diff --git "a/EGMO/segmented/en-2025-solutions.jsonl" "b/EGMO/segmented/en-2025-solutions.jsonl"
--- "a/EGMO/segmented/en-2025-solutions.jsonl"
+++ "b/EGMO/segmented/en-2025-solutions.jsonl"
@@ -4,22 +4,22 @@
 {"year": "2025", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "EGMO", "problem": "An infinite increasing sequence \\(a_{1}< a_{2}< a_{3}< \\dots\\) of positive integers is called central if for every positive integer \\(n\\) , the arithmetic mean of the first \\(a_{n}\\) terms of the sequence is equal to \\(a_{n}\\) .  \n\nShow that there exists an infinite sequence \\(b_{1},b_{2},b_{3},\\ldots\\) of positive integers such that for every central sequence \\(a_{1},a_{2},a_{3},\\ldots\\) , there are infinitely many positive integers \\(n\\) with \\(a_{n} = b_{n}\\) .", "solution": "We give an alternative proof that the sequence \\(b_{i} = 2i - 1\\) works. This proof is by contradiction, so we assume that there are only finitely many \\(a_{i}\\) such that \\(a_{i} = 2i - 1\\) .  \n\nLet \\(S(n) = \\sum_{i = 1}^{n}a_{i}\\) . We have \\(S(a_{n}) = a_{n}^{2}\\) and \\(S(a_{n + 1}) = a_{n + 1}^{2}\\) . If \\(a_{n + 1} = a_{n} + 1\\) , then it follows that  \n\n\\[S(a_{n + 1}) - S(a_{n}) = a_{n + 1}^{2} - a_{n}^{2} = a_{n + 1}^{2} - (a_{n + 1} - 1)^{2} = 2a_{n + 1} - 1.\\]  \n\nOn the other hand, if \\(a_{n + 1} = a_{n} + 1\\) , then \\(S(a_{n + 1}) - S(a_{n})\\) is \\(a_{a_{n + 1}}\\) , so it follows that \\(a_{a_{n + 1}} = 2a_{n + 1} - 1\\) . By assumption, this can only happen finitely many times, so for all sufficiently large \\(n\\) we must have \\(a_{n + 1}\\geqslant a_{n} + 2\\) .  \n\nFor large enough \\(n\\) , we now know that \\(a_{n} > 2n - 1\\) implies \\(a_{n + 1} > (2n - 1) + 2 = 2(n + 1) - 1\\) . This means that there are two cases possible:  \n\n(A) For all sufficiently large \\(n\\) (say \\(n\\geqslant N_{A}\\) ) we have \\(a_{n} > 2n - 1\\)  \n\n(B) For all sufficiently large \\(n\\) (say \\(n\\geqslant N_{B}\\) ) we have \\(a_{n}< 2n - 1\\)  \n\nIn case (A), we know for \\(m > N_{A}\\) that  \n\n\\[S(m) = S(N_{A}) + \\sum_{i = N_{A} + 1}^{m}a_{i}\\geqslant S(N_{A}) + \\sum_{i = N_{A} + 1}^{n}2i = S(N_{A}) + m(m + 1) - N_{A}(N_{A} + 1)\\] \\[\\qquad = m^{2} + m + S(N_{A}) - N_{A}(N_{A} + 1).\\]  \n\nFor \\(m\\) large enough (e.g. \\(m > N_{A}(N_{A} + 1)\\) ), this expression is always larger than \\(m^{2}\\) , contradicting \\(S(a_{n}) = a_{n}^{2}\\) for all \\(n\\) .  \n\nSimilarly, in case (B), we similarly know for \\(m > N_{B}\\) that  \n\n\\[S(m) = S(N_{B}) + \\sum_{i = N_{B} + 1}^{m}a_{i}\\leqslant S(N_{B}) + \\sum_{i = N_{B} + 1}^{n}2(i - 1) = S(N_{B}) + m(m - 1) - N_{B}(N_{B} - 1)\\] \\[\\qquad = m^{2} - m + S(N_{B}) - N_{B}(N_{B} - 1).\\]  \n\nFor \\(m\\) large enough (e.g. \\(m > S(N_{B})\\) ), this expression is always smaller than \\(m^{2}\\) , again contradicting \\(S(a_{n}) = a_{n}^{2}\\) for all \\(n\\) .", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P2.", "solution_match": "\nSolution 2. "}}
 {"year": "2025", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "EGMO", "problem": "An infinite increasing sequence \\(a_{1}< a_{2}< a_{3}< \\dots\\) of positive integers is called central if for every positive integer \\(n\\) , the arithmetic mean of the first \\(a_{n}\\) terms of the sequence is equal to \\(a_{n}\\) .  \n\nShow that there exists an infinite sequence \\(b_{1},b_{2},b_{3},\\ldots\\) of positive integers such that for every central sequence \\(a_{1},a_{2},a_{3},\\ldots\\) , there are infinitely many positive integers \\(n\\) with \\(a_{n} = b_{n}\\) .", "solution": "We claim that the sequence \\(b_{1}\\) , \\(b_{2}\\) , \\(b_{3}\\) , ... defined by \\(b_{i} = 2i - 1\\) has this property.  \n\nLemma. If there are no terms \\(a_{j}\\) such that \\(a_{j} - a_{j - 1} = 1\\) , then \\(a_{j} = a_{j - 1} + 2\\) for all \\(j\\) .  \n\nProof. Let \\(c\\) be such that \\(a_{d} = c\\) for some \\(d\\) . Now  \n\n\\[a_{1} + a_{2} + \\dots +a_{c} = c^{2}.\\]  \n\nEquality holds for \\(a_{i} = 2i - 1\\) for \\(1\\leq i\\leq c\\) , so if any difference between two consecutive terms is greater, the left- hand side of the equation is greater than \\(c^{2}\\) , a contradiction. \\(\\square\\)  \n\nLemma. If both \\(d\\) and \\(d + 1\\) are terms of the sequence, i.e. \\(a_{c} = d\\) and \\(a_{c + 1} = d + 1\\) for some \\(c\\) , then \\(a_{d + 1} = 2d + 1 = b_{d + 1}\\) .  \n\nProof. We have \\(a_{1} + a_{2} + \\dots +a_{d} = d^{2}\\) and \\(a_{1} + a_{2} + \\dots +a_{d + 1} = (d + 1)^{2}\\) . Hence \\(a_{d + 1} = (d + 1)^{2} - d^{2} = 2d + 1\\) . \\(\\square\\)  \n\nFrom the observations above, we see that we are done if there are infinitely many gaps of size 1. The only remaining case is one with finitely many gaps of size 1. This will be the subject of the following lemma.  \n\nLemma. If there are only finitely many indices \\(j\\) such that \\(a_{j + 1} - a_{j} = 1\\) , then there is an index \\(n_{0}\\) such that for all \\(k > n_{0}\\) , we have \\(a_{k} = 2k - 1\\) .  \n\nProof. Let \\(r\\) and \\(s\\) be indices such that for all the \\(j\\) satisfying \\(a_{j + 1} - a_{j} = 1\\) , we have \\(j < r\\) , \\(s\\) . Furthermore, assume \\(s > r\\) and that there are \\(i_{1}\\) and \\(i_{2}\\) such that \\(a_{i_{1}} = r\\) and \\(a_{i_{2}} = s\\) . The first goal is to show that \\(a_{s} \\geq 2s - 1\\) . If \\(a_{r} \\geq 2r - 1\\) , this is clearly the case. Assume now \\(a_{r} < 2r - 1\\) . Now \\(a_{r} \\geq 2r - 1 - m\\) , where \\(m\\) is the number of indices \\(j\\) with \\(a_{j + 1} - a_{j} = 1\\) . Denote \\(a_{r + 1} = 2r + 1 - m + \\theta_{1}\\) , \\(a_{r + 2} = 2r + 3 - m + \\theta_{2}\\) , etc. Remember that \\(a_{r + j + 1} - a_{r + j} \\geq 2\\) always. Now \\(0 \\leq \\theta_{1} \\leq \\theta_{2} \\leq \\dots\\) . Furthermore, write \\(s = r + h\\) . Now  \n\n\\[(r + h)^{2} - r^{2} = a_{r + 1} + a_{r + 2} + \\dots +a_{r + h} = \\sum_{j = 1}^{h}2r - 1 + 2j - m + \\theta_{j}.\\]  \n\nFrom this we deduce  \n\n\\[2r h + h^{2} = 2r h - h - m h + h(h + 1) + \\sum_{j = 1}^{h}\\theta_{j}.\\]  \n\nSo we obtain \\(\\sum_{j = 1}^{h}\\theta_{j} = m h\\) . Since the sequence \\(\\theta_{j}\\) is increasing, we have \\(\\theta_{h} \\geq m\\) . Hence, \\(a_{s} = a_{r + h} \\geq 2r - 1 - m + 2h + m = 2r + 2h - 1 = 2s - 1\\) .  \n\nNow \\(a_{s}\\) is exactly the desired shape. If for any \\(t > s\\) , we have \\(a_{t} - a_{t - 1} > 2\\) , then  \n\n\\[a_{s} + a_{s + 1} + \\dots +a_{t} > t^{2} - s^{2},\\]  \n\nagain a contradiction.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P2.", "solution_match": "\nSolution 3. "}}
 {"year": "2025", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "EGMO", "problem": "An infinite increasing sequence \\(a_{1}< a_{2}< a_{3}< \\dots\\) of positive integers is called central if for every positive integer \\(n\\) , the arithmetic mean of the first \\(a_{n}\\) terms of the sequence is equal to \\(a_{n}\\) .  \n\nShow that there exists an infinite sequence \\(b_{1},b_{2},b_{3},\\ldots\\) of positive integers such that for every central sequence \\(a_{1},a_{2},a_{3},\\ldots\\) , there are infinitely many positive integers \\(n\\) with \\(a_{n} = b_{n}\\) .", "solution": "Note that \\(a_{1} = 1\\) because if it is not the case, then \\(a_{1}^{2} = a_{1} + \\dots +a_{a_{1}} > a_{1} + a_{1} + \\dots +a_{1} = a_{1}^{2}\\) .  \n\nAssume by contradiction that there are only finitely many indices \\(k\\) such that \\(a_{k} = 2k - 1\\) . Set \\(i\\) to be the largest integer such that \\(a_{i} = 2i - 1\\) (which must exist as \\(a_{1} = 1\\) ). Assume that there exists \\(j \\geqslant i\\) such that \\(a_{j + 1} - a_{j} = 1\\) . Then \\(2a_{j + 1} - 1 = a_{j + 1}^{2} - a_{j}^{2} = a_{a_{j + 1}}\\) and since \\(a_{k} \\geqslant k\\) for all \\(k\\) , we have \\(a_{j + 1} \\geqslant j + 1 > i\\) , which contradicts the definition of \\(i\\) . Thus for all \\(j \\geqslant i\\) , we have \\(a_{j + 1} \\geqslant a_{j} + 2\\) , which implies by induction that \\(a_{j} \\geqslant 2j - 1\\) for \\(j \\geqslant i\\) , and even \\(a_{j} \\geq 2j\\) if \\(j > i\\) .  \n\nThere are two ways to finish the solution from here.  \n\n## First way to finish the solution  \n\nFor all \\(n\\) such that \\(a_{n} \\geqslant i\\) , we have  \n\n\\[a_{n + 1}^{2} - a_{n}^{2} = a_{a_{n + 1}} + a_{a_{n + 1} - 1} + \\dots +a_{a_{n + 1}}\\geqslant 2a_{n + 1} + 2(a_{n + 1} - 1) + \\dots +2(a_{n} + 1)\\] \\[\\qquad = (a_{n + 1} - a_{n})(a_{n + 1} + a_{n} + 1)\\] \\[\\qquad >a_{n + 1}^{2} - a_{n}^{2}.\\]  \n\nThis gives a contradiction.  \n\n## Second way to finish the solution  \n\nFor all \\(n\\) such that \\(a_{n} \\geqslant i\\) , we introduce \\(x_{n} = a_{n + 1} - a_{n}\\) . We have  \n\n\\[x_{n}^{2} + 2x_{n}a_{n} = a_{n + 1}^{2} - a_{n}^{2} = a_{a_{n + 1}} + a_{a_{n + 1} - 1} + \\dots +a_{a_{n} + 1}\\geqslant \\sum_{j = 1}^{x_{n}}(a_{a_{n}} + 2j)\\geq x_{n}a_{a_{n}} + x_{n}(x_{n} + 1).\\]  \n\nBy simplifying, we get \\(a_{a_{n}} \\leq 2a_{n} - 1\\) , which gives a contradiction.  \n\nComment. Proving that \\(a_{1} = 1\\) is not necessary for this solution. If there exists no \\(i\\) such that \\(a_{i} = 2i - 1\\) , then the same argument implies that there exists no \\(j\\) such that \\(a_{j + 1} - a_{j} = 1\\) , thus \\(a_{j} \\geqslant 2j - 1\\) for \\(j \\geqslant 1\\) .", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P2.", "solution_match": "\nSolution 4. "}}
-{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Denote by \\(B^{\\prime}\\) and \\(C^{\\prime}\\) the reflections of \\(B\\) and \\(C\\) in \\(M\\) and \\(N\\) , respectively. Points \\(C^{\\prime}\\) , \\(A\\) , \\(B^{\\prime}\\) are clearly collinear and \\(D E B^{\\prime}A\\) is a parallelogram. Since \\(E H\\perp A D\\) , we have \\(E H\\perp E B^{\\prime}\\) . Also \\(H A\\perp A B^{\\prime}\\) , so points \\(H,E,B^{\\prime},A\\) are concyclic. This gives  \n\n\\[\\angle C^{\\prime}Q H = \\angle N Q H = \\angle N E H = \\angle A E H = \\angle A B^{\\prime}H = \\angle C^{\\prime}B^{\\prime}H,\\]  \n\nand so points \\(C^{\\prime}\\) , \\(B^{\\prime}\\) , \\(Q\\) , \\(H\\) are concyclic. Analogously points \\(C^{\\prime}\\) , \\(B^{\\prime}\\) , \\(P\\) , \\(H\\) are concyclic, and so all points \\(B^{\\prime}\\) , \\(C^{\\prime}\\) , \\(P\\) , \\(Q\\) , \\(H\\) are. Now we have  \n\n\\[\\angle N M B^{\\prime} = \\angle A B^{\\prime}M = \\angle C^{\\prime}B^{\\prime}P = \\angle C^{\\prime}Q P = \\angle N Q P,\\]  \n\nwhich proves that \\(P,Q,N,M\\) are also concyclic.  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)", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 1. "}}
+{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Denote by \\(B^{\\prime}\\) and \\(C^{\\prime}\\) the reflections of \\(B\\) and \\(C\\) in \\(M\\) and \\(N\\) , respectively. Points \\(C^{\\prime}\\) , \\(A\\) , \\(B^{\\prime}\\) are clearly collinear and \\(D E B^{\\prime}A\\) is a parallelogram. Since \\(E H\\perp A D\\) , we have \\(E H\\perp E B^{\\prime}\\) . Also \\(H A\\perp A B^{\\prime}\\) , so points \\(H,E,B^{\\prime},A\\) are concyclic. This gives  \n\n\\[\\angle C^{\\prime}Q H = \\angle N Q H = \\angle N E H = \\angle A E H = \\angle A B^{\\prime}H = \\angle C^{\\prime}B^{\\prime}H,\\]  \n\nand so points \\(C^{\\prime}\\) , \\(B^{\\prime}\\) , \\(Q\\) , \\(H\\) are concyclic. Analogously points \\(C^{\\prime}\\) , \\(B^{\\prime}\\) , \\(P\\) , \\(H\\) are concyclic, and so all points \\(B^{\\prime}\\) , \\(C^{\\prime}\\) , \\(P\\) , \\(Q\\) , \\(H\\) are. Now we have  \n\n\\[\\angle N M B^{\\prime} = \\angle A B^{\\prime}M = \\angle C^{\\prime}B^{\\prime}P = \\angle C^{\\prime}Q P = \\angle N Q P,\\]  \n\nwhich proves that \\(P,Q,N,M\\) are also concyclic.  \n\n![md5:ed729686b0882a22087a6fc0e270efb8](ed729686b0882a22087a6fc0e270efb8.jpeg)", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 1. "}}
