TITLE: Is there a graph that is Ramsey for $P_{2n}$ but is $C_{2n+1}-$free
QUESTION [6 upvotes]: Write $F\to G$ to mean that for every two coloring of the edges of $F$, there exists a monochromatic copy of $G$. Nešetřil  and Rödl proved that for every graph $G$, there exists a graph $F$ such that $F\to G$ and $\omega (F)=\omega (G)$ (where $\omega$ is the size of the largest clique in the graph).
My question is an extension of that. Is there a graph $F$ so that $F \to P_{2n}$ but so $F$ is $C_{2n+1}-$free? This is trivial for $n=1$. For $n=2$, the graph $F:= C_6\cup \{(x,v_i)|i=1,3,5\}$ works (where $v_1,\dots,v_6$ are the vertices of the $C_6$ and $x\not \in C_6$).
The motivation is that if this is true, I can construct another graph with a special property (using this lemma) that will help me solve a much more interesting problem (depending on how interesting this itself is!) 
So my question is, does this generalize to all $n$? It seems that it should.

REPLY [6 votes]: A simple proof of much more: for each $n$ and $g$, there is a graph $G$ with girth at least $g$ that is Ramsey for $P_n$. As shown by Erdos, there are graphs of arbitrarily large girth and chromatic number. Let $G$ be a graph with girth at least $g$ and chromatic number at least $4n+1$, so it contains a subgraph of minimum degree at least $4n$. Any two-coloring of the edges of $G$ thus contains a monochromatic subgraph of average degree at least $2n$ and hence a subgraph of minimum degree at least $n$. One can then greedily build a monochromatic path of length $n$, showing that $G$ is Ramsey for $P_n$.