Difference between revisions of "2018 AIME II Problems/Problem 14"
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Let the incircle of <math>ABC</math> be tangent to <math>AB</math> and <math>AC</math> at <math>M</math> and <math>N</math>. By Brianchon's theorem on tangential hexagons <math>QNCBMP</math> and <math>PYQCXB</math>, we know that <math>MN,CP,BQ</math> and <math>XY</math> are concurrent at a point <math>O</math>. Let <math>PQ \cap BC = Z</math>. Then by La Hire's <math>A</math> lies on the polar of <math>Z</math> so <math>Z</math> lies on the polar of <math>A</math>. Therefore, <math>MN</math> also passes through <math>Z</math>. Then projecting through <math>Z</math>, we have | Let the incircle of <math>ABC</math> be tangent to <math>AB</math> and <math>AC</math> at <math>M</math> and <math>N</math>. By Brianchon's theorem on tangential hexagons <math>QNCBMP</math> and <math>PYQCXB</math>, we know that <math>MN,CP,BQ</math> and <math>XY</math> are concurrent at a point <math>O</math>. Let <math>PQ \cap BC = Z</math>. Then by La Hire's <math>A</math> lies on the polar of <math>Z</math> so <math>Z</math> lies on the polar of <math>A</math>. Therefore, <math>MN</math> also passes through <math>Z</math>. Then projecting through <math>Z</math>, we have | ||
<cmath> -1 = (A,O;Y,X) \stackrel{Z}{=} (A,M;P,B) \stackrel{Z}{=} (A,N;Q,C).</cmath>Therefore, <math>\frac{AP \cdot MB}{MP \cdot AB} = 1 \implies \frac{3 \cdot MB}{MP \cdot 7} = 1</math>. Since <math>MB+MP=4</math> we know that <math>MB = \frac{6}{5}</math> and <math>MB = \frac{14}{5}</math>. Therefore, <math>AN = AM = \frac{21}{5}</math> and <math>NC = 8 - \frac{21}{5} = \frac{19}{5}</math>. Since <math>(A,N;Q,C) = -1</math>, we also have <math>\frac{AQ \cdot NC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1</math>. Solving for <math>AQ</math>, we obtain <math>AQ = \frac{168}{59} \implies m+n = \boxed{227}</math>. | <cmath> -1 = (A,O;Y,X) \stackrel{Z}{=} (A,M;P,B) \stackrel{Z}{=} (A,N;Q,C).</cmath>Therefore, <math>\frac{AP \cdot MB}{MP \cdot AB} = 1 \implies \frac{3 \cdot MB}{MP \cdot 7} = 1</math>. Since <math>MB+MP=4</math> we know that <math>MB = \frac{6}{5}</math> and <math>MB = \frac{14}{5}</math>. Therefore, <math>AN = AM = \frac{21}{5}</math> and <math>NC = 8 - \frac{21}{5} = \frac{19}{5}</math>. Since <math>(A,N;Q,C) = -1</math>, we also have <math>\frac{AQ \cdot NC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1</math>. Solving for <math>AQ</math>, we obtain <math>AQ = \frac{168}{59} \implies m+n = \boxed{227}</math>. | ||
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-Vfire | -Vfire | ||
Revision as of 23:51, 23 February 2019
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 the 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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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