Three spheres of radii a, b, and c are tangent to the same plane at points A, B, and C respectively. Each sphere is externally tangent to the other two.
If a < b < c, then which internal angle of triangle ABC is largest and what is its value in terms of a, b, and c?

Take the vertical plane through points A and B, which therefore goes through the centers of those points' respective spheres.  Make the trapezoid with the centers of those spheres and the points A and B. The angles at A and B are right angles, so the trapezoid consists of a right triangle with a horizontal leg we'll call length d and a vertical leg of length b - a, on top of a rectangle of length d and height a.

We wish to find d, and this can be done from the triangle, as the hypotenuse of that triangle is a + b, so

d^2 + (b-a)^2 = (a+b)^2

therefore

d^2 = a^2 + b^2 + 2ab - b^2 - a^2 + 2ab

d^2 = 4ab

d = 2*sqrt(ab)

This can be done for each of the pairs of spheres.  The largest will use the largest two radii: b and c, and the distance, between B and C, will be 2*sqrt(bc).  The largest angle will be opposite this largest side and can be found with the law of cosines:

4bc = 4ac + 4ab - 8a sqrt(bc) cos A

cos A = (ac + ab - bc) / ( 2 a sqrt(bc))

A = arccos ((ac + ab - bc) / ( 2 a sqrt(bc)))

 

Posted by Charlie on 2005-09-14 18:55:04