## IMO 2016 – Problem 1

**IMO 2016, Problem 1.** Triangle has a right angle at . Let be the point on line such that and lies between and . Point is chosen such that and is the bisector of . Point is chosen such that and is the bisector of . Let be the midpoint of . Let be the point such that is a parallelogram (where and ). Prove that the lines and are concurrent.

*Solution:* When trying to see what’s going on, by drawing the figure associated to this problem you can easily get lost in the details… Let’s just stop after drawing points and . If drawn properly one immediately notices that seems to be the circumcenter of … Let be the circumcenter of . Since and it follows that so is cyclic. If we denote the common measure of by we find that and since we see that indeed is isosceles and thus .

We go on and look at point which seems to lie on . How do we prove this? First we note that , the midpoint of is the circumcenter of and thus .

Furthermore, and . Therefore triangles and are congruent. As a consequence . The properties of imply immediately that so is a parallelogram, and moreover a rhombus, since .

Now we ca finish by noting that is an axis of symmetry for the rhombus and triangles and are congruent. The symmetry of the figure implies that the lines and are concurrent.