$. Dividing into $PQ$, we get a nontrivial common factor.\r\nIt's also trivially false over any field that's not algebraically closed; $x^{2}+1$ and $x^{2}+1$ have no common roots over $\\mathbb{R}$." } { "Tag": [], "Problem": "Where would I send an e-mail to ask a clarification question?", "Solution_1": "usamts@usamts.org, but be warned in advance that we almost never \"clarify\" the problem questions while a round is in progress.", "Solution_2": "So I should probably state that I am assuming a property at the beginning, right?\r\n\r\nSay I stated that I assume that property #X is assumed for the problem because it seemed ambiguous. If the actual solution states that property #X is false, what would happen?" } { "Tag": [ "linear algebra", "matrix", "trigonometry", "geometry", "geometric transformation", "rotation", "trig identities" ], "Problem": "A transformation carries a triangle with vertices (0,0), (13,0), (3,2) to the triangle with vertices (0,0), (12,5), (2,3).\r\n\r\n(a) Describe the transformation.\r\n(b) Where does the transformation send the point (6,0)?\r\n\r\nThanks.", "Solution_1": "[hide=\"(a)\"]Since the transformation sends $ (0,0)$ to itself, we know that it does not involve a translation. As such, we can express the transformation as follows, and solve for $ a,b,c,d$:\n\\[ \\begin{bmatrix}\nx\\\\\ny\\end{bmatrix}\\begin{bmatrix}\na & b \\\\\nc & d\\end{bmatrix} \\equal{} \\begin{bmatrix} x'\\\\\ny'\\end{bmatrix}\\]\nWhen $ (x,y)\\equal{}(13,0)$ and $ (x',y')\\equal{}(12,5)$, we have $ 13a\\equal{}12$ and $ 13c\\equal{}5$, implying that $ a\\equal{}\\frac{12}{13}$ and $ c\\equal{}\\frac{5}{13}$.\n\nAlso, when $ (x,y)\\equal{}(3,2)$ and $ (x',y')\\equal{}(2,3)$, we have $ \\frac{36}{13}\\plus{}2b\\equal{}2$ and $ \\frac{15}{13}\\plus{}2d\\equal{}3$. Solving yields $ b\\equal{}\\minus{}\\frac 5{13}$ and $ d\\equal{}\\frac{12}{13}$. So our transformation matrix is then\\[ \\begin{bmatrix}\n12/13 & \\minus{}5/13 \\\\\n5/13 & 12/13\\end{bmatrix}.\\] Note that if we let $ \\theta\\equal{}\\sin^{\\minus{}1} \\frac 5{13}$, then the matrix is in the form\\[ \\begin{bmatrix}\n\\cos\\theta & \\minus{}\\sin\\theta \\\\\n\\sin\\theta & \\cos\\theta\\end{bmatrix},\\]so this transformation is simply a counterclockwise rotation by the angle $ \\theta$.\n[/hide]\n\n[hide=\"(b)\"]\nNow just simply plug in $ (x,y)\\equal{}(6,0)$ and we get:\\[ \\begin{bmatrix}\n6\\\\\n0\\end{bmatrix}\\begin{bmatrix}\n12/13 & \\minus{}5/13 \\\\\n5/13 & 12/13\\end{bmatrix} \\equal{} \\begin{bmatrix} 72/13\\\\\n30/13\\end{bmatrix},\\]so the resulting point is $ \\left(\\frac{72}{13},\\frac{30}{13}\\right)$.[/hide]", "Solution_2": "Thank you. That solution's beautiful. Is there an alternative solution not involving matrices?", "Solution_3": "If you draw a diagram, it's hard not to notice that there are circles around the origin, and it's easy to show: $ \\sqrt{2^2\\plus{}3^2}\\equal{}\\sqrt{3^2\\plus{}2^2}$ and $ \\sqrt{5^2\\plus{}12^2}\\equal{}\\sqrt{13^2}$. So now it seems like it is a rotation, so to prove it we just need to show that the the two given triangles are congruent, which is easily done with distance formula bashing. To find the angle of rotation, note that it is the vertex angle of the isosceles triangle $ (0,0)$, $ (13,0)$, and $ (12,5)$, which has sides $ 13,13,\\sqrt{26}$, so by the Law of Cosines, $ 13^2\\plus{}13^2\\minus{}2\\cdot13^2\\cos\\theta\\equal{}26\\implies\\cos\\theta\\equal{}\\frac{12}{13}\\implies\\theta\\equal{}\\cos^{\\minus{}1}\\frac{12}{13}\\equal{}\\sin^{\\minus{}1}\\frac5{13}$.\r\n\r\nWithout matrices for the second part, note that $ A\\equal{}(0,0)$, $ B\\equal{}(6,0)$, and $ C\\equal{}(13,0)$ are on the same line, and that $ AB\\equal{}\\frac6{13}AC$. Since rotations of lines preserve the relative positions of the points on the lines, and $ C$ rotates to $ (12,5)$, $ B$ is just $ \\left(\\frac6{13}\\cdot12,\\frac6{13}\\cdot5\\right)\\equal{}\\boxed{\\left(\\frac{72}{13},\\frac{30}{13}\\right)}$.", "Solution_4": "Thanks to both of you. Your solutions are crystal clear.", "Solution_5": "The reason you should suspect that rotation is the answer is that the double-angle formulas $ \\sin 2 \\theta \\equal{} \\frac {2 \\tan \\theta}{1 \\plus{} \\tan^2 \\theta}$ and $ \\cos 2 \\theta \\equal{} \\frac {1 \\minus{} \\tan^2 \\theta}{1 \\plus{} \\tan^2 \\theta}$ tell you that twice one of the acute angles in a $ 2 \\minus{} 3 \\minus{} \\sqrt {13}$ triangle is one of the acute angles in a $ 5 \\minus{} 12 \\minus{} 13$ triangle. Equivalently this is because $ (3 \\plus{} 2i)^2 \\equal{} 5 \\plus{} 12i$." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let three equal circles $C_{1},\\ C_{2},\\ C_{3}$ externally tangent ($C_{1}$ with $C_{2}$,$C_{2}$ with $C_{3}$,$C_{3}$with $C_{1}$) and these circles are internally tangent to onother circle $C_{4}$ .\r\nLet a point $S$ in $C_{4}$ . Let the tangents from $S$ to $C_{1}$,$C_{2}$,$C_{3}$ at the points $A,\\ B,\\ C$ .\r\nProve that $SA=SB+SC$ or $SB=SC+SA$ or $SC=SA+SB$ .", "Solution_1": "this easily follows from the generalised Ptolemy theorem!\r\nlet r be the length of the common tangent of two of the circles then we have (if for example we assume S to be in the arc AB) r*SA +r*SB=r*SC\r\ngiving the required result!", "Solution_2": "[color=darkblue]This problem is a particular case of the [b][u]Casey's theorem[/u][/b] which is similarly with the particular case of the [b][u]Ptolemeus' theorem[/u][/b] for an equilateral triangle.[/color]", "Solution_3": "http://www.mathlinks.ro/Forum/viewtopic.php?highlight=casey&t=14443 :)", "Solution_4": "What theorem called <> ?", "Solution_5": "Generalized Ptolemy = Casey's theorem. See [url=http://www.google.ro/search?q=casey+theorem&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official]here[/url]." } { "Tag": [ "geometry", "rectangle", "combinatorics unsolved", "combinatorics" ], "Problem": "I want a general solution to problem such as these, the question is to find out how many paths of the shortest possible length there are between A and B , how do I think and what is the fastest way to solve it?\r\n\r\nAny help would be greatly appriciated.", "Solution_1": "It is the standard combinatorics problem with just one rectangle but with some subcases.", "Solution_2": "For any small situation more complicated than just a rectangle, the fastest way is often just to do it by hand, as if you were building Pascal's triangle. In this case, that should take under a minute.\r\n\r\nOne can describe a general formula for the number of paths from one corner to another of a rectangular grid that doesn't pass through a given set of points. It's just inclusion-exclusion taken over paths passing through those points in all of the possible orders. It's not practical to implement in any but the most simple cases, unfortunately.", "Solution_3": "Denote $A(0,0), B(5,6)$\r\n\r\nThen the path can be\r\n\r\n(1) $(0,0)\\to (2,2)\\to (5,6)$, the number of the path is $\\binom{4}{2}\\binom{7}{3}$\r\n\r\n(2) $(0,0)\\to (2,3)\\to (5,6)$, the number of the path is $\\binom{5}{2}\\binom{6}{3}$\r\n\r\n(3) $(0,0)\\to (2,4)\\to (5,6)$, the number of the path is $\\binom{6}{2}\\binom{5}{3}$", "Solution_4": "I don't think your solution is correct libra_gold or I made a mistake evuluating it :D But I don't get the correct value.", "Solution_5": "Is this what u mean, libra_gold ??/\r\n(1) $(0,0)\\to(2,2)\\to(5,6)$, the number of path is $\\binom{4}{2}\\binom{7}{3}= 210$\r\n(2) $(0,0)\\to(1,3)\\to(5,6)$, the number of path is $\\binom{4}{3}\\binom{7}{4}= 140$\r\n(3) $(0,0)\\to(0,4)\\to(5,6)$, the number of path is $\\binom{4}{4}\\binom{7}{5}= 21$\r\n\r\ngives the total of $371$", "Solution_6": "Sorry, I found something wrong.\r\n\r\nIn fact we should delete some paths which are calculated twice.\r\n\r\nI correct my solution as following:\r\n\r\nDenote $A(0,0), B(5,6)$\r\n\r\nThen the path can be\r\n\r\n(1) $(0,0)\\to (2,2)\\to (5,6)$, the number of the path is $\\binom{4}{2}\\binom{7}{3}$\r\n\r\n(2) $(0,0)\\to (1,3)\\to (2,3)\\to (5,6)$, the number of the path is $\\binom{4}{1}\\binom{6}{3}$\r\n\r\n(3) $(0,0)\\to (1,4)\\to (2,4)\\to (5,6)$, the number of the path is $\\binom{5}{1}\\binom{5}{3}$\r\n\r\nThe correct answer should be $340$.", "Solution_7": "Ooppsss... Yeah, you're right...\r\nI make double paths at $(1,3) \\to (2,3)$ and $(2,2) \\to (2,3)$\r\nAlso at $(1,3) \\to (1,4)$ and $(0,4) \\to (1,4)$", "Solution_8": "LOL sorry I am an IDIOT" } { "Tag": [ "group theory", "abstract algebra", "search", "superior algebra", "superior algebra theorems" ], "Problem": "Someone knows the book/article regards the theory of groups\r\n above all the subgroup promormal and the T-groups?:( \r\n many thanks! :) \r\nsastra", "Solution_1": "I usually use follow book to search defines:\r\n\r\nhttp://www.amazon.com/Introduction-Theory-Groups-Joseph-Rotman/dp/0387942858\r\n\r\nTo hope it help you so :wink:" } { "Tag": [ "geometry", "rectangle", "trigonometry", "parallelogram" ], "Problem": "Let $ABCD$ be a rectangle with $BC=3AB$. Show that if $P,Q$ are the points on side $BC$ with $BP = PQ = QC$, then \\[\\angle DBC+\\angle DPC = \\angle DQC.\\]", "Solution_1": "[hide]Let $\\angle DBC = \\alpha$, $\\angle DPC = \\beta$, $\\angle DQC = \\gamma$. Then $\\alpha+\\beta = \\gamma$ iff $\\cos (\\alpha+\\beta) = \\cos \\gamma$ because $0 < \\gamma < \\frac{\\pi}{2}$ and $0 < \\alpha+\\beta < \\frac{3\\pi}{2}$. \nHence iff $\\frac{BC}{BD}\\cdot\\frac{PC}{PD}-\\frac{AB}{BD}\\cdot\\frac{AB}{PD}= \\frac{AB}{DQ}$ or after substitution $\\frac{5AB^{2}}{BD\\cdot PD}= \\frac{AB}{DQ}$. After squaring becomes $25AB^{2}\\cdot DQ^{2}= BD^{2}\\cdot PD^{2}$ and by Pythagoras $50AB^{4}= 50AB^{4}$.[/hide]", "Solution_2": "[hide=\"Easy...\"]\nSame angle labeling as [b]Andreas[/b] ( $\\angle DBC = \\alpha, \\ \\angle DPC = \\beta, \\ \\angle DQC = \\gamma$). \nThen, $\\alpha = \\arctan \\left( \\frac{1}{3}\\right), \\ \\beta = \\arctan \\left( \\frac{1}{2}\\right), \\ \\gamma= \\frac{\\pi}{4}$.\nSince $0 < \\alpha, \\beta, \\gamma \\leq \\frac{\\pi}{4}, \\ \\ \\alpha+\\beta = \\gamma \\iff \\tan(\\alpha+\\beta) = \\tan(\\gamma) \\iff \\frac{ \\frac{1}{3}+\\frac{1}{2}}{1-\\left(\\frac{1}{2}\\right) \\left( \\frac{1}{3}\\right) }= 1$, which is true. [/hide]\r\n\r\nThis was an Olympiad problem??", "Solution_3": "[hide]Let $\\angle DBC=\\alpha, \\angle DPC=\\beta, \\angle DQC=\\gamma$.\n$\\sin \\gamma=\\frac{1}{\\sqrt{2}}$. \n$\\sin(\\alpha+\\beta)=\\sin\\alpha\\cos\\beta+\\sin\\beta\\cos\\alpha=(1/\\sqrt{10})(2/\\sqrt{5})+(1/\\sqrt{5})(3\\sqrt{10})=\\frac{1}{\\sqrt{2}}$.\nThus, $\\sin\\gamma=\\sin(\\alpha+\\beta)$.\nTaking arcsin of both sides, $\\gamma+2k\\pi=\\alpha+\\beta$, for integer k. Since $0<\\gamma,\\alpha,\\beta<180$, $\\gamma=\\alpha+\\beta$.[/hide]", "Solution_4": "yeah, with trig. is all easy.\r\n\r\nis there a way to prove it without trig.?", "Solution_5": "[hide=\"ha! no trig!\"]\nPlace points $X$ and $Y$ on $\\overline{AD}$ such that $AX=XY=YD$, and let $O$ be the intersection of $\\overline{BD}$ and $\\overline{PY}$. Let $l=BP$; then $XY=l$ and simple isosceles right triangles give us $BX=PY=l\\sqrt{2}$. Also, the diagonals of parallelogram $BPDY$ bisect each other, so that $OP=\\frac{1}{2}PY=\\frac{l}{\\sqrt{2}}$. But this means that $\\triangle BXY$ and $\\triangle BPO$ are similar (by SAS similarity; also note that $\\angle BPO=\\angle BXY=135^\\circ$). Now we get congruent angles all over the place:\n\\[\\angle DBC=\\angle OBP=\\angle XBY=\\angle BYP=\\angle YPD\\]\nso that\n\\[\\angle DBC+\\angle DPC=\\angle YPC=\\angle DQC\\]\nas desired.\n[/hide]", "Solution_6": "I don't get where is $P$....\r\n\r\ngive a image please", "Solution_7": "[color=darkred]Short and nice solution[/color][hide]\nSee attached diagram!$AE=DF=AB$ and in $\\Delta EPD$ $EP=DP$ and $\\angle EPD=90$ [/hide]", "Solution_8": "[hide=\"Natural first attempt for me; ugly compared to others after checking theirs\"]\nLet first angle be called 1, second called 2, third 3.\nThen $\\cos 1=\\frac{1}{\\sqrt{10}}~\\cos 2+\\frac{1}{\\sqrt{5}}~\\cos 3=\\frac{1}{\\sqrt{2}}$\n$\\cos (1+2)=\\frac{1}{\\sqrt{10}}\\frac{1}{\\sqrt{5}}-\\frac{3}{\\sqrt{10}}\\frac{2}{\\sqrt{5}}=-\\frac{5}{50}=-\\frac{1}{\\sqrt{2}}$\nconsidering angle bounds, this implies that 1+2=3[/hide]", "Solution_9": "[hide]excellent proof \"delta\" :idea: [/hide]" } { "Tag": [ "AMC", "AIME", "AMC 12", "AMC 10" ], "Problem": "I already have aops vol. 1 and vol. 2 and introduction the counting and probability. \r\nI want to qualify for the AIME (I'm taking the AMC 12 this year) and I am not sure if the books I already have are adequate enough. Is there anything in The first steps for math Olympians that I cannot find in the books I already have?", "Solution_1": "AOPS 1 and AOPS 2 will teach you in a much better way than first steps for math olympians.", "Solution_2": "Probably not. The books you already have cover more than the first steps book. If you do some practice AMC 10s as well, you should be fine.", "Solution_3": "FSMO does a good job preparing for the AMCs but thats about it. \r\n\r\n\r\nit has a pretty good coverage of topics and you can finish the book pretty quickly, (one chapter wont take more than 1-1.5 hrs)\r\n\r\nthe problems are pretty nice i guess. the first few chapters are very elementary and the last 4 are geared towards topics on amc12.\r\n\r\n aops v1 and v2 definitely have more material and they are much more involved than FSMO, but the only thing is, v1 and v2 will take more time to properly finish because you have to reallly think deeply about many of the problems. while fsmo might only have 1 or 2 \"deep\" problems per section.\r\n\r\n\r\nonly get FSMO if you need to do really well on AMC but you dont have alot of time to study.", "Solution_4": "The Math Olympiad books will help you more on AMC 8, while the Aops Vol. 1 and 2 are great for all AMCs", "Solution_5": "its more at V1 level but its problems come from AMC12, AMC10(some of them) and AHSME.", "Solution_6": "[quote=\"giratina150\"]The Math Olympiad books will help you more on AMC 8, while the Aops Vol. 1 and 2 are great for all AMCs[/quote]\r\n\r\nThis is to some extent, false. FSFMO is quite helpful in preparing for the AMC10 & 12. Although the way the topics are covered in FSMO is somewhat not to my liking, it has great exercises and it can be helpful if you're running low on time.", "Solution_7": "I think that FSMO and the AoPS v1 and v2 are excellent. But I would start with AoPS v1 if you have time. For review, use FSMO." } { "Tag": [ "trigonometry", "trig identities", "Law of Cosines", "geometry", "Law of Sines" ], "Problem": "Prove that $\\cos\\frac{8\\pi}{35}+\\cos\\frac{12\\pi}{35}+\\cos\\frac{18\\pi}{35}=\\frac{1}{2}\\cdot\\left(\\cos\\frac{\\pi}{5}+\\sqrt7\\cdot\\sin\\frac{\\pi}{5}\\right).$", "Solution_1": "Are we allowed to brute force this. :rotfl:", "Solution_2": "[hide=\"General solution outline\"] We can write out $\\frac{8\\pi}{35}$ as $\\frac{15\\pi}{35}-\\frac{7\\pi}{35}$. We can use the trig identity $\\cos(\\pi-x)=-\\cos x$ to rewrite the angle as a negative cosine of $\\frac{30\\pi}{35}-\\frac{7\\pi}{35}$. We can write the last one out also as a subtraction of $\\frac{25\\pi}{35}-\\frac{7\\pi}{35}$. We can simplify the fractions and use cosine subtraction identities to expand. We can group together the terms with $\\cos{\\frac{\\pi}{5}}$ and $\\sin{\\frac{\\pi}{5}}$. We notice that, after some usage of the cosine identity and its corresponding (not the same) identity for sine, the sum of the arguments of the cosine and sine fuctions is $\\pi$ for each group. Therefore, these angles are parts of a triangle with special angles. We can use trig identities and the Law of Cosines and Law of Sines to find the cosines and sines of each of the angles. We can substitute in the values to see that the identity is true. (I haven't done the last part :oops: )[/hide]" } { "Tag": [ "conics", "hyperbola" ], "Problem": "I have to do a presentation about the conic section hyperbola. My schoolbook has just a little information since it is not in the school program, i. e., it is not officially taught, it is optional. So, although I am searching in google, I would like to see your sugestions not only about sites to seek information but also about methods of presentation to make it more interesting. Thank you and sorry my english.", "Solution_1": "How about trying Wikipedia? Google, in my opinion, [i]is[/i] a pretty good place to search.", "Solution_2": "http://mathworld.wolfram.com/ \r\n\r\nand do a search." } { "Tag": [ "ratio" ], "Problem": "The sums of N terms of two arithmetic progressions are in the ratio (7N+2)/(N+4). Find the ratio of their 5th term. \r\n\r\n\r\n\r\n\r\nThanks!", "Solution_1": "You've posted this problem in two forums, when it need only be placed in one. It has been answered in the other forum so I'm locking this post, next time, however, please post problems in only one forum." } { "Tag": [ "trigonometry", "geometry unsolved", "geometry" ], "Problem": "This problem has been driving me nuts for a couple of days.\r\n\r\n[img]http://www.gamepsychos.com/squarecircle.gif[/img]\r\n\r\nThe Blue Lines are 24cm long and the red one is 16cm long. The square is in the exact center of the circle.\r\n\r\nWhat is the radius of the circle?", "Solution_1": "Sorry for your inconvenience but please delete it as a substitute for somebody because I cannot delete this post from me.", "Solution_2": "I don't think the side of the square is $12\\sqrt{2}$ because the point where the blue line meets the circle isn't collinear with two of the vertices of the square.\r\n\r\nSolution:\r\n\r\nIf the blue lines are parallel to the red line, they all incline at 45 degrees.\r\nLet $2a$ be the length of the diagonal. We get $r=a+16$.\r\nDraw lines from the centre to the both ends of the blue line.\r\nThese form a right-angled triangle with sides of length $a,24,r$.\r\nSo $a^{2}+24^{2}=r^{2}\\Rightarrow a^{2}+24^{2}=(a+16)^{2}\\Rightarrow a=10$.\r\nTherefore, $r=26$.", "Solution_3": "Sorry.\r\nWhen a right-angled isosceles triangle was made, I misunderstood it.\r\n\r\nI introduce a solution in a thing with much effort.\r\n\r\n$\\cos \\angle FGB = \\frac{GB}{GF}= \\frac{16}{8\\sqrt{13}}=\\frac{2}{\\sqrt{13}}$\r\n$\\cos \\angle OGH = \\frac{GH}{OG}= \\frac{4\\sqrt{13}}{OG}$\r\n\r\nThus, $\\cos \\angle FGB = \\cos \\angle OGH \\iff \\frac{2}{\\sqrt{13}}= \\frac{4\\sqrt{13}}{OG}\\iff OG=26$\r\n\r\n\r\nThe answer that I made a mistake in not to misunderstand it deleted it." } { "Tag": [ "vector", "analytic geometry", "linear algebra", "linear algebra unsolved" ], "Problem": "S={ x e R3 / x1 + 2x2 - x3 = 0}. Find a base B from R3 that contains a base of S and a base of S orthogonal, and that the vector (0,5,-2) hast coordinates (0,1,1) in the base B.\r\n\r\nI have come till here:\r\n\r\nS=<(1,0,1)(0,1,2)>\r\nS orthogonal=(-1,-2,1)\r\n\r\nPutting these two together i get a Base, not B, lets call it Z.\r\n\r\nZ={(1,0,1)(0,1,2)(-1,-2,1)}\r\nCoordinates of (0,5,-2) in Z are (-2,1,-2).\r\n\r\nHow can i go on? Please help.", "Solution_1": "The only choice for the part of the basis orthogonal to $S$ is size- since our first try gave a coefficient of $-2$, let's multiply that one by $-2$; the third vector in the basis is now $(2,4,-2)$. We can find the second basis vector by subtracting the third from $v=(0,5,-2)$, since we know the coefficients of $v$ with respect to this basis. That makes the second basis vector $(-2,1,0)$. We have plenty of choice now for the first basis vector; anything in $S$ and not a multiple of $(-2,1,0)$ will work." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Okay, obviously I know my rank, but I have no idea what my score was at states and chapter both. How do you people know your scores? Is there a website you can find it at? Forgive me if I'm ignorant. :lol:", "Solution_1": "Don't they post the answers up at your state competition?\r\nThey should also mail the scores to your school.\r\nChapter, I don't know. Didn't get my scores there.", "Solution_2": "The scores should have been given to your coach for both state and chapter. If you want to know your scores, just ask him/her.", "Solution_3": "The scores were told to us by our mathcounts coach, and also, we got our test back too!", "Solution_4": "[quote=\"236factorial\"]The scores were told to us by our mathcounts coach, and also, we got our test back too![/quote]\r\n\r\nGetting the test back was quite unusual. They do not give the tests back. It is not that I do not believe you 236factorial :) !\r\n\r\nEvery coach gets a scoresheet back. Your coach should be able to tell you your score on the sprint, target and team rounds. You can also know which problems you missed from this sheet.", "Solution_5": "I agree. usually caoches only get their scoresheets back, and tests are basically never given back, and if they did, it wold probably be given back after all the other state competitions have been finished. Prevents cheating, see?", "Solution_6": "Okay, thanks everybody. I'll have to talk to my coach and see if he received anything." } { "Tag": [ "function", "integration", "calculus", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f: R\\minus{}>R$ be a continuous function with $ \\int_x^y f(t)dt<\\equal{}f(x\\plus{}y)$ for any real $ x,y,$. Prove $ f\\equal{}0$.", "Solution_1": "$ 0 \\equal{} \\int_x^x f(t)\\; dt \\leq f(2x)$ for all real $ x$,\r\nthus $ f(x) \\geq 0$ for all $ x$.\r\nSuppose $ f$ has a minimum in $ x$,\r\nthen $ \\int_{\\frac{x}{2}\\minus{}1}^{\\frac{x}{2}\\plus{}1}f(t)\\; dt \\leq f(x)$,\r\nbut the integration interval has length 2, and the function values in that interval are at least $ f(x)$,\r\nthus the inequalty can only be fullfilled when $ f(t) \\equal{} 0$ for $ t \\in [\\frac{x}{2}\\minus{}1,\\frac{x}{2}\\plus{}1]$.\r\nNow it is easy to see that the function is 0 for all arguments, by moving the $ x$-value to the left and the right.\r\nIf $ f$ does not have a minimum, then it has an infinum $ \\alpha$. Choose a very small $ \\epsilon>0$.\r\nThen the reasoning remains the same, when you start with a $ x$-value for which $ |f(x)\\minus{}\\alpha| < \\epsilon$." } { "Tag": [ "calculus", "derivative", "calculus computations" ], "Problem": "A man 5 ft tall walks at the rate 4 ft/sec directly away from a street light which is 20 ft above the street. At what rate is the length of this shadow changing? I don't understand like What's the difference between if the question asks you the rate of the tip of his shadow changing and the length changing? Hope you can give me some hint. thanks.", "Solution_1": "It's based on related rates, which obviously requires derivatives. You need to set up an equation that describes the scenario (looks like a triangle will do), and then take the derivative for each term with respect to time. By plugging in what you know, you can solve for what you don't." } { "Tag": [ "analytic geometry", "geometry", "3D geometry", "sphere", "number theory", "relatively prime", "complex numbers" ], "Problem": "All triples of relatively prime positive integers $(a,b,c)$ (with $a$ even) that are solutions to the equation $a^{2}+b^{2}=c^{2}$ are given by the formula $(2rs,r^{2}-s^{2},r^{2}+s^{2})$ where $r$ and $s$ are positive integers with $r>s$.\r\n\r\nFind a similar formula for the integer solutions to the equation $a^{2}+b^{2}+c^{2}=d^{2}$.", "Solution_1": "The first statement is a little off. First, we can't represent $(3, 4, 5)$ by that formula; we need to allow the first two coordinates to be switched. Second, even with that fix, we can't represent $(9, 12, 15)$; we need to handle nonprimitive triples with an extra parameter.", "Solution_2": "Well, I think I have fixed it, but either way I am referring to primitive Pythagorean triples.", "Solution_3": "Hmm now that I think about it maybe it should be moved to the Pre-Olympiad forum. If a moderator could move it please... :)", "Solution_4": "[hide=\"Cheap\"] We will instead find the integer solutions to $a^{2}+b^{2}+c^{2}+d^{2}= e^{2}$ and then consider those solutions where one of $a, b, c, d$ is zero.\n\nWe will be making use of the Lipschitz quaternions, which are the quaternions with integer components (there is another name for them, but I don't recall). \n\nWe desire to find a Lipschitz quaternion whose norm is the square of an integer. Hence, we square an arbitrary Lipschitz quaternion (because the norm is multiplicative):\n\n$(x+yi+zj+wk)^{2}= (x^{2}-y^{2}-z^{2}-w^{2})+(2xy) i+(2xz) j+(2xw) k$\n\n(I will spare you the details; quaternion multiplication is just a hassle.) The norm of this quaternion is the square of an integer, that is,\n\n$(x^{2}-y^{2}-z^{2}-w^{2})^{2}+(2xy)^{2}+(2xz)^{2}+(2xw)^{2}= \\left( x^{2}+y^{2}+z^{2}+w^{2}\\right)^{2}$\n\nWe want to set each of these to zero in turn. \n\nThe first term is troublesome, and suggests that we may not be able to generate all solutions in this way, since setting it to zero is equivalent to solving our original problem...\n\nThe other terms are symmetric, so WLOG we let $w = 0$. Then\n\n$(x^{2}-y^{2}-z^{2})^{2}+(2xy)^{2}+(2xz)^{2}= \\left( x^{2}+y^{2}+z^{2}\\right)^{2}$\n\nWhich should generate (if not all, at least a lot of) primitive solutions. [/hide]", "Solution_5": "[quote=\"t0rajir0u\"][hide=\"Cheap\"] We will instead find the integer solutions to $a^{2}+b^{2}+c^{2}+d^{2}= e^{2}$ and then consider those solutions where one of $a, b, c, d$ is zero.\n\nWe will be making use of the Lipschitz quaternions, which are the quaternions with integer components (there is another name for them, but I don't recall). \n\nWe desire to find a Lipschitz quaternion whose norm is the square of an integer. Hence, we square an arbitrary Lipschitz quaternion (because the norm is multiplicative):\n\n$(x+yi+zj+wk)^{2}= (x^{2}-y^{2}-z^{2}-w^{2})+(2xy) i+(2xz) j+(2xw) k$\n\n(I will spare you the details; quaternion multiplication is just a hassle.) The norm of this quaternion is the square of an integer, that is,\n\n$(x^{2}-y^{2}-z^{2}-w^{2})^{2}+(2xy)^{2}+(2xz)^{2}+(2xw)^{2}= \\left( x^{2}+y^{2}+z^{2}+w^{2}\\right)^{2}$\n\nWe want to set each of these to zero in turn. \n\nThe first term is troublesome, and suggests that we may not be able to generate all solutions in this way, since setting it to zero is equivalent to solving our original problem...\n\nThe other terms are symmetric, so WLOG we let $w = 0$. Then\n\n$(x^{2}-y^{2}-z^{2})^{2}+(2xy)^{2}+(2xz)^{2}= \\left( x^{2}+y^{2}+z^{2}\\right)^{2}$\n\nWhich should generate (if not all, at least a lot of) primitive solutions. [/hide][/quote]\r\n\r\nlol i was thinking that too...i love using difficult results to trivialize problems...anyway, i don't think there is any one solution, see:\r\n[url]http://mathworld.wolfram.com/PythagoreanQuadruple.html[/url]", "Solution_6": "In retrospect, [hide=\"this identity...\"] $(r_{0}^{2}-(r_{1}^{2}+r_{2}^{2}+...+r_{n}^{2}))^{2}+\\sum_{k=1}^{n}\\left( 2r_{0}r_{k}\\right)^{2}= (r_{0}+(r_{1}^{2}+r_{2}^{2}+...+r_{n}^{2}))^{2}$ [/hide] is somewhat obvious from the given, and I really didn't need to go into the quaternions and stuff. :|\n\nBy the way, an idea: if we instead use the Hurwitz quaternions, which allow for $x, y, z, w$ to be half-integers, and then multiply by $2$, we may be able to generate some more solutions.\n\nEdit: [hide=\"Hmm, it is troublesome...\"] The solutions that the Lipschitz quaternions miss occur when we let $x = \\frac{a}{2}, y = \\frac{b}{2}, z = \\frac{c}{2}, w = \\frac{d}{2}$ where $a, b, c, d$ are odd integers. This produces, after the appropriate multiplication,\n\n$\\left( \\frac{a^{2}-b^{2}-c^{2}-d^{2}}{2}\\right)^{2}+(ab)^{2}+(ac)^{2}+(ad)^{2}= \\left( \\frac{a^{2}+b^{2}+c^{2}+d^{2}}{2}\\right)^{2}$\n\nVerify that when $a, b, c, d$ are all odd every component is an integer.\n\nHowever, we run into a problem: because of the odd condition, we cannot set any of the terms to zero except the first.\n\nThis is quite unsatisfactory. All it tells us is that given an entirely odd solution $a, b, c, d$ to\n\n$a^{2}= b^{2}+c^{2}+d^{2}$\n\nWe can generate another solution where at least two of the squares being added are odd (this is a case not covered by the Lipschitz-only formulae).\n\nUnfortunately, entirely odd solutions don't exist. Odd squares are always congruent to $1 \\bmod 4$, so where the LHS is congruent to $1 \\bmod 4$, the RHS would be congruent to $3 \\bmod 4$. [/hide]", "Solution_7": "Just a small remark: in any dimension, you can get all solutions by solving $x_{1}^{2}+...+x_{n}^{2}=1$ in rational numbers, thus look for rational points on the unit sphere. One can parametrise all solutions as the intersections of the sphere with lines through $(1,0,0,0,...,0)$ and the points $(0,a_{2},...,a_{n})$ with all $a_{i}$ rational.", "Solution_8": "[quote=\"chess64\"]All triples of relatively prime positive integers $(a,b,c)$ (with $a$ even) that are solutions to the equation $a^{2}+b^{2}=c^{2}$ are given by the formula $(2rs,r^{2}-s^{2},r^{2}+s^{2})$ where $r$ and $s$ are positive integers with $r>s$.\n\nFind a similar formula for the integer solutions to the equation $a^{2}+b^{2}+c^{2}=d^{2}$.[/quote]\r\nEquation--->\r\nd^2-c^2=a^2+b^2 --->\r\n(d-c)(d+c)=a^2+b^2\r\nSince if (a+bi)(c+di)=e+fi, then (a^2+b^2)(c^2+d^2)=e^2+f^2 by some basic binomial expansion. \r\nSince complex numbers are closed over multiplication, that means that if d-c and d+c are the sum of two squares, then its cool.\r\nThe only restrictions on d-c and d+c is that they have the same parity and sum to 2d.\r\nBut we know that 2d is the sum of four squares by four square theorem because 2d is an integer.\r\nSo if d is any number, therefore, it can be done unless 2d is the sum of two squares.\r\n\r\nUm, as for a generalized solution..... that looks like messy/ugly quaternion expansions because for some a and b for d that can be represented as such, GCD(a,b)>sqrt(2d/3) which lends itself to massive ugliness.... \r\nOh and there should be no explicit thing for d besides that it includes everything except 2d=x^2+y^2 which implies d not being x^2+y^2 because 2 is the sum of two squares for some x and y and I doubt there exists an explicit formula for such numbers.", "Solution_9": "[quote=\"Scrambled\"]So if d is any number, therefore, it can be done.[/quote]\r\n\r\nThis is true (without the \"unless...\") because $d^{2}= 0^{2}+0^{2}+d^{2}$.", "Solution_10": "oops\r\nwell i'm trying to ignore the solutions with 0's in them", "Solution_11": "Different problem. Also much harder. :|", "Solution_12": "sometimes there are compounded quadruples\r\n\r\nlike 3^2+4^2=5^2\r\n\r\n5^2+12^2=13^2\r\n\r\ntherefore 3^2+4^2+12^2=13^2 and the result is a quadruple.", "Solution_13": "[hide=\"a full solution\"]My Diophanine equations book gives this solution:\n\nAt least two of $a,b,c$ must be even; otherwise $d^{2}\\equiv 2, 3 \\mod 4$, which is impossible. So let $a = 2x$ and $b = 2y$ for positive integers $x$ and $y$. Also let $d = c+z$. Then we have\n\n$4x^{2}+4y^{2}+c^{2}= (c+z)^{2}$\n\n$\\Rightarrow z^{2}= 4(x^{2}+y^{2})-2cz$\n\nWe conclude that $z = 2t$ for some positive integer $t$. Plugging this result into the above equation and solving for $c$ gives\n\n$c = \\frac{x^{2}+y^{2}-t^{2}}{t}.$\n\nNow we conclude from $d = c+z$ that\n\n$d = \\frac{x^{2}+y^{2}+t^{2}}{t}.$\n\nwith the condition that $t | x^{2}+y^{2}, \\ t \\leq \\sqrt{x^{2}+y^{2}}$. We now have our full solution to the equation:\n\n$a = 2x, b = 2y, c = \\frac{x^{2}+y^{2}-t^{2}}{t}, d = \\frac{x^{2}+y^{2}+t^{2}}{t}$\n\nwith $t | x^{2}+y^{2}, \\ t \\leq \\sqrt{x^{2}+y^{2}}.$\n\nWe know this is the full solution to the system because we made no specific assumptions about the form of $a,b,c,d$ to come to our answer. The solution can be verified by plugging into the original equation.[/hide]", "Solution_14": "Cool solution.\r\n\r\n[hide=\"Observation\"] This is just the solution I gave divided by $x$, which implies that the quaternion solutions give [i]multiples[/i] of each primitive solution, and that all there is left to do is to divide out by a greatest common factor. [/hide]", "Solution_15": "[quote=\"t0rajir0u\"]Cool solution.\n\n[hide=\"Observation\"] This is just the solution I gave divided by $x$, which implies that the quaternion solutions give [i]multiples[/i] of each primitive solution, and that all there is left to do is to divide out by a greatest common factor. [/hide][/quote]\r\nYeah, I noticed that too when I looked back at yours and compared it to my book's - it's always fun to see two very different solution methods turn up the same answer!" } { "Tag": [ "religion", "Theology" ], "Problem": "This is basically a split-off of the thread about religion, because we were arguing about a whole bunch of different issues at the same time. :D\r\n\r\nHappy debating! :)", "Solution_1": "I don't care :no:, because i just want to live my life without debate :rotfl: \r\ndebate=pointless (just kidding) \r\n\r\nHistorians have never proved god true or not, how would you expect to come close?", "Solution_2": "Okay, nvm, a mod can just delete this.", "Solution_3": "no, there is no evidence for a supernatural spirit, why need to believe in something for which there is no evidence? Thats a indirect idea of my thuoghts.", "Solution_4": "[quote=\"hello\"]no, there is no evidence for a supernatural spirit, why need to believe in something for which there is no evidence? Thats a indirect idea of my thuoghts.[/quote]\r\n\r\nthat's what I think. If there is a god, we can only know it in silence. People say praying is \"talking to god\" sometimes, but he never replies. :(", "Solution_5": "Hehehe that reminds me of a joke I heard, \r\n\r\n\"When we talk to God, it's called praying. When God talks to us, it's called crazy.\"", "Solution_6": "I hope chess64 isn't offended by that :D", "Solution_7": "[quote=\"236factorial\"][quote=\"hello\"]no, there is no evidence for a supernatural spirit, why need to believe in something for which there is no evidence? Thats a indirect idea of my thuoghts.[/quote]\n\nthat's what I think. If there is a god, we can only know it in silence. People say praying is \"talking to god\" sometimes, but he never replies. :([/quote]\r\n\r\n*sigh* yes he does. God doesn't talk in the way that we do. :noo: When you pray for something to happen, then he makes it happen. But you have to realize, God isn't stupid. If you pray for something that might not benefit you, God will do something that is better for you. For example, if you pray for a car, and you get a bike instead, that might have been better. Maybe you would have died in a car accident the next day.\r\n\r\nLet me give you an example. In 5th grade I was the worst player in my chess club. In 6th grade, I hadn't improved at all. In 7th grade, I started praying a lot for success in chess. That year, I won 2nd in the championships. In 8th grade, I continued praying and I got 1st place. Nobody would have thought that I could win 1st place, or even place in the championships before I started praying. :)", "Solution_8": "[quote=\"chess64\"]. For example, if you pray for a car, and you get a bike instead, that might have been better. Maybe you would have died in a car accident the next day.[/quote]\r\n\r\nYou could've crashed into a tree in your bike or get ran over by a car :rotfl:", "Solution_9": "[quote=\"236factorial\"][quote=\"chess64\"]. For example, if you pray for a car, and you get a bike instead, that might have been better. Maybe you would have died in a car accident the next day.[/quote]\n\nYou could've crashed into a tree in your bike or get ran over by a car :rotfl:[/quote]\r\n\r\n*sigh* you obviously don't understand. Maybe if you would have gotten your car then you would die in a car accident; only God knows. Then He will give you a bike instead b/c He knows what will happen.\r\n\r\nThe religious state of AoPSers is in a hopeless state. :noo:", "Solution_10": "[quote=\"chess64\"]The religious state of AoPSers is in a hopeless state. :noo:[/quote]\r\nEh, if you look at the poll on religion, most people here are religious with the most number of Christians.", "Solution_11": "[quote=\"stupidkid\"][quote=\"chess64\"]The religious state of AoPSers is in a hopeless state. :noo:[/quote]\nEh, if you look at the poll on religion, most people here are religious with the most number of Christians.[/quote]\r\n\r\nstupidkid is right, but in america, there is that freedom of religion, right?", "Solution_12": "[quote=\"236factorial\"][quote=\"stupidkid\"][quote=\"chess64\"]The religious state of AoPSers is in a hopeless state. :noo:[/quote]\nEh, if you look at the poll on religion, most people here are religious with the most number of Christians.[/quote]\n\nstupidkid is right, but in america, there is that freedom of religion, right?[/quote]\r\n\r\nFrom experience, I can say that most Christians know less about Christianity than I do. ;)\r\n\r\nedit: And I'm Muslim. :)", "Solution_13": "[quote=\"236factorial\"]I hope chess64 isn't offended by that :D[/quote]\r\n\r\nhehehe I hope not, I'm a christain too and I find it highly amusing.", "Solution_14": "[quote=\"Sunny\"]Hehehe that reminds me of a joke I heard, \n\n\"When we talk to God, it's called praying. When God talks to us, it's called crazy.\"[/quote]\r\n\r\nI also remember someone saying something like, \"praying is talking to God, and meditation is listening to God.\"\r\n\r\nPersonally I don't believe in the existence of a higher being, but who knows? I'll acknowledge the possibility and wait to find out for sure when the time comes, lol. I also tried telling someone once that there's no proof of a God, and therefore there's no reason to believe in one. The reply I got was, \"that's why it's called faith.\"", "Solution_15": "[quote=tacowizard][quote]How do you know nothing can be absolutely good?\n[/quote]\n\nI'm so tired of my arguments being ignored, so here you go\n\n[quote=tacowizard]Nope. I didn't say that God exists but is 'morally messed up'. The requirements for being God are being absolutely good and perfect. Even if someone or something is strong enough to be as strong as God is described, if they aren't absolutely good, they can't be God.\n\nAnd since decisions can't be absolutely good and perfect as they always benefit one side and hurt others, and so are only good or bad relative to one side(eg. losing a war is bad for one country, but good for another) for the universe as a whole, God can't exist even if there is someone with as much strength as him because that person wouldn't be morally perfect as that can't exist as good and evil are all relative to individuals(literally my entire moral relativism argument) and so they can't be God.\n\nIf absolute good can't exist by moral relativism, then one of the requirements of God simply can't be fulfilled.\n\nOnce again, selfishness as a whole isn't a very large component of human nature in politics, as those who care about themselves care not for power but a luxurious life. Therefore, those who actually care about power want to do something with it. What those people want to do they believe is good, and what they hate, they think is evil. Everyone thinks that they are good, and that the enemy is evil, or at the very least less good than them.\n\nHenceforth, good and evil are subjective relative to the universe absolutely. They can only be applied to groups. For example, if a country won a war, then that was good for the country. But to the country that lost, that was bad. We also can't use absolute good or evil in this case as some groups are benefited whereas others are harmed, and multiple frameworks exist in this case. Some argue for the greatest majority, whereas others argue for the least well off. We have no way of determining which framework is better here, because we can't determine 'betterness'. \n\nThus, absolute good can't exist, and as being absolutely good is a requirement of being God, God can't exist.[/quote][/quote]\n\nRelative to who? To sides of the universe? To material effects? Those are still material sides and material actions, which God is separate from. Yes, some things might be good or bad momentarily, but ultimately, God is the only enjoyer, so if it all goes to God, it is still considered good.\n\nAlso, it wasn't that I ignored you. It was that I wasn't participating in the debate back when you posted that.", "Solution_16": "So you're saying that because God benefits from these things, God is absolutely good?\n\nGood is based on the outcomes of decisions on groups and individuals. As every decision harms some groups and benefits others, no decision can be absolutely good, only good to some groups and evil to others. Frameworks like utilitarianism, Rawls's Theory, etc. aren't accurate as we have no way of determining whether or not which system is better than other systems as 'better' isn't defined, as its relative to good and bad.\n\nEssentially, every hero is only a hero to those that they benefit, and every evildoer is only evil to those that they harm. Harm isn't limited to physical harm. By harm, I mean to conflict with the ideals of the individual or state(eg. a man reading a newspaper is harmed morally when he reads of a murderer, because murder is against his morals and a woman who was physically hurt is harmed morally as being hurt is against her morals), and benefiting is what benefits those morals.\n\nIn order for something to be absolutely good, there can't be any ambiguity. As no action can be universally good due to groups being hurt or benefited by the action, and if we accept the inaction of this 'God' as actions, this 'God' 's actions simply can' t be absolutely good as whatever his decisions are, they would always benefit some groups and hurt others.\n\nA decision absolutely good simply isn't possible due to morally exclusive moral goals. For example, let's say that in the start of the Spanish Civil War in 1936, Francisco Franco decides that the democratic government is ineffective, and is composed of poor leadership that creates immense wealth gaps and alienates Spain. He then starts a coup to overthrow this democratic government, which is legitamately democratic. Franco's goal is to increase the material wealth of Spain, whereas the loyalist government's goal was to preserve the ideals of democracy. There obviously can't be any compromise here due to the value of government legitamacy. By the loyalist government's moral code, Franco having part of Spain means that part of Spain would have an illegitamate ruler, which by their legal code is injust. By Franco's code, the monarchy ought to be restored as it is the legitamate government, and so any trace of the Republic, which is incredibly ineffective at ruling, must be abolished as the Spanish are one people and so can't be seperated. Compromises are only created when both sides are benefited although possibly to lesser amounts, not when both sides are harmed when one side could benefit. And even if compromise can occur, it would be injust by both groups as their views are mutually exclusive.\n\nAs being absolutely good is a requirement of being God, even if some godlike being existed, they wouldn't be God even if they made the universe and were omnipotent.\n\n[quote]Relative to who? To sides of the universe? To material effects?[/quote]\n\nNot sure what you mean by this, but good and evil, as I already said, are relative to the in and out groups, or who were benefited(the in) and who are harmed(the out). There isn't any absolute good unless it benefits everyone's moral framework, which is impossible as mutually exclusive groups exist.", "Solution_17": "[quote=Doodles250]@SilverLightning Then what caused the Big Bang? The universe just randomly started existing with nothing to start it?[/quote]\n\nThe Big Bang is the moment that space-time came into existence. Before the Big Bang there was no space or time. So, it is actually meaningless to ask what caused the Big Bang to happen \u2013 there was no Universe in which that cause could have existed.\n\nQuantum physics has shown us that some events have no cause at all. Things can happen randomly for no particular reason. This unpredictable and nature of the Universe is experimentally verified but it is a fundamental property of the Universe. ", "Solution_18": "[quote=pezzeepenguin69]\n\nOkay... that's the thing... you're viewing it from the stance that something must be sustaining the universe right now... If it isn't God, it is nothing? Make that make sense. I don't believe that we can definitively say that it's God, but that's what separates theists from atheists. Theists believe that God holds the energy that sustains the universe while atheists believe that nothing holds the universe in place. It just sustains itself in a circle of life. Idk. Idc. It doesn't matter.[/quote]\n\nNo I am not. Nothing is sustaining the universe. The universe is going to rip itself apart in around a googol years. The universe is inherently chaotic. \n\nWe have also found no life forms outside of Earth so how does a circle of life hold the universe in place???", "Solution_19": "To be honest it doesn't even matter if something made the Big Bang as there's a million different possible fictional and real deities that represent a 'first god' in their corresponding religions, and God only refers to 3 of them.\n\nAnd then the Big Bang could have just been made literally by itself because the Big Bang could have been 'above reality' and then part of it leaked into the material universe-who knows.", "Solution_20": "noooo now i have to read 30 messages", "Solution_21": "[quote=tacowizard]To be honest it doesn't even matter if something made the Big Bang as there's a million different possible fictional and real deities that represent a 'first god' in their corresponding religions, and God only refers to 3 of them.\n\nAnd then the Big Bang could have just been made literally by itself because the Big Bang could have been 'above reality' and then part of it leaked into the material universe-who knows.[/quote]\n\nI'm not really sure. If a deity does not exist, can morals exist? ", "Solution_22": "[quote=shadowking2001][quote=tacowizard]To be honest it doesn't even matter if something made the Big Bang as there's a million different possible fictional and real deities that represent a 'first god' in their corresponding religions, and God only refers to 3 of them.\n\nAnd then the Big Bang could have just been made literally by itself because the Big Bang could have been 'above reality' and then part of it leaked into the material universe-who knows.[/quote]\n\nI'm not really sure. If a deity does not exist, can morals exist?[/quote]\n\nMorals are an entirely evolutionary construct.", "Solution_23": "[quote=nsd08]\n\nMorals are an entirely evolutionary construct.[/quote]\n\n\nHow so? How could morals evolve into existence? ", "Solution_24": "[quote=shadowking2001][quote=nsd08]\n\nMorals are an entirely evolutionary construct.[/quote]\n\n\nHow so? How could morals evolve into existence?[/quote]\n\nFor example, consider the moral that murder is wrong. Some people think murder is wrong and some think it is justified. However, those who think it is justified are at an evolutionary disadvantage, as people would not want to be murdered no matter whether they think they are justified in murder. Thus, they have less kids and spread their ideas less. Thus, evolution by natural selection explains why everyone thinks murder is wrong.", "Solution_25": "But its just as easy to say that God designed the world that way, so the morals would be there anyway. \n\nTo counter, why would people value human life? ", "Solution_26": "[quote=shadowking2001]But its just as easy to say that God designed the world that way, so the morals would be there anyway. \n\nTo counter, why would people value human life?[/quote]\n\nThat is indeed a possible theory. However, it is not the only theory.\n\nPeople value human life because they value their own life (due to evolutionary pressure). Thus, it is advantageous for everyone if people value life.", "Solution_27": "[quote=shadowking2001]But its just as easy to say that God designed the world that way, so the morals would be there anyway. \n\nTo counter, why would people value human life?[/quote]\n\nNo, because people thinking that murder is good is ultimately disadvantageous to the survival as a human race as a whole.", "Solution_28": "[quote=shadowking2001][quote=tacowizard]To be honest it doesn't even matter if something made the Big Bang as there's a million different possible fictional and real deities that represent a 'first god' in their corresponding religions, and God only refers to 3 of them.\n\nAnd then the Big Bang could have just been made literally by itself because the Big Bang could have been 'above reality' and then part of it leaked into the material universe-who knows.[/quote]\n\nI'm not really sure. If a deity does not exist, can morals exist?[/quote]\n\nPlease read his arguments...he posted numerous arguments.", "Solution_29": "[quote=shadowking2001]But its just as easy to say that God designed the world that way, so the morals would be there anyway. \n\nTo counter, why would people value human life?[/quote]\n\nGod couldn't make the world that way because God has to be absolutely good and having human life valued over other things causes suffering relative to groups such as animals, the environment, and saving certain groups of humans over others isn't absolutely good as one thing(human lives) are benefited whereas other aspects(another amount of human lives) are harmed. Compromise might occur, and a middle ground can be established sometimes, but there are other times when two groups want mutually exclusive things, such as if one group wants to abolish a dictatorship, whereas others see a monarchy as a legitamate government. To have a 'middle ground' would be unjust by both sides as half the country would be ruled by an illegitamate government. In this situation, its impossible to make an absolutely good decision as whatever decision made would help one group and hurt the other. As these situations exist, and there can't be absolutely good decisions made during them, God can't make absolutely good decisions and can only make decisions good for some groups and harmful to others, and therefore isn't absolutely good.\n\nIf God isn't absolutely good, then God isn't the 'God' as a requirement of being God is being the embodiment of good, which means God has to be absolutely good.\n\nAnd no, we can't decide to simply look at the majority. For example, it's obviously wrong for a doctor to kill 1 person, steal all of his organs, and use them to save 3 patients. \n\nAnd we can't decide to simply benefit the worst well off first. For example, it's obviously wrong for all rich people to be forced to donate all of their wealth to the poor, as they worked for that wealth and if that happened then nobody would want to work hard to become rich anymore.\n\n" } { "Tag": [ "algebra", "polynomial", "number theory proposed", "number theory" ], "Problem": "Let $ \\tau (k)$ denote the number of positive divisors of $ k$ and let $ \\tau (0) \\equal{} \\infty$. \r\n\r\nLet two nonconstant polynomials with integer coefficients be $ f(x)$ and $ g(x)$. If $ \\tau (f(n)) \\equal{} \\tau (g(n))$ for all integers, $ n$, then prove that $ f \\equal{} \\pm g$.\r\n\r\nEDIT: You're right - sorry. Problem edited.", "Solution_1": "As stated, $ f \\equal{} \\minus{}g$ works, as does $ f, g$ constant and equal to different integers with the same number of positive divisors.", "Solution_2": "Someone know a solution for this one?", "Solution_3": "Sorry for the delay; here's the solution:\r\n\r\n[hide=\"Solution\"]\nAssume that $ f(n) \\ne cg(n)$ for a constant $ c$. Let $ f(n) \\equal{} \\prod_{i \\equal{} 1}^k f_i (n)^{d_i}$ and $ g(n) \\equal{} \\prod_{i \\equal{} 1}^l g_i (n)^{e_i}$, where the $ f_i$ and $ g_i$ are irreducible polynomials. Note that, for an irreducible polynomial $ p$ and an arbitrary polynomial $ q$ so that $ p(x) \\not | q(x)$, there are polynomials $ r(x)$ and $ s(x)$ so that $ pr (x) \\plus{} qs (x)$ is constant. Therefore, the of primes, $ K \\equal{} \\{k_1, k_2, k_3, \\cdots , k_x \\}$, so that some member of $ K$, say $ k$, dividing both numbers in the following possibilities is finite:\n\na) $ f_i (m), f_i' (m)$\nb) $ f_i (m), g_j (m)$\nc) $ f_i (m), f_j (m)$\nd) $ g_i (m), g_j (m)$\ne) $ g_i (m), g_i'(m)$ \nf) $ g_i (n)$ for all $ n$\ng) $ f_i (n)$ for all $ n$\n\nfor some $ i, j$ (the solutions to f) and g) are clearly finite since $ f(0)$ and $ g(0)$ are finite). \n\n\n\n[b]Lemma:[/b] Take an irreducible polynomial $ t(n) | g(n)$; there exists a prime $ q$ so the number of $ 1\\le n\\le N$ with the property that there exists a prime not in $ K$ that divides $ t(n)$ at least $ q \\minus{} 1$ times is less than $ \\frac {N}{l}$. \n\n[i]Proof: [/i]For any prime, $ r$ sufficiently large, that divides $ t(n)$, there are at most $ \\deg g$ solutions to $ t(n) \\equiv 0\\bmod r$ since $ t (n)$ is irreducible. By Hensel's Lemma, the \"lifted\" solution to $ t (rt \\plus{} n) \\equiv 0\\bmod r^2$ has precisely one solution for each solution of $ t (n) \\equiv 0\\bmod r$. Therefore, there are at most $ \\deg g_i$ solutions to $ t (n) \\equiv 0\\bmod r^{q \\minus{} 1}$, and so for an extremely large $ N$, there are at most $ \\sum_{q} \\frac {N\\deg g_i}{r^{q \\minus{} 1}}$ solutions for $ n$ to $ g_i (n) \\equiv 0\\bmod r^{q \\minus{} 1}$. With $ q$ large enough, this sum will be less than $ \\frac {N}{l}$. \n\n\n\nFor any $ k_i$, increase $ q$ even more so that $ g (n) \\not \\equiv 0\\bmod k_i^{q \\minus{} 1}$ has at least a solution. Use the Chinese Remainder Theorem to obtain an $ n$ so that $ g (n) \\equiv 0\\bmod \\prod_{i \\equal{} 1}^x k_i^{q \\minus{} 1}$ has no solution. For all $ i$, take $ t_i(n) \\equal{} g_i(Mn \\plus{} y)$ so that $ Mn \\plus{} y \\equiv n\\bmod \\prod_{i \\equal{} 1}^x k_i^{q \\minus{} 1}$ and $ Mn \\plus{} y \\equiv m\\bmod p^{r \\plus{} 1}$, where $ p$ is a prime to be chosen later (but $ p > \\max {k_i}$ so that the Chinese Remainder Theorem works) and $ r$ is a prime greater than $ q\\max \\{d_i \\}\\max \\{e_i \\}$. Let $ g (Mn \\plus{} y) \\equal{} t(n)$. \n\n\nBy the Lemma, for any $ N$, less than $ \\frac {N}{l}$ numbers $ n$ between $ 1$ and $ N$ let a $ t_i (n)$ have a prime factor that divides it at least $ q \\minus{} 1$ times. Hence, there is an $ n$ so that $ g(Mn \\plus{} y)$ has no prime factor dividing it more than $ r \\minus{} 1$ times. \n\nSince $ f(m)$ and $ g(m)$ are not off by a constant, there exists an $ f_m (n)$ with $ 1\\le m\\le k$ that divides $ f$ a different number of times than it divides $ g$. Let $ f_m (n)$ be $ h(n)$; say that $ h(n)^a \\parallel{} f(n)$ and $ h(n)^b \\parallel{} g(n)$ with $ a\\ne b$. Since there are infinitely many primes dividing $ h$, we can pick a prime $ p$ be larger than any of the $ k_i$ so that $ p$ divides $ h(n)$, but is not a factor of $ h'(n)$ for some $ n$. By Hensel's Lemma, for any large integer $ k$, we may pick an $ m$ so that let $ p^k \\parallel{} h(m)$, and it follows that $ p^{ak} \\parallel{} f(m)$ and $ p^{bk} \\parallel{} g(m)$. Pick $ k$ so that $ ak\\equiv \\minus{} 1\\bmod r$. This implies that $ bk\\not \\equiv \\minus{} 1 \\bmod r$ as long as $ r > a \\minus{} b$ (which is true since $ r > \\max \\{d_i\\}\\max \\{e_i\\})$. Note that since $ p^{ak} \\parallel{} f(n)$, $ r | (ak \\plus{} 1) | \\tau (f(m \\plus{} p^{r \\plus{} 1}\\cdot n))$ for all $ n$. Therefore, $ r | \\tau (g(m \\plus{} p^{r \\plus{} 1}\\cdot n))$ for all $ n$, so $ r | \\tau (t(n))$ for all $ n$. However, this means that there is a prime that divides $ t(n)$ at least $ r \\minus{} 1$ times for all $ n$, which is a contradiction, so $ f(n) \\equal{} cg(n)$ for some constant $ c$ and all integers $ n$.\n\n\nWithout loss of generality, let $ |c| \\ge 1$. Then, $ g(n) | f(n)$ for all $ n$. Hence, if $ |g(n)| < |f(n)|$ for some $ n$, we would get that $ \\tau (g(n)) < \\tau (f(n))$ (indeed, $ f(n)$ contains all factors of $ g(n)$ and $ f(n)$ itself), which is impossible. Hence, $ |c| \\equal{} 1\\implies c \\equal{} \\pm 1\\implies f \\equal{} \\pm g$ as long as $ f$ is nonconstant. \n\n\n[/hide]" } { "Tag": [ "algebra", "polynomial", "function", "calculus", "integration", "Functional Analysis", "real analysis" ], "Problem": "Hi all\r\n\r\nI have been working on this problem for a while now, but haven't made too much progress:\r\n\r\nConsider the Hilbert space L^2([0,1]) with usual inner product norm. We wish to apply the G-S process to the set {1,x,x^2,...} to get polynomials {P_0, P_1, P_2,...}. Determine a recurrence relation between P_n, P_(n+1) and P_(n+2) and then determine a general formula for P_n which can be expressed in terms of elementary functions and derivatives.\r\n\r\nNow I haven't even got past the first part. I calculated the first few P_n, and hypothesized a recurrence relation which seems to fit: P_(n+1)' / sqrt(2n+3) = ((n-1)*4 +2)*P_n / sqrt(2n+1) + P_(n-1)' / sqrt(2n-1).\r\n\r\nBut I have been unable to prove this, and I have no idea how to start on the 2nd part. I'm hoping that someone has seen this before and can give me an idea of how to proceed. Please do not give a full solution as I want to try this myself (Or at least hide your solution), but if you could give me an idea on how to prove the recurrence relation or how to find the \"closed\" formula for part 2 that would be greatly appreciated, as I am quite stuck!\r\n\r\nThanks very much.", "Solution_1": "That recursion looks right.\r\n\r\nThese are Legendre polynomials. With the nice formula, you can prove orthogonality using integration by parts. The formula will also help to prove the recursion.", "Solution_2": "Thanks very much. I knew I had seen these polys somewhere before." } { "Tag": [ "induction", "algebra unsolved", "algebra" ], "Problem": "A bounded sequence $ (x_n)_{n\\ge 1}$ of real numbers satisfies $ x_n \\plus{} x_{n \\plus{} 1} \\ge 2x_{n \\plus{} 2}$ for all $ n \\ge 1$. Prove that this sequence has a finite limit.", "Solution_1": "$ x_n \\plus{} x_{n \\plus{} 1} \\ge 2x_n \\plus{} 2\\to x_{n \\plus{} 1}\\ge x_n \\plus{} 2\\to x_n\\ge 2n \\plus{} x_0$ is not bounded.", "Solution_2": "I think the condition should be $ x_{n}\\plus{}x_{n\\plus{}1}>\\equal{}2x_{n\\plus{}2}$.", "Solution_3": "Thank you for pointed out my mistake. Yes, it must be $ x_n\\plus{}x_{n\\plus{}1}\\ge 2x_{n\\plus{}2}$. Thanks", "Solution_4": "We can prove by induction that : $ x_{n\\plus{}m}\\leq \\frac{x_m\\plus{}2x_{m\\plus{}1}}3\\plus{}\\frac{x_{m\\plus{}1}\\minus{}x_m}3 (\\minus{}\\frac 12)^{n\\minus{}1}$.The sequence $ x_n$ is bounded . Let $ I\\leq S$ be the [b]Inferior,Superior limits[/b] of $ x_n$\r\nTherefore there are increasing sequences $ p(n),q(n)$ of natural numbers such that $ x_{p(n)}\\rightarrow I, x_{q(n)}\\rightarrow S$.\r\nThe sequence $ x_{p(n)\\plus{}1}$ is bounded , so there is $ u(n)$ such that $ x_{p(u(n))\\plus{}1}$ is convergent to $ L \\leq S$\r\nWe can write $ x_{q(2p(u(n)))}\\equal{}x_{q(2p(u(n)))\\minus{}p(u(n)) \\plus{}p(u(n))}\\leq$$ \\frac{x_{p(u(n))}\\plus{}2x_{p(u(n))\\plus{}1}}3\\plus{}\\frac{x_{p(u(n))\\plus{}1}\\minus{}x_{p(u(n))}}3 (\\minus{}\\frac 12)^{q(2p(u(n)))\\minus{}p(n)\\minus{}1}$\r\nIt's easy to see that $ q(n) \\geq n$ , so $ q(2p(u(n)))\\geq 2p(u(n))$, so for $ n\\rightarrow \\plus{}\\infty$ we obtain $ S \\leq \\frac{I\\plus{}2L}3 \\leq \\frac{I\\plus{}2S}3\\Longrightarrow S\\leq I$\r\nFinally $ I\\equal{}S$, and the sequence $ x_n$ is convergent .\r\n :cool:" } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "I think it's nice problem please help me to solve it.", "Solution_1": "no one has an idea ore solution?", "Solution_2": "No one will open a doc file. Post it here." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "let x,y,z>0\r\n 1\\3<=yx+ y z +z x <=3\r\n\r\ndetermine the range of values for\r\nx y z\r\nand\r\nx+z+y", "Solution_1": "We know that:\r\n$x^{2}+y^{2}+z^{2}\\ge xy+yz+zx \\implies$\r\n$(x+y+z)^{2}\\ge 3(xy+yz+zx) \\ge 3\\cdot\\frac{1}{3}= 1$\r\nBut i can't get the maximum. If you let $x,y,z\\ge 0$ then this maximum doesn't exist. If we let $xy+yz+zx$ fixed. $x+y+z$ goes to a maximum when $x^{2}+y^{2}+z^{2}$is maximal. \r\n$1= \\left(\\frac{3}{3}\\right)^{\\frac{3}{2}}\\ge \\left(\\frac{xy+yz+zx}{3}\\right)^{\\frac{3}{2}}\\ge xyz$\r\nI can't find a minimum.", "Solution_2": "Assume the maximum value of $x+y+z$ is $k$, so $x+y+z \\leq k$ for every $x,y,z$ that satisfy the property from the problem.\r\n\r\nSet $x=k+z, y=z=\\frac{\\sqrt{k^{2}+9}-k}{3}$ \r\n$xy+yz+zx=3z^{2}+2kz=3$\r\n\r\nThen,$x+y+z=3z+k>k$. Contradiction. So $x+y+z$ has no maximum value.\r\n\r\n\r\nAssume the minimum value of $xyz$ is $k$, so $xyz \\geq k$ for every $x,y,z$ that satisfy the property from the problem.\r\n\r\nSet $x=\\frac{k}{3}, y=1, z=\\frac{3-xy}{x+y}=\\frac{4}{x+1}-1 < 4-1=3$. [$xy+yz+zx=3$]\r\n\r\nThen, $xyzx$ so it has no bound. \r\nAnd $xyz=x\\cdot\\frac{1}{x^{2}}=\\frac{1}{x}$ so for a very big $x$ we have that $xyz \\rightarrow 0$ and this has it has no bound." } { "Tag": [ "probability" ], "Problem": "In a standard 52 card deck, I draw two cards, withcout replacement. What is the probability that I draw atleast one spade and one king (the king of spades accounts for both)?", "Solution_1": "I think it is [hide]$1-(3/4*12/13)=4/13$[/hide]", "Solution_2": "[quote=\"Art of Owna\"]In a standard 52 card deck, I draw two cards. What is the probability that I draw atleast one spade and one king (the king of spades accounts for both)?[/quote]\r\n[hide]There are 13 spade cards. There are 4 king cards. Thats a total of 17. But we counted the king of spades twice, so subtract one. Then we have 16 to get. Then the probability is $\\frac{16}{52}\\cdot\\frac{15}{51}$. Simplifying $\\frac{4}{13}\\cdot\\frac{5}{17}$. The final answer is $\\frac{20}{221}$. [/hide]\r\nIs there any replacements?", "Solution_3": "[quote=\"ckck\"][quote=\"Art of Owna\"]In a standard 52 card deck, I draw two cards. What is the probability that I draw atleast one spade and one king (the king of spades accounts for both)?[/quote]\n[hide]There are 13 spade cards. There are 4 king cards. Thats a total of 17. But we counted the king of spades twice, so subtract one. Then we have 16 to get. Then the probability is $\\frac{16}{52}\\cdot\\frac{15}{51}$. Simplifying $\\frac{4}{13}\\cdot\\frac{5}{17}$. The final answer is $\\frac{20}{221}$. [/hide]\nIs there any replacements?[/quote]\r\n\r\nNo replacements. Ill change that. I actually dont know the answer, I was just wondering what some solutions were. \r\n\r\nEDIT: But wait, the question said atleast one king and atleast one spade, isnt your way only counting the ways without both a kind and a spade and subtracing that from one (beucase if I get a kind of hearts and a nine of hearts, that wouldnt count towards the required)", "Solution_4": "[hide]\nI will solve by finding the opposite probability\n\n16 of the cards are spades or kings\n36 are neither so the number of ways of drawing neither spades nor kings is\n\n$36*35$\n\nWays to draw spades, but no kings\n$12*11$\n\nWays to draw kings, but no spades\n$3*2$\n\nProbability of not getting at least 1 spade and 1 king:\n$\\frac{(36)(35)+(12)(11)+(3)(2)}{(52)(51)}$\n\nProbability of getting at least 1 spade and 1 king\n$1-\\frac{(36)(35)+(12)(11)+(3)(2)}{(52)(51)}$\n\ndon't feel like simplifying\n\nI probably made a mistake somewhere.\n[/hide]" } { "Tag": [ "percent" ], "Problem": "Express the positive difference between $ \\frac{3}{4}$ of $ \\frac{3}{5}$ and 20 percent of 30 percent as a decimal.", "Solution_1": "$ \\frac{3}{4}$ of $ \\frac{3}{5}$ is $ \\frac{9}{20}$.\r\n\r\nand 20 % of 30% is 6 %.\r\n\r\n$ \\frac{9}{20}$ is 0.45, and 6% is 0.06.\r\n\r\n0.45-0.06=0.39\r\n\r\nanswer : 0.39" } { "Tag": [ "LaTeX" ], "Problem": "\u0388\u03c3\u03c4\u03c9 $ a, b, c$ \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03af \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03af \u03ce\u03c3\u03c4\u03b5 $ a \\plus{} b \\plus{} c \\equal{} 3$\r\n\r\n\u039d\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03ba\u03ac\u03c4\u03c9 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1:\r\n\r\n$ \\frac{a(a \\plus{} 1)}{b \\plus{} 1} \\plus{} \\frac{b(b \\plus{} 1)}{c \\plus{} 1} \\plus{} \\frac{c(c \\plus{} 1)}{a \\plus{} 1}\\geq 3$\r\n\r\n\u0398\u03ad\u03bb\u03c9 \u03bd\u03b1 \u03b4\u03c9 \u03ad\u03be\u03c5\u03c0\u03bd\u03b5\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2... :wink:\r\n\r\n[color=red][Moderator edit: \u03a1\u03b5 \u03b1\u03b4\u03b5\u03c1\u03c6\u03ad \u03bc\u03ac\u03b8\u03b5 latex \u03ba\u03b1\u03bc\u03b9\u03ac \u03c6\u03bf\u03c1\u03ac :D ][/color]", "Solution_1": "\u039c\u03b9\u03b1 \u03c3\u03c7\u03b5\u03c4\u03b9\u03ba\u03ac \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b7\u03c4\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7 \r\n\r\n\u0391\u03c0\u03cc \u03c4\u03bf\u03bd \u03b1\u03b3\u03b1\u03c0\u03b7\u03bc\u03ad\u03bd\u03bf \u03bc\u03b1\u03c2 BCS \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \r\n$ \\sum\\frac{a^2}{b\\plus{}1}\\geq\\frac{(a\\plus{}b\\plus{}c)^2}{3\\plus{}a\\plus{}b\\plus{}c}\\equal{}\\frac{3}{2}$ \r\n\r\n\u03ba\u03b1\u03b9 $ \\sum\\frac{a}{b\\plus{}1}\\equal{}\\sum\\frac{a^2}{ab\\plus{}a}\\geq\\frac{(a\\plus{}b\\plus{}c)^2}{ab\\plus{}bc\\plus{}ca\\plus{}a\\plus{}b\\plus{}c}\\geq\\frac{3}{2}$ \r\n\r\n\u03a0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c4\u03bf\u03bd\u03c4\u03b1\u03c2 we get a big QED :)", "Solution_2": "Andreescu \u03b4\u03b7\u03bb\u03b1\u03b4\u03b7 :P\r\n\r\n\u0394\u03b5\u03bd \u03bc\u03bf\u03c5 \u03bb\u03b5\u03c2, \u03b5\u03b9\u03c3\u03b1\u03b9 \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03bf\u03c4\u03b9 $ a\\plus{}b\\plus{}c\\geq ab\\plus{}bc\\plus{}ca$ ?", "Solution_3": "\u03a9\u03c1\u03b1\u03af\u03bf\u03c2 \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad \u03b1\u03c5\u03c4\u03cc \u03b5\u03af\u03c7\u03b1 \u03c3\u03c4\u03bf \u03bc\u03c5\u03b1\u03bb\u03cc \u03bc\u03bf\u03c5. Mailo \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $ (a\\plus{}b\\plus{}c)^2>\\equal{}3(ab\\plus{}bc\\plus{}ca)$ :wink:", "Solution_4": "\u0392\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03b4\u03b9\u03ba\u03cc \u03bc\u03bf\u03c5 \u03b1\u03b3\u03b1\u03c0\u03b7\u03bc\u03ad\u03bd\u03bf \u03c4\u03c1\u03cc\u03c0\u03bf:\r\n\r\n\u03b1\u03c0\u03cc \u03b1\u03bd\u03b1\u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7 (\u03b8\u03ad\u03c4\u03bf\u03bd\u03c4\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03cc\u03c4\u03b9 $ 0 \\le a \\le b \\le c$) \u03b5\u03af\u03bd\u03b1\u03b9\r\n\r\n\r\n$ \\frac {a(a \\plus{} 1)}{b \\plus{} 1} \\plus{} \\frac {b(b \\plus{} 1)}{c \\plus{} 1} \\plus{} \\frac {c(c \\plus{} 1)}{a \\plus{} 1} \\geq$\r\n\r\n$ \\frac {a(a \\plus{} 1)}{a \\plus{} 1} \\plus{} \\frac {b(b \\plus{} 1)}{b \\plus{} 1} \\plus{} \\frac {c(c \\plus{} 1)}{c \\plus{} 1} \\equal{}$\r\n\r\n$ a \\plus{} b \\plus{} c \\equal{} 3$ q.e.d. :P", "Solution_5": "maniopa den vgainei me anisotita tis anadiataksis kathws oi triades pou exeis an ypotheseis oti $ a<\\equal{}b<\\equal{}c$ den einai omoia diatetagmenes.\r\ndiladi exeis men $ a(a\\plus{}1)<\\equal{} b(b\\plus{}1) <\\equal{}c(c\\plus{}1)$ \r\nalla gia tin alli triada exeis \r\n$ 1/(c\\plus{}1) <\\equal{} 1/(b\\plus{}1) <\\equal{} 1/(a\\plus{}1)$ \r\nkai oxi \r\n$ 1/(b\\plus{}1) <\\equal{}1/(c\\plus{}1)<\\equal{} 1/(a\\plus{}1)$ pou xrisimopoieis.", "Solution_6": "[quote=\"Athinaios\"]maniopa den vgainei me anisotita tis anadiataksis kathws oi triades pou exeis an ypotheseis oti $ a < \\equal{} b < \\equal{} c$ den einai omoia diatetagmenes.\ndiladi exeis men $ a(a \\plus{} 1) < \\equal{} b(b \\plus{} 1) < \\equal{} c(c \\plus{} 1)$ \nalla gia tin alli triada exeis \n$ 1/(c \\plus{} 1) < \\equal{} 1/(b \\plus{} 1) < \\equal{} 1/(a \\plus{} 1)$ \nkai oxi \n$ 1/(b \\plus{} 1) < \\equal{} 1/(c \\plus{} 1) < \\equal{} 1/(a \\plus{} 1)$ pou xrisimopoieis.[/quote]\r\n\r\n\r\n\u03a7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce \u03bc\u03b9\u03b1 \u03cc\u03c7\u03b9 \u03c4\u03cc\u03c3\u03bf \u03b3\u03bd\u03c9\u03c3\u03c4\u03ae \u03bc\u03bf\u03c1\u03c6\u03ae \u03c4\u03b7\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2:\r\n\r\n\u03b1\u03bd \u03bf\u03b9 n-\u03ac\u03b4\u03b5\u03c2 $ a,b$ \u03ad\u03c7\u03bf\u03c5\u03bd \u03b1\u03bd\u03c4\u03af\u03b8\u03b5\u03c4\u03b7 \u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7 \u03ba\u03b1\u03b9 $ c$ \u03b1\u03bd\u03b1\u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7 \u03c4\u03bf\u03c5 b \u03b5\u03af\u03bd\u03b1\u03b9 $ a_1b_1 \\plus{} a_2b_2... \\le a_1c_1 \\plus{} a_2c_2...$\r\n\r\n\r\n\u03c4\u03ce\u03c1\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 $ \\frac {1}{a \\plus{} 1} \\ge \\frac {1}{b \\plus{} 1} \\ge \\frac {1}{c \\plus{} 1}$\r\n\r\n\u03ba\u03b1\u03b9 $ a(a \\plus{} 1) \\le b(b \\plus{} 1) \\le c(c \\plus{} 1)$ \u03ac\u03c1\u03b1 \u03b5\u03c6\u03b1\u03c1\u03bc\u03cc\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03b1\u03bd\u03b1\u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7 \u03cc\u03c0\u03c9\u03c2 \u03b1\u03bd\u03ad\u03c6\u03b5\u03c1\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9:\r\n\r\n$ \\frac {a(a \\plus{} 1)}{a \\plus{} 1} \\plus{} \\frac {b(b \\plus{} 1)}{b \\plus{} 1} \\plus{} \\frac {c(c \\plus{} 1)}{c \\plus{} 1} \\le \\frac {a(a \\plus{} 1)}{b \\plus{} 1} \\plus{} \\frac {b(b \\plus{} 1)}{c \\plus{} 1} \\plus{} \\frac {c(c \\plus{} 1)}{a \\plus{} 1}$", "Solution_7": "\u03a3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c0\u03bd\u03b5\u03cd\u03bc\u03b1 \u03ba\u03b9\u03bd\u03b5\u03af\u03c4\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03b7 \u03c0\u03b1\u03c1\u03b1\u03ba\u03ac\u03c4\u03c9 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7:\r\n\r\n\u03b1\u03bd $ a,b,c,d > 0$ \u03bd\u03b1 \u03b4\u03b5\u03b9\u03c7\u03c4\u03b5\u03af \u03cc\u03c4\u03b9\r\n\r\n$ (\\frac {(a \\plus{} b)(c \\plus{} d)}{ab \\plus{} cd})^2 \\plus{} (\\frac {(b \\plus{} c)(d \\plus{} a)}{bc \\plus{} ad})^2 \\plus{} (\\frac {(a \\plus{} c)(b \\plus{} d)}{ac \\plus{} bd})^2 \\ge 4$\r\n\r\n[color=red]Edit: \u03ad\u03ba\u03b1\u03bd\u03b1 \u03ad\u03bd\u03b1 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03c0\u03c1\u03b9\u03bd.[/color]" } { "Tag": [ "logarithms", "LaTeX", "algebra unsolved", "algebra" ], "Problem": "Please Help :blush: \r\n\r\nevaluate and prove log(2)^log(2)^log(2).......\r\n\r\nthanks :)", "Solution_1": "$ \\left( \\left( \\left( \\left( \\log 2 \\right)^\\log 2 \\right)^\\log 2 \\right)^\\log 2 \\right)^\\log 2^{\\cdot^{\\cdot^\\cdot}}$ ?\r\nor\r\n$ \\log2^\\left( \\log2^\\left( \\log2^\\left( \\log2^\\left( \\log 2 \\right) \\right) \\right) \\right)^{\\cdot^{\\cdot^\\cdot}}$ ?", "Solution_2": "it's the second one. by the way, how did you type that equation", "Solution_3": "[quote=\"seraphim\"]how did you type that equation[/quote]\r\nThis site can use $ \\text{\\LaTeX}$ :)\r\n\r\n$ \\log2^\\left( \\log2^\\left( \\log2^\\left( \\log2^\\left( \\log 2 \\right) \\right) \\right) \\right)^{\\cdot^{\\cdot^\\cdot}}$\r\n$ \\Rightarrow x = (\\log 2)^x$", "Solution_4": "i got the x- log(2)^x = 0 part, just that how would you evaluate the equation algeraically and prove the solution other than by using a calculator?" } { "Tag": [ "modular arithmetic", "number theory" ], "Problem": "What is the remainder when $5^{999,999}$ is divided by 7?", "Solution_1": "Use Fermat.", "Solution_2": "[hide]These kinds of problems often come up on mathcounts.\n\nWe want to find $5^{999,999} (\\text{ mod }7)$\n\nSo, if we look at the cycles of $5^x$ (mod 7),\n\n5 *5=25=\n4 *5=20=\n6 *5=30=\n2 *5=10=\n3 *5=15=\n1 *5=5=\n5\n\nSo there is a cycle of 6 remainders, 546231.\n\nso: $5^{999,999}$ is the same as $5^{999,999\\text{ (mod 6)}}$\n\nand $999,999 \\text{ (mod 6)} = 3$\n\nso: the remainder is the same as that of $5^3$ which is $\\boxed{6}$\n\n[/hide]\r\n\r\nEDIT: solution hidden", "Solution_3": "I did it that way too and found no pattern. must've messed up in my calculations :? \r\nthis is #29 on the '92-'93 National sprint round.\r\n\r\nWhat's Fermat?", "Solution_4": "i think he means fermat's little theroem\r\n\r\nlook it up on mathworld or...\r\n\r\n$A^{p-1}=1\\bmod p$ where $p$ is a prime and $A$ is a number not divisible by $p$", "Solution_5": "I see, how's that supposed to help?", "Solution_6": "that helps but it doesn't solve the problem on its own. The only thing that that can tell us (as far as i can tell) is that there are 6 different remainders possible.", "Solution_7": "[hide]We need to find $5^{999,999} \\pmod {7}$.\n\nFrom Fermat's little theorem, we have $5^{6} \\equiv 1 \\pmod {7}$.\n\n$999999 \\equiv 3 \\pmod {6}$, so $5^3 \\equiv 6 \\pmod {7}$[/hide]", "Solution_8": "That's an interesting theorem.", "Solution_9": "[hide]\n5^1 (mod 7)=5\n5^2 (mod 7)=4\n5^3 (mod 7)=6\n5^4 (mod 7)=2\n5^5 (mod 7)=3\n5^6 (mod 7)=1\n--------------------\nThis is the repeating pattern.\n\n999999=3 (mod 6),\nthus the remainder when 5^999999 is divided by 7 is: 6\n[/hide]", "Solution_10": "I still think my way is the best. :D", "Solution_11": "Okay... watch this awesomeness.\r\n\r\n$5^{999,999}\\equiv (-2)^{999,999}\\bmod{7}$\r\n$(-2)^{999,999}=(-8)^{333,333}\\equiv (-1)^{333,333}\\bmod{7} \\Rightarrow (-1)^{333,333}=-1$, and we have $-1\\equiv \\boxed{6}\\bmod{7}$. :lol:", "Solution_12": "That's nice too...\r\n[size=67]but I still like my way better. :D [/size]", "Solution_13": "I realized after posting it that it's basically the same thing you did. $5^3\\equiv 6\\bmod {7}$, so we have $5^{999,999}\\equiv 6^{333,333}\\bmod{7}$, and go from there. ;)", "Solution_14": "[quote=\"chess64\"]That's nice too...\n[size=67]but I still like my way better. :D [/size][/quote]Let's vote. I think JesusFreak's is little nicer, sorry. :P After all you just applied Fermat in the normal fashion.", "Solution_15": "guys...[hide=\"stop spamming\"]STOP SPAMMING!!![/hide]\r\n\r\nplease dont get off topic or make useless posts like \"interesting problem\"" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "geometry", "USAMTS", "Support" ], "Problem": "I'm sure a lot of people will disagree with me for saying this, or for being so blunt, but I've never been more embarrassed about the USA's top mathematics students.\r\n\r\nFirst, it was \"Boo hoo. I should get to take the AIME twice, because I had a bad day.\"\r\n\r\nNow, it's \"Boo hoo. I shouldn't have to take the AMC or the AIME. I'm too good for them. I already qualified for MOSP, once.\"\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138675]Fifteen forum pages of whining[/url], and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=139901]growing[/url]! Good grief!\r\n\r\nIt's not all MOSPers, but to those who have made such claims, you are an embarrassment.\r\n\r\nYou are given a free summer math program to learn advanced mathematics. (MOSP) Then you are given a free seven-month online program to learn more advanced mathematics. (WOOT) Then you think you are so much better than everyone else that you can sit on your lazy derrieres.\r\n\r\nExcuse me. But if you can't qualify for the AIME, you are NOT a well-rounded mathematician, and you don't deserve to be considered for one of the mathematicians that will receive free training or one of the six student-mathematicians that will represent the United States of America at an International Mathematical Olympiad.\r\n\r\nExcuse me again. But if you can't qualify for the USAMO, you are NOT a well-rounded mathematician, and you don't deserve to be considered for one of the mathematicians that will receive free training or one of the six mathematicians that will represent the USA at an IMO.\r\n\r\nDon't get me wrong. I understand the AIME and the USAMO are different kinds of tests. I happen to think you should be able to do well on *both* in order to qualify for MOSP and to represent the USA at the IMO. I also have been supportive of the USAMO expansion. However, the main effect of the expansion seems to have been to increase the number of whiners on the bubble. To those who are whining. Stop whining. You're an embarrassment.\r\n\r\nTo those who have gone to MOSP and are whining now, stop whining. If you can't qualify for USAMO, even with all the advantages you've been given, step aside gracefully, and let the more talented and well-rounded students have a crack at it.\r\n\r\nTo all those involved with writing and administering and scoring the AMC contests, I say \"Thank you!\" I am a coach who has had a few students qualify for the AIME, and one student qualify for MOSP. I have known, and perhaps assisted in small ways, others who have qualified for AIME, USAMO, and MOSP. Thank you for spending your time and your energy and your money to produce the AMC series of tests and beyond. You don't deserve the unprecedented negative karma that's been directed toward your series of contests by some of the very students you seek to serve.", "Solution_1": "[quote=\"rcv\"]\nExcuse me. But if you can't qualify for the AIME, you are NOT a well-rounded mathematician, and you don't deserve to be considered for one of the mathematicians that will receive free training or one of the six student-mathematicians that will represent the United States of America at an International Mathematical Olympiad.[/quote]\r\n\r\nWhat about if you're a very deep thinker, but not so fast?", "Solution_2": "Is ANY of this based on actual reason/evidence, or is everything just an opinion, which you base on to insult a bunch of random people?", "Solution_3": "Cheers.\r\n\r\nYes, there is actual reason/evidence. This is a problem that I've pointed out to several of my friends and we all had a good laugh about it. There is WAY TOO MUCH whining on this forum nowadays.\r\n\r\nI'll quote two people (not whiners) who described the situation the best, though it may be strange taking it out of context. Also, I believe the evidence takes the form of those 15 pages in that link.\r\n\r\n[quote=\"white_horse_king88\"]Aka: Compile a master list of all the ways people \"ought\" to be allowed to qualify for the USAMO.\n\nI agree with the USAMTs thing.\nAgreed with the Blue/Black Mop thing.\nAlso Red Mop. That shouldn't be a problem.\nAlso those that placed individually on the HMMT. Those problems are almost like AIME problems.\nAlso people who think the AIME screwed them over.\nAnd like ARML top ten and Mu Alpha Theta and stuff. \nThere's probably more I missed.[/quote]\n\nIt case you missed it, that was heavy sarcasm. You know it's bad when Mu Alpha Theta is called as a selection factor for the USAMO :lol: \n\n[quote=\"Ignite168\"]\nI dislike the fact that when the AMC writers set an obstacle, even if it's plain annoying computation, that people try to change that obstacle, rather than working to overcome it by simple methods like checking your work extremely closely and such. It seems to go against the spirit of \"Art of Problem Solving.\"[/quote]\r\n\r\nThus, if you're a deep thinker, but you're not fast, get fast. You don't need to be that fast for the AMC/AIME anyway. You just need to not be slow.\r\n\r\n-end of quotes-\r\n\r\n\r\nIf you want to take the USAMO or go to MOSP, you should have to take the AMC/AIME because it is only fair for everyone. Also, the system has shown itself to work, with very few outliers, year after year. Like I mentioned earlier in another thread, the only major problems we've seen have only began occurring after the expansion, which was a noble effort but too many of you have taken it for granted quickly and are now expecting more.", "Solution_4": "[quote]Fifteen forum pages of whining, and growing! Good grief! [/quote]\r\nWell half of it is people refuting the whining but yeah ok.", "Solution_5": "[quote=\"rcv\"]Don't get me wrong. I understand the AIME and the USAMO are different kinds of tests. I happen to think you should be able to do well on *both* in order to qualify for MOSP and to represent the USA at the IMO. [/quote]\r\n\r\nSince you seem to be attacking me personally in claiming that I wish to sit upon my \"lazy derriere\" (interesting accusation, considering my work ethic), I will take the time to respond once by pointing out the obvious flaw in your logic.\r\n\r\nOlympiad mathematics is tested by the USAMO, not the AIME. The AIME is simply easier to grade. Your second statement above is wholly unjustified.", "Solution_6": "Eh you don't need to be a \"well-rounded\" mathematician to do well on IMO, since being able to solve problems very quickly and make no computation error is not tested on IMO.\r\n\r\nAlso, no MoPpers whined about the system because they didn't qualify for USAMO this year; in fact, most people who like this idea are non-MoPpers.", "Solution_7": "Dude, I agree with the \"what if you're a deep thinker, but not fast\" point, and I'd also like to add \"what if you're bad at really computation heavy math, but can do the really deep stuff,\" because both describe me.\r\n\r\nI'm not saying that I wouldv'e made MOSP or IMO this year, but I'd feel fairly confident in saying that if I had the opportunity to take USAMO this year, I'd at least come close (unless all 6 problems were geometry or something, which wouldn't happen anyways).\r\n\r\nThe point is not that they wanna be lazy, but that they wanna spend their time doing real math, and practicing for silly AMC and AIME stuff, which matters more whether you're good at taking them than whether you can do deep math.\r\n\r\nAlso, notice that they're not acting \"better than everyone\" by asking for a free pass to MOSP/IMO again. Rather, they want to be guaranteed to have the chance to COMPETE FOR THEIR SPOT AGAIN, among all of the others trying out.", "Solution_8": "This feels like more of a rant, opinion, or personal thought post than that 15 page thread that you mentioned.\r\n\r\nAs Arnav said earlier, you seem to be directly attacking certain individuals. Might I mention, these individuals are discussing ways of possibly changing the AMC structure. This discussion is quite furtile, and I think it is much more substantial and meaningful than your personal embarrassment.", "Solution_9": "I personally think all these problems would have been avoided had the problems not sucked.", "Solution_10": "[quote=\"PenguinIntegral\"]I personally think all these problems would have been avoided had the problems not sucked.[/quote]\r\n\r\nAmen to that, my brother.", "Solution_11": "[quote=\"PenguinIntegral\"]I personally think all these problems would have been avoided had the problems not sucked.[/quote]\r\n\r\nIs it just me, or is saying you did badly because of bad problems just more complaining? The problems were written to test if you could do them. Therefore they cannot possibly be bad, if you got it right, you can do them, if you didn't, you can't, and that's what the test is for in the first place.", "Solution_12": "This is probably due to the fact that I didn't fail the AIME as badly a most people, but I'm getting very annoyed with the complaining as well. Thank you rcv for bringing this up!", "Solution_13": "Yeah, this completely ignores the fact that the person who brought up the possibility of automatically allowing MOPpers back into the USAMO in the first place (me) is not a MOPper and will almost certainly never be a MOPper. This is not about MOPpers being better than everyone else. As I mentioned before, if you don't like saying that they automatically \"qualify\" for the USAMO that one year, you can always have a separate list of MOPpers who didn't \"qualify\" but are \"invited\" to take the USAMO. Such a rule just very slightly increases the reward for people who score extremely high on the USAMO, and I think it's a justified increase. It's not like it would be \"once in, always in\". A MOPper would only be automatically allowed into the USAMO for one year--if they want to get in automatically again, they have to make MOP again. It would not hurt non-MOP-level contestants at all--we have as many USAMO spots to compete for as ever. The AMC people would only need to add maybe 5 spots at the most to deal with the very, very few people who have proven that they are great at the USAMO and weak on the AIME.\r\n\r\nAlso, I want to clarify that I, for one, do not blame the test, the writers, or the organizers for anything. I'm very, very thankful that they make the effort to give the AMC tests every year, and I don't think the issues this particular AIME raised could have been predicted in advance or prevented. But I also don't think that should stop me or anyone else from suggesting possible rule changes if we really feel that they would make the tests more fair.", "Solution_14": "Thank you thank you thank you rcv for taking a stand on this! But guys, as you seem to insist on complaining wherever you get a chance, please note what rcv pointed out at the top of this thread. There is already a 15-page haven of whining, so do it there.", "Solution_15": "I agree with the topic of this thread, but not rcv's reason for choosing that topic. It is an embarassment. I am not saying that the supposed \"whining\" is an embarassment, but that we have a SIX-PAGE THREAD THAT CONSISTS ALMOST ENTIRELY OF TWO FACTIONS SLINGING MUD AT EACH OTHER. Art of Problem Solving should be a haven of young mathematicians, not of a mudslinging contest to rival recent elections. Maybe this AIME was worse than normal, maybe not. My personal inclination is toward the former, but that is not the point here. The point is that we should be able to hold reasonable discussions, with out dividing into two factions bashing each other over six pages. In less than 48 hours, I might add. Please, everyone, grow up. I expect better of America's best future mathematicians.", "Solution_16": "All right, we've gone through the mudslinging. This is the [i]American[/i] Invitational Math Exam, so we should abide by the democratic principles of the United States. [Please don't argue that the US isn't a democracy :P.] Let's hold an election to determine whether we should ask for a change or not!\r\n\r\n[That's only half in jest; we need some solution that everyone will abide by on here.]", "Solution_17": "I vote no.", "Solution_18": "It seems to me that the central augument of the complaining is that AIME is for selecting USAMO, and since USAMO focuses on proof type problems, therefore any significant computation in AIME would make a bad AIME. However, that does not seem to be what the AMC says. Here is a copy of what AMC says on their website:\r\n\r\nThe American Mathematics Competitions (AMC) is dedicated to the goal of strengthening the mathematical capabilities of our nation\u2019s youth. We believe that one way to meet this goal is to identify, recognize and reward excellence in mathematics through a serics of national contests called the:\r\n\r\n * American Mathematics Contest 8 (AMC 8),\r\n * American Mathematics Contest 10 (AMC 10),\r\n * American Mathematics Contest 12 (AMC 12),\r\n * American Invitational Mathematics Examination (AIME), and\r\n * United States of America Mathematical Olympiad (USAMO).\r\n\r\nMy understanding of this is that either AIME or USAMO is one in a series of exams that serve the same purpose: strengthing the mathematical capabilities of our nations's youth. It goes without saying that certain amount of computational skill is very important. So either intentionally or unintentionally, the relatively more significant computational skill this AIME requires might be just the right kind of wakeup call. I've been a scientist and an engineer, and my own experience told me that the ability of dealing with numbers in various ways without much error is very very important even in our computer-dominated era, unless probably you'll be a pure theretical mathematician. How many of you will be pure theretical mathematician? I don't think there'll be many.\r\n\r\nI hope I understand it correctly.", "Solution_19": "I think that merely acting considerately suffices. We are entitled to express our opinions politely.\r\n\r\nI also note that few of last year's MOPpers commented on the offending fifteen-page thread, fewer of whom were complaining.\r\n\r\n\r\nI also think that it is difficult for those not accustomed to olympiad problems to realize the relation between olympiad and AIME problems, but I surmise that most such people have been to MOP. However, I have been to MOP, so I might be biased.", "Solution_20": "[quote]There's no good response to: you are a conservative bigot who does not favor change.[/quote]\nWhat...? I didn't say that. I thought your argument sounded like you thought the tweaks people are suggesting are bad partly because they're tweaks. If that's not what you meant, I apologize...but I didn't call you conservative, a bigot, or anything remotely similar to either of the two.\n[quote]Alright... we have lots of people here to like change. That's cool. And by the way, slippery slope is when the outcome does not match the trend. In this case, the fact that the trend in opinion shows complete favoring of expansion of qualification criteria and reduction of competitiveness for the... uh... \"ideologically correct purpose of mathematics\"... is definitely a trend. Not slippery slope.[/quote]\r\nOkay. Do you dislike the trend? Personally, I don't. Nobody's going to say, \"Oh, it's not impressive that you made the USAMO anymore because 500ish other people in the entire nation did too!\"\r\n\r\nBy the way, I loved #10. My reaction was basically, \"I SOLVED A PROBLEM THAT THE 2-TIME BLUE MOPPER AT MY SCHOOL DIDN'T!\" I wish the AIME tested nothing but combinatorics :lol:", "Solution_21": "[quote=\"solafidefarms\"]All right, we've gone through the mudslinging. This is the [i]American[/i] Invitational Math Exam, so we should abide by the democratic principles of the United States. [Please don't argue that the US isn't a democracy :P.] Let's hold an election to determine whether we should ask for a change or not!\n\n[That's only half in jest; we need some solution that everyone will abide by on here.][/quote]\r\nWhat's a democracy if <1% of the people will know about the vote?\r\n\r\nI'm sorry to burst everyone's bubbles but AoPS does not represent the American Math World.", "Solution_22": "I think the fact that this thread is now six pages long only adds to the embarrassment that rcv mentioned in the very first post. And now I'm contributing to it!\r\n\r\nYou know what would be REALLY funny... if the AMC people wrote this AIME specifically to weed out people who (1) don't read carefully and (2) can't compute quickly and accurately. Maybe that's what they wanted; it's their competition, so they get whatever they want. Sure you can say it wasn't good, but maybe they think it's a good test, so how can you argue with that?\r\n\r\nPersonally, I did reasonably well on the AIME. So maybe I'm just saying this because of that. But really, this AIME was not that bad of a test. A fair number of the questions had merit, and a couple of them didn't. It happens.", "Solution_23": "um. . . to penguin. . . \"counting\" is not the same as \"computation\". . . i thought the most confusing part of question 10 was finding a way to count every solution without overcounting anything, but that's not computation. after the test, my father (university math professor) went through the test, did all the problems, and he thought the test as a whole was [b]fine[/b].", "Solution_24": "Eh, I thought the test was actually pretty good. Perhaps it wasn't to your taste, but I'm sure that for the most part, it was about the same as the AIMEs in the past. Since it made it past the rigorous editing, I'm sure we can safely assume that the test was well-written.\r\n\r\nThis isn't Mu Alpha Theta guys, there aren't so many problems and these tests are as close as you will ever get to \"good\" math tests. Be thankful that you have it and don't take it for granted.", "Solution_25": "You guys are acting like this is the first AIME I've ever taken. I've taken the AIME before, and AIME 2007I was not like [i]any[/i] of the previous exams, all the way back to 1983.", "Solution_26": "Past tests aren't always the same as the current year's tests.\r\n\r\nIt will be okay guys.\r\n\r\nJust look at the difference from HMMT '05 to HMMT '06", "Solution_27": "[quote=\"chess64\"]You guys are acting like this is the first AIME I've ever taken. I've taken the AIME before, and AIME 2007I was not like [i]any[/i] of the previous exams, all the way back to 1983.[/quote]\r\n\r\nWell, what do you expect? They're not going to throw the same problems at you again (actually, #11 is nearly identical to AIME 1995 # 13).", "Solution_28": "[quote=\"Phelpedo\"]Well, what do you expect? They're not going to throw the same problems at you again (actually, #11 is nearly identical to AIME 1995 # 13).[/quote]\r\n\r\nWow... #10 was also extremely similar to an old AIME question (except it was $4 \\times 4$ before). I have to say, that's kind of lame.", "Solution_29": "mysticterminator, what a baller\r\n\r\nhe sums it up so well:\r\n\r\n[quote]and it's really really hot in my room So take off all your clothes I am gettin so hot,(uh,uh,uh,OO) I wanna take my clothes off (oo)[/quote]\n\n[/quote]" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "$ \\frac {1}{x \\minus{} y} \\plus{} \\frac {1}{y \\minus{} z} \\plus{} \\frac {1}{z \\minus{} x} \\equal{} \\frac {3}{2}$ is true for some real numbers $ x,y,z$. What value may expression $ \\frac {1}{(x \\minus{} y)^2} \\plus{} \\frac {1}{(y \\minus{} z)^2} \\plus{} \\frac {1}{(z \\minus{} x)^2}$ have?", "Solution_1": "$ \\frac{1}{x\\minus{}y}\\plus{}\\frac{1}{y\\minus{}z}\\plus{}\\frac{1}{z\\minus{}x}\\equal{}A$\r\n$ \\frac{1}{{x\\minus{}y}^2}\\plus{}\\frac{1}{{y\\minus{}z}^2}\\plus{}\\frac{1}{{z\\minus{}x}^2}\\equal{}B$\r\n$ x\\minus{}y\\equal{}a, y\\minus{}z\\equal{}b, z\\minus{}x\\equal{}c$\r\n$ A^2\\equal{}B\\plus{}2(\\frac{1}{ab}\\plus{}\\frac{1}{ac}\\plus{}\\frac{1}{bc})\\equal{}B\\plus{}\\frac{2}{abc}(a\\plus{}b\\plus{}c)\\equal{}B$ because $ a\\plus{}b\\plus{}c\\equal{}0$.\r\nSo $ B\\equal{}\\frac{9}{4}$" } { "Tag": [ "geometry", "algebra", "convex polygon", "perimeter", "geometric inequality", "IMO", "IMO 1984" ], "Problem": "Let $ d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $ n$ vertices (where $ n>3$). Let $ p$ be its perimeter. Prove that:\r\n\\[ n\\minus{}3<{2d\\over p}<\\Bigl[{n\\over2}\\Bigr]\\cdot\\Bigl[{n\\plus{}1\\over 2}\\Bigr]\\minus{}2,\\]\r\nwhere $ [x]$ denotes the greatest integer not exceeding $ x$.", "Solution_1": "Consider all the pairs of non-adjacent sides in the polygon. Each pair uniquely determines two diagonals which intersect in an interior point. The sum of these two diagonals is greater than the sum of the two sides (triangle inequality). Each side will appear in a pair $ n\\minus{}3$ times. And each time the side appears it also includes two diagonals. Thus, $ (n\\minus{}3)p < 2d$.\r\n\r\nNote that for any diagonal, its length is less than the sum of the sides on either side of it (triangle inequality). Consider all the diagonals from a fixed vertex and using the least number of sides in each application of the triangle inequality. We can sum everything up over all the diagonals considering the two cases (even or odd) to get the result." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "A quite difficult inequality", "Solution_1": "[quote=\"duc 95\"]A quite difficult inequality[/quote]\r\n\r\n\r\nsorry", "Solution_2": "What do you mean?", "Solution_3": "I cannot solve it now . can you help me?", "Solution_4": "[quote=\"duc 95\"]I cannot solve it now . can you help me?[/quote]\r\n\r\n\r\ndifficult inequality" } { "Tag": [ "induction", "LaTeX", "number theory proposed", "number theory" ], "Problem": "Let $ p > 3$ be positive integer number. $ x_1,x_2$ be roots of an equation $ x^2 \\minus{} px \\plus{} 1 \\equal{} 0$.Denote $ a_n \\equal{} x_1^n \\plus{} x_2^n$\r\n\r\nProve that $ (p \\minus{} 1)|a_n$ is not true", "Solution_1": "\\[ x_1^2 \\plus{} x_2^2 \\equal{} (x_1 \\plus{} x_2)^2 \\minus{} 2x_1x_2 \\equal{} p^2 \\minus{} 2\\]\r\n\r\n\\[ \\implies p \\minus{} 1\\nmid p^2 \\minus{} 2\\]\r\nwhich is true for ($ p\\in N$)", "Solution_2": "u mean $ a_n \\equal{} x_1^n \\plus{} x_2^n$,no?\r\nwe have $ a_n \\equal{} a_{n \\minus{} 1}.(x_1 \\plus{} x_2) \\minus{} x_1x_2(a_{n \\minus{} 2}) \\implies a_n \\equal{} a_{n \\minus{} 1}(p) \\minus{} a_{n \\minus{} 2},a_1 \\equal{} p,a_0 \\equal{} 2$\r\nwe prove it by induction,\r\nlet's suppose that \r\n$ p \\minus{} 1 | a_n \\implies p \\minus{} 1| a_{n \\minus{} 1} \\minus{} a_{n \\minus{} 2} \\equal{} (p \\minus{} 1)a_{n \\minus{} 2} \\minus{} a_{n \\minus{} 3} \\implies p \\minus{} 1 | a_{n \\minus{} 3}$ \r\nbut we know that $ p \\minus{} 1 \\nmid a_{n \\minus{} 3}$.\r\nso we have a contradiction.\r\n(note that at first we must prove $ p \\minus{} 1 \\nmid a_1,a_2,a_3$. which is easy)", "Solution_3": "nice solution shoki. i guess thats what victory.US meant. else its trivial.", "Solution_4": "[quote=\"shoki\"]u mean $ a_n \\equal{} x_1^n \\plus{} x_2^n$,no?\nwe have $ a_n \\equal{} a_{n \\minus{} 1}.(x_1 \\plus{} x_2) \\minus{} x_1x_2(a_{n \\minus{} 2}) \\implies a_n \\equal{} a_{n \\minus{} 1}(p) \\minus{} a_{n \\minus{} 2},a_1 \\equal{} p,a_0 \\equal{} 2$\nwe prove it by induction,\nlet's suppose that \n$ p \\minus{} 1 | a_n \\implies p \\minus{} 1| a_{n \\minus{} 1} \\minus{} a_{n \\minus{} 2} \\equal{} (p \\minus{} 1)a_{n \\minus{} 2} \\minus{} a_{n \\minus{} 3} \\implies p \\minus{} 1 | a_{n \\minus{} 3}$ \nbut we know that $ p \\minus{} 1 \\nmid a_{n \\minus{} 3}$.\nso we have a contradiction.\n(note that at first we must prove $ p \\minus{} 1 \\nmid a_1,a_2,a_3$. which is easy)[/quote]\r\n\r\nyes, nice solution :) :) , it seem like my way. [hide]i had a mistake in latex and i edited :oops:[/hide]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "From numbers $ 1,2,3,...,37$ we randomly choose 10 numbers. Prove that among these exist four distinct numbers, such that sum of two of them equals to the sum of other two.", "Solution_1": "There are $ \\binom{10}{2}\\equal{}45$ pairs of elements from the drawn set (let's call it $ S$).\r\nSince there are 36 possible non-negative differences of numbers between 1 and 37, there are 9 coincidences of the form $ a\\minus{}b\\equal{}c\\minus{}d$ with $ a,b,c,d\\in S$.\r\nIf all four numbers in such a coincidence are different, then (adding $ b$,$ d$ to both sides) we are done with the proof already.\r\nSo let us assume that all of these coincidences are of the form $ a\\minus{}b\\equal{}c\\minus{}a$ for some $ a,b,c\\in S$\r\nLet $ M$ be the maximum and $ m$ be the minimum numbers from $ S$.\r\nNotice that $ c\\minus{}M$ and $ m\\minus{}b$ cannot yield positive differences when $ c,b\\in S$, hence numbers $ a$ in $ a\\minus{}b\\equal{}c\\minus{}d$ must be all from the remaining 8 elements of $ S$.\r\nSince there are 9 coincidences, for some two of them we have the same $ a$, i.e. $ a\\minus{}b_1\\equal{}c_1\\minus{}a$ and $ a\\minus{}b_2\\equal{}c_2\\minus{}a$.\r\nSubstracting both equalities we have $ b_2\\plus{}c_2\\equal{}c_1\\plus{}b_1$. It is east to notice that $ b_1,b_2,c_1,c_2$ must all be different.\r\n\r\nbest regards,\r\nWojtek" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let a trianlge $ABC$ , the bisector $AD$ of its angle $A$ , the midpoint $M$ of side $BC$ and $E$ the intersection point of the circumcircles of triangles $ABC$ and $ADM$. Prove that $EA \\perp AD$.\r\n\r\n [u] Babis [/u]\r\n\r\n [i]I think it deserves to have all solutions of this classical problem.I have seen this problem as a subquestion in a Mandelbrot Competition.[/i]", "Solution_1": "This is an easy, but cute problem. Call G the intersection of $AD$ with the circumcircle of $\\Delta ABC$. Call $E'$ the intersection point of $GM$ with the same circumcircle.\r\n\r\n[u][b]Solution 1:[/b][/u] Now, it's obvious that $\\angle ACG = \\angle ADB$, so $AE'MD$ is a cyclic quadrangle. So it follows that $E'=E$, and $EA \\perp AD$\r\n\r\n[u][b]Solution 2:[/b][/u] $\\Delta DMG$ and $\\Delta E'AG$ are similar, so $\\left|GM\\right|.\\left|GE'\\right| = \\left|GD\\right|.\\left|GA\\right|$ and $AE'MD$ is a cyclic quadrangle. Again, $E=E'$ and we are done." } { "Tag": [ "algebra", "polynomial", "inequalities", "triangle inequality" ], "Problem": "Given a quartic polynomial $ f(x)=x^{4}+ax^{3}+bx^{2}+cx+d$ with $ 4$ real roots and $ |f(i)|=1$, prove that $ a=b=c=d=0$.", "Solution_1": "[hide=\"Well...\"] $ |f(i)| = |(i-p)(i-q)(i-r)(i-s)| \\ge 1$ by Triangle Inequality, where $ p, q, r, s$ are the roots. The minimum occurs when $ p = q = r = s = 0$. [/hide]", "Solution_2": "[quote=\"t0rajir0u\"][hide=\"Well...\"] $ |f(i)| = |(i-p)(i-q)(i-r)(i-s)| \\ge 1$ by Triangle Inequality, where $ p, q, r, s$ are the roots. The minimum occurs when $ p = q = r = s = 0$. [/hide][/quote]\n\nVery nice proof, but\n\n[hide]is that really Triangle Inequality? It looks like you're just using $ |ab|=|a\\parallel b|$.[/hide]", "Solution_3": "Yeah, I thought triangle inequality is just |a+b|<=|a|+|b|\r\n\r\nBut the proof still works.", "Solution_4": "Hmm. It's not actually triangle inequality, sorry. $ |i-p| \\ge |i| = 1$ follows because $ i-p$ is the hypotenuse of a right triangle with sides $ i, p$, so I suppose in some sense it would just be trivial inequality." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$ x\\geq y\\geq 1$ prove: \r\n\r\n$ \\frac{x\\minus{}y}{\\sqrt{x\\plus{}y}}\\plus{}\\frac{y\\minus{}1}{\\sqrt{y\\plus{}1}}\\plus{}\\frac{1\\minus{}x}{\\sqrt{x\\plus{}1}}\\geq 0$", "Solution_1": "$ \\to (x \\minus{} 1) \\frac {\\sqrt {x \\plus{} 1} \\minus{} \\sqrt {x \\plus{} y}}{\\sqrt {(x \\plus{} 1)(x \\plus{} y)}} \\plus{} (y \\minus{} 1) \\frac {\\sqrt {x \\plus{} y} \\minus{} \\sqrt {y \\plus{} 1}}{\\sqrt {(x \\plus{} y)(y \\plus{} 1)}} \\geq 0$\r\nor $ \\frac {(x \\minus{} 1)(1 \\minus{} y)}{\\sqrt {(x \\plus{} 1)(x \\plus{} y)}.(\\sqrt {x \\plus{} 1} \\plus{} \\sqrt {x \\plus{} y})} \\plus{} \\frac {(x \\minus{} 1)(y \\minus{} 1)}{\\sqrt {(y \\plus{} 1)(x \\plus{} y)}.(\\sqrt {y \\plus{} 1} \\plus{} \\sqrt {x \\plus{} y})} \\geq 0$\r\nlater\r\n$ \\frac {(x \\minus{} 1)(y \\minus{} 1)}{\\sqrt {x \\plus{} y}} . \\frac {x \\minus{} y \\plus{} \\sqrt {x \\plus{} y}(\\sqrt {x \\plus{} 1} \\minus{} \\sqrt {y \\plus{} 1})}{\\sqrt {(x \\plus{} 1)(y \\plus{} 1)}.(\\sqrt {x \\plus{} y} \\plus{} \\sqrt {x \\plus{} 1}).(\\sqrt {x \\plus{} y} \\plus{} \\sqrt {y \\plus{} 1})} \\geq 0$\r\nobvious true", "Solution_2": "[quote=\"zaya_yc\"]$ x\\geq y\\geq 1$ prove: \n\n$ \\frac {x \\minus{} y}{\\sqrt {x \\plus{} y}} \\plus{} \\frac {y \\minus{} 1}{\\sqrt {y \\plus{} 1}} \\plus{} \\frac {1 \\minus{} x}{\\sqrt {x \\plus{} 1}}\\geq 0$[/quote]\r\nSetting $ 2a^2 \\equal{} y \\plus{} 1,2b^2 \\equal{} x \\plus{} 1,2c^2 \\equal{} x \\plus{} y$ $ (a,b,c > 0)$, then $ a \\le b \\le c$ and\r\n\\[ x \\equal{} b^2 \\plus{} c^2 \\minus{} a^2,y \\equal{} c^2 \\plus{} a^2 \\minus{} b^2,1 \\equal{} a^2 \\plus{} b^2 \\minus{} c^2\r\n\\]\r\nAnd the inequality becomes\r\n\\[ \\sum \\frac {(b^2 \\plus{} c^2 \\minus{} a^2) \\minus{} (c^2 \\plus{} a^2 \\minus{} b^2)}{\\sqrt {2c^2}} \\ge 0\r\n\\]\r\n\r\n\\[ \\Leftrightarrow \\sum \\frac {b^2 \\minus{} a^2}{c} \\ge 0\r\n\\]\r\n\r\n\\[ \\Leftrightarrow \\sum ab(b^2 \\minus{} a^2) \\ge 0\r\n\\]\r\n\r\n\\[ \\Leftrightarrow (c \\minus{} b)(c \\minus{} a)(b \\minus{} a) (a \\plus{} b \\plus{} c)\\ge 0\r\n\\]\r\nwhich is true. :)", "Solution_3": "$ \\frac{x-y}{\\sqrt{x+y}}+\\frac{y-1}{\\sqrt{y+1}}+\\frac{1-x}{\\sqrt{x+1}}\\geq 0$ ---->\r\n\r\n$ \\frac{x-y}{\\sqrt{x+y}}+\\frac{y-1}{\\sqrt{y+1}}\\geq\\frac{x-1}{\\sqrt{x+1}}$\r\n\r\n\r\n$ ((x-y)\\sqrt{x+y}+(y-1)\\sqrt{y+1})(\\frac{x-y}{\\sqrt{x+y}}+\\frac{y-1}{\\sqrt{y+1}})\\geq (x-y+y-1)^{2}=(x-1)^{2}$\r\n \r\n--->\r\n\r\n$ \\frac{x-y}{\\sqrt{x+y}}+\\frac{y-1}{\\sqrt{y+1}}\\geq\\frac{(x-1)^{2}}{(x-y)\\sqrt{x+y}+(y-1)\\sqrt{y+1}}=\n\\frac{(x-1)^{2}}{\\sqrt{x-y}*\\sqrt{(x-y)(x+y)}+\\sqrt{y-1}*\\sqrt{(y-1)(y+1)}}\\geq\n\\frac{(x-1)^{2}}{\\sqrt{(x-y+y-1)(x^{2}-y^{2}+y^{2}-1)}}=\\frac{(x-1)^{2}}{(x-1)\\sqrt{x+1}}=\\frac{x-1}{\\sqrt{x+1}}$" } { "Tag": [], "Problem": "just wondering. i graduate (hopefully heh heh) in 2008", "Solution_1": "pinkbubblez, the one in eighth grade, u would be either a=\r\n\r\neighty\r\nor\r\neighter\r\n\r\nsounds funky to me!", "Solution_2": "heehee...I am in 8th grade as well, though the school year is drawing to a close. Only 42 days of school left. Then, I will be a big bad freshman...lol. I can't wait to get out of the middle school. I will graduate in 2008 as well.", "Solution_3": "hmm, 7th for me. I will graduate as it is right now in 2009, but I hope to skip (heheheh).", "Solution_4": "8th 2008 yay highschool next year yay!", "Solution_5": "Go 8th graders!", "Solution_6": "eighth grade pride!!! =D and we get to vote for u.s. president when we're a freshman in college........ not that it's necessarily a good thing. X__x", "Solution_7": "funny how u contradict urself, pifreak", "Solution_8": "how come sixth graders and seventh graders are referred to as weird names and eigth graders arent?", "Solution_9": "I'm a freshman but since I skipped I get teased a lot about still being an eighth grader and stuff...mrph\r\n\r\nI'm not planning to graduate anytime soon so no year. (there is no such system in Japan)", "Solution_10": "[quote=\"Scrambled\"]how come sixth graders and seventh graders are referred to as weird names and eigth graders arent?[/quote]\r\n\r\nThat is probably because the majority of us talking about this issue are 8th graders or the age of an 8th grader. It also could be because we are like the seniors of junior high and we rule!!!!!!!!!!", "Solution_11": "go 8th graders! i'm one too\r\nbut i'll never be in high school, i'm just going to transition school next year. then college", "Solution_12": "I'm a junior...and all alone among the younger ones...", "Solution_13": "[quote=\"Scrambled\"]how come sixth graders and seventh graders are referred to as weird names and eigth graders arent?[/quote]\r\n\r\nhow would i know? its just natural. at miller sixers are also called \"pixies\"...weirdness.", "Solution_14": "yes, all wiedness....\r\n\r\nthey also can be called sixies... or shorties......\r\n\r\n8th graders cant be called anything cuz nothing goes w/ their name.", "Solution_15": "[quote=\"mathfanatic\"]9th grade. Go class of 2007![/quote]\r\n\r\nGo us =)", "Solution_16": "[quote=\"Caroline\"][quote=\"mathfanatic\"]9th grade. Go class of 2007![/quote]\n\nGo us =)[/quote]\r\nay I can't believe I am now going to graduate with mathfanatic...oh wait Japanese educational system ends in March :P Go [b]us[/b] not in the us! :lol:", "Solution_17": "[quote=\"Caroline\"][quote=\"mathfanatic\"]9th grade. Go class of 2007![/quote]\n\nGo us =)[/quote]\r\n\r\nOn that note, my history teacher calls the freshman class \"007\", as in James Bond. It's kinda odd.\r\n\r\nBut yes, go me, you, Mystic, Miths, and whoever else is a freshman.", "Solution_18": "[quote=\"mathfanatic\"]\n\nBut yes, go me, you, Mystic, Miths, and whoever else is a freshman.[/quote]\r\n\r\nlots of AASTers...chahah, lunmu, divran...etc...\r\n\r\ni'm sure there's quite a few others that dont post quite as much...", "Solution_19": "I'll be graduating in 2007 too most likely.", "Solution_20": "that's because you're a freshman...", "Solution_21": "Well, I'm not really a freshman. I'm actually a junior.", "Solution_22": "[quote=\"MithsApprentice\"][quote=\"mathfanatic\"]\n\nBut yes, go me, you, Mystic, Miths, and whoever else is a freshman.[/quote]\n\nlots of AASTers...chahah, lunmu, divran...etc...\n\ni'm sure there's quite a few others that dont post quite as much...[/quote]\r\n\r\nYeah, Mr. Holbrook had a whole bunch of us sign up... Most people whose username is the first three letters of their first name and the first three letters of their last name are from AAST. Plus some others, like me ;) (We're not all freshmen though..)", "Solution_23": "[quote=\"ComplexZeta\"]Well, I'm not really a freshman. I'm actually a junior.[/quote]\r\n\r\nhow does it work that you're a junior that graduates in 2007? just asking", "Solution_24": "i'm a sophomore but i skipped a grade. i don't get teased like tare though, ha", "Solution_25": "I'm a junior because my college bases things on units rather than number of years. I'm a junior in terms of units, but this is my first year here, and I don't plan to graduate early.", "Solution_26": "clats of 2008\r\n\r\nI eat you.", "Solution_27": "what school are you going to, simon?", "Solution_28": "University of California at Santa Barbara, College of Creative Studies.", "Solution_29": "yasss 14 year bump\n4th one so far" } { "Tag": [], "Problem": "let $ x$ ,$ y$ and $ z$ are reals numbers such that :\r\n$ \\frac{x}{y\\minus{}z}\\plus{}\\frac{y}{z\\minus{}x}\\plus{}\\frac{z}{x\\minus{}y}\\equal{}0$\r\nprove that:\r\n$ \\frac{x}{(y\\minus{}z)^2}\\plus{}\\frac{y}{(z\\minus{}x)^2}\\plus{}\\frac{z}{(x\\minus{}y)^2}\\equal{}0$", "Solution_1": "[quote=\"greatestmaths\"]let $ x$ ,$ y$ and $ z$ are reals numbers such that :\n$ \\frac {x}{y \\minus{} z} \\plus{} \\frac {y}{z \\minus{} x} \\plus{} \\frac {z}{x \\minus{} y} \\equal{} 0$\nprove that:\n$ \\frac {x}{(y \\minus{} z)^2} \\plus{} \\frac {y}{(z \\minus{} x)^2} \\plus{} \\frac {z}{(x \\minus{} y)^2} \\equal{} 0$[/quote]\r\n[hide=\"Solution\"]\nMultiply the first equation by $ \\frac {1}{z \\minus{} x} \\plus{} \\frac {1}{y \\minus{} z} \\plus{} \\frac {1}{x \\minus{} y}$ to get that\n\\[ [\\frac {x}{(y \\minus{} z)^2} \\plus{} \\frac {y}{(z \\minus{} x)^2} \\plus{} \\frac {z}{(x \\minus{} y)^2}] \\plus{} [\\frac {x}{(y \\minus{} z)(x \\minus{} y)} \\plus{} \\frac {x}{(y \\minus{} z)(z \\minus{} x)} \\plus{} \\frac {y}{(z \\minus{} x)(x \\minus{} y)} \\plus{} \\frac {y}{(z \\minus{} x)(y \\minus{} z)} \\plus{} \\frac {z}{(x \\minus{} y)(z \\minus{} x)} \\plus{} \\frac {z}{(x \\minus{} y)(y \\minus{} z)}] \\equal{} 0\n\\]\nWe divided the left hand side into two brackets and we will prove that the second bracket is equal to $ 0$. We create a common denominator and expand to get that\n\\[ [\\frac {x}{(y \\minus{} z)(x \\minus{} y)} \\plus{} \\frac {x}{(y \\minus{} z)(z \\minus{} x)} \\plus{} \\frac {y}{(z \\minus{} x)(x \\minus{} y)} \\plus{} \\frac {y}{(z \\minus{} x)(y \\minus{} z)} \\plus{} \\frac {z}{(x \\minus{} y)(z \\minus{} x)} \\plus{} \\frac {z}{(x \\minus{} y)(y \\minus{} z)}] \\equal{} \\frac {zx \\minus{} x^2 \\plus{} x^2 \\minus{} xy \\plus{} yx \\minus{} y^2 \\plus{} y^2 \\minus{} zy \\plus{} z^2 \\minus{} zx \\minus{} z^2 \\plus{} zy}{(x \\minus{} y)(y \\minus{} z)(z \\minus{} x)} \\equal{} 0\n\\]\nThus, $ [\\frac {x}{(y \\minus{} z)^2} \\plus{} \\frac {y}{(z \\minus{} x)^2} \\plus{} \\frac {z}{(x \\minus{} y)^2}] \\equal{} 0$ and we are done. [/hide]", "Solution_2": "notice that $ \\frac{x}{y\\minus{}z}\\plus{}\\frac{y}{z\\minus{}x}\\equal{}\\frac{z}{y\\minus{}x}$\r\nso $ \\frac{xz\\minus{}x^2\\plus{}y^2\\minus{}yz}{(y\\minus{}z)(z\\minus{}x)(y\\minus{}x)} \\equal{} \\frac{z}{(y\\minus{}x)^2}$\r\nsimilarly , we have \r\n$ \\frac{xy\\minus{}y^2\\plus{}z^2\\minus{}xz}{(z\\minus{}y)(z\\minus{}x)(x\\minus{}y)} \\equal{} \\frac{x}{(y\\minus{}z)^2}$\r\n$ \\frac{\\minus{}xy\\plus{}x^2\\minus{}z^2\\plus{}yz}{(y\\minus{}z)(x\\minus{}z)(x\\minus{}y)} \\equal{} \\frac{y}{(x\\minus{}z)^2}$\r\nand now let's sum them up and we are done" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "$ a,b,c$ are three real strictly positive numbers such that $ abc\\equal{}1$\r\nprove that \r\n$ \\frac{1}{\\sqrt{b\\plus{}\\frac{1}{a}\\plus{}\\frac{1}{2}}}\\plus{}\\frac{1}{\\sqrt{c\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{2}}}\\plus{}\\frac{1}{\\sqrt{a\\plus{}\\frac{1}{c}\\plus{}\\frac{1}{2}}}>\\equal{}\\sqrt{2}$", "Solution_1": "try : $ a\\equal{}\\frac{1}{2}$ $ b\\equal{}\\frac{1}{2}$ $ c\\equal{}\\frac{1}{4}$ :wink:", "Solution_2": "what do you want to say ? the problem is correct :P because 1/2 x 1/2 x 1/4 = 1/16 not 1", "Solution_3": "yes sorry : i wanted to write :\r\n$ a\\equal{}\\frac{1}{2}$ and $ b\\equal{}\\frac{1}{2}$ and $ c\\equal{}4$ :wink: \r\nnow i think the inequality is not correct :P", "Solution_4": "no you can calculate it you have a mistake ths probleme is true\r\n(i just calculate it :wink: )", "Solution_5": "Let $ a\\equal{}\\frac yx,b\\equal{}\\frac zy,c\\equal{}\\frac xz$. Then our inequality is equivalent to\r\n\\[ \\sum\\frac{1}{\\sqrt{\\frac zy\\plus{}\\frac xy\\plus{}\\frac 12}}\\ge \\sqrt 2\\iff \\sum\\sqrt{\\frac{y}{2x\\plus{}y\\plus{}2z}}\\ge 1\\]\r\nVerify that \r\n\\[ \\sqrt{\\frac{y}{2x\\plus{}y\\plus{}2z}}\\ge \\frac{y}{x\\plus{}y\\plus{}z}\\iff (x\\plus{}y\\plus{}z)^2\\ge y(2x\\plus{}y\\plus{}2z)\\]\r\nWhich is clearly true. Thus \r\n\\[ \\sum\\sqrt{\\frac{y}{2x\\plus{}y\\plus{}2z}}\\ge \\sum\\frac{y}{x\\plus{}y\\plus{}z}\\equal{}1\\]\r\nSince $ x,y,z>0$ equality cannot hold.", "Solution_6": "yes good :lol: \r\nsorry i have a mystake :P" } { "Tag": [ "trigonometry", "inequalities proposed", "inequalities" ], "Problem": "$A+B+C=\\pi$. Prove that :\r\n$\\cos(A-B)\\cos(B-C)\\cos(C-A)\\geq8\\cos(A)\\cos(B)\\cos(C)$", "Solution_1": "Use the formula : $\\frac{\\cos(B-C)}{\\cos A}= \\frac{b+c}{a}$ and AM-GM ineq", "Solution_2": "[quote=\"jarod\"]Use the formula : $\\frac{\\cos(B-C)}{\\cos A}= \\frac{b+c}{a}$ and AM-GM ineq[/quote]\r\njarod, $\\{A,B,C\\}\\subset\\mathbb R$ and what is this a,b,c?:)", "Solution_3": "$\\frac{\\cos(B-C)}{\\cos A}=\\frac{\\sin(B+C) \\cos(B-C)}{\\sin A \\cos A}=\\frac{\\sin 2B+\\sin 2C}{\\sin 2A }$\r\nI think A.B.C must be acute-angel", "Solution_4": "[quote=\"zhaobin\"]\nI think A.B.C must be acute-angel[/quote]\r\nNo,not necessarily:\r\n$cos(A-B)cos(B-C)cos(C-A)\\geq8cosAcosBcosC\\Leftrightarrow$\r\n$\\Leftrightarrow p^2+(4q^3+5q)p+8q^2\\geq0,$ where $p=cos(A-B),q=cos(A+B).$\r\nLet $\\{p,q\\}\\subset[0,1].$ Then $p^2-(4q^3+5q)p+8q^2\\geq0.$\r\nProof:\r\n$(4q^3+5q)^2-4\\cdot8q^2\\leq0\\Rightarrow p^2-(4q^3+5q)p+8q^2\\geq0.$\r\nIf $(4q^3+5q)^2-4\\cdot8q^2>0$ then \r\n$(4q^3+5q)^2-4\\cdot8q^2>0\\Leftrightarrow q>\\frac{\\sqrt{\\sqrt32-5}}{2}.$ Hence, $\\frac{4q^3+5q}{2}>1.$\r\nAnd $1^2-(4q^3+5q)\\cdot1+8q^2\\geq0\\Leftrightarrow(1-q)(2q-1)^2\\geq0.$\r\nHence, $\\forall\\{p,q\\}\\subset[0,1]$ $p^2-(4q^3+5q)p+8q^2\\geq0.$\r\nHence, $\\forall\\{p,q\\}\\subset[-1,1]$ $p^2+(4q^3+5q)p+8q^2\\geq0.$\r\nHence, $cos(A-B)\\cos(B-C)\\cos(C-A)\\geq8\\cos(A)\\cos(B)\\cos(C).$ :)", "Solution_5": "nice proof.i think it is a useful skill", "Solution_6": "[quote=\"jarod\"]Use the formula : $\\frac{\\cos(B-C)}{\\cos A}= \\frac{b+c}{a}$ and AM-GM ineq[/quote]\r\nvery nice ;)" } { "Tag": [ "Putnam" ], "Problem": "Solve\r\n$ x^3 \\plus{} y^3 \\plus{} 1 \\equal{} 3 xy$\r\nWhere $ x,y$ are reals", "Solution_1": "[hide]\n$ x^3\\plus{}y^3\\plus{}1^3\\minus{}3xy1 \\equal{} (x\\plus{}y\\plus{}1)(x^2\\plus{}y^2\\plus{}1^2\\minus{}xy\\minus{}y\\minus{}x)$\nSolutions are either $ x\\plus{}y\\plus{}1\\equal{}0$ or $ x\\equal{}y\\equal{}1$.\n[/hide]", "Solution_2": "Up to a change of sign in the variables, this is essentially identical to Putnam 2006 B1. (Actually, the Putnam problem was a bit trickier.)", "Solution_3": "[quote=\"JBL\"]Up to a change of sign in the variables, this is essentially identical to Putnam 2006 B1. (Actually, the Putnam problem was a bit trickier.)[/quote]\r\n\r\nI made up this problem from my mind thinking that equality in AM-GM holds only iif the variables are equal.", "Solution_4": "You can only apply AM-GM if both $ x, y$ are nonnegative.", "Solution_5": "[quote=\"FOURRIER\"][quote=\"JBL\"]Up to a change of sign in the variables, this is essentially identical to Putnam 2006 B1. (Actually, the Putnam problem was a bit trickier.)[/quote]\n\nI made up this problem from my mind thinking that equality in AM-GM holds only iif the variables are equal.[/quote]\r\nYou can actually find this problem in the first section of Mathematical Olympiad Treasures.\r\n\r\nP.S. I am not claiming that you did not create this problem." } { "Tag": [ "quadratics" ], "Problem": "After writing the AMC and getting 104.5 (and realizing I was almost entirely lost on how to do most questions of 15-25) I really had to wonder, \"How do people do this!?\"\r\n\r\nThe thing I always hear is, \"You have the tools to solve this, it's just a matter of applying them.\" Yet, I'm very often lost on how to apply everything I know to higher-level problems.\r\n\r\nMy question is, what are common methods of approaching higher-level problems? When I look at a difficult question, how should I start myself, what should I ask myself, etc.?\r\n\r\nAny tricks or common methods you use in solving problems, or approaching problems, would be helpful. Thanks.", "Solution_1": "At where I am now, I can usually solve questions on the level of, say, 1-20 on the AMC in less than a minute, so it's hard to say what goes through my mind when I'm doing them. On harder questions though, this is how it might go (I just solved Q25 on the Fermat earlier today, so I'll try and walk you through my thought process). My apologies if this is too difficult of a question, but hopefully you can take something out of it.\r\n\r\nSee this thread for the question:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=27914\r\n\r\n1st thing: We have a line that passes through $(p,q)$. Well, that's not too difficult, we know that the equation of this line is just $y=m(x-p)+q$. This doesn't really get us anywhere, but it gives us an equation, which is good.\r\n\r\n2nd thing: Well, lets look at this triline thingy. It's easy to find x,y intercepts:\r\n\r\n$y=mx+(q-mp)$ and $y=m(x-p+q/m)$ Add em. We get: $3m=q-mp-p+q/m$\r\n\r\n3rd thing: Lets look at this last equation. It has linear expressions AND reciprocal expressions, which are awkward to deal with; lets multiply both sides by $m$.\r\n\r\n4th thing: Rearranging, we get a quadratic equation. What do quadratic equations have ? ........\r\n\r\n5th thing: TWO SOLUTIONS! (unless the discriminant is zero). So the discriminant has to be zero, because the problem says that we need EXACTLY ONE triline.\r\n\r\n6th thing: Now we have a nifty algebraic expression, which we rearrange to $(p-q)^2=12q$. This looks really simple, so we can see that we're almost done.\r\n\r\n7th thing: We have restrictions on $q$. The biggest it can be is something like $120000$. The other side is a perfect square.....\r\n\r\n8th thing: which, upon appropriate choice of $p$, can be ANY perfect square we want. So we just need the number of $x^2\\leq 120000$ which gives us that $x\\leq 346$\r\n\r\n9th thing: $12=2^2\\cdot 3$. So we need $x$ to have a factor of at LEAST $2$ and $3$; or equivalently, to be divisible by $6$. $346/6$ is just larger than $57$, so $57$ is our answer.", "Solution_2": "You have a similar score to what I had last year...and I had wondered about the same thing too (and still do). But the thing I found that really helped was to PRACTISE! Yea, yea...you're probably thinking...everyone tells me that...but when I say practise...I mean really practise. Set aside some time each day to work through past contest problems, etc. Also highly recommended is the Art of Problem Solving series (both Vol.1 and 2). They explain all the techniques really well and have hundreds of exercises. That's what I did and I've improved a lot...though definitely not at the same level as being able to solve the question above. Still, one step at a time, huh? \r\n\r\nCheers!" } { "Tag": [ "probability", "real analysis", "real analysis solved" ], "Problem": "I don't know if this is the right subforum, but I think it involves some limits, so I thought I should post it here.\r\n\r\nGiven any $p\\in [0,1]$, devise an experiment using only a fair coin which has success probability $p$. \r\n\r\nThis looks really awkward. I think it's not hard to find an experiment for $p=\\frac k{2^t}$, but what puzzles me is the fact that we regard infinite processes as \"experiments\". Some infinite processes must be involved because it can't be done in a finite number of steps for $p\\in R-Q$, for example. \r\n\r\nI think the fact that the set ${\\{\\frac k{2^t}|k\\geq 0,t\\geq 0,k\\leq 2^t}$ is dense in $[0,1]$ helps.", "Solution_1": "Yes it does involve some limits, or better said some infinite experiments.\r\n\r\nConsider the following experiment:\r\nAt step n, denote by $a_n$ the number which is 1 if we got tails and 0 if we got heads.\r\n\r\nThe probability that the number, written in base 2, $0,a_1a_2\\ldots a_n \\ldots$ is smaller that p is p." } { "Tag": [ "trigonometry", "modular arithmetic" ], "Problem": "Source: 1989 NYSML\r\nI-6. Compute the smallest positive angle $x$, in degrees, such that $\\tan (4x)=\\frac{\\cos x-\\sin x}{\\cos x+\\sin x}$.", "Solution_1": "Based on the above problem, I created my own pbm!\r\n\r\na) Compute the smallest positive angle $x$, such that \r\n$\\tan(2x)=\\frac{\\cos x - \\sin x}{\\cos x + \\sin x}$.\r\n\r\nb) Compute the smallest positive angle $x$, such that \r\n$\\cot(2x)=\\frac{\\cos x - \\sin x}{\\cos x + \\sin x}$.\r\n\r\n\r\nIt's interesting that my own pbms requires (i think) a different solving tech. from I-6.", "Solution_2": "[hide=\"I-6\"]$\\frac{\\cos x-\\sin x}{\\cos x+\\sin x}=\\frac{\\cos^2 x-\\sin^2 x}{\\sin^2 x+2\\sin x\\cos x+\\cos^2 x}$\n $=\\frac{\\cos 2x}{1+\\sin 2x}$\n $=\\frac{2\\sin 2x\\cos 2x}{2\\sin 2x+2\\sin^2 2x}$\n $=\\frac{\\sin 4x}{2\\sin 2x+2\\sin^2 2x}$\n$\\tan 4x=\\frac{\\sin 4x}{\\cos 4x}$\n $=\\frac{\\sin 4x}{1-2\\sin^2 2x}$\nthen,\n$2\\sin 2x+2\\sin^2 2x=1-2\\sin^2 2x$\nlet $z=\\sin 2x$ then, $4z^2+2z-1=0$\nwe get $z=\\frac{-1\\pm\\sqrt{5}}{4}$\nsince x is smaller if z is positive, take the positive solution and using a calculator, we get $x=9$[/hide]", "Solution_3": "[quote=\"Jiang\"][hide=\"I-6\"]$\\frac{\\cos x-\\sin x}{\\cos x+\\sin x}=\\frac{\\cos^2 x-\\sin^2 x}{\\sin^2 x+2\\sin x\\cos x+\\cos^2 x}$\n $=\\frac{\\cos 2x}{1+\\sin 2x}$\n $=\\frac{2\\sin 2x\\cos 2x}{2\\sin 2x+2\\sin^2 2x}$\n $=\\frac{\\sin 4x}{2\\sin 2x+2\\sin^2 2x}$\n$\\tan 4x=\\frac{\\sin 4x}{\\cos 4x}$\n $=\\frac{\\sin 4x}{1-2\\sin^2 2x}$\nthen,\n$2\\sin 2x+2\\sin^2 2x=1-2\\sin^2 2x$\nlet $z=\\sin 2x$ then, $4z^2+2z-1=0$\nwe get $z=\\frac{-1\\pm\\sqrt{5}}{4}$\nsince x is smaller if z is positive, take the positive solution and using a calculator, we get $x=9$[/hide][/quote]\r\n\r\nIf it came out so nice, you'd think you wouldn't need a calculator. \r\n\r\nThink Fibonacci.", "Solution_4": "I'm not sure what you mean by think Fibonacci but I divided by $\\ \\cos x$ on top and bottom to get $\\ \\frac{1 - \\tan x}{1 + \\tan x} = \\frac{\\tan 45 - \\tan x}{1 + \\tan 45 \\tan x}= \\ tan(45 - x)$ Then we can say $\\ 45 - x + 180k = 4x$ Now since we are trying to find the least positive angle $\\ x$ we can let $\\ k=0$ and we find $5x=45$ $\\ x=9$", "Solution_5": "Convert to $\\frac{ \\cos x - \\cos (90 - x) }{ \\cos x + \\cos (90 - x) }$ and use sum to product formulas to get\r\n$- \\tan 45 \\tan (x - 45) = \\tan(45 - x) = \\tan 4x \\implies x = 9$", "Solution_6": "[quote=\"Jiang\"][hide=\"I-6\"]$\\frac{\\cos x-\\sin x}{\\cos x+\\sin x}=\\frac{\\cos^2 x-\\sin^2 x}{\\sin^2 x+2\\sin x\\cos x+\\cos^2 x}$\n $=\\frac{\\cos 2x}{1+\\sin 2x}$\n $=\\frac{2\\sin 2x\\cos 2x}{2\\sin 2x+2\\sin^2 2x}$\n $=\\frac{\\sin 4x}{2\\sin 2x+2\\sin^2 2x}$\n$\\tan 4x=\\frac{\\sin 4x}{\\cos 4x}$\n $=\\frac{\\sin 4x}{1-2\\sin^2 2x}$\nthen,\n$2\\sin 2x+2\\sin^2 2x=1-2\\sin^2 2x$\nlet $z=\\sin 2x$ then, $4z^2+2z-1=0$\nwe get $z=\\frac{-1\\pm\\sqrt{5}}{4}$\nsince x is smaller if z is positive, take the positive solution and using a calculator, we get $x=9$[/hide][/quote]\r\nInteresting.. I tried that method, but gave up, b/c it was too much time consuming....\r\n\r\nAnother quick way of doing this problem is;\r\nconvert $\\tan (4x)$ to $\\frac{\\sin(4x)}{\\cos(4x)}$.\r\n$\\frac{\\sin(4x)}{\\cos(4x)}=\\frac{\\cos x-\\sin x}{\\cos x+\\sin x}$.\r\nNow cross-multiflying and rearranging would yield \r\n$\\sin(4x)\\cos(x)+\\cos(4x)\\sin(x)=\\cos(4x)\\cos(x)-\\sin(4x)\\sin(x)$.\r\nTherefore, $\\sin(5x)=\\cos(5x)$ and $\\tan(5x)=1$\r\n$5x=45^\\circ$ and $x=9^\\circ$.", "Solution_7": "It may seem like adding the $\\ 180k$ in my solution was superfluous, and in this example it turned out to be but take this example:\r\n\r\nFind the smallest positive integer soltuion to $\\ \\tan 19x = \\frac{\\cos96 + \\sin96}{\\cos96 - \\sin96}$", "Solution_8": "[quote=\"mna851\"]I'm not sure what you mean by think Fibonacci but I divided by $\\ \\cos x$ on top and bottom to get $\\ \\frac{1 - \\tan x}{1 + \\tan x} = \\frac{\\tan 45 - \\tan x}{1 + \\tan 45 \\tan x}= \\ tan(45 - x)$ Then we can say $\\ 45 - x + 180k = 4x$ Now since we are trying to find the least positive angle $\\ x$ we can let $\\ k=0$ and we find $5x=45$ $\\ x=9$[/quote]\r\n\r\nOh that is a rather nice way to do it. \r\n\r\nThough Golden Ration/Fibonacci angles are the ones which would appear in golden triangles or something like it.", "Solution_9": "[quote=\"mna851\"]It may seem like adding the $\\ 180k$ in my solution was superfluous, and in this example it turned out to be but take this example:\n\nFind the smallest positive integer soltuion to $\\ \\tan 19x = \\frac{\\cos96 + \\sin96}{\\cos96 - \\sin96}$[/quote]\r\n\r\n[hide]Dividing the numerator and denominator by $\\cos 96$, we have\n\n$\\tan 19x=\\frac{1+\\tan 96}{1-\\tan 96}=\\frac{\\tan 45+\\tan 96}{1-\\tan 45\\tan 96}=\\tan (45+96)=\\tan 141$.\n\nTherefore, $19x=141+180k$. Since $x$ is an integer, we have $141+180k\\equiv 0\\pmod{19}$. Therefore, $8+9k\\equiv 0\\pmod{19}$. The smallest possible value of $x$ will result from the smallest possible value of $k$, which is $16$. Then $19x=141+180\\cdot 16=3021$, and $x=\\boxed{159}$.\n\nCan anyone find a fast way to solve this from $19x=141+180k$? I had to check $k=1,2,3\\cdots 15$ before I found that $k=16$ was the smallest possible value of $k$.[/hide]", "Solution_10": "well, both $x$ and $k$ has to be integer, so it's really a diophatine equation:\r\n\r\n[hide]\nso $19x = 141 + 180k$, moding both sides by 19 gives $0 = 8 + 9k$, so $9k = 19u - 8$. Mod by 9 again we have $0 = u - 8$, so $u = 9v+8$. Substitute back we have $9k = 9*19v + 18*8$, or $k = 19v + 16$. Smallest $k$ implies that $v = 0$, so we have $k = 16$. This is my slow way of solving two variable diophatine equations, but hey, it works.\n[/hide]" } { "Tag": [ "pigeonhole principle", "analytic geometry", "USAMTS", "AMC" ], "Problem": "Given 5 lattice points, no three of which are collinear in the plane. Prove that if I draw all possible line segments, then at least one lattice point (excluding hte endpoints) will lie on those line segments.", "Solution_1": "[hide] consider parity:\n\nthere are only 4 possible combinations of even and odd for the x and y coordinates of the 5 points: (o,o)(e,e)(e,o)(o,e), so two of the points must have x and y coordinates with the same parity. The difference between numbers of the same parity is even, so one of the midpoints of the segments will always be a lattice point. [/hide]", "Solution_2": "Correct!\r\n\r\nNow, can you generalize?", "Solution_3": "generalize in what way?", "Solution_4": "Um, for example, extend this problem to 3-dimensions. What is the minimum number of distinct lattice points in space, such that there will always be lattice point on the planes connecting them. (Or am I not making any sense? Too tired...)\r\n\r\nOr....\r\n\r\nWhat if we had n lattice points on line segments drawn between distinct lattice points? Minimally, How many of those lattice points are needed?\r\n\r\n\r\nI think i'm blabbering, too tired....", "Solution_5": "[quote]Um, for example, extend this problem to 3-dimensions. What is the minimum number of distinct lattice points in space, such that there will always be lattice point on the planes connecting them. [/quote]\r\n\r\nI believe this was a USAMTS problem last yr, and the answer is 9, since there are 8 possible combinations of even and odd vertices. For n dimensions, there will always be lattice point if you have 2^n + 1 points. \r\n\r\nI haven't really thought about the other one, but it would make sense that if you need a lattice points on the line segments connecting b points in c dimensions then b must be greater than or equal to 2^c + n. I very well may be wrong about this, since I haven't really worked this, just thought about it briefly...", "Solution_6": "My answer: [hide]If it's is n-space, and we want c lattice points in between, it would be (c+1)^n+1[/hide]\n\n\n\nHint: [hide] ok, so it's not much of a hint, but just consider mods instead of plain ol' parity[/hide]\n\n\n\n i just felt like putting smileys here.", "Solution_7": "Definitely getting there.\r\n\r\nLike, 99 degrees warm" } { "Tag": [], "Problem": "Compute the sum of $ a$, $ b$, and $ c$ given that $ \\frac a2\\equal{}\\frac b3\\equal{}\\frac c5$ and $ abc\\equal{}1920$.", "Solution_1": "Let $ r\\equal{}\\frac a2\\equal{}\\frac b3\\equal{}\\frac c5\\equal{}\\frac{a\\plus{}b\\plus{}c}{10}$.\r\n\r\nThen we want to find $ 10r$, and since $ r^3\\equal{}\\frac{abc}{30}\\equal{}64\\implies r\\equal{}4$, our answer is $ 10\\cdot4\\equal{}\\boxed{40}$.\r\n\r\nProof that if $ r\\equal{}\\frac{a_1}{b_1}\\equal{}\\frac{a_2}{b_2}$, then $ r\\equal{}\\frac{a_1\\plus{}a_2}{b_1\\plus{}b_2}$.\r\n\r\nThis is pretty simple: we have $ a_1\\equal{}rb_1$ and $ a_2\\equal{}rb_2$, so adding these, $ a_1\\plus{}a_2\\equal{}r(b_1\\plus{}b_2)$, and dividing, $ r\\equal{}\\frac{a_1\\plus{}a_2}{b_1\\plus{}b_2}$.", "Solution_2": "For anyone interested, the three numbers are $ 8$, $ 12$, and $ 20$.", "Solution_3": "May you clarify your solution? I don't seem to see how what we're doing with $ r$ leads us to the solution.", "Solution_4": "hello, we get from\r\n$ \\frac{a}{2}\\equal{}\\frac{b}{3}$\r\n$ b\\equal{}\\frac{3}{2}a$\r\nand from\r\n$ \\frac{a}{2}\\equal{}\\frac{c}{5}$\r\n$ c\\equal{}\\frac{5}{2}a$\r\nand so we get\r\n$ abc\\equal{}a^3\\cdot\\frac{15}{4}$\r\nand from here you can compute $ a,b,c$.\r\nSonnhard.", "Solution_5": "@PhireKaLk6781\r\n\r\nAs math154 proved, $ r\\equal{}\\frac{a}{2}\\equal{}\\frac{b}{3}\\equal{}\\frac{c}{5}\\equal{}\\frac{a\\plus{}b\\plus{}c}{2\\plus{}3\\plus{}5}\\equal{}\\frac{a\\plus{}b\\plus{}c}{10}$. Since we are trying to find $ a\\plus{}b\\plus{}c\\equal{}(10)\\left(\\frac{a\\plus{}b\\plus{}c}{10}\\right)\\equal{}10r$, we simply need to find $ r$ and multiply it by $ 10$. \r\n\r\nWe know that $ r^3\\equal{}\\frac{a}{2}\\cdot\\frac{b}{3}\\cdot\\frac{c}{5}\\equal{}\\frac{abc}{30}$, and are given that $ abc\\equal{}1920$. Thus\r\n\\[ r^3\\equal{}\\frac{abc}{30}\\equal{}\\frac{1920}{30}\\equal{}64\\]\r\nSo we find that $ r\\equal{}4$, and $ a\\plus{}b\\plus{}c\\equal{}10r\\equal{}\\boxed{40}$." } { "Tag": [ "records", "Social", "IMO", "IMO 2007" ], "Problem": "In the tradition of the past years (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56717]here[/url] for example) let us vote and comment of the possible winners of the IMO 2007 Vietnam.\r\n\r\nAgain 10 options are available in the poll, ordered by their results in the last IMO, IMO 2006 Slovenia. Notice that near each country there is a sign, meaning what is their performance in the top 10 compared with IMO 2005 Mexico. \r\n\r\n- = Stationary \r\n^ = Going up in the top 10 list\r\n^^ = New in top 10\r\nv = Going down in the top 10", "Solution_1": "Why Vietnam is not in this list? It is so unjust! :D", "Solution_2": "BRAZIL! :D", "Solution_3": "[quote=\"hungkhtn\"]Why Vietnam is not in this list? It is so unjust! :D[/quote]It's always the previous year's top 10 countries (in the order they ranked).", "Solution_4": "Maybe \"Other\" should be an option. It's always possible...", "Solution_5": "[quote=\"jmerry\"]Maybe \"Other\" should be an option. It's always possible...[/quote]\r\n\r\nyou're right! :)", "Solution_6": "I see that China has favour in the peoples' eyes right now. :) \r\nOf course, maybe \"other\" is preparing extremely well...", "Solution_7": "China probably will pull off another one - but I'll go with Romania (and of course India :D)\r\n Ashwathj", "Solution_8": "interesting, the position of moldova's team is a very increasing sequence! 48 - 40 - 30 - 21 - 17 - 9 - (?)", "Solution_9": "Most probably china is going to win again.\r\nBut if it were like that what i am going to tell will come true...Then i would tell Bangladesh.\r\nBut probably not this year :(", "Solution_10": "[quote=\"e.lopes\"]interesting, the position of moldova's team is a very increasing sequence! 48 - 40 - 30 - 21 - 17 - 9 - (?)[/quote]\r\n\r\nBut the result is hard to be kept as Iura and Alex graduate from high-school this year and join the Boston fan group.", "Solution_11": "[quote=\"jmerry\"]Maybe \"Other\" should be an option. It's always possible...[/quote]I don't think it's ever been done (country not in top 10 previous year, winning the IMO). It's a matter of generation, so in order to beat China you have to have a really good generation that grows on 1-2 IMOs before that (like Romania in 96, or Bulgaria in 03).", "Solution_12": "[i][b][color=red][size=200]Vietnam [/size][/color][/b][/i]\r\n :first: :P", "Solution_13": "China, I think so :wink:", "Solution_14": "Korean team is the best. Because there are experience participants one of the third times and two of the second times. At least one person will get perfect score.\r\nthe other else will get gold medal.", "Solution_15": "You forgot Romania at the returning scores. We have a returning student (his 4th IMO!). His ML username is dzeta and he's running for the perfect score this year (he's got already 2 gold medals).", "Solution_16": "[quote=\"Valentin Vornicu\"]You forgot Romania at the returning scores. We have a returning student (his 4th IMO!). His ML username is dzeta and he's running for the perfect score this year (he's got already 2 gold medals).[/quote]Right you are!\r\n\r\nThere should be 1 returnee with 34 points.\r\n\r\nSince the editing time window has already expired on me, you can edit my post if you feel like it :).", "Solution_17": "[quote=\"MathChauffeur\"]Okay, those following the IMO are not going to be as passionate, obsessed, crazy, all-consumed, etc., etc., as those following the World Cup. However, that does not mean that we can\u2019t come up with potentially interesting tidbits leading up to the IMO -- just like thousands of reporters would, starting months (years?) before the World Cup.\n*snip*[/quote]\r\nSingapore has 2 returnees from '06, not 1, the total score is 19 + 16. Thanks.", "Solution_18": "[quote=\"jiahan\"]Singapore has 2 returnees from '06, not 1, the total score is 19 + 16. Thanks.[/quote]\r\nThanks for the correction :blush:.\r\n\r\nWas the second score 16 or 15? The official Slovenia site has 15 as the score.", "Solution_19": "there's also an albanian student who took 14 last year!", "Solution_20": "[quote=\"Albanian Eagle\"]there's also an albanian student who took 14 last year![/quote]\r\nI found a few more omissions :blush: :blush:. Can somebody check on Azerbaijan and Germany? I am not sure about them.\r\n\r\nCountry, # Returnees, Total Points\r\nUSA\t4\t125\r\nKorea\t3\t86\r\nMoldova\t3\t81\r\nArmenia\t4\t66\r\nHong Kong\t3\t65\r\nBosnia & Hertegozinia\t4\t63\r\nBulgaria\t3\t59\r\nMacau\t5\t55\r\nColumbia\t4\t54\r\nTaiwan\t2\t53\r\nIsrael\t3\t50\r\nJapan\t2\t48\r\nLituania\t3\t46\r\nMexico\t3\t46\r\nSerbia\t3\t46\r\nAustria\t3\t45\r\nGermany\t2\t45\r\nArgentina\t3\t44\r\nSweden\t3\t44\r\nEstonia\t3\t43\r\nMongolia\t3\t41\r\nUkraine\t2\t40\r\nLatvia\t3\t39\r\nIceland\t4\t38\r\nIreland\t4\t37\r\nBelarus\t2\t37\r\nChina\t1\t37\r\nPoland\t1\t36\r\nSingapore\t2\t35\r\nNorway\t3\t34\r\nPeru\t2\t34\r\nRomania\t1\t34\r\nGeorgia\t2\t32\r\nAzerbaijan\t2\t31\r\nFinland\t2\t31\r\nCroatia\t3\t30\r\nMalaysia\t3\t30\r\nBrazil\t2\t30\r\nItaly\t1\t30\r\nSlovakia\t1\t30\r\nNew Zealand\t3\t28\r\nSouth Africa\t3\t28\r\nIndia\t2\t28\r\nSwitzerland\t1\t28\r\nEl Salvador\t3\t27\r\nParaguay\t3\t27\r\nCosta Rica\t2\t27\r\nPanama\t3\t24\r\nMacedonia\t3\t23\r\nIran\t1\t23\r\nThailand\t1\t23\r\nUnited Kingdom\t1\t22\r\nBelgium\t2\t21\r\nEcuador\t3\t19\r\nHungary\t1\t19\r\nTurkmenistan\t1\t19\r\nUzbekistan\t1\t17\r\nCzech Republic\t1\t16\r\nFrance\t1\t15\r\nKazakhstan\t1\t15\r\nPortugal\t1\t15\r\nTurkey\t1\t15\r\nDenmark\t2\t14\r\nAlbania\t1\t14\r\nGreece\t1\t14\r\nSlovenia\t1\t14\r\nTrinidad & Tobago\t3\t11\r\nKyrgyzstan\t2\t11\r\nCypress\t2\t10\r\nLuxembourg\t1\t10\r\nNetherlands\t1\t10\r\nPakistan\t1\t10\r\nVenezuela\t1\t10\r\nTajikistan\t2\t8\r\nBangladesh\t1\t8\r\nNigeria\t3\t7\r\nPuetro Rico\t1\t2\r\nAustralia\t0\t0\r\nBolivia\t0\t0\r\nCanada\t0\t0\r\nLiechtenstein\t0\t0\r\nMorocco\t0\t0\r\nRussia\t0\t0\r\nSaudi Arabia\t0\t0\r\nSpain\t0\t0\r\nSri Lanka\t0\t0\r\nVietnam\t0\t0", "Solution_21": "Germany has two previous participants: Friedrich Feuerstein (IMO 2003) and Peter Scholze (since IMO 2004).", "Solution_22": "[quote=\"orl\"]Germany has two previous participants: Friedrich Feuerstein (IMO 2003) and Peter Scholze (since IMO 2004).[/quote]\r\nThanks! I wasn't sure about Georg Schr\u00f6ter in 2007 and Georg Sch\u00f6nherr in 2006.", "Solution_23": "[quote=\"MathChauffeur\"][quote=\"orl\"]Germany has two previous participants: Friedrich Feuerstein (IMO 2003) and Peter Scholze (since IMO 2004).[/quote]\nThanks! I wasn't sure about Georg Schr\u00f6ter in 2007 and Georg Sch\u00f6nherr in 2006.[/quote]\r\nyear it's a bit confusing because we live in the same city and are born in the same year (1989). but we aren't the same person ;)\r\n\r\nNaphthalin", "Solution_24": "Another correction:\r\n\r\nCountry, # Returnees, Total Points\r\nAustria 4 56", "Solution_25": "[quote=\"gighiuhui\"]guys guys it's [b]USA[/b]\n\nalso, valentin, the U.S. will beat Romania with a probability of 1.01[/quote]\r\n\r\nRush, two comments:\r\n\r\n(1) never over estimate yourself and your teams (which you are very good at and do a very often);\r\n\r\n(2) one more Zuming story for you to add: he is telling you to work on math instead of chat and waste time. Did you review all the MOSP materails? If not, go to work! See you in Sep.! (And I hope your math imporved a lot by then :|", "Solution_26": "Problems 3 and 6 were very Russian-style, so it might be Russia.", "Solution_27": "Now i know that, Vietnam is in top 4 ! :P", "Solution_28": "Vietnam is the 3rd team!!Congratulations!!! :roll: :roll: Bravo!!! :)", "Solution_29": "Go Russia!!!!!!!!!!!!!!!!!!!!!!!!!! :D :D" } { "Tag": [ "vector", "geometry", "3D geometry", "tetrahedron", "parameterization", "linear algebra" ], "Problem": "[b]Let A;B;C;D be four points in space, with position vectors a; b; c; d respectively. (We shall assume\nthe four points do not lie in a plane. They are called the vertices of a tetrahedron.) Show that the\npoint (1/4)(a + b + c + d) lies on the segment joining A to the centroid of triangle BCD. [/b]\r\n\r\nIs the above question the same basic question as this one (as I've solved this one):\r\n\r\n[b]Let A;B;C;D be four points on plane, with position vectors a; b; c; d respectively. Show that the\npoint (1/4)(a + b + c + d) lies on the segment joining A to the centroid A0 of triangle BCD. Let B0,\nC0 and D0 be the centroids of the triangles ACD, ABD and ABC respectively. Deduce that all\nfour lines AA0, BB0, CC0 and DD0 pass through one point.[/b]\r\n\r\nAnd please you will help with this next part please?\r\n\r\n[i]Deduce that all four lines joining a vertex of the tetrahedron to the centroid of the opposite triangle pass\nthrough one point.[/i]\r\n\r\nThank You!", "Solution_1": "A standard parametrization of a segment: The segment between $ P$ and $ Q$ is given by $ \\{tP\\plus{}(1\\minus{}t)Q: 0\\le t\\le 1\\}$.\r\n\r\nNow, can you find a position vector for the centroid of a face, and the appropriate $ t$-value for the centroid $ \\frac14(A\\plus{}B\\plus{}C\\plus{}D)$ of the tetrahedron?" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Prove if a,b,c>0 such that $ a^2\\plus{}b^2\\plus{}c^2\\equal{}3$ then\r\n$ \\sum\\frac{a^3}{\\sqrt{a^4\\minus{}b^4\\plus{}b^2}}>\\equal{}3$", "Solution_1": "The condition of this ineq didn't show $ a^4\\minus{}b^4\\plus{}b^2 \\ge 0$.But if $ a^4\\minus{}b^4\\plus{}b^2 \\ge0$(and others expression) , we can prove ^^\r\n$ (a^2 \\plus{} b^2 \\plus{} c^2)LHS^2 \\ge (a^2 \\plus{} b^2 \\plus{} c^2)^3 \\equal{} 27$\r\n=>$ LHS \\ge 3$\r\nDone" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Construct $ f: \\mathbb{R}\\mapsto \\mathbb{R}$ so that: (i) $ f$ is one-to-one (ii) $ f$ is continous (iii) $ f$ doesn't map closed sets to closed sets. What if we added the constraint (iv) $ f$ is bounded?", "Solution_1": "$ f(x)\\equal{}\\arctan x$. The image of the entire real line (which, of course, is closed) is an open interval. I strongly suspect that sylow_theory meant something else but this is an example for what was literally asked :).", "Solution_2": "[quote=\"fedja\"]$ f(x) \\equal{} \\arctan x$. The image of the entire real line (which, of course, is closed) is an open interval. I strongly suspect that sylow_theory meant something else but this is an example for what was literally asked :).[/quote]\r\nI think I meant something else too. Its a problem from many months ago that I have seen, and couldn't solve it. I am sure I considered arctan in that problem and it didn't work. I just dont remember the exact details. Maybe you know the exact problem. The only thing I remember it was an excercise trying to show that closed sets dont go to closed sets (and bounded sets) (unlike compact sets) but it has the one-to-one condition on it.", "Solution_3": "But if $ f$ is continuous,then it is very obviously that $ f$ map closed sets to closed sets.Have I missed something? :?:", "Solution_4": "While we're at it, can you find a function from $ \\mathbb{R}$ to $ \\mathbb{R}$ that maps open sets to open sets, but is not continuous?", "Solution_5": "[quote=\"zhaobin\"]\nBut if f is continuous,then [...] f map closed sets to closed sets.\n[/quote]\r\n\r\nCareful, here, zhaobin--[i]closed[/i] maps take closed sets to closed sets; \"f is continuous\" means that [i]preimages[/i] of closed sets are closed. The arctan function is a (the?) classic example of a continuous function (check it) that is not closed (as fedja explained above).", "Solution_6": "[quote=\"phils\"]While we're at it, can you find a function from $ \\mathbb{R}$ to $ \\mathbb{R}$ that maps open sets to open sets, but is not continuous?[/quote]\r\nPicard's theorem gives examples like this for $ \\mathbb{C}$, but it seems tricky for $ \\mathbb{R}$", "Solution_7": "[quote=\"RabbitTrails\"][quote=\"zhaobin\"]\nBut if f is continuous,then [...] f map closed sets to closed sets.\n[/quote]\n\nCareful, here, zhaobin--[i]closed[/i] maps take closed sets to closed sets; \"f is continuous\" means that [i]preimages[/i] of closed sets are closed. The arctan function is a (the?) classic example of a continuous function (check it) that is not closed (as fedja explained above).[/quote]\r\nSo does it mean such $ f$ is equivalent to an open and continuous map?", "Solution_8": "[quote=\"zhaobin\"][quote=\"RabbitTrails\"][quote=\"zhaobin\"]\nBut if f is continuous,then [...] f map closed sets to closed sets.\n[/quote]\n\nCareful, here, zhaobin--[i]closed[/i] maps take closed sets to closed sets; \"f is continuous\" means that [i]preimages[/i] of closed sets are closed. The arctan function is a (the?) classic example of a continuous function (check it) that is not closed (as fedja explained above).[/quote]\nSo does it mean such $ f$ is equivalent to an open and continuous map?[/quote]\r\nI don't know what you are asking exactly, but closed, open and continuous are \"independent\" conditions and do not imply each other. Only with injectivity can you start making such claims.", "Solution_9": "Thank you,I misunderstood all thing,I am sorry :blush:", "Solution_10": "heh... I've only been here a couple of days, but I'd have to say that you have nothing to apologize for--if everyone here knew everything, then the forum wouldn't be very useful!", "Solution_11": "[quote=\"phils\"]While we're at it, can you find a function from $ \\mathbb{R}$ to $ \\mathbb{R}$ that maps open sets to open sets, but is not continuous?\n[/quote]\r\nFor an extreme version, how about a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=31493]function[/url] that takes every open interval to the whole real line?" } { "Tag": [ "Vieta", "quadratics", "ratio", "algebra", "geometric sequence", "algebra unsolved" ], "Problem": "for three given real numbers a,b,c consider \r\n\\[s_{n}= a^{n}+b^{n}+c^{n}\\]\r\nsuppose that \r\n\r\n\\[s_{1}= 2\\] \\[s_{2}= 6\\] \\[s_{3}= 14\\]\r\n prove that\r\n\r\n\\[\\left|{s_{n}^{2}-s_{n-1}s_{n+1}}\\right| = 8\\] \r\nfor all integers n>1", "Solution_1": "$\\sigma_{1}=s_{1}=2,\\sigma_{2}=\\frac{s_{1}^{2}-s_{2}}{2}=-1,\\sigma_{3}=\\frac{s_{3}+3s_{1}\\sigma_{2}-s_{1}^{3}}{3}=0.$\r\nTherefore $s_{n}=(1+\\sqrt 2 )^{n}+(1-\\sqrt 2)^{n},s_{n}^{2}-s_{n-1}s_{n+1}=8(-1)^{n}.$", "Solution_2": "Rust, I don't understand how you deduced one of a,b,c was zero.\r\n\r\n\\[2(ab+bc+ca)=(a+b+c)^{2}-(a^{2}+b^{2}+c^{2})\\] \\[a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)\\]\r\nSubstituting the the given values implies 3abc=0\r\n\r\nUsing the idendity above is one way, but could you explain yours more, (I am relatively new to this olympiad maths so if you could that would be much appreciated)\r\n\r\nThanks", "Solution_3": "$a+b+c=2,\\ ab+bc+ca=\\frac{(a+b+c)^{2}-(a^{2}+b^{2}+c^{2})}{2}=\\frac{2^{2}-6}{2}=-1$. Let $abc=k\\ a,\\ b,\\ c$ are the roots of the cubic equation $x^{3}-2x^{2}-x-k=0$ that is to say, $a,\\ b,\\ c$ satisfy the following equations.\r\n\\[a^{3}=2a^{2}+a+k\\ \\cdots (1) \\]\r\n\r\n\\[b^{3}=2b^{2}+b+k\\ \\cdots (2) \\]\r\n\r\n\\[c^{3}=2c^{2}+c+k\\ \\cdots (3) \\]\r\n$(1)+(2)+(3)$ gives $a^{3}+b^{3}+c^{3}+2(a^{2}+b^{2}+c^{2})+a+b+c+3k$, yielding $14=2\\cdot 6+2+3k\\Longleftrightarrow k=abc=0$.\r\n\r\nThus $(1)*a^{n}+(1)*b^{n}+(3)*c^{n}$ gives $a^{n+3}+b^{n+3}+c^{n+3}=2(a^{n+2}+b^{n+2}+c^{n+2})+a^{n+1}+b^{n+1}+c^{n+1}$,\r\nyielding the following reccurence for $s_{n}$; $s_{n+3}=2s_{n+2}+s_{n+1}\\ \\cdots (3)$. Now when $k=0$, by Vieta's formula, $a,\\ b,\\ c$ are the roots of the cubic equation $x^{3}-2x^{2}-x=0\\Longleftrightarrow x(x^{2}-2x-1)=0$. Without loss of generality, we can set $a=0$, so $b,\\ c$ will become the roots of the quadratic equation $x^{2}-2x-1=0$, yielding $b+c=2,\\ bc=-1$. Rust seems to have solved by using the charactrristic equation for the reccurence (3). \r\nBack to the reccurence, we have $s_{n+1}=2s_{n}+s_{n-1}$, so we can obtain the following the relation of matrix.\r\n\\[\\left(\\begin{array}{cc}s_{n}& s_{n-1}\\\\ s_{n+1}&s_{n}\\end{array}\\right)=\\left(\\begin{array}{cc}0 & 1 \\\\ 1 & 2 \\end{array}\\right)\\left(\\begin{array}{cc}s_{n-1}& s_{n-2}\\\\ s_{n}&s_{n-1}\\end{array}\\right) \\]\r\nUse the property of determinant, we have $s_{n}^{2}-s_{n-1}s_{n+1}=(0\\cdot 2-1\\cdot 1)(s_{n-1}^{2}-s_{n-2}s_{n})$, yielding $s_{n}^{2}-s_{n-1}s_{n+1}=-(s_{n-1}^{2}-s_{n-2}s_{n})$, that is to say the sequence $\\{s_{n}^{2}-s_{n-1}s_{n+1}\\}$ is the geometric sequence with common ratio $-1$. Thus we have $s_{n}^{2}-s_{n-1}s_{n+1}=(-1)^{n-1}(s_{1}^{2}-s_{0}s_{2})=-8(-1)^{n-1}$, yielding $|s_{n}^{2}-s_{n-1}s_{n+1}|=8$." } { "Tag": [ "algebra", "polynomial", "inequalities", "complex numbers", "algebra unsolved" ], "Problem": "suppose that f(x) is an irreducible polynomial with integer cofficients .\r\nwe know that f(x) has a root like $ \\alpha$ such that $ |\\alpha| \\succ \\frac {3}{2}.$\r\nprove that $ \\alpha^3 \\plus{} 1$ is not a root of f(x).", "Solution_1": "sorry for mistake.\r\nprove that if $ f$ has a root with norm greater than $ \\frac{3}{2}$\r\nthen for every root like $ t$ ,$ t^3\\plus{}1$ cant be a root of $ f(x)$", "Solution_2": "The root with norm greater than 3/2 may be complex, right?", "Solution_3": "yes.it can be complex", "Solution_4": "Another fun problem. Sumita, I like your style :) \r\n\r\n[hide=\"Solution.\"]\nSo $ f(x) \\in \\mathbb{Z}[x]$ is irreducible, and suppose that $ z \\in \\mathbb{C}$ is a root of $ f(x)$ with the property that $ z^3 \\plus{} 1$ is also a root, that is $ f(z^3 \\plus{} 1) \\equal{} 0$. Now consider the polynomial $ g(x) \\equal{} f(x^3 \\plus{} 1)$, which also has integer coefficients. We have $ g(z) \\equal{} f(z) \\equal{} 0$, but $ f(x)$ being irreducible over $ \\mathbb{Z}$ is also irreducible over $ \\mathbb{Q}$ which implies that it must divide $ g(x)$ over $ \\mathbb{Q}$. In other words, $ f(x)$ is the minimal polynomial of $ z$, so any other polynomial over $ \\mathbb{Q}$ which has $ z$ as a root must be a multiple of it.\n\nSo now we know that there's a polynomial $ q(x) \\in \\mathbb{Q}[x]$ such that $ f(x^3 \\plus{} 1) \\equal{} q(x)f(x)$. It follows that, for [b]any[/b] root $ w$ of $ f(x)$, $ f(w^3 \\plus{} 1) \\equal{} 0$ and so $ w^3 \\plus{} 1$ is also a root. Starting with any given root $ w_0$ we now get a sequence of complex numbers $ w_1 \\equal{} w_0^3 \\plus{} 1$, $ w_2 \\equal{} w_1^3 \\plus{} 1$ etc., and $ w_n$ is a root of $ f(x)$ for all $ n$. \n\nLet's see what happens if you start with a root $ w_0$ satisfying $ |w_0| \\geq \\frac {3}{2}$. Then $ |w_1| \\equal{} |w_0^3 \\plus{} 1| \\geq |w_0^3| \\minus{} 1 \\geq \\frac {19}{8} > |w_0|$, and in general $ |w_{n \\plus{} 1}| \\equal{} |w_n^3 \\plus{} 1| \\geq |w_n^3| \\minus{} 1 \\equal{} |w_n|^3 \\minus{} 1 > |w_n|$, where the last inequality is easily shown to follow from $ |w_n| \\geq \\frac {3}{2}$. Therefore the elements of the sequence $ w_n$ are all distinct, which contradicts the fact that they are all roots of $ f(x)$. QED.\n[/hide]", "Solution_5": "thanks.your soloution was the same as mine." } { "Tag": [ "algebra", "function", "domain", "Ring Theory", "superior algebra", "superior algebra solved" ], "Problem": "Prove that :The ring Z[i] is the ring such that every ideal in it is main ideal?\r\n(I know it very famous but I dont know the solution.Can anyone help me?)", "Solution_1": "[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=75045[/url]\r\nthis is related\r\n\r\nyou probably meant : prove that it is a principal ideal domain" } { "Tag": [ "integration", "logarithms", "real analysis", "real analysis solved" ], "Problem": "Calculate the limes of the infinite series\r\n\r\n\\sumn=1 \\infty n-1(4n)!(2n)!-216-n.", "Solution_1": "By binomial\r\n[tex] \\sum\\limits_{n = 0}^{ + \\infty } {\\frac{{(2n)!}}{{\\left( {n!} \\right)^2 }}} x^n = \\frac{1}{{\\sqrt {1 - 4x} }} [/tex]\r\nThen \r\n[tex] f\\left( x \\right) = \\sum\\limits_{n = 1}^{ + \\infty } {\\frac{{(2n)!}}{{\\left( {n!} \\right)^2 }}} \\frac{{x^{n - 1} }}{{4^n }} = \\frac{1}{{x\\sqrt {1 - x} }} - \\frac{1}{x} = \\frac{1}{{\\sqrt {1 - x} \\left( {1 + \\sqrt {1 - x} } \\right)}} [/tex]\r\nIntegrate f,\r\n[tex] F\\left( x \\right) = \\int_0^x {f\\left( t \\right)dt = \\sum\\limits_{n = 1}^{ + \\infty } {\\frac{{(2n)!}}{{\\left( {n!} \\right)^2 }}} \\frac{{x^n }}{{n4^n }} = 2\\ln \\left( {\\frac{{1 + \\sqrt {1 - x} }}{2}} \\right)} [/tex]\r\nand\r\n[tex] F\\left( 1 \\right) + F\\left( { - 1} \\right) = \\sum\\limits_{n = 1}^{ + \\infty } {\\frac{{\\left( {4n} \\right)!}}{{n\\left( {\\left(2n \\right)!} \\right)^2 16^n }} = 4\\ln 2 - 2\\ln \\left( {1 + \\sqrt 2 } \\right)} [/tex]\r\n\r\nGood ?" } { "Tag": [], "Problem": "Problem $ 1$:\r\n\"Let $ S(n)$ represent the sum of the digits of n. For example, $ S(2007) \\equal{} 2 \\plus{} 0 \\plus{} 0 \\plus{} 7 \\equal{} 9$. Determine the value of the sum $ S(1) \\plus{} S(2) \\plus{} S(3) \\plus{} ... \\plus{} S(2007)$.\"\r\n\r\nProblem $ 2$:\r\n\"Determine the value of the sum $ \\frac {2}{1\\cdot{3}} \\plus{} \\frac {2}{2\\cdot{4}} \\plus{} \\frac {2}{3\\cdot{5}} \\plus{} ... \\plus{} \\frac {2}{2007\\cdot{2009}}$ in lowest terms.\"\r\n\r\nI know the answers, so I'll know if you're correct. However, I don't know how to solve the problems myself.", "Solution_1": "[hide=\"Problem 1\"]\nWe are looking for the sum of the digits from $ 1$ to $ 2007$\n\nFirst add up those which appear in the units position.\nWe have $ 0 \\plus{} 1 \\plus{} 2 \\plus{} 3 \\plus{} 4 \\plus{} 5 \\plus{} 6 \\plus{} 7 \\plus{} 8 \\plus{} 9 \\equal{} 45$ for every $ 10$ and then $ 0 \\plus{} 1 \\plus{} 2 \\plus{} 3 \\plus{} 4 \\plus{} 5 \\plus{} 6 \\plus{} 7 \\equal{} 28$ for the last one. This adds up to $ 45 \\times 200 \\plus{} 28 \\equal{} 9028$\n\nNow consider the 10s digit.\n\nFor each hundred there will be $ 10 \\times 0 \\plus{} 10 \\times 1 \\plus{} 10 \\times 2 \\plus{} \\cdots \\plus{} 10 \\times 9 \\equal{} 10 \\times ( 0 \\plus{} 1 \\plus{} 2 \\plus{} 3 \\plus{} 4 \\plus{} 5 \\plus{} 6 \\plus{} 7 \\plus{} 8 \\plus{} 9 ) \\equal{} 450$ There are no $ 10$s digits for out final seven. This adds up to $ 450 \\times 20 \\equal{} 9000$.\n\nNow consider the 100s digit\n\nFor each thousand there will be $ 100 \\times 0 \\plus{} 100 \\times 1 \\plus{} \\cdots \\plus{} 100 \\times 9 \\equal{} 4500$ There are two thosands giving us a total of $ 9000$ (Which isn't over 9000 :P)\n\nThe 1000 digits. There will be $ 1000$ 1s and $ 8$ 2s. Givin us $ 1016$\n\nTherefore the total is $ 9028 \\plus{} 9500 \\plus{} 9000 \\plus{} 1016 \\equal{} \\boxed{28044}$ I apologise for any arithmetic errors.[/hide]\r\n\r\nCheers for the error catch :)", "Solution_2": "[hide=\"Problem 2 Hint\"]$ \\frac{2}{a(a\\plus{}2)} \\equal{} \\frac{1}{a} \\minus{} \\frac{1}{a\\plus{}2}$[/hide]", "Solution_3": "Good job, SimonM. You are right, except that $ 450 * 20 \\equal{} 9000$, not $ 9500$. So the final answer is $ 28044$.\r\n\r\nAnd thanks a lot, MellowMelon! That hint really helps! I forgot all about partial fractions.", "Solution_4": "Consider all the numbers 000-999.\r\n\r\nClearly we have the same number of 0s,1s,2s...,9s.\r\n\r\nSo there are a total of 1000*3/10=300 0s, 1s, 2s..., 9s respectively.\r\n\r\nSo that gives us a sum of the digits between 000-999 of 300(0+1+2...+9) = 300(45).\r\nNow, consider only 1_ _ _, the last three digits of all numbers between 1000-1999. Clearly, it also must have sum of 300(45).\r\n\r\nNow we must also add the 1000 1s before it so thats 1000(1).\r\n\r\nFor 2000-2007, there are 8 2s and 1+2+3+...+7 for units digits.\r\n\r\nSo we have final sum of $ 300(45) \\plus{} 300(45) \\plus{} 1000(1) \\plus{} 2(8) \\plus{} 1 \\plus{} 2 \\plus{} 3 \\plus{} ... \\plus{} 7 \\equal{} 28044$", "Solution_5": "[hide=\"2\"]\nEach term is of the form $ \\frac{2}{(x\\plus{}1)(x\\minus{}1)}$. Do a PFD. Then each term is of the form $ \\frac{1}{x\\minus{}1} \\minus{} \\frac{1}{x\\plus{}1}$\n\nAfter all the terms cancel out, we're left with $ 1\\plus{}\\frac{1}{2}\\minus{}\\frac{1}{2008}\\minus{}\\frac{1}{2009}\\equal{}\\boxed{\\frac{6047091}{4034072}}$\n[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "A manufacturer of HDTVs must ship at least 100 TVs to its two West Coast warehouses. Each warehouse holds a maximum of 100 TVs. Northern warehouse already has 25 TVs on hand. Southern warehouse already has 20 TVs on hand. It cost $12 to ship a TV to the nothern warehouse and$10 to ship a TV to the southern warehouse. Union rules require that at least 300 workers be hired. Shipping a TV to northern warehouse requires 4 workers, while shipping a TV to southern warehouse requires 2 workers. \r\n\r\nHow many TVs should be shipped to each warehouse to minimize the cost? What is the minimum cost?", "Solution_1": "[hide=\"Answer\"]$50$ TV's to each. The cost is $1100$.[/hide]", "Solution_2": "@ interval\r\n\r\nif you want to denote currency on AoPS, don't just use a single dollar sign - it tells the computer that everything after the dollar sign should be in LaTeX until it sees another dollar sign, so, to type a dollarsign, type this\r\n\r\ndollarsign\\dollarsigndollarsign\r\nand you get \r\n$\\$$", "Solution_3": "[quote=\"sapphyre571\"]@ interval\n\nif you want to denote currency on AoPS, don't just use a single dollar sign - it tells the computer that everything after the dollar sign should be in LaTeX until it sees another dollar sign, so, to type a dollarsign, type this\n\ndollarsign\\dollarsigndollarsign\nand you get \n$\\$$[/quote]\r\n\r\nWhich, when actually typed in, looks like:[code]$ \\$ $[/code] and appears as $\\$$" } { "Tag": [ "real analysis" ], "Problem": "\u03a5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03af\u03c3\u03b9\u03bc\u03b7 \u03ba\u03b1\u03bc\u03c0\u03cd\u03bb\u03b7 $ \\vec x : I \\to \\mathbb R^2$ \u03cc\u03c0\u03bf\u03c5 $ I \\subset \\mathbb R$ \u03b4\u03b9\u03ac\u03c3\u03c4\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03b7 \u03b5\u03b9\u03ba\u03cc\u03bd\u03b1 \u03c4\u03b7\u03c2 \u03bd\u03b1 \u03c0\u03b5\u03c1\u03b9\u03ad\u03c7\u03b5\u03b9 \u03ad\u03bd\u03b1\u03bd \u03c4\u03cc\u03c0\u03bf \u03c4\u03bf\u03c5 $ \\mathbb R^2$?\r\n\u03a6\u03b1\u03bd\u03c4\u03ac\u03b6\u03bf\u03bc\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03b4\u03b5 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 :lol: \u03b1\u03bb\u03bb\u03ac \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03b5\u03af\u03b4\u03b1 \u03c3\u03c4\u03bf\u03bd Rudin \u03c0\u03bf\u03c5 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03cd\u03b1\u03b6\u03b5 \u03bc\u03af\u03b1 \u03c3\u03c5\u03bd\u03b5\u03c7\u03ae \u03ba\u03b1\u03bc\u03c0\u03cd\u03bb\u03b7 \u03c0\u03bf\u03c5 \u03b5\u03af\u03c7\u03b5 \u03c9\u03c2 \u03b5\u03b9\u03ba\u03cc\u03bd\u03b1 \u03c4\u03bf \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03b1\u03af\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03bc\u03bf\u03c5 \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03ae\u03b8\u03b7\u03ba\u03b5 \u03b1\u03c5\u03c4\u03ae \u03b7 \u03b1\u03c0\u03bf\u03c1\u03af\u03b1.\r\n\u0394\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ad\u03c2 \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03b2\u03bb\u03ad\u03c0\u03c9 \u03ba\u03ac\u03c4\u03b9 :blush: .\u039c\u03c0\u03bf\u03c1\u03b5\u03af \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03b5\u03b9;", "Solution_1": "\u0397 \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b7 \u03c3\u03c4\u03bf \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03b8\u03ad\u03c4\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c7\u03b9. \u0394\u03b5 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03c5\u03c0\u03ac\u03c1\u03be\u03b5\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03af\u03c3\u03b9\u03bc\u03b7 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03bf\u03c0\u03bf\u03af\u03b1\u03c2 \u03b7 \u03b5\u03b9\u03ba\u03cc\u03bd\u03b1 \u03bd\u03b1 \u03c0\u03b5\u03c1\u03b9\u03ad\u03c7\u03b5\u03b9 \u03b1\u03bd\u03bf\u03b9\u03ba\u03c4\u03cc. \u0397 \u03b9\u03b4\u03ad\u03b1 \u03c4\u03b7\u03c2 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b5\u03be\u03ae\u03c2. \u0391\u03bd \u03b1\u03c5\u03c4\u03cc \u03ae\u03c4\u03b1\u03bd \u03b4\u03c5\u03bd\u03b1\u03c4\u03cc, \u03c4\u03cc\u03c4\u03b5 \u03b8\u03b1 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03c3\u03b1\u03bc\u03b5 \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $ f: [0,1]\\to \\mathbb{R}^2$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 $ f([0,1])\\supset (a,b)\\times(c,d)$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 $ a,b,c,d$ \u03bc\u03b5 $ a0.$ We look at $ \\int_0^{1-\\epsilon}\\frac1{1+x^n}\\,dx+\\int_{1-\\epsilon}^1\\frac1{1+x^n}\\,dx=A_n+B_n.$\r\n\r\nAs for term $ A_n:$ we do have that $ \\frac1{1+x^n}\\to1$ uniformly on $ [0,1-\\epsilon].$ Uniform convergence on a bounded interval justifies interchanging limit and integral, so we can say that $ \\lim A_n=\\int_0^{1-\\epsilon}1\\,dx=1-\\epsilon.$\r\n\r\nAs for term $ B_n:$ we can say that $ 0<\\frac1{1+x^n}<1,$ so $ 01 = 3 and\r\n\r\nan+1 = \r\n\r\nan/2 if an is even\r\n(an+1983)/2 if an is odd.\r\n\r\nProve the sequence is periodic, and find the (minimum) period.", "Solution_1": "We have 1 <= a(n) < 1983 for all n :\r\n\r\na(n) < 1983 => a(n+1) < 1983 because a(n)/2 and (a(n)+1983)/2 < 1983.\r\n\r\nNow there are only finite integer values between 1 and 1982 so after at most 1987 values a(n) becomes periodic. Period is at most 1983. It's also easy to see that a(n) is always divisible by 3. There are only 660 multiple of 3 between 3 and 1982 so period is at most 661. \r\n\r\nI'm not sure that it's the minimum period, though.", "Solution_2": "[quote=\"belenos\"]Now there are only finite integer values between 1 and 1982 so after at most 1987 values a(n) becomes periodic. Period is at most 1983. [/quote]\r\n\r\nI have a question about the definition of \"periodic\" -- does it mean a sequence of numbers should cycle, or is it enough that one number is hit twice somewhere? \r\n\r\nFor ex: \r\n\r\n(1) 1, 2, 3, 4, 1, 2, 3, 4, ...\r\n(2) 1, 2, 3, 4, 1, 4, 2, 1, 4, 3, 2, 3, 3, 1, ...\r\n\r\nI know for sure (1) is periodic, but is (2)? Even if there are only a finite number of possible values for each term, does that mean that a sequence must cycle, eventually?", "Solution_3": "\"Periodic\" means that the sequence cycles. For sequences where a_{n+1} = f(a_n), it is sufficient to show that the sequence contains the same value twice to show that it eventually becomes periodic. (Why? If a_i = a_j, then a_{i+1} = a_{j+1}, etc..). However, depending on whom you are talking to, \"periodic\" may or may not mean the same as \"eventually becomes periodic\". The sequence 3, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ... (a_{n+1} = {(a_n)/2 if a_n is even, 3a_n + 1 if a_n is odd}) eventually becomes periodic, but we never have a_k = a_1 for some k>1. I don't think that this sequence would generally be considered \"periodic\" - the periodicity begins a distance into the sequence - but it depends on whom you ask.", "Solution_4": "It is (I think) called eventually periodic.", "Solution_5": "Wyra raised two good points !\r\n\r\n\r\nIt's because that a(n+1) = f(a(n)) for some function f that a(n) eventually become peridic. As soon as we get a previously got value, a(n) become periodic because a(n) depends only on the previous value.\r\n\r\nActually I've only proved that a(n) becomes periodic if we go far enough, not that a(n) is periodic 'stricto sensu' i.e. there exists no such as a(n0)=a(1).", "Solution_6": "I think I get it now. :) Thanks!", "Solution_7": "A good thing to try with recursive sequence problems is going backwards. From the formulas, we get a(n-1) = 2 a(n) or a(n-1) = 2 a(n)-1983. Is it uniquely determined? Yes. Since we know that 1 :le: a(n-1) < 1983, and since only one of 2 a(n) and 2 a(n)-1983 is within those bounds, going backwards is uniquely determined also, proving the sequence is [b]strictly periodic[/b].\r\n\r\nSince the numbers are between 1 and 1982, they are equal to their remainders modulo 1983 (none of the terms are divisible by 1983, so 0 is excluded). We see that a(n-1) :equiv: 2 a(n) in modulo 1983. To find the actual period, we need to find n such that a(n+1) = 3, i.e. a(n+1) :equiv: 3 mod 1983. Since a(n+1) :equiv: 2 a(n) :equiv: 2^2 a(n-1) :equiv: ... :equiv: 2^n a(1) :equiv: 3*2^n, we need to find n such that 3*2^n :equiv: 3 mod 1983, i.e. \r\n\r\n2^n :equiv: 1 mod 1983.\r\n\r\nAt this point i got lazy, and used a calculator to determine that the order of 2 in modulo 1983 is 660, so a(661) = 3 and the period is 660.", "Solution_8": "Very good, although I can say two things:\r\n\r\nFirstly, you can use the Euler phi function to calculate that number 661 easier.\r\n\r\nSecondly, you haven't proved that 660 is the minimum period.", "Solution_9": "I found a small problem with my last post. I went from 3*2^n :equiv: 3 mod 1983 to 2^n :equiv: 1 mod 1983. Since 3 is not relatively prime to the mod, we cannot cancel the three that easily. When we cancel the three, we get three possibilities:\r\n\r\n2^n :equiv: 1 mod 1983\r\n2^n :equiv: 662 mod 1983\r\n2^n :equiv: 1323 mod 1983\r\n\r\nThese can be summed up more simply as 2^n :equiv: 1 mod 661.\r\n\r\nSince :phi: (661) = 660 (because 661 is prime), the smallest n that works is a divisor of 660. Thus, \"all\" we have to do is calculate the remainders in mod 660 of 2^(divisors of 660) until we find a 1.\r\n\r\nI'll spare you the details: the smallest n turns out to be 660 anyway.", "Solution_10": "Well, all you have to do is prove that 2^330 is congruent to -1, that will prove it.", "Solution_11": "Are you sure? What if it's 1(mod 220) and -1(mod 330)? Then 220 is a smaller period than 660.", "Solution_12": "True. So there are a few values we have to check (660/each prime factor of 660)." } { "Tag": [ "geometry", "perpendicular bisector", "projective geometry", "geometry proposed" ], "Problem": "Let AB and CD be distinc parallel chords of a circle and let P be a point on theire com mon perpendicular bisector. The lines AP and CP meet the crcle again at Q and R respectively.If S is the intersection of the lines BR and QD, Prove that PS is rarallel to AB.", "Solution_1": "Let P be any point not on the given circle (otherwise the points P, Q, R, S would coincide and the line PS would be undefined), not necessarily on the common perpendicular bisector of the parallel chords $AB \\parallel CD$. By Pascal's theorem, the intersections $M \\equiv AB \\cap CD,$ $P \\equiv AQ \\cap CR,$ $S \\equiv BR \\cap DQ$ of the chord pairs (AB, CD), (AQ, CR), (BR, DQ) are collinear. Since the lines $AB \\parallel CD$ are parallel, their intersection M is at infinity, equivalent to the direction of these 2 parallel lines. Since the line PS passes through M, it has this direction as well, hence, $PS \\parallel AB$." } { "Tag": [ "geometry", "perimeter", "floor function", "trigonometry", "circumcircle", "trig identities", "Law of Cosines" ], "Problem": "In convex quadrilateral $ABCD$, $\\angle A \\cong \\angle C$, $AB = CD = 180$, and $AD \\neq BC$. The perimeter of $ABCD$ is 640. Find $\\lfloor 1000 \\cos A \\rfloor$. (The notation $\\lfloor x \\rfloor$ means the greatest integer that is less than or equal to $x$.)", "Solution_1": "[hide]\n\nWe use Law of Cosines in triangles ABD and CBD to find \n\n$BD^2=180^2+AD^2-2*180*AD\\cos A$\n\n$BD^2=180^2+BC^2-2*180*BC\\cos C$\n\nDefine $\\cos A=\\cos C=x$. We can set these two equations equal and simplify to get\n\n$AD^2-360*AD*x=BC^2-360*BC*x\\Rightarrow 360x(BC-AD)=BC^2-AD^2$\n\nFactoring the left side and dividing out $(BC-AD)$, we get $360x=BC+AD$. Since the perimeter of $ABCD$ is 640 and the other two sides are 180, the sum of $BC$ and $AD$ is $640-2(180)=280$. So we get $360x=280$, and thus \n\n$x=\\frac{280}{360}=\\frac{7}{9}=.777777...$\n\n$\\lfloor 1000 \\cos A \\rfloor$ is thus 777. [/hide]", "Solution_2": "Sorry for the revive, but can anyone give me an example of a diagram when $ AD \\neq BC $. \n\nFrom what I understand of it, the circumradius of triangle $ ABD $ has radius $ \\frac{BD}{\\sin A} $ and the circumradius of triangle $ CBD $ has radius $ \\frac{BC}{\\sin C} $. Therefore the circumradii are equal in length. \n\nFrom here, we can consider several possibilities:\n\n1. $ ABD $ and $ BCD $ have different circumcircles. In that case, angle $ ADB $ = angle $ CBD $. By $ AAS $ congruence, the triangles are congruent and so $ AD = BC $.\n\n2. $ ABD $ and $ ACD $ have same circumcircles. \n\nSubcase 1: $ A $ and $ C $ lie on opposite sides of line $ BD $. The only way angle $ A $ = angle $ C $ is if they are both right angles. Again we have congruent triangles and so $ AD = BC $.\n\nSubcase 2: $ A $ and $ C $ lie on the same side of line $ BD $. This quadrilateral is complex. Contradicts definition of convex. But it can be seen that even in this case $ AD = BC $.\n\nDid I do anything wrong here, and am I missing a case?", "Solution_3": "This was wayyy too easy for a #12. Agreement?", "Solution_4": "[hide=sketch]By LoC on BD in both triangles, we rearrange this to (AD^2-BC^2)=360(AD-BC)cosA, which is the same as 280=640-360=AD+BC=360cosA, which yields that floor[1000cosA]=$\\boxed{777.}$[/hide]\n[b]Remark.[/b] Hmm... The answer is the exact same as aime 1996/15 (yes i am doing problems now) and has the same form of floor[1000something]." } { "Tag": [ "Asymptote" ], "Problem": "Is there a link to a full PDF guide to Asymptote?", "Solution_1": "Everything you could every possibly want about Asymptote is found on their [url=http://asymptote.sourceforge.net/]home page[/url]. What you're looking for is probably [url=http://asymptote.sourceforge.net/asymptote.pdf]this[/url].", "Solution_2": "You actually already have a manual on your computer in your Program Files/Asymptote Folder that's the same as the PDF link ilp gave.\r\n\r\nFor package documentations, visit the respective package's website.\r\n\r\nI'm afraid there is no complete documentation of the basics and every package all in one document." } { "Tag": [], "Problem": "How much greater than $\\sqrt{16+9}$ is $\\sqrt{16}+\\sqrt9$?", "Solution_1": "[quote=\"236factorial\"]How much greater than $\\sqrt{16+9}$ is $\\sqrt{16}+\\sqrt9$?[/quote]\r\n\r\n[hide]$\\sqrt{16+9}=\\sqrt{25}=5$\n\n$\\sqrt{16}+\\sqrt{9}=4+3=7$\n\nIt's 2 greater.[/hide]", "Solution_2": "[hide]2[/hide]", "Solution_3": "[hide]\n$\\sqrt{16+9}=\\sqrt{25}=5\\\\\\\\\\sqrt{16}+\\sqrt{9}=4+3=7\\\\\\\\7-5=\\boxed{2}$[/hide]", "Solution_4": "[hide]2[/hide]", "Solution_5": "[quote=\"236factorial\"]How much greater than $\\sqrt{16+9}$ is $\\sqrt{16}+\\sqrt9$?[/quote]\r\n[hide]16+9=25, $\\sqrt{25}=5$ and $\\sqrt{16}+\\sqrt{9}=4+3=7$ so the answer is $7-5=\\framebox{2}$[/hide]", "Solution_6": "[quote=\"236factorial\"]How much greater than $\\sqrt{16+9}$ is $\\sqrt{16}+\\sqrt9$?[/quote]\r\n\r\n[hide]$\\sqrt{16+9} = \\sqrt{25} = 5$\n\n$\\sqrt{16} + \\sqrt{9} = 4 + 3 = 7$.\n\n$\\text{ EDIT: Difference = 2 }$[/hide]\r\n\r\nFixed. :blush:", "Solution_7": "[hide][quote=\"hwenterprise\"][quote=\"236factorial\"]How much greater than $\\sqrt{16+9}$ is $\\sqrt{16}+\\sqrt9$?[/quote]\n\n[hide]$\\sqrt{16+9} = \\sqrt{25} = 5$\n\n$\\sqrt{16} + \\sqrt{9} = 4 + 3 = 7$.\n\n$\\text{ Difference = -2 (Wait, that sounds wrong...)}$[/hide][/quote]\n\nYou did it back wards you need to find How much greater $\\sqrt{16}+\\sqrt9$ is not how much greater $\\sqrt{16+9}$ is...\nbtw the answer is 2...[/hide]" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Point $E,F,G,H$lie on sides $AB,BC,CD,DA$ of a convex quadrilateral $ABCD$ such that $\\frac{AE*BF*CG*DH}{EB*FC*GD*HA}=1$\r\nPoints $A,B,C,D$ lie on sides $H_{1}E_{1},E_{1}F_{1},F_{1}G_{1},G_{1}H_{1}$ a convex quadrilateral $H_{1}E_{1}F_{1}G_{1}$ such that $E_{1}F_{1}//EF,H_{1}E_{1}//HE,F_{1}G_{1}//FG,G_{1}H_{1}//GH$\r\nProve that $\\frac{E_{1}A}{AH_{1}}=\\frac{F_{1}C}{CG_{1}}$", "Solution_1": "Still no one? I have a solution with a little calculaitng.If no one can solve it in a week,I 'll post mine. :)", "Solution_2": "I think I can solve this problem with using threeliner corrdinates but I need to find time to think about it a litter longer than I did :blush: .So,wait a little more and don't post your solution.I will try to find free time and think on this problem,I am rather confidence that I will manage to solve it :P .", "Solution_3": "I will wait for you.\r\nBy the way ,Could you tell me what is \"threeliner corrdinates \"?I don't know more in Geometry. :blush: Thanks.", "Solution_4": "[quote=\"Hawk Tiger\"]Point $E,F,G,H$lie on sides $AB,BC,CD,DA$ of a convex quadrilateral $ABCD$ such that $\\frac{AE*BF*CG*DH}{EB*FC*GD*HA}=1$\nPoints $A,B,C,D$ lie on sides $H_{1}E_{1},E_{1}F_{1},F_{1}G_{1},G_{1}H_{1}$ a convex quadrilateral $H_{1}E_{1}F_{1}G_{1}$ such that $E_{1}F_{1}//EF,H_{1}E_{1}//HE,F_{1}G_{1}//FG,G_{1}H_{1}//GH$\nProve that $\\frac{E_{1}A}{AH_{1}}=\\frac{F_{1}C}{CG_{1}}$[/quote]\r\n\r\nSo :) ,yesterday was my last examination and now I am ready to start practicing math :lol: .At first here is my solution for this problem as I have promesed you dear Hawk :wink: .\r\n\r\n[color=blue]Solution:[/color]\r\n\r\nFrom the condition of the problem and Menelaus' theorem it is obvious that $FG$ and $EH$ intersect on line $BD$;let's denote the intersection point $X$.\r\nAlso let's note\r\n\r\n\\[P\\in{{F'G'}\\cap{E'H'}}\\]\r\n\\[M\\in{{BD}\\cap{F'G'}}\\]\r\n\\[N\\in{{BD}\\cap{E'H'}}\\]\r\n\\[X'\\in{{GX}\\cap{G'H'}}\\]\r\n\\[X''\\in{{HX}\\cap{G'H'}}\\]\r\n\\[O\\in{{BD}\\cap{AC}}\\]\r\n\r\nNow again from Menelaus' theorem\r\n\r\n\\[\\frac{CO}{AO}=\\frac{CM}{MP}\\frac{PN}{NA}\\]\r\n\r\nand \r\n\r\n\\[\\frac{G'D}{DH'}=\\frac{G'M}{MP}\\frac{PN}{NH'}\\]\r\n\r\nOn the other hand,we have that $GX\\parallel CM$ and $HX\\parallel AN$ =>\r\n\\[\\frac{MG'}{MC}=\\frac{XX'}{XG}\\]\r\n\\[\\frac{NH'}{NA}=\\frac{XX''}{XH}\\]\r\nBut we also have that $X'X''\\parallel GH$=>\r\n\\[\\frac{XX'}{XG}=\\frac{XX''}{XH}\\]\r\n=>\r\n\\[\\frac{MG'}{MC}=\\frac{NH'}{NA}\\]\r\n=>\r\n\\[\\frac{CO}{AO}=\\frac{CM}{MP}\\frac{PN}{NA}=\\frac{G'M}{MP}\\frac{PN}{NH'}=\\frac{G'D}{DH'}\\]\r\n=>\r\n\\[\\frac{CO}{AO}=\\frac{G'D}{DH'}\\]\r\n\r\nIn the end, using a very famous fact that if $\\frac{CO}{AO}=\\frac{G'D}{DH'}$ then $\\frac{F'B}{BE'}=\\frac{G'D}{DH'}$ we get the required property.You can similary prove that $\\frac{E'A}{AH'}=\\frac{F'C}{CG'}$.\r\nP.S. Just one more thing,in the solution above I replaced $E_{1}F_{1}G_{1}H_{1}$ by$E'F'G'H'$ :blush: ." } { "Tag": [ "inequalities" ], "Problem": "Let $ a$ and $ b$ are positive numbers such that $ a^{25}\\plus{}b^{25}\\equal{} 2.$ Prove that:\r\n\\[ \\frac{a^3}{b^2}\\plus{}\\frac{b^3}{a^2}\\geq2\\]", "Solution_1": "AM-GM \r\n $ \\frac {a^3}{b^2} \\plus{} \\frac {b^3}{a^2} \\ge 2\\sqrt {ab} (*)$\r\n But we know that $ \\sqrt {ab}\\le \\frac {a \\plus{} b}{2}\\le \\sqrt [25]{\\frac {a^{25} \\plus{} b^{25}}{2}} \\equal{} 1$\r\n Thus $ (*)$ is equivalent to $ \\frac {a^3}{b^2} \\plus{} \\frac {b^3}{a^2} \\ge 2$ Equality holds if and only if a=b=1. Q.E.D", "Solution_2": "Well,\r\n$ \\frac{a^3}{b^2} \\plus{} \\frac{b^3}{a^2}\\geq2\\sqrt {ab}$\r\nand we also know that :\r\n$ 1\\geq\\sqrt {ab}$.\r\nBut that doesn't mean:\r\n$ \\frac{a^3}{b^2} \\plus{} \\frac{b^3}{a^2}\\geq2$\r\nLook at this example:\r\n$ 4\\geq2$\r\n$ 6\\geq2$\r\nso,\r\n$ 4\\geq6$?", "Solution_3": "wangsacl, I agree with you! \r\nWith your permission I'll ask him in another way. \r\n[quote=\"broniran\"]AM-GM \n $ \\frac {a^3}{b^2} \\plus{} \\frac {b^3}{a^2} \\ge 2\\sqrt {ab} (*)$\n But we know that $ \\sqrt {ab}\\le \\frac {a \\plus{} b}{2}\\le \\sqrt [25]{\\frac {a^{25} \\plus{} b^{25}}{2}} \\equal{} 1$\n Thus $ (*)$ is equivalent to $ \\frac {a^3}{b^2} \\plus{} \\frac {b^3}{a^2} \\ge 2$ [/quote]\r\nWhy $ (*)$ is equivalent to $ \\frac {a^3}{b^2} \\plus{} \\frac {b^3}{a^2} \\ge 2$ $ ?$\r\nExplain please. Thank you!", "Solution_4": "Because $ \\sqrt {ab}\\le 1$ not $ \\sqrt {ab} \\ge 1$ , wangsacl. And because $ \\sqrt {ab}\\le 1$ , so $ 2 \\ge 2\\sqrt{ab}$. Perhaps, I haven't used the most appropriate word \"equvalent\".I should have used \" obtain\".", "Solution_5": "But that's exactly why you're wrong; you showed that it was at least something that is SMALLER than 2.", "Solution_6": "Oh, man :blush: I've made a great mistake :blush:", "Solution_7": "$ a^5\\plus{}b^5\\ge 2a^2b^2\\iff (a^5\\plus{}b^5)^{25}\\ge 2^{24}a^{50}b^{50}\\cdot (a^{25}\\plus{}b^{25})$\r\n\r\nIf we homogenize and let $ y\\equal{}x^5$, we must equivalently prove\r\n\r\n$ (y\\plus{}1)^{25}\\minus{}2^{24}\\cdot y^{10}(y^5\\plus{}1)\\ge 0$\r\n\r\nwhich is equivalent to: \r\n\r\n$ (\\minus{}1 \\plus{} y)^4 (1 \\plus{} y) F(y)\\ge 0$\r\n\r\nwhere [code]F(y)=(1 + 28 y + 382 y^2 + 3388 y^3 + 21997 y^4 + \n 111664 y^5 + 462440 y^6 + 1610480 y^7 + 4827410 y^8 + \n 12692360 y^9 + 12968500 y^10 + 12692360 y^11 + 4827410 y^12 + \n 1610480 y^13 + 462440 y^14 + 111664 y^15 + 21997 y^16 + \n 3388 y^17 + 382 y^18 + 28 y^19 + y^20)[/code]\r\n\r\nI do not know a better way for such inequalities...they are very sharp!", "Solution_8": "Yeah, I tried applying Mildorf's lemma, and this inequality was sharper... it reduced to $ a^6\\plus{}a^5b\\plus{}ab^5\\plus{}b^6 \\ge 13a^4b^2 \\minus{} 22a^3b^3 \\plus{} 13a^2b^4$ which unfortunately is false...", "Solution_9": "[quote=\"Altheman\"]we must equivalently prove\n\n$ (y + 1)^{25} - 2^{24}\\cdot y^{10}(y^5 + 1)\\ge 0$\n\n[/quote]\r\nI think, this inequality easier to prove like this:\r\n$ (y + 1)^{25} - 2^{24}\\cdot y^{10}(y^5 + 1)\\ge 0\\Leftrightarrow$\r\n$ \\Leftrightarrow(y^2+2y+1)^{12}-2^{24}y^{10}(y^4-y^3+y^2-y+1)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow(u+2)^{12}-2^{24}(u^2-u-1)\\geq0,$ where $ u=y+\\frac{1}{y}.$\r\nLet $ f(u)=12\\ln(u+2)-\\ln(u^2-u-1)-24\\ln2.$\r\nHence, $ f'(u)=\\frac{12}{u+2}-\\frac{2u-1}{u^2-u-1}=\\frac{5(u-2)(2u+1)}{(u+2)(u^2-u-1)}.$\r\nId est, $ f(u)\\geq f(2)=0.$\r\nWhat about the following:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=219238" } { "Tag": [ "geometry", "3D geometry", "prism" ], "Problem": "Three faces of a rectangular prism have the areas shown. What is the number of cubic centimeters in the volume of the rectangular prism?\n\n[asy]import three; size(101);\nreal x = 10, y = 7, z = 3;\n currentprojection = orthographic(1/2,-1/2,1/2); draw((0,0,0)--(x,0,0)--(x,0,z)--(0,0,z)--cycle3);\ndraw((x,0,0)--(x,y,0)--(x,y,z)--(x,0,z)); draw((0,0,z)--(0,y,z)--(x,y,z));\nlabel(rotate(-30)*\"$30\\mbox{ cm}^2$\",(x/2,0,z/2),fontsize(8pt));\nlabel(\"$70\\mbox{ cm}^2$\",(x/2,y/2,z),fontsize(8pt));\nlabel(rotate(30)*\"$21\\mbox{ cm}^2$\",(x,y/2,z/2),fontsize(8pt));[/asy]", "Solution_1": "Instead of finding the side lengths individually, express each face area as a product of two dimensions. So $ ab\\equal{}70, bc\\equal{}21, ac\\equal{}30$. Now we can multiply all of them together to get $ a^2b^2c^2\\equal{}44100$. But we have to remember what we want! The value of $ abc$. So take the square root of both sides to get $ abc\\equal{}\\boxed{210}$." } { "Tag": [ "calculus", "integration", "function", "logarithms", "complex analysis", "complex analysis unsolved" ], "Problem": "Let $H$ be a path composed by the $3$ lines $[i+\\infty,i-1],[i-1,-i-1],[-i-1,-i-\\infty]$. For $\\mathbb{C}\\setminus [0,\\infty]$ let $\\arg{z}$ be given by $0 < \\arg{z}< 2\\pi$ and let $\\log{z}= \\log{|z|}+i\\arg{z}$ be the correpsonding branch of the logarithm and $u^{z}= e^{z\\log{u}}$.Now show that\r\n$\\zeta(s) = \\frac{1}{\\Gamma(s)(e^{2\\pi i s}-1)}\\int_{H}\\frac{w^{s-1}}{e^{w}-1}dw$", "Solution_1": "Isnt there an identity?\r\n\r\n$\\zeta(z)\\Gamma(z)=\\int_{0}^{\\infty}\\frac{t^{z-1}}{e^{z}-1}dt$", "Solution_2": "[quote=\"sylow_theory\"]Isnt there an identity?\n\n$\\zeta(z)\\Gamma(z)=\\int_{0}^{\\infty}\\frac{t^{z-1}}{e^{z}-1}dt$[/quote]\r\n\r\nYes, that's true." } { "Tag": [ "ratio", "geometric sequence" ], "Problem": "What is the value of the sum $ 2^{\\minus{}1}\\plus{}2^{\\minus{}2}\\plus{}2^{\\minus{}3}\\plus{} \\ldots \\plus{}2^{\\minus{}9} \\plus{} 2^{\\minus{}10}$? Express your answer as a common fraction.", "Solution_1": "hello, after a short computation we get\r\n$ \\sum_{i\\equal{}1}^{10}2^{\\minus{}i}\\equal{}\\frac{1023}{1024}$\r\nSonnhard.", "Solution_2": "In these types of problems, the numerator will be 1 less than the denominator. Thus, the denominator for this solution will be 1024, and the numerator is 1023. Therefore, $ {%Error. \"division\" is a bad command.\n{1023}/{1024}} $", "Solution_3": "[hide=\"Geometric sequence solution\"]\n\nThe sum of geometric sequences is as follows:\n\n$S_n=a\\left({\\frac{1-r^n}{1-r}}\\right)$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.\n\nPlugging things in:\n\n$S_{10}={\\frac{1}{2}}\\left({\\frac{1-\\left(\\frac{1}{2}\\right)^{10}}{1-{\\frac{1}{2}}}}\\right)$\n\nSimplifying a lot:\n\n$S_{10}=1-{\\frac{1}{1024}} \\implies \\boxed{{\\frac{1023}{1024}}}$[/hide]", "Solution_4": "[quote=\"HYP135peppers\"]In these types of problems, the numerator will be 1 less than the denominator. Thus, the denominator for this solution will be 1024, and the numerator is 1023. Therefore, $ {%Error. \"division\" is a bad command.\n{1023}/{1024}} $[/quote]\n\n[hide=\"Proof\"]\nWe wish to find the value of $2^{-1}+2^{-2}+\\cdots+2^{-n}$.\n\n[quote=\"newb\"]\nThe sum of geometric sequences is as follows:\n$S_n=a\\left({\\frac{1-r^n}{1-r}}\\right)$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.\n[/quote]\n\nPluggin in values:\n\n\\[ S_n={\\frac{1}{2}}\\left({\\frac{1-\\left(\\frac{1}{2}\\right)^{n}}{1-{\\frac{1}{2}}}}\\right) .\\]\n\nThis becomes \\[S_n=\\dfrac{\\frac{2^n-1}{2^n}}{2\\cdot\\frac{1}{2}}.\\] The denominator of the fraction becomes $1$, so we are left with \n\\[\\boxed{S_n=\\dfrac{2^n-1}{2^n}}.\\]\n[/hide]" } { "Tag": [ "ratio" ], "Problem": "Nats 1989 question - I think answer they give is wrong, but can't figure out how to do. Please offer a detailed solution. Thanks.\r\n\r\nQuestion: If a typist produces 4 typed pages for every 5 handwritten pages and can produce 14 pages per hour, how many hrs would it take to type 105 hand pages?", "Solution_1": "I have some questions. \r\n\r\n1. What answer does it give?\r\n2. 14 what kind of pages per hour?", "Solution_2": "Let t = typed and h = handwritten\r\n\r\nWe have a ratio of $ 4t: 5h$\r\n\r\nSo if we have 105h,\r\n\r\n$ \\frac{4t}{5h}\\equal{}\\frac{xt}{105h}$\r\n\r\n$ x \\equal{} 84t$\r\n\r\nHence the typist has typed 105+84=189 pages in total, which would take $ \\frac{189}{14}\\equal{} 13.5$ hours.", "Solution_3": "to identity and others, yes, I got 13.5 hrs also, but the official ans. key said 6 - but I just went back and realized question in fact said [b]typed[/b] 14 pages, not produced - which changes how to do the problem and does, in fact, yield the ans of 6 -- (sorry about that - but btw, just saying 14 pages makes for a more challenging question, don't you think?) thanks to those whoresponded" } { "Tag": [ "analytic geometry", "search", "combinatorics proposed", "combinatorics" ], "Problem": "Let $ d > 0$ be a [i]rational[/i] number, and let $ G$ be the infinite graph with the vertices as points with rational coordinates. Two vertices of $ G$ are adjacent if and only if the distance between them is $ d$. Prove that $ G$ is bipartite. If $ d$ is irational does $ G$ still remain bipartite?\r\n\r\nIt is immediate to see that if $ G$ has its vertices in points with [i]real[/i] coordinates, then the $ G$ is not bipartite (This is a simple consequence of the fact that no matter how we colour the plane with two colours, there exists a monochromatic equilateral triangle).", "Solution_1": "For this problem and related questions about this graph\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?search_id=501047893&t=2635\r\n\r\nPierre.", "Solution_2": "Nice! I was in fact waiting unconsciously a reference from you. Thanks." } { "Tag": [ "function", "logarithms", "integration", "calculus", "calculus computations" ], "Problem": "Compute the value of each of the following series (preferably without using power series):\r\n\r\n$ S_{1}\\equal{}\\sum_{n\\equal{}1}^{\\plus{}\\infty}\\frac{1}{n(2n\\plus{}3)}$\r\n\r\n$ S_{2}\\equal{}\\sum_{n\\equal{}1}^{\\plus{}\\infty}\\frac{2n^{2}\\plus{}5n\\plus{}1}{(n\\plus{}2)!}$\r\n\r\n$ S_{3}\\equal{}\\sum_{n\\equal{}1}^{\\plus{}\\infty}\\frac{9n\\minus{}4}{3n(3n\\minus{}1)(3n\\minus{}2)}$\r\n\r\n$ S_{4}\\equal{}\\sum_{n\\equal{}1}^{\\plus{}\\infty}\\frac{1}{16n^{2}\\minus{}1}$\r\n\r\n$ S_{5}\\equal{}\\sum_{n\\equal{}1}^{\\plus{}\\infty}\\frac{(\\minus{}1)^{n\\minus{}1}}{n(3n\\minus{}1)}$\r\n\r\n$ S_{6}\\equal{}\\sum_{n\\equal{}1}^{\\plus{}\\infty}(\\minus{}1)^{n\\minus{}1}\\frac{n}{9n^{2}\\minus{}1}$", "Solution_1": "i am too tired to work out everyone of them in detail now (just finished writing tomorrow's lecture on weierstrass' and jacobi's elliptic functions a few minutes ago), so i will show the methods and the answer should then follow, OK ? :| \r\n\r\n$ S_{1}\\; \\equal{}\\;\\frac{2}{3}\\sum_{n \\equal{} 1}^{\\infty}\\;\\left(\\frac{1}{2n}\\minus{}\\frac{1}{2n\\plus{}3}\\right)\\; \\equal{}\\;\\frac{2}{3}\\left(1\\plus{}\\frac{1}{3}\\minus{}\\left(1\\minus{}\\frac{1}{2}\\plus{}\\frac{1}{3}\\minus{}\\frac{1}{4}\\plus{}\\frac{1}{5}\\minus{}\\;\\ldots\\right)\\right)$\r\n$ \\equal{}\\;\\frac{2}{3}\\left(\\frac{4}{3}\\minus{}\\ln 2\\right)$\r\n\r\n$ S_{2}\\; \\equal{}\\;\\sum_{n \\equal{} 1}^{\\infty}\\;\\frac{\\left[2(n\\plus{}1)(n\\plus{}2)\\minus{}(n\\plus{}2)\\minus{}1\\right]}{(n\\plus{}2)!}$\r\n$ \\equal{}\\;2(e\\minus{}1)\\minus{}(e\\minus{}1\\minus{}1)\\minus{}\\left(e\\minus{}1\\minus{}1\\minus{}\\frac{1}{2}\\right)$\r\n$ \\equal{}\\;\\frac{5}{2}$\r\n\r\n$ S_{3}\\; \\equal{}\\;\\sum_{n \\equal{} 1}^{\\infty}\\;\\left(\\frac{1}{3n\\minus{}2}\\plus{}\\frac{1}{3n\\minus{}1}\\minus{}\\frac{2}{3n}\\right)$\r\n$ \\equal{}\\;\\sum_{n \\equal{} 1}^{\\infty}\\;\\left(\\frac{1}{3n\\minus{}2}\\minus{}\\frac{1}{3n}\\right)\\;\\plus{}\\;\\sum_{n \\equal{} 1}^{\\infty}\\;\\left(\\frac{1}{3n\\minus{}1}\\minus{}\\frac{1}{3n}\\right)$\r\n$ \\equal{}\\;\\left(1\\minus{}\\frac{1}{3}\\plus{}\\frac{1}{4}\\minus{}\\frac{1}{6}\\plus{}\\frac{1}{7}\\minus{}\\frac{1}{9}\\plus{}\\;\\ldots\\right)\\;\\plus{}\\;\\left(\\frac{1}{2}\\minus{}\\frac{1}{3}\\plus{}\\frac{1}{5}\\minus{}\\frac{1}{6}\\plus{}\\frac{1}{8}\\minus{}\\frac{1}{9}\\plus{}\\;\\ldots\\right)$\r\n$ \\equal{}\\;\\left(\\int_{0}^{1}\\;\\frac{1\\minus{}x^{2}}{1\\minus{}x^{3}}\\;\\textbf dx\\right)\\;\\plus{}\\;\\left(\\int_{0}^{1}\\;\\frac{x\\minus{}x^{2}}{1\\minus{}x^{3}}\\;\\textbf dx\\right)$\r\n$ \\equal{}\\;\\left(\\int_{0}^{1}\\;\\frac{1\\plus{}x}{1\\plus{}x\\plus{}x^{2}}\\;\\textbf dx\\right)\\;\\plus{}\\;\\left(\\int_{0}^{1}\\;\\frac{x}{1\\plus{}x\\plus{}x^{2}}\\;\\textbf dx\\right)$\r\n$ \\equal{}\\;\\left(\\frac{\\pi}{6\\sqrt 3}\\plus{}\\frac{1}{2}\\ln 3\\right)\\;\\plus{}\\;\\left(\\frac{1}{2}\\ln 3\\minus{}\\frac{\\pi}{6\\sqrt 3}\\right)$\r\n$ \\equal{}\\;\\ln 3$\r\n\r\n$ S_{4}\\; \\equal{}\\;\\frac{1}{2}\\sum_{n \\equal{} 1}^{\\infty}\\;\\left(\\frac{1}{4n\\minus{}1}\\minus{}\\frac{1}{4n\\plus{}1}\\right)\\; \\equal{}\\;\\frac{1}{2}\\left(\\frac{1}{3}\\minus{}\\frac{1}{5}\\plus{}\\frac{1}{7}\\minus{}\\frac{1}{9}\\plus{}\\;\\ldots\\right)$\r\n$ \\equal{}\\;\\frac{1}{2}\\left(1\\minus{}\\frac{\\pi}{4}\\right)$\r\n\r\n$ S_{5}\\; \\equal{}\\;3\\sum_{n \\equal{} 1}^{\\infty}\\;(\\minus{}1)^{n\\minus{}1}\\left(\\frac{1}{3n\\minus{}1}\\minus{}\\frac{1}{3n}\\right)$\r\n$ \\equal{}\\;3\\left\\{\\left(\\frac{1}{2}\\minus{}\\frac{1}{5}\\plus{}\\frac{1}{8}\\minus{}\\;\\ldots\\right)\\minus{}\\left(\\frac{1}{3}\\minus{}\\frac{1}{6}\\plus{}\\frac{1}{9}\\minus{}\\;\\ldots\\right)\\right\\}$\r\n$ \\equal{}\\;3\\left\\{\\left(\\int_{0}^{1}\\;\\frac{x}{1\\plus{}x^{3}}\\;\\textbf dx\\right)\\minus{}\\left(\\frac{1}{3}\\ln 2\\right)\\right\\}$\r\n$ \\equal{}\\;3\\left\\{\\left(\\frac{\\pi}{3\\sqrt 3}\\minus{}\\frac{1}{3}\\ln 2\\right)\\minus{}\\left(\\frac{1}{3}\\ln 2\\right)\\right\\}$\r\n$ \\equal{}\\;3\\left\\{\\frac{\\pi}{3\\sqrt 3}\\minus{}\\frac{2}{3}\\ln 2\\right\\}$\r\n\r\n$ S_{6}\\; \\equal{}\\;\\frac{1}{6}\\sum_{n \\equal{} 1}^{\\infty}\\;(\\minus{}1)^{n\\minus{}1}\\left(\\frac{1}{3n\\minus{}1}\\plus{}\\frac{1}{3n\\plus{}1}\\right)$\r\n$ \\equal{}\\;\\frac{1}{6}\\left\\{\\left(\\frac{1}{2}\\minus{}\\frac{1}{5}\\plus{}\\frac{1}{8}\\minus{}\\;\\ldots\\right)\\plus{}\\left(\\frac{1}{4}\\minus{}\\frac{1}{7}\\plus{}\\frac{1}{10}\\minus{}\\;\\ldots\\right)\\right\\}$\r\n$ \\equal{}\\;\\frac{1}{6}\\left\\{\\left(\\int_{0}^{1}\\;\\frac{x}{1\\plus{}x^{3}}\\;\\textbf dx\\right)\\plus{}\\left(1\\minus{}\\int_{0}^{1}\\;\\frac{\\textbf dx}{1\\plus{}x^{3}}\\right)\\right\\}$\r\n$ \\equal{}\\;\\frac{1}{6}\\left\\{\\left(\\frac{\\pi}{3\\sqrt 3}\\minus{}\\frac{1}{3}\\ln 2\\right)\\plus{}\\left(1\\minus{}\\frac{1}{3}\\ln 2\\minus{}\\frac{\\pi}{3\\sqrt 3}\\right)\\right\\}$\r\n$ \\equal{}\\;\\frac{1}{6}\\left\\{1\\minus{}\\frac{2}{3}\\ln 2\\right\\}$" } { "Tag": [ "trigonometry", "calculus", "calculus computations" ], "Problem": "If f(x) =x+tan(x). If f is inverse of g. What is g\u2019(x).", "Solution_1": "Let $ f(x)\\equal{}y\\equal{}x\\plus{}\\tan{x}$.\r\n\r\n$ \\frac{dy}{dx}\\equal{}1\\plus{}\\sec^2{x}$, so $ g'(y)\\equal{}\\frac{1}{\\frac{dy}{dx}}\\equal{}\\frac{1}{1\\plus{}\\sec^2{x}}$. Where $ y\\equal{}x\\plus{}\\tan{x}$. \r\n\r\nWe cannot write more simpler, because $ y\\equal{}x\\plus{}\\tan{x}$ can't be written by the form of $ x\\equal{}h(y)$." } { "Tag": [ "Support", "IMO", "IMO 2008" ], "Problem": "Any guesses or predictions who will be the top-performers at IMO'2008?\r\n\r\n* There are about 10 participants who already got gold at IMO'2007.\r\n\r\n* South-Korea has a 14-years old child prodigy, Soo Hong Lee.\r\nhttp://www.imo-official.org/year_reg_team.aspx?code=KOR\r\nhttp://www.imo-official.org/participant_r.aspx?id=15628\r\nLast year, Soo Hong missed gold by a single point.\r\n\r\n* Last year, the most challenging problem was the third one.\r\nOf those who performed reasonably well on 2007/3,\r\nD\u00e1niel Kor\u00e1ndi\t(Hungary) \r\nLivia Alexandra Ilie (Romania)\r\nare still around.", "Solution_1": "Przemek Mazur from Poland was gold medalist twice. He scored 37 in Slovenia (5th place) and 30 in Vietnam (around 25-30th place). I hope that he will make the perfect score this year.", "Solution_2": "I'm pretty sure we'll have 2 gold medalists from Romania this year (both were in the team last year too, Livia and Mihai). The rest will get silver or even higher if the style of problems suits them.", "Solution_3": "I am a small fan of that S-Korean kid (14), maybe he will be the next mister Tao. He is also very cute. \r\nHow can a 13 year old kid win a silver on IMO (a rhetorical question :p)", "Solution_4": "[quote=\"Amazigh\"]I am a small fan of that S-Korean kid (14), maybe he will be the next mister Tao. He is also very cute. \nHow can a 13 year old kid win a silver on IMO (a rhetorical question :p)[/quote]\r\nIf I am not mistaken there was a 13 or 14 years old gold medalist from Kazakhstan at IMO 2004,his name is Andrei Kim.\r\nAnd I'm pretty sure that there was even younger gold medalists before...", "Solution_5": "[quote=\"Erken\"][quote=\"Amazigh\"]I am a small fan of that S-Korean kid (14), maybe he will be the next mister Tao. He is also very cute. \nHow can a 13 year old kid win a silver on IMO (a rhetorical question :p)[/quote]\nIf I am not mistaken there was a 13 or 14 years old gold medalist from Kazakhstan at IMO 2004,his name is Andrei Kim.\nAnd I'm pretty sure that there was even younger gold medalists before...[/quote]\r\n\r\nOfcourse there are younger gold medalists in the history of IMO, but it is special every time :)", "Solution_6": "[quote=\"Erken\"]If I am not mistaken there was a 13 or 14 years old gold medalist from Kazakhstan at IMO 2004,his name is Andrei Kim.[/quote]\r\n\r\nYes, but that's actually a sad story.\r\nAndrei Kim's performance became weaker and weaker, year by year.\r\n2004 he made gold, 2005 he made silver, 2006 he made bronze, and in 2007 he did not even manage to qualify for IMO.\r\n\r\nhttp://www.imo-official.org/participant_r.aspx?id=7857", "Solution_7": "[quote=\"test20\"][quote=\"Erken\"]If I am not mistaken there was a 13 or 14 years old gold medalist from Kazakhstan at IMO 2004,his name is Andrei Kim.[/quote]\n\nYes, but that's actually a sad story.\nAndrei Kim's performance became weaker and weaker, year by year.\n2004 he made gold, 2005 he made silver, 2006 he made bronze, and in 2007 he did not even manage to qualify for IMO.\n\nhttp://www.imo-official.org/participant_r.aspx?id=7857[/quote]\r\nActually he graduated from school at 2006,that's why he didn't participate to IMO 2007 :lol: ...And the reason of his following results was just that he was unlucky,and I know he was very strong matematician even in 11th grade(we have only 11 grade in Kazakhstan)...As he was absolutely first on Balkan Math Olympiad and Chinese Western Olympiad he \"beat\" all participants from China with unbelievable high score.\r\nAlso as you can see on the link above he solved all the problems at IMO 2004,but made a little bugs on almost every problem,but if he was more attentive i'm sure he could easily score 42 points at age 13 or 14.", "Solution_8": "Inverse of Tao :wink: ($ Tao^{\\minus{}1}$)\r\n\r\n[url]http://www.imo-official.org/participant_r.aspx?id=1581[/url]", "Solution_9": "So funny how with 40/42 you only got a silver medal back in the day :)", "Solution_10": "[quote=\"Valentin Vornicu\"]So funny how with 40/42 you only got a silver medal back in the day :)[/quote]\r\n\r\nThere even was one IMO, where 41/42 points only got you silver.\r\nMiss one point, and you lose gold...\r\n\r\nhttp://www.imo-official.org/year_individual_r.aspx?year=1987", "Solution_11": "Generally,all participants from China will get gold medals...", "Solution_12": "[quote=\"Erken\"][quote=\"test20\"][quote=\"Erken\"]If I am not mistaken there was a 13 or 14 years old gold medalist from Kazakhstan at IMO 2004,his name is Andrei Kim.[/quote]\n\nYes, but that's actually a sad story.\nAndrei Kim's performance became weaker and weaker, year by year.\n2004 he made gold, 2005 he made silver, 2006 he made bronze, and in 2007 he did not even manage to qualify for IMO.\n\nhttp://www.imo-official.org/participant_r.aspx?id=7857[/quote]\nActually he graduated from school at 2006,that's why he didn't participate to IMO 2007 :lol: ...And the reason of his following results was just that he was unlucky,and I know he was very strong matematician even in 11th grade(we have only 11 grade in Kazakhstan)...As he was absolutely first on Balkan Math Olympiad and Chinese Western Olympiad he \"beat\" all participants from China with unbelievable high score.\nAlso as you can see on the link above he solved all the problems at IMO 2004,but made a little bugs on almost every problem,but if he was more attentive i'm sure he could easily score 42 points at age 13 or 14.[/quote]\r\n\r\n\r\n\r\n\r\nI am also interested in this phenomenon....An inverse of Tao...\r\n\r\nChina Western Olympiad requires participants coming from western China which is the poorest area.", "Solution_13": "Lisa Sauermann from germany was 14 years old when she won a silver medal at IMO 2007, so maybe we can expect something great from her - we will see...\r\n\r\nNaphthalin", "Solution_14": "[quote=\"test20\"]\n* South-Korea has a 14-years old child prodigy, Soo Hong Lee.\n[/quote]\r\n\r\nI support him\r\n\r\nSo bad that in Georgia we finished high school 17 years old :( I could be in the list too, and I would bet on myself :D\r\n\r\nand this year it changed, our guys will have extra year at high school, and that surely means better results\r\n\r\nAnd I am sure we can expect 42 this year, and reasonably higher cuttoffs \r\n\r\ngood luck everyone ;)", "Solution_15": "i think morocco will surprise all the contries. i could probably be with the moroccan team but my scholl it not intersted in this cind of competition !!!!!! go on morocco and break the habit like in IMO 1995 \"kana lah ma3akom\"" } { "Tag": [], "Problem": "1.Prove that for every party (a group of people), you can find at less two persons such that, those persons have the same number of friends (INTO THE PARTY).", "Solution_1": "Ok... Since I am assuming you mean friends at the party...\n\n\n\n[hide]Contradiction:\n\nWe assume that the case where everyone has a different amount of friends is true.\n\n\n\nSuppose there are n friends at the party, then each person may have up to n-1 friends. To fufill the requirement that none have the same number of friends, we must assign 0 friends ,1 friend, 2 friends ,...,n-1 friends to each person. However, if we have one person with n-1 friends, then we cannot have a person with 0 friends. Contradiction![/hide]\n\n\n\nRight?", "Solution_2": "Good solution. I approve.", "Solution_3": "Yes, you must count only the friends at the party...." } { "Tag": [ "search", "MATHCOUNTS", "email" ], "Problem": "im just wonderin....", "Solution_1": "To be a member on AoPS, you are required to have a username.\r\n\r\n[hide] :rotfl: [/hide]", "Solution_2": "NO i mean why did you choose \"leoxnlin\" for ur username", "Solution_3": "If you search it(my username) on Google, you will probably find me...", "Solution_4": "iPod, iLord :D!", "Solution_5": "I am quoting a famous theorem derived by Botolemy.", "Solution_6": "because otherwise i would hav chosen a depressing one like angelofdeath22 or suffer22undead. (i was tempted by my friend's suggestion of xxdeathxx, tho. 'cause it sounds cool.) so be happy i didnt. (i get depressed easily, especially in summer.)\r\n\r\n@ 1=2: on this other forum about why did u pick ur numbers u said it was because u needed a short username...", "Solution_7": "His username is obviously a short username from a famous theorem :D!", "Solution_8": "i used my (very goofy) cousins favorite \"word\", yangosplat, and rearranged it.", "Solution_9": "I chose mine, because I admire Sesame Street's ernie. :rotfl:", "Solution_10": "pacman because pacman is cool\r\n\r\nguess 2812!", "Solution_11": "I am Danny Ham from Texas. What else do you expect?", "Solution_12": "@Pacman:\r\n\r\nHmm...\r\n\r\nBecause the stars aligned in the cradle of Draco, and Mars is in the house of Demeter, I think I will be able to guess the meaning of 2812 using the path of Orion's belt during the week after Midsummer's Night.\r\n\r\n/me uses complex astrology stuffs.\r\n\r\nI've got it! If you add up the digits in 2812, you get 13, which is an unlucky number, but it must be a lucky number because since you were born on :ninja: December 28th :ninja: under the house of Hades, 13 happens to be your lucky number!\r\n\r\nSo, was I right :P ?", "Solution_13": "hmm gee i wonder y i picked my username..... maybe it has to do with the fact that GUINEAPIGGIES R THE AMAZINGEST ANIMALS IN THE WORLD!!!! :wow: :yup: :gleam:", "Solution_14": "No, hamsters are. :wink: \r\n\r\nFor me? Oh, um... 'Cause I'm Alex?", "Solution_15": "My initials..duh\r\n\r\nI should have picked \"SupremeOvelordOfAllYouPunyEarthlingsWhoSuck\" But that might have been too long... :D :lol:", "Solution_16": "As well as arrogant?", "Solution_17": "hmm... Math website... Math website with MATHCOUNTS on it.\r\n\r\nWhat are mathcounters called again?\r\n\r\noh yea....\r\n\r\nExcept I spelled it wrong. >.< :rotfl:", "Solution_18": "[quote=\"chenhsi\"]If you search it(my username) on Google, you will probably find me...[/quote]\r\nYeah... You're on Wikipedia...\r\nhttp://en.wikipedia.org/wiki/User:Chenhsi\r\n(1+1=10 \tThis user likes binary.\r\n1+2=10 \tThis user likes ternary.\r\n2+3=10 \tThis user likes quinary.\r\n3+3=10 \tThis user likes senary.\r\n4+4=10 \tThis user likes octal.\r\n3x3=10 \tThis user likes nonary.\r\n5+5=10 \tThis user likes decimal.\r\n6+6=10 \tThis user likes dozenal.\r\n6\u00d79=42 \tThis user jokes in base 13.\r\n8+8=10 \tThis user likes hexadecimal.\r\n4\u00d75=10 \tThis user likes vigesimal.\r\n3\u00d74\u00d75=10 \tThis user likes sexagesimal.)\r\nAnd Kongregate...\r\nhttp://www.kongregate.com/accounts/chenhsi\r\nAnd AoPS.\r\n\r\nI just randomly thought up the phrase \"Here, Fishy, Fishy.\"\r\n\r\n(Yeah, I understand the 6*9=42 joke. Maybe The Earth was programmed in base 13?)", "Solution_19": "I simply wanted an original (A username that is 1 common word)\r\nI basically had a list of words that I wanted to be, and after trying them one by one, ProtestanT seemed to be the first one on the list that worked.", "Solution_20": "echen\r\n\r\nfirst initial, last name. Easy to remember and simple. If my name was something like Mymathskillsequalsathousandsouls53132456 I would forget it by tomorow.", "Solution_21": "[quote=\"splatyango\"]i used my (very goofy) cousins favorite \"word\", yangosplat, and rearranged it.[/quote]\r\n\r\nthat's probably the coolest word i have heard all summer\r\n\r\n\r\n\r\ni generally use different names for different things, though. the only site where i also used Lazarus as my nick is, iirc, Kongregate.", "Solution_22": "I always use my email for a username.... yoyomattnerdt@aim.. i have the \"yo yo\" part cause in 6th grade me and my friend would always say \"yo yo yo\" to the teachers and it would make them mad =) the other part,\r\nmatt=first name, nerd= duh...., t= last initial", "Solution_23": "just because it's my real name and I put the dotts because Ana. and Ana were already someone else's user names", "Solution_24": "Russian Rocket was the nickname for the AWESOME swimmer :winner_first: [size=200]Alexander Popov[/size]", "Solution_25": "DRAGONS ARE COOLER THAN GUINEA PIGS AND HAMSTERS AND YELLOW DUCKIES. 96 is for year of birth. \r\n\r\n@Ducky, what about your username?", "Solution_26": "xo angiee xo\r\n\r\n:D", "Solution_27": "CRAZY CHINESE MANIAC..need I say more?", "Solution_28": "i dunno i guess i just loved the ring and smooth flow of the name ra5249", "Solution_29": "my neighbor made it up. definition=crazy maniac yup, that's me :)", "Solution_30": "[size=150][color=violet]CRAZY ISLAND MANIAC!!!!!! HI KOOKHEAD!!![/color][/size] ur name is awesome!! ur my hero!!", "Solution_31": "This may be of interest:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79722[/url]", "Solution_32": "Or this, possibly, [url]http://www.mathlinks.ro/viewtopic.php?search_id=718014543&t=158658[/url]." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "For positive real numbers $a$, $b$, and $c$, solve the system\r\n\r\n\\begin{eqnarray*}cy+bz &=& a\\\\ az+cx &=& b\\\\ bx+ay &=& c \\end{eqnarray*}", "Solution_1": "Easy problem in matrix inversion. However, because of the answer you get, i feel there must be some geometric way of doing this." } { "Tag": [ "modular arithmetic", "arithmetic sequence", "number theory", "prime numbers", "number theory proposed" ], "Problem": "Giveing the infinite arithmetic progression $11,21,31,...$.Prove that in this arithmetic progression ,we have infinte prime numbers", "Solution_1": "Note that a number $\\equiv 1 \\pmod{5}$ is either $\\equiv 1 \\pmod{10}$ or $\\equiv 6 \\pmod{10}$. We'll show that there are infinitely many prime numbers of the form $5k+1$. The result follows, since they have to be of one of the above forms, and the latter form gives even numbers.\r\n\r\nReplace $5$ by any prime $q$. We'll show that $q+1, 2q+1, ...$ contains infinitely many primes. An observation: if a prime $p$ is such that $p \\mid x^q-1$ but $p \\nmid x-1$, then $q \\mid p-1$. Indeed, if $d_0$ is the smallest natural $d$ for which $p \\mid x^d-1$, all other numbers $d$ for which this happens are multiples of $d$. Then $q$ is a multiple of $d$. But since $q$ is a prime, $q=d$. Since $p \\mid x^{p-1}-1$, we have $d \\mid p-1 \\Rightarrow q \\mid p-1$. Done.