Let ABC be a triangle and P a point.
Denote:
Oa, Ob, Oc = the corcumcenters of PBC, PCA, PAB, resp.
OaaOabOac = the midway triangle of Oa
(ie Oaa, Oab, Oac = the midpoints of AOa, BOa, COa, resp.)
ObaObbObc = the midway triangle of Ob
OcaOcbOcc = the midway triangle of Oc.
Q1, Q2, Q3 = same points Q of OaaOabOac, ObaObbObc., OcaOcbOcc, resp.
For P on the Euler line, Q on the Euler line: The P-point of Q1Q2Q3 lies on the Euler line of ABC.
For example:
For Q1, Q2, Q3 = same points on the Euler lines of OaaOabOac, ObaObbObc., OcaOcbOcc, resp.
and
P = N: The NPC center of Q1Q2Q3 lies on the Euler line of ABC.
P = G: The centroid of Q1Q2Q3 lies on the Euler line of ABC.
P = O: The circumcenter of Q1Q2Q3 lies on the Euler line of ABC.
etc
APH [Hyacinthos 26807, Nov. 18, 2017]
[César Lozada, Nov. 18, 2017]
Algebraically proved.
If P, Q are such that OP=p*OH and OQ=q*OH then P*(P,Q), the P-center-of-Q1Q2Q3 lies on the Euler line of ABC and has coordinates:
P* = 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*p^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*p^5+(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))*(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))*p^4+(-(a^4+(b^2-2*c^2)*a^2-(b^2-c^2)*(2*b^2+c^2))*(a^4-(2*b^2-c^2)*a^2+(b^2-c^2)*(b^2+2*c^2))*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)^2*q+2*a^2*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*(a^10-2*(b^2+c^2)*a^8-(2*b^4-13*b^2*c^2+2*c^4)*a^6+(b^2+c^2)*(8*b^4-21*b^2*c^2+8*c^4)*a^4-(7*b^8+7*c^8+b^2*c^2*(5*b^4-28*b^2*c^2+5*c^4))*a^2+(b^4-c^4)*(b^2-c^2)*(2*b^4+9*b^2*c^2+2*c^4)))*p^3+((2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*(a^12-3*(3*b^4-5*b^2*c^2+3*c^4)*a^8+(b^2+c^2)*(16*b^4-31*b^2*c^2+16*c^4)*a^6-3*(b^2-c^2)^2*(3*b^4+11*b^2*c^2+3*c^4)*a^4+15*(b^4-c^4)*(b^2-c^2)*b^2*c^2*a^2+(b^4+4*b^2*c^2+c^4)*(b^2-c^2)^4)*q-a^2*(2*a^14-2*(b^2+c^2)*a^12-2*(9*b^4-17*b^2*c^2+9*c^4)*a^10+2*(b^2+c^2)*(25*b^4-49*b^2*c^2+25*c^4)*a^8-2*(25*b^8+25*c^8+b^2*c^2*(9*b^2-2*c^2)*(2*b^2-9*c^2))*a^6+(b^2+c^2)*(18*b^8+18*c^8+b^2*c^2*(68*b^4-171*b^2*c^2+68*c^4))*a^4+(b^2-c^2)^2*(2*b^8+2*c^8-b^2*c^2*(26*b^4+79*b^2*c^2+26*c^4))*a^2+(b^4-c^4)*(b^2-c^2)^3*(-2*b^4-10*b^2*c^2-2*c^4)))*p^2+(-(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*((b^2+c^2)*a^10-4*(b^4-b^2*c^2+c^4)*a^8+(b^2+c^2)*(6*b^4-11*b^2*c^2+6*c^4)*a^6-(b^2-c^2)^2*(4*b^4+13*b^2*c^2+4*c^4)*a^4+(b^4-c^4)*(b^2-c^2)*(b^4+5*b^2*c^2+c^4)*a^2+(b^2-c^2)^4*b^2*c^2)*q+2*a^2*((b^2+c