Return to Answer

No, this is not always possible. Consider the complete graph $K_{7}$. If $K_{7}$ can be drawn so that all crossings meet at a single point, then there is a planar graph with $8$ vertices and at least $\binom{7}{2}=21$ edges. But this contradicts the fact that every planar graph $G$ has at most $3|V(G)|-6$ edges.

This argument actually shows that every graph $G$ with at least $3|V(G)|-2$ edges can never be drawn so that all crossings occur at the same point.

No, this is not always possible.

Lemma. Let $G$ be an $n$-vertex graph with at least $3n-2$ edges. Then $G$ cannot be drawn in the plane so that all crossings occur at the same point.

Proof. We make the standard assumption that every pair of edges which intersect in a drawing are not 'tangent' at the point of intersection. Suppose $D$ is a drawing of $G$ where all edges cross at the same point. Let $G'$ be the graph obtained from $D$ by introducing a new vertex at the crossing point and then suppressing all parallel edges. We claim that $|E(G')| \geq |E(G)|$. Let $H$ be the subgraph of $G$ induced by the edges which pass through the crossing point. Since no two crossing edges are tangent, $H$ contains at most one cycle (which must be a triangle). Therefore, $H$ has average degree at most $2$. It follows that $|E(G')| \geq |E(G)|$, as claimed. Thus, $G'$ is a planar graph with $n+1$ vertices and at least $3n-2$ edges. But this contradicts the fact that every $n$-vertex planar graph has at most $3n-6$ edges.

In particular, the above lemma implies that $K_7$ cannot be drawn so that all edges cross at the same point.

Acknowledgement. Many thanks to bof for help in making this answer converge to its present form (see the comments below).

No, this is not always possible. Consider the complete graph $K_{6}$. If $K_{6}$ can be drawn so that all crossings meet at a single point, then there is a planar graph with $7$ vertices and at least $\binom{6}{2}+1=16$ edges. But this contradicts the fact that every planar graph $G$ has at most $3|V(G)|-6$ edges.

This argument actually shows that every graph $G$ with at least $3|V(G)|-3$ edges can never be drawn so that all crossings occur at the same point.

No, this is not always possible. Consider the complete graph $K_{7}$. If $K_{7}$ can be drawn so that all crossings meet at a single point, then there is a planar graph with $8$ vertices and at least $\binom{7}{2}=21$ edges. But this contradicts the fact that every planar graph $G$ has at most $3|V(G)|-6$ edges.

This argument actually shows that every graph $G$ with at least $3|V(G)|-2$ edges can never be drawn so that all crossings occur at the same point.

No, this is not always possible. Consider the complete graph $K_{6}$. If $K_{6}$ can be drawn so that all crossings meet at a single point, then there is a planar graph with $7$ vertices and at least $\binom{6}{2}+2=17$ edges. But this contradicts the fact that every planar graph $G$ has at most $3|V(G)|-6$ edges.

This argument actually shows that every graph $G$ with at least $3|V(G)|-4$ edges can never be drawn so that all crossings occur at the same point.

No, this is not always possible. Consider the complete graph $K_{6}$. If $K_{6}$ can be drawn so that all crossings meet at a single point, then there is a planar graph with $7$ vertices and at least $\binom{6}{2}+1=16$ edges. But this contradicts the fact that every planar graph $G$ has at most $3|V(G)|-6$ edges.

This argument actually shows that every graph $G$ with at least $3|V(G)|-3$ edges can never be drawn so that all crossings occur at the same point.