Is it possible that a trapezium with angles at one side both equal to 60° has all vertices in lattice points? Let $ABCD$ be a trapezium in a coordinates system. $\angle DAB=\angle ABC=60^\circ$ and $AB$ is parallel to $CD$. Is it possible that all $A, B, C, D$ vertices are lattice points (points with both coordinates being integers)?
 A: Suppose the trapezium $ABCD$ exists. Then it would be possible to translate the edge $BC$ such that $C$ and $D$ coincide, and we would then have an equilateral triangle in the integer lattice.
But this latter triangle cannot exist, and the proof can be found here. In short, if such a triangle exists it would be possible to make a regular hexagon on the integer lattice, but by rotating each of the hexagon's sides 90° about a vertex a smaller hexagon could be made, and we get infinite descent. Thus the hexagon, the triangle and finally the trapezium in the problem cannot exist.
A: I've found Pick's Theorem to be very powerful for this sort of proof. In my initial proof attempt, I attempted to produce the sides of the isosceles trapezium to meet at the apex of an equilateral triangle. However, the remainder of my (now deleted) prior proof depended on my being able to show that that apex was also a lattice point, something I have not yet managed.
It is still possible to apply Pick's theorem to the problem by starting with the insight that Parcly Taxel had in his answer below to construct a putative lattice equilateral triangle from the isosceles trapezium by translation of a vertex by a column vector with integer coordinates.
If we let the side of the equilateral triangle be $r$, then its area is $A = \frac 12 r^2 \sin \frac{\pi}{3} = \frac {\sqrt 3}4 r^2$. For a lattice polygon, the squares of all edge lengths are integral (by the Pythagorean theorem).  Hence $A$ is irrational.
However by Pick's theorem, the area $A$ is also rational (an integral multiple of $\frac 12$). $A$ is simultaneously rational and irrational - we've arrived at a contradiction. Therefore a lattice equilateral triangle cannot exist and the original lattice isosceles trapezium with base angles of $60^ \circ$ also cannot exist.
A: Assume that $A$, $B$, $C$, $D$ are lattice point. Let $E$ be the point on $AB$ such that $DC=EB$. Then $AED$ is an equilateral triangle and $E$ is a lattice point.
Take $A$ as the origin of an Argand plane. If $E$ and $D$ are represented by the complex numbers $e$ and $d$, then $\frac{d}{e}$ should have rational real and imaginary parts, which is impossible.
