Difference between revisions of "1999 USAMO Problems/Problem 6"
(→Solution) |
(→Solution) |
||
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
| − | ABCD is cyclic since it is an | + | 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=DE</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>. |
| − | 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. | + | 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. |
| − | Now angle GFA = angle GCA = angle DCA. | + | Now <math>\angle GFA=\angle GCA=\angle DCA</math>. |
| − | + | <math>\angle ACF=90+\frac{\angle DCA}{2}</math>. | |
| − | This means that angle AGF = 90-ACD/ | + | This means that <math>\angle AGF=90-\frac{\angle ACD}{2}</math>. (due to cyclic quadilateral <math>ACFG</math> as given). |
| − | Now angle FAG | + | Now \angle FAG 180-(\angleAFG+\angleFGA)=90-\frac{\angle ACD}{2}=\angle AGF<math>. |
| − | Therefore angle FAG = angle AGF. | + | Therefore </math>\angle FAG=\angle AGF$. |
QED. | QED. | ||
{{solution}} | {{solution}} | ||
Revision as of 20:49, 28 October 2013
Problem
Let 
be an isosceles trapezoid with 
. The inscribed circle 
of triangle 
meets 
at 
. Let 
be a point on the (internal) angle bisector of 
such that 
. Let the circumscribed circle of triangle 
meet line at 
and 
. Prove that the triangle 
is isosceles.
Solution
Quadrilateral is cyclic since it is an isosceles trapezoid. 
. Triangle 
and triangle are reflections of each other with respect to diameter which is perpendicular to 
. Let the incircle of triangle touch 
at 
. The reflection implies that 
, which then implies that the excircle of triangle is tangent to at . Since 
is perpendicular to which is tangent to the excircle, this implies that passes through center of excircle of triangle .
We know that the center of the excircle lies on the angular bisector of 
and the perpendicular line from to . This implies that is the center of the excircle.
Now 
.
.
This means that 
. (due to cyclic quadilateral 
as given).
Now \angle FAG 180-(\angleAFG+\angleFGA)=90-\frac{\angle ACD}{2}=\angle AGF$.
Therefore$ (Error compiling LaTeX. Unknown error_msg)\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. 
