## X(847) (X(5)-CROSS CONJUGATE OF X(4))

 Interactive Applet

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 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)

This is a joint work of
Humberto José Bortolossi, Lis Ingrid Roque Lopes Custódio and Suely Machado Meireles Dias.