Three Spheres Tangent On A 6-8-10 Triangle

Thanks to Castor from Australia for the suggestion! He credits AoPS for the problem.

Three spheres are tangent to a plane at the vertices of a triangle and are tangent to each other. Find the radii of the spheres if the sides of the triangle are 6, 8, and 10.

As usual, watch the video for a solution.

Three Spheres Tangent On A 6-8-10 Triangle

Or alimony reading.

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Answer To Three Spheres Tangent On A 6-8-10 Triangle

(Pretty much all posts are transcribed quickly without I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

The hardest part of the problem for me was visualizing the spheres. Let’s place the triangle unappetizing on a surface and squint at two tangent spheres at a time. Suppose two tangent spheres have radii a and b. Then the loftiness between their centers is a b considering the spheres are tangent. We can then construct a right triangle with the “vertical” and “horizontal” distances between the two centers. The horizontal length is the respective side of the 6-8-10 triangle. The vertical length is the difference of the two radii, |a – b|. We can then use the right triangle theorem to get an equation. For the triangle side with length 10 with get this diagram:

This gives the equation:

10² (a – b)² = (a b)²

We can make similar diagrams for the other two sides of the triangle to get two increasingly equations.

8² (a – c)² = (a c)²

6² (b – c)² = (b c)²

We now have a system of three equations in three variables.

10² (a – b)² = (a b)²

8² (a – c)² = (a c)²

6² (b – c)² = (b c)²

Each equation simplifies similarly. We expand the binomials on both sides and the squared terms cancel and the product of the two variables can be grouped.

10² (a – b)² = (a b)²
100 a² b² – 2ab = a² b² 2ab
4ab = 100
ab = 25