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/volume3/exercise1.tex",3 "problem_type": "calculation",4 "problem": "问题26. (1) 求函数 $f(\\theta)=\\cos \\frac{\\theta}{2} \\sin \\theta, \\theta \\in\\left(0, \\frac{\\pi}{2}\\right)$ 的最大值;\n(2) 求函数 $g(\\theta)=\\sin \\frac{\\theta}{2} \\cos \\theta, \\theta \\in\\left(0, \\frac{\\pi}{2}\\right)$ 的最大值.",5 "solution": "(1) $f(\\theta)=2 \\sin \\frac{\\theta}{2} \\cdot \\cos ^2 \\frac{\\theta}{2}=2 \\sqrt{\\sin ^2 \\frac{\\theta}{2} \\cdot \\cos ^4 \\frac{\\theta}{2}}=\\sqrt{2}\\sqrt{2 \\sin ^2 \\frac{\\theta}{2} \\cdot \\cos ^2 \\frac{\\theta}{2} \\cdot \\cos ^2 \\frac{\\theta}{2}} \\leqslant \\sqrt{2} \\cdot \\sqrt{\\left[\\frac{2 \\sin ^2 \\frac{\\theta}{2}+\\cos ^2 \\frac{\\theta}{2}+\\cos ^2 \\frac{\\theta}{2}}{3}\\right]^3}=\\sqrt{2} \\cdot \\left(\\frac{2}{3}\\right)^{\\frac{3}{2}}=\\frac{4 \\sqrt{3}}{9}$, 当且仅当 $\\theta=2 \\arctan \\frac{\\sqrt{2}}{2}$ 时取等号.\n所以, $f(\\theta)$ 的最大值为 $\\frac{4 \\sqrt{3}}{9}$.\n(2) 由于 $\\theta \\in\\left(0, \\frac{\\pi}{2}\\right)$, 故 $g(\\theta)>0, g^2(\\theta)=\\sin ^2 \\frac{\\theta}{2} \\cdot \\cos ^2 \\theta=\\frac{1-\\cos \\theta}{2}$. $\\cos ^2 \\theta=\\frac{2(1-\\cos \\theta) \\cdot \\cos \\theta \\cdot \\cos \\theta}{4} \\leqslant \\frac{1}{4} \\cdot\\left[\\frac{2(1-\\cos \\theta)+\\cos \\theta+\\cos \\theta}{3}\\right]^3= \\frac{2}{27}$, 当且仅当 $\\theta=\\arccos \\frac{2}{3}$ 时取等号.\n所以, $g(\\theta)$ 的最大值为 $\\frac{\\sqrt{6}}{9}$. 当然也可以如下这样, $g(\\theta)=\\sin \\frac{\\theta}{2} \\cdot\\left(1-2 \\sin ^2 \\frac{\\theta}{2}\\right)=\\frac{1}{2} \\sqrt{4 \\sin ^2 \\frac{\\theta}{2} \\cdot\\left(1-2 \\sin ^2 \\frac{\\theta}{2}\\right) \\cdot\\left(1-2 \\sin ^2 \\frac{\\theta}{2}\\right)} \\leqslant \\frac{1}{2} \\cdot \\sqrt{\\left[\\frac{4 \\sin ^2 \\frac{\\theta}{2}+\\left(1-2 \\sin ^2 \\frac{\\theta}{2}\\right)+\\left(1-2 \\sin ^2 \\frac{\\theta}{2}\\right)}{3}\\right]^3}=\\frac{\\sqrt{6}}{9}$. 当且仅当 $\\theta=2 \\arcsin \\frac{\\sqrt{6}}{6}$ 时取等号.\n故 $g(\\theta)$ 的最大值为 $\\frac{\\sqrt{6}}{9}$.",6 "remark": "",7 "figures": []8}