"Natural Extremum" Series --- 4. The Fermat Point Problem

I. "Optical Solution" to the Fermat Point

We find that the expression k_1 AP + k_2 BP + k_3 CP sought in the generalized Fermat point problem is very similar in form to \frac{PO}{v_1} + \frac{QO}{v_2} in the law of refraction. Thus, we consider applying Fermat’s Principle. First, we rewrite it as: \frac{AP}{(1/k_1)} + \frac{BP}{(1/k_2)} + \frac{CP}{(1/k_3)}, \quad v_1=1/k_1, v_2=1/k_2, v_3=1/k_3

Fermat’s Principle concerns the extremum of the sum of two terms, while the Fermat point involves three terms. To apply Fermat’s Principle, we first need to "fix" one length. As shown below:

Assume the length of BP is already given. We then draw a circular arc with BP as the radius and assume this arc is a special mirror (the speed of light changes after reflection from it). Thus, our problem becomes: find a point P on the arc such that k_1 AP + k_3 CP is minimized.

According to Fermat’s Principle, we must choose the light path. This path follows the "reflection-refraction law" corollary we derived in the previous section on "Fermat’s Principle," namely: \frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_3} \Rightarrow k_1 \sin \theta_1 = k_3 \sin \theta_2

Similarly, we can replace BP with AP or CP and set up the problem in the same way, yielding: \begin{aligned} k_3 \sin \theta_3 &= k_2 \sin \theta_4 \\ k_2 \sin \theta_5 &= k_1 \sin \theta_6 \end{aligned}

Based on the property that "vertical angles are equal" and "the sum of three adjacent angles is \pi," we obtain: \begin{aligned} \frac{\sin \theta_1}{k_3} = \frac{\sin \theta_5}{k_1}; \quad \frac{\sin \theta_3}{k_2} = \frac{\sin \theta_1}{k_3}; \quad \frac{\sin \theta_5}{k_1} = \frac{\sin \theta_3}{k_2} \\ \theta_1 + \theta_3 + \theta_5 = \pi \end{aligned}

It can be seen that \theta_1, \theta_3, \theta_5 are precisely the three internal angles of a triangle with sides k_3, k_2, k_1. At this point, BoJone believes the problem is solved. The remaining details are left for the readers to contemplate (one should not reveal all the mysteries, as that would be far too monotonous).

II. "Potential Solution" to the Fermat Point

Imagine placing \Delta ABC horizontally at a height h. Tie one end of three strings of length l together, and attach weights k_1, k_2, k_3 to the other ends. As shown in the figure, make them pass through points A, B, C respectively.

Under the influence of gravity, they will tend toward an equilibrium state, i.e., the state of minimum potential energy. This requires k_1 [h-(l-AP)] + k_2 [h-(l-BP)] + k_3 [h-(l-CP)] to be minimized (gravitational potential energy E_p = Gh, where the terms in brackets are the heights of each weight), which is equivalent to minimizing k_1 AP + k_2 BP + k_3 CP.

By the principle of equilibrium, the net force at the junction of the three strings must be zero once stabilized. Thus, we obtain the solution: it can be written in the same form as our "optical solution," or for computational convenience, as: k_1^2 + k_2^2 + 2k_1 k_2 \cos \angle APB = k_3^2 \quad \text{(Parallelogram law of forces)}

The other two cases are similar and will not be repeated. Thus, two perspectives, two methods, one problem—done!