HYACINTHOS 29009

[Vladimir Dubrovsky]:

 

The question below, posed by Konstantin Knop, appeared in one Russian facebook group. 

Consider regular triangles with vertices on three straight lines a, b, and c. It is not difficult to see that their centers form two parallel lines (each corresponding to one or the two possible orientations of the regular triangles). Both lines turn out to be perpendicular to the Euler line of the triangle formed by a, b, c, if it exists. 

Are there any ETC centers located on these two lines?  

 

 [César Lozada]:

 

Please allow me to re-write your configuration:

 

·         Let ABC be a triangle with Euler line (E) and let A’B’C’ any equilateral triangle inscribed in ABC and having its same orientation. Then the centers of all A’B’C’ lie on a line (L₁)⊥(E).

·         If A’B’C’ and ABC have opposite orientation then the centers of A’B’C’ lie on a line (L₂)⊥(E) (Therefore  (L₂)|| (L₁)).

 

(L₁) Is the trilinear polar of X(14) through ETCs 396, 523, 3272, 5637, 6137, 9194, 9201, 11549, 18776, 20579, 23284.

 

(L₂) Is the trilinear polar of X(13) through ETCs 395, 523, 6138, 9195, 9200, 11537, 18777, 20578, 23283.

 

If Q1 = (L₁)∩(E) and Q2 = (L₂)∩(E) then:

 

Q1 = EULER LINE INTERCEPT OF TRILINEAR POLAR OF X(14)

= -2*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*sqrt(3)*S+3*(a^2+b^2-c^2)*(2*a^2-b^2-c^2)*(a^2+c^2-b^2) : : (barys)

= 9*(4*R^2-SW)*S^2+3*SB*SC*SW+S*sqrt(3)*(S^2-3*SB*SC) : : (barys)

= As a point on the Euler line, this center has Shinagawa coefficients (-3*F+sqrt(3)*S/3, E+F-sqrt(3)*S)

= lies on these lines: {2, 3}, {13, 11657}, {14, 3166}, {15, 16319}, {110, 11092}, {396, 523}, {398, 8015}, {476, 8838}, {530, 32225}, {531, 5642}, {533, 3292}, {627, 15794}, {635, 15785}, {1525, 2777}, {2453, 16644}, {3163, 23713}, {3284, 23715}, {3642, 5651}, {5318, 30468}, {6105, 6107}, {8836, 20253}, {14704, 22510}, {18487, 23712}

= midpoint of X(i) and X(j) for these {i,j}: {110, 11092}, {16179, 16181}

= reflection of Q2 in X(468)

= isogonal conjugate of the antigonal conjugate of X(2993)

= circumcircle-inverse-of X(3130)

= inner-Napoleon circle-inverse-of X(4)

= orthoptic circle of Steiner inellipse-inverse-of X(383)

= polar circle-inverse-of X(471)

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (3, 4, 23722), (1113, 1114, 3130)

= [ 2.0230791373652770, 1.1482222806212000, 1.9120125319241900 ]

 

Q2 = EULER LINE INTERCEPT OF TRILINEAR POLAR OF X(13)

= 2*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)*sqrt(3)*S+3*(a^2+b^2-c^2)*(2*a^2-b^2-c^2)*(a^2+c^2-b^2) : : (barys)

= 9*(4*R^2-SW)*S^2+3*SB*SC*SW-S*sqrt(3)*(S^2-3*SB*SC) : : (barys)

= As a point on the Euler line, this center has Shinagawa coefficients (-3*F-sqrt(3)*S/3, E+F+sqrt(3)*S)

= lies on these lines: {2, 3}, {13, 3165}, {14, 11657}, {16, 16319}, {62, 8919}, {110, 11078}, {395, 523}, {397, 8014}, {476, 8836}, {530, 5642}, {531, 32225}, {532, 3292}, {628, 15793}, {636, 15784}, {1524, 2777}, {2453, 16645}, {3163, 23712}, {3284, 23714}, {3643, 5651}, {5321, 30465}, {6104, 6106}, {8838, 20252}, {14705, 22511}, {18487, 23713}

= midpoint of X(i) and X(j) for these {i,j}: {110, 11078}, {16180, 16182}

= reflection of Q1 in X(468)

= isogonal conjugate of the antigonal conjugate of X(2992)

= circumcircle-inverse-of X(3129)

= outer-Napoleon circle-inverse-of X(4)

= orthoptic circle of Steiner inellipse-inverse-of X(1080)

= polar circle-inverse-of X(470)

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (3, 4, 23721), (1113, 1114, 3129)

= [ 0.3941907924299769, -0.4763690157312665, 3.7885242039844980 ]

 

César Lozada