 {"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Introduce points \\(B^{\\prime}\\) , \\(C^{\\prime}\\) as above. Also define \\(A^{\\prime} = B^{\\prime}E\\cap C^{\\prime}D\\) , so that \\(E A D A^{\\prime}\\) is a parallelogram and \\(A D E\\) is the medial triangle of \\(A^{\\prime}B^{\\prime}C^{\\prime}\\) . It that follows that the orthocentre \\(H\\) of \\(A D E\\) is the circumcentre of \\(A^{\\prime}B^{\\prime}C^{\\prime}\\) , and in particular  \n\n\\[\\angle C^{\\prime}B^{\\prime}H = 90^{\\circ} - \\angle B^{\\prime}A^{\\prime}C^{\\prime} = 90^{\\circ} - \\angle D A E = \\angle A E H = \\angle N E H = \\angle N Q H = \\angle C^{\\prime}Q H.\\]  \n\nSo again we have that \\(C^{\\prime}\\) , \\(B^{\\prime}\\) , \\(Q\\) , \\(H\\) are concyclic and conclude as in Solution 1.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 1'."}}
-{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Let \\(X\\) be the second intersection of \\((D H M)\\) and \\((E H N)\\) and let \\(O^{\\prime}\\) be the circumcentre of \\((A M N)\\) Note that \\(\\angle M X N = \\angle M D H + \\angle N E H = 180^{\\circ} - 2\\angle D A E\\) and since \\(\\angle M O^{\\prime}N = 2\\angle D A E\\) we have that \\(X,M,O^{\\prime},N\\) is cyclic and since \\(\\angle M X H = \\angle N X H\\) it means that \\(H X\\) is the angle bisector of \\(\\angle M X N\\) but since \\(O^{\\prime}M = O^{\\prime}N\\) it means that \\(H,X,O^{\\prime}\\) are collinear. Let \\(B M\\) and \\(C N\\) intersect at \\(T\\) and let \\(K\\) and \\(L\\) be the midpoints of \\(M N\\) and \\(B C\\) . Note that \\(L\\) is also the midpoint of \\(D E\\) . Since \\(M N\\) is parallel to \\(B C\\) it means that \\(T\\) , \\(K\\) , and \\(L\\) are collinear, but since \\(A\\) , \\(K\\) , and \\(L\\) are collinear we get that \\(A\\) , \\(T\\) , \\(K\\) , and \\(L\\) are collinear. Now, \\(\\frac{T L}{T K} = \\frac{B C}{M N} = 6\\) . Since \\(K L = K A\\) it means that \\(\\frac{A T}{T K} = \\frac{T L - 2T K}{T K} = 4\\) so by the lemma below, \\(T\\) lies on \\(H O^{\\prime}\\) . Since \\(H O^{\\prime}\\) is the radical axis of \\((D H M)\\) and \\((E H N)\\) we finish the problem using the Radical Axes Theorem \\((T M\\cdot T P = T N\\cdot T Q)\\) .\n\n\n\nLemma. Let \\(A^{\\prime}\\) be the reflection of \\(A\\) around the orthocentre \\(H\\) of \\(\\triangle ABC\\) and \\(O\\) and \\(M\\) be the circumcentre of \\(\\triangle ABC\\) and the midpoint of \\(BC\\) , respectively. Let \\(T\\) be the intersection of \\(A^{\\prime}O\\) and \\(AM\\) . Then \\(\\frac{AT}{TM} = 4\\) .  \n\nProof. Since \\(OM \\parallel AA'\\) we have \\(\\frac{AT}{TM} = \\frac{AA'}{OM} = \\frac{2AH}{OM} = \\frac{4OM}{OM} = 4\\) . We used here that \\(AH = 2OM\\) . \\(\\square\\)  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"metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 2. "}}
-{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "As in solution 2, we will prove that \\(O'\\) is both on line \\(HT\\) and the radical axis of the circles, hence \\(T\\) is on the radical axis, from which we conclude the required concyclicity. We present alternative proofs of both facts, discovered by contestants.  \n\nLet \\(M_{1}\\) , \\(N_{1}\\) be the midpoints of \\(AM\\) , \\(AN\\) , respectively, so that \\(AM_{1}:M_{1}D = AN_{1}:N_{1}E = 1:3\\) . It is easy to verify (e.g. by applying the theorems of Ceva or Menelaus, or by computing in barycentric coordinates as in Solution 3) that \\(T\\) lies on \\(EM_{1}\\) and \\(DN_{1}\\) . Note that \\(M_{1}N_{1} \\parallel DE\\) , and also \\(M_{1}O' \\parallel HE\\) as both are perpendicular to \\(AMD\\) , and similarly \\(N_{1}O' \\parallel HD\\) . It follows that \\(DEH\\) and \\(N_{1}M_{1}O'\\) are homothetically similar triangles, and the center of their (negative) homothety is \\(T = DN_{1} \\cap EM_{1}\\) . Therefore \\(T\\) also lies on \\(HO'\\) , as claimed. (We also have that the homothety is by factor \\(\\frac{M_{1}N_{1}}{ED} = -\\frac{1}{4}\\) .)  \n\nNow, let \\(M_{2}\\) , \\(N_{2}\\) be the second intersection points of \\(O'M\\) , \\(O'N\\) with the circumcircles of \\(HMD\\) , \\(HNE\\) , respectively. To prove that \\(O'\\) is on the radical axis, it suffices to show that \\(O'M \\cdot O'M_{2} = O'N \\cdot O'N_{2}\\) . But by definition of \\(O'\\) we have \\(O'M = O'N\\) , so we must show \\(O'M_{2} = O'N_{2}\\) , which is equivalent to \\(M_{2}N_{2} \\parallel MN\\) . Angle chasing in circle \\(MM_{2}DH\\) gives  \n\n\\[\\angle OM_{2}H = \\angle MM_{2}H = \\angle MDH = \\angle ADH = 90^{\\circ} - \\angle EAD = 90^{\\circ} - \\angle NAM = \\angle O'M N,\\]  \n\nfrom which it follows that \\(M_{2}H \\parallel MN\\) . Similarly, we have \\(N_{2}H \\parallel MN\\) , and the two facts together imply that \\(M_{2}, H, N_{2}\\) are collinear and the line through them is parallel to \\(MN\\) , as claimed. 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"metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 2'."}}
-{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "We compute using linear combinations with respect to \\(A D E\\) . We have \\(B = 2D - E\\) \\(C = 2E - D\\) \\(M = \\frac{A + D}{2}\\) , and \\(N = \\frac{A + E}{2}\\) , from which we immediately obtain that the intersection of \\(T = B M\\cap C N\\) is \\(T = \\frac{3A + D + E}{5} = \\frac{6M - B}{5} = \\frac{6N - C}{5}\\) . As in solution 2, we reduce to showing that \\(T\\) is on the radical axis of \\(H D M\\) and \\(H E N\\) , whence \\(T M\\cdot T P = T N\\cdot T Q\\) proves the concyclicity of \\(P,Q,N,M\\)  \n\nSynthetic finish, similar to Solution 2. Let \\(O\\) be the circumcentre of \\(A D E\\) and \\(O^{\\prime} = \\frac{A + O}{2}\\) be the circumcentre of \\(A M N\\) . As in Solution 2, we have that \\(O^{\\prime}\\) is on the desired radical axis, so it is enough to show \\(T\\in O^{\\prime}H\\) . Let \\(G = \\frac{A + D + E}{3}\\) be the barycentre of \\(A D E\\) . By properties of the Euler line, we also have \\(G = \\frac{H + 2O}{3}\\) . Now using our known identities we find  \n\n\\[T = \\frac{3A + D + E}{5} = \\frac{2A + 3G}{5} = \\frac{2A + H + 2O}{5} = \\frac{H + 4O^{\\prime}}{5}\\]  \n\nand in particular \\(T\\in H O^{\\prime}\\) , as we wanted to show.  \n\nComputational finish. Let \\(f(T)\\) be the power difference at \\(T\\) w.r.t. \\(D H M\\) and \\(E H N\\) . We compute \\(f(A)\\) and \\(f(L)\\) where \\(L = \\frac{D + E}{2}\\) . Since \\(T = \\frac{3A + 2L}{5}\\) , it is enough to show that \\(3P(A) + 2P(L) = 0\\) . In the following all lengths are directed. We compute trigonometrically: Let \\(\\alpha ,\\beta ,\\gamma\\) be the angles of \\(A D E\\) and assume the diameter of its circumcircle is 1. We have  \n\n\\[f(A) = \\frac{A D^{2} - A E^{2}}{2} = \\frac{\\sin^{2}(\\gamma) - \\sin^{2}(\\beta)}{2}.\\]  \n\nTo compute \\(P(L)\\) , let \\(D^{\\prime},E^{\\prime}\\) be the second intersection points of \\(H M D,H N E\\) with \\(D E\\) , and let \\(M^{\\prime},N^{\\prime},F\\) be the feet of the perpendiculars from \\(H\\) to \\(A D,A E,D E\\) , respectively. Note that \\(D M^{\\prime} =\\) \\(\\sin \\alpha \\cos \\beta\\) , thus  \n\n\\[M^{\\prime}M = D M - D M^{\\prime} = \\frac{\\sin(\\alpha + \\beta)}{2} - \\sin \\alpha \\cos \\beta = \\frac{\\sin(\\beta - \\alpha)}{2}.\\]  \n\nWe also have \\(H M^{\\prime} = \\cos \\alpha \\cos \\beta\\) , \\(H F = \\cos \\beta \\cos \\gamma\\) , and \\(H M^{\\prime}M\\sim H F D^{\\prime}\\) , therefore  \n\n\\[F D^{\\prime} = \\frac{H F}{H M^{\\prime}} M^{\\prime}M = \\frac{\\cos\\gamma\\sin(\\beta - \\alpha)}{\\cos\\alpha}\\]  \n\nand similarly  \n\n\\[E^{\\prime}F = \\frac{\\cos\\beta\\sin(\\gamma - \\alpha)}{\\cos\\alpha}.\\]\n\n\n\nWe also have the standard \\(FD = \\sin \\gamma \\cos \\beta\\) and \\(EF = \\sin \\beta \\cos \\gamma\\) . We can finally compute  \n\n\\[f(L) = LD\\cdot LD^{\\prime} - LE\\cdot LE^{\\prime} = \\frac{\\sin\\alpha}{2} (LD^{\\prime} + LE^{\\prime}) = \\frac{\\sin\\alpha}{2} (FD^{\\prime} + FE^{\\prime} - FD - FE)\\] \\[\\qquad = \\frac{\\sin\\alpha}{4\\cos\\alpha} (\\cos \\gamma \\sin (\\beta -\\alpha) - \\cos \\beta \\sin (\\gamma -\\alpha) - 2\\cos \\alpha (\\sin \\gamma \\cos \\beta -\\sin \\beta \\cos \\gamma))\\] \\[\\qquad = \\frac{3\\sin\\alpha\\sin(\\beta - \\gamma)}{4} = \\frac{3}{4} (\\sin (\\beta +\\gamma)\\sin (\\beta -\\gamma)) = \\frac{3}{8} (\\cos (2\\gamma) - \\cos (2\\beta))\\] \\[\\qquad = \\frac{3}{4} (\\sin^{2}(\\beta) - \\sin^{2}(\\gamma)) = -\\frac{3}{2} f(A).\\quad \\square\\]  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"metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 3. "}}
+{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Let \\(X\\) be the second intersection of \\((D H M)\\) and \\((E H N)\\) and let \\(O^{\\prime}\\) be the circumcentre of \\((A M N)\\) Note that \\(\\angle M X N = \\angle M D H + \\angle N E H = 180^{\\circ} - 2\\angle D A E\\) and since \\(\\angle M O^{\\prime}N = 2\\angle D A E\\) we have that \\(X,M,O^{\\prime},N\\) is cyclic and since \\(\\angle M X H = \\angle N X H\\) it means that \\(H X\\) is the angle bisector of \\(\\angle M X N\\) but since \\(O^{\\prime}M = O^{\\prime}N\\) it means that \\(H,X,O^{\\prime}\\) are collinear. Let \\(B M\\) and \\(C N\\) intersect at \\(T\\) and let \\(K\\) and \\(L\\) be the midpoints of \\(M N\\) and \\(B C\\) . Note that \\(L\\) is also the midpoint of \\(D E\\) . Since \\(M N\\) is parallel to \\(B C\\) it means that \\(T\\) , \\(K\\) , and \\(L\\) are collinear, but since \\(A\\) , \\(K\\) , and \\(L\\) are collinear we get that \\(A\\) , \\(T\\) , \\(K\\) , and \\(L\\) are collinear. Now, \\(\\frac{T L}{T K} = \\frac{B C}{M N} = 6\\) . Since \\(K L = K A\\) it means that \\(\\frac{A T}{T K} = \\frac{T L - 2T K}{T K} = 4\\) so by the lemma below, \\(T\\) lies on \\(H O^{\\prime}\\) . Since \\(H O^{\\prime}\\) is the radical axis of \\((D H M)\\) and \\((E H N)\\) we finish the problem using the Radical Axes Theorem \\((T M\\cdot T P = T N\\cdot T Q)\\) .\n\n\n\nLemma. Let \\(A^{\\prime}\\) be the reflection of \\(A\\) around the orthocentre \\(H\\) of \\(\\triangle ABC\\) and \\(O\\) and \\(M\\) be the circumcentre of \\(\\triangle ABC\\) and the midpoint of \\(BC\\) , respectively. Let \\(T\\) be the intersection of \\(A^{\\prime}O\\) and \\(AM\\) . Then \\(\\frac{AT}{TM} = 4\\) .  \n\nProof. Since \\(OM \\parallel AA'\\) we have \\(\\frac{AT}{TM} = \\frac{AA'}{OM} = \\frac{2AH}{OM} = \\frac{4OM}{OM} = 4\\) . We used here that \\(AH = 2OM\\) . \\(\\square\\)  \n\n![md5:33c928cfe95e89ef9ab049d5028e7524](33c928cfe95e89ef9ab049d5028e7524.jpeg)", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 2. "}}
