Question

已知 $\mathcal{O}$ 是一个正八面体， $\mathcal{C}$ 是以 $\mathcal{O}$ 各面的中心为顶点的立方体，且 $\mathcal{O}$ 的体积与 $\mathcal{C}$ 的体积之比为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数，求 $m + n$ 。

Given that $\mathcal{O}$ is a regular octahedron, that $\mathcal{C}$ is the cube whose vertices are the centers of the faces of $\mathcal{O},$ and that the ratio of the volume of $\mathcal{O}$ to that of $\mathcal{C}$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

答案：11

知识点：立体几何体积坐标几何

难度评级：2450

解答：

将八面体的顶点放在 $(\pm 1, 0, 0)$ ， $(0, \pm 1, 0)$ ， $(0, 0, \pm 1)$ 。它由两个四棱锥沿着顶点为 $(\pm 1, 0, 0)$ 和 $(0, \pm 1, 0)$ 的正方形粘合而成；该正方形面积为 $2$ ，每个四棱锥的高为 $1$ ，所以 $V_{\mathcal{O}} = 2 \cdot \frac{1}{3} \cdot 2 \cdot 1 = \frac{4}{3}.$

每个面的重心是该面三个顶点的平均值，例如 $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$ ，因此立方体顶点为 $\left(\pm\frac{1}{3}, \pm\frac{1}{3}, \pm\frac{1}{3}\right)$ 。它的边长为 $\frac{2}{3}$ ，体积为 $\frac{8}{27}$ 。

比值为 $\frac{4/3}{8/27} = \frac{9}{2}$ ，所以 $m + n = 9 + 2 = 11$ 。

Place the octahedron's vertices at $(\pm 1, 0, 0),$ $(0, \pm 1, 0),$ $(0, 0, \pm 1).$ It is two square pyramids glued along the square with vertices $(\pm 1, 0, 0)$ and $(0, \pm 1, 0),$ which has area $2,$ and each pyramid has height $1,$ so $V_{\mathcal{O}} = 2 \cdot \frac{1}{3} \cdot 2 \cdot 1 = \frac{4}{3}.$

Each face centroid is the average of that face's three vertices, e.g. $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right),$ so the cube has vertices $\left(\pm\frac{1}{3}, \pm\frac{1}{3}, \pm\frac{1}{3}\right).$ Its edge is $\frac{2}{3}$ and its volume is $\frac{8}{27}.$

The ratio is $\frac{4/3}{8/27} = \frac{9}{2},$ so $m + n = 9 + 2 = 11.$

2005 AIME II 第 10 题

答案：11

难度评级：2450

解答：

其他年份的第 10 题