Problem:
Let be a scalene triangle with circumcircle and incenter . Ray meets at and meets again at ; the circle with diameter cuts again at . Lines and meet at , and is the midpoint of . The circumcircles of and intersect at points and . Prove that passes through the midpoint of either or .
Solution:
(Proposed by Evan Chen)
Let be the midpoint of , and let be the point on opposite . Observe that line passes through , and thus lines concur at the orthocenter of , which is . Denote by the -excenter of .
Next, let be the foot of the altitude from to ; observe that lies on the circle centered at through . Then, is the radical center of , and the circle with diameter ; hence line passes through ; accordingly is the orthocenter of ; denote by the foot of the altitude from to .
We claim that this lies on both the circumcircle of and . It lies on the circumcircle of since this circle is the nine-point circle of . For the other, note that , since they share the same angle at and . Consequently, , enough to imply that quadrilateral is cyclic. But lines , and meet at , so Power of a Point in cyclic quadrilaterals and gives , hence is cyclic as needed.
All that remains to show is that the midpoint of lies on . But this follows from the fact that , thus the problem is solved.
Alternate Solution (by Titu Andreescu and Cosmin Pohoata): We refer to the same figure as in the first solution. Let be the midpoint of arc of . A first key step in the problem is to note that is the orthocenter of triangle . This follows from the fact that , which implies that line must pass through the antipode of in , which is precisely the point . This together with the fact that implies the claim.
Next, it is essential to notice that is also the orthocenter of triangle , where denotes the -excenter of triangle . This can be argued as follows: since is the orthocenter of , we have by Power of a Point that (we are implicitly using the fact that the reflection of across line lies on the circumcircle of triangle ). However, the 4-tuple is a harmonic division and is the midpoint of , which easily implies that . By Power of a Point once again, this yields that the reflection of across line lies on the circumcircle of triangle , so must indeed be the orthocenter of triangle . This is crucial, since then the circumcircle of triangle is nothing but the nine-point circle of , so the foot of altitude from on becomes a good candidate for or . If denotes the midpoint of segment , then is a midline in triangle , so ; therefore is on the circle of diameter , which is precisely . It remains to show that also lies on the circumcircle of triangle , but this is clear: is cyclic, so ; also, is cyclic, so ; hence , which by Power of a Point means that is cyclic, thus completing the proof.
The problems on this page are the property of the MAA's American Mathematics Competitions