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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,

Thanks for your question.

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 
have to add +/-sqrt(3)*(-u,t)."

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,

Thanks for your reaction.

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,

Thanks for your question.

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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