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Martin Sleziak
Martin Sleziak
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(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 6.22Exercise 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."

(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."

(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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Pooter
Pooter
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(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."