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

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

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

intersect in ${\displaystyle P\,(BF\,.AE),}$ but that ${\displaystyle P\,(BF\,.AE)}$ is the same point as ${\displaystyle P\,(BF\,.EA).}$ This is the intrinsic difference between ${\displaystyle H}$ points and ${\displaystyle h}$ lines. The ${\displaystyle H}$ points lie in twos on ${\displaystyle 90}$ lines ${\displaystyle v_{12}}$ which pass by threes through the ${\displaystyle 60}$ ${\displaystyle H}$ points. The ${\displaystyle h}$ lines intersect in fours in ${\displaystyle 45}$ points ${\displaystyle P,}$ which lie in threes on the ${\displaystyle 60}$ ${\displaystyle h}$ lines. To a ${\displaystyle P}$ point, ${\displaystyle P\,(BF\,.AE),}$ may be said to correspond the pair of ${\displaystyle v_{12}}$ lines, ${\displaystyle v_{12}\,(BF\,.AE),}$ ${\displaystyle v_{12}\,(BF\,.EA).}$ In the Brianchon hexagon, on the other hand, the ${\displaystyle H'}$ points lie in fours in the ${\displaystyle 45}$ ${\displaystyle p'}$ lines, and the ${\displaystyle h'}$ lines intersect in twos in ${\displaystyle 90}$ points ${\displaystyle V'_{12},}$ which lie in threes on ${\displaystyle h'}$ llines and in twos on ${\displaystyle p'}$ 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	${\displaystyle 60}$	Pascal	lines	${\displaystyle h}$	correspond	${\displaystyle 60}$	Kirkman	points	${\displaystyle H.}$
"	${\displaystyle 20}$	Cayley-Salmon	"	${\displaystyle g}$	"	${\displaystyle 20}$	Steiner	"	${\displaystyle G.}$
"	${\displaystyle 15}$	Steiner-Plücker	"	${\displaystyle i}$	"	${\displaystyle 15}$	Salmon	"	${\displaystyle I.}$

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

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

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