Let ABC be a triangle, A'B'C' the pedal triangle of O and P a point on the Euler line such that OP/OH = t : number
Denote:
Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.
A", B", C" = the midpoints of AP, BP, CP, resp,
Ma, Mb, Mc = the midpoints of A'Na, B'Nb, C'Nc, resp
M1, M2, M3 = the midpoints of A"Ma, B"Mb, C"Mc, resp.
Conjecture:
The P point of M1M2M3 (ie the point P' on the Euler line of M1M2M3 such that O'P'/O'H' = t, where O', H' = the circumcenter, orthocenter of M1M2M3, resp.) lies on the Euler line of ABC.
P = N: Hyacinthos 29611
P = O: Hyacinthos 29613
[César Lozada]:
Conjecture proved.
t’ = OP’/OH =
= 1/8*(3*(a^6-(b^2+c^2)*a^4-(b^4-3*b^2*c^2+c^4)*a^2+(b^4-c^4)*(b^2-c^2))^2*t^6-4*(a^6-(b^2+c^2)*a^4-(b^4-3*b^2*c^2+c^4)*a^2+(b^4-c^4)*(b^2-c^2))^2*t^5+(7*a^12-11*(b^2+c^2)*a^10-(19*b^4-56*b^2*c^2+19*c^4)*a^8+(b^2+c^2)*(46*b^4-91*b^2*c^2+46*c^4)*a^6-(19*b^8+19*c^8+3*b^2*c^2*(15*b^4-41*b^2*c^2+15*c^4))*a^4-(b^4-c^4)*(b^2-c^2)*(11*b^4-45*b^2*c^2+11*c^4)*a^2+(7*b^4+17*b^2*c^2+7*c^4)*(b^2-c^2)^4)*t^4+(a^12-6*(b^2+c^2)*a^10+(15*b^4-b^2*c^2+15*c^4)*a^8-(b^2+c^2)*(20*b^4-27*b^2*c^2+20*c^4)*a^6+(15*b^8+15*c^8+b^2*c^2*(7*b^4-36*b^2*c^2+7*c^4))*a^4-(b^4-c^4)*(b^2-c^2)*(6*b^4+7*b^2*c^2+6*c^4)*a^2+(b^2-c^2)^6)*t^3+(-2*a^12+3*(b^2+c^2)*a^10+2*(b^2-3*c^2)*(3*b^2-c^2)*a^8-(b^2+c^2)*(14*b^4-31*b^2*c^2+14*c^4)*a^6+(6*b^8+6*c^8+b^2*c^2*(17*b^4-45*b^2*c^2+17*c^4))*a^4+(b^4-c^4)*(b^2-c^2)*(3*b^4-17*b^2*c^2+3*c^4)*a^2-(b^2-c^2)^4*(b^2+2*c^2)*(2*b^2+c^2))*t^2+(2*(b^2+c^2)*a^10-(8*b^4-5*b^2*c^2+8*c^4)*a^8+(b^2+c^2)*(12*b^4-19*b^2*c^2+12*c^4)*a^6-(8*b^8+8*c^8+b^2*c^2*(7*b^4-26*b^2*c^2+7*c^4))*a^4+(b^4-c^4)*(b^2-c^2)*(2*b^4+7*b^2*c^2+2*c^4)*a^2+2*(b^2-c^2)^4*b^2*c^2)*t-a^2*b^2*c^2*(2*a^6-2*(b^2+c^2)*a^4-(2*b^4-3*b^2*c^2+2*c^4)*a^2+2*(b^4-c^4)*(b^2-c^2)))/((2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*t-a^2*(a^2-b^2-c^2))/((a^4+(b^2-2*c^2)*a^2-(b^2-c^2)*(2*b^2+c^2))*t-b^2*(a^2-b^2+c^2))/((a^4-(2*b^2-c^2)*a^2+(b^2-c^2)*(b^2+2*c^2))*t-c^2*(a^2+b^2-c^2))
i.e,
P’(t) = 3*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*(a^6-(b^2+c^2)*a^4-(b^4-3*b^2*c^2+c^4)*a^2+(b^4-c^4)*(b^2-c^2))^2*t^6-4*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*(a^6-(b^2+c^2)*a^4-(b^4-3*b^2*c^2+c^4)*a^2+(b^4-c^4)*(b^2-c^2))^2*t^5+(7*a^12-11*(b^2+c^2)*a^10-(19*b^4-56*b^2*c^2+19*c^4)*a^8+(b^2+c^2)*(46*b^4-91*b^2*c^2+46*c^4)*a^6-(19*b^8+19*c^8+3*b^2*c^2*(15*b^4-41*b^2*c^2+15*c^4))*a^4-(b^4-c^4)*(b^2-c^2)*(11*b^4-45*b^2*c^2+11*c^4)*a^2+(7*b^4+17*b^2*c^2+7*c^4)*(b^2-c^2)^4)*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*t^4-(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*(7*a^12-10*(b^2+c^2)*a^10-(31*b^4