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A delightful recent problem about disconnecting the plane by straight lines suggested me the following further question, that I can't resist to post.

Let $\mathcal{F} $ be a countable family of straight lines, which is generic in the sense that no two of them are parallel, and no three are concurrent.

What can be said about the distribution of the areas of the polygons in which the plane is disconnected by $\mathcal{F} $?

After Yaakov Baruch's answer given in the linked question, we know that they can't be all equal. On the other hand, one can easily make examples where the areas are at least bounded below, and also examples where areas are bounded away from zero. But is there an example where all areas are between two positive constants? I tend to think there isn't any. More generally, what can we say about the distribution of these areas? (say, in an asymptotic sense, as limit of the distributions of the areas of the tiles within a ball of radius $R$).

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Apparently there has been some interest in characterizing the polygon that contains the origin, e.g., springerlink.com/content/f52880m850421qr2 –  Joseph O'Rourke Jun 2 '11 at 5:30
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up vote 3 down vote accepted

Here are some results for a particular random model of lines.

The equation for any line in the plane may be written $p=x\cos\theta+y\sin\theta$, with $0\leq\theta<\pi$.

The Poisson line tessellation is a random process where one places lines randomly according to a Poisson point process with parameter $\tau/\pi$ in this $(p,\theta)$ strip and with uniformly distributed orientation. This generates a system of generic random lines almost surely. I'm not sure what the state of the art is, but according to this 1964 paper by R.E. Miles (see also its sequel which contains sketches of proofs), the first three moments of the distribution of areas of the polygons resulting from the intersections of these lines are $\pi/\tau^2$, $\pi^4/2\tau^4$ and $4\pi^7/7\tau^6$.

Some numerical results for the first seven moments of the area (with parameter 1 so that $\tau=\pi$ in the notation above) from generating 250,000 polygons are in a 2007 paper by Michel and Paroux: 0.31813 (exact is $1/\pi\approx0.31831$), 0.5009 (exact is $1/2$), 1.8042 (exact is $4\pi/7\approx1.7952$), 11.47, 113.9, 1610, 29400 (where I've rounded off some unnecessary digits from their Fig. 8). They also have a figure showing the cumulative distribution function for the areas:

Figure 13 from Michel and Paroux

The behavior near zero area is apparently explained by the following asymptotic result for the probability density function for the areas (part of Theorem 6 in Miles's paper above): $P(A\in dx)=c\tau A^{-1/2}+O(\tau^2)$, where $c=\frac{1}{3\pi}\int_0^\pi d\phi\int_0^{\pi-\phi}d\psi \sqrt{2\sin\phi\sin\psi\sin(\phi+\psi)}$ which Mathematica tells me is about 0.32311.

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(Edit:) I think I may have misunderstood the question. Here are some results for polygons created by finite samples from an infinite family of lines:

Let each line have a point on the unit circle, at angle θ anticlockwise from the x-axis, where 0 ≤ θ < 1. Each line has another point also on the unit circle, at angle φ anticlockwise from its start angle. (Here, φ is the golden ratio, (1+√5)/2.) We get this:

http://www.flickr.com/photos/edwynn/5871622044/

This image shows lines (in blue) where the angles θ were generated by the start of the quasirandom sequence (nφ mod 1). This countable infinite sequence is guaranteed not to have any parallel lines. In this family, different angles also guarantee non-coincident intersections.

The image is marked up with the largest possible triangle (in red), with area ≈0.017. So, this is a case with an upper bound on area. In fact, all intersections (and therefore all polygons) fall inside that triangle plus the curved parts just below it.

From a random sample of 100,000 examples, this is the distribution of triangle areas: http://www.flickr.com/photos/edwynn/5871622056/ http://www.flickr.com/photos/edwynn/5871622082/

I agree that it seems unlikely that areas could all be both finite and nonzero, but I have no proof.

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