X₉₀₃

X₁-BLAIKIE TRANSFORM OF X₂

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' concur.

5, 6, 7 3, 4, 5 Kepler Kimberling

X₉₀₃ (X₁-BLAIKIE TRANSFORM OF X₂) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Central Lines: {2, 45}, {86, 99}

• Related Centers: X₉₀₂ (CROSSPOINT OF X₆ AND X₁₀₆), X₂ (CENTROID), X₄₅ (-BETH CONJUGATE OF X₁), X₈₆ (CEVAPOINT OF INCENTER AND CENTROID), X₉₉ (STEINER POINT)

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)

Search Value (6, 9, 13): \(0.781704235605\)

X₉₀₃ lies on the Steiner circumellipse.

▸ Incident Central Lines12 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}

▸ Functional Relations15 relations

• X₉₀₃ = reflection of X_i in X_j for these (i,j): (2, 1086), (190, 2)

• X₉₀₃ = isogonal conjugate of X₉₀₂

• X₉₀₃ = isotomic conjugate of X₅₁₉

• X₉₀₃ = complement of X₁₇₄₈₇

• X₉₀₃ = X_i-cross conjugate of X_j for these (i,j): (320, 86), (519, 2)

• X₉₀₃ = Steiner-circumellipse-antipode of X₁₉₀

• X₉₀₃ = anticomplement of X₄₃₇₀

• X₉₀₃ = projection from Steiner inellipse to Steiner circumellipse of X₁₀₈₆

• X₉₀₃ = antipode of X₂ in hyperbola {A, B, C,X₂,X₇}

• X₉₀₃ = trilinear pole of line X₂X₅₁₄

• X₉₀₃ = pole wrt polar circle of trilinear polar of X₈₇₅₆ (line X₄₁₂₀X_4895)

• X₉₀₃ = X₁₉-isoconjugate of X₂₂₃₅₆

• X₉₀₃ = X₄₈-isoconjugate (polar conjugate) of X₈₇₅₆

• X₉₀₃ = crossdifference of PU(99)

• X₉₀₃ = crossdifference of every pair of points on line X₁₀₁₇X₁₉₆₀

Notes

Darij Grinberg introduces the term Blaikie point of :g:`O <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 :g:`O <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)].

Seet Blaikie theorem in barycentrics. (Darij Grinberg, 12/28/02)