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           a(b + c)2/(b + c - a) : b(c + a)2/(c + a - b) : c(a + b)2/(a + b - c)
                                    = a2cos2(B/2 - C/2) : b2cos2(C/2 - A/2) : c2cos2(A/2 - B/2)

Trilinears            h(a,b,c) : h(b,c,a) : h(c,a,b), where h(a,b,c) = [r cos(A/2) + s sin(A/2)]2, s = semiperimeter, r = inradius

Barycentrics    a3cos2(B/2 - C/2) : b3cos2(C/2 - A/2) : c3cos2(A/2 - B/2)

Let O(A),O(B),O(C) be the excircles. Apollonius's Problem includes the construction of the circle O tangent to the three excircles and encompassing them. (The circle is called the Apollonius circle.) Let A' = O∩O(A), B'=O∩O(B), C'=O∩O(C). The lines AA',BB',CC' concur in X(181). Yff derived trilinears in 1992.

X(181) is the external center of similitude (or exsimilicenter) of the incircle and Apollonius circle. The internal center is X(1682). (Peter J. C. Moses, 8/22/03)

A proof of the the concurrence of lines AA',BB',CC' follows.
            A = exsimilicenter(incircle, A-excircle)
            A' = exsimilicenter(A-excircle, Apollonius circle)
            Let J = exsimilicenter(incircle, Apollonius circle).
By Monge's theorem, the points A, A', J are collinear. In particular, J lies on line AA', and cyclically, J lies on lines BB' and CC'. Therefore, J = X(181). (Darij Grinberg, Hyacinthos, 7461, 8/10/03)

See also

Clark Kimberling, Shiko Iwata, and Hidetosi Fukagawa, Problem 1091 and Solution, Crux Mathematicorum 13 (1987) 128-129; 217-218. [proposed 1985].

X(181) lies on these lines:
1,970    6,197    8,959    10,12    31,51    42,228    43,57    44,375    55,573    56,386    58,1324    171,511    373,748    553,1463    994,1361    1124,1685    1254,1425    1335,1686    1395,1843    1672,1683    1673,1684    1674,1693    1675,1694    1695,1697

X(181) = isogonal conjugate of X(261)
X(181) = X(872)-cross conjugate of X(1500)
X(181) = crosssum of X(I) and X(J) for these (I,J): (21,333), (86,1444)
X(181) = X(I)-beth conjugate of X(J) for these (I,J): (42,181), (660,181), (756,756)

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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