AMC10 2012 B

AMC10 2012 B · Q23

AMC10 2012 B · Q23. It mainly tests Triangles (properties), 3D geometry (volume).

A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?

一个实心四面体通过一个平面从实心木制单位立方体上切下，该平面经过一个面上两个不相邻的顶点和相对面上不与前两者相邻的一个顶点。四面体被丢弃，立方体剩余部分以切面朝下放置在桌子上。此物的高度是多少？

(A) $\sqrt{3}/3$ $\sqrt{3}/3$

(B) $2\sqrt{2}/3$ $2\sqrt{2}/3$

(D) $2\sqrt{3}/3$ $2\sqrt{3}/3$

(E) $\sqrt{2}$ $\sqrt{2}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The discarded tetrahedron can be viewed as having an isosceles right triangle of side 1 as its base, with an altitude of 1. Therefore its volume is $\frac{1}{6}$. It can also be viewed as having an equilateral triangle of side length $\sqrt{2}$ as its base, in which case its altitude $h$ must satisfy $\frac{1}{3}\cdot\frac{\sqrt{3}}{4}\cdot(\sqrt{2})^2\cdot h=\frac{1}{6},$ which implies that $h=\frac{\sqrt{3}}{3}$. The height of the remaining solid is the long diagonal of the cube minus $h$, which is $\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}$.

答案（D）：被丢弃的四面体可以看作以边长为 1 的等腰直角三角形为底面，高为 1。因此其体积为$\frac{1}{6}$。也可以看作以边长为$\sqrt{2}$的正三角形为底面，此时其高$h$满足 $\frac{1}{3}\cdot\frac{\sqrt{3}}{4}\cdot(\sqrt{2})^2\cdot h=\frac{1}{6},$ 由此得到$h=\frac{\sqrt{3}}{3}$。剩余立体的高度等于正方体的体对角线长度减去$h$，即$\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}$。

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