Take a convex polygon $P$ in the plane, and find its minimum area bounding rectangle, $R$. I'm interested in the ratio of the area of $R$ to the area of $P$. The ratio has a minimum of $1$ for rectangles and is $2$ for triangles. I want to say that $2$ is the maximum ratio possible.
Is this known? Is there a nice proof/counterexample for it?
I believe this is easy to prove. Assume you have at least four points and take the longest chord $C$ of the polygon. Now choose your rectangle such that two edges are parallel and the same length as $C$. Now observe that since each "half" of the polygon contains a triangle inscribed in the "half" rectangle (where "half" means cut by $C$), at least half of each side of the rectangle is filled.
Not quite an answer: it is a result of Dutkovsky that the area of a rectangle containing a curve of perimeter $2\pi$ is at most $4.$
EDIT I have just learned (from Erwin Lutwak) that the result is NOT a result of Dutkovsky, but a result of Lutwak, as in:
On isoperimetric inequalities related to a problem of Moser.
Amer. Math. Monthly 86 (1979), no. 6, 476–477.