jim@alberta.UUCP (Jim Easton) (02/03/86)
*** Here is an old one - perhaps old enough to have become new. I considered making up a story of Mrs. Peabody and the goldsmith and how much he should charge for a ring but ended up deciding that a straight statement of the problem was perhaps better (easier). So here it is without embellishments; A ring is made by drilling a cylindrical hole through the center of a sphere. The hole is 1 cm. long. What is the volume of the ring? Jim Easton (..!alberta!jim)
ins_apmj@jhunix.UUCP (Patrick M Juola) (02/07/86)
[munch munch] Sometime I'm going to have to have someone show me how to rot13.... In article <794@alberta.UUCP> jim@alberta.UUCP writes: > >A ring is made by drilling a cylindrical hole through the center of a >sphere. The hole is 1 cm. long. What is the volume of the ring? > > Jim Easton (..!alberta!jim) Assumption number 1 -- the problem is solvable. If it isn't, I would have rude words for the author. This means that there is enough information to solve the puzzle. The facts that are needed are a) the radius of the sphere, and b) the diameter of the hole -- at this point, the problem becomes one that we've all seen in integral calculus. All we know, however, is the length of the hole, not the diameter. By Assumption 1, however, that information is sufficient. In other words, the diameter of the hole is irrelevant! In that case, let d -> 0, the radius of the sphere goes to 1/2 cm, and the volume of the ring goes to the volume of the sphere less zero, or 4/3pi*(r^3) or pi/6 cm^3. q.e.d. Anyone who feels mathematical might want to prove that this is the solution. I.e. prove assumption 1. I've done so, but don't want to deprive you of the fun of so doing. Patrick Juola Hopkins Math ...jhunix!ins_apmj