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The Kosnita triangle is formed by the circumcenters Oa, Ob, Oc of the three triangles OBC, OCA, OAB. It is homothetic to several triangles namely the orthic, tangential, extangents triangles for example.

Now, let P be a point in the plane of the reference triangle ABC and denote by

cevP : the cevian triangle of P
acvP : the anticevian triangle of P
pedP : the pedal triangle of P
apdP : the antipedal triangle of P
ccvP : the circumcevian triangle of P
cacP : the circumanticevian triangle of P.

We seek the locus of P such that the Kosnita triangle is perspective, orthologic, parallelogic to one of these triangles.

The following table gives these loci. L denotes the line at infinity, C denotes the circumcircle. Obviously, concerning orthologic and parallelogic triangles, the Kosnita triangle can be replaced by any homothetic triangle.






pK(X2, X311) (*)

Euler perspector cubic K045

nK(X4, X264) = K211


pK(X571, X1994) (*)

K612 = pK(X216, X2)

circum-cubic K0


a K+

L + Euler line

L + line X186-X523 = tripolar of X275


Jerabek strophoid K039

L + C + Jerabek hyperbola

L + C + circum-conic with center X(5) through X110, X265, X1625


L + C + Napoleon cubic K005

C + Kosnita quartic Q064

C + quartic through X186


isogonal sextic

sextic through X2, X6


(*) in both cases, the locus of the perspector is K388.