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

always one vertex within the conic and two without, it follows that $15$ $P$ points are always within the conic and $30$ without, and that of the $45$ $p'$ lines $30$ cut the conic in two real and $15$ 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 $P\,(AB\,.DE),$ $P\,(BC\,.EF),$ $P\,(CD\,.FA)$ are on a line, $h\,(ABCDEF),$ their poles, $P\,(AB\,.DE)$ $P\,(AB\,.DE),$ $P\,(BE\,.FC)$ $P\,(BF\,.CE),$ $P\,(CF\,.AD)$ $P\,(CA\,.DF),$ meet in a point [the same as the point $H'\,(abcdef)$], the pole of $h\,(ABCDEF).$ From the $60$ points thus obtained may be produced all the other lines and points of the figure.

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

$k_{a})\;b+c'+a'=c+b'+a'=a+d'+a'=0,$
$k_{b})\;b+d'+b'=c+a'+b'=a+c'+b'=0,$
$k_{c})\;b+a'+c'=c+d'+c'=a+b'+c'=0,$
$k_{d})\;b+b'+d'=c+c'+d'=a+a'+d'=0,$

which form all four one and the same line. The quadrilateral $k_{a}k_{b}k_{c}k_{d}$ is also such that its four triangles are each homologous with $abc,$ and in fact in such a way that $k_{a}k_{b}k_{c},$ $a'b'c'$ and $abc$ have lines joining all three corresponding vertices coïncident. Take the triangles $k_{b}k_{c}k_{d}$ and $b'c'd';$ we have $k_{c}-k_{b}=b-c=b'-c'=0,$$-k_{c}-k_{d}=-a-b=-d'+c'=0,$$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 $AD,$ $BE,$ $CF$ belongs to the four Pascals $h\,(AECDBF),$ $h\,(AEFDBC),$ $h\,(ABFDEC),$ $h\,(ABCDEF).$ The points

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