Difference between revisions of "2001 AMC 12 Problems/Problem 14"
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== Solution == | == Solution == | ||
| − | 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>\binom{9}{2} = 36</math> pairs of vertices determines two equilateral triangles — one facing towards the center, and one outwards — 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. |
== Video Solution == | == Video Solution == | ||
Latest revision as of 01:21, 10 November 2024
Contents
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 facing towards the center, and one outwards — for a total of 72 triangles. However, the three triangles 
, 
, and 
are each counted 3 times, resulting in an overcount of 6. Thus, there are 
distinct equilateral triangles.
Video Solution
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
