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I am looking for a solution for a conjecture as follows.

In Cartesian plane, no exist an equilateral triangle such that three vertices are integer numbers.

I hope that you like the question and let me a answer.

Oai Thanh Đào
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    Are you measuring angles in degrees or radians? If degrees, take an equilateral triangle with side lengths $1$. – Qiaochu Yuan Mar 19 '16 at 15:38
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    Also the sum of angles must be equal to $\pi$ so of course no all three angles can be rational. – Loïc Teyssier Mar 19 '16 at 15:39
  • Thank to Dear Mister @QiaochuYuan , I am sorry, my original equestion in

    https://groups.yahoo.com/neo/groups/AdvancedPlaneGeometry/conversations/messages/3149

     – Oai Thanh Đào Mar 19 '16 at 15:49
  • I thank to Dr. @LoïcTeyssier for your answer , I am sorry, I edited and post my original question in

    https://groups.yahoo.com/neo/groups/AdvancedPlaneGeometry/conversations/messages/3149

     – Oai Thanh Đào Mar 19 '16 at 15:51
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    Nice question. Wrong forum. It might work for math.stackexchange. (Also, it has been proved that equilateral triangles embed in Z^3 and not Z^2.) Gerhard "This Is Not Math.StackExchange Forum" Paseman, 2016.03.19. – Gerhard Paseman Mar 19 '16 at 15:51
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    This has many nice proofs and generalizations. For example, compare the area formula for equilateral triangle with side $a$ $\sqrt{3}/4 a^2$ and the fact that double area must be integer. – Fedor Petrov Mar 19 '16 at 15:52
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    My favorite proof of this is to observe that if you can do it with an equilateral triangle, then by a few reflections you can do it with a regular hexagon, and then there's a nice visual proof that a regular hexagon cannot be put on the integer lattice: http://mathoverflow.net/a/25305/5701 – Vaughn Climenhaga Mar 19 '16 at 16:19
  • @VaughnClimenhaga, can't you just use the same visual proof directly on the equilateral triangle itself? – LSpice Oct 27 '17 at 17:11

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Let A,B,C be the vertices.Use the fact that the area is $E=\frac{1}{2}\cdot |det(\vec{AB},\vec{AC})|$
Since $det(\vec{AB},\vec{AC})$ is an integer and the area must be of the form
$AB^2\cdot \frac{\sqrt3}{4}$ which is not an integer you can see the contradiction

Konstantinos Gaitanas
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