Update (9/28/16): I got this riddle wrong, so read ahead with caution. See Laurent Lessard’s excellent solution to the general case to see where I went wrong. As Pilgrim notes in the comments, it’s possible to obtain a table with a radius of 32 inches.
Here’s this week’s “Riddler Classic” from fivethirtyeight:
You’re on a DIY kick and want to build a circular dining table which can be split in half so leaves can be added when entertaining guests. As luck would have it, on your last trip to the lumber yard, you came across the most pristine piece of exotic wood that would be perfect for the circular table top. Trouble is, the piece is rectangular. You are happy to have the leaves fashioned from one of the slightly-less-than-perfect pieces underneath it, but there’s still the issue of the main circle. You devise a plan: cut two congruent semicircles from the perfect 4-by-8-foot piece and reassemble them to form the circular top of your table. What is the radius of the largest possible circular table you can make?
First, let’s draw out the the pristine piece of exotic wood, measuring 48 inches high by 96 inches long. Using the bottom-left corner as the origin, we can see that our solution will be to have one semicircle with radius and center and to have the other semicircle also with radius and center at . We need to solve for . The largest two congruent semicircles will be the case where the two are tangent.
Conveniently, there’s an obvious right triangle here. Because both semicircles have the same radius, we know the hypotenuse is equal to . We know one leg is equal to , because that’s the height of the piece of wood. And for the second leg, we know this is equal to . With the Pythagorean Theorem, we can then solve for , finding that the largest round table that can be built with the two semicircles will have a radius of thirty inches.