Difference between revisions of "2018 AIME II Problems/Problem 14"
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The incircle <math>\omega</math> of triangle <math>ABC</math> is tangent to <math>\overline{BC}</math> at <math>X</math>. Let <math>Y \neq X</math> be the other intersection of <math>\overline{AX}</math> with <math>\omega</math>. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>\overline{PQ}</math> is tangent to <math>\omega</math> at <math>Y</math>. Assume that <math>AP = 3</math>, <math>PB = 4</math>, <math>AC = 8</math>, and <math>AQ = \dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | The incircle <math>\omega</math> of triangle <math>ABC</math> is tangent to <math>\overline{BC}</math> at <math>X</math>. Let <math>Y \neq X</math> be the other intersection of <math>\overline{AX}</math> with <math>\omega</math>. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>\overline{PQ}</math> is tangent to <math>\omega</math> at <math>Y</math>. Assume that <math>AP = 3</math>, <math>PB = 4</math>, <math>AC = 8</math>, and <math>AQ = \dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> size(200); import olympiad; defaultpen(linewidth(1)+fontsize(12)); | ||
+ | pair A,B,C,P,Q,Wp,X,Y,Z; B=origin; C=(6.75,0); A=IP(CR(B,7),CR(C,8)); path c=incircle(A,B,C); Wp=IP(c,A--C); Z=IP(c,A--B); X=IP(c,B--C); Y=IP(c,A--X); pair I=incenter(A,B,C); P=extension(A,B,Y,Y+dir(90)*(Y-I)); Q=extension(A,C,P,Y); | ||
+ | draw(A--B--C--cycle, black+1); draw(c^^A--X^^P--Q); | ||
+ | pen p=4+black; dot("$A$",A,N,p); dot("$B$",B,SW,p); dot("$C$",C,SE,p); dot("$X$",X,S,p); dot("$Y$",Y,dir(55),p); dot("$W$",Wp,E,p); dot("$Z$",Z,W,p); dot("$P$",P,W,p); dot("$Q$",Q,E,p); MA("\beta",C,X,A,0.3,black); MA("\alpha",B,A,X,0.7,black); </asy> | ||
==Solution 1== | ==Solution 1== | ||
− | Let sides <math>\overline{AB}</math> and <math>\overline{AC}</math> be tangent to <math>\omega</math> at <math>Z</math> and <math>W</math>, respectively. Let <math>\alpha = \angle BAX</math> and <math>\beta = \angle AXC</math>. Because <math>\overline{PQ}</math> and <math>\overline{BC}</math> are both tangent to <math>\omega</math> and <math>\angle YXC</math> and <math>\angle QYX</math> subtend the same arc of <math>\omega</math>, it follows that <math>\angle AYP = \angle QYX = \angle YXC = \beta</math>. By equal tangents, <math>PZ = PY</math>. Applying the Law of Sines to <math>\triangle APY</math> yields <cmath>\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.</cmath>Similarly, applying the Law of Sines to <math>\triangle ABX</math> gives <cmath>\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.</cmath>It follows that <cmath>2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,</cmath>implying <math>AZ = \tfrac{21}5</math>. Applying the same argument to <math>\triangle AQY</math> yields <cmath>2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),</cmath>from which <math>AQ = \tfrac{168}{59}</math>. The requested sum is <math>168 + 59 = \boxed{227}</math>. | + | Let the sides <math>\overline{AB}</math> and <math>\overline{AC}</math> be tangent to <math>\omega</math> at <math>Z</math> and <math>W</math>, respectively. Let <math>\alpha = \angle BAX</math> and <math>\beta = \angle AXC</math>. Because <math>\overline{PQ}</math> and <math>\overline{BC}</math> are both tangent to <math>\omega</math> and <math>\angle YXC</math> and <math>\angle QYX</math> subtend the same arc of <math>\omega</math>, it follows that <math>\angle AYP = \angle QYX = \angle YXC = \beta</math>. By equal tangents, <math>PZ = PY</math>. Applying the Law of Sines to <math>\triangle APY</math> yields <cmath>\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.</cmath>Similarly, applying the Law of Sines to <math>\triangle ABX</math> gives <cmath>\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.</cmath>It follows that <cmath>2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,</cmath>implying <math>AZ = \tfrac{21}5</math>. Applying the same argument to <math>\triangle AQY</math> yields <cmath>2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),</cmath>from which <math>AQ = \tfrac{168}{59}</math>. The requested sum is <math>168 + 59 = \boxed{227}</math>. |
==Solution 2 (Projective)== | ==Solution 2 (Projective)== | ||
