HYACINTHOS 25539

[Antreas P. Hatzipolakis]:

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

Denote:

Ab, Ac = the orthogonal projections of A' on BB', CC', resp.

La = the Euler line of A'AbAc

L1 = the reflection of La in AA'. Similarly L2, L3.

A1B1C1 = the triangle bounded by La, Lb, Lc.

A2B2C2 = the triangle bounded by L1, L2, L3

 

1. ABC, A1B1C1 are parallelogic.

The parallelogic center (BC, A1B1C1) is X80

 

2. ABC, A2B2C2 are parallelogic.

The parallelogic center (ABC, A2B2C2) lies on the OI line.

[Peter Moses]:

Hi Antreas,

 

1)

ABC, A1B1C1 at X(80).

A1B1C1, ABC at:
(b+c) (-2 a^5+a^4 b+3 a^3 b^2-a^2 b^3-a b^4+a^4 c-4 a^3 b c+b^4 c+3 a^3 c^2+2 a b^2 c^2-b^3 c^2-a^2 c^3-b^2 c^3-a c^4+b c^4)::
X[1109]-3 X[1962]
on lines {{1109,1962},{2650,3635},{3957 ,6758}}.
Searches {1.331131536390265404415626654 08,1.1386005366045863187958655 2005,2.23803417053954300345246 744058}.

.

 

2)

ABC, A2B2C2 at X(36)

A2B2C2, ABC at:
a (b+c) (a^5-2 a^3 b^2+a b^4+a^3 b c+b^4 c-2 a^3 c^2-b^3 c^2-b^2 c^3+a c^4+b c^4)::
on lines {{1,21},{517,3724},{523,663},{ 740,4511},{5844,10459}}.

Searches {2.747531506085739094744412559 64,2.1233042447210861031541025 1269,0.90259315583020914182194 0643109}.

crossdifference X(579) & X(661).

 

Best regards,

Peter Moses.

 

[César Lozada]:

1)      Parallelogic centers: Z(A->A1)=X(80) and

Z(A1->A) =

= (b+c)*(2*a^5-(b+c)*a^4-(3*b^2- 4*b*c+3*c^2)*a^3+(b^3+c^3)*a^2 +(b^2-c^2)^2*a-(b^2-c^2)*(b-c) *b*c) : : (barycentrics)

= X(1109)-3*X(1962)

= On lines: {1109,1962}, {2650,3635}, {3957,6758}

= [ 1.331131536390265, 1.13860053660459, 2.238034170539543 ]

 

2)      Parallelogic centers: Z(A->A2)=X(36) and

Z(A2->A) =

= a*(b+c)*(a^5-(2*b^2-b*c+2*c^2) *a^3+(b^4+c^4)*a+(b^2-c^2)*(b- c)*b*c) : : (barycentrics)

= On lines: {1,21}, {517,3724}, {523,663}, {740,4511}, {5844,10459}

= [ 2.747531506085739, 2.12330424472109, 0.902593155830209 ]

 

César Lozada