ADGEOM 2697 * ADGEOM 2698

#2697

 

[Tran Quang Hung]:

 

Dear geometers,

Let ABC be a triangle.

Perpendicular bisector of BC cuts CA,AB at Ab,Ac. Let da be perpendicular bisector of segment AbAc. Simiarlaly, we have the lines db,dc.

Then, triangle bound by lines da,db,dc has nine point circle which is tangent to circumcircle of triangle ABC.

Please, the tangent point is new ?

Best regards,

Tran Quang Hung.

 

#2698

[Paul Yiu]:

Dear QH,

 

[TQH]:  Let ABC be a triangle.

Perpendicular bisector of BC cuts CA,AB at Ab,Ac. Let da be perpendicular bisector of segment AbAc. Simiarlaly, we have the lines db,dc.Then, triangle bound by lines da,db,dc has nine point circle which is tangent to circumcircle of triangle ABC.

 

*** Euler reflection point X(110).

 

Best regards

Sincerely

Paul Yiu

 

HYACINTHOS 23787

 

[Antreas P. Hatzipolakis]:

 

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

Denote:

L1, L2, L3 = the Euler lines of AB'C', BC'A', CA'B', resp.

M1, M2, M3 = the parallels to L1, L2, L3 through P, resp.

N1, N2, N3 = the reflections of M1, M2, M3 in BC, CA, AB, resp.

Which is the locus of  P such that N1, N2, N3 are concurrent?

H lies on the locus.

 

[César Lozada]:

 

Which is the locus of  P such that N1, N2, N3 are concurrent at Z(P)?

Locus={a circumquartic through ETC’S 3,4,54,1147,2574,2575, the last two on the infinity }

 

Z(O) = X(110); Z(H) = X(186); Z(X(54)) = X(54)

 

Z(X(1147)) = (6*R^2-SA-SW)*SA*a : : (trilinears)

= cos(A)*(2*cos(A)*cos(B-C)-1) : : (trilinears)

= complementary conjugate of X(131)

= isogonal conjugate of X(1300)

= On the line at infinity, Gibert´s K039, K114, K339, Q097 and these lines:

(1,6238), (2,5654), (3,49), (4,52), (5,389), (6,4550), (20,6193), (26,6759),       

 (30,511), (40,6237), (51,381), (55,500), (56,1069), (69,4846), (74,323), (110,186),

 (113,403), (125,1568), (131,1516), (140,9729), (143,546), (146,7731), (156,1658),  

 (161,1498), (182,7514), (232,1625), (265,1531), (373,5055), (376,2979), (378,1993),

 (382,6243), (399,1495), (547,6688), (549,3819), (550,6101), (569,7503), (576,8548),

 (578,7526), (944,9933), (974,6699), (1151,8909), (1350,8717), (1351,1597),         

 (1352,7706), (1478,20019), (1614,7488), (1994,7527), (3091,3567),    

 (3153,3448), (3193,7414), (3269,3289), (3357,9938), (3426,6391), (3519,3521),      

 (3523,7999), (3524,7998), (3545,5640), (3574,5576), (3818,9969), (3832,9781),      

 (4549,6776), (5054,5650), (5167,6033), (5609,7575), (5691,9896), (5752,6985),      

 (5870,9930), (5871,9929), (5921,6403), (5986,5999), (6030,7512), (6153,6288),      

 (6285,9931), (6407,8912), (6642,9786), (6644,9306), (8549,9926), (9873,9923

 

César Lozada

HYACINTHOS 29110

[Antreas P. Hatzipolakis]:

 

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

Denote:

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

D = the Poncelet poin of ABCP

Da, Db, Dc = the orthogonal projections of D on AP, BP, CP, resp.

N1, N2, N3 = the reflections of Na, Nb, Nc in DDa, DDb, DDc, resp.

The parallels to DN1, DN2, DN3 through A', B', C', resp. concur on the pedal circle of P.

Point of concurrence?  

 

[Peter Moses]:


Hi Antreas.

Concurrence:

