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Difference between revisions of "2007 USAMO Problems/Problem 6"
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== Problem ==
 
== Problem ==
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(''Kiran Kedlaya, Sungyoon Kim'') 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
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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
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<cmath>8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,</cmath>
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<div style='text-align:center;'><math>
 
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8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,
 
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</math></div>
 
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with [[equal]]ity [[iff|if and only if]] triangle <math>ABC</math> is [[equilateral triangle|equilateral]].
 
with [[equal]]ity [[iff|if and only if]] triangle <math>ABC</math> is [[equilateral triangle|equilateral]].
  

Revision as of 09:26, 7 August 2014

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.

Solution

[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 $AX'*AY'=AI^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-r^{2}$

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

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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