There are two spherical balls, each of radius 100 cm., lying on a perfectly horizontal floor and touching each other.
What is the diameter of the largest ball that can pass through the gap between the spheres and the floor ?
Consider the vertical plane containing the centers of the two spheres, and the center of the small sphere, at its tightest point in the passage.
The spheres will have circular cross sections. Construct a square with a corner at the center of one of the larger spheres and another where the two larger spheres touch. It will also have corners on the floor, where the smaller circle touches it, and also directly under the center chosen for the first corner.
This square then includes 1/4 of one of the larger circles and 1/2 of the smaller circle.
Draw a line from the corner that's at the center of a large circle, to the center of the smaller circle. This is on an edge of the square. The top edge of the square is 100 cm., and the center of the smaller circle is r cm above the floor, where r is the radius of the smaller circle. This leaves 100-r as the portion of the side of the square that also forms part of a newly formed right triangle whose hypotenuse we just drew.
The length of this hypotenuse is just the total of the two radii, 100+r. Using the Pythagorean Theorem, 100^2+(100-r)^2 = (100+r)^2. Then
2(100^2) - 200r + r^2 = 100^2 + 200r + r^2
400r = 100^2
r = 25
so the diameter is 50 cm.
Posted by Charlie
on 2003-09-24 09:55:32