HMMT-Feb 2023 guts, Problem 28

Suppose $A B C D$ is a convex quadrilateral with $\angle A B D=105^{\circ}, \angle A D B=15^{\circ}, A C=7$, and $B C=C D=5$. Compute the sum of all possible values of $B D$.

Solution: Solution: Let $O$ be the cirumcenter of triangle $A B D$. By the inscribed angle theorem, $\angle A O C=90^{\circ}$ and $\angle B O C=60^{\circ}$. Let $A O=B O=C O=x$ and $C O=y$. By the Pythagorean theorem on triangle $A O C$, \[ x^{2}+y^{2}=49 \] and by the Law of Cosines on triangle $B O C$, \[ x^{2}-x y+y^{2}=25 \] It suffices to find the sum of all possible values of $B D=\sqrt{3} x$. Since the two conditions on $x$ and $y$ are both symmetric, the answer is equal to \[ \sqrt{3}(x+y)=\sqrt{9\left(x^{2}+y^{2}\right)-6\left(x^{2}-x y+y^{2}\right)}=\sqrt{291} \] It is easy to check that both solutions generate valid configurations. $\fbox{\sqrt{291}}$.

Answer: \sqrt{291}