X₂₇₁₅

SR(X₉₉, X₁₁₂)

X₂₇₁₅ = reflection of X₂₇₁₀ in X₃

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

X₂₇₁₅ (SR(X₉₉, X₁₁₂)) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Central Lines: {2710, 3}

• Related Centers: X₂₇₁₀ (SR(X₉₈, X₁₁₂)), X₃ (CIRCUMCENTER)

Trilinears: \(a/[(b^{2} - c^{2})(b^{4} + c^{4} - a^{2}b^{2} - a^{2}c^{2})] : :\) (M. Iliev, 5/13/07)

Search Value (6, 9, 13): \(-0.357586907999\)

X₂₇₁₅ lies on the circumcircle.

▸ Incident Central Lines32 lines

{2, 2857}, {3, 2710}, {6, 842}, {32, 2698}, {58, 2700}, {74, 187}, {81, 2856}, {98, 230}, {99, 249}, {103, 1326}, {107, 685}, {110, 647}, {111, 1495}, {112, 512}, {232, 1692}, {284, 2708}, {287, 2373}, {290, 2367}, {476, 2395}, {477, 2549}, {511, 1297}, {577, 2706}, {691, 2420}, {759, 1910}, {827, 1625}, {843, 1384}, {879, 935}, {933, 2623}, {1304, 2433}, {1333, 2699}, {2193, 2707}, {2407, 2855}

▸ Functional Relations18 relations

• X₂₇₁₅ = reflection of X₂₇₁₀ in X₃

• X₂₇₁₅ = isogonal conjugate of X₂₇₉₉

• X₂₇₁₅ = cevapoint of X_i and X_j for these (i,j): (512, 1692), (523, 2450), (1976, 2422)

• X₂₇₁₅ = X_i-cross conjugate of X_j for these (i,j): (512, 2065), (989, 98), (1691, 249), (2422, 1976)

• X₂₇₁₅ = reflection of X₁₁₂ in the Brocard axis

• X₂₇₁₅ = concurrence of reflections in sides of ABC of line X₄X₃₂

• X₂₇₁₅ = trilinear pole of line X₆X₁₅₇

• X₂₇₁₅ = X₉₂-isoconjugate of X₆₈₄

• X₂₇₁₅ = Λ(X_i,X_j) for these (i,j): (2, 1637), (99, 112)

• X₂₇₁₅ = Ψ(X_i, X_j) for these (i,j): (2, 98), (4, 32), (6, 157), (76, 3)

• X₂₇₁₅ = inverse-in-Moses-radical-circle of X₁₁₀

• X₂₇₁₅ = inverse-in-circle-O(15,16) of X₇₄

• X₂₇₁₅ = inverse-in-circle-{X₁₆₈₇,X₁₆₈₈,PU(1),PU(2)} of X₉₈

• X₂₇₁₅ = polar-circle-inverse of X₃₈₉₇₀

• X₂₇₁₅ = orthoptic-circle-of-Steiner-inellipse-inverse of X₃₈₉₇₅

• X₂₇₁₅ = Brocard-circle-inverse of X₃₄₂₃₅

• X₂₇₁₅ = barycentric product X₉₈*X₁₁₀ (circumcircle-X₂-antipodes)

• X₂₇₁₅ = barycentric quotient X₉₈/X₈₅₀

Notes

Let A' = BC∩(Brocard axis), and define CA and AB cyclically. The circumcircles of AB'C', BC'A', CA'B' concur in X₂₇₁₅. (Randy Hutson, February 10, 2016)