NMO Selection Tests for the Balkan and International Mathematical Olympiads

Since $AB=DE$, the triangles $NAB$ and $NED$ are congruent, so $NA=NE$ and $NB=ND$. Let $K$ and $L$ be the antipodes of $N$ in ${\gamma }_1$ and ${\gamma }_2$, respectively, and notice that they are the midpoints of the arcs $AE$ and $BD$, respectively. Notice further that the angles $KME$ and $LMD$ have equal measures, to deduce that the four arcs $KA,KE,LB$ and $LD$ all have the same measure. The arcs $EMN,AFN$ and $BMN$ have equal measures as well. Consequently, the inscribed angles $AFN$ and $DCN$ subtend arcs of equal measures on the respective circles, so they are congruent; that is, the points $A,F,C,D$ are co-cyclic.

We now show that the center of the circle through $A,F,C,D$ is the midpoint $O$ of the segment $KL$. To this end, we show that $O$ lies on the perpendicular bisector of any segment $X_1{X}_2$ through $N$, where $X_1$ is on ${\gamma }_1$ and $X_2$ is on ${\gamma }_2$; in particular, $O$ lies on the perpendicular bisectors of both segments $AC$ and $DF$, whence the conclusion.

Let $O_1$ and $O_2$ be the centers of the circles ${\gamma }_1$ and ${\gamma }_2$, respectively, and notice that $O{O}_{1}N{O}_2$ is a parallelogram to deduce that the segments $NO$ and $O_1{O}_2$ cross each other at their common midpoint $P$. Let further $X,X_1^{\prime }$ and $X_2^{\prime }$ be the midpoints of the segments $X_1{X}_2,N{X}_1$ and $N{X}_2$, respectively, and notice that the segments $NX$ and $X_1^{\prime }{X}_2^{\prime }$ have the same midpoint $X^{\prime }$, to infer that $OX$ is parallel to $P{X}^{\prime }$. Since the latter is perpendicular to $X_1^{\prime }{X}_2^{\prime }$, it follows that $OX$ is perpendicular to $X_1{X}_2$, so $OX$ is indeed the perpendicular bisector of the segment $X_1{X}_2$.