Analytical solution of the weighted Fermat-Torricelli problem

# Analytical solution of the weighted Fermat-Torricelli problem for tetrahedra: The case of two pairs of equal weights

Anastasios N. Zachos University of Patras, Department of Mathematics, GR-26500 Rion, Greece
###### Abstract.

The weighted Fermat-Torricelli problem for four non-collinear and non-coplanar points in states that:

Given four non-collinear and non-coplanar points and a positive real number (weight) which correspond to each point for find a fifth point such that the sum of the weighted distances to these four points is minimized. We present an analytical solution for the weighted Fermat-Torricelli problem for tetrahedra in for the case of two pairs of equal weights.

###### Key words and phrases:
weighted Fermat-Torricelli problem, weighted Fermat-Torricelli point, tetrahedra
###### 1991 Mathematics Subject Classification:
51E12, 52A10, 51E10

## 1. Introduction

Let be four non-collinear and non-coplanar points and a positive real number (weight) correspond to each point for

The weighted Fermat-Torricelli problem for four non-collinear points and non-coplanar points in states that:

###### Problem 1.

Find a unique (fifth) point which minimizes

 f(X)=4∑i=1Bi∥X−Ai∥,

where denotes the Euclidean distance and

The existence and uniqueness of the weighted Fermat-Torricelli point and a complete characterization of the solution of the weighted Fermat-Torricelli problem for tetrahedra has been established in [4, Theorem 1.1, Reformulation 1.2 page 58, Theorem 8.5 page 76, 77]).

###### Theorem 1.

,, Let there be given four non-collinear points and non-coplanar points with corresponding positive weights
(a) The weighted Fermat-Torricelli point exists and is unique.
(b) If for each point

 ∥4∑j=1,i≠jBj→u(Ai,Aj)∥>Bi, (1.1)

for holds, then
() the weighted Fermat-Torricelli point (weighted floating equilibrium point) does not belong to and
()

 4∑i=1Bi→u(A0,Ai)=→0, (1.2)

where is the unit vector from to for (Weighted Floating Case).
(c) If there is a point satisfying

 ∥4∑j=1,i≠jBj→u(Ai,Aj)∥≤Bi, (1.3)

then the weighted Fermat-Torricelli point (weighted absorbed point) coincides with the point (Weighted Absorbed Case).

We consider the following open problem:

###### Problem 2.

Find an analytic solution with respect to the weighted Fermat-Torricelli problem for tetrahedra in such that the corresponding weighted Fermat-Torricelli point is not any of the given points.

In this paper, we present an analytical solution for the weighted Fermat-Torricelli problem for regular tetrahedra in for and by expressing the objective function as a function of the linear segment which is formed by the middle point of the common perpendicular and and the corresponding weighted Fermat-Torricelli point (Section 2, Theorem 2). It is worth mentioning that this analytical solution of the weighted Fermat-Torricelli problem for a regular tetrahedron is a generalization of the analytical solution of the weighted Fermat-Torricelli point of a quadrangle (tetragon) in (see in ).

By expressing the angles for for as a function of and and taking into account the invariance property of the weighted Fermat-Torricelli point (geometric plasticity) in we obtain an analytical solution for some tetrahedra having the same weights with the regular tetrahedron (Section 3, Theorem 3).

## 2. The weighted Fermat-Torricelli problem for regular tetrahedra: The case B1=b2 and B3=b4.

We shall consider the weighted Fermat-Torricelli problem for a regular tetrahedron for and

We denote by the length of the linear segment by the common perpendicular of and where is the middle point of and is the middle point of by the weighted Fermat-Torricelli point of by the middle point of (), by the length of the linear segment and the angle for We set the edges of (Fig. 1).

###### Problem 3.

Given a regular tetrahedron and a weight which corresponds to the vertex for find a fifth point (weighted Fermat-Torricelli point) which minimizes the objective function

 f=B1a01+B2a02+B3a03+B4a04 (2.1)

for and

We set

 s≡−a6B121+2a6B101B24+a6B81B44−4a6B61B64+a6B41B84+2a6B21B104−a6B124+2√2 √(a12B221B24−8a12B201B44+29a12B181B64−64a12B161B84+98a12B141B104−112a12B121B124+ +98a12B101B144−64a12B81B164+29a12B61B184−8a12B41B204+a12B21B224) (2.2)

and

 t≡−a4B414s1/3+a4B21B242s1/3−a4B444s1/3−s1/34(B41−2B21B24+B44). (2.3)
###### Theorem 2.

The location of the weighted Fermat-Torricelli point of for and is given by:

 y=−√t2+ 12  ⎷a4B414s1/3−a4B21B242s1/3+a4B444s1/3+2(−8√2a3B21−8√2a3B24)√t(64B21−64B24)+s1/34(B41−2B21B24+B44)
###### Proof of Theorem 2:.

