Problem. Find the values for the weights $ \{m_j\}_{j=1}^{K} $ with the aim for the corresponding objective function $ \sum_{j=1}^{K} m_j |PP_j| $ to posses a minimum point precisely at $ P_{\ast} \not\in \{P_j\}_{j=1}^K $.

Theorem [3]. Let the vertices of the triangle $ P_{1}P_{2}P_3 $ be counted counterclockwise. Then for the choice

$$ m_1^{\ast} = |P_{\ast}P_1| \cdot \left| \begin{array}{ccc} 1 & 1 & 1 \\ x_{\ast} & x_2 & x_3 \\ y_{\ast} & y_2 & y_3 \end{array} \right|, \ m_2^{\ast} = |P_{\ast}P_2| \cdot \left| \begin{array}{ccc} 1 & 1 & 1 \\ x_1 & x_{\ast} & x_3 \\ y_1 & y_{\ast} & y_3 \end{array} \right|,\ m_3^{\ast} = |P_{\ast}P_3| \cdot \left| \begin{array}{ccc} 1 & 1 & 1 \\ x_1 & x_2 & x_{\ast} \\ y_1 & y_2 & y_{\ast} \end{array} \right| $$ the function $$ F(x,y) = \sum_{j=1}^3 m_{j}^{\ast} \sqrt{(x-x_j)^2+(y-y_j)^2} $$ has its stationary point at $ P_{\ast} $. Provided that the latter is chosen inside the triangle $ P_1P_{2}P_3 $ the values $ m_1^{\ast},m_2^{\ast},m_3^{\ast} $ are all positive and $$ F(x_{\ast},y_{\ast})=\min_{(x,y)\in \mathbb R^2} F(x,y)= \left| \begin{array}{cccc} 1 & 1 & 1 & 1\\ x_{\ast} & x_1 & x_2 & x_3 \\ y_{\ast} & y_1 & y_2 & y_3 \\ x_{\ast}^2+y_{\ast}^2 & x_1^2+y_1^2 & x_2^2+y_2^2 & x_3^2+y_3^2 \end{array} \right| \ . $$

Proof. Substitute the values for the weights into the equations for partial derivatives for $ F_{} $: $$ \left\{ \begin{array}{ll} \displaystyle \frac{\partial F}{\partial x} &= \displaystyle \frac{m_1^{\ast}(x_{\ast}-x_1)}{\sqrt{(x_{\ast}-x_1)^{2}+(y_{\ast}-y_1)^2}}+\frac{m_2^{\ast}(x_{\ast}-x_2)}{\sqrt{(x_{\ast}-x_1)^{2}+(y_{\ast}-y_1)^2}}+ \frac{m_3^{\ast}(x_{\ast}-x_3)}{\sqrt{(x_{\ast}-x_3)^2+(y_{\ast}-y_3)^2}}, \\ \displaystyle \frac{\partial F}{\partial y} &= \displaystyle \frac{m_1^{\ast}(y_{\ast}-y_1)}{\sqrt{(x_{\ast}-x_1)^{2}+(y_{\ast}-y_1)^2}}+\frac{m_2^{\ast}(y_{\ast}-y_2)}{\sqrt{(x_{\ast}-x_1)^{2}+(y_{\ast}-y_1)^2}}+ \frac{m_3^{\ast}(y_{\ast}-y_3)}{\sqrt{(x_{\ast}-x_3)^2+(y_{\ast}-y_3)^2}}. \end{array} \right. $$ We get $$ (x_{\ast}-x_1)\left| \begin{array}{lll} 1 & 1 & 1 \\ x_{\ast} & x_2 & x_3 \\ y_{\ast} & y_2 & y_3 \end{array} \right|+ (x_{\ast}-x_2) \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_{\ast} & x_3 \\ y_1 & y_{\ast} & y_3 \end{array} \right| + (x_{\ast}-x_3) \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_2 & x_{\ast} \\ y_1 & y_2 & y_{\ast} \end{array} \right| \ , $$ $$ (y_{\ast}-y_1)\left| \begin{array}{lll} 1 & 1 & 1 \\ x_{\ast} & x_2 & x_3 \\ y_{\ast} & y_2 & y_3 \end{array} \right|+ (y_{\ast}-y_2) \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_{\ast} & x_3 \\ y_1 & y_{\ast} & y_3 \end{array} \right|+ (y_{\ast}-y_3) \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_2 & x_{\ast} \\ y_1 & y_2 & y_{\ast} \end{array} \right| \ . $$ In order to demonstrate that both of these expressions equal to $ 0_{} $, let us utilize some properties of the determinant. Represent the first sum as the $ 4 $-th order determinant: $$ \left| \begin{array}{cccc} 1 & 1 & 1 & 1\\ x_{\ast} & x_1 & x_2 & x_3 \\ y_{\ast} & y_1 & y_2 & y_3 \\ 0 & x_{\ast}-x_1 & x_{\ast}-x_2 & x_{\ast}-x_3 \end{array} \right| $$ Add now the second row to the last one: $$ \left| \begin{array}{llll} 1 & 1 & 1 & 1\\ x_{\ast} & x_1 & x_2 & x_3 \\ y_{\ast} & y_1 & y_2 & y_3 \\ x_{\ast} & x_{\ast} & x_{\ast} & x_{\ast} \end{array} \right| \ . $$ In this determinant, the first row is proportional to the last one; therefore, the determinant equals just zero. The second equality can be verified in a similar manner.

