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Does every set of $n$ points in the Euclidean plane contain three points $A,B,C$ such that the two segments obtained by joining $A,B$, respectively $A,C$ form an angle at least equal to $(1-2/n)\pi$ at the point $A$? (Equality is of course achieved by the vertex set of a regular $n-$gone.)

Pietro Majer's example below can be generalized and shows that $(1-2/n)\pi$ has to be replaced by a somewhat smaller constant (at least for values of $n$ which are large enough). For his example we have to take $(1-2/6)\pi$ instead of $(1-2/7)\pi$. Is the best possible constant $(1-a(n))\pi$ asymptotically substantially better, ie. can $na(n)$ become for example arbitrarily large for $n$ large enough? (It is of course obvious that $a(n)$ is decreasing but how fast?)

Update: For $n=5$, one can get arbitrarily close to $(1-1/4)\pi$: Take a right-angled isocele triangle. Split the right-angled-vertex infinitesimally along a line parallel to the longest side of the initial triangle and add an additional point on the symmetry axis very high above the two infinitesimal points.

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2 Answers 2

up vote 4 down vote accepted

The key bound is $(1 - 1/n) \pi$, due to Erdős and Szekeres:


           Erdos Furedi Quote
The above is an excerpt from this paper:
           Erdos Furedi Title
The Erdős-Szekeres result is in their 1961 paper, "On some extremum problems in elementary geometry." (PDF download link).

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How does the configuration of $8=2^3$ points without angles greater than $2/3\pi+\epsilon$ look like? –  Roland Bacher Nov 14 '12 at 12:44
    
@Roland: I added a link to the Erdős-Szekeres paper. Their proof is in Section 4, p.59 ff. I cannot pursue it myself at the moment; sorry. –  Joseph O'Rourke Nov 14 '12 at 13:15
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Thank you. By the way, the configuration on $8$ points is easy: one can split the central point in Pietro Majer's example into two infinitesimally close points on a line parallel to a side of the initial triangle. –  Roland Bacher Nov 14 '12 at 14:55
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actually this way we make a new angle (1−1/6)π... –  Pietro Majer Nov 14 '12 at 17:39

Consider an exagon obtained from an equilateral triangle by cutting three small equilateral triangles from its vertices. Consider the configuration of the $6$ vertices of the exagon, plus the center of the initial equilateral triangle. It seems to me that with these $7$ points one can't do angles larger than $2\pi/3+\epsilon$.

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Indeed. So what is the best possible constant for a given integer $n$? –  Roland Bacher Nov 14 '12 at 9:14
    
Yes, it seems a beautiful and difficult variational problem. –  Pietro Majer Nov 14 '12 at 10:01
    
Sorry, since I can accept only one answer and since Joseph O'Rourke mention of the Erdos-Szekeres paper kills the problem completely, I have changed the accepted answer to his reply. –  Roland Bacher Nov 15 '12 at 9:50

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