AMC12 2009 A

AMC12 2009 A · Q22

AMC12 2009 A · Q22. It mainly tests Area & perimeter, 3D geometry (volume).

A regular octahedron has side length $1$. A plane parallel to two of its opposite faces cuts the octahedron into the two congruent solids. The polygon formed by the intersection of the plane and the octahedron has area $\frac {a\sqrt {b}}{c}$, 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 $a + b + c$?

一个正八面体的棱长为 $1$。一个平面平行于它的一对相对的面，并将该八面体切成两个全等的立体。该平面与八面体的交线所成多边形的面积为 $\frac {a\sqrt {b}}{c}$，其中 $a$、$b$、$c$ 为正整数，$a$ 与 $c$ 互质，且 $b$ 不被任何质数的平方整除。求 $a + b + c$。

(A) 10 10

(B) 11 11

(D) 13 13

(E) 14 14

Answer

Correct choice: (E)

正确答案：（E）

Solution

Firstly, note that the intersection of the plane must be a hexagon. Consider the net of the octahedron. Notice that the hexagon becomes a line on the net. Also, notice that, given the parallel to the faces conditions, the line must be parallel to precisely $\frac{1}{3}$ of the sides of the net. Now, notice that, through symmetry, 2 of the hexagon's vertexes lie on the midpoint of the side of the "square" in the octahedron. In the net, the condition gives you that one of the intersections of the line with the net have to be on the midpoint of the side. However, if one is on the midpoint, because of the parallel conditions, all of the vertices are on the midpoint of a side. Thus, we have a regular hexagon with a side length of the midline of an equilateral triangle with side length 1, which is $\frac{1}{2}$. Thus, the answer is$\frac {3\sqrt {3}}{8}$, and $a + b + c = 14\ \mathbf{(E)}$. (Can somebody clarify this and provide a diagram?)

首先注意该平面与八面体的交线必为一个六边形。考虑八面体的展开图。注意这个六边形在展开图上对应为一条线段。并且由“平行于面”的条件可知，这条线段必须与展开图中恰好 $\frac{1}{3}$ 的边平行。再注意到由对称性，六边形的两个顶点落在八面体中“正方形”那条边的中点上。在展开图中，这意味着该线段与展开图的某条边的交点必须在其中点处。而一旦有一个交点在中点处，由平行条件可推出所有顶点都在某条边的中点处。因此得到一个正六边形，其边长等于边长为 1 的等边三角形的中位线长度，即 $\frac{1}{2}$。于是面积为 $\frac {3\sqrt {3}}{8}$，从而 $a + b + c = 14\ \mathbf{(E)}$。 (有人能澄清并提供一张图吗？)

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