Let ABC be a triangle.
Denote:
Na, Nb, Nc = the NPC centers of NBC, NCA, NAB, resp.
Ma, Mb, Mc = the midpoints of AN, BN, CN, resp.
M1, M2, M3 = the midpoints of MaNa, MbNb, McNc, resp.
The NPC center of M1M2M3 lies on the Euler line.
Generalization (conjecture):
Let ABC be a triangle and P a point on the Euler line.
Denote:
Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.
Ma, Mb, Mc = the midpoints of AP, BP, CP, resp.
M1, M2, M3 = the midpoints of MaNa, MbNb, McNc, resp.
The P-point P* of M1M2M3 (ie the same, as the point P of ABC, point P* of M1M2M3) lies on the Euler line of ABC for P = N, O, G, X140 .... but not for P = X21.
Which is P* in terms of P or for some simple points P on the Euler line?
[Angel Montesdeoca]:
**** If P is a point on the Euler line:
OP/PH=t, then OP*/P*H = p/q
p = a^12 t^2 (-1-4 t-t^2+2 t^3+3 t^4)
-a^10 (b^2+c^2) t (-1-7 t-15 t^2-5 t^3+4 t^4+6 t^5)
-a^8 (b^4 t (4+19 t+24 t^2+11 t^3+2 t^4+3 t^5)+c^4 t (4+19 t+24 t^2+11 t^3+2 t^4+3 t^5)+b^2 c^2 (1+4 t+14 t^2+40 t^3+23 t^4-18 t^6))
+a^6 (2 b^6 t (3+13 t+13 t^2+7 t^3+4 t^4+6 t^5)+2 c^6 t (3+13 t+13 t^2+7 t^3+4 t^4+6 t^5)+b^4 c^2 (1+3 t+7 t^2+25 t^3+18 t^4+4 t^5-12 t^6)+b^2 c^4 (1+3 t+7 t^2+25 t^3+18 t^4+4 t^5-12 t^6))
-a^4 (b^8 t (4+19 t+24 t^2+11 t^3+2 t^4+3 t^5)+c^8 t (4+19 t+24 t^2+11 t^3+2 t^4+3 t^5)+b^4 c^4 (1-4 t-20 t^2+10 t^3+14 t^4+4 t^5-30 t^6)+b^6 c^2 (-1-3 t-7 t^2-25 t^3-18 t^4-4 t^5+12 t^6)+b^2 c^6 (-1-3 t-7 t^2-25 t^3-18 t^4-4 t^5+12 t^6))
-a^2 (b^2-c^2)^2 (b^2+c^2) (b^4 t (-1-7 t-15 t^2-5 t^3+4 t^4+6 t^5)+c^4 t (-1-7 t-15 t^2-5 t^3+4 t^4+6 t^5)+b^2 c^2 (1+3 t+7 t^2+25 t^3+18 t^4+4 t^5-12 t^6))
+(b^2-c^2)^4 t (b^4 t (-1-4 t-t^2+2 t^3+3 t^4)+c^4 t (-1-4 t-t^2+2 t^3+3 t^4)+b^2 c^2 (1+3 t-t^2+t^3+4 t^4+6 t^5))
q = a^12 t^2 (-3-4 t+t^2+6 t^3+t^4)
-a^10 (b^2+c^2) t (-3-13 t-13 t^2+t^3+12 t^4+2 t^5)
-a^8 (b^4 t (12+25 t+16 t^2+5 t^3+6 t^4+t^5)+c^4 t (12+25 t+16 t^2+5 t^3+6 t^4+t^5)+b^2 c^2 (3+4 t+26 t^2+48 t^3+21 t^4-32 t^5-6 t^6))
+a^6 (2 b^6 t (9+15 t+7 t^2+5 t^3+12 t^4+2 t^5)+2 c^6 t (9+15 t+7 t^2+5 t^3+12 t^4+2 t^5)+b^4 c^2 (3+t+13 t^2+35 t^3+22 t^4-20 t^5-4 t^6)+b^2 c^4 (3+t+13 t^2+35 t^3+22 t^4-20 t^5-4 t^6))
-a^4 (b^8 t (12+25 t+16 t^2+5 t^3+6 t^4+t^5)+c^8 t (12+25 t+16 t^2+5 t^3+6 t^4+t^5)+b^4 c^4 (7-20 t-28 t^2+30 t^3+34 t^4-52 t^5-10 t^6)+b^6 c^2 (-3-t-13 t^2-35 t^3-22 t^4+20 t^5+4 t^6)+b^2 c^6 (-3-t-13 t^2-35 t^3-22 t^4+20 t^5+4 t^6))
-a^2 (b^2-c^2)^2 (b^2+c^2) (b^4 t (-3-13 t-13 t^2+t^3+12 t^4+2 t^5)+c^4 t (-3-13 t-13 t^2+t^3+12 t^4+2 t^5)+b^2 c^2 (3+t+13 t^2+35 t^3+22 t^4-20 t^5-4 t^6))
+(b^2-c^2)^4 t (b^4 t (-3-4 t+t^2+6 t^3+t^4)+c^4 t (-3-4 t+t^2+6 t^3+t^4)+b^2 c^2 (3+t-3 t^2+3 t^3+12 t^4+2 t^5))
