Page:American Journal of Mathematics Vol. 2 (1879).pdf/15

always one vertex within the conic and two without, it follows that ${\displaystyle 15}$ ${\displaystyle P}$ points are always within the conic and ${\displaystyle 30}$ without, and that of the ${\displaystyle 45}$ ${\displaystyle p'}$ lines ${\displaystyle 30}$ cut the conic in two real and ${\displaystyle 15}$ in two imaginary points.

It now appears that the lines and points of the Brianchon figure can be produced without considering the Brianchon hexagon. Since the points ${\displaystyle P\,(AB\,.DE),}$ ${\displaystyle P\,(BC\,.EF),}$ ${\displaystyle P\,(CD\,.FA)}$ are on a line, ${\displaystyle h\,(ABCDEF),}$ their poles, ${\displaystyle P\,(AB\,.DE)}$ ${\displaystyle P\,(AB\,.DE),}$ ${\displaystyle P\,(BE\,.FC)}$ ${\displaystyle P\,(BF\,.CE),}$ ${\displaystyle P\,(CF\,.AD)}$ ${\displaystyle P\,(CA\,.DF),}$ meet in a point [the same as the point ${\displaystyle H'\,(abcdef)}$], the pole of ${\displaystyle h\,(ABCDEF).}$ From the ${\displaystyle 60}$ points thus obtained may be produced all the other lines and points of the figure.

3. If ${\displaystyle a,}$ ${\displaystyle b,}$ ${\displaystyle c}$ be the sides of a triangle and ${\displaystyle a',}$ ${\displaystyle b',}$ ${\displaystyle c',}$ ${\displaystyle d'}$ the lines of a quadrilateral such that the triangles ${\displaystyle b'c'd',}$ ${\displaystyle c'd'a',}$ ${\displaystyle d'a'f',}$ ${\displaystyle a'b'c'}$ are homologous with ${\displaystyle abc,}$ their respective axes of homology being ${\displaystyle k_{a},}$ ${\displaystyle k_{b},}$ ${\displaystyle k_{c},}$ ${\displaystyle k_{d},}$ then the intersections of ${\displaystyle k_{a},\,a';}$ ${\displaystyle k_{b},\,b';}$ ${\displaystyle k_{c},\,c';}$ ${\displaystyle k_{d},\,d'}$ are collinear. For, the equations of the axes may be written ${\displaystyle k_{a})\;b+c'=c+b'=a+d'=0,}$${\displaystyle k_{b})\;b+d'=c+a'=a+c'=0,}$${\displaystyle k_{c})\;b+a'=c+d'=a+b'=0,}$${\displaystyle k_{d})\;b+b'=c+c'=a+a'=0,}$ and we shall then have for lines through their respective intersections with sides of the quadrilateral

which form all four one and the same line. The quadrilateral ${\displaystyle k_{a}k_{b}k_{c}k_{d}}$ is also such that its four triangles are each homologous with ${\displaystyle abc,}$ and in fact in such a way that ${\displaystyle k_{a}k_{b}k_{c},}$ ${\displaystyle a'b'c'}$ and ${\displaystyle abc}$ have lines joining all three corresponding vertices coïncident. Take the triangles ${\displaystyle k_{b}k_{c}k_{d}}$ and ${\displaystyle b'c'd';}$ we have ${\displaystyle k_{c}-k_{b}=b-c=b'-c'=0,}$${\displaystyle -k_{c}-k_{d}=-a-b=-d'+c'=0,}$${\displaystyle k_{b}+k_{d}=a+c=d'-b'=0,}$ and the equations show that these lines meet in a point. Let us apply this property to the Pascal hexagram. We shall say, with Veronese, (p. 27), that the triangle formed by joining opposite vertices of a hexagon belongs to the Pascal obtained by taking the vertices in the same order; for instance, the triangle whose sides are ${\displaystyle AD,}$ ${\displaystyle BE,}$ ${\displaystyle CF}$ belongs to the four Pascals ${\displaystyle h\,(AECDBF),}$ ${\displaystyle h\,(AEFDBC),}$ ${\displaystyle h\,(ABFDEC),}$ ${\displaystyle h\,(ABCDEF).}$ The points

${\displaystyle P\,(AB\,.DE),}$	${\displaystyle P\,(AD\,.BE)}$	are on	the line	${\displaystyle p'\,(bd\,.ae);}$
${\displaystyle P\,(BC\,.EF),}$	${\displaystyle P\,(EB\,.FC)}$	"	"	${\displaystyle p'\,(ce\,.bf);}$
${\displaystyle P\,(FD\,.AC),}$	${\displaystyle P\,(AD\,.FC)}$	"	"	${\displaystyle p'\,(af\,.dc).}$