\r\n\r\nNow $q \\mid p-1 \\Rightarrow p-1=qk, p=qk+1$. So we want to show that the set $x^q-1$ for $x=2, 3, ...$ has infinitely many prime divisors. If this is false, there should be some $k$ such that any prime dividing $x^q-1$ for any $x$ divides $x^q-1$ for some $x \\leq k$. Now let $p_1, p_2, ..., p_n$ be the union of the prime divisors of $x^q-1$ for $x=2, 3, ..., k$, and let $z=p_1p_2...p_n$. Setting $x=z$, we get an obvious contradiction.\r\n\r\nNow there's a detail left: we haven't checked the condition of these primes not coinciding with those of $x-1$. So suppose that for any $x$, the primes dividing $x^q-1$ are either among $p_1, p_2, ..., p_n$ or coincide with those dividing $x-1$. Let $z-1=y$, and consider the binomial expansion of $z^q-1=(y+1)^q-1=y^q+...+qy$. Dividing by $yq$, it can be dealt with easily.", "Solution_2": "I believe you can find a proof (here on the forum) that given a natural $n$, there are infinitely many primes of the form $nk+1$. \r\n\r\nWe want infinitely many primes of the form $10k+1$. A number is of this form iff it's both of the form $2u+1$ and $5k+1$. Since all primes (except for $2$, of course :)) are odd anyway, we only need to show that there are infinitely many primes of the form $5k+1$. Assume $p_1,\\ldots,p_n$ are such primes of the form $5k+1$. Any prime divisor of $M=\\frac{N^5-1}{N-1}$, where $N=5p_1p_2\\ldots p_n$ is then a prime of the form $5k+1$ different from $p_i,\\ i\\in\\overline{1,n}$. This is because $\\left(M,N-1\\right)=(N-1,5)=1$, so for any prime $p|M$ we have $M\\not\\equiv 1\\pmod p,\\ M^5\\equiv 1\\pmod p$, so $M$ has order $5$ modulo $p$, which implies $5|p-1$.\r\n\r\nI see Severius was here first :).", "Solution_3": "Special case of the Dirichlet's theorem.", "Solution_4": "I think the point was that it's easier than Dirichlet's theorem... I've never seen Severius's proof before." } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "I'm supposed to [u]prove[/u] that a closed trajectory to an autonomous system is an periodic solution to it. I don't see how this is going to be proved. Isn't this a fact per definition? :huh:", "Solution_1": "The closed trajectory is in the phase space, right? The statement would be false in the configuration space, or for nonautonomous systems.\r\n\r\n\"Statement.\" Suppose that $X(t)$ is a solution of $\\.{X}(t)=f(t)$ ($t>0$) with the initial condition $X(0)=X_{0}$. If $X(T)=X(0)$, then $X(t+T)=X(t)$ for all $t\\ge 0$. \r\n\r\n\"Proof.\" The function $\\tilde X(t)=X(t+T)$ also satisfies the equation, as well as the initial condition. Therefore, $\\tilde X(t)=X(t)$ by an appropriate uniqueness theorem*. \r\n\r\n(*) Assuming $f$ is such that uniqueness holds." } { "Tag": [ "vector", "geometry", "geometric transformation", "geometry proposed" ], "Problem": "I was asked to post this problem, so here is it:\r\n\r\n[b]Original problem.[/b]\r\n\r\nFind all natural numbers n satisfying the following assertion: If, on a plane, we have some regular n-gons $P_1$, $P_2$, ..., $P_k$ not intersecting each other, then we can choose one of these n-gons - call it $P_i$ - and a vector $\\overrightarrow{v}$ (not the zero vector) such that for any positive number t, the image of the n-gon $P_i$ under the translation with translation vector $t\\overrightarrow{v}$ doesn't intersect the other n-gons $P_j$ (with $j \\neq i$).\r\n\r\n[Of course, this condition - that for any positive number t, the image of the n-gon $P_i$ under the translation with translation vector $t\\overrightarrow{v}$ doesn't intersect the other n-gons $P_j$ (with $j \\neq i$) - can be rewritten as follows: If the n-gon $P_i$ moves in the direction of the vector $t\\overrightarrow{v}$, then it doesn't collide with any of the other n-gons $P_j$.]\r\n\r\nThe proposed solution shows that every natural n fulfills the above assertion. Even a much more general result is true:\r\n\r\n[b]Generalized problem.[/b]\r\n\r\nIf $P_1$, $P_2$, ..., $P_k$ are k arbitrary convex polygons on the plane, and $\\overrightarrow{v}$ is an arbitrary vector, then there exists a number i such that for any positive number t, the image of the polygon $P_i$ under the translation with translation vector $t\\overrightarrow{v}$ doesn't intersect the other polygons $P_j$ (with $j \\neq i$).\r\n\r\n[The proposed solution is at http://kvant.mccme.ru/1989/12/resheniya_zadachnika_kvanta_ma.htm ; but even if you read Russian, you will not necessarily understand it, since it is written up in a very bad style. I needed a day to understand it.]\r\n\r\n[b]Variant of the original problem.[/b]\r\n\r\nOn the other hand, the original problem can be modified:\r\n\r\nFind all natural numbers n satisfying the following assertion: If, on a plane, we have some regular n-gons $P_1$, $P_2$, ..., $P_k$ not intersecting each other, then we can choose one of these n-gons - call it $P_i$ - and a vector $\\overrightarrow{v}$ (not the zero vector) [i]which is perpendicular to a side of the polygon $P_i$[/i] such that for any positive number t, the image of the n-gon $P_i$ under the translation with translation vector $t\\overrightarrow{v}$ doesn't intersect the other n-gons $P_j$ (with $j \\neq i$).\r\n\r\nIn this case, already in the case n = 5, we can find a configuration of regular 5-gons for which such a polygon $P_i$ and such a vector $\\overrightarrow{v}$ don't exist.\r\n\r\n Darij", "Solution_1": "[quote=\"darij grinberg\"]\n\n[b]Generalized problem.[/b]\n\nIf $P_1$, $P_2$, ..., $P_k$ are k arbitrary convex polygons on the plane, and $\\overrightarrow{v}$ is an arbitrary vector, then ...\n\nDarij[/quote]\r\n\r\nIf I well understand in this more general situation we have furthermore the arbitrary convex poligon (instead the regular ones) the imposed direction $\\overrightarrow{v}$ towards which $P_i$ can \"escape\". \r\n\r\nWhile in the original problem \"... we can choose one of these n-gons - call it Pi - and a vector v (not the zero vector) such that...\" \r\n\r\nIs it right?\r\n\r\n\r\nIn any case, I will try to give an \"intuitive\" solution.\r\n\r\nTake a generic Pj and tray to \"move\" it to the given direction $\\overrightarrow{v}$. If Pj1 is the first poligon you \"meet\", then try to \"move\" this always in the same direction and so on. As the poligons are a finite number they are a finite number in the \"trusted\" direction so at te end will be a poligon Pjh that can \"escape\".\r\n\r\nAnd after this, one after the other, all poligons can \"escape\" :) .\r\n\r\nPS\r\nthis problem seems to describe the situation of the people when they tray to get out of the crowded underground.", "Solution_2": "[quote=\"sprmnt21\"]If I well understand in this more general situation we have furthermore the arbitrary convex poligon (instead the regular ones) the imposed direction $\\overrightarrow{v}$ towards which $P_i$ can \"escape\".[/quote]\n\nYes, in the generalized problem,\n\n- the shapes of the polygons,\n- their mutual position (as long as they don't intersect each other) and\n- the vector $\\overrightarrow{v}$\n\ncan be arbitrary.\n\n[quote=\"sprmnt21\"]Take a generic Pj and tray to \"move\" it to the given direction $\\overrightarrow{v}$. If Pj1 is the first poligon you \"meet\", then try to \"move\" this always in the same direction and so on. As the poligons are a finite number they are a finite number in the \"trusted\" direction so at te end will be a poligon Pjh that can \"escape\".[/quote]\n\nExactly. All what you need to complete this solution is to show that cycles are impossible. Hereby, a \"cycle\" is a subset of the set of polygons {$P_1$, $P_2$, ..., $P_k$} such that: the first polygon of the subset cannot \"escape\" because it would \"meet\" the second one, the second one cannot \"escape\" because it would \"meet\" yet the third one, etc., and the last one cannot \"escape\" because it would \"meet\" the first one. Proving this is a bit of tricky. Hint: Project on an axis perpendicular to the vector $\\overrightarrow{v}$.\n\n[quote=\"sprmnt21\"]this problem seems to describe the situation of the people when they tray to get out of the crowded underground.[/quote]\r\n\r\nIndeed :D \r\n\r\n Darij", "Solution_3": "I think the reason becouse one cannot have cicles is the convexity of poligons. I later will provide an argument on this.", "Solution_4": "[quote=\"sprmnt21\"]I think the reason becouse one cannot have cicles is the convexity of poligons. I later will provide an argument on this.[/quote]\r\n\r\nHere is the intuitive argument I was referring to.\r\n\r\nLet's suppose that A is the point of firts \"impact\" of poligon Pj0 with poligon Pj1 (see fig. attached). Then Pj1 \"reacts\" to the \"impact\" moving towards direction V untill it impact Pj2 in B1. Then Pj2 \"reacts\" to the \"impact\" moving towards direction V untill it impact Pj3 in B2. And so on. Suppose that finilly Pj3 \"impact\" Pj0 in B3.\r\n\r\nBecause of the convexity of poligons all segments AiBi stay inside the correspondent poligon. \r\n\r\nFor this we should have that P0 impacted Pj3 (in our example) before impact Pj1. Contradiction!", "Solution_5": "Hmm, I must say I really don't understand this...\r\n\r\n Darij", "Solution_6": "[quote=\"darij grinberg\"]Hmm, I must say I really don't understand this...\n\n Darij[/quote]\r\n\r\n\r\nI will try to rephrase some sentences. \r\n\r\nIn the figure attached for each poligons Pji are highlighted only the segments involved in the \"meeteing\".\r\n\r\nLet's to start supposing that A is a point on the boundary of Pj0 and Pj1 where they have the firts \"meeting\"[actually could be that Pj0 and Pj1 have the first \"meeting\" in a set of points and segments; in this case let A be the \"lowest\", in the direction V, point of this set ].\r\n\r\nAfter this \"impact\" Pj1 try to \"escape\" alongside direction V and cover the distance AA1 untill meet Pj2 in B1. A1B1 is a segment of Pj1 when Pj1 \"meet\" Pj2.\r\n\r\nAnd so on.\r\n\r\nIf, as we suppose, some poligon of this chain would \"meet\" the first one, we could have the situation reppresented in the figure. Where AB3 is a segment of Pj0 and B2B'3 is a segment of Pj3 [B'3B2 is the translated segment of B3A3]. this configuration, as AB3 intersect B'3B2 and [i]fortiori[/i] Pj0 intersect Pj3, implies that Pj0 would have meet Pj3 before meet Pj1. \r\n\r\nThis is contradictory with our starting hypothesis." } { "Tag": [], "Problem": "$(a+\\sqrt{a^2-1})^{x^2-2x} + (a-\\sqrt{a^2-1})^{x^2-2x} = 2a$\r\n\r\nfind x (NOT IN TERMS OF a)", "Solution_1": "[hide]\n\nSince $a-\\sqrt{a^2-1}=\\frac{1}{a+\\sqrt{a^2-1}}$ , By letting $a+\\sqrt{a^2-1}=y$ and $x^2-2x=p$ we obtain\n\n$y^p+y^{-p}=2a$ $\\implies y^{2p}-2ay^p+1=0$ $\\implies y^p=a\\pm\\sqrt{a^2-1}$\n\nHence $p=\\pm 1$ which gives us $x=1,1\\pm\\sqrt{2}$\n[/hide]", "Solution_2": ":first:", "Solution_3": "[quote=\"shyong\"][hide]\n\nSince $a-\\sqrt{a^2-1}=\\frac{1}{a+\\sqrt{a^2-1}}$ , By letting $a+\\sqrt{a^2-1}=y$ and $x^2-2x=p$ we obtain\n\n$y^p+y^{-p}=2a$ $\\implies y^{2p}-2ay^p+1=0$ $\\implies y^p=a+-\\sqrt{a^2-1}$\n\nHence $p=+-1$ which gives us $x=1,1+-\\sqrt{2}$\n[/hide][/quote]\r\n\r\nBTW, you can use \\pm to give the $\\pm$ symbol in LaTeX.", "Solution_4": "[quote=\"paladin8\"][quote=\"shyong\"][hide]\n\nSince $a-\\sqrt{a^2-1}=\\frac{1}{a+\\sqrt{a^2-1}}$ , By letting $a+\\sqrt{a^2-1}=y$ and $x^2-2x=p$ we obtain\n\n$y^p+y^{-p}=2a$ $\\implies y^{2p}-2ay^p+1=0$ $\\implies y^p=a+-\\sqrt{a^2-1}$\n\nHence $p=+-1$ which gives us $x=1,1+-\\sqrt{2}$\n[/hide][/quote]\n\nBTW, you can use \\pm to give the $\\pm$ symbol in LaTeX.[/quote]\r\n\r\nawesome !Now that really makes it looks nicer ." } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "\"$ 1^k, 2^k, 3^k, 4^k, \\dots$\" 's k-th difference is k!\r\nex) \"$ 1^2, 2^2, 3^2, \\dots$\"'s 1-th diff = $ 2^2 \\minus{} 1^2$ = $ 3^2 \\minus{} 2^2$ = ...\r\n\"$ 1^2, 2^2, 3^2, \\dots$\"'s 2-th diff = $ 3^2 \\minus{} 2\\cdot 2^2 \\plus{} 1^2$ = $ 4^2 \\minus{} 2\\cdot 3^2 \\plus{} 2^2$ = ... = $ 2!$\r\n\r\nHow to prove?", "Solution_1": "[quote=\"ssibseya\"]\"$ 1^k, 2^k, 3^k, 4^k, \\dots$\" 's k-th difference is k!\nex) \"$ 1^2, 2^2, 3^2, \\dots$\"'s 2-th diff = $ 3^2 \\minus{} 2\\cdot 2^2 \\plus{} 1^2$ = $ 4^2 \\minus{} 2\\cdot 3^2 \\plus{} 2^2$ = $ 2!$\n\nHow to prove?[/quote]\r\nCould you explain a bit more. What is the k-th difference??\r\nBut it is easy to prove that $ (n \\plus{} 1)^2 \\minus{} 2n^2 \\plus{} (n \\minus{} 1)^2 \\equal{} n^2 \\plus{} 2n \\plus{} 1 \\minus{} 2n^2 \\plus{} n^2 \\minus{} 2n \\plus{} 1 \\equal{} 2 \\equal{} 2!$\r\nBut fx $ 3^3 \\minus{} 3*2^3 \\plus{} 1^3 \\equal{} 27 \\minus{} 18 \\plus{} 1 \\equal{} 10 \\neq 3!$....", "Solution_2": "I guess he means http://www.mathlinks.ro/Forum/viewtopic.php?t=70637 .\r\n\r\n darij", "Solution_3": "$ \\binom{n}{0} {(n \\plus{} 1)}^n \\minus{} \\binom{n}{1} {(n)}^n \\plus{} \\binom{n}{2} {(n \\minus{} 1)}^n \\minus{} \\ldots \\plus{} ( \\minus{} 1)^{n} \\binom{n}{n} {1}^n \\equal{} x \\ .$\r\n=> wanted\r\n$ \\binom{n \\plus{} 1}{0} {(n \\plus{} 1)}^n \\minus{} \\binom{n \\plus{} 1}{1} {(n)}^n \\plus{} \\binom{n \\plus{} 1}{2} {(n \\minus{} 1)}^n \\minus{} \\ldots \\plus{} ( \\minus{} 1)^{n} \\binom{n \\plus{} 1}{n} {1}^n \\equal{} 0 \\ .$\r\n=> because surjection(n to n+1) = 0\r\n$ \\binom{n}{0} {(n)}^n \\minus{} \\binom{n}{1} {(n \\minus{} 1)}^n \\minus{} \\ldots \\minus{} ( \\minus{} 1)^{n} \\binom{n}{n \\minus{} 1} {1}^n \\equal{} \\ x \\ \\equal{} n! \\ .$\r\n=> first-second = this (because surjection(n to n) = n! and you gave me this)\r\n\r\nbut\r\n$ \\binom{n}{0} {(n \\plus{} 1)}^n \\minus{} \\binom{n}{1} {(n)}^n \\plus{} \\binom{n}{2} {(n \\minus{} 1)}^n \\minus{} \\ldots \\plus{} ( \\minus{} 1)^{n} \\binom{n}{n} {1}^n \\equal{} n! \\ .$\r\n$ \\binom{n}{0} {(n \\plus{} 2)}^n \\minus{} \\binom{n}{1} {(n \\plus{} 1)}^n \\plus{} \\binom{n}{2} {(n)}^n \\minus{} \\ldots \\plus{} ( \\minus{} 1)^{n} \\binom{n}{n} {2}^n \\equal{} n! \\ .$\r\n$ \\binom{n}{0} {(n \\plus{} 3)}^n \\minus{} \\binom{n}{1} {(n \\plus{} 2)}^n \\plus{} \\binom{n}{2} {(n \\plus{} 1)}^n \\minus{} \\ldots \\plus{} ( \\minus{} 1)^{n} \\binom{n}{n} {3}^n \\equal{} n! \\ .$\r\nand so on\r\nwhy???", "Solution_4": "You are dealing with $ \\Delta^n((x\\plus{}c)^n)$. It's a constant polynomial (because $ \\Delta$ decreases degree by $ 1$).\r\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=90529 (and http://www.mathlinks.ro/Forum/viewtopic.php?p=657956 for a general explanation)." } { "Tag": [ "geometry", "ratio", "cyclic quadrilateral", "power of a point", "AMC" ], "Problem": "Two circles are concentric. A chord AC of the outer circle touches the inner circle at Q. P is the midpoint of AQ. A line through A intersects the inner circle at R and S. The perpendicular bisectors of PR and CS meet at T on the line AC. What is the ratio AT/TC? \r\n\r\nhelp please", "Solution_1": "I've been working on this one, but frankly, I don't have any good ideas. I'll keep trying it, but I thought about bringing it back up to the top just so other people see it. Where did it come from?\r\n\r\nActually, I shouldn't say I don't have any ideas: I know what the answer is. And I know a few other things about the setup, just from playing around with it on Geometer's Sketchpad. But I can't prove anything.", "Solution_2": "Okay, I've got it!\r\n\r\nHere we go:\r\nWe have by power of a point that AR * AS = AQ:^2:\r\nAQ:^2: = AQ * AQ = (2AP) * (AC/2) = AP*AC\r\nSo AR * AS = AP * AC\r\nAR/AP = AC/AS\r\nangle RAP = angle SAC\r\n\r\nSo triangle RAP is similar to triangle CAS. So angle ACS = angle ARP. angle ARP + angle PRS = 180 degrees (two angles on a line)\r\nangle ACS + angle PRS = 180 degrees (substitution)\r\nangle PCS + angle PRS = 180 degrees (renaming)\r\n\r\nThus PCSR is a cyclic quadrilateral, so the perpendicular bisectors of PR and CS meet at the center of the circumcircle. If this is on one of the edges, it must bisect that edge (since it is equidistant from the endpoints). Thus TC = TP = 3AC/8. \r\nAT = 5AC/8. AT/TC = 5/3.", "Solution_3": "im dont remember where this came from, but i liked the problem. thanks for the solution." } { "Tag": [ "function", "vector", "combinatorics proposed", "combinatorics" ], "Problem": "Suppose that ${\\mathcal H}_{n}$ is the set of all $n \\times n$ matrices $A=\\parallel a_{i,j}\\parallel $ with $a_{i,j}\\in \\{0,1\\}.$ Let [b]m[/b] be the number of all nonidentical functions $F:{\\mathcal H}_{n}\\to{\\mathcal H}_{n}$ such that $\\left(F\\left(F(A)\\right)\\right)= A$ for all $A \\in{\\mathcal H}_{n}. \\; \\;$ Find [b]m[/b] .\r\n(In case $n=2$ it's true that [b]m[/b]$=12392480$ ?)", "Solution_1": "Ok, as I see this problem, first of all, we have a vector with $n\\times n$ entries $\\in {0,1}$. Now, there are $2^{n\\ldots n}=N$ such vectors. If we consider the number of functions $f:A\\to A$ with $f(f(x))=x$ and $|A|=N$ to be $x_n$ we have the well known recurence $x_{n+1}=x_n + n\\cdot x_{n-1}$. The reasoning is well known and has been posted on the forum. \r\nBtw, I wonder if someone managed to find a solution for the recurence. I tried to do something with generating functions, but it reduces to a not very beautiful differential equation :(", "Solution_2": "Hello,\r\nI've foud some references of this problem in a book of Ioan Tomescu \"Probleme de combinatorica si teoria grafurilor\" , but it's written in Romanian and I don't know if there is a English version of this book.\r\n\r\nHere is demonstrated the recurence posted by Dust $\\ x_n=x_{n-1}+(n-1) x_{n-2}$and a formula for\r\n $\\ x_n$=$\\ n!$$\\displaystyle\\sum_{i+2j=n}\\frac{1}{i!j!2^j}$ and a closed formula for it's generating exponential function i.e $\\ F(x)=\\displaystyle\\sum_{n \\ge 0} x_n \\frac{t^n}{n!}=exp(t+\\frac{t^2} {2}) $. The demonstration is not very hard." } { "Tag": [ "trigonometry" ], "Problem": "What is $n$ if $(\\sin1)(\\sin3)(\\sin5)...(\\sin89)=\\frac{1}{2^{n}}$ and its in degrees.", "Solution_1": "$(sin1)(sin3)(sin5)...(sin89)=\\frac{1}{2^{n}}$\r\n\r\n$\\frac{1}{(sin1)(sin3)(sin5)...(sin89)}=2^{n}$\r\n\r\n$ln(\\frac{1}{(sin1)(sin3)(sin5)...(sin89)})=ln{2^{n}}$\r\n\r\n$n=\\frac{ln(\\frac{1}{(sin1)(sin3)(sin5)...(sin89)})}{ln2}$", "Solution_2": "Er... that completely misses the point. $n$ is nice. Find it.", "Solution_3": "[hide]The product equals $\\frac{\\sin 1 \\sin 2 \\sin 3 \\ldots \\sin 89}{\\sin 2 \\sin 4 \\sin 6 \\ldots \\sin 88}$.\nRepeated use of $\\cos x=\\sin 90-x$ and $\\sin x\\cos x=\\frac{1}{2}\\sin 2x$ reduces the numerator to $2^{-44}\\frac{\\sqrt{2}}{2}\\sin 2 \\sin 4 \\sin 6 \\ldots \\sin 88$. The $\\sin$ terms cancel out, leaving $2^{-44}\\frac{\\sqrt{2}}{2}$.\n$\\frac{\\sqrt{2}}{2}=\\sqrt{0.5}=\\sqrt{2^{-1}}=2^{-\\frac{1}{2}}$\nSo the product is $2^{-44}2^{-\\frac{1}{2}}$ or $\\frac{1}{2^{\\frac{89}{2}}}$.\n\nThe answer is $\\boxed{\\frac{89}{2}}$.[/hide]" } { "Tag": [ "function", "integration", "calculus", "real analysis", "real analysis unsolved" ], "Problem": "Let the function $ f$ have the property\r\n\r\n$ f(x)\\plus{}f^{\\minus{}1}(x)\\equal{}x$ for all $ x\\in\\mathbb{R}$\r\n\r\nthen find $ \\int^{a}_{\\minus{}a}f(x)dx$", "Solution_1": "$ x \\minus{}> f(x)$: .... $ f(f(f(x)))\\equal{}\\minus{}x$ and $ x \\minus{}> f(x)$ have $ f(\\minus{}x)\\equal{}\\minus{}f(x)$ and $ \\int^{a}_{\\minus{}a}f(x)dx\\equal{}0$", "Solution_2": "Er, show your work?\r\n\r\nI made it through your first three dots but I certainly don't find obvious the conclusion that $ f$ is an odd function.", "Solution_3": "FWIW, one function with the required property is f(x)=Cx. In this case, we have C+1/C=1, and C=(1+sqrt(3)i)/2 or (1-sqrt(3)i)/2.\r\nIn both cases the intergral is zero.\r\n\r\nEdit: fixed per LydianRain's observation.", "Solution_4": "In both cases $ f$ integrates to $ 0$, since it is an odd function." } { "Tag": [ "calculus", "integration", "function", "algorithm", "quadratics", "algebra", "rational function" ], "Problem": "find $\\int\\frac{\\sqrt{x^{2}+A}}{x^{2}+B}dx=?$ where $A,B$ is real.", "Solution_1": "hmm I kind of want to say to use Trigonometric Substitution, try to get it to fit [img]http://upload.wikimedia.org/math/a/1/e/a1e54d721e1fea9d5a641cda2944ca34.png[/img]\r\n\r\nas a start.....", "Solution_2": "The substitution x = (t - 1/t)*sqrt(A)/2 converts the integrand to a rational function of t, which is good enough for me.", "Solution_3": "[quote=\"trichoplax\"]The substitution x = (t - 1/t)*sqrt(A)/2 converts the integrand to a rational function of t, which is good enough for me.[/quote]\r\nI think it's equivalent tu sabstitute $sh(x)$. If you solved it then put full sulution", "Solution_4": "[quote=\"Extremal\"]\nI think it's equivalent tu sabstitute $sh(x)$. If you solved it then put full sulution[/quote]\r\nI did not continue. Since there is a well-known algorithm for integrating any rational function, it is my opinion that an integral is fully solved once it has been reduced to that case. Mathematica's solution is relatively short, so it must have used a smarter substitution than I did. Maple's solution is a mess.", "Solution_5": "[quote]Since there is a well-known algorithm for integrating any rational function, ... [/quote]\r\nThat comes with one \"but.\" There is an algorithm, provided you can factor the denominator into linear and quadratic factors. The existence of that factorization is assured by the Fundamental Theorem of Algebra, but the FTA is a non-algorithmic pure existence proof. There is no universal factoring algorithm, and in fact some kinds of factoring algorithm would be prohibited by the Abel-Galois non-solvability results.\r\n\r\nOf course in this case, the denominator will come in a factored form; that particular objection will not apply here.", "Solution_6": "[quote=\"Extremal\"]find $\\int\\frac{\\sqrt{x^{2}+A}}{x^{2}+B}dx=?$ where $A,B$ is real.[/quote]\r\nOk, My answer is next:\r\n$\\int\\frac{\\sqrt{x^{2}+A}}{x^{2}+B}$\r\nsubtitute $\\sqrt{A}sh(y)=x$ then we have \r\n$\\frac{\\sqrt{A}}{A}\\int\\frac{ch^{2}(y)}{ch^{2}(y)-1+\\frac{B}{A}}$\r\nthen we put $ch(y)=\\frac{e^{y}+e^{-y}}{2}$ and di new substtitute \r\n$e^{2x}=z$ It's done." } { "Tag": [ "function", "algebra", "polynomial" ], "Problem": "Consider the positive integers that such that its digits are either $ 1$, $ 2$, or $ 3$. \r\nDefine a function $ f(n)$ such that f(N) = the number of positive integers consisting of only $ 1$, $ 2$, and $ 3$ whose digits sum to $ N$.\r\n\r\nProve that $ f(N \\plus{} 3) \\equal{} f(N \\plus{} 2) \\plus{} f(N \\plus{} 1) \\plus{} f(N)$.", "Solution_1": "Well its a correspondence because the numbers inf(n+3) can be rewritten as either a number in f(n) with a 3 attached at the end, a number in f(n+1) with a 2 attached at the end, or a number in f(n+2) with a 1 attached at the end.", "Solution_2": "I'm confused.\r\n\r\nSo do you mean that (ii) is a specific case of the recursion at the bottom? Do you want us to prove the recursion, and then use that to find $ f(10)$? If so:\r\n\r\n[hide=\"Abridged Solution\"]This is the same as the number of ways of traversing a series of stepping stones, where from each stone you may move to the next, or skip either 1 or 2 stones.\n\nThe number of ways to get to stone $ N \\plus{} 3$ is thus the sum of the number of ways to get to stones $ N \\plus{} 2$, $ N \\plus{} 1$, and $ N$, since these are the possible locations immediately preceeding stone $ N \\plus{} 3$. This gives $ f(N \\plus{} 3) \\equal{} f(N \\plus{} 2) \\plus{} f(N \\plus{} 1) \\plus{} f(N)$. The initial conditions are $ f(1) \\equal{} 1$, $ f(2) \\equal{} 2$, and $ f(3) \\equal{} 4$.[/hide]\n[hide=\"Answer\"]Using the recursion and the initial conditions:\n$ \\begin{array}{ccc} f(4) & \\equal{} & 7 \\\\\nf(5) & \\equal{} & 13 \\\\\nf(6) & \\equal{} & 24 \\\\\nf(7) & \\equal{} & 44 \\\\\nf(8) & \\equal{} & 81 \\\\\nf(9) & \\equal{} & 149 \\\\\nf(10) & \\equal{} & 274 \\end{array}$[/hide]", "Solution_3": "[quote=\"unimpossible\"]Well its a correspondence because the numbers inf(n+3) can be rewritten as either a number in f(n) with a 3 attached at the end, a number in f(n+1) with a 2 attached at the end, or a number in f(n+2) with a 1 attached at the end.[/quote]\r\n\r\nTo be slightly more precise: let $ \\mathcal{F}_n$ denote the set of positive integers consisting of $ 1$s, $ 2$s, and $ 3$s whose digits sum to $ N$. Then $ f(n) \\equal{} |\\mathcal{F}_n|$ and the above argument can be made precisely by considering the last digit of any element of $ \\mathcal{F}_{n \\plus{} 3}$.\r\n\r\n[hide=\"Fun stuff\"] Note that the generating function for this sequence is given by $ \\frac {x \\plus{} x^2 \\plus{} x^3}{1 \\minus{} (x \\plus{} x^2 \\plus{} x^3)}$, which can be used to prove that its characteristic polynomial is $ x^3 \\equal{} x^2 \\plus{} x \\plus{} 1$ as desired. \n[hide=\"In particular...\"] Let the three roots of $ x^3 \\equal{} x^2 \\plus{} x \\plus{} 1$ be $ \\alpha, \\beta, \\gamma$. Then\n\n$ \\frac{x \\plus{} x^2 \\plus{} x^3}{1 \\minus{} (x \\plus{} x^2 \\plus{} x^3)} \\equal{} \\frac{x \\plus{} x^2 \\plus{} x^3}{(1 \\minus{} \\alpha x)(1 \\minus{} \\beta x)(1 \\minus{} \\gamma x)}$.\n\nPartial fraction decomposition will show you that $ f(n) \\equal{} a \\alpha^n \\plus{} b \\beta^n \\plus{} c \\gamma^n$ for suitably chosen $ a, b, c$, from which the conclusion follows (in a much more roundabout way than the simple counting argument). I include this argument merely to demonstrate the general power of the generating functions approach. :) [/hide] [/hide]" } { "Tag": [ "number theory open", "number theory" ], "Problem": "x^5+31=y^2 show that there is no solution in integers.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=184045" } { "Tag": [ "inequalities", "function", "inequalities proposed" ], "Problem": "$x,y,z \\ge 0$ and $x+y+z=1$. Prove that\r\n$\\frac{10}{27}+\\sqrt[3]{\\frac{1}{9}}\\le xyz+xy+yz+zx+\\sqrt[3]{x^{3}+y^{3}+z^{3}}\\le 1$\r\nP/s: To Albanian Eagle: I had edited it! :lol:", "Solution_1": "It's very easy at all :). Denote $x=ab+bc+ca+abc$ then the expression is exactly \\[x+\\sqrt[3]{1-3x}\\] then examine this function of $x$, notice that $0\\le x \\le 10/27$. \r\n\r\nTo NkMAsTeR. You need to change in your problem $x,y,z$ to $a,b,c$.", "Solution_2": "Oh, I'm sorry. This is a type error! :D :blush:\r\nIt don't think $1-3x=x^{3}+y^{3}+z^{3}$ because $1-x^{3}-y^{3}-z^{3}=3(xy+yz+zx-xyz)$\r\nBut if i replace 1 by 2, can you find the minimum and maximum of this?", "Solution_3": "Nobody like that? :maybe :(", "Solution_4": "to explain $Flove$'s idea: \\[1-3(ab+bc+ca+abc)=(a+b+c)^{3}-3((ab+bc+ca)(a+b+c)+abc)=a^{3}+b^{3}+c^{3}\\]", "Solution_5": "But \r\n$(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+3(x+y)(y+z)(z+x)$\r\n$=x^{3}+y^{3}+z^{3}+3(1-x)(1-y)(1-z)=...=x^{3}+y^{3}+z^{3}+3(xy+yz+zx-xyz)$\r\n :(", "Solution_6": "just to clearify the problem NkMAsTeR are the variables a,b,c or x,y,z in your problem ??? :P", "Solution_7": "An open question: Find the best constant $k$ such that\r\n$xyz+xy+yz+zx+\\sqrt[k]{x^{3}+y^{3}+z^{3}}\\le 1$", "Solution_8": "[quote=\"zhaobin\"]to explain $Flove$'s idea: \\[1-3(ab+bc+ca+abc)=(a+b+c)^{3}-3((ab+bc+ca)(a+b+c)+abc)=a^{3}+b^{3}+c^{3}\\] [/quote]\r\nIf you like,change $a,b,c$ to $x,y,z$ here,I think you will understand...0.0", "Solution_9": "[quote=\"zhaobin\"][quote=\"zhaobin\"]to explain $Flove$'s idea: \\[1-3(ab+bc+caabc)=(a+b+c)^{3}-3((ab+bc+ca)(a+b+c)+abc)=a^{3}+b^{3}+c^{3}\\] [/quote]\nIf you like,change $a,b,c$ to $x,y,z$ here,I think you will understand...0.0[/quote]\r\nIt must be \\[1-3(ab+bc+ca-abc)=(a+b+c)^{3}-3((ab+bc+ca)(a+b+c)-abc)=a^{3}+b^{3}+c^{3}\\] :lol:", "Solution_10": "But I wonder how can it solve? :P Normal way is seemingly impossible,", "Solution_11": "But with pvthuan, it is only average!So, I won't post it here! :lol:", "Solution_12": "Sorry I am wrong :blush:", "Solution_13": "\\[x^{3}+y^{3}+z^{3}-3xyz=x^{2}+y^{2}+z^{2}-(xy+yz+zx)=1-3(xy+yz+zx)\\] Let $t=xyz,r=xy+yz+zx$, so \\[f(t)=t+r+\\sqrt[3]{1+3t-3r}\\] Because $r^{2}\\ge 3t$ and $r\\ge 9t$ so \\[1+3t-3r \\le 1-3(t+r)+3(t+r)^{2}-(t+r)^{3}\\] \\[\\Leftrightarrow 6t+(t+r)^{3}\\le 3(t+r)^{2}\\] Which is obviously true because \\[(t+r)^{2}\\ge (t+r)^{3}\\mbox{and}(t+r)^{2}\\ge y^{2}\\ge 3t.\\] I think the rest is also easy\r\n\r\nTo NkMAsTeR: [hide]What do you feel about Hungkhtn's book?[/hide]", "Solution_14": "[quote=\"Flove\"] \\[x^{3}+y^{3}+z^{3}-3xyz=x^{2}+y^{2}+z^{2}-(xy+yz+zx)=1-3(xy+yz+zx)\\] Let $t=xyz,r=xy+yz+zx$, so \\[f(t)=t+r+\\sqrt[3]{1+3t-3r}\\] Because $r^{2}\\ge 3t$ and $r\\ge 9t$ so \\[1+3t-3r \\le 1-3(t+r)+3(t+r)^{2}-(t+r)^{3}\\] \\[\\Leftrightarrow 6t+(t+r)^{3}\\le 3(t+r)^{2}\\] Which is obviously true because \\[(t+r)^{2}\\ge (t+r)^{3}\\mbox{and}(t+r)^{2}\\ge y^{2}\\ge 3t.\\] I think the rest is also easy\n\nTo NkMAsTeR: [hide]What do you feel about Hungkhtn's book?[/hide][/quote]\nBut It's not complete! There are two inequalities here! :D \nTo Flove [hide]I don't like it much! Sorry to you and Hungkhtn[/hide]", "Solution_15": "Why it's incomplete? I don't understant? The right-hand inequality is ok now, but I think the left-hand inequality is not hard. Anyone could be try? I don't have so much time to enjoy it, because this problem is not totally beautiful.", "Solution_16": "[quote=\"NkMAsTeR\"]\nTo Flove: \nI don't like it much! Sorry to you and Hungkhtn[/quote]\r\n\r\nOh, in fact it is not as good as hungkhtn marketed it loudly. The best chapter is chapter 3. People and hung talked too much about it. The first two chapters look like a mess and may give readers the feeling that its a collections of problems and solutions. Many of them are not natural, problems were not arranged in a logical and systematic way. Still too many errors in typos, and writings.", "Solution_17": "Yes, of course it is not beautiful at all, but I think it make me thinking, not just applling an \"algorithm\".Do you know why I don't like Hungkhtn's book much? Because It has some mistakes and somewhere, the solutions are not creative!\r\nAnd can you prove the open question of me?", "Solution_18": "[quote=\"NkMAsTeR\"]Yes, of course it is not beautiful at all, but I think it make me thinking, not just applling an \"algorithm\".Do you know why I don't like Hungkhtn's book much? Because It has some mistakes and somewhere, the solutions are not creative!\nAnd can you prove the open question of me?[/quote]\r\n\r\nTo NkMAsTeR: Sometime, generalising a problem is not so important, also finding the best constant are mechanical. I'm very glad to study which part of Hungkhtn's book is as not creative as you say? Chapter II is the chapter where almost problem is the most creative, but in chapter III, the purpose is \"Methods\". Problems exposing this method are just important.", "Solution_19": "Let $m=xy+yz+zx$ and $t=xyz$ We have:\r\n$x^{3}+y^{3}+z^{3}=(x+y+z)^{3}-3(x+y)(y+z)(z+x)$\r\n$=1-3m+3t$\r\n$f(x,y,z)=xy+yz+zx+xyz+\\sqrt[3]{x^{3}+y^{3}+z^{3}}=m+t+\\sqrt[3]{1-3m+3t}=g(t)$ but $m$ is fixed$\\Rightarrow$ \r\n$g(t)$ get the maximum or minimum when\r\n 1 in 3 numbers $x,y,z=0$\r\nor 2in 3 numbers equa to each other\r\nAmssume $x\\leq y\\leq z$ and let $x+y=r$\r\nNow we have \r\n$f(0,r,z) =rz+\\sqrt[3]{r^{3}+z^{3}}=rz+\\sqrt[3]{1-3rz}$ ( Because $r+z=1$ )\r\nLet $rz=a \\text{ (}0\\leq a\\leq \\frac{1}{4}$\r\n$\\Rightarrow f(0,y,z)=a+\\sqrt[3]{1-3a}=F(a)$\r\n $F'(a)=1-\\frac{1}{ \\sqrt[3]{(1-3a)^{2}}}=0 \\Leftrightarrow a=0$\r\n Easy to get $F(a)\\leq 1$ Because $0\\leq a\\leq \\frac{1}{4}$\r\n$\\Rightarrow$ max$f(x,y,z) =1$ :lol: \r\nsimilaly min$f(x,y,z)= \\frac{10}{27}+\\sqrt[3]{ \\frac{1}{9}}$ :lol: \r\nWhat do you think? :oops:", "Solution_20": "Yes, but there are, may be you don't agree with me, some problems solved by S.O.S , are not creative. I'm sorry, but I think, it is really a good book, but, of course, there isn't any perfect things. And, I'm jsut a kid, so, please fogive me if I said somethings wrong.\r\n\r\nTo Do Tien Dang\r\n[quote=\"Do Tien Dang\"]Let $m=xy+yz+zx$ and $t=xyz$ We have:\n$x^{3}+y^{3}+z^{3}=(x+y+z)^{3}-3(x+y)(y+z)(z+x)$\n$=1-3m+3t$\n[/quote]\r\nI'm sorry, It must be $=1-3m+6t$ . Please check it and sorry if I wrong!", "Solution_21": "Actually, I don't really like problem using SOS method simply, but in almost case, SOS is the best for three-variable problem (at least better than mixing variables). If you say that SOS is not creative, then you will agree that mixing variables and almost other parts (involving EV, hafl-convex function, generation of Schur...) are also not creative, because your work is only applying, applying and applying. But this only shows that you're just a child. :lol: Such post as yours is only acceptable is diendantoanhoc, not in mathlinks. :lol: \r\n\r\nAs a matter of fact, sometime SOS is mechanical. But there's many extremely hard and nice problem where SOS is unique, but I say it not to negate other methods because I myself don't very like SOS. Such idea shouldn't post here, it should be placed in \r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=630199#630199\r\n\r\nTo you, NkMAster and other: If you fell that SOS is boring, the only reason is that, you are so professional about it, or you know nothing. :arrow:", "Solution_22": "Stop, stop :!: \r\n\r\nTo Vietnamese boys: It is not uncommon for people to have different points of view. We need arguments and discussions in a polite way. Otherwise, foreign friends will hate you.", "Solution_23": "Yes, I agree with you. Sorry, Flove. \r\nTo pvthuan: Is is inequality okie?" } { "Tag": [ "inequalities", "inequalities proposed", "109" ], "Problem": "[b]English Version.[/b]\r\n\r\nLet $a,b,c$ be non-negative real numbers. Prove that\r\n\r\n\\[\\sqrt{\\frac{a}{b+3c}}+\\sqrt{\\frac{b}{c+3a}}+\\sqrt{\\frac{c}{a+3b}}\\ge\\frac{3}{2}.\\]\r\n\r\nI have a very nice solution to this one.", "Solution_1": "[quote=\"hungkhtn\"][b]English Version.[/b]\n\nLet $a,b,c$ be non-negative real numbers. Prove that\n\\[\\sqrt{\\frac{a}{b+3c}}+\\sqrt{\\frac{b}{c+3a}}+\\sqrt{\\frac{c}{a+3b}}\\ge\\frac{3}{2}. \\]\n[/quote]\r\nLet $\\frac{a}{b+3c}=\\frac{x^{2}}{4},$ $\\frac{b}{c+3a}=\\frac{y^{2}}{4}$ and $\\frac{c}{a+3b}=\\frac{z^{2}}{4},$ where $x,$ $y$ and $z$ are non-negative numbers.\r\nHence, it remains to prove that $x+y+z\\geq3,$ where $16=7x^{2}y^{2}z^{2}+3(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}).$\r\nLet $x+y+z<3.$ Hence, exists $k>1$ that $k(x+y+z)=3.$\r\nLet $kx=u,$ $ky=v$ and $kz=w.$\r\nThus, $16=\\frac{7u^{2}v^{2}w^{2}}{k^{6}}+\\frac{3(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})}{k^{4}}<7u^{2}v^{2}w^{2}+3(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2}).$\r\nBut it's contradiction since, $16\\geq7u^{2}v^{2}w^{2}+3(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})$ \r\nfor all non-negative $u,$ $v$ and $w$ such that $u+v+w=3.$", "Solution_2": ":lol: So excellent!!! You solution is totally the same as mine!!! Contradiction method is the main key !!! Also, it is the key for another problem of mine here \r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=152337\r\n\r\nHowever, no one finds this key! Thank you very much, Arqady, since all the solutions I know is very terribly complicated.", "Solution_3": "I have applied this method here too:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=64122", "Solution_4": "[quote=\"arqady\"]I have applied this method here too:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=64122[/quote]\r\n\r\nActually, I also have the same idea as yours in proving the above inequality. Moreover, I think the following similar one is true, Arqady\r\n\r\n\\[ \\sqrt{\\frac{a}{a+2b+c}}+\\sqrt{\\frac{b}{b+2c+a}}+\\sqrt{\\frac{c}{c+2a+b}}\\le \\frac{3}{2}.\\]", "Solution_5": "[quote=\"hungkhtn\"]I think the following similar one is true, Arqady\n\\[ \\sqrt{\\frac{a}{a+2b+c}}+\\sqrt{\\frac{b}{b+2c+a}}+\\sqrt{\\frac{c}{c+2a+b}}\\le \\frac{3}{2}. \\]\n[/quote]\r\nI have checked it and I agree with you.\r\nBy the way, $ \\sqrt{\\frac{a}{a+7b+c}}+\\sqrt{\\frac{b}{b+7c+a}}+\\sqrt{\\frac{c}{c+7a+b}}\\geq1$ already,\r\nbut $ \\sqrt{\\frac{a}{a+5b+c}}+\\sqrt{\\frac{b}{b+5c+a}}+\\sqrt{\\frac{c}{c+5a+b}}\\le\\frac{3}{\\sqrt7}$ also. :wink:", "Solution_6": "[quote=\"arqady\"][quote=\"hungkhtn\"]I think the following similar one is true, Arqady\n\\[ \\sqrt{\\frac{a}{a+2b+c}}+\\sqrt{\\frac{b}{b+2c+a}}+\\sqrt{\\frac{c}{c+2a+b}}\\le \\frac{3}{2}. \\]\n[/quote]\nI have checked it and I agree with you.\nBy the way, $ \\sqrt{\\frac{a}{a+7b+c}}+\\sqrt{\\frac{b}{b+7c+a}}+\\sqrt{\\frac{c}{c+7a+b}}\\geq1$ already,\nbut $ \\sqrt{\\frac{a}{a+5b+c}}+\\sqrt{\\frac{b}{b+5c+a}}+\\sqrt{\\frac{c}{c+5a+b}}\\le\\frac{3}{\\sqrt7}$ also. :wink:[/quote]\r\n\r\nYes, I use the same method as above. But it is really complicated! So I really curious of how could you find out fastly two above inequalities? :maybe: Are your solutions easier?", "Solution_7": "I am curious too :D How can solve it? It is good to see your solution :)", "Solution_8": "Let $ \\frac{a}{a+\\alpha b+c}=\\frac{x^{2}}{\\alpha+2},$ $ \\frac{b}{b+\\alpha c+a}=\\frac{y^{2}}{\\alpha+2}$ and $ \\frac{c}{c+\\alpha a+b}=\\frac{z^{2}}{\\alpha+2},$\r\nwhere $ \\alpha,$ $ x,$ $ y$ and $ z$ are non-negative numbers.\r\nThen $ \\begin{cases}(x^{2}-\\alpha-2)a+\\alpha x^{2}b+x^{2}c=0\\\\ y^{2}a+(y^{2}-\\alpha-2)b+\\alpha y^{2}c=0\\\\\\alpha z^{2}a+z^{2}b+(z^{2}-\\alpha-2)c=0\\end{cases}.$\r\nThus, $ \\det\\left( \\begin{array}{ccc}x^{2}-\\alpha-2 & \\alpha x^{2}& x^{2}\\\\ y^{2}& y^{2}-\\alpha-2 & \\alpha y^{2}\\\\ \\alpha z^{2}& z^{2}& z^{2}-\\alpha-2 \\end{array}\\right)=0.$\r\nId est, $ (\\alpha+2)^{2}=(\\alpha-1)^{2}x^{2}y^{2}z^{2}+(\\alpha-1)\\cdot\\sum_{cyc}x^{2}y^{2}+(\\alpha+2)\\cdot \\sum_{cyc}x^{2}.$\r\nIf $ \\alpha=7$ then for $ \\sum_{cyc}\\sqrt{\\frac{a}{a+7b+c}}\\geq1$ enough to prove that $ 27\\geq12x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+3(x^{2}+y^{2}+z^{2}),$\r\nwhere $ x+y+z=3,$ which equivalent to the following inequality:\r\n$ \\sum_{cyc}(x^{5}y+x^{5}z+x^{4}y^{2}+x^{4}z^{2}+9x^{4}yz+16x^{3}y^{2}z+16x^{3}z^{2}y-45x^{2}y^{2}z^{2})\\geq0.$", "Solution_9": "I have found mistake in the proof of $ \\sqrt{\\frac{a}{a+5b+c}}+\\sqrt{\\frac{b}{b+5c+a}}+\\sqrt{\\frac{c}{c+5a+b}}\\le\\frac{3}{\\sqrt7}.$\r\nThe inequality is not true.\r\nThank you, hungkhtn and zaizai-hoang. \r\nProbably, $ \\sqrt{\\frac{a}{a+3b+c}}+\\sqrt{\\frac{b}{b+3c+a}}+\\sqrt{\\frac{c}{c+3a+b}}\\le\\frac{3}{\\sqrt5}$ is true,\r\n but if it's true, it's not easy. :wink:", "Solution_10": "[quote=\"arqady\"]Let $ \\frac{a}{a+\\alpha b+c}=\\frac{x^{2}}{\\alpha+2},$ $ \\frac{b}{b+\\alpha c+a}=\\frac{y^{2}}{\\alpha+2}$ and $ \\frac{c}{c+\\alpha a+b}=\\frac{z^{2}}{\\alpha+2},$\nwhere $ \\alpha,$ $ x,$ $ y$ and $ z$ are non-negative numbers.\nThen $ \\begin{cases}(x^{2}-\\alpha-2)a+\\alpha x^{2}b+x^{2}c=0\\\\ y^{2}a+(y^{2}-\\alpha-2)b+\\alpha y^{2}c=0\\\\\\alpha z^{2}a+z^{2}b+(z^{2}-\\alpha-2)c=0\\end{cases}.$\nThus, $ \\det\\left( \\begin{array}{ccc}x^{2}-\\alpha-2 & \\alpha x^{2}& x^{2}\\\\ y^{2}& y^{2}-\\alpha-2 & \\alpha y^{2}\\\\ \\alpha z^{2}& z^{2}& z^{2}-\\alpha-2 \\end{array}\\right)=0.$\nId est, $ (\\alpha+2)^{2}=(\\alpha-1)^{2}x^{2}y^{2}z^{2}+(\\alpha-1)\\cdot\\sum_{cyc}x^{2}y^{2}+(\\alpha+2)\\cdot \\sum_{cyc}x^{2}.$\nIf $ \\alpha=7$ then for $ \\sum_{cyc}\\sqrt{\\frac{a}{a+7b+c}}\\geq1$ enough to prove that $ 27\\geq12x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+3(x^{2}+y^{2}+z^{2}),$\nwhere $ x+y+z=3,$ which equivalent to the following inequality:\n$ \\sum_{cyc}(x^{5}y+x^{5}z+x^{4}y^{2}+x^{4}z^{2}+9x^{4}yz+16x^{3}y^{2}z+16x^{3}z^{2}y-45x^{2}y^{2}z^{2})\\geq0.$[/quote]\r\n\r\nYes, I have a quite same solution, but I just use the elementary method to construct the relationship between $ \\alpha, \\beta,\\gamma$.", "Solution_11": "[quote=\"hungkhtn\"]Let $ a,b,c$ be non-negative real numbers. Prove that\n\n$ \\sqrt {\\frac {a}{b \\plus{} 3c}} \\plus{} \\sqrt {\\frac {b}{c \\plus{} 3a}} \\plus{} \\sqrt {\\frac {c}{a \\plus{} 3b}}\\ge\\frac {3}{2}.$[/quote][quote=\"arqady\"]Let $ \\frac {a}{b \\plus{} 3c} \\equal{} \\frac {x^{2}}{4},$ $ \\frac {b}{c \\plus{} 3a} \\equal{} \\frac {y^{2}}{4}$ and $ \\frac {c}{a \\plus{} 3b} \\equal{} \\frac {z^{2}}{4},$ where $ x,$ $ y$ and $ z$ are non-negative numbers.\n\nHence, it remains to prove that $ x \\plus{} y \\plus{} z\\geq3,$ where $ 16 \\equal{} 7x^{2}y^{2}z^{2} \\plus{} 3(y^{2}z^{2} \\plus{} z^{2}x^{2} \\plus{} x^{2}y^{2}).$\n\nLet $ x \\plus{} y \\plus{} z < 3.$ But it's contradiction since, [/quote]$ 7x^2y^2z^2 \\plus{} 3(y^2z^2 \\plus{} z^2x^2 \\plus{} x^2y^2)$\n\n$ < 7x^2y^2z^2\\left(\\frac {3}{x \\plus{} y \\plus{} z}\\right)^6 \\plus{} 3\\left(y^2z^2 \\plus{} z^2x^2 \\plus{} x^2y^2\\right)\\left(\\frac {3}{x \\plus{} y \\plus{} z}\\right)^4$\n\n$ \\equal{} 16 \\minus{} \\sum{\\frac {(y \\minus{} z)^2}{(x \\plus{} y \\plus{} z)^6}\\left[5x^4 \\plus{} 732x^2yz \\plus{} x(y \\plus{} z)(5x^2 \\plus{} 8y^2 \\plus{} 253yz \\plus{} 8z^2) \\plus{} 8(y \\plus{} z)^2(y^2 \\plus{} 11yz \\plus{} z^2)\\right]}$\n\n$ \\leq 16.$\n\nOther elementary solutions see : http://www.mathlinks.ro/viewtopic.php?t=105343\n[quote=\"hungkhtn\"]I think the following similar one is true, Arqady\n\n$ \\sqrt {\\frac {a}{a \\plus{} 2b \\plus{} c}} \\plus{} \\sqrt {\\frac {b}{b \\plus{} 2c \\plus{} a}} \\plus{} \\sqrt {\\frac {c}{c \\plus{} 2a \\plus{} b}}\\le \\frac {3}{2}.$[/quote][quote=\"arqady\"]Let $ \\frac {a}{a \\plus{} \\alpha b \\plus{} c} \\equal{} \\frac {x^{2}}{\\alpha \\plus{} 2},$ $ \\frac {b}{b \\plus{} \\alpha c \\plus{} a} \\equal{} \\frac {y^{2}}{\\alpha \\plus{} 2}$ and $ \\frac {c}{c \\plus{} \\alpha a \\plus{} b} \\equal{} \\frac {z^{2}}{\\alpha \\plus{} 2},$\n\nwhere $ \\alpha,$ $ x,$ $ y$ and $ z$ are non-negative numbers.\n\nThen $ \\begin{cases}(x^{2} \\minus{} \\alpha \\minus{} 2)a \\plus{} \\alpha x^{2}b \\plus{} x^{2}c \\equal{} 0 \\\\\ny^{2}a \\plus{} (y^{2} \\minus{} \\alpha \\minus{} 2)b \\plus{} \\alpha y^{2}c \\equal{} 0 \\\\\n\\alpha z^{2}a \\plus{} z^{2}b \\plus{} (z^{2} \\minus{} \\alpha \\minus{} 2)c \\equal{} 0\\end{cases}.$\n\nThus, $ \\det\\left( \\begin{array}{ccc}x^{2} \\minus{} \\alpha \\minus{} 2 & \\alpha x^{2} & x^{2} \\\\\ny^{2} & y^{2} \\minus{} \\alpha \\minus{} 2 & \\alpha y^{2} \\\\\n\\alpha z^{2} & z^{2} & z^{2} \\minus{} \\alpha \\minus{} 2 \\end{array}\\right) \\equal{} 0.$\n\nId est, $ (\\alpha \\minus{} 1)^{2}x^{2}y^{2}z^{2} \\plus{} (\\alpha \\minus{} 1)\\sum{y^{2}z^{2} } \\plus{} (\\alpha \\plus{} 2)\\sum{x^{2}} \\equal{} (\\alpha \\plus{} 2)^{2} .$[/quote]If $ \\alpha \\equal{} 2$ then for $ \\sum_{cyc}\\sqrt {\\frac {a}{a \\plus{} 2b \\plus{} c}}\\leq\\frac {3}{2}$ enough to prove that $ x^{2}y^{2}z^{2} \\plus{} y^{2}z^{2} \\plus{} z^{2}x^{2} \\plus{} x^{2}y^{2} \\plus{} 4(x^{2} \\plus{} y^{2} \\plus{} z^{2})\\geq 16,$\r\n\r\nwhere $ x \\plus{} y \\plus{} z \\equal{} 3,$ which equivalent to the following inequality:\r\n\r\n$ \\sum{(y \\minus{} z)^2\\left[73x^4 \\plus{} 85x^3(y \\plus{} z) \\plus{} 30x^2(y^2 \\plus{} z^2) \\plus{} 7x(y \\plus{} z)(4y^2 \\plus{} 11yz \\plus{} 4z^2) \\plus{} 10(y \\plus{} z)^4\\right]}$\r\n\r\n$ \\geq0.$\r\n\r\nBy the way, if $ a,b,c$ be positive numbers, then\r\n\r\n$ \\sqrt {\\frac {a}{a \\plus{} \\alpha b \\plus{} c}} \\plus{} \\sqrt {\\frac {b}{b \\plus{} \\alpha c \\plus{} a}} \\plus{} \\sqrt {\\frac {c}{c \\plus{} \\alpha a \\plus{} b}}\\le \\frac {3}{\\sqrt {\\alpha \\plus{} 2}}$\r\n\r\nholds if and only if $ 0\\leq \\alpha\\leq\\alpha_1 \\equal{} 3.2953\\cdots$, which the greatest root of $ 16\\alpha^5 \\minus{} 281\\alpha^4 \\plus{} 1825\\alpha^3 \\minus{} 2291\\alpha^2 \\minus{} 4130\\alpha \\plus{} 100,$\r\n\r\nwith equality if $ \\alpha \\equal{} \\alpha_1,a \\equal{} 1,b \\equal{} 1.0853\\cdots$ which the greatest root of $ b^5 \\minus{} 3b^4 \\plus{} 148b^3 \\plus{} 677b^2 \\minus{} 903b \\minus{} 4,$\r\n\r\n$ c \\equal{} 1.3040\\cdots$ which the greatest root of $ c^5 \\plus{} 167c^4 \\plus{} 7298c^3 \\minus{} 16981c^2 \\plus{} 8575c \\plus{} 1024.$\r\n\r\n$ \\alpha \\equal{} 3$ see : http://www.mathlinks.ro/viewtopic.php?t=299281", "Solution_12": "[quote=\"arqady\"]\nProbably, $ \\sqrt {\\frac {a}{a \\plus{} 3b \\plus{} c}} \\plus{} \\sqrt {\\frac {b}{b \\plus{} 3c \\plus{} a}} \\plus{} \\sqrt {\\frac {c}{c \\plus{} 3a \\plus{} b}}\\le\\frac {3}{\\sqrt5}$ is true,\n but if it's true, it's not easy. :wink:[/quote]\r\nNow it's easy enough! \r\nBy http://www.artofproblemsolving.com/Forum/viewtopic.php?p=893715#893715\r\nfor $ \\alpha\\equal{}3$ we need to prove that \r\n\\[ 25\\leq4x^2y^2z^2\\plus{}\\sum_{cyc}(2x^2y^2\\plus{}5x^2)\\] \r\nwhere $ x,$ $ y$ and $ z$ are non-negatives such that $ x\\plus{}y\\plus{}z\\equal{}3.$\r\nLet $ x\\plus{}y\\plus{}z\\equal{}3u,$ $ xy\\plus{}xz\\plus{}yz\\equal{}3v^2$ and $ xyz\\equal{}w^3.$\r\nHence, $ 25\\leq4x^2y^2z^2\\plus{}\\sum_{cyc}(2x^2y^2\\plus{}5x^2)\\Leftrightarrow f(w^3)\\geq0,$ where\r\n$ f(w^3)\\equal{}2w^6\\minus{}6u^3w^3\\plus{}10u^6\\minus{}15u^4v^2\\plus{}9u^2v^4.$\r\nBut $ f'(w^3)\\equal{}4w^3\\minus{}6u^3\\leq0.$\r\nHence, for the proof enough to check $ f(w^3)\\geq0$ for $ y\\equal{}z\\equal{}1,$\r\nwhich gives $ (x\\minus{}1)^2(10x^4\\plus{}50x^3\\plus{}87x^2\\minus{}148x\\plus{}244)\\geq0,$ which is true." } { "Tag": [ "integration", "algebra proposed", "algebra" ], "Problem": "Simplify $ f(n) \\equal{}\\sum_{k \\equal{} 1}^{n}{2^{2^{k}\\minus{}k}}$", "Solution_1": "$ f(n)\\equal{}\\int_{[0,2]}\\sum_{k\\equal{}1}^{n}x^{2^{k}\\minus{}1}$ idon't know if it help." } { "Tag": [ "inequalities", "three variable inequality", "algebra", "IMO Shortlist" ], "Problem": "Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that \\[ \\frac{ab}{ab + a^5 + b^5} + \\frac{bc}{bc + b^5 + c^5} + \\frac{ca}{ca + c^5 + a^5} \\leq 1. \\]", "Solution_1": "I think so. The solution without typos :\r\n\r\nNote that $a^5 + b^5 \\geq a^3b^2 + a^2b^3$ by Muirhead, rearrangement, AM - GM, ... so\r\n\r\n\\begin{eqnarray*} \\sum \\frac{ab}{a^5 + b^5 + ab} &\\leq& \\sum \\frac{ab}{a^3b^2 + a^2b^3 + ab} \\\\ \\ &=& \\sum \\frac{1}{a^2b + ab^2 + 1} \\\\ \\ &=& \\sum \\frac{abc}{a^2b + ab^2 + abc } \\\\ \\ &=& \\sum \\frac{c}{a + b + c} \\\\ \\ &=& 1. \\end{eqnarray*} \r\n\r\nWe're done ! :cool:", "Solution_2": "One can easily show that $ x^5 \\plus{} y^5 \\geq x^2 y^3 \\plus{} x^3 y^2$ for every x, y positive\r\n\r\n\r\n$ ab/(a^5\\plus{}b^5\\plus{}ab) \\leq ab/(a^2 b^3 \\plus{}a^3 b^2\\plus{}ab) \\equal{} 1/(a b^2\\plus{}a^2 b\\plus{}1) \\equal{} 1/[ab(a\\plus{}b\\plus{}c)] \\equal{}c/(a\\plus{}b\\plus{}c)$ since $ abc \\equal{}1$\r\n\r\nSimilarly $ bc/(b^5\\plus{}c^5\\plus{}bc) \\leq a/(a\\plus{}b\\plus{}c)$ and $ ca/(c^5\\plus{}a^5\\plus{}ca)\\leq b/(a\\plus{}b\\plus{}c)$\r\n\r\nBy adding those three inequalities we get the result we are looking for.\r\n\r\nCute no?", "Solution_3": "$ \\frac {xy}{x^5 \\plus{} y^5 \\plus{} xy} \\equal{} \\frac {xyz}{x^5z \\plus{} y^5z \\plus{} xyz} \\equal{} \\frac {1}{x^5z \\plus{} y^5z \\plus{} 1}$\r\nNow, $ \\frac {1}{a\\plus{}1} \\plus{} \\frac {1}{b\\plus{}1} \\plus{} \\frac {1}{c\\plus{}1} \\implies a \\plus{} b \\plus{} c \\plus{} 2 \\le abc$\r\nThus, we have to show\r\n$ 2 \\plus{} \\sum {x^5y} \\equal{} x^5y \\plus{} x^5z \\plus{} y^5x \\plus{} y^5z \\plus{} z^5x \\plus{} z^5y \\plus{} 2 \\le (x^5 \\plus{} y^5)(y^5 \\plus{} z^5)(x^5 \\plus{} z^5) \\equal{} 2 \\plus{} \\sum {x^{10}y^5}$\r\nBut $ \\sum{x^5y} \\equal{} \\sum{x^8y^4z^3}$, and we have Muirhead.", "Solution_4": "After homogenizing by turning the $ab$ terms into $a^2b^2c$ and etc and multiplying by 2, straight-up expansion gives the LHS to be\n\n$[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$\n\nand the RHS to be\n\n$[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$\n\nSubtracting gives $RHS-LHS=2[10,5,0]-2[8,4,3]$ but $[10,5,0] \\geq [8,4,3]$ by Muirhead so we're done.", "Solution_5": "1998,I proof:\nSuppose that $a, b, c > 0$ such that $abc = 1$ , $n$ be positive integer. Prove that \n\\[ \\frac{a^{n-1}b^{n-1}}{a^{n-1}b^{n-1} + a^{2n+1} + b^{2n+1}} + \\frac{b^{n-1}c^{n-1}}{b^{n-1}c^{n-1} + b^{2n+1} + c^{2n+1}} \n\\[\\\\ + \\frac{c^{n-1}a^{n-1}}{c^{n-1}a^{n-1} + c^{2n+1} + a^{2n+1}} \\leq 1. \\]\nAnother generalization \uff1a\nLet $a,b,c $ be positive real numbers such that $abc=1 ,n\\le-1 $ or $ n\\ge2 ,$ Show that\\[\\frac{bc}{b^n+c^n+bc}+\\frac{ca}{c^n+a^n+ca}+\\frac{ab}{a^n+b^n+ab}\\le1\\]\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2835835\n\nSimilarly have\nLet $a,b,c>0 $ such that $abc=1 ,n\\ge1 $ or $ n\\le-2 ,$ Show that\n\\[\\frac{a^n}{a^n+b+c}+\\frac{b^n}{b^n+c+a}+\\frac{c^n}{c^n+a+b}\\ge1\\]\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2724850#p2724850", "Solution_6": "[quote=\"Arne\"]Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that \\[ \\frac{ab}{ab + a^5 + b^5} + \\frac{bc}{bc + b^5 + c^5} + \\frac{ca}{ca + c^5 + a^5} \\leq 1. \\][/quote]\n\n$a^5+b^5 = \\frac{a^5}{5}+\\frac{a^5}{5}+\\frac{a^5}{5}+\\frac{b^5}{5}+\\frac{b^5}{5}+\\frac{a^5}+{5}\\frac{a^5}{5}+\\frac{b^5}{5}+\\frac{b^5}{5}\\frac{b^5}{5} \\ge a^3b^2+a^2b^3 $ \nNow , \n$\\sum \\frac{ab}{ab + a^5 + b^5} \\le \\sum \\frac{1}{a^2b+ab^2+1} \\le \\sum \\frac{abc}{ab(a+b+c} \\le \\sum \\frac{c}{a+b+c)} = 1 \\Box$", "Solution_7": "Notice that we can rewrite given $LHS$ as \\[\\sum_{cyc}{\\frac{ab}{ab+a^5+b^5}}=\\sum_{cyc}{\\frac{1}{1+a^5c+b^5c}}.\\]\nThen using supstiution $a=\\frac{x}{y},b=\\frac{y}{z},c=\\frac{z}{x}$ we get \\[\\sum_{cyc}{\\frac{1}{1+a^5c+b^5c}}=\\sum_{cyc}\\frac{1}{1+\\frac{x^4z}{y^5}+\\frac{y^5}{z^4x}}\\leq \\sum_{cyc}\\frac{1}{2\\cdot \\sqrt\\frac{x^3}{z^3}+1}\\] which we got by $AM-GM$.\nThen we have \\[\\sum_{cyc}\\frac{1}{2\\cdot \\sqrt\\frac{x^3}{z^3}+1}\\leq\\sum_{cyc}\\frac{1}{2\\cdot \\sqrt\\frac{x^3}{x^3}+1}=\\sum_{cyc}\\frac{1}{2+1}=\\frac{1}{3}\\cdot 3=1.\\]Thats true because of rearrangement.\n$Q.E.D.$", "Solution_8": "@above: I don't see how you applied rearrangment in the final line. Also many solutions here use $a^5+b^5 \\ge a^3b^2+b^2a^3$ (*). I have seen some times before the trick where you reduce an inequality to cylcic sum of $a/(a+b+c)$ equals 1, and if you have this trick in mind then it is quite easy to find that you need to prove (*), and then the problem is solved. But it seems to me a very unnatural thing to think of the trick first, and to me its not obvious at all that the trick is applicable here at first, so it does not come to mind. My question is then: how did you think to use this inequality (*)?", "Solution_9": "Given $x_1, x_2, x_3,....,x_{2015} >0$ and $x_1+x_2+x_3+....+x_{2015}=1$. Find minimum value of $P=(1+x_1^2)(1+x_2^2)(1+x_3^2)...(1+x_{2015}^2)$", "Solution_10": "@above: your post has nothing do with the problem in this thread. you should remove it i think.", "Solution_11": "[quote=Mewto55555]After homogenizing by turning the $ab$ terms into $a^2b^2c$ and etc and multiplying by 2, straight-up expansion gives the LHS to be\n\n$[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$\n\nand the RHS to be\n\n$[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$\n\nSubtracting gives $RHS-LHS=2[10,5,0]-2[8,4,3]$ but $[10,5,0] \\geq [8,4,3]$ by Muirhead so we're done.[/quote]\n\n\nI don't know what he mean by $[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$ and $[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$\n\nCan someone explain ?", "Solution_12": "[quote=velraman][quote=Mewto55555]After homogenizing by turning the $ab$ terms into $a^2b^2c$ and etc and multiplying by 2, straight-up expansion gives the LHS to be\n\n$[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$\n\nand the RHS to be\n\n$[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$\n\nSubtracting gives $RHS-LHS=2[10,5,0]-2[8,4,3]$ but $[10,5,0] \\geq [8,4,3]$ by Muirhead so we're done.[/quote]\n\n\nI don't know what he mean by $[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$ and $[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$\n\nCan someone explain ?[/quote]\n\nHe is probably referring to the coefficients, and the ordered triplet represents the powers of a,b, and c respectively.", "Solution_13": "My Solution :\n\nObserve that $(x^3-y^3)(x^2-y^2) \\ge 0 \\implies x^5 + y^5 \\ge x^2y^2(x+y) \\ \\forall x,y \\in \\mathbb{R^+}$.\n\nThus, $\\sum_{cyc} \\dfrac {ab}{a^5+b^5+ab} \\le \\sum_{cyc} \\dfrac{ab}{a^2b^2(a+b)+ab} = \\sum_{cyc} \\dfrac{c}{a+b+c} = 1$.", "Solution_14": "\\[(ab+a^5+b^5)\\left(\\frac{c^4}{ab} + \\frac1a + \\frac1b\\right)\\ge (a^2+b^2 + c^2)^2\\]\n\\[\\iff \\frac{ab}{ab+a^5+b^5} \\le \\frac{c^4 + a + b}{(a^2 + b^2 + c^2)^2}\\]\nSumming up similar inequalities. \n\\[\\text{LHS} \\le \\frac{\\sum a^4 +2\\sum a }{(a^2 + b^2 + c^2)^2}\\]\nSo, it suffices to prove that \n\\begin{align*}\n\\frac{\\sum a^4 +2\\sum a }{(a^2 + b^2 + c^2)^2}&\\le 1\\\\\n\\iff a^2b^2 + b^2c^2 + c^2a^2 &\\ge a + b + c\\\\\n&= abc(a+b+c)\\\\\n\\iff \\sum a^2(b-c)^2 &\\ge 0\n\\end{align*}", "Solution_15": "Nice one [b]Vrangr[/b] :) ", "Solution_16": "observe that $$x^{5}+y^{5} = (x+y)(x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}+y^{4})$$\n$$=(x+y)(x^{3}(x-y) -y^{3}(x-y) +x^{2}y^{2}) = (x+y)((x-y)^{2}(x^{2}+y^{2}+xy)+x^{2}y^{2})\\ge (x+y)(x^{2}y^{2})$$\nsince $(x-y)^{2}\\ge0$\n$$\\Rightarrow \\sum_{cyc}\\frac{xy}{x^{5}+y^{5}+xy}\\le \\sum\\frac{xy}{(x+y)(x^{2}y^{2})+xy}= \\sum\\frac{1}{(x+y)xy+1}$$\n$$=\\sum\\frac{xyz}{(x+y)xy+xyz}= \\sum\\frac{z}{x+y+z}=1$$\nAnd we are done!", "Solution_17": "[quote=Arne]Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that \\[ \\frac{ab}{ab + a^5 + b^5} + \\frac{bc}{bc + b^5 + c^5} + \\frac{ca}{ca + c^5 + a^5} \\leq 1. \\][/quote]\n\n$$\\sum\\frac{ab}{ab+a^5+b^5}=\\sum\\frac{a^2b^2c}{a^2b^2c+a^5+b^5}\\leq\\sum\\frac{a^2b^2c}{a^2b^2c+a^3b^2+a^2b^3}=\\sum\\frac{c}{a+b+c}=1$$ :)", "Solution_18": "Observe that (By Muirhead) $a^5+b^5\\ge a^2b^2(a+b)$. We have\n\\begin{align*}\n\\sum_{cyc}\\frac{ab}{a^5+ab+b^5}&\\le\\sum_{cyc}\\frac{ab}{a^2b^2(a+b)+ab} \\\\\n&=\\sum_{cyc}\\frac1{ab(a+b)+1} \\\\\n&=\\sum_{cyc}\\frac1{ab(a+b)+abc} \\\\\n&=\\sum_{cyc}\\frac{1}{ab(a+b+c)} \\\\\n&=\\sum_{cyc}\\frac c{a+b+c} \\\\\n&=1\n\\end{align*}\n\nVery similar to a problem from USAMO 1998.", "Solution_19": "[quote=Arne]Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that \\[ \\frac{ab}{ab + a^5 + b^5} + \\frac{bc}{bc + b^5 + c^5} + \\frac{ca}{ca + c^5 + a^5} \\leq 1. \\][/quote]\nLet $a,b,c>0$ with $abc=1.$ [url=https://artofproblemsolving.com/community/c6h2071085p14829277]Establish that[/url]\n$$\\frac{1}{a^5+b^5+1}+\\frac{1}{b^5+c^5+1}+\\frac{1}{c^5+a^5+1}\\leq 1.$$\n$$\\iff$$\n$$\\frac{1}{a^3+b^3+1} +\\frac{1}{b^3+c^3+1} +\\frac{1}{c^3+a^3+1} \\leq 1$$\n[url=https://artofproblemsolving.com/community/c6h578290p3411216]h[/url] [url=https://artofproblemsolving.com/community/c6h1305663p6976340]h[/url] [url=https://artofproblemsolving.com/community/c4h1883433p12823952]h[/url]", "Solution_20": "By Muirhead\u2019s inequality, $b^5 + c^5 \\ge b^3c^2 + b^2c^3$. Hence,\n$$\\sum_{cyc} \\frac{ab}{ab+a^5+b^5} \\le \\sum_{cyc} \\frac{ab}{ab+a^3b^2 + a^2b^3} = \\sum_{cyc} \\frac{1}{1+a^2b+b^2a} = \\sum_{cyc} \\frac{c}{c+a+b} = 1$$\nas desired.", "Solution_21": "[quote=levinhkiet]Given $x_1, x_2, x_3,....,x_{2015} >0$ and $x_1+x_2+x_3+....+x_{2015}=1$. Find minimum value of $P=(1+x_1^2)(1+x_2^2)(1+x_3^2)...(1+x_{2015}^2)$[/quote]\n\neasy the answer is $\\frac{2016}{2015}$\n", "Solution_22": "WLOG let $a$$\\leq$$ b$$ \\leq c$. So we have, from rearrangement inequality, $\\begin{bmatrix} a^3 & b^3 \\\\ a^2 & b^2 \\end{bmatrix} \\geq \\begin{bmatrix} a^3 & b^3 \\\\ b^2 & a^3 \\end{bmatrix}$. Which in turn gives , $a^5+b^5 \\geq a^{3}b^{2} + a^{2}b^{3} $ \\\\ $\\frac {ab}{a^5+b^5 +ab} $$\\leq$$ \\frac {ab}{ a^{3}b^{2} + a^{2}b^{3}+ab}$\\\\ \\\\ Then it follows :)", "Solution_23": "Note that \n$\\frac{xy}{x^5+ xy+y^5}=\\frac{1}{z(x^5+y^5)+1} \\leq \\frac{1}{z(x^4y+xy^4)+1}=\\frac{1}{(x^3+y^3+1)} \\leq \\frac{1}{xy(x+y)+1}=\\frac{z}{x+y+z}$ and the result follows by summing this cyclically $\\Box$", "Solution_24": "$a^5+b^5\\ge a^4b+ab^4$ and $a^3+b^3\\ge a^2b+ab^2$ are direct by Rearrangement, so we have:\n$\\sum_{\\text{cyc}}\\frac{ab}{ab+a^5+b^5}\\le\\sum_{\\text{cyc}}\\frac1{1+a^3+b^3}\\le\\sum_{\\text{cyc}}\\frac c{c+a^2bc+ab^2c}=\\sum_{\\text{cyc}}\\frac a{a+b+c}=1$", "Solution_25": "For positive reals $x,y,z$ with $xyz=1$, prove or disprove that:\n$$\\frac1{x+y+1}+\\frac1{y+z+1}+\\frac1{z+x+1}\\le1.$$", "Solution_26": "The classic Muirhead for denominator inequalities!\n\nWe have that $\\sum_{\\text{cyc}} \\frac{ab}{ab+a^5+b^5}\\leq \\sum_{\\text{cyc}} \\frac{ab}{ab+a^3b^2+a^2b^3} = \\sum_{\\text{cyc}} \\frac{1}{abc+a^2b+ab^2}=\\frac{a+b+c}{a+b+c}=1.$\n\nRemarks: In the beginning, I tried to use AM-GM directly on the denominator to get that I wanted to show that $\\sum_{cyc} \\frac{1}{ab}\\leq 3\\implies a+b+c\\leq 3,$ but obviously the inequality was too weak. :(", "Solution_27": "\\begin{align*}\n\\sum_{\\text{cyc}} \\dfrac{ab}{{\\color{red}a^5+b^5}+ab} &\\le \\sum_{\\text{cyc}} \\dfrac{ab}{{\\color{blue}a^3b^2 + a^2b^3} + ab} \\\\\n&= \\sum_{\\text{cyc}} \\frac{ab}{ab(a^2b + ab^2 + {\\color{red}1})}\\\\\n&= \\sum_{\\text{cyc}} \\frac{1}{a^2b + ab^2 + {\\color{blue}abc}} \\\\\n&= \\sum_{\\text{cyc}} \\frac{1}{{\\color{red}ab}(a+b+c)} \\\\\n&= \\sum_{\\text{cyc}} \\frac{\\color{blue}c}{a+b+c}\\\\\n&= 1\n\\end{align*}", "Solution_28": "By Muirhead's Theorem: $a^5+b^5 \\geq a^2b^3+a^3b^2$\n$\\sum_{\\text{cyc}} \\frac{ab}{a^5+b^5+ab}\\leq \\sum_{\\text{cyc}} \\frac{ab}{a^3b^2+a^2b^3+ab} = \\sum_{\\text{cyc}} \\frac{1}{a^2b+ab^2+1}=\\sum_{\\text{cyc}}\\frac{c}{a+b+c}=\\frac{a+b+c}{a+b+c}=1$.$\\blacksquare$", "Solution_29": "Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that$$\\frac{ab}{2ab + a^5 + b^5} + \\frac{bc}{2bc + b^5 + c^5} + \\frac{ca}{2ca + c^5 + a^5} \\leq \\frac{3}{4}$$\n$$\\frac{ab}{ab + 2a^5 + b^5} + \\frac{bc}{bc +2b^5 + c^5} + \\frac{ca}{ca + 2c^5 + a^5} \\leq \\frac{3}{4}$$\n$$ \\frac{ab}{kab + a^5 + b^5} + \\frac{bc}{kbc + b^5 + c^5} + \\frac{ca}{kca + c^5 + a^5}\\leq \\frac{3}{k+2}$$\nWhere $k\\geq 1.$\n", "Solution_30": "[quote=Arne]Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that \\[ \\frac{ab}{ab + a^5 + b^5} + \\frac{bc}{bc + b^5 + c^5} + \\frac{ca}{ca + c^5 + a^5} \\leq 1. \\][/quote]\n\n$(a^3-b^3)(a^2-b^2) \\geq 0 \\implies a^5+b^5 \\geq a^2b^2(a+b),$ equality when $a=b$. So, $\\frac{ab}{a^5+b^5+ab} \\leq \\frac{ab}{a^2b^2(a+b)+ab} = \\frac{1}{ab(a+b)+1} =\\frac{abc}{ab(a+b+c)} =\\frac{c}{a+b+c}.$ \nSimilarly, $\\frac{bc}{b^5+c^5+bc} \\leq \\frac{a}{a+b+c}$ and $\\frac{ca}{c^5+a^5+ca} \\leq \\frac{b}{a+b+c},$ equality when $a=b=c=1.$ \nThe claim follows adding these three. $\\square$", "Solution_31": "Same as above solutions; posting for storage.\n\nBy Muirhead, $a^5+b^5\\ge a^3b^2+b^3a^2.$ Hence, $$\\sum_{\\text{cyc}}\\frac{ab}{ab+a^5+b^5}\\le\\sum_{\\text{cyc}}\\frac{ab}{ab+a^3b^2+b^3a^2}=\\sum_{\\text{cyc}}\\frac{1}{\\frac{a}{c}+\\frac{b}{c}+\\frac{c}{c}}=\\sum_{\\text{cyc}}\\frac{c}{a+b+c}=1.$$ $\\square$", "Solution_32": "Every solution is going to look the same I guess.\nWe first show that $a^5+b^5 \\ge a^3b^2++b^3a^2$. This will be just expanding and applying AM-GM inequality.\n\\begin{align*}\na^5+b^5 = (a+b)(a^4-a^3b+a^2b^2-ab^3+b^4) \\ge (a+b)(2a^2b^2+a^2b^2-ab(a^2+b^2)) \\ge (a+b)(3a^2b^2-2a^2b^2) = a^3b^2+b^3a^2\n\\end{align*}\nReturning to the problem:\n\\begin{align*}\n\\sum_{\\text{cyc}}\\frac{ab}{ab+a^5+b^5} = \\sum_{\\text{cyc}}\\frac{1}{1+c(a^5+b^5)} \\le \\sum_{\\text{cyc}}\\frac{1}{1+c \\cdot a^2b^2(a+b)} = \\sum_{\\text{cyc}}\\frac{1}{1+ab(a+b)} = \\sum_{\\text{cyc}}\\frac{c}{a+b+c} =1\n\\end{align*}\nwhich means we are done. $\\blacksquare$", "Solution_33": "[quote=jasperE3]For positive reals $x,y,z$ with $xyz=1$, prove or disprove that:\n$$\\frac1{x+y+1}+\\frac1{y+z+1}+\\frac1{z+x+1}\\le1.$$[/quote]\n\nInteresting. You can easily prove this result is true from straight-up expansion and then Muirhead. \n\nNote that this is actually equivalent to the original problem. From the substitution $x=a^5b$, $y=b^5c$, etc, the LHS turns into $\\sum_{cyc}{\\frac{1}{ab^5 + bc^5 + 1}}$, and dividing both top and bottom by $abc=1$ gives the original inequality. \n\n------\n\nI would be curious to see if anyone has a solution from Holder's/Cauchy's/Titu's (from $\\sum_{cyc}{\\frac{a^5+b^5}{ab+a^5+b^5}} \\ge 2$.) I tried but didn't get anywhere. ", "Solution_34": "By Power-Mean Inequality, we obtain\n$$\\sqrt[5]{\\frac{a^5+b^5}{2}} \\geq \\frac{a+b}{2}$$\n\\begin{align*}\n\\implies \\frac{a^5+b^5}{2} &\\geq \\left(\\frac{a+b}{2}\\right)^5 \\\\\n&\\geq \\left(\\frac{a+b}{2}\\right)^4 \\cdot \\frac{a+b}{2} \\\\\n& \\geq a^2b^2\\left(\\frac{a+b}{2}\\right) \\text{[using AM-GM]}\\\\\n\\end{align*}\n$$\\therefore \\boxed{a^5+b^5\\geq a^2b^2(a+b)}.$$\n\nHence,\n\\begin{align*}\n\\sum_{cyc} {\\frac{ab}{a^5+b^5+ab}} &\\leq \\sum_{cyc} {\\frac{ab}{a^2b^2(a+b)+ab}} \\\\\n&= \\sum_{cyc} {\\frac{1}{ab(a+b)+1}} \\\\\n&= \\sum_{cyc} {\\frac{c}{a+b+c}} \\text{[using $abc=1$]} \\\\\n&= 1.\n\\end{align*}\n$QED.$", "Solution_35": "Solved with [b]CT17[/b].\n\nNotice that $(b^3-c^3)(b^2-c^2) \\ge 0 \\implies b^5+c^5 \\ge b^3c^2+b^2c^3$, so we have \\[\\sum_\\text{cyc} \\frac{bc}{bc+b^5+c^5} \\le \\sum_\\text{cyc} \\frac{bc}{bc+b^3c^2+b^2c^3}=\\sum_\\text{cyc} \\frac{\\frac{1}{a}}{\\frac{1}{a}+\\frac{b}{a^2}+\\frac{c}{a^2}}=\\sum_\\text{cyc} \\frac{a}{a+b+c}=1,\\] as desired. $\\square$", "Solution_36": "[b]Version 1[/b]\nLet $a,b,c$ be positive reals such that $abc=2$. Then prove that\n\n\n$$\\sum_{cyc}{\\dfrac{a^2b^2}{2\\left(a^2b^2\\right)+a^{7}+b^{7}}}\\leq \\dfrac{1}{2}$$", "Solution_37": "[b]Generalization 1[/b]\nLet $a,b,c$ be positive reals such that $abc=k$. Then prove that\n\n\n$$\\sum_{cyc}{\\dfrac{(ab)^{n}}{k\\left(ab\\right)^{n}+a^{2n+3}+b^{2n+3}}}\\leq \\dfrac{1}{k}$$", "Solution_38": "From rearrangement inequality, we have: $a^5+b^5\\geq a^3b^2+a^2b^3 \\implies a^5+b^5\\geq a^2b^2(a+b)$\n$$\\implies \\sum_{cyc} {\\frac{ab}{a^5+b^5+ab}} \\leq \\sum_{cyc} {\\frac{ab}{a^2b^2(a+b)+ab}}$$\n$$\\implies \\sum_{cyc} {\\frac{ab}{a^5+b^5+ab}} \\leq \\sum_{cyc} {\\frac{1}{ab(a+b)+1}}$$\n$$\\implies \\sum_{cyc} {\\frac{ab}{a^5+b^5+ab}} \\leq \\sum_{cyc} {\\frac{abc}{ab(a+b)+abc}}$$\n$$\\implies \\sum_{cyc} {\\frac{ab}{a^5+b^5+ab}} \\leq \\sum_{cyc} {\\frac{c}{a+b+c}}$$\n$$\\implies \\boxed{\\sum_{cyc} {\\frac{ab}{a^5+b^5+ab}} \\leq 1}$$" } { "Tag": [ "geometry", "geometric transformation", "reflection", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Here is a beautiful problem. I already have a solution, but I'm looking for other (creative) solutions.\r\n\r\n[i]Find the area of the region\n\\[R=\\{(x,y)\\in\\mathbb{R}^{2}: \\sqrt{x}+\\sqrt{y}\\geq 1,\\ \\sqrt{1-x}+\\sqrt{1-y}\\geq 1\\}. \\][/i]\r\nMasoud Zargar", "Solution_1": "Here's something straightforward. The set in question is contained in the square $[0,1]\\times[0,1].$ The transformation $(x,y)\\mapsto(1-y,1-x)$ reflects the region about the line $x+y=1$ but leaves it unchanged. The region lies between two curves. What I will do is calculate the area below the lower curve:\r\n\r\n$\\int_{0}^{1}\\left(1-\\sqrt{x}\\right)^{2}\\,dx=\\int_{0}^{1}1-2\\sqrt{x}+x\\,dx=\\frac16.$\r\n\r\nThe region above the upper curve has the same area. We can find the area we want by subtraction:\r\n\r\nArea = $1-\\frac16-\\frac16=\\frac23.$" } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "Is it true that positive reals sattisfy\r\n$\\sqrt{\\frac{(x+y+z)^3}{x^2+y^2+z^2+xy+yz+zx}}\\ge \\frac{\\sqrt{x}+\\sqrt{y}+\\sqrt{z}}{\\sqrt{2}}$", "Solution_1": "[quote=\"Beat\"]Is it true that positive reals sattisfy\n$\\sqrt{\\frac{(x+y+z)^3}{x^2+y^2+z^2+xy+yz+zx}}\\ge \\frac{\\sqrt{x}+\\sqrt{y}+\\sqrt{z}}{\\sqrt{2}}$[/quote]\r\n\r\nAlas, it isn't: try x = 1, y = 1, z = 3.\r\n\r\n Darij" } { "Tag": [ "algebra", "polynomial", "rotation", "analytic geometry", "function" ], "Problem": "I was asked to post these once they were available, so here they are. The solutions are online as well, but have a bit of restraint and try to solve the problems rather than just searching for solutions.\r\n\r\n1. Consider the equation\r\nx1x2 + x2x3 + x3x4 + + xn\u22121xn + xnx1 = 0\r\nwhere xi is selected from the set {1,\u22121} for i = 1, 2, . . . , n.\r\n(a) Show that if the equation has a solution, then n is even. (2 points)\r\n(b) Suppose n is divisible by 4. Show that the equation has a solution. (2 points)\r\n(c) Show that if the equation has a solution, then n is divisible by 4. (6 points)\r\n\r\n\r\n2. (a) Find a polynomial f(x) with integer coeffcients and two distinct integers a and b such that\r\nf(a) = b and f(b) = a. (2 points)\r\n(b) Let f(x) be a polynomial with integer coeffcients and a, b, and c be three integers. Suppose f(a) = b, f(b) = c, and f(c) = a. Show that a = b = c. (8 points)\r\n\r\n\r\n3. (a) Consider the triangle with vertices M(0, 2n + 1), S(1, 0), and U(0, 1/(2n :^2:)), where n is a positive\r\ninteger. If :theta: = 6 MSU, prove that tan :theta: = 2n \u2212 1. (5 points)\r\n(b) Find positive integers a and b that satisfy the following equation. (2 points)\r\narctan 1/8 = arctan a \u2212 arctan b\r\n(c) Determine the exact value of the following infinite sum. (3 points)\r\narctan 1/2 + arctan 1/8 + arctan 1/18 + arctan 1/32+ + arctan 1/(2n :^2:) + \r\n\r\n\r\n4. (a) Prove: (55 + 12*:sqrt:21)1/3 + (55 \u2212 12*:sqrt:21)1/3 = 5. (5 points)\r\n(b) Completely factor x8 + x6 + x4 + x2 + 1 into polynomials with integer coefficients, and explain\r\nwhy your factorization is complete. (5 points)\r\n\r\n\r\n5. In this problem, we simulate a hula hoop as it gyrates about your waist. We model this situation by\r\nrepresenting the hoop with a rotating a circle of radius 2 initially centered at (\u22121, 0), and representing\r\nyour waist with a fixed circle of radius 1 centered at the origin. Suppose we mark the point on\r\nthe hoop that initially touches the fixed circle with a black dot (see the left figure).\r\nAs the circle of radius 2 rotates, this dot will trace out a curve in the plane (see the right figure).\r\nLet :theta: be the angle between the positive x-axis and the ray that starts at the origin and goes through\r\nthe point where the fixed circle and circle of radius 2 touch. Determine formulas for the coordinates\r\nof the position of the dot, as functions x(:theta:) and y(:theta:). The left figure shows the situation when :theta: = 0\r\nand the right figure shows the situation when :theta: = 2 :pi: /3. (10 points) (Sorry, guys, I can't attach the diagrams easily, so visit http://www.delta.edu/math/mmpc/graphics/exams/2003.II.problems.pdf if you need them or if there are typos in any of the previous problems)", "Solution_1": "1. [hide]a. Clearly, half the terms will have to be -1 and half the terms 1. As there are n terms, n is even. b. let those with indices congruent to 0 or 1 mod 4 be 1, and those with indices congruent to 2 or 3 mod 4 be -1. Solution follows. c. consider changing one of the \"-1\"s from -1 to 1. It will change the sum by either -4, 0, or 4, so the sum remains constant mod 4. As the sum was 0 mod 4 at the beginning, when all the terms are 1, we have n:equiv:0 mod 4.[/hide]\n\n\n\n2. [hide]a. f(x)=1-x. try x=0, 1 b. If f(a)=b and f(c)=d and f is a polynomial function, (c-a)|(d-b). In this case, we have (b-a)|(c-b), (c-b)|(a-c), (a-c)|(b-a). We thus have (b-a)=:pm:(c-b), and cyclic permutations of that equation. That equations simplifies to a=c OR a+c=2b. Anyway, it's easy enough from here (remember we also have cyclic permutations of those equations) to come up with a=b=c.[/hide]\n\n\n\n4. [hide]a. let x=that thing. x^3=55+12:sqrt:21+55-12:sqrt:21+3x[(55+12:sqrt:21)(55-12:sqrt:21)]^(1/3)=110+3x => x^3-3x-110=0. This factors into (x-5)(x^2+5x+22)=0. This has roots x=5, and some complex roots. However, that thing was obviously real. Thus x=5. b. Using fifth roots of unity, we find that if we plug in any fifth root of unity, we get 0. Thus, (x^4+x^3+x^2+x+1) must factor that. Indeed it does, and it factors into (x^4+x^3+x^2+x+1)(x^4-x^3+x^2-x+1). Now, it is easy to see that this is a complete factorization because that can be written as (x^5-1)(x^5+1)/(x-1)(x+1), and the only things that divide that are complex roots of -1 and 1. [/hide]", "Solution_2": "3. [hide]a. in the problem statement, :theta:=MSU, not 6MSU, so i'll assume the problem is right and that was just a typo. in any case, (O is the origin in case you didn't guess) :theta:=OSM-OSU, and tan:theta:=[(2n+1)-(1/2n^2)]/[1+(2n+1)(1/2n^2)]=[4n^3+2n^2-1]/[2n^2+2n+1]=2n-1. b. 8=(1+ab)/(a-b) => 8a-8b=1+ab => ab-8a+8b=-1 => (a+8)(b-8)=-65. (a,b)=(57, 7) is one solution methinks. It's easy to find all solutions but I won't. c. by (a), arctan(1/2n^2)=arctan(2n+1)-arctan(2n-1) so this whole thing is arctan(:inf:) (you know that means limit of arctan n as n goes to infinity) - arctan (1)= :pi:/2-:pi:/4=:pi:/4[/hide]", "Solution_3": "5. [hide]at :theta:, the circles intersect (are tangent, whatever) at (cos :theta:, sin :theta:). The center of the big circle will be the negative of that. Okay, now if the big circle were centered at the origin, the marked point would be at (2cos :theta:/2, 2sin :theta:/2). We now translate that by the negative of that original point/vector to get (x(:theta:), y(:theta:))=(2cos:theta:/2-cos:theta:, 2sin:theta:/2-sin:theta:)[/hide]", "Solution_4": "does everyone hate my solutions so much they don't even want to discuss them...", "Solution_5": "Mystic, I told you they looked good... and got some information out of you on AIM ;) What should I say here?", "Solution_6": "oh sorry i forgot. i was hoping people might possibly have other solutions though but whatever." } { "Tag": [], "Problem": "When the temperature goes up $ 3^\\circ$ on the Cantor scale, it goes up $ 8^\\circ$ on the Frobenius scale. On both scales, $ 18^\\circ$ is the same temperature. How many Frobenius degrees are equal to $ 30^\\circ$ Cantor?", "Solution_1": "When the temperature goes up 3on the Cantor scale, it goes up 8 on the Frobenius scale.\r\nWhen it goes up 12 on Cantor (18 to 30), it goes up 32 on the Frobenius scale.\r\n18+32=[b]50[/b]" } { "Tag": [ "function", "trigonometry", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Show that continuity of the differential is essential in the inverse function theorem by considering the function $ f: (\\minus{}1,1)\\to\\mathbb{R}$ defined by $ f(0)\\equal{}0$ and $ f(t)\\equal{}t\\plus{}2t^2\\sin(1/t)$ for $ t\\neq0$. Show that $ f$ is everywhere differentiable, and even that $ f'$ is bounded, that $ f'(0)\\equal{}1$, but that $ f$ is not injective in any neighborhood of $ 0$", "Solution_1": "Which part are you having trouble with? For example:\r\n$ f'(t) \\equal{} 1 \\plus{} 4t\\sin\\left(\\frac {1}{t}\\right) \\minus{} 2\\cos\\left(\\frac {1}{t}\\right)$ for $ t\\neq 0$. Since $ |t| < 1$ and sine and cosine are bounded anyway, $ |f'(t)|$ is bounded.\r\nAnd:\r\n$ f'(0) \\equal{} \\lim_{x\\to 0}\\frac {f(x) \\minus{} f(0)}{x} \\equal{} \\lim_{x\\to 0}\\frac {x \\plus{} 2x^2\\sin\\left(\\frac {1}{x}\\right)}{x} \\equal{} \\lim_{x\\to 0}1 \\plus{} 2x\\sin\\left(\\frac {1}{x}\\right) \\equal{} 1$. Ehm..I'm doing something wrong here apparently (it should be $ 0$)." } { "Tag": [ "geometry", "geometric transformation", "rotation" ], "Problem": "examin the relationship below. Determine whcih of the following properties each relationship satisfies: the Reflexive Property (a is a real number, a=a), the symmetric property(a=b, b=a). Transitive property(a=b, b=c, a=c). Then state wether the relationship is an equivalence relation. If a preperty is not satisfied by a given relationship, give a counterexample.\r\n\r\n1 A less than B, where A and B are real numbers\r\n\r\n2.A less or equal than B( <= ), A and B are real numbers\r\n\r\n3.A is disvisible by b, where a and b are real numbers\r\n\r\n4.Figure a is a relfection of figure b.\r\n\r\n5.Firgure a is a rotation of firgure b\r\n\r\n6. A is a sister of B, where a and b are people.\r\n\r\n53. A has the same last name as b, where a and b are people", "Solution_1": "[hide]\n1. Transitive, not equivalence\n2. Equivalence (in some cases)\n3. Transitive, reflexive, not equivalence\n4. symmetric, not equivalence\n5. Reflexive, symmetric, not equivalence\n6. Transitive, not equivalence\n53. Equivalence\n[/hide]", "Solution_2": "Based on the content of this problem, I think it should go in Puzzles and Brainteasers. I'm not a moderator, though, so I can't move it." } { "Tag": [ "inequalities" ], "Problem": "Give a,b,c is three side of a triangle and $ a\\plus{}b\\plus{}c\\equal{}2$.Prove that:\r\n$ a^2\\plus{}b^2\\plus{}c^2\\plus{}2abc<2$", "Solution_1": "Setting $ a \\equal{} x \\plus{} y,b \\equal{} y \\plus{} z,c \\equal{} z \\plus{} x$ and after homogenization the inequality becomes:\r\n\r\n$ 2 \\sum x^3 \\plus{} 6 \\sum _{sym}x^2y \\plus{} 10 xyz < 2\\sum x^3 \\plus{} 6 \\sum _{sym} x^2y \\plus{} 12xyz \\Longleftrightarrow 2xyz > 0$ which is true :wink:", "Solution_2": "Another solution:\r\n\r\nBy triangle ineq. we have $ a= b >= c then 1/(b+c) >= 1/(a+c) >= 1/(a+b);\r\nTchebytchev : a/(b+c)+b/(a+c)+c/(a+b) >= 1/3.(a+b+c)(1/(b+c)+1/(a+c)+1/(a+b));\r\nCauchy : ( (b+c)+(a+c)+(a+b)) (1/(b+c)+1/(a+c)+1/(a+b)) >= 9\r\nThen a/(b+c)+b/(a+c)+c/(a+b) >= 3/2.\r\nSo if we suppose K and l positive real ( not mentionned ?)\r\nWe just need :1) ab +ac + bc >= a 2 +b 2 +c 2 \r\n 2) ab +ac +bc >= 1/3(a 2 +b 2 +c 2 )\r\nwhich are equivalente to : (a+b+c) 2 >= 3(a 2 +b 2 +c 2 )\r\norrect me if not", "Solution_2": "[quote=\"AYMANE\"]because of the semetry we can suppose that a >= b >= c then 1/(b+c) >= 1/(a+c) >= 1/(a+b);\nTchebytchev : a/(b+c)+b/(a+c)+c/(a+b) >= 1/3.(a+b+c)(1/(b+c)+1/(a+c)+1/(a+b));\nCauchy : ( (b+c)+(a+c)+(a+b)) (1/(b+c)+1/(a+c)+1/(a+b)) >= 9\nThen a/(b+c)+b/(a+c)+c/(a+b) >= 3/2.\nSo if we suppose K and l positive real ( not mentionned ?)\nWe just need :1) ab +ac + bc >= a 2 +b 2 +c 2 \n 2) ab +ac +bc >= 1/3(a 2 +b 2 +c 2 )\nwhich are equivalente to : (a+b+c) 2 >= 3(a 2 +b 2 +c 2 )\norrect me if not[/quote]\r\n\r\n1. the question is asking you to find the best constant k and l.\r\n2. Your signs for (1) and (2) are wrong.", "Solution_3": "I forget writting $k,l$ in the question.", "Solution_4": "That's why my answer was for an anathor exarcice :D" } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Let $m$ be a positive integer. Prove that there are integers $a, b, k$, such that both $a$ and $b$ are odd, $k\\geq0$ and\n\\[2m=a^{19}+b^{99}+k\\cdot2^{1999}\\]", "Solution_1": "This is from China 1999,you can find it in Resources", "Solution_2": "Yes it's China 1999. \nHere I propose the rather more general version : Let be given $m,b\\in\\mathbb N^*$ where $b$ is odd then there exist $a$ odd and $k\\in\\mathbb N$ so that $2m=a^{19}+b^{99}+k*2^{1999}$.", "Solution_3": "[hide=\"Solution to original\"](in collaboration with hyperbolictangent)\nIt suffices to show that $a^{19}$ and $b^{99}$ range over every odd residue $\\bmod{2^{1999}}$ as $a$ and $b$ range over every odd residue $\\bmod{2^{1999}}$.\n\n$a^{19}$: Consider $z=(a+t\\cdot 2^{1000})^{19}$, where $t\\in\\mathbb{Z}$. Then $z\\equiv a^{19}+19at\\cdot 2^{1000}\\bmod{2^{1999}}$. Since $\\gcd(19a,2^{1999})=1$, we know that as $t$ varies, $19at\\cdot 2^{1000}$ takes on every residue $\\bmod{2^{1999}}$ that is also a multiple of $2^{1000}$. Hence it suffices to show that $a^{19}$ takes on every odd residue $\\bmod{2^{1000}}$. We can continue this argument until we eventually arrive at proving that $a^{19}$ takes on every odd residue $\\bmod{4}$, which is clear. This proves that $a^{19}$ takes on every odd residue $\\bmod{2^{1999}}$.\n\n\n$b^{99}$: Consider $w=(b+s\\cdot 2^{1000})^{99}$, where $s\\in\\mathbb{Z}$. Then $w\\equiv b^{99}+99bs\\cdot 2^{1000}\\bmod{2^{1999}}$. Since $\\gcd(99b,2^{1999})=1$, we know that as $s$ varies, $99bs\\cdot 2^{1000}$ takes on every residue $\\bmod{2^{1999}}$ that is also a multiple of $2^{1000}$. Hence it suffices to show that $b^{99}$ takes on every odd residue $\\bmod{2^{1000}}$. We can continue this argument until we eventually arrive at proving that $b^{99}$ takes on every odd residue $\\bmod{4}$, which is clear. This proves that $b^{99}$ takes on every odd residue $\\bmod{2^{1999}}$.\n\nThis completes the proof.[/hide]\n\n[hide=\"Proof of generalization\"]This follows quickly; once we have chosen $m$ and $b$, we can choose a sufficiently low (possibly negative) $a$ such that $2m\\equiv a^{19}+b^{99}\\bmod{2^{1999}}$. Then the choice of a positive $k$ is fixed.[/hide]", "Solution_4": "Sorry to revive. Take $b=1$. Now we apply Hensel's Lemma on $f(a)=a^{19}+1-2m$ and since $a$ is odd $f(a)\\equiv 0\\pmod 2$ has a solution and $f'(a)\\not\\equiv 0\\pmod 2$ thus for any $k$ there is $a_k$ with $f(a_k)\\equiv 0\\pmod{2^{k}}$. In particular we take $k=1999$. And thus $2m=a_{1999}^{19}+1^{99}+k\\cdot 2^{1999}$. See we can assume $k\\ge 0$ because we can otherwise vary $a_{1999}\\mod 2^{1999}$ unless $k\\ge 0$." } { "Tag": [ "function", "percent" ], "Problem": "Because of medical Technology advances the disability rates for people over 65 have been dropping rather dramatically. The function\r\nR(t)= 26.3e^-0.016 (0<=t<=18)\r\ngives the disability rate R(t), in percent, for people over age 65 from 1982 (t=0) through 2000 where t is measured in years.\r\nWhat was the disability rate in 1986?\r\n\r\nI know that your supposed to plug 4 in for t and the problem becomes\r\n26.3e^-.064\r\nand I know e= 2.7182818\r\nbut where do I go from here? I don't want the answer I just want to know what I'm doing wrong. Your help would be greatly appreciated.", "Solution_1": "You are not doing anything wrong! :) \r\n\r\nJust pull out your calculator and evaluate $ 26.3e^{\\minus{}0.064}$, and you have your answer." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all function $ f: R\\rightarrow R$ that satisfy for all $ x,y>0$ : $ f(1\\plus{}xf(y))\\equal{}yf(x\\plus{}y)$ . (National Contest \"Vranceanu Procopiu\"-Romania 2004)", "Solution_1": "Here is a proof if we assume $ f: \\mathbb {R}^ \\plus{} \\rightarrow\\mathbb {R}^ \\plus{}$. \r\nIt can be shortened if we use \r\n\"surjective\" and \"monotonicty\" imply continuity.\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=218300" } { "Tag": [ "AMC", "AIME" ], "Problem": "Can I be tied for 7th instead of being in 8th?", "Solution_1": "I MISSED #1 ONCE AGAIN!!!!!!!!!!!!", "Solution_2": "Haha Beta believe it or not so did I!\r\nWhat answer did you pick?\r\nI picked E 19.5", "Solution_3": "I WAS SOOO CLOSE!!! :( :( :( :( :(", "Solution_4": "lol missing number 1. I on;y got 2 of the ones I did and I got number 1.", "Solution_5": "I'm unhappy. I guessed on two of the problems I got wrong, if I 'dlet those be blank I would be several places higher", "Solution_6": "rep- i think there might be a small problem.\r\n\r\ni was at a friends house some days ago, and never logged off my account there. so i think this friend of mine sent in the first set of answers to you. that's the score of 97.\r\n\r\nmy real solutions were sent a bit later.\r\n\r\ni'm entirely serious. i know, it's hard to believe, but there are people who know me who are willing to back up my integrity.\r\n\r\nsorry for the incovenience =/", "Solution_7": "7 was really easy, but I forgot that 6*1=1*6=6 so my answer was off by 2", "Solution_8": "[quote=\"NightFlarer\"]Haha Beta believe it or not so did I!\nWhat answer did you pick?\nI picked E 19.5[/quote]\r\n17.125", "Solution_9": "What is an aime-qualifying score for this?", "Solution_10": "Grrr... 2 points away...", "Solution_11": "[quote=\"Rep123max\"][b]113[/b]. Genius1601-96[/quote]\r\n\r\nWhen you get a chance, you might want to fix that....should read \"13\".", "Solution_12": "[quote=\"Rep123max\"]I think the most missed one was 7, and I don't know if anyone answered #25. Perhaps Teddy did? Did you? I actually think that #25 was easier than some of the others.[/quote]\r\n\r\nYeah, I got #25. It was moderately easy for #25.", "Solution_13": "#24 was an easy one too", "Solution_14": "Rep, are you still going to make the graph?", "Solution_15": "NOOOOOOOO!!! Rosen beat me by 1!!! If I had just gotten 6 right...\r\nI thought 23 was pretty easy though, probably because I didn't really do it, lol.", "Solution_16": "Rep. I am 99.9% sure I scored the lowest and it wouldn't bug me a bit. I am quite proud I scored as high as I did.", "Solution_17": "[quote=\"Rep123max\"]Is anybody against the graph? Here it is, if you don't want to see it, don't look at it.[/quote]\r\n\r\nI am discouraged. I am suicidal now. \r\nMan, maybe I should work on math stuff again.\r\n\r\nYou guys are all geniuses.\r\n\r\npar excellence!", "Solution_18": "pkZKMbyW4UPdkRsgJ5xP+j8i+qckkW5ZznguEmtF5pAz8FyMwxsvrm2qBbia", "Solution_19": "I really don't know what to think about my score. A lot of people said that Rep's test was more of an AMC 12 level, but if that's true, my score is very very high for me. But then, I only answered the easy questions.", "Solution_20": "[quote=\"h_s_potter2002\"]I really don't know what to think about my score. A lot of people said that Rep's test was more of an AMC 12 level, but if that's true, my score is very very high for me. But then, I only answered the easy questions.[/quote]\r\n\r\nI only answered the easy questions and I still made stupid mistakes! :D", "Solution_21": "[quote=\"nr1337\"][quote=\"h_s_potter2002\"]I really don't know what to think about my score. A lot of people said that Rep's test was more of an AMC 12 level, but if that's true, my score is very very high for me. But then, I only answered the easy questions.[/quote]\n\nI only answered the easy questions and I still made stupid mistakes! :D[/quote]\r\n\r\nI think it's easier to make stupid mistakes on easy questions since you go through the problem so fast, not really thinking about it, so you're more likely to miss something.\r\n\r\nBut in hard questions, you look at the question more carefully, so you can understand it better.", "Solution_22": "Hey guys,\r\n\r\nI know i don't post much, but I do follow the board pretty closely. I was planning on taking the mock amc B, but when i woke up yesterday (wed), and tried going to the forum. there was a 400 Result Error. So i figured the AoPS server was having difficulties. 6 o clock came around and I still couldnt get onto the forum, so i figured this was happening to everyone and someone would postpone the mock amc B. \r\n\r\nI guess i shouldve read the error because it was a just a stupid cookie on my computer that was preventing me from going into the forum. SO i missed the mock amc B and was bummed out. I took the test this morning and got a 105.5. \r\n\r\ndang. tied for 5th.\r\n\r\ni know it's unofficial and won't count. but i felt like singing my sorrow.", "Solution_23": "Rep, for some reason Nosepuppy is in two places on the High Score list...why is that?", "Solution_24": "[quote=\"Nosepuppy\"]rep- i think there might be a small problem.\n\ni was at a friends house some days ago, and never logged off my account there. so i think this friend of mine sent in the first set of answers to you. that's the score of 97.\n\nmy real solutions were sent a bit later.\n\ni'm entirely serious. i know, it's hard to believe, but there are people who know me who are willing to back up my integrity.\n\nsorry for the incovenience =/[/quote]", "Solution_25": "Ah, OK.", "Solution_26": "If you still have the problems can you PM them to me...I accidentally got the calendar mixed up in my head somehow and I didn't answer any of the problems? If that's so then what is my score? lol, j/p. Anyways, do you think you could PM me the problems? Thanks.", "Solution_27": "Hmm... better be careful with that graph\r\n\r\nBy manipulating the chart I was able to see everyone's score...\r\n\r\nWon't say how, just letting you know though...", "Solution_28": "Wait - do you mean you saw everyone's names beside their scores? That's amazing - I have no idea how you did that.", "Solution_29": "[quote=\"white_horse_king88\"]Wait - do you mean you saw everyone's names beside their scores? That's amazing - I have no idea how you did that.[/quote]\r\n\r\nI do :maybe:", "Solution_30": "Rep, you should delete all the names next to the numbers, the graph will still look the same", "Solution_31": "pkZKMbyW4UPdkRsgJ5xP+j8i+qckkW5ZznguEmtF5pAz8FyMwxsvrm2qBbia", "Solution_32": "Ahhh! I would have been tied at 131 with Teddy if only I partook!", "Solution_33": "[quote=\"Philip_Leszczynski\"]Ahhh! I would have been tied at 131 with Teddy if only I partook![/quote]\r\n\r\n :| It's a mock contest...I hardly think it's worth reviving an old topic like this just to say that." } { "Tag": [], "Problem": "A party has $ 6$ attendees. Attendees shake hands with other attendees, some pairs may shake hands more than once or not at all. Each list below records the number of hands each attendee shook at the party. Which one of the lists must be incorrect?\r\n\r\n$ \\mathrm{(A)}\\ (1,\\,2,\\,2,\\,2,\\,2,\\,3)\\qquad\\mathrm{(B)}\\ (1,\\,3,\\,3,\\,3,\\,4,\\,4)\\qquad\\mathrm{(C)}\\ (3,\\,3,\\,3,\\,3,\\,3,\\,3)\\qquad\\mathrm{(D)}\\ (2,\\,2,\\,3,\\,3,\\,3,\\,4)\\qquad\\mathrm{(E)}\\ \\text{None of these}$", "Solution_1": "[hide]When people shake hands, there must be an even number of total shakes from each side, since $ 2$ people are involved in $ 1$ handshake. Only choice $ \\fbox{D}$ has an odd number of shakes: $ 2\\plus{}2\\plus{}3\\plus{}3\\plus{}3\\plus{}4\\equal{}17$ handshakes. [/hide]" } { "Tag": [ "conics", "parabola", "probability", "trigonometry", "quadratics", "algebra", "binomial theorem" ], "Problem": "11.) The parabola $y = x^2$ is reflected in the line y = 3, producing a new parabola. This new parabola is reflected in the line x = 2, producing the parabola $y = -x^2 + bx + c$. Compute the ordered pair (b,c).\r\n\r\n12.) Find the largest value of x for which $x^2 + y^2 = x + y$ has a solution, if x and y are real.\r\n\r\n13.) Let z be such that $z^7 = 1$ and $z \\neq 1$. Compute the numerical value of $z^{10} + \\frac{1}{z^{10}} + z^{30} + \\frac{1}{z^{30}} + z^{50} + \\frac{1}{z^{50}}$\r\n\r\n14.) Find, as a proper base ten fraction, the sum of the following infinite series of repeating decimals, all written in base eight: \r\nS = .07* + .007* + .0007* + ... (where * indicates a repeating digit)\r\n\r\n15.) Find all positive integer values of n, 1 < n < 100, such that $\\binom{n}{2}$ is a perfect square.\r\n\r\n16.) An \u201cunfair\u201d coin has a 2/3 probability of turning up heads. If this coin is tossed 50 times, what is the probability that the total number of heads is even?\r\n\r\n17.) Find the smallest positive integer such that when its leading digit is removed and placed at the end of the number, the resulting number is twice as large as the original.\r\n\r\n18.) Find the value of $\\binom{50}{50} - \\binom{50}{48} + \\binom{50}{46} - \\ldots - \\binom{50}{4} + \\binom{50}{2} - \\binom{50}{0}$\r\n\r\n19.) Given that $1 + \\frac{1}{2^2} + \\frac{1}{3^2} + \\frac{1}{4^2} + \\cdots = \\frac{\\pi^2}{6}$, find the value of $1 - \\frac{1}{2^2} + \\frac{1}{3^2} - \\frac{1}{4^2} + \\cdots$", "Solution_1": "[hide=\"Number 18\"]\nConsider $(1+i)^{50}$. Using the binomial theorem, we can expand this as \n\n\\begin{eqnarray*} (1+i)^{50} &=& \\binom{50}{50}+\\binom{50}{49}i+\\binom{50}{48}i^2+\\binom{50}{47}i^3+\\cdots+\\binom{50}{1}i^{49}+\\binom{50}{0}i^{50}\\\\ &=& \\binom{50}{50}+\\binom{50}{49}i-\\binom{50}{48}-\\binom{50}{47}i+\\cdots+\\binom{50}{1}i-\\binom{50}{0} \\end{eqnarray*}\n\nwhere every other term has an $i$ and the sign of the terms switch every two terms. So then we can write this as \\[ \\left[\\binom{50}{50}-\\binom{50}{48}+\\cdots+\\binom{50}{2}-\\binom{50}{0}\\right]+\\left[\\binom{50}{49}-\\binom{50}{47}+\\cdots-\\binom{50}{3}+\\binom{50}{1}\\right]i. \\]\n\nBasically we just showed that what we are trying to find is the real part of $(1+i)^{50}.$\n\nWe can convert $(1+i)^{50}$ to polar form using $1+i=\\sqrt{2}\\text{cis } 45^\\circ.$ Thus\n\n\\begin{eqnarray*} (1+i)^{50} &=& (\\sqrt{2}\\text{cis }45^\\circ)^{50}\\\\ &=& (\\sqrt{2})^{45}\\text{cis }(45\\cdot 50)^\\circ\\\\ &=& 2^{45/2}\\text{cis } 2250^\\circ\\\\ &=& 2^{45/2}\\text{cis } 90^\\circ\\\\ &=& 2^{45/2}\\cos 90^\\circ + 2^{45/2}i\\sin 90^\\circ\\\\ &=& 0 + 2^{45/2}i\\end{eqnarray*}\n\nThus we can match up real and imaginary components:\n\n\\begin{eqnarray*} 0+2^{45/2}i &=& \\left[\\binom{50}{50}-\\binom{50}{48}+\\cdots+\\binom{50}{2}-\\binom{50}{0}\\right]+\\left[\\binom{50}{49}-\\binom{50}{47}+\\cdots-\\binom{50}{3}+\\binom{50}{1}\\right]i\\\\ &\\Rightarrow & \\binom{50}{50}-\\binom{50}{48}+\\cdots+\\binom{50}{2}-\\binom{50}{0} = 0\\\\ &\\text{and}& \\binom{50}{49}-\\binom{50}{47}+\\cdots -\\binom{50}{3}+\\binom{50}{1} = 2^{45/2} \\end{eqnarray*}\n\nSo our answer is $\\boxed{0.}$[/hide]", "Solution_2": "Hints to the ones i did:\r\n\r\n[hide]12. Kinda reason through, manipulating takes more effort and time\n\n13. Look at $\\mod{7}$\n\n14. I think it's 1/7...not sure =(\n\n15. $\\frac{n(n-1)}{2}=k^2$, now look at divisibility by $2$ and trial and error because we need the other factor and the quotient to be a square.\n\n17. $199a=80b+8c$ where the original number is $abc$\n[/hide]", "Solution_3": "I would prefer you put your answers. That way I can tell you if you're right or not. Your hints are kind of vague.", "Solution_4": "[quote=\"gauss202\"]I would prefer you put your answers. That way I can tell you if you're right or not. Your hints are kind of vague.[/quote]\r\n\r\nOK, i'll post my solutions tonight, I'm a bit busy right now. So far numbers 11,19,16, and 8 have been bugging me :(", "Solution_5": "[hide=\"Number 11\"]\nBasically, reflecting $y=x^2$ over $y=3$ does two things. First, it flips $y=x^2$ over and, second, it translates it upwards 6 units. So now the vertex of the parabola is at $(0,6)$. It follows that the new equation is $y=-x^2+6.$\n\nReflecting this new parabola over $x=2$ does not change which way the parabola is facing. It just translates the parabola over 4 units, essentially. So the vertex is now at $(4,6)$. Thus we have $y=-(x-4)^2+6=-x^2+8x-10\\Rightarrow (b,c)=\\boxed{(8,-10).}$[/hide]\n\n[hide=\"Number 16\"]\nFor the number of heads to be even, we can have 0 heads, 2 heads, 4 heads, and so forth, up to 50 heads. Now we want to add up the probability of each one of these cases. First, we need to find the probability of each one. If we have $k$ heads, then we must have $50-k$ tails. There are $\\binom{50}{k}$ ways to get $k$ heads when tossing the coins. Thus our probability of getting $k$ heads is $\\binom{50}{k}\\left(\\frac 23\\right)^k\\left(\\frac 13\\right)^{50-k}.$ We want to sum this for $k=0,2,4,\\cdots,50.$ Thus we want \n\n\\[ \\binom{50}{0}\\left(\\frac 23\\right)^0\\left(\\frac 13\\right)^{50}+\\binom{50}{2}\\left(\\frac 23\\right)^2\\left(\\frac 13\\right)^{48}+\\cdots+\\binom{50}{50}\\left(\\frac 23\\right)^{50}\\left(\\frac 13\\right)^{0}=\\frac 1{3^{50}} \\left[\\binom{50}{0}+2^2\\binom{50}{2}+\\cdots+2^{50}\\binom{50}{50}\\right]. \\]\n\nThat sum looks kinda hard to evaluate. But it does remind of us some things which we can sum easily. From the binomial theorem with $(x+y)^{50}$, $(x-y)^{50}$, $x=2$ and $y=1$, we get:\n\n\\begin{eqnarray*} \\binom{50}{0}+2\\binom{50}{1}+2^2\\binom{50}{2}+\\cdots+2^{50}\\binom{50}{50} &=& 3^{50}\\\\ \\text{and} \\binom{50}{0}-2\\binom{50}{1}+2^2\\binom{50}{2}-\\cdots+2^{50}\\binom{50}{50} &=& 1 \\end{eqnarray*}\n\nAdding these two yields $2\\left[\\binom{50}{0}+2^2\\binom{50}{2}+\\cdots+2^{50}\\binom{50}{50}\\right] = 3^{50}+1.$ \n\nIt follows that our answer is thus $\\boxed{\\frac{3^{50}+1}{2\\cdot3^{50}}}.$\n\nYou can generalize this to $\\frac{3^n+1}{2\\cdot3^n}$ where $n$ is the number of coins.[/hide]\n\n[hide=\"Number 19\"]\nLet $S=1-\\frac 1{2^2}+\\frac 1{3^2}-\\cdots$ and $T=1+\\frac 1{2^2}+\\frac 1{3^2}+\\cdots$. Then \\[ S+T = 2\\left(1+\\frac 1{3^2}+\\frac 1{5^2}+\\cdots\\right). \\]\n\nWe can evaluate the RHS of the above equation(leave off the 2 out in front of the parentheses for now):\n\n\\begin{eqnarray*} 1+\\frac 1{3^2}+\\frac 1{5^2}+\\cdots &=& \\left(1+\\frac 1{2^2}+\\frac 1{3^2}+\\cdots\\right) - \\left(\\frac 1{2^2}+\\frac 1{4^2}+\\frac 1{6^2}+\\cdots\\right)\\\\ &=& \\left(1+\\frac 1{2^2}+\\frac 1{3^2}+\\cdots\\right)-\\frac 1{2^2}\\left(1+\\frac 1{2^2}+\\frac 1{3^2}+\\cdots\\right)\\\\ &=& \\left(1-\\frac 1{2^2}\\right)\\left(1+\\frac 1{2^2}+\\frac 1{3^2}+\\cdots\\right)\\\\ &=& \\frac 34\\cdot \\frac{\\pi^2}6\\\\ &=& \\frac{\\pi^2}8 \\end{eqnarray*}\n\nSo\n\n\\begin{eqnarray*} S+T &=& 2\\left(1+\\frac 1{3^2}+\\frac 1{5^2}+\\cdots\\right)\\\\ S+\\frac{\\pi^2}6 &=& 2\\left(\\frac{\\pi^2}8\\right)\\\\ \\Rightarrow S &=& \\boxed{\\frac{\\pi^2}{12}} \\end{eqnarray*}[/hide]", "Solution_6": "heh, looks like you've already started. I guess I'll just post solutions one by one then.\r\n[hide=\"13\"]\nAfter rationalizing the denominator, bringing it to one single fraction, and cancelling out $z^7$ (since it equals $1$), we have that the sum we are looking for is $z^6+z^5+z^4+z^3+z^2+z$, let us call this $S$. Now notice that $z^7-1=(z-1)(z^6+z^5+z^4+z^3+z^2+z+1)=(z-1)(S+1)=0$. Now, since $z \\not=1$, then $S+1$ must $=0$, therefore $S=\\boxed{-1}$\n[/hide]", "Solution_7": "[quote=\"Iversonfan2005\"]heh, looks like you've already started. I guess I'll just post solutions one by one then.[/quote]\r\n\r\nPosting 3 at a time is far better. Look at joml88's post, for example.", "Solution_8": "[quote=\"3\"]Posting 3 at a time is far better.[/quote]\r\n\r\nHa, ha, ha... :rotfl: \r\n\r\n3...", "Solution_9": "[quote=\"3\"]\n 3 [/quote]\r\n\r\nind33d", "Solution_10": "[quote=\"joml88\"]\n\n[hide=\"Number 19\"]\nLet $S=1-\\frac 1{2^2}+\\frac 1{3^2}-\\cdots$ and $T=1+\\frac 1{2^2}+\\frac 1{3^2}+\\cdots$. Then \\[ S+T = 2\\left(1+\\frac 1{3^2}+\\frac 1{5^2}+\\cdots\\right). \\]\n\nWe can evaluate the RHS of the above equation(leave off the 2 out in front of the parentheses for now):\n\n\\begin{eqnarray*} 1+\\frac 1{3^2}+\\frac 1{5^2}+\\cdots &=& \\left(1+\\frac 1{2^2}+\\frac 1{3^2}+\\cdots\\right) - \\left(\\frac 1{2^2}+\\frac 1{4^2}+\\frac 1{6^2}+\\cdots\\right)\\\\ &=& \\left(1+\\frac 1{2^2}+\\frac 1{3^2}+\\cdots\\right)-\\frac 1{2^2}\\left(1+\\frac 1{2^2}+\\frac 1{3^2}+\\cdots\\right)\\\\ &=& \\left(1-\\frac 1{2^2}\\right)\\left(1+\\frac 1{2^2}+\\frac 1{3^2}+\\cdots\\right)\\\\ &=& \\frac 34\\cdot \\frac{\\pi^2}6\\\\ &=& \\frac{\\pi^2}8 \\end{eqnarray*}\n\nSo\n\n\\begin{eqnarray*} S+T &=& 2\\left(1+\\frac 1{3^2}+\\frac 1{5^2}+\\cdots\\right)\\\\ S+\\frac{\\pi^2}6 &=& 2\\left(\\frac{\\pi^2}8\\right)\\\\ \\Rightarrow S &=& \\boxed{\\frac{\\pi^2}{12}} \\end{eqnarray*}[/hide][/quote]\r\nHmm... easier way?\r\n\r\n\\begin{eqnarray*} \\frac{\\pi^2}6 &=& 1+\\frac 1{2^2}+\\frac 1{3^2}+\\frac1{4^2}+\\cdots \\\\ S &=& 1-\\frac 1{2^2}+\\frac 1{3^2}-\\frac1{4^2}+\\cdots \\\\ \\text{Subtract: } \\frac{\\pi^2}6-S &=& \\frac 2{2^2}+\\frac 2{4^2}+\\frac2{6^2}+\\cdots=\\frac12\\left(1+\\frac 1{2^2}+\\frac 1{3^2}+\\cdots\\right)=\\frac{\\pi^2}{12} \\\\ S &=& \\frac{\\pi^2}6-\\frac{\\pi^2}{12}=\\boxed{\\frac{\\pi^2}{12}} \\end{eqnarray*}", "Solution_11": "[quote=\"gauss202\"]12.) Find the largest value of x for which $x^2 + y^2 = x + y$ has a solution, if x and y are real.[/quote]\r\n\r\n[hide=\"12\"]\n$x^2-x+(y^2-y)=0$\n\n$x=\\frac{1\\pm\\sqrt{1-4(y^2-y)}}{2}$\n\n$x=\\frac{1\\pm\\sqrt{-4y^2+4y+1}}{2}$\n\n$\\therefore x\\in\\mathbb{R}\\iff -4y^2+4y+1\\ge 0$\n\n$\\iff \\frac{1}{2}-\\frac{\\sqrt{2}}{2}\\le y\\le\\frac{1}{2}+\\frac{\\sqrt{2}}{2}$\n\n$\\Rightarrow \\boxed {\\frac{1}{2}+\\frac{\\sqrt{2}}{2}}$\n[/hide]", "Solution_12": "[quote=\"gauss202\"]15.) Find all positive integer values of n, 1 < n < 100, such that $\\binom{n}{2}$ is a perfect square.[/quote]\r\n\r\n[hide]$n^2-n=2k^2$\n\nThe discrimant of this quadratic equation is $8k^2+1$. If we want $n\\in\\mathbb{Z}$, then $8k^2+1=z^2$ for $z\\in\\mathbb{Z}$.\n\n$\\iff z^2-8k^2=1$\n\nPELL!\n\n$\\sqrt{8}=2+\\frac{1}{1+\\frac{1}{4+\\frac{1}{1+\\frac{1}{4+\\cdots}}}}$\n\nconvergents: $\\frac{2}{1}$, $\\frac{3}{1}$, $\\frac{14}{5}$, $\\frac{17}{6}$, $\\frac{82}{29}$, $\\frac{99}{35}$, ... (the rest are too big)\n\nperiod is 2\n\n$(z,k)=(3,1),(17,6),(99,35),\\ldots$\n\n$n=\\frac{1\\pm z}{2}$\n\n$=\\boxed {2,9,50}$[/hide]" } { "Tag": [ "AMC", "AIME" ], "Problem": "How many 4-tuples (a, b, c, d) satisfy 0 < a < b < c < d < 500, a + d = b + c, and bc - ad = 93?", "Solution_1": "[quote=\"jin thynj\"]How many 4-tuples (a, b, c, d) satisfy 0 < a < b < c < d < 500, a + d = b + c, and bc - ad = 93?[/quote]\r\n\r\nLet $ a\\equal{}b\\minus{}k$ , then $ d\\equal{}c\\plus{}k$, where $ k$ is positive integer.\r\n\r\nSo $ bc\\minus{}ad\\equal{}93 \\Leftrightarrow bc\\minus{}(b\\minus{}k)(c\\plus{}k)\\equal{}93$\r\n$ \\Leftrightarrow k(c\\minus{}b\\plus{}k)\\equal{}93\\equal{}3 \\times 31$\r\n\r\n$ k|93$ and $ k^2 GM. Here, we'll use something called contradiction which disapproves the first assumption.\n\nLet's assume $\\frac {a+b}{2} < \\sqrt {ab}$.\n\nThen we can do:\n\n$\\frac {a+b}{2} < \\sqrt {ab}\\\\\na+b < 2 \\sqrt {ab}\\\\\na^2+2ab+b^2 < 4ab\\\\\na^2+b^2 < 2ab$\n\nLet's use our simplest case, 1 and 2. When we plug a=1 and b=2 (doesn't matter which order you use), we get:\n\n$1^2+2^2 = 5\\\\\n2 \\cdot 2 \\cdot 1 = 4\\\\\n\\text {But}\\\\\n5 < 4$\n\nSo, our assumption is wrong and that $\\frac {a+b}{2} > \\sqrt {ab}$. And by the first proof we did, we can now summarize our AM-GM proof as:\n\n$\\frac {a+b}{2} \\geq \\sqrt {ab}$.[/hide]", "Solution_1": "I'm making this post because I want people to see examples on how proofs should look like:\r\n\r\n[hide=\"\"Proof by induction\"\"]Statement: $1+2+\\cdots+n=\\frac{n(n+1)}{2}$\n\nAssume that $n=1$. Obviously $1=\\frac{(1)\\cdot(1+1)}{2}=1$.\n\nSuppose the statement is true for $n$ and we can safely assume that $1+2+\\cdots+n=\\frac{n(n+1)}{2}$ holds.\n\nThen, $1+2+\\cdots+n+(n+1)=\\frac{n(n+1)}{2}+(n+1)=(n+1)\\left(\\frac{n}{2}+1\\right)=\\frac{(n+1)(n+2)}{2}$\n\nTherefore, the statement also holds for $n+1$\n\nBy induction hypothesis, the statement is true for all $n\\geq 1$.[/hide]\n\n[hide=\"\"Proof of basic 2 variable AM-GM\"\"]Proving inequalities are the best proofs you can do. It takes cleverness on how you manipulate the variables and how you 'see through' a sense of measure in an inequality. This may sound deep and confusing, but my opinion will make some sense once you have proven many inequalities. Here we got a classical inequality that by working forward and simplifying until you see something obvious will be very rewarding.\n\nSo how can you prove $\\frac{a+b}{2}\\geq \\sqrt{ab}$ when both $a,b$ are positive numbers? Note that the condition that $a,b$ are positive is [u]necessary[/u].\n\nLet's try to work on something familiar: so the above inequality is equivalent to $a+b\\geq 2\\sqrt{ab}$.\n\nLet's see what happens when we square both sides; this is a valid operation since we are working with positive numbers: $(a+b)^2\\geq 4ab$\n\nBut, $(a+b)^2=a^2+2ab+b^2\\geq 4ab$ and if we try to simplify we get $a^2-2ab+b^2\\geq 0$.\n\nRecall that $a^2-2ab+b^2=(a-b)^2$ and so we have $(a-b)^2\\geq 0$. The last part is true since the square of any real number is always greater than or equal to 0.\n\nNow, if you work backwards using the fact that the square of any real number is always greater than or equal to 0, we can form the following steps:\n\n$(a-b)^2\\geq 0$ (Initial statement)\n\n$a^2-2ab+b^2\\geq 0$ (Expand square)\n\n$a^2+2ab+b^2\\geq 4ab$ (Add 4ab to both sides, a positive quantity, preserves inequality)\n\n$(a+b)^2\\geq 4ab$ (Reduce to square form)\n\n$a+b\\geq 2\\sqrt{ab}$ (Taking square root to both sides that are positives)\n\n$\\frac{a+b}{2}\\geq \\sqrt{ab}$ (Divide both sides by 2)\n\nAM-GM is proved. QED[/hide]\r\nHopefully, I didn't bore some of you with so much details. Happy proving!\r\n3X.lich", "Solution_2": "Silverfalcon, your proof is wrong. If you want to prove AM-GM by contradiction (I can't imagine why) you need to show that $AM \\frac{13}{24}$\r\n\r\nIf you're above Getting Started or experienced in math induction, please leave these to getting starters.", "Solution_16": "[quote=\"towersfreak2006\"]Here's another way to show $1+2+3+\\ldots+n=\\frac{(n)(n+1)}{2}$:\n\nLet $S=1+2+3+\\ldots+n$.\nThen also, $S=n+n-1+n-2+\\dots+1$.\n\nAdd the two together:\n$2S=(1+n)+(2+n-1)+(3+n-2)+\\dots+(n+1)$.\n$2S=(n+1)+(n+1)+(n+1)+\\ldots+(n+1)$.\nSince $S$ had $n$ terms, then $2S$ has $n$ terms of $(n+1)$, so:\n$2S=(n+1)(n)$.\n$S=1+2+3+\\ldots+n=\\frac{(n+1)(n)}{2}$.[/quote]\r\n\r\nThat's the same thing that Moubinool did.\r\n\r\nAs for the Pigeon Hole Principle, here is the stronger version of it:\r\n\r\nGiven $n$ holes and $kn+1$ pigeons to place in them, there exists a hole with at least $k+1$ pigeons in it.", "Solution_17": "Just wondering whether proof by induction is better or proof like the Gauss one. i mean if we use induction we don't really know \"why\" , it's more like forcing it through.\r\n\r\nAnother thing is that how is one supposed to know whether to use induction or if there is a better method by just looking at the question?", "Solution_18": "[quote=\"joml88\"][quote=\"towersfreak2006\"]Here's another way to show $1+2+3+\\ldots+n=\\frac{(n)(n+1)}{2}$:\n\nLet $S=1+2+3+\\ldots+n$.\nThen also, $S=n+n-1+n-2+\\dots+1$.\n\nAdd the two together:\n$2S=(1+n)+(2+n-1)+(3+n-2)+\\dots+(n+1)$.\n$2S=(n+1)+(n+1)+(n+1)+\\ldots+(n+1)$.\nSince $S$ had $n$ terms, then $2S$ has $n$ terms of $(n+1)$, so:\n$2S=(n+1)(n)$.\n$S=1+2+3+\\ldots+n=\\frac{(n+1)(n)}{2}$.[/quote]\n\nThat's the same thing that Moubinool did.\n\nAs for the Pigeon Hole Principle, here is the stronger version of it:\n\nGiven $n$ holes and $kn+1$ pigeons to place in them, there exists a hole with at least $k+1$ pigeons in it.[/quote]\r\n\r\nJust for the sidenote, what joml88 posted is called General Pigeon Hole Principle.\r\n\r\nMy PHP description covers when $k=1$.", "Solution_19": "Pigeonhole problem:\r\n\r\nGiven 16 distinct positive integers not exceeding 100, prove that it is possible to choose 4 distinct integers a, b, c and d such that $a+c = b + d$.\r\n\r\nIs this statement still true for 15 distinct positive integers?", "Solution_20": "Oh, I didn't notice. :oops: \r\n\r\nI don't know how to use Pigeonhole for white_horse_king88's problem. :? \r\nMaybe something with the Fibonacci sequence? It seems to be happy: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89", "Solution_21": "Hmm, look at the equation as the equality of two differences. How many unique differences are there?", "Solution_22": "[quote=\"white_horse_king88\"]Pigeonhole problem:\n\nGiven 16 distinct positive integers not exceeding 100, prove that it is possible to choose 4 distinct integers a, b, c and d such that $a+c = b + d$.\n\nIs this statement still true for 15 distinct positive integers?\n\n[/quote]\r\n[hide]\nWe can rewrite this as $a-b = d-c$. If we let any pairing of 2 of the 16 numbers (in the form $x-y$ where $x > y$) have a unique difference, then we have $\\frac{15(15-1)}{2} = 105$ unique differences. This implies that there are more than 100, namely 105, positive integers less than 100 which is not true. It is, therefore, possible to choose 4 distinct integers a, b, c and d such that $a+c = b+d$ from these 16.\n\nI think another way to prove it would be to show that we have $15*14 = 210$ possibilities for differences (either x-y or y-x) and then that there are only 200 possibilities for differences.\n\n[/hide]" } { "Tag": [], "Problem": "7 times a # is the same as 12 more than 3 times the #.find the #", "Solution_1": "Call the number $ \\alpha$.\r\n\r\n$ 7 \\alpha \\equal{} 12 \\plus{} 3 \\alpha$\r\n\r\n$ 4\\alpha \\equal{} 12$\r\n\r\n$ \\alpha \\equal{} \\boxed{3}$" } { "Tag": [ "symmetry", "linear algebra", "linear algebra solved" ], "Problem": "Prove that any bilinear inner product $g: L\\times L\\to K$ where the field $K$ has characteristic different from 2 can uniquely be decomposed into a sum of symmetric and antisymmetric inner products.", "Solution_1": "the solution is just one line but it is much nicer if you think about this yourself with this hint :\r\n\r\nsuppose $g=k+l$ with k symmetric and l antisymmetric\r\n\r\nthen $g(v,w) = k(v,w)+l(v,w)$\r\n\r\nand $g(w,v)=k(w,v)+l(w,v)=k(v,w)-l(v,w)$ where in the last equation i used symmetry and antisymmetry of k and l respectively\r\n\r\nlooking at these two equations and using the fact that the characteristic is not two, should immediately help you find k and l, and thus help you prove decomposition", "Solution_2": "thank u very much.You are right, it's trivial.Just that when I read the problem I didn't quite studied for the subject :oops:" } { "Tag": [ "number theory", "prime numbers", "number theory unsolved" ], "Problem": "Does there exist infinite, growing sequence of prime numbers $(q_n)$, such that $q_{n+2}+2q_n=3q_{n+1}$ for all natural numbers $n$.", "Solution_1": "Assume there does exist such a sequence.\r\n\r\nIt would mean that we can find integers $\\alpha,\\beta$ s.t. $q_n=\\alpha\\cdot 2^n+\\beta,\\ \\forall n\\ge 0$ (use the characteristic equation of the recurrence). We easily get a contradiction from here." } { "Tag": [ "inequalities", "LaTeX", "inequalities proposed" ], "Problem": "Here's my inequalitie. :D", "Solution_1": "Why didn't you post it in LaTeX? :?: For some LaTeX help, please visit [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]this site[/url].\r\nAnyway, we usually write inequality, not inequalitie :P \r\n\r\nHis inequality is the following one:\r\n\r\nFor $a,b,c,d>0$ satisfying $a+b+c+d=5$ and $abc\\ge\\frac{3}{2},abd\\ge\\frac{3}{2},bcd\\ge\\frac{3}{2},cda\\ge\\frac{3}{2}$ prove that:\r\n\r\n$\\frac{7}{4}\\le\\frac{1}{1+\\sqrt{abc}}+\\frac{1}{1+\\sqrt{abd}}+\\frac{1}{1+\\sqrt{bcd}}+\\frac{1}{1+\\sqrt{cda}}\\le2$", "Solution_2": "Yes, that is the inequality.So, try to solve it. ;)", "Solution_3": "$\\frac{7}{4}\\le\\frac{1}{1+\\sqrt{abc}}+\\frac{1}{1+\\sqrt{abd}}+\\frac{1}{1+\\sqrt{bcd}}+\\frac{1}{1+\\sqrt{cda}}\\le2$\r\n\r\nSo, the right estimate is clear since $\\frac{1}{1+\\sqrt{t}}\\le\\frac{1}{1+\\sqrt{\\frac{3}{2}}}$ because $t\\ge\\frac{3}{2}$. So we have:\r\n\r\n$V=\\frac{1}{1+\\sqrt{abc}}+\\frac{1}{1+\\sqrt{abd}}+\\frac{1}{1+\\sqrt{bcd}}+\\frac{1}{1+\\sqrt{cda}}\\le\\frac{4}{1+\\sqrt{\\frac{3}{2}}}$. We want to prove that $V\\le2$. It is sufficient to prove that $\\frac{4}{1+\\sqrt{\\frac{3}{2}}}\\le2\\iff2\\le1+\\frac{\\sqrt{3}}{\\sqrt{2}}\\iff\\sqrt{2}\\le\\sqrt{3}$ what's OK. Note that we have never have equality. Second note that we didn't need the information about the sum..\r\n\r\n\r\nI will have a look at the first inequality later..", "Solution_4": "We need that sum for the left inequality. ;)", "Solution_5": "Simple...But a hell lot's of work for the first one", "Solution_6": "I know that is easy.But how did you say, it's lot of work.This problem I've composed for 8 grade at a contest in my district." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,$ $b$ and $c$ are non-negative numbers. Prove that\r\n$\\sqrt[3]{\\frac{(2a+b)(2b+c)(2c+a)}{27}}\\geq\\sqrt{\\frac{ab+ac+bc}{3}}.$", "Solution_1": "If k>=1/4,a,b,c>=0, we have\r\n\r\n ((k*a+b)*(k*b+c)*(k*c+a)/(k+1)^3)^(1/3)>=((a*b+b*c+c*a)/3)^(1/2).", "Solution_2": "[quote=\"fjwxcsl\"]If k>=1/4,a,b,c>=0, we have\n\n ((k*a+b)*(k*b+c)*(k*c+a)/(k+1)^3)^(1/3)>=((a*b+b*c+c*a)/3)^(1/2).[/quote]\r\nTry $k=3$ or $k=\\frac{1}{3}$ and $a=b=1,$ $c=0.$ :wink:", "Solution_3": "Solly! Correctly,If 3/7<=k<=7/3, a,b,c>=0, then\r\n\r\n ((k*a+b)*(k*b+c)*(k*c+a)/(k+1)^3)^(1/3)>=((a*b+b*c+c*a)/3)^(1/2).", "Solution_4": "Very nice Inequality, Arqady!\r\nIn my solution, we have to show that\r\n\r\n$5\\sum a^{3}b+\\sum ab^{3}\\ge 3\\sum a^{2}b^{2}+3abc\\sum a$,\r\n\r\nwhich is true.", "Solution_5": "[quote=\"fjwxcsl\"]Solly! Correctly,If $ 3/7<\\equal{}k<\\equal{}7/3, a,b,c>\\equal{}0,$ then\n\n $ ((k*a\\plus{}b)*(k*b\\plus{}c)*(k*c\\plus{}a)/(k\\plus{}1)^3)^{1/3}>\\equal{}((a*b\\plus{}b*c\\plus{}c*a)/3)^{1/2}.$[/quote]\r\nI have obtained: $ \\frac{1}{k_{0}}\\leq k\\leq k_{0},$ where $ k_{0}\\equal{}\\frac{3\\sqrt[3]{4}\\minus{}2\\plus{}\\sqrt{18\\sqrt[3]2\\minus{}12\\sqrt[3]4}}{2}\\equal{}2.333698853...$\r\nId est, indeed the following inequality is true too. \r\n\r\nLet $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\\plus{}ac\\plus{}bc\\equal{}3.$ Prove that:\r\n\\[ (3a\\plus{}7b)(3b\\plus{}7c)(3c\\plus{}7a)\\geq1000\\]", "Solution_6": "A proof of the original inequality see here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=262970", "Solution_7": "[quote=\"arqady\"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that\n$ \\sqrt [3]{\\frac {(a \\plus{} 2b)(b \\plus{} 2c)(c \\plus{} 2a)}{27}}\\geq\\sqrt {\\frac {ab \\plus{} ac \\plus{} bc}{3}}.$[/quote]\r\nWrite our inequality as\r\n$ (a \\plus{} 2b)(b \\plus{} 2c)(c \\plus{} 2a) \\ge (ab \\plus{} bc \\plus{} ca)\\sqrt {3(ab \\plus{} bc \\plus{} ca)}.$\r\nApplying AM-GM inequality, we have\r\n$ \\sqrt {3(ab \\plus{} bc \\plus{} ca)} \\le \\frac {1}{2}\\left( a \\plus{} b \\plus{} c \\plus{} \\frac {3(ab \\plus{} bc \\plus{} ca)}{a\\plus{}b\\plus{}c}\\right),$\r\nThus our inequality reduce to\r\n$ 2(a \\plus{} 2b)(b \\plus{} 2c)(c \\plus{} 2a)(a \\plus{} b \\plus{} c) \\ge (ab \\plus{} bc \\plus{} ca)[(a \\plus{} b \\plus{} c)^2 \\plus{} 3(ab \\plus{} bc \\plus{} ca)],$\r\nwhich is, after expanding, equivalent to\r\n$ \\sum a^3b \\plus{} 5\\sum ab^3 \\minus{} 3\\sum a^2b^2 \\minus{} 3abc\\sum a \\ge 0,$\r\nor\r\n$ \\sum ab(a \\minus{} 2b)^2 \\plus{} \\sum ab^3 \\plus{} \\sum a^2b^2 \\minus{} 3abc\\sum a \\ge 0.$\r\nBy Cauchy Schwarz, we can easily show that\r\n$ \\sum ab(a \\minus{} 2b)^2 \\ge abc\\sum a,$\r\nMoreover, the AM-GM yields\r\n$ \\sum ab^3 \\ge abc\\sum a$, and $ \\sum a^2b^2 \\ge abc\\sum a.$\r\nOur proof is now completed. :)\r\n\r\nI have found (and proved) the largest range of $ k$ such that the following inequality:\r\n$ (a \\plus{} kb)(b \\plus{} kc)(c \\plus{} ka) \\ge (k \\plus{} 1)^3,$\r\nholds for all $ a,b,c \\ge 0$ satisfying $ ab \\plus{} bc \\plus{} ca \\equal{} 3$ is\r\n$ \\frac {3\\sqrt [3]{4} \\minus{} 2 \\minus{} \\sqrt {18\\sqrt [3]{2} \\minus{} 12\\sqrt [3]{4}}}{2} \\le k \\le \\frac {3\\sqrt [3]{4} \\minus{} 2 \\plus{} \\sqrt {18\\sqrt [3]{2} \\minus{} 12\\sqrt [3]{4}}}{2} .$\r\n:)", "Solution_8": "NIce proof, can_hang2007! \r\n[quote=\"can_hang2007\"]\nI have found (and proved) the largest range of $ k$ such that the following inequality:\n$ (a \\plus{} kb)(b \\plus{} kc)(c \\plus{} ka) \\ge (k \\plus{} 1)^3,$\nholds for all $ a,b,c \\ge 0$ satisfying $ ab \\plus{} bc \\plus{} ca \\equal{} 3$ is\n$ \\frac {3\\sqrt [3]{4} \\minus{} 2 \\minus{} \\sqrt {18\\sqrt [3]{2} \\minus{} 12\\sqrt [3]{4}}}{2} \\le k \\le \\frac {3\\sqrt [3]{4} \\minus{} 2 \\plus{} \\sqrt {18\\sqrt [3]{2} \\minus{} 12\\sqrt [3]{4}}}{2} .$\n:)[/quote]\r\nOf cause! I also proved it by the same of the following way:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=262970" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "Prove that \r\n\r\n$\\lim_{x\\to0}(1+rx)^{\\frac{1}{x}}= e^{r}$ \r\n\r\ngiven that $\\lim_{x\\to0}(1+x)^{\\frac{1}{x}}= e$. \r\n\r\n[Do not use L'Hopital's rule.]", "Solution_1": "[quote=\"236factorial\"]Prove that \n\n$\\lim_{x\\to0}(1+rx)^{\\frac{1}{x}}= e^{r}$ \n\ngiven that $\\lim_{x\\to0}(1+x)^{\\frac{1}{x}}= e$. \n\n[Do not use L'Hopital's rule.][/quote]\r\n\r\nLet $u = rx$\r\n$x = u/r$\r\n$\\lim_{u\\to0}(1+u)^{\\frac{1}{u/r}}=\\lim_{u\\to0}((1+u)^{\\frac{1}{u})^{r}}= e^{r}$" } { "Tag": [ "AMC", "AIME", "Purple Comet" ], "Problem": "I think we should have a post about how nm did in Purple Comet.\r\n\r\nOur team, the Atomic City Deuterons, got an honorable mention in the MS Small school category, with 2 people! \r\n\r\nNew Mexico Red or Green, got an honorable mention in the MS large school category. Congrats!\r\n\r\nThese are the competitors on our team:\r\nAlexandr Wang\r\nWillie Zhao\r\n\r\nThese are the competitors i know of on NM R or G:\r\nRobert Chen\r\nDrew Brost\r\nGeelon So\r\nJoseph Griego\r\n\r\nThese are competitors i know of on New Mexico 6(MS):\r\nAndy Chen", "Solution_1": "Just like on MC, my team made many \"stupid\" mistakes (most of them were made by me :( )\r\n\r\nCongrats to all who participated, though.", "Solution_2": "my team got 11 too. We guessed on #14 and got it right, we got #12 because i didn't make a mistake with similarity (it has happened many times before), and we got #13 right because I didn't mess up in casework! :D :lol: \r\nAll the other problems were pretty trivial (except 15), but my teammate misread #2 and #3, and made an arithmetic error on #7. :wallbash_red: \r\n\r\nOn the high school one we got 10/25... pretty terrible i know. we participated in both ( :lol: ), but i guess the word we is misused. my teammate didn't do any of the problems during the high school one, and kept saying \"beef stroganoff,\" while drawing overly brawny caricatures of 18th century men on the whiteboard. Long story short, it didn't go well. I really liked the high school problems, though. They reminded me of AIME problems, but not as hard. Anyway, I think we did well! :D", "Solution_3": "Aw dang. We should have taken both MS and HS, too. I would have probably decided to try the HS myself, though. Again, Congrats to the NM Purple Comet Teams!", "Solution_4": "How many did andy's team get", "Solution_5": "9........making post long enough......", "Solution_6": "lol\r\nWe almost got 9, until i checked most of my partner's work. I'm so happy we got 14,13, and 12 right. especially 13, because my casework didn't really look right.", "Solution_7": "bleh... looked over the Purple Comet question, now I realized the I didn't need casework for the one that i'm so happy about (argh), and for the HS one, I could have gotten at least 5 more with my current knowledge (I was reeeeeeeeeeeaaaaaaaaaaaaaally bad at math before nats). urgh." } { "Tag": [], "Problem": "3.) Solve the system of equations:\r\n5x-2y+7z = 46\r\nx+5y+z = 13\r\n5y-z = 1", "Solution_1": "(1) $ 5x-2y+7z=46$\r\n(2) $ x+5y+z=13$\r\n(3) $ 5y-z=1$\r\n\r\n(2)+(3) $ x+10y=14\\implies{x=14-10y}$\r\n(1)+7(3) ${ 5x+33y=53\\implies{70-50y+33y=53}\\implies{y=1}\\implies(x,y,z)=(4,1,4)}$", "Solution_2": "Thank you very much." } { "Tag": [ "AMC", "AIME", "probability", "factorial", "Pascal\\u0027s Triangle" ], "Problem": "Three red counters, four green counters and five blue counters are placed in a row in random order. Find the probability that no two blue counters are adjacent.", "Solution_1": "[quote=\"mathkid's original post\"]\nthe denominator is clearly 12! / (5!4!3!)\n\nbut to find the number of arrangements where no blue is together..\nthis is what i started doing.\n\nletting * be a blue chip\n\n*_*_*_*_*\n\nsince there must be at least one of the seven blue chips in between each..\n\n*7*6*5*4*\nwhere the digits are the number of chips that can be place in each\n\nAfter that, the last three chips can be placed anywhere. there is a total of 10 places for the first, 11 for the next, and 12 for the last.. \n\ni don't know where i'm going with this. \n\nif possible, try to explain how i can do it like i started with the blue chips already arranged.. if you can't, an alternate solution will suffice\n\nthanks[/quote]\r\n\r\nInstead of counting the combos where no two blue are adjacent, we shall count the ones that have 2 blue adjacent, 3, 4, and 5 although this might lead to PIE.\r\n\r\n-interesting_move", "Solution_2": "Instead of counting the combos where no two blue are adjacent, we shall count the ones that have 2 blue adjacent, 3, 4, and 5 although this might lead to PIE.\r\n\r\n-interesting_move", "Solution_3": "My guess for the number of such arrangement : 8! * 7! / (3! * 4! * 5!)\n\n[hide]\n\nFirst I put the red and green counters : there are 7!/(4!*3!) way to do that. Then I insert the blue ones :\n\n\n\na1 b1 a2 b2 a3 b3 a4 b4 a5 b5 a6 where ai are the *number* of red or green counters put between the blue counters, bi the blue counters.\n\n\n\n :Sigma: ai = 7, a1, a6 >= 0, a2..a5 >= 1.\n\nai=bi+1 for 2<=i<=5, bi=ai otherwise.\n\n :Sigma: bi = 3, with bi >= 0.\n\n\n\nBy De Moivre formula the number of solutions is C(3+6-1,6-1) = 8!/3!*5!. Hence the result.\n\n\n\nCorrect ?[/hide]", "Solution_4": "I'm curious to see if PIE works... \r\n\r\nIf it did, would this be right?\r\n\r\n1-(11*C(5,2)-10*C(5,3)+9*C(5,3)-8*C(5,5))/(12!/4!*3!*5!)", "Solution_5": "Belenos, your answer appears to be correct. Another way: [hide]Continuing mathkid's line of reasoning from the *_*_*_*_*: we have 3 non-blue counters remaining, and 6 slots (count the sides) to put them in: C(8,3) ways. Then, distinguishing between red and green, we multiply by C(7,3). Same result as belenos, but a lot less work.[/hide]", "Solution_6": "[quote=\"Kamior\"]a lot less work.[/quote]\r\n\r\nAt *lot* less work ???? You're kidding !", "Solution_7": "Quick counting method for the numerator, this is what I used to do as a kid before I knew about factorials and such (Think I was trying to work out lottery odds with like choose 6 numbers out of 48):\r\n\r\nYou have 12 \"slots\" and 5 blue chips. First arrangement is--\r\nX _ X _ X _ X _ X _ _ _\r\nlocking the first 4 chips in place, there are 4 positions for the 5th chip that work (just visually). moving the 4th chip to the right one space, there are now 3 positions for the last chip, so for all positions of the last 2 (given the first 3), there are 4th triangle = 10 possible positions. Moving the 3rd chip one space to the right, there are now 3rd triangle = 6 positions for the other 2, so for all positions of the last 3, there are sum of first 4 triangles (call them tetrahedral numbers)=1+3+6+10=20 positions. Similarly, unlocking the 2nd chip gives the sum of the first 4 tetrahedral numbers=35, and first chip, the sum of the first 4 4-dimensional triangular numbers (sums of first n tetrahedrals)=56 for the numerator. All this can be read directly from Pascal's triangle if you know what you're looking for. Gives an answer of 1/495." } { "Tag": [ "search", "AMC" ], "Problem": "Each ring has 2 more squares on a side than the previous one. Since the first ring has 3^2-1^2, the second has 5^2-3^2, etc. we see that the nth ring has (2n+1)^2 - (2n-1)^2 = 8n. So for n = 100, we have 800 squares.", "Solution_1": "I too got 800, the amount of squares in a ring are multiples of 8. so 100 * 8 = 800.", "Solution_2": "Hey Max, the e-mail address on your signature is wrong, I think. You might wanna change it...", "Solution_3": "[color=cyan]I suspect that he is engaging in a common tactic to post ones e-mail without getting spammed. Many programs are designed to search the web for things that fit the (blah)@(blah).(blah) form, adding them to a long list to recieve spam. By slightly misspelling ones e-mail address in a way a human wishing to e-mail him might correct, or alternatively by writing something like repDELETETHISPART123max@(blah).(blah), he can let people know his e-mail without getting spammed. Or, it might just be a typo, and all my elaborate explanation would be gone to naught.[/color]", "Solution_4": "hmmm...I've seen this before. At LiveJournal, your e-mail address appears normal, but there's some sort of useless code in the middle of it so that spam-robots will get a garbled e-mail address that includes various nonsense letters or something...." } { "Tag": [ "MATHCOUNTS" ], "Problem": "With all these books and courses on \"speed reading,\" I sometimes...\r\n\r\nWhat is speed reading?\r\nIs \"speed reading\" a hype, a myth, ... or just a market for people to sell courses and book about? \r\n\r\nWould reading faster prevent you from understanding some of what you're reading?\r\nOr is speed reading simply a collection of techniques to keep you focused on reading? \r\n\r\nAre there comparable methods to actually better understand what you're reading rather than just \"read\" faster?", "Solution_1": "I took a speed reading course long time ago, but did not practice it since then. I suppose it may help if you are a law student and must read 10000 pages every week. But it doesn't help to learn mathematics faster. And if I read a book for pleasure, I don't try to finish it as soon as possible.", "Solution_2": "Did the speed reading course help you at all at anything useful?\r\n\r\nThe only time when I would be interested in reading faster is on the SAT verbal section... but even then I would need to read certain parts slowly to understand them better.\r\n\r\nIs there any use in speed reading or is it more or less skimming? Can you retain information well if you speed read?", "Solution_3": "The ability to \"speed read\" is just the ability to skim effectively. A good course on speed reading will teach you how to quickly recognize the important information on a page without having to read all of the words. So yes, it's just skimming. But skimming is something you could be good or bad at. Reading 1/10 of the words in a paragraph at random will still give you an idea of what the paragraph is about, but if you read the right 1/10 of the words, you might get pretty much all of the useful info out of it. \r\n\r\nThat doesn't mean I have any good recommendations for you, though. I am sure lots of the stuff people sell about speed reading is not actually that helpful, and will either tell you obvious things or useless things. Just read free stuff about speed reading online. I'm sure if you google it and look around, you'll find a decent guide. Then you can practice (perhaps with SAT reading comp questions!) :)", "Solution_4": "I speed read before taking notes on a biology chapter (usually over 20 pages per chapter).This allows me to use effective note-taking skills and discern what would be important for my notes. (Should I annotate this passage about how they concluded DNA carries genetic info or should I save that space for later?).", "Solution_5": "Thanks for responding, guys.\r\n[b]\n7h3.D3m0n.117, as you speed read the chapter, what particular things do you you look for when determining what is important?\n\nXevarion, do you consider speed reading something worth looking into?\n[/b]", "Solution_6": "Yeah. It's really useful if you want to read about stuff but you just want an overview, or you want to get an impression of a book before reading it in detail. I'm so accustomed to the mathematical style of reading (read VERY slowly because everything is intricate and complicated and you need to understand the details) that I can't efficiently read anything else... :|", "Solution_7": "Xevarion, have you tried to learn how to speed read in the past? If you have, what would you recommend? \r\nRight now, I am peeking around with google to find a guide or some general suggestions on it as you suggested.", "Solution_8": "I haven't tried. My roommate is really good at it, so I'm going to try to get him to teach me. After that happens, if you're still interested, I'll send you whatever tips I got from him. Anyone else who's also interested, PM me and I'll write to you too. But it might be a few weeks before this happens.", "Solution_9": "[quote=\"aufha\"][b]\n7h3.D3m0n.117, as you speed read the chapter, what particular things do you you look for when determining what is important?[/b][/quote]\r\n\r\nWell first of all, general concepts are definately important but details that are already in another set of chapter notes should not be repeated (unless there is an extension). I normally make a habit of not recording notes that I already know. Notes should include study questions that not only tests your memorization of facts, but also application. \r\n\r\nFor example, the word \"recombinant types\"\r\n\r\nFirst I would define what it is. Then I would write the question, \"How is this different from parental types?\" After that, I try to make a general connection. (Such as, \"How does this apply to the rest of genetics?\").", "Solution_10": "Thank you both for your responses, Xevarion and 7h3.D3m0n.117.", "Solution_11": "In my mind, it has no importance to read speedy, but it's important to do meaningly.", "Solution_12": "I think, it is important to be able to read speedy, e.g. at school exams (history, language...) you have to read and understand texts quickly to be done in time.", "Solution_13": "Speed reading has only helped me in the MathCounts State CD round '07. If I speed read a chapter in a book, I find that I have to read it again to remember/understand most of it.", "Solution_14": "Speed reading isn't just about skimming. It's about how fast you can go while achieving close to 100% comprehension. It comes in handy during mathcounts countdown rounds, but also has many other uses. You should try to increase you reading speed as much as possible without sacrificing comprehension because that's what really matters. JFK had a reading speed of over 1000 words per minute." } { "Tag": [ "function", "Functional Analysis", "real analysis", "real analysis unsolved" ], "Problem": "Given a Banach space $E$, does there always exist a differentiable function $f: E \\rightarrow \\mathbb{R}$ such that $f' \\neq 0$ ? If $E$ is finite-dimensional, the answer is yes. What about the general case ?", "Solution_1": "there always exist linear functional isn't it ? (Hahn Banach)", "Solution_2": "Obviously it reduces to Hahn-Banach ... :blush: Thanks Alekk ..." } { "Tag": [ "floor function", "logarithms", "modular arithmetic", "real analysis", "real analysis unsolved" ], "Problem": "Prove that the following series converges:\r\n\\[\r\n\\sum_{n \\geq 1} \\frac{(-1)^{\\lfloor n\\sqrt{2} \\rfloor}}{n}.\r\n\\]\r\n\r\nEstimate its value.", "Solution_1": "this serie is convergent from Leibiniz criterion,but not absolutely convergent.", "Solution_2": "[quote=\"cezar lupu\"]this serie is convergent from Leibiniz criterion[/quote] It is not that simple. Indeed, using the Abel transform, it is not hard to show that it suffices to prove that $\\sum_{n=1}^m (-1)^{\\lfloor n\\sqrt 2\\rfloor}=O(m^{1-\\delta})$ for some $\\delta>0$, but the only proof of that I know uses Fourier series and other \"non-elementary\" techniques. I wonder whether vess or someone else has a simpler approach. As to estimating the value, once we know the speed of convergence, we can give some rough bounds. I don't know if it is possible to find an exact value of the sum though.", "Solution_3": "I see a similatr problem as follow:\r\n\r\n$\\sum_{n \\geq 1} \\frac{(-1)^{\\lfloor n \\rfloor}}{n}$.\r\n\r\nand the general will be true?", "Solution_4": "Of course.", "Solution_5": "but it is easy to say,hard to do,", "Solution_6": "Be careful with generalizations here: you can easily find $\\alpha>0$ such that $\\sum\\frac{(-1)^{\\lfloor n\\alpha\\rfloor}}n$ diverges. Try $\\alpha=2/3$, for example. You'll get the series $\\frac 11-\\frac 12+\\frac 13 +\\frac 14-\\frac 15+\\frac 16 +\\frac 17 -\\frac 18+\\frac 19+\\dots$. It is important that $\\sqrt 2$ is irrational, and, morover, cannot be approximated by rationals too well.", "Solution_7": "sorry :blush: \r\nand my problem should be :\r\n$ \\sum_{n \\geq 1} \\frac{(-1)^{\\lfloor \\sqrt n \\rfloor}}{n}$ converges.\r\nthanks fedja,can you prove your solution?", "Solution_8": "I finally found a more or less elementary solution. Let $S_m=\\sum_{n=1}^m (-1)^{[n\\sqrt 2]}$. If we show that $S_m=o(m)$, we can use the Abel transform to conclude that our series converges if and only if the series $\\sum_{m\\geqslant 1}\\frac {S_m}{m(m+1)}$ converges and, in case of convergence, they converge to the same sum. We shall show that $S_m=O(\\log m)$, which is more than enough for our purposes.\r\n\r\nLet $(1+\\sqrt 2)^\\ell=A_\\ell+B_\\ell\\sqrt 2$. The numbers $A_\\ell$ and $B_\\ell$ can be easily found from the recurrence relations $A_1=B_1=1$, $A_{\\ell+1}=A_{\\ell}+2B_\\ell$, $B_{\\ell+1}=A_\\ell+B_\\ell$. What is important for us is that $A_\\ell$ are odd for all $\\ell$, $B_{\\ell+1}\\leq 3B_\\ell$ and $|B_\\ell\\sqrt 2-A_\\ell|=(\\sqrt 2-1)^\\ell=\\frac 1{A_\\ell+B_\\ell\\sqrt 2}<\\frac 1{2 B_\\ell}$.\r\n\r\nWe shall also need a well-known fact that $|A-B\\sqrt 2|\\geq \\frac1{100 B}$\r\nfor all integer $A$ and $B\\geqslant 1$. It follows from here that, for any $B\\geq 1$, any two integers $n$ for which the distance from $n\\sqrt 2$ to the nearest integer is less than $\\frac 1{2B}$ are at least $\\frac B{100}$ units apart. \r\n\r\nLet now $m$ be any positive integer. Choose the largest $B=B_\\ell$ that is less than $\\frac m2$. Let $m'=m-2B$. Note that $m'\\leqslant \\frac 23m$ (otherwise we could use $B_{\\ell+1}$). We have $|S_m|\\leqslant |S_{m'}|+\\sum_{n=m'+1}^{m'+B}\\left|(-1)^{[n\\sqrt 2]}+(-1)^{[(n+B)\\sqrt 2]}\\right|$. Note now that, unless the distance from $n\\sqrt 2$ to the nearest integer is less than $\\frac{1} {2B}$, we have $[(n+B)\\sqrt 2]=[n\\sqrt 2]+A$ and, since $A$ is odd, the corresponding term in the sum is $0$. But there may be only 101 exceptional values of $n$ for which it isn't so. Thus, $|S_m|\\leqslant |S_{m'}|+202$. Now do the same for $m'$ instead of $m$ and so on. We'll make at most $C\\log m$ steps adding a constant during each step.\r\n\r\nThe estimate we got is quite rough: a direct computation shows that, for $m<200,000$ we have $|S_m|\\leqslant 7$, so it is quite believable that the Abel transformation of the original series converges fairly fast. The sum of first 200,000 terms is $-0.5154\\dots$ and it is quite believable that all digits presented are correct for the sum of the entire series.", "Solution_9": "@fedja : using the uniform distribution of $\\lfloor n \\sqrt{2}\\rfloor \\pmod{2}$ one can prove that\r\n\\begin{eqnarray*}\\sum_{n = 1}^{m}(-1 )^{\\lfloor n \\sqrt{2}\\rfloor}& = & o \\left( m \\right) \\end{eqnarray*}\r\nIs this enough to ensure convergence ?", "Solution_10": "Unfortunately, the answer is \"No\"." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "[img]http://img241.imageshack.us/img241/913/adsz2nh3.jpg[/img]\r\n\r\n\r\nwhat is the approximete length (in feet) of the longest ladder you can carry horizontally around the corner of the corridor shown here?round your answer down to the nearest foot.(thomas calculus chapter 4 practice exercises question 72)", "Solution_1": "The same [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=24520]link[/url] I sent to someone else Tuesday." } { "Tag": [ "probability" ], "Problem": "Four distinct points, $A$, $B$, $C$, and $D$, are to be selected from $1996$ points evenly spaced around a circle. All quadruples are equally likely to be chosen. What is the probability that the chord $AB$ intersects the chord $CD$?\n\n$\\text{(A)}\\ \\frac 14 \\qquad \\text{(B)}\\ \\frac 13 \\qquad \\text{(C)}\\ \\frac 12 \\qquad \\text{(D)}\\ \\frac 23\\qquad \\text{(E)}\\ \\frac 34$", "Solution_1": "[hide=\"Solution\"]Notice that all quadruples are equally likely likely to be chosen. Thus, we need only examine the six orderings of the points $A$, $B$, $C$, and $D$:\n\n$ACBD$, $ADBC$ , $ABCD$, $ADCB$, $ABDC$, and $ACDB$. \n\nOut of these six orderings, only two satisfy the intersection condition, so our answer is $\\frac 26=\\frac 13$.[/hide]", "Solution_2": "Clever problem!\r\n\r\n[hide]Every set of four given points can be labeled A,B,C,and D in 24 different ways. In 8 of these possibilities, AB crosses CD. Therefore the probability is 1/3 and the answer is [b]B[/b] :D [/hide]", "Solution_3": "Hexaditom - you don't need to consider all 24 pairings, because it's cyclic; you only need to consider six of them (fix one of the points).", "Solution_4": "when I first looked at the problem, being lazy I wanted to say 1/2- either it crosses or it doesn't.\r\nThen I solved the problem looking at the ABCD orderings and got 1/3\r\nnow, I know the fastest solution- we split the problem into three equally likely cases,\r\n1) It intersects\r\n2) Both A and B farther to the left (or right) than C,D. (This looks like two seperately removed self contained line segments)\r\n3) B C (or AD) contained within the arc marked off by the other segment\r\nThese three are equally likely, so it is 1/3", "Solution_5": "[quote=\"me@home\"]when I first looked at the problem, being lazy I wanted to say 1/2- either it crosses or it doesn't.\nThen I solved the problem looking at the ABCD orderings and got 1/3\nnow, I know the fastest solution- we split the problem into three equally likely cases,\n1) It intersects\n2) Both A and B farther to the left (or right) than C,D. (This looks like two seperately removed self contained line segments)\n3) B C (or AD) contained within the arc marked off by the other segment\nThese three are equally likely, so it is 1/3[/quote]\r\n\r\nCases 2 and 3 are very similar. What if one chord is the diameter and the other chord does not intersect it? Is that case 2 or 3? Also, how do you know for sure that each case is equally likely? Your answer is right, but your solution seems shaky.", "Solution_6": "I was simply looking for an intuitive approach, and judging from the answers possible this seems the best; also, as to your specific concerns, we can see that overall one case can't intuitively overpower another, with cancellations etc. The case 2 and case 3 options are just two subcases of non-intersect; they are meant to show that there are twice as many; it doesn't matter which case the diameter fits in really.", "Solution_7": "you set A as one point.\nthey intersect if and only iff b is opposite q, which is equally likelyas c or d, so it is 1/3\n" } { "Tag": [], "Problem": "A uniform rope of weight W hangs between two trees. The ends of the rope are the same height, and they each make angle theta with the trees. Find\r\n\r\na) the tension at either end of the rope\r\n\r\nb) the tension in the middle of the rope\r\n\r\n\r\n\r\nThanks for the help guys.", "Solution_1": "For part (a), imagine a rod of weight W suspended by two [i]massless [/i]ropes at each end, each of which makes an angle theta with the trees. That will give you the same answer.\r\n\r\nFor part (b), the tension is zero. Intuitively, since the rope makes less of an angle with the horizontal as you approach the middle, it becomes more \"slack.\"", "Solution_2": "I think the part a) you have is correct, but the part b) answer doesn't seem to be. It is not zero because in the book I got the question from, it gave you a \"hint\" in which it gave you a definite angle (it was 45 degrees actually), and told you the tension for both parts (for the second part it was W/2). So I don't think it's zero.", "Solution_3": "Sorry, I didn't think that through very carefully.\r\n\r\nWhat I should have said, for part (b), is: now imagine an infinitesimally short segment of massless rope connecting two rods, each of weight W/2, and which are each connected to the trees with two other segments of massless rope making angles of theta with the horizontal at their other ends.", "Solution_4": "For partb,horizontal component of tension at the is equal to tension in the middle." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "On a chess board ($8*8$) there are written the numbers $1$ to $64$: on the first line, from left to right, there are the numbers $1, 2, 3, ... , 8$; on the second line, from left to right, there are the numbers $9, 10, 11, ... , 16$;etc. The $\\\"+\\\"$ and $\\\"-\\\"$ signs are put to each number such that, in each line and in each column, there are $4$ $\\\"+\\\"$ signs and $4$ $\\\"-\\\"$ signs. Then, the $64$ numbers are added. Find all the possible values of this sum.", "Solution_1": "[hide]Write each number in the form A + 8B, where A ranges from 1 to 8. A is constant along columns, B along rows. Sum As columnwise and Bs rowwise. The sum is 0. [][/hide]", "Solution_2": "On a chess board ($8\\times 8$) there are written the numbers $1$ to $64$: on the first line, from left to right, there are the numbers $1, 2, 3, ... , 8$; on the second line, from left to right, there are the numbers $9, 10, 11, ... , 16$;etc. The ''$ + $'' and ''$- $'' signs are put to each number such that, in each line and in each column, there are $4$ ''$+$'' signs and $4$ ''$-$'' signs. Then, the $64$ numbers are added. Find all the possible values of this sum." } { "Tag": [ "algebra", "polynomial", "function", "calculus", "derivative", "real analysis" ], "Problem": "Let $x_{0}.< I remember doing this on some practice test and getting Thursday...but yeah. Probably thursday then.", "Solution_4": ":rotfl: you atleat tried\n\nI don't get what you are saying.", "Solution_5": "way to have a conclusive answer on this thread guys... anyway we can tell n is a leap year because day 200 of year N+1 is divisible by seven, meaning there must be 266 not 265 days between the two dates. Counting backwards we go back 565 days = 5 (mod 7) going back 5 days from tuesday we get to Thursday (A)", "Solution_6": "I got Thursday as well (mod 7 technique)\r\nBut mathemonster, please don't post on a 5-month-old topic.", "Solution_7": "its not 5 months, its 1 year and 5 months..\r\n\r\nand i also got A by mod 7 teck", "Solution_8": "I got Thursday too.\r\nI think they give you year N+1, which is a leap year, to prove N-1 is not a leap year. But you know that they would have to give you N+1 as a leap year or N-1 could either be a leap year or a normal one, so you wouldn't have to figure it out, right? :D", "Solution_9": "I got Thursday by counting", "Solution_10": "Welcome to AoPS? Don't bump 10 year old posts...", "Solution_11": "Year N is a leap year. Then year N-1's 100th day is 565 days before 300th day of year N. Then the 100th day of N-1 is Thursday.", "Solution_12": "[quote=Skupp3]Welcome to AoPS? Don't bump 10 year old posts...[/quote]\n\nOkay ", "Solution_13": "[hide=Solution]Note that this means the 6th day of year $N$ is Tuesday and the 4th day of year $N+1$ is Tuesday, so year $N+1$ is a leap year. There are hence $565$ days between the 100th of year $N-1$ and the 300th of year $N$, and $565 \\equiv 5 \\pmod 7$. Going backward five days gives $\\text{Thursday} \\implies \\boxed{\\text{(A)}}$.[/hide]\n\n[b]End time:[/b] 4:15 PM\n\n[b]Interruptions + Solution Writing + Non-Strict Environment:[/b] 15 minutes\n\n[b]Projected End Time on Real Test:[/b] 4:00 PM (49 minutes)\n\n[b]Satisfaction:[/b] 7/10\n\n[b]Score:[/b] 145.5", "Solution_14": "[quote=OlympusHero][hide=Solution]Note that this means the 6th day of year $N$ is Tuesday and the 4th day of year $N+1$ is Tuesday, so year $N+1$ is a leap year. There are hence $565$ days between the 100th of year $N-1$ and the 300th of year $N$, and $565 \\equiv 5 \\pmod 7$. Going backward five days gives $\\text{Thursday} \\implies \\boxed{\\text{(A)}}$.[/hide]\n\n[b]End time:[/b] 4:15 PM\n\n[b]Interruptions + Solution Writing + Non-Strict Environment:[/b] 15 minutes\n\n[b]Projected End Time on Real Test:[/b] 4:00 PM (49 minutes)\n\n[b]Satisfaction:[/b] 7/10\n\n[b]Score:[/b] 144[/quote]\n\nWait...did you speed run this AMC 10?? :o (You did a great job, BTW. :))", "Solution_15": "Yes, I did. I believe the question I missed was #18 (geometric probability except not really probability) for anyone curious.\n\nOh wait oops I got an answer of 47 which was not an answer choice so I actually would've left that blank so 145.5 not 144 :rotfl:", "Solution_16": "So Pr000 :omighty:", "Solution_17": "[hide=Solution]The $100^{\\text{th}}$ day of the year $N-1$ is $565$ days before the $300^{\\text{th}}$ day of year $N.$ Hence, the $100^{\\text{th}}$ day of the year $N-1$ is the same day as the $-565\\equiv-5\\equiv2\\text{th}\\pmod7$ day after Tuesday, which is $\\boxed{\\textbf{(A) Thursday}}.$" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "We know that $ f: \\mathbb{R} \\to \\mathbb{R}$, $ f$ is continuous on $ \\mathbb{R}$ and for all $ x\\notin \\mathbb{Q} : f'(x) \\equal{} 0$. Is it true that $ f \\equal{} const$ ?", "Solution_1": "Some modification of the [url=http://en.wikipedia.org/wiki/Cantor_function]Cantor function[/url] should be an appropriate counterexample.", "Solution_2": "[quote=\"t0rajir0u\"]Some modification of the [url=http://en.wikipedia.org/wiki/Cantor_function]Cantor function[/url] should be an appropriate counterexample.[/quote]\r\n\r\nThe Cantor set is uncountable.\r\n\r\nOne can show that if $ f\\in C([a,b])$, $ f'$ exists for $ x\\in [a,b]\\setminus A$ where $ A$ is countable, and $ f'\\in L^1([a,b])$, then $ f$ is absolutely continuous (and hence can be recovered from $ f'$ by integration). In this case, one easily sees that we must have $ f$ constant." } { "Tag": [ "function", "calculus", "derivative", "algebra", "functional equation", "cyclic function" ], "Problem": "Please help me again with this one:\r\n\r\nIf F(x-3/x+1) + F(3+x/1-x)=x: x not=1; x not=-1, determine the function of F(x)", "Solution_1": "Somebody help me ,Please!", "Solution_2": "Posted in the wrong section - it's not linear algebra.", "Solution_3": "Do you mean $ F \\left( \\frac{x\\minus{}3}{1\\plus{}x} \\right) \\plus{} F \\left( \\frac{x\\plus{}3}{1\\minus{}x} \\right) \\equal{} x$", "Solution_4": "yes that's what i mean. i know it's in the wrong section and feel so sorry about it.", "Solution_5": "[hide=\"Hint\"] Let $ G(x) \\equal{} \\frac{x \\plus{} 3}{1 \\minus{} x}$. What is $ G^{\\minus{}1}(x)$? [/hide]", "Solution_6": "yes i know that.\r\nf(g) + f(inv g) = x\r\nbut i still can't get the answer. Would you please help me to solve it?", "Solution_7": "[hide=\"Second hint\"] Compute $ g(g(x))$. [/hide]", "Solution_8": ":| sorry i have been evaluating so many time, but still can't get the answer.", "Solution_9": "[hide=\"another hint\"]Try plugging in $ g(x)$ and $ g(g(x))$ as $ x$ to the original equation. Then solve for $ f(x)$.[/hide]", "Solution_10": "[quote=\"worthawholebean\"][hide=\"another hint\"]Try plugging in $ g(x)$ and $ g(g(x))$ as $ x$ to the original equation. Then solve for $ f(x)$.[/hide][/quote]\r\nNice approach, worthawholebean! Can you show me how can you think of it? Does it have any method to solve this kind of functional equation?", "Solution_11": "Plugging $ x\\rightarrow \\frac {x \\minus{} 3}{x \\plus{} 1}$ we get\r\n\\[ f(x) \\plus{} f(\\frac {x \\plus{} 3}{1 \\minus{} x}) \\equal{} \\frac {x \\minus{} 3}{x \\plus{} 1}\\ \\ \\ (i)\r\n\\]\r\nand plugging $ x\\rightarrow \\frac {3 \\plus{} x}{1 \\minus{} x}$ we get\r\n\\[ f(x) \\plus{} f(\\frac {x \\minus{} 3}{x \\plus{} 1}) \\equal{} \\frac {3 \\plus{} x}{1 \\minus{} x}\\ \\ \\ (ii)\r\n\\]\r\nadding $ (i)$ and $ (ii)$ we get $ f(x) \\equal{} \\frac {x(3 \\plus{} x)(3 \\minus{} x)}{2(x \\plus{} 1)(x \\minus{} 1)}$.", "Solution_12": "[quote=\"ghjk\"][quote=\"worthawholebean\"][hide=\"another hint\"]Try plugging in $ g(x)$ and $ g(g(x))$ as $ x$ to the original equation. Then solve for $ f(x)$.[/hide][/quote]\nNice approach, worthawholebean! Can you show me how can you think of it? Does it have any method to solve this kind of functional equation?[/quote]I used t0rajir0u's hints and got that $ g(g(g(x)))\\equal{}x$, so I tried plugging in random things for a while and eventually tried $ g(x)$ and $ g(g(x))$, and I had a linear system.", "Solution_13": "In other words, $ g(x)$ is a [[cyclic function]] of order 3. They are useful in that you can plug away end up with a usually solvable system of equations.\r\nFor a simple example, to find $ f(x)$ such that $ f(x) \\plus{} 3f(1/x) \\equal{} x\\minus{}1$, we note that $ f(x) \\equal{} 1/x$ is cyclic with order 2. Then we plug in $ 1/x$ for $ x$ to get \r\n$ f(1/x)\\plus{}3f(x)\\equal{}1/x \\minus{} 1$. Now we can solve this system.", "Solution_14": "[quote=\"archimedes1\"we note that $ f(x) \\equal{} 1/x$ is cyclic with order 2..[/quote]\r\nI think it's incorrect since depending on the definition you provide, I guess the order of a cyclic function is n in oder to have: f^n(x)=x for some integer n>1.\r\nNow we can't find any integer n>1 to satisfy:f^n(x)=1/x, so how can you state that f(x)=1/x is cyclic with oder 2? Am I wrong anywhere? :maybe:", "Solution_15": "Now I'm forgetting whether $ f^n(x)$ means $ f(f(f(...f(x))...)$ or $ (f(x))^n$. According to AoPS Volume II, the first definition is correct, but according to some WOOT instructors the second is correct. In either case, the definition in the wiki means $ f(f(...f(x))...)$, where $ f$ is iterated $ n$ times.\r\n\r\n$ f(f(x)) \\equal{} x$, so $ f(x)$ has order 2... right?", "Solution_16": "As far as I know, $ f^n(x)$ always means iteration (whenever $ f(x)$ is not a trigonometric function). Incidentally, the \"cyclic functions\" we're talking about here are elements of finite order in the [url=http://en.wikipedia.org/wiki/Projective_linear_group]projective linear group[/url].", "Solution_17": "That ambiguous notation has always bothered me, but there's not much you can do about it - $ f^n(x)$ is iteration on any non-trigonometric function.$ f^{(n)}$ is differentiation as well.", "Solution_18": "[quote=\"worthawholebean\"]That ambiguous notation has always bothered me, but there's not much you can do about it - $ f^n(x)$ is iteration on any non-trigonometric function.$ f^{(n)}$ is differentiation as well.[/quote]\r\n...at WOOT we were taught that $ f^{(n)}$ means iteration and $ f^n(x)$ is taking $ f(x)$ to the $ n$th power. I guess that the definition in a problem is usually apparent from the context.", "Solution_19": "I remember something about that in one of the WOOT lectures, but the more common standard is that $ f^{(n)}$ denotes the $ n^{th}$ derivative.", "Solution_20": "its correct the function i found or not? :(", "Solution_21": "Thank you friends, you just opened up my mind.\r\nBut the answer key is :\r\nf(x) = 4x/(1-x^2) - 0.5\r\n :maybe:\r\ni am sure it is getting so close now." } { "Tag": [ "probability", "linear algebra", "matrix", "probability and stats" ], "Problem": "As a simplified model for weather forecasting, suppose that the weather (either wet or dry) tomorrow will be the same as the weather today with probability p.\r\n\r\nIf the weather is dry on January 1st, what is the probability (in terrms of p) that the weather will be dry on February 1st?", "Solution_1": "That probability is $ \\frac{1\\plus{}(2p\\minus{}1)^{31}}{2}.$\r\n\r\nUnless $ p$ is very close to either $ 1$ or $ 0,$ that should be very close to $ \\frac12.$\r\n\r\nWhat you describe (besides bearing no resemblance to actual weather forecasting) is a two-state Markov chain whose transition matrix is\r\n\r\n$ P\\equal{}\\begin{pmatrix}p&1\\minus{}p\\\\1\\minus{}p&p\\end{pmatrix}.$\r\n\r\nThis matrix has eigenvalues $ 1$ and $ 2p\\minus{}1$ and can be readily diagonalized. Using the diagonal form, we compute that:\r\n\r\n$ P^n\\equal{}\\frac12\\begin{pmatrix}1\\plus{}(2p\\minus{}1)^n&1\\minus{}(2p\\minus{}1)^n\\\\1\\minus{}(2p\\minus{}1)^n&1\\plus{}(2p\\minus{}1)^n\r\n\\end{pmatrix}.$\r\n\r\nThe answer above is the $ (1,1)$ entry of $ P^n$ for $ n\\equal{}31.$" } { "Tag": [ "conics", "ellipse" ], "Problem": "$a$ and $b$ are positive parameters.Find the minimum value $S = \\sqrt{(x-a)^2+b^2}+\\sqrt{(x-b)^2+a^2}$.", "Solution_1": "let there be $A(a,b)$ and $B(b,-a)$ as well as $P(x,0)$.\r\n\r\nthe we want to minimize $\\overline{AP}+\\overline{PB}$ and thats exactly when $A,P,B$ are colinear. \r\n\r\nthe line through $A$ and $B$ is $y=\\frac{b+a}{a-b}(x-a)+b=\\frac{a+b}{a-b}x -\\frac{a+b}{a-b}a+b$\r\n\r\nso $S$ is the distance between $A$ and $B$: $S=\\sqrt{(a-b)^2+(b+a)^2}=\\sqrt{2a^2+2b^2}$\r\n\r\npeeta", "Solution_2": "great observation. i once did a similar problem which used the properties of ellipse!!!", "Solution_3": "What's that problem? Can you post it here?" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Each unit square of a 2 by n grid is to be colored with blue or green so that no 2by2 green square is obtained, let c_{n} be the number of different colorings. Determine max{k} so that 3^k divides c_{2001} \r\n\r\nLet n>0 be a integer, s={1,...n}. Show that the nuber of permutations of s with no fixed points and the number of permuttions of s with exactly one fixed point differ by 1.\r\n\r\nLet s={1,...n} T be the set of all subsets of s (including s and the empty set). One tries to choose 3 sets T1, T2, T3from the set T so that T1 \\in T3, T2\\in T3. I howm any ways can this be done?", "Solution_1": "I think I have solved the first one.\r\n\r\nLet $h_{n}$ be the number of good colorings, with two most-right squares green, and $g_{n}$ the numbers of good colorings, with two most-right squares colored either blue, either green-blue. (For example $h_{1}= 1, g_{1}= 3 , h_{2}= 3 , g_{2}= 12$).\r\n\r\nWe obviously have:\r\n$c_{n}= h_{n}+g_{n}$\r\nNow we think of a reccurency. If we want to create a good coloring of 2xN grid, we can put two squares colored in any way to the right end of a good colored 2x(N-1) grid, which has two most-right squares colored non-green.\r\nWe can also put two squares, which won't be both green, to the right end of the good 2x(N-1) grid, if the two most-right squares of that 2x(N-1) grid were both green. So we have:\r\n$c_{n}= 3\\cdot h_{n-1)+4 \\cdot g_{n-1}= 4 \\cdot c_{n-1}-h_{n-1}}$\r\nNow we notice, that $h_{n}= g_{n-1}$\r\nSo we have:\r\n$c_{n}= 4 \\cdot c_{n-1}-h_{n-1}= 4 \\cdot c_{n-1}-g_{n-2}= 4 \\cdot c_{n-1}-c_{n-2}+h_{n-2}$\r\nNow we continue, and we get:\r\n$c_{n}= 4 \\cdot c_{n-1}-c_{n-2}+c_{n-3}-...+(-1)^{n}c_{1}$\r\nbut there is also:\r\n$c_{n+1}= 4 \\cdot c_{n}-c_{n-1}+c_{n-2}-...+(-1)^{n+1}c_{1}$\r\nadding up we get:\r\n$c_{n+1}= 3 \\cdot ( c_{n}+c_{n-1})$\r\nWhich is absolutely sattisfying us.\r\nNow by checking that $c_{1}= 4 , c_{2}= 15$ (note that $c_{0}= 1$, so our reccurency works!!) we can easily find k =1000.", "Solution_2": "The second one is preety obvious, since from inclusion-exclusion we get, that the number of permutations with no fixed point is equal to:\r\n$n! \\cdot (\\frac{1}{0!}-\\frac{1}{1!}+\\frac{1}{2!}-...+(-1)^{n}\\cdot\\frac{1}{n!})$\r\nnow we choose the fixed point in n ways, and make a permutation with no fixed point with a set of ${1, 2, ... ,n-1}$. So we get:\r\n$n \\cdot (n-1)! \\cdot (\\frac{1}{0!}-\\frac{1}{1!}+\\frac{1}{2!}-...+(-1)^{n-1}\\cdot \\frac{1}{(n-1)!})$\r\nThoose two obiously differ by $n! \\cdot (-1)^{n}\\cdot\\frac{1}{n!}$\r\n\r\nThe third one is also not to hard. Every element can be either in $T_{3}$ and is not in any other subset, or it is in $T_{1}$ and $T_{3}$, or it is in $T_{2}$ and $T_{3}$, or in $T_{1}$ , $T_{2}$ and $T_{3}$. It can also not be an element of any of the subsets $T_{1}$ , $T_{2}$ and $T_{3}$. so there are $5^{n}$ such triples of subsets.\r\n\r\nSorry for double-posting, but I posted one solution, and after that I started thinking of the other.\r\nI hope my solutions are clear and understandable (and true of course :P )\r\nAnd sorry for my awful english :wink:\r\n\r\n\r\nTheese were nice problems, I especially like the first one.", "Solution_3": "[quote=\"jgao\"]Let n>0 be a integer, s={1,...n}. Show that the nuber of permutations of s with no fixed points and the number of permuttions of s with exactly one fixed point differ by 1.[/quote]\r\n\r\nI think I have solved this one:\r\n\r\n\r\nDefine an $n$-permutation as a permutation of the set $\\{ i | 1 \\leq i \\leq n \\}$\r\nLet the number of $n$-permutations with no fixed points be $a_{n}$\r\nLet the number of $n$-permutations with one fixed point be $b_{n}$\r\n\r\nConsider any $n$-permutation be $\\{ p_{i}| 1 \\leq i \\leq n \\}$\r\nLet $G$ be a graph where there is a directed edge from vertex $i$ to vertex $j$ iff $p_{i}=j$\r\nLet $q_{p_{i}}=i$\r\nThen fixed points correspond to vertices each having an edge to itself\r\nFor any sub-permutation of the permutation with no fixed points,\r\nEach vertex in the corresponding sub-graph has $1$ in-edge and $1$ out-edge\r\nThus the sub-graph is a set of disjoint cycles\r\n\r\nConsider the graph corresponding to an $n$-permutation with no fixed points\r\nConsider the vertex $n$\r\nIf $p_{p_{n}}\\not=n$, removing vertex $n$ and replacing directed edges $(q_{n},n)$ and $(n,p_{n})$ with directed edge ${(q_{n},p_{n}}$ will result in an $(n-1)$-permutation with no fixed point,\r\nAnd there are exactly $(n-1)$ $n$-permutations for each $(n-1)$-permutation because you can insert vertex $n$ after any of the $(n-1)$ vertices\r\nIf $p_{p_{n}}=n$, merging vertices $n,p_{n}$ will result in an $(n-1)$-permutation with one fixed point\r\nAnd there is exactly $1$ $n$-permutation for each $(n-1)$-permutation\r\n\r\nThus $a_{n}=(n-1)a_{(n-1)}+b_{(n-1)}$\r\n\r\nConsider the graph corresponding to an $n$-permutation with one fixed point\r\nRemoving the vertex corresponding to the fixed point will result in an $(n-1)$-permutation with no fixed points\r\nAnd there are exactly $n$ $n$-permutation for each $(n-1)$-permutation\r\n\r\nThus $b_{n}=na_{(n-1)}$\r\n\r\nSo we have\r\n$a_{n}=(n-1)a_{(n-1)}+(n-1)a_{(n-2)}$\r\n$a_{n}-na_{(n-1)}=-a_{(n-1)}+(n-1)a_{(n-2)}$\r\n$(a_{n}-b_{n})=-(a_{(n-1)}-b_{(n-1)})$\r\n\r\nAnd it only remains to prove that $a_{1}$ and $b_{1}$ differ by $1$\r\n\r\n\r\nEdit: Someone beat me to it :P but I still like my solution better, because in my opinion it uses intrinsic properties of the problem :lol:" } { "Tag": [ "\\/closed" ], "Problem": "How do you use the avatar you uploaded from your computer in the classroom? I looked through the faq and couldn't find anything.", "Solution_1": "If your avatar is a gif or a jpg, you should be able to change your avatar in the Classroom to your board avatar by just clicking on the little person by the input box. After clicking that, scroll down and your avatar should appear. The avatars for the classroom are refreshed every 24 hours, so if you upload something right now, for example, it won't be available tonight." } { "Tag": [ "inequalities", "absolute value", "inequalities proposed" ], "Problem": "let $x_1,x_2 ... x_n$ be real numbers not equal to $1$ and with product $1$\r\nshow that :\r\n$\\left| \\frac{x_1 x_2}{x_1-1}\\right|+\\left| \\frac{x_2 x_3}{x_2-1}\\right|+...+\\left| \\frac{x_n x_1}{x_n-1}\\right| > \\left[ \\frac{n+1}{2} \\right]$\r\n\r\nhere $\\left[ x \\right]$ denotes the greatest integer less than $x$ while $|x|$ the absolute value of $x$\r\n\r\nShow that $\\left[ \\frac{n+1}{2} \\right]$ can not be improved.", "Solution_1": "So? :oops:" } { "Tag": [ "search" ], "Problem": "I saw lots of TST tests, but I have no clue what a TST is. I tried using the search(since everyone who doesn't seems to get attacked) but I don't think I'm to good with keywords. So could someone explain what they are?", "Solution_1": "Team Selection Test - theres another thread on this in the Other Topics forum.", "Solution_2": "A simple search on \"what is tst\" reveals the magic answer: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=what+is+tst&t=22017" } { "Tag": [ "polynomial", "function", "search", "superior algebra", "superior algebra unsolved" ], "Problem": "This is a problem from a previous IMO\r\nLet $ a > b > c > d$ be positive integers and suppose that $ ac\\plus{}bd \\equal{} (b\\plus{}d\\plus{}a\\minus{}c)(b\\plus{}d\\minus{}a\\plus{}c)$. Prove that ab + cd is not prime.\r\n\r\nThere exists a solution using the Eisenstein integers.", "Solution_1": "Use function search, i have http://www.mathlinks.ro/Forum/viewtopic.php?p=119217#119217", "Solution_2": "[quote=\"N.T.TUAN\"]Use function search, i have http://www.mathlinks.ro/Forum/viewtopic.php?p=119217#119217[/quote]\r\nYes, but I am looking for the solution using Eisenstein integers ;)", "Solution_3": "[hide]\nThe given condition is equivalent to\n\\[ ac\\plus{}bd\\equal{}(b\\plus{}d)^{2}\\minus{}(a\\minus{}c)^{2}\\Rightarrow a^{2}\\minus{}ac\\plus{}c^{2}\\equal{}b^{2}\\plus{}bd\\plus{}d^{2}\\]\nLet $ p \\equal{} ab\\plus{}cd$ and suppose $ p$ prime. Multiply above equation by $ b^{2}$ and replace $ ab$ with $ p\\minus{}cd$ whenever it appears. We get\n\\[ (p\\minus{}cd)^{2}\\minus{}bc(p\\minus{}cd)\\plus{}b^{2}c^{2}\\equal{}b^{2}(b^{2}\\plus{}bd\\plus{}d^{2})\\]\n\n\\[ p(p\\minus{}2cd\\minus{}bc)\\equal{}(b^{2}\\minus{}c^{2})(b^{2}\\plus{}bd\\plus{}d^{2})\\qquad (*)\\]\nSince $ p$ is a prime dividing a product of two integers, it divides one of them. But\n\\[ 0 0$ $ X_r \\equal{} \\bigcup_{x \\in X} B_r(x)$, where $ B_r(x)$ is the ball centered at $ x$ of radius $ r$. Thus, $ X_r$ is what we get by taking a small ball around every element of $ X$, and so $ X_r$ is clearly open. I claim that $ I \\equal{} \\bigcap_{r > 0} X_r \\equal{} X$. Certainly $ X \\subset X_r$ for every $ r$ and so $ X \\subseteq I$. Suppose we have some point of $ M$ that's not in $ X$, say $ y$. For any $ x \\in X$, we know there is some $ r$ such that $ r < d(x, y)$, so $ y$ will not be in a ball around $ x$ with radius $ r$. Thus, for small enough $ r$ we have that $ y \\not\\in X_r$ and so $ y \\not \\in I$, so $ X \\equal{} I$, as desired.", "Solution_3": "[quote=\"JBL\"]$ X_r$ is what we get by taking a small ball around every element of $ X$, and so $ X_r$ is clearly open. [/quote]\r\nThis isn't my specialty, but $ B_r(x)$ need not be open if $ r$ is small and $ x$ is isolated, so I believe a necessary and sufficient condition is that $ X$ contain no isolated points. \r\n\r\nA general point like this was recently raised on Tim Gowers' blog - namely, that the weakest point in an argument is usually when a word like \"clearly\" is being used :P", "Solution_4": "$ B_r(x)$ is the open ball in $ M$ around $ X$, and $ X_r$ is a union of these open balls and so open. So that step is actually fine. (But you're right that as a stylistic matter I shouldn't have used \"clearly.\")", "Solution_5": "I believe the problem is here:[quote=\"JBL\"]For any $ x \\in X$, we know there is some $ r$ such that $ r < d(x, y)$, so $ y$ will not be in a ball around $ x$ with radius $ r$. Thus, for small enough $ r$ we have that $ y \\not\\in X_r$ and so $ y \\not \\in I$.[/quote]Though the first statement is true, the conclusion does not to follow. But the argument should work if $ X$ is nowhere dense in $ M$.", "Solution_6": "The problem with the reasoning, as JoeBlow showed, is that r depends on x so the infimum of such r's could well be 0. sos440 is right that such a statement holds in topological spaces with strong topologies. What is an example of a topological space in which it does not hold?", "Solution_7": "[quote=\"\u00a7outh\u00a7tar\"]What is an example of a topological space in which it does not hold?[/quote]Take the simplest example of a topological space $ X$ in which not every point is closed: $ X\\equal{}\\{1,2\\}$ with the topology $ \\{\\emptyset,\\{1\\},\\{1,2\\}\\}$. Then $ 1$ is not a closed point and $ \\{2\\}$ is not the intersection of open subsets.", "Solution_8": "[quote=\"JoeBlow\"]But the argument should work if $ X$ is nowhere dense in $ M$.[/quote] Yes, this is sufficient, but actually there is a simpler necessary and sufficient condition: [hide=\"Answer to my question\"]This argument works if and only if $ X$ is closed in $ M$. (So, oddly enough, this argument \"usually\" fails to write open sets as intersections of open sets. Also, a nice example of a spectacular failure of this argument (suggested by JoeBlow's comment) is when $ M \\equal{} \\bf R$ and $ X \\equal{} \\bf Q$.)[/hide]", "Solution_9": "The notion of $ G_\\delta$-sets applies here.", "Solution_10": "Not to mention the argument misses the case for empty sets :rotfl: \r\n\r\n :maybe:" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "The sequence $ S\\equal{}a_{1}a_{2}...a_{N}$, where $ a_{i}\\in \\{A, B\\}$ is not palindromic. Let $ a_{i}^{j}, i\\leq j$ denote the subsequence $ a_{i}a_{i\\plus{}1}...a_{j}$. For any subsequence $ a_{i}^{j}$, let $ F(a_{i}^{j})\\equal{}A^{k}B^{j\\minus{}i\\plus{}1\\minus{}k}$, where $ k$ is the number of A's and $ j\\minus{}i\\plus{}1\\minus{}k$ is the number of B's in the subsequence. Thus, for any length N sequence, we get $ \\frac{N*(N\\plus{}1)}{2}$ F's, one for each subsequence. Let this collection of F's be called $ D(S)$. If two sequences $ S_{1}$ and $ S_{2}$, have same D, i.e, $ D(S_{1}) \\equal{} D(S_{2})$, then show that $ S_{1}$ must be of the form $ Sx_{1}Sx_{2}Sx_{3}...Sx_{k}S$ and $ S_{2}$ in that case will equal $ Sx_{k}Sx_{k\\minus{}1}Sx_{k\\minus{}2}...Sx_{1}S$, for some fixed sequence $ S$ (same for both) and $ x_{i}\\in\\{A, B\\}$.", "Solution_1": "Sorry, a small part of the qn is missing. The last time will read - If two sequences $ S_{1}$ and $ S_{2}$, have same D, i.e, $ D(S_{1}) \\equal{} D(S_{2})$, then show that $ S_{1}$ must be of the form $ Sx_{1}Sx_{2}Sx_{3}...Sx_{k}S$ and $ S_{2}$ in that case will equal $ Sx_{k}Sx_{k\\minus{}1}Sx_{k\\minus{}2}...Sx_{1}S$, for some fixed sequence $ S$ (same for both) and $ x_{i}\\in\\{A, B\\}$, OR $ S_{1}$ is the reversed sequence $ S_{2}$. Does anyone have any ideas about this problem and how to go about it? Thanks." } { "Tag": [ "inequalities", "algebra solved", "algebra" ], "Problem": "Find the greatest positive constant A which satisfies:\r\n\r\nx/\\sqrt(y2 + z2) + y/\\sqrt(x2 + z2) + z/\\sqrt(x2 + y2) > A\r\n\r\nfor any positive reals x, y and z.\r\n\r\n[Well, I was wondering: shouldn't it be \\geq A? Or am I totally wrong?]", "Solution_1": "Agree the strict inequality is strange.\r\n\r\nA = 3/\\sqrt 2 no ?\r\n\r\n. We can suppose \\sum x = 1 since it's homogeneous.\r\n\r\n. C.S : \\sum x / \\sqrt(y+z) * \\sum \\sqrt(y+z) >= (\\sum \\sqrt x) >= 3 (last by concavity of \\sqrt)\r\n\r\n. C.S again : (\\sum \\sqrt(y+z)) <= 6 \\sum x <= 2 (last by convexity of x)", "Solution_2": "In fact, A=2 and the problem is quite simple. All you have to do is to observe that taking a=b and c very small the expression tends to 2 and to apply the Arithmetic-Harmonic mean:\r\n x/sqrt(y^2+z^2)=sqrt(x^2/(y^2+z^2))>=2x^2/(x^2+y^2+z^2). Sum up these inequalities and observe that we cannot have equality." } { "Tag": [ "\\/closed" ], "Problem": "Is there a reason why the beginner level problems are in \"Getting Started\", medium level questions are in \"Intermediate [b]Topics[/b]\", and the hard level questions are in \"Advanced [b]Problems[/b]\"?\r\nJust out of curiosity...", "Solution_1": "It's all subconscious. Heh, you noticed something that Mr. Rusczyk probably didn't when he renamed the forums...\r\n\r\nWell, it seems somewhat natural to me...\"Getting Questions\" wouldn't make much sense, right?", "Solution_2": "Yes, but shouldn't Advanced Problems be Advanced Topics since the others are called \"... Topics\"", "Solution_3": "Well, no...he can name then whatever he wants as long as they get the message across.\r\n\r\nBut why does it matter?", "Solution_4": "I actually find the deviation in the naming pleasant. But seriously, go worry about other things." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Hi. I was looking for the 2008 Nat Cd Round. They took it down from the mathcounts website. Do any of you know how to access it now? :huh:", "Solution_1": "here , u can have it 4 free. :D I'm willing to!", "Solution_2": "I think he was talking about the 2008 web cast? if so, it's just gone from the Mathcounts website forever. Every year, they take off the old webcast forever to replace it with the new one.", "Solution_3": "well... i was looking for the webcast.. :maybe: \r\nbut thank you cubex!! :lol:", "Solution_4": "Maybe we can share our answers if anyone else tried to solve these? Please post your answers here.", "Solution_5": "If you want a solution for a particular problem, it's probably best to create a new thread. :wink:\r\n\r\nEDIT: Not to be a hall monitor for MATHCOUNTS, but is it okay to post a single test on a thread (like CubeX did)?", "Solution_6": "Ok here are the solutions for the 2008 Nationals:", "Solution_7": "could u give me 2008 national mathcounts sprint.\r\nPlease", "Solution_8": "CuBeX, I'm pretty sure it's illegal to post full MATHCOUNTS tests on the Internet.", "Solution_9": "Catalog is [url=http://mathcounts.org/Document.Doc?id=294]here[/url] and order form is [url=http://mathcounts.org/Document.Doc?id=295]here[/url]." } { "Tag": [ "trigonometry", "geometry", "number theory" ], "Problem": "Hi,\r\nI guess I should start some olympiad preparation now. If someone can recommend me some nice preparation books(national olympiad level), that'll be great!\r\nThanks!", "Solution_1": "engel :lol:", "Solution_2": "? :)didn't get it", "Solution_3": "Chen Sun (if that is correct) recommends [i]Problem Solving Strategies[/i] by Arthur Engel.", "Solution_4": "[quote=\"boxedexe\"]Chen Sun (if that is correct) recommends [i]Problem Solving Strategies[/i] by Arthur Engel.[/quote]\r\n\r\nYes, that is correct :clap2: . And you, boxedexe, are?", "Solution_5": "His name is in his signature -- Masoud Zargar. ;)", "Solution_6": "Oops, just notice a typo in the title :oops:", "Solution_7": "[quote=\"sunchips\"]engel :lol:[/quote]\r\nThanks! What kind of questions does the book cover? I mean the difficulty. >=national olympiad?", "Solution_8": "ya its very good", "Solution_9": "I would call it pre-olympiad, but it has a lot of (the easier) olympiad problems.", "Solution_10": "[quote=\"Phelpedo\"]I would call it pre-olympiad, but it has a lot of (the easier) olympiad problems.[/quote]\r\n :P hmm By what percentage? roughly\r\nEasy: Medium: Hard:", "Solution_11": "anyone?\r\n :) i just want to make sure because I did mail back a book which i ordered several weeks ago and turned out to be fairly easy and \"meaningless\" when I got it. I had to pay the postage anyways:wink: \r\n\r\nThanks! i know it's hard to say because everyone is different. i just want to have a general idea.", "Solution_12": "It ranges from trivial problems to IMO level problems.", "Solution_13": "Thanks! Can I buy it somewhere in Canada?\r\nOr I may have to order online on amazon.com?", "Solution_14": "P.Solving strategies is a good book but u have to learn most of the topics from the problem its not an ideal book for learning theory.\r\nIf u want to b a good problem solver u should read the book the art nd craft of problem solving by paul zeitz\r\n>moon\r\ni love math i love bangladesh", "Solution_15": "[quote=\"andy_tok\"]Thanks! Can I buy it somewhere in Canada?\nOr I may have to order online on amazon.com?[/quote]\r\nOnline: http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191/sr=8-1/qid=1169909691/ref=sr_1_1/102-9944810-3528962?ie=UTF8&s=books\r\nIt might exist in a big bookstore like Chapters or Indigo.", "Solution_16": "Here is a web url:\r\nhttp://www.kalva.demon.co.uk\r\nhere u may get a lot of probs of different math olympiads.\r\nNd for learning better u may read some books\r\n1.103 trigonometry problems.\r\n2.The art and craft of problem solving\r\nnd u should practice probs thats d only way 2 b grt math wizard\r\n>moon", "Solution_17": "[1] Mathematical Olympiad Challenges, Second Edition by Titu Andreescu \r\n\r\n[2] Challenging Problems in Geometry by Alfred S. Posamentier \r\n\r\n[3] Mathematical Olympiad Treasures by Titu Andreescu \r\n\r\n[4] 102 Combinatorial Problems by Titu Andreescu \r\n\r\n[5] Problem-Solving Strategies (Problem Books in Mathematics) by Arthur Engel \r\n\r\n[6] 103 Trigonometry Problems: From the Training of the USA IMO Team by Titu Andreescu \r\n\r\n[7] A Path to Combinatorics for Undergraduates: Counting Strategies (Paperback) \r\nby Titu Andreescu, Zuming Feng \r\n\r\n[8] Principles and Techniques in Combinatorics (Paperback) \r\nby Chen Chuan-Chong, Koh Khee-Meng\r\n\r\n[9] Mathematical Miniatures (New Mathematical Library) (Paperback) \r\nby Svetoslav Savchev, Titu Andreescu\r\n\r\n[10] Elementary Number Theory (Hardcover) \r\nby David M. Burton", "Solution_18": "Thanks! :)", "Solution_19": "sigh... i wish Titu Andreescu had included similar commentary ahead of 102 problems in Combinatorics, just like he did for 103 problems in Trigonometry. \r\n\r\nI could really use some training for combinatorics" } { "Tag": [ "puzzles" ], "Problem": "A seven digit positive octal integer N is formed by each of the nonzero octal digits from 1 to 7 exactly once, such that N satisfies all the following conditions: \r\n\r\n(A) The sum of the digits 1 and 2 and all the digits between them is equal to the octal number 12.\r\n\r\n(B) The sum of the digits 2 and 3 and all the digits between them is equal to the octal number 23.\r\n\r\n(C) The sum of the digits 3 and 4 and all the digits between them is equal to the octal number 34.\r\n\r\nGiven that the first digit of N is greater than the last digit, what are the possible value(s) of N?", "Solution_1": "[hide=\"solution\"]1+2+3+4+5+6+7=34 (octal). Therefore 3,4 have to be the end digits. Since 4>3, 4 is the first digit and 3 is the last.\n4xxxxx3\nThe 2~3 segment excludes 4 so the maximum possible is 30. Since we are aiming for 23, we must subtract 5 additional. There are several ways to do so: (5), (4,1), (3,2), but only (5) is valid.\n\n452xxx3\n\nNow we work on the 2~1 segment. It excludes 4,5,3 so the maximum possible is 20. We are left with 1,6,7 and must subtract 6, thus 6 is excluded.\n\n4527163 is our desired number.[/hide]", "Solution_2": "[quote=\"james4l\"]The maximum possible is 30. Since we are aiming for 23, we must subtract 5 additional.[/quote]\r\n\r\nI believe you mean 7.", "Solution_3": "No, $ 30_8\\minus{}23_8\\equal{}5_8$.", "Solution_4": "I stand corrected" } { "Tag": [ "geometry", "perimeter", "calculus", "integration", "national olympiad" ], "Problem": "find the no. of all integer sided isosceles obtuse angled triangles with perimeter 2008", "Solution_1": "hi I've solved it in this way\r\nsuppose in triangle ABC ,C be the obtuse angle.\r\nnow 900 \r\n$ \\sqrt{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{a+c}}+\\sqrt{\\frac{c}{a+b}}$ find min", "Solution_1": "[quote=\"zaya_yc\"]a,b,c>0 \n$ \\sqrt{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{a+c}}+\\sqrt{\\frac{c}{a+b}}$ find min[/quote]\r\nThe minimum does not exist. :wink:", "Solution_2": "What do you mean?('The minimum does not exist?)Why?\r\nI know that $ \\sum\\sqrt{\\frac{a}{b+c}}>2$ But i don't know is this one true:$ \\sum\\sqrt{\\frac{a}{b+c}}\\geq\\frac{3}{\\sqrt{2}}$ :maybe: ?", "Solution_3": "[quote=\"Erken\"]What do you mean?('The minimum does not exist?)Why?\nI know that $ \\sum\\sqrt{\\frac{a}{b+c}}>2$ [/quote]\nYou are right and the value $ 2$ does not hold. \n[quote=\"Erken\"]But i don't know is this one true:$ \\sum\\sqrt{\\frac{a}{b+c}}\\geq\\frac{3}{\\sqrt{2}}$ :maybe: ?[/quote]\r\nIt's wrong: try $ a=b=1$ and $ c=0.07.$", "Solution_4": "But it means that the answer is $ 2+\\frac{1}{\\infty}$,isn't it?", "Solution_5": "[quote=\"Erken\"]But it means that the answer is $ 2+\\frac{1}{\\infty}$,isn't it?[/quote]\r\nI am sorry, what is this? This is not number. :wink:", "Solution_6": "[quote=\"Erken\"]What do you mean?('The minimum does not exist?)Why?\nI know that $ \\sum\\sqrt{\\frac{a}{b+c}}>2$ [/quote]\r\n$ 2$ is called infimum of that express :lol:", "Solution_7": "[quote=\"Erken\"]I know that $ \\sum\\sqrt {\\frac {x}{y\\plus{}z}}>2$. [/quote]\r\n\r\nIf $ x,y,z > 0,\\frac {49 \\minus{} 9\\sqrt {17}}{32}\\equal{} 0.371\\cdots< k<\\frac {49 \\plus{} 9\\sqrt {17}}{32}\\equal{} 2.69\\cdots$, then\r\n\r\n\\[ \\sqrt {\\frac {x}{y\\plus{}kz}}\\plus{}\\sqrt {\\frac {y}{z\\plus{}kx}}\\plus{}\\sqrt {\\frac {z}{x\\plus{}ky}}>\\frac {2}{\\sqrt [4]{ k}}\\].\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105343]I have not an elementary solution to these inequalities.[/url]", "Solution_8": "[quote=\"zaya_yc\"]a,b,c>0 \n$ \\sqrt {\\frac {a}{b \\plus{} c}} \\plus{} \\sqrt {\\frac {b}{a \\plus{} c}} \\plus{} \\sqrt {\\frac {c}{a \\plus{} b}}$ find min[/quote]\r\nIf $ a,b,c\\ge0$,we have $ Min \\equal{} 2 \\Leftrightarrow a \\equal{} b,c \\equal{} 0$and itspermutations :)\r\nIf $ a,b,c > 0$,we have a stronger result is\r\n$ \\sqrt {\\frac {a}{b \\plus{} c}} \\plus{} \\sqrt {\\frac {b}{a \\plus{} c}} \\plus{} \\sqrt {\\frac {c}{a \\plus{} b}}>\\sqrt {4 \\plus{} \\frac {(a \\minus{} b)^2} {(c \\plus{} a)(c \\plus{} b)}}$\r\nComprehend: \r\n$ \\sqrt [n]{\\frac {a}{b \\plus{} c}} \\plus{} \\sqrt [n]{\\frac {b}{a \\plus{} c}} \\plus{} \\sqrt [n]{\\frac {c}{a \\plus{} b}}> 2 \\plus{} \\sqrt [n]{\\frac {b \\plus{} c}{a \\plus{} c}}(\\sqrt [n]{\\frac {a \\plus{} c}{b \\plus{} c}} \\minus{} 1)^2$", "Solution_9": "I think that $ 2$ is the largest number such that the inequality holds for all positive numbers $ a,b,c$\r\nIndeed, when $ a\\equal{}b$, $ c$----> 0 , the left hand side of the inequality come to $ 2$ . So, we can't replace $ 2$ by any another number greater than $ 2$.\r\nOn other hand, we can't find the minimum value of given express.", "Solution_10": "[quote=\"ngtl\"]I think that $ 2$ is the largest number such that the inequality holds for all positive numbers $ a,b,c$\nIndeed, when $ a \\equal{} b$, $ c$----> 0 , the left hand side of the inequality come to $ 2$ . So, we can't replace $ 2$ by any another number greater than $ 2$.\nOn other hand, we can't find the minimum value of given express.[/quote]\r\nPlease read my discussion of this problem with arqady above." } { "Tag": [ "pigeonhole principle", "number theory", "AMC" ], "Problem": "Consider a set of 1,985 positive integers, not necessarily distinct, and none with prime factors bigger than 23. Prove that there must exist 4 integers in this set whose product is equal to the fourth power of an integer.", "Solution_1": "This should go in another forum, I think...", "Solution_2": "i think it still applies... it's still kinda under the category of american mathematics competitions.\r\n\r\nI dont know how to solve it, but here are just some details that may help....\r\n\r\nthe primes <=23 are 2,3,5,7,11,13,17,19,23... so all the 1985 numbers can be written as a product of powers of these numbers\r\n\r\ni think there was a similar problem in art and craft.... and it talked something about parity, where any of these two numbers' product had to be a perfect square, so their powers had to be both even or both odd.", "Solution_3": "PK has the correct approach. :lol:", "Solution_4": "Pigeonhole the parity, done.", "Solution_5": "[quote=\"darktreb\"]Pigeonhole the parity, done.[/quote]\r\n\r\n :?: could you go deeper into it please?", "Solution_6": "I see the approach, too, but am having trouble putting pidgeon-holing it. Then again, it is an IMO problem.", "Solution_7": "Let's see if this works...\r\n\r\nSo every number is of the form $2^{a_1}3^{a_2} \\cdots 23^{a_9}$. Considering all the $a_i \\mod{2}$ we have $2^9 = 512$ \"different\" numbers. But since $1985 > 3(512)$ there must be $4$ of the \"same\" number; aka four that have the same parity on all the exponents. Then multiplying these four together, we see that the new number has all exponents that are $0 \\mod{4}$.", "Solution_8": "Very nice", "Solution_9": "[quote=\"paladin8\"]Let's see if this works...\n\nSo every number is of the form $2^{a_1}3^{a_2} \\cdots 23^{a_9}$. Considering all the $a_i \\mod{2}$ we have $2^9 = 512$ \"different\" numbers. But since $1985 > 3(512)$ there must be $4$ of the \"same\" number; aka four that have the same parity on all the exponents. Then multiplying these four together, we see that the new number has all exponents that are $0 \\mod{4}$.[/quote]\r\n\r\nBe careful. 5, 5, 5, and 125 are all the \"same\" number under your pigeon hole, but do not have a perfect fourth-power as their product.", "Solution_10": "[quote=\"DPopov\"]Consider a set of 1,985 positive integers, not necessarily distinct, and none with prime factors bigger than 23. Prove that there must exist 4 integers in this set whose product is equal to the fourth power of an integer.[/quote]\r\n\r\nI saw this problem and decided to register :). Anyway, this solution took me several hours to produce :oops:\r\n\r\nI worked backwards (but I'll present the solution forwards for ease of reading). First, map each integer n to a 9-tuple a(n) = (a0, a1, ..., a9), where ai is the exponent of the ith prime in n's prime factorization. Considered mod 2, there are 512 possible tuples (which will be our buckets). If m and n map to the same bucket, then a(mn) = a(m)+a(n) = 2*a(m) has entries that are all 0 mod 2, and thus in {0, 2} mod 4. There are at least (1985-512)/2 = 737 such pairs. Two of them map to the same tuple (mod 4), and since their entries are all in {0, 2}, they add to the 0 tuple, which is our desired 4th power.\r\n\r\nI'm pretty sure that works.", "Solution_11": "excuse me, what is pigeon-hole(as you can see, im bad at number theory, actually i never learn about it...)", "Solution_12": "I don't think pigeon hole's quite number theory, it has a huge number of uses in all subjects.\r\nhttp://mathworld.wolfram.com/DirichletsBoxPrinciple.html" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "for a given positive integer a_0 define the squence a_n by:\r\na_{i+1}= a_{i}/2 if a_{i} even\r\na_{i+1}= 3a_{i} -1 if a_{i} odd \r\nprove that: if n is a natural number such that a_n=a_0 then 2^n>a_0", "Solution_1": "It was already posted and if I remember correctly no solution was proposed. The solution I had for that on the exam (or actually sent as it appeared during first round so the problem was to be dealt at home) took 6 pages and it was quite unappealing - I don't know any better solution.", "Solution_2": "How about thinking the binary expension system?" } { "Tag": [ "integration", "trigonometry", "calculus", "parameterization", "function", "complex analysis", "calculus computations" ], "Problem": "$\\int_{0}^\\infty \\frac{\\sin x}{x}=\\frac{\\pi}{2}$\r\n\r\nProve or disprove.", "Solution_1": "This can be done by contour integration ([url=http://en.wikipedia.org/wiki/Fresnel_integral]here[/url]), but there is another approach, by introducing a parameter into the integral... I wish I could remember it. :(", "Solution_2": "[quote=\"mathnerd314\"]$\\int_{0}^\\infty \\frac{\\sin x}{x}=\\frac{\\pi}{2}$\n\nProve or disprove.[/quote]\r\n\r\nThat's correct . A classical way is the the Fourier transformer of the gate function .\r\n :cool:", "Solution_3": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=73681]Link 1[/url]\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41288]Link 2[/url]\r\n\r\nI've got my own favorite method in link 2.\r\n\r\nThis has probably appeared in other places on Mathlinks/AoPS as well; I was just looking for my own posts.", "Solution_4": "Yesterday I was having a look in the Complex Analysis - Computations section. I found something very interesting [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=44440]there[/url].\r\n\r\nIt seems just pure luck that we stumble upon this integral." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "let $f: [0,\\infty)\\longrightarrow R$ be increasing function ,such that $f(x)-3x\\ ,\\ f(x)-x^3$ are increasing too.\r\nprove that :$f(x)-x^2-x$ is an increasing function.\r\n :oops: :blush:", "Solution_1": "For any $x,y\\in [0,+\\infty),x>y$,we need to prove $f(x)-x^2-x>f(y)-y^2-y$\r\n$f(x)-f(y)>(x-y)(x+y+1)$\r\nSince $f(x),f(x)-x^3,f(x)-3x$ incerasing.\r\n$f(x)-f(y)>0,f(x)-f(y)>x^3-y^3=(x-y)(x^2+xy+y^2),f(x)-f(y)>3(x-y)$\r\nNow we only need to prove $\\frac{1}{3}(x^2+xy+y^2)+2\\geq x+y+1$\r\nThat's because $\\frac{1}{6}[(x-1)^2+(y-1)^2+(x+y-2)^2]\\geq0$\r\nSo $f(x)-f(y)=\\frac{1}{3}(f(x)-f(y))+\\frac{2}{3}(f(x)-f(y))>\\frac{1}{3}(x-y)(x^2+xy+y^2)+2(x-y)$\r\n$=(x-y)[\\frac{1}{3}(x^2+xy+y^2)+2]\\geq(x-y)(x+y+1)$\r\n$f(x)-x^2-x$ increases", "Solution_2": "Another proof. Use the identity\r\n\\[ 3 [f(x) - x^2 - x] = 2[f(x) - 3x] + [f(x) - x^3] + (x-1)^3 + 1 \\, . \\]\r\nEach of the four terms on the right is an increasing function. Hence the left side is increasing.\r\n\r\nWe didn't need the hypothesis that $f$ itself be increasing, because it follows from $f(x) - 3x$ being increasing.", "Solution_3": "Yeah,that's a easy solution." } { "Tag": [ "geometry", "ratio", "3D geometry", "sphere" ], "Problem": "Hi, we were assigned a problem to do a poster on in Chemistry but I can't seem to figure this out.\r\n\r\nThe problem is \r\n[code]\nA mole of marbles would cover the continental united states by a depth of how much?\n\nA regular marble is .5 inch across\nThe continental united states is 2,959,064.44 square miles\nThat is 1.562386024 x 10^10 feet\nThat is 1.874863229184 x 10^11 inches.[/code]\r\n\r\nMy initial thought was to get area covered by a marble (2 dimensional area) and multiple it by 6.02 x 10^23 and divide it by the area of the USA. Is this correct or should I approach another method? :ninja:", "Solution_1": "I would find the volume of 1 mol of marbles then use volume equals area of the base multiplied by height where the area of the base is the area of the continental US.", "Solution_2": "I would look up the ratio of empty to total space in a compact arrangement of spheres, just in case we'll be covered to a high depth. \r\nHere we go. Check my math. I've been known to make mistakes.\r\nLet's use a packing density of 0.74.\r\nThe volume of a marble is:\r\n(pi/6)diameter^3=(pi/6)0.125 cu. in.=0.6545 cu. in.\r\nConsidering the density (gaps between marbles), each marble will account for\r\n0.06545/0.74 cu. in. = 0.08845 cu. in.\r\n6.02 * 10^23 (1 mole) marbles (and the gaps in between) would occupy\r\n(6.02 * 10^23)(0.08845 cu. in.) = 5.32 * 10^22 cu. in.\r\n\r\nFrom the other end, if the continental united states is 2,959,064.44 square miles\r\nThat is (2.959 * 10^6 sq mi)(5280 ft/mi)^2 = 8.25*10^13 sq ft\r\n(1 mile = 5280 ft, but 1 sq mile = 5280^2 sq ft)\r\nIt is also (8.25*10^13)*144 sq in = 1.79*10^16 sq. in.\r\n\r\n(5.32 * 10^22 cu. in.)/(1.79*10^16 sq. in.) = 4.48*10^6 in is the height of the marble covering. Dividing by 12 in/ft we find it's 3.74*10^5 ft. Further dividing by 5280 ft/mi, we find it's 70.7 miles." } { "Tag": [ "inequalities", "number theory theorems", "number theory" ], "Problem": "I'm looking for a proof of Liouville Inequality (in algebaric number theory).\r\nIs there some proof online ? At worst, can anyone post it ?", "Solution_1": "Do you mean this one: For all algebraic $\\alpha$ of degree $\\geq2$ there exists a positive constant $c$ such that for all rational $p/q$ we have \\[ \\left|\\alpha-\\frac{p}{q}\\right|\\geq \\frac{c}{q^n}? \\] because I have a proof of that one.", "Solution_2": "At the time I meant a different Liouville Inequality ... (I've seen the proof of yours)\r\nUnfortunately I forgot which :( (But I guess there is not too much of them :))\r\nAs fast as I retrieve it, I will make my question more precise." } { "Tag": [ "puzzles" ], "Problem": "Hello :) \r\n\r\nHow are you?\r\n\r\nI'm new to this board.\r\n\r\nJason", "Solution_1": ":spam:\r\n\r\nDon't spam, please. But anyway, hi. :P", "Solution_2": ".......As it is more than three months after the original post and you feel the need to respond, you yourself are spamming. Did you not see that no one answered him?", "Solution_3": ":( Sorry!" } { "Tag": [ "geometry", "power of a point", "radical axis", "geometry proposed" ], "Problem": "[quote=\"Virgil Nicula\"][color=darkred]For $k\\in\\overline {1,2}$ the circle $C_k=C(O_k,r_k)$ is interior tangent to the circle $C=C(O,r)$ in the points $T_k$ and $|r_1-r_2|\\le O_1O_2$. Denote: $A_1\\in C_1$, $A_2\\in C_2$ for which the line $A_1A_2$ is common exterior tangent of the circles $C_1$ and $C_2$; the intersection $R\\in T_1A_1\\cap T_2A_2$. Prove that:\n$1^{\\circ}.\\blacktriangleright\\ R\\in C$ and $OR\\perp A_1A_2\\ ;$\n$2^{\\circ}.\\blacktriangleright$ The point $R$ belongs to the radical axis of the circles $C_1$ and $C_2\\ ;$\n$3^{\\circ}.\\blacktriangleright\\ O_1\\in C_2\\Longrightarrow$ the line $A_2B_2$ is tangent to the circle $C_1$, where for any $k\\in \\overline {1,2}$ the point $B_k\\in C_k\\ ,$ $B_k\\ne A_k$ and the line $B_1B_2$ is the second common exterior tangent of the circles $C_1$ and $C_2$.[/color][/quote]", "Solution_1": "Consider $r_1>r_2$:\r\n1\u00ba)The common tangent at $T_1$ cut the line $A_1A_2$ at $P$, now the triangle $PA_1T_1$ is isosceles then $ MB = NC\r\n\r\nAnd, PMN and PBC is similar. From this, BM=MN also can be made. (<< I cannot do this now..)", "Solution_8": "My idea: Construct the angle bisectors of $ \\angle BEF$ and $ \\angle EFC$. These intersect at $ X$. Then $ AX$ bisects $ \\angle BAC$ and you get three congruent triangles and $ \\angle BXC\\equal{}\\frac{3}{2}(180\\minus{}A)$. So you can pretty easily construct $ X$ from these two properties and then finding $ E$ and $ F$ is easy." } { "Tag": [ "geometry", "3D geometry", "sphere", "pyramid", "algebra", "polynomial", "function" ], "Problem": "The number of spheres needed to create a triangular pyramid is a polunomial function given by $ N(x)\\equal{}\\frac{1}{6}x^3\\plus{}\\frac{1}{2}x^2\\plus{}\\frac{1}{3}x$, where $ x$ is the number of layers and $ N(x)$ is the number of spheres. How many spheres are in the eighth layer?", "Solution_1": "Eight layer = # of spheres needed to make $ 8$ layer pyramid - #of shperes needed to make $ 7$ layer pyramid\r\nPlugging into equation gives $ \\frac168^3 + \\frac128^2 + \\frac138 - \\left(\\frac167^3 + \\frac127^2 + \\frac137\\right) =$(calc)$ = 36$\r\nAlso, you can use logic and see that each layer is the traingle number.\r\n\r\n[color=blue]Fixed $ \\text{\\LaTeX}$. :) - Izzy[/color]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ x\\equal{}a\\plus{}\\frac1{b},y\\equal{}b\\plus{}\\frac1{c},z\\equal{}c\\plus{}\\frac1{a}$.Show that:\r\n$ xy\\plus{}yz\\plus{}zx\\ge 2(x\\plus{}y\\plus{}z)$", "Solution_1": "[quote=\"duongptnkvn\"]Let $ x \\equal{} a \\plus{} \\frac1{b},y \\equal{} b \\plus{} \\frac1{c},z \\equal{} c \\plus{} \\frac1{a}$.Show that:\n$ xy \\plus{} yz \\plus{} zx\\ge 2(x \\plus{} y \\plus{} z)$[/quote]\r\nIt's obviously wrong. Try $ a\\equal{}b\\equal{}5$ and $ c\\equal{}\\minus{}0.001.$ :wink:", "Solution_2": "I think, that $ a,b,c>0$ :P" } { "Tag": [ "articles", "AMC", "USA(J)MO", "USAMO" ], "Problem": "When are the definite articles [b]el[/b] and [b]la[/b] not used in spanish? \r\n\r\nHow do you say \"stop sign\" in spanish? My book says that it's \"senal de stop\", but I think that's a typo.", "Solution_1": "In most cases it is there.\r\n\r\nSome cases that it isn't:\r\nWhen stating someone's job.\r\nMi padre es bombero.\r\n\r\nOn SOME countries, but it is disputable.\r\n\r\nYo soy de Mexico.\r\n\r\n\r\nI don't realy know Spanish so don't kill me if I am wrong :nhl:", "Solution_2": "Why don't they use the article in this sentence? \r\n\r\nUsamos ingredientes puros solamente. \r\n\r\n(We only use pure ingredients. )\r\n\r\nIn addition, many books say that the definite article is used before a noun in a [i]general sense[/i]... what exactly does that mean?", "Solution_3": "I think its because you don't say \"we use the pure ingredients\"\r\n\r\nbut you have to ask a spanish dude to be sure", "Solution_4": "Sometimes, in spanish we don't use articles, because the article is obvious when you read the verbs.\r\n\r\ne.g.: Cocinamos comida (we cook food). The verb \"cocinamos\" is always for nosotros (us). \r\n\r\nIn spanish, verbs can be infinitive or \"conjugados\". In english, verbs are only in infinitive.\r\n\r\nLet me explain you.\r\n\r\nLet's put the verb love (amar). In english you say: I love. You love. He love. We love. The person or article change, but the verb is always the same (the spelling does not change). In spanish, we say: Yo amo, tu amas, el ama, nosotros amamos. The verb (spelling) change you see?? (amo, amas, ama, amamos). So, let's say again: We love our grandma. Now, let's say \"amamos a nuestra abuela\". We don't have to put \"nosotros amamos a nuestra abuela\" becuase the ver \"amamos\" is only for the person \"us\". In conclusion, in spanish, a verb can change it spelling to change the person or article we're talking about. In english, the verb is the same, but you have to include the article.", "Solution_5": "No.. I mean the definite article el or la as in \r\n\r\nel libro ([i]the[/i] book) or \r\n\r\nla almohada (the pillow)\r\n\r\nAnd I think the stuff about the verb conjugations is very cool :D", "Solution_6": "[quote=\"236factorial\"]No.. I mean the definite article el or la as in \n\nel libro ([i]the[/i] book) or \n\nla almohada (the pillow)\n\nAnd I think the stuff about the verb conjugations is very cool :D[/quote]\r\n\r\nAre you talking why do we say \"el libro\" en vez de \"la libro\", for example??", "Solution_7": "I'm talking about something like: \r\n\r\n\u00bfPor qu\u00e9 no usa usted \"el\" en esta oraci\u00f3n: Usamos ingredientes puros solamente. ?\r\n\u00bfPor qu\u00e9 no es : Usamos [i]los[/i] ingredientes puros solamente. ? \r\n\r\n[size=75]I hope that made sense...[/size]\r\n(Does \"usa\" mean \"you use\"?)", "Solution_8": "[quote=\"236factorial\"]I'm talking about something like: \n\n\u00bfPor qu\u00e9 no usa usted \"el\" en esta oraci\u00f3n: Usamos ingredientes puros solamente. ?\n\u00bfPor qu\u00e9 no es : Usamos [i]los[/i] ingredientes puros solamente. ? \n\n[size=75]I hope that made sense...[/size]\n(Does \"usa\" mean \"you use\"?)[/quote]\r\n\r\n\"usa\" means \"he/she uses\" .\r\n\r\nYo can use \"el\" in that sentence, but, when you're talking, the meaning of the sentence is obvious, so it's not necessary to put it.", "Solution_9": "[quote=\"Jos\u00e9\"]\n\"usa\" means \"he/she uses\" .\n\nYo can use \"el\" in that sentence, but, when you're talking, the meaning of the sentence is obvious, so it's not necessary to put it.[/quote]\r\n\r\nHow do you say \"you use\"?\r\n\r\nCan you give an example of a sentence where the meaning is not obvious, and \"el\", \"la\", \"los\", or \"las\" is necessary?", "Solution_10": "[quote=\"Jos\u00e9\"]\nYo can use \"el\" in that sentence[/quote]\r\n\r\nSorry, it was: you can use \"los\" in that sentence", "Solution_11": "[quote=\"236factorial\"][quote=\"Jos\u00e9\"]\n\"usa\" means \"he/she uses\" .\n\nYo can use \"el\" in that sentence, but, when you're talking, the meaning of the sentence is obvious, so it's not necessary to put it.[/quote]\n\nHow do you say \"you use\"?\n\nCan you give an example of a sentence where the meaning is not obvious, and \"el\" is necessary?[/quote]\r\n\r\n..you use=vos us\u00e1s...\r\n\r\ndo you use pen? (vos) us\u00e1s lapicera???.\r\n\r\nAbout the other thing: \u00bfEL sombrero est\u00e1 en su lugar correcto? (is the hat in it's correct place?) [you can't say:\u00bfsombrero est\u00e1 en su lugar correcto? ... Sorry 236factorial if I'm not to good to explain...", "Solution_12": "you have to put articles when they're before a noun, but it isn't necessary to put them when they're before a verb.", "Solution_13": "Do you mean after a verb? Despu\u00e9s de un verbo? Or otherwise, the sentence that I mentioned wouldn't follow your rule (ingredientes is a noun and I didn't [i]have[/i] to put an article before it). \r\n\r\nSorry if I'm making this too hard :D", "Solution_14": "[quote=\"236factorial\"]Do you mean after a verb? Despu\u00e9s de un verbo? Or otherwise, the sentence that I mentioned wouldn't follow your rule (ingredientes is a noun and I didn't [i]have[/i] to put an article before it). \n\nSorry if I'm making this too hard :D[/quote]\r\n\r\nBecuase ingredientes is plural I think... if not, you would have put ingrediente (singular)\r\n\r\nUsamos (los) ingredientes puros solamente... Putting or not LOS doesn't change the meaning, because by only looking at ingredientes you see that it's plural... Sorry if I made a very huge mess in your head.", "Solution_15": "[quote=\"Jos\u00e9\"]Becuase ingredientes is plural I think... if not, you would have put ingrediente (singular)\n\nUsamos (los) ingredientes puros solamente... Putting or not LOS doesn't change the meaning, because by only looking at ingredientes you see that it's plural... Sorry if I made a very huge mess in your head.[/quote]\r\n\r\nLos esp\u00e1rragos son saludables.\r\n\r\nThe \"los\" is for some reason necessary...", "Solution_16": "[quote=\"236factorial\"]\nLos esp\u00e1rragos son saludables.\n\nThe \"los\" is for some reason necessary...[/quote]\r\n\r\nDo you know something about \"sujeto y predicado\"?? (subject and object). The \"los\" is necessary because you need it (obligation) for studying and dividing that sentence in sujeto y predicado. Moreover, when el, la, los, las are starting a sentence, they're almost ever necessary.\r\n\r\nLos esp\u00e1rragos son saludables.\r\nEsp\u00e1rragos son saludables. (The 1\u00ba sound best).\r\n\r\nMe gusta comer esp\u00e1rragos.\r\nMe gusta comer los esp\u00e1rragos (The 2\u00ba sound best-don't you think that los here is unnecessary??).\r\n\r\nThe problem is that one get used to it when you speak every day, and it's dificult to study without talking and talking fluently every day.", "Solution_17": "Thanks for replying. By the way, what does 1\u00ba mean?", "Solution_18": "1\u00b0\r\nCan mean to things: 1 degree (un grado) or First (Primero)\r\nBy the way, You don't say \"Usamos los ingredientes puros solamente\" for the same reason why you don't say \"We use the pure ingredients only\"\r\nThe definite article is used to determinate the noun. For example:\r\n\"Voy a comprar libros\" (\"I'm going to buy books\")\r\n\"Voy a comprar los libros\" (I'm going to bue the books\")\r\nWhen you say the first one, the books that you are going to buy are not so clear, I mean, If you are in the middle of a conversation and you say that, someone will not be sure about which books are your going to buy. But if you use the second one (and in the conversation you have already mentioned the books) it is clear that the books that you are going to buy are those wich you have already said.\r\n\r\nTo make more clear the diference between \"Usamos ingredientes puros solamente\" and \"Usamos los ingredientes puros solamente\" you will have to put them into a contex, for example \"Al usar ingredientes para nuestro pastel, usamos los ingredientes puros solamente\" (At using ingredients for our cake, we use the pure ingredents only) And: \"En nuestro pastel, usamos ingredientes puros solamente\" (In our cake, we use pure ingredients only)", "Solution_19": "[quote=\"236factorial\"]Thanks for replying. By the way, what does 1\u00ba mean?[/quote]\r\n\r\n1\u00ba=First." } { "Tag": [ "calculus", "integration", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Proove that:\r\n\\[ \\lim_{n\\to\\infty }\\,\\int_{0}^{\\infty }\\left(\\sum_{i\\equal{}1}^{\\infty }\\frac{i\\plus{}x}{\\left(i^{n}\\plus{}x^{n}\\right)^{n}}\\right)\\, dx\\equal{}\\frac{3}{2}\\]\r\nOr the generalization:\r\n\\[ \\lim_{n\\to\\infty }\\,\\int_{0}^{\\infty }\\left(\\sum_{i\\equal{}1}^{\\infty }\\frac{\\alpha i^{a}\\plus{}\\beta x^{b}}{\\left(i^{n}\\plus{}x^{n}\\right)^{n\\plus{}\\kappa }}\\right)\\, dx\\equal{}\\alpha\\plus{}\\frac{\\beta }{b\\plus{}1}\\]\r\n$ b > 0$ and the others constants dont have restrition", "Solution_1": "Here are some calculations:\r\n\r\n$ I \\equal{}\\sum_{i \\equal{} 1}^{\\infty}\\int_{0}^{\\infty}\\frac{i\\plus{}x}{(i^{n}\\plus{}x^{n})^{n}}dx$\r\n\r\nUsing the substitutions $ x^{n}\\equal{} y$ and $ y \\equal{} i^{n}t$ we get that\r\n\r\n$ I \\equal{}\\sum_{i \\equal{} 1}^{\\infty}\\frac{1}{n}\\frac{1}{i^{n^{2}\\minus{}2}}\\int_{0}^{\\infty}\\left(\\frac{t^{1/n\\minus{}1}}{(1\\plus{}t)^{n}}\\plus{}\\frac{t^{2/n\\minus{}1}}{(1\\plus{}t)^{n}}\\right)dt$\r\n\r\n$ \\equal{}\\frac{1}{n}\\left(\\beta(1/n,n\\minus{}1/n)\\plus{}\\beta(2/n,n\\minus{}2/n)\\right)\\zeta(n^{2}\\minus{}2)$\r\n\r\n$ \\equal{}\\left(\\frac{\\Gamma(1/n)}{n}\\frac{\\Gamma(n\\minus{}1/n)}{\\Gamma(n)}\\plus{}\\frac{\\Gamma(2/n)}{n}\\frac{\\Gamma(n\\minus{}2/n)}{\\Gamma(n)}\\right)\\zeta(n^{2}\\minus{}2)$\r\n\r\nNow use Stirling's formula combined with $ n\\minus{}\\Gamma(1/n)\\rightarrow\\gamma$, $ \\zeta(\\infty) \\equal{} 1$ and the Legendre's formula\r\n\r\n$ \\Gamma(1/n)\\Gamma(1/n\\plus{}1/2) \\equal{}\\frac{\\sqrt{\\pi}}{2^{2/n\\minus{}1}}\\Gamma(2/n)$,\r\n\r\nand you get that the first term converges to 1 and the second one to 1/2.\r\n\r\nAs for the generalization, it follows by similar computations." } { "Tag": [ "\\/closed" ], "Problem": "See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1004242#1004242", "Solution_1": "The forum doesn't allow SVG images.\r\n:)", "Solution_2": "The wiki automatically converts to PNG for display; get the thumbnail from MediaWiki; it'll be PNG." } { "Tag": [ "college", "search", "superior algebra", "superior algebra theorems" ], "Problem": "I'm find a free version of this book, but i can't fint out. Please help me, Thanks a lot", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=707706759&t=191144" } { "Tag": [ "analytic geometry" ], "Problem": "how many shortest paths are there in space from (0,0,0) to (6,6,6)? A path consists in moves of increasex,y or z coordinate by exactly 1.", "Solution_1": "You want to find out how many paths of length exactly 18 there are given that each step can only be of length 1 in the positive $ x$, $ y$, or $ z$ direction. This is a counting problem. For any configuration of 18 steps, 6 of them will be $ x$'s, 6 of them will be $ y$'s, and 6 of them will be $ z$'s. Thus the problem is equivalent to finding the number of ways to arrange:\r\n\r\nXXXXXXYYYYYYZZZZZZ,\r\n\r\nwhich is just $ \\frac{18!}{6!6!6!}\\equal{}\\boxed{17153136}$.", "Solution_2": "To add to that, there's pretty much a general formula for counting the number of paths when moving a units among x-axis, b units among y-axis, and c units among z-axis, which is just $ \\frac{(a\\plus{}b\\plus{}c)!}{a!b!c!}$, which also works together with the similar formula for the paths on a grid formula for 2-dimensional walks (with x horizontal, y vertical) $ \\frac{(x\\plus{}y)!}{x!y!}$.\r\n\r\nSimilar formulas also work for higher dimensions." } { "Tag": [ "calculus", "integration", "logarithms", "trigonometry", "calculus computations" ], "Problem": "Answer the following questions.\r\n\r\n(1) Evaluate $\\int_e^{e^e} \\frac{\\ln (\\ln x)}{x\\ln x} dx.$\r\n\r\n(2) Let $\\alpha,\\beta$ be real numbers.Find the values of $\\alpha,\\beta$ for which the following equality holds for any real numbers $p,q.$\r\n\r\n\\[ \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}}(p\\cos x+q\\sin x)(x^2+\\alpha x+\\beta) dx=0. \\]", "Solution_1": "For 1)\r\n$\\int_{e}^{e^e} \\frac{\\ln (\\ln (x) )}{x \\ln (x)} dx = \\int_{e}^{e^e} (\\ln (\\ln (x) ) )' (\\ln (\\ln (x) ) ) dx = [\\frac{1}{2} (\\ln (\\ln (x) ) )]_{e}^{e^e}=\\frac{1}{2}$\r\n\r\nFor 2)\r\nlet's make an integration by parts : \r\n$I=\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} (p \\sin x -q \\cos x)'(x^2+\\alpha x + \\beta )dx$\r\n$=p(\\frac{\\pi^2}{2} +2\\beta ) + \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} (p \\cos x +q \\sin x)'(2x+\\alpha )dx$\r\n$=p(\\frac{\\pi^2}{2} +2\\beta ) + 2q \\alpha -2 \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} (p \\cos x -q \\sinh x)dx$\r\n$=p(\\frac{\\pi^2}{2} -4 +2 \\beta ) + q (2 \\alpha) = 0$\r\nsince this is verified for all $p,q \\in \\mathbb_R$ then :\r\n$\\beta = 2- \\frac{\\pi^2}{4}$ and $\\alpha = 0$" } { "Tag": [ "function", "limit", "algebra unsolved", "algebra" ], "Problem": "is it true that if f is monotonous and surjective, then f is continuous?", "Solution_1": "Yes, it is true.", "Solution_2": "Can you post the proof? please. thank you", "Solution_3": "It's an immediate proof. Suppose $ f$ is growing. Fix a point $ a \\in \\mathbb{R}$. Since $ f$ is growing it has a limit $ A \\leq f(a)$ as $ x \\to a^{ \\minus{} }$*. But we cannot have $ A < f(a)$ because then the range of $ f$ would not contain any number between $ A$ and $ f(a)$. This means $ \\lim_{x \\to a^{ \\minus{} }} f(x) \\equal{} f(a)$. Analogously, $ \\lim_{x \\to a^{ \\plus{} }} f(x) \\equal{} f(a)$\r\n\r\n*I looked at your previous posts, and maybe you're not so familiar with analysis. So let me explain this part. It is an important property of the real numbers that every non-empty bounded from above set $ M \\subset \\mathbb{R}$ has a least upper bound $ B \\equal{} \\sup M$. This is equivalent to the so called completeness of $ \\mathbb{R}$. To prove it you need to go down to the construction of the real numbers. Anyway, in this case we let $ A \\equal{} \\sup \\{f(x) : x < a \\}$. It's easy to verify that $ \\lim_{x \\to a^{\\minus{}}} f(x) \\equal{} A$", "Solution_4": "oh, it was kind of basic property of f. thank you for your help. :)" } { "Tag": [], "Problem": "The set {1, 2, 3, ..., 2005} is partitioned into two sets: the sets of numbers having an odd digit sum and the set of numbers having an even digit sum. Let A be the sum of the numbers in the first set and let B be the sum of the numbers in the second set. Determine A - B.", "Solution_1": "Out of 2 consecutive numbers, one must have a even sum, and one must have an odd sum.\r\n\r\nie, if n has an odd sum, n+1 has an even sum\r\nif n has an even sum, n+1 has an odd sum\r\n\r\nSo, its' (1-2)+(3-4)+...+(2003-2004)+2005=-1002+2005=1003", "Solution_2": "Hm... Don't think that works :D", "Solution_3": "Oh hm, I found a little mistake that I didn't consider, stuff like, 1999 and 2000, hehe", "Solution_4": "[hide]In the set, there are 9 numbers with only 1 digit. From those numbers, there are $5$ numbers with an odd digit sum.\n\nThere are 90 numbers of 2 digits in the set.\nTo obtain an odd sum, the number must exist out of an odd number and an even number. This is possible in two ways: OE or EO (E=even, O=odd)\nThere are 5 odd digits, and 5 even digits, so realising that the even digit in the second possibility can not be 0, the number of odd sums is $5.5+4.5=45$\n\nThere are 900 numbers of 3 digits in the set. The odd sums can be obtained like this:\nEEO, EOE, OEE, OOO. Thus the ammount of odd sums is: $4.5^{2}.2+5^{3}.2=450$\n\nThen we get the numbers with four digits, and starting with 1. There are thousand such numbers, and to obtain an odd sum, we must have 1OOE,1OEO,1EOO or 1EEE.\nThis gives us $4.5^{3}=500$\n\nThere are only 6 numbers with 4 digits and greater or equal to 2000 in the set. Of them, 3 have an odd sum.\n\nSo in the entire set, there are $1003$ numbers with an odd digit sum. That means that there are $1002$ numbers with an odd sum.\n\nTherefor, A-B$=1003-1002=1$\n\nShorter an better way:\nWe have the following possibilities to obtain an odd sum in the set: 1OOE,1OEO,1EOO,0OOO,EEO,EOE,OEE,EEE.\nThen there are the 3 odd sums that start with a 2\nSo that means $5^{3}.3+5.5^{3}+3=1003$\n\nCorrect?[/hide]", "Solution_5": "On a first glimp: for 1..1999\r\n\r\nWhen is SoD(n) = SoD(n+1) [mod 2] ?\r\nevery jump 9-->10, except when n ended in an even number of 9's.\r\n\r\nSo that means 200 - 18 = 182. Since this is 91 of each parity, same for the 18 (=9 of each parity), here we have a 0.