Let us evaluate $ F(x_{\ast},y_{\ast}) $: $$ F(x_{\ast},y_{\ast}) = $$ $$ = \left[(x_{\ast}-x_1)^2+(y_{\ast}-y_1)^2 \right] \left| \begin{array}{lll} 1 & 1 & 1 \\ x_{\ast} & x_2 & x_3 \\ y_{\ast} & y_2 & y_3 \end{array} \right| + \left[(x_{\ast}-x_2)^2+(y_{\ast}-y_2)^2 \right] \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_{\ast} & x_3 \\ y_1 & y_{\ast} & y_3 \end{array} \right| $$ $$ + \left[(x_{\ast}-x_3)^2+(y_{\ast}-y_3)^2 \right] \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_2 & x_{\ast} \\ y_1 & y_2 & y_{\ast} \end{array} \right| \ . $$ To prove the equivalence of this expression to the one from the statement of the theorem, let us split it into the $ x_{} $-part and the $ y_{} $-part. First, keep the $ x_{} $-terms in brackets of the previous formula: $$ (x_{\ast}-x_1)^2 \left| \begin{array}{lll} 1 & 1 & 1 \\ x_{\ast} & x_2 & x_3 \\ y_{\ast} & y_2 & y_3 \end{array} \right| + (x_{\ast}-x_2)^2 \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_{\ast} & x_3 \\ y_1 & y_{\ast} & y_3 \end{array} \right| + (x_{\ast}-x_3)^2 \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_2 & x_{\ast} \\ y_1 & y_2 & y_{\ast} \end{array} \right| \ . $$ Similar to the proof of the first part of the theorem, represent this linear combination as the $ 4 $-th order determinant as follows $$ \left| \begin{array}{cccc} 1 & 1 & 1 & 1\\ x_{\ast} & x_1 & x_2 & x_3 \\ y_{\ast} & y_1 & y_2 & y_3 \\ 0 & (x_{\ast}-x_1)^2 & (x_{\ast}-x_2)^2 & (x_{\ast}-x_3)^2 \end{array} \right| \ . $$ Multiply the first row by $ (-x_{\ast}^2) $, the second one by $ 2\, x_{\ast} $ and add the obtained rows to the last one: $$ \left| \begin{array}{cccc} 1 & 1 & 1 & 1\\ x_{\ast} & x_1 & x_2 & x_3 \\ y_{\ast} & y_1 & y_2 & y_3 \\ x_{\ast}^2 & x_1^2 & x_2^2 & x_3^2 \end{array} \right| \ . $$ In exactly the same manner the equality $$ (y_{\ast}-y_1)^2 \left| \begin{array}{lll} 1 & 1 & 1 \\ x_{\ast} & x_2 & x_3 \\ y_{\ast} & y_2 & y_3 \end{array} \right| + (y_{\ast}-y_2)^2 \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_{\ast} & x_3 \\ y_1 & y_{\ast} & y_3 \end{array} \right| +(y_{\ast}-y_3)^2 \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_2 & x_{\ast} \\ y_1 & y_2 & y_{\ast} \end{array} \right| $$ $$ =\left| \begin{array}{cccc} 1 & 1 & 1 & 1\\ x_{\ast} & x_1 & x_2 & x_3 \\ y_{\ast} & y_1 & y_2 & y_3 \\ y_{\ast}^2 & y_1^2 & y_2^2 & y_3^2 \end{array} \right| $$ can be proven. The linear property of determinant with respect to its rows completes the proof of the last statement of the theorem. ♦

