Question

一个金字塔有边长为 $1$ 的正方形底面，侧面都是等边三角形。一个立方体放在金字塔内，使其一个面在金字塔底面上，而相对的面所有边都位于金字塔侧面上。这个立方体的体积是多少？

A pyramid has a square base with sides of length $1$ and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?

$5\sqrt{2} - 7$

$7 - 4\sqrt{3}$

$\dfrac{2\sqrt{2}}{27}$

$\dfrac{\sqrt{2}}{9}$

$\dfrac{\sqrt{3}}{9}$

答案：A

知识点：棱锥正方体体积

难度评级：2150

解答：

设立方体边长为 $x$ 。

取经过正方形底面对角线的垂直截面。该截面外轮廓斜边为 $\sqrt2$ ，立方体截面高为 $x$ ，两侧各有直角边为 $x$ 的三角形，中间宽为 $\sqrt2x$ 。

因此 $\sqrt2=\sqrt2x+2x$ ，所以 $x=\sqrt2-1$ ，立方体体积为 $x^3=(\sqrt2-1)^3=5\sqrt2-7$ 。

所以正确答案是 A。

Let the cube have side length $x$ . Take a vertical diagonal cross-section of the pyramid through opposite vertices of the square base.

This cross-section is an isosceles right triangle with hypotenuse $\sqrt2$ . The cube appears as a rectangle of height $x$ and width $\sqrt2x$ , leaving two congruent right isosceles triangles of leg $x$ .

Thus $\sqrt2=\sqrt2x+2x$ , so $x=\sqrt2-1$ . The cube volume is $x^3=(\sqrt2-1)^3=5\sqrt2-7$ .

Thus, A is the correct answer.

2011 AMC 10B 第 22 题

答案：A

难度评级：2150

解答：

其他年份的第 22 题