Property of Darboux cubic related to 2016 USA TSTST
by TelvCohl, Jul 13, 2016, 11:45 AM
Property : Given a 
and a point 
lying on the Darboux cubic of 
Let 
be the pedal triangle of WRT and let 
be the perspector of 
Let 
be the intersection of 
and let 
be the circumcircle of 
(define 
similarly). Then is the radical center of 
Proof :
Lemma (well-known) : Given a and two points 
Let 
be the cevian triangle of WRT and let 
cuts 
at 
respectively. Then 
are concurrent.
Proof : From Cevian nest theorem we get
is the cevian triangle of 
so notice is the cevian triangle of 
WRT 
we conclude that (by Cevian nest theorem) are concurrent.
Back to the main problem :
Let
cuts 
at 
respectively. Clearly, 
is the radical center of 
and the pedal circle 
of WRT 
lies on the radical axis 
of 
so if 
cuts again at then 
Analogously, we can prove 
lies on 
and 
lies on 
so the radical axis 
of passes through and the intersection of 
hence from Lemma we conclude that 
Similarly, we can prove lies on the radical axis of 
and so is the radical center of
and a point 
lying on the Darboux cubic of 
Let 
be the pedal triangle of WRT and let 
be the perspector of 
Let 
be the intersection of 
and let 
be the circumcircle of 
(define 
similarly). Then is the radical center of 
Proof :
Lemma (well-known) : Given a
and two points 
Let 
be the cevian triangle of WRT and let 
cuts 
at 
respectively. Then 
are concurrent.Proof : From Cevian nest theorem we get
is the cevian triangle of 
so notice is the cevian triangle of 
WRT 
we conclude that (by Cevian nest theorem) are concurrent.Back to the main problem :
Let
cuts 
at 
respectively. Clearly, 
is the radical center of 
and the pedal circle 
of WRT 
lies on the radical axis 
of 
so if 
cuts again at then 
Analogously, we can prove 
lies on 
and 
lies on 
so the radical axis 
of passes through and the intersection of 
hence from Lemma we conclude that 
Similarly, we can prove lies on the radical axis of 
and so is the radical center of 
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