\r\n\r\nAbove, we have 2005-2004,...,2001-2000, which makes me believe the answer is 3? Why do you all say 1003?", "Solution_6": "I don't, I know my answer's wrong, hehe", "Solution_7": "I think $0$ because consider $\\overline{xa}$ which $x \\in \\{0,1,...,200\\}$ and ${a \\in\\{0,1,...,9}$ .if x is even so $A-B$ which $A,b \\in \\overline{xa}$ is $5$ and it is $-5$ if it is odd and we know sum of the digit of $x$ is even in half cases. so $A-B=0$.\r\nI think I explain it badly :blush:", "Solution_8": "I woke up at 4 am this night, realising how easily and nice this problem could be solved:\r\n\r\nFor 1..1999, we have that SoD(k) and SoD(2000-k) always have the same parity, and since all parities appear equally much, we have 0 for 1..1999\r\n\r\nAnd from there on the other 6 terms give a trivial 3 indeed :)", "Solution_9": "[quote=\"a_vakilian\"]I think $0$ because consider $\\overline{xa}$ which $x \\in \\{0,1,...,200\\}$ and ${a \\in\\{0,1,...,9}$ .if x is even so $A-B$ which $A,b \\in \\overline{xa}$ is $5$ and it is $-5$ if it is odd and we know sum of the digit of $x$ is even in half cases. so $A-B=0$.\nI think I explain it badly :blush:[/quote]well, in your example you go from 0 to 200, so you have more even. ;) so you have part of a +5. That part is {0,...,5} and not {6,...,9}, which is 3/5 of 5 = 3. You again confirm my claim. I am very sure now it is 3.", "Solution_10": "Yes Peter I want to say there is little wrong withme because I consider $\\{1,2,..,2009\\}$.So It is $3$.", "Solution_11": "But for 2009 it is 5! :P" } { "Tag": [ "inequalities" ], "Problem": "Let $a,b,c$ are positive number such that:\r\n$a+b+c=\\frac1a+\\frac1b+\\frac1c$\r\nwe have $a^{2}+b^{2}+c^{2}\\geq \\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}$\r\n :)", "Solution_1": "After homogenizing, it's enough to prove that \\[\\sum \\frac{a^{2}}{bc}\\geq \\sum \\frac{ac}{b^{2}}\\] But this is just Muirhead; $[2,-1,-1]\\succ[1,1,-2]$.", "Solution_2": "I have proved the general inequality but certainly, I doubt its valid. May you justify it?\r\nLet $a,b,c$ are positive number such that $a+b+c=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$\r\nProve that\r\n$a^{r}+b^{r}+c^{r}\\geq \\frac{1}{a^{r}}+\\frac{1}{b^{r}}+\\frac{1}{c^{r}}$ for all of positive real $r > 1$.", "Solution_3": "It is equivalent to \r\n$\\frac{\\sqrt[r]{a^{r}+b^{r}+c^{r}}}{a+b+c}\\ge \\frac{\\sqrt[r]{(ab)^{r}+(bc)^{r}+(ac)^{r}}}{ab+bc+ac}$\r\n\r\nLet $(a, b, c) = v$ and $(b, c, a) = w$. It is \r\n$\\frac{\\|v\\|_{r}}{\\|v\\|_{1}}\\ge \\frac{\\langle v, w\\rangle_{r}}{\\langle v, w \\rangle_{1}}$\r\n\r\nIn other words, Cauchy-Schwartz introduces more error when the powers are higher. I conjecture that if $1 \\le q \\le r$, we can replace 1 by $q$ in the denominators and it's still true. I suspect this should be obvious to me, but some dude hit me really hard in the head today at frisbee, and I have a terrible headache now and can't think." } { "Tag": [], "Problem": "In triangle ABC, AX=XY=YB=BC, and the measure of angle ABC=120 degrees. What is the measure of angle BAC?\r\n\r\nHide your answers!\r\n\r\nBilly", "Solution_1": "[hide]We can make angle YBC = x, and angle XBY = 120 - x. Since XY = YB, then it is an isoscoles triangle which means that the base angles are equal. BXY must then also be 120 - x. Angle AXY = 180 - (120 - x) = 60 + x. Angle A = [180 - (60 + x)]/2 = (120 - x)/2. Angle C = (180 - x)/2. Angle A + B + C = 180 = (120 - x)/2 + (180 - x)/2 + 120. Solving this makes x = 90. Plugging it into the equation for angle A, we get 15 degrees.[/hide]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "A permutation $ \\{x_1, \\ldots, x_{2n}\\}$ of the set $ \\{1,2, \\ldots, 2n\\}$ where $ \\mbox{n}$ is a positive integer, is said to have property $ T$ if $ |x_i \\minus{} x_{i \\plus{} 1}| \\equal{} n$ for at least one $ i \\in \\{1,2, \\ldots, 2n \\minus{} 1\\}$. Show that, for each $ \\mbox{n}$, there are more permutations with property T than without.", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?p=372274" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "If n=1(mod 2),n>1 prove it is impossible that m^(n-1)=-1 (mod n)", "Solution_1": "Just note that if the order of $ m$ modulo $ n$ is $ d$ then $ d$ divides $ 2(n\\minus{}1)$ and $ d$ not divides $ n\\minus{}1$ (because $ n\\minus{}1$ is even. Then $ 2(n\\minus{}1)$ divides $ \\phi{n}$, which is impossible because $ 2(n\\minus{}1)>n>\\phi{n}$ fo $ n>1$", "Solution_2": "[quote=\"gilcu3\"]Just note that if the order of $ m$ modulo $ n$ is $ d$ then $ d$ divides $ 2(n \\minus{} 1)$ and $ d$ not divides $ n \\minus{} 1$ (because $ n \\minus{} 1$ is even. Then $ 2(n \\minus{} 1)$ divides $ \\phi{n}$, which is impossible because $ 2(n \\minus{} 1) > n > \\phi{n}$ fo $ n > 1$[/quote]\r\nbut why $ 2(n \\minus{} 1)$ divides $ \\phi{n}$ ? i don't understand :blush: :blush:", "Solution_3": "It is the good solution.\r\nLet $ 2^c$ the maximum power of $ 2$ that divides $ n\\minus{}1$. Note that like $ d$ divides $ 2(n \\minus{} 1)$ and $ d$ not divides $ n \\minus{} 1$ then $ d\\equal{}2^{c\\plus{}1}q$ for some $ q$ odd.\r\nThen for each prime $ p$ (not $ 2$ because $ n$ is odd) that divides $ n$ the order $ s$ of $ m$ modulo $ p$ is of the form $ s\\equal{}2^{c\\plus{}1}r$ for some $ r$ odd. And then like $ s|p\\minus{}1$. $ p\\equal{}2^{c\\plus{}1}v$ for some $ v$ odd.\r\nThen $ n$ is of the form $ 2^{c\\plus{}1}j\\plus{}1$ and this contradict the choice of $ c$", "Solution_4": "[quote=\"gilcu3\"]It is the good solution.\nThen for each prime $ p$ (not $ 2$ because $ n$ is odd) that divides $ n$ the order $ s$ of $ m$ modulo $ p$ is of the form $ s \\equal{} 2^{c \\plus{} 1}r$ for some $ r$ odd.\n[/quote]\r\n\r\nI didn't understand that either. Why does it hold for every prime divisor of n?", "Solution_5": "For each prime $ p$ that divides $ n$ \r\n$ m^{n\\minus{}1} \\equal{} \\minus{}1 (mod p)$\r\nThen if $ s$ is the order, $ s$ divides $ 2(n\\minus{}1)$ and $ s$ don't divides $ n\\minus{}1$." } { "Tag": [ "inequalities", "logarithms" ], "Problem": "Prove, for b > a > 0: (a + b)/2 > (b^b/a^a)^(1/(b-a))/e.", "Solution_1": "Here's a go at reducing it to something solvable:\r\nmultiplying through by e and taking the natural logarithm of both sides, and letting b = (k + 1)a, k > 0, and simplifying, we get \r\n1 + ln((k+2)a/2) >? ln(a) + (k + 1)ln(k)/k\r\n1 + ln(a) + ln((k + 2)/2) >? ln(a) + ln(k) + ln(k)/k\r\n1 + ln((k + 2)/2) >? ln(k) + ln(k)/k\r\n\r\nWe might perhaps reduce this to two cases: where 1 - ln(2) > ln(k)/k (which is for large k, and since we know ln(k + 2) > ln(k), we will be done) and alternatively the alternate case (in which case k is small and so ln(k + 2) - ln(k) is sufficiently large to overcome the differences in the other terms.) I'm not sure about this, particularly whether or not it will work.", "Solution_2": "Here is the inequality in a better form:\r\n\r\n:sqrt:((1/3)(a^2+ab+b^2)) > (a+b)/2 > (1/e)(b^b/a^a)^(1/(b-a)) > (b-a)/ln(b/a)" } { "Tag": [ "geometry", "3D geometry", "prism", "Gauss" ], "Problem": "Can you guys help me with this problem? Thanks in advance. :lol: \r\n\r\n A rectangular fish tank has a base 2 feet wide and 3 feet long. When the tank is partially filled with water, a solid cube with an edge length of 1 foot is placed in the tank. If no overflow of water from the tank is assumed, by how many inches will the level of the water in the tank rise when the cube becomes completely submerged? \r\n \r\nA. 1/6 \r\n \r\nB. 1/2 \r\n \r\nC. 2 \r\n \r\nD. 3 \r\n \r\nE. 4", "Solution_1": "Well when the cube goes in, assuming that it's a paper cube it wouldn't be able to sink below the surface so the answer is 0. :lol: \r\nNah just joking, this is how you do it.\r\nThe cube has a volume of 1 cubic foot.\r\nThe fish tank has a base of 6 square feet.\r\nThat means that in order to get 1 cubic feet of water in the tank, you need an rectangular prism with a base of 6 and a height of $ \\frac{1}{6}$\r\nTherefore, the answer is $ \\frac{1}{6}$.", "Solution_2": "However, the problem asked for inches, so it would be 1/6(12)=2 inches.", "Solution_3": "Ah smart.\r\n :rotfl:", "Solution_4": "This should have been in the AMC forum. It came from the AMC 10 and 12.", "Solution_5": "@Z-coli and Zhangpeijin: thnx for helping me solve it :) \r\n\r\n@ gauss: really? it was a homework problem for me." } { "Tag": [ "probability", "expected value" ], "Problem": "A deck of cards is shuffled. Then you draw from the top until you draw an ace. What is the average amount of cards you draw until the ace comes up?", "Solution_1": "[quote=\"Wumbate\"]A deck of cards is shuffled. Then you draw from the top until you draw an ace. What is the average amount of cards you draw until the ace comes up?[/quote]\r\n[hide]..I suppose it's not 4 is it..because expected value is the same for all numbers, so...52/13=4[/hide]", "Solution_2": "I think the average cards drawn before an ace is between the 6th and 7th card (52/4/2=6.5)", "Solution_3": "[hide]I think the answer is $\\frac{53}{5}$[/hide]", "Solution_4": "number of cards left, percentage that it's an ace, number of aces out of 100 attempts\r\n52 7.6% 7\r\n51 7.8% 13\r\n50 8.0% 19\r\n49 8.1% 25\r\n48 8.3% 30\r\n47 8.5% 35\r\n46 8.6% 40\r\n\r\nEdit: This is probably wrong, I forgot to take into account that the more aces you find the lower the chances of aces in the future become. The more I look at this problem the more I agree with the 10th card. If you only have 13 cards (all of one suit) then you will find the ace on average around 6th or 7th draw. If you increase the base set to 52 cards (four of each suit) then the chances probably decrease but not very much.", "Solution_5": "If \"average amount\" means expected value, then SplashD is correct. More generally, \r\n\r\n[hide]if you shuffle a deck of $m$ cards, $n$ of which are aces, the first ace is on average the $\\frac{m+1}{n+1}$th card from the top of the deck.\n\nHere's one way of showing that:\n\nHow many ways can an ace be first? Well, there are $m-1 \\choose n-1$ ways to choose the places for the other aces. What if the first ace is second? There are $m-2 \\choose n-1$ ways to arrange the other aces. What if the first ace is $m-n+1$st (that is, $n$th from the end)? Well, there are only $n-1 \\choose n-1$ ways to arrange the other $n-1$ aces (they have to be the last $n-1$ cards).\n\nSo, the expected value is just $E = \\frac{1}{{\\binom m n}}\\sum_{i = 1}^{m-n+1}i \\cdot{m-i \\choose n-1}$. How to evaluate this? Well, one way is peel this sum apart, the same way one can peel a sum like $r+2r^{2}+3r^{3}+4r^{4}+\\ldots$ into many geometric sums. Equivalently, we can re-write the sum as $\\sum_{i = 1}^{m-n+1}i \\cdot{m-i \\choose n-1}= \\sum_{i = 1}^{m-n+1}(m+1) \\cdot{m-i \\choose n-1}-\\sum_{i = 1}^{m-n+1}(m+1-i) \\cdot{m-i \\choose n-1}= (m+1)\\sum_{i = 1}^{m-n+1}{m-i \\choose n-1}-n\\sum_{i = 1}^{m-n+1}{m-i+1\\choose n}$.\n\nFrom here, you apply the hockey-stick identity and everything works out nicely and you get the answer that I gave.\n\n[/hide]", "Solution_6": "I got [hide]$\\frac{53}{5}$[/hide] with many combinations and summations like JBL mentioned. I concur with SplashD :wink:", "Solution_7": "A slight generalization: if there are $m$ cards, $n$ of which are aces ($1 \\leq n \\leq m$), what is the expected value of the position of the $k$th ace ($1 \\leq k \\leq n$) in the deck?", "Solution_8": "should be [hide]$\\frac{m+1}{n+1}$[/hide]", "Solution_9": "That formula gives the answer in the case $k = 1$, which I gave in my first post. The question I just asked is more general.", "Solution_10": "oh sorry. i didn't see that.\r\n\r\ni found that vendermonte's identity comes into the mix but i'm not sure of how to evaluate the sum since you have to worry about multiplying respective probabilities by the corresponding number of cards pulled.", "Solution_11": "[hide]There are $n$ aces and $m-n$ non-aces. \nEach of the $m-n$ non-aces can be inserted either above the top ace, below the bottom ace, or between any two consecutive aces. \nSo, there are $n+1$ space to insert a non-ace card. \nSo, the average number of non-ace cards in each space is $\\frac{m-n}{n+1}$. \nThe $k$th ace has $k$ spaces above it. \nSo, on average, there are $\\frac{k(m-n)}{n+1}$ non-ace cards above the $k$th ace. \nThere are always $k$ aces that are above the $k$th ace or are the $k$th ace. \nSo, the expected value of the position of the $k$th ace is $\\frac{k(m-n)}{n+1}+k=\\boxed{\\frac{k(m+1)}{n+1}}$[/hide]\r\n1st post in Intermediate Topics", "Solution_12": "wow that was a nice solution :lol:" } { "Tag": [ "Duke", "college" ], "Problem": "Okay. We need to get our act together for DMM. (which happens to be on Halloween this year... coincidence?) Let's not get our butts kicked by TJ again and let's make sure Hamster has a miserable time if he goes.\r\n\r\nWell, not really, but you know what I mean.", "Solution_1": "Exactly who is this directed at?", "Solution_2": "to North Carolina, looks like. can't let those Virginians or Georgians or South Carolinians win.\r\n\r\nbtw good luck making Hamster feel \"miserable\"", "Solution_3": "Besides Enloe, what NC schools are sending teams?\r\n\r\nEdit: what NC schools are sending teams that have a legit shot at being top ten?", "Solution_4": "uh, NCSSM? lol", "Solution_5": "Woohoo, I'm glad I left an impression on North Carolina. But you know guys, maybe you should just accept defeat. Knowing when to gg is a good skill to have. So is making carriers. GLHF\r\n\r\nJust kidding. TJ's team isn't as good this year as it was last year.", "Solution_6": "[quote=\"Hamster1800\"]Woohoo, I'm glad I left an impression on North Carolina. But you know guys, maybe you should just accept defeat. Knowing when to gg is a good skill to have. So is making carriers. GLHF\n\nJust kidding. TJ's team isn't as good this year as it was last year.[/quote]\r\n\r\nbm is also a good skill to have\r\n\r\nalso [img]http://www.ghetto-overlord.com/Blog/Pictures/ICCUP/idracup.JPG[/img] YEAH\r\n\r\n\r\nbut dude we'll see", "Solution_7": "[quote=\"Hamster1800\"]Woohoo, I'm glad I left an impression on North Carolina. But you know guys, maybe you should just accept defeat. Knowing when to gg is a good skill to have. So is making carriers. GLHF\n\nJust kidding. TJ's team isn't as good this year as it was last year.[/quote]\r\n\r\nwe pwned you peoples xP i wonder if this is a beginning of a new era, hmmm", "Solution_8": "Darn I lost to the Chinese Taylor. That was pretty sad. Also the first two tiebreakers weren't that bad....although I forgot the (2,q,q) cases at first.", "Solution_9": "[quote=\"Hamster1800\"]Darn I lost to the Chinese Taylor. That was pretty sad. Also the first two tiebreakers weren't that bad....although I forgot the (2,q,q) cases at first.[/quote]\r\n\r\ndarn I lost to myself\r\n\r\nthat was even sadder\r\n\r\nand the first tiebreak was pretty easy except I kinda went an extra step\r\nand the second I was pretty dumb or something\r\nand the third I was screwing around and got something like x+y+z=8 1/x+1/y+1/z = -3/4 but then I was like oh noes I probably want another 3var guy so I am not dead or something\r\n\r\nalso darn we lost to TJ B? what? darn [though someone on TJ B got #10 hmm]", "Solution_10": "How do you do number 10?", "Solution_11": "Darn... Our team lost to AAST B by 0.5 points... It was our first time at Duke, though.\r\nIndividual finals was so fun-although slightly intimidating.\r\n\r\nI spent most of the time in power explaining the orbit-stabilizer theorem and Burnside's Lemma to my teammates.", "Solution_12": "[quote=\"33286\"]How do you do number 10?[/quote]\r\n\r\nhmm... $ \\left(a,b,c\\right)\\implies\\left(\\frac{a\\plus{}b\\plus{}c}{3},b,c\\right)\\implies\\left(\\frac{a\\plus{}b\\plus{}c}{3},\\frac{a\\plus{}4b\\plus{}4c}{9},c\\right)\\implies\\left(\\frac{a\\plus{}b\\plus{}c}{3},\\frac{a\\plus{}4b\\plus{}4c}{9},\\frac{4a\\plus{}7b\\plus{}16c}{27}\\right)\\equal{}\\left(a_1,b_1,c_1\\right)$. It's easy to show that $ a\\plus{}2b\\plus{}3c\\equal{}a_1\\plus{}2b_1\\plus{}3c_1$, and so we have found an invariant.", "Solution_13": "[quote=\"bpgbcg\"]Darn... Our team lost to AAST B by 0.5 points[/quote]\r\n\r\nthat was just too close. -phew-\r\n\r\nwhen they said that we got 5th by a sliver, mu A thought that they got edged out XD", "Solution_14": "Good job, Allen!!!! :P", "Solution_15": "[quote=\"Hamster1800\"]Woohoo, I'm glad I left an impression on North Carolina. But you know guys, maybe you should just accept defeat. Knowing when to gg is a good skill to have. So is making carriers. GLHF\n\nJust kidding. TJ's team isn't as good this year as it was last year.[/quote]\r\n\r\nOops i didn't accept defeat.\r\n\r\nP.S. Your cousin goes to my school.\r\n\r\nDude, i just want to make it clear that the answer for #10 is $ \\frac{40}{6}$, not $ \\frac{20}{3}.$\r\n\r\nAlso, I failed mandelbrot, so i should be in good shape for PuMaC :)", "Solution_16": "AAST finally beat TJ at something. :)", "Solution_17": "[quote=\"not_trig\"][quote=\"Hamster1800\"]Darn I lost to the Chinese Taylor. That was pretty sad. Also the first two tiebreakers weren't that bad....although I forgot the (2,q,q) cases at first.[/quote]\n\ndarn I lost to myself\n\nthat was even sadder\n\nand the first tiebreak was pretty easy except I kinda went an extra step\nand the second I was pretty dumb or something\nand the third I was screwing around and got something like x+y+z=8 1/x+1/y+1/z = -3/4 but then I was like oh noes I probably want another 3var guy so I am not dead or something\n\nalso darn we lost to TJ B? what? darn [though someone on TJ B got #10 hmm][/quote]\r\n\r\nWHAT FAIL #3 was just newton sums", "Solution_18": "So does anyone have the full results? My team failed. But we did beat some teams :P", "Solution_19": "I guessed 48/7 ((1a+2b+4c)/7) for 10. I guess I wasn't far off...\r\n\r\nI'd like to see full results as well.", "Solution_20": "[quote=\"PI-Dimension\"][quote=\"not_trig\"][quote=\"Hamster1800\"]Darn I lost to the Chinese Taylor. That was pretty sad. Also the first two tiebreakers weren't that bad....although I forgot the (2,q,q) cases at first.[/quote]\n\ndarn I lost to myself\n\nthat was even sadder\n\nand the first tiebreak was pretty easy except I kinda went an extra step\nand the second I was pretty dumb or something\nand the third I was screwing around and got something like x+y+z=8 1/x+1/y+1/z = -3/4 but then I was like oh noes I probably want another 3var guy so I am not dead or something\n\nalso darn we lost to TJ B? what? darn [though someone on TJ B got #10 hmm][/quote]\n\nWHAT FAIL #3 was just newton sums[/quote]\r\n\r\nyes i was aware but i don't know that crap" } { "Tag": [ "MATHCOUNTS" ], "Problem": "I heard you guys talking about it earlier, how can I get involved?", "Solution_1": "Wait until February. I'll probably send you something or get Lucian to. Basically, there's going to be a test at a school in your area. If you do well (even with only MathCounts-level knowledge you'd be fine) you make the team. The actual meet is in late May/early June; see [url]http://www.arml.com[/url]." } { "Tag": [ "limit", "logarithms", "integration", "real analysis", "real analysis unsolved" ], "Problem": "find :) \r\n\r\n$ \\lim_{n\\to\\infty}\\;\\left\\{\\;\\sum_{k=1}^{n}\\;\\frac{k^{3}+n^{3}}{k^{4}+n^{4}}\\;\\right\\},\\;\\;\\;\\quad\\text{i.e.}$ \r\n\r\n$ \\lim_{n\\to\\infty}\\;\\left\\{\\;\\frac{1+n^{3}}{1+n^{4}}+\\frac{8+n^{3}}{16+n^{4}}+\\frac{27+n^{3}}{81+n^{4}}+\\frac{64+n^{3}}{256+n^{4}}+\\cdots\\;\\;\\;\\cdots+\\frac{n^{3}+n^{3}}{n^{4}+n^{4}}\\;\\right\\}$", "Solution_1": "$ \\frac{1}{4}\\ln 2+\\int_{0}^{1}\\frac{1}{x^{4}+1}dx$" } { "Tag": [ "logarithms", "limit", "number theory unsolved", "number theory" ], "Problem": "for each number $ N$, suppose that Bob chooses a random number $ n$ in the interval $ \\frac{1}{2}\\leq n\\leq \\frac{3}{2}N$. If he repeats this process many times, prove that approximately $ \\frac{1}{\\ln{N}}$ of his numbers will be prime.\r\n\r\nIn other words, given\r\n\r\n$ P(N)\\equal{}($prob that an integer $ n$ in the interval $ \\frac{1}{2}\\leq n\\leq \\frac{3}{2}N$ is a prime number)\r\n\r\nthen prove that\r\n\r\n$ \\lim_{N\\to\\infty}\\frac{P(N)}{1/\\ln{N}}\\equal{}1$\r\n\r\n\r\nMore generally, given\r\n\r\n$ P(c_1,c_2,N)\\equal{}($Prob that an integer $ n$ in the interval $ c_1N\\leq N\\leq c_2N$ is a prime number)\r\n\r\nGive the correct term for $ X$ for the following equation:\r\n\r\n$ \\lim_{N\\to\\infty}\\frac{P(c_1,c_2,N)}{X}\\equal{}1.$", "Solution_1": "Am I wrong, or is it the same as\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=170399\r\n\r\nPierre.", "Solution_2": "so does anyone know what should go in $ X$?", "Solution_3": "Same answer : $ \\frac 1 {ln(N)}$, and same way as in the given link.\r\n\r\nPierre." } { "Tag": [ "probability" ], "Problem": "In a envelope, there are 2 one dollar bills, 2 two dollar bills, 2 five dollar bills, and 5 ten dollar bills. What is the probability that if i draw any 2 bills, without replacement, that i get a sum of more than 10 dollars.", "Solution_1": "[quote=\"usaha\"]In a envelope, there are 2 one dollar bills, 2 two dollar bills, 2 five dollar bills, and 5 ten dollar bills. What is the probability that if i draw any 2 bills, without replacement, that i get a sum of more than 10 dollars.[/quote]\r\n\r\n[hide]You can count and get 40 sums that are more than 10. There are a total of $\\binom{11}{2}=55$ ways to choose 2 bills. So the answer is $\\frac{40}{55}=\\boxed{\\frac{8}{11}}$[/hide]", "Solution_2": "I got the same answer as SplashD, but used a different method [hide]there is a 5/11 probability that you will pick a ten first, which guarantees success. if you pick a one, two, or five first, which has a 6/11 chance , you will only win if you then pick a ten, which has a probability of 5/10=1/2, so if you pick anything but a ten first, you have a 6/11*1/2=3/11 chance of succeeding. So, the answer is 5/11+3/11=[color=red]8/11[/color][/hide]", "Solution_3": "[quote=\"SplashD\"][quote=\"usaha\"]In a envelope, there are 2 one dollar bills, 2 two dollar bills, 2 five dollar bills, and 5 ten dollar bills. What is the probability that if i draw any 2 bills, without replacement, that i get a sum of more than 10 dollars.[/quote]\n\n[hide]You can count and get 40 sums that are more than 10. There are a total of $\\binom{11}{2}=55$ ways to choose 2 bills. So the answer is $\\frac{40}{55}=\\boxed{\\frac{8}{11}}$[/hide][/quote]\n[hide]You are aware that you have to pick a 10 aren't you?[/hide]", "Solution_4": "[quote=\"bpms\"][quote=\"SplashD\"][quote=\"usaha\"]In a envelope, there are 2 one dollar bills, 2 two dollar bills, 2 five dollar bills, and 5 ten dollar bills. What is the probability that if i draw any 2 bills, without replacement, that i get a sum of more than 10 dollars.[/quote]\n\n[hide]You can count and get 40 sums that are more than 10. There are a total of $\\binom{11}{2}=55$ ways to choose 2 bills. So the answer is $\\frac{40}{55}=\\boxed{\\frac{8}{11}}$[/hide][/quote]\n[hide]You are aware that you have to pick a 10 aren't you?[/hide][/quote]\r\n\r\nUmm... yeah. Why? Is my answer wrong or something?", "Solution_5": "[quote=\"SplashD\"][/quote][quote=\"bpms\"][quote=\"SplashD\"][quote=\"usaha\"]In a envelope, there are 2 one dollar bills, 2 two dollar bills, 2 five dollar bills, and 5 ten dollar bills. What is the probability that if i draw any 2 bills, without replacement, that i get a sum of more than 10 dollars.[/quote]\n\n[hide]You can count and get 40 sums that are more than 10. There are a total of $\\binom{11}{2}=55$ ways to choose 2 bills. So the answer is $\\frac{40}{55}=\\boxed{\\frac{8}{11}}$[/hide][/quote]\n[hide]You are aware that you have to pick a 10 aren't you?[/hide][/quote][quote=\"SplashD\"]\n\nUmm... yeah. Why? Is my answer wrong or something?[/quote]\r\nNo your answer is right. (I think)" } { "Tag": [ "MIT", "college" ], "Problem": "Does someone know when it will be released??? I know it's from mit to late march, but does someone know the specific day?", "Solution_1": "Uh, it says mid to late March on the MIT admissions website. So that's the best you're going to get. It depends on how quickly they can get everything done...", "Solution_2": "Didn't they get it out pi day last year (3/14) ? I figure it's probably around the same time this year", "Solution_3": "it's usually around pi day, if not exactly pi day. last year was a couple days late due to some weird shipping error they had with the physical letters themselves. \r\n\r\nGood luck to everyone who applied (myself included)!" } { "Tag": [ "number theory", "least common multiple", "induction", "inequalities", "number theory proposed" ], "Problem": "Consider $k$ positive integers $a_{1},a_{2},...,a_{k}$ satisfying $1 \\leq a_{1}1$. Then :\r\n$a_{i}-a_{i-1}\\geq \\gcd (a_{i}, a_{i-1}) = \\frac{a_{i}a_{i-1}}{[a_{i}, a_{i-1}]}\\geq \\frac{a_{i}a_{i-1}}n$.\r\nThus $a_{i-1}\\leq \\frac n{1+\\frac n{a_{i}}}\\leq \\frac n{k+1-(i-1)}$, as desired (the last inequality follows from the induction hypothesis).\r\n\r\nFrom the inequality, it easily follows that if $i \\leq k-[ \\sqrt n ]$ then $a_{i}\\leq [\\sqrt n ]$. Hence, since the given sequence is increasing, we have $k-[ \\sqrt n ] \\leq [ \\sqrt n ]$, which leads to the result.\r\n\r\nPierre." } { "Tag": [ "induction" ], "Problem": "prove that $\\frac{\\sqrt{6}}{5}+\\frac{\\sqrt{20}}{9}+\\frac{\\sqrt{42}}{13}+..............+\\frac{\\sqrt{2n(n+1)}}{4n+1}<\\frac{n}{2}$ for all $n\\in {N^{*}}$", "Solution_1": "$\\frac{\\sqrt{2n(n+1)}}{4n+1}$ or rather $\\frac{\\sqrt{2n(2n+1)}}{4n+1}$ ?\r\n\r\nBecause $\\frac{\\sqrt{2n(n+1)}}{4n+1} < 1/2$ which makes the problem obvious.", "Solution_2": "What's the 'easy way' to see that it is <1/2?", "Solution_3": "Use induction", "Solution_4": "I will use induction:\r\nFirstly we check if it's true for $n=1$\r\n$\\frac{\\sqrt{6}}{5}<\\frac{1}{2}$\r\nIt's true. \r\nNow I'll show that if it's true for $n$, then it's true for $n+1$\r\nAssumption :\r\n$\\frac{\\sqrt{6}}{5}+\\frac{\\sqrt{20}}{9}+\\frac{\\sqrt{42}}{13}+..............+\\frac{\\sqrt{2n(n+1)}}{4n+1}<\\frac{n}{2}$\r\nThesis:\r\n$\\frac{\\sqrt{6}}{5}+\\frac{\\sqrt{20}}{9}+\\frac{\\sqrt{42}}{13}+..............+\\frac{\\sqrt{2n(n+1)}}{4n+1}+\\frac{\\sqrt{2(n+1)(n+2)}}{4n+5}<\\frac{n+1}{2}$\r\n\r\nNow I add $\\frac{\\sqrt{2(n+1)(n+2)}}{4n+5}$ to the both sides of our thesis, and I get this:\r\n\r\n$\\frac{\\sqrt{6}}{5}+\\frac{\\sqrt{20}}{9}+\\frac{\\sqrt{42}}{13}+..............+\\frac{\\sqrt{2n(n+1)}}{4n+1}+\\frac{\\sqrt{2(n+1)(n+2)}}{4n+5}<\\frac{n}{2}+\\frac{\\sqrt{2(n+1)(n+2)}}{4n+5}$\r\n\r\nNow I'm going to prove, that:After I do so I'll have the thesis\r\n\r\n$\\frac{n}{2}+\\frac{\\sqrt{2(n+1)(n+2)}}{4n+5}<\\frac{n+1}{2}$\r\nAfter I do so I'll have the thesis\r\n\r\n$2\\sqrt{2(n+1)(n+2)}<4n+5$\r\n$(n+1)(n+2)<(4n+5)^2$\r\n$8n^2+16n+9>0$, which is true for all $n$, because $\\Delta<0$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that if $ a, b, c > 0 $ then\r\n$ 8(a^2+b^2)(b^2+c^2)(c^2+a^2)(a+b+c)^2 \\ge 3(a+b)^2(b+c)^2(c+a)^2(a^2+b^2+c^2)$", "Solution_1": "I've gotten a fun!\r\n\r\nYour inequality is equivalent to\r\n\\[5\\sum a^6b^2+10\\sum a^5b^3+10\\sum a^4b^4\\geq 6\\sum a^6bc+2\\sum a^5b^2c+4\\sum a^4b^2c^2+8\\sum a^4b^3c +10\\sum a^3b^3c^2.\\]\r\nApplying several times AM-GM we obtain result:\r\n1) $\\sum a^6b^2 \\geq 2\\sum a^6bc$;\r\n2) $\\sum a^5b^3 \\geq \\sum a^5b^2c$;\r\n3) $\\sum a^4b^4\\geq \\sum a^4b^3c$;\r\n4) $\\sum a^4b^4\\geq 2\\sum a^4b^2c^2$;\r\n5) $\\sum a^6b^2 \\geq \\sum a^3b^3c^2$;\r\n6) $\\sum a^6b^2 \\geq \\sum a^3b^3c^2$.", "Solution_2": "I've also got a fun! Prove the more general inequality:\r\n For all positive real numbers $ x_1,x_2,...,x_n$ we have \r\n $\\frac{x_2^2+x_3^2+...+x_n^2}{n_1}\\cdot\\frac{x_1^2+x_3^2+...+x_n^2}{n-1}\\cdot...\\cdot\\frac{x_1^2+x_2^2+...+x_{n-1}^2}{n-1}\\geq\\frac{n(x_1^2+x_2^2+...+x_n^2)}{(x_1+x_2+...+x_n)^2}\\cdot\\(\\frac{x_2+...+x_n}{n-1})^2\\cdot...\\cdot (\\frac{x_1+x_2+...+x_{n-1}}{n-1})^2 $", "Solution_3": "And here is my solution (with Mr.Nam Dung 's help )for it.The inequality is equal to:\r\n $ \\prod \\frac{(n-1)(x_1^2+...+x_{n-1}^2)}{(x_1+...+x_{n-1})^2} \\geq \\frac{n(x_1^2+...+x_n^2)}{(x_1+...+x_n)^2} $\r\nWe have:\r\n $ LHS = \\prod (\\sum \\frac{(x_i-x_j)^2}{(x_1+...+x_{n-1})^2}+1) \\geq $ \r\n $ \\sum \\frac{(x_i-x_j)^2}{(x_1+...+x_n)^2}+1 =\\frac{n(x_1^2+...+x_n^2)}{(x_1+...+x_n)^2} $ :) .", "Solution_4": "Yes, that is!" } { "Tag": [ "limit", "advanced fields", "advanced fields unsolved" ], "Problem": "Let A be a c*-algebra. I tried to show that given a positive a in A, the closure of the set aAa is the smallest hereditary C*-subalgebra containing a, I was able to show that it contains a^2, but not a... and why is it hereditary?", "Solution_1": "it's not easy at all :-). the proof can be found in\r\nB. Blackadar, Operator Algebras: Theory of C*-Algebras and von Neumann Algebras, Proposition II.3.4.2 (ii)", "Solution_2": "Thanks a lot!\r\nI actually got the book and understood the proofe, but now I don't get another thing -\r\nin the same proposition - part (iii) - why these the element h =$ \\sum_{n \\equal{} 1}^{\\infty}{2^{ \\minus{} n} x_{n}^{*} x_n}$\r\nmake $ B \\equal{} \\overline{hAh}$?\r\n\r\n(I assume it has something to do with the fact that: $ b \\in \\overline{x^{*}Ax}$ if and only if\r\n$ lim_{\\alpha \\rightarrow 0} {(xx^{*})^{\\alpha}b(xx^{*})^{\\alpha}} \\equal{} b$, is that correct?)", "Solution_3": "Take $ \\{u_{n}\\}$ to be any countable approximate unit in B (there is such an approximate unit as B is separable)\r\nand take $ a=\\sum 2^{-n}u_{n}$, so $ a \\in B^{+}$ and as $ \\overline{aAa}$ is the smallest hereditary subalgebra containning a, $ \\overline{aAa} \\subseteq B$.\r\nAs $ 2^{-n}u_{n} \\leq a$, and $ \\overline{aAa}$ is hereditary, $ u_{n} \\in \\overline{aAa}$,\r\nand that is why if $ b \\in B$ we get ${ b=\\lim{u_{n}bu_{n}}}$ and ${{ u_{n}bu_{n}}} \\in \\overline{aAa}$, so $ b \\in \\overline{aAa}$, so fimally $ B = \\overline{aAa}$." } { "Tag": [], "Problem": "Seven students in a class compare their marks in 12 subjects studied and observe that no two of the students have identical marks in all 12 subjects. Prove that we can choose 6 subjects such that any two of the students have different marks in at least one of these subjects.", "Solution_1": "Let $\\{X_i\\}_{i=1}^{12}$ be the subjects, and $(a,b,c,d,e,f,g)$ be the students.\r\n\r\nIf (in example) $(a,b) \\in X_1$, then one of $(a,j), (b,j)$ is in $X_1$, for j = c,d,e,f,g.\r\n\r\nOur strategy is as follows. Pick a subject containing (a,b) [there is atleast one]. From our lemma above, we guarantee 5 other unique pairs in that subject. Now pick a subject containing a pair we havent got to yet. We guarantee 4 other unique pairs, because only one can repeat (it is easy to show **) Continuing in this fashion, we guarantee 6+5+4+3+2+1 = 21 unique pairs, and we are done.\r\n\r\n** a short proof with handwaving: if in example we used (a,b), then we have also used either (x,a) or (x,b). So when we use (i, j), we use either (y,i) or (y,j). So if (x,a) = (y,i) in example, we have a,b,i,j,x fixed, so that when y varies it can only match up with (x,a) at most once.", "Solution_2": "Clarification -- what do the ordered pairs represent? And why would \"only one of $(a,j),(b,j)$\" be in $X_1$ if $(a,b) \\in X_1$?" } { "Tag": [ "integration", "function", "calculus", "real analysis", "real analysis unsolved" ], "Problem": "let $f: [0,1]\\to\\mathbb{R}$ be integrable. let $f(x)=0$ everywhere except the cantor set. then show that $\\int_{0}^{1}f(x)dx=0$. Also my friend said that if $f=0$ everywhere except $\\mathbb{Q}\\cap [0,1]$ then this is not true. i think it might be something to do with the fact that the characteristic function of $\\mathbb{Q}$ not being integrable, but i'm not able to prove both these statements. can someone help? perhaps a hint or something. thanks in advance", "Solution_1": "Riemann integral, right?", "Solution_2": "yep, riemann integral. i dont know lebesgue :)", "Solution_3": "See [b]Theorem 11[/b] [url=http://www.math.cuhk.edu.hk/~mat2060/mat2060b/Notes/notes2.pdf]here[/url].\r\n\r\nEDIT: jmerry's approach is indeed better.", "Solution_4": "ya i know this theorem but how is it going to help me with the problem i posted in post no 1? sorry if i didnt get the entire meaning you intended to convey\r\n\r\nso let me state more clearly, i'd like to have hints for the problem in post 1, the $\\int_{0}^{1}f(x)dx=0$ thing", "Solution_5": "For this sort of problem, you're probably best off working with upper and lower sums, which depend only on the partition. The function is integrable if and only if the upper sums can be made arbitrarily close to each other.\r\n\r\nFor the Cantor set, try a partition with $3^{n}$ parts of equal length (assign the endpoints to the interval with members of the Cantor set). The lower sums are all zero. The upper sums aren't, but the function is zero on most of the pieces. Calculate the upper sums and show they go to zero.\r\n\r\nFor the characteristic function of $\\mathbb{Q}$, the upper sums are all 1 and the lower sums are all zero for any partition.", "Solution_6": "thanks jmerry" } { "Tag": [ "inequalities", "trigonometry", "calculus", "calculus computations" ], "Problem": "Prove that $ 2\\plus{}2x\\sin\\frac{1}{x}>\\cos\\frac{1}{x}\\ \\forall x\\in\\mathbb{R}\\setminus\\{0\\}$. Find the minimum value of $ M\\in\\mathbb{R}$ for which $ M\\plus{}2x\\sin\\frac{1}{x}>\\cos\\frac{1}{x}\\ \\forall x\\in\\mathbb{R}\\setminus\\{0\\}$.", "Solution_1": "Can this be shown analytically? The behavior around 0 is highly irregular.", "Solution_2": "For $ |x| < \\frac {1}{2}$ we have $ 2 \\plus{} 2x\\sin \\frac {1}{x} > 1$ and $ \\cos \\frac {1}{x}\\leq 1$. For $ |x|\\geq \\frac {1}{2} > \\frac {1}{\\pi}$ we have $ x\\sin \\frac {1}{x} > 0$ so the inequality clearly holds.\r\n\r\nTo find $ M$, let $ x_1 > x_2 > ...$ be the positive points at which $ f(x) \\equal{} \\cos \\frac {1}{x} \\minus{} 2x\\sin \\frac {1}{x}$ obtains local maxima and show that $ f(x_1) > f(x_2) > ...$. Then $ M \\equal{} f(x_1)\\approx 1.05498$, where $ x_1\\approx 0.16834$ is the second greatest solution of $ f'(x) \\equal{} 0\\Leftrightarrow \\left(1 \\plus{} 2x^2\\right)\\sin \\frac {1}{x} \\equal{} 2x\\cos \\frac {1}{x}$. (though strictly speaking, no $ M$ satisfies the condition as stated; the $ >$ should be $ \\geq$)." } { "Tag": [], "Problem": "Let $n$ be a positive integer,prove (1) to (3) \r\n\r\n(1) $1111^{n}-1109^{n}$ can be divided by $2$\r\n\r\n(2) $11^{n}-8^{n}-3^{n}$ can be divided by $24$\r\n\r\n(3) $2450^{n}-1370^{n}+1150^{n}-250^{n}$ can be divided by $1980$\r\n\r\n\r\n\r\nThis is a Japan Medical University entrance exam in [b]1980[/b] . not so difficult :P\r\n\r\n[hide=\"Hint\"]\n$a,b$ ; positive integer$(a>b)$,then $a^{n}-b^{n}$can be divided by $a-b$ :lol:\n[/hide]", "Solution_1": "[hide=\"1\"]$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}+.......+y^{n-1})$\n\nSo $1111^n-1109^n=(1111-1109)(........)$\n $1111^n-1109^n=(2)(......)$[/hide]\n\n\n[hide=\"2\"]$11^n-8^n-3^n=(3)[11^{n-1}+8(............)-3^{n-1}]$\n\nNow $11^{n-1}-3^{n-1}$ is divisible by $8$\n\nso we can take a factor of $8$ out of $[11^{n-1}+8(............)-3^{n-1}]$\n\n$\\Longrightarrow 11^n-8^n-3^n=(24)(......)$[/hide]", "Solution_2": "[quote=\"porte\"]Let $n$ be a positive integer,prove (1) to (3) \n\n(1) $1111^{n}-1109^{n}$ can be divided by $2$\n\n(2) $11^{n}-8^{n}-3^{n}$ can be divided by $24$\n\n(3) $2450^{n}-1370^{n}+1150^{n}-250^{n}$ can be divided by $1980$\n\n\n\nThis is an Japan Medical University entrance exam in [b]1980[/b] . not so difficult :P\n\n[hide=\"Hint\"]\n$a,b$ ; positive integer$(a>b)$,then $a^{n}-b^{n}$can be divided by $a-b$ :lol:\n[/hide][/quote]\n\n[hide=\"3\"]\n$180|(2450^n-1370^n)+(1150^n-250^n)$ and $11|2450^n-250^n)-(1370^n-1150^n)$\n\n[/hide]", "Solution_3": "thank you for reply , Both are best answer !shadysaysurspammed and shyong :P", "Solution_4": "[quote=\"porte\"](1) $1111^{n}-1109^{n}$ can be divided by $2$[/quote]\r\nIt's obvious that the difference of two odd numbers is an even number. :P", "Solution_5": "[quote=\"ArcticMonkey\"][quote=\"porte\"](1) $1111^{n}-1109^{n}$ can be divided by $2$[/quote]\nIt's obvious that the difference of two odd numbers is an even number. :P[/quote]\r\n\r\nYeah that's right but I gave that identity beacuse it is helpful for the other questions.", "Solution_6": "[quote=\"shadysaysurspammed\"]\nNow $11^{n-1}-3^{n-1}$ is divisible by $8$\n\n[/quote]\r\n\r\n\r\nCan anyone prove that $11^n-3^n$ is divisible by $8$?", "Solution_7": "[quote=\"smellyfart\"]\nCan anyone prove that $11^n-3^n$ is divisible by $8$?[/quote]\r\n\r\n$11^n-3^n=(11-3)(11^{n-1}+\\cdots+3^{n-1})$ :lol:", "Solution_8": "Or look mod 8.", "Solution_9": "[quote=\"porte\"]Let $n$ be a positive integer,prove (1) to (3) \n\n(1) $1111^{n}-1109^{n}$ can be divided by $2$\n\n(2) $11^{n}-8^{n}-3^{n}$ can be divided by $24$\n\n(3) $2450^{n}-1370^{n}+1150^{n}-250^{n}$ can be divided by $1980$\n\n\n\nThis is a Japan Medical University entrance exam in [b]1980[/b] . not so difficult :P\n\n[hide=\"Hint\"]\n$a,b$ ; positive integer$(a>b)$,then $a^{n}-b^{n}$can be divided by $a-b$ :lol:\n[/hide][/quote]\r\n\r\nWhat does this have to do with medicine? :huh:", "Solution_10": "Many Japanese universities offer their own entrance exams in addition to the standarized test. Students usually need to take Japanese, math, English and one more subject (either social studies or science, depending on the major). Although math (and English, Japanese etc.) has nothing to do with medicine, the students still need to take the exams for all these different subjects." } { "Tag": [ "search", "calculus", "LaTeX", "limit", "trigonometry", "geometry", "derivative" ], "Problem": "How to show $ lim_{x \\longrightarrow 0} \\frac {sinx}{x} \\equal{} 1$ without L'Hospital rule?", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1450063994&t=242582\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1450063994&t=211925\r\n\r\n(I found these by searching for the word \"circular\" in the Calculus & Analysis forum.)\r\n\r\nSome [[LaTeX]] suggestions: the code you should have used is \\lim_{x \\to 0} \\frac{\\sin x}{x} = 1.\r\n\r\nKent Merryfield's posts [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229706]here[/url] has some very good general comments, and there's more related discussion of L'Hopital and how you might want to use it or why you might want to avoid it in certain circumstances [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=256605]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190611]here[/url].", "Solution_2": "Actually we can use the seris expansion of [b]sin x[/b]\r\n[img]http://upload.wikimedia.org/math/2/9/c/29cb648e96291f52707c7225630c1a17.png[/img]\r\nand then we can divide by x.as x tends to 0 we see that substituting o in the series gives us the ans as 1", "Solution_3": "Unless that series is the definition you're using, that's as circular as anything else.", "Solution_4": "There is a proof of the limit using the Squeeze theorem. Shouldn't be too hard to find ;)", "Solution_5": "the one using the squeez theorem is quite easy . its an elegant proof using geometry . \r\n\r\nit given in all ncert textbooks followed in india here is the link for the proof..\r\n\r\n[url]http://ncert.nic.in/textbooks/testing/Index.htm[/url] jus go here and in the top select the class as X1 and subject mathematics,a pdf opens and the go to chapter limits and derivatives" } { "Tag": [ "calculus", "integration", "geometry", "3D geometry", "parameterization", "analytic geometry", "calculus computations" ], "Problem": "Crunch time finals next week...oy. I got this and not sure on second step.\r\n4) Evaluate the surface integral $\\int\\int_{\\sigma}(y^2+z^2)dS$ where $\\sigma$ is the portion of the cone $x=\\sqrt{3(y^2+z^2)}$ for $0\\leq x\\leq3$.\r\nMy book says that a special form of the surface integral can be simplified if you use x=x y=y and z=g(x,y).\r\nI used this with: $y=y$ $z=z$ $x=\\sqrt{3(y^2+z^2)}$\r\nSo I could use the formula: $\\int\\int_{D}f(y,z,g(y,z))\\sqrt{\\frac{\\partial x}{\\partial y}^2+\\frac{\\partial x}{\\partial z}^2+1}\\,dA\\Rightarrow$\r\n$\\frac{\\partial x}{\\partial y}=\\frac{\\sqrt{3}y}{\\sqrt{y^2+z^2}}$ and $\\frac{\\partial x}{\\partial z}= \\frac{\\sqrt{3}z}{\\sqrt{y^2+z^2}}$\r\nPlug in the variables and you get: $2\\int\\int_{D}y^2+z^2\\,dA$\r\nMy problem is I am not sure how to find the area D.", "Solution_1": "The set $D$ is the projection of your cone onto the $yz$ plane, and your $dA$ is $dy\\,dz.$ You've made the minor adjustments needed to turn a standard formula for a surface given by $z=h(x,y)$ into formulas that apply to a surface given by $x=h(y,z).$ Since you're using $y$ and $z$ as your parameters, you need to find out what values $(y,z)$ can take on.\r\n\r\n$x=\\sqrt{3(y^2+z^2)}\\le3$ so $\\sqrt{y^2+z^2}\\le\\frac3{\\sqrt{3}}=\\sqrt{3}$ or $y^2+z^2\\le3.$ Your set $D$ is a circular disk of radius $\\sqrt{3}.$\r\n\r\nYour integral is $2\\int\\int_{y^2+z^2\\le3}y^2+z^2\\,dy\\,dz.$\r\n\r\nOf course, now you're going to use polar coordinates. OK, it's not quite standard looking polar coordinates, since this is $y$ and $z,$ but given what you did at the beginning of the problem, you should be able to see what I mean.", "Solution_2": "Not quite sure how I could keep this as x=x y=x and z=g(x,y)....\r\n\r\nMaybe x=z y=y z=g(x,y) ???\r\n\r\nor is it legitimate to just continue by making it:\r\n$2\\int_{0}^{2\\pi}\\int_{0}^{3}r\\cdot r\\,dr\\,d\\theta$", "Solution_3": "Just don't worry about the names of the coordinates - work by analogy.\r\n\r\nAnd it is, as you say, \"legtimate to just continue by making it ...\" only please get it right:\r\n\r\n$2\\int_0^{2\\pi}\\int_0^{\\sqrt{3}}r^2\\cdot r\\,dr\\,d\\theta$", "Solution_4": "doh! :blush: wish I could say that was a typo.............just typing without thinking...oy." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Prove that if A is a Jordan region and f and g are integrable functions on A, then fg is also integrable on A.", "Solution_1": "Riemann integrable, of course. Since $fg=\\frac14((f+g)^{2}-(f-g)^{2})$, it suffices to prove that $f^{2}$ is integrable whenever $f$ is. Being Riemann integrable, $f$ is bounded: $|f|\\le M$, say. Therefore, $|f(x)^{2}-f(x')^{2}|\\le 2M|f(x)-f(x')|$ for any $x,x'\\in A$. This implies, after some routine work, that Riemann sums for $f^{2}$ have a limit." } { "Tag": [ "LaTeX" ], "Problem": "A3. n > 2. x1, x2, ... , xn are real numbers such that 2 \u2264 xi \u2264 3. Show that (x12 + x22 - x32)/(x1 + x2 - x3) + (x22 + x32 - x42)/(x2 + x3 - x4) + ... + (xn-12 + xn2 - x12)/(xn-1 + xn - x1) + (xn2 + x12 - x22)/(xn + x1 - x2) \u2264 2(x1 + x2 + ... + xn) - 2n.", "Solution_1": "man ke hichi nafahmidam.\r\nva khodaavand latex aafarid... :chief:", "Solution_2": "hads mizanam ke masalat ghashange vali hichish maloom nis", "Solution_3": "kheili tablooye ke az ye jaei copy kardi. ye jaei mesle kava ,ye keshvari ,ye sali ,soale 3 jabr(fekrkonam shortlist)\r\n\r\nfekr konam soorate doroste soal ine\r\n\r\n$n>2\\ ,\\ x_1,x_2,\\dots,x_n$ are real numbers such that$2\\le x_i\\le 3$.show that:\r\n\r\n$\\frac{x_1^2+x_2^2-x_3^2}{x_1+x_2-x_3}+\\frac{x_2^2+x_3^2-x_4^2}{x_2+x_3-x_4}+\\dots+\\frac{x_{n-1}^2+x_n^2-x_1^2}{x_{n-1}+x_n-x_1}+\\frac{x_n^2+x_1^2-x_2^2}{x_n+x_1-x_2}\\le 2(x_1+x_2 + \\dots+x_n)-2n$", "Solution_4": "hahahahah belakhare fahmidam soalo az koja cop zadi :trampoline: :clap: \r\n\r\ninham ye link:\r\nhttp://www.kalva.demon.co.uk/short/sh95.html", "Solution_5": "khob bashe ghanool mikonam\r\nmanam rahehalesho cop mizanam", "Solution_6": "[quote=\"amir2\"]akhe adam ingadr tablo cop mizane[/quote]\r\nbaba hala mage chiye... khaaste ye maseleye ghashang matrah kone... :lol:", "Solution_7": "The ith term on the lhs is (xi2 + xi+12 - xi+22)/(xi + xi+1 - xi+2) (with cyclic subscripts). We can write this as (xi + xi+1 + xi+2) - 2xixi+1/(xi + xi+1 - xi+2). Now (xi - 2)(xi+1 - 2) \u2265 0, so -2xixi+1 \u2264 -4(xi + xi+1 - 2) = -4( (xi + xi+1 - xi+2) + (xi+2 - 2) ). Hence the ith term \u2264 (xi + xi+1 + xi+2) - 4(1 + (xi+2 - 2)/(xi + xi+1 - xi+2) ). But the constraints on xi imply that (xi + xi+1 - xi+2) and (xi+2 - 2) are positive and (xi + xi+1 - xi+2) \u2264 4. So the ith term <= (xi + xi+1 + xi+2) - 4(1 + (xi+2 - 2)/4) = xi + xi+1 - 2. Hence lhs \u2264 2s - 2n.", "Solution_8": "[quote=\"amir2\"]The ith term on the lhs is (xi2 + xi+12 - xi+22)/(xi + xi+1 - xi+2) (with cyclic subscripts). We can write this as (xi + xi+1 + xi+2) - 2xixi+1/(xi + xi+1 - xi+2). Now (xi - 2)(xi+1 - 2) \u2265 0, so -2xixi+1 \u2264 -4(xi + xi+1 - 2) = -4( (xi + xi+1 - xi+2) + (xi+2 - 2) ). Hence the ith term \u2264 (xi + xi+1 + xi+2) - 4(1 + (xi+2 - 2)/(xi + xi+1 - xi+2) ). But the constraints on xi imply that (xi + xi+1 - xi+2) and (xi+2 - 2) are positive and (xi + xi+1 - xi+2) \u2264 4. So the ith term <= (xi + xi+1 + xi+2) - 4(1 + (xi+2 - 2)/4) = xi + xi+1 - 2. Hence lhs \u2264 2s - 2n.[/quote]\r\nlet me guess....kalva? :D", "Solution_9": "ramtin az babate del garmit mamnoon vali man natoonestam hal konam va be hamin dalil matrah karmam albate soal male hamon jaee ke amir goft bood\r\n\r\nlot fan ye rahe hale ghashang bedid" } { "Tag": [ "summer program", "Mathcamp", "geometry", "MATHCOUNTS", "AMC", "USA(J)MO", "USAMO" ], "Problem": "Is mathcamp more about contest math or about learning difficult math?", "Solution_1": "If all you want to do is do contest math all day, go to AwesomeMath.\r\n\r\nMathcamp will give you a deep understanding in topics you've never heard of. The classes are in areas you might not see again until your graduate years (though, because of the 1-week timeframe, they're often very specific).\r\n\r\nOf course, we certainly do do contest math at Mathcamp, but it's not the main focus (though you can make it your main focus if you really want). Every week there are problem-solving classes (usually two -- one \"intro class,\" which last year was taught by a different visitor every week and ranged in difficulty from Mathcounts to USAMO with them), and an advanced problem class). Every Wednesday there's Team Problem Solving, where teams of about 8 people (the same team every week -- scores are cumulative) are given 2 hours to collaborate on a USAMO-style test. Finally, every Saturday there are Relays, where teams work on two questions at a time to compete for a bag of candy. Relays can get quite wacky -- questions usually just ask you to find some number like an ARML individual round, but other times they might ask your team to stand with 3 feet total on the ground, or play a game blindfolded. The last Relay included a competition to make up the best rap on a mathematical topic for bonus points (Laura Zehender has a video of this on Facebook). One Relay was done in the swimming pool area, where a team member had to swim across to turn in an answer.", "Solution_2": "The video mentioned above is here: http://www.facebook.com/video/video.php?v=73908337943 :)", "Solution_3": "I really would say that in general, the answer to your question is \"about learning difficult math.\" (Or I might replace it with \"abstract\" or \"advanced.\")", "Solution_4": "then i think I will definately attend because i don't like contest math very much and am not very good at it :) \r\nwill people look down on me if im not good at it? (right now i cud prb get 7 on the AIME)\r\nAs for the video no offense but i could freestyle ALOT better than that, i'd post link to some vids but real names scare me :P im PROMYS this summer but maybe ill do mathcamp next summer if I get in", "Solution_5": "Last year I got a 4 on the AIME, and I did not feel \"looked down on\" at all.", "Solution_6": "I believe some people last year did not make AIME. I know at least one of this years' campers did not make AIME. And no, we don't really care. It's not something that comes up much.", "Solution_7": "Personally, if someone can't get at least a 12 or 13 on the AIME, I don't find it worth trying to have an intelligent conversation with them. Math ability is a good indicator of intelligence; and besides, the best mathematicians are usually the ones who do contests in high school.", "Solution_8": "12 or 13 on the AIME? Why aim so low!? I personally refuse to talk to anyone in real life who isn't either a member of my immediate family or a perfect scorer on the IMO.\r\n\r\n[hide]To be honest, when I first saw benzi's post the only reason I took it for sarcasm was because it was kinda outlandish; hence, I made it even worse to emphasize that we are, in fact, being sarcastic. And then I made this hidden message because I realize that sarcasm can be hard to detect on the internet sometimes <_< (and actually in real life, apparently. Only my group of friends can recognize when I'm being sarcastic; random people I talk to think I'm being serious usually)[/hide]", "Solution_9": "I too am curious to know if Benzi's post was sarcastic.", "Solution_10": "I'm fairly sure it is.", "Solution_11": "Hmm is associating AIME score with math ability also sarcastic? :P", "Solution_12": "I sure hope so :P .\r\n\r\n(And I'm one of those people who's pretty oblivious to sarcasm. Once I actually thought my friend was deaf because of that...heh :roll: . but I was able to catch on with Zeph's post, at least.)", "Solution_13": "Lol. Thank you for your help, Zeph.\r\n[quote=\"solafidefarms\"]I too am curious to know if Benzi's post was sarcastic.[/quote]\nIt was.\n\n[quote=\"perfect628\"]Hmm is associating AIME score with math ability also sarcastic? :P[/quote]\nYes. So there were a few points of sarcasm. First of all, I cannot get a 12 or 13 on the AIME. Second, math ability is often no indicator of intelligence, and AIME score is often no indicator of math ability. Third, mathematicians are often not the ones who do contests in high school.\n\n[quote=\"LadyKn1ght\"](And I'm one of those people who's pretty oblivious to sarcasm. [...])[/quote]\r\nLol. I'm going to need to work on that.\r\n\r\nYeah, after I posted that message I got to thinking that maybe I'd trolled too hard. But hey, it's all in good fun; and besides, a lot of the time I'm the guy who pushes the limit on that anyway. :lol: :spam: :rotfl:", "Solution_14": "[quote=\"benzi455\"] Third, mathematicians are often not the ones who do contests in high school. [quote]\r\nIs that really true? *cough Terence Tao\r\nwhile I kno wjust b/c one mathematician did well on contests that is not the rule i am interested to know if most top mathematicians did math contests in HS", "Solution_15": "I think he's saying that while there is a correlation between contest skill and higher-level math success, it's not as prominent as some people make it out to be. I'd say it also depends on the contest; for instance, the ARML Power Round is probably a better indicator than Mathcounts Countdown." } { "Tag": [ "trigonometry" ], "Problem": "If a little car with initial velocity $v_{0}= 2 \\sqrt{gR}$ rides through a loop with radius $R$, at which height does the car leave the track. You can neglect friction.", "Solution_1": "Hint: Using the Newton's second law and the law of conservation of energy to find the answer.\r\n\r\n[hide]\nNewton's second law:\n\n$\\large N+mg\\cos\\theta=\\frac{mv^{2}}{R}$\n\n$\\Longrightarrow N=\\frac{mv^{2}}{R}-mg\\cos\\theta$\n\nWhen the car leaves the track\n\n$\\large N=0 \\Longrightarrow \\frac{mv^{2}}{R}=g\\cos\\theta_{0}$\n\n$\\large \\cos\\theta_{0}=\\frac{h-R}{R}$ where $h$ is the height we need.\n\nThe law of conservation of energy:\n\n$\\large \\frac{mv_{0}^{2}}{2}=mgh+\\frac{mv^{2}}{2}$\n\nWith $\\large v_{0}=2\\sqrt{gR}$ we have $\\boxed{h=\\frac{5R}{3}}$\n\n[/hide]", "Solution_2": "Thanks alot.\r\n\r\nThat was more or less what I found. A pitty that I forgot the cosine on my exam :)", "Solution_3": "[quote=\"phucnv87\"]\n\nNewton's second law:\n\n$\\large N+mg\\cos\\theta=\\frac{mv^{2}}{R}$\n\n$\\Longrightarrow N=\\frac{mv^{2}}{R}-mg\\cos\\theta$\n\n[/quote]\r\n\r\nThis is not quite correct.\r\n\r\nThe correct expression is:\r\n\r\n$ma_{R}-m(\\frac{v}{R})^{2}R+mg(1-\\cos\\theta) = 0$\r\n\r\n$\\Rightarrow a_{R}= \\frac{v^{2}}{R}-g(1-\\cos\\theta)$\r\n\r\nThe car leaves the track if $a_{R}< 0$\r\n\r\n$\\Rightarrow \\frac{v^{2}}{R}-g(1-\\cos\\theta) < 0$\r\n\r\n$\\Rightarrow \\frac{v^{2}}{R}< g(1-\\cos\\theta)$\r\n\r\n$\\Rightarrow \\frac{v^{2}}{gR}-1 <-\\cos\\theta$\r\n\r\n$\\Rightarrow 1-\\frac{v^{2}}{gR}> \\cos\\theta$\r\n\r\nFrom the conservation of energy:\r\n\r\n$\\frac12 mv_{0}^{2}= \\frac12 mv^{2}+mgh$\r\n\r\n$v^{2}= v_{0}^{2}-2gh$\r\n\r\n$\\Rightarrow 1-\\frac{v_{0}^{2}-2gh}{gR}> \\cos\\theta$\r\n\r\n$\\Rightarrow 1-\\frac{4gR-2gh}{gR}> \\cos\\theta$\r\n\r\n$\\Rightarrow \\frac{-3R+2h}{R}> \\cos\\theta$\r\n\r\n$\\Rightarrow \\frac{-3R+2h}{R}> \\cos\\theta \\ge-1$\r\n\r\n$\\Rightarrow-3R+2h >-R$\r\n\r\n$\\Rightarrow 2h > 2R$\r\n\r\n$\\therefore$ If $h > R$ then the car leaves the track." } { "Tag": [ "LaTeX" ], "Problem": "Evaluate: $ \\frac{2}{7} \\cdot \\frac{5}{9} \\plus{} \\frac{5}{7} \\cdot \\frac{5}{9}$. Express your answer as a common fraction.", "Solution_1": "Instead of multiplying it out, we see that we can factor out $ \\frac{5}{9}$. This gives us $ \\frac{5}{9}(\\frac{2}{7}\\plus{}\\frac{5}{7})\\equal{}\\frac{5}{9}(1)\\equal{}\\boxed{\\frac{5}{9}}$", "Solution_2": "hello, we get $ \\frac{10}{63}\\plus{}\\frac{25}{63}\\equal{}\\frac{35}{63}\\equal{}\\frac{5}{9}$\r\nSonnhard.", "Solution_3": "[quote=\"jxl28\"]Instead of multiplying it out, we see that we can factor out $ \\frac {5}{9}$. This gives us $ \\frac {5}{9}(\\frac {2}{7} + \\frac {5}{7}) = \\frac {5}{9}(1) = \\boxed{\\frac {5}{9}}$[/quote]\r\n\r\nreally random $ \\text{\\LaTeX}$ hint. Use [code]$\\left(\\right)$[/code] to make the parentheses bigger and actually cover the fractions." } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO" ], "Problem": "Some Mock AIME questions are sufficiently poorly worded that I would like to re-word them. I obviously am not going to go around doing this for \"real\" contests, but is it okay to do this for the Mock AIME?", "Solution_1": "Yeah, go ahead. I don't see why anyone would have problems with that. :lol:", "Solution_2": "On a similar note, would it be acceptable to enter paraphrased USAMO questions from before 1989 (since I don't think we have the original wording from those years), with a disclaimer that the question is paraphrased?", "Solution_3": "I guess it would be ok if the original wording is unavailable.", "Solution_4": "[quote=\"Boy Soprano II\"]On a similar note, would it be acceptable to enter paraphrased USAMO questions from before 1989 (since I don't think we have the original wording from those years), with a disclaimer that the question is paraphrased?[/quote]Unavailable? :huh: I don't think you should post the problems if you don't have the original version.\r\n\r\nI have 1972-1986. If I have time I'll try to post some, although I'm going to be very busy this year." } { "Tag": [], "Problem": "A plane has a speed of \"p\" km/hr. when flying in still air. The plane flies from New York City to Washington D.C., and then back to New York again, with and against a wind of speed \"w\" km/hr. find the average speed of the plane.", "Solution_1": "To solve this, we must use [url=http://www.artofproblemsolving.com/Wiki/index.php/Harmonic_mean]harmonic mean[/url].\r\n\\[ \\dfrac{2}{\\frac {1}{p \\plus{} w} \\plus{} \\frac {1}{p \\minus{} w}} \\equal{} \\frac {p^2 \\minus{} w^2}{p}\r\n\\]\r\nThus, our answer is $ \\frac {p^2 \\minus{} w^2}{p}$ km/hr.", "Solution_2": "You should hide your solution so as to give others the opportunity to solve it. It's $ \\frac {p^2 \\minus{} w^2}{p}$, not \\frac {p^2 - w^2}{2p}. The 2's in the numerator and denominator cancel out.", "Solution_3": "You should've used the \"*\" symbol if you already knew the answer. Then, you are supposed to hide your solution as it is of no use to anyone else. I assumed you didn't know the answer which is why I didn't hide my solution.", "Solution_4": "Wait,is it against regulations to put problems you made on the Middle School section. In that case, I should stop.", "Solution_5": "I don't think it's against the regulations... :ninja:" } { "Tag": [ "LaTeX" ], "Problem": "How to type the dollar signs in the forums (not in LaTeX)?", "Solution_1": "It would be nice if you would read the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=51330]announcements[/url] in this forum and then ask questions (if there are any left :P)." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $n$ be a positive integer. An [i]a-stump[/i] of $n$ is the integer gained by removing one or more digits from the right hand side of the base $a$ representation of $n$. Let $T_{a}(n)$ be the sum of all the [i]a-stumps[/i] of $n$. If $S_{a}(n)$ is the sum of the digits in the base $a$ representation of $n$, prove that $n=S_{a}(n)+(a-1)T_{a}(n)$", "Solution_1": "Let the digits on $n$ in base $a$ be $n_{k}, n_{k-1},\\cdots,n_{2},n_{1},n_{0}$. Then $n=\\sum_{i=0}^{k}a^{i}n_{i}$.\r\nWe say that the jth a-stump of $n$ is the one which removes the last $j$ digits. Then the jth a-stump is $\\sum_{i=j}^{k}a^{i-j}n_{i}$. So we have\r\n\\[T_{a}(n)=\\sum_{j=1}^{k}\\sum_{i=j}^{k}a^{i-j}n_{i}=\\sum_{i=1}^{k}n_{i}\\sum_{j=1}^{i}a^{i-j}=\\sum_{i=1}^{k}n_{i}\\left(\\frac{a^{i}-1}{a-1}\\right).\\]\r\nSo finally,\r\n\\[(a-1)T_{a}(n)=\\sum_{i=1}^{k}\\left(a^{i}-1\\right)n_{i}=\\sum_{i=0}^{k}a^{i}n_{i}-\\sum_{i=0}^{k}n_{i}=n-S_{a}(n).\\]", "Solution_2": "exactly. it can be solved with just plain manipulations. so isn't the prob too easy? i mean, for APMO...and in the exam it was to be proven only for base 10." } { "Tag": [], "Problem": "Say you have a huge speech due on Monday (wow reminds me of something that happened last year). You just finished writing it on Sunday and now you have to memorize it. How would you go about doing this?\r\n\r\nI really don't know what I would do. Since palm cards don't help me much. I guess I will just try and memorize as much as possible (after all I am the one who wrote it), get an early night sleep and hope for the best. :P", "Solution_1": "If I wasn't winging it (which is what I normally do) I would almost certainly read it line by line and repeat it until I was confident on that section. Then I'd move onto the next section (although still saying the first section out loud again)\r\n\r\nHowever, if you are pushed for time. Learn the flow of the argument and talk naturally following the flow.", "Solution_2": "Huh. Well...that's a problem.\r\nJust go with the flow, remember what your main points and statistics/facts and go for it. \r\nThat's what I usually do, if it's not complete improv. I manage to fool the teachers...:D", "Solution_3": "Another thing I tend to do is use humour so I remember it better. :D", "Solution_4": "Hmm, in reading and study skills (yes, it's a class) we discussed ways to memorize things. Apparently, the best way is to start with what it begins with, create a weird/interesting (so you remember it) picture in your head of the next thing, do it again for the next, and keep doing that for all the major parts.\r\n\r\nAlso, if you just say it over and over, it gets recorded in your first level memory, if someone keeps saying it to you, it gets recorded in your second level memory, if you write it down in addition to the others, it gets recorded in your third level memory, and I forgot what to do to keep it in your fourth level memory :roll:" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find n<>0 natural if 210 divides 6+6^2+ .......+6^n.", "Solution_1": "The problem is equivalent to determining $n$ such that\r\n\\[ 5^2 \\cdot 7 | 6^n - 1 \\]\r\nObserve that\r\n\\[ 6^n - 1 \\equiv ( - 1 )^n - 1 ( \\mbox{mod } 7 ) \\]\r\nSo $n$ has to be even. Moreover\r\n\\[ 6^n \\equiv 1 ( \\mbox{mod } 25 ) \\Longleftrightarrow 5| n \\]\r\nTherefore $n$ must be a multiple of $10$." } { "Tag": [ "function", "algebra", "domain", "complex analysis", "topology", "complex analysis unsolved" ], "Problem": "Hi all,\r\n\r\nBelow are two problems in which one is asked to prove the constancy of a given function, here we go:\r\n\r\n(1) Suppose the function f is analytic on and inside a simple closed curve C, f(z) is never zero for all z on and inside C, and |f(z)| is constant on C. Prove that f is a constant function.\r\n\r\n(2) Suppose that f(z) is an entire function such that |f(z)| = 1 for all |z| = 1, and f(z) is never 0 inside the open unit disc centred at the origin. Prove that f is a constant function.\r\n\r\nI was thinking if it was possible to come up with functions (or composite of functions) to show that in fact f is bounded on the whole complex plane and then conclude by Liouville. But I haven't managed to solve these problems so far. It's not hard to see that the two problems are in fact very similar. Another problem I encountered and solved is as follows, though I'm not sure if the solution is correct:\r\n\r\n(3) Suppose that g is an entire function and Im g(z) + Re g(z) > 1 for all z in the complex plane. Prove that g is a constant function.\r\n\r\nMy solution runs as follows: Im g + Re g > 1 implies that g maps the complex plane to the half plane above the line y = -x + 1, which we shall denote by U. As U is a simply connected domain, we can find an analytic isomorphism from U to the open unit disc centred at the origin, say h.\r\n\r\nThus h(g(C)) is contained in h(U), the open unit disc. Observe that h(g(z)) (viewed here as a composite function) is entire and bounded. We conclude by using Liouville's Theorem that g is in fact constant throughout the complex plane.\r\n\r\nCan anyone help me with these 3 problems?", "Solution_1": "(2) is a special case of (1). Let's consider (1). If $ |f(z)| \\equal{} K$ on $ C$, then $ |f(z)|\\le K$ inside $ C$, by the maximum principle. But $ 1/f$ is analytic too, so $ |1/f(z)|\\le 1/K$ inside $ C$. Thus $ |f(z)| \\equal{} K$ everywhere inside $ C$. However, the image of an open set under a nonconstant analytic function is open. (Either that, or use the form of maximum principle with an equality case.)", "Solution_2": "Hi,\r\n\r\n I got the first part too, but I do not quite get your last part of your argument. Could you please clarify this a little?" } { "Tag": [ "abstract algebra", "superior algebra" ], "Problem": "Let G be a finite group and let I ={g in G: g^2 = e} \\ {e} be its subset of involutions. Show that G is abelian if card(I) => (3/4)card(G).", "Solution_1": "A solution is given at http://mathforum.org/library/drmath/view/66879.html when you take T to be the identity automorphism.\r\n\r\nThis is a theorem of G.A.Miller from around 1906.\r\n\r\nG. A. Miller. \"Note on the Possible Number of Operators of Order 2 in a Group of Order 2^m\"\r\nThe Annals of Mathematics, Second Series, Vol. 7, No. 2 (Jan., 1906), pp. 55-60\r\nhttp://www.ams.org/mathscinet-getitem?mr=1502317\r\nhttp://dx.doi.org/10.2307/1967420\r\n\r\nMiller's proof is: Such a G is generated by its involutions. Suppose by way of contradiction that H is not abelian. Then G must have a non-central involution, t. Consider H=C_G(t). If g an element of order 2 in G-H, then gt cannot have order 2, since is abelian. Hence for every element of order 2 in G-H, there is an element of order higher than 2 in G-H. Since G-H contains at least |G|/2 elements, at least 1/4 of the elements of |G| are of order larger than 2, but this is a contradiction.", "Solution_2": "See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=27046]this thread[/url]." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "give 0 0$ and $ a \\plus{} b \\plus{} c \\equal{} 3$; prove that: \r\n$ (3a \\plus{} \\sqrt {9a^2 \\plus{} 16})(3b \\plus{} \\sqrt {9b^2 \\plus{} 16})(3c \\plus{} \\sqrt {9c^2 \\plus{} 16})\\leq 512 \\equal{} 8^3$", "Solution_1": "let $ 3a\\equal{}x,3b\\equal{}y$ and $ 3c\\equal{}z$.\r\nconsider the function $ f$ defined for every positive real number $ t$ by $ f(x) \\equal{}ln(t \\plus{} \\sqrt {t^2 \\plus{} 16})$ we have $ f''(x)\\equal{}\\minus{}1/2(x^2\\plus{}16)^{\\minus{}\\frac{3}{2}}2x<0$ so $ f$ is concave and by jensen's inequality $ f(x)\\plus{}f(y)\\plus{}f(z) \\le 3f(3)\\equal{}3ln(8).$", "Solution_2": "[quote=\"tchebytchev\"]let $ 3a \\equal{} x,3b \\equal{} y$ and $ 3c \\equal{} z$.\nconsider the function $ f$ defined for every positive real number $ t$ by $ f(x) \\equal{} ln(t \\plus{} \\sqrt {t^2 \\plus{} 16})$ we have $ f''(x) \\equal{} \\minus{} 1/2(x^2 \\plus{} 16)^{ \\minus{} \\frac {3}{2}}2x < 0$ so $ f$ is concave and by jensen's inequality $ f(x) \\plus{} f(y) \\plus{} f(z) \\le 3f(3) \\equal{} 3ln(8).$[/quote]\r\n\r\n\r\n\r\nI thinks we should have a solution by Cauchy or Bunhiacopxki; ...", "Solution_3": "I think if the stronger is trues, it'll be easier \r\n$ \\sum\\ (3a \\plus{} \\sqrt {9a^2 \\plus{} 16})(3b \\plus{} \\sqrt {9b^2 \\plus{} 16}) \\leq 675$\r\nIs it trues ? :wink:", "Solution_4": "[quote=\"nguoivn\"]I think if the stronger is trues, it'll be easier \n$ \\sum\\ (3a \\plus{} \\sqrt {9a^2 \\plus{} 16})(3b \\plus{} \\sqrt {9b^2 \\plus{} 16}) \\leq 675$\nIs it trues ? :wink:[/quote]\r\nI think 192 not 675 because of a=b=c=1\r\ncan you write your solution? \r\nthank you", "Solution_5": "Sorry, I posted wrong. 192, not 675.\r\nI don't have solution for it. I try to expand and uue Am-Gm but I don't obtain any result.\r\nHowever, there are many ways to use Am-Gm which I haven't to try. So, I think somebody 'll find a solution for it :)" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "geometric transformation", "dilation" ], "Problem": "Let $ \\mathcal Q$ be a unit cube. We say that a tetrahedron is [b]good[/b] if all its edges are equal and all of its vertices lie on the boundary of $ \\mathcal Q$. Find all possible volumes of good tetrahedra.", "Solution_1": "This has nothing to do with combinatorics at all. Also, it's very easy: any such tetrahedron has side length at most $ \\sqrt{2}$, by comparison of circumradii. There is such a tetrahedron, $ T$: take 4 vertices of the cube such that no two are adjacent. Then we can get tetrahedra of any smaller volume by a dilation of $ T$ centered at any of its vertices." } { "Tag": [ "USAMTS", "Ross Mathematics Program", "induction", "quadratics", "AMC", "AIME", "ratio" ], "Problem": "What were your round 2 scores?\r\n\r\nMine were 455[b]55[/b]. I lost a point for #1 because I didn't explicitly say why $ n\\leq 7$ doesn't work.", "Solution_1": "Got 25 again, but no commended this time.", "Solution_2": "553-5\r\n\r\nGot a \"good execution\" for number 1 but no commended.\r\n\r\nIs the following a valid lemma for 3?\r\n\r\n\"Two 4-row remarkable triangular arrays with the same top row are identical.\"", "Solution_3": "5 5 5 5 [b]5[/b]\r\n\r\nMy solution to number one was a pure program. lol", "Solution_4": "5 5 5 5 4\r\n\r\nhad a nice grader for #5, I should try to get commends...\r\n\r\ntotal: 24+24=48\r\n\r\nI could get gold if I try hard enough...", "Solution_5": "5/5/5/5/3\r\n\r\nmy friend got a 4 for #5, and it turns out we made the same mistake...=(\r\n\r\n25+23=48. Still in contention for gold =D", "Solution_6": "I got 21 on the first round, so I have 45 now. I'm hoping that I can pull off two 25s in the last two rounds in order to make gold.", "Solution_7": "woohoo\r\n\r\n5 5 [b]5 5[/b] 5\r\n\r\ndidn't take off for typos on #5 :)", "Solution_8": "[quote=\"lingomaniac88\"]I got 21 on the first round, so I have 45 now. I'm hoping that I can pull off two 25s in the last two rounds in order to make gold.[/quote]\n\nThe cutoff for gold is 96.\n\n[quote=\"worthawholebean\"]\"Is the following a valid lemma for 3?\n\nTwo 4-row remarkable triangular arrays with the same top row are identical.\"[/quote]\r\n\r\nWell yeah, isn't this trivial?", "Solution_9": "[b]5[/b]5[b]55[/b]4\r\n\r\nWow... 6 commends .. (guess who I am!) :lol:", "Solution_10": "[b]5[/b] 5 [b]5[/b] 5 [b]5[/b] !\r\n\r\nWow...I didn't expect so many (to me) commends, especially for no.1...it's sheerly bruteforce...", "Solution_11": "555[b]5[/b]5", "Solution_12": "[quote=\"CatalystOfNostalgia\"][quote=\"worthawholebean\"]\"Is the following a valid lemma for 3?\n\nTwo 4-row remarkable triangular arrays with the same top row are identical.\"[/quote]\n\nWell yeah, isn't this trivial?[/quote]\r\nGrader said it was invalid.", "Solution_13": "[b]5[/b] 5 5 [b]5[/b] [b]5[/b]\r\n\r\nNow I have to look at the third round...", "Solution_14": "5 5 5 [b]5 5[/b]\r\n\r\nAmazingly, my four page solutions both got commended, which I thought was kind of strange.", "Solution_15": "[quote=\"buzzer11\"]I don't know, I could be wrong as well. I don't know what ordered/unordered counting is, either.[/quote]\r\n\r\nConsidering the talk about complex sums that are hard to calculate, it looks as if your proof would qualify as ordered counting.", "Solution_16": "[quote=\"mountain.dew\"]Does anyone know the difference between unordered counting and ordered counting? :huh:[/quote]\nazjps's solution, and case 2 of buzzer11's solution are ordered counting. The nature of ordered counting is that the exact position of each point matters. Each point is indexed with a number that describes its position, and the number of triangles is expressed as a nested summation that sums over the indicies.\n\nCase 1 of buzzer11's solution is unordered counting. The nature of unordered counting is to look at all the points involved in forming a triangle and see how many other triangles use the same set of points in different roles. Then the problem of counting triangles reduces to the problem of counting sets of points on the arcs.\n\n[quote=\"buzzer11\"]Here's what I did: Diagram is attached.[/quote]\r\nI can stop talking about about the rubric and how a typical grader would grade buzzer11's solution. It looks familiar, so I think I graded that one myself. (Note: I always initial my comments \"EJS\", so people can tell what I graded.) That problem was so close to earning four points that I started to expand the triple summation at the end so that I cound spot the computational error and take only one point off. But due to the quadratic terms in the summation, the expansion was getting quite messy and I realized that duplicating the calculations would take at least 15 minutes. I could not afford 15 minutes. So I stuck to the rubric and took two points off.\r\n\r\nErin Schram\r\nUSAMTS grader", "Solution_17": "Yes, you did grade my solution (\"EJS\" was written in the comment box).\r\nI think I understand unordered vs. ordered counting now.\r\nAnd oh well about the three points. :) I'll just try and do well enough this round to make the AIME cutoff.", "Solution_18": "You could just take AMC to make AIME. I think 100 in AMC 12 or 120 on AMC 10 is a lot easier than 68 on USAMTS.", "Solution_19": "Actually that depends. Considering the time allotment for USAMTS, everyone has enough time to do serious research if necessary, and some people just work better if there's a lot of time. Also, it's good too have a safety net just in case, say, you get the flu on both AMC test dates. (Highly unlikely, just a hypothetical example) The cutoffs of both competitions are pretty equivalent. Someone who can get a 100 on the AMC 12 should be able to get 68 on USAMTS.", "Solution_20": "I tend to think that scoring a 100 on the AMC 12 is much easier than scoring a 68 on the USAMTS. After all, about 10000 people a year make it through the AMC 12, versus a couple hundred (at most) for the USAMTS. Naturally that statistic is skewed due to the difference in the total number of total participants, but still.. USAMTS requires effort that many students wouldn't be willing to give during their free time but the AMCs only take up an hour and half of time.", "Solution_21": "Yes, I agree that the AMC 12 is easier. However, considering that I screwed up badly on last year's AMC 12 and missed the cutoff by a few points, I want to have a back up way into the AIME.", "Solution_22": "[quote=\"azjps\"]After all, about 10000 people a year make it through the AMC 12, versus a couple hundred (at most) for the USAMTS. Naturally that statistic is skewed due to the difference in the total number of total participants[/quote]\r\n\r\nConsider the ratios of students making it from both competitions. Assuming a low estimate of 100 people from USAMTS as you stated, and considering about 500 USAMTS participants, we've got around 20% of people making it to AIME. However, from the AMC 12, it's only supposed to be the top 5% or so, so from this statistic, a much higher percentage of people make it from USAMTS. I still think both are comparable when considering both difficulty and time allotment.", "Solution_23": "There's a really big difference though between the average person who takes the AMC 12 and the average person who does the USAMTS.", "Solution_24": "Hmm, looks like we're not talking about Round 2 anymore...\r\n\r\nYou can't really compare percent of qualifiers from the two tests. The people that do AMC are from all around the nation and probably aren't as bright as those who do USAMTS. \r\n\r\nAnd while we're at it...What happens if you take the AMC 10 and AMC 12?", "Solution_25": "[quote=\"le00327146\"]And while we're at it...What happens if you take the AMC 10 and AMC 12?[/quote]\r\n\r\nIt's possible to qualify through both ways. But you can only take one of each (like 10A and 12B). Obviously, you need to earn a higher score on the 10 compared to the 12 in order to qualify for AIME.", "Solution_26": "[quote=\"vishalarul\"][quote=\"le00327146\"]And while we're at it...What happens if you take the AMC 10 and AMC 12?[/quote]\n\nIt's possible to qualify through both ways. But you can only take one of each (like 10A and 12B). Obviously, you need to earn a higher score on the 10 compared to the 12 in order to qualify for AIME.[/quote]\r\n\r\nThey take your better score right?", "Solution_27": "They'll take whichever score makes you qualify more easily into the upper rounds, since you can't exactly find a correlation between an AMC 10 and AMC 12 score.", "Solution_28": "5,3,3,3,2\r\n\r\ni'm so horrible at this contest, but it's my first year doing this -.-;", "Solution_29": "Don't worry, you're not horrible. It's just that all the people who post on AOPS are the ones who get all the high scores. There's a bunch of people with lower scores who don't bother with (or don't know about) the forum." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$a,b,c>0,\\ abc=1\\Longrightarrow 27\\le \\overline {\\underline {\\left|\\ 3\\sum a\\cdot \\sum \\frac 1a\\le 21+\\sum \\left(a^3+\\frac{1}{a^3}\\right)\\ \\right| }}.$", "Solution_1": "$3\\sum a\\cdot \\sum \\frac{1}{a}\\le 21+\\sum \\left(a^{3}+\\frac{1}{a^{3}}\\right)\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(a^{4}bc+a^{3}b^{3}-3a^{3}b^{2}c-3a^{3}c^{2}b+4a^{2}b^{2}c^{2})\\geq0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(c^{3}a^{3}-c^{3}a^{2}b-c^{3}b^{2}a+c^{3}b^{3})+\\sum_{cyc}(a^{4}bc-a^{3}b^{2}c-b^{3}a^{2}c+b^{4}ac)-$\r\n$-4\\cdot\\sum_{cyc}(a^{3}c^{2}b-2a^{2}b^{2}c^{2}+b^{3}c^{2}a)\\geq0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(a-b)^{2}c((a+b)c^{2}-4abc+(a+b)ab)\\geq0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(a-b)^{2}c\\left((a+b)(c-\\sqrt{ab})^{2}+2\\sqrt{ab}c(\\sqrt a-\\sqrt b)^{2}\\right)\\geq0.$ :)" } { "Tag": [ "AMC", "AIME", "ARML", "USAMTS" ], "Problem": "Other than old AIME problems and mock AIME's, what are some good practice problems around the level of AIME problems 7-12? any suggestions for books or other problems?\r\n\r\nthanks. what are you guys doing to prep for AIME?", "Solution_1": "AoPS v2 if you haven't done it already is great practice for AIME.\r\n\r\nThe Intermediate/Pre-Olympiad forums have plenty of problems all the time to do, too.\r\n\r\nThe AIME Problem Series (and Seminar in the spring I think) give good practice, but you might encounter problems you have already solved (not sure about this).\r\n\r\n... and I'm sure there are plenty of other things to practice with that I can't think of right now.", "Solution_2": "From what I've seen, ARML problems are good too. But trust me, just doing the old AIME problems especially with recent years having I and II, you will have lots of practice, especially if you work on most of them as if you were taking the actual test.", "Solution_3": "[quote=\"probability1.01\"]From what I've seen, ARML problems are good too. But trust me, just doing the old AIME problems especially with recent years having I and II, you will have lots of practice, especially if you work on most of them as if you were taking the actual test.[/quote]\r\n\r\nAgree. Emphasis on \"as if you were taking the actual test\".", "Solution_4": "What about Thomas and other people's awesome Mock AIME?!!\r\n\r\nAlso, USAMTS types are very good because they usually require thinking. Some problems may not be hard as later AIME problems but again, if they were to be solved in 3 hours, who knows? (assuming the calculation works out nice way in integers)\r\n\r\nThere are many contests out there in different nations as well. You can even use them to prepare for AIME. Well, at least that's what I would do except I'm mainly concentrating on scoring high on AMC-12 now. :lol:", "Solution_5": "[quote=\"Silverfalcon\"]Also, USAMTS types are very good because they usually require thinking. Some problems may not be hard as later AIME problems but again, if they were to be solved in 3 hours, who knows? (assuming the calculation works out nice way in integers)[/quote]\r\n\r\nI disagree. USAMTS problems are usually much harder than AIME problems. If you just want to do the actual problem, not the proof, then maybe you could use some USAMTS. But in general, I don't think USAMTS are very good for AIME practice. Just stick to AIME. There are so many of them that you can do one every week and still not run out by the time you take the real AIME. (If you start now.)" } { "Tag": [ "algebra", "polynomial", "function", "continued fraction", "algebra proposed" ], "Problem": "The question is to find the value of this continued fraction:\r\n\r\n$1+\\frac{1}{2+\\frac{1}{3+\\frac{1}{4+\\frac{1}{5+...}}}}$\r\n\r\nIt goes on forever in this arithmetic sequence.\r\n\r\nI need someone's help to find an exact value. I suspect it's either the solution of an irreducible polynomial of degree 3 or greater, or transcendental.\r\n\r\nI was able to get the following approximation:\r\n$\\frac{7489051}{5225670}$", "Solution_1": "Wait, according to the site:\r\n[url]http://mathworld.wolfram.com/ContinuedFractionConstant.html[/url]\r\nthis is equal to:\r\n$\\frac{I_0(2)}{I_1(2)}$.\r\n\r\nAlthough I don't understand modified Bessel functions (can someone explain?), and mathematica refuses to simplify this result, I suppose this may be as exact an answer as I'll get. Too bad.", "Solution_2": "OK, looks like that is equal to:\r\n\r\n$\\frac{\\sum_{k=0}^\\infty \\frac{1}{(k!)^2}}{\\sum_{k=0}^\\infty \\frac{1}{(k)!(k+1)!}}$\r\n\r\nOf course, this is not evaluable in exact form. I'll just leave it at that... Ugh." } { "Tag": [ "geometry", "perimeter", "geometry unsolved" ], "Problem": "Let $ P$ be a convex polygon with sides of integer lenght and odd perimeter. Prove that the area of $ P$ is greater than or equal to $ \\frac{\\sqrt{3}}{4}$.", "Solution_1": "The $ P$ of minimal area is the equilateral triangle of side 1, perimeter 3 and area $ {{\\sqrt 3} \\over 4}$.\r\n\u00c9 isso ?", "Solution_2": "Could you show your work?" } { "Tag": [ "floor function", "combinatorics proposed", "combinatorics" ], "Problem": "There are several scientists collaborating in Niichavo. During an $ 8$-hour working day, the scientists went to cafeteria, possibly several times.It is known that for every two scientist, the total time in which exactly one of them was in cafeteria is at least $ x$ hours ($ x>4$).\r\nWhat is the largest possible number of scientist that could work in Niichavo that day,in terms of $ x$?", "Solution_1": "I guess the answer is $ \\left\\lfloor\\frac{16}x\\right\\rfloor$.\r\n\r\nAssume that $ \\frac{16}3\\frac{16}3,a\\plus{}d\\plus{}c\\plus{}e>\\frac{16}3,b\\plus{}d\\plus{}c\\plus{}f>\\frac{16}3$. Sum up the three equations to get $ 2(a\\plus{}b\\plus{}c\\plus{}d\\plus{}e\\plus{}f)>16$, or $ a\\plus{}b\\plus{}c\\plus{}d\\plus{}e\\plus{}f>8$. On the other hand, we must have $ a\\plus{}b\\plus{}c\\plus{}d\\plus{}e\\plus{}f\\plus{}g\\le8$, contradiction. Therefore, there are only 2 scientists at most. But $ \\left\\lfloor\\frac{16}x\\right\\rfloor\\equal{}2$ for $ \\frac{16}3i$ represent the time for which exactly one of the scientists $i$ or $j$ is in the cafeteria. Let $T=\\sum t_{i,j}$. Now for each interval we assign a weight based on its length of time. For every interval with $k$ scientists and weight $w$, this interval contributes $wn(n-k)$ to the sum $T$. For example, suppose there is the set $\\{1,2\\}$ consisting of $2$ elements and taking time $w$. Then this contributes $w$ to each of $t_{1,j}$ where $j\\ne 2$ and $t_{i,2}$ where $i\\ne 1$ so it is $w(2)(n-2)$ overall. Since we are trying to maximize the time for the given number of scientists, we are really trying to maximize $\\min(t_{i,j})$. Therefore, the way to maximize is to first maximize the sum $T$, then make all of the $t_{i,j}$ equal so the minimum is maximized too. However to maximize $T$, since the total time has to be $8$ hours and an interval with $k$ scientists contributes $wk(n-k)$, the best we can do is maximize $k(n-k)$ by setting $k= \\left\\lfloor \\frac {n}{2}\\right\\rfloor$ and using no other intervals, so essentially being greedy. Then the sum will be $8\\left\\lfloor \\frac {n}{2}\\right \\rfloor(n-\\left\\lfloor \\frac {n}{2} \\right\\rfloor)$. So $\\min(t_{i,j})=\\frac {16\\left\\lfloor \\frac {n}{2}\\right \\rfloor(n-\\left\\lfloor \\frac {n}{2} \\right\\rfloor)}{(n)(n-1)}$ which is the maximum number of hours for a given number of scientists. The construction is simply to divide the $8$ hours into $n\\choose {\\left\\lfloor \\frac {n}{2} \\right\\rfloor}$ equal intervals and do all possible sets of $\\left\\lfloor \\frac {n}{2} \\right\\rfloor$, this guarantees that the $t_{i,j}$ will all be equal. \n\nNow the original problem asks for $n$ in terms of $x$. For $\\frac{8k+8}{2k+1}4$ condition is given." } { "Tag": [ "calculus", "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "1^1+2^2+3^3+4^4+....+n^n = ? mod ( n ) \r\n\r\nfind the solution for n=1000....\r\n\r\nIs there any general case for this problem ?", "Solution_1": "[quote=\"epsilon07\"]1^1+2^2+3^3+4^4+....+n^n = ? mod ( n ) \n\nfind the solution for n=1000....\n\nIs there any general case for this problem ?[/quote]\r\n\r\nI mean n=1000 \r\n\r\nCan anybody solve this? I need the solution...", "Solution_2": "[quote=\"epsilon07\"]1^1+2^2+3^3+4^4+....+n^n = ? mod ( n ) \n\nfind the solution for n=1000....\n\nIs there any general case for this problem ?[/quote]\r\n\r\nI dont think there is a general method.\r\n\r\nDirect calculus gives the result : $ 700$\r\n\r\nYou can improve the calculus speed writing $ (100n\\plus{}a)^{100n\\plus{}a}\\equal{}a^aa^{100n}(100n\\plus{}1)\\pmod{1000}$", "Solution_3": "[quote=\"pco\"][quote=\"epsilon07\"]1^1+2^2+3^3+4^4+....+n^n = ? mod ( n ) \n\nfind the solution for n=1000....\n\nIs there any general case for this problem ?[/quote]\n\nI dont think there is a general method.\n\nDirect calculus gives the result : $ 700$\n\nYou can improve the calculus speed writing $ (100n \\plus{} a)^{100n \\plus{} a} \\equal{} a^aa^{100n}(100n \\plus{} 1)\\pmod{1000}$[/quote]\r\n\r\n\r\n\u0131 got it, thanks ... is there any other solution?" } { "Tag": [ "calculus", "integration", "geometry", "function", "derivative", "analytic geometry", "real analysis" ], "Problem": "Let $f$ be a continuous function on the plane with real values such that the integral of $f$ over each square of area 1 in the plane is 0. What can we say about $f$?", "Solution_1": "Let $g(x,y)=\\int_x^{x+1}\\int_y^{y+1}f(u,v)\\, dv\\, du$.\r\nBy the Fundamental Theorem, $g_{yx}(x,y)=f(x+1,y+1)-f(x,y+1)-f(x+1,y)+f(x,y)$. Since we know that $g=0$, its derivatives are zero.\r\n\r\nLet $p(x,y)$ be any continuous function on $[0,1]^2$ with integral 0 and $\\int_0^1 p(0,y)-p(1,y)\\, dy=0=\\int_0^1 p(x,0)-p(x,1)\\, dx.$\r\nLet $h(x,y)$ be any continuous function on $[0,1]^2$ with $\\int_0^1 h(x,y)\\, dy=0$ for all $x$, and $h(0,a)=h(1,a)=p(1,a)-p(0,a), h(b,0)=h(b,1)$ for all $a,b.$\r\nLet $v(x,y)$ be any continuous function on $[0,1]^2$ with $\\int_0^1 v(x,y)\\, dx=0$ for all $y$, and $v(0,a)=v(1,a), v(b,0)=v(b,1)=p(b,1)-p(b,0)$ for all $a,b.$\r\n\r\nNow, for $x,y\\in [0,1)$, let $f(x+m,y+n)=p(x,y)+mh(x,y)+nv(x,y)$. This $f$ satisfies the conditions, and any $f$ that satisfies the conditions can be uniquely expressed in this form.\r\n\r\nThis still allows for some rather strange examples.", "Solution_2": "Hmmm,... My understanding was that the sides of square are not necessarily parallel to the coordinate axes, which makes the problem completely different...", "Solution_3": "I considered an analogous problem some years ago (with disks instead of squares), but I didn't find a final answer. My progress with the disk was the following: the assumption means that the convolution of $f$ with the characteristic function of a disk is zero, and so its Fourier transform is zero. But the Fourier transform of a convolution is just a product and so the Fourier transform of $f$ must be zero (you need to prove the the Fourier transform of the characteristic function is non-zero). The problem is that some summability assumptions on $f$ are needed in order to compute the transform. I would also appreciate a more elementary approach.", "Solution_4": "There is a beautiful (but not really elementary) theorem that says that any shift-invariant closed subspace $E$ of $C(\\mathbb R)$ is just the closure of all exponential polynomials contained in $E$. Unfortunately, I do not remember whether the result holds in $\\mathbb R^2$. :blush: :(" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that the sides $a,b,c$ of a triangle satisfythe inequality\r\n\r\n$5(ab^2+bc^2+ca^2)\\ge 3(a^2b+b^2c+c^2a)+6abc$", "Solution_1": "Suppose $a\\ge b,c$. Rewrite the inequality as\\[3(a-b)(a-c)(c-b)+2(\\sum ab(b-c))\\ge0\\]\r\nIf $c\\ge b$, then we are done.\r\nAssume $a\\ge b\\ge c$. Let a=c+x+y,b=c+x,$x,y\\ge0$. Since a,b,c are sidelengths of a triangle, $c\\ge y$.\r\nNow substitue this into the inequality, we obtain \\[(c+x)(c+x+y)x+cy(x+y)\\ge3xy(x+y)\\]This is true since $(c+y)(c+x+y)x+cy(x+y)\\ge(x+y)(y+x+y)x+y^2(x+y)=(x+y)(x^2+y^2+2xy)\\ge4xy(x+y)$.\r\n\r\nP.S. Probably there are some computational errors in the proof." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "[color=darkblue][size=150]Let equation: $ (x\\plus{}1).lnx\\minus{}x.ln(x\\plus{}1)\\equal{}0$.\n\nProve that this equation have only one root?[/size][/color]", "Solution_1": "I can prove it ^^ .\r\n$ x \\ge 0$\r\n$ f(x) \\equal{} (x \\plus{} 1)lnx \\minus{} xln(x \\plus{} 1)$\r\n$ f'(x) \\equal{} ln(\\frac {x}{x \\plus{} 1} ) \\plus{} \\frac {x \\plus{} 1}{x} \\minus{} \\frac {x}{x \\plus{} 1}$\r\nSetting :$ t \\equal{} \\frac {x}{x \\plus{} 1} < 1$\r\n$ g(x) \\equal{} lnt \\plus{} \\frac {1}{t} \\minus{} t$\r\n$ g'(x) \\equal{} \\frac {1}{t} \\minus{} \\frac {1}{t^2} \\minus{} 1 \\le 0 \\forall t \\in R$ \r\n=>$ g(x) \\ge g(1) \\equal{} 0$\r\n=>$ f'(x) \\ge 0$\r\n=>...\r\nBut we have :$ f(...) \\equal{} ... > 0$\r\nDone ;)\r\n[hide=\"@thay_Quy\"] \u0110\u01a1n \u0111i\u1ec7u vi\u1ebft sao h\u1ea3 th\u1ea7y ?[/hide]", "Solution_2": "Thank [b]Mitdac123[/b]. You are very good stupid boy.\r\n\r\n[hide=\"For only mitdac123\"] \u0110\u01a1n \u0111i\u1ec7u ti\u1ebfng Anh l\u00e0: Monotone [/hide]", "Solution_3": "[quote=\"thanhnam2902\"]Thank [b]Mitdac123[/b]. You are very good [b]stupid[/b] boy.\n\n[hide=\"For only mitdac123\"] \u0110\u01a1n \u0111i\u1ec7u ti\u1ebfng Anh l\u00e0: Monotone [/hide][/quote]Is this the kind of language allowed in AoPS? Very degrading...\r\n\r\nEDIT: Nevermind, he called himself that way too :what?:", "Solution_4": "Thank you very much." } { "Tag": [ "function", "logarithms", "calculus", "calculus computations" ], "Problem": "Is there a curve that bounds the nth prime function from the left? For instance, $ P_n \\geq 2n\\plus{}1$, would be the right bounds.", "Solution_1": "Are you looking for something like $ 2^n>p_n$?", "Solution_2": "A consequence of the [url=http://en.wikipedia.org/wiki/Prime_number_theorem]prime number theorem[/url] is that $ p_n\\leq \\frac {3}{2}n\\log_2 n$ for $ n\\geq 2$. Albanian Eagle's assertion also follows from the more elementary [url=http://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate]Bertrand's postulate[/url]." } { "Tag": [ "inequalities", "function", "calculus", "derivative", "algebra", "polynomial", "inequalities proposed" ], "Problem": "If a,b,c is positive numbers with abc=1, then \\[ \\frac{ab}{1\\plus{}c^4}\\plus{}\\frac{bc}{1\\plus{}a^4}\\plus{}\\frac{ca}{1\\plus{}b^4}\\ge \\frac{3}{2}\\]", "Solution_1": "After using Am-Gm and Cauchy Schwart, it's becomes:\r\n$ \\frac{a^2\\plus{}b^2\\plus{}c^2}{ab\\plus{}bc\\plus{}ca} \\plus{} \\frac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2} \\geq\\ 2$ (always trues)\r\n :)", "Solution_2": "$ 2^{nd}$ proof:\r\nLet $ x\\equal{} \\frac{a}{b}$...\r\nBy Cauchy Schwart, we only need to prove:\r\n$ \\frac{2}{3} (a^3\\plus{}b^3\\plus{}c^3)^2 \\geq\\ \\sum\\ ab(a^4\\plus{}b^4)$\r\nThere are many ways to prove it :)", "Solution_3": "Your proofs are really magnificent :rotfl: . Remember, not every mathlinkers is as smart as you are.", "Solution_4": "[quote=\"nguoivn\"]$ 2^{nd}$ proof:\nLet $ x \\equal{} \\frac {a}{b}$...\nBy Cauchy Schwart, we only need to prove:\n$ \\frac {2}{3} (a^3 \\plus{} b^3 \\plus{} c^3)^2 \\geq\\ \\sum\\ ab(a^4 \\plus{} b^4)$\nThere are many ways to prove it :)[/quote]\r\nBut it's wrong. :( Try $ a \\equal{} 2$ and $ b \\equal{} c \\equal{} 1.$ :wink:\r\n\r\nWith same conditions $ \\frac {ab}{1 \\plus{} c^7} \\plus{} \\frac {bc}{1 \\plus{} a^7} \\plus{} \\frac {ca}{1 \\plus{} b^7}\\ge \\frac {3}{2}$ is true too.\r\nIt seems that $ \\frac {ab}{1 \\plus{} c^{14}} \\plus{} \\frac {bc}{1 \\plus{} a^{14}} \\plus{} \\frac {ca}{1 \\plus{} b^{14}}\\ge \\frac {3}{2}$ is true .\r\nBut $ \\frac {ab}{1 \\plus{} c^{15}} \\plus{} \\frac {bc}{1 \\plus{} a^{15}} \\plus{} \\frac {ca}{1 \\plus{} b^{15}}\\ge \\frac {3}{2}$ is wrong already: $ a \\equal{} \\frac {1}{1.15^2}$ and $ b \\equal{} c \\equal{} 1.15.$", "Solution_5": "Sorry, arquady. The $ 2^{nd}$ proof was false.\r\nI 'll post full first proof:\r\nLenma: $ 1 \\plus{} x^4 \\geq\\ x(x^2 \\plus{} 1)$\r\n$ LHS \\equal{} ab \\plus{} bc \\plus{} ca \\minus{} \\sum\\ \\frac {c^3}{1 \\plus{} c^4} \\geq\\ ab \\plus{} bc \\plus{} ca \\minus{} \\sum\\ \\frac {c^2}{c^2 \\plus{} 1} \\geq\\ \\frac {3}{2}$\r\n$ \\Leftrightarrow ab \\plus{} bc \\plus{} ca \\plus{} \\sum\\ \\frac {1}{1 \\plus{} a^2} \\geq\\ \\frac {9}{2}$\r\nLet $ a \\equal{} \\frac {x}{z}$...\r\nIt's becomes: $ \\frac {x}{y} \\plus{} \\frac {y}{z} \\plus{} \\frac {z}{x} \\plus{} \\sum\\ \\frac {x^2}{x^2 \\plus{} y^2} \\geq\\ \\frac {9}{2}$\r\nUsing Cauchy Schwart: \r\n$ LHS \\geq\\ \\frac {(x \\plus{} y \\plus{} z)^2}{xy \\plus{} yz \\plus{} zx} \\plus{} \\frac {(x \\plus{} y \\plus{} z)^2}{2(x^2 \\plus{} y^2 \\plus{} z^2)} \\equal{} \\frac {x^2 \\plus{} y^2 \\plus{} z^2}{xy \\plus{} yz \\plus{} zx} \\plus{} \\frac {xy \\plus{} yz \\plus{} zx}{x^2 \\plus{} y^2 \\plus{} z^2 } \\plus{} \\frac {5}{2} \\geq\\ \\frac {9}{2}$ \r\n\r\n[quote=\"Nguyen Ngoc Linh\"]Your proofs are really magnificent :rotfl: . Remember, not every mathlinkers is as smart as you are.[/quote]\nNo, I am only a lazy man :)\n\n[quote=\"arqady\"]\nWith same conditions $ \\frac {ab}{1 \\plus{} c^7} \\plus{} \\frac {bc}{1 \\plus{} a^7} \\plus{} \\frac {ca}{1 \\plus{} b^7}\\ge \\frac {3}{2}$ is true too.\nIt seems that $ \\frac {ab}{1 \\plus{} c^{14}} \\plus{} \\frac {bc}{1 \\plus{} a^{14}} \\plus{} \\frac {ca}{1 \\plus{} b^{14}}\\ge \\frac {3}{2}$ is true .\nBut $ \\frac {ab}{1 \\plus{} c^{15}} \\plus{} \\frac {bc}{1 \\plus{} a^{15}} \\plus{} \\frac {ca}{1 \\plus{} b^{15}}\\ge \\frac {3}{2}$ is wrong already: $ a \\equal{} \\frac {1}{1.15^2}$ and $ b \\equal{} c \\equal{} 1.15.$[/quote]\r\n\r\nWe can use the following way to prove :maybe:\r\nBesides, with $ k \\equal{} 7$ , if using Cauchy Schwart, we 'll have to prove:\r\n$ \\frac {2}{3}(a^4 \\plus{} b^4 \\plus{} c^4)^2 \\geq\\ \\sum\\ ab(a^6 \\plus{} b^6)$\r\nIs it trues ? :wink:", "Solution_6": "This is also true.\r\n\r\nIf a,b,c are positive real numbers with abc=1, then\r\n\r\n\\[ \\frac {ab}{2 \\plus{} c^4} \\plus{} \\frac {bc}{2 \\plus{} a^4} \\plus{} \\frac {ca}{2 \\plus{} b^4}\\ge 1\r\n\\]", "Solution_7": "[quote=\"nguoivn\"]\nWe can use the following way to prove :maybe:\nBesides, with $ k \\equal{} 7$ , if using Cauchy Schwart, we 'll have to prove:\n$ \\frac {2}{3}(a^4 \\plus{} b^4 \\plus{} c^4)^2 \\geq\\ \\sum\\ ab(a^6 \\plus{} b^6)$\nIs it trues ? :wink:[/quote]\nWrong. Try $ a \\equal{} 2$ and $ b \\equal{} c \\equal{} 1.$\n[quote=\"Vasc\"]This is also true.\n\nIf a,b,c are positive real numbers with abc=1, then\n\\[ \\frac {ab}{2 \\plus{} c^4} \\plus{} \\frac {bc}{2 \\plus{} a^4} \\plus{} \\frac {ca}{2 \\plus{} b^4}\\ge 1\n\\]\n[/quote]\r\nI have proved that with same conditions and for all $ k\\geq0$ holds:\r\n\\[ \\frac {ab}{k \\plus{} c^4} \\plus{} \\frac {bc}{k \\plus{} a^4} \\plus{} \\frac {ca}{k \\plus{} b^4}\\ge \\frac{3}{k\\plus{}1}\r\n\\]\r\n :maybe:", "Solution_8": "[quote=\"arqady\"]\nI have proved that with same conditions and for all $ k\\geq0$ holds:\n\\[ \\frac {ab}{k \\plus{} c^4} \\plus{} \\frac {bc}{k \\plus{} a^4} \\plus{} \\frac {ca}{k \\plus{} b^4}\\ge \\frac {3}{k \\plus{} 1}\n\\]\n:maybe:[/quote]\r\nMe,too. But for $ k \\equal{} 2$ there is a nicer solution. :)", "Solution_9": "[quote=\"Vasc\"]This is also true.\n\nIf a,b,c are positive real numbers with abc=1, then\n\\[ \\frac {ab}{2 \\plus{} c^4} \\plus{} \\frac {bc}{2 \\plus{} a^4} \\plus{} \\frac {ca}{2 \\plus{} b^4}\\ge 1\n\\]\n[/quote]\r\nhi Vasc :) \r\nyour inequality is equivalent to : $ \\sum{\\frac{x^3}{2x^2\\plus{}y^2z^2}} \\geq 1$ ( with the same conditions!)\r\nnow by cauchy-shwarz : $ LHS (\\sum{x(2x^2\\plus{}y^2z^2)}) \\geq (x^2\\plus{}y^2\\plus{}z^2)^2$\r\nhence it sufficies to prove that: $ (x^2\\plus{}y^2\\plus{}z^2)^2 \\geq 2(x^3\\plus{}y^3\\plus{}z^3) \\plus{}xyz(x\\plus{}y\\plus{}z)$\r\nor : $ (x^2\\plus{}y^2\\plus{}z^2)^2 \\geq 2(x^3\\plus{}y^3\\plus{}z^3)(xyz)^{1/3} \\plus{} xyz(x\\plus{}y\\plus{}z)$\r\nit sufficies to show the stronger : $ (x^2\\plus{}y^2\\plus{}z^2)^2 \\geq \\frac{2(x^3\\plus{}y^3\\plus{}z^3)(x\\plus{}y\\plus{}z)}{3} \\plus{} x^2y^2\\plus{}x^2z^2\\plus{}y^2z^2$\r\nthat is : $ \\frac{(x^2\\plus{}y^2\\plus{}z^2\\minus{}xy\\minus{}yz\\minus{}xz)^2}{3} \\geq 0$ , which is obvious \r\nfor the general case : notice that the inequality is equivalent to : $ \\sum{\\frac{x^5}{kx^4\\plus{}1} \\minus{} \\frac{3}{k\\plus{}1}} \\geq 0$ , we put $ x\\equal{}e^{a}$ ,and Jensen finishes the proof :) \r\nI hope that I didn't any mistake :(", "Solution_10": "[quote=\"anas\"] your inequality is equivalent to : $ \\sum{\\frac {x^3}{2x^2 \\plus{} y^2z^2}} \\geq 1$ ( with the same conditions!)([/quote]\r\nWhy ?\r\n\r\nAnd why Jensen, since $ f$ is not convex ?", "Solution_11": "[quote=\"Vasc\"][quote=\"anas\"] your inequality is equivalent to : $ \\sum{\\frac {x^3}{2x^2 \\plus{} y^2z^2}} \\geq 1$ ( with the same conditions!)([/quote]\nWhy ?\n\nAnd why Jensen, since $ f$ is not convex ?[/quote]\r\n\r\nfor the first one , put $ x\\equal{}ab$ etc.......\r\nfor the second one , after putting $ x\\equal{}e^{a}$ , consider the function $ f(a)\\equal{}\\frac{e^{5a}}{ke^{4a}\\plus{}1} \\minus{} \\frac{1}{k\\plus{}1}$ the second derivative is : $ \\frac{e^{5a}(k^2(e^{4a})^2\\minus{}6ke^{4a}\\plus{}25)}{(ke^{4a}\\plus{}1)^3}$ :wink:", "Solution_12": "Indeed, you are right Anas. The inequality is trivial by Jensen. :wink:", "Solution_13": "But we can't apply Jensen for the following :\r\n\r\n$ \\frac {ab}{2 \\plus{} c^5} \\plus{} \\frac {bc}{2 \\plus{} a^5} \\plus{} \\frac {ca}{2 \\plus{} b^5}\\ge 1$, \r\n\r\nwhere $ abc\\equal{}1$.", "Solution_14": "[quote=\"anas\"][quote=\"Vasc\"]This is also true.\n\nIf a,b,c are positive real numbers with abc=1, then\n\\[ \\frac {ab}{2 \\plus{} c^4} \\plus{} \\frac {bc}{2 \\plus{} a^4} \\plus{} \\frac {ca}{2 \\plus{} b^4}\\ge 1\n\\]\n[/quote]\nhi Vasc :) \nyour inequality is equivalent to : $ \\sum{\\frac {x^3}{2x^2 \\plus{} y^2z^2}} \\geq 1$ ( with the same conditions!)\nnow by cauchy-shwarz : $ LHS (\\sum{x(2x^2 \\plus{} y^2z^2)}) \\geq (x^2 \\plus{} y^2 \\plus{} z^2)^2$\n[quote]hence it sufficies to prove that: $ (x^2 \\plus{} y^2 \\plus{} z^2)^2 \\geq 2(x^3 \\plus{} y^3 \\plus{} z^3) \\plus{} xyz(x \\plus{} y \\plus{} z)$[/quote]\nor : $ (x^2 \\plus{} y^2 \\plus{} z^2)^2 \\geq 2(x^3 \\plus{} y^3 \\plus{} z^3)(xyz)^{1/3} \\plus{} xyz(x \\plus{} y \\plus{} z)$\nit sufficies to show the stronger : $ (x^2 \\plus{} y^2 \\plus{} z^2)^2 \\geq \\frac {2(x^3 \\plus{} y^3 \\plus{} z^3)(x \\plus{} y \\plus{} z)}{3} \\plus{} x^2y^2 \\plus{} x^2z^2 \\plus{} y^2z^2$\nthat is : $ \\frac {(x^2 \\plus{} y^2 \\plus{} z^2 \\minus{} xy \\minus{} yz \\minus{} xz)^2}{3} \\geq 0$ , which is obvious \nfor the general case : notice that the inequality is equivalent to : $ \\sum{\\frac {x^5}{kx^4 \\plus{} 1} \\minus{} \\frac {3}{k \\plus{} 1}} \\geq 0$ , we put $ x \\equal{} e^{a}$ ,and Jensen finishes the proof :) \nI hope that I didn't any mistake :([/quote]\n\n[quote]hence it sufficies to prove that: $ (x^2 \\plus{} y^2 \\plus{} z^2)^2 \\geq 2(x^3 \\plus{} y^3 \\plus{} z^3) \\plus{} xyz(x \\plus{} y \\plus{} z)$[/quote]\r\n\r\n Why?", "Solution_15": "I made a stupid error :blush: ,I'm always hurry :mad: , I've corrected my proof :\r\n\r\nyour inequality is equivalent to : $ \\sum{\\frac {x^3}{2x^2 \\plus{} y^2z^2}} \\geq 1$ ( with the same conditions!)\r\nnow by cauchy-shwarz : $ LHS (\\sum{x(2x^2 \\plus{} y^2z^2)}) \\geq (x^2 \\plus{} y^2 \\plus{} z^2)^2$\r\nhence it sufficies to prove that: $ (x^2 \\plus{} y^2 \\plus{} z^2)^2 \\geq 2(x^3 \\plus{} y^3 \\plus{} z^3) \\plus{} yz\\plus{}xz\\plus{}xy$ ( since $ xyz\\equal{}1$)\r\nit sufficies to show the stronger : $ (x^2\\plus{}y^2\\plus{}z^2)^2 \\geq 2(x^3\\plus{}y^3\\plus{}z^3)\\frac{x\\plus{}y\\plus{}z}{3} \\plus{} (yz\\plus{}xz\\plus{}xy)\\frac{ (x\\plus{}y\\plus{}z)^2}{9}$ , which is easy to prove by mixing variables , because $ f(x,y,z)\\minus{} f(x ,\\frac{y\\plus{}z}{2} , \\frac{y\\plus{}z}{2}) \\geq 0$ is true for $ x\\equal{}max(x,y,z)$", "Solution_16": "[quote=\"Vasc\"]But we can't apply Jensen for the following :\n\n$ \\frac {ab}{2 \\plus{} c^5} \\plus{} \\frac {bc}{2 \\plus{} a^5} \\plus{} \\frac {ca}{2 \\plus{} b^5}\\ge 1$, \n\nwhere $ abc \\equal{} 1$.[/quote]\r\nThe Cauchy Schwarz still works here. here is it :)", "Solution_17": "Very nice solution, Can_hang2007. :lol:", "Solution_18": "Using anas' method, we shall prove :\r\n\r\nif $ x,y,z$ be positive numbers such that $ xyz \\equal{} 1$ and $ k\\geq 0$, then\r\n\r\n$ \\frac {yz}{k \\plus{} x^p} \\plus{} \\frac {zx}{k \\plus{} y^p} \\plus{} \\frac {xy}{k \\plus{} z^p}\\geq \\frac {3}{k \\plus{} 1}$\r\n\r\nholds for $ 2 \\minus{} 2\\sqrt {2}\\leq p\\leq 2 \\plus{} 2\\sqrt {2}$.\r\n[hide=\"Proof.\"]Substitute $ x \\equal{} e^a,y \\equal{} e^b,z \\equal{} e^c$, the original inequality is equivalent to\n\n$ \\frac {3}{k \\plus{} 1}\\leq \\sum{\\frac {e^{b \\plus{} c}}{k \\plus{} e^{pa}}} \\equal{} \\sum{\\frac {e^{ \\minus{} a}}{k \\plus{} e^{pa}}}\\equiv \\sum{f(a)}$\n\nwith $ a \\plus{} b \\plus{} c \\equal{} 0$. It's trivial by Jensen inequality since\n\n$ f''(a) \\equal{} \\frac {k^2 \\minus{} \\left(p^2 \\minus{} 2p \\minus{} 2\\right)ke^{pa} \\plus{} (p \\plus{} 1)^2e^{2pa}}{e^a\\left(k \\plus{} e^{pa}\\right)^3}$\n\n$ \\equal{} \\frac {\\left[2k \\minus{} \\left(p^2 \\minus{} 2p \\minus{} 2\\right)e^{pa}\\right]^2 \\plus{} p^2\\left(4 \\plus{} 4p \\minus{} p^2\\right)e^{2pa}}{4e^a\\left(k \\plus{} e^{pa}\\right)^3}\\geq 0$\n\nwhen $ 2 \\minus{} 2\\sqrt {2}\\leq p\\leq 2 \\plus{} 2\\sqrt {2}$.[/hide]\r\nLet $ x,y,z$ be positive numbers such that $ xyz \\equal{} 1$ and $ p\\geq 2 \\minus{} 2\\sqrt {2}$, prove or disprove :\r\n\r\n$ \\frac {yz}{k \\plus{} x^p} \\plus{} \\frac {zx}{k \\plus{} y^p} \\plus{} \\frac {xy}{k \\plus{} z^p}\\geq \\frac {3}{k \\plus{} 1}$\r\n\r\nholds if and only if $ 0\\leq k\\leq \\frac {1}{2}$.\r\n\r\nIf $ x,y,z$ be positive numbers such that $ xyz \\equal{} 1$, then\r\n\r\n$ \\frac {yz}{k \\plus{} x^5} \\plus{} \\frac {zx}{k \\plus{} y^5} \\plus{} \\frac {xy}{k \\plus{} z^5}\\geq \\frac {3}{k \\plus{} 1}$\r\n\r\nholds if and only if $ 0\\leq k \\leq k_1\\equal{}3.8193725239\\cdots$ or $ k\\geq k_2\\equal{}9.6722267758\\cdots$,\r\n\r\nwhere $ k_1$ and $ k_2$ are two roots of the irreducible polynomial \r\n\r\n$ 32000000000000000k^{33}\\minus{}4164907500000000000k^{32}$\r\n\r\n$ \\plus{}108670789214853515625k^{31}\\minus{}1256797627930842690625k^{30}$\r\n\r\n$ \\plus{}7493064794466948225000k^{29}\\minus{}25179783180321850391250k^{28}$\r\n\r\n$ \\plus{}69998100809896006842966k^{27}\\minus{}166282369027994424494619k^{26}$\r\n\r\n$ \\plus{}248104421159487398937774k^{25}\\minus{}400710378866130892958873k^{24}$\r\n\r\n$ \\plus{}396257666090594064385257k^{23}\\minus{}509491530874008328479924k^{22}$\r\n\r\n$ \\minus{}346738950771159102749972k^{21}\\minus{}2408825170503114333783615k^{20}$\r\n\r\n$ \\minus{}4189941102747476571660963k^{19}\\minus{}6765634615405559909195838k^{18}$\r\n\r\n$ \\minus{}8362769665280304962816235k^{17}\\minus{}8896513381195451506877313k^{16}$\r\n\r\n$ \\minus{}7632962071922521639315144k^{15}\\minus{}4837463042034085759524150k^{14}$\r\n\r\n$ \\minus{}1695616682343142863021315k^{13}\\plus{}395117094548260011146381k^{12}$\r\n\r\n$ \\plus{}927599047555212875891916k^{11}\\plus{}634580917305149169698544k^{10}$\r\n\r\n$ \\minus{}24826238628013847693808k^9\\minus{}1021031023301135926285248k^8$\r\n\r\n$ \\minus{}1866256679324080330596864k^7\\minus{}1987337463714132235990272k^6$\r\n\r\n$ \\minus{}1436664498839673863823360k^5\\minus{}735887569963796015874048k^4$\r\n\r\n$ \\minus{}267609651620487080214528k^3\\minus{}66701786894907139031040k^2$\r\n\r\n$ \\minus{}10326857903621496373248k\\minus{}755623749045475344384$." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "$ 100$ positive real numbers $ x_1,\\;x_2,\\;...,\\;x_{100}$ are such that\r\n\r\n$ \\textbf{(1)}\\; x_{1}^2 \\plus{} x_{2}^2 \\plus{} ... \\plus{} x_{100}^2 > 10000$\r\n$ \\textbf{(2)}\\; x_{1} \\plus{} x_{2} \\plus{} ... \\plus{} x_{100} < 300$\r\n\r\nProve that there exist numbers $ x_i,\\; x_j,\\; x_k$ such that $ x_i \\plus{} x_j \\plus{} x_k > 100$ for some $ 1\\leq i,\\;j,\\;k\\leq 100$.", "Solution_1": "I think you can use Drichlet to done. I will try it and post soon.", "Solution_2": "[quote=\"thanhnam2902\"]I think you can use Drichlet to done. I will try it and post soon.[/quote]\n\nI have a feeling this will be tough... :P If I'm not getting this wrong, the opening post says:\n[quote=\"Yuriy Solovyov\"]$ 100$ positive real numbers\n(...)\n$ \\textbf{(2)}\\; x_{1} \\plus{} x_{2} \\plus{} ... \\plus{} x_{100} < 100$\n(...)\nProve that there exist numbers $ x_i,\\; x_j,\\; x_k$ such that $ x_i \\plus{} x_j \\plus{} x_k > 100$ for some $ 1\\leq i,\\;j,\\;k\\leq 100$.[/quote]\r\n\r\nHowever, that's not really possible, since all x's are positive and, if $ x_{1} \\plus{} x_{2} \\plus{} ... \\plus{} x_{100} < 100$ then $ x_i\\plus{}x_j\\plus{}x_k R. Suppose that this sequence converges uniformly to a function f: I -> R. Then f also is integrable, and \r\n\\[\\int_{I}{f = \\mathop{\\lim }_{j \\to \\infty }}\\int_{I}{f_{j}}\\]", "Solution_1": "Riemann or Lebesgue integration? I assume it's Riemann. For any partition of $I$ and any choice of sample points, the Riemann sums you get for $f$ and $f_{j}$ are different by at most $\\mathrm{vol}(I)\\sup_{x\\in I}|f_{j}(x)-f(x)|$, which tends to $0$.", "Solution_2": "R is for all reals and it is riemann integration" } { "Tag": [ "function" ], "Problem": "A positive integer which contains no prime factors except 2 and 3 can be divided\r\nby exactly 24 different numbers and its square can be divided by exactly 77\r\ndifferent numbers. What is the integer?\r\n\r\nI have no idea where to even begin on thins problem without using my calculator to multiply 100 numbers.", "Solution_1": "Let the number be $ 2^x \\cdot 3^y$.\r\n\r\nThus:\r\n\r\n$ (x\\plus{}1)(y\\plus{}1)\\equal{}24$\r\n$ (2x\\plus{}1)(2y\\plus{}1)\\equal{}77$\r\n\r\nExpanding gives\r\n\r\n$ xy\\plus{}x\\plus{}y\\equal{}23$\r\n$ 4xy\\plus{}2x\\plus{}2y\\equal{}76 \\implies 2xy\\plus{}x\\plus{}y\\equal{}38$\r\n\r\nThus, $ xy\\equal{}15$.\r\n\r\nThis gives $ x\\plus{}y\\equal{}8$. Clearly, $ x,y\\equal{}5,3$ or $ 3,5$.\r\n\r\nSo there are 2 different possibilites. $ 2^5 \\cdot 3^3 \\equal{} 32 \\cdot 27 \\equal{} 864$, or $ 2^3 \\cdot 3^5 \\equal{} 8 \\cdot 243 \\equal{} 1944$.", "Solution_2": "What you need here is the [i]number of divisors[/i] function, commonly denoted $ d(n)$.\r\n\r\nIf $ n \\equal{} \\prod p_i^{k_i}$, we have $ d(n) \\equal{} \\prod (k_i \\plus{} 1)$ (Why?).\r\n\r\nThis fancy notation doesn't really make sense until you see an example, so here it is:\r\n$ d(12) \\equal{} d(2^23^1) \\equal{} (2 \\plus{} 1)(1 \\plus{} 1) \\equal{} 3\\cdot 2 \\equal{} 6$. These are $ 1,2,3,4,6,12$.\r\n\r\nEdit: Oh, also, try the High School Basics forum next time, this problem belongs there.\r\n\r\nEdit: Sorry, used \u2211 instead of\u00a0\u220f :huh: And yes, each $ p_i$ is a prime (I thought that was obvious but maybe not)", "Solution_3": "[quote=\"grn_trtle\"]What you need here is the [i]number of divisors[/i] function, commonly denoted $ d(n)$.\n\nIf $ n \\equal{} \\prod p_i^{k_i}$, we have $ d(n) \\equal{} \\sum (k_i \\plus{} 1)$ (Why?).\n\nThis fancy notation doesn't really make sense until you see an example, so here it is:\n$ d(12) \\equal{} d(2^23^1) \\equal{} (2 \\plus{} 1)(1 \\plus{} 1) \\equal{} 3\\cdot 2 \\equal{} 6$. These are $ 1,2,3,4,6,12$.\n\nEdit: Oh, also, try the High School Basics forum next time, this problem belongs there.[/quote] \r\n\r\nShould that summation after the d(n) be a product? \r\nAlso, i know it is probably assumed, but you might want to say that the p_i's are primes :wink:" } { "Tag": [ "algorithm", "function" ], "Problem": "U have function $f: \\mathbb{N}\\times \\mathbb{N}\\to \\mathbb{N}$; $f(m,n)=m+\\frac{(m+n)(m+n+1)}{2}$ (bijection).\r\n\r\nHow to find $(m,n) \\in \\mathbb{N}\\times \\mathbb{N}$, which $f(m,n)=6666$?\r\n\r\nCan be this solved with some algoritm :?: \r\n\r\n\r\ngoska", "Solution_1": "Hmmm, what you have there is a bijection from the rationals to the integers. Look at the first few values and see if you can spot a pattern...", "Solution_2": "[hide=\"at first I thought this would be too ugly\"]\nbut it wasnt\n\n$m+\\frac{m^{2}+2mn+n^{2}+m+n}{2}=6666$\n\n$m^{2}+2mn+n^{2}+3m+n-13332=0$\n\n$m^{2}+(2n+3)m+(n^{2}+n-13332)=0$\n\n$m=\\frac{-(2n-3)\\pm\\sqrt{(2n+3)^{2}-(4)(n^{2}+n-13332)}}{2}$\n\n$m=\\frac{-(2n-3)\\pm\\sqrt{8n+53337}}{2}$\n\n\n(111,3) is one integer solution I got fairly quick after my calculator told me that $231^{2}=53361$ \n[/hide]", "Solution_3": "[hide]\nLet $S=m+n$. Then $f(m,n)=m+\\frac{S(S+1)}{2}$. But $0\\leq m\\leq S$ so $\\frac{S(S+1)}{2}\\leq f(m,n)\\leq \\frac{(S+1)(S+2)}{2}-1$. Wow, that proves that $f$ is a bijection and gives an easy way to find $S$ and $m$.\n\nIn order for $f(m,n)$ to equal 6666, $S$ must be the greatest integer that satisfies $\\frac{S(S+1)}{2}\\leq 6666$, or $S(S+1)\\leq 13332$. $S$ will be somewhere between 110 and 120, so write $S=110+a$. Then\n\\[S(S+1)=(110+a)(111+a)=12210+221a+a^{2}\\]\nThis is less than or equal to 13332 when $a=4$, but not when $a=5$, so $S=114$. In this case, $\\frac{S(S+1)}{2}=6555$, so $m=111$ and $n=3$ as a result: $\\boxed{(m,n)=(111,3)}$.\n[/hide]" } { "Tag": [ "logarithms", "calculus", "calculus computations" ], "Problem": "Determine a closed form for the sum $f(y)=\\sum_{k=1}^{\\infty}\\frac1{y^{2}+k^{2}}$.", "Solution_1": "I think this is the formula you are looking for:\r\n\r\n$\\pi\\frac{e^{\\pi x}+e^{-\\pi x}}{e^{\\pi x}-e^{-\\pi x}}=\\frac{1}{x}+2x\\sum_{k=1}^{\\infty}\\frac{1}{x^{2}+k^{2}}$.", "Solution_2": "[quote=\"didilica\"]I think this is the formula you are looking for:\n\n$\\pi\\frac{e^{\\pi x}+e^{-\\pi x}}{e^{\\pi x}-e^{-\\pi x}}=\\frac{1}{x}+2x\\sum_{k=1}^{\\infty}\\frac{1}{x^{2}+k^{2}}$.[/quote]\r\n\r\nYes, brilliant! I had done this before, I used:\r\n\r\n\\[f(y)=\\sum_{k=1}^{\\infty}\\frac1{y^{2}+k^{2}}= \\frac1{2y}\\sum_{k=1}^{\\infty}\\frac{2y}{y^{2}+k^{2}}= \\frac1{2y}\\sum_{k=1}^{\\infty}\\frac{d}{dy}\\left[ \\log \\left(y^{2}+k^{2}\\right)-\\log k^{2}\\right]\\]\r\n\r\nand this is premissable because $\\frac{d}{dy}\\log k^{2}=0$, so we have \r\n\r\n\\[f(y) =\\frac1{2y}\\sum_{k=1}^{\\infty}\\frac{d}{dy}\\log\\left(1+\\frac{y^{2}}{k^{2}}\\right) =\\frac1{2y}\\frac{d}{dy}\\log\\prod_{k=1}^{\\infty}\\left(1+\\frac{y^{2}}{k^{2}}\\right)\\]\r\n\r\nand since $\\mbox{sinh}z = z\\prod_{k=1}^{\\infty}\\left(1+\\frac{z^{2}}{\\pi^{2}k^{2}}\\right)$ we have\r\n\r\n\\[f(y)=\\frac1{2y}\\frac{d}{dy}\\log\\left(\\frac{\\mbox{sinh}\\left(\\pi y\\right)}{\\pi y}\\right) =\\frac1{2y}\\cdot \\frac{\\pi y}{\\mbox{sinh}\\left(\\pi y\\right)}\\left(\\frac{\\pi^{2}y\\mbox{cosh}\\left(\\pi y\\right)-\\pi\\mbox{sinh}\\left(\\pi y\\right)}{\\pi^{2}y^{2}}\\right)\\]\r\n\\[=\\frac1{2y^{2}}\\left(\\pi y\\mbox{coth}\\left(\\pi y\\right)-1\\right)\\]" } { "Tag": [], "Problem": "What is the positive difference, in inches, between the greatest cabin\r\nheight and the least cabin height of the following business jets?\r\n\\[ \\begin{tabular}{|c|c|} \\hline\r\n\\textbf{Business Jet} & \\textbf{Cabin Height} \\\\ \\hline\r\nLearjet 31A & 4 feet, 5 inches \\\\ \\hline\r\nLearjet 60 & 5 feet, 8 inches \\\\ \\hline\r\nCessna Citation V & 4 feet, 7 inches \\\\ \\hline\r\nGulfstream V & 6 feet, 2 inches \\\\ \\hline\r\n\\end{tabular}\\]", "Solution_1": "The smallest height is 4 feet, 5 inches, which = 53 inches. The largest height (I crave a better word) is 6 feet, 2 inches, which is 74 inches. 74 - 53 = [b]21[/b]. \r\nqed" } { "Tag": [], "Problem": "find all positive integers $x,y,z$ s.t. $x+y+z=xyz$", "Solution_1": "[quote=\"hydro\"]find all positive integers $x,y,z$ s.t. $x+y+z=xyz$[/quote]\r\n\r\nI think ordering helps here.", "Solution_2": "[hide] Assume wlog that $x \\geq y \\geq z$ \nClearly if x = 1 then the equation is false\n if x = 2 then the LHS = 4,5 or 6. However the RHS = 2,4 or 8. (the RHS is 4 when y=2 z=1, but the LHS when y=1) Hence no solutions.\n if x = 3 then (3,2,1) is a solution.\n If y=3 then the RHS $\\geq 9$ but the LHS is < 9 unless z=3 (and no solutions here)\n If y=2 then we only need condsider z=2 which does not yield a solution \n If y=1 then z=1 and we do not have a solution\n if x > 3 then: if y>2 , the LHS $\\geq 3x$ (equality only occurring when z=1) and the LHS $\\leq 3x$ (equality only occuring when x=y=z)\n Therefore y=2 or 1\n if y=2: The LHS $\\leq x+4$ (equality only when z=2) and the RHS $\\geq 2x$ but since x is at least 4 this is at least x+4, and equality only occurs when z=1, so y cannot equal 2\n if y=1 then z=1 and we have x+2=x which is of course absurd\n\nThe only solution is x=3 y=2 z=1. (then also y=3 x=2 z=1 and all other permutations)[/hide]\r\n\r\nHopefully there is a neater solution", "Solution_3": "There you go :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=52572 .\r\nAnd here for the generalization :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=63671 ." } { "Tag": [ "Mafia", "pre100" ], "Problem": "Ok, game 18 will start as soon as I send out roles and stuff.\n\nPeople Playing:\nPeter VDD\nKlebian\nSuperNova\nLadyKn1ght\nlyra\nlasagnaman\nperfectnumber628\nArs\nCaptain Sugar\nsamath\nAdunakhor\ndaine\nchesspro\nMelissa\nbubka\nIgnite168\nGoBraves\nchuckzeduck\nslamchess\nb-flat\nrandomdragoon\nmatt276eagles\ndts\nSly Si\npoetryinmath\n\nRoles (this is a Mafia vs. Werewolf):\n4 Mafia\n4 Werewolves\n1 Vigilante\n\n1 Inspector\n1 Angel\n1 Seer\n1 Sorcerer\n1 Fairy\n\n3 Mason (have the ability to induct on one night)\n2 Miller\n4 Civilian\n\n1 Assassin\n1 Mathematician (for one night, he/she can choose 3 or more people to inspect, and will get a return saying how many of the group are mafia or werewolf. he/she must specify mafia or werewolf)\n\nAccusations are a follows: Accusation and Second do not count as aye. Once seconded, person on block has a maximum of 24 hours to give a defense. Votes begin as soon as defense is given, or 24 hours is up.\n\nAnyone who seems inactive (no post for more than 5 days), and has not given a prior reason as to why, will get one and only one warning and will be subject to replacement if they do not respond.\n\nI'm sending roles out now. DAY BEGINS:", "Solution_1": "May I be the first to remark that while this game certainly looks fun (Mafia v Werewolves always is), 25 people may be a bit too much.\r\nWatch for around 8 people not posting at all. Anyway, it's strange that we have 2 millers, but whatever. 3 Mason will be interesting, makes for a lot more behind the scenes action.\r\n\r\nAnyway, I hope that this post won't get me killed for being the first to speak. If it does, so be it, that's how most Mafia games turn out anyway -_- \r\n\r\nNote that I am making no implications in this post, this line being a message to the Mafia, Werewolves, Innocents, and other.", "Solution_2": "[quote]Watch for around 8 people not posting at all. [/quote]\r\nNot me.\r\n\r\n\r\nOh, and can someone give me a quick rundown of how the roles work? Specifically the ones that didn't exist last game, because I read through that one.", "Solution_3": "mafia vs werewolves! should be fun.....\r\n\r\nand besides, if there are 8 inactives, that makes 8 easy lynches right? =P so we don't have to play the \"most suspicious\" game like we did in 17....", "Solution_4": "This topic should help:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=38004\r\n\r\nBut I'll explain some too.\r\n\r\nWerewolves are similar to mafia, but basically they have a different set of corresponding roles.\r\n\r\nWerewolves --- Mafia\r\nSeer --- Inspector\r\nSorcerer --- Angel (Note that both can save from vigilante kills)\r\n\r\nOther than that, it's fairly basic, or was in the last game. \r\nMathmetician is explained also.", "Solution_5": "I'm here. My first Mafia/Werewolves game. :) \r\n\r\nI may not be very active (but I'll try to be) until Thursday, since I have a test then.", "Solution_6": "ok, just showing i'm alive", "Solution_7": "[quote=\"poetryinmath\"]ok, just showing i'm alive[/quote]\r\nSame here.", "Solution_8": "I'm ready to play.", "Solution_9": "ready!!!\r\nwhoopeee!", "Solution_10": "ready!!!\r\nwhoopeee!", "Solution_11": "Ready to play.", "Solution_12": "i'm here.\r\n\r\nminor detail: i will be gone this friday-sunday", "Solution_13": "Let's lynch the bird first.", "Solution_14": "posting to show I am here but Randomdragon wait until everbodyy posts before we start random accusations", "Solution_15": "I'm back from camp finally, but you can replace me if you want.", "Solution_16": "I hereby declare that I am no longer playing in this game.", "Solution_17": "I think we're just gonna have to end this game now, or maybe postpone it until summer is over...", "Solution_18": "I'll replace.", "Solution_19": "whom though? :huh:", "Solution_20": "how about this:\r\n\r\nWe restart the whole game in October. That's the start of fall. I'll sign up now.", "Solution_21": "Signup list:\r\n\r\nnutz\r\nakalra1\r\n\r\nbuild off this", "Solution_22": "okeydokey. I'll pm the past modder of this game and tell him that we are making a signup list.", "Solution_23": "Hey don't start spamming here now. I'll tell pulak to make a final effort. No registration is open, so there's no ppint in signing up. Each player must be pmed individually, and a huge new set must be taken in fro mthe signup thread. So sign up there if you like.", "Solution_24": "We are not starting this game over, this is Pulak's game and it's up to him.", "Solution_25": "ok. If you're talking pianoforte, I pmed him.", "Solution_26": "Yeah I am now back from Russia so I can rejoin the game", "Solution_27": "How about this:\r\n\r\nwe get a whole new signup for game 18(including moderator) So we can actually play.\r\n\r\nIs that okay?", "Solution_28": "How about this: just stop reviving this thread and get over it... this one failed.", "Solution_29": "last post" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that for nonnegative reals $x, y, z,$\r\n\r\n\\[\\sum \\sqrt{\\frac{2x}{y+z}} \\geq 2\\sqrt{2}.\\]", "Solution_1": "you correct it very quickly :D \r\nI'll post my solution.\r\nwe prove $\\sum \\sqrt{\\frac{x}{y+z}} \\geq 2$.\r\nbecause $\\sqrt{ \\frac{x}{y+z}}=\\frac{x}{\\sqrt {x(y+z)}} \\geq \\frac{2x}{x+y+z}$\r\nwe have proved :)", "Solution_2": "wt about positive x,y,z, is the minimum 3?", "Solution_3": "for $x=y=1,z \\longrightarrow 0$\r\n$\\sqrt{\\frac{2x}{y+z}} \\longrightarrow 2\\sqrt{2}$.", "Solution_4": "[quote=\"zhaobin\"]you correct it very quickly :D \nI'll post my solution.\nwe prove $\\sum \\sqrt{\\frac{x}{y+z}}\\geq 2$.\nbecause $\\sqrt{ \\frac{x}{y+z}}=\\frac{x}{\\sqrt{x(y+z)}}\\geq \\frac{2x}{x+y+z}$\nwe have proved :)[/quote]\r\n\r\n$\\sqrt{ \\frac{x}{y+z}}=\\frac{x}{\\sqrt{x(y+z)}}\\geq \\frac{2x}{x+y+z}$\r\nWhat inequality does this come from?\r\nedit: oh wait nv I see it know, it is just AM-GM (but why the wierd form you wrote it in?)\r\n\r\n$\\frac{x}{\\sqrt{x(y+z)}}\\geq \\frac{2x}{x+y+z}\\leftrightarrow \\frac{x+(y+z)}{2}\\geq \\sqrt{x(y+z)}$ or AM-GM on x, y+z", "Solution_5": "Posted some day ago, http://www.mathlinks.ro/Forum/viewtopic.php?t=108180" } { "Tag": [ "ARML", "MIT", "college", "Stanford", "Princeton", "CMU", "Putnam" ], "Problem": "I'm a couple days early with the X vs Y threads still going on, but maybe some last minute people can get ideas from here. \r\n\r\nI'm 95% likely going to Harvey Mudd. Is anyone else? My main concern is that it's so far from home and anyone I know...\r\n\r\nAnyway, it's been fun, guys. Not that anyone knows me because I either lurk or spam southeast state boards. I think I'm burnt out on high school math, but maybe I'll head back to the College Playground sometime if my interest returns. \r\n\r\nOh yeah... there's still ARML and Mu Alpha Theta. It never ends!!", "Solution_1": "Caltech.", "Solution_2": "I'm going to Caltech as well. \r\n\r\ne: Also, I'm from Chicago, so I understand a little bit of how you're feeling about going so far away from home. In the end, I just decided I had to go to the school I liked the best, and use this as a chance to learn to deal with living far away from home. And of course, you'll meet new people at HM and make new friends. It'll probably be fine.", "Solution_3": "heading off to MIT :)", "Solution_4": "That is funny, I just wanted to open up that thread to figure out which college you guys are finally joining as I think for some colleges the deadlines is on 1st May. Maybe you can also give the main reasons what made you accept the offer, e.g. I want to take courses X, Y, Z and thus thought it is the curriculum that fits me best, I want to take courses under professor A, Stanford has the best weather, most of my friends are going to Caltech, MIT made the most successful advertisement on AoPS, Princeton hold the best college preview session, Berkeley has the best statistics department, at CMU I can happily play with robots etc.", "Solution_5": "I picked HMC because the only other school to accept me and give significant money was University of Alabama. UA is not so great with math and CS, and I think I'll fit in better at Mudd.\r\n\r\nCool things about HMC: Putnam & pizza night, undergrad focus, lots of research opportunities and Clinics, which let you do real word problem solving in your field.", "Solution_6": "Princeton for me. Strong analysis program and the undergraduate focus were what attracted me.", "Solution_7": "Stanford - cs program/classes, people, job opportunities, weather, campus, location, etc.", "Solution_8": "Stanford- Math + {Physics, Econ, or CS} (i.e., double major)", "Solution_9": "University of Chicago! Gonna be a huge change in climate, but I'm looking forward to this awesome, new experience! :) Going there because.. they're the best school that accepted me? :) And I really enjoyed my time when I visited, and their staff were some of the friendliest out of the universities I visited.", "Solution_10": "[quote=\"Cicatriz\"]II'm 95% likely going to Harvey Mudd. Is anyone else? My main concern is that it's so far from home and anyone I know...[/quote]. That's where I'm going. There were 3 schools I really wanted to go to, but Caltech rejected me and MIT put me on the waitlist, so that leaves Harvey Mudd (and they offered a chunk of money, which is always nice).\r\n\r\nBut it was one of my favorite schools in the first place because its a great math and science oriented community with an incredible math department surrounded by liberal arts colleges for a nice balance. I'll be able to do research as an undergrad, and even if I end up in engineering or chemistry, I'll still get to do a lot of math along the way. I get the feeling I'm going to get a great education in a great environment (I'm talking about the people and the climate, I need to get away from northeastern weather).", "Solution_11": "MIT as well. :D", "Solution_12": "Princeton. They offered more aid than MIT.", "Solution_13": "I applied to two colleges, Michigan and MIT. I got into both. While Michigan offered me a scholarship, MIT's financial aid was about the same amount. So I chose MIT.", "Solution_14": "I ended up choosing Harvard. MIT and Harvard's financial offers were about the same. Both are great schools. I loved Harvard's pre-frosh and did cool stuff. I was scared at first when I saw quite the number of drunks Friday evening (including my freshman host friend and his buddies). I didn't attend MIT's CPW because of interview conflicts, so didn't get to see MIT firsthand. \r\n\r\nI chose Harvard mostly because of the strong alumni (they had two dinner gatherings for us-one in December and one in April) and close friends there. Pretty sweet.", "Solution_15": "hey mathfanatic where you at that scholarship dinner during the caltech preview weekend?", "Solution_16": "I'll be at Princeton next year... for math/finance", "Solution_17": "MIT, probably Course 6 (Electrical Engineering and Computer Science).", "Solution_18": "I'm probably the only one ever from this forum lol but i chose Purdue (I actually live in the same town).\r\nIt was between there and Caltech; there were some things I didn't like about Caltech, but overall I liked it ok.\r\nSo it was mostly because of the money.\r\n\r\nAnyone going to or know anyone else going to purdue too?", "Solution_19": "[quote=\"d343seven\"]I'm probably the only one ever from this forum lol but i chose Purdue (I actually live in the same town).\nIt was between there and Caltech; there were some things I didn't like about Caltech, but overall I liked it ok.\nSo it was mostly because of the money.\n\nAnyone going to or know anyone else going to purdue too?[/quote]\r\n\r\nI think I met some people from West Layafette HS at the Caltech Prefrosh Weekend who were also deciding between Caltech and Purdue.", "Solution_20": "[quote=\"1234567890\"][quote=\"d343seven\"]I'm probably the only one ever from this forum lol but i chose Purdue (I actually live in the same town).\nIt was between there and Caltech; there were some things I didn't like about Caltech, but overall I liked it ok.\nSo it was mostly because of the money.\n\nAnyone going to or know anyone else going to purdue too?[/quote]\n\nI think I met some people from West Layafette HS at the Caltech Prefrosh Weekend who were also deciding between Caltech and Purdue.[/quote]\r\n\r\nThen that was me and 2 of my friends from here haha. Whats your name? I might remember you. The other two are going to Caltech. Are you going to Caltech?", "Solution_21": "I'm Saurabh (I think I met you guys playing mini golf at Ruddock). And, yup, I'm going to Caltech.", "Solution_22": "Although the one campus which has enrolled the largest number of Southern California ARML team members over our history has been Caltech, this year was an exception: no new Caltech students. There will be SoCal ARML team members as freshmen this year at MIT, Yale, Cornell, Princeton, Penn, UC Berkeley, Stanford, USC, and UCI.", "Solution_23": "Massive bump but I just have to say that these people probably have kids now", "Solution_24": "[quote=peelybonehead]Massive bump but I just have to say that these people probably have kids now[/quote]\n\nfor real", "Solution_25": "Bro every single person in this thread go into a t-10 uni, ", "Solution_26": "everyone pretty much went to either princeton, standford, MIT, or caltech\n", "Solution_27": "[quote=Turtle09]everyone pretty much went to either princeton, standford, MIT, or caltech[/quote]\n\nor harvery mudd", "Solution_28": "\n[quote name=\"peelybonehead\" url=\"https://artofproblemsolving.com/community/p25726575\"]Massive bump but I just have to say that these people probably have kids now[/quote]\n\nWe're like the new generation of AoPS-ers\n\nOne day we'll have kids and we'll stop being active here and then the kids of the next generation will bump our threads just like you just did", "Solution_29": "[quote=ktm]UChicago =) where the squirrels rule -.-[/quote]\n\nNah mate Vandy is where the squirrels rule :rotfl:" } { "Tag": [ "logarithms", "integration", "calculus", "calculus computations" ], "Problem": "prove that: $ 1\\minus{}\\frac{1}{4}\\plus{}\\frac{1}{7}\\minus{}\\frac{1}{10}\\plus{}\\cdots\\plus{}\\frac{(\\minus{}1)^{n\\plus{}1}}{3n\\minus{}2}\\plus{}\\cdots\\equal{}\\frac{1}{3}\\Big(\\frac{\\pi}{3}\\minus{}\\ln 2\\Big)$", "Solution_1": "Just note that $ \\sum_{n\\equal{}0}^{\\infty} \\frac{(\\minus{}1)^n}{3n\\plus{}1} \\equal{} \\int_{0}^{1} \\sum_{n\\equal{}0}^{\\infty} (\\minus{}1)^n x^{3n} \\, dx \\equal{} \\int_{0}^{1} \\frac{dx}{1\\plus{}x^3}$, which is justified by Abel's theorem. The remaining is just a calculation involving partial fraction technique, so we have the answer.\r\n\r\np.s. But the answer is $ \\frac{\\pi}{9}(\\sqrt{3}\\pi \\plus{} 3 \\log 2)$. :lol:", "Solution_2": "[quote=\"sos440\"]p.s. But the answer is $ \\frac {\\pi}{9}(\\sqrt {3}\\pi \\plus{} 3 \\log 2)$. :lol:[/quote]Actually it's $ \\frac {1}{9}\\left(\\pi \\sqrt {3} \\plus{} 3\\log 2\\right)$. ;)", "Solution_3": "[quote=\"JoeBlow\"]Actually it's $ \\frac {1}{9}\\left(\\pi \\sqrt {3} \\plus{} 3\\log 2\\right)$. ;)[/quote]\r\n\r\nSorry for typo :blush:" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Find all limit points of the sequence $ x_n\\equal{}cosn$.", "Solution_1": "It should be all of $ [\\minus{}1,1]$ by the Weyl equidistribution theorem, since $ 2\\pi$ is irrational.", "Solution_2": "Dear blahblahblah thanks for your help.Can you give me information about \"Weyl equidistribution theorem\"!" } { "Tag": [], "Problem": "[asy]draw((0,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--(0,0)--cycle);\ndraw((1,0)--(6,0)--(6,1)--(5,1)--(5,2)--(4,2)--(4,3)--(3,3)--(3,2)--(2,2)--(2,1)--(1,1)--(1,0)--cycle);\ndraw((2,0)--(5,0)--(5,1)--(4,1)--(4,2)--(3,2)--(3,1)--(2,1)--cycle);\ndraw((3,0)--(3,1)--(4,1)--(4,0)--cycle);\nfill((1,0)--(1,1)--(2,1)--(2,0)--cycle,black);\nfill((3,0)--(3,1)--(4,1)--(4,0)--cycle,black);\nfill((5,0)--(5,1)--(6,1)--(6,0)--cycle,black);\nfill((2,1)--(2,2)--(3,2)--(3,1)--cycle,black);\nfill((4,1)--(4,2)--(5,2)--(5,1)--cycle,black);\nfill((3,2)--(3,3)--(4,3)--(4,2)--cycle,black);[/asy]\r\n\r\nA \"stair-step\" figure is made up of alternating black and white squares in each row. Rows $ 1$ through $ 4$ are shown. All rows begin and end with a white square. The number of black squares in the $ 37$th row is\r\n\r\n\\[ \\textbf{(A)}\\ 34 \\qquad\r\n\\textbf{(B)}\\ 35 \\qquad\r\n\\textbf{(C)}\\ 36 \\qquad\r\n\\textbf{(D)}\\ 37 \\qquad\r\n\\textbf{(E)}\\ 38\r\n\\]", "Solution_1": "[quote=\"AIME15\"][asy]draw((0,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--(0,0)--cycle);\ndraw((1,0)--(6,0)--(6,1)--(5,1)--(5,2)--(4,2)--(4,3)--(3,3)--(3,2)--(2,2)--(2,1)--(1,1)--(1,0)--cycle);\ndraw((2,0)--(5,0)--(5,1)--(4,1)--(4,2)--(3,2)--(3,1)--(2,1)--cycle);\ndraw((3,0)--(3,1)--(4,1)--(4,0)--cycle);\nfill((1,0)--(1,1)--(2,1)--(2,0)--cycle,black);\nfill((3,0)--(3,1)--(4,1)--(4,0)--cycle,black);\nfill((5,0)--(5,1)--(6,1)--(6,0)--cycle,black);\nfill((2,1)--(2,2)--(3,2)--(3,1)--cycle,black);\nfill((4,1)--(4,2)--(5,2)--(5,1)--cycle,black);\nfill((3,2)--(3,3)--(4,3)--(4,2)--cycle,black);[/asy]\n\nA \"stair-step\" figure is made up of alternating black and white squares in each row. Rows $ 1$ through $ 4$ are shown. All rows begin and end with a white square. The number of black squares in the $ 37$th row is\n\\[ \\textbf{(A)}\\ 34 \\qquad \\textbf{(B)}\\ 35 \\qquad \\textbf{(C)}\\ 36 \\qquad \\textbf{(D)}\\ 37 \\qquad \\textbf{(E)}\\ 38\n\\]\n[/quote]\r\n\r\n[hide]Row $ n$ contains $ n\\minus{}1$ black squares. Row $ 37$ contains $ \\boxed{36}$ black squares.[/hide]", "Solution_2": "Answer is,In nth row there will be n-1 black squares so in 37 th row there will be 36 rows ", "Solution_3": "[hide=Solution 1]So in each row the number of blocks is twice the row number minus 1 so there will will 73 blocks in it. The number of blacks is always half of one less than the number of the blocks. So 73-1 = 72 and 72/2 =36.[/hide]\n[hide=Solution 2]An alternative and easier solution is that the number of blacks is always 1 less than the row number. This is because in each row the number of blocks is twice the row number minus 1. The number of blacks is always half of one less than the number of the blocks. Applying this there will be 36 blacks in the 37th row [/hide] \n\nMy second solution is easier however the first is more logic based." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Do there exist two coprime posivite integers a,b such that the sequence:\r\nL(0)=a,L(1)=b and L(n+2)=3*L(n+1)-L(n) for n>=0 does not contain any prime?", "Solution_1": "This problem hasnot a classic solution", "Solution_2": "But is there any solution for this one? If someone has any, then please share it with us!", "Solution_3": "The answer is yes.\r\nSet $\\lambda = \\frac {3 + \\sqrt {5}} {2} $\r\n$ F_n $ = $ \\lambda ^ {n} + \\lambda ^ {-n} $\r\n\r\n$ L_n $ = $ \\frac{ \\lambda ^ {n} - \\lambda ^ {-n} } {\\sqrt{5} } $\r\n\r\nIt's easy to see that both $ F_n $ and $ L_n $ are integers and satisfy the reccurence of the problem. Moreover, it's easy to see that $ L_n $ and $ L_{n+1} $ are coprime for all positive integers n.\r\nWe have $ F_n ^ 2 - 5 L_n ^ 2 = 4 $ \r\nNow suppose $ L_n $ is a prime.\r\nThus $ ( F_n - 2 ) ( F_n + 2 ) = 5 p ^ 2 $ with $ p \\in P $\r\nIt's easy to see that this equation can't happen for sufficently large n, so there exist a $ N_0 $ such that $ L_n $ isn't a prime for all $ n \\geq N_0 $.\r\nChoose $ a= L_{N_0} , b = L_{N_0 + 1 } $ and we are done.", "Solution_4": "After such a bad day, thanks for this magnificient proof. Now I'm happy I've seen a truly wonderful problem and solution." } { "Tag": [ "geometry", "3D geometry", "dodecahedron", "sphere", "vector", "symmetry", "geometry unsolved" ], "Problem": "A regular dodecahedron is inscribed in a unit sphere and vectors are drawn from the centre of the sphere to the vertices. What is the sum of these vectors?", "Solution_1": "By symmetry, each vector has a vector pointing the other way, so the sum is 0. Hope this is right :)" } { "Tag": [], "Problem": "in the figure below we have AP=2PB, find the angule BOP\r\n\r\nhttp://img194.imageshack.us/img194/3549/imagemdwk.png\r\n\r\nsorry,but I couldn\u00b4t post the image here,", "Solution_1": "[quote=\"wanderson\"]in the figure below we have AP=2PB, find the angle BOP\n[/quote]\r\n\r\nAre you referring to the arcs $ AP$ and $ PB$ or the line segments $ AP$ and $ PB$ ?", "Solution_2": "probably the line segments AP and PB(I just copied the question),but can be the arc AP and PB too, \r\nI do not know" } { "Tag": [ "inequalities", "Pythagorean Theorem", "geometry" ], "Problem": "Let $ ABCD$ be a quadrilateral inscribed in a semicircle with diameter $ AD\\equal{}x$. If $ AB\\equal{}a$, $ BC\\equal{}b$, and $ CD\\equal{}c$, then prove \\[ x^3\\minus{}(a^2\\plus{}b^2\\plus{}c^2)x\\minus{}2abc\\equal{}0\\]", "Solution_1": "Anybody solve this problem?\r\n\r\nHint:[hide]ptolemy[/hide]", "Solution_2": "[hide=\"solution\"] Draw in the other semicircle so ABCD is inscribed in a circle. Notice that, as AD is a diameter, angles ABD and ACD are right.\n\nAs that implies ABCD is cyclic, we use Ptolemy's Inequality and the Pythagorean Theorem to get \n$ ac \\plus{} bx \\equal{} \\sqrt {(x^2 \\minus{} a^2)(x^2 \\minus{} c^2)}$.\n$ \\implies{(ac)^2 \\plus{} (bx)^2 \\plus{} 2abcx \\equal{} x^4 \\minus{} (cx)^2 \\minus{} (ax)^2 \\plus{} (ac)^2}$\n$ \\implies{x^4 \\minus{} (a^2 \\plus{} b^2 \\plus{} c^2)x^2 \\minus{} 2abcx \\equal{} 0}$.\n\nAs $ x > 0$, we may safely divide by x to get the desired result.[/hide]\r\n\r\nA bit easy for a #10? I assume the problems are supposed to fairly increase in difficulty the farther you go...", "Solution_3": "I am not sure. But this is an interesting problem. A few questions to evaluate what happened:\r\n\r\nFirst, it seems a bit strange to be that when you have the equation for ptolemy, it is clear that $ a$ and $ c$ are interchangeable. But it seems that you couldn't switch say $ a$ and $ b$ without changing the expression. But infact when you expand, the equation is \"symmetric\" in $ a$, $ b$, $ c$, (that is you can switch them around and the question is unchanged). Can you give a geometric reason for why this is so. As a side note, it is kind of cool that a geometrical argument can give rise to such an algebraic oddity.\r\n\r\nSecond, geometrically, what does the $ x\\equal{}0$ solution correspond to? Why did it appear in our equation?" } { "Tag": [], "Problem": "How many ordered pairs of positive integers $ (x,y)$ are solutions of $ 5x\\plus{}3y\\equal{}29$?", "Solution_1": "The pair of numbers with the largest x is $ (4, 3)$. Then, we can decrease the x by multiples of three. The only other pair using this method would be $ (1, 8)$. Therefore, there are $ \\boxed{2}$ pairs." } { "Tag": [], "Problem": "1) find the base 2 decimal expansion of: 11/16 and 4/7\r\n\r\n2) find the base 3 decimal expansion of: 11/81 and 7/39\r\n\r\n3) in base 10, 9-1=2 :^3: , 9 2 -1 = 2:^4: *5 , \r\n9 :^4: -1=2 :^3:*5*41\r\nusing the above, determine the length of the period in the base [i]nine [/i]expansion of each of the fractions 1/5, 1/(2 :^5: ) and 1/(5*14) in which all of the digits are base 9", "Solution_1": "[quote=\"random_mathematician\"]... 9 :^3: -1=2 :^3:*5*41 ...\n[/quote]\r\nAre you sure that is right? 9 :^3: = 729 :?", "Solution_2": "oops!! :blush: \r\n\r\nyeh fixed it now\r\n\r\nits meant to be 9 :^4: not 9:^3:!!\r\n\r\nalready edited 4 ur convenience...", "Solution_3": "Quote:1) find the base 2 decimal expansion of: 11/16 and 4/7\n\n[hide]\n\nNote: an underline denotes a repeating cycle of digits.\n\n\n\nSince 11 = 8 + 2 + 1 = 10112\n\ndividing by 16 = 24 moves the (binary?) point 4 places to the left.\n\n\n\nAnswer: 11/16 = 0.10112\n\n\n\n\n\nSince 7 = 1112, I used long division in base 2 and found that 1/7 = 0.0012\n\n\n\nMultiplying by 4 = 22 moves the point 2 places to the right.\n\n\n\nAnswer: 4/7 = 0.1002[/hide]", "Solution_4": "I got half of #2 . . .\n\n\n\nQuote:2) Find the base 3 decimal expansion of: 11/81\n\n[hide]Since 11 = 9 + 2, we have: 11/81 = 9/81 + 2/81\n\n\n\n which translates to: 1/9 + 2/81 = 0/31 +1/32 + 0/33 + 2/34\n\n\n\nTherefore: 11/81 = 0.01023[/hide]" } { "Tag": [], "Problem": "Like it? Or was the old one better?", "Solution_1": "Oh my God that's TUPAC.Please keep it,or if you don't want it just give it to me.I'm a huge $2Pac$ fan. :coolspeak:", "Solution_2": "[quote=\"cezar lupu\"]Oh my God that's TUPAC.Please keep it,or if you don't want it just give it to me.I'm a huge $2Pac$ fan. :coolspeak:[/quote]\r\nwhat? Tupac?", "Solution_3": "[quote=\"Iversonfan2005\"]Like it? Or was the old one better?[/quote]\r\n\r\n\r\n my god, tupac was the most annoying person to ever enter rap, all he did was cry about how african americans were being treated unfairly, when he himslef, killed numerous african americans (excluding the hundreds of cops, he planned on eliminating). all this person did was cry for attention. And to think people thought of him as a visionary, more like mockery. In my opinion, stick with the nba snapshot with iverson or whoever he is.", "Solution_4": "Hey,G-UNIT like it or not $2PAC$ was,is and will be the greatest rapper of all time. :D :D :D I think you don't know nothing about Tupac's life and about Rap.... Anyway if you don't know at least one of Pac's songs it means that you know nothing about Rap Music as far as I'm concerned :starwars:", "Solution_5": "i like the 2pac better. :lol:", "Solution_6": "[quote=\"cezar lupu\"]Hey,G-UNIT like it or not $2PAC$ was,is and will be the greatest rapper of all time. :D :D :D I think you don't know nothing about Tupac's life and about Rap.... Anyway if you don't know at least one of Pac's songs it means that you know nothing about Rap Music as far as I'm concerned :starwars:[/quote]\r\n\r\nTrue, True, Biggie and Eminem weren't far behind though", "Solution_7": "[quote=\"cezar lupu\"]Oh my God that's TUPAC.Please keep it,or if you don't want it just give it to me.I'm a huge $2Pac$ fan. :coolspeak:[/quote]\r\n\r\nIf you want a tupac avatar, I could find you one, or just look for it on Google Images.", "Solution_8": "Yes,but I don't know how to find one that little to put it in my profile image.Can you please send me one? :help:", "Solution_9": "I see you got one :)", "Solution_10": "Yes,thanks anyway.", "Solution_11": "it seems to be in a rap club...can i join it? \r\ni was just wondering about my avatar...what do you think about an EMINEM avatar? :lol:\r\n\r\ni like TUPAC", "Solution_12": "[quote=\"EUCLA\"]it seems to be in a rap club...can i join it? \ni was just wondering about my avatar...what do you think about an EMINEM avatar? :lol:\n\ni like TUPAC[/quote]\r\n\r\nSure, get an Eminem one", "Solution_13": "[quote=\"Iversonfan2005\"][quote=\"cezar lupu\"]Hey,G-UNIT like it or not $2PAC$ was,is and will be the greatest rapper of all time. :D :D :D I think you don't know nothing about Tupac's life and about Rap.... Anyway if you don't know at least one of Pac's songs it means that you know nothing about Rap Music as far as I'm concerned :starwars:[/quote]\n\nTrue, True, Biggie and Eminem weren't far behind though[/quote]\n[quote=\"G-UNIT\"][quote=\"Iversonfan2005\"]Like it? Or was the old one better?[/quote]\n\n\n my god, tupac was the most annoying person to ever enter rap, all he did was cry about how african americans were being treated unfairly, when he himslef, killed numerous african americans (excluding the hundreds of cops, he planned on eliminating). all this person did was cry for attention. And to think people thought of him as a visionary, more like mockery. In my opinion, stick with the nba snapshot with iverson or whoever he is.[/quote]\r\n\r\nhey cool it off people. you dont have to make the other person what he believes and you two should cool it. otherwise i have to pm the admins to lock this topic", "Solution_14": "[quote=\"Palytoxin\"][quote=\"Iversonfan2005\"][quote=\"cezar lupu\"]Hey,G-UNIT like it or not $2PAC$ was,is and will be the greatest rapper of all time. :D :D :D I think you don't know nothing about Tupac's life and about Rap.... Anyway if you don't know at least one of Pac's songs it means that you know nothing about Rap Music as far as I'm concerned :starwars:[/quote]\n\nTrue, True, Biggie and Eminem weren't far behind though[/quote]\n[quote=\"G-UNIT\"][quote=\"Iversonfan2005\"]Like it? Or was the old one better?[/quote]\n\n\n my god, tupac was the most annoying person to ever enter rap, all he did was cry about how african americans were being treated unfairly, when he himslef, killed numerous african americans (excluding the hundreds of cops, he planned on eliminating). all this person did was cry for attention. And to think people thought of him as a visionary, more like mockery. In my opinion, stick with the nba snapshot with iverson or whoever he is.[/quote]\n\n\n \nhey cool it off people. you dont have to make the other person what he believes and you two should cool it. otherwise i have to pm the admins to lock this topic[/quote]\r\n\r\n Palytoxin, please use complete and logical sentences when speaking, otherwise; no one understands what you are trying to say.", "Solution_15": "I don't think the conversation is heated enough to ask the administrators to close this topic. They've only posted their opinions, strong as they might be. They're not trying to indoctrinate the other person. ;)", "Solution_16": "[quote=\"G-UNIT\"][/quote][quote=\"Palytoxin\"][quote=\"Iversonfan2005\"][quote=\"cezar lupu\"]Hey,G-UNIT like it or not $2PAC$ was,is and will be the greatest rapper of all time. :D :D :D I think you don't know nothing about Tupac's life and about Rap.... Anyway if you don't know at least one of Pac's songs it means that you know nothing about Rap Music as far as I'm concerned :starwars:[/quote]\n\nTrue, True, Biggie and Eminem weren't far behind though[/quote]\n[quote=\"G-UNIT\"][quote=\"Iversonfan2005\"]Like it? Or was the old one better?[/quote]\n\n\n my god, tupac was the most annoying person to ever enter rap, all he did was cry about how african americans were being treated unfairly, when he himslef, killed numerous african americans (excluding the hundreds of cops, he planned on eliminating). all this person did was cry for attention. And to think people thought of him as a visionary, more like mockery. In my opinion, stick with the nba snapshot with iverson or whoever he is.[/quote]\n\n\n \nhey cool it off people. you dont have to make the other person what he believes and you two should cool it. otherwise i have to pm the admins to lock this topic[/quote][quote=\"G-UNIT\"]\n\n Palytoxin, please use complete and logical sentences when speaking, otherwise; no one understands what you are trying to say.[/quote]\r\n\r\nI don't really see the non-logicality of what he wrote. :?", "Solution_17": "chinese rap is better! go Jay Chow :D", "Solution_18": "The admins shoot you, fool\r\nBetter believe it, they rule", "Solution_19": "I changed it again and put 50 Cent", "Solution_20": "tupac no more :o :?: \r\n\r\n\r\nnow you have 50 :oops: not bad :!: ;)", "Solution_21": "Jay Chow is chinese pop not rap!!!!\r\n\r\nOn a side note, I don't like rap. No offense to any rapper homey's.\r\nI liked your iverson one better.", "Solution_22": "Yeah, I'll just switch between Iverson, Tupac, and 50 Cent, like every week or something :)", "Solution_23": "[quote=\"Iversonfan2005\"]Yeah, I'll just switch between Iverson, Tupac, and 50 Cent, like every week or something :)[/quote]\r\n\r\nI don't like a lot of the mainstream rappers because all they only sing about \"hot girls\" (candy shop? wtf? worst song, [b]ever[/b]) to make themselves seem cool. tupac is okay (was he the one that dedicated the song to his mom?) and so are a few others...\r\n\r\nthere are some underground rappers that manage to sound nice and have meaningful lyrics at the same time... can't recall their names right now, though.", "Solution_24": "[quote=\"xdl615x\"]Jay Chow is chinese pop not rap!!!!\n\nOn a side note, I don't like rap. No offense to any rapper homey's.\nI liked your iverson one better.[/quote]\r\n\r\nlol I thought it was Jay Chou??", "Solution_25": "[quote=\"elvenchamp777\"]chinese rap is better! go Jay Chow :D[/quote]\r\nJay Chow is rap+pop. Anyway, I don't quite like him.\r\n\r\nI got my first avatar, how's it? :P", "Solution_26": "[quote=\"shobber\"][quote=\"elvenchamp777\"]chinese rap is better! go Jay Chow :D[/quote]\nJay Chow is rap+pop. Anyway, I don't quite like him.\n\nI got my first avatar, how's it? :P[/quote]\r\ni like it, exept for the pink, i don't like pink :D", "Solution_27": "Thanks! :D", "Solution_28": "LOL IT'S JAY CHOU. obviously you are not chinese. :P", "Solution_29": "[quote=\"mathclass\"]LOL IT'S JAY CHOU. obviously you are not chinese. :P[/quote]\r\nWell I am a Chinese. I just do not know how to spell his lastname. We call his Chinese name all Jay in our daily chats. And the first time I heard of the name Jay I thought of Gay...", "Solution_30": "[quote=\"mathclass\"]I don't like a lot of the mainstream rappers because all they only sing about \"hot girls\" (candy shop? wtf? worst song, [b]ever[/b]) to make themselves seem cool. .[/quote]\r\nThis is why Eminem is, and will always be, the greatest rapper ever. Every one of his songs has meaning for himself or someone close to him. Unlike meaningless crap such as... \"the sweat drips from my balls...\"." } { "Tag": [ "trigonometry", "calculus", "calculus computations" ], "Problem": "I do get the basic aspects of it, but when it comes converting, I am a bit 'shaky'.\r\n1.lim x->0 (x-sinx/tan x-x). I know it is a indeterminate prob. and it is a 0/0 problem. I tried to split into two different equation -sin x/-x and something else to get the right answer. The answer is 1/2 by the way.\r\n2. lim x->pi (3 sec x/2+tan x) the answer is 3. What I did was I put sec x/tan x on one, 3/1 on the other and needed one more for it to make sense. those would equal 1 and 3 respectively. But it is the 2 that is the problem. Please help.", "Solution_1": "Stop trying to split up fractions, especially when the addition or subtraction is in the denominator.\r\n\r\nOn the first: there's a lot of cancellation, and using L'Hopital's rule will require more than one application. It is still the only way you hknow that will work (if you haven't learned power series yet).\r\n\r\nOn the second: $ \\frac{3\\sec x}{2+\\tan x}=\\frac{3}{2\\cos x+\\sin x}$. It's not indeterminate anymore." } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let $ ABCD$ be a square. Denote $ O\\in AC\\cap BD$ . For a point $ G\\in [OB]$ define \r\n\r\nthe points $ E\\in AG\\cap BC$ and $ F\\in[CD]$ so that $ GF\\perp \\overline {AGE}$ . Prove that :\r\n\r\n$ 1\\blacktriangleright$ For any $ G\\in [OB]$ exists the point $ K\\in [GF]$ for which $ AK\\equal{}EF$ and $ m(\\widehat {EKF})\\equal{}135^{\\circ}$ .\r\n\r\n$ 2\\blacktriangleright$ The lines $ BD$ , $ AF$ , $ EK$ are concurrently ( [b]I added this point ![/b] ).", "Solution_1": "This is my solution (sorry, my English is bad)\r\nNote H is the intersection of AF and BD. I'll prove that K is the intersection of GF and HE.\r\nWe have ADFG is cyclic so 1$, $x_k\\in R$, $k\\in\\overline {1,n}$, $\\sum x_k=na$, $\\sum x^2_k=nb$\nthen for any $k\\in \\overline {1,n}$, $|x_k-a|\\le \\sqrt{(n-1)(b-a^2)}$.[/color][hide=\"Some aplications.\"] $1^{\\circ}\\blacktriangleright \\sum x_k=\\frac{n+1}{2}\\ ,\\ \\sum x^2_k=1+\\left(\\frac{n-1}{2}\\right)^2\\Longrightarrow$$\\left(\\forall\\right) k\\in \\overline {1,n}\\ ,\\ |2nx_k-(n+1)|\\le (n-1)^2\\ .$\n\n$2^{\\circ}\\blacktriangleright 2+\\sum x_k=\\sum x^2_k=n\\Longrightarrow \\left(\\forall\\right) k\\in \\overline {1,n}\\ ,\\ x_k\\in \\left[-1\\ ,\\ 3-\\frac 4n\\right]\\ .$\n\n$3^{\\circ}\\blacktriangleright$ In any triangle $ABC$ there is the relations\n\n$|r_a+r_b-2r_c|\\le \\sqrt{(4R+r)^2-3p^2}\\ ;\\ \\max_{x\\in \\{a,b,c\\}} \\left|p-\\frac 32 \\cdot x\\right|\\le \\sqrt{p^2-3r(4R+r)}\\ .$[/hide]", "Solution_1": "For 1:\r\n\r\nAssume $(\\forall k)|2nx_k-(n+1)|>(n-1)^2$, then square it and expand it, sum up all the inequalities, substitute the given sums and get a contradiction.\r\n\r\nFor 2:\r\n\r\nAssume $(\\forall k)(x_k+1)(x_k-3+{4\\over n})>0$. Expand it, sum up all the inequalities, substitute the given sums and get a contradiction.\r\n\r\nFor 3:\r\n\r\nWhat's $p$? Perimeter? Semiperimeter?", "Solution_2": "The $p$ is the semiperimeter and $P$ is the perimeter (in Romania) !. At other times $p\\equiv s$." } { "Tag": [ "calculus", "integration", "algebra", "function", "domain", "Ring Theory", "superior algebra" ], "Problem": "What is the difference between the quotient field of Q[x] and the quotient field of Z[x]?\r\n\r\nI dont just want an answer though.. I want someone to help explain the whole reasoning and thought process behind it so that next time I'll be able to reason through it...\r\n\r\nThanks!!!", "Solution_1": "look at the definitions, these things are just defined differently.\r\nbut there is a canonical isomorphism. more generally, if $ R$ is an integral domain, we may identity $ Q(Q(R)[X])$ with $ Q(R[X])$ by taking a common denominator." } { "Tag": [], "Problem": "Evaluate $ (a \\plus{} b)(a^2 \\minus{} ab \\plus{} b^2)$ for $ a \\equal{} 5$ and $ b \\equal{} 4$.", "Solution_1": "We have two solutions - we could substitute $ 5$ for $ a$ and $ 4$ for $ b$ to get the value, but that takes a while, and of course we want to do this quickly. Note that $ (a \\plus{} b)(a^2 \\minus{} ab \\plus{} b^2)$ is the factorized form of $ a^3 \\plus{} b^3$. Therefore, all we need to do is compute $ 5^3 \\plus{} 4^3 \\equal{} 125 \\plus{} 64 \\equal{} \\boxed{189}$." } { "Tag": [], "Problem": "Can the second reference (that one that's not a math teacher) be a parent? I remember reading that it could be a parent, but I just want to make sure.\r\n\r\nAlso, do I mail the application test and personal information stuff separately or together?", "Solution_1": "We'd strongly prefer the second reference [i]not[/i] come from a parent. If you need a little extra time to solicit a second reference from someone else, we can work with you--send us an e-mail.\r\n\r\nIf your parents would like to write to us about you, they may submit a parent's statement (details near the bottom of the [url=http://www.awesomemath.org/application.shtml]application page[/url].)\r\n\r\nOh, and mail the stuff together. No sense in using two envelopes.\r\n\r\n-Zac", "Solution_2": "[quote=\"zacacox\"]We'd strongly prefer the second reference [i]not[/i] come from a parent. If you need a little extra time to solicit a second reference from someone else, we can work with you--send us an e-mail.\n\nIf your parents would like to write to us about you, they may submit a parent's statement (details near the bottom of the [url=http://www.awesomemath.org/application.shtml]application page[/url].)\n\nOh, and mail the stuff together. No sense in using two envelopes.\n\n-Zac[/quote]\r\n\r\nim assuming the \"stuff\" would just be the personal info and the parent letter, correct?\r\n\r\nbecause i think it says for the recommendations to be mailed/emailed separately.", "Solution_3": "Correct. And the test, too, if you're sending it right at the deadline." } { "Tag": [ "AMC", "AIME" ], "Problem": "I'd like to introduce some easy AIME to the people who are new to AMC so they can see what AIMEs are like.\r\n\r\nIt would be solved by me first, of course.\r\n\r\nWhat is the sum of all two-digit numbers that each is divisible by the both of its digits?\r\n\r\nChanged the wording from Kalva, who changed wording from the Official AMC Problem. :P", "Solution_1": "What is the sum of all two digit numbers where each is divisible by both of its digits?\r\n[hide]Well, we obviously have all the double-digits: 11+22+33+44+55+66+77+88+99.\nThen there are the even number where the second digit is double the first: 12, 24, 36, 48 (3*4, 3*8, 3*12, 3*16). Finally there is 15 . 11+22+33+44+55+66+77+88+99+12+24+36+48+15=630. So my answer is 630.[/hide]", "Solution_2": "[hide]\n\\[S=\\sum^{9}_{k=1}11k+\\sum^{4}_{k=1}12k+15\\]\n\\[S=495+120+15\\]\n\\[S=630\\]\n\n\n\n[/hide]", "Solution_3": "Oh, I see." } { "Tag": [ "probability" ], "Problem": "Two six-sided dice are fair in the sense that each face is equally likely to turn up. However, one of the dice has the $ 4$ replaced by $ 3$ and the other die has the $ 3$ replaced by $ 4$. When these dice are rolled, what is the probability that the sum is an odd number?\r\n\r\n$ \\textbf{(A)}\\ \\frac{1}{3}\\qquad \r\n\\textbf{(B)}\\ \\frac{4}{9}\\qquad \r\n\\textbf{(C)}\\ \\frac{1}{2}\\qquad \r\n\\textbf{(D)}\\ \\frac{5}{9}\\qquad \r\n\\textbf{(E)}\\ \\frac{11}{18}$", "Solution_1": "For the first dice the sides are $ {1,2,3,3,5,6}$.\r\nFor the second dice the sides are $ {1,2,4,4,5,6}$. \r\nFor the sum to be odd, one of the numbers had to be even and the other has to be odd. So, we have two cases.\r\n\r\nCase 1 (The first dice rolls to be odd and the other to be even)\r\n$ \\frac{4*4}{6*6}\\equal{}\\frac{4}{9}$\r\nCase 2 (The first dice rolls to be even and the other to be odd)\r\n$ \\frac{2*2}{6*6}\\equal{}\\frac{1}{9}$\r\n\r\nAdding these up we get $ \\frac{5}{9}$" } { "Tag": [ "calculus", "derivative", "function", "vector", "calculus computations" ], "Problem": "For the function $ f(x,y,z) \\equal{} x^2 \\plus{} y^2 \\plus{} z^3 \\minus{} 3z$, find $ \\nabla f$.\r\n\r\n(i) What is the rate of change of $ f(x,y,z)$ in the direction normal to the cylinder $ x^2 \\plus{} y^2 \\equal{} 25$ at the point $ (3, \\minus{}4, 4)$?\r\n(ii) At which points does $ \\nabla f$ have no component in the $ z$-direction?\r\n(iii) Find and classify the stationary points of $ f$.\r\n(iv) Sketch the contours of $ f$ and add to the sketch a few arrows showing the directions of $ \\nabla f$.", "Solution_1": "I'll get you started:\r\n$ \\nabla f\\equal{}\\left(f_{x},f_{y},f_{z}\\right)\\equal{}(2x,2y,3z^{2}\\minus{}3)$. Then a stationary point would be where $ \\nabla f\\equal{}(0,0,0)$, so that $ 2x\\equal{}2y\\equal{}3z^{2}\\minus{}3\\equal{}0$. Then $ (x,y,z)\\equal{}(0,0,\\pm 1)$. If you want to classify them, you can use a Hessian matrix. The second derivatives are easy to compute.", "Solution_2": "Okay, this looks like a natural extension of the two-variable case. But what does \"normal to the cylinder\" mean? I thought we can only find the direction in which the rate of change is the greatest, which is $ \\nabla f$", "Solution_3": "I'm assuming normal to the cylinder means perpendicular to it. The gradient gives you the tangent vector, from which you can find the normal vector (do you remember how to find normal lines?).", "Solution_4": "I see, in that case is the normal vector $ <3, \\minus{}4, 0>$? Then the answer to the first part would be $ <3, \\minus{}4, 0> \\cdot <6, \\minus{}8, 45>$ for a value of $ 50$. But the how do you interpret this value?\r\n\r\nPart (iv) doesn't seem possible because it's like asking us to plot a 4D contour..", "Solution_5": "No, it's in $ \\mathbb{R}^{3}$. You hold one value fixed and see what the function does for other values. [url=http://www.math.hmc.edu/calculus/tutorials/surfacecontours/]Here's[/url] a little tutorial.", "Solution_6": "Alright, thanks for the tip!", "Solution_7": "I can't understand \"The gradient gives you the tangent vector, from which you can find the normal vector\", but for (i) we have to project $ f$'s gradient into the normal of the cylinder. whose versor is $ ({3\\over5},\\minus{}{4\\over5}).$", "Solution_8": "Yes, I forgot to normalize there :maybe:" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let $n$ be a positive integer. Consider all permutations of $\\{1,2,...,n\\}$. Let $A$ denote the set of those permutations such that each number is either greater than all of the numbers to its left or less than all of the numbers to its right. Let $B$ denote the set of those permutations $a_{1},a_{2},...,a_{n}$ such that for $1\\leq i\\leq n-1$, there is a $j>i$ with $|a_{j}-a_{i}|=1$. Show that $|A|=|B|$.", "Solution_1": "I do not think it is true:\r\n\r\nFor $n = 3$, $A$ consists of all permutations except $\\{3, 2, 1\\}$, but $B$ consists of all permutations except $\\{2, 3, 1\\}$ and $\\{2, 1, 3\\}$. Did I misunderstand or make a mistake?", "Solution_2": "[quote=\"JBL\"]I do not think it is true:\n\nFor $n = 3$, $A$ consists of all permutations except $\\{3, 2, 1\\}$, but $B$ consists of all permutations except $\\{2, 3, 1\\}$ and $\\{2, 1, 3\\}$. Did I misunderstand or make a mistake?[/quote]\r\n\r\nWhen $n=3$ then $A=\\{(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2)\\}$ and $B=\\{(1,2,3),(1,3,2),(3,1,2),(3,2,1)\\}$ therefore $5=|A|\\not = |B|=4$ !!!! , problem is false! I think so! \r\n\r\nThis is a problem in ''A Path To Combinatorics For Undergraduates'' of Tittu and Zum." } { "Tag": [ "FTW" ], "Problem": "WAIT!!!!\r\n\r\nHow exactly does a FTW tourney work?????\r\n\r\nPlease tell me Where, When, Who, and How.", "Solution_1": "...people play against each other as scheduled, results are recorded, a winner is decided.", "Solution_2": "....in the FTW site?", "Solution_3": "Yes, on FTW. Hence the name [b]FTW tourney[/b].", "Solution_4": "In the future, please read the stickies. One of them pointed to a thread that described in great detail the workings of an FTW tournament. \r\n\r\nLocked." } { "Tag": [ "trigonometry", "inequalities proposed", "inequalities" ], "Problem": "Prove that for all real numbers $\\alpha_1,\\alpha_2,\\ldots,\\alpha_n$ we have \\[ \\sum_{i=1}^n \\sum_{j=1}^n ij \\cos {(\\alpha_i - \\alpha_j )} \\geq 0. \\]\r\n[i]Octavian Stanasila[/i]", "Solution_1": "[i]Dear V.V., [/i]this Stanasila's problem is a amusement:\r\n\r\n$ S\\equiv \\sum\\limits_{i=1}^n \\sum\\limits_{j=1}^n ij\\cos(x_1-x_j )=\\sum\\limits_{i=1}^n \\sum\\limits_{j=1}^n ij(\\cos x_i\\cos x_j +\\sin x_i\\sin x_j)= $\r\n$ \\sum\\limits_{i=1}^n i\\cos x_i\\cdot\\sum\\limits_{j=1}^n j\\cos x_j+\\sum\\limits_{i=1}^n i\\sin x_i\\cdot \\sum\\limits_{j=1}^n j\\sin x_j=\\left( \\sum\\limits_{i=1}^n i\\cos x_i \\right)^2+ $ $ \\left( \\sum\\limits_{i=1}^n i\\sin x_i \\right)^2\\ge 0. $\r\n\r\n[b][u]Remark.[/u] Two interesting problems:[/b]\r\n\r\n[b]1. When[/b] $ S=0 $ ?\r\n\r\n[b]2. Ascertain sum[/b] $ S\\ if\\ \\left(\\forall\\right)k\\in \\overline {1,n},\\ x_{k+1}=x_k+r $ [b](arithmetical progression).[/b]", "Solution_2": "[quote=\"levi\"][i]Dear V.V., [/i]this Stanasila's problem is a amusement:[/quote]It is easy, but it's not the only easy problem given at a Romanian TST :)" } { "Tag": [], "Problem": "We have been given a number $ a\\equal{}152855965207715 / 248623746255151$. Determine the rational numbers $ b\\equal{}n_1/m_1, c\\equal{}n_2/m_2$ such that $ 00$ are as close to $ 0$ as possible but they are not $ 0$.", "Solution_1": "Solved by the Euclidean algorithm. 484929707720781/788749333008502 and 585062048733224/951616890777555." } { "Tag": [], "Problem": "In three bowling games, Alice scores 139, 143, and 144. What score will alice need in a fourth game in order to have an average score of 145 for all four games?", "Solution_1": "[hide=\"Answer\"]$141$[/hide]", "Solution_2": "just by looking at the answer i think its wrong because your answer is not over 145 so the average can't be 145.", "Solution_3": "Oops... I meant [hide=\"this\"]$154$[/hide]", "Solution_4": "[hide] 139 + 143 + 144= 426, An average of 145 on 4 games would bring a total of 580. 580-426=[b]154[/b][/hide]", "Solution_5": "[hide]She needs 580 points and she currently only has 426 pts. So she needs 154 points. [/hide]", "Solution_6": "Its 154 :blush: :rotfl:", "Solution_7": "[quote=\"math92\"]In three bowling games, Alice scores 139, 143, and 144. What score will alice need in a fourth game in order to have an average score of 145 for all four games?[/quote]\r\n\r\n[hide]\n\nAdd them up. Get 426. 145*4=580. Then, subtract 426 from 580, to get $\\boxed{154}$.[/hide]", "Solution_8": "[hide] you add $139,143,144$. Then, you multiply $145$ by 4. subtract the two sums to get 154. The answer.\n$\\boxed{154}$\n[/hide]", "Solution_9": "[quote=\"math92\"]In three bowling games, Alice scores 139, 143, and 144. What score will alice need in a fourth game in order to have an average score of 145 for all four games?[/quote]\r\n\r\n[hide]139 is 6 under 145. 143 is 2 less than 145. 144 is one less than 145. 6+2+1=9 which means that Alice is 9 less than where she needs to be. 145+9=154.\n\n$154$[/hide]", "Solution_10": "[hide][quote=\"math92\"]In three bowling games, Alice scores 139, 143, and 144. What score will alice need in a fourth game in order to have an average score of 145 for all four games?[/quote]\n\n145*4-139-143-144\n580-282-144\n580-426\n154[/hide]", "Solution_11": "$145*4 = 580$\r\n$580-(139+143+144) =$\r\n$580-426 =$\r\n$154$\r\n :D She will need to score $154$ points.", "Solution_12": "[hide] 145*4=580\n580-139-143-144=\n154 [/hide]", "Solution_13": "please don't revive month old topics. :wink: :wink: :wink:" } { "Tag": [ "inequalities" ], "Problem": "Let reals $x_{1} g$ set $g$ to $x_{j}$, $x_{j}$ to $y_{k}$, and choose a different value for $y_{k}$ (This is where the permutations kick in). Eventually, if you never run into the $y_{k}\\le g$ condition, you will have $g = x_{n}$, $\\Rightarrow x_{n}+x_{j}< x_{j}+y_{k}$, contradiction since $x_{n}$ is the largest real according to our given. So if $y_{1}= x_{j}$ with $j \\ne 1$, we get a contradiction. Therefore, we must have $x_{1}= y_{1}$. You'll notice that for each determined value of $y_{k}$, $y_{k+1}$ must take a value from the ever-shrinking set $S-\\{y_{k}\\}$ until we've completely exhausted $S$. So if we assume $x_{i}\\ne y_{i}$ for the remainder of the $i$'s, using a similar argument as above we do get $x_{i}= y_{i}$ $\\forall$ $i = 1,2,...,n$, $Q.E.D.$[/hide]", "Solution_6": "Nice Problem, here's my solution :D:\r\n[hide]\nLet $A={x_{1},x_{2},x_{3}...x_{n}}$\nConsider $S={x_{1}+y_{1},x_{2}+y_{2},...x_{n}+y_{n}}={x_{1}+x_{\\in A},x_{2}+x_{\\in A},...x_{n}+x_{\\in A}}$.\nWe wish to pick $x_{1}$ and $y_{1}$ so that \n$x_{1}+y_{1}0.$ \r\n\r\nkunny's original question asks for the value $F(1).$\r\n\r\nHint: one can make useful connections between Laplace transforms of periodic functions and Fourier series.", "Solution_2": "Well, this definately isn't the best way, but here's what I did: \\[I=\\lim_{n\\to\\infty}\\int_{0}^{2n}e^{-x}f(x) \\ dx\\] \\[=\\sum_{j=0}^\\infty\\left(\\int_{2j}^{2j+1}xe^{-x}\\ dx+\\int_{2j+1}^{2j+2}(2-x)e^{-x}\\ dx\\right)\\] \\[=\\sum_{j=0}^\\infty \\left(2j(e-1)^{2}+e^{2}-2e+1\\right)e^{-2(j+1)}\\] \\[=\\sum_{j=0}^\\infty (-4j-2)e^{-2j-1}+\\sum_{j=0}^\\infty (2j+1)e^{-2j-2}+\\sum_{j=0}^\\infty (2j+1)e^{-2j}\\] \\[=\\frac{e^{2}+1}{(e+1)^{2}}\\] Edit: I'm not too sure about my answer because when I tried the solution Kent was referring this is what I got:\r\n\r\nFor a periodic function $f(x)$ with period $p$, we have $F(s) = \\frac{1}{1-e^{-ps}}\\int_{0}^{p}f(t)e^{-st}\\ dt$, so in our case we have \\[F(s) = \\frac{1}{1-e^{-2s}}\\left(\\int_{0}^{1}te^{-st}\\ dt+\\int_{1}^{2}(2-t)e^{-ts}\\ dt\\right) = \\frac{\\tanh{s/2}}{s^{2}}\\] and then $F(1) = \\tanh 1/2 \\not = \\frac{e^{2}+1}{(e+1)^{2}}$", "Solution_3": "You're better off taking the derivative first before doing the Laplace transform, and using that \r\n\r\n$\\mathcal{L}(f')(s)=sF(s)-f(0).$\r\n\r\nWorking that way, I get that $F(s)=\\frac{e^{s}-1}{s^{2}(e^{s}+1)}$ meaning the integral in this problem is\r\n\r\n$F(1)=\\frac{e-1}{e+1}.$", "Solution_4": "OK, Jimmy and I agree with each other - or at least I agree with his second answer.", "Solution_5": "Any idea where I went wrong in the first one?" } { "Tag": [ "integration", "calculus", "calculus computations" ], "Problem": "Find the whole portion of:\r\n\r\n$A=\\sum_{n=1}^{10^{9}}n^{\\frac{-2}{ 3}}$", "Solution_1": "If by \"whole portion\" you mean \"integer part\", the answer is 2997.\r\n\r\nThis sum can be compared to an integral. The comparison is quite visual - draw boxes both above and below the curve $y=x^{-2/3}.$ We can show that\r\n\r\n$\\int_1^{m+1}x^{-2/3}dx<\\sum_{n=1}^mn^{-2/3}<1+\\int_1^mx^{-2/3}dx$\r\n\r\n$3(m+1)^{1/3}-3<\\sum_{n=1}^mn^{-2/3}<3m^{1/3}-2.$\r\n\r\nFor $m=10^9$ this clearly places the sum between 2997 and 2998." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "hi ! $ x,y,z\\geq0$ prove that \r\n[img]http://www.texify.com/img/%5CLARGE%5C%21%28xy%2Byz%2Bxz%29%5Csqrt%5B%5D%7B%28x%2By%2Bz%29%5E2%2B4xyz-%28x%2By%2Bz%29%2B1%7D%5Cgeq6xyz.gif[/img]", "Solution_1": "let $ r\\equal{}\\sqrt[3]{xyz}$ its obious that $ xy\\plus{}yz\\plus{}zx\\geq 3r^2$ and by AM-GM $ \\dfrac{(x\\plus{}y\\plus{}z)^2}{4}\\plus{}1\\geq (x\\plus{}y\\plus{}z)$ and $ \\dfrac{3(x\\plus{}y\\plus{}z)^2}{4}\\plus{}4xyz\\geq \\dfrac{27r^2}{4}\\plus{}4r^3$ thus $ (xy\\plus{}yz\\plus{}zx)\\sqrt{(x\\plus{}y\\plus{}z)^2\\plus{}4xyz\\minus{}(x\\plus{}y\\plus{}z)\\plus{}1}\\geq 3r^2\\sqrt{\\dfrac{27r^2}{4} \\plus{}4r^3}\\geq (3r^2)(\\sqrt{4r^2})\\equal{}6r^3\\equal{}6xyz$" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ n$ be a positive integer. Given a sequence of $ n$ elements, each 1 or -1. In one step it is allowed to transform sequence in to another one such that every element becomes equal to the product of his neighbours (each element have 2, except first and last element having 1 neighbour). Find all $ n$ such that for every possible arrangement of 1 and -1 in first sequence after finite number of steps we will get sequence $ (1,1,1,\\ldots,1)$ (every element is 1).\r\n\r\nExample of transformation:\r\n\\[ (1, \\minus{} 1,1, \\minus{} 1,1) \\mapsto ( \\minus{} 1, 1, 1, 1, \\minus{} 1) .\r\n\\]\r\nSorry for my BAD English. :blush:", "Solution_1": "anyone, please! :)", "Solution_2": "Sort of a sketch: Let us go backwards. From $ (1,1,1,1,\\dots,1,1,1,1)$ we may reach either $ (1,1,1,1,\\dots,1,1,1,1)$ itself or $ ( \\minus{} 1,1, \\minus{} 1,1, \\minus{} 1,\\dots,1, \\minus{} 1,1, \\minus{} 1)$. We consider the latter case. Here the string $ \\minus{} 1,1$ appears $ \\frac {n \\plus{} 1}{2}$ times (in the last appearance a $ 1$ is missing). Now from $ ( \\minus{} 1,1, \\minus{} 1,1, \\minus{} 1,\\dots,1, \\minus{} 1,1, \\minus{} 1)$ we can reach $ (x, \\minus{} 1, x, 1, x, \\minus{} 1, x,\\dots, x, \\minus{} 1, x)$, where $ x$ can be replaced with either $ 1$ or $ \\minus{} 1$. Note that the string $ x, \\minus{} 1,x,1$ appears $ \\frac {n \\plus{} 1}{4}$ times (in the last appearance a $ 1$ is again missing). Again, from here we can go to $ (y, x, \\minus{} y, 1, \\minus{} y, x, y, 1, \\dots, \\minus{} y, x, y)$, where $ x,y\\in\\{1, \\minus{} 1\\}$. Here the string $ y, x, \\minus{} y, 1, \\minus{} y, x, y, 1$ appears $ \\frac {n \\plus{} 1}{8}$ times (in the last appearance of the string a $ 1$ is again missing). We have something like the following:\r\n\r\n. . . . . . . . . . . . \r\n. . . . . . . . . . . . \r\ny, x, -y, 1, -y, x, y, 1, . . ., -y, x, y \r\nx, -1, x, 1, x, -1, x, 1, . . ., x, -1, x \r\n-1, 1, -1, 1, -1, 1, -1, . . ., -1, 1, -1\r\n1, 1, 1, 1, 1, 1, 1, . . ., 1, 1, 1 \r\n\r\nProceeding on like this finally we are left with a single string, where the last $ 1$ will be missing, and we have freedom of choice for all the other terms of the string. This implies $ \\frac {n \\plus{} 1}{2^k} \\equal{} 1$ for some $ k$, and thus $ n \\equal{} 2^k \\minus{} 1$.\r\n\r\nBut we have a little problem. If there are some $ 1$'s or $ \\minus{} 1$'s in the last string, for which we cannot apply our freedom of choice, we can yet go backwards until they are deleted and reach a state where all of the terms can take arbitrary values.\r\n\r\nA second approach: Since before reaching $ (1,1,1,1,\\dots,1,1,1,1)$ we must reach $ (\\minus{}1,1,\\minus{}1,1,\\dots,\\minus{}1,1,\\minus{}1)$, we must have $ n\\equal{}2n_1\\plus{}1$ for some $ n_1$. Now we delete all the $ \\minus{}1$'s in our last sequence to get a new sequence of $ n_1$ $ 1$'s. Applying the same argument for $ n_1$ we find that $ n_1\\equal{}2n_2\\plus{}1$ for some $ n_2$. If we keep going like this we will eventually have $ n_k\\equal{}0$ for some $ k$. That means $ n\\equal{}2^k\\minus{}1$." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all functions $f,g,h: \\mathbb{R} \\mapsto \\mathbb{R}$ such that $f(x) - g(y) = (x-y) \\cdot h(x+y)$ for $x,y \\in \\mathbb{R}.$", "Solution_1": "Take $x=y$ and you find $f=g$. Next, take $y=0$ and you find $ xh(x)=f(x)-f(0)$. Put $ t(x)=f(x)-f(0)$. Then $ (x-y)t(x+y)=(x+y)(t(x)-t(y))$. Now, take $ y=1$ and you find $ t(x+1)=\\frac{x+1}{x-1} (t(x)-t(1))$. Substitute here $x+1$ instead of $x$ and obtain $ t(x+2)=\\frac{x+2}{x} (\\frac{x+1}{X-1}(t(x)-t(1))-t(1))$. But $ t(x+2)=\\frac{x+2}{x-2} (t(x)-t(2))$. From the two relations we find that $t$ is of the form $ ax+b$ and since $ t(0)=0$ we have $t(x)=ax$ and then $ f(x)=ax+b$.", "Solution_2": "Thats not all the possible functions, note for example $ h(x) \\equal{} ax$, $ f(x) \\equal{} g(x) \\equal{} ax^2 \\plus{} b$\r\n\r\n[quote=\"harazi\"] From the two relations we find that $ t$ is of the form $ ax \\plus{} b$ [/quote]\r\n\r\nIn fact from that tow little guys we get $ t(x)\\equal{}x^2\\times (\\frac{1}{2} f(2)\\minus{}f(1))\\plus{} x (2 f(1) \\minus{} \\frac{1}{2} f(2) )$ for $ x\\not\\equal{} 0, 1, 2$ :D\r\n\r\nMake the substitucion $ a\\equal{}\\frac{1}{2} f(2)\\minus{}f(1) \\wedge b\\equal{}( 2f(1) \\minus{} \\frac{1}{2} f(2) \\Leftrightarrow f(2)\\equal{}4a\\plus{}b \\wedge f(1)\\equal{}a\\plus{}b$ and we get $ t(x)\\equal{}ax^2\\plus{}bx$ for all reals, because it is true for $ x\\equal{}0$ by defenition of $ t(x)$, and $ f(2)\\equal{}4a\\plus{}b\\equal{}a2^2\\plus{}b2$, $ f(1)\\equal{}a\\plus{}b\\equal{}a1^2\\plus{}1b$\r\n\r\nRecall $ f(0)\\equal{}c$ and we get $ f(x)\\equal{}ax^2\\plus{}bx\\plus{}c\\equal{}g(x)$ and $ h(x)\\equal{}ax\\plus{}b$\r\n\r\nNow check that $ ax^2\\plus{}bx\\plus{}c\\minus{}ay^2\\minus{}by\\minus{}c\\equal{}(x\\minus{}y) \\times (a(x\\plus{}y) \\plus{} b)$ occur for $ x, y$ reals.", "Solution_3": "Let $P(x,y)$ be the assertion and let $g(0)=c,kf(0)=k$.Then $P(x,0) \\Rightarrow f(x)-c=xh(x) \\Rightarrow f(x)=xh(x)+c$.$P(0,y) \\Rightarrow k-g(y)=-yh(y) \\Rightarrow g(y)=k+yh(y)$.Substituting these in the original equation we get $xh(x)-yh(y)+f=(x-y)h(x+y)$ where $f=c-k$.$y=0 \\Rightarrow f=0 \\Rightarrow c=k$.So the equation may be rewritten as $xh(x)-yh(y)=(x-y)h(x+y)$\nNow $P(x,-x) \\Rightarrow h(x)+h(-x)=2h(0) \\forall x \\neq 0$.Substituting $y=-y$ we get\n$xh(x)+yh(-y)=(x+y)h(x-y)$\n$\\Rightarrow xh(x)+y(2h(0)-h(y))=(x+y)h(x-y)$\n$\\Rightarrow (x-y)h(x+y)+2h(0)y=(x+y)h(x-y)$\n\nSubstituting $y=(x-1)$ in this equation yeilds $h(2x-1)=(2x-1)(h(1)-2h(0))$.In other words $h(x)=dx$ for some $d \\in \\mathbb{R}$.\nWe may hence conclude that \n$f(x)=g(x)=dx^2+c$ for some $d \\in \\mathbb{R}$\n\nThese system clearly satisfies the original FE." } { "Tag": [ "graph theory", "combinatorics solved", "combinatorics" ], "Problem": "An infinite tree graph is given in the plane such that every lattice point is a vertex of the graph, and every edge of the graph connects two lattice points whose Euclidean distance is at most 1998.\r\nProve that there exist two points in the plane, a unit distance apart, such that the path which connects them in the tree is at least 10^1998 long !", "Solution_1": "So many Combinatorics Problems. And so few solutions. Where do you find them , ORL? Do you have any solutions for them?", "Solution_2": "Hi Dust,\r\n\r\nat first I cannot refer each problem I posed to a certain olympiad. Sometimes I found it in any journal. I have some of the solutions but I want to make the forum members try to solve it. I cannot evaluate whether the forum members dislike my Combinatorics, especially graph theory, problems (meaning they are not sufficiently appealing) or maybe they regard them as challenging and did not find any approach yet.", "Solution_3": "I solved it, but all combinatoric problems are very hard to expound in foreign language. It is the main reason why many combinatoric problems haven't solution on this forum, I think.", "Solution_4": "I think there are only few native English speakers on the forum who might complain. Nobody should try to raise his pedagogical index finger in order to indicate that one's English is superb. As long as the forum members know what you try to explain I feel there is no problem about it. Nobody will kill you ! Simply try !!! :)", "Solution_5": "Suppose it is false.\r\n\r\nLet T be the tree. Let us appoint (0,0) as the root of T.\r\nLet l be 1998, L be 101998.\r\n\r\nConsider square Q with center (0,0) and side-lenght l*L.\r\nLet V be the set of vertex on border of Q.\r\nAt initial moment V has 4l*L points with unit distance between neighboring vertexes, and if X,Y are \r\n\r\nneighbor then |PathInTree(X,Y)| |V|-1.\r\n\r\nStep 6) Go to Step 1).\r\n\r\nWe will stop process 1)-6) when set V will first time have two neighboring vertex X and Y whose Euclidian distance is at least l*L ==> |PathInTree(X,Y)| \\geq l*L/l (because each rib (branch) of graph is at most l) = L !!!! Contradictory with property |PathInTree(X,Y)|$ is dense in $G$, then $G$ is abelian.\r\n\r\n2) Let $H$ be a closed subgroup of a compact group $G$ such that $H \\neq G$. Then there exists a character $f: G \\to S$ such that $f \\not \\equiv 1$ and $f(H)=1$, where $S=\\{z : |z|=1\\}$.\r\n\r\nRemembering that a character is just a continuous homomorphism $f: G \\to S$", "Solution_1": "the first one you have to use the fact that a cyclic group is abelian and the operation $G \\times G \\to G : (x,y) \\mapsto xy$ is continuous.\r\n\r\n\r\nFS", "Solution_2": "In 2) do you suppose that $H$ is normal? If the conjugates of $H$ generate $G$ (e.g., $G=S_{3}$ and $H=\\{e,(12)\\}$), then there is no such $f$. \r\n\r\nIf $H$ is normal, then compose the canonical homomorphism $G\\to G/H$ with a non-principal character of $G/H$.", "Solution_3": "THANKS FOR YOUR HELP. \r\n\r\njust two things:\r\n\r\n1) $H$ is always normal, since by definition $G$ is Hausdorff (it is a topological group, well i am working with that definition :P ) , so $H$ is, and by compactness of $G$ , $H$ is compact , then $H$ is a compact hausdorff space, therefore normal :) .\r\n\r\n\r\n2)sorry , but I am not very familiar with characters, so why can i suppose that there exists such a non-principal character? .\r\n\r\n\r\n\r\nAdriana", "Solution_4": ":) I meant to ask if $H$ is a normal subgroup (as in $x\\in H\\implies gxg^{-1}\\in H$ for any $g\\in G$), not a normal topological space. \r\nNon-principal characters exist on any locally compact [b]abelian[/b] group; in fact the characters separate the points of the group by the [url=http://en.wikipedia.org/wiki/Pontryagin_duality]Pontryagin duality theorem[/url]. Using Pontryagin duality may be an overkill, but I just can't think of a simple direct proof of the existence of non-principal characters. :( For a non-abelian group you take its quotient by the commutant and hope that the quotient is nontrivial.", "Solution_5": "well thank u so much. i really appreacite ur help\r\n\r\n\r\n\r\nAdriana", "Solution_6": "[quote=\"mlok\"]In 2) do you suppose that $H$ is normal? If the conjugates of $H$ generate $G$ (e.g., $G=S_{3}$ and $H=\\{e,(12)\\}$), then there is no such $f$. \n\nIf $H$ is normal, then compose the canonical homomorphism $G\\to G/H$ with a non-principal character of $G/ H$.[/quote]\r\n\r\n\r\nI was thinking about this, and it's not necessary that $H$ be normal, we can consider the left (right) coset $G/H$ which is a group and under the canonical epimorphism is a topological space.", "Solution_7": "I don't see how $G/H$ is a group... And consider the above example with $S_{3}$: if a homomorphism $f\\colon S_{3}\\to \\mathbb C$ sends $H$ to $1$, then $f$ is identically equal to $1$.", "Solution_8": "[quote=\"mlok\"]I don't see how $G/H$ is a group... And consider the above example with $S_{3}$: if a homomorphism $f\\colon S_{3}\\to \\mathbb C$ sends $H$ to $1$, then $f$ is identically equal to $1$.[/quote]\r\n\r\n\r\nYes, you are right, it was a stupid mistake, of course $H$ has to be normal, otherwise the operation, in $G/H$ , is not well defined.\r\n\r\n\r\n\r\nAdriana" } { "Tag": [ "conics", "hyperbola", "geometry", "geometric transformation", "reflection" ], "Problem": "Prove that for every positive integer n , the difference $ s_n = (\\sum_{k=1}^{n}[\\frac{n}{k}]) - [\\sqrt{n}] $ is an even integer , where [x] denotes the integer part of x", "Solution_1": "Let $A = \\{1,2, \\cdots , n \\}$ and $S = \\sum_{k=1}^n [\\frac n k ].$\r\n\r\nFor each $k$, the number $[\\frac n k]$ denotes the number of multiples of $k$ in $A$. Therefore, each element $i$ from $A$ contributes for $1$ in $S$ exactly as many times as its number of positive divisors, say $d(i).$\r\nIt follows that $S = \\sum_{k=1}^n d(k).$\r\nBut, it is well known that $d(k)$ is odd if and only if $k$ is a perfect square. Thus $S$ has the same parity than the number of squares in $A$, which is $[\\sqrt n ]$, and the desired result now follows easily.\r\n\r\nPierre.", "Solution_2": "I thought of this geometrically (but in essence it's the same solution as pbornsztein's).\r\n\r\nThe summation in the problem is the number of lattice points in the first quadrant that are on or below the hyperbola $xy=n$. Of these lattice points, $[\\sqrt{n}]$ of them lie on the line $y=x$. The remaining lattice points can be paired by reflection across the line $y=x$, so there are an even number of them.\r\n\r\nYakko" } { "Tag": [ "geometry", "geometric transformation", "reflection", "geometry unsolved" ], "Problem": "this is 220 exercise [b]a ra rb [/b]of Prof Lu\u00eds Lopes Thesis (Handbook of triangles construction)\r\n\r\ni made [b]hc= 2rarb/ (ra+rb)[/b] and i have the line of c side\r\n\r\nbut 220 exercise says that ra rb hc is a datum we have hc so the problem is now 164 with [b]a hc ra[/b] but this is a mistake because 164 problem says \r\nsin(gama)=[b]hb/a[/b] isn't [b]hc[/b]\r\n\r\ni want the A point construction with i've got now \r\n\r\n :(", "Solution_1": "It is very easy to finish. $ F \\in c$ is foot of the C-altitude $ h_c.$ Having the sidelines $ a, c$ interseting at $ B,$ construct the internal and external bisectors $ BI_b$ and $ BI_a$ of the $ \\angle B.$ (The excenters $ I_b, I_a$ are unknown, but you can draw the angle bisectors.) Draw 2 parallels $ p, q$ to $ a$: $ p$ at the distance $ r_a$ from $ a$ on the opposite side of $ a$ than $ F,$ $ q$ at the distance $ r_b$ from $ a$ on the same side of $ a$ as $ F.$ The parallel line $ p$ cuts the external bisector $ BI_a$ of the angle $ \\angle B$ at $ I_a,$ the parallel line $ q$ cuts the internal bisector $ BI_b$ of the $ \\angle B$ at $ BI_b$ at $ I_b.$ Draw the excircles $ (I_a), (I_b)$ tangent to $ a.$ The center line $ I_aI_b$ should cut $ a$ at $ C.$ Reflection of $ a$ in the line $ I_aI_b$ is the common internal tangent of $ (I_a), (I_b)$ other than $ a,$ i.e., the sideline $ b$ cutting the sideline $ c$ at $ A.$", "Solution_2": ":thumbup: :clap:" } { "Tag": [], "Problem": "Three machines P, Q, R work together and can do a job in x hours. P needs an extra 6 hours when working alone; Q needs an extra 1 hour when working alone; R needs an extra x hours when working alone. Find x.\r\n\r\nBlah, anyone know how to do this? I hate these sort of problems.", "Solution_1": "remember that their individual rates sum up to the total rate.", "Solution_2": "Ah! There we go. Thanks.", "Solution_3": "I believe there are a couple of these in AoPS 1 Chapter 3 if you want practice.", "Solution_4": "I think that x is 2/3 hour" } { "Tag": [ "modular arithmetic" ], "Problem": "Show that $a^2\\equiv1\\pmod8$ for any odd $a$.\r\n\r\nShow that $a\\equiv{b}\\pmod{m}$ and $0 b$\nThis implies that because $m,b > 0$, $1-k>0 \\Rightarrow k<1 \\Rightarrow k=0$ because it must be a nonnegative integer. \n\nTherefore $a=m(0)+b \\implies a=b \\ \\ \\ \\ \\mathbb{W}^5$[/hide]", "Solution_6": "[quote=\"chess64\"][hide=\"1\"]\n\nSince $a$ is odd, $a \\equiv 1, 3, 5, \\text { or } 7 \\pmod {8}.$ Squaring, we see $a^2 \\equiv 1, 9, 25, 49 \\equiv 1, 1, 1, 1, \\equiv 1 \\pmod {8} \\ \\ \\ \\ \\ \\mathbb{W}^5$[/hide]\n\n[hide=\"2\"]\n\n$a \\equiv b \\pmod {m} \\implies 0 < a=mk+b < m$\n$mk+b < m$\n$m(k-1)<-b$\n$m(-1)(1-k) < -b$\n$m(1-k) > b$\nThis implies that because $m,b > 0$, $1-k>0 \\Rightarrow k<1 \\Rightarrow k=0$ because it must be a nonnegative integer. \n\nTherefore $a=m(0)+b \\implies a=b \\ \\ \\ \\ \\mathbb{W}^5$[/hide][/quote]Looks good. Here's my solution for 2.\n[hide=\"2\"]Let $a=b+km$. Then $b+km-1.$ Therefore $-10, a\\plus{}b\\plus{}c\\equal{}1$ . Prove that:\r\n$ (1 \\plus{} a)^{\\frac{b}{a}} (1 \\plus{} b)^{\\frac{c}{b}} (1 \\plus{} c)^{\\frac{a}{c}} \\ge \\frac{{64}}{{27}}$\r\n-----\r\nEasy?", "Solution_1": "[quote=\"ML2008\"]Let $ a,b,c > 0, a + b + c = 1$ . Prove that:\n$ (1 + a)^{\\frac {b}{a}} (1 + b)^{\\frac {c}{b}} (1 + c)^{\\frac {a}{c}} \\ge \\frac {{64}}{{27}}$\n-----\nEasy?[/quote]\r\n$ Ln(LHS)= b \\frac{ln(1+a)}{a}+c \\frac{ln(1+b)}{b}+a \\frac{ln(1+c)}{c}= af(c)+bf(a)+cf(b)$ where $ f(x)= \\frac{ln(1+x)}{x}$, $ x > 0$;\r\n$ f'(x)= \\frac{1}{x^2}(\\frac{x}{1+x}-ln(1+x)) < 0 \\Longrightarrow$ f(x) - decrease for $ x>0$\r\n$ f''(x)= \\frac{1}{x^3}(\\frac{x}{(1+x)^2}-\\frac{3x}{1+x}+2ln(1+x)) > 0 \\Longrightarrow$ f(x) - concave for $ x>0$\r\nSo by Jensen :\r\n$ b \\frac{ln(1+a)}{a}+c \\frac{ln(1+b)}{b}+a \\frac{ln(1+c)}{c} \\ge \r\n(a+b+c)^2 \\frac{ ln(1+\\frac{ab+bc+ca}{a+b+c} )} {ab+bc+ca}=$$ \\frac{ ln(1+ab+bc+ca )} {ab+bc+ca}\\ge 3 ln(1+\\frac{1}{3} )=ln(\\frac{64}{27})$;\r\n$ (ab+bc+ca \\le \\frac{1}{3})$" } { "Tag": [ "analytic geometry", "trigonometry", "conics", "parabola" ], "Problem": "How would you write $y = x$ in polar form?\r\n\r\n$r\\sin x = r\\cos x$\r\n\r\nHow would you isolate the $r$?", "Solution_1": "Polar equations don't always have to have r, just like cartesian equations don't always have both y and x. In this case, either you can just draw the solution and see, or do what you did, and solve for x to get the equation to be $\\theta = \\frac{\\pi}4$ to be the equation.", "Solution_2": "So not all $f(x,y)$ can be converted to polar form?\r\n\r\nOr would that denote a line going through $45^\\circ$?", "Solution_3": "btw, how do I convert the equation into polar form if there is a coefficient? Like $y=x^2 +1$", "Solution_4": "Do you mean coefficient, exponent, or constant? Since you added an exponent and a constant I'm guessing you used the wrong term. The equation you gave (a parabola), being a conic, would be graphed in polar coordinates following the equation of a conic in polar. Problem is, conics in polar form need to have the focus at the pole (as far as I know) to be written in this form.", "Solution_5": "Hmm, so you can't write a polar form of an equation with a coefficient? In coefficient I mean an $x^0$ term like a 1 or a -2 in the equation.", "Solution_6": "[quote=\"breez\"]So not all $f(x,y)$ can be converted to polar form?\n\nOr would that denote a line going through $45^\\circ$?[/quote]\n\nJust as not all $f(x, y)$ can be written explicitly in $y =$ form, so all polar $p(r, \\theta)$ cannot be written explicitly in $r =$ form, but all $f(x, y)$ can be written in some sort of implicit polar form by simple substitution.\n\nYes, the equation $\\theta = \\frac{\\pi}{4}$ denotes the set of all points $(r, \\theta)$ such that $\\theta = \\frac{\\pi}{4}$, hence $r$ ranges across the reals.\n\n[quote=\"breez\"]btw, how do I convert the equation into polar form if there is a coefficient? Like $y=x^2 +1$[/quote]\r\n\r\n$r \\sin \\theta = r^2 \\cos^2 \\theta + 1$\r\n$r^2 \\cos^2 \\theta - r \\sin \\theta + 1 = 0$\r\n$r = \\frac{\\sin \\theta \\pm \\sqrt{\\sin^2 \\theta - 4 \\cos^2 \\theta}}{2 \\cos^2 \\theta}$", "Solution_7": "How exactly did you substitute the 1 into that polar equation?\r\n\r\nIs there a generalized method that works for all coefficients?", "Solution_8": "You simply add one; no need to convert it. \r\n\r\nTo what was said earlier -- conics of the form $\\frac{ep}{1\\pm p\\sin\\theta}$ (and the corresponding one with $\\cos$) have to have a focus at the center, but there are conics of other forms which do not have to have a focus at the center.", "Solution_9": "Isn't the 1 adding to the length of r? In $x^2 + 1$, the 1 adds to the y-axis value. If you added 1 in the polar equation, it would lengthen the r and not increase the y values by 1. \r\n\r\n$r \\sin \\theta$ must increase by 1.", "Solution_10": "[quote=\"breez\"]Isn't the 1 adding to the length of r? In $x^2 + 1$, the 1 adds to the y-axis value. If you added 1 in the polar equation, it would lengthen the r and not increase the y values by 1. \n\n$r \\sin \\theta$ must increase by 1.[/quote]\r\n\r\nI didn't add $1$ to the polar equation, I merely kept the $1$ that was already there. If you plug my solution (both signs) into your calculator it should produce the same graph. I don't understand what you're confused about :huh:\r\n\r\nEdit: As has already been said, [b]not all polar graphs are explicitly $r =$ form.[/b] Any implicit Cartesian graph $f(x, y) = 0$ can be transformed into an implicit polar graph $p(r, \\theta) = 0$ by using $p(r, \\theta) = f(r \\cos \\theta, r \\sin \\theta)$.\r\n\r\nExamples:\r\n\r\n$y = x^3 + x^2 + 1 \\implies r \\sin \\theta = r^3 \\cos^3 \\theta + r^2 \\cos^2 \\theta + 1$\r\n\r\n$x^2 y + y^2 x + 3 = 0 \\implies r^3 \\cos^2 \\theta \\sin \\theta + r^3 \\sin^2 \\theta \\cos \\theta + 3= 0$\r\n\r\n$y = e^x + 17 \\implies r \\sin \\theta = e^{r \\cos \\theta} + 17$", "Solution_11": "Thank you. I get it." } { "Tag": [ "trigonometry", "geometry solved", "geometry" ], "Problem": "let ABC a right-angled triangle in C and I medium of [ BC ]. show that sin(\\alpha>0$, where $\\ell(\\cdot)$ denotes the length of the interval, but I'm not sure. And how do you define the notion for unmeasurable $S$?", "Solution_4": "I suppose you could get around it by saying that any purely random countable sequence of points in $S$ is uniformly distributed in $I$ (in the normal way, say using Weyl's Criterion). But I'm still a little unsure of that, and I should be wary of talking about random things when I haven't really studied probability theory at all. \r\n\r\nYou're right, this is a tricky problem to state the way I want to. This is pretty much the same as a question I had on homework, but it was worded much differently (it was really just an exercise on points of Lebesgue density). I noticed that the result we proved actually said something about \"uniformly\" distributed subsets of intervals, so I thought I'd post it here. Perhaps I should have thought about it a bit more carefully rather than toss it into the ring so casually. \r\n\r\nBut I my meaning is clear, at least -- that you can't make a measurable set that's uniformly distributed as grobber defined it unless its measure is 0 or differs by a set of measure 0 from the smallest interval containing it.", "Solution_5": "Well, in that case, it follows immediately from the fact that a measurable set can be approximated arbitrarily well with open sets. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=poorly&t=68528]this[/url], for example.\r\n\r\nEdit:\r\n\r\nSorry; I erased that stupid comment of mine.", "Solution_6": "[quote]you can always choose a convergent subsequence[/quote]\r\n\r\nI won't challenge that, but if you did, then it wouldn't be random, would it? Or are you saying that there are enough convergent subsequences that if I looked at a random sequence it might be a convergent one? That seems unlikely to me. (heh.)\r\n\r\nBut on the other hand, does the fact that such sequences exist already screw me up, even if the probability of such a sequence being chosen is infinitesimal?" } { "Tag": [ "search", "function", "\\/closed" ], "Problem": "I have a question. In this website, you have your username, and on the bottom of it it says your rank in stars. How do I improve my rank, since my rank says New Member and no stars?", "Solution_1": "Well, you're a P versus NP now. :D You can improve your rank by the number of posts you have (I think 50 will get you the next higher rank, Hodge Conjecture).\r\nbtw, this belongs in games and fun factory or something.", "Solution_2": "ty easyas3.14159...", "Solution_3": "Correction for easyasPI... according to my sources after 15 posts you gain your first rank, but you can ask an admin to be sure. But if you don,t want to take advice from a new user, thats okay. :starwars:", "Solution_4": "You should try searching for these kinds of things before posting. I posted this question before and the post got locked :blush: ;)", "Solution_5": "Use the search function, Luke... :yoda:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Rank+Stars&t=79143[/url]\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Rank+Stars&t=73553[/url]\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Rank+Stars&t=28375[/url]\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Rank+Stars&t=18966[/url]\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Rank+Stars&t=13045[/url]\r\n\r\nOkay, I'll stop.", "Solution_6": "[quote=\"dogseatcheese\"]You should try searching for these kinds of things before posting. I posted this question before and the post got locked :blush: ;)[/quote]Exactly :)" } { "Tag": [ "email", "search" ], "Problem": "Hello,\r\n\r\nI hope this will not be considered spam. I found this trick (that worked) for Hotmail. Since a few months ago they are offering 250 MB, but that only work for the new accounts, and they start first with 25 and afterwards 250. But all of us who are old member don't have that nice feature and still have our simple 2MB of storage (what is ridiculus if you compare with 1GB of Yahoo and 2+GB of Gmail). Then, here there is a trick to get first your 25 and afterwards the other 250 (after ten days)\r\n\r\nLogin to your account\r\nGo to Option\r\nClick My Profile\r\nChange Country to United States\r\nWait for browser to Load US Settings\r\nChange state to Florida and Zip Code to 33332 (I tried with mine, Georgia and 30332 and didn't work, but with that one it does work)\r\nClick Update\r\nClick Continue\r\n\r\nGo to Language and Make sure it is English. Paste this link in Same Browser: http://by17fd.bay17.hotmail.msn.com/cgi-bin/AccountClose\r\n\r\nWait until the screen says \"Your Hotmail account is closed and ready to be deleted.\" Click Close Account.\r\n\r\nGo Back to Login and relogin to ur Account. Just reactivate ur account. No email will be deleted from ur account and now ur account size limit increase to 25MB. (Really!)\r\n\r\nNow, it is important to do it inmediately. It shouldn't be that you do half of it one day and the other half the next day, because as you see, YOUR ACCOUNT IS CLOSED AND SCHEDULED TO DELETION, then, if you wait a day, may be it will be deleted!\r\n\r\nAny way, I hope it helps!", "Solution_1": "mine worked. THANK YOU!", "Solution_2": "awesome\r\n :lol:", "Solution_3": "who still uses hotmail? :o \r\ngmail is way better. i have 50 invites if anyone wants one.", "Solution_4": "[quote=\"furious\"]who still uses hotmail? :o \ngmail is way better. i have 50 invites if anyone wants one.[/quote]\r\n\r\nI use hotmail... and gmail... I actually like hotmail more, but that's just me.", "Solution_5": "[quote=\"djimenez\"] \nSince a few months ago they are offering 250 MB, but that only work for the new accounts\n[/quote]\r\n\r\nI started my account ~5 years ago, and it automatically got upgraded to 250 MB.", "Solution_6": "[quote=\"furious\"]who still uses hotmail? :o \ngmail is way better. i have 50 invites if anyone wants one.[/quote]\r\n\r\ni use yahoo for regular email.\r\ni use gmail to send myself projects and things.\r\ni use hotmail for livejournal stuff, and when i fill out college things for standardized tests.", "Solution_7": "yeah im with paladin, i use both hotmail and gmail, but im just more used to hotmail. so yeah. i use hotmail more.", "Solution_8": "i like hotmail too :) didn't even consider getting gmail (i guess it's also the fact that i don't exchange that many messages) this trick was very cool though thanks alot djimenez", "Solution_9": "GMAIL>any thing else.\r\nGmail is the only free email that doesn't attach a stupid auto generated signature at the end of each email.\r\nThat gets really annoying some times.", "Solution_10": "Gmail owns you.\r\n\r\nRecently they changed their style of displaying their ad's in Gmail.\r\n\r\nLook at the spam section for some recipes. :)", "Solution_11": "OR another way to boost your HOTMAIL MB up:\r\n\r\n1. Find a friend who has MSN Internet(Broadband)\r\n2. Tell them to add their address to their network of registered users.\r\n3. you should have it", "Solution_12": "[quote=\"themonster\"][quote=\"djimenez\"] \nSince a few months ago they are offering 250 MB, but that only work for the new accounts\n[/quote]\n\nI started my account ~5 years ago, and it automatically got upgraded to 250 MB.[/quote]You are lucky, most of the people didn't got that (well, at least most of the people who opened hotmail from out of the country).\r\n\r\nAny way, I use several emails for different things. Hotmail mainly for email lists, walla for the replys from ML only, yahoo for personal emails, and recently gmail too for personal emails, but not too many people have the gmail account!", "Solution_13": "[quote=\"djimenez\"]and recently gmail too for personal emails, but not too many people have the gmail account![/quote]\r\n\r\nactually that is not true. With all the account holders having 50 invites, and refilled each time they use some, there are more than you might expect. I mean a lot more.", "Solution_14": "[quote=\"xxreddevilzxx\"][quote=\"djimenez\"]and recently gmail too for personal emails, but not too many people have the gmail account![/quote]\n\nactually that is not true. With all the account holders having 50 invites, and refilled each time they use some, there are more than you might expect. I mean a lot more.[/quote]Well, let me rephrase though. Not too many people who use to write me have my gmail address. But for example, I usually send all my invitations to a place where they give to several people, and twice the receiver has emailed me saying... \"Hey, do I know you?\". Usually, the people who have my gmail account is the people who usually send me files attached.", "Solution_15": "THAT would explain why mine didn't auto-upgrade. I put my country of origin as Zimbabwe.", "Solution_16": "[quote=\"bleumoose\"]THAT would explain why mine didn't auto-upgrade. I put my country of origin as Zimbabwe.[/quote]May be this could help to clarify this,\n\n[quote=\"Hotmail Main Page\"]250MB inbox [b]available only in the 50 United States, District of Columbia, and Puerto Rico[/b]. Eligible Hotmail users will first receive 25MB at sign-up. Please allow at least 30 days for activation of your 250MB storage to verify your e-mail account and help prevent abuse. Microsoft Corporation reserves the right to provide 250MB inbox to free Hotmail accounts at its discretion.[/quote]That is written in very small letter, and at the end. That means that non-american people didn't have that automatic update.", "Solution_17": "wow. didn't know so many ppl liked hotmail.\r\nalso, when i had a hotmail account, it automatically updated to 250MB. but since i like gmail better, i let the account expire. and in hotmail i was getting constant junk email. it was messages with just symbols in it. like \"$%(@#(*~\\\". that annoyed me greatly.", "Solution_18": "What is G-Mail and how do I get it?\r\nThanks", "Solution_19": "[quote=\"lightspeed\"]What is G-Mail and how do I get it?\nThanks[/quote]\r\n\r\nsearch the words \"gmail\" and \"invite\" on aops.", "Solution_20": "Or go here: http://isnoop.net/gmail/ :)", "Solution_21": "[quote=\"lightspeed\"]What is G-Mail and how do I get it?\nThanks[/quote]\r\n\r\ngmail is a mail started by google.", "Solution_22": "[quote=\"lightspeed\"]What is G-Mail and how do I get it?\nThanks[/quote]Give me an email address and I can send you an invitation.", "Solution_23": "ok, i've already got the 25mb account. And for 250's, do i have to do it again after ten days?", "Solution_24": "[quote=\"shobber\"]ok, i've already got the 25mb account. And for 250's, do i have to do it again after ten days?[/quote]No, in a few days it will automatically upgrade itself. In my case that took two days!", "Solution_25": "[quote=\"djimenez\"][quote=\"shobber\"]ok, i've already got the 25mb account. And for 250's, do i have to do it again after ten days?[/quote]No, in a few days it will automatically upgrade itself. In my case that took two days![/quote]\r\nThat's cool!\r\nThanks a lot!" } { "Tag": [ "calculus", "integration", "logarithms", "function", "calculus computations" ], "Problem": "Evaluate $ \\int_0^1\\frac {\\ln\\left(x\\right)}{1 \\minus{} x^2}\\,dx$", "Solution_1": "There are three ways that I have done on my own. I don't want to do anymore for I didn't come up with them on my own.\r\n\r\n[hide]\n\n\n[b]Way one[/b]\n\nWe know that\n\n$ \\forall{x}\\in(0,1)\\quad\\frac {1}{1 - x^2} = \\sum_{n = 0}^{\\infty}x^{2n}$\n\n$ \\therefore\\qua\\int_0^1\\frac {\\ln(x)}{1 - x^2}\\,dx$\n\n$ = \\int_0^1\\ln(x)\\sum_{n = 0}^{\\infty}x^{2n}\\,dx$\n\nNow assuming uniform convergence\n\n$ = \\sum_{n = 0}^{\\infty}\\int_0^1\\ln(x)x^{2n}\\,dx$\n\n$ = - \\sum_{n = 0}^{\\infty}\\frac {1}{(2n + 1)^2}$\n\n$ = \\frac { - \\pi^2}{8}$\n\n\n[b]Way two[/b]\n\nWe know that\n\n$ \\ln(x) = \\int_1^x\\frac {dy}{y}$\n\nSo we can see that\n\n$ \\int_0^1\\frac {\\ln(x)}{1 - x^2}\\,dx$\n\n$ = \\int_0^1\\int_1^x\\frac {dy}{y(1 - x^2)}\\,dx$\n\nNow switching the integration order\n\n$ = - \\int_0^1\\int_0^y\\frac {dx}{y(1 - x^2)}\\,dy$\n\nEvaluation on the inner integral leads to\n\n$ = - \\int_0^1\\frac {\\tanh^{ - 1}(y)}{y}\\,dy$\n\nNow we know that\n\n$ \\forall{y}\\in(0,1)\\quad\\tanh^{ - 1}(y) = \\sum_{n = 0}^{\\infty}\\frac {y^{2n + 1}}{2n + 1}$\n\nSo\n\n$ - \\int_0^1\\frac {\\tanh^{ - 1}(y)}{y}\\,dy$\n\n$ = - \\int_0^1\\frac {1}{y}\\sum_{n = 0}^{\\infty}\\frac {y^{2n + 1}}{2n + 1}\\,dt$\n\n$ = - \\int_0^1\\sum_{n = 0}^{\\infty}\\frac {y^{2n}}{2n + 1}$\n\nNow assuming uniform convergence \n\n$ = - \\sum_{n = 0}^{\\infty}\\frac {1}{2n + 1}\\int_0^1{y^{2n}}\\,dy$\n\n$ = - \\sum_{n = 0}^{\\infty}\\frac {1}{(2n + 1)^2}$\n\n$ = \\frac { - \\pi^2}{8}$\n\n\n[b]Way three[/b]\n\n$ \\int_0^1\\frac {\\ln(x)}{1 - x^2}\\,dx$\n\n$ = \\frac {1}{2}\\int_0^1\\frac {\\ln(x)}{1 - x}\\,d + \\int_0^1\\frac {\\ln(x)}{1 + x}$\n\nNow we know that\n\n$ \\forall{x}\\in(0,1)\\quad\\frac {1}{1 + x} = \\sum_{n = 0}^{\\infty}( - 1)^nx^n\\,\\wedge\\, \\frac {1}{1 - x} = \\sum_{n = 0}^{\\infty}x^n$\n\nSo we have that\n\n$ = \\frac {1}{2}\\int_0^1\\frac {\\ln(x)}{1 - x}\\,d + \\int_0^1\\frac {\\ln(x)}{1 + x}$\n\n$ = \\frac {1}{2}\\int_0^1\\ln(x)\\sum_{n = 0}^{\\infty}( - 1)^nx^n\\,dx + \\frac {1}{2}\\int_0^1\\ln(x)\\sum_{n = 0}^{\\infty}x^n\\,dx$\n\nNow assuming uniform convergence\n\n$ = \\frac {1}{2}\\sum_{n = 0}^{\\infty}( - 1)^n\\int_0^1\\ln(x)x^n\\,dx + \\frac {1}{2}\\sum_{n = 0}^{\\infty}\\int_0^1\\ln(x)x^n\\,dx$\n\n$ = - \\frac {1}{2}\\sum_{n = 0}^{\\infty}\\frac {( - 1)^n}{(n + 1)^2} - \\frac {1}{2}\\sum_{n = 0}^{\\infty}\\frac {1}{(n + 1)^2}$\n\n$ = - \\frac {1}{2}\\sum_{n = 1}^{\\infty}\\frac {( - 1)^{n - 1}}{n^2} - \\frac {1}{2}\\sum_{n = 1}^{\\infty}\\frac {1}{n^2}$\n\n$ = \\frac { - 1}{2}\\left(1 - 2^{1 - 2}\\right)\\zeta(2) - \\frac {1}{2}\\zeta(2)$\n\n$ = \\frac { - \\pi^2}{24} - \\frac {\\pi^2}{12}$\n\n$ = \\frac { - \\pi^2}{8}$\n\n\nBy any means\n\n$ \\boxed{\\int_0^1\\frac {\\ln(x)}{1 - x^2}\\,dx = \\frac { - \\pi^2}{8}}\\,\\,\\blacksquare$[/hide]", "Solution_2": "Thanks.\r\nI too have solved the problem using series. I was looking for any other method if possible.\r\n\r\nThanks again", "Solution_3": "[quote=\"vidyamanohar\"]Thanks.\nI too have solved the problem using series. I was looking for any other method if possible.\n\nThanks again[/quote]\r\nJust out of curiosity, which of the three (if any) did you use?", "Solution_4": "[quote=\"Mathstud28\"][quote=\"vidyamanohar\"]Thanks.\nI too have solved the problem using series. I was looking for any other method if possible.\n\nThanks again[/quote]\nJust out of curiosity, which of the three (if any) did you use?[/quote]\r\n\r\nThe first one.", "Solution_5": "Sorry for reviving the thread, but I wanted to show another way of solving the integral (inspired by one of Kouichi Nakagawa's posts)\r\n\r\nUsing the substitution $ x \\equal{} e^{t}$, we obtain:\r\n\r\n$ \\minus{} \\int_{0}^{\\infty}\\frac {te^tdt}{e^2t \\minus{} 1}$\r\n\r\nNow, lets substitute t with $ \\frac {u}{2}$:\r\n\r\n$ \\minus{} \\int_{0}^{\\infty}\\frac {\\frac {u}{4}e^{u/2}du}{e^u \\minus{} 1}$\r\n\r\nwhich is $ \\minus{} \\int_{0}^{\\infty}\\frac {u}{4}e^{u/2}\\sum_{n \\equal{} 1}^{\\infty}e^{ \\minus{} nu}du$\r\n\r\nAs the interchange of sum-integral order can be justified, we have\r\n\r\n$ \\minus{} \\sum_{n \\equal{} 1}^{\\infty}\\int_{0}^{\\infty}\\frac {u}{4}e^{u/2}e^{ \\minus{} nu}du$\r\n\r\nor\r\n\r\n$ \\minus{} \\sum_{n \\equal{} 1}^{\\infty}\\mathcal{L}\\left[\\frac {u}{4}e^{u/2}\\right](n)$\r\n\r\nwhere $ \\mathcal{L}[f(x)](z)$ denotes the Laplace Transform of the function $ f$. Finding the transform is easy, and we have:\r\n\r\n$ \\minus{} \\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{(2n \\minus{} 1)^2} \\equal{} \\minus{} \\frac {\\pi^2}{8}$.", "Solution_6": "Or something in between:\r\n\r\n$ \\int^1_0 \\frac{x ^ \\alpha}{1\\minus{}x^2}\\,dx\\equal{}\\int^1_0 x ^ \\alpha \\sum ^\\infty_{n\\equal{}0} x^{2n}\\,dx\\equal{}\\int^1_0 \\sum ^\\infty_{n\\equal{}0} x^{2n\\plus{}\\alpha}\\,dx\\equal{}$\r\n\r\n$ \\equal{} \\sum ^\\infty_{n\\equal{}0} \\int^1_0x^{2n\\plus{}\\alpha}\\,dx \\equal{} \\sum ^\\infty_{n\\equal{}0} \\frac{x^{2n\\plus{}\\alpha\\plus{}1}}{2n\\plus{}\\alpha\\plus{}1}\\Big|^1_0\\equal{}\\sum ^\\infty_{n\\equal{}0} \\frac{1}{2n\\plus{}\\alpha\\plus{}1}\\equal{}F(\\alpha)$\r\n\r\n$ I\\equal{}\\frac{d F}{d\\alpha}\\Big|_{\\alpha \\equal{} 0} \\equal{}\\minus{}\\sum ^\\infty_{n\\equal{}0} \\frac{1}{(2n\\plus{}1)^2}\\equal{}\\minus{}\\frac{\\pi}{8}$" } { "Tag": [], "Problem": "ok so the problem is (10n-1)(10n+1)= 999,999,999,999 \r\nand the answer is 6, which doesn't make any sense at all.\r\n\r\nI got 100,000 as the answer doing 100n^2-1= 999,999,999,999 and then dividing by 100 and then taking the square root. Could anybody corroborate this answer or tell me what's wrong?