2018 AIME II Problems/Problem 14
Contents
Problem
The incircle of triangle
is tangent to
at
. Let
be the other intersection of
with
. Points
and
lie on
and
, respectively, so that
is tangent to
at
. Assume that
,
,
, and
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let sides and
be tangent to
at
and
, respectively. Let
and
. Because
and
are both tangent to
and
and
subtend the same arc of
, it follows that
. By equal tangents,
. Applying the Law of Sines to
yields
Similarly, applying the Law of Sines to
gives
It follows that
implying
. Applying the same argument to
yields
from which
. The requested sum is
.
Solution 2 (Projective)
Let the incircle of be tangent to
and
at
and
. By Brianchon's theorem on tangential hexagons
and
, we know that
and
are concurrent at a point
. Let
. Then by La Hire's
lies on the polar of
so
lies on the polar of
. Therefore,
also passes through
. Then projecting through
, we have
Therefore,
. Since
we know that
and
. Therefore,
and
. Since
, we also have
. Solving for
, we obtain
.
-Vfire
Solution 3 (Combination of Law of Sine and Law of Cosine)
Let the center of the incircle of be
. Link
and
. Then we have
Let the incircle of be tangent to
and
at
and
, let
and
.
Use Law of Sine in and
, we have
therefore we have
Solve this equation, we have
As a result, ,
,
,
,
So,
Use Law of Cosine in and
, we have
And we have
So
Solve this equation, we have
As a result,
So, the final answer of this question is
~Solution by (Frank FYC)
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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