|Information from Kimberling's Encyclopedia of Triangle Centers|
Trilinears f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a [cos2(B - C)][b cos(C - A) + c cos (A - B)]
Barycentrics af(a,b,c) : bf(b,c,a) : cf(c,a,b)
Let A'B'C' be the cevian triangle of a point X = x : y : z (trilinears), let L(A) be the line through A parallel to B'C', and define L(B) and L(C) cyclically. The lines L(A), L(B), L(C) form a triangle homothetic to A'B'C', with homothetic center
ax2(by + cz) : by2(cz + ax) : cz2(ax + by).
We denote this point by D(X) and call it the Danneels point of X. (Eric Danneels, Hyacinthos, Jan. 28, 2005).
The formula is simpler in barycentrics: if U = u : v : w, then
D(U) = u2(v + w) : v2(w + u) : w2(u + v).
If X is line the Euler line, then D(X) is on the Euler line. A proof follows. Let a1, b1, c1 be cos A, cos B, cos C, respectively, so that the Euler line is given parametrically as x(t) : y(t) :z (t) by
x = b1c1 + ta1
y = c1a1 + tb1
z = a1b1 + tc1,
where t is an arbitrary function homogenous of degree 0 in a, b, c. Then D(X) is the point U = u : v : w given by
u = ax2(by + cz)
v = by2(cz + ax)
w = cz2(ax + by).
A point P = p : q : r is on the Euler line if
a(b2 - c2)(b2 + c2 - a2)p
+ b(c2 - a2)(c2 + a2 - b2)q
+ c(a2 - b2)(a2 + b2 -c2)r = 0.
It is easy to check that the point U satisfies this equation. (This sort of algebraic proof is more inclusive than a geometric proof, because here, a,b,c are indeterminates or variables. They can, for example, take values that are not sidelengths of a triangle. Here, cos A is defined as (b2 + c2 - a2)/(2bc), so that no dependence on a geometric angle is necessary.)
The appearance of (I, J) in the following list means that D(X(I)) = X(J):
(1,42) (2,2) (3,418) (4,25) (6,3051) (7,57) (8,200) (69,394) (75,321) (100,55) (110,84) (264,324) (366,367) (651,222) (653,196) (1113,25) (1114,25) (1370,455)
X(3078) lies on this line: (2,3)
X(3078) = X(5)-Ceva conjugate of X(233)