HYACINTHOS 25300

[Antreas P. Hatzipolakis]:

 

Now, let O'a, O'b, O'c be the reflections of Oa,Ob,Oc [ = the circumcenters of IBC, ICA, IAB, resp.] in BC, CA, AB, resp. and N'a, N'b, N'c the reflections of Na,Nb,Nc [ = the NPC centers of IBC, ICA, IAB, resp.] in BC, CA, AB, resp. and P a point.

Which are the loci of P such that the NPCs of :

1. IPO'a, IPO'b, IPO'c
2 IPN'a, IPN'b, IPN'c

are coaxial ?

Too complicated?

O lies on the locus 1. Second point of intersection (other than the midpoint of OI)?

 

[César Lozada]:

 

>> Excellent as always ! Thanks !!

Thank YOU!

 

>> Second point of intersection (other than the midpoint of OI)?

Did you mean: Second point of intersection (other than midpoint IP) ?

 

Note: O’aO’bO’c = Fuhrmann triangle of ABC.

 

1)      Which are the loci of P such that the NPCs of : IPO'a, IPO'b, IPO'c are coxial?

The entire plane.

 

The 2^nd point intersection Q1(P) lies on the circle c1 {N, sqrt(R^2-2*R*r)/2}, through ETC’s 10, 502, 946.

Q1(P) is a fixed point on c1 for all points P on the conic q1={O’a, O’b, O’c, I, P}. Moreover, Q1(P)=center of q1.

 

ETC-pairs (P,Q1(P)): (3,10), (4,946), (8,10), (191,10), (355,10), (1158,10), (2475,10), (3307,10), (3308,10), (5220,10), (5880,10), (6256,10), (9782,10), (10093,10), (10094,10), (10912,10), (10940,10), (11023,10), (11524,10)

 

Example:

Q1(X(11)) = Q1(X(1769))=

= 16*p^6*q*(-2*q+p)+8*(2*q^2+1)* q*p^5-8*(2*q^4-2*q^2+1)*p^4+( 16*q^4-12*q^2-3)*q*p^3+(1-q^2) *((16*q^2+2)*p^2+(4*q^2-11)*q* p-2*q^2+3) : : (trilinears), where p=sin(A/2), q=cos((B-C)/2)

= On lines: {10,2804}, {108,10058}, {946,1387}, {952,10271}

=  [ 2.062923504958657, 2.40796703285916, 1.021491841485582 ]

 

2)      Which are the loci of P such that the NPCs of : IPN'a, IPN'b, IPN'c are coaxial?

The conic q2 with trilinear equation:

CyclicSum[ (b-c)*(a^2*(a-b-c)*(a^5-(b+c)* a^4-2*(b^2+c^2)*a^3+(b+c)*(2* b^2-b*c+2*c^2)*a^2+(b^4+b^2*c^ 2+c^4)*a-(b+c)*(b^4+c^4-b*c*( b+c)^2))*u^2+b*c*((b+c)*a^5-( b^2+c^2)*a^4-(b+c)*(2*b^2-b*c+ 2*c^2)*a^3+2*(b^4+c^4-b*c*(b^ 2-b*c+c^2))*a^2+(b^3-c^3)*(b^ 2-c^2)*a-(b^2-c^2)^2*(b-c)^2)* w*v) ] = 0

through ETC’s 1, 65, 11696 and having center = midpoint(65, 11696).

 

For P on q2, the 2^nd point of intersection Q2(P) is always:

Q2(P) = (b+c)*a^11-(4*(b^2+b*c+c^2))* a^10+7*b*c*(b+c)*a^9+(15*b^2- 22*b*c+15*c^2)*(b^2+b*c+c^2)* a^8-(b+c)*(10*b^4-3*b^2*c^2+ 10*c^4)*a^7-(20*b^6+20*c^6-( 41*b^4+41*c^4-2*b*c*(19*b^2- 11*b*c+19*c^2))*b*c)*a^6+(b+c) *(20*b^6+20*c^6-(42*b^4+42*c^ 4-b*c*(35*b^2-17*b*c+35*c^2))* b*c)*a^5+(2*b^2+3*b*c+2*c^2)*( 5*b^4+5*c^4-b*c*(18*b^2-29*b* c+18*c^2))*(b-c)^2*a^4-(b^2-c^ 2)*(b-c)*(15*b^6+15*c^6-(26*b^ 4+26*c^4-b*c*(12*b^2+11*b*c+ 12*c^2))*b*c)*a^3+(b^2-c^2)^2* (7*b^4+7*c^4-15*b*c*(2*b^2-3* b*c+2*c^2))*b*c*a^2+(b^2-c^2)^ 3*(b-c)*(4*b^4+4*c^4-b*c*(13* b^2-19*b*c+13*c^2))*a-(b^2-c^ 2)^4*(b-c)^4 :: (trilinears)

= center of q2

= On line: {65,11696}

= midpoint of X(65) and X(11696)

= [ 0.118317762735891, -0.11528256852149, 3.665867292698074 ]

 

Regards,

César Lozada