2001 AMC 12 Problem 14

Each of the $\binom{9}{2} = 36$ pairs of vertices is a side of exactly two equilateral triangles, giving $72$ triangles counted with multiplicity. The triangles $A_1 A_4 A_7,$ $A_2 A_5 A_8,$ and $A_3 A_6 A_9$ have all three vertices in the set, so each is counted three times instead of once, an overcount of $2$ apiece. The number of distinct triangles is $72 - 3 \cdot 2 = 66.$ Thus, the correct answer is D.