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Difference between revisions of "2007 USAMO Problems/Problem 6"
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== Problem ==
 
== Problem ==
 +
(''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
 +
<cmath>8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,</cmath>
 +
with [[equal]]ity [[iff|if and only if]] triangle <math>ABC</math> is [[equilateral triangle|equilateral]].
  
−
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
+
== Solutions ==
  
−
<math>
+
=== Solution 1 ===
−
8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,
+
<center><asy>
−
</math>
+
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></center>
  
−
with equality if and only if triangle <math>ABC</math> is equilateral.
+
'''Lemma.'''
 +
<cmath>P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}</cmath>
  
−
== Solution ==
+
''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>\omega</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>\omega</math> is [[orthogonal]] to the inversion circle, so <math>\mathcal{I}(\omega)=\omega</math>. Consider <math>\mathcal{I}(\omega_{A})=\omega_{A}'</math>. Note that <math>\omega_{A}</math> passes through <math>A</math> and is tangent to <math>\omega_{A}</math>, hence <math>\omega_{A}'</math> is a line that is tangent to <math>\omega</math>. Furthermore, <math>\omega_{A}'\perp AO</math> because <math>\omega_{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>\omega</math> and [[perpendicular]] to <math>AO</math>.
 +
 
 +
Let <math>\omega_{A} \cap AO=X</math> and <math>\omega_{A}' \cap AO=X'</math> (second intersection).
 +
 
 +
Let <math>\Omega_{A} \cap AO=Y</math> and <math>\Omega_{A}' \cap AO=Y'</math> (second intersection).
 +
 
 +
Evidently, <math>AX=2AP_{A}</math> and <math>AY=2AQ_{A}</math>. We want:
 +
<cmath>\star=AQ_{A}-AP_{A}=\frac{1}{2}(AY-AX)=\frac{(s-a)^{2}}{2}\left(\frac{1}{AY'}-\frac{1}{AX'}\right)</cmath>
 +
by inversion. Note that <math>\omega_{A}' || \Omega_{A}'</math>, and they are tangent to <math>\omega</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', AY' = AI\cos\frac{1}{2}|\angle B-\angle C|\pm r</math>. So <math>AX'\cdot AY' = AI^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-r^{2}</math>.
 +
<cmath>\begin{align*}
 +
\star &= \frac{(s-a)^{2}}{2}\cdot\frac{AX' - AY'}{AY'\cdot 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}.
 +
\end{align*}</cmath>
 +
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
 +
<cmath>\begin{align*}
 +
\star &= \frac{\frac{(s-a)^{2}}{r}}{\frac{1+\cos (\angle B-\angle C)}{1-\cos A}+1} \\
 +
&= \frac{\frac{(s-a)^{2}}{r}\cdot (1-\cos \angle A)}{\cos(\angle B-\angle C)+\cos(\pi-\angle B-\angle C)} \\
 +
&= \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{align*}</cmath>
 +
 
 +
'''End Lemma'''
 +
 
 +
The problem becomes:
 +
<cmath>\begin{align*}
 +
8\prod \frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}} &\leq R^{3} \\
 +
\frac{2^{9}R^{6}\left[\prod(s-a)\right]^{4}}{r^{3}a^{4}b^{4}c^{4}} &\leq R^{3} \\
 +
\left(\frac{4AR}{abc}\right)^{4}\cdot\left(\frac{A}{rs}\right)^{4}2rR^{2} &\leq R^{3} \\
 +
2r &\leq R,
 +
\end{align*}</cmath>
 +
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>.
 +
 
 +
{{alternate solutions}}
 +
 
 +
== See also ==
 +
* <url>viewtopic.php?t=145852 Discussion on AoPS/MathLinks</url>
  
 
{{USAMO newbox|year=2007|num-b=5|after=Last Question}}
 
{{USAMO newbox|year=2007|num-b=5|after=Last Question}}
 +
 +
[[Category:Olympiad Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 14:11, 31 August 2017

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.

Solutions

Solution 1

[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 $AX'\cdot AY' = AI^{2}\cos^{2}\frac{1}{2}|\angle B-\angle C|-r^{2}$. \begin{align*} \star &= \frac{(s-a)^{2}}{2}\cdot\frac{AX' - AY'}{AY'\cdot 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}. \end{align*} 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 \begin{align*} \star &= \frac{\frac{(s-a)^{2}}{r}}{\frac{1+\cos (\angle B-\angle C)}{1-\cos A}+1} \\ &= \frac{\frac{(s-a)^{2}}{r}\cdot (1-\cos \angle A)}{\cos(\angle B-\angle C)+\cos(\pi-\angle B-\angle C)} \\ &= \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{align*}

End Lemma

The problem becomes: \begin{align*} 8\prod \frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}} &\leq R^{3} \\ \frac{2^{9}R^{6}\left[\prod(s-a)\right]^{4}}{r^{3}a^{4}b^{4}c^{4}} &\leq R^{3} \\ \left(\frac{4AR}{abc}\right)^{4}\cdot\left(\frac{A}{rs}\right)^{4}2rR^{2} &\leq R^{3} \\ 2r &\leq R, \end{align*} 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 $w$.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

  • <url>viewtopic.php?t=145852 Discussion on AoPS/MathLinks</url>
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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