HYACINTHOS 25553

[Antreas P. Hatzipolakis]:

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

Denote:

Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.

R1 = the radical axis of the circles (Nb, NbA'), (Nc, NcA')

R2 = the radical axis of the circles (Nc, NcB'), (Na, NaB')

R3 = the radical axis of the circles (Na, NaC'), (Nb, NbC')

1. R1, R2, R3 are concurrent.

Which is the point of concurrence as P moves on the Euler line?

2. The parallels to R1, R2, R3 through A,B,C resp. are concurrent.

Which is the point of concurrence as P moves on the Euler line?

3. Which is the locus of P such that the reflections of R1, R2, R3 in BC, CA, AB are concurrent?

N lies on the locus.

[César Lozada]:

 

1)      For P=u:v:w (trilinears), the point of concurrence Z(P) is:

Z(P) = Complement(AntigonalConjugate( P))

= a*u*(-b*c*u^2*(v^2+w^2)+v*w*( b^2*v^2+c^2*w^2))*(b^2+c^2-a^ 2)+(a^2-b^2)*b^2*c*u^2*v^3-b*( 3*a^4-(4*b^2+3*c^2)*a^2-b^2*c^ 2+(b^2+c^2)^2)*w*u^2*v^2-c*(3* a^4-(3*b^2+4*c^2)*a^2-b^2*c^2+ (b^2+c^2)^2)*w^2*u^2*v+(a^2-c^ 2)*c^2*b*w^3*u^2-w^2*a^2*b*v^ 2*c*(b*v+c*w)-a*u*v*(2*a^2-b^ 2-c^2)*w*(-(b^2+c^2-a^2)*u^2+ 2*b*c*v*w) : :

 

Z(P) = P for P on the line-at-infinity

Z(P) = O for P on de circumcircle of ABC

 

Other ETC pairs (P,Z(P)): (1,214), (2,2482), (3,1511), (5,6592), (6,6593), (7,10427), (8,1145), (9,6594), (13,619), (14,618), (20,3184), (23,187), (54,11597), (64,11598), (67,141), (69,5181), (76,5976), (80,10), (83,8290), (265,5), (316,858), (671,2), (895,6), (1117,10264), (1156,9), (1177,206), (1263,140), (1320,1), (1325,36), (1337,16), (1338,15), (1916,39), (2070,6150), (3065,3647), (3254,142), (3407,10291), (3484,6760), (3557,2029), (3558,2028), (5011,1155), (5057,5074), (5203,5159), (5503,11165), (5504,1147), (5505,8542), (5508,2887), (5523,468), (5962,2072), (6328,620), (8781,6337), (9180,1649), (9513,11672), (10152,4), (10290,10335), (10693,960), (11599,1125), (11600,623), (11601,624), (11602,629), (11603,630), (11604,442), (11605,427), (11606,6292), (11607,100), (11608,5745), (11609,2092), (11610,32), (11611,3666), (11703,3628), (11744,2883)

 

When P moves on the Euler line of ABC, Z(P) moves on the cubic K038 !!!

 

Off topic: Here is a barycentric parametrization of cubic K038:

x(t) = f(a, b, c, t)

y(t) = f(b, c, a, t)

z(t) = f(c, a, b, t)

 

where

f(a, b, c, t) = (-a^4*(b^2+c^2-a^2)-(b^4-3*b^ 2*c^2+c^4)*a^2+(b^4-c^4)*(b^2- c^2))*t-a^2*((b^2+c^2-a^2)^2- b^2*c^2))*((2*a^4-(b^2+c^2)*a^ 2-(b^2-c^2)^2)*(a^6-(b^2+c^2)* a^4-(b^4-3*b^2*c^2+c^4)*a^2+( b^4-c^4)*(b^2-c^2))*t^2-2*a^2* ((b^2+c^2)*a^6-(3*b^4-2*b^2*c^ 2+3*c^4)*a^4+(b^2+c^2)*(3*b^4- 5*b^2*c^2+3*c^4)*a^2-(b^4+3*b^ 2*c^2+c^4)*(b^2-c^2)^2)*t+a^2* b^2*c^2*(2*a^4-(b^2+c^2)*a^2-( b^2-c^2)^2)

 

2)      Indicated parallel lines concur at:

Z2(P) = AntigonalConjugate(P)

= u/(-a^2*(b^2+c^2-a^2)*u^2+(a^ 2-b^2)*a*b*u*v+(a^2-c^2)*a*c* u*w+a^2*b*c*v*w) : :

 

Z(P) = P for P on the line-at-infinity

Z(P) = H for P on de circumcircle of ABC

 

When P moves on the Euler line of ABC, Z(P) moves on the cubic K025.

