2002 AIME I 第 5 题

Each pair of vertices determines three squares: two having the pair as a side and one having it as a diagonal

Only squares with all four vertices among the $A_i$ are overcounted, and each such square arises from $6$ of its vertex pairs

Each of the $\binom{12}{2} = 66$ pairs of vertices determines exactly three squares: two having the pair as a side (one on each side of the segment) and one having it as a diagonal. That counts $3 \cdot 66 = 198$ squares.

A square is overcounted only if it has more than two vertices among the $A_i.$ If three vertices of a square lie on the circumcircle, the square’s own circumcircle shares three points with it and hence coincides with it, and an inscribed square’s vertices are spaced $90^\circ$ apart — three steps of the dodecagon — so the fourth vertex is also an $A_i.$ The fully inscribed squares are exactly $A_1A_4A_7A_{10},$ $A_2A_5A_8A_{11},$ and $A_3A_6A_9A_{12},$ and each is generated by all $\binom{4}{2} = 6$ of its vertex pairs, so each is counted $6$ times instead of once.

The number of distinct squares is $198 - 3 \cdot 5 = 183.$