X₃₆₈₃

X₃₂₁₉com(EXTOUCH TRIANGLE)

X₃₆₈₃ = excentral-to-ABC barycentric image of X₃₂₁₉

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

X₃₆₈₃ (X₃₂₁₉com(EXTOUCH TRIANGLE)) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Central Lines: {9, 55}, {31, 37}

• Related Centers: X₉ (MITTENPUNKT), X₅₅ (INSIMILICENTER), X₃₁ (2nd POWER POINT), X₃₇ (CROSSPOINT OF INCENTER AND CENTROID)

Barycentrics: \(f(a,b,c) : f(b,c,a) : f(c,a,b)\), where \(f(a,b,c) = a(-a + b + c)(2a + b + c)\)

Search Value (6, 9, 13): \(3.361198331077\)

▸ Incident Central Lines33 lines

{2, 1155}, {3, 1709}, {6, 968}, {9, 55}, {21, 60}, {31, 37}, {38, 1279}, {42, 44}, {45, 612}, {63, 354}, {65, 405}, {144, 3475}, {191, 942}, {212, 1212}, {238, 846}, {381, 1698}, {392, 993}, {430, 1213}, {452, 1837}, {518, 1621}, {553, 1125}, {756, 902}, {940, 1707}, {958, 3057}, {1011, 2182}, {1100, 1962}, {1104, 2292}, {1214, 1456}, {1376, 3305}, {2187, 2267}, {2223, 3294}, {2293, 2318}, {2328, 2361}

▸ Functional Relations2 relations

• X₃₆₈₃ = complement of X₂₀₂₉₂

• X₃₆₈₃ = excentral-to-ABC barycentric image of X₃₂₁₉

Notes

In the plane of a triangle ABC, let

• A'B'C' = circumcevian triangle of X₁;

• DEF = cevian triangle of X₁₀;

• L_a = line tfhrough D perpendicular to BC;

• A″ = L_a∩AA', and define B″ and C″ cyclically.

Then X(36;83) is the finite fixed point of the affine transformation that carries A'B'C' onto A″B″C'. (Angel Montesdeoca, November 21, 2022)