# Concurrent bitangents of a quartic curve

What is the maximum number of concurrent bitangents, i.e. all intersecting at the same point, of a smooth complex projective quartic curve? Can the number of concurrent bitangents be six?

• Can the number of concurrent bitangents be six?
– bog
Sep 20, 2017 at 13:03
• OK. Edited the question to make it clearer.
– R.P.
Sep 20, 2017 at 13:09

I think the maximum number of concurrent bitangents is at most 4. Consider the double cover $S\rightarrow \mathbb{P}^2$ branched along your quartic curve. Fix one point $q$ of $S$ above the point of intersection; each bitangent lift to a "line" in $S$ passing through $q$. The description of these lines is well-known. We can represent $S$ as the blown-up of $\mathbb{P}^2$ along 7 points $p_1,\ldots ,p_7$, and assume that one of the lines is the exceptional divisor above $p_1$. Since any pair of our "lines" in $S$ must meet, we don't have much choice: we can have the strict transform of the line in $\mathbb{P}^2$ joining $p_1$ to another point, say $p_2$, then the strict transform of the conic through $p_1$ and 4 other points $\neq p_2$, say $p_3,\ldots ,p_6$, and finally the the strict transform of the cubic passing through all the $p_i$ and doubly through $p_7$.
So the maximum is $\leq 4$; I do not know if 4 can be actually realized. $3$ is easy, by projecting a cubic surface with 3 concurrent lines.