11
$\begingroup$

The following question was asked last year at KoMal (May 2015):

Do there exist infinitely many (not necessarily convex) 2015-gons in the plane such that every three of them have a common interior point, but no four have a common point?

I don't have any idea in how to tackle this problem.
I believe the answer is 'no' and it's related to Helly's theorem.
Any help is welcome, thanks.

$\endgroup$
0
$\begingroup$

I've managed to figure out that, if it is possible, you must use arbitrarily large polygons. Here's the argument:

Suppose you don't have arbitrarily large polygons. That means that the maximal diameter (longest chord within a single polygon) is bounded above by some constant $C$. Take any two polygons. They must intersect, for any three polygons to have a common point, which means that they are no more than $2C$ apart. Pick some polygon to be the "center" one. All other polygons are within $2C$ of it, so all polygons are contained in a circle of radius no more than $3C$ centered at that polygon.

Now split the set of polygons into groups of threes. The area covered by each group is finite and positive, since you must have some intersection area by the definition for any three groups. You also know that there is no intersection between the area intersected by any two groups, since then you would have a common point between four polygons. There are infinitely many groups, and each contributes a finite and distinct amount of area, so the total area covered is infinite. But the area of our circle is $9C^2\pi$, which is finite. Therefore our polygons couldn't possibly be bounded in size, and therefore must be allowed to be arbitrarily large.

Now to prove that it's not possible: Take any pair of polygons. Every other polygon must intersect with both in some finite area, and that area must be distinct from the area provided by any other polygon, as otherwise you would have four common points. Add up all of these intersection areas for each other polygon. The total area is infinite. But the intersection areas were all within those two polygons, and the total area they cover is finite, so you couldn't have gotten infinite intersection area. So by contradiction, the answer is "no", if the argument holds.

However, there is a problem with both of my arguments, I've realized. It's a common misconception/mistake related to infinity, see if you can spot it.

The problem:

$\sum_{i = 1}^{\infty} 2^{-i} = 1$, an infinite number of finite terms, but the sum is finite.

$\endgroup$
  • $\begingroup$ "There are infinitely many groups, and each contributes a finite and distinct amount of area, so the total area covered is infinite." This is not true. $\endgroup$ – san Jan 16 '16 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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