• In a projective plane a set ${\displaystyle {\mathfrak {o}}}$  of points is called oval, if:

(1) Any line ${\displaystyle g}$  meets ${\displaystyle {\mathfrak {o}}}$  in at most two points.

If ${\displaystyle |g\cap {\mathfrak {o}}|=0}$  the line ${\displaystyle g}$  is an exterior (or passing) line; in case ${\displaystyle |g\cap {\mathfrak {o}}|=1}$  a tangent line and if ${\displaystyle |g\cap {\mathfrak {o}}|=2}$  the line is a secant line.

(2) For any point ${\displaystyle P\in {\mathfrak {o}}}$  there exists exactly one tangent ${\displaystyle t}$  at P, i.e., ${\displaystyle t\cap {\mathfrak {o}}=\{P\}}$ .

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 ${\displaystyle {\mathfrak {o}}}$  of points is an oval if and only if ${\displaystyle |{\mathfrak {o}}|=n+1}$  and no three points are collinear (on a common line).

Theorem

Let be ${\displaystyle {\mathfrak {o}}}$  an oval in a pappian projective plane of characteristic ${\displaystyle \neq 2}$ .
${\displaystyle {\mathfrak {o}}}$  is a nondegenerate conic if and only if statement (P3) holds:

(P3): Let be ${\displaystyle P_{1},P_{2},P_{3}}$  any triangle on ${\displaystyle {\mathfrak {o}}}$  and ${\displaystyle {\overline {P_{i}P_{i}}}}$  the tangent at point ${\displaystyle P_{i}}$  to ${\displaystyle {\mathfrak {o}}}$ , then the points

${\displaystyle P_{4}:={\overline {P_{1}P_{1}}}\cap {\overline {P_{2}P_{3}}},\ P_{5}:={\overline {P_{2}P_{2}}}\cap {\overline {P_{1}P_{3}}},\ P_{6}:={\overline {P_{3}P_{3}}}\cap {\overline {P_{1}P_{2}}}}$ 

are collinear.^[1]

Proof

Let the projective plane be coordinatized inhomogeneously over a field ${\displaystyle K}$  such that ${\displaystyle P_{3}=(0),\;g_{\infty }}$  is the tangent at ${\displaystyle P_{3},\ (0,0)\in {\mathfrak {o}}}$ , the x-axis is the tangent at the point ${\displaystyle (0,0)}$  and ${\displaystyle {\mathfrak {o}}}$  contains the point ${\displaystyle (1,1)}$ . Furthermore, we set ${\displaystyle P_{1}=(x_{1},y_{1}),\;P_{2}=(x_{2},y_{2})\ .}$  (s. image)
The oval ${\displaystyle {\mathfrak {o}}}$  can be described by a function ${\displaystyle f:K\mapsto K}$  such that:

${\displaystyle {\mathfrak {o}}=\{(x,y)\in K^{2}\;|\;y=f(x)\}\ \cup \{(\infty )\}\;.}$ 

The tangent at point ${\displaystyle (x_{0},f(x_{0}))}$  will be described using a function ${\displaystyle f'}$  such that its equation is

${\displaystyle y=f'(x_{0})(x-x_{0})+f(x_{0})}$ 

Hence (s. image)

${\displaystyle P_{5}=(x_{1},f'(x_{2})(x_{1}-x_{2})+f(x_{2}))}$  and ${\displaystyle P_{4}=(x_{2},f'(x_{1})(x_{2}-x_{1})+f(x_{1}))\;.}$ 

I: if ${\displaystyle {\mathfrak {o}}}$  is a non degenerate conic we have ${\displaystyle f(x)=x^{2}}$  and ${\displaystyle f'(x)=2x}$  and one calculates easily that ${\displaystyle P_{4},P_{5},P_{6}}$  are collinear.

II: If ${\displaystyle {\mathfrak {o}}}$  is an oval with property (P3), the slope of the line ${\displaystyle {\overline {P_{4}P_{5}}}}$  is equal to the slope of the line ${\displaystyle {\overline {P_{1}P_{2}}}}$ , that means:

${\displaystyle f'(x_{2})+f'(x_{1})-{\frac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}}}={\frac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}}}}$  and hence

(i): ${\displaystyle (f'(x_{2})+f'(x_{1}))(x_{2}-x_{1})=2(f(x_{2})-f(x_{1}))}$  for all ${\displaystyle x_{1},x_{2}\in K}$ .

With ${\displaystyle f(0)=f'(0)=0}$  one gets

(ii): ${\displaystyle f'(x_{2})x_{2}=2f(x_{2})}$  and from ${\displaystyle f(1)=1}$  we get

(iii): ${\displaystyle f'(1)=2\;.}$ 

(i) and (ii) yield

(iv): ${\displaystyle f'(x_{2})x_{1}=f'(x_{1})x_{2}}$  and with (iii) at least we get

(v): ${\displaystyle f'(x_{2})=2x_{2}}$  for all ${\displaystyle x_{2}\in K}$ .

