# A fundamental property of the Fermat-Torricelli point for tetrahedra in the
three dimensional Euclidean Space

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

We prove the following fundamental property for the Fermat-Torricelli point
for four non-collinear and non-coplanar points forming a tetrahedron in
$\mathbb{R}^{3},$ which states that: The three bisecting lines having as a
common vertex the Fermat-Torricelli point formed by each pair of equal angles,
which are seen by the opposite edges of the tetrahedron meet perpendicularly
at the Fermat-Torricelli point. Furthermore, we give an alternative proof,
which is different from the one obtained by Bajaj and Mehlhos for the
unsolvability of the Fermat-Torricelli problem for tetrahedra in
$\mathbb{R}^{3}$ using only algebraic computations for some angles, which have
as a common vertex the Fermat-Torricelli point of the tetrahedron.

###### Key words and phrases:

Fermat-Torricelli point, tetrahedra

###### 2010 Mathematics Subject Classification:

Primary 51M14,51M20; Secondary 51M16

##  1\. Introduction

The Fermat Problem for four non-collinear and non-coplanar points
$A_{i}(x_{i},y_{i},z_{i})$ forming a tetrahedron $A_{1}A_{2}A_{3}A_{4}$ in
$\mathbb{R}^{3}$ states that ([2],[6], [8], [18]):

######  Problem 1 (The Fermat problem for $A_{1}A_{2}A_{3}A_{4}$ in
$\mathbb{R}^{3}$).

Find $A_{0}(x,y,z)$ in $\mathbb{R}^{3},$ such that:

$f(\\{A_{0}\\})=\sum_{i=1}^{4}\sqrt{(x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}}\to
min.$ (1.1)

The unsolvability of the Fermat-Torricelli problem for tetrahedra in
$\mathbb{R}^{3}$ has been proved by Bajaj, Mehlhos, Melzak and Cockane in
[3],[11], [12], by applying Galois theory in some specific examples.
Therefore, there is no Euclidean construction to locate the Fermat-Torricelli
point $A_{0}$ of $A_{1}A_{2}A_{3}A_{4}$ in $\mathbb{R}^{3}.$

It is worth mentioning that Synge ([17]) was the first who gave a non-
Euclidean construction for the Fermat-Torricelli point using some spindles. He
considered around the two opposite edges of the tetrahedron and created two
isosceles triangles containing two angles $\pi-\alpha_{102}$ and
$\pi-\alpha_{304}$ and by rotating two circular arcs having as chords these
skew edges he showed that there is a common value such that
$\alpha_{102}=\alpha_{304},$ whuch yields a unique touching point (Fermat-
Torricelli point) of the two spindles. Rubinstein, Thomas and Weng ([13]) use
a more specific construction to find a Steiner tree having two Fermat-
Torricelli points for a tetrahedron in $\mathbb{R}^{3},$ which is based on the
Simpson line formed by the two vertices of two equilateral triangles with side
lengths the two skew edges and located at the exterior of the tetrahedron.
Recently, we use a similar construction with Synge ([20, Theorem 5]) by
constructing some isosceles triangles at the exterior of the skew edges of the
tetrahedron, in order locate the Fermat-Torricelli point inside a tetrahedron
in $\mathbb{R}^{3}.$

Kupitz, Martini, Abu-Saymeh, Hajja proved various properties of the Fermat-
Torricelli point for some specific classes of tetrahedra having their opposite
edges equal (isosceles tetrahedra) ([6],[2]).

It is well known that the existence and uniqueness of the Fermat point $A_{0}$
in $\mathbb{R}^{3}$ is derived by the convexity of the Euclidean norm
(distance) and compactness arguments.

Sturm and Lindelof gave a complete characterization of the solutions of the
Fermat problem for $m$ given points in $\mathbb{R}^{n}$ ([16],[9]). Kupitz and
Martini gave an alternative proof by using subdifferential calculus ([7],
[8]). Eriksson and Noda Sakai Morimoto discovered some new characterizations
for the Fermat-Torricelli point for tetrahedra in $\mathbb{R}^{3}$ (([5],
[10]).

We shall focus on the characterization of solutions for $m=4,$ $n=3$ (([7],
[8])) Let $\\{A_{1},A_{2},A_{3},A_{4}\\}$ be a tetrahedron and $A_{0}$ be a
point in $\mathbb{R}^{3}.$ We denote by $\vec{u}(A_{j},A_{i})$ the unit vector
from $A_{j}$ to $A_{i}$ for $i,j=0,1,2,3,4.$

Two cases may occur:

(I) If for each point $A_{i}\in\\{A_{1},A_{2},A_{3},A_{4}\\}$

$\|\sum_{j=1,j\neq i}^{4}\vec{u}(A_{j},A_{i})\|>1,$

for $i,j=1,2,3,4,$ then

(a) $A_{0}$ does not belong to $\\{A_{1},A_{2},A_{3},A_{4}\\},$

(b) $\sum_{i=1}^{4}\vec{u}(A_{0},A_{i})=\vec{0}$

(Fermat-Torricelli solution).

