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 The Five Touching Spheres 2 (Posted on 2019-09-15)
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?

 See The Solution Submitted by Brian Smith No Rating

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 iterative soln | Comment 1 of 4
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)

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
 Posted by Steven Lord on 2019-09-16 02:06:06

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