We have an angle $\angle AOB$ and two segments of lengths $x$ and $y$. Construct a point $P$ inside $\angle AOB$ such that $$\frac{\mathrm{dist}(P, OA)}{\mathrm{dist}(P,OB)}=\frac{x}y. $$

My idea is to generalise the construction of the angle-bisector: we draw a circle with center $O$ intersecting $OA$ at $A'$ and $OB$ at $B'$ and then two circles with equal radiuses and centers at $A'$ and $B'$ and any of their intersection will be on the angle bisector. This might useful since this is the case $x=y$. Probably at some point we have to draw a circle with radius $x$ and one with radius $y$.

Another useful idea is given a segment of length $a$, the construction of a segment of length $b$ such that $a/b=x/y$ is easily done: let $MN=x$, $P\in MN$ such that $NP=y$ and $XY\parallel MN$ such that $XY=a$ (these can be easily constructed). Then we have to consider $Q\in XM\cap YN$ and $Z\in QP\cap XY$ and we will have $YZ=b$.


1 Answer 1


The distance of a point to a line is being measured perpendicularily to the given line.

Thence the way to solve this construction is to draw a parallel to OA with distance x and a further parallel to OB with distance y. The intersection point P of these parallels obviously does bow to the requested ratio. Finally draw the line OP, which provides all other points P', which would solve the same ratio too.

--- rk


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