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?
(In reply to
Solution (I'd like confirmation) by Jer)
I used the same triangle setup as you did, but I used Pythagorean theorem to solve: once for triangle ADC and then once for triangle BDC.
Let AD=x then the two equations are x^2 + 16/3 = (r+2)^2 and (x+r+6)^2 + 16/3 = 64.
Solve the first equation for x and substitute into the second to get:
(sqrt[r^2+4r-4/3]+r+6)^2 + 16/3 = 64.
Expand the the squaring and simplify to get:
(r+6) * sqrt[r^2+4r-4/3] = -r^2 - 8r + 12
Squaring each side and simplifying again gets the same quadratic 2r^2+15r-9=0.
re: Solution (I'd like confirmation)
