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

(New page: == Problem == Given the nine-sided regular polygon <math>A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9</math>, how many distinct equilateral triangles in the plane of the polygon have at least two ...)
 
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== Solution ==
 
== Solution ==
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Each of the <math>\binom{9}{2} = 36</math> pairs of vertices determines <math>2</math> equilateral triangles — one facing towards the center, and one outwards — for a total of <math>2 \cdot 36 = 72</math> triangles. However, the <math>3</math> triangles <math>A_1A_4A_7</math>, <math>A_2A_5A_8</math>, and <math>A_3A_6A_9</math> are each counted <math>3</math> times (once for each of the <math>\binom{3}{2} = 3</math> possible pairs of vertices, all of which are vertices of the <math>9</math>-gon), whereas they should of course only be counted once, resulting in an overcount of <math>(3-1) \cdot 3 = 6</math>. Thus, there are <math>72-6 = \boxed{\text{(D) }66}</math> distinct equilateral triangles.
  
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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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== Video Solution ==
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https://youtu.be/gMWJisI9Ulk
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2001|num-b=13|num-a=15}}
 
{{AMC12 box|year=2001|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 06:00, 6 June 2025

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 $2$ equilateral triangles — one facing towards the center, and one outwards — for a total of $2 \cdot 36 = 72$ triangles. However, the $3$ triangles $A_1A_4A_7$, $A_2A_5A_8$, and $A_3A_6A_9$ are each counted $3$ times (once for each of the $\binom{3}{2} = 3$ possible pairs of vertices, all of which are vertices of the $9$-gon), whereas they should of course only be counted once, resulting in an overcount of $(3-1) \cdot 3 = 6$. Thus, there are $72-6 = \boxed{\text{(D) }66}$ distinct equilateral triangles.

Video Solution

https://youtu.be/gMWJisI9Ulk

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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