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

4 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

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

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:

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