\r\nThanks in advance :lol:", "Solution_1": "i agree with you\r\ngo report this question\r\n\r\nif it said how many digits, then 6 is correct", "Solution_2": "Someone entered the problem incorrectly. It is now fixed. Thanks for the report." } { "Tag": [ "geometry", "circumcircle", "trigonometry", "trig identities", "Law of Cosines" ], "Problem": "A triangle inscribed in a circle of radius 5 has two sides of length 5 and 6. Find the two possible lengths of the third side of the triangle.", "Solution_1": "Call the triangle $ \\triangle ABC$, with $ AB \\equal{} 5$ and $ BC \\equal{} 6$. Then we want to find $ AC$. Let the circumcenter of the triangle be point $ E$. Then we have $ \\triangle ABE$ is regular, so $ \\angle ABE \\equal{} 60^\\circ$. In $ \\triangle BCE$, we use the law of cosines to find $ \\angle EBC \\equal{} \\cos^{ \\minus{} 1}\\bigg(\\frac {5^2 \\minus{} 5^2 \\minus{} 6^2}{ \\minus{} 2*5*6}\\bigg) \\equal{} \\cos^{ \\minus{} 1} \\frac {3}{5}$. Then $ \\angle ABC \\equal{} 60^\\circ \\pm \\cos^{ \\minus{} 1}\\frac {3}{5}$. Now, by law of cosines:\r\n\r\n$ AC^2 \\equal{} BC^2 \\plus{} AB^2 \\minus{} 2AB*BC*\\cos \\angle ABC$\r\n$ \\implies \\equal{} 5^2 \\plus{} 6^2 \\minus{} 2*5*6\\cos \\bigg(60 \\pm \\cos^{ \\minus{} 1}\\frac {3}{5}\\bigg)$\r\n$ AC \\equal{} \\sqrt {61 \\minus{} 60 \\bigg(\\frac {3}{5}\\cos 60 \\mp \\frac {4}{5}\\sin 60\\bigg)}$\r\n$ \\implies \\equal{} \\sqrt {61 \\minus{} 60 (3/10 \\mp 4/10 \\sqrt{3})}$\r\n$ \\implies \\equal{} \\sqrt {43 \\pm 24 \\sqrt {3}}$\r\n\r\nSo $ AC \\approx 9.196152423$ or $ AC \\approx 1.196152423$" } { "Tag": [ "floor function", "AMC", "number theory" ], "Problem": "How many positive integers $ N$ less than $ 1000$ are there such that the equation $ x^{\\lfloor x\\rfloor} \\equal{} N$ has a solution for $ x$? (The notation $ \\lfloor x\\rfloor$ denotes the greatest integer that is less than or equal to $ x$.)", "Solution_1": "[hide=\"solution\"] We divide this problem into some cases.\n\nCase 1: x is an integer-\n\nClearly, this can only be $ 1^1, 2^2, 3^3$, and $ 4^4$, so four here.\n\nCase 2: $ \\lfloor{x}\\rfloor\\equal{}2$\n\nThen let $ x\\equal{}\\sqrt{a}$ and a ranges from 5 to 8, inclusive, so four more.\n\nCase 3: $ \\lfloor{x}\\rfloor\\equal{}3$\n\nThen let $ x\\equal{}\\sqrt[3]{a}$. a ranges from 28 to 63, so 36 choices.\n\nCase 4: $ \\lfloor{x}\\rfloor\\equal{}4$\n\nThen let $ x\\equal{}\\sqrt[4]{a}$. a ranges from 257 to 624, which gives 368.\n\nAfter this, it is easy to see that $ \\lfloor{x}\\rfloor\\equal{}5$ will not work as then N would not be in the given range.\n\nIn all, we have 412.[/hide]", "Solution_2": "One line:\r\n\\[ (2^1 \\minus{} 1^1) \\plus{} (3^2 \\minus{} 2^2) \\plus{} (4^3 \\minus{} 3^3) \\plus{} (5^4 \\minus{} 4^4)\\equal{}412\r\n\\]", "Solution_3": "Which totally explains how it works :roll:", "Solution_4": "Yes, in fact it does.\r\n\r\nTake some positive integer $ k$. Then for any $ j$ in the interval $ [k, k \\plus{} 1)$, we have $ \\lfloor j \\rfloor \\equal{} k$, so $ j^k \\equal{} N$. The interval for $ N$ is $ [k^k, (k \\plus{} 1)^k)$. Thus we have $ (k \\plus{} 1)^k \\minus{} k^k$.\r\n\r\n :| I think my one-line solution explains it better.", "Solution_5": "[quote=\"AIME15\"]Which totally explains how it works :roll:[/quote]\r\n\r\nIt does, but if you were retarded like me, you would say $ 4^4\\equal{}64$ and get the wrong answer.", "Solution_6": "Yeah, but my solution prevents the retardedness. You see, you would've calculated $ 4^3\\equal{}64$ before $ 4^4$, and if you got $ 4^4\\equal{}64$, then you would know something was amiss since you already know $ 4^3\\equal{}64$.", "Solution_7": "[hide=\"i got this one wrong\"]\nI put 788 =(\nLet's graph this thing.\n\nSo we have 1 as a solution, right? So that's 1.\n\nThen case 2: floor x = 2. We know $ x^2$ ranges from 4 to 9, including 4 but not 9...\n\nThen case 3: floor x = 3 => similar method, [27, 64)\n\nCase 4: floor x = 4 => similar method, [256, 625)\n\n1 + 5 + 37 + 369 = 412[/hide]", "Solution_8": "[quote=\"Yongyi781\"]One line:\n\\[ (2^1 \\minus{} 1^1) \\plus{} (3^2 \\minus{} 2^2) \\plus{} (4^3 \\minus{} 3^3) \\plus{} (5^4 \\minus{} 4^4) \\equal{} 412\n\\]\n[/quote]\r\nI did it the same way, but since I can't write, I almost thought $ 2^1 \\equal{} 2^2$, $ 3^2 \\equal{} 3^3$, and $ 4^3 \\equal{} 4^4$ and calculated $ 5^4 \\minus{} 1^1 \\equal{} 624$ before I did a\r\n\r\nSANITY CHECK", "Solution_9": "[quote=\"abacadaea\"][quote=\"AIME15\"]Which totally explains how it works :roll:[/quote]\n\nIt does, but if you were retarded like me, you would say $ 4^4 \\equal{} 64$ and get the wrong answer.[/quote]\r\n\r\nThat makes 2 of us.\r\nI triple-checked all my arithmetic on that one too.\r\n :(", "Solution_10": "[quote=\"abacadaea\"][quote=\"AIME15\"]Which totally explains how it works :roll:[/quote]\n\nIt does, but if you were retarded like me, you would say $ 4^4 \\equal{} 64$ and get the wrong answer.[/quote]\r\n\r\nYou know this is very interesting: I made the [b]exact[/b] same mistake the first time I attempted the problem. Luckily I checked my work and caught it (motivated by the similar $ 2^6$ mistake on the AMC 12A), but it's just amazing that all around the country people were making the exact same mistake. :o", "Solution_11": "[hide=\"epic failure\"]\nI added wrong at he end and got 410. :( \n[/hide]", "Solution_12": "I got 413 because I counted when N = 1 twice. I found the same thing as what Yongyi said first, but I also added the case when it's like (1/2)^0...which is the same as 1^1. I was kind of annoyed when I realized afterwards.", "Solution_13": "[quote=\"brightzhu\"][hide=\"epic failure\"]\nI added wrong at he end and got 410. :( \n[/hide][/quote]\r\n\r\nI failed more. I added correctly to get 412. I even wrote it on my scantron. Then while checking my work I added wrong and changed it to 422.", "Solution_14": "[quote=\"Walk Around The River\"][quote=\"abacadaea\"][quote=\"AIME15\"]Which totally explains how it works :roll:[/quote]\n\nIt does, but if you were retarded like me, you would say $ 4^4 \\equal{} 64$ and get the wrong answer.[/quote]\n\nThat makes 2 of us.\nI triple-checked all my arithmetic on that one too.\n :([/quote]\r\n\r\nHeh.\r\n\r\nUm, I kinda failed by adding wrong to get 411. A friend of mine added badly and got 410. Fail?\r\n\r\nI also checked my arithmetic...", "Solution_15": "I can't complain though. I decided to check my work, and in doing so I changed 6's right answer to a wrong answer but I also changed 8's wrong answer to the right one. Funny, huh?\r\n\r\nno not really... =(", "Solution_16": "Somehow I ended up with 410. Is this explainable?", "Solution_17": "during the actual contest, i had no idea what to do with this question\r\n\r\nafterwards, I play with some numbers on the calculator..and realize you just have to sum\r\n\r\n(n+1)^n - n^n \r\n\r\ncool problem anyway", "Solution_18": "Why is our first case $x>0$? In other words why can\u2019t we have negative $x$ that satisfy $x^{\\lfloor x\\rfloor} = N$ for an integer $N$ that satisfies the conditions of the problem?", "Solution_19": "Also HMMO Team #1\n\nThis problem was pretty nice but my team and I didnt notice in time that $x$ doesn't have to be an integer :wallbash: \n\nGood problem overall :)", "Solution_20": "We take cases on $\\lfloor x \\rfloor$. If $\\lfloor x \\rfloor = 1$, we can only have $N = 1$, for $1$ way. If $\\lfloor x \\rfloor = 2$, we can have $N$ ranging from $2^2$ to $3^2-1$, for $3^2-2^2$ ways. Similarly, $\\lfloor x \\rfloor = 3$ gives $4^3 - 3^3$ ways and $\\lfloor x \\rfloor = 4$ gives $5^4 - 4^4$ ways. But $\\lfloor x \\rfloor = 5$ is not possible, so the answer is $1+3^2-2^2+4^3-3^3+5^4-4^4 = \\boxed{412}$." } { "Tag": [ "function", "trigonometry", "algebra unsolved", "algebra" ], "Problem": "Hi guys,\r\n\r\nI would like to know the solution of\r\n\r\nx^x = k, where k is a defined constant\r\n\r\n\r\nI've thought about it for a long time, but... nothing to do :( \r\n\r\nThanks in advance,\r\n\r\nGabriele :)", "Solution_1": "hello, here we have $ x\\equal{}\\frac{\\ln(k)}{\\text{LambertW}(\\ln(k))}$.\r\nSonnhard.", "Solution_2": "Thank you very much, Sonnhar...\r\n\r\n\r\nI watched what LambertW is...\r\n\r\nIt is the Lambert W function... but... really... what is it???\r\n\r\nI didn't understand anything, except that it is the inverse function of $ f(x)\\equal{}we^w$...\r\nIs it a sort of \"sine\", a transcendental function? Otherwise, how can I calculate its values?\r\n\r\n\r\nThank you again\r\nGabriele :)", "Solution_3": "hello, here you will found some informations about this function\r\nhttp://mathworld.wolfram.com/LambertW-Function.html\r\nSonnhard." } { "Tag": [ "AMC", "AIME" ], "Problem": "What would be a good book to help me get something better than 4 or 5 on an AIME?", "Solution_1": "AoPS I & II & Subject books\r\nArt and Craft of Problem Solving\r\nOld AIME's", "Solution_2": "I heard of a book called Engel. Would that help? \r\n\r\nWhat kind of topics does Art and Craft of Problem Solving cover?", "Solution_3": "[quote=\"game.guy\"]I heard of a book called Engel. Would that help? [/quote]\r\nI think you want, problem solving strategies by Engel :wink: . I am studying from that book right now, and it 'can' help you with the harder questions on the AIME, but it is mostly at an Olympiad level. On the few practice AIMES i did recently i got 6,7.. and still Engel's book is hard to comprehend... Maybe its just me", "Solution_4": "What topics do those two books cover?" } { "Tag": [ "probability", "function", "conditional probability", "\\/closed" ], "Problem": "I was wondering if subforums could be created for each class so that the problem sets would be more organized. \r\n\r\nFor example, in the Intermediate Counting/Probability class, make subforums by week or by subject. (Week is less work :D)\r\n\r\nWeek 1, Week 2, Week 3,...., Week 12\r\n\r\nor\r\n\r\nConditional Probability, 1-1 Correspondences, PIE, Generating Functions, etc\r\n\r\nAnyway, just a suggestion.", "Solution_1": "Not going to happen I'm afraid. That would mean creating too many subforums, and making proper permissions for each." } { "Tag": [ "function", "inequalities", "integration", "calculus", "calculus computations" ], "Problem": "Suppose f is a differentiable real function such that f(x) + f'(x)<=1 for all x, and\r\nf(0) = 0. What is the largest possible value of f(1)?", "Solution_1": "Faced with a differential inequality, the first place to look would be techniques from differential equations. In particular, the multiplying factor method for first order linear equations suggests that we multiply both sides of the inequality by $ e^x.$ Try that, and see what develops.\r\n\r\n[hide=\"Short answer.\"]$ f(1)\\le 1\\minus{}e^{\\minus{}1}$[/hide]", "Solution_2": "Let $ h(x)\\equal{}f(x)\\plus{}f'(x)\\le 1$. Then $ f$ is a solution of $ y'(x)\\plus{}y(x)\\equal{}h(x)$. This can be written $ \\left(e^xy(x)\\right)'\\equal{}h(x)e^x$ and so $ f(x)\\equal{}e^{\\minus{}x}\\int_{0}^xe^th(t)dt$. for $ x\\equal{}1$ we have $ f(1)\\equal{}e^{\\minus{}1}\\int_0^1e^th(t)dt\\le e^{\\minus{}1}\\int_0^1e^tdt\\equal{}1\\minus{}e^{\\minus{}1}$. Equality is obtained when $ h(t)\\equiv 1$, i.e for $ f(x)\\equal{}1\\minus{}e^{\\minus{}x}$.", "Solution_3": "Solution...\r\n[hide]\nLet $ f(x) \\equal{} e^{ \\minus{} x}v(x)$, $ v(0) \\equal{} 0$. We have $ e^{ \\minus{} x}v(x) \\minus{} e^{ \\minus{} x}v(x) \\plus{} e^{ \\minus{} x}v'(x)\\leq 1$, so $ v'(x)\\leq e^{x}$. \n\nTaking integral, we will have $ \\int_0^x v'(t) dt \\leq \\int_0^x e^{t} dt$, or $ v(x) \\minus{} v(0)\\leq e^x \\minus{} 1$. Putting x=1: $ v(1)\\leq e \\minus{} 1$, so $ f(1)\\leq 1 \\minus{} e^{ \\minus{} 1}$, and $ f(1) \\equal{} 1 \\minus{} e^{ \\minus{} 1}$ if $ v(x) \\equal{} e^x \\minus{} 1$, or $ f(x) \\equal{} 1 \\minus{} e^{ \\minus{} x}$.\n\nOttem, sorry, my solution is same to your's...\n[/hide]", "Solution_4": "Ottem's solution, Kirill's solution, and the solution I hinted at but didn't write up differ in organization and notation, but are all essentially the same solution, and are motivated by the same idea." } { "Tag": [ "USAMTS", "geometry", "geometric transformation", "rotation", "linear algebra", "matrix", "analytic geometry" ], "Problem": "Post scores for USAMTS here and your grade. For example (here are my statistics):\r\n\r\nScore: 20, Grade: 7\r\n\r\n\r\nYou may also post what you got off on. For example (here is what I got off on):\r\n\r\nI got a 2 on problem 4 for not getting the 40 case.\r\nI got a 3 on problem 5 for my proof of part b being incomplete.", "Solution_1": "Darn I got a 22... missed 3 points on #4 for forgetting about the 40 case and not constructing the 60 case... oh well :(", "Solution_2": "25.\r\n\r\nI'm glad my ridiculous \"1489+3=1502\" was excused...", "Solution_3": "5/5/5/4/5.\r\n\r\nOn #4 for whatever reason I didn't justify that my constructions implied convexity.", "Solution_4": "Lol 17. Which is pretty good considering how much time+effort I actually put into it. Yay this means I don't have to try for gold anymore and can be content with 15+ :)", "Solution_5": "Score: 17 Grade: 7\r\n\r\nDidn't put in a lot of work, 5/5/2/0/5\r\n\r\n#4 had no idea what to do (I don't like working on hard geometry)\r\n\r\n#3 my PPW was rejected...", "Solution_6": "5/5/5/5/5\r\n\r\nComment for #3:\r\n\r\n$ 1488\\plus{}4 \\neq 1490$\r\nOtherwise nice.\r\n\r\n :rotfl:", "Solution_7": "5/5/5/3/4 Grade 11\r\nI wonder how I got full points for 3, I needlessly complicated the question with a rotation matrix in order to avoid slope arguments.\r\nI'm mad at myself for number 4. I considered including an argument about being able to construct both the 40 and 60 (or whatever they were), but decided against it because it would increase my page count and wanted the beautiful short proof, not the legit long one.\r\nNumber 5 I am also mad with myself because I lost a point for cutting out an algebra step in part (b)\r\nGL to everyone on round 2.", "Solution_8": "5/5/5/5/5. Apparently I satisfied their criteria for number 4, even though I didn't show that the hexagons were constructable. Also, my proof for #3 was not as concise as the grader would have liked. Hi five for that! :lol: \r\n\r\nI think it's funny how of the 15 people on the leaderboard who got 25's, 10 of them are in 11th grade.", "Solution_9": "In 8th grade, and it was my first time writing formal proofs. I got a 3/5/2/2/2=14, which I am satisfied with.\r\n\r\nI lost 2 points for screwing up casework and getting 72. Also, I didn't notice the slick hexagon thing in the posted solution.\r\n\r\nThis was easy for me, and I got full credit.\r\n\r\nOnly got 8 points, and failed to show that they were the only ones. I can't understand the connection to number theory in this problem. Lost 3 points.\r\n\r\nAt the last second, I noticed the flipping over. I didn't see the 40 case, and forgot to prove that they could be constructed (didn't use the fact that the hexagon was convex...). Lost 3 points\r\n\r\nI assumed something when doing (a), and my proof wasn't rigorous in part (b).\r\n\r\nI learned a lot though. I typed these in Word, but the ugliness motivated me to learn $ \\text{\\LaTeX}$ for this round. Also, it was a very good experience on proofs, since this was my first time. After really understanding the problems and seeing other people's solution, I learned a lot.[hide=Index][/hide]", "Solution_10": "3/5/5/0/4 - 17 :lol: \r\nwhich is good pretty for me, since i got 7 last year :(", "Solution_11": "Oh well seeing that everyone did better than me um I am not going to post mine. I never wrote proofs and I kinda didn't write them right.", "Solution_12": "Blegh. I've never done proofs before...but I still did so horrible...\r\n\r\n3/3/0/1/0=7. :( :( :(", "Solution_13": "25 for me: I guess the limited rigor of my hexagon constructability argument was enough. (I thought I would get 4 on #4)", "Solution_14": "3/4/5/2/5 = 19.\r\n\r\nNot as good as I hoped but I didn't spend much time on it.\r\nI got a point off on 2 for not considering 1*11111 as a possible factorization, but I figured it was clear that it wouldn't work in that case... should I protest it? :(\r\n\r\nEDIT: looking at the rubric, I did earn a 4... :( too bad. I thought I wouldn't have to explain that case. oh well.", "Solution_15": "5/5/5/5/5, 12th grade", "Solution_16": "4/5/5/0/4\r\nfail @ geometry :(", "Solution_17": "I'm so proud that three of my solutions were programmed :) way too lazy to solve them by hand.", "Solution_18": "5/4/5/5/5=24, even though i didn't prove existence for either case on #4 (maybe just because my diagram was so good). But i still don't know why i lost a point on #2...my grader said i needed either to check or dismiss the case x+y=11111, x-y=1 in the 5-digit part (where $ (x^2, y^2)$ is the pair), except i clearly said that in this case, x=5556 and so $ x^2$ has more than 5 digits, and therefore this case is clearly absurd. Um.", "Solution_19": "5/5/0/0/0\r\nBasically I was like FIDDLESTICKS, I DONT SEE AN IMMEDIATELY TRIVIAL AND STRAIGHTFORWARD SOLUTION TO #3-5 and didn't spend the time figure out more. Maybe I'll try harder on round 2?", "Solution_20": "2/0/0/0/0... LOL. I'm in 8th grade and this is my first time writing proofs. And I only had a week to work on the problems. I managed to screw up the casework on the only one I did, though. >.<", "Solution_21": "oh ok well mine was 1/1/0/0/1", "Solution_22": "I'm definitely not going to protest these scores that I really don't deserve because the solutions were written on the weekend before (along with college apps...). \r\n\r\n1: 3 - missed cases\r\n2: 5 - yay!\r\n3: 4 - yeah over counted, like I expected, but only one point off??\r\n4: 1 - I went nowhere\r\n5: 5 - really? I didn't know I could write good proofs!\r\n\r\nTotal: 18", "Solution_23": "5/5/5/2/5 can only miss one more point to get a gold >_<", "Solution_24": "2/5/5/.../... darn now my best is an 87", "Solution_25": "9th grade\r\nI got 5/5/5/3/5 cuz i apparently failed to show that the hexagon was convex...\r\n\r\n[quote=\"dnkywin\"]Darn I got a 22... missed 3 points on #4 for forgetting about the 40 case and not constructing the 60 case... oh well :([/quote]\r\ngood job, and now you are going to score worse than last year... :P", "Solution_26": "3/5/5/2/5 = 20. 9th grade\r\n\r\nOvercounted/Messed up casework for #1. Completely forgot about 40 case for #4. I'm really surprised that I got full credit for 5b, since I made up a lot of stuff about sequences for my proof...", "Solution_27": "3/5/4/4/5=21\r\n\r\nhmmm maybe i should not be up til 4am the night before doing these, but i am unfortunately procrastinating on round 2 more so than i did i round one as of right now.\r\n\r\nsomehow got 120 on number 1, was too sleepy at 4am to finish number 3 ( i am very surprised that my claim of clearly the answer is 1488 (which i guessed instead of 1492) in the end got a 4), and wasn't sure if constructing the diagrams was necessary on number 4 and decided to keep it short.\r\n\r\ni guess i'll have to try for gold next year.", "Solution_28": "8th grade...like others, this was my first real attempt at writing proofs. Luckily, USAMTS graders are not very harsh...\r\n\r\n5/5/4/0/5=19\r\n\r\nEh, only 4 grade 8 or younger on the leaderboard, so I don't feel too bad about this.\r\n\r\nFor #3 I just stated the answer and said to use analytic geometry, and somehow I got a 4...?\r\n\r\n#4 I trig-bashed unsuccessfully because I had an error in, I think, the third line in a three page solution.\r\n\r\nOn pace for silver :)", "Solution_29": "Voila.\r\n5/4/2/0/1 = 12 (10th grade)\r\nA list of excuses for my peers :| :\r\nVery first USAMTS. Very first time I wrote formal proofs, ever. Spent too much time figuring out LATEX. I overslept.\r\n#2: one small, small error in counting cases and I lost a point. Guess that's what happens when you write it the night before. :!: #3: thought it was only one point for putting the answer only? Apparently the scoring criterion modified that. #5: WHAT?? No points for second part?\r\nAnyways, enough viewing the forums for me. Time to get back to finishing Round 2!", "Solution_30": "5/5/5/5/5 11th grade\r\n\r\na Bit surprised I got full points on 4 as I did not prove both were constructable", "Solution_31": "5/4/5/2/5 = 21 (8th)\r\n\r\n#2: I found a solution for the 6-digit case. Go me.\r\n#4: Same as a lot of other people, omitted the 40 case.", "Solution_32": "9th grade,\r\n5/5/5/3/5\r\n\r\nUgh I did them all 11 o'clock at night before the due date.\r\nI didn't know you had to construct the hexagons :(" } { "Tag": [ "geometry", "rotation", "geometry proposed" ], "Problem": "A piece of cardboard has the shape of a pentagon $ABCDE$ in which $BCDE$ is a square and $ABE$ is an isosceles triangle with a right angle at $A$. Prove that the pentagon can be divided in two different ways in three parts that can be rearranged in order to recompose a right isosceles triangle.", "Solution_1": "Let a = BC be the square side. Area $|ABCDE| = \\frac{5a^2}{4}.$ Let x be the leg of the isosceles right triangle with the area $\\frac{x^2}{2} = \\frac 5 4 a^2.$ Then $x = a\\ \\sqrt{\\frac 5 2.}$ Since $AC = AD = a\\ \\sqrt{\\frac 1 4 + \\frac 9 4} = a\\ \\sqrt{\\frac 5 2}$, one of these is the 1st cut, say AD. Rotate the triangle $\\triangle ADE$ by $90^\\circ$ around A, so that AE coincides with AB and D goes into $D' \\in BE$. Let M be the midpoint of BC. Since the triangles $\\triangle MCD \\cong \\triangle MBD'$ are congruent, DM is the 2nd cut." } { "Tag": [ "AoPS Books" ], "Problem": "i had some free time so i looked up how many people actually post on AOPS. out of 19361, only 7625 have ever posted anything. that's about 40%. that also includes the people that posted 1 thing, or 2 things. that's not very good. :( does anyone have any opinions on this?", "Solution_1": "they are awesome people", "Solution_2": "There are lots of people that are guests or just look around, reading threads.", "Solution_3": "Maybe some people join just to take classes.", "Solution_4": "You'll usually get that with most online forums. Only about 10% of the people are active.", "Solution_5": "Well, for one, there are a lot of people, like my dad, who don't have very many things to post about, but read through the forums, and do some things on AoPSWiki.\r\n\r\nBesides, I'm sure that a lot of people are searching for forums and join lots of them, but find that they aren't that interested in math after all, so they stop coming.", "Solution_6": "Yeah, I've signed up to tons of websites but I end up not posting. I wonder if any of the moderators or administrators have any histograms that would give us a good idea of member participation.", "Solution_7": "Well i know alot of people at my school who buy the AoPS books to learn and have an account in the fourm. But they don't really like math that much and find the fourm quite boring.", "Solution_8": "Some people sign up because they thing this forum is really nice and all, but they get bored really fast and do not visit the website for the rest of their lives.", "Solution_9": "Also, some might sign up only for the Classes, or to browse Resources and such. I think the level of activity here is pretty astounding and 40% is not low at all.", "Solution_10": "[quote=\"drunner2007\"]Yeah, I've signed up to tons of websites but I end up not posting. I wonder if any of the moderators or administrators have any histograms that would give us a good idea of member participation.[/quote]We do, but they are not for public use ;)", "Solution_11": "in an other forum members that had 0 posts for a year were deleted... but it was only forum", "Solution_12": "Another reason why some ppl have so little posts is because they want to register to see various ppl's profiles or talk to them by private messaging; they think that the opportunity of posting something so that many ppl around the world will see it is really cool, but then they either get bored or have nothing to post.\r\nI still think that this aops -mathlinks forum is much more active and has much more posts than any other forum where you need to be a regusterd user to post. :!:", "Solution_13": "[quote=\"rem\"]Another reason why some ppl have so little posts is because they want to register to see various ppl's profiles or talk to them by private messaging; they think that the opportunity of posting something so that many ppl around the world will see it is really cool, but then they either get bored or have nothing to post.\nI still think that this aops -mathlinks forum is much more active and has much more posts than any other forum where you need to be a regusterd user to post. :!:[/quote]\r\nThat's true. Some users don't post because they rather just read or they don't know what to post." } { "Tag": [ "Princeton", "college" ], "Problem": "Anybody got thei PSAT scores yet? If you have, post it here...if you wish", "Solution_1": "got them today, 235: 80v, 77m, 78w, -1, -1, -3, respectively", "Solution_2": "very nice good sir or ma'am", "Solution_3": "impressive...\r\n\r\nwhat grade are you in?\r\n\r\nshame that they deduct so many points for so few errors on the math...no?", "Solution_4": "that would be \"sir,\" im in the 11th grade, Tuesday test, from last year, verbal up 3 points, math down 3, writing up 10; overall: up 10 points from my score.", "Solution_5": "you should get national merit with ease, but then you have to write an essay", "Solution_6": "did the rest of you get scores yet? i have not, but i think we should within the week, or maybe next week even.", "Solution_7": "My friends got theirs today, but I'm sick so I'll get my Mom to pick them up tomorrow.", "Solution_8": "Are the cutoff scores in the Princeton Review the cutoff for Commended or Semi-finalist?", "Solution_9": "I got mine, but...yeah...they're not as good as I wanted (nowhere near *******'s) :D. But it's ok, cuz I'm a sophomore. *Frantically buys SAT books off ebay*", "Solution_10": "i got mine back, since my counselor was so nice to tell me early. my score is 232, with 77 writing, 75 math, 80 writing. a little ironic, since i dont do anything for writing, but i spend a lot of time on math. even so, im in 10th grade, so im pretty happy with my results.", "Solution_11": "Do they mail the scores to you?\r\n\r\n-interesting_move", "Solution_12": "238...", "Solution_13": "[quote=\"mathnovice\"]Are the cutoff scores in the Princeton Review the cutoff for Commended or Semi-finalist?[/quote]\r\nThe cutoff scores change each year. Last year, the cutoff for semifinalist was 209, and the cutoff for commended was 200 or 201.", "Solution_14": "222, Am worried that the cutoff for semifinalist will be 223 or something like that, though...", "Solution_15": "221 --- I should have had higher, but I when I boxed in my answer I wrote \"92\" instead of \"192.\" Note to self- never do that again. : ( I hope I make the cutoff though.", "Solution_16": "Edit: That was a bit too whiny.", "Solution_17": "mathnovice arent you from MA? I'll pray that it's 221 too. That'd be horrible to be so close and not make it.", "Solution_18": "It'll probably be a lot lower than 221.", "Solution_19": "Matt -- I'm pretty sure it varies from state to state. MA probably has a higher cutoff than most places, although I strongly doubt it will be 220's.", "Solution_20": "I did so badly...[hide]191[/hide].\n\n\n\n-interesting_move", "Solution_21": "i'm positive that it's been ~222 meaning =/- 1 or 2 points for many years now. My guidance counselor says that in his ~10 years it's gone from 220 to 222, and I know for sure that last year's cutoff was 222. And it is the highest in the nation... I read that NJ's was 221 for this year's seniors, not positive that it's true though.", "Solution_22": "Wait... the cutoff for NATIONAL merit varies between states? That doesn't seem right.", "Solution_23": "it is right, oklahoma's was like 207 last year, when other states were in the 220's and suches", "Solution_24": "[quote=\"confuted\"]Wait... the cutoff for NATIONAL merit varies between states? That doesn't seem right.[/quote]\r\n\r\nThe statement is correct. And I think the statement that MA is usually one of the two or three most competitive states is also correct. I was comfortably above the cutoff, decades ago, in Minnesota. The states of the upper Midwest, where many high school students score well on all manner of standardized tests, historically have NOT been very competitive for getting above the National Merit Scholarship cutoff, because many students in that region self-select to take the ACT rather than the SAT and its associated PSAT/NMSQT. What makes a state have a high cutoff score is BOTH lots of well-prepared students AND a high percentage of the well-prepared students showing up to take the PSAT/NMSQT, which is, after all, a voluntary test. \r\n\r\nI suppose the policy underlying the differing cutoff scores around the county is to ensure that young people from all states of the Union have a chance at getting a [url=http://www.nationalmerit.org/entering.html]National Merit Scholarship[/url]. (After edit: yes, that's it, and the link is below.) \r\n\r\nhttp://www.nationalmerit.org/nmsp.html#semifinalists \r\n\r\n\"To ensure that academically able young people from all parts of the United States are included in this talent pool, Semifinalists are designated on a state representational basis.\"\r\n\r\nIt's not a perfect system (the scholarships are terribly underfunded, as they long have been), but it's a decades-old first attempt at helping bright young people from families of modest means go to college. State university systems with low list prices have undoubtedly had a much bigger impact on access to higher education than has the National Merit Scholarship program--the former worked better for me than the latter, given my family circumstances at the time.", "Solution_25": "Yeah, my mother was one of the top scorers in Maine back in the day -- odds are not good that she would have made it in New York or Massachusetts. Many of the scholarships are distributed with a large helping of arbitraryness, though -- look at the business-backed scholarships (one of which my mom recieved, because my grandmother worked for Worldbook at the time, who happened to sponsor a National Merit Scholarship.)", "Solution_26": "When do you find out if you qualified for semi-finalist? If I had gotten one more question right, I probably would not have to worry. : (", "Solution_27": "231, soph. Stupid careless mistakes.", "Solution_28": "[quote=\"mathnovice\"]When do you find out if you qualified for semi-finalist? If I had gotten one more question right, I probably would not have to worry. : ([/quote]\r\nYou'll find out at the beginning of your senior year. It'll be a long wait.", "Solution_29": "238", "Solution_30": "227(80W 77M 70V) but i 1590ed the SAT so its all good : )", "Solution_31": "I'm a sophomore, I got my scores a while back, 206(61V, 78M, 67W) This was my first time taking the PSATs.", "Solution_32": "240 but i took them before as a sophomore also and got 220 (80 M, 72 V, 68 W)" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all functions $ f: \\mathbb{R}\\to\\mathbb{R}$ that satisfy $ f((x\\minus{}y)^2)\\equal{}f^2(x)\\minus{}2xf(y)\\plus{}y^2,\\forall x,y\\in\\mathbb{R}$", "Solution_1": "[hide]\nPlug in $ x\\equal{}y\\equal{}0$ to get $ f(0)\\equal{}f(0)^2$, so $ f(0)$ is either 0 or 1.\n\nPlug in $ x\\equal{}y$ to get $ f(0)\\equal{}f(x)^2\\minus{}2xf(x)\\plus{}x^2\\equal{}(f(x)\\minus{}x)^2$. If $ f(0)\\equal{}0$, then $ f(x)\\equal{}x$ for all $ x$. This function works.\n\nIf $ f(0)\\equal{}1$, then for each $ x$, $ f(x)\\equal{}x\\plus{}1$ or $ f(x)\\equal{}x\\minus{}1$. Suppose there exists $ a$ with $ f(a)\\equal{}a\\minus{}1$ Then $ x\\equal{}a,y\\equal{}0$ gives\n\\[ a^2\\pm 1\\equal{}f(a^2)\\equal{}(a\\minus{}1)^2\\minus{}2a\\equal{}a^2\\minus{}4a\\plus{}1\\]\nIf $ f(a^2)\\equal{}a^2\\plus{}1$, then $ a\\equal{}0$, which we know is false. Otherwise, $ a\\equal{}1/2$ but then $ f(1/4)\\equal{}1/4\\minus{}1$, repeating the argument gives $ 1/4\\equal{}1/2$, a contradiction.\n\nSo the only other solution is $ f(x)\\equal{}x\\plus{}1$ for all $ x$. This solution turns out to work, too, since\n\\[ (x\\minus{}y)^2\\plus{}1\\equal{}(x\\plus{}1)^2\\minus{}2x(y\\plus{}1)\\plus{}y^2\\]\nfor all $ x,y$.\n[/hide]", "Solution_2": "$ x\\equal{}0$ gives $ f(y^2)\\equal{}f(0)\\plus{}y^2$ so $ \\forall x\\ge 0: \\ f(x)\\equal{}x\\plus{}a\\ (f(0)\\equal{}a)$\r\nso if $ x\\ge 0$ gives $ (x\\minus{}y)^2\\plus{}a\\equal{}(x\\plus{}a)^2\\minus{}2xf(y)\\plus{}y^2\\iff 2xy\\minus{}a\\equal{}\\minus{}2x(a\\minus{}f(y))\\minus{}a^2$\r\nthen $ \\forall y\\in R: \\ f(y)\\equal{}y\\plus{}a$ because $ x\\equal{}y\\equal{}0$ gives $ a^2\\equal{}a$\r\n\r\n$ f((x\\minus{}y)^2)\\equal{}x^2\\plus{}y^2\\minus{}2xy\\plus{}a$\r\n$ f^2(x)\\plus{}y^2\\minus{}2xf(y)\\equal{}x^2\\plus{}y^2\\minus{}2xy\\plus{}a^2$\r\n\r\nthen $ f(x)\\equal{}x$ or $ f(x)\\equal{}x\\plus{}1$", "Solution_3": "[quote=\"scorpius119\"][hide]\n\nIf $ f(0) \\equal{} 1$, then for each $ x$, $ f(x) \\equal{} x \\plus{} 1$ or $ f(x) \\equal{} x \\minus{} 1$. Suppose there exists $ a$ with $ f(a) \\equal{} a \\minus{} 1$ Then $ x \\equal{} a,y \\equal{} 0$ gives\n\\[ a^2\\pm 1 \\equal{} f(a^2) \\equal{} (a \\minus{} 1)^2 \\minus{} 2a \\equal{} a^2 \\minus{} 4a \\plus{} 1\n\\]\nIf $ f(a^2) \\equal{} a^2 \\plus{} 1$, then $ a \\equal{} 0$, which we know is false. Otherwise, $ a \\equal{} 1/2$ but then $ f(1/4) \\equal{} 1/4 \\minus{} 1$, repeating the argument gives $ 1/4 \\equal{} 1/2$, a contradiction.\n\n[/quote]\r\n\r\nIn my opinion, there is a intricacy in above argument. Because there exist [b]some[/b] values $ a$ such that $ f(a) \\equal{} a \\minus{} 1$. With $ a \\equal{} \\frac{1}{2}$ then $ f(1/2)\\equal{}1/2\\minus{}1$ and $ f(1/4)\\equal{}1/4\\minus{}1$, we can't conclude $ \\frac{1}{2}\\equal{}\\frac{1}{4}$", "Solution_4": "Well, I was able to conclude that if there is an $ a$ that satisfies $ f(a)\\equal{}a\\minus{}1$, then $ a$ must equal $ 1/2$. But if we let $ b\\equal{}1/4$, we must also have $ f(b)\\equal{}b\\minus{}1$, and then $ b$ must equal $ 1/2$. Since $ b\\equal{}1/4$, this is not possible.\r\n\r\nAlternatively, plug in $ x\\equal{}1/4,y\\equal{}0$ to get that $ f(1/16)$ is something other than $ (1/16)\\pm 1$.", "Solution_5": "Yes, i see, my stupid question." } { "Tag": [ "trigonometry", "geometry", "trig identities", "Law of Sines", "Law of Cosines" ], "Problem": "The skyhawk in Cedar point?\r\nhttp://www.cedarpoint.com/public/park/rides/thrill/skyhawk/index.cfm\r\n\r\nI just need to make trig questions like something that could be put on a worksheet with the skyhawk. Does anyone have any ideas.", "Solution_1": "Just putting out random ideas:\r\n\r\n1) The length of the skyhawk arm is ______ ft and they form an isosceles triangle with angle ____degrees. Find the length of the third side.\r\n\r\n2) Find the area of the triangle spanned by the arms of the skyhawk.", "Solution_2": "If you can create some sort of triangle, you can make a trig problem revolving around the Law of Sines or Law of Cosines (or anything else if I'm leaving them out)." } { "Tag": [ "geometry", "trigonometry" ], "Problem": "The area of triangle $ ABC$ is $ 1$, $ AC \\equal{} 2BC$, point $ K$ is middle of $ AC$. The circle with center $ K$ crosses side $ AB$ in points $ M$ and $ N$, where $ AM \\equal{} MN \\equal{} NB$. Find the area of that part of triangle $ ABC$, which is inside the circle.", "Solution_1": "wat do u mean by circle with circumcentre K? K is the circumcentre of which triangle? Please be clear.", "Solution_2": "[quote=\"hell_ever\"]wat do u mean by circle with circumcentre K? K is the circumcentre of which triangle? Please be clear.[/quote]\r\n\r\nSorry for my bad english. I mean K- is center of circle", "Solution_3": "How to solve it? What to use?", "Solution_4": "Meatman, your problems are too easy for the olympiad level. If you continue to post easy problems here, you will probably not get useful replies. I am moving this one to the high school section as well.\r\n\r\nRegards, yetti.\r\n________________________________________\r\n\r\nDenote $ b \\equal{} AC, c \\equal{} AB.$ $ AK \\equal{} \\frac {b}{2},$ $ AM \\equal{} \\frac {c}{3},$ $ AN \\equal{} \\frac {2c}{3}.$ By the cosine theorem for the $ \\triangle AKM, \\triangle AKN,$\r\n\r\n$ KM^2 \\equal{} AK^2 \\plus{} AM^2 \\minus{} 2 AK \\cdot AM \\cos A \\equal{} \\frac {b^2}{4} \\plus{} \\frac {c^2}{9} \\minus{} \\frac {bc}{3}\\cos A$\r\n\r\n$ KN^2 \\equal{} AK^2 \\plus{} AN^2 \\minus{} 2 AK \\cdot AN \\cos A \\equal{} \\frac {b^2}{4} \\plus{} \\frac {4c^2}{9} \\minus{} \\frac {2bc}{3}\\cos A$\r\n\r\nFrom $ KM \\equal{} KN,$\r\n\r\n$ KN^2 \\minus{} KM^2 \\equal{} \\frac {c^2}{3} \\minus{} \\frac {bc}{3} \\cos A \\equal{} \\frac {c}{3} \\left(c \\minus{} b \\cos A \\right) \\equal{} 0$\r\n\r\nIt follows that $ \\cos A \\equal{} \\frac {c}{b},$ which means that the angle $ \\angle B \\equal{} 90^\\circ$ is right, $ B$ is on a circle $ (O)$ with diameter $ AC.$ From the condition $ AC \\equal{} 2 BC,$ $ B$ is also on a circle $ (C)$ centered at $ C$ and with radius $ CB \\equal{} \\frac {AC}{2},$ congruent to the circle $ (O).$ $ B$ is one of the equivalent intersections of the circles $ (O), (C)\\ \\Longrightarrow$ the right angle $ \\triangle ABC$ has angles $ \\angle B \\equal{} 90^\\circ, \\angle C \\equal{} 60^\\circ, \\angle A \\equal{} 30^\\circ$ and $ \\cos A \\equal{} \\frac {c}{b} \\equal{} \\frac {\\sqrt 3}{2}.$ Substituting $ b \\equal{} \\frac {2c}{\\sqrt 3}, b^2 \\equal{} \\frac {4c^2}{3}$ into the cosine formula for $ KM$ yields $ KM^2 \\equal{} \\frac {c^2}{9}, KN \\equal{} KM \\equal{} \\frac {c}{3} \\equal{} MN,$ the $ \\triangle KMN$ is equilateral. The desired area is then, using $ KM^2 \\equal{} \\frac {c^2}{9} \\equal{} \\frac {b^2}{12},$\r\n\r\n$ S \\equal{} \\frac {\\pi \\cdot KM^2}{3} \\plus{} \\frac {KM^2 \\sqrt 3}{4} \\equal{} b^2\\ \\frac {4 \\pi \\plus{} 3 \\sqrt 3}{144}.$", "Solution_5": "Oh, sorry :blush: I won't post such kind of problems there.\r\nThanks for solution." } { "Tag": [ "geometry", "rectangle", "calculus", "integration" ], "Problem": "I found this problem in Harold Reiter's mathcount's workshops: http://www.math.uncc.edu/~hbreiter/problems/HawaiiIndex.htm , but I have no clue where to even begin. Can someone help? Thanks.\r\n\r\nHow many rectangles are defined by the grid lines below? How many of them are squares?", "Solution_1": "Uh look at this\r\n\r\n[hide]You know the largest rectangle can at most have a width of 2 squares. So break it up into 4, 6*2, rectangles count how many squares/rectangles in each one and times by 4. I count 63 total rectanges 17 of which are squares... that means 252 rectanles and 68 squares. There are more rectangles however.\n\nNow you also have to account for rectangles that overlap the boundaries that you created. There are only 4 squares total that cross those borders. I count 24 extra rectangles on each segment *4=96. Now just add them all together...\n\nI get 352 total squares and 72 of those are squares. I'm quite sure about that answer... I used to know a formula that is applied to counting rectangles... but I can't remember it.[/hide]\r\n\r\nhope that helps.", "Solution_2": "[color=darkred]Don't forget to count rectangles with non-integral side lengths...[/color]", "Solution_3": "[quote=\"Treething\"][color=darkred]Don't forget to count rectangles with non-integral side lengths...[/color][/quote]\r\n\r\nBut those aren't defined by the gridlines.", "Solution_4": "Oh. The gridlines. Sorry, I thought it meant intersections :blush: . I gotsta learn how to read...", "Solution_5": "[quote] You know the largest rectangle can at most have a width of 2 squares. [/quote]\r\n\r\nSorry if I am not reading the problem carefully .. but there are rectangles ( may be with \"holes\" in the middle , but formed by lines of the grid) which are larger...", "Solution_6": "I think your best bet may be to determine the number of rectangles if there were grid lines where the 'hole' is, then subtract out the ones that don't exist as a result of the hole. \r\n\r\n[hide=\"Doing so yields...\"]If there were no hole, there are $\\binom{9}{2}$ ways to choose the two vertical sides and $\\binom{9}{2}$ ways to choose the two horizontal sides, so a total of 36^2.\n\nThe rectangle will not exist as a result of the hole iff it has at least one vertical and one horizontal side that passes through the hole. So how many ways are there to choose two vertical lines such that at least one passes through the hole? There are 3 choices for the line that will pass through the hole, times 8 choices for the second line, minus 3 choices for repetitions (if we number the lines from 1 to 9, then 45, 46, and 56 are each counted twice). So 21 ways. Similarly, there are 21 ways for the horizontal lines. So in all 21^2 are bad.\n\nThus the answer is 36^2 - 21^2 = 855.[/hide]\n\nAs for how many squares there are,\n[hide]Just counting them may work best. There are 48 1x1s, 24 2x2s, 1 4x4, 16 5x5s, 9 6x6s, 4 7x7s, and 1 8x8 (note that for the 5x5s through 8x8s, there are the usual (8-n)^2 n x ns because the hole is too small to take any away. In all, 103.[/hide]", "Solution_7": "Or, you can actually count the rectangles directly. I did it by breaking them down into height and witdth and making a chart:\r\n\r\n\r\n[code]\n\\W 1 2 3 4 5 6 7 8\nH\\ \n1 48 36 24 20 16 12 8 4\n2 36 24 12 10 8 6 4 2\n3 24 12 0 0 0 0 0 0\n4 20 10 0 1 2 3 2 1\n5 16 8 0 2 4 6 4 2\n6 12 6 0 3 6 9 6 3\n7 8 4 0 2 4 6 4 2\n8 4 2 0 1 2 3 2 1[/code]\r\nAdding up all of the entries, I get 477 rectangles and 91 squares -- anyone want to chime in on why this is so much smaller than Keone's answer? I notice that he made a definite mistake in counting the number of 5x5 squares -- did he do something similar on the rectangles as a whole, or is it my mistake?\r\n\r\nActually, I do see one mistake in his work -- you don't necessarily need a vertical and a horizontal side to cross the hole. Consider, for example, some 3x8 rectangles.", "Solution_8": "I get 477 total rectangles. 91 of these are square.\r\n\r\nThere are 6 full-height vertical lines and 6 full-width horizontal lines. Choose one pair of horizontal lines plus a pair of vertical lines: 6C2x6C2=225.\r\n\r\nConsider the top section directly above the empty square. There are 3 full-width horizontal lines and 3 short vertical lines. One pair from each: 3C2x3C2=9.\r\n\r\nContinuing the entire top section above the empty square (and extending the width of the large square). There are 3 full-width horizontal lines. Choose 2. There are 6 full-height vertical lines. Choose 1. There are 3 short vertical lines. Choose 1. 3C2x6x3=54.\r\n\r\nRepeat for the section below, left, and right of empty square.\r\n\r\nTotal: $225+4(9+54)=225+252=477$\r\n\r\n[quote=\"JBL\"]Adding up all of the entries, I get 447 rectangles and 91 squares [/quote]\r\nJBL: I think your table is correct, so there may be an arithmetic error.", "Solution_9": "Fixed -- no arithmetic error, actually, just a typo. Thanks.", "Solution_10": "Thanks guys! I learned a lot." } { "Tag": [ "geometry", "perpendicular bisector", "geometric transformation", "geometry unsolved" ], "Problem": "On the exterior of a triangle $ABC$ we draw two similar right triangles $ABD , ACE$ , with right angles at $D , E$ and $\\angle BAD = \\angle CAE$. If $M$ is the midpoint of $BC$ , prove that $MD = ME$.\r\n\r\n [u]Babis[/u]\r\n\r\n [i]Mery Cristmas !!![/i]", "Solution_1": "That's easy.Just let F is the midpoint of AB,and G is the midpoint of AC,DF=AB/2=MG,GE=AC/2=FM,and \u2220DFB=\u2220EGC,\u2220BFM=\u2220A=\u2220MGC,so \u2220DFM=\u2220MGE,so triangle DFM and MGE is congruent,so MD=ME. :)", "Solution_2": "Two additional solutions\n\nM1) Take $K$ and $L$ the midpoints of $AB$ and $AC$, see that $\\triangle DKM \\equiv \\triangle MLE$ (s.a.s), done.\n\nM2) Take $P$ on the perpendicular bisector of $BC$ so that $\\angle BCP = \\angle ABD$, then see that there is a spiral similarity centered at $B$, which takes $P$ to $M$ and $A$ to $D$, and a spiral similarity centered at $C$, taking $P$ to $M$ and $A$ to $E$, hence $\\frac{DM}{AP}=\\frac{ME}{AP}$ , consequently $ME=MD$ and, as it has been seen, $\\widehat {(DM, AP)} = \\widehat {(ME, AP)} = \\angle ABD$, so $\\angle DME = 2\\angle ABD$ . \n\nBest regards,\nsunken rock" } { "Tag": [ "trigonometry", "function", "limit", "symmetry", "algebra proposed", "algebra" ], "Problem": "1. Suppose $ f: \\left[ \\minus{} \\frac {\\pi}{2},\\frac {\\pi}{2}\\right ]\\to \\mathbb{R}$ is a continuous function. Prove that $ f(x) \\minus{} \\tan x \\equal{} 0$ has at least one solution in $ \\left( \\minus{} \\frac {\\pi}{2},\\frac {\\pi}{2}\\right)$.\r\n\r\n2. Suppose $ f: [ \\minus{} \\pi,\\pi]\\to \\mathbb{R}$ is a continuous function and $ f(x) \\equal{} f( \\minus{} x)$. Prove that $ f(x) \\minus{} \\tan x \\equal{} 0$ has at least two solutions in $ \\left[ \\minus{} \\pi, \\pi\\right]$", "Solution_1": "the first one is obvious throug graph :) but dont know how to do it rigorously :oops:", "Solution_2": "[quote=\"v.dinesh\"]the first one is obvious throug graph :) but dont know how to do it rigorously :oops:[/quote]\r\n\r\nIt is obvious throug the graph and it also follows from the Bolzano Theorem, since g(x) = f(x) - tanx is continious in [m, n] for $ m,n \\in (\\minus{}\\frac{\\pi}{2}, \\frac{\\pi}{2})$ with g(m) < 0 and g(n) > 0 (these m,n exists because $ \\lim_{x \\rightarrow \\frac {\\pi}{2}}g(x) \\equal{} \\plus{}\\infty$, and $ \\lim_{x \\rightarrow \\minus{}\\frac {\\pi}{2}}g(x) \\equal{} \\minus{}\\infty$, because f is continious in $ [\\minus{}\\frac{\\pi}{2}, \\frac{\\pi}{2}]$ and so the limits exists)", "Solution_3": "Another way for 1 is to use the fact that continuous functions have maxima and minima on closed and bounded intervals. Take the maximum $ M$ of $ f$, and note that if $ \\tan x > M$ (which it is somewhere on $ (\\minus{}\\pi/2,\\pi/2)$, then $ f(x) \\minus{} \\tan x < 0$. Likewise, the same reasoning on the minimum $ m$ gives an $ x$ such that $ f(x) \\minus{} \\tan x > 0$. Then the intermediate value theorem finishes.\r\n\r\nFor 2, reducing it showing that $ |f(x)| \\equal{} \\tan x$ has a root in $ [0,\\pi/2)$. Suppose WLOG that $ f(0) > 0$, and then use the maximum argument to show there is an $ x$ such that $ f(x) < \\tan x$.", "Solution_4": "[quote=\"MellowMelon\"]For 2, reducing it showing that $ |f(x)| \\equal{} \\tan x$ has a root in $ [0,\\pi/2)$. Suppose WLOG that $ f(0) > 0$, and then use the maximum argument to show there is an $ x$ such that $ f(x) < \\tan x$.[/quote]How does this give us two roots in $ [ \\minus{} \\pi ,\\pi]$? Note the function satisfies $ f(x)\\equal{}f(\\minus{}x)$, not $ f(x)\\equal{}\\minus{}f(\\minus{}x)$.", "Solution_5": "In $ [0,\\pi]$, if $ f(x) \\equal{} \\tan x$ we have a solution on that interval, obviously. If $ \\minus{}f(x) \\equal{} \\tan x$, then $ \\minus{}f(\\minus{}x) \\equal{} \\tan x \\Rightarrow f(\\minus{}x) \\equal{} \\tan \\minus{}x$. Then we can reduce that to $ [0,\\pi/2]$ by symmetry and find one solution there, giving two on the larger one.", "Solution_6": "I still must not be understanding.\r\n[quote=\"MellowMelon\"]In $ [0,\\pi]$, if $ f(x) \\equal{} \\tan x$ we have a solution on that interval, obviously. If $ \\minus{} f(x) \\equal{} \\tan x$, then $ \\minus{} f( \\minus{} x) \\equal{} \\tan x \\Rightarrow f( \\minus{} x) \\equal{} \\tan \\minus{} x$.[/quote]So far you have proven that the existence of one solution to $ f(x) \\equal{} \\tan x$ in $ [ \\minus{} \\pi,\\pi]$ is equivalent to the existence of one solution of $ |f(x)| \\equal{} \\tan x$ in $ [0,\\pi]$. I agree, at least one such solution certainly exists.[quote=\"MellowMellon\"]Then we can reduce that to $ [0,\\pi/2]$ by symmetry and find one solution there, giving two on the larger one.[/quote]Reduce what? Find one solution where?\r\n\r\nI really don't think symmetry is important here. For example, if I were to replace $ \\tan x$ by something non-symmetric like, say, $ \\tan x \\plus{} x^2 \\plus{} 7x \\plus{} 1$, the conclusion that $ f(x) \\equal{} \\tan x \\plus{} x^2 \\plus{} 7x \\plus{} 1$ has at least two roots in $ [ \\minus{} \\pi, \\pi]$ for any continuous even $ f$ would still hold. And in fact it would still hold for a much wider class of functions than just continuous even ones.", "Solution_7": "(fix: when I talked about $ |f(x)| \\equal{} \\tan x$, I probably meant $ |f(x)| \\equal{} |\\tan x|$, which changes things when going to other intervals)\r\n\r\nThe symmetry is just a way of simplifying the solution. If it were $ \\tan x$ plus some asymmetric function, you'd probably be right that the conclusion would remain the same, and I admit it would take a stronger proof than mine.\r\n\r\nThe symmetry involved is that $ \\tan x \\equal{} \\minus{}\\tan (\\pi \\minus{} x)$. If any $ f(x)$ defined on $ [0,\\pi/2$ has a solution to $ |f(x)| \\equal{} |\\tan x|$, then any $ f(x)$ defined on $ [\\pi/2,\\pi]$ has a solution to that equation because we let $ g(x) \\equal{} f(\\pi \\minus{} x)$ be defined on $ [0,\\pi/2]$, and the solution we know exists for $ |g(x)| \\equal{} |\\tan x|$ implies that \\[ |g(x)| \\equal{} |\\tan (x)|\\]\r\n\\[ |f(\\pi \\minus{} x)| \\equal{} |\\minus{}\\tan (\\pi \\minus{} x)|\\]\r\n\\[ |f(\\pi \\minus{} x)| \\equal{} |\\tan ( \\pi \\minus{} x )|\\]" } { "Tag": [ "function", "probability", "calculus", "integration", "quadratics" ], "Problem": "The new wave function for a particle is made by adding two functions of particle in states $ a$ and $ b$ ,such that the function is so: \r\n\r\n$ \\psi\\equal{}C_a\\psi_a\\plus{}C_b\\psi_b$\r\n\r\nIn this equation,$ C_a$ and $ C_b$ are constants.$ C_a\\equal{}\\frac{1}{2}$ is given.What is the probability of particle being in state b?", "Solution_1": "Don't we just evaluate the integral $ \\int_{\\minus{}L}^{L}\\Psi\\Psi^{*}\\ \\text{d}x$?", "Solution_2": "I need the solution for checking whether mine is correct or not :) .", "Solution_3": "I could be wrong, but I'd say that the probability of the particle being in state b is just 3/4, due to the normalization condition for wavefunctions. Namely, I think $ C_b \\equal{} \\sqrt(3)/2$, since $ |C_a|^2 \\plus{} |C_b|^2 \\equal{} 1$.\r\n\r\nI conclude this because we already know that\r\n\\[ \\int_{ \\minus{} \\infty}^\\infty |\\psi_a|^2 d(\\mbox{whatever}) \\equal{} \\int_{ \\minus{} \\infty}^\\infty |\\psi_b|^2 d(\\mbox{whatever}) \\equal{} 1.\r\n\\]\r\nOh wow, I'm a \"Hodge Conjecture\" now.\r\n\r\nOh wait, nevermind, it's not so simple. That is, we want to find $ |C_b|^2$ given that\r\n\\[ \\int_{ \\minus{} \\infty}^\\infty |\\psi|^2 d(\\mbox{whatever}) \\equal{} \\int_{ \\minus{} \\infty}^\\infty |C_a\\psi_a \\plus{} C_b\\psi_b|^2 d(\\mbox{whatever}) \\equal{} 1.\r\n\\]\r\nThen\r\n\\[ 1 \\equal{} \\int_{ \\minus{} \\infty}^\\infty |C_a\\psi_a \\plus{} C_b\\psi_b|^2 \\equal{} \\int_{ \\minus{} \\infty}^\\infty (C_a\\psi_a \\plus{} C_b\\psi_b)(C_a\\psi_a^* \\plus{} C_b\\psi_b^*) \\equal{} \\\\\r\nC_a^2 \\int_{ \\minus{} \\infty}^\\infty |\\psi_a|^2 \\plus{} C_a C_b \\int_{ \\minus{} \\infty}^\\infty \\psi_a \\psi_b^* \\plus{} C_a C_b \\int_{ \\minus{} \\infty}^\\infty \\psi_b \\psi_a^* \\plus{} C_b^2 \\int_{ \\minus{} \\infty}^\\infty |\\psi_b|^2 \\equal{} \\\\\r\nC_a^2 \\plus{} C_b^2 \\plus{} C_a C_b \\int_{ \\minus{} \\infty}^\\infty \\psi_a \\psi_b^* \\plus{} \\psi_b \\psi_a^*.\r\n\\]\r\nAnd that's a quadratic in $ C_b$, assuming you know the last integral.", "Solution_4": "yes,I also got 3/4.", "Solution_5": "The answer is $ \\frac{3}{4}$ if $ \\psi_1$ and $ \\psi_2$ are orthogonal. If they\r\nare not, the answer may be different, as genericme pointed out." } { "Tag": [ "calculus", "inequalities", "function" ], "Problem": "Find $y_{max}$ and $y_{min}$ of $y=\\frac{x^{2}-3x+4}{x^{2}+3x+4}$. (hint - do not use calculus -- theres a faster way with inequality :) )", "Solution_1": "You can use the discriminant and A.M-G.M. :D", "Solution_2": "\\[\\frac{x^{2}-3x+4}{x^{2}+3x+4}=y\\] \\[x^{2}-3x+4=yx^{2}+3xy+4y\\] \\[x^{2}(y-1)+x(3y+3)+4y-4=0\\] \\[y=1 \\Rightarrow x=0\\] \\[y\\not =1 \\Rightarrow\\] \\[\\Delta \\geq 0\\] \\[(3y+3)^{2}-(4y-4)^{2}\\geq 0\\] \\[7y^{2}-50y+7 \\leq 0\\] \\[y_{min}=\\frac{2}{17}\\] \\[y_{max}=\\frac{98}{17}\\]", "Solution_3": "Why can you use discriminant immediately?\r\nYou should answer the values of $x,\\ y$ for which the given equation is minimized or maximized.", "Solution_4": "I consider function\r\n$f(x)=x^{2}(y-1)+x(3x+3)+4y-4$ This function has at least one root,so it's discriminant is greater that zero or equal", "Solution_5": "You should classify the case of $y=1$ and $y\\neq 1.$" } { "Tag": [ "logarithms" ], "Problem": "Simplify the following equations.\r\n\r\n(1) $\\log_{2}{18}+\\frac{1}{2}\\log_{2}{\\frac{1}{3}}-\\frac{3}{2}\\log_{2}{\\sqrt[3]{12}}$\r\n\r\n(2) $(\\log_{10}2)^{3}+\\log_{10}{8}\\cdot \\log_{10}{5}+(\\log_{10}5)^{3}$\r\n\r\n(3) $\\log_{4}{3}\\cdot \\log_{27}25\\cdot \\log_{5}{16}$\r\n\r\n(4) $3^{2\\log_{3}2}$", "Solution_1": "1.\r\n\r\n[hide] $\\log_{2}18+\\frac{1}{2}\\log_{2}\\frac{1}{3}-\\frac{3}{2}\\log_{2}\\sqrt[3]{12}= \\log_{2}18+\\log_{2}\\sqrt{\\frac{1}{3}}-\\log_{2}\\sqrt{12}= \\log_{2}\\frac{18\\sqrt{\\frac{1}{3}}}{\\sqrt{12}}= \\log_{2}\\frac{18\\sqrt{3}}{6\\sqrt{3}}= \\log_{2}3$ [/hide]", "Solution_2": "(1)\r\n$\\log_{2}{18}+\\frac{1}{2}\\log_{2}{\\frac{1}{3}}-\\frac{3}{2}\\log_{2}{\\sqrt[3]{12}}$\r\n$=\\log_{2}{18}+\\log_{2}{(\\frac{1}{3})^{\\frac{1}{2}}}-\\log_{2}{12^{(\\frac{1}{3})(\\frac{3}{2})}}$\r\n$=\\log_{2}{18}+\\log_{2}{(\\frac{1}{3})^{\\frac{1}{2}}}-\\log_{2}{12^{\\frac{1}{2}}}$\r\n$=\\log_{2}{(18\\cdot(\\frac{1}{3})^{\\frac{1}{2}}(\\frac{1}{12^{\\frac{1}{2}}}))}$\r\n$=\\log_{2}{3}$", "Solution_3": "[quote=\"vishalarul\"](1)\n$=\\log_{2}{18}+\\log_{2}{(\\frac{1}{3})^{\\frac{1}{2}}}-\\log_{2}{12^{\\frac{1}{2}}}$\n$=\\log_{2}{(18\\cdot(\\frac{1}{3})^{\\frac{1}{2}}\\cdot12^{\\frac{1}{2}})}$\n[/quote]\r\n\r\nThe sign in front of the $\\log_{2}12^{\\frac{1}{2}}$ is negative, so I think you divide by $12^{\\frac{1}{2}}$ and get $\\log_{2}3$, like E^(pi*i)=-1 got.", "Solution_4": "[quote=\"vishalarul\"](1)\n$\\log_{2}{18}+\\frac{1}{2}\\log_{2}{\\frac{1}{3}}-\\frac{3}{2}\\log_{2}{\\sqrt[3]{12}}$\n$=\\log_{2}{18}+\\log_{2}{(\\frac{1}{3})^{\\frac{1}{2}}}-\\log_{2}{12^{(\\frac{1}{3})(\\frac{3}{2})}}$\n$=\\log_{2}{18}+\\log_{2}{(\\frac{1}{3})^{\\frac{1}{2}}}-\\log_{2}{12^{\\frac{1}{2}}}$\n$=\\log_{2}{(18\\cdot(\\frac{1}{3})^{\\frac{1}{2}}\\cdot12^{\\frac{1}{2}})}$\n$=\\log_{2}{36}$\n$=2\\log_{2}{6}$\n$=2(1+\\log_{2}{3})$[/quote]\r\n\r\nYou have an error: The minus sign in row 3 gave a dot in row 4, and it should've been a column or a slash.", "Solution_5": "(2)\r\n$(\\log_{10}2)^{3}+\\log_{10}{8}\\cdot\\log_{10}{5}+(\\log_{10}5)^{3}$\r\n$=(\\log_{10}2+\\log_{10}5)((\\log_{10}2)^{2}-(\\log_{10}2)(\\log_{10}5)+(\\log_{10}5)^{2})+3(\\log_{10}{2})(\\log_{10}{5})$\r\n$=(\\log_{10}2)^{2}-(\\log_{10}2)(\\log_{10}5)+(\\log_{10}5)^{2})+3(\\log_{10}{2})(\\log_{10}{5})$\r\n$=(\\log_{10}2)^{2}+2(\\log_{10}2)(\\log_{10}5)+(\\log_{10}5)^{2}$\r\n$=(\\log_{10}2+\\log_{10}5)^{2}$\r\n$=1$\r\n\r\n(3)\r\n$\\log_{4}{3}\\cdot \\log_{27}25\\cdot \\log_{5}{16}$\r\n$=\\frac{\\log_{10}3}{\\log_{10}{4}}\\cdot\\frac{\\log_{10}25}{\\log_{10}27}\\cdot\\frac{\\log_{10}16}{\\log_{10}5}$\r\n$=\\frac{\\log_{10}3}{2\\log_{10}{2}}\\cdot\\frac{2\\log_{10}5}{3\\log_{10}3}\\cdot\\frac{4\\log_{10}2}{\\log_{10}5}$\r\n$=\\frac{4}{3}$\r\n\r\n(4)\r\nLet $u=3^{2\\log_{3}2}$\r\n$\\log_{3}u=2\\log_{3}2=\\log_{3}4$\r\n$u=4$\r\n$3^{2\\log_{3}2}=4$", "Solution_6": "Answers are all right. :) \r\n\r\nComment:\r\n\r\nFor (2), let $a=\\log_{10}2,\\ b=\\log_{10}5,$ since $a+b=1,$\r\n\r\n$(\\log_{10}2)^{3}+\\log_{10}{8}\\cdot\\log_{10}{5}+(\\log_{10}5)^{3}$\r\n$=a^{3}+3ab+b^{3}=a^{3}+b^{3}+3ab(a+b)=(a+b)^{3}=1.$\r\n\r\nFor (4), by definition $log_{3}4$ is the root of the equation $3^{x}=4.$" } { "Tag": [ "function" ], "Problem": "$ (\\exists )\\ f: \\mathcal R\\rightarrow\\mathcal R$ so that $ f(1)\\equal{}2$ and $ (\\forall )\\ x\\in\\mathcal R$ , $ f(f(x))\\equal{}x^2\\minus{}2x\\plus{}2\\ \\ ?$", "Solution_1": "No. Put x=1 in the equation given to get $ f(2) \\equal{} 1$. Since $ f(1) \\equal{} 2$, by continuity there exists $ 1 1. In particular we have convergence for $z=1$. But the sum telescopes and we get $\\lim_{n}a_{n}= g(1) \\neq 0$.\r\n\r\nThis can easily be extended by induction to show that $a_{n}$ never converges to 0 (for any order pole). I don't know why I didn't realize this when I posted originally.", "Solution_3": "Does it matter that it was a pole and not some other kind of singularity? Yes.\r\n\r\n$f(z)=\\sum_{n=1}^{\\infty}\\frac{z^{n}}n=-\\log(1-z)$ has a branch point at $1.$ The power series converges everywhere else on the unit circle.\r\n\r\nNote that for this branch point, you cannot employ jmerry's proof; you can't subtract off the singularity.\r\n\r\n(jmerry didn't address the issue of poles of higher order, but you would simply subtract off the principal part of the Laurent expansion about 1, and it should work out the same.)", "Solution_4": "One thing I kept thinking about in doing this problem is this: We must have that $\\sum a_{n}$ diverges because otherwise we would by Abel's theorem have that the sum is continuous on the ray between 0 and 1 which would contradict having a pole. However, is it possible to extend this to say that the partial sums diverge to infinity?\r\n\r\nIf we instead had an essential singularity at 1 I don't see how the continuity is contradictory. Is it?" } { "Tag": [ "function", "group theory", "abstract algebra", "number theory", "prime numbers", "superior algebra", "superior algebra solved" ], "Problem": "Prove that the in group of bijections (one to one functions) from $\\mathbb{N}$ to $\\mathbb{N}$, denote it by $(G,\\circ )$, there are three subgroups, each isomorphic with one of the the three folowing groups:\r\na) $(\\mathbb{Z},+)$\r\nb) $(\\mathbb{Q},+)$\r\nc) $(\\mathbb{Q}_+^*,\\cdot)$", "Solution_1": "You can get every countably infinite group as a subgroup: given a countably infinite group $G$, identify the countable set $\\mathbb{N}$ with the set of elements of $G$, and take the bijections on this set induced by left action of the group ($g(x)=gx$). That's obviously isomorphic to $G$. Since each of the groups we care about is a countably infinite group, we can do this for them.", "Solution_2": "The original solution was quite ugly. They constructed some particular examples of izomophimes for every desired group.\r\n-for the first they chosen a bijections of infinite order\r\n-for the third they partitioned $\\mathbb{N}$ into a infinite number of infinite subsets. and on each subset they find a bijection ,$f_{p_i}$, of infinite order. Then using the fact that $(\\mathbb{Q}_+^*,\\cdot)$ is generated by prime numbers $p_i$, the conclusion folows. \r\n\r\nThinking about this I got to the same aproach as jmerry. thank you for your time." } { "Tag": [ "geometry", "3D geometry", "induction" ], "Problem": "I recently figured out something. (maybe some one has figured it out before me)\r\n\r\nI was looking at the animation on the side that said The sum of the first n odd natural numbers is n^2. Then I though what the sum of the first n even natural numbers are. After about 30 seconds, I figured out what it is. The sum of the first n even natural numbers is n(n+1). How would you prove this though?", "Solution_1": "If you accept that the sum of the first $ n$ odd natural numbers is $ n^2$, then you get the sum of the first $ n$ even natural numbers by adding $ 1$ to each of the first $ n$ odd natural numbers; in other words, you get the sum $ n^2 \\plus{} n \\equal{} n(n \\plus{} 1)$. :)\r\n\r\nThis also allows you to prove that the sum of the first $ n$ natural numbers (which is half the sum of the first $ n$ even natural numbers) is $ \\frac {n(n \\plus{} 1)}{2}$. This is the $ n^{th}$ triangular number $ T_n$.\r\n\r\nThere are well-known techniques for a lot of different summations; in particular, it is possible to find formulas for the sum of the first $ n$ squares, cubes, fourth powers, ...", "Solution_2": "We prove $ 2\\plus{}4\\plus{}6\\plus{}...\\plus{}2n\\equal{}n(n\\plus{}1)$ using induction:\r\n\r\n[b]Base Case ($ n\\equal{}1$)[/b]:\r\nThe sum of the first $ 1$ even natural number is $ 2$, and the formula yields $ 1(1\\plus{}1)\\equal{}2$, so our base case is proven.\r\n\r\n[b]Show that $ n$ implies $ n\\plus{}1$[/b]:\r\nAssume that our statement is true for $ n\\equal{}m$:\r\n$ 2\\plus{}4\\plus{}6\\plus{}...\\plus{}2m\\equal{}m(m\\plus{}1)$\r\nClearly, since the next even natural number is $ 2(m\\plus{}1)$, we add the quantity to both sides:\r\n$ 2\\plus{}4\\plus{}6\\plus{}...\\plus{}2m\\plus{}2(m\\plus{}1)\\equal{}m(m\\plus{}1)\\plus{}2(m\\plus{}1)$\r\nBy algebraically manipulating the RHS, we get:\r\n$ 2\\plus{}4\\plus{}6\\plus{}...\\plus{}2(m\\plus{}1)\\equal{}(m\\plus{}1)[(m\\plus{}1)\\plus{}1]$\r\n\r\nAnd this looks just like our original formula, so we are done." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "1) Let $ A,B \\in S_n(R)$ slove matrix equations :\r\n $ \\begin{cases} e^A \\equal{} I_n \\plus{} B \\\\\r\ne^B \\equal{} I_n \\plus{} A \\end{cases}$ \r\n $ I_n$ is identity matrix\r\n 2) Let $ A,B \\in S^{ \\plus{} }_n(R)$ slove matrix equations :\r\n $ \\begin{cases} Ae^A \\equal{} Be^B \\\\\r\nA^3 \\equal{} B \\plus{} 6I_n\\end{cases}$\r\n My solution 1) ..2) the same \r\n From 1) we have $ e^A \\plus{} A \\equal{} e^B \\plus{} B$\r\n By $ A \\in S_n(R)$ the exists $ T \\in O_n(R)$ ,$ D \\equal{} diag(\\lambda_1,\\lambda_2,...,\\lambda_n) \\in D_n(R)$ ,$ A \\equal{} TDT^{ \\minus{} 1}$ then \r\n $ e^A \\plus{} A \\equal{} T(e^D \\plus{} D)T^{ \\minus{} 1}$\r\nBy $ \\lambda_1,\\lambda_2,...,\\lambda_n \\in R$ ,$ \\forall (i,j) \\in$ {1,2,...,n} $ (e^{\\lambda_i} \\plus{} \\lambda_i \\equal{} e^{\\lambda_j} \\plus{} \\lambda_j$ then $ \\lambda_i \\equal{} \\lambda_j$ for all $ 1 \\leq i,j \\leq n$\r\n Show that the exists $ P \\in R[X]$ Such that : $ \\forall i \\in$ {1,2,...,n} $ P(e^{\\lambda_i} \\plus{} \\lambda_i) \\equal{} \\lambda_i$\r\n (by interpolation for $ e^{\\lambda_i}\\plus{}\\lambda_i$ distinct ) then We have :\r\n $ Tdiag(P(e^{\\lambda_1} \\plus{} \\lambda_1),...,P(e^{\\lambda_n} \\plus{} \\lambda_n))T^{ \\minus{} 1} \\equal{} Tdiag(\\lambda_1,...,\\lambda_n)T^{ \\minus{} 1}$ Or $ P(e^{A} \\plus{} A) \\equal{} A$ the same the exists $ Q \\in R[X]$ such that $ Q(e^B \\plus{} B) \\equal{} B$ \r\nand $ AB \\equal{} P(e^{A} \\plus{} A)Q(e^{B} \\plus{} B) \\equal{} P(e^{A} \\plus{} A)Q(e^{A} \\plus{} A) \\equal{} Q(e^A \\plus{} A)P(e^{A} \\plus{} A) \\equal{} Q(e^{B} \\plus{} B)P(e^{A} \\plus{} A) \\equal{} BA$\r\n Then the exists $ \\omega \\in O_n(R)$ Such that :\r\n $ \\omega^{ \\minus{} 1}A \\omega \\equal{} diag(a_1,...,a_n)$ ({$ a_1,...,a_n$}={$ \\lambda_1,...,\\lambda_n$})\r\n $ \\omega^{ \\minus{} 1}B \\omega \\equal{} diag(b_1,...,b_n)$\r\n $ \\omega^{ \\minus{} 1}(e^A \\plus{} A) \\omega \\equal{} diag(e^{a_1} \\plus{} a_1,...,e^{a_n} \\plus{} a_n)$\r\n $ \\omega^{ \\minus{} 1}(e^B \\plus{} B) \\omega \\equal{} diag(e^{b_1} \\plus{} b_1,...,e^{b_n} \\plus{} b_n)$\r\n Since $ e^A \\plus{} A \\equal{} e^B \\plus{} B$ then $ e^{a_i} \\plus{} a_i \\equal{} e^{b_i} \\plus{} b_i$ for all $ 1 \\leq i \\leq n$ then $ a_i \\equal{} b_i$ Or $ A \\equal{} B$ \r\nand need solution $ e^A \\equal{} A \\plus{} I_n$ eassy slove $ e^{a_i} \\equal{} a_i \\plus{} 1$ have unique root $ a_i \\equal{} 0$ for all $ 1 \\leq i \\leq n$ and $ A \\equal{} B \\equal{} 0$ :P \r\n :)", "Solution_1": "hi...[b]QuyBac[/b] \r\n You poof that $ AB\\equal{}BA$ is complicated too . Imples as : $ B\\equal{}e^{A}\\minus{}I_n$ then $ AB\\equal{}A(e^A\\minus{}I_n)\\equal{}(e^A\\minus{}I_n)A\\equal{}BA$\r\n Imples for equation : $ e^{A}\\plus{}A\\equal{}e^B\\plus{}B$" } { "Tag": [ "induction" ], "Problem": "http://www.UploadYourImages.com/view/597759induction.jpg\r\n\r\npretty sure i got the second part wrong but i do not understand why, anyone help?", "Solution_1": "[hide=\"My try at this\"]\nAssume that $S_{k}$ is true then for $S_{k+1}$\nit is $5+25+125+...+5^{k}+5^{k+1}$ or\n$({5/4})(5^{k}-1)+5^{k+1}$\nthen\n$5(5^{k}/4-1/4+5^{k})$\n$5(5(5^{k})/4-1/4)$\n$({5/4})(5^{k+1}-1)$ \nwhich is what we are trying to get at [/hide]\r\n\r\nSo that is my whack at it", "Solution_2": "Yes, that's correct, but you lack a base case. :lol: \r\n\r\nWell just do n=1 and it's obvious so yeah.", "Solution_3": "[hide]\nsay k=4. Then, let k+1=u. u=5. Then, substituting u for k, we know that it is true. \nIt's kind of a deep intuition that all of us have.[/hide]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let try this one. :D (I thinks it is very weak but anyway it is nice)\r\n\r\n$a,b,c$ are positive number such that $a+b+c=3$ then\r\n$\\sum \\frac{a^3(2a^2+bc)}{(b+c)^2} \\geq \\frac32 \\sum \\frac{a^4}{b^2+c^2}$\r\nHave fun", "Solution_1": "Use the ineq $(b+c)^2 \\leq 2(b^2+c^2)$ , we need to prove :\r\n $\\sum\\frac{a^5+a^3bc-a^4b-a^4c}{b^2+c^2} \\geq 0$ \r\n$\\iff \\sum\\frac{a^3(a-b)(a-c)}{b^2+c^2} \\geq 0$ , which is right by Vornicu-Schur .", "Solution_2": "Wow, it is very nice. But I don't know what is Vornicu-Schur. I have looked it up in Mathlink and found that $\\sum f(a)(a-b)(a-c) \\geq 0$ But what is a condition of $f(a)$ in order to satisfy this inequality is a little difficult for me to understand. Can anyone explain it for me please? I think it is a very interesting equality. \r\n By the way, from here, we can find that $\\sum \\frac{a^3(2a^2+bc)}{(b+c)^2} \\geq \\frac{9}{4}$ easily .Right?" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "email", "AMC 10", "geometry", "trapezoid" ], "Problem": "Hey. . .\r\n\r\nI got invited to take the APMO, you guys got any info on it? I'd appreciate any help you may give me.\r\n\r\nThanks.", "Solution_1": "Whoa, how did you get invited?", "Solution_2": "I think all of the people who went to MOP last year, plus anyone not at MOP who got a perfect score on the AMC-12 this year, were invited to do APMO. So maybe 20-30 people.\r\n\r\nEveryone who takes it submits their papers to AMC, which grades them all and then chooses the top 10 (by their grading) to be the official US entries. People don't actually travel anywhere to compete in this.", "Solution_3": "Yeah I think that dudes right. . . but like, the problems don't seem to be as hard as the USAMO, I dunno though.", "Solution_4": "[quote=\"Idiot_without_a_village\"]Yeah I think that dudes right. . . but like, the problems don't seem to be as hard as the USAMO, I dunno though.[/quote]\r\n\r\nWait..you went to MOP last year?", "Solution_5": "Naw dawg, I only got a perfect score on dis yer's AMC 12. . . Mebbe MOP dis yer, but I dun think it'll happen man.", "Solution_6": "[quote=\"Idiot_without_a_village\"]Naw dawg, I only got a perfect score on dis yer's AMC 12. . . Mebbe MOP dis yer, but I dun think it'll happen man.[/quote]\r\n\r\nSo you're one of the 12 who got the perfect score? Nice!", "Solution_7": "Wait, are you kidding? Is this the first year they've invited perfect AMCer's? I feel so cheated! :)", "Solution_8": "Yeah, I got some email or something. . . I dunno, it all happened so fast, my math teacher dude was like, you wanna do this APMO shiznit, and I was like, fo shizzle. . . so, it's all good.", "Solution_9": "Got in through the AMC 10A Perfect score.", "Solution_10": "AMC10? That's pretty tite homie. . . yeah man, I guess we just need to plow through some of their old problems or something, the early ones were really easy and then a lil' later ya know they got tougher. . . aight man, I dunno about this thing but I guess we'll see what happens, it's all about the AIME/USAMO this year mon.", "Solution_11": "Did anyone here take this? How did it go?", "Solution_12": "Yep, I think I got 20-25 points.", "Solution_13": "#1,#2, #4 are almost trivial... #3 and #5 just seems impossible", "Solution_14": "Are the problems posted somewhere? If not everyone has taken it yet (I don't know how this works), please don't post them here.", "Solution_15": "[quote=\"ComplexZeta\"]Are the problems posted somewhere? If not everyone has taken it yet (I don't know how this works), please don't post them here.[/quote]\r\n\r\nYes, please wait until the problems are posted on http://cms.math.ca/Competitions/APMO/ to discuss them publicly. In past years, the offical results have been nullified because the problems were posted online before all participants had taken the exam. The organizers are rather slow to post the official problems, but please respect their rules.", "Solution_16": "The questions are finally up. http://www.cms.math.ca/Competitions/APMO/exam/apmo2004.html\n\n\n\nThe only one I'm 100% sure about my answer is #1.\n\nI got:\n\n[hide]\n\n{2}\n\n[/hide]", "Solution_17": "#1, not quite complete: [hide]assume you have at least 2 numbers: if you have two relatively prime numbers, you lose, where winning is defined as having finitely many terms. Thus, all terms must have a greatest common divisor, k. So our numbers can be written as k*a1, ..., k*an. Then, using our law, k | ai + aj for all i, j. That means that all of the ai are congruent to 0 mod k, or all divisible by k. This is a contradiction unless k = 1, which also makes us lose. Otherwise, we must have either (a) k even and all numbers congruent to k/2, in which case we lose unless k = 2, from which we get the solution {2} by going back and reapplying the process or (b) we only have two numbers. I didn't work these last two cases through at all, but I'm sure that the first leads to a contradiction if the solution isn't {2}, and the second leads to a contradiction.[/hide]\n\n\n\nProblem 5 I can reduce to proving (abc)^2 + 2 :ge: ab + bc + ca, but I don't actually know if that is true or not.\n\n\n\nNumber 2 I can do uglily with abSinC/2 area formula. Looking for a nicer way.\n\n\n\nI think number 3 can be solved by somehow coloring the plane or setting up an isomorphism (I don't know if this is exactly the right word) between the points and sections of the plane, or something. Basically, I know that it's easy to two-color a plane cut up by any number of lines so that each section of the plane borders only different-colored sections. I'm wondering if there's some way to use that fact to do this question.", "Solution_18": "Oh, and I think I just got number 4:\n\n\n\n[hide]Cases: 1) n = 1, 2, 3, 4 : the result follows by calculation. \n\nFrom now on, n > 4. 2) n and n + 1 are composite. In that case, we have that they are both > 4, and as such, all of their prime factors are represented in the numerator of the fraction. There are clearly more than enough 2's to go around. Thus, it is even.\n\n3) n odd prime > 3. (n - 1)! :equiv: -1 (mod n), and (n + 1) | (n - 1)! because, as above, there are plenty of factors to go around. (The only problem could possibly be n + 1 = 8, but even there, we're fine.) (n + 1) :equiv: 1 (mod n), thus (n - 1)! = (n + 1)*(k*n - 1). We get that k must be odd, since (n - 1)!/(n + 1) is even and n is odd, so our expression is just less than an odd number and thus floor of it is even.\n\n4) (n + 1) odd prime > 5. Everything basically works the same way. The n disappears into the (n - 1)!, leaving us with k*(n + 1) - 1, and k must be odd for the same reasons, so everything again works.\n\n[/hide]\n\n\n\nThis was kind of sketchy, but I'm confident of it.", "Solution_19": "number 1 is kinda easy, but it might be a slightly more interesting problem if i is not allowed to equal j.", "Solution_20": "If i is not allowed to equal j, I think any proof will still work with the only added affect of removing the one solution.", "Solution_21": "right, you just get slightly more cases to examine", "Solution_22": "Then, you'll get infinitely many sets starting with two member sets we get (a+b)/(a,b)=a. It follows that a and b aren't relatilevly prime, the solution I have for 2 element sets is \r\n{a, a(a-1)}\r\nMost likely you'll be able to getb infinitely many sets for any given number of elements.", "Solution_23": "Demon... Which one are you on the AMC website?", "Solution_24": "[quote=\"Demon\"]Then, you'll get infinitely many sets starting with two member sets we get (a+b)/(a,b)=a. It follows that a and b aren't relatilevly prime, the solution I have for 2 element sets is \n{a, a(a-1)}\nMost likely you'll be able to getb infinitely many sets for any given number of elements.[/quote]\r\n\r\nactually i have proved that if i is not allowed equal to j, then |S|<3. so S={a,a(a+1)} or S={a}. The proof is quiet complicated and involve some case work so i am interested in seeing a shorter proof.\r\nthis would make a nice question 4 and the original question four can be bumped to 1. \r\n\r\nJBL, for number 2, since OH is consistent in all three triangles, you gotta prove the average of the two heights (WLOG A to OH and B to OH) is equal to C to OH. but the average of the heights is equilvalent to the midpoint height of the right trapezoid, which is the perpendicular bisector of AB. Then using the fact that the centroid lies on the Euler line and dividing the median into the ratio of 1:2, we complete our geometric proof", "Solution_25": "2004APMO #2 can be solved by vector easily.\r\n\r\nLet X be the vector from O to X. Then H=A+B+C. The cross product\r\n0=HxH=(A+B+C)xH=AxH+BxH+CxH. Note AxH,BxH,CxH are perpendicular\r\nto the plane of ABC and their magnitudes are twice the area of triangles AOH,\r\nBOH,COH. So the magnitude part of the equation AxH+BxH+CxH=0 gives\r\nthe desired conclusion." } { "Tag": [ "topology", "real analysis", "real analysis unsolved" ], "Problem": "Let S be a subset of R. Show that S is an inverval if and only if for all a,b in S, we have [a,b] contained in S.", "Solution_1": "How, in particular, are you defining interval? Some books take it to be only sets of the form $ \\{a\\leq x\\leq b\\}$, others allow open, and half-open. And are we allowed to take $ a\\equal{}\\minus{}\\infty$, $ b\\equal{}\\infty$?", "Solution_2": "we're allowing open and closed and unbounded (infinitis). But he actually said it doesn't matter because you can prove it for any form.", "Solution_3": "Still, what is you definition of 'interval'? In our analysis course, the definition was exactly what you want to prove.", "Solution_4": "This result only states that the only connected sets in $ \\mathbb{R}$ with the usual topology (since here connected = connected by arcs) are the intervals. One implication is trivial. For the other one, embed $ \\mathbb{R}$ in $ \\overline{\\mathbb{R}}$ and for that set $ S$ take $ \\alpha \\equal{} \\inf S$ and $ \\beta \\equal{} \\sup S$. It follows that $ (\\alpha, \\beta) \\subseteq S$ and we are done." } { "Tag": [ "geometry" ], "Problem": "Yup. He's my current professor for discrete mathematics.\r\n\r\nwhat should I ask him...he's so approachable yet..he also shines in a way too bright for me.", "Solution_1": "That depends on what you are interested in yourself and how you are doing in his course. If you like the subject and are getting straight A's, it may be a good idea just to come and say that you want some additional extracurricular reading to do and proceed from there but if you are not interested in his area in the slightest or if your typical midterm grade is between C and D, then all you'll be really able to ask is questions about homeworks. \r\n\r\nAs for being \"too bright\", that shouldn't prevent you from talking to him. Bright professors usually do not bite their students :P." } { "Tag": [ "geometry", "rectangle" ], "Problem": "i have no idea what that \"proof without words\" flash animation explains?\r\ncan someone explain to me how they prove that the sum of squares equal that equation?", "Solution_1": "Thats a good question. I will try to figure it out over the next day so check back in a few.", "Solution_2": "At the beginning of the animation, we see all the squares laid out. Then they use all of the squares three times to make a rectangle. The height of the rectangle is 2n+1. The width of the rectangle is 1+2+3+...+n=(n)(n+1)/2, which is proven in another animation. So we have $1^2+2^2+3^2+\\cdots+n^2=\\frac{(2n+1)\\frac{n(n+1)}{2}}{3}=\\frac{n(n+1)(2n+1)}{6}$", "Solution_3": "ohhhh! nice.. thanks" } { "Tag": [ "geometry", "incenter", "geometric transformation", "homothety", "ratio", "radical axis", "angle bisector" ], "Problem": "Prove that in triangle $ABC$, radical center of its excircles lies on line $GI$, which $G$ is Centroid of triangle $ABC$, and $I$ is the incenter.", "Solution_1": "According to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=20029#20029]http://www.mathlinks.ro/Forum/viewtopic.php?t=5958 post #2[/url], the radical center of the excircles of triangle ABC is the Spieker point of triangle ABC, that is, the incenter of triangle A'B'C', where A', B', C' are the midpoints of the sides BC, CA, AB of triangle ABC. Thus, it remains to show that the incenter of triangle A'B'C' lies on the line GI. Now, it is well-known that the homothety with center G and ratio $-\\frac12$ maps the points A, B, C to the points A', B', C'. Hence, this homothety maps the incenter I of triangle ABC to the incenter of triangle A'B'C'. Since the center of the homothety is G, this yields that the incenter of triangle A'B'C' lies on the line GI. Proof complete.\r\n\r\nActually, this problem is way too well-known for an exam imho...\r\n\r\n darij", "Solution_2": "this was posted also on the italian forum [url=http://olimpiadi.dm.unipi.it/oliForum/viewtopic.php?t=6470]here[/url], this is the edriv's solution:\r\n\r\nLet the A-excircle touch AB on L an the B-excircle touch AB on K. Then as well-know the radical axes of $ \\Gamma_a$ and $ \\Gamma_b$ pass through the point M that is the midpoint of LK that is the midpoint of AB. Also we have that the radical axes is perpendicular to $ I_aI_b$ that is parallel to the internal angle bisector of $ \\angle CAB$ and the radical axes of $ \\Gamma_a$ and $ \\Gamma_b$ is the internal M-angle bisector of the medial triangle and equally for the others, from which the radical center P of the excircles is the incenter of the medial triangle (the Spieker point of ABC) so for an homothety with center G it lies on IG with $ IG \\equal{} 2 GP$.\r\n\r\nok It's the same solution of darij :|", "Solution_3": "[quote=\"Omid Hatami\"]Prove that in triangle $ ABC$, radical center of its excircles lies on line $ GI$, which $ G$ is Centroid of triangle $ ABC$, and $ I$ is the incenter.[/quote]\r\nI have a solution :D Maybe it is same of you :D \r\n[img]http://i209.photobucket.com/albums/bb172/Ineq/excircles.jpg[/img]\r\nLet $ D$ is midpoint of $ AC$\r\n$ a$ is radical axis of $ O_1)$ and $ (O_3)$ \r\nEasy to prove $ I$ is orthocenter of $ O_1O_2O_3$\r\nand $ AE\\equal{}CF$\r\nSo $ DE\\equal{}DF \\Rightarrow P_{D|(O_3)}\\equal{}P_{D|(O_1)}$\r\nso $ D \\in a$ and $ a // BI$\r\n$ K\\equal{}GI\\cap a$ then $ 2GK\\equal{}IG$\r\nSimilar we have desire" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Give a, b, c>0, and:\u2211(a).^2=3 ,Find The minimum of this product:\r\n (1-3(a.^3)*b)(1-3(b.^3)*c)(1-3(c.^3)*a)\r\n\r\n\r\n\r\n\r\n Have a good time.", "Solution_1": "[quote=\"navid\"]Give $a,\\,b,\\,c>0$ and $\\sum a^{2}=3$. Find the minimum of product: \\[\\left(1-3a^{3}b\\right)\\left(1-3b^{3}c\\right)\\left(1-3c^{3}a\\right)\\]\n[/quote]\r\n[b]navid[/b], Is it true?", "Solution_2": "Yes it is tru and is there any one who solve it?" } { "Tag": [ "function", "algebra", "polynomial", "induction", "LaTeX", "quadratics", "Putnam" ], "Problem": "A certain 2005-degree polynomial f(x) has this property:\r\nFor any n, the numbers n, f(n), f(f(n)), f(f(f(n))), ... are all relatively prime.\r\nFind f(x).", "Solution_1": "does it have integer coefficients? i think that might make it easier...\r\nf(x)=1 works but it is not exactly of degree 2005. :lol:", "Solution_2": "is natural number 1 considered 'relatively prime' to every integer?", "Solution_3": "Quite frankly, I don't know the answer. I think 1 is relatively prime to itself, but is 0 relatively prime to itself? And how does relatively prime work with negative numbers?\r\n\r\nI thought that this function was impossible at first, but it may be possible afterall.\r\n\r\nBy the way, n must be an integer.", "Solution_4": "[quote=\"Wumbate\"]...is 0 relatively prime to itself? And how does relatively prime work with negative numbers?\n\n...\nBy the way, n must be an integer.[/quote]\r\n\r\n0 has no common factor with any integer becuz any number times 0 is 0, which can't possibly be relatively prime to any number, maybe to itself since 1*0=0. but im not sure if they have a convention for 1. yes 'relatively prime' works with negative numbers because it is just a matter of sign, it doesn't affect the divisors. so $n$ must be a [b]nonzero[/b] integer :D", "Solution_5": "[quote=\"Palytoxin\"][quote=\"Wumbate\"]...is 0 relatively prime to itself? And how does relatively prime work with negative numbers?\n\n...\nBy the way, n must be an integer.[/quote]\n\n0 has no common factor with any integer becuz any number times 0 is 0, which can't possibly be relatively prime to any number, maybe to itself since 1*0=0. but im not sure if they have a convention for 1. yes 'relatively prime' works with negative numbers because it is just a matter of sign, it doesn't affect the divisors. so $n$ must be a [b]nonzero[/b] integer :D[/quote]\r\n\r\nAh, I never knew that negative integers could be relatively prime to each other.\r\nAccording to Mathworld, two integers that are relatively prime to each other if their greatest common divisor is 1. So, in fact, 1 is relatively prime to any nonzero integer.\r\n\r\nHere's the link: http://mathworld.wolfram.com/RelativelyPrime.html", "Solution_6": "1 is relatively prime to all other integers. 0 is [i]not[/i] relatively prime with any integer other than 1 (or -1, if we include the negatives), as for any $p \\geq 2$, $p | p$ and $p | 0$. Thus, the greatest common divisior of 0 and anything else is that thing (or its absolute value, if we include negatives), with the exception of 0 and 0, for which there is no greatest common division (as every integer divides both 0 and 0).\r\n\r\nFor the purposes of this problem, however, it is best to assume (1) that n is a positive number and (2) that none of the numbers f(n), f(f(n)), ... are 0.", "Solution_7": "my guess is... $f(x)=x^{2005}+1$ but i dunno how to prove it :( :wallbash:", "Solution_8": "That is wrong, since f(2)=2^2005+1=odd\r\nand f(odd)=odd^2005+1=even, so 2 and the even number share 2 as a factor.\r\n\r\nI'm trying to work considerations like that into a general polynomial.", "Solution_9": "hmm i see your point\r\n\r\nhow about $f(x)=2x^{2005}+1$ ? :D", "Solution_10": "Wumbate, how do you know such a polynomial exists?\r\n\r\nPaly, you should test before you guess. For any multiple of 3, that second polynomial fails after 2 interations.", "Solution_11": "I don't know it exists, in fact my hunch is that it doesn't.\r\nHowever, this problem is one from a pack of problems I got, without the answers, so it would be weird if this problem was actually a trick question, since all the others weren't.\r\nI don't know where the problem comes from originally.", "Solution_12": "Well, my challenge then: solve the problem where the degree of f is replaced by (a) 0 and (b) 1. Then let's think about 2005.\r\n\r\n\r\nActually, there's almost certainly a modular argument of some sort. I just don't know what it would be :-)", "Solution_13": "How about degree 2? Let $f(x) = x^2 - x + 1$. It's well known that the sequence will be relatively prime in that case. Can you guys prove it?", "Solution_14": "For degree 0: f(x)=1\r\nFor degree 1: I feel like I'm missing something obvious. Impossible?\r\n\r\nProof for degree 2:\r\n$f(x)=x^2-x+1=(x)(x-1)+1$\r\n\r\nLet $x_0$ be the first of the iteration.\r\n\r\nFirst, note that $x_{n+1}-1=(x_n)*(x_n-1)$ which, by induction backwards, is simply the product of all the previous n's times $x_0-1$ (don't know how to latex this nicely).\r\n\r\nThen $f(x_{n+1})=(x_n)*(x_n-1)+1$. The value $(x_n-1)$, by the above, has every previous value multiplied into it. So taking this mod any previous x will = 1, i.e. they are relatively prime.\r\n\r\n\r\nThis function allows me to solve the 2005 degree one:\r\n$f(x)=x^{2004}*(x-1)+1=x^{2005}-x^{2004}+1$\r\nThe proof would be the same as above.", "Solution_15": "Pretty good. The quadratic problem was half of a Putnam problem from 1956. \r\n\r\nLet's generalize. Let $f$ be an integer polynomial such that $f(0) = 1$ and $f(1) = 1$. Then, believe it or not, the given sequence will be relatively prime. One way to prove it is to use your \"product\" method. Another way (perhaps slicker) is to use mods (JBL was right).", "Solution_16": ":? and what makes this problem intermediate topics? am i missing something? :help:", "Solution_17": "Yes, the 2005-degree polynomial I gave isn't the only one. It can be any polynomial of the form:\r\n$f(x)=(x)^a(x-1)^b+1$\r\nwhere a+b=2005 and a, b>0\r\n\r\nI thought it was intermediate because it came from a set of problems which were easier than this one.", "Solution_18": "Even more general than that: $f(x) = 1 + x(x-1) g(x)$, where $g$ is an arbitrary polynomial (of degree say 2003).", "Solution_19": "Oh, that's right, thanks for the correction.", "Solution_20": ":help: intermediate topics too hard for me, i better stay in getting started... :help: :oops_sign:", "Solution_21": "Is this the only polynomial like that? I somehow doubt it", "Solution_22": "Here are 6 general forms that I can think of:\r\n\r\n1. $f(x) = x(x-1) g(x) + 1$.\r\n2. $f(x) = x(x+ 1) g(x) - 1$.\r\n3. $f(x) = (x-1) x (x + 1) g(x) - 2x^2 + 1$. \r\n4. $f(x) = (x-1) x (x+1) g(x) + 2x^2 - 1$.\r\n5. $f(x) = (x - 1) x (x + 1) g(x) - x (x + 1) + 1$. \r\n6. $f(x) = (x - 1) x (x + 1) g(x) + x(x - 1) - 1$.\r\n\r\nI think I can prove that those 6 forms cover all such polynomials, but I would have to think some more to be sure.\r\n\r\nThose 6 forms may look mysterious, but they can be stated more simply by looking at their values:\r\n\r\n1. $f(0) = 1$, $f(1) = 1$.\r\n2. $f(0) = -1$, $f(-1) = -1$.\r\n3. $f(0) = 1$, $f(1) = -1$, $f(-1) = -1$.\r\n4. $f(0) = -1$, $f(-1) = 1$, $f(1) = 1$.\r\n5. $f(0) = 1$, $f(1) = -1$, $f(-1) = 1$.\r\n6. $f(0) = -1$, $f(-1) = 1$, $f(1) = -1$.\r\n\r\nIn each, we get a cycle of values starting from 0 and going through -1 or 1.\r\n\r\n(By the way, notice that even the quadratic example gives a infinite sequence of relatively prime numbers. As a corollary, we get an amusing proof that there are infinitely many primes.)", "Solution_23": "[quote]How about degree 2? Let $f(x) = x^2 - x + 1$. It's well known that the sequence will be relatively prime in that case.[/quote]\r\n\r\nAre you thinking of $x^2 - 2x + 2$, which generates the sequence $1 + a^{2^k}$ with $a=x-1$?", "Solution_24": "No, I meant $x^2 - x + 1$. The function $x^2 - 2x + 2$, while interesting, doesn't work for this problem. (Just set $n = 4$ say.)\r\n\r\nPerhaps \"well known\" was an exaggeration. What I really meant was \"I have seen it before\" (probably from the Putnam 1956 problem)." } { "Tag": [ "modular arithmetic", "search", "Diophantine Equations", "pen" ], "Problem": "Solve the equation $x^2 +7=2^n$ in integers.", "Solution_1": "[quote=\"Peter\"]Solve the equation $ x^2 \\plus{} 7 \\equal{} 2^n$ in integers.[/quote]\r\nHere is my solution\r\nWe have:\r\n$ x^2\\equal{}2^n\\minus{}7 \\equiv (\\minus{}1)^n\\minus{}1(mod 3)$\r\nIf $ n\\equal{}2k\\plus{}1;k \\in Z$ then $ x^2 \\equiv \\minus{}2(mod 3)$,wrong\r\nSo $ n\\equal{}2k$ with $ k \\in Z$\r\nThen $ 2^{2k}\\minus{}x^2\\equal{}7$\r\n<=>$ (2^k\\minus{}x^2)(2^k\\plus{}x^2)\\equal{}7$\r\n=>$ 2^k\\plus{}x^2\\equal{}7 and 2^k\\minus{}x^2\\equal{}1$\r\nSo $ 2.2^k\\equal{}k\\equal{}>k\\equal{}2$\r\n=>$ n\\equal{}4$=>$ x^2\\equal{}9$=>$ x \\equal{}\\pm 3$", "Solution_2": "What's wrong with $ x^2\\equiv \\minus{}2\\pmod3$?\r\n\r\nSchmu", "Solution_3": "Nothing is wrong with it. There are actually such solutions: $ x\\equal{}1, x\\equal{}5, ...$. So please try again.", "Solution_4": "[quote=\"Peter\"]Solve the equation $ x^2 \\plus{} 7 \\equal{} 2^n$ in integers.[/quote]\r\n\r\nThis one is famous,but it seems there are no elementary proof. :(", "Solution_5": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=98196791&t=192321]Posted[/url]. This equation goes back to Ramanujan and the [url=http://www.math.sc.edu/~filaseta/gradcourses/TheMath784Notes.pdf]standard solution[/url] due to Nagell involves a fair amount of work in $ \\mathbb{Z} \\left[ \\frac{1 \\plus{} \\sqrt{\\minus{}7}}{2} \\right]$.", "Solution_6": "Hmm... not really a PEN-problem if that's what it needs to solve this.", "Solution_7": "[quote=\"Peter\"]Hmm... not really a PEN-problem if that's what it needs to solve this.[/quote]\r\nAs one can see in 8th post in the link above,I said that this problem was given at a Preparation olympiad,so it seems to me that this problem must have almost elementary solution,furthermore,in my opinion it should not be very hard to find a such proof,as even 9th graders participated in it :roll: ...\r\nAlso [url=http://www.mathlinks.ro/viewtopic.php?t=150421]this problem[/url] was used at that Olympiad,although nobody was able to find a proof without using Gaussian numbers during olympiad and as I can see at ML elementary solution wasn't discovered for a long time too.\r\nThis remark gives us hope that there must exist \"almost elementary\" proof and let's just wait until somebody...\r\nUnfortunately Teacher who gave us this problem can not find the place where he saw elementary solution and I have a sense that he maybe didn't even try to search :mad: \r\nBest regards.", "Solution_8": "[quote=\"Erken\"][quote=\"Peter\"]Hmm... not really a PEN-problem if that's what it needs to solve this.[/quote]\nAs one can see in 8th post in the link above,I said that this problem was given at a Preparation olympiad,so it seems to me that this problem must have almost elementary solution,furthermore,in my opinion it should not be very hard to find a such proof,as even 9th graders participated in it :roll: ...[/quote]\r\nMaybe the person putting this in the test didn't know it was so difficult, or had a wrong solution in mind (like the few posted on ML)?\r\n\r\nThe fact that this is a famous problem that Ramanujan couldn't solve, strongly suggests that this is NOT a pre-olympiad level problem.", "Solution_9": "[quote=\"Peter\"]Solve the equation $x^2 +7=2^n$ in integers.[/quote]\n \n$X^2 +7=2^n$\n$X^2+16-9 =2^n$\n$(X+3)(X-3)=2^4(2^q-1)$ for $n=4+q$\nWe see $x =2k+1$\n$=>(k+2)(k-1)=4(2q-1)$\nObserve $(k+2,k-1) =1$ for being odd or even,\nAnd also $(4,2q-1) =1$\n$k+2=4$ or $k-1=4$\n$=> k=2$ or $k=5$\n$X=5$ or $x=11$\nFor $n=4, x=3$\nFor $n<4$, we may check the values of n.\n$(x,n) =(1,3),(3,4),(5,5),(11,7)$", "Solution_10": "$(X+3)(X-3)=2^4(2^q-1)$ for $n=4+q$\nWe see $x =2k+1$\n$=>(k+2)(k-1)=4(2q-1)$\n\nIt should be $8(2^q-1)$ . How did you come up with $4(2q-1)$?", "Solution_11": "[quote=reveryu]$(X+3)(X-3)=2^4(2^q-1)$ for $n=4+q$\nWe see $x =2k+1$\n$=>(k+2)(k-1)=4(2q-1)$\n\nIt should be $8(2^q-1)$ . How did you come up with $4(2q-1)$?[/quote]\n\nConsider (X+3)(X-3)=2(k+2)2(k-1)=4(k+2)(k-1)\nBut you're right about 2q-1.it should be 2^q-1\nThat was a perfect solution", "Solution_12": "[quote=venkats]\n \n\n$=>(k+2)(k-1)=4(2q-1)$\nObserve $(k+2,k-1) =1$ for being odd or even,\nAnd also $(4,2q-1) =1$\n$k+2=4$ or $k-1=4$\n[/quote]\nCounter-example: $2 \\cdot 15= 6\\cdot 5$?\n\n", "Solution_13": "[quote=venkats][quote=\"Peter\"]Solve the equation $x^2 +7=2^n$ in integers.[/quote]\n \n$X^2 +7=2^n$\n$X^2+16-9 =2^n$\n$(X+3)(X-3)=2^4(2^q-1)$ for $n=4+q$\nWe see $x =2k+1$\n$=>(k+2)(k-1)=4(2q-1)$\nObserve $(k+2,k-1) =1$ for being odd or even,\nAnd also $(4,2q-1) =1$\n$k+2=4$ or $k-1=4$\n$=> k=2$ or $k=5$\n$X=5$ or $x=11$\nFor $n=4, x=3$\nFor $n<4$, we may check the values of n.\n$(x,n) =(1,3),(3,4),(5,5),(11,7)$[/quote]\n\n$(k+2,k-1) =3$ for k=1 so k-1=4*3 or 4 at most but anyways nice solution." } { "Tag": [ "geometry", "college", "Berkeley Math Circle" ], "Problem": "Okay so we need a list of Bay Area MOPpers so we can get together at MOP and overwhelm everyone else muwahahahaha.\r\n\r\nHere's what I know so far:\r\nAlbert Gu - Saratoga\r\nDavid Zeng - Saratoga\r\nGay - Saratoga\r\nWendy Mu - Saratoga\r\nTaylor Han - Gunn\r\nJohn Boyle - Paly\r\nLynnelle Ye - Paly\r\nColleen Lee - Paly\r\nVishal Arul - Homestead\r\nPaul Christiano - Harker\r\nEvan O'Dorney - Berkeley Math Circle\r\n\r\nYay that's like 15% of the people at MOP.\r\n\r\nSo should I add what color MOP and/or score to the list?", "Solution_1": "[quote=\"Ubemaya\"]Okay so we need a list of Bay Area MOPpers so we can get together at MOP and overwhelm everyone else muwahahahaha.\n\nHere's what I know so far:\nAlbert Gu - Saratoga\nDavid Zeng - Saratoga\nGay - Saratoga\nTaylor Han - Gunn\nJohn Boyle - Gunn\nLynnelle Ye - Paly\nVishal Arul - Homestead\n\nHmm uh I think we need more people.[/quote]\r\nboyle is paly\r\ncolleen lee - paly for cgmo", "Solution_2": "[quote=\"Ubemaya\"]Okay so we need a list of Bay Area MOPpers so we can get together at MOP and overwhelm everyone else muwahahahaha.\n\nHere's what I know so far:\nAlbert Gu - Saratoga\nDavid Zeng - Saratoga\nGay - Saratoga\nTaylor Han - Gunn\nJohn Boyle - Paly\nLynnelle Ye - Paly\nColleen Lee - Paly\nVishal Arul - Homestead\n\nHmm uh I think we need more people.[/quote]\r\ndude that's a pretty awesome third person you got on the list\r\n\r\nAND UHH YOU SEEM TO BE FORGETTING A VERY PARTICULAR WINNER FROM THE BAY AREA COUGH COUGH", "Solution_3": "2 winners in fact...Paul Christiano from Harker (his username on AoPS is a string of binary...), Evan O'Dorney.", "Solution_4": "dude bay area pwnd pretty awesomely huh", "Solution_5": "Whos the other one? I believe one was the award winner, no?", "Solution_6": "whoops yeah forgot Christiano...", "Solution_7": "It totally should've had me, Brian Zhang, and a few others on there...:( MUHAHA RED MOP CHEAPNESS", "Solution_8": "Man, you guys are seriously pwn. Have fun at MOP, and I want to see some blue/black/IMO team members in a few years.\r\nAnd CGMO is possibly THE least legit way to get into MOP. Just sayin.", "Solution_9": "Yay first person from Homestead to make MOP! :wow: \r\nHey, Ubemaya, how many points did you improve from last year?\r\n\r\nI think you should specify which color each person made...", "Solution_10": "dude both red and CGMO are now fake mop\r\n\r\nblue and black are real mop", "Solution_11": "red mop is 'mop lite'\r\n\r\ni'm not sure what to call cgmo mop", "Solution_12": "[quote=\"vishalarul\"]Yay first person from Homestead to make MOP! :wow: \n[/quote]\r\ndude no fair i dont get to say things like this", "Solution_13": "[quote=\"Altheman\"]red mop is 'mop lite'\n\ni'm not sure what to call cgmo mop[/quote]\r\n\r\nthey are both fake mop\r\n\r\nunless of course a girl happens to legitimately make real mop and just incidentally happens to be on the cgmo team", "Solution_14": "[quote=\"Ubemaya\"]\nGay - Saratoga\n[/quote]\r\n\r\nlol.\r\nguess quatto didnt make mop this year\r\n[hide] \nsarcasm\nQuatto's gay :o \n[/hide]", "Solution_15": "[quote=\"MysticTerminator\"][quote=\"Altheman\"]red mop is 'mop lite'\n\ni'm not sure what to call cgmo mop[/quote]\n\nthey are both fake mop\n\nunless of course a girl happens to legitimately make real mop and just incidentally happens to be on the cgmo team[/quote]\r\nmehh could we consider red MOPers MOPlets?", "Solution_16": "did bowei not make it?", "Solution_17": "[quote=\"MysticTerminator\"][quote=\"Altheman\"]red mop is 'mop lite'\n\ni'm not sure what to call cgmo mop[/quote]\n\nthey are both fake mop\n\nunless of course a girl happens to legitimately make real mop and just incidentally happens to be on the cgmo team[/quote]\r\n\r\ndude\r\n\r\ni definitely made 'real mop' last year\r\n\r\ni blame the higher ups", "Solution_18": "[quote=\"mathclass\"]did bowei not make it?[/quote]\r\nHmmm...did you make it?\r\nCurrently we don't know if bowei made it I think...judging from his beastly skills he had to have made it.", "Solution_19": "I made 'fake mop' with a score of 14, but i'm going to rsi instead.", "Solution_20": "[quote=\"Hamster1800\"][quote=\"MysticTerminator\"][quote=\"Altheman\"]red mop is 'mop lite'\n\ni'm not sure what to call cgmo mop[/quote]\n\nthey are both fake mop\n\nunless of course a girl happens to legitimately make real mop and just incidentally happens to be on the cgmo team[/quote]\n\ndude\n\ni definitely made 'real mop' last year\n\ni blame the higher ups[/quote]\r\n\r\nyeah I know\r\n\r\nso you made real MOP\r\n\r\ngood job" } { "Tag": [], "Problem": "Josh and Mike live 13 miles apart. Yesterday, Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?\r\n\r\n$ \\textbf{(A)}\\ 4\\qquad\r\n\\textbf{(B)}\\ 5\\qquad\r\n\\textbf{(C)}\\ 6\\qquad\r\n\\textbf{(D)}\\ 7\\qquad\r\n\\textbf{(E)}\\ 8$", "Solution_1": "[quote=\"Silverfalcon\"]Josh and Mike live 13 miles apart. Yesterday, Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?\n\nA. 4\nB. 5\nC. 6\nD. 7\nE. 8[/quote]\r\n\r\n[hide]\n$d=rt$\nSince Josh rode for twice as long as Mike, and 4/5 the rate, we could say that $13=2\\cdot(\\frac{4a}{5})+1\\cdot a)$, where a is some value (we don't know the exact time or rate). Solving, $a=5$. That would mean that Mike rode for $d=rt=5\\cdot 1=5$ miles.\nB\n[/hide]", "Solution_2": "[hide=\"Answer\"]Josh had gone $2\\cdot \\frac{4}{5}=\\frac{8}{5}$ of the distance that Mike had, and $8+5=13$, so Mike had ridden $5$ miles. $\\boxed{B}$[/hide]", "Solution_3": "[hide]\nLet $M$=distance travelled by Mike; $(2/3)D$=amount travelled by Josh; m=Mike's speed.\n$\\\\13-M=(2/3)D=(4/3)mt2\\\\13-mt=8mt/5\\\\13=13mt/5\\\\mt=5\\\\M=mt=5$.\n[/hide]\r\n\r\nMasoud Zargar" } { "Tag": [ "FTW" ], "Problem": "Another poll for writing instruments... :)\r\n\r\nI use a pentel p207 and a rotring tikky ii. (oh yeah, I forgot to add pen to the poll, can a mod add it for me?)", "Solution_1": "I use a Pilot G2 0.5 mechanical pencil. I have yet to encounter anything better than it in terms of writing, and it's great for spinning as well. :)", "Solution_2": "Why is there two of each option in the poll? You can still edit it because it is less than 24 hours. I like to use mechanical pencils.", "Solution_3": "I don't know why its like that...", "Solution_4": "pencils ftw", "Solution_5": "[youtube]zMvdpfN8utw[/youtube]\r\n\r\n32 seconds in.", "Solution_6": "how come no one has voted for the second copy on either of the options??", "Solution_7": "[quote=\"eggylv999\"]how come no one has voted for the second copy on either of the options??[/quote]\r\n\r\nI voted for woodcase pencil (the second one) and it transferred the vote to the first one. Glitch.", "Solution_8": "The mods don't like you, no adding pen option today\r\n\r\n\r\n:( die", "Solution_9": "pilot G2 pens. ;]\r\ngreat for spinning too...", "Solution_10": "Why is it called a writing instrument if you can't make music from it?", "Solution_11": "You can, bash it against a table and you'll see what I mean.\r\n :ninja: :ninja: :ninja:" } { "Tag": [ "logarithms" ], "Problem": "$ (\\frac{1}{3})^{\\log _9 \\left( x^2 \\plus{} 2x \\plus{} 4 \\right)} \\equal{} 6^{\\log _\\frac{1}{6} \\left( x\\plus{}2 \\right)}$", "Solution_1": "hello, taking the logarithm of both sides we get\r\n$ \\log_9 (x^2 \\plus{} 2x \\plus{} 4)\\cdot \\ln(3^{ \\minus{} 1})\\equal{}\\ln(6)\\log_{\\frac {1}{6}}(x \\plus{} 2)$\r\n$ \\frac {\\ln(x^2 \\plus{} 2x \\plus{} 4)}{2\\ln(3)}\\ln(3^{ \\minus{} 1}) \\equal{} \\frac {\\ln(x \\plus{} 2)}{\\ln(6^{ \\minus{} 1})}\\ln(6)$\r\n$ \\ln(x^2 \\plus{} 2x \\plus{} 4) \\equal{} 2\\ln(x \\plus{} 2)$\r\n$ x^2 \\plus{} 2x \\plus{} 4 \\equal{} x^2 \\plus{} 4x \\plus{} 4$\r\n$ x \\equal{} 0$\r\n$ x \\equal{} 0$ fulfilles our original equation.\r\nSonnhard.", "Solution_2": "Dr Sonnhard Graubner, you missed an equals sign\r\nin the first line of your calculations. I placed it in:\r\n\r\n[quote=\"Dr Sonnhard Graubner\"]hello, taking the logarithm of both sides we get\n$ \\log_9 (x^2 \\plus{} 2x \\plus{} 4)\\cdot \\ln(3^{ \\minus{} 1})\\equal{}\\ln(6)\\log_{\\frac {1}{6}}(x \\plus{} 2)$\n$ \\frac {\\ln(x^2 \\plus{} 2x \\plus{} 4)}{2\\ln(3)}\\ln(3^{ \\minus{} 1}) \\equal{} \\frac {\\ln(x \\plus{} 2)}{\\ln(6^{ \\minus{} 1})}\\ln(6)$\n$ \\ln(x^2 \\plus{} 2x \\plus{} 4) \\equal{} 2\\ln(x \\plus{} 2)$\n$ x^2 \\plus{} 2x \\plus{} 4 \\equal{} x^2 \\plus{} 4x \\plus{} 4$\n$ x \\equal{} 0$\n$ x \\equal{} 0$ fulfilles our original equation.\nSonnhard.[/quote]", "Solution_3": "hello, yes, you have right, i inserted it, thank you.\r\nSonnhard." } { "Tag": [ "quadratics", "induction", "function", "limit", "algebra", "polynomial", "quadratic formula" ], "Problem": "The real sequence $x_{0},x_{1},x_{2},...$ is defined by $x_{0}=1,x_{1}=k,x_{n+2}=x_{n}-x_{n+1}$. Show that there is only one value of $k$ such that $x_{n}>0\\forall n\\geq 0$.", "Solution_1": "[hide=\"Solution\"] Setting $x_{n}= 0$ and solving for $k$ produces a convergent of $\\boxed{ \\phi-1 }$. [/hide]", "Solution_2": "What is $\\phi$ ?", "Solution_3": "The root of the equation $x^{2}-x-1=0$\r\n\r\nby the quadratic formula, that's $\\frac{1\\pm\\sqrt{5}}{2}$, and the symbol $\\phi$ usually denotes the positive one. Phi has many interesting properties.\r\n\r\nMost notably,\r\n$\\phi+1=\\phi^{2}$ (from the equation above)\r\n$\\phi^{-1}=\\phi-1$", "Solution_4": "[hide=\"Somewhat of a further explanation\"] It's not hard to see by induction that we have [b]something like[/b] $x_{n}= (-1)^{n}( F_{n}k-F_{n-1})$ (too lazy to work out the exact form). [/hide]", "Solution_5": "so if, me too! :wink: \r\n[hide]hint: use $u_{n}=c_{1}(\\frac{-1-\\sqrt{5}}{2})^{n}+c_{2}(\\frac{-1+\\sqrt{5}}{2})^{n}$.[/hide]", "Solution_6": "Not quite necessary; and you still have to prove the relevant lemma about the convergents, which is that the difference between the convergents and the number alternate signs.", "Solution_7": "[quote=\"N.T.TUAN\"]The real sequence $x_{0},x_{1},x_{2},...$ is defined by $x_{0}=1,x_{1}=k,x_{n+2}=x_{n}-x_{n+1}$. Show that there is only one value of $k$ such that $x_{n}>0\\forall n\\geq 0$.[/quote]\r\n\r\nIt is necessary and sufficient for the function to satisfy: \\[x_{n}>x_{n+1}\\qquad (\\forall n)\\]\r\nNotice that, since $x_{n+1}=x_{n-1}-x_{n}$, this is equivalent to: \\[2x_{n}>x_{n-1}\\qquad (\\forall n)\\]\r\nLet $F_{M}$ be Fibonaccis, as usual. Suppose that \\[F_{m}x_{q}>F_{m+1}x_{q+1}\\]\r\nThen by applying the Fibonacci recursion $F_{m+2}=F_{m}+F_{m+1}$ as well as the $x$-sequence recursion $x_{q+2}=x_{q}-x_{q+1}$ we derive:\r\n\\begin{eqnarray*}F_{m}x_{q}&>&F_{m+1}x_{q+1}\\\\ \\iff F_{m+2}x_{q}&>&F_{m+1}x_{q-1}\\\\ \\iff F_{m+2}x_{q-2}&>&F_{m+3}x_{q-1}\\\\ \\iff F_{m+4}x_{q-2}&>&F_{m+3}x_{q-3}\\end{eqnarray*}\r\nIt follows by induction (since the previous two observations serve as base cases, and the result holds for any $n$) that \\[F_{m}x_{q}>F_{m+1}x_{q+1}\\] for $m$ even and that \\[F_{m}x_{q}\\frac{x_{1}}{x_{0}}>\\frac{F_{m+1}}{F_{m+2}}\\]\r\nfor even $m$. Taking $m$ to $\\infty$, it is clear that $\\lim_{n\\to \\infty}\\frac{F_{m}}{F_{m+1}}=x_{1}=\\phi$ (a well known and easy to prove result)", "Solution_8": "Wow :wink: \r\nuse $u_{n}=c_{1}(\\frac{-1-\\sqrt{5}}{2})^{n}+c_{2}(\\frac{-1+\\sqrt{5}}{2})^{n}$.\r\nIF $u_{n}>0$ we have $0<\\frac{u_{n}}{(\\frac{1+\\sqrt{5}}{2})^{2k}}=c_{1}+c_{2}(\\frac{\\sqrt{5}-1}{\\sqrt{5}+1})^{2k}$ therefore $c_{1}\\geq 0$. Similary $c_{1}\\leq 0$, and we obtain $c_{1}=0$, therefore $c_{2}=1$.\r\n\r\nFinal, $k=u_{1}=\\phi$.", "Solution_9": "All this \"characteristic polynomial\" stuff relied on in your proofs can be incorporated into easier elementary proofs, thats really all my proof did.", "Solution_10": "characteristic polynomials are elementary. :wink:" } { "Tag": [], "Problem": "\u0391\u03bd \u03ad\u03bd\u03b1 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03ad\u03c7\u03b5\u03b9 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03bc\u03b5\u03bc\u03ae\u03ba\u03b7 \u03b1\u03ba\u03b5\u03c1\u03b1\u03af\u03bf\u03c5\u03c2 \u03ba\u03b1\u03b9 \u03b7 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 \u03c4\u03bf\u03c5 \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03ad\u03bd\u03bf\u03c5 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03cc \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 1, \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9\u03bf.\r\n\r\n[hide]\n\u0394\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03bf\u03b9 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03c4\u03bf\u03c5 \u03b9\u03c3\u03bf\u03cd\u03bd\u03c4\u03b1\u03b9 \u03bc\u03b5 3, 4, 5.\n[/hide]", "Solution_1": "\u0397 \u03bb\u03cd\u03c3\u03b7 \u03ba\u03c1\u03c5\u03c6\u03ae \u03b3\u03b9\u03b1 \u03cc\u03c3\u03bf\u03c5\u03c2 \u03b8\u03ad\u03bb\u03bf\u03c5\u03bd \u03bd\u03b1 \u03c4\u03bf \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ae\u03c3\u03bf\u03c5\u03bd \u03b1\u03ba\u03cc\u03bc\u03b1. \u03a0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03c0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03cc\u03c4\u03b9 \u03b7 \u03b5\u03bd\u03b1\u03c3\u03c7\u03cc\u03bb\u03b7\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03bf \u03b1\u03c0\u03cc \u03bc\u03b9\u03c3\u03ae \u03ce\u03c1\u03b1 \u03bc\u03b5 \u03c4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bd\u03c4\u03ac\u03b5\u03b9 \u03b1\u03bd\u03bf\u03cd\u03c3\u03b9\u03b1.\r\n\r\n[hide]\n\u03a3\u03c5\u03bc\u03b2\u03bf\u03bb\u03af\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 $ r$ \u03c4\u03b7\u03bd \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 \u03c4\u03bf\u03c5 \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5, $ S$ \u03c4\u03bf \u03b5\u03bc\u03b2\u03b1\u03b4\u03cc\u03bd, $ p$ \u03c4\u03b7\u03bd \u03b7\u03bc\u03b9\u03c0\u03b5\u03c1\u03af\u03bc\u03b5\u03c4\u03c1\u03bf, $ a\\le b \\le c$ \u03c4\u03b9\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9:\n\n$ S^2 \\equal{} p(p \\minus{} a)(p \\minus{} b)p(p \\minus{} c)$\n$ S \\equal{} pr \\Leftrightarrow (r \\equal{} 1) S \\equal{} p$\n\n\u03ac\u03c1\u03b1\n\n$ p^2 \\equal{} p(p \\minus{} a)(p \\minus{} b)p(p \\minus{} c) \\Leftrightarrow p \\equal{} (p \\minus{} a)(p \\minus{} b)(p \\minus{} c) (I)$\n\n\u03ad\u03c3\u03c4\u03c9 $ x \\equal{} p \\minus{} a, y \\equal{} p \\minus{} b, z \\equal{} p \\minus{} c$ ($ \\Leftrightarrow x \\ge y \\ge z$) \u03c4\u03cc\u03c4\u03b5\n\n$ x, y, z$ \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03b9 \u03ae \u03bc\u03b9\u03c3\u03ac \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03ce\u03bd \u03b1\u03ba\u03b5\u03c1\u03b1\u03af\u03c9\u03bd \u03c4\u03b1\u03c5\u03c4\u03cc\u03c7\u03c1\u03bf\u03bd\u03b1\n\u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03c3\u03ac \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03ce\u03bd \u03b1\u03ba\u03b5\u03c1\u03b1\u03af\u03c9\u03bd \u03c4\u03cc\u03c4\u03b5 $ (I) \\Leftrightarrow x \\plus{} y \\plus{} z \\equal{} \\frac {m}{8}$ \u03bc\u03b5 $ m \\equal{}$ \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc\u03c2, \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c4\u03bf\u03c0\u03bf\n\u03ac\u03c1\u03b1 \u03bf\u03b9 $ x, y, z$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03b9\n\n$ (I) \\Leftrightarrow x \\plus{} y \\plus{} z \\equal{} xyz \\Rightarrow 3x \\ge xyz \\Leftrightarrow yz \\le 3$\n\n\u03ac\u03c1\u03b1 \n\u03b5\u03af\u03c4\u03b5 $ y \\equal{} 3$ \u03ba\u03b1\u03b9 $ z \\equal{} 1$ \u03ac\u03c1\u03b1 $ 4 \\plus{} x \\equal{} 3x \\Leftrightarrow x \\equal{} 2 < y \\Leftrightarrow$ \u03ac\u03c4\u03bf\u03c0\u03bf\n\u03b5\u03af\u03c4\u03b5 $ y \\equal{} 2$ \u03ba\u03b1\u03b9 $ z \\equal{} 1$ \u03ac\u03c1\u03b1 $ 3 \\plus{} x \\equal{} 2x \\Leftrightarrow x \\equal{} 3$\n\u03b5\u03af\u03c4\u03b5 $ y \\equal{} 1$ \u03ba\u03b1\u03b9 $ z \\equal{} 1$ \u03ac\u03c1\u03b1 $ x \\equal{} x \\plus{} 2 \\Leftrightarrow$ \u03ac\u03c4\u03bf\u03c0\u03bf\n\n\u03ac\u03c1\u03b1 $ x \\equal{} 3, y \\equal{} 2, z \\equal{} 1$ \u03ac\u03c1\u03b1 $ p \\equal{} x \\plus{} y \\plus{} z \\equal{} 6 \\Leftrightarrow a \\equal{} p \\minus{} x \\equal{} 3, b \\equal{} p \\minus{} y \\equal{} 4, c \\equal{} p \\minus{} z \\equal{} 5$\n\n\u03b1\u03bb\u03bb\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 $ a^2 \\plus{} b^2 \\equal{} 3^2 \\plus{} 4^2 \\equal{} 5^2 \\equal{} c^2$ \u03ac\u03c1\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b1 $ a, b, c$ \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c4\u03bf \u03c0\u03c5\u03b8\u03b1\u03b3\u03cc\u03c1\u03b5\u03b9\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \n\n\u03ac\u03c1\u03b1 \u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9\u03bf\n[/hide]", "Solution_2": "\u03a9\u03c1\u03b1\u03af\u03bf\u03c2 :wink:" } { "Tag": [ "geometry", "ratio" ], "Problem": "Suppose the cevians $AP, BQ, CR$ meet at $T$. Prove that\r\n\r\n$\\frac{TP}{AP} + \\frac{TQ}{BQ} + \\frac{TR}{CR} = 1$.", "Solution_1": "A Hint : Use areas. We have \r\n \r\n $\\frac{TR}{AR} $ = $\\frac{(TBC)}{(ABC)} $\r\n\r\n where $(TBC) , (ABC) $ means areas of the corespondin triangles. Write now two more relations for the other ratios and you are done.If problem , please ask me again.\r\n\r\nBabis" } { "Tag": [ "geometry" ], "Problem": "Let $x,y,z$ be positve real numbers satifying $\\left\\{\\begin{array}{c}x3x^{2}+3xy+y^{2}=75y^{2}+3z^{2}=27x^{2}+xz+z^{2}=16\\end{array}\\right$\r\nFind value of $A=xy+2yz+3zx$", "Solution_1": "You can imagine the areas of three triangles.", "Solution_2": "Yes, this is a nice approach. The rezult is $24\\sqrt{3}$" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Hi, I am having trouble with the follow question regarding to n x n matrices.\r\n\r\nGiven that A and B are both n x n matrices, prove that\r\n1. det(AB) = det(A)det(B)\r\n2. adj(AB) = adj(B)adj(A)\r\n3. rank(AB) <= min[rank(A), rank(B)]\r\n\r\nI hope I can get help with these questions. Thank you very much.", "Solution_1": "What do you mean by $adj(A)$?", "Solution_2": "He means the adjungated matrix, which you find by replacing each element with its cofactor, and then taking the transpose\r\n\r\nI would like to add a question, does the adjungated matrix have geometric meaning, i mean when \r\n$C^{-1}AC=B$ is \r\n$C^{-1}Adj(A)C=Adj(B)$?" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Show that there exists exactly $\\binom{\\left[ \\frac{k}{2} \\right]}{k}$ sequences $\\ a_{1}, a_{2}, \\ldots, a_{k+1}$ of integer numbers $\\geq 0,$ for which $a_{1}=0$ and $|a_{i} - a_{i+1}|=1$ for all $i = 0, \\ldots, k.$", "Solution_1": "First of all, it should be $\\binom{k}{\\left[ \\frac{k}{2} \\right]}$ not $\\binom{\\left[ \\frac{k}{2} \\right]}{k}$ (the latter always equals $0$).\r\n\r\nLet $f(k,m)$ be the number of such sequences $a_{1}, a_{2}, \\ldots, a_{k+1}$ that $a_{k+1} = m$. Then the original problem is equivalent to computing $F(k)=\\sum_{m=0}^k f(k,m)$. \r\n\r\nNote that $f(k,k)=1$, $f(2n+1,0)=0$ and $f(2n,0)=C_n$, the $n$-th [url=http://mathworld.wolfram.com/CatalanNumber.html]Catalan number[/url]. We can also let $f(k,-1)=0$ and $f(k,m)=0$ for $m>k$. Then $f(k+1,m) = f(k,m-1) + f(k,m+1)$ for all $m\\geq 0$. Summation over $m=0..k+1$ gives $F(k+1) = 2F(k) - f(k,0)$ implying\r\n$F(2n) = 2F(2n-1) = 4F(2(n-2)) - 2C_{n-1}$\r\n$F(2n+1) = 2F(2n) - C_n = 4F(2n-1) - C_n$\r\nthat together with $F(0)=F(1)=1$ give rise to\r\n$F(2n) = 4^n - 2 \\sum_{i=0}^{n-1} 4^{n-1-i} C_i$\r\n$F(2n+1) = 4^n - \\sum_{i=0}^{n-1} 4^{n-1-i} C_{i+1}$\r\n\r\nThe sum $\\sum_{i=0}^{n-1} 4^{n-1-i} C_i$ equals the coefficient of $x^{n-1}$ in the expansion of $\\frac{1}{1-4x}\\cdot\\frac{1-\\sqrt{1-4x}}{2x} = \\frac{(1-4x)^{-1} - (1-4x)^{-1/2}}{2x}$ that is \r\n\\[ 2^{2n-1}(-1)^n\\left(\\binom{-1}{n} - \\binom{-1/2}{n}\\right) = 2^{2n-1} - \\frac{1}{2}\\binom{2n}{n} \\]\r\nimplying $F(2n)=\\binom{2n}{n}$.\r\n\r\nSimilarly, \r\n\\[ \\sum_{i=0}^{n-1} 4^{n-1-i} C_{i+1} = \\sum_{i=0}^n 4^{n-i} C_i - 4^n = 4^n - \\frac{1}{2}\\binom{2(n+1)}{n+1} = 4^n - \\binom{2n+1}{n} \\]\r\nimplying $F(2n+1)=\\binom{2n+1}{n}$." } { "Tag": [], "Problem": "A snake of mass $m$ moves from one end of a boat (of mass $M$) to the other. What's the distance the boat travels in the mean time if its length is $L$?", "Solution_1": "One starts with the normal assumptions, frictionless interfaces between the boat and the water, the snake has negible length the snake has a smaller mass than the boat and nothing interacts with any outside objects.\r\n\r\nThe balance point of the snake/boat system doesn't change and to counter act the snake's movement. The boat moves a distance equal to m*L/M." } { "Tag": [], "Problem": "The set {2, 3, 4, 5, 9} contains five of the factors of which one of the four integers? \r\n\r\n480, 1260, 1890, 3888", "Solution_1": "[hide]i did trial and error and came up with 1260.[/hide]", "Solution_2": "[quote=\"math92\"][hide]i did trial and error and came up with 3888.[/hide][/quote]\r\n\r\n5 isn't a factor of that :roll:", "Solution_3": "[hide]\n1260\n\nObviously divisible by 2\n1+2+6+0=9 which is a multiple of 3 which means its divisible by 3\n1200 is divisible by 4. 60 is divisible by 4.\nObviously divisible by 5.\n1+2+6+0=9 which is a multiple of 9 which mean its divisble by 9.[/hide]", "Solution_4": "[quote=\"236factorial\"][quote=\"math92\"][hide]i did trial and error and came up with 3888.[/hide][/quote]\n\n5 isn't a factor of that :roll:[/quote]\r\n :wallbash: \r\ni keep messing up every problem i do!!!!! :(", "Solution_5": "[hide]they are all divisible by 2\nthey are all divisible by 3\nthe first 2 and the last 1 are divisible by 4\nthe first 3 are divisible by 5\nthe middle 2 are divisible by 9\nthis leaves the second one [b]1260[/b] the only one with all the factors\n[/hide]", "Solution_6": "[hide]\nSince there's a five, 3888 doesn't work.\nSince there's a 9, 480 doesn't work.\n1890 doesn't work because there's a 4.\nSo, the answer is 1260.[/hide]" } { "Tag": [], "Problem": "Moi nguoi oi\r\nsap thi quoc gia 2006 roi\r\nDoi tuyen Khtn manh lam :)", "Solution_1": "[color=blue]\u0110\u1ed9i tuy\u1ec3n c\u1ee7a [b]Khoa h\u1ecdc t\u1ef1 nhi\u00ean[/b] khi n\u00e0o ch\u1eb3ng m\u1ea1nh m\u00e0 b\u1ea1n l\u1ea1i n\u00f3i v\u1eady? [/color]", "Solution_2": "see here:\r\nhttp://www.nxbgd.com.vn/toanhoctuoitre/?p=5&id=26&vtopicID=2", "Solution_3": "Nh\u01b0ng m\u00e0 em th\u1ea5y d\u00f9 sao th\u00ec c\u0169ng \u0111\u1eebng b\u1ecb \u1ea3nh h\u01b0\u1edfng b\u1edfi ng\u01b0\u1eddi ta w\u00e1!!!n\u1ebfu kh\u00f4ng m\u00ecnh s\u1ebd kh\u00f4ng l\u00e0m b\u00e0i dc!!ch\u1ec9 c\u1ea7n m\u00ecnh c\u1ed1 g\u1eafng h\u1ebft s\u1ee9c th\u00ec k\u1ebft w\u1ea3 th\u1ebf n\u00e0o c\u0169ng th\u1ea5y vui m\u00e0!!! :roll:" } { "Tag": [ "function" ], "Problem": "find all functions like $ f: N \\to N \\minus{} {1}$ that $ \\forall n \\in N$:\r\n\\[ f(n)\\plus{}f(n\\plus{}1)\\equal{}f(n\\plus{}2)f(n\\plus{}3)\\minus{}168\\]", "Solution_1": "Why post this twice?\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?p=1329198#1329198[/url]" } { "Tag": [ "trigonometry", "inequalities unsolved", "inequalities" ], "Problem": "For a triangle , prove that :\r\n$ a^2\\,\\preceq\\,\\frac{b^2}{\\cos^2x}\\plus{}\\frac{c^2}{\\sin^2x}$\r\nwith x any angle in the triangle", "Solution_1": "[quote=\"memath\"]For a triangle , prove that :\n$ a^2\\,\\preceq\\,\\frac {b^2}{\\cos^2x} \\plus{} \\frac {c^2}{\\sin^2x}$\nwith x any angle in the triangle[/quote]\r\n\r\n\r\n$ \\frac {b^2}{\\cos^2x} \\plus{} \\frac {c^2}{\\sin^2x}\\equal{}(cos^{2}x\\plus{}sin^{2}x)(\\frac {b^2}{\\cos^2x} \\plus{} \\frac {c^2}{\\sin^2x})\\geq (b\\plus{}c)^{2}\\geq a^{2}$", "Solution_2": "[quote=\"zaya_yc\"][quote=\"memath\"]For a triangle , prove that :\n$ a^2\\,\\preceq\\,\\frac {b^2}{\\cos^2x} \\plus{} \\frac {c^2}{\\sin^2x}$\nwith x any angle in the triangle[/quote]\n\n\n$ \\frac {b^2}{\\cos^2x} \\plus{} \\frac {c^2}{\\sin^2x} \\equal{} (cos^{2}x \\plus{} sin^{2}x)(\\frac {b^2}{\\cos^2x} \\plus{} \\frac {c^2}{\\sin^2x})\\geq (b \\plus{} c)^{2}\\geq a^{2}$[/quote]\r\n\r\ncorrect :)", "Solution_3": "memath : a,b,c triangle side ? ---> $ b \\plus{} c > a$ $ \\frac{c^{2}}{sin^{2}x}\\plus{}\\frac{b^{2}}{cos^{2}x}>a^{2}$", "Solution_4": "[quote=\"zaya_yc\"]memath : a,b,c triangle side ? ---> $ b \\plus{} c > a$ $ \\frac {c^{2}}{sin^{2}x} \\plus{} \\frac {b^{2}}{cos^{2}x} > a^{2}$[/quote]\r\nyes we have a,b,c triangle side. why ? you're answer is correct" } { "Tag": [], "Problem": "Does anyone know the divisibility rule for 31?", "Solution_1": "[hide=\"Test for divisibility by 31\"]\nSubtract three times the last digit from the remaining leading truncated number. If the result is divisible by 31, then so was the first number. Apply this rule over and over again as necessary.\nExample: 7998-->799-3*8=775-->77-3*5=62 which is twice 31, so 7998 is also divisible by 31.\n[/hide]" } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Please help me to resolve this :\r\n\r\n$ \\lim_{ n->{\\infty}} \\prod_{k=1}^{n} \\frac{(k+1)^{2}}{k (k+2) }$", "Solution_1": "We can write:\r\n\\[\\prod_{k=1}^n\\dfrac{(k+1)^2}{k(k+2)}=\\prod_{k=1}^n\\dfrac{k+1}{k}\\prod_{k=1}^n\\dfrac{k+1}{k+2}=\\\\\r\n\\qquad\\\\\\qquad\\\\ \\dfrac{\\displaystyle\\prod_{k=1}^n(k+1)}{\\displaystyle\\prod_{k=1}^nk}\\dfrac{\\displaystyle\\prod_{k=1}^n(k+1)}{\\displaystyle\\prod_{k=1}^n(k+2)}=(n+1)\\dfrac2{n+2}\\to 2 \\text\\quad{if}\\quad n\\to+\\infty\\]", "Solution_2": "But what about [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=45150]www.mathlinks.ro/Forum/viewtopic.php?t=45150[/url]", "Solution_3": "Since \\[\\dfrac{k^3+1}{k^3-1}=\\dfrac{(k+1)(k^2-k+1)}{(k-1)((k+1)^2-(k+1)+1)}\\]\r\nwe get\\[\\prod_{k=2}^n\\dfrac{k^3+1}{k^3-1}=\\frac32\\dfrac{n^2+n}{n^2+n+1}\\xrightarrow[n\\to+\\infty]{}\\frac32\\]" } { "Tag": [ "trigonometry", "function", "calculus", "calculus computations" ], "Problem": "How would you go about solving the following.\r\nMaximize $ \\cos(a)\\plus{}\\cos(b)\\plus{}\\cos(c)$ given that $ a\\plus{}b\\plus{}c\\equal{}S$ and $ S$ is a positive integer. The original problem also assumes that $ a,b,c$ are also positive integers but I would like to solve this \"simpler\" case first.", "Solution_1": "[url=http://en.wikipedia.org/wiki/Lagrange_multipliers]Lagrange multipliers[/url]. The computation is very straightforward depending on the value of $ S$ and whether $ a, b, c$ are restricted to being positive.", "Solution_2": "I understand that I have to use Lagrange multipliers, however, I run into problems when trying to solve the system that follows.", "Solution_3": "Lagrange multipliers tells you the extrema occur when $ \\sin a \\equal{} \\sin b \\equal{} \\sin c$. This means that $ \\cos^2 a \\equal{} \\cos^2 b \\equal{} \\cos^2 c \\equal{} r^2$ for some positive real $ r$ and hence $ \\cos a \\plus{} \\cos b \\plus{} \\cos c$ can take on the values $ 3r, r, \\minus{}r, \\minus{}3r$. To maximize, we want the value $ 3r$, and picking the value of $ \\cos a$ and $ \\sin a$ therefore uniquely determines the angles $ a, b, c$ up to addition of $ 2 \\pi$; that is, $ a \\equiv b \\equiv c \\bmod 2 \\pi$. From here there are only a few choices that sum up to $ S$, depending on the value of $ S$." } { "Tag": [], "Problem": "Prove that $ \\sqrt {m}$, if m is not a perfect square, is not rational.\r\n\r\nProve that $ \\sqrt{p}$,where p is a prime, cannot be rational.", "Solution_1": "We can clearly see that 2 is a consequence of 1 :wink:\r\n\r\n[hide=\"to prove 1\"]\nLet $ \\sqrt{m} \\equal{} \\frac {p}{q}$ where the fraction is in its simplest form \n$ \\implies m \\equal{} \\frac {p^2}{q^2} \\implies p^2 \\equal{} m \\cdot q^2$\n$ \\implies p \\equiv 0 \\mod m \\implies q \\equiv 0 \\mod m$\nBut this is a contradiction!\n\nNote that the problem doesnt arise if m is a perfect square (Why?) :)\n\nAs i said earlier, 2 is a consequence of 1 :)\n[/hide]", "Solution_2": "[quote=\"madness\"]$ \\implies p \\equiv 0 \\mod m \\implies q \\equiv 0 \\mod m$[/quote]There is an important step missing within this implication. If $ p\\equiv0\\mod m$, $ p\\equal{}mk$ for some integer $ k$. Then, $ m^2k^2\\equal{}mq^2\\implies q^2\\equal{}mk^2$. The rest follows.", "Solution_3": "sorry i thought that was understood :oops: :)", "Solution_4": "[quote=\"madness\"]\n\nNote that the problem doesnt arise if m is a perfect square (Why?) :)\n\n[/quote]\r\n\r\nIs that because if $ m$ is a perfect square, then when $ \\sqrt{m}$ is written as $ \\frac{p}{q}$, the integer $ q$ is 1? So that $ m \\equal{} \\frac {p^2}{q^2} \\implies p^2 \\equal{} m \\cdot q^2$ becomes \r\n$ p^2 \\equal{} m \\cdot (1^2)$\r\n$ p^2 \\equal{} m$\r\nAnd we can no longer derive the contradiction?", "Solution_5": "The proof is not correct. $ m|p^2\\not\\implies m|p$. For example, $ 4|6^2$, but $ 4\\not|6$." } { "Tag": [ "USAMTS", "search" ], "Problem": "In problem 3, what is the term \"disjoint squares\" supposed to mean?", "Solution_1": "You could search on google.", "Solution_2": "DPatrick said, in another thread, that correctly interpreting the problem is part of the problem, so interpretation help should NOT be given.", "Solution_3": "yeah, I was about to tell him that.\r\n\r\n :|", "Solution_4": "We cannot clarify problem statement while the round is in progress, and it is against the USAMTS rules to ask for help." } { "Tag": [ "conics", "hyperbola" ], "Problem": "It is part of the definition of being a hyperbola. b :^2: =c :^2: -a :^2: .\r\n\r\n-interesting_move", "Solution_1": "Erm ... I think you can prove it based upon the definition of the hyperbola (using difference of distances between two points) and the equation ... it should follow fairly directly from that.\r\n\r\nInteresting_move, that isn't definitional -- it's something you prove. a, b and c all have meaning of their own.", "Solution_2": "i think you're both right [quote]t is part of the definition of being a hyperbola. b =c -a [/quote] from the algabraic side, that's true but from the geometric side you can and do have to prove it.\r\nSo you're both right." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "[url=http://www.imagehosting.com][img]http://www.imagehosting.com/out.php/i1811033_TTinequality1.gif[/img][/url]\r\n\r\nHave A Great Day!", "Solution_1": "Homogenize to abc=1\r\n\r\ncyclic sum of $ \\frac{1}{c\\sqrt{1/c\\plus{}2c^2}}$\r\n\r\nNow we make a change of variables a,b,c-->1/a,1/b,1/c\r\n\r\n$ \\frac{c}{\\sqrt{c\\plus{}2/c^2}}\\equal{}\\frac{c^2}{\\sqrt{c^3\\plus{}2}}$\r\n\r\nwhich we have to show is $ \\ge\\sqrt{{a\\plus{}b\\plus{}c}}$\r\n\r\nIt's too late to figure this out for me, anyone else want to figure out this seemingly apparent inequality.", "Solution_2": "From Holder's inequality we have \r\n\\[ \\left(\\sum\\frac{ab}{\\sqrt{ab+2c^2}}\\right)^2\\left(\\sum ab(ab+2c^2)\\right)\\&\\ge (ab+bc+ca)^3\\]\r\nwhich implies\r\n\\begin{align*}\\left(\\sum\\frac{ab}{\\sqrt{ab+2c^2}}\\right)^2&\\ge \\frac{(ab+bc+ca)^3}{\\sum ab(ab+2c^2)}\\\\\r\n&=\\frac{(ab+bc+ca)^3}{(ab+bc+ca)^2}\\\\\r\n&=ab+bc+ca\\end{align*}\r\nwhich was what we wanted.", "Solution_3": "Thanks Nayel. Your solution is very beautiful!\r\n\r\nHave A Great Day!", "Solution_4": "I think you should write the author's name of this problem. \r\nIt's my ine80 in my topic :) \r\nhttp://www.toanthpt.net/forums/showthread.php?t=9244&page=11\r\n@can_hang, mitdac, honey_suck: My last nick is nhocnhoc but I can't \"k\u00edch ho\u1ea1t\" it .So I must to use this ugly name :D (sr because I speak English very bad)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Does anybody know the winning strategy of the Super Nim game? \r\n\r\nhttp://www.gameintellect.com/java-games/super-nim/#menu\r\n------------------------------------------------------------------------------------------------------------------------------------------\r\nSuper Nim was created by GameIntellect.com\r\n\r\nAt the beginning of the game a few chips are placed randomly on the board. The players remove chips in turns. A move consists in removing all the chips in a row or in a column. The player to remove the very last piece wins.\r\n\r\nTo make a move select any column or row and click on corresponding letter or number. For example, to remove all of the pieces from column \"b\" click on the letter \"b\".\r\n\r\nPress \"Comp. starts\" to give computer the first turn. When you make it to the next level, your current score will be cleared and you'll have to start from 0 all over again.\r\n\r\n---------------------------------------------------------------------------------------------------------------------------------------------\r\n\r\nPlease help~", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293868" } { "Tag": [ "geometry" ], "Problem": "This is really just talking about what could be improved from '08, and in case you left something off the list. Please read through all posts. That will make it easier for the next person. \r\n\r\nin response to the question it \"Will you go next year\" by thing1\r\nThere seems to be less time to ask the teacher questions and, for me and at least 3 others, were afraid to stop the class and ask questions. They should (encourage/have 5 minute breaks in the middle of lectures for) questions. \r\n\r\nWho else went to the math forums at night? If there were more of those, and maybe a rating of difficulty for each one and what is expected of you to know then that would probably be adequate. \r\n\r\nAs a camp:\r\nThere was a wonderful amount of activities scheduled on Saturday, but a rather noticable lack of such on Sundays. \r\n\r\nA more specific food complaint: A lack of breakfast on Sundays, a lack of Vitamin C in the food and drinks, and a lack of constantly available vegetarian options (although 80-90% isn't bad).", "Solution_1": "i felt like having free time on sunday was okay... maybe there could be one or two small activities. but i liked the free time. \r\n\r\nbut yea, definitely time for questions, cuz i felt so intimidated by the profs. X_X and i felt scared to ask questions/look stupid. >< priorityyy. a lotta the time i felt lost ... then i just zoned out. =_=''\r\n\r\nfooodddd.... -.- its edible. \"this is math camp not food camp! lol\" but if improved, itd be better.", "Solution_2": "Suggestions:\r\n\r\nMake the food better (though I didn't really eat very much at all...)\r\n\r\nAbout the classes:\r\n\r\nHmm... question times might be helpful?\r\n\r\nAlso, why was Olympiad Combinatorics problem set same as Combinatorics class... some days everybody just did Team Contest instead of the work...\r\n\r\nAlso, the team contest problems were way too favored for those who were in Combinatorics class...\r\n\r\nLagrange Interpolation-You put it on team contest right after it is taught in class...\r\nThe nice combinatorial argument with the inscribed 3/8 area triangle: You put it on team contest right after its solution is shown in class (a not-exactly-the-same version though...)\r\n\r\nQuestion times... might be helpful? \r\n\r\nMore math forums would be helpful... (though admittedly I didn't go... oh well... I should be less lazy?)\r\n\r\nI didn't care at all that there weren't any activities on sundays... I like spare time...", "Solution_3": "Hmm my experience was that I was like: ASK QUESTIONS PEOPLE. I think that it went pretty well in olympiad geometry (I'd go around while people did the problem sets). The kids in my dorm asked me questions occasionally (I don't think that they were intimidated anyway). I also got questions from some of the 'standout kids' in the class. I also visited a few classes and I think that I got some questions out of people. \r\n\r\nI think that it is a good idea to have question times. But just so you know, once you get to college, you won't get babied like that. You have to be proactive in asking questions...\r\n\r\nYou could always order something from Walmart or buy food at that court that was open in the student union.\r\n\r\nI would say that some classes were better prepared than others. I do remember that happening in the Olympiad combo class one or two days. I think I had a new problem set for the geometry class everyday.\r\n\r\nI guess one of the things that I picked up on was that there were some assistants/teachers that were hard to understand because of their English.", "Solution_4": "[quote=\"leoxnlin\"]\nI'd say rather that your money goes to the whole experience of living, learning, and playing, and especially the social interactions, rather than the amount of problem-solving skill gained.\n\nI'd like more math forums though... Altheman made geometry actually seem reasonable (I am a former geometry-hater)... :P[/quote]\r\nI really think AMSP should be more intense like with more lectures and problem solving sessions like in the 2006 schedule. I mean we could fit so many more lectures in and yes I know we should spend more time solving problems on our own but we could just do that at home and like I probably learn more in a one hour lecture than I do in like four hours just doing random problems. Also it might be nice to leave some of the problems unsolved at the end of the problem solving session so we have something to think about after class. Overall I don't think I learned a lot from the team contests but I really improved my problem solving ability by attending the lectures and math forums. The schedule probably can't be changed since they have already determined which classes to teach but surely we could have more math forums even if they are like a repeat of some class for people than didn't take it. There was tons of time to do team contest problems but I didn't get many of them or learn much. Like if you look on the will you go next year post it seems that several of people agree with me.", "Solution_5": "Was I the only one who thought the food was actually pretty good?", "Solution_6": "[quote=\"leoxnlin\"]Was I the only one who thought the food was actually pretty good?[/quote]\r\n\r\nseriouslyy??", "Solution_7": "[quote=\"leoxnlin\"]Was I the only one who thought the food was actually pretty good?[/quote]\r\n\r\n... I am so glad that I'm learning to cook...", "Solution_8": "but u cant cook at amsp.... \r\nunless ur like alexhhmun who got that HUGE box of ramen [that he wouldnt share!!! >< ] ahahaha jk. xP", "Solution_9": "I thought the food tasted pretty good...and it was better than I expected...\r\nWell you could cook...you just needed a counselor watching you...", "Solution_10": "I liked the food too...although I think supper was too early, so I got hungry by bedtime.", "Solution_11": "[quote=\"shelly32494\"]unless ur like alexhhmun who got that HUGE box of ramen [that he wouldnt share!!! >< ][/quote]\r\n\r\nyeah it was yummy.\r\n\r\nhe shared with me because we are best buddies.\r\n\r\nAlso stock up on snacks. You are gonna need them.\r\n\r\nAlso, I'm used to good homemade food, so it might be just me.\r\n\r\nGuacamole was good at a restaurant in Fort Worth, but HORRIBLE at AMSP for some reason.\r\n\r\nas you can see, i'm picky even when i'm hungry.\r\nexcept when it's mom's cooking. :)\r\n\r\ni ate sushi for lunch the second half of the camp. It was DELICIOUS.\r\n\r\nNext time i'ma use wasabi.", "Solution_12": "Yeah...just get snacks...\r\n\r\nOr you could order pizza or chinese food at like 11 PM, like we did :lol:", "Solution_13": "so, all in all, \r\nbring extra food money!! lol XP", "Solution_14": "student uniuon stores/pub arent open all the time! :/\r\n\r\nmy dorm ended up ordering pizzas...hahaha at like midnight[i think a lot of dorms did actually LOL]", "Solution_15": ":wallbash: Yeah more math jams would be great and if possible more classes too. :spam: \r\nOne thing that I didn't like about class was that it was way too rigorous and we didn't go through the steps in solving the problem. Like It would be nice if the lectures were more like WOOT lectures in that they outline problem solving strategies more. Also I felt that the mentors just ignored my questions and were like figure it out. But mostly yeah it probably would be better to make the camp more intense and have like lectures on weekends because there was really nothing to do on weekends except work on team contest problems which got boring after a while. Yeah so on weekends it would be really nice to have some lectures. :starwars: Like overall organized classes are a lot more effective than random problems.", "Solution_16": "The place to buy food on campus was closed at dinner. And I DID share the ramen... I ate 4 or 5 of 12...\r\n\r\nMost days the food wasn't bad. Just that it was really inconsistent.", "Solution_17": "Yes, alexhhmun DID share his ramen. :) I tried to eat it in the dorm but we had nothing to cook it with. So I just used hot tap water from the sink :stink: but it still tasted pretty good.\r\n\r\nC'mon guys. It was cafeteria food. What did you expect? I thought it was fine. Although the lack of plenty of vegetarian choices was a problem, I know a couple of people who \"starved\" since they didnt always serve vegetarian food and ate stuff shipped from home.", "Solution_18": "its impossible to starve... I ate practically nothing and didn't starve.\r\n\r\nOf course, I took some food from the peopple next door...", "Solution_19": "pythag, the food wasn't bad, but I maintain that the drinks lacked vitamin C, which is why we all got sick. I got sick 2 days after leaving camp. \r\n\r\nI noticed but wasn't allowed to get orange juice from wal-mart. :(", "Solution_20": "I was first to get sick. And for some reason all Zeb did during that class was ask me if I was going to die. WHY DOES EVERYONE WANT ME TO DIE...?\r\n\r\nSo everyone in my room blames me for them getting sick. :)", "Solution_21": "GOSH DARN IT JUST DRINK OJ! :P\r\n\r\nThe food needs to improve and people(not just counselors, STUDENTS!) need to start organizing stuph to do on weekends...\r\nThen it's probably perfect.", "Solution_22": ":mad: to everyone in my dorm who ate all of the food that I brought...", "Solution_23": "[quote=\"alexhhmun\"]pythag, the food wasn't bad, but I maintain that the drinks lacked vitamin C, which is why we all got sick. I got sick 2 days after leaving camp. \n\nI noticed but wasn't allowed to get orange juice from wal-mart. :([/quote]\r\nThe problem is almost all non-preserved (and some preserved) fruit, vegetable, AND meat has vitamin C...", "Solution_24": "[quote=\"sooozyy\"]student uniuon stores/pub arent open all the time! :/\n\nmy dorm ended up ordering pizzas...hahaha at like midnight[i think a lot of dorms did actually LOL][/quote]\r\n\r\nlol pizza after 9 dearrr XP \r\nhahhaa. \r\n\r\nwhattttt. alexhhmun how could u betrayyy me??!!? i didnt get ramennn. hahaha jkjk. XP \r\n\r\nhmm. i was homesick when i got back from amsp. lol XP hahaha. \r\nbut srlsly, being tired from camp may have something to do with getting sick and stuff.", "Solution_25": "The problem sessions sometimes weren't so good like some people were noisy. Also during the problem session I think the mentors should focus more on helping us become better at problem solving rather than putting all their emphasis on proof writing. Lectures on weekends would be REALLY good like overall I don't see the point in working on team contest problems by yourself at camp when you can just do it from home or having competitions because we are not going there so we can get the satisfaction of beating someone (at least I'm not). Man is the food all everybody complains about seriously." } { "Tag": [ "geometry", "rotation" ], "Problem": "In Triangle $ABC$, $AC=BC=AC$. Three lines, each drawn from each of the separate verticies of the triangle, meet at point $D$ inside the triangle so that $AD=7$, $BD=5$, and $CD=8$. What is the area of triangle $ABC$?", "Solution_1": "rotate 60 around point B" } { "Tag": [ "geometry", "geometric transformation", "search" ], "Problem": "thanks :lol:", "Solution_1": "If you see this site please say how can we join at?", "Solution_2": "Buddy, I had pass for lib.homelinux once, but it is changed now!! Could you share it with me?! :oops:", "Solution_3": "what is in this site? when trying to access this site it opened a dialog box asking for some login name and password.? and there was written \"wooden boy with long nose\" :rotfl: what is this man?", "Solution_4": "its asks for passwords and usernames :rotfl:", "Solution_5": "What is this site? How do you enter it?\u201c What does this mean!?!??!?!?!?!?!?!?!?!? :wallbash_red: \u00c4\u00e5\u00f0\u00e5\u00e2\u00ff\u00ed\u00ed\u00fb\u00e9 \u00ec\u00e0\u00eb\u00fc\u00f7\u00e8\u00ea \u00f1 \u00e4\u00eb\u00e8\u00ed\u00ed\u00fb\u00ec \u00ed\u00ee\u00f1\u00ee\u00ec (\u00f0\u00f3\u00f1\u00f1\u00ea\u00e8\u00e9 \u00e2\u00e0\u00f0\u00e8\u00e0\u00ed\u00f2) (\u00e8\u00e7 8 \u00e1\u00f3\u00ea\u00e2)? (\u00e8 \u00e8\u00ec\u00ff, \u00e8 \u00ef\u00e0\u00f0\u00ee\u00eb\u00fc, \u00eb\u00e0\u00f2\u00e8\u00ed\u00e8\u00f6\u00e5\u00e9, \u00ec\u00e0\u00eb\u00e5\u00ed\u00fc\u00ea\u00e8\u00ec\u00e8 \u00e1\u00f3\u00ea\u00e2\u00e0\u00ec\u00e8). \u00cf\u00f0\u00ee\u00f1\u00fc\u00e1\u00e0 \u00e8\u00ec\u00ff \u00e8 \u00ef\u00e0\u00f0\u00ee\u00eb\u00fc \u00ed\u00e8\u00e3\u00e4\u00e5 \u00ed\u00e5 \u00ef\u00f3\u00e1\u00eb\u00e8\u00ea\u00ee\u00e2\u00e0\u00f2\u00fc, \u00e8\u00ed\u00e0\u00f7\u00e5 \u00ef\u00f0\u00e8\u00e4\u00e5\u00f2\u00f1\u00ff \u00f7\u00e0\u00f8\u00e5 \u00ec\u00e5\u00ed\u00ff\u00f2\u00fc.\u201d on lib.homelinux.org.", "Solution_6": "[quote=\"spider_boy\"]what is in this site? when trying to access this site it opened a dialog box asking for some login name and password.? and there was written \"wooden boy with long nose\" :rotfl: what is this man?[/quote]\r\nPinocchio", "Solution_7": "[quote=\"Xevarion\"][quote=\"spider_boy\"]what is in this site? when trying to access this site it opened a dialog box asking for some login name and password.? and there was written \"wooden boy with long nose\" :rotfl: what is this man?[/quote]\nPinocchio[/quote]\r\n\r\nno I understood that part.But what is inside this site?I am very curious!! :D :lol:", "Solution_8": "me too...\r\nHas anyone searched in google how to enter?", "Solution_9": "I again tried to open this site but it again asked for password etc..\r\nBut the text in the dialog box that my browser opens is in russian( maybe it is in russian in your compter also) and it says: \u0414\u0435\u0440\u0435\u0432\u044f\u043d\u043d\u044b\u0439 \u043c\u0430\u043b\u044c\u0447\u0438\u043a \u0441 \u0434\u043b\u0438\u043d\u043d\u044b\u043c \u043d\u043e\u0441\u043e\u043c (\u0440\u0443\u0441\u0441\u043a\u0438\u0439 \u0432\u0430\u0440\u0438\u0430\u043d\u0442) (\u0438\u0437 8 \u0431\u0443\u043a\u0432)? (\u0438 \u0438\u043c\u044f, \u0438 \u043f\u0430\u0440\u043e\u043b\u044c, \u043b\u0430\u0442\u0438\u043d\u0438\u0446\u0435\u0439, \u043c\u0430\u043b\u0435\u043d\u044c\u043a\u0438\u043c\u0438 \u0431\u0443\u043a\u0432\u0430\u043c\u0438). \u041f\u0440\u043e\u0441\u044c\u0431\u0430 \u0438\u043c\u044f \u0438 \u043f\u0430\u0440\u043e\u043b\u044c \u043d\u0438\u0433\u0434\u0435 \u043d\u0435 \u043f\u0443\u0431\u043b\u0438\u043a\u043e\u0432\u0430\u0442\u044c, \u0438\u043d\u0430\u0447\u0435 \u043f\u0440\u0438\u0434\u0435\u0442\u0441\u044f \u0447\u0430\u0448\u0435 \u043c\u0435\u043d\u044f\u0442\u044c.\r\ntranslation: Wooden boy with long nose(russian version)(8 letters)? (and username, and password, in latin, with little letters).It would be good if you don't publish your username and password anywhere.If not you will have to change it frequently.\r\n\r\nFirst of all are they joking?would someone publish his/her username and password anywhere?\r\nSecondly And I did some google search and learned this information: it is Belorussian science library. and it is semibroken and unstable.And it includes tones of ebooks. :) \r\nSo can our belorussian users help us?\r\n\r\nedit:hey guys I have found the password and login.so i edited this message because it revealed the solution to this great mystery!!I will not post the information here.you can pm me for the information.I will answer only the first 10 pms. :D :) :P", "Solution_10": "I did some google search and in a site there was a riddle: the name of a large Russian river is the username and password.\r\nIt says in latin, right?", "Solution_11": "[quote=\"thatin\"]I did some google search and in a site there was a riddle: the name of a large Russian river is the username and password.\nIt says in latin, right?[/quote]\r\n\r\nI searched in some russian sites and found ::\u0414\u043b\u0438\u043d\u043d\u0435\u0439\u0448\u0430\u044f \u0440\u0435\u0447\u043d\u0430\u044f \u0441\u0438\u0441\u0442\u0435\u043c\u0430 - \u041e\u0431\u044c-\u0418\u0440\u0442\u044b\u0448 - \u043f\u0440\u043e\u0442\u044f\u043d\u0443\u043b\u0430\u0441\u044c \u043d\u0430 5410 \u043a\u043c \u043e\u0442 \u0437\u0430\u043f\u0430\u0434\u043d\u043e\u0433\u043e \u041a\u0438\u0442\u0430\u044f \u0434\u043e \u0421\u0435\u0432\u0435\u0440\u043d\u043e\u0433\u043e \u041b\u0435\u0434\u043e\u0432\u0438\u0442\u043e\u0433\u043e \u043e\u043a\u0435\u0430\u043d\u0430. \u0412\u0442\u043e\u0440\u0430\u044f \u043f\u043e \u0434\u043b\u0438\u043d\u0435 \u0440\u0435\u0447\u043d\u0430\u044f \u0441\u0438\u0441\u0442\u0435\u043c\u0430 - \u0410\u043c\u0443\u0440-\u0428\u0438\u043b\u043a\u0430-\u041e\u043d\u043e\u043d (4416 \u043a\u043c) - \u0442\u0435\u0447\u0435\u0442 \u0438\u0437 \u0441\u0435\u0432\u0435\u0440\u043d\u043e\u0439 \u041c\u043e\u043d\u0433\u043e\u043b\u0438\u0438, \u0432\u0434\u043e\u043b\u044c \u0440\u043e\u0441\u0441\u0438\u0439\u0441\u043a\u043e-\u043a\u0438\u0442\u0430\u0439\u0441\u043a\u043e\u0439 \u0433\u0440\u0430\u043d\u0438\u0446\u044b \u0432 \u0422\u0438\u0445\u0438\u0439 \u043e\u043a\u0435\u0430\u043d. \u0421\u0440\u0435\u0434\u0438 \u043e\u0442\u0434\u0435\u043b\u044c\u043d\u044b\u0445 \u0440\u0435\u043a \u0441\u0430\u043c\u043e\u0439 \u0434\u043b\u0438\u043d\u043d\u043e\u0439 \u044f\u0432\u043b\u044f\u0435\u0442\u0441\u044f \u041b\u0435\u043d\u0430 - 4269 \u043a\u043c, \u0434\u0430\u043b\u0435\u0435 \u0438\u0434\u0443\u0442 \u0418\u0440\u0442\u044b\u0448 \u0438 \u041e\u0431\u044c,\r\n\r\nit says the longest river system in Russia is \"\u041e\u0431\u044c-\u0418\u0440\u0442\u044b\u0448\" and the longest river is \"\u041b\u0435\u043d\u0430\".so... the first one can help.\r\noh I love mysteries :roll: :roll:\r\n\r\nedit: I tried but none of \"\u041e\u0431\u044c-\u0418\u0440\u0442\u044b\u0448\" and \"\u041b\u0435\u043d\u0430\" worked.", "Solution_12": "hey guys I have found the login and password.I will not post the information here.you can pm me for the information.I will answer only the first 10 pms.", "Solution_13": "I think that the password and username change frequently so... hurry :D", "Solution_14": "I got in but there was some russian site :maybe:", "Solution_15": "[quote=\"anirudh\"]I got in but there was some russian site :maybe:[/quote]\r\n\r\nno everyone can use it easily.Just click _djvu(\u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430,...)(if you want math materials).Then it will open a huge page where there are many books. :D (In my comp the page opens approx for 1 hour because my internet speed is slow. :( )The page contains books in many languages such as german ,english,russian, and french.", "Solution_16": "no, their internet speed is slow. In my pc and in my father's pc I waited for 2 hours and it didn't opened.", "Solution_17": "dear users.I will not answer your questions on this issue from now on.(i am very sorry) :(", "Solution_18": "you can just use http://lookforbook.com/ to find the book you want and then use the password when necessary..." } { "Tag": [ "geometry", "3D geometry", "symmetry", "linear algebra", "matrix", "graph theory" ], "Problem": "Everyone probably know the famous problem consisting of a circuit having 12 resistors in the shape of a cube such that the resistors make up the edges. We pass current from one vertex and are required to find the effective resistance between this vertex and another one (the difficulty of the problem depends upon which vertexes are considered!!)\r\n\r\nIn Physics, we usually solve this problem by assuming some symmetry considerations like equal partition of current or equal voltages etc. While studying Graph Theory in Alan Tucker's book, I cam across a similar example where he asked whether 2 graphs, one like the cubical circuit above and another circular graph were isomorphic. While his question ended in a negative there, I thought, could it be possible to solve the problem without any assumptions by finding a suitable, simple isomorphic graph that can be easily solved! Before getting down to some brainstorming, I just want to know, has anyone of you (or somebody you know) considered this problem before. Are you aware of a solution to the above problem??? Please post what you think and your possible solutions!! \r\n\r\nOne possible graph of the original circuit , with arrows depicting directions of currents,(of course the actual directions could be different!) , is \r\n\r\n[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=358[/img]", "Solution_1": "Would this be better in the Combinatorics Section ???? There isn't much 'Physics' in my question. Can the moderator kindly move the topic! :lol:", "Solution_2": "ok sinec i became a mod now ...so i am moving this", "Solution_3": "Index vertices of the bottom cube face 0, 1, 2, 3 counter-clockwise and vertices of the top cube face 4, 5, 6, 7 clockwise, the vertices 3, 4 being connected by a cube edge. Ground vertex 0 and connect grounded DC voltage supply $ V$ to one of the remaining vertices. Vertices 1, 3, 7 are equivalent, vertices 2, 4, 6 are equivalent, it is sufficient to test vertices 1, 2, 5 (on a cube edge, face diagonal, and body diagonal form vertex 0). Let $ v_1, v_2, ..., v_7$ be voltages at the vertices 1, 2, ..., 7 and $ I$ current supplied by the voltage source. Denote the conductance $ G \\equal{} 1/R.$ The node analysis equation for the circuit is\r\n\r\n\r\n$ \\left( \\begin{array}{cccccccc} 3G & \\minus{} G & 0 & 0 & 0 & \\minus{} G & 0 & 1 \\\\\r\n\\minus{} G & 3G & \\minus{} G & 0 & \\minus{} G & 0 & 0 & 0 \\\\\r\n0 & \\minus{} G & 3G & \\minus{} G & 0 & 0 & 0 & 0 \\\\\r\n0 & 0 & \\minus{} G & 3G & \\minus{} G & 0 & \\minus{} G & 0 \\\\\r\n0 & \\minus{} G & 0 & \\minus{} G & 3G & \\minus{} G & 0 & 0 \\\\\r\n\\minus{} G & 0 & 0 & 0 & \\minus{} G & 3G & \\minus{} G & 0 \\\\\r\n0 & 0 & 0 & \\minus{} G & 0 & \\minus{} G & 3G & 0 \\\\\r\n1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\\r\n\\end{array}\\right) \\cdot \\left( \\begin{array}{cccccccc} v_1 \\\\\r\nv_2 \\\\\r\nv_3 \\\\\r\nv_4 \\\\\r\nv_5 \\\\\r\nv_6 \\\\\r\nv_7 \\\\\r\n\\minus{} I \\\\\r\n\\end{array}\\right) \\equal{} \\left( \\begin{array}{cccccccc} 0 \\\\\r\n0 \\\\\r\n0 \\\\\r\n0 \\\\\r\n0 \\\\\r\n0 \\\\\r\n0 \\\\\r\nV \\\\\r\n\\end{array}\\right)$\r\n\r\nThe first 7 diagonal elements $ g_{ii}$ of the conductance matrix $ \\mathbf G$ are sums of conductances connected to the i-the node, while the off-diagonal elements $ g_{ij}, i \\neq j$ are negative conductances connected between the i-th and j-th node. If 2 nodes are not connected, $ g_{ij} \\equal{} 0.$ Each voltage source (one in this case) has assigned an extra column at the right and an extra row at the bottom, filled with $ g_{i8} \\equal{} g_{8i} \\equal{} 1,$ if the positive terminal of the voltage supply is connected to the i-th node, $ g_{i8} \\equal{} g_{8i} \\equal{} \\minus{} 1,$ if the negative terminal of the voltage supply is connected to the i-th node (none in this case), and $ g_{i8} \\equal{} g_{8i} \\equal{} 0$ otherwise. Solving for the current $ I$ by Cramer rule,\r\n\r\n$ I \\equal{} \\minus{} \\frac {\\det \\mathbf G_I}{\\det \\mathbf G} \\equal{} \\frac {384}{224} VG \\equal{} \\frac {12}{7}\\ \\frac {V}{R} \\equal{} \\frac {V}{R_{01}},\\ \\ R_{01} \\equal{} \\frac {7}{12} R.$\r\n\r\nTo calculate $ R_{02}, R_{05},$ just move the ones in the 8-th column and 8-th row of the conductance matrix $ \\mathbf G$ from the 1st to the 2nd resp. 5th position. In all cases, $ \\det \\mathbf G_I \\equal{} 384 VG^7$ while $ \\det \\mathbf G \\equal{} \\minus{} 288 G^6, \\minus{} 320 G^6,$ when connecting the voltage supply to the 2nd, 5th node, respectively, resulting in $ R_{02} \\equal{} \\frac {3}{4} R, R_{05} \\equal{} \\frac {5}{6} R.$ The node analysis will work for any resistance / capacitance network, the cube symmetry and resistor equality just make the determinants simpler and easier to calculate. Would someone move this thread back to physics ?", "Solution_4": "Sorry for bumping this about 9 years later, but do we have a closed form for the answer of the following problem (seems I saw it somewhere but I can't find it):\n\nLet a set of vertices of a connected graph be given and each edge corresponds to $1 \\Omega$. If the set of vertices is fixed find the expected value of the equivalent resistance over all connected graphs with those vertices.\n\nI mean the two vertices across which we have to find the resistance is fixed.", "Solution_5": "This is known as Foster's theorem. The sum of all equivalent resistances across each resistor in the graph is $n-1$, where $n$ is the number of vertices. So I guess expected value is $\\frac{n-1}{m}$, where $m$ is the number of resistors.", "Solution_6": "No, that theorem doesn't answer the question. For example, consider the graph for $n-1$ resistors connected in series. The average resistance between two randomly chosen vertices in that graph is $\\frac{n+1}{3}$.\n\nBasically, we're dealing with vertices that aren't connected directly all the time, and they're going to drive the averages up. The set of connected graphs isn't exactly all that nice combinatorially, so I doubt that there will be a closed form for the average over all of them.", "Solution_7": "Interestingly, this is the exact same question on this year's Sir Isaac Newton Physics Contest from Waterloo." } { "Tag": [], "Problem": "Please see [url=http://www.cyneer.com/phpBB3/viewtopic.php?f=35&t=15]here[/url]\r\n\r\nTo start this up, I will need the help of a banker and 2 college deans, one for C.I.T. (city instute of technology) and one for G.E.C. (General Education College).\r\n\r\nPlease sign up for forum and post on here:\r\n\r\nName on cyneer:\r\nPosition trying out for:\r\nWhy you are resposible enough:\r\n\r\n[color=red][size=200]PLEASE IGNORE[/size][/color]", "Solution_1": "STOP making these stupid city games! The planet one was creative, and revamped and our town were both dumb! STOP! :mad: :mad: :mad:", "Solution_2": "lol\r\n\r\nThis time though, it is not congested to the space of one thread.\r\n\r\nPlus the design of the other ones could use some... work.\r\n\r\n :|\r\n\r\nEdit:\r\nI also thought:\r\nThere will be not spam in the way of our precious eyes if it is on a different forum, now is there?", "Solution_3": "Note: CIT=Caltech Institute of Technology" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all pairs $(x,y)$ of positive integers such that $\\frac{x^{2}+y^{2}}{x-y}$ is an integer and divides 1995.", "Solution_1": "Well, a little bit of synthetic division tells us that\r\n\r\n$\\frac{x^{2}+y^{2}}{x-y}= 1+y+\\frac{2y^{2}}{x-y}$\r\n\r\nBecause we can choose $x$ so that $x-y$ is any divisor of $2y^{2}$, it follows that $\\frac{2y^{2}}{x-y}$ can be any divisor of $2y^{2}$.\r\n\r\nNow, $1995 = 3 \\cdot 5 \\cdot 7 \\cdot 19$. We can set our expression equal to every factor of $1995$ in turn, all of which should yield a solution. For example, letting $x = 2y^{2}+y$ we have for a given divisor $d$ of $1995$ the solution\r\n\r\n$\\left( 2(d-2)^{2}+(d-2), d-2 \\right)$\r\n\r\nThere are many others that I don't really want to list out :|" } { "Tag": [ "inequalities", "geometry", "circumcircle", "geometry solved" ], "Problem": "$ABC$ a triangle ( all angles $< 90$) and $P$ a point inside the triangle. Denote by $L$, $M$, $N$ the orthogonal projections of $P$ on the sides $BC$, $CA$, $AB$ of triangle $ABC$.\r\n\r\nDetermine the place of $P$ for which $BL^2+CM^2+AN^2$ is minimal.\r\n\r\n\r\n\r\n((PLEAZ SEE ATTACHMENT))", "Solution_1": "It is well known (Carnot's relation) that\r\n\\[\r\nBL^2+CM^2+AN^2=LC^2+MA^2+NB^2.\r\n\\]\r\nTherefore, in order to minimize the sum\r\n\\[BL^2+CM^2+AN^2+LC^2+MA^2+NB^2\r\n\\]\r\nobserve that\r\n\\[BL^2+LC^2\\ge \\frac12 \\left( BL+LC \\right)^2 =\\frac12 BC^2 \\]\r\nwith equality iff $BL=LC$.\r\nUsing the similar inequalities, we deduce that the minimal value is reached when $P$ is the circumcenter of the triangle." } { "Tag": [ "inequalities", "Euler", "geometry", "integration", "calculus" ], "Problem": "This should be easy:\r\nProve 1+1/2+1/3+1/4+1/5...+1/1278576<20", "Solution_1": "Actually... It isn't really easy.\r\n\r\nI guess you have to know the formula which is ln(x)+0.5773 or the Euler-Mascheroni number.\r\n\r\nln(1278576)+0.5773=14.63.\r\n\r\nI haven't tried it yet, but someone can try 1+(1/2+1/2)+(1/4+1/4+1/4+1/4)+(1/8....)+.... and see if it is under 20.", "Solution_2": "[quote=\"Ragingg\"]Actually... It isn't really easy.\n\nI guess you have to know the formula which is ln(x)+0.5773 or the Euler-Mascheroni number.\n\nln(1278576)+0.5773=14.63.\n\nI haven't tried it yet, but someone can try 1+(1/2+1/2)+(1/4+1/4+1/4+1/4)+(1/8....)+.... and see if it is under 20.[/quote]\r\n\r\nln? Just group it into 1, 1/2+1/3+1/4(teeny bit more), 1/5+1/6+1/7+1/8+1/9+1/10(less), 1/9+1/10+1/11...+1/18(less) and so on.", "Solution_3": "? I don't really understand your reasoning.\r\n\r\nln=natural log or log base e.", "Solution_4": "i hate to break it to you but i dont think ur formula is right...\r\n\r\nlets take for example 1+1/2+1/3+1/4+1/5=2.2833333333333333...\r\nand ln(5)+0.5773=1.609437912+0.5773=2.186737912...", "Solution_5": "[quote=\"bob123\"]i hate to break it to you but i dont think ur formula is right...\n\nlets take for example 1+1/2+1/3+1/4+1/5=2.2833333333333333...\nand ln(5)+0.5773=1.609437912+0.5773=2.186737912...[/quote]\r\n\r\nIt's correct as n-> :inf:", "Solution_6": "[quote=\"Ragingg\"]Actually... It isn't really easy.\n\nI guess you have to know the formula which is ln(x)+0.5773 or the Euler-Mascheroni number.\n\nln(1278576)+0.5773=14.63.\n\nI haven't tried it yet, but someone can try 1+(1/2+1/2)+(1/4+1/4+1/4+1/4)+(1/8....)+.... and see if it is under 20.[/quote]\n\nYou must prove the forumla you speak of... merely saying \"you have to know it\" doesn't cut it in a proof... you also cannot make the assumption that we are at infinity when in truth 1278576 is nowhere close to infinity therefore you're method does not hold...\n\nAttempt to prove it without using a calculator to calculate ln(blah)... there is a way.\n\n\n[5th edit] As a matter of fact, using the ln(n) approx to prove this is above the level of this forum for sure. If anyone wants, I will post a ln(n) proof so you can see what I'm talking about... The other method of adding 1 + 1/2 + 1/3 +1/3 + 1/5 + 1/5 + 1/5 + 1/5 unfortunately will show the upper bound, however adding these up all the way to the requested value is almost impossible to do by hand, it would be tedious... I think the scope of this proof is well out of the reaches of this forum.\n\n\n[4th edit]\n[quote=\"Scrambled\"]ln? Just group it into 1, 1/2+1/3+1/4(teeny bit more), 1/5+1/6+1/7+1/8+1/9+1/10(less), 1/9+1/10+1/11...+1/18(less) and so on.[/quote]\r\n\r\nThis is not a proof in any way, and I'm not quite sure I see where you are going with this at all... Could you please explain?", "Solution_7": "Scrambled wrote:ln? Just group it into 1, 1/2+1/3+1/4(teeny bit more), 1/5+1/6+1/7+1/8+1/9+1/10(less), 1/9+1/10+1/11...+1/18(less) and so on.\n\nThis is not a proof in any way, and I'm not quite sure I see where you are going with this at all... Could you please explain?[/quote]\n\n\n\n[hide]unless i am quite mistaken: add up all the terms from 1 to 1/18. How far is that from 4? Should be about .23 smaller\n\nWe'll assume that 1/x+1/x+1...+1/2x which in all cases is smaller than 1 except the 2nd one. so we assume all are one. this turns out to be 1/1048576. .23 = 23/100 which is approximatley equal to 23000/1048576. So it is approximately 23000/1000000 which gives about that much leeway. Ad 23000 to the bottom[/hide]\n\n\n\nmaybe i seriously screwed up", "Solution_8": "well I just don't see what you're trying to convey... what do you mean by a bit less and a tiny bit more?", "Solution_9": "[quote=\"Spoon\"]well I just don't see what you're trying to convey... what do you mean by a bit less and a tiny bit more?[/quote]\r\n\r\nsorry i meant less than 1, more than 1...", "Solution_10": "you would be adding groups for a while if you did it that way..", "Solution_11": "I will post my proof (using the ln method):\r\n\r\nFirst, I will show that [tex]\\displaystyle{1+\\frac{1}{2}+\\ldots +\\frac{1}{n} < \\ln(n) + 1}[/tex].\r\n\r\nThe area under the curve of [tex]\\displaystyle{\\int_1^{n} \\frac{1}{x}\\; dx = \\ln(n)}[/tex]. If we want to approximate this area, we could use the right side approximation, with width of one. The right side approximation of [tex]\\displaystyle{\\int_1^\\infty \\frac{1}{x}\\; dx}[/tex] will be less than the exact value, because it's monotonically deacreasing... therefore the right side sum with width of 1 would be [tex]\\displaystyle{\\sum_{k=2}^n \\frac{1}{k}}[/tex].\r\n\r\nSo [tex]\\displaystyle{\\frac{1}{2}+\\ldots +\\frac{1}{n} < \\ln(n)}[/tex]. We simply have to add 1 to both sides and the conjecture is complete.\r\n\r\nNow we must prove that [tex]\\ln(1278576) < 19[/tex].\r\n\r\nFactor 1278576, which can be done by hand and you get: [tex]\\ln(2^4\\cdot 3^2\\cdot 13\\cdot 683) < 19[/tex].\r\n\r\n[tex]3^2<13<2^4>d^{2}$\r\n\r\nwell the first one is obvious but is the second one valid(yes it's true i know for all books can't be wrong at a time :wink: )\r\nbut what's the justification...since the plate itself is infinite is this a valid assumption?", "Solution_1": "a) The surface charge density induced on the plane must be finite, but the surface is infinite;\r\nb) The field generated by the induced charges will decay as $z$ increases.\r\n\r\nThese two facts should be sufficient to support your second claim. (They are sufficient to me, at least ;)). Btw, what is the first uniqueness theorem?", "Solution_2": "yse that's what i wanted to know whether this was an approximation(well almost everything is :wink: )or does it have something to do with things i don't know :( \r\nso anyways it's intutively obvious\r\n\r\n\r\n[b]Uniqueness theorem[/b]:The solution to laplace equation is uniquely specified in a given volume if it's value is given everywhere in the boundary.", "Solution_3": "[quote=\"Djole\"]a) The surface charge density induced on the plane must be finite, but the surface is infinite;\nb) The field generated by the induced charges will decay as $z$ increases.\n\n[/quote]\r\nstill the plate is also infinitely long then how can one argue that the potential will be approximately 0", "Solution_4": "The potential is constant everywhere, and since the potential is defined up to an additive constant, you may always choose this constant so that the potential is zero.", "Solution_5": "sorry i didn't get that can u be a bit more clear :maybe:", "Solution_6": "I meant that every point on the metal plane has the same potential. Now, the potential is any function $\\varphi$ which satisfies $\\nabla \\varphi =-\\vec E$, so if $\\varphi$ satisfies this, $\\varphi+C$ also satisfies it. \r\n\r\nBut looking back, I'm not sure which potential you're talking about -- I was talking about the potential of the plane -- which one did you mean?", "Solution_7": "no i was talking about the potential about a point thatis infinitely away from the point charge but may be close to the infinite sheet", "Solution_8": "It will also have a zero potential, because there is a somewhat lengthy argument proving that the potential of the uncharged infinite plane be zero at each of its points, or something like that...", "Solution_9": "Actually, the plane is charged with $-q$. Anyhow, it doesn't matter: $V = 0$ is only a reference; the physical importance lies in $\\Delta V$.", "Solution_10": "yes immanuel is right when he says that the plate will have charge $-q$....\r\ni thinka n intutive argument(if someone can give a better arguement please do give :) ) that at far off distanves from the charge the induced charges are negligible hence potential duw to them is also..." } { "Tag": [ "Princeton", "college" ], "Problem": "Did princeton review change their AP books' cover art for the 06-07 editions?", "Solution_1": "Uh, yes they did. I can't imagine why you would want to know though..." } { "Tag": [ "percent", "geometry", "geometric transformation", "reflection" ], "Problem": "Each time a sound wave bounces off a reflector it loses 20$ \\%$ of its strength. What percent of its original strength has it lost after three reflections? Express your answer as a decimal percent.", "Solution_1": "After each reflection, it has $ .8$ of its strength left, so after three, it has $ (0.8)^3\\equal{}.512$ of its strength left. As a percent, it has lost $ 100\\% \\minus{}51.2 \\%\\equal{}\\boxed{ 48.8\\% }$", "Solution_2": "we have $ (.8)^3$, or $ .512. \n100(1-.512)=100\\cdot.488$, or $ \\boxed{48.8}$\r\n\r\nEDIT: beaten." } { "Tag": [ "trigonometry", "conics", "hyperbola", "geometry", "circumcircle", "perpendicular bisector" ], "Problem": "Prove that if the inscribed quadrilateral with perpendicular diagonals is also a circumscribed one, then it is symmetric with respect to one of its diagonals.\r\n\r\nIt's easy with trig, but i haven't find a geometric proof.", "Solution_1": "If I'm not mistaken (which is possible), there are too many conditions here.\r\n\r\nSee the attached diagram. Assume $ABCD$ admits an incircle. Then $AB+CD=AD+BC$. If we denote that sum by $K$, we can denote the sides as shown. But then $BA-BC=a+b-K=DA-DC$. Assuming $a+b-K\\neq 0$, that means that $B$ and $D$ lie on a hyperbola whose foci are $A$ and $C$, hence $BD$ is a hyperbolic chord which is perpendicular to hyperbola's focal axis, and that, because of the geometric properties of a hyperbola, means that $B$ and $D$ are symmetric with respect to $AC$.\r\n\r\nIf on the other hand $a+b-K=0$, then $b=K-a\\iff BA=BC$ and $a=K-b\\iff DA=DC$, hence both $B$ and $D$ lie on the perpendicular bisector of $AC$, which means that $BD$ is the symmetrial axis of $ABCD$.\r\n\r\nHence it's proven that a kite is the only type of quadrilateral which admits an incircle and has perpendicular diagonals.\r\n\r\nNow, a kite can also admit a circumcircle if two of its non-adjacent angles are right (but that's just a special case of the previous statement).", "Solution_2": "Another proof, for those who are not fluent in conics, relies on Pythagorean theorem.\r\n\r\nLet $O$ be the intersection of the diagonals. Denote $x=OA, y=OC, p=OB, q=OD$. Then, assuming $ABCD$ admits an incircle, we have\r\n\r\n$\\sqrt{x^{2}+p^{2}}+\\sqrt{y^{2}+q^{2}}=\\sqrt{x^{2}+q^{2}}+\\sqrt{y^{2}+p^{2}}$\r\n\r\nSquare it, cancel the equal terms, square it again, expand, cancel the equal terms once more to get\r\n\r\n$x^{2}q^{2}+y^{2}p^{2}=x^{2}p^{2}+y^{2}q^{2}$\r\n\r\nwhich is equivalent to $(x^{2}-y^{2})(p^{2}-q^{2})=0$\r\n\r\nTherefore $x=y$ or $p=q$, which means that one of the diagonals is the symmetrial axis of the quadrilateral." } { "Tag": [ "advanced fields", "advanced fields theorems" ], "Problem": "The diagonal of the Cartesian product $ X \\times X$ is the set $ \\Delta\\equal{}\\{(x,x): x \\in X \\}$.\r\nEvery set $ V \\subset X \\times X$ that contains $ \\Delta$ and satisfies the condition $ V\\equal{}V^{\\minus{}1}$ is called an entourage of the diagonal. Let $ D_X$ be the family of all entourage of the diagonal $ \\Delta \\subset X \\times X$.\r\nA uniformity on a set $ X$ is a subfamily $ U$ of $ D_X$ which satisfies the following conditions:\r\n(1) if $ V \\in U$ and $ V \\subset W \\in D_X$ then $ W \\in U$;\r\n(2) if $ V_1,V_2 \\in U$, then $ V_1 \\cap V_2 \\in U$;\r\n(3) For every $ V \\in U$ there exists a $ W \\in U$ such that $ W \\circ W \\subset V$;\r\n(4) $ \\bigcap U\\equal{}\\Delta.$\r\n\r\nI don't understand why from (3) follows that for every $ V \\in U$ there exists $ W \\in U$ such that $ W \\circ W \\circ W \\subset V$.", "Solution_1": "I think that applying (3) twice, we get \r\n$ W \\subset W \\circ W \\subset Z \\subset Z \\circ Z \\subset V$.\r\nThen $ (W \\circ W) \\circ W \\subset Z \\circ Z \\subset V$." } { "Tag": [ "limit", "calculus", "inequalities" ], "Problem": "Se considera ecuatia :$x^n-nx^{n-1}+p=0,p \\in (0,1)$.\r\nAratati ca oricare ar fi $n \\in 2N+1, n \\ge 3$ ecuatia are 3 radacini reale $a_n,b_n,c_n.$ si:\r\na)$a_n<00$).\r\nFor $x>0$ we have $f'(x)=(x^n-nx^{n-1}+p)'$ has a unique root and $f(0)>0$, $f(1)<0$. It follows the equation has exactly two positive roots and $b_n\\in (0,1)$ and $a_n>1$.\r\nBut, in such case, it is clear that $\\lim \\sqrt[n]{b_n+1}=1$.", "Solution_2": "[quote=\"Myth\"]Hmm... There is something wrong...\nWe have $x^n-nx^{n-1}+p$ is increasing for $x<0$. So there is a unique negative root (since $p>0$).\nFor $x>0$ we have $f'(x)=(x^n-nx^{n-1}+p)'$ has a unique root and $f(0)>0$, $f(1)<0$. It follows the equation has exactly two positive roots and $b_n\\in (0,1)$ and $a_n>1$.\nBut, in such case, it is clear that $\\lim \\sqrt[n]{b_n+1}=1$.[/quote]\r\n\r\nIf I understood, we wish to solve:\r\n\r\n$g(x) = x^{n-1} (n - x) = p > 0$.\r\n\r\nThere is one negative root, and two positive roots in $[0,n]$. It is easy using calculus or AM-GM inequality to see that $g(x)$ has a maximum at $x = n-1$, where it is quite large.\r\nSo one positive solution will be very close to $n$ and the other is close to 1. \r\nIf the solution near $1$ is $1 - \\alpha/(n-1)$ we need $e^\\alpha \\sim (n-1)/p$, i.e.\r\n$b_n = 1 - \\frac{log(n-1)-log(p)}{n-1} + O(1/n^2)$.", "Solution_3": "Please, fix your reply ;)", "Solution_4": "[quote=\"Myth\"]Please, fix your reply ;)[/quote]\r\n\r\nTeX or mathematics? I fixed the TeX...", "Solution_5": "I said \"something wrong\" in my reply, because conclusion $\\lim ...=1$ is quite meaningless. Why we need to prove it, if we know that $b_n\\in[0,1]$?", "Solution_6": "I see that you are learning Romanian Mikhail :) :D", "Solution_7": "[quote=\"Myth\"] $\\lim ...=1$ is quite meaningless. Why we need to prove it, if we know that $b_n\\in[0,1]$?[/quote]\r\n\r\nYes, but one can determine asymptotics of the roots regardless of any wrong or un-natural statements in the original problem.", "Solution_8": "$f\\prime=nx^{(n-2)}(x-n+1)$\r\n obtinem din tabloul de variatie urmatoarele \r\n$a_n<0 , 0n-1$\r\n$10$ and any $c\\in\\mathbb R$. \r\n\r\nHint #2: it suffices to prove that $f(x)+\\epsilon x^{2}$ is convex for any $\\epsilon>0$.", "Solution_2": "[quote=\"mlok\"]I think you meant $g_{s}(t)\\le f(t)$ for $t\\in\\mathbb R$. Think on the graph of $g_{s}$ as a \"generalized tangent\" to the graph of $f$ at $s$. [/quote]You're right, it's been fixed." } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "Hi,\r\n\r\nCan anybody direct me to the right path for the 3rd problem from the following problem set:\r\nhttp://www.math.ubc.ca/~rfroese/math267/hw5.pdf\r\n\r\nI got the Fourier Series for g(x) in complex exp. form, but do not really understand the rest. e.g. \"then expanding y(t) in a series with unknown coefficients and substituting it in the equation, and then determining the values of the coefficients that make the equation hold.\" what is this sentence saying? Like I find the characteristic eq. and then find its roots to get y(t)?\r\n\r\nThanks in advance.", "Solution_1": "You are supposed to write $y(t)=\\sum_k y_k e^{\\pi ikt}$ where $y_k\\in\\mathbb C$. Then you are supposed to formally differentiate the series with respect to $t$ and to write the resulting expression for the left hand side. In your case you'll get $\\sum_k (-\\pi^2 k^2+i\\alpha\\pi k+\\beta)y_k e^{\\pi ikt}$. This expansion should be the same as the Fourier expansion $g(t)=\\sum_k g_k e^{\\pi ikt}$ for the right hand side you found. It remains to equate the coefficients at each exponent $e^{\\pi ikt}$ to find $y_k$. I'll let you continue from here.", "Solution_2": "Right, thanks! I found g_k to be 2/(i*k*pi) with k being any real odd numbers. Now I am having difficulties equating this g_k with y_k (I always had troubles with that). Could you possibly provide a quick hint or something?\r\n\r\nBy the way your response totally clarified this problem for me. Much appreciations.", "Solution_3": "All you need to do is to solve $(-\\pi^2 k^2+i\\alpha\\pi k+\\beta)y_k =g_k$ for $y_k$.", "Solution_4": "Thanks. I was thinking too complicating I guess." } { "Tag": [], "Problem": "This is easy:\r\n\r\nCan we label each vertex of a 45-gon with one of the digits {0, 1, ..., 9}, \r\nso that for each pair of distinct digits i, j one of the 45 sides has vertices labeled i, j?", "Solution_1": "Hrmm... this may not be very clear, but here goes.\r\n\r\nThe trick is for every number to be next to every other number. Since each vertice is next to two other vertices, if there are n vertices with a label i, then the number i can be next to 2n different numbers. Since there are nine numbers other than i, we would have to have 5 of each number (so that 2n=10>9). But that would be a total of 50 vertices, so no.\r\n\r\n.dzhonatan", "Solution_2": "hmz\r\n\r\nI think mine is easier (or if it is what you wrote, at least more clear):\r\n\r\n10 numbers on 45 corners\r\nso some numbers will be on at least 5 places.\r\nso those numbers need 10 different numbers as neighbours\r\n\r\nbut there are only 9 differents for each number \r\nso solved", "Solution_3": "No, you aren't solving the problem...\r\n\r\nthere's nothing in the problem that says that you can't have the same two vertices next to each other twice, which is what you are saying.\r\n\r\n.dzhonatan", "Solution_4": "there are 45 pairs possible and you need 45 different ones.\r\n\r\nI don't think I have to explain why there cannot be double ones then? :?", "Solution_5": "Ok, that works then.\r\n\r\nBut you didn't say that in your first post.\r\n\r\n.dzhonatan", "Solution_6": "ya I know I thought that was too obvious, the art is to write them as short as possible :)\r\n\r\nSorry." } { "Tag": [ "algebra", "polynomial", "combinatorics unsolved", "combinatorics" ], "Problem": "My book isnt exactly student friendly \r\n\r\nbut this is what i came up with after solving the problem 3 times\r\n\r\nthe problem is $\\ ( 1+x+x^{2}\\ ) \\ ( 1+x \\ )^{n}$ find coefficient of $x^{k}$\r\n\r\ni am looking in for $x^{k-n}$ in $\\ (1+x+x^{2}\\ )$\r\n\r\n$\\ (1+x+x^{2}\\ ) = \\frac{1-x^{2}}{1-x}= \\ ( 1-x^{2}\\ ) \\ (1-x \\ )^{-1}$\r\n\r\ncoefficient of $x^{k-n}$ would then be $\\binom{1+k-n-1}{k-n}= \\binom{k-n}{k-n}= 1$\r\n\r\nand i dunno if i'm right or not :(\r\n\r\ncan someone explain how to do this", "Solution_1": "[quote=\"Lena\"]My book isnt exactly student friendly \n\nbut this is what i came up with after solving the problem 3 times\n\nthe problem is $\\ ( 1+x+x^{2}\\ ) \\ ( 1+x \\ )^{n}$ find coefficient of $x^{k}$\n\ni am looking in for $x^{k-n}$ in $\\ (1+x+x^{2}\\ )$\n\n$\\ (1+x+x^{2}\\ ) = \\frac{1-x^{2}}{1-x}= \\ ( 1-x^{2}\\ ) \\ (1-x \\ )^{-1}$\n\ncoefficient of $x^{k-n}$ would then be $\\binom{1+k-n-1}{k-n}= \\binom{k-n}{k-n}= 1$\n\nand i dunno if i'm right or not :(\n\ncan someone explain how to do this[/quote]\r\n\r\nOne sloppy error putting $1+x+x^{2}$ into another form. More importantly, $1+x+x^{2}$ [i]is a polynomial in standard form[/i]: you know the coefficients of every term already. The coefficient of $x^{0}, x^{1}$ and $x^{2}$ is 1, and all other coefficients are 0. The other factor is also a polynomial. I have no idea where $x^{k-n}$ came from -- either you've mistyped or you have misread.", "Solution_2": "well i dont exactly understand it yet\r\n\r\nanother try is to open the brackets\r\n\r\nmainly write something like $\\ ( 1+x \\ )^{n}+x \\ (1+x \\ )^{n}+x^{2}\\ ( 1+x \\ )^{n}$", "Solution_3": "And then?", "Solution_4": "$\\binom{n}{k}+\\binom{n+k+1-1}{k}+\\binom{n+k+2-1}{k}$\r\n\r\n$\\binom{n+k+1-1}{k}+\\binom{n+k+2-1}{k}$ this part i am most definitely confusing\r\n\r\ni'm not sure if to add 1 or subtract 1", "Solution_5": "Why don't you explicitly state what you're trying to do? Once you know exactly what it is that you want, then you can jump on the formulas that do it.", "Solution_6": "I want to find the coefficient of $x^{k}$ in the phrase $\\ ( 1+x+x^{2}\\ ) \\ ( 1+x \\ )^{n}$\r\n\r\nwhich means i want to find the coefficient of $x^{k}$ in $\\ ( 1+x \\ )^{n}+x \\ (1+x \\ )^{n}+x^{2}\\ ( 1+x \\ )^{n}$\r\n\r\nmeaning find the coefficient of $x^{k}$ in $\\ ( 1+x \\ )^{n}$ and add to that the coefficient of $x^{k-1}$ in $x \\ (1+x \\ )^{n}$ and add to that the coefficient of $x^{k-2}$ in $x^{2}\\ ( 1+x \\ )^{n}$\r\n\r\nso the answer would be $\\binom{n}{k}+\\binom{n+k-1-1}{k-1}+\\binom{n+k-2-1}{k-2}$", "Solution_7": "Just as a note for future: You'll get more responses for such problems in the Intermediate forum. See if your formula works by testing a couple of values. Also, consider the numbers you add to powers by multiplying the binomial by $1,x,x^{2}$.", "Solution_8": "[gee i've been going about this all wrong\r\n\r\nthe coefficient of $x^{k}$ in $\\ ( 1+x \\ )^{n}$ is $\\binom{n}{k}$ \r\nthe coefficient of $x^{k}$ in $x \\ (1+x \\ )^{n}$ is the same as $x^{k-1}$ in $\\ ( 1+x \\ )^{n}$ \r\nthe coefficient of $x^{k}$ in $x^{2}\\ ( 1+x \\ )^{n}$ is the same as the coefficient of $x^{k-2}$ in $\\ ( 1+x \\ )^{n}$\r\nso the correct answer would be \r\n\r\n$\\binom{n}{k}+\\binom{n}{k-1}+\\binom{n}{k-2}$", "Solution_9": "Yes. (You actually essentially had it last time, but you were too eager to use a formula which didn't apply.)", "Solution_10": "thx JBL \r\n\r\nmainly for not giving me the answer\r\n\r\nonce i understood where i had to derive the coefficient from, it was easy to solve\r\n\r\nas i said, my book isnt student friendly, it didnt explain that in so many words... i read almost every page of the book betweeen each and every letter" } { "Tag": [ "trigonometry", "geometry", "algebra open", "algebra" ], "Problem": "We have\r\n1) 0 < a, b, c, x, y, z < pi;\r\n2) a + b + c + x + y + z = pi;\r\n3) sin a * sin b * sin c = sin x * sin y * sin z;\r\n4) sin (y/2) * sin (c/2 + z) = sin (c/2) * sin (y/2 + b);\r\n5) sin (z/2) * sin (a/2 + x) = sin (a/2) * sin (z/2 + c).\r\nProve that sin (x/2) * sin (b/2 + y) = sin (b/2) * sin (x/2 + a). :D", "Solution_1": "Try to use this geometrical interpretation:\r\nTake the triangle ABC with the chevians AX, BY, CZ such, that the angle BAX = a; the angle CBY = b; the angle ACZ = c; the angle CAX = x; the angle ABY = y; the angle BCZ = z.", "Solution_2": "Can anyone solve it???", "Solution_3": "But it is rather nice problem" } { "Tag": [ "integration", "function", "inequalities", "calculus", "derivative", "vector", "topology" ], "Problem": "Consider the functional $F(u)=\\dfrac12\\int\\limits_a^b u'^2(x)dx-\\dfrac1p\\int\\limits_a^b|u(x)|^pdx-\\int\\limits_a^bh(x)u(x)dx$ with $\\mathrm{dom}\\, F=W: =C^1[a,b]$, $p>2$.\r\n\r\nDenote\r\n$f(t,s): =F(tu+sv)$ ($\\|u\\|=\\|v\\|=1$ and $\\left\\langle u,v\\right\\rangle=0$ where $\\left\\langle u,v\\right\\rangle=\\int\\limits_a^bu'v'dx$ and $\\|u\\|=\\sqrt{\\left\\langle u,u\\right\\rangle}$) and\r\n$\\hat F(u,v): =\\max\\limits_{t,s\\ge0}f(t,s)$.\r\n\r\nIs is not hard to prove that $\\hat F(u,v)$ exists, and $\\hat F(u,v)>0$ implies $t=t(u,v)$ and $s=s(u,v)$ are both strictly positive functions.\r\n\r\nProve that $\\hat F(u,v)>0$ implies that $t=t(u,v)$ and $s=s(u,v)$ are both $C^1$.\r\n\r\n[b]P.S.[/b] My idea is as follows. As $f(t(u,v),s(u,v))=\\hat F(u,v)$, it is necessary that\r\n\r\n$\\frac{\\partial f}{\\partial t}(t(u,v),s(u,v))=0$,\r\n$\\frac{\\partial f}{\\partial s}(t(u,v),s(u,v))=0$.\r\n\r\nSo if denoting $A$ the Hessian of $f$ at $t=t(u,v),s=s(u,v)$, we prove that $\\det A\\neq0$, we will be able to apply Implicit Function Theorem and thus prove smoothness of $t$ and $s$.\r\n\r\nThe trouble is that I cannot prove that $\\det A\\neq0$; maybe you offer another approach.", "Solution_1": "[quote=\"Vladimir Lubyshev\"]Consider the functional $F(u)=\\dfrac12\\int\\limits_a^b u'^2(x)dx-\\dfrac1p\\int\\limits_a^b|u(x)|^pdx-\\int\\limits_a^bh(x)u(x)dx$ with $\\mathrm{dom}\\, F=W: =C^1[a,b]$, $p>2$.\n\nDenote\n$f(t,s): =F(tu,sv)$ [/quote]\r\nWait a minute; $F$ had only one argument just a moment ago :?. And what is $h(x)$? Is it also smooth or just integrable?", "Solution_2": "Oh, I am sorry. I have corrected the mistake, here is the right version: $f(t,s): =F(tu+sv)$. Suppose $h$ to be \"good enought\", but it would be better if we consider more general case for $h$ (just integrable).", "Solution_3": "[quote=\"Vladimir Lubyshev\"]\nIs is not hard to prove that $\\hat F(u,v)$ exists and $\\hat F(u,v)>0$ implies $t=t(u,v)$ and $v=v(u,v)$ are both strictly positive functions.\n[/quote]\r\nNow, I have very serious doubts about this statement. Indeed, since $\\|u\\|=\\|v\\|=1$ and $u$ and $v$ are orthogonal and since $h$ is arbitrary, we can write \r\n\\[ f(t,s)=\\frac 1 2 (t^2+s^2)-\\frac 1 p\\int_a^b |tu+sv|^p - At-Bs \\]\r\nwhere $A$ and $B$ are some constants that may be chosen to be any two real numbers for fixed $u,v$ if we choose $h$ in an appropriate way. Now choose $u$ and $v$ strictly positive, $A=0$ and $B$ big positive. Then, using the trivial inequality $(x+y)^p\\ge x^p+y^p$ for positive $x,y$, we conclude that \r\n\\[ f(t,s)-f(t,0)\\le \\frac{s^2}2-s^p\\frac 1p\\int_a^b v^p- Bs\\,. \\]\r\nIf $B$ is large and the integral is not too small, this difference is strictly negative for positive $s$. On the other hand, it is clear that $f(t,0)>0$ for small positive $t$. So the maximum is attained at a point $(t_*,0)$ for some $t_*>0$. :?", "Solution_4": "Suppose $\\hat F(u,v)>0$. It it obvious that one of the $t,s$ must be strictly positive, let it be $t$. Let's prove that $s$ must be strictly positive as well. Suppose the contrary. It is clear that\r\n\r\n$f(t,s)\\ge f(t,0)+f(0,s)$\r\n\r\nfor $\\ge0$ (in our case = 0). Thus $f(t,t)\\ge2f(t,0)=2\\hat F(u,v)>\\hat F(u,v)$. We have come to a contradiction.\r\n\r\nSo your reasoning contains mistakes, they are embodied in (some of) the following items:\r\n\r\n1) It seems not mandatory that $A=\\int\\limits_a^b hu$ and $B=\\int\\limits_a^b hv$ can be made arbitrary reals due to the restriction $\\langle u,v\\right angle\\rangle=0$ and $\\|u\\|=\\|v\\|=1$.\r\n\r\n2) You have made too many presuppositions: apart from 1), you supposed $\\int\\limits_a^bv^p$ to be not too small. All this can easily turn into the contradiction with our $\\langle u,v\\rangle=0$ and $\\|u\\|=\\|v\\|=1$.\r\n\r\n3) You cannot state that $f(t,0)>0$ for small $t>0$.\r\n\r\n[b]P.S.[/b] Having received the problem, I thought that $\\hat F(u,v)>0$ doesn't imply that $t$ and $s$ are both $>0$ so I asked the professor and he responded it was known that in this case $t$ and $s$ [i]must[/i] be $>0$. It may happen to be wrong in the case $\\hat F(u,v)=0$ only. Yet, it is known that in the case $\\hat F(u,v)=0$ $t$ and $s$ are not mandatory $C^1$. So the condition $\\hat F(u,v)>0$ is crucial. In my proof of strict positiveness of $t$ and $s$, I essentially used that condition: $2\\hat F(u,v)>\\hat F(u,v)$.", "Solution_5": "[quote=\"Vladimir Lubyshev\"]\n\n$f(t,s)\\ge f(t,0)+f(0,s)$\n\n[/quote]\nThis is completely unclear to me and, moreover, obviously false because, under your assumptions it is equivalent to\n\\[ \\int_a^b |tu+sv|^p\\le \\int_a^b|tu|^p+\\int_a^b |sv|^p\\,, \\]\nwhich is very far from being always true...\nAnother mistake you make is the assumption that $f(t,0)+f(0,t)=2f(t,0)$, i.e., that $f(0,t)=f(t,0)$, which doesn't follow from anywhere.\n[quote]So your reasoning contains mistakes [/quote]\nI'm afraid not (unless I misunderstood something in your original post). I looked at all three of your objections and they are all invalid.\n\n1)if $u$ is not a constant multiple of $v$, then, for all $A$ and $B$, we can always find $h$ such that $\\int uh=A$ and $\\int vh=B$.\n\n2) Since adding a constant to $v$ doesn't change $\\|v\\|$ or $\\langle u,v\\rangle$ (both depend only on the derivative of $v$), we can make $\\int|v|^p$ as large as we want.\n\n3) If $A=0$, $f(t,0)=\\frac 1 2 t^2- t^p \\frac 1p\\int_a^b |u|^p=\\frac 1 2 t^2-Const\\, t^p$, which is strictly positive for $t^{p-2}<\\frac{Const}{2}$ \n\n[quote]I asked the professor and he responded it was known that in this case t and s must be >0.[/quote]\r\nHe(she?) is only human and, therefore, may make mistakes too :).", "Solution_6": "Yes, it seems you are right. My \"proof\" contain outrage mistakes! :blush: I have to think on this problem again. Thank you very much. :oops:\r\n\r\nCan you say something on smoothness of $t$ and $s$? :?", "Solution_7": "Now I absolutely confirmed in your correctness! You founded a counter example. So that professor was wrong or I misunderstood something. :D \r\n\r\nI hastened stating my \"objections\" 1) - 3) :wallbash:. As to the first \"objection\", one can even prove that for any two linearly independent vectors $u$ and $v$ of a locally convex space $X$ and any $A,B\\in\\mathbb R$ there exists $x^*\\in X^*$ with $\\left\\langle u,x^*\\right\\rangle=A$ and $\\left\\langle v,x^*\\right\\rangle=B$. So the fact that $\\forall\\,A,B\\in\\mathbb R\\;\\;\\exists\\,h\\in L^p(a,b)$ ($p\\ge2$) with $\\int_a^buh=A$ and $\\int_a^bvh=B$ follows from this general case. Did you mean $h\\in L^p(a,b)$ ($p\\ge2$)?", "Solution_8": "[quote=\"Vladimir Lubyshev\"]\nCan you say something on smoothness of $t$ and $s$? :?[/quote]\r\nI'm afraid we first have to figure out whether they are correctly defined. I mean, since the function $f(t,s)$ is not concave, it seems to me that it is quite feasible that it may have 2 or more different points where the maximum is attained. :roll: Of course, if we assume a priori that $t,s$ are unique and positive, we may try to ask about their smoothness in a neighborhood of the corresponding pair $(u,v)$. But then, since we've piled up so many assumptions already, why not to add the assumption that the Hessian is non-degenerate as well :P ? You see what I'm driving at: this problem is hardly of much interest by itself; I assume that you or your professor need it for something else, in which case it would be wise to check first whether what we figured out already doesn't spell the end of this approach to that \"something else\". :)", "Solution_9": "I should stop here. When saying about existence of $h$ with $\\int_a^b uh=A$ and $\\int_a^b vh=B$, did you mean $h\\in L^p(a,b)$ ($p\\ge2$) or one can even suppose $h\\in C^r([a,b])$?", "Solution_10": "You can even find $h\\in C^\\infty$. :)", "Solution_11": "How to prove it?", "Solution_12": "Just take two points $x,y\\in [a,b]$ such that the vectors $(u(x),u(y))$ and $(v(x), v(y))$ are not proportional to each other and choose an appropriate linear combination of two positive $C^\\infty$ functions one of which is supported by a small meighborhood of $x$ and another by a small neighborhood of $y$. Many other constructions are possible too.", "Solution_13": "What about the following non-constructive approach. $X: =C^1[a,b]$, consider the space $Y$ consisting of linear functionals on $X$ of the form $f\\mapsto\\int_a^b fh$ as $h$ vary over $C^\\infty[a,b]$ then $Y$ separates points of $X$ by virtue of the Dubois Raymond theorem. Now endow $X$ with the topology $\\sigma(X,Y)$, then $X^*=Y$. It remained only to apply the statement about locally convex spaces from one of my previous posts (about existence of $x^*\\in X^*$ with ...).\r\n\r\nAre there any mistakes?", "Solution_14": "That seems fine too. The only catch is that one has to prove that your topology separates points, i.e., that for every $u$ that is not identically $0$, there exists $h\\in C^\\infty$ such that $\\int uh\\ne 0$ and when proving it, you may have to do something very similar to the construction in my post. Otherwise, what you said is perfectly correct. :)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ G= $ { $ a_1, a_2 ,... , a_{100} $ } be a subset of the set {1,2, ... , 200} such that : with $ 1 \\leq i < j \\leq 100 $ , $ \\sum_{i=1}^{100}a_i = 10080 $ , $ a_i + a_j $ is different from 201 \r\n a) Prove that : the number of the odd numbers of G is a multiple of 4 \r\n b) Calculate $ a_1^2 + a_2^2 +... +a_{100}^2 $", "Solution_1": "Nasty computations, I got wrong 4 four times;... :( \r\n\r\nFirst, divide the integers in $\\{1,2, \\cdots, 200 \\}$ by pairs of the form $(a,201-a)$ where $a=1,2 \\cdots, 100$.\r\nSince any two of the $a_i$'s do not add up to $201$ it follows that $G$ contains at most one term from each pair. In another hand, since $|G| = 100$ we then have to choose exactly one term from each pair.\r\nThat means $a \\in G$ if and only if $201-a \\notin G.$ (1)\r\n\r\n1) Let $r$ be the number of odd terms in $G$.\r\nLet $S = 10080 = \\sum_{a_i \\in G} a_i = P+I$, where $P$ is the same sum but only over the even terms and $I$ is only the odd ones (odd = Impair and even = Pair, in french :P ).\r\nNote that there are exactly $r$ even integers which do not appear in $P$.\r\nThus, \r\n$I= \\sum_{200-2a \\notin G} (2a+1) = r - \\sum_{200-2a \\notin G} (200-2a) + 200r = 201r - \\sum_{i=1}^{100} (2i) + P$.\r\nIt follows that \r\n$201r = S -2P + \\sum_{i=1}^{100} (2i) =10080 + 10100 -2P = 0\\mod[4]$\r\nsince $P$ is even as a sum of even integers.\r\nThus $r = 0 \\mod[4]$.\r\n\r\n2) Let $S = S_1 + S_2$ where $S_1$ (resp $S_2$) is the same sum but only over the $a_i$'s which are not greater than 100, and which form the set $G_1$, (resp. not less than 101 and which form $G_2$).\r\nLet $y$ be the number of terms in the sum $S_2$. From (1), we deduce that :\r\n$S_2 = \\sum (201-a_i)$ where the sum is over the $a_i$'s not greater than 100 and which do not appear in $S_1$.\r\nThus $S-S_1 = S_2 = 201y+S_1 - \\sum_{i=1}^{100} i$\r\nso that $201y + 2S_1 = 10080 + \\sum_{i=1}^{100} i = 15130.$ (2)\r\n\r\nTherefore :\r\n$S' = \\sum_{a_i \\in G} a_i^2 = \\sum_{a_i \\in G_1} a_i^2 + \\sum_{a_i \\in G_2} a_i^2 = \\sum_{a_i \\in G_1} a_i^2 + \\sum_{a_i \\notin G_1,a_i \\leq 100} (201-a_i)^2 = \\sum_{i =1}^{100} i^2 + 201(201y+2S_1 - 2\\sum_{i =1}^{100} i )$\r\nUsing (2), we deduce (hope there is not mistake in my computations :? ) that $S' = 1349380.$\r\n\r\nPierre." } { "Tag": [], "Problem": "Let A and B be disjoint sets whose union is the set of natural numbers. Show that for every natural number n there exist distinct $ a,b>n$ such that $ \\{a,,b,a\\plus{}b\\}$ is either a subset of A or B.\r\n\r\n(By the way when I type \\subset it only gives $ \\subset$, which is propert subset; what's the correct code then?)", "Solution_1": "Since $ A\\cup B\\equal{}\\mathbb N$, one of $ A$ and $ B$ contain two elements $ a$ and $ b$ such that $ a>b>n$ and $ a\\minus{}b>n$, for each $ n\\in\\mathbb N$. WLOG let $ a,b\\in A$. Assume the contrary. Then $ a,b\\in A\\Rightarrow a\\plus{}b,a\\minus{}b\\in B\\Rightarrow 2a,2b\\in A\\Rightarrow 2(a\\plus{}b),2(a\\minus{}b)\\in B$. Now $ a,b,2a,2b\\in A\\Rightarrow 2a\\plus{}b,a\\plus{}2b,3a,3b\\in B$. But this implies $ 2(a\\minus{}b)\\in B, a\\plus{}2b\\in B$ and $ 3a\\equal{}2(a\\minus{}b)\\plus{}a\\plus{}2b\\in B$, contradiction. Hence the conclusion.", "Solution_2": "[quote=\"mathwizarddude\"]\n(By the way when I type \\subset it only gives $ \\subset$, which is propert subset; what's the correct code then?)[/quote]\r\n\r\nYou can use subseteq to get $ \\subseteq$. Notice, however, that usage of $ \\subset$ varies - some authors use it to mean \"proper subset\", others use it to mean any subset (proper or not). It's not quite as standardized as $ <$ and $ \\leq$.", "Solution_3": "I think you have a small typo:[quote=\"nayel\"]$ 3a \\equal{} 2(a \\minus{} b) \\plus{} a \\plus{} 2b\\in B$[/quote]should be $ 3a \\equal{} 2(a \\minus{} b) \\plus{} a \\plus{} 2b\\in A$\r\nNice solution!" } { "Tag": [ "real analysis", "real analysis solved" ], "Problem": "We consider the sequences of real numbers (x_n)n>=1 and (s_k)k>=1 which verify the conditions:\r\n1)x_n is in (0,1) for all n>=1;\r\n2)sum(n=1,k) (x_n)^(s_k)=1 for any k>=1;\r\n3)there exists s real, so that lim(k-->oo) sum(n=1,k) (x_n)^s=1.\r\na)Prove that the sequence(s_k)converges\r\nb)prove that lim(s_k)=s.", "Solution_1": "a) \r\n We have\r\n1)\r\n s(k) \r\n 1= x(1)^s(k) +x(2)^s(k) +.....+x(k)^s(k) \r\n \\leq x(1)^s(k+1) +x(2)^s(k+1)+.......+x(k)^s(k+1)\r\n < x(1)^s(k+1)+x(2)^s(k+1)+.........+x(k)^s(k+1) +x(k+1)^s(k+1) =1\r\n Nonsense.\r\n --> x(k+1)>s(k) for all k\r\n \r\n2)\r\nFor all k , s(k) \\leq s\r\n {If s(k)>s ,we deduce \r\n 1= x(1)^s(k) +x(2)^s(k) +.....+x(k)^s(k) \r\n \\leq x(1)^s+x(2)^s+.........+x(k)^s +x(k+1)^s\r\n oo} [x(i)^s] =1 }\r\n \r\nFor {s(k)} is an increase and bounded sequence , there exists finite limit c of {s(k)} \r\n \r\nb) lim s(k)=c --> c \\leq s \r\n Assume coo}[x(n)^c]=lim sum{n-->oo}[x(n)^s] \r\n So \r\n lim sum{n-->oo}[x(n)^c-x(n)^s] =0 (*)\r\n But , for 0 (s-c).ln(1/x(1)).x(1)^s = M\r\n So \r\n lim sum{n-->oo}[x(n)^c - x(n)^s] \\geq M >0 ,contradict with (*)\r\n -->\r\n c=s \r\n (q.e.d)" } { "Tag": [], "Problem": "How many numbers are in the list $ \\minus{}36, \\minus{}29, \\minus{}22, \\ldots, 41, 48?$", "Solution_1": "The list contains the numbers $ \\minus{}36, \\minus{}29, \\minus{}22, \\ldots, 41, 48$\r\n\r\nAdding 36 to every number in the list gives:\r\n\r\n$ 0, 7, 14, \\ldots, 77, 84$\r\n\r\nWe want to count consecutive integers, so we divide each term by 7:\r\n\r\n$ 0, 1, 2, \\ldots, 11, 12$\r\n\r\nThus, there are $ 12\\minus{}0\\plus{}1 \\equal{} 13$ numbers in the list.", "Solution_2": "$ a_n - a_1 = (a + (n - 1)d) - a = (n - 1)d$.\r\nAnd $ a_2 - a_1 = (a + d) - a = d$.\r\nThus,\r\n\\begin{eqnarray*} n & = & \\frac {a_n - a_1}{a_2 - a_1} + 1 \\\\\r\n& = & \\frac {48 - (-36)}{(-29) - (-36)} + 1 \\\\\r\n& = & \\frac {84}{7} + 1 \\\\\r\n& = & 12 + 1 \\\\\r\n& = & \\boxed{13} \\end{eqnarray*}\r\n\r\ncf. http://www.mathlinks.ro/viewtopic.php?t=260977", "Solution_3": "[hide=My solution]Add 43 to every number. 7,14,21...84,91. 91/7=[u]13[/u] numbers." } { "Tag": [], "Problem": "Rays $ BA$ and $ BC$ form an acute angle. Including angle $ ABC$, how many acute angles are formed by the rays in the diagram?\n\n[asy]draw((0,0)--10dir(0),linewidth(1),Arrow);\ndraw((0,0)--10dir(88),linewidth(1),Arrow);\ndraw((0,0)--9dir(15),linewidth(1),Arrow);\ndraw((0,0)--11dir(35),linewidth(1),Arrow);\ndraw((0,0)--7dir(55),linewidth(1),Arrow);\ndraw((0,0)--11dir(70),linewidth(1),Arrow);\n\ndot(8.5dir(88));\ndot((0,0));\ndot(8.5dir(0));\ndot(7dir(15));\ndot(9dir(35));\ndot(5.5dir(55));\ndot(9.5dir(70));\n\n\nlabel(\"A\",8.5dir(88),W);\nlabel(\"B\",(0,0),SW);\nlabel(\"C\",8.5dir(0),S);\nlabel(\"D\",7dir(15),SE);\nlabel(\"E\",9dir(35),SE);\nlabel(\"F\",5.5dir(55),NW);\nlabel(\"G\",9.5dir(70),SE);[/asy]", "Solution_1": "The answer is the fifth triangular number, or 15.", "Solution_2": "We can pick any two rays, and there are 6 total. so 6C2=15.\r\n\r\nIn case you didnt understand how mew got 5th triagnular numb.", "Solution_3": "What about the 90degree angle formed by rays AB and BC? so shouldn't it be 14?", "Solution_4": "[quote=\"JSGandora\"]What about the 90degree angle formed by rays AB and BC? so shouldn't it be 14?[/quote]\n\nThe problem specifically says that the angle ABC is acute.\n\n[quote=\"AIME15\"]\n\nIn case you didnt understand how mew got 5th triagnular numb.[/quote]\r\n\r\nIts 5+4+3+2+1...\r\n\r\n5 little angles\r\n\r\n4 combination of 2 little angles\r\n\r\n3 combinations of 3 little angles\r\n\r\n...\r\n\r\nand so on.", "Solution_5": "[quote=JSGandora]What about the 90degree angle formed by rays AB and BC? so shouldn't it be 14?[/quote]\nDude. Read the question properly\n", "Solution_6": "Rays $BA$ and $BC$ form an acute angle." } { "Tag": [ "trigonometry", "inequalities", "geometry", "inequalities proposed" ], "Problem": "A triangle $ ABC$ is given with side lengths $ a, b$ and $ c$. A point $ P$ lies inside $ ABC$, and the distances from to three sides are $ x, y$ and $ z$ , respectively. Prove that \r\n\r\n$ a\\sin A \\plus{} b\\sin B \\plus{} c\\sin C\\geq9\\sqrt[3]{xyz}$", "Solution_1": "Note that $ 9\\sqrt[3]{xyz}\\leq 3(x\\plus{}y\\plus{}z) \\leq 3(h_a\\plus{}h_b\\plus{}h_c)$\r\nWhere equality occurs if and only if the triangle is equilateral and the point is the fermat point.\r\nTherefore it suffices to show that $ \\sum a \\sin A\\geq 3(h_a\\plus{}h_b\\plus{}h_c)$\r\nBut, from Chebyshev we have that \r\n$ \\sum a \\sin A\\geq \\frac 13 (a\\plus{}b\\plus{}c)(\\sin A\\plus{} \\sin B\\plus{}\\sin C)$\r\n$ \\equal{} \\frac 13 (a\\plus{}b\\plus{}c)\\left(\\frac{a\\plus{}b\\plus{}c}{2R} \\right) \\equal{} \\frac{(a\\plus{}b\\plus{}c)^2}{3abc}\\cdot 2\\Delta$\r\nWhere $ \\Delta$ is the area of the triangle.\r\nSince $ h_a \\equal{} \\frac {2\\Delta }{a}$ therefore it suffices to show that \r\n$ (a\\plus{}b\\plus{}c)^2\\geq 3abc\\cdot \\left(\\frac 1a\\plus{}\\frac 1b\\plus{}\\frac 1c\\right) \\equal{} 3(ab\\plus{}bc\\plus{}ca)$\r\nWhich is perfectly true from the trivial inequality.\r\nEquality occurs if and only if $ P$ is the Fermat Point of the triangle $ ABC$ which is equilateral." } { "Tag": [ "arithmetic series" ], "Problem": "How do I set up to find the sum of all multiples of 12 between 100 and 1000?", "Solution_1": "Use sum of arithmetic series formula. The sum of an arithmetic series with least term a, difference d, and n terms has sum:\r\n\r\n$ \\frac {n}{2}(2a \\plus{} (n \\minus{} 1)d)$\r\n\r\nAn alternate formula would be: leas term a, greatest term b, and has n terms:\r\n\r\n$ \\frac{n}{2}(a\\plus{}b)$\r\n\r\n[hide=\"solution\"]\n\nThe first term is 108, the difference is 12, and there are $ (996\\minus{}108)/12\\plus{}1\\equal{}75$ terms.\n\nSo the sum is:\n\n$ \\frac{75}{2}(216\\plus{}(74)(12))$\n$ \\equal{}41400$[/hide]" } { "Tag": [ "combinatorics solved", "combinatorics", "graph theory", "Tournament of Towns" ], "Problem": "Given a graph with $ 50$ vertices. Show that it has two vertices $ A$ and $ B$ such that the number of vertices joined to both $ A$ and $ B$ is even.", "Solution_1": "[quote=\"Number1\"]Given a graph with $ 50$ vertices. Show that it has two vertices $ A$ and $ B$ such that the number of vertices joined to both $ A$ and $ B$ is even.[/quote]\r\n\r\nMore generally, if n is a positive integer, then any graph S with 2n vertices has two vertices which have an even number of common neighbours.\r\n\r\nThis is problem 14.10 in Arthur Engel's [i]Problem-Solving Strategies[/i]. Here is my solution (more or less the same as Engel's one):\r\n\r\nThe [i]degree[/i] of a vertex in a graph will mean the number of its neighbours, or, equivalently, the number of edges starting at this vertex.\r\n\r\nWe distinguish between two cases:\r\n\r\n[i]Case 1:[/i] Some vertex C of the graph S has an odd degree. This means that it has an odd number of neigbours. Consider the subgraph of S whose vertices are these neighbours (and whose edges are only those edges of S whose both endpoints are neighbours of C). Then, this subgraph has an odd number of vertices; thus, by a well-known theorem (which states that if a graph has an odd number of vertices, then it has a vertex with even degree), it must have a vertex of even degree. In other words, in our graph S, there exists a neighbour D of the vertex C which is connected with an even number of other neighbours of C. In other words, the number of common neighbours of the vertices C and D is even. Problem solved.\r\n\r\n[i]Case 2:[/i] Every vertex of the graph S has an even degree.\r\n\r\nLet A be an arbitrary vertex of the graph S. Construct a subgraph S' of S as follows:\r\n- The subgraph S' should contain all the 2n vertices of the graph S except of the vertex A.\r\n- The subgraph S' should contain only those edges of the graph S which pass through a neighbour of A (in other words, an edge of S is taken as an edge of S' if and only if at least one of the endpoints of this edge is a neighbour of A; the other endpoint can be arbitrary, but it cannot be A since A is not a vertex of S').\r\n\r\nThen, the graph S' has an odd number of vertices (2n - 1 vertices, to be precise), and thus, by the well-known theorem we used above, it contains a vertex D with even degree. This vertex D cannot be a neighbour of A in the graph S (in fact, since we are in Case 2, the vertex D has an even degree in the graph S, i. e. there is an even number of edges of the graph S starting at the vertex D; now, if D would be a neighbour of A, then all of these edges would be edges of the subgraph S', except of the edge joining D to A, and thus, there would be an odd number of edges of the subgraph S' starting at the vertex D, so that the degree of the vertex D in the subgraph S' would be odd). Hence, this vertex D is not a neighbour of A. Its degree in the subgraph S' is the number of common neighbours of the vertices A and D in the graph S. Since this degree is even, we thus see that the vertices A and D have an even number of common neighbours. Problem solved.\r\n\r\n[b]EDIT:[/b] Other solutions can be found at http://mathoverflow.net/questions/17269/let-g-be-a-graph-such-that-for-all-u-v-v-g-u-no-equal-to-v-n-u-n-v .\r\n\r\n Darij" } { "Tag": [ "inequalities", "trigonometry" ], "Problem": "If inequality $ \\frac {\\sin ^{3} x}{\\cos x} \\plus{} \\frac {\\cos ^{3} x}{\\sin x} \\ge k$ is hold for every $ x\\in \\left(0,\\frac {\\pi }{2} \\right)$, what is the largest possible value of $ k$?\n\n$\\textbf{(A)}\\ \\frac {1}{2} \\qquad\\textbf{(B)}\\ \\frac {3}{4} \\qquad\\textbf{(C)}\\ 1 \\qquad\\textbf{(D)}\\ \\frac {3}{2} \\qquad\\textbf{(E)}\\ \\text{None}$", "Solution_1": "From Holder\r\n\r\n$ LHS^2 \\left( cos^2 \\plus{} sin^2 \\right) \\ge \\left( sin^2 \\plus{} cos^2x \\right) ^3$\r\n\r\nAnswer is $ \\boxed{C}$", "Solution_2": "By rearrengement inequality\r\n\r\n$ \\dfrac{\\sin^3(x)}{\\cos(x)}\\plus{}\\dfrac{\\cos^3(x)}{\\sin(x)}\\ge \\dfrac{\\sin^3(x)}{\\sin(x)}\\plus{}\\dfrac{\\cos^3(x)}{\\cos(x)}\\equal{}\\sin^2(x)\\plus{}\\cos^2(x)\\equal{}1$ \r\n\r\nwith equality when $ \\cos(x)\\equal{}\\sin(x)\\Leftrightarrow x\\equal{}\\dfrac{\\pi}{4}$\r\n\r\nSo the answer is $ C$.\r\n\r\n:)", "Solution_3": "$ \\frac{(\\sin^2 x \\plus{} \\cos^2 x)^2 \\minus{} 2\\sin^2 x \\cos^2 x}{\\sin x \\cos x} \\equal{} \\frac{2 \\minus{} \\sin^2 2x} {\\sin 2x}$\n\nWhile $ \\sin 2x$ is increasing, above expression will decrease. So it has its minima at $ \\sin 2x \\equal{} 1$.\nThen $ k\\equal{}1$ and $ 2x \\equal{} \\frac{\\pi}{2} \\Rightarrow x \\equal{} \\frac{\\pi}{4}$", "Solution_4": "we can also prove it easily by Cauchy-Shcwarz.", "Solution_5": "$ \\frac {\\sin ^{3} x}{\\cos x} \\plus{} \\frac {\\cos ^{3} x}{\\sin x} \\equal{} \\frac {\\sin^4x \\plus{} \\cos^4x}{\\sin x\\cos x}\\ge \\frac {(\\sin^2x \\plus{} \\cos^2x)^2}{2\\sin x\\cos x} \\equal{} \\frac {1}{\\sin 2x}\\ge 1$ for any $ x\\in\\left(0\\ ,\\ \\frac {\\pi}{2}\\right)$ , where $ \\sin 2x\\in(0\\ ,\\ 1]$ .\r\n\r\nI used the simple inequality $ a^2\\plus{}b^2\\ge \\frac 12\\cdot (a\\plus{}b)^2$ for any real numbers $ a$ and $ b$ in particular case $ a: \\equal{}\\sin^2x$ and $ b: \\equal{}\\cos^2x$ ." } { "Tag": [], "Problem": "AP (bad teacher) vs. regular (good teacher)...\r\nWhich should I go for?\r\n\r\nWell, a good teacher will definitely make class more enjoyable, but in this particular case, the grading is supposedly harder in the regular class! Huge drop in GPA expected.\r\nAnd unfortunately, the quality of teachers doesn't show up on the transcript.", "Solution_1": "What subject? How is your transcript as it stands?", "Solution_2": "Both US History and govt/politics. (required)\r\n\r\nMy transcript for last semester:\r\nAP Econ 100\r\nEnglish II Honors 92\r\nWorld History 90\r\nAP Calc 100\r\nPhysics 98\r\nFrench 99\r\nAP Chem 98\r\nGraphic Design 100.", "Solution_3": "I wouldn't obsess over GPA or how you'll look to colleges or anything else. Choose the course that you'll learn the most from and do your best.", "Solution_4": "If the teacher is as good as you say, you might be able to still take the AP exam. Plus if you work hard in the class, you might be able to get a recommendation. Consider this: I know people who took AP Environmental for their science instead of regular physics...guess which class is easier and less useful (and in my opinion less interesting).", "Solution_5": "I should have remembered from chem this year that 200+ horrible class hours each week isn't really worth the AP designation. And since a lot of people in my school are going to take IB next year, the people in the class won't be fabulous anyway.\r\nI guess I'll just have to work harder to keep up the transcript. *sigh*\r\n\r\nThanks everyone for the advice! :)" } { "Tag": [ "calculus", "trigonometry", "calculus computations" ], "Problem": "Hi, today i've been handed a calculus assignment and Im stuck on a particular question and need some help. Here it goes:\r\n\r\nProve that $ f(x) \\equal{} xSin(x)$ and $ f(x) \\equal{} Cos(x)$ intersect at right angles to eachother.\r\n\r\nMy work:\r\n\r\n$ xSin(x) \\equal{} Cos(x)$\r\n$ x \\equal{} \\frac{Cos(x)}{Sin(x)}$\r\n$ x \\equal{} Cot(x)$\r\n\r\n$ f'(x) \\equal{} \\frac{\\minus{}1}{f'(x)}$\r\n$ xCos(x) \\plus{} Sin(x) \\equal{} \\frac{1}{Sin(x)}$\r\n$ Cot(x)Cos(Cot(x)) \\plus{} Sin(Cot(x)) \\equal{} \\frac{1}{Sin(Cot(x))}$\r\n\r\nNow Im stuck, I dont know how to prove that this is true!!!", "Solution_1": "Your notation is horrible. What you really want to show is that if $ f(x) = x\\sin x$, $ g(x) = \\cos x$, and $ f(a) = g(a)$, then $ f'(a)g'(a) = 1$.\r\n\r\nAs you noted, $ f(a) = g(a)$ implies $ a = \\cot a$.\r\n\r\nNow it's straightforward to compute\r\n\r\n\\begin{eqnarray*}\r\nf'(a)g'(a) &=& (a\\cos a+\\sin a)(-\\sin a) \\\\\r\n&=& (\\cot a \\cos a + \\sin a)(-\\sin a) \\\\\r\n&=&\\left(\\frac{1-\\sin^2 a}{\\sin a} + \\sin a\\right)(-\\sin a) \\\\\r\n&=&\\left(\\frac{1}{\\sin a}\\right)(-\\sin a) \\\\\r\n&=& -1\r\n\\end{eqnarray*}", "Solution_2": ":o never thought of it that way, thank you!!!!!!" } { "Tag": [ "algebra", "polynomial", "integration", "complex analysis", "complex analysis unsolved" ], "Problem": "let $ P(z)\\equal{}a_{0}\\plus{}...\\plus{}a_{n}z^{n}$ be a polynomial of order $ n\\geq 1$\r\n $ C_{r}$ be the circle with radius $ r>0$, counterclockwise oriented\r\nprove that \\[ \\frac{1}{2\\pi i}\\int_{C_{r}}z^{n\\minus{}1}|P(z)|^2dz\\equal{}a_{0}\\overline{a_{n}}r^{2n}\\]\r\n\r\nAny idea will be highly appreciated", "Solution_1": "Write $ |P(z)|^2\\equal{}P(z)\\overline{P(z)}$. Split into terms. What is $ \\int_{C_r}z^a\\overline{z}^b$? Well, on $ C_r$, $ \\overline{z}\\equal{}\\frac{r^2}{z}$", "Solution_2": "thanks, that was I though :lol:" } { "Tag": [ "percent", "percent decrease" ], "Problem": "A number is increased by $ 50\\%$ and then the result is decreased by $ 50\\%$. What is the percent of decrease from the original number to the final number?", "Solution_1": "let's assume the original number is $ 100$\r\n\r\nafter increasing it by $ 50\\%$, our number would be $ 150$\r\n\r\nafter decreasing it by $ 50\\%$, our number would be $ 75$\r\n\r\n\r\nthus, the percent of decrease from the original number to the final number is $ \\frac{100\\minus{}75}{100}\\equal{}\\frac{25}{100}\\equal{}25\\%$\r\n\r\n\r\nPS the percent decrease is calculated with this formula: $ \\frac{\\text{original} \\minus{} \\text{new}}{\\text{original}}$", "Solution_2": "[quote=\"GameBot\"]A number is increased by $ 50\\%$ and then the result is decreased by $ 50\\%$. What is the percent of decrease from the original number to the final number?[/quote]\r\n\r\n25 :)\r\n\r\nI totally agree with vallon22's response. No need for supposing that the number is 100, it could just be x*1.5*.5 = .75x, but it doesn't really matter x or 100, really... :lol:", "Solution_3": "I think somewhere izzy said saying i agree, and like stuff like that are like spam?\r\nBut it's ok seeing your a new member." } { "Tag": [], "Problem": "Sa se arate ca:\r\n$ (m^2 \\plus{} 9mn \\plus{} n^2)\\vdots 11\\Rightarrow (m^2 \\minus{} n^2)\\vdots 11; \\ (\\forall) m;n\\in \\mathbb{Z}.$", "Solution_1": "$ m^2\\plus{}9mn\\plus{}n^2 \\vdots 11 \\Rightarrow (m\\minus{}n)^2 \\plus{} 11mn \\vdots 11 \\Rightarrow (m\\minus{}n)^2 \\vdots 11 \\Rightarrow m\\minus{}n \\vdots 11$\r\n\r\n$ \\Rightarrow m$ si $ n$ dau acelasi rest la impartirea cu $ 11$ de unde rezulta concluzia", "Solution_2": "In aceleasi conditii, demonstrati ca:\r\n$ (m^2\\plus{}9mn\\plus{}n^2)\\vdots 77 \\Rightarrow (m^2\\minus{}n^2)\\vdots 77.$", "Solution_3": "Ramane de demonstrat divizibilitatea cu $ 7$, care se face in mod analog ca si in cazul divizibilitatii cu $ 11$" } { "Tag": [ "limit", "real analysis", "integration", "probability", "function", "probability and stats" ], "Problem": "Let $ (X_n: n\\in\\mathbb{N})$ sequence of independent equally distributed random variables such that $ \\mathbb{E}[X_1]\\equal{}0$. Prove that $ \\lim_{n\\rightarrow \\infty}\\frac{1}{n}\\mathbb{E}[|X_1\\plus{}\\dots\\plus{}X_n|]\\equal{}0$.", "Solution_1": "This would be easy if each $ X_i$ had finite variance $ \\sigma^2.$ If that were the case,\r\n\r\n$ \\mathbb{E}\\left(\\sum X_i\\right)^2=n\\sigma^2,$ and by Cauchy-Schwarz,\r\n\r\n$ \\mathbb{E}\\left(\\left|\\sum X_i\\right|\\right)\\le \\left(\\mathbb{E}\\left(\\sum X_i\\right)^2\\right)^{\\frac12}=\\sqrt{n}\\,\\sigma.$\r\n\r\nBut you didn't say anything about finite variance, so we'll have to find another proof.", "Solution_2": "what you are trying to prove is the \"modified\" law of large number where the almost sure convergence is replaced by the $ L^1$ convergence.\r\nFor $ \\epsilon > 0$, take $ a$ large enough such that:\r\n$ |E[ X 1_{|X|n - s1 xn-1 + s2 xn-1 + ...... + (-1)n sn\n\n\n\nWith \n\ns1 = :Sigma: ai\n\ns2 = :Sigma: ai aj\n\nsn = prod ai\n\nif ai are the roots.\n\n\n\nThen :Sigma: ai = s1 - 2s2 = 1 - 2s2\n\nAll the roots are real so :Sigma: ai >= 0 and so s2 = -1 and :Sigma: ai = 3 !\n\n\n\nLet bi = ai\n\n\n\nWe have :Sigma: bi = 3, prod bi = 1.\n\n\n\nAM-GM : :Sigma: bi / n >= prod bi or n <= 3 :mrgreen: \n\n\n\nx-1, x-x-1, x3+x-x-1 are examples of such polynoms for n <= 3.[/hide]", "Solution_2": "Thats exactly the same solution I had!\r\n\r\nI kept wanting to use AM-GM but couldn't because of negatives.. then I thought of using the squares, and then that was it.. quite nice." } { "Tag": [ "probability", "articles", "geometry" ], "Problem": "Okay, Here's my query, and it is -- It's late at night, you're walking in a street (There are very few people on the street), it's raining pretty heavily (a heavy downpour), you can hear loud thunderstorms and see intermittent glimpses of lighting. Now, In the prevailing circumstances, what's the probability of you getting struck by lightning (or a lightning bolt)? \r\n\r\nAssume you forgot to carry an umbrella with you -- The downpour was not expected at all -- sort of impromptu.\r\n\r\nConsider another case (case no 2) in which you're walking on a road on the outskirts of the city (which is almost completely devoid of people,trees, buildings,etc). Does the probability of you getting hit by lightning increase here when compared with first case? \r\n\r\nNow, it does appear obvious intuitively (in general) that the probability of getting struck by lightning is very low. (In the presence of buildings, people, trees, etc), but, I would also like to know whether the probability of getting hit by lightning increase in the absence of the aforementioned things (trees,etc)?\r\n\r\nBut I'm slightly more interested in seeking a scientific, or a coherent/logical explanation rather than an intuitive one. (Both are welcomed)\r\n\r\nThank you!", "Solution_1": "The National Safety Council estimates that 70 to 120 people per year are killed by lightning strikes. Say 100, just to be even. The US population is 308 million, maybe a little more. So the probability of being killed by a lightning strike is roughly $ 3.25\\times 10^{\\minus{}7}$, pretty small.\r\n\r\nNow not everyone who is struck by lightning dies of course, so your numerator rises by a little. But it certainly won't alter the order of magnitude by more than 1.", "Solution_2": "If I remember correctly, lightning tends to strike higher objects more often, so if you're in a street the lightning will hit a house more often.", "Solution_3": "A month or two ago, a woman was killed by lightning in Southern California - I think it was in San Bernardino. It was extensively covered in the local news, in part because the storm was quite unusual, especially for the time of year. She was in a residential neighborhood with plenty of trees and houses around. Apparently, the lightning bolt hit a tree and then jumped sideways from the tree to hit her.", "Solution_4": "[quote=\"worthawholebean\"]If I remember correctly, lightning tends to strike higher objects more often, so if you're in a street the lightning will hit a house more often.[/quote]\r\n\r\nYes, you're right. \r\n\r\nAssume the following scenario:You're walking in a completely deserted land -- there are no buildings, trees, etc. Now, if there are no \"higher\" objects -- The probability of the lightning directly striking you will increase-- but does lightning \"directly\" strike people? (It shouldn't strike people as in the \"news article\" that [b]Kent Merryfield [/b] had cited) \r\n\r\nI'm just asking the probability of getting struck by lightning as in case (1) -- When you're in a busy street, and in case (2) -- When you're in a completely deserted area as aforementioned. Certainly, the probability of getting struck by lightning in the second case will be more as and when compared with the first case.", "Solution_5": "Fontana, rather than San Bernardino (close enough), and it appears she was standing under the tree. Here are two news stories:\r\n\r\n[url=http://lethalapp.com/news/2009/06/rare-california-lightning-death/]Link.[/url]\r\n[url=http://www.pe.com/localnews/sbcounty/stories/PE_News_Local_S_wea04.4b9641e.html]Link.[/url]", "Solution_6": "[quote=\"Kent Merryfield\"]Fontana, rather than San Bernardino (close enough), and it appears she was standing under the tree. Here are two news stories:\n\n[url=http://lethalapp.com/news/2009/06/rare-california-lightning-death/]Link.[/url]\n[url=http://www.pe.com/localnews/sbcounty/stories/PE_News_Local_S_wea04.4b9641e.html]Link.[/url][/quote]\r\n\r\nYes, I felt sad :( after reading the article. Maybe it drives home the point:\"Death could sometimes occur in the weirdest possible manner.\"", "Solution_7": "doesn't lightning go upward, correct me if im wrong\r\nsomething about massive static electricity" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c > 0$\r\n\r\n$ \\sqrt {\\dfrac{{2a}}{{a \\plus{} b}}} \\plus{} \\sqrt {\\dfrac{{2b}}{{b \\plus{} c}}} \\plus{} \\sqrt {\\dfrac{{2c}}{{c \\plus{} a}}} \\ge \\dfrac{{\\left( {a \\plus{} b \\plus{} c} \\right)\\left( {ab \\plus{} bc \\plus{} ca} \\right)}}{{\\sqrt {\\left( {{a^2} \\plus{} {b^2} \\plus{} {c^2}} \\right)\\left( {{a^2}{b^2} \\plus{} {b^2}{c^2} \\plus{} {c^2}{a^2}} \\right)} }}$", "Solution_1": "[quote=\"leviethai\"]Let $ a,b,c > 0$\n\n$ \\sqrt {\\dfrac{{2a}}{{a \\plus{} b}}} \\plus{} \\sqrt {\\dfrac{{2b}}{{b \\plus{} c}}} \\plus{} \\sqrt {\\dfrac{{2c}}{{c \\plus{} a}}} \\ge \\dfrac{{\\left( {a \\plus{} b \\plus{} c} \\right)\\left( {ab \\plus{} bc \\plus{} ca} \\right)}}{{\\sqrt {\\left( {{a^2} \\plus{} {b^2} \\plus{} {c^2}} \\right)\\left( {{a^2}{b^2} \\plus{} {b^2}{c^2} \\plus{} {c^2}{a^2}} \\right)} }}$[/quote]\r\n\r\n\r\n\r\nUsing Holder's Inequality\r\n\r\n$ (\\sum \\sqrt {\\frac {2a}{a \\plus{} b}})^2(\\sum {4a^2(a \\plus{} b)})\\geq (2a \\plus{} 2b \\plus{} 2c)^3$\r\n\r\nwe have \r\n\r\n$ (\\sum \\sqrt {\\frac {2a}{a \\plus{} b}})^2\\geq \\frac {(2a \\plus{} 2b \\plus{} 2c)^3}{\\sum {4a^2(a \\plus{} b)}}$\r\n\r\nwe only prove \r\n\r\n$ \\frac {(2a \\plus{} 2b \\plus{} 2 c)^3}{\\sum {4a^2(a \\plus{} b)}}\\geq \\frac {(a \\plus{} b \\plus{} c)^2(ab \\plus{} bc \\plus{} ca)^2}{(a^2 \\plus{} b^2 \\plus{} c^2)(a^2b^2 \\plus{} b^2 c^2 \\plus{} c^2a^2)}$\r\n\r\nor \r\n\r\n$ \\frac {2(a \\plus{} b \\plus{} c)}{a^2(a \\plus{} b) \\plus{} b^2(b \\plus{} c) \\plus{} c^2(c \\plus{} a)}\\geq \\frac {(ab \\plus{} bc \\plus{} ca)^2}{(a^2 \\plus{} b^2 \\plus{} c^2)(a^2b^2 \\plus{} b^2 c^2 \\plus{} c^2a^2)}$\r\n\r\n\r\n\r\nThe inequality holds,but i can not prove. :lol:", "Solution_2": "[quote=\"leviethai\"]Let $ a,b,c > 0$\n\n$ \\sqrt {\\dfrac{{2a}}{{a \\plus{} b}}} \\plus{} \\sqrt {\\dfrac{{2b}}{{b \\plus{} c}}} \\plus{} \\sqrt {\\dfrac{{2c}}{{c \\plus{} a}}} \\ge \\dfrac{{\\left( {a \\plus{} b \\plus{} c} \\right)\\left( {ab \\plus{} bc \\plus{} ca} \\right)}}{{\\sqrt {\\left( {{a^2} \\plus{} {b^2} \\plus{} {c^2}} \\right)\\left( {{a^2}{b^2} \\plus{} {b^2}{c^2} \\plus{} {c^2}{a^2}} \\right)} }}$[/quote]\r\n\r\nby Holder, we have\r\n\r\n\\[ \\left( {\\sum {\\sqrt {\\frac {{2a}}{{a \\plus{} b}}} } } \\right)^2 \\sum {a\\left( {a \\plus{} b} \\right)} \\sum {a^2 } \\ge 2\\left( {a \\plus{} b \\plus{} c} \\right)^4 ,\\]\r\nwe get\r\n\r\n\\[ \\left( {\\sum {\\sqrt {\\frac {{2a}}{{a \\plus{} b}}} } } \\right)^2 \\ge \\frac {{2\\left( {a \\plus{} b \\plus{} c} \\right)^4 }}{{\\sum {a\\left( {a \\plus{} b} \\right)} \\sum {a^2 } }},\\]\r\nso we just need to prove that\r\n\r\n\\[ \\frac {{2\\left( {a \\plus{} b \\plus{} c} \\right)^4 }}{{\\sum {a\\left( {a \\plus{} b} \\right)} \\sum {a^2 } }} \\ge \\frac {{\\left( {a \\plus{} b \\plus{} c} \\right)^2 \\left( {ab \\plus{} bc \\plus{} ca} \\right)^2 }}{{\\left( {a^2 \\plus{} b^2 \\plus{} c^2 } \\right)\\left( {a^2 b^2 \\plus{} b^2 c^2 \\plus{} c^2 a^2 } \\right)}},\\]\r\nor\r\n\r\n\\[ 2\\left( {a \\plus{} b \\plus{} c} \\right)^2 \\left( {a^2 b^2 \\plus{} b^2 c^2 \\plus{} c^2 a^2 } \\right) \\ge \\left( {ab \\plus{} bc \\plus{} ca} \\right)^2 \\left( {a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca} \\right),\\]\r\nthat's easy to prove by Muirhead.\r\n\r\n :D", "Solution_3": "@hedeng123: I hope you will prove it soon :)\r\n@kuing: I don't like Murihead so much, so can you give me another solution? Thank you :) \r\nI have 1 proof, using only AM-GM and CAuchy - Schwartz\r\n\r\nP/S: my english is not good......", "Solution_4": "[quote=\"leviethai\"]@hedeng123: I hope you will prove it soon :)\n@kuing: I don't like Murihead so much, so can you give me another solution? Thank you :) \nI have 1 proof, using only AM-GM and CAuchy - Schwartz\n\nP/S: my english is not good......[/quote]\r\n\r\n\r\n\r\n\r\n$ 2(a \\plus{} b \\plus{} c)^2(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2) \\minus{} (ab \\plus{} ca \\plus{} bc)^2(a^2 \\plus{} bc \\plus{} b^2 \\plus{} ca \\plus{} c^2 \\plus{} ab)\\geq 0$\r\n\r\n\r\n $ \\Longleftrightarrow \\sum_{sym} (2a^4b^2 \\minus{} 2a^4bc \\plus{} 3a^3b^3 \\minus{} 2a^3b^2c \\minus{} a^2b^2c^2)\\geq 0$\r\n\r\n\r\nor \r\n\r\n$ \\Longleftrightarrow (a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2 \\plus{} 5\\sum b^2c^2(a \\minus{} b)(a \\minus{} c) \\plus{} 2abc\\sum(b \\plus{} c)(a \\minus{} b)(a \\minus{} c)\\geq 0$", "Solution_5": "[quote=\"hedeng123\"][quote=\"leviethai\"]@hedeng123: I hope you will prove it soon :)\n@kuing: I don't like Murihead so much, so can you give me another solution? Thank you :) \nI have 1 proof, using only AM-GM and CAuchy - Schwartz\n\nP/S: my english is not good......[/quote]\n\n\n\n\n$ 2(a \\plus{} b \\plus{} c)^2(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2) \\minus{} (ab \\plus{} ca \\plus{} bc)^2(a^2 \\plus{} bc \\plus{} b^2 \\plus{} ca \\plus{} c^2 \\plus{} ab)\\geq 0$\n\n\n $ \\Longleftrightarrow \\sum_{sym} (2a^4b^2 \\minus{} 2a^4bc \\plus{} 3a^3b^3 \\minus{} 2a^3b^2c \\minus{} a^2b^2c^2)\\geq 0$\n\n\nor \n\n$ \\Longleftrightarrow (a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2 \\plus{} 5\\sum b^2c^2(a \\minus{} b)(a \\minus{} c) \\plus{} 2abc\\sum(b \\plus{} c)(a \\minus{} b)(a \\minus{} c)\\geq 0$[/quote]\r\nGreat, I think this is computer's power :) \r\nSo we have another Inequality. Thank you.\r\nI try to prove this inequality more elementary.", "Solution_6": "[quote=\"leviethai\"]@hedeng123: I hope you will prove it soon :)\n@kuing: I don't like Murihead so much, so can you give me another solution? Thank you :) \nI have 1 proof, using only AM-GM and CAuchy - Schwartz\n\nP/S: my english is not good......[/quote]\r\n\r\nthen.......how about the $ uvw$ method? use it to prove that ineq is so easy.\r\n\r\ndo you like it?", "Solution_7": "[quote=\"kuing\"][quote=\"leviethai\"]@hedeng123: I hope you will prove it soon :)\n@kuing: I don't like Murihead so much, so can you give me another solution? Thank you :) \nI have 1 proof, using only AM-GM and CAuchy - Schwartz\n\nP/S: my english is not good......[/quote]\n\nthen.......how about the $ uvw$ method? use it to prove that ineq is so easy.\n\ndo you like it?[/quote]\r\n\r\nYou can.........\r\n\r\nI don't like it because I can't use it :blush: , I want everything can be solved in elementary solutions :) .\r\n\r\nP/S: I'm Grade 10 Student", "Solution_8": "[quote=\"kuing\"]\nor\n\\[ 2\\left( {a \\plus{} b \\plus{} c} \\right)^2 \\left( {a^2 b^2 \\plus{} b^2 c^2 \\plus{} c^2 a^2 } \\right) \\ge \\left( {ab \\plus{} bc \\plus{} ca} \\right)^2 \\left( {a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca} \\right),\\]\nthat's easy to prove by Muirhead.\n\n :D[/quote]\r\nAM-GM solved it easily, my friend :)", "Solution_9": "[quote=\"nguoivn\"][quote=\"kuing\"]\nor\n\\[ 2\\left( {a \\plus{} b \\plus{} c} \\right)^2 \\left( {a^2 b^2 \\plus{} b^2 c^2 \\plus{} c^2 a^2 } \\right) \\ge \\left( {ab \\plus{} bc \\plus{} ca} \\right)^2 \\left( {a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca} \\right),\\]\nthat's easy to prove by Muirhead.\n\n :D[/quote]\nAM-GM solved it easily, my friend :)[/quote]\r\n\r\n :blush: \r\n\r\nhow to solved... :what?:", "Solution_10": "My solution :) ( This is a reason why I said this inequality wasn't hard)\r\n\r\nFrom AM-GM inequality, we have\r\n\r\n$ \\sqrt {\\dfrac{{2a}}{{a \\plus{} b}}} \\equal{} \\dfrac{{2a}}{{\\sqrt {2a} .\\sqrt {a \\plus{} b} }} \\ge \\dfrac{{4a}}{{3a \\plus{} b}}$\r\n\r\n$ \\sqrt {\\dfrac{{2b}}{{b \\plus{} c}}} \\ge \\dfrac{{4b}}{{3b \\plus{} c}},\\sqrt {\\dfrac{{2c}}{{c \\plus{} a}}} \\ge \\dfrac{{4c}}{{3c \\plus{} a}}$\r\n\r\nSo that,\r\n\r\n$ \\sqrt {\\dfrac{{2a}}{{a \\plus{} b}}} \\plus{} \\sqrt {\\dfrac{{2b}}{{b \\plus{} c}}} \\plus{} \\sqrt {\\dfrac{{2c}}{{c \\plus{} a}}} \\ge 4\\left( {\\dfrac{a}{{3a \\plus{} b}} \\plus{} \\dfrac{b}{{3b \\plus{} c}} \\plus{} \\dfrac{c}{{3c \\plus{} a}}} \\right)$\r\n\r\nFrom Schwartz inequality and the following well-known inequality $ {{a^2} \\plus{} {b^2} \\plus{} {c^2}} \\ge {ab \\plus{} bc \\plus{} ca}$\r\n\r\n$ \\dfrac{a}{{3a \\plus{} b}} \\plus{} \\dfrac{b}{{3b \\plus{} c}} \\plus{} \\dfrac{c}{{3c \\plus{} a}} \\equal{} \\dfrac{{{a^2}}}{{3{a^2} \\plus{} ab}} \\plus{} \\dfrac{{{b^2}}}{{3{b^2} \\plus{} bc}} \\plus{} \\dfrac{{{c^2}}}{{3{c^2} \\plus{} ca}}$\r\n\r\n$ \\ge \\dfrac{{{{\\left( {a \\plus{} b \\plus{} c} \\right)}^2}}}{{3\\left( {{a^2} \\plus{} {b^2} \\plus{} {c^2}} \\right) \\plus{} ab \\plus{} bc \\plus{} ca}} \\ge \\dfrac{{{{\\left( {a \\plus{} b \\plus{} c} \\right)}^2}}}{{4\\left( {{a^2} \\plus{} {b^2} \\plus{} {c^2}} \\right)}}$\r\n\r\nTherefore,\r\n\r\n$ \\sqrt {\\dfrac{{2a}}{{a \\plus{} b}}} \\plus{} \\sqrt {\\dfrac{{2b}}{{b \\plus{} c}}} \\plus{} \\sqrt {\\dfrac{{2c}}{{c \\plus{} a}}} \\ge \\dfrac{{{{\\left( {a \\plus{} b \\plus{} c} \\right)}^2}}}{{{a^2} \\plus{} {b^2} \\plus{} {c^2}}}$\r\n\r\nWe also have, (just prove analogously)\r\n\r\n$ \\sqrt {\\dfrac{{2a}}{{a \\plus{} b}}} \\plus{} \\sqrt {\\dfrac{{2b}}{{b \\plus{} c}}} \\plus{} \\sqrt {\\dfrac{{2c}}{{c \\plus{} a}}} \\equal{} \\sqrt {\\dfrac{{2ac}}{{ac \\plus{} bc}}} \\plus{} \\sqrt {\\dfrac{{2ba}}{{ba \\plus{} ca}}} \\plus{} \\sqrt {\\dfrac{{2cb}}{{cb \\plus{} ab}}}$\r\n\r\n$ \\ge \\dfrac{{{{\\left( {ab \\plus{} bc \\plus{} ca} \\right)}^2}}}{{{a^2}{b^2} \\plus{} {b^2}{c^2} \\plus{} {c^2}{a^2}}}$\r\n\r\nSo that,\r\n\r\n$ \\sqrt {\\dfrac{{2a}}{{a \\plus{} b}}} \\plus{} \\sqrt {\\dfrac{{2b}}{{b \\plus{} c}}} \\plus{} \\sqrt {\\dfrac{{2c}}{{c \\plus{} a}}} \\ge \\dfrac{1}{2}\\left( {\\dfrac{{{{\\left( {a \\plus{} b \\plus{} c} \\right)}^2}}}{{{a^2} \\plus{} {b^2} \\plus{} {c^2}}} \\plus{} \\dfrac{{{{\\left( {ab \\plus{} bc \\plus{} ca} \\right)}^2}}}{{{a^2}{b^2} \\plus{} {b^2}{c^2} \\plus{} {c^2}{a^2}}}} \\right)$\r\n\r\n$ \\ge \\dfrac{{\\left( {a \\plus{} b \\plus{} c} \\right)\\left( {ab \\plus{} bc \\plus{} ca} \\right)}}{{\\sqrt {\\left( {{a^2} \\plus{} {b^2} \\plus{} {c^2}} \\right)\\left( {{a^2}{b^2} \\plus{} {b^2}{c^2} \\plus{} {c^2}{a^2}} \\right)} }}$ (from AM-GM inequality)\r\n\r\nThe proof is completed. Equality holds if and only if $ a \\equal{} b \\equal{} c$.", "Solution_11": "nice proof, leviethai ... :)", "Solution_12": "[quote=\"kuing\"][quote=\"nguoivn\"][quote=\"kuing\"]\nor\n\\[ 2\\left( {a \\plus{} b \\plus{} c} \\right)^2 \\left( {a^2 b^2 \\plus{} b^2 c^2 \\plus{} c^2 a^2 } \\right) \\ge \\left( {ab \\plus{} bc \\plus{} ca} \\right)^2 \\left( {a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca} \\right),\\]\nthat's easy to prove by Muirhead.\n\n :D[/quote]\nAM-GM solved it easily, my friend :)[/quote]\n\n :blush: \n\nhow to solved... :what?:[/quote]\r\nMy proof is also similar to you, Kuing. But with this last ineq, we can prove it more simpler:\r\nThe first, rewrite it to: \r\n$ \\frac {2(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2)}{(ab \\plus{} bc \\plus{} ca)^2} \\ge \\frac {a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca}{(a \\plus{} b \\plus{} c)^2}$\r\nOr $ \\sum (\\frac {4a^2}{(ab \\plus{} bc \\plus{} ca)^2} \\minus{} \\frac {1}{(a \\plus{} b \\plus{} c)^2})(b \\minus{} c)^2 \\ge 0$\r\nAssume that $ a \\ge b \\ge c$, then easily to see that $ Sa, Sb \\ge 0$\r\nIn other hand, by AM-GM, we also have: \r\n$ Sb \\plus{} Sc \\equal{} \\frac {2(b^2 \\plus{} c^2)}{(ab \\plus{} bc \\plus{} ca)^2} \\minus{} \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\ge \\frac {(b \\plus{} c)^2}{(ab \\plus{} bc \\plus{} ca)^2} \\minus{} \\frac {1}{(a \\plus{} b \\plus{} c)^2}$\r\nSo, we can prove $ Sb \\plus{} Sc \\ge 0$ if show that: $ (b \\plus{} c)(a \\plus{} b \\plus{} c) \\ge ab \\plus{} bc \\plus{} ca$ (obviously trues).\r\nThe proof completed :) \r\nPS: Nice proof, leviethai :)", "Solution_13": "[quote=\"nguoivn\"]\nMy proof is also similar to you, Kuing. But with this last ineq, we can prove it more simpler:\nThe first, rewrite it to: \n$ \\frac {2(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2)}{(ab \\plus{} bc \\plus{} ca)^2} \\ge \\frac {a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca}{(a \\plus{} b \\plus{} c)^2}$\nOr $ \\sum (\\frac {4a^2}{(ab \\plus{} bc \\plus{} ca)^2} \\minus{} \\frac {1}{(a \\plus{} b \\plus{} c)^2})(b \\minus{} c)^2 \\ge 0$\nAssume that $ a \\ge b \\ge c$, then easily to see that $ Sa, Sb \\ge 0$\nIn other hand, by AM-GM, we also have: \n$ Sb \\plus{} Sc \\equal{} \\frac {2(b^2 \\plus{} c^2)}{(ab \\plus{} bc \\plus{} ca)^2} \\minus{} \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\ge \\frac {(b \\plus{} c)^2}{(ab \\plus{} bc \\plus{} ca)^2} \\minus{} \\frac {1}{(a \\plus{} b \\plus{} c)^2}$\nSo, we can prove $ Sb \\plus{} Sc \\ge 0$ if show that: $ (b \\plus{} c)(a \\plus{} b \\plus{} c) \\ge ab \\plus{} bc \\plus{} ca$ (obviously trues).\nThe proof completed :) \nPS: Nice proof, leviethai :)[/quote]\r\n\r\noh...S.O.S method...\r\nthank you :)" } { "Tag": [ "IMO 2003", "inequalities", "IMO", "arithmetic sequence", "IMO Shortlist" ], "Problem": "Let $n$ be a positive integer and let $x_1\\le x_2\\le\\cdots\\le x_n$ be real numbers.\r\nProve that\r\n\r\n \\[\r\n \\left(\\sum_{i,j=1}^{n}|x_i-x_j|\\right)^2\\le\\frac{2(n^2-1)}{3}\\sum_{i,j=1}^{n}(x_i-x_j)^2.\r\n \\]\r\nShow that the equality holds if and only if $x_1, \\ldots, x_n$ is an arithmetic sequence.", "Solution_1": "We can rewrite the given expression as\r\n\\( \\sum_i (2i-n-1)x_i \\)^2 \r\n\\leq (n^2 - 1)/3 \\( n\\sum x_i^2 - (\\sum x_i )^2\\).\r\nAs the expression is translation invariant (obvious from the original form), we can assume that \\sum x_i =0. The rest follows by Cauchy-Schwartz.", "Solution_2": "The crucial idea is that we may assume that Sum x_k = 0. I had tried to solve this problem, but I did not succeed. I came up with the first identity in Jayanta's message, but I was not smart enough to make his clever assumption. Let me complete the proof\r\n\r\n(a) Since sum_{i,j} |x_j x_i|= 2 sum_{i2 and x_1 <=x_2<=...<=x_n and y_1<=y_2<=...<=y_n. Then\r\n\r\n(sum_{i,j} |x_i-x_j|) * (sum_{i,j} |y_i-y_j|) <= 2(n^2-1)/3 sum_{i,j} |x_i-x_j)(y_i-y_j)| .\r\n\r\nThis would be a generalization of the IMO problem. \r\n\r\n\r\nWe could assume that sum x_k=sum y_k=0 and use the identity\r\n\r\nn*(sum x_k * y_k) = (sum x_k)*(sum y_k) + sum_{i=0, then the rhs (except the (n^2-1)/3 factor) is\r\n[a_1^2+a_2^2...+a_{n-1}^2 + \r\n(a_1+a_2)^2+(a_2+a_3)^2+...+(a_{n-2}+a_{n-1})^2+\r\n...\r\n+(a_1+a_2+...+a_{n-1})^2]\r\nwrite a k-sum square as \r\n(a_{i+1}+...+a_{i+k})^2 as k^2 [(a_{i+1}+...+a_{i+k})/k]^2, so altogether we have\r\n(n-1)*1^2+(n-2)*2^2+...+1*(n-1)^2 terms, then use the power mean inequality to get it\r\ncheers,", "Solution_7": "Elinor, I hope this is the way you did it..\r\n\r\nThe number of terms is \r\n\r\n(n-1)*1^2+(n-2)*2^2+...+1*(n-1)^2 = sum (n-k)*k^2 for k=1,2,...,n-1 = n^2(n^2-1)/12. \r\n\r\nSo the rhs of the reduced inequality (i.e. the sum for i= (n/2)*S' , where S' is the sum=(a_1+...+a_{n-1} +\r\n(a_1+a_2) + (a_2+a_3)+...+(a_{n-2}+a_{n-1})+..+(a_1+....+a_{n-1}))\r\n\r\nIn order to do this we observe that after making the operations in the two sums we see that for k=1,2,..,n-1, the coefficient of a_k in S is nk(n-k)/2 and in S' is k(n-k). (of course, this needs proof!!) Thus we actually have equality\r\n\r\nS= (n/2)*S' , \r\n\r\nand the result follows.", "Solution_8": "[quote=\"Achilleas Sinefakopoulos\"]Question: Is the following inequality true?\n\nLet n>2 and x_1 <=x_2<=...<=x_n and y_1<=y_2<=...<=y_n. Then\n\n(sum_{i,j} |x_i-x_j|) * (sum_{i,j} |y_i-y_j|) <= 2(n^2-1)/3 sum_{i,j} |x_i-x_j)(y_i-y_j)| .\n\n...(Of course, the inequality could be false)[/quote]\r\n\r\nIt is false in general, (as will be shown later). This can also be easily seen by trying n=3. However under certain restrictions for the x_i, y_i the attack given by Elinor and Achilles can work by replacing the Cauchy-Schwartz (or AM-RMS) inequality with Chebyshev's inequality. However this gives the very awkward restriction that if f(i, j) = |x_i-x_j|/|i-j| and g(i,j) = |y_i-y_j|/|i-j|, f(i, j)g(k,l), showing that the originally proposed inequality is in general false.", "Solution_9": "Alison, thanks for your reply!", "Solution_10": "[b]Lemma:[/b] $\\sum_{i,j=1}^n |x_i - x_j| |i - j| = \\frac{n \\sum_{i,j=1} |x_i - x_j|}{2}$.\n[i]Proof:[/i] Let $a_1 = x_1$, and let $a_{i} = x_i - x_{i-1}$ for all $i$ with $2 \\leq i \\leq n$. The lemma we wish to prove can easily be seen to be equivalent to\n$\\sum_{1 \\leq i < j \\leq n} (i - j)(a_{i+1} + a_{i+2} + \\cdots + a_j) = \\frac{n}{2} \\sum_{1 \\leq i < j \\leq n} (a_{i+1} + a_{i+2} + \\cdots + a_j).$\n\nWhen expanded, both sides are a linear combination of $a_1, a_2, \\cdots, a_n$. We will determine the coefficients of each variable on both sides of the equation.\n\nOn the left-hand side, $a_k$ appears if and only if $i+1 \\leq k \\leq j$, in which case it comes with a factor if $(j-i)$. Hence, the coefficient of $a_k$ on the left-hand side is $\\sum_{i=1}^{k-1} \\sum_{j=k}^n (j-i) = \\frac{n(k-1)(n-k+1)}{2}$. On the right-hand side, $a_k$ appears iff $i+1 \\leq k \\leq j$, in which case it comes with a factor if $\\frac{n}{2}$. Hence, the coefficient of $a_k$ on the right-hand side is $\\frac{n(k-1)(n-k+1)}{2}$, which is equal to the coefficient on the left. $\\blacksquare$\n\nIt can easily be seen that $\\sum_{i=1}^n \\sum_{j=1}^n |i - j|^2 = \\frac{2(n^2-1)}{3} \\cdot \\frac{n^2}{4}$. By the Cauchy-Schwarz inequality,\n\\begin{align*}\n\\frac{2(n^2 - 1)}{3} \\left( \\sum_{i,j=1}^n |x_i - x_j|^2 \\right)\n&= \\frac{\\left( \\sum_{i,j=1}^n |x_i - x_j|^2 \\right) \\left( \\sum_{j=1}^n |i-j|^2 \\right)}{\\frac{n^2}{4}} \\\\\n&\\geq \\left (\\frac{ \\sum_{i,j=1}^n |x_i - x_j| \\, |i - j| }{ \\frac{n}{2} } \\right)^2 \\\\\n&= \\left( \\sum_{i,j=1}^n |x_i - x_j| \\right)^2, \n\\end{align*}\nas desired.\n\nThe cases of equality easily follow from the above application of Cauchy-Schwarz.", "Solution_11": "I have known that this problem can be solved using Cauchy-Schwarz inequality, but even with this information I haven't known how to solve it. It can be solved by using derivatives in a very simple way. Just compute derivatives on each variable and it can be easily seen that $x_2-x_1=x_3-x_2$ and so on.", "Solution_12": "We first make use of symmetry to rewrite the inequality as\n\\[\\left(\\sum_{1\\le ij\\ge1}^{n}(x_i-x_j)^2$ since $(x_i-x_j)^2 = (x_j-x_i)^2$. Next, let $J = \\sum_{i>j\\ge1}^{n}(x_i-x_j)^2$ and $K = \\sum_{i = 1}^n\\sum_{i \\ge j \\ge 1}^nx_j^2$. Also, let $c_n = \\binom{n}{2}, c_{n-1} = \\binom{n}{2}-n, \\ldots c_1 = \\binom{n}{2}-(n-1)n = -\\binom{n}{2}$. By Cauchy-Schwarz, $$JK = \\left((x_2-x_1)^2+(x_3-x_1)^2+(x_3-x_2)^2+\\ldots+(x_n^2-x_{n-1}^2)\\right)\\left((1^2)+(2^2+1^2)+(3^2+\\ldots+2^2+1^2)\\right)$$ $$\\ge \\left(1(x_2-x_1)+2(x_3-x_1)+1(x_3-x_2)+\\ldots+2(x_n-x_{n-2})+1(x_n-x_{n-1})\\right)^2$$ $$= \\left(c_nx_n+c_{n-1}x_{n-1}+c_{n-2}x_{n-2}+\\ldots+c_1x_1\\right)^2 = I^2$$ where the last step can be proven by considering specific values of $x_i$. \n\nNext, I will find the specific value of $K$. We have the following lemmas: $\\sum_{i = 1}^ni^3 = \\frac{n^2(n+1)^2}{4}, \\sum_{i = 1}^ni^2 = \\frac{n(n+1)(2n+1)}{6}, \\sum_{i = 1}^ni = \\frac{n(n+1)}{2}.$ Therefore, $$K = \\sum_{i = 1}^{n-1} \\frac{i(i+1)(2i+1)}{6} = \\sum_{i = 1}^n \\frac{2i^3+3i^2+i}{6} = \\frac{\\frac{(n-1)^2n^2}{2}+\\frac{(n-1)n(2n-1)}{2}+\\frac{(n-1)n}{2}}{6} = \\frac{(n-1)n^2(n+1)}{12}.$$ Therefore, we have $$R = \\frac{4(n^2-1)}{3K}JK \\ge \\frac{4(n-1)(n+1)}{\\frac{3(n-1)n^2(n+1)}{12}}I^2 = \\frac{16}{n^2}I^2.$$\n\nFinally, it suffices to prove $L = \\frac{16}{n^2}I^2$. Note that $L = 4\\left(\\sum_{i \\ge 1 \\ge 1}|x_i-x_j|\\right)^2$. Also, let $d_i = 2i-(n+1)$. For specific $x_p$, notice that if $i > p$, then $|x_i-x_p|$ will result in a negative $x_p$ term. If $i < p$, then $|x_p-x_i|$ will result in a positive $x_p$ term, and otherwise nothing. We now see that the coefficient of $x_p$ will be $$\\left((p-1)-1+1\\right)-\\left(n-(p+1)+1)\\right) = 2p-n-1$$ as desired. However, $\\frac{nd_i}{2} = pn-\\frac{n^2}{2}-\\frac{n}{2} = \\binom{n}{2}-n(n-p) = xc_i$. This means $L = 4\\left(\\frac{2}{n}\\right)^2I^2 = \\frac{16}{n^2}I^2$, which proves the inequality. Finally, we see that equality holds when $\\frac{|x_i-x_j|}{|i-j|}$ is a constant which gives equality in the Cauchy-Schwarz. Since $i < j \\implies x_i < x_j$, this means $x_1, x_2, \\ldots x_n$ must be an arithmetic sequence. $\\blacksquare$\n\n[b][color=#f00]Remarks: [/color][/b] If this problem did not tell you the equality condition, it would be quite harder because you have to guess it, which makes the Cauchy-Schwarz part much more contrived. I actually enjoyed bashing this problem unlike some other problems I did.", "Solution_17": "Shift so that \\(\\sum_{i=1}^nx_i=0\\). By Cauchy-Schwarz, \\[ \\left[\\sum_{i=1}^n\\left(i-\\frac{n+1}2\\right)x_i\\right]^2 \\le\\frac{n(n+1)(n-1)}{12}\\cdot\\sum_{i=1}^nx_i^2;\\] that is, \\[\\left[4\\sum_{i=1}^n\\left(i-\\frac{n+1}2\\right)x_i\\right]^2 \\le\\frac{2(n^2-1)}3\\left[2n\\sum_{i=1}^nx_i^2-2\\left(\\sum_{i=1}^nx_i\\right)^2\\right],\\] which is equivalent to the desired inequality. Equality holds iff \\(x_1\\), \\ldots, \\(x_n\\) is an arithmetic sequence.", "Solution_18": "Since shifting the sequence $(x_1,x_2,\\dots, x_n)$ does not affect the problem, we can assume that the sum is $0$. We have\n\\begin{align*}\\left(\\sum_{i,j=1}^{n}|x_i-x_j|\\right)^2 &=\\left(2\\sum_{1\\le i\\le j\\le n}(x_j-x_i)\\right)^2 \\\\\n&= \\left((2n-2)x_n+(2n-6)x_{n-1}+\\dots +(2-2n)x_1\\right)^2 \\\\\n&\\le ((2n-2)^2+(2n-6)^2+(2n-10)^2+\\dots + (2-2n)^2)(x_1^2+x_2^2+\\dots + x_n^2) \\\\\n&= \\frac{4(n-1)(n)(n+1)}{3}(x_1^2+x_2^2+\\dots + x_n^2) \\\\\n&= \\frac{2(n^2-1)}{3}\\cdot 2(nx_1^2 + nx_2^2 + \\dots + nx_n^2) \\\\\n&= \\frac{2(n^2-1)}{3}\\cdot 2\\left((n-1)\\left(\\sum_{i=1}^{n}{x_i^2}\\right) + \\left(\\sum_{i=1}^{n}{x_i}\\right)^2 - 2\\sum_{1\\le i m^2 + mn + n^2 \\ \\forall m,n >0$, thus we can let $m^2 + mn + n^2 = {(m+n-k)}^2$ for some $k \\in \\mathbb{Z}^+$. Expanding out, we get the equality $mn - 2km - 2kn + k^2 = 0$, which is equivalent to $(m-2k)(n-2k) = 3k^2$.\r\n\r\nFrom our original assumption that $\\gcd (m, n) = 1$, we have $\\gcd (m-2k, n-2k) = 1$, thus $\\exists \\ q_1 \\in \\mathbb{Z}^+ , q_1 \\mid k$, such that letting $q_2 = \\frac{k}{q_1}$, we have $\\gcd(q_1, q_2) = 1$, satisfying the condition that either $3{q_1}^2 \\mid m-2k, {q_2}^2 \\mid n-2k$ or ${q_1}^2 \\mid m-2k, 3{q_2}^2 \\mid n-2k$. By symmetry of $m, n$, we can assume WLOG that $3{q_1}^2 \\mid m-2k, {q_2}^2 \\mid n-2k$.\r\n\r\nNote that we can say even further more, that $\\mid m-2k \\mid = 3{q_1}^2$, and $\\mid n-2k \\mid = {q_2}^2$, thus we have either $m = 2k + 3{q_1}^2, n = 2k + {q_2}^2$ or $m = 2k - 3{q_1}^2, n = 2k - {q_2}^2$. This is equivalent to $m = q_1 (2q_2 \\pm 3q_1), n = q_2 (2q_1 \\pm q_2)$.\r\n\r\nFinally, dropping the condition that $m, n$ must be coprime, we thus have the final solution of the form $m = d q_1 (2q_2 \\pm 3q_1), n = d q_2 (2q_1 \\pm q_2)$, where $d, q_1, q_2 \\in \\mathbb{Z}^+, \\gcd (q_1, q_2) = 1$, with the additional checking to be done that $m, n$ are positive in the case of $m = d q_1 (2q_2 - 3q_1), n = d q_2 (2q_1 - q_2)$. Indeed, checking back into the expression $m^2 + mn + n^2$, we have $m^2 + mn + n^2 = {d^2 (3q_1^2 \\pm 3q_1q_2 + q_2^2)}^2$, which is a perfect square. Computing, some of the first few solutions are $(3, 5), (5, 16), (7, 8), (7, 33), (11, 24), \\ldots$", "Solution_2": "[quote]From our original assumption that \\gcd (m, n) = 1, we have \\gcd (m-2k, n-2k) = 1[/quote]\r\n\r\nit's not true, take n=5, m=8, k=1. \r\ncan you make it right?" } { "Tag": [ "AMC", "AIME", "probability", "rotation", "AIME II" ], "Problem": "Couldn't understand the solution for AIME II, 2003, #12 and #13. I'd really really appreciate a step by step and carefully written solution.\r\n\r\n12. The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates. For each candidate, the exact precentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate. What is the smallest possible number of members in the committee?\r\n\r\n13. A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is m/n, where m and n are relatively positive integers, find m+n.\r\n\r\nI have some headway on #12, it seems, but #13 is completely mindboggling.\r\nThanks a lot.", "Solution_1": "[hide=\"3\"]\nPlace the bug and the triangle on the complex plane with the vertices as the third roots of unity and $ 1$, i.e. $ 1$, $ \\omega$, and $ \\omega^{2}$ where $ w^{3}=1$ is not purely real.\n\nThen, the probability will be the coefficient on $ 1$, when powers are reduced as far as possible, in the expansion of:\n\\[ \\left( \\omega+\\frac{1}{\\omega}\\right)^{10}\\]\ndivided by $ 2^{10}$, as each time we multiply a complex number by $ \\omega$ it is rotated around the complex plane $ 120^\\circ$ in the counterclockwise direction (by DeMoivre's Theorem), and similarly dividing by $ \\omega$ rotates in the clockwise direction. (Why?)\n\n This is equal to:\n\\[ \\frac{ \\binom{10}{2}+\\binom{10}{5}+\\binom{10}{8}}{2^{10}}= \\frac{171}{512}\\implies m+n= \\boxed{683}\\]\n[/hide]", "Solution_2": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=713219#p713219[/url]\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=713221#p713221[/url]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "assume that $G=(V,E)$ is a connected graph with minimum degree $\\delta$ prove that if $|V|>2\\delta$ then $G$ contains a path with length at least $2\\delta$ :?:", "Solution_1": "take the longest path, if it's shorter than $2d$ then this path is a cycle, since it is a cycle any vertex cannot be adjacent to any vertex which is not in that cycle, so this graph cannot be connected", "Solution_2": "yeah correct your solution is the same as mine...\r\nany other solutions?? :?:" } { "Tag": [ "geometry", "inradius", "incenter", "circumcircle", "geometric transformation", "reflection", "Euler" ], "Problem": "Given a triangle $ ABC$ with an interior point $ P$ in it, suppose the distances from $ P$ to $ BC,CA, AB$ are $ p,q,r$ respectively. \r\nAlso, suppose that $ pqr\\ge R^3$ where $ R$ is the inradius of triangle $ ABC$.\r\nProve that the point $ P$ lies inside or on the circle with centre $ S$ and radius $ SI$ where $ I$ is the incenter and $ S$ circumcenter of triangle $ ABC$.", "Solution_1": "This problem was posted on Mathematical reflections with the solution. As soon as I get the link I post it. The clue to solve this problem is using the Euler pedal theorem plus some AM-GM inequalities" } { "Tag": [ "vector", "function", "algebra", "polynomial" ], "Problem": "I've studied Diophantine Equations in two variables, but I have not found any resources for those in three or more variables. Consider, for example, the equation $ ax\\plus{}by\\plus{}cz\\equal{}n$, where $ a$, $ b$, $ c$, and $ n$ are positive integers and $ x$, $ y$, and $ z$ are the variables, also constrained to be positive integers.\r\n\r\nPerhaps what I am most interested in is a way to count the number of nonnegative solutions to an equation like this. For an equation like the above where $ a$, $ b$, and $ c$ are $ 1$, we can use balls and urns, but I don't know what to do otherwise. A method for counting solutions would come in handy on coin problems; for example, counting the number of ways to make change for a dollar using pennies, nickels, and dimes would be equivalent to counting the solutions to $ x\\plus{}5y\\plus{}10z$ for integers $ x$, $ y$, and $ z$. Any suggestions?", "Solution_1": "$ ax\\plus{}by\\plus{}cz\\equal{}n$ defines a plane in 3-space with normal vector $ $. The number of solutions is, of course, the number of lattice points in quadrant 1 on the plane.\r\n\r\nA quick Googling found [url=http://adsabs.harvard.edu/abs/2008arXiv0805.1702Z]this[/url].", "Solution_2": "If you're familiar with the generating functions approach to the balls-and-urns problem, the corresponding generating function for pennies, nickels, and dimes is\r\n$ \\frac {1}{(1 \\minus{} x)(1 \\minus{} x^5)(1 \\minus{} x^{10})} \\equal{} s_0 \\plus{} s_1 x \\plus{} ...$\r\n\r\nwhere $ s_n$ is the number of ways to make change for $ n$ cents. In theory, it is possible to solve this problem by finding the partial fraction decomposition of the LHS, and the answer is some sum of exponentials with polynomial coefficients involving the fifth and tenth roots of unity. \r\n\r\nNote that this is very different from asking the same question about integers alone; compare the [url=http://en.wikipedia.org/wiki/Coin_problem]coin problem[/url] with [url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity]Bezout's identity[/url]. Without the positive restriction, the three-variable problem reduces to the two-variable problem. The difficulty of this problem is not in the number of variables, but in the positivity restriction." } { "Tag": [ "geometry", "rotation", "integration", "calculus", "calculus computations" ], "Problem": "I know this problem is very easy for all of you but I would like to know the answer:\r\n\r\nThe line segment connecting the point (1,2) to (5,5) is rotated around the line x= -3.\r\nFind the surface area of the resulting solid.", "Solution_1": "anyone?\r\n\r\nI just want the answer... it doesnt matter if you dont want to post the solution", "Solution_2": "that does not have a finite surface area, you need to have more restrictions because the line continues on forever, making it infinite surface area, but if you do give other restrictions, one thing that will help would be moving everything up 3 units, so you would rotate about the y-axis", "Solution_3": "No, it has finite area. qweretyq's problem did say the line segment, not the line.\r\n\r\nLet's find an equation for that line segment: $-3x+4y=5,$ or $x=\\frac{4y-5}3$ for $2\\le y\\le 5.$\r\n\r\nTake an infinitesimal piece of this segment. Its length is $ds=\\sqrt{dx^2+dy^2}=\\sqrt{1+\\left(\\frac{dx}{dy}\\right)^2}\\,dy= \\sqrt{1+\\left(\\frac43\\right)^2}=\\frac53\\,dy.$ When you rotate it around the line $x=-3,$ it forms a slanted ribbon whose length is the circumference of a circle, $2\\pi(x+3),$ and whose width is $ds.$\r\n\r\nArea = $\\int2\\pi(x+3)ds=\\int_2^52\\pi\\left(\\frac53\\right)\\left(\\frac{4y-5}3+3\\right)dy =\\frac{700\\pi}9.$\r\n\r\nThe rotated object is a portion of a cone. The cone's vertex lies on the $(x,y)$ plane at the point $(-3,-1).$" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "Let $a 0$ there is a $\\delta >0$such that if $P$ and $Q$ are partitions of $[a,b]$ with $\\| P \\|<\\delta$ and $\\| Q \\| <\\delta$ then $| R(f,P)- R(f,Q) | <\\epsilon$ for any Riemann sums $R(f,P)$ and $R(f,Q)$ of $f$ with respect to the partitions $P$ and $Q$. ( the two Riemann sums do not ncecssarily involve the same sample points) \r\n\r\nShow that $f$ is integrable", "Solution_1": "Note that most of this is the same as the definition of integrability. Mimic the proof that a Cauchy sequence of real numbers is convergent (or use the fact to produce a candidate integral $I$).", "Solution_2": "Could you explain in detail? Please help. Thx alot." } { "Tag": [ "modular arithmetic", "number theory", "least common multiple" ], "Problem": "If $ n$ is any whole number, $ n^2(n^2 \\minus{} 1)$ is always divisible by\r\n\r\n$ \\textbf{(A)}\\ 12 \\qquad\\textbf{(B)}\\ 24 \\qquad\\textbf{(C)}\\ \\text{any multiple of }12 \\qquad\\textbf{(D)}\\ 12 \\minus{} n \\qquad\\textbf{(E)}\\ 12\\text{ and }24$", "Solution_1": "Is $ 0$ divisible by $ 12$?\r\nWhat happens if $ n\\equal{}1$?\r\nWhat is AHSME, and why do these problems look so easy?", "Solution_2": "[hide=\"Solution\"]\n\n[b](A)[/b]. \n\nConsider the case where $ n \\equal{} 2$. \n\n$ 2^2(2^2 \\minus{} 1) \\equal{} 12$, which clearly eliminates choices [b]B-E[/b].\n[/hide] \r\n\r\nHere's more information on the [url=http://www.artofproblemsolving.com/Wiki/index.php/AHSME]AHSME[/url].", "Solution_3": "I understood that the hidden solution would be the official solution, and I might have checked that answer if I was taking the test and had something to gain by doing so.\r\nHowever, I do not like any of the choices because when $ n\\equal{}\\minus{}1$, or $ n\\equal{}0$, or $ n\\equal{}1$, $ n^2(n^2 \\minus{} 1)\\equal{}0$, and I have never called $ 0$ a multiple of anything.\r\nIf $ 0$ can be called a multiple of any natural number, I would have to agree with the official solution.\r\nIf $ 0$ cannot be called a multiple of any natural number, the expression may sometimes be divisible by that one answer, but is not always divisible by any of the choices. So it all hinges on what the meaning of the word \"is\" is (or rather on what the meaning of the word \"divisible\" is).", "Solution_4": "By definition, $ n|0$ for all $ n\\in\\mathbb{R}$.\r\nWe can claim this because $ 0\\equiv 0\\pmod{n}$ for any $ n$. By definition, $ n|k\\iff k\\equiv 0\\pmod n$.", "Solution_5": "[hide=\"My Solution\"]\nWe have $ n^2(n^2 \\minus{} 1) \\equal{} n\\cdot (n \\minus{} 1)n(n \\plus{} 1)$. Clearly, this quantity is divisible by $ 3$ and $ 2$ because there are 3 consecutive integers. Also, if $ n$ is even, then the whole quantity is divisible by $ 4$. If $ n$ is odd, then the whole quantity will still be divisible by $ 4$. We have covered all the cases so $ n^2(n^2 \\minus{} 1)$ is therefore divisible by $ \\mathrm{lcm}{(2,3,4)} \\equal{} \\boxed{12 \\text{(A)}}$.\n[/hide]", "Solution_6": "factor n(n)(n+1)(n-1). if n is odd then due to n+1 and n-1 it is automatically divisble by 8. when n is even then it is divisile by 4. So we already have it being divisible by 4. but since there is n(n+1)(n-1) 3 consecutive numbers then it is divisible by 3. So hence it is divisible by A)12", "Solution_7": "This is a reply to a months-old discussion, but the formal definition of divisibility is \"$ a \\mid b$ if and only if there exists a $ c$ such that $ ac \\equal{} b$.\" (not mods, that's reversed; mods are defined by divisibility) Here if $ b \\equal{} 0$, then $ c \\equal{} 0$ for any $ a$ works, so we say everything divides $ 0$. The definition of divisibility is not something in dispute, by the way." } { "Tag": [ "calculus", "trigonometry" ], "Problem": "Hi everybody, I'm an 8th grade, I'm doing in my math class good so far. I got for my first marking period A+ and for second one, its comming probably A+ too. I'm little bit worried about my high school math, I looked in those problems under high school math they look scary. Maybe most of you guys been like that?Is it easy to learn high level math? Took you more time and research to figure out problems? and how you did in 8th grade, and high school math?Thanks\r\n\r\n\r\nSorry for my bad English, I'm new in country so....", "Solution_1": "I've moved this thread to this forum, where it will hopefully get more replies.\r\n\r\nNeedless to say, nice job on your math results so far! \r\n\r\nMy two cents:\r\n\r\n(1) Pretty much everything looks scary until you learn how to do it. For a baby, standing on two feet and walking is pretty scary, and babies bump their heads a lot, but eventually they learn to walk. The first time I drove a car, it was terrifying. Then, after a few times, it became the coolest thing ever. (Now I live in Southern California, so it's terrifying again, for a different reason.) \r\n\r\n(2) Most things worth doing aren't easy. That's why they're worth doing.\r\n\r\nI suspect that if you're getting A+'s in 8th grade math at your school, that you will continue to do excellent in your school classes. For very good students (like yourself), the type of math taught in most schools is pretty easy. It's the type of math -- and problem solving! -- that we do on this site that's hard. See (2) above.", "Solution_2": "[quote=\"might\"]I'm little bit worried about my high school math, I looked in those problems under high school math they look scary. Maybe most of you guys been like that?Is it easy to learn high level math? Took you more time and research to figure out problems? and how you did in 8th grade, and high school math?Thanks[/quote]\r\n\r\nWhile DPatrick's points still stand, please note that the problems in the high school section of AoPS are *nothing* like what you'll be doing in a normal high school. AoPS is for interested, bright students. High school is not. So relax, if you enjoy math, it'll all come together for you :)", "Solution_3": "well, I'm kind of thinking I wanna become a computer engineer, so I would need take I think calculus or trignomerty or something like that for it.", "Solution_4": "You will do trigonometry, eventually. You will do calculus, eventually. There's no need to be worrying about them in 8th grade.\r\n\r\nCalculus is normally done in the last year of high school or the first year of college, for a student interested in math/science/engineering.", "Solution_5": "If you are interested in learning mathematics, in really learning it than I would recommend taking the hardest math classes your school has to offer. But, also get involved in extracurricular activies related to different types of math, Summer programs are probably your best bet.\r\n\r\nDon't fall into the trap of thinking that high school math is all you will need (as I did and am now regretting)." } { "Tag": [ "inequalities open", "inequalities" ], "Problem": "Let $ a;b;c;d$ be positive real numbers.Prove that :\r\n\r\n$ (\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d})(\\frac{1}{1+a}+\\frac{1}{1+b}+\\frac{1}{1+c}+\\frac{1}{d}) \\geq \\frac{16}{1+abcd}$", "Solution_1": "[quote=\"chien than\"]Let $ a;b;c;d$ be positive real numbers.Prove that :\n\n$ (\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d})(\\frac{1}{1+a}+\\frac{1}{1+b}+\\frac{1}{1+c}+\\frac{1}{d}) \\geq \\frac{16}{1+abcd}$[/quote]\r\nMaybe $ \\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}\\right)\\left(\\frac{1}{1+a}+\\frac{1}{1+b}+\\frac{1}{1+c}+\\frac{1}{1+d}\\right) \\geq \\frac{16}{1+abcd}$ $ ?$\r\nIf so, try $ a=b=c=d=0.8.$ :wink:" } { "Tag": [ "geometry", "3D geometry", "prism" ], "Problem": "Amy wants to design a rectangular prism with a volume of 32 cubic units such that one edge is twice as long as another and each dimension is an integer. How many unique prisms could she make?", "Solution_1": "We have sides x, 2x, and y. So then 2*x^2*y=32. \r\nx^2*y =16\r\nThen we have ordered pairs (1,16),(2,4), and (4,1) for a total of 3 prisms." } { "Tag": [], "Problem": "What is the digit in the thousandths place of the decimal\nequivalent of $ \\frac{3}{16}$?", "Solution_1": "What is there to say? Dividing reveals the thousands digit as $ \\boxed{7}$.", "Solution_2": "$ \\frac {3}{16} \\equal{} 3(\\frac {1}{16}) \\equal{} 3\\cdot0.0625 \\equal{} .1875$\r\n\r\nHence, the answer is $ \\boxed{7}$" } { "Tag": [], "Problem": "What is the sum of the solutions to $ (y \\plus{} 2)^2 \\equal{} 25$?", "Solution_1": "If $ (y \\plus{} 2)^2 \\equal{} 25$, then y+2=5 or -5. These possibilities give 3 and -7 for y, so the answer is 3+(-7)=-4." } { "Tag": [ "abstract algebra", "group theory", "invariant", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ G$ be a finite abelian group and $ x$ is an element of maximal order.Let $ H$ be a maximal subgroup of $ G$ such that $ H \\cap \\langle x\\rangle\\equal{}\\{0\\}$(which means if $ K$,such that $ H 0$, define $ a_0$ and $ a_1$ two positive integers each of the form $ 2p$, where $ p$ is a arbitrary prime much and much greater of $ t$.\r\n\r\nNow consider the recurrence $ a_{n \\plus{} 2} \\equal{} \\frac {a_{n \\plus{} 1} \\plus{} a_n}{2} \\plus{} 2t$ forall $ n \\in \\mathbb{N}$.\r\n\r\nThere exist a index $ i$ (explicit or in function of t) such that $ a_j$ is not a integer of the form $ 2p$, for some $ 2 \\le j \\le i$?\r\n\r\n[size=75](Ask me if the question is not clear)[/size]", "Solution_1": "[quote=\"bboypa\"]Fixed a positive integer $ t > 0$, define $ a_0$ and $ a_1$ two positive integers each of the form $ 2p$, where $ p$ is a arbitrary prime much and much greater of $ t$.\n\nNow consider the recurrence $ a_{n \\plus{} 2} \\equal{} \\frac {a_{n \\plus{} 1} \\plus{} a_n}{2} \\plus{} 2t$ forall $ n \\in \\mathbb{N}$.\n\nThere exist a index $ i$ (explicit or in function of t) such that $ a_j$ is not a integer of the form $ 2p$, for some $ 2 \\le j \\le i$?\n\n[size=75](Ask me if the question is not clear)[/size][/quote]\r\nWe have $ a_{n\\plus{}2}\\minus{}a_{n\\plus{}1}\\equal{}2t\\minus{}\\frac{a_{n\\plus{}1}\\minus{}a_n}{2}$\r\n\r\nAnd so $ a_{n\\plus{}2}\\minus{}a_{n\\plus{}1}\\equal{}t\\plus{}\\frac{a_{n}\\minus{}a_{n\\minus{}1}}{4}$\r\n\r\nAnd so $ a_{2k\\plus{}1}\\minus{}a_{2k}\\equal{}t\\plus{}\\frac{t}{4}\\plus{}\\frac{t}{4^2}\\plus{}$...$ \\plus{}\\frac{t}{4^{k\\minus{}1}}\\plus{}\\frac{a_1\\minus{}a_0}{4^{k}}$\r\n\r\n$ a_{2k\\plus{}1}\\minus{}a_{2k}\\equal{}\\frac{4t}{3}(1\\minus{}\\frac{1}{4^k})\\plus{}\\frac{a_1\\minus{}a_0}{4^k}$\r\n\r\nAnd obviously, there exist an index $ k$ such that RHS is not an integer." } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "linear algebra unsolved" ], "Problem": "Solve the equation\r\n\r\n$X^{2006}=\\left(\\begin{array}{cc} 9&3\\\\\\ 15&5\\end{array}\\right)$\r\n\r\n :)", "Solution_1": ":roll: : t = Tr(T), d = det(T), $T^2 - t*T + d*I_2 = O_2$, d = 0, t = 14, $X^2 = t*X, X^{2006} = t^{2005}*X, X = ...$", "Solution_2": "[quote=\"mareleG\"]:roll: : t = Tr(T), d = det(T), $T^2 - t*T + d*I_2 = O_2$, d = 0, t = 14, $X^2 = t*X, X^{2006} = t^{2005}*X, X = ...$[/quote]\r\n\r\nNIce,nice.... :) \r\n\r\nBut how did you find that $T^2 - t*T + d*I_2 = O_2$?? :?:", "Solution_3": "it is the caracteristic polynomial. But if you don't know caracteristic polynomial theory, you can check this out : this is just calculations ..", "Solution_4": "[quote=\"alekk\"]it is the caracteristic polynomial. But if you don't know caracteristic polynomial theory, you can check this out : this is just calculations ..[/quote]\r\n\r\nOh,yeah.\r\nI finally unterstood this. It was an easy one..... :D", "Solution_5": "Actually, Socrates one can easily solve the following general equation:\r\n\r\n \\[ X^{2}=A, \\]\r\n where $X\\in M_{2}(\\mathbb{C})$ and $det A=0$. :) ;)" } { "Tag": [ "linear algebra" ], "Problem": "Compute the determinant:\r\n\r\n$ \\triangle_n\\equal{} \\left| \\begin{array}{ccccc}\r\n1 & a_1 & a_2 & \\dots & a_{n\\minus{}1} \\\\\r\n1 & \\minus{} \\frac{1}{a_1} & 0 & \\dots & 0\\\\\r\n1 & 0 & \\minus{}\\frac{1}{a_2} & \\dots & 0 \\\\\r\n\\dots & \\dots & \\dots & \\dots & \\dots \\\\\r\n1 & 0 & 0 & \\dots & \\minus{} \\frac{1}{a_{n\\minus{}1}} \\end{array} \\right|$ , where $ a_1,a_2, \\dots a_{n\\minus{}1} \\in \\mathbb{C}^*$.", "Solution_1": "$ \\Delta_{n} = \\left|\\begin{array}{ccccc} 1 & {a_1} & {a_2} & {\\dots} & {a_{n - 1}} \\\\\r\n1 & { - \\frac {1}{a_1}} & 0 & {\\dots} & 0 \\\\\r\n1 & 0 & { - \\frac {1}{a_2}} & {\\dots} & 0 \\\\\r\n{\\dots} & {\\dots} & {\\dots} & {\\dots} & {\\dots} \\\\\r\n1 & 0 & 0 & {\\dots} & { - \\frac {1}{a_{n - 1}}}\\end{array}\\right| = \\\\\r\n= \\frac {1}{a_1a_2.\\dots a_{n - 1}}.\\left|\\begin{array}{ccccc} 1 & {a_1} & {a_2} & {\\dots} & {a_{n - 1}} \\\\\r\n{a_1} &{ { - 1}} & 0 & {\\dots} & 0 \\\\\r\n{a_2} & 0 &{ { - 1}} & {\\dots} & 0 \\\\\r\n{\\dots} & {\\dots} & {\\dots} & {\\dots} & {\\dots} \\\\\r\n{a_{n - 1}} & 0 & 0 & {\\dots} & { - 1}\\end{array}\\right| = \\\\\r\n \\\\\r\n= \\frac {1}{a_1^2a_2^2.\\dots a_{n - 1}^2}.\\left|\\begin{array}{ccccc} 1 & {a_1^2} & {a_2^2} & {\\dots} & {a_{n - 1}^2} \\\\\r\n{a_1} &{ { - a_1}} & 0 & {\\dots} & 0 \\\\\r\n{a_2} & 0 &{ { - a_2}} & {\\dots} & 0 \\\\\r\n{\\dots} & {\\dots} & {\\dots} & {\\dots} & {\\dots} \\\\\r\n{a_{n - 1}} & 0 & 0 & {\\dots} & { - a_{n - 1}}\\end{array}\\right| = \\\\\r\n\\\\\r\n= \\frac {1}{a_1^2a_2^2.\\dots a_{n - 1}^2}.\\left|\\begin{array}{ccccc} {1 + a_1^2 + a_2^2 + \\dots + a_{n - 1}^2} & {a_1^2} & {a_2^2} & {\\dots} & {a_{n - 1}^2} \\\\\r\n0 &{ { - a_1}} & 0 & {\\dots} & 0 \\\\\r\n0 & 0 &{ { - a_2}} & {\\dots} & 0 \\\\\r\n{\\dots} & {\\dots} & {\\dots} & {\\dots} & {\\dots} \\\\\r\n0 & 0 & 0 & {\\dots} & { - a_{n - 1}}\\end{array}\\right| = \\\\\r\n= \\frac {( - 1)^{n - 1}.(1 + a_1^2 + a_2^2 + \\dots + a_{n - 1}^2)}{a_1a_2.\\dots a_{n - 1}}.$" } { "Tag": [ "geometry", "trigonometry", "area of a triangle" ], "Problem": "Find the largest possible area of a triangle that has two sides of lengths 30 and 40.", "Solution_1": "The triangle with the major area is the right and he area is 600", "Solution_2": "Hey, uh...could you elaborate on why the right triangle always has the most area?", "Solution_3": "The area of a triangle can be determined by two sides and their included angle, and is equal to $ \\frac{1}{2} ab \\sin \\theta$. Since we know $ a$ and $ b$, we would like to maximize $ \\sin \\theta$, which is done by letting $ \\theta \\equal{} \\frac{\\pi}{2}$." } { "Tag": [ "analytic geometry", "USAMTS", "search", "inequalities", "geometry", "Gauss" ], "Problem": "How hard did you think round 1 was in comparison to round 1 of Year 19.\r\n\r\nI personally think that this year's was harder than last year's but not by much. first of all, last year, in 8th grade, i got a 19, and now, since i have gotten better, and i am able to solve one of the problems that i got a 1 on last year, i could probly get a 23 or 24 maybe 25. on this year's test, #1 was very tricky and required alot of precise casework, which is not good for a #1. So this year, i see myself struggling to get a 21 or 22....", "Solution_1": "I found them sort of the same dificulty...\r\n\r\nHowever, I think that this round was easier to make a stupid mistake on.", "Solution_2": "I thought this year was easier, I could solve all of them, but perhaps I just put in more time this year.", "Solution_3": "I've never done this contest before, but I looked at a few past rounds before (not first rounds in particular), and this round was easier than I expected. Just now I looked at the round 1 / year 19 problems, and the last one seems kind of hard. Then again, it would probably be possible to brute-force it with coordinates.", "Solution_4": "Hmm... solutions.tex > Properties > Modified: Tuesday, August 05, 2008. I'm weird.\r\n\r\nI suppose the questions were a bit trickier, though I didn't find them as difficult as before (or there's the unlikely possibility that I might have improved..).", "Solution_5": "I find these problems are getting too bashable: 1 (CASEWORK) 4 (COORDINATES) 2 and 3 were pretty easy (unless I messed up... ugh) In my opinion, 5 was the only nice problem. Although it did get a little bit bashy with the algebra, I actually like the problem.\r\n\r\nAnyway, I also think that the non-bashable problems are getting harder...", "Solution_6": "Maybe it's just me, but I thought the problems were a bit too easy. Like last year, I couldn't solve a single problem, but this year, I looked at the problems and was like...yeah, too simple. Then again, I was reeeeeeeeeeeeaaaaaaaaaaaaaaaaaalllllllllllllllly stupid last year, but then again, I might still be this year.", "Solution_7": "its proabbly because you're smarter this year....\r\n\r\ni looked at both years probs and their all hard :D", "Solution_8": "I think I partially screwed up #1, kinda did good on #2, totally messed up #3 and #4, and did well on #5.\r\nMeh. Epoch Phail... :(", "Solution_9": "If this year is easy, I don't know how I'm gonna survive next year... :P", "Solution_10": "I think this year's USAMTS was hard. Pretty hard, compared to older ones.\r\nSo i guess they'd just be getting harder and harder as the years go by...", "Solution_11": "I think this round was harder, but it could just have been me being stupid (especially on 2 and 5). #2... I tried to use $ lcm(x,y,z)|x \\plus{} y \\plus{} z \\minus{} 1$ and got stuck from there. After a while of getting nowhere, I tried to write a program to search for solutions based on this, and this gave me nothing because I messed up the order of operations in the program...\r\n\r\n#5, I started with the same idea as the official, but I tried to flat out expand for the $ r$ odd case, rearrange into two square roots, and show the numbers we were square-rooting had a difference of 1. After pages of work, I just got stuck. epic failure, hoping for a 3.\r\n\r\n[i]Don't discuss the current round!! -- Admin[/i]", "Solution_12": "Hey, I did the exact same epic failure for that number five too! I had the same idea of splitting into odd and even, and epically failed the odd case by expanding!\r\n\r\n :ninja:", "Solution_13": "I think this year's first round was harder but the problems were not as well developed. If you know what I mean..", "Solution_14": "the problems in general were easier, but #1 on this year's test was like WHAT? i missed it i was off by 1, so that kinda stinx. oh well. so yeah, they should kinda cut down on the bashing problems.", "Solution_15": "I think this year was much easier.\r\nLast year, I looked at the problems, stared at them for like another hour, and all I was thinking was Say Whaaat?\r\nAnd then chickened out after a couple days.\r\nBut this year I solved 1,2,4,and 5. 3 turned into an inequality bash for me that didn't get anywhere.", "Solution_16": "last year #1 < this year #1\r\nlast year #2 > this year #2\r\nlast year #3 < this year #3\r\nlast year #4 < this year #4 ( i like 2-d geometry more than 3-d geometry)\r\nlast year #5 < this year #5 \r\n\r\nso yeah..... this is my opinion.....", "Solution_17": "really?\r\n\r\nI found that \r\n\r\nthis year #1 < last year #1.\r\n\r\nActually, it is pretty hard to mess up #1 with Burnside's Lemma actually.\r\n\r\nI made several mistakes but whenever I made a mistake, my answer was not an integer. So I knew I had made a mistake.", "Solution_18": "how exactly do you use burnside's lemma anyway? I don't quite get what your are supposed to count...\r\n\r\nSo I missed number 1 (got 23) by stupidly overcounting. :-<\r\n\r\nAs for #2, last year's was easier. by far. The angles you could find from the diagram easily, and the rest was playing with the angles.\r\n\r\n#1 this year... idk. I think its not quite comparable to #1 round 1 last year, but some of the other rounds had similar problems. But I'd rate it a bit harder than most geometric counting problems, but it can still be done by casework.\r\n\r\n#2 - so i tried to use $ lcm(x,y,z)|x\\plus{}y\\plus{}z\\minus{}1$. found a few useless facts, overall got nowhere, and then tried writing a program to find the answers. But i Phailed that too... just figured out I messed up stinking order of operations there. I just thought there were no solutions.\r\n\r\noverall :\\", "Solution_19": "[quote=\"facis\"]how exactly\n\nSo I missed number 1 (got 23) by stupidly overcounting. :-<\n\n[/quote]\r\n\r\ni'm with you.... :(", "Solution_20": "I don't think #1 is hard as long as you count your casework carefully and avoid stupid mistakes.", "Solution_21": "wow we're still talking about round 1?\r\n\r\ngauss are you done with round 2??\r\n\r\nEDIT: hey 1181 posts", "Solution_22": "yes, it's not like we can talk about any other rounds...*cough*" } { "Tag": [ "integration", "limit", "logarithms", "calculus", "real analysis", "real analysis unsolved" ], "Problem": "a question i asked in class yesterday... :D took nearly 30 mins to get it done completely... students got both confused and excited at the same time... \r\n\r\nwe already know from riemann,\r\n$ \\int_a^b\\;f(x)\\;dx\\;\\equal{}\\;\\lim_{n\\to\\infty}\\left\\{\\left(\\dfrac{b\\minus{}a}{n}\\right)\\sum_{m\\equal{}1}^n\\;f\\left(a\\plus{}\\dfrac{m(b\\minus{}a)}{n}\\right)\\right\\}$\r\n\r\nnow show\r\n$ \\int_a^b\\;f(x)\\;dx\\;\\equal{}\\;\\sum_{n\\equal{}1}^\\infty\\;\\sum_{m\\equal{}1}^{2^n\\minus{}1}\\;\\left\\{(\\minus{}1)^{m\\plus{}1}\\cdot\\left(\\dfrac{b\\minus{}a}{2^n}\\right)\\cdot\\,f\\left(a\\plus{}\\dfrac{m(b\\minus{}a)}{2^n}\\right)\\right\\}$\r\n\r\nand hence show for $ 0\\leq p<\\infty$,\r\n\r\n$ \\sum_{n\\equal{}1}^\\infty\\;\\sum_{m\\equal{}1}^{2^n\\minus{}1}\\;\\dfrac{(\\minus{}1)^{m\\plus{}1}}{2^n}\\;\\left(\\ln\\,\\left(\\dfrac{2^n}{m}\\right)\\right)^p\\;\\equal{}\\;\\int_0^1\\;\\left(\\ln\\,\\left(\\frac{1}{x}\\right)\\right)^p\\;dx\\;\\equal{}\\;\\boxed{p!}$\r\n\r\napparently, this double series summing to $ p!$ was known to archimedes using this method ! what a genius !! :o", "Solution_1": "[quote=\"misan\"]\nwe already know from riemann,\n$ \\int_a^b\\;f(x)\\;dx\\; \\equal{} \\;\\lim_{n\\to\\infty}\\left\\{\\left(\\dfrac{b \\minus{} a}{n}\\right)\\sum_{m \\equal{} 1}^n\\;f\\left(a \\plus{} \\dfrac{m(b \\minus{} a)}{n}\\right)\\right\\}$[/quote]\r\nHow do you prove this?", "Solution_2": "[quote=\"Field\"][quote=\"misan\"]\nwe already know from riemann,\n$ \\int_a^b\\;f(x)\\;dx\\; \\equal{} \\;\\lim_{n\\to\\infty}\\left\\{\\left(\\dfrac{b \\minus{} a}{n}\\right)\\sum_{m \\equal{} 1}^n\\;f\\left(a \\plus{} \\dfrac{m(b \\minus{} a)}{n}\\right)\\right\\}$[/quote]\nHow do you prove this?[/quote]\r\n\r\nWell, when we know that the integral exists, it's obvious, since the RHS is just a sequence of Riemann sums. But the limit can exist even though the integral doesn't, so I wouldn't call it \"true\" exactly." } { "Tag": [ "inequalities", "linear algebra", "matrix", "complex numbers", "linear algebra unsolved" ], "Problem": "Let $ A,B \\in M_n{(\\mathbb{C})}$ and $ I$ is $ n^2$x$ n^2$ identity matrix.\r\nProve inequality Kronecker product Matrix :\r\n 1)$ \\det(A\\otimes{\\overline{A}} \\plus{} I) \\geq 0$\r\n 2)If $ AB \\equal{} BA$ then $ \\det(A\\otimes{\\overline{A}} \\plus{} B\\otimes{\\overline{B}}) \\geq 0$\r\n Here $ \\overline{A}$ is the conjugate matrix of $ A$ \r\n[hide=\"Answer of me\"]Implies but Very interesting[/hide] :P", "Solution_1": "[hide=\"Solution 1)\"]Let $ \\lambda_1,\\lambda_2,...,\\lambda_n$ are eigenvalues of $ A$ matrix we can similar ,are real or complex numbers .\n $ \\beta_1,\\beta_2,...,\\beta_n$ are eigenvalues of $ B$ matrix we can similar ,are real or complex numbers \n The exist $ T \\in GL_n{(\\mathbb{C})}$ Such that :\n $ A = T\\begin{pmatrix}\\lambda_1 & * & * & * & * & * & * \\\\\n0 & \\lambda_2 & * & * & * & * & * \\\\\n0 & 0 & \\lambda_3 & * & * & * & * \\\\\n... & ... & .... & ... & ... & ... & ... \\\\\n0 & 0 & 0 & ... & ... & ... & \\lambda_n\\end{pmatrix}T^{ - 1}$\nPut $ T_A = \\begin{pmatrix}\\lambda_1 & * & * & * & * & * & * \\\\\n0 & \\lambda_2 & * & * & * & * & * \\\\\n0 & 0 & \\lambda_3 & * & * & * & * \\\\\n... & ... & .... & ... & ... & ... & ... \\\\\n0 & 0 & 0 & ... & ... & ... & \\lambda_n\\end{pmatrix}$\nThen $ A = TT_AT^{ - 1}$\nUse $ \\forall X,Y \\in M_n{(\\mathbb{C})}$ then $ \\overline{XY} = (\\overline{X})(\\overline{Y})$ We have \n $ \\overline{A} = (\\overline{T})(\\overline{T_A})(\\overline{T})^{ - 1}$\nWe have $ A\\otimes{\\overline{A}} = (TT_AT^{ - 1})\\otimes{((\\overline{T})(\\overline{T_A})(\\overline{T})^{ - 1})} = (T\\otimes{\\overline{T}})(T_A\\otimes{\\overline{T_A}})(T^{ - 1}\\otimes{(\\overline{T})^{ - 1}}) = (T\\otimes{\\overline{T}})(T_A\\otimes{\\overline{T_A}})(T\\otimes{\\overline{T}})^{ - 1}$\nThen we have :\n $ \\det(A\\otimes{\\overline{A}} + I) = \\det(T_A\\otimes{\\overline{T_A}} + I) = \\det\\begin{pmatrix}\\lambda_1\\overline{T_A} + I_n & * & * & * & * & * & * \\\\\n0 & \\lambda_2\\overline{T_A} + I_n & * & * & * & * & * \\\\\n0 & 0 & \\lambda_3\\overline{T_A} + I_n & * & * & * & * \\\\\n... & ... & .... & ... & ... & ... & ... \\\\\n0 & 0 & 0 & ... & ... & ... & \\lambda_n\\overline{T_A} + I_n\\end{pmatrix} = \\prod_{i = 1}^{n}{(\\lambda_i\\overline{T_A} + I_n)} = \\prod_{i = 1}^{n}{(|\\lambda_i|^2 + 1)}\\prod_{1\\leq i < j \\leq n}{(\\lambda_i\\overline{\\lambda_j} + 1)(\\lambda_j\\overline{\\lambda_i} + 1)} = \\prod_{i = 1}^{n}{(|\\lambda_i|^2 + 1)}\\prod_{1\\leq i < j\\leq n}{|\\lambda_i\\overline{\\lambda_j} + 1|^2}\\geq o$[/hide]\n[hide=\"Solution 2)\"]$ AB = BA$ Then The exist $ T \\in GL_n{(\\mathbb{C})}$ Such that :\n $ A = TT_AT^{ - 1},B = TT_BT^{ - 1}$\nHere $ T_B = \\begin{pmatrix}\\beta_1 & * & * & * & * & * & * \\\\\n0 & \\beta_2 & * & * & * & * & * \\\\\n0 & 0 & \\beta_3 & * & * & * & * \\\\\n... & ... & .... & ... & ... & ... & ... \\\\\n0 & 0 & 0 & ... & ... & ... & \\beta_n\\end{pmatrix}$\nAnd $ A\\otimes{\\overline{A}} + B\\otimes{\\overline{B}} = (T\\otimes{\\overline{T}})(T_A\\otimes{\\overline{T_A}} + T_B\\otimes{\\overline{T_B}})(T\\otimes{\\overline{T}})^{ - 1}$\nHence \n $ \\det(A\\otimes{\\overline{A}} + B\\otimes{\\overline{B}}) = \\det(T_A\\otimes{\\overline{T_A}} + T_B\\otimes{\\overline{T_B}}) ==\\prod_{i=1}^{n}{(\\lambda_k\\overline{T_A}+\\beta_k\\overline{T_B})}= \\prod_{i = 1}^{n}{(|\\lambda_i|^2 + |\\beta_i|^2)}\\prod_{1\\leq i < j \\leq n}{(\\lambda_i\\overline{\\lambda_j} + \\beta_i\\overline{\\beta_j})(\\lambda_j\\overline{\\lambda_i} + \\beta_j\\overline{\\beta_i})} = \\prod_{i = 1}^{n}{(|\\lambda_i|^2 + |\\beta_i|^2)}\\prod_{1\\leq i < j \\leq n}{|\\lambda_i\\overline{\\lambda_j} + \\beta_i\\overline{\\beta_j}|^2} \\geq 0$[/hide]\r\n Sorry if I wrong :blush:", "Solution_2": "2) implies 1)\r\nIf $ spectrum(A)\\equal{}(\\lambda_i)$ then $ spectrum(A\\otimes\\bar{A})\\equal{}(\\lambda_i\\bar{\\lambda_j})_{ij}$.\r\nIf $ AB\\equal{}BA$ then $ A,B$ are simultaneously triangularizable ; thus $ A\\otimes\\bar{A},B\\otimes\\bar{B}$ are simultaneously triangularizable. Then there exists an ordering of $ spectrum(A)\\equal{}(\\lambda_i)$ and $ spectrum(B)\\equal{}(\\mu_j)$ such that $ spectrum(A\\otimes\\bar{A}\\plus{}B\\otimes\\bar{B})\\equal{}(\\lambda_i\\bar{\\lambda_j}\\plus{}\\mu_i\\bar{\\mu_j})_{ij}$. If $ i\\equal{}j$ then these values are non negative ; if $ i\\not\\equal{}j$ then these values are pairwise conjugate.", "Solution_3": "I add inequality as : \r\n$ A,B$ are Hermitian matricies (or $ SP_c(A) \\subset R,SP_c(B) \\subset R$) $ AB \\equal{} BA$ then :\r\n1)$ \\det(A\\otimes{A} \\plus{} I) \\geq O$\r\n2)$ \\det(A\\otimes{A} \\plus{} B\\otimes{B}) \\geq O$\r\n3)$ \\det(A\\otimes{A} \\plus{} B\\otimes{B} \\plus{} I) \\geq O$", "Solution_4": "Let $ A,B,C \\in M_n{\\mathbb{(C)}}$ $ AB \\equal{} BA,BC \\equal{} CB,CA \\equal{} AC$ then $ \\det(A\\otimes{\\overline{A}} \\plus{} B\\otimes{\\overline{B}} \\plus{} C\\otimes{\\overline{C}}) \\geq 0$\r\n If $ AB \\equal{} BA,BC \\equal{} CB,CA \\equal{} AC$ Then The exist $ T \\in GL_n{(\\mathbb{C})}$ Such that :\r\n $ A \\equal{} TT_AT^{ \\minus{} 1},B \\equal{} TT_BT^{ \\minus{} 1},C \\equal{} TT_CT^{ \\minus{} 1}$\r\nHere $ T_C \\equal{} \\begin{pmatrix}\\phi_1 & * & * & * & * & * & * \\\\\r\n0 & \\phi_2 & * & * & * & * & * \\\\\r\n0 & 0 & \\phi_3 & * & * & * & * \\\\\r\n... & ... & .... & ... & ... & ... & ... \\\\\r\n0 & 0 & 0 & ... & ... & ... & \\phi_n\\end{pmatrix}$\r\nAnd $ A\\otimes{\\overline{A}} \\plus{} B\\otimes{\\overline{B}} \\plus{} C\\otimes{\\overline{C}} \\equal{} (T\\otimes{\\overline{T}})(T_A\\otimes{\\overline{T_A}} \\plus{} T_B\\otimes{\\overline{T_B}} \\plus{} T_C\\otimes{\\overline{T_C}})(T\\otimes{\\overline{T}})^{ \\minus{} 1}$\r\nHence \r\n $ \\det(A\\otimes{\\overline{A}} \\plus{} B\\otimes{\\overline{B}} \\plus{} C\\otimes{\\overline{C}}) \\equal{} \\det(T_A\\otimes{\\overline{T_A}} \\plus{} T_B\\otimes{\\overline{T_B}} \\plus{} T_C\\otimes{\\overline{T_C}}) \\equal{} \\equal{} \\prod_{i \\equal{} 1}^{n}{(\\lambda_k\\overline{T_A} \\plus{} \\beta_k\\overline{T_B} \\plus{} \\phi_k\\overline{T_C})} \\equal{} \\prod_{i \\equal{} 1}^{n}{(|\\lambda_i|^2 \\plus{} |\\beta_i|^2 \\plus{} |\\phi_i|^2)}\\prod_{1\\leq i < j \\leq n}{(\\lambda_i\\overline{\\lambda_j} \\plus{} \\beta_i\\overline{\\beta_j} \\plus{} \\phi_i\\overline{\\phi_j})(\\lambda_j\\overline{\\lambda_i} \\plus{} \\beta_j\\overline{\\beta_i} \\plus{} \\phi_j\\overline{\\phi_i})} \\equal{} \\prod_{i \\equal{} 1}^{n}{(|\\lambda_i|^2 \\plus{} |\\beta_i|^2 \\plus{} |\\phi_i|^2)}\\prod_{1\\leq i < j \\leq n}{|\\lambda_i\\overline{\\lambda_j} \\plus{} \\beta_i\\overline{\\beta_j} \\plus{} \\phi_i\\overline{\\phi_j}|^2} \\geq 0$\r\n case ) Let $ A,B,C \\in M_n{\\mathbb{(R)}}$ $ AB \\equal{} BA,BC \\equal{} CB,CA \\equal{} AC$ then $ \\det(A\\otimes{{A}} \\plus{} B\\otimes{{B}} \\plus{} C\\otimes{{C}}) \\geq 0$\r\n Generral is correct , I hope that You are complete it .... :play_ball: .............", "Solution_5": "[quote=\"QuyBac\"]\n Let $ A,B,C \\in M_n{\\mathbb{(R)}}$ $ AB \\equal{} BA,BC \\equal{} CB,CA \\equal{} AC$ then $ \\det(A\\otimes{{A}} \\plus{} B\\otimes{{B}} \\plus{} C\\otimes{{C}}) \\geq 0$\n [/quote]\r\n It is same :oops: \r\nIf $ AB \\equal{} BA,BC \\equal{} CB,CA \\equal{} AC$ Then The exist $ T \\in GL_n{(\\mathbb{C})}$ Such that :\r\n $ A \\equal{} TT_AT^{ \\minus{} 1},B \\equal{} TT_BT^{ \\minus{} 1},C \\equal{} TT_CT^{ \\minus{} 1}$ ,note $ \\overline{T^{ \\minus{} 1}} \\equal{} (\\overline{T})^{ \\minus{} 1}$\r\nThen $ A \\equal{} \\overline{A} \\equal{} (\\overline{T})( \\overline{T_A})( \\overline{T})^{ \\minus{} 1}$ ,$ B \\equal{} \\overline{B} \\equal{} (\\overline{T})( \\overline{T_B})( \\overline{T})^{ \\minus{} 1}$ ,$ C \\equal{} \\overline{C} \\equal{} (\\overline{T})( \\overline{T_C})( \\overline{T})^{ \\minus{} 1}$ .....\r\n and $ \\det(A\\otimes{{A}} \\plus{} B\\otimes{{B}} \\plus{} C\\otimes{{C}}) \\equal{} \\det(A\\otimes{\\overline{A}} \\plus{} B\\otimes{\\overline{B}} \\plus{} C\\otimes{\\overline{C}}) \\geq 0$\r\nPs: [color=red]If You think single that $ A \\in M_n{\\mathbb{(R)}}$ , $ \\det(A\\otimes{A} \\plus{} I) \\geq 0$[/color] [b]Is very hard [/b] [color=red]If you don't thinks to [/color] $ \\det(A\\otimes{\\overline{A}} \\plus{} I)\\geq 0$ ???\r\nIf you see It is easy , I for example Let $ A \\in M_n{\\mathbb{(C)}},B \\in M_n{\\mathbb{(R)}}$ such that $ AB \\equal{} BA$ then $ \\det(A\\otimes{\\overline{A}} \\plus{} B\\otimes{B}) \\geq 0$ It is is very hard If you don't thinks to $ \\det(A\\otimes{\\overline{A}} \\plus{} B\\otimes{\\overline{B}}) \\geq 0$.............." } { "Tag": [ "geometry", "3D geometry", "sphere", "incenter", "quadratics", "pyramid", "modular arithmetic" ], "Problem": "Recently, I've noticed an increase in number of people questioning, \"How can I study for AMC?\" or \"My score is ### and I can't seem to improve,\" and so on. So, I'm bringing back the old idea that I used sometime ago. Hopefully, this can help others and I know it will help me as well (I don't get to practice math as I used to nowadays).\r\n\r\nEveryday (this is tentative), I'm going to post two problems that I wrote on specific category. These problems will have ideas taken from actual AMC questions. They will be [b]very[/b] similar to the original one, and I'll cite the original ones. But, problems are purely MINE (not copied off from some internet site) so please respect them and if there's an error, please correct or let me know. :) \r\n\r\nAnother thing... To make this more fun, you'll be able to gain points on answering the questions. I'm going to take FIRST TWO CORRECT responses. Each question will worth 1 point so maximum you can get on each day is 2 points. Here are brief rules (on hidden text):\r\n\r\n[hide]Question Answering and Score Rule\n\nThere are just some rules to follow.\n\n1. When posting your answer, answer itself does NOT count. You must have your own solution typed in your own words so that I can see that you have solved the problem on your own. If you take previous poster's answer and change it, I may or may not know it. But it is disgrace to yourself.\n\n2. You cannot win everyday. If you win on one day, you cannot win the next day but the day after. It's only possible to win every other day. This is to even out the scores and give other people some chance. So, if you have gotten the question right, hold your answers until at least two people with correct answers post. Then post or elaborate or so on.\n\n3. I'm not native speaker so don't argue with me about English rule in these problems. You're here to do math, not English. :D \n\n4. When you have questions to past problems, contact me FIRST. I'll probably make a new thread dedicated just for that problem so new problems don't get mixed up with old ones.\n\n5. There is no prize folks. Just the fact that you might be one step closer to 150. Who knows. :roll: \n\n6. I'll try to do this everyday. I will really try but I have millions of things to do now. I don't usually get home till 7 so if I'm not posting problems, just wait until I post new one. Don't send millions of PMs about this. \n\n[/hide]\n\nPLEASE READ THOSE RULES!\n\nFirst Set of Problems\n\n1. (Modified from AMC 12A 2004 #22) Three mutually tangent unit spheres (radius 1) are placed on the surface. A soccer ball of unknown radius is placed on top of them. The distance from the surface to top of a soccer ball is in form $ \\displaystyle 8 \\plus{} \\frac{2 \\sqrt{m}}{3}$ where $ m$ is not divisible by any perfect square greater than 1. What is the remainder when $ m$ is divided by 23?\nA) 3\nB) 11\nC) 17\nD) 19\nE) 22\n\n2. (Modified from AMC 12B 2004 #19) Roskolnikov makes a slice in a cone-shaped lumber so that it is now a truncated cone with two different radii. Inside the lumber, he positions a sphere so that it touches top, bottom, and lateral surfaces of it. If one of the radii is 2008, what is the minimum length of other radii so that the sphere has an integer radius?\nA) 411\nB) 502\nC) 636\nD) 675\nE) 1004\n\n*Just for fun, where is this character from?*\n[hide][i]Crime and Punishment[/i][/hide]", "Solution_1": "Still working on solution 1, but here's #2.\r\n\r\n[hide=\"Solution 2\"]\nTaking a radial cross-section of the cone before the cut, it is obvious that the center of the sphere lies at the incenter of this radial cross-section. Drawing in the angle bisectors and the radii of the sphere to the tangential points in this cross-section, we notice that we have determined several right triangles. Those that are of interest to us:\n\nTriangle 1: \nSide a = r\nSide b = r_1\nAngle b = x/2\n\nTriangle 2:\nSide a = r\nSide b = 2008\nAngle a = x/2\n\nNow, since r/2008 must equal r_1/r, we end with the quadratic r^2 = 2008*r_1, which leads us to the obvious conclusion that r_1=502 at minimum. (2008=2^3*251, for a perf. square, we need another 2 and another 251.)\n\nAnswer: B, 502.[/hide]", "Solution_2": "[hide=\"1?\"]Connect the centers of the balls to form a pyramid. Let $ x\\equal{}1\\plus{}r$ where $ r$ is the radius of the soccerball. The pyramid has a 2 x 2 x 2 base and a vertex that has a distance of x from the other three points. Then $ h^2\\plus{}\\frac{4}{3}\\equal{}(x)^2$\n\nThe total distance is $ x\\plus{}h\\equal{}8\\plus{}\\frac{2}{3}\\cdot \\sqrt{m}$\n\nSo okay...lets analyze $ \\sqrt{h^2\\plus{}\\frac{4}{3}}\\plus{}h$...taking h=0, it is root 4/3...taking h arbtrarily large, that expression is arbitrarily large. Hence for any prime p=m, we have a solution for h. But these all satisfy the problem (to be extra ridiculous, the sequence 23n+b where n is an integer has infinitely many primes by dirchlet's theorem so there cannot be a unique answer...)\n\nIf I missed something, my bad. But I read the problem like ten times.[/hide]", "Solution_3": "@Altheman: Try to use number 8. It's actually a lot simpler than what you're thinking (honestly, I think you've gotten the hard part).", "Solution_4": "The question for #1 needs to be slightly clarified as Altheman noted (its sort of confusing whether or not you want $ m$ to always have a certain residue $ \\pmod{23}$ or for a certain specific case).\r\n\r\n[hide=\"1\"]\n[img]http://www.artofproblemsolving.com/Wiki/images/7/76/2004_AMC12A-22a.png[/img]\n\n(Ignore those 2s and pretend they are $ r$s, because the original problem statement had 2 instead of unknown. Its from the wiki [I made it])\n\nThe heights of the bottom parts (eg below the center) of the bottom spheres and the top part of the top sphere is $ 1 \\plus{} r$. The height of the centers of the bottom sphere to that of the top sphere can be calculated by drawing a triangular pyramid between those four points. The distance from any of the three bottom centers to the center of the equilateral triangle face of the pyramid is $ \\frac {2}{\\sqrt {3}}$ using $ 30 \\minus{} 60 \\minus{} 90 \\triangle$s. The other edges are of length $ r \\plus{} 1$. Thus by Pythagorean thm $ h \\equal{} \\sqrt {(r \\plus{} 1)^2 \\minus{} 4/3}$, and the total height is $ 1 \\plus{} r \\plus{} \\sqrt {(r \\plus{} 1)^2 \\minus{} 4/3}$.\n\nHere comes the ambiguous part, we assume $ r \\equal{} 7$ just to make our answer nice. Then the height becomes $ 8 \\plus{} \\sqrt {64 \\minus{} 4/3} \\equal{} 8 \\plus{} \\frac {2\\sqrt {141}}{3}$, and the residue upon division by $ 23$ is $ 3\\ \\mathrm{(A)}$.\n[/hide]\r\n\r\nEdit: h_s_potter2002, I noticed that and fixed it before I saw your post. Thanks anyway!", "Solution_5": "[hide=\"azjps\"]$ \\sqrt{64\\minus{}4/3} \\equal{} \\sqrt{\\frac{188}{3}} \\equal{} \\frac{2\\sqrt{47}}{\\sqrt{3}} \\equal{} \\frac{2\\sqrt{141}}{3}$[/hide]", "Solution_6": "[hide=\"1\"]Draw lines from the center of the soccer ball to each of the three unit spheres and connect the centers of the three spheres. Thus, we have a tetrahedron with an equilateral triangle base of side length 2. Let the radius of the soccer ball be $ r$ and the height of the tetrahedron be $ h$. The distance to the top of the soccer ball is just $ 1 \\plus{} r \\plus{} h$.\n\nNow, we find that $ h^2 \\plus{} \\left(\\frac {2\\sqrt {3}}{3}\\right)^2 \\equal{} (r \\plus{} 1)^2$. Hence, $ h \\equal{} \\sqrt {r^2 \\plus{} 2r \\minus{} \\frac {1}{3}}$. The distance to the top of the soccer ball is just $ 1 \\plus{} r \\plus{} \\sqrt {r^2 \\plus{} 2r \\minus{} \\frac {1}{3}}$. Since there is an 8, let $ r \\equal{} 7$. Then, we have $ 8 \\plus{} \\sqrt {63 \\minus{} \\frac {1}{3}} \\equal{} 8 \\plus{} \\sqrt {\\frac {564}{9}} \\equal{} 8 \\plus{} \\frac {2\\sqrt {141}}{3}$. Thus, the residue is 3, or A.[/hide]\r\n\r\nedit: too late :roll:", "Solution_7": "My bad on #1. I didn't really consider the problem in full case. Thanks for pointing out this.\r\n\r\nAnyway:\r\n\r\n[hide=\"Correct Answer\"]\n1. A\nThe expression was indeed $ \\displaystyle 8 \\plus{} \\frac{2 \\sqrt{141}}{3}$. \n\n2. B\nI solved it using trapezoid and circle though. It's probably similar logic as other person solved.\n[/hide]\r\n\r\nSecond Set of Problems (Just one problem; Lack of time)\r\n\r\n3. (Inspired by AMC 12B 2002 #23) Quadrilateral DACB is drawn such that AD = AC, and BC = 20. A line with length of 20 was drawn from A so that it touches BC at its midpoint, E. If ABD is a right triangle, then find BD.\r\nA) 28\r\nB) $ 21 \\sqrt{2}$\r\nC) 30\r\nD) $ 10 \\sqrt{10}$\r\nE) $ 15 \\sqrt{7}$", "Solution_8": "[hide=\"Solution #3\"]\nBy the Law of Cosines on $ \\triangle AEC, AEB$, we have\n\\begin{eqnarray*} AC^2 & = & 20^2 + 10^2 - 2\\cdot 20 \\cdot 10 \\cos \\angle AEC \\\\\nAB^2 & = & 20^2 + 10^2 - 2\\cdot 20 \\cdot 10 \\cos (180 - \\angle AEC) \\end{eqnarray*}\nSumming these give $ AB^2 + AC^2 = AB^2 + AD^2 = 1000$ (alternatively, use Stewarts). In triangle $ DAB$, Pythagorean Thm gives $ AB^2 + AD^2 = BD^2$, so $ BD = \\sqrt {1000} = 10\\sqrt {10}\\ \\mathrm{(D)}$.\n[/hide]\n\n\n\n[hide=\"Edit history\"]*edits* I just read the rules ... sorry. (I'm not so much interested in winning than in practicing though)\n\nRequest for clarification: Which angle in $ \\triangle ABD$ is right? I don't think it was specified in the problem statement, but I believe its necessary for the answer. Below is how I drew my diagram.\n\nEdit again: Thanks\n\nEdit again: Nobody's posting a full solution, so I did. \n[/hide]", "Solution_9": "Yes, I remembered early this morning that I didn't specify that BD was the hypotenuse.\r\n\r\nSo yes, your left diagram is correct.", "Solution_10": "[hide]using stewart theorem[/hide]", "Solution_11": "is it just me, or does 1 look like an old AIME problem... one with balls aligned so that their centers form octagon, etc... if so, would a similar approach work?", "Solution_12": "I think silverfalcon mentioned in the rules to post questions regarding old problems on a separate thread, but one hasn't been created yet so I guess its safe to answer here for now.\r\n\r\nI'm guessing the AIME problem in question is [url=http://www.artofproblemsolving.com/Wiki/index.php/1998_AIME_Problems/Problem_10]1998 #10[/url]; and the answer to your question is to a degree, yes (since both solutions involve connecting the radii and then Pythagorean).", "Solution_13": "[hide=\"Correct Answer\"]\n3. D\nNice job, azjps, especially with neat drawing. [/hide]\r\n\r\nI'll keep total scores once we hit like 10 problems. I don't want to do it every set.\r\n\r\nAgain, lack of time (I have a big test tomorrow and Tuesday) so here's one.\r\n\r\n4. (My own) Orthodiagonal trapezoid PQRS (PQ and RS are bases) has median MN, and it is given that ON = 19 where O is the midpoint of MN. If RS = 59, then the sum of the squares of diagonals is $ \\displaystyle S_d$. What is the remainder when $ \\displaystyle S_d$ is divided by 2008?\r\nA) 1760\r\nB) 1761\r\nC) 1762\r\nD) 1763\r\nE) 1764", "Solution_14": "what i have so far: we can find PQ because of what we know about SR and ON (and thus MN). i was wondering if you could use ptolemy's theorem.\r\nhere's my general question: are orthodiagonal trapezoids cyclic?", "Solution_15": "[quote=\"ghjk\"]You meant the solution of \"The Zuton Force\" is wrong, Teneroez?? :o . Can you show me your solution?? Maybe you also get it wrong :lol:.Anyway, I think solution of Zuton Force is right! Brute force won't work in this case! :wink:[/quote]\r\n\r\nNo I also agree that his solution is wrong. You cannot set one plate because some plates have multiples (more than just 1).\r\n\r\nLike firecricket91 said, try the case with 2 blue plates and 2 red plates.\r\n\r\nAccording to Zuton Force, he did it with the idea of $ \\frac {(4 - 1)!}{2!2!}$ (generically of $ \\frac {(n - 1)!}{k!l!m!...}$ where $ k + l + m + .... = n$, and each is just how many times each different item appears which is $ \\frac {6}{4}$ in this case, which isnt even an integer so clearly his idea is wrong. \r\nHe just did the normal $ (n - 1)!$ in a circle then put it over how many of each appears to account for overcounts. \r\n${ (n - 1)}!$ only works for if all things are different, but not in this case. It is very hard to account for overcounts using this case (since objects can be the same).\r\n\r\nThats why in my earlier post, I said this problem is a lot harder than AMC--this can be like 10+ on AIME.\r\n\r\nHeres an old topic about the exact same type of question. You will see the only way to do this [i]is[/i] brute force.\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1662713791&t=57731\r\nbut even that \"simplification\" does not work when there are like 4 different colored ones", "Solution_16": "Wait, but the original problem said that there was 1 white plate, so isn't Zuton Force correct?", "Solution_17": "Okay, try it with 1 white plate 2 red plates and 2 blue plates.\r\n\r\nList them out. you wont get the same answer.\r\n\r\nWhy does Zuton only set the 1 white plate?\r\n\r\nWHy cant he set one of the two red plates.\r\n\r\nSetting one of the two red plates would give a completely differnt answer than setting one of the 1 white plates.\r\n\r\nThat shouldnt be the case because both answers should be the same.", "Solution_18": "Uh, can you list them out for me?\r\n\r\nI have W B B R R, W R B B R, W R R B B, W B R R B, W B R B R , W R B R B.\r\n\r\nisn't that it?\r\n\r\nIsn't WBRBR different from WRBRB because one is a mirror image of the other?\r\n\r\nMirror images are not superimposable, right?\r\n\r\nUh...to answer your question, white is chosen over the other colors because it has only $ \\emph{one}$. If you choose the red, you have to consider the case where where the rotation makes it indistinguishable. (IE, RRBBW, RBBWR)", "Solution_19": "Uh, here's a problem (not sure if it's AMC level)\r\n\r\n$ \\cos \\frac{\\pi}{7} \\cos \\frac{2\\pi}{7} \\cos \\frac{3\\pi}{7} \\cos \\frac{4\\pi}{7} \\cos \\frac{5\\pi}{7} \\cos \\frac{6\\pi}{7} \\equal{} ?$", "Solution_20": "[quote=\"Zellex\"]Uh, here's a problem (not sure if it's AMC level)\n\n$ \\cos \\frac {\\pi}{7} \\cos \\frac {2\\pi}{7} \\cos \\frac {3\\pi}{7} \\cos \\frac {4\\pi}{7} \\cos \\frac {5\\pi}{7} \\cos \\frac {6\\pi}{7} \\equal{} ?$[/quote]\r\n[hide=\"Solution\"]\nLet $ k \\equal{} \\frac {\\pi}{7}$. Therefore, we wish to find the value of $ \\cos k\\cos 2k\\cos 3k\\cos 4k\\cos 5k\\cos 6k$. Notice that $ \\cos ik \\equal{} \\minus{} \\cos k(7 \\minus{} i)$. Hence, the value that we seek is $ \\minus{} \\cos^2k\\cos^22k\\cos^23k \\equal{} \\minus{} \\cos^2k\\cos^22k\\cos^24k$. Let $ P \\equal{} \\cos k\\cos 2k\\cos 4k\\implies \\minus{} P^2 \\equal{} \\cos k\\cos 2k\\cos 3k\\cos 4k\\cos 5k\\cos 6k$. Thus,\n\\[ P\\sin k \\equal{} \\cos k\\sin k\\cos 2k\\cos 4k\n\\]\nNoticing that $ \\cos k\\sin k \\equal{} \\frac {\\sin 2k}{2}$, we have that\n\\[ P\\sin k \\equal{} \\frac {\\sin 2k\\cos 2k\\cos 4k}{2}\n\\]\nAgain, we notice that $ \\cos 2k\\sin 2k \\equal{} \\frac {\\sin 4k}{2}$, so\n\\[ P\\sin k \\equal{} \\frac {\\sin 4k\\cos 4k}{4} \\equal{} \\frac {\\sin 8k}{8} \\equal{} \\frac { \\minus{} \\sin k}{8}\\implies P \\equal{} \\frac { \\minus{} 1}{8}\\implies \\minus{} P^2 \\equal{} \\boxed{\\frac { \\minus{} 1}{64}}\n\\]\n[/hide]", "Solution_21": "Let $ 1 < log_{10}({log_{10}{a}}) < 100$. There are $ N$ integer solutions for $ a$. Let $ S$ be the sum of the digits of $ N$. Let $ S'$ be the sum of the digits of $ S$. Find $ S'$", "Solution_22": "[quote=\"firecricket91\"]Let $ 1 < log_{10}({log_{10}{a}}) < 100$. There are N integer solutions for $ a$. Let $ S$ be the sum of $ N$. Let $ S'$ be the sum of the digits of $ S$. Find $ S'$[/quote]\r\n\r\nWhen you say \"Let $ S$ be the sum of $ N$\", do you mean digits of $ N$, or all possible integer solutions for $ a$?", "Solution_23": "[quote=\"123456789\"][quote=\"firecricket91\"]Let $ 1 < log_{10}({log_{10}{a}}) < 100$. There are N integer solutions for $ a$. Let $ S$ be the sum of $ N$. Let $ S'$ be the sum of the digits of $ S$. Find $ S'$[/quote]\n\nWhen you say \"Let $ S$ be the sum of $ N$\", do you mean digits of $ N$, or all possible integer solutions for $ a$?[/quote]\r\nyea sorry.", "Solution_24": "Which one? Digits?", "Solution_25": "Solving the inequality, we have $ 10^{10} < a < 10^{10^{100}}$\r\n\r\nThus, $ N \\equal{} 10^{10^{100}} \\minus{} 10^{10} \\minus{} 1$\r\nThe number $ 10^{10^{100}}$ in base-10 form is a single $ 1$ followed by $ 10^{100}$ zeros.\r\nThe number $ 10^{10}$ in base-10 form is a single $ 1$ followed by $ 10$ zeros.\r\nThe number $ 1$ in base-10 form is a single $ 1$ followed by $ 0$ zeros.\r\n\r\nThus, $ N$ consists (in digits) of $ (10^{100} \\minus{} 1)$ nines and $ 1$ eight.\r\n\r\nSo $ S \\equal{} 9(10^{100}) \\minus{} 1$.\r\n\r\nThen, $ S$ consists of $ 100$ nines and $ 1$ eight.\r\nSo $ S' \\equal{} 100(9) \\plus{} 8 \\equal{} \\boxed{908}$.", "Solution_26": "[quote=\"123456789\"][hide=\"Solution\"]Solving the inequality, we have $ 10^{10} < a < 10^{100^{10}}$ or $ 10^{10} < a < 10^{10^{20}}$.\n\nThus, $ N \\equal{} 10^{10^{20}} \\minus{} 10^{10} \\minus{} 1$\nThe number $ 10^{10^{20}}$ in base-10 form is a single $ 1$ followed by $ 10^{20}$ zeros.\nThe number $ 10^{10}$ in base-10 form is a single $ 1$ followed by $ 10$ zeros.\nThe number $ 1$ in base-10 form is a single $ 1$ followed by $ 0$ zeros.\n\nThus, $ N$ consists (in digits) of $ 10^{20} \\minus{} 1$ nines and $ 1$ eight.\n\nSo $ S \\equal{} 9(10^{20}) \\minus{} 1$.\n\nThen, $ S'$ consists of $ 20$ nines and $ 1$ eight.\nSo $ S' \\equal{} 20(9) \\plus{} 8 \\equal{} \\boxed{188}$.[/hide][/quote]\n\nIts not 10^{10^{20}} , it would be 10^{10^{100}}, also you cannot go from \n\n[quote]Thus, $ N$ consists (in digits) of $ 10^{20} \\minus{} 1$ nines and $ 1$ eight.\n\nSo $ S \\equal{} 9(10^{20}) \\minus{} 1$.[/quote]\r\n\r\nbecause some digits can carry forming a new number.", "Solution_27": "[quote=\"firecricket91\"][quote=\"123456789\"][hide=\"Solution\"]Solving the inequality, we have $ 10^{10} < a < 10^{100^{10}}$ or $ 10^{10} < a < 10^{10^{20}}$.\n\nThus, $ N \\equal{} 10^{10^{20}} \\minus{} 10^{10} \\minus{} 1$\nThe number $ 10^{10^{20}}$ in base-10 form is a single $ 1$ followed by $ 10^{20}$ zeros.\nThe number $ 10^{10}$ in base-10 form is a single $ 1$ followed by $ 10$ zeros.\nThe number $ 1$ in base-10 form is a single $ 1$ followed by $ 0$ zeros.\n\nThus, $ N$ consists (in digits) of $ 10^{20} \\minus{} 1$ nines and $ 1$ eight.\n\nSo $ S \\equal{} 9(10^{20}) \\minus{} 1$.\n\nThen, $ S'$ consists of $ 20$ nines and $ 1$ eight.\nSo $ S' \\equal{} 20(9) \\plus{} 8 \\equal{} \\boxed{188}$.[/hide][/quote]\n\nIts not 10^{10^{20}} , it would be 10^{10^{100}}, also you cannot go from \n\n[quote]Thus, $ N$ consists (in digits) of $ 10^{20} \\minus{} 1$ nines and $ 1$ eight.\n\nSo $ S \\equal{} 9(10^{20}) \\minus{} 1$.[/quote]\n\nbecause some digits can carry forming a new number.[/quote]\r\n\r\nI edited my post. Is it right now?", "Solution_28": "yea thats what I got.\r\n\r\nyou can post next one if you want.", "Solution_29": "lol. I don't know any problems. You can post one if you want." } { "Tag": [], "Problem": "do any of u have any recommendation on honor chemistry text book? cuz i wanna learn i honor chemistry at home, so i can take chemistry AP in 10th grade", "Solution_1": "as a homeschooler, i use A Beka for chemistry I...very good book in my opinion, short, precise, and to-the-point...", "Solution_2": "Our school uses Zhumdahl (sp?), but my school is extremely inferior compared to many of those on the forums.", "Solution_3": "but what text book does most high school use?", "Solution_4": "prentice hall and mcgraw hill are typically very common and popular among public schools...\r\n\r\nhowever, they are not necessarily always the best..." } { "Tag": [ "inequalities", "function", "Cauchy Inequality", "inequalities unsolved" ], "Problem": "Let $ x,y,z>0$ satisfy $ xyz\\equal{}1$. Solve that:\r\n$ \\sqrt{4\\plus{}9x^{2}}\\plus{}\\sqrt{4\\plus{}9y^{2}}\\plus{}\\sqrt{4\\plus{}9z^{2}}\\leq\\sqrt{13}(x\\plus{}y\\plus{}z)$", "Solution_1": "It's equivalent to:\r\n$ \\sum(\\sqrt{4+9x^{2}}-\\sqrt{13}x+(\\sqrt{13}-\\frac{9}{\\sqrt{13}})\\cdot ln(x))\\leq 0$\r\nLet $ f: (0,\\infty)\\to\\mathbb R$ ,$ f(x)=\\sqrt{4+9x^{2}}-\\sqrt{13}x+(\\sqrt{13}-\\frac{9}{\\sqrt{13}})\\cdot ln(x)$\r\nBecause $ f''(x)<0$ and $ f'(1)=0$ we are done.\r\n\r\n[hide=\"PS\"]\nI've seen this technique used by Arqady.Very nice\n[/hide]", "Solution_2": "Who can solve this problems by the Cauchy-Schwarz inequality?", "Solution_3": "How do you choose $ \\sqrt{13}\\minus{}\\frac{9}{\\sqrt{13}}$", "Solution_4": "[quote=\"bibobeo\"]How do you choose $ \\sqrt{13}\\minus{}\\frac{9}{\\sqrt{13}}$[/quote]\r\n\r\nConsider the function $ f(x) \\equal{}\\sqrt{4\\plus{}9x^{2}}\\minus{}\\sqrt{13}\\plus{}C\\cdot ln(x)$\r\nBecause our inequality has the equality case when $ x \\equal{} y \\equal{} z \\equal{} 1$,we must find $ C$ in such a way that $ f(1)\\geq f(x),\\forall x\\in (0,\\infty)$ If we managed to find such a constant it would mean that the inequality is true ,since $ 0 \\equal{} 3f(1)\\geq f(x)\\plus{}f(y)\\plus{}f(z)$ ,which is equivalent to the initial inequality.\r\nIf $ f(1)\\geq f(x),\\forall x\\in (0,\\infty)$ it means that $ f(1)$ must be the maximum value of $ f$ $ \\Leftrightarrow$ $ f'(1) \\equal{} 0$ $ (*)$ \r\nFrom $ (*)$ we get the value of $ C$.\r\nIt remains only to check if $ f$ really has a maximum point (check if $ f$ is concave ).", "Solution_5": "OK! It is a special technique!", "Solution_6": "Sorry, but I don't see why $ f''(x)<0$ with $ x\\geq 0$", "Solution_7": "I got $ f''(x)\\equal{}\\frac{36}{(4\\plus{}9x^{2})\\sqrt{4\\plus{}9x^{2}}}\\minus{}\\frac{4}{\\sqrt{13}x^{2}}$\r\nI don't see why $ f''(x)<0$ when $ x>0$\r\nIt's easy to see when $ x\\geq 1$, what about $ 0 0$\nIt's easy to see when $ x\\geq 1$, what about $ 0 < x < 1$?[/quote]\r\n\r\nWe have $ 4(\\frac{9}{\\sqrt{(4+9x^{2})^{3}}}-\\frac{1}{\\sqrt{13}x^{2}})<0$ $ \\Leftrightarrow$ \r\n$ \\frac{9}{\\sqrt{(4+9x^{2})^{3}}}<\\frac{1}{\\sqrt{13}x^{2}}$ $ \\Leftrightarrow$\r\n$ 81\\cdot 13 x^{4}<4^{3}+9^{3}\\cdot x^{6}+27\\cdot 4 x^{2}(4+9x^{2})$ $ \\Leftrightarrow$\r\n$ 81x^{4}<64+729x^{6}+432x^{2}$ $ \\Leftrightarrow$ \r\n$ 0<64+729x^{6}+x^{2}(432-81x^{2})$ ,which it's clearly true if $ x\\in (0,1)$\r\n\r\nIf $ x\\geq 1$ we may write it as :\r\n$ 0<64+x^{4}(729x^{\\frac{6}{4}}-81)+432x^{2}$ ,which it's clearly true since $ x\\geq 1$", "Solution_9": "This is my homework and my teacher say that it can be solved by Cauchy-Schwarz inequality but he don't say clearly! But I can't do it by that way", "Solution_10": "[quote=\"nktp\"]Let $ x,y,z$ are positive real numbers such that $ xyz = 1$. Prove that\n\n$ \\sqrt {4 + 9x^{2}} + \\sqrt {4 + 9y^{2}} + \\sqrt {4 + 9z^{2}}\\leq\\sqrt {13}(x + y + z)$.[/quote]Using Cauchy inequality,\r\n\r\n$ \\sum\\sqrt {4 + 9x^{2}} = \\sum\\sqrt {4\\sqrt [3]{x^{2}y^{2}z^{2}} + 9x^{2}}\\equiv\\sum\\sqrt {4u^{2}v^{2}w^{2} + 9u^{6}}$\r\n\r\n$ = \\sum{u\\sqrt {4v^{2}w^{2} + 9u^{4}}\\leq\\sqrt {\\sum{u^2}}\\sqrt {\\sum{\\left(4v^{2}w^{2} + 9u^{4}\\right)}}\\leq\\sqrt {13}\\sum{u^3}\\equiv\\sqrt {13}\\sum{x}}$,\r\n\r\nthe last inequality holds for\r\n\r\n$ 13\\left(\\sum{u^3}\\right)^2 - \\sum{u^2}\\sum{\\left(4v^{2}w^{2} + 9u^{4}\\right)}$\r\n\r\n$ = 4\\sum{u^6} - 13\\sum{u^4\\left(v^2 + w^2\\right)} + 26\\sum{v^3w^3} - 12u^2v^2w^2$\r\n\r\n$ = 2\\sum{\\left(v^6 + w^6 - v^4w^2 - v^2w^4\\right)} - 11\\sum{\\left(v^4w^2 + v^2w^4 - 2v^3w^3\\right)}$\r\n\r\n$ + 4\\left(\\sum{v^3w^3} - 3u^2v^2w^2\\right)$\r\n\r\n$ = 2\\sum{(v - w)^2(v + w)^2\\left(v^2 + w^2\\right)} - 11\\sum{v^2w^2(v - w)^2}$\r\n\r\n$ + 2\\sum{vw}\\sum{u^2(v - w)^2}$\r\n\r\n$ = \\sum{(v - w)^2\\left[2v^4 + 4v^3w - 7v^2w^2 + 4vw^3 + 2w^4 + 2u^2(vw + wu + uv)\\right]}\\geq 0$.\r\n\r\nThe generalization see: http://www.mathlinks.ro/Forum/viewtopic.php?t=165627\r\n\r\nBy the way, if $ x,y,z$ be nonnegative real numbers such that $ 2xyz + yz + zx + xy = 5$, then\r\n\r\n$ \\sqrt {4 + 9x^{2}} + \\sqrt {4 + 9y^{2}} + \\sqrt {4 + 9z^{2}}\\leq\\sqrt {13}(x + y + z)$,\r\n\r\nwith equality if and only if $ x = y = z = 1$." } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "I noticed that everyone seems to expect the AMC B cutoff score to drop after the A cutoff score did. That means people don't really have a score of 120 to aim for any more, they just do the best they can, and maybe it will be enough.\r\n\r\nThis seems to show a way of satisfying the original goal of the scoring changes in a different way than the one originally given.\r\n\r\nWhat if the target score of 120 was not given, and they just take the top 5%. That way people would not have a score to shoot for, and would be forced to do their best and try all the problems.", "Solution_1": "Perhaps they should get rid of a number cutoff and go to the percentages directly. Doesn't that make more sense? Or, better yet, go by strictly number of people qualifying - like how they do it for USAMO qualification.", "Solution_2": "LOL, I just edited that idea into my original post, and then I saw your post. So perhaps it is a good idea.", "Solution_3": "This year was weird because of the scoring change. Usually having an absolute cutoff helps people rather than hurts them, when using the 5% rule might push the cutoff up to 106 or higher.", "Solution_4": "If they were to go by a percentage cutoff - is there enough of a difference between the skill levels between the A testers and the B testers to potentially skew the results?", "Solution_5": "Right, and also it sucks when you can't even tell what score you're shooting for to qualify. Having an absolute cutoff just makes it so much easier, especially for the people who don't care or aren't savvy enough to dig up statistics to estimate a relative cutoff. Besides, AMC at the lower levels should be (and I think is) more about individual achievement than competition, and a relative cutoff sort of violates that spirit." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $ \\lambda_1 \\leq \\lambda_2 \\leq...$ be a positive sequence and let $ K$ be a constant such that \\[ \\sum_{k=1}^{n-1} \\lambda^2_k < K \\lambda^2_n \\;(n=1,2,...).\\] Prove that there exists a constant $ K'$ such that \\[ \\sum_{k=1}^{n-1} \\lambda_k < K' \\lambda_n \\;(n=1,2,...).\\] \r\n\r\n[i]L. Leindler[/i]", "Solution_1": "fourth problem ,here : [url]http://www.mathematik.uni-muenchen.de/~yarotsky/jarnik/Jarnik_daniel_harrer.pdf[/url]" } { "Tag": [], "Problem": "What is conduction, convection and radiation and what states of matter does it occur in?\r\n\r\nI also need to know how to draw a diagram of each type of heat transfer. \r\n\r\nthanks :)", "Solution_1": "Conduction is direct transfer of heat due to physical contact. It can occur in all states of matter. Convection is heat transfer due to convection currents. It is based on the idea that things at higher temperature tend to be less dense than the same thing at a lower temperature. Therefore, when something is heated, the heated substance tends to move in convection currents throughout the substance and transfer heat to the rest of the substance. This is typical of liquid and gaseous substances. Radiation is simply an electromagnetic wave emitted by a substance. I believe radiation exists for all states of matter.", "Solution_2": "Dragon was right as far as he went.\r\n\r\nAllow me to add a bit.\r\n\r\nConduction is energy transfer between molecules in contact. Think of a stove in contact with a pot. (It is part of heat transfer in all phases.)\r\n\r\nConvection is energy transfer by conduction, when supplemented by movement of molecules to bring these moving molecules in contact to conduct energy. Think of the water inside the pot, particularly when boiling. (This occurs in mobile phases -- liquid, gas, plasma.)\r\n\r\nRadiation is energy transfer by emitting/absorbing photons. Think of the pot inside a normal oven. (They don't sell as well if they are called \"radiation ovens.\") (This occurs in all phases.)" } { "Tag": [ "conics", "hyperbola", "ellipse", "parabola", "geometry", "3D geometry", "trigonometry" ], "Problem": "How is the eccentricity values determined for the conic sections?\r\n\r\nie: if e>1 it is a hyperbola\r\n if 1>e>0 it is an ellipse\r\n\r\nI understand why a parabola has an eccentricity of 1 and why a circle has one of 0, but I don't understand how they derived the e for hyperbolas and ellipses.\r\n\r\nAlso, how do you derive the directrix for the conic sections?", "Solution_1": "Edit: http://mathworld.wolfram.com/Eccentricity.html\r\n\r\nI have actually derived an expression for the eccentricity in terms of the angle of the plane that cuts the conic section and the angle of the cone itself :D\r\n\r\n$e = 1 - \\cos \\theta + \\cot \\alpha \\sin \\theta$\r\n\r\nWhere $\\theta$ is the tilt of the plane away from the plane perpendicular to the axis of the cone (that is, $\\theta = 0$ is a circle) and $\\alpha$ is the tilt of the cone away from the plane perpendicular to the axis of the cone (that is, small $\\alpha$ is a wide cone, and $\\alpha \\approx \\frac{\\pi}{2}$ is almost a stick).\r\n\r\nFor a parabola, $\\theta = \\alpha$ (can you see why?) and the expression becomes $e = 1 - \\cos \\theta + \\cos \\theta = 1$.\r\n\r\nFor a circle, $\\theta = 0$ and the expression becomes $e = 1 - 1 + 0 = 0$.\r\n\r\nFor a hyperbola, $\\theta > \\alpha$ (can you see why?) and it can be shown either graphically or with angle summation formulas that $e > 1$.\r\n\r\nFinally, for an ellipse, $0 < \\theta < \\alpha$ and it can again be shown that $0 < e < 1$.\r\n\r\n\r\n\r\nI can tell you that the eccentricity of an ellipse is a measure of how far it deviates from being a circle. In fact, the polar equation for an ellipse is $r = \\frac{k}{1 - e \\cos t}$ (I have avoided $\\theta$ because of my use of it above) and you can clearly see that $e = 0$ describes a circle, and $0 < e < 1$ describes a wobbly sort of circle - in fact, an ellipse!\r\n\r\n(I worked for awhile on describing the polar equation of a conic section solely in terms of the geometry of the cone and the plane that cuts it, so that is why I have derived such an odd equation for $e$ - I could also tell you my expression for $k$ if you would like :D )", "Solution_2": "Well, e=c/a (c=distance from center to focus, a=distance from center to vertex). If you think about the ellipse, the vertex is farther away from the center than is the focus. It is the opposite on the hyperbola. This makes e<1 is an ellipse and e>1 a hyperbola.", "Solution_3": "If $e = c/a$, why is e also equal to the ratio $PF: PD$ where $PF$ is the distance from focus to a point and PF is distance from directrix to focus?", "Solution_4": "This is what Wikipedia has to say about your question: \r\n\r\n\"The four defining conditions above can be combined into one condition that depends on a fixed point F (the focus), a line L (the directrix) not containing F and a positive number e (the eccentricity). The corresponding conic section consists of all points whose distance to F equals e times their distance to L. For 0 < e < 1 we obtain an ellipse, for e = 1 a parabola, and for e > 1 a hyperbola.\r\n\r\n\"For an ellipse and a hyperbola, two focus-directrix combinations can be taken, each giving the same full ellipse or hyperbola. The distance from the center to the directrix is a/e, where a is the semi-major axis of the ellipse, or the distance from the center to the tops of the hyperbola. The distance from the center to a focus is ae.\r\n\r\n\"In the case of a circle e = 0 and one imagines the directrix to be infinitely far removed from the center. However, the statement that the circle consists of all points whose distance is e times the distance to L is not useful, because we get zero times infinity.\r\n\r\n\"The eccentricity of a conic section is thus a measure of how far it deviates from being circular.\"\r\n\r\n\r\nI think it answers your question perfectly. You can read the entire article at http://en.wikipedia.org/wiki/Conic_sections.", "Solution_5": "That's a good question. I don't have the answer as to how to prove it, but those two distances you mentioned are used to distinguish conics. If the sum of distances from a given point on the conic to the two foci is equal to a constant, it is an ellipse. If the difference of the distances is equal to a constant, it is a hyperbola. (The constant is always equal to 2a, or the length of major/transverse axis in a ellipse/hyperbola, respectively).", "Solution_6": "This is what Wikipedia has to say about your question: \r\n\r\n\"The four defining conditions above can be combined into one condition that depends on a fixed point F (the focus), a line L (the directrix) not containing F and a positive number e (the eccentricity). The corresponding conic section consists of all points whose distance to F equals e times their distance to L. For 0 < e < 1 we obtain an ellipse, for e = 1 a parabola, and for e > 1 a hyperbola.\r\n\r\n\"For an ellipse and a hyperbola, two focus-directrix combinations can be taken, each giving the same full ellipse or hyperbola. The distance from the center to the directrix is a/e, where a is the semi-major axis of the ellipse, or the distance from the center to the tops of the hyperbola. The distance from the center to a focus is ae.\r\n\r\n\"In the case of a circle e = 0 and one imagines the directrix to be infinitely far removed from the center. However, the statement that the circle consists of all points whose distance is e times the distance to L is not useful, because we get zero times infinity.\r\n\r\n\"The eccentricity of a conic section is thus a measure of how far it deviates from being circular.\"\r\n\r\n\r\nI think it answers your question perfectly. You can read the entire article at http://en.wikipedia.org/wiki/Conic_sections.", "Solution_7": "I don't get why the formulas are the way they are. I'm looking for a proof or derivation of the eccentricity formulas of the conic shapes.\r\n\r\nMy textbook says eccentricity is $PF: PD$ where P is a point on the section, F is focus, and D is directrix.\r\n\r\nThe wikipedia article says it is equal to$c: a = F_1F_2 : 2A$", "Solution_8": "I think you are looking for a simple explaination? I'm not exactly sure. As weiyan said earlier, eccentricity measures how much the graph deviates from being a circle. I don't know much about trig/polar but think of it this way. A cirlce is made by intersecting the two cones with a plane, and having the plane parallel to the base of the cone. Now slightly tilt the plane. The intersection is an ellipse. Tilt a little more so that exactly one of the bases of the cone is interected. This creates a parabola. Tilt still more, such that both bases are intersected by and the plane, and it has been nearly a 90 degree tilt from the way you formed the circle. The intersection is now a hyperbola. Notice, the eccentricity is increasing as the new conic sections are formed. This also relates the conic's graph to its eccentricity." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a\\ge b\\ge c>0$ and $ x,y,z$ be real numbers such that $ x\\plus{}y\\plus{}z\\equal{}0,\\,x\\ge 0,\\,z\\le 0$. Show that\r\n\\[ a^xb^yc^z\\ge 1.\\]", "Solution_1": "[quote=\"nayel\"]Let $ a\\ge b\\ge c > 0$ and $ x,y,z$ be real numbers such that $ x \\plus{} y \\plus{} z \\equal{} 0,\\,x\\ge 0,\\,z\\le 0$. Show that\n\\[ a^xb^yc^z\\ge 1.\n\\]\n[/quote]\r\n\r\n1)if $ y\\ge0$\r\n$ a^xb^yc^z\\ge a^x c^{y \\plus{} z} \\equal{} a^xc^{ \\minus{} x} \\equal{} (\\frac {a}{c})^x\\ge1$.\r\n\r\n2)if $ y\\le0$\r\n$ a^xb^yc^z\\ge a^{x\\plus{}y}c^z\\equal{}a^{\\minus{}z}c^z\\equal{}(\\frac {a}{c})^{\\minus{}z}\\ge1$", "Solution_2": "Very nice NextPeace! :)\r\n\r\nAn interesting corollary of the above inequality: let $ a\\ge b\\ge c>0$ and $ x,y,z$ be real numbers such that $ x\\ge \\frac{x\\plus{}y\\plus{}z}{3}\\ge z$. Then the following inequality holds\r\n\\[ a^xb^yc^z\\ge (abc)^\\frac{x\\plus{}y\\plus{}z}{3}.\\]" } { "Tag": [ "function", "induction", "number theory unsolved", "number theory" ], "Problem": "[i]A given funtion $ f \\ : \\ N \\ \\to \\ N$ satisfies three conditions :\n\n $ 1 / f(1) \\ \\equal{} \\ 1$\n\n$ 2/ 3f(n). f( 2n\\plus{}1 ) \\ \\equal{} \\ f(2n) \\left( 1 \\plus{} 3f(n) \\right) \\ \\forall \\ n \\ \\in \\ N$\n\n $ 3/ f(2n) < 6f(n) \\ \\forall \\ n \\ \\in \\ N$\n\nFind all pairs of positive integers $ \\left( k ; m \\right)$ such that :\n\n $ f(k) \\ \\plus{} \\ f(m) \\ \\equal{} \\ 293$[/i]", "Solution_1": "From 2) we conclude $ 3f(n)|f(2n)$ and this yields a stronger version of 3) given by $ f(2n)=3f(n)$. After substitution into 2) we get $ f(2n+1)=1+3f(n)$. Switching to the dual system we observe that for all $ N>0$ and $ a_k\\in\\{\\pm 1\\}, a_N=1$ we get the recursion $ f(\\sum_{k=0}^Na_k2^k)=a_0+3f(\\sum_{k=0}^{N-1}a_{k+1}2^k)$. This together with 1) can be used in a straightforward manner to prove (by induction on $ N\\geq 0$) that $ f(\\sum_{k=0}^Na_k2^k)=\\sum_{k=0}^Na_k3^k$ holds.\r\n\r\nFor $ N\\leq M$ assume that $ f(\\sum_{k=0}^Na_k2^k)+f(\\sum_{k=0}^Mb_k2^k)=293$ with $ a_k,b_k\\in\\{0,1\\}$ and $ a_N=b_M=1$ holds. This is equivalent to $ 1\\cdot 3^5+1\\cdot 3^3 +2\\cdot 3^2+1\\cdot 3^1 +2\\cdot 3^0=\\sum_{k=0}^M(a_k+b_k)3^k$, where we have formally set $ a_k=0$, if $ k>N$. Since $ 0\\leq a_k+b_k<3$ holds, this is equivalent to the equations $ a_0=b_0=a_2=b_2=1$ and $ a_1+b_1=a_4+b_4=a_5+b_5=1$. Therefore, all pairs $ (A,B)$ with $ f(A)+f(B)=293$ are given by the set $ \\{(5+2a+16b+32c\\,,\\,55-2a-16b-32c)\\,|\\,a,b,c\\in\\{0,1\\}\\}$" } { "Tag": [ "floor function", "inequalities", "combinatorics unsolved", "combinatorics" ], "Problem": "find the minimum value of k>1 that: for any k subsets of X={1,2,..,n} ,there are 2 set A,B that A C B", "Solution_1": "Your minimum value is $\\binom{n}{\\left\\lfloor\\frac{n}{2}\\right\\rfloor}+1$. In fact, Sperner's theorem states that if we have k subsets $X_1$, $X_2$, ..., $X_k$ of an n-element set Y such that $X_i\\not\\subseteq X_j$ for any $i\\neq j$, then $k\\leq\\binom{n}{\\left\\lfloor\\frac{n}{2}\\right\\rfloor}$; and indeed, this inequality is strict, since we can set $k=\\binom{n}{\\left\\lfloor\\frac{n}{2}\\right\\rfloor}$ and take as $X_1$, $X_2$, ..., $X_k$ all possible $\\left\\lfloor\\frac{n}{2}\\right\\rfloor$-element subsets of Y, and these k subsets will obviously satisfy $X_i\\not\\subseteq X_j$ for any $i\\neq j$.\r\n\r\nA proof of Sperner's theorem can be found at http://www.mathlinks.ro/Forum/viewtopic.php?t=14689 .\r\n\r\n[Edited to make it more clear.]\r\n\r\n Darij" } { "Tag": [ "MATHCOUNTS" ], "Problem": "In the game of Bing Bong, there are two ways to score. A smash is worth 7 points, and a bash is worth 9 points. What is the greatest score that cannot be attained in a game of Bing Bong?", "Solution_1": "The question is not clear. Please state the exact rules of the game.", "Solution_2": "[hide] By the Chicken McNugget Theorem, $ 7\\times{9}\\minus{}(7\\plus{}9)\\equal{}\\boxed{47}$.[/hide]", "Solution_3": "[hide]So, you do (7*9)-(7+9)=47[/hide]" } { "Tag": [ "function", "number theory solved", "number theory" ], "Problem": "Let $f$ be a bijective function from $A = \\{ 1, 2, \\ldots, n \\}$ to itself. Show that there is a positive integer $M$ such that $f^{M}(i) = f(i)$ for each $i$ in $A$, where $f^{M}$ denotes the composition $f \\circ f \\circ \\cdots \\circ f$ $M$ times.", "Solution_1": "Let $k \\in A$ be given.\r\nSince $A$ is finite the set $\\{k,f(k), \\cdots, f^p(k), \\cdots, \\}$ is finite, thus there exist $a,b$ such that $b>a$ and $f^a(k) = f^b(k).$\r\nThus $f^a(f^{b-a}(k)) = f^a(k)$. Since $f$ is bijective it follows that $f^{b-a}(k) = k.$\r\nTherefore, for each $k \\in A$ there exists a positive integer $M_k$ such that $f^{M_k}(k) = k.$\r\nIt is easy to deduce that for every positive integer $p$, we have $f^{pM_k} (k) =k.$\r\n\r\nNow, to conlude, just choose $M = \\text{lcm} (M_1, \\cdots, M_n).$\r\n\r\nPierre.", "Solution_2": "Graph Theory?", "Solution_3": "Honestly, this is a pretty cool problem.\n\nEssentially we must find a $M$ so that $f^{M-1}(k)=k$.\nSince we have that $f$ is a bijection that means that after some turns, there exists some $B_k \\in \\mathbb{N}$ so that $f^{B_k}(k) = k$.\nSo now let's have a graph $G_k$, with $n$ vertices. We connect the pairs of vertices of the form $(v_{f^{t}(k)},v_{f^{t+1}(k)})$, where $t$ is a non-negative integer number less than $B_k$ (note that $f^{0}(k)=k$), but now notice that we must have a cycle in the graph $G_k$, and essentially we must have that $B_k$ is the length of the cycle.\n\nSo we must have that $B_t \\mid M-1$, where $t$ is a positive integer number less than or equal to $n$.\nThus we can set $M-1=B_1B_2\\dots B_n$, this implies that $M=B_1B_2\\dots B_n +1$." } { "Tag": [ "analytic geometry", "vector", "trigonometry", "Pythagorean Theorem", "geometry" ], "Problem": "How fast something is traveling south and/or east when you're given how fast it is when you're traveling southeast? Example: A car is traveling southeast at $ \\frac{\\sqrt{3}}{3}$. How fast is the car traveling going south?\r\n\r\nIs it any different for northeast, northwest, or southwest?\r\n\r\nThanks!!", "Solution_1": "Take a Cartesian plane with a right angled coordinate system in it, and let the positive direction of y axis denote north. Now, positive direction of x axis denotes east, negative y south, and negative x west. \r\nAssign a vector to the car's velocity using this system-if you put the car in the origin of the system, then the given vector is a radius vector whose modulus is $ \\frac {\\sqrt {3}}{3}$, in fourth quadrant with a 45 degrees angle between it and the positive x (hence, the angle between negative y and the vector is also 45 degrees).\r\nThe south component is its projection on the y axis, i.e. $ \\frac {\\sqrt {3}}{3}\\frac {\\sqrt {2}}{2} \\equal{} \\frac {\\sqrt {6}}{6}$.\r\n\r\nThis was a complicated manner of showing something rather simple, but when put in this manner, it will be easier for you to move to harder problems of this type (i.e. Car is traveling 60 degrees southeast at the given velocity)\r\n\r\nSimpler way would consist of taking the vector $ v$ into its components $ v_S$ and $ v_E$ denoting the speed toward south and east, respectively (they're orthogonal). Knowing that SE is \"right in the middle of it\", you know that your task is reduced to a simple application of Pythagorean theorem in an isosceles right triangle with a know hypotenuse.", "Solution_2": "Could you explain how you got $ \\frac {\\sqrt {2}}{2}$? Thanks again!", "Solution_3": "Basic trigonometry: $ \\cos 45^0\\equal{}\\frac{\\sqrt{2}}{2}$. How did it appear in our calculations? Well, in the first approach, you have a vector (whose modulus is $ v\\equal{}\\frac{\\sqrt{3}}{3}$) and want to know the modulus of its orthogonal projection on an axis. If the angle between the vector and the axis is $ \\alpha\\equal{}45^0$, then the projection's modulus is (it's pretty much obvious, just draw a sketch) $ v_S\\equal{}v\\cos\\alpha\\equal{}\\frac{\\sqrt{3}}{3}\\frac{\\sqrt{2}}{2}$.\r\n\r\nIn the other approach, the same formula is obtained, applying Pythagoras' theorem:\r\n\r\n$ v^2\\equal{}v_S^2\\plus{}v_E^2$\r\n\r\nnoting that $ v_S\\equal{}v_E$, we have $ 2v_S^2\\equal{}v^2$\r\n\r\nor $ v_S\\equal{}\\frac{\\sqrt{2}}{2}v\\equal{}\\frac{\\sqrt{2}}{2}\\frac{\\sqrt{3}}{3}$.", "Solution_4": "Oh okay, thanks alot!" } { "Tag": [], "Problem": "What number is $ 10\\%$ of $ 20\\%$ of $ 30\\%$ of 40? Express your answer as a decimal to the nearest hundredth.", "Solution_1": "$ \\frac{1}{10} \\cdot \\frac{1}{5} \\cdot \\frac{3}{10} \\cdot 40 \\equal{} \\frac{6}{25} \\equal{} 0.24$" } { "Tag": [ "geometry", "geometric transformation", "homothety", "ratio", "combinatorics proposed", "combinatorics" ], "Problem": "$A$ is a compact convex set in plane. Prove that there exists a point $O \\in A$, such that for every line $XX'$ passing through $O$, where $X$ and $X'$ are boundary points of $A$, then\n\\[ \\frac12 \\leq \\frac {OX}{OX'} \\leq 2.\\]", "Solution_1": "For all $M$ on the boundary of $A$, let $A_M$ be the homothetic of $A$ with respect to the homothty $h_M$ of pole $M$ and ratio $\\frac 2 3 $.\r\nUsing the convexity of $A$, we deduce that $A_M$ is convex and contained in $A$ for all such point $M$.\r\n\r\nUsing the convexity of $A$ and $A_M$, it is easy to verify that is $M,N,P$ are three points on the boundary of $A$ and $G$ is the center of gravity of $MNP$ then $G$ belongs to each of the sets $A_M,A_N,A_P$.\r\nIt follows that the family of the sets $A_M$ is a set of bounded convex sets such that any three have a common point.\r\nFrom Helly's theorem, we deduce that there exists a point $O$ which belongs to all the sets $A_M$ where $M$ is a point of the boundary of $A$.\r\n\r\nNow, let's consider any line $d$ which passes through $O$. This line meet the boundary of $A$ in two points $X,X'$.\r\nThen, from above $O$ belongs to the segment $XX''$ where $X'' = h_X(X')$, and to the segment $XX'$. Thus, $OX \\leq XX'' = \\frac 2 3 XX'$ so that $OX' \\geq \\frac 1 3 XX'$. Therefore $\\frac {OX } {OX'} \\leq 2$.\r\nThe other inequality is obtained by interchanging $X$ and $X'$.\r\n\r\nPierre.", "Solution_2": "It's very similar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16474]this[/url] problem. In fact, in order to prove that, I proved a very similar lemma (although weaker).", "Solution_3": "what is HELLYS THEOREM?", "Solution_4": "Given a collection of convex sets in $\\mathbb R^{n}$ s.t. each $n+1$ have non-void intersection, all of them have non-void intersection.", "Solution_5": "Another solution that is very beautiful.\r\nIf A and B are two convex figures. Then there is atranslation that takes A in B if and only if Every 3 points of A could be taken in B with a translation.\r\nProof of this is easily with Helly theorem.", "Solution_6": "A is a convex set in plane so it is compact. :)\nSo, we can set a triangle which the size is maximum. If we set $ O $ the centroid of it, we are done. :)", "Solution_7": "Here is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=520132]another thread dedicated to this problem[/url]. One can see that the center of gravity(the centroid) of $A$ satisfies the requirement.", "Solution_8": "a nice problem\n here's my solution:\nwe will use $S$ to refer to the convex set \ntake a triangle $ABC$ with vertices in the boundary of $S$ with maximum area\nnote that if we construct the parallel from $ A$ to $BC$ let it $\\ell_a$ similarly $\\ell_c$ and $\\ell_b$\nthen the triangle from $\\ell_a$ $\\ell_c$ $\\ell_b$ contains $S$ (let it $T$)\ntake $O$ the centroid of $ABC$\ntake a line throw $O$ cuts $S$ in $X,Y$, $ABC$ at $X',Y'$ and $T$ at $X'',Y''$ (with $X',X''$ closer to $X$)\n$2 \\ge \\frac{OX''}{OY'} \\ge \\frac{OX}{OY} \\ge \\frac{OX'}{OY''} \\ge \\frac{1}{2}$\nand we win :D ", "Solution_9": "It can be proved immediately from Minkovski's theorem. It states that for convex sets in $R^d$ we can find a point $O$ such that for every line $XOX'$ we have $\\dfrac{XO}{X'O} \\leq d$" } { "Tag": [ "geometry", "circumcircle" ], "Problem": "Source: Rocket City Math League Round 1, Apollo Division\r\n\r\n9. If $\\overline{AB}$ and $\\overline{CD}$ are perpendicular chords to a circle that intersect at E where $\\overline{AE}=6$, $\\overline{EB}=2$,$\\overline{CE}=4$, and $\\overline{ED}=3$. What is the area of the circle?\r\n\r\n(BTW, no calculators are allowed on RCML's)", "Solution_1": "[hide]An inscribed triangle in this circle has sides of $8$, $2 \\sqrt {13}$, and $2 \\sqrt 5$. Since $\\frac {abc}{4A}$ gives us the radius of the circumscribed circle of a triangle with sides $a$, $b$, $c$, and area $A$, we plug in the values to get $\\frac {8 \\cdot 2 \\sqrt {13} \\cdot 2 \\sqrt 5}{\\frac 12 \\cdot 4 \\cdot 8 \\cdot 4} = \\frac {\\sqrt {65}}{2}$. The area of the circle is then $\\left(\\frac {\\sqrt {65}}{2}\\right)^2 \\cdot \\pi = \\frac {65\\pi}{4}$.[/hide]", "Solution_2": "[quote=\"tarquin\"][hide]An inscribed triangle in this circle has sides of $8$, $2 \\sqrt {13}$, and $2 \\sqrt 5$. Since $\\frac {abc}{4A}$ gives us the radius of the circumscribed circle of a triangle with sides $a$, $b$, $c$, and area $A$, we plug in the values to get $\\frac {8 \\cdot 2 \\sqrt {13} \\cdot 2 \\sqrt 5}{\\frac 12 \\cdot 4 \\cdot 8 \\cdot 4} = \\frac {\\sqrt {65}}{2}$. The area of the circle is then $\\left(\\frac {\\sqrt {65}}{2}\\right)^2 \\cdot \\pi = \\frac {65\\pi}{4}$.[/hide][/quote]\r\n\r\nThanks, but how did you figure out the lengths of the sides?", "Solution_3": "The chords are perpendicular so they make right triangles if you connect different points.", "Solution_4": "There is a simple formula for perpendicular chords.\r\nLet\r\n$AE=a$\r\n$BE=b$\r\n$CE=c$\r\n$DE=d$\r\n\r\n$\\text{Radius}=\\displaystyle \\frac{\\sqrt{a^2+b^2+c^2+d^2}}{2}$\r\n\r\nIn this case\r\n\r\n$\\text{Radius}=\\displaystyle \\frac{\\sqrt{2^2+3^2+4^2+6^2}}{2}=\\displaystyle \\frac{\\sqrt{65}}{2}$\r\n\r\n$\\text{Area}=\\left({\\displaystyle \\frac{\\sqrt{65}}{2}\\right)}^2\\cdot{\\pi}=\\displaystyle \\frac{65\\pi}{4}}$", "Solution_5": "I believe that that's called the root-mean-square", "Solution_6": "thanks.", "Solution_7": "If you don't like formulas, here is a graphical approach to the problem.", "Solution_8": "[quote=\"NoSoupForYou\"]If you don't like formulas, here is a graphical approach to the problem.[/quote]\r\n\r\nwow, did you make that using paint?", "Solution_9": "Yeah, it took me a while." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "If p and q are different odd primes, then, if p - 1|q-1, a is an integer, an (a,pq) = 1, prove that $a^{q-1}$ is congruent to 1 mod pq.", "Solution_1": "If $q-1=k(p-1)$, then $a^{q-1}=(a^{p-1})^k\\equiv1\\bmod p$. Since also $a^{q-1}\\equiv1\\bmod q$, the number $a^{q-1}-1$ is divisible by p and q, i.e. $a^{q-1}\\equiv1\\bmod pq$." } { "Tag": [ "\\/closed" ], "Problem": "Can people who've taken classes in something you have not share information from their transcripts and what they have learned or is that against the rules?", "Solution_1": "They cannot share the transcripts." } { "Tag": [ "limit", "trigonometry", "calculus", "calculus computations" ], "Problem": "(1) Calculate $ \\lim_{n\\to\\infty}n\\left(\\sec^{n}\\frac{a}{n}\\minus{}1\\right)$\r\n\r\n(2) Let $ \\theta \\equal{}\\theta(n) \\equal{}\\frac{2\\pi}{n}$. Calculate $ \\lim_{n\\to\\infty}\\frac{\\tan\\theta (\\sec^{n}\\theta\\minus{}1)}{\\sec\\theta\\minus{}1}$. This limit has a simple geometrical interpretation.", "Solution_1": "$ \\sec x\\equal{}1\\plus{}\\frac{1}2 x^{2}\\plus{}O(x^{4})\\quad ,\\quad\\left(1\\plus{}\\frac{a^{2}}{2}x^{2}\\right)^{\\frac{1}{x}}\\equal{}1\\plus{}\\frac{a^{2}}{2}x\\plus{}O(x^{2})$\r\n\r\nTherefore $ \\sec^{n}\\left(\\frac{a}{n}\\right)\\equal{}\\left(1\\plus{}\\frac{a^{2}}{2}\\frac{1}{n^{2}}\\plus{}O\\left(\\frac{1}{n^{4}}\\right)\\right)^{n}\\equal{}1\\plus{}\\frac{a^{2}}{2}\\frac{1}{n}\\plus{}O\\left(\\frac{1}{n}\\right)$\r\n\r\n$ \\Rightarrow n\\left(\\sec^{n}\\left(\\frac{a}{n}\\right)\\minus{}1\\right)\\to\\frac{a^{2}}{2}$\r\n\r\n\r\n$ \\frac{\\tan\\theta\\left(\\sec^{n}\\theta\\minus{}1\\right)}{\\sec\\theta\\minus{}1}\\equal{}\\frac{\\tan\\frac{2\\pi}{n}}{n\\left(\\sec\\frac{2\\pi}{n}\\minus{}1\\right)}n\\left(\\sec^{n}\\frac{2\\pi}{n}\\minus{}1\\right)\\sim\\frac{\\frac{2\\pi}{n}}{n\\frac{1}2\\left(\\frac{2\\pi}{n}\\right)^{2}}\\frac{(2\\pi)^{2}}{2}\\equal{}2\\pi$", "Solution_2": "[quote=\"Dr. Doe\"]$ \\sec x \\equal{} 1\\plus{}\\frac{1}{2} x^{2}\\plus{}O(x^{4})\\quad ,\\quad\\left(1\\plus{}\\frac{a^{2}}{2}x^{2}\\right)^{\\frac{1}{x}}\\equal{} 1\\plus{}\\frac{a^{2}}{2}x\\plus{}O(x^{2})$\n\nTherefore $ \\sec^{n}\\left(\\frac{a}{n}\\right) \\equal{}\\left(1\\plus{}\\frac{a^{2}}{2}\\frac{1}{n^{2}}\\plus{}O\\left(\\frac{1}{n^{4}}\\right)\\right)^{n}\\equal{} 1\\plus{}\\frac{a^{2}}{2}\\frac{1}{n}\\plus{}O\\left(\\frac{1}{n}\\right)$\n\n$ \\Rightarrow n\\left(\\sec^{n}\\left(\\frac{a}{n}\\right)\\minus{}1\\right)\\to\\frac{a^{2}}{2}$\n\n\n$ \\frac{\\tan\\theta\\left(\\sec^{n}\\theta\\minus{}1\\right)}{\\sec\\theta\\minus{}1}\\equal{}\\frac{\\tan\\frac{2\\pi}{n}}{n\\left(\\sec\\frac{2\\pi}{n}\\minus{}1\\right)}n\\left(\\sec^{n}\\frac{2\\pi}{n}\\minus{}1\\right)\\sim\\frac{\\frac{2\\pi}{n}}{n\\frac{1}{2}\\left(\\frac{2\\pi}{n}\\right)^{2}}\\frac{(2\\pi)^{2}}{2}\\equal{} 2\\pi$[/quote]\nHi everyone. \nSome correction to Dr.Doe\n\n[quote=\"Dr. Doe\"]$ \\quad\\left(1\\plus{}\\frac{a^{2}}{2}x^{2}\\right)^{\\frac{1}{x}}\\equal{} 1\\plus{}\\frac{a^{2}}{2}x\\plus{}O(x^{2})$[/quote]\nhere must be \n$ \\quad\\left(1\\plus{}\\frac{a^{2}}{2}x^{2}\\right)^{\\frac{1}{x}}\\equal{} 1\\plus{}\\frac{a^{2}}{2}x\\plus{}O(x^{3})$\nand also here \n[quote=\"Dr. Doe\"]\nTherefore $ \\sec^{n}\\left(\\frac{a}{n}\\right) \\equal{}\\left(1\\plus{}\\frac{a^{2}}{2}\\frac{1}{n^{2}}\\plus{}O\\left(\\frac{1}{n^{4}}\\right)\\right)^{n}\\equal{} 1\\plus{}\\frac{a^{2}}{2}\\frac{1}{n}\\plus{}O\\left(\\frac{1}{n}\\right)$[/quote]\r\n here must be \r\n$ ...\\equal{}\\left(1\\plus{}\\frac{a^{2}}{2}\\frac{1}{n^{2}}\\plus{}O\\left(\\frac{1}{n^{4}}\\right)\\right)^{n}\\equal{} 1\\plus{}\\frac{a^{2}}{2}\\frac{1}{n}\\plus{}O\\left(\\frac{1}{n^{3}}\\right)$", "Solution_3": "I don't understand your first correction.\r\n\r\n$ \\left(1\\plus{}\\frac{a^{2}}{2}x^{2}\\right)^{\\frac{1}{x}}\\equal{}1\\plus{}\\frac{a^{2}}{2}x\\plus{}\\frac{a^{4}}{8}x^{2}\\plus{}...$ and that's not $ 1\\plus{}\\frac{a^{2}}{2}x\\plus{}O(x^{3})$ .\r\n\r\nBut in your second correction actually a small Landau symbol $ o\\left(\\frac{1}{n}\\right)$ would be better than a big one.", "Solution_4": "my first correction was not correct :(" } { "Tag": [ "inequalities" ], "Problem": "GIVEN x,y,z are positive reals such that x+y+z>=xyz . prove that x^2+y^2+z^2>= xyz :blush:", "Solution_1": "hmm.. it's a bit weird \r\nsetting $ a\\equal{}\\frac{1}{x} ; b\\equal{}\\frac{1}{y} ; c\\equal{} \\frac{1}{z}$\r\nwe have $ \\sum ab \\geq 1$\r\nwe need to prove that $ \\sum \\frac{ab}{c} \\geq 1$\r\nwhich is equivalent to $ abc (\\sum \\frac{1}{c^2} ) \\geq 1$\r\nnote that $ abc (\\sum \\frac{1}{c^2}) \\geq abc (\\sum \\frac{1}{ab}) \\equal{} \\sum a$\r\nso it's suffice to prove that $ \\sum a \\geq 1$\r\non the other hand , $ \\sum a \\geq \\sqrt{3(\\sum ab)} \\geq \\sqrt{3} > 1$ ..... a bit weird", "Solution_2": "$ \\sum {x^2} \\ge \\sum {xy}$\r\n$ \\sum\\frac{1}{x} \\sum {x^2} \\ge \\sum {x} \\ge xyz$\r\n\r\n$ \\text {since }\\sum\\frac{1}{x}\\equal{}\\frac{\\sum{xy}}{xyz}$\r\n\r\n$ \\implies \\sum {x^2} \\ge \\frac{( xyz)^2}{\\sum {xy}} \\ge \\frac{( xyz)^2}{\\sum {x^2}}$\r\n\r\n\r\n$ \\implies \\sum {x^2} \\ge xyz$" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "for m;n is two integer number such that : (m;n)=d \r\n prove that : $m^{phi(n)}+n^{phi(m)}=1+kmn$", "Solution_1": "It is not true. For example n=m=2.\r\nObviosly $d=(m,n)=1$. If d=1, then for $x=m^{\\phi(n)}+n^{\\phi(m)}-1$ we have $n|x$ and $m|x$, therefore $x=kmn$.", "Solution_2": "thanks you \r\nfirt ; i think it true :lol:" } { "Tag": [ "function", "calculus", "derivative", "inequalities theorems", "inequalities" ], "Problem": "how to use it?\r\n\r\ncan somebody show me how to find the minimum of\r\n\r\nthis simple example $ x \\plus{} 1/{x}$\r\n\r\nby AM-GM its minimum is 2\r\n\r\nbut how by lagrange. can we use lagrange?\r\nplease :)", "Solution_1": "Lagrange multipliers are used when we have some multi-variable function subjected to some constraints, and we want to find the extrema.\r\n\r\nI don't think we need to use Lagrange for this problem. If you use the derivative tests, we get\r\n$ 1\\minus{}\\frac{1}{x^{2}}\\equal{}0\\Rightarrow 1\\equal{}\\frac{1}{x^{2}}\\Rightarrow x\\equal{}\\pm1$\r\nPlugging in,\r\n$ 1\\plus{}1\\equal{}2$ and $ \\minus{}1\\minus{}1\\equal{}\\minus{}2$\r\nThen 2 is a local minimum and -2 is a local maximum.", "Solution_2": "[quote=\"JRav\"]Lagrange multipliers are used when we have some multi-variable function subjected to some constraints, and we want to find the extrema.\n\nI don't think we need to use Lagrange for this problem. If you use the derivative tests, we get\n$ 1 \\minus{} \\frac {1}{x^{2}} \\equal{} 0\\Rightarrow 1 \\equal{} \\frac {1}{x^{2}}\\Rightarrow x \\equal{} \\pm1$\nPlugging in,\n$ 1 \\plus{} 1 \\equal{} 2$ and $ \\minus{} 1 \\minus{} 1 \\equal{} \\minus{} 2$\nThen 2 is a local minimum and -2 is a local maximum.[/quote]\r\n\r\nthanks,\r\nactually i dont know alot about calculus. :blush: \r\n :D \r\nwhy $ 1 \\minus{} \\frac{1}{x^{2}} \\equal{} 0$", "Solution_3": "Take the derivative of each term and set it equal to zero.", "Solution_4": "oh ya thanks\r\n\r\nthe gradient equal to $ 0$" } { "Tag": [ "geometry" ], "Problem": "Please differentiate these:\r\n\r\nDifferential Geometry, Algebraic Geometry and Projective Geometry. It is my elective, and I do not know which is which. \r\n\r\nThanks in advance", "Solution_1": "Information on all of these topics is easy to find on [url=http://www.math.niu.edu/~rusin/known-math/welcome.html]Dave Rusin's site[/url] or wikipedia. However, this is the wrong subforum to be asking this question in (advanced fields in the college playground would be better, for example)" } { "Tag": [ "function", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ \\mathcal{A}$ unital $ C^*$-algebra, $ x\\in\\mathcal{A}$ normal element and $ f\\in C(\\sigma(x))$. Prove that \\[ \\sigma(f(x))\\equal{}f(\\sigma(x))\\equal{}\\{f(\\lambda): \\lambda\\in \\sigma(x)\\}\\].\r\n\r\nI would be grateful for any useful hints.", "Solution_1": "everything takes place in the commutative, unital C*-algebra which is generated by $ x$, and the spectrum does not change when we make the C*-algebra smaller. then, by gelfand-neymark, the claim may be translated to ordinary functions where the spectrum is just the image." } { "Tag": [], "Problem": "Given that the midpoints of the sides of a triangle are connected, how many triangles are in the resulting diagram?", "Solution_1": "[asy]pair A=(-5,0), B=(5,0), C=(0, 5sqrt(3)), D=(-2.5, 2.5sqrt(3)), E=(2.5, 2.5sqrt(3)), F=(0,0);\ndraw(A--B--C--cycle);\ndraw(D--E--F--cycle);\nlabel(\"$A$\", A, SW);\nlabel(\"$B$\", B, SE);\nlabel(\"$C$\", C, N);\nlabel(\"$D$\", D, NW);\nlabel(\"$E$\", E, NE);\nlabel(\"$F$\", F, S);[/asy]\r\nWe can see, from the diagram, that the triangles are $ \\triangle ADF,\\triangle DEF,\\triangle EFB,\\triangle DEC,$ and $ \\triangle ABC$. Therefore, we have a total of $ \\boxed{5}$ triangles." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "For real numbers $ x_1, x_2, \\cdots x_n$, we have $ x_1 > 1, x_2 > 2, \\cdots x_n > n$, prove that \r\n\r\n$ \\frac {(x_1 \\plus{} x_2 \\plus{} \\cdots \\plus{} x_n)^2}{\\sqrt {x_1^2 \\minus{} 1^2} \\plus{} \\sqrt {x_2^2 \\minus{} 2^2} \\plus{} \\cdots \\plus{} \\sqrt {x_n^2 \\minus{} n^2}}$.\r\n\r\n---\r\n\r\nPositive integers $ x_1, x_2, \\cdots x_n$ satisfy $ x_1\\plus{}x_2\\plus{}\\cdots\\plus{}x_n\\equal{}1$.\r\n\r\nProve that \r\n\r\n$ \\frac{1}{1\\plus{}x_1}\\plus{}\\frac{1}{1\\plus{}x_1\\plus{}x_2}\\plus{}\\cdots\\plus{}\\frac{1}{1\\plus{}x_1\\plus{}x_2\\plus{}\\cdots\\plus{}x_n}<$\r\n\r\n$ \\sqrt{\\frac{2}{3}\\left(\\frac{1}{x_1}\\plus{}\\frac{1}{x_2}\\plus{}\\cdots\\plus{}\\frac{1}{x_n}\\right)}$.\r\n\r\n---\r\n\r\nReal numbers $ x_1, x_2, \\cdots x_n$ satisfy $ x_1^2\\plus{}x_2^2\\plus{}\\cdots\\plus{}x_n^2\\equal{}1$.\r\n\r\nProve that $ \\frac{x_1}{1\\plus{}x_1^2}\\plus{}\\frac{x_2}{1\\plus{}x_1^2\\plus{}x_2^2}\\plus{}\\cdots\\plus{}\\frac{x_n}{1\\plus{}x_1^2\\plus{}x_2^2\\plus{}\\cdots\\plus{}x_n^2}<\\sqrt{\\frac{n}{2}}$.", "Solution_1": "What are we supposed to prove in no 1? :huh:", "Solution_2": "For some reason I can't edit anymore...\r\n\r\n#1 should be:\r\n\r\nFor real numbers $ x_1, x_2, \\cdots x_n$, we have $ x_1 > 1, x_2 > 2, \\cdots x_n > n$, find the minimal value of \r\n\r\n$ \\frac {(x_1 \\plus{} x_2 \\plus{} \\cdots \\plus{} x_n)^2}{\\sqrt {x_1^2 \\minus{} 1^2} \\plus{} \\sqrt {x_2^2 \\minus{} 2^2} \\plus{} \\cdots \\plus{} \\sqrt {x_n^2 \\minus{} n^2}}$.\r\n\r\nThanks for pointing out. :)", "Solution_3": "Let $ y_{i} = \\sqrt{x_{i}^{2} - i^{2}}$ > 0 for all $ 1 \\leq i \\leq n$\r\n\r\nThen we have to minimise $ S$ = $ \\frac{(\\sqrt{y_{1}^{2} + 1^{2}} + \\sqrt{y_{2}^{2} + 2^{2}} + \\cdots + \\sqrt{y_{n}^{2} + n^{2}})^{2}}{y_{1} + y_{2} + \\cdots + y_{n}}$\r\n\r\n= $ \\frac{y_{1}^{2} + y_{2}^{2} + \\cdots + y_{n}^{2} + 2\\sum_{1\\leq i a^2$\r\n\r\n$ a^3 > n^2 > (1/4) a^4$ so $ a < 4$ and $ n < 2$\r\n\r\nn = 1 does not work, so n = 0 is only solution" } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Give a sequence\r\n$ A_k \\equal{} \\frac {1}{32}( (17 \\plus{} 12\\sqrt {2})^k \\plus{} (17 \\minus{} 12\\sqrt {2})^k \\minus{} 2 )$\r\n\r\n$ k$ is a natural number.\r\nProve that $ A_k$ is triangular number.\r\n\r\nthanks!", "Solution_1": "[hide=\"idea\"]use pell equation.note that $ 17^2 \\minus{} 12^2.2 \\equal{} 1$.[/hide]\nif u want a full solution...\n[hide=\"solution\"]\ndefine these sequences:\n$ X_k \\equal{} 17X_{k \\minus{} 1} \\plus{} 24Y_{k \\minus{} 1},Y_k \\equal{} 12X_{k \\minus{} 1} \\plus{} 17Y_{k \\minus{} 1}$.we can easily prove that $ X_k \\plus{} Y_k\\sqrt2 \\equal{} (17 \\plus{} 12\\sqrt2)^k$.\nand also we can prove that $ A_k \\equal{} \\frac {X_k \\minus{} 1}{16}$.note that $ X_k \\equiv 1 \\pmod {16}$.\nbut we have \n$ X_k^2 \\minus{} 2.Y_k^2 \\equal{} 1 \\implies (X_k \\minus{} 1)(X_k \\plus{} 1) \\equal{} 2.Y_k^2,16 \\mid X_k \\minus{} 1 \\implies \\gcd(\\frac {X_k \\minus{} 1}{16},\\frac {X_k \\plus{} 1}{2}) \\equal{} 1 \\implies X_k \\equal{} 2.l_k^2 \\minus{} 1,\\frac {X_k \\minus{} 1}{16} \\equal{} m_k^2 \\implies A_k \\equal{} \\frac {l_k^2 \\minus{} 1}{8},l_k \\equal{} 2.t_k \\minus{} 1 \\implies A_k \\equal{} \\frac {t_k(t_k \\minus{} 1)}{2}$\nDONE![/hide]", "Solution_2": "define $ x \\equal{} \\sqrt{2} \\minus{} 1$ and $ y \\equal{} \\sqrt{2} \\plus{} 1$\r\n\r\n notice that $ xy \\equal{} 1$ and $ x^4 \\equal{} 17 \\minus{} 12\\sqrt{2}$ and $ y^4 \\equal{} 17 \\plus{} 12\\sqrt{2}$\r\n\r\n $ \\implies 32A_k \\equal{} x^{4k} \\plus{} y^{4k} \\minus{} 2 (xy)^{4k}$\r\n $ \\equal{} (y^{2k} \\minus{} x^{2k})^2$\r\n $ \\equal{} (y^k \\minus{} x^k)^2 (y^k \\plus{} x^k)^2$\r\n $ \\equal{} (y^{2k} \\plus{} x^{2k} \\minus{} 2)( y^{2k} \\plus{} x^{2k} \\plus{} 2)$\r\n\r\n so if $ N_k \\equal{} \\frac{1}{4}(y^{2k} \\plus{} x^{2k} \\minus{} 2)$\r\n then $ A_k \\equal{} \\frac{1}{2}N_k(N_k \\plus{} 1)$\r\n\r\n so it remains to show that $ N_k$ is an integer which can be done in the usual way either by expanding the powers using the binomial expression or by forming a linear recurrence relation for $ N_k$", "Solution_3": "[quote=\"Lomisedu\"]Give a sequence\n$ A_k \\equal{} \\frac {1}{32}( (17 \\plus{} 12\\sqrt {2})^k \\plus{} (17 \\minus{} 12\\sqrt {2})^k \\minus{} 2 )$\n\n$ k$ is a natural number.\nProve that $ A_k$ is triangular number.\n\nthanks![/quote]\r\nIf $ A_k\\equal{}\\frac{x_k(x_k\\plus{}1)}2$ then $ x_k\\equal{}\\frac{\\sqrt{1\\plus{}8A_k}\\minus{}1}2$\r\n\r\n$ 8A_k\\plus{}1\\equal{}\\frac 14((3\\plus{}2\\sqrt 2)^{2k}\\plus{}\\frac 1{(3\\plus{}2\\sqrt 2)^{2k}}\\plus{}2)$ $ \\equal{}\\frac 14((3\\plus{}2\\sqrt 2)^k\\plus{}\\frac 1{(3\\plus{}2\\sqrt 2)^k})^2$\r\n\r\nSo $ x_k\\equal{}\\frac 14(2\\plus{}(3\\plus{}2\\sqrt 2)^k\\plus{}\\frac 1{(3\\plus{}2\\sqrt 2)^k})$\r\n\r\nAnd we just have to show that this $ x_k\\equal{}\\frac {2\\plus{}(3\\plus{}2\\sqrt 2)^k\\plus{}(3\\minus{}2\\sqrt 2)^k}4$ is an integer, which is easy :\r\n\r\n$ x_0\\equal{}1$\r\n$ x_1\\equal{}2$\r\n$ x_{k\\plus{}2}\\equal{}6x_{k\\plus{}1}\\minus{}x_{k}\\minus{}2$\r\n\r\nAnd $ A_k\\equal{}\\frac{x_k(x_k\\plus{}1)}2$\r\n\r\n\r\n(a bit too late :) )" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "The vertices of a regular 2005- gon are colored red, white and blue. When ever two vertices of different colors stand next to each other, we are allowed to recolor them into the third color.\r\n(a) Prove that there is a \ufb01nite sequence of allowed recolorings after which all the vertices are of the same color.\r\n(b) Is that color uniquely determined by the initial coloring?", "Solution_1": "[hide=\"b\"] Yes. Instead of red, white, and blue, put 0's, 1's and 2's. Under the given operation, the sum of all numbers around this 2005-gon does not change modulo 3. In order for us to have all numbers around the circle to be 0, 1 or 2, then, we would have to have an initial sum congruent to 0, 1 or 2, respectively, modulo 3. Thus, the initial configuration determines a final color.[/hide]\r\n\r\nFor part a, I think you can create a process by which you can strictly increase the number of occurrences of whichever color you want. But I'm too lazy to outline it.", "Solution_2": "Hi. Now I'm at the first variant which is [i]a[/i]. Let's begin with I and II vertices. The first is blue. Anyway, we must choose one. And we chosen. Now we have two options. First is that we can select [i]same[/i] or [i] different[/i] colors. ... After some operations such that we're already doing, will establish the same situation we'd had at the outset. It's time now to use advantages of amateaurity. We can easily assume that in doing the requested operation, quite same thing will happen and wherever in a place, the all will go replaying.\r\nI will then think of the [i]b[/i] variant of the problem.\r\nThanks.", "Solution_3": "Drive an injection one of the corner to another i.e., make an injection from the first to the last number (and these numbers are what are used in counting the operations while you're solving the problem). Then you'll see whether or not there exists unique way to do.\r\nOnce thanks![/i]", "Solution_4": "[quote=\"K81o7\"][hide=\"b\"] Yes. Instead of red, white, and blue, put 0's, 1's and 2's. Under the given operation, the sum of all numbers around this 2005-gon does not change modulo 3. In order for us to have all numbers around the circle to be 0, 1 or 2, then, we would have to have an initial sum congruent to 0, 1 or 2, respectively, modulo 3. Thus, the initial configuration determines a final color.[/hide]\n\nFor part a, I think you can create a process by which you can strictly increase the number of occurrences of whichever color you want. But I'm too lazy to outline it.[/quote]\r\n\r\nYes.. I'm not sure if there is an easier way.. but here's mine:\r\nLet $ \\{a_{i}\\}$ be the vertices' colour and $ b$ be the final colour\r\nLet $ a_{i+kn}=a_{i}$ $ \\forall k\\in\\mathbb{Z}$\r\nLet the operation on $ (a_{i},a_{i+1})$ be $ O(i)$\r\nIf $ \\exists a_{i}\\neq a_{i+1}$, either keep or do $ O(i)$ so that $ a_{i}=b$ or $ a_{i+1}=b$\r\nStart with any contiguous block of $ \\{a_{i}=b| j\\leq i\\leq k\\}$, $ a_{k+1}\\neq b$\r\nLet $ x\\geq k+2$ be the smallest such that $ a_{x}\\neq a_{k+1}$\r\nIf $ a_{x}\\neq a_{k}$ then do $ O(x-1)$ ($ x$ decreases by 1)\r\nIf $ a_{x}=a_{k}$ and $ x=k+2$ do $ O(k),O(k+1),O(k)$ ($ k$ increases)\r\nIf $ a_{x}=a_{k}$ and $ x>k+2$ do $ O(k+1),O(k)$ ($ x$ decreases by 2)\r\nSince $ x$ cannot decrease indefinitely without $ k$ increasing, the process will terminate with the desired outcome.\r\n\r\n(In a more easy-to-see form, WLOG 0 is the final colour so I use the operations 01[color=red]0[/color] -> 220 -> 211 -> 0[color=red]0[/color]1 and 11[color=red]0[/color] -> 122 -> [color=red]0[/color]02 to bunch zeros together to the left)" } { "Tag": [ "analytic geometry", "probability", "calculus", "integration", "function", "symmetry", "algebra" ], "Problem": "For [i]n [/i]distinct points chosen randomly and independently in the unit interval, what is the expected length of the largest sub-interval.\r\n\r\nWhere \"randomly\" means uniform distribution and the \"largest subinterval\" is the maximum of {$x_1,x_2-x_1,x_3-x_2,\\dots,x_n-x_{n-1},1-x_n$} where $0\\leq x_1\\leq x_2\\leq \\dots\\leq x_n\\leq 1$.", "Solution_1": "The differences are distributed uniformly on the simplex $u_i>0,\\sum u_i=1$. This is not obvious, but it can be routinely proven by induction.\r\n\r\nNow, the largest coordinate- that's ugly. It comes down to finding the center of mass of an irregular box-like region (quadrilateral when $n=3$, hexahedral with no two faces parallel when $n=4$, etc.)", "Solution_2": "[quote=\"jmerry\"]The differences are distributed uniformly on the simplex $u_i>0,\\sum u_i=1$. This is not obvious, but it can be routinely proven by induction. [/quote]\r\n\r\nAll that is needed is that the probability distribution on the differences is cyclically invariant.\r\nThus, one can assume the longest interval is the last one, $[x,1]$. The probability density of such configurations (among all $n$-point configurations) is proportional to $x^{n-1}$. The integral of that density is $1/n$ so that the correctly normalized probability density is $n x^{n-1}$. The answer to the problem is therefore $\\int_0^1 (1-x) n x^{n-1} dx = \\frac{1}{n+1}$.", "Solution_3": "[quote=\"fleeting_guest\"]\nAll that is needed is that the probability distribution on the differences is cyclically invariant.\n[/quote]\r\nYou mean that for example $1-x_n$ and $x_{n}-x_{n-1}$ has the same distribution??? ($x_{n-1}0 , xyz=1$ \r\nProve that :\r\n\\[ x+y+z+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\leq \\frac{x}{y}+\\frac{y}{z}+\\frac{z}{x} +3 \\]\r\n\r\nI try so hard and i only solve that :\r\n\\[ x+y+z+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\leq \\frac{x^n}{y^n}+\\frac{y^n}{z^n}+\\frac{z^n}{x^n} +3 \\]\r\n\\[ n \\geq 2 \\]\r\nI think it's so hard and interesting .\r\nAnybody help me ???", "Solution_2": "Yes, Galois, your inequality is true. Here is my proof:\r\n\r\nIn fact, just as Tung Lam, I substitute $x=a^3$, $y=b^3$ and $z=c^3$. Then your problem can be rewritten as follows:\r\n\r\n[i]If x, y, z are arbitrary positive numbers with xyz = 1, then prove that\n\n $\\displaystyle \\frac{\\left( x-1\\right) \\left( y-1\\right) }{y}+\\frac{\\left( y-1\\right) \\left( z-1\\right) }{z}+\\frac{\\left( z-1\\right) \\left( x-1\\right) }{x}\\leq 0$.[/i]\r\n\r\nNow here is my proof of this inequality: We can rewrite the inequality in the form\r\n\r\n $\\displaystyle \\left( x-1\\right) \\left( 1-\\frac{1}{y}\\right) +\\left( y-1\\right) \\left( 1-\\frac{1}{z}\\right) +\\left( z-1\\right) \\left( 1-\\frac{1}{x}\\right) \\leq 0$,\r\n\r\nor, equivalently,\r\n\r\n $\\displaystyle \\left( x-1\\right) \\left( \\frac{1}{y}-1\\right) +\\left( y-1\\right) \\left( \\frac{1}{z}-1\\right) +\\left( z-1\\right) \\left( \\frac{1}{x}-1\\right) \\geq 0$.\r\n\r\nSince xyz = 1, we can use Harazi's substitution $\\displaystyle x=\\frac{u}{v}$, $\\displaystyle y=\\frac{v}{w}$, $\\displaystyle z=\\frac{w}{u}$, where u, v, w are three positive reals. Then, the inequality we have to prove simplifies to\r\n\r\n $\\displaystyle \\left( \\frac{u}{v}-1\\right) \\left( \\frac{w}{v}-1\\right) +\\left( \\frac{v}{w}-1\\right) \\left( \\frac{u}{w}-1\\right) +\\left( \\frac{w}{u}-1\\right) \\left( \\frac{v}{u}-1\\right) \\geq 0$.\r\n\r\nWe expand this and get\r\n\r\n $\\displaystyle \\frac{v^{3}w^{3}+w^{3}u^{3}+u^{3}v^{3}+3u^{2}v^{2}w^{2}-vu^{3}w^{2}-u^{3}wv^{2}-v^{3}u^{2}w-uv^{3}w^{2}-uv^{2}w^{3}-w^{3}vu^{2}}{u^{2}v^{2}w^{2}}\\geq 0$.\r\n\r\nThus, it remains to prove that\r\n\r\n$v^{3}w^{3}+w^{3}u^{3}+u^{3}v^{3}+3u^{2}v^{2}w^{2}-vu^{3}w^{2}-u^{3}wv^{2}-v^{3}u^{2}w-uv^{3}w^{2}-uv^{2}w^{3}-w^{3}vu^{2}\\geq 0$.\r\n\r\nBut if we now define X = vw, Y = wu and Z = uv, then we see that\r\n\r\n$v^{3}w^{3}+w^{3}u^{3}+u^{3}v^{3}+3u^{2}v^{2}w^{2}-vu^{3}w^{2}-u^{3}wv^{2}-v^{3}u^{2}w-uv^{3}w^{2}-uv^{2}w^{3}-w^{3}vu^{2}=X^{3}+Y^{3}+Z^{3}+3XYZ-\\left( YZ\\left( Y+Z\\right) +ZX\\left( Z+X\\right) +XY\\left( X+Y\\right) \\right)$,\r\n\r\nand this is $\\geq 0$ after Schur. Hence, the inequality is proven.\r\n\r\nNice inequality, Galois! Although the \"^3\"'s were a bit needless...\r\n\r\n Darij", "Solution_3": "Schur's Again!!! :) :) \r\n\r\n[quote=\"darij grinberg\"]\n $\\displaystyle \\left( \\frac{u}{v}-1\\right) \\left( \\frac{w}{v}-1\\right) +\\left( \\frac{v}{w}-1\\right) \\left( \\frac{u}{w}-1\\right) +\\left( \\frac{w}{u}-1\\right) \\left( \\frac{v}{u}-1\\right) \\geq 0$.\n[/quote]\n\nMay we declare victory here since it's equivalent to \n\n$\\displaystyle v^{-2}(v-u)(v-w) + w^{-2}(w-u)(w-v) + u^{-2}(u-v)(u-w) \\geq 0$\n\nwhich is the Schur's inequality for $r=-2$.\n\nSee (and pay attention to the first reply by Moubinool) \nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=1166\n[quote=\"Moubinool\"]\nFor any t \\in R, (a,b,c)\\in R+3 \n\nat(a-b)(a-c) + bt(b-c)(b-a)+\nct(c-a)(c-b) \\geq 0\n\nwith equality iff a=b=c\n[/quote]\r\n\r\nOr you can prove a more generalized Schur's inequality:\r\n\r\nIf $f(x)$ is a monotone function (increasing or decresing, not necessary strict) and $f(x)\\geq 0$ in it's domain, then for any $a,b,c$ in it's domain:\r\n\r\n$f(a) (a-b)(a-c) + f(b)(b-c)(b-a) + f(c)(c-a)(c-b) \\geq 0$.\r\n\r\nMake sure you prove this as a lemma if you ever use it in a Olympiad contest and don't forget to quote my name if you use it in the case $f(x)$ is a decresing function. :D :D :D", "Solution_4": "[quote=\"fuzzylogic\"]Or you can prove a more generalized Schur's inequality:\n\nIf $f(x)$ is a monotone function (increasing or decresing, not necessary strict) and $f(x)\\geq 0$ in it's domain, then for any $a,b,c$ in it's domain:\n\n$f(a) (a-b)(a-c) + f(b)(b-c)(b-a) + f(c)(c-a)(c-b) \\geq 0$.[/quote]\r\n\r\nThis is so cool! And the proof is really the same as for the \"standard\" Schur inequality, just with a little modification. Thanks for this generalization!\r\n\r\n Darij", "Solution_5": "hmm... I knew that it's for increasing or convex functions:\r\n\r\nif $f:\\mathbb{R}\\to\\mathbb{R}_+$ and its an increasing or a convex function we have \r\n$f(a)(a-b)(a-c)+f(b)(b-a)(b-c)+f(c)(c-a)(c-b)\\ge 0$", "Solution_6": "[quote=\"Tung Lam\"]\nLet $ x,y,z>0 , xyz=1$ \nProve that :\n\\[ x+y+z+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\leq \\frac{x}{y}+\\frac{y}{z}+\\frac{z}{x} +3 \\]\n[/quote]\r\n\r\nHehe, in his post \"Inequality - abc = 1\", [b]Arne[/b] noticed that this is actually equivalent to IMO2000/2. :D \r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=848" } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "$ \\lim\\limits_{ n \\to \\infty} \\frac{1.3.5...(2n\\minus{}1)}{2.4.6....2n}$", "Solution_1": "0.\r\nFirst method: Stirling's approximation. The expression is just $ \\frac {\\binom{2n}{n}}{4^n}$, and from Stirling's approximation, we have: $ \\lim \\frac {n!}{(n/e)^n\\sqrt {2\\pi n}} \\equal{} 1$, thus $ \\lim \\frac {\\binom{2n}{n}\\sqrt {\\pi n}}{4^n} \\equal{} 1$, and together with $ \\lim \\frac {1}{\\sqrt {\\pi n}} \\equal{} 0$ it implies that $ \\lim \\frac {\\binom{2n}{n}}{4^n} \\equal{} 0$\r\n\r\nA more elementary approach: Let $ a_{n} \\equal{} \\frac {1}{2}\\cdot\\frac {3}{4}\\cdots\\frac {2n \\minus{} 1}{2n}$. We can show that $ a_n \\le \\frac {1}{\\sqrt {3n\\plus{}1}}$, and this is enough (this problem was found in Titu's \"50 Introductory Problems in Algebra\"). It is just induction." } { "Tag": [ "rotation", "Support", "geometry", "geometric transformation" ], "Problem": "Can anybody explain the direction of friction on the front and rear wheels of a bicycle with a proper argument. I was actually analyzing the way my cycle chain rotates today so let me check if my analysis was correct. (Proper Reasoning obviously required from your side)", "Solution_1": "it is towards the front(the direction u r moving) on the back wheel and toward back on the front wheel.\r\nwell basically the reason is the construstion of bicycle...wehen u push the pedals of the bicycle u apply both torque and force on the back bicycle wheels but the impulse due to torque is more than due to force hence the wheel tends to rotate witha speed $ \\omega$ such that $ \\omega r > v$ hence a frictional force acts wich supports translation that is forward motion...\r\nIn case of the front wheel due to the additional condition of the cycle being a rigid system as u pedal along with the back wheel sthe front ones are bound to move forward but not rotate...hence friecction acts backward to provide torque and support rotation and oppose translation", "Solution_2": "[quote=\"pardesi\"] the impulse due to torque is more than due to force [/quote]\r\n\r\nThis is what I was pondering about. Any specific reason or is just standard theory. :D", "Solution_3": "let me assume the biscycle tyres to be discs...\r\nSo $ \\omega r \\equal{} \\frac {2Fdt}{M} > \\frac {Fdt}{M} \\equal{} v$ ...and we are done..\r\nand yes i don't beleive in standard theories..i beleive in change( i hate hillary :D )" } { "Tag": [ "email" ], "Problem": "I'm trying to find one that does not use java (flash preferred), has free chat enabled, and isn't a \"club\" that you have to join.\r\n\r\nEDIT: And my dad doesn't let me install anything on his computer (his is the one with internet access.)\r\n\r\nhttp://www.jagex.com has all of these, but java doesn't work on my compy for whatever reason.", "Solution_1": "You might want to look at [url=http://www.freechess.org]FICS[/url]. It has all the things you mention. You will need to download some graphical chess program though - eg winboard for windows, which is free, as FICS is entirely textbased. Its easy to set up something like winboard to run your games though.", "Solution_2": "[quote=\"TripleM\"]You might want to look at [url=http://www.freechess.org]FICS[/url]. It has all the things you mention. You will need to download some graphical chess program though - eg winboard for windows, which is free, as FICS is entirely textbased. Its easy to set up something like winboard to run your games though.[/quote]\r\n\r\nProblem with FICS is that it won't let you on if you have a free email. Which kinda sucks.", "Solution_3": "[quote=\"Fermatprime\"][quote=\"TripleM\"]You might want to look at [url=http://www.freechess.org]FICS[/url]. It has all the things you mention. You will need to download some graphical chess program though - eg winboard for windows, which is free, as FICS is entirely textbased. Its easy to set up something like winboard to run your games though.[/quote]\n\nProblem with FICS is that it won't let you on if you have a free email. Which kinda sucks.[/quote]\r\n\r\nThey used to let u sign (i'm not sure if they still do) in as a guest and then talk to a moderator and he'll set up an account with your free email. That's what I did, but for some reason my email wasn't free it was @msn.com which costs money.", "Solution_4": "Playchess is free and pretty good, but I think you have to get fritz.", "Solution_5": "Yeah, I play there too. Although people seem to get annoyed at me because I have dialup. But I suppose that happens everywhere. I don't think you need Fritz - I'm pretty sure theres something else (free) that you can download that lets you play there, but Fritz has apparently a nicer interface.", "Solution_6": "[quote=\"Fermatprime\"]Problem with FICS is that it won't let you on if you have a free email. Which kinda sucks.[/quote]\r\nBe clever. Go to http://www.mailinator.com and read the blurb on their front page. The site is rather useful, and FICS accepts a @mailinator.com \"address\" for your email.", "Solution_7": "[url]http://www.chesskids.com[/url] has some nice links to chess-playing sites.", "Solution_8": "Thanx for the info on Mailinator. I may sign up for FICS that way later on. It's probably worth it, from what I've heard...\r\n\r\nOf course, I just played an absolutely horrible game on Yahoo!, so I'm not in the mood right now. I got to a point where I was four pawns up, and then basically went insane and ended up losing...first a pawn, then the exchange, then a rook, then my queen...", "Solution_9": "A number of people on Yahoo use chess programs to cheat. A word to the wise. :D", "Solution_10": "[quote=\"Kyptonite\"]A number of people on Yahoo use chess programs to cheat. A word to the wise. :D[/quote]\r\n\r\nYeah, but I doubt he did. Look at this position:\r\n\r\nWhite: Kd2, Qb7, Ra1, Re1, Ne3, Pa2, Pc5, Pe2\r\n\r\nBlack: Kg8, Qg5, Ra8, Re8, Bc4, Pa6, Pc7, Pd6, Pf3, Pg6, Ph5\r\n\r\nThe game continued: 1.exf3 Rxe3! (a sound exchange sac) 2. Qxa8+?? (a line I had nearly overlooked in my calculations) 2...Kg7????\r\n\r\n2... Re8+! would have won the queen and the game.", "Solution_11": "Im not a chess addict, but ill play someone if they want to play. If you go to funorb.com and make an account, add me to your friends list, my name is hyperhavoc5. they have lots of games, one including chess", "Solution_12": "http://www.chessvariants.org doesnt have [i]normal[/i] chess but it has many very fun chess variants free on the internet that dont require download\r\n\r\nunfortunately, the computer in the chess games is very stupid and has very little sense of strategy", "Solution_13": "nowadays we have chess.com", "Solution_14": "[quote=mathleticguyyy]nowadays we have chess.com[/quote]\n\n(-_-)\nI broke chess.com computer mode by letting 2 level 1 computers play each other", "Solution_15": "I think that you should go with chesskid.com", "Solution_16": "chess.com", "Solution_17": "chess kids is a good site", "Solution_18": "Yes it teaches u too!!!", "Solution_19": "chesskids has very funny animations for the videos" } { "Tag": [ "inequalities", "pigeonhole principle", "geometry", "perpendicular bisector", "combinatorics unsolved", "combinatorics" ], "Problem": "Fifteen boys are standing on a field, and each of them has a ball. No two distances between two of the boys are equal. Each boy throws his ball to the boy standing closest to him. \r\n\r\n(a) Show that one of the boys does not get any ball. \r\n(b) Prove that none of the boys gets more than five balls.", "Solution_1": "a) is classic: Consider the pair of boys standing closest to each other; they will throw to each other. If none of the other boys throw to them either of them, the result follows by induction. Otherwise, there will not be enough throws among the remaining boys for everyone to get a ball, so the result follows. \r\n\r\nb) is pretty interesting: Suppose that a boy gets more than five balls. Then there will be at least six boys who are closer to him than anyone else. Consider the boy he is closest to, and draw a circle with the original boy as center and the radius as the distance between the two boys. Now note that any of the remaining five boys must lie closer to the original boy than the other one, so they must lie on the original boy's side of the perpendicular bisector of the segment connecting the two boy, which cuts off a 2pi/3 arc of the circle - we'll drop the line restriction for this weaker restriction, because it turns out this suffices. \r\n\r\nRepeat this logic for the second-closest boy, the third, etc. up to the fifth, and note that each arc extends to 2pi/3 on the largest circle of the fifth boy because the circles are concentric. Now note that the sixth boy has to be outside the fifth boy's circle by assumption, but also that because the 2pi/3 arcs cannot contain any other boys, the projections of the boys onto the largest circle are at least pi/3 apart, so they take up at least 5pi/6 of the circle. If they take up more than 5pi/6, then the sixth boy cannot possibly be closer to the first boy than each of the other boys by the logic. If equality holds, however, we bring back the perpendicular bisector restriction and note that this requires all of the first five boys to be equidistant from the first boy; contradiction.\r\n\r\nThat was rather longwinded for a fairly simple and intuitive problem; does anyone have a better solution?", "Solution_2": "[hide=\"Solutions\"]\n(a) Suppose the opposite case. If all the boys get a ball, each one would get just one ball (obvious). Now, take the pair whose distance has the shortest length. It's easy to see that each boy will throw his ball in the other, because this is the least distance in all the configuration. So these boys will get a ball. Analogously, taking the other $ 13$ boys, and since no boy throw his ball in that \"isolated pair\", we can apply the same reasoning. Thus only one boy will remain, a contradiction.\n(b) If there exist a boy which get at least six balls (call this boy \"$ A$\"), there would be two balls' trajectories which form a $ \\pi /3$ angle, and using some inequalities in the triangle, you can notice that there's two boys which throw his ball in $ A$ but the distance between them is smaller than the distance to $ A$, contradiction.\n[/hide]", "Solution_3": "[quote=\"raf92\"]\n(b) If there exist a boy which get at least six balls (call this boy \"$ A$\"), there would be two balls' trajectories which form a $ \\pi /3$ angle, and using some inequalities in the triangle, you can notice that there's two boys which throw his ball in $ A$ but the distance between them is smaller than the distance to $ A$, contradiction.\n[/quote]\r\n\r\nThis seems more like handwaving than an actual proof. It's the same idea I used, but you can't just say \"using some inequalities in the triangle\".", "Solution_4": "[quote=\"fwolth\"]\nThis seems more like handwaving than an actual proof. [/quote]\r\nOk, forgive me. I wanted to be brief, and maybe my solution lost clarity. The inequality I mean is, using the conventional notation, given a triangle $ \\Delta ABC$, if $ \\angle BAC>\\angle ABC$, therefore $ a>b$.", "Solution_5": "Oh, I understand now. That's indeed a very nice proof. By Pigeonhole, there must be an angle of at most $ \\pi/3$ which must be the largest angle in that triangle, contradiction.", "Solution_6": "Yes, fwolth, that's exactly it.", "Solution_7": "Also another problem is to show that the lines formed by balls travelling don't form a closed polygon.", "Solution_8": "Suppose $ A_1,A_2,\\ldots,A_n$ are arranged so that $ A_i$ throws to $ A_{i\\plus{}1}$ where $ A_{n\\plus{}1}\\equal{}A_1$. Then we must have $ A_1A_2>A_2A_3>\\ldots>A_{n\\minus{}1}A_n>A_nA_1>A_1A_2$, contradiction.", "Solution_9": "Alternately, we can apply pure logic, assuming that this process continues finitely....\r\n\r\nSuppose $ C$ is the closest person to both $ A$ and $ B$, Suppose A gets the ball, passes it on to C, and finally the ball reaches $ B$ , the cycle repeats, the process continues infinitely, thus to make the opp. happen $ B$ should not receive the ball.....thus proved.[hide][/hide]", "Solution_10": "You appear to have badly misunderstood the question. (Each person has their own ball, and each person throws once and only once. If everyone threw whenever they got a ball, the process would continue indefinitely no matter what: the two closest-together people would toss back-and-forth forever.)", "Solution_11": "[quote=\"reza\"]Fifteen boys are standing on a field, and each of them has a ball. No two distances between two of the boys are equal. Each boy throws his ball to the boy standing closest to him. \n\n(a) Show that one of the boys does not get any ball. \n(b) Prove that none of the boys gets more than five balls.[/quote]\r\n\r\na)suppose that all the boys stay with 1 ball.Look at then as a graph.consider the boy $ b_1$ whose shortest edge is the longest edge, when compared to other boys.This boy $ b_1$ throws to boy $ b_2$, and so on.But boy $ b_{15}$ throws to $ b_1$. But due to the fact that $ b_1$ has the propertie that his shortest edge is thr longest edge when compared to others, $ b_{15}$ couldn\u00b4t possibly throw to him and the result follows.", "Solution_12": "[quote]You appear to have badly misunderstood the question. (Each person has their own ball, and each person throws once and only once. If everyone threw whenever they got a ball, the process would continue indefinitely no matter what: the two closest-together people would toss back-and-forth forever.)\n[/quote]\r\n\r\nOOps...thanx JBL..." } { "Tag": [], "Problem": "$a,b,c$ are positive reals that are not equal to 1. And satisfy: $a^{x}=b^{y}=c^{z}$, $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=0$.\r\n\r\nFind $abc$.", "Solution_1": "[hide]$a^{\\frac{1}{x}}a^{\\frac{1}{y}}a^{\\frac{1}{z}}=1=a^{yz}a^{xz}a^{xy}=a^{yz}b^{yz}c^{yz}=\\left(abc\\right)^{yz}\\implies \\boxed{abc=1}$[/hide]" } { "Tag": [ "articles" ], "Problem": "Who,in your opinion, is more likely to become the next winner of the Champions League 2008-2009. As for me, it is Liverpool, though, I hope Barca will bash them in the semi-final.\r\n\r\nP.S: for those who are still unaware of what champions league is:\r\n\r\nhttp://en.wikipedia.org/wiki/UEFA_Champions_League", "Solution_1": "My favourite is LIverpool too.They show fascinating games nowadays(Especially their game with MU).\r\nBut apparently Barca has the best staff in League.So it would be interesting to watch the match between these 2 teams :D", "Solution_2": "Barcaaaaaaa !\r\nI hope :p", "Solution_3": "So eventually Barcelona and Chelsea qualified for the semi-finals. I'd say their chances are about 50-50.\r\nBy the way the game between Liverpool - Chelsea was awesome, probably, the best match ever, well, at least it deserves to be among the most memorable matches of Champions League.", "Solution_4": "Talking about soccer, you should've seen the Corinthians-Santos match for the Brazilian Cup the other day. There were tons of amazing plays, and the Corinthians scored a goal with 15 seconds left in extra time to win.", "Solution_5": "[quote=\"Erken\"]So eventually Barcelona and Chelsea qualified for the semi-finals. I'd say their chances are about 50-50.\nBy the way the game between Liverpool - Chelsea was awesome, probably, the best match ever, well, at least it deserves to be among the most memorable matches of Champions League.[/quote]\r\nI think Barca will own Chelsea easily!\r\nI'm quite disappointed with Liverpool's defense on that \"crazy\" match.Steven Gerrald should be on that match, not in the audience.", "Solution_6": "[quote=\"ghjk\"]\nI think Barca will own Chelsea easily!\nI'm quite disappointed with Liverpool's defense on that \"crazy\" match.Steven Gerrald should be on that match, not in the audience.[/quote]\r\n\r\nI sincerely hope so. \r\n\r\nbtw, goalkeepers quite sucked that night as well, both Cech and Reina. \r\n\r\nSo finally, all semi-finalists are known. MU is not as impressive as it was last year, so I am more inclined to think that the winner of the couple Barcelona - Chelsea will win the CS cup.", "Solution_7": "[color=red]Manchester United [/color]WILL WIN THE CL\r\n\r\nbtw, barca 0-0 chelsea.", "Solution_8": "btw, Manchester United 1-0 Arsenal. hope we get the away goal against arsenal.\r\n\r\n1-1 my prediction at the emirates", "Solution_9": "So the second day of the semi-finals is coming. I'd like to see MU and Barca in the final, and the Red Devils, in my opinion, are already in the final. But Barcelona will have to do a harder job tomorrow.\r\n\r\nMy prediction for today is $ 1\\minus{}1$.", "Solution_10": "[quote=\"Erken\"]So eventually Barcelona and Chelsea qualified for the semi-finals. I'd say their chances are about 50-50.\nBy the way the game between Liverpool - Chelsea was awesome, probably, the best match ever, well, at least it deserves to be among the most memorable matches of Champions League.[/quote]\r\n\r\nI agree about Barcelona and Chelsea. The game tonight is pretty much unpredicatble, having in mind the key players that won't play for Barcelona (including the injured T. Henry) and Chelsea, with their new coach. :P\r\n\r\nAbout Arsenal and Manchester, I believe that Manchester will qualify for the final, but they are not as good as they were in the beginning of the year, so I am not sure if they are going to be able to win CL. :huh: \r\n\r\nMy personal favourite is Chelsea, but all I want is to watch some nice soccer, as most of us watched during the game Liverpool - Chelsea. :)", "Solution_11": "Manchester won 1 - 3:) It was a nice game\r\nI am looking forward to the match tomorrow, to see the other finalist :)", "Solution_12": "arsenal 1-3 manchester united aka CHAMPIONS OF ENGLAND, EUROPE, and WORLD\r\n\r\nhere we come rome!!!!!!!!!!!!!!!", "Solution_13": "[b]Can Barcelona find a way past a Premier League defence?[/b]\r\n\r\nBarcelona may be dominant domestically, but defeating Premier League opposition in Europe is another matter\r\n\r\nFor a club with the St George's cross on its badge, in a city where Sant Jordi's day is celebrated with thousands of improvised stalls draped in the Catalan flag selling roses for her and books for him, FC \u00adBarcelona must be sick of the sight of the English. Last season the roses were yet to wilt when Bar\u00e7a's Champions League hopes withered at the hands of \u00adManchester United; this season another English \u00advisitor to Camp Nou and another 0-0 draw \u00adthreatens a familiar end.\r\n\r\nWhen Gerard Piqu\u00e9 squeezed the ball past Real Madrid's Iker Casillas on \u00adSaturday night, it completed a 6-2 \u00adhammering that has been declared the greatest \u00adperformance in the club's history. It also took Barcelona's total to a staggering 100 league goals. In all competitions their front three alone have 94. Barcelona have scored 146 times in 55 games and are the Champions League's top scorers, with 10 goals more than Chelsea.\r\n\r\nThe thing is, of course, that only one of their matches has been against an \u00adEnglish team. Dominating domestically is one thing; defeating Premier League \u00adopposition another. Last Tuesday was the first time Barcelona had failed to score at home this season.\r\n\r\nThe last time they had been held in Camp Nou was a year earlier \u2013 against Manchester United at the same stage. A 0-0 draw then, followed by a 1-0 defeat at Old Trafford, put them out. \u00adBarcelona's bitter reaction to Chelsea's tactics spoke of frustration \u2013 the English again the \u00adspanner in Bar\u00e7a's works.\r\n\r\nBarcelona have won only 20 of their 50 games against English clubs in European competition. In knockout ties they have emerged defeated as often as victorious and they have won only one of their last five European semi-finals against English teams \u2013 against Tottenham in the 1981-82 Cup Winners' Cup. United, Liver\u00adpool and Leeds all saw them off. United beat them in the 1991 Cup \u00adWinners' Cup final, Mark Hughes scoring the winner.\r\n\r\nRecent Champions League campaigns have been more striking still. It is six years since Barcelona were knocked out by a non-English side, when an extra-time \u00adbreakaway saw Juventus beat 10 men at Camp Nou. Last season they were knocked out by United, the season before \u00adby Liverpool and two seasons before that by Chelsea. In between they knocked out Chelsea and defeated Arsenal in the 2006 final but there were \u00adextenuating \u00adcircumstances: Arsenal had their \u00adgoalkeeper sent off and Chelsea, too, saw red. Barcelona have not beaten a Chelsea side with 11 men.\r\n\r\nBarcelona's record over the last five games against English teams at Camp Nou reads: drawn three, lost two, won none. It is these last two draws that most stick in the throat, chiefly for the manner in which they were achieved.\r\n\r\nManchester United arrived feted as a devastating attacking force but spent 90 minutes defending, with Wayne Rooney employed as an auxiliary full-back. \u00adChelsea came with Guus Hiddink promising attack but they parked the bus, the goalkeeper Petr Cech providing more \"passes\" than any of his team-mates.\r\n\r\nThe style is not unusual but its \u00adsuccess is. Most teams have attempted to pack players behind the ball this season. \u00ad\"Chelsea did the same as we did at the Camp Nou,\" said the Real Madrid coach, Juande Ramos. But Barcelona \u00adeventually found a way through their defences, as they have against other sides who have sought to frustrate them.\r\n\r\nThey should have scored against \u00adChelsea too, Samuel Eto'o, Bojan Krkic and Alexander Hleb missing excellent chances that would have put a different spin on the supposed tactical \"genius\" of Hiddink's approach.\r\n\r\nBut breaking down a team that comes to defend is difficult; breaking down an English team that has come to defend is harder still. They are physically stronger, better organised and faster. Beating them requires qualities that some Spanish sides do not have.\r\n\r\nAs Robert Pires admitted after Villarreal were defeated by Arsenal: \"They played with an intensity La Liga teams cannot live with.\" Barcelona's deep commitment to a footballing philosophy is laudable but some criticise the lack of a plan B.\r\n\r\nChelsea assumed Barcelona's technical superiority but not the inevitability of defeat. Barcelona did not like it but it was licit \u2013 especially with European \u00adreferees. The Spanish game permits far less \u00adcontact and, as the Athletic Bilbao coach, Joaquin Caparros, says, it is causing teams \u00adproblems beyond the country's borders. \u00adDaniel Alves described Chelsea as employing \u00ad\"excessive force\" in the first leg, whereas John Terry said he saw \"only one bad foul\".\r\n\r\nIn Europe \u00adBarcelona have discovered that the \u00adreferees are not the same as they are in Spain. And nor are the opponents, especially the ones from the other land of St George.\r\n\r\n[i]- written by Sid Lowe[/i]", "Solution_14": "A fun trivia fact concerning the article Erken quoted: today's Saint George's day in the old (Ortodox) calendar :lol: \r\n\r\nAnd concerning the tonight's semifinal: I believe Bar\u00e7a has what it takes to get through. After the Saturday Real demolition at Bernabeu, they're confident, and Guardiola has probably prepared a tactical surprise for tonight. We'll just have to wait and see the final outcome (it's the match between Total Football (something like the Dutch in the past) and Destructive football (something like Italians in the past) )", "Solution_15": "Chelsea 1 - 1 Barcelona! It's not fair! :( The referee is not good enough. I am not satisfied with his decisions :mad: Anyway, congratulations, Barca! :|", "Solution_16": "2day the ref was really, really bad. abidal shouldnt had a red card (what a dive by anelka!), maluda foul should have been a pk, not a fk, pique handball should be a pk (he even admitted!!!), and the last minute hand ball by etoo should have been a pk.\r\n\r\nthough the reaction of ballack and drogba to the ref was priceless.", "Solution_17": "Yep, you are right. We could understand Michael Ballack and Didier Drogba's disappointment and their reaction to the decisions of the ref. After Barcelona's performance today, I think that Manchester United will defend their Euro crown successful.", "Solution_18": "no, guys, you are wrong.\r\n\r\nFristly, probably, we'll see Barcelona in the final, because they've been showing a great perfomance every time they were on a field, well maybe except for this match.\r\n\r\nYes, someone might argue that Barcelona didn't deserve to tie the match, but that's not the case, as long as, they are the only team, in my opinion, who is still trying to play entertainment football. \r\n\r\nAnd we'll see two strongest teams of the world in the final, and that's the only thing matters.", "Solution_19": "That was really some bad refereeing. I honestly believe that the last 20 minutes the referee was 'compensating' Abidal for Bar\u00e7a by taking their side in some dubious situations, but it still doesn't approve his actions (and what about first 70 minutes and all those dives in Bar\u00e7a's penalty box, wasn't at least one a penalty?). Two beautiful goals, and total domination of science (tactics) over art (technique)- but Bar\u00e7a goes to the finals. I was rooting for Bar\u00e7a last night, so I'm satisfied :lol:", "Solution_20": "Probably tomorrow, despite the result, some people will say that Hiddink totally overplayed Guardiola, and only some crazy case prevented him from the triumph. Maybe they are right, maybe not. Maybe only this crazy case helps Essien to score an unbelievable goal in the very beginning of the match. Afterwards, the match went under the scenario of Hiddink, who knows what would've happened, if Barcelona opened the score in the beginning? There was a moment, when Xavi shoted from the distance, and the series of ricochets might've led to goal, who knows?\r\nThen, Chelsea would need to play open football, and I am pretty sure - Barcelona would've just bashed them. And what would you say then?\r\n\r\nThat Guardiola is a genius? And that Barcelona is the best fotball team and that they play celestial football?\r\n\r\nIt is not about the bad refereeing , it's not about the tactics, it's not about who desreved it more. It is football, and the case plays a great role in it.", "Solution_21": "[quote=\"Erken\"]no, guys, you are wrong.\n\nFristly, probably, we'll see Barcelona in the final, because they've been showing a great perfomance every time they were on a field, well maybe except for this match.\n\nYes, someone might argue that Barcelona didn't deserve to tie the match, but that's not the case, as long as, they are the only team, in my opinion, who is still trying to play entertainment football. \n\nAnd we'll see two strongest teams of the world in the final, and that's the only thing matters.[/quote]\r\n\r\nYes, I agree that Barca always performs great performance and indeed they are a strong team. But you see when they meet an English FC, they didn't (or can't?) show their strength. This is why I guess and believe that Man Utd will defend their Euro crown successful, not only because of the way Barca played last night. :)\r\n\r\nAs you said, it's football, and we don't know what will happen. We will have to wait to see who is the new king of Europe. :wink:", "Solution_22": "may the best one win, but where was messi or etoo at the 2 games (barc-chels)? i didn't know either of them was playing untill the game ended in standford bridge (i probably spelled that wrong), celebrating", "Solution_23": "so, any predictions who's going to win the final?\r\n\r\nheres mine: Man Utd 2-1 Barca\r\n\r\nits going to be an great match. both teams r best know for their goals. attack vs attack.", "Solution_24": "[quote=\"hjoon0125\"]so, any predictions who's going to win the final?\n\nheres mine: Man Utd 2-1 Barca\n\nits going to be an great match. both teams r best know for their goals. attack vs attack.[/quote]\r\n\r\nare u kidding? There is no chance MU will play an attacking football. I expect somewhat a repetition of the last Camp Nou game. But since Barcelona is way stronger now, my prediction is MU 1 - 2 Barcelona. Besides, Barcelona has scored more goals than MU and managed to beat more serious opponents than MU did.", "Solution_25": "I hope Manchester United wins in a similar manner like they had won in 1999.", "Solution_26": "Hopes: Barca wins\r\n\r\nReality: MU ALWAYS scores in last minute (especially in Premier League), and I'm afraid that something similar will happen now...\r\n\r\nPredict: Barca-MU 1:2", "Solution_27": "Reality is that MU last year played way better than now, while Barca was way weaker. Nevertheless, MU had to do they best to overcome Barcelona in the semi - final last year. So if Iniesta and Henry will play - I believe Barcelona will win. If english clubs played open attacking football, nobody would be able to confront the Barcelona.", "Solution_28": "im pretty sure iniesta and henry will play. no abidal nor alves 4 the final, but captain puyol's back so that doesnt hurt there chances.\r\n\r\nimo, we need to play attacking football. big teams like barca know how to score even if the opponet puts 10 or 11 men behind the ball.", "Solution_29": "Looks like Barca's close to the win...", "Solution_30": "Oh nooooo!!!!! :( :ewpu: :o", "Solution_31": "Barca 2-0 ManUtd\r\n\r\ngah, now i look like a fool. barca were way better than manutd. deserved at the least well done. congratulations barcelona fans. they were just the better team. congratulations once again. that was not the manutd i expected.", "Solution_32": "also congratulations to erken and bugi. man i watched the game and none, yes NONE of our players showed up. though give credit to puyol, OUTSTANDING performance.", "Solution_33": "It takes a special team to make United look ordinary, let alone ugly, but Barcelona, a truly beautiful team, had done so and no one, not least Ferguson, denied the best side had won." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all primes $ p,q$ such that $ pq | 2^p \\plus{} 2^q$.", "Solution_1": "If $ p \\equal{} q$ then $ p^2 | 2^{p \\plus{} 1}$, then $ p \\equal{} q \\equal{} 2$.\r\n\r\n$ p \\equal{} 2$ implies $ 2^{q} \\plus{} 4 \\equiv 0 (mod 2q) \\Rightarrow 2^q \\plus{} 4 \\equiv 0 (mod q)$ \r\n\r\n$ \\Leftrightarrow 2 \\plus{} 4 \\equiv 0 (mod q) \\Leftrightarrow 6 \\equiv 0 (mod q)$ and, since $ (q,2) \\equal{} 1$, $ q \\equal{} 3$. \r\n\r\nIf $ (p,2) \\equal{} 1$ and $ (q,2) \\equal{} 1$ then $ 2^p \\plus{} 2^q \\equiv 0 (mod pq) \\Rightarrow 2 \\plus{} 2^q \\equiv 0 (mod p) \\Rightarrow 2^q$ \r\n\r\n$ \\equiv \\minus{} 2 (mod p) \\Leftrightarrow 2^{q \\minus{} 1} \\equiv \\minus{} 1 (mod p)$ $ (1)$ and $ 2^{2(q \\minus{} 1)} \\equal{} 1 (mod p)$ $ (2)$.\r\n\r\nLet $ K$ be the order of $ 2$ m\u00f3d $ p$. By $ (1)$ $ K$ doesn't divide $ (q \\minus{} 1)$, but by $ (2)$ $ K$ divides $ 2(q \\minus{} 1)$. Then the greatest \r\n\r\npower of $ 2$ that divides $ K$ is the gratest power of $ 2$ that divides $ 2(q \\minus{} 1)$. Let $ 2^{m \\plus{} 1}$ be such power.Then $ 2^{m \\plus{} 1}$ \r\n\r\ndivides $ p \\minus{} 1$ (because it divides the order and the order divides $ p \\minus{} 1$).\r\n\r\nSimilarly, let $ 2^n$ be the greatest power that divides $ p \\minus{} 1$, then $ 2^{n \\plus{} 1}$ divides $ q \\minus{} 1$.\r\n\r\nThen $ 2^{n \\plus{} 1} | 2^m$ and $ 2^{m \\plus{} 1} | 2^n$. Contradiction, so there is not solution for $ p$ and $ q$ coprimes with $ 2$." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $ a,b,c$ be integers such that none is zero and $ abc$ divides both $ a^2b\\plus{}b^2c\\plus{}c^2a$ and $ ab^2\\plus{}bc^2\\plus{}ca^2$. Prove that $ |a|\\equal{}|b|\\equal{}|c|$.", "Solution_1": "We can resume that $ \\gcd(a,b,c) \\equal{} 1$\r\nNow we will prove that $ \\gcd(a,b) \\equal{} 1$ ,if not there exist a prime p such that $ p^n\\parallel{}\\gcd(a,b)$\r\nTherefore exist $ a$ or $ b$ such that $ p^n\\parallel{}a$ or $ p^n\\parallel{}b$ \r\nSuppose $ p^n\\parallel{}a$( other case is similar ) \r\nBecause $ ab|ac^2 \\plus{} b^2c \\plus{} a^2b$ so $ p^{2a}|ac^2 \\plus{} b^2c \\plus{} a^2b$\r\nFrom condition we have $ p^{2n}|a^2b \\plus{} b^2c$ therefore $ p^{2n}|c^2a$\r\nBecause $ p^n\\parallel{}a$ so $ p|c$ (contradiction )\r\nSo $ \\gcd(a,b) \\equal{} gcd(b,c) \\equal{} \\gcd(c,a) \\equal{} 1$\r\nBut from condition then $ a|b^2c$ so $ |a| \\equal{} 1$\r\nIt is similar to prove that $ |b| \\equal{} |c| \\equal{} 1$\r\nProblem claim." } { "Tag": [ "calculus", "integration", "function", "topology", "calculus computations" ], "Problem": "Suppose that $\\vec{F}(\\vec{r})$ is continue in a connected and open set $D$. Then line integral $\\int_{C} \\vec{F}(\\vec{r}) . d \\vec{r}$ independent of path If and only if $\\vec{F}(\\vec{r}) = \\nabla f(\\vec{r})$ (conservative) .\r\nany proof?", "Solution_1": "Say the integral is independent of path. You need to define a function $\\phi$ such that $F=\\nabla \\phi$.\r\n\r\nThe idea is that we pick some $a\\in D$ and define\r\n\r\n$\\phi(x)=\\int^x_a F\\cdot dr$\r\n\r\nIt remains to show that $\\phi$ has the property we want of it. Restrict consideration to a path along the $x-$axis and show that the $x-$component of $F$ is $\\frac{\\partial\\phi}{\\partial x}$. You'll need the mean value theorem there. Proceeding similarly, you get $F=\\nabla\\phi$.\r\n\r\nNow for the other direction. Say that $F=\\nabla\\phi$. Since $C$ is smooth, we can parametrize it with respect to $t$, using the relations $dx=\\frac {dx}{dt}dt$ via the chain rule and so on. Hence you should get that\r\n\r\n$\\int_C F\\cdot dr=\\int_C \\frac {d\\phi}{dt}dt=\\phi(b)-\\phi(a)$, where $a,b$ are the endpoints of $C$. But this is just independence of path." } { "Tag": [ "AMC", "AIME", "logarithms" ], "Problem": "Determine the number of ordered pairs $ (a,b)$ of integers such that $ \\log_a b \\plus{} 6\\log_b a\\equal{}5, 2 \\leq a \\leq 2005$, and $ 2 \\leq b \\leq 2005$.\r\n\r\nWhy is the below method incorrect?\r\n\r\n[b]Incorrect Method:[/b]\r\n$ \\log_a b\\equal{}\\frac{1}{\\log_b a}$\r\nSubstituting $ x\\equal{}\\log_a b$, we have $ x \\plus{} 6\\times\\frac{1}{x}\\equal{}5$. Multiplying both sides by $ x$ and factoring, we get $ (x\\minus{}6)(x\\plus{}1)\\equal{}0$ so either\r\n\r\n$ 1) x\\equal{}\\log_a b\\equal{}6$\r\nSo these ordered pairs are of the form $ (a,a^6)$.\r\n\r\n$ 1) x\\equal{}\\log_a b\\equal{}\\minus{}1$\r\nThis doesn't yield integer solutions within the given range.\r\n\r\nPlugging in 2, 3, and 4, we find that 2 and 3 works so our answer is 2.\r\n\r\nExcept not...", "Solution_1": "when you multiply through by x, you should get $ x^2 \\minus{} 5x \\plus{} 6 \\equal{} 0$, not $ (x\\minus{}6)(x\\plus{}1) \\equal{} x^2 \\minus{} 5x \\minus{} 6 \\equal{} 0$.", "Solution_2": "Anyways to finish off what max said.\r\n\r\nIt would be a^3=b and a^2=b,\r\n\r\nsince the solutions are 2 and 3.\r\n\r\nFor a^3=b, we find that a can any integer from 2 to 12 inclusive \r\nFor a^2=b, we find that a can be any integer from 2 to 44 inclusive\r\n\r\nIt doesnt matter what b is because there will be no reapeating (a,b) since a and b are greater than 1.\r\n\r\nSo first one has 11, second has 43, 11+43=054" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "find $xy+2yz+3zx$ given that \r\n $x^{2}+xy+ \\frac{y^{2}}{3}=25$\r\n $\\frac{y^{2}}{3}+ z^{2}=9$ \r\n $x^{2} +xz+ z^{2}=16$", "Solution_1": "Maybe $x^2+xz+z^2=16$?", "Solution_2": "[quote=\"jin\"]Maybe $x^2+xz+z^2=16$?[/quote]\r\n oh you are right,I am very sorry,I will edit it", "Solution_3": "[quote=\"quangtrung\"]find $xy+2yz+3zx$ given that \n $x^{2}+xy+ \\frac{y^{2}}{3}=25$\n $\\frac{y^{2}}{3}+ z^{2}=9$ \n $x^{2} +xz+ z^{2}=16$[/quote]\r\n\r\nIt is an old USSR math olympiad problem.\r\nYou can use geometrical interpretation.It is easy.", "Solution_4": "$3^2=x^2+(\\frac{y}{\\sqrt3})^2+2x(\\frac{y}{\\sqrt3})sin{\\frac{5\\pi}{6}}$\r\n$4^2=z^2+(\\frac{y}{\\sqrt3})^2$\r\n$5^2=x^2+z^2+2xzsin{\\frac{2\\pi}{3}}$\r\nAnd $\\frac{5\\pi}{6}+\\frac{2\\pi}{3}+\\frac{\\pi}{2}=2\\pi$\r\nSo we can found a triangle ABC,in which contains a point O.And $OC=x,OB=\\frac{y}{\\sqrt3},OA=z,\\angle{AOB}=\\frac{\\pi}{2},\\angle{AOC}=\\frac{2\\pi}{3}$,then we can have:$AB=3,AC=4,BC=5$,so $\\angle{BAC}=\\frac{\\pi}{2}$\r\nFor one thing, $S_{\\triangle{ABC}}=6$\r\nFor another thing,$S_{\\triangle{ABC}}=\\frac{1}{4\\sqrt3}(xy+2yz+3xz)$\r\nso $xy+2yz+3xz=24\\sqrt3$", "Solution_5": "Yeah,the same solution I got." } { "Tag": [ "quadratics", "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "$X,Y,Z$ E N .If $(x,y)=1$ and $x^{2}+y^{2}=z^{4}$ show that $7$ divides $xy$ :)\r\n \r\n edited; :wink:", "Solution_1": "$x=y=1, z=2$ :huh:", "Solution_2": "[quote=\"Maruf\"]$X,Y,Z$ E N .If $(x,y)=1$ and $x^{2}+Y^{2}=z^$ show that $7$ divides $xy$ :)[/quote]\r\nmaybe the problem is:\r\n$X,Y,Z$ E N .If $(x,y)=1$ and $x^{2}+Y^{2}=z^{2}$ show that $7$ divides $xy$ :)", "Solution_3": "$x=3,y=4,z=5$.", "Solution_4": "maybe the problem is $x^{2}+y^{2}=z^{4}$ then $7|xy$ :P", "Solution_5": "We may suppose y is even. Then\r\n\\[x=m^{2}-n^{2}, y=2mn,z^{2}=m^{2}+n^{2}\\]\r\nQuadratic residues $\\pmod7$ are 0,1,2,4...", "Solution_6": "[quote=\"RIP\"]We may suppose $(x,y,z)=1$ and y is even. Then\n\\[x=m^{2}-n^{2}, y=2mn,z^{2}=m^{2}+n^{2}\\]\nQuadratic residues $\\pmod7$ are 0,1,2,4...[/quote]\r\n \r\n\r\n can you explain your solution,thanks!!", "Solution_7": "That's wrong solution.\r\nNo, that's right.", "Solution_8": "It is easy to get $x=(a^{2}-b^{2})-4a^{2}b^{2},y=4ab(a^{2}-b^{2})$. Sufficiently chek case $a=1,b=2,3,4,5$ (another cases a=c,b=2c,3c,4c,5c) when $7\\not |y$. \r\na=1 give $x=(b^{2}-1)^{2}-4b^{2}=b^{4}-6b^{2}+1=b^{4}+b^{2}+1(mod 7)$.\r\n$b^{2}=4(mod 7), b^{4}=2(mod 7)$ or $b^{2}=2(mod 7),b^{4}=4(mod 7)$.\r\nFor both cases $x=b^{4}+b^{2}+1=0(mod 7)$. :D\r\nEasy $x=a^{4}+a^{2}b^{2}+b^{4}(mod 7)$, therefore $xy=4ab(a^{6}-b^{6})(mod 7)$. If $7|ab$ obviosly $7|xy$, else $a^{6}=b^{6}=1$, therefore $7|xy$." } { "Tag": [ "number theory", "modular arithmetic" ], "Problem": "1. What is mod, how to find it easily, and give me some applications/problems with it.\r\n\r\n2. How to find a number in base of another number easily and give me some applications/problems.\r\n\r\n3. What is the thing with the parantheses with one number on top of the other? What is the concept in words, formula, and what are applications at high levels? A problem using it would be nice too.\r\n\r\nThanks.", "Solution_1": "[quote=\"neelnal\"]1. What is mod, how to find it easily, and give me some applications/problems with it.\n\n2. How to find a number in base of another number easily and give me some applications/problems.\n\n3. What is the thing with the parantheses with one number on top of the other? What is the concept in words, formula, and what are applications at high levels? A problem using it would be nice too.\n\nThanks.[/quote]\r\n\r\nOkay, one at a time.\r\n\r\n\"Mod\" is a shorthand way to write \"modulo\". The best example of modular arithmetic is the clock. For example, if it's midnight right now and we want to find what time it will be 100 hours from now, we would take $100 \\mod 12$. This is the same thing as saying \"the remainder that you get when you divide 100 by 12\".\r\n\r\nThe thing with parentheses is the \"choose\" function. For example, if you want to find the number of ways you can \"choose\" 2 objects out of a bag with 5 objects in it, you would take $5 \\choose 2$. This assumes that the order doesn't matter. It is equivalent to saying $\\frac{5!}{2!3!}$.\r\n\r\nAnd I don't know much about converting one base to another.", "Solution_2": "[quote=\"neelnal\"]3. What is the thing with the parantheses with one number on top of the other? What is the concept in words, formula, and what are applications at high levels? A problem using it would be nice too.[/quote]\r\n$\\binom{n}{k}\\equiv_{n}C_{k}\\equiv\\frac{n!}{k!(n-k)!}$\r\nIt is how many ways there are to choose, without replacement and without order, $k$ objects from $n$ objects.", "Solution_3": "So, does anyone know a quick way to do the base thingy?\r\n\r\nLike what is 500 in base 17?\r\n\r\nIs there a quick way? Or do I just have to divide and stuf?", "Solution_4": "In bases, each digit signifies $x*y^{z}$ where x is the digit, y is the number base that you are in, and z is the number of digits to the left of the units digit that you are. For instance, in base 10, the number 4239 is equal to $9*10^{0}+3*10^{1}+2*10^{2}+4*10^{3}$.\r\n\r\nThe base 2 number 1001 in base 10 would be equal to $1*2^{0}+1*2^{3}$ which is equal to 9. \r\n\r\nThe base 10 number 283 in base 5 would be equal to $2*5^{3}+1*5^{2}+1*5^{1}=3*5^{0}$.\r\n\r\n500 in base 17 is equal to $2*17^{2}+B*17^{1}+7*17^{0}$\r\n\r\nNote that when there are bases higher than base 10, letters are used to represent higher numbers, so that $A=11, B=12, C=13,$ and so on. The most common number base higher than 10 is 16.\r\n\r\nTo get a number from base x to base y where neither x nor y are equal to 10, I normally change the number in base x to base 10, and then I change that number to base y.", "Solution_5": "I think you mean A = 10, B = 11, C = 12 etc", "Solution_6": "[quote=\"mustafa\"]The base 10 number 283 in base 5 would be equal to $2*5^{3}+1*5^{2}+1*5^{1}=3*5^{0}$.[/quote]\r\nThe equal sign is supposed to be a plus. \r\n\r\n$283=2\\cdot5^{3}+1\\cdot5^{2}+1\\cdot5^{1}+3\\cdot5^{0}=2113_{5}$", "Solution_7": "Some info...\r\n\r\nThe reason base ten is the one we use is that we have ten fingers and count by tens...\r\n\r\nThe reason base sixteen is the most common base above ten is that it is used in computer programming...\r\n\r\n$\\binom{n}{k}$ is almost always read as \"n choose k\" but in some cases it is read as \"n take k\"...\r\n\r\nYou probably know this, but $n!=n(n-1)(n-2)...(3)(2)(1)$...\r\n\r\nIMPORTANT\r\nThe reason $\\binom{n}{k}=\\frac{n!}{k!(n-k)!}$ is because think about it. If order doesn't matter then you have to pick k things out of a set of n. You have n choices for the first thing, n-1 for the second, n-2 for the third, ..., n-k+1 for the kth item. So the number of choices is $(n)(n-1)(n-2)...(n-k+1)$. This quantity can be rewritten as $\\frac{n!}{(n-k)!}$. The thing is, we made this formula not counting the fact that order matters. So we need to factor in that order doesn't matter. There are k numbers you multiplied in making our answer. There are $k!$ ways to make the combination, when we really only have 1. $\\frac{n!}{(n-k)!k!}$." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Two players, A and B, play the following game: they retire coins of a pile which contains initially 2006 coins. The players play removing alternatingly, in each move, from 1 to 7 coins, each player keeps the coins that retires. If a player wishes he can pass(he doesn't retire any coin), but to do that he must pay 7 coins from the ones he retired from the pile in past moves. These 7 coins are taken to a separated box and don't interfere in the game any more. The winner is the one who retires the last coin, and A starts the game. Determine which player can win for sure, it doesn't matter how the other one plays. Show the winning strategy and explain why it works.", "Solution_1": "[quote=\"carlosbr\"]Two players, A and B, play the following game: they retire coins of a pile which contains initially 2006 coins. The players play removing alternatingly, in each move, from 1 to 7 coins, each player keeps the coins that retires. If a player wishes he can pass(he doesn't retire any coin), but to do that he must pay 7 coins from the ones he retired from the pile in past moves. These 7 coins are taken to a separated box and don't interfere in the game any more. The winner is the one who retires the last coin, and A starts the game. Determine which player can win for sure, it doesn't matter how the other one plays. Show the winning strategy and explain why it works.[/quote]\n\nIf they won't may pass at that way:\n$A$ starts with taking $6$ coins and after $B$ he does $8-i$ or if $B$ and he can win by take the last after $251$ moves.\n\nAt the other way:\n$A$ takes always $7$ coins until they reach a number, after at least $143$ moves each, he has to bring the number on $8$, or B did, but $A$ can pass at least once more, (142*14<1998), (143*14>1998), so B has to take sth and then A do $8-i$ and win.\nIf B did earlier less then $8$, A win very simply.", "Solution_2": "[hide=My solution]$A$ removes always $7$ coins. If $B$ removes always $7$ coins (without passing) then after $143$ moves each one, there are $2006-143\\times 14=2006-2002=4$ coins, so $A$ wins. If $B$ removes always $7$ coins (without passing), in his last move he can pass and then $A$ removes $3$ coins, or remove at most $3$ coins, and $A$ removes the remaining coins (may be $0$); but now $A$ can pass more times than $B$, so $B$ will have to remove at least one coin, then $A$ removes the remaining coins and win.\nIf $B$ pass or remove less than $7$ coins in some turn, then we can get again to $8$ coins and $A$ can pass more times than $B$, so $B$ will have to remove at least one coin, then $A$ removes the remaining coins and win.[/hide]", "Solution_3": "Let $n(X)$ be the total of coins of player $X$. Because every number is a lose number or a win number, the better strategy to $A$ is to maximize $n(A)$. Indeed, if $n(A)=7q+r$ and $n(B)= 7l+r$ and $q>l$, $A$ will win because he can pass in a lose number and win in a win point.\nThus, the better strategy to $A$ is always choose $7$. Because $B$ is not interesting in lose (=0), by symmetry he has to choose always 7.\nThen, after several time, the pile will have $18$ coins and $n(A)=n(B)$. Thus, $A$ can choose $7$ again, and $11$ is a lose number, because if $B$ chooses $t>3$, $A$ can easily win and if $B$ choose $x$, $A$ can choose $3-x$ putting $B$ in number $8$, and because in this time $n(A)>n(B)$, $B$ will have to choose a number $v$ and of course $8-v$ less equal to $7$. Thus $A$ always wins." } { "Tag": [ "geometry", "number theory" ], "Problem": "A teacher wants to arrange 3 copies of $ \\emph{Introduction to Geometry}$ and 4 copies of $ \\emph{Introduction to Number Theory}$ on a bookshelf. In how many ways can he do that?", "Solution_1": "hello\r\nno. of ways=$ \\frac{(3\\plus{}4)!}{3!4!}\\equal{}35$\r\nthank u", "Solution_2": "Additionally, we can pick $ \\binom{7}{3}\\equal{}\\boxed{35}$ spots to put the $ \\emph{Introduction to Geometry}$ books.", "Solution_3": "Or $ \\binom{7}{4}=\\boxed{35} $ for $ Introduction $ $ to $ $ Number Theory $", "Solution_4": "Wait, are the books distinguishable?", "Solution_5": "I don't think so if they are the same kind...", "Solution_6": "They should have specified distinguishable ways or not", "Solution_7": "All copies of both books are indistinguishable, so I would assume that's why they didn't specify that.", "Solution_8": "What they said all copies are identical so they are indistinguishable. (or maybe that was just added later)" } { "Tag": [ "geometry", "3D geometry", "prism" ], "Problem": "a) Prove that we can assign one of the numbers $1$ or $-1$ to the vertices of a cube such that the product of the numbers assigned to the vertices of any face is equal to $-1$.\r\n\r\nb) Prove that for a hexagonal prism such a mapping is not possible.", "Solution_1": "For (a): we put $-1$ in two opposite corners and $1$ in the other corners.\r\nI'll think later about (b).\r\n\r\nValentin, will you post grades 9/10? I'm interested in those...", "Solution_2": "[hide=\"Major Hint\"]For b, consider the products on the opposite hexagons and prove they are different.[/hide]" } { "Tag": [ "IMC", "college contests" ], "Problem": "As I've heard, IMC 2004 will take place in Skopje, Macedonia. They have announced last year that Bulgaria will organise the 2004 IMC, but I guess they have changed their mind.\r\nFor us, the romanians, this only means higher transport cost, since Bulgaria is our neighbour wile Macedonia is not.", "Solution_1": "upgraded the forum's profile :D" } { "Tag": [ "calculus", "integration", "search", "real analysis", "real analysis theorems" ], "Problem": "I'm sorry if this already has been asked. But my search on this forum did not pay off...\r\n\r\nAnyways, I was wondering if there is a book out there containing challenging integrals (problems, solutions and hopefully some hints on how to reason)? So I'm basically want integrals, nothing else. A book with the same philosophy as Problems and Theorems in Analysis by George Polya, Gabor Szeg\u00f6.\r\n\r\nThank you.", "Solution_1": "You might try [url=https://www.cambridge.org/us/catalogue/catalogue.asp?isbn=0521796369]Irresistible Integrals[/url] (or try [url=http://www.amazon.com/Irresistible-Integrals-Symbolics-Experiments-Evaluation/dp/0521796369]Amazon[/url].)" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f$ and $ g$ be measurable functions on $ E$.\r\na) If $ f$ and $ g$ are finite a.e., show that $ f\\plus{}g$ is measurable no matter how we define it at the points when it has the form $ \\infty\\minus{}\\infty$ or $ \\minus{}\\infty\\plus{}\\infty$.\r\nb) Show that $ fg$ is measurable without restriction on the finiteness of $ f$ and $ g$. Show that $ f\\plus{}g$ is measurable if it is defined to have the same value at every point where it has the form $ \\infty\\minus{}\\infty$ or $ \\minus{}\\infty\\plus{}\\infty$.\r\n\r\nFor a), I am thinking that since $ m(\\left|f\\right|\\equal{}\\infty)\\equal{}m(\\left|g\\right|\\equal{}0)$, we can write $ \\{f\\plus{}g>\\alpha\\}\\equal{}\\bigcup_{r\\in\\mathbb{Q}}\\left(\\{f\\equal{}\\infty\\}\\cup\\{\\infty>f>r\\}\\right)\\cap\\left(\\{g\\equal{}\\infty\\})\\cup\\{\\infty>g>\\alpha\\minus{}r\\}\\right)$, but the sets where $ f,g\\equal{}\\pm\\infty$ are null sets, they are measurable so the entire union\\intersection above is measurable.\r\n\r\nNot quite as sure about b).", "Solution_1": "I am sure you mean this: ;)\r\n[quote=\"JRav\"]$ m(\\left|f\\right| \\equal{} \\infty) \\equal{} m(\\left|g\\right| \\equal{} 0)$[/quote]\n$ m(\\left|f\\right| \\equal{} \\infty) \\equal{} m(\\left|g\\right| \\equal{} \\infty)\\equal{}0$\n[quote]$ \\{f \\plus{} g > \\alpha\\} \\equal{} \\bigcup_{r\\in\\mathbb{Q}}\\left(\\{f \\equal{} \\infty\\}\\cup\\{\\infty > f > r\\}\\right)\\cap\\left(\\{g \\equal{} \\infty\\})\\cup\\{\\infty > g > \\alpha \\minus{} r\\}\\right)$[/quote]$ \\{f \\plus{} g > \\alpha\\} \\equal{} \\bigcup_{r\\in\\mathbb{Q}}\\left(\\{f \\equal{} \\infty\\}\\cup\\{\\infty > f > r\\}\\right)\\cap\\left(\\{g \\equal{} \\infty\\}\\cup\\{\\infty > g > \\alpha \\minus{} r\\}\\right)$ - here clearing one right bracket.\n\nFor (b) use the decomposition \\[ f\\equal{}f 1_{\\{0 a\\} \\equal{} \\{<\\infty>f \\plus{} g > a\\}\\cup\\{f\\plus{}g\\equal{}\\infty\\}\\cup\\{f \\plus{} g \\equal{} \\infty \\minus{} \\infty\\}\\cup\\{f \\plus{} g \\equal{} \\minus{} \\infty \\plus{} \\infty\\}$, where the last two sets are null sets, so the whole thing is measurable. Correct?", "Solution_5": "Hmm, do you mean $ \\{f\\plus{}g\\equal{}\\infty\\minus{}\\infty\\}\\equal{}\\{f\\equal{}\\infty\\}\\cup\\{g\\equal{}\\minus{}\\infty\\}$? I would reduce to the finite case by restricting to the complement of the null-set $ \\{f\\equal{}\\pm\\infty\\}\\cup\\{g\\equal{}\\pm\\infty\\}$. Then everything is fine. Otherwise you need to run through too many cases what I usually try to avoid. :o", "Solution_6": "I'm not sure if that is what I \"mean,\" but it seems that $ \\{f\\plus{}g\\equal{}\\infty\\minus{}\\infty\\}$ and $ \\{f\\equal{}\\infty\\}\\cap\\{g\\equal{}\\minus{}\\infty\\}$ are the same thing. It seems like I just need to consider that case, and $ \\{f\\plus{}g\\equal{}\\minus{}\\infty\\plus{}\\infty\\}\\equal{}\\{f\\equal{}\\minus{}\\infty\\}\\cap\\{g\\equal{}\\infty\\}$, in which case I just get one big union and intersection.", "Solution_7": "Apology for the late reply. Of course there was a typo in my last posting - and you have corrected it in your reply. But with that everything seems to be fine with the solutions to your questions. :)", "Solution_8": "Just one more question: you said that my final question (part (c)) was just a special case of part (a). Why is that true since I am not necessarily assuming $ f$ and $ g$ to be finite a.e. in the third part?", "Solution_9": "Right, got me! ;) I think we need to amend your proof of (a) a bit - since, e.g. what happens to \\[ \\{f+g >\\alpha\\}=\\bigcup_{r\\in\\mathbb{Q}}\\left(\\{f =\\infty\\}\\cup\\{\\infty > f > r\\}\\right)\\cap\\left(\\{g =\\infty\\})\\cup\\{\\infty > g >\\alpha-r\\}\\right)\\] when we set $ \\infty-\\infty=\\infty$? It is still correct? I missed this nasty detail, to be honest. And there will be another as we will see below.\r\n\r\nWhile writing the first reply I had my proof of (a) in mind which I will briefly sketch: Let $ A$ be the set of nasty points, i.e. $ A=A_1\\cup A_2$ with $ A_1=\\{f=+\\infty\\}\\cap\\{g=-\\infty\\}$ and $ A_2=\\{f=-\\infty\\}\\cap\\{g=+\\infty\\}$. We know that $ A_1$ and $ A_2$ are measurable sets of measure zero. The function $ f+g$ restricted to $ E\\setminus A$ is measurable; we could use your proof for this. Here I assume that $ \\infty+\\infty=\\infty$ and $ -\\infty+(-\\infty)=-\\infty$.\r\n\r\nNow the problem comes in when we have a look at the measurability of $ f+g$ restricted to $ A_i$, $ i=1,2$. If $ A_i$ does not contain any proper measurable subset, then $ f+g$ must be constant on $ A_i$. This does not happen if the $ \\sigma$-algebra is complete with respect to the measure on $ E$, i.e. that every subset of a null-set is measurable. If every subset of $ A$ is measurable we can certainly define $ f+g$ on $ A$ by any functions with values in $ \\overline{\\mathbb R}$.\r\n\r\nNow you might follow my comment on (c) which can be solved [i]like[/i] the addition of (a). Hope that this makes sense to you. I think with this you can amend your proof." } { "Tag": [ "geometry" ], "Problem": "[quote]Let $ P$ be the common point of three circles $ \\|\\begin{array}{cccc}C_1(O_1,r_1)\\\\\n\\\\\nC_2(O_2,r_2)\\\\\n\\\\\nC_3(O_3,r_3)\\end{array}$ and let $ \\{\\begin{array}{cccc} P_1\\in C_1\\cap C_2\\\\\n\\\\\nP_2\\in C_2\\cap C_3\\\\\n\\\\\nP_3 \\in C_3\\cap C_1\\end{array}$\nThe tangent in $ P$ at $ C_1$ meets $ C_2$ in $ N_1$ and $ C_3$ in $ M_3$,the tangent in $ P$ at $ C_2$ meets $ C_3$ in $ N_2$and $ C_1$ in $ M_1$,and the tangent in $ P$ at $ C_3$ meets $ C_1$ in $ N_3$ and $ C_2$ in $ M_2$.Prove that \\[ P_1M_1 \\cdot P_2M_2 \\cdot P_3M_3=P_1N_1 \\cdot P_2N_2 \\cdot P_3N_3\\][/quote]", "Solution_1": "here is a proof [url]http://www.mathlinks.ro/weblog_entry.php?p=1126659#1126659[/url][/url]" } { "Tag": [], "Problem": "What is the final temperature in degrees centigrade of 100.0 g of water at 30.0 degrees C if it is mixed with 50.0g of water at 0.0 degrees C?", "Solution_1": "First, we must assume that:\r\n\r\nI. The system in study is isolated.\r\n\r\nII. There is transfer of energy between the two parts of the system only under the form of heat.\r\n\r\nWith these assumptions made, we can write\r\n\r\n$ Q_1 \\plus{} Q_2 \\equal{} 0$\r\n\r\nwhere $ Q_1$ is the heat transfered by the water at 30.0\u00baC and $ Q_2$ is the heat transfered to the water at 0.0\u00baC.\r\n\r\nNow we are going to make two more assumptions:\r\n\r\nIII. The water at 0.0\u00baC is in the liquid state.\r\n\r\nIV. The heat capacity of water is constant (in the temperature range of this problem, this assumption is valid).\r\n\r\nTherefore:\r\n\r\n$ m_1\\,\\,c \\,\\, (T \\minus{} 30.0) \\plus{} m_2 \\,\\, c \\,\\, (T \\minus{} 0.0) \\equal{} 0$,\r\n\r\nwhere T is the final temperature of the system. Now just substitute $ m_1 \\equal{} 100\\,\\,g$, $ m_2 \\equal{} 50.0\\,\\,g$, and $ c \\equal{} 4.18\\,\\,J/g/K$." } { "Tag": [ "algorithm", "email", "search" ], "Problem": "In my country people use max flow or Hungarian method for Bipartite Matching(BPM). But I know hoffman algorithm is the best for BPM. But I don\u2019t know its implementation. And as far I know in abroad Hoffman is popular. So if any one have the implementation of Hoffman Algo can u post it here or send me by email?\r\n\r\nMahbub", "Solution_1": "Really? What's the time bounds for it?\r\n\r\nHopcroft-Karp is usually the one people refer to the fastest BM algo... Time: O(E sqrt(V))", "Solution_2": "well what do u mean by time bound?\r\n\r\nas far i know, BPM is better than maxflow, and Hopman (or known as hopkroft Karp) is better than BPM.\r\n\r\ndo u have implimentation of HK? can u plz send that?\r\n\r\nMahbub", "Solution_3": "Nope, I've never written any HK implementation... Never understood it really, and never had the need for it either. ;)\r\n\r\nWhat I do know about it, is that it's basicly the same as the normal augmenting path algorithm, but finds multiple augmenting paths each \"cycle\", instead of just one...", "Solution_4": "hmm.. I have to learn it later :P", "Solution_5": "Hopcroft-Karp is actually almost the same as the \"normal\" augmenting-path Bipartite Matching algorithm, but you search for many paths at once. I attach a paper on it. I have coded it once, but I think on a competition like IOI it's better to know how to code normal BPM quickly and to make it efficient, the improvement of sqrt(N) is not very impressing in practice.", "Solution_6": "Great!