Let ABC be a triangle and AaBbCc the orthic triangle.
The circle (Aa,AaA) intersects AB,AC at Ab,Ac, resp. (other than A)
The circle (Bb,BbB) intersects BC,BA at Bc,Ba, resp. (other than B)
The circle (Cc,CcC) intersects CA,CB at Ca,Cb, resp. (other than C)
The Euler lines of the triangles:
1. AAbAc, BBcBa, CCaCb
2. AaAbAc, BbBcBa, CcCaCb
[=perp. bisectors of AbAc, BcBa, CaCb, resp.]
are concurrent.
Points ?
APH, 21 February 2012
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1) (-s1*s2^2+2*s1^2*s3+s2*s3)/(2*s1*s3)
2) (3*s1*s3-s2^2)/(2*s3)
It is the orhocenter of the orthic triangle.
Where I use Moley'trick and s1=a+b+c, s2=ab+bc+ca and s3=abc
Etienne Rousee, Hyacinthos #20850
The second part of the problem is obvious. Here is a proof for the first part of the problem:-
ΑπάντησηΔιαγραφήClearly, the Euler line of AAbAc passes through Aa and anti-parallel to OH wrt \angle BAC. Suppose, A1B1C1 is the circumcevian triangle of H wrt ABC. Isogonal conjugate of the point at infinity on OH wrt ABC be P. Note that, HO passes through the Kosnita point of HB1C1. So the isogonal conjugate of HO wrt \angle HB1C1 passes through the reflection of A on B1C1, let that point be A'. If we reflect HA' on the midpoint of B1C1, then suppose, H goes to A2, so HA' goes to A2A. Clearly, A2B2C2 is homothetic to AaBbCc and A2A is anti-parallel to OH wrt \angle BAC and A2A,B2B,C2C concurr at P. So the Euler lines of AAbAc,BBcBa,CCbCa are concurrent.