Page:American Journal of Mathematics Vol. 2 (1879).pdf/16

10 These three $$p'$$ lines meet in a point, namely, the Brianchom point $$H'\,(aecdbf),$$ hence the two triangles formed by the vertical rows of $$P$$ points are homologous. The sides of the first are the Pascals $$h\,(ABFDEC),$$ $$h\,(ABCDEF),$$ $$h\,(CAEFDB),$$ and the corresponding sides of the second are $$AD,$$ $$BE,$$ $$CF,$$ hence these three pairs of sides intersect in a line which, as it is an axis of homology corresponding to the centre of homology $$H'\,(aecdbf),$$ we shall call the line $$k\,(AECDBF).$$ In the same way it may be shown that the triangle formed by any three of the four Pascals to which the triangle $$AD,$$ $$BE,$$ $$CF$$ belongs are homologous therewith, therefore the intersections of the four axes of homology, $$k\,(AECDBF),$$ $$k\,(AEFDBC),$$ $$k\,(ABFDEC),$$ $$k\,(ABCDEF)$$ with the four Pascal lines $$h\,(AECDBF),$$ $$h\,(AEFDBC),$$ $$h\,(ABFDEC),$$ $$h\,(ABCDEF)$$ respectively, are four points on one straight line. As this line is obtained by means of the triangle $$AD,$$ $$BE,$$ $$CF,$$ we shall call it the line $$l\,(AD\,.BE\,.CF).$$ To each triangle formed by three fundamental lines, no two of which pass through the same point on the conic, corresponds a line $$l$$ of the same notation; there are $$15$$ such triangles, hence there are $$15$$ lines $$l.$$ To each $$H'$$ point corresponds a $$k$$ line, hence there are $$60$$ lines $$k,$$ divided into $$15$$ groups of four each, which intersect corresponding $$h$$ lines on the $$15$$ $$l$$ lines.

4. The triangles $$ABC,$$ $$abc,$$ are homologous. Let us call their centre of homology $$C\,(ABC\,.abc),$$ their axis $$a\,(ABC\,.abc).$$ Let us say that the points $$C\,(ABC\,.abc),$$ $$C\,(ADC\,.adc)$$ are joined by the line $$c\,(\overline{ac}\,.bd)$$ and that the lines $$a\,(ABC\,.abc),$$ $$a\,(ADC\,.adc)$$ intersect in $$A\,(\overline{ac}\,.bd),$$ where the bar is drawn over the letters that are repeated. I have shown (Educational Times, Question 5698,) that $$c\,(\overline{ac}\,.bd),$$ $$c\,(ac\,.\overline{bd})$$ intersect in $$P\,(AC\,.BD),$$ and that $$A\,(\overline{ac}\,.bd),$$ $$A\,(ac\,.\overline{bd})$$ are connected by $$p'\,(ac\,.bd).$$ There are $$20$$ points $$C$$ and $$20$$ lines $$a.$$ Each $$C$$ point is joined to $$9$$ other $$C$$ points by $$c$$ lines, hence there are $$\tfrac{1}{2}(9\,.\,20)=90$$ lines $$c,$$ which pass by twos through the $$45$$ points $$P,$$ and $$90$$ points $$A$$ which lie in twos on the $$45$$ lines $$p'.$$ The six $$c$$ lines intersect in pairs in three points on one straight line, viz., the $$P$$ points on $$h\,(ACEBDF),$$ hence they form the sides of a Pascal hexagon; and for a similar reason the six $$A$$ points of the same notation are the vertices of a Brianchon hexagon.

5. The Brianchon hexagon formed by joining alternate vertices of $$ABCDEF$$ has for its sides $$AC,$$ $$BD,$$ $$CE,$$ $$DF,$$ $$EA,$$ $$FB.$$ The conic inscribed