Three balls of radius 2 are all tangent to one another as well as to a ball of radius 6. A fifth, smaller ball is tangent to all four balls. What is its radius?
Call the 5 balls: 3 x b2, b6, and bs (bs, for ball: small).
The centers of the 3 b2's make a horizontal equilateral triangle which will be the base of two different tetrahedrons, one inscribed within the other. Each tetrahedron has three identical isosceles triangle sides, with the second tetrahedron situated within the first, symmetrically. The first (outer) tetrahedron has as its top vertex the center of b6 and as its upward rising edges (of length 8) the combined radii of one b2 ball and the b6 ball, with the edge passing through the b2-b6 tangent point.
Similarly, the inner tetrahedron has as its top vertex the center of bs, and edges of length of 2+ radii(bs).
The trick (I think) is to locate the point on the vertical line between the two vertices that is the tangent point of bs with b6 (which is also the lowest point of b6 and highest point of bs).
I am also thinking that using an iterative numerical method might be the easiest route...
Edited on August 26, 2019, 11:22 am
just a start
