[Tran Quang Hung]:
Let ABC be a triangle with NPC center N.
The reflection of the circle (NBC) in the lines CA, AB meets BC at Cb, Bc, resp.
Define similarly the points Ac, Ca and Ba, Ab.
Let Oa, Ob, Oc be the circumenters of the triangles AAbAc, BBcBa, CCaAb, resp.
Then X(186) of the triangle OaObOc lies on the Euler of ABC. Which is this point?
Let Ka, Kb, Kc be the circumcenters of the triangles ABaCa, BCbAb, CAcBc, resp.
Then X(186) of the triangle KaKbKc lies on the Euler line of ABC. Which is this point?
Also, if H1 and H2 be the orthocenters of the triangles OaObOc and KaKbKc, resp. then the line H1H2 is parallel to the Euler line of ABC. Which is this line?
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[Ercole Suppa]
Dear Tran Quang Hung,
X(186) of the triangle OaObOc is the point:
P1 = (name pending)
= 2 a^22-9 a^20 (b^2+c^2)+4 a^18 (3 b^4+8 b^2 c^2+3 c^4)-(b^2-c^2)^8 (b^6+b^4 c^2+b^2 c^4+c^6)+a^16 (3 b^6-41 b^4 c^2-41 b^2 c^4+3 c^6)+a^14 (-20 b^8+28 b^6 c^2+74 b^4 c^4+28 b^2 c^6-20 c^8)+2 a^2 (b^2-c^2)^6 (2 b^8-b^6 c^2-b^2 c^6+2 c^8)+2 a^10 b^2 c^2 (11 b^8+3 b^6 c^2+26 b^4 c^4+3 b^2 c^6+11 c^8)-2 a^8 (b^2-c^2)^2 (b^10+3 b^8 c^2-6 b^6 c^4-6 b^4 c^6+3 b^2 c^8+c^10)-a^4 (b^2-c^2)^4 (5 b^10-17 b^8 c^2-b^6 c^4-b^4 c^6-17 b^2 c^8+5 c^10)+a^12 (14 b^10-24 b^8 c^2-71 b^6 c^4-71 b^4 c^6-24 b^2 c^8+14 c^10)+2 a^6 (b^2-c^2)^2 (b^12-10 b^10 c^2+b^8 c^4-6 b^6 c^6+b^4 c^8-10 b^2 c^10+c^12) : : (barys)
= R^2 S^4 + (-92 R^6-21 R^2 SB SC+99 R^4 SW+4 SB SC SW-35 R^2 SW^2+4 SW^3) S^2 + 132 R^6 SB SC-157 R^4 SB SC SW+63 R^2 SB SC SW^2-8 SB SC SW^3 : : (barys)
As a point on the Euler line, X() has Shinagawa coefficients {92 R^6-99 R^4 SW-4 SW^3-R^2 (S^2-35 SW^2), -132 R^6+157 R^4 SW-4 S^2 SW+8 SW^3+21 R^2 (S^2-3 SW^2)}
= lies on this line: {2,3}
= (6-8-13) search numbers [-10.7997916222310703, -11.6408214221380734, 16.6842139074943085]
***************
X(186) of the triangle KaKbKc is the point:
P2 = (name pending)
