HYACINTHOS 27119

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and P a point.

Denote:

La, Lb, Lc = the Euler lines of PBC, PCA, PAB, resp.

 

La1, Lb1, Lc1 = the reflections of La, Lb, Lc in BC, CA, AB, resp..

La2, Lb2, Lc2 = the parallels to La1, Lb1, Lc1 through A, B, C, resp.

La3, Lb3, Lc3 = the reflections of La2, Lb2, Lc2 in BC, CA, AB, resp.

La4, Lb4, Lc4 = the parallels to La3, Lb3, Lc3 through A, B. C, resp.

 

For P = I ==> La4, Lb4. Lc4 are concurrent.
or, if A*B*C* is the triangle bounded by La3, Lb3, Lc3, then ABC, A*B*C* are parallelogic.

Which is the locus of P such that ABC, A*B*C* are parallelogic?
Circumcircle + Linf + Neuberg cubic?



[ César Lozada ]:

> Which is the locus of P such that ABC, A*B*C* are parallelogic?
> Circumcircle + Linf + Neuberg cubic?

 
Your assumption is exact.

 
Denote Q = A->A* and Q*=A*->A


For P on the circumcircle:

Q=reflection of O in the midpoint point of P and H.

ETC-pairs (P,Q): (74,265), (98,6321), (99,6033), (100,10742), (101,10741), (102,10747), (103,10739), (104,10738), (105,15521), (106,15522), (109,10740), (110,7728), (112,12918), (1291,14980), (1292,10743), (1293,10744), (1294,10745), (1295,10746), (1296,10748), (1297,10749), (1300,13556)


For P on the line at infinity:

Q = Isogonal(Midpoint(Reflection(Isogonal(P), O), O))

ETC-pairs (P,Q): (30,5627)

 
For other P:

ETC-pairs (P,Q): (1,79), (3,4), (4,5), (399,14451), (484,14452), (1138,30), (1263,1141), (3065,80), (8431,6761)

 

General expression for Q and Q* are complicated.

 
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Examples:

Q*(X(1)) = X(65)X(267) ∩ X(517)X(11524)

= (16*q*p^4+(4*q^2-3)*p*(4*p^2-3)+4*q*(4*q^2-9)*p^2-3*q*(4*q^2-5))*p : : (trilinears), where p=sin(A/2), q=cos((B-C)/2)

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

= on lines: {65, 267}, {517, 11524}, {5697, 13744}

= [ -5.0645337171986260, -7.6000262569020610, 11.2396982215467100 ]

 

Q*(X(3)) = X(4)

Q*(X(4)) = X(10263)

Q*(X(74)) = X(399)

 

César Lozada