^2)*a^12-(5*b^4-2*b^2*c^2+5*c^4)*a^10+5*(b^2+c^2)*(2*b^4-3*b^2*c^2+2*c^4)*a^8-2*(5*b^8+5*c^8+3*b^2*c^2*(b^4-4*b^2*c^2+c^4))*a^6+(b^2+c^2)*(5*b^8+5*c^8+8*(b^4-3*b^2*c^2+c^4)*b^2*c^2)*a^4-(b^2-c^2)^2*(b^8+c^8+6*b^2*c^2*(b^2+c^2)^2)*a^2-(b^4-c^4)*(b^2-c^2)^3*b^2*c^2))*p-(a^2+b^2-c^2)*(a^2-b^2+c^2)*(-a^2+b^2+c^2)*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*a^2*b^2*c^2*q-(2*a^8-4*(b^2+c^2)*a^6+6*b^2*c^2*a^4+(b^2+c^2)*(4*b^4-7*b^2*c^2+4*c^4)*a^2-(2*b^4+3*b^2*c^2+2*c^4)*(b^2-c^2)^2)*a^4*b^2*c^2 : : (barycentrics)
ETC triads (P, Q, P*(P,Q)) for P,Q in { G, O, H, N, X(20) }:
(2, 2, 549), (2, 3, 3), (2, 4, 5), (2, 5, 140), (2, 20, 550), (3, 4, 13406), (3, 5, 10020), (4, 2, 549), (4, 3, 3), (4, 4, 5), (4, 5, 140), (4, 20, 550), (5, 5, 10289)
It can also be proved that for each point P and any Q there exist an unique point P” which is the reflection of Q-of-ABC in P*(P,Q), i.e, P*(P,Q) is the midpoint of Q-of-ABC and a fixed point P” depending on P only. Coordinates of P” are:
P” = 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*p^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*p^5+(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))*(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))*p^4+a^2*(a^10-2*(b^2+c^2)*a^8-(2*b^4-17*b^2*c^2+2*c^4)*a^6+(b^2+c^2)*(8*b^4-25*b^2*c^2+8*c^4)*a^4-(7*b^8+7*c^8+b^2*c^2*(9*b^4-40*b^2*c^2+9*c^4))*a^2+(b^4-c^4)*(b^2-c^2)*(2*b^4+13*b^2*c^2+2*c^4))*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*p^3-a^2*(a^14-(b^2+c^2)*a^12-(3*b^2+b*c-3*c^2)*(3*b^2-b*c-3*c^2)*a^10+(b^2+c^2)*(25*b^4-52*b^2*c^2+25*c^4)*a^8-(25*b^8+25*c^8+4*b^2*c^2*(5*b^4-23*b^2*c^2+5*c^4))*a^6+(b^2+c^2)*(9*b^8+9*c^8+b^2*c^2*(38*b^4-93*b^2*c^2+38*c^4))*a^4+(b^2-c^2)^2*(b^8+c^8-b^2*c^2*(13*b^4+43*b^2*c^2+13*c^4))*a^2-(b^4-c^4)*(b^2-c^2)^3*(b^4+6*b^2*c^2+c^4))*p^2+a^2*((b^2+c^2)*a^12-(5*b^4-2*b^2*c^2+5*c^4)*a^10+5*(b^2+c^2)*(2*b^4-3*b^2*c^2+2*c^4)*a^8-2*(5*b^8+5*c^8+b^2*c^2*(b^2-3*c^2)*(3*b^2-c^2))*a^6+(b^2+c^2)*(5*b^8+5*c^8+2*b^2*c^2*(4*b^4-11*b^2*c^2+4*c^4))*a^4-(b^4-c^4)^2*(b^4+4*b^2*c^2+c^4)*a^2-(b^4-c^4)*(b^2-c^2)^3*b^2*c^2)*p-b^2*c^2*a^4*(a^8-2*(b^2+c^2)*a^6+2*b^2*c^2*a^4+(b^2+c^2)*(2*b^4-3*b^2*c^2+2*c^4)*a^2-(b^6-c^6)*(b^2-c^2)) : : (barys)
ETC-pairs (P, P”): (2, 3), (3, 1658), (4, 3), (5, 10126), (1113, 3), (1114, 3)
This means that:
* As for P=X(2) we have P”=X(3), then P*(X(2),Q) = midpoint(P”=X(3),Q)
* As for P=X(3) we have P”=X(1658), then P*(X(3),Q) = midpoint(X(1658), Q)
and so on.