(a^2 c^6 p^4 q^4-b^2 c^6 p^4 q^4-c^8 p^4 q^4+a^2 c^6 p^3 q^5-b^2 c^6 p^3 q^5+c^8 p^3 q^5-2 a^4 c^4 p^4 q^3 r+6 a^2 b^2 c^4 p^4 q^3 r-4 b^4 c^4 p^4 q^3 r+4 a^2 c^6 p^4 q^3 r-2 b^2 c^6 p^4 q^3 r-2 c^8 p^4 q^3 r+a^4 c^4 p^3 q^4 r+2 a^2 b^2 c^4 p^3 q^4 r-3 b^4 c^4 p^3 q^4 r-a^2 c^6 p^3 q^4 r+3 b^2 c^6 p^3 q^4 r+3 a^4 c^4 p^2 q^5 r-4 a^2 b^2 c^4 p^2 q^5 r+b^4 c^4 p^2 q^5 r+2 a^2 c^6 p^2 q^5 r-2 b^2 c^6 p^2 q^5 r+c^8 p^2 q^5 r+a^6 b^2 p^4 q^2 r^2-3 a^4 b^4 p^4 q^2 r^2+3 a^2 b^6 p^4 q^2 r^2-b^8 p^4 q^2 r^2+a^6 c^2 p^4 q^2 r^2-6 a^4 b^2 c^2 p^4 q^2 r^2+6 a^2 b^4 c^2 p^4 q^2 r^2-b^6 c^2 p^4 q^2 r^2-3 a^4 c^4 p^4 q^2 r^2+6 a^2 b^2 c^4 p^4 q^2 r^2-8 b^4 c^4 p^4 q^2 r^2+3 a^2 c^6 p^4 q^2 r^2-b^2 c^6 p^4 q^2 r^2-c^8 p^4 q^2 r^2+a^8 p^3 q^3 r^2-a^6 b^2 p^3 q^3 r^2-3 a^4 b^4 p^3 q^3 r^2+5 a^2 b^6 p^3 q^3 r^2-2 b^8 p^3 q^3 r^2-5 a^6 c^2 p^3 q^3 r^2+6 a^4 b^2 c^2 p^3 q^3 r^2-6 a^2 b^4 c^2 p^3 q^3 r^2+5 b^6 c^2 p^3 q^3 r^2+4 a^4 c^4 p^3 q^3 r^2-4 a^2 b^2 c^4 p^3 q^3 r^2-9 b^4 c^4 p^3 q^3 r^2+3 a^2 c^6 p^3 q^3 r^2+9 b^2 c^6 p^3 q^3 r^2-3 c^8 p^3 q^3 r^2+2 a^8 p^2 q^4 r^2-5 a^6 b^2 p^2 q^4 r^2+3 a^4 b^4 p^2 q^4 r^2+a^2 b^6 p^2 q^4 r^2-b^8 p^2 q^4 r^2-6 a^6 c^2 p^2 q^4 r^2+9 a^4 b^2 c^2 p^2 q^4 r^2-9 a^2 b^4 c^2 p^2 q^4 r^2+6 b^6 c^2 p^2 q^4 r^2+4 a^4 c^4 p^2 q^4 r^2+a^2 b^2 c^4 p^2 q^4 r^2-8 b^4 c^4 p^2 q^4 r^2-a^2 c^6 p^2 q^4 r^2+2 b^2 c^6 p^2 q^4 r^2+c^8 p^2 q^4 r^2+a^8 p q^5 r^2-3 a^6 b^2 p q^5 r^2+3 a^4 b^4 p q^5 r^2-a^2 b^6 p q^5 r^2-3 a^4 b^2 c^2 p q^5 r^2+3 a^2 b^4 c^2 p q^5 r^2+4 a^4 c^4 p q^5 r^2-3 a^2 b^2 c^4 p q^5 r^2+a^2 c^6 p q^5 r^2-2 a^4 b^4 p^4 q r^3+4 a^2 b^6 p^4 q r^3-2 b^8 p^4 q r^3+6 a^2 b^4 c^2 p^4 q r^3-2 b^6 c^2 p^4 q r^3-4 b^4 c^4 p^4 q r^3+a^8 p^3 q^2 r^3-5 a^6 b^2 p^3 q^2 r^3+4 a^4 b^4 p^3 q^2 r^3+3 a^2 b^6 p^3 q^2 r^3-3 b^8 p^3 q^2 r^3-a^6 c^2 p^3 q^2 r^3+6 a^4 b^2 c^2 p^3 q^2 r^3-4 a^2 b^4 c^2 p^3 q^2 r^3+9 b^6 c^2 p^3 q^2 r^3-3 a^4 c^4 p^3 q^2 r^3-6 a^2 b^2 c^4 p^3 q^2 r^3-9 b^4 c^4 p^3 q^2 r^3+5 a^2 c^6 p^3 q^2 r^3+5 b^2 c^6 p^3 q^2 r^3-2 c^8 p^3 q^2 r^3+2 a^8 p^2 q^3 r^3-8 a^6 b^2 p^2 q^3 r^3+9 a^4 b^4 p^2 q^3 r^3-2 a^2 b^6 p^2 q^3 r^3-b^8 p^2 q^3 r^3-8 a^6 c^2 p^2 q^3 r^3+8 a^4 b^2 c^2 p^2 q^3 r^3-4 a^2 b^4 c^2 p^2 q^3 r^3+10 b^6 c^2 p^2 q^3 r^3+9 a^4 c^4 p^2 q^3 r^3-4 a^2 b^2 c^4 p^2 q^3 r^3-18 b^4 c^4 p^2 q^3 r^3-2 a^2 c^6 p^2 q^3 r^3+10 b^2 c^6 p^2 q^3 r^3-c^8 p^2 q^3 r^3+2 a^8 p q^4 r^3-5 a^6 b^2 p q^4 r^3+4 a^4 b^4 p q^4 r^3-a^2 b^6 p q^4 r^3-8 a^6 c^2 p q^4 r^3+3 a^4 b^2 c^2 p q^4 r^3+3 a^2 b^4 c^2 p q^4 r^3+5 a^4 c^4 p q^4 r^3-3 a^2 b^2 c^4 p q^4 r^3+a^2 c^6 p q^4 r^3+a^8 q^5 r^3-2 a^6 b^2 q^5 r^3+a^4 b^4 q^5 r^3-2 a^4 b^2 c^2 q^5 r^3+a^4 c^4 q^5 r^3+a^2 b^6 p^4 r^4-b^8 p^4 r^4-b^6 c^2 p^4 r^4+a^4 b^4 p^3 q r^4-a^2 b^6 p^3 q r^4+2 a^2 b^4 c^2 p^3 q r^4+3 b^6 c^2 p^3 q r^4-3 b^4 c^4 p^3 q r^4+2 a^8 p^2 q^2 r^4-6 a^6 b^2 p^2 q^2 r^4+4 a^4 b^4 p^2 q^2 r^4-a^2 b^6 p^2 q^2 r^4+b^8 p^2 q^2 r^4-5 a^6 c^2 p^2 q^2 r^4+9 a^4 b^2 c^2 p^2 q^2 r^4+a^2 b^4 c^2 p^2 q^2 r^4+2 b^6 c^2 p^2 q^2 r^4+3 a^4 c^4 p^2 q^2 r^4-9 a^2 b^2 c^4 p^2 q^2 r^4-8 b^4 c^4 p^2 q^2 r^4+a^2 c^6 p^2 q^2 r^4+6 b^2 c^6 p^2 q^2 r^4-c^8 p^2 q^2 r^4+2 a^8 p q^3 r^4-8 a^6 b^2 p q^3 r^4+5 a^4 b^4 p q^3 r^4+a^2 b^6 p q^3 r^4-5 a^6 c^2 p q^3 r^4+3 a^4 b^2 c^2 p q^3 r^4-3 a^2 b^4 c^2 p q^3 r^4+4 a^4 c^4 p q^3 r^4+3 a^2 b^2 c^4 p q^3 r^4-a^2 c^6 p q^3 r^4-2 a^6 b^2 q^4 r^4+2 a^4 b^4 q^4 r^4-2 a^6 c^2 q^4 r^4-4 a^4 b^2 c^2 q^4 r^4+2 a^4 c^4 q^4 r^4+a^2 b^6 p^3 r^5+b^8 p^3 r^5-b^6 c^2 p^3 r^5+3 a^4 b^4 p^2 q r^5+2 a^2 b^6 p^2 q r^5+b^8 p^2 q r^5-4 a^2 b^4 c^2 p^2 q r^5-2 b^6 c^2 p^2 q r^5+b^4 c^4 p^2 q r^5+a^8 p q^2 r^5+4 a^4 b^4 p q^2 r^5+a^2 b^6 p q^2 r^5-3 a^6 c^2 p q^2 r^5-3 a^4 b^2 c^2 p q^2 r^5-3 a^2 b^4 c^2 p q^2 r^5+3 a^4 c^4 p q^2 r^5+3 a^2 b^2 c^4 p q^2 r^5-a^2 c^6 p q^2 r^5+a^8 q^3 r^5+a^4 b^4 q^3 r^5-2 a^6 c^2 q^3 r^5-2 a^4 b^2 c^2 q^3 r^5+a^4 c^4 q^3 r^5) (-a^2 c^6 p^6 q^3+c^8 p^6 q^3-2 a^2 c^6 p^5 q^4-a^2 c^6 p^4 q^5+a^4 c^4 p^6 q^2 r-a^2 b^2 c^4 p^6 q^2 r+2 b^4 c^4 p^6 q^2 r-2 a^2 c^6 p^6 q^2 r+c^8 p^6 q^2 r+4 a^2 b^2 c^4 p^5 q^3 r+2 b^4 c^4 p^5 q^3 r-2 a^2 c^6 p^5 q^3 r-4 b^2 c^6 p^5 q^3 r+2 c^8 p^5 q^3 r-a^4 c^4 p^4 q^4 r+7 a^2 b^2 c^4 p^4 q^4 r-6 a^2 c^6 p^4 q^4 r+2 a^2 b^2 c^4 p^3 q^5 r-2 a^2 c^6 p^3 q^5 r+a^4 b^4 p^6 q r^2-2 a^2 b^6 p^6 q r^2+b^8 p^6 q r^2-a^2 b^4 c^2 p^6 q r^2+2 b^4 c^4 p^6 q r^2+2 a^6 b^2 p^5 q^2 r^2-2 a^4 b^4 p^5 q^2 r^2-2 a^2 b^6 p^5 q^2 r^2+2 b^8 p^5 q^2 r^2+2 a^6 c^2 p^5 q^2 r^2-8 a^4 b^2 c^2 p^5 q^2 r^2+6 a^2 b^4 c^2 p^5 q^2 r^2-4 b^6 c^2 p^5 q^2 r^2-2 a^4 c^4 p^5 q^2 r^2+6 a^2 b^2 c^4 p^5 q^2 r^2+4 b^4 c^4 p^5 q^2 r^2-2 a^2 c^6 p^5 q^2 r^2-4 b^2 c^6 p^5 q^2 r^2+2 c^8 p^5 q^2 r^2+a^8 p^4 q^3 r^2+2 a^6 b^2 p^4 q^3 r^2-6 a^4 b^4 p^4 q^3 r^2+2 a^2 b^6 p^4 q^3 r^2+b^8 p^4 q^3 r^2-4 a^4 b^2 c^2 p^4 q^3 r^2+2 a^2 b^4 c^2 p^4 q^3 r^2-4 b^6 c^2 p^4 q^3 r^2+6 a^2 b^2 c^4 p^4 q^3 r^2+6 b^4 c^4 p^4 q^3 r^2-2 a^2 c^6 p^4 q^3 r^2-4 b^2 c^6 p^4 q^3 r^2+c^8 p^4 q^3 r^2+2 a^8 p^3 q^4 r^2-2 a^6 b^2 p^3 q^4 r^2-2 a^4 b^4 p^3 q^4 r^2+2 a^2 b^6 p^3 q^4 r^2-2 a^6 c^2 p^3 q^4 r^2+4 a^4 b^2 c^2 p^3 q^4 r^2-6 a^2 b^4 c^2 p^3 q^4 r^2-2 a^4 c^4 p^3 q^4 r^2+10 a^2 b^2 c^4 p^3 q^4 r^2-6 a^2 c^6 p^3 q^4 r^2+a^8 p^2 q^5 r^2-2 a^6 b^2 p^2 q^5 r^2+a^4 b^4 p^2 q^5 r^2-a^2 b^4 c^2 p^2 q^5 r^2+2 a^2 b^2 c^4 p^2 q^5 r^2-a^2 c^6 p^2 q^5 r^2-a^2 b^6 p^6 r^3+b^8 p^6 r^3-2 a^2 b^6 p^5 q r^3+2 b^8 p^5 q r^3+4 a^2 b^4 c^2 p^5 q r^3-4 b^6 c^2 p^5 q r^3+2 b^4 c^4 p^5 q r^3+a^8 p^4 q^2 r^3-2 a^2 b^6 p^4 q^2 r^3+b^8 p^4 q^2 r^3+2 a^6 c^2 p^4 q^2 r^3-4 a^4 b^2 c^2 p^4 q^2 r^3+6 a^2 b^4 c^2 p^4 q^2 r^3-4 b^6 c^2 p^4 q^2 r^3-6 a^4 c^4 p^4 q^2 r^3+2 a^2 b^2 c^4 p^4 q^2 r^3+6 b^4 c^4 p^4 q^2 r^3+2 a^2 c^6 p^4 q^2 r^3-4 b^2 c^6 p^4 q^2 r^3+c^8 p^4 q^2 r^3+4 a^8 p^3 q^3 r^3-2 a^6 b^2 p^3 q^3 r^3-2 a^2 b^6 p^3 q^3 r^3-2 a^6 c^2 p^3 q^3 r^3+2 a^2 b^4 c^2 p^3 q^3 r^3+2 a^2 b^2 c^4 p^3 q^3 r^3-2 a^2 c^6 p^3 q^3 r^3+5 a^8 p^2 q^4 r^3-4 a^6 b^2 p^2 q^4 r^3-a^2 b^6 p^2 q^4 r^3-4 a^6 c^2 p^2 q^4 r^3+4 a^4 b^2 c^2 p^2 q^4 r^3-a^4 c^4 p^2 q^4 r^3+3 a^2 b^2 c^4 p^2 q^4 r^3-2 a^2 c^6 p^2 q^4 r^3+2 a^8 p q^5 r^3-2 a^6 b^2 p q^5 r^3-2 a^2 b^6 p^5 r^4-a^4 b^4 p^4 q r^4-6 a^2 b^6 p^4 q r^4+7 a^2 b^4 c^2 p^4 q r^4+2 a^8 p^3 q^2 r^4-2 a^6 b^2 p^3 q^2 r^4-2 a^4 b^4 p^3 q^2 r^4-6 a^2 b^6 p^3 q^2 r^4-2 a^6 c^2 p^3 q^2 r^4+4 a^4 b^2 c^2 p^3 q^2 r^4+10 a^2 b^4 c^2 p^3 q^2 r^4-2 a^4 c^4 p^3 q^2 r^4-6 a^2 b^2 c^4 p^3 q^2 r^4+2 a^2 c^6 p^3 q^2 r^4+5 a^8 p^2 q^3 r^4-4 a^6 b^2 p^2 q^3 r^4-a^4 b^4 p^2 q^3 r^4-2 a^2 b^6 p^2 q^3 r^4-4 a^6 c^2 p^2 q^3 r^4+4 a^4 b^2 c^2 p^2 q^3 r^4+3 a^2 b^4 c^2 p^2 q^3 r^4-a^2 c^6 p^2 q^3 r^4+4 a^8 p q^4 r^4-2 a^6 b^2 p q^4 r^4-2 a^6 c^2 p q^4 r^4+a^8 q^5 r^4-a^2 b^6 p^4 r^5-2 a^2 b^6 p^3 q r^5+2 a^2 b^4 c^2 p^3 q r^5+a^8 p^2 q^2 r^5-a^2 b^6 p^2 q^2 r^5-2 a^6 c^2 p^2 q^2 r^5+2 a^2 b^4 c^2 p^2 q^2 r^5+a^4 c^4 p^2 q^2 r^5-a^2 b^2 c^4 p^2 q^2 r^5+2 a^8 p q^3 r^5-2 a^6 c^2 p q^3 r^5+a^8 q^4 r^5) : :
On the pedal circle of P.

