Constructing fourth collinear point satsfying the mentioned condition

Question

Given three collinear points $A,B,C$ such that $B$ is in between $A$ and $C$ . If $AB = m$ and $BC = n$ construct a point $X$ along $BC$ such that $CX = \frac{BC×AC}{3AB-BC}$ and $C$ lies between $B$ and $X$ . (Using only straightedge and compass).

My try : i thought of that maybe similar triangles might be helping here. $3AB$ can be easily made using compass $BC$ can be subtracted from it too by just using a compass . Let that point be $Z$ , i am not able to think of now how to use $Z$ point to determine that $X$ point.

Answer 1

I try to follow the idea of @aschepler:

To fit $BC\times AC$ format, we need to draw a circle O pass AB.

First draw a circle O' with center C and radius $(3AB-BC)$. Then draw suitable circle O pass AB such that O cut O'(Some O with too small radius may never cut O'), call one of the intersect point D. Then CD should cut circle O at another point X'(X' may overlap with X). Since $CA\times CB=CD\times CX'$, Then we find X on line ABC such that $CX=CX'$ and C lies between B and X.

Answer 2

You need to somehow make some $2$-d figure instead of work on $1$-d line. I try right angled triangle. It is easy to draw.

You can Draw $A'C\perp$ where $A'C=AC$

Draw $DB\perp AB$ where $DB=3AB-BC$ use your mentioned method or else. (D is at the same side of AC as A')

So $\bigtriangleup ACX \cong \bigtriangleup DBC$ since $\frac{AC}{CX}=\frac{DB}{BC}$

Show $DC\| AX$ and then you can easily find X then by ruler.