The value $$ \left| \begin{array}{ccc} 1 & 1 & 1 \\ x_{\ast} & x_2 & x_3 \\ y_{\ast} & y_2 & y_3 \end{array} \right| $$ equals twice the area of the triangle $ P_{\ast} P_2P_3 $.

The equalities $$ (x_{\ast}-x_1)\left| \begin{array}{lll} 1 & 1 & 1 \\ x_{\ast} & x_2 & x_3 \\ y_{\ast} & y_2 & y_3 \end{array} \right|+ (x_{\ast}-x_2) \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_{\ast} & x_3 \\ y_1 & y_{\ast} & y_3 \end{array} \right|+ (x_{\ast}-x_3) \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_2 & x_{\ast} \\ y_1 & y_2 & y_{\ast} \end{array} \right|=0 \ , $$ $$ (y_{\ast}-y_1)\left| \begin{array}{lll} 1 & 1 & 1 \\ x_{\ast} & x_2 & x_3 \\ y_{\ast} & y_2 & y_3 \end{array} \right|+ (y_{\ast}-y_2) \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_{\ast} & x_3 \\ y_1 & y_{\ast} & y_3 \end{array} \right|+ (y_{\ast}-y_3) \left| \begin{array}{lll} 1 & 1 & 1 \\ x_1 & x_2 & x_{\ast} \\ y_1 & y_2 & y_{\ast} \end{array} \right|=0 \ , $$ appeared in the proof of the theorem, are equivalent to the known geometric property [1]: $$ \overrightarrow{P_{\ast}P_1} \cdot S_{_{\triangle P_{\ast}P_2P_3}}+\overrightarrow{P_{\ast}P_2} \cdot S_{_{\triangle P_{\ast}P_3P_1}}+\overrightarrow{P_{\ast}P_3} \cdot S_{_{\triangle P_{\ast}P_1P_2}}=\overrightarrow{\mathbb O} \ . $$ Geometrical meaning of the determinant $$ \left| \begin{array}{cccc} 1 & 1 & 1 & 1\\ x_{\ast} & x_1 & x_2 & x_3 \\ y_{\ast} & y_1 & y_2 & y_3 \\ x_{\ast}^2+y_{\ast}^2 & x_1^2+y_1^2 & x_2^2+y_2^2 & x_3^2+y_3^2 \end{array} \right| $$ is connected with $$ h=-\frac{1}{S} \left| \begin{array}{cccc} 1 & 1 & 1 & 1\\ x_{\ast} & x_1 & x_2 & x_3 \\ y_{\ast} & y_1 & y_2 & y_3 \\ x_{\ast}^2+y_{\ast}^2 & x_1^2+y_1^2 & x_2^2+y_2^2 & x_3^2+y_3^2 \end{array} \right| \quad for \quad S= \left| \begin{array}{cccc} 1 & 1 & 1\\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{array} \right| , $$ which is known as the power of the point $ P_{\ast} $ with respect to the circle through the points $ P_1, P_2, $ and $ P_3 $ (the circumscribed circle of the triangle) [2]. If one denotes the circumcenter of the triangle $ P_1P_2P_3 $ by $ C_{} $ then $$ h=|CP_{\ast}|^2-|CP_j|^2 \quad \mbox{ for } \ j \in \{1,2,3\} \ , $$ and, provided that $ P_{\ast} $ lies inside this triangle, the value $ h_{} $ is negative.

