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?


2 Answers 2


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.

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    $\begingroup$ the rest of the argument being that the minimum area bounding rectangle has area no bigger than the one that you specify. $\endgroup$ Commented Apr 9, 2012 at 3:36
  • $\begingroup$ Right, in other words that the rectangle actually contains the polygon. $\endgroup$ Commented Apr 9, 2012 at 16:34

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: Lutwak, E. On isoperimetric inequalities related to a problem of Moser. Amer. Math. Monthly 86 (1979), no. 6, 476–477.


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