| {"year": "2009", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(n,k\\geq 2\\) be positive integers and let \\(a_{1}\\) , \\(a_{2}\\) , \\(a_{3}\\) , ..., \\(a_{k}\\) be distinct integers in the set \\(\\{1,2,\\ldots ,n\\}\\) such that \\(n\\) divides \\(a_{i}(a_{i + 1} - 1)\\) for \\(i = 1,2,\\ldots ,k - 1\\) . Prove that \\(n\\) does not divide \\(a_{k}(a_{1} - 1)\\) .", "solution": "We proceed indirectly and assume that \n\n\\[a_{i}(a_{i + 1} - 1)\\equiv 0\\pmod {n}\\] \n\nfor \\(i = 1,\\ldots ,k\\) (indices taken modulo \\(k\\) ). We claim that this implies all the \\(a_{i}\\) are equal modulo \\(n\\) . \n\nLet \\(q = p^{e}\\) be any prime power dividing \\(n\\) . Then, \\(a_{1}(a_{2} - 1)\\equiv 0\\) (mod \\(q\\) ), so \\(p\\) divides either \\(a_{1}\\) or \\(a_{2} - 1\\) . \n\n- If \\(p \\mid a_{1}\\) , then \\(p \\nmid a_{1} - 1\\) . Then \n\n\\[a_{k}(a_{1} - 1)\\equiv 0\\pmod {q}\\implies a_{k}\\equiv 0\\pmod {q}.\\] \n\nIn particular, \\(p \\mid a_{k}\\) . So repeating this argument, we get \\(a_{k - 1} \\equiv 0\\) (mod \\(q\\) ), \\(a_{k - 2} \\equiv 0\\) (mod \\(q\\) ), and so on. \n\n- Similarly, if \\(p \\mid a_{2} - 1\\) then \\(p \\nmid a_{2}\\) , and \n\n\\[a_{2}(a_{3} - 1)\\equiv 0\\pmod {q}\\implies a_{3}\\equiv 1\\pmod {q}.\\] \n\nIn particular, \\(p \\mid a_{3} - 1\\) . So repeating this argument, we get \\(a_{4} \\equiv 0\\) (mod \\(q\\) ), \\(a_{5} \\equiv 0\\) (mod \\(q\\) ), and so on. \n\nEither way, we find \\(a_{i}\\) (mod \\(q\\) ) is constant (and either 0 or 1). \n\nSince \\(q\\) was an arbitrary prime power dividing \\(n\\) , by Chinese remainder theorem we conclude that \\(a_{i}\\) (mod \\(n\\) ) is constant as well. But this contradicts the assumption of distinctness.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2009/1, proposed by Ross Atkins (AUS) \n"}} |
| {"year": "2009", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with circumcenter \\(o\\) . The points \\(P\\) and \\(Q\\) are interior points of the sides \\(C A\\) and \\(A B\\) respectively. Let \\(K\\) , \\(L\\) , \\(M\\) be the midpoints of \\(\\overline{{B P}}\\) \\(\\overline{{C Q}}\\) , \\(\\overline{{P Q}}\\) . Suppose that \\(\\overline{{P Q}}\\) is tangent to the circumcircle of \\(\\triangle K L M\\) . Prove that \\(O P = O Q\\) .", "solution": "a point, we have \\(- AQ \\cdot QB = OQ^2 - R^2\\) and \\(- AP \\cdot PC = OP^2 - R^2\\) . Therefore, it suffices to show \\(AQ \\cdot QB = AP \\cdot PC\\) . \n\n\n \n\nAs \\(\\overline{ML} \\parallel \\overline{AC}\\) and \\(\\overline{MK} \\parallel \\overline{AB}\\) we have that \n\n\\[\\angle APQ = \\angle LMP = \\angle LKM\\] \\[\\angle PQA = \\angle KMQ = \\angle MLK\\] \n\nand consequently we have the (opposite orientation) similarity \n\n\\[\\triangle APQ \\sim \\triangle MKL.\\] \n\nTherefore \n\n\\[\\frac{AQ}{AP} = \\frac{ML}{MK} = \\frac{2ML}{2MK} = \\frac{PC}{QB}\\] \n\nid est \\(AQ \\cdot QB = AP \\cdot PC\\) , which is what we wanted to prove.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2009/2, proposed by Sergei Berlov (RUS) \n"}} |
