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The family of rectangles has the cover property, i.e.:

For every $R\geq 1$, $k\geq 1$: every rectangle with aspect ratio $kR$ can be exactly covered by $\lceil k\rceil$ (possibly overlapping) rectangles of aspect ratio at most $R$.

I am trying to generalize this property to arbitrary convex shapes, using the following natural generalization of the aspect ratio concept:

(side length of smallest containing square) / (side length of largest contained square)

This factor can be termed slimness factor, as many papers define an "$R$-fat shape" as a shape for which this factor is at most $R$. So, for example, the slimness factor of a square is 1, the slimness factor of a circle is $\sqrt 2$, etc.

The slimness factor of a rectangle is identical to its aspect ratio, so this is indeed a proper generalization of the aspect ratio concept. However, the cover property is not true when the slimness factor is used instead of the aspect ratio.

For example, taking $R=1$, the only shape with slimness factor 1 is a square, but obviously most convex shapes cannot be exactly covered by a finite number of squares.

So, my question is: Does there exist an $R$ such that every convex shape with slimness factor $kR$ can be exactly covered by $\lceil k\rceil$ convex shapes of slimness factor at most $R$?

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    $\begingroup$ A bit related: mathoverflow.net/questions/134407/… $\endgroup$
    – Per Alexandersson
    May 20, 2014 at 10:22
  • $\begingroup$ Are the areas of all the shapes supposed to be the same? $\endgroup$
    – Ben Barber
    May 20, 2014 at 12:32
  • $\begingroup$ @BenBarber not necessarily. $\endgroup$
    – Erel Segal-Halevi
    May 20, 2014 at 12:37
  • $\begingroup$ Then I'm afraid I'm having trouble understanding. For example, what does "obviously most convex shapes cannot be covered by a finite number of squares" mean? $\endgroup$
    – Ben Barber
    May 20, 2014 at 12:38
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    $\begingroup$ @BenBarber For example, how can you cover a circle with a finite number of squares? Maybe it was unclear that the cover must be exact - i.e. the union of the covering shapes must be equal to the covered shape (not larger than it). I added the word "exactly" to make this clear. $\endgroup$
    – Erel Segal-Halevi
    May 20, 2014 at 12:40

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