Difference between revisions of "1999 USAMO Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | ABCD is cyclic since it is isosceless trapezoid.AD=BC. | + | ABCD is cyclic since it is an isosceless 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=DE, 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+angle DCA/2. | ||
| + | This means that angle AGF = 90-ACD/2 (due to cyclic quadilateral ACFG as given). | ||
| + | Now angle FAG = 180-(AFG+FGA) = 90-ACD/2 = angle AGF. | ||
| + | |||
| + | Therefore angle FAG = angle AGF. | ||
| + | QED. | ||
{{solution}} | {{solution}} | ||
Revision as of 20:38, 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 an isosceless 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=DE, 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+angle DCA/2. This means that angle AGF = 90-ACD/2 (due to cyclic quadilateral ACFG as given). Now angle FAG = 180-(AFG+FGA) = 90-ACD/2 = angle AGF.
Therefore angle FAG = angle AGF. QED. 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. 
