The Weighted Fermat Triangle Problem

Abstract

We completely solve the generalized Fermat problem: given a triangle 𝑃₁, 𝑃₂, 𝑃₃ and three positive numbers 𝜆₁, 𝜆₂, 𝜆₃, find a point 𝑃 for which the sum 𝜆₁𝑃₁𝑃+𝜆₂𝑃₂𝑃+𝜆₃𝑃₃𝑃 is minimal. We show that the point always exists and is unique, and indicate necessary and sufficient conditions for the point to lie inside the triangle. We provide geometric interpretations of the conditions and briefly indicate a connection with dynamical systems.

1. Introduction

Pierre Fermat (1601–1665) formulated the following problem:

Given a triangle 𝐴𝐵𝐶, find a point 𝑃 such that the sum of the three distances from 𝑃 to the vertices 𝐴,𝐵,𝐶 be minimal.

In the literature, one can find various beautiful ways to solve the problem (see, e.g., [1–4]). In short, the answer is as follows. If every angle of 𝐴𝐵𝐶 measures less than 120^∘, then the point 𝑃 in the interior of the triangle such that ∠𝐴𝑃𝐵=∠𝐵𝑃𝐶=∠𝐶𝑃𝐴=120^∘ minimizes the sum of the three distances. If one of the angles of 𝐴𝐵𝐶 measures 120^∘ or more, then the vertex corresponding to this angle minimizes the sum of the three distances to the vertices.

An important application is the shortest network problem, used in the construction of telephone, pipeline, and roadway networks; see for example [5].

In this paper, we consider a weighted Fermat triangle problem:

Given a triangle 𝑃₁𝑃₂𝑃₃ and given three positive numbers 𝜆₁, 𝜆₂, 𝜆₃, find a point 𝑃 on the triangle such that the weighted sum of the distances to the three vertices 𝜆₁𝑃₁𝑃+𝜆₂𝑃₂𝑃+𝜆₃𝑃₃𝑃 be the least possible.

A possible application is the problem of constructing a consumer center servicing three given cities in such a way as to minimize the total distance to all three, but also making the distance to a given city inversely proportional to the population of that city. As we found out after working out a solution, this problem had been previously formulated by Greenberg and Robertello in 1965 [6] as the three factory problem and solved using trigonometry; two subsequent papers by van de Lindt [7] and Tong and Chua [8] offered geometric solutions. There is also a higher-dimensional generalization in [9]. However, we think our approach is still of interest, firstly, because of the geometric connections explored throughout the paper, and secondly because of its accessibility. Except possibly for the last section, the paper can be understood by students who have completed the calculus sequence.

Using calculus, we obtain necessary and sufficient conditions for 𝜆₁𝑃₁𝑃+𝜆₂𝑃₂𝑃+𝜆₃𝑃₃𝑃 to attain its absolute minimum in the interior of the triangle 𝑃₁𝑃₂𝑃₃. We show uniqueness of such minimizing point, and present an elegant geometric construction of this point.

In the event that the absolute minimum of 𝜆₁𝑃₁𝑃+𝜆₂𝑃₂𝑃+𝜆₃𝑃₃𝑃 does not occur in the interior of the triangle, we show that one and only one vertex of the triangle minimizes this sum, and we locate that vertex.

In the last section, we briefly show a connection between the Fermat problem and a gradient dynamical system.

2. Existence of a Minimum inside the Triangle

Assuming our plane has Cartesian coordinates (𝑥,𝑦), let 𝑃_𝑖 have coordinates (𝑥_𝑖,𝑦_𝑖), 𝑖=1,2,3, and let 𝑃 have coordinates (𝑥,𝑦). Call 𝑑_𝑖 the distance between 𝑃_𝑖 and 𝑃, 𝑖=1,2,3 (see Figure 1).

Figure 1 

Then, the problem is to minimize the function

𝑓(𝑃)=𝑓(𝑥,𝑦)=𝜆₁𝑑₁+𝜆₂𝑑₂+𝜆₃𝑑₃. (2.1)

This function is continuous on the whole plane ℝ², so it must attain an absolute minimum on the closed triangle 𝑃₁𝑃₂𝑃₃.

Let us find the gradient of 𝑓(𝑃). First of all, we have𝑑²_𝑖=(𝑥−𝑥_𝑖)²+(𝑦−𝑦_𝑖)²,𝑖=1,2,3, (2.2)and therefore, 𝜕𝜕𝑥(𝑑_𝑖)²=2(𝑥−𝑥_𝑖). (2.3) If 𝑃≠𝑃_𝑖, then 𝑑_𝑖 itself is differentiable, in which case we have2𝑑_𝑖𝜕𝑑𝜕𝑥^𝑖=2(𝑥−𝑥_𝑖). (2.4) It follows that 𝜕𝑑𝜕𝑥^𝑖=𝑥−𝑥_𝑖𝑑_𝑖,if𝑃≠𝑃_𝑖. (2.5) Similarly, we get𝜕𝑑𝜕𝑦^𝑖=𝑦−𝑦_𝑖𝑑_𝑖,if𝑃≠𝑃_𝑖. (2.6) Therefore, the gradient ∇𝑑_𝑖 of 𝑑_𝑖 is equal to ∇𝑑_𝑖=1𝑑_𝑖(𝑥−𝑥_𝑖,𝑦−𝑦_𝑖),if𝑃≠𝑃_𝑖,where𝑖=1,2,3. (2.7) Let us call 𝐮_𝑖=∇𝑑_𝑖=1𝑑_𝑖(𝑥−𝑥_𝑖,𝑦−𝑦_𝑖1)=𝑑_𝑖𝑃_𝑖𝑃. (2.8) This is a unit vector, defined for every 𝑃≠𝑃_𝑖, 𝑖=1,2,3.

Getting back to our function 𝑓(𝑃), we conclude that 𝑓 is differentiable on the open domainΩ=ℝ²⧵{𝑃₁,𝑃₂,𝑃₃}, (2.9) and its gradient is equal to∇𝑓=𝜆₁𝐮₁+𝜆₂𝐮₂+𝜆₃𝐮₃onΩ. (2.10) Let us see when 𝑓 can have stationary points.

Lemma 2.1. A necessary condition for ∇𝑓 to be zero (at some point in Ω) is that 𝜆₁<𝜆₂+𝜆₃,𝜆₂<𝜆₃+𝜆₁,𝜆₃<𝜆₁+𝜆₂. (2.11)
(Geometrically, this means that we can construct a nondegenerate triangle with sides 𝜆₁, 𝜆₂, 𝜆₃.)
Indeed, ∇𝑓=0 is equivalent to 𝜆₁𝐮₁+𝜆₂𝐮₂+𝜆₃𝐮₃=0. This, in turn, means that the polygonal curve with sides 𝜆₁𝐮₁, 𝜆₂𝐮₂, 𝜆₃𝐮₃ must be a triangle; see Figures 2 and 3.
Moreover, in our case the triangle cannot be degenerate, since this would imply that the three vectors 𝐮₁, 𝐮₂, 𝐮₃ are parallel, which is impossible, for the points 𝑃₁, 𝑃₂, 𝑃₃ do not lie on one line.
Since 𝐮₁,𝐮₂, and 𝐮₃ are unit vectors, this triangle has sides 𝜆₁, 𝜆₂, 𝜆₃, so these three numbers must satisfy the triangle inequalities (2.11).

Figure 2 

Figure 3 

Lemma 2.2. If ∇𝑓(𝑃)=0 at some 𝑃∈Ω, then at this point one has𝐮₁⋅𝐮₂=𝜆²₃−𝜆²₁−𝜆²₂𝜆₁𝜆₂,𝐮₂⋅𝐮₃=𝜆²₁−𝜆²₂−𝜆²₃2𝜆₂𝜆₃,𝐮₃⋅𝐮₁=𝜆²₂−𝜆²₁−𝜆²₃𝜆₁𝜆₃. (2.12)
Indeed, if ∇𝑓(𝑃)=0, then at 𝑃 we have𝜆₁𝐮₁+𝜆₂𝐮₂+𝜆₃𝐮₃=0. (2.13)
Let us dot multiply this equality successively by 𝐮₁, 𝐮₂, and 𝐮₃. Recalling that 𝐮_𝑖⋅𝐮_𝑖=‖𝐮_𝑖‖²=1, we get𝜆₁+𝜆₂𝐮₁⋅𝐮₂+𝜆₃𝐮₁⋅𝐮₃𝜆=0,₁𝐮₂⋅𝐮₁+𝜆₂+𝜆₃𝐮₂⋅𝐮₃𝜆=0,₁𝐮₃⋅𝐮₁+𝜆₂𝐮₃⋅𝐮₂+𝜆₃=0. (2.14)To simplify matters, let us call for a moment 𝑣₃=𝐮₁⋅𝐮₂,𝑣₂=𝐮₁⋅𝐮₃,𝑣₁=𝐮₂⋅𝐮₃. (2.15)Then, the previous system looks like𝜆₁+𝜆₂𝑣₃+𝜆₃𝑣₂𝜆=0,₁𝑣₃+𝜆₂+𝜆₃𝑣₁𝜆=0,₁𝑣₂+𝜆₂𝑣₁+𝜆₃=0. (2.16) Let us multiply the first equation by 𝜆₁, the second by 𝜆₂, and subtract the second from the first: 𝜆²₁+𝜆₁𝜆₃𝑣₂−𝜆²₂−𝜆₂𝜆₃𝑣₁=0. (2.17) This can be rewritten as𝜆₁𝜆₃𝑣₂−𝜆₂𝜆₃𝑣₁=𝜆²₂−𝜆²₁. (2.18) Let us adjoin to this equation the last equation in (2.16), previously multiplied by 𝜆₃ :𝜆₁𝜆₃𝑣₂−𝜆₂𝜆₃𝑣₁=𝜆²₂−𝜆²₁,𝜆₁𝜆₃𝑣₂+𝜆₂𝜆₃𝑣₁+𝜆²₃=0. (2.19)Adding both equations in this system, and solving for 𝑣₂, we get𝑣₂=𝜆²₂−𝜆²₁−𝜆²₃𝜆₁𝜆₃. (2.20) Similarly one gets𝑣₁=𝜆²₁−𝜆²₂−𝜆²₃2𝜆₂𝜆₃,𝑣₃=𝜆²₃−𝜆²₁−𝜆²₂2𝜆₁𝜆₂. (2.21) Our result follows if we recall the notation (2.15).