Examples:
P = X(1) -> X(13756).
P = X(3) -> X(25641).

Best regards,
Peter Moses.

HYACINTHOS 26850

[Antreas P. Hatzipolakis]:

Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P.

Denote:

Pab, Pac = the orthogonal projections of Pa on OB,OC, resp.

(N1) = the NPC of PaPabPac. Similarly (N2),(N3)

Ra = the radical axis of (N2),(N3. Similarly Rb, Rc.

Sa = the parallel to Ra through A. Similarly Sb, Sc

1. Which is the locus of P such that Sa,Sb,Sc are concurrent? The Euler Line + + ??

2. Let P be a point on the Euler Line.

2.1. Which is the locus of the radical center P' of (N1),(N2),(N3) [point of concurrence of Ra,Rb,Rc] as P moves on the Euler line?

2.2. Which is the locus of the point of concurrence P" of Sa,Sb,Sc (if concur) as P moves on the Euler line?

[...]


[Peter Moses]:

1: Euler line and this conic:
2 a^2 (a^2 - b^2 - c^2) (a^6 b^2 - 3 a^4 b^4 + 3 a^2 b^6 - b^8 + a^6 c^2 + 2 a^2 b^4 c^2 - 3 b^6 c^2 - 3 a^4 c^4 + 2 a^2 b^2 c^4 + 8 b^4 c^4 + 3 a^2 c^6 - 3 b^2 c^6 - c^8) x^2 + (3 a^12 - 11 a^10 b^2 + 15 a^8 b^4 - 10 a^6 b^6 + 5 a^4 b^8 - 3 a^2 b^10 + b^12 - 11 a^10 c^2 + 21 a^8 b^2 c^2 + 6 a^6 b^4 c^2 - 33 a^4 b^6 c^2 + 19 a^2 b^8 c^2 - 2 b^10 c^2 + 15 a^8 c^4 + 6 a^6 b^2 c^4 + 56 a^4 b^4 c^4 - 16 a^2 b^6 c^4 - b^8 c^4 - 10 a^6 c^6 - 33 a^4 b^2 c^6 - 16 a^2 b^4 c^6 + 4 b^6 c^6 + 5 a^4 c^8 + 19 a^2 b^2 c^8 - b^4 c^8 - 3 a^2 c^10 - 2 b^2 c^10 + c^12) y z + cyclic
 
2.1: A nasty cubic.
 
2.2: circumconic thru X(3519)
(b^2-c^2) (a^2-b^2-c^2) (a^8-3 a^6 b^2+4 a^4 b^4-3 a^2 b^6+b^8-3 a^6 c^2-11 a^4 b^2 c^2+3 a^2 b^4 c^2-4 b^6 c^2+4 a^4 c^4+3 a^2 b^2 c^4+6 b^4 c^4-3 a^2 c^6-4 b^2 c^6+c^8) y z + cyclic

****

Suppose we parameterize a point on the Euler lines as a^2 SA + k SB SC::, then the concurrence is
1 / (a^2 SA (S^2 + 5 SA^2) + k (3 S^2 - SA^2) SB SC)::
 
Thus:
1): L, k = -1
concurrence = 1/(a^2 SA (S^2+5 SA^2)-(3 S^2-SA^2) SB SC)::
 
2): O, k = 0
concurrence = 1/(a^2 SA (S^2+5 SA^2)):: on lines {{4,3521},{93,403},...}
 
3): G, k = 1
concurrence = 1/(a^2 SA (S^2+5 SA^2)+(3 S^2-SA^2) SB SC)::
 
4): N, k = 2
concurrence = 1/(a^2 SA (S^2+5 SA^2)+2 (3 S^2-SA^2) SB SC)::
 
5): H, k = Infinity
concurrence = 1/((3 S^2-SA^2) SB SC):: = X(3519).
 
6): Schiffler, k = R/(r+R)
concurrence = 1/(a^2 (r+R) SA (S^2+5 SA^2)+R (3 S^2-SA^2) SB SC)::
 
Best regards,
Peter Moses.

[Peter Moses]:


Hi Antreas,

Some details for those, unfortunately mostly uninspiring, points ... plus one other:

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

L, k=-1; (a^8 - 3*a^6*b^2 + 4*a^4*b^4 - 3*a^2*b^6 + b^8 - 4*a^6*c^2 - 6*a^4*b^2*c^2 - 6*a^2*b^4*c^2 - 4*b^6*c^2 + 6*a^4*c^4 + 13*a^2*b^2*c^4 + 6*b^4*c^4 - 4*a^2*c^6 - 4*b^2*c^6 + c^8)*(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8 - 3*a^6*c^2 - 6*a^4*b^2*c^2 + 13*a^2*b^4*c^2 - 4*b^6*c^2 + 4*a^4*c^4 - 6*a^2*b^2*c^4 + 6*b^4*c^4 - 3*a^2*c^6 - 4*b^2*c^6 + c^8):: 

on lines {}.
alternative barycentrics: 1/(a^2 SA (S^2+5 SA^2)-(3 S^2-SA^2) SB SC).

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

O, k=0; b^2*c^2*(-a^2 + b^2 - c^2)*(a^2 + b^2 - c^2)*(a^4 + 3*a^2*b^2 + b^4 - 2*a^2*c^2 - 2*b^2*c^2 + c^4)*(a^4 - 2*a^2*b^2 + b^4 + 3*a^2*c^2 - 2*b^2*c^2 + c^4):: 

on lines {{4, 3521}, {93, 403}, {186, 1105}, {235, 6344}, {1217, 7505}, {1300, 3518}, {8884, 11815}}.
X(i)-isoconjugate of X(j) for these (i,j): {{255, 3520}, {2169, 11591}}.
barycentric product X(i)X(j) for these {i,j}: {{324, 11815}, {2052, 3521}}.
barycentric quotient X(i)/X(j) for these {i,j}: {{53, 11591}, {393, 3520}, {3521, 394}, {11815, 97}}.
major center: Tan[A] / (2 Cos[2 A] + 3).
alternative barycentrics: 1/(a^2 SA (S^2+5 SA^2)).

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

G, k=1; (a^8 - 3*a^6*b^2 + 4*a^4*b^4 - 3*a^2*b^6 + b^8 + 10*a^4*b^2*c^2 + 10*a^2*b^4*c^2 - 6*a^4*c^4 - 15*a^2*b^2*c^4 - 6*b^4*c^4 + 8*a^2*c^6 + 8*b^2*c^6 - 3*c^8)*(a^8 - 6*a^4*b^4 + 8*a^2*b^6 - 3*b^8 - 3*a^6*c^2 + 10*a^4*b^2*c^2 - 15*a^2*b^4*c^2 + 8*b^6*c^2 + 4*a^4*c^4 + 10*a^2*b^2*c^4 - 6*b^4*c^4 - 3*a^2*c^6 + c^8):: 

on lines {}.
alternative barycentrics: 1/(a^2 SA (S^2+5 SA^2)+(3 S^2-SA^2) SB SC).

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

N, k=2; (a^8 - 3*a^6*b^2 + 4*a^4*b^4 - 3*a^2*b^6 + b^8 - a^6*c^2 + 6*a^4*b^2*c^2 + 6*a^2*b^4*c^2 - b^6*c^2 - 3*a^4*c^4 - 8*a^2*b^2*c^4 - 3*b^4*c^4 + 5*a^2*c^6 + 5*b^2*c^6 - 2*c^8)*(a^8 - a^6*b^2 - 3*a^4*b^4 + 5*a^2*b^6 - 2*b^8 - 3*a^6*c^2 + 6*a^4*b^2*c^2 - 8*a^2*b^4*c^2 + 5*b^6*c^2 + 4*a^4*c^4 + 6*a^2*b^2*c^4 - 3*b^4*c^4 - 3*a^2*c^6 - b^2*c^6 + c^8):: 

on lines {}.
alternative barycentrics: 1/(a^2 SA (S^2+5 SA^2)+2 (3 S^2-SA^2) SB SC).

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

 

X(3519).

H, k = Infinity; (a^2 - b^2 - c^2)*(a^4 - a^2*b^2 + b^4 - 2*a^2*c^2 - 2*b^2*c^2 + c^4)*(a^4 - 2*a^2*b^2 + b^4 - a^2*c^2 - 2*b^2*c^2 + c^4):: 

on lines {{2, 1493}, {3, 539}, {4, 93}, {5, 1173}, {6, 17}, {20, 13452}, {49, 343}, {54, 140}, {64, 1657}, {65, 2962}, {68, 12606}, {69, 12363}, {70, 14791}, {74, 550}, {185, 14861}, {265, 5562}, {290, 7768}, {340, 8795}, {394, 15317}, {399, 14862}, {511, 13433}, {524, 5576}, {542, 13564}, {567, 12899}, {895, 11585}, {1176, 3564}, {1503, 9935}, {1899, 11577}, {2889, 3448}, {2914, 14643}, {2918, 5898}, {2979, 12291}, {3426, 5073}, {3431, 3523}, {3521, 13754}, {3522, 11270}, {3527, 3574}, {3850, 14483}, {4857, 6286}, {5056, 13565}, {5059, 11738}, {5068, 14491}, {5449, 15002}, {5504, 12359}, {6101, 12226}, {6145, 7574}, {6515, 9827}, {7517, 15069}, {10018, 11597}, {10625, 14864}, {10628, 11744}, {11138, 11600}, {11139, 11601}, {11225, 15047}, {11412, 12280}, {12936, 15232}}.
anticomplement X(1493).
complement X(11271).
midpoint of X(i) and X(j) for these {i,j}: {{2888, 12325}, {11412, 12280}}.
reflection of X(i) in X(j) for these {i,j}: {{195, 1209}, {6243, 6152}, {6288, 2888}, {11271, 1493}, {12226, 6101}, {12316, 3574}, {13431, 12242}, {13432, 13431}}.
{X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (2, 11271, 1493), (17, 18, 2963), (93, 562, 14111), (195, 1656, 12242), (195, 13432, 13431), (1209, 12242, 1656), (1209, 13431, 12242), (1656, 13432, 195), (12242, 13431, 195).
on K618.
on Jerabek.
isogonal conjugate of X(3518).
anticomplement of the isogonal of X(1487).
X(i)-anticomplementary conjugate of X(j) for these (i,j): {{1487, 8}, {2962, 2889}}.
X(11140)-Ceva conjugate of X(2963).
X(i)-isoconjugate of X(j) for these (i,j): {{1, 3518}, {4, 2964}, {19, 1994}, {49, 158}, {92, 2965}, {143, 2190}, {162, 1510}, {1973, 7769}, {2148, 14129}, {2167, 14577}, {2216, 6152}}.
cevapoint of X(i) and X(j) for these (i,j): {{633, 634}, {14813, 14814}}.
barycentric product X(i)X(j) for these {i,j}: {{3, 11140}, {63, 2962}, {69, 2963}, {93, 394}, {252, 343}, {525, 930}}.
barycentric quotient X(i)/X(j) for these {i,j}: {{3, 1994}, {5, 14129}, {6, 3518}, {17, 472}, {18, 473}, {48, 2964}, {51, 14577}, {69, 7769}, {93, 2052}, {184, 2965}, {216, 143}, {252, 275}, {562, 14165}, {570, 6152}, {577, 49}, {647, 1510}, {930, 648}, {2962, 92}, {2963, 4}, {8603, 10633}, {8604, 10632}, {11140, 264}, {14111, 11547}}.
major center: 1/(Cot[A] - 3 Tan[A]).
alternative barycentrics: 1/((3 S^2-SA^2) SB SC).

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

Schiffler, k = R/(r+R);  b*c*(-(a^9*b) + a^8*b^2 + 4*a^7*b^3 - 4*a^6*b^4 - 6*a^5*b^5 + 6*a^4*b^6 + 4*a^3*b^7 - 4*a^2*b^8 - a*b^9 + b^10 + a^9*c + a^8*b*c - 2*a^7*b^2*c - 2*a^6*b^3*c + 2*a^3*b^6*c + 2*a^2*b^7*c - a*b^8*c - b^9*c - 3*a^7*b*c^2 + 5*a^6*b^2*c^2 + 8*a^5*b^3*c^2 - 14*a^4*b^4*c^2 - 7*a^3*b^5*c^2 + 13*a^2*b^6*c^2 + 2*a*b^7*c^2 - 4*b^8*c^2 - 3*a^7*c^3 + 3*a^6*b*c^3 + 2*a^5*b^2*c^3 - a^3*b^4*c^3 - 7*a^2*b^5*c^3 + 2*a*b^6*c^3 + 4*b^7*c^3 + 8*a^4*b^2*c^4 - 14*a^2*b^4*c^4 + 6*b^6*c^4 + 4*a^5*c^5 + 2*a^3*b^2*c^5 + 8*a^2*b^3*c^5 - 6*b^5*c^5 + 3*a^3*b*c^6 + 5*a^2*b^2*c^6 - 2*a*b^3*c^6 - 4*b^4*c^6 - 3*a^3*c^7 - 3*a^2*b*c^7 - 2*a*b^2*c^7 + 4*b^3*c^7 + a*b*c^8 + b^2*c^8 + a*c^9 - b*c^9)*(-(a^9*b) + 3*a^7*b^3 - 4*a^5*b^5 + 3*a^3*b^7 - a*b^9 + a^9*c - a^8*b*c + 3*a^7*b^2*c - 3*a^6*b^3*c - 3*a^3*b^6*c + 3*a^2*b^7*c - a*b^8*c + b^9*c - a^8*c^2 + 2*a^7*b*c^2 - 5*a^6*b^2*c^2 - 2*a^5*b^3*c^2 - 8*a^4*b^4*c^2 - 2*a^3*b^5*c^2 - 5*a^2*b^6*c^2 + 2*a*b^7*c^2 - b^8*c^2 - 4*a^7*c^3 + 2*a^6*b*c^3 - 8*a^5*b^2*c^3 - 8*a^2*b^5*c^3 + 2*a*b^6*c^3 - 4*b^7*c^3 + 4*a^6*c^4 + 14*a^4*b^2*c^4 + a^3*b^3*c^4 + 14*a^2*b^4*c^4 + 4*b^6*c^4 + 6*a^5*c^5 + 7*a^3*b^2*c^5 + 7*a^2*b^3*c^5 + 6*b^5*c^5 - 6*a^4*c^6 - 2*a^3*b*c^6 - 13*a^2*b^2*c^6 - 2*a*b^3*c^6 - 6*b^4*c^6 - 4*a^3*c^7 - 2*a^2*b*c^7 - 2*a*b^2*c^7 - 4*b^3*c^7 + 4*a^2*c^8 + a*b*c^8 + 4*b^2*c^8 + a*c^9 + b*c^9 - c^10):: 

on lines {}.
alternative barycentrics: 1/(a^2 (r+R) SA (S^2+5 SA^2)+R (3 S^2-SA^2) SB SC).

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

 

X(13597).

X(30), k=-2; (a^8 - 3*a^6*b^2 + 4*a^4*b^4 - 3*a^2*b^6 + b^8 - 3*a^6*c^2 - 2*a^4*b^2*c^2 - 2*a^2*b^4*c^2 - 3*b^6*c^2 + 3*a^4*c^4 + 6*a^2*b^2*c^4 + 3*b^4*c^4 - a^2*c^6 - b^2*c^6)*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 - 3*a^6*c^2 - 2*a^4*b^2*c^2 + 6*a^2*b^4*c^2 - b^6*c^2 + 4*a^4*c^4 - 2*a^2*b^2*c^4 + 3*b^4*c^4 - 3*a^2*c^6 - 3*b^2*c^6 + c^8):: 

on lines {{4,11792},{25,13507},{30, 11703},{99,1232},{110,140},{ 112,6748},{476,5899},{953, 5957},{2687,5959},{2699,5958}} .
reflection of X(4) in X(11792).
on circumcircle.
isogonal conjugate of X(13391).
Collings transform of X(11792).
alternative barycentrics: 1/(a^2 SA (S^2+5 SA^2)-2 (3 S^2-SA^2) SB SC)).

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

Best regards,
Peter Moses.

 

HYACINTHOS 26373

[Le Viet An]:

Let ABC be a triangle and HaHbHc the orthic triangle.

The line IH intersects HbHc, HcHa, HaHb at Ka,Kb,Kc, resp.

The perpendiculars to IA, IB, IC  from Ka, Kb, Kc, resp. bound a triangle A'B'C'.

Then the circumcircle of the triangle A'B'C' touches the NPC of ABC.

Which is

1. the center of the circle ?

2. the touchpoint ?


