Problem:
Let A1β,A2β,A3β,β¦,A12β 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 {A1β,A2β,A3β,β¦,A12β}?
Solution:
Each pair {Aiβ,Ajβ} of the (212β)=66 pairs of vertices generates three squares, one having diagonal AiβAjβ, and the other two having AiβAjβ as a side. However, each of the three squares A1βA4βA7βA10β,A2βA5βA8βA11β, and A3βA6βA9βA12β is counted six times. The total number of squares is therefore 3β
66β15=183β.
The problems on this page are the property of the MAA's American Mathematics Competitions