CoolFace
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math-ai/BlueMO

BlueMO 🚀 BlueMO: A Comprehensive Collection of Challenging Mathematical Olympiad Problems from the Little Blue Book Series   BlueMO is a comprehensive and challenging dataset comprising mathematical olympiad problems paired with detailed solutions, meticulously curated from the esteemed "Little Blue Book" (小蓝书) series (Second Edition)—a vital resource for Chinese students training for national and international olympiad math competitions. Designed to… See the full description on the dataset page: https://huggingface.co/datasets/math-ai/BlueMO.

sourceHugging Facecc-by-nd-4.0updated 8mo agoView on Hugging Face
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1{2    "source_file": "./raw_volume-zh/volume9/exercise2.tex",3    "problem_type": "calculation",4    "problem": "问题2. $\\triangle A B C$ 的边 $B C$ 和边 $A C$ 分别取定长 $a$ 和 $b$, 而边 $A B$ 的长度可变动.\n以边 $A B$ 作为正方形的一边向三角形外作正方形.\n设 $O$ 是所作正方形的中心, 并设 $B C$ 和 $A C$ 的中点分别为 $M$ 和 $N$. 试求 $O M+O N$ 的最大值.",5    "solution": "设正方形为 $A B D E$, 则 $O M=\\frac{C E}{2}, O N=\\frac{C D}{2}$, 所以 $O M+O N= \\frac{1}{2}(C D+C E)$. 设 $A B=c$, 则 $B D=A E=c, A D=B E=\\sqrt{2} c$. 对四边形 $A C B D$ 和 $A C B E$ 分别应用广义 Ptolemy 定理可得 $O M+O N=\\frac{1}{2}(C D+ C E) \\leqslant \\frac{\\sqrt{2}+1}{2}(a+b)$. 所以 $O M+O N$ 的最大值为 $\\frac{\\sqrt{2}+1}{2}(a+b)$.",6    "remark": "",7    "figures": []8}