## X(3035) (COMPLEMENT OF X(11))

 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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Download all construction files and macros: tc.zip (10.1 Mb).
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[(a - b + c)(a - c)2 + (a + b - c)(a - b)2]
Barycentrics    af(a,b,c) : bf(b,c,a) : cf(c,a,b)

X(3035) lies on the Spieker circle and these lines:
1,1145    2,11    3,119    8,1317    9,1768    10,140    12,404    36,529    40,1537    80,1698    104,631    153,2551    230,1575    405,2932    468,1861    474,498    516,1638    549,993    620,2787    632,1484    676,2804    899,1818    908,1155    960,2800    1125,1387    1532,2077    2801,3041    2826,3039    2827,3038

X(3035) = midpoint of X(I) and X(J) for these I,J: 1,1145    3,119    8,1317    10,214    11,100    40,1537    908,1155    1532,2077
X(3035) = reflection of X(I) in X(J) for these I,J: 1387,1125    3036,10
X(3035) = complement of X(11)
X(3035) = X(885)-Ceva conjugate of X(518)

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