HYACINTHOS 27715

[Antreas P.  Hatzipolakis]:

 

Let ABC be a triangle and P a point.

The perpendicular bisectors of BC, CA, AB intersect the circumcircle at (A1, A2), (B1, B2), (C1, C2) ,resp.

 

The circumcircles of PA1A2, PB1B2, PC1C2 are coaxial.

 

Now, let (A'1, A'2), (B'1, B'2), (C'1, C'2) be the reflections of  (A1, A2), (B1, B2), (C1, C2) in BC, CA, AB, resp.

Which is the locus of P such that the circumcircles of  PA'1A'2, PB'1B'2, PC'1C'2 are coaxial ?
N lies on the locus.

 

[César Lozada]:
 

> Now, let (A'1, A'2), (B'1, B'2), (C'1, C'2) be the reflections of  (A1, A2), (B1, B2), (C1, C2) in BC, CA, AB, resp.

> Which is the locus of P such that the circumcircles of  PA'1A'2, PB'1B'2, PC'1C'2 are coaxial ?
> N lies on the locus.

The locus is {Linf} ∪ {Gibert’s K800, through ETC’s 1, 3, 5, 30, 191, 399, 5127, 5535, 6326, 7701, 13513 and vertices of triangles ABC, EXCENTRAL, FUHRMANN, FUHRMANN2}

 

Let P ∈ K800 and P’ be the other point of intersection.

ETC pairs (P,P’): (1, 6326), (5, 399), (191, 5535), (399, 5), (5535, 191), (6326, 1), (7701, 13513), (13513, 7701)

 

Missing point in the above list:

P’(X(5127)) = X(4)X(5535) ∩ X(5)X(580)

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

= 2*X(502)-3*X(5587)

= on the Fuhrmann circle, the cubic K800 and these lines: {4, 5535}, {5, 580}, {355, 13514}, {502, 5587}

= antipode of X(13514) in the Fuhrmann circle

= X(6798) of excentral triangle

= [ 4.8220713398357920, 1.0085543351807640, 0.7168631699350233 ]

 

P->P’ seems to map K800 -> K800, but I have not proved it.

 

César Lozada