X₁₄₉₀

X₁X_4∩X₃X_9

X₁₄₉₀ = reflection of X₈₄ in X₃

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

X₁₄₉₀ (X₁X_4∩X₃X_9) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

• Central Lines: {84, 3}

• Related Centers: X₈₄ (ISOGONAL CONJUGATE OF X₄₀), X₃ (CIRCUMCENTER)

Trilinears: \(1 + \sec A - \sec B - \sec C : :\)

Search Value (6, 9, 13): \(66.335101010823\)

X₁₄₉₀ lies on the Darboux cubic.

▸ Incident Central Lines17 lines

{1, 4}, {3, 9}, {20, 78}, {40, 64}, {63, 411}, {165, 191}, {224, 908}, {386, 990}, {910, 1192}, {975, 991}, {1045, 1047}, {1210, 1467}, {1498, 3347}, {2130, 3473}, {2131, 3353}, {3182, 3348}, {3183, 3354}

▸ Functional Relations15 relations

• X₁₄₉₀ = isogonal conjugate of X₃₃₄₅

• X₁₄₉₀ = reflection of X₈₄ in X₃

• X₁₄₉₀ = X_i-Ceva conjugate of X_j for these (i,j): (20, 40), (78, 1), (329, 9)

• X₁₄₉₀ = X₂₀₇-cross conjugate of X₁

• X₁₄₉₀ = homothetic center of hexyl and 2nd extouch triangles

• X₁₄₉₀ = X₁₅₅-of-hexyl-triangle

• X₁₄₉₀ = X₁₅₅-of-2nd-extouch-triangle

• X₁₄₉₀ = perspector of hexyl triangle and cevian triangle of X₂₀

• X₁₄₉₀ = perspector of hexyl triangle and anticevian triangle of X₄₀

• X₁₄₉₀ = perspector of ABC and the reflection in X₂₈₂ of the antipedal triangle of X₂₈₂

• X₁₄₉₀ = {X₁,X₁₇₄₅}-harmonic conjugate of X₂₂₃

• X₁₄₉₀ = excentral-isogonal conjugate of X₄₆

• X₁₄₉₀ = excentral-isotomic conjugate of X₁₇₂₁

• X₁₄₉₀ = pedal antipodal perspector of X₁ wrt excentral triangle

• X₁₄₉₀ = ABC-to-excentral barycentric image of X₄

Notes

X₁₄₉₀ is X₆₈-of-the-excentral triangle and the reflection of X₈₄ in X₃. These and other properties itemized here were reported by Darij Grinberg, May 19, 2003.

Let U(A) be the circle with center A having the radius of the A-excircle, and define U(B) and U(C) cyclically. Then X₁₄₉₀ is the radical center of the three circles. (Hauke Reddmann, Hyacinthos, Jan. 8, 2009)

Let O_a be the circle centered at A and passing through the A-excenter, and define O_b and O_c cyclically. The radical center of O_a, O_b, O_c = X₁₄₉₀. (Randy Hutson, September 14, 2016)

Let A'B'C' be the cevian triangle of X₁₀₃₄. Let A″ be the orthocenter of AB'C', and define B″ and C″ cyclically. The lines AA″, BB″, CC″ concur in X₁₄₉₀. (Randy Hutson, September 14, 2016)