Published online by Cambridge University Press: 24 October 2008
The theorem of Wallace that if four arbitrary lines he taken in a plane the circumcircles of the four triangles formed by them in threes meet in a point—in which for circles we may equally well take conies drawn through two arbitrary points I, J of the plane —may be proved very easily from three dimensions. Through the four given lines draw four arbitrary planes, forming a tetrahedron X, Y, Z, T. A unique cubic curve can be drawn through I, J and the four vertices. Projecting this curve from the point T upon the original plane, we get a conic which passes through I, J and the vertices of the triangle formed by the lines in which the plane meets the planes TYZ, TZX, TXY, that is, by three.of the given lines. This conic, which is one of the conics of the theorem, also passes through the third point K, other than I, J, in which the original plane meets the cubic curve. Projecting in turn from the other three vertices of the tetrahedron, we get the other three conics of the theorem, which also pass through K.
* Given by Wallace, W. (“Scoticus”) in 1804Google Scholar in Leybourn's, Mathematical Repositor:New Series, p. 22Google Scholar: see Mackay, , Proc. Edin. Math. Soc. 9 (1891), p. 87.Google Scholar
† Leipzig Berichte, 65 (1914), p. 329.Google Scholar The method of proof given here was suggested by Prof. H. F. Baker, whose attention I had drawn to Meyer's theorem.
‡ Mathematical Papers, pp. 38–54: “Synthetic Proof of Miguel's Theorem.” The date of the paper is 1870.Google Scholar
* See Segre, in Encyk. Math. Wiss., III. c. 7, p. 894, footnote 368.Google Scholar
* Compare the statement (for n = 2) in Clifford, , Common Sense of the Exact Sciences (1885), p. 80.Google Scholar