| {"year": "2007", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Real numbers \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) are fixed. For each \\(1 \\leq i \\leq n\\) we let \\(d_{i} = \\max \\{a_{j} : 1 \\leq j \\leq i\\} - \\min \\{a_{j} : i \\leq j \\leq n\\}\\) and let \\(d = \\max \\{d_{i} : 1 \\leq i \\leq n\\}\\) . \n\n(a) Prove that for any real numbers \\(x_{1} \\leq \\dots \\leq x_{n}\\) we have \n\n\\[\\max \\left\\{|x_{i} - a_{i}|:1\\leq i\\leq n\\right\\} \\geq \\frac{1}{2} d.\\] \n\n(b) Moreover, show that there exists some choice of \\(x_{1} \\leq \\dots \\leq x_{n}\\) which achieves equality.", "solution": "Note that we can dispense of \\(d_{i}\\) immediately by realizing that the definition of \\(d\\) just says \n\n\\[d = \\max_{1 \\leq i \\leq j \\leq n} (a_{i} - a_{j}).\\] \n\nIf \\(a_{1} \\leq \\dots \\leq a_{n}\\) are already nondecreasing then \\(d = 0\\) and there is nothing to prove (for the equality case, just let \\(x_{i} = a_{i}\\) ), so we will no longer consider this case. \n\nOtherwise, consider any indices \\(i < j\\) with \\(a_{i} > a_{j}\\) . We first prove (a) by applying the following claim with \\(p = a_{i}\\) and \\(q = a_{j}\\) : \n\nClaim — For any \\(p \\leq q\\) , we have either \\(|p - a_{i}| \\geq \\frac{1}{2} (a_{i} - a_{j})\\) or \\(|q - a_{j}| \\geq \\frac{1}{2} (a_{i} - a_{j}) \n\nProof. Assume for contradiction both are false. Then \\(p > a_{i} - \\frac{1}{2} (a_{i} - a_{j}) = a_{j} + \\frac{1}{2} (a_{i} - a_{j}) > q\\) , contradiction. \\(\\square\\) \n\nAs for (b), we let \\(i < j\\) be any indices for which \\(a_{i} - a_{j} = d > 0\\) achieves the maximal difference. We then define \\(x_{\\bullet}\\) in three steps: \n\nWe set \\(x_{k} = \\frac{a_{i} + a_{j}}{2}\\) for \\(k = i, \\ldots , j\\) . We recursively set \\(x_{k} = \\max (x_{k - 1}, a_{k})\\) for \\(k = j + 1, j + 2, \\ldots\\) . We recursively set \\(x_{k} = \\min (x_{k + 1}, a_{k})\\) for \\(k = i - 1, i - 2, \\ldots\\) . \n\nBy definition, these \\(x_{\\bullet}\\) are weakly increasing. To prove this satisfies (b) we only need to check that \n\n\\[|x_{k} - a_{k}| \\leq \\frac{a_{i} - a_{j}}{2} \\qquad (\\star)\\] \n\nfor any index \\(k\\) (as equality holds for \\(k = i\\) or \\(k = j\\) ). \n\nWe note \\((\\star)\\) holds for \\(i < k < j\\) by construction. For \\(k > j\\) , note that \\(x_{k} \\in \\{a_{j}, a_{j + 1}, \\ldots , a_{k}\\}\\) by construction, so \\((\\star)\\) follows from our choice of \\(i\\) and \\(j\\) giving the largest possible difference; the case \\(k < i\\) is similar.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2007/1, proposed by Michael Albert (NZL) \n"}} |
