X₁₄₂

COMPLEMENT OF X₉

X₁₄₂ is the {X₂,X₇}-harmonic conjugate of X₉. For a list of other harmonic conjugates, click Tables at the top of this page.

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

X₁₄₂ (COMPLEMENT OF X₉) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Central Lines: {1001, 1125}

• Related Centers: X₁₀₀₁ (MIDPOINT OF X₁ AND X₉), X₁₁₂₅ (COMPLEMENT OF X₁₀)

Trilinears: \(b + c - [(b - c)^{2}]/a : c + a - [(c - a)^{2}]/b : a + b - [(a - b)^{2}]/c\)

Barycentrics: \(ab + ac - (b - c)^{2} : bc + ba - (c - a)^{2} : ca + cb - (a - b)^{2}\)

Search Value (6, 9, 13): \(2.158066839118\)

▸ Linear Combinations1 combination

• X₁₄₂ = \(3\) X₂ + X₇ = \(3\) X₂ - X₉

▸ Incident Central Lines13 lines

{1, 277}, {2, 7}, {3, 516}, {5, 971}, {10, 141}, {37, 1086}, {86, 284}, {116, 119}, {214, 528}, {269, 948}, {354, 3059}, {377, 950}, {474, 954}

▸ Functional Relations17 relations

• X₁₄₂ = midpoint of X_i and X_j for these (i,j): (7, 9), (8, 3243), (100, 3254)

• X₁₄₂ = reflection of X₁₀₀₁ in X₁₁₂₅

• X₁₄₂ = isogonal conjugate of X₁₁₇₄

• X₁₄₂ = isotomic conjugate of X₃₂₀₀₈

• X₁₄₂ = complement of X₉

• X₁₄₂ = X₁₀₀-Ceva conjugate of X₅₁₄

• X₁₄₂ = crosspoint of X₂ and X₈₅

• X₁₄₂ = crosssum of X₆ and X₄₁

• X₁₄₂ = X₁₉₀-beth conjugate of X₁₄₂

• X₁₄₂ = X₉-of- medial triangle

• X₁₄₂ = polar conjugate of isogonal conjugate of X₂₂₀₅₃

• X₁₄₂ = centroid of the set {X₁, X₄, X₇, X₄₀}

• X₁₄₂ = perspector of circumconic centered at X₁₂₁₂

• X₁₄₂ = X₂-Ceva conjugate of X₁₂₁₂

• X₁₄₂ = centroid of ABCX₇

• X₁₄₂ = center of circumconic that is locus of trilinear poles of lines passing through X₁₂₁₂

• X₁₄₂ = X₉₉₆₉-of-excentral-triangle

Notes

Let A' be the midpoint between A and X₇, and define B' and C' cyclically; the triangle A'B'C' is homothetic to the medial triangle, and the center of homothety is X₁₄₂. (Randy Hutson, 9/23/2011)

Let A' be the intersection of these three lines:

1. through midpoint of CA perpendicular to BX₁

2. through midpoint of AB perpendicular to CX₁

3. through midpoint of AX₁ perpendicular to BC.

Define B' and C' cyclically. Then X₁₄₂ = X₆-of-A'B'C'. The triangle A'B'C' is the complement of the excentral triangle, and also the extraversion triangle of X₁₀. (Randy Hutson, September 14, 2016)