## X(1131) (ARCTAN(2) KIEPERT POINT)

 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.

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This applet was built with the free and multiplatform dynamic geometry software C.a.R..

 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) = 1/(sin A + 2 cos A)
Barycentrics    (sin A)f(A,B,C) : (sin B)f(B,C,A) : (sin C)f(C,A,B)

To construct the Vecten point, X(485), squares are erected outward on the sides of ABC. If A', B', C' are the centers of these squares, then triangle A'B'C' is perspective to ABC with perspector X(485). Now let A" be the midpoint of the side of the A-square that does not touch line BC, and define B" and C" cyclically. Then triangle A"B"C" is perspective to ABC with perspector X(1131). Angle(A"BC) = angle(B"CA) = angle(C"AB), so that X(1131) lies on the Kiepert hyperbola. Here, the common angle is arctan(2). (Darij Grinberg 9/22/02)

X(1131) lies on these lines: 2,490    6,1132    20,485    175,226

X(1131) = isogonal conjugate of X(1151)
X(1131) = isotomic conjugate of X(1270)

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