Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2007 USAMO Problems/Problem 6

Problem

(Kiran Kedlaya, Sungyoon Kim) Let $ABC$ be an acute triangle with $\omega$, $\Omega$, and $R$ being its incircle, circumcircle, and circumradius, respectively. Circle $\omega_A$ is tangent internally to $\Omega$ at $A$ and tangent externally to $\omega$. Circle $\Omega_A$ is tangent internally to $\Omega$ at $A$ and tangent internally to $\omega$. Let $P_A$ and $Q_A$ denote the centers of $\omega_A$ and $\Omega_A$, respectively. Define points $P_B$, $Q_B$, $P_C$, $Q_C$ analogously. Prove that \[8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,\] with equality if and only if triangle $ABC$ is equilateral.

Solutions

Solution 1

[asy] size(400); defaultpen(fontsize(8)); pair A=(2,8), B=(0,0), C=(13,0), I=incenter(A,B,C), O=circumcenter(A,B,C), p_a, q_a, X, Y, X1, Y1, D, E, F; real r=abs(I-foot(I,A,B)), R=abs(A-O), a=abs(B-C), b=abs(A-C), c=abs(A-B), x=(((b+c-a)/2)^2)/(r^2+4*r*R+((b+c-a)/2)^2), y=((b+c-a)/2)^2/(r^2+((b+c-a)/2)^2); p_a=x*(O-A)+A; q_a=y*(O-A)+A; X=intersectionpoint(Circle(p_a,x*abs(O-A)), A+(O-A)*0.01--O); X1=intersectionpoint(intersectionpoint(incircle(A,B,C),I--I+O-A)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+O-A)-abs(A-O)*expi(angle(A-O)-pi/2), q_a--q_a+O-A); Y=intersectionpoint(Circle(q_a,y*abs(O-A)), O--2*O-A); Y1=intersectionpoint(intersectionpoint(incircle(A,B,C),I--I+A-O)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+A-O)-abs(A-O)*expi(angle(A-O)-pi/2), O--A); draw(intersectionpoint(incircle(A,B,C),I--I+A-O)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+A-O)-abs(A-O)*expi(angle(A-O)-pi/2),blue+0.7); draw(intersectionpoint(incircle(A,B,C),I--I+O-A)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+O-A)-0.5*abs(A-O)*expi(angle(A-O)-pi/2),red+0.7); draw(A--B--C--A); draw(Circle(A,abs(foot(I,A,B)-A))); draw(incircle(A,B,C)); draw(I--A--O--Y); draw(Circle(p_a,x*abs(O-A)),red+0.7); draw(Circle(q_a,y*abs(O-A)),blue+0.7); label("$A$",A,(-1,1)); label("$I$",I,(-1,0));label("$O$",O,(-1,-1)); label("$P_A$",p_a,(0.5,1));label("$Q_A$",q_a,(1,0)); label("$X$",X,(1,0));label("$Y$",Y,(1,-1)); label("$X'$",X1,(1,0));label("$Y'$",Y1,(0,1)); label("$\omega$",I+r*expi(pi/12), (1,0)); label("$\omega_A$",p_a+x*abs(O-A)*expi(pi/6), (1,1)); label("$\Omega_A$",q_a+y*abs(O-A)*expi(pi/6), (1,0)); label("$\Omega_A'$",intersectionpoint(incircle(A,B,C),I--I+A-O)-abs(A-O)*expi(angle(A-O)-pi/2), (0,2)); label("$\omega_A'$",intersectionpoint(incircle(A,B,C),I--I+O-A)-0.5*abs(A-O)*expi(angle(A-O)-pi/2), (0,2)); dot(p_a^^q_a^^A^^I^^O^^X^^Y^^X1^^Y1); [/asy]

Lemma. \[P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}\]

Proof. Note $P_{A}$ and $Q_{A}$ lie on $AO$ since for a pair of tangent circles, the point of tangency and the two centers are collinear.

