HMMT-Nov 2023 gen, Problem 10

Let $A B C D$ be a convex trapezoid such that $\angle A B C=\angle B C D=90^{\circ}, A B=3, B C=6$, and $C D=12$. Among all points $X$ inside the trapezoid satisfying $\angle X B C=\angle X D A$, compute the minimum possible value of $C X$.

Solution: Solution: Let $P=A D \cap B C$. Then, we see that the locus if $X$ is the $\operatorname{arc} \widehat{B D}$ of $\odot(P B D)$. Let $O$ be its center, and $R$ be the radius of $\odot(P B D)$, then the answer is $C O-R$. Let $T$ be the second intersection of $\odot(P B D)$ and $C D$. We can compute $B P=2$, so by Power of Point, $C T \cdot C D=C P \cdot C B=48$, so $C T=4$. Since the projections of $O$ onto $C D$ and $C B$ are midpoints of $B P$ and $D T$, we get by Pythagorean theorem that $C O=\sqrt{8^{2}+7^{2}}=\sqrt{113}$. By Pythagorean theorem, we also get that $R=\sqrt{8^{2}+1^{2}}=\sqrt{65}$, hence the answer. $\fbox{\sqrt{113}-\sqrt{65}.}$.

Answer: \sqrt{113}-\sqrt{65}.