X₁₄₉₅

CROSSSUM OF X₂ AND X₃₀

X₁₄₉₅ = midpoint of X₂₃ and X₁₁₀

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

X₁₄₉₅ (CROSSSUM OF X₂ AND X₃₀) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Central Lines: {23, 110}

• Related Centers: X₂₃ (FAR-OUT POINT), X₁₁₀ (FOCUS OF KIEPERT PARABOLA)

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

Search Value (6, 9, 13): \(-1.073001771887\)

X₁₄₉₅ lies on the Walsmith rectangular hyperbola.

▸ Incident Central Lines12 lines

{3, 3426}, {6, 25}, {23, 110}, {24, 185}, {30, 113}, {52, 156}, {74, 186}, {125, 468}, {182, 373}, {187, 237}, {263, 1383}, {1204, 1498}

▸ Functional Relations19 relations

• X₁₄₉₅ = midpoint of X₂₃ and X₁₁₀

• X₁₄₉₅ = reflection of X_i in X_j for these (i,j): (125, 468), (1531, 113)

• X₁₄₉₅ = isogonal conjugate of X₁₄₉₄

• X₁₄₉₅ = isotomic conjugate of isogonal conjugate of X₉₄₀₇

• X₁₄₉₅ = X_i-Ceva conjugate of X_j for these (i,j): (4, 39176), (78, 1), (329, 9)

• X₁₄₉₅ = crosspoint of X₆ and X₇₄

• X₁₄₉₅ = crosssum of X_i and X_j for these (i,j): (74, 6), (1304, 647)

• X₁₄₉₅ = crossdifference of every pair of points on line X₂X₅₂₅

• X₁₄₉₅ = inverse-in-Moses-radical-circle of X₁₈₇

• X₁₄₉₅ = intersection of tangents to Moses-Jerabek conic at X₆ and X₇₄

• X₁₄₉₅ = pole of Brocard axis wrt Moses radical circle

• X₁₄₉₅ = {X₃,X₅₆₅₁}-harmonic conjugate of X₅₆₅₀

• X₁₄₉₅ = trilinear pole of PU(87)

• X₁₄₉₅ = inverse-in-Parry-isodynamic-circle of X₃₅₆₉; see X₂

• X₁₄₉₅ = X₃₂₁₈-of-orthic-triangle if ABC is acute

• X₁₄₉₅ = antipode of X₁₂₅ in Walsmith rectangular hyperbola

• X₁₄₉₅ = orthocenter of X₁₁₀X₁₄₉₅X₃₅₆₉

• X₁₄₉₅ = orthocenter of X₁₂₅X₃₅₆₉X₃₅₈₀

• X₁₄₉₅ = orthic-isogonal conjugate of X₃₉₁₇₆

Notes

Let L_a be the line through A parallel to the de Longchamps line, and define L_b and L_c cyclically. Let M_a be the reflection of BC in L_a, and define M_b and M_c cyclically. Let A' = M_b∩M_c, and define B', C' cyclically. The triangle A'B'C' is inversely similar to, and 3 times the size of, ABC. Let A″B″C″ be the reflection of A'B'C' in the de Longchamps line. The triangle A″B″C″ is homothetic to ABC, with center of homothety X₁₄₉₅. (See Hyacinthos #16741/16782, Sep 2008.) (Randy Hutson, December 26, 2015)