# Maximum area of a rectangle inside a triangle

I recently came across a problem where it gave a triangle with integer side lengths, and it asked you to find the maximum area of a rectangle of a triangle.

I solved the problem correctly, but it took me about 5-6 min to do so. I looked at the solution they gave for it and it stated that the maximum area is always:

(1/2)*(Area of Circumscribed Triangle)

Could someone please prove this? The solution didn't provide one for this problem.

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Let the triangle be $ABC$. I think we may take for granted that an optimal rectangle must have all its corners on the side of the triangle, so two of them must lie on a single side, which we may take to be $AB$. Picture this with $AB$ horizontal and $C$ on top.

Now fold the triangle with one corner at $C$ and its opposite side being the top side of the rectangle across the rectangle, with $C$ ending up at a new position $C'$. Fold the other two small triangles, with corners at $A$ and $B$ respectively, similarly over the left and right edges of the rectangle. Note carefully that the sides of the folded top triangle align with the slanted sides of the other two folded triangles.

If $C'$ lies above $AB$, then the other two triangles will overlap, and all three will cover the rectangle, so the combined area of the small triangles is greater than the area of the rectangle.

If $C'$ lies below $AB$, the small triangles will cover the rectangle with a bit sticking outside, so again, theire combined area is greater than that of the rectangle.

If $C'$ lies on $AB$, the triangles will exactly cover the rectangle. This is the optimum.

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Nice and “Elementarily, Watson!” –  Michael Hoppe Jan 2 '14 at 13:41
Nice explanation. Thanks for the proof. –  Varun Iyer Jan 2 '14 at 15:59