# Question on a concrete example of n points

Can anyone give a concrete example of n points in the unit square (for instance, n runs from 3 through a large number) that can be generated by the algorithm here or any other algorithm or any construction such that the areas of the smallest triangles, multiplied by n^2, form a sequence that is non-decreasing for large n or tends to be unbounded?

• Example of what? Your question provides no information about what you are asking! Aug 30, 2013 at 3:56
• I'm sure that for any concrete set of $n$ points, there is an algorithm or construction that will generate that set of points, so I suppose you can take any $n$ points you like as an example. Aug 30, 2013 at 4:34
• @GerryMyerson the link points to an algorithm for finding sets that are counterexamples to a conjecture of Heilbronn. I agree, though, that in its current form the question should be closed. Aug 30, 2013 at 4:37
• If the comment above is meant to be a clarification of the question, please edit it into the question. People shouldn't have to wade through the comments to understand the question. Aug 30, 2013 at 4:40
• The comment mentions the unit square. If that's an important part of the problem, that should be in the body of the question, too. Aug 30, 2013 at 10:05

Unbounded might not be too hard if I understand the question correctly, although my argument is very much vaguer than I would like. For a set $S$ of points in the unit square let $n=|S|$ and $f(S)$ be $n^2$ times the smallest area of any of the $\binom{n}{3}$ triangles determined by three of the points. You link to an article which constructs for each $n \ge 3$ an explicit (in some sense) set $T_n$ with $f(T_n) \gt c\log(n)$ for some fixed $c$. Are you asking for a sequence of points so that the sequence $a_n=f(\lbrace p_1,p_2,\cdots,p_n\rbrace)$ is increasing or at least unbounded?
Here is a possible attack using only that information. Considering the actual construction in the linked article might give better results. It seems possible that the following could work, i.e. have $\limsup(a_n)=\infty$, although it could at the same time have $\liminf(a_n)=0$: First let me comment that we can assume that the points in the set $T_n$ have the form $(\frac{a}{n^k},\frac{b}{n^k})$ for $k=4$, I don't know that that helps but it makes things seem more controlled (proof at end).
Consider the sets $T_n$ above (about which I know nothing beyond what is stated). Perhaps one can always find a sequence of indices $n_1,n_2,\cdots$ so that $T_{n_i}$ is "almost a subset" of $T_{n_{i+1}}$ in the sense that we can perturb the points of the smaller one by a sufficiently small amount (perhaps none by more than $1/2^{n_i}$) such that they coincide with a subset $T'_{n_i}$ of the larger one having $f(T'_{n_i}) \gt \frac{c}{2}\log(n_i).$ Then (skipping some routine details) there would be a sequence so that $a_n \gt \frac{c}{2}\log(n)$ for $n=n_1,n_2,\cdots.$
The question of $a_n$ increasing or at least having a positive lower bound might be harder.
My argument that the points of $T_n$ can have coordinates of the form $\frac{\cdot}{n^k}$ for $k=4$ is that moving each point to make this happen will change the area of each little triangle by no more than $\frac{1}{n^k}$. I don't know if we really need $k$ as big as $4$.