Hellenic Mathematical Olympiad

Let $K$ be the intersection point of $B\Delta$, $\Gamma E$ and $N$ be the second intersection point of $c_1$, $c_2$. Let $Z$ be the point of intersection of the tangents at $B,\Gamma$ of the circle $c$. We will prove that the points $K,N,Z$ are collinear. Let $T$ be the intersection of $BZ$ with $c_1$ and $P$ the intersection of $Z\Gamma$ with $c_2$, then: $${\hat{E}}_1={\hat{T}}_1:(\text{inscribed in }{c}_1\text{ at the same arc }BK)$$ $${\hat{E}}_1={\hat{B}}_1:(\text{chord and tangent at }c)$$ $${\hat{\Delta }}_1={\hat{P}}_1:(\text{inscribed in }{c}_2\text{ at the same arc }\Gamma K)$$ $${\hat{\Delta }}_1={\hat{\Gamma }}_1:(\text{chord and tangent at }c)$$ $${\hat{B}}_1={\hat{\Gamma }}_1:(\text{ZB and ZGammaare tangents at }c)$$ From the above inequalities we have: $${\hat{B}}_1={\hat{T}}_1,\text{ so }KT//B\Gamma \text{ and }{\hat{P}}_1={\hat{\Gamma }}_1,\text{ so }KP//B\Gamma .$$ Therefore $B\Gamma PT$ is an isosceles trapezium and thus it is cyclic (let $c_3$ its circle). This means that the common chord $KN$ of the circles $c_1$ and $c_2$ will pass through the radical center $Z$ of the circles $c_1,c_2,c_3$. Suppose first that $A,K,N$ are collinear. Then, since $K,N,T$ are collinear, we have that $A,K,N,T$ are collinear, so $A,K,N$ are on the symmedian $AT$. Conversely, if $K$ is on the symmedian $AT$, then $A,K,T$ are collinear and since $K,N,T$ are collinear, we conclude that $A,K,N$ are collinear.

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Alternative solution.

Let $K$ be the intersection point of $B\Delta$, $\Gamma E$ and $N$ the second common point of $c_1,c_2$. Let $Z$ be the point of intersection of the tangents at $B,\Gamma$ of the circle $c$. If $H$ is the intersection of $EB$ and $\Delta \Gamma$, we will prove that $K,N,Z,H$ are collinear. Using Pascal's theorem at the degenerate hexagon $BB\Gamma \Gamma \Delta E$, we have that $H,K,Z$ are collinear. The point $H$ has power with respect to $c$, $HE\cdot HB=H\Delta \cdot H\Gamma$. Therefore $K,N,H$ are collinear. Therefore $K,N,Z,H$ are collinear.

Now it is clear that $A,K,N$ are collinear if and only if $K$ is on the symmedian $AZ$.