Question

一个四棱锥的底面是边长为 $1$ 的正方形，侧面都是等边三角形。一个立方体放在该四棱锥内，使它的一个面在四棱锥底面上，而相对的那个面所有边都在四棱锥的侧面上。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：棱锥正方体立体几何

难度评级：2030

解答：

设顶点为 $A$ ，底面正方形为 $BCDE$ 。于是 $AB=AD=1$ ，且 $BD=\sqrt2$ ，所以 $\triangle BAD$ 是等腰直角三角形。

设立方体边长为 $x$ 。它与 $\triangle BAD$ 所在平面的交线是一个高为 $x$ 、宽为 $\sqrt2\,x$ 的长方形，其上方两个顶点在 $AB$ 与 $AD$ 上。因为 $AB$ 与 $AD$ 与底边成 $45^\circ$ ，长方形外侧的两段 $BD$ 各长 $x$ ，所以得 $x=\dfrac{\sqrt2}{2+\sqrt2}=\sqrt2-1$ 。 $\sqrt2=BD=\sqrt2\,x+2x,$

体积为 $(\sqrt2-1)^3=5\sqrt2-7.$

所以正确答案是 A。

Let the apex be $A$ and the base be square $BCDE.$ Then $AB=AD=1$ and $BD=\sqrt2,$ so $\triangle BAD$ is an isosceles right triangle.

Let the cube have edge length $x.$ Its intersection with the plane of $\triangle BAD$ is a rectangle of height $x$ and width $\sqrt2\,x,$ whose top corners lie on $AB$ and $AD.$ Because the legs $AB$ and $AD$ meet the base at $45^\circ,$ each portion of $BD$ outside the rectangle has length $x,$ so $\sqrt2=BD=\sqrt2\,x+2x,$ which reduces to $x=\dfrac{\sqrt2}{2+\sqrt2}=\sqrt2-1.$

The volume is $(\sqrt2-1)^3=5\sqrt2-7.$

Thus, the correct answer is A.

2011 AMC 12B 第 18 题

答案：A

难度评级：2030

解答：

其他年份的第 18 题