X3078

DANNEELS POINT OF X5

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
< A B C >: Reference Triangle △ABC [ A B ]: Sideline AB (Base c = 1) [ B C ]: Sideline BC (Side a) [ C A ]: Sideline CA (Side b) { 2, 3 }: Euler Line A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_5 = NINE-POINT CENTER (0.5561, 0.2489) X_233 = -CEVA CONJUGATE OF X(140) (0.5475, 0.2218) X_2 = CENTROID (0.5374, 0.2000) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_3078 = DANNEELS POINT OF X(5) (0.5651, 0.2724) A B C X5 X233 X2 X3 X3078 (a, b, c) = (5, 6, 7) | AB = 1

X3078 (DANNEELS POINT OF X5) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(a[\cos^{2}(B - C)][b \cos(C - A) + c \cos (A - B)] : :\)
Barycentrics: \(a^{2}[\cos^{2}(B - C)][b \cos(C - A) + c \cos (A - B)] : :\)
Search Value (6, 9, 13): \(2.017201680077\)
▸ Incident Central Lines1 line

{2, 3}

▸ Functional Relations2 relations

Notes

\(ax^{2}(by + cz)\) : \(by^{2}(cz + ax)\) : \(cz^{2}(ax + by)\).

We denote this point by D(X) and call it the Danneels point of X. (Eric Danneels, Hyacinthos, Jan. 28, 2005). In addition to the discussion below, see the preamble to X8012 and the Danneels points beginning at X8012.

If X lies on the line at infinity, then D(X) = X2 of cevian triangle of X. (Randy Hutson, Jul 23, 2015)

The formula is simpler in barycentrics: if U =\(u : v : w\), then

D(U) = \(u^{2}(v + w)\) : \(v^{2}(w + u)\) : \(w^{2}(u + v)\).

If X is line the Euler line, then D(X) is on the Euler line. A proof follows. Let \(a_{1}\), \(b_{1}\), \(c_{1}\) 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 = \(b_{1}\)\(c_{1}\) + \(ta_{1}\)

y = \(c_{1}\)\(a_{1}\) + \(tb_{1}\)

z = \(a_{1}\)\(b_{1}\) + \(tc_{1}\),

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 = \(ax^{2}(by + cz)\)

v = \(by^{2}(cz + ax)\)

w = \(cz^{2}(ax + by)\).

A point P =\(p : q : r\) is on the Euler line if

\(a(b^{2}\) - \(c^{2})\)(\(b^{2}\) + \(c^{2}\) - \(a^{2})\)p

  • \(b(c^{2}\) - \(a^{2})\)(\(c^{2}\) + \(a^{2}\) - \(b^{2})\)q

  • \(c(a^{2}\) - \(b^{2})\)(\(a^{2}\) + \(b^{2}\) -\(c^{2})\)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 (\(b^{2}\) + \(c^{2}\) - \(a^{2})\)/(2bc), so that no dependence on a geometric angle is necessary.)

The appearance of (i,j) in the following list means that D(Xi) = Xj:

  • (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) For an extension of this list, see the preamble to X8012.