In triangle ABC, it is known that AB = 9 and BC/AC = 40/41.
Find the maximum possible area of the triangle ABC.
(In reply to Nice problem and nice solutions.
by Ady TZIDON)
Actually, Ady, 820 = 40*41/2, so the 5 is really incidental.
I am struck that 41^2 = 40^2 + 9^2. Is this why things simplified so nicely?