AMC12 2011 B

AMC12 2011 B · Q18

AMC12 2011 B · Q18. It mainly tests Coordinate geometry, 3D geometry (volume).

A pyramid has a square base with side 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，且侧面为等边三角形。在金字塔内放置一个立方体，使其中一个面在金字塔底面上，其对面的所有边都在金字塔的侧面上。求此立方体的体积。

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

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

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

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

Answer

Correct choice: (A)

正确答案：（A）

Solution

We can use the Pythagorean Theorem to split one of the triangular faces into two 30-60-90 triangles with side lengths $\frac{1}{2}, 1$ and $\frac{\sqrt{3}}{2}$. Next, take a cross-section of the pyramid, forming a triangle with the top of the pyramid and the midpoints of two opposite sides of the square base. This triangle is isosceles with a base of 1 and two sides of length $\frac{\sqrt{3}}{2}$. The height of this triangle will equal the height of the pyramid. To find this height, split the triangle into two right triangles, with sides $\frac{1}{2}, \frac{\sqrt2}{2}$ and $\frac{\sqrt{3}}{2}$. The cube, touching all four triangular faces, will form a similar pyramid that sits on top of the cube. If the cube has side length $x$, the small pyramid has height $\frac{x\sqrt{2}}{2}$ (because the pyramids are similar). Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid. $x +\frac{x\sqrt{2}}{2} = \frac{\sqrt2}{2}$. $x\left(1+\frac{\sqrt{2}}{2} \right) =\frac{\sqrt{2}}{2}$ $x\left(2+\sqrt{2}\right) = \sqrt{2}$ $x = \frac{\sqrt{2}}{2+\sqrt{2}} \cdot \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2}-2}{4-2} = \sqrt{2}-1 =$side length of cube. $\left(\sqrt{2}-1\right)^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(-1) + 3(\sqrt{2})(-1)^2 + (-1)^3 = 2\sqrt{2} - 6 +3\sqrt{2} - 1 =\textbf{(A)} 5\sqrt{2} - 7$

我们可以用勾股定理将其中一个三角形侧面分成两个 $30$-$60$-$90$ 三角形，其边长分别为 $\frac{1}{2}, 1$ 和 $\frac{\sqrt{3}}{2}$。接着取金字塔的一个截面，该截面形成一个三角形，其顶点为金字塔顶点，底边为正方形底面两条对边的中点连线。该三角形是等腰三角形，底边长为 $1$，两腰长为 $\frac{\sqrt{3}}{2}$。该三角形的高等于金字塔的高。为求此高，将该三角形分成两个直角三角形，其边长为 $\frac{1}{2}, \frac{\sqrt2}{2}$ 和 $\frac{\sqrt{3}}{2}$。立方体与四个三角形侧面都相切，会在立方体上方形成一个相似的小金字塔。若立方体的棱长为 $x$，则小金字塔的高为 $\frac{x\sqrt{2}}{2}$（因为两个金字塔相似）。因此，立方体的高加上小金字塔的高等于大金字塔的高： $x +\frac{x\sqrt{2}}{2} = \frac{\sqrt2}{2}$。 $x\left(1+\frac{\sqrt{2}}{2} \right) =\frac{\sqrt{2}}{2}$ $x\left(2+\sqrt{2}\right) = \sqrt{2}$ $x = \frac{\sqrt{2}}{2+\sqrt{2}} \cdot \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2}-2}{4-2} = \sqrt{2}-1 =$ 立方体的棱长。 $\left(\sqrt{2}-1\right)^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(-1) + 3(\sqrt{2})(-1)^2 + (-1)^3 = 2\sqrt{2} - 6 +3\sqrt{2} - 1 =\textbf{(A)} 5\sqrt{2} - 7$。

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