Let $\omega$ touch $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. Note $AE=AF=s-a$. Consider an inversion, $\mathcal{I}$, centered at $A$, passing through $E$, $F$. Since $IE\perp AE$, $\omega$ is orthogonal to the inversion circle, so $\mathcal{I}(\omega)=\omega$. Consider $\mathcal{I}(\omega_{A})=\omega_{A}'$. Note that $\omega_{A}$ passes through $A$ and is tangent to $\omega_{A}$, hence $\omega_{A}'$ is a line that is tangent to $\omega$. Furthermore, $\omega_{A}'\perp AO$ because $\omega_{A}$ is symmetric about $OA$, so the inversion preserves that reflective symmetry. Since it is a line that is symmetric about $AO$, it must be perpendicular to $AO$. Likewise, $\mathcal{I}(\Omega_{A})=\Omega_{A}'$ is the other line tangent to $\omega$ and perpendicular to $AO$.

Let $\omega_{A} \cap AO=X$ and $\omega_{A}' \cap AO=X'$ (second intersection).

Let $\Omega_{A} \cap AO=Y$ and $\Omega_{A}' \cap AO=Y'$ (second intersection).

Evidently, $AX=2AP_{A}$ and $AY=2AQ_{A}$. We want: \[\star=AQ_{A}-AP_{A}=\frac{1}{2}(AY-AX)=\frac{(s-a)^{2}}{2}\left(\frac{1}{AY'}-\frac{1}{AX'}\right)\] by inversion. Note that $\omega_{A}' || \Omega_{A}'$, and they are tangent to $\omega$, so the distance between those lines is $2r=AX'-AY'$. Drop a perpendicular from $I$ to $AO$, touching at $H$. Then $AH=AI\cos\angle OAI=AI\cos\frac{1}{2}|\angle B-\angle C|$. Then $AX', AY' = AI\cos\frac{1}{2}|\angle B-\angle C|\pm r$. So $AX'\cdot AY' = AI^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-r^{2}$. \begin{align*} \star &= \frac{(s-a)^{2}}{2}\cdot\frac{AX' - AY'}{AY'\cdot AX'} \\ &= \frac{r(s-a)^{2}}{AI^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-r^{2}} \\ &= \frac{\frac{(s-a)^{2}}{r}}{\left(\frac{AI}{r}\right)^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-1}. \end{align*} Note that $\frac{AI}{r} = \frac{1}{\sin\frac{A}{2}}$. Applying the double angle formulas and $1-\cos\angle A=\frac{2(s-b)(s-c)}{bc}$, we get \begin{align*} \star &= \frac{\frac{(s-a)^{2}}{r}}{\frac{1+\cos (\angle B-\angle C)}{1-\cos A}+1} \\ &= \frac{\frac{(s-a)^{2}}{r}\cdot (1-\cos \angle A)}{\cos(\angle B-\angle C)+\cos(\pi-\angle B-\angle C)} \\ &= \frac{(s-a)^{2}(1-\cos \angle A)}{2r\sin \angle B\sin \angle C} \\ &= \frac{(s-a)^{2}(s-b)(s-c)}{rbc\sin\angle B\sin\angle C} \\ P_{A}Q_{A} &= \frac{4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}} \end{align*}

End Lemma

The problem becomes: \begin{align*} 8\prod \frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}} &\leq R^{3} \\ \frac{2^{9}R^{6}\left[\prod(s-a)\right]^{4}}{r^{3}a^{4}b^{4}c^{4}} &\leq R^{3} \\ \left(\frac{4AR}{abc}\right)^{4}\cdot\left(\frac{A}{rs}\right)^{4}2rR^{2} &\leq R^{3} \\ 2r &\leq R, \end{align*} which is true because $OI^{2}=R(R-2r)$, equality is when the circumcenter and incenter coincide. As before, $\angle OAI=\frac{1}{2}|\angle B-\angle C|=0$, so, by symmetry, $\angle A=\angle B=\angle C$. Hence the inequality is true iff $\triangle ABC$ is equilateral.

Comment: It is much easier to determine $AP_{A}$ by considering $\triangle IAP_{A}$. We have $AI$, $\angle IAO$, $IP_{A}=r+r(P_{A})$, and $AP_{A}=r(P_{A})$. However, the inversion is always nice to use. This also gives an easy construction for $w_{A}$ because the tangency point is collinear with the intersection of $w_{A}'$ and $w$.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

  • <url>viewtopic.php?t=145852 Discussion on AoPS/MathLinks</url>
2007 USAMO (Problems • Resources)
Preceded by
Problem 5 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_USAMO_Problems/Problem_6&oldid=87293"