2020 AIME I Problems/Problem 6
A flat board has a circular hole with radius and a circular hole with radius such that the distance between the centers of the two holes is 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 where and are relatively prime positive integers. Find
Solution 1 (Pythagorean Theorem)
Set the common radius to . First, take the cross section of the sphere sitting in the hole of radius . 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 and base . Therefore, the height of this circle outside of the hole is .
The other circle follows similarly for a height (outside the hole) of . Now, if we take the cross section of the entire board, essentially making it two-dimensional, we can connect the centers of the two spheres, then form another right triangle with base , 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 . Now we can set up an equation in terms of with the Pythagorean theorem: Simplifying a few times, Therefore, our answer is .
Solution 2 (Tangential Distance)
Let and be the centers of the holes, let be the point of crossing and radical axes of the circles. So has equal tangential distance to any point of both spheres. In particular to the circles (https://en.wikipedia.org/wiki/Radical_axis.)
Let be the point of tangency of the spheres common radius centered at and Let be the angle between and flat board. In the plane, perpendicular to board
Distance between and midpoint of is email@example.com, vvsss
Video solution (With Animation)
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