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

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== Solution ==
 
== Solution ==
 
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Each of the <math>\binom{9}{2} = 36</math> pairs of vertices determines two equilateral triangles, for a total of 72 triangles. However, the three triangles <math>A_1A_4A_7</math>, <math>A_2A_5A_8</math>, and <math>A_3A_6A_9</math> are each counted 3 times, resulting in an overcount of 6. Thus, there are <math>\boxed{66}</math> distinct equilateral triangles.
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 ==

Latest revision as of 13:51, 29 March 2020

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 $\binom{9}{2} = 36$ pairs of vertices determines two equilateral triangles, for a total of 72 triangles. However, the three triangles $A_1A_4A_7$, $A_2A_5A_8$, and $A_3A_6A_9$ are each counted 3 times, resulting in an overcount of 6. Thus, there are $\boxed{66}$ distinct equilateral triangles.

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
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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