-73*b^2*c^2+31*c^4)*a^8+(b^2+c^2)*(84*b^4-163*b^2*c^2+84*c^4)*a^6-(71*b^8+71*c^8+b^2*c^2*(15*b^4-164*b^2*c^2+15*c^4))*a^4+(b^4-c^4)*(b^2-c^2)*(22*b^4+47*b^2*c^2+22*c^4)*a^2-(b^2-c^2)^6)*t^3+(4*a^16-(61*b^4-70*b^2*c^2+61*c^4)*a^12+(b^2+c^2)*(163*b^4-280*b^2*c^2+163*c^4)*a^10-(180*b^8+180*c^8+(65*b^4-448*b^2*c^2+65*c^4)*b^2*c^2)*a^8+(b^2+c^2)*(86*b^8+86*c^8+(118*b^4-409*b^2*c^2+118*c^4)*b^2*c^2)*a^6-(b^2-c^2)^2*(5*b^8+5*c^8+7*(16*b^4+29*b^2*c^2+16*c^4)*b^2*c^2)*a^4-(b^4-c^4)*(b^2-c^2)^3*(9*b^4+10*b^2*c^2+9*c^4)*a^2+(b^2-c^2)^6*(b^2+2*c^2)*(2*b^2+c^2))*t^2+(-4*(b^2+c^2)*a^14+2*(11*b^4-5*b^2*c^2+11*c^4)*a^12-(b^2+c^2)*(50*b^4-81*b^2*c^2+50*c^4)*a^10+4*(15*b^8+15*c^8+b^2*c^2*(2*b^4-33*b^2*c^2+2*c^4))*a^8-4*(b^2+c^2)*(10*b^8+10*c^8+b^2*c^2*(2*b^4-25*b^2*c^2+2*c^4))*a^6+2*(b^2-c^2)^2*(7*b^8+7*c^8+6*b^2*c^2*(4*b^4+5*b^2*c^2+4*c^4))*a^4-(b^4-c^4)*(b^2-c^2)^3*(2*b^4+b^2*c^2+2*c^4)*a^2-2*(b^2-c^2)^6*b^2*c^2)*t+a^2*b^2*c^2*(4*a^10-10*(b^2+c^2)*a^8+2*(2*b^4+9*b^2*c^2+2*c^4)*a^6+(b^2+c^2)*(8*b^4-17*b^2*c^2+8*c^4)*a^4-(b^2-c^2)^2*(8*b^4+9*b^2*c^2+8*c^4)*a^2+2*(b^4-c^4)*(b^2-c^2)^3) : :
ETC pairs (P,P’): (2,3628), (4,3850), (5,34420)
P’( X(3) ) = P’(t=0) = MIDPOINT OF X(140) AND X(5498)
= 4*a^10-10*(b^2+c^2)*a^8+2*(2*b^4+9*b^2*c^2+2*c^4)*a^6+(b^2+c^2)*(8*b^4-17*b^2*c^2+8*c^4)*a^4-(b^2-c^2)^2*(8*b^4+9*b^2*c^2+8*c^4)*a^2+2*(b^4-c^4)*(b^2-c^2)^3 : : (barys)
= (49*R^2-12*SW)*S^2-(19*R^2-4*SW)*SB*SC : : (barys)
= 3*X(2)+X(10226), 15*X(2)+X(34350), 3*X(3)+X(18567), X(3)+3*X(34331)
= As a point on the Euler line, this center has Shinagawa coefficients (E-48*F, -3*E+16*F)
= lies on these lines: {2, 3}, {15311, 32415}
= midpoint of X(i) and X(j) for these {i,j}: {140, 5498}, {461, 11343}, {10125, 23336}, {18420, 25647}
= reflection of X(i) in X(j) for these (i,j): (3536, 33001), (3628, 12043)
= complement of the complement of X(10226)
= [ 4.2238135009577640, 3.3431510513947140, -0.6232770925770827 ]
P’( X(20) ) = P’(t=-1) = MIDPOINT OF X(3) AND X(15948)
= 9*S^4+(16*R^2*(80*R^2-31*SW)-11*SB*SC+44*SW^2)*S^2-4*(4*R^2-SW)*(112*R^2-17*SW)*SB*SC : : (barys)
= lies on these lines: {2, 3}
= midpoint of X(3) and X(15948)
= reflection of X(25450) in X(10691)
= [ 9.1279702749931410, 8.2343705360318430, -6.2729629391883490 ]
P’( X(550) ) = P’(t=-1/2) = MIDPOINT OF X(3530) AND X(15949)
= 540*S^4+3*(25*R^2*(805*R^2-284*SW)-156*SB*SC+520*SW^2)*S^2-5*(5*R^2*(2125*R^2-764*SW)+296*SW^2)*SB*SC : : (barys)
= lies on these lines: {2, 3}
= midpoint of X(3530) and X(15949)
= reflection of X(26028) in X(21518)
= [ 6.0041159705419130, 5.1187570381120540, -2.6742208385740800 ]
César Lozada
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