2002 AIME I Problem 5

Question

Let

A_1,

A_2,

A_3,

\ldots,

A_{12}

be the vertices of a regular dodecagon. How many distinct squares in the plane of the dodecagon have at least two vertices in the set

\{A_1, A_2, A_3, \ldots, A_{12}\}?

Answer

Answer: 183

Concepts:counting pairs regular polygon counting shapes in figures

Difficulty rating: 2480

Small hint:

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

Big hint:

Only squares with all four vertices among the

A_i

are overcounted, and each such square arises from

6

of its vertex pairs

Solution:

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.