2001 AMC 12 Problem 14

Attempt Problem 14 of the 2001 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 12 solutions, or check the answer key.

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14.

Given the nine-sided regular polygon A1A2A3A4A5A6A7A8A9,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 {A1,A2,…,A9}?\{A_1, A_2, \ldots, A_9\}?
3030
3636
6363
6666
7272
Answer: D
Concepts:counting shapes in figuresregular polygondouble counting
Small hint:
Each pair of vertices is a side of exactly two equilateral triangles
Big hint:
This counts 2(92)2\binom{9}{2} triangles, but the three all-vertex triangles get counted three times each
Solution:
Each of the (92)=36\binom{9}{2} = 36 pairs of vertices is a side of exactly two equilateral triangles, giving 7272 triangles counted with multiplicity.
The triangles A1A4A7,A_1 A_4 A_7, A2A5A8,A_2 A_5 A_8, and A3A6A9A_3 A_6 A_9 have all three vertices in the set, so each is counted three times instead of once, an overcount of 22 apiece.
The number of distinct triangles is 72−3⋅2=66.72 - 3 \cdot 2 = 66.
Thus, the correct answer is D.
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