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 sec A sec 2A : sec B sec 2B : sec C sec 2C
Barycentrics tan A sec 2A : tan B sec 2B : tan C sec 2C
This point solves a problem posed by Antreas Hatzipolakis in Hyacinthos #3130, June 25, 2001; see also Jean-Pierre Ehrmann, #3135, June 26, 2001. The problem and solution may be stated as follows. Let ABC be a triangle, La, Lb, Lc the perpendicular bisectors of sides BC, CA, AB, and AA', BB', CC' the altitudes of ABC, respectively. Let Ab be the point of intersection of AA' and Lb, and let Ac be the point of intersection of AA' and Lc. Let A" be the point of intersection of BAb and CAc. Define B" and C" cyclically. Then triangle A"B"C" is perspective to triangle ABC, with perspector X(847).
X(847) lies on the ; see McCay orthic cubic.
X(847) lies on these lines: 2,254 3,925 4,52 24,96 91,225 378,1105 403,1093
X(847) = isogonal conjugate of X(1147)
X(847) = X(5)-cross conjugate of X(4)
X(847) = cevapoint of X(485) and X(486)