HJB --- GMA --- UFF


Click here to access the list of all triangle centers.

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)

This is a joint work of
Humberto José Bortolossi, Lis Ingrid Roque Lopes Custódio and Suely Machado Meireles Dias.

If you have questions or suggestions, please, contact us using the e-mail presented here.

Departamento de Matemática Aplicada -- Instituto de Matemática -- Universidade Federal Fluminense

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