It is very easy problem. Solution: Denote $ P = AK\cap BC$.Using Menelaus theorem for $ \triangle ACP,\triangle ABP$ and transversals $ \overline{X_{4}X_{3}K}$ and $ \overline{X_{1}X_{2}K}$: $ PX_{3}= PX_{2}\Longleftrightarrow\frac{BX_{2}}{BX_{1}}\cdot\frac{CX_{4}}{CX_{3}}\cdot\frac{AX_{1}}{AX_{4}}= 1$.The last one is obvious,because $ BX_{1}= BX_{2},CX_{3}= CX_{4},AX_{4}= AX_{1}$.(use that the center of $ \Gamma$ is incenter of $ \triangle ABC$.)Thus we are done.

Let $ D$, $ E$,$ F$ be the points of contact of the incircle with the sides $ BC$, $ AC$ and $ AB$ and let $ K_{1}$ and $ K_{2}$ be the points where $ X_{1}X_{2}$ and $ X_{3}X_{4}$ cut $ AD$. Notice that $ D$ is the midpoint of $ X_{2}X_{3}$ and that $ FX_{1}= EX_{4}$. It is easy to prove that $ X_{1}X_{2}\parallel DF$ and that $ X_{3}X_{4}\parallel DE$ so $ \Delta AFD\sim AX_{1}K_{1}$ and $ AED\sim AX_{4}K_{2}$, then $ AK_{1}= AD\frac{AF+FX_{1}}{AF}= AD\frac{AE+EX_{4}}{AE}= AK_{2}$ so $ K_{1}= K_{2}= K$, which completes the proof.

To dM@gGo+: Did you take participant in this olympiad?

No. David is Panama's deputy leader, not a contestant. By the way, nice solution, David!

Thanks

It's easy to note that $ CX_{3}=CX_{4}$, $ BX_{2}=BX_{1}$ and $ AX_{4}=AX_{1}$. Let $ L$ the line parallel to $ BC$ through A, let $ P=X_{3}X_{4}\cap L$ and $ Q=X_{2}X_{1}\cap L$ then, $ AP=AX_{4}=AX_{1}=AQ$ so $ A$ is the midpoint of $ PQ$, since $ X_{2}X_{3}$ is parallel to $ PQ$, we conclude that $ AK$ passes through the midpoint of $ X_{2}X_{3}$. $ Tipe$

i have a trigonometric solution (as ussual ), i'll post it later when i have a little more time... it uses that $ BX_{1}=BX_{2}$, $ CX_{3}=CX_{4}$, some power or a point and some law of sines for the gergonne cevian (that is what we want to prove) combined with briggs formulas... to david: how did the contestants do at this problem? it seems much more easier than last year's q5..

campos wrote: i have a trigonometric solution (as ussual ), i'll post it later when i have a little more time... There's no solution if we can't see it... Just kidding! Well, I have to say it was a 50/50: near 50% of the contestants solved it throughly. Problems 3 and 6 of this Iberoamerican Math Olympiad were more interesting cases: nobody solved them.

Erken wrote: It is very easy problem. last one is obvious,because $ BX_{1}= BX_{2},CX_{3}= CX_{4},AX_{4}= AX_{1}$.(use that the center of $ \Gamma$ is incenter of $ \triangle ABC$.)Thus we are done. proof?

This is probably much more roundabout and complicated than needed but here's an alternative method. Continue labeling the points counterclockwise around the circle as $ X_{5}, X_{6}$. Clearly $ X_{2}X_{3}=X_{4}X_{5}=X_{6}X_{1}$, because $ I$ is equidistant from the sides of the triangle. Then those three arcs have equal measure in this circle. From angle chasing, we get $ m\angle{AX_{1}X_{2}}+m\angle{KX_{4}X_{1}}=180^{\circ}$, and so the circumcircle of $ \triangle{KX_{1}X_{4}}$ is tangent to $ AB$ and $ AC$. Then $ KA$ is the symmedian of $ \triangle{KX_{1}X_{4}}$, and so it's the median in $ \triangle{KX_{2}X_{3}}$. The conclusion follows.

Is easy to see that $AX_1X_4, BX_2X_3, CX_2X_3$ are isosceles triangles, thus $\angle{X_1KX_4}=\angle{AX_1X_4}=\angle{AX_4X_1}=\frac{\pi-A}{2}$ so $AX_1$ and $AX_4$ are tangents to the circumcirle of $X_1X_4K$. Finally $KA$ is symmedian and $X_2X_3$ is antiparallel to $X_1X_4$; hence done.

Let the incircle touch $BC,CA,AB$ at $D,E,F$,respectively. Then $DEF$ and $X_1X_4K$ are homothetic with homothety center $A$. So $A,D,K$ are collinear since $D$ is the midpoint of $X_2X_3$ we get the desired result.