HMMT-Feb 2021 geo, Problem 7

Let $O$ and $A$ be two points in the plane with $O A=30$, and let $\Gamma$ be a circle with center $O$ and radius $r$. Suppose that there exist two points $B$ and $C$ on $\Gamma$ with $\angle A B C=90^{\circ}$ and $A B=B C$. Compute the minimum possible value of $\lfloor r\rfloor$.

Solution: Solution: Let $f_{1}$ denote a $45^{\circ}$ counterclockwise rotation about point $A$ followed by a dilation centered $A$ with scale factor $1 / \sqrt{2}$. Similarly, let $f_{2}$ denote a $45^{\circ}$ clockwise rotation about point $A$ followed by a dilation centered $A$ with scale factor $1 / \sqrt{2}$. For any point $B$ in the plane, there exists a point $C$ on $\Gamma$ such that $\angle A B C=90^{\circ}$ and $A B=B C$ if and only if $B$ lies on $f_{1}(\Gamma)$ or $f_{2}(\Gamma)$. Thus, such points $B$ and $C$ on $\Gamma$ exist if and only if $\Gamma$ intersects $f_{1}(\Gamma)$ or $f_{2}(\Gamma)$. So, the minimum possible value of $r$ occurs when $\Gamma$ is tangent to $f_{1}(\Gamma)$ and $f_{2}(\Gamma)$. This happens when $r / \sqrt{2}+r=30 / \sqrt{2}$, i.e., when $r=\frac{30}{\sqrt{2}+1}=30 \sqrt{2}-30$. Therefore, the minimum possible value of $\lfloor r\rfloor$ is $\lfloor 30 \sqrt{2}-30\rfloor=12$. $\fbox{12}$.

Answer: 12