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Difference between revisions of "2007 USAMO Problems/Problem 6"
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== Solution ==
 
== Solution ==
 +
 +
Lemma:
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<math>
 +
P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}
 +
</math>
 +
 +
Proof:
 +
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.
 +
 +
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 w_{A}<math>\cap</math> AO=X and w_{A}'<math>\cap</math> AO=X' [second intersection].
 +
 +
Let \Omega_{A}<math>\cap</math> AO=Y and \Omega_{A}'<math>\cap</math> AO=Y' [second intersection].
 +
 +
Evidently, <math>AX=2AP_{A}</math> and <math>AY=2AQ_{A}</math>. We want:
 +
 +
<math>
 +
\star=AQ_{A}-AP_{A}=\frac{1}{2}(AY-AX)=\frac{(s-a)^{2}}{2}(\frac{1}{AY'}-\frac{1}{AX'})
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</math>
 +
 +
by inversion. Note that <math>w_{A}' // \Omega_{A}'</math>, and they are tangent to <math>w</math>, so the distance between those lines is <math>2r=AX'-AY'</math>. Drop a perpendicular from <math>I</math> to <math>AO</math>, touching at <math>H</math>. Then <math>AH=AI\cos\angle OAI=AI\cos\frac{1}{2}|\angle B-\angle C|</math>. Then <math>AX'</math>, <math>AY'</math>=<math>AI\cos\frac{1}{2}|\angle B-\angle C|\pm r</math>. So <math>AX'*AY'=AI^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-r^{2}</math>
 +
 +
<math>
 +
\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}
 +
</math>
 +
 +
Note that <math>\frac{AI}{r}=\frac{1}{\sin\frac{A}{2}}</math>. Applying the double angle formulas and <math>1-\cos\angle A=\frac{2(s-b)(s-c)}{bc}</math>, we get
 +
 +
<math>
 +
\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 }
 +
</math>
 +
 +
<math>
 +
\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}
 +
</math>
 +
 +
<math>
 +
P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}
 +
</math>
 +
 +
End Lemma--
 +
 +
The problem becomes:
 +
<math>
 +
8\prod \frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}\le R^{3}
 +
</math>
 +
 +
<math>
 +
\frac{2^{9}R^{6}\left[\prod(s-a)\right]^{4}}{r^{3}a^{4}b^{4}c^{4}}\le R^{3}
 +
</math>
 +
 +
<math>
 +
\left(\frac{4AR}{abc}\right)^{2}\cdot\left(\frac{A}{rs}\right)^{4}2rR^{2}\le R^{3}
 +
</math>
 +
 +
<math>
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2r\le R
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</math>
 +
 +
which is true because <math>OI^{2}=R(R-2r)</math>, equality is when the circumcenter and incenter coincide. As before, <math>\angle OAI=\frac{1}{2}|\angle B-\angle C|=0</math>, so, by symmetry, <math>\angle A=\angle B=\angle C</math>. Hence the inequality is true iff <math>\triangle ABC</math> is equilateral.
 +
 +
 +
Comment: It is much easier to determine <math>AP_{A}</math> by considering <math>\triangle IAP_{A}</math>. We have <math>AI</math>, <math>\angle IAO</math>, <math>IP_{A}=r+r(P_{A})</math>, and <math>AP_{A}=r(P_{A})</math>. However, the inversion is always nice to use. This also gives an easy construction for <math>w_{A}</math> because the tangency point is collinear with the intersection of <math>w_{A}'</math> and <math>w</math>.
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 +
Solution by AoPS user Altheman
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 +
 +
  
 
{{USAMO newbox|year=2007|num-b=5|after=Last Question}}
 
{{USAMO newbox|year=2007|num-b=5|after=Last Question}}

Revision as of 00:08, 26 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

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)
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