Trilinears f(a,b,c) : f(b,c,a) : f(c,a,b), where
f(a,b,c) = p(a,b,c)y(a,b,c)/a, polynomials p and y as given below
Barycentrics g(a,b,c) : g(b,c,a) : g(c,a,b), where
g(a,b,c) = p(a,b,c)y(a,b,c), polynomials p and y as given below
In A History of Mathematics, Florian Cajori wrote, "H. C. Gossard of the University of Oklahoma showed in 1916 that the three Euler lines of the triangles formed by the Euler line and the sides, taken by twos, of a given triangle, form a triangle . . . perspective with the given triangle and having the same Euler line." Let ABC be the given triangle and A'B'C' the Gossard triangle - that is, the triangle perspective with the given triangle and having the same Euler line. The lines AA', BB', CC' concur in X(402), named the Gosssard perspector by John Conway (1998).
Actually, X(402) dates back to an article by Christopher Zeeman in Wiskundige Opgaven 8 (1899-1902) 305. For details, see Paul Yiu's Hyacinthos message #7536 and others with Gossard in the subject line. (In ETC, the change of name from Gossard Perspector to Zeeman-Gossard Perspector was made on Oct. 15, 2003.) Further details are given by Wilson Stothers in Hyacinthos #8383, Oct. 21, 2003.
Barycentrics for X(402) were received from Paul Yiu (2/20/99); the polynomials p and y referred to above are given as follows:
p(a,b,c) = 2a4 - a2b2 - a2c2 - (b2 - c2)2
y(a,b,c) = a8 - a6(b2 + c2) + a4(2b2 - c2)(2c2 - b2) + [(b2 - c2)2][3a2(b2 + c2) - b4 - c4 - 3b2c2]
X(402) lies on this line: 2,3
X(402) = complement of X(1650)
, select the center name from the list and, then, click on the vertices A, B and C successively.