Here's a simple question I've been struggling with for weeks. Draw a row of n adjacent squares (so 2(n+1) boundary points). Join every pair of boundary points by a line segment. The resulting graph G_n is a K_{n+1, n+1} with two extra boundary lines. Here is a colored illustration of G_3: https://oeis.org/A331452/a331452_8.png (A331452 has many other pictures.) Thanks to the work of Legendre (2009), Griffiths (2010), and Max Alekseyev (2010, 2015),we know the numbers of regions (or cells) in G_n, which is A306302: 4, 16, 46, 104, 214, 380, 648, ... (and has an elegant formula), and also the number of vertices (A331755) and edges (A331757). My question is, why are all the cells in G_n triangles or quadrilaterals? Why are there no pentagons or higher-order cells? Max gives an indirect proof of this fact referring back to an old Russian theorem about training sets for threshold functions. But there ought to be a proof just from geometry. Once we know this fact, we can deduce the numbers of triangular cells (A324042) and quadrilaterals (A324043).