Difference between revisions of "1999 USAMO Problems/Problem 6"
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Let <math>ABCD</math> be an isosceles trapezoid with <math>AB \parallel CD</math>. The inscribed circle <math>\omega</math> of triangle <math>BCD</math> meets <math>CD</math> at <math>E</math>. Let <math>F</math> be a point on the (internal) angle bisector of <math>\angle DAC</math> such that <math>EF \perp CD</math>. Let the circumscribed circle of triangle <math>ACF</math> meet line <math>CD</math> at <math>C</math> and <math>G</math>. Prove that the triangle <math>AFG</math> is isosceles. | Let <math>ABCD</math> be an isosceles trapezoid with <math>AB \parallel CD</math>. The inscribed circle <math>\omega</math> of triangle <math>BCD</math> meets <math>CD</math> at <math>E</math>. Let <math>F</math> be a point on the (internal) angle bisector of <math>\angle DAC</math> such that <math>EF \perp CD</math>. Let the circumscribed circle of triangle <math>ACF</math> meet line <math>CD</math> at <math>C</math> and <math>G</math>. Prove that the triangle <math>AFG</math> is isosceles. | ||
| − | == Solution ==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. | + | == Solution == |
| + | 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. | ||
tri here means triangle. | tri here means triangle. | ||
{{solution}} | {{solution}} | ||
Revision as of 21:34, 28 October 2013
Problem
Let 
be an isosceles trapezoid with 
. The inscribed circle 
of triangle 
meets 
at 
. Let 
be a point on the (internal) angle bisector of 
such that 
. Let the circumscribed circle of triangle 
meet line at 
and 
. Prove that the triangle 
is isosceles.
Solution
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. tri here means triangle. This problem needs a solution. If you have a solution for it, please help us out by adding it.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
