HYACINTHOS 28916

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, P a point and A'B'C' the pedal tiangle of P.

Denote:

A", B", C" = the midpoints of AP, BP, CP, resp.
Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.
Ma, Mb, Mc = the midpoints of A"Na, B"Nb, C"Nc, resp.

Which is the locus of P such that A'B'C', MaMbMc are perspective?


[César Lozada]: 

Locus = {Linf} ∪ {circumcircle} ∪  {q5: excentral-circumquintic to ETC’s 1, 5, 110, 1113, 1114, 2041, 2042}

q5 =∑ [y*z*((b^2-c^2)*(9*a^2-5*b^2-5*c^2)*b^2*c^2*x^3+(b^2-c^2)*(5*a^4-7*(b^2+c^2)*a^2+2*b^4+10*b^2*c^2+2*c^4)*a^2*x*y*z-a^2*c^2*(3*a^2-b^2+c^2)*(a^2-c^2)*y^3+a^2*(4*a^6-(8*b^2+c^2)*a^4+(5*b^4+3*b^2*c^2-5*c^4)*a^2-(b^2-c^2)*(b^4+2*c^4))*z*y^2-a^2*(4*a^6-(b^2+8*c^2)*a^4-(5*b^4-3*b^2*c^2-5*c^4)*a^2+(b^2-c^2)*(2*b^4+c^4))*z^2*y+a^2*b^2*(3*a^2+b^2-c^2)*(a^2-b^2)*z^3) ] = 0 (barys)

 

ETC pairs (P, Q(P)=perspector): (1, 15950), (5, 5)

 

Some others:

Q( X(1113) ) = EULER LINE INTERCEPT OF X(2100)X(5886)

= OH*(3*S^2-SB*SC)-2*R*(S^2-3*SB*SC) : : (barys)

= X(2100)+3*X(5886), X(2102)-3*X(10283), 3*X(14561)+X(15162)

= lies on these lines: {2, 3}, {2100, 5886}, {2102, 10283}, {2574, 10272}, {12041, 14500}, {14374, 15806}, {14561, 15162}

= midpoint of X(i) and X(j) for these {i,j}: {550, 10751}, {12041, 14500}

= reflection of X(13627) in X(10124)

= complement of the complement of X(15154)

= [ 0.4719735317143392, -0.3987914692771249, 3.6989169460003860 ]

 

Q( X(1114) ) = EULER LINE INTERCEPT OF X(2101)X(5886)

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

= X(2101)+3*X(5886), X(2103)-3*X(10283), 3*X(14561)+X(15163)

= lies on these lines: {2, 3}, {2101, 5886}, {2103, 10283}, {2575, 10272}, {12041, 14499}, {14375, 15806}, {14561, 15163}

= midpoint of X(i) and X(j) for these {i,j}: {550, 10750}, {12041, 14499}

= reflection of X(13626) in X(10124)

= complement of the complement of X(15155)

= [ 6.8632611041169110, 5.9756357251315450, -3.6639730681606490 ]

 

César Lozada