HYACINTHOS 23900

-

Antreas P. Hatzipolakis

 

 

[APH]:

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

Denote:

Ab, Ac = the NPC centers of AB'P, AC'P, resp.

Bc, Ba = the NPC centers of BC'P, BA'P, resp.

Ca, Cb = the NPC centers of CA'P,CB'P, resp.

Na, Nb, Nc = the NPC centers of AB'C', BC'A', CA'B', resp.

For P = O

The circumcircles of NaAbAc, NbBcBa, NcCaCb concur at the midpoint of ON.

For P = N

The circumcircles of NaAbAc, NbBcBa, NcCaCb are concurrent.

Point?

Note: The locus is quite complicated, I think !

[César Lozada]:

 

Locus for concurrence of the three circles:

A quintic through X(3), X(5), X(523), X(2070) with barycentric equation:

a^2*((b^2-c^2)^2*b^4*c^4*x^5-((a^8*c^2-(3*b^2+4*c^2)*a^6*c^2+3*(b^4+b^2*c^2+2*c^4)*a^4*c^2-(b^6-b^4*c^2-3*b^2*c^4+4*c^6)*a^2*c^2+(b^6-4*b^4*c^2-3*b^2*c^4+c^6)*c^4)*y^3+(a^8*b^2-(4*b^2+3*c^2)*a^6*b^2+3*(2*b^4+b^2*c^2+c^4)*a^4*b^2-(4*b^6-3*b^4*c^2-b^2*c^4+c^6)*a^2*b^2+(b^6-3*b^4*c^2-4*b^2*c^4+c^6)*b^4)*z^3)*a^2*y*z+((2*a^6+(-5*b^2-8*c^2)*a^4+(4*b^4-b^2*c^2+10*c^4)*a^2-(b^2+4*c^2)*(b^4-4*b^2*c^2+c^4))*y+(2*a^6+(-8*b^2-5*c^2)*a^4+(10*b^4-b^2*c^2+4*c^4)*a^2-(4*b^2+c^2)*(b^4-4*b^2*c^2+c^4))*z)*a^2*b^2*c^2*y^2*z^2-(a^8+(-6*b^2-6*c^2)*a^6+(12*b^4+17*b^2*c^2+12*c^4)*a^4-(b^2+c^2)*(10*b^4-11*b^2*c^2+10*c^4)*a^2+(3*b^4-2*b^2*c^2+3*c^4)*(b^2-c^2)^2)*b^2*c^2*x^3*y*z-(a^12+(-6*b^2-6*c^2)*a^10+(15*b^4+17*b^2*c^2+15*c^4)*a^8-(b^2+c^2)*(20*b^4-7*b^2*c^2+20*c^4)*a^6+(15*b^8-6*b^6*c^2+18*b^4*c^4-6*b^2*c^6+15*c^8)*a^4-(b^2+c^2)*(6*b^8-19*b^6*c^2+23*b^4*c^4-19*b^2*c^6+6*c^8)*a^2+(b^4-b^2*c^2+c^4)*(b^2-c^2)^4)*x*y^2*z^2) + … = 0

Points of concurrence Z(P) (trilinears):

Z(O) = X(140)

 

Z(N) = a*(SA^2+5*S^2)*(S^2+SB*SC) : :

= X(5)+3*X(51)

= midpoint of X(i),X(j) for these {i,j}: {5,143}, {140,5446}, {389,546}                                                     

= On lines:  (3,5640), (4,3521), (5,51), (6,156), (30,5462), (110,1173), (140,5446), (185,3845), (373,632), (381,3567), (389,546), (403,6746), (511,3628), (547,1216), (548,5892), (567,3518), (568,3091), (1112,1594), (1614,7545), (1656,3060), (2979,5070), (3090,6243), (3527,6642), (3627,9730), (3843,5890), (3851,5889), (3861,6000), (5066,5907), (5422,7517), (5447,6688), (7529,9777)

= [ 0.228687122439348, -0.19664680952800, 3.671256678147139 ]

 

Z(X(2070)) = (cos(2*A)-3/2)*cos(B-C)+cos(A)*cos(2*(B-C))+cos(A)-cos(3*A) : : (on Euler line)

= (23*R^2-6*SW)*X(3)+(13*R^2-2*SW)*X(4)

= Inverse of X(6143) in Polar circle

= midpoint of X(i),X(j) for these {i,j}: {5,2070}, {403,7575}                                                               

= reflection of X(i) in X(j) for these (i,j): (546,403), (2071,3530)                                                             

= On lines:  (2,3), (1287,5966)

= [ 0.388974180558294, -0.48157186604911, 3.794533844299168 ]

 

César Lozada