Let ABC be a triangle and P a point on the Euler line (such that PO/OH = t : number?).
1. Denote:
N1, N2, N3 = the NPC centers of OBC, OCA, OAB, resp.
Aa, Ab, Ac = the midpoints of AN1, BN1, CN1, resp.
Ba, Bb, Bc = the midpoints of AN2, BN2, CN2, resp.
Pa, Pb, Pc = same to P points of the triangles AaAbAc, BaBbBc, CaCbCc, resp.
(ie Pa lies on the Euler line of AaAbAc such that PaOa / OaHa = t and similarly Pb, Pc)
Conjecture:
The NPC center of PaPbPc lies on the Euler line of ABC.
Which is it in terms of t?
2. Denote:
N1, N2, N3 = the NPC centers of NBC, NCA, NAB, resp.
Aa, Ab, Ac = the midpoints of AN1, BN1, CN1, resp.
Ba, Bb, Bc = the midpoints of AN2, BN2, CN2, resp.
Pa, Pb, Pc = same to P points of the triangles AaAbAc, BaBbBc, CaCbCc, resp.
Conjecture:
The Circumcenter of PaPbPc lies on the Euler line of ABC.
Case 1: circumcenter O, the NPC center lies on the Euler line
1)
Your conjecture is true for any t.
Let Np be the NPC center of PaPbPc.
Np(P) = midpoint of P and X(5498)
ETC pairs (P,Np(P)): (3,10212), (5498,5498), (10125,140), (15761,12010), (18282,10125)
Some others:
Np( X(2) ) = midpoint of X(2) and X(5498)
= 10*a^10-27*(b^2+c^2)*a^8+2*(7*b^4+23*b^2*c^2+7*c^4)*a^6+(b^2+c^2)*(20*b^4-43*b^2*c^2+20*c^4)*a^4-(b^2-c^2)^2*(24*b^4+23*b^2*c^2+24*c^4)*a^2+7*(b^4-c^4)*(b^2-c^2)^3 : : (barys)
= As a point on the Euler line, this center has Shinagawa coefficients (3*E-136*F, -9*E+24*F)
= on the line {2, 3}
= midpoint of X(2) and X(5498)
= reflection of X(2) in X(12043)
= reflection of X(2) in the line X(523)X(12043)
= [ 3.7046892382443110, 2.8253962515940150, -0.0252356483857741 ]
Np( X(4) ) = midpoint of X(4) and X(5498)
= 2*a^10+(b^2+c^2)*a^8-2*(5*b^4-3*b^2*c^2+5*c^4)*a^6+(b^2+c^2)*(4*b^4-7*b^2*c^2+4*c^4)*a^4+(b^2-c^2)^2*(8*b^4-3*b^2*c^2+8*c^4)*a^2-5*(b^4-c^4)*(b^2-c^2)^3 : : (barys)
= As a point on the Euler line, this center has Shinagawa coefficients (E-24*F, -3*E-56*F)
= on lines: {2, 3}, {6153, 20584}, {13851, 15806}
= midpoint of X(i) and X(j) for these {i,j}: {4, 5498}, {10125, 18567}, {18282, 18377}
= reflection of X(3) in X(12043)
= reflection of X(3) in the line X(523)X(12043)
= X(5498)-of-Euler triangle
= X(12043)-of-X3-ABC reflections triangle
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (5, 3627, 14940), (5, 18403, 140), (5, 18567, 10125), (546, 1885, 3861), (1656, 18566, 15332), (3850, 3860, 13487), (3856, 12811, 23411), (12102, 22249, 6240)
= [ -0.4483048634633169, -1.3166421468115770, 4.7590959051446940 ]
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2)
Your conjecture is true for any t.
Let Op be the NPC center of PaPbPc.
Op(P) = midpoint of P and X(5501)
ETC pairs (P,Op): (3,15327), (5,15957), (546,19940), (5501,5501), (10126,3628), (10289,13469)
Op( X(2) ) = midpoint of X(2) and X(5501)
= 6*a^16-19*(b^2+c^2)*a^14-(9*b^4+10*b^2*c^2+9*c^4)*a^12+3*(b^2+c^2)*(41*b^4+8*b^2*c^2+41*c^4)*a^10-(235*b^8+235*c^8+2*b^2*c^2*(59*b^4+52*b^2*c^2+59*c^4))*a^8+(b^2+c^2)*(219*b^8+219*c^8-5*b^2*c^2*(80*b^4-67*b^2*c^2+80*c^4))*a^6-(b^2-c^2)^2*(111*b^8+111*c^8-b^2*c^2*(108*b^4+139*b^2*c^2+108*c^4))*a^4+(b^4-c^4)*(b^2-c^2)^3*(29*b^4-100*b^2*c^2+29*c^4)*a^2-(3*b^4-20*b^2*c^2+3*c^4)*(b^2-c^2)^6 : : (barys)
= 22*S^4+(R^2*(251*R^2-192*SW)+30*SB*SC+34*SW^2)*S^2+3*(5*R^4-2*SW^2)*SB*SC : : (barys)
= on lines: {2, 3}
= midpoint of X(i) and X(j) for these {i,j}: {2, 5501}, {12100, 20030}
= reflection of X(i) in X(j) for these (i,j): (2, 12056), (10109, 13469), (15334, 11540)
= [ -0.2791624620081381, -1.1479459472822270, 4.5642405047989070 ]
Op( X(4) ) = midpoint of X(4) and X(5501)
= 2*a^16-17*(b^2+c^2)*a^14+5*(9*b^4+10*b^2*c^2+9*c^4)*a^12-(b^2+c^2)*(39*b^4-8*b^2*c^2+39*c^4)*a^10-(25*b^8+25*c^8+2*b^2*c^2*(b^4+12*b^2*c^2+c^4))*a^8+(b^2+c^2)*(73*b^8+73*c^8-b^2*c^2*(144*b^4-133*b^2*c^2+144*c^4))*a^6-(b^2-c^2)^2*(53*b^8+53*c^8-3*b^2*c^2*(12*b^4+19*b^2*c^2+12*c^4))*a^4+(b^4-c^4)*(b^2-c^2)^3*(15*b^4-44*b^2*c^2+15*c^4)*a^2-(b^4-12*b^2*c^2+c^4)*(b^2-c^2)^6 : : (barys)
= 2*S^4+(R^2*(41*R^2-32*SW)+26*SB*SC+6*SW^2)*S^2+(R^2*(133*R^2-96*SW)+14*SW^2)*SB*SC
= on lines: {2, 3}
= midpoint of X(i) and X(j) for these {i,j}: {4, 5501}, {546, 20030}
= reflection of X(i) in X(j) for these (i,j): (3, 12056), (3530, 13469), (10289, 12811), (15327, 15957), (15335, 3856), (15336, 12108), (15957, 19940)
= [ -4.4321565637157640, -5.2899843456878170, 9.3485720583293760 ]
César Lozada
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