+{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "As in solution 2, we will prove that \\(O'\\) is both on line \\(HT\\) and the radical axis of the circles, hence \\(T\\) is on the radical axis, from which we conclude the required concyclicity. We present alternative proofs of both facts, discovered by contestants.  \n\nLet \\(M_{1}\\) , \\(N_{1}\\) be the midpoints of \\(AM\\) , \\(AN\\) , respectively, so that \\(AM_{1}:M_{1}D = AN_{1}:N_{1}E = 1:3\\) . It is easy to verify (e.g. by applying the theorems of Ceva or Menelaus, or by computing in barycentric coordinates as in Solution 3) that \\(T\\) lies on \\(EM_{1}\\) and \\(DN_{1}\\) . Note that \\(M_{1}N_{1} \\parallel DE\\) , and also \\(M_{1}O' \\parallel HE\\) as both are perpendicular to \\(AMD\\) , and similarly \\(N_{1}O' \\parallel HD\\) . It follows that \\(DEH\\) and \\(N_{1}M_{1}O'\\) are homothetically similar triangles, and the center of their (negative) homothety is \\(T = DN_{1} \\cap EM_{1}\\) . Therefore \\(T\\) also lies on \\(HO'\\) , as claimed. (We also have that the homothety is by factor \\(\\frac{M_{1}N_{1}}{ED} = -\\frac{1}{4}\\) .)  \n\nNow, let \\(M_{2}\\) , \\(N_{2}\\) be the second intersection points of \\(O'M\\) , \\(O'N\\) with the circumcircles of \\(HMD\\) , \\(HNE\\) , respectively. To prove that \\(O'\\) is on the radical axis, it suffices to show that \\(O'M \\cdot O'M_{2} = O'N \\cdot O'N_{2}\\) . But by definition of \\(O'\\) we have \\(O'M = O'N\\) , so we must show \\(O'M_{2} = O'N_{2}\\) , which is equivalent to \\(M_{2}N_{2} \\parallel MN\\) . Angle chasing in circle \\(MM_{2}DH\\) gives  \n\n\\[\\angle OM_{2}H = \\angle MM_{2}H = \\angle MDH = \\angle ADH = 90^{\\circ} - \\angle EAD = 90^{\\circ} - \\angle NAM = \\angle O'M N,\\]  \n\nfrom which it follows that \\(M_{2}H \\parallel MN\\) . Similarly, we have \\(N_{2}H \\parallel MN\\) , and the two facts together imply that \\(M_{2}, H, N_{2}\\) are collinear and the line through them is parallel to \\(MN\\) , as claimed. \\(\\square\\)\n\n\n![md5:a1349e2b0d6dbddf473be279a244f481](a1349e2b0d6dbddf473be279a244f481.jpeg)", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 2'."}}
+{"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "We compute using linear combinations with respect to \\(A D E\\) . We have \\(B = 2D - E\\) \\(C = 2E - D\\) \\(M = \\frac{A + D}{2}\\) , and \\(N = \\frac{A + E}{2}\\) , from which we immediately obtain that the intersection of \\(T = B M\\cap C N\\) is \\(T = \\frac{3A + D + E}{5} = \\frac{6M - B}{5} = \\frac{6N - C}{5}\\) . As in solution 2, we reduce to showing that \\(T\\) is on the radical axis of \\(H D M\\) and \\(H E N\\) , whence \\(T M\\cdot T P = T N\\cdot T Q\\) proves the concyclicity of \\(P,Q,N,M\\)  \n\nSynthetic finish, similar to Solution 2. Let \\(O\\) be the circumcentre of \\(A D E\\) and \\(O^{\\prime} = \\frac{A + O}{2}\\) be the circumcentre of \\(A M N\\) . As in Solution 2, we have that \\(O^{\\prime}\\) is on the desired radical axis, so it is enough to show \\(T\\in O^{\\prime}H\\) . Let \\(G = \\frac{A + D + E}{3}\\) be the barycentre of \\(A D E\\) . By properties of the Euler line, we also have \\(G = \\frac{H + 2O}{3}\\) . Now using our known identities we find  \n\n\\[T = \\frac{3A + D + E}{5} = \\frac{2A + 3G}{5} = \\frac{2A + H + 2O}{5} = \\frac{H + 4O^{\\prime}}{5}\\]  \n\nand in particular \\(T\\in H O^{\\prime}\\) , as we wanted to show.  \n\nComputational finish. Let \\(f(T)\\) be the power difference at \\(T\\) w.r.t. \\(D H M\\) and \\(E H N\\) . We compute \\(f(A)\\) and \\(f(L)\\) where \\(L = \\frac{D + E}{2}\\) . Since \\(T = \\frac{3A + 2L}{5}\\) , it is enough to show that \\(3P(A) + 2P(L) = 0\\) . In the following all lengths are directed. We compute trigonometrically: Let \\(\\alpha ,\\beta ,\\gamma\\) be the angles of \\(A D E\\) and assume the diameter of its circumcircle is 1. We have  \n\n\\[f(A) = \\frac{A D^{2} - A E^{2}}{2} = \\frac{\\sin^{2}(\\gamma) - \\sin^{2}(\\beta)}{2}.\\]  \n\nTo compute \\(P(L)\\) , let \\(D^{\\prime},E^{\\prime}\\) be the second intersection points of \\(H M D,H N E\\) with \\(D E\\) , and let \\(M^{\\prime},N^{\\prime},F\\) be the feet of the perpendiculars from \\(H\\) to \\(A D,A E,D E\\) , respectively. Note that \\(D M^{\\prime} =\\) \\(\\sin \\alpha \\cos \\beta\\) , thus  \n\n\\[M^{\\prime}M = D M - D M^{\\prime} = \\frac{\\sin(\\alpha + \\beta)}{2} - \\sin \\alpha \\cos \\beta = \\frac{\\sin(\\beta - \\alpha)}{2}.\\]  \n\nWe also have \\(H M^{\\prime} = \\cos \\alpha \\cos \\beta\\) , \\(H F = \\cos \\beta \\cos \\gamma\\) , and \\(H M^{\\prime}M\\sim H F D^{\\prime}\\) , therefore  \n\n\\[F D^{\\prime} = \\frac{H F}{H M^{\\prime}} M^{\\prime}M = \\frac{\\cos\\gamma\\sin(\\beta - \\alpha)}{\\cos\\alpha}\\]  \n\nand similarly  \n\n\\[E^{\\prime}F = \\frac{\\cos\\beta\\sin(\\gamma - \\alpha)}{\\cos\\alpha}.\\]\n\n\n\nWe also have the standard \\(FD = \\sin \\gamma \\cos \\beta\\) and \\(EF = \\sin \\beta \\cos \\gamma\\) . We can finally compute  \n\n\\[f(L) = LD\\cdot LD^{\\prime} - LE\\cdot LE^{\\prime} = \\frac{\\sin\\alpha}{2} (LD^{\\prime} + LE^{\\prime}) = \\frac{\\sin\\alpha}{2} (FD^{\\prime} + FE^{\\prime} - FD - FE)\\] \\[\\qquad = \\frac{\\sin\\alpha}{4\\cos\\alpha} (\\cos \\gamma \\sin (\\beta -\\alpha) - \\cos \\beta \\sin (\\gamma -\\alpha) - 2\\cos \\alpha (\\sin \\gamma \\cos \\beta -\\sin \\beta \\cos \\gamma))\\] \\[\\qquad = \\frac{3\\sin\\alpha\\sin(\\beta - \\gamma)}{4} = \\frac{3}{4} (\\sin (\\beta +\\gamma)\\sin (\\beta -\\gamma)) = \\frac{3}{8} (\\cos (2\\gamma) - \\cos (2\\beta))\\] \\[\\qquad = \\frac{3}{4} (\\sin^{2}(\\beta) - \\sin^{2}(\\gamma)) = -\\frac{3}{2} f(A).\\quad \\square\\]  \n\n![md5:74efbddd0a29cbc1175d0f3b5702823a](74efbddd0a29cbc1175d0f3b5702823a.jpeg)", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 3. "}}
 {"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "Let \\(T = BM\\cap CN\\) , let \\(F\\) be the foot of the altitude from \\(A\\) to \\(BC\\) , let \\(O\\) be the circumcentre of \\((ADE)\\) , let \\(D^{\\prime}\\neq D\\) and \\(E^{\\prime}\\neq E\\) be the second intersections of \\((DHMP)\\) and \\((EHNQ)\\) with line \\(BC\\) and let \\(U\\) and \\(V\\) be the antipodes of \\(D\\) and \\(E\\) on \\((DHMP)\\) and \\((EHNQ)\\) , respectively.  \n\nWe begin with a bit of barycentric coordinates. Set barycentric coordinates in \\(\\triangle ABC\\) , set so that \\(A = (1,0,0)\\) , \\(B = (0,1,0)\\) , and \\(C = (0,0,1)\\) . The definitions of \\(D\\) and \\(E\\) give \\(D = (0,2 / 3,1 / 3)\\) and \\(E = (0,1 / 3,2 / 3)\\) , whence \\(M = (1 / 2,1 / 3,1 / 6)\\) and \\(N = (1 / 2,1 / 6,1 / 3)\\) . This means that line \\(BM\\) is given by \\((1 / 2:y:1 / 6)\\) while line \\(CN\\) is given by \\((1 / 2:1 / 6:z)\\) . So their intersection \\(T\\) is \\((1 / 2:1 / 6:1 / 6) = (3:1:1)\\) , giving \\(T = \\frac{3A + B + C}{5} = \\frac{3A + D + E}{5}\\) .  \n\nOur next tool is the linearity of the power of a point. Let \\(f:\\mathbb{R}^{2}\\to \\mathbb{R}\\) be defined by  \n\n\\[f(Z) = \\mathrm{Pow}_{(DHMP)}(Z) - \\mathrm{Pow}_{(EHNQ)}(Z).\\]  \n\nIt suffices to show that \\(f(T) = 0\\) ; from there, the required concyclicity will follow from \\(TM\\cdot TP =\\) \\(TN\\cdot TQ\\) . The key observation is that \\(f\\) is a linear function. In particular, \\(f(T) = \\frac{3f(A) + f(D) + f(E)}{5}\\) So, we need only show that \\(3f(A) + f(D) + f(E) = 0\\) . Pick an arbitrary one- dimensional coordinate system on the line \\(BC\\) and let \\(\\tau\\) be the map from a point on \\(BC\\) to its coordinate. We compute  \n\n\\[f(A) = AM\\cdot AD - AN\\cdot AE = \\frac{AD^2 - AE^2}{2} = \\frac{FD^2 - FE^2}{2}\\] \\[\\qquad = (\\tau (E) - \\tau (D))\\left(\\tau (F) - \\tau \\left(\\frac{D + E}{2}\\right)\\right),\\] \\[f(D) = -DE\\cdot DE^{\\prime} = (\\tau (E) - \\tau (D))(\\tau (D) - \\tau (E^{\\prime})),\\] \\[f(E) = ED\\cdot ED^{\\prime} = (\\tau (E) - \\tau (D))(\\tau (E) - \\tau (D^{\\prime})).\\]  \n\nRearranging, it suffices to show that \\(3\\tau (F) = \\tau (D^{\\prime}) + \\tau (E^{\\prime}) + \\tau ((D + E) / 2)\\) . This can be rewritten as \\(3F = D^{\\prime} + E^{\\prime} + (D + E) / 2\\) . By projecting down to line \\(BC\\) , it suffices to show that the displacement vector \\(H + 2A - (O + U + V)\\) is perpendicular to line \\(BC\\) .\n\n\n\nWe do this using complex numbers. Let \\((A D E)\\) be the unit circle with \\(A = a\\) , \\(D = d\\) , and \\(E = e\\) , so that \\(O = 0\\) and \\(H = h = a + d + e\\) . Note that \\(U\\) satisfies \\(U M\\perp A D\\) and \\(U H\\perp D H\\perp A E\\) . Translating these conditions into equations, we have \\(u = a d\\overline{{u}}\\) and \\(u + a e\\overline{{u}} = h + a e\\overline{{h}}\\) . Rearranging gives  \n\n\\[(d + e)u = d u + e(a d\\overline{{u}}) = d(h + a e\\overline{{h}}) \\Rightarrow u = \\frac{d h + a d e\\overline{{h}}}{d + e}.\\]  \n\nComputing \\(v\\) similarly, we get that  \n\n\\[v:= H + 2A - (O + U + V) = h + 2a - \\frac{(d + e)h + 2a d e\\overline{{h}}}{d + e} = 2\\left(a - \\frac{a d e\\overline{{h}}}{d + e}\\right) = -\\frac{2d e}{d + e}.\\]  \n\nThis displacement vector \\(v\\) satisfies \\(v = d e\\overline{{v}}\\) and so it is orthogonal to line \\(D E\\) , as desired.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 4. "}}
 {"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "This solution uses almost exclusively complex numbers. As in other solutions, we reduce to showing that \\(H\\) , \\(S:= (D H M)\\cap (E H N)\\) , and \\(T:= B M\\cap C N\\) are collinear; this is all of the synthetic information we shall use. (If one computes \\(T = \\frac{3A + D + E}{5}\\) using means other than complex numbers, the solution becomes much shorter.)  \n\nWe use complex numbers with \\(A = a\\) , \\(D = d\\) , and \\(E = e\\) on the unit circle. Note that \\(H = a + d + e\\) , \\(M = \\frac{a + d}{2}\\) , and \\(B = 2d - e\\) . We will make great use of the \"arbitrary line intersection formula,\" which says that the intersection between lines \\(W X\\) and \\(Y Z\\) can be written explicitly as  \n\n\\[\\frac{(\\overline{{u}} x - w\\overline{{x}})(y - z) - (w - x)(\\overline{{y}} z - y\\overline{{z}})}{(\\overline{{w}} - \\overline{{x}})(y - z) - (w - x)(\\overline{{y}} - \\overline{{z}})}.\\]  \n\nWe first use this to find \\(T = t\\) . We compute  \n\n\\[b - m = (2d - e) - \\frac{a + d}{2} = \\frac{3d - a - 2e}{2}\\] \\[\\overline{{b}} - m = \\frac{3a e - d e - 2a d}{2a d e}\\] \\[\\overline{{b}} m - b\\overline{{m}} = \\frac{2e - d}{d e}\\cdot \\frac{a + d}{2} -(2d - e)\\cdot \\frac{a + d}{2a d}\\] \\[\\qquad = \\frac{a(a + d)(2e - d) - (2d - e)(a + d)e}{2a d e} = \\frac{(a + d)(2a e + e^{2} - a d - 2d e)}{2a d e}.\\]  \n\nIf \\(\\mathcal{E}\\) is some expression, we use the notation \\(\\mathcal{E} - \\{\\sim \\}\\) to denote \\(\\mathcal{E}\\) minus the expression formed by swapping \\(d\\) and \\(e\\) in \\(\\mathcal{E}\\) . We now have  \n\n\\[t = \\frac{(\\overline{{b}}m - b\\overline{{m}})(c - n) - \\{\\sim\\}}{(\\overline{{b}} - \\overline{{m}})(c - n) - \\{\\sim\\}}\\] \\[\\quad = \\frac{(a + d)(2a e + e^{2} - a d - 2d e)(3e - a - 2d) - \\{\\sim\\}}{(3a e - d e - 2a d)(3e - a - 2d) - \\{\\sim\\}}\\] \\[\\quad = \\frac{[a^{3}(d - 2e) + a^{2}(3d^{2} - 7d e + 5e^{2}) + a(2d^{3} - d^{2}e - 3d e^{2} + 3e^{3}) + d e(4d^{2} - 8d e + 3e^{2})] - \\{\\sim\\}}{[a^{2}(2d - 3e) + a(4d^{2} - 11d e + 9e^{2}) + d e(2d - 3e)] - \\{\\sim\\}}\\] \\[\\quad = \\frac{a^{3}(3(d - e)) - a^{2}(2(d^{2} - e^{2})) + a(-(d^{3} - e^{3}) + 2d e(d - e)) + d e(d^{2} - e^{2})}{a^{2}(5(d - e)) - a(5(d^{2} - e^{2})) + d e(5(d - e))}\\] \\[\\quad = \\frac{3a^{3} - 2a^{2}(d + e) - a(d^{2} - d e + e^{2}) + d e(d + e)}{5(a^{2} - a(d + e) + d e)}\\] \\[\\quad = \\frac{(a - d)(a - e)(3a + d + e)}{5(a - d)(a - e)} = \\frac{3a + d + e}{5}.