Difference between revisions of "2010 AMC 12A Problems/Problem 8"
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+ | ==Video Solution== | ||
+ | https://youtu.be/kU70k1-ONgM?t=785 | ||
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+ | ~IceMatrix | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}} | {{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}} |
Latest revision as of 06:30, 28 May 2020
Problem
Triangle has . Let and be on and , respectively, such that . Let be the intersection of segments and , and suppose that is equilateral. What is ?
Solution
Let .
Since , triangle is a triangle, so .
Solution 2(Trig and Angle Chasing)
Let . Let . Because is equilateral, we get , so . Because is equilateral, we get . Angles and are vertical, so . By triangle , we have , and because of line , we have . Because Of line , we have , and by line , we have . By quadrilateral , we have .
By the Law of Sines, we have . By the sine addition formula(which states by the way), we have . Because cosine is an even function, and sine is an odd function, we have . We know that , and , hence . The only value of that satisfies (because is an angle of the triangle) is . We seek to find , which as we found before is , which is . The answer is
-vsamc
Video Solution
https://youtu.be/kU70k1-ONgM?t=785
~IceMatrix
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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.