finding a point that has specific distances from 2 other points and a line using only a compass and straight-edge

Question

Is there a way to find said point, which lies on half-plane, such that it has a distance of n from line L, n-m from point A, and n-2m from point B, with both points A and B located on the same half-plane?

Answer 1

I don't think, that it always works. Intuitively, because we have three conditions on $P$ but the plane is only two dimensional.

Let the distances $m,n$ be given. Then $P$ must lie on a circle centered at $A$ with radius $n-m$ and simultaneously on a circle centered at $B$ with radius $n-2m$. So $P$ must lie on the intersection of these circles.

Depending on the distance between $A$ and $B$ and depending on $m,n$ the circles intersect (as usual) in zero, one or two points. This does not leave enough freedom of choice to ensure that the distance between $P$ and $L$ is $n$. This only happens in certain cases.