Three circles of radius 6, 7, and 8 are externally tangent to each other. There exists a smaller circle tangent to all three (in the space created between the three original circles).

What is the radius of this smallest circle?

Using the center of the small circle as a vertex for three triangles with the bases formed by each of the center-to-center lines connecting the three given circles, the sides of, say the one with base 13, are 6+r, 7+r, 13, where r is the radius of the new small circle.  The others work the same way.

Then the angle at the center of the circle that's part of the given triangle is arccos(((6+r)^2 + (7+r)^2 - 13^2)/(2(6+r)(7+r))).  Similarly for the other two formed triangles.  The three angles at the center must add to 2 pi radians.

The following spreadsheet, in which the Solver function has been used to set the SUM in column D (there's no column C) equal to a 10-digit approximation of 2 pi by changing cell B2, which represents the radius of the small circle, shows that to be 1.070063694. (I wouldn't trust the places after that, as 2 pi was not entered to more than 10 places--and the solver doesn't allow setting a different cell to be the goal of a given cell.)

R  1.07006369440565    =ACOS(((6+B$2)^2+(7+B$2)^2-13^2)/(2*(6+B$2)*(7+B$2)))
                       =ACOS(((7+B$2)^2+(8+B$2)^2-15^2)/(2*(7+B$2)*(8+B$2)))
                       =ACOS(((6+B$2)^2+(8+B$2)^2-14^2)/(2*(6+B$2)*(8+B$2)))

                       =SUM(D2:D4)                                             =2*PI()

Posted by Charlie on 2004-07-28 15:47:59