\\]  \n\n(The factorization in the last line can be motivated by noting that the expression, while cubic in \\(a\\) , is only quadratic in \\(d\\) . When written out as a polynomial in \\(d\\) , each coefficient is divisible by \\(a - e\\) ; by symmetry, the numerator is divisible by \\(a - d\\) as well, and the factorization follows.)\n\n\n\nComputing \\(S = s\\) is slightly harder, as it is the intersection of two circles rather than of two lines. We get around this by noting that \\(\\{h,d,m,s\\}\\) are concyclic if and only if \\(\\{0,h - d,h - m,h - s\\}\\) are concyclic, which happens if and only if \\(\\{\\frac{1}{h - d},\\frac{1}{h - m},\\frac{1}{h - s}\\}\\) are collinear. (One can see this by inversion, or just by writing out the cross- ratio in the special case when one of the points is zero.) Thus \\(\\frac{1}{h - s}\\) is the intersection of the line through \\(w:= \\frac{1}{h - d}\\) and \\(x:= \\frac{1}{h - m}\\) and the line through \\(y:= \\frac{1}{h - e}\\) and \\(z:= \\frac{1}{h - n}\\) . We compute  \n\n\\[w - x = \\frac{1}{a + e} -\\frac{1}{\\frac{a + d}{2} + e} = -\\frac{a - d}{(a + e)(a + d + 2e)}\\] \\[\\overline{w - x} = \\frac{ae^{2}(a - d)}{(a + e)(2ad + ae + de)}\\] \\[\\overline{w} x - w\\overline{x} = \\frac{ae}{a + e}\\cdot \\frac{2}{a + d + 2e} -\\frac{1}{a + e}\\cdot \\frac{2ade}{2ad + ae + de}\\] \\[\\qquad = 2ae\\frac{(2ad + ae + de) - d(a + d + 2e)}{(a + e)(a + d + 2e)(2ad + ae + de)}\\] \\[\\qquad = \\frac{2ae(a - d)(d + e)}{(a + e)(a + d + 2e)(2ad + ae + de)}.\\]  \n\nUsing the line intersection formula, we have  \n\n\\[\\frac{1}{h - s} = \\frac{(\\overline{w} x - w\\overline{x})(y - z) - \\{\\sim\\}}{(\\overline{w} - \\overline{x})(y - z) - \\{\\sim\\}}\\] \\[= \\frac{[\\frac{2ae(a - d)(d + e)}{(a + e)(a + d + 2e)(2ad + ae + de)}] - \\{\\sim\\}}{[\\frac{ae^{2}(a - d)}{(a + e)(2ad + ae + de)}] - \\{\\sim\\}}\\] \\[= 2(d + e)\\frac{[e(ad + 2ae + de)] - \\{\\sim\\}}{[e^{2}(a + d + 2e)(ad + 2ae + de)] - \\{\\sim\\}}\\] \\[= 2(d + e)\\frac{[a(de + 2e^{2}) + de^{2}] - \\{\\sim\\}}{[a^{2}(de^{2} + 2e^{3}) + a(d^{2}e^{2} + 5de^{3} + 4e^{4}) + de(de^{2} + 2e^{3})] - \\{\\sim\\}}\\] \\[= \\frac{2(d + e)(a(2(e^{2} - d^{2})) + de(e - d))}{(a^{2} + de)(2(e^{3} - d^{3}) + de(e - d)) + a(4(e^{4} - d^{4}) + 5de(e^{2} - e^{2}))}\\] \\[= \\frac{2(d + e)(2a(d + e) + de)}{(a^{2} + de)(2d^{2} + 3de + 2e^{2}) + a(d + e)(4d^{2} + 5de + 4e^{2})}.\\]  \n\nSince \\(h - t = \\frac{2a + 4d + 4e}{5}\\) , this gives us  \n\n\\[\\frac{h - s}{h - t} = \\frac{5}{4}\\cdot \\frac{(a^{2} + de)(2d^{2} + 3de + 2e^{2}) + a(d + e)(4d^{2} + 5de + 4e^{2})}{(d + e)(2ad + 2ae + de)(a + 2d + 2e)}.\\]  \n\nIt is easy to see that this is real by using the symmetry of the expressions (both the numerator and denominator satisfy \\(\\mathcal{E} = a^{2}d^{3}e^{3}\\overline{\\mathcal{E}}\\) . We conclude that \\(H\\) , \\(S\\) , and \\(T\\) are collinear, as desired.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 5. "}}
 {"year": "2025", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle. Points \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) lie on a line in this order and satisfy \\(B D =\\) \\(D E = E C\\) . Let \\(M\\) and \\(N\\) be the midpoints of \\(A D\\) and \\(A E\\) , respectively. Let \\(H\\) be the orthocentre of triangle \\(A D E\\) . Let \\(P\\) and \\(Q\\) be points on lines \\(B M\\) and \\(C N\\) , respectively, such that \\(D\\) , \\(H\\) , \\(M\\) , and \\(P\\) are concyclic and \\(E\\) , \\(H\\) , \\(N\\) , and \\(Q\\) are concyclic. Prove that \\(P\\) , \\(Q\\) , \\(N\\) , and \\(M\\) are concyclic.  \n\nThe orthocentre of a triangle is the point of intersection of its altitudes.", "solution": "As usual, we reduce to proving \\(T = (3A + D + E) / 5\\) is on the radical axis and compute; this time in Cartesian coordinates.  \n\nLet \\(A(0,h)\\) , \\(D(b,0)\\) , \\(E(c,0)\\) be coordinates for \\(ADE\\) . Then \\(H(0, - \\frac{b c}{h})\\) , \\(M(\\frac{b}{2},\\frac{h}{2})\\) , \\(N(\\frac{c}{2},\\frac{h}{2})\\) and \\(T(\\frac{b + c}{5},\\frac{3h}{5})\\) . We compute \\(O_{D}(x_{D},y_{D})\\) the circumcentre of \\(DMH\\) and obtain \\(O_{E}\\) by symmetry. We then have to verify that \\(O_{D}O_{E} \\perp HT\\) , which can be done by comparing slopes.  \n\nThe centre \\(O_{D}\\) can be given by the intersection of perpendicular bisectors of \\(DM\\) and \\(DH\\) . This gives the following system of equations on \\(x_{D},y_{D}\\) :\n\n\n\n\\[h\\cdot x_{D} + c\\cdot y_{D} = \\frac{b h}{2} -\\frac{b c^{2}}{2 h}\\] \\[-b\\cdot x_{D} + h\\cdot y_{D} = -\\frac{3 b^{2}}{4} +\\frac{h^{2}}{4}\\]  \n\nSolving the system gives  \n\n\\[(h^{2} + b c)x_{D} = \\frac{3 b^{2}c}{4} -\\frac{b c^{2}}{2} +\\frac{h^{2}b}{2} -\\frac{h^{2}c}{4}\\] \\[(h^{2} + b c)y_{D} = -\\frac{h b^{2}}{4} +\\frac{h^{3}}{4} -\\frac{b^{2}c^{2}}{2 h}\\]  \n\nThe formulas for \\(x_{E}, y_{E}\\) will be the same, swapping \\(b \\leftrightarrow c\\) by symmetry; thus \\(y_{D} - y_{E}\\) and \\(x_{D} - x_{E}\\) will be antisymmetric in \\(b, c\\) and divisible by \\(b - c\\) , and explicitly:  \n\n\\[(h^{2} + b c)(x_{D} - x_{E}) = \\frac{b - c}{4} (5b c + 3h^{2})\\] \\[(h^{2} + b c)(y_{D} - y_{E}) = -\\frac{b - c}{4} h(b + c)\\]  \n\nSo \\(\\frac{y_{D} - y_{E}}{x_{D} - x_{E}} = -\\frac{h(b + c)}{5b c + 3h^{2}}\\) .  \n\nThe other slope is more immediate:  \n\n\\[\\frac{y_{T} - y_{H}}{x_{T} - x_{H}} = \\frac{3h / 5 + b c / h}{(b + c) / 5} = \\frac{5b c + 3h^{2}}{h(b + c)} = -\\frac{x_{D} - x_{E}}{y_{D} - y_{E}}\\]  \n\nso indeed the two slopes correspond to perpendicular lines.\n\n\n\n## Day 2", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P3.", "solution_match": "\nSolution 6. "}}
-{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "We will prove that \\(\\triangle B I R\\sim \\triangle C I S\\) , since the statement then follows from \\(\\angle T R I = \\angle B R I = \\angle C S I =\\) \\(\\angle T S I\\) .  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 \n\nStep 1. Let us prove \\(\\angle R B I = \\angle S C I\\) . We will use directed angles:  \n\n\\[(B R,B I) = (B R,A B) + (A B,B I) = (A Q,A B) + (B I,B C)\\] \\[\\qquad = (C Q,B C) + (B I,B C) = (C I,C B) + (B I,B C),\\]  \n\nwhich is symmetric in \\(B,C\\) . Therefore, analogously we would obtain the same expression for \\((C S,C I)\\)  \n\nStep 2. Let us prove \\(B R / B I = C S / C I\\) . Clearly \\(B R = A Q\\) and \\(C S = A P\\) . Angle chasing gives \\(\\angle I C B = \\angle Q C B = \\angle A P Q\\) , and similarly \\(\\angle P Q A = \\angle C B I\\) , and so \\(\\triangle I B C\\sim \\triangle A Q P\\) , from which the desired \\(A Q / B I = A P / C I\\) follows. This finishes the solution.  \n\nRemark. In the alternative solutions below, the notation \\((A B C)\\) refers to the circle passing through points \\(A\\) , \\(B\\) , and \\(C\\) . \\(D\\) will refer to the midpoint of arc \\(B C\\) not containing point \\(A\\) , unless stated otherwise.  \n\nLet \\(I_{A}\\) be the \\(A\\) - excenter of \\(\\triangle A B C\\) . It is well- known that points \\(B\\) , \\(C\\) , \\(I\\) , and \\(I_{A}\\) lie on a circle with diameter \\(I I_{A}\\) . The center of this circle is point \\(D\\) . It is clear that \\(A\\) , \\(I\\) and \\(D\\) are collinear.  \n\nIn some of the solutions, some of the facts below may be used:\n\n\n\n\\(\\cdot B,T,I,C\\) are concyclic. Indeed, \\(\\angle B T C = \\angle Q A P = A + \\frac{B + C}{2} = 90^{\\circ} + \\frac{A}{2} = \\angle B I C\\) , so \\(B,T,I,C\\) are concyclic.  \n\n\\(\\angle Q A P = \\angle B T C = \\angle R T S\\) . Indeed, since \\(Q A\\parallel B T\\) and \\(P A\\parallel C T\\) , then \\(\\angle Q A P = \\angle B T C =\\) \\(\\angle R T S\\)  \n\n\\(\\cdot R Q\\) and \\(P S\\) are tangents to \\((A B C)\\) . Indeed, \\(\\angle R Q B = \\angle Q B A = \\angle Q A B\\) and similarly for \\(P S\\)  \n\n\\(\\cdot \\triangle A Q P\\sim I B C\\) (see Solution 1 for the proof).  \n\n\\(\\cdot Q P\\) is the perpendicular bisector of \\(A I\\) . Indeed, \\(Q A = Q I\\) and \\(P A = P I\\) so \\(P Q\\perp A I\\) and \\(P Q\\) bisects \\(A I\\) .", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 1. "}}
+{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "We will prove that \\(\\triangle B I R\\sim \\triangle C I S\\) , since the statement then follows from \\(\\angle T R I = \\angle B R I = \\angle C S I =\\) \\(\\angle T S I\\) .  \n\n![md5:fde912fbdc79cac01c462235f3a9f56d](fde912fbdc79cac01c462235f3a9f56d.jpeg)\n  \n\nStep 1. Let us prove \\(\\angle R B I = \\angle S C I\\) . We will use directed angles:  \n\n\\[(B R,B I) = (B R,A B) + (A B,B I) = (A Q,A B) + (B I,B C)\\] \\[\\qquad = (C Q,B C) + (B I,B C) = (C I,C B) + (B I,B C),\\]  \n\nwhich is symmetric in \\(B,C\\) . Therefore, analogously we would obtain the same expression for \\((C S,C I)\\)  \n\nStep 2. Let us prove \\(B R / B I = C S / C I\\) . Clearly \\(B R = A Q\\) and \\(C S = A P\\) . Angle chasing gives \\(\\angle I C B = \\angle Q C B = \\angle A P Q\\) , and similarly \\(\\angle P Q A = \\angle C B I\\) , and so \\(\\triangle I B C\\sim \\triangle A Q P\\) , from which the desired \\(A Q / B I = A P / C I\\) follows. This finishes the solution.  \n\nRemark. In the alternative solutions below, the notation \\((A B C)\\) refers to the circle passing through points \\(A\\) , \\(B\\) , and \\(C\\) . \\(D\\) will refer to the midpoint of arc \\(B C\\) not containing point \\(A\\) , unless stated otherwise.  \n\nLet \\(I_{A}\\) be the \\(A\\) - excenter of \\(\\triangle A B C\\) . It is well- known that points \\(B\\) , \\(C\\) , \\(I\\) , and \\(I_{A}\\) lie on a circle with diameter \\(I I_{A}\\) . The center of this circle is point \\(D\\) . It is clear that \\(A\\) , \\(I\\) and \\(D\\) are collinear.  \n\nIn some of the solutions, some of the facts below may be used:\n\n\n\n\\(\\cdot B,T,I,C\\) are concyclic. Indeed, \\(\\angle B T C = \\angle Q A P = A + \\frac{B + C}{2} = 90^{\\circ} + \\frac{A}{2} = \\angle B I C\\) , so \\(B,T,I,C\\) are concyclic.  \n\n\\(\\angle Q A P = \\angle B T C = \\angle R T S\\) . Indeed, since \\(Q A\\parallel B T\\) and \\(P A\\parallel C T\\) , then \\(\\angle Q A P = \\angle B T C =\\) \\(\\angle R T S\\)  \n\n\\(\\cdot R Q\\) and \\(P S\\) are tangents to \\((A B C)\\) . Indeed, \\(\\angle R Q B = \\angle Q B A = \\angle Q A B\\) and similarly for \\(P S\\)  \n\n\\(\\cdot \\triangle A Q P\\sim I B C\\) (see Solution 1 for the proof).  \n\n\\(\\cdot Q P\\) is the perpendicular bisector of \\(A I\\) . Indeed, \\(Q A = Q I\\) and \\(P A = P I\\) so \\(P Q\\perp A I\\) and \\(P Q\\) bisects \\(A I\\) .", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 1. "}}
 {"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "We use complex numbers, with \\((A B C)\\) as the unit circle. Set \\(D = d\\) , \\(P = p\\) , \\(Q = q\\) , so that \\(A = a = - \\frac{p q}{d}\\) , \\(B = b = - \\frac{d q}{p}\\) , and \\(C = c = - \\frac{d q}{q}\\) . Write \\(z\\sim w\\) if \\(z / w\\) is a nonzero real number. We observe that  \n\n\\[\\frac{R - T}{S - T}\\sim \\frac{R - B}{S - C} = \\frac{Q - A}{P - A} = \\frac{q + \\frac{p q}{d}}{p + \\frac{p q}{d}} = \\frac{(d + p)q}{(d + q)p}.\\]  \n\nSo, it suffices to show that \\(\\frac{I - R}{I - S}\\sim \\frac{(d + p)q}{(d + q)p}\\) . Indeed,  \n\n\\[I - R = (d + p + q) - (Q + B - A) = d + p + \\frac{d q}{p} - \\frac{p q}{d} = (d + p)\\left(1 + \\frac{(d - p)q}{d p}\\right) = \\frac{(d + p)(d p + d q - p q)}{d p},\\]  \n\nso  \n\n\\[\\frac{I - R}{I - S} = \\frac{\\frac{d + p}{d p}}{\\frac{d + q}{d q}} = \\frac{(d + p)q}{(d + q)p}.\\]", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 2. "}}
