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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.
|Information from Kimberling's Encyclopedia of Triangle Centers|
Trilinears bc/(b + c - 2a) : ca/(c + a - 2b) : ab/(a + b - 2c)
Barycentrics 1/(b + c - 2a) : 1/(c + a - 2b) : 1/(a + b - 2c) (Darij Grinberg, 12/28/02)
James Blaikie (1847-1929) proposed the following problem. Let O be any point in the plane of triangle ABC, and let any straight line g through O meet BC in P, CA in Q, AB in R; then, if points P', Q', R' be taken on the line so that
PO = OP', QO = OQ', RO = OR',
prove that AP', BQ', CR' are concurrent.
Darij Grinberg introduces the term Blaikie point of O and g for the point Z of concurrence. If
O = x : y : z and g = [k : l : m] (barycentric coordinates),
then Z has first barycentric 1/[k(y-z) - (ly-mz)]. Given a point S = u : v : w, Grinberg then defines the S-Blaikie transform of O as the Blaikie point of O and OS. The first barycentric of Z can be written as
1/[yw(y+x) + zv(z+x) - yz(2u+v+w)].
Visit Blaikie theorem in barycentrics. (Darij Grinberg, 12/28/02)
X(903) lies on these lines:
2,45 7,528 27,648 75,537 86,99 310,670 320,519 335,536 350,889 527,666 675,901 812,1022
X(903) = reflection of X(I) in X(J) for these (I,J): (2,1086), (3,190)
X(903) = isogonal conjugate of X(902)
X(903) = isotomic conjugate of X(519)
X(903) = X(I)-cross conjugate of X(J) for these (I,J): (320,86), (519,2)