Let the radius and the center of the original circle be `R ` and `O_1 ` and of the new circle — `r ` and `O_2 . ` Also denote the distance `O_1 O_2 ` as `x .`

Please look at the picture (not drawn to a scale). The angle BAF is an angle of a regular pentagon, so it is equal to `108^@ . ` From the symmetry of the problem the points `A , ` `O_1 , ` `O_2 ` and the point of the circles' tangency lie on one line, and also the angle `O_1 A B ` is equal to `54^@ . ` There is also a right angle between the tangent line AB and the corresponding radius of the smaller circle.

Now we easily get two equations: `r = R - x ` and `r / ( R + x ) = sin ( 54^@ ) . ` From them we can eliminate `x ` and infer that

`r = R * ( 2 sin ( 54^@ ) ) / ( 1 + sin ( 54^@ ) ) ,` so the diameter is `20 * ( 2 sin ( 54^@ ) ) / ( 1 + sin ( 54^@ ) ) .`

There is a separate fact that `sin ( 54^@ ) = ( sqrt ( 5 ) + 1 ) / 4 , ` so we can rationalize the denominator and get the answer in the desired form. It is `sqrt ( 320 ) , ` so `m = 320 , n = 0 , ` and `m + n = 320 .`

**Further Reading**