Equilateral Triangle Vertices on Lattice PointsDate: 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. |
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