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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.

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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.

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It's also clear that $k \ge 2$, since the largest rectangle in a triangle of area 2 has area 1. Perhaps $k = 2$? –  Dave Radcliffe Mar 25 '13 at 17:28
    
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. –  Ross Millikan Mar 26 '13 at 23:50
    
Is it necessary that the polygon of $(n-1)$ sides is within the $n$-sided polygon? –  RicardoCruz Jun 14 '13 at 14:57

1 Answer 1

up vote 2 down vote accepted

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.

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