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
Solution 1
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
is the midpoint of
and
. Similarly, we can find from angle chasing that
. Therefore,
is the angle bisector of
. From the angle bisector theorem, we have
, so
and
.
Lastly, we apply power of a point from points and
with respect to
and have
and
, so we can compute that
and
. From the Pythagorean Theorem, we result in
, so
Solution 2
From solution 1, we have CG = 5/2 and =
=
. Therefore,
is isosceles with EF = EA.
Let EF = x, then DE = \sqrt{2}x. Therefore
.
Using Cosine rule on
(14x - 25)(2x + 25) = 0 and x = \frac{25}{14}. Hence,
, so
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
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