Difference between revisions of "2014 USAMO Problems/Problem 5"
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But <math>YO=OA</math> as well, and <math>\angle YOX=\angle AOO_1</math>, so <math>\triangle OYX\cong \triangle OAO_1</math>. Thus <math>XY=AO_1=AO</math>. | But <math>YO=OA</math> as well, and <math>\angle YOX=\angle AOO_1</math>, so <math>\triangle OYX\cong \triangle OAO_1</math>. Thus <math>XY=AO_1=AO</math>. | ||
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+ | ==See also== | ||
+ | {{USAMO newbox|year=2014|num-b=4|num-a=6}} |
Revision as of 10:28, 17 June 2020
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
.
See also
2014 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |