$14$ points in a regular hexagon of side $2$ Can we place $14$ points in a regular hexagon of side $2$ such that the minimal distance between  points is $>1$?
Background:
We can place $13$ points in a regular hexagon of side $2$ so that the minimal distance between $2$ points is $\frac{2}{\sqrt{3}}$. For this, divide the hexagon into $6$ equilateral triangles and consider the points: $6$ -- vertices of the hexagon, $6$ the center of the triangles, and $1$ the center of the hexagon, $13$ in total. 
We can place $19$ points in the hexagon such that the minimal distance between them is $1$. For this, divide the hexagon into equilateral triangles of side $1$ and consider all the possible vertices, $19$ in number. 
From this answer, whenever we take $20$ points in the hexagon, the minimal distance between them is $\le 1$. 
From this answer, whenever we take $19$ points in the hexagon, the minimal distance between them is $\le 1$. ( I believe the result, not totally sure about the proof).
Connected results would be great too. Thanks for your interest! 
 A: Here are 15 points with minimal distance $1$ and all minimal distances marked.

Now, move the blue points towards the centre a bit (e.g., until they form a smaller equilateral triangle and with blue-blue distances equal to blue-red -- that is, with these distances $=3-\sqrt3=1.2679\ldots$). That allows you to move the green dots inwards as well, thereby making all distances slightly $>1$.

(Finally, drop one of the points in order to reach the desired number of $14$ points.)
A: Can this way work?
We can at most put 11 points on the side. The gray part means the area in which points will have <= 1 distance with those 11 points. So there will still be an acceptable area in the middle for the rest of 3 points.

A: Iris had a nice idea to draw the allowed curved region.

Now we use @Hagen von Eitzen: method but instead of placing in the curved region a triangle, we place a quadrilateral with all sides $>1$. Wiggling a bit ( same method) gets us $16$ points all separated by more than $1$.
It turns out that we can place a pentagon with vertices in the curved region and all sides of length $>1$. The same method now gives us $17$ points  in the hexagon with parwise distances $>1$. 

Calculation that the top vertex of the regular pentagon is below the top arc: we have the height of the pentagon $=\frac{\sqrt{5 + 2 \sqrt{5}}}{2}\cdot 1$, so we need to check that 
$$2 \sqrt{3} - (1 +\frac{\sqrt{5 + 2 \sqrt{5}}}{2} + \frac{\sqrt{3}}{2})= 0.0592\ldots>0$$ 
A: Here is a way to fit 16 points inside, such that the minimum distance is greater than $1$.



*

*Begin the construction with the regular pentagon in the middle.

*Construct the equilateral triangles below it.

*Construct a line parallel to the base line thus far, slightly below it.

*Extend the edges of the outer triangles to meet the line from the previous step.

*Construct the hexagon using the points found in the previous step. Declare this hexagon's edge to be 2 units, and it follows that all point-to-point distances so far are greater than 1.

*Construct $A'$ and $B'$ using reflection.

*Equilateral triangles for $R$ and $S$.

*There are lots of way to put in the last two points at the top. The illustration uses one last equilateral triangle.


Note that the points could all be shifted up a tiny bit, and then they would all fit even in the interior of the hexagon. 
Take a look at @Iris's answer. I believe it is similar to this. Mine also has 11 points "around the edge". They could probably be distributed evenly like in @Iris's answer. @Iris put three points in the middle, but perhaps did not realize a pentagon could fit there too.
A: Alternative solution:
Consider a regular hexagon of side $2$ given by the linear inequalities (denote by $H$):
\begin{align}
\pm \frac{x}{2} \pm \frac{y}{2\sqrt{3}} &\le 1, \\
\pm y &\le \sqrt{3}.
\end{align}
We need to find $(x_i, y_i) \in H, i = 1, 2, \cdots, 14$ such that
$\sqrt{(x_i - x_j)^2 + (y_i - y_j)^2} > 1$ for any $1\le i < j \le 14$.
Using Matlab global optimization solver, we find a feasible solution as follows:
\begin{align}
(299/739, \sqrt{3}),\\
(-53/345, 751/982),\\
(-411/577, \sqrt{3}),\\
(-243/185, 333/421),\\
(-2 + 109\sqrt{3}/2157, -109/719),\\
(-2 + 263\sqrt{3}/705, -263/235),\\
(-253/603, -\sqrt{3}), \\
(101/547, -236/301),\\
(87/112, -\sqrt{3}),\\
(2-261\sqrt{3}/895, -783/895),\\
(2-38\sqrt{3}/693, 38/231),\\
(2-266\sqrt{3}/705, 266/235),\\
(682/867, 137/862),\\
(-224/297, -66/373).
\end{align}
(Use Maple to check its feasibility)
It holds that $\sqrt{(x_i - x_j)^2 + (y_i - y_j)^2} > 1.116$ for any $1\le i < j \le 14$.
Remarks: This numerical approach works for $n \le 17$. 
Anybody knows better global optimization solvers?