A consequence of (ii) and (v) is

${\displaystyle f(x_{2})=x_{2}^{2},\;x_{2}\in K}$ .

Hence ${\displaystyle {\mathfrak {o}}}$  is a nondegenerate conic.

Remark: Property (P3) is fulfilled for any oval in a pappian projective plane of characteristic 2 with a nucleus (all tangents meet at the nucleus). Hence in this case (P3) is also true for non-conic ovals.^[2]

Theorem

Any oval ${\displaystyle {\mathfrak {o}}}$  in a finite pappian projective plane of odd order is a nondegenerate conic section.

Proof

^[3]

For the proof we show that the oval has property (P3) of the 3-point version of Pascal's theorem.

Let be ${\displaystyle P_{1},P_{2},P_{3}}$  any triangle on ${\displaystyle {\mathfrak {o}}}$  and ${\displaystyle P_{4},P_{5},P_{6}}$  defined as described in (P3). The pappian plane will be coordinatized inhomogeneously over a finite field ${\displaystyle K}$ , such that${\displaystyle P_{3}=(\infty ),\;P_{2}=(0),\;P_{1}=(1,1)}$  and ${\displaystyle (0,0)}$  is the common point of the tangents at ${\displaystyle P_{2}}$  and ${\displaystyle P_{3}}$ . The oval ${\displaystyle {\mathfrak {o}}}$  can be described using a bijective function ${\displaystyle f:K^{*}:=K\cup \setminus \{0\}\mapsto K^{*}}$ :

${\displaystyle {\mathfrak {o}}=\{(x,y)\in K^{2}\;|\;y=f(x),\;x\neq 0\}\;\cup \;\{(0),(\infty )\}\;.}$ 

For a point ${\displaystyle P=(x,y),\;x\in K\setminus \{0,1\}}$ , the expression ${\displaystyle m(x)={\tfrac {f(x)-1}{x-1}}}$  is the slope of the secant ${\displaystyle {\overline {PP_{1}}}\;.}$  Because both the functions ${\displaystyle x\mapsto f(x)-1}$  and ${\displaystyle x\mapsto x-1}$  are bijections from ${\displaystyle K\setminus \{0,1\}}$  to ${\displaystyle K\setminus \{0,-1\}}$ , and ${\displaystyle x\mapsto m(x)}$  a bijection from ${\displaystyle K\setminus \{0,1\}}$  onto ${\displaystyle K\setminus \{0,m_{1}\}}$ , where ${\displaystyle m_{1}}$  is the slope of the tangent at ${\displaystyle P_{1}}$ , for ${\displaystyle K^{**}:=K\setminus \{0,1\}\;:}$  we get

${\displaystyle \prod _{x\in K^{**}}(f(x)-1)=\prod _{x\in K^{**}}(x-1)=1\quad {\text{und}}\quad m_{1}\cdot \prod _{x\in K^{**}}{\frac {f(x)-1}{x-1}}=-1\;.}$ 

(Remark: For ${\displaystyle K^{*}:=K\setminus \{0\}}$  we have: ${\displaystyle \displaystyle \prod _{k\in K^{*}}k=-1\;.}$ )
Hence

${\displaystyle -1=m_{1}\cdot \prod _{x\in K^{**}}{\frac {f(x)-1}{x-1}}=m_{1}\cdot {\frac {\displaystyle \prod _{x\in K^{**}}(f(x)-1)}{\displaystyle \prod _{x\in K^{**}}(x-1)}}=m_{1}\;.}$ 

Because the slopes of line ${\displaystyle {\overline {P_{5}P_{6}}}}$  and tangent ${\displaystyle {\overline {P_{1}P_{1}}}}$  both are ${\displaystyle -1}$ , it follows that ${\displaystyle {\overline {P_{1}P_{1}}}\cap {\overline {P_{2}P_{3}}}=P_{4}\in {\overline {P_{5}P_{6}}}}$ . This is true for any triangle ${\displaystyle P_{1},P_{2},P_{3}\in {\mathfrak {o}}}$ .

So: (P3) of the 3-point Pascal theorem holds and the oval is a non degenerate conic.

• B. Segre: Ovals in a finite projective plane, Canadian Journal of Mathematics 7 (1955), pp. 414–416.
• G. Järnefelt & P. Kustaanheimo: An observation on finite Geometries, Den 11 te Skandinaviske Matematikerkongress, Trondheim (1949), pp. 166–182.
• Albrecht Beutelspacher, Ute Rosenbaum: Projektive Geometrie. 2. Auflage. Vieweg, Wiesbaden 2004, ISBN 3-528-17241-X, p. 162.
• P. Dembowski: Finite Geometries. Springer-Verlag, 1968, ISBN 3-540-61786-8, p. 149

• Simeon Ball and Zsuzsa Weiner: An Introduction to Finite Geometry [1]