\r\n\r\nWell, HK is a great algorithm to have for competitions where \"prewritten code\" is allowed. :)", "Solution_7": "[quote=\"kubus\"]to code normal BPM quickly[/quote]\r\n\r\nWhat's the normal algorithm?", "Solution_8": "normal algo is Hungerian Algo or u can also use Maxflow-BPM", "Solution_9": "Just using \"maxflow\" is enough?", "Solution_10": "Well, if you use the maxflow and bear in mind that what you are trying to do is just a matching, where capacities are all 1 and so on, you can make it quite simple - just keep finding an augmenting path using simple dfs.", "Solution_11": "actually the method (DFS) kubus is statis is nothing but hungerian tree algortihm. but th difference is that we mainly understand that hangerian is BFS algorithm... but kubus is just using DFS instead of DFS." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "I would like to know (using some illusionary example) the following:\r\n\r\n$A$ is said to be complete [b]if[/b] it contains $B$.\r\n\r\nIs it the same as saying\r\n\r\n$A$ is said to be complete [b]if and only if[/b] it contains $B$?\r\n\r\nI know the usage of [b]if[/b] and [b]if and only if[/b] have different meaning in general, but in the case of the example above, are they different?\r\n\r\nHere is how I see the first statement is equivalent to the second statement:\r\nIf $A$ is complete, then it must then contain $B$. Suppose $B$ is not contained in $A$, then $A$ wouldn't have been complete.", "Solution_1": "Yes, the definitional meaning of \"if\" is different from the logical meaning of \"if\". I prefer to use other words to avoid this confusion.\r\nOn the other hand, I don't like using \"if and only if\" in a definition either. That phrase is much better suited to a theorem.\r\n\r\nAlso, be careful with your examples- you've given a statement and its contrapositive, not the statement and its converse.", "Solution_2": "[quote=\"jmerry\"]Also, be careful with your examples- you've given a statement and its contrapositive, not the statement and its converse.[/quote]\r\n\r\n :huh: I am not entire clear. Examples would be appreciated.", "Solution_3": "I personally would see the top statement as misleading, because it still implies (albeit only to the pedant) that there could exist an entity $A$ that does not contain $B$, but could still be said to be complete. It could be equivalently rephrased as \"any $A$ that contains $B$ must be called complete\".\r\n\r\nPersonally I would use the \"[b]if and only if[/b]\" definition, hence eliminating this ambiguity entirely while maintaining the integrity of the original definition. It could be rephrased as above as \"any $A$ that contains $B$ must be called complete and any $A$ that does not contain $B$ cannot be called complete\".\r\n\r\nI see no reason for this situation to be different to that of theorems. Why not have a consistent usage of \"if\" and \"iff\"?", "Solution_4": "To make myself clear, my interest is about knowing proper usage of [b]if[/b] and not to make the original statement unambiguous. Here is the reason why I raise this question. Many times I encounter statements using only [b]if[/b] to describe definitions. Although it is one directional, the author seems to assume the other direction also to be valid in subsequent theorems (Yes, the author could have strengthen his definition like what you just gave to avoid ambiguity). It has led me to believe that [b]if[/b] and [b]if and only if[/b] can be used interchangebly in some cases.\r\n\r\n[quote=\"Ilthigore\"]there could exist an entity $A$ that does not contain $B$, but could still be said to be complete.[/quote]\r\nI think you have a point there. The statement\r\n\r\n$A$ is complete [b]if[/b] it contains $B$\r\n\r\nleaves out some [i]possibility[/i] that $A$ can be complete if it contains $C$ (with or without $B$). However the other statement\r\n\r\n$A$ is complete [b]if and only if[/b] it contains $B$\r\n\r\ndoes not leave out such possibility." } { "Tag": [], "Problem": "IN THE NAME OF GOD\r\n\r\nhiiiiiiii\r\n\r\nIf a man can walk 4km in 1 hour and his wife can walk 5km in 1 hour, how many minutes will it take them to meet if they begin walking at the same time at opposite ends of a 12km road, and have agreed to meet at the half way point? \r\n \r\nMay God help you to all the best \r\n\r\nSALAH", "Solution_1": "EDIT: Sorry I didnt see it said \"halfway there\" .", "Solution_2": "This is not the correct solution\r\nis there another solution??????\r\n\r\nregards\r\n\r\nsalah", "Solution_3": "[hide]The man will make it to the halfway point in $ \\frac {6 km}{4 \\frac {km}{hr}} \\equal{} 1.5 hr \\equal{} 90 minutes.$\nHis wife will make it in $ \\frac {6 km}{5 \\frac {km}{hr}} \\equal{} 1.2 hr \\equal{} 72 minutes.$\n\nThe wife will make it first, and spend 18 minutes waiting for her husband. The answer is thus $ 90 minutes$. Is that correct?[/hide]\r\n\r\nSilly miscalculation, I edited.", "Solution_4": "Yes, I agree with -Elixer-'s answer of $ 90$ minutes (except that the wife makes it in $ 72$ minutes (your $ 1.2$ hours) and waits for $ 18$ minutes).", "Solution_5": "I did it like this:\r\n[hide]\nFirst I found the rates that each person can walk 1 km at.\nWife: 5 km in an hour. 60 minutes \u00f7 5 km = 12 minutes to walk 1 km.\nMan: 4 km in an hour. 60 minutes \u00f7 4 km = 15 min to walk 1 km.\nThe wife will get to the halfway point in 12*6 = 72 min, and the man will get there in 15*6 = 90 minutes.\nTherefore, they will meet at the halfway point in [b]90[/b] minutes, and the wife will just have to wait for 18 minutes. :no: [/hide]" } { "Tag": [], "Problem": "\u03a4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03b9\u03b1 $ K,L,M,N,J,O$ \u03ba\u03b9\u03bd\u03bf\u03c5\u03bd\u03c4\u03b1\u03b9 \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03b1 \u03c4\u03c9\u03bd \u03c4\u03bc\u03b7\u03bc\u03b1\u03c4\u03c9\u03bd $ AB,BC,CD,DE,EZ$ \u03b5\u03bd\u03bf\u03c2 \u03ba\u03b1\u03bd\u03bf\u03bd\u03b9\u03ba\u03bf\u03c5 \u03b5\u03be\u03b1\u03b3\u03c9\u03bd\u03bf\u03c5 $ ABCDEZ$ \u03bc\u03b5 \u03c0\u03b5\u03c1\u03b9\u03bc\u03b5\u03c4\u03c1\u03bf $ 2$.\r\nN\u03b1 \u03b2\u03c1\u03b5\u03b8\u03b5\u03b9 \u03c4\u03bf \u03b5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf \u03c4\u03b7\u03c2 \u03c0\u03b5\u03c1\u03b9\u03bc\u03ad\u03c4\u03c1\u03bf\u03c5 \u03c4\u03bf\u03c5 \u03b5\u03be\u03b1\u03b3\u03ce\u03bd\u03bf\u03c5 $ KLMNJO$", "Solution_1": "\u03a9\u03c1\u03b1\u03af\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1!\r\n\r\n\u0397 \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 $ \\sqrt{3}$, \u03c3\u03c9\u03c3\u03c4\u03ac?\r\n\r\n\u0397 \u03b3\u03bd\u03c9\u03c3\u03c4\u03ae \u03c4\u03b5\u03c7\u03bd\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 \u03cc\u03c4\u03b9 \u03b7 \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03b7 \u03b1\u03c0' \u03c4\u03b7\u03bd \u03c4\u03b5\u03b8\u03bb\u03b1\u03c3\u03bc\u03ad\u03bd\u03b7 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03c0\u03b9\u03ac\u03bd\u03b5\u03b9 \u03c3\u03b5 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03c5 \u03b5\u03af\u03b4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1 :wink:", "Solution_2": "\u03c3\u03c9\u03c3\u03c4\u03bf\u03c2 \u03b1\u03bd \u03b5\u03c7\u03b5\u03b9\u03c2 \u03c7\u03c1\u03bf\u03bd\u03bf \u03ba\u03b1\u03bd\u03c4\u03bf \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03b1", "Solution_3": "\u039c\u03b9\u03b1 \u03c0\u03b5\u03c1\u03b9\u03b3\u03c1\u03b1\u03c6\u03b9\u03ba\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03bf\u03c0\u03bf\u03b9\u03bf\u03bd \u03c4\u03b7\u03bd \u03b8\u03b5\u03bb\u03b5\u03b9. \r\n\u0391\u03c0\u03bf \u039d.\u03c3\u03c5\u03bd\u03b9\u03bc\u03b7\u03c4\u03bf\u03bd\u03c9\u03bd ....\u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03c0\u03bb\u03b5\u03c5\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03bd\u03b5\u03bf\u03c5 \u03b5\u03be\u03b1\u03b3\u03c9\u03bd\u03bf\u03c5 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 $ \\sqrt{x^2+y^2+xy}\\geq{\\sqrt{3}/2(x+y)}$ .\r\n\u0395\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03b7 \u03b3\u03b9\u03b1 \u03bf\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03b5\u03c2 , \u03c0\u03c1\u03bf\u03c3\u03b8\u03b5\u03c3\u03b7 \u03ba\u03b1\u03c4\u03b1 \u03bc\u03b5\u03bb\u03b7.....\u03ba\u03b1\u03b9 \u03c0\u03b1\u03b9\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 $ \\sqrt{3}$ \u03bf\u03c0\u03c9\u03c2 \u03b1\u03bd\u03b1\u03c6\u03b5\u03c1\u03b8\u03b7\u03ba\u03b5 \u03b1\u03c0 \u03c4\u03bf\u03bd bilot4", "Solution_4": "\u03a3\u03c5\u03b3\u03b3\u03bd\u03ce\u03bc\u03b7 \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b1 \u03bd\u03c9\u03c1\u03af\u03c4\u03b5\u03c1\u03b1. \u0395\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03c9\u03c1\u03b1\u03af\u03b1 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03b1\u03c2:wink: \r\n\u0395\u03b3\u03ce \u03c4\u03bf \u03ad\u03ba\u03b1\u03bd\u03b1 \u03c4\u03b5\u03bb\u03b5\u03af\u03c9\u03c2 \u03b1\u03bb\u03bb\u03af\u03c9\u03c2 (\u03ba\u03b1\u03b8\u03b1\u03c1\u03ac \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7)... \u0395\u03c7\u03c9 \u03c3\u03c5\u03bd\u03ac\u03c8\u03b5\u03b9 \u03c4\u03bf \u03c3\u03c7\u03ae\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b8\u03ad\u03bb\u03b5\u03b9 \u03c0\u03bf\u03bb\u03cd \u03c0\u03b5\u03c1\u03b1\u03b9\u03c4\u03ad\u03c1\u03c9 \u03b5\u03be\u03ae\u03b3\u03b7\u03c3\u03b7... \u0391\u03c0\u03bb\u03ac \u03ad\u03c7\u03c9 \u03c0\u03ac\u03c1\u03b5\u03b9 $ CD \\equal{} CD', CM \\equal{} CM', \\angle{BCD'} \\equal{} 120$ \u03ba\u03b1\u03b9 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03ac\u03bb\u03bb\u03b1. \u0391\u03c5\u03c4\u03cc \u03bc\u03b1\u03c2 \u03b4\u03af\u03bd\u03b5\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03cc\u03c4\u03b9 $ LM \\equal{} LM'$ \u03ba\u03c4\u03bb. \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03c3\u03b7\u03bc\u03b1\u03af\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b7 \u03c0\u03b5\u03c1\u03af\u03bc\u03b5\u03c4\u03c1\u03bf\u03c2 \u03b9\u03c3\u03bf\u03cd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf $ KL \\plus{} LM' \\plus{} M'N' \\plus{} N'O' \\plus{} O'P' \\plus{} P'K'$. \u03a4\u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf \u03cc\u03bc\u03c9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf \u03ae \u03af\u03c3\u03bf \u03c4\u03bf\u03c5 $ KK' \\equal{} AA'$ (\u03c9\u03c2 \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b7\u03bb\u03cc\u03b3\u03c1\u03b1\u03bc\u03bc\u03bf). \u03a4\u03b5\u03bb\u03b9\u03ba\u03ce\u03c2 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 $ AA' \\equal{} 3AB \\equal{} \\sqrt {3}$.\r\n\r\n\r\nY.\u0393. \u0398\u03b1 \u03ad\u03b2\u03b1\u03b6\u03b1 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bd\u03c9\u03c1\u03af\u03c4\u03b5\u03c1\u03b1 \u03b1\u03bb\u03bb\u03ac \u03b5\u03af\u03c7\u03b1 \u03ad\u03bd\u03b1 \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03bc\u03b5 \u03c4\u03bf \u03c3\u03c5\u03bd\u03ac\u03c8\u03c9 \u03c4\u03bf \u03c3\u03c7\u03ae\u03bc\u03b1. \u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c4\u03bf \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03cc \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03ac \u03c4\u03bf\u03c5 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03cc :)", "Solution_5": "\u0391\u03c0\u03bb\u03b1 \u03b5\u03ba\u03c0\u03bb\u03b7\u03ba\u03c4\u03b9\u03ba\u03b7 \u03bb\u03c5\u03c3\u03b7 ............", "Solution_6": "[quote=\"xp\"]\u0391\u03c0\u03bb\u03b1 \u03b5\u03ba\u03c0\u03bb\u03b7\u03ba\u03c4\u03b9\u03ba\u03b7 \u03bb\u03c5\u03c3\u03b7 ............[/quote]\r\n\r\n\u039d\u03b1\u03b9, \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03c5\u03c0\u03ad\u03c1\u03bf\u03c7\u03b7 \u03bb\u03cd\u03c3\u03b7! :coolspeak:\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_7": "\u03a3\u03b1\u03c2 \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03bf\u03bb\u03cd :)", "Solution_8": "\u03a0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9 \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae \u03bb\u03cd\u03c3\u03b7 :) \u03a4\u03ce\u03c1\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b5\u03af\u03c3\u03c4\u03b5 \u03c4\u03bf \u03b3\u03b5\u03bd\u03b9\u03ba\u03cc \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b3\u03b9\u03b1 \u03bd-\u03b3\u03c9\u03bd\u03bf :wink:", "Solution_9": "E\u03b9\u03c3\u03b1\u03b9 \u03b1\u03bd\u03b7\u03c3\u03c5\u03c7\u03bf \u03c0\u03bd\u03b5\u03c5\u03bc\u03b1 .\u039c\u03b9\u03b1 \u03ba\u03bf\u03c5\u03b2\u03ad\u03bd\u03c4\u03b1 \u03b5\u03b9\u03c0\u03b1\u03bc\u03b5 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c8\u03b1\u03be\u03b5\u03c2.....\u0391\u03bd \u03c3\u03c4\u03bf \u03b4\u03b9\u03b1\u03bb\u03bb\u03b5\u03b9\u03bc\u03b1 9-9.10 \u03b4\u03b5\u03bd \u03b5\u03ba\u03b1\u03bd\u03b1 \u03bb\u03b1\u03b8\u03bf\u03c2 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5:\r\nA\u03bd $ S$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03b5\u03c1\u03af\u03bc\u03b5\u03c4\u03c1\u03bf\u03c2 \u03c4\u03bf $ n$ \u03b3\u03c9\u03bd\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 $ S'_{min}\\equal{}Scos(\\frac{\\pi}{n})$" } { "Tag": [], "Problem": "Determine the number of perfect squares that there are between $ 40000$ and $ 640000$ that are multiples of $ 3, 4$ and $ 5$, simultaneously.", "Solution_1": "[hide]\n$ 3|n^{2}\\implies 3|n$\n$ 4|n^{2}\\implies 2|n$\n$ 5|n^{2}\\implies 5|n$\n$ 3,4,5 | n^{2}\\implies 3 \\times 2 \\times 5 | n \\iff 30 | n$\n\nAssuming between means exclusive, we want numbers between $ \\sqrt{40000}=200$ and $ \\sqrt{640000}=800$ that are multiples of $ 30$. We can transform our range by shifting it by multiples of $ 30$:\n\n$ \\#(200,800) \\equiv \\#(200-180, 800-180) \\equiv \\#(20, 620) \\equiv \\#(20, 20 \\times 30+20)$\n\n$ 20 < 30 \\times 1 < 30 \\times 20 < 30 \\times 20+20$\n\nThus, there are $ 20$ such numbers.\n\n[/hide]" } { "Tag": [ "ARML" ], "Problem": "From the mailing I received today from Tom Kilkelly:\r\n\r\n1. 40 Sofia Math Circle (Bulgaria)\r\n2T. 37 Lynbrook High School\r\n3T. 37 Phillips Exeter Academy \r\n4. 36 Illinois Math and Science Academy\r\n5T. 35 Thomas Jefferson HS \r\n5T. 35 Western Washington \r\n7T. 34 Georgia ARML\r\n7T. 34 Mongomery Blair HS \r\n9T. 33 Colorado ARML\r\n9T. 33 Long Beach Math Circle\r\n9T. 33 AAST \r\n12T. 32 New Mexico Math League\r\n12T. 32 Phillips Academy \r\n12T. 32 Western MASS A\r\n12T. 32 Iowa City West HS\r\n16T. 31 Naperville North HS \r\n16T. 31 Wayzata HS \r\n16T. 31 Daderyds Gymnasium (Sweeden)\r\n16T. 31 Stuyvesant HS", "Solution_1": "35 is an [b]amazing[/b] score for us considering we didn't seriously answer the part worth 5 points :D", "Solution_2": "There were two 5-point parts; #3 and #4e. Our 33 comes from getting a 1 out of 5 on #3 and 2 out of 5 on #4e, and all of the other points.\r\n\r\n[\"Long Beach Math Circle\" is a large fraction of the Southern California ARML team - as I assume that \"Western Washington\" is a large fraction of the Washington ARML team - right?]", "Solution_3": "[quote=\"Kent Merryfield\"]There were two 5-point parts; #3 and #4e. Our 33 comes from getting a 1 out of 5 on #3 and 2 out of 5 on #4e, and all of the other points.\n\n[\"Long Beach Math Circle\" is a large fraction of the Southern California ARML team - as I assume that \"Western Washington\" is a large fraction of the Washington ARML team - right?][/quote]\r\n\r\nWestern Washington is a large fraction, but there are also a number of strong students who couldn't make it (3 hour drives and such). I think we lost a majority of points on #3; basically, nobody could do it and in the last 3 minutes we scrambled to write random numbers down, and we didn't even finish that.", "Solution_4": "[quote=\"paladin8\"]basically, nobody could do it[/quote]\r\n\r\nProbably because only two people were working on it until the last few minutes :| \r\n\r\nWe should probably divvy up the work more efficiently next time :)", "Solution_5": "[quote=\"t0rajir0u\"][quote=\"paladin8\"]basically, nobody could do it[/quote]\n\nProbably because only two people were working on it until the last few minutes :| \n\nWe should probably divvy up the work more efficiently next time :)[/quote]\r\nYou and us both. That's our story too.", "Solution_6": "A belated solution to #3:\r\n\r\n[quote=\"Mij\"]Problem #3: Place the numbers from 0 to 80 in a 9 x 9 square array so that the sum of the numbers in each of its 49 3x3 subsquares is the same.\n\nHere's a solution in ternary:\n$\\begin{bmatrix}0000&0110&0220&0001&0111&0221&0002&0112&0222 \\\\ 1001&1111&1221&1002&1112&1222&1000&1110&1220 \\\\ 2002&2112&2222&2000&2110&2220&2001&2111&2221 \\\\ 0010&0120&0200&0011&0121&0201&0012&0122&0202 \\\\ 1011&1121&1201&1012&1122&1202&1010&1120&1200 \\\\ 2012&2122&2202&2010&2120&2200&2011&2121&2201 \\\\ 0020&0100&0210&0021&0101&0211&0022&0102&0212 \\\\ 1021&1101&1211&1022&1102&1212&1020&1100&1210 \\\\ 2022&2102&2212&2020&2100&2210&2021&2101&2211 \\\\ \\end{bmatrix}$\nEssentially, to generate this table, you follow a certain algorithm (hinted at in the problem itself). First, separate this 9 x 9 square into 9 3x3 squares, in which the upper left hand corners of each read from $0000$ to $0022$. Then, in each of the 3 x 3's, to find the number below a certain number, add $1001$ WITHOUT carrying (for example, $1102+1001=2100$). To find the number to the right of a certain number, add $0110$ WITHOUT carrying. By doing this you can find out the rest of the table.\n\nHere is the array in base 10:\n$\\begin{bmatrix}0&12&24&1&13&25&2&14&26 \\\\ 28&40&52&29&41&53&27&39&51 \\\\ 56&68&80&54&66&78&55&67&79 \\\\ 3&15&18&4&16&19&5&17&20 \\\\ 31&43&46&32&44&47&30&42&45 \\\\ 59&71&74&57&69&72&58&70&73 \\\\ 6&9&21&7&10&22&8&11&23 \\\\ 34&37&49&35&38&50&33&36&48 \\\\ 62&65&77&60&63&75&61&64&76 \\\\ \\end{bmatrix}$\nEvery 3 x 3 subsquare will add up to the same number, $360$. Also, if you rotate the whole matrix or transpose it, you will still have a viable solution :) \n\nI in fact figured out this algorithm during the Nov. meeting, but this revelation came too late for me to finish #3 in time :([/quote]", "Solution_7": "man...i wish my school would do these types of things...sadly, there is not enough interest\r\n\r\np.s. is there anywhere i can see the contest?", "Solution_8": "You live in Illinois, but I take it you don't go to either Naperville North or IMSA, right?", "Solution_9": "Grrrrr, I was almost sure we had closer to a perfect than that... But we counted the later parts more important than the early ones and we didn't have a great proof for 4e, so I guess that's why we got but a 34. :(", "Solution_10": "[quote=\"Kent Merryfield\"]You live in Illinois, but I take it you don't go to either Naperville North or IMSA, right?[/quote]\r\n\r\nyes...although my school is well-rounded, the math program only teaches basic proficiency, (my school is new trier...)", "Solution_11": "I would like to bring the ARML Power Contest and/or the Mandelbrot to my school by next year, but there might not be enough interest." } { "Tag": [ "algorithm", "algebra", "polynomial" ], "Problem": "Hi,\r\n\r\nYou are given an undirected graph.Now the problem is to find number of loop less paths between two given nodes in this graph.\r\n\r\nAre there any algorithms for solving this problem.\r\n\r\nAre there any special cases where it can be solved ??\r\n\r\nThank You.\r\nAbhilash I.", "Solution_1": "For small n (up to 20 something-ish) a simple 2^n*n DP would work. I can't think of anything for larger n though.\r\n\r\nThis paper might be of interest: http://maths.uq.edu.au/~kroese/ps/robkro_rev.pdf", "Solution_2": "Well, we would need to define some more specific rules for the program before we can create the algorithm. Can links be traveled only once? Once in each direction? An arbitrary number of passes (would frequently result in infinite paths)?\r\n\r\nBut regardless, the algorithm should be fairly simple. You create a stack or queue (same result) containing all the nodes in the graph, and begin by adding the first of the two source nodes. Then for each node in the stack, you add all the linked nodes to the stack/queue (or all the linked nodes that have not already been added, if you want each node to only be usable once). If you want to avoid back-and-forth (you probably do, otherwise there would almost always be infinite paths) then give the new node a parent (there may be more than one) which is the identifier of the old node which led to that node, and add the additional condition that new nodes must not be the parent of the node currently being processed. Finally, when the ending node is reached, increment a paths_found variable by one. (Depending on the rules of the system, you probably don't want to add the ending node to the path, but it would be neither correct nor incorrect to do so.) It's a lot like A* pathfinding - store nodes in a stack/queue, and when each node is processed add each adjacent node to the stack/queue if certain conditions are met.", "Solution_3": "You must have missed the 'loopless' criterion he mentioned ;)\r\n\r\nAnyway, of course a simple traversal would give you the right answer, but thats easily going to be exponential. I'm pretty sure we're looking for an algorithm that would actually run in a limited amount of time (the brute force method won't work as high as the DP method will).", "Solution_4": "[quote=\"TripleM\"]You must have missed the 'loopless' criterion he mentioned ;)[/quote]\nSorry - I did.\n\n[quote=\"TripleM\"]Anyway, of course a simple traversal would give you the right answer, but thats easily going to be exponential. I'm pretty sure we're looking for an algorithm that would actually run in a limited amount of time (the brute force method won't work as high as the DP method will).[/quote]\r\nWell, the problem there is that there is no such algorithm. Given an arbitrary number of nodes and an arbitrary number of connections of arbitrary complexity, there is no method of finding the number of paths between two nodes without actually traversing the path. There are certain optimizations that can be made to avoid redundant calculations, but attempting the problem without traversal is akin to searching for an object before identifying it - it's simply a necessary part of processing the given input.\r\n\r\nIn any case, it wasn't stated or implied in the original post that a DP is what we're looking for. If someone asks how to find the shortest path between two points with arbitrary obstacles, it seems reasonable to point to A*, given the lack of equivalent alternatives. I don't see the difference here.", "Solution_5": "Can you actually prove its NP complete? There are very similar problems in graphs which do have polynomial-time solutions, so I wouldn't go saying that there is no algorithm without being very sure. \r\nOf course, the link I gave above probably does prove it is NP complete. Then again, some exponential algorithms are far faster than others.\r\n\r\nAnyway, he asked if there were special cases the problem could be solved on. The DP gives a way of solving it for small n which a brute force approach does not.\r\n\r\n(edit - by the way, you said a shortest path algorithm with arbitrary obstacles should use A* - but of course, Dijkstra is a simple polynomial algorithm for that - you must have meant something else).", "Solution_6": "[quote](edit - by the way, you said a shortest path algorithm with arbitrary obstacles should use A* - but of course, Dijkstra is a simple polynomial algorithm for that - you must have meant something else).[/quote]\r\nI said arbitrary obstacles, not arbitrary nodes and links. Sorry if I wasn't clear.", "Solution_7": "Hi,\r\nProblem Statement:\r\nGiven an undirected graph find the number of edge disjoint paths between \r\ntwo given nodes from that graph ?\r\n\r\nIs there any polynomial algorithm or is it NP-complete??\r\n\r\nThank You.\r\nAbhilash I." } { "Tag": [ "AMC", "AMC 10", "AMC 12", "AIME", "calculus", "trigonometry", "logarithms" ], "Problem": "From the AMC12 Discussion Thread:\r\n[quote=\"DPopov\"]AMC 10 is harder to qualify with in my opinion. There is greater stress on not making stupid mistakes.[/quote]\r\nAnd I agree.\r\n\r\nIs it time to turn AMC10/12 back into one, again? \r\n\r\nOr, just phase out AMC10!\r\n\r\nI may be influenced by the skill level of the AoPS members, but it seems like AMC12 is now appropriate for many middle schoolers, that is, as long as they are guided correctly. Also, by keeping some problems sufficiently challenging (certainly the 20s), it will remain appropriate for high schoolers as well.", "Solution_1": "No, no, no. Obviously, AoPS is nowhere near an accurate sample of AMC-takers. Most of my friends who took it did 10-15 problems on the 10 and would only have answered 8-10 on the 12. The 12 is far too advanced for most freshmen and below. With that few questions answered, you just get bored with the remaining time.", "Solution_2": "[quote=\"MathChauffeur\"]\nI may be influenced by the skill level of the AoPS members, but it seems like AMC12 is now appropriate for many middle schoolers, that is, as long as they are guided correctly. Also, by keeping some problems sufficiently challenging (certainly the 20s), it will remain appropriate for high schoolers as well.[/quote]\r\n\r\nWhen I was in middle school, I could barely qualify for AIME via the AMC 10 (120.5). So the AMC 12 would have been IMPOSSIBLE.", "Solution_3": "[quote=\"mysmartmouth\"][quote=\"MathChauffeur\"]\nI may be influenced by the skill level of the AoPS members, but it seems like AMC12 is now appropriate for many middle schoolers, that is, as long as they are guided correctly. Also, by keeping some problems sufficiently challenging (certainly the 20s), it will remain appropriate for high schoolers as well.[/quote]\n\nWhen I was in middle school, I could barely qualify for AIME via the AMC 10 (120.5). So the AMC 12 would have been IMPOSSIBLE.[/quote]\r\n\r\nYeah. I qualified last year (8th grade) with a 124.", "Solution_4": "9th grade, took the 12, and did respectably. 118.5.", "Solution_5": "Comments, scores, and opinions on this forum are not representative of the AMC 10.\r\n\r\nAbout 4000 teachers believe the AMC 10 is appropriate for at least 100,000 students. Scores, averages, number of problems answered, and the number of students participating, along with teacher comments we receive all convince the Committee on the American Mathematics Competitions that the AMC 10 has a place in our contest sequence.\r\n\r\nRead the letter from the AMC 10 Chair, Doug Faires, in the 2006 Summary of Results and Awards, page 12 and 13. (All contest managers from last year received this book in June or July 2006.)\r\n\r\n\r\nSteve Dunbar\r\nAMC Director", "Solution_6": "At our school, we've given only the AMC-12, to all grades 9-12 for the past five years. \r\n\r\nLast year our top three freshman scores were 102, 101.5, and 95. Our top three sophomore scores last year were 122, 114.5, and 103.\r\n\r\nWe will probably give the AMC-10 and AMC-12 next year due to the change in scoring. Under the old system, a freshman could find eleven problem she could do correctly, do them, and qualify for the AIME. Now, she'd have to find fourteen, which may be next to impossible for a student who hasn't been practicing on AMC tests to know the style of questions.\r\n\r\nIdeally, I'd still give the AMC-12 exclusively, as I believe it sets the bar higher for the underclassmen. I don't like the idea of giving an intentionally easier test to sophomores (many of them are in pre-calculus at our school) just because they're a year younger than the juniors. But, I want to get them exposed to the AIME under contest environments (even when freshmen get one or two right, it's still an accomplishment to be proud of.)\r\n\r\nTDP.", "Solution_7": "Which contest (AMC 10 or AMC 12) to give to a student is a question best left to the discretion of the teacher who knows the student well. If you have sophomores in pre-calculus, then maybe the AMC 12 is the proper contest to administer to those students. Your experience and judgment is important in helping students. That's wh we often refer to the teachers and contest managers as coaches.\r\n\r\nI don't think the word \"easier\" is the right word to distinguish between the AMC 10 and the AMC 12. The AMC 10 contest does not contain trigonometry, logarithms, complex numbers, some advanced algebra of polynomials and their roots, and does not use function notation. As we say in the brochures, the \"AMC 10 covers material normally associated with grades 9 and 10.\" For a student who has covered only that material, the AMC 10 may not be easier but it certainly is less advanced.\r\n\r\nI have written comments on the Certification Form from many teachers who appreciate the distinction between the AMC 10 and the AMC 12. The number who have written those comments is many more than have written in this forum. Again, I don't think you can take the comments of this forum as representative. Remember, your experience may be different, and our rules accommodate both experiences.\r\n\r\n\r\nSteve Dunbar\r\nAMC Director", "Solution_8": "I guess it depends on the purpose of the AMC-10 test.\r\n\r\nIf the purpose of the test is to determine participation on the AIME or USAMO, where trigonometry, logarithms, complex numbers, advanced algebra of polynomials and their roots, and functional notation all appear, then no, I don't think it's fulfilling it's purpose. If a student designs to take the AIME or USAMO, I think they should take the AMC-12 test, even if they are a freshman, and even if they haven't had the requisite mathematics. Exposure to the AMC-12 will be more of help experience-wise than the AMC-10 for those level of exams. \r\n\r\nThe AIME floor value last year was eight. I find it difficult to believe that a student who could get eight right on the AIME would not be able to get at least eleven right on the AMC-12 last year. (Per http://www.unl.edu/amc/b-registration/b1-archive/2003-2004/faq2003-04.html, there are about a dozen problems in common between AMC-10 and AMC-12)\r\n\r\nIf, on the other hand, the purpose of the AMC-10 test is to challenge students with \"outside the box\" mathematics at a 9th or 10th grade \"appropriateness level\" (read, Algebra I, II, and Geometry) then, yes it is fulfilling it's purpose.\r\n\r\nI'm not saying these goals need be mutually exclusive, but it should be noted that they are not very similar. Obviously, students participating in this forum are going to be more desirous of an AIME qualification, or even an USAMO qualification, and so make comments to that effect.\r\n\r\nAnd my choice of \"easier\" in the prior post may not have been appropriate. Less rigorous wouldn't be appropriate either, but I do want an exam that will push beyond currently known content. If one is able to reach the stars, one will be content with that.\r\n\r\nTDP." } { "Tag": [ "geometry", "algebra", "polynomial", "conics", "inequalities", "function", "trigonometry" ], "Problem": "Since I finished summer school Geometry, I would now like to know what material will be covered in Algebra 2......Anything will be helpful :)", "Solution_1": "Matrices\r\nPolynomial Factorization (Just a little harder then algebra I?)\r\nBinomial Theorem\r\nNon-Linear Systems of Equations\r\nConics\r\nSynthetic Division/Algebraic Long Division (Think that's what it is called)\r\nInequalities/Systems of Inequalities\r\nExponential Function\r\nRational Functions\r\nLogarithms/Natural Log\r\nA little basic trig\r\n\r\nThose are just a few of the topics..." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $x$,$y$,$z$ $\\in$ $(0,1)$ and $\\sum x$ $\\leq$ $\\frac{3}{2}$.Prove that\r\n$2$$\\sum x$+$\\sum \\frac{(4y^2z^2-1)x}{y^2+z^2-1}$ $\\leq$ $\\frac{3}{4}$", "Solution_1": "take $x=y=z=\\frac{1}{2}$ and u get a contradiction", "Solution_2": "Sorry,I posted carelessly.Let $x$,$y$,$z$>0 and $\\sum x$$\\leq$ $\\frac{3}{2}$.Prove that\r\n$2\\sum x-\\sum \\frac{(4y^2z^2-1)x}{y^2+z^2-1}$ $\\leq$ $\\frac{3}{4}$" } { "Tag": [ "inequalities", "calculus", "inequalities proposed" ], "Problem": "Let $ a,b,c,d$ be real numbers with sum 0. Prove the inequality:\r\n\\[ (ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd)^2 \\plus{} 12\\geq 6(abc \\plus{} abd \\plus{} acd \\plus{} bcd).\r\n\\]", "Solution_1": "Posted here \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=71330", "Solution_2": "After applying Rolle's theorem to $(x-a)(x-b)(x-c)(x-d)$ we get an inequality in $u,v,w$ s.t. $u+v+w=0$. (technique similar with the one used for proving Schur for several variables) I solved that through $w=-(u+v)$; then I considered $LHS-RHS$ as $f(u)$; $f' \\left( - \\frac{v}{2} \\right)=0$; the rest is easy. :D\r\n\r\nCan anyone find a nice solution for the inequality with $u+v+w=0$: \r\n\r\n\\[ (uv+vw+wu)^2 + 3 \\geq 6uvw \\ ? \\]", "Solution_3": "Perfect_Radio, can you post your solution in detail? :) ;)", "Solution_4": "We have $P(x) = x^4+(ab+ac+ad+bc+bd+cd)x^2-x(abc+abd+acd+bcd)+abcd$, so $P'(x) = 4x^3 + 2(ab+ac+ad+bc+bd+cd)x-(abc+abd+acd+bcd)$.\r\n\r\nBy Rolle's Theorem, there are $u,v,w \\in \\mathbb{R}$ s.t. $P'(x) = 4(x-u)(x-v)(x-w)$. (in case we have $a=b$ for example, then $a$ is also a root $P'(x)$; else we apply the Rolle argument)\r\n\r\nNow equate the two formulas and substitute in the inequality. As I said in my previous post, my solution for the $u+v+w=0$ inequality is pretty ugly (you can do it yourself). Try to solve this inequality. Maybe you'll find a nice solution.", "Solution_5": "[quote=\"perfect_radio\"]\n\nCan anyone find a nice solution for the inequality with $u+v+w=0$: \n\n\\[ (uv+vw+wu)^2 + 3 \\geq 6uvw \\ ? \\][/quote]\r\nYou can easy see that we only need to prove this ineq.\r\n $x^2y^2+(x^2+y^2)(x+y)^2+3\\geq 6xy(x+y)$wher $x,y\\geq 0$. :) \r\nLet's use Caush's ineq. for 12 numbers\r\n$x^2y^2+8*\\frac{(x^2+y^2)(x+y)^2}{8}+1+1+1\\geq \\sqrt[12]{x^2y^2*(\\frac{(x^2+y^2)(x+y)^2}{8})^8}\\geq 6xy(x+y)$ \r\nMybe not wery nice :( but not wery ugly too :D .", "Solution_6": "[quote=\"perfect_radio\"]Can anyone find a nice solution for the inequality with $u+v+w=0$: \n\n\\[ (uv+vw+wu)^2 + 3 \\geq 6uvw \\ ? \\][/quote]\r\n Suppose $u \\geq 0 \\geq v \\geq w$ \r\n Put $m=-v$ ; $n=-w$ \r\n So $u =m+n$ (1) \r\n Replace (1) into the ineq , we need to prove :\r\n $(m^n+n^2+mn)^2 +3 \\geq 6mn(m+n)$ , where $n \\geq m \\geq 0$ \r\n ;) Maybe this ineq can be solved by AM-GM , Cauchy ,... but I used calculus to solve this ineq . It's not nice but easy .", "Solution_7": "[quote=\"nttu\"][quote=\"perfect_radio\"]Can anyone find a nice solution for the inequality with $u+v+w=0$: \n\n\\[ (uv+vw+wu)^2 + 3 \\geq 6uvw \\ ? \\][/quote]\n Suppose $u \\geq 0 \\geq v \\geq w$ \n Put $m=-v$ ; $n=-w$ \n So $u =m+n$ (1) \n Replace (1) into the ineq , we need to prove :\n $(m^n+n^2+mn)^2 +3 \\geq 6mn(m+n)$ , where $n \\geq m \\geq 0$ \n ;) Maybe this ineq can be solved by AM-GM , Cauchy ,... but I used calculus to solve this ineq . It's not nice but easy .[/quote]\r\n :rotfl: :blush: I think you have a typo ther must be $(m^2+n^2+mn)^2 +3 \\geq 6mn(m+n)$ :P :D ,and for this ineq. see my laste post.(change m,n->x,y :rotfl: :oops: )", "Solution_8": "\\displaystyle d=-a-b-c\r\n\\displaystyle abc+abd+acd+cbd=abc-(a+b+c)(ab+bc+ac)=-(-abc+\\sum_{sym}^{}a^2b+3abc)=-(\\sum_{sym}^{}a^2b+2abc)=-(a+b)(a+c)(b+c)\r\n\\displaystyle ab+ac+bc+ad+cd+bd=ab+bc+ac-(a+b+c)^2=-\\frac{1}{2}((a+b)^2+(a+c)^2+(b+c)^2)\r\nLet \\displaystyle x=a+b y=a+c z=b+c \r\nWe need \\displaystyle \\frac{1}{4}(x^2+y^2+z^2)^2+12\\ge -6xyz\r\n\\displaystyle x^2+y^2+z^2\\ge 3|xyz|^{\\frac{2}{3}}\r\nSo, \\displaystyle \\frac{1}{4}(x^2+y^2+z^2)^2+12\\ge \\frac{9}{4}|xyz|^{\\frac{4}{3}}+12=\\frac{3}{4}|xyz|^{\\frac{3}{4}}+\\frac{3}{4}|xyz|^{\\frac{3}{4}}+\\frac{3}{4}|xyz|^{\\frac{3}{4}}+12\\ge 4\\times(\\frac{12\\times3^3}{4^3})^{\\frac{1}{4}}|xyz|=6|xyz|\\ge-6xyz", "Solution_9": ":oops_sign: \r\n$ d\\equal{}\\minus{}a\\minus{}b\\minus{}c$\r\n$ abc\\plus{}abd\\plus{}acd\\plus{}cbd\\equal{}abc\\minus{}(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ac)\\equal{}\\minus{}(\\minus{}abc\\plus{}\\sum_{sym}^{}a^2b\\plus{}3abc)\\equal{}\\minus{}(\\sum_{sym}^{}a^2b\\plus{}2abc)\\equal{}\\minus{}(a\\plus{}b)(a\\plus{}c)(b\\plus{}c)$\r\n$ ab\\plus{}ac\\plus{}bc\\plus{}ad\\plus{}cd\\plus{}bd\\equal{}ab\\plus{}bc\\plus{}ac\\minus{}(a\\plus{}b\\plus{}c)^2\\equal{}\\minus{}\\frac{1}{2}((a\\plus{}b)^2\\plus{}(a\\plus{}c)^2\\plus{}(b\\plus{}c)^2)$\r\nLet $ x\\equal{}a\\plus{}b$\r\n $ y\\equal{}a\\plus{}c$ \r\n $ z\\equal{}b\\plus{}c$\r\nWe need $ \\frac{1}{4}(x^2\\plus{}y^2\\plus{}z^2)^2\\plus{}12\\ge \\minus{}6xyz$\r\n$ x^2\\plus{}y^2\\plus{}z^2\\ge 3|xyz|^{\\frac{2}{3}}$\r\nSo, $ \\frac{1}{4}(x^2\\plus{}y^2\\plus{}z^2)^2\\plus{}12\\ge$ $ \\frac{9}{4}|xyz|^{\\frac{4}{3}}\\plus{}12\\equal{}\\frac{3}{4}|xyz|^{\\frac{4}{3}}\\plus{}\\frac{3}{4}|xyz|^{\\frac{4}{3}}\\plus{}\\frac{3}{4}|xyz|^{\\frac{4}{3}}\\plus{}12\\ge$ $ 4\\times(\\frac{12\\times3^3}{4^3})^{\\frac{1}{4}}|xyz|\\equal{}6|xyz|\\ge\\minus{}6xyz$", "Solution_10": "As a three-year late side note, we can prove the inequality in $ u,v,w$ by using the fact $ x^3 \\plus{} ax \\plus{} b$ has three real roots iff\r\n\\[ 4 a^3 \\plus{} 27 b^2 \\leq 0 .\r\n\\]", "Solution_11": "[quote=\"Valentin Vornicu\"]Let $ a,b,c,d$ be real numbers with sum 0. Prove the inequality:\n\\[ (ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd)^2 \\plus{} 12\\geq 6(abc \\plus{} abd \\plus{} acd \\plus{} bcd).\\]\n[/quote]\r\nMy soloution is here [url]http://vimf.freeforums.org/m-t-s-bdt-olympiad-t162-20.html[/url]", "Solution_12": "[quote=\"litbon\"][quote=\"Valentin Vornicu\"]Let $ a,b,c,d$ be real numbers with sum 0. Prove the inequality:\n\\[ (ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd)^2 \\plus{} 12\\geq 6(abc \\plus{} abd \\plus{} acd \\plus{} bcd).\\]\n[/quote]\nMy soloution is here [url]http://vimf.freeforums.org/m-t-s-bdt-olympiad-t162-20.html[/url][/quote]\n\nNext time just post the solution here. it seems your forum is out of service now..", "Solution_13": "Can anyone find a nice solution for the inequality with $u+v+w=0$: \n\n\\[ (uv+vw+wu)^2 + 3 \\geq 6uvw \\ ? \\][/quote]\n\nSubstituting $w=-(u+v)$, we have the inequality\n $(u^2+uv+v^2)^2+3\\geq6uv(u+v)$\nAfter writing $u=z+t, v=z-t$, the inequality becomes\n$(3z^2+t^2)^2+ 3 \\geq 12z(z^2-t^2)$\nAs $t^2 \\geq 0$ , it is sufficient to prove that\n$(3z^2)^2+3 \\geq 12z^3$, which is obvious from AM-GM\n\n", "Solution_14": "Does anyone find the a,b,c,d when LHS and RHS are equal?", "Solution_15": "hi can any one help me with this problem i ve seen seen the aops solution but didnt understand \nwe have 0<=x,y,z<=1 prove that\n(x/(y+z+1))+(y/(x+z+1))+(z/(x+y+1))+(1-x)(1-y)(1-z)<=1", "Solution_16": "[quote=Valentin Vornicu]Let $ a,b,c,d$ be real numbers with sum 0. Prove the inequality:\n\\[ (ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd)^2 \\plus{} 12\\geq 6(abc \\plus{} abd \\plus{} acd \\plus{} bcd).\n\\][/quote]\n\nLet $4u=a+b+c+d$, $6v^2=ab+cd+ac+bd+ad+bc$ ($v^2$ can be negative), $4w^3=abc+bcd+cda+dab$ and $P(x)=(x-a)(x-b)(x-c)(x-d)=x^4-4ux^3+6v^2x^2-4w^3x+abcd$.\nThus by Rolle's theorem $P'(x)=4(x^3-3ux^2+3v^2x-w^3)$ has a three real roots, which gives $$\\Delta=6912(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=-6912(4v^6+w^6)\\geq 0\\text{.}$$ \nSo we need to prove\n$$3v^4+1\\geq 2w^3$$\n,after squaring it's equivalent to\n$$9v^8+6v^4+1\\geq 4w^6$$\nor\n$$9v^8+6v^4-4w^6+1\\geq 9v^8+16v^6+6v^4+1=(3v^2+1)^2(9v^4-2v^2+1)\\geq 0$$\n, which is true." } { "Tag": [], "Problem": "What are some of the scholarship competitions in MAth??\r\n(LIke essays/researches)", "Solution_1": "Try searching on a website such as Fastweb.com or Collegeboard's scholarship search. I doubt if there's math scholarship competitions out there, but it's worth a try." } { "Tag": [ "number theory", "prime numbers", "number theory unsolved" ], "Problem": "Find all prime numbers $p$ which satisfy the following condition: For any prime $q < p$, if $p = kq + r, 0 \\leq r < q$, there does not exist an integer $q > 1$ such that $a^{2} \\mid r$.", "Solution_1": "I guess this is supposed to be \"...there does not exist an integer $ a>1$ such that $ a^2 | r$\"?", "Solution_2": "I'm still new to this writing a proof thing, so I'd appreciate it if people could comment on how I could improve, thx\r\n\r\n[hide]The only such p are $ 2$, $ 3$, $ 5$, $ 7$, and $ 13$.\nProof:\nI will divide up our search in to two cases.\n\nCase 1: p is a prime NOT in the form $ 3^k \\plus{} 4$.\nIn this case, we are guaranteed that as long as $ p$ is greater than $ 5$, $ p\\minus{}4$ will be divisible by a prime larger than $ 3$.\nThus we see that we can have $ p \\equal{} kq \\plus{} 2^2$, and $ q > 2^2$. This rules out any prime NOT of the form $ 3^k \\plus{} 4$ that is greater than 5. By inspection we see that every prime less than or equal to 5 works, thus 2, 3, and 5 are numbers that we are looking for.\n\nCase 2: p is a prime of the form $ 3^k \\plus{} 4$.\nIn this case, let us consider $ k \\ge 4$. Then $ p \\minus{} 7^2 \\equal{} 3^k \\minus{} 45 \\equal{} 9(3^{k\\minus{}2} \\minus{} 5)$. Since $ k \\ge 4$, the expression is positive, and is divisible by the square of a prime. Trying cases for $ k \\equal{} 1$ to $ k \\equal{} 3$, we see that $ 7$ and $ 13$ are also numbers we are looking for.\n\nThus the numbers we are looking for are $ 2$, $ 3$, $ 5$, $ 7$, and $ 13$.\n[/hide]", "Solution_3": "Case 1 seems fine, however:\r\n[quote=\"Math4tots\"]\nCase 2: p is a prime of the form $ 3^k \\plus{} 4$.\nIn this case, let us consider $ k \\ge 4$. Then $ p \\minus{} 7^2 \\equal{} 3^k \\minus{} 45 \\equal{} 9(3^{k \\minus{} 2} \\minus{} 5)$. Since $ k \\ge 4$, the expression is positive, and is divisible by the square of a prime. Trying cases for $ k \\equal{} 1$ to $ k \\equal{} 3$, we see that $ 7$ and $ 13$ are also numbers we are looking for.\n[/quote]\r\nThis is not very clear to me. What do you take $ q$ and $ r$ to be in this case? If not the proof itself, at least the exposition seems to be flawed.", "Solution_4": "This is not a problem too difficult.\r\nI will try to make a resolution in the shortest time(by the busy work :lol: )", "Solution_5": "math10 may have given up, so here's my solution.\r\n\r\nWe prove that if $ p > 13$ is prime, there is a prime $ q < p$ such that the least residue of $ p$ modulo $ q$ is not square-free, which is the property described in the problem. In fact, we prove a stronger statement: if $ p>13$ is odd (not necessarily prime), there is a prime $ q$ such that the least residue of $ p$ modulo $ q$ is either $ 4$, $ 8$ or $ 9$. \r\n\r\nSuppose that some odd $ p>13$ is a counterexample: there's no prime $ q < p$ for which the remainder upon dividing $ p$ by $ q$ belongs to $ \\{4,8,9\\}$. \r\n\r\nIf there is a prime $ q > 3$ diving $ p\\minus{}4$, then $ 4$ is the least residue of $ p$ modulo $ q$, which contradicts our assumption on $ p$. Since $ p$ is odd, $ p\\minus{}4$ is also not divisible by $ 2$, so the only prime divisor of $ p\\minus{}4$ must be $ 3$. In other words, $ p\\minus{}4 \\equal{} 3^k$ for some natural number $ k$.\r\n\r\nConsider now $ p\\minus{}8$. If it has any prime factor greater than $ 8$, we once again contradict our assumption on $ p$. Thus we may assume that all of its prime factors belong to $ \\{2,3,5,7\\}$. But $ 2$ is impossible since $ p\\minus{}8$ is odd, and $ 3$ is impossible since $ p\\minus{}4$ is divisible by $ 3$. Thus $ p\\minus{}8$ is divisible only by $ 5$, $ 7$ or both. \r\n\r\nNext consider $ p\\minus{}9$. For the same reason as above, it too can only have prime factors among $ \\{2,3,5,7\\}$. Since $ p\\minus{}4$ is divisible by $ 3$ and not by $ 5$, $ p\\minus{}9$ is divisible by neither $ 3$ nor $ 5$. It may or may not be divisible by $ 7$, but we claim that in either case our previous friend $ p\\minus{}8$ cannot be divisible by $ 7$. Here's why: If $ p\\minus{}9$ is divisible by $ 7$, then $ p\\minus{}8$ certainly isn't. If $ p\\minus{}9$ is not divisible by $ 7$, then it must be a power of $ 2$ (since no other prime factors are available). But powers of $ 2$ can only leave remainders of $ 1, 2, 4$ upon division by $ 7$, so $ p\\minus{}8$ which is one greater than a power of $ 2$ cannot be divisible by $ 7$. \r\n\r\nRecall that the only possible prime factors of $ p\\minus{}8$ were $ 5$ and $ 7$, and we just ruled out $ 7$. We may therefore conclude that $ p\\minus{}8$ is a power of $ 5$: $ p\\minus{}8 \\equal{} 5^m$. Comparing this with $ p\\minus{}4 \\equal{} 3^k$ we obtain:\r\n\r\n(*) $ 5^m \\plus{} 4 \\equal{} 3^k$\r\n\r\nWe conclude the proof by showing that (*) only has the trivial solution $ 5\\plus{}4 \\equal{} 3^2$, which corresponds to $ p\\equal{}13$. Reducing mod $ 4$ we see that $ k$ must be even, so $ k \\equal{} 2t$ for some $ t$. Now $ 5^m \\equal{} 3^{2t} \\minus{} 4 \\equal{} (3^t\\minus{}2)(3^t\\plus{}2)$. But these factors cannot both be divisible by $ 5$ (they differ by $ 4$), so the only way their product can be a power of $ 5$ is if one of them is $ 1$ and the other is a power of $ 5$. This forces $ t\\equal{}1$, $ k\\equal{}2$ as required. QED", "Solution_6": "[quote=randomgraph]Case 1 seems fine, however:\n[quote=\"Math4tots\"]\nCase 2: p is a prime of the form $ 3^k \\plus{} 4$.\nIn this case, let us consider $ k \\ge 4$. Then $ p \\minus{} 7^2 \\equal{} 3^k \\minus{} 45 \\equal{} 9(3^{k \\minus{} 2} \\minus{} 5)$. Since $ k \\ge 4$, the expression is positive, and is divisible by the square of a prime. Trying cases for $ k \\equal{} 1$ to $ k \\equal{} 3$, we see that $ 7$ and $ 13$ are also numbers we are looking for.\n[/quote]\nThis is not very clear to me. What do you take $ q$ and $ r$ to be in this case? If not the proof itself, at least the exposition seems to be flawed.[/quote]\n\nThis finish is somewhat similar.\nTake $r=5^2$, $q =$ a prime that divides $3^{k \\minus{} 1} \\minus{} 7 = \\frac{p-5^2}{3}$ that is greater than $5$. Since $k \\geq 4$ the number $3^{k-1}-7 >1$ and $3 \\not |5$, so it is not a power of 3, and if $5|3^{k-1}-7$ then $5|p$, contradiction. Thus if $q$ does not exist, then $3^{k-1}-7$ is a power of 2. However $3^{k-1}-7 \\equiv 4,2 (mod 8),$ so $3^{k-1}-7 \\leq 4 \\implies k<4$, contradiction. Now finish as in the quote by considering cases $k\\in [1,3]\\cup \\mathbb{N}$." } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ G \\equal{} \\langle g_1, \\ldots, g_r \\rangle$ be a finitely generated (not necessarily finite) group, en let $ N$ be an abelian normal subgroup of $ G$. Show that every element of $ [G,N]$ can be written in the form\r\n\r\n\\[ g \\equal{} [g_1,n_1]\\ldots [g_r,n_r]\\]\r\n\r\nfor certain elements $ n_1, \\ldots, n_r \\in N$. Note that $ g_1, \\ldots, g_r$ are fixed.", "Solution_1": "[quote=\"Jan\"]Let $ G \\equal{} \\langle g_1, \\ldots, g_r \\rangle$ be a finitely generated (not necessarily finite) group, en let $ N$ be an abelian normal subgroup of $ G$. Show that every element of $ [G,N]$ can be written in the form\n\\[ g \\equal{} [g_1,n_1]\\ldots [g_r,n_r]\n\\]\nfor certain elements $ n_1, \\ldots, n_r \\in N$. Note that $ g_1, \\ldots, g_r$ are fixed.[/quote]\r\n\r\nLet $ [a,b] \\equal{} aba^{ \\minus{} 1}b^{ \\minus{} 1}.$ Forget for now that G is finitely generated. Show that $ [a_1a_2,n_1] \\equal{} [a_1,a_2n_1a_2^{ \\minus{} 1}][a_2,n_1],$ and $ [a_1, n_1][a_1,n_2] \\equal{} [a_1,n_1n_2],$ for all $ a_i \\in G, n_i \\in N.$ \r\n\r\nNow your problem should be easy." } { "Tag": [ "Gauss", "induction", "modular arithmetic", "algebra", "polynomial", "pen", "number theory" ], "Problem": "prove that for every natural $n$ \r\n\r\n$x^2+y^2+1 \\equiv 0$ \r\n\r\nhas solution.", "Solution_1": "Didn't you mean $x^2+y^2+1 \\equiv 0$ $mod$ $n$ ?", "Solution_2": "I think so!!", "Solution_3": "what if n=4? :?", "Solution_4": "[quote=\"amir2\"]prove that for every natural $n$ \n\n$x^2+y^2+1 \\equiv 0$ \n\nhas solution.[/quote]\r\n\r\nWell it's a very nice question. I think the result should be \r\n\r\nProblem: Let $n$ be a positive integer. Prove that $4\\nmid n$ if and only if there exist $x,y$ integers so that $n\\mid x^2+y^2+1$\r\n\r\nHere's my very very ugly solution:\r\n\r\nIf $4\\mid n$, as a comment of al. M.V , no such pair $(x,y)$ exist!\r\n\r\nNow let me consider the case $4\\nmid n$ we will so that $n|x^2+y^2+1$ for some $x,y$.\r\n\r\n1) $n=p$ where $p$ is an odd prime : then $p$ or $2p$ is not in the form $4^k(8r+7)$. So by Gauss Theorem $p$ or $2p$ can be writen in the form $x^2+y^2+z^2$. Of cause there one of $x,y,z$; namely $z$ is relative to $p$ ie... $\\gcd(p,z)=1$. By muliple the conveser of $z$ modulo $p$ we get the problem is true in that case.\r\n\r\n2) $n=p^a$ for any $a$ is positive integer. Now use induction on $a$; just put $x_{a+1}=x_a+tp^a$ for $t$ is integer we get: the problem is true in the case\r\n\r\n3) if $n=2$ then chose $x=0$ and $y=z=1$ so problem is true in that case\r\n\r\n4) At last use Chines Remainder Theorem so if $4\\nmid n$ then such pair $(x,y)$ exists. \r\n\r\n\r\nIt's ugly but i hope it's clear :D", "Solution_5": "It's not ugly, but you can avoid aplying Gauss' Theorem. If we want to show that we can find $x,y$ s.t. $p|x^2+y^2+1$, it suffices to notice that the sets $\\{x^2\\pmod p|x\\in\\overline{0,p-1}\\}$ and $\\{-1-y^2\\pmod p|y\\in\\overline{0,p-1}\\}$ together have $p+1$ elements, so they must have an element in common.", "Solution_6": "All you need is Chevalley's theorem and you are done.\r\n\r\nBomb", "Solution_7": "In LIV Polish Math Olympiad we had a problem:\r\nProve that for every prime $p>3$ there exist integers $x,y,k$ with $0<2k by angle chasing $ \\angle ABB_1\\equal{}\\angle B_1A_1F$ and $ \\angle CBB_1\\equal{}\\angle C_1B_1B\\equal{}\\angle C_1AG\\equal{}\\angle AA_1B_1$ hence result.\r\n2. ABC=AA1F => again by angle chasing we prove that AB1GC1 is cyclic (prove is similar to 1 so I leave it to you)." } { "Tag": [ "Pascal\\u0027s Triangle", "geometric sequence" ], "Problem": "I've been looking at pascal's triangle and the binomial expansion theorem and all that kind of stuff, and I've been trying to figure out if something works. But I got stuck with this expression-\r\n\r\ny:^3:/x:^2: - y:^4:/x:^3: + y:^5:/x:^4: - y^6/x:^5: + .... \r\n\r\n- because I don't think I know how to solve things like this. Except I might, I got an answer, but it isn't the one I hoped I would get. So if I\"m right, then I'm wrong, but if I'm wrong then I might be right :lol: if that makes any sense. So is this simplifiable to a non \"...-ing\" expression?", "Solution_1": "Think geometric progression.\r\n\r\na + ar + ar^2 + ar^3... = a/(1-r) as long as |r| < 1.", "Solution_2": "[quote=\"TripleM\"]Think geometric progression.\n\na + ar + ar^2 + ar^3... = a/(1-r) as long as |r| < 1.[/quote]\r\n\r\nIt's indeed an Infinite geometric progression with a = y^3/x^2\r\n\r\nand r= -y/x.\r\n\r\nThe series can be summed only if |-y/x| <1\r\n\r\ni.e if |y| < |x| and x !=0\r\n\r\nunder the above restriction, the sum of the series , as stated by TripleM is\r\n\r\ny^3/x^2/(1 + y/x) = y^3/x(x+y)" } { "Tag": [ "puzzles" ], "Problem": "There are four grades A, B, C, D. A is the best and D is the worst. Four students Jim, Jane, Peter and Lilian appeared in an examination. After the results were published , they made the following statements to a common acquaintance:\r\n\r\n Jim\u2019s Statement: : If I have got A then Lilian has got D. \r\n\r\n Lilian\u2019s Statement: : (i) If I got C then Jim has got D; (ii) Jim\u2019s grade is better than Peter\u2019s grade. \r\n\r\nJane\u2019s Statement: If I didn\u2019t get A then Jim\u2019s grade is not A. \r\n\r\nPeter\u2019s Statement : If Jim\u2019s grade is A, then Jane\u2019s grade is not B, Lilian \u2019s grade is C, but my grade is neither A nor B.\r\n\r\nCan you help the acquaintance to determine the grades of each of the four students, given that all the above statements are true?", "Solution_1": "I think there are a lot of possibilities.", "Solution_2": "A possibility is\r\n\r\nJim:C\r\nJane:A\r\nLillian:B\r\nPeter:D", "Solution_3": "[hide]If Jim has an A, then Jane does not have an A. By Jane's statement, this means that Jim doesn't have an A, which leads to contradiction. This means that Jim does not have an A.\n\nBecause of Lilian's second statement, Jim doesn't have a D.\n\nThus, Jim can have either a B or a C.\n\nLet's assume that Jim has a C. This means that Peter has a D. Jane and Lilian got the A and B. Thus, two solutions are:\nJim: C\nLilian: A\nJane: B\nPeter: D\n\nJim: C\nLilian: B\nJane: A\nPeter: D\n\nLet's assume that Jim has a B. This means Peter can have a C or a D. Let's assume that Peter has a C. This means that Lilian and Jane have the A and D. Thus, here are two more solutions.\n\nJim: B\nLilian: A\nJane: D\nPeter: C\n\nJim: B\nLilian: D\nJane: A\nPeter: C\n\nLet's assume that Peter has a D. This means that Lilian and Jane have the A and C. If Jane has the C, then Jim has the D, which is a contradiction. So, the final solution is:\n\nJim: B\nLilian: A\nJane: C\nPeter: D\n\nThere are 5 solutions altogether.[/hide]", "Solution_4": "OR...\r\n\r\n[hide]They all are enemies and they lie about their grades. :rotfl: [/hide]", "Solution_5": "[quote=\"K Sengupta\"]There are four grades A, B, C, D. A is the best and D is the worst. Four students Jim, Jane, Peter and Lilian appeared in an examination. After the results were published , they made the following statements to a common acquaintance:\n\n Jim\u2019s Statement: : If I have got A then Lilian has got D. \n\n Lilian\u2019s Statement: : (i) If I got C then Jim has got D; (ii) Jim\u2019s grade is better than Peter\u2019s grade. \n\nJane\u2019s Statement: If I didn\u2019t get A then Jim\u2019s grade is not A. \n\nPeter\u2019s Statement : If Jim\u2019s grade is A, then Jane\u2019s grade is not B, Lilian \u2019s grade is C, but my grade is neither A nor B.\n\nCan you help the acquaintance to determine the grades of each of the four students, given that all the above statements are true?[/quote] they can't all be true.\r\njim says that if he gets an A then lilian gets a d and peter says that if jim gets an a then lilian got a c they're contidicting eachother..." } { "Tag": [], "Problem": "A slug climbs ten inches in ten minutes. It then rests two minutes. It continues climbing at a constant rate and rests for two minutes after climbing ten minutes. How many minutes will it take the slug to reach the top of a twenty-foot tower?", "Solution_1": "20 ft. = 240 inches. Therefore, it will take the slug 24 periods of climbing, which means 23 breaks in between, making it $ 240\\plus{}46\\equal{}\\boxed{286}$" } { "Tag": [ "trigonometry" ], "Problem": "simplify: ((sin x)^4+4((cos x)^2))^.5 +((cos x)^4 + 4((sin x)^2))^.5 \r\n(i dont have the answer)", "Solution_1": "Well it can be writen as $ \\sin^{4}{x}\\plus{}\\cos^{4}{x}\\plus{}4(\\sin{x}\\plus{}\\cos{x})$. I'm not sure how much farther though.", "Solution_2": "It might be helpful to also post the entire problem as well..", "Solution_3": "$ \\sqrt{\\sin^4 x \\plus{} 4 \\cos^2 x} \\plus{} \\sqrt{\\cos^4 x \\plus{} 4 \\sin^2 x}$\r\n$ \\equal{} \\sqrt{\\sin^4 x \\plus{} 4(1 \\minus{} \\sin^2 x)} \\plus{} \\sqrt{\\cos^4 x \\plus{} 4(1 \\minus{} \\cos^2 x)}$\r\n$ \\equal{} \\sqrt{(\\sin^2 x)^2 \\minus{} 4 \\sin^2 x \\plus{} 4} \\plus{} \\sqrt{(\\cos^2 x)^2 \\minus{} 4 \\cos^2 x \\plus{} 4}$\r\n$ \\equal{} \\sqrt{(\\sin^2 x \\minus{} 2)^2} \\plus{} \\sqrt{(\\cos^2 x \\minus{} 2)^2}$\r\n$ \\equal{} |\\sin^2 x \\minus{} 2| \\plus{} |\\cos^2 x \\minus{} 2|$\r\n$ \\equal{} 2 \\minus{} \\sin^2 x \\plus{} 2 \\minus{} \\cos^2 x$\r\n$ \\equal{} 4 \\minus{} (\\sin^2 x \\plus{} \\cos^2 x)$\r\n$ \\equal{} 4 \\minus{} 1$\r\n$ \\equal{} 3$.\r\n\r\nNote that for all real $ x$, $ |\\sin x| \\le 1$ and $ |\\cos x| \\le 1$, from which it follows that $ |\\sin^2 x \\minus{} 2| \\equal{} 2 \\minus{} \\sin^2 x$, and so forth." } { "Tag": [], "Problem": "$ a) \\frac {6.19}{6.24 \\minus{} 3.12\\times 2}$\r\n\r\n$ b) \\frac {7.8}{ \\minus{} 5.64 \\minus{} 3.1233}$\r\n\r\n$ c) \\frac {2.4\\div 3}{0.6 \\minus{} 1.8\\div 3}$\r\n\r\n$ c) \\frac {3.4\\times 1.4}{ \\minus{} 1.8 \\minus{} 3\\frac {2}{3}\\times ( \\minus{} 2)}$\r\n\r\n :wink: No calculator !", "Solution_1": "This is just a matter of plugging it into your calculator, I'm guessing.\r\n\r\n[hide=\"answers\"]\na) undefined, because $ 6.24\\minus{}3.12\\cdot2\\equal{}0$\nb)$ \\minus{}.8901$\nc) again, undefined\nd) $ .7843$[/hide]", "Solution_2": "I thought it would be a good idea to bring up why A and C are undefined. You can [b]never[/b] divide a number by 0 because multiplication and divsion are interchangable meaning that the product of the divisor and quotient will equal the dividend, and in mutliplication of 2 numbers the product divided by one of te number will always equal the other number.", "Solution_3": "SOLUTION OF a is\r\n6.19/0 =infinity :rotfl:", "Solution_4": "[quote=\"salahcool\"]SOLUTION OF a is\n6.19/0 =infinity :rotfl:[/quote]\r\nYes. The positive infinity.\r\nSee the endless ascending sequence $ \\frac{1}{2}, \\frac{1}{1}, \\frac{1}{\\frac{1}{2}}, \\frac{1}{\\frac{1}{3}}, \\frac{1}{\\frac{1}{4}}, \\frac{1}{\\frac{1}{10}}, \\frac{1}{\\frac{1}{100}}, \\frac{1}{\\frac{1}{1000}} ... \\frac{1}{0}$\r\nThen fewer is a discriminant then bigger is a fraction." } { "Tag": [ "integration", "trigonometry", "calculus", "calculus computations" ], "Problem": "Find the volume of the solid enclosed by the cylinder $ {{x}^{2}}\\plus{}{{y}^{2}}\\equal{}9$ and the planes y + z = 5 and z = 1. Help me, please.", "Solution_1": "$ \\begin{gathered}\r\n z \\plus{} y \\equal{} 5,{\\text{ }}z \\equal{} 1,{\\text{ }}{x^2} \\plus{} {y^2} \\equal{} 9. \\hfill \\\\\r\n D \\equal{} \\left\\{ {\\left. {\\left( {x,y} \\right)} \\right|{\\text{ }} \\minus{} \\sqrt {9 \\minus{} {x^2}} \\leqslant y \\leqslant \\sqrt {9 \\minus{} {x^2}} ,{\\text{ }} \\minus{} 3 \\leqslant x \\leqslant 3} \\right\\}. \\hfill \\\\\r\n V \\equal{} \\iint\\limits_D {dxdy}\\int\\limits_1^{5 \\minus{} y} {dz} \\equal{} \\iint\\limits_D {\\left( {4 \\minus{} y} \\right)dxdy} \\equal{} \\int\\limits_{ \\minus{} 3}^3 {dx} \\int\\limits_{ \\minus{} \\sqrt {9 \\minus{} {x^2}} }^{\\sqrt {9 \\minus{} {x^2}} } {\\left( {4 \\minus{} y} \\right)dy} \\equal{} \\hfill \\\\\r\n \\equal{} 2\\int\\limits_0^3 {\\left. {\\left( {4y \\minus{} \\frac{{{y^2}}}\r\n{2}} \\right)} \\right|_{ \\minus{} \\sqrt {9 \\minus{} {x^2}} }^{\\sqrt {9 \\minus{} {x^2}} }dx} \\equal{} 16\\int\\limits_0^3 {\\sqrt {9 \\minus{} {x^2}} dx} \\equal{} \\left\\{ \\begin{gathered}\r\n x \\equal{} 3\\sin t,{\\text{ }}dx \\equal{} 3\\cos tdt, \\hfill \\\\\r\n 0 \\leqslant x \\leqslant 3,{\\text{ }}0 \\leqslant t \\leqslant \\frac{\\pi }\r\n{2} \\hfill \\\\ \r\n\\end{gathered} \\right\\} \\equal{} \\hfill \\\\\r\n \\equal{} 48\\int\\limits_0^{\\pi /2} {\\sqrt {9 \\minus{} 9{{\\sin }^2}t} \\cos tdt} \\equal{} 144\\int\\limits_0^{\\pi /2} {{{\\cos }^2}tdt} \\equal{} 72\\int\\limits_0^{\\pi /2} {\\left( {1 \\plus{} \\cos 2t} \\right)dt} \\equal{} \\hfill \\\\\r\n \\equal{} \\left. {72\\left( {t \\plus{} \\frac{1}\r\n{2}\\sin 2t} \\right)} \\right|_0^{\\pi /2} \\equal{} 36\\pi {\\text{ }}\\left( {{\\text{cubic units}}} \\right). \\hfill \\\\ \r\n\\end{gathered}$" } { "Tag": [ "limit", "integration", "inequalities", "calculus", "calculus computations" ], "Problem": "Find $ \\lim_{n\\to\\infty} \\frac{1}{n^2} \\sum_{k\\equal{}1}^n \\sqrt[4]{k(k\\plus{}1)(k\\plus{}2)(k\\plus{}3)}$.", "Solution_1": "[color=darkred] \\lim_{n\\to\\infty} \\frac{1}{n^2} \\sum_{k=1}^n \\sqrt[4]{k(k+1)(k+2)(k+3)}=\\frac{1}{2}[/color]", "Solution_2": "[quote=\"shubin\"]$ \\lim_{n\\to\\infty} \\frac{1}{n^2} \\sum_{k\\equal{}1}^n \\sqrt[4]{k(k\\plus{}1)(k\\plus{}2)(k\\plus{}3)}\\equal{}\\frac{1}{2}$[/quote]\r\n\r\nHow did you get it?", "Solution_3": "[quote=\"kunny\"][quote=\"shubin\"]$ \\lim_{n\\to\\infty} \\frac {1}{n^2} \\sum_{k \\equal{} 1}^n \\sqrt [4]{k(k \\plus{} 1)(k \\plus{} 2)(k \\plus{} 3)} \\equal{} \\frac {1}{2}$[/quote]\n\nHow did you get it?[/quote]\r\n$ \\int\\limits_{0}^{1} \\sqrt[4]{(x^2\\plus{}3x\\plus{}1)^2\\minus{}1}dx \\equal{}\\frac{1}{2}$", "Solution_4": "[quote=\"Hong Quy\"]$ \\int\\limits_{0}^{1} \\sqrt [4]{(x^2 \\plus{} 3x \\plus{} 1)^2 \\minus{} 1}dx \\equal{} \\frac {1}{2}$[/quote]\r\n\r\n$ \\int_{0}^{1} x \\, dx \\equal{} \\frac {1}{2}$.\r\n\r\nOne may prove like this:\r\n\r\n$ \\frac {1}{n^2} \\sum_{k \\equal{} 1}^n \\sqrt [4]{k(k \\plus{} 1)(k \\plus{} 2)(k \\plus{} 3)}$ $ \\equal{} \\frac {1}{16} \\sum_{k \\equal{} 1}^n \\sqrt [4]{\\left(\\frac {4k}{n}\\right)\\left(\\frac {4k \\plus{} 1}{n}\\right)\\left(\\frac {4k \\plus{} 2}{n}\\right)\\left(\\frac {4k \\plus{} 3}{n}\\right)} \\cdot \\frac {4}{n}$ $ \\equal{} \\frac {1}{16} \\sum_{k \\equal{} 1}^n \\frac {4k \\plus{} c_k}{n} \\cdot \\frac {4}{n}$\r\n\r\nfor some $ 0 < c_k < 3$, by intermediate value theorem applied to the inequality\r\n\r\n$ \\frac {4k}{n} < \\sqrt [4]{\\left(\\frac {4k}{n}\\right)\\left(\\frac {4k \\plus{} 1}{n}\\right)\\left(\\frac {4k \\plus{} 2}{n}\\right)\\left(\\frac {4k \\plus{} 3}{n}\\right)} < \\frac {4k \\plus{} 3}{n}.$\r\n\r\nThen it converges to $ \\frac {1}{16}\\int_{0}^{4} x \\, dx \\equal{} \\int_{0}^{1} t \\, dt$." } { "Tag": [ "induction", "graph theory", "combinatorics proposed", "combinatorics" ], "Problem": "There is a complete Graph $K_n$ and all its edges are colored in two colors, prove there exists a Hamilton Cycle such that going arround it one only change of color once.", "Solution_1": "Ok, I hope it's ok.....\r\nSo, just do it by induction...I mean, consider a bicolored $K_{n+1}$. Take some $K_n$ of it, apply induction, and the vertex left is $A$. If the cycle is of the form $A-X_1-B-X_2-C$, where all the edges in the $A$ and $C$ chains are of the same color, and $X_i$ are \"intermediate\" vertices. Now just consider the color of $A-X_i$ and look to the left or right of it.....\r\nI hope I didn't miss any cases..... :oops:" } { "Tag": [ "geometry", "parallelogram", "circumcircle", "analytic geometry", "function" ], "Problem": "Does anyone know any general theorems that give you information as to how a line can bisect the area of a polygon? I think I have seen such a thm. before, but I thought it was too difficult to learn at the time.", "Solution_1": "I found one. Can anyone find the proof?\r\nIf a line bisects the area of a parallelogram, then the line passes through the intersection of the diagonals of the parallelogram.", "Solution_2": "if it is a regular polygon, a line through the circumcenter suffices, one exercise related exercise that you should work on is: prove that such a line exists\r\n[hide=\"hint\"]continuity[/hide]", "Solution_3": "I'll followed your hint Altheman. Tell me if you see anything incorrect.\r\nThm: There exists atleast one line that bisects the area of a concave polygon.\r\nProof: Consider a concave polygon $A$ and an arbitrary line $l_k$ situated in the coordinate plane. Furthermore, let $l_k$ be defined by the equation $y=mx+k$ where $m$ is a constant. Finally, note that $A$ and $l_k$ are infinite sets of pts.\r\nNow define $A(k)$ to equal the lower area bounded by $A$ and $l_k$. We see that, for different values of k, $A(k)$ ranges from $0$ to $A$'s entire area. It is also easy to see that $A(k)$ is a continuous function of $k$ since $A$ is concave.\r\nThus, if we let $k_0$ be the value of $k$ that makes $A(k)=0$, and $k_t$ be the value of $k$ that makes $A(k)$ equal the total area of $A$, then there exists some value $k'$ such that $k_02, O2, CO2, Ar, and other trace elements. Anyway, that stuff is all solutions because if you heat/cool it enough (depending on what it is), it will be a liquid with a bunch of components. Well, that's a nice way of thinking of it at least.\r\n\r\nchicken: a) density b) \"broth\" that's cooled down enough to congeal c) The fat and broth will have different melting points. Figure it out from there. ...or use a spoon.", "Solution_1": "The first part looks right. For the second part, just use what you know about homogeneous mixtures -- in air, is the nitrogren one big sea with little blobs of oxygen floating in it, or is it all mixed in together?\r\n\r\nFor the third part, a you should figure out yourself, b I'm not sure what they want, and c you should be able to do by thinking about liquids and solids.", "Solution_2": "Very standard questions of science...intro to chemistry...\r\n\r\nFor your first question regarding why the air we breathe and steel is homogeneous, think about the definition of homogeneous. \r\n\r\nI'll show ou an example:\r\n\r\nIf you took a salad, I bet you could pick out the lettuce and cabbage for me. This is a heterogeneous mixture because you could easily seperate the parts by physical means. Now, suppose you put that salad into a blender. (Not that you would want to) When it comes out, all the parts will look the same. At this point, I bet you couldn't diffrentiate between lettuce and cabbage (even though before they entered the blender, they were totally different). This is a homogeneous mixture.\r\n\r\nThe same thing applies to air and steel like your teacher put it. If I gave you a sample of our atmosphere (and you could see its parts) you would not be able to reach out and grab the carbon dioxide or the oxygen. It's mixed so well that you'd have to be close to the atomic level to see different parts. This is why it's a homogeneous mixture. Same thing with the steel. It is so well mixture that you could not tell what is what.\r\n\r\nThe only difference between a homogeneous mixture and a solution is that in a solution, there is a solvent and a solute. And one or more substances must be completely dissolved into another. Solutions usually refer to liquids.\r\n\r\nFor your other problem,\r\n\r\na) Pretty much common sense. The fat rose to the top b/c it is the least dense of the substances originally in the chicken soup (aka. the mixture).\r\n\r\nb) The gelatinous mixture is basically the rest of the mixture that the fat seperated out of. There may even be some bits of chicken in it still.\r\n\r\nc) This is also common sense. Just take a knife (or a fork even) and cut off the fat at the top and remove it. It is basically already seperated. All you have to do is seperate the layers and remove the fat. The more professional way would be to take a tiny tube and suck the fat into it. This is how most chemists remove certain substances from a mixture. There are other ways which you will learn in high school chemistry", "Solution_3": "Question: For b in the first part if the coffee is cold enough then you couldn't you argue it's heterogeneous because the sugar content near the bottom is higher than the top and thus the sugar is not spread throughout the coffee equally?", "Solution_4": "But it's not really seperable -- you couldn't get the sugar out. (I don't remember exactly the definitions of homogeneous and heterogeneous . . . I think I just went on the basis of \"if it looks homogeneous, it probably is.\")", "Solution_5": "Well, what do [i]you[/i] think will happen?", "Solution_6": "[quote=\"Tare\"]Question: For b in the first part if the coffee is cold enough then you couldn't you argue it's heterogeneous because the sugar content near the bottom is higher than the top and thus the sugar is not spread throughout the coffee equally?[/quote]\r\n\r\nwell sure...you could argue it that way...but ppl would still call it homogeneous...\r\n\r\nYou could also argue that if you were to raise the temperature of the mixture enough, then the coffee would evaporate and you could collect the gas and cool is again. This way you will definitely have seperated coffee and sugar since the boiling point of sugar is much much higher than that of water.\r\n\r\nHowever, I think we all know what he meant by coffee, cream, and sugar.", "Solution_7": "Maybe you could do a little experimenting with sugar and water.", "Solution_8": "hint- when you make coffee, when you make tea, and you put sugar into it, the coffee and the tea... they are hot. so you can put lots and lots of sugar into it. wow that was a really obvious hint. sry.", "Solution_9": "[quote=\"Rep123max\"]it starts to turn into a solid (im scientifically challenged)[/quote]\r\nMmm I was going to say the same thing until I realized that in order to put the maximum amount of sugar in water and melt it it has to be hot/warm so it's probably just going to become cold and thus some of the sugar will not dissolve anymore and sink to the bottom.", "Solution_10": "Tare, sugar does not melt in hot water. It dissolves.\r\n\r\nSpeaking of which, I should try the supersaturation experiment with sugar, since it didn't work with salt (fairly predictable, it turns out)", "Solution_11": ":oops: just keep in mind all of this I just know and that I didn't learn any of this...\r\n\r\nYou actually conducted an experiment? how?", "Solution_12": "Mmmm, take some water and dissolve a cystalline solid of some sort in it. Salt is a bad choice. Keep adding and stirring until the solid will not dissolve anymore and stays on the bottom [i]in solid form[/i]. Heat the mixture while stirring until all the crystals are completely dissolved. If you add too much in the first step, you won't be able to get it to dissolve. If you use some exotic substance, solubility may not increase with heat, which would also be a problem. Anyway, once it's all dissolved, allow the solution to cool slowly. In theory, the solute should stay dissolved, creating a supersaturated solution. It doesn't work with salt because the solubility of salt in water doesn't increase enough with temperature. Once you have the supersaturated solution and it's cooled back down to room temperature, if you add a [i]single grain[/i] of the solid crystal to the mixture, all of the excess crystals that were originally laying on the bottom should drop out of solution. I've seen pictures in the chemistry book, but haven't gotten it to work yet. I need to try it with sugar, but first I should make sure we have enough that we won't run out.", "Solution_13": "Ooooh...I did something like that for my fourth or third grade science report for Japanese school over at San Jose and I did solutability testing of sugar and water within different temperatures of water but I've never tried that...*looks in pantry* mmm, must ransack Cooking Class :twisted:", "Solution_14": "homogeneous -> cannot be separated by physical methods, i.e. filtering.", "Solution_15": "*nods rippledance*...thats a very textbooky definition though", "Solution_16": "Basically, the solubility of liquids increases as the temperature increases. That's a basic textbook principle. See what you can do with that.\r\n\r\nBTW, does IPC stand for \"introduction to physical chemistry\"? I can't imagine what else it stands for.", "Solution_17": "Hmm, this reminds me of the first few weeks of Earth Science class when we were defining basic stuff like solution.\r\nMixture: Can be separated by physical means.\r\nSolution: Can not be separated by physical means.\r\n\r\nSo Mixture=heterogeneous and solution=homogeneous, si?" } { "Tag": [ "geometry open", "geometry" ], "Problem": "Hello everyone,\r\nWhen can a regular polygon have n vertices,i.e., for which n there exists a regular polyhedra of n vertices.\r\nThis question came up due to a question posed by the ALBANIAN EAGLE in reference to the question 'What are the number of ways of arranging n objects on a polyhedra with n vertices',which I posted in combinatorics open problem section.\r\nThank You\r\nHave a nice time. :?: :)", "Solution_1": "Let me just refer you to http://en.wikipedia.org/wiki/Regular_polyhedra . From an elementary viewpoint, your question is pretty hard to formalize - what exactly is a regular polyhedron? What is a polyhedron? The simplest definition that comes into my mind uses some basic linear optimization theory.\r\n\r\n darij", "Solution_2": "[quote=\"darij grinberg\"]Let me just refer you to http://en.wikipedia.org/wiki/Regular_polyhedra . From an elementary viewpoint, your question is pretty hard to formalize - what exactly is a regular polyhedron? What is a polyhedron? The simplest definition that comes into my mind uses some basic linear optimization theory.\n\n darij[/quote]\r\n\r\nWell, got the answer for the number of symmetries of a regular polygon from http://mathworld.wolfram.com/RegularPolyhedron.html\r\n\r\nAnyway Thanks for refering to the http://en.wikipedia.org/wiki/Regular_polyhedra .\r\n :) :)" } { "Tag": [], "Problem": "Let $x,y,z>0$ such that $xyz=1$. Prove that: \\[x+y^{2}+z^{3}> 2.5.\\]", "Solution_1": "[hide]\n$x+y^{2}+z^{3}=\\frac{1}{6}x+\\frac{1}{6}x+\\frac{1}{6}x+\\frac{1}{6}$ $x+\\frac{1}{6}x+\\frac{1}{6}x+\\frac{1}{3}y^{2}+\\frac{1}{3}y^{2}+\\frac{1}{3}y^{2}+\\frac{1}{2}z^{3}+\\frac{1}{2}z^{3}$$\\ge 11 \\sqrt[11]{\\frac{x^{6}y^{6}z^{6}}{6^{6}\\cdot 3^{3}\\cdot 2^{2}}}>2.5$[/hide]", "Solution_2": "[quote=\"hydro\"][hide]\n$x+y^{2}+z^{3}=\\frac{1}{6}x+\\frac{1}{6}x+\\frac{1}{6}x+\\frac{1}{6}$ $x+\\frac{1}{6}x+\\frac{1}{6}x+\\frac{1}{3}y^{2}+\\frac{1}{3}y^{2}+\\frac{1}{3}y^{2}+\\frac{1}{2}z^{3}+\\frac{1}{2}z^{3}$$\\ge 11 \\sqrt[11]{\\frac{x^{6}y^{6}z^{6}}{4 \\cdot 9 \\cdot 36}}>2.5$[/hide][/quote]\r\n\r\n$6^{6}\\cdot3^{3}\\cdot2^{2}\\neq 4\\cdot9\\cdot36$. (It is still good enough to give the answer, though.)", "Solution_3": "sorry, edited :wink:" } { "Tag": [ "geometry", "LaTeX" ], "Problem": "I am taking a class from AoPS and need 5 things in a header, but I only know how to make 3 (rhead, chead, lhead). How do you add multiple lines in it :?:\r\nThanks,\r\nminicon", "Solution_1": "Just put \\\\ in your text eg \\chead{this is on \\\\ 2 lines}. You may need to increase the headheight using the geometry package.", "Solution_2": "That worked. Thanks!\r\n:lol: :)" } { "Tag": [ "modular arithmetic", "number theory", "number theory unsolved" ], "Problem": "Show that there exists a positive interger n such as all it's digit are 1 and it is divisible by 2003!", "Solution_1": "I mean 2003 .", "Solution_2": "Yes definitely.. there are infinitely many such numbers, the number $n = 2003$ can be replaced with any positive integer coprime with $10$ and $9$, and if you do not restrict yourself to the decimal system, the number can just be replaced with any positive integer coprime to the base, and another suitable number. (E.g. in base $5$, to get a number consisting of a string of ones that is divisible by $n$, $n$ must be coprime with $4$ as well.) In fact, he 'suitable number' condition can also be dropped, as we can always construct a number that is also divisible by that 'suitable number' as well. All these can be illustrated with the case of $n=2003$.\r\n\r\nFor the case of $n=2003$, since $n$ is prime, by Fermat's Little Theorem, $10^{2002} \\equiv 1 \\pmod{2003}$. The integer $\\frac{10^{2002}-1}{9}$ then is our desired number. In particular, the sequence $\\frac{10^{2002k}-1}{9}$ for $k = 1, 2, 3, \\ldots$ generates an infinite number of integers divisible by $2003$.", "Solution_3": "Indeed you don't need it to be coprime with something other than the base (like the $9$ here).\r\nAnd also indeed, any number coprime to the base works, but all these stuff was posted before.", "Solution_4": "[quote=\"dblues\"]$\\frac{10^{2002k}-1}{9}$ for $k = 1, 2, 3, \\ldots$ generates an infinite number of integers divisible by $2003$.[/quote] \r\nBecause $(\\frac{2}{2003})=-1 (2003=3(mod 8)),(frac{5}{2003})=(\\frac 35 )=-1\\Longrightarrow (\\frac{10}{2003})=1$.\r\nTherefore $2003| \\frac{10^{1001k}-1}{9}$ for $k=1,2,3,...$" } { "Tag": [], "Problem": "\u0393\u03b5\u03bd\u03af\u03ba\u03b5\u03c5\u03c3\u03b7 \u03bc\u03b9\u03b1\u03c2 \u03b1\u03c0\u03bf \u03c0\u03b1\u03bb\u03b9\u03bf \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\r\n\r\nA\u03bd $ a_{1}^{2}\\plus{}a_{2}^{2}\\plus{}....\\plus{}a_{n}^2\\equal{}c_1$ \u03ba\u03b1\u03b9 $ a_{1}\\plus{}a_{2}\\plus{}....\\plus{}a_{n}\\equal{}c_2$ \u03bc\u03b5 $ a_i>0, n>2$ \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b8\u03b5\u03af \u03c4\u03bf $ (a_1)_{max}$ \u03ba\u03b1\u03b9 $ (a_1)_{min}$ (\u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 \u03c5\u03c0\u03b1\u03c1\u03be\u03b7\u03c2 \u03b3\u03b9\u03b1 min)", "Solution_1": "Efarmozwntas tin anisotita Cauchy-Scharz gia tis $ n \\minus{} ades$ $ (1,1,...,1),(a_1,a_2,...,a_n)$ pernoume oti:\r\n\r\n $ n(a_1^2 \\plus{} a_2^2 \\plus{} ... \\plus{} a_n^2)\\geq(a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n)^2$\r\n or\r\n $ nc_1 \\geq c_2^2$ $ (1)$\r\n\r\n Twra,efarmozwntas tin Cauchy-Scharw gia tis $ (n \\minus{} 1) \\minus{} ades$ $ (1,1,...,1),(a_2,a_3,...,a_n)$ pernoume:\r\n\r\n $ (n \\minus{} 1)(a_2^2 \\plus{} ... \\plus{} a_n^2)\\geq(a_2 \\plus{} ... \\plus{} a_n)^2$ or\r\n $ (n \\minus{} 1)(c_1 \\minus{} a_1^2)\\geq(c_2 \\minus{} a_1)^2$ or\r\n $ na_1^2 \\minus{} 2c_2a_1 \\plus{} (c_2^2 \\minus{} nc_1 \\plus{} c_1)\\leq0$\r\n\r\nTheoroume to aristero melos tis anisotitas polyonymo ws pros $ a_1$.\r\n\r\nI diakrinousa tou polyonymou aftou einai:\r\n\r\n $ D \\equal{} 4c_2^2 \\minus{} 4n(c_2^2 \\minus{} nc_1 \\plus{} c_1)$ or\r\n $ D \\equal{} 4(1 \\minus{} n)(c_2^2 \\minus{} nc_1) > 0$ logw tis $ (1)$ kai epeidi $ n > 2$ \r\n\r\nEpeidi $ D > 0$ to polyonymo exei dyo anises pragmatikes rizes oi opoies kai antistoixoun me tin elaxisti kai megisti timi tou $ a_1$.\r\n\r\n Oi rizes tou polyonymou einai:\r\n\r\n\r\n $ \\frac {c_2 \\plus{} \\minus{} \\sqrt {(1 \\minus{} n)(c_2^2 \\minus{} nc_1)}}{n}$, oi opoies kai\r\neinai i megisti kai elaxisti timi antoistoixws tou $ a_1$.\r\n\r\n~Kostas", "Solution_2": "\u038c\u03bc\u03bf\u03c1\u03c6\u03b1, \u03b1\u03bb\u03bb\u03ac \u03b8\u03c5\u03bc\u03af\u03c3\u03bf\u03c5 \u03cc\u03c4\u03b9 $ a_i > 0$...\r\n\r\n\u0393\u03b9\u03b1 \u03c4\u03b1 $ c_1, c_2$ \u03b8\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $ c_1 > 0, c_2 > 0$ \u03ba\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 $ \\frac {c_2^2}{n} \\leq c_1 < c_2^2$ (\u03cc\u03c7\u03b9 \u03bc\u03cc\u03bd\u03bf $ nc_1\\geq c_2^2$).\r\n\r\n\u0393\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ac, \u03b7 \u03c0\u03c1\u03ce\u03c4\u03b7 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03af\u03b1 \u03c3\u03c6\u03b1\u03af\u03c1\u03b1 \u03bc\u03b5 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 $ \\sqrt {c_1}$ \u03b5\u03bd\u03ce \u03b7 \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03b1\u03bd\u03c4\u03b9\u03c0\u03c1\u03bf\u03c3\u03c9\u03c0\u03b5\u03cd\u03b5\u03b9 \u03ad\u03bd\u03b1 \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf \u03ba\u03ac\u03b8\u03b5\u03c4\u03bf \u03c3\u03c4\u03bf $ (1,1\\ldots 1)$ \u03ba\u03b1\u03b9 \u03c0\u03bf\u03c5 \u03c3\u03c5\u03bd\u03b1\u03bd\u03c4\u03ac\u03b5\u03b9 \u03c4\u03bf\u03c5\u03c2 \u03ac\u03be\u03bf\u03bd\u03b5\u03c2 \u03c3\u03c4\u03bf $ c_2$. \u0397 \u03c4\u03bf\u03bc\u03ae \u03c4\u03bf\u03c5\u03c2 \u03ac\u03c1\u03b1 \u03c3\u03c4\u03bf \u03b8\u03b5\u03c4\u03b9\u03ba\u03cc $ 2^n$-\u03bc\u03cc\u03c1\u03b9\u03bf \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03b1\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 (ok, \u03c3\u03c6\u03b1\u03af\u03c1\u03b1 :P) \u03c0\u03bf\u03c5 \u03bf\u03b9 \u03b1\u03ba\u03c1\u03b1\u03af\u03b5\u03c2 \u03b1\u03c0\u03bf\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5 \u03b1\u03c0\u03cc \u03c4\u03bf\u03bd \u03ac\u03be\u03bf\u03bd\u03b1 $ x_1$ \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03af\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03b1\u03bd \u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b5\u03c2 \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03ad\u03c2 \u03af\u03c3\u03b5\u03c2 \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03bf\u03c5\u03c2 (\u03bb\u03cc\u03b3\u03c9 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03af\u03b1\u03c2). \u03a4\u03cc\u03c4\u03b5 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03b7\u03bd \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b5\u03af\u03c7\u03b5\u03c2 \u03b3\u03c1\u03ac\u03c8\u03b5\u03b9 \u03ba\u03b9 \u03b5\u03c3\u03cd \u03c0\u03c1\u03b9\u03bd: $ (n \\minus{} 1)x^2 \\plus{} (c_2 \\minus{} x)^2 \\equal{} (n \\minus{} 1)c_1$.\r\n\r\n\u038c\u03c7\u03b9 \u03cc\u03c4\u03b9 \u03b7 \u03bf\u03c5\u03c3\u03af\u03b1 \u03b1\u03c5\u03c4\u03bf\u03cd \u03c0\u03bf\u03c5 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ae, \u03b1\u03c0\u03bb\u03ce\u03c2 \u03b8\u03b1 \u03ae\u03c4\u03b1\u03bd \u03ba\u03c1\u03af\u03bc\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c4\u03bf\u03bd\u03b9\u03c3\u03c4\u03b5\u03af \u03b7 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae \u03b5\u03c1\u03bc\u03b7\u03bd\u03b5\u03af\u03b1 \u03b5\u03b4\u03ce :)\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_3": "Hey,i geometriki lysi einai i \"swsti\" lysi gia to provlima kata tin gnwmi mou aplws den mporousa na to kataferw \r\ngeometrika gia afto kai anatreksa se algevrikes methodous.\r\nDystyxws den mporw na dw/katalavw giati isxyei i afstiri anisotita $ \\frac {c_2^2}{n} 0$ such that $ xyz \\equal{} 1$ . Prove that \r\n\r\n$ \\frac {2 \\minus{} x}{x^2 \\plus{} 1} \\plus{} \\frac {2 \\minus{} y}{y^2 \\plus{} 1} \\plus{} \\frac {2 \\minus{} z}{z^2 \\plus{} 1}\\geq\\frac {3}{2}$", "Solution_1": "[quote=\"silouan\"]Let $ x,y,z > 0$ such that $ xyz \\equal{} 1$ . Prove that \n\n$ \\frac {2 \\minus{} x}{x^2 \\plus{} 1} \\plus{} \\frac {2 \\minus{} y}{y^2 \\plus{} 1} \\plus{} \\frac {2 \\minus{} z}{z^2 \\plus{} 1}\\geq\\frac {3}{2}$[/quote]\r\n\r\nIt's true since $ \\frac {2 \\minus{} x}{x^2 \\plus{} 1}\\ge\\frac1{2}\\minus{}\\frac3{4}lnx$ :)", "Solution_2": "[quote=\"dduclam\"][quote=\"silouan\"]Let $ x,y,z > 0$ such that $ xyz \\equal{} 1$ . Prove that \n\n$ \\frac {2 \\minus{} x}{x^2 \\plus{} 1} \\plus{} \\frac {2 \\minus{} y}{y^2 \\plus{} 1} \\plus{} \\frac {2 \\minus{} z}{z^2 \\plus{} 1}\\geq\\frac {3}{2}$[/quote]\n\nIt's true since $ \\frac {2 \\minus{} x}{x^2 \\plus{} 1}\\ge\\frac1{2} \\minus{} \\frac3{4}lnx$ :)[/quote]\r\nTry $ x\\equal{}0.1.$ :wink:", "Solution_3": "[quote=\"dduclam\"] \nIt's true since $ \\frac {2 \\minus{} x}{x^2 \\plus{} 1}\\ge\\frac1{2} \\minus{} \\frac3{4}lnx$ :)[/quote]\r\n\r\nThis is not true ... Take for example $ x\\equal{}\\frac{1}{10}$ :wink:", "Solution_4": "[quote=\"arqady\"][quote=\"dduclam\"][quote=\"silouan\"]Let $ x,y,z > 0$ such that $ xyz \\equal{} 1$ . Prove that \n\n$ \\frac {2 \\minus{} x}{x^2 \\plus{} 1} \\plus{} \\frac {2 \\minus{} y}{y^2 \\plus{} 1} \\plus{} \\frac {2 \\minus{} z}{z^2 \\plus{} 1}\\geq\\frac {3}{2}$[/quote]\n\nIt's true since $ \\frac {2 \\minus{} x}{x^2 \\plus{} 1}\\ge\\frac1{2} \\minus{} \\frac3{4}lnx$ :)[/quote]\nTry $ x \\equal{} 0.1.$ :wink:[/quote]\r\n\r\nMy mistake :( I'm sorry.", "Solution_5": "RCF-Theorem (corollary) works here.", "Solution_6": "[quote=\"Vasc\"]RCF-Theorem (corollary) works here.[/quote]\r\n\r\nVasc that was also my idea . But I am wondering if there is something elementary ... Note also that the inequality is equivalent to \r\n$ \\sum\\frac{3a^2\\minus{}2ab\\minus{}b^2}{a^2\\plus{}b^2}\\geq 0$", "Solution_7": "[quote=\"silouan\"]Let $ x,y,z > 0$ such that $ xyz \\equal{} 1$ . Prove that \n\n$ \\frac {2 \\minus{} x}{x^2 \\plus{} 1} \\plus{} \\frac {2 \\minus{} y}{y^2 \\plus{} 1} \\plus{} \\frac {2 \\minus{} z}{z^2 \\plus{} 1}\\geq\\frac {3}{2}$[/quote]\r\nWe are not need RCF here. ;) Just AM-GM. :)\r\nSee my solution at: http://canhang2007.wordpress.com/2008/12/13/nice-inequality-of-silouan/ :)", "Solution_8": "Does anyone have another solution? I have\r\n$ b^4c^4\\minus{}2b^4c^3\\minus{}2b^3c^4\\minus{}2b^3c^3\\plus{}5b^4c^2\\plus{}5b^2c^4\\minus{}2b^3c^2\\minus{}2b^2c^3\\plus{}6b^2c^2\\minus{}2b^3c\\minus{}2bc^3\\minus{}2b^2c\\minus{}2bc^2\\plus{}c^2\\plus{}b^2\\minus{}2bc\\minus{}2b\\minus{}2c\\plus{}5 \\ge 0$. Can someone write this as a sum of squares?", "Solution_9": "Can-Hang, you have amazingly simple proof!\r\nI have some ideas about this inequality too. However, i cannot finish :noo: \r\nInequality is equivalent to proving that if $ xyz = 1$\r\n\r\n$ 2\\sum_{cyc}{\\frac {1}{x^2 + 1}} \\geq \\frac {3}{2} + \\sum_{cyc}{\\frac {x}{x^2 + 1}}$\r\n\r\nAfter simple transformation and from Cauchy\u2013Schwarz inequality , we have:\r\n\r\n${ \\frac {x}{x^2 + 1} = \\frac {x}{x^2 + xyz} = \\frac {1}{x + yz} \\leq \\frac {1}{4}(\\frac {1}{x} + \\frac {1}{yz}) = \\frac {1}{4}(\\frac {1}{x} + x)}$\r\n\r\nTherefore, it is enough to prove stronger inequality:\r\n\r\n$ 2\\sum_{cyc}{\\frac {1}{x^2 + 1}} \\geq \\frac {3}{2} + \\frac {1}{4}\\sum_{cyc}{(\\frac {1}{x} + x)}$ or maybe we use the fact that for $ xyz = 1$ we have $ \\sum_{cyc}{\\frac {1}{x}} = \\sum_{cyc}{xy}$ and maybe from there........\r\nAnyone can finish it?\r\nThanks for attention. :)", "Solution_10": "Can_hang2007 , your solution is great and purely elementary :lol: Thanks for sharing", "Solution_11": "[quote=\"silouan\"][quote=\"Vasc\"]RCF-Theorem (corollary) works here.[/quote]\n\nVasc that was also my idea . But I am wondering if there is something elementary ... Note also that the inequality is equivalent to \n$ \\sum\\frac {3a^2 \\minus{} 2ab \\minus{} b^2}{a^2 \\plus{} b^2}\\geq 0$[/quote]\r\n$ \\iff$\r\n$ \\sum_{cyc} (a\\minus{}b)^2 \\left ( \\frac{9b\\minus{}3a}{a^2\\plus{}b^2} \\plus{} \\frac{\\minus{}3b\\minus{}c}{b^2\\plus{}c^2} \\plus{} \\frac{a\\plus{}3c}{c^2\\plus{}a^2} \\right ) \\geqslant 0$\r\n\r\n(If my calculations are right)", "Solution_12": "What about this? Is it nice enough for you, silouan?" } { "Tag": [ "number theory", "greatest common divisor", "modular arithmetic" ], "Problem": "Let $a,b,c$ and $n$ be positive integers such that $(a+bc)(b+ca)=5^{n}$. Prove that $n$ is even.", "Solution_1": "[hide] We know that the GCD of the two terms divide their difference, $(a+bc)-(b+ca) = (a-b)(1-c) \\equiv 0 \\pmod{5}$. Then $a \\equiv b \\pmod{5}$ or $c \\equiv 1 \\pmod{5}$. In each case, the expression $\\pmod{5}$ is a square, which I believe is sufficient in determining that $n$ is even. [/hide]" } { "Tag": [ "LaTeX" ], "Problem": "How do I get Texnic Center out of overwrite mode? i.e. if I want to insert stuff back into a paragraph it replaces the other stuff.", "Solution_1": "press insert on your keyboard. :D", "Solution_2": "Thanks. :lol:", "Solution_3": "Ok, I have another unrelated question. I don't think this affects my $ TeX $ document but how do you get rid of the bad boxes? It says that I have \"underfull hboxes\". this is just annoying me a little bit.", "Solution_4": "[quote=\"joml88\"]Ok, I have another unrelated question. I don't think this affects my $ TeX $ document but how do you get rid of the bad boxes? It says that I have \"underfull hboxes\". this is just annoying me a little bit.[/quote]\r\nThat's a tricky question; many things can cause underfull (or overfull) hboxes. They don't matter unless your document doesn't look right." } { "Tag": [ "calculus", "integration", "function", "LaTeX", "Putnam", "real analysis", "real analysis unsolved" ], "Problem": "For a continuous function f:[0,1]\u2192R, let I = \u2320((x)^2)f(x) \u2013 x(f(x))^2 dx with limits 1 and zero respectively.\r\n\r\nFind the maximum value of I and corresponding f(x).\r\n\r\nPS: How do you use those math symbols in posts?[/hide]", "Solution_1": "The pretty math symbols are $ \\text{\\LaTeX}$. There's a subforum with various resources [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]here[/url].\r\n\r\nThis was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122456]Putnam 2006 B5[/url].", "Solution_2": "Beat me to the punch - I was about to link to Putnam 2006 B5.\r\n\r\nThe only thing I have to add is the FAQ link: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX.[/url]" } { "Tag": [ "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "The sequence $\\{x_{n},\\, n\\ge1\\}$ is defined as follows: $x_{1}=a$ and $x_{n+1}=3x_{n}-x_{n}^{3},\\, n\\ge1.$ Determine the set of real numbers $a$ for which the sequence is convergent. \r\n(A. Kukush)", "Solution_1": "[hide=\"ignore\"]we claim that $a\\in [-2,2]$\n\nLemma 1:if $|x_{n}|\\geq \\sqrt{2}$ $\\implies |x_{n}|\\leq |x_{n+1}|$\nProof : \n$|x_{n+1}|=|x_{n}\\parallel 3-x_{n}^{2}|\\geq |x_{n}\\parallel 3-2|=|x_{n}|$\n\nLemma 2: if $|x_{n}|>2$ $\\implies |x_{n+1}|\\geq 2$\nProof : let $f(x)=3x-x^{3}$ then $f'(x)=3(1+x)(1-x)$ . So it is monotonic when $|x|>2$ . and $f(2)=-2$ , $f(-2)=2$ so it is obvious that $|f(x)|>2$ when $|x|>2$ and hence implies above lemma .\n\nTo prove the question , we observed that for $x=|a|\\leq 2$ $\\implies |f(x)|\\leq 2$ where $f(x)=3x-x^{3}$ so the sequence $x_{n}$ converges in the interval .\n\nwhen $x=|a|>2>\\sqrt{2}$ , from lemma 1,2 we know that this interval is increasing , so $f(x)$ diverges.[/hide]", "Solution_2": "I can't understand your proof.\r\n\r\nBut If $a \\in [-2,2]$,we let $a=2\\sin x$\r\nthen $x_{n}=2\\sin 3^{n}x$\r\nwill it always be convergent?", "Solution_3": "OK,I find the answer is:\r\n$a=2\\sin x$,For every $x=\\frac{\\pi}{4\\cdot 3^{m}}+\\frac{k\\pi}{\\cdot 3^{m}},k,m \\in Z$ or $a=0$\r\nAnd if Moubinool agree on my answer.\r\nI will post my solution :P", "Solution_4": "i think that i only prove that $|a|>2$ then the series divergent . :blush: \r\n\r\nif i followed your hint , i got\r\n\r\n$a=2\\sin x$ where $x=\\frac{k\\pi}{3^{m}}$ , $k,m\\in\\mathbb{Z}$", "Solution_5": "but what about $\\sqrt 2$?\r\nI think the answer may combine yours and mine :lol:" } { "Tag": [ "trigonometry", "geometry", "similar triangles", "inequalities unsolved", "inequalities" ], "Problem": "A, B, C are the angles of a triangle such that:\r\n\r\n$ \\sin A+\\sin\\frac{B}{2}+\\sin\\frac{C}{3}= \\frac{3}{2}$.\r\n\r\nFind value of A,B,C.\r\n\r\n\r\nSorry, my english is poor...", "Solution_1": "[quote=\"nkht-tk14\"]A, B, C are the angles of a triangle such that:\n$sinA+\\sin\\frac{B}{2}+\\sin\\frac{C}{3} = \\frac{3}{2}$.\nFind value of A,B,C.\n[/quote]\r\nFor example: $A=35.42842...^{\\circ},$ $B=42^{\\circ}$ and $C=102.57157...^{\\circ}.$ ;)", "Solution_2": "he wants to prove that: A = pi/6, B = pi/3, and C = pi/2.", "Solution_3": "Who can prove that: $A = \\frac{\\pi}{6}, B = \\frac{\\pi}{3}$, and $C = \\frac{\\pi}{2}$ ???", "Solution_4": "Shouldn`t this be moved..?", "Solution_5": "Admins, this topic should be at Geometry. ;)", "Solution_6": "At the moment there is nothing to move. There are infinitely many non-similar triangles ABC satisfying $\\sin A+\\sin\\frac{B}{2}+\\sin\\frac{C}{3}=\\frac32$, so the problem makes no sense. (This is pretty much what Arqady wanted to say, too.)\r\n\r\n Darij" } { "Tag": [ "Euler", "inequalities", "logarithms", "function", "inequalities unsolved" ], "Problem": "Hi. This is interesting:\r\n\r\nLet us denote\r\n$\\displaystyle \\Phi=\\{(x, y)\\in\\Re^+\\times\\Re^+\\mid x^y\\ge y^x\\}$.\r\nThe program Derive6 generated the following graph:\r\nhttp://www.petocerno.host.sk/FILES/flower.jpg\r\nThere we can see two main boundaries in this graph:\r\nThe first boundary is: $y=x$ for $x>0$.\r\nThe second boundary I could not make out. It looks like hyperbola.\r\nIt is approximately:\r\n$\\displaystyle y=\\frac{e}{x-(e-\\sqrt e)}+(e-\\sqrt e)$.\r\nWhere $e$ is an Euler number.\r\n\r\nBut the most interesting thing is that these\r\nboundaries (in the graph of course) meets at point $(e,e)$.\r\n\r\nAnd from this graph many interesting inequalities can be made out.\r\n\r\nFor instance prove that for every $x\\in\\Re^+$ the following\r\ninequality holds:\r\n$e^x\\ge x^e$.\r\n\r\nAnd for every $a\\in\\Re^+, a\\neq e$ there are such numbers $b,c\\in\\Re^+$ that:\r\n$a^b\\ge b^a$,\r\n$a^c < c^a$.\r\n\r\nI would be grateful for any clear ideas or proofs you can post. :) \r\n\r\nPeter Cerno, SLOVAKIA", "Solution_1": "A proof for $e^x\\geq x^e$ proceeds as follows:\r\n\r\n$\\frac {d}{dx}\\left(\\frac {x^e}{e^x}\\right)=\\frac {e^x(ex^{e-1})-e^x(x^e)}{(e^x)^2}$\r\n\r\nSo we need to consider $e^x(ex^{e-1})-e^x(x^e)$, which factors as $e^xx^{e-1}(e-x)$, which is strictly negative for $x>e$, hence we have the inequality for $x\\geq e$. So it remains to prove that $e^x\\geq x^e$ for $x\\geq e$. Put $g(x)=x^e/e^x$. Then $g(x)=0$, $g(e)=1$, and $g$ is differentiable on $(0,\\infty)$. If $g(x)\\geq 1$, then we need to have $g(x)=1$ somewhere (by continuity). If this is to happen, it must happen in $(0,e)$. But then we can apply Rolle's theorem to $g$ over $[x,e]$ to conclude that $g'$ vanishes somewhere in $(x,e)$. But $g'$ does not vanish at all, a contradiction.\r\n\r\nThe second part should be similar, so I'll leave it.", "Solution_2": "Hi, blahblahblah.\r\n\r\nThank you for your proof.\r\nI am not sure, but I think that the simpler proof can be based on the\r\nobservation, that:\r\n$e^x\\ge x^e$\r\nis equivalent with\r\n$x\\ge e\\ln x$.\r\nIf we set $f(x) = x - e\\ln x$, then\r\n$f'(x) = 1 - \\frac{e}{x}$ and\r\n$f''(x) = \\frac{e}{x^2} > 0$ for all $x\\in \\Re^+$\r\nso for $x=e$ we have local (and also global) minimum.\r\nMoreover $f(e) = 0$.\r\n\r\nNote: This problem was inspired by another problem:\r\nDecide which number is greater: $e^\\pi$ or $\\pi^e$.\r\n\r\nBut can you give me an exact formula for the second\r\nboundary of the set $\\Phi$ discussed above?\r\n(with proof of course) ;)", "Solution_3": "First consider the equation\r\n$x^{x^{x^{x^{x^{x...\\infty}}}}}=2$. This was published in VTM (that's a czech magazine, Peete, u probably know it). The solution is simple. With repacing into itself we get $x^2=2$, so $x=\\sqrt 2$. In more general case\r\n$x^{x^{x^{x^{x^{x...\\infty}}}}}=n$ we get $x=\\sqrt[n] n$. This is correct for ne the upper boundary is y=x and the lower $z^{z^{z^{z^{z^{z...\\infty}}}}}$, whwere\r\n$z=\\sqrt[x]x$. For 1e, for which x' and x are boundaries. I think for x<1 there is only one boundary. I know this \"proove\" doesn't look very useful, but by evaluating the powers you will find it's true.", "Solution_4": "Thanks, Kondr.\r\n\r\nSo I suppose there is not an easy way how to\r\ncompute $z^{z^{z^{...}}}$ for $z=x^{1/x}$ ... :) \r\n\r\nFor those who want to see a graph of\r\nthe function $f(x)=x^{1/x}$ just click\r\non the link below:\r\nhttp://www.petocerno.host.sk/FILES/flower2.jpg\r\n\r\nIt is not so hard to show, that $f(x)$ reaches\r\nit's global maximum for $x=e$ ...", "Solution_5": "Using Maple I found the second bound for x>e:\r\n$-x \\frac {LambertW(-\\frac {ln(x)} x)} {ln(x)}$. I wanted to find more about Lambert function and found\r\nhttp://mathworld.wolfram.com/LambertW-Function.html , where a lot of information about the aproximations of it can be found.", "Solution_6": "Another proof for $e^x\\geq x^e$\r\nSince $e^t=\\sum_{k=0}^{\\infty}\\frac{t^k}{k!}$, we have $e^t\\geq 1+t$. (Equality holds when t=0.)\r\nLet $t=\\frac{x-e}{e}$, then $e^\\frac{x-e}{e}\\geq 1+\\frac{x-e}{e}$.\r\nThen $e^\\frac{x}{e}\\cdot e^{-1}\\geq \\frac{x}{e}$, or $e^x\\geq x^e$" } { "Tag": [ "quadratics", "inequalities" ], "Problem": "Two astronauts A and B, each of mass M ,float freely in space at relative rest.\r\n\r\nA throws a spanner of mass m to B ,who catches it and throws it back to A. Each throw results in the same relative velocity of spanner and thrower.\r\n\r\nShow that spanner does not reach A unless 2m +M < M.$\\sqrt5$", "Solution_1": "[quote]Show that spanner does not reach A unless 2m +M 0$\n\n$M > \\frac{m \\pm \\sqrt{m^2 + 4 m^2}}{2}$\n\n$M > \\frac{m + \\sqrt{m^2 + 4 m^2}}{2}$\n\n$M > \\frac{1 + \\sqrt{5}}{2} m$\n[/hide]", "Solution_3": "I\u2019ll consider scalar variables, momentum and velocity, positive in the astronaut A\u2019s direction of movement (that is also the spanner\u2019s final velocity direction, $\\;\\;v\\;\\;$) and negative in the opposite direction, except for the spanner's relative velocity,$\\;\\;v_{s}\\;,\\;$ that I considered as absolute value.\r\n \r\n When astronaut A throws the spanner, momentum conservation of the two body system tells us that \r\n$\\qquad Mv_{A}+m(v_{A}-v_{s})\\,=\\,0\\;\\;\\;\\Rightarrow\\;\\;v_{A}\\,=\\,{m \\over{M+m}}\\,v_{s}\\;\\;$, where$\\;\\;v_{A}\\;\\;$ is astronaut A\u2019s velocity \r\n\r\nThe three body system , the two astronauts and the spanner, conserves its momentum, equal to zero all the time, so\r\n\r\n$\\qquad Mv_{B}+Mv_{A}+mv\\,=\\,0\\,\\,$ after the spanner was thrown by astronaut B, i.e., $\\;\\;v_{B}\\;\\;$ and $\\;v\\;$ are the final velocities of astronaut B and of the spanner.\r\n\r\nPutting $\\;\\;\\;v_{B}\\,=\\,v+v_{s},\\;\\;\\;$ together with $\\;\\;v_{A}\\;\\;$ from the beginig, in the expression above we get\r\n\r\n$\\qquad v\\,=\\,{\\bigg{(}{M \\over{M+m}}\\bigg{)}}^{2}\\cdot v_{s}\\;\\;\\;$ and dividing it by $\\;v_{A}\\;,$ results\r\n\r\n$\\qquad{v \\over v_{A}}\\,=\\,{M^{2}\\over{(M+m)m}},\\;\\;$ that must be greater than 1. Then\r\n\r\n$\\qquad m^{2}+Mm-M^{2}\\,<\\,0\\;\\;$\r\n\r\nConsidering it as a quadratic trinomial in m, we find the positive root\r\nas $\\;\\;m\\,={M\\over2}(\\sqrt5\\,-\\,1)\\;\\;$ so, as the coefficient of the first degree of m is positive, $\\;\\;m\\,<\\,{M\\over2}(\\sqrt5\\,-\\,1)\\;\\;$ that leads to the desired answer.", "Solution_4": "For Whatever its worth, Here is a \"physicist way\" :) \r\n(The problem is not hard in any case, but still a physicist way is the lazy one)\r\n \r\nFrame of reference is the space station (where everyone is at rest in the initial condition) and consider the limiting case (Final velocity of spanner = Final velocity of A) (Direction of velocities will not be given as they would be obvious) \r\n\r\nLet the spanner be of mass $1$ and is thrown with a velocity $1$ (We have freedom to choose our units :) ) and man A of mass $M$ will move with a velocity ($1/M$)\r\nIn the end, we have velocity of spanner=velocity of A= $1/M$ \r\nRelative velocity of A and (spanner initially thrown) = $1+1/M$ = relative velocity of B and final velocity of spanner($=1/M$) this implies velocity of B=1.\r\nConservation of momentum gives:\r\n $M (1/M)+1(1/M) = M$ this gives the required result.\r\n$M^{2}-M-1=0 \\implies M = \\frac{1+\\sqrt 5}2$ \r\n(After changing M to (M/m) and only taking the sign which makes sense to get the inequality etc...)" } { "Tag": [ "conics", "ellipse", "AMC", "AIME", "analytic geometry", "special factorizations" ], "Problem": "I forgot how to graph them, but I just took a Mock AIME that had me draw an ellipse; I got it right. Is this the right method?\r\n\r\n[hide=\"in here\"]Let's take the ellipse defined by $ 16x^2\\minus{}320x\\plus{}25y^2\\minus{}1500y\\minus{}15900\\equal{}0$. This is equivalent to $ (4x\\minus{}40)^2\\plus{}(5y\\minus{}150)^2\\equal{}200^2$. Now take the circle defined by $ (x\\minus{}40)^2\\plus{}(y\\minus{}150)^2\\equal{}200^2$ and \"squash\" the radii down; the vertical radii are now 1/4 of what they used to be and the horizontal radii are now 1/5 of what they used to be. This is the ellipse.[/hide]\r\n\r\nIf this is the right method, then do I have my vertical/horizontal radii backwards?", "Solution_1": "[quote=\"1=2\"]I forgot how to graph them, but I just took a Mock AIME that had me draw an ellipse; I got it right. Is this the right method?\n\n[hide=\"in here\"]Let's take the ellipse defined by $ 16x^2 \\minus{} 320x \\plus{} 25y^2 \\minus{} 1500y \\minus{} 15900 \\equal{} 0$. This is equivalent to $ (4x \\minus{} 40)^2 \\plus{} (5y \\minus{} 150)^2 \\equal{} 200^2$. Now take the circle defined by $ (x \\minus{} 40)^2 \\plus{} (y \\minus{} 150)^2 \\equal{} 200^2$ and \"squash\" the >> radii << down; the vertical radii are now 1/4 of what they used to be and the horizontal radii are now 1/5 of what they used to be. This is the ellipse.[/hide]\n\nIf this is the right method, then do I have my vertical/horizontal radii backwards?[/quote]\r\n\r\nFor an ellipse, they are not called \"radii;\" they have semiaxes instead.\r\n\r\nI don't see your method working out.\r\n\r\nAmong different types of solutions:\r\n\r\nSet 4x - 40 and 5y - 150, respectively, equal to 0 to find this ellipse's center: (10,30).\r\n\r\nSet (4x - 40)^2 = 200^2, with y = 30, and solve for the two values of x to get \r\nthe two coordinates of the ellipse at the extreme at the sides.\r\n\r\nSet (5y - 150)^2 = 200^2, with x = 10, and solve for the two values of y to get\r\nthe two coordinates of the ellipse at the extreme at its top and bottom.", "Solution_2": "You can also graph this ellipse by rewriting it into standard form.\r\nThe standard equation of an ellipse is $ \\frac{(x\\minus{}h)^2}{a^2}\\plus{}\\frac{(y\\minus{}k)^2}{b^2}\\equal{}1$\r\n[hide]\n$ 16x^2\\minus{}320x\\plus{}25y^2\\minus{}1500y\\minus{}15900 \\equal{}0 \\\\\n16x^2\\minus{}320x\\plus{}25y^2\\minus{}1500y \\equal{}15900 $\nFactoring out the 16 and 25 so the coefficients of $ x^2$ and $ y^2$ are 1:\n$ 16(x^2\\minus{}20x) \\plus{} 25(y^2\\minus{}60y) \\equal{} 15900 $\nCompleting the square:\n$ 16(x^2\\minus{}20x\\plus{}100)\\plus{}25(y^2\\minus{}60y\\plus{}900) \\equal{}15900 \\plus{} (16)(100) \\plus{} (25)(900) \\\\\n16(x\\minus{}10)^2 \\plus{} 25(y\\minus{}30)^2 \\equal{}40,000 $\nDividing both sides by 40,000 so the equation is set equal to one:\n$ \\frac{16(x\\minus{}10)^2}{40,000} \\plus{} \\frac{25(y\\minus{}30)^2}{40,000} \\equal{}\\frac{40,000}{40,000} \\\\\n\\frac{(x\\minus{}10)^2}{2500}\\plus{}\\frac{(y\\minus{}30)^2}{1600} \\equal{}1$\nNow the equation is in standard form. \nIn standard form, the center is located at (h,k), so the center of this ellipse is (10,30).\n$ a^2$ corresponds to the horizontal axis, and $ b^2$ corresponds to the vertical axis. \n$ a\\equal{}50$ and is the distance from the center to the left and right edges of the ellipse, and $ b\\equal{}40$ and is the distance from the center to the top and bottom edges.\nUsing the values of $ a$ and $ b$, you can find four points to graph the ellipse. With $ a$, you can find one point directly to the left of the center and one point directly to the right. These points are $ (10\\pm50,30)$, or (-40,30) and (60,30).\nWith $ b$, you can find one point directly above the center and one directly below.\nThese points are $ (10,30\\pm40)$, or (10,70) and (10,-10).\n\nSince $ a^2 > b^2$, the ellipse is stretched horizontally.\n\nSo the ellipse is centered at (10,30) and passes through the points (-40,30), (60,30), (10,70), and (10,-10).\n[/hide]", "Solution_3": "tornado was correct. :lol:", "Solution_4": "[quote=\"maximos\"]tornado was correct. :lol:[/quote]\r\n\r\nIt's not no longer true.\r\n\r\nOnce tornado.adv4 got it correct, that user remained correct.\r\n\r\ntornado.adv4 is correct.", "Solution_5": "yuh .. right! thanks for correcting my grammar. It [u]TOTALLY[/u] changed my life a lot . :w00tb: \r\n\r\n\r\nJust kidding [b]Arrange your tan[/b]. :lol:\r\n\r\n\r\nP.S. [b]Arrange your tan[/b] I didn't mean you are wrong or anything on showing your solution. I'm so sorry if ever you got me wrong. Peace men. message me if you want, let's both promote world peace. :omighty:", "Solution_6": "the way tornado did is the standard way. although the way you did it is pretty interesting.", "Solution_7": "Actually you can also think of this in terms of affine transformation... (say taking the circle $ (x \\minus{} 40)^2 \\plus{} (y \\minus{} 150)^2 \\equal{} 200^2$ and apply the transformation $ x\\mapsto 4x',y\\mapsto5y'$) :)\r\n(in fact this was what you basically did)" } { "Tag": [ "function", "complex analysis", "complex analysis unsolved" ], "Problem": "Prove that any entire function that tends to infinity at infinity is a polynomial. ($lim_{z \\to \\infty} |f(z)|= + \\infty$)", "Solution_1": "When you develop $g(z)=f\\left(\\frac 1z\\right)$ as a Laurent series around $0$, you get something of the form $\\frac{a_n}{z^n}+\\ldots+\\frac{a_1}z+a_0$ (and the result follows, of course). \r\n\r\nThis is because, on the one hand, the seris must be finite to the left, since $g$ has a pole at $0$, and, on the other hand, it cannot contain any positive values of $z$, because $g$ minus its principal part is a bounded entire function (so it must be constant)." } { "Tag": [ "analytic geometry", "geometry", "3D geometry" ], "Problem": "How many lines in a three dimensional rectangular coordiante system pass through four distinct points of the form $(i,j,k)$ where $i,j,$ and $k$ are positive integers not exceeding four?\r\n\r\n$\\text{(A)} \\ 60 \\qquad \\text{(B)} \\ 64 \\qquad \\text{(C)} \\ 72 \\qquad \\text{(D)} \\ 76 \\qquad \\text{(E)} \\ 100$", "Solution_1": "(D) 76\r\n16 lines perpendicular to each plane parallel to one of the side of the cube for 3*16 plus the lines that pass through parallel edges separated by the cube's volume for 4*6 lines, plus the 4 lines passing through the cube's volume that connect opposite vertices for a total of 3*16 + 4*6 + 4 = 76.", "Solution_2": "We have four colinear points of the form $(i,j,k)$, where each of the three coordinates is between 1 and 4. \r\n\r\nCase 1: Two of the three coordinates are the same for all four points. The third coordinate goes from 1 to 4. This can happen in 3*4*4 = 48 ways.\r\n\r\nCase 2: One of the three coordinates is the same for all four points. The other two coordinates can either go in the same direction, or go in opposite directions. This can happen in 3*4*2 = 24 ways.\r\n\r\nCase 3: No coordinate remains the same for all four points. Order the points in terms of the $i$ coordinate. Then the $j$ and $k$ coordinates will both be either increasing or decreasing. So this can happen in 2*2 = 4 ways.\r\n\r\nSo the total number of lines is 48 + 24 + 4 = 76.", "Solution_3": "TheAmazingOne's method can be extended to harder versions of this question. For instance, what is the answer if we replace the second \"four\" in the question by \"five?\" We could keep going like that, replacing it first with 5, then 6 and so on. When we get to 8, however, it will get more complicated -- why?", "Solution_4": "When we get to 8 we'll have to start considering lines where the $i$ coordinate increases by two each time the $j$ coordinate increases by 1, etc, rather than only having to consider lines where coordinates increase at the same rate or remain constant. Actually, for 7 we would also have to consider some of those, such as the line through (1,1,1),(1,1,3),(1,1,5),(1,1,7).", "Solution_5": "Yes, you're right, I meant 7 instead of 8 -- $2n - 1$, in fact, where $n$ is the \"first\" 4 in the question.\r\n\r\nWhile I'm here: what is the answer to the original question if we replace [i]both[/i] instances of \"four\" in the question with $n$? What if we replace the first with $n$ and the second with $n + 1$?", "Solution_6": "If we replace them both with $n$,\r\n\r\nWe have n colinear points of the form (i,j,k), where each of the three coordinates is between 1 and n.\r\n\r\nCase 1: Two of the three coordinates are the same for all n points. The third coordinate goes from 1 to n. This can happen in 3*n*n = 3n\u00b2 ways.\r\n\r\nCase 2: One of the three coordinates is the same for all n points. The other two coordinates can either go in the same direction, or go in opposite directions. This can happen in 3*n*2 = 6n ways.\r\n\r\nCase 3: No coordinate remains the same for all n points. Order the points in terms of the i coordinate. Then the j and k coordinates will both be either increasing or decreasing. So this can happen in 2*2 = 4 ways.\r\n\r\nSo the total number of lines is 3n\u00b2 + 6n + 4." } { "Tag": [ "geometry", "MATHCOUNTS", "blogs" ], "Problem": "Just wanted to know. (im on mathlinks.ro)", "Solution_1": "Im on Aops....but shouldnt this be on F&G or maybe The Round Table??", "Solution_2": "aops", "Solution_3": "good old google", "Solution_4": "lol, you found this from google!?", "Solution_5": "What were you searching for? \"math\" \"message board\" \"art\" \"the area of a regular dodecagon = 3R 2 \"? :lol:", "Solution_6": "[quote=\"Phelpedo\"]good old google[/quote]\r\n\r\nI didn't know you could see this message directly from the google site :D :D .\r\n\r\nIn my case, http://www.MathLinks.ro\r\n\r\nI would say that this topic should be move to F&G.", "Solution_7": "If this post were in F&G or The Round Table, it would not measure how users of the MATHCOUNTS Forum accessed this site. Therefore, I don't intend to move this poll.\r\n\r\nAt the time of this writing, the poll is 13 AoPS, 3 Mathlinks. I think the poll confirms the common wisdom is that most US users and most pre-Olympiad users probably use the AoPS site. (I am actually a little surprised there are so many mathlinks users lurking in this forum.)", "Solution_8": "i was searching for MC results and just found the site", "Solution_9": "[quote=\"Phelpedo\"]i was searching for MC results and just found the site[/quote]\r\nI did the exact same thing.", "Solution_10": "I get on AoPS. For the .ro at the end of mathlinks, what country is that?", "Solution_11": "[quote=\"mathgeek2006\"]I get on AoPS. For the .ro at the end of mathlinks, what country is that?[/quote]Romania (where site creator was from)", "Solution_12": "That's what I thought--just wanted to make sure.", "Solution_13": "[quote=\"biffanddoc\"][quote=\"mathgeek2006\"]I get on AoPS. For the .ro at the end of mathlinks, what country is that?[/quote]Romania (where site creator was from)[/quote]*caugh* is from :P :lol:", "Solution_14": "But as I hear from Mr. Rusczyk's blog, you are coming over here as soon as you can get a H1B visa? That way you will learn not to wander into the way of paintball-shooting thugs ;) !\r\n\r\nBilly", "Solution_15": "[quote=\"solafidefarms\"]But as I hear from Mr. Rusczyk's blog, you are coming over here as soon as you can get a H1B visa?[/quote]Probably ;) However that doesn't change the \"is\" into \"was\" :D", "Solution_16": "I use mathlinks because i think it looks way cooler. :cool:", "Solution_17": "So the sites both use the same database?", "Solution_18": "[quote=\"NoSoupForYou\"]So the sites both use the same database?[/quote]Actually they are one and the same site (since last summer). Just that there are two ways to access it, and two skins which look differently but offer (almost) the same functionality. \r\nThe MathLinks skin is optimized for people interested in advanced maths and olympiads. The AoPS skin is optimized for the gifted math students or their parents/teachers searching for addition resources, not necessarily USAMO/MOSP-level.", "Solution_19": "[quote=\"Valentin Vornicu\"][quote=\"NoSoupForYou\"]So the sites both use the same database?[/quote]Actually they are one and the same site (since last summer). Just that there are two ways to access it, and two skins which look differently but offer (almost) the same functionality. \nThe MathLinks skin is optimized for people interested in advanced maths and olympiads. The AoPS skin is optimized for the gifted math students or their parents/teachers searching for addition resources, not necessarily USAMO/MOSP-level.[/quote]\r\n\r\nI suppose that MathLinks and Art of Problem Solving started as two separated sites (I suppose, I have been here just a few months and I haven't find out yet), nevertheless, the truth is that I have some idea of the amount of work that to build, update and administrate a site like this may be. I just wanted to make a stop in the way and say: [i]Guys, you really do a good job[/i]. I suppose the creators of the sites are the same current administrators, those are, Valentin, DPatrick, MCrawford, rrusczyk and vRusczyk (even when most of the people usually \"blame\" just Valentin for such a great job, actually I don't know exactly what part of the job is done by each of the Administrators) but anyway, I just wanted to say, speaking by myself, and I suppose that speaking by most of the users of the site: [b]We do appreciate all the work you do! Keep going[/b].\r\n\r\nBest regards,", "Solution_20": "i totally agree with djimenez, thank you to the administrators :D", "Solution_21": "im on aops", "Solution_22": "same here...\r\nand kingof21penguins, you said that there was nothing good about Aops forums but apparently you joined too! :)", "Solution_23": "magix for the last time i am not mark\r\nwho do u think i am", "Solution_24": "i see... you confused me the whole time", "Solution_25": "you would think that it would be obvious that i wouldnt mock my self and ill give u a nickel if ...." } { "Tag": [ "calculus", "geometry", "3D geometry" ], "Problem": "Here is a very challenging calc problem I was given... If anyone could help or post a solution to it that would be great!!\r\n\r\nYou have a metal cube that is 1 meter on each side and you drill a hole through\r\nthe center of the cube, starting at one corner and ending at the corner opposite. If the radius of\r\nthe drill is significantly less than 1 meter, what is the volume of the remaining material?\r\n(\u201csignificantly less than\u201d means \u201ca very small fraction of.\u201d)", "Solution_1": "Uh...1? Depending on how big the drill is...Can mod move this to some other forum?" } { "Tag": [ "inequalities solved", "inequalities" ], "Problem": "I proved the following ineq:(but I don't know the author?) \r\na)Let $a,b,c,d\\in R$ such that :$\\sum a^2=1$.Prove that \r\n$\\sum \\frac{1}{1-ab}\\le 8$\r\n And I have a generalization question:\r\nb)Let $a_1,a_2,...,a_n\\in R$ such that $\\sum a_i^2=1$.Prove that :\r\n$\\sum \\frac{1}{1-a_ia_j}\\le \\frac{n^2}{2}$\r\n ;)", "Solution_1": "Here it is\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=leq8&t=55410", "Solution_2": "Your link does not provide any solution pvthuan. The general solution is here:\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=299899&start=500[/url]\r\nPost 509." } { "Tag": [ "induction", "number theory", "greatest common divisor", "Putnam", "combinatorics proposed", "combinatorics" ], "Problem": "Let $ S $ be a finite set of integers , each greater than 1 . Suppose that , for each integer n , there is some $ s \\in S $ such that the greatest common divisor of $ s $ and $ n $ equals $ 1 $ or $ s $ . Show that there exist $ s,t \\in S $ whose greatest common divisor is prime .", "Solution_1": "It's almost the same as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29019]this one[/url].\r\n\r\nWe can prove by induction on $n$ that given a set of $n$ elements, no two of which have a prime gcd, we can find a natural $N=N(n)$ which has gcd$>1$ with each one of the $n$ numbers, but, at the same time, is not divisible by any of them:\r\n\r\nSuppose we've shown it for $n-1$ numbers $a_i,\\ i\\in\\overline{1,n-1}$, and we add another number $a_n$ to the collection. \r\n\r\nIf $a_n$ has at least one prime factor which does not appear in the decomposition of the previous $n-1$ numbers, then we can choose either to multiply $N(n-1)$ with that prime factor to get $N(n)$ or to take $N(n)=N(n-1)$ unchanged, according to the situation. Either way we're done in this case. Assume, on the other hand, that all the prime factors of $a_n$ appear in the other $n-1$ numbers. If $N(n-1)$ is divisible by the number, then divide $N(n-1)$ by any prime factor of $a_n$ to get $N(n)$. If $(N(n-1),a_n)=1$, multiply $N(n-1)$ with any prime factor of $a_n$ to get $N(n)$. It's easy to see that these operations give the desired result due to our hypotheses.", "Solution_2": "This is some Putnam B6 question from a few years ago (1999?).", "Solution_3": "This is one the problem in last issue of the Iran's mathematic magazin :lol: \r\n\r\n[i]Solution:[/i]\r\n\r\nSuppose that $n$ be the smallest number $\\in \\mathbb{Z}$ s.t. $gcd (n,s)>1$. It's obviously that for every $s\\in S$ ,$n$ hasn't the repeated factor in prime divisor.It's obviously that there exist $s\\in S$ s.t. $s\\mid n$.In the other way if $p$ be the prime number s.t. $p\\mid s$ and according to choosing $n$ , $\\frac np$ is coprime some $t\\in S$ and we know that $n$ is not coprime to $t$ ,so that $p\\mid t$ but is not divisibe by the another prime divisor of $n$ then $gcd(s,t)=p$ and the proof is complete. :D", "Solution_4": "what if S is a finite set of prime numbers?", "Solution_5": "I wanted to ask this as well, when I first saw the problem, but then I thought that maybe by \"two elements whose gcd is prime\" nttu didn't necessarily mean that the two elements must be distinct. So if the set has a prime member, then we're done, since it has a prime gcd with itself." } { "Tag": [ "MATHCOUNTS" ], "Problem": "A company makes stadium seats. The company attaches the seats to a steel rod. The length of each rod needed depends on the # of seats attached to it. The distance from the center of 1 seat to the center of the next seat is 2.5 feet. The distance from the end of the steel rod to the center of the nearest seat is 2 feet.\r\n\r\nA.\tThe company received an order to make 3 seats of stadium seats. The 1st set will contain 2 seats, 2nd set will contain 3 seats, and 3rd set will contain 4 seats. What is the length, in feet of each steel rod the company will use to make each set of stadium seats? Explain your steps.\r\nB.\tWrite an expression that could be used to determine the length of a steel rod used to make a set of n stadium seats. Explain how you know your expression is right?", "Solution_1": "you can figure out that a seat is one foot long...\r\n\r\nthat gives you 2.5s-1.5, s is the number of seats...where did this question come from?\r\n\r\noops forgot the end of the rod, I thought it looked like this s-s-s-s-s\r\n\r\nthis doesn't look like mathcounts, what is it?", "Solution_2": "It doesn't matter how long the seat is\r\n[hide]\n6.5 feet for set 1\n9 feet for set 2\n11.5 feet for set 3\nOne seat must have two feet in each direction from the end of the steel rod making $4 feet$. Two must have another two and a half feet of spacing making $6.5 feet$ and so on.\nYour expression is:\n1.5 + 2.5s\n[/hide]\r\nEdit: This problem would be probably more suited for: http://www.artofproblemsolving.com/Forum/index.php?f=299" } { "Tag": [], "Problem": "i need help proving that if quadrilateral ABCD is orthodiagonal and circumscribed around a circle, then (AB)(CD)=(BC)(AD).", "Solution_1": "[hide]\nProve \n\n1. $AB+CD=BC+AD$\n\n2. $(AB)^{2}+(CD)^{2}=(BC)^{2}+(AD)^{2}$\n\nThe result follows. \n[/hide]" } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "If $ x$ is a real number such that $ x^2\\minus{}x$ and $ x^n\\minus{}x$ are integers for some $ n \\ge 3$, prove that $ x$ is an integer.", "Solution_1": "If x is rational then it is trivially integer (just substitute x in lowest term in x\u207f-x). Now p(x)=k in Z, deg(p(x))=2 admit always solution of the form x=a+bt, with a,b in Z and t a real number with minimum polynomial quadratic(by definition).\r\nIf x is in R/Q then x in not in {0,1}, and since (x\u207f-x)/(x\u00b2-x) is in Q and is expressible by sum of power of x, we conclude that x is rational(t=0)." } { "Tag": [ "AMC", "AIME", "trigonometry", "AIME II", "AIME I" ], "Problem": "Hey guys,\r\n\r\nI am in the process of making a mock AIME and I thought some of you guys who still have yet to take the AIME II or anyone might be interested. The test should be finished tonight or tomorrow so it should be posted sometime then. \r\n\r\nPlease post here if you are interested.", "Solution_1": "I want to take it.", "Solution_2": "Yeah I'll take it.", "Solution_3": "Sure, I'll take it.", "Solution_4": "I'll do it (and fail).", "Solution_5": "i may do it (depending on how much time i have).", "Solution_6": "I'll read over the problems for sure. I can check if you want (I'm not sure that I can find a three hour block but I still enjoy looking at these tests).", "Solution_7": "[quote=\"Altheman\"]I'll read over the problems for sure. I can check if you want (I'm not sure that I can find a three hour block but I still enjoy looking at these tests).[/quote]\r\n\r\nThat'll be great. I was hoping I could find someone to take a look at it before I give it out. I'll pm you about it when I get it finished.", "Solution_8": "I might, it depends on the time.", "Solution_9": "ill take it", "Solution_10": "I'll take it too!!", "Solution_11": "I've already taken the AIME I, but I would love to take it.", "Solution_12": "I'll probably fail miserably, but I'll take it if I can find some time", "Solution_13": "ILL TAKE IT!!\r\nyay... phail", "Solution_14": "I guess I'll take it.", "Solution_15": "I'll take it.", "Solution_16": "I might take it just for the hell of it, even though I took the AIME I.", "Solution_17": "I'll take it\r\n\r\n(btw make it really hard)", "Solution_18": "I'll take it. Sounds like fun.", "Solution_19": "k, i'll take it", "Solution_20": "sign me up, too!", "Solution_21": "Oh, cool, I'll take it.\r\nThanks.", "Solution_22": "I'll take it, especially because I am taking the AIME II.", "Solution_23": "[quote=\"davidyko\"]I might take it just for the hell of it, even though I took the AIME I.[/quote]\r\n\r\nsame here\r\n\r\nalso, make it hard, really hard. harder than aime i", "Solution_24": "[quote=\"mchoi815\"][quote=\"davidyko\"]I might take it just for the hell of it, even though I took the AIME I.[/quote]\n\nsame here\n\nalso, make it hard, really hard. harder than aime i[/quote]\r\n\r\nhehe i'll try :P", "Solution_25": "I'll join the list of people who'll take it!\r\n\r\n(Want to mix it up some, after reading 20 \"I'll take it!\" posts)", "Solution_26": "Okay everyone, the Mock AIME for 2008 is just undergoing some edits by Altheman and not_trig so it should be out soon! (probably 1~2 hours)", "Solution_27": "Well, if it's not up yet, I guess I can still jump and an take it\r\n\r\nIt must be coming out right about now", "Solution_28": "Really good test. Though I failed badly...\r\nIt seems a lot more difficult than the normal AIME, though..." } { "Tag": [], "Problem": "Do any of you have any advise for very good ways of making money (non-members) as i am new :(", "Solution_1": "get mining lvl up to 30 and mine coal and sell it for 100-200 gp each piece.\r\n\r\nelvenchamp777", "Solution_2": "get mining lvl up to 30 and mine coal and sell it for 100-200 gp each piece.\r\n\r\nelvenchamp777", "Solution_3": "get mining and smithing up to high lvls, then mine for gold, mithril, coal etc, make stuff out of it, and sell it to other players or the merchants even to get some good cash", "Solution_4": "I hate runescape, the tutorial takes too long( I never got past the tutorial becuse i died to a rat)", "Solution_5": "[quote=\"Guard 125\"]Do any of you have any advise for very good ways of making money (non-members) as i am new :([/quote]\r\n\r\nSimple: find some market nest that is empty, and fill it. There are a bunch of them open, specially on the web, the hard part is to find them, because they are not that evident untill something start making money with them (Examples: Ebay, Google, E-Diet, E-Harmony, and other good bussiness in the web) unfortunatelly those examples are already filled, try to find one that is not.\r\n\r\nBest,", "Solution_6": "[quote=\"GoBraves\"]I hate runescape, the tutorial takes too long( I never got past the tutorial becuse i died to a rat)[/quote]\r\nhm...\r\ni thought u culdn't die on tutorial island....\r\ni tried to but once i got to 1 hp the rat kept doing 0's for a loooooong time till i gave up.\r\nanother way to make money is to sell nats (nature runes) they go to about 300gp each\r\nor you can smith or craft and enchant ammy (amulets) and sell.\r\nI'm a member and i sell ranaar weeds and unidentified herbs above that for lots of money :D \r\ngrey wolf furs also help too, and pickpocketing gaurds is good too.\r\nthat's all i can think of...", "Solution_7": "[quote=\"djimenez\"][quote=\"Guard 125\"]Do any of you have any advise for very good ways of making money (non-members) as i am new :([/quote]\n\nSimple: find some market nest that is empty, and fill it. There are a bunch of them open, specially on the web, the hard part is to find them, because they are not that evident untill something start making money with them (Examples: Ebay, Google, E-Diet, E-Harmony, and other good bussiness in the web) unfortunatelly those examples are already filled, try to find one that is not.\n\nBest,[/quote]\r\nsrry about the double post people...\r\nbut are you talking about runescape, cuz it doesn't sound like it....", "Solution_8": "[quote=\"Jeffreywang\"][quote=\"GoBraves\"]I hate runescape, the tutorial takes too long( I never got past the tutorial becuse i died to a rat)[/quote]\nhm...\ni thought u culdn't die on tutorial island....\ni tried to but once i got to 1 hp the rat kept doing 0's for a loooooong time till i gave up.\nanother way to make money is to sell nats (nature runes) they go to about 300gp each\nor you can smith or craft and enchant ammy (amulets) and sell.\nI'm a member and i sell ranaar weeds and unidentified herbs above that for lots of money :D \ngrey wolf furs also help too, and pickpocketing gaurds is good too.\nthat's all i can think of...[/quote]\r\n\r\nI left in on and left cuz i had to do something, and when i came back, well i was dead.", "Solution_9": "[quote=\"GoBraves\"][quote=\"Jeffreywang\"][quote=\"GoBraves\"]I hate runescape, the tutorial takes too long( I never got past the tutorial becuse i died to a rat)[/quote]\nhm...\ni thought u culdn't die on tutorial island....\ni tried to but once i got to 1 hp the rat kept doing 0's for a loooooong time till i gave up.\nanother way to make money is to sell nats (nature runes) they go to about 300gp each\nor you can smith or craft and enchant ammy (amulets) and sell.\nI'm a member and i sell ranaar weeds and unidentified herbs above that for lots of money :D \ngrey wolf furs also help too, and pickpocketing gaurds is good too.\nthat's all i can think of...[/quote]\n\nI left in on and left cuz i had to do something, and when i came back, well i was dead.[/quote]\r\n\r\nWhere did you appear at?", "Solution_10": "U sure you weren't just logged out? cuz u can't die on tutorial island", "Solution_11": "[quote=\"entropywins\"]U sure you weren't just logged out? cuz u can't die on tutorial island[/quote]\r\nthat's wat i was thinking too\r\nanyone here buying essense? (rune essense)", "Solution_12": "[quote=\"Jeffreywang\"][quote=\"djimenez\"][quote=\"Guard 125\"]Do any of you have any advise for very good ways of making money (non-members) as i am new :([/quote]\n\nSimple: find some market nest that is empty, and fill it. There are a bunch of them open, specially on the web, the hard part is to find them, because they are not that evident untill something start making money with them (Examples: Ebay, Google, E-Diet, E-Harmony, and other good bussiness in the web) unfortunatelly those examples are already filled, try to find one that is not.\n\nBest,[/quote]\nsrry about the double post people...\nbut are you talking about runescape, cuz it doesn't sound like it....[/quote]\r\n\r\nSorry, I didn't know runscape is a game. Of course, I wasn't talking about it.\r\n\r\nBest,", "Solution_13": "Ok i dont know what really happened that day, it was like 2 years ago or something like that, but runescape is old, there's no point in the game, and has terrible graphics, so its not really my idea of fun. I find ragnarok, and gunbound, much better, but even those two dont compare to halo 2 or any nfl or nba games.", "Solution_14": "Well, I don't know the game, I have never played it, but even when I am not a fan of computer games, I prefer the old ones. I still have my 486 Laptop with Doom, Pacman, Prince of Persia, Digger and many more (all versions from 1994 or before :P :D )\r\n\r\nBest,", "Solution_15": "[quote=\"GoBraves\"]Ok i dont know what really happened that day, it was like 2 years ago or something like that, but runescape is old, there's no point in the game, and has terrible graphics, so its not really my idea of fun. I find ragnarok, and gunbound, much better, but even those two dont compare to halo 2 or any nfl or nba games.[/quote]\r\nRunescape is much better now, they made the graphics like 1000x better. They added a ton of new stuff, and if you've only played it 2 years ago when they are always making it better you shouldn't diss it...you can diss the version from 2 years ago all you want, but never diss the current version. Also, in response to post #1, i would suggest getting 40 crafting and good mining to go with it (at least 20, over 30 recommended) then mine silver in crafting guild-sell for 100-300gp each (the more at a time you have, the more per piece people sell.", "Solution_16": "Runescape is really repetative. The graphics still suck. There isn't really a point to the game, as gobraves said. Not only can it not even compare to games like Halo 2, Fable, and Morrowind, I don't even consider it that great of a free online game. I'd rather have a nice game of Texttwist any day.", "Solution_17": "I PLAYED TEXT TWIST IN 4th grade. Pretty entertaining until your eyes hurt. YEs RUNESCAPE is a piece of junk. Worse than all other XBOX games" } { "Tag": [ "geometry", "incenter", "circumcircle", "vector" ], "Problem": " 0$ and $xyz=1$ [edit]( $x=\\frac{a}{b},y=\\frac{b}{c},z=\\frac{c}{a}$) prove that is this inequlity true\r\n\r\n$9(ax+by+cz) > = (a+b+c)(x+y+z)^2$\r\nIt might not be true but if it is that it is very helpfull.I'll now try to solve it :?", "Solution_1": "I can't understand what you mean here.\r\nBut I have substituted $a=b=1$, $x=1$, $y=1/c$, $z=c$. And the inequlity becomes wrong for sufficiently large $c$." } { "Tag": [ "geometry", "rectangle", "combinatorics unsolved", "combinatorics" ], "Problem": "Prove that for $ 5\\times16$ table and we colored all of its cells by 3 colors : Blue Green Red.\r\nProve that there are rectangles with vertices of the same colors and with side parallel to the sides of the rectangle.", "Solution_1": "Is it too hard? Anyone can solve it? Help me! Thanks", "Solution_2": "It is an immediate result of Pigeonhole.", "Solution_3": "Count the number of pairs of grids with same color in each of the 16 columns. For example, bbrrg gives 2 such pairs and gggbr gives 3 such pairs. It is easy to see that in each of the 16 columns there are at least 2 such pairs, give a total of at least 32 pairs.\r\n\r\nAccording to the row number (1 to 5) of grids in each pair we can assign each pair into one of the 10 sets. (row 12, row 13,..., row 45) So there will be a set contains at least 4 such pairs. Two of the 4 pairs must have same color, thus gives a desired rectangle.", "Solution_4": "How do we colour each cell in 5x15 table by R,G and B such that there are no rectangle with vertices of the same colors and with side parallel to the sides of the rectangle? :) Thanks." } { "Tag": [ "calculus", "derivative", "algebra unsolved", "algebra" ], "Problem": "Let $ w$ be a primitive $ 2007$th root of unity.\r\n\r\nCalculate $ \\sum_{k \\equal{} 1}^{2006}\\frac {1}{2 \\minus{} w^k}$.", "Solution_1": "let $ f(x) \\equal{} \\prod_{i \\equal{} 1}^{2006}(x \\minus{} \\omega ^i) \\equal{} \\frac {x^{2007} \\minus{} 1}{x \\minus{} 1}$\r\nthen $ g(x) \\equal{} \\frac {f'(x)}{f(x)} \\equal{} \\sum_{i \\equal{} 1}^{2006}\\frac {1}{x \\minus{} \\omega ^i}$\r\nSo you are looking for $ g(2)$ which is a straightforward derivative computation.", "Solution_2": "The derivative is even simpler if we use the definition\r\n\\[ f(x) \\equal{} \\prod_{k\\equal{}0}^{2006} (x \\minus{} w^k) \\equal{} x^{2007} \\minus{} 1.\r\n\\]\r\nWe can always subtract the $ k \\equal{} 0$ term from the sum at the end." } { "Tag": [ "function", "limit", "irrational number", "real analysis", "real analysis unsolved" ], "Problem": "A cute problem :):\r\n\r\nProve or disprove the existence of a differentiable function $f:\\mathbb R\\to\\mathbb R$ which takes rationals to rationals and irrationals to irrationals, but which is not linear on any interval.", "Solution_1": "Since there is a rational number between any two irrational numbers and there is an irrational number between any two rational numbers, I think we can just let $f(x) = x$ if $x$ is rational and $f(x) = -x$ if $x$ is irrational.\r\n\r\nDoes this work?", "Solution_2": "I would like to see you differentiate that. :D", "Solution_3": "[quote=\"blahblahblah\"]I would like to see you differentiate that. :D[/quote]\r\n\r\nWhoops...I missed that part... :oops:", "Solution_4": "[hide=\"Nevermind\"]If $f$ is differentiable the it is continous. So if we have a sequence of converging rationals to a irrational number $\\lim f(r_{n})=f(\\lim r_{n})$. Does this mean that a rational equals an irational?[/hide]\r\n\r\nAny solutions?", "Solution_5": "@spix: *I thinhk that is not true...It only means that u have a sequence of rational numbers converging to an irrational one :wink:\r\n@grobber: That site requires special access? I can not open it, but i think the link is broken or something :(", "Solution_6": "[quote=\"grobber\"]A cute problem :):\n\nProve or disprove the existence of a differentiable function $f: \\mathbb R\\to\\mathbb R$ which takes rationals to rationals and irrationals to irrationals, but which is not linear on any interval.[/quote]\r\n\r\nf(x) = x/(1+|x|)", "Solution_7": "[quote=\"trichoplax\"][quote=\"grobber\"]A cute problem :):\n\nProve or disprove the existence of a differentiable function $f: \\mathbb R\\to\\mathbb R$ which takes rationals to rationals and irrationals to irrationals, but which is not linear on any interval.[/quote]\n\n$f(x) = x/(1+|x|)$[/quote]How about a $C^\\infty$-function?" } { "Tag": [], "Problem": "Hi everyone,\r\n\r\nCould someone please tell me if this is correct?\r\n\r\nThe number x is prime only if x is odd.\r\n\r\nP: X is odd\r\nQ: The number x is prime\r\n\r\nP arrow Q\r\nThe negation form of this is \r\n :huh: \r\n\r\nTherefore, it would be read as \"The number x is prime and x is not odd\"\r\n\r\nThank you very much", "Solution_1": "Yes I think it is correct.\r\n\r\nThe negation for $ p\\Rightarrow q$ is $ p\\wedge\\sim q$.\r\n\r\nYou can verify it by constructing the tables for both.\r\n\r\n\r\n :)" } { "Tag": [ "conics", "parabola", "symmetry", "search" ], "Problem": "An arch is in the shape of a parabola with a vertical axis. The arch is 15 ft high at the center and the 40 ft wide at the base. At which height above the base is the width 20 ft?\r\n\r\n$ V(0,15)$ so my equation becomes: $ (x\\minus{}0)^2\\equal{}\\minus{}4p(y\\minus{}15)\\rightarrow x^2\\equal{}\\minus{}4p(y\\minus{}15)$\r\n\r\n$ F(0,15\\minus{}p)$ & $ D: y\\equal{}15\\plus{}p$\r\n\r\nBase = 40 ft. So at the axis of symmetry, it's 20 ft to the left & right. So that means the x value we're at for a total of 20 ft is actually x = 10.\r\n\r\nI can substitute 10 into my equation, but I still can't grasp how to find p.\r\n\r\nJust a hint please :)", "Solution_1": "You can plot the point (20,0) to get p.", "Solution_2": "[quote=\"Carbon57\"]You can plot the point (20,0) to get p.[/quote]Keep going :p", "Solution_3": "First off, your 10 is wrong: you've split 40 into two halves, and then you inexplicably split it again. You've correctly figured out what value of $ x$ you want to be plugging in -- what you haven't figured out is what the associated $ y$ value is. Once you figure that out, you'll be able to plug both in and solve for $ p$.\r\n\r\nA similar topic was http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=544884669&t=146085", "Solution_4": "[quote=\"JBL\"]First off, your 10 is wrong: you've split 40 into two halves, and then you inexplicably split it again. You've correctly figured out what value of $ x$ you want to be plugging in -- what you haven't figured out is what the associated $ y$ value is. Once you figure that out, you'll be able to plug both in and solve for $ p$.\n\nA similar topic was http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=544884669&t=146085[/quote]Hmm, I don't see why my x value is wrong.\r\n\r\nThe x-intercepts are $ I_x(\\pm20,0)$, which gives my Arch a base of 40 ft. Wouldn't I have to split it again? I need to have a base of 20 ft, so my points for a 20 ft base are $ (\\pm10,y)$.\r\n\r\nNow let me go look at that thread you linked, thanks.", "Solution_5": "sorry, but what is the equation of a parabola when its easy to find the vertex and the axis of symmetry?\r\nisnt it something like $ y \\equal{} a(q \\minus{} p)^2$ ?\r\nalso is the answer? \r\n[hide]\n15\n[/hide]", "Solution_6": "The equation for a parabola is:\r\n\r\n$ y\\minus{}k\\equal{}\\frac{1}{4p}(x\\minus{}h)^2$\r\n\r\nRight?", "Solution_7": "lol! I GOT IT :) Thanks! I have my x-intercepts but yet I wasn't using them :p", "Solution_8": "ok i found it, isnt the equation of a parabola:\r\n$ y \\equal{} a(x\\minus{}p)^2\\plus{}q$ ?", "Solution_9": "[quote=\"neurosurgeon1118\"]sorry, but what is the equation of a parabola when its easy to find the vertex and the axis of symmetry?\nisnt it something like $ y \\equal{} a(q \\minus{} p)^2$ ?\nalso is the answer? \n[hide]\n15\n[/hide][/quote]No. I'm using the equation of the Parabola: Foci version, lol ...\r\n\r\n$ x^2 \\equal{} 4py$ etc\r\n\r\nAnswer is 11.25 ft" } { "Tag": [ "geometric series" ], "Problem": "Let\r\n\\[ w \\equal{} \\frac { \\minus{} 1 \\plus{} i\\sqrt {3}}{2}\r\n\\]\r\nwhere $ i \\equal{} \\sqrt { \\minus{} 1}$.\r\n\r\nCompute the product $ (1 \\minus{} w \\plus{} w^2)(1 \\plus{} w \\plus{} w^2)$\r\n\r\nis there an easier way to do this than just pluggin' and chuggin'?\r\n\r\ni got 0 by just working it out, but i was wondering if there is something that i should have noticed and just circled 0. :ninja: .", "Solution_1": "[quote=\"DaReaper\"]Let\n\\[ w = \\frac { - 1 + i\\sqrt {3}}{2}\n\\]\nwhere $ i = \\sqrt { - 1}$.\n\nCompute the product $ (1 - w + w^2)(1 + w + w^2)$\n\nis there an easier way to do this than just pluggin' and chuggin'?[/quote]\r\n[hide=\"Cis!\"]Simplifying that out we now have $ w^{4} + w^{2} + 1$. And notice that $ \\frac { - 1 + i\\sqrt {3}}{2}$ is the same as ${ - \\frac {1}{2} + \\frac {\\sqrt {3}}{2}}i$, which is the same as $ cis\\frac {2\\pi}{3}$. Then $ (cis\\frac {2\\pi}{3})^{4} + (cis\\frac {2\\pi}{3})^{2} + 1 = cis\\frac {8\\pi}{3} + cis\\frac {4\\pi}{3} + 1 = - \\frac {1}{2} + \\frac {\\sqrt{3}}{2}i - \\frac {1}{2} - \\frac {\\sqrt{3}}{2}i + 1 = -1 + 1=0$[/hide]", "Solution_2": "[hide=\"Hint\"]\nRecognize that $ w$ is a third root of unity, so $ w^3\\equal{}1$. Try some manipulation with this equation.\n[hide=\"Giveaway\"]\n$ (w\\minus{}1)(w^2\\plus{}w\\plus{}1)\\equal{}0$, and $ w\\ne1$, so...\n[/hide]\n[/hide]", "Solution_3": "You should immediately recognize $ \\omega\\equal{}\\frac{\\minus{}1\\plus{}i\\sqrt{3}}{2}$ as a third root of unity (at least I do after having seen it so many times).\r\n\r\nIn general, [hide]\nif $ \\omega$ is an $ n^{\\text{th}}$ root of unity, $ \\sum_{k\\equal{}0}^{n\\minus{}1}\\omega^{k}\\equal{}0$. Why? Use the geometric series closed form to simplify that and it should be obvious. Or look at the post above mine ;)\n[/hide]" } { "Tag": [ "Mafia" ], "Problem": "Seriously, what were the results? Also who got what role?", "Solution_1": "It was a game in a *Missouri room*...you expect it to be legit?", "Solution_2": "*cough* I'd still like results, you know.\r\n\r\nMASS ROLECLAIM EVERYBODY!\r\n\r\nI was townie game 1 and doctor game 2 and would-be mafia game e (which never happened?)", "Solution_3": "I ownt when I was jester. Tytia.", "Solution_4": "Yeah, I remember that one. I only played that one game, in which I was mafia. I recall Calvin kept kicking me in the back, so I moved from the foot of his bed to the couch.\r\n\r\nAround five minutes into the game.\r\nMe: *taps Max on the shoulder for water bottle* \r\nMax: Huh? Oh. (hands me water bottle)\r\nMe: Max is jester. (tosses water bottle. calvin catches.)\r\nMax: (WITHOUT the water bottle, I must note) No, we, uh, decided Bob was jester, remember?\r\nEveryone else: Yeah.\r\n(five seconds later)\r\nEveryone else: We vote to lynch max.\r\nMax: Haha, losers, I was jester. I win.\r\n\r\nBleh? Max was being pretty obvious about it - he claimed doctor (no one counterclaimed, surprisingly. I think calvin was doctor?) and was in general talked more than anyone else about who was who - a \"clear\" sign of mafia guilt. I knew he wasn't mafia, but also knew he wasn't doctor (I heard John come over to my side of the room to pick doctor, and Max was on the other side), so he had to be jester.\r\n\r\nThat was a fail game.", "Solution_5": "Well that was cuz I thought I was jester since [i]someone[/i] tapped me when he was like \"I am tapping the jester\" (ahem John...)", "Solution_6": "aaw now i'm too old for this sort of thing *cries*", "Solution_7": "srsly yeah, time flies" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "$\\sum_{i=2}^{n}\\frac{1}{1+i^{2}+i^{4}}=?$", "Solution_1": "Hint: $x^{4}+x^{2}+1 = (x^{2}+x+1)(x^{2}-x+1)$, $(x+1)^{2}-(x+1)+1 = x^{2}+x+1$.", "Solution_2": "$\\sum_{i=1}^{n}\\frac{1}{1+i^{2}+i^{4}}=\\sum_{i=1}^{n}(\\frac{1}{2i}(\\frac{1}{i^{2}-i+1}-\\frac{1}{i^{2}+i+1}))=\\sum_{i=1}^{n}(\\frac{1}{2i}(\\frac{1}{i^{2}-i+1}-\\frac{1}{(i+1)^{2}-(i+1)+1})$\r\nand then what?", "Solution_3": "\\[\\sum_{k=1}^{n}\\frac1{k^{4}+k^{2}+1}=\\sum_{k=1}^{n}\\frac1{2k}\\left(\\frac1{k^{2}-k+1}-\\frac1{k^{2}+k+1}\\right)=\\frac12\\left(\\sum_{k=1}^{n}\\frac1{k^{2}+k+1}\\left(\\frac1{k+1}-\\frac1k\\right)+1-\\frac1{(n+1)(n^{2}+n+1)}\\right)=\\]\r\n\\[=\\frac12\\left(\\sum_{k=1}^{n}\\left(\\frac1{k^{2}+k+1}+\\frac1{k+1}-\\frac1k\\right)+1-\\frac1{(n+1)(n^{2}+n+1)}\\right)=\\]\r\n\\[=\\frac12\\left(\\sum_{k=1}^{n}\\frac1{k^{2}+k+1}+\\frac1{n+1}-\\frac1{(n+1)(n^{2}+n+1)}\\right)=\\]\r\n\\[=\\frac12\\left(\\sum_{k=1}^{n}\\frac1{k^{2}+k+1}+\\frac n{n^{2}+n+1}\\right)\\]\r\nBut I have no idea what to do with the sum $\\sum_{k=1}^{n}\\frac1{k^{2}+k+1}$ :( . Does anybody? Maybe this is well known :maybe: ?", "Solution_4": "One way is using the residue theorem from complex analysis...\r\n$\\sum_{k=1}^{+\\infty}\\frac{1}{k^{4}+k^{2}+1}=\\frac{\\pi\\sqrt{3}}{6}\\tanh\\left(\\frac{\\pi\\sqrt{3}}{2}\\right)-\\frac{1}{2}$\r\n\r\nEDIT: I see now that a finite sum was intended, but in view of the above result, I suspect no nice closed form for the finite form..." } { "Tag": [], "Problem": "how many people do they take to nationals?", "Solution_1": "4 per state", "Solution_2": "there is also 7 other territories/d.c. and department of defense. so 228 people + coaches go to nationals.", "Solution_3": "*FYI, coach of winning team goes." } { "Tag": [ "limit", "integration", "inequalities", "function", "real analysis", "real analysis unsolved" ], "Problem": "Compute $ \\underset{n\\to \\infty }{\\mathop{\\lim }}\\,\\int_{0}^{1}{\\frac {dx}{\\left( 1 \\plus{} x^{n} \\right)\\sqrt [n]{1 \\plus{} x^{n}}}}.$\r\n\r\nI think we can apply the DCT here but don't know how to find a bound for the integrand.", "Solution_1": "$ g(x)\\equal{}1$ on $ [0,1]$ makes a perfectly good dominating function.", "Solution_2": "Okay, then do you imply that is $ f\\le g,$ since the integrand is positive.\r\n\r\nBut my huge question is how do you find that bound!! :(", "Solution_3": "Bernoulli's inequality implies:\r\n$ \\left(1 \\plus{} x^n\\right)^{1 \\plus{} \\frac{1}{n}} > 1 \\plus{} x^n \\left(1 \\plus{} \\frac{1}{n} \\right) > 1$\r\nNow take the reciprocals and the inequalities reverse, giving you the dominant function you were after.", "Solution_4": "You don't have to be precise about that sort of thing- just that 1+something is bigger than 1, an nth root of something bigger than 1 is bigger than 1, multiplying keeps us there, and then dividing by the whole thing flips it to being less than 1.\r\nIt should be obvious; it took me about a second to see the bound. Be bold and make crude estimates; you can always refine later if they aren't good enough.\r\n\r\n[Edit- Bernoulli's inequality? Serious overkill.]", "Solution_5": "Ahh well, reading it as you phrased it, then I have $ 1\\plus{}x^n\\ge1$ and $ \\sqrt[n]{1\\plus{}x^n}\\ge1$ and then that produces faster our result.\r\n\r\nMy problem was that we have $ 0\\le x\\le1$ and then I was complicating myself on finding a bound with that requirement, but now it's clear." } { "Tag": [ "calculus", "derivative", "quadratics", "algebra", "polynomial", "geometry", "perpendicular bisector" ], "Problem": "You are walking north on a straight path. Ahead of you and on the east side of the path is\r\na beautiful billboard 54 feet across. The billboard is aligned perpendicularly to the path,\r\nand the west edge of the billboard is 18 feet from the path. At what point on the path\r\nwould you see the billboard from the widest possible angle?", "Solution_1": "This is for computations and tutorials. \r\n\r\nLet $x$ be the distance between you and the point on the path that is closest to the billboard. The angle in question is $\\mathrm{arccot}(x/72)-\\mathrm{arccot}(x/18)$ (why?) -- take derivative and equate it to zero.", "Solution_2": "I am still stuck on isolating x once I set it equal to 0.", "Solution_3": "Don't you get a quadratic equation for $x$? And of the simplest kind, without linear term.", "Solution_4": "In this case, the near end of the billboard is 18 feet from the path and the far end is 72 feet from the path. The correct answer should be $x=36$ feet.\r\n\r\nFor the generalized version, in which the near end is distance $a$ from the path and the far end distance $b,$ the $x$ that maximizes the viewing angle works out to be $\\sqrt{ab},$ the geometric mean of $a$ and $b.$ (Among other things, that implies that from that point, if you look at a 45 degree angle, you'll always be looking into the interior of the billboard.)\r\n\r\nIt's a fairly staightforward calculus problem, assuming you know how to differentiate arccot (or arctan, if you do the labeling a little differently) and can handle the resulting polynomial algebra. But I'm interested in the non-calculus solution. Can you arrive at this conclusion using only concepts in Euclidean geometry?", "Solution_5": "[quote=\"Kent Merryfield\"]But I'm interested in the non-calculus solution. Can you arrive at this conclusion using only concepts in Euclidean geometry?[/quote]\r\n[hide]\nLet $A,B$ the endpoints of the billboard.\nTake $P$ be a point on the line $d$ (the road that is). Consider the circle of center $O_{P}$ which passes through $A,B,P.$\nSince $\\measuredangle A O_{P}B = 2 \\measuredangle A P B$, it suffices to maximize $\\measuredangle A O_{P}B.$\nWe know that $O_{P}$ moves on the perpendicular bisector of $AB$ as $P$ varies. The closer $O_{P}$ is to $AB$, the bigger the angle $\\measuredangle A O_{P}B$ is. But keep in mind the fact that the circle $\\left( O_{P}\\right)$ must cut $d.$\nA bit of intuition will help you see that the minimum corresponds exactly to the case when the circle is tangent to $d$, in a point which we'll call $X$. Now let $AB \\cap D = \\left\\{ W \\right\\}.$ Then, by the power of a point theorem, $WA \\cdot WB = WX^{2}$, so that\n\\[WX = \\sqrt{WA \\cdot WB}= \\sqrt{ab}. \\]\n[/hide]", "Solution_6": "perfect_radio's post is exactly the argument I was looking for." } { "Tag": [ "inequalities", "algebra" ], "Problem": "If $ a,b,c$ are three positive real numbers, prove that $ \\frac {a^{2}\\plus{}1}{b\\plus{}c}\\plus{}\\frac {b^{2}\\plus{}1}{c\\plus{}a}\\plus{}\\frac {c^{2}\\plus{}1}{a\\plus{}b}\\ge 3$", "Solution_1": "[quote=\"makar\"]If $ a,b,c$ are three positive real numbers, prove that $ \\frac {a^{2} \\plus{} 1}{b \\plus{} c} \\plus{} \\frac {b^{2} \\plus{} 1}{c \\plus{} a} \\plus{} \\frac {c^{2} \\plus{} 1}{a \\plus{} b}\\ge 3$[/quote]\r\n$ \\sum \\frac{a^2\\plus{}1}{b\\plus{}c} \\ge \\frac{2a}{b\\plus{}c} \\ge 3$ (nebsit inequality)", "Solution_2": "it's equivalent to \r\n\r\n$ \\frac{(a\\minus{}1)^2}{b\\plus{}c}\\plus{}\\frac{(b\\minus{}1)^2}{a\\plus{}c}\\plus{}\\frac{(c\\minus{}1)^2}{a\\plus{}b}\\plus{}\\frac{2(a\\minus{}b)^2}{(a\\plus{}c)(b\\plus{}c)}\\plus{}\\frac{2(b\\minus{}c)^2}{(a\\plus{}c)(a\\plus{}b)}\\plus{}\\frac{2(a\\minus{}c)^2}{(a\\plus{}b)(b\\plus{}c)}\\ge0$", "Solution_3": "[quote=\"makar\"]If $ a,b,c$ are three positive real numbers, prove that $ \\frac {a^{2} \\plus{} 1}{b \\plus{} c} \\plus{} \\frac {b^{2} \\plus{} 1}{c \\plus{} a} \\plus{} \\frac {c^{2} \\plus{} 1}{a \\plus{} b}\\ge 3$[/quote]\r\nThe inequality is obviously true because $ (a^2 \\plus{} 1)(b^2 \\plus{} 1)(c^2 \\plus{} 1) \\geq (a \\plus{} b)(b \\plus{} c)(a \\plus{} c)$ holds for all positive real numbers $ a,b$ and $ c$.", "Solution_4": "rmo 2006 guys were too lucky!", "Solution_5": "This is very easy problem.", "Solution_6": "That is easily solved by AM-GM and cs \r\n \u2211(a^2+1)/(b+c)\u2265\u22112a/(b+c)\u2265\u22112a^2/(ab+ac)\u22652(\u2211a)^2/2\u2211bc\u22653", "Solution_7": "can also be done very easily using titu's lemma\n", "Solution_8": "[quote=Zarif]This is very easy problem.[/quote]\nApplying AM-GM and Nesbitt`s inequality we have ", "Solution_9": "If $ a,b,c$ are three positive real numbers, then $ \\frac {a^{2}+1}{b+c}+\\frac {b^{2}+1}{c+a}+\\frac {c^{2}+1}{a+b}\\ge \\frac {a^{2}+1}{a+b}+\\frac {b^{2}+1}{b+c}+\\frac {c^{2}+1}{c+a}\\ge 3\\sqrt[3]{\\frac {(a^{2}+1)(b^{2}+1)(c^{2}+1)}{(a+b)(b+c)(c+a)}}\\ge 3.$", "Solution_10": "Take $x = b + c, y = c + a, z = a + b$.\\\\\nThus, $a = \\dfrac{y+z-x}{2}, b = \\dfrac{z+x-y}{2}, z = \\dfrac{x+y-z}{2}$.\\\\\n\nTherefore, $\\dfrac{a}{b+c} + \\dfrac{b}{c+a} + \\dfrac{c}{a+b} = \\dfrac{y+z-x}{2x}+ \\dfrac{z+x-y}{2y}+ \\dfrac{x+y-z}{2z}$\\\\\n$\\hspace{5.5cm} = \\dfrac{1}{2}\\left[ \\left(\\dfrac{x}{y} + \\dfrac{y}{x}\\right) + \\left(\\dfrac{y}{z} + \\dfrac{z}{y}\\right) + \\left(\\dfrac{z}{x} + \\dfrac{x}{z}\\right)\\right] $\\\\\n$\\hspace{5.5cm}\\geq \\dfrac{1}{2}[2 + 2 +2] = 3.$", "Solution_11": "[quote=makar]If $ a,b,c$ are three positive real numbers, prove that $ \\frac {a^{2}\\plus{}1}{b\\plus{}c}\\plus{}\\frac {b^{2}\\plus{}1}{c\\plus{}a}\\plus{}\\frac {c^{2}\\plus{}1}{a\\plus{}b}\\ge 3$[/quote]\n\nSorry for reviving this thread for such a mere solution of mine, but its a little different.\n\nBy Titu's Lemma,\n$ \\frac {a^{2}\\plus{}1}{b\\plus{}c}\\plus{}\\frac {b^{2}\\plus{}1}{c\\plus{}a}\\plus{}\\frac {c^{2}\\plus{}1}{a\\plus{}b}$\n$\\geq \\frac{(a+b+c)^2}{2(a+b+c)} + \\frac{3^2}{2(a+b+c)}$\nWrite $x=\\sum_{a,b,c} a$\nTherefore, we have to show that $\\frac{x^2+9}{2x} \\geq 3$\n$\\implies x^2+9 \\geq 6x$\n$\\implies (x-3)^2+6x \\geq 6x$\n$\\implies (x-3)^2 \\geq 0$ and that's trivial as any square is $\\geq 0$.", "Solution_12": "Suppose $a\\geq{b}\\geq{c}$,then $a^2+1\\geq{b^2+1}\\geq{c^2+1}$ and $\\frac{1}{b+c}\\geq\\frac{1}{c+a}\\geq\\frac{1}{a+b}$.By chebyshev inequality,we have\n$$LHS\\geq{\\frac{1}{3}(a^2+1+b^2+1+c^2+1)(\\frac{1}{b+c}+\\frac{1}{c+a}+\\frac{1}{a+b})}\\geq{\\frac{1}{3}(b+c+c+a+a+b)(\\frac{1}{b+c}+\\frac{1}{c+a}+\\frac{1}{a+b})}\\geq{3}$$", "Solution_13": "[quote=makar]If $ a,b,c$ are three positive real numbers, prove that $ \\frac {a^{2}\\plus{}1}{b\\plus{}c}\\plus{}\\frac {b^{2}\\plus{}1}{c\\plus{}a}\\plus{}\\frac {c^{2}\\plus{}1}{a\\plus{}b}\\ge 3$[/quote]\n$x,y,z>0$,prove that:\n\n\\[{\\frac {1+x}{\\sqrt {y+z}}}+{\\frac {1+y}{\\sqrt {z+x}}}+{\\frac {z+1}{\n\\sqrt {x+y}}}\\geq 3\\,\\sqrt {2}\\]\n", "Solution_14": "By Titu inequality and AM-GM inequality,\n\n$\\sum_{cyc}\\frac{a^2+1}{b+c}\\geq \\frac{(a+b+c)^2}{2(a+b+c)}+\\frac{3^2}{2(a+b+c)}\\geq 2\\sqrt{\\frac{a+b+c}{2}\\cdot \\frac{9}{2(a+b+c)}}\\geq 3$\n\n:D", "Solution_15": "[quote=xzlbq][quote=makar]If $ a,b,c$ are three positive real numbers, prove that $ \\frac {a^{2}\\plus{}1}{b\\plus{}c}\\plus{}\\frac {b^{2}\\plus{}1}{c\\plus{}a}\\plus{}\\frac {c^{2}\\plus{}1}{a\\plus{}b}\\ge 3$[/quote]\n$x,y,z>0$,prove that:\n\n\\[{\\frac {1+x}{\\sqrt {y+z}}}+{\\frac {1+y}{\\sqrt {z+x}}}+{\\frac {z+1}{\n\\sqrt {x+y}}}\\geq 3\\,\\sqrt {2}\\][/quote]\n\n[b]Proof[/b] [b]:[/b] Since $$\\sum \\frac{2x}{\\sqrt{y+z}}\\ge \\sum \\sqrt{y+z}$$ $$\\Leftrightarrow \\sum \\frac{(x-y)^2}{\\sqrt{(x+z)(y+z)}(\\sqrt{x+z}+\\sqrt{y+z})}\\ge 0$$ thus $$\\sum \\frac{x+1}{\\sqrt{y+z}}\\ge $$ $$\\ge \\frac{1}{2}\\sum \\sqrt{y+z}+\\sum \\frac{1}{\\sqrt{y+z}}\\ge 3\\sqrt{2}$$", "Solution_16": "$ \\frac {a^{2}+1}{b+c}+\\frac {b^{2}+1}{c+a}+\\frac {c^{2}+1}{a+b}\\ge 2\\left(\\frac {a}{b+c}+\\frac {b}{c+a}+\\frac {c}{a+b}\\right)\\ge 3$\nFirst by AM-GM, second by Nesbitt's\nEquality iff $a=b=c=1$", "Solution_17": "[quote=WolfusA]$ \\frac {a^{2}+1}{b+c}+\\frac {b^{2}+1}{c+a}+\\frac {c^{2}+1}{a+b}\\ge 2\\left(\\frac {a}{b+c}+\\frac {b}{c+a}+\\frac {c}{a+b}\\right)\\ge 3$\nFirst by AM-GM, second by Nesbitt's\nEquality iff $a=b=c=1$[/quote]\n\nBrilliant, oh no, there was analogous proof before.", "Solution_18": "[quote=makar]If $ a,b,c$ are three positive real numbers, prove that $ \\frac {a^{2}\\plus{}1}{b\\plus{}c}\\plus{}\\frac {b^{2}\\plus{}1}{c\\plus{}a}\\plus{}\\frac {c^{2}\\plus{}1}{a\\plus{}b}\\ge 3$[/quote]\n\nA different solution:\n\\[\\sum \\frac{a^2+1}{b+c} \\geq \\sum \\frac{a^2+1}{\\sqrt{(b^2+1)(c^2+1)}} \\geq 3 \\sqrt[3]{\\prod \\frac{a^2+1}{\\sqrt{(b^2+1)(c^2+1)}}}=3\\]", "Solution_19": "[quote=janitor]\n\nBrilliant, oh no, there was analogous proof before.[/quote]\n\nI didn't read it though.", "Solution_20": "Apply AM GM and nesbitt's inequality ", "Solution_21": "This can be solved by Jensen's inequality taking the function as f(x)=(x\u00b2+1)/((a+b+c)-x)\n", "Solution_22": "$LHS=\\frac{a^2}{b+c}+\\frac{1}{b+c}+\\frac{b^2}{a+c}+\\frac{1}{a+c}+\\frac{c^2}{a+b}+\\frac{1}{a+b}\\geq\\frac{(a+b+c+3)^2}{4(a+b+c)}$ by Titu lemma\nNow, let $a+b+c=k$\nWe need to prove $\\frac{(k+3)^2}{4k}\\geq 3$ \nOr, $k^2+6k+9\\geq 12k$ \nOr, $(k-3)^2\\geq 0$ which is trivial [s]by trivial inequality[/s]", "Solution_23": "[quote=Leooooo][quote=xzlbq][quote=makar]If $ a,b,c$ are three positive real numbers, prove that $ \\frac {a^{2}\\plus{}1}{b\\plus{}c}\\plus{}\\frac {b^{2}\\plus{}1}{c\\plus{}a}\\plus{}\\frac {c^{2}\\plus{}1}{a\\plus{}b}\\ge 3$[/quote]\n$x,y,z>0$,prove that:\n\n\\[{\\frac {1+x}{\\sqrt {y+z}}}+{\\frac {1+y}{\\sqrt {z+x}}}+{\\frac {z+1}{\n\\sqrt {x+y}}}\\geq 3\\,\\sqrt {2}\\][/quote]\n\n[b]Proof[/b] [b]:[/b] Since $$\\sum \\frac{2x}{\\sqrt{y+z}}\\ge \\sum \\sqrt{y+z}$$ $$\\Leftrightarrow \\sum \\frac{(x-y)^2}{\\sqrt{(x+z)(y+z)}(\\sqrt{x+z}+\\sqrt{y+z})}\\ge 0$$ thus $$\\sum \\frac{x+1}{\\sqrt{y+z}}\\ge $$ $$\\ge \\frac{1}{2}\\sum \\sqrt{y+z}+\\sum \\frac{1}{\\sqrt{y+z}}\\ge 3\\sqrt{2}$$[/quote]\n\nNice!", "Solution_24": "$a,b,c,d>0$,prove\n\n\\[{\\frac {1+a}{\\sqrt {b+c+d}}}+{\\frac {1+b}{\\sqrt {a+c+d}}}+{\\frac {1+c}\n{\\sqrt {d+a+b}}}+{\\frac {1+d}{\\sqrt {a+b+c}}}\\geq \\frac{8}{3}\\,\\sqrt {3}\\]", "Solution_25": "2-var hold.\n\n$x,y>0$,prove that\n\n\\[{\\frac {1+x}{\\sqrt {y}}}+{\\frac {1+y}{\\sqrt {x}}}\\geq 4\\]", "Solution_26": "\n$a,b,c,d,e>0$,prove that\n\n\\[{\\frac {1+a}{\\sqrt {b+c+d+e}}}+{\\frac {1+b}{\\sqrt {c+d+e+a}}}+{\\frac {\n1+c}{\\sqrt {d+e+a+b}}}+{\\frac {1+d}{\\sqrt {e+a+b+c}}}+{\\frac {1+e}{\n\\sqrt {a+b+c+d}}}\\geq 5\\]\n", "Solution_27": "$n$-var hold\n\n$x_i>0,i=1,...,n,n\\geq 2,S=\\sum_{i=1}^n{x_i}$,prove that\n\n\\[\\sum_{i=1}^n{\\frac{(1+x_i)}{\\sqrt{S-x_i}}}\\geq 2\\,{\\frac {n}{\\sqrt {n-1}}}\\]", "Solution_28": "[quote=xzlbq]$a,b,c,d>0$,prove\n\n\\[{\\frac {1+a}{\\sqrt {b+c+d}}}+{\\frac {1+b}{\\sqrt {a+c+d}}}+{\\frac {1+c}\n{\\sqrt {d+a+b}}}+{\\frac {1+d}{\\sqrt {a+b+c}}}\\geq \\frac{8}{3}\\,\\sqrt {3}\\][/quote]\n\nBy Holder,need to prove\n\n${\\frac { \\left( \\left( 1+a \\right) ^{2} \\left( b+c+d+1 \\right) +\n \\left( 1+b \\right) ^{2} \\left( c+d+a+1 \\right) + \\left( 1+c \\right) ^\n{2} \\left( d+a+b+1 \\right) + \\left( 1+d \\right) ^{2} \\left( a+b+c+1\n \\right) \\right) ^{3}}{ \\left( b+c+d \\right) \\left( 1+a \\right) ^{4}\n \\left( b+c+d+1 \\right) ^{3}+ \\left( a+c+d \\right) \\left( 1+b\n \\right) ^{4} \\left( c+d+a+1 \\right) ^{3}+ \\left( d+a+b \\right) \n \\left( 1+c \\right) ^{4} \\left( d+a+b+1 \\right) ^{3}+ \\left( a+b+c\n \\right) \\left( 1+d \\right) ^{4} \\left( a+b+c+1 \\right) ^{3}}}$\n$\\geq {\n\\frac {64}{3}}$\n", "Solution_29": "$x,y,z>0$,prove that\n\n\\[{\\frac {yz+1}{\\sqrt { \\left( z+x \\right) \\left( x+y \\right) }}}+{\n\\frac {xz+1}{\\sqrt { \\left( x+y \\right) \\left( y+z \\right) }}}+{\n\\frac {xy+1}{\\sqrt { \\left( y+z \\right) \\left( z+x \\right) }}}\\geq 2\\,\n\\sqrt {2+2\\,{\\frac {xyz}{ \\left( y+z \\right) \\left( z+x \\right) \n \\left( x+y \\right) }}}\\]\n", "Solution_30": "[quote=xzlbq]$x,y,z>0$,prove that\n\n\\[{\\frac {yz+1}{\\sqrt { \\left( z+x \\right) \\left( x+y \\right) }}}+{\n\\frac {xz+1}{\\sqrt { \\left( x+y \\right) \\left( y+z \\right) }}}+{\n\\frac {xy+1}{\\sqrt { \\left( y+z \\right) \\left( z+x \\right) }}}\\geq 2\\,\n\\sqrt {2+2\\,{\\frac {xyz}{ \\left( y+z \\right) \\left( z+x \\right) \n \\left( x+y \\right) }}}\\][/quote]\n\nBy Holder,need to prove\n\n\\[{\\frac { \\left( {\\it \\sum} \\left( \\left( yz+1 \\right) ^{2} \\left( 3\\,\nx+2\\,y+2\\,z \\right) \\right) \\right) ^{3}}{{\\it \\sum} \\left( \\left( \nz+x \\right) \\left( x+y \\right) \\left( yz+1 \\right) ^{4} \\left( 3\\,x+\n2\\,y+2\\,z \\right) ^{3} \\right) }}\\geq 8+8\\,{\\frac {xyz}{ \\left( y+z\n \\right) \\left( z+x \\right) \\left( x+y \\right) }}\\]", "Solution_31": "Use Tittu Lemma, after separating 1 from each numerator and we can apply it on 3 terms for 2 time and at last apply AM-GM... ", "Solution_32": "Let $ a,b,c$ are three positive real numbers. Prove that\n$$ \\frac {a^{2}+1}{a+b}+\\frac {b^{2}+1}{b+c}+\\frac {c^{2}+1}{c+a}\\ge 3$$\n[quote=makar]If $ a,b,c$ are three positive real numbers, prove that $$ \\frac {a^{2}\\plus{}1}{b\\plus{}c}\\plus{}\\frac {b^{2}\\plus{}1}{c\\plus{}a}\\plus{}\\frac {c^{2}\\plus{}1}{a\\plus{}b}\\ge 3$$[/quote]\n$$\\sum \\frac {a^2+1}{a+b}\\ge \\frac {1}{2}\\sum \\frac {(a+1)^2}{a+b}\\ge \\frac {1}{2} \\frac {(\\sum a+3)^2}{2 \\sum a}\\ge 3$$\n$$\\sum \\frac {a^2+1}{b+c}\\ge \\frac {1}{2}\\sum \\frac {(a+1)^2}{b+c}\\ge \\frac {1}{2} \\frac {(\\sum a+3)^2}{2 \\sum a}\\ge 3$$\n[url]https://math.stackexchange.com/questions/2452783/prove-that-a2-1b2-1c2-1-ge-a-bb-cc-a-for-a-b-c?noredirect=1[/url]\n$$(a^2 + 1)(1+b^2) \\ge (a + b)^2$$", "Solution_33": "Let $ a,b,c$ are three positive real numbers. Prove that \n$$(a^2+1)(b^2+1)(c^2+1)\\ge \\frac {1}{27} (5a+b)(5b+c)(5c+a) $$\n$$(a^2+1)(b^2+1)(c^2+1)\\ge \\frac {8}{27} (2a+b)(2b+c)(2c+a) $$\n$$(a^2+1)(b^2+1)(c^2+1)\\ge \\frac {8}{(k+1)^3} (ka+b)(kb+c)(kc+a) $$\nWhere $k=1,2,3,4,5.$", "Solution_34": "Let $ a,b,c$ are real numbers. [url=https://math.stackexchange.com/questions/2452783/prove-that-a2-1b2-1c2-1-ge-a-bb-cc-a-for-a-b-c?noredirect=]Prove that[/url]\n$$(a^2 + 1)(b^2 + 1)(c^2 + 1) \\ge (a + b)(b + c)(c + a)$$\n[url=https://math.stackexchange.com/questions/1955141/if-a-b-and-c-are-three-positive-real-numbers-prove-that-fraca21b?rq=1]h[/url] [url=https://artofproblemsolving.com/community/c6h383100p2124305]h[/url]\n$$\\implies$$\nLet $a,b,c>0 $. Prove that$$\\frac{a^2+1}{b+c} +\\frac{(b^2+1)(c^2+1)}{(c+a)(a+b)}\\geq 2$$\n$$\\frac{a^2+1}{a+b} +\\frac{(b^2+1)(c^2+1)}{(b+c)(c+a)}\\geq 2$$" } { "Tag": [ "function", "geometry", "geometric transformation", "algebra", "domain", "complex numbers", "AMC" ], "Problem": "A function f(z) is defined on the complex numbers by f(z)=(a+bi)z where a and b are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin.\r\n\r\nThis isnt the whole entire question, but is the image of each point f(z) and when it says its equidistant, is it equidistant to z and (0+0i)?\r\n\r\nIf it is, why is f(z) an image?", "Solution_1": "That's what it means. Or another way,\r\n\r\n$ |f(z) \\minus{} z| \\equal{} |f(z) \\minus{} 0|$", "Solution_2": "Yes, the number 0 is viewed as the origin in the complex plane. In a general setting, \"zero\" and \"the origin\" are usually interchangeable terms. Sometimes, we denote the complex number $ a\\plus{}bi$ as $ (a,b)$, emphasizing that the complex plane really is the Euclidean plane.\r\n\r\nIt's a standard mathematical convention to speak of functions as \"mappings\" which take objects to their \"images\" under the mapping. You may be more familiar with this language in a geometric setting---if you apply a translation to a circle, then your original circle's [i]image[/i] under this translation (a certain type of function) is a congruent circle in a different location. So in general, if $ f$ is a function, and $ z$ is an object in the domain of $ f$, we speak of $ f(z)$ as the image of $ z$ under the mapping $ f$." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "prove that\r\n\r\n$ (\\forall a,b \\in \\mathbb{R}^{*\\plus{}});(1\\plus{}a^2)(1\\plus{}b^2) \\ge a(1\\plus{}b^2)\\plus{}b(1\\plus{}a^2)$", "Solution_1": "Hi :) \r\n\r\n$ (1 \\plus{} a^{2})(1 \\plus{} b^{2}) \\equal{} \\frac {(1 \\plus{} a^{2})(1 \\plus{} b^{2})}{2} \\plus{} \\frac {(1 \\plus{} a^{2})(1 \\plus{} b^{2})}{2} \\geq a(1 \\plus{} b^{2}) \\plus{} b(1 \\plus{} a^{2})$\r\n\r\nbecause\r\n\r\n$ (1 \\minus{} x)^2 \\geq 0 \\equal{} >$\r\n$ 1 \\plus{} x^2 \\geq 2x$", "Solution_2": "See here : http://www.mathlinks.ro/Forum/viewtopic.php?t=318272 :wink:" } { "Tag": [ "IMC", "college contests" ], "Problem": "For a permutation $ \\sigma\\in S_n$ with $ (1,2,\\dots,n)\\mapsto(i_1,i_2,\\dots,i_n)$, define\r\n\\[ D(\\sigma) \\equal{} \\sum_{k \\equal{} 1}^n |i_k \\minus{} k|\r\n\\]\r\nLet\r\n\\[ Q(n,d) \\equal{} \\left|\\left\\{\\sigma\\in S_n : D(\\sigma) \\equal{} d\\right\\}\\right|\r\n\\]\r\nShow that when $ d \\geq 2n$, $ Q(n,d)$ is an even number.", "Solution_1": "[hide=\"Hint\"]A very interesting idea is to consider the determinant \n$ \\Delta(x)\\equal{}|a_{ij}|$ where $ a_{ij}\\equal{}x^{|i\\minus{}j|}$. \nTherefore $ \\Delta(x)\\equal{}\\sum_{\\sigma \\in S_{n}}(\\minus{}1)^{Inv(\\sigma)}x^{D(\\sigma)}$. So $ Q(n,d)$ has the same parity as the coefficient of $ x^d$ in $ \\Delta(x)$.\nComputing $ \\Delta(x)$ we get $ (1\\minus{}x^2)^{n\\minus{}1}$ so for $ d\\geq 2n$ the coefficient of $ x^d$ is 0 in $ \\Delta(x)$ so $ Q(n,d)$ is even.[/hide]", "Solution_2": "An alternative method:\r\n\r\nThe involution $ \\sigma \\leftrightarrow \\sigma^{\\minus{}1}$ preserves $ D(\\sigma)$, so the parity of $ Q(n,d)$ is the same as that of $ R(n,d)$, where $ R(n,d)$ only counts permutations which are their own inverse (the remaining permutations all pair off). Every permutation counted in $ R(n,d)$ is the product of disjoint 2-cycles. We next perform the following involution: \r\n\r\nGiven a permutation $ \\sigma$, let $ (i,j)$ be the smallest pair lexicographically (assuming one exists) satisfying $ \\max\\{i,j\\}<\\min\\{\\sigma(i), \\sigma(j)\\}$. We replace the cycles $ (i,\\sigma(i))$ and $ (j,\\sigma(j))$ by $ (i,\\sigma(j))$ and $ (j, \\sigma(i))$. Again this preserves $ D(\\sigma)$, so again we can pair off and reduce to computing the parity of $ S(n,d),$ where $ S$ is the collection of permutations that are their own inverses such that no such $ (i,j)$ exists. \r\n\r\nConsider such a permutation, and let $ i<\\sigma(i)$ be any point which is not fixed by it. If $ i \\frac{5}{4}$\r\nEquality doesn't occur. Indeed, I don't like it :wink:" } { "Tag": [ "invariant", "Functional Analysis", "combinatorial geometry", "advanced fields", "advanced fields unsolved" ], "Problem": "$C$ a convex, closed, bounded subset of a Hilbert space (not compact ..). Suppose that $f: C\\to C$ \r\nsatisfies $|f(x)-f(y)| \\leq |x-y|$ for any $x,y \\in C$. Prove that $f$ has at least \r\na fixed point.", "Solution_1": "I've recently learned about a generalization. First some terminology:\r\n\r\nWe say that a point $x$ of a bounded subset $K$ of a normed space is [i]diametral[/i] if $\\sup_{y\\in K}\\|x-y\\|=\\rho(K)$, the diameter of $K$. Naturally, a non-diametral point will simply be a point which is not diametral.\r\n\r\nLet $K$ be a closed, convex, bounded subset of a normed space. We say that $K$ has [i]normal structure[/i] if every closed, convex subset of $K$ which has at least two points has a non-diametral point.\r\n\r\nNotice that every closed, convex, bounded subset $K$ containing at least two points of a uniformly convex Banach space $B$ has a non-diametral point, and hence every closed, convex, bounded subset of such a Banach space has normal structure. Indeed, let $x,y$ be two distinct points of $K$. The definition of uniform convexity easily shows that $\\frac{x+y}2$ cannot be diametral. \r\n\r\n\r\nThe generalization I was talking about is this:\r\n\r\n$(*)$ Let $K$ be a weakly compact convex subset of a Banach space $B$, having normal structure. Then every map $f: K\\to K$ such that $\\|f(x)-f(y)\\|\\le\\|x-y\\|,\\ \\forall x,y\\in K$ (i.e. a so-called non-expansive map) has a fixed point.\r\n\r\nBecause a Hilbert space is reflexive, a closed, convex, bounded subset is weakly compact. On the other hand, because a Hilbert space is uniformly convex, by the observation above, a closed, convex, bounded subset has normal structure. Hence the hypotheses of $(*)$ are satisfied, so the problem posted by alekk is indeed a consequence of $(*)$. \r\n\r\n\r\nNow for a proof of $(*)$:\r\n\r\nConsider the collection $\\mathcal C$ of all $f$-invariant closed, convex subsets of $K$. All such sets are weakly compact, so a descending chain has non-empty inetrsection. This means that $\\mathcal C$, partially ordered by inclusion, satisfies the hypotheses of Zorn's Lemma, and must thus have minimal elements. Let $\\widetilde K$ be such a minimal element. $\\widetilde K$ either consists of one point only, in which case we are done, or has at least two points. The aim will now be to show that the latter cannot occur.\r\n\r\nSince $K$ has normal structure, $\\widetilde K$ has a non-diametral point $x_{0}$. From now on we forget about $K$ and only work inside $\\widetilde K$. \r\n\r\nThere is a $t>0$ such that $\\sup_{y\\in\\widetilde K}\\|x_{0}-y\\|\\le\\rho\\left(\\widetilde K\\right)-t$. Now consider the (non-empty, because it contains $x_{0}$) set $L=\\left\\{x\\in\\widetilde K\\ |\\ sup_{y\\in\\widetilde K}\\|x-y\\|\\le\\rho\\left(\\widetilde K\\right)-t\\right\\}$. $L$ is closed, because the map $x\\mapsto\\sup_{y\\in\\widetilde K}\\|x-y\\|$ is continuous. Also, notice that $L$ is convex: if $x_{i}\\in L,\\ i=1,2$ and $y\\in\\widetilde K$, and $\\lambda\\in[0,1]$, then $\\|\\lambda x_{1}+(1-\\lambda)x_{2}-y\\|\\le\\lambda\\|x_{1}-y\\|+(1-\\lambda)\\|x_{2}-y\\|\\le\\rho\\left(\\widetilde K\\right)-t$. Finally, $L$ is $f$-invariant: $\\widetilde K$ is minimal $f$-invariant, so an arbitrary $y\\in\\widetilde K$ is in the closed convex hull of $f\\left(\\widetilde K\\right)$. This means that for an arbitrary $x\\in L$ we have $\\|f(x)-y\\|\\le\\sup_{z\\in\\widetilde K}\\|f(x)-f(z)\\|$, which, in turn, by the non-expansivity of $f$, is at most $\\sup_{z\\in\\widetilde K}\\|x-z\\|\\le\\rho\\left(\\widetilde K\\right)-t$. This proves that $f(x)\\in L$, as desired. The definition of $L$ implies that it cannot be equal to $\\widetilde K$ so we have found a closed, convex $f$-invariant set strictly smaller than $\\widetilde K$. This contradicts the minimality of $\\widetilde K$, and finishes the proof.", "Solution_2": "beautiful proof! :)" } { "Tag": [], "Problem": "What are all ordered triples of real numbers (x,y,z) which satisfy \r\n\r\n(x+y)(x+y+z) = 120 \r\n(y+z)(x+y+z) = 96 \r\n(x+z)(x+y+z) = 72? \r\n\r\nI'll also post this problem in other forums.", "Solution_1": "Summing these equations :\r\n\r\n[(x + y) + (y + z) + (z + x)](x + y + z) = 288\r\n\r\n=> x + y + z = 144 or x + y + z = -144\r\n\r\nIt is trivial from here on.", "Solution_2": "Good thinking. You're on the right track so far. That's what I did, but I want to see if anybody else here can get this correct. It's not that hard anyways.", "Solution_3": "Arne wrote:Summing these equations :\n\n[(x + y) + (y + z) + (z + x)](x + y + z) = 288\n\n=> x + y + z = 144 or x + y + z = -144\n\nIt is trivial from here on.\n\n\n\n[hide]Don't we have\n\n(2x+2y+2z)(x+y+z) = 288\n\nso\n\n(x+y+z)(x+y+z) = 144\n\n(x+y+z) = 12 or -12[/hide] (as opposed to 144 or -144)?\n\n\n\n[hide]If x+y+z = 12, then\n\nx+y=10\n\ny+z=8\n\nx+z=6\n\nSo z=2, x=4, y=6.\n\n\n\nIf x+y+z = -12, then\n\nx+y=-10\n\ny+z=-8\n\nx+z=-6\n\nSo z=-2, x=-4, y=-6\n\n\n\nHence the ordered triples are (4,6,2) or (-4, -6, -2).[/hide]" } { "Tag": [ "calculus", "integration", "trigonometry", "limit", "calculus computations" ], "Problem": "does \r\n$ \\int_0^\\infty sin x dx$ exist?\r\nif so, what is it's value and the method to solve it?", "Solution_1": "yes it is equal to 1.", "Solution_2": "i think that it will become a limit of n tending to infinity of the integral $ \\int_0^n sinx dx$and then we evaluate the integral and take $ \\cos\\infty$ = 0 as its a random number :maybe:", "Solution_3": "No, the it cannot exist, as the integral doesn't converge.", "Solution_4": "The integral doesn't converge in the classical sense. On the other hand, we can try some summation method. The easiest one to use will be Abel-Poisson, which amounts to finding the limit $ \\lim_{r\\to 0\\plus{}}\\int_0^\\infty e^{\\minus{}rx}\\sin x\\,dx$. Writing $ \\sin$ as a difference of exponents, integrating and taking the limit of the resulting expression, you get $ 1$.", "Solution_5": "[quote=\"fedja\"]The integral doesn't converge in the classical sense. On the other hand, we can try some summation method. The easiest one to use will be Abel-Poisson, which amounts to finding the limit $ \\lim_{r\\to 0 \\plus{} }\\int_0^\\infty e^{ \\minus{} rx}\\sin x\\,dx$. Writing $ \\sin$ as a difference of exponents, integrating and taking the limit of the resulting expression, you get $ 1$.[/quote]\r\nyes that is what i did. and integral of cosx =0." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Hi \r\n\r\nSuppose p>2 is a prime and $x^{2}+1$ is reducible in $\\mathbb{F}_{p}$[x]. Prove that the multiplicative group of $\\mathbb{F}_{p}$ contains an element of order 4. Deduce that p is congruent to 1 modulo 4. \r\n\r\nWhat exactly do we have to do with this question? \r\n\r\nThnx!", "Solution_1": "Show that any zero of $x^{2}+1$ has order 4.\r\n\r\nNow, what's the order of the multiplicative group $F_{p}^{*}$? What must be true if there is an element of order 4?\r\n\r\nThe converse is also true; whenever $p\\equiv 1\\mod 4$, there is an element of order 4 and $x^{2}+1$ factors. This follows from the fact that any finite multiplicative subgroup of a field is cyclic.", "Solution_2": "we can use Legendre's symbol, and euler's criterion (sorry i'm not that specific, i'm in a hurry this morning :oops: )" } { "Tag": [], "Problem": "Co bac nao biet cho nao co tai lieu toan tren dai hoc khong cho minh xin voi, Cam on nhieu!!!!!", "Solution_1": "[quote=\"langtupanda\"]Co bac nao biet cho nao co tai lieu toan tren dai hoc khong cho minh xin voi, Cam on nhieu!!!!![/quote]\r\n hinh` nhu ban o nuoc' ngoai`\r\n ban co' the vao` day\r\n http://vietnamnet.vn/giaoduc/tuyensinh/" } { "Tag": [], "Problem": "How many positive integers less than 1000 have a 1 in their base 16 expansion?\r\nPlease include everything you used to sole this problem.\r\n\r\nThanks", "Solution_1": "[hide]\nFor integers between 1 and 15 (1 \"digit\" in base 16), there is only one: 1.\n\nFor integers between 16 and 255 (2 \"digits\" in base 16): there are\n16 that are 1x and 15 that are y1, but 11 is counted twice, so there are only 30.\n\nFor integers between 256 and 1000 (3 \"digits\" in base 16): there are 256 that are 1xy. For numbers in the 200-2FF range, there are 31 that contain a 1. (It is the same as the combined number for 1 and 2-digit base 16 numbers) For numbers in the 300-3FF range, there are also 31 numbers that contain a 1, but 1000 in base 10 is less than 3FF. Since 400 (base 16) is 1024 (base 10), we counted one extra number in base 16: 3F1. It is pretty evident that 3E1 is the last number containing a 1 that is less than 1000. \n\nSo:\n1 + 30 + 256 + 31 + 31 - 1 = 348.\n\n[/hide]\n\nThere is probably an inclusion/exclusion method for this problem as well going something like this:\ntotal = {# that have 1 in the one's digit} + {# that have 1 in the 16's digit} + {# that have 1 in the 256's digit} - {# that have 1 in the one's digit and 16's digit} - {# that have 1 in the one's digit and 256's digit} - {# that have 1 in the 16's digit and 256's digit} + {# that have 1 in all three digits}[/hide]\r\n\r\nDo you have access to the SDMC sub-forum? We could just make a new topic to discuss solutions." } { "Tag": [ "inequalities" ], "Problem": "\u039d\u0394\u039f\r\n$ a^2 b \\plus{} b^2 c \\plus{} c^2 a \\geq ab \\plus{} bc \\plus{} ca$\r\n\u03bf\u03c4\u03b1\u03bd $ abc\\equal{}1$ \u03ba\u03b1\u03b9 $ a,b,c >0$", "Solution_1": "\u03a0\u03b1\u03c1\u03b1\u03bb\u03af\u03b3\u03bf \u03b3\u03b9\u03b1\u03c5\u03c4\u03ae\u03bd \u03bd\u03b1 \u03c0\u03ac\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03c4\u03bf \u0392\u03b1\u03c6\u03b5\u03af\u03b4\u03b7 10 \u03b5\u03c5\u03c1\u03ce \u03c3\u03c4\u03b9\u03c2 \u0392\u03c1\u03c5\u03be\u03ad\u03bb\u03bb\u03b5\u03c2 :P \u0391\u03c0\u03cc \u0391\u039c-\u0393\u039c \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \r\n$ 2a^2b\\plus{}b^2c\\geq 3\\sqrt[3]{a^4b^4c}\\equal{}3\\sqrt[3]{a^3b^3}\\equal{}3ab$ \u03a0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c4\u03bf\u03bd\u03c4\u03b1\u03c2 \u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ac \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b7 \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03b7 :)", "Solution_2": "\u03a3\u03c4\u03b1 \u03c5\u03c0' \u03cc\u03c8\u03b9\u03bd .\u0394\u03bf\u03c5\u03bb\u03b5\u03cd\u03b5\u03c4\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03b5 Lagrange \u03cc\u03c0\u03bf\u03c5 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 a=b=c=1 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9\u03c2 min", "Solution_3": "\u039b\u03cd\u03bd\u03b5\u03c4\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03bc\u03b5 Muirhead \u03b1\u03bc\u03ad\u03c3\u03c9\u03c2, \u03c0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03b9\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03bf \u03b4\u03b5\u03be\u03af \u03bc\u03ad\u03bb\u03bf\u03c2 \u03bc\u03b5 $ \\sqrt[3]{abc}$", "Solution_4": "\u039a\u03b1\u03b9 Holder" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "let $ a_1,a_2,a_3,... and b_1,b_2,b_3,...$ are 2 sequnce such that :\r\n \\[ \\forall n \\geq 2 : (a_n \\minus{} a_(n\\minus{}1))(a_(n\\minus{}1) \\minus{} a_(n\\minus{}2))\\plus{} (b_n \\minus{} b_(n\\minus{}1))(b_(n\\minus{}1) \\minus{} b_(n\\minus{}2))\\equal{}0\\]\r\nprove that there is a natural number like k such that : \\[ a_k \\equal{} a_(k\\plus{}2008)\\]", "Solution_1": "stvs_f , You might have made some typing mistakes. \r\nFirstly, does a(k+2008) on the last line mean a_(k+2008)\r\n If it does , then your statement is wrong because a_k can be made any number as b_k varies .\r\n\r\nAlso, do the two sequences contain natural numbers only?", "Solution_2": "the question is for USATST 2008.\r\nall of themare index\r\nand $ a_i and b_i \\in Z$", "Solution_3": "You reposted it even knowing the source?\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=1247503#p1247503\r\n\r\n[i]Please stop[/i] reposting problems that can already be found on this forum; all the discussion for one problem should stay collected in one topic. If you want someone to take another look at it or have a question, revive the topic yourself." } { "Tag": [ "FTW", "MATHCOUNTS" ], "Problem": "People take pride in different accomplishments. Some prefer having the ability of Countdowning well, possibly because they like that they can show their talents off. Others are proud of being able to score well on written tests. Which holds more value with you?\r\n\r\nPersonally, I valued countdown a lot more last year. I looooved FTW (I still do) but my skills didn't show in written. This year, my last year, I feel like solid written scores is a better attribute, I think because it's my last shot at this and I want to make it as far as possible. And, It seems whichever I enjoy more I do better at (Chapter last year: 5th written, 1st countdown. This year: 1st written, 2nd countdown.)", "Solution_1": "I used to be a more countdown oriented person as well, but now I've lost most interest for speed (which I should probably change :maybe: ).", "Solution_2": "Hmm. Are you referring to importance to somebody or the importance according to the rules? Either way, for me, it's written.", "Solution_3": "I'm wayy better at countdown, but written definitely is more important. Countdown trophies are bigger, though :)", "Solution_4": "Written for sure. Probably because I stink at countdown... I focus a lot more of my time to written as well.", "Solution_5": "I think written is more important... besides cd is like 2 or 3 questions.\r\nAnd our written trophies are bigger than cd ones :P", "Solution_6": "Written is definitely more important, but countdown is definitely more fun :P \r\n\r\nPoint is, you first have to do well to even MAKE it to countdown, but from then on it's just a popularity show. \r\n\r\nAnd countdown trophies aren't much bigger than regular trophies in my chapter. At states, we didn't even get countdown trophies -- we got these cheap copper medals with stickers on them that didn't even say \"MathCounts\" :( ...", "Solution_7": "For me, definitely written. I used that to gauge how well I was doing in the math problems. The countdown tested how awake and lucky I was.", "Solution_8": "Although my chapter/state's countdown is unofficial, I somehow got 2nd at countdown chapter.\r\nSo I kinda value both of them the same even though I suck at CD...but fortunately I'm realizing that I'm actually getting faster at solving problems mentally.", "Solution_9": "I generally value written more since it gives you a score to aim for. But still, I get pretty upset when I do badly at CD.", "Solution_10": "I haven't done a single MC test ALL YEAR!!! woohooo! anyway, both for me.", "Solution_11": "Actually, in all perspective, i think they are equally valued.. I'm kind of an average guy, like i get 3rd place in chapter and then i move up to countdown and get 1st.. This is a concept one has too learn!!! The only way to get into countdown is with written!!! I know this sixth grader that is totally beastly at countdown and possibly may be better than people at nationals, but his written test is not that good... that's why he doesn't get better. My mom is always telling me, \"SHOW ME YOUR BALANCE!!!\" and I know she's right.. So I improved. And now I'm pretty leveled out. It's really helpful to be balanced! :D", "Solution_12": "Countdown, much more intense. Also, at state, if you pwn, everyones like OMG :O" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all integers $x,y,z,t$ such that $x^y+y^z+z^t=x^{2005}$.\r\nCan we replase 2005 by n?", "Solution_1": "somebody? help", "Solution_2": "[quote=\"Beat\"]Find all integers $x,y,z,t$ such that $x^{y}+y^{z}+z^{t}=x^{2005}$.\nCan we replase 2005 by n?[/quote]\r\nOne solution $y=1, t=1, z=x^{n}-x-1$, were $x$ is any integer.", "Solution_3": "Can we find all the solutions? :(" } { "Tag": [], "Problem": "How many different integers can be expressed as the sum of \ntwo different numbers in the set $ \\{1,2,3,4,5,6\\}$?", "Solution_1": "The minimum is $ 3$, and the maximum is $ 11$, so our answer is $ 11 \\minus{} 3 \\plus{} 1 \\equal{} \\boxed9$." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let f,g be two functions from the positive integers to the positive integers.\r\nProve there exist distinct p,q naturals such that \r\n\r\nf(p)=\nBabel and hyphenation patterns for english, dumylang, nohyphenation, ge\nrman, ngerman, german-x-2008-06-18, ngerman-x-2008-06-18, french, loaded.\n(\"C:\\Program Files\\MiKTeX 2.7\\tex\\latex\\base\\latex209.def\"\nFile: latex209.def 1998/05/13 v0.52 Standard LaTeX file\n\n\n Entering LaTeX 2.09 COMPATIBILITY MODE\n *************************************************************\n !!WARNING!! !!WARNING!! !!WARNING!! !!WARNING!! \n \n This mode attempts to provide an emulation of the LaTeX 2.09\n author environment so that OLD documents can be successfully\n processed. It should NOT be used for NEW documents!\n \n New documents should use Standard LaTeX conventions and start\n with the \\documentclass command.\n \n Compatibility mode is UNLIKELY TO WORK with LaTeX 2.09 style\n files that change any internal macros, especially not with\n those that change the FONT SELECTION or OUTPUT ROUTINES.\n \n Therefore such style files MUST BE UPDATED to use\n Current Standard LaTeX: LaTeX2e.\n If you suspect that you may be using such a style file, which\n is probably very, very old by now, then you should attempt to\n get it updated by sending a copy of this error message to the\n author of that file.\n *************************************************************\n\n\\footheight=\\dimen102\n\\@maxsep=\\dimen103\n\\@dblmaxsep=\\dimen104\n\\@cla=\\count79\n\\@clb=\\count80\n\\mscount=\\count81\n(\"C:\\Program Files\\MiKTeX 2.7\\tex\\latex\\base\\tracefnt.sty\"\nPackage: tracefnt 1997/05/29 v3.0j Standard LaTeX package (font tracing)\n\\tracingfonts=\\count82\nLaTeX Info: Redefining \\selectfont on input line 101.\n)\n\\symbold=\\mathgroup4\n\\symsans=\\mathgroup5\n\\symtypewriter=\\mathgroup6\n\\symitalic=\\mathgroup7\n\\symsmallcaps=\\mathgroup8\n\\symslanted=\\mathgroup9\nLaTeX Font Info: Redeclaring math alphabet \\mathbf on input line 293.\nLaTeX Font Info: Redeclaring math alphabet \\mathsf on input line 294.\nLaTeX Font Info: Redeclaring math alphabet \\mathtt on input line 295.\nLaTeX Font Info: Redeclaring math alphabet \\mathit on input line 301.\nLaTeX Info: Redefining \\em on input line 311.\n\n(\"C:\\Program Files\\MiKTeX 2.7\\tex\\latex\\base\\latexsym.sty\"\nPackage: latexsym 1998/08/17 v2.2e Standard LaTeX package (lasy symbols)\n\\symlasy=\\mathgroup10\nLaTeX Font Info: Overwriting symbol font `lasy' in version `bold'\n(Font) U/lasy/m/n --> U/lasy/b/n on input line 47.\n)\nLaTeX Font Info: Redeclaring math delimiter \\lgroup on input line 375.\nLaTeX Font Info: Redeclaring math delimiter \\rgroup on input line 377.\nLaTeX Font Info: Redeclaring math delimiter \\bracevert on input line 379.\n)\n(\"C:\\Program Files\\MiKTeX 2.7\\tex\\latex\\base\\letter.cls\"\nDocument Class: letter 1999/04/29 v1.2z Standard LaTeX document class\n(\"C:\\Program Files\\MiKTeX 2.7\\tex\\latex\\base\\size10.clo\"\nFile: size10.clo 2005/09/16 v1.4f Standard LaTeX file (size option)\n)\n\\longindentation=\\dimen105\n\\indentedwidth=\\dimen106\n\\labelcount=\\count83\nCompatibility mode: definition of \\rm ignored.\nCompatibility mode: definition of \\sf ignored.\nCompatibility mode: definition of \\tt ignored.\nCompatibility mode: definition of \\bf ignored.\nCompatibility mode: definition of \\it ignored.\nCompatibility mode: definition of \\sl ignored.\nCompatibility mode: definition of \\sc ignored.\nLaTeX Info: Redefining \\cal on input line 391.\nLaTeX Info: Redefining \\mit on input line 392.\n)\n\n! LaTeX Error: LaTeX2e command \\usepackage in LaTeX 2.09 document.\n\nSee the LaTeX manual or LaTeX Companion for explanation.\nType H for immediate help.\n ... \n \nl.3 \\usepackage\n {amsfonts}\n? x\n \nHere is how much of TeX's memory you used:\n 233 strings out of 95305\n 2564 string characters out of 1183058\n 46191 words of memory out of 1500000\n 3513 multiletter control sequences out of 110000\n 3640 words of font info for 14 fonts, out of 1200000 for 2000\n 14 hyphenation exceptions out of 8191\n 19i,0n,19p,211b,36s stack positions out of 5000i,500n,10000p,200000b,5000s\nNo pages of output.\n[\\code][/code]", "Solution_1": "The problem is caused by using an older version of LaTeX which doesn't support the packages:\r\n[quote] Entering LaTeX 2.09 COMPATIBILITY MODE \n ************************************************************* \n !!WARNING!! !!WARNING!! !!WARNING!! !!WARNING!! [/quote]\r\nAre you using \\documentstyle{letter} which causes this instead of \\documentclass{letter}?\r\nIf it's not that please post a minimal example of your tex document.", "Solution_2": "Hi,\r\n\r\nthelog file contains the answers:\r\n\r\n[quote=\"1/(ln x)\"]\n New documents should use Standard LaTeX conventions and start\n with the \\documentclass command.\n...\n\n! LaTeX Error: LaTeX2e command \\usepackage in LaTeX 2.09 document.\n[/quote]\r\n\r\nPerhaps have a look also at the modern alternative letter class scrlttr2, I'm using it and I'm very satisfied with it.\r\n\r\nStefan" } { "Tag": [ "linguistics" ], "Problem": "We have just started to encourage students at the Metroplex Math Circle to look into the Linguistics Olympiads.\r\n\r\nhttp://metroplexmathcircle.wordpress.com/2008/11/28/naclo-2009/\r\n\r\nAre these contests very popular outside the US, and do the problem solving skills of math competitions lend themselves to success in these contests?", "Solution_1": "Hi! I'm a linguistics graduate student, and even though I've never participated in any linguistics competitions, I can certainly address your questions about linguistics and mathematical problem solving.\r\n\r\nSo, I looked over some problems from past linguistics competitions (NAMCLO 2007 and IOL6) and the questions seem to range from logic problems having to do with words to problems that are along the lines of \"real linguistics\" (in fact, one of my professors gave us a problem he'd written for the ILO as an exercise the first day of class). I imagine that strong mathematical problem solving skills are as good a preparation as any for the \"logic problems having to do with words\" kinds of problems. For the \"real linguistics\" types of problems, again, I think that mathematical problem solving would be as good a preparation as any for the problem-solving skills, but it would also be important to have a good familiarity with linguistics to know the framework within which linguistics problems work (eg, so you'll know how phonology problems in general work). It's not completely false that the best linguists are the ones that have looked at the most languages.\r\n\r\nAnyway, I'm not too much help on how linguistics competitions themselves work, but if you'd like a reading list for real linguistics, or have any questions about linguistics in general, feel free to PM me.", "Solution_2": "Thank you, Osud! I will PM you but I'll post here as well in case anyone else has a similar interest. \r\n\r\nI am very much a lay person but I have always been fascinated by linguists and by the fact that so many linguists make great contributions outside of their field (like Pinkner, Chomsky, Tolkien, Nietzche, etc.) I'm sure it was very basic, but I also enjoyed the Teaching Company series by McWhorter (http://www.teach12.com/ttcx/CourseDescLong2.aspx?cid=2270)\r\n\r\nWhile I'm too old to be anything but a linguistics fan, computational linguistics might be something that would combine my son's interests. He certainly enjoys all of his AOPS classes and Latin, but the one thing he would do all day long is his Logic class (http://www.eimacs.com).\r\n\r\nI guess that by the fact you are on these forums you have more than a passing interest in mathematics :) and any resources or advice you could give would be greatly appreciated.", "Solution_3": "I would definitely agree with Osud, though I would emphasize even more the connection between linguistics and mathematics. Many of the \"real linguistics\" problems, as far as I've found, really only require mathematical reasoning; the \"good familiarity with linguistics\" is mainly what one gets after doing many of these problems, not by studying linguistics from outside sources. Just to note, I'm a high school student who participated in the NACLO. The website http://webscript.princeton.edu/~ahesterb/index.php is extremely good for linguistics problems." } { "Tag": [ "invariant", "combinatorics", "IMO Shortlist", "game", "termination" ], "Problem": "$ 1994$ girls are seated at a round table. Initially one girl holds $ n$ tokens. Each turn a girl who is holding more than one token passes one token to each of her neighbours.\r\n \r\na.) Show that if $ n < 1994$, the game must terminate. \r\nb.) Show that if $ n \\equal{} 1994$ it cannot terminate.", "Solution_1": "A harder variant of this problem can be discussed [url=http://www.mathlinks.ro/viewtopic.php?p=7623#7623]here.[/url]", "Solution_2": "Denote the girl originally holding all coins $A$. If $n<1994$, we just remove the girl opposite $A$ and consider the girls sitting in a row with A in the center. We use a string to represent the coins in each girl's hand. I'll just write out the numbers to the right of $A$.\n\n0 ...\n20...\n02...\n101...\n201...\n011...\n111...\n211...\n021...\n202...\n\n\nClearly, the string cycles between 20202020.... and 1111111... (in each cycle, another 20 and 11 appear)\n\nThus, if $n<1994$, we get 111..... ($A$ is left with no coins if $n$ is even, and one coin if $n$ is odd), so the girl opposite $A$ never gets a coin.\n\nIf n=1994, the 0's and 2's switch back and forth, i.e.\n\n20202020.....\n02020202.....\n\nso the game does not end.", "Solution_3": "@Above Sorry, your proof is probably very wrong - a person can have way more than $2$ cards. \n\n[hide=A \"symmetric\" invariant for (a) ] Start from any girl, and number them $1, \\cdots, 1994$, and denote $C[j]$ by the number of tokens of $j^{th}$ girl. The quantity $$ \\sum_{j \\; odd} C[j] \\mod 2$$ is always $0$ [/hide]\n\n[hide= IMO Compendium \"assymetric\" invariant for (a) ] The quantity $$\\sum_{00$, $x^2-x-6>0$ The solution set for this is described by $\\boxed{A}$[/hide]", "Solution_5": "So we agree...[/hide]" } { "Tag": [ "linear algebra", "matrix", "function", "geometry", "geometric transformation", "integration", "trigonometry" ], "Problem": "Problem statement: Consider the graph $ U$ where the vertices are points in $ \\mathbb{R}^2$ and two points are joined if they are at distance $ 1$ from each other in $ \\mathbb{R}^2$. What is the chromatic number $ \\chi$ of this graph?\r\n\r\nKnown results: $ 4 \\leq \\chi \\leq 7$\r\n\r\nProof: See the diagram here which has the $ 7$-coloring as well as a subgraph with chromatic number $ 4$: http://en.wikipedia.org/wiki/Image:Hadwiger-Nelson.svg. The hexagons all have diameter slightly less than $ 1$.\r\n\r\nIdea: Let $ A$ be the \"adjacency matrix\" for this graph (in this case not actually a matrix but rather a linear operator). Let $ \\lambda_{max}$ and $ \\lambda_{min}$ be the largest and smallest eigenvalues of $ A$. Then we can show that\r\n\\[ \\chi \\geq 1 \\minus{} \\frac {\\lambda_{max}}{\\lambda_{min}}\r\n\\]\r\nProof: Let $ c_1,\\ldots,c_{\\chi}$ be the color classes. Let $ f$ be the largest eigenfunction of $ A$, and let $ f_1,\\ldots,f_{\\chi}$ be functions such that $ f_i(x) \\equal{} f(x)$ if $ x \\in c_i$ and $ 0$ otherwise. Then clearly $ f \\equal{} f_1 \\plus{} \\ldots \\plus{} f_{\\chi}$. Then we have\r\n\\[ 2\\lambda_{min}(c \\minus{} 1)\\parallel{}f\\parallel{}^2 \\equal{} \\sum_{i,j} \\lambda_{min}\\parallel{}f_i \\minus{} f_j\\parallel{}^2\\]\r\n\\[ \\equal{} \\sum_{i,j} \\lambda_{min} < f_i \\minus{} f_j,f_i,f_j > \\leq \\sum_{i,j} < f_i \\minus{} f_j,A(f_i \\minus{} f_j) >\\]\r\nBut $ < f_i,Af_i > \\equal{} 0$ if we have a proper coloring, since no two two points in $ c_i$ can be connected. Thus (combined with the fact that $ A$ is symmetric) the above reduces to\r\n\\[ \\sum_{i,j} \\minus{} 2 < f_i,Af_j > \\equal{} \\minus{} 2 < f,Af > \\equal{} \\minus{} 2\\lambda_{max}\\parallel{}f\\parallel{}^2\r\n\\]\r\nThus $ 2\\lambda_{min}(c \\minus{} 1)\\parallel{}f\\parallel{}^2 \\leq \\minus{} 2\\lambda_{max}\\parallel{}f\\parallel{}^2$\r\n\r\nThen, since $ \\lambda_{min}$ is negative, when we divide by $ 2\\lambda_{min}\\parallel{}f\\parallel{}^2$ we get $ c \\minus{} 1 \\geq \\frac {\\lambda_{max}}{\\lambda_{min}}$, which is what we wanted.\r\n\r\nNote that this logic holds for any linear operator $ B$ where $ A_{ij} \\equal{} 0 \\implies B_{ij} \\equal{} 0$. Now, assuming that $ B$ is an operator that commutes with translation operators, i.e. $ B(f(x \\plus{} t)) \\equal{} (Bf)(x \\plus{} t)$, then the eigenvectors of $ B$ are given by the Fourier coefficients over $ \\mathbb{R}^2$ corresponding to $ Bf$, where $ f$ is such that $ f(0) \\equal{} 1$ and $ f(x) \\equal{} 0$ if $ x \\neq 1$. Working this out, we see that they can be parameterized as\r\n\\[ \\mu(r,s) \\equal{} \\int_{ \\minus{} \\pi}^{\\pi} f(\\theta) e^{i(r\\cos(\\theta) \\plus{} s \\sin(\\theta))} d\\theta\r\n\\]\r\nfor any $ f$ we choose, provided that $ f(x) \\equal{} f(x \\plus{} \\pi)$ so that the integral evaluates to a real number always. If $ f$ is the constant function, then using this integral to give us a bound on the chromatic number is not quite good enough to show that $ \\chi \\geq 5$, but it is possible that a better $ f$ will yield a better bound. Can we find such an $ f$, or prove that no such $ f$ exists? Also note that we can re-paramterize with a proper set of substitutions as\r\n\\[ \\mu(t,\\phi) \\equal{} \\int_{ \\minus{} \\pi}^{\\pi} f(\\theta \\plus{} \\phi) e^{itcos(\\theta)} d\\theta\r\n\\]\r\nSo, the basic question is, can we find an $ f$ to get a better bound than the current one on the chromatic number?\r\n\r\nComment: I've used this same idea with seemingly more success on the [i]odd-distance graph[/i], where two points are connected if their distance is an odd integer. There we just need to show that a certain integral is finite, and if I can get the asymptotics down right I think I can show that. I'd prefer not to type the work for that problem up for now, since I think I'm close to solving it. If I get stuck or end up solving it, I'll put it on arXiv and link to it from here.\r\n\r\nAlso, I apologize if the exposition is a bit terse here. I will try to make the explanations more detailed later.", "Solution_1": "Quick question: what vector space is $ A$ as a linear operator operating on?", "Solution_2": "We can associate the vertex set $ V$ with $ \\mathbb{R}^2$, so $ A$ acts on $ L^2(\\mathbb{R}^2)$. One thing to note is that this approach actually only tells us about measurable colorings of the plane, but this question is also interesting (I believe that the current best bound for measurable colorings is $ 5$ colors)." } { "Tag": [], "Problem": "a nice one :)", "Solution_1": "[hide]\nNo such sequence exists. Suppose one did. Then $P(n_{k})\\neq 0$ for all $k$, so no $n_{k}$ has any of its digits 0. In particular, letting $a$ be the number of digits of $n_{1}$, for each $d\\geq a$, the sequence must go from a number with at most $d$ digits to one with at least $d+1$ digits.\n\nIf $n_{k}$ has at most $d$ digits and $n_{k+1}$ has at least $d+1$, then $n_{k}\\leq 10^{d}-1$ and $n_{k+1}\\geq 10^{d}+10^{d-1}+\\cdots+1=\\frac{10^{d+1}-1}{9}$ (no digit can be 0). In particular, this means $P(n_{k})\\geq \\frac{10^{d}+8}{9}$.\n\nHowever, we also have $P(n_{k})\\leq 9^{d}$ (since at most $d$ digits, each at most 9). Therefore, we must have $9^{d+1}\\geq 10^{d}+8$ for each $d\\geq a$, but this is false for sufficiently large $d$ (actually, $d\\geq 21$ is enough), a contradiction.\n[/hide]" } { "Tag": [ "MATHCOUNTS", "Ring Theory", "FTW", "AMC", "USA(J)MO", "USAMO", "AMC 8" ], "Problem": "what time will it be aired tomorowed????\r\nis it after 4:00 eastern time? if not, i will be mad because i will have to skip school to watch it and i cant skip school(although i would like to)", "Solution_1": "Don't worry its 5:00p.m. eastern, 3:00 mountain. And it says on their website.\r\n\r\nEdit:@ajain, i didn't feel like posting another post so hopefully your looking at this. It is to cover countdown. I can't wait. I can't wait. Oh shoot, I have track practice. :(", "Solution_2": "Is this regarding the webcast of the Nation MathCounts Countdown round?\r\n\r\nHere is the webpage:\r\n\r\n[url]http://mathcounts.org/NETCOMMUNITY/Page.aspx?pid=1290&srcid=1290[/url]\r\n\r\nBut, if you do miss it, it will be archived in the same link i provided.", "Solution_3": "Message from Denver!!!\r\n\r\nBoth replies are right. Sprint starts in like 13 hours. Shentang, Alan is still using your lucky pencil at all our practices and doing well!!\r\n\r\nWe'll be sure to post results as soon as we can, but hopefully you catch the webcast.\r\n\r\nFrost", "Solution_4": "Its 30 minutes before the announcing of the winners. From what I have heard, 43 was the top score.", "Solution_5": "i think it was bobby shen..\r\nmy coach told me that a 6th grader from the team he proctored got 41", "Solution_6": "Dang, people are doing good. I wish I was there. Guess what? My track practice was canceled because of the rain, now I can watch the web cast. @frost13, yep, thats the pencil I used to get the cookie and win bingo. Ah, good times.", "Solution_7": "ahh!\r\ni am using a crap computer right now. i have 2 right next to me. the one i am typing from is my fast one and there is no link showing up.", "Solution_8": "Should we be able to see it right now? It's not working for me :( .", "Solution_9": "Yes, the webcast is working for me right now.", "Solution_10": "and now my sound isnt working...", "Solution_11": "Is the webcast really live? According to tinytim's post, the thing should have started an hour and a half earlier...", "Solution_12": "Shoot, I really hate this, so many downloads, I am missing all the fun.....even tried my friends computer and my dad's. :(", "Solution_13": "kyyuan or whoever is whatching can you give me an update. i cant hear a thing. i just know theyre anouncing some teams", "Solution_14": "Me too please, cause I don't see or hear anything. Only thing I will be doing-refreshing this page over and over again till someone posts.", "Solution_15": "You admitted Darryl's luck, didn't you? So why were you expecting maximized credit for Darryl? Just because he's a sixth grader, and he made USAMO, and perfect score on AMC 8, and perfect 800 SAT math score, and a USCF top-rated chess player, or something like those?", "Solution_16": "How did Maryland get 3 people in top 6 and not win 1st team.... hmmm... they must have done (relatively) poorly on the team round... but still, Texas only had one person in cd. Was Texas written something like 1, 13, 14, 15?\r\n\r\nAnd Indiana must be pretty disappointed.\r\n\r\nBut congrats to all. (I feel brave congratulating people who are 1000000 times better than me...)", "Solution_17": "[quote=\"Fanatic\"]Could anyone clarify \"by luck\"? I think we're not giving Darryl Wu nearly enough credit here... or did I miss something?[/quote]\r\n\r\nWell, I have to admit Darryl is really beast, but there were a lot of things that made it more \"by luck\". For example, Lyndon Ji, Kevin Tian, Kevin Li, and many other really good people who should've qualified for countdown did not. Darryl got in the \"easier\" brackets (I think). He almost did not make finals. Bobby made 4 careless mistakes or something like that. Also, you could say that anyone could have won in Darryl's place in the final round--because he won by getting each of the questions he had 44.99999999 (emphasizing Bobby's speed) seconds on. Again, Darryl is really good, but I think many people are still better than him.\r\n\r\n@leoxnlin: They screwed up on the team round. Indiana screwed up too. According to tinytim who heard from Kevin Tian, I heard Lyndon Ji and the Kevins missed \"every other\" question on the Sprint by misreading.\r\n\r\nAnyway, I think nationals was really screwed up this year...", "Solution_18": "What do you mean Darryl almost failed to make finals? Was it because of Evan Miller?", "Solution_19": "[quote=\"mathcrazed\"]\nThings of note:\n\nYeah... ermm... sorry.\nExcuse list: Bloody nose at 2:30 AM - 3:00 AM day of competition. Bloody nose when I woke up around 7. Took advil before test (which isn't supposed to do anything anyway XD). Bloody nose the day before and the night before (something with bloody noses in spring). Braces adjusted on Tuesday, but then my bond broke on wednesday and started poking into my cheek. Uber stomachache the night before.\n\n[/quote]\r\n\r\nhey i also had a bloody nose that morning. only i woke up three times during the night and my pillow had some stains of blood. :oops:", "Solution_20": "Hmm, it appears that Denver is a really dry place, with all the bloody noses...\r\nDidn't you have a humidifier in your hotel rooms?\r\nAnyways, I guess most of you are just surprised that a 6th grader won and are placing it on luck...\r\n(Just like me)\r\nAnyways, Darryl seemed sort of... unhappy, when he won... (Hunched back, starting at the floor, shuffling his feet, etc.)", "Solution_21": "[quote=\"math154\"]\n\n@leoxnlin: They screwed up on the team round. Indiana screwed up too. According to tinytim who heard from Kevin Tian, I heard Lyndon Ji and the Kevins missed \"every other\" question on the Sprint by misreading.\n\n[/quote]\r\n\r\nHmm.... and Texas had a perfect team round?", "Solution_22": "[quote=\"ZhangPeijin\"]Hmm, it appears that Denver is a really dry place, with all the bloody noses...\nDidn't you have a humidifier in your hotel rooms?\nAnyways, I guess most of you are just surprised that a 6th grader won and are placing it on luck...\n(Just like me)\nAnyways, Darryl seemed sort of... unhappy, when he won... (Hunched back, starting at the floor, shuffling his feet, etc.)[/quote]\r\nUnhappy? :huh: No way, you should have said...unenthusiastic. :o", "Solution_23": "Ha, I DOUBT he was unenthusiastic. I bet it was just the shock of suddenly becoming National Champion and that the reality of the situation didn't hit him at that very moment. I'll bet he's feeling EXTREMELY happy, and lucky right now :D", "Solution_24": "I have a feeling it's way more complicated than that...", "Solution_25": "[quote=\"math154\"][quote=\"Fanatic\"]Could anyone clarify \"by luck\"? I think we're not giving Darryl Wu nearly enough credit here... or did I miss something?[/quote]\n\nWell, I have to admit Darryl is really beast, but there were a lot of things that made it more \"by luck\". For example, Lyndon Ji, Kevin Tian, Kevin Li, and many other really good people who should've qualified for countdown did not. Darryl got in the \"easier\" brackets (I think). He almost did not make finals. Bobby made 4 careless mistakes or something like that. Also, you could say that anyone could have won in Darryl's place in the final round--because he won by getting each of the questions he had 44.99999999 (emphasizing Bobby's speed) seconds on. Again, Darryl is really good, but I think many people are still better than him.\n\n@leoxnlin: They screwed up on the team round. Indiana screwed up too. According to tinytim who heard from Kevin Tian, I heard Lyndon Ji and the Kevins missed \"every other\" question on the Sprint by misreading.\n\nAnyway, I think nationals was really screwed up this year...[/quote]\r\n\r\numm when i used to lose at something i always thought the other person/people ot luckier than me and that was the reason why I lost and they won, but now i realize maybe they are actually better...