HYACINTHOS 26775

[Antreas P. Hatzipolakis]:

 

 

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

 

Denote:

 

A", B", C" = the reflections of P in the midpoints Ma, Mb, Mc of AA', BB', CC', resp.

Ab, Ac = the intersections of B"C" and AB,AC, resp.
Bc, Ba = the intersections of C"A" and BC,BA, resp.
Ca, Cb = the intersections of A"B" and CA,CB, resp.

The perpendicular bisectors of AbAc, BcBa, CaCb are concurrent.

The point of concurrence is the circumcenter O" of the triangle A"B"C".

1. Which is the locus of P such that O" lies on the Euler line of ABC?

2. Which is the locus of P such that A"B"C", ABC are orthologic?

 

[César Lozada]:

 

1. This quartic:

∑ [ (b^2-c^2)*a^2*x^2*(x^2+y*z)-y* z*((a^4+b^2*a^2-b^4-c^4)*y^2-( a^4+c^2*a^2-b^4-c^4)*z^2) ] = 0 (barys)

 

through vertices of anticomplementary triangle and ETC’s: 2 , 30

 

ETC pairs (P,Za): (2, 549), (30,30).

 

2. L_∞ ∪ { Lucas cubic K007}

The orthologic center Za=A->A” lies on the Darboux cubic K004

 

ETC pairs (P, Za): (2,4), (4,64), (7,84), (8,1), (20,20), (69,3), (189,3345), (253,3346), (329,40), (1032,3348), (1034,3347), (5932,1490), (14361,1498), (14362,3183), (14365,2131)

 

Orthologic center Za”=A”->A:

ETC pairs (P,Za”): (2,381), (4,4), (7,11372), (8,1), (20,5895), (69,6776)

 

Some others:

 

Za”( X(189) ) = X(1)X(4) ∩ X(189)X(3345)

= a*(a^12-2*(3*b^2-8*b*c+3*c^2)* a^10-4*b*c*(b+c)*a^9+(15*b^2+ 2*b*c+15*c^2)*(b-c)^2*a^8-4*( 5*b^4+5*c^4+2*b*c*(5*b^2+9*b* c+5*c^2))*(b-c)^2*a^6+8*(b^2- c^2)*(b-c)*b*c*(3*b^2+2*b*c+3* c^2)*a^5+(b^2-c^2)^2*(15*b^4+ 15*c^4+2*b*c*(4*b^2+b*c+4*c^2) )*a^4-32*(b^2-c^2)*(b-c)*b*c*( b^4+c^4)*a^3-2*(b^2-c^2)^4*(3* b^2-8*b*c+3*c^2)*a^2+4*(b^2-c^2)^3*(b-c)*b*c*(3*b^2-2*b*c+3* c^2)*a+(b^4+c^4-6*b*c*(2*b^2- b*c+2*c^2))*(b^2-c^2)^4) : : (barys)

= on lines: {1, 4}, {189, 3345}

= reflection of X(i) in X(j) for these (i,j): (189, 6245), (1490, 223)

= [ -67.936621609123460, -76.01316799636480, 87.620529222063150 ]

 

Za”( X(253) ) = X(4)X(6) ∩ X(253)X(3346)

= a^18+2*(b^2+c^2)*a^16-4*(7*b^ 4-12*b^2*c^2+7*c^4)*a^14+72*( b^4-c^4)*(b^2-c^2)*a^12-2*(b^ 2-c^2)^2*(41*b^4+98*b^2*c^2+ 41*c^4)*a^10+4*(b^4-c^4)*(b^2- c^2)*(11*b^4+26*b^2*c^2+11*c^ 4)*a^8-4*(b^2-c^2)^2*(b^4+c^4) *(3*b^2+c^2)*(b^2+3*c^2)*a^6+ 8*(b^4-c^4)*(b^2-c^2)^3*(b^4+ 6*b^2*c^2+c^4)*a^4-(b^2-c^2)^ 4*(7*b^8+7*c^8+2*b^2*c^2*(14* b^4+29*b^2*c^2+14*c^4))*a^2+2* (b^2+c^2)*(b^2-c^2)^8 : : (barys)

= on lines: {4, 6}, {253, 3346}

= reflection of X(i) in X(j) for these (i,j): (253, 6247), (1498, 1249)

= [ 19.402861174972030, 23.62848935293965, -21.672687150884010 ]

 

Za”( X(329) ) = X(4)X(57) ∩ X(9)X(119)

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

= on lines: {4, 57}, {9, 119}, {40, 329}, {381, 3358}, {517, 1490}, {971, 2095}, {1005, 3576}, {1158, 5177}, {1706, 5777}, {1750, 2093}, {2829, 3586}, {3452, 6908}, {5436, 10269}, {5437, 6913}, {5440, 6282}, {5709, 6259}, {6692, 6846}, {6843, 8257}, {7680, 9612}, {7962, 7966}, {7994, 11500}

= reflection of X(i) in X(j) for these (i,j): (84, 57), (329, 6260) , (7994, 11500)

= [ 53.531285805465490, 63.26128721529533, -64.862358577357990 ]

 

César Lozada