8
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Dec 2, 2014 at 22:28
  • 2
    $\begingroup$ @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. $\endgroup$ Dec 3, 2014 at 8:14
  • $\begingroup$ OK, after computing I agree. $\endgroup$ Dec 3, 2014 at 9:23

1 Answer 1

3
$\begingroup$

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 $R>1$.

$\endgroup$
3
  • 1
    $\begingroup$ This is a beautiful proof, thanks! I created a GeoGebra worksheet to illustrate it: tube.geogebra.org/material/show/id/1536593 $\endgroup$ Aug 29, 2015 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. $\endgroup$ Aug 29, 2015 at 18:44
  • $\begingroup$ Yes, I have already realizd that, sorry. $\endgroup$ Aug 29, 2015 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.