Difference between revisions of "2020 AIME II Problems/Problem 7"
m (→Problem) |
|||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Two congruent right circular cones each with base radius <math>3</math> and height <math>8</math> have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance <math>3</math> from the base of each cone. A sphere with radius <math>r</math> lies withing both cones. The maximum possible value of <math>r^2</math> is <math>\frac{m}{n}</math>, where <math>m</math> | + | Two congruent right circular cones each with base radius <math>3</math> and height <math>8</math> have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance <math>3</math> from the base of each cone. A sphere with radius <math>r</math> lies withing both cones. The maximum possible value of <math>r^2</math> is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
==Solution== | ==Solution== |
Revision as of 16:39, 8 June 2020
Contents
Problem
Two congruent right circular cones each with base radius and height
have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance
from the base of each cone. A sphere with radius
lies withing both cones. The maximum possible value of
is
, where
and
are relatively prime positive integers. Find
.
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, , and the bases form the positive
and
axes. Then label the vertices of the region enclosed by the two triangles as
in a clockwise manner. We want to find the radius of the inscribed circle of
. By symmetry, the center of this circle must be
.
can be represented as
Using the point-line distance formula,
~mn28407
Video Solution
https://youtu.be/bz5N-jI2e0U?t=44
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
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.