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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
Copyright © 1996-2008 Alexander Bogomolny
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