HYACINTHOS 27185

 

[Antreas P. Hatzipolakis]:

 

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

Denote:

A", B", C" = the reflections of A', B', C' in L, resp.

A*,B*, C* = the reflections of A, B, C in PA", PB", PC", resp.

P lies on the Euler line of A*B*C*.

Which is the point P wrt triangle A*B*C*?


[César Lozada]:

Suppose OP/OH=t and let O*, H* be the circumcenter, orthocenter of A*B*C*. Then 

O*P/O*H* = R^2/((9*R^2-2*SW)*t)

 

A faster way to find P w/r to A*B*C*: it is equivalently the ABC-circumcircle-inverse of P. 

Examples: 

- the circumcircle-inverse of X(2) is X(23), then P=X(2)-of-ABC = X(23)-of-A*B*C*

- the circumcircle-inverse of X(4) is X(186), then  P=X(4)-of-ABC = X(186)-of-A*B*C*

 

ETC pairs (P,circumcircle-inverse-of-P):  (2,23), (4,186), (5,2070), (20,2071), (21,1325), (22,858), (23,2), (24,403), (25,468), (26,2072), (27,2073), (28,2074), (29,2075), (30,3), (140,5899), (186,4), (237,1316), (376,7464), (378,10295), (381,7575), (382,15646), (403,24), (468,25), (858,22), (859,3109), (1113,1113), (1114,1114), (1316,237), (1325,21), (1995,7426), (2070,5), (2071,20), (2072,26), (2073,27), (2074,28), (2075,29), (3109,859), (3148,5112), (3153,7488), (3515,10151), (3520,13619), (3658,7477), (4184,5196), (4225,7424), (4226,7468), (4228,7469), (4230,7473), (4234,7481), (4235,7482), (4236,7475), (4238,7476), (4240,7480), (4243,7479), (5004,5005), (5005,5004), (5055,12105), (5112,3148), (5159,9909), (5189,6636), (5196,4184), (5899,140), (5999,15915), (6636,5189), (6644,11799), (6660,11007), (7387,10257), (7419,7478), (7424,4225), (7426,1995), (7464,376), (7468,4226), (7469,4228), (7471,15329), (7472,11634), (7473,4230), (7475,4236), (7476,4238), (7477,3658), (7478,7419), (7479,4243), (7480,4240), (7481,4234), (7482,4235), (7488,3153), (7492,10989), (7502,7574), (7574,7502), (7575,381), (9909,5159), (10151,3515), (10257,7387), (10295,378), (10296,10298), (10297,14070), (10298,10296), (10989,7492), (11007,6660), (11634,7472), (11799,6644), (12083,15122), (12105,5055), (13473,15750), (13619,3520), (14070,10297), (15122,12083), (15329,7471), (15646,382), (15750,13473), (15915,5999)

 

More:

Triangles ABC and A*B*C* are directly similar and their center of similitude lies on the circumcircle of ABC. The ratio A*B*C*/ABC = (OH/R)*|t|. The center os similitude is:

 

Zs(t) = (SB+SC)*(SA-SB)*(SA-SC)*(OH^ 2*t+3*R^2+SB-SW)*(OH^2*t+3*R^ 2+SC-SW) : : (barys)

 

ETC pairs (P, Zs): (2,9060), (3,476), (4,1304), (20,10420), (23,1302), (186,107), (376,691), (378,935), (403,1301), (468,9064), (550,1291), (1113,1113), (1114,1114), (2071,925), (3651,1290), (7414,2766), (7417,10102), (7418,2770), (7421,2689), (7422,842), (7425,2752), (7429,2687), (7430,2690), (7440,2688), (7444,2758), (7454,2695), (7464,99), (10295,112), (13619,933), (14127,12030)

 

Some others:
Zs (N) = X(5)X(477) ∩ X(74)X(2070)

= (SB+SC)*(SA-SB)*(SA-SC)*(15* R^2+2*SB-4*SW)*(15*R^2+2*SC-4* SW) : : (barys)

= on the circumcircle and on these lines: {5, 477}, {74, 2070}, {842, 13595}, {930, 7471}, {933, 7480}, {1141, 10096}, {1291, 15329}, {1294, 3153}, {2693, 7488}, {7426, 9076}, {14670, 14979}

= trilinear pole of the line {6, 11559}

= [ 0.3850728401490158, -0.4734119563332962, 3.7906852946848090 ]

 

Zs(X(22)) = X(20)X(841) ∩ X(22)X(477)

= a*(a^10-(3*b^2-2*c^2)*a^8+( 2*b^4-3*b^2*c^2-3*c^4)*a^6+(2* b^6-3*c^6-b^2*c^2*(3*b^2-8*c^ 2))*a^4-(b^2-c^2)^3*(3*b^2+2* c^2)*a^2+(b^4-c^4)*(b^2-c^2)^ 3)*b*c*(a^10+(2*b^2-3*c^2)*a^ 8-(3*b^4+3*b^2*c^2-2*c^4)*a^6- (3*b^6-2*c^6-b^2*c^2*(8*b^2-3* c^2))*a^4+(b^2-c^2)^3*(2*b^2+ 3*c^2)*a^2+(b^4-c^4)*(b^2-c^2) ^3)*(a^2-c^2)*(a^2-b^2) : : (barys)

on the circumcircle and on these lines: {20, 841}, {22, 477}, {23, 1300}, {74, 858}, {112, 7471}, {468, 1299}, {842, 7493}, {915, 7469}, {1289, 7480}, {1370, 2693}, {2373, 3260}, {3563, 7426}, {3658, 10100}, {4226, 10098}, {4240, 10423}

= trilinear pole of the line {6, 7706}

= [ 0.7213001480848624, -0.7689936025441857, 3.8401369076296470 ]

 

César Lozada