Re: Open geometry problem?

On 30 Apr 2006 23:51:00 +0100, Thomas Mautsch <mautsch@xxxxxxx> wrote:

In news:<dr0852122dtdb65125ia66fiold5j0far8@xxxxxxx>
schrieb quasi <quasi@xxxxxxxx>:
On 29 Apr 2006 16:05:53 +0100, Thomas Mautsch wrote:
quasi wrote:

Prove or disprove:

Given a closed disk in the plane and n distinct points, n>2, in the
disk. No special restrictions are imposed on the n points (for example
-- collinearity is allowed, and boundary points are also allowed).

Then the disk can be partitioned into n closed convex
subregions of equal areas (shared boundaries are allowed but the
subregions must have pairwise non-overlapping interiors) such that
each subregion contains exactly one of the n given points.

It might be better, first to investigate
the weaker version of this conjecture:

Given a closed, bounded, convex set in the plane D,
and n (distinct) points p1,p2,...,pn in D;
is it possible to represent D as the union of
n closed, convex sets A1, A2,..., An with disjoint interiours,
^^^^^^^^^^
such that each point pi is an element of Ai for i=1, 2,...,n.
...
As quasi has not given his prove of the case n=3 yet,
here is my proof of the weaker version of the conjecture:

... [proof of weakened conjecture for the case n=3 snipped]

This ends the proof of the case n=3.

So, up till now, the weak form of the conjecture
is proven to be true for n = 0, 1, 2, 3, 4, 6, 8.

The first open case that remains is n = 5.

Any ideas how to do this or the general case?

For the weakened conjecture (with no requirement for equal areas), the
general case is no problem:

Only because I forgot to write that I want
the convex sets A1, A2,..., An to be of *equal area*.

[ As if such a thing had not happened to you before. ;-) ]

The only difference to your conjecture is that
the points p(i) are not required to be in the *interior* of the sets A(i).

Any new ideas on this problem?

Ah, ok -- thanks for the clarification -- I'll play with it.

quasi
.

Relevant Pages

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