Results of the previous section can evidently be extended to the case of $ K=4 $ points in $ \mathbb R^{3} $. Let the points $ \{P_j=(x_j,y_j,z_j)\}_{j=1}^4 $ be noncoplanar, and be counted in such a manner that the value of the determinant $$ V=\left| \begin{array}{cccc} 1 & 1 & 1 & 1 \\ x_1 & x_2 & x_3 & x_4 \\ y_1 & y_2 & y_3 & y_4 \\ z_1 & z_2 & z_3 & z_4 \end{array} \right| $$ is positive.

Теорема [3]. Set

$$ m_1^{\ast}= |P_{\ast}P_1|\cdot \left| \begin{array}{cccc} 1 & 1 & 1 & 1 \\ x_{\ast} & x_2 & x_3 & x_4 \\ y_{\ast} & y_2 & y_3 & y_4 \\ z_{\ast} & z_2 & z_3 & z_4 \end{array} \right|,\ m_2^{\ast}= |P_{\ast}P_2|\cdot \left| \begin{array}{cccc} 1 & 1 & 1 & 1 \\ x_1 & x_{\ast} & x_3 & x_4 \\ y_1 & y_{\ast} & y_3 & y_4 \\ z_1 & z_{\ast} & z_3 & z_4 \end{array} \right|,\ \dots ; $$ i.e., $ m_j^{\ast}=|P_{\ast}P_j| V_j $ where $ V_j $ equals the determinant obtained on replacing the $ j_{} $-th column of $ V_{} $ by the column¹⁾ $ \left[1,x_{\ast},y_{\ast},z_{\ast} \right]^{\top} $. For these values of weights, the function $$ F(x,y,z) = \sum_{j=1}^4 m_{j}^{\ast}\sqrt{(x-x_j)^2+(y-y_j)^2+(z-z_j)^2} $$ has its stationary point at $ P_{\ast}=(x_{\ast},y_{\ast},z_{\ast}) $. If $ P_{\ast} $ lies inside the tetrahedron $ P_1P_2P_3P_4 $, then the values $ \{m_{j}^{\ast} \} $ are all positive, and $$ F(x_{\ast},y_{\ast},z_{\ast})=\min_{(x,y,z)\in \mathbb R^3} F(x,y,z) $$ $$ =-\ \left| \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ x_{\ast} & x_1 & x_2 & x_3 & x_4 \\ y_{\ast} & y_1 & y_2 & y_3 & y_4 \\ z_{\ast} & z_1 & z_2 & z_3 & z_4 \\ x_{\ast}^2+y_{\ast}^2+z_{\ast}^2 & x_1^2+y_1^2+z_1^2 & x_2^2+y_2^2+z_2^2 & x_3^2+y_3^2+z_3^2 & x_4^2+y_4^2+z_4^2 \end{array} \right| \ . $$

Proof is similar to that of the planar case.

