John Randall wrote: > Basic idea: Over time, the numbers become closer. Since their sum is > constant and positive, they end up nonnegative. It suffices to show that > the sum of squares eventually decreases. It does on one step if x+y+z>0, > but I cannot yet do the rest.
I was on the right track, but did not quite have the right quantity. Spoiler follows. Suppose the values on the pentagon are (in order) p,q,r,s,t. Let S=(p-r)^2+(q-s)^2+(r-t)^2+(s-p)^2+(t-q)^2 (the sum of squares of differences 2 apart). Now replace p=p+r r=-r s=s+r to get a new sum S'. A calculation shows S'-S=2r(p+q+r+s+t). The parenthesized sum is positive by hypothesis, and we only do the indicated move if r is negative. Thus S is nonnegative, and decreases on each move. The process terminates. Best wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