Geometric Interpretation of Equalities (2.12)
Assume that conditions (2.11) hold, so that we can construct the triangle in Figure 3. Call 𝜃_𝑖 the angle opposite to the side 𝜆_𝑖 in this triangle (see Figure 4).
Then, for example,𝐮₁⋅𝐮₂=cos(𝜋−𝜃₃)=−cos𝜃₃ (2.22) (recall that the 𝐮_𝑖 are unit vectors), and the first equality in (2.12) follows from the law of the cosine. The other two equalities in (2.12) have analogous geometric interpretations.

Figure 4 

Lemma 2.3. If 𝑃 lies on one of the sides of the triangle 𝑃₁𝑃₂𝑃₃, but does not coincide with one of the vertices, then ∇𝑓(𝑃)≠0.
Indeed, assume, for example, that 𝑃 lies on the side 𝑃₁𝑃₂ (see Figure 5).
Then 𝐮₁ and 𝐮₂ are parallel. Moreover, the vectors 𝜆₁𝐮₁+𝜆₂𝐮₂ and 𝜆₃𝐮₃ are linearly independent, and at least the second one is nonzero. Therefore,∇𝑓(𝑃)=(𝜆₁𝐮₁+𝜆₂𝐮₂)+𝜆₃𝐮₃≠0. (2.23)

Figure 5 

Lemma 2.4. If 𝑃 lies outside the triangle 𝑃₁𝑃₂𝑃₃, then ∇𝑓(𝑃)≠0.
Indeed, if 𝑃 lies outside this triangle, then it must lie on one of the half-planes whose boundary is the line joining two vertices, which does not contain the third vertex. Assume, for example, that 𝑃 lies on the half-plane with the boundary through 𝑃₁𝑃₂ which does not contain 𝑃₃ (see Figures 6 and 7).
Then, if we draw the vectors 𝜆₁𝐮₁, 𝜆₂𝐮₂, and 𝜆₃𝐮₃ starting at a common origin 𝑃, all three will lie on the same half-plane with boundary being the line parallel to 𝑃₁𝑃₂ passing through 𝑃. Since all three vectors are nonzero, so is their sum𝜆₁𝐮₁+𝜆₂𝐮₂+𝜆₃𝐮₃=∇𝑓(𝑃). (2.24)

Figure 6 

Figure 7 

The gradient ∇𝑓 does not exist at the vertices of the triangle. However, we can compute one-sided directional derivatives at these points.

Let us start by analyzing the behavior of 𝑑₁(𝑃)=𝑑₁(𝑥,𝑦) near the singular point 𝑃₁. Let us fix an arbitrary unit vector 𝐧 and a nonzero number ℎ. Then,𝑑₁(𝑃₁+ℎ𝐧)−𝑑₁(𝑃₁)ℎ=𝑑₁(𝑃₁+ℎ𝐧)ℎ=‖ℎ𝐧‖ℎ=|ℎ|ℎ. (2.25)

Therefore,𝐷⁺_𝐧𝑑₁(𝑃₁)=^deflim_ℎ→0+𝑑₁(𝑃₁+ℎ𝐧)−𝑑₁(𝑃₁)ℎ𝐷=1,_−𝐧𝑑₁(𝑃₁)=^deflim_ℎ→0−𝑑₁(𝑃₁+ℎ𝐧)−𝑑₁(𝑃₁)ℎ=−1. (2.26)

Let us denote by 𝛼_𝑖 the angle of the triangle 𝑃₁𝑃₂𝑃₃ at 𝑃_𝑖, where 𝑖=1,2,3 (see Figure 8).