Geometrical meanings of the values $ V,\{V_j\}, F(x_{\ast},y_{\ast},z_{\ast}) $ are similar to their counterparts from the planar case. The value $ V_{} $ equals six times the volume of tetrahedron $ P_1P_2P_3P_4 $, while the value $$ -\frac{1}{V}\ \left| \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ x_{\ast} & x_1 & x_2 & x_3 & x_4 \\ y_{\ast} & y_1 & y_2 & y_3 & y_4 \\ z_{\ast} & z_1 & z_2 & z_3 & z_4 \\ x_{\ast}^2+y_{\ast}^2+z_{\ast}^2 & x_1^2+y_1^2+z_1^2 & x_2^2+y_2^2+z_2^2 & x_3^2+y_3^2+z_3^2 & x_4^2+y_4^2+z_4^2 \end{array} \right| $$ is known as the power of the point $ P_{\ast} $ with respect to a sphere circumscribed to that tetrahedron [2]; it is equivalent to $$ h=|CP_{\ast}|^2-|CP_j|^2 \quad \mbox{ for } \ j\in \{1,2,3,4 \} \ , $$ where $ C_{} $ stands for the circumcenter of the tetrahedron. If $ P_{\ast} $ lies inside the tetrahedron, then $ h_{} $ is negative.

Results of the previous section can evidently be extended to the case of $ K=n+1 $ points in $ \mathbb R^{n} $.

Theorem [4]. Let the points $ \{P_j=(x_{j1},\dots,x_{jn})\}_{j=1}^{n+1} $ be noncoplanar and be counted in such a manner that the value

$$ V= \left| \begin{array}{llll} 1 & 1 & \dots & 1\\ x_{11} & x_{21} & \dots & x_{n+1,1} \\ \vdots & \vdots & & \vdots \\ x_{1n} & x_{2n} & \dots & x_{n+1,n} \end{array} \right| $$ is positive. Denote by $ V_j $ the determinant obtained on replacing the $ j_{} $-th column of $ V_{} $ by the column²⁾ $ \left[1,x_{\ast 1},\dots,x_{\ast n} \right]^{\top} $. Then for the choice $$ \left\{m_j^{\ast} = |P_{\ast}P_j| V_j \right\}_{j=1}^{n+1} $$ the function $$ F(P) = \sum_{j=1}^{n+1} m_{j}^{\ast} |PP_j| $$ has its stationary point at $ P_{\ast}=(x_{\ast 1},\dots,x_{\ast n}) $. If $ P_{\ast} $ lies inside the simplex $ P_1P_2\dots P_{n+1} $, then the values $ \{ m_j^{\ast} \} $ are all positive, and $$ F(P_{\ast})= \displaystyle \sum_{j=1}^{n+1} V_j |P_{\ast}P_j|^2 =- \left| \begin{array}{lllll} 1 & 1 & \dots & 1 & 1 \\ x_{11} & x_{21} & \dots & x_{n+1,1} & x_{\ast 1} \\ \vdots & \vdots & & \vdots & \vdots \\ x_{1n} & x_{2n} & \dots & x_{n+1,n} & x_{\ast n} \\ \displaystyle \sum_{\ell=1}^n x_{1\ell}^2 & \displaystyle \sum_{\ell=1}^n x_{2\ell}^2 & \dots & \displaystyle \sum_{\ell=1}^n x_{n+1,\ell}^2 & \displaystyle \sum_{\ell=1}^n x_{\ast\ell}^2 \end{array} \right| \ . $$

[1].Prasolov V.V. Problems in Plane Geometry. Vol. 2. Nauka. Moscow, 1991, p. 10 (in Russian)

[2]. Uspensky J.V. Theory of Equations. New York. McGraw-Hill. 1948

[3]. Uteshev A.Yu. Analytical Solution for the Generalized Fermat-Torricelli Problem. Amer.Math.Monthly. V. 121, N 4, 318-331, 2014. Text ☞ HERE

[4]. Uteshev A.Yu., Yashina M.V. Stationary Points for the Family of Fermat-Torricelli-Coulomb-like potential functions. Proc. 15th Workshop CASC (Computer Algebra in Scientific Computing), Berlin 2013. Springer. LNCS. V.8136 , 2013, P. 412-426.