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 | + | 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 | + | with [[equal]]ity [[iff|if and only if]] triangle <math>ABC</math> is [[equilateral triangle|equilateral]]. |
== Solution == | == Solution == | ||
+ | {{image}} | ||
+ | |||
+ | '''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>. | ||
− | |||
− | Let \ | + | <!--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''' | ||
− | |||
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> | ||
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<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>. | ||
− | + | == See also == | |
+ | {{USAMO newbox|year=2007|num-b=5|after=Last Question}} | ||
− | + | [[Category:Olympiad Geometry Problems]] | |
− | |||
− | |||
− |
Revision as of 15:59, 26 April 2007
Problem
Let be an acute triangle with , , and being its incircle, circumcircle, and circumradius, respectively. Circle is tangent internally to at and tangent externally to . Circle is tangent internally 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.
Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
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 inversions map a circle's center collinear with the center of inversion. Likewise, is the other line tangent to and perpendicular to .
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 .
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 |