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. | ||
− | + | == Video Solution == | |
+ | 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}} | ||
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:26, 30 November 2021
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, 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 |
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