Can an equilateral triangle have vertices at integral lattice points?
Integral lattice points are such points as (101, 254) or (3453, 12), but not points such as (123.4, 1) or (√2, 5)
If you can't find a solution in the 2D Cartesian plane, can you find one in a 3 (or more) dimensional space?
For 2D the answer is no. Assume true with s the length of an edge. The area of the triangle is s^2*sqrt(3)/4. By Pick's theorem the area is k/2 for some integer k. Therefore,
sqrt(3) = 2*k/s^2
s^2 is an integer for vertices lying on latice points. Hence, we have a contradiction since sqrt(3) is not rational.
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Posted by Jerry
on 2004-07-17 13:49:49 |