2010 AMC 12A Problems/Problem 8
Contents
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 and the angle between the hypotenuse and the shorter side is , 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
Solution 3 (Similar Triangles)
Notice that and . Hence, triangle AEB is similar to triangle CFA. Since , , as triangle CFE is equilateral. Therefore, , and since ,. Thus, the measure of equals to -HarryW
Video Solution by the Beauty of Math
https://youtu.be/kU70k1-ONgM?t=785
See also
2010 AMC 10A (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 10 Problems and Solutions |
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 |
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