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. The circle <math>\omega_A</math> is tangent internally to <math>\Omega</math> at <math>A</math> and externally tangent to <math>\omega</math>. Circle <math>\Omega_A</math> is internally tangent 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 | ||
+ | <cmath>8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,</cmath> | ||
+ | with equality [[iff|if and only if]] triangle <math>ABC</math> is equilateral. | ||
− | + | == Solutions == | |
− | < | + | === Solution 1 === |
− | + | <center><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></center> | ||
− | + | '''Lemma.''' | |
+ | <cmath>P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}</cmath> | ||
− | + | ''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>. |
− | <math> | + | Let <math>\omega_{A} \cap AO=X</math> and <math>\omega_{A}' \cap AO=X'</math> (second intersection). |
− | |||
− | </math> | ||
− | + | Let <math>\Omega_{A} \cap AO=Y</math> and <math>\Omega_{A}' \cap AO=Y'</math> (second intersection). | |
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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: | ||
− | + | <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>. |
− | \star=AQ_{A}-AP_{A}=\frac{1}{2}(AY-AX)=\frac{(s-a)^{2}}{2}\left(\frac{1}{AY'}-\frac{1}{AX'}\right) | + | <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}} \\ | |
− | by inversion. Note that <math> | + | &= \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 |
− | \star=\frac{(s-a)^{2}}{2} | + | <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)} \\ | |
− | 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 | + | &= \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}} |
− | \star= \frac{ \frac{(s-a)^{2}}{r}}{ \frac{1+\cos \angle B-\angle C}{1-\cos A}+1 }=\frac{ \frac{(s-a)^{2}}{r} | + | \end{align*}</cmath> |
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− | P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}} | ||
− | </ | ||
'''End Lemma''' | '''End Lemma''' | ||
− | |||
The problem becomes: | 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>. | |
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− | </math> | ||
− | + | {{alternate solutions}} | |
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− | + | == See also == | |
− | + | * <url>viewtopic.php?t=145852 Discussion on AoPS/MathLinks</url> | |
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− | < | ||
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{{USAMO newbox|year=2007|num-b=5|after=Last Question}} | {{USAMO newbox|year=2007|num-b=5|after=Last Question}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 05:52, 27 August 2024
Contents
Problem
(Kiran Kedlaya, Sungyoon Kim) Let be an acute triangle with
,
, and
being its incircle, circumcircle, and circumradius, respectively. The circle
is tangent internally to
at
and externally tangent to
. Circle
is internally tangent to
at
and tangent internally to
. Let
and
denote the centers of
and
, respectively. Define points
,
,
,
analogously. Prove that
with equality if and only if triangle
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]](http://latex.artofproblemsolving.com/d/c/7/dc7462ed555c973a14140b29ca11f239711d0576.png)
Lemma.
Proof. Note and
lie on
since for a pair of tangent circles, the point of tangency and the two centers are collinear.
Let touch
,
, and
at
,
, and
, respectively. Note
. Consider an inversion,
, centered at
, passing through
,
. Since
,
is orthogonal to the inversion circle, so
. Consider
. Note that
passes through
and is tangent to
, hence
is a line that is tangent to
. Furthermore,
because
is symmetric about
, so the inversion preserves that reflective symmetry. Since it is a line that is symmetric about
, it must be perpendicular to
. Likewise,
is the other line tangent to
and perpendicular to
.
Let and
(second intersection).
Let and
(second intersection).
Evidently, and
. We want:
by inversion. Note that
, and they are tangent to
, so the distance between those lines is
. Drop a perpendicular from
to
, touching at
. Then
. Then
. So
.
Note that
. Applying the double angle formulas and
, we get
End Lemma
The problem becomes:
which is true because
, equality is when the circumcenter and incenter coincide. As before,
, so, by symmetry,
. Hence the inequality is true iff
is equilateral.
Comment: It is much easier to determine by considering
. We have
,
,
, and
. However, the inversion is always nice to use. This also gives an easy construction for
because the tangency point is collinear with the intersection of
and
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.