X₁₄

2nd ISOGONIC CENTER

Construct the equilateral triangle BA'C having base BC and vertex A' on the positive side of BC; similarly construct equilateral triangles CB'A and AC'B based on the other two sides. The lines AA',BB',CC' concur in X₁₄. The antipedal triangle of X₁₄ is equilateral.

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

X₁₄ (2nd ISOGONIC CENTER) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Equilateral Side Triangles: △BCA', △CAB', △ABC' erected on sidelines with apexes A', B', C'

• Concurrence Lines: A–A', B–B', C–C' concurring at X₁₄

Trilinears: \(\csc(A - \pi /3) : \csc(B - \pi /3) : \csc(C - \pi /3)\)

Barycentrics: \(f(a,b,c) : f(b,c,a) : f(c,a,b)\), where \(f(a,b,c) = a^{4} - 2(b^{2} - c^{2})^{2} + a^{2}(b^{2} + c^{2} - 4*\)\(\sqrt{3}\)*Area(ABC))

Tripolars: \(1/(a^{4}+a^{2} b^{2}-2 b^{4}+a^{2} c^{2}+4 b^{2} c^{2}-2 c^{4}-2 \sqrt{3} a^{2} S) : :\)

Search Value (6, 9, 13): \(4.373431772013\)

▸ Alternate Coordinates2 forms

• Trilinears: \(\sec(A + \pi /6) : \sec(B + \pi /6) : \sec(C + \pi /6)\)

• Barycentrics: \((S_{A} - \sqrt{3} S) (S_{B} + S_{C}) + 4 S_{B} S_{C} : :\)

X₁₄ lies on the Neuberg cubic.

Related Collection 1 Topic Guides

T_13: The Isogonic & Napoleon Central Lines • T_18: The Apollonian Circles • T_24: Equilateral, Isogonic & Napoleon Triangles

▸ Incident Central Lines21 lines

{2, 15}, {3, 18}, {4, 62}, {5, 17}, {6, 13}, {11, 203}, {16, 30}, {76, 298}, {98, 383}, {99, 302}, {148, 616}, {202, 1478}, {226, 554}, {262, 1080}, {275, 473}, {299, 533}, {397, 546}, {484, 1276}, {530, 671}, {532, 622}, {633, 636}

▸ Functional Relations32 relations

• X₁₄ = \(3^{1/2}(r^{2} + 2rR + s^{2})\)*X₁ - 6\(r(3^{1/2}\)R + 2s)*X₂ - 2\(r(3^{1/2}\)r - 3s)*X₃ (Peter Moses, April 2, 2013)

• X₁₄ = reflection of X_i in X_j for these (i,j): (13, 115), (16, 395), (99, 618), (299, 624), (617, 619)

• X₁₄ = isogonal conjugate of X₁₆

• X₁₄ = isotomic conjugate of X₂₉₉

• X₁₄ = complement of X₆₁₇

• X₁₄ = anticomplement of X₆₁₉

• X₁₄ = circumcircle-inverse of X₆₁₀₅

• X₁₄ = orthocentroidal-circle-inverse of X₁₃

• X₁₄ = cevapoint of X₁₆ and X₆₁

• X₁₄ = X_i-cross conjugate of X_j for these (i,j): (16, 17), (30, 13), (395, 2)

• X₁₄ = trilinear pole of line X₃₉₆X₅₂₃ (polar of X₄₇₁ wrt polar circle)

• X₁₄ = pole wrt polar circle of trilinear polar of X₄₇₁

• X₁₄ = X₄₈-isoconjugate (polar conjugate) of X₄₇₁

• X₁₄ = antigonal image of X₁₃

• X₁₄ = reflection of X₁₃ in line X₁₁₅X₁₂₅

• X₁₄ = X₁₆-of-4th-Brocard triangle

• X₁₄ = X₁₆-of-orthocentroidal-triangle

• X₁₄ = orthocorrespondent of X₁₄

• X₁₄ = homothetic center of inner Napoleon triangle and antipedal triangle of X₁₄

• X₁₄ = inner-Napoleon-isogonal conjugate of X₃

• X₁₄ = outer-Napoleon-to-inner-Napoleon similarity image of X₁₆

• X₁₄ = inner-Napoleon-to-outer-Napoleon similarity image of X₁₃

• X₁₄ = orthocenter of X₁₃X₉₈X₂₃₉₄

• X₁₄ = X₁₆-of-pedal-triangle of X₁₄

• X₁₄ = {X₂₆₅,X₁₉₈₉}-harmonic Conjugate of X₁₃

• X₁₄ = homothetic center of (equilateral) antipedal triangle of X₁₄ and triangle formed by circumcenters of BCX₁₄, CAX₁₄, ABX₁₄

• X₁₄ = homothetic center of triangle formed by circumcenters of BCX₁₃, CAX₁₃, ABX₁₃ and triangle formed by nine-point centers of BCX₁₄, CAX₁₄, ABX₁₄

• X₁₄ = Cundy-Parry Phi transform of X₁₈

• X₁₄ = Cundy-Parry Psi transform of X₆₂

• X₁₄ = Kosnita(X₁₄,X₁) point

• X₁₄ = Kosnita(X₁₄,X₁₄) point

• X₁₄ = Thomson-isogonal conjugate of X₃₄₃₁₈

Notes

Let N_aN_bN_c, N_a'N_b'N_c' be the outer and inner Napoleon triangles, resp. Let A' be the isogonal conjugate of N_a, wrt N_a'N_b'N_c', and define B' and C' cyclically. The lines N_a'A', N_b'B', N_c'C' concur in X₁₄. (Randy Hutson, January 29, 2015)

Let Nap_aNap_bNap_c be the outer Napoleon triangle and Nap'_aNap'_bNap'_c the inner Napoleon triangle. Let A* be the isogonal conjugate, wrt Nap'_aNap'_bNap'_c, of Nap_a, and define B* and C* cyclically. The lines Nap'_aA*, Nap'_bB*, Nap'_cC* concur in X₁₄. (Randy Hutson, December 2, 2017)

The line X₁₄X₁₆ is parallel to the Euler line, and the distance between the two lines is \((S_{B} - S_{C})(S_{C} - S_{A})(S_{A} - S_{B})/|(\cot(\omega ) - 3^{1/2})|((E - 8F)S^{2})^{1/2}\). (Kiminari Shinagawa, February 20, 2018)

Let O be a point (not necessarily X₃), let H=ABCDEF be a regular hexagon with center O, and let P be a point. Define 6 triangles T₁=ABP, T₂=BCP, … , T₆=FAP.

Claim 1: The points X₁₄-of-Ti lie on OP, i=1…6.

Next, define 6 reflection triangles T₁' = reflection of T₁ in AB, T₂' = reflection of T₂ in BC, …, T₆' = reflection of T₆ in FA.

Claim 2: If P is interior to H, the points X₁₄-of-Ti’ lie on a rectangular hyperbola centered at O: Figure. (Dan Reznik, December 10, 2021)

X₁₄ is the {X₆,X₃₈₁}-harmonic conjugate of X₁₃.