In particular:
P = X(2), P* = X(547) = X(2)+X(5) = 7 X(3) + 5 X(4)
P = X(3), P* = X(5498)
X(5498) = 6th HATZIPOLAKIS-MONTESDEOCA POINT
Let ABC be a triangle, let Na be the nine-point center of the triangle BCO, where O = X(3), and define Nb and Nc cyclically. The nine-point center of the triangle NaNbNc is X(5498), which lies on the Euler line of ABC. (Antreas Hatzipolakis, May 30, 2013)
P = X(4), P* =X(546) = X(4) + X(5) = X(3) + 3 X(4)
P = N= X(5),
P* = (2 a^16
-5 a^14 (b^2+c^2)
-a^12 (17 b^4+26 b^2 c^2+17 c^4)
+a^10 (91 b^6+115 b^4 c^2+115 b^2 c^4+91 c^6)
-3 a^8 (55 b^8+28 b^6 c^2+28 b^4 c^4+28 b^2 c^6+55 c^8)
+3 a^6 (b^2+c^2) (51 b^8-92 b^6 c^2+79 b^4 c^4-92 b^2 c^6+51 c^8)
-a^4 (b^2-c^2)^2 (75 b^8-68 b^6 c^2-93 b^4 c^4-68 b^2 c^6+75 c^8)
+a^2 (b^2-c^2)^4 (b^2+c^2) (17 b^4-64 b^2 c^2+17 c^4)
-(b^2-c^2)^6 (b^4-14 b^2 c^2+c^4) : .... : ...) =
(a^12+14 a^10 (b^2+c^2)+(b^2-c^2)^4 (b^4+18 b^2 c^2+c^4)-a^8 (65 b^4+64 b^2 c^2+65 c^4)+50 a^6 (2 b^6+b^4 c^2+b^2 c^4+2 c^6)+2 a^2 (b^2-c^2)^2 (7 b^6-18 b^4 c^2-18 b^2 c^4+7 c^6)+a^4 (-65 b^8+50 b^6 c^2+39 b^4 c^4+50 b^2 c^6-65 c^8)) X(3)
- (a^12-18 a^10 (b^2+c^2)+(b^2-c^2)^4 (b^4-14 b^2 c^2+c^4)+a^8 (63 b^4+64 b^2 c^2+63 c^4)-46 a^6 (2 b^6+b^4 c^2+b^2 c^4+2 c^6)-2 a^2 (b^2-c^2)^2 (9 b^6-14 b^4 c^2-14 b^2 c^4+9 c^6)+a^4 (63 b^8-46 b^6 c^2-25 b^4 c^4-46 b^2 c^6+63 c^8)) X(4),
On lines: {2,3}
with (6 - 9 - 13) - search number (0.396416771170110, -0.474148909179255, 3.78595983233766).
P* = Midpoint of X(5501) and X(10289)
X(5501) = 9th HATZIPOLAKIS-MONTESDEOCA POINT
Let N be a the nine-point center of triangle ABC. Let Na be the nine-point center of NBC, and define Nb and Nc cyclically. The circumcenter of NaNbNc is X(5501), which lies on the Euler line of ABC. (Antreas Hatzipolakis, June 2, 2013)
X(10289) = 7th HATZIPOLAKIS-MOSES-EULER POINT
Let A'B'C' the pedal triangle of the nine-point center, N = X(5), of a triangle ABC. Let
Oa = circumcenter of NB'C', and define Ob and Oc cyclically
Ooa = circumcenter of NObOc, and define Oob and Ooc cyclically.
Then X(10289) = nine-point center of OoaOobOoc; this point lies on the Euler line of ABC. See Antreas Hatzipolakis and Peter Moses, Hyacinthos 24670.
P* = Reflection of X(12057) in X(3628)
X(12057) = MIDPOINT OF X(140) AND X(10289) (Antreas Hatzipolakis and César Lozada, Hyacinthos 25504)
X(3628) is the centroid of the set {A', B', C', X(5)}, where A'B'C' is the medial triangle; more generally, H-1(X; M, 2) is the centroid of the set {A', B', C', X}. (Angel Montesdeoca, 12/20/2011)
Angel Montesdeoca
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