Figure 8 

Lemma 2.5. Ifcos𝛼₁>𝜆²₁−𝜆²₂−𝜆²₃2𝜆₂𝜆₃, (2.27)then the absolute minimum of the function 𝑓(𝑃) given by (2.1) on the triangle 𝑃₁𝑃₂𝑃₃ is not attained at 𝑃₁.
To show this, let us compute the one-sided directional derivative 𝐷⁺_𝐧𝑓(𝑃₁). Now, only the first term of 𝑓(𝑃), that is, 𝜆₁𝑑₁, is not differentiable at 𝑃₁; for the other two terms, we can compute the directional derivative in the usual way. Therefore, we have𝐷⁺_𝐧𝑓(𝑃₁)=𝜆₁𝐷⁺_𝐧𝑑₁(𝑃₁)+𝜆₂∇𝑑₂(𝑃₁)⋅𝐧+𝜆₃∇𝑑₃(𝑃₁)⋅𝐧=𝜆₁+[𝜆₂𝐮₂(𝑃₁)+𝜆₃𝐮₃(𝑃₁)]⋅𝐧. (2.28)The smallest value of this derivative will happen when 𝐧 is parallel to the vector 𝜆₂𝐮₂(𝑃₁)+𝜆₃𝐮₃(𝑃₁) and has the opposite direction. For such 𝐧 we get𝐷⁺_𝐧𝑓(𝑃₁)=𝜆₁−‖𝜆₂𝐮₂(𝑃₁)+𝜆₃𝐮₃(𝑃₁)‖. (2.29) Let us denote𝐯₂=𝐮₂(𝑃₁),𝐯₃=𝐮₃(𝑃₁); (2.30) these are unit vectors directed along the sides 𝑃₂𝑃₁ and 𝑃₃𝑃₁, respectively, as shown in Figure 9. With this notation, we have
𝐷⁺_𝐧𝑓(𝑃₁)=𝜆₁−‖𝜆₂𝐯₂+𝜆₃𝐯₃‖. (2.31) Notice that the vector 𝐧 we have chosen, directed opposite to 𝜆₂𝐯₂+𝜆₃𝐯₃, points towards the interior of the triangle 𝑃₁𝑃₂𝑃₃.
If this derivative is negative, this means that when we move from 𝑃₁ in the direction of 𝐧, the function 𝑓(𝑃) will decrease, so that 𝑃₁ cannot be a minimum. For this derivative to be negative, we must have‖𝜆₂𝐯₂+𝜆₃𝐯₃‖²>𝜆²₁ (2.32)or(𝜆₂𝐯₂+𝜆₃𝐯₃)⋅(𝜆₂𝐯₂+𝜆₃𝐯₃)>𝜆²₁, (2.33)or still𝜆²₂+2𝜆₂𝜆₃𝐯₂⋅𝐯₃+𝜆²₃>𝜆²₁. (2.34)This implies that𝐯₂⋅𝐯₃>𝜆²₁−𝜆²₂−𝜆²₃2𝜆₂𝜆₃. (2.35)Finally, notice that 𝐯₂⋅𝐯₃=cos𝛼₁ (see Figure 9).

Figure 9 

In a totally similar way, one can prove that ifcos𝛼₂>𝜆²₂−𝜆²₃−𝜆²₁2𝜆₃𝜆₁, (2.36) then the absolute minimum 𝑓(𝑃) on 𝑃₁𝑃₂𝑃₃ cannot be attained at 𝑃₂, and ifcos𝛼₃>𝜆²₃−𝜆²₁−𝜆²₂2𝜆₁𝜆₂, (2.37) then the absolute minimum 𝑓(𝑃) on 𝑃₁𝑃₂𝑃₃ cannot be attained at 𝑃₃.

