[Antreas P. Hatzipolakis]:
Let ABC be a triangle, P, Q two points on the Euler line and PaPbPc, QaQbQc the pedal triangles of P,Q, resp.
Denote:
Na, Nb, Nc = the NPC centers of PPbPc, PPcPa, PPaPb, resp.
N1, N2, N3 = the reflections of Na, Nb, Nc in QQa QQb, QQc, resp.
The circumcenter O* of N1N2N3 lies on the Euler line
Which is the ratio O*P/O*Q ?
For P = H, Q = N, the O* is the midpoint of HN.
[Angel Montesdeoca]:
** For P = H, Q = N, the O* is the midpoint of HN.
More in general, if P, Q are two points on the Euler line such that OP/PH=m, OQ/QH=n=m/(m+2), then O* is the midpoint of PQ.
Examples: P=N (m=1), Q=X(140) (n=1/3) , O*=X(3628) = midpoint of X(5) and X(140).
P=G (m=1/2), Q=X(549) (n=1/5) , O*=X(140)= midpoint of X(2) and X(549).
P=X(20), Q=X(550) , O*=(10 a^4 -3(b^2-c^2)^2-7a^2 (b^2+c^2):..:...:)= midpoint of X(20) and X(550).
On lines {2,3}....
P=X(21), Q=X(5428) , O*=(a (4 a^6-4 a^5 (b+c)+a^4 (-8 b^2+2 b c-8 c^2)-3 b c (b^2-c^2)^2+8 a^3 (b^3+c^3)+a^2 (4 b^4+b^3 c+8 b^2 c^2+b c^3+4 c^4)-4 a (b^5-b^4 c-b c^4+c^5)):..:...:)= midpoint of X(21) and X(5428).
On lines {2,3}....
P=X(22), Q=X(7502) , O*=X(7555)= midpoint of X(22) and X(7502).
P=X(23), Q=X(7575) , O*=(a^2 (4 a^8+2 a^4 b^2 c^2-8 a^6 (b^2+c^2)-(b^2-c^2)^2 (4 b^4-b^2 c^2+4 c^4)+a^2 (8 b^6-3 b^4 c^2-3 b^2 c^4+8 c^6)):..:...:)= midpoint of X(23) and X(7575).
On lines {2,3}....
P=X(25), Q=X(6644) , O*=(a^2 (a^8+2 a^4 b^2 c^2-2 a^6 (b^2+c^2)-(b^2-c^2)^2 (b^4-b^2 c^2+c^4)+a^2 (2 b^6-3 b^4 c^2-3 b^2 c^4+2 c^6)):..:...:)= midpoint of X(25) and X(6644).
On lines {2,3}....
P=X(26), Q=X(1658) , O*=(a^2 (2 a^8+2 a^4 b^2 c^2-4 a^6 (b^2+c^2)-(b^2-c^2)^2 (2 b^4+b^2 c^2+2 c^4)+a^2 (4 b^6-b^4 c^2-b^2 c^4+4 c^6)):..:...:)= midpoint of X(26) and X(1658).
On lines {2,3}....
P=X(140), Q=X(3530) , O*=(10 a^4+3 (b^2-c^2)^2-13 a^2 (b^2+c^2):..:...:)= midpoint of X(140) and X(3530).
On lines {2,3}....
Angel Montesdeoca
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