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Three Circles (Posted on 2004-07-28) Difficulty: 4 of 5
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?

No Solution Yet Submitted by ThoughtProvoker    
Rating: 3.0000 (3 votes)

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Solution numerical answer | Comment 2 of 7 |

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
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