2020 AIME II Problems/Problem 7

Problem

Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies withing both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$, where $m$n and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Take the cross-section of the plane of symmetry formed by the two cones. Let the point where the bases intersect be the origin, $O$, and the bases form the positive $x$ and $y$ axes. Then label the vertices of the region enclosed by the two triangles as $O,A,B,C$ in a clockwise manner. We want to find the radius of the inscribed circle of $OABC$. By symmetry, the center of this circle must be $(3,3)$. $\overline{OA}$ can be represented as $8x-3y=0$ Using the point-line distance formula, $$r^2=\frac{15^2}{(\sqrt{8^2+3^2})^2}=\frac{225}{73}$$ This implies our answer is $225+73=\boxed{298}$. ~mn28407