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### Equilateral Triangle Vertices on Lattice Points

```Date: 11/26/2002 at 19:20:57
From: Carol Awu
Subject: Equilateral Triangle Vertices on Lattice Points

Can lattice points be the vertices of an equilateral triangle? And can
you explain how and why?

So far, I know that the triangle has to be tilted because if it were
flat on an x,y grid, it wouldn't work because:

2a = each side of the triangle
a = half of the side of triangle, also the base of the triangle
is split into 2 right triangles
2a = hypotenuse
h = height

(2a)^2 = a^2+h^2
4a^2 = a^2+h^2
3a^2 = h^2
3 = (a^2)/(h^2)
Square root of 3 = a/h

But that's not possible because the square root of 3 is irrational,
while a/h is a fraction, therefore rational.

So all I know for now is that the triangle has to be slanted on the
grid. However, I'm not sure how to do it. In addition, I've drawn out
3 congruent circles forming a triangle with their radius. I'm not sure
how that helps...
```

```
Date: 11/27/2002 at 03:40:55
From: Doctor Floor
Subject: Re: Equilateral Triangle Vertices on Lattice Points

Hi, Carol,

We may assume that A(0,0) is one of the lattice points, and B(2t,2u)
is one of the others. Then the midpoint of AB is M(t,u). Now from M
to the third vertex C we have to go along a line perpendicular to AB
through a distance of sqrt(3)*sqrt(t^2+u^2).

Since the slope of AB is u/t, and the product of slopes of
perpenicular lines must be -1, we know that the slope of MC must be
-t/u. Or stated in another way: we have to add to the coordinates of
M a multiple of (-u,t).

Since the distance from M to C has to be sqrt(3)*sqrt(t^2+u^2), we
see that in fact we have to add +/-sqrt(3)*(-u,t). And now we see
that the coordinates of C cannot be rational, and thus C cannot be a
lattice point.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 11/27/2002 at 13:07:11
From: Carol Awu
Subject: Equilateral Triangle Vertices on Lattice Points

Hi Doctor,

I got confused in the last part of what you said, "Since the distance
from M to C has to be sqrt(3)*sqrt(t^2+u^2), we see that in fact we

Where did the sqrt(3)*sqrt(t^2+u^2) as distance of MC come from? And
at the next thing you said I am more than lost.
```

```
Date: 11/27/2002 at 14:53:01
From: Doctor Floor
Subject: Re: Equilateral Triangle Vertices on Lattice Points

Hi, Carol,

The sqrt(3)*sqrt(t^2+u^2) was mentioned earlier in the first
paragraph. You can see by the Pythagorean theorem that the length of
segment AM is equal to sqrt(t^2+u^2), and MC (the altitude) has to be
sqrt(3) times as long as AM.

Now if we calculate the distance betweeen M and the point Q, which is
M plus (-u,t), then this distance is sqrt(t^2+u^2), the same as AM. We
had derived that instead of sqrt(t^2+u^2) the distance from M to C
had to be sqrt(3) times as much. This means that we also have to add
sqrt(3) times as much.

Since B was a lattice point, we see that t and u are rational numbers.
But the coordinates of C have become

(t -/+ sqrt(3)*u, u +/- sqrt(3)*t)

and these coordinates cannot be rational, since sqrt(3) is irrational.

For the reasoning to achieve the irrationality of these coordinates,
see for instance, from the Dr. Math library:

Prove that 7 + 17*sqrt(17) is Irrational
http://mathforum.org/library/drmath/view/61250.html

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 11/29/2002 at 18:30:51
From: Carol Awu
Subject: Equilateral Triangle Vertices on Lattice Points

Hi,

Thanks for your reply, but where did point Q come from? That's where I
got lost again. And the whole (-u,t) thing.
```

```
Date: 11/30/2002 at 08:07:26
From: Doctor Floor
Subject: Re: Equilateral Triangle Vertices on Lattice Points

Hi, Carol,

1. Suppose we have a point T(x,y) and a point U, which is T + (-u,t),
thus U(x-u,y+t). Then the slope of TU is -t/u, and thus line TU is
perpendicular to AB.

If instead of (-u,t) we add 2*(-u,t) = (-2u,2t) to T, then we get a
point U'(x-2u,y+2t), and still the slope of TU' is -t/u.

We may take other multiples of (-u,t) as well. That is why I say that
we have to add a multiple of (-u,t).

2. From M we had to go through the right distance in the direction
indicated by adding multiples of (-u,t). I just defined a point Q as M
+ (-u,t), thus this point Q has coordinates (t-u,t+u). I then
calculated the distance MQ to see how much distance more or less I had
to travel from M to get in the point C that would make ABC
equilateral.

I hope this helps.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 11/30/2002 at 20:14:13
From: Carol Awu
Subject: Thank you (Equilateral Triangle Vertices on Lattice Points)

Hi,

Thank you SO much for the reply! You have no idea how
happy you've made me. Thanks again.
```
Associated Topics:
High School Coordinate Plane Geometry

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