# LAttice Points

1. Feb 1, 2005

Hello all

In a ordinary syatem of rectangular coordinates, the points for which both coordinates are integers are called lattice points . Prove that a triangle whose vertices are lattice points cannot be equilateral. Ok so I know that in a equilateral triangle the angle measures are $$\frac{\pi}{3}$$.Assuming that we do have an equilateral triangle then we know that $$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$ which is irrational. Hence we cannot have lattice points.

Is this sufficient enough to qualify as a proof?

Thanks

Last edited: Feb 1, 2005
2. Feb 1, 2005

### hypermorphism

Nope. The isosceles right triangle formed by the points (0,0), (1,0), (0,1) has a hypotenuse of length $$\sqrt{2}$$.

3. Feb 1, 2005

but thats a isoceles right triangle. I am talking about only an equilateral triangle. The definition of an equilateral triangle is that: all angles and all sides are equal

Thanks

4. Feb 1, 2005

### hypermorphism

Your proof rested on the fact that the sine of one of the angles is irrational. I showed that this does not prove your statement since sines of angles of triangles exist that are irrational in the lattice, not that an equilateral triangle exists. In other words, you need to find another approach or further your statement into one about the sides of the triangle. :)

Last edited: Feb 1, 2005
5. Feb 1, 2005

### Hurkyl

Staff Emeritus
No. You've arrived at a key step, but you haven't finished the proof. You haven't even talked about the vertices of the triangle yet!

6. Feb 1, 2005

but you are only trying to prove it for only equilateral triangles .

7. Feb 1, 2005

### hypermorphism

Let's try a direct approach. What are you implying by the fact that the sine of pi/3 is irrational ?

8. Feb 1, 2005

### Hurkyl

Staff Emeritus
Nothing in your argument distinguishes between equalateral, isosceles, or even scalene triangles. If it was valid, you would have disproven the existence of the triangle hypermorphism demonstrated. :tongue2:

9. Feb 1, 2005

ok so because $$\frac{\sqrt{3}}{2}$$ is irrational, this implies that the distance from the vertex to the midpoint of the base is irrational. But this should be rational because of the laattice points?

Thanks

10. Feb 1, 2005

### Hurkyl

Staff Emeritus
Nope. Consider hypermorphism's example!

11. Feb 1, 2005

I am not sure if this is correct but:

If we have an equilateral triangle of sides of length $$a$$ then the altitude is $$\sqrt{a^2 - (\frac{a}{2})^2} = a\sqrt{\frac{3}{2}}$$. So the area is $$a^2\sqrt{\frac{3}{2}}$$ which is irrational. This is irrational area, but any triangle with lattice points has a rational area.

Is this correct?

PS: How would I use the irrationality of $$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$ to prove this because my text (Courant) suggests it?

Thanks

Last edited: Feb 1, 2005
12. Feb 1, 2005

I am not sure whether to assume that the area of a triangle with lattice points is rational.

13. Feb 1, 2005

### learningphysics

No you can't. It's generally not true.

Try a proof by contradiction. Assume the triangle is equilateral with lattice points.

A couple of hints:

1) Translate the triangle so that one point is at the origin. If the original triangle had lattice points, then so does the translated triangle.

2) Write the coordinates of the translated triangle in polar form.

Last edited: Feb 2, 2005
14. Feb 2, 2005

### Hurkyl

Staff Emeritus
Actually, I think this one's right -- I think the area of a polygon whose vertices lie on a lattice is a rational function of the number of points lying on the polygon, and the number lying inside the polygon.

In any case, you could use vector geometry to prove the area of a triangle with vertices on the lattice must be rational. (which is how I would have done the problem)

And no, you should not assume it's rational, you have to prove it, or reference a theorem. :tongue2:

15. Feb 2, 2005

### learningphysics

Oops! Yes, you're right. Sorry about that!

16. Feb 2, 2005

### Curious3141

You'll find Pick's theorem intriguing.

17. Feb 2, 2005

So I take it that If I use Pick's Theorem, then I have a formal proof? Also I would I convert the cartesian coordinates into polar coordinates?

Thanks

18. Feb 2, 2005

### learningphysics

If you're allowed to use Pick's theorem, then you should be fine. Like Hurkyl
said you can also use vector geometry to prove the area must be rational.

Don't worry about the polar coordinates. I was referring to a different way
to solve the problem, not related to the area of the triangle. I didn't realize that the area had to be rational.

I think the area method might be easier. But if you're interested I've described my idea below.

Translate to the origin, then write the cartesian coordinates of the triangle vertices like this:
$$(0,0)$$
$$(Acos\alpha, Asin\alpha)$$
$$(Acos(\alpha+60),Asin(\alpha+60))$$

A is length of the triangle.

If $$Acos\alpha$$ and $$Asin\alpha$$ are integers then using the cosine sum formula you can prove that $$Acos(\alpha+60)$$ is not an integer (proving the third point is not a lattice point leading to a contradiction). Or you can prove that $$Asin(\alpha+60)$$ (the y coordinate instead) is not an integer using the sine sum formula.

19. Feb 2, 2005

### Hurkyl

Staff Emeritus
Pick, that's it! I couldn't remember its name.

20. Feb 2, 2005

### Curious3141

With Pick's it's a beautiful and elegant proof.

Let one side of the triangle be $l$. Now $l^2$ has to be integral because all the vertices lie on lattice points.

The area of the triangle is given by

$$\frac{1}{2}l^2\sin{\frac{\pi}{3}} = \frac{1}{2}l^2\frac{\sqrt{3}}{2}$$

which is irrational. But the area is rational by Pick's theorem, and we have a contradiction. Hence no lattice point triangle can be equilateral.