= 2 a^22-11 a^20 (b^2+c^2)+23 a^18 (b^2+c^2)^2-(b^2-c^2)^8 (b^6+c^6)-a^16 (19 b^6+67 b^4 c^2+67 b^2 c^4+19 c^6)+a^14 (-6 b^8+28 b^6 c^2+62 b^4 c^4+28 b^2 c^6-6 c^8)+a^2 (b^2-c^2)^6 (5 b^8-3 b^6 c^2-5 b^4 c^4-3 b^2 c^6+5 c^8)-a^4 (b^2-c^2)^4 (9 b^10-11 b^8 c^2-7 b^6 c^4-7 b^4 c^6-11 b^2 c^8+9 c^10)+a^8 (b^2-c^2)^2 (12 b^10+13 b^8 c^2+29 b^6 c^4+29 b^4 c^6+13 b^2 c^8+12 c^10)+a^12 (28 b^10+18 b^8 c^2-13 b^6 c^4-13 b^4 c^6+18 b^2 c^8+28 c^10)+2 a^6 (b^2-c^2)^2 (2 b^12-5 b^10 c^2-6 b^6 c^6-5 b^2 c^10+2 c^12)-a^10 (28 b^12+7 b^10 c^2+13 b^8 c^4-24 b^6 c^6+13 b^4 c^8+7 b^2 c^10+28 c^12) : : (barys)
= (5 R^2-2 SW) S^4 + (-160 R^6-51 R^2 SB SC+164 R^4 SW+14 SB SC SW-55 R^2 SW^2+6 SW^3) S^2 + 192 R^6 SB SC - 212 R^4 SB SC SW+81 R^2 SB SC SW^2-10 SB SC SW^3 : : (barys)
As a point on the Euler line, X() has Shinagawa coefficients {(5 R^2 - 2 SW) (32 R^4 - S^2 - 20 R^2 SW + 3 SW^2), -192 R^6 + 212 R^4 SW - 14 S^2 SW + 10 SW^3 + R^2 (51 S^2 - 81 SW^2)}
= lies on this line: {2,3}
= (6-8-13) search numbers [-7.19335336272051906, -8.04389703781748783, 12.5295255216521102]
*************** 
The line H1H2 is the trilinear polar of the point
P3 = (a-b) (a+b) (a-c) (a+c) (a^2+b^2-c^2) (a^2-b^2+c^2) (a^16 (b^2-c^2)+b^2 (b^2-c^2)^6 (b^2+c^2)^2-a^14 (4 b^4+b^2 c^2-5 c^4)+a^12 (4 b^6+10 b^4 c^2-b^2 c^4-9 c^6)-a^2 (b^2-c^2)^4 (4 b^8+6 b^6 c^2+4 b^4 c^4+5 b^2 c^6+c^8)+a^10 (4 b^8-16 b^6 c^2-14 b^4 c^4-3 b^2 c^6+5 c^8)+a^8 (-10 b^10+15 b^8 c^2+10 b^6 c^4+8 b^4 c^6+8 b^2 c^8+5 c^10)+a^4 (b^2-c^2)^2 (4 b^10+4 b^8 c^2-b^6 c^4-b^4 c^6+9 b^2 c^8+5 c^10)+a^6 (4 b^12-9 b^10 c^2+5 b^8 c^4+4 b^6 c^6+8 b^4 c^8-3 b^2 c^10-9 c^12)) (a^16 (b^2-c^2)-c^2 (b^2-c^2)^6 (b^2+c^2)^2+a^14 (-5 b^4+b^2 c^2+4 c^4)+a^12 (9 b^6+b^4 c^2-10 b^2 c^4-4 c^6)+a^10 (-5 b^8+3 b^6 c^2+14 b^4 c^4+16 b^2 c^6-4 c^8)+a^2 (b^2-c^2)^4 (b^8+5 b^6 c^2+4 b^4 c^4+6 b^2 c^6+4 c^8)-a^8 (5 b^10+8 b^8 c^2+8 b^6 c^4+10 b^4 c^6+15 b^2 c^8-10 c^10)-a^4 (b^2-c^2)^2 (5 b^10+9 b^8 c^2-b^6 c^4-b^4 c^6+4 b^2 c^8+4 c^10)+a^6 (9 b^12+3 b^10 c^2-8 b^8 c^4-4 b^6 c^6-5 b^4 c^8+9 b^2 c^10-4 c^12)) : : (barys)
= (6-8-13) search numbers [1.08939964531861435, 3.18120003404388610, 0.935495391268482101]
Best regards
Ercole Suppa
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