 {"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "In the following, all segment notations denote vectors.  \n\nAs mentioned above, we find \\(\\triangle A Q P\\sim \\triangle I B C\\) , and by definitions of the parallelograms we have \\(B R = A Q\\) and \\(C S = A P\\) as well as \\(\\angle R T S = \\angle Q A P\\) , so it suffices to show \\(\\angle R I S = \\angle Q A P\\) . From the similarity \\(\\triangle A Q P\\sim \\triangle I B C\\) , we have a spiral map \\(\\lambda\\) such that \\(I B = \\lambda A Q\\) and \\(I C = \\lambda A P\\) . It follows that \\(I R = I B + B R = (\\lambda +1)A Q\\) and \\(I S = I C + C S = (\\lambda +1)A P\\) . Because \\(\\lambda +1\\) is also a spiral map, we have \\(\\triangle I R S\\sim \\triangle A Q P\\) and in particular \\(\\angle R I S = \\angle Q A P\\) , as we wanted to show.  \n\nRemark. This solution is deeply related to the complex numbers solution; indeed, the vectors can be interpreted as complex numbers and the spiral map as a complex scalar multiplication. But it only relies on the additive structure of the complex numbers as a real plane and the linear map acting on them (rather than, e.g., multiplying two points together), making vectors a slightly more natural language for the claims.  \n\nRemark. The number 1 in the solution above represents the identity map.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 3. "}}
-{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(E\\) , \\(F\\) , \\(G\\) be the midpoints of \\(A I\\) , \\(B Q\\) , \\(C P\\) . As in Solution 1, angle chase shows that \\(\\triangle A Q P\\sim\\) \\(\\triangle I B C\\) .  \n\nNote that by the Mean Geometry Theorem we have that \\(\\frac{1}{2} A Q P + \\frac{1}{2} I B C = E F G\\) is similar to \\(\\triangle I B C\\) . Homothety with center \\(A\\) and scale- factor 2 maps \\(E F G\\) to \\(I R S\\) . Hence \\(\\angle R I S = \\angle F E G = \\angle Q A P =\\) \\(\\angle B T C = \\angle R T S\\) , so \\(R,T,I,S\\) are concyclic.  \n\nRemark. As shown above, \\(E\\) lies on \\(Q P\\) and \\(A I\\perp P Q\\) . One can prove that \\(\\angle F E G = \\angle B I C\\) in another way. Let \\(J\\) be the midpoint of \\(P Q\\) . Then \\(\\angle B I C = \\angle F J G\\) by midlines and \\(\\angle F J G = \\angle F E G\\) by the lemma below applied in \\(B C P Q\\) .  \n\nLemma. Let \\(A B C D\\) is a cyclic quadrilateral and \\(E\\) is the intersection of its diagonals. Then the midpoints of \\(A B\\) , \\(B C\\) , \\(C D\\) and the foot of the perpendicular from \\(E\\) to \\(B C\\) are concyclic.\n\n\n![](data:image/jpeg;base64,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"metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 4. "}}
-{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(O\\) be the circumcenter of \\((A B C)\\) . Let \\(M\\) , \\(N\\) , and \\(L\\) be the midpoints of \\(O D\\) , \\(P C\\) , and \\(Q B\\) respectively.  \n\nClaim 1. \\(\\triangle O P Q\\) and \\(\\triangle D C B\\) are directly similar.  \n\nProof. Clearly \\(D B = D C\\) and \\(O Q = O P\\) . Also note that \\(\\angle Q O P = 2\\angle Q D P = 2\\angle Q D A + 2\\angle P D A =\\) \\(\\angle B D A + \\angle C D A = \\angle B D C\\) . So the two triangles are directly similar by SAS.  \n\nClaim 2. \\(M L = M N\\) and \\(\\angle L M N = 180^{\\circ} - \\angle B A C\\)  \n\nProof. Note that since \\(\\triangle O Q P\\sim \\triangle D B C\\) by the Mean Geometry Theorem, we have that the average of the two triangles is also similar to them, therefore \\(\\triangle M L N\\sim \\triangle D B C\\Rightarrow M L = M N\\) and \\(\\angle L M N =\\) \\(\\angle B D C = 180^{\\circ} - \\angle B A C\\)  \n\nLet \\(K\\) be the reflection of \\(A\\) over \\(M\\)  \n\nClaim 3. \\(K\\) is the circumcenter of \\(\\triangle R T S\\)  \n\nProof. Note that since \\(A Q R B\\) and \\(A P S C\\) are parallelograms we have that \\(A - L - R\\) are collinear and that \\(A - N - S\\) are collinear. The homothety centered at \\(A\\) with scale- factor 2 maps \\(\\triangle L M N\\) to \\(\\triangle R K S\\) , therefore \\(K R = K S\\) and \\(\\angle R K S = \\angle L M N = \\angle B D C = 2(180^{\\circ} - \\angle R T S)\\) (and \\(K\\) and \\(T\\) are in opposite sides of \\(R S\\) ), implying that \\(K\\) is the circumcenter of \\(\\triangle R T S\\)  \n\nClaim 4. \\(K T = K I\\)  \n\nProof. Note that \\(A O K D\\) is a parallelogram. Let \\(B T\\) intersect the \\((A B C)\\) again at point \\(G\\) . Since \\(\\angle A B G = \\angle A B T = \\angle Q A B = \\angle Q C A\\Rightarrow A Q = A G\\) and also \\(O Q = O G\\) hence \\(A O\\perp Q G\\) . Then by Reim's theorem we have that \\(Q G\\parallel T I\\) and also that \\(A O\\parallel D K\\) , so \\(D K\\perp T I\\) . Since \\(D I = D T\\) , it means that \\(K D\\) is the perpendicular bisector of \\(T I\\) , therefore \\(K T = K I\\) .  \n\nThis means that \\(R T I S\\) is cyclic with center \\(K\\) .  \n\nRemark. When \\(K\\) and \\(T\\) are on the same side of \\(R S\\) , it can be shown that \\(\\angle R K S = 2\\angle R T S\\) .\n\n\n![](data:image/jpeg;base64,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"metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 5. "}}
-{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "As shown above, we have that \\(BTIC\\) is cyclic. Let \\(D\\) and \\(E\\) be the second intersections of \\(AC\\) and \\(AB\\) with this circle, respectively. Since the center of this circle lies on \\(AI\\) (by symmetry about \\(AI\\) ), we have that \\(AB = AD\\) and \\(AC = AE\\) , therefore \\(BE = CD\\) . Note that since \\(C - I - Q\\) and \\(A - B - E\\) are collinear, by Reim's theorem we have that \\(AQ \\parallel EI\\) and since \\(AQ \\parallel BT\\) , we have that \\(BT \\parallel EI\\) . Similarly, we get \\(CT \\parallel DI\\) . Let \\(F\\) and \\(G\\) be the intersections of \\(ID\\) and \\(IE\\) with \\(PS\\) and \\(QR\\) , respectively. Clearly, \\(RGEB\\) and \\(FSCD\\) are parallelograms. Since \\(RGEB\\) is parallelogram and \\(BEIT\\) is isosceles trapezoid, we have that \\(RGIT\\) is isosceles trapezoid. Similarly, \\(SFIT\\) is isosceles trapezoid. Hence, both of them are cyclic. Note also that \\(QR = AB = AD = PF\\) and \\(QG = AE = AC = PS\\) . Since \\(QR\\) and \\(PS\\) are tangents to the circumcircle of \\(\\triangle ABC\\) we have that \\(R\\) and \\(F\\) are symmetric (reflections) about the perpendicular bisector of \\(PQ\\) . Similarly, \\(G\\) and \\(S\\) are symmetric about the perpendicular bisector of \\(PQ\\) . This gives us that \\(QP \\parallel RF \\parallel GS\\) and that \\(RFSG\\) is an isosceles trapezoid, hence a cyclic quadrilateral with \\(\\angle RGS = 180^{\\circ} - \\angle GQP = 180^{\\circ} - \\angle QAP = 180^{\\circ} - \\angle RTS \\Rightarrow R, G, S, T\\) are concyclic. Combining all the facts about the cyclic quadrilaterals we proved above, we have that \\(R, G, S, F, I, T\\) are concyclic. Therefore \\(R, T, I, S\\) lie on a circle.\n\n\n![](data:image/jpeg;base64,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"metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 6. "}}
-{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(E\\) be the \\(A\\) - excenter of \\(\\triangle ABC\\) . Let the midpoints of \\(AQ, QB, CP, PA\\) be the points \\(F, G, H, J\\) , respectively. Both \\(PD\\) and \\(EC\\) are perpendicular to \\(CI\\) , hence \\(PD \\parallel CE\\) .  \n\nSince \\(PA = PC\\) we have that \\(AJHC\\) is an isosceles trapezoid so it is cyclic. Let \\(K\\) be the second intersection of \\((AJHC)\\) and \\(AI\\) . Then \\(\\angle ADP = \\angle ACP = \\angle ACH = \\angle AKH \\Rightarrow DP \\parallel KH\\) . So \\(KH\\) is a line passing through the midpoint of the side \\(CP\\) of trapezoid \\(DPCE\\) and parallel to the bases, hence \\(K\\) is the midpoint of \\(DE\\) . Similarly, we show that the circle \\((AFGB)\\) passes through the midpoint of \\(DE\\) . Homothety centered at \\(A\\) with scale- factor 2 maps \\((AJH)\\) to \\((APS)\\) , \\((AFG)\\) to \\((AQR)\\) , and line \\(AI\\) to line \\(AI\\) . This means that the circles \\((AQR)\\) and \\((APS)\\) intersect on \\(AI\\) , call it point \\(L\\) .  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\\(\\angle I L R = 180^{\\circ} - \\angle A Q R = \\angle Q A B = \\angle Q C B = 180^{\\circ} - \\angle I T B = 180^{\\circ} - \\angle I T R\\) , therefore \\(R,L,I,T\\) are concyclic. Similarly, we get that \\(S,L,T,I\\) are concyclic. Combining these, it means that \\(R\\) and \\(S\\) belong to the circle \\((L I T)\\) . The conclusion follows.  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"metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 7. "}}
-{"year": "2025", "tier": "T2", "problem_label": "5", "problem_type": null, "exam": "EGMO", "problem": "Let \\(n > 1\\) be an integer. In a configuration of an \\(n \\times n\\) board, each of the \\(n^{2}\\) cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate \\(90^{\\circ}\\) counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of \\(n\\) , the maximum number of good cells over all possible starting configurations.", "solution": "We will show that the maximum number of good cells over all possible starting configurations is  \n\n\\[\\frac{n^{2}}{4} \\quad \\text{if} n \\text{ is even and}\\] \\[0 \\quad \\text{if} n \\text{ is odd.}\\]  \n\n## Odd \\(n\\)  \n\nFirst, we will prove that there are no good cells if \\(n\\) is an odd number.  \n\nFor Turbo to reach her goal, she must return to her initial cell after visiting every cell exactly once. Consider the chessboard coloring of the board. Without loss of generality, we assume that Turbo starts in a black cell. Since, at every step, Turbo moves to a cell of a different color; she will be in a white cell after \\(n^{2} \\equiv 1 \\mod 2\\) moves. Thus, it is impossible for Turbo to come back to her initial black cell on her \\(n^{2}\\) - th move, which is a contradiction. Thus there are no good cells.  \n\n## Lower bound for even \\(n\\)  \n\nWe will now construct a starting configuration with \\(\\frac{n^{2}}{4}\\) good cells for even \\(n\\) .  \n\nLet \\((i,j)\\) denote the cell in row \\(i\\) and column \\(j\\) . Consider the following cycle  \n\n\\[\\begin{array}{r l} & {(1,1)\\rightarrow (1,2)\\rightarrow (1,3)\\rightarrow \\ldots \\rightarrow (1,n)}\\\\ & {\\qquad \\rightarrow (2,n)\\rightarrow (2,n - 1)\\rightarrow \\ldots \\rightarrow (2,2)}\\\\ & {\\qquad \\ldots}\\\\ & {\\qquad \\rightarrow (2i - 1,2)\\rightarrow (2i - 1,3)\\rightarrow \\ldots \\rightarrow (2i - 1,n)}\\\\ & {\\qquad \\rightarrow (2i,n)\\rightarrow (2i,n - 1)\\rightarrow \\ldots \\rightarrow (2i,2)}\\\\ & {\\qquad \\ldots}\\\\ & {\\qquad \\rightarrow (n,n)\\rightarrow (n,n - 1)\\rightarrow \\ldots \\rightarrow (n,2)}\\\\ & {\\qquad \\rightarrow (n,1)\\rightarrow (n - 1,1)\\rightarrow \\ldots \\rightarrow (2,1)\\rightarrow (1,1).} \\end{array} \\quad (1,1)\\]  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)\n  \n\nNote that the cycle returns to the initial cell after visiting every cell exactly once. To prove that \\((1,1)\\) is good, we need to find a starting configuration such that Turbo traverses this cycle.  \n\nLet \\(c_{i}\\) be the \\((i - 1)\\) - th cell on the cycle: so we have \\(c_{0} = (1,1)\\) , \\(c_{2} = (1,2)\\) , ..., \\(c_{n^{2} - 1} = (2,1)\\) . For every \\(i\\) , we draw an arrow in cell \\(c_{i}\\) pointing towards cell \\(c_{i + 1}\\) (or pointing towards \\(c_{0}\\) if \\(i = n^{2} - 1\\) ) and then rotate this arrow \\(i\\) times \\(90^{\\circ}\\) in the clockwise direction. After \\(i\\) moves, the arrow in \\(c_{i}\\) will have rotated \\(i\\) times \\(90^{\\circ}\\) counterclockwise and be in the same direction as on the path defined above. Thus, Turbo will traverse the cycle \\(c_{0}, c_{1}, c_{2}, \\ldots , c_{n^{2} - 1}, c_{0}\\) and \\((1,1)\\) is good.  \n\nEvery four moves, all arrows point in the same direction as in the beginning. Moreover, the board will return to its initial configuration after traversing the full cycle, since \\(n^{2}\\) , the length of the cycle, is divisible by 4. Therefore Turbo can also start at any \\(c_{i}\\) with 4 \\(| i\\) and follow the same route. Hence the cells \\(c_{0}, c_{4}, c_{8}, \\ldots , c_{n^{2} - 4}\\) are good and there are \\(\\frac{n^{2}}{4}\\) of such cells.\n\n\n\n## Upper bound for even \\(n\\)  \n\nWe will prove that for even \\(n\\) and any start configuration there are at most \\(\\frac{n^{2}}{4}\\) good cells.  \n\nLet \\(a_{0}\\) be a good cell. Let \\(a_{0},a_{1},a_{2},\\ldots ,a_{n^{2} - 1},a_{n^{2}} = a_{0}\\) be the sequence of cells that Turbo visits when she starts at \\(a_{0}\\) . Now suppose there is another good cell \\(b_{0}\\) and let \\(b_{0},b_{1},b_{2},\\ldots ,b_{n^{2} - 1},b_{n^{2}} = b_{0}\\) be the sequence of cells that Turbo visits when she starts at \\(b_{0}\\) .  \n\nNote that, since \\(4\\mid n^{2}\\) , the arrows are back to their initial configuration after \\(n^{2}\\) steps. Thus, if Turbo keeps walking after returning to her initial cell, she would just traverse the same cycle over and over again.  \n\nConsider the upper left corner of the board. With standard row and column numbering, the corner cell is \\((1,1)\\) . This cell has only two neighbours, so both the \\(a\\) - route and the \\(b\\) - route must have cells \\((2,1),(1,1),(1,2)\\) in that order or \\((1,2),(1,1),(2,1)\\) in that order. Without loss of generality, \\(a_{i - 1} =\\) \\((2,1)\\) , \\(a_{i} = (1,1)\\) and \\(a_{i + 1} = (1,2)\\) for some \\(i\\) . Let \\(j\\) be such that \\(b_{j} = (1,1)\\) . If \\(b_{j - 1} = (2,1) = a_{i - 1}\\) then the arrow in cell \\((2,1)\\) must be pointed in the same direction after \\(i - 1\\) steps and after \\(j - 1\\) steps, so \\(i\\equiv j\\) mod 4. But then the arrow in cell \\(b_{j} = (1,1) = a_{i}\\) must also be pointed in the same direction after \\(i\\) and after \\(j\\) steps, so Turbo moves to \\(b_{j + 1} = a_{i + 1}\\) in both cases, and again finds the arrow pointed in the same direction in both cases. Continuing, we find that the \\(b\\) - route is actually identical to \\(a_{4t},a_{4t + 1},\\ldots ,a_{n^{2}} = a_{0},a_{1},\\ldots ,a_{4t - 1},a_{4t}\\) for some \\(t\\) , as any other starting point would have the arrows in the wrong direction initially.  \n\nNow suppose instead that \\(b_{j + 1} = (2,1) = a_{i - 1}\\) . Considering the \\(a\\) - route, the arrows in the upper left corner after \\(i - 1\\) steps must be like this:  \n\n![](data:image/jpeg;base64,/9j/4AAQSkZJRgABAQAAAQABAAD/2wBDAAgGBgcGBQgHBwcJCQgKDBQNDAsLDBkSEw8UHRofHh0aHBwgJC4nICIsIxwcKDcpLDAxNDQ0Hyc5PTgyPC4zNDL/2wBDAQkJCQwLDBgNDRgyIRwhMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjL/wAARCABWAVsDASIAAhEBAxEB/8QAHwAAAQUBAQEBAQEAAAAAAAAAAAECAwQFBgcICQoL/8QAtRAAAgEDAwIEAwUFBAQAAAF9AQIDAAQRBRIhMUEGE1FhByJxFDKBkaEII0KxwRVS0fAkM2JyggkKFhcYGRolJicoKSo0NTY3ODk6Q0RFRkdISUpTVFVWV1hZWmNkZWZnaGlqc3R1dnd4eXqDhIWGh4iJipKTlJWWl5iZmqKjpKWmp6ipqrKztLW2t7i5usLDxMXGx8jJytLT1NXW19jZ2uHi4+Tl5ufo6erx8vP09fb3+Pn6/8QAHwEAAwEBAQEBAQEBAQAAAAAAAAECAwQFBgcICQoL/8QAtREAAgECBAQDBAcFBAQAAQJ3AAECAxEEBSExBhJBUQdhcRMiMoEIFEKRobHBCSMzUvAVYnLRChYkNOEl8RcYGRomJygpKjU2Nzg5OkNERUZHSElKU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6goOEhYaHiImKkpOUlZaXmJmaoqOkpaanqKmqsrO0tba3uLm6wsPExcbHyMnK0tPU1dbX2Nna4uPk5ebn6Onq8vP09fb3+Pn6/9oADAMBAAIRAxEAPwD3+iszxDdT2egXk9tIY5lTCuACVJIGRnjIz3qD+w73/oZtX/75tv8A4zQBtUVg3GlXVrbS3EnibWNkSF2wltnAGT/yxog0m7ubeKePxNrGyRA65S2zgjP/ADxoA3qKxf7Dvf8AoZtX/wC+bb/4zWfoVjqWp6HaXs/iTVBLMm5giW4Gc9v3VAHVUVi/2He/9DNq/wD3zbf/ABmj+w73/oZtX/75tv8A4zQBtUVi/wBh3v8A0M2r/wDfNt/8Zo/sO9/6GbV/++bb/wCM0AbVFctPYalFr1jYr4l1Typ4JpGJS2yChjAx+6/2z+lX/wCw73/oZtX/AO+bb/4zQBtUVi/2He/9DNq//fNt/wDGat2Gn3FnIzzate3oYYCXAiAX3GxFNAF+iiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKxtae6kvtMsre9ms1uJJPMkgVC+FQkAb1YYz7UAbNFYv9h3v/Qzav/3zbf8Axmq93YXFl5Hm+JtZ/fzLCmEtj8xzjP7npxQB0VFYv9h3v/Qzav8A9823/wAZrP1yx1LTdImu4fEmqGRGQAOluRy4B/5ZehoA6qisX+w73/oZtX/75tv/AIzUF9YTadYXF7deKtWjt7eNpZHK23yqoyT/AKn0FAHQ0Vz1lp82o2Nve2virVpLe4jWWNwtthlYZB/1Poan/sO9/wChm1f/AL5tv/jNAG1RXK6DY6lqehWV9P4k1QSzxB2CJbgAn0/dVof2He/9DNq//fNt/wDGaANqisX+w73/AKGbV/8Avm2/+M1sRqUiRGdpCqgF2xlvc4AGfoKAMnxV/wAi1efRf/QhWxWP4q/5Fq8+i/8AoQrYoApaz/yA9Q/69pP/AEE0uk/8gax/694//QRWf4m0+/utLupdMuZEu1t3VbdjmKfIPysOxPZhgg4zkcF3huwvrbTbebU7iRrtoEVrcHEUGAPlUdyO7HJJz0HAANqsXwj/AMinp3/XL+prarF8I/8AIp6d/wBcv6mgDR1C8/s/T57v7PcXPkoX8m2TfI+Oyr3Ncx4e8SyXOnNqE1hqkzX0hnCxwl44lwFVFOewUZ6ZYse9dhVSy0+Kwe7MLNsuZzOUONqMQA232JBY+7E96AKS+IAzBf7J1YZOMm1OB+tbFFFAGPef8jfpP/Xpdf8AoUNbFY95/wAjfpP/AF6XX/oUNbFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABWPqf/Iw6H/10m/9FmtisfU/+Rh0P/rpN/6LNAGxWN4h66T/ANhGL+TVs1yXiXS9cN1Yy6PcedG99G8sNz8y2/X94h64HdCcHjBXnIB1tYviz/kWrn/fi/8ARi1o2No1nbCJ7me5fJZpZ2BZifoAAPYACs7xZ/yLVz/vxf8AoxaANqsTxJpcF9ptzNcvK6Q28jLBvxGXAJDMo+8R2zkAgHGea26qarG8uj3scalne3kVVHUkqcCgCh4c0q30/T7eW1aWOOaBGa335jDEAllU/dJJOcYBznGea2qq6ZG8WlWccilXWBFZT1BCjirVAGL4Q/5FHS/+vda2qxfCH/Io6X/17rW1QAUUUUAY/ir/AJFi+HcoAD6HcMGk/s3W/wDoYP8AyTT/ABpfFX/ItXn0X/0IVsUAYU1nq9vBJNN4jCRRqXdms0woAySeaWKy1eeFJovEQeORQysLNMEHkHrSeKdNfUtIvElunWyW2kMlsigecdp4Zuu3/ZGM9yRkUeF9MbTNJtY4bp3sTboYreRQfJOAcK3Xb6Kc47EDAoAk/s3W/wDoYP8AyTT/ABrI8MWGrSeGrB4da8qMx/Kn2VG28nuTzXYVi+Ef+RT07/rl/U0AL/Zut/8AQwf+Saf40f2brf8A0MH/AJJp/jWzRQBjf2brf/Qwf+Saf40f2brf/Qwf+Saf41s0UAcjdWGrjxPpqHW8yG1uSsn2VPlAaLIxnnOR+Vav9m63/wBDB/5Jp/jS3n/I36T/ANel1/6FDWxQBjf2brf/AEMH/kmn+NWrC11C3kZrzUvtakYVfIWPB9cir9FABRRRQAUUUUAFFFFABRRWNr2tnSX061hRZLzUroW1uHPyqdpZnb1AVScdzgcZyADZornm1m9h8QnQJmthd3Fo11Z3AibYdrBWVk3ZyNynhuQT0xzX8Ha3rfiHStN1W7trKKzu7UyOsbMJElDADAOQVI3Hrkcdc0AdTRRRQAUUUUAFFFFABWDr0c02raNHbXH2eYyylZdgfaPLORg8c1vVj6n/AMjDof8A10m/9FmgBP7N1v8A6GD/AMk0/wAaguINTtfK8/xKkfnSCKPdaINznoBz14roK47xXoCalc6fJqNzLOh1CMQRJ+7W34PzLjnf/tE8dgMnIBsf2brf/Qwf+Saf41keJbDV49AuGm1rzUDR5T7Ki5+de4NdPYwXFtbCK5uzdupOJWjCMR23beCfcAfSs7xZ/wAi1c/78X/oxaAF/s3W/wDoYP8AyTT/ABo/s3W/+hg/8k0/xrZooAxv7N1v/oYP/JNP8aP7N1v/AKGD/wAk0/xrZooA4/wxYatJ4Y054da8qMwgrH9lRto9Mk81r/2brf8A0MH/AJJp/jSeEP8AkUdL/wCvda2qAMb+zdb/AOhg/wDJNP8AGteNWWJFd97hQGbGNx9cU6igDH8Vf8i1efRf/QhWxWN4r/5Fi+x12DafQ7hjPtmk8nxN/wA/+kf+AUn/AMdoAu6wCdEvwBkm2kwB/uml0oEaPYgjBFvHx/wEVR8nxN/z/wCkf+AUn/x2jyfE3/P/AKR/4BSf/HaANqsXwj/yKenf9cv6mjyfE3/P/pH/AIBSf/HayfC8PiE+GrAwXumLF5fyiSzkLAZPUiUfyoA7CisXyfE3/P8A6R/4BSf/AB2jyfE3/P8A6R/4BSf/AB2gDaorF8nxN/z/AOkf+AUn/wAdo8nxN/z/AOkf+AUn/wAdoAdef8jfpP8A16XX/oUNbFchdQ+If+Eo00NfaWZfstzsIs5NoG6LOR5vJ6Y59a1fJ8Tf8/8ApH/gFJ/8doA2qKxfJ8Tf8/8ApH/gFJ/8dq3YR6skjHULmyljx8ot7d4yD7ku1AF+iiigAooooAKKKKACsrXNDTWFspFmMF3Y3AubWbbuCuAVIIyMqVYgjI69Qa1aKAMJdBuH1t9buLuF9QS1NrbYgPlQqWDMSu7LEkLk5HAAHcl3hTQ7jw34dtNInvYrtbVPLjlSAxEr7gs2T+VbdFABRRRQAUUUUAFFFFABWPqf/Iw6H/10m/8ARZrYrA19bptW0UWUkMdz5su1poy6AeWc5AIP60Ab9Y3iBSx0rAJxqEROB0GGpPJ8Tf8AP/pH/gFJ/wDHaPJ8Tf8AP/pH/gFJ/wDHaANqsXxZ/wAi1c/78X/oxaPJ8Tf8/wDpH/gFJ/8AHayfE0PiEaBcGe90xo90eQlnID/rFxyZT3oA7CisXyfE3/P/AKR/4BSf/HaPJ8Tf8/8ApH/gFJ/8doA2qKxfJ8Tf8/8ApH/gFJ/8do8nxN/z/wCkf+AUn/x2gA8If8ijpf8A17rW1XH+F4fEJ8MacYL3TFi8hdoks5CwHuRKP5VreT4m/wCf/SP/AACk/wDjtAG1RWL5Pib/AJ/9I/8AAKT/AOO1sRhxEglKmTaNxUYBPfA5wKAMnxV/yLV59F/9CFbFY/ir/kWrz6L/AOhCtigAooooAKxfCP8AyKenf9cv6mtqsXwj/wAinp3/AFy/qaANqiiigAooooAx7z/kb9J/69Lr/wBChrYrHvP+Rv0n/r0uv/Qoa2KACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACsfU/wDkYdD/AOuk3/os1sVj6n/yMOh/9dJv/RZoA2KKKKACsXxZ/wAi1c/78X/oxa2qxfFn/ItXP+/F/wCjFoA2qKKKACiiigDF8If8ijpf/XutbVYvhD/kUdL/AOvda2qACiiigCnqth/aemTWfmeX5gA37c4wQen4VcoooAKKKKACqWkaf/ZWk21j5vm+Sm3ft25/DJoooAu0UUUAFFFFAFOaw83V7S/8zH2eGWLZt+9vKHOe2Nn61coooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKp3Vh9p1Cxu/M2/ZWdtu3O7cpXr260UUAXKKKKACqWraf/AGppstn5vlbyp37d2MMG6ZHpRRQBdooooAKKKKAKWkaf/ZWkWth5vm+RGE37du7HfGTirtFFABRRRQB//9k=)\n  \n\nConsidering the \\(b\\) - route instead, the arrows after \\(j - 1\\) steps must be like this:  \n\n![](data:image/jpeg;base64,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)\n  \n\nFrom the arrows in cell \\((1,1)\\) we see that \\(i\\equiv j + 1\\) mod 4. However, for the cells \\((2,1)\\) and \\((1,2)\\) this gives a contradiction.  \n\nWe conclude that the only possible good cells are \\(a_{4t}\\) for \\(t = 0,1,\\ldots ,\\frac{n^{2}}{4} - 1\\) , which gives at most \\(\\frac{n^{2}}{4}\\) good cells.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P5.", "solution_match": "\nSolution."}}
+{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(E\\) , \\(F\\) , \\(G\\) be the midpoints of \\(A I\\) , \\(B Q\\) , \\(C P\\) . As in Solution 1, angle chase shows that \\(\\triangle A Q P\\sim\\) \\(\\triangle I B C\\) .  \n\nNote that by the Mean Geometry Theorem we have that \\(\\frac{1}{2} A Q P + \\frac{1}{2} I B C = E F G\\) is similar to \\(\\triangle I B C\\) . Homothety with center \\(A\\) and scale- factor 2 maps \\(E F G\\) to \\(I R S\\) . Hence \\(\\angle R I S = \\angle F E G = \\angle Q A P =\\) \\(\\angle B T C = \\angle R T S\\) , so \\(R,T,I,S\\) are concyclic.  \n\nRemark. As shown above, \\(E\\) lies on \\(Q P\\) and \\(A I\\perp P Q\\) . One can prove that \\(\\angle F E G = \\angle B I C\\) in another way. Let \\(J\\) be the midpoint of \\(P Q\\) . Then \\(\\angle B I C = \\angle F J G\\) by midlines and \\(\\angle F J G = \\angle F E G\\) by the lemma below applied in \\(B C P Q\\) .  \n\nLemma. Let \\(A B C D\\) is a cyclic quadrilateral and \\(E\\) is the intersection of its diagonals. Then the midpoints of \\(A B\\) , \\(B C\\) , \\(C D\\) and the foot of the perpendicular from \\(E\\) to \\(B C\\) are concyclic.\n\n\n![md5:17c822e365e079d4dd97dbebf5fb270d](17c822e365e079d4dd97dbebf5fb270d.jpeg)", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 4. "}}