P*( X(3), X(2) ) = MIDPOINT OF X(2) AND X(1658)
= (37*R^2-10*SW)*S^2-3*(5*R^2-2* SW)*SB*SC : : (barycentrics)
= X(4)-4*X(12010)
= As a point on the Euler line, this center has Shinagawa coefficients (-3*E-40*F, 9*E+24*F)
= On lines: {2, 3}, {1154, 10182}
= midpoint of X(i) and X(j) for these {i,j}: {2, 1658}, {12100, 13383}
= reflection of X(i) in X(j) for these (i,j): (2, 10125), (10224, 2), (10226, 12100)
= {X(1658), X(10125)}-Harmonic conjugate of X(10224)
= [2.48108111117190, 1.60501603366136, 1.38438517651991]
P*( X(3), X(3) ) = MIDPOINT OF X(3) AND X(1658)
= (SB+SC)*(S^2+(26*R^2+SA-8*SW)* SA) : : (barycentrics)
= 3*X(3)+X(26) = 7*X(3)+X(7387) = 13*X(3)+3*X(9909) = 3*X(3)-X(11250) = 5*X(3)-X(12084) = 9*X(3)-X(12085) = 2*X(3)+X(12107) = 5*X(3)+3*X(14070) = X(20)+3*X(10201) = X(156)-3*X(11202) = 3*X(182)-X(11255) = X(5448)-3*X(10182) = X(7689)+3*X(11202)
= As a point on the Euler line, this center has Shinagawa coefficients (-E-16*F, 3*E+16*F)
= On lines: {2, 3}, {143, 11430}, {156, 7689}, {182, 11255}, {185, 5944}, {541, 14862}, {1154, 12038}, {1493, 14831}, {1511, 5562}, {2883, 13289}, {5010, 8144}, {5448, 10182}, {5663, 10282}, {6102, 13367}, {6200, 11266}, {6396, 11265}, {7691, 11597}, {8718, 15055}, {9682, 13925}, {9730, 10610}, {10540, 11440}, {10575, 12041}, {10627, 11561}, {10645, 11268}, {10646, 11267}, {11424, 13451}, {12359, 12893}, {13352, 14449}, {13353, 15053}, {14805, 15043}
= midpoint of X(i) and X(j) for these {i,j}: {3, 1658}, {26, 11250}, {156, 7689}, {548, 13383}, {10226, 12107}
= reflection of X(i) in X(j) for these (i,j): (5, 10125), (10224, 140), (10226, 3), (12107, 1658), (13371, 5498), (13406, 10020)
= [ 4.557578162027791, 3.67603523286139, -1.007780600240179 ]
P*( X(3), X(20) ) = MIDPOINT OF X(20) AND X(1658)
= (-23*R^2+6*SW)*S^2+(37*R^2-10* SW)*SB*SC : : (barycentrics)
= 3*X(3)-2*X(5498) = 5*X(3)-4*X(10212)
= As a point on the Euler line, this point has Shinagawa coefficients (-E-24*F, 3*E+40*F)
= On lines: {2, 3}, {10264, 12289}, {11750, 12041}, {12281, 14677}
= midpoint of X(20) and X(1658)
= reflection of X(i) in X(j) for these (i,j): (4, 10125), (10224, 3), (10226, 548)
= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (3, 382, 6143), (3, 1657, 3153)
= [ 10.787069314589230, 9.88909283046978, -8.184277930535881 ]
P*( X(5), X(2) ) = MIDPOINT OF X(2) AND X(10126)
= 16*S^4+(245*R^4-180*R^2*SW-24* SA^2+24*SA*SW+28*SW^2)*S^2+3*( 11*R^4-12*R^2*SW+4*SW^2)*SB*SC : : (barycentrics)
= 5*X(2)-X(10285)
= on line {2, 3}
= midpoint of X(2) and X(10126)
= reflection of X(2) in X(12057)
= [ 5.018775784272121, 4.13601620413315, -1.539089406002628 ]
P*( X(5), X(3) ) = MIDPOINT OF X(3) AND X(10126)
= (103*R^4+12*SW^2-4*(19*R^2-4* SA)*SW-16*SA^2)*S^2-(53*R^4- 36*R^2*SW+4*SW^2)*SB*SC : : (barycentrics)
= on line {2, 3}
= midpoint of X(3) and X(10126)
= reflection of X(5) in X(12057)
= [ 7.095272835125935, 6.20703540333595, -3.931255182767862 ]
P*( X(5), X(4) ) = MIDPOINT OF X(4) AND X(10126)
= 16*S^4+(4*SW^2-4*(7*R^2+2*SA)* SW+39*R^4+8*SA^2)*S^2+(139*R^ 4-108*R^2*SW+20*SW^2)*SB*SC : : (barycentrics)
= 3*X(5)-2*X(13469) = 5*X(5)-X(14142)
= on line {2, 3}
= midpoint of X(4) and X(10126)
= reflection of X(3) in X(12057)
= [ 0.865781682564494, -0.00602219427244, 3.245242147527840 ]
P*( X(5), X(20) ) = MIDPOINT OF X(20) AND X(10126)
= 16*S^4-(20*SW^2-4*(31*R^2-10* SA)*SW+167*R^4-40*SA^2)*S^2+( 245*R^4-180*R^2*SW+28*SW^2)* SB*SC : : (barycentrics)
= on line {2, 3}
= midpoint of X(20) and X(10126)
= reflection of X(4) in X(12057)
= [ 13.324763987687380, 12.42009300094433, -11.107752513063560 ]
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