2005 AIME I 第 9 题

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9.

二十七个单位立方体各有四个面被涂成橙色,使得两个未涂色的面共用一条边。然后将这 2727 个立方体随机排列成一个 3×3×33 \times 3 \times 3 的立方体。已知整个大立方体表面全为橙色的概率为 paqbrc\frac{p^a}{q^b r^c}, 其中 ppqq, 和 rr 是不同的质数,aabb, 和 cc 是正整数,求 a+b+c+p+q+ra + b + c + p + q + r

Twenty-seven unit cubes are each painted orange on a set of four faces so that the two unpainted faces share an edge. The 2727 cubes are then randomly arranged to form a 3×3×33 \times 3 \times 3 cube. Given that the probability that the entire surface of the larger cube is orange is paqbrc,\frac{p^a}{q^b r^c}, where p,p, q,q, and rr are distinct primes and a,a, b,b, and cc are positive integers, find a+b+c+p+q+r.a + b + c + p + q + r.

答案:74
知识点:基本概率独立事件正方体分类讨论
难度评级:2920
解答:

每个单位立方体都有一条“坏边”:即两个未涂色面共用的那条边。大立方体的表面全为橙色,当且仅当每个单位立方体的坏边不接触任何可见面。一个均匀随机的朝向会使坏边均匀地落在该立方体的 1212 条棱的位置中,所以对每个单位立方体,只需数出两侧面都被隐藏的棱位置数。

一个角块显示 33 个共顶点的面;安全棱是相对顶点处的 33 个隐藏面所共用的棱,所以概率为 312=14\frac{3}{12} = \frac{1}{4}。一个棱块显示 22 个相邻面,它们接触 4+41=74 + 4 - 1 = 7 条棱,留下 55 条安全棱,概率为 512\frac{5}{12}。一个面心块显示 11 个面,接触 44 条棱,留下 88 条安全棱,概率为 812=23\frac{8}{12} = \frac{2}{3}。中心块总是符合条件。

88 个角块、1212 个棱块和 66 个面心块,所以概率为 (14)8(512)12(23)6=512234318, \begin{aligned} &\left(\frac{1}{4}\right)^{8}\left(\frac{5}{12}\right)^{12}\left(\frac{2}{3}\right)^{6} \\ &= \frac{5^{12}}{2^{34} \cdot 3^{18}}, \end{aligned} 因此 a+b+c+p+q+ra + b + c + p + q + r =12+34+18= 12 + 34 + 18 +5+2+3+ 5 + 2 + 3 =74= 74

Each unit cube has one "bad edge": the edge shared by its two unpainted faces. The larger cube's surface is entirely orange exactly when every unit cube's bad edge touches no visible face. A uniformly random orientation places the bad edge uniformly among the cube's 1212 edge positions, so for each unit cube we count the edge positions both of whose faces are hidden.

A corner cube shows 33 faces meeting at a vertex; the safe edges are those of the 33 hidden faces meeting at the opposite vertex, so the probability is 312=14.\frac{3}{12} = \frac{1}{4}. An edge cube shows 22 adjacent faces, which touch 4+41=74 + 4 - 1 = 7 edges, leaving 55 safe: probability 512.\frac{5}{12}. A face-center cube shows 11 face touching 44 edges, leaving 88 safe: probability 812=23.\frac{8}{12} = \frac{2}{3}. The center cube is always fine.

With 88 corner, 1212 edge, and 66 face-center cubes, the probability is (14)8(512)12(23)6=512234318, \begin{aligned} &\left(\frac{1}{4}\right)^{8}\left(\frac{5}{12}\right)^{12}\left(\frac{2}{3}\right)^{6} \\ &= \frac{5^{12}}{2^{34} \cdot 3^{18}}, \end{aligned} so a+b+c+p+q+ra + b + c + p + q + r =12+34+18= 12 + 34 + 18 +5+2+3+ 5 + 2 + 3 =74.= 74.

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