Question

在立方体 $ABCDEFGH$ 中， $C$ 与 $E$ 是相对顶点， $J$ 和 $I$ 分别是棱 $\overline{FB}$ 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？

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 $R^2?$

$\dfrac{5}{4}$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\dfrac{25}{16}$

$\dfrac{9}{4}$

答案：C

知识点：正方体菱形勾股定理

难度评级：1910

小提示：

四边形 $EJCI$ 是菱形，可以用对角线求面积

The quadrilateral $EJCI$ is a rhombus, so use its diagonals.

大提示：

用立方体边长表示对角线 $IJ$ 和 $CE$

Express the diagonals $IJ$ and $CE$ in terms of the cube side length.

视频讲解：

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文字解答：

设立方体边长为 $s$ 。注意截面四边形每条边长度相同，所以 $EJCI$ 是菱形。菱形面积等于两条对角线乘积的一半，因此它的面积为 $\frac12 IJ\cdot CE$ 。由勾股定理， $IJ=FH=s\sqrt{2}\text{。}$ 再次使用勾股定理可得 $\begin{align*}CE&=\sqrt{AC^2+AE^2}\\&=\sqrt{(s\sqrt{2})^2+s^2}\\&=\sqrt{2s^2+s^2}\\&=s\sqrt{3}\end{align*}$ 因此 $\begin{align*}R&=\dfrac{\frac12 IJ\cdot CE}{s^2}\\&=\dfrac{\frac12 s^2\sqrt{2}\sqrt{3}}{s^2}\\&=\sqrt{\dfrac32}\end{align*}$ 所以 $R^2=\dfrac32$ ，正确答案是 C。

Allow $s$ to represent the length of an edge of the cube. Noting that each side of the cross section is equal in length, we conclude that $EJCI$ is a rhombus. The area of this rhombus can be calculated as $\frac12 IJ\cdot CE,$ as the area of a rhombus is equal to half the product of its diagonals. Using the Pythagorean Theorem: $IJ=FH=s\sqrt{2}.$ Similarly, using the Pythagorean Theorem again lets us see that: $\begin{align*}CE&=\sqrt{AC^2+AE^2}\\&=\sqrt{(s\sqrt{2})^2+s^2}\\&=\sqrt{2s^2+s^2}\\&=s\sqrt{3}\end{align*}$ Therefore, $\begin{align*}R&=\dfrac{\frac12 IJ\cdot CE}{s^2}\\&=\dfrac{\frac12 s^2\sqrt{2}\sqrt{3}}{s^2}\\&=\sqrt{\dfrac32}\end{align*}$ Thus, $R^2=\dfrac32,$ and the correct answer is C.

2018 AMC 8 第 24 题