Placing exactly six points along the boundary or inside of a square, so that each pair of points are equally spaced away from each other?

Question

This would be perhaps a good brain teaser at the elementary levels of Geometry, and I am trying to get my head around it.

How can we space exactly six points anywhere along the boundary or inside of a square, so that each pair of points are equally spaced away from each other?

I started by placing the four points on the four corners, and fifth one in the middle, but obviously that was to no avail. I thought to space them along two-thirds distance away along each edge, starting with the first at a vertex, but then realized that because of crossover along the edges, I am again not getting the same distance.

After reading the answers below, I realize that the problem in my head was not written in the intended manner. What I had on my mind was

If six points are placed either along the boundary or inside of a unit square, such that the minimum distance of each of these points from the remaining five points has the same value, then what is the numerical value of this distance?

I realize that my previous post was quite misleading. Does this make better sense, and is there a way out for it?

Answer 1

Alternative approach to prove impossibility.

Assumed that all 6 points are required to be distinct from each other.

Any $3$ points that are equidistant from each other must form an equilateral triangle.

Therefore, without loss of generality, points $P_1, P_2, P_3$ form an equilateral triangle.

However, since $P_4$ is also to be this same distance from both $P_1$ and $P_2$, points $P_1, P_2, P_4$ must also form an equilateral triangle.

Similarly, points $P_1, P_2, P_5$ must also form an equilateral triangle.

This is impossible because:

given points $P_1, P_2$, in order for a third point to form an equilateral triangle with $P_1$ and $P_2$ it must be on the perpendicular bisector of $P_1, P_2$, with an altitude $= \frac{\sqrt{3}}{2} \times $ the length of $\overline {P_1, P_2}$.

Therefore, there are only two distinct vertices possible that can be used to form an equilateral triangle with $P_1, P_2$.

Therefore you can not have $P_1, P_2, P_3$ and $P_1, P_2, P_4$ and $P_1, P_2, P_5$ all equilateral triangles.

Edit
The impossibility could have been proven, without consideration of point $P_5$, by noting that if $\overline{P_1P_2}$ has length $= 1$ unit, then, per the above analysis, $\overline{P_3P_4}$ must have a lenth of $\sqrt{3}$ units, which is not equal to $1$ unit.

Answer 2

The only possible case where this can happen is if the points are all a distance zero from each other. Otherwise, the set of points would, without loss of generality, be a unit-embedding of $K_6$ (the connected graph on $6$ vertices) into the plane $\mathbb{R}^2$, which is impossible, since the smallest Euclidean dimension that $K_6$ can be embedded into is $5$. See here for a more general and detailed proof of this claim.