\r\ni dont get how other good people who shouldve qualified has anything to do with wheter he won or not...maybe you think they are better than him or something just because youve heard of them before and you never heard of darryl...\r\nok he made 4 careless mistakes, thats kinda the point of the competition, not to make mistakes so you cant say he got lucky just because his opponent made careless msitakes, you have to give him the credit for not making the careless mistakes...and the 45 seconds, thats actually really smart...you want to use up all your time to check your work so you dont make carless mistakes, im sure he couldve gotten some earlier...and the point about he couldve lost earlier and might not have made the finals, well, maybe had he lost, the person that wouldve beat him might of still beat bobby in the finals and you would be making the same argument against that guy...anyways doesnt matter what i think, the fact of the matter is he won so hes the best...besides he also made usamo as a 6th grader so you cant say he just got \"lucky\" on one day (and no disrespect, but the guy you are trying to defend did not make usamo in 7th grade).", "Solution_26": "[quote=\"the future\"](and no disrespect, but the guy you are trying to defend did not make usamo in 7th grade).[/quote]\r\n\r\nthis is when i get ticked off when people spout off facts about people when they don't know the true situation...\r\n\r\nBobby Shen, like Kevin Chen before him, only took the AMC 8 in 6th grade and 7th grade. So, he didn't make the USAMO because he didn't try to make the USAMO.", "Solution_27": "I just feel like a lot of people wouldn't be saying the things they are if they knew Darryl posted on AoPS. If I was Darryl and I just won Nats, I log onto AoPS and everyone's spouting off about how lucky I got... I'd be pissed. Down 3-0, 3-1, etc. approaching the 45th second of the countdown question... it's insanely hard to keep your cool.", "Solution_28": "true, the fact is that Darryl beat Bobby, and one can't say that Bobby didn't win because of stress, because whatever bobby was experiencing, darryl was too. Anyway nice job to both of you. i have a bad post count.", "Solution_29": "OK Guys, I'm gonna go ahead and lock this.\r\n\r\nThey both did a great job. Congratulations to both Bobby and Darryl. Special super congratulations for beating out the 8th graders. Maybe we'll have a classic rematch next year!" } { "Tag": [ "absolute value" ], "Problem": "Solve for $ x$ in terms of $ a,b$: $ |x\\minus{}a|\\equal{}|x\\minus{}b|$.\r\n\r\n[hide=\"solution1\"]\nBy the number line, the distance from $ x$ to $ a$ equals the distance from $ x$ to $ b$ (in terms of magnitude). Thus $ x\\equal{}\\frac{a\\plus{}b}{2}$.\n[/hide]\r\n\r\nHow do we do this the \"normal way\" (not by number line)?", "Solution_1": "You mean algebraically?\r\n\r\n$ x\\minus{}a\\equal{}\\minus{}(x\\minus{}b)\\equal{}\\minus{}x\\plus{}b\\Rightarrow 2x\\equal{}a\\plus{}b\\Rightarrow x\\equal{}\\frac{a\\plus{}b}{2}$\r\n\r\nJust use the rules of absolute value arithmetic.", "Solution_2": "This is as clearly as I can put it:\r\n$ |x - a| = |x - b|$\r\nDivide both sides by $ |x - a|$:\r\n$ 1 = \\frac {|x - b|}{|x - a|}$\r\nThis is the same as:\r\n$ 1 = {|}\\frac {x - b}{x - a}{|}$\r\nWhich means:\r\n$ 1 = \\frac {\\pm(x - b)}{x - a}$\r\nSo, solve the resulting equations:\r\n$ 1 = \\frac {x - b}{x - a}$\r\n$ x - a = x - b$\r\n$ \\boxed{a = b}$\r\nThat's not very informative, but the next equation:\r\n$ 1 = \\frac {b - x}{x - a}$\r\n$ x - a = b - x$\r\n$ 2x = b + a$\r\n$ \\boxed{x = \\frac{b + a}{2}}$\r\nWhich is also equal to $ 2b/2$ or $ 2a/2$ by substitution, which simplifies to $ x = a$ or $ x = b$.\r\nThat makes the original equation $ x - x = x - x$ by more substitution, which leads to $ 0 = 0$, which signifies that there are infinitely many solutions." } { "Tag": [ "geometry", "rhombus", "perimeter", "analytic geometry", "quadratics", "geometry unsolved" ], "Problem": "A circle intersects each side of a rhombus twice, dividing each side into three parts. Starting from any vertex, go around the rhombus in a fixed direction, color the segments red, white, blue, red, white, blue, red, white, blue, red, white, blue. Prove that the sum of the lengths of the red segments is equal to that of the blue ones.", "Solution_1": "Let $ XYZT$ be a rhombus, let a circle $ (P)$ with radius $ r$ intersect the rhombus sides $ XY, YZ, ZT, TX$ at $ A, B, C, D, E, F, G, H,$ so that the points $ X, A, B, Y, C, D, Z, E, F, T, G, H$ follow on the rhombus perimeter in this order. We have to show (using directed line segments)\r\n \r\n$ \\Sigma_{RED} \\minus{} \\Sigma_{BLUE} \\equal{} (\\overline{AX} \\plus{} \\overline{CY} \\plus{} \\overline{ZE} \\plus{} \\overline{TG}) \\minus{} (\\overline{YB} \\plus{} \\overline{ZD} \\plus{} \\overline{FT} \\plus{} \\overline{HX}) \\equal{} 0$ \r\n\r\nPut the coordinate origin at the intersection $ O$ of the rhombus diagonals, x-axis along $ OX,$ y-axis along $ OY.$ Equations of the rombus sidelines and of the circle $ (P)$ are\r\n\r\n$ XY: \\ \\ \\ \\ y \\equal{} \\minus{} m(x \\minus{} q)$\r\n$ YZ: \\ \\ \\ \\ y \\equal{} \\plus{} m(x \\plus{} q)$\r\n$ ZT: \\ \\ \\ \\ y \\equal{} \\minus{} m(x \\plus{} q)$\r\n$ TX: \\ \\ \\ \\ y \\equal{} \\plus{} m(x \\minus{} q)$\r\n\r\n$ (P): \\ \\ (x \\minus{} a)^2 \\plus{} (y \\minus{} b)^2 \\equal{} r^2$\r\n\r\nwhere $ \\pm q$ are the x-coordinates of $ X, Z.$ Substituting the 4 line equations into the circle equation yields 4 quadratic equations for the intersections:\r\n\r\n$ A, B: \\ \\ (1 \\plus{} m^2) x^2 \\minus{} 2(a \\minus{} mb \\plus{} m^2q)x \\plus{} ... \\equal{} 0$\r\n$ C, D: \\ \\ (1 \\plus{} m^2) x^2 \\minus{} 2(a \\plus{} mb \\minus{} m^2q)x \\plus{} ... \\equal{} 0$\r\n$ E, F: \\ \\ (1 \\plus{} m^2) x^2 \\minus{} 2(a \\minus{} mb \\minus{} m^2q)x \\plus{} ... \\equal{} 0$\r\n$ G, H: \\ \\ (1 \\plus{} m^2) x^2 \\minus{} 2(a \\plus{} mb \\plus{} m^2q)x \\plus{} ... \\equal{} 0$\r\n\r\nPutting $ k \\equal{} \\frac {1}{1 \\plus{} m^2},$ sums of the roots are \r\n\r\n$ x_{A} \\plus{} x_{B} \\equal{} 2k(a \\minus{} mb \\plus{} m^2q)$\r\n$ x_{C} \\plus{} x_{D} \\equal{} 2k(a \\plus{} mb \\minus{} m^2q)$ \r\n$ x_{E} \\plus{} x_{F} \\equal{} 2k(a \\minus{} mb \\minus{} m^2q)$\r\n$ x_{G} \\plus{} x_{H} \\equal{} 2k(a \\plus{} mb \\plus{} m^2q)$ \r\n\r\nIt is sufficient to prove the proposition for projections of the line segments to the x-axis. $ O$ is projection of $ Y, T$ on the x-axis. Let $ A', B', ..., H'$ be projections of $ A, B, ..., H$ on the x-axis.\r\n \r\n$ (\\overline{A'X} \\plus{} \\overline{C'O} \\plus{} \\overline{ZE'} \\plus{} \\overline{OG'}) \\minus{} (\\overline{OB'} \\plus{} \\overline{ZD'} \\plus{} \\overline{F'O} \\plus{} \\overline{H'X}) \\equal{}$ \r\n\r\n$ \\equal{} \\left[(q \\minus{} x_{A}) \\plus{} (0 \\minus{} x_{C}) \\plus{} (x_{E} \\plus{} q) \\plus{} (x_{G} \\minus{} 0)\\right]\\ \\minus{}$\r\n\r\n$ \\minus{} \\ \\left[(x_{B} \\minus{} 0) \\plus{} (x_{D} \\plus{} q) \\plus{} (0 \\minus{} x_{F}) \\plus{} (q \\minus{} x_{H}) \\right] \\equal{}$\r\n\r\n$ \\equal{} \\left[(x_{E} \\plus{} x_{F}) \\plus{} (x_G \\plus{} x_{H})\\right] \\minus{} \\left[(x_{A} \\plus{} x_{B}) \\plus{} (x_{C} \\plus{} x_{D})\\right] \\equal{} 0$\r\n\r\nQ.E.D." } { "Tag": [ "ratio", "analytic geometry", "rotation", "geometry", "power of a point", "Pythagorean Theorem" ], "Problem": "If circular arcs $ AC$ and $ BC$ have centers at $ B$ and $ A$, respectively, then there exists a circle tangent to both $ \\stackrel{\\frown}{AC}$ and $ \\stackrel{\\frown}{BC}$, and to $ \\overline{AB}$. If the length of $ \\stackrel{\\frown}{BC}$ is $ 12$, then the circumference of the circle is\n[asy]unitsize(4cm);\ndefaultpen(fontsize(8pt)+linewidth(.8pt));\ndotfactor=3;\n\npair O=(0,.375);\npair A=(-.5,0);\npair B=(.5,0);\npair C=shift(-.5,0)*dir(60);\ndraw(Arc(A,1,0,60));\ndraw(Arc(B,1,120,180));\ndraw(A--B);\ndraw(Circle(O,.375));\ndot(A);\ndot(B);\ndot(C);\nlabel(\"$A$\",A,SW);\nlabel(\"$B$\",B,SE);\nlabel(\"$C$\",C,N);[/asy]$ \\textbf{(A)}\\ 24 \\qquad \\textbf{(B)}\\ 25 \\qquad \\textbf{(C)}\\ 26 \\qquad \\textbf{(D)}\\ 27 \\qquad \\textbf{(E)}\\ 28$", "Solution_1": "Cutting a string the length of $ \\stackrel{\\frown}{BC}$, we find that it is pretty much half the circumference of the circle. $ 12\\cdot 2 \\equal{} 24$; the answer is $ \\boxed{A}$.", "Solution_2": "[hide]From Power of a Point and substitution, you can find the ratio of the radius as well as circumference of the smaller circle to AB is $ \\frac{3}{8}$. ABC is an equilateral triangle as all sides are radii of congruent circles so arc BC is a sixth of the circumference of circle A. This means the circumference of the smaller circle is $ \\frac{3}{8}*72\\equal{}27$ or D.[/hide]", "Solution_3": "I bashed with coordinates. Not that quick, but it works.", "Solution_4": "[quote=\"Lazarus\"]Cutting a string the length of $ \\stackrel{\\frown}{BC}$, we find that it is pretty much half the circumference of the circle. $ 12\\cdot 2 \\equal{} 24$; the answer is $ \\boxed{A}$.[/quote]\r\nHuman error is a terrible thing. \"Pretty much\" is not the same as \"exactly,\" especially when the answer choices are rather close to each other numerically.", "Solution_5": "Nevermind got it.", "Solution_6": "My solution:\r\n\r\nFrom the equilateral triangle, we have $ 12 \\equal{} 1/6 * 2\\pi r \\implies r \\equal{} 36/\\pi$\r\n$ (36\\pi \\minus{} R)^2 \\equal{} R^2 \\plus{} 18^2/\\pi^2\\implies R \\equal{} 972/(72\\pi)$\r\n$ 2\\pi R \\equal{} 972/36 \\equal{} 27$", "Solution_7": "how is it equilateral if AB is not an arc but rather a line? is it not isoceles?\r\n\r\nnvm", "Solution_8": "GlenthemanN:\r\nSince B is the center of the arc AC, we have AB=BC. Similarly, since A is the center of the arc BC, we have BA=AC. This gives us AB=BC=CA, or ABC being an equilateral triangle.", "Solution_9": "mathwiz, could you explain how you get $ (36\\pi \\minus{} R)^2 \\equal{} R^2 \\plus{} 18^2/\\pi^2\\implies R \\equal{} 972/(72\\pi)$\r\n\r\nive tried going the coordinate route to no avail, I have\r\n$ (x\\minus{}r)^2 \\plus{}y^2 \\equal{} r^2$\r\n$ (x\\plus{}r)^2 \\plus{}y^2 \\equal{} r^2$\r\nand for the circle\r\n$ x^2 \\plus{} (y\\minus{}R)^2 \\equal{} R^2$\r\n\r\nbig thanks to anyone who helps out :blush:", "Solution_10": "If we let M be the midpoint of AB and O the center of the circle, then it's just an application of Pythagorean theorem on triangle OMB.", "Solution_11": "If one circle is internally tangent to another, then the centers of the two circles and the point of tangency are collinear. This is harder to see in my opinion than power of a point is.\r\n\r\nIn this case, draw a segment from A to the point of tangency opposite it (T), let it cut the circle at Q, let M be the midpoint of AM, O the center of the inscribed circle. Then A,O, and T are collinear, so by power of a point, $ AQ \\cdot AT \\equal{} AM^2 \\equal{} (\\frac {AQ}{2})^2$. Then the problem is easy, since we know AQ.", "Solution_12": "yes thats the issue I had. Cant believe I didn't see it, guess the two arcs threw me off. I wasnt sure if the line BO extended would pass through the tangent or not :ninja: got it not. makes a lot more sense", "Solution_13": "[quote=\"GlenthemanN\"]yes thats the issue I had. Cant believe I didn't see it, guess the two arcs threw me off. I wasnt sure if the line BO extended would pass through the tangent or not :ninja: got it not. makes a lot more sense[/quote]\r\nThe diagram is a bit misleading in this regard, but it is a fact that if you have an circle internally tangent to another, then the centers of both circles and the point and tangency will be collinear. It's a bit more obvious if you try a few diagrams yourself, and maybe rotate the paper around when you look at it...", "Solution_14": "Quick question-- why isn't the triangle made by A, the midpoint of AB, and the center of the circle a 30-60-90? Or is it? After finding AB I tried using ratios to find r but it doesn't work.", "Solution_15": "It's not 30-60-90. Can you elaborate why you think it is?", "Solution_16": "Denote $AB=r$ and $\\stackrel{\\frown}{BC}=l$ . Observe that $\\pi r=3l\\implies \\boxed{r=\\frac {3l}{\\pi}}\\ (*)$ . Denote the length $x$ of the radius for the circle $w$ which\n\n is tangent to $AB$ and which is interior tangent to the circles $C(A,r)\\ ,\\ C(B,r)$ . Hence $(r-x)^2=x^2+\\left(\\frac r2\\right)^2$ $\\implies $ $x=\\frac {3r}{8}$ . \n\nIn conclusion, the length of the circumference $L$ for the circle $w$ is $2\\pi\\cdot\\frac {3r}{8}=\\frac {3\\pi }{4}\\cdot r\\stackrel{(*)}{=}\\frac {3\\pi}{4}\\cdot\\frac {3l}{\\pi}=\\frac {9l}{4}$ . For $l:=12$ obtain that $L=27$ .", "Solution_17": "I thought that it would be since triangle ABC is equilateral.", "Solution_18": "[quote=\"Ignite168\"][hide]From Power of a Point and substitution, you can find the ratio of the radius as well as circumference of the smaller circle to AB is $ \\frac{3}{8}$. [/quote]\nCould you elaborate on where this particular result came from? I'm not seeing where power of a point comes in.", "Solution_19": "[quote=\"SalsaOnStrings\"][quote=\"Ignite168\"][hide]From Power of a Point and substitution, you can find the ratio of the radius as well as circumference of the smaller circle to AB is $ \\frac{3}{8}$. [/quote]\nCould you elaborate on where this particular result came from? I'm not seeing where power of a point comes in.[/quote]\n\nI dont understand that either. Where does the PPT come from?\n\nEdit: Nvm. I figured it out.\n\n[asy]\nunitsize(2cm);\ndefaultpen(fontsize(8pt)+linewidth(.8pt));\ndotfactor=3;\npair O=(0,.375);\npair A=(-.5,0);\npair B=(.5,0);\npair C=shift(-.5,0)*dir(60);\ndraw(Arc(A,1,0,60));\ndraw(Arc(B,1,120,180));\ndraw(A--B);\ndraw(Circle(O,.375));\ndot(A);\ndot(B);\ndot(C);\nlabel(\"$A$\",A,SW);\nlabel(\"$B$\",B,SE);\nlabel(\"$C$\",C,N);\ndraw(C--B,blue);\ndraw(C--(0,0),blue);[/asy]", "Solution_20": "[quote=mathwizarddude]My solution:\n\nFrom the equilateral triangle, we have $ 12 \\equal{} 1/6 * 2\\pi r \\implies r \\equal{} 36/\\pi$\n$ (36\\pi \\minus{} R)^2 \\equal{} R^2 \\plus{} 18^2/\\pi^2\\implies R \\equal{} 972/(72\\pi)$\n$ 2\\pi R \\equal{} 972/36 \\equal{} 27$[/quote]\n\nCould someone explain the $(36\\pi \\minus{} R)^2$ term?", "Solution_21": "[quote=macandcheese][quote=mathwizarddude]My solution:\n\nFrom the equilateral triangle, we have $ 12 \\equal{} 1/6 * 2\\pi r \\implies r \\equal{} 36/\\pi$\n$ (36\\pi \\minus{} R)^2 \\equal{} R^2 \\plus{} 18^2/\\pi^2\\implies R \\equal{} 972/(72\\pi)$\n$ 2\\pi R \\equal{} 972/36 \\equal{} 27$[/quote]\n\nCould someone explain the $(36\\pi \\minus{} R)^2$ term?[/quote]\nIt was a typo. It's supposed to be $(36/\\pi-R)^2 = R^2 + 18^2/\\pi^2$, where $R$ denotes the radius of the circle.\n\nIf we let M be the midpoint of AB and O the center of the circle, then it's just an application of Pythagorean theorem on triangle OMB.", "Solution_22": "Since $\\Delta ABC$ is equilateral, $\\angle CAB = 60$ degrees, so the circle with center $A$ containing points $B$ and $C$, has circumference $\\frac{360}{60} \\cdot 12=72$ and radius $\\frac{36}{\\pi}$.\nDraw the altitude from $C$ to $AB$ and let the intersection with $AB$ be point $M$. Let the radius of the circle we are trying to find be $r$ and let it's center be $O$. Then, extend $AO$ until it intersects circle $A$; call this point of intersection $N$. Then, since $AN=\\frac{36}{\\pi}$, $AO=\\frac{36}{\\pi}-r$. We also know that $OM=r$ and $AM=\\frac{18}{\\pi}$, so by the Pythagorean theorem, $\\left(\\frac{36}{\\pi}-r\\right)=r^2+\\left( \\frac{18}{\\pi} \\right)^2 \\implies$ $r=\\frac{27}{2\\pi}$. Hence, our answer is $\\boxed{27\\pi}$." } { "Tag": [ "puzzles" ], "Problem": "two guys were thrown in jail and as they were sitting they wanted to know each other's sentence. All they know was that their sentences differ by one year. they had the following coversation:\r\n\r\nPerson1: I don't know how much you got\r\nPerson2: I know you don't know how much I got\r\nperson1: Then I know how much you got\r\nperson2: Then I also know how much you got\r\n\r\n\r\nwhat are the sentences of both of these men?", "Solution_1": "I like this, but it's not calculus. :)\r\n\r\nWhen person A says he doesn't know what the other got, A can't have 1 year. If he did B would have 2 and A would be able to figure it out.\r\n\r\nWhen B says he is aware of this we know that B can't have 2 either. If B had 2 there would be a possibility that A had 1.\r\n\r\nWhen A then says he knows how much B got, we know that A must have 3. (So B is 2 or 4 but we know that B can't be 2). Again, A can't have 1.\r\n\r\nSo A has 3 and B has 4. I believe.", "Solution_2": "The jail sentence are as follows:\r\n\r\nPerson 1 --> n years\r\nPerson 2 --> (n + 1) or (n-1) years\r\n\r\nPerson1: I don't know how much you got \r\n[b]Person1 is not having a sentence for 1 year or else he could have known that Person2 is having it for 2 years (there cannot be a sentence for 0 years)[/b]. \r\n\r\nPerson2: I know you don't know how much I got \r\n[b]This means Person2 is not having the sentence for 2 years. Then, he/she could have been so sure[/b]. \r\n\r\nperson1: Then I know how much you got \r\n[b]Person1 can interpret this message and he/she comes to know how much Person2 has got. This implies that Person1 is having a sentence for 3 years. And, he/she comes to know that Person2 is not having it for 2 years. Thus, Person2 has it for 4 years[/b].\r\n\r\nperson2: Then I also know how much you got\r\n\r\n[b]Person1 ==> 3 years\nPerson2 ==> 4 years[/b]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "For which integers $n\\geq 2$, the number $(n-1)^{n^{n+1}}+(n+1)^{n^{n-1}}$ is divisible by $n^{n}$ ?", "Solution_1": "It is obviosly and posted before.", "Solution_2": "How is it obvious? Can you show?", "Solution_3": "http://www.mathlinks.ro/viewtopic.php?t=150522" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Let $ A,B,C,Q$ be fixed points on plane. $ M,N,P$ are intersection points of $ AQ,BQ,CQ$ with $ BC,CA,AB$. $ D',E',F'$ are tangency points of incircle of $ ABC$ with $ BC,CA,AB$. Tangents drawn from $ M,N,P$ (not triangle sides) to incircle of $ ABC$ make triangle $ DEF$. Prove that $ DD',EE',FF'$ intersect at $ Q$.", "Solution_1": "I am surprised because of it was a contest problem !\r\n\r\nAny way, we can get a solution of the wanted result, applying the proof of the generalization of a familiar problem in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=140191]Interesting and hard[/url], of [b][size=100]cvix[/size][/b].\r\n\r\nKostas Vittas.", "Solution_2": "Dear Kostas,\r\nWhy are you surprised?", "Solution_3": "Can someone write out a solution?I can't continue [b]vittasko[/b]'s comment...", "Solution_4": "[quote=\"Omid Hatami\"]Let $ A,B,C,Q$ be fixed points on plane. $ M,N,P$ are intersection points of $ AQ,BQ,CQ$ with $ BC,CA,AB$. $ D',E',F'$ are tangency points of incircle of $ ABC$ with $ BC,CA,AB$. Tangents drawn from $ M,N,P$ (not triangle sides) to incircle of $ ABC$ make triangle $ DEF$. Prove that $ DD',EE',FF'$ intersect at $ Q$.[/quote]\n\nwe just need prove $CP,NB$ and $D'D$ are concurrent at $Q$\n\n Tangents drawn from $ N,P$ (not triangle sides) to incircle of $ ABC$ intersection $AB, AC at J , K$ and intersection incircle of $ ABC$ at $O,V$\n\n*) $PKCB$ is a special quadrilateral ( i don't know this name :( ) and we know following lemma:\n$F'O$ intersection $D'E'$ at $X$ then $X$ is pole of $PC.$\n\n*) $NJBC$ is a special quadrilateral and $E'V$ intersection $D'F'$ at $Y$ then $Y$ is pole of $BN$\n\n*) $OV$ intersection Tangents drawn from $D'$ to incircle of $ ABC$ at $Z$ then $Z$ is pole of $DD'$\n\nand use Pascal theorem for $(D'D'F'VOE')$ we have $XYZ$ therefore $DD',PC,BN$ are concurrent\n[img]http://ns0.upanh.com/b1.s33.d1/eae7bc59240a3ffea1a834c2d9a151b9_50500680.untitled.700x0.jpg[/img]", "Solution_5": "The idea of using pole-polar and Pascal is good, but additionally I think the whole problem is rested on nothing else than two well known properties of pole-polar: La Hire\u2019s Theorem ([i]Let x and y be the polars of X and Y, respectively; then X is on line y \u21d4 Y is on line x[/i]) and this: [i]Let x, y, z be the polars of distinct points X, Y, Z, respectively; then Z = x\u2229y \u21d4 z = XY[/i] (what can be proved easily using La Hire).\n\nLet $E''F''$ meet $BC$ at $G$, $E'E''$ meet $F'D'$ at $H$, $F'F''$ meet $D'E'$ at $K$. From now on, while talking about poles and polars we do so with respect to the incircle of $ABC$.\nFirst notice that $E''F''$ is the polar of $D$, as $G$ lies on the polar of $D$, $D$ must lie on the polar of $G$, but we know that the tangent point $D'$ is also on the polar of $G$, so $DD'$ is the polar of $G$. Then to show that $DD'$ passes through $Q$, it suffices to show that $Q$ lies on the polar of $G$.\nNow notice that $F'F''$ is the polar of $P$ and $D'E'$ is the polar of $C$, then $PC$ is the polar of $K\\equiv F'F''\\cap D'E'$.\n$E'E''$ is the polar of $N$ and $F'D'$ is the polar of $B$, then $NB$ is the polar of $H$. Then $HK$ is the polar of $Q\\equiv PN\\cap BN$.\nUsing Pascal for $E''F''F'D'D'E'$ we get $G\\equiv E''F''\\cap D'D', K\\equiv F''F'\\cap D'E', H\\equiv E''E'\\cap F'D'$ collinear.\nAs $G$ lies on the polar of $Q$, $Q$ must lie on the polar of $G$. q.e.d.\nBy the same way we get $EE'$ and $FF'$ pass through $Q$.\n\nNote: Having a look back at the [b]cvix[/b]'s [i]interesting and hard[/i] [b]vittasko[/b] mentioned above, we see that the problem can be solved in the same way, of course with the aid of a well known lemma (instead of Pascal in our problem).", "Solution_6": "We can use brianchon's theorem on $BD'CNDP$ then we get Q, D, D' are collinear, it finishes the proof. :D" } { "Tag": [ "inequalities", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Let $p,q$ are positive real numbers.\r\nProve the following inequality.\r\n\r\n\\[\\frac{p^2+q^2}{2}\\ln \\frac{p^2+q^2}{2}\\geqq -\\frac{1}{2}(p-q)^2+\\frac{p^2\\ln p^2+q^2\\ln q^2}{2}\\]", "Solution_1": "Can anyone solve this problem? :?", "Solution_2": "I will. ;) \r\nlet $f(p)=\\frac{p^2+q^2}{2}\\ln \\frac{p^2+q^2}{2} +\\frac{1}{2}(p-q)^2-\\frac{p^2\\ln p^2+q^2\\ln q^2}{2} $\r\nwe have $f'(p)=\\frac{p^2+q^2}{2} \\frac{2}{p^2+q^2}p+p ln \\frac{p^2+q^2}{2} +(p-q)-pln p^2-p=p ln \\frac{p^2+q^2}{2} +(p-q)-pln p^2$\r\ntry to show $f'(p) \\leq 0,p \\geq q$,$f'(p) \\geq 0,p \\leq q$(by$f\\\"(p)$\r\nthen we are done :) \r\nthank you for your question", "Solution_3": "Thank you for your reply,zhaobin.\r\n\r\nYou are right.\r\n\r\nkunny" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Let there be a polynomial such that\r\n$ P(x)\\equal{}x^{4}\\plus{}ax^{3}\\plus{}bx^{2}\\plus{}cx\\plus{}d$.\r\nIf $ P(1)\\equal{}10$,\r\n$ P(2)\\equal{}20$, and\r\n$ P(3)\\equal{}30$, find $ \\frac{P(10)\\plus{}P(\\minus{}8)}{10}$.", "Solution_1": "Let $ Q(x) \\equal{} P(x) \\minus{} 10x$, which is a monic quartic with roots $ x \\equal{} 1,2,3,p$. Then $ Q(x) \\equal{} (x \\minus{} p)(x \\minus{} 1)(x \\minus{} 2)(x \\minus{} 3)$, and $ \\frac {P(12) \\plus{} P( \\minus{} 8)}{10}$ $ \\equal{} \\frac {Q(12) \\plus{} 120 \\plus{} Q( \\minus{} 8) \\minus{} 80}{10}$ $ \\equal{} \\frac {[12 \\minus{} p]\\cdot 9 \\cdot 10 \\cdot 11 \\plus{} 120 \\plus{} [ \\minus{} 8 \\minus{} p] \\cdot ( \\minus{} 1) \\cdot 9 \\cdot 10 \\cdot 11 \\minus{} 80}{10}$ $ \\equal{} \\frac {20 \\cdot 9 \\cdot 10 \\cdot 11 \\plus{} 40}{10} \\equal{} 1984$. \r\n\r\n- edited.", "Solution_2": "Oh..OOPS I made a mistake in the question. \r\nYou didn't make a stupid mistake...my question was wrong. \r\n\r\nIt's supposed to be\r\nFind $ \\frac{P(12)\\plus{}P(\\minus{}8)}{10}$. \r\nSorry =\\" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Let $A$ be a square matrix of order 3. If $A\\acute{A}= I$ and $det.A = 1$, then $det.(A-I) = ?$\r\n\r\nAlso can anyone help me with some important properties of orthogonal and hermitian matrices ?", "Solution_1": "Sorry, but what does $\\acute{A}$ means?", "Solution_2": "I think he meant $A A^{T}= I_{n}$. That's the definition of an orthogonal matrix.", "Solution_3": "$det(A-I)=0$ . we can prove this as follows. $det(A-I)=detA^{T}.det(A-I)$ since$detA^{T}=1$. therefore $det(A-I)=det(A^{T}A-A^{T})=det(I-A^{T})=det(I-A)^{T}=det(I-A)$. since A is of order 3, $det(I-A)=-det(A-I)$ which gives the answer that the value of the determinant is 0 :wink:" } { "Tag": [ "counting", "distinguishability" ], "Problem": "A mathematical prodigy wishes to put 2 of his IMO gold medals and 2 of his IPhO gold medals in one row. How many distinct arrangements are possible?", "Solution_1": "Assuming the medals are indistinguishable, the answer is $ \\frac{4!}{2!2!}$, or $ \\boxed{6}$.", "Solution_2": "Or we could do $ \\binom{4}{2}$ to pick the two spots to put the IMO medals and the rest must go into their slots.", "Solution_3": "This kid has 2 IMO gold medals and 2 IPhO gold medals. He can't figure this out himself?", "Solution_4": "Bruh, he probably already knows how to do it and the problem is not asking if he knows it or not. Plus, he could've stolen them people.", "Solution_5": "[quote=huili2010]Bruh, he probably already knows how to do it and the problem is not asking if he knows it or not. Plus, he could've stolen them people.[/quote]\n\ngood point. ", "Solution_6": "He is a [quote]...mathematical prodigy...[/quote] He should be able to figure it out. \n\n[hide = Solution if distinguishable]There are $ 4! = 24$ ways to order the medals. [/hide]\n[hide = Solution if indistinguishable]There are $ \\frac{4!}{2!2!}=6 $ ways to order the medals. We divide by $ 2 $ twice because the IMO and IPhO medals are indistinguishable, but IMO and IPhO medals are probably different. [/hide]", "Solution_7": "[hide=my dumb solution 1: Write down the options]I write them down. There are $\\boxed{6}$[/hide]\n[hide=my dumb solution 2]We see that there are $4!$ ways if they were distinguishable. We divide by $2\\cdot2$ to account for indistinguishability, for an answer of $\\boxed{6}$[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "saabet konid ke be ezaaye har adade tabiii mese $n$,\r\n$\\sum_{k=0}^{n} \\left(\\begin{array}{c} {n-k} \\\\ {[\\frac{n-k}{2}]}\\end {array} \\right)2^k\\left(\\begin{array}{c} n\\\\k\\end{array} \\right)=\\left(\\begin{array}{c} 2n+1\\\\n \\end{array} \\right)$\r\n\r\n\r\n\r\nkhodayish poostam kande shod ta neveshtamesh :stretcher:\r\n\r\n\r\n[color=red]Moderator Edit : LaTeX Formula Corrected[/color]", "Solution_1": "Man ke nafahmidam shayad manzooret in bashe: $\\sum_{k=0}^{n}\\binom{n-k}{[\\frac{n-k}2]} 2^k\\binom nk=\\binom {2n+1}n$\r\n\r\nDar zemn mitoonid be jaye in hame array neveshtan az [color=red]\\binom{m}{n}[/color] baraye $\\binom mn$ estefade konid." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Can someone give me the answer to this question?\r\n\r\n(X,M) is a measurable space. Show that \r\nif f:X -> [-oo, oo] and f^-1((r,oo]) in M for all rational r,\r\nthen f is measurable.", "Solution_1": "The definition of 'measurable function' is that $f^{-1}(B)$ is measurable for every Borel set $B$, right?\r\n\r\nSo consider $X = \\{X\\text{ is a subset of }\\mathbb{R} \\mid f^{-1}(X)\\text{ is measurable }\\}$. Show that $X$ is an $\\sigma$-algebra. If $X$ contains all $(a,\\infty]$, we can conclude that all Borel sets are in $X$, and $f$ is a measurable function.\r\n\r\nNow we have only that $(r,\\infty]\\in X$ for all rationals $r$. For a real $a$, we can find a sequence of rationals $r_n\\searrow a$, $(a,\\infty] = \\bigcup (r_n,\\infty]$ then $f^{-1}(a,\\infty] = \\bigcup f^{-1}(r_n,\\infty]$. Hence $(a,\\infty]\\in X$." } { "Tag": [ "function", "calculus", "derivative", "inequalities" ], "Problem": "Show whether the function $ f(x)\\equal{}2x^3\\minus{}x\\plus{}1$ is convex or concave.", "Solution_1": "I think that $ f\"(x)\\equal{}12x$ is enough to solve the problem!! no?", "Solution_2": "Adding to the post above, that implies that the function is concave on $ ( \\minus{} \\infty,0)$ and convex on $ (0,\\infty)$.", "Solution_3": "How would you do it without the second derivative?", "Solution_4": "You could prove it with the converse of Jensen's inequality:\r\n\r\nProve that for all nonnegative $ x,y$,\r\n\r\n$ \\frac{f(x)\\plus{}f(y)}{2}\\ge f\\left(\\frac{x\\plus{}y}{2}\\right)$.\r\n\r\nAlthough taking the second derivative is without a question the easiest way.", "Solution_5": "[quote=\"Yongyi781\"]Adding to the post above, that implies that the function is concave on $ ( \\minus{} \\infty,0]$ and convex on $ [0,\\infty)$.[/quote]\r\n\r\nThose aren't disjoint partitions of the continuum. $ x \\equal{} 0$ is an inflection point of the function - you could describe it as nonconvex and nonconcave, but I believe concave and convex are terms limited to a positive second derivative.", "Solution_6": "Indeed, you are right. I was being absent-minded when I posted that :blush:", "Solution_7": "[quote=\"modularmarc101\"]How would you do it without the second derivative?[/quote]\r\nwe can say that $ f$ is convex if:\r\n\r\n$ \\forall(x,y)\\in{I^{2}} \\forall(\\alpha,\\beta)\\in{[0,1]^{2}} \\alpha\\plus{}\\beta\\equal{}1\\Rightarrow f(\\alpha x\\plus{}\\beta y)\\le\\alpha f(x)\\plus{}\\beta f(y)$", "Solution_8": "Are there any times when the second derivative doesn't work?" } { "Tag": [ "trigonometry", "geometry", "trig identities", "Law of Sines" ], "Problem": "Prove the Law Of Tangents\r\n\r\n$ \\frac{a \\minus{}b}{a\\plus{}b}$ = $ \\frac{tan\\frac{(A\\minus{}B)}{2}}{tan\\frac{(A\\plus{}B)}{2}}$", "Solution_1": "[hide=\"Proof\"]Since $ \\tan x \\equal{} \\frac {\\sin x}{\\cos x}$, we know that \n\n$ \\frac {\\tan (\\frac {A \\minus{} B}{2})}{\\tan ( \\frac {A \\plus{} B}{2})} \\equal{} \\frac {\\sin (\\frac {A \\minus{} B}{2}) \\cos (\\frac {A \\plus{} B}{2}) } {\\sin (\\frac {A \\plus{} B}{2}) \\cos (\\frac {A \\minus{} B}{2})}$\n\nBy the Product-to-Sum identity, we can rewrite this as \n\n$ \\frac {\\sin A \\minus{} \\sin B}{\\sin A \\plus{} \\sin B}$\n\nNow notice that by the Law of Sines, \n\n$ \\sin M \\equal{} \\frac {m}{2R}$ so, \n\n$ \\frac {\\sin A \\minus{} \\sin B}{\\sin A \\plus{} \\sin B} \\equal{} \\frac {a/2R \\minus{} b/2R}{a/2R \\plus{} b/2R} \\equal{} \\frac {a \\minus{} b}{a \\plus{} b}$ [/hide]", "Solution_2": "Good job, just pulled it from vol2." } { "Tag": [ "probability", "combinatorics proposed", "combinatorics" ], "Problem": "In the weekly State Lottery, a sequence of seven numbers from the set $ \\{ 0,1,...,9 \\}$ is picked at random. Compute the probability that the sequence contains exactly five distinct numbers.", "Solution_1": "[quote=\"moldovan\"]In the weekly State Lottery, a sequence of seven numbers from the set $ \\{ 0,1,...,9 \\}$ is picked at random. Compute the probability that the sequence contains exactly five distinct numbers.[/quote]\r\n\r\nis it {10c5 * 7c5*10 }/10^7 ?" } { "Tag": [ "number theory", "greatest common divisor" ], "Problem": "Are two null sets disjoint?", "Solution_1": "Sure, their intersection is the null set. :) It's sorta like $ \\text{GCD}[1,1]\\equal{}1$.\r\n\r\nNeither of them have any prime factors, but the GCD still exists." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be real positive numbers such that absolute difference between any two of them is less than $ 2$. Prove that: $ a \\plus{} b \\plus{} c < \\sqrt {ab \\plus{} 1} \\plus{} \\sqrt {ac \\plus{} 1} \\plus{} \\sqrt {bc \\plus{} 1}$", "Solution_1": "[b][color=darkblue]This problem is from Russia Olympiad 2004 Grade 9.\n\nWe have \n\n$ \\begin{array}{cc} \\\\\n \\\\\n \\\\\n|a \\minus{} b| < 2 \\\\\n \\\\\na^2 \\minus{} 2ab \\plus{} b^2 < 4 \\\\\n \\\\\na^2 \\plus{} 2ab \\plus{} b^2 < 4(1 \\plus{} ab) \\\\\n \\\\\na \\plus{} b < 2\\sqrt {1 \\plus{} ab} \\\\\n \\\\\n \\\\\n \\\\\n\\end{array}$. \n\nBy adding them we obtain:\n $ a \\plus{} b \\plus{} c < \\sqrt {ab \\plus{} 1} \\plus{} \\sqrt {ac \\plus{} 1} \\plus{} \\sqrt {bc \\plus{} 1}$[/color][/b]", "Solution_2": "Cool :cool:" } { "Tag": [ "inequalities", "search", "function", "rearrangement inequality" ], "Problem": "These are 3 inequalities that I was not able to prove:\r\n\r\n$1) \\frac{a^2}{b^5} + \\frac{b^2}{c^5} + \\frac{c^2}{d^5} + \\frac{d^2}{a^5} \\geq \\frac{1}{a^3} + \\frac{1}{b^3} + \\frac{1}{c^3} +\\frac{1}{d^3}$\r\n\r\n$2) \\frac{1}{a^2+bc} + \\frac{1}{b^2+ac} + \\frac{1}{c^2+ab} \\leq \\frac{a+b+c}{2abc}$\r\n\r\n$3) \\frac{1}{a^3+b^3+abc} + \\frac{1}{b^3+c^3+abc} + \\frac{1}{a^3+c^3+abc} \\leq \\frac{1}{abc}$\r\n\r\nUpdate: I forgot to say that $a,b,c > 0$", "Solution_1": "The first inequality is not true for all real $a,b,c,d$. (Take $a=-1, b=-1, c=-1, d=-2$, for example.)\r\n\r\nFor positive $a,b,c,d$ though, you can use Rearrangement on $\\{a^2, b^2, c^2, d^2\\}$ and $\\left\\{\\frac{1}{a^5},\\frac{1}{b^5},\\frac{1}{c^5},\\frac{1}{d^5}\\right\\}$.", "Solution_2": "[quote=\"towersfreak2006\"]you can use Rearrangement on $\\{a^2, b^2, c^2, d^2\\}$ and $\\left\\{\\frac{1}{a^5},\\frac{1}{b^5},\\frac{1}{c^5},\\frac{1}{d^5}\\right\\}$.[/quote]\r\n\r\nCan you show me how to rearrangment them?", "Solution_3": "If $a^2$ is the greatest element in the first set, then $\\frac{1}{a^5}$ is the least element in the second set, since $a>0$. Hence, the right hand side, which is equivalent to $a^2\\cdot\\frac{1}{a^5}+b^2\\cdot\\frac{1}{b^5}+c^2\\cdot\\frac{1}{c^5}+d^2\\cdot\\frac{1}{d^5}$, is the sum of the product of corresponding elements in oppositely sorted sets. From Rearrangement Inequality, we know that the sum of the product of corresponding elements in any other orderings of the two sets is greater than or equal to the sum of the product of corresponding elements in oppositely sorted sets. The left hand side is one such sum.", "Solution_4": "Rearrangement Inequality, thank you towersfreak2006 I just learn that.\r\n\r\nSo there're 2 questions left. :roll:", "Solution_5": "[quote]$\\frac{1}{a^2+bc} + \\frac{1}{b^2+ac} + \\frac{1}{c^2+ab} \\leq \\frac{a+b+c}{2abc}$[/quote]\r\n\r\nNote that the given inequality is homogenous in its variables. Thus, we can normalize to $abc=1$. But first, rewrite as \\[ \\frac{1}{a^2+\\frac{abc}{a}} + \\frac{1}{b^2+\\frac{abc}{b}} + \\frac{1}{c^2+\\frac{abc}{c}} \\leq \\frac{a+b+c}{2abc}, \\] \\[ \\frac{2}{a^2+\\frac{1}{a}}+\\frac{2}{b^2+\\frac{1}{b}}+\\frac{2}{c^2+\\frac{1}{c}}\\le a+b+c. \\] From GM-HM on each of the three terms, we have \\[ \\frac{2}{a^2+\\frac{1}{a}}+\\frac{2}{b^2+\\frac{1}{b}}+\\frac{2}{c^2+\\frac{1}{c}}\\le \\sqrt{\\frac{1}{a}}+\\sqrt{\\frac{1}{b}}+\\sqrt{\\frac{1}{c}}. \\] Since $abc=1$, then $\\sqrt{abc}=1$. Hence, multiply the given inequality by $\\sqrt{abc}$, \\[ \\frac{2}{a^2+\\frac{1}{a}}+\\frac{2}{b^2+\\frac{1}{b}}+\\frac{2}{c^2+\\frac{1}{c}}\\le\\sqrt{bc}+\\sqrt{ca}+\\sqrt{ab}. \\] From Rearrangment Inequality, we know that \\[ \\frac{2}{a^2+\\frac{1}{a}}+\\frac{2}{b^2+\\frac{1}{b}}+\\frac{2}{c^2+\\frac{1}{c}}\\le\\sqrt{bc}+\\sqrt{ca}+\\sqrt{ab}\\le\\sqrt{a^2}+\\sqrt{b^2}+\\sqrt{c^2}=a+b+c, \\] as desired. :)", "Solution_6": "3 can be found a million times on the forum (use the search function)\r\n\r\nHere is a solution to 2.\r\n\r\nWe want to prove\r\n\\[ \\sum \\frac{1}{a^2+bc}\\leq \\frac{1}{2}\\sum \\frac 1{bc} \\]\r\nBy summing cyclically, it is enough to prove that\r\n\\[ \\frac{1}{a^2+bc}\\leq \\frac{1}{4}\\left(\\frac 1{ab}+\\frac 1{ac}\\right) \\]\r\nThis rearranges to\r\n\\[ 4abc\\leq (a^2+bc)(b+c) \\]\r\nBy AM-GM, we get\r\n\\[ (a^2+bc)(b+c)\\geq (2a\\sqrt{bc})(2\\sqrt{bc})=4abc \\]\r\nso the result follows.", "Solution_7": "Here is a solution to 3\r\n\r\nhttp://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27&year=1997&p=185470", "Solution_8": "I've got a question to towersfreak2006. [quote]Note that the given inequality is homogenous in its variables. Thus, we can normalize to $abc=1$ . But first, rewrite as...[/quote].\r\nWhy can we assume, that $abc=1$. Could you explain?", "Solution_9": "If $abc \\neq 1$, simply replace $a$ with $ad$, $b$ with $bd$ and $c$ with $cd$. The inequality will still look exactly the same (if it was true before, it will be true now), and we have changed the product $abc$ to $d^3$ times the previous product -- in particular, we can change it to 1 with a good choice of $d$. We can do this because every term has the same total power in the variables (it is homogeneous).", "Solution_10": "sorry... This was rubbish", "Solution_11": "Since $a^2+1/a\\geq 2\\sqrt{a}$, it follows that\r\n\\[ \\frac{2}{a^2+a^{-1}}\\leq \\frac{2}{2\\sqrt{a}}=\\sqrt{\\frac{1}{a}} \\]\r\nIn any event, I feel that my solution is more natural, since I avoid dehomogenizing, which is an odd step when presented with a homogeneous inequality (unless, of course, you want to apply Jensen's)", "Solution_12": "[quote=\"Laputa\"]$2) \\frac{1}{a^2+bc} + \\frac{1}{b^2+ac} + \\frac{1}{c^2+ab} \\leq \\frac{a+b+c}{2abc}$\n\nI forgot to say that $a,b,c > 0$[/quote]\r\n\r\n$\\frac{a^2+bc}{2}\\geq \\sqrt{a^2bc}=a\\sqrt{bc}$ $\\Rightarrow \\frac1{a^2+bc}\\leq \\frac{1}{2a\\sqrt{bc}}=\\frac{\\sqrt{bc}}{2abc}$\r\n\r\n$\\frac1{a^2+bc}+\\frac1{b^2+ac}+\\frac1{c^2+ab}\\leq \\frac{\\sqrt{bc}+\\sqrt{ac}+\\sqrt{ab}}{2abc}\\leq \\frac{a+b+c}{2abc}$\r\n\r\nThe last one is true since Cauchy's yields that $(a+b+c)(b+c+a)\\geq (\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca})^2$", "Solution_13": "[quote=\"Laputa\"]$3) \\frac{1}{a^3+b^3+abc} + \\frac{1}{b^3+c^3+abc} + \\frac{1}{a^3+c^3+abc} \\leq \\frac{1}{abc}$\n\nI forgot to say that $a,b,c > 0$[/quote]\r\n\r\nUsing your favourite inequality (for instance, rearrangement), we have $a^3+b^3\\geq a^2b+ab^2$\r\n\r\n$a^3+b^3+abc\\geq a^2b+ab^2+abc=ab(a+b+c)\\Rightarrow \\frac1{a^3+b^3+abc}\\leq \\frac1{ab(a+b+c)}$\r\n\r\n$\\frac1{a^3+b^3+abc}+\\frac1{b^3+c^3+abc}+\\frac1{c^3+a^3+abc}\\leq \\frac{a+b+c}{abc(a+b+c)}=\\frac1{abc}$", "Solution_14": "Thank you to all of you. Very nice solutions! :) \r\n\r\n[quote=\"blahblahblah\"]We want to prove\n\\[ \\sum \\frac{1}{a^2+bc}\\leq \\frac{1}{2}\\sum \\frac 1{bc} \\]\nBy summing cyclically, it is enough to prove that\n\\[ \\frac{1}{a^2+bc}\\leq \\frac{1}{4}\\left(\\frac 1{ab}+\\frac 1{ac}\\right) \\]\n[/quote]\r\n\r\nWhat is the sigma for? I never see such notation before. And what is summing cyclically?", "Solution_15": "The $\\sum$, in this context, is a kind of shorthand for writing symmetric sums, e.g, we use\r\n\r\n$\\sum a^2b$ to denote $a^2b+b^2c+c^2a$.\r\n\r\nBy summing cyclically, I mean that we sum that inequality and the two similar ones", "Solution_16": "see here a solution for 3\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=58140" } { "Tag": [ "inequalities", "geometry unsolved", "geometry" ], "Problem": "Let ABC be a triangle, BM; CN are medians. Assume that $ \\angle{B}\\leq\\angle{C}$. Prove that: $ BM+CM\\geq CN+BN$", "Solution_1": "If $\\hat C \\leqslant {90^o}$ , then: (see fig.)\n$\\begin{gathered}\n \\left. \\begin{gathered}\n {\\rm A}{\\rm E}|| = {\\rm B}C\\mathop {}\\limits_{} \\hfill \\\\\n AZ||BC\\mathop {}\\limits_{} \\hfill \\\\\n ZL \\bot BC\\mathop {}\\limits_{} \\hfill \\\\\n KZ = KL\\mathop {}\\limits_{} \\hfill \\\\\n ZI = EC \\hfill \\\\ \n\\end{gathered} \\right\\}\\mathop {}\\limits_{}^{} \\mathop {}\\limits_{}^{} \\Rightarrow \\mathop {}\\limits_{}^{} \\left\\{ \\begin{gathered}\n AB = IL = 2BN\\mathop {}\\limits_{}^{} ,\\mathop {}\\limits_{}^{} IE = LC = ZC = 2CN \\hfill \\\\\n BZ = BL = 2MC\\mathop {}\\limits_{}^{} ,\\mathop {}\\limits_{}^{} BE = 2BM \\hfill \\\\ \n\\end{gathered} \\right. \\hfill \\\\\n \\hfill \\\\\n So:\n\\mathop {}\\limits_{}^{} IL + IE < BL + BE\\mathop {}\\limits_{}^{} \\Rightarrow \\mathop {}\\limits_{}^{} 2BN + 2CN < 2MC + 2BM\\mathop {}\\limits_{}^{} \\Rightarrow \\mathop {}\\limits_{}^{} BN + CN < MC + BM \\hfill \\\\ \n\\end{gathered} $.\n\n\u03a4he same conclusion when $\\hat C > {90^o}$" } { "Tag": [ "probability", "geometry", "analytic geometry", "similar triangles", "absolute value" ], "Problem": "I hated that target about the elevator...", "Solution_1": "Me too! :D They have the solutions up there, but even though I know how to do it, I still hate those kinds of problems.", "Solution_2": "I figured out it was 7th floor FOR EACH TIME about a minute before and then i put 5/32 and then my brain told me it was 3/16 and i was trying to choose with 10 seconds left but i didn't change it... so i got it wrong... even if i did get that target right, i still would be 2nd but still...", "Solution_3": "Can anyone post a *coherent* solution for the elevator target (#7), the tile target (#6), and the Fibonacci Sprint Problem (#28)?\r\n\r\nI stumbled upon the answer for #6, but it was one of those questions where you kind of intuit the answer, so I am looking for a good solution. Also, I guessed correctly on number 28, but I didn't really solve it either. Thanks a bunch.", "Solution_4": "Well first know that there are $2^5 = 32$ ways to flip the coin. \r\nThen the only combinations that work are,\r\nHTHHH\r\nHHTHH\r\nHHTTH\r\nHHHTH\r\nHHHTT\r\nWhich makes 6 combinations so the answer is $\\frac 6{32} = \\frac 3{16}$.", "Solution_5": "28\r\nThere's a 6-number pattern of remainders 1, 1, 2, 3, 1, 0, so you divide 100 by 6 for remainder of 4 making the answer $3$.", "Solution_6": "28. 1 1 2 3 5 8 13 21\r\nWrite that in mod 4.\r\n1 1 2 3 1 0 1 1 2 3 1 0 You see the pattern 112310 and you can find that 100 is 4 mod 6, so it's the 4th one, with is 3.\r\n\r\nEdit: Tarquin beat me to it", "Solution_7": "6. You know that the number of bricks can be expressed as 7x2 (7x*x) where x is the length of the width. Then you start adding layers until you get a number like that 34, 65, 93, 118, 140, 159, 175... 175 is 7*52 so the answer is 7 rows of bricks.", "Solution_8": "[quote=\"bob123\"]6. You know that the number of bricks can be expressed as 7x2 (7x*x) where x is the length of the width. Then you start adding layers until you get a number like that 34, 65, 93, 118, 140, 159, 175... 175 is 7*52 so the answer is 7 rows of bricks.[/quote]\r\n\r\nYeah, but is there any way to just get 7, or do you have to add? Also, is there any formula that would allow you to find the number of bricks if there were, say, 14 rows, without adding?", "Solution_9": "[quote=\"tarquin\"]Well first know that there are $2^5 = 32$ ways to flip the coin. \nThen the only combinations that work are,\nHTHHH\nHHTHH\nHHTTH\nHHHTH\nHHHTT\nWhich makes 6 combinations so the answer is $\\frac 6{32} = \\frac 3{16}$.[/quote]\r\n\r\nThe first one (HTHHH) doesn't work because the problems says \"what is the probability that [b]each[/b] of her next five stops is on the 7th floor or higher.\" So if it was HTHHH, it would be 7,6,7,8,9, but they all have to be above 6. Also, the third one (HHTTH) doesn't work for the same reason (you wind up on the 6th floor after 4 flips). I think the ones you are missing are HHTHT and HHHHH.", "Solution_10": "Whatever", "Solution_11": "[quote=\"Phelpedo\"][quote=\"bob123\"]6. You know that the number of bricks can be expressed as 7x2 (7x*x) where x is the length of the width. Then you start adding layers until you get a number like that 34, 65, 93, 118, 140, 159, 175... 175 is 7*52 so the answer is 7 rows of bricks.[/quote]\n\nYeah, but is there any way to just get 7, or do you have to add? Also, is there any formula that would allow you to find the number of bricks if there were, say, 14 rows, without adding?[/quote]\r\n\r\nSum the first and last number, divide by 2, multiply by the number of terms... That's a way you could do it... Then you have to find two numbers that are both 1 mod 3, and their sum can be expressed as 2x2... I just said the way I did it at state because I had lots of time there.", "Solution_12": "[quote=\"chess64\"]Me too! :D They have the solutions up there, but even though I know how to do it, I still hate those kinds of problems.[/quote]\r\n\r\nWhere can you get the solutions?", "Solution_13": "http://www.mathcounts.org/webarticles/articlefiles/510-05SolutionSetS.pdf", "Solution_14": "The first two flips must be heads. The next two flips must not be TT, and the last flip can be anything.\r\n\r\n(1/4)(3/4)\r\n\r\n3/16", "Solution_15": "Yes that's a much better way.", "Solution_16": "Wanna hear some thing really bad?\r\nI got EVERYSINGLETARGETPROBLEMWRONG (exept for 8)\r\nStupid runny nose...ran out of tissues after the sprint.\r\n\r\nWell, at least this thread got some of my questions done.\r\n\r\nHow do you do # 29 and 30 on the sprint?", "Solution_17": "#29[hide]\nCall BE and FC both x. Lable the lengths of AB as 28 and CD as 58 as given.\n\nThen using similar triangles make the proportion $\\frac{28}{x} = \\frac{x}{58}$ Cross multiply to get $x^2 = (28)(58) = 1624$ since $x^2$ is the area of the square and what you are looking for, the answer is 1624.[/hide]", "Solution_18": "#30\r\n\r\n[hide]\nCall the radius of the small circle 2x and the radius of the larger cirlce 4x.\n\nThe combined areas of the two semicircles is then \n\n$\\frac{16x^2 \\pi}{2} + \\frac{4x^2 \\pi }{2} = 10x^2 \\pi$\n\nSince the sector ACP needs to be half that, its area will be $5x^2 \\pi$\n\nTo find the measure of the angle ACP, call it y, use the equation \n\n$\\frac{y}{360} (16x^2)\\pi = 5x^2 \\pi$ which simplifies to \n$\\frac{16x^2y\\pi}{360} = 5x^2\\pi$ \nSimplify the $x^2$ and $\\pi$ out, and reduce to get\n$\\frac{2y}{45} = 5$\nMultiply by 45 \n$2y = 225$\n$y=112.5$[/hide]", "Solution_19": "[quote=\"frost13\"]#30\nand I apologize ahead of time for having to use pi instead of the symbol, everytime I clicked on the Emoticon I got 10 lines of nonsense in my post???\n[/quote]\r\n\r\nTry $\\pi$. \"\\pi.\"", "Solution_20": "#30\r\nJust know that the little semi circle is 1/4 of the bigger semicircle so that makes a total of 5/4, and all you need is 5/8. 5/8 of 180 is 112.5 or something.", "Solution_21": "[quote=\"chess64\"][quote=\"frost13\"]#30\nand I apologize ahead of time for having to use pi instead of the symbol, everytime I clicked on the Emoticon I got 10 lines of nonsense in my post???\n[/quote]\n\nTry $\\pi$. \"\\pi.\"[/quote]\r\n\r\nThanks Chess64, I edited it", "Solution_22": "[quote=\"dwx314\"]Wanna hear some thing really bad?\nI got EVERYSINGLETARGETPROBLEMWRONG (exept for 8)\nStupid runny nose...ran out of tissues after the sprint.\n\nWell, at least this thread got some of my questions done.\n\nHow do you do # 29 and 30 on the sprint?[/quote]\r\n\r\nWas mathlover happy :? ?\r\n\r\n\r\nI didn't get number 29. In fact, I guessed 2005 because it looked like around 2005 :P .\r\n\r\n30 was easy, though. I set the area of the smaller semicircle as 1, so the area of the larger one must be 4. 2.5/4 = 5/8. 5/8 = x/180.", "Solution_23": "[quote=\"Treething\"][quote=\"dwx314\"]Wanna hear some thing really bad?\nI got EVERYSINGLETARGETPROBLEMWRONG (exept for 8)\nStupid runny nose...ran out of tissues after the sprint.\n\nWell, at least this thread got some of my questions done.\n\nHow do you do # 29 and 30 on the sprint?[/quote]\n\nWas mathlover happy :? ?\n[/quote]\r\ntreething, since I don't expect mathlover will be visiting here anytime soon :\r\n\r\nMathlover was happy for her team who took 1st, 4th, 6th, 7th, and 10th at state, but she wasn't too fond of the target round herself :(", "Solution_24": "[quote=\"frost13\"]Mathlover was happy for [b]her team who took 1st, 4th, 6th, 7th, and 10th at state[/b], but she wasn't too fond of the target round herself :([/quote]\r\n\r\nUmm... There's only 4 people in a team.", "Solution_25": "I didnt like number 24 the one with all the equations of the lines that took way too long!\r\nAnyone know a quick and painless way of doing that?", "Solution_26": "find points of intersection, use distance formula, and then heron's formula? hmm... that'd be a lot longer, wouldn't it....", "Solution_27": "[quote=\"NoSoupForYou\"][quote=\"frost13\"]Mathlover was happy for [b]her team who took 1st, 4th, 6th, 7th, and 10th at state[/b], but she wasn't too fond of the target round herself :([/quote]\n\nUmm... There's only 4 people in a team.[/quote]\r\n\r\nOne of the \"members\" must have been an individual not on the team.", "Solution_28": "Yes, one of those 5, #7 was an individual who was 10th at chapter. I consider my \"team\" to be the 12 kids I take to the competition.", "Solution_29": "# 24\r\n\r\n[hide]Find the points of intersection. \nFor any polygon on a coordinate plane, there is a way to find the area that is nice and simple. Triangle:\npoint (a,b)\npoint (c,d)\npoint (e,f)\n\nList them in order and then put the first point again at the end.\n(a,b)\n(c,d)\n(e,f)\n(a,b)\n\nThen you do a shoe string like multiplication and addition.\nad + cf + eb and the other is\nbc +de + fa\nThen take the absolute value of the difference of these quantities and divide by 2. \n\nThis works on any sided polygon, just list the points in order[/hide]", "Solution_30": "Nice solution I forgot about shoestring! :D", "Solution_31": "[quote=\"nat mc\"]# 24\n\n[hide]Find the points of intersection. \nFor any polygon on a coordinate plane, there is a way to find the area that is nice and simple. Triangle:\npoint (a,b)\npoint (c,d)\npoint (e,f)\n\nList them in order and then put the first point again at the end.\n(a,b)\n(c,d)\n(e,f)\n(a,b)\n\nThen you do a shoe string like multiplication and addition.\nad + cf + eb and the other is\nbc +de + fa\nThen take the absolute value of the difference of these quantities and divide by 2. \n\nThis works on any sided polygon, just list the points in order[/hide][/quote]\r\n\r\nThis was actually #25.", "Solution_32": "i hated number 6 i miss interpreted the problem horrible grammer", "Solution_33": "Which number 6, because neither one seems to have bad grammar to me?" } { "Tag": [], "Problem": "A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length $2000$. What is the length of each side of the octagon?\r\n\r\n\r\nA. $1/3(2000)$\r\n\r\nB. $2000 (\\sqrt{2}-1)$\r\n\r\nC. $2000(2-\\sqrt{2})$\r\n\r\nD. $1000$\r\n\r\nE. $1000\\sqrt{2}$", "Solution_1": "Dude, the correct answer is not up there.\r\n\r\n[hide]$x=\\frac{(2000-x)\\sqrt{2}}{2}\\Longrightarrow \\boxed{x=2000(\\sqrt{2}-1)}$.[/hide]", "Solution_2": "I think he just mistyped. I got the same thing, although I went through a few more steps... Is that a formula you have memorized, or am I missing something obvious?", "Solution_3": "Let $x$ be a side of the octagon. We can also write it as $\\frac{(2000-x)}{2}\\cdot\\sqrt{2}$ since we were told that it is a regular octagon and a side of the isosceles triangle is just $\\frac{(2000-x)}{2}$. :wink:", "Solution_4": "Okay, now that I look at it, that's exactly what I did. I just used two variables and didn't substitute soon enough.", "Solution_5": "[hide]$x+x\\sqrt{2}=2000\\implies x=\\frac{2000}{1+\\sqrt{2}}\\implies x=2000(\\sqrt{2}-1), \\boxed{B}$.[/hide]\r\n\r\n1000th post :D !", "Solution_6": "ok. Here's my solution.\r\n\r\n\r\n[hide=\"solution\"]$x\\sqrt{2}+x=2000$ Since x/rt2 is 1/2 of the complement of the side.\n\nSo $x=2000(\\sqrt{2}-1)$[/hide]\r\n\r\nedit: same as pianofote's." } { "Tag": [ "abstract algebra", "algebra", "function", "domain", "real analysis", "real analysis theorems" ], "Problem": "Let $ \\equiv$ and $ \\sim$ be two equivalence relations on the same set $ A$. If $ a\\equiv a_{1}$ implies that $ a\\sim a_{1}$, prove that each $ \\sim$ equivalence class is partitioned by the $ \\equiv$ equivalence classes it contains.\r\n\r\nHow does one handle partitions of equivalence classes? I know the theorem for equivalence classes under $ \\equiv$ partitioning $ A$, but what to do for equivalence classes of other equivalence classes...\r\n\r\nAnother question: let $ \\alpha: A\\rightarrow B$ be a map, $ \\equiv$ denotes the kernel equivalence ($ a\\equiv a_{1}$ iff $ \\alpha(a)\\equal{}\\alpha(a_{1})$ for $ a,a_{1}\\in A$), and let $ \\varphi: A\\rightarrow A_{\\equiv}$ denote the mapping $ \\varphi(a)\\equal{}[a]$. Finally, let $ \\sigma([a])\\equal{}\\alpha(a)$. If $ \\alpha(A)$ is a finite set, prove that $ \\left|A_{\\equiv}\\right|\\equal{}\\left|\\alpha(A)\\right|$. I already proved that $ \\sigma$ is one-to-one, and that $ \\sigma$ is onto if $ \\alpha$ is. This is a bijection then, because $ \\alpha: A\\rightarrow\\alpha(A)$ is obviously onto, so $ \\sigma$ is onto, correct?", "Solution_1": "For your first question, what you have says that $ \\equiv$ is finer than $ \\sim$ (or $ \\sim$ is coarser than $ \\equiv$). The same translates to the partitions by equivalence classes, just follow the definition.\r\n\r\nFor your second, you are again right. What you have as equivalence classes are the fibres $ \\alpha^{\\minus{}1}(b)$ of the map $ \\alpha$, and of course $ \\sigma$ is only onto when $ \\alpha$ is, when considering co-domain $ B$. But clearly what matters is the restriction to $ \\alpha(A)$, the range of $ \\alpha$, so $ \\sigma$ is then a bijection between $ A_{\\equiv}$ and $ \\alpha(A)$.", "Solution_2": "Maybe it's that the wording is confusing me, but could you elaborate on the first paragraph? Thanks.", "Solution_3": "Assume $ a \\equiv b$ always implies $ a \\sim b$. Take now an element $ y$ in the class $ [x]_{\\sim}$. Clearly $ [y]_{\\equiv} \\subseteq [y]_{\\sim} \\equal{} [x]_{\\sim}$, since $ z \\equiv y$ implies $ z \\sim y$. If there exists $ y' \\in [y]_{\\sim} \\setminus [y]_{\\equiv}$ do the same, and so on. Another way to see it, is to check that $ \\equiv$ is an equivalence relation on the set $ [x]_{\\sim}$, and partition it accordingly.", "Solution_4": "This is probably getting annoying, but could you clarify the statement \"do the same, and so on\"? Based on what is there, I am thinking that the partition is all of the elements in $ \\equiv$ equivalence classes and those just in $ \\sim$, but I'm not sure.\r\n\r\nThanks.", "Solution_5": "Look at one of the $ \\sim$ equivalence classes, be it $ [x]_{\\sim}$. \r\n\r\n1. Take an element $ y \\in [x]_{\\sim}$. Clearly $ [y]_{\\equiv} \\subseteq [y]_{\\sim} \\equal{} [x]_{\\sim}$, since $ z \\equiv y$ implies $ z \\sim y$. If $ [y]_{\\sim} \\equal{} [y]_{\\equiv}$, we are done, having partitioned $ [x]_{\\sim}$ in $ \\equiv$ equivalence classes (albeit just one in this case!).\r\n\r\n2. If not, there exists $ y' \\in [y]_{\\sim} \\setminus [y]_{\\equiv}$. Do the same as step one for $ y'$. If $ [y]_{\\sim} \\equal{} [y]_{\\equiv} \\cup [y']_{\\equiv}$, we are done, having partitioned $ [x]_{\\sim}$ in $ \\equiv$ equivalence classes (two in this case!).\r\n\r\n3. If not, there exists $ y'' \\in [y]_{\\sim} \\setminus ([y]_{\\equiv} \\cup [y']_{\\equiv})$. Do the same as step one for $ y''$. And so on.\r\n\r\nAt some point, there are no more elements left, so we succeed partitioning $ [x]_{\\sim}$ in $ \\equiv$ equivalence classes. Then we do the same for the other $ \\sim$ equivalence classes.", "Solution_6": "Thank you!", "Solution_7": "[quote=\"mavropnevma\"]Assume $ a \\equiv b$ always implies $ a \\sim b$. Take now an element $ y$ in the class $ [x]_{\\sim}$. Clearly $ [y]_{\\equiv} \\subseteq [y]_{\\sim} \\equal{} [x]_{\\sim}$, since $ z \\equiv y$ implies $ z \\sim y$. If there exists $ y' \\in [y]_{\\sim} \\setminus [y]_{\\equiv}$ do the same, and so on.[/quote]\nWe must pay attention here. This does not work if a $ \\sim$-equivalence class decomposes into [u]uncountably[/u] many $ \\equiv$-equivalence classes. E.g., on $ \\mathbb R$ take $ x\\equiv y$ iff $ x\\equal{}y$ and $ [x]_\\sim\\equal{}\\mathbb R$ for every $ x\\in \\mathbb R$. Then this constructive method can never exhaust $ \\mathbb R$.\n[quote]Another way to see it, is to check that $ \\equiv$ is an equivalence relation on the set $ [x]_{\\sim}$, and partition it accordingly.[/quote]I believe only an abstract argument like that can be given as a valid proof, since the cardinality of the searched decomposition can exceed every constructive, hence countable, approach." } { "Tag": [ "search", "calculus", "complex analysis", "complex numbers" ], "Problem": "Hey\r\n\r\nI posted this in the Pre-Olympiad Forum but the answer I received didn't make sense and I was hoping for a clarification. Sorry if I am being pushy, if necessary, please delete one of the double posts....\r\n\r\n\r\nAnyways, my question is to prove\r\n\r\nProve the following formula\r\n\r\n\r\n$\\sum_{k=0}^{n}{\\binom{n}{k}}^{2}= \\binom{2n}{n}$\r\n\r\n\r\nThe answer that was shown to me used Lemma (Vandermonde's Identity), but I have never seen that nor do I understand how it works. Is there a simpler method?\r\n\r\n\r\nAny help is appreciated, thanks.", "Solution_1": "Both this problem and the more general Vandermonde result (of which this is a special case) are on this forum dozens of times. I suggest you do a search of the forum. There are doubtless many explanations available elsewhere on the internet, as well.", "Solution_2": "Good idea. I guess sometimes my brain just doesn;t work.", "Solution_3": "In fact, there are two additional proofs of it currently near the top of one of the Olympiad combinatorics forums, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121075\r\n\r\n\r\nAs an aside, \"Complex Analysis\" refers to analysis (formal calculus) over the complex numbers, so this was a poor choice of forum for your question. :wink:" } { "Tag": [ "inequalities" ], "Problem": "\\[ \\begin{array}{l} k \\ge 1,prove: \\\\\r\n\\frac {{a_1^k \\plus{} a_2^k \\plus{} ...a_n^k }}{n} \\ge (\\frac {{a_1 \\plus{} a_2 \\plus{} ...a_n }}{n})^k \\\\\r\nk < 1,prove: \\\\\r\n\\sqrt [m]{{\\frac {{a_1 \\plus{} a_2 \\plus{} .... \\plus{} a_n }}{n}}} \\ge \\frac {{\\sqrt [m]{{a_1 }} \\plus{} \\sqrt [m]{{a_2 }}...... \\plus{} \\sqrt [m]{{a_n }}}}{n} \\\\\r\n\\end{array}\r\n\\]\r\nCan you tell me how to make this post bigger , it's too small .", "Solution_1": "Just use the power mean inequalities: http://en.wikipedia.org/wiki/Power_mean", "Solution_2": "I don't know how to use power mean inequalities in this problem . Can you show me in details ?", "Solution_3": "If you use \"Jensen's inequality\", it is easy for you to prove it.\r\nplease refer to http://en.wikipedia.org/wiki/Jensen%27s_inequality or http://mathworld.wolfram.com/JensensInequality.html." } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Find all real polynomials that\r\n\r\n$p(x+p(x))=p(x)+p(p(x))$\r\n\r\n\r\n :wink: \r\n\r\nFitim Dika", "Solution_1": "continuos here, please http://www.mathlinks.ro/Forum/viewtopic.php?p=632796#632796 :wink:" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Give an example of a function f:R->R s.t. f(x3)=f(x) for all x \\in R and which isn't continuous at any point.\r\n\r\nThis might be kind of easy, but it sure is nice :).", "Solution_1": "f(x) = 1 if x is algebraic, else f(x) = 0.", "Solution_2": "Right. I did exactly the same thing :).", "Solution_3": "sharpened question: find a function f: R->R that does not only satisfy\r\nthe original requirements, but is moreover surjective." } { "Tag": [ "inequalities", "trigonometry", "vector", "inequalities unsolved" ], "Problem": "As Vasc said, it is right to dedicate an own topic to this nice inequality \r\n\r\n[quote=\"Vasc\"]This is the Fan's Inequality:\n\nIf $x_1+...+x_n=0$, then $\\frac{x_1x_2+...+x_nx_1}{x_1^2+...+x_n^2} \\leq cos{\\frac{2\\pi}{n}}$[/quote]", "Solution_1": "Equality occurs for $x_k=A\\cos{\\frac{2k\\pi}{n}}+B\\sin{\\frac{2k\\pi}{n}}$, where A and B are arbitrary real numbers. That means for $x_k+x_{k+2}=2cos{\\frac{2\\pi}{n}}x_{k+1}$.", "Solution_2": "This isn't a proof \"per se\", since I'm using a strong result, but I thought the approach was nice enough to post :).\r\n\r\nFor every $i\\in\\overline{1,n}$, let $v_i\\subset\\mathbb R^n$ be the vector $(x_i,x_{i+1},\\ldots,x_{i+n-1})$ (where $x_{n+1}=x_1$). What we want to prove is that the angle $\\theta$ between $v_i$ and $v_{i+1}$ (which is the same for all $i$) is $\\ge\\frac{2\\pi}n$. \r\n\r\nNow consider the piecewise linear curve with vertices $0,v_1,v_1+v_2,\\ldots,v_1+\\ldots+v_{n-1},v_1+\\ldots+v_n=0$. We can approximate this curve arbitrarily well with smooth closed curves, which, by Fenchel's Theorem (the result I mentioned above), have total curvature $\\ge2\\pi$. The total curvature of these curves, however, tends to the sum of the external angles of our curve, which is $n\\theta$, so the conclusion is that $n\\theta\\ge2\\pi$, i.e. exactly what we wanted.", "Solution_3": "[quote=\"grobber\"]This isn't a proof \"per se\", since I'm using a strong result, but I thought the approach was nice enough to post :).\n\nFor every $i\\in\\overline{1,n}$, let $v_i\\subset\\mathbb R^n$ be the vector $(x_i,x_{i+1},\\ldots,x_{i+n-1})$ (where $x_{n+1}=x_1$). What we want to prove is that the angle $\\theta$ between $v_i$ and $v_{i+1}$ (which is the same for all $i$) is $\\ge\\frac{2\\pi}n$. \n\nNow consider the piecewise linear curve with vertices $0,v_1,v_1+v_2,\\ldots,v_1+\\ldots+v_{n-1},v_1+\\ldots+v_n=0$. We can approximate this curve arbitrarily well with smooth closed curves, which, by Fenchel's Theorem (the result I mentioned above), have total curvature $\\ge2\\pi$. The total curvature of these curves, however, tends to the sum of the external angles of our curve, which is $n\\theta$, so the conclusion is that $n\\theta\\ge2\\pi$, i.e. exactly what we wanted.[/quote]\r\nVery Very nice idea(I think it is a proof).Can you show me a proof for fenchel's theorem or a link?I can't find it on the internet.\r\nThanks" } { "Tag": [ "geometry", "area of a triangle", "Heron\\u0027s formula" ], "Problem": "Find the length of the altitude to the longest side of a triangle whose sides have lengths 13, 20, and 21.", "Solution_1": "By [url=http://mathworld.wolfram.com/HeronsFormula.html]Heron's Formula[/url], the area of the triangle is $ \\sqrt {27\\times{14}\\times{7}\\times{6} } = \\sqrt { {9}\\times{3}\\times{7}\\times{2}\\times{7}\\times{3}\\times{2} } = 3\\times{3}\\times{7}\\times{2} = 126$\r\n\r\nLet $ h$ be the altitude to the longest side (21) of the triangle, then we have $ \\frac {21h}{2} = 126\\implies{h = \\boxed{12}}$", "Solution_2": "This is a very common problem." } { "Tag": [], "Problem": "How many zeros are at the end of (31!)/(16!*8!*4!*2!*1!)", "Solution_1": "[quote=\"Art of Owna\"]How many zeros are at the end of (31!)/(16!*8!*4!*2!*1!)[/quote]\r\n\r\n[hide=\"solution\"]There are 6+1=7 multiples of 5 in 31!, 3 multiples of 5 in 16!, 1 multiple of 5 in 8! and none in the others. \n\n7-3-1=3 so $\\boxed{3}$ zeroes.[/hide]", "Solution_2": "Are you sure?\r\n\r\nFor example, using your way, ur inplying there is 1 zero at the end of (5!)/(4!)", "Solution_3": "Yeah. JLI is right.", "Solution_4": "No shes not... just think about it \r\nLets give an example.. 7!/(3!*4!), using her way, there is 1 multiple of 5 in 7!, and none in 3! or 4!, 1-0 is 1 so thats the correct answer... Nope.. type it in ur calculator..", "Solution_5": "Actually, since $10=2\\cdot 5$, you must also consider whether or not there is an ample number of two's to combine with the fives to make zeros. \r\n\r\n[hide]For speed, we can only consider even numbers and multiples of 5. At the end, we get $3$ factors of 5 but only $1$ factor of 2. \nedit: scratch that, no factors of 2. \nThus, there is $\\boxed{0}$ zeros. [/hide]", "Solution_6": "[hide=\"hmmmm\"]The number of zeroes in the numerator is 7. THe number in the denominator is 4. So there are 3. That's what I got. :| \n\nAtleast it was right in a previous MC question :huh: [/hide]", "Solution_7": "kyyuanmathcount i think ur reasoning is correct but I think if u calculate it it would be no zeros \r\n\r\n#H34N1, you cant just subtract zeros.. from the top and bottom.. whats 400/8- does that have 2-0=2 zeros?", "Solution_8": "Oh.\r\n\r\nBut don't the zeroes cancel out...", "Solution_9": "[quote]Oh. \n\nBut don't the zeroes cancel out...[/quote]\r\nNot necessarily..., when you cancel out the zeros, ur just really subtracting the number of 2's and 5s in the numerator with the number of 2's and 5s in the denominator, so that only works if there is equal amounts of 2s and 5s in both the demoninator and numerator\r\n\r\nFor example, 10^3/10^2, you can cancel out zeros because there are 3 2s and 5s in the numerator and 2 2s and 5s in the denominator\r\n\r\nbut like 400/8, you cannot do that", "Solution_10": "The reason $\\frac{7!}{3!4!}$ doesn't work is because $3!4!$ has as many zeroes as $7!$.\r\n\r\n$31!$ has $15+7+3+1 = 26$ zeroes, while $16!8!4!2!1!$ has $8+4+2+1+4+2+1+2+1+1 = 26$ zeroes. So the expression has $\\boxed{\\text{no}}$ zeroes!\r\n\r\nCan someone prove for me that an expression of the form $\\frac{n!}{a_{1}!a_{2}!a_{3}!...a_{k}!}$ is always an integer where $a_{1}+a_{2}+... a_{k}= n$? I think it's true but I'm not sure...", "Solution_11": "What are you talking about? 31! does not have 26 zeros, it hsa 26 factors of 2, infact 31! has 7 zeros, the denominator has 4 zeros BUT when you divide them it will have 0 zeros", "Solution_12": "because you don't have enough 2's\r\n\r\n(You need 2's and 5's to make 10) :D", "Solution_13": "[quote=\"Art of Owna\"]No shes not... just think about it \nLets give an example.. 7!/(3!*4!), using her way, there is 1 multiple of 5 in 7!, and none in 3! or 4!, 1-0 is 1 so thats the correct answer... Nope.. type it in ur calculator..[/quote]\r\nJli is a guy.\r\nI'm getting 0 for the answer", "Solution_14": "[quote=\"Art of Owna\"]kyyuanmathcount i think ur reasoning is correct but I think if u calculate it it would be no zeros \n\n#H34N1, you cant just subtract zeros.. from the top and bottom.. whats 400/8- does that have 2-0=2 zeros?[/quote]\r\n\r\nWhoops, my bad, i'll edit that.", "Solution_15": "[quote=\"usaha\"]What are you talking about? 31! does not have 26 zeros, it hsa 26 factors of 2, infact 31! has 7 zeros, the denominator has 4 zeros BUT when you divide them it will have 0 zeros[/quote]\r\n\r\nWhoops, my bad... I meant multiples of 2... and you should have been able to figure that out by yourself.", "Solution_16": "[quote=\"Ubemaya\"]Can someone prove for me that an expression of the form $\\frac{n!}{a_{1}!a_{2}!a_{3}!...a_{k}!}$ is always an integer where $a_{1}+a_{2}+... a_{k}= n$? I think it's true but I'm not sure...[/quote]\r\n\r\n[hide=\"hmm\"]Wouldn't that just be ${n\\choose a_{1},a_{2},\\ldots,a_{k}}={n\\choose a_{1}}{n-a_{1}\\choose a_{2}}\\cdots{a_{k}\\choose a_{k}}$, which is always an integer?[/hide]", "Solution_17": "[quote=\"13375P34K43V312\"][quote=\"Ubemaya\"]Can someone prove for me that an expression of the form $\\frac{n!}{a_{1}!a_{2}!a_{3}!...a_{k}!}$ is always an integer where $a_{1}+a_{2}+... a_{k}= n$? I think it's true but I'm not sure...[/quote]\n\n[hide=\"hmm\"]Wouldn't that just be ${n\\choose a_{1},a_{2},\\ldots,a_{k}}={n\\choose a_{1}}{n-a_{1}\\choose a_{2}}\\cdots{a_{k}\\choose a_{k}}$, which is always an integer?[/hide][/quote]\r\nWell for example, can you prove that if $n \\geq k$ then $\\frac{n!}{k!(n-k)!}$ is an integer without combinatorics?", "Solution_18": "there are 3 multiples of 5 in the end. i hope you guys know that.\r\n\r\nnow count the number of powers of 2's. the top has $15+7+3+1=26$. The bottom has $8+4+2+1+4+2+1+2+1+1=26$. So ther are no multiples of 2. i feel really stupid now. :wallbash: \r\n\r\nthe answer is 0\r\n\r\noh, and by the way, im a guy. :D (j=johnny li=last name)" } { "Tag": [], "Problem": "To all who may be confused: Compiz and Beryl are window managers which run in the Gnome/KDE Desktop Environments. The Gnome window manager is metacity and the KDE window manager is kwin.\r\n\r\nI'm curious as to who uses which window manager. Ubuntu comes with metacity (Gnome) by default, so I'm expecting that to be the clear majority. Of course, KDE 4 has completely blown it out of the water. Personally, I like the minimalism of fluxbox, but if you go as far as evilwm, it's a bit too little (although the source for evilwm is nice to read to see how you would make a window manager at all).", "Solution_1": "I'd vote for Compiz Fusion, but I have an ATI card, which kinda sucks..." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "$ \\sum_{n\\equal{}1}^{\\infty}\\frac{n^2}{e^n}\\equal{}?$", "Solution_1": "at first find $ \\sum x^{n}n^{2}$ for $ |x|<1$ you will get some differential equation :)", "Solution_2": "hello, after the hint above we get\r\n$ \\sum_{n\\equal{}1}^{\\infty}\\frac{n^2}{e^n}\\equal{}\\frac{e(e\\plus{}1)}{(e\\minus{}1)^3}$.\r\nSonnhard." } { "Tag": [ "algebra", "polynomial", "Vieta" ], "Problem": "Source: ACoPS\r\n(Hungary, 1899) Let $r$ and $s$ be the roots of \r\n$x^{2}-(a+d)x+(ad-bc)=0$.\r\nProve that $r^{3}$ and $s^{3}$ are the roots of \r\n$y^{2}-(a^{3}+d^{3}+3abc+3bcd)y+(ad-bc)^{3}=0$.", "Solution_1": "From ?vietta's? sums $r+s=a+d$ and $rs=ad-bc$. Thus $r^{3}+s^{3}=(r+s)^{3}-3rs(r+s)=(a+d)^{3}-3(ad-bc)(a+d)=a^{3}+3a^{2}d+3ad^{2}+d^{3}-3a^{2}d+3abc-3ad^{2}+3bcd=a^{3}+d^{3}+3abc+3bcd$. Also, $(rs)^{3}=(ad-bc)^{3}$. Therefore the roots of $y^{2}-(a^{3}+d^{3}+3abc+3bcd)y+(ad-bc)^{3}$ are $r^{3}$ and $s^{3}$.", "Solution_2": "[quote=\"drunner2007\"]From ?vietta's? sums $r+s=a+d$ and $rs=ad-bc$. Thus $r^{3}+s^{3}=(r+s)^{3}-3rs(r+s)=(a+d)^{3}-3(ad-bc)(a+d)=a^{3}+3a^{2}d+3ad^{2}+d^{3}-3a^{2}d+3abc-3ad^{2}+3bcd=a^{3}+d^{3}+3abc+3bcd$. Also, $(rs)^{3}=(ad-bc)^{3}$. Therefore the roots of $y^{2}-(a^{3}+d^{3}+3abc+3bcd)y+(ad-bc)^{3}$ are $r^{3}$ and $s^{3}$.[/quote]\r\nYes, that's how i did..this was pretty easy, since all u have to do is just applying Vieta's theorem.", "Solution_3": "yeah its a little too easy for acops" } { "Tag": [ "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "$ x\\in R$\r\n\r\n$ u_{n}=\\frac{(-1)^{n}}{\\sqrt{n}+e^{inx}ln(n)}$ \r\n\r\nstudy the convergence of $ \\sum u_{n}$", "Solution_1": "[quote=\"Moubinool\"]$ x\\in R$\n\n$ u_{n}=\\frac{(-1)^{n}}{\\sqrt{n}+e^{inx}ln(n)}$ \n\nstudy the convergence of $ \\sum u_{n}$[/quote]\r\n\r\n$ |\\frac{(-1)^{n}}{\\sqrt{n}+e^{inx}\\ln n}| \\geq \\frac{1}{|\\sqrt{n}|+|e^{inx}\\ln n|}\\geq \\frac{1}{\\sqrt{n}+\\ln n}\\geq \\frac{1}{2\\sqrt{n}}$\r\n\r\nNow, $ \\sum_{n=1}^{\\infty}\\frac{1}{2\\sqrt{n}}=+\\infty$." } { "Tag": [], "Problem": "Ten points are marked on a circle. How many distinct convex polygons of three or more sides can be drawn using some (or all) of the ten points as vertices?", "Solution_1": "[hide]All polygons are distinct unless the have the same vertices. Thus, we simply need to count the number of ways to choose distinct vertices, for a total of $\\binom{10}{3}+\\binom{10}{4}+\\cdots+\\binom{10}{10}$ polygons. This expression can be simplified as $2^{10}-\\binom{10}{0}-\\binom{10}{1}-\\binom{10}{2}=1024 - 1 - 10 - 45=\\boxed{968}$. [/hide]", "Solution_2": "[hide=\"solution\"]$\\sum_{k=3}^{10}\\binom{10}{k}=x$\n$\\sum_{k=1}^{10}\\binom{10}{k}=\\binom{10}{0}+\\binom{10}{1}+\\binom{10}{2}+x=2^{10}$\n$x=1024-10-45-1=\\boxed{968}$[/hide]", "Solution_3": "Altheman: 2^{10} ;)", "Solution_4": "[quote=\"Hokkage\"]Altheman: 2^{10} ;)[/quote]\r\n\r\nedited...i did not bother to preview that message", "Solution_5": "first of all 2^10, is more than 999. and 2^10 includes polygons with 0, 1, and 2 sides, and u can have polygons with 0,1, and 2 sides", "Solution_6": "He subtracted off the aforementioned polygons...\r\nLook at altheman's hidden post.", "Solution_7": "What if the ten points are evenly spaced, and polygons are considered the same if they are congruent instead of sharing the same vertices?\nHow would you approach?", "Solution_8": "Combo AIME grind: Use complementary counting. The total amount of polygons is $2^{10}$, and the undesirables give $\\binom{10}{0}+\\binom{10}{1}+\\binom{10}{2}=56$ ways. The answer is $2^{10}-56=\\boxed{968}$.", "Solution_9": "[quote=Hokkage]He subtracted off the aforementioned polygons...\nLook at altheman's hidden post.[/quote]\n\nBecause nC0+nC1+...+nCn =2^n", "Solution_10": "$1024-\\binom{10}{0}-\\binom{10}{1}-\\binom{10}{2}=\\boxed{968}.$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "k,l,m,n are Natural numbers, k=k+2", "Solution_1": "It's well-known that there exist natural numbers $ a,b,c,d$ such that $ k \\equal{} ab, n \\equal{} cd, l \\equal{} ac, m \\equal{} bd$.\r\nAs $ k < l < m < n$, $ ab < ac < bd < cd$, i.e. $ a < d, b < c$, so $ d\\geq a \\plus{} 1, c \\geq b \\plus{} 1$.\r\nSo $ cd \\geq (a \\plus{} 1)(b \\plus{} 1) \\equal{} ab \\plus{} a \\plus{} b \\plus{} 1$, i.e. $ (cd \\minus{} ab)^2 \\geq (a \\plus{} b \\plus{} 1)^2$.\r\n\r\nNow back to the original inequality, $ (\\frac {n}{2} \\minus{} \\frac {k}{2})^2 \\geq k \\plus{} 2 \\Leftrightarrow (cd \\minus{} ab)^2 \\geq 4(ab \\plus{} 2)$ \r\nDivide two cases, \r\nCase 1) $ a \\neq b$, as $ a \\geq 1, a \\plus{} b \\geq 1 \\plus{} 2 \\equal{} 3$\r\nAs $ (cd \\minus{} ab)^2 \\geq (a \\plus{} b \\plus{} 1)^2$, it's left to prove that $ (a \\plus{} b \\plus{} 1)^2 \\geq 4(ab \\plus{} 2)$\r\nSimiplify to get $ (a \\minus{} b)^2 \\plus{} 2(a \\plus{} b \\minus{} 3) \\geq 0$, this is true as $ (a \\minus{} b)^2 \\geq 0$ and $ a \\plus{} b \\geq 3$.\r\n\r\nCase 2) $ a \\equal{} b$\r\nAs $ ac < bd$, so $ c \\neq d$, WLOG let $ c < d$.\r\nSo $ d > c \\geq a \\plus{} 1$, i.e. $ d \\geq a \\plus{} 2$\r\nSo $ cd \\geq (a \\plus{} 1)(a \\plus{} 2) \\equal{} a^2 \\plus{} 3a \\plus{} 2$, then $ (cd \\minus{} a^2)^2 \\geq (3a \\plus{} 2)^2$.\r\n\r\nSo left to prove $ (3a \\plus{} 2)^2 \\geq 4(a^2 \\plus{} 2)$\r\n$ (3a \\plus{} 2)^2 \\minus{} 4(a^2 \\plus{} 2) \\equal{} 5a^2 \\plus{} 12a \\minus{} 4 \\equal{} (a \\minus{} 1)(5a \\plus{} 17) \\plus{} 13 > 0$ \r\nSo we are done.\r\n\r\nSeems the equality can't hold..." } { "Tag": [ "inequalities" ], "Problem": "How many integer solutions are there for: $-6 < 3x-2 < 7$.", "Solution_1": "[hide]$-4<3x<9$\n\n$-\\frac{4}{3}< x < 3$\n\n$-1\\leq x\\leq 2$\n\n$-1,0,1,2$\n\n$4$ solutions[/hide]", "Solution_2": "[quote=\"vishalarul\"][hide]$-4<3x<9$\n\n$-\\frac{4}{3}< x < 3$\n\n$-1\\leq x\\leq 2$\n\n$-1,0,1,2$\n\n$4$ solutions[/hide][/quote]\r\n\r\ncould you explain how you got: $-1\\leq x\\leq 2$ ?", "Solution_3": "well the easiest way would be from\r\n\r\n$\\frac{4}{3}< x <3 =$\r\n$\\frac{4}{3}< x$ and $x<3$. \r\n\r\nGraph or imagine the number line and you can find your solutions.\r\n\r\nedit: actually to recieve full credit for that problem the answer should be in the form\r\n\r\n$\\{x|-1\\leq x \\leq 2 \\}$", "Solution_4": "[quote=\"xxxxx\"][quote=\"vishalarul\"][hide]$-4<3x<9$\n\n$-\\frac{4}{3}< x < 3$\n\n$-1\\leq x\\leq 2$\n\n$-1,0,1,2$\n\n$4$ solutions[/hide][/quote]\n\ncould you explain how you got: $-1\\leq x\\leq 2$ ?[/quote]\n[hide]Well, you need the integer solutions, so the last two numbers at the end do not count... So we can ignore the $3$, and since $-\\frac{4}{3}$ is a fraction anyway, we can get the bounds $-1\\leq x\\leq 2$[/hide]", "Solution_5": "[quote=\"vishalarul\"][quote=\"xxxxx\"][quote=\"vishalarul\"][hide]$-4<3x<9$\n\n$-\\frac{4}{3}< x < 3$\n\n$-1\\leq x\\leq 2$\n\n$-1,0,1,2$\n\n$4$ solutions[/hide][/quote]\n\ncould you explain how you got: $-1\\leq x\\leq 2$ ?[/quote]\n[hide]Well, you need the integer solutions, so the last two numbers at the end do not count... So we can ignore the $3$, and since $-\\frac{4}{3}$ is a fraction anyway, we can get the bounds $-1\\leq x\\leq 2$[/hide][/quote]\r\n\r\ni see." } { "Tag": [ "analytic geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "You have a chessboard whose dimensions are 15 x 15 (225 squares), and a chess piece that can move a distance of 8 or 9 squares, horizontally or vertically. You place the piece anywhere on the board. How many moves can you make without visiting the same square twice?", "Solution_1": "Not hard. Labell the squares of the chessboard with ordered pairs $(x,y)$, $1\\leq x,y\\leq 15$\r\nFirst of all, a path that contains at least one cell of the type $(8,y)$ or $(x,8)$ must contain at most 15 squares, since there is a direction along which you cannot move from or to it. On the other hand you cannot reach a cell of one of those two tipes from a cell not belonging to it, so if the longest non self-intersecting path consistent of allowed moves and containing cells $(x,y)$ with $x,y\\neq8$ has length greater than 15, then that will be maximal. There are $14^{2}=296$ such squares; we will show we can visit all of them. Consider the sequence $a_{0}=9,\\;\\;a_{1}=9-8=1,\\;\\;a_{2}=1+9=10,\\;\\;a_{3}=10-8=2,\\;\\;a_{4}=2+9=11 \\;\\ldots\\; a_{13}=15,\\; \\; a_{14}=15-8=7$; all moves of the type $(a_{x},a_{y})\\mapsto(a_{x\\pm1},a_{y})$ or $(a_{x},a_{y})\\mapsto(a_{x},a_{y\\pm1})$, with $1\\leq x,y\\leq 14$ and such that for the new coordinates $(a_{x'},a_{y'})$, $x'$ and $y'$ continue to belong to that interval are allowed moves inside our $15\\times 15$ chessboard. So just take a nonintersecting path $(x_{i},y_{i})$, $1\\leq i\\leq 196$ inside a $14\\times 14$ chessboard in which all consecutive cells in the path are adyacent, and which goes through all of them (for example, an s-like one) and the path $(a_{x_{i}},a_{y_{i}})$, $1\\leq i\\leq 196$ inside the initial $15\\times15$ chessboard will achieve maximality within the restrictions initially imposed. Since it consists of $196$ cells, the number of steps will be $195$.", "Solution_2": "Very nice, your solution is absolutely correct :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For every positive integer $n$ prove the inequality for all $n \\geq 1$\r\n\r\n$\\sqrt{1+\\sqrt{3+\\sqrt{5+\\sqrt{...+\\sqrt{2n-1}}}}} < 2$", "Solution_1": "This inequality holds iff \\[ \\sqrt{3+\\sqrt{5+\\sqrt{...+\\sqrt{2n-1}}}} < 3 \\]\r\nAnd again, this inequality holds iff \\[ \\sqrt{5+\\sqrt{...+\\sqrt{2n-1}}} < 6 \\]\r\nAnd this one holds iff \\[ \\sqrt{7+\\sqrt{...+\\sqrt{2n-1}}} < 31 \\]\r\nAnd this one holds iff \\[ \\sqrt{9+\\sqrt{...+\\sqrt{2n-1}}} < 954 \\]\r\n...\r\n\r\nIn the end, we will have to prove that $\\sqrt{2n-1} < (((((2^2-1)^2-3)^2-5)^2-7)^2-...)^2- 2n-3$\r\n\r\nAnd that's where I'm stuck :(", "Solution_2": "I used induction. There is another way ?", "Solution_3": "use $\\sqrt{2n-1}<2\\sqrt{2n-3}+1$\r\nso we have \r\n$\\sqrt{1+\\sqrt{3+\\sqrt{5+\\sqrt{...+\\sqrt{2n-1}}}}} <\\sqrt{1+\\sqrt{3+\\sqrt{5+\\sqrt{...+\\sqrt{2n-3}+1}}}}$\r\nthen use $\\sqrt{2n-3}<2\\sqrt{2n-5},n \\ge 3$\r\nso $\\sqrt{1+\\sqrt{3+\\sqrt{5+\\sqrt{...+\\sqrt{2n-3}+1}}}}<\\sqrt{1+\\sqrt{3+\\sqrt{5+\\sqrt{...+\\sqrt{2n-5}+1}}}}<\\cdots<\\sqrt{1+\\sqrt{3+\\sqrt{5}+1}}<\\sqrt{1+\\sqrt{3}+1}<1+1=2$" } { "Tag": [ "probability" ], "Problem": "All of the students in a history class are juniors or seniors. There are twice as many girls as boys in the class, and there are 3 times as many senior girls as junior girls. If a student is selected at random, what is the probability that the student is a junior girl?", "Solution_1": "[hide]Let x be the total number of boys in the class. There are twice as many girls as boys, so the number of girls is 2x. Now the total number of students is 3x. Since there are 3 times as many senior girls as junior girls, the number of sen. girls is (3/4)(2x)=3x/2 and the number of junior girls is (1/4)(2x)=x/2. The probablity of choosing a junior girl is x/2 over 3x=[b]1/6[/b][/hide]" } { "Tag": [], "Problem": "From 6 am to noon the temperature rose $8^\\circ$ C. From noon to 3 pm, it rose another $3^\\circ$ C, but then dropped $5^\\circ$ C to $-3^\\circ$ C at 6 pm. How many degrees Celsius was the temperature at 6 am?", "Solution_1": "[hide]You have temperature x, + 8 + 3 -5 = -3. The answer is -9. If you work backwards, you would have -3 + 5 - 3 -8 = x. \n\n\nTherefore, the answer is -9. [/hide]", "Solution_2": "[hide]\nIn total you have:\n$x+8+3-5=-3\\\\\nx=-3-8-3+5\\\\\nx=-9$[/hide]", "Solution_3": "[hide]-9[/hide]", "Solution_4": "[hide]let x= temp at 6 am\n\nx+8+3-5=-3\nx+11-5 =-3\nx+6 =-3\nx =-3-6\nx = -9\n\ntherefore, the temp at 6 am is -9 degrees.[/hide]\r\n\r\n[size=84][color=green]Yes, I agree. Please hide your answers.[/color][/size]", "Solution_5": "[hide=\"answer\"]x+8+3-5=-3\nx+6=-3\nx=-9[/hide]\r\n\r\nby the way jsc hide your answer next time ;)", "Solution_6": "[hide]$x+8+3-5=-3$\n$x+6=-3$\n$x=-9$[/hide]", "Solution_7": "[hide]-9[/hide]" } { "Tag": [ "inequalities", "calculus", "quadratics", "function", "number theory", "inequalities proposed" ], "Problem": "$a,b,c\\ge0$. Prove \r\n$2(1+abc)+\\sqrt{2(1+a^2)(1+b^2)(1+c^2)}\\ge (1+a)(1+b)(1+c)$. :)\r\nWanted: solutions with as little calculus as possible :D", "Solution_1": "[quote=\"spanferkel\"]$a,b,c\\ge0$. Prove \n$2(1+abc)+\\sqrt{2(1+a^2)(1+b^2)(1+c^2)}\\ge (1+a)(1+b)(1+c)$. :)\nWanted: solutions with as little calculus as possible :D[/quote]\r\n\r\nAs you wish:\r\n $\\sqrt{2(1+a^2)(1+b^2)(1+c^2)} \\geq (1+a)(1+b)(1+c)-2(1+abc)$\r\n<---> $\\sqrt{2(1+a^2)(1+b^2)(1+c^2)} \\geq a+b+c-1+ab+bc+ca-abc$\r\n<---> $2(1+a^2)(1+b^2)(1+c^2) \\geq (a+b+c-1+ab+bc+ca-abc)^2$\r\n<---> $(-1-a-b-c+ab+bc+ca+abc)^2 \\geq 0$ :D", "Solution_2": "$2(1+abc)+\\sqrt{2(1+a^2)(1+b^2)(1+c^2)}\\geq (1+a)(1+b)(1+c) \\leftrightarrow 2(1+a^2)(1+b^2)(1+c^2)\\geq [a+b+c+ab+bc+ca-1-abc]^2 \\leftrightarrow (1+a^2)(1+b^2)(1+c^2)\\geq 2(a+b+c)(ab+bc+ca)+2abc-2(a+b+c)-2abc(ab+bc+ca) \\leftrightarrow (ab+bc+ca-a-b-c)^2+(abc-1)^2\\geq 2(abc-1)(a+b+c-ab-bc-ca)$\r\nwhich is true\r\n\r\nso,does this solution satisfy you? :D", "Solution_3": "This is neat. A nice identity:\r\n\r\n$\\boxed{(-1+ab+ bc+ca)^2+(a+b+c-abc)^2=(1+a^2)(1+b^2)(1+c^2)}$ :)", "Solution_4": "yes its really nice \r\nand its application to this problem was straightforward.\r\n\r\nI didnt notice Gibbenergy s post when I was posting. Our solutions are identical!\r\nIs it also your solution Spanferkel", "Solution_5": "no :blush: I didn't want to expand it all and used some calculus :blush:", "Solution_6": "[quote=\"spanferkel\"]This is neat. A nice identity:\n\n$\\boxed{(-1+ab+ bc+ca)^2+(a+b+c-abc)^2=(1+a^2)(1+b^2)(1+c^2)}$ :)[/quote]\r\n\r\n Very nice identity. \r\n Your problem is solve by quadratic form. Since after expand, I treat it as the function with a and using discrimination to access that forms. So, my sols and Albani Eagle is quite different.\r\n But anyway, Very nice problems", "Solution_7": "[quote=\"Gibbenergy\"] So, my sols and Albani Eagle is quite different.\n But anyway, Very nice problems[/quote] Well maybe your approach was different in the beginning but the solution is exactly the same. Add one more step in yours (putting it all on the LHS) and you have the same as Albanian Eagle :)\r\nAnyway maybe using that identity in a hidden way can help constructing other ineqs harder to solve :D :D", "Solution_8": "Spanferkel !\r\n\r\n Your identity is very very nice.\r\n Indeed. Here is my conjecture:\r\n$(-1+ab+bc+ca)^k + (a+b+c-abc) ^k$\r\n 1/ $k$ is odd . It can be factor . I mean it can divided by $(ab+b-abc+bc-1+a+c+ac)$\r\n 2/ $k$ is even. \r\n If $k$ is not the power of $2$ , I mean $k$ not $2^2,2^3,2^4,...2^n$ it can be factor to $(a^2+1)(b^2+1)(c^2+1)$\r\n If $k$ is the power of $2$. Irreducible form.\r\nI will try to prove them. :D", "Solution_9": "Nice idea, Gibbs. Obvious for odd $k$.\r\nBut the second one doesn't hold for $k=12$.\r\nIn fact, $x^k+y^k$ is generally divisible by $x^2+y^2$ only of $k=2r$, with odd $r$. :( \r\n\r\nWhat I meant was rather finding nice difficult [i]inequalities [/i]that have an elegant solution based on some (hidden) use of the above identity... Less straightforward ones than my original one. :D", "Solution_10": "[quote=\"spanferkel\"]Nice idea, Gibbs. Obvious for odd $k$.\nBut the second one doesn't hold for $k=12$.\nIn fact, $x^k+y^k$ is generally divisible by $x^2+y^2$ only of $k=2r$, with odd $r$. :( \n\n :D[/quote]\r\n\r\n In case $k=12$ and also $k=20$ ...you are right, it doesn't have the factor $(a^2+1)(b^2+1)(c^2+1)$ but it still can be factor out. I think it should be a very nice number theory problem. \r\n :(", "Solution_11": "So, good luck!! And especially for proving that for $2^n$, it's irreducible... :P\r\n\r\nAnd I hope you have a computer algebra system for trying :play_ball: :whistling:", "Solution_12": "Now let's see if we can do something more general:\r\nIf $a_i, b_i, c_i \\ge0$ for $i=1,...,n$, prove or disprove\r\n\r\n$\\sum_{i=1}^n a_ib_ic_i+\\sqrt{\\frac{\\sum a_i^2\\sum b_i^2\\sum c_i^2}n}\\ge \\frac2{n^2} \\sum a_i\\sum b_i\\sum c_i$ :lol:", "Solution_13": "As to the identity, I've just found a generalization of it and a *beautiful* proof. Well, once proven it looks so easy that I don't think it is brand new. I've posted it in Combinatorics [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=474387]here.[/url]", "Solution_14": "Thanks Albnian Eagle for providing me the link of this topic. The following is my solution:\r\n\r\nBecause $ 2(1\\plus{}a^2)\\equal{}(1\\plus{}a)^2\\plus{}(1\\minus{}a)^2,$ and $ (1\\plus{}b^2)(1\\plus{}c^2)\\equal{}(b\\plus{}c)^2 \\plus{}(bc\\minus{}1)^2,$ the Cauchy Schwarz Inequality gives us\r\n$ \\sqrt{2(1\\plus{}a^2)(1\\plus{}b^2)(1\\plus{}c^2)} \\ge (b\\plus{}c)(1\\plus{}a)\\plus{}(1\\minus{}a)(bc\\minus{}1),$\r\nand hence, we deduce that\r\n$ 2(abc\\plus{}1) \\plus{}\\sqrt{2(1\\plus{}a^2)(1\\plus{}b^2)(1\\plus{}c^2)} \\ge 2(abc\\plus{}1)\\plus{}(b\\plus{}c)(1\\plus{}a)\\plus{}(1\\minus{}a)(bc\\minus{}1) \\equal{}1\\plus{}a\\plus{}b\\plus{}c\\plus{}ab\\plus{}bc\\plus{}ca\\plus{}abc\\equal{}(1\\plus{}a)(1\\plus{}b)(1\\plus{}c),$\r\nas desired." } { "Tag": [ "limit", "function", "real analysis", "real analysis unsolved" ], "Problem": "Let $ \\tau : \\mathbb{N}\\to \\mathbb{N}$ strictly increasing, $ (a_n) \\in (\\mathbb{R}_{\\plus{}})^{\\mathbb{N}}$ decreasing and $ \\varepsilon : \\mathbb{N}\\to \\{1,2\\}$. If $ \\sum (\\minus{}1)^{\\varepsilon (n)} a_n$ converge,\r\nDoes $ \\sum (\\minus{}1)^{\\varepsilon (n)} a_{\\tau (n)}$ converge?", "Solution_1": "$ \\lim_{N\\to \\infty} \\sum_{n\\equal{}1}^N (\\minus{}1)^{\\varepsilon (n)} a_n \\equal{} L < \\infty$\r\n\r\nfix $ \\delta > 0$, then $ \\exists M_\\delta$ s.t.\r\n\r\n$ \\frac{\\delta}{C}>|\\sum_{n\\equal{}1}^{M_\\delta} (\\minus{}1)^{\\varepsilon (n)} a_n \\minus{} L | \\equal{}|\\sum_{n\\equal{}M_\\delta\\plus{}1}^{\\infty} (\\minus{}1)^{\\varepsilon (n)} a_n|$\r\n\r\n\r\n$ |\\sum_{n\\equal{}M_\\delta\\plus{}1}^\\infty (\\minus{}1)^{\\varepsilon (n)} a_{\\tau (n)}| \\le C|\\sum_{n\\equal{}M_\\delta\\plus{}1}^{\\infty} (\\minus{}1)^{\\varepsilon (n)} a_n|$\r\n\r\nand we are done? just having problems with that C.... i want to use 2...", "Solution_2": "I believe there is not enough information to answer. Did you just make this problem up?\r\n\r\nObserve what happens when the sign function is constant and we choose different terms of the harmonic series", "Solution_3": "[quote=\"\u00a7outh\u00a7tar\"]I believe there is not enough information to answer. [/quote]\nDo you mean that you can find $ \\tau : \\mathbb{N}\\to \\mathbb{N}$ strictly increasing, $ (a_n) \\in (\\mathbb{R}_{ \\plus{} })^{\\mathbb{N}}$ decreasing and $ \\varepsilon : \\mathbb{N}\\to \\{1,2\\}$ s.t. $ \\sum ( \\minus{} 1)^{\\varepsilon (n)} a_n$ converges and $ \\sum ( \\minus{} 1)^{\\varepsilon (n)} a_{\\tau (n)}$ diverges?\n\n[quote=\"\u00a7outh\u00a7tar\"]Did you just make this problem up?[/quote]\r\nYes I did", "Solution_4": "Related question: can you show that for a decreasing sequence $ (a_n)$, $ a_n > 0$ and $ a_n \\to 0$, if $ \\varepsilon^{\\minus{}1}(2)$ has density $ \\frac{1}{2}$ in $ \\bf N$ then $ \\sum (\\minus{}1)^{\\varepsilon(n)} a_n$ converges? (This is close to the alternating series test.)", "Solution_5": "[quote=\"JBL\"]... if $ \\varepsilon^{ \\minus{} 1}(2)$ has density $ \\frac {1}{2}$ in $ \\bf N$ ...[/quote]\r\nIf $ \\left|\\sum_{k\\equal{}1}^n(\\minus{}1)^{\\varepsilon(k)}\\right|$ is bounded, then there's an easy proof - but you can have density $ \\frac12$ without that sum being bounded.", "Solution_6": "Joel, are you sure about that result? What about 1 copy of $ \\frac {1}{1}$, minus 2 copies of $ \\frac {1}{2}$, plus 3 copies of $ \\frac {1}{3}$, and so on? That series does not converge, but I believe that the plus signs have density $ \\frac {1}{2}$. I hope I haven't miscalculated.", "Solution_7": "[quote=\"Ravi B\"]Joel, are you sure about that result?[/quote] Sorry, I wasn't claiming it as true: I played around with it far enough that I could have made Kent Merryfield's comment, suspected that it might not be true if the partial sums weren't bounded, and then didn't find a counter-example immediately so put it up here. That is a counter-example. I think my original thought was that a counter-example might be used to construct a counter-example to the original question, but my initial attempts based off your example haven't been successful. I.e., I'm trying to do the problem in reverse by finding a supersequence that does converge when signed in the same way.\r\n\r\nEdit: More broadly, I think the statement should be true if $ \\varepsilon^{\\minus{}1}(2)$ exists but is not $ \\frac{1}{2}$, and probably also if $ \\varepsilon^{\\minus{}1}(2)$ does not exist, so this looked like the best place to look for a counter-example.", "Solution_8": "I believe the answer to the original question is \"not necessarily\". Here is a counterexample. First, define\r\n\\[ x_k \\equal{} \\frac {1}{k \\log(k \\plus{} 1)}.\r\n\\]\r\nNote that the sum of the $ x_k$ diverges to infinity.\r\n\r\nConsider the series consisting of 1 copy of $ x_1$, minus 1 copy of $ x_1$, plus 2 copies of $ x_2$, minus 2 copies of $ x_2$, and so on. That series converges (to zero).\r\n\r\nNow consider the \"subseries\" consisting of 1 copy of $ x_1$, minus 1 copy of $ x_2$, plus 2 copies of $ x_2$, minus 2 copies of $ x_3$, and so on. That subseries diverges to infinity, by comparing to the sum of the $ x_k$.", "Solution_9": "Nice conterexample Ravi B.", "Solution_10": "Thanks. Joel's side question was useful in pointing the way to the counterexample." } { "Tag": [], "Problem": "My AP chemistry class is going at a very slow pace, so I am attempting some self-study on my own. I have Chemistry by Zumdahl, and Chemistry by Mortimer. Would studying solely from these two books and doing practice tests be enough preparation for me to have a decent chance at making it past the local level?", "Solution_1": "Indeed.\r\n\r\nAnd, as always, I recommend [u]General Chemistry [/u]by Linus Pauling. Besides those three books, studying from past tests is very likely to put you on the road towards camp.", "Solution_2": "Thanks, inteluser\r\nI ordered the book just now on Amazon. It looks like a very interesting book, =)", "Solution_3": "question: is Mortimer- Robert G or Charles E?\r\n\r\nAlso, first two books are somewhat expensive so I don't think I'll get both. Which do you think is more suitable for someone with a pretty solid understanding of AP chem topics (so like, I wouldn't need a complete intro to those topics.. maybe which is harder? unless they both are pretty well beyond AP.. then which is more \"user-friendly?\"). thanks", "Solution_4": "Charles Mortimers book is my all time favorite chem book. i think it is the best. :)", "Solution_5": "[quote=\"saturnlife\"]My AP chemistry class is going at a very slow pace, so I am attempting some self-study on my own. I have Chemistry by Zumdahl, and Chemistry by Mortimer. Would studying solely from these two books and doing practice tests be enough preparation for me to have a decent chance at making it past the local level?[/quote]\r\n\r\nprinciples of chemistry, raymond chang" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "If $ H$ and $ K$ are subgroup of a finite group then $ |HK| \\equal{} \\frac {|H\\parallel K|} {|H\\cap K|}$", "Solution_1": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=232063\r\n\r\nSee the 10th post. (I think this has been done in other threads as well)." } { "Tag": [ "puzzles" ], "Problem": "This is directly copied from one of local competitions around my area.\r\n\r\nThe Protagoras Paradox\r\n\r\nPerhaps one of the earliest known paradoxes is about the Greek law teacher Protagoras, who took a poor but talented student and agreed to teach him out a fee on condition that after the student completed his studies and won his first law case, he would pay Protagoras a certain sum. The student agreed to do this. Well, the student completed his studies but did not take any law cases. Some time elasped and Protagoras sued the student for the sum. Here are the arguments they gave in court.\r\n\r\nStudent: If I win the case, then by definition, I don't have to pay. If I lose the case, then I will not yet have won my first case, and I have not contracted to pay Protagoras until after I have won my first case. So whether I win the case or lose the case, I don't have to pay.\r\n\r\nProtagoras: If he loses the case, then by definition he has to pay me (after all, this is what the case is about). If he wins the case, then he will have won his first case, hence he has to pay me. In either case, he has to pay me.\r\n\r\n[From Raymond Smullyan's [i]What is the Name of this Book[/i]]", "Solution_1": "See http://en.wikipedia.org/wiki/Paradox_of_the_Court .", "Solution_2": "Thanks.\r\n\r\nP.S. It seems like every time I post, you have something to teach me. :lol:", "Solution_3": "In my opinion, Protagoras was right.\r\n\r\nRegarding the student's first statement: \"If I win the case, then by definition, I don't have to pay\"\r\nWinning the case obviosly does not grant the student exemption to any payment to Protagoras in the future, but it is after he wins the case that he should pay the fee. \r\n\r\nRegarding the student's first statement: \"If I lose the case, then I will not yet have won my first case, and I have not contracted to pay Protagoras until after I have won my first case.\"\r\nNote that losing the case => paying the fee, but the opposite is not true.\r\nIf the court decided that the student should pay, then although it was unreasonable, the student must pay.", "Solution_4": "the student is right because he only has to pay his teacher if he doesnt win his first case", "Solution_5": "double post!!! XD", "Solution_6": "[quote=\"Nightmare124\"]the student is right because he only has to pay his teacher if he doesnt win his first case[/quote]\r\nIt is if, not only if, as the problem states. Signing a contract that says \"I cannot give money to somebody if I have not win any cases\" is completely nonsense. He may give the money to Protagoras if he wants to, or someone force him to.\r\n\r\nIn fact Protagoras could just sue the student for murder :D", "Solution_7": "Maybe you guys are both right. Or, both wrong. That's why it's paradox. :)", "Solution_8": "I think they're both right, and they should just forget about the whole thing because they both owe each other. :D\r\n\r\n[quote=\"Nightmare124\"]double post!!! XD[/quote]\r\nBeyond unnecessary. :|", "Solution_9": "I cannot see where the paradox is\r\nWhat the student says makes no sense at all :mad: \r\nI have explained it in the forth post\r\nCan someone please explain why this is a paradox?", "Solution_10": "Okay, let's see...\r\n\r\nStudent: Sued for money owed. If he wins, he does not have to pay the fine he is being sued for. If he loses, since he hasn't won his first law case, he does not have to pay. Either way, he doesn't pay. [b]Both arguements are valid.[/b]\r\n\r\nProtagoras: Suing for money owed. If he wins, he gets the money he sued for. If he loses, his student has won his first law case, and must pay. [b]Both arguements are valid.[/b] \r\n\r\nNow you get it?", "Solution_11": "[quote=\"erabel\"]Okay, let's see...\n\nStudent: Sued for money owed. If he wins, he does not have to pay the fine he is being sued for. If he loses, since he hasn't won his first law case, he does not have to pay. Either way, he doesn't pay. [b]Both arguements are valid.[/b]\n\nProtagoras: Suing for money owed. If he wins, he gets the money he sued for. If he loses, his student has won his first law case, and must pay. [b]Both arguements are valid.[/b] \n\nNow you get it?[/quote]\r\n\r\nThe first statement of the student is invalid. Of course he need not pay the sum Protagoras sue him for, but his victory triggers another payment, which does not exist at the time he is sued. Unless the court says, \"The contract is annulled. You don't need to pay Protagoras any more.\" which sounds really weird.\r\n\r\nThe court should let the student win, then the student still have to pay Protagoras the sum. The student may sue Protagoras back, but this is another case, nothing related to the first case. What effect will a verdict bring should not be a factor affecting the court's decision.", "Solution_12": "Okay basically we have:\r\n\r\nIf the student loses the case then he needs to pay Protagoras. But the student claims that he won't have to pay Protagoras because he hasn't won a case yet. But winning a case is not the only reason he would need to pay Portagoras. (aka \"student winning case $ \\implies$ he pays\" but NOT \"student winning case $ \\leftrightarrow$ he pays\"). In this case, he would have to pay as a result of the court commanding him to pay.\r\n\r\nNow if the student wins, then he will NOT have to pay as a result of the court commanding him to. But he WILL have to pay as a result of the original contract.\r\n\r\nThe key is noticing the difference between paying money as a result of losing the case and paying money as a result of the original contract.", "Solution_13": "[quote=\"tjhance\"]Okay basically we have:\n\nIf the student loses the case then he needs to pay Protagoras. But the student claims that he won't have to pay Protagoras because he hasn't won a case yet. But winning a case is not the only reason he would need to pay Portagoras. (aka \"student winning case $ \\implies$ he pays\" but NOT \"student winning case $ \\leftrightarrow$ he pays\"). In this case, he would have to pay as a result of the court commanding him to pay.\n\nNow if the student wins, then he will NOT have to pay as a result of the court commanding him to. But he WILL have to pay as a result of the original contract.\n\nThe key is noticing the difference between paying money as a result of losing the case and paying money as a result of the original contract.[/quote]\r\n\r\nTotally agree :D \r\nThe student should hire another lawyer :P", "Solution_14": "I think I should sue them both on the grounds of mental harrasment\r\n:D lol" } { "Tag": [ "symmetry" ], "Problem": "How many lines of symmetry exist for a regular heptagon (a regular seven-sided polygon)?", "Solution_1": "This is simply one for each vertex, since the same lines go through the opposing side, thus $ \\fbox{7}$.", "Solution_2": "there will be a symmetry for each side of the vertex so the answer is 7\n\nso the answer is 7 \n\nrate my post thanks", "Solution_3": "There will be 1 line of symmetry for each vertex. So the answer would be 7" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "for each integer n>2 find the smallest natural number f(n) such that for any subset of A={1,2,...,n} like B with f(n) elements , there is 3 elements of B like\r\n \r\nx,y,z that are pairwise coprime.", "Solution_1": "This won't be a very useful post...\r\n\r\nThe problem was posted here: http://www.mathlinks.ro/viewtopic.php?t=85233\r\n\r\nAnd the solution from Kalva is here:\r\n\r\nhttp://web.archive.org/web/20040810234501/www.kalva.demon.co.uk/balkan/bsoln/bsol924.html" } { "Tag": [ "analytic geometry", "geometry", "quadratics", "graphing lines", "slope", "perpendicular bisector", "geometry solved" ], "Problem": "The problem is:-\r\n\r\nFind the coordinates of point $P$ lying inside triangle $ABC$ such that $\\angle APB = \\angle BPC = \\angle CPA$,\r\n\r\n[b]where A =(0,0) B =(8,0) and C =(5,4).[/b]\r\n\r\nPlease give the whole numerical soln. and not just the method.I know the method but want a sol. which doesn't involve solving Complicated eqns.\r\n\r\n[hide]NOTE:- The answer is quite complicated involving surds. An elegant soln with minimum of algebra will be preferred.[/hide]", "Solution_1": "The 1st Fermat point P is the intersection of 3 circles centered at the orthocenters of the equilateral triangles constructed on the triangle sides c = AB, a = BC, b = CA and passing through the adjacent triangle vertices. We could get the equations of 2 of these circles and find their intersection other than the common vertex. On the other hand, the 1st Fermat point P is also the intersection of the lines connecting the outward vertices of the equilateral triangles with the opposite vertices of the triangle $\\triangle ABC$. Using the circles would lead to quadratic equations, but using these lines would lead to linear equations, hence, it is easier. We need 2 of these lines to get the common intersection of the 3. The triangle side lengths are\r\n\r\n$c = AB = 8$\r\n\r\n$a = BC = \\sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} = \\sqrt{9 + 16} = 5$\r\n\r\n$b = CA = \\sqrt{(x_A - x_C)^2 + (y_A - y_C)^2} = \\sqrt{25 + 16} = \\sqrt{41}$\r\n\r\nLet D, E be the midpoints of the sides AB, BC and G, H the outward vertices of the equilateral triangles $\\triangle ABG, \\triangle BCH$. The coordinates of the midpoints D, E are\r\n\r\n$D = (x_D, y_D) = \\left(\\frac{x_A + x_B}{2}, \\frac{y_A + y_B}{2}\\right) = (4, 0)$\r\n\r\n$E = (x_E, y_E) = \\left(\\frac{x_B + x_C}{2}, \\frac{y_B + y_C}{2}\\right) = \\left(\\frac{13}{2}, 2\\right)$\r\n\r\nThe coordinates of the vertex G of the equilateral triangle $\\triangle ABG$ is\r\n\r\n$G = (x_G, y_G) = \\left(x_E, -\\frac{c \\sqrt 3}{2}\\right) = \\left(4, -4 \\sqrt 3\\right)$\r\n\r\nThe equation of the line BC is\r\n\r\n$BC:\\ \\ y - y_B = \\frac{y_C - y_B}{x_C - x_B}\\ (x - x_B),\\ \\ y = -\\frac 4 3\\ (x - 8) = -\\frac 4 3\\ x + \\frac{32}{3}$\r\n\r\nThe slope of this line is $m = -\\frac 4 3$. The vertex H of the equilateral triangle $\\triangle BCH$ lies on the perpendicular bisector of the segment BC. The perpendicular bisector of the segment BC passes through the midpoint E and it has the slope $k = -\\frac 1 m = \\frac 3 4$, because the product of slopes of 2 perpendicular lines is equal to -1. Hence, the equation of this perpendicular bisector is\r\n\r\n$EH:\\ \\ y - y_E = k (x - x_E),\\ \\ y = \\frac 3 4 \\left(x - \\frac{13}{2}\\right) + 2 = \\frac 3 4\\ x - \\frac{23}{8}$\r\n\r\nThe distance EH is equal to\r\n\r\n$EH = \\sqrt{(x_H - x_E)^2 + (y_H - y_E)^2} = \\frac{a \\sqrt 3}{2} = \\frac{5 \\sqrt 3}{2}$\r\n\r\nSince the point H lies on the line EH, $y_H = \\frac 3 4\\ x_H - \\frac{23}{8}$ and\r\n\r\n${\\left(x_H - \\frac{13}{2}\\right)^2 + \\left(\\frac 3 4\\ x_H - \\frac{39}{8}\\right)^2} = \\frac{75}{4}$\r\n\r\n$x_H^2 - 13x_H + \\frac{169}{4} + \\frac{9}{16}\\ x_H^2 - \\frac{117}{16}\\ x_H + \\frac{1521}{64} = \\frac{75}{4}$\r\n\r\n$64x_H^2 - 832x_H + 2704 + 36x_H^2 - 468x_H + 1521 = 1200$\r\n\r\n$100x_H^2 - 1300x_H + 3025 = 0$\r\n\r\n$4x_H^2 - 52x_H + 121 = 0$\r\n\r\n$x_H = \\frac{52 + \\sqrt{2704 - 1936}}{8} = \\frac{13}{2} + 2 \\sqrt 3$\r\n\r\n$y_H = \\frac 3 4\\ x_H - \\frac{23}{8} = 2 + \\frac{3 \\sqrt 3}{2}$\r\n\r\nFor the outward vertex H of the equilateral triangle $\\triangle BCH$, we had to take the root with the plus sign. Having the coordinates of the points G, H, we can find the equations of the lines AH, CG.\r\n \r\n$AH:\\ \\ y - y_A = \\frac{y_H - y_A}{x_H - x_A}\\ (x - x_A)$\r\n\r\n$y = \\frac{2 + \\frac{3 \\sqrt 3}{2}}{\\frac{13}{2} + 2 \\sqrt 3}\\ x = \\frac{4 + 3 \\sqrt 3}{13 + 4 \\sqrt 3}\\ x = \\frac{16 + 23 \\sqrt 3}{121}\\ x$\r\n\r\n$CG:\\ \\ y - y_C = \\frac{y_C - y_G}{x_C - x_G}\\ (x - x_C)$\r\n\r\n${y = \\frac{4 + 4 \\sqrt 3}{5 - 4}\\ (x - 5)} + 4 = 4(1 + \\sqrt 3)x - 4(4 + 5 \\sqrt 3)$\r\n\r\nSolving the set of 2 equations for the lines AH, CG, we get their intersection P\r\n\r\n$\\frac{16 + 23 \\sqrt 3}{121}\\ x = 4(1 + \\sqrt 3)x - 4(4 + 5 \\sqrt 3)$\r\n\r\n$(16 + 23 \\sqrt 3) x = 484(1 + \\sqrt 3)x - 484(4 + 5 \\sqrt 3)$\r\n\r\n$(468 x + 461 \\sqrt 3) x = 484(4 + 5 \\sqrt 3)$\r\n\r\n\r\n$x_P = \\frac{484(4 + 5 \\sqrt 3)}{468 + 461 \\sqrt 3} = \\frac{2^2 \\cdot 11^2 (3 \\cdot 41^2 - 2^4 \\cdot 31 \\sqrt 3)}{3 \\cdot 11^2 \\cdot 1153} = \\frac{2^2 (3 \\cdot 41^2 - 2^4 \\cdot 31 \\sqrt 3)}{3 \\cdot 1153}$\r\n\r\n$y_P = \\frac{16 + 23 \\sqrt 3}{121}\\ x_P = \\frac{16 + 23 \\sqrt 3}{121} \\cdot \\frac{2^2 (3 \\cdot 41^2 - 2^4 \\cdot 31 \\sqrt 3)}{3 \\cdot 1153} =$\r\n\r\n$= \\frac{2^2 (2^7 \\cdot 3 \\cdot 11^2 + 11^2 \\cdot 19 \\cdot 47 \\sqrt 3)}{3 \\cdot 11^2 \\cdot 1153} = \\frac{2^2 (2^7 \\cdot 3 + 19 \\cdot 47 \\sqrt 3)}{3 \\cdot 1153}$\r\n\r\nYetti" } { "Tag": [ "conics", "hyperbola", "geometry", "geometric transformation", "reflection", "Asymptote", "geometry unsolved" ], "Problem": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14640[/img]\r\n\r\n :(", "Solution_1": "This problem is similar to the problem [url]http://www.mathlinks.ro/viewtopic.php?t=209728[/url], click the link for details. $ J$ is the intersection of the directrix $ d_1$ with the line $ a_1 \\parallel s_1,$ where $ s_1$ is the unknown hyperbola asymptote. Perpendicular to the directrix $ d_1$ from $ P$ cuts $ d_1, a_1$ at $ D, A,$ respectively. Perpendicular to the directrix $ d_1$ from $ Q$ cuts $ d_1, a_1$ at $ E, B,$ respectively. $ \\frac{AJ}{AD} \\equal{}\\frac{BJ}{BE} \\equal{} \\epsilon$ is the hyperbola excentricity. Reflect $ a_1$ into $ a_1'$ in the line $ PDA.$ Parallels to $ a_1'$ through $ P, Q$ cut the directrix $ d_1$ at $ S, T$ and $ \\frac{PD}{PS} \\equal{} \\frac{QE}{QT} \\equal{} \\epsilon$ $ \\Longrightarrow$ the hyperbola focus $ F_1$ correcponding to the directrix $ d_1$ is intersection of circles $ (P), (Q)$ with centers $ P, Q$ and radii $ PS, QT,$ respectively. If the 2 circles intersect at all, we get 2 possible foci $ F_1, F_1'$ and again, 2 different hyperbolas exist with the given directrix $ d_1,$ the given asymptote direction $ a_1,$ and passing through the 2 given points $ P, Q.$\r\n\r\nProceeding with one possible focus $ F_1$: Perpendicular $ p$ to $ d_1$ through $ F_1$ is the hyperbola major axis. Perpendicular to $ a_1$ through $ F_1$ is also perpendicular to the unknown asymptote $ s_1 \\parallel a_1,$ it cuts $ d_1$ at $ K.$ This point lies on the hyperbola pedal circle and perpendicular to $ F_1K$ through $ K$ (parallel to $ a_1$) cuts the hyperbola major axis $ p$ at the hyperbola center $ C,$ identical with the center of the pedal circle $ (C)$ with radius $ CK.$ The other hyperbola focus is the reflection $ F_2$ of $ F_1$ in $ C,$ one hyperbola asymptote is $ s_1 \\equiv CK$ and the other asymptote is reflection $ s_2$ of $ s_1$ in the major axis $ p \\equiv F_1CF_2.$ Reflections $ P', Q'$ of $ P, Q$ in $ C$ also lie on the hyperbola. The pedal circle $ (C)$ cuts the hyperbola major axis $ p$ at the hyperbola vertices $ V_1, V_2,$ so that we have 6 points $ P, Q, P', Q', V_1, V_2.$ See the above link on how to construct more hyperbola points.", "Solution_2": ":thumbup: :wink:" } { "Tag": [ "function", "absolute value", "algebra proposed", "algebra" ], "Problem": "once I've heard a question :\r\n\r\nIs it possible that $(a_1+a_2+...+a_n) -> g \\in R$ and \r\n$b_n=a_1^3+a_2^3+...+a_n^3$ is not congruent ?", "Solution_1": "Just a silly question...\r\n\r\nIs it congruent or .... convergent?\r\n\r\ncheers! :D :D", "Solution_2": "in fact you can prove the following statement. If $f$ is a function such that for every series $\\sum_{n=1}^{\\infty}u_n$ converges then $\\sum_{n=1}^{\\infty}f(u_n)$ also converges then\r\n$f(x)=kx$ for every $x$ in $[a,b]$ where $a<02+b2. Try taking (a+bi)2002=a-bi and multiplying by a+bi on both sides and see what happens.[/hide]", "Solution_2": "(a + bi)^2002 = a - bi \r\ntimes a+bi\r\n(a+bi)^2003=a^2+b^2,\r\n(0,0) works...", "Solution_3": "it says find the number of ordered pairs, and 1 is not the answer... there are more.", "Solution_4": "[hide]\n\nif (a+bi)^2002 = a-bi, we know |(a+bi)^2002| = |a-bi| =>. But |a+bi| = |a-bi|, so |a+bi| = 1. Therefore we can write z = a + bi = cis :theta:. Furthermore, a-bi = cis (- :theta: ). cis :theta: ^ 2002 = cis (2002 * :theta: ). Therefore 2002 * :theta: = - :theta: (mod 2pi) or 2003 * :theta: = 0 (mod 2pi). So we should get 2003 unique solutions for :theta: ( 2pi * i / 2003 for 0 :le: i :le: 2002 ). and therefore 2003 ordered pairs (a,b). (I might be wrong, just guessing)[/hide]", "Solution_5": "[hide]I get 2004 since we have (0,0) as well. But other than that small detail, I agree with everything Nukular did.[/hide]", "Solution_6": "[hide]\n\nyeah i forgot (0,0)... \n\n\n\nalways forget the trivial solution \n\n\n\n[/hide][/hide]", "Solution_7": "[quote]So we should get 2003 unique solutions for ( 2pi * i / 2003 for 0 i 2002 ). and therefore 2003 ordered pairs (a,b).[/quote]\r\n\r\nCould you explain that?", "Solution_8": "Well, if we have 2003 :theta: = 0 (mod 2pi), then \r\n\r\n2003 :theta: = 2pi * k for some integer k right?. Since 0 <= :theta: < 2*pi, (thats the standard for dealing with unique angles), any integer k from 0 to 2002 will give us a unique :theta:. Therefore we will have 2003 unique :theta: and therefore 2003 unique pairs (a,b). Add in the (0,0) case and you get 2004 solutions." } { "Tag": [], "Problem": "Hmm. I went to see this with some of my friends and wasted a good $5.75 on a very dumb movie. There was no plot, and it wasn't even that scary, because when the monsters came you knew it was all fake. I was actually laughing really hard about the movie.", "Solution_1": "Don't spoil the movie man.\r\nThe beginning and middle were good, but I hated the ending.", "Solution_2": "what's the village hmm", "Solution_3": "I have no idea what this is but I'm the only person who voted in the poll\n\nI also have no idea what this is about so I chose randomly" } { "Tag": [ "trigonometry" ], "Problem": "hi there \r\nprove this plz \r\ni need hlp\r\ntan(a)+tan(b)+tan(c)=tan(z)tan(b)tan(c)\r\na,b,c <>90 :oops:\r\nalso \r\na,b,c are the angel of a triangle", "Solution_1": "Simple use $c = \\pi-a-b$ and trig rules for tangent and a little bit of algebra.\r\n\r\ntry it", "Solution_2": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143021" } { "Tag": [ "limit", "logarithms", "calculus", "calculus computations" ], "Problem": "we all know \r\n\r\n$\\lim_{p\\rightarrow\\infty}\\;\\sum_{n=1}^{p}\\;\\frac{1}{n}\\quad\\;\\;\\;$ diverges.\r\n\r\nwhat about\r\n\r\n$\\lim_{p\\rightarrow\\infty}\\;\\sum_{n=1}^{p}\\;\\frac{1}{n^{1+\\imath}}$ ? \r\n\r\ndoes it also diverge ? or converge ? :D \r\n\r\nnote: $\\;\\;\\;\\imath\\;\\;=\\;\\;\\sqrt{-1}$", "Solution_1": "Since $n^{i}=e^{i\\log n}$ is essentially constant on any interval of the form $N\\le n\\le 1.01N$, the sum of $\\frac{1}{n^{1+i}}$ over such an interval is about $0.01$ in absolute value. The series diverges." } { "Tag": [ "trigonometry", "geometry unsolved", "geometry" ], "Problem": "Let be given a right triangle $ABC$ at $A$. Denote: $O$ be the midpoint of the side $BC;\\,M\\in AB,\\,N\\in AC: \\,BM^{2}+CN^{2}=MN^{2}$\r\nProve that: $\\angle{MON}=90^{\\circ}$", "Solution_1": "$ON^{2}={a^{2}\\over 4}+CN^{2}-aCN\\cos\\gamma={a^{2}\\over 4}+CN^{2}-bCN$\r\n\r\n$OM^{2}={a^{2}\\over 4}+BM^{2}-cBM$\r\n\r\nAdding those two we get\r\n\r\n$OM^{2}+ON^{2}={a^{2}\\over 2}+MN^{2}-(bCN+cBM)\\qquad (*)$\r\n\r\nFrom $(b-CN)^{2}+(c-BM)^{2}=MN^{2}$, after simplification, we get $bCN+cBM={a^{2}\\over 2}$, hence $(*)$ becomes $OM^{2}+ON^{2}=MN^{2}$. QED", "Solution_2": "Through vertex $B$ of $\\bigtriangleup ABC,$ we draw a line perpendicular to $AB$ and we denote as $L,$ it\u2019s intersection point, from the segment line $NO.$ So, we have $BL = CN$ $($ because of $OB = OC$ $).$\r\n\r\nFrom the right triangle $\\bigtriangleup BLM$ $($ $\\angle LBM = 90^{o}$ $),$ $\\Longrightarrow$ $(BM)^{2}+(BL)^{2}= (ML)^{2}$ and so, we have that $ML= MN.$\r\n\r\nHence, because of the point $O,$ is also the midpoint of the segment $LN,$ we conclude that the segment line $MO,$ is midperpenticular of the segment $LN$ and the proof is completed.\r\n\r\nKostas Vittas." } { "Tag": [ "trigonometry", "geometry proposed", "geometry" ], "Problem": "Let $ABC$ be an equilateral triangle and $\\Gamma$ its incircle. If $D$ and $E$ are points on the segments $AB$ and $AC$ such that $DE$ is tangent to $\\Gamma$, show that $\\frac{AD}{DB}+\\frac{AE}{EC}=1$.", "Solution_1": "Let $M$ be the tangency point of the circle with $AB$, we define the same way $N(N \\in BC)$,$P(P \\in AC)$,$Q(Q \\in DE)$. Also $MB=BN=NC=CP=c$, $MD=DQ=a$, $QE=EP=b$, $AD=c-a$, $AE=c-b$.\r\n\r\nThen our statement is equivalent to show:$3ab=c^{2}-ac-bc$\r\n\r\nBy cosine law in triangle $AED$: $(b+c)^{2}=(c-a)^{2}+(c-b)^{2}-2cos60^\\circ(c-a)(c-b)$, and the result follows.", "Solution_2": "With the last notations. $BE$ cut $CD$ at $J$ and $AJ$ cut $BC$ at $K$, then: $\\frac{AD}{DB}+\\frac{AE}{EC}=\\frac{AJ}{JK}$ \r\nBut $BE, CD, MP$ concur at $J$, now $J$ is the midpoint of $AK$,i.e. $AJ=JK$, therefore: $\\frac{AD}{DB}+\\frac{AE}{EC}=1$", "Solution_3": "Dear Mathlinkers,\nrevisiting the last proof, we have\n1. the relation of van Aubel\n2. the Newton theorem\n3. the midline.\n\nSincerely\nJean-Louis", "Solution_4": "This problem is a special case of the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=615952]Modern Sangaku[/url] ;) :)" } { "Tag": [ "topology", "symmetry" ], "Problem": "Ampere's Law applies when we have a contour around a wire with a current on it. My question is, does this current have to be infinitely long?\r\n\r\nBecause we can do it for an infinitely long wire. If $ B$ is the magnetic field at radius $ R$, then we have $ 2 \\pi R B \\equal{} \\mu_0 i$ or $ B \\equal{} \\frac{\\mu_0 i}{2 \\pi R}$, which is the correct formula.\r\n\r\nHowever, what if we said that Ampere's Law applied for a small segment of wire, like the one in the diagram? The magnetic field should still be symmetric around the wire, and so we should get $ 2 \\pi R B \\equal{} \\mu_0 i$ and arrive at the same formula. But by the Biot-Savart law, the field strength should be $ \\frac{\\mu_0 i dl}{4 \\pi R^2}$. It thus seems to me like Ampere's Law only applies when the wire is infinitely long.\r\n\r\nHowever, Ampere's Law also works for a solenoid. So this doesn't make sense. (Do the solenoid and the infinite wire supposed be similar because they have the same topology or something?? That is, if we embed the infinite wire in a projective space...)\r\n\r\nSo where did I go wrong?", "Solution_1": "Ampere's law applies only when you have steady state currents, and these occur only in closed loops.\r\n\r\nIf the current is infinitely long, you can put the rest of the loop infinitely far away so that its effect on the magnetic field is negligible. The circuit can still be closed, so Ampere's Law still applies.\r\n\r\nOn the other hand, if the current is finitely long, either (a) you do not have a closed loop or (b) the rest of the loop has a non-negligible effect on the magnetic field.\r\nIn case (a) Ampere's Law no longer applies. You need Maxwell's correction, which deals with time rates of change of electric flux caused by the accumulation of charge at one end of the wire.\r\nIn case (b) the symmetry of the magnetic field is broken. $ B$ is no longer constant over the line integral." } { "Tag": [ "topology", "probability and stats" ], "Problem": "Fro two prob. measures $ P$ and $ Q$ on $ (\\mathbb{R}, \\mathcal{B})$, the total variation distance is defined as $ \\parallel P\\minus{}Q\\parallel \\equal{}\\sup_{B\\subset\\mathcal{B}}|P(B)\\minus{}Q(B)|$\r\nThe sup metric is defines as $ d(P,Q) \\equal{}\\sup_{x\\in \\mathbb{R}}|P((\\minus{}\\infty, x])\\minus{}Q((\\minus{}\\infty, x])|$\r\nAren't the topologies generated by these two the same? Isn't there any connection between these two?", "Solution_1": "What about $ P_n\\equal{}\\sum_1^n \\delta_{\\frac{k}{n}}$ ?", "Solution_2": "I don't understand your point alekk. Could you please be more precise?", "Solution_3": "$ P_n \\equal{} \\sum_1^n \\frac{1}{n} \\delta_{\\frac {k}{n}}$ converges towards the uniform distribution on $ [0;1]$ for the distance $ d$ but not for $ \\| \\cdot \\|$, isn't it?", "Solution_4": "Well that's true, because $ P_n$ converges in distribution to the uniform distribution on $ [0,1]$. \r\nWe have $ d(P,Q)\\le \\parallel P\\minus{}Q\\parallel$. So we maynot expect convergence with respect to $ \\|\\cdot\\|$.\r\nCan we say that the topology generated by $ d$ is finer than that by $ \\|\\cdot\\|$?", "Solution_5": "In fact, if $ P_{n}\\equal{}\\sum_{1}^{n}\\frac{1}{n}\\delta_{\\frac{k}{n}}$ and $ P$ corresponds to the uniform distribution on $ [0,1]$, it can be shown that $ \\parallel P_n\\minus{}P\\parallel \\equal{}1$ by choosing $ B_n\\equal{} \\displaystyle\\cup_{i\\equal{}0}^{n\\minus{}1} \\left(\\frac{i}{n}, \\frac{i\\plus{}1}{n}\\right)$. Am I right?", "Solution_6": "yes, this is what I meant in the previous post :lol:" } { "Tag": [ "quadratics", "complex numbers", "algebra", "algebra solved" ], "Problem": "Hey, Ive just been introduced to complex numbers and I'm having real trouble working out the question below....\r\n\r\nThe question is solve:\r\n\r\n(1 + i)z2 + (3 - 2i)z - (21 - 7i) = 0\r\n\r\nanswer. z = 3 - i OR -7/2 + 7/2i\r\n\r\nCan someone please explain how to work out the roots for the equation, I have been looking through textbooks with no luck, any help is greatly appreciated!\r\n\r\nCheers,\r\nRoo", "Solution_1": "You should use ordinar formulae for roots of a quadratic equation:\r\n\\[z_{\\pm}=\\frac{-(3-2i)\\pm \\sqrt{(3-2i)^2+4(1+i)(21-7i)}}{2(1+i)}=\\frac{-(3-2i)\\pm\\sqrt{107+44i}}{2(1+i)}=\\] \\[=\\frac{-(3-2i)\\pm \\sqrt{(11+2i)^2}}{2(1+i)}=\\frac{-(3-2i)\\pm(11+2i)}{2(1+i)}=\\] \\[=(-(3-2i)\\pm(11+2i))\\cdot (1-i)/4=3-i\\ OR\\ -\\frac{7}{2}+\\frac{7}{2}i.\\]", "Solution_2": "aaaaaaaaahh right, :D cheers" } { "Tag": [ "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "Let $n\\geq 2$ be an integer and denote by $H_{n}$ the set of column vectors $^{T}(x_{1},\\ x_{2},\\ \\ldots, x_{n})\\in\\mathbb{R}^{n}$, such that $\\sum |x_{i}|=1$. \r\n\r\nProve that there exist only a finite number of matrices $A\\in\\mathcal{M}_{n}(\\mathbb{R})$ such that the linear map $f: \\mathbb{R}^{n}\\rightarrow\\mathbb{R}^{n}$ given by $f(x)=Ax$ has the property $f(H_{n})=H_{n}$.\r\n\r\n[hide=\"Comment\"]In the contest, the problem was given with a) and b):\n\na) Prove the above for $n=2$;\n\nb) Prove the above for $n\\geq 3$ as well.[/hide]", "Solution_1": "The standard basis vectors $e_{n}$ (and also $-e_{n}$) are the [url=http://en.wikipedia.org/wiki/Extreme_point]extreme points[/url] of $H_{n}$. It follows that $A$ permutes $\\{\\pm e_{n}\\}$. Therefore, there are exactly $2^{n}n!$ such matrices." } { "Tag": [ "limit", "algebra", "polynomial", "function", "calculus" ], "Problem": "What is the value of $ 0^0$? \r\n\r\nGoogle results in $ 1$. [url=http://www.google.com/search?client=opera&rls=en&q=0%5E0&sourceid=opera&ie=utf-8&oe=utf-8]link[/url]\r\n\r\nWolfram Alpha results in \"indeterminate\". [url=http://www51.wolframalpha.com/input/?i=0%5E0]link[/url]", "Solution_1": "$ 0^0$ means $ \\lim_{(x, y) \\to (0, 0)} x^y$.\r\n$ 0^0$ [b]don't[/b] mean $ \\lim_{x \\to 0} x^x$.\r\n\r\nCan you image $ z\\equal{}x^y$?", "Solution_2": "In some cases it can be defined as 1 or 0 for some purpose, but generally it is indeterminate.", "Solution_3": "[quote=\"Kouichi Nakagawa\"]$ 0^0$ means $ \\lim_{(x, y) \\to (0, 0)} x^y$.\n$ 0^0$ [b]don't[/b] mean $ \\lim_{x \\to 0} x^x$.\n\nCan you image $ z \\equal{} x^y$?[/quote]Um, may I ask what is the difference? :blush:", "Solution_4": "What's the difference?\r\n\r\n$ \\lim_{x\\to0^\\plus{}}x^x\\equal{}1.$\r\n\r\n$ \\lim_{x\\to0^\\plus{}}\\left(e^{\\minus{}1/x}\\right)^x\\equal{}e^{\\minus{}1},$ which is neither one nor zero.\r\n\r\n$ \\lim_{x\\to0^\\plus{}}\\left(e^{\\minus{}1/x}\\right)^{\\sqrt{x}}\\equal{}0.$\r\n\r\nActually the main reason I posted was to mention another interpretation:\r\n\r\nWhen using summation notation to represent polynomials or power series, [b]by convention,[/b] the symbol $ x^0$ stands for the the constant everywhere-defined function $ 1.$ Hence within that particular convention, $ 0^0\\equal{}1.$ In other words:\r\n\r\n$ f(x)\\equal{}\\sum_{k\\equal{}0}^na_kx^k\\equal{}a_0\\plus{}a_1x\\plus{}a_2x^2\\plus{}\\cdots\\plus{}a_nx^n$ and by convention, $ f(0)\\equal{}a_0.$", "Solution_5": "Most calculators are programmed to spit out \"undefined\" on $ 0^0$.", "Solution_6": "[b]Proposition:[/b] Let $ A, B$ be finite sets, and let $ A^B$ denote the set of functions from $ B$ to $ A$. Then $ |A^B| \\equal{} |A|^{|B|}$.\r\n\r\nThere is one function from the empty set to itself: it does nothing. Therefore $ 0^0 \\equal{} 1$ in [b]combinatorics.[/b] (This is the same function that makes $ 0! \\equal{} 1$, the empty permutation.)", "Solution_7": "What t0rajir0u just said is also essentially a convention. If you didn't accept what he said, mathematics would not be destroyed - but it's mildly more convenient and consistent to accept it.", "Solution_8": "[quote=\"FantasyLover\"][quote=\"Kouichi Nakagawa\"]$ 0^0$ means $ \\lim_{(x, y) \\to (0, 0)} x^y$.\n$ 0^0$ [b]don't[/b] mean $ \\lim_{x \\to 0} x^x$.\n\nCan you image $ z \\equal{} x^y$?[/quote]Um, may I ask what is the difference? :blush:[/quote]\r\n[b]Remark 1[/b]\r\n$ \\lim_{(x, y) \\to (0, 0)} x^y \\neq \\lim_{x \\to 0} \\left(\\lim_{y \\to 0} x^y \\right)\\equal{}1$.\r\n$ \\lim_{(x, y) \\to (0, 0)} x^y \\neq \\lim_{y \\to 0} \\left(\\lim_{x \\to 0} x^y \\right)\\equal{}0$.\r\n$ \\lim_{(x, y) \\to (0, 0)} x^y \\neq \\lim_{y \\to 0} \\left(\\lim_{x \\to (1/2)^{1/y}} x^y \\right)\\equal{}\\frac{1}{2}$.\r\nand more...\r\n\r\n[b]Remark 2[/b]\r\n$ \\lim_{x \\to 0} \\left(\\lim_{y \\to 0} f(x, y) \\right)\\equal{}0, \\ \\lim_{y \\to 0} \\left(\\lim_{x \\to 0} f(x, y) \\right)\\equal{}0 \\nRightarrow \\lim_{(x, y) \\to (0, 0)} f(x, y) \\ \\text{is undefined}$.\r\nFor example, $ f(x, y)\\equal{}\\frac{x^2y^2}{x^2y^2\\plus{}(x\\minus{}y)^2}, \\ x^2y^2\\plus{}(x\\minus{}y)^2\\neq0$.", "Solution_9": "[quote=\"Kent Merryfield\"]it's mildly more convenient and consistent to accept it.[/quote]\r\nRight. In combinatorics one often finds that it is possible to extrapolate the value of a sequence defined on the positive integers to a value at $ 0$, and usually (but not always!) this value is consistent with a certain way of thinking about functions on the empty set, or empty permutations, or empty products. For example, if you define the Fibonacci numbers $ F_n$ as the number of ways to tile a board of size $ n$ with tiles of size $ 1$ and $ 2$, then you get $ F_1 \\equal{} 1, F_2 \\equal{} 2$, etc. (note that this is the \"combinatorialist's indexing\" instead of the \"number theorist's indexing,\" which is the one consistent with $ F_n | F_{mn}$). Interpolating the Fibonacci recursion backwards gives $ F_0 \\equal{} 1$, which suggests that the number of ways to tile the empty board is $ 1$ - the empty tiling. The reason this makes sense is that the Fibonacci recursion is based on deleting the last tile, but if a tiling only has one tile then deleting the last tile gives an empty tiling.\r\n\r\n(The really curious thing is that often it is possible to extrapolate the values of a sequence defined on the positive integers to the [b]negative[/b] integers, and moreover that these numbers have combinatorial significance; write down the negative Fibonacci numbers, for example.)\r\n\r\nNote that $ 0^0 \\equal{} 1$ is also the most convenient convention for writing down power series, which is not an accident; power series and combinatorics are closely related.", "Solution_10": "0^0=0^x/0^x=0/0=indeterminate.\r\nBut on one of the tests i took the choices for 0^0 were 1)complex 2)irrational 3)rational 4)real 5) undefined.\r\nSo I think it should be undefined even though when you write it it is indeterminate.", "Solution_11": "[b]Everything depends on context.[/b] In a calculus class $ 0^0$ is a classic example of an \"indeterminate form,\" and what this means is essentially what Kent Merryfield outlined: it means it's an example of a situation in which a standard arithmetic operation isn't continuous. In a combinatorial context, these concerns are irrelevant.", "Solution_12": "[quote=\"t0rajir0u\"][b]Everything depends on context.[/b] In a calculus class $ 0^0$ is a classic example of an \"indeterminate form,\" and what this means is essentially what Kent Merryfield outlined: it means it's an example of a situation in which a standard arithmetic operation isn't continuous. In a combinatorial context, these concerns are irrelevant.[/quote]\r\n\r\nSo, we can generally say that in DISCREET contexts, $ 0^0$ is often taken to be $ 1$. In CONTINUOUS contexts, though, it's not determined.", "Solution_13": "This is rather simple compared to the rigorous reasons involving limits given above, but I'll post it anyway because it's kind of interesting. :) \r\n\r\nI have read that mathematicians originally defined the zeroth power of a number because of the rules they had proved for dividing powers with like bases. They had come up with and proved rules for simplifying $ \\frac{x^a}{x^b}=x^{a-b}$ when $ a>b$ and $ a0$\r\n\r\n$ \\Leftrightarrow 2t^3\\plus{}3t^2\\minus{}3t\\minus{}2\\plus{}\\frac{3\\sqrt 3}{2} \\ge 0$ , ($ t >0$ ) - which is true .\r\n\r\nEquality holdes for : $ x_1\\equal{} \\minus{}1$ , $ x_2\\equal{}0$ , $ x_3 \\equal{} \\frac{\\minus{}1\\plus{}\\sqrt 3}{2}$ ;", "Solution_2": "Why is it correct to assume values for $ x_1$ and $ x_2$ ?", "Solution_3": "[quote=\"modularmarc101\"]Why is it correct to assume values for $ x_1$ and $ x_2$ ?[/quote]\r\n\r\n$ \\Leftrightarrow (x_2 \\minus{} x_1)^3 \\ge \\frac {2}{3\\sqrt 3}(x_1 \\plus{} x_2 \\minus{} 2x_3)(x_1 \\minus{} 2x_2 \\plus{} x_3)( \\minus{} 2x_1 \\plus{} x_2 \\plus{} x_3)$ ; ($ x1 < x_2 < x_3$)\r\n\r\n1. Substitution : $ x_1\\plus{}a$ , $ x_2\\plus{}a$ , $ x_3\\plus{}a$ don\"t change this inecuality , so we can assume : $ x_2\\equal{}0$\r\n\r\n2. It is homogenous inequality => we can assume : $ x_1\\equal{}\\minus{}1$ , ($ x_1 we can assume : $ x_1 \\equal{} \\minus{} 1$ , ($ x_1 < x_2$)[/quote]\r\n\r\nwho can help me with other solution. :)" } { "Tag": [], "Problem": "What are the differences between incandescence, fluorescence, and phosphorescence? How do these relate to quantum mechanics and wave-particle duality?", "Solution_1": "[quote=\"ffdbzathf\"]What are the differences between incandescence, fluorescence, and phosphorescence? How do these relate to quantum mechanics and wave-particle duality?[/quote]\r\n\r\nThing one: A photon is a quantum of light. Its energy is given by the expression $h\\nu$. Quantum mechanics says that physical things, energy, momentum and so on come in discrete bits; i.e. particles, called quanta. So then a particle of light is a photon. Since a photon travels at speed $c=\\lambda\\nu$, light - a group of photons - can be viewed as a wave having wavelength, $\\lambda$.\r\n\r\nIncandescence: the emission of light (energy); i.e. photons, due to a rise in the temperature of the source of the energy.\r\n\r\nPhosphoresence: luminescence which persists after the energy source which produced it is removed.\r\n\r\nFluorescence: luminescence which is not phosphoresence.\r\n\r\nLuminesence: the emission of light other than by incandescence." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c$ be the positive real numbers.Prove that:\r\n$\\sum a(a+2b+2)^{2}\\geq (a+b+c)(a+b+c+2)^{2}$", "Solution_1": "[quote=\"NguyenDungTN\"]Let $a,b,c$ be the positive real numbers.Prove that:\n$\\sum a(a+2b+2)^{2}\\geq (a+b+c)(a+b+c+2)^{2}$[/quote]\r\nThe inequality is equivalent to\r\n\\[\\sum a(a+2b)^{2}+4\\sum a(a+2b)+4\\sum a \\ge (a+b+c)^{3}+4(a+b+c)^{2}+4(a+b+c) \\]\r\nor\r\n\\[\\sum a(a+2b)^{2}\\ge (a+b+c)^{3}\\]\r\nor\r\n\\[\\sum c(a-b)^{2}\\ge 0 \\]\r\nwhich is trivial. We have done. :)" } { "Tag": [], "Problem": "i want to talk to ppl but i can't in 'for the win'", "Solution_1": "what do you mean?", "Solution_2": "seriously,what do u mean? :huh:", "Solution_3": "I think you need to play around 40 games to chat on FTW.\r\n\r\nI thinks that what you asked, right?" } { "Tag": [], "Problem": "Solve the equation : \r\n $ (13+ \\sqrt{x})^{\\frac{1}{3}} + (13- \\sqrt{x})^{\\frac{1}{3}} = y $ \r\n $ x \\in R+ , y \\in Z $", "Solution_1": "Let's try to solve it !", "Solution_2": "(x,y) is either (20188/27,1), (196,2), (123200,3), or (29645/216).\r\n\r\nThe maximum of the left side occurs at x=0 and it is always decreasing. The only integers it passes through are 1, 2, 3, and 4. Setting y equal to these values and solving for x produces those four points." } { "Tag": [ "geometry" ], "Problem": "The length of each side of the larger square is 12 cm. The remaining figures are an inscribed square and two inscribed circles.What is the number of square centimeters in the area of the smaller circle? Express your answer in terms of $ \\pi$.\n\n[asy]size(100); import graph;\ndraw(Circle((0,0),1)); draw((sqrt(2),0)--(0,sqrt(2))--(-sqrt(2),0)--(0,-sqrt(2))--cycle); draw(Circle((0,0),sqrt(2))); draw((sqrt(2),sqrt(2))--(-sqrt(2),sqrt(2))--(-sqrt(2),-sqrt(2))--(sqrt(2),-sqrt(2))--cycle);[/asy]", "Solution_1": "Since the side length of the large square is 12, the diameter of the large circle is 12. 12 is also the diagonal of the small square, so the side length of the small square is $ 6\\sqrt{2}$. That means the radius of the small circle is $ 3\\sqrt{2}$, so the area of the sqmall circle is $ 18\\pi$." } { "Tag": [ "induction" ], "Problem": "At a turnament between $n$ persons, everyone playes exactly one time against everyone else, and at one game there is everytime a winner and a looser.\r\nProve that one can arrange the participants in a chain$P_1 \\to P_2 \\to ... \\to P_n$ such that the $i$-th person has won against the $(i+1)$-th person.", "Solution_1": "Hmm maybe I'm missing something, but I this problem seems rather trivial...\r\n\r\nUsing induction,\r\n\r\nwith only 2 players, if player 1 beats player 2, then we have P1->P2 , otherwise we have P2->P1.\r\n\r\nNow, our induction hypothesis will be that for n players, there is a way of ordering them P1->P2->...Pn-1->Pn .\r\n\r\nSo, now we add 1 more player to these n players, producing an additional n matches. \r\nWe start at the beginning of our ordering. If Pn+1 beat P1, then he will go to the front and the ordering will be Pn+1->P1->P2...->Pn. If he lost to P1, then we see how he did against P2. If he beat P2, then the ordering will be P1->Pn+1->P2->...->Pn. We keep on going in this manner until we find a player that he lost to who is pointing to a player that he beat, and we can put Pn+1 exactly in between these 2 players. If it turns out that he didn't beat anyone, then he just goes to the end, leaving a final ordering of P1->P2->...Pn->Pn+1.\r\n\r\n\r\nThis feels a little bit too easy for this forum, maybe I missed something ? :o" } { "Tag": [], "Problem": "Na deiksete oti den einai dinaton na diameriseis ena kivo se mikra kivakia ola ek ton opoio exoun diaforetiko megethos.\r\n\r\nStergios", "Solution_1": "\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03c0\u03c9\u03c2 \u03ad\u03c7\u03c9 \u03c4\u03b7\u03bd \u03b5\u03b9\u03ba\u03cc\u03bd\u03b1 \u03b1\u03bb\u03bb\u03ac \u03b7 \u03b1\u03bb\u03ae\u03b8\u03b5\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03b5\u03cd\u03bf\u03bc\u03b1\u03b9 \u03c3\u03c4\u03b7 \u03b4\u03b9\u03b1\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7 \u03bb\u03cc\u03b3\u03c9 3 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03c9\u03bd. \r\n\r\n\u03a4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c4\u03bf \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03b5\u03af \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03c0\u03bf\u03c5 \u03b4\u03b5 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03c4\u03bf \u03ba\u03b1\u03bb\u03cd\u03c8\u03b5\u03b9\u03c2 \u03bc\u03b5 \u03ba\u03cd\u03b2\u03bf\u03c5\u03c2 (\u03b1\u03bd\u03b1\u03b3\u03ba\u03b1\u03c3\u03c4\u03b9\u03ba\u03ac \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf\u03c5\u03c2) \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 6 \u03bc\u03b5\u03c1\u03b9\u03ad\u03c2. \u0398\u03b1 \u03bc\u03b5\u03af\u03bd\u03b5\u03b9 \u03ba\u03b5\u03bd\u03cc. \r\n\u038c\u03c4\u03b1\u03bd \u03ba\u03bf\u03bb\u03bb\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c3\u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03ad\u03bd\u03b1 \u03c0\u03b9\u03bf \u03bc\u03b5\u03b3\u03ac\u03bb\u03bf \u03c4\u03cc\u03c4\u03b5 \u03b1\u03bd \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03ac\u03ba\u03b9 \u03c0\u03b5\u03c1\u03b9\u03ad\u03c7\u03b5\u03c4\u03b1\u03b9 \u03bc\u03ad\u03c3\u03b1 \u03c3\u03c4\u03bf \u03bc\u03b5\u03b3\u03ac\u03bb\u03bf \u03c4\u03cc\u03c4\u03b5 \u03c4\u03bf \u03bc\u03b5\u03b3\u03ac\u03bb\u03bf \u03be\u03b5\u03c7\u03b5\u03b9\u03bb\u03af\u03b6\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 4 \u03b1\u03ba\u03bc\u03ad\u03c2. \u0391\u03bd \u03b2\u03ac\u03bb\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03b1\u03c0\u03ad\u03bd\u03b1\u03bd\u03c4\u03b9 \u03b4\u03b5 \u03c7\u03c9\u03c1\u03ac\u03b5\u03b9 \u03bc\u03b5\u03c4\u03ac \u03bd\u03b1 \u03bc\u03c0\u03b5\u03b9 \u03ac\u03bb\u03bb\u03bf \u03c3\u03c4\u03bf \u03ba\u03b5\u03bd\u03cc.\r\n\r\n\u03a4\u03bf \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf \u03c0\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b1\u03c5\u03be\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03c9\u03bd \u03b3\u03b5\u03b9\u03c4\u03cc\u03bd\u03c9\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03ba\u03bf\u03bb\u03bb\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c4\u03b9\u03c2 \u03b3\u03c9\u03bd\u03af\u03b5\u03c2 (\u03bc\u03b5\u03b3\u03ac\u03bb\u03bf\u03c5-\u03bc\u03b9\u03ba\u03c1\u03bf\u03cd \u03ba\u03cd\u03b2\u03bf\u03c5) \u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 \u03c4\u03bf \u03bc\u03b5\u03b3\u03ac\u03bb\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03bd\u03b1 \u03be\u03b5\u03c7\u03b5\u03b9\u03bb\u03af\u03b6\u03b5\u03b9 \u03c4\u03bf \u03bc\u03b5\u03b3\u03ac\u03bb\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b1\u03c0\u03cc 2 \u03b1\u03ba\u03bc\u03ad\u03c2. \u039a\u03b1\u03b9 \u03c0\u03ac\u03bb\u03b9 \u03cc\u03bc\u03c9\u03c2 \u03cc\u03c4\u03b9 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9\u03c2 \u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03af\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03ba\u03b5\u03bd\u03ac \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03c7\u03c9\u03c1\u03ac\u03b5\u03b9 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9. \r\n\r\n\u0394\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03b1\u03bd \u03b2\u03b3\u03ac\u03b6\u03b5\u03b9\u03c2 \u03c4\u03bf \u03bd\u03cc\u03b7\u03bc\u03b1 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03ac \u03c0\u03bf\u03c5 \u03b3\u03c1\u03ac\u03c6\u03c9. \u0395\u03af\u03bd\u03b1\u03b9 \u03bc\u03c0\u03ad\u03c1\u03b4\u03b5\u03bc\u03b1 \u03c7\u03c9\u03c1\u03af\u03c2 \u03c3\u03c7\u03ae\u03bc\u03b1, \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03b1 \u03bd\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03ad\u03bd\u03b1 \u03b6\u03ac\u03c1\u03b9 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03c3\u03b7\u03bc\u03b5\u03b9\u03ce\u03bd\u03b5\u03b9 \u03c0\u03ac\u03bd\u03c9 \u03c4\u03bf\u03c5. \r\n\r\n\u0395\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03c3\u03c4\u03bf \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c3\u03c4\u03b9\u03c2 2 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03bf \u03b5\u03c0\u03b9\u03c7\u03b5\u03af\u03c1\u03b7\u03bc\u03b1 \u03bc\u03b5 \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03b4\u03b5\u03bd \u03b5\u03c6\u03b1\u03c1\u03bc\u03cc\u03b6\u03b5\u03c4\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03ad\u03bd\u03b1 \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03bd\u03b1 \u03c4\u03bf \u03c0\u03b5\u03c1\u03b9\u03bb\u03c5\u03ba\u03bb\u03ce\u03c3\u03b5\u03b9\u03c2 \u03b1\u03c0\u03cc \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03b1 \u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 \u03c4\u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1 \u03bd\u03b1 \u03be\u03b5\u03c7\u03b5\u03b9\u03bb\u03af\u03b6\u03b5\u03b9 \u03b1\u03c0\u03cc \u03bc\u03b9\u03b1 \u03bc\u03cc\u03bd\u03bf \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac", "Solution_2": "Nomizo pos katalaba tin perigrafi sou. Alla einai entiposiako oti auto pou perigrafeis mporei na ginei. Diladi, mporeis na kalipseis ena mikro kivaki me mi-temnomenous kivous olous diaforetikou megethous kai megaliterous apo to arxiko kivaki, xoris na afinoun kanena keno. Einai diskolo na to ferei kapoios sta matia tou os eikona, alla einai mia poli oraia askisi xeirotexnias (Ego basika to dokimasa me xarti :-) )\r\n\r\nEpiseis sosti i paratirisi sou me ta tetragona, gia tin akriveia o antistoixos diamerismos einai dinatos se 2 diastaseis. Paratheto to epomeno link : http://www.srcf.ucam.org/tms/sqsq.php\r\n\r\nStergios", "Solution_3": "\u039c\u03bf\u03c5 \u03b4\u03ce\u03c3\u03b1\u03c4\u03b5 \u03bc\u03b9\u03b1 \u03b9\u03b4\u03ad\u03b1, \u03b5\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c9\u03c3\u03c4\u03ae...\r\n[hide]\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bf \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03b2\u03bf\u03c5 \u03c0\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03b3\u03c1\u03ac\u03c6\u03b5\u03b9 \u03bf Anto.\n\n\u03a0\u03ac\u03c1\u03c4\u03b5 \u03c4\u03b7\u03bd \u03ba\u03ac\u03c4\u03c9 \u03ad\u03b4\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03b2\u03bf\u03c5 \u03ba' \u03ad\u03c3\u03c4\u03c9 $ K_1$ \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03c4\u03b7\u03c2 \u03ad\u03b4\u03c1\u03b1\u03c2.\n\n\u03a4\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03b1\u03bd\u03b1\u03b3\u03ba\u03b1\u03c3\u03c4\u03b9\u03ba\u03ac \u03b4\u03b5 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03ac\u03ba\u03c1\u03b7 \u03b3\u03b9\u03b1\u03c4\u03af \u03b7 \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03b7 \u03ad\u03b4\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03b7 \u03b4\u03c5\u03bd\u03b1\u03c4\u03ae (\u03c3\u03ba\u03b5\u03c6\u03c4\u03b5\u03af\u03c4\u03b5 \u03ad\u03bd\u03b1 \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03b9\u03c3\u03bc\u03ad\u03bd\u03bf a la Anto: \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03ac\u03ba\u03c1\u03b7). \u0386\u03c1\u03b1, \u03c4\u03bf $ K_1$ \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03c0\u03c9\u03c3\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03c0\u03b5\u03c1\u03b9\u03ba\u03c5\u03ba\u03bb\u03c9\u03bc\u03ad\u03bd\u03bf \u03b1\u03c0\u03cc \u03b1\u03c5\u03c3\u03c4\u03b7\u03c1\u03ac \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf\u03c5\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03af\u03b6\u03bf\u03c5\u03bd \u03ad\u03bd\u03b1 \"\u03c4\u03b5\u03af\u03c7\u03bf\u03c2\" \u03b3\u03cd\u03c1\u03c9 \u03c4\u03bf\u03c5. \u0388\u03c4\u03c3\u03b9, \u03c0\u03ac\u03bd\u03c9 \u03b1\u03c0\u03cc \u03c4\u03bf $ K_1$ \u03b8\u03b1 \u03bc\u03c0\u03b1\u03af\u03bd\u03bf\u03c5\u03bd \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03bc\u03cc\u03bd\u03bf \u03b1\u03c5\u03c3\u03c4\u03b7\u03c1\u03ac \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03b1 \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03c3\u03c5\u03bd\u03b1\u03c1\u03bc\u03bf\u03bb\u03bf\u03b3\u03bf\u03cd\u03bd \u03ad\u03bd\u03b1 \u03c4\u03ad\u03bb\u03b5\u03b9\u03b1 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03b9\u03c3\u03bc\u03ad\u03bd\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03c3\u03c4\u03b7\u03bd \u03c0\u03ac\u03bd\u03c9 \u03ad\u03b4\u03c1\u03b1 \u03c4\u03bf\u03c5 $ K_1$.\n\n\u0388\u03c3\u03c4\u03c9 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd $ K_2$ \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03c0\u03bf\u03c5 \u03c3\u03c4\u03ad\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03bf $ K_1$. \u038c\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03c1\u03b9\u03bd, \u03b7 \u03c0\u03ac\u03bd\u03c9 \u03ad\u03b4\u03c1\u03b1 \u03c4\u03bf\u03c5 $ K_2$ \u03b8\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c4\u03ad\u03bb\u03b5\u03b9\u03b1 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03b9\u03c3\u03bc\u03ad\u03bd\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf.\n\nProceed ad nauseum :D\n[/hide]\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_4": "Poli oraia :-)\r\n\r\nAuti einai i idea !\r\n\r\nStergios", "Solution_5": "Sorry \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03ba\u03b1\u03b8\u03c5\u03c3\u03c4\u03ad\u03c1\u03b7\u03c3\u03b7.\r\n\u03a3\u03c4\u03ad\u03c1\u03b3\u03b9\u03bf \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b4\u03af\u03ba\u03b9\u03bf \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b4\u03b9\u03cc\u03c1\u03b8\u03c9\u03c3\u03b7. \u039d\u03b1 \u03ad\u03bd\u03b1\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 \u03bd\u03b1 \u03b4\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03c0\u03c9\u03c2 \u03ad\u03bd\u03b1\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c2 \u03c0\u03b5\u03c1\u03b9\u03ba\u03c5\u03ba\u03bb\u03ce\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf\u03c5\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c5\u03c2 :\r\n\u0392\u03ac\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03bf \u03c0\u03ac\u03c4\u03c9\u03bc\u03b1 \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc \u03ba\u03cd\u03b2\u03bf \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03c4\u03bf\u03bd \u03ba\u03b1\u03bb\u03cd\u03c0\u03c4\u03bf\u03c5\u03bc\u03b5 \u03b3\u03cd\u03c1\u03c9 \u03b3\u03cd\u03c1\u03c9 \u03b1\u03c0\u03cc \u03c0\u03b9\u03bf \u03bc\u03b5\u03b3\u03ac\u03bb\u03bf\u03c5\u03c2 \u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 \u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1\u03c2 \u03bd\u03b1 \u03be\u03b5\u03c7\u03b5\u03b9\u03bb\u03af\u03b6\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b1\u03c0\u03cc \u03bc\u03b9\u03b1 \u03b1\u03ba\u03bc\u03ae.\r\n\u0392\u03bb\u03ad\u03c0\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03bf \u03b1\u03c0\u03cc \u03c0\u03ac\u03bd\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03b1\u03bd \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03ad\u03bd\u03b1 \u03c7\u03b1\u03bc\u03b7\u03bb\u03cc \u03ba\u03c4\u03af\u03c1\u03b9\u03bf \u03c0\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03ba\u03c5\u03ba\u03bb\u03ce\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c8\u03b7\u03bb\u03cc\u03c4\u03b5\u03c1\u03b5\u03c2 \u03c0\u03bf\u03bb\u03c5\u03ba\u03b1\u03c4\u03bf\u03b9\u03ba\u03af\u03b5\u03c2.\r\n\r\n\u0396\u03bf\u03c5\u03bc\u03c0\u03ac\u03bc\u03b5 \u03bc\u03ad\u03c3\u03b1 \u03c3\u03c4\u03b7 \u03b3\u03b7 2 \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ad\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03c0\u03bf\u03bb\u03c5\u03ba\u03b1\u03c4\u03bf\u03b9\u03ba\u03af\u03b5\u03c2 \u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03c6\u03c4\u03ac\u03c3\u03bf\u03c5\u03bd \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03b1\u03c1\u03ac\u03c4\u03c3\u03b1 \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03cd\u03c8\u03bf\u03c2 \u03bc\u03b5 \u03c4\u03bf \u03ba\u03c4\u03af\u03c1\u03b9\u03cc \u03bc\u03b1\u03c2. \u03a4\u03ce\u03c1\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c7\u03ce\u03c1\u03bf\u03c2 \u03b3\u03b9\u03b1 \u03bd\u03b1\u03ba\u03bf\u03bb\u03bb\u03ae\u03c3\u03b5\u03b9 \u03b1\u03c0\u03cc \u03c0\u03ac\u03bd\u03c9 \u03ba\u03b1\u03b9 \u03ad\u03bd\u03b1\u03c2 \u03b1\u03ba\u03cc\u03bc\u03b1 \u03ba\u03cd\u03b2\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf \u03c5\u03c0\u03cc\u03b3\u03b5\u03b9\u03bf \u03c0\u03bf\u03c5 \u03be\u03b5\u03c7\u03c9\u03c1\u03af\u03b6\u03bf\u03c5\u03bd \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b1 \u03ba\u03ac\u03c4\u03c9 \u03bf\u03b9 \u03b2\u03c5\u03b8\u03b9\u03c3\u03bc\u03ad\u03bd\u03bf\u03b9 \u03ba\u03cd\u03b2\u03bf\u03b9. \r\n\r\n\u03a4\u03bf\u03c5 \u03a0\u03b1\u03bd\u03b1\u03b3\u03b9\u03ce\u03c4\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u039f\u039a. \u038a\u03c3\u03c9\u03c2 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b1\u03c6\u03bf\u03cd \u03b1\u03c5\u03c4\u03cc \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03c6\u03bf\u03c1\u03ac \u03ba\u03b1\u03c4\u03b5\u03b2\u03b1\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c4\u03ac\u03be\u03b7 \u03bc\u03b5\u03b3\u03ad\u03b8\u03bf\u03c5\u03c2 \u03b5\u03c0\u2019 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf, \u03c4\u03cc\u03c4\u03b5 \u03bc\u03b1\u03c2 \u03b2\u03ac\u03b6\u03b5\u03b9 \u03c6\u03c1\u03ac\u03b3\u03bc\u03b1 \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03b1\u03c0\u03cc \u03cc\u03bb\u03b1 \u03b1\u03c6\u03bf\u03cd \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c0\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03ac\u03c4\u03c9 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03cc. \u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03c0\u03c9\u03c2 \u03b1\u03bd \u03bf\u03b9 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03af\u03c3\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b5\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03cc\u03bd\u03c4\u03c9\u03c2 \u03c4\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9. \u0391\u03bd \u03b5\u03c0\u03b9\u03c4\u03c1\u03ad\u03c0\u03b5\u03c4\u03b1\u03b9 \u03bf\u03b9 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03af\u03c3\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b5\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03c0\u03c9\u03c2 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c4\u03bf \u03b5\u03c0\u03b9\u03c7\u03b5\u03af\u03c1\u03b7\u03bc\u03b1.", "Solution_6": "[quote=\"gpapargi\"]\u03a3\u03c4\u03ad\u03c1\u03b3\u03b9\u03bf \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b4\u03af\u03ba\u03b9\u03bf \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b4\u03b9\u03cc\u03c1\u03b8\u03c9\u03c3\u03b7. \u039d\u03b1 \u03ad\u03bd\u03b1\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 \u03bd\u03b1 \u03b4\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03c0\u03c9\u03c2 \u03ad\u03bd\u03b1\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c2 \u03c0\u03b5\u03c1\u03b9\u03ba\u03c5\u03ba\u03bb\u03ce\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf\u03c5\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c5\u03c2 :\n\u0392\u03ac\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03bf \u03c0\u03ac\u03c4\u03c9\u03bc\u03b1 \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc \u03ba\u03cd\u03b2\u03bf \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03c4\u03bf\u03bd \u03ba\u03b1\u03bb\u03cd\u03c0\u03c4\u03bf\u03c5\u03bc\u03b5 \u03b3\u03cd\u03c1\u03c9 \u03b3\u03cd\u03c1\u03c9 \u03b1\u03c0\u03cc \u03c0\u03b9\u03bf \u03bc\u03b5\u03b3\u03ac\u03bb\u03bf\u03c5\u03c2 \u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 \u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1\u03c2 \u03bd\u03b1 \u03be\u03b5\u03c7\u03b5\u03b9\u03bb\u03af\u03b6\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b1\u03c0\u03cc \u03bc\u03b9\u03b1 \u03b1\u03ba\u03bc\u03ae.\n\u0392\u03bb\u03ad\u03c0\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03bf \u03b1\u03c0\u03cc \u03c0\u03ac\u03bd\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03b1\u03bd \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03ad\u03bd\u03b1 \u03c7\u03b1\u03bc\u03b7\u03bb\u03cc \u03ba\u03c4\u03af\u03c1\u03b9\u03bf \u03c0\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03ba\u03c5\u03ba\u03bb\u03ce\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c8\u03b7\u03bb\u03cc\u03c4\u03b5\u03c1\u03b5\u03c2 \u03c0\u03bf\u03bb\u03c5\u03ba\u03b1\u03c4\u03bf\u03b9\u03ba\u03af\u03b5\u03c2.\n\n\u0396\u03bf\u03c5\u03bc\u03c0\u03ac\u03bc\u03b5 \u03bc\u03ad\u03c3\u03b1 \u03c3\u03c4\u03b7 \u03b3\u03b7 2 \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ad\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03c0\u03bf\u03bb\u03c5\u03ba\u03b1\u03c4\u03bf\u03b9\u03ba\u03af\u03b5\u03c2 \u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03c6\u03c4\u03ac\u03c3\u03bf\u03c5\u03bd \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03b1\u03c1\u03ac\u03c4\u03c3\u03b1 \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03cd\u03c8\u03bf\u03c2 \u03bc\u03b5 \u03c4\u03bf \u03ba\u03c4\u03af\u03c1\u03b9\u03cc \u03bc\u03b1\u03c2. \u03a4\u03ce\u03c1\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c7\u03ce\u03c1\u03bf\u03c2 \u03b3\u03b9\u03b1 \u03bd\u03b1\u03ba\u03bf\u03bb\u03bb\u03ae\u03c3\u03b5\u03b9 \u03b1\u03c0\u03cc \u03c0\u03ac\u03bd\u03c9 \u03ba\u03b1\u03b9 \u03ad\u03bd\u03b1\u03c2 \u03b1\u03ba\u03cc\u03bc\u03b1 \u03ba\u03cd\u03b2\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf \u03c5\u03c0\u03cc\u03b3\u03b5\u03b9\u03bf \u03c0\u03bf\u03c5 \u03be\u03b5\u03c7\u03c9\u03c1\u03af\u03b6\u03bf\u03c5\u03bd \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b1 \u03ba\u03ac\u03c4\u03c9 \u03bf\u03b9 \u03b2\u03c5\u03b8\u03b9\u03c3\u03bc\u03ad\u03bd\u03bf\u03b9 \u03ba\u03cd\u03b2\u03bf\u03b9. \n[/quote]\n\n\u03a9\u03c1\u03b1\u03af\u03b1 \u03c0\u03b5\u03c1\u03b9\u03b3\u03c1\u03b1\u03c6\u03ae. \u039b\u03af\u03b3\u03bf \u03bc\u03c0\u03ad\u03c1\u03b4\u03b5\u03bc\u03b1, eh? ;)\n\n[quote]\n\u03a4\u03bf\u03c5 \u03a0\u03b1\u03bd\u03b1\u03b3\u03b9\u03ce\u03c4\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u039f\u039a. \u038a\u03c3\u03c9\u03c2 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b1\u03c6\u03bf\u03cd \u03b1\u03c5\u03c4\u03cc \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03c6\u03bf\u03c1\u03ac \u03ba\u03b1\u03c4\u03b5\u03b2\u03b1\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c4\u03ac\u03be\u03b7 \u03bc\u03b5\u03b3\u03ad\u03b8\u03bf\u03c5\u03c2 \u03b5\u03c0\u2019 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf, \u03c4\u03cc\u03c4\u03b5 \u03bc\u03b1\u03c2 \u03b2\u03ac\u03b6\u03b5\u03b9 \u03c6\u03c1\u03ac\u03b3\u03bc\u03b1 \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03b1\u03c0\u03cc \u03cc\u03bb\u03b1 \u03b1\u03c6\u03bf\u03cd \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c0\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03ac\u03c4\u03c9 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03cc. \u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03c0\u03c9\u03c2 \u03b1\u03bd \u03bf\u03b9 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03af\u03c3\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b5\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03cc\u03bd\u03c4\u03c9\u03c2 \u03c4\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9. \u0391\u03bd \u03b5\u03c0\u03b9\u03c4\u03c1\u03ad\u03c0\u03b5\u03c4\u03b1\u03b9 \u03bf\u03b9 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03af\u03c3\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b5\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03c0\u03c9\u03c2 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c4\u03bf \u03b5\u03c0\u03b9\u03c7\u03b5\u03af\u03c1\u03b7\u03bc\u03b1.[/quote]\r\n\r\n\u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03c3\u03b5 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b7 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c1\u03b9\u03c3\u03b7 \u03b1\u03bd\u03b1\u03c6\u03b5\u03c1\u03cc\u03bc\u03bf\u03c5\u03bd, \u03c4\u03bf ad nauseum \u03b5\u03ba\u03b5\u03af \u03c0\u03ae\u03b3\u03b1\u03b9\u03bd\u03b5 :P\r\n\r\nThat said... think you can do an infinite partition? :)\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_7": "\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c1\u03b9\u03c3\u03b7 \u03bc\u03b5 \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03c5\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c5\u03c2. \r\n\u03a6\u03b1\u03bd\u03c4\u03ac\u03b6\u03bf\u03bc\u03b1\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf:\r\n\u03a0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9\u03c2 \u03c4\u03bf \u03bc\u03b5\u03b3\u03ac\u03bb\u03bf \u03ba\u03cd\u03b2\u03bf \u03ba\u03b1\u03b9 \u03c7\u03c9\u03c1\u03af\u03b6\u03b5\u03b9\u03c2 \u03ba\u03ac\u03b8\u03b5 \u03b1\u03ba\u03bc\u03ae \u03c3\u03b5 2 \u03af\u03c3\u03b1 \u03bc\u03ad\u03c1\u03b7. \u0388\u03c4\u03c3\u03b9 \u03b2\u03ac\u03b6\u03b5\u03b9\u03c2 \u03bc\u03ad\u03c3\u03b1 \u03c3\u03c4\u03bf \u03bc\u03b5\u03b3\u03ac\u03bb\u03bf \u03ba\u03cd\u03b2\u03bf $ 2^3$ \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1. \u03a4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03c4\u03bf \u03ba\u03c1\u03b1\u03c4\u03ac\u03c2. \u03a4\u03b1 \u03ac\u03bb\u03bb\u03b1 7 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c4\u03b1 \u03b1\u03bb\u03bb\u03ac\u03be\u03b5\u03b9\u03c2 \u03b3\u03b9\u03b1\u03c4\u03af \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c5\u03c2 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03bf\u03cd \u03bc\u03b5\u03b3\u03ad\u03b8\u03bf\u03c5\u03c2.\r\n\r\n\u03a4\u03bf\u03bd \u03b5\u03c0\u03cc\u03bc\u03b5\u03bd\u03bf \u03b8\u03b1 \u03c4\u03bf\u03bd \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03bc\u03b5 \u03ac\u03bb\u03bb\u03b1 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03cc\u03bb\u03b1 \u03af\u03c3\u03b1 \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03bf\u03c5\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7 \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 \u03c4\u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ac \u03b8\u03b1 \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03b1\u03b8\u03b5\u03af \u03bc\u03b5 \u03ac\u03bb\u03bb\u03b1 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03b1 \u03cc\u03bb\u03b1 \u03af\u03c3\u03b1 \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03bf\u03c5\u03c2. \u039a\u03ac\u03b8\u03b5 \u03c6\u03bf\u03c1\u03ac \u03c4\u03b5\u03bb\u03b5\u03b9\u03ce\u03bd\u03b5\u03b9\u03c2 \u03ad\u03bd\u03b1 \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf (\u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03c4\u03b7\u03c2 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c1\u03b9\u03c3\u03b7\u03c2 \u03c4\u03bf\u03c5 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5) \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03ba\u03b1\u03c4\u03b5\u03b2\u03b1\u03af\u03bd\u03b5\u03b9\u03c2 \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf (\u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03c0\u03c1\u03ce\u03c4\u03b1 \u03b7 \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae 7\u03ac\u03b4\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03af\u03b6\u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd \u03ba\u03ac\u03b8\u03b5 \u03bc\u03b9\u03b1 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ad\u03c2). \u0391\u03c5\u03c4\u03cc \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c0\u03b5\u03c1\u03ac\u03c3\u03b5\u03b9\u03c2 \u03b1\u03c0\u03cc \u03cc\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03be\u03b5\u03bc\u03b5\u03af\u03bd\u03b5\u03b9 \u03ba\u03ac\u03c4\u03b9. \r\n\u039a\u03ac\u03b8\u03b5 \u03c6\u03bf\u03c1\u03ac \u03c7\u03c9\u03c1\u03af\u03b6\u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd \u03b1\u03ba\u03bc\u03ae \u03c3\u03b5 p \u03af\u03c3\u03b1 \u03ba\u03bf\u03bc\u03bc\u03ac\u03c4\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9\u03c2 $ p^3$ \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1.\r\n\u0393\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03af\u03b4\u03b9\u03bf \u03bc\u03ad\u03b3\u03b5\u03b8\u03bf\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c5 \u03ba\u03cd\u03b2\u03bf\u03c5 \u03bc\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03bd \u03ac\u03bb\u03bb\u03bf (\u03ad\u03c3\u03c4\u03c9 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf \u03bc\u03ad\u03bb\u03bb\u03bf\u03bd \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03af\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5\u03c2), \u03ba\u03ac\u03b8\u03b5 \u03c6\u03bf\u03c1\u03ac \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 p \u03c0\u03c1\u03ce\u03c4\u03bf. \u0391\u03c6\u03bf\u03cd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03b9 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9. \r\n\r\n\u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03bc\u03b7 \u03bc\u03bf\u03c5 \u03be\u03ad\u03c6\u03c5\u03b3\u03b5 \u03ba\u03ac\u03c4\u03b9", "Solution_8": "gpapargi, \u03c6\u03bf\u03b2\u03ac\u03bc\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03b4\u03b5\u03bd \u03b4\u03bf\u03c5\u03bb\u03b5\u03cd\u03b5\u03b9 \u03b7 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03ae \u03c3\u03bf\u03c5. \u03a4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03ac \u03c4\u03b1 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03b4\u03b5\u03bd \u03b1\u03c1\u03ba\u03bf\u03cd\u03bd \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03af\u03c3\u03b5\u03b9\u03c2 \u03cc\u03bb\u03bf \u03c4\u03bf\u03bd \u03ba\u03cd\u03b2\u03bf.\r\n\r\n\u0391\u03bd \u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03bc\u03ae\u03ba\u03bf\u03c2 1, \u03c4\u03cc\u03c4\u03b5 \u03c4\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03c3\u03bf\u03c5 \u03ad\u03c7\u03bf\u03c5\u03bd \u03bc\u03ae\u03ba\u03b7 1/2,1/3,1/4,.... \u0391\u03bb\u03bb\u03ac \u03c4\u03cc\u03c4\u03b5, \u03ba\u03b1\u03bb\u03cd\u03c0\u03c4\u03bf\u03c5\u03bd \u03c3\u03c5\u03bd\u03bf\u03bb\u03b9\u03ba\u03cc \u03cc\u03b3\u03ba\u03bf $ (1/2)^3 \\plus{} (1/3)^3 \\plus{} \\ldots \\equal{} 0.202... < 1$.", "Solution_9": "[quote=\"Demetres\"]gpapargi, \u03c6\u03bf\u03b2\u03ac\u03bc\u03b1\u03b9 \u03c0\u03c9\u03c2 \u03b4\u03b5\u03bd \u03b4\u03bf\u03c5\u03bb\u03b5\u03cd\u03b5\u03b9 \u03b7 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03ae \u03c3\u03bf\u03c5. \u03a4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03ac \u03c4\u03b1 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03b4\u03b5\u03bd \u03b1\u03c1\u03ba\u03bf\u03cd\u03bd \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03af\u03c3\u03b5\u03b9\u03c2 \u03cc\u03bb\u03bf \u03c4\u03bf\u03bd \u03ba\u03cd\u03b2\u03bf.\n\n\u0391\u03bd \u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03bc\u03ae\u03ba\u03bf\u03c2 1, \u03c4\u03cc\u03c4\u03b5 \u03c4\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03c3\u03bf\u03c5 \u03ad\u03c7\u03bf\u03c5\u03bd \u03bc\u03ae\u03ba\u03b7 1/2,1/3,1/4,.... \u0391\u03bb\u03bb\u03ac \u03c4\u03cc\u03c4\u03b5, \u03ba\u03b1\u03bb\u03cd\u03c0\u03c4\u03bf\u03c5\u03bd \u03c3\u03c5\u03bd\u03bf\u03bb\u03b9\u03ba\u03cc \u03cc\u03b3\u03ba\u03bf $ (1/2)^3 \\plus{} (1/3)^3 \\plus{} \\ldots \\equal{} 0.202... < 1$.[/quote]\r\n\r\nH \u03b1\u03bb\u03ae\u03b8\u03b5\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03ba\u03b9 \u03b5\u03b3\u03ce \u03b4\u03b5\u03bd \u03ba\u03b1\u03c4\u03ac\u03bb\u03b1\u03b2\u03b1 \u03c0\u03ce\u03c2 \u03b3\u03b5\u03bc\u03af\u03b6\u03b5\u03b9 \u03cc\u03bb\u03bf\u03c2 \u03bf \u03ba\u03cd\u03b2\u03bf\u03c2...\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_10": "\u03a0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b4\u03af\u03ba\u03b9\u03bf Demetres. \u038c\u03c6\u03b5\u03b9\u03bb\u03b1 \u03bd\u03b1 \u03b4\u03c9 \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b1\u03bd \u03c6\u03c4\u03ac\u03bd\u03b5\u03b9 \u03c3\u03c4\u03bf 1. \u0389\u03c4\u03b1\u03bd \u03c4\u03cc\u03c3\u03bf \u03b1\u03c0\u03bb\u03cc \u03c4\u03b5\u03c3\u03c4 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03b2\u03b9\u03b1\u03c3\u03cd\u03bd\u03b7 \u03c4\u03bf \u03c0\u03b1\u03c1\u03ac\u03bb\u03b5\u03b9\u03c8\u03b1. \u0394\u03c5\u03c3\u03c4\u03c5\u03c7\u03ce\u03c2 \u03c4\u03bf \u03bc\u03ad\u03b3\u03b5\u03b8\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03b2\u03bf\u03c5 \u03b5\u03bb\u03b1\u03c4\u03c4\u03ce\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c0\u03ac\u03c1\u03b1 \u03c0\u03bf\u03bb\u03cd \u03b3\u03c1\u03ae\u03b3\u03bf\u03c1\u03b1. \u0398\u03c5\u03bc\u03af\u03b6\u03b5\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03b1\u03c0\u03cc \u0396\u03ae\u03bd\u03c9\u03bd\u03b1.\r\n\r\n\u0395\u03b4\u03ce \u03c0\u03bf\u03c5 \u03c4\u03b1 \u03bb\u03ad\u03bc\u03b5, \u03b1\u03bd \u03c0\u03b5\u03c1\u03b9\u03b3\u03c1\u03b1\u03c6\u03b5\u03af \u03bc\u03b9\u03b1 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c1\u03b9\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b4\u03af\u03bd\u03b5\u03b9 \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c0\u03ac\u03bd\u03c9 \u03b1\u03c0\u03cc 1 \u03b1\u03c5\u03c4\u03cc \u03bc\u03b1\u03c2 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03ae \u03b8\u03ad\u03bb\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b4\u03af\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 1; \u039c\u03ac\u03bb\u03bb\u03bf\u03bd \u03c4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf. \u0388\u03c7\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03c5\u03c0\u03cc\u03c8\u03b7 \u03c4\u03bf\u03c5 \u03b1\u03bd \u03bb\u03cd\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c1\u03b9\u03c3\u03b7 \u03ae \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03b5\u03c4\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 (\u03bb\u03c7 \u03bc\u03b9\u03b1 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03bc\u03b5 \u03ac\u03c4\u03bf\u03c0\u03bf \u03c0\u03c7 \u03cc\u03c4\u03b9 \u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b7 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c1\u03b9\u03c3\u03b7 \u03c4\u03cc\u03c4\u03b5 \u03b8\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b7 \u03c0\u03bf\u03c5 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03c7\u03c4\u03b7\u03ba\u03b5 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9); \u03a1\u03c9\u03c4\u03ac\u03c9 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03be\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03c4\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03bc\u03c0\u03bb\u03ad\u03be\u03b5\u03b9. :)", "Solution_11": "\u03a3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03b8\u03b5\u03c2 \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03bd\u03b1 \u03b9\u03c3\u03bf\u03cd\u03c4\u03b1\u03b9 \u03bc\u03b5 1. (\u0391\u03ba\u03cc\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03cc. \u03a0\u03ce\u03c2 \u03b8\u03b1 \u03b5\u03af\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03b9 \u03b3\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c0\u03b1\u03c1\u03b1\u03bb\u03b5\u03af\u03c8\u03b5\u03b9 \u03bd\u03b1 \u03ba\u03b1\u03bb\u03cd\u03c8\u03bf\u03c5\u03bc\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf (\u03bc\u03ae \u03ba\u03b5\u03bd\u03cc) \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd \u03bc\u03b5 \u03c3\u03c5\u03bd\u03bf\u03bb\u03b9\u03ba\u03cc \u03cc\u03b3\u03ba\u03bf 0; )\r\n\r\n\u039c\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03b1\u03c1\u03ba\u03b5\u03c4\u03ae \u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b5\u03b9\u03b1 \u03ad\u03c7\u03c9 \u03bb\u03cd\u03c3\u03b5\u03b9 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1. \u0394\u03b5\u03bd \u03b8\u03b1 \u03c0\u03c9 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf \u03c0\u03b1\u03c1\u03cc\u03bd \u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c1\u03b9\u03c3\u03b7 \u03ae \u03cc\u03c7\u03b9, \u03b4\u03b9\u03cc\u03c4\u03b9 \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03bc\u03b5\u03b3\u03ac\u03bb\u03b7 \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03b1. \u0398\u03b1 \u03b1\u03c1\u03ba\u03b5\u03c3\u03c4\u03ce \u03bd\u03b1 \u03c0\u03c9 \u03cc\u03c4\u03b9 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b1\u03c0\u03bb\u03ae.\r\n\r\n------------------------\r\n\r\n\u0391\u03c0\u03bf\u03c1\u03af\u03b1 formatting: \u03a3\u03c4\u03bf \u03c4\u03ad\u03bb\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03c0\u03c1\u03ce\u03c4\u03b7\u03c2 \u03c0\u03b1\u03c1\u03b1\u03b3\u03c1\u03ac\u03c6\u03bf\u03c5, \u03b8\u03ad\u03bb\u03c9 \u03bd\u03b1 \u03ba\u03bb\u03b5\u03af\u03c3\u03c9 \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03ad\u03bd\u03b8\u03b5\u03c3\u03b7 \u03b1\u03bc\u03ad\u03c3\u03c9\u03c2 \u03bc\u03b5\u03c4\u03ac \u03c4\u03bf \u03b5\u03c1\u03c9\u03c4\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc, \u03b1\u03bb\u03bb\u03ac \u03bc\u03bf\u03c5 \u03b2\u03b3\u03ac\u03b6\u03b5\u03b9 ;) . \u03a0\u03ce\u03c2 \u03c4\u03bf \u03b1\u03c0\u03bf\u03c6\u03b5\u03cd\u03b3\u03c9 \u03b1\u03c5\u03c4\u03cc;", "Solution_12": "[quote=\"Demetres\"]\u03a3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03b8\u03b5\u03c2 \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03bd\u03b1 \u03b9\u03c3\u03bf\u03cd\u03c4\u03b1\u03b9 \u03bc\u03b5 1. (\u0391\u03ba\u03cc\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03cc. \u03a0\u03ce\u03c2 \u03b8\u03b1 \u03b5\u03af\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03b9 \u03b3\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c0\u03b1\u03c1\u03b1\u03bb\u03b5\u03af\u03c8\u03b5\u03b9 \u03bd\u03b1 \u03ba\u03b1\u03bb\u03cd\u03c8\u03bf\u03c5\u03bc\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf (\u03bc\u03ae \u03ba\u03b5\u03bd\u03cc) \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd \u03bc\u03b5 \u03c3\u03c5\u03bd\u03bf\u03bb\u03b9\u03ba\u03cc \u03cc\u03b3\u03ba\u03bf 0; )\n\n\u039c\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03b1\u03c1\u03ba\u03b5\u03c4\u03ae \u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b5\u03b9\u03b1 \u03ad\u03c7\u03c9 \u03bb\u03cd\u03c3\u03b5\u03b9 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1. \u0394\u03b5\u03bd \u03b8\u03b1 \u03c0\u03c9 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf \u03c0\u03b1\u03c1\u03cc\u03bd \u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c1\u03b9\u03c3\u03b7 \u03ae \u03cc\u03c7\u03b9, \u03b4\u03b9\u03cc\u03c4\u03b9 \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03bc\u03b5\u03b3\u03ac\u03bb\u03b7 \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03b1. \u0398\u03b1 \u03b1\u03c1\u03ba\u03b5\u03c3\u03c4\u03ce \u03bd\u03b1 \u03c0\u03c9 \u03cc\u03c4\u03b9 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b1\u03c0\u03bb\u03ae.[/quote]\r\n[hide=\"Hint\"]\u03a3\u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 quote \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ae\u03b4\u03b7 \u03ad\u03bd\u03b1 hint ;)[/hide]\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_13": "[quote=\"Demetres\"]\u03a3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03b8\u03b5\u03c2 \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03bd\u03b1 \u03b9\u03c3\u03bf\u03cd\u03c4\u03b1\u03b9 \u03bc\u03b5 1. (\u0391\u03ba\u03cc\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03cc. \u03a0\u03ce\u03c2 \u03b8\u03b1 \u03b5\u03af\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03b9 \u03b3\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c0\u03b1\u03c1\u03b1\u03bb\u03b5\u03af\u03c8\u03b5\u03b9 \u03bd\u03b1 \u03ba\u03b1\u03bb\u03cd\u03c8\u03bf\u03c5\u03bc\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf (\u03bc\u03ae \u03ba\u03b5\u03bd\u03cc) \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd \u03bc\u03b5 \u03c3\u03c5\u03bd\u03bf\u03bb\u03b9\u03ba\u03cc \u03cc\u03b3\u03ba\u03bf 0; )[/quote]\r\n\r\n\u039a\u03b1\u03bd\u03b5\u03af\u03c2 \u03b2\u03c1\u03b5 guys?\r\n\r\nTo\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b5\u03c5\u03ba\u03bf\u03bb\u03cc\u03c4\u03b5\u03c1\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03b8\u03af\u03b3\u03b5\u03b9 \u03bf Demetres, \u03c4\u03b9 \u03bb\u03ad\u03c4\u03b5? \u0388\u03c7\u03b5\u03b9 \u03b2\u03c1\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03bc\u03af\u03b1 \u03bc\u03b7 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b7 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c1\u03b9\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03b2\u03bf\u03c5 \u03c3\u03b5 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ce\u03bd \u03bc\u03b5\u03b3\u03b5\u03b8\u03ce\u03bd \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03ba\u03b1\u03bb\u03cd\u03c0\u03c4\u03bf\u03c5\u03bd \u03cc\u03bb\u03bf\u03bd \u03c4\u03bf\u03bd [i]\u03cc\u03b3\u03ba\u03bf[/i] \u03c4\u03bf\u03c5 \u03b1\u03c1\u03c7\u03b9\u03ba\u03bf\u03cd \u03ba\u03cd\u03b2\u03bf\u03c5?\r\n\r\nDemetres, \u03b1\u03bd \u03b4\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03c0\u03ac\u03b5\u03b9 \u03bd\u03b1 \u03b8\u03b1\u03c6\u03c4\u03b5\u03af \u03c4\u03bf thread, \u03b2\u03ac\u03bb\u03b5 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 :)\r\n\r\nCheerio,\r\n\r\nDurandal 1707\r\n\r\nEdit: \u03bd\u03b1 \u03b2\u03ac\u03bb\u03c9 \u03ad\u03bd\u03b1 [hide=\"hint.\"]Just do it :)[/hide]", "Solution_14": "\u039a\u03b1\u03bb\u03b7\u03c3\u03c0\u03b5\u03c1\u03b1 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03bf \u03b5\u03bc\u03b5\u03bd\u03b1 \u03bc\u03b5\u03c4\u03b1 \u03b1\u03c0\u03bf \u03b1\u03c1\u03ba\u03b5\u03c4\u03bf \u03ba\u03b1\u03b9\u03c1\u03bf. \u0394\u03c5\u03c3\u03c4\u03c5\u03c7\u03c9\u03c2 \u03bf\u03b9 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03b5\u03c2 \u03c5\u03c0\u03bf\u03c7\u03c1\u03b5\u03c9\u03c3\u03b5\u03b9\u03c2 \u03b4\u03b5\u03bd \u03bc\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c4\u03c1\u03b5\u03c0\u03bf\u03c5\u03bd \u03bd\u03b1 \u03b5\u03c0\u03b9\u03c3\u03ba\u03b5\u03c0\u03c4\u03bf\u03bc\u03b1\u03b9 \u03c4\u03b7\u03bd \u03bc\u03b9\u03ba\u03c1\u03b7 \u03bc\u03b1\u03c2 \u03ba\u03bf\u03b9\u03bd\u03bf\u03c4\u03b7\u03c4\u03b1 \u03bf\u03c3\u03bf \u03c3\u03c5\u03c7\u03bd\u03b1 \u03b8\u03b1 \u03b7\u03b8\u03b5\u03bb\u03b1.\r\n\u03a4\u03b5\u03c3\u03c0\u03b1 \u03b5\u03b9\u03b4\u03b1 \u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03c3\u03ba\u03b5\u03c6\u03c4\u03b7\u03ba\u03b1 \u03b5\u03bd\u03b1 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03bf \u03c4\u03bf\u03c5, \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b5\u03c7\u03c9 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b7\u03c3\u03b5\u03b9 \u03b1\u03ba\u03bf\u03bc\u03b7 \u03bd\u03b1 \u03bb\u03c5\u03c3\u03c9. \u03a4\u03bf \u03b3\u03c1\u03b1\u03c6\u03c9 \u03bc\u03c0\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b8\u03b5\u03bb\u03b5\u03c4\u03b5 \u03bd\u03b1 \u03c4\u03bf \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b7\u03c3\u03b5\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03b5\u03c4\u03c3\u03b9.\r\n\r\n\u039d\u03b1 \u03b4\u03b5\u03b9\u03be\u03b5\u03c4\u03b5 \u03bf\u03c4\u03b9 \u03b1\u03bd \u03c0\u03b1\u03c1\u03bf\u03c5\u03bc\u03b5 \u03ba\u03c5\u03b2\u03bf\u03c5\u03c2 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03bf\u03c5 \u03bc\u03b5\u03b3\u03b5\u03b8\u03bf\u03c5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03b5\u03bd\u03c9\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03c4\u03b5\u03c4\u03bf\u03b9\u03bf \u03c4\u03c1\u03bf\u03c0\u03bf \u03c9\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03b7\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b5\u03bd\u03b1 \u03bd\u03b5\u03bf \u03ba\u03c5\u03b2\u03bf \u03bf\u03c0\u03bf\u03b9\u03bf\u03bd\u03b4\u03b7\u03c0\u03bf\u03c4\u03b5 \u03b4\u03b9\u03b1\u03c4\u03b1\u03c3\u03b5\u03c9\u03bd. \r\n\r\n\u03a5.\u0393. \u039f\u03b9 \u03ba\u03c5\u03b2\u03bf\u03b9 \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03c0\u03b1\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bc\u03c0\u03bf\u03c1\u03bf\u03c5\u03bd \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03c3\u03bf\u03b9 \u03b8\u03b5\u03bb\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03b9 \u03bf\u03c0\u03bf\u03b9\u03bf\u03c5 \u03bc\u03b5\u03b3\u03b5\u03b8\u03bf\u03c5\u03c2 \u03b8\u03b5\u03bb\u03bf\u03c5\u03bc\u03b5(\u03b1\u03c1\u03ba\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03bf\u03c5 \u03bc\u03b5\u03b3\u03b5\u03b8\u03bf\u03c5\u03c2 \u03bc\u03b5\u03c4\u03b1\u03be\u03c5 \u03c4\u03bf\u03c5\u03c2 \u03bf\u03c0\u03c9\u03c2 \u03bb\u03b5\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bf \u039d\u03c4\u03bf\u03c5\u03c1\u03b1\u03bd\u03c4\u03b1\u03bb).", "Solution_15": "[quote=\"Anto\"]Na deiksete oti den einai dinaton na diameriseis ena kivo se mikra kivakia ola ek ton opoio exoun diaforetiko megethos.\n\nStergios[/quote]\r\n\r\nSygnwmh pou grafw se greeklish alla o sygkekrimenos ypologisths de mou parexei ellhnika. Den exw diabasei thn ypoloiph koubenta, alla me mia grhgorh matia, den einai to teleftaio 8ewrhma tou Ferma gia n=3?", "Solution_16": "\u039a\u03cd\u03c1\u03b9\u03bf\u03b9, \u03c0\u03c1\u03bf\u03c3\u03bf\u03c7\u03ae:\r\n\r\n[b]@basted[/b]\r\n[quote=\"basted\"][quote=\"Anto\"]Na deiksete oti den einai dinaton na diameriseis ena kivo se mikra kivakia ola ek ton opoio exoun diaforetiko megethos.\n\nStergios[/quote]\n\nSygnwmh pou grafw se greeklish alla o sygkekrimenos ypologisths de mou parexei ellhnika. Den exw diabasei thn ypoloiph koubenta, alla me mia grhgorh matia, den einai to teleftaio 8ewrhma tou Ferma gia n=3?[/quote]\n\n1. \u0394\u03b5 \u03b3\u03c1\u03ac\u03c6\u03bf\u03c5\u03bc\u03b5 \u03ad\u03bd\u03b1\u03bd \u03ba\u03cd\u03b2\u03bf \u03c9\u03c2 \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b4\u03cd\u03bf \u03ba\u03cd\u03b2\u03c9\u03bd \u03b1\u03bb\u03bb\u03ac \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03bf\u03c4\u03ad\u03c1\u03c9\u03bd.\n2. \u03a0\u03b9\u03bf \u03c3\u03b7\u03bc\u03b1\u03bd\u03c4\u03b9\u03ba\u03cc: \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03b4\u03b5 \u03bc\u03b1\u03c2 \u03c0\u03b5\u03c1\u03b9\u03bf\u03c1\u03af\u03b6\u03b5\u03b9 \u03c3\u03b5 \u03c1\u03b7\u03c4\u03ac \u03bc\u03ae\u03ba\u03b7 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ce\u03bd.\n\n\n[b]@filosofimenos[/b]\n[quote=\"filosofimenos\"]\u039d\u03b1 \u03b4\u03b5\u03b9\u03be\u03b5\u03c4\u03b5 \u03bf\u03c4\u03b9 \u03b1\u03bd \u03c0\u03b1\u03c1\u03bf\u03c5\u03bc\u03b5 \u03ba\u03c5\u03b2\u03bf\u03c5\u03c2 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03bf\u03c5 \u03bc\u03b5\u03b3\u03b5\u03b8\u03bf\u03c5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03b5\u03bd\u03c9\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03c4\u03b5\u03c4\u03bf\u03b9\u03bf \u03c4\u03c1\u03bf\u03c0\u03bf \u03c9\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03b7\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b5\u03bd\u03b1 \u03bd\u03b5\u03bf \u03ba\u03c5\u03b2\u03bf \u03bf\u03c0\u03bf\u03b9\u03bf\u03bd\u03b4\u03b7\u03c0\u03bf\u03c4\u03b5 \u03b4\u03b9\u03b1\u03c4\u03b1\u03c3\u03b5\u03c9\u03bd. [/quote]\r\n\r\n\u03a4\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03bb\u03b5\u03c2 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 (\u03b1\u03c6\u03bf\u03cd \u03ad\u03bd\u03b1\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c2 \u03b4\u03b5 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03ba\u03cd\u03b2\u03c9\u03bd). \u03a0.\u03c7. \u03b1\u03bd \u03c3\u03bf\u03c5 \u03b4\u03ce\u03c3\u03c9 7 \u03ba\u03cd\u03b2\u03bf\u03c5\u03c2 \u03bc\u03b5 \u03bc\u03ae\u03ba\u03bf\u03c2 $ 1,2\\ldots 7$ \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c6\u03c4\u03b9\u03ac\u03be\u03b5\u03b9\u03c2 \u03ad\u03bd\u03b1\u03bd \u03ba\u03b1\u03b9\u03bd\u03bf\u03cd\u03c1\u03b9\u03bf \u03ba\u03cd\u03b2\u03bf. \u0395\u03c0\u03af\u03c3\u03b7\u03c2, \u03b1\u03bd \u03c3\u03bf\u03c5 \u03b4\u03bf\u03b8\u03bf\u03cd\u03bd \u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03b9 \u03ba\u03cd\u03b2\u03bf\u03b9 $ C_n$ \u03bc\u03b5 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03c0.\u03c7. $ \\frac {1}{\\sqrt [3]{n}}$, \u03ba\u03b1\u03b9 \u03c0\u03ac\u03bb\u03b9 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c3\u03c5\u03bd\u03b1\u03c1\u03bc\u03bf\u03bb\u03bf\u03b3\u03ae\u03c3\u03b5\u03b9\u03c2 \u03ad\u03bd\u03b1\u03bd \u03ba\u03cd\u03b2\u03bf \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03cc\u03bb\u03bf\u03c5\u03c2.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_17": "[quote=\"Durandal\"]\nDemetres, \u03b1\u03bd \u03b4\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03c0\u03ac\u03b5\u03b9 \u03bd\u03b1 \u03b8\u03b1\u03c6\u03c4\u03b5\u03af \u03c4\u03bf thread, \u03b2\u03ac\u03bb\u03b5 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 :)\n[/quote]\r\n\r\n\u0398\u03b1 \u03b2\u03ac\u03bb\u03c9 \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03c3\u03b5 2-3 \u03bc\u03ad\u03c1\u03b5\u03c2. \u0393\u03b9\u03b1 \u03cc\u03c3\u03bf\u03c5\u03c2 \u03b8\u03ad\u03bb\u03bf\u03c5\u03bd \u03bd\u03b1 \u03be\u03b1\u03bd\u03b1\u03b4\u03bf\u03ba\u03b9\u03bc\u03ac\u03c3\u03bf\u03c5\u03bd, \u03bd\u03b1 \u03c0\u03c9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03b9\u03c3\u03c4\u03b5\u03af \u03ad\u03bd\u03b1\u03c2 \u03ba\u03c5\u03b2\u03bf\u03c2 \u03c3\u03b5 (\u03ac\u03c0\u03b5\u03b9\u03c1\u03b1) \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03cc\u03bb\u03b1 \u03b5\u03ba \u03c4\u03c9\u03bd \u03bf\u03c0\u03bf\u03af\u03c9\u03bd \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bd \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ad\u03c2 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2.\r\n\r\n\u0397 \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae \u03bc\u03bf\u03c5 \u03bb\u03cd\u03c3\u03b7 \u03b5\u03af\u03c7\u03b5 \u03ad\u03bd\u03b1 \u03bb\u03ac\u03b8\u03bf\u03c2. \u0397 \u03ba\u03b1\u03b9\u03bd\u03bf\u03cd\u03c1\u03b3\u03b9\u03b1 \u03bc\u03bf\u03c5 \u03bb\u03cd\u03c3\u03b7 (\u03b5\u03bb\u03c0\u03af\u03b6\u03c9 \u03c0\u03c9\u03c2 \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03bb\u03ac\u03b8\u03bf\u03c2) \u03b8\u03ad\u03bb\u03b5\u03b9 \u03bb\u03af\u03b3\u03b7 \u03b1\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7. (\u03a7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce \u03c3\u03c5\u03bc\u03c0\u03ac\u03b3\u03b5\u03b9\u03b1.) \u038c\u03c3\u03bf\u03b9 \u03b4\u03b5\u03bd \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03bf\u03c5\u03bd \u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c3\u03c5\u03bc\u03c0\u03ac\u03b3\u03b5\u03b9\u03b1 (compactness) \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03bf\u03c5\u03bd... \r\n\r\n[hide]\n\u03a5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03ad\u03bd\u03b1 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03bc\u03b5 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 $ 0 \\times 0 \\times 0$. (\u03a6\u03c5\u03c3\u03b9\u03ba\u03ac, \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b5\u03ba\u03c6\u03ce\u03bd\u03b7\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2, \u03c3\u03c5\u03bc\u03c0\u03b5\u03c1\u03b1\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03ad\u03bd\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9.)\n[/hide]", "Solution_18": "[quote=\"Demetres\"][quote=\"Durandal\"]\nDemetres, \u03b1\u03bd \u03b4\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03c0\u03ac\u03b5\u03b9 \u03bd\u03b1 \u03b8\u03b1\u03c6\u03c4\u03b5\u03af \u03c4\u03bf thread, \u03b2\u03ac\u03bb\u03b5 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03b1\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 :)\n[/quote]\n\n\u0398\u03b1 \u03b2\u03ac\u03bb\u03c9 \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03c3\u03b5 2-3 \u03bc\u03ad\u03c1\u03b5\u03c2. \u0393\u03b9\u03b1 \u03cc\u03c3\u03bf\u03c5\u03c2 \u03b8\u03ad\u03bb\u03bf\u03c5\u03bd \u03bd\u03b1 \u03be\u03b1\u03bd\u03b1\u03b4\u03bf\u03ba\u03b9\u03bc\u03ac\u03c3\u03bf\u03c5\u03bd, \u03bd\u03b1 \u03c0\u03c9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03b9\u03c3\u03c4\u03b5\u03af \u03ad\u03bd\u03b1\u03c2 \u03ba\u03c5\u03b2\u03bf\u03c2 \u03c3\u03b5 (\u03ac\u03c0\u03b5\u03b9\u03c1\u03b1) \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03cc\u03bb\u03b1 \u03b5\u03ba \u03c4\u03c9\u03bd \u03bf\u03c0\u03bf\u03af\u03c9\u03bd \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bd \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ad\u03c2 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2.\n\n\u0397 \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae \u03bc\u03bf\u03c5 \u03bb\u03cd\u03c3\u03b7 \u03b5\u03af\u03c7\u03b5 \u03ad\u03bd\u03b1 \u03bb\u03ac\u03b8\u03bf\u03c2. \u0397 \u03ba\u03b1\u03b9\u03bd\u03bf\u03cd\u03c1\u03b3\u03b9\u03b1 \u03bc\u03bf\u03c5 \u03bb\u03cd\u03c3\u03b7 (\u03b5\u03bb\u03c0\u03af\u03b6\u03c9 \u03c0\u03c9\u03c2 \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03bb\u03ac\u03b8\u03bf\u03c2) \u03b8\u03ad\u03bb\u03b5\u03b9 \u03bb\u03af\u03b3\u03b7 \u03b1\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7. (\u03a7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce \u03c3\u03c5\u03bc\u03c0\u03ac\u03b3\u03b5\u03b9\u03b1.)[/quote]\r\n\r\n\u039c\u03ac\u03bb\u03bb\u03bf\u03bd \u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03bf \u03bc\u03c5\u03b1\u03bb\u03cc \u03bc\u03b1\u03c2 :)\r\n\r\nCheerio,\r\n\r\nDurandal 1707", "Solution_19": "\u0394\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03af\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ad\u03bd\u03b1 \u03ba\u03cd\u03b2\u03bf \u03c3\u03b5 \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ce\u03bd \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03c9\u03bd.\r\n\r\n\u038c\u03c0\u03c9\u03c2 \u03b5\u03af\u03c0\u03b1 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf post, \u03b7 \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae \u03bc\u03bf\u03c5 \u03bb\u03cd\u03c3\u03b7 (\u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03c5\u03c1\u03af\u03c3\u03c4\u03b7\u03ba\u03b1 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03b1\u03c0\u03bb\u03ae) \u03ba\u03b1\u03c4\u03ad\u03c1\u03c1\u03b5\u03c5\u03c3\u03b5. \u0399\u03b4\u03bf\u03cd \u03bc\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b8\u03ad\u03bb\u03b5\u03b9 \u03bb\u03af\u03b3\u03b7 \u03b1\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7:\r\n\r\n\u0391\u03c2 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf $ A \\equal{} [0,1] \\times [0,1] \\times [0,1]$. \u039c\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03c9 \u03cc\u03c4\u03b9 \u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03c9\u03bd $ 0 \\times 0 \\times 0$, \u03c4\u03cc\u03c4\u03b5 \u03b1\u03c5\u03c4\u03cc \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf $ B \\equal{} [0,1] \\times [0,1] \\times [1/2,1]$. \u039a\u03bf\u03b9\u03c4\u03ac\u03b6\u03c9 \u03c4\u03b7\u03bd \u03ba\u03ac\u03c4\u03c9 \u03ad\u03b4\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03b2\u03bf\u03c5. \u0395\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03b9\u03c3\u03bc\u03ad\u03bd\u03bf \u03c3\u03b5 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03ac\u03ba\u03b9\u03b1. \u0399\u03c3\u03c7\u03c5\u03c1\u03af\u03b6\u03bf\u03bc\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03ac\u03ba\u03b9\u03b1. \u03a0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9, \u03b1\u03bd \u03c5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1, \u03c4\u03cc\u03c4\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c3\u03c5\u03bc\u03c0\u03ac\u03b3\u03b5\u03b9\u03b1 \u03c4\u03bf\u03c5 $ [0,1] \\times [0,1]$, \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03ac\u03ba\u03b9 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03c9\u03bd $ 0 \\times 0$, \u03ac\u03c4\u03bf\u03c0\u03bf. \u03a3\u03c5\u03bd\u03b5\u03c7\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c0\u03c9\u03c2 \u03c3\u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03c4\u03bf\u03c5 Durandal \u03b3\u03b9\u03b1 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b1 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1: \u0391\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 $ K_1$ \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03c0\u03bf\u03c5 \u03b5\u03c6\u03ac\u03c0\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03ba\u03ac\u03c4\u03c9 \u03ad\u03b4\u03c1\u03b1, $ K_2$ \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03c0\u03bf\u03c5 \u03b5\u03c6\u03ac\u03c0\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf $ K_1$ \u03ba.\u03c4.\u03bb. \u039c\u03ad\u03c7\u03c1\u03b9 \u03c0\u03b9\u03bf \u03cd\u03c8\u03bf\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c6\u03c4\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5; \u0399\u03c3\u03c7\u03cd\u03c1\u03af\u03b6\u03bf\u03bc\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03be\u03b5\u03c0\u03b5\u03c1\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf 1/2. \u0393\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03b4\u03b5\u03af\u03be\u03c9 \u03b1\u03c5\u03c4\u03cc, \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03c9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03bc\u03ae\u03ba\u03bf\u03c2 \u03c4\u03bf\u03c5 $ K_1$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03c4\u03bf\u03c5 1/3. (\u0395\u03c0\u03b5\u03b9\u03b4\u03ae 1/3 + 1/9 + ... = 1/2.) \u0391\u03c5\u03c4\u03cc \u03b1\u03c6\u03ae\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b1\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c3\u03c4\u03bf\u03bd \u03b1\u03bd\u03b1\u03b3\u03bd\u03ce\u03c3\u03c4\u03b7 :D . \r\n\r\n\u0393\u03b9\u03b1 \u03bd\u03b1 \u03b5\u03af\u03bc\u03b1\u03b9 \u03c4\u03b5\u03bb\u03b5\u03af\u03c9\u03c2 \u03c3\u03c9\u03c3\u03c4\u03cc\u03c2, \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03c0\u03c9\u03c2 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03c9 \u03cc\u03c4\u03b9 \u03c4\u03b1 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ae\u03c3\u03b9\u03bc\u03b1 \u03b1\u03bb\u03bb\u03ac \u03b1\u03c5\u03c4\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 \u03b3\u03bd\u03c9\u03c3\u03c4\u03ae \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c3\u03c4\u03b7\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03b7\u03c3\u03b9\u03bc\u03cc\u03c4\u03b7\u03c4\u03b1.", "Solution_20": "[quote=\"Demetres\"]\u0391\u03c2 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03bf \u03b1\u03c1\u03c7\u03b9\u03ba\u03cc\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf $ A \\equal{} [0,1] \\times [0,1] \\times [0,1]$. \u039c\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03c9 \u03cc\u03c4\u03b9 \u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03c9\u03bd $ 0 \\times 0 \\times 0$, \u03c4\u03cc\u03c4\u03b5 \u03b1\u03c5\u03c4\u03cc \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf $ B \\equal{} [0,1] \\times [0,1] \\times [1/2,1]$. \u039a\u03bf\u03b9\u03c4\u03ac\u03b6\u03c9 \u03c4\u03b7\u03bd \u03ba\u03ac\u03c4\u03c9 \u03ad\u03b4\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03b2\u03bf\u03c5. \u0395\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03b4\u03b9\u03b1\u03bc\u03b5\u03c1\u03b9\u03c3\u03bc\u03ad\u03bd\u03bf \u03c3\u03b5 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03ac\u03ba\u03b9\u03b1. \u0399\u03c3\u03c7\u03c5\u03c1\u03af\u03b6\u03bf\u03bc\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03ac\u03ba\u03b9\u03b1. \u03a0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9, \u03b1\u03bd \u03c5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd \u03ac\u03c0\u03b5\u03b9\u03c1\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1, \u03c4\u03cc\u03c4\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c3\u03c5\u03bc\u03c0\u03ac\u03b3\u03b5\u03b9\u03b1 \u03c4\u03bf\u03c5 $ [0,1] \\times [0,1]$, \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03ac\u03ba\u03b9 \u03b4\u03b9\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03c9\u03bd $ 0 \\times 0$, \u03ac\u03c4\u03bf\u03c0\u03bf. \u03a3\u03c5\u03bd\u03b5\u03c7\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c0\u03c9\u03c2 \u03c3\u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7 \u03c4\u03bf\u03c5 Durandal \u03b3\u03b9\u03b1 \u03c0\u03b5\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03ad\u03bd\u03b1 \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9\u03b1: \u0391\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 $ K_1$ \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03c0\u03bf\u03c5 \u03b5\u03c6\u03ac\u03c0\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03ba\u03ac\u03c4\u03c9 \u03ad\u03b4\u03c1\u03b1, $ K_2$ \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03c0\u03bf\u03c5 \u03b5\u03c6\u03ac\u03c0\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf $ K_1$ \u03ba.\u03c4.\u03bb. \u039c\u03ad\u03c7\u03c1\u03b9 \u03c0\u03b9\u03bf \u03cd\u03c8\u03bf\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c6\u03c4\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5; \u0399\u03c3\u03c7\u03cd\u03c1\u03af\u03b6\u03bf\u03bc\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03be\u03b5\u03c0\u03b5\u03c1\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf 1/2. \u0393\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03b4\u03b5\u03af\u03be\u03c9 \u03b1\u03c5\u03c4\u03cc, \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03c9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03bc\u03ae\u03ba\u03bf\u03c2 \u03c4\u03bf\u03c5 $ K_1$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03c4\u03bf\u03c5 1/3. (\u0395\u03c0\u03b5\u03b9\u03b4\u03ae 1/3 + 1/9 + ... = 1/2.) \u0391\u03c5\u03c4\u03cc \u03b1\u03c6\u03ae\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b1\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c3\u03c4\u03bf\u03bd \u03b1\u03bd\u03b1\u03b3\u03bd\u03ce\u03c3\u03c4\u03b7 :D.[/quote]\r\n\r\n\u0388\u03be\u03bf\u03c7\u03b1, well done! :clap2:\r\n\r\n\u0391\u03c5\u03c4\u03ae \u03ae\u03c4\u03b1\u03bd \u03b7 \u03b9\u03b4\u03ad\u03b1 \u03bc\u03bf\u03c5, \u03cc\u03c4\u03b9 \u03c0\u03b7\u03b3\u03b1\u03af\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03c3\u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03ba\u03c5\u03b2\u03ac\u03ba\u03b9 \u03b4\u03b5 \u03c6\u03c4\u03ac\u03bd\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c4\u03ad \u03c3\u03c4\u03bf \u03ae\u03bc\u03b9\u03c3\u03c5 \u03c4\u03bf\u03c5 \u03cd\u03c8\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03b2\u03bf\u03c5 :)\r\n\r\n\u03a9\u03c3\u03c4\u03cc\u03c3\u03bf, \u03b8\u03b1 \u03b5\u03c0\u03b9\u03bc\u03b5\u03af\u03bd\u03c9 \u03c3\u03c4\u03bf \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03cc \u03bc\u03bf\u03c5 \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03c0\u03b9\u03bf \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc (\u03ba\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03bf):\r\n\r\n[b]\u03a5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03bc\u03af\u03b1 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c1\u03b9\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03b2\u03bf\u03c5 \u03c3\u03b5 (\u03ac\u03c0\u03b5\u03b9\u03c1\u03bf\u03c5\u03c2) \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf\u03c5\u03c2 \u03ba\u03cd\u03b2\u03bf\u03c5\u03c2 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03bf\u03cd \u03bc\u03b5\u03b3\u03ad\u03b8\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03b3\u03b5\u03bc\u03af\u03b6\u03bf\u03c5\u03bd \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03cc\u03bb\u03bf\u03bd \u03c4\u03bf\u03bd \u03cc\u03b3\u03ba\u03bf \u03c4\u03bf\u03c5 \u03b1\u03c1\u03c7\u03b9\u03ba\u03bf\u03cd \u03ba\u03cd\u03b2\u03bf\u03c5?[/b]\r\n\r\nCheerio,\r\n\r\nDurandal 1707" } { "Tag": [ "Euler" ], "Problem": "i've a nice question for all...\r\nwrite a program in c++ such that when a name (with multiple constituents) for ex:JOHN SMITH,GEORGE THOMAS,etc., the output shud be J.SMITH, G.THOMAS,...... \r\nALSO PLZ RATE THIS QUESTION AND SHARE YOUR FEELINGS ABOU IT..........", "Solution_1": "I'm desperately waiting for a reply...", "Solution_2": "[b] @ shibin.george [/b]: Okay, I have developed the code for it, and here it is:\r\n\r\n[code] \n\n#include \n#include \n#include \nusing namespace std;\nvoid main()\n{\nstring a= \"John\";\nstring b= \"Smith\";\na.insert(4,\".\"); // This statement inserts \".\" at the 4th position in string a //\na.erase(1,3); // This erases three characters of \"string a\" starting from 'O'//\nstring c=a+b; // concatenation of Strings 'a' and 'b' //\nputs(c); // Printing the final string, you can also use \"cout\" too //\ngetch();\n}\n\n[/code]\n\n[b] Output [/b]: J.Smith\n\nThis is a pretty easy programming problem, but you need to be aware of \"insert\",\"erase\" operations that are part of the C++ string class (Which is a part of the std namespace that allows you to manipulate strings safely) \n\nIf you intend to enter the strings at run-time, you need to be careful while using the \"insert\" and \"erase\" operations. (For example , if you enter the strings as-- Michael , and Slater, and you desire an output as -- M.Slater, you need to use insert and erase operation as follows:\n\n[code]\n\na.insert(7,\".\");\na.erase(1,6);\nstring c=a+b;\ncout<\n#include \n\nusing namespace std;\n\nvoid printShortName(string firstName, string lastName)\n{\n cout << firstName[0] << \".\" << lastName << endl;\n}\n\nvoid printShortName(string fullName)\n{\n int index = fullName.find(' ');\n printShortName(\n fullName.substr(0, index - 1),\n fullName.substr(index + 1, fullName.length() - index));\n}\n\nint main()\n{\n printShortName(\"John Smith\");\n printShortName(\"Leonhard Euler\");\n // etc...\n return 0;\n}[/code]", "Solution_4": "[b] @Yongyi781 [/b] : Actually, I was developing the C++ code for the given problem, so that when we enter a string at run-time , say, something like -- David West, the output would be -- D.West, before you came up with your idea.\r\n\r\nI really can't understand how you could so [b] rudely [/b] (I'm construing your remark -- \"No Good\" to be unintentional though) dismiss my program as being \"No Good\", I know for sure , my code works really fine if the length of the string is static. But since the OP didn't explicitly mention whether the string should be entered before run-time or after, I thought of first posting the code assuming that the \"string\" is mentioned before run-time.\r\n\r\nI think you had developed the code for \"strings of arbitrary size\" which I feel is good, but it would have been much better had you desisted from using that obnoxious, snooty remark.(i.e.-- regarding my program)", "Solution_5": "sorry @gen8, but i just gave JOHN SMITH as an example....\r\nmy intention is that this code should sould work for all types of names of any length and any characters......\r\npleeaase,make the necessary corrections......", "Solution_6": "[quote=\"Gen8\"](I'm construing your remark -- \"No Good\" to be unintentional though)[/quote]\r\n\r\nNo problem. It was unintentional that it would be construed as rude :)\r\n\r\nBut I believe that a good program should be generalizable, like a good math problem should be, unless there is a compelling reason not to (such as performance, etc).\r\n\r\nshibin.george: I have posted my code. Does it work?" } { "Tag": [ "Harvard", "college", "Yale", "Princeton", "geometry", "geometric transformation", "reflection" ], "Problem": "I do not have too much money but I don't think I will recieve all that much financial aid plus I am looking to go through grad school so I was wondering what the best colleges for your money were?\r\nI am looking for great academics as well as the money...\r\nso colleges like Harvard and Yale are out because of the money", "Solution_1": "[quote=\"qweretyq\"]I do not have too much money but I don't think I will recieve all that much financial aid plus I am looking to go through grad school so I was wondering what the best colleges for your money were?\nI am looking for great academics as well as the money...\nso colleges like Harvard and Yale are out because of the money[/quote]\r\n\r\nI have been said (by Harvard Alumni) that if you happen to be admited in Harvard, the money is not really a problem, that a big percentage (greater than 90%) receive BIG waivers, otherwise they couldn't study there. I haven't checked the facts, but otherwise, I doubt that a person I know, with a \"Descent Latin American Salary\" (~$30K a year) could have afford to have both of his sons at Harvard at once!", "Solution_2": "qweretyq says, \"...but I don't think I will recieve all that much financial aid plus I am looking to go through grad school...\"\r\n\r\nIf you're planning to go to grad school in mathematics or almost any of the sciences, then grad school (Ph.D. programs and a few types of masters' programs) is essentially free. Usually a program which admits you will supply a tuition waiver and some sort of stipend; the stipend may be in exchange for part-time teaching/grading/etc. or for doing research, depending on whether you're a US citizen or not and how much money the department has. Stuff like health insurance is also often covered, or at least subsidized.", "Solution_3": "On the other hand, if you're talking about professional (business, law, med) school, you're going to end up paying a bundle.\r\n\r\n\r\nOne thing to point out, though: if you don't have much money, you'd almost certainly be [i]better[/i] off going to Harvard or Yale than to schools that cost, say, USD25,000 instead of USD40,000 per year. This is because they actually have the endowments to give financial aid packages to many, many students that other schools can't necessarily afford. (For Harvard, at least, all financial aid is need-based rather than academic scholarships, so you don't have to compete with the rest of the student body for the money.)\r\n\r\nThat said, of course, not everyone who needs a good financial aid package will get into Harvard or Yale. Depending where in the country you are (or, I suppose, what country you are in), local state schools often have programs that make them very economically attractive to bright students, so you can get simultaneously a really cheap education (some will pay you to come if you are a nat'l merit finalist, for example) and a collection of bright kids in the same program as you.", "Solution_4": "I agree that you should not allow financial aid concerns to deter you from applying to top colleges. Many offer generous packages. Harvard, Princeton and Yale, for example, all recently eliminated the parental contribution for families earning less than $40,000 a year, and greatly reduced the contribution for families earning between $40,000 and $60,000. \r\n\r\nFinancial aid offices are usually flexible; if the financial aid application does not accurately reflect your family's situation, write them a letter. If you're accepted into your dream school but they offer less aid than another school, call up your dream school. They might be able to match a competing offer.\r\n\r\nYou should consider your state school, especially if it offers an honors program. Many selective colleges offer merit scholarships to top applicants. Harvey Mudd and University of Chicago come to mind. Rice University's base tuition is significantly lower than comparable schools; they're also generous with financial aid. There are also many private scholarships you can apply for.", "Solution_5": "Pulling out my 2005 Fiske guide to colleges, it lists the following schools as \"Bangs for the buck\":\r\n\r\nAmong private colleges and universities:\r\nAntioch\r\nBard\r\nBYU\r\nCentre College\r\nClark U\r\nCooper Union (well, duh, it's free)\r\nDeep Springs\r\nEarlham\r\nElon U\r\nGrinnell\r\nHendrix\r\nKenyon\r\nLawrence U\r\nMacalester\r\nU. of Redlands\r\nRhodes College\r\nRice U\r\nSt. Olaf\r\nUniversity of the South (Sewanee)\r\nTrinity U\r\nWake Forest U\r\nWarren Wilson College\r\nWashington and Lee U\r\n\r\nand among public colleges and universities:\r\nU. of British Columbia *\r\nU. of Florida\r\nU. of Kansas\r\nMary Washington U.\r\nUMass\r\nMcGill *\r\nUNC Asheville\r\nUNC Chapel Hill\r\nU. of Oregon\r\nQueens University *\r\nSUNY Bing +\r\nSUNY Albany +\r\nSUNY Geneseo +\r\nSUNY Stony Brook +\r\nU. of Texas\r\nTexas A&M\r\nU. of Toronto *\r\nCollege of William and Mary\r\nU. of Wisconsin - Madison\r\n\r\n(*: Canadian schools)\r\n(+: SUNY = State University of New York)\r\n\r\nNotably missing from the public colleges are U. Michigan - Ann Arbor and U. California - many, which provide really really good educations and fall in a range of very cheap by private school standards but very pricy by public school standards.", "Solution_6": "cost don't matter bc when you get a nice job you can pay back the debt\r\nif you don't go to a good school (you can save money) you will not get bigger salaries", "Solution_7": "[quote=\"juicybooty911\"]if you don't go to a good school (you can save money) you will not get bigger salaries[/quote]This is most assuredly [b]not[/b] true.", "Solution_8": "[quote=\"DPatrick\"][quote=\"juicybooty911\"]if you don't go to a good school (you can save money) you will not get bigger salaries[/quote]This is most assuredly [b]not[/b] true.[/quote]\r\n\r\nI suppose you have counterexamples in mind, but what are they?", "Solution_9": "[quote=\"qweretyq\"]I do not have too much money but I don't think I will recieve all that much financial aid.[/quote]\r\n\r\nIf you don't have much money, you should get good financial aid. As JBL and others in the thread have pointed out, sometimes the very top schools, if you can get into them, actually have the best resources for providing financial aid. You should definitely apply to the schools you strongly desire to attend, after preparing well to be a strong candidate for admission, and then let the school show you what it can offer in a financial aid package. Don't ASSUME you won't get enough financial aid; apply to various schools and find out what you'll get.", "Solution_10": "[quote=\"tokenadult\"][quote=\"DPatrick\"][quote=\"juicybooty911\"]if you don't go to a good school (you can save money) you will not get bigger salaries[/quote]This is most assuredly [b]not[/b] true.[/quote]\n\nI suppose you have counterexamples in mind, but what are they?[/quote]\r\nIt's not that I have specific counterexamples in mind, it's more the certitude of juicybooty911's statement that I objected to: \"If you don't go to a good school...you [b]will not[/b] get bigger salaries\" (emphasis added).\r\n\r\nMy main point is that (in my opinion, of course) the difference between attending a so-called \"top-tier\" school (like Harvard/MIT/Caltech) and a so-called \"second-tier\" school (like many in the list that JBL posted) is much, much smaller than many people think. I'm not saying it's nonexistant, but you can go to a state school and be rich and/or successful and/or happy (whichever of these is most important to you), and you can go to Harvard and get nothing out of it.\r\n\r\nIf you want a counterexample, how about Paul Allen?", "Solution_11": "How did Paul Allen meet Bill Gates? \r\n\r\nBill Gates dropped out of Harvard because he had a one-million-dollar trust fund from his grandfather already (from birth), so he had enough capital to found a business. He had also attended the top private school in the Seattle area, which at that time had a higher annual tuition bill than Harvard. So he is adequately prepared to self-educate as an adult, without a need for a college degree. But I don't know as many biographical details about Paul Allen.", "Solution_12": "You might want to take a look at the article [url=http://www.theatlantic.com/doc/200410/easterbrook]Who Needs Harvard?