2007 USAMO Problems/Problem 6

Problem

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.

Solution

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 $w$ 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 $IE\perp AE$ , $w$ is orthogonal to the inversion circle, so $\mathcal{I}(w)=w$ . Consider $\mathcal{I}(w_{A})=w_{A}'$ . Note that $w_{A}$ passes through $A$ and is tangent to $w_{A}$ , hence $w_{A}'$ is a line that is tangent to $w$ . Furthermore, $w_{A}'\perp AO$ because inversions map a circle's center collinear with the center of inversion. Likewise, $\mathcal{I}(\Omega_{A})=\Omega_{A}'$ is the other line tangent to $w$ and perpendicular to $AO$ .

Let w_{A} $\cap$ AO=X and w_{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:

$\star=AQ_{A}-AP_{A}=\frac{1}{2}(AY-AX)=\frac{(s-a)^{2}}{2}(\frac{1}{AY'}-\frac{1}{AX'})$

by inversion. Note that $w_{A}' // \Omega_{A}'$ , and they are tangent to $w$ , 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 $AX'*AY'=AI^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-r^{2}$

$\star=\frac{(s-a)^{2}}{2}*\frac{AX'-AY'}{AY'*AX'}=\frac{r(s-a)^{2}}{AI^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-r^{2}}=\frac{ \frac{(s-a)^{2}}{r}}{ \left(\frac{AI}{r}\right)^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-1}$

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

$\star= \frac{ \frac{(s-a)^{2}}{r}}{ \frac{1+\cos \angle B-\angle C}{1-\cos A}+1 }=\frac{ \frac{(s-a)^{2}}{r}*(1-\cos \angle A) }{ \cos \angle B-\angle C+\cos \pi-\angle B-\angle C }$

$\star=\frac{ (s-a)^{2}(1-\cos \angle A)}{2r\sin \angle B\sin \angle C}=\frac{ (s-a)^{2}(s-b)(s-c)}{rbc\sin\angle B\sin\angle C}$

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

End Lemma--

The problem becomes: $8\prod \frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}\le R^{3}$

$\frac{2^{9}R^{6}\left[\prod(s-a)\right]^{4}}{r^{3}a^{4}b^{4}c^{4}}\le R^{3}$

$\left(\frac{4AR}{abc}\right)^{2}\cdot\left(\frac{A}{rs}\right)^{4}2rR^{2}\le R^{3}$

$2r\le R$

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 $w$ .

Solution by AoPS user Altheman

2007 USAMO (Problems • Resources)
Preceded by Problem 5	Followed by Last Question
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

Page

Toolbox

Search

2007 USAMO Problems/Problem 6

Problem

Solution