HYACINTHOS 27614

[Tran Quang Hung]:

Let ABC be a triangle.
Incenter I.
The reflection of Hatzipolakis axis of triangle IBC in line BC is
da.

Similarly, we have db and dc.

The lines da,db,dc bound triangle A'B'C'.

Circumcenter of triangle A'B'C' lies on OI line of ABC.
Which is this point?

 

[Angel Montesdeoca]:

*** The circumcenter of triangle A'B'C' lies on OI line of ABC and
is

W = R (3 R^2-4 s^2) X(1) + 2 (r R^2+2 R^3+4 r s^2) X(3),

with first barycentric coordinate:

a^2 (a^11+a^10 (b+c)
+a^9 (-5 b^2+b c-5 c^2)
-a^8 (5 b^3+4 b^2 c+4 b c^2+5 c^3)
+a^7 (10 b^4-4 b^3 c+11 b^2 c^2-4 b c^3+10 c^4)
+a^6 (10 b^5+6 b^4 c+9 b^3 c^2+9 b^2 c^3+6 b c^4+10 c^5)
+a^5 (-10 b^6+6 b^5 c-4 b^4 c^2+9 b^3 c^3-4 b^2 c^4+6 b c^5-10 c^6)
-a^4 (10 b^7+4 b^6 c+2 b^5 c^2+5 b^4 c^3+5 b^3 c^4+2 b^2 c^5+4 b c^6+10
c^7)
+a^3 (5 b^8-4 b^7 c-5 b^6 c^2-b^5 c^3+2 b^4 c^4-b^3 c^5-5 b^2 c^6-4 b
c^7+5 c^8)
+a^2 (b-c)^2 (5 b^7+11 b^6 c+10 b^5 c^2+11 b^4 c^3+11 b^3 c^4+10 b^2
c^5+11 b c^6+5 c^7)
-a (b^2-c^2)^4 (b^2-b c+c^2)
-(b-c)^4 (b+c)^5 (b^2-b c+c^2) )

(6 - 9 - 13) - search numbers of W: (-183.709240207533,
-151.398945410382, 193.244199092572).

Angel Montesdeoca