HYACINTHOS 29250

[Antreas P. Hatzipolakis]:

Let ABC be an acute angled triangle.

The internal angle bisectors of A, B, C  intersect OH (Euler line) at A', B', C', resp,
(If the triangle is not acute angled with A> 90, then we take the external angle bisector of A instead) (*)

Denote:

Ao, Ah  = the orthogonal projections of A' on AO, AH, resp.
Bo, Bh = the orthogonal projections of B' on BO, BH, resp.
Co, Ch = the orthogonal projections of C' on CO, CH, resp.

A*B*C* = the triangle bounded by AoAh, BoBh, CoCh.

1. The orthocenter of A*B*C* lies on the Euler line of ABC.

 

2. ABC, A*B*C* are orthologic.

 

(*) Another description holding for all triangles:
Take the internal angle bisectors of the angles A, B. C of the triangles HAO, HBO, HCO, resp. instead of the triangle ABC.
(if the triangle is acute angled then they coincide)

[César Lozada]:

Please allow me to re-write this configuration.

Let ABC be a triangle.

Denote:

-          A’, B’, C’ the intersections of the Euler line of ABC and the internal angle bisectors of ABC

-          A”, B”, C” the intersections of the Euler line of ABC and the external angle bisectors of ABC

-          A’o, A’h  = the orthogonal projections of A’ on AO, AH, resp., and cyclically (B’o, B’h), (C’o, C’h)

-          A”o, A”h  = the orthogonal projections of A” on AO, AH, resp., and cyclically (B”o, B”h), (C”o, C”h)

-          A₁B₁C₁ = the triangle bounded by A’o A’h, B’o B’h, C’o C’h

-          A₂B₂C₂ = the triangle bounded by A”o A”h, B”o B”h, C”o C”h

Then:

1)      The orthocenters H1, H2 of A₁B₁C₁, A₂B₂C₂ lie on the Euler line of ABC

2)      ABC and A₁B₁C₁ are orthologic

3)      ABC and A₂B₂C₂ are parallelogic

 

-----------------------------------

 

Notes:  

-           A’o A’h ⊥  AI   and  A”o A”h ∥  AI

-          A₁B₁C₁, A₂B₂C₂ are parallelogic, directly similar and perspective with perspector X(185).

-          A₁B₁C₁ is homothetic to 29 triangles defined here (circumperps included) and parallelogic to the following triangles with the given centers: (Fuhrmann, 0, 4), (K798i, 0, 5), (1st Parry, Q1, 9810), (2nd Parry, Q1, 9811), (2nd Sharygin, H1, 2254) (0 means a not calculated center. For H1, Q1, see below). Also, it is orthologic to 71 triangles in the above list.

-          The only triangle in the list which is perspective to A₂B₂C₂ is the 2^nd Sharygin triangle. Listed orthologic triangles to A₂B₂C₂ with orthologic centers: (Fuhrmann, 0, 4), (K798i, 0, 5), (1st Parry, Q2, 9810), (2nd Parry, Q2, 9811), (2nd Sharygin, H2, 2254) (0 means a not calculated center. For H2, Q2, see below). There are 71 triangles in the list which are parallelogic to A₂B₂C₂.

 

 

Orthologic center ABC à A₁B₁C₁ = X(1)

 

Orthologic center A₁B₁C₁ à  ABC

Q1 = X(5)X(113) ∩ X(10574)X(24904)

= a^2*((b^2+c^2)*a^12-4*(b^4+c^4)*a^10+5*(b^6+c^6)*a^8-(b+c)*(b^2+c^2)*b^2*c^2*a^7-(11*b^4-18*b^2*c^2+11*c^4)*b^2*c^2*a^6+(b+c)*(b^2+b*c+c^2)*(3*b^2-4*b*c+3*c^2)*b^2*c^2*a^5-(b^2+c^2)*(5*b^6+5*c^6-(10*b^4+10*c^4+(8*b^2-25*b*c+8*c^2)*b*c)*b*c)*(b+c)^2*a^4-(b^2-c^2)*(b-c)*(3*b^4+3*c^4+4*(b^2+b*c+c^2)*b*c)*b^2*c^2*a^3+(4*b^8+4*c^8-(b^4+c^4+2*(b^2+6*b*c+c^2)*b*c)*b^2*c^2)*(b^2-c^2)^2*a^2+(b^3-c^3)*(b^2-c^2)^3*b^2*c^2*a+(b^2-c^2)^3*(b-c)*(b^3+c^3)*(-b^4-c^4-(b^2+3*b*c+c^2)*b*c)) : : (barys)