[Peter Moses]:


Hi Antreas,


1).
(a+b-c) (a-b+c) (a^2+b^2-c^2) (a^2-b^2+c^2) (2 a^12 b-2 a^11 b^2-6 a^10 b^3+7 a^9 b^4+4 a^8 b^5-8 a^7 b^6+4 a^6 b^7+2 a^5 b^8-6 a^4 b^9+2 a^3 b^10+2 a^2 b^11-a b^12+2 a^12 c-8 a^11 b c+4 a^10 b^2 c+16 a^9 b^3 c-19 a^8 b^4 c-2 a^7 b^5 c+14 a^6 b^6 c-10 a^5 b^7 c+2 a^4 b^8 c+2 a^3 b^9 c-2 a^2 b^10 c+2 a b^11 c-b^12 c-2 a^11 c^2+4 a^10 b c^2-2 a^9 b^2 c^2-a^8 b^3 c^2+8 a^7 b^4 c^2-11 a^6 b^5 c^2-2 a^5 b^6 c^2+9 a^4 b^7 c^2-2 a^3 b^8 c^2-a^2 b^9 c^2-6 a^10 c^3+16 a^9 b c^3-a^8 b^2 c^3-28 a^7 b^3 c^3+17 a^6 b^4 c^3+10 a^5 b^5 c^3-11 a^4 b^6 c^3+8 a^3 b^7 c^3-3 a^2 b^8 c^3-6 a b^9 c^3+4 b^10 c^3+7 a^9 c^4-19 a^8 b c^4+8 a^7 b^2 c^4+17 a^6 b^3 c^4-16 a^5 b^4 c^4+6 a^4 b^5 c^4-11 a^2 b^7 c^4+9 a b^8 c^4-b^9 c^4+4 a^8 c^5-2 a^7 b c^5-11 a^6 b^2 c^5+10 a^5 b^3 c^5+6 a^4 b^4 c^5-20 a^3 b^5 c^5+15 a^2 b^6 c^5+4 a b^7 c^5-6 b^8 c^5-8 a^7 c^6+14 a^6 b c^6-2 a^5 b^2 c^6-11 a^4 b^3 c^6+15 a^2 b^5 c^6-16 a b^6 c^6+4 b^7 c^6+4 a^6 c^7-10 a^5 b c^7+9 a^4 b^2 c^7+8 a^3 b^3 c^7-11 a^2 b^4 c^7+4 a b^5 c^7+4 b^6 c^7+2 a^5 c^8+2 a^4 b c^8-2 a^3 b^2 c^8-3 a^2 b^3 c^8+9 a b^4 c^8-6 b^5 c^8-6 a^4 c^9+2 a^3 b c^9-a^2 b^2 c^9-6 a b^3 c^9-b^4 c^9+2 a^3 c^10-2 a^2 b c^10+4 b^3 c^10+2 a^2 c^11+2 a b c^11-a c^12-b c^12)::
on lines {{4, 109}, {65, 7649}}.

2).
(a-b-c) (b-c)^2 (a^2+b^2-c^2) (a^2-b^2+b c-c^2) (a^2-b^2+c^2) (a^6-a^5 b-a^4 b^2+2 a^3 b^3-a^2 b^4-a b^5+b^6-a^5 c+a^4 b c+a^3 b^2 c-a^2 b^3 c-a^4 c^2+a^3 b c^2+a b^3 c^2-b^4 c^2+2 a^3 c^3-a^2 b c^3+a b^2 c^3-a^2 c^4-b^2 c^4-a c^5+c^6)::
on lines {{4, 2222}, {11, 7649}, {117, 1737}, {119, 1877}, {650, 5190}, {867, 10017}, {5089, 5513}}.
On the NP circle.
inverse of X(5190) in the Stevanovic circle.
inverse of X(2222) in the polar circle.


Best regards,
Peter Moses.

HYACINTHOS 26370

[Le Viet An]:

Let ABC be a triangle.

Denote:

J = the reflection of I in the Feuerbach point Fe 

[ J = X(80)].

Then the cevian circle of J passes through Fe.

Which point is its center (lying on the OI line)?

 

 

[Angel Montesdeoca]:

The center of cevian circle of X(80) is W = (9r(r+R)-s^2) X(1) - 9r(2r-R) X(3).

W = ( a (4 a^6
               -a^5 (b+c)
                -a^4 (13 b^2-8 b c+13 c^2)
                +a^3 (2 b^3+11 b^2 c+11 b c^2+2 c^3)
               +2 a^2 (7 b^4-8 b^3 c-3 b^2 c^2-8 b c^3+7 c^4)
               -a (b-c)^2 (b^3+12 b^2 c+12 b c^2+c^3)
               -(b^2-c^2)^2 (5 b^2-8 b c+5 c^2) ) : ... : ... ).
               
(6 - 9 - 13) - search numbers  of W: (81.2652311218832, 67.3974022848629, -80.5261053100206).


The cevian circle of X(80) intersects the incircle in X(11) and X(3025).

Angel Montesdeoca

HYACINTHOS 26369

[Le Viet An]:

Let ABC be a triangle.

Denote:

J = the reflection of I in the Feuerbach point Fe
[ J = X(80)].

Then the cevian circle of J passes through Fe.

Which point is its center (lying on the OI line)?


[Peter Moses]:


Hi Antreas,

The circumcircle of the cevian of a point P passes through X(11) for P on the Feuerbach circumhyperbola and on a (b-c) y^2 z^2 + cyclic = 0.

Specifically for P = X(80), the circle's center is

a (4 a^6-a^5 b-13 a^4 b^2+2 a^3 b^3+14 a^2 b^4-a b^5-5 b^6-a^5 c+8 a^4 b c+11 a^3 b^2 c-16 a^2 b^3 c-10 a b^4 c+8 b^5 c-13 a^4 c^2+11 a^3 b c^2-6 a^2 b^2 c^2+11 a b^3 c^2+5 b^4 c^2+2 a^3 c^3-16 a^2 b c^3+11 a b^2 c^3-16 b^3 c^3+14 a^2 c^4-10 a b c^4+5 b^2 c^4-a c^5+8 b c^5-5 c^6:: 

on lines {{1, 3}, {88, 12515}}.
(9 r (r + R) - s^2) X[1] + 9 r (R - 2 r) X[3].

Best regards,
Peter Moses.

 

HYACINTHOS 26367

[Le Viet An]:

Dear Mr Rodinos

Let ABC be a triangle and L the Euler line.

Denote:

E = the point of concurrence of the reflections of L in BC, CA, AB, resp. [ =X(110)]
P = the reflection of E in O [ = X(74)]

The lines passing through P and parallels to OA,OB,OC intersect L at Pa,Pb,Pc, resp.
The perpendiculars to APa, BPb, CPc, at Pa,Pb, Pc, resp. bound a triangle A'B'C'.

1. The NPC of A'B'C' passes through O.
Which is its center?
2.The orthocenter of A'B'C' lies on the circumcircle of ABC.
Which point is it?

Thank you very much.
Best regards,
Le Viet An.

 

[César Lozada ]:

 

1)       

N’ = (a^26-4*(b^2+c^2)*a^24+28*b^2* c^2*a^22+(b^2+c^2)*(21*b^4-68* b^2*c^2+21*c^4)*a^20-(23*b^8+ 23*c^8+b^2*c^2*(68*b^4-235*b^ 2*c^2+68*c^4))*a^18-(b^2+c^2)* (45*b^8+45*c^8-4*b^2*c^2*(84* b^4-149*b^2*c^2+84*c^4))*a^16+ (120*b^12+120*c^12-(219*b^8+ 219*c^8+b^2*c^2*(479*b^4-1162* b^2*c^2+479*c^4))*b^2*c^2)*a^ 14-(b^4-c^4)*(b^2-c^2)*(78*b^ 8+78*c^8+b^2*c^2*(351*b^4-889* b^2*c^2+351*c^4))*a^12-(b^2-c^ 2)^2*(45*b^12+45*c^12-(495*b^ 8+495*c^8+2*b^2*c^2*(98*b^4- 615*b^2*c^2+98*c^4))*b^2*c^2)* a^10+(b^4-c^4)*(b^2-c^2)*(98* b^12+98*c^12-(259*b^8+259*c^8+ b^2*c^2*(511*b^4-1328*b^2*c^2+ 511*c^4))*b^2*c^2)*a^8-(b^2-c^ 2)^4*(56*b^12+56*c^12+(205*b^ 8+205*c^8-3*b^2*c^2*(57*b^4+ 242*b^2*c^2+57*c^4))*b^2*c^2)* a^6+(b^4-c^4)*(b^2-c^2)^3*(9* b^12+9*c^12+(95*b^8+95*c^8+16* b^2*c^2*(b^2-4*b*c-c^2)*(b^2+ 4*b*c-c^2))*b^2*c^2)*a^4+(b^2- c^2)^6*(3*b^12+3*c^12-b^2*c^2* (7*b^4+61*b^2*c^2+7*c^4)*(b^2+ c^2)^2)*a^2-(b^2-c^2)^8*(b^2+ c^2)*(b^4+c^4+b*c*(b^2+3*b*c+ c^2))*(b^4+c^4-b*c*(b^2-3*b*c+ c^2)))*a : :

= On lines: {74,186}, {520,13293}

= [ 12.969555486717660, 13.03416707640744, -11.368938334090460 ]

 

2)      H’ =  X(1304)

 

César Lozada

 

HYACINTHOS 26366

[Le Viet An]:

Dear Mr Rodinos

Let ABC be a triangle and L the Euler line.

Denote:

E = the point of concurrence of the reflections of L in BC, CA, AB, resp. [ =X(110)]
P = the reflection of E in O [ = X(74)]

The lines passing through P and parallels to OA,OB,OC intersect L at Pa,Pb,Pc, resp.
The perpendiculars to APa, BPb, CPc, at Pa,Pb, Pc, resp. bound a triangle A'B'C'.

1. The NPC of A'B'C' passes through O.
Which is its center?
2.The orthocenter of A'B'C' lies on the circumcircle of ABC.
Which point is it?

Thank you very much.
Best regards,
Le Viet An.

