Page:American Journal of Mathematics Vol. 2 (1879).pdf/9

Salmon line, the line $g\,(ACE\,.BFD).$ It passes through the point $G\,(ACE\,.BDF).$ Through the two conjugate points $G\,(ACE\,.BDF),$ $G\,(ACE\,.BFD),$ pass respectively the lines $g\,(ACE\,.BFD),$ $g\,(ACE\,.BDF).$ These two $g$ lines we shall call, for the present, corresponding $g$ lines. They are not conjugate with respect to the conic $S.$ (Veronese, Nuovi Teoremi sul Hexagrammum Mysticum, p. 26.) The $H$ points on $G\,(ACE\,.BDF)$ correspond to the $h$ lines through $g\,(ACE\,.BDF);$ hence we shall say that $g\,(ACE\,.BDF)$ corresponds to $G\,(ACE\,.BDF),$ while it passes through $G\,(ACE\,.BFD).$

The symbol for the Salmon point in which four $g$ lines intersect is obtained in the same way as that of the Steiner-Plücker line through four $G$ points. In fact, the lines $g\,(BDA\,.ECF),$ $g\,(EDF\,.BCA),$ $g\,(BCF\,.EDA),$ $g\,(BDF\,.ECA),$ intersect in the Salmon point $I\,(BE\,.CD\,.AF);$ and the $I$ points on $g\,(ACE\,.BDF),$ are $I\,(AB\,.CD\,.EF),$ $I\,(AD\,.CF\,.EB),$ $I\,(AF\,.CB\,.ED).$

Professor Cayley (Quarterly Journal, Vol. IX,) gives a table to show in what kind of a point each Pascal line meets every one of the $59$ other Pascal lines. By attending to the notation of Pascal lines such a table may be dispensed with. His $90$ points, "$m,$" $360$ points "$r,$" 360 points "$t,$" $360$ points "$z,$" and $90$ points "$w$" are the intersections each of two Pascals whose symbols can easily be derived one from another. For instance,

${h\,(ABCDEF)} \atop {h\,(ABCFED)}$ $>$ "$m,$"	${h\,(DEFABC)} \atop {h\,(DEFBCA)}$ $>$ "$r,$"	${h\,(ACEDBF)} \atop {h\,(ABCDEF)}$ $>$ "$t,$"
${h\,(ACFBED)} \atop {h\,(AFEDCB)}$ $>$ "$z,$"		${h\,(ACEBFD)} \atop {h\,(ABCDEF)}$ $>$ "$w.$"

By producing the lines and points of the Brianchon hexagon, as we may call the corresponding circumscribed hexagon, we should find occasion for the same symbols, in small letters, for the $H',\,G',\,I'$ points, which are the poles of the $h,\,g,\,i$ lines, and for the $h',\,g',\,i'$ lines, which are the poles of the $H,\,G,\,I$ points.

It was shown by Kirkman that the two Kirkman points

$H\,(BFCEAD),$$H\,(BFDEAC),$

are on a line through the point $P\,(AB\,.FE).$ I shall call this line $v_{12}\,(BF\,.EA)$ (and it happens that my notation here coïncides with that of Veronese, p. 43). So the points

$H\,(BFCAED),$$H\,(BFDAEC),$

are on the line $v_{12}\,(BF\,.AE),$ which passes through $P\,(EB\,.FA)$ and which does not coïncide with $v_{12}\,(BF\,.EA).$ Through each point $P$ pass two $v_{12}$ lines,