− | Let the incircle of <math>ABC</math> be tangent to <math>AB</math> and <math>AC</math> at <math> | + | Let the incircle of <math>ABC</math> be tangent to <math>AB</math> and <math>AC</math> at <math>Z</math> and <math>W</math>. By Brianchon's theorem on tangential hexagons <math>QWCBZP</math> and <math>PYQCXB</math>, we know that <math>ZW,CP,BQ</math> and <math>XY</math> are concurrent at a point <math>O</math>. Let <math>PQ \cap BC = M</math>. Then by La Hire's <math>A</math> lies on the polar of <math>M</math> so <math>M</math> lies on the polar of <math>A</math>. Therefore, <math>ZW</math> also passes through <math>M</math>. Then projecting through <math>M</math>, we have |
− | <cmath> -1 = (A,O;Y,X) \stackrel{ | + | <cmath> -1 = (A,O;Y,X) \stackrel{M}{=} (A,Z;P,B) \stackrel{M}{=} (A,W;Q,C).</cmath>Therefore, <math>\frac{AP \cdot ZB}{MP \cdot AB} = 1 \implies \frac{3 \cdot ZB}{ZP \cdot 7} = 1</math>. Since <math>ZB+ZP=4</math> we know that <math>ZP = \frac{6}{5}</math> and <math>ZB = \frac{14}{5}</math>. Therefore, <math>AW = AZ = \frac{21}{5}</math> and <math>WC = 8 - \frac{21}{5} = \frac{19}{5}</math>. Since <math>(A,W;Q,C) = -1</math>, we also have <math>\frac{AQ \cdot WC}{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>. |
+ | 😃 | ||
-Vfire | -Vfire | ||
Line 62: | Line 69: | ||
~Solution by <math>BladeRunnerAUG</math> (Frank FYC) | ~Solution by <math>BladeRunnerAUG</math> (Frank FYC) | ||
+ | |||
+ | ==Solution 4 (Projective geometry)== | ||
+ | [[File:2018 AIME II 14.png|500px|right]] | ||
+ | <i><b>Claim</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> | ||
+ | [[File:2018 AIME II 14a.png|300px|right]] | ||
+ | <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>P'Q'||B'C', \angle B'P'Q' = \angle B'C'Q' = 90^\circ \implies B'C'Q'P'</math> be a square. | ||
+ | |||
+ | Therefore <math>M'N'</math> be the diameter <math>\implies P'C', B'Q',</math> be diagonals of the square. <math>M'N'</math> and <math>X'Y'</math> be midlines which crossing in the center <math>I.</math> Therefore lines <math>PC, MN, AX,</math> and <math>BQ</math> are concurrent. | ||
+ | |||
+ | Lines <math>P'Q'||M'N' ||B'C' \implies PQ, MN</math> and <math>BC</math> are concurrent. | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | The cross-ratio associated with a list of four collinear points <math>A,P,M,D</math> is defined as <cmath>(A,P;M,B)={\frac {AP\cdot MB}{AB\cdot PM}}.</cmath> | ||
+ | The cross-ratio be <i><b>projective invariant</b></i> of a quadruple of collinear points, so | ||
+ | [[File:2018 AIME II 14b.png|300px|right]] | ||
+ | <cmath>(A,P; M,B) = {\frac {A'P'\cdot M'B'}{A'B'\cdot P'M'}} = \frac {M'B'}{P'M'} = 1.</cmath> | ||
+ | <cmath>(A,P; M,B)={\frac {3\cdot (7 - AM)}{7\cdot (AM - 3)}} = 1 \implies AM = \frac {21}{5} \implies AN = AM = \frac {21}{5}.</cmath> | ||
+ | [[File:2018 AIME II 14c.png|300px|right]] | ||
+ | <cmath>(A,Q;N,C)={\frac {AQ\cdot NC}{AC\cdot QN}} = \frac {AQ\cdot (AC- AN)}{AC\cdot (AN-AQ)} = 1.</cmath> | ||
+ | <cmath>AQ \cdot (8 - \frac{21}{5}) = 8 \cdot (\frac{21}{5} – AQ) \implies AQ = \frac{168}{59}.</cmath> | ||
+ | |||
+ | For visuals only, I will show how one can find the perceptor <math>D</math> and the image’s plane. | ||
+ | <math>E_0</math> is image of inversion <math>E</math> with respect <math>\omega.</math> | ||
+ | <math>UW</math> is the diameter of <math>\omega, E,E_0,U,W</math> are collinear. | ||
+ | <math>DU \perp \omega, DE_0\perp WD, UV \perp WD, UV</math> is diameter of <math>\omega'</math>. | ||
+ | |||
+ | Plane of images is perpendicular to <math>WD.</math> | ||
+ | |||
+ | Last diagram shows the result of transformation. Transformation is possible. The end. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Video Solution by Mop 2024== | ||
+ | https://youtu.be/SIs1JFLFzyw | ||
+ | |||
+ | ~r00tsOfUnity | ||
+ | |||
+ | ==See Also== | ||
{{AIME box|year=2018|n=II|num-b=13|num-a=15}} | {{AIME box|year=2018|n=II|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:32, 28 January 2024
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)
Claim
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.
and
be midlines which crossing in the center
Therefore lines
and
are concurrent.
Lines and
are concurrent.
Solution
The cross-ratio associated with a list of four collinear points is defined as
The cross-ratio be projective invariant of a quadruple of collinear points, so
For visuals only, I will show how one can find the perceptor and the image’s plane.
is image of inversion
with respect
is the diameter of
are collinear.
is diameter of
.
Plane of images is perpendicular to
Last diagram shows the result of transformation. Transformation is possible. The end.
vladimir.shelomovskii@gmail.com, vvsss
Video Solution by Mop 2024
~r00tsOfUnity
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.