2018 AIME II Problems/Problem 14

Revision as of 08:22, 31 August 2022 by Vvsss (talk | contribs) (Solution 4 (Projective geometry))


The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$. Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$. Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$, respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$. Assume that $AP = 3$, $PB = 4$, $AC = 8$, and $AQ = \dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.


[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

Let the sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $Z$ and $W$, respectively. Let $\alpha = \angle BAX$ and $\beta = \angle AXC$. Because $\overline{PQ}$ and $\overline{BC}$ are both tangent to $\omega$ and $\angle YXC$ and $\angle QYX$ subtend the same arc of $\omega$, it follows that $\angle AYP = \angle QYX = \angle YXC = \beta$. By equal tangents, $PZ = PY$. Applying the Law of Sines to $\triangle APY$ yields \[\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.\]Similarly, applying the Law of Sines to $\triangle ABX$ gives \[\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.\]It follows that \[2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,\]implying $AZ = \tfrac{21}5$. Applying the same argument to $\triangle AQY$ yields \[2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),\]from which $AQ = \tfrac{168}{59}$. The requested sum is $168 + 59 = \boxed{227}$.

Solution 2 (Projective)

Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $M$ and $N$. By Brianchon's theorem on tangential hexagons $QNCBMP$ and $PYQCXB$, we know that $MN,CP,BQ$ and $XY$ are concurrent at a point $O$. Let $PQ \cap BC = Z$. Then by La Hire's $A$ lies on the polar of $Z$ so $Z$ lies on the polar of $A$. Therefore, $MN$ also passes through $Z$. Then projecting through $Z$, we have \[-1 = (A,O;Y,X) \stackrel{Z}{=} (A,M;P,B) \stackrel{Z}{=} (A,N;Q,C).\]Therefore, $\frac{AP \cdot MB}{MP \cdot AB} = 1 \implies \frac{3 \cdot MB}{MP \cdot 7} = 1$. Since $MB+MP=4$ we know that $MP = \frac{6}{5}$ and $MB = \frac{14}{5}$. Therefore, $AN = AM = \frac{21}{5}$ and $NC = 8 - \frac{21}{5} = \frac{19}{5}$. Since $(A,N;Q,C) = -1$, we also have $\frac{AQ \cdot NC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1$. Solving for $AQ$, we obtain $AQ = \frac{168}{59} \implies m+n = \boxed{227}$. 😃 -Vfire

Solution 3 (Combination of Law of Sine and Law of Cosine)

Let the center of the incircle of $\triangle ABC$ be $O$. Link $OY$ and $OX$. Then we have $\angle OYP=\angle OXB=90^{\circ}$

$\because$ $OY=OX$

$\therefore$ $\angle OYX=\angle OXY$

$\therefore$ $\angle PYX=\angle YXB$

$\therefore$ $\sin \angle PYX=\sin \angle YXB=\sin \angle YXC=\sin \angle PYA$

Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $M$ and $N$, let $MP=YP=x$ and $NQ=YQ=y$.

Use Law of Sine in $\triangle APY$ and $\triangle AXB$, we have

$\frac{\sin \angle PAY}{PY}=\frac{\sin \angle PYA}{PA}$

$\frac{\sin \angle BAX}{BX}=\frac{\sin \angle AXB}{AB}$

therefore we have


Solve this equation, we have $x=\frac{6}{5}$

As a result, $MB=4-x=\frac{14}{5}=BX$, $AM=x+3=\frac{21}{5}=AN$, $NC=8-AN=\frac{19}{5}=XC$, $AQ=\frac{21}{5}-y$, $PQ=\frac{6}{5}+y$

So, $BC=\frac{14}{5}+\frac{19}{5}=\frac{33}{5}$

Use Law of Cosine in $\triangle BAC$ and $\triangle PAQ$, we have

$\cos \angle BAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}$

$\cos \angle PAQ=\frac{AP^2+AQ^2-PQ^2}{2\cdot AP\cdot AQ}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$

And we have

$\cos \angle BAC=\cos \angle PAQ$


$\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$

Solve this equation, we have $y=\frac{399}{295}=QN$

As a result, $AQ=AN-QN=\frac{21}{5}-\frac{399}{295}=\frac{168}{59}$

So, the final answer of this question is $168+59=\boxed {227}$

~Solution by $BladeRunnerAUG$ (Frank FYC)

Solution 4 (Projective geometry)

2018 AIME II 14.png


Let the sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $M$ and $N$, respectively. Then lines $PQ, MN,$ and $BC$ are concurrent and lines $PC, MN, AX,$ and $BQ$ are concurrent.


Let $E$ be point of crossing $AX$ and $BQ.$ We make projective transformation such that circle $\omega$ maps into the circle and point $E$ maps into the center of new circle point $I.$ We denote images using notification $X \rightarrow X'.$

2018 AIME II 14a.png

$BCQP$ maps into $B'C'Q'P'$, so lines $B'Q'$ and $A'X'$ be the diameters. This implies $P'Q'||B'C', \angle B'P'Q' = \angle B'C'Q' = 90^\circ \implies B'C'Q'P'$ be a square.

Therefore $M'N'$ be the diameter $\implies P'C', B'Q',$ be diagonals of the square. $M'N'$ and $X'Y'$ be midlines which crossing in the center $I.$ Therefore ines $PC, MN, AX,$ and $BQ$ are concurrent.

Lines $P'Q'||M'N' ||B'C' \implies PQ, MN$ and $BC$ are concurrent.

Solution The cross-ratio associated with a list of four collinear points $A,P,M,D$ is defined as \[(A,P;M,B)={\frac {AP\cdot MB}{AB\cdot PM}}.\] The cross-ratio be projective invariant of a quadruple of collinear points, so

2018 AIME II 14b.png

\[(A,P; M,B)  = {\frac {A'P'\cdot M'B'}{A'B'\cdot P'M'}} = \frac {M'B'}{P'M'} = 1.\] \[(A,P; M,B)={\frac {3\cdot (7 - AM)}{7\cdot (AM - 3)}} = 1 \implies AM = \frac {21}{5} \implies AN = AM =  \frac {21}{5}.\]

2018 AIME II 14c.png

\[(A,Q;N,C)={\frac {AQ\cdot NC}{AC\cdot QN}} = \frac {AQ\cdot (AC- AN)}{AC\cdot (AN-AQ)} = 1.\] \[AQ \cdot (8 - \frac{21}{5}) = 8 \cdot (\frac{21}{5} – AQ) \implies  AQ = \frac{168}{59}.\]

For visuals only, I will show how one can find the perceptor $D$ and the image’s plane. $E_0$ is image of inversion $E$ with respect $\omega.$ $UW$ is the diameter of $\omega, E,E_0,U,W$ are collinear. $DU \perp \omega, DE_0\perp WD, UV \perp WD, UV$ is diameter of $\omega'$.

Plane of images is perpendicular to $WD.$

Last diagram shows the result of transformation. Transformation is possible. The end.

vladimir.shelomovskii@gmail.com, vvsss

See Also

2018 AIME II (ProblemsAnswer KeyResources)
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

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