Let $\Pi$ be a projective plane of even order $2n$. Let $\mathcal{H}_1,\mathcal{H}_2$ be two hyperovals in $\Pi$. Show that if $|\mathcal{H}_1 \cap \mathcal{H}_2|> n+1$ then $\mathcal{H}_1 =\mathcal{H}_2$.
We know that in a hyperoval no three points are collinear, and there are no tangent lines, namely any line intersects a hyperoval in $2$ points or it is disjoint from it. Also, a hyperoval contains $2n+2$ points. Suppose $|\mathcal{H}_1 \cap \mathcal{H}_2|> n+1$. We look at $\mathcal{H}_1 \setminus(\mathcal{H}_1 \cap \mathcal{H}_2)$. Denote $|\mathcal{H}_1 \setminus(\mathcal{H}_1 \cap \mathcal{H}_2)|=x$. Then we have $$2n+2 =|\mathcal{H}_1|=|\mathcal{H}_1 \setminus(\mathcal{H}_1 \cap \mathcal{H}_2)|+ |\mathcal{H}_1 \cap \mathcal{H}_2|> x + n+1$$ so $x < n+1$. I am stuck at this point, any ideas?
Consider the points that are not in $\mathcal{H}_{1} \cap \mathcal{H}_{2}$. If such a point $P$ is in $\mathcal{H}_{1} \cup \mathcal{H}_{2}$, then any line on $P$ meets at most one point of $\mathcal{H}_{1} \cap \mathcal{H}_{2}$; on the other hand, if $P \not\in \mathcal{H}_{1} \cup \mathcal{H}_{2}$, then $P$ lies on $n+1$ secant lines to $\mathcal{H}_{1}$, and so at least one of these secant lines must meet $\mathcal{H}_{1} \cap \mathcal{H}_{2}$ in two points (since $P$ must be joined to every point of $\mathcal{H}_{1} \cap \mathcal{H}_{2}$, which has more than $n+1$ points, by an $\mathcal{H}_{1}$-secant). By a similar argument, this same property holds for $\mathcal{H}_{2}$.
Thus if we put $$\mathcal{S} = \{ P \ : \ P \in (\mathcal{H}_{1} \cap \mathcal{H}_{2})^{c} \mid \mbox{ every line on $P$ meets $\mathcal{H}_{1} \cap \mathcal{H}_{2}$ in at most one point}\}$$ then we can see that $$\mathcal{H}_{1} = \mathcal{H}_{2} = (\mathcal{H}_{1} \cap \mathcal{H}_{2}) \cup \mathcal{S}.$$