AMC8 2018

AMC8 2018 · Q24

AMC8 2018 · Q24. It mainly tests 3D geometry (volume).

In the cube ABCDEFGH with opposite vertices C and E, J and I are the midpoints of edges $\overline{FB}$ and $\overline{HD}$, respectively. Let $R$ be the ratio of the area of the cross-section EJCI to the area of one of the faces of the cube. What is $R^2$?

在立方体ABCDEFGH中，相对顶点为C和E，J和I分别是边$\overline{FB}$和$\overline{HD}$的中点。设$R$为截面EJCI的面积与立方体一个面的面积之比。$R^2$是多少？

(A) $\frac{5}{4}$ $\frac{5}{4}$

(B) $\frac{4}{3}$ $\frac{4}{3}$

(D) $\frac{25}{16}$ $\frac{25}{16}$

(E) $\frac{9}{4}$ $\frac{9}{4}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $s$ denote the length of an edge of the cube. Now $EJCI$ is a non-square rhombus whose area is $\frac{1}{2}EC\cdot JI$, because the area of a rhombus is half the product of the lengths of its diagonals. By the Pythagorean Theorem, $JI=FH=s\sqrt{2}$, and using the Pythagorean Theorem twice, $EC=s\sqrt{3}$. Thus $R=\frac{\frac{1}{2}EC\cdot JI}{s^2}=\frac{\frac{1}{2}(s\sqrt{3})(s\sqrt{2})}{s^2}=\frac{\sqrt{6}}{2}$ and $R^2=\frac{6}{4}=\frac{3}{2}$.

答案（C）：设 $s$ 表示立方体棱长。$EJCI$ 是一个非正方形的菱形，其面积为 $\frac{1}{2}EC\cdot JI$，因为菱形面积等于两条对角线长度乘积的一半。由勾股定理可得 $JI=FH=s\sqrt{2}$；再两次使用勾股定理可得 $EC=s\sqrt{3}$。因此 $R=\frac{\frac{1}{2}EC\cdot JI}{s^2}=\frac{\frac{1}{2}(s\sqrt{3})(s\sqrt{2})}{s^2}=\frac{\sqrt{6}}{2}$，并且 $R^2=\frac{6}{4}=\frac{3}{2}$。

Topics

GE.013 3D geometry (volume)