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.
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 86,99 310,670 320,519 335,536 350,889 527,666 675,901 812,1022X(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)