# Intersecting Family of Triangulations

Let $\cal T_n$ be the family of all triangulations on an $n$-gon using $(n-3)$ non-intersecting diagonals. The number of triangulations in $\cal T_n$ is $C_{n-2}$ the $(n-2)$th Catalan number. Let $\cal S \subset \cal T_n$ be a subfamily of triangulations with the property that every two triangulations of $\cal S$ have a common diagonal.

Problem: Show that $|\cal S| \le |\cal T_{n-1}|$.

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Have you thought about using a cluster algebra mutation argument? Triangulations having all but one diagonal in common correspond to adjacent clusters in a cluster algebra of type $A$, so maybe this helps? –  Jan Grabowski Nov 27 '12 at 11:09