2014 AIME I Problems/Problem 15
Problem 15
In ,
,
, and
. Circle
intersects
at
and
,
at
and
, and
at
and
. Given that
and
, length
, where
and
are relatively prime positive integers, and
is a positive integer not divisible by the square of any prime. Find
.
Solution
First we note that is an isosceles right triangle with hypotenuse
the same as the diameter of
. We also note that
since
is a right angle and the ratios of the sides are
.
From congruent arc intersections, we know that , and that from similar triangles
is also congruent to
. Thus,
is an isosceles triangle with
, so G is the midpoint of
and
. Similarly, we can find from angle chasing that
is the angle bisector of
. From the angle bisector theorem, we have
AF = 15/7
CF = 20/7$.
Lastly, we apply power of a point from points$ (Error compiling LaTeX. Unknown error_msg)AC
\omega
AE \times AB=AF \times AG
CD \times CB=CG \times CF
EB = \frac{17}{14}
DB = \frac{31}{14}
DE = \frac{25 \sqrt{2}}{14}
a+b+c=25+2+14= \boxed{041}$
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
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