| {"year": "2007", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Consider five points \\(A\\) , \\(B\\) , \\(C\\) , \\(D\\) and \\(E\\) such that \\(ABCD\\) is a parallelogram and \\(BCED\\) is a cyclic quadrilateral. Let \\(\\ell\\) be a line passing through \\(A\\) . Suppose that \\(\\ell\\) intersects the interior of the segment \\(DC\\) at \\(F\\) and intersects line \\(BC\\) at \\(G\\) . Suppose also that \\(EF = EG = EC\\) . Prove that \\(\\ell\\) is the bisector of angle \\(DAB\\) .", "solution": "Let \\(M\\) , \\(N\\) , \\(P\\) denote the midpoints of \\(\\overline{CF}\\) , \\(\\overline{CG}\\) , \\(\\overline{AC}\\) (noting \\(P\\) is also the midpoint of \\(\\overline{BD}\\) ). \n\nBy a homothety at \\(C\\) with ratio \\(\\frac{1}{2}\\) , we find \\(\\overline{MNP}\\) is the image of line \\(\\ell \\equiv \\overline{AGF}\\) . \n\n\n \n\nHowever, since we also have \\(\\overline{EM} \\perp \\overline{CF}\\) and \\(\\overline{EN} \\perp \\overline{CG}\\) (from \\(EF = EG = EC\\) ) we conclude \\(\\overline{PMN}\\) is the Simson line of \\(E\\) with respect to \\(\\triangle BCD\\) , which implies \\(\\overline{EP} \\perp \\overline{BD}\\) . In other words, \\(\\overline{EP}\\) is the perpendicular bisector of \\(\\overline{BD}\\) , so \\(E\\) is the midpoint of arc \\(\\overline{BCD}\\) . \n\nFinally, \n\n\\[\\angle (\\overline{AB}, \\ell) = \\angle (\\overline{CD}, \\overline{MNP}) = \\angle CMN = \\angle CEN\\] \\[\\qquad = 90^{\\circ} - \\angle NCE = 90^{\\circ} + \\angle ECB\\] \n\nwhich means that \\(\\ell\\) is parallel to a bisector of \\(\\angle BCD\\) , and hence to one of \\(\\angle BAD\\) . (Moreover since \\(F\\) lies on the interior of \\(\\overline{CD}\\) , it is actually the internal bisector.)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2007/2, proposed by Charles Leytem (LUX) \n"}} |
| {"year": "2007", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "In a mathematical competition some competitors are (mutual) friends. Call a group of competitors a clique if each two of them are friends. Given that the largest size of a clique is even, prove that the competitors can be arranged into two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.", "solution": "Take the obvious graph interpretation \\(G\\) . We paint red any vertices in one of the maximal cliques \\(K\\) , which we assume has \\(2r\\) vertices, and paint the remaining vertices green. We let \\(\\alpha (\\bullet)\\) denote the clique number. \n\nInitially, let the two rooms \\(A = K\\) , \\(B = G - K\\) . \n\nClaim — We can move at most \\(r\\) vertices of \\(A\\) into \\(B\\) to arrive at \\(\\alpha (A) \\leq \\alpha (B) \\leq \\alpha (A) + 1\\) . \n\nProof. This is actually obvious by discrete continuity. We move one vertex at a time, noting \\(\\alpha (A)\\) decreases by one at each step, while \\(\\alpha (B)\\) increases by either zero or one at each step. \n\nWe stop once \\(\\alpha (B) \\geq \\alpha (A)\\) , which happens before we have moved \\(r\\) vertices (since then we have \\(\\alpha (B) \\geq r = \\alpha (A)\\) ). The conclusion follows. \\(\\square\\) \n\nSo let's consider the situation \n\n\\[\\alpha (A) = k \\geq r \\qquad \\text{and} \\qquad \\alpha (B) = k + 1.\\] \n\nAt this point \\(A\\) is a set of \\(k\\) red vertices, while \\(B\\) has the remaining \\(2r - k\\) red vertices (and all the green ones). An example is shown below with \\(k = 4\\) and \\(2r = 6\\) . \n\n\n \n\nNow, if we can move any red vertex from \\(B\\) back to \\(A\\) without changing the clique number of \\(B\\) , we do so, and win. \n\nOtherwise, it must be the case that every \\((k + 1)\\) - clique in \\(B\\) uses every red vertex in \\(B\\) . For each \\((k + 1)\\) - clique in \\(B\\) (in arbitrary order), we do the following procedure.