| {"year": "2012", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle and \\(J\\) the center of the \\(A\\) -excircle. This excircle is tangent to the side \\(B C\\) at \\(M\\) , and to the lines \\(A B\\) and \\(A C\\) at \\(K\\) and \\(L\\) , respectively. The lines \\(L M\\) and \\(B J\\) meet at \\(F\\) , and the lines \\(K M\\) and \\(C J\\) meet at \\(G\\) . Let \\(S\\) be the point of intersection of the lines \\(A F\\) and \\(B C\\) , and let \\(T\\) be the point of intersection of the lines \\(A G\\) and \\(B C\\) . Prove that \\(M\\) is the midpoint of \\(\\overline{S T}\\) .", "solution": ". \n\nWe employ barycentric coordinates with reference \\(\\triangle A B C\\) . As usual \\(a = B C\\) , \\(b = C A\\) , \\(c = A B\\) , \\(s = \\frac{1}{2} (a + b + c)\\) . \n\nIt's obvious that \\(K = (- (s - c):s:0)\\) , \\(M = (0:s - b:s - c)\\) . Also, \\(J = (- a:b:c)\\) . We then obtain \n\n\\[G = \\left(-a:b: - \\frac{as + (s - c)b}{s - b}\\right).\\] \n\nIt follows that \n\n\\[T = \\left(0:b: - \\frac{-as + (s - c)}{s - b}\\right) = (0:b(s - b):b(s - c) - as).\\] \n\nNormalizing, we see that \\(T = \\left(0, - \\frac{b}{a},1 + \\frac{b}{a}\\right)\\) , from which we quickly obtain \\(M T = s\\) . Similarly, \\(M S = s\\) , so we're done.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2012/1, proposed by Evangelos Psychas (HEL) \n"}} |
| {"year": "2012", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{2}\\) , \\(a_{3}\\) , ..., \\(a_{n}\\) be positive reals with product 1, where \\(n \\geq 3\\) . Show that \n\n\\[(1 + a_{2})^{2}(1 + a_{3})^{3}\\dots (1 + a_{n})^{n} > n^{n}.\\]", "solution": "L \n\nTry the dumbest thing possible: by AM- GM, \n\n\\[(1 + a_{2})^{2}\\geq 2^{2}a_{2}\\] \\[(1 + a_{3})^{3} = \\left(\\frac{1}{2} +\\frac{1}{2} +a_{3}\\right)^{3}\\geq \\frac{3^{3}}{2^{2}}a_{3}\\] \\[(1 + a_{4})^{4} = \\left(\\frac{1}{3} +\\frac{1}{3} +\\frac{1}{3} +a_{4}\\right)^{4}\\geq \\frac{4^{4}}{3^{3}}a_{4}\\] \\[\\qquad \\vdots\\] \n\nand so on. Multiplying these all gives the result. The inequality is strict since it's not possible that \\(a_{2} = 1\\) , \\(a_{3} = \\frac{1}{2}\\) , et cetera.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2012/2, proposed by Angelo di Pasquale (AUS) \n"}} |
| {"year": "2012", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "The liar's guessing game is a game played between two players \\(A\\) and \\(B\\) . The rules of the game depend on two fixed positive integers \\(k\\) and \\(n\\) which are known to both players. \n\nAt the start of the game \\(A\\) chooses integers \\(x\\) and \\(N\\) with \\(1 \\leq x \\leq N\\) . Player \\(A\\) keeps \\(x\\) secret, and truthfully tells \\(N\\) to player \\(B\\) . Player \\(B\\) now tries to obtain information about \\(x\\) by asking player \\(A\\) questions as follows: each question consists of \\(B\\) specifying an arbitrary set \\(S\\) of positive integers (possibly one specified in some previous question), and asking \\(A\\) whether \\(x\\) belongs to \\(S\\) . Player \\(B\\) may ask as many questions as he wishes. After each question, player \\(A\\) must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any \\(k + 1\\) consecutive answers, at least one answer must be truthful. \n\nAfter \\(B\\) has asked as many questions as he wants, he must specify a set \\(X\\) of at most \\(n\\) positive integers. If \\(x\\) belongs to \\(X\\) , then \\(B\\) wins; otherwise, he loses. Prove that: \n\n(a) If \\(n \\geq 2^{k}\\) , then \\(B\\) can guarantee a win. \n(b) For all sufficiently large \\(k\\) , there exists an integer \\(n \\geq (1.99)^{k}\\) such that \\(B\\) cannot guarantee a win.", "solution": "Call the players Alice and Bob. \n\nPart (a): We prove the following. \n\nClaim — If \\(N\\geq 2^{k} + 1\\) , then in \\(2k + 1\\) questions, Bob can rule out some number in \\(\\{1,\\ldots ,2^{k} + 1\\}\\) form being equal to \\(x\\) . \n\nProof. First, Bob asks the question \\(S_{0} = \\{2^{k} + 1\\}\\) until Alice answers \"yes\" or until Bob has asked \\(k + 1\\) questions. If Alice answers \"no\" to all of these then Bob rules out \\(2^{k} + 1\\) . So let's assume Alice just said \"yes\". \n\nNow let \\(T = \\{1,\\ldots ,2^{k}\\}\\) . Then, he asks \\(k\\) - follow up questions \\(S_{1}\\) , ..., \\(S_{k}\\) defined as follows: \n\n- \\(S_{1} = \\{1,3,5,7,\\ldots ,2^{k} - 1\\}\\) consists of all numbers in \\(T\\) whose least significant digit in binary is 1. \n- \\(S_{2} = \\{2,3,6,7,\\ldots ,2^{k} - 2,2^{k} - 1\\}\\) consists of all numbers in \\(T\\) whose second least significant digit in binary is 1. \n- More generally \\(S_{i}\\) consists of all numbers in \\(T\\) whose \\(i\\)th least significant digit in binary is 1. \n\nWLOG Alice answers these all as \"yes\" (the other cases are similar). Among the last \\(k + 1\\) answers at least one must be truthful, and the number \\(2^{k}\\) (having zeros in all relevant digits) does not appear in any of \\(S_{0}\\) , ..., \\(S_{k}\\) and is ruled out. \\(\\square\\)\n\n\n\nThus in this way Bob can repeatedly find non- possibilities for \\(x\\) (and then relabel the remaining candidates \\(1, \\ldots , N - 1\\) ) until he arrives at a set of at most \\(2^{k}\\) numbers. \n\nPart (b): It suffices to consider \\(n = \\lceil 1.99^{k}\\rceil\\) and \\(N = n + 1\\) for large \\(k\\) . At the \\(t\\) th step, Bob asks some question \\(S_{t}\\) ; we phrase each of Alice's answers in the form \" \\(x \\notin B_{t}\\) \", where \\(B_{t}\\) is either \\(S_{t}\\) or its complement. (You may think of these as \"bad sets\"; the idea is to show we can avoid having any number appear in \\(k + 1\\) consecutive bad sets, preventing Bob from ruling out any numbers.) \n\nMain idea: for every number \\(1 \\leq x \\leq N\\) , at time step \\(t\\) we define its weight to be \n\n\\[w(x) = 1.998^{e}\\] \n\nwhere \\(e\\) is the largest number such that \\(x \\in B_{t - 1} \\cap B_{t - 2} \\cap \\dots \\cap B_{t - e}\\) . \n\nClaim — Alice can ensure the total weight never exceeds \\(1.998^{k + 1}\\) for large \\(k\\) . \n\nProof. Let \\(W_{t}\\) denote the sum of weights after the \\(t\\) th question. We have \\(W_{0} = N < 1000n\\) . We will prove inductively that \\(W_{t} < 1000n\\) always. \n\nAt time \\(t\\) , Bob specifies a question \\(S_{t}\\) . We have Alice choose \\(B_{t}\\) as whichever of \\(S_{t}\\) or \\(\\overline{S_{t}}\\) has lesser total weight, hence at most \\(W_{t} / 2\\) . The weights of for \\(B_{t}\\) increase by a factor of 1.998, while the weights for \\(\\overline{B_{t}}\\) all reset to 1. So the new total weight after time \\(t\\) is \n\n\\[W_{t + 1} \\leq 1.998 \\cdot \\frac{W_{t}}{2} + \\# \\overline{B_{t}} \\leq 0.999W_{t} + n.\\] \n\nThus if \\(W_{t} < 1000n\\) then \\(W_{t + 1} < 1000n\\) . \n\nTo finish, note that \\(1000n < 1000 \\left(1.99^{k} + 1\\right) < 1.998^{k + 1}\\) for \\(k\\) large. \n\nIn particular, no individual number can have weight \\(1.998^{k + 1}\\) . Thus for every time step \\(t\\) we have \n\n\\[B_{t} \\cap B_{t + 1} \\cap \\dots \\cap B_{t + k} = \\emptyset .