2019 AMC 12A Problems/Problem 22
Contents
[hide]Problem
Circles and , both centered at , have radii and , respectively. Equilateral triangle , whose interior lies in the interior of but in the exterior of , has vertex on , and the line containing side is tangent to . Segments and intersect at , and . Then can be written in the form for positive integers , , , with . What is ?
Solution
Let be the point of tangency between and , and be the midpoint of . Note that and . This implies that , and . Thus, .
If we let be the side length of , then it follows that and . This implies that , so . Furthermore, (because ) so this gives us the equation to solve for the side length , or . Thus, The problem asks for .
Video Solution
https://www.youtube.com/watch?v=2eASfdhEyUE
Video Solution by Richard Rusczyk - https://www.youtube.com/watch?v=CaYgfNEUBwA&list=PLyhPcpM8aMvLgfgbaTLDaV_jRYfn1A-x_&index=2 - AMBRIGGS
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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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