Why is a+b-c least here?

Question

This part is from the well-known book What is Mathematics.

Chapter VII (Maxima and Minima) - Section 5 (Steiner’s Problem) - Article 3 (A Complementary Problem).

The problem: a, b, c is the distance of arbitrary point P to three different points A、B、C in the plane respectively, and we seek the least value of a+b-c.

The theorem at bottom states that: when one of the angle for example C is greater than 120°, then a+b-c is least at the point P where the two shorter sides of the triangle (AC and BC) subtend (from point P) angles of 60° and the longest side (AB) subtends an angle of 120°.

The figure (which is accurate) demonstrates this situation at P'.

But it seems to me that if P’ in the figure is say at the symmetric point about C or simply coincides with A then the value of a+b-c (after some measure) will be smaller. Or is it talking about a relative minimum rather than a global minimum?

Very much appreciate it if any one who has the book at hand bother to take a look at it.

Answer 1

Here's a contour plot, made with Mathematica, of $a+b-c$ for $A=(1,0)$, $B=(-1,0)$ and $C=(0.2,0.2)$. The minimum is clearly at $B$. Point $P'$ defined in the book (black dot) looks like a saddle point of the function.

EDIT.

A simple construction shows that $P'$ cannot be a minimum (see figure below). The ellipse with foci $A$ and $B$, passing through $P'$, is tangent to the circle of centre $C$ and radius $CP'$. For any other point $P''$ on the ellipse, near to $P'$, the sum $AP''+BP''$ is the same as $AP'+BP'$, whereas $CP''>CP'$, because the circle is inside the ellipse. Hence $AP''+BP''-CP''<AP'+BP'-CP'$ and $P'$ is thus a local maximum for $a+b-c$ along the ellipse.

On the other hand, the hyperbola with foci $A$ and $C$, passing through $P'$, is tangent to the circle of centre $B$ and radius $BP'$. With a reasoning analogous to that shown above one can prove that $P'$ is instead a local minimum along the hyperbola. Hence $P'$ is a saddle point.

Answer 2

In the figure, the green curves are the $2$ branches of the hyperbola with foci $A$ and $C$. Minimum $a+b-c$ is attained when the circle with center $B$ touches the hyperbola at $P'$.

Note that $\alpha = \beta$ and $\alpha = \delta$

In the second figure, the foci of the green hyperbola are $B$ and $C$ and the center of the circle is $A$.

Note that $\gamma = \delta$.

Since $\alpha = \beta = \gamma = \delta$

and $\alpha + \beta + \gamma = 180^o$

therefore $\alpha = \beta = \gamma = 60^o$