Difference between revisions of "2020 AIME II Problems/Problem 7"
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==Solution== | ==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, <math>O</math>, and the bases form the positive <math>x</math> and <math>y</math> axes. Then label the vertices of the region enclosed by the two triangles as <math>O,A,B,C</math> in a clockwise manner. We want to find the radius of the inscribed circle of <math>OABC</math>. By symmetry, the center of this circle must be <math>(3,3)</math>. <math>\overline{OA}</math> can be represented as <math>8x-3y=0</math> Using the point-line distance formula, <cmath>r^2=\frac{15^2}{(\sqrt{8^2+3^2})^2}=\frac{225}{73}\ | + | Take the cross-section of the plane of symmetry formed by the two cones. Let the point where the bases intersect be the origin, <math>O</math>, and the bases form the positive <math>x</math> and <math>y</math> axes. Then label the vertices of the region enclosed by the two triangles as <math>O,A,B,C</math> in a clockwise manner. We want to find the radius of the inscribed circle of <math>OABC</math>. By symmetry, the center of this circle must be <math>(3,3)</math>. <math>\overline{OA}</math> can be represented as <math>8x-3y=0</math> Using the point-line distance formula, <cmath>r^2=\frac{15^2}{(\sqrt{8^2+3^2})^2}=\frac{225}{73}\implies225+73=\boxed{298}</cmath>~mn28407 |
==Video Solution== | ==Video Solution== | ||
Revision as of 13:27, 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 
n 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, ![\[r^2=\frac{15^2}{(\sqrt{8^2+3^2})^2}=\frac{225}{73}\implies225+73=\boxed{298}\]](http://latex.artofproblemsolving.com/d/2/2/d22afd3408a0b21a08bf2cac0c4902ba9be8f749.png)
~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. 
