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?