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](https://math.stackexchange.com/questions/793151/is-it-always-possible-to-partition-a-fat-shape-to-fat-shapes), currently with no reply.



Fixed point of fatness
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 $R>1$.


