IMO Team Selection Test 2

Because of the requirement on the length, the configuration is fixed: $M$ lies on the ray $CB$ past $B$, and $N$ lies on the ray $ED$ past $D$. See the figure. Let $\alpha =\angle BAC$ and $\beta =\angle ABC$. Moreover, let $H$ be the orthocentre of the triangle (in other words: the intersection of $AD$, $BE$, and $CF$). Thales's theorem yields that $AFHE$, $BDHF$, $CEHD$, $ABDE$, $BCEF$, and $CAFD$ are cyclic. Because of the cyclic quadrilateral $ABDE$, we get $\angle CED=180^{\circ }-\angle AED=\angle ABD=\beta$ and because of the cyclic quadrilateral $BCEF$, we get $\angle AEF=180^{\circ }-\angle CEF=\angle CBF=\beta$. Analogously, $\angle CDE$ and $\angle BDF$ equal $\alpha$. From $\angle CED=\beta =\angle AEF$ it follows that $\angle DEH=90^{\circ }-\beta =\angle FEH$. Hence, $EH$ is the angular bisector of $\angle DEF$. Because $DE\parallel FM$, we get that $\angle MFE=180^{\circ }-\angle FED=180^{\circ }-2(90^{\circ }-\beta )=2\beta$. As $FN$ is the angular bisector of $\angle MFE$, we have $\angle EFN=\frac{1}{2}\cdot 2\beta =\beta$. Because $\angle FEH=90^{\circ }-\beta$, we also see that $FN$ and $EH$ are perpendicular, hence $EH$ is not only the angular bisector in $△FEN$, but it is also an altitude. Therefore, this line is also the perpendicular bisector of $FN$. As $B$ lies on this line, we get $|BF|=|BN|$. We already saw that $\angle CDE=\alpha =\angle BDF$. Because $DE\parallel FM$, we also have $\angle BMF=\angle CDE=\alpha$, hence $\angle DMF=\angle BMF=\angle BDF=\angle MDF$. Thus, $|FM|=|FD|$. Let $S$ be the intersection of $AC$ with $MF$. Then we have $\angle BFM=\angle AFS$ and because $DE\parallel FM$, we get $\angle CED=\angle CSF$. The exterior angle theorem in triangle $AFS$ yields that $\angle CSF=\angle SAF+\angle AFS=\alpha +\angle AFS$.

Combining everything, we obtain $\angle CED=\alpha +\angle BFM$. On the other hand, we knew that $\angle CED=\beta$, hence $\angle BFM=\beta -\alpha$. Moreover, we know that $\angle BMF=\alpha$. We conclude that $|BF|=|BM|$ if and only if $\beta -\alpha =\alpha$, or if and only if $\beta =2\alpha$. Because we already know that $|BF|=|BN|$, we get: $B$ is the circumcentre of $△FMN$ if and only if $\beta =2\alpha$. Before, we saw that $EH$ is the perpendicular bisector and altitude in triangle $EFN$, hence this triangle is isosceles with top angle $E$, which yields that $\angle DNF=\angle ENF=\angle EFN=\beta$. Moreover, we know that $\angle CDE=\alpha =\angle BDF$, from which it follows that $\angle NDF=\angle NDB+\angle BDF=\angle CDE+\angle BDF=2\alpha$. Hence, $|FD|=|FN|$ if and only if $\beta =2\alpha$. Because we already knew that $|FM|=|FD|$, we now get: $F$ is the circumcentre of $△DMN$ if and only if $\beta =2\alpha$. We conclude that $F$ is the circumcentre of $△DMN$ if and only if $B$ is the circumcentre of $△FMN$, as both properties are equivalent to $\beta =2\alpha$. $\square$