= lies on these lines: {5, 113}, {517, H1}, {10574, 24904}

= [ -155.7636168728372000, -181.1376122836620000, 200.9345269272906000 ]

 

Parallelogic center ABC à A₂B₂C₂ = X(1)

 

Parallelogic center A₂B₂C₂ à  ABC

Q2 = X(5)X(31847) ∩ X(185)X(526)

= a^2*(a^2-b^2+b*c-c^2)*((b^2+c^2)*a^10-2*(b+c)*b*c*a^9-(3*b^4+3*c^4-(b+c)^2*b*c)*a^8+4*(b^3+c^3)*b*c*a^7+2*(b^6+c^6-(b^3+c^3)*(b+c)*b*c)*a^6+5*(b^2-c^2)*(b-c)*b^2*c^2*a^5+(b^2-c^2)^2*(2*b^2-c^2)*(b^2-2*c^2)*a^4-2*(b^2-c^2)*(b-c)*(2*b^4+2*c^4+3*(b^2+c^2)*b*c)*b*c*a^3-(b^2-c^2)^2*(b+c)^2*(3*b^4+3*c^4-8*(b^2-b*c+c^2)*b*c)*a^2+(b^2-c^2)^3*(b-c)*(2*b^2+b*c+2*c^2)*b*c*a+(b^4+c^4-(b^2-b*c+c^2)*b*c)*(b^2-c^2)^4) : : (barys)

= lies on these lines: {5, 31847}, {185, 526}, {513, H2}

= [ 15.7632537278549000, 9.9378665840844040, -10.5147448737763100 ]

 

Orthocenters:

H1 = EULER LINE INTERCEPT OF X(517)X(Q1)

= 2*a^13-2*(b+c)*a^12-(5*b^2+2*b*c+5*c^2)*a^11+(b+c)*(5*b^2-4*b*c+5*c^2)*a^10+(b^4+c^4+4*(b^2+4*b*c+c^2)*b*c)*a^9-(b+c)*(b^4+c^4-(7*b^2-16*b*c+7*c^2)*b*c)*a^8+(6*b^6+6*c^6-(13*b^2+10*b*c+13*c^2)*b^2*c^2)*a^7-(b+c)*(6*b^6+6*c^6-(b^4+c^4+2*(7*b^2-11*b*c+7*c^2)*b*c)*b*c)*a^6-(b^2-c^2)^2*(4*b^4+4*c^4+(4*b^2+13*b*c+4*c^2)*b*c)*a^5+(b^2-c^2)*(b-c)*(4*b^6+4*c^6+(3*b^4+3*c^4+2*(2*b^2+7*b*c+2*c^2)*b*c)*b*c)*a^4-(b^2-c^2)^2*(b^6+c^6-2*(b^4+c^4+(4*b^2+3*b*c+4*c^2)*b*c)*b*c)*a^3+(b^2-c^2)^3*(b-c)^3*(b^2+3*b*c+c^2)*a^2+(b^6-c^6)*(b^2-c^2)^3*a-(b^2-c^2)^5*(b-c)*(b^2+c^2) : : (barys)

= lies on these lines: {2, 3}, {517, Q1}

= [ -300.9278480782251000, -301.0035127395635000, 350.9174877992477000 ]

 

H2 = EULER LINE INTERCEPT OF X(513)X(Q2)

