Why is maximum number of joints of 6 lines is 4?

Question

The following is considered in Lary Guth's Polynomial Methods in Combinatorics, page 14. Let $L$ be a set of lines in $\mathbb R^3$. A point $x$ which lies in some set of three non-co-planar lines of $L$ is called a joint of $L$. Suppose $L$ has $6$ lines. Then, why is it that $L$ has at most $4$ joints?

This has been my approach so far:

1. Note that the tetrahedron has 6 edges and 4 vertices. If we take our $L$ to be the set of 6 lines containing each of the six edges of a tetrahedron, we get that each of its vertex is a joint as the three lines intersecting any vertex are non-co-planar.

2. Now, I want to argue that if one wants to maximize the number of joints possible for any set of six lines, the configuration of a tetrahedron is the best possible one.

The problem is that I don't know why or how to prove 3. Any suggestions will be really helpful :)

Answer 1

Let $G = (V,E)$ be a graph, and we're interested in the vertices which fulfill $deg(v) \geq 3$. We notice that $|E| = 6$ from the assignment, and we can use your earlier fact that a tetrahedron is a lower bound.

By the handshaking lemma, $\sum_{v \in V} deg(v) = 2 |E| = 12$. This is immediately fulfilled by $4$ joints. Therefore, a tetrahedron is an upper bound as well.

Answer 2

Note that $2$ intersecting lines define a plane. Thus, a joint is formed by $3$ intersecting planes. Let's add one more plane to these three. Note that $4$ planes have at most $6$ distinct lines as intersections so $4$ is the minimum number of planes to contain $6$ intersecting lines that satisfy the requirement that no three of them are co-planar. Finally, there are $4$ ways to pick $3$ planes out of the set of $4$ so we can have at most $4$ joints. Tetrahedron is the shape with $4$ planes and $6$ lines.