2020 AIME II Problems/Problem 7
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, ![\[r^2=\frac{15^2}{(\sqrt{8^2+3^2})^2}=\frac{225}{73}\implies225+73=\boxed{298}\]](http://latex.artofproblemsolving.com/d/2/2/d22afd3408a0b21a08bf2cac0c4902ba9be8f749.png)
~mn28407
Solution 2 (Official MAA)
Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the intersection of their axes of symmetry. Let 
be the point in the cross section where the bases of the cones meet, and let 
be the center of the sphere. Let the axis of symmetry of one of the cones extend from its vertex, 
, to the center of its base, 
. Let the sphere be tangent to 
at 
. The right triangles 
and 
are similar, implying that the radius of the sphere is![\[CE = AD \cdot\frac{BC}{AB} = AD \cdot\frac{BD-CD}{AB} =3\cdot\frac5{\sqrt{8^2+3^2}} = \frac{15}{\sqrt{73}}=\sqrt{\frac{225}{73}}.\]](http://latex.artofproblemsolving.com/8/3/6/8361df2c618af0b5ace94849fd5ccfb7eecbc0d0.png)
The requested sum is 
.
![[asy] unitsize(0.6cm); pair A = (0,0); pair TriangleOneLeft = (-6,0); pair TriangleOneDown = (-3,-8); pair TriangleOneMid = (-3,0); pair D = (0,-3); pair TriangleTwoDown = (0,-6); pair B = (-8,-3); pair C = IP(TriangleOneMid -- TriangleOneDown, B--D); pair EE = foot(C, A, B); real radius = arclength(C--EE); path circ = Circle(C, radius); draw(A--B--TriangleTwoDown--cycle); draw(B--D); draw(A--TriangleOneLeft--TriangleOneDown--cycle); draw(circ); draw(C--EE); draw(TriangleOneMid -- TriangleOneDown, gray); dot("$B$", B, W); dot("$E$", EE, NW); dot("$A$", A, NE); dot("$D$", D, E); dot("$C$", C, SE); [/asy]](http://latex.artofproblemsolving.com/1/6/1/1614cf232770744f98f625145dfcb7dd2ef79acb.png)
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. 