\n\n\n\n- If all \\(k + 1\\) vertices are still green, pick one and re-color it blue. This is possible since \\(k + 1 > 2r - k\\) . \n\n- Otherwise, do nothing. \n\nThen we move all the blue vertices from \\(B\\) to \\(A\\) , one at a time, in the same order we re- colored them. This forcibly decreases the clique number of \\(B\\) to \\(k\\) , since the clique number is \\(k + 1\\) just before the last blue vertex is moved, and strictly less than \\(k + 1\\) (hence equal to \\(k\\) ) immediately after that. \n\nClaim — After this, \\(\\alpha (A) = k\\) still holds. \n\nProof. Assume not, and we have a \\((k + 1)\\) - clique which uses \\(b\\) blue vertices and \\((k + 1) - b\\) red vertices in \\(A\\) . Together with the \\(2r - k\\) red vertices already in \\(B\\) we then get a clique of size \n\n\\[b + ((k + 1 - b)) + (2r - k) = 2r + 1\\] \n\nwhich is a contradiction. \n\nRemark. Dragomir Grozev posted the following motivation on his blog: \n\nI think, it's a natural idea to place all students in one room and begin moving them one by one into the other one. Then the max size of the cliques in the first and second room increase (resp. decrease) at most with one. So, there would be a moment both sizes are almost the same. At that moment we may adjust something. \n\nTrying the idea, I had some difficulties keeping track of the maximal cliques in the both rooms. It seemed easier all the students in one of the rooms to comprise a clique. It could be achieved by moving only the members of the maximal clique. Following this path the remaining obstacles can be overcome naturally.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2007/3, proposed by Vasily Astakhov (RUS) \n"}} |
| {"year": "2007", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "In triangle \\(ABC\\) the bisector of \\(\\angle BCA\\) meets the circumcircle again at \\(R\\) , the perpendicular bisector of \\(\\overline{BC}\\) at \\(P\\) , and the perpendicular bisector of \\(\\overline{AC}\\) at \\(Q\\) . The midpoint of \\(\\overline{BC}\\) is \\(K\\) and the midpoint of \\(\\overline{AC}\\) is \\(L\\) . Prove that the triangles \\(RPK\\) and \\(RQL\\) have the same area.", "solution": "We first begin by proving the following claim. \n\nClaim — We have \\(CQ = PR\\) (equivalently, \\(CP = QR\\) ). \n\nProof. Let \\(O = \\overline{LQ} \\cap \\overline{KP}\\) be the circumcenter. Then \n\n\\[\\angle OPQ = \\angle KPC = 90^{\\circ} - \\angle PCK = 90^{\\circ} - \\angle LCQ = \\angle \\angle CQL = \\angle PQO.\\] \n\nThus \\(OP = OQ\\) . Since \\(OC = OR\\) as well, we get the conclusion. \n\nDenote by \\(X\\) and \\(Y\\) the feet from \\(R\\) to \\(\\overline{CA}\\) and \\(\\overline{CB}\\) , so \\(\\triangle CXR \\cong \\triangle CYR\\) . Then, let \\(t = \\frac{CQ}{CR} = 1 - \\frac{CP}{CR}\\) . \n\n\n \n\nThen it follows that \n\n\\[[RQ L] = [X Q L] = t(1 - t)\\cdot [X R C] = t(1 - t)\\cdot [Y C R] = [Y K P] = [R K P]\\] \n\nas needed. \n\nRemark. Trigonometric approaches are very possible (and easier to find) as well: both areas work out to be \\(\\frac{1}{8} ab \\tan \\frac{1}{2} C\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2007/4, proposed by Marek Pechal (CZE) \n"}} |
