HYACINTHOS 25254

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the cevian triangle of H.

Denote:

Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.

Mab, Mac = the orthogonal projections of Ma on BB', CC', resp.

Mbc, MBa = the orthogonal projections of Mb on CC', AA', resp.

Mca, Mcb = the orthogonal projections of Mc on AA', BB', resp.

La, Lb, Lc = the Euler lines of HMabMac, HMbcMba, HMcaMcb, resp.

Na, Nb, Nc = the NPC centers of HMabMac, HMbcMba, HMcaMcb, resp.

L1, L2, L3 = the Euler lines of MaMabMac, MbMbcMba, McMcaMcb, resp.

N1, N2, N3 = the NPC centers of MaMabMac, MbMbcMba, McMcaMcb, resp.

 

1. La, Lb, Lc are concurrent. Point?

2. The parallels to La, Lb, Lc through A, B, C concur on the circumcircle (at the antipode of the Euler line reflection point = X74)

3. ABC, NaNbNc are orthologic. Orthologic centers?

 

4. L1, L2, L3 are concurrent. Point?

5. The parallels to L1, L2, L3 through A, B, C concur on the circumcircle (at the antipode of the Euler line reflection point = X74)

6. ABC, N1N2N3 are orthologic. Orthologic centers.

7. NaNbNc, N1N2N3 are bilogic (perspective and orthologic). Perspector, orthologic centers ?

 

[César Lozada]:

 

1)       

Z1(La/\Lb/\Lc) = midpoint of X(4) and X(973)

= a*((b^2+c^2)*a^12-4*(b^2+c^2)^ 2*a^10+(b^2+c^2)*(5*b^4+b^2*c^ 2+5*c^4)*a^8+12*b^2*c^2*(b^4+ c^4)*a^6-(b^4-c^4)*(b^2-c^2)*( 5*b^4+12*b^2*c^2+5*c^4)*a^4+4* (b^2-c^2)^2*(b^8+c^8-b^2*c^2*( b^2+c^2)^2)*a^2-(b^4-c^4)*(b^ 2-c^2)^3*(b^4-5*b^2*c^2+c^4)) : : (trilinears)

= (-9*cos(2*A)+cos(4*A))*cos(B- C)+(4*cos(A)-cos(3*A))*cos(2*( B-C))+4*cos(A)+cos(3*A) : : (trilinears)

= On lines: {4,973}, {54,154}, {235,1843}, {546,1154}, {1209,7403}, {3091,9971}, {6756,10110}, {10301,10619}

= midpoint of X(i) and X(j) for these {i,j}: {4,973}, {3574,11576}

= [ -1.264793210508239, -3.91060310490440, 6.931755805537384 ]

 

2)      Z2 = X(54), not X(74)

 

3)      Orthologic centers:

Z3(A->Na) = reflection of X(64) in X(125)

= 1/(a*(a^8-2*(b^2+c^2)*a^6+7*b^ 2*c^2*a^4+2*(b^4-3*b^2*c^2+c^ 4)*(b^2+c^2)*a^2-(b^4+3*b^2*c^ 2+c^4)*(b^2-c^2)^2)) : : (trilinears)

= 1/((2*cos(2*A)+3)*cos(B-C)-6* cos(A)-cos(3*A))

= On Jerabek hyperbola, K255, K495 and these lines: {2,11598}, {3,113}, {4,974}, {6,1562}, {30,5504}, {54,10721}, {64,125}, {68,5663}, {69,146}, {72,2778}, {73,9627}, {74,403}, {110,2883}, {265,6000}, {542,6391}, {690,2435}, {895,1503}, {1539,4846}, {1853,3426}, {3448,6225}, {3519,10628}, {3521,9730}, {3532,10990}, {6145,11381}, {6699,10606}, {7505,11270}

= midpoint of X(i) and X(j) for these {i,j}: {3448,6225}, {5895,10117}

= reflection of X(i) in X(j) for these (i,j): (64,125), (110,2883), (2935,113)

= trilinear pole of the line {647,800}

= antigonal conjugate of X(64)

= isogonal conjugate of X(2071)

= anticomplement of X(11598)

= antipode of X(64) in Jerabek hyperbola

= 2^nd Droz-Farny circle-inverse-of-X(122)

= [ -24.646780662675290, -27.05650836430864, 33.747530578432800 ]

 

Z3[Na->A)  = midpoint of X(389) and X(6756)

= 2*a^10-(b^2+c^2)*a^8-8*(b^4+c^ 4)*a^6+10*(b^4-c^4)*(b^2-c^2)* a^4-2*(b^2-c^2)^4*a^2-(b^4-c^ 4)*(b^2-c^2)^3 :: (barycentrics)

= (5*cos(2*A)-1)*cos(B-C)-cos(A) *cos(2*(B-C))-2*cos(A)-cos(3* A) : : (trilinears)

= (4*R^2+SW)*X(4)+(4*R^2-SW)*X( 64)

= On lines: {3,3589}, {4,64}, {6,7487}, {20,10601}, {30,5462}, {51,3575}, {52,524}, {141,7401}, {185,428}, {343,7544}, {389,1503}, {468,3574}, {511,9825}, {546,5449}, {569,6329}, {1192,3088}, {1204,1907}, {1216,10127}, {1350,6803}, {1498,6995}, {1595,6696}, {1596,5893}, {1597,5894}, {1598,2883}, {1990,8884}, {2777,11566}, {3066,6816}, {3517,10192}, {3567,6146}, {3629,6193}, {6240,9781}, {6759,7715}, {7553,9730}, {8550,9833}, {9973,11387}

