## X(1654) (1ST HATZIPOLAKIS PARALLELIAN 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) = bc(-a2 + b2 + c2 + bc + ca + ab)
Barycentrics    g(a,b,c) : g(b,c,a) : g(c,a,b), where g(a,b,c) = -a2 + b2 + c2 + bc + ca + ab

Let P be a point in the plane of, but not on a sideline of, triangle ABC. Let Ba be the point where the line through P parallel to line BC meets line BA, and let Ca be the point where the line through P parallel to line BC meets line CA. Define Cb, Ab, Ac, and Bc cyclically. If P = X(1654), then

|ABa| + |ACa| = |BCb| + |BAb| = |CAc| + |CBc|

(Antreas Hatzipolakis, Anopolis #20, 1/20/02)

X(1654) lies on these lines: 2,6    8,192    10,894    37,319    71,1762    190,594

X(1654) = reflection of X(86) in X(1213)
X(1654) = anticomplement of X(86)
X(1654) = X(I)-Ceva conjugate of X(J) for these (I,J): (10,2), (894,192)

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