X₁₆₅₇

{X₃,X₄}-HARMONIC CONJUGATE OF X₁₆₅₆

X₁₆₅₇ is the {X₃,X₄}-harmonic conjugate of X₁₆₅₆.

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

X₁₆₅₇ ({X₃,X₄}-HARMONIC CONJUGATE OF X₁₆₅₆) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Central Lines: {3, 4} (Euler line)

• Related Centers: X₁₆₅₆ (INTERSECTION OF EULER LINE AND LINEX), X₃ (CIRCUMCENTER), X₄ (ORTHOCENTER)

Trilinears: \(3 \cos A - 4 \cos B \cos C : :\)

Barycentrics: \(5a^{4} - 2b^{4} - 2c^{4} - 3a^{2}b^{2} - 3a^{2}c^{2} + 4b^{2}c^{2} : :\)

Search Value (6, 9, 13): \(31.700327504442\)

▸ Linear Combinations1 combination

• X₁₆₅₇ = \(3\) X₃₈₁ - \(2\) X₃₈₂

▸ Incident Central Lines4 lines

{2, 3}, {195, 1181}, {399, 1498}, {516, 1482}

▸ Functional Relations6 relations

• X₁₆₅₇ = reflection of X_i in X_j for these (i,j): (3, 20), (4, 550), (382, 3)

• X₁₆₅₇ = complement of X₃₃₇₀₃

• X₁₆₅₇ = {X₃₈₁,X₃₈₂}-harmonic conjugate of X₅₀₇₆

• X₁₆₅₇ = Ehrmann-mid-to-ABC similarity image of X₃₈₂

• X₁₆₅₇ = circumcircle inverse of X₃₄₁₅₂

• X₁₆₅₇ = endo-homothetic center of Ehrmann mid-triangle and ABC-X3 reflections triangle; the homothetic center is X₃₈₂

Notes

As a point on the Euler line, X₁₆₅₇ has Shinagawa coefficients \((3, -7)\).

Let L_a be the polar of X₃ wrt the A-power circle, and define L_b and L_c cyclically. Let A' = L_b∩L_c, B' = L_c∩L_a, C' = L_a∩L_b. Triangle A'B'C' is homothetic to ABC, and the orthocenter of A'B'C' is X₁₆₅₇. (Randy Hutson, December 2, 2017)

Let O_A be the circle centered at the A-vertex of the Ehrmann mid-triangle and passing through A; define O_B and O_C cyclically. X₁₆₅₇ is the radical center of O_A, O_B, O_C. (Randy Hutson, August 28, 2020)