부산대학교 도서관

A projective oval is defined as any nonempty set ${\bf K}$ of points of aprojective plane, $\pi$ say, no three of which are collinear and throughevery one of which there is a unique line (the tangent) containing no otherpoint of ${\bf K}$. Points of such a plane can then be divided intoexterior points (two tangents meeting) or interior points (no tangentsmeeting). If $A_i$, $B_i\ (i=1,2,3)$ are six not necessarily distinctpoints of an oval ${\bf K}$ such that $A_iB_j\neq A_jB_i$ for $i\neqj\ (i,j=1,2,3)$, then the sextuple $(A_1,A_2,A_3;B_1,B_2,B_3)$ is a hexagoninscribed in ${\bf K}$. For $i\neq j\ (i,j=1,2,3)$ the line $A_iB_j$ is adiagonal and the point $A_iB_j\cap A_jB_i$ is a diagonal point of thehexagon. 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 planeover some commutative field and ${\bf K}$ is a conic in $\pi$. In thispaper the following extension of this result is proved. Suppose that thefinite projective plane $\pi$ of order $n$, $n$ odd, contains an oval ${\bfK}$ and that every hexagon ${\bf E}$ inscribed in ${\bf K}$ satisfies thefollowing requirement: Let $R$, $S$, $T$ be the diagonal points of ${\bfE}$ and let $d$ be a diagonal of $E$ through $T$. If $R$, $S$ and $d\capRS$ are exterior points, then $T=d\cap RS$. Then $\pi$ is the Desarguesianplane over some commutative field, and ${\bf K}$ is a conic in $\pi$.