The Smiths, the Andrings and the Cliffords all hold a big party. Everyone shakes hands with every member of the other two families (no one shakes hands with members of their own family), 142 handshakes in all.

Assuming that there at least as many Andrings as Smiths, and at least as many Cliffords as Andrings, how many of each family are present?

(In reply to re: 1/2 Surface of a Rectangular Solid... by Larry)

Larry wrote: So if shaking hands is equivalent to finding surface area, a triple handshake (ie one member of each family) is equivalent to finding volume. 

Not quite.  The handshaking problem is as area problem with the number of families, N, denoting the number of dimensions of the rectangular object.

Between 1 family, the result is always 0 (we're talking area, remember?)

Between 2 families it's a simple 2D surface: number in Family A x number in Family B.

Between 3 families it's 1/2 the total surface of a 3D rectangular object: AxB + BxC + CxA.

Between 4 families it becomes 1/4 the total surface of a hyper-rectangle: AB+AC+AD+BC+BD+CD

Posted by Erik O. on 2004-06-14 13:26:09