1999 USAMO Problems/Problem 6

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

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.

1999 USAMO (Problems • Resources)
Preceded by Problem 5	Followed by Last Question
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

Page

Toolbox

Search

1999 USAMO Problems/Problem 6

Problem

Solution

See Also