2020 AIME II Problems/Problem 7
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 within both cones. The maximum possible value of is , where and are relatively prime positive integers. Find .
Solution (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 isThe requested sum is .
Not part of MAA's solution, but this: https://www.geogebra.org/calculator/xv4nm97a is a good visual of the cones in GeoGebra.
Solution (Clean analytic geometry)
Using the diagram above, we notice that the desired length is simply the distance between the point and . We can mark as since it is units away from each of the bases. Point is . Thus, line is . We can use the distance from point to line formula , where and are the coordinates of the point, and A, B, and C are the coefficients of the line in form . Plugging everything in, we get .
Solution 1: Graph paper coordbash
We graph this on graph paper, with the scale of . So, we can find then divide by to convert to our desired units, then square the result. With 5 minutes' worth of coordbashing, we finally arrive at .
Video Solution 2
https://www.youtube.com/watch?v=0XJddG43pIk ~ MathEx
Video Solution 3
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