2018 AMC 8 第 24 题

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.