| {"year": "2009", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Suppose that \\(s_{1},s_{2},s_{3},\\ldots\\) is a strictly increasing sequence of positive integers such that the sub-sequences \\(s_{s_{1}}\\) , \\(s_{s_{2}}\\) , \\(s_{s_{3}}\\) , ...and \\(s_{s_{1} + 1}\\) , \\(s_{s_{2} + 1}\\) , \\(s_{s_{3} + 1}\\) , ...are both arithmetic progressions. Prove that the sequence \\(s_{1}\\) , \\(s_{2}\\) , \\(s_{3}\\) , ...is itself an arithmetic progression.", "solution": "We present two solutions. \n\n\\(\\P\\) First solution (Alex Zhai). Let \\(s(n):= s_{n}\\) and write \n\n\\[s(s(n)) = Dn + A\\] \\[s(s(n) + 1) = D^{\\prime}n + B.\\] \n\nIn light of the bounds \\(s(s(n))\\leq s(s(n) + 1)\\leq s(s(n + 1))\\) we right away recover \\(D = D^{\\prime}\\) and \\(A\\leq B\\) \n\nLet \\(d_{n} = s(n + 1) - s(n)\\) . Note that \\(\\sup d_{n}< \\infty\\) since \\(d_{n}\\) is bounded above by \\(A\\) \n\nThen we let \n\n\\[m:= \\min d_{n},\\qquad M:= \\max d_{n}.\\] \n\nNow suppose \\(a\\) achieves the maximum, meaning \\(s(a + 1) - s(a) = M\\) . Then \n\n\\[\\underbrace{d_{s(s(a))} + \\cdot\\cdot\\cdot + d_{s(s(a + 1)) - 1}}_{D\\mathrm{~terms}} = \\underbrace{\\left[s(s(s(a + 1))) - s(s(s(a)))\\right]}_{D\\mathrm{~terms}}\\] \\[\\qquad = (D\\cdot s(a + 1) + A) - (D\\cdot s(a) + A) = DM.\\] \n\nNow \\(M\\) was maximal hence \\(M = d_{s(s(a))} = \\dots = d_{s(s(a + 1)) - 1}\\) . But \\(d_{s(s(a))} = B - A\\) is a constant. Hence \\(M = B - A\\) . In the same way \\(m = B - A\\) as desired. \n\n\\(\\P\\) Second solution. We retain the notation \\(D\\) , \\(A\\) , \\(B\\) above, as well as \\(m = \\min_{n}s(n+\\) \\(1) - s(n)\\geq B - A\\) . We do the involution trick first as: \n\n\\[D = \\left[s(s(s(n) + 1)) - s(s(s(n)))\\right] = s(Dn + B) - s(Dn + A)\\] \n\nand hence we recover \\(D\\geq m(B - A)\\) \n\nThe edge case \\(D = B - A\\) is easy since then \\(m = 1\\) and \\(D = s(Dn + B) - s(Dn + A)\\) forces \\(s\\) to be a constant shift. So henceforth assume \\(D > B - A\\) \n\nThe idea is that right now the \\(B\\) terms are \"too big\", so we want to use the involution trick in a way that makes as many \" \\(A\\) minus \\(B\\) \" shape expressions as possible. This motivates considering \\(s(s(s(n + 1))) - s(s(s(n) + 1) + 1) > 0\\) , since the first expression will have all \\(A\\) 's and the second expression will have all \\(B\\) 's. Calculation gives: \n\n\\[s(D(n + 1) + A) - s(Dn + B + 1) = \\left[s(s(s(n + 1))) - s(s(s(n) + 1) + 1)\\right]\\] \\[\\qquad = (D s(n + 1) + A) - (D(s(n) + 1) + B)\\] \\[\\qquad = D(s(n + 1) - s(n)) + A - B - D.\\]\n\n\n\nThen by picking \\(n\\) achieving the minimum \\(m\\) , \n\n\\[\\underbrace{m(D + A - B - 1)}_{>0}\\leq s(s(s(n + 1))) - s(s(s(n) + 1) + 1)\\leq Dm + A - B - D\\] \n\nwhich becomes \n\n\\[(D - m(B - A)) + ((B - A) - m)\\leq 0.\\] \n\nSince both of these quantities were supposed to be nonnegative, we conclude \\(m = B - A\\) and \\(D = m^{2}\\) . Now the estimate \\(D = s(Dn + B) - s(Dn + A)\\geq m(B - A)\\) is actually sharp, so it follows that \\(s(n)\\) is arithmetic.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2009/3, proposed by Gabriel Carroll (USA) \n"}} |