= 2*a^13-2*(b+c)*a^12-(5*b^2-6*b*c+5*c^2)*a^11+(b+c)*(5*b^2-4*b*c+5*c^2)*a^10+(b^4+c^4-2*(5*b^2-8*b*c+5*c^2)*b*c)*a^9-(b+c)*(b^4+c^4-(7*b^2-16*b*c+7*c^2)*b*c)*a^8+(6*b^6+6*c^6-(4*b^4+4*c^4+15*(b-c)^2*b*c)*b*c)*a^7-(b^2-c^2)*(b-c)*(6*b^4+6*c^4+11*(b^2+c^2)*b*c)*a^6-(4*b^6+4*c^6-(4*b^4+4*c^4+(11*b^2+2*b*c+11*c^2)*b*c)*b*c)*(b-c)^2*a^5+(b^2-c^2)*(b-c)*(4*b^6+4*c^6+(3*b^2+c^2)*(b^2+3*c^2)*b*c)*a^4-(b^2-c^2)^2*(b^6+c^6+2*(b^4+c^4-(3*b^2-7*b*c+3*c^2)*b*c)*b*c)*a^3+(b^2-c^2)^3*(b-c)*(b^4+c^4+(b-c)^2*b*c)*a^2+(b^4+c^4-(2*b^2-b*c+2*c^2)*b*c)*(b^2-c^2)^4*a-(b^2-c^2)^5*(b-c)*(b^2+c^2) : : (barys)

= lies on these lines: {2, 3}, {513, Q2}

= [ 13.9745257026850700, 13.0681406281315900, -11.8562908926536800 ]

 

Parallelogic center A₁B₁C₁ à  A₂B₂C₂ = H1

Parallelogic center A₂B₂C₂ à  A₁B₁C₁  = H2

 

Homothetic center ( A₂B₂C₂ , 2^nd Sharygin):

Q3 = X(659)X(Q2) ∩ X(2254)X(H2)

= a(2*b*c*a^15-(b+c)^3*a^14+(b^4+c^4-(3*b-c)*(b-3*c)*b*c)*a^13+(b+c)*(4*b^4+4*c^4+(b-c)^2*b*c)*a^12-(4*b^6+4*c^6+(3*b^4+3*c^4+(15*b^2-14*b*c+15*c^2)*b*c)*b*c)*a^11-(b+c)*(5*b^6+5*c^6-(11*b^4+11*c^4-4*(b^2+c^2)*b*c)*b*c)*a^10+(5*b^8+5*c^8+(4*b^6+4*c^6-(5*b^2+21*b*c+5*c^2)*(b-c)^2*b*c)*b*c)*a^9-(b+c)*(14*b^6+14*c^6-(26*b^4+26*c^4-(17*b^2-2*b*c+17*c^2)*b*c)*b*c)*b*c*a^8+(4*b^6+4*c^6-(2*b^4+2*c^4+3*(2*b^2-3*b*c+2*c^2)*b*c)*b*c)*(b+c)^2*b*c*a^7+(b^2-c^2)*(b-c)*(5*b^8+5*c^8+(6*b^6+6*c^6-(13*b^4+13*c^4+3*(b^2+6*b*c+c^2)*b*c)*b*c)*b*c)*a^6-(5*b^10+5*c^10+(b^2+b*c+c^2)*(13*b^6+13*c^6-(4*b^4+4*c^4+7*(b-c)^2*b*c)*b*c)*b*c)*(b-c)^2*a^5-(b^2-c^2)*(b-c)*(4*b^10+4*c^10-(5*b^8+5*c^8+(9*b^6+9*c^6-(b^4+c^4-15*(b^2+c^2)*b*c)*b*c)*b*c)*b*c)*a^4+(b^2-c^2)^2*(4*b^10+4*c^10-(3*b^8+3*c^8-(b^6+c^6-2*(b^4+c^4+(7*b^2-10*b*c+7*c^2)*b*c)*b*c)*b*c)*b*c)*a^3+(b^2-c^2)^3*(b-c)*(b^8+c^8-(3*b^6+3*c^6+4*(b^2-b*c+c^2)*b^2*c^2)*b*c)*a^2-(b^8+c^8-(2*b^2-3*b*c+2*c^2)*(b^4+c^4-(b^2-b*c+c^2)*b*c)*b*c)*(b^2-c^2)^4*a+(b^2-c^2)^5*(b-c)*(b^2+c^2)*b^2*c^2 ) : : (barys)

= lies on these lines: {659, Q2}, {2254, H2}

= [ 49.2735064796446900, 25.5183921771737300, -36.7675330928949900 ]

 

César Lozada