IMO 2016 Shortlisted Problems

Denote the common angle in (1) by $\theta$. As $△ABF\sim △ACD$, we have $\frac{AB}{AC}=\frac{AF}{AD}$ so that $△ABC\sim △AFD$. From $EA=ED$, we get $$\angle AFD=\angle ABC=90^{\circ }+\theta =180^{\circ }-\frac{1}{2}\angle AED.$$ Hence, $F$ lies on the circle with centre $E$ and radius $EA$. In particular, $EF=EA=ED$. As $\angle EFA=\angle EAF=2\theta =\angle BFC$, points $B,F,E$ are collinear. As $\angle EDA=\angle MAD$, we have $ED\parallel AM$ and hence $E,D,X$ are collinear. As $M$ is the midpoint of $CF$ and $\angle CBF=90^{\circ }$, we get $MF=MB$. In the isosceles triangles $EFA$ and $MFB$, we have $\angle EFA=\angle MFB$ and $AF=BF$. Therefore, they are congruent to each other. Then we have $BM=AE=XM$ and $BE=BF+FE=AF+FM=AM=EX$. This shows $△EMB\cong △EMX$. As $F$ and $D$ lie on $EB$ and $EX$ respectively and $EF=ED$, we know that lines $BD$ and $XF$ are symmetric with respect to $EM$. It follows that the three lines are concurrent.

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Alternative solution.

From $\angle CAD=\angle EDA$, we have $AC\parallel ED$. Together with $AC\parallel EX$, we know that $E,D,X$ are collinear. Denote the common angle in (1) by $\theta$. From $△ABF\sim △ACD$, we get $\frac{AB}{AC}=\frac{AF}{AD}$ so that $△ABC\sim △AFD$. This yields $\angle AFD=\angle ABC=90^{\circ }+\theta$ and hence $\angle FDC=90^{\circ }$, implying that $BCDF$ is cyclic. Let ${\Gamma }_1$ be its circumcircle. Next, from $△ABF\sim △ADE$, we have $\frac{AB}{AD}=\frac{AF}{AE}$ so that $△ABD\sim △AFE$. Therefore, $$\angle AFE=\angle ABD=\theta +\angle FBD=\theta +\angle FCD=2\theta =180^{\circ }-\angle BFA.$$ This implies $B,F,E$ are collinear. Note that $F$ is the incentre of triangle $DAB$. Point $E$ lies on the internal angle bisector of $\angle DBA$ and lies on the perpendicular bisector of $AD$. It follows that $E$ lies on the circumcircle ${\Gamma }_2$ of triangle $ABD$, and $EA=EF=ED$. Also, since $CF$ is a diameter of ${\Gamma }_1$ and $M$ is the midpoint of $CF$, $M$ is the centre of ${\Gamma }_1$ and hence $\angle AMD=2\theta =\angle ABD$. This shows $M$ lies on ${\Gamma }_2$. Next, $\angle MDX=\angle MAE=\angle DXM$ since $AMXE$ is a parallelogram. Hence $MD=MX$ and $X$ lies on ${\Gamma }_1$. We now have two ways to complete the solution. - Method 1. From $EF=EA=XM$ and $EX\parallel FM$, $EFMX$ is an isosceles trapezoid and is cyclic. Denote its circumcircle by ${\Gamma }_3$. Since $BD$, $EM$, $FX$ are the three radical axes of ${\Gamma }_1$, ${\Gamma }_2$, ${\Gamma }_3$, they must be concurrent. - Method 2. As $\angle DMF=2\theta =\angle BFM$, we have $DM\parallel EB$. Also, $$\angle BFD+\angle XBF=\angle BFC+\angle CFD+90^{\circ }-\angle CBX=2\theta +(90^{\circ }-\theta )+90^{\circ }-\theta =180^{\circ }$$ implies $DF\parallel XB$. These show the corresponding sides of triangles $DMF$ and $BEX$ are parallel. By Desargues' Theorem, the two triangles are perspective and hence $DB$, $ME$, $FX$ meet at a point.

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Alternative solution.

Let the common angle in (1) be $\theta$. From $△ABF\sim △ACD$, we have $\frac{AB}{AC}=\frac{AF}{AD}$ so that $△ABC\sim △AFD$. Then $\angle ADF=\angle ACB=90^{\circ }-2\theta =90^{\circ }-\angle BAD$ and hence $DF\perp AB$. As $FA=FB$, this implies $△DAB$ is isosceles with $DA=DB$. Then $F$ is the incentre of $△DAB$. Next, from $\angle AED=180^{\circ }-2\theta =180^{\circ }-\angle DBA$, points $A,B,D,E$ are concyclic. Since we also have $EA=ED$, this shows $E,F,B$ are collinear and $EA=EF=ED$. Note that $C$ lies on the internal angle bisector of $\angle BAD$ and lies on the external angle bisector of $\angle DBA$. It follows that it is the $A$-excentre of triangle $DAB$. As $M$ is the midpoint of $CF$, $M$ lies on the circumcircle of triangle $DAB$ and it is the centre of the circle passing through $D,F,B,C$. By symmetry, $DEFM$ is a rhombus. Then the midpoints of $AX$, $EM$ and $DF$ coincide, and it follows that $DAFX$ is a parallelogram. Let $P$ be the intersection of $BD$ and $EM$, and $Q$ be the intersection of $AD$ and $BE$. From $\angle BAC=\angle DCA$, we know that $DC$, $AB$, $EM$ are parallel. Thus we have $\frac{DP}{PB}=\frac{CM}{MA}$. This is further equal to $\frac{AE}{BE}$ since $CM=DM=DE=AE$ and $MA=BE$. From $△AEQ\sim △BEA$, we find that $$\frac{DP}{PB}=\frac{AE}{BE}=\frac{AQ}{BA}=\frac{QF}{FB}$$ by the Angle Bisector Theorem. This implies $QD\parallel FP$ and hence $F,P,X$ are collinear, as desired.