ok im stumped. im looking for an equilateral triangle on the cartesian plane such that the coordinates of the vertices are all integers .
what are the vertices?
they are not given, they are for you to find
i guess it can be (0,0) (0,0) (0,0) an equilateral triangle wid side length=0
no thats a trivial case
ok, non trivial ?
hmm, maybe it is imposslbe.
hi
The fact that the altitude of the triangle is going to fall within the midpoint of one of the sides, and the midpoint must be half of the distance from a given set of points, I dont think it is possible to have all 6 points being integers in normal euclidian space... I don't know a good way of articulating this point... lol
, there are 3 points,
yes, but six co-ordinate points.
, right, the midpoint is not a problem, for instance you can have 2 as your base, and half of it is 1
here is what i did , ok i realised that (0,0) (x,0) and (1/2 x , sqrt3 / 2 x ) wont work
Yes, but then your altitude will give that other point at 1/2 sqrt 3 * a which is irrational...
so lets look at rotating this equilateral triangle
right, i agree , so lets rotate it
http://mathworld.wolfram.com/EquilateralTriangle.html the first set of equations here kind of make the point im trying to.
so imagine we have a circle, one vertex will be the center, the other two will land on the circle, and make 60 degrees .
(0,0), ( x1,y1) (x2,y2)
, you know i would hate to find out this is impossible problem,
here Given the distances of a point from the three corners of an equilateral triangle, , , and , the length of a side is given by (16) (Gardner 1977, pp. 56-57 and 63). There are infinitely many solutions for which , , and are integers. In these cases, one of , , , and is divisible by 3, one by 5, one by 7, and one by 8 (Guy 1994, p. 183).
I'm fairly sure that because of the law of sines and the fact that they want this in normal euclidian space that it is unable to be done... but I cant think of a good proof for you.
looks like wolfram says there are infintie solutions
No, that is Napoleon's theory... it deals with creating eq triangles from the sides of any given trigangle..
That is a rather different question.
hmmm, well hero posted it
http://www.ams.org/samplings/feature-column/fcarc-taxi about half way down they have a solution... but it does seem to violate a few of euclids axioms.
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