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Difference between revisions of "1999 USAMO Problems/Problem 6"
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== Solution ==
 
== Solution ==
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{{solution}}
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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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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)
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Problem 5 		Followed by
Last Question
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All USAMO Problems and Solutions

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