X₃₄₄₇

X₅₂₃-VERTEX CONJUGATE OF X₅₂₃

X₃₄₄₇ = isogonal conjugate of X₃₄₄₈

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

X₃₄₄₇ (X₅₂₃-VERTEX CONJUGATE OF X₅₂₃) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Central Lines: {50, 232}, {23, 325}

• Related Centers: X₁₅₇₆ (ISOGONAL CONJUGATE OF X₈₅₀), X₆ (SYMMEDIAN POINT), X₅₀ (-CEVA CONJUGATE OF X₁₈₄), X₂₃₂ (-CEVA CONJUGATE OF X₁₃₂), X₂₃ (FAR-OUT POINT), X₃₂₅ (-HIRST INVERSE OF X₆₉)

Trilinears: \(a/[a^{2}vwyz - ux(bw + cv)(bz + cy)] : :\), where \(u : v : w\) =\(x : y : z\) = X₅₂₃

Search Value (6, 9, 13): \(-8.228842683765\)

▸ Alternate Coordinates1 form

• Trilinears: \(a/[a^{6} - a^{4}(b^{2} + c^{2}) + a^{2}(b^{4} - b^{2}c^{2} + c^{4}) - (b^{2} + c^{2})(b^{2} - c^{2})^{2}] (E. Danneels)\)

▸ Incident Central Lines5 lines

{23, 325}, {50, 232}, {264, 2453}, {511, 2070}, {2079, 2485}

▸ Functional Relations3 relations

• X₃₄₄₇ = isogonal conjugate of X₃₄₄₈

• X₃₄₄₇ = X₁₅₇₆-cross conjugate of X₆

• X₃₄₄₇ = perspector of ABC and reflection of tangential triangle in Euler line

Notes

Let E be the Euler line and T_AT_BT_C the tangential triangle of ABC. Let D_A = E∩T_BT_C, and define D_B and D_C cyclically. Let A' = D_BT_B∩D_CT_C, and define B' and C' cyclically. Then A'B'C' is perspective to ABC, and the perspector is X₃₄₄₇. For a sketch, click X₃₄₄₇andX(7669) (X(3447)andX(7669).gif). (Angel Montesdeoca, April 22, 2016)