HYACINTHOS 29027

 

[Antreas P. Hatzipolakis]:

 

VARIATION

Let ABC be a triangle and P a point.

Denote:

A', B', C' = the midpoints of AP, BP, CP, resp.

(Na), (Nb), (Nc) = the NPCs of PBC, PCA, PAB, resp.

The perpendicular from A' to PB intersects again (Nb) at Ab
The perpendicular from A' to PC intersects again (Nc) at Ac

 

The perpendicular from B' to PC intersects again (Nc) at Bc
The perpendicular from B' to PA intersects again (Na) at Ba  

 

The perpendicular from C' to PA intersects again (Na) at Ca
The perpendicular from C' to PB intersects again (Nb) at Cb

Ma, Mb, Mc = the midpoints of AbAc, BcBa, CaCb, resp.

Which is the locus of P such that 

 

1. ABC, MaMbMc are orthologic?

 

2. The perpendicular bisectors of AbAc, BcBa, CaCb are concurrent?. 
The entire plane?

 

[César Lozada]:

 

1)       Locus = {sidelines} ∪ { Linf} ∪ {q9: circum-nonic through vertices of orthic triangle and ETCs 4, 13, 14, 1113, 1114 } 

 

 

Q9: ∑ [ y*z*((-c^4*(S^2-(2*SA-SC)*SA)*y^2+2*SA*(S-SA)*(S+SA)*(b^2-c^2)*z*y+b^4*(S^2-(2*SA-SB)*SA)*z^2)*x^5-2*((S^4-(SA^2-3*SA*SB+3*SA*SC-SB^2)*S^2-(SA*SB-3*SA*SC+3*SB^2)*SA^2)*z*y^2+2*(S^4-(SA^2+3*SA*SB-3*SA*SC-SC^2)*S^2+(3*SA*SB-SA*SC-3*SC^2)*SA^2)*z^2*y)*x^4+a^4*SA*(b^2-c^2)*z^3*y^3*x+a^6*y^3*z^3*(SB*y-SC*z)) ] = 0 (barys)

Orthologic centers:

ABC->MaMbMc = Qa( X(4) ) = X(4);   MaMbMc ->ABC = Qm( X(4) ) = X(5)

For P=X(13): (Qa, Qm  ) = { X(2), X(6669) )

For P=X(14): (Qa, Qm  ) = { X(2), X(6670) )

 

 Some related centers:

 

Qm( X(1113) ) = X(5)X(2883) ∩ X(546)X(14374)

= SA*(a*b*c*(8*S^2+(SB+SC)*(6*R^2-7*SA-3*SW))+2*(SB+SC)*(10*R^2-SA-SW)*OH*S) : : (barys)

= lies on these lines: {5, 2883}, {546, 14374}, {1113, 32110}, {1312, 13754}, {1313, 14915}, {1346, 9730}, {1347, 16194}, {1531, 10751}, {2575, 6699}

= [ -3.9079893834448600, -5.2235415630161940, 9.0606498948162950 ]

 

Qm( X(1114) ) = X(5)X(2883) ∩ X(546)X(14375)

= SA*(a*b*c*(8*S^2+(SB+SC)*(6*R^2-7*SA-3*SW))-2*(SB+SC)*(10*R^2-SA-SW)*OH*S) : : (barys)

= lies on these lines: {5, 2883}, {546, 14375}, {1114, 32110}, {1312, 14915}, {1313, 13754}, {1346, 16194}, {1347, 9730}, {1531, 10750}, {2574, 6699}

 

= [ 5.0137328667146680, 4.5873282212622280, -1.8492086866808570 ]

 

2) The entire plane.

For P=x:y:z (barys) the point of concurrence is:

Q(P) = ((3*S^2-SA*SB)*y^2*x*z+(3*S^2-SA*SC)*z^2*y*x-4*b^2*c^2*x^2*y*z+2*(S^2+SB*SC)*y^2*z^2+(S^2-SB^2)*y^3*z+(S^2-SC^2)*y*z^3+(SA+SC)*SA*x^2*z^2-(SA+SB)*SB*y^3*x-(SA+SC)*SC*z^3*x+(SA+SB)*SA*x^2*y^2)*x : :

 

If P lies on the circumcircle then Q(P)=X(5). If P lies at the infinity then Q(P) = P.

Other ETC pairs (P,Q(P)): (1,10), (2,16509), (3,5449), (4,5), (6,16511), (13,2), (14,2), (67,16511), (80,10), (265,5449), (671,16509)

 

Some related centers:

 

Q( X(5) ) = COMPLEMENT OF X(15345)

= (4*S^2+R^2*(R^2-5*SA-SW)+2*SA^2-2*SB*SC)*(S^2+SB*SC) : : (barys)

= 3*X(2)+X(25043)

= lies on these lines: {2, 3459}, {5, 51}, {140, 6592}, {252, 6150}, {1493, 19553}, {1510, 20327}, {3628, 13856}, {5501, 20414}, {8254, 24385}, {10285, 14143}, {14051, 24306}, {14142, 18400}, {15957, 31879}, {18282, 32428}

= midpoint of X(i) and X(j) for these {i,j}: {10285, 14143}, {15345, 25043}

= reflection of X(i) in X(j) for these (i,j): (13856, 3628), (18016, 140), (20414, 5501), (31879, 15957)

= complement of X(15345)

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (2, 25043, 15345), (5, 21230, 16336)

= [ -0.7501175371962464, 0.1700747164410257, 3.8691285415388580 ]

 

César Lozada