## X(3039) (COMPLEMENT OF X(1358))

 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 - c)2/(a - b + c) + (a - b)2/(a + b - c)]
Barycentrics    af(a,b,c) : bf(b,c,a) : cf(c,a,b)

X(3039) lies on the Spieker circle and these lines:
2,1358    8,3021    9,80    960,2809    1292,1376    2814,3042    2826,3035    2832,3038    2835,3040

X(3039) = midpoint of X(8) and X(3021)
X(3039) = complement of X(1358)
X(3039) = crosssum of X(6) and X(1357)

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