X₂₈₈₃

COMPLEMENTARY CONJUGATE OF X₄

X₂₈₈₃ = midpoint of X₄ and X₁₄₉₈

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

X₂₈₈₃ (COMPLEMENTARY CONJUGATE OF X₄) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Central Lines: {4, 1498}

• Related Centers: X₄ (ORTHOCENTER), X₁₄₉₈ (REFLECTION OF X₆₄ IN X₃)

Trilinears: \(f(a,b,c) : f(b,c,a) : f(c,a,b)\), where f(a,b,c) is as described just before X₂₈₈₃

Barycentrics: \([3a^{4} - 2a^{2}(b^{2} + c^{2}) - (b^{2} - c^{2})^{2}] [a^{4}(b^{2} + c^{2}) - 2a^{2}(b^{2} - c^{2})^{2} + (b^{2} - c^{2})^{2}(b^{2} + c^{2})] : :\)

Search Value (6, 9, 13): \(-11.869608844287\)

▸ Incident Central Lines17 lines

{2, 64}, {3, 1661}, {4, 6}, {20, 154}, {30, 156}, {51, 1906}, {184, 1885}, {185, 235}, {221, 497}, {388, 2192}, {389, 1596}, {417, 1624}, {459, 1192}, {468, 1204}, {496, 942}, {550, 1511}, {1593, 1619}

▸ Functional Relations8 relations

• X₂₈₈₃ = midpoint of X₄ and X₁₄₉₈

• X₂₈₈₃ = reflection of X₃₃₅₇ in X₁₄₀

• X₂₈₈₃ = complement of X₆₄

• X₂₈₈₃ = complementary conjugate of X₄

• X₂₈₈₃ = X₂-Ceva conjugate of X₈₀₀

• X₂₈₈₃ = X₁₂₅₁₃-of-orthic-triangle if ABC is acute

• X₂₈₈₃ = perspector of circumconic centered at X₈₀₀

• X₂₈₈₃ = center of circumconic that is locus of trilinear poles of lines passing through X₈₀₀

Notes

In the plane of a triangle ABC, let

• H = X₄;

• L = X₂₀;

• A_b = AL∩BH;

• A_c = AL∩CH;

• U = line through A_b parallel to AC;

• V = line through A_c parallel to AB;

• A₁ = U∩V;

• A₂ = U∩BC;

• A₃ = V∩BC;

• O = circle {{A₁, A₂, A₃}};

• A' = pole of BC wrt O, and define B' and C' cyclically.

Then A'B'C' = reflection of ABC in X₂₈₈₃. (Angel Montesdeoca, July 30, 2021)