[Peter Moses]:

Hi Antreas,

1) 

a^2 (a^26-4 a^24 b^2+21 a^20 b^6-23 a^18 b^8-45 a^16 b^10+120 a^14 b^12-78 a^12 b^14-45 a^10 b^16+98 a^8 b^18-56 a^6 b^20+9 a^4 b^22+3 a^2 b^24-b^26-4 a^24 c^2+28 a^22 b^2 c^2-47 a^20 b^4 c^2-68 a^18 b^6 c^2+291 a^16 b^8 c^2-219 a^14 b^10 c^2-273 a^12 b^12 c^2+585 a^10 b^14 c^2-357 a^8 b^16 c^2+19 a^6 b^18 c^2+68 a^4 b^20 c^2-25 a^2 b^22 c^2+2 b^24 c^2-47 a^20 b^2 c^4+235 a^18 b^4 c^4-260 a^16 b^6 c^4-479 a^14 b^8 c^4+1318 a^12 b^10 c^4-839 a^10 b^12 c^4-350 a^8 b^14 c^4+655 a^6 b^16 c^4-251 a^4 b^18 c^4+12 a^2 b^20 c^4+6 b^22 c^4+21 a^20 c^6-68 a^18 b^2 c^6-260 a^16 b^4 c^6+1162 a^14 b^6 c^6-967 a^12 b^8 c^6-1127 a^10 b^10 c^6+2196 a^8 b^12 c^6-964 a^6 b^14 c^6-128 a^4 b^16 c^6+149 a^2 b^18 c^6-14 b^20 c^6-23 a^18 c^8+291 a^16 b^2 c^8-479 a^14 b^4 c^8-967 a^12 b^6 c^8+2852 a^10 b^8 c^8-1587 a^8 b^10 c^8-943 a^6 b^12 c^8+1075 a^4 b^14 c^8-199 a^2 b^16 c^8-20 b^18 c^8-45 a^16 c^10-219 a^14 b^2 c^10+1318 a^12 b^4 c^10-1127 a^10 b^6 c^10-1587 a^8 b^8 c^10+2578 a^6 b^10 c^10-773 a^4 b^12 c^10-220 a^2 b^14 c^10+75 b^16 c^10+120 a^14 c^12-273 a^12 b^2 c^12-839 a^10 b^4 c^12+2196 a^8 b^6 c^12-943 a^6 b^8 c^12-773 a^4 b^10 c^12+560 a^2 b^12 c^12-48 b^14 c^12-78 a^12 c^14+585 a^10 b^2 c^14-350 a^8 b^4 c^14-964 a^6 b^6 c^14+1075 a^4 b^8 c^14-220 a^2 b^10 c^14-48 b^12 c^14-45 a^10 c^16-357 a^8 b^2 c^16+655 a^6 b^4 c^16-128 a^4 b^6 c^16-199 a^2 b^8 c^16+75 b^10 c^16+98 a^8 c^18+19 a^6 b^2 c^18-251 a^4 b^4 c^18+149 a^2 b^6 c^18-20 b^8 c^18-56 a^6 c^20+68 a^4 b^2 c^20+12 a^2 b^4 c^20-14 b^6 c^20+9 a^4 c^22-25 a^2 b^2 c^22+6 b^4 c^22+3 a^2 c^24+2 b^2 c^24-c^26):: 

on lines {{74, 186}, {520, 13293}}.

2). X(1304).

Best regards,
Peter Moses.

HYACINTHOS 26363

[Le Viet An]:

 

Let ABC be a triangle and DaDbDc the pedal triangle of I (intouch triangle).

Denote: 

 

Ab,Ac = the reflections of Da in B, C, resp.
Similarly Bc, Ba,Ca,Cb.
The lines BaCa, CbAb, AcBc bound triangle A'B'C'

[A' = AbCb /\ AcBc, B' = BaCa /\ BcAc, C' = CaBa /\ CbAb]

Then the Euler lines of A'AbAc, B'BcBa, C'CaCb are concurrent.

Which point is the point of concurrence?

 

[Angel Montesdeoca]

The Euler lines of A'AbAc, B'BcBa, C'CaCb are concurrent at W= 4 X(10) - 3 X(11)

W = (4 a^3 (b+c)-a^2 (b^2+14 b c+c^2)-4 a (b^3-2 b^2 c-2 b c^2+c^3)+(b^2-c^2)^2 : ... : ....)

W is the reflection of X(i) in X(j), for these {i, j}: {11,1145}, {149,3036}, {1317,100}, {1320,3035}, {5183,4394}, {6154,5541}, {7972,9945}, {7993,13226}, {12653,1387}.

W lies on lines X(i)X(j) for these {i, j}: {1, 6174}, {8, 190}, {10, 11}, {12, 10129}, {40, 550}, {56, 100}, {65, 10427}, {80, 4668}, {149, 2551}, {214, 3635}, {519, 1155}, {1000, 4413}, {1018, 4534}, {1320, 3035}, {1376, 13279}, {1387, 3624}, {2183, 3943}, {2254, 6366}, {2800, 3962}, {2829, 6361}, {3434, 13272}, {3617, 10707}, {3649, 10956}, {3679, 4679}, {3885, 8256}, {3893, 11362}, {3922, 12736}, {4701, 12732}, {4746, 12572}, {5082, 10953}, {5087, 6735}, {5252, 5856}, {5433, 10912}, {5434, 12648}, {5660, 11531}, {7091, 12641}, {7173, 13463}, {7972, 9945}, {7993, 13226}, {11500, 12245}, {12247, 12249}.

 (6 - 9 - 13) - search numbers  of W: (4.46619037228263, -2.98139348497599, 3.64338749199116).


Angel Montesdeoca

HYACINTHOS 26360

[Antreas P. Hatzipolakis]:

 

 

Let ABC be a triangle.

Denote:

Ab,Ac = the orthogonal projections of A on IB, IC, resp.

La = the Euler line of AAbAc

Laa, Lab, Lac = the reflections of La in IA,IB,IC, resp.

A1B1C1 = the triangle bounded by Laa, Lab, Lac

L1= the Euler line of A1B1C1. Similarly L2 ,L3.


1. L1, L2, L3 are concurrent.

2. the reflections of L1, L2 ,L3 in IA, IB,IC, resp. are concurrent.


[Peter Moses]:

Hi Antreas,

1).
a^2 (a^8 b^3+a^7 b^4-3 a^6 b^5-3 a^5 b^6+3 a^4 b^7+3 a^3 b^8-a^2 b^9-a b^10-2 a^9 b c+a^8 b^2 c+2 a^7 b^3 c-2 a^6 b^4 c+6 a^5 b^5 c-10 a^3 b^7 c+2 a^2 b^8 c+4 a b^9 c-b^10 c+a^8 b c^2-4 a^7 b^2 c^2+a^6 b^3 c^2+5 a^5 b^4 c^2-3 a^4 b^5 c^2+3 a^3 b^6 c^2-4 a b^8 c^2+b^9 c^2+a^8 c^3+2 a^7 b c^3+a^6 b^2 c^3-14 a^5 b^3 c^3+2 a^4 b^4 c^3+10 a^3 b^5 c^3-5 a^2 b^6 c^3+2 a b^7 c^3+b^8 c^3+a^7 c^4-2 a^6 b c^4+5 a^5 b^2 c^4+2 a^4 b^3 c^4-14 a^3 b^4 c^4+4 a^2 b^5 c^4+5 a b^6 c^4-3 b^7 c^4-3 a^6 c^5+6 a^5 b c^5-3 a^4 b^2 c^5+10 a^3 b^3 c^5+4 a^2 b^4 c^5-12 a b^5 c^5+2 b^6 c^5-3 a^5 c^6+3 a^3 b^2 c^6-5 a^2 b^3 c^6+5 a b^4 c^6+2 b^5 c^6+3 a^4 c^7-10 a^3 b c^7+2 a b^3 c^7-3 b^4 c^7+3 a^3 c^8+2 a^2 b c^8-4 a b^2 c^8+b^3 c^8-a^2 c^9+4 a b c^9+b^2 c^9-a c^10-b c^10):: 

on lines {{104, 5663}, {109, 3028}, {513, 3109}, {6789, 10176}}.
reflection of X(3109) in the OI line.

2).
2 a^7-2 a^5 b^2-a^4 b^3-3 a^3 b^4+a^2 b^5+3 a b^6+a^4 b^2 c-b^6 c-2 a^5 c^2+a^4 b c^2+8 a^3 b^2 c^2-a^2 b^3 c^2-3 a b^4 c^2-b^5 c^2-a^4 c^3-a^2 b^2 c^3+2 b^4 c^3-3 a^3 c^4-3 a b^2 c^4+2 b^3 c^4+a^2 c^5-b^2 c^5+3 a c^6-b c^6:: 

on lines {{1, 523}, {30, 944}, {265, 952}, {405, 2452}, {6741, 11735}}.
midpoint X(6742) and X(7984).
reflection of X(i) in X(j) for these {i,j}: {{3109, 1}, {6741, 11735}].

Best regards,
Peter Moses.

 

 

HYACINTHOS 26358

[Le Viet An] (*):

Let ABC be a triangle and L a line passing through O.

Denote:

A',B',C' = the orthogonal projections of A,B,C on L, resp. 

 

La, Lb, Lc = the parallels though A', B', C' to BC,CA,AB, resp.

A*B*C* = the triangle bounded by La, Lb, Lc

 

The NPC of A*B*C* touches the circumcircle of ABC.

For L = Euler line, the touchpoint is X110.

For L = X74OX110 which is the touchpoint?

For L = OK (Brocard axis) which is it?

(*) 9-circle touches circumcircle  

[Peter Moses]:

Hi Antreas,

For an line L = OP{p,q,r}, the touchpoint, T, is
T = a^2(b^2 (p - r) + (a^2 - c^2) q) (c^2 (p - q) + (a^2 - b^2) r):: = the isogonal conjugate of the orthopoint of (L / \ infinity).

For P = X(74), T = X(476),
P = K (Brocard), T = X(99).
P = I, T = X(100).

 

Best regards,
Peter Moses.

