## X(3042) (COMPLEMENT OF X(1364))

 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[(a - b + c)(a - c)2cos2B + (a + b - c)(a - b)2cos2C]
Barycentrics    af(a,b,c) : bf(b,c,a) : cf(c,a,b)

X(3042) lies on the Spieker circle and these lines:
2,1364    8,1361    10,2818    72,1845    102,958    109,1376    117,1329    118,124    151,2551    474,1795    960,2817    2814,3039    2815,3039

X(3042) = midpoint of X(I) and X(J) for these I,J: 8,1361    72,1845
X(3042) = reflection of X(3040) in X(10)
X(3042) = complement of X(1364)

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