As $\angle DAE=\angle AFE$, we obtain that $EA$ is tangent to the circumcircle of triangle $ADF$. Hence, $AE^2=ED*EF$. As triangle $AEF$ is right angled, line $EM$ is $E$-symmedian of triangle $AEF$ and thus also perpendicular to $AF$. Then $FE^2-FX^2=AE^2-AX^2$ or $(FE^2-FX^2)-(AE^2-AX^2)=0$ $(1)$. Also $\angle EXF = 90^\circ - \angle CBA = \angle FDC$, so quadrilateral $EDCX$ is cyclic and $FD*FE=FC*FX$ or $FD*FE-FC*FX=0$. Subtracting this equation from equation $(1)$, we get $(AE^2-ED*EF)-(AX^2-XC*XF)=0$. Finally, as $AE^2-ED*EF=0$ we also get $AX^2-XC*XF=0$ or $AX^2=XC*XF$ which implies that $AX$ is tangent to $\Gamma$.

This problem was proposed by me . Solution: By simple angle chasing $\measuredangle ADF+ \measuredangle DCF=180^{\circ}$ $\Longrightarrow$ $AD$ is tangent to $\odot (CFD)$, since $$MA^2=MD^2 \Longrightarrow \text{$M$ belongs in the radical axis of $\odot (A)$ and $\odot (CDF).$} $$On the other hand, so from $\measuredangle EAD=\measuredangle AFE$ we get $$EA^2=ED.EF \Longrightarrow \text{$E$ belongs in the radical axis of $\odot (A)$ and $\odot (CDF).$} $$hence $EM$ is the radical axis of $\odot (A)$ and $\odot (CDF).$ Finally by radical axis in $\odot (A)$, $\odot (ABC)$ and $\odot (CDF)$ we get, the tangent by $A$ to $\odot (ABC)$, $EM$ and $CF$ are concurrent in $X.$

We do not need $DE \perp AB$. Let $DE$ meet $\Gamma$ at $P \neq F$, and let $Q \equiv AP \cap BC$. Notice that $\angle DAE= \angle AFE= \angle ABP$ thus $BP \parallel AD$. Furhermore $\tfrac{BE}{EA} = \tfrac{BP}{AD} = \tfrac{BQ}{DQ}$. By Menelau's $M$, $E$ and $Q$ are collinear. By Pascal on $AABCFP$ we get $Q$, $E$ and $AA \cap CF$ are collinear, thus $E$, $M$ and $AA \cap CF$ are collinear.

Easy for P2. $M$ is the circumcenter of $\bigtriangleup EAD$ so $\angle AEM=\angle EAM=\angle EAD$. Let $\{N\}=\overline{EX}\cap \overline{AF}$. We have, $$\angle NEF=90^\circ-\angle MEA=90^\circ-\angle EAD=90^\circ-\angle AFD$$then $\angle ANE=90^\circ$. Clearly, it is enough to show that $\angle XAF=\angle ABF$. Since $\angle BEF=90^\circ=\angle ANX$, we only need to prove that $\bigtriangleup BEF\sim \bigtriangleup ANX$. So, we must show that $\frac{AN}{EB}=\frac{NX}{EF}\iff AN\cdot EF=NX\cdot EB$. We have $\angle XFN=\angle CFA=\angle CBA=\angle DEB$ and $\angle BED=\angle XNF$ so $\bigtriangleup BED\sim \bigtriangleup FNX$, hence, $\frac{EB}{NF}=\frac{ED}{NX}\therefore EB\cdot NX=ED\cdot NF$. So it suffices to prove that $AN\cdot EF=ED\cdot NF\iff \frac{AN}{NF}=\frac{ED}{EF}$. But, $$\frac{AN}{NF}=\frac{AE}{EF}\cdot \frac{\sin \angle AEM}{\sin (90^\circ-\angle AEM)}=\frac{AE}{EF}\cdot \tan \angle AEM$$so $ \frac{AN}{NF}=\frac{ED}{EF}\iff \frac{ED}{AE}=\tan \angle AEM=\tan \angle DAE$, which is obviously true in $\bigtriangleup EAD$.

What were the notes for Gold, Silver and Bronze of this competition?

JJ44e wrote: What were the notes for Gold, Silver and Bronze of this competition? They were 36, 27 and 14, respectively.

