Given a convex set $S\subset\mathbb{R}^2$ of unit area, how small of a quadrilateral can we inscribe it in?

Question

Every bounded convex planar body $S$ of area $A$ can be inscribed in a triangle of area at most $2A$, with the upper bound attained by the parallelograms. See Wolfram MathWorld for citations.

What constant do we obtain for quadrilaterals? Obviously, we can guarantee an upper bound of $2$ by the above result. In fact, one can bound the area by $2$ just in the case of rectangles, by placing the sides parallel or orthogonal to the diameter of $S$.

Conversely, the circle provides a lower bound of $4/\pi\approx 1.273$. I think the regular hexagon may give $A=4/3\approx 1.333$ by lining up the sides of the quadrilateral with four sides of the hexagon, but I haven't proven this is optimal for the hexagon.

The answer in the worst case has to be strictly less than $2$, as the only shapes that attain the bound in the triangular case allow for $A=1$ in the quadrilateral case. One might worry about a series of shapes approaching but never attaining this bound, but the Blaschke selection theorem guarantees this will not happen (because we can use affine transformations to restrict our attention to a bounded region).

In the event that the quadrilateral case is solved, I'm curious about the general case of $k$-gons for all $k\ge3$ (both our best proven upper bounds, and the hardest-to-inscribe known shapes that approach this bound - I would be surprised if these coincide for large $k$).

Answer 1

Even for quadrilateral, it is probably still an open problem.

For any convex region $K$ with unit area, let

• $C_n(K)$ be a circumscried $n$-gon of $K$ with minimum area.
• $c_n = \sup\limits_{K}\verb/Area/(C_n(K))$

According to a 2009 paper Circumscribed Polygons of Small Area by Dan Ismailescu, $$\frac{3}{\sqrt{5}} \le c_4 \le \sqrt{2}$$ The lower bound $\frac{3}{\sqrt{5}}$ is attained by a regular pentagon. In $1983$, Kuperberg has asked the question whether $c_4 = \frac{3}{\sqrt{5}}$. As least up to $2013$, this has not been settled.

According to above paper, we also have

• $c_n \le \sec\frac{\pi}{n}$ for $n \ge 3$
• $c_n \le \frac{n-2}{\pi} \tan\frac{\pi}{n-2}$ for $n \ge 5$

The first bound was proved in above paper, it beat the second bound for $5 \le n \le 11$. The second bound was proved by Fejes Tóth in 1940s.