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.

  • $\begingroup$ This might help: the sum of the interior angles of a convex pentagon is ... $\endgroup$ – Ethan Bolker Aug 9 '17 at 14:12
  • $\begingroup$ @EthanBolker .. $3\pi$. How would this help? $\endgroup$ – user261263 Aug 9 '17 at 14:18
  • $\begingroup$ One of the five angles is more than a right angle. $\endgroup$ – Ethan Bolker Aug 9 '17 at 15:55
  • $\begingroup$ @EthanBolker But you assume the points are the vertices of a convex pentagon. $\endgroup$ – user261263 Aug 9 '17 at 16:03
  • $\begingroup$ Isn't that the case you need help with? $\endgroup$ – Ethan Bolker Aug 9 '17 at 18:46

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 $n\le 4$.

  • $\begingroup$ Thank you for the answer. I have a question: you said "Each of the $k$ arcs determined by these points is $≤180 \circ$, or otherwise we could shrink the circle". If $k \ge 3$, no matter the measure of the arcs, we can't shrink the circle anymore since $3$ points or more determine a circle. $\endgroup$ – user261263 Aug 9 '17 at 14:42

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