Beautiful problem - and a new solution! Let $F'$ be the second intersection of $FE$ with the circumcircle of $\triangle ABC$. $\textbf{Claim:}$ $AF', EM, BC$ concur(or are parallel). $\textbf{Proof)}$ We know that $$\angle F'BA = \angle F'FA = \angle DFA = \angle BAD$$meaning that $F'B$ is parallel to $AD$. Now let $T$ be the exsimilicenter of the circumcircles of $\triangle F'BE$ and $\triangle AED$. If we let $EM \cap F'B = M'$, then $E, M, M'$ are collinear as $AD \parallel F'B$. Then $T,M,M',E$ are collinear. Taking the homothety centered at $T$ sending one of the circumcircles to the other and noticing that $F'B$ and $AD$ are parallel diameters we get that $T \in AF'$ and $T \in BD$. $\square$ Now using Pascal on degenerate hexagon $AABCFF'$ we get that the tangent at $A$ to $\Gamma$, $CF$ and $TE$ concur meaning that $AX$ is indeed tangent to $\Gamma$ $\square$

Remarks

Attachments:

We use directed angles$\mod 180^\circ$. Let $Q$ be the other intersection of line $\overline{EF}$ with $\Gamma$ and $P=\overline{QA}\cap\overline{BC}$. First, note that \[ \measuredangle ABQ = \measuredangle AFQ = \measuredangle AFE = \measuredangle EAD = \measuredangle BAD, \]so $\overline{BQ}\parallel\overline{AD}$. Then, since $E=\overline{AB}\cap\overline{DQ}$ and $P=\overline{QA}\cap\overline{BC}$, we know that $P,E,M$ are collinear provable by homothety . Let $X'$ be the intersection of $\overline{CF}$ with the tangent through $A$. Now, by Pascal's theorem on cyclic hexagon $AABCFQ$, we have $X', E, P$ collinear. Thus, $X'$ is on line $\overline{EM}$, implying $X'=X$ which gives the desired result. $\blacksquare$

Claim1 : $EDCX$ is cyclic. Proof : $\angle XED = \angle MED = \angle ADE = \angle 90 - \angle DAE = \angle 90 - \angle AFE = \angle FAE = \angle FCD$. Let $EM$ meet $AF$ at $H$. Claim2 : $EM \perp AF$. Proof : $\angle HEA = \angle 90 - \angle ADE = \angle 90 - \angle HAE$. Note that we have $AE^2 = ED.EF$ and $FC.FX = FD.FE$ so $FC.FX = EF^2 - AE^2$ so we have $XC.XF = FX^2 - EF^2 + EA^2 = FX^2 - EH^2 - HF^2 + AH^2 + EH^2 = FX^2 - HF^2 + AH^2 = HX^2 + AH^2 = XA^2$ so $XA^2 = XC.XF$ so $XA$ is tangent.

Let $L=DE\cap (ABC)\neq F$, $N=AD\cap BC\neq A$ and let $Z=AL\cap BC$. Let $AA\cap CF=X'$; we want to show that $X'=X$. Claim 1 : $X-E-Z$ colinear Pf: By PASCAL's theorem on $AABCFL$, we know that the points $AA\cap CF=X'$, $AB\cap FL=E$ and $BC\cap LA=Z$ are colinear. $\blacksquare$ Claim 2: Z-E-M colinear Pf: By the angle condition, $BL$ is parallel to $AD$. Hence in $\triangle DZA$, line $X-E-M$ is the $Z-midline$. $\blacksquare$ Together, Claim 1 and Claim 2 solve the problem.

Nice! Since $M$ is midpoint of $AD$ and $\angle AED=90$, we have $ME=MD=MA$. Then $\angle XED=\angle MED=\angle MDE=90-\angle EAD=90-\angle AFE=\angle EAF=\angle BCF$ which means that $(E D C X)$ is cyclic, then $FC*FX=FD*FE$. And since $\angle EAD=\angle AFE$, we have $AE$ is tangent to $(A D F)$, then $AE^2=ED*EF$ $(2)$. Then we have $\angle EAF=90-\angle AFE=90-\angle EAD=90-\angle MEA$, which means that $EX \perp AF$. Then it is well known that $AE^2+FX^2=EF^2+AX^2$ $(3)$. Then using $(1)$,$(2)$ and $(3)$, we conclude that $AE^2+FC*FX+CX*FX=AE^2+FX^2=FE^2+AX^2=ED*EF+FD*FE+AX^2$ --> $AX^2=XC*XF$, then $AX$ is tangent to $(A B C)$, as desired.