Difference between revisions of "2018 AIME II Problems/Problem 14"
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==Solution 4 (Projective geometry)== | ==Solution 4 (Projective geometry)== | ||
+ | [[File:2018 AIME II 14.png|400px|right]] | ||
+ | <i><b>Lemma</b></i> | ||
+ | |||
+ | Let the sides <math>\overline{AB}</math> and <math>\overline{AC}</math> be tangent to <math>\omega</math> at <math>M</math> and <math>N</math>, respectively. Then | ||
+ | lines <math>PQ, MN,</math> and <math>BC</math> are concurrent and lines <math>PC, MN, AX,</math> and <math>BQ</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>E</math> be point of crossing <math>AX</math> and <math>BQ.</math> We make projective transformation such that circle <math>\omega</math> maps into the circle and point <math>E</math> maps into the center of new circle point <math>I.</math> We denote images using notification <math>X \rightarrow X'.</math> | ||
+ | |||
+ | <math>BCQP</math> maps into <math>B'C'Q'P'</math>, so lines <math>B'Q'</math> and <math>A'X'</math> be the diameters. | ||
+ | This implies <math>PQ||BC, \angle BPQ = \angle BCQ = 90^o \implies BCQP</math> be a square. | ||
+ | Therefore <math>MN</math> be the diameter <math>\implies PC, BQ, MN,</math> be diagonals of the square which crossing in the center <math>I.</math> | ||
+ | So <math>PQ||MN ||BC \implies PQ, MN</math> and <math>BC</math> are concurrent. | ||
+ | |||
+ | Lines PC, MN, AX and BQ are concurrent. | ||
+ | |||
+ | |||
~vvsss, www.deoma-cmd.ru | ~vvsss, www.deoma-cmd.ru | ||
Revision as of 12:27, 12 June 2022
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
.
Diagram
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)
Solution 4 (Projective geometry)
Lemma
Let the sides and
be tangent to
at
and
, respectively. Then
lines
and
are concurrent and lines
and
are concurrent.
Proof
Let be point of crossing
and
We make projective transformation such that circle
maps into the circle and point
maps into the center of new circle point
We denote images using notification
maps into
, so lines
and
be the diameters.
This implies
be a square.
Therefore
be the diameter
be diagonals of the square which crossing in the center
So
and
are concurrent.
Lines PC, MN, AX and BQ are concurrent.
~vvsss, www.deoma-cmd.ru
See Also
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.