Interactive Applet |
You can move the points A, B and C (click on the point and drag it).
Press the keys “+” and “−” to zoom in or zoom out the visualization window and use the arrow keys to translate it.
You can also construct all centers related with this one (as described in ETC) using the “Run Macro Tool”. To do this, click on the icon , select the center name from the list and, then, click on the vertices A, B and C successively.
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) = bc/[16D2 + (a2+c2-b2)(a2+b2-c2)][16D2 - 3(b2+c2-a2)2],
where D = area(ABC)
= sec(B - C)/(1 - 4 cos2A) : sec(C - A)/(1 - 4 cos2B) : sec(A - B)/(1 - 4 cos2C) (Eric Weisstein, Nov. 17, 2005)Barycentrics af(a,b,c) : bf(b,c,a) : cf(c,a,b)
X(1141) was first noted (Hyacinthos #1498, September 25, 2000) by Bernard Gibert as a point of intersection of the circumcircle and certain cubic, denoted Kn. To define Kn, note first that the Neuberg cubic is the locus of a point M such that the reflections of M in the sidelines of triangle ABC are the vertices of a triangle perspective to ABC. The locus of the perspector is the cubic Kn, and X(1141) is the point, other than A,B,C, in which Kn meets the circumcircle. Also, X(1141) is the perspector when M = X(1157). In
Jean-Pierre Ehrmann and Bernard Gibert, "Special Isocubics," downloadable from
Cubics in the Triangle Plane,
the point X(1141) is labeled E, barycentrics are given, and it is established that this point also lies on the line X(5)-to-X(110) [listed below as (5,49)], two other cubics, and the hyperbola that passes through the points A, B, C, X(4), X(5).
Let A' be the reflection of A in line BC, and define B' and C' cyclically. Let Ab be the reflection of A' in AB, and define Ac, Bc, Ba, Ca, Cb cyclically. Let
A1 = BAb∩CAc, and define B1 and C1 cyclically,
A2 = BAc∩CAb, and define B2 and C2 cyclically,
A3 = BBa∩CCa, and define B3 and C3 cyclically,
A4 = BBc∩CCb, and define B4 and C4 cyclically,
A5 = BCa∩CBa, and define B5 and C5 cyclically,
A6 = BCb∩CBc, and define B6 and C6 cyclically.Then triangle AnBnCn is perspective to ABC, for n = 1,2,3,4,5,6. The six perspectors are
X(1141), X(186), X(4), X(54), X(265), X(5), respectively. (Keith Dean, #4953, 3/12/02; coordinates by Paul Yiu, #4963; summary by Dean, #4971)
X(1141) lies on the conic of {A, B, C, X(3), X(49)}, the conic of {A, B, C, X(6), X(567)}, and the conic of {A, B, C, X(70), X(253), X(254)}.
X(1141) is the antipode of X(930) on the circumcircle, and X(1141) lies on the line of the
nine-point center, X(5), and its isogonal conjugate, X(54).
X(1141) lies on these lines:
2,128 3,252 4,137 5,49 53,112 79,109 94,96 95,99X(1141) = reflection of X(I) in X(J) for these (I,J): (4,137), (930,3)
X(1141) = isogonal conjugate of X(1154)
X(1141) = isotomic conjugate of X(1273)
X(1141) = anticomplement of X(128)
X(1141) = isogonal conjugate of (1154)
X(1141) = X(231)-cross conjugate of X(2)