1999 USAMO Problems/Problem 6
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 .
Therefore . 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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