2014 USAMO Problems/Problem 5
Contents
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 1
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 .
Solution 2
Since is a cyclic quadrilateral, . and , we find . That is, is a cyclic quadrilateral. Let be mid-point of . are collinear and . Let be second intersection of with circumcircle of the triangle . Let , . Since is mid-point of the arc , . Since is a cyclic quadrilateral, . Since is the orthocenter of triangle , . Thus, and is a cyclic quadrilateral. So, and . We will prove that is a parallelogram.
https://wiki-images.artofproblemsolving.com//7/7b/Usamo2014-5.png (figure link)
We see that is an isosceles triangle and . Also and . Then, . By spiral similarity, and . Hence, , . Since , we get is a parallelogram. As a result, .
(Lokman GÖKÇE)
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 |