Linkage and Inversion of Two Points

Question

I am having a little bit of trouble proving the following question from "Experiencing Geometry: Euclidean and Non-Euclidean With History":

Show that for the linkage in Figure 16.7 the points $P$ and $Q$ are the inversions of each other through the circle of inversion with center at $C$ and radius $r = \sqrt{s^2 - d^2}$.

I know that by definition of inversion pairs, $P$ and $Q$ are only inversions of each other if $CP \cdot CQ = r^2$, so I believe that I need to show this. I also know that the points $C$, $P$ and $Q$ are collinear, and so the lines from $CQ$ and $RS$ must be perpendicular to each other.

The textbook does not really give much insight about how to approach the problem, so some advice on how to prove this would be helpful.

Answer 1

Below is the top part of your diagram with a point $T$ and several line lengths (e.g., $\lvert CP\rvert=a$ and $\lvert TR\rvert=c$) added:

All $4$ sides of $PRQS$ being equal means it's a rhombus, so the diagonals are perpendicular bisectors of each other at $T$ (with $\lvert PT\rvert = \lvert TQ\rvert = b$), and we also have

$$\measuredangle RPQ = \measuredangle SPQ \tag{1}\label{eq1A}$$

In addition, using the SSS criteria (i.e., all $3$ side lengths are equal) of determining congruent triangles, then $\triangle RPC \cong SPC$ and

$$\measuredangle RPC = \measuredangle SPC \tag{2}\label{eq2A}$$

From adding \eqref{eq1A} and \eqref{eq2A}, and using that the sum of angles about a point is $360^{\circ}$, we get $\measuredangle RPQ + \measuredangle RPC = \measuredangle SPQ + \measuredangle SPC = 180^{\circ}$, i.e., as you already stated, the points $C$, $P$ and $Q$ are collinear. Next, using the Pythagorean theorem with $\triangle CTR$ gives

$$(a+b)^2 + c^2 = s^2 \tag{3}\label{eq3A}$$

and using that theorem with $\triangle PTR$ results in

$$b^2 + c^2 = d^2 \tag{4}\label{eq4A}$$

Next, \eqref{eq3A} minus \eqref{eq4A} becomes

$$a^2 + 2ab = s^2 - d^2 \tag{5}\label{eq5A}$$

As you already stated, $P$ and $Q$ being circular inversions of each other with center $C$ means that circle has a radius $r$ with

$$\lvert CP\rvert\cdot\lvert CQ\vert = r^2 \;\;\to\;\; a(a+2b) = r^2 \;\;\to\;\; a^2 + 2ab = r^2 \tag{6}\label{eq6A}$$

Comparing \eqref{eq6A} to \eqref{eq5A} gives that

$$r^2 = s^2 - d^2 \;\;\to\;\; r = \sqrt{s^2-d^2} \tag{7}\label{eq7A}$$

Answer 2

Apologies for the crude hand sketch.

When hinges $(R,S)$ are fully taken apart they assume new positions $(U,V)$ respectively. (Length $UV=2d,$ double the rhombus link side.)

Point $P$ goes to $P1$, point $Q$ to $Q1$, center C is fixed.

Points $P1,Q1$ merge, Points $(U,P1,Q1, V )$, are aligned straight, line $UV$ still perpendicular to $CPQ.$

On a circle of inversion a point and its inverse coincide.

i.e., when they do coincide, they are on the boundary of the inversion circle radius $r$.

By Pythagoras theorem its radius is

$$ CP1= CQ1=r=\sqrt{s^2-d^2}.$$