What is the least $k > 0$ such that every convex polygon of area $k$ contains a rectangle of area 1?

I can prove that $k \le 8$, but surely this can be improved. Let $\mathcal{C}$ be a convex polygon of area 8, and let $\overline{PQ}$ be a diameter of $\mathcal{C}$. There is a bounding rectangle $ABCD$ such that $\overline{AB}$ is parallel to $\overline{PQ}$. The line segment $\overline{PQ}$ divides $ABCD$ into two rectangles, at least one of which has area 4 or greater.

Assume without loss of generality that the area of $PQBA$ is at least 4, and let $R$ be a point where $\mathcal{C}$ meets $\overline{AB}$. Then the area of the triangle $PQR$ is at least 2, and the largest rectangle inscribed in $PQR$ has area at least 1.

enter image description here

Update: Andrés Koropecki pointed out the following theorem of W. Blaschke. Let $K$ is a convex body in $\mathbb{E}^2$, and let $T$ be a triangle with maximum area among all triangles contained in $K$. Then $\frac{\mathrm{Area}(T)}{\mathrm{Area}(K)} \ge \frac{3\sqrt{3}}{4\pi}$ with equality iff $K$ is an ellipse. This implies that my constant $k$ is at most $\frac{8\pi}{3\sqrt3} \approx 4.837$.

Update 2: Bertram Felgenhauer has kindly shown me a proof that $k \le 4$. I will post it later.

  • 1
    $\begingroup$ It's also clear that $k \ge 2$, since the largest rectangle in a triangle of area 2 has area 1. Perhaps $k = 2$? $\endgroup$ Mar 25 '13 at 17:28
  • $\begingroup$ I believe $k \gt 2$. In the figure above, take a point X on CQ and consider PXQR. The rectangles in PQR and PXQ don't line up. I believe this gets at least $2+\epsilon$, but it is not a proof. $\endgroup$ Mar 26 '13 at 23:50
  • $\begingroup$ Is it necessary that the polygon of $(n-1)$ sides is within the $n$-sided polygon? $\endgroup$ Jun 14 '13 at 14:57

The main result of Marek Lassak, "Approximation of convex bodies by rectangles", Geom. Dedicata 47 (1993), 111–117, doi:10.1007/BF01263495 is:

Let $C$ be a convex body in the plane. We can inscribe a rectangle $R$ in $C$ such that a homothetic copy $S$ of $R$ is circumscribed about $C$. The positive homothety ratio is at most 2 and $\frac12|S|\le|C|\le 2|R|$.

($|\cdot|$ denotes area.) In particular, $k\le 2$, which is optimal as noted in comments. According to Lassak, the fact that every convex body contains a rectangle $R$ with $|C|\le 2|R|$ was shown in K. Radziszewski, "Sur une problème extrémal relatif aux figures inscrites et circonscrites aux figures convexes", Ann. Univ. Mariae Curie-Sklodowska, Sect. A 6 (1952), 5–18.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.