Given that this is a pretty obvious and trivial problem, I'm not sure why it hasn't been solved yet. [asy][asy] size(300); pair A = (1, 2), B = (3, 1), C = (3, 2), D = (1, 1); draw(A--C--B--D--cycle); draw(circle(A, 2)); draw(circle(A, 1)); draw(circle(B, 1)); draw(circle(B, 2)); dot("$A$", A, NW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$D$", D, SW); [/asy][/asy] The problem is equivalent to constructing a rectangle with diagonal endpoints $A$ and $B$. Since we know the side lengths of the rectangle (1 and 2), this is trivialized by considering the intersection points of a circle centered at $A$ with radius 2 and a circle centered at $B$ with radius 1 and cyclic permutations, as shown above. The construction trivially works.

Quote: I'm not sure why it hasn't been solved yet. I think I was sleeping at 3:16 AM but I am not entirely sure...

Just reflect $A$ across $B$ to get $(5,0)$ then rotate $B$ around $A$ $90^\circ$ clockwise, then you can get the unit vector by pentisecting the distance from $(0,0)$ to $(5,0)$.

@hamstpan I dont think the measurements are given, which is what makes it difficult. How do you know what length 1 or 2 is?

Zorger74 wrote: @hamstpan I dont think the measurements are given, which is what makes it difficult. How do you know what length 1 or 2 is? We're given the length of $AB$ which is a perfect square root. Constructing lengths of 1 and 2 should be trivial enough.

@hamstpan really? I guess you could get 2sqrt5, then 5, but how do you go from 5 to 1?

You can get $1$ and $2$ by finding the apollonius circle that is the locus of all points whose distances to $A$ and $B$ are in ratio 1:2 (it's not too bad to construct this circle). Then find this circle's intersection with the circle whose diameter has endpoints on $A$ and $B$.