2017 USAJMO Problems/Problem 5
Contents
Problem
Let and be the circumcenter and the orthocenter of an acute triangle . Points and lie on side such that and . Ray intersects the circumcircle of triangle in point . Prove that .
Solution 1
It's well known that the reflection of across , , lies on . Then is just the reflection of across , which is equivalent to the reflection of across . Reflect points and across to points and , respectively. Then is the midpoint of minor arc , so are collinear in that order. It suffices to show that .
Claim: . The proof easily follows.
Proof: Note that . Then we have . So, it suffices to show that Notice that , so that Therefore, it suffices to show that But it is easy to show that , implying the result.
Solution 2
Suppose ray intersects the circumcircle of at , and let the foot of the A-altitude of be . Note that . Likewise, . So, . is cyclic, so . Also, . These two angles are on different circles and have the same measure, but they point to the same line ! Hence, the two circles must be congruent. (This is also a well-known result)
We know, since is the midpoint of , that is perpendicular to . is also perpendicular to , so the two lines are parallel. is a transversal, so . We wish to prove that , which is equivalent to being cyclic.
Now, assume that ray intersects the circumcircle of at a point . Point must be the midpoint of . Also, since is an angle bisector, it must also hit the circle at the point . The two circles are congruent, which implies NDP is isosceles. Angle ADN is an exterior angle, so . Assume WLOG that . So, . In addition, . Combining these two equations, .
Opposite angles sum to , so quadrilateral is cyclic, and the condition is proved.
-william122
See also
2017 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |