X₁₁₉

FEUERBACH ANTIPODE

X₁₁₉ = complement of X₁₀₄

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

X₁₁₉ (FEUERBACH ANTIPODE) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Circle: Nine-Point Circle (nine_pointcircle) — Passes through side midpoints M_a, M_b, M_c and altitude feet H_a, H_b, H_c; centered at X₅

• Central Lines: {5, 11}

• Related Centers: X₅ (NINE-POINT CENTER), X₁₁ (FEUERBACH POINT)

Trilinears: \((\csc A)(-1 + \cos B + \cos C)[\sin 2B + \sin 2C + 2(-1 + \cos A)(\sin B + \sin C)] : :\)

Barycentrics: \((-1 + \cos B + \cos C)[\sin 2B + \sin 2C + 2(-1 + \cos A)(\sin B + \sin C)] : :\)

Search Value (6, 9, 13): \(-1.537391637975\)

▸ Incident Central Lines12 lines

{1, 5}, {2, 104}, {3, 123}, {4, 100}, {10, 124}, {116, 142}, {125, 442}, {135, 431}, {136, 429}, {214, 515}, {381, 528}, {517, 908}

▸ Functional Relations15 relations

• X₁₁₉ = midpoint of X_i and X_j for these (i,j): (4, 100), (104, 153)

• X₁₁₉ = reflection of X_i in X_j for these (i,j): (11, 5), (3, 3035)

• X₁₁₉ = isogonal conjugate of X₁₅₃₈₁

• X₁₁₉ = complement of X₁₀₄

• X₁₁₉ = complementary conjugate of X₅₁₇

• X₁₁₉ = X₄-Ceva conjugate of X₅₁₇

• X₁₁₉ = nine-point-circle-antipode of X₁₁

• X₁₁₉ = X₁₀₄-of-medial triangle

• X₁₁₉ = X₂₀₇₂-of-Fuhrmann-triangle

• X₁₁₉ = inverse-in-polar-circle of X₉₁₅

• X₁₁₉ = anticenter of cyclic quadrilateral ABCX₁₀₀

• X₁₁₉ = Λ(X₁, X₃), wrt orthic triangle

• X₁₁₉ = Λ(X₄, X₈), wrt orthic triangle

• X₁₁₉ = Spieker-radical-circle-inverse of X₃₄₄₅₉

• X₁₁₉ = center of rectangular circumhyperbola passing through isogonal and isotomic conjugates of X₃₆₅₇

Notes

Let N_a = X₅-BCX₁, N_b = X₅-of-CAX₁, N_c = X₅-of-ABX₁. Then X₁₁₉ = X₂₀₇₁-of-N_aN_bN_c. (Randy Hutson, January 29, 2018)

Let H_aH_bH_c be the orthic triangle. Let L_a be the antiorthic axis of AH_bH_c, and define L_b and L_c cyclically. Let A″ = L_b∩L_c, B″ = L_c∩L_a, C″ = L_a∩L_b. Then triangle A″B″C″ is inversely similar to ABC, with similicenter X₉. The lines H_aA″, H_bB″, H_cC″ concur in X₁₁₉. Also, X₁₁₉ = X₃ of A″B″C″. (Randy Hutson, January 29, 2018)