HYACINTHOS 25490

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and A'B'C' the pedal triangle of N.

Denote:

P, Pa, Pb, Pc = same points on the Euler lines of ABC, AB'C', A'BC', A'B'C, resp.

Pab, Pac = the orthogonal projections of Pa on AC, AB, resp.

Pbc, Pba = the orthogonal projections of Pb on BA, BC, resp.

Pca, Pcb = the orthogonal projections of Pc on CB, CA, resp.

La, Lb, Lc = the Euler lines of APabPac, BPbcPba, CPcaPcb, resp.

A*B*C* = the triangle bounded by La, Lb, Lc

 

Conjecture:

ABC, A*B*C* are parallelogic.

The parallelogic center (ABC, A*B*C*) is the P.

For P = H and Pa, Pb, Pc = the H's of AB'C', A'BC', A'B'C, resp., it is true:

La, Lb, Lc are concurrent and perpendicular to BC,CA,AB, resp.

(Hyacinthos #23820)

 

If it is true then which is the locus of the other parallelogic center (A*B*C*, ABC) as P moves on the Euler line?

Or, if it is too complicated, which is the parallelogic center (A*B*C*, ABC) for some simple, other than H, points on the Euler line (P = O,N,G,X140...)

 

[César Lozada]:

 

 

 

> ABC, A*B*C* are parallelogic.

> The parallelogic center (ABC, A*B*C*) is the P.

Right.

 

When P moves on the Euler line the other parallelogic center Z*(P)  moves on a parabola p through ETCs: 3530, 5663, 10110.

 

Center of p: X(5663)

Focus:

F = (78*cos(2*A)-4*cos(4*A)+83)*co s(B-C)+(-52*cos(A)+8*cos(3*A)) *cos(2*(B-C))+(2*cos(2*A)+11)* cos(3*(B-C))+6*cos(5*A)-94*cos (A)-22*cos(3*A) : : (trilinears)

= X(20)-9*X(476) = 9*X(3258)-13*X(5067)

= On lines: {20,476}, {3258,5067}

= [ -2.435001993518502, -2.24466174148195, 6.318508146095808 ]

 

Vertex of p:

V = long coordinates =

= On lines {}

= [ -2.536870780780805, -2.36249872664846, 6.447103960716758 ]

 

Axis of p = trilinear pole of:

Ap = long coordinates

= on lines {}

= [ 0.496663710256132, -0.68505275259020, 3.885702367428609 ]

 

Directrix of p = trilinear pole of:

Dp = csc(A)/(8*cos(2*A)+8*cos(2*B)+ 8*cos(2*C)-7*cos(2*(B-C))+cos( 2*B+4*C)+cos(4*B+2*C)+17) : : (trilinears)

= on lines {}

= [ 1.869204017796796, 1.23516791842294, 1.922838684016129 ]

 

ETC pairs (P,Z*(P)): (3,3530), (4,10110)

 

Z*(G) = a*((b^2+c^2)*a^2-b^4+24*b^2*c^ 2-c^4) : : (trilinears)

= (cos(2*A)-12)*cos(B-C)-11*cos( A) : : (trilinears)

= 11*X(2)+X(51) = 3*X(2)+X(373) = 7*X(2)-X(3819) = 13*X(2)-X(3917) = 7*X(2)+X(5640) = 5*X(2)+X(5943) = 9*X(2)-X(7998) = 15*X(2)+X(11002) = 8*X(3628)+X(9729) = X(6102)+5*X(10170) = X(6102)-10*X(11695)

= On lines: {2,51}, {575,6090}, {576,5544}, {3589,9027}, {3848,9026}, {5663,6723}, {6102,10170}, {8705,9822}

= midpoint of X(i) and X(j) for these {i,j}: {3819,5640}, {5650,5943}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (2,10219,6688), (373,5650,11002), (373,11002,5943)

= [ 2.420306858340283, 1.59382313520880, 1.420183761375079 ]

 

Z*(H) = X(10110)

 

Z*(N) = a*((b^2+c^2)*a^6-(3*b^4-2*b^2* c^2+3*c^4)*a^4+3*(b^2+c^2)*(b^ 2+3*b*c+c^2)*(b^2-3*b*c+c^2)*a ^2-(b^4-13*b^2*c^2+c^4)*(b^2-c ^2)^2) : : (trilinears)

= (2*cos(2*A)-13)*cos(B-C)-4*cos (A) : : (trilinears)

= 3*X(2)-X(11592) = 15*X(5)+X(185) = 7*X(5)+9*X(373) = 13*X(5)+3*X(9730) = 3*X(5)-X(11017) = 3*X(5)+X(12006) = X(143)+7*X(3090) = X(185)+5*X(11017) = X(185)-5*X(12006) = 27*X(373)-7*X(12006)

= On lines: {2,11592}, {5,113}, {143,3090}, {156,11484}, {1216,10095}, {2979,9781}, {3567,11591}, {5447,10110}, {5876,11451}

= midpoint of X(11017) and X(12006)

= complement of X(11592)

= {X(5), X(12006)}-Harmonic conjugate of X(11017)

= [ 1.250511980906196, 0.48889049416131, 2.725042456300612 ]

 

 

César Lozada