Each of conditions (2.27), (2.36), (2.37) implies the corresponding one in (2.11) (condition (2.27) implies the first one, etc.). Indeed, assume, for example, that (2.27) holds. Then, since cos𝛼₁<1, we have𝜆²₁−𝜆²₂−𝜆²₃2𝜆₂𝜆₃<1. (2.38)This is equivalent to 𝜆²₁−𝜆²₂−𝜆²₃<2𝜆₂𝜆₃ (2.39)or𝜆²₁<𝜆²₂+𝜆²₃+2𝜆₂𝜆₃=(𝜆₂+𝜆₃)². (2.40)Since all 𝜆_𝑖 are positive, this in turn is equivalent to𝜆₁<𝜆₂+𝜆₃. (2.41) The other two inequalities are proved in a similar way.

Hence, if (2.27), (2.36), (2.37) hold, we can construct a nondegenerate triangle with sides 𝜆₁, 𝜆₂, 𝜆₃, as in Figure 4. Calling, as before, 𝜃_𝑖 the angle opposite to 𝜆_𝑖 on this triangle, and recalling that the cosine function decreases on [0,𝜋], we can rewrite conditions (2.27), (2.36), (2.37) in the following very natural way:𝛼_𝑖<𝜋−𝜃_𝑖,for𝑖=1,2,3. (2.42)Lemma 2.6. Conditions (2.27), (2.36), (2.37) are necessary for the existence of the absolute minimum of 𝑓(𝑃) in the interior of the triangle 𝑃₁𝑃₂𝑃₃.
Indeed, assume, for example, thatcos𝛼₁≤𝜆²₁−𝜆²₂−𝜆²₃2𝜆₂𝜆₃. (2.43)Pick any point 𝑃 in the interior of the triangle 𝑃₁𝑃₂𝑃₃ (see Figure 10).
Then the angle ∠𝑃₂𝑃𝑃₃ is strictly bigger than 𝛼₁. Therefore,cos𝛼₁>cos∠𝑃₂𝑃𝑃₃=𝐮₂⋅𝐮₃, (2.44)and our assumption implies that𝜆²₁−𝜆²₂−𝜆²₃2𝜆₂𝜆₃>𝐮₂⋅𝐮₃. (2.45) Then, we cannot have ∇𝑓(𝑃)=0 at this point, for otherwise we would get a contradiction with the second equality in (2.12).

Figure 10 

Theorem 2.7. The function 𝑓(𝑃) attains its absolute minimum in the interior of the triangle 𝑃₁𝑃₂𝑃₃ if, and only if, conditions (2.27), (2.36), (2.37) hold, or, equivalently, if conditions (2.42) hold.
Indeed, we know, by Lemma 2.6, that these conditions are necessary. Conversely, assume that the conditions hold.
Consider a circle 𝐶_𝑅, with center anywhere on the triangle, and with a radius 𝑅 so large that the whole triangle lies in its interior and, moreover, on the boundary of 𝐶_𝑅 the minimum of 𝑓(𝑃) is larger than, say, 𝑓(𝑃₁). (This can be achieved because 𝑓(𝑃) tends to infinity as 𝑃 tends to infinity in any direction.)
Now, the continuous function 𝑓(𝑃) must attain a minimum on the compact set 𝐶_𝑅. By our choice of 𝑅, this minimum is not on the boundary of 𝐶_𝑅. Further, by Lemma 2.5 and the remark following it, this minimum is not attained on either vertex 𝑃₁,𝑃₂,𝑃₃. Since these are the only singular points of ∇𝑓(𝑃), it follows that the minimum must occur at a point 𝑃 inside 𝐶_𝑅 at which ∇(𝑃)=0. (This also proves that the gradient must vanish somewhere.) By Lemma 2.4, we conclude that the minimum must lie on the triangle 𝑃₁𝑃₂𝑃₃ (vertices excepted). Therefore, by Lemma 2.3, the minimum must lie in the interior of the triangle 𝑃₁𝑃₂𝑃₃, at a point for which ∇𝑓(𝑃)=0.

3. Uniqueness of the Minimum

As in the classical case, the function 𝑓(𝑃) attains its absolute minimum value exactly at one point.

