2014 USAJMO Problems/Problem 6
Let be a triangle with incenter , incircle and circumcircle . Let be the midpoints of sides , , and let be the tangency points of with and , respectively. Let be the intersections of line with line and line , respectively, and let be the midpoint of arc of .
(a) Prove that lies on ray .
(b) Prove that line bisects .
Solution 1: We will prove this via contradiction: assume that line intersects line at and line and , with and not equal to . Let and . We know that because is a midsegment of triangle ; thus, by alternate interior angles (A.I.A) , because triangle is isosceles. Also by A.I.A, . Furthermore, because is an angle bisector of triangle , it is also an altitude of the triangle; combining this with from the Exterior Angle Theorem gives . Also, because they are vertical angles. This completes part (a).
Solution 2: First we show that the intersection of with the internal angle bisector of is the same as the intersection of with the internal angle bisector of Let denote the intersection of with the internal angle bisector of and let denote the side lengths of By Menelaus on with respect to Similarly, Since and divide in the same ratio, they must be the same point. Now, since lies on ray
Solution 3: By the Iran Lemma, we know concur, so Part follows easily.
Solution 1: Using a similar argument to part (a), point U lies on line . Because , triangle is isosceles. Similarly, triangle is isosceles, from which we derive that . Hence, triangle is isosceles.
Note that lies on both the circumcircle and the perpendicular bisector of segment . Let be the midpoint of ; our goal is to prove that points , , and are collinear, which equates to proving lies on ray .
Because is also an altitude of triangle , and and are both perpendicular to , . Furthermore, we have because is a parallelogram. (incomplete)
Solution 2: Let , , and be the excenters of . Note that the circumcircle of is the nine-point circle of . Since is the external angle bisector of , is the midpoint of . Now and are parallel since both are perpendicular to the internal angle bisector of . Since bisects , it bisects as well.
Solution 3: Let be the antipode of with respect to the circumcircle of triangle . Then, by the Incenter-Excenter lemma, is the center of a circle containing , , and . Because is a diameter, and are tangent to the aforementioned circle; thus by a well-known symmedian lemma, coincides with the -symmedian of triangle . From part (a); we know that is cyclic (we can derive a similar argument for point ); thus coincides with the median of triangle , and we are done.
Solution 4: Let be the midpoints of arcs respectively, and let be the tangency point between the -mixtilinear incircle of and . It's well-known that , bisects , , and .
Now, it's easy to see , so and are homothetic at . But bisects , so Part follows directly from the homothety. ~ ike.chen