X₃₅₃₀

EULER LINE INTERCEPT OF LINE X₁₅X_3411

X₃₅₃₀ = complement of X₅₄₆

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

X₃₅₃₀ (EULER LINE INTERCEPT OF LINE X₁₅X_3411) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Derived Triangle: Medial Triangle △M_aM_bM_c — vertices M_a, M_b, M_c (Formed by the midpoints of sides BC, CA, AB)

• Central Lines: {2, 546}, {2, 3} (Euler line)

• Related Centers: X₂ (CENTROID), X₅₄₆ (MIDPOINT OF X₄ AND X₅), X₃ (CIRCUMCENTER)

Trilinears: \(5 \cos A + 2 \sin B \sin C : :\)

Search Value (6, 9, 13): \(5.224990106056\)

▸ Alternate Coordinates1 form

• Trilinears: \(6a^{4} + b^{4} + c^{4} - 2b^{2}c^{2} - 7a^{2}b^{2} - 7a^{2}c^{2} : :\)

▸ Linear Combinations1 combination

• X₃₅₃₀ = \(3\) X₂ + \(5\) X₃ = \(21\) X₂ - \(5\) X₄ = \(7\) X₃ + X₄ = \(3\) X₃ + X₅

▸ Incident Central Lines4 lines

{2, 3}, {15, 3411}, {16, 3412}, {1385, 3244}

▸ Functional Relations9 relations

• X₃₅₃₀ = midpoint of X_i in X_j for these (i,j): (3, 140), (5, 548), (546, 550)

• X₃₅₃₀ = complement of X₅₄₆

• X₃₅₃₀ = anticomplement of X₃₅₀₁₈

• X₃₅₃₀ = circumcircle-inverse of X₃₇₉₄₉

• X₃₅₃₀ = orthocentroidal-circle-inverse of X₅₀₇₉

• X₃₅₃₀ = {X₂,X₃}-harmonic conjugate of X₅₅₀

• X₃₅₃₀ = {X₂,X₄}-harmonic conjugate of X₅₀₇₉

• X₃₅₃₀ = {X₂,X₂₀}-harmonic conjugate of X₃₈₅₅

• X₃₅₃₀ = {X₃,X₅}-harmonic conjugate of X₅₄₈

Notes

As a point on the Euler line, X₃₅₃₀ has Shinagawa coefficients \((7, -5)\).

Let P₇' and U₇' be the bicentric pair constructed as with PU(7), but with circles half the size. The midpoint of P₇' and U₇' is X₃₅₃₀. (Randy Hutson, December 26, 2015)