BMO 2017

WLOG, assume $AB<AC$. Let $M$ be the midpoint of side $BC$ and let the circumcircle of $DFE$ intersect $AF$ again at $K$. Since $$90^{\circ }+\angle MED=180^{\circ }-\angle MDE=\angle ABC+\frac{\angle BAC}{2}=\angle AFE=\angle DFE+\angle AFD=90^{\circ }+\angle AFD$$ it follows that $$\angle AFD=\frac{\angle ABC-\angle ACB}{2}=\angle MED$$ Because $$\angle DKE=\angle DME=90^{\circ }$$ and $$\angle KED=\angle KFE=\angle MED$$ we get $△KDE\cong △MDE$ from which it follows that $DE$ is the perpendicular bisector of $MK$ and here we get $$\angle FAD=\angle KAD=\angle MAD$$ It is obvious that $P$ is the intersection of $ME$ and $AF$. Let $ME$ intersect $\Gamma$ again at $L$. From the angle bisector theorem in triangle $△AMP$ we get $$\frac{PE}{ME}=\frac{AP}{AM}=\frac{LP}{LM}(1)$$ ($LA$ is the external angle bisector of $\angle MAP$ since $LE$ is the diameter of $\Gamma$). Now we prove that $CE$ and $CL$ are angle bisectors of $\angle MCP$. Let $M^{\prime }$ be the point on $LE$ such that $\angle M^{\prime }CE=\angle ECP$. From the angle bisector theorem we get $$\frac{M^{\prime }E}{PE}=\frac{C{M}^{\prime }}{CP}=\frac{M^{\prime }L}{PL}(2)$$ Multiplying (1) and (2) we get $\frac{ME}{LM}=\frac{M^{\prime }E}{M^{\prime }L}$ adding 1 on both sides we get $LM=L{M}^{\prime }$ from which it follows that $M=M^{\prime }$ and thus $CE$ and $CL$ are the bisectors of $\angle MCP$. Now we have $$\angle MPC=90^{\circ }-\angle MCP=90^{\circ }-2\angle MCE=90^{\circ }-2\angle EAC=90^{\circ }-\angle BAC$$

Since $X$ and $Y$ are perpendicular to $AB$ and $AC$ we have $BXPM$ and $CYPM$ are concyclic. Here we get $$\angle MYC=\angle MPC=90^{\circ }-\angle BAC$$ and it follows that $YM\perp AX$. Similarly we get $XM\perp AY$ and so $M$ is the ortocenter of $△AXY$ giving us $M=H^{\prime }$. Since $ATHS$ and $ATQF$ are both concyclic it is enough to prove that $HSFQ$ is concyclic. Since $$\angle BQC=180^{\circ }-\angle BAC=\angle BHC$$ and $HQ\perp BC$ it follows that $BC$ is the perpendicular bisector of $HQ$. It is enough to prove that $BC$ is the perpendicular bisector of $SF$. Let $AM$ and $TH$ meet $\Gamma$ again at points $A^{\prime }$ and $N$ respectively. Since $HN$ passes through the midpoint of side $BC$ and $$\angle BHC=180^{\circ }-\angle BAC=\angle BNC$$ it follows that $BNCH$ is a parallelogram. From here we get that $$\angle NCB=\angle HBC=90^{\circ }-\angle ACB$$ giving us $\angle NCA=90^{\circ }$ and similarly $\angle NBA=90^{\circ }$. This means $AN$ is the diameter of $\Gamma$, so $$\angle N{A}^{\prime }{A}^{\prime }=\angle N{A}^{\prime }A=90^{\circ }=\angle HSA=\angle HS{A}^{\prime }$$ and from here we have $HS\parallel A^{\prime }N$. Now since $HS\parallel A^{\prime }N$ and $M$ is the midpoint of $HN$ (because $BHCN$ is a parallelogram) we get

that $HSN{A}^{\prime }$ is a paralelogram. Since $$\angle FAE=\angle EAM=\angle EA{A}^{\prime }$$ HS, AT are concurrent. we get that $F{A}^{\prime }BC$ is an isosceles trapezoid which means that $ME$ is the perpendicular bisector of $F{A}^{\prime }$ (since it is the perpendicular bisector of $BC$). This gives us $BF=C{A}^{\prime }=BS$ and $CF=B{A}^{\prime }=CS$ giving us that $SBFC$ is a deltoid, meaning that $BC$ is the perpendicular bisector of $FS$. This means that $HSFQ$ is an isosceles trapezoid. Now from the radical axis theorem of the circumcircles of $HSFQ$, $HSAT$ and $ATQF$ we get that $QF$,