2010 AIME II Problems/Problem 14
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 I (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 |