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/volume14/exercise5.tex",3    "problem_type": "calculation",4    "problem": "问题4. 将 $a \\times b \\times c\\left(a, b, c \\in \\mathbf{N}^*\\right)$ 的长方体分成 $a b c$ 个 $1 \\times 1 \\times 1$ 的小正方体.\n求长方体的一条对角线穿过的小正方体的个数.",5    "solution": "设长方体占据的空间区域为 $\\{(x, y, z) \\mid 0 \\leqslant x \\leqslant a, 0 \\leqslant y \\leqslant b, 0 \\leqslant z \\leqslant c\\}$.\n将质点沿对角线从 $(0,0,0)$ 穿行到 $(a, b, c)$ 的路径用参数 $t$ 表示为\n$$\nx=a t, y=b t, z=c t, t \\in[0,1] .\n$$\n在 $t$ 递增的过程中, 当且仅当 $x, y, z$ 中至少有一个为整数时, 质点将穿人一个新的小正方体,这里 $t \\in[0,1)$. 设\n$$\n\\begin{aligned}\n& A_x=\\{t \\mid t \\in[0,1), \\text { at } \\in \\mathbf{Z}\\}, \\\\\n& A_y=\\{t \\mid t \\in[0,1), b t \\in \\mathbf{Z}\\}, \\\\\n& A_z=\\{t \\mid t \\in[0,1), c t \\in \\mathbf{Z}\\},\n\\end{aligned}\n$$\n则有 $\\left|A_x\\right|=a,\\left|A_y\\right|=b,\\left|A_z\\right|=c$.\n又记 $d_1=(a, b), d_2=(b, c), d_3=(c, a), d=(a, b, c)$, 则\n$$\nA_x \\cap \\dot{A_y}=\\left\\{t \\mid t \\in[0,1), d_1 t \\in \\mathbf{Z}\\right\\},\n$$\n从而\n$$\n\\left|A_x \\cap A_y\\right|=d_1 .\n$$\n同理得 $\\left|A_y \\cap A_z\\right|=d_2,\\left|A_z \\cap A_x\\right|=d_3,\\left|A_x \\cap A_y \\cap A_z\\right|=d$.\n故由容斥原理得, 质点一共穿过的小正方体个数为\n$$\n\\begin{aligned}\n\\left|A_x \\cup A_y \\cup A_z\\right|= & \\left|A_x\\right|+\\left|A_y\\right|+\\left|A_z\\right|-\\left|A_x \\cap A_y\\right|-\\left|A_y \\cap A_z\\right|- \\\\\n& \\left|A_z \\cap A_x\\right|+\\left|A_x \\cap A_y \\cap A_z\\right| \\\\\n= & a+b+c-(a, b)-(b, c)-(c, a)+(a, b, c) .\n\\end{aligned}\n$$",6    "remark": "",7    "figures": []8}