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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
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