TITLE: Maximum degree in maximal triangle free graphs
QUESTION [10 upvotes]: It's easy to see that in bipartite maximal triangle free graphs (n vertices), the maximum degree is at least $\lceil n/2 \rceil$.  What about mtf graphs in general?  Must there always be some vertex of high degree?  Before I jump in with both feet, is there an obvious reason why there cannot be an absolute constant $c$ such that for every mtf $G$, $\Delta(G) > cn$?  A handy counterexample to c = 1/4?

REPLY [7 votes]: A maximal triangle-free graph has diameter two, so its breadth-first spanning tree has a vertex with a number of children that is at least proportional to $\sqrt n$.
I have the feeling that it should be possible to cook up a matching $O(\sqrt n)$ upper bound on the maximum degree, by using incidence graphs of the points and lines in a projective plane, but that doesn't quite work: it is triangle-free (and bipartite), has maximum degree $O(\sqrt n)$, and prevents adding any more point-point or line-line edges, but there is no obstacle to adding more point-line edges.
(Edit to add, after Greg left a comment on an earlier version of this answer): See the Kim 1995 reference from http://en.wikipedia.org/wiki/Triangle-free_graph:
Kim, J. H. (1995), "The Ramsey number $R(3,t)$ has order of magnitude $\frac{t^2}{\log t}$", Random Structures and Algorithms 7: 173–207 .
He describes a construction for a triangle-free graph in which every independent set has size $O(\sqrt{n\log n})$. If one makes the graph maximal, this doesn't increase the size of its independent sets. But in a triangle-free graph, the neighbors of every vertex form an independent set, so the maximalized version of Kim's graph has degree $O(\sqrt{n\log n})$.