Trilinears           bcf(a,b,c) : caf(b,c,a): abf(c,a,b), where f(a,b,c) is the 1st barycentric given below

Trilinears           g(A,B,C) : g(B,C,A) : g(C,A,B),
                                    where g(A,B,C) = (J - 1)cos A + 2(J + 1)cos B cos C, where J = |OH|/R; see X(1113)

Barycentrics    f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = (J - 1)a²SA + 2(J + 1) SB SC, where
                                    S_A = (b² + c² - a²)/2, and S_B and S_C are defined cyclically.

X(1312) is a point of intersection of the Euler line and the nine-point circle. Its antipode on the nine-point circle
is X(1313). Of the two points, X(1312) is the one closer to X(4). The asymptotes of the Jerabek hyperbola meet at X(125) on the nine-point circle. One of the asymptotes meets the circle again at X(1312), and the other, at X(1313). Thus, the points X(125), X(1312), X(1313) form a right triangle of which the midpoint of the hypotenuse X(1312)-to-X(1313) is X(5). (Peter J. C. Moses, 3/14/2003)

X(1312) lies on this line: 2,3

X(1312) lies on the nine-point circle

X(1312) = {X(2),X(1113)}-harmonic conjugate of X(468)
X(1312) = {X(2),X(858)}-harmonic conjugate of X(1313)
X(1312) = {X(4),X(403)}-harmonic conjugate of X(1313)
X(1312) = {X(427),X(468)}-harmonic conjugate of X(1313)

For a list of harmonic conjugates, click More at the top of this page.

X(1312) = midpoint of X(4) and X(1113)
X(1312) = reflection of X(1313) in X(5)
X(1312) = complement of X(1114)
X(1312) = X(1113)-Ceva conjugate of X(523)