Difference between revisions of "2010 AIME II Problems/Problem 14"
m |
m (→See also) |
||
Line 13: | Line 13: | ||
== See also == | == See also == | ||
− | {{AIME box|year=2010|num-b=14|num-a=15|n= | + | {{AIME box|year=2010|num-b=14|num-a=15|n=II}} |
Revision as of 16:14, 3 April 2010
Problem 14
Triangle with right angle at , and . Point on $\overbar{AB}$ (Error compiling LaTeX. Unknown error_msg) is chosen such that and . The ratio can be represented in the form , where , , are positive integers and is not divisible by the square of any prime. Find .
Solution
Label the center of the circumcircle of as and the intersection of with the circumcircle as . It now follows that . Hence is isosceles and .
Denote the projection of onto . Now . By the pythagorean theorem, . Now note that . By the pythagorean theorem, . Hence it now follows that,
This gives that the answer is .
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |