Interactive Applet |
You can move the points A, B and C (click on the point and drag it).
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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 2CThis 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, L_{a}, L_{b}, L_{c} the perpendicular bisectors of sides BC, CA, AB, and AA', BB', CC' the altitudes of ABC, respectively. Let A_{b} be the point of intersection of AA' and L_{b}, and let A_{c} be the point of intersection of AA' and L_{c}. Let A" be the point of intersection of BA_{b} and CA_{c}. 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)