Δευτέρα 11 Νοεμβρίου 2024

ETC

X(66180) = CENTER OF THE 2nd HATZIPOLAKIS-VAN TIENHOVEN EQUILATERAL TRIANGLE

Trilinears    sqrt(3)*cos(A) - 2*(cos((A-2*Pi)/3) + 2*cos((B-2*Pi)/3)*cos((C - 2*Pi)/3))*(cos(2*A/3 - 7*Pi/6) + cos(2*B/3 - 7*Pi/6) + cos(2*C/3 - 7*Pi/6)) : :

See César Lozada, euclid 7201.

X(66180) lies on these lines: {3, 3276}, {4, 65155}

X(66180) = reflection of X(3276) in X(41109)
X(66180) = Cundy-Parry-Psi-transform of the anticomplement of X(41111)


X(66181) = CENTER OF THE 3rd HATZIPOLAKIS-VAN TIENHOVEN EQUILATERAL TRIANGLE

Trilinears    sqrt(3)*cos(A) - 2*(cos((A-4*Pi)/3) + 2*cos((B-4*Pi)/3)*cos((C - 4*Pi)/3))*(cos(2*A/3 - 15*Pi/6) + cos(2*B/3 - 15*Pi/6) + cos(2*C/3 - 15*Pi/6)) : :

See César Lozada, euclid 7201.

X(66181) lies on this line: {3, 3277}

X(66181) = reflection of X(3277) in X(41110)
X(66181) = Cundy-Parry-Psi-transform of the anticomplement of X(41109)


X(66182) = HOMOTHETIC CENTER OF THESE TRIANGLES: 2nd MORLEY AND 2nd HATZIPOLAKIS-VAN TIENHOVEN EQUILATERAL TRIANGLE

Trilinears    cos(A)+(cos((A-2*Pi)/3)+2*cos((B-2*Pi)/3)*cos((C-2*Pi)/3))*(1-2*cos((2*A-3*Pi)/3)-2*cos((2*B-3*Pi)/3)-2*cos((2*C-3*Pi)/3)) : :

See César Lozada, euclid 7201.

X(66182) lies on this line: {3, 3276}


X(66183) = HOMOTHETIC CENTER OF THESE TRIANGLES: 3rd MORLEY AND 3rd HATZIPOLAKIS-VAN TIENHOVEN EQUILATERAL TRIANGLE

Trilinears    cos(A)+(cos((A-4*Pi)/3)+2*cos((B-4*Pi)/3)*cos((C-4*Pi)/3))*(1-2*cos((2*A-7*Pi)/3)-2*cos((2*B-7*Pi)/3)-2*cos((2*C-7*Pi)/3)) : :

See César Lozada, euclid 7201.

X(66183) lies on this line: {3, 3277}


Δεν υπάρχουν σχόλια:

Δημοσίευση σχολίου

REGULAR POLYGONS AND EULER LINES

Let A1A2A3 be an equilateral triangle and Pa point. Denote: 1, 2, 3 = the Euler lines of PA1A2,PA2A3, PA3A1, resp. 1,2,3 are concurrent. ...