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

re: Nice problem and nice solutions. by Steve Herman)

If a decimal point was missing, than it is there now.

As to 41^2 = 40^2 + 9^2

....if k=1 then the sides are 40 and 41, the area = .5*40*41 *sin C.

The maximum is achieved at C=90deg, so ABC is a right-angled triangle.

Makes sense.