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Difference between revisions of "1999 USAMO Problems/Problem 6"

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== Solution ==
 
== Solution ==
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ABCD is cyclic since it is isosceless trapezoid.AD=BC.tri ADC and tri BCD are reflections of each other with refect to diameter which is perpendicular to AB.Let incircle of tri ADC touches DC at K.Reflection implies that Dk=DE.This implies that excircle of tri ADC is tangent to DC at E.Since EF is perpendicular to DC which is tangent to excircle this implies EF passes through center of excircle of tri ADC.We know center of excirle lies on angular bisector of DAC and line perpendicular to DC at E,this implies that F is the centre of excirlce.Now angle GFA=angle GCA=angle DCA.angle ACF=90+angle DCA/2.This mean that angle AGF=90-ACD/2(due to cyclic quadilateral ACFG as given).Now angle FAG=180-(AFG+FGA)=90-ACD/2 =angle AGF.thereforeangle FAG=angle AGF.This completes the proof.
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Quadrilateral <math>ABCD</math> is cyclic since it is an isosceles trapezoid. <math>AD=BC</math>. Triangle <math>ADC</math> and triangle <math>BCD</math> are reflections of each other with respect to diameter which is perpendicular to <math>AB</math>. Let the incircle of triangle <math>ADC</math> touch <math>DC</math> at <math>K</math>. The reflection implies that <math>DK=CE</math>, which then implies that the excircle of triangle <math>ADC</math> is tangent to <math>DC</math> at <math>E</math>. Since <math>EF</math> is perpendicular to <math>DC</math> which is tangent to the excircle, this implies that <math>EF</math> passes through center of excircle of triangle <math>ADC</math>.
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tri here means triangle.
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{{solution}}
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We know that the center of the excircle lies on the angular bisector of <math>DAC</math> and the perpendicular line from <math>DC</math> to <math>E</math>. This implies that <math>F</math> is the center of the excircle.  
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Now <math>\angle GFA=\angle GCA=\angle DCA</math>.  
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<math>\angle ACF=90^\circ+\frac{\angle DCA}{2}</math>.
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This means that <math>\angle AGF=90^\circ-\frac{\angle ACD}{2}</math>. (due to cyclic quadilateral <math>ACFG</math> as given).
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Now <math>\angle FAG =180^\circ- (\angle AFG + \angle FGA)=90^\circ-\frac{\angle ACD}{2}=\angle AGF</math>.
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Therefore <math>\angle FAG=\angle AGF</math>.  
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QED.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 10:11, 27 September 2024

Problem

Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$. The inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$. Let $F$ be a point on the (internal) angle bisector of $\angle DAC$ such that $EF \perp CD$. Let the circumscribed circle of triangle $ACF$ meet line $CD$ at $C$ and $G$. Prove that the triangle $AFG$ is isosceles.

Solution

Quadrilateral $ABCD$ is cyclic since it is an isosceles trapezoid. $AD=BC$. Triangle $ADC$ and triangle $BCD$ are reflections of each other with respect to diameter which is perpendicular to $AB$. Let the incircle of triangle $ADC$ touch $DC$ at $K$. The reflection implies that $DK=CE$, which then implies that the excircle of triangle $ADC$ is tangent to $DC$ at $E$. Since $EF$ is perpendicular to $DC$ which is tangent to the excircle, this implies that $EF$ passes through center of excircle of triangle $ADC$.

We know that the center of the excircle lies on the angular bisector of $DAC$ and the perpendicular line from $DC$ to $E$. This implies that $F$ is the center of the excircle.

Now $\angle GFA=\angle GCA=\angle DCA$. $\angle ACF=90^\circ+\frac{\angle DCA}{2}$. This means that $\angle AGF=90^\circ-\frac{\angle ACD}{2}$. (due to cyclic quadilateral $ACFG$ as given). Now $\angle FAG =180^\circ- (\angle AFG + \angle FGA)=90^\circ-\frac{\angle ACD}{2}=\angle AGF$.

Therefore $\angle FAG=\angle AGF$. QED.

See Also

1999 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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