Page:American Journal of Mathematics Vol. 2 (1879).pdf/10

viz., through $P\,(EB\,.FA)$ pass $v_{12}\,(EF\,.AB)$ and $v_{12}\,(EA\,.FB).$ There is but one $P$ point on each $v_{12}$ line. Through each $H$ point pass three $v_{12}$ lines; through $H\,(BCDEFA)$ pass $v_{12}\,(BC\,.EF),$ $v_{12}\,(CD\,.FA),$ $v_{12}\,(DE\,.AB).$ There are therefore ${\tfrac {3.60}{2}}$ or $90$ $v_{12}$ lines in all. If we look for the corresponding property of $h$ lines, we find that

$h\,(BFCEAD),$$h\,(BFDEAC),$

intersect in $P\,(BF\,.EA),$ and that

$h\,(BFCAED),$$h\,(BFDAEC),$

intersect in $P\,(BF\,.AE),$ but that $P\,(BF\,.AE)$ is the same point as $P\,(BF\,.EA).$ This is the intrinsic difference between $H$ points and $h$ lines. The $H$ points lie in twos on $90$ lines $v_{12}$ which pass by threes through the $60$ $H$ points. The $h$ lines intersect in fours in $45$ points $P,$ which lie in threes on the $60$ $h$ lines. To a $P$ point, $P\,(BF\,.AE),$ may be said to correspond the pair of $v_{12}$ lines, $v_{12}\,(BF\,.AE),$ $v_{12}\,(BF\,.EA).$ In the Brianchon hexagon, on the other hand, the $H'$ points lie in fours in the $45$ $p'$ lines, and the $h'$ lines intersect in twos in $90$ points $V'_{12},$ which lie in threes on $h'$ llines and in twos on $p'$ lines. Not even in a hexagon which can be inscribed in one conic and circumscribed about another is there entire correspondence between Kirkman points and Pascal lines.

To resume:

To	$60$	Pascal	lines	$h$	correspond	$60$	Kirkman	points	$H.$
"	$20$	Cayley-Salmon	"	$g$	"	$20$	Steiner	"	$G.$
"	$15$	Steiner-Plücker	"	$i$	"	$15$	Salmon	"	$I.$

On	each	$h$	line	lie three $H$'s, and one $G.$
"	"	$g$	"	lie three $H$'s, three $I$'s and one $G.$
"	"	$i$	"	lie four $G$'s.

Through	each	$H$	point	pass three $h$'s, and one $g.$
"	"	$G$	"	pass three $h$'s, three $i$'s and one $g.$
"	"	$I$	"	pass four $g$'s.

The whole arrangement can be diagrammatically represented by a simple figure: