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For the purposes of this question, define the following properties of convex sets in the plane:

  • A set is $R$-fat (for $R\geq 1$) if it contains a disc of side-length $x$ and is contained in a disc of side-length $R\cdot x$, for some positive $x$.
  • A set is $R$-cuttable if it can be cut (using a straight line) to two non-empty convex $R$-fat shapes.

For example:

  • A disc is 1-fat. It is not 1-cuttable since it cannot be cut to two discs. But it is 2-cuttable since it can be cut to two semi-discs, and a semi-disc is 2-fat.
  • A $1\times 10$ rectangle is $\sqrt{101}$-fat. It is $\sqrt {26}$-cuttable since it can be cut to two $1\times 5$ rectangles. Hence it is also $\sqrt{101}$-cuttable,

MY QUESTION IS: what is the smallest value $R_0$ such that every convex $R_0$-fat shape is $R_0$-cuttable?

The first example shows that $R_0 \geq 2$.

NOTE: I asked a similar question in Math.SE, currently with no reply.

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  • $\begingroup$ I suspect there is no smallest $R$. A rhombus with diagonals $1$ and $d$ (large) is $R$-fat, with $R:=(d^2+1)/2d$. The best way to cut it, it seems, is into two isosceles triangles, but they are not $R$ fat. Is there a better way to cut it? $\endgroup$ – Pietro Majer Dec 2 '14 at 22:28
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    $\begingroup$ @PietroMajer: They are. The critical situation is a rhombus with angles $\pi/3$ and $2\pi/3$; then it is 2-fat, and it can be cut into two isosceles triangles, which are also 2-fat. $\endgroup$ – Ilya Bogdanov Dec 3 '14 at 8:14
  • $\begingroup$ OK, after computing I agree. $\endgroup$ – Pietro Majer Dec 3 '14 at 9:23
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Still, Pietro Majer is right --- there is no such $R_0$.

Let $S$ be the convex hull of the unit circle centered at the origin and the point $A=(2R-1,0)$. Then it is $R$-fat, with the unit disk inside it and the smallest enclosing disk with radius $R$ and center $(R-1,0)$.

Now assume that $S$ is cut into two convex sets $T$ and $T'$, with $T$ containing $A$. Let $O$ and $r$ be the center and the radius of a largest disk $D$ contained in $T$ (we have $r<1$, otherwise $T=S$). Then the abscissa of $O$ is at most $(2R-1)(1-r)$. Now, $T$ contains the point $B$ on the ray $AO$ with $AB=AO+r\geq (2R-1)r+r=2Rr$. Notice that $B$ lies strictly inside $S$; since the cut is straight and $B$ lies on the boundary of $D$, $T$ also contains some other point $C$ on the line perpendicular to $AB$ and passing through $B$. Then $AC>AB=2Rr$, so the smallest disr enclosing $T$ has the radius greater than $Rr$, and $T$ is not $R$-fat.

REMARK, edited. Notice that in this example, one may still cut $S$ into two $R$-fat non-convex but simply connected sets by a non-straight cut (namely, by a circular arc). As Erel mentions in the comment, this is always possible for every $R>1$.

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    $\begingroup$ This is a beautiful proof, thanks! I created a GeoGebra worksheet to illustrate it: tube.geogebra.org/material/show/id/1536593 $\endgroup$ – Erel Segal-Halevi Aug 29 '15 at 18:39
  • $\begingroup$ If the pieces don't have to be convex, then, I think, for every $R>1$, you can always cut a very small disk, which is not contained in the internal disk. The disk is 1-fat, and the remainder is still R-fat. $\endgroup$ – Erel Segal-Halevi Aug 29 '15 at 18:44
  • $\begingroup$ Yes, I have already realizd that, sorry. $\endgroup$ – Ilya Bogdanov Aug 29 '15 at 19:15

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