Perceived angular bisection

Given collinear points A, B, C in the plane, the locus of points P such that $\angle \mathrm{APB}=\angle \mathrm{BPC}\ne 0$ is the unique generalised circle through B which inverts A to C.

Those angles are also trivially equal (both zero) for all P collinear with A, B, C and not between A and C. The locus of other points P is a true circle, except when $AB\; =\; BC$, in which case it is the straight line that is the perpendicular bisector of AC through B; "generalised" circle includes this case as a notional "circle of zero curvature".

Equivalently, the locus is the circle through B centred at a point O chosen such that $\mathrm{OA}.\mathrm{OC}={\mathrm{OB}}^2$ (again with the exception when $AB\; =\; BC$).

This result answers the question: given three points in a line and no other information — three lamp-posts in an otherwise-dark landscape, for instance — whereabouts are the viewpoints from which the middle one appears, through perspective, to be exactly halfway in-between the other two?

Proof

Consider such a point P. Without loss of generality, assume AB < BC (the AB=BC case is trivial, and the AB > BC case is symmetrical with this one).

Draw the unique circle through P and B whose centre is collinear with A, B, and C (e.g. by constructing the perpendicular bisector of PB and marking its centre where that line intersects with the line through A and C). Call this centre O.

Label the angles as shown: $$\begin{array}{cccc}\angle \mathrm{OPA}& =& c& \\ \angle \mathrm{APB}& =& d& \text{which is also}\angle BPC\\ \angle \mathrm{OCP}& =& e& \\ \angle \mathrm{OBP}& =& g& \end{array}$$

Then consider:

$$\begin{array}{cccc}\text{in △CPB,}& e+d+(\mathrm{180°}-g)& =& \mathrm{180°}\\ & e+d& =& g\\ \text{△POB is isosceles,}& c+d& =& g\\ \text{therefore,}& e& =& c\\ \text{i.e.,}& \angle \mathrm{OPA}& =& \angle \mathrm{OCP}\end{array}$$

But the triangles OAP and OPC also have the angle at O in common, so they have two angles in common, therefore they are similar, therefore the ratios of corresponding sides are equal: $\mathrm{OA}/\mathrm{OP}=\mathrm{OP}/\mathrm{OC}$.

And $\mathrm{OP}=\mathrm{OB}$, because both are radii of the circle, so $\mathrm{OA}/\mathrm{OB}=\mathrm{OB}/\mathrm{OC}$ or $\mathrm{OA}.\mathrm{OC}={\mathrm{OB}}^2$, which is one definition of the circle of inversion.

Note that the position of O is the same whichever P we pick; it doesn't depend on any of the angles. (This is easier to see if we rearrange $\mathrm{OA}.\mathrm{OC}={\mathrm{OB}}^2$ into $\mathrm{OA}=\frac{{\mathrm{AB}}^2}{\mathrm{BC}-\mathrm{AB}}$ — it's uniquely specified by the positions of A, B, and C.) So the circle does represent the locus of all P.

Bit off-topic for this blog though eh

Yes. But it is a problem I've been thinking about for a while — in fact, ever since I read that, given the original formulation (three lamp-posts on a darkened plain) you can't even tell from a photograph whether or not they're evenly-spaced, because perspective means that they could be anywhere. But if your photograph has four lamp-posts in, you can tell whether or not they're evenly-spaced, because the cross-ratio is projection-invariant.

So I wondered whether, given three lamp-posts, there was always a point from which they appear evenly-spaced. It felt sort-of plausible that there were such points, but I had no intuition what their locus looked like. The first time I attacked this problem I plotted the points by brute force, and was surprised to see an apparently perfect circle. With the cosine rule and lots of ghastly slog (and Wolfram Alpha) I found the equation of the circle and the position of O. But that was clearly not The Book's proof, so I shelved the draft blog post — for some years. Then I happened to be reading about circle inversion, and suddenly realised it was talking about the three-lamp-posts circle. Eventually I was able to use circle-inversion techniques to come up with a geometrical, not algebraic proof, which I'm much happier with.

And if you're dying to see how bad the ghastly brute-force proof looked, here's a part of it ($p$ is $\mathrm{AB}$ and $q$ is $\mathrm{BC}$):

$p^2{q}^2{x}^2-2p^{2}q{x}^3+p^2{x}^4+p^2{x}^2{y}^2+2p{q}^2{x}^3+2p{q}^{2}x{y}^2-4pq{x}^4-4pq{x}^2{y}^2+2p{x}^5+4p{x}^3{y}^2+2px{y}^4+q^2{x}^4+2q^2{x}^2{y}^2+q^2{y}^4-2q{x}^5-4q{x}^3{y}^2-2qx{y}^4+x^6+3x^4{y}^2+3x^2{y}^4+y^6=p^2{q}^2{x}^2-2p^{2}q{x}^3-2p^{2}qx{y}^2+p^2{x}^4+2p^2{x}^2{y}^2+p^2{y}^4+2p{q}^2{x}^3-4pq{x}^4-4pq{x}^2{y}^2+2p{x}^5+4p{x}^3{y}^2+2px{y}^4+q^2{x}^4+q^2{x}^2{y}^2-2q{x}^5-4q{x}^3{y}^2-2qx{y}^4+x^6+3x^4{y}^2+3x^2{y}^4+y^6$

(I didn't set that by hand; there used to be an online Latex-to-MathML converter, but it seems to have since been retired.)