# Qvist's theorem

In projective geometry, Qvist's theorem, named after the Finnish mathematician Bertil Qvist [de], 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:
* Any line l meets Ω in at most two points, and
* 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
### Qvist's theorem

Let Ω be an oval in a finite projective plane of order n.
    (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.

### Proof

(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.
(b) Let s be a secant, s ∩ Ω = {P$_0$, P$_1$} and s= {P$_0$, P$_1$,...,P$_n$}. 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.

### Example in a pappian plane of even order

Using inhomogeneous coordinates over a field K, |K| = n even, the set
    Ω$_1$ = {(x, y) | y = x$^2$} ∪ {(∞)},
the projective closure of the parabola y = x$^2$, is an oval with the point N = (0) as nucleus (see image), i.e., any line y = c, with c ∈ K, is a tangent.

## Definition and property of hyperovals

* Any oval Ω in a finite projective plane of even order n has a nucleus N.
    The point set Ω := Ω ∪ {N} is called a hyperoval or (n + 2)-arc. (A finite oval is an (n + 1)-arc.)
One easily checks the following essential property of a hyperoval:
* For a hyperoval Ω and a point R ∈ Ω the pointset Ω \ {R} is an oval.
This property provides a simple means of constructing additional ovals from a given oval.

### Example

For a projective plane over a finite field K, |K| = n even and n > 4, the set
    Ω$_1$ = {(x, y) | y = x$^2$} ∪ {(∞)} is an oval (conic section) (see image),
    Ω$_1$ = {(x, y) | y = x$^2$} ∪ {(0), (∞)} is a hyperoval and
    Ω$_2$ = {(x, y) | y = x$^2$} ∪ {(0)} is another oval that is not a conic section. (Recall that a conic section is determined uniquely by 5 points.)