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Satellite page for ☞ Fermat-Torricelli problem and its development

**Problem.** Find the values for the
weights
with the aim for the corresponding objective function to posses a minimum point precisely at .

Th

**Theorem** [3]. *Let the vertices of the triangle* *be counted counterclockwise. Then for the choice*

*the function*

*has its stationary point at* . *Provided that the latter is chosen inside the triangle*
*the values* *are all positive and*

**Proof**. Substitute the values for the weights into the equations for partial derivatives for :

We get

In order to demonstrate that both of these expressions equal to , let us utilize some properties of the determinant. Represent the first sum as the -th order determinant:

Add now the second row to the last one:

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 :

To prove the equivalence of this expression to the one from the statement of the theorem, let us split it into the -part and the -part. First, keep the -terms in brackets of the previous formula:

Similar to the proof of the first part of the theorem, represent this linear combination as the -th order determinant as follows

Multiply the first row by , the second one by and add the obtained rows to the last one:

In exactly the same manner the equality

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

equals twice the area of the triangle . The equalities

appeared in the proof of the theorem, are equivalent to the known geometric property [1]:

Geometrical meaning of the determinant

is connected with

which is known as **the power of the point** **with respect to the circle** through the points and (the circumscribed circle of the triangle)
[2]. If one denotes the circumcenter of the triangle by then

for

and, provided that lies inside this triangle, the value is negative.

Results of the previous section can evidently be extended to the case of points in . Let the points be noncoplanar, and be counted in such a manner that the value of the determinant

is positive.

Т

**Теорема** [3]. *Set*

*i.e.,* *where*
*equals the determinant obtained on replacing the* *-th column of * *by the column*1) . *For these values of weights, the function*

*has its stationary point at * .
*If* *lies inside the tetrahedron* , *then
the values * * are all positive, and*

**Proof** is similar to that of the planar case.

Geometrical meanings of the values are similar to their counterparts from the planar case. The value equals six times the volume of tetrahedron , while the value

is known as **the power of the point** **with respect to a sphere circumscribed to that tetrahedron** [2]; it is equivalent to

for

where stands for the circumcenter of the tetrahedron. If lies inside the tetrahedron, then is negative.

Results of the previous section can evidently be extended to the case of points in .

Th

**Theorem** [4]. *Let the points* *be noncoplanar and
be counted in such a manner that the value*

*is positive. Denote by* *the determinant obtained on replacing the* -*th column of* *by the column*2) .
*Then for the choice*

*the function*

*has its stationary point at* . *If* *lies inside the simplex* , *then
the values * *are all positive, and*

[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.