# Maximum of the minimal distance of a set of points in an equilateral triangle

In this question, a closed triangle on a plane is a set of all points in its area and on its boundary, while an open triangle excludes its boundary.

Now, the problems: Let $T$ be an equilateral triangle with sides of $1$ on a plane, $S = \{ A_1, ..., A_s \} \subset T$ be a set of some distinct points.

The general problem: Determine $$M_s = \max_{S \subset T} \left [ \min_{1 \le i < j \le s} d(A_i,A_j) \right ],$$ if it exists; otherwise, determine $$U_s = \sup_{S \subset T} \left [ \min_{1 \le i < j \le s} d(A_i,A_j) \right ].$$ Assume that:

$i) \; T \text{ is closed}, \; s = n^2+1 \; (n \ge 4)$.

$ii) \; T \text{ is closed}, \; s = n^2 \; (n \ge 4)$.

$iii) \; T \text{ is open}, \; s = n^2 \; (n \ge 3)$.

$(n \in \mathbb{Z^{+}})$

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This problem is posed by myself out of curiosity when I considered $ii, \; iii$ in the case of $n = 3$. I haven't solved even this particular case so far. Therefore, we can start with this case, which is called The specific problem from now on. Also, FYI, the common situation (?) of the problems is originated from another problem which can be easily proved by the Pigeonhole Principle.

P.S: I don't know exactly what to title and what to tag. So, if I had it wrong, please help me fix it, and thank you for that!

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Update: I don't know why but when I made up The general problem, I just totally forgot about the potential sub-problem of

$iv) \; T \text{ is open}, \; s = n^2+1 \; (n \ge 3)$.

Furthermore, I find it necessary to add some sub-results, which can be easily proved by the Pigeonhole Principle, as followed:

$a) \; \text{If } s = n^2+1$, $$U_s = \sup_{S \subset T} \left [ \min_{1 \le i < j \le s} d(A_i,A_j) \right ] \le \dfrac{1}{n}$$ $b) \; \text{If } s = n^2$, $$U_s = \sup_{S \subset T} \left [ \min_{1 \le i < j \le s} d(A_i,A_j) \right ] \le \dfrac{\sqrt{3}}{n}$$

$(n \in \mathbb{Z^+}\backslash \{1,2\})$.

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Earlier i said value for $s= 3m+1$ is $\frac{1}{\sqrt{3}m}$, this is only true for $m=1$, for higher $m$ other orientation(i wish i can explain it by figures, but i am not familiar with figure plotting) of points exceeds this value. so my final answer is:
Value of expression is: $\dfrac{1}{\left [ \dfrac{s}{3} \right ]}$ if $s\neq 4$, and for $s = 4$ value is $\frac{1}{\sqrt{3}}$, where $[x]$ denotes the maximum integer greater or equal to $x$.
Note that the answer remains the same for both open and closed triangles. If triangle is open one can take a triangle of length $1-\epsilon$ inside the original triangle with sides parallel, then find optimum orientation in smaller triangle and then make $\epsilon\rightarrow 0$, to conclude expression of value remains the same.
Explanation: For $s= 4$, it turns out that optimum orientation is one in which three points lie on vertexes and one on orthocenter. For $s = 7$, optimum orientations consist of one with one point at orthocenter while other points at boundary, while other orientations with all points at boundary, but the answer is same $\frac{1}{3}$. For $s\neq 4$ and $s\neq 7$ and $s>2$, optimum orientation consists of points at boundary, with three points on vertexes.
Your results are really helpful in case $T$ is closed (as well as in case $T$ is open, if they're improved just a little). I just have two wonders as followed: $A.$ In case of $s = 3m$, why can't the result be $\frac{1}{\sqrt{3}m}$ (the optimum scenario is the same scenario of the case of $s = 3m+1$ excluding the orthocenter) ? $B.$ Can you prove your results? – Vincent J. Ruan Feb 2 '15 at 4:29
About your second answer, I have some improvements and questions as followed (from now on, we just consider the cases of closed $T$ and $s \in \mathbb{Z^+} \backslash \{ 1,2,3 \}$): $C.$ In case of $s \in \{ 4,5,6,7 \}$, please provide a proof or a hint of one for your results. $D.$ In case of $s \ge 8$, your result can be improved to $\left ( \left \lceil \dfrac{s-1}{3} \right \rceil \right )^{-1}$. This is originated from your optimum scenario with the orthocenter added. – Vincent J. Ruan Feb 9 '15 at 13:45