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|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)
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)