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.

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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.

Also Read :

Geometry Shortlist Problem in IMO 2015


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