\\] \n\nThen once Bob stops, if he declares a set of \\(n\\) positive integers, and \\(x\\) is an integer Bob did not choose, then Alice's question history is consistent with \\(x\\) being Alice's number, as among any \\(k + 1\\) consecutive answers she claimed that \\(x \\in \\overline{B_{t}}\\) for some \\(t\\) in that range. \n\nRemark (Motivation). In our \\(B_{t}\\) setup, let's think backwards. The problem is equivalent to avoiding \\(e = k + 1\\) at any time step \\(t\\) , for any number \\(x\\) . That means \n\n- have at most two elements with \\(e = k\\) at time \\(t - 1\\) ,- thus have at most four elements with \\(e = k - 1\\) at time \\(t - 2\\) ,- thus have at most eight elements with \\(e = k - 2\\) at time \\(t - 3\\) ,- and so on. \n\nWe already exploited this in solving part (a). In any case it's now natural to try letting \\(w(x) = 2^{e}\\) , so that all the cases above sum to \"equally bad\" situations: since \\(8 \\cdot 2^{k - 2} = 4 \\cdot 2^{k - 1} = 2 \\cdot 2^{k}\\) , say. \n\nHowever, we then get \\(W_{t + 1} \\leq \\frac{1}{2} (2W_{t}) + n\\) , which can increase without bound due to contributions from numbers resetting to zero. The way to fix this is to change the weight to \\(w(x) = 1.998^{e}\\) , taking advantage of the little extra space we have due to having \\(n \\geq 1.99^{k}\\) rather than \\(n \\geq 2^{k}\\) .\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2012/3, proposed by David Arthur (CAN) \n"}} |
| {"year": "2012", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(f: \\mathbb{Z} \\to \\mathbb{Z}\\) such that, for all integers \\(a\\) , \\(b\\) , \\(c\\) that satisfy \\(a + b + c = 0\\) , the following equality holds: \n\n\\[f(a)^{2} + f(b)^{2} + f(c)^{2} = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).\\]", "solution": "arbitrary \\(k\\in \\mathbb{Z}\\) , we have \n\n(i) \\(f(x) = k x^{2}\\) \n\n(ii) \\(f(x) = 0\\) for even \\(x\\) , and \\(f(x) = k\\) for odd \\(x\\) , and \n\n(iii) \\(f(x) = 0\\) for \\(x\\equiv 0\\) (mod 4), \\(f(x) = k\\) for odd \\(x\\) , and \\(f(x) = 4k\\) for \\(x\\equiv 2\\) (mod 4). \n\nThese can be painfully seen to work. (It's more natural to think of these as \\(f(x) = x^{2}\\) , \\(f(x) = x^{2}\\) (mod 4), \\(f(x) = x^{2}\\) (mod 8), and multiples thereof.) \n\nSet \\(a = b = c = 0\\) to get \\(f(0) = 0\\) . Then set \\(c = 0\\) to get \\(f(a) = f(- a)\\) , so \\(f\\) is even. Now \n\n\\[f(a)^{2} + f(b)^{2} + f(a + b)^{2} = 2f(a + b)\\left(f(a) + f(b)\\right) + 2f(a)f(b)\\] \n\nor \n\n\\[(f(a + b) - (f(a) + f(b)))^{2} = 4f(a)f(b).\\] \n\nHence \\(f(a)f(b)\\) is a perfect square for all \\(a,b\\in \\mathbb{Z}\\) . So there exists a \\(\\lambda\\) such that \\(f(n) = \\lambda g(n)^{2}\\) , where \\(g(n)\\geq 0\\) . From here we recover \n\n\\[\\boxed{g(a + b) = \\pm g(a)\\pm g(b).}\\] \n\nAlso \\(g(0) = 0\\) \n\nLet \\(k = g(1)\\neq 0\\) . We now split into cases on \\(g(2)\\) \n\n- \\(g(2) = 0\\) . Put \\(b = 2\\) in original to get \\(g(a + 2) = \\pm g(a) = +g(a)\\) \n\n- \\(g(2) = 2k\\) . Cases on \\(g(4)\\) : \n\n- \\(g(4) = 0\\) , then we get \\((g(n))_{n\\geq 0} = (0,1,2,1,0,1,2,1,\\ldots)\\) . This works. \n\n- \\(g(4) = 4k\\) . This only happens when \\(g(1) = k\\) , \\(g(2) = 2k\\) , \\(g(3) = 3k\\) , \\(g(4) = 4k\\) . Then \n\n\\[*g(5) = \\pm 3k\\pm 2k = \\pm 4k\\pm k.\\] \n\n\\[*g(6) = \\pm 4k\\pm 2k = \\pm 5k\\pm k.\\] \n\n\\\\*... \n\nand so by induction \\(g(n) = nk\\)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2012/4, proposed by Liam Baker (SAF) \n"}} |
