Difference between revisions of "2002 AIME I Problems/Problem 5"
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== Solution == | == 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 <math>66(3)=198</math> 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 <math>9+6=15</math>, and <math>198-15=\fbox{183}</math>. |
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== See also == | == See also == | ||
{{AIME box|year=2002|n=I|num-b=4|num-a=6}} | {{AIME box|year=2002|n=I|num-b=4|num-a=6}} | ||
Revision as of 09:07, 19 August 2008
Problem
Let 
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 
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 
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 
, and 
.
See also
| 2002 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
