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

Rh 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 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

which form all four one and the same line. The quadrilateral $$k_ak_bk_ck_d$$ is also such that its four triangles are each homologous with $$abc,$$ and in fact in such a way that $$k_ak_bk_c,$$ $$a'b'c'$$ and $$abc$$ have lines joining all three corresponding vertices coïncident. Take the triangles $$k_bk_ck_d$$ 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