+{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(O\\) be the circumcenter of \\((A B C)\\) . Let \\(M\\) , \\(N\\) , and \\(L\\) be the midpoints of \\(O D\\) , \\(P C\\) , and \\(Q B\\) respectively.  \n\nClaim 1. \\(\\triangle O P Q\\) and \\(\\triangle D C B\\) are directly similar.  \n\nProof. Clearly \\(D B = D C\\) and \\(O Q = O P\\) . Also note that \\(\\angle Q O P = 2\\angle Q D P = 2\\angle Q D A + 2\\angle P D A =\\) \\(\\angle B D A + \\angle C D A = \\angle B D C\\) . So the two triangles are directly similar by SAS.  \n\nClaim 2. \\(M L = M N\\) and \\(\\angle L M N = 180^{\\circ} - \\angle B A C\\)  \n\nProof. Note that since \\(\\triangle O Q P\\sim \\triangle D B C\\) by the Mean Geometry Theorem, we have that the average of the two triangles is also similar to them, therefore \\(\\triangle M L N\\sim \\triangle D B C\\Rightarrow M L = M N\\) and \\(\\angle L M N =\\) \\(\\angle B D C = 180^{\\circ} - \\angle B A C\\)  \n\nLet \\(K\\) be the reflection of \\(A\\) over \\(M\\)  \n\nClaim 3. \\(K\\) is the circumcenter of \\(\\triangle R T S\\)  \n\nProof. Note that since \\(A Q R B\\) and \\(A P S C\\) are parallelograms we have that \\(A - L - R\\) are collinear and that \\(A - N - S\\) are collinear. The homothety centered at \\(A\\) with scale- factor 2 maps \\(\\triangle L M N\\) to \\(\\triangle R K S\\) , therefore \\(K R = K S\\) and \\(\\angle R K S = \\angle L M N = \\angle B D C = 2(180^{\\circ} - \\angle R T S)\\) (and \\(K\\) and \\(T\\) are in opposite sides of \\(R S\\) ), implying that \\(K\\) is the circumcenter of \\(\\triangle R T S\\)  \n\nClaim 4. \\(K T = K I\\)  \n\nProof. Note that \\(A O K D\\) is a parallelogram. Let \\(B T\\) intersect the \\((A B C)\\) again at point \\(G\\) . Since \\(\\angle A B G = \\angle A B T = \\angle Q A B = \\angle Q C A\\Rightarrow A Q = A G\\) and also \\(O Q = O G\\) hence \\(A O\\perp Q G\\) . Then by Reim's theorem we have that \\(Q G\\parallel T I\\) and also that \\(A O\\parallel D K\\) , so \\(D K\\perp T I\\) . Since \\(D I = D T\\) , it means that \\(K D\\) is the perpendicular bisector of \\(T I\\) , therefore \\(K T = K I\\) .  \n\nThis means that \\(R T I S\\) is cyclic with center \\(K\\) .  \n\nRemark. When \\(K\\) and \\(T\\) are on the same side of \\(R S\\) , it can be shown that \\(\\angle R K S = 2\\angle R T S\\) .\n\n\n![md5:475715ec425df4654345ac12f68d35c0](475715ec425df4654345ac12f68d35c0.jpeg)", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 5. "}}
+{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "As shown above, we have that \\(BTIC\\) is cyclic. Let \\(D\\) and \\(E\\) be the second intersections of \\(AC\\) and \\(AB\\) with this circle, respectively. Since the center of this circle lies on \\(AI\\) (by symmetry about \\(AI\\) ), we have that \\(AB = AD\\) and \\(AC = AE\\) , therefore \\(BE = CD\\) . Note that since \\(C - I - Q\\) and \\(A - B - E\\) are collinear, by Reim's theorem we have that \\(AQ \\parallel EI\\) and since \\(AQ \\parallel BT\\) , we have that \\(BT \\parallel EI\\) . Similarly, we get \\(CT \\parallel DI\\) . Let \\(F\\) and \\(G\\) be the intersections of \\(ID\\) and \\(IE\\) with \\(PS\\) and \\(QR\\) , respectively. Clearly, \\(RGEB\\) and \\(FSCD\\) are parallelograms. Since \\(RGEB\\) is parallelogram and \\(BEIT\\) is isosceles trapezoid, we have that \\(RGIT\\) is isosceles trapezoid. Similarly, \\(SFIT\\) is isosceles trapezoid. Hence, both of them are cyclic. Note also that \\(QR = AB = AD = PF\\) and \\(QG = AE = AC = PS\\) . Since \\(QR\\) and \\(PS\\) are tangents to the circumcircle of \\(\\triangle ABC\\) we have that \\(R\\) and \\(F\\) are symmetric (reflections) about the perpendicular bisector of \\(PQ\\) . Similarly, \\(G\\) and \\(S\\) are symmetric about the perpendicular bisector of \\(PQ\\) . This gives us that \\(QP \\parallel RF \\parallel GS\\) and that \\(RFSG\\) is an isosceles trapezoid, hence a cyclic quadrilateral with \\(\\angle RGS = 180^{\\circ} - \\angle GQP = 180^{\\circ} - \\angle QAP = 180^{\\circ} - \\angle RTS \\Rightarrow R, G, S, T\\) are concyclic. Combining all the facts about the cyclic quadrilaterals we proved above, we have that \\(R, G, S, F, I, T\\) are concyclic. Therefore \\(R, T, I, S\\) lie on a circle.\n\n\n![md5:eecb68b2c8a14a8f09cb4dd29102e21f](eecb68b2c8a14a8f09cb4dd29102e21f.jpeg)", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 6. "}}
+{"year": "2025", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "EGMO", "problem": "Let \\(A B C\\) be an acute triangle with incentre \\(I\\) and \\(A B \\neq A C\\) . Let lines \\(B I\\) and \\(C I\\) intersect the circumcircle of \\(A B C\\) at \\(P \\neq B\\) and \\(Q \\neq C\\) , respectively. Consider points \\(R\\) and \\(S\\) such that \\(A Q R B\\) and \\(A C S P\\) are parallelograms (with \\(A Q \\parallel R B\\) , \\(A B \\parallel Q R\\) , \\(A C \\parallel S P\\) , and \\(A P \\parallel C S\\) ). Let \\(T\\) be the point of intersection of lines \\(R B\\) and \\(S C\\) . Prove that points \\(R\\) , \\(S\\) , \\(T\\) , and \\(I\\) are concyclic.", "solution": "Let \\(E\\) be the \\(A\\) - excenter of \\(\\triangle ABC\\) . Let the midpoints of \\(AQ, QB, CP, PA\\) be the points \\(F, G, H, J\\) , respectively. Both \\(PD\\) and \\(EC\\) are perpendicular to \\(CI\\) , hence \\(PD \\parallel CE\\) .  \n\nSince \\(PA = PC\\) we have that \\(AJHC\\) is an isosceles trapezoid so it is cyclic. Let \\(K\\) be the second intersection of \\((AJHC)\\) and \\(AI\\) . Then \\(\\angle ADP = \\angle ACP = \\angle ACH = \\angle AKH \\Rightarrow DP \\parallel KH\\) . So \\(KH\\) is a line passing through the midpoint of the side \\(CP\\) of trapezoid \\(DPCE\\) and parallel to the bases, hence \\(K\\) is the midpoint of \\(DE\\) . Similarly, we show that the circle \\((AFGB)\\) passes through the midpoint of \\(DE\\) . Homothety centered at \\(A\\) with scale- factor 2 maps \\((AJH)\\) to \\((APS)\\) , \\((AFG)\\) to \\((AQR)\\) , and line \\(AI\\) to line \\(AI\\) . This means that the circles \\((AQR)\\) and \\((APS)\\) intersect on \\(AI\\) , call it point \\(L\\) .  \n\n![md5:9e3f4eae8a85d6f55293984efee47f24](9e3f4eae8a85d6f55293984efee47f24.jpeg)\n\n\n\n\nNow, \\(\\angle I L R = 180^{\\circ} - \\angle A Q R = \\angle Q A B = \\angle Q C B = 180^{\\circ} - \\angle I T B = 180^{\\circ} - \\angle I T R\\) , therefore \\(R,L,I,T\\) are concyclic. Similarly, we get that \\(S,L,T,I\\) are concyclic. Combining these, it means that \\(R\\) and \\(S\\) belong to the circle \\((L I T)\\) . The conclusion follows.  \n\n![md5:94d8549749b3f283915bfb6e310b329a](94d8549749b3f283915bfb6e310b329a.jpeg)", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P4.", "solution_match": "\nSolution 7. "}}
+{"year": "2025", "tier": "T2", "problem_label": "5", "problem_type": null, "exam": "EGMO", "problem": "Let \\(n > 1\\) be an integer. In a configuration of an \\(n \\times n\\) board, each of the \\(n^{2}\\) cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate \\(90^{\\circ}\\) counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of \\(n\\) , the maximum number of good cells over all possible starting configurations.", "solution": "We will show that the maximum number of good cells over all possible starting configurations is  \n\n\\[\\frac{n^{2}}{4} \\quad \\text{if} n \\text{ is even and}\\] \\[0 \\quad \\text{if} n \\text{ is odd.}\\]  \n\n## Odd \\(n\\)  \n\nFirst, we will prove that there are no good cells if \\(n\\) is an odd number.  \n\nFor Turbo to reach her goal, she must return to her initial cell after visiting every cell exactly once. Consider the chessboard coloring of the board. Without loss of generality, we assume that Turbo starts in a black cell. Since, at every step, Turbo moves to a cell of a different color; she will be in a white cell after \\(n^{2} \\equiv 1 \\mod 2\\) moves. Thus, it is impossible for Turbo to come back to her initial black cell on her \\(n^{2}\\) - th move, which is a contradiction. Thus there are no good cells.  \n\n## Lower bound for even \\(n\\)  \n\nWe will now construct a starting configuration with \\(\\frac{n^{2}}{4}\\) good cells for even \\(n\\) .  \n\nLet \\((i,j)\\) denote the cell in row \\(i\\) and column \\(j\\) . Consider the following cycle  \n\n\\[\\begin{array}{r l} & {(1,1)\\rightarrow (1,2)\\rightarrow (1,3)\\rightarrow \\ldots \\rightarrow (1,n)}\\\\ & {\\qquad \\rightarrow (2,n)\\rightarrow (2,n - 1)\\rightarrow \\ldots \\rightarrow (2,2)}\\\\ & {\\qquad \\ldots}\\\\ & {\\qquad \\rightarrow (2i - 1,2)\\rightarrow (2i - 1,3)\\rightarrow \\ldots \\rightarrow (2i - 1,n)}\\\\ & {\\qquad \\rightarrow (2i,n)\\rightarrow (2i,n - 1)\\rightarrow \\ldots \\rightarrow (2i,2)}\\\\ & {\\qquad \\ldots}\\\\ & {\\qquad \\rightarrow (n,n)\\rightarrow (n,n - 1)\\rightarrow \\ldots \\rightarrow (n,2)}\\\\ & {\\qquad \\rightarrow (n,1)\\rightarrow (n - 1,1)\\rightarrow \\ldots \\rightarrow (2,1)\\rightarrow (1,1).} \\end{array} \\quad (1,1)\\]  \n\n![md5:f72c11b17e739eadf5ff82c9c5828a0d](f72c11b17e739eadf5ff82c9c5828a0d.jpeg)\n  \n\nNote that the cycle returns to the initial cell after visiting every cell exactly once. To prove that \\((1,1)\\) is good, we need to find a starting configuration such that Turbo traverses this cycle.  \n\nLet \\(c_{i}\\) be the \\((i - 1)\\) - th cell on the cycle: so we have \\(c_{0} = (1,1)\\) , \\(c_{2} = (1,2)\\) , ..., \\(c_{n^{2} - 1} = (2,1)\\) . For every \\(i\\) , we draw an arrow in cell \\(c_{i}\\) pointing towards cell \\(c_{i + 1}\\) (or pointing towards \\(c_{0}\\) if \\(i = n^{2} - 1\\) ) and then rotate this arrow \\(i\\) times \\(90^{\\circ}\\) in the clockwise direction. After \\(i\\) moves, the arrow in \\(c_{i}\\) will have rotated \\(i\\) times \\(90^{\\circ}\\) counterclockwise and be in the same direction as on the path defined above. Thus, Turbo will traverse the cycle \\(c_{0}, c_{1}, c_{2}, \\ldots , c_{n^{2} - 1}, c_{0}\\) and \\((1,1)\\) is good.  \n\nEvery four moves, all arrows point in the same direction as in the beginning. Moreover, the board will return to its initial configuration after traversing the full cycle, since \\(n^{2}\\) , the length of the cycle, is divisible by 4. Therefore Turbo can also start at any \\(c_{i}\\) with 4 \\(| i\\) and follow the same route. Hence the cells \\(c_{0}, c_{4}, c_{8}, \\ldots , c_{n^{2} - 4}\\) are good and there are \\(\\frac{n^{2}}{4}\\) of such cells.\n\n\n\n## Upper bound for even \\(n\\)  \n\nWe will prove that for even \\(n\\) and any start configuration there are at most \\(\\frac{n^{2}}{4}\\) good cells.  \n\nLet \\(a_{0}\\) be a good cell. Let \\(a_{0},a_{1},a_{2},\\ldots ,a_{n^{2} - 1},a_{n^{2}} = a_{0}\\) be the sequence of cells that Turbo visits when she starts at \\(a_{0}\\) . Now suppose there is another good cell \\(b_{0}\\) and let \\(b_{0},b_{1},b_{2},\\ldots ,b_{n^{2} - 1},b_{n^{2}} = b_{0}\\) be the sequence of cells that Turbo visits when she starts at \\(b_{0}\\) .  \n\nNote that, since \\(4\\mid n^{2}\\) , the arrows are back to their initial configuration after \\(n^{2}\\) steps. Thus, if Turbo keeps walking after returning to her initial cell, she would just traverse the same cycle over and over again.  \n\nConsider the upper left corner of the board. With standard row and column numbering, the corner cell is \\((1,1)\\) . This cell has only two neighbours, so both the \\(a\\) - route and the \\(b\\) - route must have cells \\((2,1),(1,1),(1,2)\\) in that order or \\((1,2),(1,1),(2,1)\\) in that order. Without loss of generality, \\(a_{i - 1} =\\) \\((2,1)\\) , \\(a_{i} = (1,1)\\) and \\(a_{i + 1} = (1,2)\\) for some \\(i\\) . Let \\(j\\) be such that \\(b_{j} = (1,1)\\) . If \\(b_{j - 1} = (2,1) = a_{i - 1}\\) then the arrow in cell \\((2,1)\\) must be pointed in the same direction after \\(i - 1\\) steps and after \\(j - 1\\) steps, so \\(i\\equiv j\\) mod 4. But then the arrow in cell \\(b_{j} = (1,1) = a_{i}\\) must also be pointed in the same direction after \\(i\\) and after \\(j\\) steps, so Turbo moves to \\(b_{j + 1} = a_{i + 1}\\) in both cases, and again finds the arrow pointed in the same direction in both cases. Continuing, we find that the \\(b\\) - route is actually identical to \\(a_{4t},a_{4t + 1},\\ldots ,a_{n^{2}} = a_{0},a_{1},\\ldots ,a_{4t - 1},a_{4t}\\) for some \\(t\\) , as any other starting point would have the arrows in the wrong direction initially.  \n\nNow suppose instead that \\(b_{j + 1} = (2,1) = a_{i - 1}\\) . Considering the \\(a\\) - route, the arrows in the upper left corner after \\(i - 1\\) steps must be like this:  \n\n![md5:c6fb9df05a7b0617c31ee0581f632de0](c6fb9df05a7b0617c31ee0581f632de0.jpeg)\n  \n\nConsidering the \\(b\\) - route instead, the arrows after \\(j - 1\\) steps must be like this:  \n\n![md5:f60d351055b547df6566fbcbe9ff1e35](f60d351055b547df6566fbcbe9ff1e35.jpeg)\n  \n\nFrom the arrows in cell \\((1,1)\\) we see that \\(i\\equiv j + 1\\) mod 4. However, for the cells \\((2,1)\\) and \\((1,2)\\) this gives a contradiction.  \n\nWe conclude that the only possible good cells are \\(a_{4t}\\) for \\(t = 0,1,\\ldots ,\\frac{n^{2}}{4} - 1\\) , which gives at most \\(\\frac{n^{2}}{4}\\) good cells.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P5.", "solution_match": "\nSolution."}}