Taking into account the symmetry of the weights and for and the symmetries of the regular tetrahedron the objective function (2.13) of the weighted Fermat-Torricelli problem (Problem 3) could be reduced to an equivalent Problem: Find a point which belongs to the midperpendicular of and and minimizes the objective function

 f2=B1a01+B4a04. (2.5)

We express and as a function of

 a201=(a2)2+⎛⎜⎝a√222−y⎞⎟⎠2, (2.6)
 a202=(a2)2+⎛⎜⎝a√222−y⎞⎟⎠2, (2.7)
 a203=(a2)2+⎛⎜⎝a√222+y⎞⎟⎠2, (2.8)
 a204=(a2)2+⎛⎜⎝a√222+y⎞⎟⎠2, (2.9)

where the length of is

By replacing (2.6) and (2.9) in (2.5) we get:

 B1   ⎷(a2)2+⎛⎜⎝a√222−y⎞⎟⎠2+B4   ⎷(a2)2+⎛⎜⎝a√222+y⎞⎟⎠2→min. (2.10)

By differentiating (2.10) with respect to and by squaring both parts of the derived equation, we get:

 B21(a√222−y)2(a2)2+(a√222−y)2=B24(a√222+y)2(a2)2+(a√222+y)2 (2.11)

which yields

 64y4(B21−B24)−8√2a3y(B21+B24)+3a4(B21−B24)=0. (2.12)

By solving the fourth order equation (2.12) with respect to , we derive two complex solutions and two real solutions (see in  for the general solution of a fourth order equation with respect to ) which depend on and One of the two real solutions with respect to is (2). The real solution (2) gives the location of the weighted Fermat-Torricelli point at the interior of (see fig. 1).

We shall state the Complementary weighted Fermat-Torricelli problem for a regular tetrahedron ([2, pp. 358]), in order to explain the second real solution which have been obtained by (2.12) with respect to (see also in  for the case of a quadrangle).

###### Problem 4.

Given a regular tetrahedron and a weight (a positive or negative real number) which corresponds to the vertex for find a fifth point (weighted Fermat-Torricelli point) which minimizes the objective function

 f=B1a01+B2a02+B3a03+B4a04 (2.13)

for and

###### Proposition 1.

The location of the complementary weighted Fermat-Torricelli point (solution of Problem 4) of the regular tetrahedron for and is the exactly same with the location of the corresponding weighted Fermat-Torricelli point of for and

###### Proof of Proposition 1:.

Taking into account theorem 2, for we derive:

 →B1+→B2+→B3+→B4=→0 (2.14)

or

 (−→B1)+(−→B2)+(−→B3)+(−→B4)=→0. (2.15)

From (2.14) and (2.15), we derive that the complementary weighted Fermat-Torricelli point coincides with the weighted Fermat-Torricelli point We note that the vectors may change direction from to simultaneously, for

###### Proposition 2.

The location of the complementary weighted Fermat-Torricelli point (solution of Problem 4) of the regular tetrahedron for or and is given by:

 y=√t2+ +12  ⎷a4B414s1/3−a4B21B242s1/3+a4B444s1/3−2(−8√2a3B21−8√2a3B24)√t(64B21−64B24)+s1/34(B41−2B21B24+B44)
###### Proof of Proposition 2:.

Considering (2.10) for or and and differentiating (2.10) with respect to and by squaring both parts of the derived equation, we obtain (2.12) which is a fourth order equation with respect to The second real solution of gives (2). Taking into account the real solution (2) and the weighted floating equilibrium condition we obtain that the complementary weighted Fermat-Torricelli point for coincides with the complementary weighted Fermat-Torricelli point for ( Fig. 2 ). From (2), we derive that the complementary is located outside the regular tetrahedron (Fig. 2). Figure 2. The complementary weighted Fermat-Torricelli point A′0 of a regular tetrahedron A1A2A3A4 for B1=B2>0 and B3=B4<0 or B1=B2<0 and B3=B4>0 for ∥B1∥>∥B4∥

###### Example 1.

Given a regular tetrahedron in from (2) and (2) we get and respectively, with six digit precision. The weighted Fermat-Torricelli point and the complementary weighted Fermat-Torricelli point for and corresponds to The complementary weighted Fermat-Torricelli point for and or and lies outside the regular tetrahedron and corresponds to

We proceed by calculating the angles for

###### Proposition 3.

The angles for are given by:

 α102=arccos⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝1−a22⎛⎜⎝(a2)2+(a√222−y)2⎞⎟⎠⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠, (2.17)
 α304=arccos⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝1−a22⎛⎜⎝(a2)2+(a√222+y)2⎞⎟⎠⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠, (2.18)

and

 α104=α203=α103=α204=arccos(a√222−y)2+(a√222−y)2−a222 ⎷(a2)2+(a√222−y)2 ⎷(a2)2+(a√222+y)2. (2.19)
###### Proof of Proposition 3:.

