2009 AMC 12A 第 22 题

Question

一个正八面体的边长为

1

。一个平行于其中两张相对面的平面把该八面体切成两个全等的立体。该平面与八面体相交形成的多边形面积为

\dfrac{a\sqrt{b}}{c}

，其中

a

，

b

，和

c

是正整数，

a

与

c

互质，且

b

不被任何质数的平方整除。求

a + b + c

？

A regular octahedron has side length

1.

A plane parallel to two of its opposite faces cuts the octahedron into two congruent solids. The polygon formed by the intersection of the plane and the octahedron has area

\dfrac{a\sqrt{b}}{c},

where

a,

b,

and

c

are positive integers,

a

and

c

are relatively prime, and

b

is not divisible by the square of any prime. What is

a + b + c?

10

11

12

13

14

答案：E

知识点：立体几何正多边形面积

难度评级：2270

小提示：

切割平面经过不属于那两张平行面的六条边的中点。

The cutting plane passes through the midpoints of the six edges that do not belong to the two parallel faces

大提示：

这些中点形成一个边长为

\dfrac{1}{2}

的正六边形；求它的面积。

Those midpoints form a regular hexagon with side

\dfrac{1}{2};

find its area

解答：

设那两张平行面为三角形。该平面经过不在这些面上的六条边的中点，形成一个边长为

\dfrac{1}{2}

的等边六边形；由对称性它也等角，因此是正六边形。

正六边形由六个等边三角形组成，所以面积为

6\cdot\frac{\sqrt{3}}{4}\left(\frac{1}{2}\right)^2 = \frac{3\sqrt{3}}{8}\text{。}

因此

a = 3

，

b = 3

，

c = 8

，且

a + b + c = 14

。

因此，正确答案是 E。

Let the two parallel faces be triangles. The plane passes through the midpoints of the six edges not on those faces, forming an equilateral hexagon of side

\dfrac{1}{2},

which by symmetry is also equiangular and hence regular.

A regular hexagon is six equilateral triangles, so its area is

6\cdot\frac{\sqrt{3}}{4}\left(\frac{1}{2}\right)^2 = \frac{3\sqrt{3}}{8}.

Thus

a = 3,

b = 3,

c = 8,

and

a + b + c = 14.

Thus, the correct answer is E.

2009 AMC 12A 第 22 题

答案：E

难度评级：2270

小提示：

大提示：

解答：

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