AMC10 2011 B

AMC10 2011 B · Q22

AMC10 2011 B · Q22. It mainly tests Coordinate geometry, 3D geometry (volume).

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?

一个金字塔具有边长为 1 的正方形底面，其侧面为等边三角形。在金字塔内放置一个立方体，使一面在金字塔底面上，其相对面所有的边都在金字塔的侧面上。这个立方体的体积是多少？

(A) $5\sqrt{2}-7$ $5\sqrt{2}-7$

(B) $7-4\sqrt{3}$ $7-4\sqrt{3}$

(D) $\frac{\sqrt{2}}{9}$ $\frac{\sqrt{2}}{9}$

(E) $\frac{\sqrt{3}}{9}$ $\frac{\sqrt{3}}{9}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Let $A$ be the apex of the pyramid, and let the base be the square $BCDE$. Then $AB=AD=1$ and $BD=\sqrt{2}$, so $\triangle BAD$ is an isosceles right triangle. Let the cube have edge length $x$. The intersection of the cube with the plane of $\triangle BAD$ is a rectangle with height $x$ and width $\sqrt{2}x$. It follows that $\sqrt{2}=BD=2x+\sqrt{2}x$, from which $x=\sqrt{2}-1$. Hence the cube has volume $$(\sqrt{2}-1)^3=(\sqrt{2})^3-3(\sqrt{2})^2+3\sqrt{2}-1=5\sqrt{2}-7.$$

答案（A）：设 $A$ 为金字塔的顶点，底面为正方形 $BCDE$。则 $AB=AD=1$ 且 $BD=\sqrt{2}$，所以 $\triangle BAD$ 为等腰直角三角形。设立方体的棱长为 $x$。立方体与 $\triangle BAD$ 所在平面的交截面是一个高为 $x$、宽为 $\sqrt{2}x$ 的矩形。由此得 $\sqrt{2}=BD=2x+\sqrt{2}x$，因此 $x=\sqrt{2}-1$。因此立方体体积为 $$(\sqrt{2}-1)^3=(\sqrt{2})^3-3(\sqrt{2})^2+3\sqrt{2}-1=5\sqrt{2}-7.$$

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