Howell, Leonard W. (MSFC-VP62) wrote: > I didn't have much luck solving this one on my own but found the > solution at mathworld. > > Can you explain how it was solved, hopefully with some J thinking? >
Here's my solution, with no J thinking. Spoiler below.... This method gives the result: maybe it does not show it is actually the largest square. For some 0<x<1, take vertices (x,0,0) (0,x,0) (1,1-x,1) (1-x,1,1) Calculating the square of the side in two ways gives 2x^2=2(1-x)^2+1, so x=3/4, and the side of the square is 3%2*%:2=1.06066. Best wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
