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2002 AIME I Problems/Problem 5

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Problem

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}\} ?

Solution

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}.

See also

2002 AIME I (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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