Παρασκευή 10 Φεβρουαρίου 2012

Perspective


Let ABC be a triangle, A'B'C' the orthic triangle, A1B1C1 the cevian triangle of G and A2B2C2 the circumcevian triangle of G with respect A1B1C1.


Denote:

A* = A2O /\ A'N

B* = B2O /\ B'N

C* = C2O /\ C'N

The triangles ABC, A*B*C* are perspective (?)
(perspector on the Euler line?)

Variation:

A** = A2N /\ A'O

B** = B2N /\ B'O

C** = C2N /\ C'O

Are the triangles:

ABC, A**B**C**

A*B*C*, A**B**C**

perspective?

APH, 10 February 2012

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X(73160) = X(2)X(6120)∩X(15859)X(66554) Barycentrics    Sin[B/3]*Sin[(2*B)/3]*Sin[B] + (Sin[A]*Sin[B] + Sin[C/3]*Sin[(2*C)/3])*Sin[C] : : ...