Theorem 3.1. Assume that conditions (2.27), (2.36), (2.37) hold. Then 𝑓(𝑃) attains its absolute minimum value in the interior of the triangle 𝑃₁𝑃₂𝑃₃ at exactly one point.
We already know, by Theorem 2.7, that 𝑓(𝑃) attains its minimum at some point 𝑃 inside the triangle 𝑃₁𝑃₂𝑃₃.
Arguing by contradiction, assume that the minimum is also attained at some other point 𝑃^′ inside the triangle. Then 𝑃^′ must lie on one of the triangles 𝑃𝑃₁𝑃₂, 𝑃𝑃₂𝑃₃, or 𝑃𝑃₃𝑃₁ (possibly on one of the sides 𝑃𝑃_𝑖). Assume that 𝑃 lies on 𝑃𝑃₂𝑃₃ as in Figure 11.
Since we have both ∇𝑓(𝑃)=0 and ∇𝑓(𝑃^′)=0, by Lemma 2.2 we must have𝐮₂⋅𝐮₃=𝐮^′₂⋅𝐮^′₃=𝜆²₁−𝜆²₂−𝜆²₃2𝜆₂𝜆₃. (3.1) On the other hand, we have𝐮₂⋅𝐮₃=cos∠𝑃₂𝑃𝑃₃,𝐮^′₂⋅𝐮^′₃=cos∠𝑃₂𝑃𝑃₃. (3.2) But ∠𝑃₂𝑃^′𝑃₃ is strictly bigger than ∠𝑃₂𝑃𝑃₃, so we must have 𝐮^′₂⋅𝐮^′₃<𝐮₂⋅𝐮₃, a contradiction.

Figure 11 

4. Construction of the Interior Minimizing Point

Let us assume that conditions (2.27), (2.36), (2.37), or, equivalently, conditions (2.42), are satisfied. Then, by Theorem 2.7, there is a point 𝑃 at which the function 𝑓(𝑃) attains its minimum; moreover, 𝑃 lies in the interior of the triangle 𝑃₁𝑃₂𝑃₃. Also, by Theorem 3.1, this point is unique.

To actually find the point, we can use a construction inspired by the one for the classical case (see, e.g., [4]).

Taking 𝑃₁𝑃₂ as one of the sides, let us construct a triangle 𝑃₁𝑃₂𝑃^′₃, as in Figure 12, which is similar to the triangle in Figure 4, with sides 𝜆₁, 𝜆₂, 𝜆₃. Moreover, let us choose the angles so that the angle at 𝑃₁ is 𝜃₁, the angle at 𝑃₂ is 𝜃₂, and the angle at 𝑃^′₃ is 𝜃₃. Further, let us draw the circumcircle to this triangle, and let 𝑂 be its center.

Figure 12 

The arc 𝑃₁𝑃^′₃𝑃₂ of this circle spans the angle 𝜃₃, so the complementary arc will span 𝜋−𝜃₃.

Similarly, let us construct 𝑃₁𝑃^′₂𝑃₃, also similar to the triangle with sides 𝜆₁, 𝜆₂, 𝜆₃, as in Figure 12, so that the angle at 𝑃₁ is 𝜃₁, the angle at 𝑃^′₂ is 𝜃₂, and the angle at 𝑃₃ is 𝜃₃. Let us also draw the circumcircle to 𝑃₁𝑃^′₂𝑃₃, and let 𝑂^′ be its center.

Now, the formula for the radius of the circumscribed circle (see, e.g., [1], page 13), applied to the triangle 𝑃₁𝑃^′₃𝑃₂, yields𝑃₁𝑃₂sin𝜃₃=2𝑃₁𝑃𝑂,whence¹𝑃₂𝑃₁𝑂=2sin𝜃₃. (4.1) Applying the same result to the triangle 𝑃₁𝑃₃𝑃^′₂, we obtain𝑃₁𝑃^′₂sin𝜃₃=2𝑃₁𝑂^′𝑃,whence¹𝑃^′₂𝑃₁𝑂^′=2sin𝜃₃. (4.2) We conclude that𝑃₁𝑃₂𝑃₁𝑂=𝑃₁𝑃^′₂𝑃₁𝑂^′. (4.3) Hence, the isosceles triangles 𝑃₁𝑂𝑃₂ and 𝑃₁𝑂^′𝑃^′₂ are similar. Therefore, the angle ∠𝑂𝑃₁𝑃₂ is equal to the angle ∠𝑂^′𝑃₁𝑃^′₂. Hence,∠𝑂𝑃₁𝑂^′=∠𝑂𝑃₁𝑃₂+∠𝑃₂𝑃₁𝑃₃+∠𝑃₃𝑃₁𝑂^′=∠𝑂^′𝑃₁𝑃^′₂+𝛼₁+∠𝑃₃𝑃₁𝑂^′=𝛼₁+𝜃₁. (4.4)Now, by our assumption (2.42), 𝛼₁<𝜋−𝜃₁, whence ∠𝑂𝑃₁𝑂^′<𝜋. This guarantees that, firstly, the two circles are not tangent, and secondly, the other point 𝑃 of intersection of these circles, besides 𝑃₁, will occur inside the triangle 𝑃₁𝑃₂𝑃₃. Indeed, from our construction it follows that ∠𝑃₂𝑃𝑃₁=𝜋−𝜃₃ and ∠𝑃₁𝑃𝑃₃=𝜋−𝜃₂. Consequently,2𝜋−(𝜋−𝜃₂)−(𝜋−𝜃₂)=𝜃₃+𝜃₂=𝜋−𝜃₁, (4.5) which is less than 𝜋, so 𝑃 cannot lie below the line 𝑃₂𝑃₃.

