Difference between revisions of "1999 USAMO Problems/Problem 6"
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<math>\angle ACF=90+\frac{\angle DCA}{2}</math>. | <math>\angle ACF=90+\frac{\angle DCA}{2}</math>. | ||
This means that <math>\angle AGF=90-\frac{\angle ACD}{2}</math>. (due to cyclic quadilateral <math>ACFG</math> as given). | This means that <math>\angle AGF=90-\frac{\angle ACD}{2}</math>. (due to cyclic quadilateral <math>ACFG</math> as given). | ||
− | Now <math>\angle FAG - (\ | + | Now <math>\angle FAG - (\angle AFG + \angle FGA)=90-\frac{\angle ACD}{2}=\angle AGF</math>. |
Therefore <math>\angle FAG=\angle AGF</math>. | Therefore <math>\angle FAG=\angle AGF</math>. |
Revision as of 22:17, 17 May 2015
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.