Really? Nobody? I guess everybody is on vacation. Ok, then.
The n outer circles have their centers at a distance of R + r from the center circle. Pick an outer circle. Form a triangle by joining the center C of the selected outer circle with the center of the inner circle, and with one of its points of tangency with an outer circle. This is a right triangle, with the angle at C = (2pi/n)/2, opposite side r and hypotenuse (R+r).
So, sin(pi/n) = r/(R+r)
Inverting,
1/sin(pi/n) = (R+r)/r = R/r + 1
R/r = -1 + 1/sin(pi/n)
r/R = 1/(-1 + 1/sin(pi/n)) = sin(pi/n)/(1-sin(pi/n))
Final answer.
Hope I haven't made a mistake.
Checking. Let n = 6. Then r/R = sin(30 degrees)/(1-sin(30 degrees)) = 1. That checks out.
Not on vacation (spoiler)
