Given A=(1,0), B=(0,0), and C=(0,1)
find all unit length equilateral triangles XYZ such that the lengths AX, BY, and CZ are all one unit.

let X,Y,Z be (x1,x2) (y1,y2) (z1,z2) respectively
then this boils down to solving these 6 equations
(x1-1)^2+x2^2=1
y1^2+y2^2=1
z1^2+(z2-1)^2=1
(x1-y1)^2+(x2-y2)^2=1
(y1-z1)^2+(y2-z2)^2=1
(x1-z1)^2+(x2-z2)^2=1

Tried for an analytical solution, still working it out, but for now I have Mathematica's solutions which it found these 12 (in order x,y,z)
1 (1/2,sqrt(3)/2) (1,0) (0,0)
2 (1/2,sqrt(3)/2) (-1/2,sqrt(3)/2) (0,0)
3 (1/2,-sqrt(3)/2) (1,0) (0,0)
4 (1/2,-sqrt(3)/2) (-1/2,sqrt(3)/2) (0,0)
5 (0,0) (0,1) (sqrt(3),1/2)
6 (0,0) (0,1) (-sqrt(3),1/2)
7 (0,0) (sqrt(3)/2,-1/2) (sqrt(3)/2,1/2)
8 (0,0) (-sqrt(3)/2,-1/2) (-sqrt(3)/2,1/2)
9 (1,1) (1,0) (1/2,(2-sqrt(3))/2)
10 (1,1) (1,0) (1/2,(2+sqrt(3))/2)
11 ((2-sqrt(3))/2,1/2) (1,0) (1,1)
12 ((2+sqrt(3))/2,1/2) (1,0) (1,1)

Posted by Daniel on 2009-09-09 14:58:36