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Difference between revisions of "2001 AMC 12 Problems/Problem 14"

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== Solution ==
 
== Solution ==
  
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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>.
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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 two 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>.
  
 
== See Also ==
 
== See Also ==

Revision as of 17:05, 18 August 2016

Problem

Given the nine-sided regular polygon $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$, how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set $\{A_1,A_2,\dots,A_9\}$?

$\text{(A) }30 \qquad \text{(B) }36 \qquad \text{(C) }63 \qquad \text{(D) }66 \qquad \text{(E) }72$

Solution

Each of the ${9\choose 2}=36$ pairs of vertices determines two equilateral triangles, one on each side of the segment. This would give us $72$ triangles. However, note that there are three equilateral triangles that have all three vertices among the vertices of the polygon. These are the triangles $A_1A_4A_7$, $A_2A_5A_8$, and $A_3A_6A_9$. We counted each of these two times (once for each side). Hence we overcounted by $6$, and the correct number of equilateral triangles is $72-6=\boxed{66}$.

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

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