| {"year": "2009", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with \\(A B = A C\\) . The angle bisectors of \\(\\angle C A B\\) and \\(\\angle A B C\\) meet the sides \\(B C\\) and \\(C A\\) at \\(D\\) and \\(E\\) , respectively. Let \\(K\\) be the incenter of triangle \\(A D C\\) . Suppose that \\(\\angle B E K = 45^{\\circ}\\) . Find all possible values of \\(\\angle C A B\\) .", "solution": "Here is the solution presented in my book EGM0. \n\nLet \\(I\\) be the incenter of \\(A B C\\) , and set \\(\\angle D A C = 2x\\) (so that \\(0^{\\circ}< x< 45^{\\circ}\\) ). From \\(\\angle A I E = \\angle D I C\\) , it is easy to compute \n\n\\[\\angle K I E = 90^{\\circ} - 2x, \\angle E C I = 45^{\\circ} - x, \\angle I E K = 45^{\\circ}, \\angle K E C = 3x.\\] \n\nHaving chased all the angles we want, we need a relationship. We can find it by considering the side ratio \\(\\frac{I K}{K C}\\) . Using the angle bisector theorem, we can express this in terms of triangle \\(I D C\\) ; however we can also express it in terms of triangle \\(I E C\\) . \n\n\n \n\nBy the law of sines, we obtain \n\n\\[\\frac{I K}{K C} = \\frac{\\sin 45^{\\circ}\\cdot\\frac{E K}{\\sin(90^{\\circ} - 2x)}}{\\sin(3x)\\cdot\\frac{E K}{\\sin(45^{\\circ} - x)}} = \\frac{\\sin 45^{\\circ}\\sin(45^{\\circ} - x)}{\\sin(3x)\\sin(90^{\\circ} - 2x)}.\\] \n\nAlso, by the angle bisector theorem on \\(\\triangle I D C\\) , we have \n\n\\[\\frac{I K}{K C} = \\frac{I D}{D C} = \\frac{\\sin(45^{\\circ} - x)}{\\sin(45^{\\circ} + x)}.\\]\n\n\n\nEquating these and cancelling \\(\\sin \\left(45^{\\circ} - x\\right) \\neq 0\\) gives \n\n\\[\\sin 45^{\\circ}\\sin \\left(45^{\\circ} + x\\right) = \\sin 3x\\sin \\left(90^{\\circ} - 2x\\right).\\] \n\nApplying the product- sum formula (again, we are just trying to break down things as much as possible), this just becomes \n\n\\[\\cos \\left(x\\right) - \\cos \\left(90^{\\circ} + x\\right) = \\cos \\left(5x - 90^{\\circ}\\right) - \\cos \\left(90^{\\circ} + x\\right)\\] \n\nor \\(\\cos x = \\cos \\left(5x - 90^{\\circ}\\right)\\) \n\nAt this point we are basically done; the rest is making sure we do not miss any solutions and write up the completion nicely. One nice way to do this is by using product- sum in reverse as \n\n\\[0 = \\cos \\left(5x - 90^{\\circ}\\right) - \\cos x = 2\\sin \\left(3x - 45^{\\circ}\\right)\\sin \\left(2x - 45^{\\circ}\\right).\\] \n\nThis way we merely consider the two cases \n\n\\[\\sin \\left(3x - 45^{\\circ}\\right) = 0 \\mathrm{and} \\sin \\left(2x - 45^{\\circ}\\right) = 0.\\] \n\nNotice that \\(\\sin \\theta = 0\\) if and only \\(\\theta\\) is an integer multiple of \\(180^{\\circ}\\) . Using the bound \\(0^{\\circ} < x < 45^{\\circ}\\) , it is easy to see that the permissible values of \\(x\\) are \\(x = 15^{\\circ}\\) and \\(x = \\frac{45^{\\circ}}{2}\\) . As \\(\\angle A = 4x\\) , this corresponds to \\(\\angle A = 60^{\\circ}\\) and \\(\\angle A = 90^{\\circ}\\) , which can be seen to work.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2009/4, proposed by Hojoo Lee, Peter Vandendriessche, Jan Vonk (BEL) \n"}} |
| {"year": "2009", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(f\\colon \\mathbb{Z}_{>0}\\to \\mathbb{Z}_{>0}\\) such that for positive integers \\(a\\) and \\(b\\) , the numbers \n\n\\[a, f(b), f(b + f(a) - 1)\\] \n\nare the sides of a non-degenerate triangle.", "solution": ". \n\nThe only function is the identity function (which works). We prove it is the only one. Let \\(P(a,b)\\) denote the given statement. \n\nClaim — We have \\(f(1) = 1\\) , and \\(f(f(n)) = n\\) . (In particular \\(f\\) is a bijection.) \n\nProof. Note that \n\n\\[P(1,b)\\Rightarrow f(b) = f(b + f(1) - 1).