HMMT-Feb 2022 guts, Problem 16

Let $A B C$ be an acute triangle with $A$-excircle $\Gamma$. Let the line through $A$ perpendicular to $B C$ intersect $B C$ at $D$ and intersect $\Gamma$ at $E$ and $F$. Suppose that $A D=D E=E F$. If the maximum value of $\sin B$ can be expressed as $\frac{\sqrt{a}+\sqrt{b}}{c}$ for positive integers $a, b$, and $c$, compute the minimum possible value of $a+b+c$.

Solution: Solution: First note that we can assume $A B<A C$. Suppose $\Gamma$ is tangent to $B C$ at $T$. Let $A D=$ $D E=E F=x$. Then, by Power of a Point, we have $D T^{2}=D E \cdot D F=x \cdot 2 x=2 x^{2} \Longrightarrow D T=x \sqrt{2}$. Note that $C T=s-b$, and since the length of the tangent from $A$ to $\Gamma$ is $s$, we have $s^{2}=A E \cdot A F=6 x^{2}$, so $C T=x \sqrt{6}-b$. Since $B C=B D+D T+T C$, we have $B D=B C-x \sqrt{2}-(x \sqrt{6}-b)=a+b-x(\sqrt{2}+\sqrt{6})$. Since $a+b=2 s-c=2 x \sqrt{6}-c$, we have $B D=x(\sqrt{6}-\sqrt{2})-c$. Now, by Pythagorean Theorem, we have $c^{2}=A B^{2}=A D^{2}+B D^{2}=x^{2}+[x(\sqrt{6}-\sqrt{2})-c]^{2}$. Simplifying gives $x^{2}(9-4 \sqrt{3})=x c(2 \sqrt{6}-2 \sqrt{2})$. This yields \[ \frac{x}{c}=\frac{2 \sqrt{6}-2 \sqrt{2}}{9-4 \sqrt{3}}=\frac{6 \sqrt{2}+10 \sqrt{6}}{33}=\frac{\sqrt{72}+\sqrt{600}}{33} \] $\fbox{705}$.

Answer: 705