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?

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

2 Answers 2


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.


(This is really a comment to the good answer of abx, but I don't have the reputation for that.) Indeed 4 concurrent bitangents seems to be realisable, for example by the Klein quartic (what else?).

In the book Classical Algebraic Geometry by Dolgachev you will find Exercise 6.22 (in the version I have, at least) which reads:

"Show that the set of 28 bitangents of the Klein quartic contains 21 subsets of four concurrent bitangents and each bitangent has 3 concurrency points."


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