2020 AIME II Problems/Problem 7

Revision as of 03:30, 8 June 2020 by Mn28407 (talk | contribs) (Solution Added)

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, $$ (Error compiling LaTeX. ! Missing $ inserted.)r^2=$\frac{15^2}{(\sqrt{8^2+3^2})^2}=\frac{225}{73}$$This implies our answer is$225+73=\boxed{298}$. ~mn28407

See Also

2020 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See Also

2020 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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