 

 

HYACINTHOS 26354

 [Antreas P. Hatzipolakis]:

   
Let ABC be a triangle and A'B'C',IaIbIc the pedal, antipedal triangles of I, resp.

Denote:

A",B",C" = the antipodes of A', B', C', in the incircle, resp.

The circumcircles of IA"Ia, IB"Ib, IC"Ic are coaxial.

2nd intersection (other than I)?

 

[Angel Montesdeoca]:

***  2nd intersection (other than I) of the circumcircles of IA"Ia, IB"Ib, IC"Ic is

W = (a (2 a^9
               -15 a^8 (b+c)
              +10 a^7 (b^2+12 b c+c^2)
              +2 a^6 (21 b^3-61 b^2 c-61 b c^2+21 c^3)
               -2 a^5 (21 b^4+70 b^3 c-62 b^2 c^2+70 b c^3+21 c^4)
                -4 a^4 (9 b^5-67 b^4 c+22 b^3 c^2+22 b^2 c^3-67 b c^4+9 c^5)
               +2 a^3 (b-c)^2 (23 b^4+6 b^3 c-74 b^2 c^2+6 b c^3+23 c^4)
               +2 a^2 (b-c)^2 (3 b^5-49 b^4 c-2 b^3 c^2-2 b^2 c^3-49 b c^4+3 c^5)
               -4 a (b^2-c^2)^2 (4 b^4-25 b^3 c+10 b^2 c^2-25 b c^3+4 c^4)
               3 (b-c)^4 (b+c)^3 (b^2-6 b c+c^2)) : .... : ....).
              
The coaxal axis is X(1)X(5806)
 
 (6 - 9 - 13) - search numbers  of W: (8.16415864716562, 8.77789505718487, -6.20441301022005).
            
              
 *** If Ra is the the radical center of incircle and circumcircle  of IA"Ia,  and define Rb, Rc cyclically.
 
   Ra, Rb , Rc concur in a point
  
   Z = (a (a^5 (b+c)
           -a^4 (b^2+14 b c+c^2)
          +a (b-c)^2 (b^3-9 b^2 c-9 b c^2+c^3)
          -2 a^3 (b^3-5 b^2 c-5 b c^2+c^3)
          +2 a^2 (b^4+4 b^3 c+14 b^2 c^2+4 b c^3+c^4)
           -(b^2-c^2)^2 (b^2-6 b c+c^2)) : ... : ....).
          
 Z lies on the coaxal axis X(1)X(5806) of the circumcircles of IA"Ia, IB"Ib, IC"Ic,  and  on lines X(i)X(j) for these {i, j}:  {1,5806}, {8,3740}, {65,390}, {354,4297}, {950,8581}, {2136,3303}, {2951,11518}, {3057,6738}, {3488,12675}, {3698,10389}, {3893,4423}, {4321,5665}, {5836,8236}, {8275,9957}, {9848,12672}.
 
  (6 - 9 - 13) - search numbers  of Z: (1.98657016904173, 2.01465647989418, 1.32902376396145).
 
  Angel Montesdeoca

 

 

HYACINTHOS 25346

[Le Viet An]:


Let ABC be a triangle and Fe, Fa, Fb, Fc the Feuerbach points.

Denote:

PaPbPc = the pedal triangle of I.

Qa, Qb, Qc  = the orthogonal projections of the excenters Ia,Ib, Ic on BC, CA, AB, resp.

The lines through Pa, Pb, Pc and parallels to FaQa, FbQb, FcQc, resp. bound a triangle A'B'C'.

The NPC of A'B'C' passes through the Feuerbach point Fe.

Which point is its center?

Le Viet An

***********

Note: Let A"B"C" be the triangle bounded by FaQa, FbQb, FcQc.

The homothetic center of A'B'C', A"B"C" is the Feuerbach point Fe.

APH

 

[Peter Moses]:


Hi Antreas,


FaFbFc = Feuerbach triangle.
PaPbPc = intouch triangle.
QaQbQc = extouch triangle.

A' = {-4 a^3+6 a^2 b+a b^2-3 b^3+6 a^2 c-2 a b c+3 b^2 c+a c^2+3 b c^2-3 c^3,(a-c) (a-2 b-c) (a-b+c),(a-b) (a-b-2 c) (a+b-c)}.

>The NPC of A'B'C' passes through the Feuerbach point Fe.

>Which point is its center?

2 a^6 b-a^5 b^2-5 a^4 b^3+2 a^3 b^4+4 a^2 b^5-a b^6-b^7+2 a^6 c+6 a^5 b c-5 a^4 b^2 c-7 a^3 b^3 c+2 a^2 b^4 c+a b^5 c+b^6 c-a^5 c^2-5 a^4 b c^2-10 a^3 b^2 c^2-6 a^2 b^3 c^2+a b^4 c^2+3 b^5 c^2-5 a^4 c^3-7 a^3 b c^3-6 a^2 b^2 c^3-2 a b^3 c^3-3 b^4 c^3+2 a^3 c^4+2 a^2 b c^4+a b^2 c^4-3 b^3 c^4+4 a^2 c^5+a b c^5+3 b^2 c^5-a c^6+b c^6-c^7::

on lines {{1, 30}, {442, 3833}, {758, 12267},...}.
(OI^2 - 8 R^2)  X[1]+ OI^2 X[79].

Best regards,
Peter Moses.

 

 

HYACINTHOS 26344

[Le Viet An]:

 

Let ABC be a triangle and P a point.

Denote:

Ab, Ac = the orthogonal projections of A on PB, PC, resp.
Similarly Bc, Ba and Ca, Cb.

La, Lb, Lc = the Euler lines of AAbAc, BBcBa, CCaCb, resp.

A'B'C' = the triangle bounded by La, Lb, Lc.

For P = G, the centroid of A'B'C' lies on the NPC of ABC.

Which point is it?

 

[César Lozada]:

 

Q = reflection of X(11569) in X(5)

= (SB-SC)^2*(3*S^2-9*SB*SC-2*SW^ 2)*(3*S^2-(3*SA-2*SW)*(3*SA+2* SW)) : : (barycentrics)

= on the nine-points circle and these lines: {4,11568}, {5,11569}, {114,6032}, {126,9771}, {543,13234}, {2793,12494}, {3849,9127}, {6094,13377}

= midpoint of X(i) and X(j) for these {i,j}: {4,11568}, {6094,13377}

= reflection of X(11569) in X(5)

= antipode of X(11569) in the nine-points circle

= [ -0.786976228359526, 2.06720654330792, 2.572741288090982 ]

 

César Lozada

HYACINTHOS 26332

[Antreas P. Hatzipolakis]:

Let ABC be a triangle.

Denote:

(Ia), (Ib), (Ic) = the A-, B-,C- excircle, resp.

(Na), (Nb), (Nc) = the NPCs of IBC, ICA, IAB, resp.

Ra = the radical axis of (Ia) and (Na)
Rb = the radical axis of (Ib) and (Nb)
Rc = the radical axis of (Ic) and (Nc)

A*B*C* = the triangle bounded by Ra, Rb, Rc

1. The NPC of A*B*C* passes through the Feuerbach point.

Center of the circle?

2. ABC, A*B*C* are orthologic.

Orthologic centers?



[Angel Montesdeoca]:

*** 1. Center of the NPC of A*B*C* :

W = ( a^8 (b-c)^2
         -2 a^7 (b^3+5 b^2 c+5 b c^2+c^3)
          -a^6 (2 b^4+11 b^3 c+30 b^2 c^2+11 b c^3+2 c^4)
          +a^5 (6 b^5+45 b^4 c+41 b^3 c^2+41 b^2 c^3+45 b c^4+6 c^5)
           +2 a^4 b c (15 b^4+94 b^3 c-14 b^2 c^2+94 b c^3+15 c^4)
           -6 a^3 (b^7+10 b^6 c-11 b^4 c^3-11 b^3 c^4+10 b c^6+c^7)
           +a^2 (b^2-c^2)^2 (2 b^4-19 b^3 c-158 b^2 c^2-19 b c^3+2 c^4)
           +a (b-c)^4 (b+c)^3 (2 b^2+27 b c+2 c^2)
            -(b-c)^6 (b+c)^4 : .... : ....).
 
  (6 - 9 - 13) - search numbers  of W: (0.777426283641197, 0.0129765810797222, 3.27286856409479)
  
 
 *** 2. ABC, A*B*C* are orthologic.
 
The orthologic center (ABC, A*B*C) is  X(5557)=Garcia-Feuerbach Point GF(1/3),   where GF is the mapping defined at X(5550).     (Emmanuel José Garcia; September 11, 2013).

The orthologic center (A*B*C*, ABC) is:

V =  2(r+10R) X(5) - 5R X(40).

V = (a^5 (b^2-10 b c+c^2)
         -a^4 (b^3+11 b^2 c+11 b c^2+c^3)
        -2 a^3 (b-c)^2 (b^2+6 b c+c^2)
       +2 a^2 (b-c)^2 (b^3+7 b^2 c+7 b c^2+c^3)
       +a (b^2-c^2)^2 (b^2+18 b c+c^2)
        -(b-c)^4 (b+c)^3 : ... : ....).
          
 V = X(5)X(40) /\ X(946)X(12680)
         
  (6 - 9 - 13) - search numbers  of  V: (-3.10967432619234, -3.68825662960864, 7.62930722218683)
         
  Angel Montesdeoca  

HYACINTHOS 26330

[Antreas P. Hatzipolakis]:

General Theorem (*)

Let ABC be a triangle, L a line and P a point on L.

 

Denote:

A1, B1, C1 = the other than P intersections of the circumcircles of PBC, PCA, PAB and L, resp. 

