HYACINTHOS 25436

[Tran Quang Hung]:

 

Let ABC be a triangle with incenter I.

A line d cuts IA,IB,IC at A',B',C'.

P is reflection of I through d.

 Then the circles (PAA'),(PBB'),(PCC') are coaxial and the second intersection lies on circumcircle (O) of ABC.

 

[César Lozada]:

 

Let d = [L, M, N] (trilinears). Then the 2^nd point of intersection Z(d) is:

 

Z(d) = f(a, b, c, A, B, C, L, M, N)  :  f(b, c, a, B, C, A, M, N, L) : f(c, a, b, C, A, B,  N, L, M)

 

where f(a, b, c, A, B, C, L, M, N) = a/((2*a+b+c)*L^2-2*L*a*(M*cos( C)+N*cos(B))-(b+c)*(M^2-2*M*N* cos(A)+N^2))

 

The distance OZ(d) is R, then Z(d) lies on the circumcircle of ABC.  

------------------------

Specifically, if d is the trilinear polar of Q=u:v:w (trilinears). Then

 

Z(Q) =  a/((b+c)*u^2*(v^2-2*cos(A)*v* w+w^2)+2*a*u*v*w*(cos(B)*v+ cos(C)*w)-(2*a+b+c)*v^2*w^2) : :

 

ETC pairs:

(Q, Z(Q))=(X(I),X(J)) for these (I,J):

(27,741), (57,106), (81,759), (88,2718), (89,105), (514,106), (651,2718), (2988,2733), (2990,2716), (4622,11635), (6650,2700), (8378,9070), (9250,2728)

 

(Z(Q),Q)=(X(I),X(J)) for these (I,J):

(105,89), (106,57), (106,514), (741,27), (759,81), (2700,6650), (2716,2990), (2718,88), (2718,651), (2728,9250), (2733,2988), (9070,8378), (11635,4622)

---------------------------

Some examples (all coordinates are trilinears):

 

d =Line IO = PolarTrilineal( X(651) )

Z(d) = X(2718)

 

d = Nagel line {1,2} = PolarTrilineal( X(190) )

Z(d) = a/((b+c)*(a^3+5*b*c*a-b*c*(b+ c))+(b^2-10*b*c+c^2)*a^2) : :

= on the circumcircle and these lines:  {1,6079}, {100,1149}, {901,3915}, {995,2748} , {7292,9059}

= [ -0.383640795837512, 0.77045217676459, 3.284339496072360 ]

 

d = Euler line {2,3} = PolarTrilineal( X(648) )

Z(d) = 1/((a^5-(b^2+c^2)*a^3+(b^2-c^ 2)*(b-c)*a^2+b^2*c^2*a-(b^4-c^ 4)*(b-c))*(b+c)) : :

= on the circumcircle and these lines: {12,2222}, {21,1290}, {23,9070}, {28,2766}, {30,6011}, {74,6003}, {100,1325}, {101,4053}, {108,2074}, {109,5127}, {110,758}, {476,6757}, {523,759}, {842,7427}, {2651,4588}, {2691,4221}, {2701,4653}, {4227,10100}, {6012,7481}, {7469,9058}

= Trilinear pole of the line {6, 2610}

= [ 11.840688873989800, 3.57702652298284, -4.300671821999037 ]

 

d = Brocard axis {3,6} = PolarTrilineal( X(110) )

Z(d) = a/((a^4-(b^2+c^2)*a^2-(b^2-c^ 2)*(b-c)*a+b^2*c^2)*(b+c)) : :

= on the circumcircle and these lines:  {58,2702}, {98,6002}, {99,740}, {100,1931}, {101,1326}, {110,3747}, {511,6010}, {512,741}, {789,5209}, {813,1500}, {825,5006}, {2703,3736}

= [ 2.755617944032597, -1.50226613010891, 3.408871213198577 ]

 

d = Soddy line {1,7} = PolarTrilineal( X(658) )

Z(d) = a/((b+c)*a^4-2*(b^2+c^2)*a^3+( b^3+c^3)*a^2-(b^2-c^2)*(b-c)* b*c) : :

= on the circumcircle and these lines:  {1,927}, {3,813}, {41,919}, {100,2340}, {101,7193}, {103,9320}, {105,663}, {108,1429}, {109,2223}, {112,5009}, {741,7254}, {929,990}, {934,1458}, {991,1308}, {1305,3100}, {2222,5091}, {2704,11012}, {2737,5732}

= reflection of X(813) in X(3)

= antipode of X(813) in the circumcircle

= [ 13.312754611302090, 5.95856889793745, -6.628846883419437 ]

 

César Lozada