IMO 2019 Shortlisted Problems

Denote the centres of ${\omega }_B$ and ${\omega }_C$ by $O_B$ and $O_C$, let their radii be $r_B$ and $r_C$, and let $BC$ be tangent to the two circles at $T$ and $U$, respectively.



From the cyclic quadrilaterals $AFDC$ and $ABDE$ we have $$\angle MD{O}_B=\frac{1}{2}\angle FDB=\frac{1}{2}\angle BAC=\frac{1}{2}\angle CDE=\angle O_{C}DN,$$ so the right-angled triangles $DM{O}_B$ and $DN{O}_C$ are similar. The ratio of similarity between the two triangles is $$\frac{DN}{DM}=\frac{O_{C}N}{O_{B}M}=\frac{r_C}{r_B}.$$ Let $\phi =\angle DMN$ and $\psi =\angle MND$. The lines $FM$ and $EN$ are tangent to ${\omega }_B$ and ${\omega }_C$, respectively, so $$\angle MTP=\angle FMP=\angle DMN=\phi \text{and}\angle QUN=\angle QNE=\angle MND=\psi .$$ (It is possible that $P$ or $Q$ coincides with $T$ or $U$, or lie inside triangles $DMT$ or $DUN$, respectively. To reduce case-sensitivity, we may use directed angles or simply ignore angles $MTP$ and $QUN$.)

In the circles ${\omega }_B$ and ${\omega }_C$ the lengths of chords $MP$ and $NQ$ are $$MP=2r_B\cdot \sin \angle MTP=2r_B\cdot \sin \phi \text{and}NQ=2r_C\cdot \sin \angle QUN=2r_C\cdot \sin \psi .$$ By applying the sine rule to triangle $DNM$ we get $$\frac{DN}{DM}=\frac{\sin \angle DMN}{\sin \angle MND}=\frac{\sin \phi }{\sin \psi }.$$ Finally, putting the above observations together, we get $$\frac{MP}{NQ}=\frac{2r_B\sin \phi }{2r_C\sin \psi }=\frac{r_B}{r_C}\cdot \frac{\sin \phi }{\sin \psi }=\frac{DM}{DN}\cdot \frac{\sin \phi }{\sin \psi }=\frac{\sin \psi }{\sin \phi }\cdot \frac{\sin \phi }{\sin \psi }=1,$$ so $MP=NQ$ as required.