 

Off topic: Here is a barycentric parametrization of cubic K025:

x(t) = f(a, b, c, t)

y(t) = f(b, c, a, t)

z(t) = f(c, a, b, t)

 

where

f(a, b, c, t) = ((2*a^4-(b^2+c^2)*a^2-(b^2-c^ 2)^2)*t+a^2*(b^2+c^2-a^2))/(( a^4*(b^2+c^2-a^2)+(b^4-3*b^2* c^2+c^4)*a^2-(b^4-c^4)*(b^2-c^ 2))*t+a^2*((b^2+c^2-a^2)^2-b^ 2*c^2))

 

3)      Locus = {circumcircle} \/ { Euler line of ABC } \/ {circum-quartic CyclicSum(y*z*(2*(S^2-SA^2)*x^ 2-(SB+SC)^2*y*z))=0, through H}

 

For P on the Euler line such that OP=t*OH, the point of concurrence Z3(P) is:

Z3(P) = (((b^2+c^2-a^2)^2-b^2*c^2)*t- b^2*c^2)*((2*a^4-(b^2+c^2)*a^ 2-(b^2-c^2)^2)*(a^6-(b^2+c^2)* a^4-(b^4-3*b^2*c^2+c^4)*a^2+( b^4-c^4)*(b^2-c^2))*t^2-2*a^2* ((b^2+c^2)*a^6-(3*b^4-2*b^2*c^ 2+3*c^4)*a^4+(b^2+c^2)*(3*b^4- 5*b^2*c^2+3*c^4)*a^2-(b^4+3*b^ 2*c^2+c^4)*(b^2-c^2)^2)*t+a^2* b^2*c^2*(2*a^4-(b^2+c^2)*a^2-( b^2-c^2)^2))*a : : (trilinears)

= on cubic K038

 

ETC pairs (P,Z3(P)): (2,187), (3,30), (5,6150), (21,36), (23,2482), (186,131), (1113,3), (1114,3), (2070,6592), (2071,3184)

 

Others:

Z3( H ) = circumcircle-inverse-of-X(155)

=  (a^4-2*(b^2+c^2)*a^2+b^4+c^4) *(-a^2+b^2+c^2)*(2*a^8-3*(b^2+ c^2)*a^6+(b^2+c^2)^2*a^4-(b^4- c^4)*(b^2-c^2)*a^2+(b^2-c^2)^ 4)*a : : (trilinears)

= (cos(A)+cos(3*A))*(2*cos(B)* cos(C)*cos(B-C)+cos(2*A)) : : (trilinears)

= on cubic K038 and lines: {2,5962}, {3,49}, {30,136}, {186,1299}

= midpoint of X(186) and X(10420)

= complement of X(5962)

= circumcircle-inverse-of-X(155)

= [ 2.880260812083525, 0.59572312888983, 1.898889633252870 ]

 

Z3(X(20)) = circumcircle-inverse-of-X( 1498)

= (-a^2+b^2+c^2)^2*(2*a^10-(b^2+ c^2)*a^8-8*(b^2-c^2)^2*a^6+10* (b^4-c^4)*(b^2-c^2)*a^4-2*(b^ 2-c^2)^2*(b^4+6*b^2*c^2+c^4)* a^2-(b^4-c^4)*(b^2-c^2)^3)*a : : (trilinears)

= cos(A)^2*((5*cos(2*A)+7)*cos( B-C)-cos(A)*cos(2*(B-C))-10* cos(A)-cos(3*A)) : : (trilinears)

= on cubic K038 and lines: {3,64}, {30,122}, {131,10257}, {520,4091}, {631,6761}, {1304,2071}, {2060,3346}

= midpoint of X(i) and X(j) for these {i,j}: {3,6760}, {1304,2071}

= reflection of X(11589) in X(3)

= circumcircle-inverse-of-X( 1498)

= [ 5.304040466174815, 4.26929744483353, -1.763021656596442 ]

 

César Lozada