How to find the point on a line which has the minimum length to three points?

Question

I've learnt some ways of finding a point on a line which can minimize the sum of length to other points.

I want to generalize this to three points using geometric methods.

Question:

There're three points $(A,B,C)$ on same side of a line, finding a point on that line to minimize $PA+PB+PC$.

I know that the Lagrange Multiplier Method can solve such problems, but I want to find some geometric meaning.

My Attempt

Just like the two point case, let point $B$ go to another side.

$P$ maybe lies on the intersection of $BA, BC$ with the x-axis.

But I suddenly realized that the point of minimum total distance from the three vertices of the triangle is called Fermat–Torricelli Point ($F$ on picture).

But that point may not on the line. But we can get an intersection by connecting $BF$.

I can't prove which point is smaller, or there are other smaller points.

Answer 1

This is not something you can solve with geometric constructions for more than two points.

Suppose the $n$ points are $(x_i,y_i)$ and the target point is constrained to the $x$-axis, then the problem is equivalent to minimising the following function: $$\sum_i\sqrt{(x-x_i)^2+y_i^2}$$ When considering the unconstrained problem, the set of points whose sum of distances to the $n$ points is constant is an $n$-ellipse, and the constrained problem is indirectly asking for that constant for which the $n$-ellipse is tangent to the $x$-axis. For two points, this shape is a normal ellipse, and the depicted reflection method works because the shortest distance between two points is a straight line (and the ellipse is of algebraic degree 2).

For three points, however, the 3-ellipse is of degree 8 – far too large to admit an exact solution in all cases.

I took your last diagram and assigned coordinates $(0,2),(2,-2),(3,3)$ to $A,B,C$ respectively. What is $P$ in your diagram – the $x$-intercept of the line between the Fermat point of $\triangle ABC$ and $B$ – lies at $(1.61509982\dots,0)$ where the sum of distances to $A,B,C$ is $$7.911\mathbf64173\dots$$ This is quite good, but not optimal. The optimal point is $(1.59405306\dots,0)$ whose corresponding sum-distance is $$7.911\mathbf42964\dots$$

That this differs only after the fourth decimal place shows how delicate this problem is, and the unchallenged superiority of numerical methods for tackling such problems.

Answer 2

For the special case of the three points $(0,0)$, $(1,1)$, $(2,2)$, I find that the minimal $x$ is a root (approximately $0.5473905291$) of the polynomial $$3 x^8-36 x^7+206 x^6-708 x^5+1567 x^4-2280 x^3+2128 x^2-1152 x+256$$ The Galois group of this polynomial, according to Maple, is $S_8$, which is not solvable. This means that not only is there no "ruler and compasses" geometric construction of the root, there is no expression for it using radicals.