Let ΔA’B’C’ be the
medial triangle of ΔABC.
Let X be the
X(n)-ETC-center-of- ΔABC. Let A” be the X(n)-ETC-center-of-ΔAB’C’, B” be the
X(n)-ETC-center-of-ΔBC’A’ and C” be the X(n)-ETC-center-of-ΔCA’B’.
The triangles ΔA’B’C’
and ΔA”B”C” are cyclologic (*)
(*) i.e., {a’}
= circumcircle(ΔA’B”C”), {b’} = circumcircle(ΔB’C”A”) and {c’} = circumcircle(ΔC’A”B”)
concur at Z’
and
{a”} = circumcircle(ΔA”B’C’), {b”} = circumcircle(ΔB”C’A’) and {c”} = circumcircle(ΔC”A’B’) concur at Z”.
If X= u : v :w (trilinears) then:
1)
Z’
= u*(v*cos(B)-w*cos(C))*(u*(w*sin(B)-v*sin(C))+v*w*sin(B-C)) : :
2)
Z”
= f(a,b,c,u,v,w)*g(a,b,c,u,v,w) : :
where:
f(a,b,c,u,v,w)= (a*b*c*v*w+c*(a^2-c^2)*w*u+b*(a^2-b^2)*u*v+a*(-b^2-c^2+a^2)*u^2)
g(a,b,c,u,v,w)= (-b*c *(v^2+w^2)*u
+a*(b*v+c*w) *v*w +(2*a^2-b^2-c^2)*u*v*w)
3)
Z’
lies on circumcircle of ΔA’B’C’ and
Z” lies on circumcircle of ΔA”B”C”
4)
Z’
and Z” are symmetrical w/r to the radical trace of circumcircles of ΔA’B’C’ and ΔA”B”C”
5)
Z”=midpoint(X,
anticomplement(Z’)) or Z’=complement(reflection(X,Z”)).
Note: anticomplement(Z’) lies on the circumcircle of
ΔABC.
6)
If
{h(T)} is the rectangular circum-hyperbola with center T on the NPC of ΔABC,
then for all P in {h(T)} we have Z’(P) = T, i.e.:
Z’(P)=X(11) for all P on Feuerbach hyperbola
Z’(P)=X(115)
for all P on Kiepert hyperbola
Result (6) above let us deduce
the existence of some rectangular circum-hyperbolas:
Center
|
Trilinear equation
[Through ETC-centers]
|
Perspector's 1st coordinate
Properties
ETC 6-9-13 search
|
X(116)
|
Σ a*(b-c)*(-b^2+a*b-b*c+c*a-c^2)*v*w
[4, 103, 947, 1002, 1126, 1174, 2141, 3681, 3730, 4184]
|
a*(b-c)*(-b^2+a*b-b*c+c*a-c^2)
Complement of X(3261)
-6.381084332890005
|
X(120)
|
Σ (c^3+c*a^2-c*b^2-2*a*b*c+b*a^2+b^3-b*c^2)*v*w
[4, 668, 1292, 4236]
|
X(3290)
|
X(122)
|
Σ b*c*(b^2-c^2)*(3*a^4-(b^2-c^2)^2-2*a^2*(b^2+c^2))*v*w
[4, 20, 253, 1249, 1294, 3346, 3668, 5930]
|
(3*a^4-(b^2-c^2)^2-2*a^2*(b^2+c^2))*(b^2-c^2)/a
Complement of X(3265)
-3.600719569463247
|
X(123)
|
Σ(c-b)*(a^4+2*b*c*a^2-2*a*b*c*(b+c)-(b^2-c^2)^2)*v*w
[4, 961, 998, 1295, 1766, 2995, 3436]
|
(b-c)*(a^4+2*b*c*a^2-2*b*c*(b+c)*a-(b^2-c^2)^2)
(6,2431)∩(230,231)
-3.012216446340097
|
X(124)
|
Σ (b-c)*(-b^3+b*a^2-a*b*c-c^3+c*a^2)*v*w*a
[4, 58, 102, 573, 959, 994, 3417, 3869, 4225]
|
a*(b-c)*(-b^3-a*b*c-c^3+(b+c)*a^2)
(6,652)∩(230,231)
-3.443760727600679
|
X(5514)
|
Σ (c-b)*(a^3+a^2*(b+c)-(b-c)^2*(b+c)-a*(b+c)^2)*v*w
[4, 40, 57, 189, 196, 223, 329, 937, 972, 1817, 2184, 3194, 3345]
|
X(6129)
|
X(5515)
|
Σ b*c*(b-c)*(a^2+(b+c)^2)*v*w
[4, 75, 388, 1010, 1065, 1220, 2345, 4385]
|
(b-c)*(a^2+(b+c)^2)/a
(230,231)∩(513,3700)
|
X(5521)
|
Σ (b-2)*(a^2-b^2+c^2)*(a^2+b^2-c^2)*v*w
[4, 19, 28, 34, 286, 915, 1118, 1119, 5317]
|
(b-c)/(b^2+c^2-a^2)
Isogonal conjugate of X(1332)
-0.772000939862511
|
Some triads of
(N, I, J) such that Z’=X(I) and Z”=X(J) for n=N:
(1, 11, 214),
(2, 115, 2482), (3, 125, 1511), (5, 137, Q5), (6, 125, Q6), (8, 11, 1145), (9,
11, Q9)
where:
Q5= (1+2*cos(2*B)+2*cos(2*C)-2*cos(2*A))*
(2*cos(A)*(cos(2*B-2*C)-2+4*cos(A)^2)+(- 1+4*cos(A)^2)*cos(B-C)) : :
= On lines (2,1263),
(3,2888), (5,930), (30,128), (137,3628) (at least)
= On loci K038, K067
= Complement of
X(1263)
= Inverse of X(5898)
in circumcircle
= Midpoint of:
(5,930) (at least)
= Reflection of
(137/3628) (at least)
=
[7.346237636455233551025, 4.968703721274385706387, -3.189778541954149563316]
Q6=
a*(a^4-b^4-c^4+b^2*c^2)*(2*a^2-b^2-c^2)::
= On lines (2,67),
(3,1177), (5,542), (6,110), (74,5085), (113,1503), (125,3589) (at least)
= On loci: K042,
K043, K565
=
Complementary conjugate of X(858)
= Complement of X(67)
= Midpoint of: (6,110) (at least)
= Reflection of: (125/3589), (141/5972) (at
least)
= [0.9502357345836802528899,
0.2943117751284885726792, 2.998339837010649974545]
Q9 = (2*a^2-a*(b+c)-(b-c)^2)*(a^2-2*a*(b+c)+b^2+b*c+c^2)::
= On lines (1,3939),
(2,3254), (3,2801), (9,100), (10,528), (119,516), (142,3035) (at least)
= Complement of
X(3254)
= Midpoint of:
(9,100) (at least)
= Reflection of:
(142/3035) (at least)
=
[3.086904417102852112698, 0.3728597156591479833209,
1.957805717019037216497]
