Difference between revisions of "2014 USAMO Problems/Problem 5"
Bluelinfish (talk | contribs) m (→Solution) |
|||
Line 12: | Line 12: | ||
</cmath> | </cmath> | ||
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>. | ||
+ | |||
+ | ==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 |