Qvist's theorem

In projective geometry, Qvist's theorem, named after the Finnish mathematician Bertil Qvist, is a statement on ovals in finite projective planes. Standard examples of ovals are non-degenerate (projective) conic sections. The theorem gives an answer to the question How many tangents to an oval can pass through a point in a finite projective plane? The answer depends essentially upon the order (number of points on a line −1) of the plane.

Definition of an oval

 * In a projective plane a set $Ω$ of points is called an oval, if:
 * 1) Any line $l$ meets $Ω$ in at most two points, and
 * 2) For any point $P ∈ Ω$ there exists exactly one tangent line $t$ through $P$, i.e., $t ∩ Ω = {P}$.

When $|l ∩ Ω | = 0$ the line $l$ is an exterior line (or passant), if $|l ∩ Ω| = 1$ a tangent line and if $|l ∩ Ω| = 2$ the line is a secant line.

For finite planes (i.e. the set of points is finite) we have a more convenient characterization:
 * For a finite projective plane of order $n$ (i.e. any line contains $n + 1$ points) a set $Ω$ of points is an oval if and only if $|Ω| = n + 1$ and no three points are collinear (on a common line).

Statement and proof of Qvist's theorem
Let $Ω$ be an oval in a finite projective plane of order $n$.
 * Qvist's theorem
 * (a) If $n$ is odd,
 * every point $P ∉ Ω$ is incident with 0 or 2 tangents.
 * (b) If $n$ is even,
 * there exists a point $N$, the nucleus or knot, such that, the set of tangents to oval $Ω$ is the pencil of all lines through $N$.

(a) Let $t_{R}$ be the tangent to $Ω$ at point $R$ and let $P_{1}, ..., P_{n}$ be the remaining points of this line. For each $i$, the lines through $P_{i}$ partition $Ω$ into sets of cardinality 2 or 1 or 0. Since the number $|Ω| = n + 1$ is even, for any point $P_{i}$, there must exist at least one more tangent through that point. The total number of tangents is $n + 1$, hence, there are exactly two tangents through each $P_{i}$, $t_{R}$ and one other. Thus, for any point $P$ not in oval $Ω$, if $P$ is on any tangent to $Ω$ it is on exactly two tangents.
 * Proof:

(b) Let $s$ be a secant, $s ∩ Ω = {P_{0}, P_{1}}|undefined$ and $s= {P_{0}, P_{1},...,P_{n}}|undefined$. Because $|Ω| = n + 1$ is odd, through any $P_{i}, i = 2,...,n$, there passes at least one tangent $t_{i}$. The total number of tangents is $n + 1$. Hence, through any point $P_{i}$ for $i = 2,...,n$ there is exactly one tangent. If $N$ is the point of intersection of two tangents, no secant can pass through $N$. Because $n + 1$, the number of tangents, is also the number of lines through any point, any line through $N$ is a tangent.

Using inhomogeneous coordinates over a field $K, |K| = n$ even, the set
 * Example in a pappian plane of even order:

the projective closure of the parabola $Ω_{1} = {(x, y) | y = x^{2}} ∪ {(∞)}|undefined$, is an oval with the point $y = x^{2}$ as nucleus (see image), i.e., any line $N = (0)$, with $y = c$, is a tangent.

Definition and property of hyperovals

 * Any oval $c ∈ K$ in a finite projective plane of even order $n$ has a nucleus $N$.


 * The point set $Ω$ is called a hyperoval or ($\overline{Ω} := Ω ∪ {N}$)-arc. (A finite oval is an ($n + 2$)-arc.)

One easily checks the following essential property of a hyperoval:
 * For a hyperoval $n + 1$ and a point $\overline{Ω}$ the pointset $R ∈ \overline{Ω}$ is an oval.

This property provides a simple means of constructing additional ovals from a given oval.

For a projective plane over a finite field $\overline{Ω} \ {R}$ even and $Ω_{1}$, the set
 * Example:
 * $K, |K| = n$ is an oval (conic section) (see image),
 * $n > 4$ is a hyperoval and
 * $Ω_{1} = {(x, y) | y = x^{2}} ∪ {(∞)}|undefined$ is another oval that is not a conic section. (Recall that a conic section is determined uniquely by 5 points.)