International Mathematical Olympiad

Denote by $\omega$ and ${\omega }^{\prime }$ the incircles of $△ABC$ and $△A^{\prime }{B}^{\prime }{C}^{\prime }$ and let $I$ and $I^{\prime }$ be the centres of these circles. Let $N$ and $N^{\prime }$ be the second intersections of $BI$ and $B^{\prime }{I}^{\prime }$ with $\Omega$, the circumcircle of $A^{\prime }BC{C}^{\prime }{B}^{\prime }A$, and let $O$ be the centre of $\Omega$. Note that $ON\perp AC$, $O{N}^{\prime }\perp A^{\prime }{C}^{\prime }$ and $ON=O{N}^{\prime }$ so $N{N}^{\prime }$ is parallel to the angle bisector $I{I}^{\prime }$ of $AC$ and $A^{\prime }{C}^{\prime }$. Thus $I{I}^{\prime }\Vert N{N}^{\prime }$ which is antiparallel to $B{B}^{\prime }$ with respect to $BI$ and $B^{\prime }{I}^{\prime }$. Therefore $B,I,I^{\prime },B^{\prime }$ are concyclic.

Further define $P$ as the intersection of $AC$ and $A^{\prime }{C}^{\prime }$ and $M$ as the antipode of $N^{\prime }$ in $\Omega$. Consider the circle ${\Gamma }_1$ with centre $N$ and radius $NA=NC$ and the circle ${\Gamma }_2$ with centre $M$ and radius $M{A}^{\prime }=M{C}^{\prime }$. Their radical axis passes through $P$ and is perpendicular to $MN\perp N{N}^{\prime }\Vert IP$, so $I$ lies on their radical axis. Therefore, since $I$ lies on ${\Gamma }_1$, it must also lie on ${\Gamma }_2$. Thus, if we define $Z$ as the second intersection of $MI$ with $\Omega$, we have that $I$ is the incentre of triangle $Z{A}^{\prime }{C}^{\prime }$. (Note that the point $Z$ can also be constructed directly via Poncelet's porism.)

Consider the incircle ${\omega }_c$ with centre $I_c$ of triangle $C^{\prime }{B}^{\prime }Z$. Note that $\angle ZI{C}^{\prime }=90^{\circ }+\frac{1}{2}\angle Z{A}^{\prime }{C}^{\prime }=90^{\circ }+\frac{1}{2}\angle Z{B}^{\prime }{C}^{\prime }=\angle Z{I}_c{C}^{\prime }$, so $Z,I,I_c,C^{\prime }$ are concyclic. Similarly $B^{\prime },I^{\prime },I_c,C^{\prime }$ are concyclic.

The external centre of dilation from $\omega$ to ${\omega }_c$ is the intersection of $I_c$ and $C^{\prime }Z$ ( $D$ in the picture), that is the radical centre of circles $\Omega ,C^{\prime }{I}_{c}IZ$ and $I{I}^{\prime }{I}_c$. Similarly, the external centre of dilation from ${\omega }^{\prime }$ to ${\omega }_c$ is the intersection of $I^{\prime }{I}_c$ and $B^{\prime }{C}^{\prime }$ ( $D^{\prime }$ in the picture), that is the radical centre of circles $\Omega ,B^{\prime }{I}^{\prime }{I}_c{C}^{\prime }$ and $I{I}^{\prime }{I}_c$. Therefore the Monge line of $\omega ,{\omega }^{\prime }$ and ${\omega }_c$ is line $D{D}^{\prime }$, and the radical axis of $\Omega$ and circle $I{I}^{\prime }{I}_c$ coincide. Hence the external centre $T$ of dilation from $\omega$ to ${\omega }^{\prime }$ is also on the radical axis of $\Omega$ and circle $I{I}^{\prime }{I}_c$.

Now since $B,I,I^{\prime },B^{\prime }$ are concyclic, the intersection $T^{\prime }$ of $B{B}^{\prime }$ and $I{I}^{\prime }$ is on the radical axis of $\Omega$ and circle $I{I}^{\prime }{I}_c$. Thus $T^{\prime }=T$ and $T$ lies on line $B{B}^{\prime }$. Finally, construct a circle ${\Omega }_0$ tangent to $A^{\prime }{B}^{\prime },B^{\prime }{C}^{\prime },AB$ on the same side of these lines as ${\omega }^{\prime }$. The centre of dilation from ${\omega }^{\prime }$ to ${\Omega }_0$ is $B^{\prime }$, so by Monge's theorem the external centre of dilation from ${\Omega }_0$ to $\omega$ must be on the line $TB{B}^{\prime }$. However, it is on line $AB$, so it must be $B$ and $BC$ must be tangent to ${\Omega }_0$ as desired.