Construction of a point with resprect to a triangle For a given  triangle $ABC$, how to construct a point $P$ such that $PA \colon PB \colon PC = 1 \colon 2 \colon4$?

Here is a solution: 
Orient the triangle such that AC lies on the x-axis and B lies on the y-axis. Then, we have that $A = (-a,0), B = (0,b), C = (0,c)$. Draw circles around each of these points with a radius $r$, $2r$ and $4r$. 
If these circles are to meet at a point $(x, y)$, then we have the following simultaneous equations: 
$$(x+a)^2 + y^2 = r^2$$
$$x^2 + (y-b)^2 = (2r)^2$$ 
$$ (x-c)^2 + y^2 = (4r)^2$$
There are three equations and three variables. You can now solve it to obtain, $x$ and $y$ (and $r$ too which will be helpful if you want to physically construct it with a compass). 
