Difference between revisions of "2001 AMC 12 Problems/Problem 14"
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| − | Each of the <math>{9\choose 2}=36</math> pairs of vertices determines two equilateral triangles, one on each side of the segment. This would give us <math>72</math> triangles. However, note that there are three equilateral triangles that have all three vertices among the vertices of the polygon. These are the triangles <math>A_1A_4A_7</math>, <math>A_2A_5A_8</math>, and <math>A_3A_6A_9</math>. We counted each of these three times (once for each side). Hence we overcounted by <math>6</math>, and the correct number of equilateral triangles is <math>72-6=\boxed{66}</math>. | + | Each of the <math>{9\choose 2}=36</math> pairs of vertices determines two equilateral triangles, one on each side of the segment. This would give us <math>72</math> triangles. However, note that there are three equilateral triangles that have all three vertices among the vertices of the polygon. These are the triangles <math>A_1A_4A_7</math>, <math>A_2A_5A_8</math>, and <math>A_3A_6A_9</math>. We counted each of these three times (once for each side). Hence we overcounted by <math>2</math> for each of these triangles for a total of <math>6</math> overcounted, and the correct number of equilateral triangles is <math>72-6=\boxed{66}</math>. |
== See Also == | == See Also == | ||
Revision as of 17:11, 18 August 2016
Problem
Given the nine-sided regular polygon 
, how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set 
?
Solution
Each of the 
pairs of vertices determines two equilateral triangles, one on each side of the segment. This would give us 
triangles. However, note that there are three equilateral triangles that have all three vertices among the vertices of the polygon. These are the triangles 
, 
, and 
. We counted each of these three times (once for each side). Hence we overcounted by 
for each of these triangles for a total of 
overcounted, and the correct number of equilateral triangles is 
.
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
