Let $A_1,A_2,A_3,\cdots,A_{12}$ 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 $\{A_1,A_2,A_3,\cdots,A_{12}\} ?$
There are 66 ways of picking two vertices. Note with any two vertices one can draw three squares (two with the vertices forming a side, another with the vertices forming the diagonal). So so far we have $66(3)=198$ squares, but we have overcounted since some squares have their other two vertices in the dodecagon as well. All 12 combinations of two distinct vertices that form a square side only form 3 squares, and all 12 combinations of two vertices that form a square diagonal only form 6 squares. So in total, we have overcounted by $9+6=15$, and $198-15=\fbox{183}$.