X₃₇₄₀

X₂₁₀com[n(INCENTRAL)*n(MEDIAL)]

X₃₇₄₀ = reflection of X₃₇₄₂ in X₂

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

X₃₇₄₀ (X₂₁₀com[n(INCENTRAL)*n(MEDIAL)]) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Central Lines: {3742, 2}

• Related Centers: X₃₇₄₂ (com[n(INCENTRAL)*n(MEDIAL)]), X₂ (CENTROID)

Barycentrics: \(a(b^{2} + c^{2} - ab - ac + 4bc) : :\)

Search Value (6, 9, 13): \(3.292329299997\)

▸ Incident Central Lines22 lines

{2, 210}, {5, 10}, {9, 165}, {37, 43}, {44, 171}, {55, 3305}, {72, 1698}, {120, 125}, {200, 1001}, {374, 966}, {375, 511}, {405, 2900}, {612, 1386}, {748, 3246}, {756, 899}, {936, 958}, {1051, 1100}, {1107, 2664}, {1155, 3219}, {2801, 3035}, {3057, 3617}, {3555, 3624}

▸ Functional Relations4 relations

• X₃₇₄₀ = reflection of X₃₇₄₂ in X₂

• X₃₇₄₀ = isotomic conjugate of X₃₂₀₂₁

• X₃₇₄₀ = complement of X₃₅₄

• X₃₇₄₀ = anticomplement of X₃₈₄₈

Notes

In the plane of a triangle ABC, let

• A_aA_bA_c = A-extouch triangle, and define B_aB_bB_c and C_aC_bC_c cyclically

• A_b' = A_bA_c∩B_bB_c, and define B_c' and C_a' cyclically

• A_c' = A_bA_c∩C_bC_c, and define B_a' and C_b' cyclically

Then X₃₇₄₀ = centroid-of-{A_b', A_c', B_c', B_a', C_a', C_b'}. These 6 points lie on the radical circle of the excircles. (Randy Hutson, September 14, 2016)

Let J_aJ_bJ_c = excentral triangle. Let

• A_b” = J_bJ_c∩B_bB_c, and define Bc’’ and Ca’’ cyclically

• Ac’’ = J_bJ_c∩C_bC_c, and define Ba’’ and Cb’’ cyclically

Then X₃₇₄₀ = centroid-of-{A_b”,A_c”, B_c”, B_a”, C_a”, C_b”}. These 6 points lie on an ellipse. (Randy Hutson, September 14, 2016)

Continuing, let A_b* = B'C'∩B_cB_a, and define B_c* and C_a* cyclically

A_c* = B'C'∩C_aC_b, and define B_a* and C_b* cyclically

Then X₃₇₄₀ = centroid of {A_b*,A_c*,B_c*,B_a*,C_a*,C_b*}. These 6 points lie on the radical circle of the excircles. (Randy Hutson, September 14, 2016)