We claim that 𝑃 is the desired minimizing point.

Indeed, geometrically, the fact that ∠𝑃₂𝑃𝑃₁=𝜋−𝜃₃, ∠𝑃₁𝑃𝑃₃=𝜋−𝜃₂, and ∠𝑃₁𝑃𝑃₃=𝜋−𝜃₁ guarantees that at 𝑃 one can arrange the vectors 𝜆_𝑖𝐮_𝑖 as in Figure 4, and therefore, 𝜆₁𝐮₁+𝜆₂𝐮₂+𝜆₃𝐮₃=0, (4.6)that is, ∇𝑓(𝑃)=0.

Here is an algebraic proof of the same fact. At 𝑃, we have‖∇𝑓(𝑃)‖²=(𝜆₁𝐮₁+𝜆₂𝐮₂+𝜆₃𝐮₃)⋅(𝜆₁𝐮₁+𝜆₂𝐮₂+𝜆₃𝐮₃)=𝜆²₁+𝜆²₂+𝜆²₃+2𝜆₁𝜆₂𝐮₁⋅𝐮₂+2𝜆₁𝜆₃𝐮₁⋅𝐮₃+2𝜆₂𝜆₃𝐮₂⋅𝐮₃=𝜆²₁+𝜆²₂+𝜆²₃+2𝜆₁𝜆₂cos(𝜋−𝜃₃)+2𝜆₁𝜆₃cos(𝜋−𝜃₂)+2𝜆₂𝜆₃cos(𝜋−𝜃₁)=𝜆²₁+𝜆²₂+𝜆²₃−2𝜆₁𝜆₂cos𝜃₃−2𝜆₁𝜆₃cos𝜃₂−2𝜆₂𝜆₃cos𝜃₁. (4.7) Now, by the cosine law, 𝜆²₁+𝜆²₂−2𝜆₁𝜆₂cos𝜃₃=𝜆²₃, (4.8)so the above expression simplifies to‖∇𝑓(𝑃)‖²=𝜆²₃+𝜆²₃−2𝜆₁𝜆₃cos𝜃₂−2𝜆₂𝜆₃cos𝜃₁. (4.9)Now we add and subtract 𝜆²₁ and apply the cosine law twice again: ‖∇𝑓(𝑃)‖²=(𝜆²₃+𝜆²₁−2𝜆₁𝜆₃cos𝜃₂)+𝜆²₃−2𝜆₂𝜆₃cos𝜃₁−𝜆²₁=(𝜆²₂+𝜆²₃−2𝜆₂𝜆₃cos𝜃₁)−𝜆²₁=𝜆²₁−𝜆²₁=0. (4.10)Note. As for the classical case, it is not hard to show that actually the points 𝑃, 𝑃₃, and 𝑃^′₃ lie on the same line, and so do the points 𝑃, 𝑃₂, and 𝑃^′₂; this provides another geometric way of constructing the minimizing point 𝑃; see [8]. Moreover, generalizing the situation in the classical case, one can see that 𝑃₃𝑃^′₃=𝑑/𝜆₃ and 𝑃₂𝑃^′₂=𝑑/𝜆₂, where 𝑑 is the minimum of our function 𝑓(𝑃) (attained at the point 𝑃 we just constructed).

5. Degenerate Cases

Case 1. When one of the triangle inequalities (2.11) fails to hold then, by Lemma 2.1, the absolute minimum cannot occur in Ω, so it must happen at one of the vertices of our triangle 𝑃₁𝑃₂𝑃₃.

Assume, for example, that 𝜆₁≥𝜆₂+𝜆₃. (5.1) Then, we claim that the minimum is attained at 𝑃₁.

Indeed, we have𝑓(𝑃₂)=𝜆₁𝑃₁𝑃₂+𝜆₃𝑃₂𝑃₃≥(𝜆₂+𝜆₃)𝑃₁𝑃₂+𝜆₃𝑃₂𝑃₃=𝜆₂𝑃₁𝑃₂+𝜆₃(𝑃₁𝑃₂+𝑃₂𝑃₃). (5.2) By the triangle inequality, applied to 𝑃₁𝑃₂𝑃₃, we have 𝑃₁𝑃₂+𝑃₂𝑃₃>𝑃₁𝑃₃. Therefore, the last expression is strictly greater than𝜆₂𝑃₁𝑃₂+𝜆₃𝑃₁𝑃₃=𝑓(𝑃₁). (5.3) This shows that 𝑓(𝑃₂)>𝑓(𝑃₁). One shows analogously that 𝑓(𝑃₃)>𝑓(𝑃₁).

