Find the maximum number of points $P$ in a plane such that all the triangles having vertices in $P$ are not obtuse. 
Find the maximum number of points $P$ in a plane such that all the
  triangles having vertices in $P$ are not obtuse. (Degenerate triangles are also considered)


Obviously the vertices of a rectangle satisfy the condition, and I suspect that $4$ is the maximum. Supposing otherwise, there are $5$ points satisfying the property. If there are $4$ points, let's say $A_1,A_2,A_3,A_4$ such that the convex envelope of them contains the $5$th point $A_5$, then since none of the angles $A_iA_5A_{i+1}$ is obtuse and their sum is $2\pi$, all are right angles. But in this case we have degenerate triangles - which have obtuse angle - therefore we get a contradiction. It remains open the case when none of the points is contained in the convex envelope of the other four and this is the case I need help with.
 A: Let $P$ be a set, consisting of $n$ points. Among all circles having none of the points in its exterior, pick one of minimal radius. Then some points $p_1,\ldots, p_k$ are on the circle and the rest, $p_{k+1},\ldots, p_n$ are inside. 
Certainly, $k\ge2$ as otherwise we can shringk the circle. 
Each of the $k$ arcs determined by these points  is $\le180^\circ$, or otherwise we could shrink the circle.


*

*Any $p_i$ outside the convex polygon spanned by $p_1,\ldots,p_k$ leads to an obtuse triangle with the two vertices determining the arc that $p_i$ is closest to.

*Any $p_i$ ony an edge of said convex polygon leads to a degenerate obtuse triangle

*For such any point $p_i$ in the inside of said convex polygon, the line $p_1p_i$ leaves the polygon by ntersecting some edge $p_ap_b$. Then $p_i$ is in the triangle $p_1p_ap_b$ and hence one of the triangle $p_1p_ap_i$, $p_ap_bp_i$, $p_bp_1p_i$ is obtuse (their angles at $p_i$ add up to $360^\circ$).


We conclude that $n=k$, i.e., the points of $P$ are concircular.
If the points are labelled in circular order, we must have that $p_ip_j$ is a diameter whenever $i,j$ are not concsecutive. In particulyr, $n>4$ would imply that $p_1p_3$ as well as $p_1p_4$ is a diameter, i.e., $p_3=p_4$, contradiction.
We conclude that
$n\le 4$.
