2015 AIME II Problems/Problem 11
Contents
[hide]Problem
The circumcircle of acute has center . The line passing through point perpendicular to intersects lines and and and , respectively. Also , , , and , where and are relatively prime positive integers. Find .
Solution
Call the and foot of the altitudes from to and , respectively. Let and let . Notice that because both are right triangles, and . Then, . However, since is the circumcenter of triangle , is a perpendicular bisector by the definition of a circumcenter. Hence, . We can use the Pythagorean theorem to find , so we have Likewise, because both are right triangles, and . Hence, since as well, we have that . It follows that . We add this to to get , so . Our answer is .
Solution 2
Notice that , so . From this we get that . So , plugging in the given values we get , so , and .
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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