Question

对一个三角形，它的中点三角形由连接三边中点得到。递归定义一列多面体 $\mathcal{P}_i$ 如下： $\mathcal{P}_0$ 是体积为 $1$ 的正四面体。为了得到 $\mathcal{P}_{i+1}$ ，将 $\mathcal{P}_i$ 的每个面上的中点三角形替换为一个向外突出的正四面体，该中点三角形作为这个正四面体的一个面。 $\mathcal{P}_3$ 的体积为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Given a triangle, its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra $\mathcal{P}_i$ is defined recursively as follows: $\mathcal{P}_0$ is a regular tetrahedron whose volume is $1.$ To obtain $\mathcal{P}_{i+1},$ replace the midpoint triangle of every face of $\mathcal{P}_i$ by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of $\mathcal{P}_3$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：101

知识点：体积长度、面积与体积的缩放关系递推等比数列

难度评级：2990

解答：

在某个面上的中点三角形处贴上一个四面体，会把原来的面替换成 $6$ 个边长为一半的等边三角形： $3$ 个角上的三角形，加上新四面体暴露出的 $3$ 个面。因此 $\mathcal{P}_i$ 的所有面全等，边长为原来的 $\left(\frac{1}{2}\right)^i$ 倍，且 $\mathcal{P}_i$ 有 $4 \cdot 6^i$ 个面。

从 $\mathcal{P}_i$ 到 $\mathcal{P}_{i+1}$ 时，每个面上粘上一个正四面体；每个新四面体都与 $\mathcal{P}_0$ 相似，比例为 $\left(\frac{1}{2}\right)^{i+1}$ ，因而体积为 $\left(\frac{1}{8}\right)^{i+1}$ 。增加的体积为 $4 \cdot 6^i \left(\frac{1}{8}\right)^{i+1} = \frac{1}{2}\left(\frac{3}{4}\right)^i.$

因此 $\mathcal{P}_3$ 的体积为 $1 + \frac{1}{2} + \frac{3}{8} + \frac{9}{32} = \frac{69}{32}$ ，所以 $m + n = 69 + 32 = 101$ 。

Attaching a tetrahedron over the midpoint triangle of a face replaces that face by $6$ equilateral triangles of half the side length: the $3$ corner triangles plus $3$ exposed faces of the new tetrahedron. So all faces of $\mathcal{P}_i$ are congruent, with side $\left(\frac{1}{2}\right)^i$ times the original, and $\mathcal{P}_i$ has $4 \cdot 6^i$ faces.

Passing from $\mathcal{P}_i$ to $\mathcal{P}_{i+1}$ glues one regular tetrahedron onto each face; each is similar to $\mathcal{P}_0$ with ratio $\left(\frac{1}{2}\right)^{i+1},$ hence has volume $\left(\frac{1}{8}\right)^{i+1}.$ The volume added is $4 \cdot 6^i \left(\frac{1}{8}\right)^{i+1} = \frac{1}{2}\left(\frac{3}{4}\right)^i.$

Therefore the volume of $\mathcal{P}_3$ is $1 + \frac{1}{2} + \frac{3}{8} + \frac{9}{32} = \frac{69}{32},$ and $m + n = 69 + 32 = 101.$

2001 AIME II 第 12 题

答案：101

难度评级：2990

解答：

其他年份的第 12 题