Finite set of non-collinear points on plane with every point having ≥ 3 equidistant neighbors? Does there exist a finite set of points on the Euclidean plane, such that:

*

*No 3 points are collinear, and

*Every one of the points has (at least) three other points in the set at the same distance from it?

It seems to me that the answer should be No, but my naïve attempts to prove it have failed.
 A: Yes.
Place five points $P_1,P_2,P_3,P_4,P_5$ in a regular pentagon inscribed in the unit circle centered at the origin. For each of these points $P$, we're going to add another point $Q$ somewhere on the circle centered at $P$ and passing through the neighbors of $P$ on the pentagon:

If we choose consistent relative positions on each arc for the new $Q$'s, we will have five new points $Q_1, Q_2, Q_3, Q_4, Q_5,$ arranged in another pentagon centered at the origin. Our construction currently makes all the $P_i$'s circumcenters, but not all the $Q_i$'s. We want each $Q_i$ to be equidistant from $Q_{i+2}, Q_{i+3},$ and $P_i$ (where we take indices mod $5$).
If we choose each $Q_i$ to be the midpoint of our arc, then the distance $\overline{Q_iQ_{i+2}}$ will be less than $\overline{Q_iP_i}$. On the other hand, if each $Q_i$ all the way to the side approaching $P_{i+1}$, then the distance will be greater. So by continuity, there is a point at which they are equal, and each $Q_i$ is a circumcenter.
A: While there are 6-point solutions, and 6 is the minimum, nevertheless there is a 7-point solution that feels to me to be the simplest, and it's very symmetric, namely
the regular 6-gon (hexagon) together with it's center.
A: I can't say much about colinearity here, but the current record holder in the Polymath Project for the "Hadwiger-Nelson problem" lists a 510-vertex unit distance graph of chromatic number 5, so in particular, an arrangement of 510 points where every point has at least four other points at distance 1.
A: 
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A: This text is an extended Answer that I have posted some 17h ago and removed some 16h ago.
THE NEW PART: I've removed my answer because @PietroMajer has posted earlier (some 19h ago) an elegant 9-point example that had a richer and more impressive structure (and Pietro has posted a comment under the OP too). This 9-point solution contained a 6-point solution.
However, I've decided to mention the existence of that 6-point solution explicitly. Of the Pietro's 9 points one may take only 6-points
$$ \{\xi^k+c\cdot\xi^j: k=0,1,2\ \ \text{and}\ \ j=0,1\} $$
THE OLD PART (posted about 3h after @PietroMajer's solution):
In $\ \mathbb C,\ $ let $a$ and $b$ and $c$ be the vertices of a triangle such that $\ |a-c|=|b-a|=|c-a|=1.\ $ Let $\ v\in\mathbb C\ $ be such that
$\ |v|=1,\ $ and $\ \pm v\ $ be different from any $\ a-c\ $ or $\ b-a\ $ or
$\ c-a.\ $ Then the $\ 6$-point set consisting of points $\ x+v,\ $ where $\ x\ $ is any of the points $\ a\ $ or $\ b\ $ or $\ c\ $ together with these three points, provides an example of a $\ 6$-point planar set such that each point has distance $\ 1\ $ from at least three other selected points.
Is $\ 6\ $ the record or is it $\ 5?$ — I feel lazy :) but this should not be a difficult question to settle.
Oh, $\ 5\ $ is impossible. This is a simple exercise. Thus, the record is $\ \mathbf 6$.


PS. 5-point solution is impossible even if collinearity was allowed.
A: The figure below has all line segments shown congruent. The quadrilaterals sharing edges with the central triangle are squares.

A: Example: If $1,\xi,\xi^2$ are the three cubic roots of unity, (and $c$ is  generic, with $|c|=1$) $\xi^k+c\xi^j$ gives a set of $9$ points, no three of which collinear, where each point has $4$ points at the same distance.
