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Mathematics 22 Online
OpenStudy (anonymous):

ok im stumped. im looking for an equilateral triangle on the cartesian plane such that the coordinates of the vertices are all integers .

OpenStudy (sriram):

what are the vertices?

OpenStudy (anonymous):

they are not given, they are for you to find

OpenStudy (sriram):

i guess it can be (0,0) (0,0) (0,0) an equilateral triangle wid side length=0

OpenStudy (anonymous):

no thats a trivial case

OpenStudy (anonymous):

ok, non trivial ?

OpenStudy (anonymous):

hmm, maybe it is imposslbe.

OpenStudy (anonymous):

hi

OpenStudy (agreene):

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

OpenStudy (anonymous):

, there are 3 points,

OpenStudy (agreene):

yes, but six co-ordinate points.

OpenStudy (anonymous):

, right, the midpoint is not a problem, for instance you can have 2 as your base, and half of it is 1

OpenStudy (anonymous):

here is what i did , ok i realised that (0,0) (x,0) and (1/2 x , sqrt3 / 2 x ) wont work

OpenStudy (agreene):

Yes, but then your altitude will give that other point at 1/2 sqrt 3 * a which is irrational...

OpenStudy (anonymous):

so lets look at rotating this equilateral triangle

OpenStudy (anonymous):

right, i agree , so lets rotate it

OpenStudy (agreene):

http://mathworld.wolfram.com/EquilateralTriangle.html the first set of equations here kind of make the point im trying to.

OpenStudy (anonymous):

so imagine we have a circle, one vertex will be the center, the other two will land on the circle, and make 60 degrees .

OpenStudy (anonymous):

(0,0), ( x1,y1) (x2,y2)

OpenStudy (anonymous):

, you know i would hate to find out this is impossible problem,

OpenStudy (anonymous):

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).

OpenStudy (agreene):

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.

OpenStudy (anonymous):

looks like wolfram says there are infintie solutions

OpenStudy (agreene):

No, that is Napoleon's theory... it deals with creating eq triangles from the sides of any given trigangle..

OpenStudy (agreene):

That is a rather different question.

OpenStudy (anonymous):

hmmm, well hero posted it

OpenStudy (agreene):

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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