# Solution problem 2 of shortlist problem geometry in IMO 2015

Problem 2.

Let ABC be a triangle inscribed into a circle $\Omega$ with center O. A circle $\Gamma$ with center A meets the side BC at points D and E such that D lies between B and E. Moreover, let F and G be the common points of $\Gamma$ and $\Omega$. We assume that F lies on the arc AB of $\Omega$ not containing C, and G lies on the arc AC of $\Omega$ not containing B. The circumcircles of the triangles BDF and CEG meet the sides AB and AC again at K and L, respectively. Suppose that the lines FK and GL are distinct and intersect at X. Prove that the points A, X, and O are collinear.

Solution 1.

It suﬃces to prove that the lines FK and GL are symmetric about AO. Now the segments AF and AG, being chords of $\Omega$ with the same length, are clearly symmetric with respect to AO. Hence it is enough to show

$\angle KFA=\angle AGL$ …………..(1)

Let us denote the circumcircles of BDF and CEG by $\omega_B\text{ and }\omega_c$, respectively. To prove (1), we start from

$\angle KFA=\angle DFG+\angle GFA-\angle DFK$.

In view of the circles $\omega_b,\Gamma,\text{ and }\Omega$, this may be rewritten as

$\angle KFA=\angle CEG+\angle GBA-\angle DBK=\angle CEG-\angle CBG$.

Due to the circles $\omega_c\text{ and }\Omega$, we obtain $\angle KFA=\angle CLG-\angle CAG=\angle AGL$. Thereby the problem is solved.