# Graph with 3 blocks which any two of them are connected.

For a graph with 3 blocks like $$B_1, B_2, B_3$$ if every two of them are common in a vertex like $$v,u,w$$ then non of these vertices are cut-vertex. Because $$G-\{v\}$$ is still a connected graph. Am I right?

• I don't think it is possible to have such 3 blocks, because if you define block to be the maximal biconnected subgraph of $G$, then if the above is satisfied, then $B_1\cup B_2 \cup B_3$ will be a larger biconnected subgraph, hence $B_i$ are not blocks. Nov 25 '21 at 9:10

Let $$B_1$$, $$B_2$$ and $$B_3$$ be distinct pairwise-intersecting blocks.
Let $$u \in B_1 \cap B_2$$, $$v \in B_2 \cap B_3$$ and $$w \in B_1 \cap B_3$$. Let $$P_1$$ be a path in $$B_1$$ from $$w$$ to $$u$$, $$P_2$$ be a path in $$B_2$$ from $$u$$ to $$v$$ and $$P_3$$ be a path in $$B_3$$ from $$v$$ to $$w$$. These three paths are edge-disjoint by $$(a)$$ and node-disjoint (that is they do not have common inner nodes) by $$(b)$$. Thus if $$u$$, $$v$$ and $$w$$ are distinct, the concatenation of $$P_1$$, $$P_2$$, $$P_3$$ is a circuit containing $$u$$, $$v$$ and $$w$$, from which we get that $$w \in B_2$$, contradicting $$(b)$$. Hence $$u = v = w$$.
So it is not that $$u$$, $$v$$ and $$w$$ are not cut-vertices, but instead that they are the same cut-vertex.