\\] \n\nOtherwise, the function \\(f\\) is periodic modulo \\(N = f(1) - 1\\geq 1\\) . This is impossible since we can fix \\(b\\) and let \\(a\\) be arbitrarily large in some residue class modulo \\(N\\) . \n\nHence \\(f(1) = 1\\) , so taking \\(P(n,1)\\) gives \\(f(f(n)) = n\\) . \n\nClaim — Let \\(\\delta = f(2) - 1 > 0\\) . Then for every \\(n\\) , \n\n\\[f(n + 1) = f(n) + \\delta \\quad \\mathrm{or}\\quad f(n - 1) = f(n) + \\delta\\] \n\nProof. Use \n\n\\[P(2,f(n))\\Rightarrow n - 2< f(f(n) + \\delta)< n + 2.\\] \n\nLet \\(y = f(f(n) + \\delta)\\) , hence \\(n - 2< y< n + 2\\) and \\(f(y) = f(n) + \\delta\\) . But, remark that if \\(y = n\\) , we get \\(\\delta = 0\\) , contradiction. So \\(y\\in \\{n + 1,n - 1\\}\\) and that is all. \\(\\square\\) \n\nWe now show \\(f\\) is an arithmetic progression with common difference \\(+\\delta\\) . Indeed we already know \\(f(1) = 1\\) and \\(f(2) = 1 + \\delta\\) . Now suppose \\(f(1) = 1,\\ldots ,f(n) = 1 + (n - 1)\\delta\\) . Then by induction for any \\(n\\geq 2\\) , the second case can't hold, so we have \\(f(n + 1) = f(n) + \\delta\\) , as desired. \n\nCombined with \\(f(f(n)) = n\\) , we recover that \\(f\\) is the identity.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2009/5, proposed by Bruno Le Floch (FRA) \n"}} |
| {"year": "2009", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) be distinct positive integers and let \\(M\\) be a set of \\(n - 1\\) positive integers not containing \\(s = a_{1} + \\dots +a_{n}\\) . A grasshopper is to jump along the real axis, starting at the point 0 and making \\(n\\) jumps to the right with lengths \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in \\(M\\) .", "solution": "The proof is by induction on \\(n\\) . Assume \\(a_{1} < \\dots < a_{n}\\) and call each element of \\(M\\) a mine. Let \\(x = s - a_{n}\\) . We consider four cases, based on whether \\(x\\) has a mine and whether there is a mine past \\(x\\) . \n\n- If \\(x\\) has no mine, and there is a mine past \\(x\\) , then at most \\(n - 2\\) mines in \\([0, x]\\) and so we use induction to reach \\(x\\) , then leap from \\(x\\) to \\(s\\) and win. \n\n- If \\(x\\) has no mine but there is also no mine to the right of \\(x\\) , then let \\(m\\) be the maximal mine. By induction hypothesis on \\(M \\setminus \\{m\\}\\) , there is a path to \\(x\\) using \\(\\{a_{1}, \\ldots , a_{n - 1}\\}\\) which avoids mines except possibly \\(m\\) . If the path hits the mine \\(m\\) on the hop of length \\(a_{k}\\) , we then swap that hop with \\(a_{n}\\) , and finish. \n\n- If \\(x\\) has a mine, but there are no mines to the right of \\(x\\) , we can repeat the previous case with \\(m = x\\) . \n\n- Now suppose \\(x\\) has a mine, and there is a mine past \\(x\\) . There should exist an integer \\(1 \\leq i \\leq n - 1\\) such that \\(s - a_{i}\\) and \\(y = s - a_{i} - a_{n}\\) both have no mine. By induction hypothesis, we can then reach \\(y\\) in \\(n - 2\\) steps (as there are two mines to the right of \\(y\\) ); then \\(y \\to s - a_{i} \\to s\\) finishes. \n\nRemark. It seems much of the difficulty of the problem is realizing induction will actually work. Attempts at induction are, indeed, a total minefield (ha!), and given the position P6 of the problem, it is expected that many contestants will abandon induction after some cursory attempts fail.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2009/6, proposed by Dmitry Khramtsov (RUS) \n"}} |
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