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：

其他年份的第 22 题