AMC12 2001 A

AMC12 2001 A · Q14

AMC12 2001 A · Q14. It mainly tests Basic counting (rules of product/sum), Inclusion–exclusion (basic).

Given the nine-sided regular polygon $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$, how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set $\{A_1,A_2,\dots,A_9\}$?

给定正九边形 $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$，在该多边形所在平面内，有多少个不同的等边三角形至少有两个顶点属于集合 $\{A_1,A_2,\dots,A_9\}$？

(A) 30 30

(B) 36 36

(D) 66 66

(E) 72 72

Answer

Correct choice: (D)

正确答案：（D）

Solution

Each of the $\binom{9}{2} = 36$ pairs of vertices determines $2$ equilateral triangles — one facing towards the center, and one outwards — for a total of $2 \cdot 36 = 72$ triangles. However, the $3$ triangles $A_1A_4A_7$, $A_2A_5A_8$, and $A_3A_6A_9$ are each counted $3$ times (once for each of the $\binom{3}{2} = 3$ possible pairs of vertices, all of which are vertices of the $9$-gon), whereas they should of course only be counted once, resulting in an overcount of $(3-1) \cdot 3 = 6$. Thus, there are $72-6 = \boxed{\text{(D) }66}$ distinct equilateral triangles.

每一对顶点（共有 $\binom{9}{2} = 36$ 对）都确定 $2$ 个等边三角形——一个朝向中心，另一个朝向外侧——因此共有 $2 \cdot 36 = 72$ 个三角形。然而，$3$ 个三角形 $A_1A_4A_7$, $A_2A_5A_8$, 和 $A_3A_6A_9$ 各被计算了 $3$ 次（对应于 $\binom{3}{2} = 3$ 种可能的顶点对，这些顶点对都在该正九边形上），但它们当然只应计数一次，因此多计了 $(3-1) \cdot 3 = 6$。所以不同的等边三角形共有 $72-6 = \boxed{\text{(D) }66}$ 个。

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