29-th Balkan Mathematical Olympiad

Let $l\cap AO=\{K\}$, and $G$ be the other end point of the diameter of $\Gamma$ through $A$. Then $D$, $C$, $G$ are collinear. Moreover, $E$ is the orthocenter of triangle $ADG$. Therefore $GE\perp AD$ and $G$, $E$, $B$ are collinear.

As $\angle CDF=\angle GDK=\angle GAC=\angle GFC$, $FG$ is tangent to the circumcircle of triangle $CFD$ at $F$. As $\angle FBE=\angle FBG=\angle FAG=\angle GFK=\angle GFE$, $FG$ is also tangent to the circumcircle of $BFE$ at $F$. Hence the circumcircles of the triangles $CFD$ and $BFE$ are tangent at $F.<br/><br/>---<br/><br/>\ast \ast Alternativesolution.\ast \ast <br/><br/>Let$O_1$and$O_2$bethecircucentreoftriangles$BEF$and$CDF$respectively.Wewishtoprovethat$O_1O_2$and$F$arecollinear.Since$\angle O_1FE = 90^\circ - \angle EBF$and$\angle O_2FD = 90^\circ - \frac{1}{2}\angle DO_2F = 90^\circ - \angle DCF = \angle FCE$itsufficestoshowthat$\angle ABE = 90^\circ$,whichisequivalenttoprovingthequadrilateral$ABEK$iscyclic.Therefore,itremainstoprovethat$ABEK$iscyclic.Lettheline$l$cutthecircumscribedcircleofthetriangle$ABC$atthepoint$H \neq F$.Since$AHCF$isacyclicquadrilateral,fromthepowerofpointweobtain$\overline{AE} \cdot \overline{EC} = \overline{FE} \cdot \overline{EH}$.Moreover,sincethequadrilateral$AKCD$iscyclic(because$\angle AKD = \angle ACD = 90^\circ$),wehave$\overline{AE} \cdot \overline{EC} = \overline{ED} \cdot \overline{EK}$.Thus,fromthelasttworelations,wehave$\overline{FE} \cdot \overline{EH} = \overline{ED} \cdot \overline{EK}$.Let$\overline{DF} = a$,$\overline{FC} = b$,$\overline{EK} = c$.Since$OH \perp FH$,$K$isthemidpointof$FH$,thus$\overline{KH} = b+c$.Therefore,therelation$\overline{FE} \cdot \overline{EH} = \overline{ED} \cdot \overline{EK}$isrewrittenas<br/><br/>$B$,$E$,$G$arecollinear.Let$t$bethetangentlinetothecircumcircleofthetriangle$BFE$at$F$.Since$B$,$F$,$C$,$G$areconcyclic$K$and$L$betwopointson$t$suchthattheyareondifferentsidesoftheline$l$.Thensince$t$istangenttothecircumcircleof$BFE$at$F$wehave$\angle FBE = \angle LFE = \angle KFB = \angle FCD$using(\ast ).So$t$isalsotangenttothecircumcircleofthetriangle$FCD$at$F$.Hence,wearedone.$\triangle GFH \sim \triangle GAF$.Hence$\overline{DF} \cdot \overline{DH} = \overline{DE} \cdot \overline{DK}$.Butsince$\overline{DF} \cdot \overline{DH} = \overline{DB} \cdot \overline{DA}$,fromthepowerofpoint,weobtain$\overline{DE} \cdot \overline{DK} = \overline{DB} \cdot \overline{DA}$,therefore$ABEK$isacyclicquadrilateral,whichiswhatwewantedtoprove.<br/><br/>---<br/><br/>\ast \ast Alternativesolution.\ast \ast <br/><br/>Let$AO$intersect$\Gamma$at$G$($G \neq A$).Since$\angle ACG = 90^\circ$,$D$,$C$,$G$arecollinear.Hence$E$istheorthocenterofthetriangle$ADG$so$\overline{GF}^2 = \overline{GE} \cdot \overline{GB} - \overline{GC} \cdot \overline{GD}$,whichshowsthat$GF$istangenttothecircumcirclesof$BFE$and$FCD$at$F$. Hence, we are done.