HMMT-Nov 2017 team, Problem 9

Let $A, B, C, D$ be points chosen on a circle, in that order. Line $B D$ is reflected over lines $A B$ and $D A$ to obtain lines $\ell_{1}$ and $\ell_{2}$ respectively. If lines $\ell_{1}, \ell_{2}$, and $A C$ meet at a common point and if $A B=4, B C=3, C D=2$, compute the length $D A$.

Solution: Let the common point be $E$. Then since lines $B E$ and $B D$ are symmetric about line $B A, B A$ is an exterior bisector of $\angle D B E$, and similarly $D A$ is also an exterior bisector of $\angle B D E$. Therefore $A$ is the $E$-excenter of triangle $B D E$ and thus lie on the interior bisector of $\angle B E D$. Since $C$ lies on line $A E$, and by the fact that $A, B, C, D$ are concyclic, we get that $\angle A B C+\angle A D C=180^{\circ}$, which implies $\angle D B C+\angle B D C=\frac{1}{2}(\angle D B E+\angle B D E)$, so $C$ is the incenter of triangle $B D E$. Thus $\angle A B C=\angle C D A=\frac{\pi}{2}$, and thus $D A^{2}=A C^{2}-C D^{2}=A B^{2}+B C^{2}-C D^{2}=3^{2}+4^{2}-2^{2}=21$. The length of $D A$ is then $\sqrt{21}$. $\fbox{\sqrt{21}}$.

Answer: \sqrt{21}