[/url] in the [i]Atlantic Monthly[/i]. It discusses a study that tracked students who were accepted into Ivy League schools but chose to attend less prestigious colleges. Twenty years after graduation, these students were earning the same salaries as Ivy League grads.", "Solution_13": "I've heard of that conclusion, but I think that study badly needs replication before I'd bet my children's future on its conclusion.", "Solution_14": "New HP CEO Mark Hurd had a BS degree from Baylor. Stanford must be steaming mad that one of his graduates was not selected for a job in its own backyard. :D", "Solution_15": "ok, here's a kicker, see if anyone can solve THIS one:\r\n1. your parents make upper-middle class amounts of money.\r\n2. they are only going to pay four years of state school tuition for you.\r\n3. you would LIKE to go to a top-tier school.\r\n4. you have potential to be accepted at said school.\r\n5. however, because of #1, your financial aid is squat.\r\n6. you also have SOME money saved up, which you are required to report when applying for financial aid, but have plans for this money to be spent on a trip you have been saving for for four years.\r\n7. oh yes, you also want to attend a professonal school.\r\n8. and your job will not pay large sums of money after you get out. so loans are not a good idea.\r\n9. but you REALLY need/want to go to said professional school, because therein lies your deepest passion.\r\n\r\n\r\nbtw, tokenadult- i recognize your argument but i really don't think its valid at all. well, maybe it is, but i don't think it matters. do you think your children are really going to be ALL THAT much happier if they are making, oh, i dunno, $10k more than \"very-well-off\" amounts? because i think regardless of where your children (or anyones) go to school, as long as it is a decent, respectable school, they are going to be making enough money to be happy. i just beg you not to make them think that ivies/top-tiers are the ONLY good schools out there. because [b]that[/b] is not true. :) \r\n\r\n[ [i]edited by moderator because double dollar sign was read as LaTeX code--no changes in substance of post[/i] ]", "Solution_16": "Well, always the toughest situation is to have a higher EFC (estimated family contribution) than your parents are willing to pay at a suitable school. Young people in that situation get my greatest sympathy, because I had to deal with that situation when I was college age. I couldn't obtain financial aid based on my own (meager) income, because my parents' income had to be counted, by law, to determine what *I* could afford to pay for college, but my parents weren't actually making their resources available to me for college expenses after a certain age. \r\n\r\nMaybe in your situation you need to work backwards from what you will do when you have your professional school degree to pursue your passion. Is there any kind of loan forgiveness program or other way to erase student loan debt for someone who pursues that occupation? Does it matter, in that occupation, where a person attended college before starting professional school? Maybe you do need to minimize expense (and debt) up-front in your case to make achieving your goal possible. I would interview people who pursue that occupation now (something, by the way, that I stupidly DIDN'T do when I was your age, thinking about my dream occupation at the time) and find out what they would suggest. \r\n\r\nDo any of the other participants here have suggestions for this situation?", "Solution_17": "I understand the desire to attend an elite college like H/Y/P/S/M/C. Ears do perk up when I mention I attended one of these schools but ultimately you will be judged by who you are and what you accomplish. Admittance to one of these schools is not an automatic ticket to anywhere. At best they offer opportunities. Such opportunities are also available at less selective schools.\r\n\r\nTake a look at the most recent US Presidents. Six of the last eight (the GB's being the exceptions) did not attend elite colleges as undergraduates : [list]\nBill Clinton (Georgetown U)\nRonald Reagan (Eureka College)\nJimmy Carter (US Naval Academy)\nGerald Ford (U Michigan)\nRichard Nixon (Whittier College)\nLyndon Johnson (Southwest Texas State Teachers College)\n[/list]\n\nTake a look at the Forbes Top 10 list of richest Americans. Yes, Bill Gates, the Harvard dropout, is at the top of the list but the rest did not attend elite colleges: \n[list]\nWarren Buffett (U Nebraska)\nPaul Allen (Washington St U dropout)\nMichael Dell (U Texas dropout)\nLawrence Ellison (U Illinois dropout)\n[/list]\r\n(The remaining slots are filled by five non-Ivy heirs of WalMart founder Sam Walton, who graduated from the University of Missouri.)\r\n\r\nEven in academia, if you examine the CV's of math professors at Princeton and MIT, many attended Ivies as undergraduates, and many graduated from foreign universities. The rest attended schools like U Chicago, U Maryland, U Colorado, U Arizona, Brandeis, and Purdue. For an academic career, it's more important which graduate school you attend than which college.\r\n\r\nIt is a misconception to think that your Ivy League roommate's megarich dad will offer you a high-paying job upon graduation. Today's Ivy admissions process is much more of a meritocracy than it was in the old days when prep schools fed their graduates straight into the elite schools. This is especially true if you hang out with math and science majors who typically come from middle class backgrounds. \r\n\r\nI met intelligent, creative people at both the liberal arts college and big name university I attended. The teaching was definitely superior at the small college. Professors who excel at research do not necessarily excel at teaching. In fact there's often an inverse correlation.\r\n\r\nElite schools have a lot to offer but there are many other excellent schools where you can get a quality education. It's up to you to make the most of your environment wherever you are.\r\n\r\nBleumoose, if you end up attending your state school, it's not the end of the world. If you want to shoot for your dream school, there are private scholarships available. Books like [url=http://www.amazon.com/exec/obidos/ASIN/0060937653/qid=1112478928/sr=2-1/ref=pd_bbs_b_2_1/104-3755691-1763125]How to Go to College Almost for Free[/url] give tips on how to win merit-based scholarships. I wouldn't worry about graduate/professional school for now. Your interests may very well change.", "Solution_18": "So it would probably be best for me to excel as an honors student at the state school, which also offers a very good professional school in that field, instead of (maybe) entering an ivy league school which also offers a great professional school in that field, but costs 10x as much (no exaggeration) and is harder to get into (it is easier for a state resident to get into a professional school in THIS field)? because that is what all of the people that i have talked to did.\r\n\r\ni guess i halfway know that's the right thing to do but when people you have topped academically and extracurricularly for so long are getting into top universities, and you are going to State U., it's still a little discouraging. one feels that one must be at the top of the game to \"succeed.\"\r\n\r\non another note: does anyone know of any cheap, small-medium (15,000-) liberal arts school that have good science/biology programs? esp in the southeast? (sorry, off-topic?)", "Solution_19": "[quote=\"bleumoose\"]ok, here's a kicker, see if anyone can solve THIS one:\n1. your parents make upper-middle class amounts of money.\n2. they are only going to pay four years of state school tuition for you.\n3. you would LIKE to go to a top-tier school.\n4. you have potential to be accepted at said school.\n5. however, because of #1, your financial aid is squat.\n6. you also have SOME money saved up, which you are required to report when applying for financial aid, but have plans for this money to be spent on a trip you have been saving for for four years.\n7. oh yes, you also want to attend a professonal school.\n8. and your job will not pay large sums of money after you get out. so loans are not a good idea.\n9. but you REALLY need/want to go to said professional school, because therein lies your deepest passion.\n[/quote]\r\n\r\nMy family was upper-middle class (roughly six figures annual income in canadian dollars) and I still got about 32000 in aid from princeton per year. Trust me, (4) is by far the biggest hurdle on your list. Of the people I know who got into Ivy League caliber universities last year (mainly the big three and one at MIT and one at Caltech), almost all got truly massive amounts of financial aid. Mine was the smallest!", "Solution_20": "Wow. Financial aid is more generous than I thought. I recommend checking the college discussion websites to get a feel for which schools are generous with aid (both merit-based and need-based). Many students share information about their awards.", "Solution_21": "Bleumoose,\r\n\r\nI sense that the college decision choice for you is a bit more complicated than the money. You mentioned science/biology... that sounds like you'd like to do pre-med? And you also mentioned that your eventual job wouldn't make tons of money. Just as an example, let's suppose you wanted to get an M.D. and then work in rural areas (which doesn't make much money). You could go to an undergraduate institution which has an excellent acceptance rate for med school (there are lots, and they don't all appear to be top-tier), and then do a program (such as at Michigan State, or at least they used to have this) where your tuition etc. are cut if you do a special program which concentrates on doing medicine in rural area. \r\n\r\nIf you're willing to disclose a few more of the details of your dream (pm me if you'd like), we might be able to give you some more specific ideas on what to do.\r\n\r\nBut very generally, my philosophy is that you should always apply to the places you really want to go, and see what happens. Maybe you'll get in, and maybe they'll give you lots of financial aid! But if you don't apply, then they can't possibly give you financial aid. Best not to cut off your options if you can avoid it.", "Solution_22": "it's veternary medicine. esp at a zoo/circus (exotics). they have programs for vets that cut tuition if you work in rural areas but i am unaware of any for zoo/circus vets. either way, however, one is still looking at a salary of <$50k, which wouldnt pay off large school loans. it is complicated and scary. i appreciate EVERYONE for your help and advice!", "Solution_23": "(People in the math department here sometimes have very strange conversations, so that's where I got this information.) If you get a student loan, you will have to pay back a certain amount of it each year until either it is paid off or a certain amount of time has expired (I think it's 20 years, but I'm not sure). The amount you pay each year depends on your salary. Therefore if you take out a big loan and get a low-paying job, you won't end up paying back the full loan. Furthermore, the loan will be interest-free (I think), and so it won't adjust for inflation.", "Solution_24": "In the US at least I don't think you will be able to find loans that are interest-free or forgivable after a certain number of years. Student loans are generally interest-free during the time you're a student and then for a grace period after graduation. After that the interest starts to accrue.\r\n\r\nDuring the repayment period you can claim hardship (forbearance) to temporarily postpone or lessen payments but you still have to repay the full amount in the future.\r\n\r\nSometimes loans (up to a certain amount) will be forgiven if you choose to enter jobs like teaching, medicine or law in underserved areas.", "Solution_25": "Wow -- veterinary medicine with a zoo/exotic specialization is pretty difficult. While in graduate school, I had a housemate for a year who was applying to veterinary schools; in fact, she was partially considering a zoo specialization! As I recall, not all states have veterinary schools, and few veterinary schools have zoo/exotic specializations. My impression at the time was that veterinary schools didn't seem to care much about the prestige of one's undergraduate institution. They were much more interested in whether you were a resident of the state they were in, and in how well you'd done in your undergrad program, and in how much vet-tech volunteer experience you'd had. Things might be different now, of course. But to deal with the money factor, start looking at state residency requirements for the vet schools that have the zoo/exotic specialization; the way to make this affordable might be to move to such a state after college and work there as a vet assistant until you achieve residency. (In-state tuition is usually significantly cheaper.)", "Solution_26": "[quote=\"ComplexZeta\"]People in the math department here sometimes have very strange conversations, so that's where I got this information. . . . [ details omitted] [/quote]\r\n\r\nMost federal student loans in the United States have to be paid back in full, and I'm sure the same is true of most state loans and private college loans. Nor is low income in itself a basis for loan extension. The best reason to borrow money for higher education is to obtain work that allows a higher income, income sufficient to pay off the loan. Otherwise it is probably more shrewd to find a less expensive way to go to college." } { "Tag": [ "inequalities", "combinatorics unsolved", "combinatorics" ], "Problem": "How many subsets of $\\{1, 2, ... , 2n\\}$ do not contain two numbers with sum $2n+1$?", "Solution_1": "We have $n$ pairs of the form $(k,2n + 1 - k)$ with $k \\in \\{1,2,\\dots,n\\}$. For each of these pairs, either $k$ belongs to the \"chosen\" subset, or $2n + 1 - K$ does, or neither $k$ nor $2n + 1 - k$ do. So, there are 3 \"possibilities\" for each pair, giving $3^n$ subsets :)", "Solution_2": "^very nice, but since I am bad at combo I use algebra. Partition $\\{1, 2, ... , 2n\\}$ into $A=\\{1, 2, ... , n\\}$ and $B=\\{n+1, n+2, ... , 2n\\}$ and let $S$ be a subset of $\\{1, 2, ... , 2n\\}$. There are $\\dbinom{n}{k}$ ways to pick $k$ elements from $A$ to be contained in $S$, where $0 \\le k \\le n$. Let them be $a_1,a_2,...,a_k$. Now we cannot have in $S$ the elements $(2n+1)-a_1,(2n+1)-a_2,...,(2n+1)-a_k$, which are all contained in $B$. Choose to include the remaining $n-k$ elements in $B$ in $2^{n-k}$ ways. Thus there are $\\dbinom{n}{k}2^{n-k}$ ways total and summing over all $k$ gives\n\n\\[\\sum_{k=0}^{n}\\dbinom{n}{k}2^{n-k}=\\sum_{i=0}^{n}\\dbinom{n}{i}2^i=(1+2)^n=3^n\\]\n\nwhere we used the substitutions $n-k=i$ and $\\dbinom{n}{k}=\\dbinom{n}{n-k}$, and in the second to last inequality we used the binomial theorem." } { "Tag": [ "AMC" ], "Problem": "Call a positive real number special if it has a decimal representation that consists entirely of 0 and 7. For example, $\\frac{700}{99}$=7.070707070707... and 77.007 are special numbers. What is the smallest n such that 1 can be written as a sum of n special numbers?", "Solution_1": "[hide]since 7 is not as easy to work with, you can divide everything by 7, making the problem with 0's and 1's summing up to 1/7.\n\n1/7 = $0.\\overline{142857}$\n\nthen we can subtract repeating sequences of 0's and 1's until we get 0. you want to subtract sequences with as many 1's as possible to minimize the n.\n\nfirst, subtract $0.\\overline{111111}$ (that makes 1 number) to get $0.\\overline{031746}$.\n\nthen subtract $0.\\overline{011111}$ (that makes 2 numbers) to get $0.\\overline{020635}$.\n\nthen subtract $0.\\overline{010111}$ twice (4 numbers so far) to get $0.\\overline{000413}$.\n\nthen subtract $0.\\overline{000111}$ (5 numbers so far) to get $0.\\overline{000302}$.\n\nthen subtract $0.\\overline{000101}$ twice (7 numbers) to get $0.\\overline{000100}$.\n\nfinally, subtract $0.\\overline{000100}$ to get 0.\n\nwe subtracted 8 numbers (meaning the sum of the 8 numbers = 1/7), so the answer is [b]8[/b]?[/hide]", "Solution_2": "That's a very interesting way to do it :)" } { "Tag": [ "algebra", "polynomial", "LaTeX", "real analysis", "real analysis unsolved" ], "Problem": "In the following $T_0,T_1,...,T_n,...$ are [b] Chebychev polynomials[/b] of first kind, that is $T_0(x)=1 , T_1(x)=x$ and\r\n \\[ T_{n+1}(x)=2xT_n(x)-T_{n-1}(x) \\; \\; \\; \\mbox{for}\\; \\; n\\in {\\mathbb N}^* . \\]\r\n$l_0(x),l_1(x),...,l_n(x),...$ are [b] Lucas polynomials [/b] defined as $l_0(x)=2 , l_1(x)=x ,$ and for $n \\ge 1$\r\n\\[ l_{n+1}(x)=x l_n(x)+l_{n-1}(x) \\; . \\]\r\nIt is supposed that a polynomial $f$ [b] is given [/b], and moreover $\\displaystyle f(x) =\\displaystyle \\sum\\limits_{k=0}^{n} c_kT_k(x) \\; .$\r\n[b] Find a closed form for the sum [/b] $(Tf)(x,y): = \\sum\\limits_{k=0}^{n} c_k l_k(x)l_k(y) .$", "Solution_1": "$Tf(x,y)=2f\\Big(\\frac{xy}{4}+\\frac{1}{4}\\sqrt{(x^2+4)(y^2+4)}\\Big)+ 2f\\Big(\\frac{xy}{4}-\\frac{1}{4}\\sqrt{(x^2+4)(y^2+4)}\\Big)$\r\n\r\n[this is a silly problem anyway..]", "Solution_2": "[quote=\"cimabue\"]$Tf(x,y)=....$[/quote] Please take care how you use LaTex: I do'nt understand !", "Solution_3": "Here my proof \r\n$T_n(x)=1/2(x+\\sqrt{x^2-1})^n+1/2(x-\\sqrt{x^2-1})^n$\r\n$L_n(x)=2i^{n}T_n(-ix/2)$\r\n\r\n$L_k(x)L_k(y)=4(-1)^kT_k(-ix/2)T_k(-iy/2)$ (1)\r\n\r\nUse the definition of $T_k(x)=\\cos(n\\arccos x))$\r\n\r\n$T_k(x)T_k(y)=1/2(\\cos(n(\\arccos x + \\arccos y))+\\cos(n(\\arccos x - \\arccos y))= 1/2(T_k(xy-\\sqrt{1-x^2}\\sqrt{1-y^2})+T_k(xy+\\sqrt{1-x^2}\\sqrt{1-y^2})$ (2)\r\n\r\n(1) and (2) gives \r\n\r\n$L_k(x)L_k(y)=2T_k(xy/4+1/4\\sqrt{4+x^2}\\sqrt{4+y^2})+T_k(xy/4-1/4\\sqrt{4+x^2}\\sqrt{4+y^2})$ (2)\r\n\r\nand you get the formulae of cimabue" } { "Tag": [], "Problem": "[b][size=100][color=DarkBlue]\u0394\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf $ \\bigtriangleup ABC$ \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 $ D,$ \u03c4\u03bf\u03c5 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03bc\u03ae\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 $ (O),$ \u03b1\u03c0\u03cc \u03c4\u03b7 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03bf\u03b4\u03b9\u03ac\u03bc\u03b5\u03c3\u03bf \u03b4\u03b9\u03b1 \u03c4\u03b7\u03c2 \u03ba\u03bf\u03c1\u03c5\u03c6\u03ae\u03c2 $ A.$ \u0395\u03ac\u03bd $ D'$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b1\u03bd\u03c4\u03b9\u03b4\u03b9\u03b1\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03c5 $ D,$ \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b7 \u03c3\u03b7\u03bc\u03b5\u03b9\u03bf\u03c3\u03b5\u03b9\u03c1\u03ac $ K\\equiv BC\\cap AD',\\ B,\\ L\\equiv BC\\cap DD',\\ C,$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03ae.[/color][/size][/b]\r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.", "Solution_1": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03c4\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03c0\u03bb\u03b5\u03c5\u03c1\u03bf $ ABDC$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03cc \u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 $ (D'A,D'B,D'D,D'C)\\equal{}\\minus{}1$ \u03c0\u03ac\u03c1\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03c4\u03bf\u03bc\u03ae \u03b1\u03c5\u03c4\u03ae\u03c2 \u03c4\u03b7\u03c2 \u03b4\u03ad\u03c3\u03bc\u03b7\u03c2 \u03bc\u03b5 \u03c4\u03b7\u03bd $ BC$ \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf . \r\n\r\n**\u03a3\u03c5\u03b3\u03c7\u03c9\u03c1\u03ad\u03c3\u03c4\u03b5 \u03bc\u03b5 \u03b3\u03b9\u03b1 \u03c4\u03c5\u03c7\u03cc\u03bd \u03bb\u03ac\u03b8\u03b7 \u03b3\u03b9\u03b1\u03c4\u03af\u03b5\u03af\u03bc\u03b1\u03b9 \u03c3\u03b5 internet cafe \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03c3\u03c7\u03ae\u03bc\u03b1" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "hi\r\n\r\nwhat's the dual of $ T$, the order 12 rotational subgroup of $ T_d$, the tetrahedral group? dual group is defined as the group of all homomorphisms from $ T$ to $ \\mu(\\mathbb{C})$.\r\n\r\nthanks", "Solution_1": "It's $ Hom(T,\\mathbb{C}^x)\\equal{}Hom(T/T',\\mathbb{C}^x)\\cong T/T'\\cong \\mathbb{Z}/3$.", "Solution_2": "i'm really sorry i cannot understand. what's $ T/T'$? and why is it isomorphic to $ \\mathbb{Z}/3\\mathbb{Z}$?" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Consider the pair of four-digit positive integers\r\n\r\n$ (M, N) \\equal{} (3600, 2500)$.\r\n\r\nNotice that $ M$ and $ N$ are both perfect squares, with equal digits in two places, and differing digits in the remaining two places. Moreover, when the digits differ, the digit in $ M$ is exactly one greater than the corresponding digit in $ N$.\r\n\r\nFind all pairs of four-digit positive integers $ (M, N)$ with these properties.", "Solution_1": "[quote=\"AndrewTom\"]Consider the pair of four-digit positive integers\n\n$ (M, N) \\equal{} (3600, 2500)$.\n\nNotice that $ M$ and $ N$ are both perfect squares, with equal digits in two places, and differing digits in the remaining two places. Moreover, when the digits differ, the digit in $ M$ is exactly one greater than the corresponding digit in $ N$.\n\nFind all pairs of four-digit positive integers $ (M, N)$ with these properties.[/quote]\r\nSo $ x^2\\minus{}y^2\\in\\{11,101,110,1001,1010,1100\\}$. from which we'll only keep $ \\{11,101,1001,1100\\}$ since $ x^2\\minus{}y^2$ cant be $ 2$ mod 4.\r\n\r\n$ x^2\\minus{}y^2\\equal{}11$ $ \\implies$ $ (x\\plus{}y,x\\minus{}y)\\in\\{(11,1)\\}$ and so $ (x,y)\\in\\{(6,5)\\}$ so no solution (no 4 digits squares)\r\n$ x^2\\minus{}y^2\\equal{}101$ $ \\implies$ $ (x\\plus{}y,x\\minus{}y)\\in\\{(101,1)\\}$ and so $ (x,y)\\in\\{(51,50)\\}$ which is a solution\r\n$ x^2\\minus{}y^2\\equal{}1001$ $ \\implies$ $ (x\\plus{}y,x\\minus{}y)\\in\\{(1001,1),(143,7),(91,11),(77,13)\\}$ and so $ (x,y)\\in\\{(501,500),(75,68),(51,40),(45,32)\\}$ from which only $ \\{(75,68),(51,40),(45,32)\\}$ are solutions.\r\n$ x^2\\minus{}y^2\\equal{}1100$ $ \\implies$ $ (x\\plus{}y,x\\minus{}y)\\in\\{(550,2),(110,10),(50,22)\\}$ (I kept only couple where elements are both odd or both even) and so $ (x,y)\\in\\{(276,274),(60,50),(36,14)\\}$ from which only $ \\{(60,50)\\}$ is solution.\r\n\r\nHence the answer : $ \\boxed{(M,N)\\in\\{2601,2500),(5625,4624),(2601,1600),(2025,1024),(3600,2500)\\}}$" } { "Tag": [ "AMC", "AIME" ], "Problem": "(Me) For some positive real $ x,k$,\r\n\\[ x^5 \\plus{} \\frac {1}{x^5} \\equal{} k^5 \\minus{} 5k^3 \\plus{} 5k \\equal{} \\frac {33}{4 \\sqrt {2}}\r\n\\]\r\nIf $ k \\equal{} x \\plus{} \\frac {1}{x}$ and $ x \\equal{} \\sqrt {a}$ where $ a$ is a positive integer not divisible by square of any prime, find $ a^{a^a}$.\r\n\r\nLevel: Medium Hidden Discoveries Mathematics Challenge (equivalent to AIME)", "Solution_1": "Erm, it looks like this problem is missing something. Like a way to determine $ a$. :P", "Solution_2": "When I plug in $ k\\equal{}x\\plus{}\\frac{1}{x}$ on the RHS, I get $ x^5\\plus{}\\frac{1}{x^5}\\equal{}x^5\\plus{}\\frac{1}{x^5}$, so the equation works for all values of $ x$...?", "Solution_3": "Ah, I forgot to put the last part to this question (I had it but I forgot).\r\n\r\nI fixed it now.", "Solution_4": "[hide=\"The magic of lucky guessing...\"]$ \\dfrac {33}{4 \\sqrt {2}}\\equal{}\\dfrac {33}{\\sqrt {32}}\\equal{}\\dfrac {32}{\\sqrt {32}}\\plus{}\\dfrac {1}{\\sqrt {32}}\\equal{} \\sqrt {32}\\plus{}\\dfrac {1}{\\sqrt {32}}\\\\\nx^5\\plus{}\\dfrac {1}{x^5}\\equal{}\\sqrt {32}\\plus{}\\dfrac {1}{\\sqrt {32}}\\\\\nx^5\\equal{}\\sqrt {32}\\\\\nx\\equal{}\\sqrt {2}\\\\\na\\equal{}2\\\\\n2^{2^2}\\equal{}2^4\\equal{}\\boxed{16}$\n[/hide]", "Solution_5": "[hide=\"Easy bounding solution\"]\nWe can calculate that $ \\frac {33}{4 \\sqrt {2}} \\approx 5.8 < 6$. So\n\\[ a^2\\sqrt {a} \\plus{} \\frac {1}{a^2 \\sqrt {a}} < 6\n\\]\nBut the only integers $ a$ such that $ a^2 < 6$ are $ a \\equal{} 1$ and $ 2$. We check and see that $ (\\sqrt {2})^5 \\plus{} (1/\\sqrt {2})^5 \\equal{} \\frac {33}{4 \\sqrt {2}}$. So 2^(2^2) = 16.[/hide]\n\n[hide=\"Even easier\"]\n3^(3^3) is too big to calculate without a calculator, so 2^(2^2) = 16. QED :P[/hide]\n\n[hide=\"Or since a is squarefree...\"]\n\\[ a^2 \\sqrt {a} \\plus{} \\frac {\\sqrt {a}}{a^3} \\equal{} \\frac {33}{8} \\cdot \\sqrt {2}\n\\]\n\n\\[ \\frac {a^2 \\plus{} 1/a^3}{33/8} \\equal{} \\sqrt {2/a}\n\\]\nThe left hand side is rational. Therefore $ 2/a$ must be the square of a rational number, so $ a \\equal{} 2\\cdot b^2$ where $ b$ is an integer. However, $ a$ is squarefree, and so $ a \\equal{} 2$.[/hide]", "Solution_6": "What is the purpose of $ k$ in this problem?", "Solution_7": "[quote=\"Altheman\"]What is the purpose of $ k$ in this problem?[/quote]\r\n\r\nRed herring, I guess? Though it's pretty hard to get led off track by it; probably why he's not using the question! :)", "Solution_8": "\\[ \\left(x^5 \\plus{} x^{ \\minus{} 5}\\right)^2 \\equal{} a^5 \\plus{} a^{ \\minus{} 5} \\plus{} 2 \\equal{} \\frac {33^2}{32}\r\n\\]\r\n\r\n\\[ a^{10} \\minus{} \\frac {1025}{32}\\cdot a^5 \\plus{} 1 \\equal{} 0\r\n\\]\r\n\r\n\\[ a^5(32a^5 \\minus{} 1025) \\equal{} \\minus{} 32\r\n\\]\r\nSince $ a$ is an integer, a fifth power of some integer divides the LHS, so it must divide the RHS. It is clear then that, necessarily, $ a\\in \\{ \\minus{} 2, \\minus{} 1,1,2\\}$. Checking gives $ a \\equal{} 2$.", "Solution_9": "[quote=\"Altheman\"]What is the purpose of $ k$ in this problem?[/quote]\r\n\r\nWell, I pondered on whether to put $ x^5 \\plus{} \\frac{1}{x^5}$ or not. Obviously, having that made the problem very easy.\r\n\r\nBut, if you were to just solve the problem like this:\r\n\r\n$ k^5 \\minus{} 5k^3 \\plus{} 5k \\equal{} \\frac {33}{4 \\sqrt {2}}$\r\n\r\nIt is not possible to \"get lucky\" on this. But then, only with this form, it is practically impossible to solve this problem. The only way is if you're familiar with going up higher exponents from $ x \\plus{} \\frac{1}{x}$ (which is how I came up with $ k^5 \\minus{} 5k^3 \\plus{} 5k$) but still, this is not what AMC/AIME will likely to test people. It is not just a very approachable problem as most AMC problems are so I decided to not put on the contest." } { "Tag": [], "Problem": "What is the thousands digit of $ 9\\cdot37^{2}\\cdot77\\cdot91\\cdot143$?\r\n\r\n\r\nYes, there's an easier way to do this.", "Solution_1": "EDIT: I added my signature\r\n\r\n[hide=\"Umm...lets see\"]\n\nAnswer: 4\n\nYeah, all i did was type the numbers into my TI-89... :rotfl: :blush: \n\n[/hide]", "Solution_2": "[hide=\"Hint 1\"] The given quantity is a square. [/hide]\n[hide=\"Hint 2\"] $ 1001 = 7 \\cdot 11 \\cdot 13$ [/hide]\n[hide=\"Hint 3\"] $ 111 = 3 \\cdot 37$ [/hide]\n[hide=\"HInt 4\"] $ (1+x+...+x^{n})^{2}= 1+2x+3x^{2}+...$ [/hide]", "Solution_3": "Note that that whole thing is equal to $ (3\\cdot37\\cdot7\\cdot11\\cdot13)^{2}$. Then you pull out your 1337 bag and notice that $ 3\\cdot37=111$ and $ 7\\cdot11\\cdot13=1001$ like we all have memorized, then we reach further into our 1337 bag, and remember that when we multiply a three digit number $ \\overline{abc}$ by 1001 the product is $ \\overline{abcabc}$, which we also all have memorized. Then we find find that the quantity we want is just $ (111111)^{2}$, which, by another part of our 1337 bag, is just 12345654321, which we, also have memorized. But if we don't have that memorized, we note that if we were to long multiply $ 111111\\cdot111111$ the traditional way, when we add at the end, we would have one 1, then two one's, then a stack of three one's, then four, then five, then six, then five, back down to one 1. \r\n\r\nHence the thousands digit is 4.\r\n\r\nAs you can see, then 1337 bag was very helpful in this problem.", "Solution_4": "blah, use the indianamath bag." } { "Tag": [ "calculus", "derivative", "analytic geometry", "graphing lines", "slope", "trigonometry", "calculus computations" ], "Problem": "an equation of the line tangent to the graph of y=cos(2x) at x=pi/4 is?", "Solution_1": "What are you having trouble with? Use a derivative to get your slope, then you have a slope and a point, so you can find the equation.", "Solution_2": "$ y'\\equal{}\\minus{}2\\sin{2x}$. Plug in $ x\\equal{}\\frac{\\pi}{4}$ to get $ m\\equal{}\\minus{}2\\sin{\\pi}/{2}\\equal{}\\minus{}2$. Then find the equation from\r\n$ y_{1}\\minus{}y_{0}\\equal{}m(x_{1}\\minus{}x_{0})$\r\nTake $ (x_{1},y_{1})\\equal{}(\\pi/4,0)$ so we get\r\n$ \\minus{}y\\equal{}\\minus{}2(\\pi/4\\minus{}x)\\Rightarrow y\\equal{}\\frac{\\pi}{2}\\minus{}2x$" } { "Tag": [ "function", "algebra", "polynomial", "integration", "calculus", "derivative", "quadratics" ], "Problem": "Determine all continous functions $f: R\\to\\mathbb R$ such that \r\n $F(f(x))+f(F(x))=x^2$, $\\forall$ $x$ $\\in$ $R$, where $F'(x)=f(x)$ ,$F(0)=0$.", "Solution_1": "$f(x)=x$ and $f(x)=0$ work. not sure about proving these are the only two functions.", "Solution_2": "Any ideeas? :lol:", "Solution_3": "My be there are general method to solve this equation , but l got simple idea ,\r\n if we write $f(x) = \\sum\\limits_{i = 0}^n {a_i x^i }$ as a polynomial , n must be $\\le 1$ cause if n > 1 $\\deg (F(f(x)) +(f(F(x)) \\ge 6$ contradict $\\deg (x^2 ) = 2$\r\nthere are tow cases :\r\n(1) $f(x) = ax + b$ substitute in the equation ,we get a = 1 and b = o and one of solution is $f(x) = x$\r\n(2) $f(x) = m$ and with substituting we get m= 0 and the other solution is $f(x) = 0$", "Solution_4": "[quote=\"franck\"]My be there are general method to solve this equation , but l got simple idea ,\n if we write $f(x) = \\sum\\limits_{i = 0}^n {a_i x^i }$ as a polynomial , n must be $\\le 1$ cause if n > 1 $\\deg (F(f(x)) +(f(F(x)) \\ge 6$ contradict $\\deg (x^2 ) = 2$\nthere are tow cases :\n(1) $f(x) = ax + b$ substitute in the equation ,we get a = 1 and b = o and one of solution is $f(x) = x$\n(2) $f(x) = m$ and with substituting we get m= 0 and the other solution is $f(x) = 0$[/quote]\r\nOk.If f is not polynomials function :D ??", "Solution_5": "csabafarkas, can you post the solution please?", "Solution_6": "ok! :D", "Solution_7": "[quote=\"franck\"]ok! :D[/quote]\r\n\r\nWhat is OK!! :D", "Solution_8": "This problem looks really hard to me :)\r\n\r\nI don't know why $f(x) = 0$ has been mentioned, though -- it's clearly not a solution of the original equation.", "Solution_9": "What about this trying , :D \r\n\r\nWith checking we found that $f(x)=0$ is a solution , otherwise ,\r\n\r\nLet $g(x)=F(f(x))+f(F(x))=x^2$ ( * ) g is even , yielding $g(x)=g(-x)$\r\n\r\n$F(f(x))+f(F(x))=F(-f(x))+f(-F(x))$\r\n\r\n$F(f(x)) - F(-f(x)) = - (f(F(x))-f(-F(x)))$ --- ( ** )\r\n\r\nAny function , so f , is odd or even or ( neither odd or even )\r\n\r\nFrom ( ** ) We conclude that ( 1 ) : $F(f(x)) = f(F(x)$ or ( 2 ) : $F(f(x)) = - f(F(x)$ ( the last case doesn't exist ( * ) \r\n\r\nSub ( 1 ) in ( * ) we get $F(f(x)) = \\frac{{x^2 }}{2}$ or \r\n\r\n$\\int\\limits_0^{f(x)} {f(t)dt = \\frac{{x^2 }}{2}} = \\int\\limits_0^x {tdt}$\r\n\r\n$\\int\\limits_0^0 {f(t)dt} = \\int\\limits_{f(x)}^x {tdt} \\Rightarrow 0 = \\frac{{x^2 }}{2} - \\frac{{f^2 (x)}}{2}$\r\n\r\n$f^2 (x) = x^2 \\Rightarrow f(x) = \\pm x$ ( by checking , we found the other solution is $f(x) = x$ ).", "Solution_10": "[quote=\"franck\"]What about this trying , :D \n\nWith checking we found that $f(x)=0$ is a solution , otherwise ,\n\nLet $g(x)=F(f(x))+f(F(x))=x^2$ ( * ) g is even , yielding $g(x)=g(-x)$\n\n$F(f(x))+f(F(x))=F(-f(x))+f(-F(x))$\n\n$F(f(x)) - F(-f(x)) = - (f(F(x))-f(-F(x)))$ --- ( ** )\n\nAny function , so f , is odd or even or ( neither odd or even )\n\nFrom ( ** ) We conclude that ( 1 ) : $F(f(x)) = f(F(x)$ or ( 2 ) : $F(f(x)) = - f(F(x)$ ( the last case doesn't exist ( * ) \n\nSub ( 1 ) in ( * ) we get $F(f(x)) = \\frac{{x^2 }}{2}$ or \n\n$\\int\\limits_0^{f(x)} {f(t)dt = \\frac{{x^2 }}{2}} = \\int\\limits_0^x {tdt}$\n\n$\\int\\limits_0^0 {f(t)dt} = \\int\\limits_{f(x)}^x {tdt} \\Rightarrow 0 = \\frac{{x^2 }}{2} - \\frac{{f^2 (x)}}{2}$\n\n$f^2 (x) = x^2 \\Rightarrow f(x) = \\pm x$ ( by checking , we found the other solution is $f(x) = x$ ).[/quote]\r\n\r\nI think is not ok!My opinion is $g(-x)=F(f(-x))+f(F(-x))$ :)", "Solution_11": "Let $g(x)=F(f(x))+f(F(x))=x^2$ ( * ) g is even , yielding $g(x)=g(-x)$\r\n\r\n$F(f(x))+f(F(x))=F(f(-x))+f(F(-x))$\r\n\r\n$F(f(x)) - F(f(-x)) = - (f(F(x))-f(F(-x)))$ --- ( ** )\r\n\r\nAny function , so f , is odd or even or ( neither odd or even )\r\n\r\nFrom ( ** ) We conclude that ( 1 ) : $F(f(x)) = f(F(x)$ or ( 2 ) : $F(f(x)) = -f(F(x)$ ( the last case doesn't exist ( * ) \r\n\r\nSub ( 1 ) in ( * ) we get $F(f(x)) = \\frac{{x^2 }}{2}$ or ( complete as above )\r\n\r\n[quote=\"csabafarkas\"]\n\nI think is not ok!My opinion is $g(-x)=F(f(-x))+f(F(-x))$ :)[/quote]\r\n\r\n*******************************************************************\r\nthe idea in my last solution is : ;) \r\n( $g$ is even , $g(x)=g(-x) \\forall x \\in R$ , \r\nSINCE $f(x),F(x) \\in R$ , we can write $g(f(x))=g(-f(x)) , g(F(x))=g(-F(x))$ ) ;) ,\r\n so $g(f(x)), F(x)) = g ( f(-x)) , F(-x)))$ and $f(F(x))+F(f(x))=f(-F(x))+F(-f(x))$\r\n*******************************************************************\r\n\r\ncsabafarkas, can you post the solution please? :)", "Solution_12": "If f is even $f(-x)=f(x)$ then $F'(-x)=F'(x)$ and $F(x)=F(-x)+c ,F(0)=0$ so $F(-x)=F(x)$ so $F(f(x))=F(f(-x))$ so $F(f(x))-F(f(-x))=0$.\r\nWhy $F(f(x))=f(F(x))$? :)", "Solution_13": "[quote=\"franck\"]From ( ** ) We conclude that ( 1 ) : $F(f(x)) = f(F(x)$ or ( 2 ) : $F(f(x)) = -f(F(x)$ ( the last case doesn't exist ( * )[/quote]\nWhere does this conclusion come from?\n\n[quote=\"csabafarkas\"]$F'(-x)=F'(x)$ and $F(x)=F(-x)+c ,F(0)=0$ so $F(-x)=F(x)$[/quote]\r\nThis is wrong -- if $F'(-x)=F'(x)$ then $F(x)=-F(-x)+c ,F(0)=0$ so $F(-x)=-F(x)$ -- the derivative of an odd function is even, and vice versa.", "Solution_14": "[quote=\"JBL\"][quote=\"franck\"]From ( ** ) We conclude that ( 1 ) : $F(f(x)) = f(F(x)$ or ( 2 ) : $F(f(x)) = -f(F(x)$ ( the last case doesn't exist ( * )[/quote]\nWhere does this conclusion come from?\n\n[quote=\"csabafarkas\"]$F'(-x)=F'(x)$ and $F(x)=F(-x)+c ,F(0)=0$ so $F(-x)=F(x)$[/quote]\nThis is wrong -- if $F'(-x)=F'(x)$ then $F(x)=-F(-x)+c ,F(0)=0$ so $F(-x)=-F(x)$ -- the derivative of an odd function is even, and vice versa.[/quote]\r\nyes Thanks JBL\r\n :blush: :D :!:", "Solution_15": "[quote=\"JBL\"][quote=\"franck\"]From ( ** ) We conclude that ( 1 ) : $F(f(x)) = f(F(x)$ or ( 2 ) : $F(f(x)) = -f(F(x)$ ( the last case doesn't exist ( * )[/quote]\nWhere does this conclusion come from?[/quote]\r\nI have conclude the last two solutions by notice symmetrical between both sides of the equation .\r\n\r\n $F(f(x)) - F(-f(x)) = - (f(F(x))-f(-F(x)))$ --- ( ** )\r\n\r\n $F(f(x)) = f(F(x))$ exist (**)\r\n\r\n sub in both sides $f(F(x))-F(-f(x))= - ( F(f(x)-f(-F( x))$\r\n\r\n so $F(f(x))-F(-f(x)) = -f(F(x)+f(-F(x))$\r\n\r\n $F(f(x))-F(-f(x))=-(f(F(x)-f(-F(x))$ as required\r\n\r\n, another symmetric $F(f(x)) = - f(F(x))$ doesn't exist ( ** ) \r\n\r\nJbl , l appreciate your opinion that you mentioned above ,Please tell me if there is falses , May be there are another solutions through another ideas but i achieved this . :)", "Solution_16": "[quote=\"franck\"] $F(f(x)) - F(-f(x)) = - (f(F(x))-f(-F(x)))$ --- ( ** )\n\n $F(f(x)) = f(F(x))$ exist (**)[/quote]\r\n\r\nI am confused about this step. Perhaps I am missing something obvious. Also, it's not clear to me: you are saying that there exists some value $x_0$ such that $f(F(x_0)) = F(f(x_0))$, right? Then, you make a substitution into the equation (**) and end up pulling out the equation (**) again ... I don't think I know what you're proving. Perhaps you can add more explanation to your steps?\r\n\r\n\r\nA suggestion that this question may be very very hard:\r\nIf we drop the condition $F(0) = 0$ and try to solve instead the equation $f(F(x)) + F(f(x)) = x^2 + (3 + 3\\sqrt5)x -\\frac{15+21\\sqrt5}2$, we would find not only a quadratic solution for $F$ but also a cubic. Now, it is possible that dropping $F(0) = 0$ increases the wildness of the solution, and it's possible that changing the quadratic we're trying to get increases the wildness of the solution set, but I think this suggests that this question could have a great many solutions we won't find.", "Solution_17": "[quote=\"JBL\"]This problem looks really hard to me :)\n\nI don't know why $f(x) = 0$ has been mentioned, though -- it's clearly not a solution of the original equation.[/quote]\r\n\r\nAs JBL said : $f(x) = 0$ doesn't exist the equation because let $f(x)=0 \\Rightarrow F(x) = c \\,\\, \\forall x \\in R$ , But $F(0)=0$ , So $F(x)=0 \\,\\, \\forall x \\in R$ , substitute in ( * ) we find that $0=x^2 \\,\\,\\forall x \\in R$ which isn't true .", "Solution_18": "Franck teh solution is not correct... :(", "Solution_19": "[quote=\"franck\"][quote=\"JBL\"]This problem looks really hard to me :)\nI don't know why $f(x) = 0$ has been mentioned, though -- it's clearly not a solution of the original equation.[/quote]\nAs JBL said : $f(x) = 0$ doesn't exist the equation because let $f(x)=0 \\Rightarrow F(x) = c \\,\\, \\forall x \\in R$ , But $F(0)=0$ , So $F(x)=0 \\,\\, \\forall x \\in R$ , substitute in ( * ) we find that $0=x^2 \\,\\,\\forall x \\in R$ which isn't true .[/quote]\r\nFrom this result , we see that $f(x) = 0 \\Rightarrow x = 0 , f(0)=0$\r\nI will (try) to proof that $f(x)=x$ is the only solution on $R$ ,\r\nLet ${f(x)}>x$ in some interval $( a , b ) \\subset R^+$ , $\\frac{{f(x)}}{x} = \\frac{{f(x) - f(0)}}{{x - 0}} > 1$ on ( a,b) , take the limit when $x \\to 0$ , we obtain $f'(x) > 1 > 0$ , So $f$ is strictly increasing on ( a, b) .\r\nNow \\[ f(x) > x \\Rightarrow f(f(x)) > f(x) \\Rightarrow f(f(x))f'(x) > f(x)f'(x) \\] on ( a,b)\r\nIntegrate both sides on any interval $\\subset ( a,b)$ , then :\r\n\r\n\\[ F(f(x)) > \\frac{{f^2 (x)}}{2} > \\frac{{x^2 }}{2} ........ ( 1 ) \\]\r\n\r\nIntegrate both sides in $f(x)>x$ on ( a,b) we get $F(x) > \\frac{{x^2 }}{2}$ , since $f$is strictly increasing on ( a , b ) , then \\[ f(F(x)) > f(\\frac{{x^2 }}{2}) > \\frac{{x^2 }}{2} ......... ( 2 ) \\]\r\nPlug ( 1 ) , ( 2 ) on the equation , we get $f(F(x)) + F(f(x)) > x^2$ , a contradiction ,\r\nSame with $R^-$ , and the other case $f(x)x :)", "Solution_21": "Franck is not correct :( \r\n1) we can take only one x where f(x)>x\r\n2) f is not certanly derivable\r\n ;)", "Solution_22": "[quote=\"csabafarkas\"]Determine all continous functions $f: R\\to\\mathbb R$ such that \n $F(f(x))+f(F(x))=x^2$, $\\forall$ $x$ $\\in$ $R$, where $F'(x)=f(x)$ ,$F(0)=0$.[/quote]\n\n[quote=\"csabafarkas\"]I have a problem:\n1)f is not certanly dervable\n2) we can take only one x where f(x)>x :)[/quote]\r\n\r\n1)\r\nwe have given $F'(x) = f(x) \\forall x\\in R$ , that is $F(x)$ is differentiable $\\forall x\\in R$ , now suppose for contrary that $F(f(x))$ isn't differentiable at $x = x_0 \\in R$ , since $f(x) : R \\to R , \\exists y_0 \\in R .s.t . f(x_0 ) = y_0$ implies that $F(x)$ isn't differentiable at $y_0=f(x_0)$ ( contradict that $F(x)$ is differentiable $\\forall x\\in R$ ) \r\nwe deduce that $F(f(x))$ is differentiable $\\forall x\\in R$ , so $\\left( {F(f(x))} \\right)^\\prime$ is defined as well $\\forall x\\in R$ , in other words , $f(f(x)).f'(x)$ is exist $\\forall x\\in R .....(I)$.\r\nnow , if we say that ( f is not certanly direvable ) then there are $x_0 \\in R$ .s.t. $f'(x_0 )$ doesn't exist ,which contradict .(I).\r\n\r\n2) if $f(x) \\ne x$ , let $f(x_0 ) > x_0$ , for one point $x = x_0 \\in R$ , we claim that $f(x) > x$ in some neighborhood around $x_0$ , say $(x_0 - \\varepsilon ,x_0 + \\varepsilon ) = (a,b)$ \r\nIf isn't then $f(x_0 ) =y_0> x_0$ , while $\\mathop {\\lim }\\limits_{x \\to x_0 } f(x)\\ne y_0$ , which contradict that $f$ is continuous on R as we search for .", "Solution_23": "csabafarkas , ? :lol:", "Solution_24": "[quote=\"franck\"]csabafarkas , ? :lol:[/quote] \r\n:P", "Solution_25": "[quote=\"csabafarkas\"][quote=\"franck\"]csabafarkas , ? :lol:[/quote] \n:P[/quote] is not correct\r\nby taking the limit with x tends to 0, we obtain the derivative in 0, and this doen't imply the monotonicity on the interval (a,b) doesn't", "Solution_26": "[quote=\"csabafarkas\"][quote=\"csabafarkas\"][quote=\"franck\"]csabafarkas , ? :lol:[/quote] \n:P[/quote] is not correct\nby taking the limit with x tends to 0, we obtain the derivative in 0, and this doen't imply the monotonicity on the interval (a,b) doesn't[/quote]\r\n\r\nfranck? :D", "Solution_27": "[quote=\"csabafarkas\"][/quote][quote=\"csabafarkas\"][quote=\"csabafarkas\"][quote=\"franck\"]csabafarkas , ? :lol:[/quote] \n:P[/quote] is not correct\nby taking the limit with x tends to 0, we obtain the derivative in 0, and this doen't imply the monotonicity on the interval (a,b) doesn't[/quote][quote=\"csabafarkas\"]\n\nfranck? :D[/quote]\r\n\r\nby taking the limit with x tends to 0 ( taking limit for what , what is the function you have limit for ? ) \r\nand i dont know what are you talking about :evilgrin:\r\ncan you give more details ? :roll:", "Solution_28": "[quote=\"franck\"][/quote][quote=\"csabafarkas\"][/quote][quote=\"csabafarkas\"][quote=\"csabafarkas\"][quote=\"franck\"]csabafarkas , ? :lol:[/quote] \n:P[/quote] is not correct\nby taking the limit with x tends to 0, we obtain the derivative in 0, and this doen't imply the monotonicity on the interval (a,b) doesn't[/quote][quote=\"csabafarkas\"]\n\nfranck? :D[/quote][quote=\"franck\"]\n\nby taking the limit with x tends to 0 ( taking limit for what , what is the function you have limit for ? ) \nand i dont know what are you talking about :evilgrin:\ncan you give more details ? :roll:[/quote]\r\nSorry is not correct :!:" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If $ a,b,c>0$, then\r\n\r\n$ \\frac{a^2}{a^2\\plus{}b^2}\\plus{}\\frac{b^2}{b^2\\plus{}c^2}\\plus{}\\frac{c^2}{c^2\\plus{}a^2}\\leq\\frac{1}{2}(\\frac{a^2}{b^2}\\plus{}\\frac{b^2}{c^2}\\plus{}\\frac{c^2}{a^2})$", "Solution_1": "[quote=\"filomen\"]If $ a,b,c > 0$, then\n\n$ \\frac {a^2}{a^2 \\plus{} b^2} \\plus{} \\frac {b^2}{b^2 \\plus{} c^2} \\plus{} \\frac {c^2}{c^2 \\plus{} a^2}\\leq\\frac {1}{2}(\\frac {a^2}{b^2} \\plus{} \\frac {b^2}{c^2} \\plus{} \\frac {c^2}{a^2})$[/quote]\r\nUsing the AM-GM Ineq, we have:\r\n\\[ \\sum \\frac{a^2}{a^2\\plus{}b^2} \\le \\frac{1}{2}(\\sum \\frac{a}{b})\\]\r\nSo that we need to prove:\r\n\\[ \\sum \\frac{a}{b} \\le \\sum \\frac{a^2}{b^2}\\]\r\nThis's a known result with $ a,b,c \\in R$\r\n :lol:" } { "Tag": [ "topology", "limit", "logarithms" ], "Problem": "Find a countable normal T1- Spape but not metrizable. \r\n\r\nthanks in advance", "Solution_1": "Does countable mean countably infinite? If not, then the standard example of a $ T_1$ but not $ T_2$ will work, since every metrizable space is $ T_2$.\r\n\r\n$ X\\equal{}\\{a, b\\}$ and $ \\mathcal{T}\\equal{}\\{\\emptyset, \\{a\\}, \\{a, b\\}\\}$.", "Solution_2": "it need to be [b]NORMAL[/b], your example is not", "Solution_3": "Oh sorry, I misread that. I'd usually call normal + T_1 the $ T_4$ axiom.\r\n\r\nThis countable condition makes this quite hard. I have tons of good uncountable examples.", "Solution_4": "thanks for your attention\r\n\r\nMy idea is : make a normal space but not metrizable , and then take a countable subspace but I have a difficulty is : subspace of normal space [b]may not[/b] be normal . I am still getting stuck", "Solution_5": "I got one!\r\n\r\nMy first instinct was to try the \"Fort space.\" Unfortunately, the uncountable one works but the countable one does not. Here is a variant on that space that I think works. It is sort of complicated, though. \r\n\r\nLet $ X\\equal{}\\mathbb{Z}^\\plus{}$ (positive integers). Let $ N(n, E)$ be the number of integers in a set $ E\\subset X$ that are less than or equal to $ n$. So for $ E\\equal{}\\{1, 3, 7, 27\\}$, we have $ N(8, E)\\equal{}3$.\r\n\r\nDefine a topology by declaring a set $ U$ open if it excludes 1, or if $ \\lim_{n\\to\\infty}N(n, U)/n\\equal{}1$. \r\n\r\nSo note that $ E$ is closed iff $ 1\\in E$ or $ 1\\notin E$ and $ \\lim_{n\\to\\infty}N(n, E)/n\\equal{}0$.\r\n\r\nTo get normal, break it into cases. If $ A$ and $ B$ are closed disjoint sets, and $ 1$ is not in either set, then both are open so we are done. So assume $ 1\\in A$. Then since $ 1\\notin B$, we have $ \\lim_{n\\to\\infty} N(n, B)/n\\equal{}0$, so $ B\\supset B$ and $ X\\setminus B\\supset A$ are both open and we have normal.\r\n\r\n$ T_1$ is trivial since if neither of your points are 1, the are both open sets. And if one of the points is 1, the other is not and it is open.\r\n\r\nThe tricky part is that $ X$ is not first countable, and hence not metrizable. So the claim is that $ 1$ has no countable local basis. Suppose it does, say $ \\{B_n\\}$. Since each $ B_n$ contains 1, it must be infinite. Choose $ x_n\\in B_n$ such that $ x_n>e^n$. Then define $ U\\equal{}X\\setminus \\{x_n\\}$. Clearly $ U$ does not contain any $ B_n$. but it is a neighborhood of $ 1$ since $ N(n, U)>n\\minus{}\\ln n$, so $ \\lim_{n\\to\\infty}N(n, U)/n\\equal{}1$. \r\n\r\nHopefully that works :)", "Solution_6": "thanks, I am trying to understand your example" } { "Tag": [ "geometry", "perimeter", "incenter", "circumcircle", "AMC", "AIME", "search" ], "Problem": "1.Three parallel chords of a circle have lengths 2,3,4 and subtend angles x,y,x+y at the center (x+y <180). Find cos x\r\n\r\n\r\n2.A hexgon is inscribed in a circle. Five sides have length 81, and the other side length 31. Find the sum of the three diagonals form a vertex on the short side\r\n\r\n\r\n3.The triangle ABC has AB=1995, bc=1993, CA=1994. CX is an altitude. Find the distance between the points at which the incircles of ACX and BCX touch CX\r\n\r\n4. The incircle of ABC touches AB at P and has radius 21. If AP=23 and PB=27 find the perimeter of ABC\r\n\r\n5. A circle radius r is inscribed in ABCD. It touches AM at P and CD at Q. AP=19, PB=26, CQ=37,QD=23. Find r.\r\n\r\n6. A triangle has sides 20,21,22. The line through the incenter parallel to the shortest side meets the other two sides at X and Y. Find XY\r\n\r\n7. The triangle ABC has AB=24. The median CE is extended to meet te circumcircle at F. CE=27, and the median AD=18. Find area of ABF", "Solution_1": "a couple of these problems have been posted before on this forum:\r\n\r\nfor example, heres 5\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=385895#p385895\r\n\r\nEDIT: Actually, now that i think of it, these are like all AIME problems....use the search function or check Contests" } { "Tag": [ "logarithms", "floor function", "algebra proposed", "algebra" ], "Problem": "Powers of $2$ in base $10$ start with $3$ or $4$ more frequently? What is their state in base $3$? First write down an exact form of the question.", "Solution_1": "You have to find whether the density of those $n$ for which $\\left\\{n\\log_{10}2\\right\\}\\in(\\log_{10}3,\\log_{10}4)$ (the necessary and sufficient condition for $2^n$ to begin with $3$) is larger than that of those $n$ for which $\\left\\{\\log_{10}2\\right\\}\\in(\\log_{10}4,\\log_{10}5)$ (the necessary and sufficient condition for $2^n$ to begin with $4$). Use Weyl's Theorem: the first density is $\\log_{10}\\frac 43$, while the second is $\\log_{10}\\frac 54$, smaller than the first. In conclusion, more powers of $2$ begin with $3$ than with $4$. \r\n\r\nI'm not sure what the second question means though. In base $3$, which numbers begin with $3$ in base $3$? Maybe the question turns into: are there more numbers of the form $2^n$ that begin with $10$ or with $11$ in base $3$? Anyway, an analysis like that above can be performed.", "Solution_2": "[quote=\"grobber\"]the density of those $n$ for which $\\{\\left\\lfloor\\log_{10}2\\right\\rfloor\\}\\in(\\log_{10}3,\\log_{10}4)$ (the necessary and sufficient condition for $2^n$ to begin with $3$)[/quote]\r\nI think you mistyped something. The expression $\\{\\left\\lfloor\\log_{10}2\\right\\rfloor\\}\\in(\\log_{10}3,\\log_{10}4)$ doesn't make sense! (There is no $n$ in it and the left side is always $0$)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Given $ a,b,c,d > 0$,and $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2 \\equal{} 4$.Prove that:\r\n$ \\frac {a^3}{b^2 \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\plus{} d^2} \\plus{} \\frac {c^3}{d^2 \\plus{} a^2} \\plus{} \\frac {d^3}{a^2 \\plus{} b^2} \\ge\\ 2$", "Solution_1": "$ \\frac {a^3}{b^2 \\plus{} c^2}\\geq\\ a^2 \\minus{} \\frac {1}{4}a(b^2 \\plus{} c^2)$\r\n$ \\frac {b^3}{c^2 \\plus{} d^2}\\geq\\ b^2 \\minus{} \\frac {1}{4}b(c^2 \\plus{} d^2)$\r\n$ \\frac {c^3}{d^2 \\plus{} a^2}\\geq\\ c^2 \\minus{} \\frac {1}{4}c(d^2 \\plus{} a^2)$\r\n$ \\frac {d^3}{a^2 \\plus{} b^2}\\geq\\ d^2 \\minus{} \\frac {1}{4}d(a^2 \\plus{} b^2)$\r\n$ \\frac {a^3}{b^2 \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\plus{} d^2} \\plus{} \\frac {c^3}{d^2 \\plus{} a^2} \\plus{} \\frac {d^3}{a^2 \\plus{} b^2} \\ge\\ 4 \\minus{} \\frac {1}{4}(a(b^2 \\plus{} c^2) \\plus{} b(c^2 \\plus{} d^2) \\plus{} c(d^2 \\plus{} a^2) \\plus{} d(a^2 \\plus{} b^2))$\r\n$ a(b^2 \\plus{} c^2) \\plus{} b(c^2 \\plus{} d^2) \\plus{} c(d^2 \\plus{} a^2) \\plus{} d(a^2 \\plus{} b^2)\\leq$$ \\frac {a^2 \\plus{} 1}{2}(b^2 \\plus{} c^2) \\plus{} \\frac {b^2 \\plus{} 1}{2}(c^2 \\plus{} d^2) \\plus{} \\frac {c^2 \\plus{} 1}{2}(d^2 \\plus{} a^2) \\plus{} \\frac {d^2 \\plus{} 1}{2}(a^2 \\plus{} b^2)$=$ 4 \\plus{} \\frac {1}{2}(a^2(b^2 \\plus{} c^2) \\plus{} b^2(c^2 \\plus{} d^2) \\plus{} c^2(d^2 \\plus{} a^2) \\plus{} d^2(a^2 \\plus{} b^2))\\leq$\r\n$ \\frac {1}{2}(a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2)^2 \\equal{} 8$", "Solution_2": "The following inequality is still easy.\r\nLet $ a,$ $ b,$ $ c,$ and $ d$ are positive numbers such that $ a^3 \\plus{} b^2 \\plus{} c^3 \\plus{} d^3 \\equal{} 4.$ Prove that\r\n\\[ \\frac {a^3}{b^2 \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\plus{} d^2} \\plus{} \\frac {c^3}{d^2 \\plus{} a^2} \\plus{} \\frac {d^3}{a^2 \\plus{} b^2} \\ge\\ 2\r\n\\]\r\nWhat about the following? :) \r\nLet $ a,$ $ b,$ $ c,$ and $ d$ are positive numbers such that $ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} d^4 \\equal{} 4.$ Prove that\r\n\\[ \\frac {a^3}{b^2 \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\plus{} d^2} \\plus{} \\frac {c^3}{d^2 \\plus{} a^2} \\plus{} \\frac {d^3}{a^2 \\plus{} b^2} \\ge\\ 2\r\n\\]", "Solution_3": "The following inequality is still easy.\r\nLet $ a,$ $ b,$ $ c,$ and $ d$ are positive numbers such that $ a^3 \\plus{} b^2 \\plus{} c^3 \\plus{} d^3 \\equal{} 4.$ Prove that\r\n\\[ \\frac {a^3}{b^2 \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\plus{} d^2} \\plus{} \\frac {c^3}{d^2 \\plus{} a^2} \\plus{} \\frac {d^3}{a^2 \\plus{} b^2} \\ge\\ 2\r\n\\]\r\n$ \\frac {a^3}{b^2 \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\plus{} d^2} \\plus{} \\frac {c^3}{d^2 \\plus{} a^2} \\plus{} \\frac {d^3}{a^2 \\plus{} b^2} \\ge\\ \\frac {(\\sum{a^3})^2}{a^3(b^2 \\plus{} c^2) \\plus{} b^3(c^2 \\plus{} d^2) \\plus{} c^3(d^2 \\plus{} a^2) \\plus{} d^3(a^2 \\plus{} b^2)}\\ge\\ \\frac {16}{a^3(\\frac {b^3 \\plus{} b^3 \\plus{} 1 \\plus{} c^3 \\plus{} c^3 \\plus{} 1}{3}) \\plus{} b^3(\\frac {c^3 \\plus{} c^3 \\plus{} 1 \\plus{} d^3 \\plus{} d^3 \\plus{} 1}{3}) \\plus{} c^3(\\frac {d^3 \\plus{} d^3 \\plus{} 1 \\plus{} a^3 \\plus{} a^3 \\plus{} 1}{3}) \\plus{} d^3(\\frac {a^3 \\plus{} a^3 \\plus{} 1 \\plus{} b^3 \\plus{} b^3 \\plus{} 1}{3})}$\r\n= $ \\frac {16}{\\frac {8}{3} \\plus{} \\frac {2}{3}(a^3b^3 \\plus{} a^3c^3 \\plus{} b^3c^3 \\plus{} b^3d^3 \\plus{} c^3d^3 \\plus{} c^3a^3 \\plus{} d^3a^3 \\plus{} d^3b^3)}\\ge$\r\n$ \\frac {16}{\\frac {8}{3} \\plus{} \\frac {1}{3}(a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} d^3)^2} \\equal{} 2$", "Solution_4": "[quote=\"arqady\"]What about the following? :) \nLet $ a,$ $ b,$ $ c,$ and $ d$ are positive numbers such that $ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} d^4 \\equal{} 4.$ Prove that\n\\[ \\frac {a^3}{b^2 \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\plus{} d^2} \\plus{} \\frac {c^3}{d^2 \\plus{} a^2} \\plus{} \\frac {d^3}{a^2 \\plus{} b^2} \\ge\\ 2\n\\]\n[/quote]\r\n$ \\frac {a^3}{b^2 \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\plus{} d^2} \\plus{} \\frac {c^3}{d^2 \\plus{} a^2} \\plus{} \\frac {d^3}{a^2 \\plus{} b^2}\\ge\\ \\frac {(\\sum{a^4})^2}{a^5(b^2 \\plus{} c^2) \\plus{} b^5(c^2 \\plus{} d^2) \\plus{} c^5(d^2 \\plus{} a^2) \\plus{} d^5(a^2 \\plus{} b^2)}$\r\nwith $ 4a^5b^2\\leq\\ 2a^6b^2 \\plus{} a^4b^4 \\plus{} a^4$\r\nand $ \\sum_{cyc}{a^6(b^2 \\plus{} c^2)}\\leq\\ \\frac {1}{2}(\\sum{a^4})^2$\r\n$ \\sum_{cyc}{a^4b^4}\\leq\\ \\frac {1}{4}(\\sum{a^4})$\r\nso $ LHS\\ge\\ \\frac {16}{\\frac {1}{4}16 \\plus{} 2 \\plus{} 2} \\equal{} 2$", "Solution_5": "[quote=\"ye109\"]\n$ \\frac {(\\sum{a^3})^2}{a^3(b^2 \\plus{} c^2) \\plus{} b^3(c^2 \\plus{} d^2) \\plus{} c^3(d^2 \\plus{} a^2) \\plus{} d^3(a^2 \\plus{} b^2)}\\ge\\ \\frac {16}{a^3(\\frac {b^3 \\plus{} 1 \\plus{} c^3 \\plus{} 1}{2}) \\plus{} b^3(\\frac {c^3 \\plus{} 1 \\plus{} d^3 \\plus{} 1}{2}) \\plus{} c^3(\\frac {d^3 \\plus{} 1 \\plus{} a^3 \\plus{} 1}{2}) \\plus{} d^3(\\frac {a^3 \\plus{} 1 \\plus{} b^3 \\plus{} 1}{2})}$\n[/quote]\nId est, $ \\frac {a^3 \\plus{} 1}{2}\\geq a^2$ $ ?$ :maybe:\n[quote=\"ye109\"][quote=\"arqady\"]What about the following? :) \nLet $ a,$ $ b,$ $ c,$ and $ d$ are positive numbers such that $ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} d^4 \\equal{} 4.$ Prove that\n\\[ \\frac {a^3}{b^2 \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\plus{} d^2} \\plus{} \\frac {c^3}{d^2 \\plus{} a^2} \\plus{} \\frac {d^3}{a^2 \\plus{} b^2} \\ge\\ 2\n\\]\n[/quote]\n\nand $ \\sum_{cyc}{a^6(b^2 \\plus{} c^2)}\\leq\\ \\frac {1}{2}(\\sum{a^4})^2$\n[/quote]\r\nWhy it's true?", "Solution_6": "[quote=\"arqady\"][quote=\"ye109\"]\n$ \\frac {(\\sum{a^3})^2}{a^3(b^2 \\plus{} c^2) \\plus{} b^3(c^2 \\plus{} d^2) \\plus{} c^3(d^2 \\plus{} a^2) \\plus{} d^3(a^2 \\plus{} b^2)}\\ge\\ \\frac {16}{a^3(\\frac {b^3 \\plus{} 1 \\plus{} c^3 \\plus{} 1}{2}) \\plus{} b^3(\\frac {c^3 \\plus{} 1 \\plus{} d^3 \\plus{} 1}{2}) \\plus{} c^3(\\frac {d^3 \\plus{} 1 \\plus{} a^3 \\plus{} 1}{2}) \\plus{} d^3(\\frac {a^3 \\plus{} 1 \\plus{} b^3 \\plus{} 1}{2})}$\n[/quote]\nId est, $ \\frac {a^3 \\plus{} 1}{2}\\geq a^2$ $ ?$ :maybe:[/quote]\r\no ,i got a foolish,i will edit it :wink:", "Solution_7": "i have corrected it,arqady .", "Solution_8": "OK. But with $ a^4\\plus{}b^4\\plus{}c^4\\plus{}d^4\\equal{}4$ it's still unsolving.", "Solution_9": "[quote=\"arqady\"]OK. But with $ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} d^4 \\equal{} 4$ it's still unsolving.[/quote]\r\n why\uff1fi think the proof is right :maybe:", "Solution_10": "[quote=\"ye109\"]\nand $ \\sum_{cyc}{a^6(b^2 \\plus{} c^2)}\\leq\\ \\frac {1}{2}(\\sum{a^4})^2$\n[/quote]\r\nWhy it's true?", "Solution_11": "[quote=\"arqady\"][quote=\"ye109\"]\nand $ \\sum_{cyc}{a^6(b^2 \\plus{} c^2)}\\leq\\ \\frac {1}{2}(\\sum{a^4})^2$\n[/quote]\nWhy it's true?[/quote]\r\n :wink: i find it's wrong,i think another proof." } { "Tag": [ "trigonometry", "inequalities" ], "Problem": "In triangle $ABC$, prove that: \\[\\cot\\frac{A}{2}+\\cot\\frac{B}{2}+\\cot\\frac{C}{2}\\ge 3\\left(\\tan\\frac{A}{2}+\\tan\\frac{B}{2}+\\tan\\frac{C}{2}\\right)\\]", "Solution_1": "[hide]we have $\\tan{\\frac{A}{2}}=\\frac{2r}{b+c-a}$ and so on.\nSo the inequality becomes\n$\\frac{b+c-a}{2r}+\\frac{a+c-b}{2r}+\\frac{a+b-c}{2r}\\geq3\\left(\\frac{2r}{b+c-a}+\\frac{2r}{a+c-b}+\\frac{2r}{a+b-c}\\right)$\n$\\frac{a+b+c}{12r^{2}}\\geq\\frac{1}{b+c-a}+\\frac{1}{a+c-b}+\\frac{1}{a+b-c}$\n\nbut we have $r=\\sqrt{\\frac{(p-a)(p-b)(p-c)}{p}}$,so the inequality becomes\n$\\frac{(a+b+c)^{2}}{3(a+b-c)(a+c-b)(b+c-a)}\\geq\\frac{1}{b+c-a}+\\frac{1}{a+c-b}+\\frac{1}{a+b-c}$\n\nand with the ravi substitution\n$b+c-a=x$\n$a+c-b=y$\n$a+b-c=z$\nit becomes\n\n$\\frac{(x+y+z)^{2}}{3xyz}\\geq\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$\n$(x+y+z)^{2}\\geq3(xy+yz+zx)$\n$x^{2}+y^{2}+z^{2}\\geq xy+yz+zx$\nthat is the simpler of the inequality :lol: [/hide]\r\n\r\nbye", "Solution_2": "[hide=\"alternatively you could\"]\nsubstitute $x=\\tan \\frac{A}{2}$ etc. so $x,y,z$ positive. Then use trig addition to prove $xy+yz+zx=1$.\n[/hide]", "Solution_3": "Wow yeah, since I stink at trig, this is what I think he's saying.\r\n\r\n[hide]Since for triangle $ABC$, $\\cot{A/2}\\cot{B/2}\\cot{C/2}= \\cot{A/2}+\\cot{B/2}+\\cot{C/2}$, we can see that for our equation, equating\n$tan{A/2}= x$...etc as Scorpius said,\n\nwe have $\\frac{1}{xyz}= \\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}= \\frac{xy+yz+zx}{xyz}$...\n\nEdit: I think this works.\n\nEdit: Sigh... does not work... what's the next step?[/hide]", "Solution_4": "[hide=\"finishing the solution\"]\nSo after the substitution, the inequality becomes\n\\[\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\geq 3(x+y+z)\\]\nwhich simplifies to\n\\[(xy+yz+zx)^{2}\\geq 3xyz(x+y+z)\\]\n\\[\\iff x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}\\geq x^{2}yz+xy^{2}z+xyz^{2}\\]\nThis is true just by AM-GM's such as:\n\\[\\frac{x^{2}y^{2}+x^{2}z^{2}}{2}\\geq x^{2}yz\\]\n[/hide]" } { "Tag": [], "Problem": "When placing each of the digits 2, 4, 6, 8 and 9 in exactly one\nof the boxes in the addition problem shown, what is the greatest\npossible sum?\n\n[asy]size(100);\npath box = (0,0)--(1,0)--(1,1)--(0,1)--cycle; real shifter = 1 + .2; draw(box ^^ shift(shifter*right)*box ^^ shift(2shifter*right)*box ^^ shift(shifter*down + shifter*right)*box ^^ shift(shifter*down + 2shifter*right)*box); label(\"$+$\",shift(shifter*down + shifter*left)*(.5,.5));\ndraw(shift(1.15shifter*down)*((-1,0)--(3shifter,0)));[/asy]", "Solution_1": "We want to have 9xx added to 8x, in order to have the biggest sum.\r\n\r\n$ 942 \\plus{} 86 \\equal{} 1010$\r\n\r\nLet's see if we can get anything bigger.\r\n\r\n$ 962 \\plus{} 84 \\equal{} \\boxed{1046}$", "Solution_2": "We want the largest numbers in the largest places, so we have a 9 in the hundreds place. We have the 8 and the 6 in the tens place, and 4 and 2 in the ones place. Adding yields 9*100+14*10+6=1046.\r\n\r\nAlso, the diagram doesn't show up in the game." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Here's a good problem I wanna share with you;\r\nLet $S$ be the set of real numbers, $x\\in [0,1]$, such that for all $\\epsilon>0$ there exists some rational number $\\frac{p}{q}\\ \\text{st}\\ |x-\\frac{p}{q}|<\\frac{\\epsilon}{q^{3}}$. Prove that $S$ is infinite and that it does not contain any interval. What more can you say about $S$?", "Solution_1": "Let $x_{n}= \\sum_{k=1}^{n}a^{-b^{k}}$ for $a\\geq 2$ and $b \\geq 4$ and $x$ be the limit of $(x_{n})$. Then\r\n\r\n\\begin{eqnarray*}x-x_{n}& = & a^{-b^{n+1}}+a^{-b^{n+2}}+a^{-b^{n+3}}+\\cdots\\\\ & \\leq & a^{-b^{n+1}}+a^{-b^{n+1}(1+(b-1))}+a^{-b^{n+1}(1+2(b-1))}+\\cdots\\\\ & = & a^{-b^{n+1}}\\sum_{k = 1}^{\\infty}\\left( a^{-b^{n+1}(b-1)}\\right)^{k-1}\\\\ & = & a^{-b^{n+1}}\\frac{1}{1-a^{-b^{n+1}(b-1)}}\\\\ & < & 2\\cdot a^{-b^{n+1}}\\\\ & = & 2\\cdot \\left( a^{-b^{n}}\\right)^{3}a^{-(b-3)b^{n}}\\\\ \\end{eqnarray*}\r\n\r\nAnd for any $\\epsilon$, we can choose $n$ such that $2\\cdot a^{-(b-3)b^{n}}< \\epsilon$. So reversing this process would show that $x$ is in $S$. But there are infinitely many choice of $(a, b)$, $S$ is infinite.\r\n\r\n\r\n\r\nYou can use Liouville's theorem to show that $\\frac{r}{\\sqrt{2}}$ for nonzero rational $r$ is always not in $S$, so for any two distinct $x, y \\in S$, we can always find a number $z\\notin S$ that lies between $x$ and $y$. So any interval (except for an empty set or a singleton) cannot be a subset of $S$.", "Solution_2": "If you don't require that $|x-\\frac{p}{q}| > 0$ then $S$ is infinite just because it contains all rational numbers (between 0 and 1). :) \r\n\r\nNow I will give an alternative proof that $S$ doesn't contain any interval. It doesn't use Liouville's theorem.\r\nLet us compute the measure of the set $S$. The set $S$ is covered by intervals $(\\frac{p}{q}-\\frac\\varepsilon{q^{3}}, \\frac{p}{q}+\\frac\\varepsilon{q^{3}})$. The total measure of all intervals is\r\n\\[\\sum_{q=1}^\\infty \\sum_{p=0}^{q}\\frac{2\\varepsilon}{q^{3}}= 2\\varepsilon \\cdot \\sum_{q=1}^\\infty \\frac{q+1}{q^{3}}\\leq C \\varepsilon. \\]\r\nSince $\\varepsilon$ can be arbitrarily small, $S$ has measure $0$. Therefore, $S$ does not contain any interval." } { "Tag": [ "integration", "real analysis", "calculus", "algebra", "function", "domain", "trigonometry" ], "Problem": "Friends....\r\nThis question may be silly for many of you. \r\nWe define $ X$ to be integrable if $ E|X|$ is finite.\r\nMy question is, does $ E(X)$ finite imply $ X$ is integrable? Because I think $ E(X)$ could be finite when $ E|X|$ is infinite.", "Solution_1": "NO: $ E[X]$ finite does not imply $ E[|X|] < \\infty$. Try to find a counter example when $ X$ oscilate a lot.", "Solution_2": "But for my question in Analysis forum, Professor Kent Merryfield gave the following reply. That time I accepted, but now I am realizing.\r\n\r\n\r\nThe hypothesis was that $ \\int_Xf\\ d\\mu$ is finite. This only makes sense as a Lebesgue integral if $ \\int_X|f|\\ d\\mu$ is finite.", "Solution_3": "And I'm sticking with that answer. \r\n\r\nalekk, the problem with what you just said is that \"oscillate\" implies that you're taking the domain in some particular order. This can be true for integrals over $ \\mathbb{R}.$ For instance, we can say that $ \\int_0^{\\infty}\\frac{\\sin x}{x}\\,dx$ converges as an improper integral despite not being absolutely integrable - but we can [i]only[/i] say that because we've agreed to take the real line in a particular order - and that's because $ \\mathbb{R}$ (or $ [0,\\infty)$) has some structure [i]other[/i] that that of a measure space.\r\n\r\nWhat about a probability sample space? The basic issue is that the [i]only[/i] structure such a space has is that of a measure space - and in a measure space, and in the definition of the Lebesgue integral, we must be free to select and rearrange measurable subsets in any order whatsoever. Hence the definition of the Lebesgue integral - and the Lebesgue integral is the only definition we have that makes any sense for what $ E(X)$ is - must be insensitive to rearrangements, from which we arrive at the requirement for absolute integrability.\r\n\r\nExample: Suppose $ U$ is uniformly distributed on $ \\left(\\minus{}\\frac{\\pi}2,\\frac{\\pi}2\\right)$ and let $ X\\equal{}\\tan U.$ This gives a reasonably well-known distribution on the line. We can say that $ X$ has a probability density function of $ f(x)\\equal{}\\frac{1}{\\pi(1\\plus{}x^2)}$ on $ \\mathbb{R}.$ Despite the symmetry of this distribution, despite the fact that $ 0$ is clearly the median (and mode) of this distribution, we must say that $ E(X)$ is undefined.", "Solution_4": "Yes sir, you are right. We define $ EX\\equal{}EX^\\plus{}\\minus{}EX^\\minus{}$. So if $ EX$ is finite then both $ EX^\\plus{}$ and $ EX^\\minus{}$ have to be finite and hence $ E|X|\\equal{}EX^\\plus{}\\plus{}EX^\\minus{}$ is also finite. Am I right sir?", "Solution_5": "Yes, precisely." } { "Tag": [ "calculus", "derivative", "logarithms", "function", "integration", "calculus computations" ], "Problem": "If $ y \\equal{} x^{\\ln x}$, then $ y'$ is _____.\r\n\r\nI got $ \\frac {x^{\\ln x} \\ln x}{x}$, but the answer key says $ \\frac {2x^{\\ln x} \\ln x}{x}$. Could someone please explain what I did wrong? Thanks in advance!", "Solution_1": "[quote=\"314.math\"]Could someone please explain what I did wrong?[/quote] No, because you only showed what answer you got, not how you got it. Show your workings, then we can discuss what's wrong with them. (I personally would have approached this problem by rewriting $ x^{\\ln x} \\equal{} e^{(\\ln x)^2}$ before taking the derivative, but probably you tried to use some formula or other instead.)", "Solution_2": "check again, actually your mistake is quite obvious, your derivative for (ln x)^2 is wrong...", "Solution_3": "[quote=\"JBL\"][quote=\"314.math\"]Could someone please explain what I did wrong?[/quote] No, because you only showed what answer you got, not how you got it. Show your workings, then we can discuss what's wrong with them. (I personally would have approached this problem by rewriting $ x^{\\ln x} \\equal{} e^{(\\ln x)^2}$ before taking the derivative, but probably you tried to use some formula or other instead.)[/quote]\r\n\r\nHere's what I used:\r\n\r\n$ \\frac {d}{dx}a^u \\equal{} a^u \\ln a \\frac {du}{dx} \\Rightarrow \\frac {d}{dx} \\equal{} x^{\\ln x} \\cdot \\ln x \\cdot \\frac {1}{x}$", "Solution_4": "hello, writing your term in the form\r\n$ y\\equal{}e^{x\\ln(x)}$ we get for the first derivative\r\n$ y'\\equal{}e^{x\\ln(x)}(\\ln(x)\\plus{}1)$\r\nSonnhard.", "Solution_5": "[quote=\"Dr Sonnhard Graubner\"]hello, writing your term in the form\n$ y \\equal{} e^{x\\ln(x)}$ we get for the first derivative\n$ y' \\equal{} e^{x\\ln(x)}(\\ln(x) \\plus{} 1)$\nSonnhard.[/quote]\r\n\r\nI don't get how you rewrote my original problem. Where did the $ e$ come from? Also, what is incorrect about my method?\r\n\r\nThanks!", "Solution_6": "hello, sorry, i have done a mistake, lets make it clear\r\n$ y\\equal{}x^{\\ln(x)}$ taking the logarithm of both sides we get\r\n$ \\ln(y)\\equal{}\\ln(x)\\ln(x)\\equal{}(\\ln(x))^2$\r\nor\r\n$ y\\equal{}e^{(\\ln(x))^2}$\r\nfrom here we get\r\n$ y'\\equal{}e^{(\\ln(x))^2}(2\\ln(x)\\cdot1/x)$\r\nSonnhard.", "Solution_7": "[quote=\"Dr Sonnhard Graubner\"]hello, sorry, i have done a mistake, lets make it clear\n$ y \\equal{} x^{\\ln(x)}$ taking the logarithm of both sides we get\n$ \\ln(y) \\equal{} \\ln(x)\\ln(x) \\equal{} (\\ln(x))^2$\nor\n$ y \\equal{} e^{(\\ln(x))^2}$\nfrom here we get\n$ y' \\equal{} e^{(\\ln(x))^2}(2\\ln(x)\\cdot1/x)$\nSonnhard.[/quote]\r\n\r\nOk, I follow what you did, but why is my way wrong? What is the motivation to take the ln of both sides?", "Solution_8": "[quote=\"314.math\"]Here's what I used:\n\n$ \\frac {d}{dx}a^u \\equal{} a^u \\ln a \\frac {du}{dx} \\Rightarrow \\frac {d}{dx} \\equal{} x^{\\ln x} \\cdot \\ln x \\cdot \\frac {1}{x}$[/quote] Now, go look in your textbook or notes and find under what assumptions the rule $ \\frac {d}{dx} a^x \\equal{} a^x \\ln a$ is valid. (Hint: there is some restriction on $ a$.)\r\n\r\n(I really like the way this rule applies to the function $ g(x) \\equal{} x^2$, with $ a \\equal{} x$ and $ u \\equal{} 2$: we have $ \\frac {d}{dx} x^2 \\equal{} x^2 \\ln x \\cdot \\frac {du}{dx} \\equal{} 0$. :wink:)\r\n\r\n\r\nGeneral principle: when you remember a theorem (one special case is a computational rule of this sort), you must remember not only the [i]conclusion[/i] of the theorem but also the [i]conditions under which the conclusion is valid[/i]. In this case, you are applying the theorem in a circumstance in which it is not valid, so the fact that you get a wrong answer is not terribly surprising.", "Solution_9": "[color=green][Moderator says: please do not quote the preceding post for no good reason.][/color]\r\n\r\nHmmm... My sheet only says that $ \\frac {d}{dx} (a^x) \\equal{} a^x \\ln a$. The only restriction that I'm aware of is that for $ \\int u^n du \\equal{} \\frac {u^{n \\plus{} 1}}{n \\plus{} 2} \\plus{} C, n\\neq \\minus{} 1$. Can you please tell me the restriction? I know one exists, since $ \\frac {d}{dx} x^2 \\equal{} 2x$, not 0. :)", "Solution_10": "What JBL is trying to say it that $ \\frac{d}{dx}a^x \\equal{} a^x \\ln a$ only if $ a$ is constant and does not depend on $ x$. Since this is not your case, you cannot use this formula. What you are dealing with is a non-standard function (non-standard is the sense that you can't find it in most derivation tables) so you need to use a somewhat more complicated method than simply reading it off from a table. In the same sense you could say that since $ \\frac{d}{dx} x^n \\equal{} n x^{n\\minus{}1}$ then $ \\frac{d}{dx}x^{\\ln x} \\equal{} \\ln x \\, x ^{\\ln x \\minus{} 1}$. This is, of course, wrong for the same reason - the formula holds only if $ n$ is constant / does not depend on $ x$.", "Solution_11": "[quote=\"milin\"]What JBL is trying to say ...[/quote] Well, yes and no: I was trying not to say it but instead to get 314.math to discover it for him or herself. :)", "Solution_12": "I'm aware of that and would not have saved him the trouble had he not asked for it as directly as he did. :)" } { "Tag": [ "modular arithmetic", "number theory", "greatest common divisor", "number theory unsolved" ], "Problem": "Find all integer solutions to $2 x^4 + 1 = y^2.$", "Solution_1": "For $|y|<2$ the solutions $(x,y)$ are $(0,-1),(0,1)$.\r\nLet's assume $y>1$, the equation is equivalent to $(y+1)(y-1)=2x^4$. Because $y+1,y-1$ are coprime for odd divisors, then there exist such odd, coprime numbers $u,v$ that $x=2^tuv$ and $y-1=2^mu^4$ , $y+1=2^nv^4$, where $m,n$ are postive integers and $t$ is a non-negative positive integer.\r\nLet's note that $m+n = 4t+1= 1$ $(\\mod 4)$.\r\nSo we have $2^nv^4-2^mu^4=2$, so $min\\{m,n\\}=1$ and $max\\{m,n\\}>min\\{m,n\\}$:\r\n1) $m=1$ thus $2^{n-1}v^4-u^4=1$ ($n>1$), but because $m+n = 1$ $(\\mod 4)$ so $n-1=3$ $(\\mod 4)$,so $n \\geq 3$. However, $u^4+1=2$ $(\\mod 4)$, so we have a contradiction because $8|2^{n-1}$ so $2^{n-1}=0$ $(\\mod 4)$.\r\n2)$n=1$ thus $v^4-2^{m-1}u^4=1$ ($n>1$), so $(v^2-1)(v^2+1)=2^{m-1}u^4$. Because $v^2+1,v^2-1$ ($v>1$) are coprime for odd divisors, then there exist such odd, coprime numbers $a,b$ that $u=ab$ and $v^2-1=2^{m-2}a^4$ , $v^2+1=2b^4$. Thus\r\n$(v+2^{\\frac{m-2}{2}}a^2)(v-2^{\\frac{m-2}{2}}a^2)=1$ ( because $m+n=1$ $(\\mod 4)$ and $n=1$ then $m-2$ is even), but that is impossible.\r\n1) and 2) haven't given any solutions, so there aren't any solutions for $y>1$ and also for $y<-1$- the equality is symmetric, so the only solutions are: $(0,1),(0,-1)$.", "Solution_2": "if we don't consider trivial answers,then\r\nno solution,\r\nwe know that $ x$ is even so let's replace $ x$ by $ 2x$ and we get,\r\n$ 32x^4 \\equal{} (y \\minus{} 1)(y \\plus{} 1)$\r\nso we have 2 cases :\r\n1-$ y \\equiv \\minus{} 1 \\pmod {4}$ which implies \r\n$ y \\minus{} 1 \\equal{} 2t^4,y \\plus{} 1 \\equal{} 16s^4,(t,s) \\equal{} 1,ts \\equal{} x$\r\nso we get\r\n$ 16s^4 \\minus{} 2t^4 \\equal{} 2 \\implies 8s^4 \\equal{} t^4 \\plus{} 1$\r\nbut we obviously have \r\n$ t^4 \\plus{} 1 \\equiv 2 \\pmod {8}$.\r\nso this case is ommitted. \r\n2-$ y \\equiv 1 \\pmod {4}$ which implies \r\n$ y \\minus{} 1 \\equal{} 16s^4,y \\plus{} 1 \\equal{} 2t^4,(t,s) \\equal{} 1,ts \\equal{} x$\r\nso we get \r\n$ 2t^4 \\minus{} 16s^4 \\equal{} 2 \\implies t^4 \\minus{} 8s^4 \\equal{} 1$.\r\nor in other words,\r\n$ t^4 \\minus{} 1^4 \\equal{} 2(2s^2)^2$.\r\nu can easily prove that the equation $ x^4 \\minus{} y^4 \\equal{} 2z^2$ has no integer solutions(by infinite descent).\r\nso we r done!", "Solution_3": "x^4 = 1 mod 5\r\n2*x^4+1 = 3 mod 5\r\nno y that y^2 = 3 mod 5", "Solution_4": "[quote=\"skytin\"]x^4 = 1 mod 5\n2*x^4+1 = 3 mod 5\nno y that y^2 = 3 mod 5[/quote]\r\nwhat if $ 5 | x$ ?", "Solution_5": "We may first assume $ x,y \\geq 0$.\r\n\r\n$ 2x^{4}\\equal{}y^{2}\\minus{}1\\equal{}(y\\plus{}1)(y\\minus{}1)$.\r\n$ 2|(y\\plus{}1)(y\\minus{}1) \\Rightarrow 4|(y\\plus{}1)(y\\minus{}1) \\Rightarrow 2|x$.\r\nLet $ x\\equal{}2x_{1}$, then $ 32x_{1}^{4}\\equal{}(y\\plus{}1)(y\\minus{}1) \\Rightarrow y\\equal{}16y_{1}\\plus{}1$ or $ 16y_{1}\\minus{}1$.\r\nThen $ x_{1}^{4}\\equal{}y_{1}(8y_{1}\\plus{}1)$ or $ y_{1}(8y_{1}\\minus{}1)$.\r\nBut since $ gcd(y_{1},8y_{1}\\minus{}1)\\equal{}gcd(y_{1},8y_{1}\\plus{}1)\\equal{}1$, both $ y_{1}$ and $ 8y_{1}\\plus{}1$ (or $ \\minus{}1$) must be $ 4^{th}$ powers.\r\nBut $ 8y_{1}\\minus{}1$ cannot be a $ 4^{th}$ power, so we can only have $ x_{1}^{4}\\equal{}y_{1}(8y_{1}\\plus{}1)$ (i.e. $ y\\equal{}16y_{1}\\plus{}1$).\r\n\r\nNote that $ y_{1},8y_{1}\\plus{}1$ are still $ 4^{th}$ powers, so let $ y_{1}\\equal{}a^{4}$, $ 8y_{1}\\plus{}1\\equal{}b^{4}$.\r\nThen $ b^{4}\\minus{}1\\equal{}(b^{2}\\plus{}1)(b\\plus{}1)(b\\minus{}1)\\equal{}8a^{4}$.\r\n\r\nClearly b is odd.\r\nConsider then the integers $ \\frac{b^{2}\\plus{}1}{2},\\frac{b\\plus{}1}{2},\\frac{b\\minus{}1}{2}$.\r\nTaken pairwise, each gcd is 1 ($ \\frac{1}{2}(b\\plus{}1)(b\\minus{}1)\\equal{}\\frac{b^2\\minus{}1}{2}\\equal{}\\frac{b^2\\plus{}1}{2}\\minus{}1$).\r\nHowever, their product = $ a^{4}$, so each of them must be a $ 4^{th}$ power.\r\nIn particular, $ \\frac{b\\plus{}1}{2},\\frac{b\\minus{}1}{2}$, 2 consecutive integers, must both be $ 4^{th}$ powers, possible only when $ \\frac{b\\minus{}1}{2}\\equal{}0, \\frac{b\\plus{}1}{2}\\equal{}1 \\Rightarrow b\\equal{}1 \\Rightarrow y\\equal{}16y_{1}\\plus{}1\\equal{}2(8y_{1}\\plus{}1)\\minus{}1\\equal{}2b^{4}\\minus{}1\\equal{}1$, and $ x\\equal{}0$.\r\n\r\n$ \\therefore (x,y)\\equal{}(0,1),(0,\\minus{}1)$ are the only solutions in integers.", "Solution_6": "im not sure if my ideas are correct cuz i wasnt sure how to proceed \r\n\r\nbut this is what i had:\r\n\r\n$ 2x^4\\equal{}(y\\minus{}1)(y\\plus{}1)$\r\n\r\nwe take this mod 4 and find that $ x$ is even and $ y\\equal{}4k\\plus{}1$ or $ y\\equal{}4k\\plus{}3$\r\n\r\nwith $ y\\equal{}4k\\plus{}3$, we find a contradiction. so, $ y\\equal{}4k\\plus{}1$\r\n\r\nlet $ x\\equal{}2a$\r\n\r\n\r\n$ 32a^4\\plus{}1\\equal{}(4k\\plus{}1)^2$\r\n\r\nthen we have\r\n$ 8a^4\\equal{}(2k)(2k\\plus{}1)$\r\n\r\ni think we could use infinite descent as mentioned before to show that either a or k must be 0 but im not sure how to. does anyone know how?", "Solution_7": "Here is my solution\n\nI will assume $y$ is positive.\n$2x^4+1=y^2\\implies 2x^4=(y-1)(y+1)$, so replace $y-1$ with $2z$. This gives $2x^4=4z(z+1)$, and again we will replace $x$ with $2x_1$ to get $8x_1^4=z(z+1)$. Now $\\gcd(z,z+1)=1$, hence either $z=8a^4$ and $z+1=b^4$ or that $z=a^4$ and $z+1=8b^4$ for some integers $a,b$ (since $y$ is positive- we can ignore the case $z=-8a^4$ and $z+1=-b^4$ etc).\n\n[b]Case 1[/b]\n$z=8a^4$ and $z+1=b^4$\nThen $8a^4+1=b^4$ which rearranges to $8a^4=(b-1)(b+1)(b^2+1)$. Since $b$ must be odd, substitute $b-1=2c\\implies b+1=2(c+1)$ and $b^2+1=2(2c^2+2c+1)$. Then $a^4=c(c+1)(2c^2+2c+1)$.\n\n$\\gcd(c,c+1)=1$\n$\\gcd(c, 2c^2+2c+1)=\\gcd(c,(c+1)^2+c)=\\gcd(c,(c+1)^2)=1$\n$\\gcd(c+1, 2c^2+2c+1)=\\gcd(c+1,(c+1)^2+c)=\\gcd(c+1,c)=1$\n\nHence all of $c$, $c+1$ and $2c^2+2c+1$ are fourth powers, i.e. $c$ and $c+1$ are consecutive squares. This implies $c=0\\implies a=0 \\implies z=0 \\implies y=1$. This gives the solution $(0,1)$.\n\n[b]Case 2[/b]\n$z=a^4$ and $z+1=8b^4$\nThen $a^4+1=8b^4$. Looking modulo $5$, the LHS is one of $1,2$ while the RHS is one of $0,3$, a contradiction.\n\nThus the only solution is $(0,1)$, however since we assumed $y$ is positive, there is also the solution $(0,-1)$.", "Solution_8": "How can one miss this solution?\nConsider the pell's equation $y^2-2z^2=1$ . We need to find when $z$ is a perfect square. Note that $(y,z)=(3,2)$ is a solution and we have all solutions from this.\n $Y+\\sqrt2 Z=(3+2\\sqrt2)^N=(1+\\sqrt2)^{2N}$ this gives $Z= 2^{N-1}+2^{N-2}+....+2+1=2^N-1$\nand $2^N-1$ is a perfect square only when Z=0,1 so the solution are $(0,1) (0-1)$" } { "Tag": [ "geometry", "3D geometry", "ratio", "rectangle", "probability" ], "Problem": "Hi everyone, I just practiced with the 2001 State Sprint Round, and there's a few I'm struggling with.\r\n\r\n8. Sixty-four unit cubes are placed together to create a large cube. How many cubes with integer dimensions are in the 4x4x4 cube?\r\n\r\nAnswer: 100\r\n\r\nI'm trying to figure out a way to do this that will work with all problems similar to this...I tried dividing 64 by 1, 8, 27, and 64 (four types of cubes that will fit), getting rid of the remainder, and adding the results together, but that came out to be 75.\r\n\r\n13. What is the ratio of the width to length of a floded sheet of peper if the rectangle formed when the sheet is folded in half as shown is similar to the original rectangular shset? Express you answer as a common fraction in simplest radical form.\r\n\r\nAnswer: (:sqrt:2)/2\r\n\r\nNo idea how that came to be...\r\n\r\n21. A standard die is tossed twice. What is the probability of obtaining exactly one 5? Express you answer as a common fraction.\r\n\r\nAnswer: 5/18\r\n\r\nI got 5/6 by doing 1/6 (probability of getting a 5 on the first roll) times 5/6 (probability of gettin anything but 5 on the second roll). Then, I multiplied that by 2 because it could also come out vice versa (getting anything but a 5 and then getting a 5).\r\n\r\nThere are a few more, but I'll post them later. Can I just get help on these three for now? Thank you to whoever responds! :)", "Solution_1": "I just did this sprint yesterday, so these answers are fresh in my mind...\r\n8. For these problems, I always start by looking for a pattern. There are 64 unit cubes, so you get 64 there. Notice that the unit cubes are 1*1*1 and 64 is 4*4*4. There is also 1 4*4*4 cube, and there are 8 (2*2*2) 3*3*3 cubes. If you notice this pattern, you realize that you have to just sum up all of the cubes less than 4^3, and that gives you 100. THe reason this works is because the numbers corresponds to what I call its opposite cube.. at least for this figure. $64+27+8+1=100$ This problem is easier to do than to explain...\r\n\r\n13. In this problem, the width of the paper is the one being folded, so I drew a diagram of an unfolded and folded paper. I labeled the width of the folded paper w, the length l, and the width of the unfolded paper 2w. I then set up the proportion $l/w=2w/l$ By cross multiplying, I got $2w^{2}=l^2$}. By finding the roots of these two, I got $w\\sqrt{2}=l$. Which therefore gave me the ratio of $w/l = 1/\\sqrt{2}$ (as in the equation, if the width was 1, then the width was sqrt(2). By simplifying that, you get the answer $\\sqrt{2}/2$. Again, this problem sounds much more complex than it really is, as it is just a simple algebra problem. \r\n\r\n21. For the last problem, you just multiplied wrong. 5/6*1/6*2 is 5/18 not 5/6... the way I did it is list all ten combinations (or just think them out) and put the denominator as 36 which also simplifies to 5/18.", "Solution_2": "Wow, thank you! I totally understand the first one now, and silly me on the third one.\r\n\r\nThere's just one thing I don't understand on the second one...my friend was trying to explain it, and I don't completely understand him, neither.\r\n\r\nBut my question is, why is the porportion l/w = 2w/l instead of l/w = l/2w? I thought that since in the first ratio, you were comparing the length to the width, you had to do the same in the second ratio.\r\n\r\nThanks again! :)", "Solution_3": "[quote=\"SoccerBrainy40\"]Wow, thank you! I totally understand the first one now, and silly me on the third one.\n\nThere's just one thing I don't understand on the second one...my friend was trying to explain it, and I don't completely understand him, neither.\n\nBut my question is, why is the porportion l/w = 2w/l instead of l/w = l/2w? I thought that since in the first ratio, you were comparing the length to the width, you had to do the same in the second ratio.\n\nThanks again! :)[/quote]\r\n\r\nNo prob... On th second one, you have to remember that the question states that the rectangles are similar, and if you fold the paper, then the [b]width gets smaller[/b], and the only way the rectangle can remain similar is if the length is proportional to the width in the same way the width was proportional to the length in the unfolded paper. Also, if you cross-multiply l/w=2l/w, then you get 2lw=lw, which will get you nowhere. Sprint problems are (usually) pretty straightforward... I think this and number 24 are the hardest if you don't know how to solve them.", "Solution_4": "The only one I can grasp is the first one, where cubes are same as squares. E.G. if you have an 8x8 checkerboard you will have 64+49+36+25 etc. squares, so cubes its the same.", "Solution_5": "cubes are in 3 dimensions", "Solution_6": "[quote=\"SoccerBrainy40\"]Hi everyone, I just practiced with the 2001 State Sprint Round, and there's a few I'm struggling with.\n\n8. Sixty-four unit cubes are placed together to create a large cube. How many cubes with integer dimensions are in the 4x4x4 cube?\n\nAnswer: 100\n\nI'm trying to figure out a way to do this that will work with all problems similar to this...I tried dividing 64 by 1, 8, 27, and 64 (four types of cubes that will fit), getting rid of the remainder, and adding the results together, but that came out to be 75.\n\n13. What is the ratio of the width to length of a floded sheet of peper if the rectangle formed when the sheet is folded in half as shown is similar to the original rectangular shset? Express you answer as a common fraction in simplest radical form.\n\nAnswer: (:sqrt:2)/2\n\nNo idea how that came to be...\n\n21. A standard die is tossed twice. What is the probability of obtaining exactly one 5? Express you answer as a common fraction.\n\nAnswer: 5/18\n\nI got 5/6 by doing 1/6 (probability of getting a 5 on the first roll) times 5/6 (probability of gettin anything but 5 on the second roll). Then, I multiplied that by 2 because it could also come out vice versa (getting anything but a 5 and then getting a 5).\n\nThere are a few more, but I'll post them later. Can I just get help on these three for now? Thank you to whoever responds! :)[/quote]\r\n\r\n8.\r\nfind a pattern\r\n\r\n1X1X1=1=1^3\r\n2X2X2=9=1^3+2^3\r\n3X3X3=36=1^3+2^3+3^3\r\n4X4X4=100=1^3+2^3+3^3+4^3\r\n\r\nso the generic formula for this is $1^3+2^3+3^3...n^3$\r\n\r\n13.\r\nI need the picture..I dont know what it looks like :blush: \r\n\r\n21.\r\none case is rolling 5 on the first and then rolling anything other than a 5 the second time\r\nanother is rolling anything but a 5 and then rolling a 5\r\n\r\nthe probability of them happening for both is 5/36\r\n\r\n5/36+5/36=10/36=5/18" } { "Tag": [ "trigonometry" ], "Problem": "(tan 115+tan 55 +tan 15)/(tan 75+tan 70+tan 35)=?", "Solution_1": "Where did you get this problem? It doesn't seem to have a nice answer.\r\n\r\n$\\frac{\\tan {115}+\\tan {55}+\\tan {15}}{\\tan {75} + \\tan {70} + \\tan {35}}$", "Solution_2": "The denominator equals $\\tan 75\\tan 70\\tan 35$...if that helps any.", "Solution_3": "In other words, if $\\alpha +\\beta +\\gamma =180^{\\circ}$, then $\\tan\\alpha +\\tan\\beta +\\tan\\gamma=\\tan\\alpha\\tan\\beta\\tan\\gamma$.\r\n\r\nMasoud Zargar" } { "Tag": [ "counting", "derangement", "combinatorics unsolved", "combinatorics" ], "Problem": "How many biyectios from {1, 2, 3, 4...n } to {1, 2, 3, 4...n } are that follow that f( x) is different from x... for any value of x?", "Solution_1": "If you mean the number of bijections from $ \\{1, 2, \\ldots, n\\}$ that have no fixed points, I believe that's just the number of [url=http://en.wikipedia.org/wiki/Derangement]derangements[/url]..." } { "Tag": [ "videos", "percent", "ratio" ], "Problem": "Two months ago, Javier Heredia found himself in a different country without any of the local money in his wallet. Not a penny, and he had some big purchases to make.\r\n\r\nHeredia arranged to meet a currency broker at a busy city landmark, and they agreed that Heredia would pay $130 for the new money. As they met, sketchy characters with weapons gathered nearby. The broker handed Heredia the cash without problems.\nBut the smooth transaction didn't happen here on Earth - it took place in the fictional city of Freeport on the make-believe planet of Norrath, inside the Internet video game EverQuest. Heredia wasn't even himself at time; he was Gretk, a 5-foot-tall talking rat.\nIn fact, the only thing that was real was the money.\nVideo games are nothing new, but the latest generation of games, so-called MMORPGs, or massively multiplayer online role-playing games, has spawned an interesting new wrinkle in electronic entertainment: the virtual economy.\nAt auction houses and on eBay, players can use real money from the real world to buy game money, implements and weapons of their game's universe. Brokers take orders for fictional currency, make-believe swords, even entire characters (a Ranger Wood Elf male with very nice gear for $269).\r\n\r\nThe practice, while increasingly common as online games grow in popularity, raises interesting questions about intellectual property rights - shouldn't gamemakers get a percentage since they invented all this stuff? - and whether gamers like Heredia are being true to the rules of play.\r\n\r\nSoftware and computer companies produce a wide variety of MMORPGs with an array of fantasy worlds to explore, from outer space to Tolkeinesque lands. Yet for all their differences these games share a form of meritocracy; through experience (and hours of play) gamers can build characters, called \"avatars\" among gamers, with ever greater strengths and ever greater goods, permitting an ever more interesting level of play.\r\n\r\nBecause the level of interaction in the games is so high, and because the social rewards can be so great, it has become attractive for players to pump the hard currency of the real world into these fake universes, all in the name of achieving status and prestige in the virtual society.\r\n\r\n* * *\r\n\r\nEdward Castronova is an economist at Indiana University who specializes in the video game industry. He estimates that revenues for online gaming were $1.9-billion in 2003 and will grow to $9.8-billion by 2009. He also figures that up to $100-million in real-world money is captured annually by dealers in virtual currency and goods, and that number will keep growing.\nAccording to Castronova's survey of players, the average gamer is a man in his 20s, with a full-time job and disposable income. The average time spent within the virtual worlds is 20 hours per week, he said, adding that 20 percent of the EverQuest gamers he surveyed declared, I live in Norrath but I travel outside of it regularly.\nIn EverQuest, like other MMORPGs, players create a character, or avatar, that they will play each time they return to the game. This character has special powers, such as strength or agility, based on its race and class. An ogre, for instance, has more raw strength than a half-elf, and a barbarian is better in combat than a Vah Shir (a feline humanoid).\nAvatars take on tasks in exchange for experience, game money or items. Castronova found that fighting magical beasts yielded friends, currency and prestige.\nBut in between killing magical beasts and accompanying lizards on adventures, Castronova noticed that Norrath's economy was booming.\nHe witnessed virtual bakers making virtual bread. Virtual merchants selling virtual boots and hats, along with weapons and armor. Even virtual homes. Castronova realized Norrath is a consumer society. Everyone wants more game money (platinum is the currency) to gain higher status and, well, amass more stuff.\nHe did some research, and the results were startling: The average player was generating 319 platinum pieces, about $3.42, for each hour spent \"working\" in the game. Castronova tallied the wealth all the players created in one year in Norrath: $2,266 per capita.\r\n\r\nThe exchange rate between Norrath's currency and the U.S. dollar is determined in a highly liquid (if illegal)\r\n currency market. According to the daily exchange rate published at [url=http://www.gameusd.com/everquest-plat.htm]Everquest Plat Research[/url] ,its value exceeds that of the Japanese yen and the Italian lira. The creation of dollar-valued items in Norrath occurs at a rate such that Norrath's GNP per capita easily exceeds that of dozens of countries, including India and China.\r\n\r\nCastronova discovered that EverQuest is the 77th richest country in th", "Solution_1": "i don't think you finished typing lol...\r\n\r\nbut i know Everquest is richer than ukraine is lol\r\n\r\nit's kinda pathetic :P", "Solution_2": "everquest is not richer than ukraine. it has a more affluent economy in terms of sales and trade, and has a higher ratio-ed income per capita.\r\n\r\nif you dont believe that the virtual world has gone to hell yet, here's a little light reading\r\n\r\n[url=http://www.msnbc.msn.com/id/7341316/]Gamer stabs gamer over virtual sword[/url]", "Solution_3": "[quote=\"L_Li\"]everquest is not richer than ukraine. it has a more affluent economy in terms of sales and trade, and has a higher ratio-ed income per capita.\n\nif you dont believe that the virtual world has gone to - yet, here's a little light reading\n\n[url=http://www.msnbc.msn.com/id/7341316/]Gamer stabs gamer over virtual sword[/url][/quote]\r\nLet me see if I can dig up the link where a duper made thousands of dollars...\r\n\r\nMMO's are rediculous. Everyone and their mom want to make one, the whole industry is full of low quality clones, the players are all 12-year olds who curse at you, and the servers are always down.\r\n\r\nI prefer something like diablo over a LAN.", "Solution_4": "One lady made 16 dollar bills and tried to use 10 of them at Target :o" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c>0$.Prove the following inequality:\r\n\r\n $\\frac{a^2}{b+c}+\\frac{b^2}{c+a}+\\frac{c^2}{a+b}\\geq\\frac{2a^2}{b+c+2a}+\\frac{2b^2}{c+a+2b}+\\frac{2c^2}{a+b+2c}$ ;)", "Solution_1": "The way to proceed is clear: $\\frac{2a^2}{b+c+2a}\\leq\\frac{1}{2}(\\frac{a^2}{b+a}+\\frac{a^2}{c+a})$. Then it is Chebyshev inequality." } { "Tag": [ "function", "integration", "geometry", "abstract algebra", "limit", "calculus", "inequalities" ], "Problem": "For open sets $ U \\subset \\mathbb{C}$, let $ L^p_a(U)$ be the space of all analytic functions $ f$ in $ U$ such that $ \\int_U |f|^p \\, dA < \\infty$, where $ dA$ is the usual area measure.\r\n\r\ni) Construct a set such that $ L^1_a(U)$ has finite (non-zero) dimension\r\nii) Can $ L^2_a(U)$ have finite (non-zero) dimension?", "Solution_1": "If $ U\\subset V$ then $ L_a^p(V)\\subset L_a^p(U)$. So we are looking for \"big\" open sets. The entire plane $ \\mathbb C$ is, obviously, too big. How about $ \\mathbb C$ minus a point? Nope, the spaces are still empty... Two points don't seem to help either... But without three points you have $ \\frac{1}{(z\\minus{}1)(z\\minus{}2)(z\\minus{}3)}$ which is in $ L^1$. \r\n\r\nThe problem with $ L_a^2$ seems that it does not allow the function to have poles...", "Solution_2": "For i), I can prove that $ f \\in L^1_a(\\mathbb{C} \\setminus \\{1,2,3\\})$ has at most simple poles in 1, 2 and 3. But how do I finish to show that $ f$ is a multiple of the function mlok wrote down?\r\n\r\nFor ii), I have no idea. I tried to look at reproducing kernels in $ L^2_a(U)$, but there doesn't really seem to be a contradiction in $ L^2_a(U)$ being finite-dimensional there.", "Solution_3": "For i): Since $ f$ has to be $ L^1$, $ \\lim_{z\\to\\infty} f(z) \\equal{} 0$, thus $ f$ can be holomorphically continued to $ \\infty$. To be in $ L^1$, $ f$ has to have a zero of order at least $ 3$ in $ z \\equal{} \\infty$.\r\n\r\nFor ii): Not yet a full solution but two seemingly crucial thoughts: \r\n(1) If $ U^c \\equal{} \\mathbb{C}\\setminus U$ has nonempty interior, then $ L^2_\\alpha(U)$ has infinite dimension.\r\n(2) If $ z\\in U^c$ is an isolated point of $ U^c$, then $ L^2_\\alpha(U) \\cong L^2_\\alpha(U\\cup\\{z\\})$.", "Solution_4": "Ok, I can prove i) completely now. Although I don't feel I have a satisfactory proof of the fact that $ f$ can't be integrable around an essential singularity (in 0, let's say). My argument is to make the change of variable $ z \\equal{} \\frac {1}{w}$, so that we are considering the convergence of $ \\int_{\\mathbb{C}} \\left | \\frac{a_{\\minus{}2}}{w^2} \\plus{} \\frac {a_{\\minus{}1}}{w} \\plus{} g(w) \\right | \\, dA$ where $ g$ is entire. But $ \\int_{|w| < R} \\left | \\frac{a_{\\minus{}2}}{w^2} \\plus{} \\frac {a_{\\minus{}1}}{w} \\right | \\, dA$ grows like $ R$ and unless $ g \\equal{} 0$ then $ \\int_{|w| < R} | g(w) | \\, dA$ grows faster than $ R^2$, by the subharmonicity of $ |g|$.. It just feels like I don't completely understand what is going on with this argument. How fast does the integral of a transcendental entire function really grow? \r\n\r\n\r\nFor property (1), what did you have in mind? When I saw of your post I thought of the following: If $ a$ is in the interior of $ \\mathbb{C} \\setminus U$, then $ g(z) \\equal{} \\frac {1}{z \\minus{} a}$ is analytic and bounded in $ U$. Hence if $ 0 \\neq f \\in L^2_a(U)$, then $ g^n f \\in L^2_a(U)$ are linearly independent.", "Solution_5": "To estimate the integral of $ |f|$ near an isolated singular point $ 0$, just note that if $ f(z)\\equal{}\\sum_{k\\equal{}\\minus{}\\infty}^\\infty a_k z^k$, then the integral of $ f$ over the circumference of radius $ r>0$ is at least $ 2\\pi |a_k|r^k$ for every $ k$. Integrating this inequality with respect to $ r$, we see that $ f\\notin L^1_a$ unless $ a_k\\equal{}0$ for $ k\\le \\minus{}2$. Similarly, $ f\\notin L^2_a$ unless $ a_k\\equal{}0$ for $ k\\le \\minus{}1$.\r\n\r\nThe $ L^2_a$ question is more interesting. I have an idea of how to solve it but I do not really like my approach. Let $ F\\equal{}\\mathbb C\\setminus U$. Note that $ F$ does not contain isolated points (we can extend any function from $ L^2_a$ analitically to those points). Also, F cannot have bounded connected closed subsets (otherwise we can consider the conformal map $ g$ of the unbounded component of the complement of such a subset to the unit disk mapping $ \\infty$ to $ 0$ and $ g^2,g^3,g^4,\\dots$ will be in $ L^2_a$. In each bounded component, we can just take the identically $ 0$ function. This (if I do not mistake) should imply that connected components of $ F$ are just points. Now just apply the separation of singularities lemma to $ f$ to build functions whose singularities lie in arbitrarily small neighborhoods of $ n\\plus{}1$ distinct points of $ f$. This will give you $ n\\plus{}1$ linearly independent functions $ g_k$ that are as good as $ f$ locally but may decay like $ \\frac 1z$ at $ \\infty$. To improve the decay at infinity to $ z^{\\minus{}2}$ just subtract $ g_1$ from $ g_k$ with an appropriate coefficients (assuming that it is $ g_1$ that fails to decay as $ z^{\\minus{}2}$ at infinity from the very beginning).", "Solution_6": "What do you mean, exactly, by a singularity here? And what if the $ f$ you happen to pick only has finitely many?", "Solution_7": "By \"singularity\" I mean a point $ z$ to which the function cannot be analytically extended from $ U$ (i.e., there is no open $ V\\supset U\\cup\\{z\\}$ such that $ f$ is analytic in $ V$). Note that it is enough to consider the case when $ F$ has Lebesgue measure $ 0$ (otherwise we can create a bounded function decaying at infinity by just taking the Cauchy integral of the characteristic function of some bounded portion of $ F$), so such analytic extension, if it is there, is just the extension by continuity and we do not need to go into the tricky business of checking whether 2 different extensions agree here.\r\n\r\nThere cannot be finitely many singularities because then they would be isolated, which is impossible for an $ L^2_a$ function.\r\n\r\nAnyway, I found a better idea that works in $ L^p_a$ with $ p>2$. If $ f$ is a not identically zero function in $ L^p_a$, then $ g(z)\\equal{}\\frac {f(z)\\minus{}f(w)}{z\\minus{}w}$ is another such function. It seems that applying this operation many times, we should get linearly independent functions unless $ f$ is rational. For $ L^2_a$, we can try $ g(z)\\equal{}f(w_2)\\frac {f(z)\\minus{}f(w_1)}{z\\minus{}w_1}\\minus{}f(w_1)\\frac {f(z)\\minus{}f(w_2)}{z\\minus{}w_2}$. See if you can make this trick work. :)", "Solution_8": "Well, as you indicated, the coefficients have to be adjusted so that $ f$ decays fast enough at infinity. It seems easiest to assume then that $ U$ contains a neighborhood of $ \\infty$ (the change of variables $ z \\equal{} \\frac {1}{w}$ does not change the dimension). \r\n\r\nBut in fact, once you have a function $ f$ that decays like $ \\frac {1}{z^n}$ at $ \\infty$, it is easy to choose coefficients $ c_j$ so that $ \\sum c_j \\frac {f(z) \\minus{} f(w_j)}{z \\minus{} w_j}$ decays like $ \\frac{1}{z^{n\\plus{}1}}$. Just calculate the coefficients in the laurent expansion around $ \\infty$ explicitly to get a system of linear equations that is not over-determined. So if $ f$ has infinitely many non-zero coefficients in its laurent expansion, then we can construct countably many linearly independent functions in $ L^2_a(U)$. But what if $ f$ does not have an infinite expansion?\r\n\r\nAbout this:\r\n[quote=\"fedja\"] Note that it is enough to consider the case when $ F$ has Lebesgue measure $ 0$ (otherwise we can create a bounded function decaying at infinity by just taking the Cauchy integral of the characteristic function of some bounded portion of $ F$)[/quote]\r\nIsn't such a Cauchy integral zero?", "Solution_9": "[quote=\"Kalle\"]the change of variables $ z \\equal{} \\frac {1}{w}$ does not change the dimension[/quote]\nWhy? \n[quote]\nBut what if $ f$ does not have an infinite expansion?\n[/quote]\nIf we consider the case when $ F$ has measure $ 0$, then no rational function can be in $ L^p_a(U)$ with $ p\\ge 2$.\n[quote]\nIsn't such a Cauchy integral zero?[/quote]\r\nI mean the integral $ \\int_S \\frac {dm_2(w)}{z\\minus{}w}$ where $ S$ is a bounded subset of $ F$ of positive measure, which behaves like $ \\frac{m_2(S)}z$ at $ \\infty$. ($ m_2$ is the Lebesgue measure on the plane, as usual).", "Solution_10": "[quote=\"fedja\"][quote=\"Kalle\"]the change of variables $ z \\equal{} \\frac {1}{w}$ does not change the dimension[/quote]\nWhy?\n[/quote]\nIf $ V$ is the image of $ U$ under $ 1/z$, then $ f(z) \\leftrightarrow \\frac{f(w)}{w^2}$ is a unitary correspondence between $ L^2_a(U)$ and $ L^2_a(V)$. \n\n[quote=\"fedja\"][quote=\"Kalle\"]But what if $ f$ does not have an infinite expansion?[/quote]\nIf we consider the case when $ F$ has measure $ 0$, then no rational function can be in $ L^p_a(U)$ with $ p\\ge 2$.\n[/quote]\r\nOoops. I am stupid :P \r\n\r\nDo you agree that the problem is solved now?", "Solution_11": "[quote=\"Kalle\"]Do you agree that the problem is solved now?[/quote]\r\nSeems like that :). Does it mean that it is time to post the next one? :P" } { "Tag": [ "geometry", "AMC", "AIME", "email", "ratio", "perimeter", "circumcircle" ], "Problem": "Has anyone of you cleared the Regional Mathematics Olympiad ???\r\nCan you please help out, i am in 11th now and i would be appearing for the first time in RMO (2009). I don't want to miss this opportunity.can you please tell me , how much number thoery and geometry must i do...in order to be able to solve RMO Problems .....(actually i am able to solve many of the previous years' RMO Problems and very few INMO Problems ,but not all)\r\n\r\nPlease help me out", "Solution_1": "hey hi check this out [u]www.artofproblemsolving.com/forum/index.php?f=513[/u]", "Solution_2": "Hey !! Would it help if i solve AIME nad Junior Balkan Question Papers coz' i frequently solve them and am able to crack some of those problems.....????", "Solution_3": "how does one apply for RMO? im in 10th and wish to apply", "Solution_4": "hey its easy u just see se who ur regional coordintor is and then contact him:\r\ncheck this http://www.iapt.in\r\nbtw where do u live?\r\n :D", "Solution_5": "through which medium should i contact? email, post, telephone....which one?", "Solution_6": "To Analytic: \r\nYou need to tell us where you live.\r\nAnd you better hurry.\r\nEnrollment might have been already over by now.\r\n\r\nHere's some info:\r\nhttp://www.isical.ac.in/~rmo/\r\n\r\nIn W.B. the procedure is:\r\nhttp://www.isical.ac.in/~rmo/rmo08.html\r\n\r\nHere's the list of regional coordinators:\r\nhttp://olympiads.hbcse.tifr.res.in/enrollment/list-of-rmo-coordinators", "Solution_7": "well, i live in delhi. it will be sad if the enrollment has ended. i know that professor tripathi is the my regional co-ordinator and ways of contacting. normally, how do you contact your regional coordinator? phone or email?\r\ni found email as a more convenient way but he has not responded.", "Solution_8": "C'mon you all don't panic, last date for registration is 24th october. so there is quite a lot of time you see!! :oops:\r\n\r\nRegisteration is quite easy: \r\n1) Either the forms are made available by the school and sent to the cordinator\r\nOR\r\n2) You can individually register yourself by sending a DD of 50 rupees in the name of regional cordinator, delhi Region to Vasant Valley school , New delhi with the necesary info_ --> Name, Class, age, Address, Father's Name, School's Name and 2 passport size Photographs\r\n\r\n\r\nI hope you would find this info useful!!!\r\n\r\n :idea: :wink:", "Solution_9": "this information seems to be useful, but i'll still confirm with the coordinator.but thanks :)", "Solution_10": "also send ur class tenth marksheet as a conformation and it should be [b]CERTIFIED[/b]", "Solution_11": "[quote=\"amangupta2\"]C'mon you all don't panic, last date for registration is 24th october. so there is quite a lot of time you see!! :oops:\n\nRegisteration is quite easy: \n1) Either the forms are made available by the school and sent to the cordinator\nOR\n2) You can individually register yourself by sending a DD of 50 rupees in the name of regional cordinator, delhi Region to Vasant Valley school , New delhi with the necesary info_ --> Name, Class, age, Address, Father's Name, School's Name and 2 passport size Photographs\n\n\nI hope you would find this info useful!!!\n\n :idea: :wink:[/quote]\r\n\r\n\r\nI am sorry but the last date was 26th septempber", "Solution_12": "hell_ever is correct \r\ni am also sorry!", "Solution_13": "oh!\r\nnoooooooooooooo :( :mad: \r\n:'(\r\nvery bad.\r\ndarn darn!!!!", "Solution_14": "C'mon ths is not possible , coz' when i filed the form, it cleary mentioned that the registration has to be done before 24th pctober. and no markshet is required..(maybe it s required if you register individually)\r\n\r\n\r\nAnyways i need not worry coz' i have already registered for RMO through my school(DPS Noida) [And not markshet copy was required]\r\n\r\nHope that you are all able to register for it.(coz'there is still time)\r\n\r\nALL THE BEST !!", "Solution_15": "i duuno about other :) regions but yes in uttar pradeshthe date of rmo is over!@", "Solution_16": "well, there is hope. today, professor just responded. so, maybe last date is 24th october.", "Solution_17": "hey delhi guys!!\r\n\r\nRMO registration isn't over yet.\r\nyes!! its 24 th oct....\r\n\r\ni dunno why many of u have some confusoion\r\n\r\ni had also sent a mail i got the reply.. i'm in XI as well \r\n\r\ncan the genius one who is able to solve past RMO problems let me which ones u r able to???let me try and see if a stupid one like me has any hope of even having a hope!! \r\nreply fast dude!!!", "Solution_18": "hey solve these i was able to solve only three (with partial solution s............... :P )\r\n\r\n1. Two boxes contain between them 65 balls of several different sizes. Each ball is white, black,\r\nred or yellow. If you take any 5 balls of the same colour at least two of them will always be\r\nof the same size (radius). Prove that there are at least 3 balls which lie in the same box have\r\nthe same colour and have the same size (radius).\r\n\r\n.\r\n3. A square sheet of paper ABCD is so folded that B falls on the mid-point M of CD. Prove\r\nthat the crease will divide BC in the ratio 5 : 3.\r\n4. Find the remainder when 21990 is divided by 1990.\r\n5. P is any point inside a triangle ABC. The perimeter of the triangle AB + BC + CA = 2s.\r\nProve that\r\ns < AP + BP + CP < 2s.\r\n6. N is a 50 digit number (in the decimal scale). All digits except the 26th digit (from the left)\r\nare 1. If N is divisible by 13, find the 26th digit.\r\n7. A censusman on duty visited a house which the lady inmates declined to reveal their individual\r\nages, but said \u2014 \u201cwe do not mind giving you the sum of the ages of any two ladies you may\r\nchoose\u201d. Thereupon the censusman said \u2014 \u201cIn that case please give me the sum of the ages of\r\nevery possible pair of you\u201d. The gave the sums as follows : 30, 33, 41, 58, 66, 69. The censusman\r\ntook these figures and happily went away. How did he calculate the individual ages of the ladies\r\nfrom these figures.\r\n8. If the circumcenter and centroid of a triangle coincide, prove that the triangle must be equilateral.\r\n1", "Solution_19": "RMO Problems are quite easy , so even if i have solved may, it doesn't make me a genius anyways (as you mentioned)\r\n\r\nactually i have solved many of them almost 50% of all RMO's Problems and about 10% of all INMO's Problems :\r\n\r\nBut the Problem is that I am still weak in Combinatorics and set theory and Algebra part\r\n\r\nAs far as Number Theory and Geometry and General Questions (Tricky) are concerned I have ben able to solve most of them.....\r\n\r\nAnyways!! Let's Look forward to RM0 2009..I would be appearing for RMO for the first time this year!!!\r\nBye the way which region are you all appearing from??, I am Appearing from Delhi Region.\r\n\r\nALL THE BEST TO YOU ALL TOO !!!\r\n] :P :idea: :roll: :blush: :wink:", "Solution_20": "i am from uttar pradesh region!", "Solution_21": "Have you People heard of Terence Tao ??? The Best Mathematician at Present . If not then go to google and Surery read about it !! It is truely Inspiring....", "Solution_22": "[quote=\"amangupta2\"]Have you People heard of Terence Tao ??? The Best Mathematician at Present . If not then go to google and Surery read about it !! It is truely Inspiring....[/quote]\r\nAnother senseless post...well,even if it had such great influence on you, it doesn't yet mean that everyone wants to hear or to read about it.If you really want to argue about this theme then go to the Round Table section or somewhere else and discuss who is the best mathematician in the world, as much as you wish.\r\nP.S:\r\nFurthermore I doubt that Terence Tao is the best mathematician...Perelman is much better, even Terence Tao admitted this fact.", "Solution_23": "Sir\r\nI am really sorry , If I hurt you in anyways through that post. \r\nActually, my aim of creating that post was not to prove whether Terence Tao is the best mathematician or not ,but i just wanted to let my Indian friends know about his childhood and his World Record at IMO and that is all . \r\n\r\n :( :huh: :!:", "Solution_24": "same region.. let's see what happens....\r\n\r\nthe thing which made me call u a genius...that u cud spare time for atlest knowing what combi... is all abt i dont know even that!!!lolzzz\r\nso forget abt what i do..that will not pose any kind of trouble 2 u...\r\nbye\r\n\r\n\r\nps: sorry didn't read the other page..", "Solution_25": "hey y so many processes\r\nlast yr there was no proof--roll no etc that the one appearing 4 RMO was me!!!! :rotfl: :rotfl:" } { "Tag": [], "Problem": "Was anyone at the SEIBA honor band tryouts this Saturday? I didn't even get recalled, but I swear there were like a 100 trumpets trying out for 20 spots...\r\n\r\nCurrently I'm working on the jazz band guitar part. As soon as my director gets an electric guitar for the school I can practice for the jazz competition in March. I'm thinking about \"borrowing\" some Hendrix solos if I ever need a solo part...", "Solution_1": "No, I wasn't. I'm not in band, but I play piano for Ames Jazz II.\r\n\r\nDid you tryout for all-state?", "Solution_2": "Nah, I suck too much for all-state...", "Solution_3": "Penguin, are you a high school trumpeter? \r\n\r\nI auditioned and made the band, again. I've made it three of my possible four years.", "Solution_4": "[quote=\"PathofNeo\"]Penguin, are you a high school trumpeter? \n[/quote]\r\nYes (8th grade), though I suck. By suck I mean I can't even manage an A above the staff consistantly...", "Solution_5": "Oh, you've got plenty of time. A good regiment of practice will straighten almost anything out. If you want to increase your range, try ascending long tones (12-20 counts) until you reach your limit. When you're there, hold it as long as you can while producing a good tone. Then do it again. And again. And again. :lol: (Hooray for being a trumpet major)" } { "Tag": [ "function", "geometry", "geometric transformation", "reflection", "quadratics", "algebra", "quadratic formula" ], "Problem": "Here's another problem i need help with, hope this is the right section for it\r\n\r\nFind an equation that defines the inverse of function f(x)= 2x-3", "Solution_1": "[hide]\n$f(x)=y=2x-3$.\n\nTo find $f^{-1}(x)$, we swap the roles of $x$ and $y$. Note that this reflects the graph through the line $y=x$.\n\n$x=2y-3$.\n\nSolving for $y$, we get $y=\\frac{x+3}{2}$. \n\nThis is the inverse of the original function.\n\n$\\boxed{f^{-1}(x)=\\frac{x+3}{2}}$.[/hide]", "Solution_2": "now try the inverse of the function $f(x)=(2x-3)^{2}$\r\n$\\textbf{Hint}:$ Its similar to yours.", "Solution_3": "Would you do the same thing?\r\n[hide]$f(x)=y=(2x-3)^{2}$\n\nSwitch x and y:\n\n$x=(2y-3)^{2}$\n$x=4y^{2}-12y+9$\n$4y^{2}-12y+(9-x)=0$\n\nNow, use the quadratic formula:\n\n$y=\\frac{12\\pm\\sqrt{144-16(9-x)}}{8}$\n$y=\\frac{3\\pm\\sqrt{9-9+x}}{2}$\n\n$\\boxed{y=\\frac{3\\pm\\sqrt{x}}{2}}$\n\nNow that I've done all that work, I realize I could have just taken the square root in the first step. :mad: [/hide]" } { "Tag": [ "algebra", "polynomial", "function" ], "Problem": "Try this problem! \r\n\r\n Find all polynomials f(x) such that f(x)+f(x+2)=2f(x+1)\r\n for all real x.\r\n\r\nI don't find it difficult, nevertheless interesting! :)", "Solution_1": "[quote=\"socrates\"]Try this problem! \n\n Find all polynomials f(x) such that f(x)+f(x+2)=2f(x+1)\n for all real x.\n\nI don't find it difficult, nevertheless interesting! :)[/quote]\r\n\r\nI love math problems in the F&G forum. :D \r\n\r\n[hide]f(x)+f(x+2)=2f(x+1)\nf(x+1)=(f(x)+f(x+2))/2\nThis means f(x+1) is the arithmetic mean of f(x) and f(x+2), which indicates that f(x) is a linear function. \nSo, any linear function would satisfy f(x)[/hide]", "Solution_2": "Very well! \r\nPlease tell me the general rule for this.! :)", "Solution_3": "[quote=\"socrates\"]Very well! \nPlease tell me the general rule for this.! :)[/quote]\r\n\r\nWell, you posted it in at least 2 other forums, check there! ;) :rotfl: :rotfl:", "Solution_4": "[quote=\"socrates\"]Very well! \nPlease tell me the general rule for this.! :)[/quote]The rule is simple: you don't post serious math problems in the Games & Fun Factory forums, and you don't post a problem twice on the forum! \r\n\r\nDiscussion about the problem is carried on in this topic: \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=271412" } { "Tag": [], "Problem": "The origin is the limiting point of a system of co-axial circles of which $ x^2 \\plus{} y^2 \\plus{} 2gx \\plus{} 2fy \\plus{} c \\equal{} 0$ is a member. The equation of the orthogonal system is given by-\r\nA] $ (x^2 \\plus{} y^2)(g \\plus{} \\lambda.f) \\plus{} c(x \\plus{} \\lambda y) \\equal{} 0$ \r\nB] $ (x^2 \\plus{} y^2)(g \\plus{} \\lambda.f) \\plus{} c(\\lambda x \\plus{} y) \\equal{} 0$\r\nC] $ (x^2 \\plus{} y^2)(\\lambda g \\plus{} f) \\plus{} c(x \\plus{} \\lambda y) \\equal{} 0$\r\nD] none of these\r\n\r\nwhere $ \\lambda$ belongs to R\r\n\r\nAns-A", "Solution_1": "IIT jee math syllabus includes the following for circles:\r\n\r\nEquation of a circle in various forms, equations of tangent, normal and chord.\r\n\r\nParametric equations of a circle, intersection of a circle with a straight line or a circle, equation of a circle through the points of intersection of two circles and those of a circle and a straight line. \r\n\r\nI dont think coaxial and other things are necessary.:) :blush: \r\nMay come in comprehension though but I am ready to risk it. :D", "Solution_2": "i am ready to risk everything in co- geo. :rotfl:", "Solution_3": "what do u mean by\r\n1. coaxial circles (any 2 circles can be said to be coaxial, if axis is considered to be a diameter)\r\n2. limiting point\r\n??", "Solution_4": "[quote=\"achu_182\"]what do u mean by\n1. coaxial circles (any 2 circles can be said to be coaxial, if axis is considered to be a diameter)\n2. limiting point\n??[/quote]\r\n\r\n1. same radical axis.\r\n2. members of the family with zero radius.\r\n\r\nmethod... x2+y2=0 is also amember. s1+ts2=0. diffrntiate. replacce dy/dx by -dx/dy. int." } { "Tag": [ "algebra", "function", "domain" ], "Problem": "Let $ x,y,z>0$ and $ x\\plus{}y\\plus{}z\\equal{}\\sqrt{3}$. Find maximum of $ P\\equal{}\\sqrt{x^2\\plus{}xy\\plus{}y^2}\\plus{}\\sqrt{y^2\\plus{}yz\\plus{}z^2}\\plus{}\\sqrt{z^2\\plus{}zx\\plus{}x^2}$", "Solution_1": "Is the maximum achievable, or is it an upper bound?\r\n\r\nEdit: I agree :) , I thought it was weird though that $ P$ was defined, which at first glance suggested that the value would be attainable.", "Solution_2": "Upper bound -- it occurs at $ (\\sqrt{3},0,0)$ and permutations, which are not in the domain space.\r\n\r\nOn the other hand, the minimum [i]is[/i] attainable.", "Solution_3": "You can post full solution?" } { "Tag": [], "Problem": "find the product and suggest a mechanism", "Solution_1": "Isn't the product just 1,1-dimethylcyclohexane?", "Solution_2": "that is the minor product", "Solution_3": "Interesting. So, what is the major product?", "Solution_4": "Is the major product 3,3,5-trimethylcyclopentanone? I am surprised that you have to study such a specific issue.", "Solution_5": "yes that is the right answer and the mechanism?", "Solution_6": "Well, to your information the mechanism of the Clemmensen reduction is not fully understood at present, as far as I know: the best we (or I) have is one or another [i]proposal[/i] for the mechanism. Also, I see no interest in knowing the mechanism of such a transformation - there are plenty of mechanisms much more important than this in an organic chemistry course. However, I will leave you the proposal I know - you can then figure out by yourself the mechanism of the particular Clemmensen reduction of 1,3-diketones.\r\n\r\nThe reaction of a carbonyl compound RCOR' occurs at the surface of the zinc catalyst as follows:\r\n\r\nStep 1. The pi-bond of the C=O linkage undergoes homolitic cleavage to form the radical species $ Zn\\minus{}O\\minus{}C(RR')\\cdot$.\r\n\r\nStep 2. There is elimination of zinc oxide to form $ Zn\\equal{}CRR'$.\r\n\r\nStep 3. Addition of $ H^{\\plus{}}$ to form $ Zn^{\\plus{}}\\minus{}CHRR'$.\r\n\r\nStep 4. Further addition of $ H^{\\plus{}}$ to give the final product, $ RCH_{2}R'$ and $ Zn^{2\\plus{}}$." } { "Tag": [ "logarithms", "algebra proposed", "algebra" ], "Problem": "Solve the system equation: $ \\{\\begin{array}{l}x^2 + y^2 = 1 \\\\\r\ne^\\frac{1}{e}(e^{x\\ln x}+e^{y\\ln y})=(x+y)^2 \\\\\r\n\\end{array} $", "Solution_1": "Nobody like this problem?\r\n[hide=\"Hint\"]$ x\\ln x \\geq \\minus{} \\frac {1}{e}$[/hide]" } { "Tag": [ "logarithms" ], "Problem": "a physical quantity is related to four variables as\r\n \u03a9 =( a^3.b^2 ) / \u221ay . n\r\n\r\nthe percentage errors in measurement of a,b,y,n are 1%,3%,4% and 2% respectively. if value of \u03a9 is calculated as 3.763 unit from the given values...to what value should you round off the result ??", "Solution_1": "[quote=\"eureka-123\"]a physical quantity is related to four variables as\n \u03a9 =( a^3.b^2 ) / \u221ay . n\n\nthe percentage errors in measurement of a,b,y,n are 1%,3%,4% and 2% respectively. if value of \u03a9 is calculated as 3.763 unit from the given values...to what value should you round off the result ??[/quote]\r\n[hide]\nAssuming you meant $ \\Omega \\equal{}\\frac{a^3b^2}{\\sqrt{y}\\cdot n}$\n\nthen $ \\ln\\Omega \\equal{} 3\\ln a \\plus{} 2\\ln b \\minus{}\\frac{1}{2}\\ln y \\minus{}\\ln n$.\n\nDifferentiating the above expression:\n\n\\[ \\frac{d\\Omega}{\\Omega} \\equal{} 3\\frac{d a}{a} \\plus{} 2\\frac{d b}{b} \\minus{}\\frac{1}{2}\\frac{d y}{y} \\minus{}\\frac{d n}{n}\\]\n\nTherefore,\n\n\\[ \\frac{\\sigma_{\\Omega}}{\\bar{\\Omega}} \\equal{} 3\\frac{\\sigma_a}{\\bar{a}} \\plus{} 2\\frac{\\sigma_b}{\\bar{b}} \\plus{}\\frac{1}{2}\\frac{\\sigma_y}{\\bar{y}} \\plus{}\\frac{\\sigma_n}{\\bar{n}}.\\]\n\nThis implies $ \\frac{\\sigma_{\\Omega}}{\\bar{\\Omega}} \\equal{} 3(1\\%) \\plus{} 2(3\\%) \\plus{}\\frac{1}{2}(4\\%) \\plus{} 2\\% \\equal{} 13\\%$\n\n\\[ \\sigma_{\\Omega} \\equal{} 0.13\\cdot \\bar{\\Omega} \\\\ \\\\ \\equal{} 0.13(3.763) \\equal{} 0.49\\quad \\text{to two significant figures.}\\]\n\n\\[ \\therefore \\Omega \\in [3.3, 4.3)\\][/hide]" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all real functions f if f(xy+((f(x))^2)=f(f(x))f(x+y).\r\n Harazi", "Solution_1": "Just the beginning:\r\n\r\nWe put x=0 and find f(f 2 (0))=f(f(0))f(y) for all real y. \r\n\r\nCase 1: f is constant. We put f(x)=k for all x, so k 2 =k, so f(x)=1 for all x or f(x)=0 for all x. Both of these verify the equation, so they're both solutions.\r\n\r\nCase 2: f isn't constant. From f(f 2 (0))=f(f(0))f(y) we find f(f(0))=0. In the initial eqn we make x=f(0), so we get f(yf(0))=0 for all reals y, so f(0)=0, because otherwise yf(0) would cover all reals, so f would be constant. \r\n\r\nLet's take an a s.t. f(a)=0. We put x=a and find f(ay)=0 for all reals y, so a=0 (by the same argument as above). This means that the only soln of f(x)=0 is x=0. \r\n\r\nWe take y=-x, so f(f 2 (x)-x 2 )=0, so, by the previous observation, f 2 (x)-x 2, which means that for every real x we have f(x)=x or f(x)=-x. \r\n\r\nI haven't continued.", "Solution_2": "Another step:\r\n\r\nIf we make y=0 we get f(x2)=f(f2(x))=f(f(x))*f(x) (*)\r\n\r\nWe can write the initial relation like this:\r\n\r\nf(x(x+y))=f(f(x))f(x+y). Let a=x and b=x+y. We have f(ab)=f(f(a))f(b) for all real a and b. Since f(ab) is symmetric in a and b we also have f(ab)=f(a)f(f(b)). We multiply these and we combine the result with (*) to get (ab)2 = f(a2)*f(b2), so f(x)/x is constant for all positive x, which means that we either have f(x)=x for all x>0 or f(x)=-x for all x>0.", "Solution_3": "I'm doing this step by step :D \r\n\r\nLet's assume f(x)=x for all x>0. Let's take a,b<0. We have f(f(a))f(b)=f(ab)=ab, so we either have f(a)/a=f(b)/b=1 or f(a)/a=f(b)/b=-1, so for all x<= f(x)/x is constant, so we either have f(x)=x for all x<= or f(x)=-x for all x<0. Since we assumed f(x)=x for x>0, we have 2 cases: \r\n\r\nf(x)=x if x>0 and -x if x<0\r\nf(x)=x for all reals x.\r\n\r\nI think that if we assume f(x)=-x for all x<0 we get the other 2 choices:\r\n\r\nf(x)=-x if x>0 and f(x)=x if x<0\r\nf(x)=-x for all reals x.\r\n\r\nI think these 4 and the constant solutions f(x)=1 for all x and f(x)=0 for all x are the solutions of the problem. Is it right? I haven't checked the case f(x)=-x for all x>0, so I might be wrong.", "Solution_4": "Yes, grobber, you are right., but the function f(x)=-x if x>0 and x if x<0 is not solution. I think that this problem solves itself, you have just to work on it. Let's remember the IMO 2002 function, another example of this type.", "Solution_5": "Any continuation?" } { "Tag": [ "algebra", "polynomial", "function", "real analysis", "real analysis unsolved" ], "Problem": "let $f(x,y)$ be a continuous function from $R^2$ \r\nto $R$ such that:\r\n*for all $x_0 \\in R$, $y -> f(x_0, y)$ is a polynomial\r\n*for all $y_0 \\in R$, $x -> f(x , y_0)$ is a polynomial\r\n\r\nprove that $f \\in R[X,Y]$.", "Solution_1": "It has been proved many times on forum, Alekk. By the way, as far as I remember, the problem works without the assumption of continuity. I remember anyway to have seen it at ENS without this condition. I will check.", "Solution_2": "It's true, we don't need the hypothesis of continuity. I do believe it's been discussed before, and I sort of remember a proof by Myth, but I'm definitely not sure, and I can't find it.. :?", "Solution_3": "and what was the idea of the proof ?", "Solution_4": "I will post a solution tomorow, I remember to have solved this one some time ago. I will try to rediscover the idea.", "Solution_5": "Here is what I did. Hope it's good. First of all, by the fact the the reals are not countable, one can find two numbers $m,n$ and two infinite sets $A,B$ such that there exist functions $ g_0,..., g_n$ from $A$ to the reals and functions $ h_0,...,h_m$ from $B$ to the reals such that $ f(x,y)=g_0(x)+g_1(x)y+...+g_n(x) y^n$ for all $x$ in $A$ and real $y$ and $ f(x,y)=h_0(y)+...+h_m(y) x^m$ for all $y$ in $B$ and real $x$ (I just wrote that there exist $m,n$ such that for infinitely many reals $x$ the function which takes $y$ to $ f(x,y)$ is a polynomial of degree at most $n$ and the same for $m$). Now, take $ y_0,...,y_n$ different from $B$ and write the condition $ h_0(y)+...+h_m(y)x^m=g_0(x)+...+g_n(x) y^n$ for $y_0,...,y_n$ and all $x$ in $A$. It will give you a linear system in the unknowns $ g_0(x),..., g_n(x)$ and you will easily see that we can find $P_0,...,P_n$ polynomials such that $ g_i(x)=P_i(x)$ when $x\\in A$. Thus we can find a two-variable polynomial $P_1(X,Y)$ such that $ f(x,y)=P_1(x,y)$ whenever $x\\in A$ and $ y\\in R$. The same argument shows that we can find a two variable polynomial $ P_2(X,Y)$ such that $ f(x,y)=P_2(X,Y)$ whenever $x\\in R$ and $ y\\in B$. It follows that $ P_1-P_2$ vanishes on the infinite set $AxB$ and thus it vanishes everywhere, hence $ f(x,y)=P_1(x,y)$ whenever $x$ is in $A$ or $y$ is in $B$. Now, take two reals $x,y$. I want to show that $ f(x,y)=P_1(x,y)$. If $x\\in A$ it's clear. Assume the contrary. The function $ t(x)= f(x,y)$ is a polynomial $ S(x)$. Moreover $ S(a)=P_1(a,y)$ whenever $a\\in A$, which is infinite, thus in fact the polynomials are identical and thus $ f(x,y)=S(x)=P_1(x,y)$.", "Solution_6": "very nice :) \r\nThe next question asked if the conclusion holds if we only suppose that $f(x_0,y)$\r\nand $f(x,y_0)$ are polynomial when $x_0 \\in Q$ or $y_0 \\in Q$ ?", "Solution_7": "This one is easier :), and I definitely remember it being solved by Myth. The answer is no. I don't know if this was the exact same example that Myth probided, but it was along the same lines:\r\n\r\nWe take $f(x,y)=\\sum_{n\\ge 1}\\alpha_n(x-q_1)(y-q_1)\\ldots(x-q_n)(y-q_n)$, where $(q_n)_n$ is an enumeration of the rationals, and $\\alpha_n$ are rationals chosen to be so small that the series converges absolutely for all reals $x,y$.", "Solution_8": "Does the same stay true if we replace polynomials by rational functions?" } { "Tag": [], "Problem": "In the following figure, $CD$ is a half chord perpendicular to the diameter $AB$ of the semicircle with center $O$. A circle with center $P$ is inscribed, toucher $AB$ at $E$ and arc $BD$ at $F$. Prove that $\\Delta EDA$ is isosceles.", "Solution_1": "I've got a proof using trigometry and bisector theorem to show $\\measuredangle CDE= \\measuredangle EDB$. If you want I can post it. :D", "Solution_2": "please do.", "Solution_3": "This is gonna be a little long :P \r\n\r\n$WLOG$:$OP=1$\r\nAlso:$\\measuredangle POE = \\alpha$\r\n\r\n$PE=CE=sin \\alpha$, $OF=1+sin\\alpha$\r\n$\\Rightarrow OC=cos \\alpha-sin \\alpha$\r\n\r\n$(DA)^{2}=(AC)^{2}+(OD)^{2}-(OC)^{2}$\r\n$(DA)^{2}=(1+cos\\alpha)^{2}+(1+sin\\alpha)^{2}-(cos\\alpha-sin\\alpha)^{2}$\r\n$(DA)^{2}= 2(1+sin\\alpha)(1+cos\\alpha)=(1+sin\\alpha+cos\\alpha)^{2}$ $\\Rightarrow$ $DA=1+sin\\alpha+cos\\alpha$\r\n\r\n$AE=1+sin\\alpha+cos\\alpha$\r\n\r\n$\\Rightarrow$ $\\boxed{DA=AE}$ $\\mathbb{QED}$\r\n\r\nWow, I found a better solution :!: .", "Solution_4": "[quote=\"M4RI0\"]This is gonna be a little long :P \n\n$WLOG$:$OP=1$\nAlso:$\\measuredangle POE = \\alpha$\n\n$PE=CE=sin \\alpha$, $OF=1+sin\\alpha$\n$\\Rightarrow OC=cos \\alpha-sin \\alpha$\n\n$(DA)^{2}=(AC)^{2}+(OD)^{2}-(OC)^{2}$\n$(DA)^{2}=(1+cos\\alpha)^{2}+(1+sin\\alpha)^{2}-(cos\\alpha-sin\\alpha)^{2}$\n$(DA)^{2}= 2(1+sin\\alpha)(1+cos\\alpha)=(1+sin\\alpha+cos\\alpha)^{2}$ $\\Rightarrow$ $DA=1+sin\\alpha+cos\\alpha$\n\n$AE=1+sin\\alpha+cos\\alpha$\n\n$\\Rightarrow$ $\\boxed{DA=AE}$ $\\mathbb{QED}$\n\nWow, I found a better solution :!: .[/quote]\r\n\r\nNice. See if you can find a way with massive Pythagoras.", "Solution_5": "Maybe if someone could show in a geometric way that $\\measuredangle CDE=\\measuredangle EDB$, it would be really nice." } { "Tag": [ "probability" ], "Problem": "The intergers from 1 to 10 inclusive, are partitioned at random into two sets of five elements each. What is the probability that 1 and 2 are in the same set?\r\n\r\nmy method seems right, but I'm getting the wrong answer. Would anyone mind checking my work? Thank you:)\r\n[hide]\nThere are $\\frac{10!}{5!5!}=252$ ways to choose the elements in the two sets. There are $\\frac{8!}{3!5!}=56$ ways to choose the other three elements in set 1( the set that 1 and 2 are in). so the my answer is $\\frac{56}{252}$. Do I have to multiply $\\frac{56}{252}$ by 2, since there are two sets, 1 and 2 can be in set 1 or set 2. [/hide][/hide]", "Solution_1": "[quote=\"myc\"]The intergers from 1 to 10 inclusive, are partitioned at random into two sets of five elements each. What is the probability that 1 and 2 are in the same set?\n\nmy method seems right, but I'm getting the wrong answer. Would anyone mind checking my work? Thank you:)\n[hide]\nThere are $\\frac{10!}{5!5!}=252$ ways to choose the elements in the two sets. There are $\\frac{8!}{3!5!}=56$ ways to choose the other three elements in set 1( the set that 1 and 2 are in). so the my answer is $\\frac{56}{252}$. Do I have to multiply $\\frac{56}{252}$ by 2, since there are two sets, 1 and 2 can be in set 1 or set 2. [/hide][/hide][/quote]\r\nActually, you do have to multiply by 2, and then you can simplify $2 \\times \\frac{56}{252}$ to $\\frac{4}{9}$, which I believe is the answer.\r\n\r\nAnother way to do the problem:\r\n1 is in one of the sets. 2 can go in any of the other 9 spaces, 4 of which are in the same set as 1, 5 of which are not. Each space has equal probability. Thus our probability is $\\frac49$.", "Solution_2": "[hide] This is my solution:\nThere are $\\frac{10!}{2^{5}}$ ways to create the 5 sets.\nThink of it as ordering the numbers in a list, and every 2 are in a set. For example, $abcdefghij$ would have the sets $\\{a,b\\}$, $\\{c,d\\}$, etc. Each pair of 2 could also be flipped. For example, $abcdefghij\\equiv bacdefghij$. There are $10!$ ways to list the numbers and $2^{5}$ ways to represent the same 5-set solution in a list.\n\nThere $\\frac{8!}{2^{4}}$ ways to create the 5 sets, given that one of the sets is $\\{1,2\\}$, in the same way as above.\nThus, the answer is $\\frac{\\frac{8!}{2^{4}}}{\\frac{10!}{2^{5}}}=\\frac{2}{10\\cdot9}=\\boxed{\\frac{1}{45}}$ [/hide]\r\nEDIT: Oops, I thought it was 5 sets of 2 elements. :blush: \r\nIs my answer correct, though?", "Solution_3": "[quote=\"lotrgreengrapes7926\"][hide] This is my solution:\nThere are $\\frac{10!}{2^{5}}$ ways to create the 5 sets.\nThink of it as ordering the numbers in a list, and every 2 are in a set. For example, $abcdefghij$ would have the sets $\\{a,b\\}$, $\\{c,d\\}$, etc. Each pair of 2 could also be flipped. For example, $abcdefghij\\equiv bacdefghij$. There are $10!$ ways to list the numbers and $2^{5}$ ways to represent the same 5-set solution in a list.\n\nThere $\\frac{8!}{2^{4}}$ ways to create the 5 sets, given that one of the sets is $\\{1,2\\}$, in the same way as above.\nThus, the answer is $\\frac{\\frac{8!}{2^{4}}}{\\frac{10!}{2^{5}}}=\\frac{2}{10\\cdot9}=\\boxed{\\frac{1}{45}}$ [/hide]\nEDIT: Oops, I thought it was 5 sets of 2 elements. :blush: \nIs my answer correct, though?[/quote]\r\nWell, actually, if that had been the problem, the answer would be $\\frac19$ because 1 is in one of the sets, and 2 has to be placed in one particular spot out of 9 possible ones.\r\n\r\nYour solution was almost correct. You just forgot to multiply by 5 because {1,2} could be any of {a,b}, {c,d}, etc.", "Solution_4": "[quote=\"nebula42\"]Well, actually, if that had been the problem, the answer would be $\\frac19$ because 1 is in one of the sets, and 2 has to be placed in one particular spot out of 9 possible ones.[/quote] That was easy. :|", "Solution_5": "yeah, this problem is incredibly easy if you think about it - the 1 has to be in a certain spot, and there are 9 spots for the 2 to be in, and only four work, making the probability $\\frac49$." } { "Tag": [ "inequalities", "function", "trigonometry", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Prove that $\\sqrt[2006]{\\cos x}\\leq \\cos \\frac{x}{2006}$ for $0\\leq x \\leq \\frac{\\pi}{2}.$", "Solution_1": "Let $f(t)=t\\ln \\cos \\frac{x}t,t \\ge 1$\r\n$f'(t)=\\ln \\cos \\frac{x}t+\\frac{\\sin \\frac{x}{t}}{\\cos \\frac{x}{t}}\\frac{x}{t}$\r\nLet:\r\n$g(a)=ln \\cos a+a\\tan a,00$\r\nand $g(0)=0$\r\nthen we have:$f'(t)=g(\\frac{x}{t})>0$\r\nso $f(2006)>f(1)$\r\nWhich comes the desired inequality :)" } { "Tag": [ "geometry", "geometric transformation", "reflection", "projective geometry", "geometry open" ], "Problem": "Could anybody please show me some results about the antipole and the perspector PLEASE :(\r\n\r\nFriendly \r\n\r\nJos\u00e9 Carlos", "Solution_1": "Antipole? Never heard about it. If you wanted to say \"antipode\", then the [i]antipode[/i] of a point P with respect to a circle k is the reflection of this point P in the center of the circle k. Usually, this term is only used if the point P lies on the circle k; then, the antipode of P is also called the [i]point diametrically opposite[/i] to the point P on the circle k.\r\n\r\nAs for the perspector, there is really not much to say. If $A_1A_2...A_n$ and $B_1B_2...B_n$ are two polygons such that the lines $A_1B_1$, $A_2B_2$, ..., $A_nB_n$ all concur at one point, then this point is called the [i]perspector[/i] of the polygons $A_1A_2...A_n$ and $B_1B_2...B_n$, and the polygons themselves are called [i]perspective[/i] (or [i]in perspective[/i]). The most important result on perspectivity is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30323]the Desargues theorem[/url].\r\n\r\n Darij", "Solution_2": "[quote=\"darij grinberg\"]As for the perspector, there is really not much to say. If $A_1A_2...A_n$ and $B_1B_2...B_n$ are two polygons such that the lines $A_1B_1$, $A_2B_2$, ..., $A_nB_n$ all concur at one point, then this point is called the [i]perspector[/i] of the polygons $A_1A_2...A_n$ and $B_1B_2...B_n$, and the polygons themselves are called [i]perspective[/i] (or [i]in perspective[/i]). The most important result on perspectivity is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30323]the Desargues theorem[/url].\n\n Darij[/quote]\r\n\r\n :? I see... let me ask you something... If you talk about two polygons such that the lines join the homologous vertices are concurrent at $O$, then we say that the polygons are in [i]perspective from [/i]$O$... the dual defintion says that if two polygons such that the intersetion points of the homologous sides lie in $l$, then we say that the polygons are in [i]perspective from [/i]$l$... What about the poligons (even more the n-line complete) wich are perspective from a point and from a line?? (I think it could be better if it's possible to prove that if two $n$-line complete or $n$-point complete are perspective from a line, then they are perspective from a point, because... What's the definition of perspective... How can u identify if two figures are in perspective froma point and from a line... It have to use perspective froma point and perspective from a line??...)\r\n\r\n[b]Theorem. [/b](Desargues) Two triangles are in perspective from a point iff they are in perspective from a line.\r\n\r\nSo it follows in...\r\n\r\n[b]Theorem.[/b] Given two complete quadrangles such that 5 points of intersection of their correspondent sides lie in a line, the the 6 intersection points of the correspondent sides lie in a line and the two quadrangles are perspective from a point and from a line.\r\n\r\nPD. Have you took [b]Measure Theory??[/b]... I need some concepts about areas... :?:" } { "Tag": [ "geometry", "3D geometry", "dodecahedron" ], "Problem": "An edgy spider walks only along the edges from $A$ to $B$ of the dodecahedron formed by the net shown. What is the number of edges in the shortest path that the spider could take?\r\n\r\n[img]7605[/img]", "Solution_1": "Uh, what net?", "Solution_2": "[quote=\"bpms\"]Uh, what net?[/quote]\r\n\r\nThat shape in the figure is supposed to be a net of a dodecahedron. Am I missing something here? :maybe:", "Solution_3": "Is it\r\n\r\n[hide] 3 edges?[/hide]", "Solution_4": "I think the answer is [hide]5[/hide] edges.\r\n\r\nJorian", "Solution_5": "I think my answer might be right, but I visualized it, so I also wouldn't be surprised if it's wrong.", "Solution_6": "[quote=\"jhredsox\"]I think the answer is [hide]5[/hide] edges.\n\nJorian[/quote]\r\n\r\n\r\nno, youre considering him walking on the net, instead of the dodecahedron...\r\n\r\n\r\n\"An edgy spider walks only along the edges from A to B of the [b]dodecahedron formed by the net shown. [/b]", "Solution_7": "[hide]It's 5 on the net, but when it becomes a dodecahedron, it's less since the edges come closer and new connections are made.\n\nI visualized it and got 3... not sure though, I suck at visualizing things :P\n[/hide]", "Solution_8": "try putting some of the net together... \r\n\r\nI still got my answer.", "Solution_9": "How does [i]that[/i] make a dodecahedron?\r\n\r\n-jorian", "Solution_10": "dodecahedron means twelve sides i think (?)", "Solution_11": "[quote=\"jhredsox\"]How does [i]that[/i] make a dodecahedron?\n\n-jorian[/quote]\r\nDodeca = 12 right?\r\n\r\nThat net forms a 12 faced polyhedron.", "Solution_12": "[hide=\"Solution\"]By putting the net together, you get a dodecahedron. The \"V\" shaped gaps between the A and the dot next to it become one spot. Thus, the answer becomes $\\boxed{3\\ \\text{edges}}$, which is our answer.[/hide]", "Solution_13": "[quote=\"mwpl11\"]dodecahedron means twelve sides i think (?)[/quote]\r\n\r\nno, 12 faces.", "Solution_14": "oh yeah. oops :blush:\r\n\r\nI forgot it was 3D. my bad.", "Solution_15": "[quote=\".:Logic:.\"][quote=\"bpms\"]Uh, what net?[/quote]\n\nThat shape in the figure is supposed to be a net of a dodecahedron. Am I missing something here? :maybe:[/quote]\r\nIt wasn't there when I posted.", "Solution_16": "[quote=\".:Logic:.\"][quote=\"jhredsox\"]How does [i]that[/i] make a dodecahedron?\n\n-jorian[/quote]\nDodeca = 12 right?\n\nThat net forms a 12 faced polyhedron.[/quote]\r\n\r\noh, i thought you guys meant putting lines between each of the sides\r\n\r\nwhoops\r\n\r\njorian", "Solution_17": "I believe it is\r\n\r\n[hide] $3$ [/hide]", "Solution_18": "Could someone post a solution and explain this? I don't get it at all.", "Solution_19": "I already posted a solution :roll:", "Solution_20": "It's more visualization than \"solution\".... :P", "Solution_21": "Here's how I did it\r\n\r\n[img]http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=7638[/img]\r\n\r\nI kind of saw that was connected\r\n\r\nso that's the path (correct me if it isn't)\r\n\r\n-jorian\r\n\r\nedit: why isn't the pic showing??!!", "Solution_22": "[img]7638[/img]\r\n\r\nthat's what I did.\r\n\r\n(picture jhredosx)", "Solution_23": "Just a note: that was the picture I made, anirudh just posted it because I had trouble posting it.\r\n\r\nSo yah, that's how I did it.\r\n\r\n-jorian", "Solution_24": "Please do not call me by my name; rather #H34N1 is fine :)", "Solution_25": "Sorry #H34N1, I'm so used to anirudh, plus I can't remember #H34N1. Please don't reply as this topic has served its purpose (unless you have a question) and we're getting off-track.\r\n\r\n-Jorian" } { "Tag": [ "calculus" ], "Problem": "sorry i know i shouldnt have posted this here, but..bcz.. i dont know where to post this question\r\n\r\ncan someone tell me this , does ti 89 titanium solve questions in step by step or does it just give the answer?\r\n thanks, adm u can delete this after the answr sorry", "Solution_1": "Just the answer.", "Solution_2": "Just the answer, but you can easily program the calculator to show at least a few steps with a basic knowledge of the TI programming language.", "Solution_3": "any one knows the calculus made easy key for ti 89 titanium?", "Solution_4": "[quote=\"binomial_4eva\"]any one knows the calculus made easy key for ti 89 titanium?[/quote]\r\n\r\nUm... You have to buy it to get the key.", "Solution_5": "[quote=\"vishalarul\"][quote=\"binomial_4eva\"]any one knows the calculus made easy key for ti 89 titanium?[/quote]\n\nUm... You have to buy it to get the key.[/quote]\r\n\r\nBuy what? :huh:", "Solution_6": "Do you guys mean an app? B/c I know one app for the calculus is one you must buy. But if by \"key\" you mean something else...please explain.\r\n\r\nAnyway since it is sort of trivial in my mind to make an 89 show some steps for you, has anyone done this and be willing to post the program?\r\n\r\nAlso can you do both dec->frac and frac->dec?" } { "Tag": [ "geometry", "circumcircle", "trigonometry" ], "Problem": "The distances from a point on the circumcircle of an equilateral triangle to the two nearest vertices are 3 and 6. Find the distance from the point to the furthest vertex. (Please explain.)\r\n\r\nThank you.", "Solution_1": "Let $ABC$ be the equilateral triangle and $P$ given point in the circumcircle. Let $PA=3$ and $PB=6$ Now angle $APB=120^\\circ$ since $ABCD$ is cyclic and $C=60^\\circ$. Now $AB^2=3^2+6^2-2\\cdot 3\\cdot 6\\cos 60^\\circ=45-18=27$ so $AB=3\\sqrt{3}$. It follows from Ptolemy's theorem that $3\\sqrt{3}\\cdot 3+3\\sqrt{3}\\cdot 6=PC\\cdot 3\\sqrt{3}$ so that $PC=3+6=9$.", "Solution_2": "thank you, but can u please tell me what Ptolemys Theorem says.", "Solution_3": "Mathworld.Wolfram is a really good site for this. [url]http://mathworld.wolfram.com/PtolemysTheorem.html[/url]\r\n\r\nAside: does anyone know why it doesn't find the server unless you include the /index.html? I get a server not found when just typing in mathworld.wolfram.com.", "Solution_4": "http://www.mathlinks.ro/Forum/viewtopic.php?t=34094 ;)\r\n\r\n\r\n\r\n[quote=\"Hokkage\"]Aside: does anyone know why it doesn't find the server unless you include the /index.html? I get a server not found when just typing in mathworld.wolfram.com.[/quote]\r\nI don't... which browser are you using?" } { "Tag": [ "real analysis", "topology", "real analysis unsolved" ], "Problem": "Suppose continuum hypothesis is wrong. \r\nLet S be a set of real numbers and |N| < |S| < |R|. Is it true mes(S) = 0?", "Solution_1": "Sure. The measurable set of positive measure contains a closed subset of positive measure. But the closed set is either not-more-than-countable or have capacity continuum (it's classical result of, I think, Lindelof. It may be proved as follows: 1)call the point a from the set A to be a point of condensation, if any neighborhood of a contains more than countably many points of A. 2) All points of A except countably many are points of condensation. 3) The set of points of condensation of closed set A is perfect set (i.e. the closed set, which contains all its limit points) 4) not-empty perfect set has capacity of continuum; it's easy to construct injection from the set of sequences {a1,a2,...}, a[i]=0 or 1, to it, as it is made for Cantor set: choose a complementary interval, and consider points left to it, if a[1]=0 and right to it, if a[1]=1. Then choose a complementary interval left (or right) to the first choosen and do the same.By a principle of embeeded segments, you get a point of A from a sequence and it's clearly an injection.", "Solution_2": "You must not suppose S is measurable. Maybe there exists nonmeasurable set S with property |N| < |S| < |R|.\r\n\r\n[quote]perfect set (i.e. the closed set, which contains all its limit points)[/quote]\r\nIs it exact definition of perfect set? I mean closed set always contains its limit points.", "Solution_3": "Ops, sorry. The perfect set is such a close set, that each its point is a limit point (i.e. the set coinciding with its limit set). \r\n\r\nAs for non-measurable set, I do not know indeed, what to do. \r\n\r\nWhere is this problem from? It may appear to be hopeless.", "Solution_4": "I made lectures about continuum hipothesis (cardinal numbers, ordinal numbers, Erdos theorem and so on). It is very interesting theme. And now I have some questions related to it. So this is my problem and don't know solution at the moment.", "Solution_5": "Lets move it to Analysis section.", "Solution_6": "This problem is independent of ZFC.", "Solution_7": "Any detailed comment?", "Solution_8": "If we assume Martin's Axiom (MA) and the negation of Continuum Hypothesis (CH) then every set of cardinality less than continuum has measure zero. The proof of this fact is easy, and I can post it tomorrow if anybody is interested. (And it is known that MA + $\\neg$ CH is consistent with ZFC.)\r\n\r\nThe other direction is trickier. It is consistent with ZFC that there is a set of cardinality less than continuum that is not a null set. However, I do not know a simple argument for this. :( If I don't come up with a short proof, I'll find a reference and post it here.", "Solution_9": "We will prove that MA implies that every set of cardinality less than continuum is a null set.\r\n\r\nConsider a set $A\\subset [0,1]$ s.t. $|A| < |{\\mathbb R}|$. We are going to cover $A$ with null sets $S_{k}$, $k\\in \\omega$.\r\n\r\nWe consider sentences of two types\r\n1) \"$x\\in \\mathbf{S_{k}}$\" for $x\\in A$, $k\\in\\omega$. Here $\\mathbf{S_{k}}$ is a variable name;\r\n2) \"$[s, t] \\not\\subset \\mathbf{S_{k}}$\" for $s,t \\in{\\mathbb Q}$ ($s 0$ let \\[D_{k, \\varepsilon}= \\{p\\in P: \\exists \\text{ intervals }[s_{1},t_{1}], \\dots, [s_{k}, t_{k}] \\text{ s.t. their union has measure at least }1-\\varepsilon \\text{ and }([s_{i}, t_{i}]\\not\\subset S_{k})\\in p\\}.\\] The set $D_{k, \\varepsilon}$ is dense: Let $p$ be a condition. The set $F$ of $x$ mentioned in $p$ is finite. Therefore, we can pick (finitely many) disjoint intervals in the complement of $F$ whose total length is greater than $1-\\varepsilon$.\r\n\r\nNow we prove that $P$ satisfies the countable chain condition. That is, every antichain (a set of pairwise inconsistent conditions) is at most countable. Observe that if two conditions $p$ and $q$ have the same sentences of the second type, then they are consistent. But there are only countably many different combinations of sentences of the second type. Hence every antichain is countable.\r\n\r\nWe are now ready to use MA. Recall that Martin's Axiom states that given\r\n1) a partial ordered set $P$, that satisfies the countable chain condition,\r\n2) a family of dense sets $D_\\alpha$; the cardinality of the family is less than continuum,\r\nthere exists a (\"generic\") set $G$ s.t. $G$ is consistent, and $G$ intersects every $D_\\alpha$.\r\nWe apply MA to our set $P$ and dense sets $D_{x}$ (for $x\\in A$) and $D_{k, \\varepsilon}$ ($k\\in \\omega$, $\\varepsilon>0$, $\\varepsilon\\in{\\mathbb Q}$). Define \\[S_{k}= \\{x: (x\\in \\mathbf{S_{k}}) \\in p, p\\in G\\}.\\] We prove that $A\\subset\\bigcup_{k\\in \\omega}S_{k}$. Let $x\\in A$. Since $D_{x}$ is dense, $D_{x}\\cap G \\neq \\emptyset$. Let $p\\in D_{x}\\cap G$. By the definition of $D_{x}$, there exists $k$ s.t. $(x\\in \\mathbf{S_{k}}) \\in p$. Since $p\\in G$, $x\\in S_{k}$.\r\nFinally, we prove that $\\mu(S_{k}) = 0$. Let $p\\in D_{k,1/2^{m}}\\cap G$. The condition $p$ contains sentences $[s_{i}, t_{i}]\\not\\subset \\mathbf{S_{k}}$, s.t. $\\mu(\\bigcup_{i=1}^{m}[S_{i}, t_{i}]) \\geq 1-1/2^{m}$. Since $p\\in G$, and all conditions in $G$ are consistent, no condition $q\\in G$ contains \"$(x\\in \\mathbf{S_{k}})$\" for $x\\in[s_{i}, t_{i}]$. We conclude $\\bigcup_{i=1}^{m}[s_{i}, t_{i}] \\cap S_{k}= \\emptyset$. So $\\mu(S_{m}) \\leq 1/2^{m}$ for every $m$.\r\n\r\nWe have proved that $A \\subset \\bigcup_{k\\in\\omega}S_{k}$, where each $S_{k}$ is a null set.", "Solution_10": "Here is a reference to a good article on the subject:\r\n\r\nH. Judah and S. Shelah, \"The Kunen-Miller Chart (Lebesgue Measure, the Baire Property, Laver Reals and Preservation Theorems for Forcing)\", The Journal of Symbolic Logic, Vol. 55, N 3, Sept. 1990.", "Solution_11": "Thanks. It should be studied. Too far away from clarity :)", "Solution_12": "But will the more general conclusion be true:\r\nIf $|A|<|B|$,then $m^{*}(A)\\le m^{*}(B)$ :?:", "Solution_13": "What is $m*$? Exterior measure?", "Solution_14": "Yes,that's what I mean,And sorry I forgot to type ^.\r\nAnd if this is true,the original problem will be easy to solve,consider the Cantor-Set.", "Solution_15": "If there exists a set $S$ s.t. $|S| < \\mathbf{c}$ and $\\mu^{*}(S) > 0$, then the answer to your question is \"no\": when $A = S$, $B$ is the Cantor set, we have $|A|< |B|$, but $\\mu^{*}(A) >\\mu^{*}(B)$.\r\n\r\nOn the other hand, if for every set $S$ s.t. $|S| < \\mathbf{c}$, $\\mu^{*}(S) = 0$, then the answer is trivially \"yes\".\r\n\r\nSo your question is equivalent to Myth's original question. Hence the answer to it is also independent of ZFC." } { "Tag": [ "logarithms", "calculus", "calculus computations" ], "Problem": "I need to find out if the following series convergers or divergers to infinity. Please show work\r\n\r\n(cos (n*pi))/(1+sqrt n) n runs from 1 to infinity\r\n\r\n1/(ln (n!) n runs from 2 to infinity\r\n\r\nNeed some help fast", "Solution_1": "For 1 note: $\\sum_{n=1}^\\infty \\frac{\\cos(n\\pi)}{1+\\sqrt{n}}= \\sum_{n=1}^\\infty \\frac{(-1)^{n}}{1+\\sqrt{n}}$. Do you have it now?\r\n\r\nMy gut says diverge for 2, but don't see a way to show it right off. Maybe compare to $\\sum\\frac{1}{n\\ln n}$ ?\r\n\r\nEdit: Yeah, just tried it. $\\sum_{n=2}^\\infty \\frac{1}{\\ln n!}> \\sum_{n=2}^\\infty \\frac{1}{n\\ln n}$ and the latter diverges. Not sure what I was thinking since $n\\ln n=\\ln n^{n}$ and $n^{n}$ grows faster than $n!$... Oh well.", "Solution_2": "Thank you very much. And yes it helped. I see now that cos n*pi is just a funny way to write (-1)^n :)", "Solution_3": "$\\sum_{2}^{n}\\frac{1}{k\\ln{k}}=\\ln{\\ln{n}+C'+o(1).}$\r\nwhere $C'$ is constant\r\n :wink:" } { "Tag": [], "Problem": "\u03c7\u03c4\u03b5\u03c2 \u03c0\u03ae\u03b3\u03b1\u03bc\u03b5 \u03bc\u03b5 \u03c4\u03bf \u03a1\u03ac\u03bb\u03bb\u03b7 (RAKAR) \u03bd\u03b1 \u03c0\u03b1\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03c3\u03ba\u03ac\u03ba\u03b9...\r\n\u03ba\u03b1\u03b8\u03ce\u03c2 \u03c4\u03bf\u03c0\u03bf\u03b8\u03b5\u03c4\u03bf\u03cd\u03c3\u03b1\u03bc\u03b5 \u03c4\u03b1 \u03c0\u03bf\u03cd\u03bb\u03b9\u03b1 \u03b5\u03af\u03c7\u03b1 \u03c4\u03b7\u03bd \u03b5\u03be\u03ae\u03c2 \u03b9\u03b4\u03ad\u03b1:\r\n\u03c0o\u03b9\u03b1 \u03b7 \u03c0\u03b9\u03b8\u03b1\u03bd\u03cc\u03c4\u03b7\u03c4\u03b1 \u03bd\u03b1 \u03c4\u03c1\u03b1\u03b2\u03ae\u03be\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b1 \u03c0\u03bf\u03cd\u03bb\u03b9\u03b1 \u03b1\u03c0\u03cc \u03bc\u03b9\u03b1 \u03b1\u03b4\u03b9\u03b1\u03c6\u03b1\u03bd\u03ae\u03c2 \u03c3\u03b1\u03ba\u03bf\u03cd\u03bb\u03b1 \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03ba\u03b1\u03b8\u03bf\u03c1\u03b9\u03c3\u03bc\u03ad\u03bd\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac.\r\n\u039c\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03ce\u03c1\u03b1 \u03b1\u03c1\u03c7\u03af\u03c3\u03b1\u03bc\u03b5 \u03bd\u03b1 \u03c0\u03b1\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c3\u03ba\u03ac\u03ba\u03b9... \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf \u03c4\u03ad\u03bb\u03bf\u03c2 \u03b1\u03c6\u03bf\u03cd \u03b5\u03af\u03c7\u03b1\u03bc\u03b5 \u03bc\u03b5\u03af\u03bd\u03b5\u03b9 \u03bc\u03b5 1\u039a \u03ba\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c0\u03b9\u03cc\u03bd\u03b9 \u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1\u03c2, \u03b2\u03b3\u03ac\u03bb\u03b1\u03bc\u03b5 2 Q (\u03c4\u03b1\u03c5\u03c4\u03cc\u03c7\u03c1\u03bf\u03bd\u03b1!) \u03ba\u03b1\u03b9 \u03bd\u03af\u03ba\u03b7\u03c3\u03b5 \u03bf \u03a1\u03ac\u03bb\u03bb\u03b7\u03c2. \u0395\u03be\u03b5\u03c4\u03ac\u03c3\u03c4\u03b5 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03c0\u03bf\u03b9\u03b5\u03c2 \u03b4\u03b9\u03b1\u03c4\u03ac\u03be\u03b5\u03b9\u03c2 \u039a-Q \u03b5\u03c0\u03b9\u03c4\u03c1\u03ad\u03c0\u03bf\u03c5\u03bd \u03c4\u03b7 \u03bd\u03af\u03ba\u03b7 \u03c4\u03bf\u03c5 \u03b5\u03bd\u03cc\u03c2 \u03c0\u03b1\u03af\u03ba\u03c4\u03b7 \u03bf\u03c0\u03c9\u03c3\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03c0\u03bf\u03b9\u03b1 \u03b7 \u03c0\u03b9\u03b8\u03b1\u03bd\u03cc\u03c4\u03b7\u03c4\u03b1 \u03bd\u03b1 \u03c3\u03c5\u03bc\u03b2\u03b5\u03af \u03b1\u03c5\u03c4\u03cc. :P \r\n\r\n\r\n\r\n\r\n\r\n\r\n[color=#f8f8f8]PS: 2 ill S6[/color]:stretcher:", "Solution_1": "\u039c\u03b7\u03bd \u03b4\u03b5\u03b9\u03c4\u03b5 \u03c4\u03bf hint(\u03b5\u03b9\u03bd\u03b1\u03b9 \u03b7 \u03bb\u03c5\u03c3\u03b7...)\u03c0\u03c1\u03b9\u03bd \u03c4\u03bf \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b7\u03c3\u03b5\u03c4\u03b5....\r\n[hide]\u0395\u03c0.....\u03b3\u03b9\u03b1\u03c4\u03b9 \u03b2\u03bb\u03b5\u03c0\u03b5\u03b9\u03c2 \u03c4\u03bf hint?\u0395\u03bb\u03b1 \u03c4\u03c9\u03c1\u03b1..,\u03b1\u03c6\u03bf\u03c5 \u03b4\u03b5\u03bd \u03c4\u03b7\u03bd \u03b5\u03c7\u03b5\u03b9\u03c2 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b7\u03c3\u03b5\u03b9 \u03ba\u03b1\u03b8\u03bf\u03bb\u03bf\u03c5...\u039b\u03bf\u03b9\u03c0\u03bf\u03bd \u03b5\u03bb\u03b1 ,\u03ba\u03bb\u03b5\u03b9\u03c3\u03c4\u03bf \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c7\u03b9\u03c3\u03b5....[/hide]\n\n[hide]\u039a\u03b1\u03bb\u03b1 \u03c0\u03b1\u03bb\u03b9 \u03c4\u03b1 \u03b9\u03b4\u03b9\u03b1??\u03a0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b7\u03c3\u03b5 \u03bd\u03c4\u03b5....[/hide]\n\n[hide]\u0395,\u03b4\u03b5\u03bd \u03b1\u03bd\u03c4\u03b5\u03c7\u03b5\u03c3\u03b1\u03b9 \u03c1\u03b5 \u03c0\u03b1\u03b9\u03b4\u03b9 \u03bc\u03bf\u03c5....\u03a4\u03b5\u03bb\u03bf\u03c2 \u03c0\u03b1\u03bd\u03c4\u03c9\u03bd,\u03b5\u03c0\u03b9\u03b4\u03b7 \u03b4\u03b5\u03bd \u03b2\u03b1\u03b6\u03b5\u03b9\u03c2 \u03bc\u03c5\u03b1\u03bb\u03bf \u03b8\u03b1 \u03c3\u03bf\u03c5 \u03c0\u03c9 \u03c4\u03bf \u03b1\u03c0\u03bf\u03c4\u03b5\u03bb\u03b5\u03c3\u03bc\u03b1... (2(8!)^2)/31! \u039a\u03b1\u03b9 \u03c4\u03b9 \u03b5\u03b3\u03b9\u03bd\u03b5 \u03c4\u03c9\u03c1\u03b1 \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03b7\u03b8\u03b7\u03ba\u03b5\u03c2??[/hide]\r\n\r\n :rotfl: :rotfl: :rotfl:.......", "Solution_2": "\u03a0\u03bf\u03bb\u03cd \u03c9\u03c1\u03b1\u03af\u03bf \u03ba\u03b9 \u03c7\u03b9\u03bf\u03c5\u03bc\u03bf\u03c1\u03b9\u03c3\u03c4\u03b9\u03ba\u03cc \u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03b9\u03b4\u03ac\u03ba\u03b9 \u03c3\u03bf\u03c5 \u03b1\u03b3\u03b1\u03c0\u03b7\u03c4\u03ad \u03a1\u03ac\u03bb\u03bb\u03b7\r\n\u03bc\u03b5 \u03bc\u03b5\u03b3\u03ac\u03bb\u03b7 \u03b4\u03cc\u03c3\u03b7 \u03c7\u03b9\u03bf\u03cd\u03bc\u03bf\u03c1 \r\n\u03b1\u03c5\u03c4\u03bf \u03b4\u03b5\u03b9\u03c7\u03bd\u03b5\u03b9 \u03ba\u03b1\u03bb\u03cc \u03c7\u03b1\u03c1\u03b1\u03ba\u03c4\u03b7\u03c1\u03b1\r\n\u03b1\u03bb\u03bb\u03ac \u03ba\u03b9 \u03bc\u03b5\u03b3\u03ac\u03bb\u03bf \u03bc\u03c5\u03b1\u03bb\u03bf :D \r\n\u03a7\u03b1\u03b9\u03c1\u03b5\u03c4\u03ce" } { "Tag": [ "factorial" ], "Problem": "If a! divided by b! is a multiple of four but not a multiple of 8, then what is the maximum value of a-b?", "Solution_1": "I think it's 3. Any group of four consecutive integers, when multiplied together, will be divisible by 8.", "Solution_2": "This is because you have a multiple of $ 2$ and a multiple of $ 4$, and $ 2 \\times 4 \\equal{} 8$." } { "Tag": [ "ratio", "analytic geometry", "arithmetic sequence" ], "Problem": "So, that name doesn't really mean anything, but here's the question:\r\n\r\nConsider the infinite grid in the plane of points arranged in a square lattice with each point a distance 1 from its 4 nearest neighbors. Say we tile it with a finite number $n$ of other square lattices such that the set of vertices of the original grid is equal to the disjoint union of the vertices of the $n$ new grids. If the shortest distance between two points in the $i^{th}$ of the $n$ grids is $l_n$, prove that $\\frac{1}{l_1^2}+\\frac{1}{l_2^2}+\\ldots+\\frac{1}{l_n^2} = 1$. Also prove that this does not necessarily hold for a tiling of a square lattice with infinitely many other such lattices.", "Solution_1": "Maybe I'm just tired.. I'm not understanding the problem :blush: :blush: \r\n\r\nDo you mind clarifying it a bit? (Maybe a diagram would help)\r\n\r\nWhat does your bathroom floor look like anyways? :P", "Solution_2": "I can give you a brief description of one example, to which I will add a drawing tonight when I'm not running to class:\r\n\r\nIt's easy to dissect a grid into two grids: take the two new grids at 45 degree angles to the old grid with side lengths 1/2. Or, in other words, on the cartesian plane take the set {(a, b) s.t. a + b odd} and the set {(a, b) s.t. a + b even}. Each of these is a square lattice. Their disjoint (meaning no over-lap) union is all the lattice points in the plane. We can do this lots of other ways, as well.", "Solution_3": "A cute problem! The underlying idea is very easy, but it's expressed in an interesting way.\r\n\r\nHere's a generalization: Let $S$ denote the set of lattice points, which is partitioned into the sets $S_1, S_2, \\dots, S_n$. Each set $S_i$ is periodic, that is, there exist non-zero integers $e_i$, $f_i$, $g_i$, and $h_i$, such that if $(x,y) \\in S_i$, then $(x + e_i, y + f_i) \\in S_i$ and $(x + g_i, y + h_i) \\in S_i$.\r\n\r\nThen each set $S_i$ has a well-defined density $d_i$. What does that mean? Suppose we take a very large region $R$ in the plane; this will contain many lattice points, and many points in $S_i$. Then the ratio of points in $S_i$ in $R$ to lattice points in $R$ goes to a limit as the region $R$ increases unboundedly. This limit is the density $d_i$.\r\n\r\nNow, since the $S_i$ partition $S$, the sum of the densities must be 1. This is just the analogy of the result that if we have a number of mutually exclusive events that cover all possibilities, then the sum of their probabilities is 1.\r\n\r\nIn JBL's result, the density of the $i$th grid is $1/l_i^2$, and the result follows.\r\n\r\nThis idea also solves the following problem: Let $S$ denote the set of integer, which is partitioned into sets $S_1, S_2, \\dots, S_n$, such that each $S_i$ is an arithmetic progression with common difference $d_i$. Prove that\r\n\\[\\frac{1}{d_1} + \\frac{1}{d_2} + \\dots + \\frac{1}{d_n} = 1.\\]" } { "Tag": [ "vector", "calculus", "calculus computations" ], "Problem": "Hi this was shown in my book, but I do not understand how they deduced it. Can someone please explain this to me? Thanks.\r\n\r\n[b]My book said this:[/b]\r\nFrom the equations\r\n$ XdX\\minus{}YdY\\plus{}cdc\\equal{}0$\r\n$ YdX\\plus{}XdY\\minus{}bdc\\equal{}0$,\r\nHence,\r\n$ \\displaystyle\\frac{dc}{X^2\\plus{}Y^2}\\equal{}\\frac{dX}{bY\\minus{}cX}\\equal{}\\frac{Xdc\\minus{}cdX}{X(X^2\\plus{}c^2)}$. How did they come up with this? :huh:", "Solution_1": "Rewrite the condition as $ (X, \\minus{} Y,\\ c)\\cdot (dX,\\ dY,\\ dc) \\equal{} 0$ and $ (Y,\\ X, \\minus{} b)\\cdot (dX,\\ dY,\\ dc) \\equal{} 0$,\r\n\r\n$ (dX,\\ dY,\\ dc)$ is perpendicular to two vectors $ (X, \\minus{} Y,\\ c),\\ (Y,\\ X,\\ \\minus{}b).$\r\n\r\n$ \\therefore (dX,\\ dY,\\ dc)\\parallel{}(bY \\minus{} cX,\\ cY \\plus{} bX,\\ X^2 \\plus{} Y^2).$", "Solution_2": "Okay kunny, thank you very much for the explanation.\r\n\r\nSo [b](dX, dY, dc)[/b] is perpendicular to [b](X, -Y, c)[/b] and [b](Y, X, -b)[/b] which implies [b](dX, dY, dc)[/b] is a scalar multiple of [b](X, -Y, c)[/b] x [b](Y, X, -b)[/b].", "Solution_3": "Yes. :oops:" } { "Tag": [ "trigonometry", "inequalities proposed", "inequalities" ], "Problem": "Let $a;b;c>0$ satisfying $a+b+c=1$.Prove that:\r\n$\\frac{a}{a+bc}+\\frac{b}{b+ca}+\\frac{\\sqrt{abc}}{c+ab}\\leq 1+\\frac{3\\sqrt{3}}{4}$", "Solution_1": "[hide=\"A hint\"]Put $x=ab,y=bc,z=ca$ then $a=\\tan{\\frac{X}{2}}$, e.t.c. Where $X,Y,Z$ are angles of a triangle.[/hide]" } { "Tag": [], "Problem": "The reaction; 4Ag(s) + O2 (g) yields 2Ag2O (s) is exothermic. Which statement about the reaction is correct?\r\n\r\nA. It is spontaneous at all temperatures.\r\nB. It is spontaneous only at low temperatures.\r\nC. It is spontaneous only at high temperatures\r\nD. It is non-spontaneous at all temperatures.\r\n\r\nCan someone answer this question and explain why the correct answer is the correct answer?\r\nthanks![/code]", "Solution_1": "The criteria of spontaneity at constant T and P is $ \\Delta G \\equal{} \\Delta H \\minus{} T \\Delta S \\leq 0$. For this reaction, since it is exothermic, we already have $ \\Delta H < 0$; however, a solid and [i]a gas[/i] are converted into [i]a solid[/i]: so $ \\Delta S$ must be negative. Since the (thermodynamic) temperature T is always positive, the free energy variation will be negative only if $ |T \\Delta S| < |\\Delta H|$. Therefore, the correct answer seems to be B.", "Solution_2": "thanks carcul." } { "Tag": [ "integration", "trigonometry", "limit", "calculus", "calculus computations" ], "Problem": "I was solving some sample problems for final exam, and i saw these three , they are really cute :D :\r\n\r\n\r\n$1.$ Evaluate:\r\n\r\n$I=\\int xe^{2x}\\cos3xdx$\r\n\r\n$2.$ Prove that:\r\n\r\n$\\sqrt{3}\\int^{\\frac{1}{\\sqrt{3}}}_{0}x\\arctan xdx=\\sum^{\\infty}_{n=1}\\frac{(-1)^{n-1}}{(4n^{2}-1)3^{n}}$\r\n\r\nand then evaluate the sum of the series $\\sum^{\\infty}_{n=1}\\frac{(-1)^{n-1}}{(4n^{2}-1)3^{n}}$\r\n\r\n$3.$ For $n>1$ prove that:\r\n\r\n$I=\\int^{\\infty}_{0}\\frac{dx}{\\left(x+\\sqrt{1+x^{2}}\\right)^{n}}=\\frac{n}{n^{2}-1}$", "Solution_1": "For the first one\r\n\r\n$\\int x e^{ax}\\cos bx \\cdot dx = \\int x e^{ax}\\frac{e^{ibx}+e^{-ibx}}{2}\\cdot dx$\r\n\r\nand integrate by parts", "Solution_2": "For the second one, we have:\r\n\r\n$\\arctan x = x-\\frac{x^{3}}{3}+\\frac{x^{5}}{5}-\\frac{x^{7}}{7}+... = \\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}x^{2n-1}}{2n-1}, for |x|<1$\r\n\r\nand so \r\n\r\n$x\\arctan x = x^{2}-\\frac{x^{4}}{3}+\\frac{x^{6}}{5}-\\frac{x^{8}}{7}+...$\r\n\r\n$\\int x\\arctan x = \\frac{x^{3}}{3}-\\frac{x^{5}}{3\\cdot5}+\\frac{x^{7}}{5\\cdot7}-... = \\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}x^{2n+1}}{(2n-1)(2n+1)}= \\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}x^{2n+1}}{4n^{2}-1}$\r\n\r\nMaking $x = \\frac{1}{\\sqrt{3}}$, we get the sum\r\n\r\n$\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{3^{n}\\sqrt{3}(4n^{2}-1)}$\r\n\r\nwhich multiplicating by $\\sqrt{3}$ becomes\r\n\r\n$\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{3^{n}(4n^{2}-1)}$\r\n\r\nOn the other hand,\r\n\r\n$\\int x\\arctan x = \\frac{\\arctan x}{2}(1+x^{2})-\\frac{x}{2}$\r\n\r\nand so \r\n\r\n$\\int^\\frac{1}{\\sqrt{3}}_{0}x\\arctan x = \\frac{\\pi}{9}-\\frac{1}{2\\sqrt{3}}$\r\n\r\nand multiplicating by $\\sqrt{3}$ we get $\\frac{\\pi\\sqrt{3}}{9}-\\frac{1}{2}$\r\n\r\nThe final conclusion is that\r\n\r\n$\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{3^{n}(4n^{2}-1)}= \\frac{\\pi\\sqrt{3}}{9}-\\frac{1}{2}$.", "Solution_3": "[quote=\"sinajackson\"]$3.$ For $n>1$ prove that:\n\n$I=\\int^{\\infty}_{0}\\frac{dx}{\\left(x+\\sqrt{1+x^{2}}\\right)^{n}}=\\frac{n}{n^{2}-1}$[/quote]\r\n\r\n\\[I=\\int^{\\infty}_{0}\\frac{dx}{\\left(x+\\sqrt{1+x^{2}}\\right)^{n}}=\\int^{\\infty}_{0}\\frac{dx}{\\left(x+\\sqrt{1+x^{2}}\\right)^{n}}\\cdot\\frac{\\left(x-\\sqrt{1+x^{2}}\\right)^{n}}{\\left(x-\\sqrt{1+x^{2}}\\right)^{n}}\\]\r\n\\[=\\int^{\\infty}_{0}\\frac{\\left(x-\\sqrt{1+x^{2}}\\right)^{n}}{\\left[x^{2}-(1+x^{2})\\right]^{n}}dx=\\int^{\\infty}_{0}\\frac{\\left(x-\\sqrt{1+x^{2}}\\right)^{n}}{(-1)^{n}}dx=\\int^{\\infty}_{0}\\left(\\sqrt{1+x^{2}}-x\\right)^{n}dx\\]\r\n\r\nlet $u=\\sqrt{1+x^{2}}-x$ solving this for $x$ gives $x=\\frac{1}{2}\\left(\\frac{1}{u}-u\\right)$ so that $dx=-\\frac{1}{2}\\left(\\frac{1}{u^{2}}+1\\right) du$, also we have that $u=1$ when $x=0$ and that $u=0$ when $x=\\infty$ since \r\n\r\n\\[\\lim_{x\\to\\infty}\\left(\\sqrt{1+x^{2}}-x\\right)=\\lim_{x\\to\\infty}\\left(\\sqrt{1+x^{2}}-x\\right)\\cdot\\frac{\\sqrt{1+x^{2}}+x}{\\sqrt{1+x^{2}}+x}\\]\r\n\\[=\\lim_{x\\to\\infty}\\frac{1}{\\sqrt{1+x^{2}}+x}=0\\]\r\n\r\nand the integral becomes\r\n\r\n\\[I=\\frac{1}{2}\\int^{1}_{0}u^{n}\\left(\\frac{1}{u^{2}}+1\\right)du =\\frac{1}{2}\\int_{0}^{1}\\left(u^{n-2}+u^{n}\\right)du\\]\r\n\\[=\\frac{1}{2}\\left[\\frac{u^{n-1}}{n-1}+\\frac{u^{n+1}}{n+1}\\right]_{u=0}^{1}=\\frac{1}{2}\\left(\\frac{1}{n-1}+\\frac{1}{n+1}\\right)=\\frac{n}{n^{2}-1}\\]" } { "Tag": [ "function", "trigonometry", "calculus", "vector", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f,g: \\mathbb{R} \\to \\mathbb{R}$ satisfy: for all $ x,y \\in \\mathbb{R}$\r\n1- $ f(x \\plus{} y) \\equal{} f(x)f(y) \\minus{} g(x)g(y)$.\r\n2- $ g(x \\plus{} y) \\equal{} g(x)f(y) \\plus{} g(y)f(x)$.\r\n3- $ g'(0) \\equal{} 0$.\r\nProve that $ f^2(x) \\plus{} g^2(x) \\equal{} 1$.", "Solution_1": "You mean, besides the trivial case $ f(x)\\equiv g(x)\\equiv 0$?", "Solution_2": "So.. it's pretty clear these are just the sine and cosine functions, but are we supposed to prove that those are the only ones satisfying this?", "Solution_3": "i dont think this problem goes here since it has to do with derivatives. i think it means that g(x) is tangent to the x-axis at the origin but it says that there shouldnt be calculus probs here. anyways...how do u solve this?", "Solution_4": "The problem is incorrect as stated. Something is wrong - probably the initial condition (3). Perhaps that was supposed to be $ g'(0)\\equal{}1?$\r\n\r\nIt's clear to me that it is incorrect as stated because everything in statements (1), (2), and (3) is linear homogeneous, meaning that the set of all functions satisfying it is a vector space, and we could multiply any solutions by a fixed scalar to have another solution. Given that, there's no way to draw the nonhomogeneous conclusion $ f^2(x)\\plus{}g^2(x)\\equal{}1.$\r\n\r\nFarenhajt's objection is really part of this observation." } { "Tag": [ "geometry" ], "Problem": "Does anyone know of some good english sites besides amazon to get books?", "Solution_1": "are you asking for math books? then there is [url=higherpreception.com]this[/url]", "Solution_2": "That url doesn't seem to exist...\r\n(at least for me) :)", "Solution_3": "Me neither...\r\n\r\n[url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php]This[/url] is much much better...", "Solution_4": "[url]http://www.highperception.com/(S(a0ewui55zzle1gfmn5msoq45))/Default.aspx[/url][/url]", "Solution_5": "Is that a european site? Because americans will have to pay uber-shipping for those...", "Solution_6": "he asked for a european site dood! !!! ! ! ! ! !", "Solution_7": "Apparently we interpreted \"english\" differently...i'm very very sorry...\r\n\r\nJeez...", "Solution_8": "[quote=\"Poincare\"]he asked for a european site dood! !!! ! ! ! ! ![/quote]\r\nNo amazon doesn't have some books like olympiad geometry and an introduction to diophantine equations do you know any north american sites that sell these?" } { "Tag": [ "modular arithmetic" ], "Problem": "In how many different ways can change be made for $ \\$10$ if only quarters or dimes are used?", "Solution_1": "We want $ 25q \\plus{} 10d \\equal{} 1000\\implies 5q \\plus{} 2d \\equal{} 200\\implies q\\equiv0\\pmod2$\r\n$ \\implies q \\equal{} 2m,m\\in\\mathbb Z\\ge0\\implies d \\equal{} 100 \\minus{} 5m$. Also, $ 2m,100 \\minus{} 5m\\ge0\\implies0\\le m\\le20\\implies\\boxed{21}$.", "Solution_2": "Or, we know we can make it with 100 dimes, and we can substitute 5 dimes for 2 quarters, 100/5+1=21.", "Solution_3": "Another way:\r\n\r\nLet's just figure out how to make each half dollar. Each half dollar can be 2 quarters or 5 dimes. So there are 2 ways to make each half dollar.\r\n\r\nWe are simply choosing 20 objects, the half dollars, from two possible choices (quarters or dimes).\r\n\r\nLet the 0 be a half-dollar. We have 00000000000000000000. We add in a |. All 0's to the left of this divider will be made from quarters, and all 0's to the right will be made of dimes. There are now 21 symbols, 20 0's and 1 |. There are therefore $ \\binom{21}{1}\\equal{}21$ ways to do this.", "Solution_4": "Third way, sort of similar to AIME's. \r\n\r\nwe see that 10 dollars is 40 quarters. every two quarters can be replaced by dimes, so we have 40/2=20. but, we're forgetting the scenario where all of them are dimes, so $ 20\\plus{}1\\equal{}\\boxed{21}$" } { "Tag": [ "geometry", "incenter", "geometry unsolved" ], "Problem": "Given $\\triangle ABC$ with incenter $I$,$d$ is line through $I$ and cuts two segment $AB,AC$, with any $M,N \\in d$ prove that\r\n$b.S_{BMN}+c.S_{CMN}=a.S_{AMN}$", "Solution_1": "Let MN meet AB and AC at P and Q. Then $\\frac{\\triangle{BMN}}{\\triangle{AMN}}=\\frac{BF}{AF}$ and $\\frac{\\triangle{CMN}}{\\triangle{AMN}}=\\frac{CE}{AE}$, thus the equality becomes theorem 1 of: http://www.mathlinks.ro/Forum/viewtopic.php?t=100964", "Solution_2": "yes you are right, if I change incenter $I$ by excenter $J_{a}$ of angle $A$and $d$ isline thruogh $J_{a}$ and cuts ray $\\vec{AB},\\vec{AC}$ this problem is still true!\r\nAnd if I change $I$ by $L$ Lemoine,$d$is line through $L$,with any $M,N\\in d$ then in three number $a^{2}S_{AMN},b^{2}S_{BMN},c^{2}S_{CMN}$ there is one is sum other! \r\nCan it be done by your menthod ? :) :)" } { "Tag": [ "linear algebra", "matrix", "algorithm", "ratio", "superior algebra", "superior algebra solved" ], "Problem": "If we start with a matrix with integer entries,of type (2,2),wich has the determinant 1, and make transformations by adding a line(colon) multiplied by an integer to other line(colon)in all possible ways, can we obtain all the matrixes\r\nwith integer entries wich have also the determinant 1?", "Solution_1": "Take A=(a b\r\n c d)\r\ndetA=1, so ad-bc=1. So all pairs (a,c), (a,b), (b,d), (c,d) are coprime.\r\nApply Euclid's algorithm to the lines 1 and 2. For instance suppose a=cq+r with 0<=r<|c|. So add line 2 multiplied by -q to line 1. Notice that the new matrix is of the form \r\n(r b'\r\n c d)\r\nNow repeat this (possible by Euclid's algorithm) until a matrix of the form\r\n(1 x\r\n 0 y)\r\nThe new matrix has y for it's deterninant, but the transformations we have made don't change the det, so y=1. Now add line 2 multiplied by -x to line 1 and get (1 0\r\n 0 1)\r\nSo we can bring the matrix A to I_2=(1 0\r\n 0 1)\r\nNow I must prove the converse: we can turn I_2 into any matrix A with determinant 1. Get back to it later.", "Solution_2": "Sorry for the delay. I almost thought I missed this problem.\r\nSo back to my proof. I must prove that the matrix I_2 can be turned into any matrix with integer entries\r\na b\r\nc d\r\nof determinant 1 using only the following operation:\r\nadd a line (column) multiplied by an integer to the other line (column).\r\nIt is obvious that a,b are coprime or else det(a,b;c,d) is divisible by their gcd.\r\n\r\nI now work only with lines and show that I can obtain a matrix of the form\r\na x\r\nc y. For now I don't care about the elements x,y. Practicaly I want to reach from (1,0) to (a,c) only by one of the following operation repeated a certain number of times:\r\n(q,w) -> (q+kw) or (q,w)->(kq+w) for some integer k.\r\n\r\nTo do this I walk back the path of Euclid's algorithm.\r\n(a,c)<-(r1,c)<-(r1,r2)<-(r3,r2)<-...<-(1,0) where\r\na=cq1+r1\r\nc=r1q2+r2\r\n...\r\nI think this part of the solution is complete.\r\n\r\nI turned I_2 into\r\na x\r\nc y\r\nBecause the transformations used don't modify the det ay-cx=1, but ad-bc=1 and by Chinese Remainder Theorem or smth simpler we get the existance of an integer m such that y=d+mc and x=b+ma.\r\nAdd column 1 multiplied by -m to column 2 and get\r\na b\r\nc d\r\n\r\nHope you don't ask me to write my soln again.\r\n\r\nP.S. There was one (among others :D ) grammar mistake in my previous proof:\r\n\"Now repeat this (possible by Euclid's algorithm) until a matrix of the form\r\n(1 x\r\n0 y)\"\r\nAnd it stops here. I should have ended it in \"is obtained.\r\n\r\nP.P.S. There is a big grammar error in the problem text: xxxxtt writes colon instead of column. \r\nFrom Collins English Dictionary:\r\ncolon1(`k?l?n) (pl, -lons) the punctuation mark : , usually preceding an explanation or an example of what has gone before, a list, or an extended quotation(pl, -lons) this mark used for certain other purposes, such as expressions of time, as in 2:45 p.m., or when a ratio is given in figures, as in 5:3(pl. -la (-l?)) (in classical prosody) a part of a rhythmical period with two to six feet and one principal accent or ictus[C16: from Latin, from Greek kolon limb, hence part of a strophe, clause of a sentence] colon2(`k?l?n) , pl -lons or -la (-l?)the part of the large intestine between the caecum and the rectum[C16: from Latin: large intestine, from Greek kolon] colon3(k?`ln; French kl) a colonial farmer or plantation owner, esp. in a French colony[French: colonist, from Latin colonus, from colere to till, inhabit] coln (k?`l?n; Spanish ko`lon) , pl -lons or -lones (Spanish -`lones)the standard monetary unit of Costa Rica, divided into 100 cntimosthe standard monetary unit of El Salvador, divided into 100 centavos[C19: American Spanish, from Spanish, after Cristbal Coln Christopher Columbus] Coln (k`ln; Spanish ko`ln) a port in Panama, at the Caribbean entrance to the Panama Canal. Chief Caribbean port. Pop.: 68 688 (1987). Former name: AspinwallArchipilago de (arti`pjelao e). the official name of the Galpagos Islands.", "Solution_3": "I found out it was easier to prove the second part.\r\nThe transformations allowed act as the multiplication of A to the left or right by a matrix of the form:\r\n1 0, 1 k, where k is an integer.\r\nk 1, 0 1,\r\nNotice that these matrices are invertible in Z and their inverses are\r\n1 -k and 1 0. They are still allowed transformations.\r\n0 1 -k 1\r\nIf in the first I proved that ant A can be transformed into I_2, I proved that there exist E_1, E_2...E_n and F_1...F_m matrices of order 2 which are allowed tranformations such that E_1*...E_n*A*F_1*...*F_m=I_2.\r\nAll we have to do now to go from I_2 to A is to simplify the matrices E_i and F_j by their inverses and get A=E_n'...E_1'I_2F_m'...F_1'." } { "Tag": [ "geometry", "analytic geometry", "rectangle", "combinatorics proposed", "combinatorics" ], "Problem": "Given 3n points in the plane such that distance between any two point is not greater than 1 and no three are collinear. Prove that we can select n triangles satisfy:\r\n1. Each of 3n points is a vertex of one triangle of n triangles.\r\n2. Any two of n triangles have no common point in their area.\r\n3. Sum of area of n triangles is less than 1/2\r\n--------------------------------------", "Solution_1": "Well, 3) is incorrect because it does not holds for $n=1$ and the three vertices of an equilateral triangle with side-length 1.\r\n\r\nIn another hand, since there is a finite number of points, we may find an orthogonal system of coordinates such that the $n$ points have pairwise distinct x-coordinate.\r\nLet $A_1,...,A_{3n}$ be the points, where the ordering follows the ordering of their x-coordinate. Now choose the triangles $A_{3i+1}A_{3i+2}A_{3i}$ for $i=0,...,n-1$ and you will have 1) and 2).\r\n\r\nPierre.", "Solution_2": "[quote=\"pbornsztein\"]Well, 3) is incorrect because it does not holds for $n=1$ and the three vertices of an equilateral triangle with side-length 1.\n\nIn another hand, since there is a finite number of points, we may find an orthogonal system of coordinates such that the $n$ points have pairwise distinct x-coordinate.\nLet $A_1,...,A_{3n}$ be the points, where the ordering follows the ordering of their x-coordinate. Now choose the triangles $A_{3i+1}A_{3i+2}A_{3i}$ for $i=0,...,n-1$ and you will have 1) and 2).\n\nPierre.[/quote]\r\nyou re wrong. Because n=1, three points are $A_1 ,A_2 ,A_3$, so the area of triangle $A_1 A_2 A_3$ is ${\\sqrt3 / 4} < 1/2$", "Solution_3": "We can find two perpendicular directions s.t. none of the lines determined by two points among the $3n$ has one of the two directions. Choose a rectangle having sides in the two directions s.t. it's minimal. Then both its sides are $\\le 1$. We choose triangles as Pierre said, according to one of the directions. The triangles are covered by a set of disjoint rectangles (a rectangle for each triangle), and these rectangles are covered by the large rectangle, which has area $\\le 1$. On the other hand, a triangle covered by a rectangle has area $\\le \\frac 12\\times$(the area of the rectangle), so we can now obtain our conclusion." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "$ \\frac{1}{2}\\geq a,b,c \\geq 0$\r\n$ a\\plus{}b\\plus{}c\\equal{}1$\r\nprove that \r\n$ \\frac{9}{32}\\geq a^3\\plus{}b^3\\plus{}c^3\\plus{}4abc$\r\n :lol:", "Solution_1": "Since $ a\\plus{}b\\plus{}c\\equal{}1$ it follows that there exist the numbers $ x,y,z \\ge 0$, $ x\\plus{}y\\plus{}z \\not\\equal{} 0$, such that:\r\n$ a\\equal{}\\frac {x} {x\\plus{}y\\plus{}z}, b\\equal{}\\frac {y} {x\\plus{}y\\plus{}z}, c\\equal{}\\frac {z} {x\\plus{}y\\plus{}z}$\r\nFrom $ \\frac {1} {2} \\ge a, b, c \\ge 0$, we deduce that $ x, y, z$ are the sides of a triangle, maybe degenerate. \r\nSo, $ x\\equal{}n\\plus{}p$, $ y\\equal{}m\\plus{}p$, $ z\\equal{}m\\plus{}n$ with $ m, n, p \\ge 0$ and $ m\\plus{}n\\plus{}p \\not\\equal{} 0$.\r\nThe inequality is equivalent with the following inequality:\r\n$ \\frac {9} {32}\\ge\\frac {(n\\plus{}p)^3} {8(m\\plus{}n\\plus{}p)^3}\\plus{}\\frac {(m\\plus{}p)^3} {8(m\\plus{}n\\plus{}p)^3}\\plus{}\\frac {(m\\plus{}n)^3} {8(m\\plus{}n\\plus{}p)^3}\\plus{}\\frac {4(n\\plus{}p)(m\\plus{}p)(m\\plus{}n)} {8(m\\plus{}n\\plus{}p)^3}$\r\nAfter reductions, the inequality is equivalent with:\r\n$ m^3\\plus{}n^3\\plus{}p^3\\plus{}22mnp \\ge m^2n\\plus{}m^2p\\plus{}n^2p\\plus{}n^2m\\plus{}p^2m\\plus{}p^2n$\r\n\r\nIf $ m,n,p>0$ then the last one results from Schur:\r\n$ m(m\\minus{}n)(m\\minus{}p)\\plus{}n(n\\minus{}m)(n\\minus{}p)\\plus{}p(p\\minus{}m)(p\\minus{}n) \\ge 0$ \r\n\r\nIf one of $ m,n,p$ is zero then the inequality is trivial: \r\n$ m^3\\plus{}n^3 \\ge m^2n\\plus{}mn^2 \\Leftrightarrow (m\\plus{}n)(m\\minus{}n)^2 \\ge 0$\r\n\r\nThe equality case is for $ p\\equal{}0$ and $ m\\equal{}n$ and the permutations.\r\nIn other words, when: $ (a,b,c)\\equal{}\\{ (\\frac {1} {2}, \\frac {1} {4}, \\frac {1} {4}),(\\frac {1} {4}, \\frac {1} {2}, \\frac {1} {4}),(\\frac {1} {4}, \\frac {1} {4}, \\frac {1} {2}) \\}$." } { "Tag": [ "LaTeX", "articles" ], "Problem": "\\footnote command are really helpful, but is troublesome if you have a great number in the same page.\r\n\r\nI don\u2019t know what kind of interactivity we can use in adobe, but I was thinking in something like \u201cput your mouse on the Word and it will show a small box with the \u2018footnote\u2019\u201d. But I think this kind of thing are not in the libraries of latex.\r\n\r\nJust think in this kind of situation. You have 30 links in the same page going to another page with a explanation. You cannot make a footnote with this much of information. But for the reader will be troublesome (to not say boring) to go to the page, go back and make this process 30 times for each page!\r\n\r\nI can make some solutions for this kind of problem. Like make a small explanation below each paragraph, make a lateral box.\r\n\r\nI was just hoping that there were another\u2019s solutions. You don\u2019t need to make a complete code here, just give your idea, if it is sophisticated, you can put your code. \r\n\r\nBut the main point is that I don\u2019t want to change PDF to another format.", "Solution_1": "The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=cooltooltips]cooltooltips package[/url] will do exactly this [i]put your mouse on the Word and it will show a small box with the \u2018footnote\u2019\u201d.[/i]. The [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/cooltooltips/cooltooltips.pdf]documentation[/url] shows what's possible. Cooltooltips is supported by MiKTeX.", "Solution_2": "Really Cool. But...It do not make a PDF in LED but in anothers programs it do (TexMaker etc. (but with some errors))\r\n\r\nTrying the code:\r\n\r\n[code]\\cooltooltip[0 0 1]{Example}{This is an example of a cool tooltip. Pretty cool, eh?}{http://www.ctan.org/}{Visit CTAN on the Web}{This text\\strut}[/code]\n\nResult from LED\n\n[code]-----------------------\n)\n! Missing number, treated as zero.\n \n \\pdfannot \nl.4 ...g//}{Visit CTAN on the Web}{This text\\strut}\n \n?[/code]\r\n\r\neh... I think I will not use this package.\r\n\r\nThe old footnote were more simple, useful and old-fashioned.", "Solution_3": "It cannot depend on the editor unless you've misconfigured how it's compiled.\r\nThis complete minimal example works well:\r\n[code]\\documentclass{article}\n\\usepackage{cooltooltips}\n\\begin{document}\n\\cooltooltip[0 0 1]{Example}{This is an example of a cool tooltip. Pretty cool, eh?}{http://www.ctan.org/}{Visit CTAN on the Web}{This text\\strut}\n\\end{document}[/code]and should work fine for you using pdflatex on any system." } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "geometry unsolved" ], "Problem": "Let $ ABC$ be a triangle with $ A$ a right angle and $ \\angle B<\\angle C$. The tangent at $ A$ to the circumcircle $ k$ of $ \\triangle ABC$ meets the line $ BC$ at $ D$. Let $ E$ be the reflection of $ A$ in $ BC$, $ X$ be the foot of the perpendicular from $ A$ to $ BE$, and $ Y$ the midpoint of segment $ AX$. Line $ BY$ intersect $ k$ again at $ Z$. Prove that $ BD$ is tangent to the circumcircle of $ \\triangle ADZ$.", "Solution_1": "let $ AE,BC$ intersect at $ H$ and let $ AZ,BC$ intersect each other at point $ K$,so it sufficies to show that $ KD^2\\equal{}KZ.KA$.\r\n\r\nnote that $ AZBE$ is cyclic so $ \\angle ZBX\\equal{}\\angle ZAH$.\r\n\r\nalso in $ \\triangle AEX$ points $ Y,H$ are the midpoints of sides $ AX,AE$ respectively so $ HY\\parallel EX$ hence:\r\n\r\n$ \\angle HYB\\equal{}\\angle YBX\\equal{}\\angle ZAH$\r\n\r\nthus $ YZAH$ is cyclic which means that $ \\angle HYA\\equal{}\\angle HZA\\equal{}90^\\circ$.\r\n\r\nbut on the other hand we know that $ \\angle AHB\\equal{}90^\\circ$ thus $ \\angle ZHB\\equal{}\\angle ZAH$ therefore we have:\r\n\r\n$ \\triangle KZH\\sim\\triangle KAH\\Rightarrow KH^2\\equal{}KZ.KA\\equal{}KB.KC$.\r\n\r\nso in order to complete the proof,it's sufficient to show that $ KH\\equal{}KD$,now let $ D'$ be the reflection of point $ H$ wrt point $ K$,now we have to show that $ D'\\equiv D$.\r\n\r\nbut note that according to [b]newton formula[/b] we have $ \\boxed{(D'H,BC)\\equal{}\\minus{}1} (I)$\r\n\r\non the other hand we know that $ \\boxed{(DH,BC)\\equal{}\\minus{}1} (II)$\r\n\r\nnow according to (I),(II) we get that $ D\\equiv D'$.\r\n\r\nQED", "Solution_2": "seems that your problem is from shortlist 1998:\r\n[url]http://www.mathlinks.ro/viewtopic.php?p=123012#123012[/url]\r\n\r\nanyway here's another solution:\r\n\r\nlet $ G$ be a point on circle $ k$ such that $ AG$ is a diameter and let $ BD,AE$ intersect at $ H$.\r\n\r\nfirst of all we prove that $ Z,H,G$ are collinear,note that we have:\r\n\r\n$ \\angle AEG \\equal{} 90^\\circ \\equal{} \\angle AXB$ and also $ \\angle AGE \\equal{} \\angle ABX$ thus $ \\triangle AGE\\sim\\triangle ABX$.\r\n\r\nbut on the other hand points $ H,Y$ are the midpoints of $ AE,AX$ respectively,hence $ \\triangle AHG\\sim\\triangle AGE$ therefor:\r\n\r\n$ \\angle ABY \\equal{} \\angle AGH$ thus $ Z,H,G$ are collinear.\r\n\r\nnow in order to prove the tangency of $ BD$ and the circumcircle of $ \\triangle AZD$,it's sufficient to show that $ \\angle ZAD \\equal{} \\angle ZDC$. (*)\r\n\r\nnow note that $ AD$ is tangent to circle $ k$ hence $ \\angle AGZ \\equal{} \\angle ZAD$,thus for proving (*) we have to show that $ \\angle AGZ \\equal{} \\angle ZDC$ or equivalently we have to prove that $ GOZD$ is cyclic:\r\n\r\n$ \\boxed{OH.HD \\equal{} AH^2 \\equal{} AH.HE} (I)$\r\n\r\n$ \\boxed{GH.HZ \\equal{} AH.HE} (II)$\r\n\r\n$ (I),(II)\\Rightarrow OH.HD \\equal{} GH.HZ$ thus $ GOZD$ is cyclic.\r\n\r\nQED", "Solution_3": "Thank BaBaK Ghalebi very much.", "Solution_4": "See also\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=198944\nSincerely\nJean-Louis" } { "Tag": [ "inequalities", "LaTeX", "inequalities proposed" ], "Problem": "If $ a,b,c$ are nonnegative real numbers, then\r\n\r\n$ \\sum \\frac {3a^2\\minus{}bc}{2a^2\\plus{}b^2\\plus{}c^2} \\le \\frac 3{2}$.", "Solution_1": "[quote=\"Vasc\"]If $ a,b,c$ are nonnegative real numbers, then\n\n$ \\sum \\frac {3a^2 \\minus{} bc}{2a^2 \\plus{} b^2 \\plus{} c^2} \\le \\frac 3{2}$.[/quote]\r\nAfter expansion we obtain:\r\n$ \\sum \\frac {3a^2 \\minus{} bc}{2a^2 \\plus{} b^2 \\plus{} c^2} \\le \\frac 3{2}\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{sym}(4a^5b \\minus{} 9a^4b^2 \\plus{} 5a^3b^3 \\plus{} a^4bc \\plus{} 6a^3b^2c \\minus{} 7a^2b^2c^2)\\geq0,$ which obviously true.\r\nI have not found nothing without expansion. :(", "Solution_2": "I think there is at least a solution without expansion. :wink:", "Solution_3": "I found only the SOS form .It is easy to see that the inequality is equivalent to \r\n\r\n$ \\sum \\frac{(a\\minus{}b)^2(a^2\\minus{}4ab\\plus{}b^2\\minus{}2ac\\minus{}2bc\\plus{}2c^2)}{(2a^2\\plus{}b^2\\plus{}c^2)(2b^2\\plus{}c^2\\plus{}a^2)}\\leq 0$ \r\n\r\nBut the $ 2a^2\\plus{}b^2\\plus{}c^2$ reminds me some problems of Vasc that solved with Cauchy Swartz .\r\n\r\nFor example $ \\sum\\frac{(b\\plus{}c)^2}{2a^2\\plus{}b^2\\plus{}c^2}\\leq 3$ :wink:", "Solution_4": "Yes, silouan. It can be rewritten as\r\n\\[ \\frac12\\sum_{\\rm cyclic}\\frac{(b\\minus{}c)^2}{2a^2\\plus{}b^2\\plus{}c^2}\\plus{}\\sum_{\\rm cyclic}\\frac{(b\\plus{}c)^2}{2a^2\\plus{}b^2\\plus{}c^2}\\geq3.\\]", "Solution_5": "[quote=\"Vasc\"]If $ a,b,c$ are nonnegative real numbers, then\n\n$ \\sum \\frac {3a^2 \\minus{} bc}{2a^2 \\plus{} b^2 \\plus{} c^2} \\le \\frac 3{2}$.[/quote]\r\n$ \\sum \\frac {3a^2 \\minus{} bc}{2a^2 \\plus{} b^2 \\plus{} c^2} \\le \\frac 3{2} \\\\ \\Leftrightarrow \\sum (\\frac {3a^2 \\minus{} bc}{2a^2 \\plus{} b^2 \\plus{} c^2} \\minus{} \\frac 1{2}) \\le 0 \\\\ \\Leftrightarrow \\sum \\frac {4a^2 \\minus{}2 bc\\minus{}b^2\\minus{}c^2}{2a^2 \\plus{} b^2 \\plus{} c^2} \\le 0 \\\\ \\Leftrightarrow 2\\sum \\frac {b^2\\plus{}c^2\\minus{}2a^2}{2a^2 \\plus{} b^2 \\plus{} c^2} \\ge \\sum \\frac {(b\\minus{}c)^2}{2a^2 \\plus{} b^2 \\plus{} c^2}$\r\nNotice that\r\n$ \\sum \\frac {b^2\\plus{}c^2\\minus{}2a^2}{2a^2 \\plus{} b^2 \\plus{} c^2} \\\\ \\equal{}\\sum (a^2\\minus{}b^2)(\\frac{1}{2b^2\\plus{}a^2\\plus{}c^2}\\minus{}\\frac{1}{2a^2\\plus{}b^2\\plus{}c^2}) \\\\ \\equal{} \\sum \\frac{(a^2\\minus{}b^2)^2}{(2a^2\\plus{}b^2\\plus{}c^2)(2b^2\\plus{}a^2\\plus{}c^2)}$ \r\nthe inequality is equivalent to\r\n\\[ \\sum (a\\minus{}b)^2 \\left(\\frac{2(a\\plus{}b)^2}{(2a^2\\plus{}b^2\\plus{}c^2)(2b^2\\plus{}a^2\\plus{}c^2)}\\minus{} \\frac{1}{2c^2\\plus{}a^2\\plus{}b^2}\\right) \\ge 0\r\n\\]\r\nThis inequality can be rewritten as\r\n\\[ S_a(b\\minus{}c)^2\\plus{}S_b(c\\minus{}a)^2\\plus{}S_c(a\\minus{}b)^2 \\ge 0\r\n\\]\r\nwhere\r\n$ S_a\\equal{}2(b\\plus{}c)^2(2a^2\\plus{}b^2\\plus{}c^2)\\minus{}(2b^2\\plus{}a^2\\plus{}c^2)(2c^2\\plus{}a^2\\plus{}b^2) \\\\ \\equal{} a^2(b^2\\plus{}c^2\\plus{}8bc)\\plus{}4bc(b^2\\plus{}c^2)\\minus{}b^2c^2\\minus{}a^4 \\\\ \\ge a^2b^2\\minus{}a^4$\r\n$ S_b \\ge a^2b^2\\minus{}b^4$\r\n$ S_c \\equal{}c^2(a^2\\plus{}b^2\\plus{}8ab)\\plus{}4ab(a^2\\plus{}b^2)\\minus{}a^2b^2\\minus{}c^4 \\ge 0$\r\n(we assume that $ a \\ge b \\ge c$)\r\nIt suffits to prove \r\n$ a^2(b^2\\minus{}a^2)(b\\minus{}c)^2\\plus{}b^2(a^2\\minus{}b^2)(a\\minus{}c)^2 \\ge 0$\r\n, which is clearly true because $ a \\ge b \\ge 0$ and $ ab\\minus{}bc \\ge ab\\minus{}ac \\ge 0$.\r\nThe inequality is proved.", "Solution_6": "Nice solution, ehku. :lol: \r\nAnother solution is as follows: \r\nSince\r\n$ S_a \\equal{} 4bc(2a^2 \\plus{} b^2 \\plus{} c^2) \\minus{} (a^2 \\minus{} b^2)(a^2 \\minus{} c^2)$,\r\nwe may write the inequality as\r\n$ 4\\sum bc(2a^2 \\plus{} b^2 \\plus{} c^2)(b \\minus{} c)^2 \\ge (a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2$,\r\nwhich is easy.\r\n\r\nMuch harder than this is the following:\r\n\r\n$ \\sum \\frac {5a^2\\plus{}4bc}{a^2\\plus{}2b^2\\plus{}2c^2} \\ge \\frac {27}{5}$,\r\n\r\nbecause equality holds for (1,1,1) and (4,1,1).", "Solution_7": "[quote=\"Vasc\"]Nice solution, ehku. :lol: \n\n\n$ \\sum \\frac {5a^2 \\plus{} 4bc}{a^2 \\plus{} 2b^2 \\plus{} 2c^2} \\ge \\frac {27}{5}$,\n\nbecause equality holds for (1,1,1) and (4,1,1).[/quote]\r\nTry for $ (a,b,c)\\equal{}(1,0,0)$.", "Solution_8": "[quote=\"ehku\"][quote=\"Vasc\"]Nice solution, ehku. :lol: \n\n\n$ \\sum \\frac {5a^2 \\plus{} 4bc}{a^2 \\plus{} 2b^2 \\plus{} 2c^2} \\ge \\frac {27}{5}$,\n\nbecause equality holds for (1,1,1) and (4,1,1).[/quote]\nTry for $ (a,b,c) \\equal{} (1,0,0)$.[/quote]\r\nThanks, ehku. The reverse inequality holds; that is\r\n$ \\sum \\frac {5a^2 \\plus{} 4bc}{a^2 \\plus{} 2b^2 \\plus{} 2c^2} \\le \\frac {27}{5}$.\r\n\r\nMore general,\r\n\r\n$ \\sum \\frac {p(3 \\minus{} p)a^2 \\plus{} 2(1 \\minus{} p)bc}{pa^2 \\plus{} b^2 \\plus{} c^2} \\le \\frac {3(1 \\plus{} p)(2 \\minus{} p)}{2 \\plus{} p}$,\r\n\r\nfor $ 0 < p < 1$ and any nonnegative $ a,b,c$. Equality holds for $ (1,1,1)$ and again for $ (\\frac 2{p},1,1)$ or any cyclic permutation. For $ p > 1$, the reverse inequality holds. \r\nI have no solution to these inequalities. :oops: \r\nFor $ p \\equal{} 2$ and $ p \\equal{} 3$ we get two known inequalities.", "Solution_9": "[quote=\"Vasc\"][quote=\"ehku\"][quote=\"Vasc\"]Nice solution, ehku. :lol: \n\n\n$ \\sum \\frac {5a^2 \\plus{} 4bc}{a^2 \\plus{} 2b^2 \\plus{} 2c^2} \\ge \\frac {27}{5}$,\n\nbecause equality holds for (1,1,1) and (4,1,1).[/quote]\nTry for $ (a,b,c) \\equal{} (1,0,0)$.[/quote]\nThanks, ehku. The reverse inequality holds; that is\n$ \\sum \\frac {5a^2 \\plus{} 4bc}{a^2 \\plus{} 2b^2 \\plus{} 2c^2} \\le \\frac {27}{5}$.\n\nMore general,\n\n$ \\sum \\frac {p(3 \\minus{} p)a^2 \\plus{} 2(1 \\minus{} p)bc}{pa^2 \\plus{} b^2 \\plus{} c^2} \\le \\frac {3(1 \\plus{} p)(2 \\minus{} p)}{2 \\plus{} p}$,\n\nfor $ 0 < p < 1$ and any nonnegative $ a,b,c$. Equality holds for $ (1,1,1)$ and again for $ (\\frac 2{p},1,1)$ or any cyclic permutation. For $ p > 1$, the reverse inequality holds. \nI have no solution to these inequalities. :oops: \nFor $ p \\equal{} 2$ and $ p \\equal{} 3$ we get two known inequalities.[/quote]\r\nIt is very interesting your inequality.\r\nIt seems be difficult to create and prove inequalities of this type. I have found some nice results, for example the following:\r\n\r\nIf $ a^2\\plus{}b^2\\plus{}c^2 \\le 2(ab\\plus{}bc\\plus{}ca)$ then \r\n\\[ \\frac{a^2\\minus{}bc}{3a^2\\plus{}b^2\\plus{}c^2}\\plus{}\\frac{b^2\\minus{}ca}{3b^2\\plus{}c^2\\plus{}a^2}\\plus{}\\frac{c^2\\minus{}ab}{3c^2\\plus{}a^2\\plus{}b^2} \\le 0\r\n\\]\r\nFor $ (a,b,c)\\equal{}(1,1,0)$ or $ (1,1,1)$ or $ (1,1,4)$ it becomes equality. \r\nPlease try it, Vasc, there is a very interesting solution. \r\nI am trying to find a general inequality for all above inequalities.", "Solution_10": "[quote=\"ehku\"] If $ a^2 \\plus{} b^2 \\plus{} c^2 \\le 2(ab \\plus{} bc \\plus{} ca)$ then\n\\[ \\frac {a^2 \\minus{} bc}{3a^2 \\plus{} b^2 \\plus{} c^2} \\plus{} \\frac {b^2 \\minus{} ca}{3b^2 \\plus{} c^2 \\plus{} a^2} \\plus{} \\frac {c^2 \\minus{} ab}{3c^2 \\plus{} a^2 \\plus{} b^2} \\le 0\n\\]\nFor $ (a,b,c) \\equal{} (1,1,0)$ or $ (1,1,1)$ or $ (1,1,4)$ it becomes equality. [/quote]\r\nThis is problem 18 in Algebraic Inequalities (p. 345 and 369), by Nguyen Anh Tuan. \r\nEquality holds for $ a^2 \\plus{} b^2 \\plus{} c^2\\equal{}2(ab \\plus{} bc \\plus{} ca)$.\r\n\r\nAs a consequence, in any triangle\r\n\\[ \\sum \\frac {a^4 \\minus{} b^2c^2}{3a^4 \\plus{} b^4 \\plus{} c^4} \\le 0.\r\n\\]", "Solution_11": "[quote=\"Vasc\"][quote=\"ehku\"] If $ a^2 \\plus{} b^2 \\plus{} c^2 \\le 2(ab \\plus{} bc \\plus{} ca)$ then\n\\[ \\frac {a^2 \\minus{} bc}{3a^2 \\plus{} b^2 \\plus{} c^2} \\plus{} \\frac {b^2 \\minus{} ca}{3b^2 \\plus{} c^2 \\plus{} a^2} \\plus{} \\frac {c^2 \\minus{} ab}{3c^2 \\plus{} a^2 \\plus{} b^2} \\le 0\n\\]\nFor $ (a,b,c) \\equal{} (1,1,0)$ or $ (1,1,1)$ or $ (1,1,4)$ it becomes equality. [/quote]\nThis is problem 18 in Algebraic Inequalities (p. 345 and 369), by Nguyen Anh Tuan. \nEquality holds for $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 2(ab \\plus{} bc \\plus{} ca)$.\n\nAs a consequence, in any triangle\n\\[ \\sum \\frac {a^4 \\minus{} b^2c^2}{3a^4 \\plus{} b^4 \\plus{} c^4} \\le 0.\n\\]\n[/quote]\r\n :o Is that right ?\r\nI hope the following identiy is not in your book:\r\nFor all reals $ a,b,c,p$ we have \r\n\\[ \\frac{a^2\\minus{}bc}{(p\\plus{}1)a^2\\plus{}b^2\\plus{}c^2}\\plus{}\\frac{b^2\\minus{}ca}{(p\\plus{}1)b^2\\plus{}c^2\\plus{}a^2}\\plus{}\\frac{c^2\\minus{}ab}{(p\\plus{}1)c^2\\plus{}a^2\\plus{}b^2} \\\\ \\equal{}\\frac{(\\sum a^2\\minus{}p\\sum ab)[(\\sum a^2)(\\sum a^2\\minus{}\\sum ab)\\plus{}p(\\sum a^2b^2\\minus{}abc \\sum a)]}{((p\\plus{}1)a^2\\plus{}b^2\\plus{}c^2)((p\\plus{}1)b^2\\plus{}c^2\\plus{}a^2)((p\\plus{}1)c^2\\plus{}a^2\\plus{}b^2)}\r\n\\]", "Solution_12": "[quote=\"Vasc\"][quote=\"ehku\"][quote=\"Vasc\"]Nice solution, ehku. :lol: \n\n\n$ \\sum \\frac {5a^2 \\plus{} 4bc}{a^2 \\plus{} 2b^2 \\plus{} 2c^2} \\ge \\frac {27}{5}$,\n\nbecause equality holds for (1,1,1) and (4,1,1).[/quote]\nTry for $ (a,b,c) \\equal{} (1,0,0)$.[/quote]\nThanks, ehku. The reverse inequality holds; that is\n$ \\sum \\frac {5a^2 \\plus{} 4bc}{a^2 \\plus{} 2b^2 \\plus{} 2c^2} \\le \\frac {27}{5}$.\n\nMore general,\n\n$ \\sum \\frac {p(3 \\minus{} p)a^2 \\plus{} 2(1 \\minus{} p)bc}{pa^2 \\plus{} b^2 \\plus{} c^2} \\le \\frac {3(1 \\plus{} p)(2 \\minus{} p)}{2 \\plus{} p}$,\n\nfor $ 0 < p < 1$ and any nonnegative $ a,b,c$. Equality holds for $ (1,1,1)$ and again for $ (\\frac 2{p},1,1)$ or any cyclic permutation. For $ p > 1$, the reverse inequality holds. \nI have no solution to these inequalities. :oops: \nFor $ p \\equal{} 2$ and $ p \\equal{} 3$ we get two known inequalities.[/quote]\r\nThis inequality is true for all reals $ a,b,c$.\r\n\r\nI have proved it in the small case $ \\frac{1}{2} \\le p \\le 1$ (that also means the inequality $ \\sum \\frac {5a^2 \\plus{} 4bc}{a^2 \\plus{} 2b^2 \\plus{} 2c^2} \\le \\frac {27}{5}$ was proved, of course, if I didn't make any mistake).\r\nTo prove the original inequality in the case $ 0 \\le p \\le 1$, it reduces to prove\r\n\\[ (pb^2\\plus{}c^2\\plus{}a^2)(pa\\minus{}2c)^2\\plus{}(pc^2\\plus{}a^2\\plus{}b^2)(pa\\minus{}2b)^2 \\ge (1\\minus{}p)(p\\plus{}2)(b^2\\minus{}c^2)^2\r\n\\]\r\nfor all $ a \\ge b\\ge c \\ge 0$.\r\nI think it is true for all $ 1 \\ge p \\ge 0$ but I can only prove it for $ \\frac{1}{2} \\le p \\le 1$.\r\n\r\nThe case $ p \\ge 1$ is much more difficult.", "Solution_13": "[quote=\"Vasc\"] \nAnother solution is as follows: \nSince\n$ S_a \\equal{} 4bc(2a^2 \\plus{} b^2 \\plus{} c^2) \\minus{} (a^2 \\minus{} b^2)(a^2 \\minus{} c^2)$,\nwe may write the inequality as\n$ 4\\sum bc(2a^2 \\plus{} b^2 \\plus{} c^2)(b \\minus{} c)^2 \\ge (a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2$,\nwhich is easy.\n[/quote]\r\n\r\nVasc, why do you say it is easy? How do you finish your proof?\r\n\r\nThank you very much.", "Solution_14": "[quote=\"manlio\"][quote=\"Vasc\"] \nAnother solution is as follows: \nSince\n$ S_a \\equal{} 4bc(2a^2 \\plus{} b^2 \\plus{} c^2) \\minus{} (a^2 \\minus{} b^2)(a^2 \\minus{} c^2)$,\nwe may write the inequality as\n$ 4\\sum bc(2a^2 \\plus{} b^2 \\plus{} c^2)(b \\minus{} c)^2 \\ge (a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2$,\nwhich is easy.\n[/quote]\n\nVasc, why do you say it is easy? How do you finish your proof?\n\nThank you very much.[/quote]\r\nWell, to prove this one, we will asume $ a \\ge b \\ge c$, then we have\r\n\\[ 4\\sum bc(2a^2\\plus{}b^2\\plus{}c^2)(b\\minus{}c)^2 \\ge 4ab(a^2\\plus{}b^2\\plus{}2c^2) (a\\minus{}b)^2 \\ge 4ab(a^2\\plus{}b^2)(a\\minus{}b)^2 \\ge 8a^2b^2(a\\minus{}b)^2\r\n\\]\r\nMoreover, \r\n\\[ (a\\minus{}b)^2(b\\minus{}c)^2(c\\minus{}a)^2 \\le a^2b^2(a\\minus{}b)^2\r\n\\]\r\nhence the inequality is true. :)", "Solution_15": "[quote=\"can_hang2007\"][quote=\"manlio\"][quote=\"Vasc\"] \nAnother solution is as follows: \nSince\n$ S_a \\equal{} 4bc(2a^2 \\plus{} b^2 \\plus{} c^2) \\minus{} (a^2 \\minus{} b^2)(a^2 \\minus{} c^2)$,\nwe may write the inequality as\n$ 4\\sum bc(2a^2 \\plus{} b^2 \\plus{} c^2)(b \\minus{} c)^2 \\ge (a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2$,\nwhich is easy.\n[/quote]\n\nVasc, why do you say it is easy? How do you finish your proof?\n\nThank you very much.[/quote]\nWell, to prove this one, we will asume $ a \\ge b \\ge c$, then we have\n\\[ 4\\sum bc(2a^2 \\plus{} b^2 \\plus{} c^2)(b \\minus{} c)^2 \\ge 4ab(a^2 \\plus{} b^2 \\plus{} 2c^2) (a \\minus{} b)^2 \\ge 4ab(a^2 \\plus{} b^2)(a \\minus{} b)^2 \\ge 8a^2b^2(a \\minus{} b)^2\n\\]\nMoreover,\n\\[ (a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2 \\le a^2b^2(a \\minus{} b)^2\n\\]\nhence the inequality is true. :)[/quote]\r\n\r\nVery nice and very simple :) \r\n\r\nThank you very much.", "Solution_16": "[quote=\"ehku\"][/quote][quote=\"Vasc\"][quote=\"ehku\"][quote=\"Vasc\"]Nice solution, ehku. :lol: \n\n\n$ \\sum \\frac {5a^2 + 4bc}{a^2 + 2b^2 + 2c^2} \\ge \\frac {27}{5}$,\n\nbecause equality holds for (1,1,1) and (4,1,1).[/quote]\nTry for $ (a,b,c) = (1,0,0)$.[/quote]\nThanks, ehku. The reverse inequality holds; that is\n$ \\sum \\frac {5a^2 + 4bc}{a^2 + 2b^2 + 2c^2} \\le \\frac {27}{5}$.\n\nMore general,\n\n$ \\sum \\frac {p(3 - p)a^2 + 2(1 - p)bc}{pa^2 + b^2 + c^2} \\le \\frac {3(1 + p)(2 - p)}{2 + p}$,\n\nfor $ 0 < p < 1$ and any nonnegative $ a,b,c$. Equality holds for $ (1,1,1)$ and again for $ (\\frac 2{p},1,1)$ or any cyclic permutation. For $ p > 1$, the reverse inequality holds. \nI have no solution to these inequalities. :oops: \nFor $ p = 2$ and $ p = 3$ we get two known inequalities.[/quote][quote=\"ehku\"]\nThis inequality is true for all reals $ a,b,c$.\n\nI have proved it in the small case $ \\frac {1}{2} \\le p \\le 1$ (that also means the inequality $ \\sum \\frac {5a^2 + 4bc}{a^2 + 2b^2 + 2c^2} \\le \\frac {27}{5}$ was proved, of course, if I didn't make any mistake).\nTo prove the original inequality in the case $ 0 \\le p \\le 1$, it reduces to prove\n\\[ (pb^2 + c^2 + a^2)(pa - 2c)^2 + (pc^2 + a^2 + b^2)(pa - 2b)^2 \\ge (1 - p)(p + 2)(b^2 - c^2)^2\n\\]\nfor all $ a \\ge b\\ge c \\ge 0$.\nI think it is true for all $ 1 \\ge p \\ge 0$ but I can only prove it for $ \\frac {1}{2} \\le p \\le 1$.\n\nThe case $ p \\ge 1$ is much more difficult.[/quote]\r\nIt was proved for all $ p \\ge 3$.\r\nThe useful lemma posted here http://www.mathlinks.ro/viewtopic.php?p=1042081#1042081 can help us to prove it for $ 0 < p < \\frac {1}{2}$ or $ 1 < p < 3$, then the above inequality can be proved completely.\r\nI will try prove two these cases by the same way for $ 1 \\ge p \\ge \\frac {1}{2}$ or $ p \\ge 3$.", "Solution_17": "[quote=\"ehku\"] The useful lemma posted here http://www.mathlinks.ro/viewtopic.php?p=1042081#1042081 can help us to prove it [/quote]\r\nThis lemma is a particular case of EV-Theorem. For this case, the method is too laborious.\r\n\r\nSome ideas for Mathlinkers.\r\nThis is an inequality of six degree. There is an inequality of four degree with the same equality conditions:\r\n\r\n$ p^2\\sum a^4 \\plus{} 4(p \\plus{} 1)\\sum a^2b^2 \\plus{} (p^2 \\minus{} 4)\\sum a^2bc \\minus{} (p^2 \\plus{} 2p)\\sum ab(a^2 \\plus{} b^2) \\ge 0$,\r\n\r\nwhich is equivalent to $ \\frac 1{2}\\sum(pb \\plus{} pc \\minus{} (p \\plus{} 2)a)^2(b \\minus{} c)^2 \\ge 0$.\r\nCan we use this to prove our difficult inequality ?\r\n\r\nThe inequality becomes identity for $ p \\equal{} 1$. Dividing by $ p \\minus{} 1$ :o , we find an inequality for any $ p > 0$. It is interesting to prove this inequality.", "Solution_18": "The inequality was proved for all $ p \\ge \\frac{1}{2}$ and all reals $ a,b,c$.\r\n\r\n\r\n[quote]This lemma is a particular case of EV-Theorem. For this case, the method is too laborious.[/quote] \nMaybe it is not nice, but it is not laborious, because we needn't expand it.\n\n\n[hide=\"Dear Vasc\"][quote=\"ehku\"]\nI hope the following identiy is not in your book:\nFor all reals $ a,b,c,p$ we have\n\\[ \\frac {a^2 \\minus{} bc}{(p \\plus{} 1)a^2 \\plus{} b^2 \\plus{} c^2} \\plus{} \\frac {b^2 \\minus{} ca}{(p \\plus{} 1)b^2 \\plus{} c^2 \\plus{} a^2} \\plus{} \\frac {c^2 \\minus{} ab}{(p \\plus{} 1)c^2 \\plus{} a^2 \\plus{} b^2} \\\\\n\\equal{} \\frac {(\\sum a^2 \\minus{} p\\sum ab)[(\\sum a^2)(\\sum a^2 \\minus{} \\sum ab) \\plus{} p(\\sum a^2b^2 \\minus{} abc \\sum a)]}{((p \\plus{} 1)a^2 \\plus{} b^2 \\plus{} c^2)((p \\plus{} 1)b^2 \\plus{} c^2 \\plus{} a^2)((p \\plus{} 1)c^2 \\plus{} a^2 \\plus{} b^2)}\n\\]\n[/quote]\n :blush: \n\nThe inequality\n\\[ \\frac {a^2 \\minus{} bc}{3a^2 \\plus{} b^2 \\plus{} c^2} \\plus{} \\frac {b^2 \\minus{} ca}{3b^2 \\plus{} c^2 \\plus{} a^2} \\plus{} \\frac {c^2 \\minus{} ab}{3c^2 \\plus{} a^2 \\plus{} b^2} \\le 0\n\\]\nfor $ a^2 \\plus{} b^2 \\plus{} c^2 \\le 2(ab \\plus{} bc \\plus{} ca)$ can be proved without the above indentity:\nWe need only consider two case $ a \\ge b \\ge c \\ge 0$ or $ a\\ge b \\ge 0 > c$. \n If $ a\\ge b \\ge 0 \\ge c$ then $ a^2 \\plus{} b^2 \\plus{} c^2 \\le 2(ab \\plus{} bc \\plus{} ca) \\le 2ab$, contradiction. Then $ a \\ge b \\ge c \\ge 0$.\nBecause\n\\[ (a^2 \\minus{} bc)(b\\plus{}c) \\ge (b^2 \\minus{} ca)(c\\plus{}a) \\ge (c^2 \\minus{} ab)(a\\plus{}b) \\\\ (3a^2 \\plus{} b^2 \\plus{} c^2)(b\\plus{}c) \\ge (3b^2 \\plus{} c^2 \\plus{}a^2)(c\\plus{}a) \\ge (3c^2 \\plus{} a^2 \\plus{} b^2)(a\\plus{}b) >0\n\\]\n, by Chebysev inequality we get\n\\[ LHS \\le (\\sum (a^2 \\minus{} bc)(b\\plus{}c))\\left(\\sum \\frac{1}{(3a^2 \\plus{} b^2 \\plus{} c^2)(b\\plus{}c)}\\right)\\equal{}0\n\\]\nbecause $ \\sum (a^2 \\minus{} bc)(b\\plus{}c)\\equal{}0$.[/hide]", "Solution_19": "...and now, it was proved for all positive real $ p$ ! :!:", "Solution_20": "[b]Problem.[/b] Let $ a,b,c,p$ be real numbers. We denote \r\n\\[ F(a,b,c)\\equal{}\\sum \\frac {p(3 \\minus{} p)a^2 \\plus{} 2(1 \\minus{} p)bc}{pa^2 \\plus{} b^2 \\plus{} c^2} \\minus{} \\frac {3(1 \\plus{} p)(2 \\minus{} p)}{2 \\plus{} p}\r\n\\].\r\nProve that $ (p\\minus{}1).F(a,b,c) \\ge 0$ for all real numbers $ a,b,c$ and all positive real number $ p$ with equality if and only if $ p\\equal{}1$ or $ (a,b,c)\\equal{}(1,1,1)$ or $ (a,b,c)\\equal{}(1,1,\\frac{2}{p})$ and their permutations.\r\n[b]Solution.[/b]\r\nBecause \\[ (2\\plus{}p)[p(3\\minus{}p)a^2\\plus{}2(1\\minus{}p)bc]\\minus{}(1\\plus{}p)(2\\minus{}p)(pa^2\\plus{}b^2\\plus{}c^2) \\\\ \\equal{}2p(2a^2\\minus{}b^2\\minus{}c^2)\\plus{}(p\\minus{}1)(p\\plus{}2)(b\\minus{}c)^2\r\n\\]\r\n, we have\r\n\\[ F(a,b,c)\\equal{}\\sum \\left(\\frac {p(3 \\minus{} p)a^2 \\plus{} 2(1 \\minus{} p)bc}{pa^2 \\plus{} b^2 \\plus{} c^2}\\minus{}\\frac {(1 \\plus{} p)(2 \\minus{} p)}{2 \\plus{} p}\\right) \\\\ \\equal{} \\frac{2p}{2\\plus{}p} \\sum \\frac{2a^2\\minus{}b^2\\minus{}c^2}{pa^2\\plus{}b^2\\plus{}c^2}\\plus{}(p\\minus{}1)\\sum \\frac{(b\\minus{}c)^2}{pa^2\\plus{}b^2\\plus{}c^2}.\r\n\\]\r\nNotice that\r\n\\[ \\sum \\frac{2a^2\\minus{}b^2\\minus{}c^2}{pa^2\\plus{}b^2\\plus{}c^2} \\equal{} \\sum(a^2\\minus{}b^2)\\left(\\frac {1}{pa^2 \\plus{} b^2 \\plus{} c^2}\\minus{}\\frac {1}{pb^2\\plus{}c^2\\plus{}a^2}\\right) \\\\ \\equal{}(1\\minus{}p) \\sum \\frac{(a^2\\minus{}b^2)^2}{(pa^2 \\plus{} b^2 \\plus{} c^2)(pb^2\\plus{}c^2\\plus{}a^2)}\r\n\\]\r\n, we have\r\n\\[ F(a,b,c)\\equal{}\\frac{2p(1\\minus{}p)}{2\\plus{}p}\\sum \\frac{(a^2\\minus{}b^2)^2}{(pa^2 \\plus{} b^2 \\plus{} c^2)(pb^2\\plus{}c^2\\plus{}a^2)}\\plus{} \\\\ \\plus{}(p\\minus{}1)\\sum \\frac{(b\\minus{}c)^2}{pa^2\\plus{}b^2\\plus{}c^2} \\\\ \\equal{} (p\\minus{}1) \\left(\\sum \\frac{(a\\minus{}b)^2}{pc^2\\plus{}a^2\\plus{}b^2} \\minus{}\\frac{2p}{2\\plus{}p} \\sum \\frac{(a^2\\minus{}b^2)^2}{(pa^2 \\plus{} b^2 \\plus{} c^2)(pb^2\\plus{}c^2\\plus{}a^2)}\\right).\r\n\\]\r\nHence the original inequality is equivalent to\r\n\\[ \\sum \\frac{(a\\minus{}b)^2}{pc^2\\plus{}a^2\\plus{}b^2} \\ge \\frac{2p}{2\\plus{}p} \\sum \\frac{(a^2\\minus{}b^2)^2}{(pa^2 \\plus{} b^2 \\plus{} c^2)(pb^2\\plus{}c^2\\plus{}a^2)}\r\n\\]\r\nfor all reals $ a,b,c$ and positive real $ p$.\r\nFrom the inequality $ (x\\minus{}y)^2 \\ge (|x|\\minus{}|y|)^2 \\ \\forall x,y$, we see that it suffits to prove the above inequality for $ a,b,c \\ge 0$ and $ p > 0$.\r\nThis inequality is equivalent to\r\n\\[ \\sum (a\\minus{}b)^2\\left(\\frac{2\\plus{}p}{pc^2\\plus{}a^2\\plus{}b^2} \\minus{}\\frac{2p(a\\plus{}b)^2}{(pa^2 \\plus{} b^2 \\plus{} c^2)(pb^2\\plus{}c^2\\plus{}a^2)}\\right) \\ge 0.\r\n\\]\r\n, which can be rewritten as\r\n\\[ S_a(b\\minus{}c)^2\\plus{}S_b(c\\minus{}a)^2\\plus{}S_c(a\\minus{}b)^2 \\ge 0\r\n\\]\r\nwith\r\n\\[ S_c\\equal{}(p\\plus{}2)(pa^2 \\plus{} b^2 \\plus{} c^2)(pb^2\\plus{}c^2\\plus{}a^2)\\minus{}2p(a\\plus{}b)^2(pc^2\\plus{}a^2\\plus{}b^2)\r\n\\]\r\nand $ S_a, S_b$ are determined similarly.\r\nWe can rewritten $ S_c$ as two forms\r\n\\[ S_c\\equal{}(pc^2\\plus{}a^2\\plus{}b^2)(pb\\minus{}2a)^2\\plus{}(p\\plus{}2)(c^2\\minus{}a^2)[pc^2\\plus{}a^2\\plus{}b^2\\plus{}(1\\minus{}p)(pb^2\\plus{}a^2\\plus{}c^2)] \\\\ S_c\\equal{}(pc^2\\plus{}a^2\\plus{}b^2)(pa\\minus{}2b)^2\\plus{}(p\\plus{}2)(c^2\\minus{}b^2)[pc^2\\plus{}a^2\\plus{}b^2\\plus{}(1\\minus{}p)(pa^2\\plus{}b^2\\plus{}c^2)]\r\n\\]\r\nand similarly for $ S_a$ and $ S_b$.\r\n\r\nIt is very useful we rewrite in both two above forms.\r\n\r\n(see my next post).", "Solution_21": "Now, we assume $ a \\ge b \\ge c \\ge 0$ and consider three cases of $ p$:\r\n\r\n[b]First case : $ 0