HJB --- GMA --- UFF


Click here to access the list of all triangle centers.

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 Run Macro Tool, 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: (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            sec A sec 2A : sec B sec 2B : sec C sec 2C
Barycentrics    tan A sec 2A : tan B sec 2B : tan C sec 2C

This point solves a problem posed by Antreas Hatzipolakis in Hyacinthos #3130, June 25, 2001; see also Jean-Pierre Ehrmann, #3135, June 26, 2001. The problem and solution may be stated as follows. Let ABC be a triangle, La, Lb, Lc the perpendicular bisectors of sides BC, CA, AB, and AA', BB', CC' the altitudes of ABC, respectively. Let Ab be the point of intersection of AA' and Lb, and let Ac be the point of intersection of AA' and Lc. Let A" be the point of intersection of BAb and CAc. Define B" and C" cyclically. Then triangle A"B"C" is perspective to triangle ABC, with perspector X(847).

X(847) lies on the ; see McCay orthic cubic.

X(847) lies on these lines: 2,254    3,925    4,52    24,96    91,225    378,1105    403,1093

X(847) = isogonal conjugate of X(1147)
X(847) = X(5)-cross conjugate of X(4)
X(847) = cevapoint of X(485) and X(486)

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

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