Taking into account the cosine law in and (2), we obtain (2.17), (2.18) and (2.19), respectively.

###### Corollary 1.

[5, Theorem 4.3, p. 102] If then

 αi0j=arccos(−13), (2.20)

for and

###### Proof.

By setting in (2.17), (2.18) and (2.19), we obtain (2.20). ∎

## 3. The weighted Fermat-Torricelli problem for tetrahedra in the three dimensional Euclidean Space: The case B1=b2 and B3=b4.

We consider the following lemma which gives the invariance property (geometric plasticity) of the weighted Fermat-Torricelli point for a given tetrahedron in ([9, Appendix AII,pp. 851-853])

###### Lemma 1.

[9, Appendix AII,pp. 851-853] Let be a regular tetrahedron in and each vertex has a non-negative weight for Assume that the floating case of the weighted Fermat-Torricelli point occurs:

 ∥4∑j=1,i≠jBj→u(Ai,Aj)∥>Bi. (3.1)

If is connected with every vertex for and a point is selected with corresponding non-negative weight on the ray that is defined by the line segment and the tetrahedron is constructed such that:

 ∥4∑j=1,i≠jBj→u(A′i,A′j)∥>Bi, (3.2)

then the weighted Fermat-Torricelli point of is identical with

We consider a tetrahedron which has as a base the equilateral triangle with side and the vertex is located on the ray with corresponding non-negative weights at the vertices and at the vertices

Assume that we choose and non negative weights which satisfy the inequalities (3.1), (3.2) and which correspond to the weighted floating case of and

We denote by the length of the linear segment the angle for by the height of from to by the dihedral angle between the planes and and by the dihedral angle between the planes and and by the corresponding weighted Fermat-Torricelli point of the regular tetrahedron

###### Theorem 3.

The location of the weighted Fermat-Torricelli point of a tetrahedron which has as a base the equilateral triangle with side and the vertex is located on the ray for and under the conditions (3.1), (3.2) and is given by:

 a04′=√a220+a224′−2a24′(√a202−h20,12cosα124′+h0,12sinα124′cos(αg4′−α)) (3.3)

where

 α=arccosa202+a223−a2032a23−√a22−h20,12cosα123h0,12sinα123 (3.4)

and

 h0,12= ⎷4a201a202−(a201+a202−a212)24a212. (3.5)
###### Proof of Theorem 3:.

From lemma 1, we get Therefore, we get the relations (3.3) and (3.4) from a generalization of the cosine law in which has been introduced for tetrahedra in [8, Solution of Problem 1, Formulas (2.14) and (2.20),p. 116].

###### Remark 1.

We may consider a tetrahedron by placing on the ray defined by and is the corresponding weighted Fermat-Torricelli point. Taking into account lemma 1, we get

The author is sincerely grateful to Professor Dr. Vassilios G. Papageorgiou for his very valuable comments, many fruitful discussions and for bringing my attention to this particular problem.

## References

•  V. Boltyanski, H. Martini, V. Soltan, Geometric Methods and Optimization Problems, Kluwer, Dordrecht-Boston-London, 1999.
•  R. Courant and H. Robbins, What is Mathematics? Oxford University Press, New York, 1951.
•  Y. Kupitz and H. Martini, The Fermat-Torricelli point and isosceles tetrahedra. J. Geom. 49 (1994), no. 1-2, 150–162.
•  Y.S. Kupitz and H. Martini, Geometric aspects of the generalized Fermat-Torricelli problem, Bolyai Society Mathematical Studies. 6, (1997) , 55-127.
•  R. Noda, T. Sakai and M. Morimoto, Generalized Fermat’s problem, Canad. Math. Bull., 34, no.1, (1991) 96–104.
•  S. L. Shmakov, A universal method of solving quartic equations, International Journal of Pure and Applied Mathematics, 71, no. 2 (2011), 251-259.
•  E. Weiszfeld, Sur le point lequel la somme des distances de n points donnes est minimum, Tohoku Math. J.43, (1937), 355–386.
•  A.N. Zachos and G. Zouzoulas, The weighted Fermat-Torricelli problem for tetrahedra and an ”inverse” problem, J. Math. Anal. Appl, 353, (2009), 114–120.
•  A. Zachos and G. Zouzoulas, An evolutionary structure of pyramids in the three dimensional Euclidean Space , J. Convex Anal. 18, no. 3 (2011), 833–853.
•  A.N. Zachos, Analytical solution of the weighted Fermat-Torricelli problem for convex quadrilaterals in the Euclidean plane: The case of two pairs of equal weights ,arXiv:1406.2947v2, Submitted 11-06-2014, pp. 13.
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