Let ABC be a triangle and BCA', CAB', ABC' equilateral triangles erected out/in - wardly of ABC.
Denote:
Bc, Cb, Ca, Ac, Ab, Ba = the circumcenters of CC'B, BB'C, AA'C, CC'A, BB'A, AA'B, resp.
Oa, Ob, Oc = the circumcenters of BcCbA, CaAcB, AbBaC, resp.
1. OaObOc is equilateral.
Which is its center?
2. The circumcircle of OaObOc touches the circumcircle of ABC.
Which is the touchpoint?
3. The Euler lines of BcCbA, CaAcB, AbBaC are concurrent.
Point of concurrence?
[César Lozada]:
· For BCA’, CAB’ and ABC’ erected outwardly ABC:
Triangle OaObOc is equilateral and has sides-length:
ObOc = 4*S^2*R*|(SW+sqrt(3)*S)/((sqrt (3)*SA+S)*(sqrt(3)*SB+S)*(sqrt (3)*SC+S))|
Oa has coordinates:
Oa = (a^2+b^2+c^2+2*sqrt(3)*S)*a : -2*(b^2-c^2)*b : 2*(b^2-c^2)*c (trilinears)
1)
The center of OaObOc is:
Zo = X(3)X(74) ∩ X(15)X(23)
= (3*a^4-2*(b^2+c^2)*a^2-b^4-c^4 -2*sqrt(3)*(b^2+c^2-a^2)*S)*a : : (trilinears)
= (5*SW-18*R^2+sqrt(3)*S)*X(3)+( 9*R^2-2*SW)*X(74)
= On lines: {2, 9749}, {3, 74}, {4, 8838}, {15, 23}, {16, 11003}, {61, 3457}, {184, 9736}, {511, 11127}, {1495, 13350}, {3130, 5640}, {3132, 5012}, {3170, 5237}, {5463, 9143}, {5615, 11422}, {6770, 11078}, {7712, 10645}, {11142, 11624}
= {X(3), X(110)}-Harmonic conjugate of X(11131)
= [ 83.659926140168850, 94.82327257766553, -100.618490136554300 ]
2) X(110)
3)
E = (3*a^4-3*(b^2+c^2)*a^2+2*(b^2- c^2)^2-2*sqrt(3)*S*a^2)*(a^2-b ^2)*(a^2-c^2) : : (barycentrics)
= (SW+sqrt(3)*S)*X(99)-(9*R^2-2* SW)*X(110)
= On lines: {2, 9735}, {99, 110}, {476, 9202}, {1316, 11130}, {6321, 8838}
= [ 36.561544488050230, -9.57902997708241, -6.602258374597138 ]
· For BCA’, CAB’ and ABC’ erected inwardly ABC:
Triangle OaObOc is equilateral and has sides-length:
ObOc = 4*S^2*R*|(SW-sqrt(3)*S)/((sqrt (3)*SA-S)*(sqrt(3)*SB-S)*(sqrt (3)*SC-S))|
Oa has coordinates:
Oa = (a^2+b^2+c^2-2*sqrt(3)*S)*a : -2*(b^2-c^2)*b : 2*(b^2-c^2)*c (trilinears)
1)
The center of OaObOc is:
Zo = X(3)X(74) ∩ X(16)X(23)
= (3*a^4-2*(b^2+c^2)*a^2-b^4-c^4 +2*sqrt(3)*(b^2+c^2-a^2)*S)*a : : (trilinears)
= (-18*R^2+sqrt(3)*S+5*SW)*X(3)+ (9*R^2-2*SW)*X(74)
= On lines: {2, 9750}, {3, 74}, {4, 8836}, {15, 11003}, {16, 23}, {62, 3458}, {184, 9735}, {511, 11126}, {1495, 13349}, {3129, 5640}, {3131, 5012}, {3171, 5238}, {5464, 9143}, {5611, 11422}, {6773, 11092}, {7712, 10646}, {11141, 11626}
= {X(3), X(110)}-Harmonic conjugate of X(11130)
= [ -1.515109111358711, -3.70317144141322, 6.903679685051399 ]
2) X(110)
3)
E = (3*a^4-3*(b^2+c^2)*a^2+2*(b^2- c^2)^2+2*sqrt(3)*S*a^2)*(a^2-b ^2)*(a^2-c^2) : : (barycentrics)
= (SW-sqrt(3)*S)*X(99)-(9*R^2-2* SW)*X(110)
= On lines: {2, 9736}, {99, 110}, {476, 9203}, {1316, 11131}, {6321, 8836}
= [ -0.136248978144573, -0.64667209347805, 4.151244690382059 ]
· Relations between both triangles:
- They are not perspective.
- They are orthologic. Centers (barys):
T1=((15*S^2-3*SA*(36*R^2-9*SA- 2*SW)-SW^2)*S^2+(3*SA-2*SW)*SA *SW^2
-2*sqrt(3)*S*(3*(-SA-2*SW+6*R^ 2)*S^2-(6*SA-5*SW)*SA*SW))*(SB +SC)::
= On line {15, 110}
= [ -2.256709079343813, -6.66073353669564, 9.293499582393888 ]
T2 =((15*S^2-3*SA*(36*R^2-9*SA -2*SW)-SW^2)*S^2+(3*SA-2*SW)* SA*SW^2
+2*sqrt(3)*S*(3*(-SA-2*SW+6*R^ 2)*S^2-(6*SA-5*SW)*SA*SW))*(SB +SC)::
= On line {16, 110}
= [ 162.984692386448900, 73.44022077808081, -122.426269465894100 ]
- They are parallelogic. Centers (barys):
L1 = a^2*((4*a^2-2*b^2-2*c^2) *S+sqrt(3)*((b^2+c^2)*a^2+b^4- 4*b^2*c^2+c^4))* (a^2-b^2)*(a^2-c^2) : :
= On lines: {15, 1337}, {22, 11629}, {110, 9162}, {691, 5467}, {805, 5994} , {10409, 10411}
= [ -0.773509143373610, -0.74560934613080, 4.513859787708909 ]
L2 = a^2*(-(4*a^2-2*b^2-2*c^2)*S+sq rt(3)*((b^2+c^2)*a^2+b^4-4*b^2 *c^2+c^4))* (a^2-b^2)*(a^2-c^2) : :
= On lines: {16, 1338}, {22, 11630}, {110, 9163}, {691, 5467}, {805, 5995} , {10410, 10411}
= [ 4.335159893888770, 116.20632437725030, -78.810710807214460 ]
· Relations with other triangles* :
- Both triangles are perspective to the anti-orthocentroidal triangle
- The outer equilateral triangle is orthologic to these triangles: 1st Ehrmann, inner-Napoleon
- The inner equilateral triangle is orthologic to these triangles: 1st Ehrmann, 1st Morley, 2nd Morley, 3rd Morley, outer-Napoleon, Roussel, Stammler
- The outer equilateral triangle and the inner-Napoleon triangle are parallelogic
- The inner equilateral triangle is parallelogic to these triangles: 1st Ehrmann, 1st Morley, 2nd Morley, 3rd Morley, outer-Napoleon, Roussel, Stammler
*Centers: later.
Note that the lists of orthologic triangles to both triangles do not coincide. It is dued to the fact that some Morleys triangles are not catalogued, for example, the interior trisectors of the external angles of a triangle meet at the vertices of another equilateral triangle.
César Lozada
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