= midpoint of X(389) and X(6756)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (4,9786,6247), (1595,11438,6696), (3567,7576,6146), (9833,11432,8550)

= [ 0.851669104062236, 0.42469172989728, 2.953569082411384 ]

 

4)       

Z4(L1/\L2/\L3) = midpoint of X(4) and X(974)

= a^2*((b^2+c^2)*a^8-2*(b^4+c^4) *a^6+b^2*c^2*(b^2+c^2)*a^4+2*( b^2-c^2)^2*(b^4-b^2*c^2+c^4)* a^2-(b^4-c^4)*(b^2-c^2)*(b^4- 3*b^2*c^2+c^4)) : : (barycentrics)

= (cos(2*A)+cos(4*A)-2)*cos(B-C) +(2*cos(A)-cos(3*A))*cos(2*(B- C))-cos(3*A) : : (trilinears)

= On lines: {4,974}, {6,110}, {51,125}, {74,9781}, {143,10224}, {389,546}, {399,11432}, {511,5159}, {568,7723}, {578,1511}, {1593,11598}, {1598,9934}, {1986,3567}, {2777,10110}, {3448,7394}, {4232,9973}, {5446,6699}, {5462,9826}, {5504,6642}, {5622,10117}, {5943,5972}, {5946,7706}

= midpoint of X(i) and X(j) for these {i,j}: {4,974}, {125,1112}, {389,7687}, {5446,6699}

= reflection of X(9826) in X(5462)

= {X(51), X(125)}-Harmonic conjugate of X(1112)

= [ 0.963037112451180, 0.65281445535612, 2.744237345529598 ]

 

5)      Z5 = X(74)

 

6)       

Z6(A->N1) = X(6145) = ABC-TO-HATZIPOLAKIS-MOSES- TRIANGLE ORTHOLOGY CENTER

 

Z6[N1->A)  = midpoint of X(389) and X(6756) = Z3(Na->A)

 

7)

Perspector: Z7(Na, N1) = Z3[Na->A)  = midpoint of X(389) and X(6756)

Orthologic centers:

Z7(Na->N1) = 2*a^22-6*(b^2+c^2)*a^20-(b^2- 3*c^2)*(3*b^2-c^2)*a^18+3*(b^ 2+c^2)*(9*b^4-8*b^2*c^2+9*c^4) *a^16-(20*b^8+20*c^8+b^2*c^2*( b^4-20*b^2*c^2+c^4))*a^14-2*( b^2+c^2)*(14*b^8+14*c^8-b^2*c^ 2*(15*b^4-26*b^2*c^2+15*c^4))* a^12+(b^2-c^2)^2*(42*b^8+42*c^ 8+b^2*c^2*(35*b^4+38*b^2*c^2+ 35*c^4))*a^10-2*(b^4-c^4)*(b^ 2-c^2)*(b^8+c^8-19*b^2*c^2*(b^ 4+c^4))*a^8-(b^2-c^2)^2*(22*b^ 12+22*c^12+(7*b^8+7*c^8-2*b^2* c^2*(2*b^4+9*b^2*c^2+2*c^4))* b^2*c^2)*a^6+2*(b^4-c^4)*(b^2- c^2)^3*(5*b^8+5*c^8-7*b^2*c^2* (b^4+c^4))*a^4+(b^2-c^2)^6*(b^ 2+c^2)^2*(b^4+7*b^2*c^2+c^4)* a^2-(b^2+c^2)^3*(b^2-c^2)^8 : : (barycentrics)

= on lines {}

= [ -2.897926231821713, -2.90183539331030, 6.987131861193836 ]

 

Z7(N1->Na) = 2*a^22-4*(b^2+c^2)*a^20-(13*b^ 4-18*b^2*c^2+13*c^4)*a^18+(b^ 2+c^2)*(43*b^4-54*b^2*c^2+43* c^4)*a^16-(20*b^8+20*c^8+7*( 11*b^4-16*b^2*c^2+11*c^4)*b^2* c^2)*a^14-8*(b^2+c^2)*(7*b^8+ 7*c^8-(29*b^4-42*b^2*c^2+29*c^ 4)*b^2*c^2)*a^12+(b^2-c^2)^2*( 70*b^8+70*c^8+(39*b^4-98*b^2* c^2+39*c^4)*b^2*c^2)*a^10-2*( b^4-c^4)*(b^2-c^2)*(b^8+c^8+2* b^2*c^2*(25*b^4-41*b^2*c^2+25* c^4))*a^8-(b^2-c^2)^2*(38*b^ 12+38*c^12-(85*b^8+85*c^8+2*b^ 2*c^2*(4*b^4-39*b^2*c^2+4*c^4) )*b^2*c^2)*a^6+4*(b^4-c^4)*(b^ 2-c^2)^3*(5*b^8+5*c^8-2*b^2*c^ 2*(b^2+c^2)^2)*a^4-(b^2-c^2)^ 6*(b^2+c^2)^2*(b^4+5*b^2*c^2+ c^4)*a^2-(b^2+c^2)^3*(b^2-c^2) ^8 : : (barycentrics)

= on lines: {2777,10095}

= [ 0.654365820206404, 0.24965024769701, 3.165814701098879 ]

 

César Lozada