2014 USAMO Problems/Problem 5
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 .
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 .
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.
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, .
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