bitangent of smooth quartic plane curve Suppose $C \subset \mathbb{P}^2$ is a smooth plane quartic curve over an algebraically closed field, where $C$ is the zero set of a homogeneous polynomial of degree 4.
What is the quickest way to prove that it has at least one bitangent?
One way to prove it is to prove the Plucker formulas, which immediately give the exact number 28. But if I want to prove just the existence of one bitangent, is there a shortcut possible?
 A: I'm not sure about the quickest, but you can give a very elementary argument using an incidence correspondence. (If you've seen the proof e.g. in Shafarevich that every cubic surface contains at least 1 line, this one is exactly analogous.)
Let $\mathbf P^{14}$ be the space of plane quartics, and $(\mathbf P^2)^\ast$ the space of lines in the plane. We consider the incidence correspondence
$$ I = \left\{ (Q,L) \mid L \text{ is bitangent to } Q \right\} \subset \mathbf P^{14} \times (\mathbf P^2)^\ast$$.
We want to prove the first projection $\pi_1$ is surjective. For that it is enough to show 
(i) $I$ is irreducible of dimension 14, and
(ii) the general fibre of $\pi_1$ has dimension 0; equivalently, some fibre of $\pi_1$ has dimension 0.
For (i), look at the second projection $\pi_2: I \rightarrow (\mathbf P^2)^\ast$. The fibre over a point $[L]$ is the set of all quartics bitangent to the line $L$. It's easy to see that in the space $k^5$ of binary quartic polynomials, the subset of polynomials of the form $q(x,y)^2$ for $q$ a quadratic is irreducible of codimension 2. Dividing by scalars, that proves (i).
For (ii), just choose your favourite smooth quartic and find all its bitangents.  
A: One way is to use the fact that a smooth projective curve of $C$ genus $g$ has exactly $2^{g-1}(2^g-1)$ odd theta-characteristics (i.e., divisor classes $\eta$ such that $2\eta = K_C$ and $\dim H^0(C,\eta)$ is odd). Any odd theta-characteristic can be represented by an effective divisor. When $C$ is a plane quartic (so $g = 3$) such a divisor has form $P_1 + P_2$, and the equality 
$$2(P_1 + P_2) = K_C$$ 
means that the line through $P_1$ and $P_2$ is a bitangent.
Another way is to observe that the double covering $X$ of $\mathbb{P}^2$ branched along $C$ is a del Pezzo surface, hence has many (in fact 56) lines (i.e., curves $L$ such that $(-K_X)\cdot L = 1$), and the image of any such curve in $\mathbb{P}^2$ is a bitangent.