The other two possibilities of failure of (2.11) are discussed analogously; this leads to the following result.

Theorem 5.1. If 𝜆₁≥𝜆₂+𝜆₃, then the absolute minimum of 𝑓(𝑃) is attained at 𝑃₁ and only at 𝑃₁. Similarly, if 𝜆₂≥𝜆₁+𝜆₃, the minimum is attained at 𝑃₂, and if 𝜆₃≥𝜆₁+𝜆₂, the minimum is attained at 𝑃₃.

Case 2. Assume now that the triangle inequalities (2.11) hold, but one of the conditions (2.42) fails to hold. We claim that only one of these conditions can fail.
Indeed, if we had, say, both𝛼₁≥𝜋−𝜃₁,𝛼₂≥𝜋−𝜃₂, (5.4)then we would have both𝛼₁+𝜃₁𝛼≥𝜋,₂+𝜃₂≥𝜋. (5.5)Adding up, we would get (𝛼₁+𝛼₂)+(𝜃₁+𝜃₂)≥2𝜋, (5.6) which is impossible, since 𝛼₁+𝛼₂+𝛼₃<𝜋 and 𝜃₁+𝜃₂+𝜃₃<𝜋. So, only one of the inequalities in (2.42) can fail to hold.
If we have, for example,𝛼₁≥𝜋−𝜃₁, (5.7) then we will have both 𝛼₂<𝜋−𝜃₂ and 𝛼₃<𝜋−𝜃₃. This implies, by Lemma 2.5, the remark following it, and Theorem 2.7, that the minimum must be attained at 𝑃₁.
The other two possibilities of failure of (2.11) are discussed similarly. The following result summarizes our discussion.

Theorem 5.2. If conditions (2.11) hold and 𝛼₁≥𝜋−𝜃₁, then the absolute minimum of 𝑓(𝑃) is attained at 𝑃₁. Similarly, if 𝛼₂≥𝜋−𝜃₂, then the minimum is attained at 𝑃₂, and if 𝛼₃≥𝜋−𝜃₃, the minimum is attained at 𝑃₃.

6. The Classical Case

The classical Fermat triangle problem happens when𝜆₁=𝜆₂=𝜆₃. (6.1) Then, the triangles in Figures 3 and 4 are equilateral, and therefore𝜃₁=𝜃₂=𝜃₃=60^∘. (6.2)Also, all the right-hand sides in (2.12) are equal to −1/2. Conditions (2.42) become𝛼_𝑖<120^∘for𝑖=1,2,3. (6.3) From here one can easily deduce the classical result stated in the introduction, especially as discussed in [3].

7. The Fermat Gradient System

Assume conditions (2.42) hold. As we observed before, the gradient (2.10) of the weighted distance sum 𝑓(𝑥,𝑦) given by (2.1) is defined in all of Ω=ℝ²⧵{𝑃₁,𝑃₂,𝑃₃}. Since the function 𝑓(𝑥,𝑦) has a global minimum at the optimal point 𝑃, the trajectories in Ω of the gradient system(𝑥̇,𝑦)̇=−∇𝑓(𝑥,𝑦) (7.1) will converge to the asymptotically stable equilibrium 𝑃. This follows immediately from the fact that 𝑉(𝑥,𝑦)=−‖∇𝑓(𝑥,𝑦)‖² is a global Lyapunov function for the system on Ω (see, e.g., [10, Section 9.3]). Moreover, the trajectories of (7.1) are orthogonal to the level curves of the weighted sum 𝑓(𝑥,𝑦). Figure 13 illustrates the situation for a more or less randomly chosen triangle, for the classical Fermat problem, when all the weights 𝜆_𝑖 coincide. We have depicted the direction field of the gradient system plus several trajectories. The closed lines are the level curves of 𝑓.

Figure 13 

Several intriguing questions arise. For example, when the triangle is equilateral, a single level curve will be tangent to all three sides as shown in Figure 14.

Figure 14 

In general, under which condition(s) is there a level curve simultaneously tangent to two sides? To all three sides? For a generic triangle, and for the generalized Fermat problem, is it always possible to pick the weights 𝜆_𝑖 so that a single level curve of 𝑓 will be tangent to all three sides?