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

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 $$k$$).

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.

• Thanks for the information! What’s the 2013 source for the problem being open as of then? Nov 13, 2020 at 18:00
• W Kuperberg himself? asks the question on MO in 2013. Nov 13, 2020 at 18:10