| {"year": "2012", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with \\(\\angle B C A = 90^{\\circ}\\) , and let \\(D\\) be the foot of the altitude from \\(C\\) . Let \\(X\\) be a point in the interior of the segment \\(C D\\) . Let \\(K\\) be the point on the segment \\(A X\\) such that \\(B K = B C\\) . Similarly, let \\(L\\) be the point on the segment \\(B X\\) such that \\(A L = A C\\) . Let \\(M = \\overline{A L} \\cap \\overline{B K}\\) . Prove that \\(M K = M L\\) .", "solution": "Let \\(\\omega_{A}\\) and \\(\\omega_{B}\\) be the circles through \\(C\\) centered at \\(A\\) and \\(B\\) ; extend rays \\(A K\\) and \\(B L\\) to hit \\(\\omega_{B}\\) and \\(\\omega_{A}\\) again at \\(K^{*}\\) , \\(L^{*}\\) . By radical center \\(X\\) , we have \\(K L K^{*}L^{*}\\) is cyclic, say with circumcircle \\(\\omega\\) . \n\n\n \n\nBy orthogonality of \\((A)\\) and \\((B)\\) we find that \\(\\overline{A L}\\) , \\(\\overline{A L^{*}}\\) , \\(\\overline{B K}\\) , \\(\\overline{B K^{*}}\\) are tangents to \\(\\omega\\) (in particular, \\(K L K^{*}L^{*}\\) is harmonic). In particular \\(\\overline{M K}\\) and \\(\\overline{M L}\\) are tangents to \\(\\omega\\) , so \\(M K = M L\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2012/5, proposed by Josef Tkadlec (CZE) \n"}} |
| {"year": "2012", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Find all positive integers \\(n\\) for which there exist non-negative integers \\(a_{1}, a_{2}, \\ldots , a_{n}\\) such that \n\n\\[\\frac{1}{2^{a_{1}}} + \\frac{1}{2^{a_{2}}} + \\dots + \\frac{1}{2^{a_{n}}} = \\frac{1}{3^{a_{1}}} + \\frac{2}{3^{a_{2}}} + \\dots + \\frac{n}{3^{a_{n}}} = 1.\\]", "solution": "The answer is \\(n\\equiv 1,2\\) (mod 4). To see these are necessary, note that taking the latter equation modulo 2 gives \n\n\\[1 = \\frac{1}{3^{a_{1}}} +\\frac{2}{3^{a_{2}}} +\\dots +\\frac{n}{3^{a_{n}}}\\equiv 1 + 2 + \\ldots +n\\pmod {2}.\\] \n\nNow we prove these are sufficient. The following nice construction was posted on AOPs by the user cfheolipixin. \n\nClaim — If \\(n = 2k - 1\\) works then so does \\(n = 2k\\) \n\nProof. Replace \n\n\\[\\frac{k}{3^{r}} = \\frac{k}{3^{r + 1}} +\\frac{2k}{3^{r + 1}}. \\quad (*)\\] \n\nClaim — If \\(n = 4k + 2\\) works then so does \\(n = 4k + 13\\) \n\nProof. First use the identity \n\n\\[\\frac{k + 2}{3^{r}} = \\frac{k + 2}{3^{r + 2}} +\\frac{4k + 3}{3^{r + 3}} +\\frac{4k + 5}{3^{r + 3}} +\\frac{4k + 7}{3^{r + 3}} +\\frac{4k + 9}{3^{r + 3}} +\\frac{4k + 11}{3^{r + 3}} +\\frac{4k + 13}{3^{r + 3}}\\] \n\nto fill in the odd numbers. The even numbers can then be instantiated with \\((\\ast)\\) too. \n\nThus it suffices to construct base cases for \\(n = 1\\) , \\(n = 5\\) , \\(n = 9\\) . They are \n\n\\[1 = \\frac{1}{3^{0}}\\] \\[\\quad = \\frac{1}{3^{2}} +\\frac{2}{3^{2}} +\\frac{3}{3^{2}} +\\frac{4}{3^{3}} +\\frac{5}{3^{3}}\\] \\[\\quad = \\frac{1}{3^{2}} +\\frac{2}{3^{3}} +\\frac{3}{3^{3}} +\\frac{4}{3^{3}} +\\frac{5}{3^{3}} +\\frac{6}{3^{4}} +\\frac{7}{3^{4}} +\\frac{8}{3^{4}} +\\frac{9}{3^{4}}.\\]", "metadata": {"resource_path": "IMO/segmented/en-IMO-2012-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2012/6, proposed by Dusan Djukic (SRB) \n"}} |
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