HYACINTHOS 29607

[Kadir Atintas]:

 

Let ABC be a triangle and P be a point. 

Denote:

Oa, Ob, Oc = the circumcenters of PBC, PCA, PAB, resp. 

Oaa, Obb, Occ = the  circumcenters of PObOc, POcOa, POaOb, resp. 

Which is the locus of P such that ABC, OaaObbOcc are perspective?

The Napoleon-Feuerbach cubic  + + ? 

Some perspectors?

 

[César Lozada]:

 

Locus = The Napoleon-Feuerbach cubic K005

 

ETC pairs {P, Q(P)=perspector}: {1, 104}, {3, 186}, {4, 4}, {5, 33565}, {17, 3479}, {18, 3480}, {61, 3439}, {62, 3438}, {195, 18212}

 

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Q( X(54) ) = ISOGONAL CONJUGATE OF X(19552)

= (SB+SC)*(3*S^2-SA^2)*(S^2+(3*R^2+SB-2*SW)*SB)*(S^2+(3*R^2+SC-2*SW)*SC) : : (barys)

= lies on the cubics K039, K466 and on these lines: {3, 2888}, {4, 3432}, {26, 11671}, {49, 15345}, {137, 3518}, {184, 13505}, {186, 1141}, {378, 15960}, {570, 14586}, {930, 7512}, {1147, 13504}, {1263, 2070}, {2914, 8157}, {3520, 15620}, {5944, 11273}, {6143, 23319}, {7488, 25150}, {7502, 13512}, {13367, 20574}, {21394, 24144}

= anticomplement of the complementary conjugate of X(6150)

= isogonal conjugate of X(19552)

= antigonal conjugate of the isogonal conjugate of X(21394)

= [ 8.0353884953189140, 3.8366720672165050, -2.7241339470819330 ]

 

Q( X(3336) ) = X(3)X(3648) ∩ X(48)X(16553)

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

= (-1+cos(B-C)+cos(2*A)+cos(2*B)+2*cos(2*B+C)-cos(B+2*C))*(-1+cos(B-C)+cos(2*A)+cos(2*C)+2*cos(B+2*C)-cos(2*B+C))  : : (trilinears)

= lies on these lines: {3, 3648}, {48, 16553}

= [ -12.5610071527344900, 20.0158926044868800, -4.4191040199367690 ]

 

César Lozada