| {"year": "2007", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(a\\) and \\(b\\) be positive integers. Show that if \\(4ab - 1\\) divides \\((4a^{2} - 1)^{2}\\) , then \\(a = b\\) .", "solution": "As usual, \n\n\\[4ab - 1\\mid (4a^{2} - 1)^{2}\\iff 4ab - 1\\mid (4ab\\cdot a - b)^{2}\\iff 4ab - 1\\mid (a - b)^{2}.\\] \n\nThen we use a typical Vieta jumping argument. Define \n\n\\[k = \\frac{(a - b)^{2}}{4ab - 1}.\\] \n\nNote that \\(k = 0 \\iff a = b\\) . So we will prove that \\(k > 0\\) leads to a contradiction. \n\nIndeed, suppose \\((a,b)\\) is a minimal solution with \\(a > b\\) (we have \\(a\\neq b\\) since \\(k\\neq 0\\) - By Vieta jumping, \\((b,\\frac{b^{2} + k}{a})\\) is also such a solution. But now \n\n\\[\\frac{b^{2} + k}{a}\\geq a\\Rightarrow k\\geq a^{2} - b^{2}\\] \\[\\Rightarrow \\frac{(a - b)^{2}}{4ab - 1}\\geq a^{2} - b^{2}\\] \\[\\Rightarrow a - b\\geq (4ab - 1)(a + b)\\] \n\nwhich is absurd for \\(a,b\\in \\mathbb{Z}_{>0}\\) . (In the last step we divided by \\(a - b > 0\\) -", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2007/5, proposed by Kevin Buzzard, Edward Crane (UNK) \n"}} |
| {"year": "2007", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\) be a positive integer. Consider \n\n\\[S = \\{(x,y,z) \\mid x,y,z \\in \\{0,1,\\ldots ,n\\} , x + y + z > 0\\}\\] \n\nas a set of \\((n + 1)^{3} - 1\\) points in the three- dimensional space. Determine the smallest possible number of planes, the union of which contains \\(S\\) but does not include \\((0,0,0)\\) .", "solution": "The answer is \\(3n\\) . Here are two examples of constructions with \\(3n\\) planes: \n\n\\(\\cdot x + y + z = i\\) for \\(i = 1,\\ldots ,3n\\) \n\n\\(\\cdot x = i\\) \\(y = i\\) \\(z = i\\) for \\(i = 1,\\ldots ,n\\) \n\nSuppose for contradiction we have \\(N< 3n\\) planes. Let them be \\(a_{i}x + b_{i}y + c_{i}z + 1 = 0\\) for \\(i = 1,\\ldots ,N\\) . Define the polynomials \n\n\\[A(x,y,z) = \\prod_{i = 1}^{n}(x - i)\\prod_{i = 1}^{n}(y - i)\\prod_{i = 1}^{n}(z - i)\\] \\[B(x,y,z) = \\prod_{i = 1}^{N}(a_{i}x + b_{i}y + c_{i}z + 1).\\] \n\nNote that \\(A(0,0,0) = (- 1)^{n}(n!)^{3}\\neq 0\\) and \\(B(0,0,0) = 1\\neq 0\\) , but \\(A(x,y,z) =\\) \\(B(x,y,z) = 0\\) for any \\((x,y,z)\\in S\\) . Also, the coefficient of \\(x^{n}y^{n}z^{n}\\) in \\(A\\) is 1, while the coefficient of \\(x^{n}y^{n}z^{n}\\) in \\(B\\) is 0. \n\nNow, define \n\n\\[P(x,y,z):= A(x,y,z) - \\lambda B(x,y,z).\\] \n\nwhere \\(\\lambda = \\frac{A(0,0,0)}{B(0,0,0)} = (- 1)^{n}(n!)^{3}\\) . We now have that \n\n\\(\\cdot P(x,y,z) = 0\\) for any \\(x,y,z\\in \\{0,1,\\ldots ,n\\}^{3}\\) \n\nBut the coefficient of \\(x^{n}y^{n}z^{n}\\) is 1. \n\nThis is a contradiction to Alon's combinatorial nullstellensatz.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2007/6, proposed by Gerhard Woeginger (NLD) \n"}} |
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