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Difference between revisions of "2007 USAMO Problems/Problem 6"

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Note <math>P_{A}</math> and <math>Q_{A}</math> lie on <math>AO</math> since for a pair of tangent circles, the point of tangency and the two centers are [[collinear]].  
 
Note <math>P_{A}</math> and <math>Q_{A}</math> lie on <math>AO</math> since for a pair of tangent circles, the point of tangency and the two centers are [[collinear]].  
  
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Let <math>w</math> touch <math>BC</math>, <math>CA</math>, and <math>AB</math> at <math>D</math>, <math>E</math>, and <math>F</math>, respectively. Note <math>AE=AF=s-a</math>. Consider an [[inversion]], <math>\mathcal{I}</math>, centered at <math>A</math>, passing through <math>E</math>, <math>F</math>. Since <math>IE\perp AE</math>, <math>w</math> is [[orthogonal]] to the inversion circle, so <math>\mathcal{I}(w)=w</math>. Consider <math>\mathcal{I}(w_{A})=w_{A}'</math>. Note that <math>w_{A}</math> passes through <math>A</math> and is tangent to <math>w_{A}</math>, hence <math>w_{A}'</math> is a line that is tangent to <math>w</math>. Furthermore, <math>w_{A}'\perp AO</math> because inversions map a circle's center collinear with the center of inversion. Likewise, <math>\mathcal{I}(\Omega_{A})=\Omega_{A}'</math> is the other line tangent to <math>w</math> and [[perpendicular]] to <math>AO</math>.  
+
Let <math>w</math> touch <math>BC</math>, <math>CA</math>, and <math>AB</math> at <math>D</math>, <math>E</math>, and <math>F</math>, respectively. Note <math>AE=AF=s-a</math>. Consider an [[inversion]], <math>\mathcal{I}</math>, centered at <math>A</math>, passing through <math>E</math>, <math>F</math>. Since <math>IE\perp AE</math>, <math>w</math> is [[orthogonal]] to the inversion circle, so <math>\mathcal{I}(w)=w</math>. Consider <math>\mathcal{I}(w_{A})=w_{A}'</math>. Note that <math>w_{A}</math> passes through <math>A</math> and is tangent to <math>w_{A}</math>, hence <math>w_{A}'</math> is a line that is tangent to <math>w</math>. Furthermore, <math>w_{A}'\perp AO</math> because <math>w_{A}</math> is symmetric about <math>OA</math>, so the inversion preserves that reflective symmetry. Since it is a line that is symmetric about <math>AO</math>, it must be perpendicular to <math>AO</math>. Likewise, <math>\mathcal{I}(\Omega_{A})=\Omega_{A}'</math> is the other line tangent to <math>w</math> and [[perpendicular]] to <math>AO</math>.  
  
  

Revision as of 21:46, 28 April 2007

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

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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 $w_{A}$ 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 $w$ and perpendicular to $AO$.


Let $\displaystyle \Omega_{A} \cap AO=Y$ and $\displaystyle \Omega_{A}' \cap AO=Y'$ (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 $w_{A}' // \Omega_{A}'$, 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 $w$.

See also

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