Graphs with non-disjoint odd cycles satisfies $\chi \le 5$ Given a graph $G$ with the property that any two odd cycles in the graph share a  vertex, then show that the chromatic index of the graph, $\chi$, is at most $5$.
 A: Let $C$ be the smallest odd cycle in $G$.  Note that $C$ has no chords (i.e. edges between its vertices that aren't part of the cycle), as otherwise there would be a smaller odd cycle.  Hence we can color $C$ with $3$ colors, and it will be a valid partial coloring of G.
Now if we throw all the vertices in $C$ from $G$, the remaining graph is bipartite because any odd cycle in $G$ must share a vertex with $C$ by assumption. Now we can color all of those vertices with two new colors. We have colored $G$ with $5$ colors, so $\chi(G) \le 5$.
A: We can do better. In fact, we will have $\chi(G) \le 3$ for such a graph. 
First, a lemma.
Lemma: Let $G$ be a graph with no odd cycles. Then $G$ is $2$-colorable.
Proof: Pick an arbitrary vertex $v\in G$. Color all vertices an even distance away from $v$ one color, and an odd distance away from $v$ another color. Note that this coloring is well defined precisely because there are no odd cycles. $\square$
Now, suppose $G$ is a graph such that all odd cycles intersect in one and only one vertex. Call it $w$. Then the graph $G'$ obtained from $G$ by removing $w$ is a graph with no odd cycles. Hence $G'$ is $2$-colorable. Add $w$ back and color it a third color. This is  $3$-coloring for $G$. Hence $\chi(G) \le 3$.
