AMC12 2009 B

AMC12 2009 B · Q20

AMC12 2009 B · Q20. It mainly tests Basic counting (rules of product/sum), Geometry misc.

A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$, and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$. In addition, no two planes intersect inside or on $Q$. The cuts produce $n$ pyramids and a new polyhedron $R$. How many edges does $R$ have?

一个凸多面体 $Q$ 有顶点 $V_1,V_2,\ldots,V_n$，并且有 $100$ 条棱。用平面 $P_1,P_2,\ldots,P_n$ 切割该多面体，使得平面 $P_k$ 只切割那些在顶点 $V_k$ 处相交的棱。此外，任意两平面都不在 $Q$ 的内部或表面相交。切割产生 $n$ 个棱锥和一个新的多面体 $R$。求 $R$ 有多少条棱。

(A) 200 200

(B) 2n 2n

(D) 400 400

(E) 4n 4n

Answer

Correct choice: (C)

正确答案：（C）

Solution

Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Three edges of $R$ meet at each vertex, so $R$ has a total of $\frac 12 \cdot 3 \cdot 200 = \boxed {300}$ edges.

$Q$ 的每条棱都被两个平面切割，因此 $R$ 有 $200$ 个顶点。$R$ 的每个顶点处有三条棱相交，所以 $R$ 的棱总数为 $\frac 12 \cdot 3 \cdot 200 = \boxed {300}$。

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