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.
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1{2 "source_file": "./raw_volume-zh/volume8/chapter1.tex",3 "problem_type": "calculation",4 "problem": "例3. 已知非零复数 $a 、 b 、 c$ 满足 $\\frac{a}{b}=\\frac{b}{c}=\\frac{c}{a}$, 试求 $\\frac{a+b-c}{a-b+c}$ 的一切可能值.",5 "solution": "分析:与解设 $\\frac{a}{b}=\\frac{b}{c}=\\frac{c}{a}=k$, 则 $a=b k, b=c k, c=a k$, 也就有 $c= a k, b=a k \\cdot k=a k^2, a=a k^2 \\cdot k=a k^3$. 因为 $a \\neq 0$, 所以有 $k^3=1$, 解得 $k=1$ 或 $k=-\\frac{1}{2} \\pm \\frac{\\sqrt{3}}{2} \\mathrm{i}$.\n所以 $\\frac{a+b-c}{a-b+c}=\\frac{a+a k^2-a k}{a-a k^2+a k}=\\frac{1+k^2-k}{1-k^2+k}$.\n若 $k=1$, 则原式 $=1$;\n若 $k=-\\frac{1}{2}+\\frac{\\sqrt{3}}{2} \\mathrm{i}$, 则原式 $=-\\frac{1}{2}-\\frac{\\sqrt{3}}{2} \\mathrm{i}$ ;\n若 $k=-\\frac{1}{2}-\\frac{\\sqrt{3}}{2} \\mathrm{i}$, 则原式 $=-\\frac{1}{2}+\\frac{\\sqrt{3}}{2} \\mathrm{i}$.\n综上所述, $\\frac{a+b-c}{a-b+c}$ 的一切可能值为 $1 、-\\frac{1}{2}-\\frac{\\sqrt{3}}{2} \\mathrm{i}$ 和 $-\\frac{1}{2}+\\frac{\\sqrt{3}}{2} \\mathrm{i}$.",6 "remark": "注:连等式设 $k$ 是常用的解题技巧.",7 "figures": []8}