Two balls of radius 2 and two balls of radius 6 are all externally tangent to each other. A fifth, smaller ball is tangent to all four balls. What is its radius?

I get

**r = 0.727925** +/- 0.000005

Call the 4 balls: A6, B6, A2, B2, and the 5th: C, with corresponding labeled centers. We place A6 and B6 on either side of the origin on the x-axis at (-6,0,0) and (6,0,0), and we place A2 at (0, sqrt(28), 0) so that the three centers make a 12, 8, 8 triangle in the x-y plane of with a y-height=sqrt(28). We place B2 atop these three. By symmetry, the x coordinate of B2 is 0.

Being lazy, I avoided further trig and iterated by hand B2 = (0,y,z) to get the center of B2 that yielded distances of (6+2), (6+2) and (2+2) to the other 3 centers.

I found:

B2 = (0, 3.77965, 3.7033)

which gave lengths:

8.0000115501235669

8.0000115501235669

4.0000161646277368

Now, ball C in the middle is also at x=0 by symmetry, and so once again i skipped trig class and iterated C = (0, *, *) to get:

C = (0, 2.81805, 1.15048)

Subtracting the known radii gave radius estimates for C:

CA6 = CB6 = 6 + 0.72792761798906369

CA2 = 2 + 0.72792450487870397

CB2 = 2 + 0.72792310309804797

And finally, checking by subtracting the average C_radius from the distance between C's center and the other centers, I found all is OK to 3 sig figs:

CA6-C_r CB6-C_r CA2-C_r CB2-C_r

6.000 6.000 2.000 2.000

*Edited on ***September 16, 2019, 8:38 pm**