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

Given points A and B, the locus of points P such that PA:PB = 1:2 (P is twice as far from B as it is from A) is a circle, and similarly for locus PA:PC = 1:4. So draw the two circles, and P is at an intersection of the circles. Note that you may get 0, 1, or 2 possible points for P. In the diagram below I show a situation giving 2 points $P_1$ and $P_2$.

Here is an overview of the construction:
1) Construct point $H$ on $\overline {AB}$ such that $AH = \frac{1}{3}AB$.
2) Construct point $J$ on ray $\overrightarrow {BA} $ (not on $\overline {AB}$) such that $AJ = AB$.
3) Construct the circle that has $\overline {HJ}$ as a diameter.
4) Construct point $U$ on $\overline {AC}$ such that $AU = \frac{1}{5}AC$.
5) Construct point $W$ on ray $\overrightarrow {CA} $  (not on $\overline {AC}$) such that $AW = \frac{1}{3}AC$.
6) Construct the circle that has $\overline {UW}$ as a diameter.
7) Mark point $P$ at an intersection of the circles.
ADDED:
I just realized that it would be easier to construct the 1:2 circle between A and B and the 1:2 circle between B and C. Oh, well, what I showed above also works.
ADDED LATER:
I just had to add a diagram of the better construction using two 1:2 circles. In the diagram below, $AH = \frac13AB$, $AI=AB$, $BQ=\frac13BC$, $BM=BC$, and $\overline {HI}$ and $\overline {MQ}$ are diameters of circles $p$ and $u$ respectively.

