3
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The slimness factor of a geometric 2-dimensional shape is defined (for this question) as the ratio of the side-length of its smallest enclosing square to the side-length of its largest enclosed square. For example, the slimness of a square is 1, of a right-angled isosceles triangle is 2, of a 30-by-10 rectangle is 3, etc.

A shape is called $R$-fat if its slimness factor is at most $R$. (See Fat object on Wikipedia).

Given a convex $R$-fat shape, is it always possible to partition it to two or more convex $R$-fat shapes?

The sketch below (drawn with GeoGebra) shows some shapes for which I managed to find such a partition. The number inside each shape is its slimness factor, which is also the maximum slimness factor of the pieces.

Note that without the convexity constraint there is a trivial solution: just cut a very small square that is not contained in the inner square. The square is of course R-fat, and the remaining (non-convex) shape is also R-fat (see the flag-shape at the left).

My guess is that such a partition is always possible even with the convexity constraint, but I find it difficult to make general assertions about convex shapes. Here I need your help.

enter image description here

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    $\begingroup$ By partition do you mean with a single linear cut? If not I have a trivial general solution... $\endgroup$ – Rahul May 14 '14 at 0:51
  • $\begingroup$ @Rahul do you mean to cut a tiny square that is not contained in the inner square? This is interesting. I should add a convexity constraint. $\endgroup$ – Erel Segal-Halevi May 14 '14 at 6:51
  • $\begingroup$ Breaking news: the question has been answered in MathOverflow: mathoverflow.net/q/188644/34461 $\endgroup$ – Erel Segal-Halevi Aug 31 '15 at 5:18

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