Date: Fri, 22 Oct 2000 13:14:10 -0300 (GMT)

From: Ing. Isi Haim

Alexander, hello:

With all my excuses, I want to remark that your argument about "re:trigonometry inequality" in your answer to "François" of 10 Nov 1997 is not correct: When you say that p/R attains its minimum for the equilatral triangle, you are making a mistake: for the equilateral triangle, p/R = 3sqroot(3)/2, while for the rectangular isosceles tiangle, p/R = sqroot(2)+1. In the first case, p/R is greater than in the second case.

So, to prove the trigonometry inequality, another way must be taken.

Best regards.

Isi HAIM