# Thread: Is there an equilateral triangle in the plane all of whose vertices have integer coor

1. ## Is there an equilateral triangle in the plane all of whose vertices have integer coor

Is there an equilateral triangle in the plane all of whose vertices have integer coordinates? If you can, answer the same question for vertices required to have rational coordinates.

Extend this to a tretahedron in the 3-D Cartesian coordinates system.

2. Originally Posted by mathwizard
Is there an equilateral triangle in the plane all of whose vertices have integer coordinates? If you can, answer the same question for vertices required to have rational coordinates.
If you can do it for rational coordinates, then by a dilation (multiplying the coordinates of the vertices by the l.c.m. of their denominators) you can do it for integer coordinates. On the other hand, if it's impossible for rational coordinates then obviously it's impossible with integer coordinates.

Now suppose it can be done with rational coordinates. Then all three sides of the triangle must have rational slope. But if one side has slope t, then (from the formula for tan(θ+120°)) one of the other sides will have slope $\displaystyle \frac{t-\sqrt3}{1+t\sqrt3} = \frac{1+3t^2-(1+t^2)\sqrt3}{1-3t^2},$ which is irrational. Therefore no such triangle exists.

3. Originally Posted by Opalg
If you can do it for rational coordinates, then by a dilation (multiplying the coordinates of the vertices by the l.c.m. of their denominators) you can do it for integer coordinates. On the other hand, if it's impossible for rational coordinates then obviously it's impossible with integer coordinates.

Now suppose it can be done with rational coordinates. Then all three sides of the triangle must have rational slope. But if one side has slope t, then (from the formula for tan(θ+120°)) one of the other sides will have slope $\displaystyle \frac{t-\sqrt3}{1+t\sqrt3} = \frac{1+3t^2-(1+t^2)\sqrt3}{1-3t^2},$ which is irrational. Therefore no such triangle exists.
Exactly. Have you tried to crack the case with the tretahedron?

4. Originally Posted by mathwizard
Have you tried to crack the case with the tretahedron?
The points (1,1,1), (–1,–1,1), (–1,1,–1), (1,–1,–1) are the vertices of a regular tetrahedron. See here.