Difference between revisions of "2014 USAMO Problems/Problem 5"
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\angle AOY=2\angle ACY=2(90-\angle PAC)=2(90-\frac{A}{2})=180-\angle A = \angle XOO_1 | \angle AOY=2\angle ACY=2(90-\angle PAC)=2(90-\frac{A}{2})=180-\angle A = \angle XOO_1 | ||
</cmath> | </cmath> | ||
− | But <math>YO=OA</math> as well, and <math>\angle YOX=\angle AOO_1</math>, so <math>\triangle OYX\ | + | But <math>YO=OA</math> as well, and <math>\angle YOX=\angle AOO_1</math>, so <math>\triangle OYX\sim \triangle OAO_1</math>. Thus <math>XY=AO_1=AO</math>. |
Revision as of 11:59, 10 July 2019
Problem
Let be a triangle with orthocenter
and let
be the second intersection of the circumcircle of triangle
with the internal bisector of the angle
. Let
be the circumcenter of triangle
and
the orthocenter of triangle
. Prove that the length of segment
is equal to the circumradius of triangle
.
Solution
Let be the center of
,
be the center of
. Note that
is the reflection of
across
, so
. Additionally
so
lies on
. Now since
are perpendicular to
and their bisector,
is isosceles with
, and
. Also
But
as well, and
, so
. Thus
.