# Does this iterated Delaunay triangulation process always “explode”?

Let $P$ be a set of three noncollinear points in $\mathbb{R}^2$. Iteratively form the Delaunay triangulation $\cal T$ of $P$, and then augment $P$ by the circumcircle centers of all triangles in $\cal T$, repeating these two steps:

(1) ${\cal T} = \operatorname{DelTri}( P )$.

(2) $P = P \cup \{ x : x \; \textrm{is center of circumcircle of}\; \triangle \in {\cal T} \}$.

It appears that, after a few iterations, $P$ contains either an exactly zero-area triangle of three collinear points, or three nearly collinear points whose circumcircle center lies far outside the initial $P$. For example: After $5$ iterations, subsets of $3$ nearly collinear points occur.
Starting with an equilateral triangle leads to collinearity in four steps: Initial $P$ consists of corners of equilateral triangle.

Q. Is there any initial triangle that does not lead to a $3$-points-collinear or nearly collinear triangle?

By nearly collinear I mean that, as the number of iterations increases, the diameter of $P$ grows without bound: it grows because some triangles become very flat and their circumcenters are cast far away. Truly $3$-points-collinear would throw the circumcenter to $\infty$. I'm asking whether these conditions eventually necessarily occur for any initial triangle.

Added. Just a small amount of data, but $25$ random start triangles with coordinates uniform in $[-1,1]$ each reach a triangle of area smaller than $10^{-10}$ within $10$ iterations.

• What does "nearly collinear" mean? – Igor Rivin Oct 9 '17 at 19:44
• Since at the very first step if any of your triangles is acute you are going to get a triangle with an angle $\geq 2\pi/3,$ which, in my book, is pretty close to "nearly collinear"... – Igor Rivin Oct 9 '17 at 19:57
• @IgorRivin: Good question, I will edit. – Joseph O'Rourke Oct 9 '17 at 20:00