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

(Solution)
Line 22: Line 22:
  
 
{{AMC12 box|year=2001|num-b=13|num-a=15}}
 
{{AMC12 box|year=2001|num-b=13|num-a=15}}
 +
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:13, 13 June 2017

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 three times (once for each side). Hence we overcounted by $2$ for each of these triangles for a total of $6$ overcounted, and the correct number of equilateral triangles is $72-6=\boxed{66}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS