Interactive Applet |
You can move the points A, B and C (click on the point and drag it).
Press the keys “+” and “−” to zoom in or zoom out the visualization window and use the arrow keys to translate it.
You can also construct all centers related with this one (as described in ETC) using the “Run Macro Tool”. To do this, click on the icon _{}, select the center name from the list and, then, click on the vertices A, B and C successively.
Information from Kimberling's Encyclopedia of Triangle Centers |
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)a^{2}SA + 2(J + 1) SB SC, where
S_{A} = (b^{2} + c^{2} - a^{2})/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)