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

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== Problem ==
 
== Problem ==
  
−
Let <math>ABC</math> be an acute triangle with <math>\omega</math>, <math>\Omega</math>, and <math>R</math> being its incircle, circumcircle, and circumradius, respectively.  Circle <math>\omega_A</math> is tangent internally to <math>\Omega</math> at <math>A</math> and tangent externally to <math>\omega</math>.  Circle <math>\Omega_A</math> is tangent internally to <math>\Omega</math> at <math>A</math> and tangent internally to <math>\omega</math>.  Let <math>P_A</math> and <math>Q_A</math> denote the centers of <math>\omega_A</math> and <math>\Omega_A</math>, respectively.  Define points <math>P_B</math>, <math>Q_B</math>, <math>P_C</math>, <math>Q_C</math> analogously.  Prove that
+
Let <math>ABC</math> be an [[acute triangle]] with <math>\omega</math>, <math>\Omega</math>, and <math>R</math> being its [[incircle]], [[circumcircle]], and circumradius, respectively.  [[Circle]] <math>\omega_A</math> is [[tangent]] internally to <math>\Omega</math> at <math>A</math> and [[externally tangent|tangent externally]] to <math>\omega</math>.  Circle <math>\Omega_A</math> is [[internally tangent|tangent internally]] to <math>\Omega</math> at <math>A</math> and tangent internally to <math>\omega</math>.  Let <math>P_A</math> and <math>Q_A</math> denote the [[center]]s of <math>\omega_A</math> and <math>\Omega_A</math>, respectively.  Define points <math>P_B</math>, <math>Q_B</math>, <math>P_C</math>, <math>Q_C</math> [[analogous]]ly.  Prove that
  
−
<math>
+
<div style='text-align:center;'><math>
 
8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,
 
8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,
−
</math>
+
</math></div>
  
−
with equality if and only if triangle <math>ABC</math> is equilateral.
+
with [[equal]]ity [[iff|if and only if]] triangle <math>ABC</math> is [[equilateral triangle|equilateral]].
  
 
== Solution ==
 
== Solution ==
 +
{{image}}
 +
 +
'''Lemma''':
  
−
Lemma:
 
 
<math>
 
<math>
 
P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}
 
P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}
 
</math>
 
</math>
  
−
Proof:
+
'''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>.  
+
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].  
+
<!--Let <math>w_{A} \cap AO=X</math> and <math>w_{A}'<math>\cap</math> AO=X'</math> (second intersection).-->
 +
 
 +
Let <math>\displaystyle \Omega_{A} \cap AO=Y</math> and <math>\displaystyle \Omega_{A}' \cap AO=Y'</math> (second intersection).  
  
 
Evidently, <math>AX=2AP_{A}</math> and <math>AY=2AQ_{A}</math>. We want:
 
Evidently, <math>AX=2AP_{A}</math> and <math>AY=2AQ_{A}</math>. We want:
  
−
<math>
+
<div style='text-align:center;'><math>
−
\star=AQ_{A}-AP_{A}=\frac{1}{2}(AY-AX)=\frac{(s-a)^{2}}{2}(\frac{1}{AY'}-\frac{1}{AX'})
+
\star=AQ_{A}-AP_{A}=\frac{1}{2}(AY-AX)=\frac{(s-a)^{2}}{2}\left(\frac{1}{AY'}-\frac{1}{AX'}\right)
−
</math>
+
</math></div>
  
−
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>
+
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>
+
<div style='text-align:center;'><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}
 
\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>
+
</math></div>
  
 
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
 
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>
+
<div style='text-align:center;'><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 }
 
\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>
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<math>
 
<math>
 
P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}
 
P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}
−
</math>
+
</math></div>
 +
 
 +
'''End Lemma'''
  
−
End Lemma--
 
  
 
The problem becomes:
 
The problem becomes:
−
<math>
+
 
 +
<div style='text-align:center;'><math>
 
8\prod \frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}\le R^{3}
 
8\prod \frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}\le R^{3}
 
</math>
 
</math>
Line 69: Line 74:
 
<math>
 
<math>
 
2r\le R
 
2r\le R
−
</math>
+
</math></div>
  
 
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.
 
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>.
 
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>.
  
−
Solution by AoPS user Altheman
+
== See also ==
 +
{{USAMO newbox|year=2007|num-b=5|after=Last Question}}
  
−
 
+
[[Category:Olympiad Geometry Problems]]
−
 
 
−
 
 
−
{{USAMO newbox|year=2007|num-b=5|after=Last Question}}
 

Revision as of 16:59, 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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $\displaystyle \Omega_{A} \cap AO=Y$ 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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