X(97) = ISOGONAL CONJUGATE OF X(53)

Trilinears

\(cot A sec(B - C) : cot B sec(C - A) : cot C sec(A - B)\)

Barycentrics

\(cos A sec(B - C) : cos B sec(C - A) : cos C sec(A - B) In the plane of a triangle ABC, let\)

Notes

In the plane of a triangle ABC, let Ia = line through X(3) parallel to BC, and define Ib and Ic cyclically. Ac = Ia∩AB, and define Ba and Cb cyclically. Ab = Ia∩AC, and define Bc and Ca cyclically. Oa = circumcircle of A, Bc, Cb, and define Ob and Oc cyclically. Then

X(97) = radical center of Oa, Ob, Oc. See also X(77). (Ivan Pavlov, April 1, 2022)

X(97) lies on these lines: 2,95 3,54 110,418 216,288 276,401

X(97) = isogonal conjugate of X(53)

X(97) = isotomic conjugate of X(324)

X(97) = anticomplement of X(34836)

X(97) = complement of isogonal conjugate of X(34433)

X(97) = X(95)-Ceva conjugate of X(54)

X(97) = X(3)-cross conjugate of X(95)

X(97) = cevapoint of X(3) and X(577)

X(97) = X(51)-isoconjugate of X(92)

X(97) = Cundy-Parry Phi transform of X(7592)

Let Y = Λ(P,X), let Q = isogonal conjugate of P, and let Y and Z be the points where line YQ meets the circumcircle; then Ψ(P,X) = Z.