 

The tangents to circumcircles of PBC, PCA, PAB at A1, B1, C1, resp. bound a triangle A'B'C'.

The circumcircles of ABC and A'B'C' are tangent. 

(*)  AoPS

Question: 
If L = the Euler line and P = O, H, N, ...., then which are the touchpoints of the circumcircles of ABC and A'B'C' ?

 

[Peter Moses]:

Hi Antreas,

 

P = O

X(1304).

 

P = H

X(10420).

 

P = N

a^2 (a-b) (a+b) (a-c) (a+c) (a^2+b^2-c^2) (a^2-b^2+c^2) (a^10-4 a^8 b^2+4 a^6 b^4+2 a^4 b^6-5 a^2 b^8+2 b^10-3 a^8 c^2+9 a^6 b^2 c^2-3 a^4 b^4 c^2+2 a^2 b^6 c^2-5 b^8 c^2+2 a^6 c^4-10 a^4 b^2 c^4-3 a^2 b^4 c^4+2 b^6 c^4+2 a^4 c^6+9 a^2 b^2 c^6+4 b^4 c^6-3 a^2 c^8-4 b^2 c^8+c^10) (a^10-3 a^8 b^2+2 a^6 b^4+2 a^4 b^6-3 a^2 b^8+b^10-4 a^8 c^2+9 a^6 b^2 c^2-10 a^4 b^4 c^2+9 a^2 b^6 c^2-4 b^8 c^2+4 a^6 c^4-3 a^4 b^2 c^4-3 a^2 b^4 c^4+4 b^6 c^4+2 a^4 c^6+2 a^2 b^2 c^6+2 b^4 c^6-5 a^2 c^8-5 b^2 c^8+2 c^10):: 

on line {{1141,13619},...}.

isoconjugate of X(656) and X(10096).

barycentric quotient X(112)/X(10096).

 

P = G

X(10098).

 

Best regards,

Peter Moses.

 

 

HYACINTHOS 26323

[Antreas P. Hatzipolakis]:

 

In the McCay's cubic entry K003 in CTC , are listed the following triangle centers lying on the cubic:

X(1), X(3), X(4), X(1075), X(1745), X(3362), E(412)=X(1075)*

The last one is the isogonal conjugate of X(1075) not listed in ETC.

In order to be included in ETC: Which are its properties (lines it is lying on, etc) ?

[César Lozada]:

X(1075)* = Isogonal conjugate of X(1075)

= (S^2-SB*SC)/(16*R^4-8*(SB+SC)* R^2-S^2-2*SA^2+SW^2) : :  (barycentrics)

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

= On McCay cubic(K003) and these lines: {3,1075}, {577,6759}, {1092,2055}

= Isogonal conjugate of X(1075)

= X(4)-crossconjugate-of-X(3)

= [ 1.303070197580847, 1.35587259452337, 2.100566440661658 ]

 

César Lozada

HYCINTHOS 26322

[Antreas P. Hatzipolakis]:

 

 

Let ABC be a triangle and P a point.

Denote:

P* = the Poncelet point of (ABCP)
(ie the point the NPCs of ABC, PBC, PCA, PAB concur at)

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

A1 = the other than P* intersection of P*A* and the NPC of PBC
B1 = the other than P* intersection of P*B* and the NPC of PCA
C1 = the other than P* intersection of P*C* and the NPC of PAB

A2 = the antipode of A1 in the NPC of PBC
B2 = the antipode of B1 in the NPC of PCA
C2 = the antipode of C1 in the NPC of PAB

P*, A2, B2, C2 are concyclic.

Which is the center of the circle for P = 

 

1. I
2. O
3. N

?

[Peter Moses]:

Hi Antreas,

 

Center of the circle for P{p,q,r} is:

p (c^2 (a^2-b^2-c^2) p^2 q^2+c^2 (a^2-b^2+c^2) p q^3-2 (a^4-2 a^2 b^2+b^4-2 a^2 c^2+c^4) p^2 q r-(2 a^4-4 a^2 b^2+2 b^4-3 a^2 c^2-3 b^2 c^2+c^4) p q^2 r+2 a^2 c^2 q^3 r+b^2 (a^2-b^2-c^2) p^2 r^2-(2 a^4-3 a^2 b^2+b^4-4 a^2 c^2-3 b^2 c^2+2 c^4) p q r^2-2 a^2 (a^2-b^2-c^2) q^2 r^2+b^2 (a^2+b^2-c^2) p r^3+2 a^2 b^2 q r^3)::

 

1). X(1).

 

2). X(12038).

 

3). (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) (a^12-6 a^10 b^2+14 a^8 b^4-16 a^6 b^6+9 a^4 b^8-2 a^2 b^10-6 a^10 c^2+18 a^8 b^2 c^2-14 a^6 b^4 c^2-3 a^4 b^6 c^2+6 a^2 b^8 c^2-b^10 c^2+14 a^8 c^4-14 a^6 b^2 c^4-3 a^4 b^4 c^4-4 a^2 b^6 c^4+4 b^8 c^4-16 a^6 c^6-3 a^4 b^2 c^6-4 a^2 b^4 c^6-6 b^6 c^6+9 a^4 c^8+6 a^2 b^2 c^8+4 b^4 c^8-2 a^2 c^10-b^2 c^10):: 

on lines {{2,11016},{5,128},{140,389},{ 195,252},{930,7604},{1487, 1656}}.

 

4). P -> X(n).

{{1,1},{2,12040},{3,12038},{4, 5},{6,12039},{13,5459},{14, 5460},{74,12041},{80,11},{98, 12042},{110,1511},...}.

 

Best regards,

Peter Moses.

 

HYACINTHOS 26317

[Antreas P. Hatzipolakis]:

 

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

The line B'C' intersects the circumcircle at Ba, Ca
The line C'A' intersects the circumcircle at Cb, Ab
The line A'B' intersects the circumcircle at Ac, Bc

Denote:

A1, B1, C1 = the midpoints of AA', BB', CC', resp.
A2, B2, C2 = the midpoints of BaCa, CbAb, AcBc, resp.

A*, B*, C* = the reflections of A2, B2, C2 in B1C1, C1A1, A1B1, resp.

Which is the locus of P such that ABC, A*B*C* are:

1. perspective

2. orthologic
?

Note: The A*B*C* is inscribed in the NPC. See

APH

 

 

[César Lozada]:

 

Algebraically simpler:

 

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

The line B'C' intersects the circumcircle at Ba, Ca
The line C'A' intersects the circumcircle at Cb, Ab
The line A'B' intersects the circumcircle at Ac, Bc
These intersections lead generally to 2nd degree equations with squares roots.

Denote:

A1, B1, C1 = the midpoints of AA', BB', CC', resp.
A2, B2, C2 = the midpoints of BaCa, CbAb, AcBc, resp.
A2B2C2 = the pedal triangle of O w/r to A’B’C’
A*, B*, C* = the reflections of A2, B2, C2 in B1C1, C1A1, A1B1, resp.

1)      Perspective:

Locus = {Gibert’s Q066=Stammler quartic through ETC’s 1, 2, 4, 254, 1113, 1114, 1138, 2184, 3223, 3346, 3459, 8049, 9510, 13483, 13484, 13574, 13575)}

 

ETC-pairs (P, Z1(P)=perspector): (1,12), (2,2), (4,4), (254,68), (1113,1313), (1114,1312), (1138,5627), (3346,6526), (3459,252), (13574,10415)

 

Z1( X(2184) ) = (b+c)^2*(a-b+c)^2*(a+b-c)^2//( a^3+(b+c)*a^2-(b+c)^2*a-(b^2- c^2)*(b-c)) : : (barycentrics)

= On lines: {2,7367}, {4,6611}, {11,1435}, {84,5715}, {225,1427}, {226,1439}, {1422,2006}, {1436,7490}, {1440,6612}, {2184,5514}, {3772,7129}, {6356,6358}

= {X(226), X(8808)}-Harmonic conjugate of X(1903)

= [ -0.214603731338937, -0.32118572451519, 3.962071705651325 ]

 

Z1( X(13575) ) = (4)X(251) ∩ X(6)X(66)

= (S^2-2*SA*SC+SB^2)*(S^2-2*SA* SB+SC^2)*SB*SC : : (barycentrics)

= On the cubic K701 and these lines: {2,1235}, {4,251}, {6,66}, {22,5523}, {25,2353}, {111,1289}, {112,7391}, {127,13575}, {232,2165}, {468,8770}, {1383,6995}, {1400,2156}, {2395,6753}, {2987,6515}, {3172,5064}, {5133,8743}, {7735,8882}

= polar conjugate of X(315)

= [ 0.771108835007703, 1.28820741641366, 2.392932192848292 ]

 

2)      Orthologic:

Locus = {sidelines} \/ {  circum-nonic  q9 through ETC’s 2,4,7, vertices of triangles anticomplementary and cevian-of-X(264)}

q9: CyclicSum[ y*z*((SB*b^2*y-SC*c^2*z)*a^4* y^3*z^3+2*(-b^2*z^2+c^2*y^2)* SA^2*a^2*x^5-(b^2-c^2)*(3*S^2- SB*SC)*a^2*x*y^3*z^3-2*(-(SC* S^2+(2*SA*SB+SA*SC-2*SC^2)*SA) *c^2*y+(SB*S^2+(SA*SB+2*SA*SC- 2*SB^2)*SA)*b^2*z)*x^4*y*z) ] = 0 (barys)

 

ETC triads: (2,4,3), (4,3,4), ( 7,1,5)

 

César Lozada