X(67) = ISOGONAL CONJUGATE OF X(23)

Trilinears

\(bc/(b4 + c4 - a4 - b2c2) : :\)

Barycentrics

\(1/(b4 + c4 - a4 - b2c2) : : Let A' be the reflection in BC of the A-vertex of the antipedal triangle of :ref:`X(6) <X(6)>\), and define B’ and C’ cyclically. The circumcircles of A’BC, B’CA, C’AB concur at X(67). Also, let A’ be the reflection of X(6) in BC, and define B’ and C’ cyclically. The circumcircles of A’BC, B’CA, C’AB concur in X(67). Note: the above 2 sets of circumcircles are identical. (Randy Hutson, November 18, 2015) X(67) lies on these lines: 3,542 4,338 6,125 50,248 74,935 110,141 265,511 290,340 524,858 526,879 X(67) = midpoint of X(69) and X(3448) <X(3448)>`

Notes

Let A’ be the reflection in BC of the A-vertex of the antipedal triangle of X(6), and define B’ and C’ cyclically. The circumcircles of A’BC, B’CA, C’AB concur at X(67). Also, let A’ be the reflection of X(6) in BC, and define B’ and C’ cyclically. The circumcircles of A’BC, B’CA, C’AB concur in X(67). Note: the above 2 sets of circumcircles are identical. (Randy Hutson, November 18, 2015)

X(67) lies on these lines: 3,542 4,338 6,125 50,248 74,935 110,141 265,511 290,340 524,858 526,879

X(67) = midpoint of X(69) and X(3448)

X(67) = reflection of X(i) in X(j) for these (i,j): (6,125), (110,141)

X(67) = isogonal conjugate of X(23)

X(67) = isotomic conjugate of X(316)

X(67) = circumcircle-inverse of X(3455)

X(67) = cevapoint of X(141) and X(524)

X(67) = X(187)-cross conjugate of X(2)

X(67) = antigonal image of X(6)

X(67) = trilinear pole of line X(39)X(647)

X(67) = polar conjugate of X(37765)

X(67) = pole wrt polar circle of trilinear polar of X(37765) (line X(9517)X(9979))

X(67) = X(63)-isoconjugate of X(8744)

X(67) = orthocenter of X(3)X(74)X(879)

X(67) = perspector of ABC and X(2)-Ehrmann triangle; see X(25)

X(67) = X(19)-isoconjugate of X(22151)