IMO 2016 Shortlisted Problems

Let $AM$ meet $\Gamma$ again at $Y$ and $XY$ meet $BC$ at $D^{\prime }$. It suffices to show $D^{\prime }=D$. We shall apply the following fact.

- Claim. For any cyclic quadrilateral $PQRS$ whose diagonals meet at $T$, we have $$\frac{QT}{TS}=\frac{PQ\cdot QR}{PS\cdot SR}$$ Proof. We use $[W_1{W}_2{W}_3]$ to denote the area of $W_1{W}_2{W}_3$. Then $$\frac{QT}{TS}=\frac{[PQR]}{[PSR]}=\frac{\frac{1}{2}PQ\cdot QR\sin \angle PQR}{\frac{1}{2}PS\cdot SR\sin \angle PSR}=\frac{PQ\cdot QR}{PS\cdot SR}$$ Applying the Claim to $ABYC$ and $XBYC$ respectively, we have $1=\frac{BM}{MC}=\frac{AB\cdot BY}{AC\cdot CY}$ and $\frac{B{D}^{\prime }}{D^{\prime }C}=\frac{XB\cdot BY}{XC\cdot CY}$. These combine to give $$\begin{array}{c}\frac{B{D}^{\prime }}{C{D}^{\prime }}=\frac{XB}{XC}\cdot \frac{BY}{CY}=\frac{XB}{XC}\cdot \frac{AC}{AB}\end{array}\text{\hspace{1em}(1)}$$ Next, we use directed angles to find that $\measuredangle XBF=\measuredangle XBA=\measuredangle XCA=\measuredangle XCE$ and $\measuredangle XFB=\measuredangle XFA=\measuredangle XEA=\measuredangle XEC$. This shows triangles $XBF$ and $XCE$ are directly similar. In particular, we have $$\begin{array}{c}\frac{XB}{XC}=\frac{BF}{CE}\end{array}\text{\hspace{1em}(2)}$$ In the following, we give two ways to continue the proof.

- Method 1. Here is a geometrical method. As $\angle FIB=\angle AIB-90^{\circ }=\frac{1}{2}\angle ACB=\angle ICB$ and $\angle FBI=\angle IBC$, the triangles $FBI$ and $IBC$ are similar. Analogously, triangles $EIC$ and $IBC$ are also similar. Hence, we get $$\begin{array}{c}\frac{FB}{IB}=\frac{BI}{BC}\text{ and }\frac{EC}{IC}=\frac{IC}{BC}\end{array}\text{\hspace{1em}(3)}$$ Next, construct a line parallel to $BC$ and tangent to the incircle. Suppose it meets sides $AB$ and $AC$ at $B_1$ and $C_1$ respectively. Let the incircle touch $AB$ and $AC$ at $B_2$ and $C_2$ respectively. By homothety, the line $B_{1}I$ is parallel to the external angle bisector of $\angle ABC$, and hence $\angle B_{1}IB=90^{\circ }$. Since $\angle B{B}_{2}I=90^{\circ }$, we get $B{B}_2\cdot B{B}_1=B{I}^2$, and similarly $C{C}_2\cdot C{C}_1=C{I}^2$. Hence, $$\begin{array}{c}\frac{B{I}^2}{C{I}^2}=\frac{B{B}_2\cdot B{B}_1}{C{C}_2\cdot C{C}_1}=\frac{B{B}_1}{C{C}_1}\cdot \frac{BD}{CD}=\frac{AB}{AC}\cdot \frac{BD}{CD}.\end{array}\text{\hspace{1em}(4)}$$ Combining (1), (2), (3) and (4), we conclude $$\frac{B{D}^{\prime }}{C{D}^{\prime }}=\frac{XB}{XC}\cdot \frac{AC}{AB}=\frac{BF}{CE}\cdot \frac{AC}{AB}=\frac{B{I}^2}{C{I}^2}\cdot \frac{AC}{AB}=\frac{BD}{CD}$$ so that $D^{\prime }=D$. The result then follows.

- Method 2. We continue the proof of Solution 1 using trigonometry. Let $\beta =\frac{1}{2}\angle ABC$ and $\gamma =\frac{1}{2}\angle ACB$. Observe that $\angle FIB=\angle AIB-90^{\circ }=\gamma$. Hence, $\frac{BF}{FI}=\frac{\sin \angle FIB}{\sin \angle IBF}=\frac{\sin \gamma }{\sin \beta }$. Similarly, $\frac{CE}{EI}=\frac{\sin \beta }{\sin \gamma }$. As $FI=EI$, we get $$\frac{B{D}^{\prime }}{C{D}^{\prime }}=\frac{AC}{AB}\cdot {\left(\frac{\sin \gamma }{\sin \beta }\right)}^2=\frac{\sin 2\beta }{\sin 2\gamma }\cdot {\left(\frac{\sin \gamma }{\sin \beta }\right)}^2=\frac{\tan \gamma }{\tan \beta }=\frac{ID/CD}{ID/BD}=\frac{BD}{CD}.$$ This shows $D^{\prime }=D$ and the result follows.

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Alternative solution.

Let ${\omega }_A$ be the $A$-mixtilinear incircle of triangle $ABC$. From the properties of mixtilinear incircles, ${\omega }_A$ touches sides $AB$ and $AC$ at $F$ and $E$ respectively. Suppose ${\omega }_A$ is tangent to $\Gamma$ at $T$. Let $AM$ meet $\Gamma$ again at $Y$, and let $D_1,T_1$ be the reflections of $D$ and $T$ with respect to the perpendicular bisector of $BC$ respectively. It is well-known that $\angle BAT=\angle D_{1}AC$ so that $A,D_1,T_1$ are collinear. We then show that $X,M,T_1$ are collinear. Let $R$ be the radical centre of ${\omega }_A,\Gamma$ and the circumcircle of triangle $AEF$. Then $R$ lies on $AX,EF$ and the tangent at $T$ to $\Gamma$. Let $AT$ meet ${\omega }_A$ again at $S$ and meet $EF$ at $P$. Obviously, $SFTE$ is a harmonic quadrilateral. Projecting from $T$, the pencil ($R,P;F,E$) is harmonic. We further project the pencil onto $\Gamma$ from $A$, so that $XBTC$ is a harmonic quadrilateral. As $T{T}_1\parallel BC$, the projection from $T_1$ onto $BC$ maps $T$ to a point at infinity, and hence maps $X$ to the midpoint of $BC$, which is $M$. This shows $X,M,T_1$ are collinear. We have two ways to finish the proof.

- Method 1. Note that both $AY$ and $X{T}_1$ are chords of $\Gamma$ passing through the midpoint $M$ of the chord $BC$. By the Butterfly Theorem, $XY$ and $A{T}_1$ cut $BC$ at a pair of symmetric points with respect to $M$, and hence $X,D,Y$ are collinear. The proof is thus complete.

- Method 2. Here, we finish the proof without using the Butterfly Theorem. As $DT{T}_1{D}_1$ is an isosceles trapezoid, we have $$\measuredangle YTD=\measuredangle YT{T}_1+\measuredangle T_{1}TD=\measuredangle YA{T}_1+\measuredangle A{D}_{1}D=\measuredangle YMD$$ so that $D,T,Y,M$ are concyclic. As $X,M,T_1$ are collinear, we have $$\measuredangle AYD=\measuredangle MTD=\measuredangle D_1{T}_{1}M=\measuredangle A{T}_{1}X=\measuredangle AYX$$ This shows $X,D,Y$ are collinear.