Least sum of distances

Question

• Problem:

Let $A, B, C, D$ be points in a $3$-dimensional space.

Find the point $X$ that minimizes the sum of the distances $AX+ BX + CX + DX$.

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• Context:

During a course, I was assigned a problem about minimizing the sums of the squares of the distances of one point from other points in a plane.

Now, I am curious about what methods I can use to tackle and make some progresses with this more general question.

Answer 1

In the beautiful book The Mathematical Mechanic. Using Physical Reasoning to Solve Problems [1], professor Mark Levi offered the following mechanical proof of the fact that $ \angle{AXB} = \angle{CXD} $ and that the two angles are bisected by the same line (N.B.: the same result holds for any different pairing of these points).

Let us take four identical constant tension springs (i.e., whose tension is independent of their elongation) of tension $ T = 1 $ and connect them as shown in the picture ([1], p. 62).

It is clear that the potential energy of each spring is equal to its length, and that the potential energy of the system is equal to the combined length of the springs. It follows that the minimal-length configuration has minimal potential energy. Consequently, it corresponds to an equilibrium, and thus to the vanishing of the sum of the forces $ \mathbf{a}, \mathbf{b}, \mathbf{c} $, and $ \mathbf{d} $ acting upon the point $ X $:

\begin{equation} \tag{1} \mathbf {a} + \mathbf{b}= - (\mathbf{c} + \mathbf{d}). \end{equation}

The springs have identical constant tension: $|\mathbf{a}|= |\mathbf{b}| = |\mathbf{c}| = |\mathbf{d}|=1$. Hence, the vector $ \mathbf {a} + \mathbf{b} $ lies on the bisector of $ \angle{AXB} $ and the vector $ \mathbf{c} + \mathbf{d} $ lies on the bisector of $ \angle{CXD} $. From (1), it follows that these bisectors lie on the same line.

Furthermore, the fact that $|\mathbf{a}|= |\mathbf{b}| = |\mathbf{c}| = |\mathbf{d}|=1$ implies that $\mathbf {|a + b|} = 2\cos(\angle{AXB})$ and $\mathbf{|c + d|} = 2\cos(\angle{CXD})$. From (1), we have $ \angle{AXB} = \angle {CXD} $.

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Reference:

1 Levi, M., 2009: The Mathematical Mechanic. Using Physical Reasoning to Solve Problems. Princeton University Press, pp. 61-63, 161-162.

Answer 2

• Two-dimensional space

Let $A=(x_1,y_1),B=(x_2,y_2),C=(x_3,y_3),D = (x_4,y_4),X = (x,y)$.

Then, $$AX+ BX + CX + DX=\big((x-x_1)^2+(y-y_1)^2\big)^{\frac12}+\big((x-x_2)^2+(y-y_2)^2\big)^{\frac12}+\big((x-x_3)^2+(y-y_3)^2\big)^{\frac12}+\big((x-x_4)^2+(y-y_4)^2\big)^{\frac12}$$

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