부산대학교 도서관

A projective oval is defined as any nonempty set ${\bf K}$ of points of a projective plane, $\pi$ say, no three of which are collinear and through every one of which there is a unique line (the tangent) containing no other point of ${\bf K}$. Points of such a plane can then be divided into exterior points (two tangents meeting) or interior points (no tangents meeting). If $A_i$, $B_i\ (i=1,2,3)$ are six not necessarily distinct points of an oval ${\bf K}$ such that $A_iB_j\neq A_jB_i$ for $i\neq j\ (i,j=1,2,3)$, then the sextuple $(A_1,A_2,A_3;B_1,B_2,B_3)$ is a hexagon inscribed in ${\bf K}$. For $i\neq j\ (i,j=1,2,3)$ the line $A_iB_j$ is a diagonal and the point $A_iB_j\cap A_jB_i$ is a diagonal point of the hexagon. A beautiful theorem of F. Buekenhout [Arch. Math. (Basel) {\bf 17} (1966), 89--93; MR0190840 (32 \#8250)] says that if every hexagon inscribed in ${\bf K}$ satisfies Pascal's theorem, then $\pi$ is the Desarguesian plane over some commutative field and ${\bf K}$ is a conic in $\pi$. In this paper the following extension of this result is proved. Suppose that the finite projective plane $\pi$ of order $n$, $n$ odd, contains an oval ${\bf K}$ and that every hexagon ${\bf E}$ inscribed in ${\bf K}$ satisfies the following requirement: Let $R$, $S$, $T$ be the diagonal points of ${\bf E}$ and let $d$ be a diagonal of $E$ through $T$. If $R$, $S$ and $d\cap RS$ are exterior points, then $T=d\cap RS$. Then $\pi$ is the Desarguesian plane over some commutative field, and ${\bf K}$ is a conic in $\pi$.