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Is there a triangle with integer sides as well as integer co ordinates when none of the angles is $90$? I tried to solve the general case but I am stuck with it.


Let the Triangle be $T$ whose vertics are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ such that $x_i\neq x_j\neq y_i \neq y_j$ and angles such that $A_i\neq \frac{\pi}{2}$

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Note that if the coordinates are all integral then the area is an integer or a half-integer (by Pick's Theorem, q.v.). So you are asking for triangles with integer side lengths and integer (or half-integer) area. These are "Heronian triangles" as discussed at and elsewhere. – Gerry Myerson Jul 7 '14 at 12:54

The triangle with vertexes $(-3,0),(3,0),(0,4)$

Note 1: Consider any Pythagorician triple $(a,b,c)$, then $a,b,c \in \mathbb{N}$ and $a^2+b^2=c^2$. Now consider the triangle with vertexes of coordinates $(0,0),(b,0),(0,a)$. Finally to avoid the right angle consider the triangle with vertexes $(-b,0),(b,0),(0,a)$. Clearly all coordinates are integers and $$\|(b,0) - (-b,0)\|=2b, \quad \|(\pm b,0) - (0,a)\| = c,$$ which shows that all the edges have integer lengths.

Note 2: Exploiting the same idea, we may also construct non isosceles triangles. Therefore consider $(a,b,c)$ a Pythagorician triple and $(b,d,e)$ another Pythaforician triple (e.g. $(5,12,13)$ and $(12,35,37)$), then the triangle $(-a,0), (d,0), (0,b)$ has integer side lengths and vertexes coordinates.

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Is there a solution in general? – HashimKhan Jul 7 '14 at 10:37
@HashimKhan I don't know if this represent all the triangle respecting the condition, but at least you already have infinitely many different ones with this technique. – Surb Jul 7 '14 at 10:45
Note that you may also consider the triangles of the form $(b,0),(0,-a),(0,a)$. – Surb Jul 7 '14 at 11:32
@HashimKhan Observe that by a simple translation of vector $(0,1)$ of the triangle constructed in NOTE 2, you get a triangle with all the properties of the update except that there are $2$ coordinates that are identical in the vertexes $(-a,1),(d,1)$. – Surb Jul 7 '14 at 12:31

Using complex numbers for the rational coordinates, take an initial triangle with $$A=a+0\ i,\ B=b+0\ i,\ C=0+c\ i.$$ This triangle works provided each of $a^2+c^2,\ b^2+c^2$ is a rational square. This triangle could be rescaled by multiplying all by a common denominator to make the coordinates all integers. However for now just keep the coordinates rational. [For a general triangle in the plane, one can pick a side and show that the foot of the perpendicular to that side from the third vertex has rational coordinates. Such a triangle has rational area using a determinant area formula based on the coordinates. This implies the length of a perpendicular from one vertex to the extended opposite side is rational, so that the general triangle may be repositioned as in the above "initial triangle" description.]

The next question is how the initial triangle may be rotated (and perhaps stretched) and keep rational sides. To do this we use a complex number $z=x+i\ y$ as a multiplier, where $x,y$ are rational to keep the new coordinated rational, and also $|z|=\sqrt{x^2+y^2}$ should be rational, in order to keep sidelengths rational, since each length gets multiplied by $|z|.$ Then the rotated/stretched triangle is $Az,Bz,Cz,$ and has rational sidelengths.

This construction then gives all of those for which one side has the foot of the perpendicular from the third vertex at the origin. Then one may move that by a rational translation so the perpendicular foot may be at any rational point, and finally multiply everything by some integer in order to make the sidelengths and coordinated all integers.

Here is one obtained by the above method, with all sides and coordinates integers, for which there are no horizontal or vertical sides or right angle. $$A=(288,-540),\ B=(225,120),\ C=(-160,300)$$ I used a multiplier $z$ of norm $17$, and the sides are $AB=39\cdot 17,$ $AC=56\cdot 17,$ and $BC=25 \cdot 17.$

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