HYACINTHOS 23129

Antreas P. Hatzipolakis

 

[APH]:

Let ABC be a triangle with orthocenter H, incenter I and excenters Ia,Ib,Ic.
A',B',C' lie on BC,CA,AB, resp. such that  B'C' _|_ IaH, C'A' _|_ IbH, A'B' _|_ IbH.
Prove that the NPC center of the triangle A'B'C' lies on the line IH.

References:
Buratino Giggle [=Tran Quang Hung]
https://www.facebook.com/groups/439719556180075/permalink/444311072387590/
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=625851

Variation

 

The perpendiculars to HIa, HIb, HIc from A,B,C bound triangle A"B"C".

Which point is the circumenter of A"B"C" ?

Is it lying on the IH line?

 

[Peter Moses]:

Hi Antreas,

 

Yes, it is on IH.

 

The point is:

 

2 a^10+a^9 b-4 a^8 b^2-3 a^7 b^3+a^6 b^4+3 a^5 b^5+a^4 b^6-a^3 b^7+a^2 b^8-b^10+a^9 c-2 a^8 b c-4 a^7 b^2 c+a^6 b^3 c+4 a^5 b^4 c+3 a^4 b^5 c-a^2 b^7 c-a b^8 c-b^9 c-4 a^8 c^2-4 a^7 b c^2+4 a^6 b^2 c^2+a^5 b^3 c^2-a^4 b^4 c^2+2 a^3 b^5 c^2-2 a^2 b^6 c^2+a b^7 c^2+3 b^8 c^2-3 a^7 c^3+a^6 b c^3+a^5 b^2 c^3-6 a^4 b^3 c^3-a^3 b^4 c^3+a^2 b^5 c^3+3 a b^6 c^3+4 b^7 c^3+a^6 c^4+4 a^5 b c^4-a^4 b^2 c^4-a^3 b^3 c^4+2 a^2 b^4 c^4-3 a b^5 c^4-2 b^6 c^4+3 a^5 c^5+3 a^4 b c^5+2 a^3 b^2 c^5+a^2 b^3 c^5-3 a b^4 c^5-6 b^5 c^5+a^4 c^6-2 a^2 b^2 c^6+3 a b^3 c^6-2 b^4 c^6-a^3 c^7-a^2 b c^7+a b^2 c^7+4 b^3 c^7+a^2 c^8-a b c^8+3 b^2 c^8-b c^9-c^10::

 

Searches {-0.0268745477834600894141063274825,-0.226063404646283397563799394890,3.80957355333186457346903529626}

 

Best regards,

Peter Moses.