Circle problem

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A flat board has a circular hole with radius 1 and a circular hole with radius 2 such that the distance between the centers of the two holes is 7 Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find m+n

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a156f24285b03d26c18745c310c33b11767325d5

Set the common radius to b55ca7a0aa88ab7d58f4fc035317fdac39b17861. First, take the cross section of the sphere sitting in the hole of radius 1. If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenuse b55ca7a0aa88ab7d58f4fc035317fdac39b17861 and base dce34f4dfb2406144304ad0d6106c5382ddd1446. Therefore, the height of this circle outside of the hole is 4b162b52a624076f304ee4a250639394cda413a8.

The other circle follows similarly for a height (outside the hole) of 85aaa9f6788ce44bec07b0a8d5e87dec1c71f9a1. Now, if we take the cross section of the entire board, essentially making it 2-D, we can connect the centers of the two spheres, then form another right triangle with base e0a0db32027a732ac57d37ef2ae9bb150f65b108, as given by the problem. The height of this triangle is the difference between the heights of the parts of the two spheres outside the holes, which is 52c41bc5747b9f89f6614fa0e61c46b015fcba43. Now we can set up an equation in terms of b55ca7a0aa88ab7d58f4fc035317fdac39b17861 with the Pythagorean theorem:

cbd1a428780203111cbd1b284d308f73f78fdad9

Simplifying a few times,

243f6f0f03b8d804532e636ab2a4bd0ad282e84e

ec578873f2f75ba83a8745a7948aeb7035976398

2e4094ce8a7b36802c3abd62a846026e4e5a5756

94f541428b127ae0c127fb0b550f69a709891a70

0ccb966bb82a5cbd6a5efd0bcdede596d4f4f27d

c663410726dc7c944f1964a8d34788dabacb75cb

Therefore, our answer is 952cf60fbe678f4ff0dc12dae6e235b9e33ca121.

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