# Can this be bounded [closed]

Let us denote by $a_{k}$ the supremum of the minimum of the angle formed by the line segments between any $k$ points in the unit triangle. Is $\sqrt{k}a_{k}$ bounded as $k\rightarrow\infty$?

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Erm... You have about $k^2$ segments, so at least one angle is $O(k^{-2})$. Apparently either I miss something, or what you wrote is not what you meant. –  fedja Sep 22 '13 at 17:34
I share @fedja's confusion: there are $O(k^3)$ angles, and their sum is is $O(k),$ (since the sum of the angles around any one point is at most $2\pi.$) So, the smallest anglesfor any configuration is at most $c/k^2.$ No? –  Igor Rivin Sep 22 '13 at 17:51
@mfan I look only at those made by the lines "adjacent" in the "direction ordering" and ignore the rest. –  fedja Sep 22 '13 at 17:52
@IgorRivin and fedja: The fact is that the angles at the same vertex can have overlaps. To be more clear, the sum of all the angles is $k(k-1)(k-2)\pi/6$. Thanks. –  mfan Sep 22 '13 at 17:56
What @fedja said (and this does cut down the number of angles you look at to $O(k^2),$ my mistake, so the smallest angle is at least $c/k.$) –  Igor Rivin Sep 22 '13 at 18:03
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## closed as off-topic by Ricardo Andrade, Andres Caicedo, Andy Putman, Andrey Rekalo, Will JagySep 22 '13 at 18:15

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