constructing segments with equal cross ratio

I was puzzeling again and had the following problem:

Given:

• two segments $AD$ and $PS$
• on $AD$ there are points $B$ and $C$ so that $AD \gt AC \gt AB$ (so they are in order A, B , C, D )
• on $PS$ there is a point Q

Question:

How to construct (geometricly, with compass and straightedge) point R on $QS$ so that $$ \frac{AC \times BD} {AB \times CD} = \frac{PR \times QS} {PQ \times RS} $$

(so the crossratio of ABCD is the cross ratio of PQRS )

PS you may NOT assume that the lines $AP$, $BQ$ and $DS$ meet in a single point. (that would be to easy)

Solution 1:

Here is a construction that is not "tricky" or perhaps elegant but is straightforward.

In your desired formula, let's replace the segment length $RS$ with the equivalent $(PS-PR)$, so we have only one segment length that is dependent on point $R$. Then we get

$$ \frac{AC \cdot BD} {AB \cdot CD} = \frac{PR \cdot QS} {PQ \cdot (PS-PR)} $$

The only unknown in that equation is $PR$. Solving that equation for $PR$ we get

$$\begin{align} PR &= \frac{AC \cdot BD \cdot PQ \cdot PS}{AC \cdot BD \cdot PQ + AB \cdot CD \cdot QS} \\[2 ex] &= \frac{PQ \cdot PS}{PQ + \frac{AB}{AC}\left(\frac{CD}{BD}\cdot QS\right)} \end{align}$$

There are many other ways to express that, but I chose an expression that uses only segments on $\overline{AD}$ in the ratios. We can use the standard Euclidean construction methods of ratios and of addition to construct length $PR$. For clarity, I broke the construction into three phases, but this could all be done in one place.

On any angle, construct points $QS$, $BD$, and $CD$ so they have the distances appropriate to their names from vertex $O$. Then construct point $E$ so the line through points $E$ and $QS$ is parallel to the line through points $CD$ and $BD$. Then the distance between points $O$ and $E$ is $\frac{CD}{BD}\cdot QS$.

(In this diagram I copied points $O$ and $E$ to $O'$ and $E'$.) Now construct points $AB$ and $AC$ so they have the distances appropriate to their names from vertex $O'$. Then construct point $F$ so the line through points $F$ and $E'$ is parallel to the line through points $AB$ and $AC$. Then the distance between points $O'$ and $F$ is $\frac{AB}{AC}\left(\frac{CD}{BD}\cdot QS\right)$. Then extend ray $\overrightarrow{O'F}$ to point $FPQ$ so the distance between points $F$ and $FPQ$ equals the distance between points $P$ and $Q$. Then the distance between points $O'$ and $FPQ$ is $PQ+\frac{AB}{AC}\left(\frac{CD}{BD}\cdot QS\right)$.

(In this diagram I copied points $O'$ and $FPQ$ to $O''$ and $FPQ'$.) Now construct points $PQ$ and $PS$ so they have the distances appropriate to their names from vertex $O''$. Then construct point $PR$ so the line through points $PR$ and $PS$ is parallel to the line through points $PQ$ and $FPQ'$.

The distance between points $O''$ and $PR$ is the desired distance between points $P$ and $R$ in the original diagram. Construct point $R$ appropriately and we are done. I tested this construction with Geogebra and it works.