 {"year": "2025", "tier": "T2", "problem_label": "6", "problem_type": null, "exam": "EGMO", "problem": "In each cell of a \\(2025 \\times 2025\\) board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 1, and the sum of the numbers in each column is equal to 1. Define \\(r_{i}\\) to be the largest value in row \\(i\\) , and let \\(R = r_{1} + r_{2} + \\dots + r_{2025}\\) . Similarly, define \\(c_{i}\\) to be the largest value in column \\(i\\) , and let \\(C = c_{1} + c_{2} + \\dots + c_{2025}\\) .What is the largest possible value of \\(\\frac{R}{C}\\) ?  \n\nWhat is the largest possible value of \\(\\frac{R}{C}\\) ?", "solution": "Answer: \\(\\frac{2025}{89}\\) .  \n\nIn general, if the table is \\(m^{2} \\times m^{2}\\) , the answer is \\(\\frac{m^{2}}{2m - 1}\\) .  \n\nThe example is as follows: label rows and columns from 1 to \\(m^{2}\\) , from top to bottom and left to right. For the first \\(m\\) columns, write \\(\\frac{1}{m}\\) in all squares whose coordinates have the same residue modulo \\(m\\) and place 0 everywhere else. For the remaining \\(m^{2} - m\\) columns, place \\(\\frac{1}{m^{2}}\\) everywhere. Then \\(R = m^{2} \\cdot \\frac{1}{m} = m\\) , and \\(C = m \\cdot \\frac{1}{m} + (m^{2} - m) \\cdot \\frac{1}{m^{2}} = 2 - \\frac{1}{m}\\) . So the ratio is as claimed.  \n\n<table><tr><td>1/2</td><td>0</td><td>1/4</td><td>1/4</td></tr><tr><td>0</td><td>1/2</td><td>1/4</td><td>1/4</td></tr><tr><td>1/2</td><td>0</td><td>1/4</td><td>1/4</td></tr><tr><td>0</td><td>1/2</td><td>1/4</td><td>1/4</td></tr></table>  \n\nIn particular, when \\(n := m^{2} = 2025\\) , we get \\(\\frac{2025}{89}\\) . Now we need to show that \\(\\frac{R}{C} \\leq \\frac{n}{2\\sqrt{n} - 1}\\) .  \n\nFor each row, select one cell having the largest value appearing in said row and colour it red. Then, without loss of generality, we may rearrange the columns such that the red cells appear in the first \\(k\\) columns from the left, and each column contains at least one red cell, for some \\(k \\leq n\\) .  \n\nFor the \\(j^{\\mathrm{th}}\\) column, for all \\(1 \\leq j \\leq k\\) , let \\(p_{j}\\) and \\(n_{j}\\) denote the sum and number of red cells in it, respectively. We observe that \\(c_{j}\\) , the biggest number in the \\(j^{\\mathrm{th}}\\) column, is at least \\(\\frac{p_{j}}{n_{j}}\\) , for all \\(1 \\leq j \\leq k\\) . For all other columns, the largest value they contain is at least \\(\\frac{1}{n}\\) , as their sum is 1. Thus, \\(C \\geq \\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\dots + \\frac{p_{k}}{n_{k}} + \\frac{n - k}{n}\\) .  \n\nWe can also observe that \\(R = p_{1} + p_{2} + \\dots + p_{k}\\) .  \n\nTherefore we have to show that:  \n\n\\[p_{1} + p_{2} + \\dots +p_{k} \\leq \\frac{n}{2\\sqrt{n} - 1} \\cdot \\left(\\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\dots + \\frac{p_{k}}{n_{k}} + \\frac{n - k}{n}\\right). \\quad (*)\\]  \n\nBy construction, \\(n_{1} + n_{2} + \\dots + n_{k} = n\\) , and, as the numbers in every column are nonnegative, we see that \\(p_{j} \\leq 1\\) for every \\(j\\) . Also, since each number in a red cell is at least \\(\\frac{1}{n}\\) , we also have \\(p_{j} \\geq \\frac{n_{j}}{n}\\) .  \n\nSince our inequality is linear in each \\(p_{j}\\) , it suffices to prove it when each variable equals one of its two critical values. By relabeling, we may assume that \\(p_{j} = \\frac{n_{j}}{n}\\) for \\(1 \\leq j \\leq t\\) , and \\(p_{j} = 1\\) for \\(t + 1 \\leq j \\leq k\\) , for an integer \\(0 \\leq t \\leq k\\) .  \n\nFirst, if \\(t = k\\) , we observe that \\(p_{1} + p_{2} + \\dots +p_{k} = \\frac{n_{1}}{n} + \\frac{n_{2}}{n} + \\dots + \\frac{n_{k}}{n} = 1\\) , and that \\(\\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\dots + \\frac{p_{k}}{n_{k}} = \\frac{k}{n}\\) , so the inequality becomes \\(1 \\leq \\frac{n}{2\\sqrt{n} - 1}\\) , which is true.  \n\nFrom now on we may assume that \\(t < k\\) . We need to show that:  \n\n\\[\\frac{n_{1} + \\dots + n_{t}}{n} +k - t \\leq \\frac{n}{2\\sqrt{n} - 1} \\cdot \\left(\\frac{t}{n} + \\frac{1}{n_{t + 1}} + \\dots + \\frac{1}{n_{k}} + \\frac{n - k}{n}\\right).\\]  \n\nBy Cauchy- Schwarz inequality we have that:  \n\n\\[\\frac{1}{n_{t + 1}} + \\dots + \\frac{1}{n_{k}} \\geq \\frac{(k - t)^{2}}{n_{t + 1} + \\dots + n_{k}} = \\frac{(k - t)^{2}}{n - (n_{1} + \\dots + n_{t})}. \\quad (CS)\\]\n\n\n\nLet \\(n_{1} + \\dots +n_{t} = n\\cdot q\\) , where \\(0\\leq q< 1\\) . Thus, it is now enough to show that:  \n\n\\[q + k - t\\leq \\frac{n}{2\\sqrt{n - 1}}\\left(\\frac{t}{n} +\\frac{(k - t)^{2}}{n - nq} +\\frac{n - k}{n}\\right).\\]  \n\nLet \\(k - t = \\ell \\geq 1\\) . The inequality becomes:  \n\n\\[q + \\ell \\leq \\frac{1}{2\\sqrt{n} - 1}\\cdot \\left(n - \\ell +\\frac{\\ell^{2}}{1 - q}\\right).\\]  \n\nRearranging this we get:  \n\n\\[n + q + \\frac{\\ell^{2}}{1 - q}\\geq 2(q + \\ell)\\sqrt{n}.\\]  \n\nIf \\(q = 0\\) the inequality is trivially true by \\(A M - G M\\) . Suppose now that \\(0< q< 1\\) . Then, by Cauchy- Schwarz, we have:  \n\n\\[q + \\frac{\\ell^{2}}{1 - q} = \\frac{q^{2}}{q} +\\frac{\\ell^{2}}{1 - q}\\geq (q + \\ell)^{2}.\\]  \n\nIt is therefore enough to show that \\(n + (q + \\ell)^{2}\\geq 2(q + \\ell)\\sqrt{n}\\) , which is true by \\(A M - G M\\) , completing the proof.  \n\nRemark. Another way to show \\(n + q + \\frac{\\ell^{2}}{1 - q}\\geq 2(q + \\ell)\\sqrt{n}\\) is to split it into two \\(A M - G M\\) 's, as:  \n\n\\[\\left(n(1 - q) + \\frac{\\ell^{2}}{1 - q}\\right) + (nq + q)\\geq 2\\ell \\sqrt{n} +2q\\sqrt{n}.\\]", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P6.", "solution_match": "\nSolution 1. "}}
 {"year": "2025", "tier": "T2", "problem_label": "6", "problem_type": null, "exam": "EGMO", "problem": "In each cell of a \\(2025 \\times 2025\\) board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 1, and the sum of the numbers in each column is equal to 1. Define \\(r_{i}\\) to be the largest value in row \\(i\\) , and let \\(R = r_{1} + r_{2} + \\dots + r_{2025}\\) . Similarly, define \\(c_{i}\\) to be the largest value in column \\(i\\) , and let \\(C = c_{1} + c_{2} + \\dots + c_{2025}\\) .What is the largest possible value of \\(\\frac{R}{C}\\) ?  \n\nWhat is the largest possible value of \\(\\frac{R}{C}\\) ?", "solution": "We prove the main inequality \\((\\ast)\\) in a slightly different manner. Instead of the strong lower bound \\(p_{j}\\geq \\frac{n_{j}}{n}\\) , we use the weaker, simpler and more immediate lower bound \\(p_{j}\\geq 0\\) (thus proving the inequality in a larger regime).  \n\nAs in Solution 1, suppose \\(p_{j} = 0\\) for \\(1\\leq j\\leq t\\) and \\(p_{j} = 1\\) for \\(t + 1\\leq j\\leq k\\) , with \\(\\ell = k - t\\) . We also denote by \\(m = n_{t + 1} + \\dots +n_{k}\\) and note that \\(m\\leq n - t\\) , since \\(n_{j}\\geq 1\\) for each \\(i\\leq t\\) . We need to prove that:  \n\n\\[\\ell \\leq \\frac{n}{2\\sqrt{n} - 1}\\cdot \\left(\\frac{1}{n_{t + 1}} +\\dots +\\frac{1}{n_{k}} +\\frac{n - k}{n}\\right).\\]  \n\nRearranging and using the same Cauchy- Schwarz (CS) as in Solution 1, we see it suffices to show that:  \n\n\\[(2\\sqrt{n} -1)\\ell \\leq n - k + \\frac{n\\ell^{2}}{m},\\]  \n\nor equivalently, that:  \n\n\\[2\\sqrt{n}\\ell \\leq n - t + \\frac{n\\ell^{2}}{m}.\\]  \n\nBut since \\(n - t\\geq m\\) this immediately follows from \\(2\\sqrt{n}\\ell \\leq m + \\frac{n\\ell^{2}}{m}\\) , which is a simple application of \\(A M - G M\\) .  \n\n(Note that in Solution 1 the case \\(t = k\\) , which is equivalent to \\(m = \\ell = 0\\) , was dealt with separately, to avoid the appearance of \\(\\frac{0}{0}\\) terms such as \\(\\frac{n\\ell^{2}}{m}\\) . It is easy to verify that the corresponding term should in fact be 0 and the final \\(A M - G M\\) replaced with \\(0\\leq 0\\) , and all transitions are valid. Alternatively, the case can be argued directly by simply noting that \\((\\ast)\\) evaluates to \\(0\\leq \\frac{n - k}{2\\sqrt{n} - 1}\\) , which is obvious.)", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P6.", "solution_match": "\nSolution 1'."}}
 {"year": "2025", "tier": "T2", "problem_label": "6", "problem_type": null, "exam": "EGMO", "problem": "In each cell of a \\(2025 \\times 2025\\) board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 1, and the sum of the numbers in each column is equal to 1. Define \\(r_{i}\\) to be the largest value in row \\(i\\) , and let \\(R = r_{1} + r_{2} + \\dots + r_{2025}\\) . Similarly, define \\(c_{i}\\) to be the largest value in column \\(i\\) , and let \\(C = c_{1} + c_{2} + \\dots + c_{2025}\\) .What is the largest possible value of \\(\\frac{R}{C}\\) ?  \n\nWhat is the largest possible value of \\(\\frac{R}{C}\\) ?", "solution": "1\". This is an alternative way of getting the upper bound on \\(\\frac{R}{C}\\) from  \n\n\\[\\frac{R}{C}\\leq \\frac{p_{1} + p_{2} + \\ldots + p_{k}}{\\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\ldots + \\frac{p_{k}}{n_{k}} + \\frac{n - k}{n}}.\\]  \n\nUsing the fact that \\(\\sum_{j = 1}^{k}n_{j} = n\\) , we can rewrite the above right hand side as follows:\n\n\n\n\\[\\frac{\\sum_{j = 1}^{k}p_{j}}{\\sum_{j = 1}^{k}\\left(\\frac{p_{j}}{n_{j}} +\\frac{n_{j} - 1}{n}\\right)}.\\]  \n\nWe notice that this is a quotient of affine functions in the \\(p_{j}\\) 's, for which the denominator does not vanish over the set defined by \\(0 \\leq p_{j} \\leq 1\\) . Therefore the maximum of this function is attained when a certain number of \\(p_{j}\\) 's are 1 and the others are 0. Without loss of generality we may assume that the first \\(t\\) are equal to 1 and the other \\(k - t\\) are 0 for some \\(0 \\leq t \\leq k\\) . Then one has that the previous expression is at most  \n\n\\[\\frac{t}{\\sum_{1 \\leq j \\leq t} \\left(\\frac{1}{n_{j}} + \\frac{n_{j} - 1}{n}\\right) + \\sum_{t < j \\leq k} \\frac{n_{j} - 1}{n}}.\\]  \n\nWe now lower bound the denominator by observing that the second sum is non negative, while each term of the first sum can be bounded by \\(AM - GM\\) as follows:  \n\n\\[\\frac{1}{n_{j}} + \\frac{n_{j} - 1}{n} \\geq \\frac{2}{\\sqrt{n}} - \\frac{1}{n}.\\]  \n\nWe therefore have  \n\n\\[\\frac{R}{C} \\leq \\max_{1 \\leq t \\leq k} \\frac{t}{\\sum_{1 \\leq j \\leq t} \\left(\\frac{2}{\\sqrt{n}} - \\frac{1}{n}\\right)} = \\frac{n}{2\\sqrt{n} - 1},\\]  \n\nwhich finishes the proof.", "metadata": {"resource_path": "EGMO/segmented/en-2025-solutions.jsonl", "problem_match": "P6.", "solution_match": "\nSolution "}}