HMMT-Feb 2022 guts, Problem 19

In right triangle $A B C$, a point $D$ is on hypotenuse $A C$ such that $B D \perp A C$. Let $\omega$ be a circle with center $O$, passing through $C$ and $D$ and tangent to line $A B$ at a point other than $B$. Point $X$ is chosen on $B C$ such that $A X \perp B O$. If $A B=2$ and $B C=5$, then $B X$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$. Compute $100 a+b$.

Solution: Solution: Note that since $A D \cdot A C=A B^{2}$, we have the tangency point of $\omega$ and $A B$ is $B^{\prime}$, the reflection of $B$ across $A$. Let $Y$ be the second intersection of $\omega$ and $B C$. Note that by power of point, we have $B Y \cdot B C=B B^{2}=4 A B^{2} \Longrightarrow B Y=\frac{4 A B^{2}}{B C}$. Note that $A X$ is the radical axis of $\omega$ and the degenerate circle at $B$, so we have $X B^{2}=X Y \cdot X C$, so \[ B X^{2}=(B C-B X)(B Y-B X)=B X^{2}-B X(B C+B Y)+B C \cdot B Y \] This gives us \[ B X=\frac{B C \cdot B Y}{B C+B Y}=\frac{4 A B^{2} \cdot B C}{4 A B^{2}+B C^{2}}=\frac{80}{41} \] $\fbox{8041}$.

Answer: 8041