(II) If there is a point $A_{i}\in\\{A_{1},A_{2},A_{3},A_{4}\\}$ satisfying

$\|{\sum_{j=1,j\neq i}^{4}\vec{u}(A_{j},A_{i})}\|\leq 1.$

for $i,j=1,2,3,4,$ then $A_{0}\equiv A_{i}$

(Fermat-Cavallieri solution).

Hence, we get two characterization of solutions for the Fermat problem for
$A_{1}A_{2}A_{3}A_{4}$ in $\mathbb{R}^{3}.$

The Fermat-Torricelli solution is a tree, which consists of the quadruple of
line segments $\\{A_{0}A_{1},A_{0}A_{2},A_{0}A_{3},A_{0}A_{4}\\}.$

The Fermat-Cavallieri tree solution is a tree, which consists of the triad of
line segments $\\{A_{i}A_{j},A_{i}A_{k},A_{i}A_{l}\\},$ for
$i,j,k,l=1,2,3,4,i\neq j\neq k\neq l.$

Abu-Abas, Abu-Saymeh and Hajja proved the non-isogonal property of the Fermat-
Torricelli point in $\mathbb{R}^{3}$ ([1], [2])

In this paper, we prove a fundamental property of the Fermat-Torricelli point
for tetrahedra in $\mathbb{R}^{3},$ by using basic algebra of vectors, which
are transformed in spherical coordinates in $\mathbb{R}^{3}$ and we give an
alternative proof for the unsolvability of the Fermat-Torricelli point for
tetrahedra in $\mathbb{R}^{3},$ by obtaining an implicit expression for two
angles having as a common vertex the Fermat-Torricelli point. Our main results
are:

Main Result 1.

The three bisecting lines having as a common vertex the Fermat-Torricelli
point $A_{0}$ and formed by each pair of equal angles, which are seen by the
opposite edges of $A_{1}A_{2}A_{3}A_{4}$ meet perpendicularly at $A_{0}$
(Section 2, Theorem 1)

Main Result 2.

The Fermat-Torricelli problem for four non-collinear and non-coplanar points
forming a tetrahedron in $\mathbb{R}^{3}$ is not in general solvable by
Euclidean constructions (Section 3, Theorem 2).

##  2\. A fundamental property of the Fermat-Torricelli point for tetrahedra
in $\mathbb{R}^{3}$

Let $A_{1}A_{2}A_{3}A_{4}$ be a tetrahedron and $A_{0}$ be the Fermat-
Torricelli point inside $A_{1}A_{2}A_{3}A_{4}$ in $\mathbb{R}^{3}$ We denote
by $a_{i,j0k}$ the angle that is formed by the line segment that connects
$A_{0}$ with the trace of the orthogonal projection $A_{i}$ to the plane
defined by the triangle $\triangle A_{j}A_{0}A_{k}$ with the line segment
$A_{i}A_{0}.$ We set $\alpha_{i0j}\equiv\angle A_{i}A_{0}A_{j}.$

We need the following well known lemma ([11],[17]), in order to prove the main
result (Theorem 1):

######  Lemma 1.

If

$\|\sum_{j=1,j\neq i}^{4}\vec{u}(A_{j},A_{i})\|>1,$

then

$\cos\alpha_{102}=\cos\alpha_{304},$ (2.1)

$\cos\alpha_{203}=\cos\alpha_{104},$ (2.2)

$\cos\alpha_{103}=\cos\alpha_{204}$ (2.3)

and

$1+\cos\alpha_{102}+\cos\alpha_{103}+\cos\alpha_{104}=0.$ (2.4)

###### Proof.

The inner product of the unit vectors $\vec{u}(A_{0},A_{i}),$
$\vec{u}(A_{0},A_{j})$ yields:

$\vec{u}(A_{0},A_{i})\cdot\vec{u}(A_{0},A_{j})=\cos\alpha_{i0j},$ (2.5)

for i,j=1,2,3,4.

Taking into account the balancing condition of unit vectors
$\vec{u}(A_{0},A_{i}),$ for $i=1,2,3,4,$ we get:

$\vec{u}(A_{0},A_{1})+\vec{u}(A_{0},A_{2})=-(\vec{u}(A_{0},A_{3})+\vec{u}(A_{0},A_{4})),$
(2.6)

$\vec{u}(A_{0},A_{2})+\vec{u}(A_{0},A_{3})=-(\vec{u}(A_{0},A_{1})+\vec{u}(A_{0},A_{4}))$
(2.7)

$\vec{u}(A_{0},A_{1})+\vec{u}(A_{0},A_{3})=-(\vec{u}(A_{0},A_{2})+\vec{u}(A_{0},A_{4}))$
(2.8)

$\vec{u}(A_{0},A_{4})=-\vec{u}(A_{0},A_{1})+\vec{u}(A_{0},A_{2})+\vec{u}(A_{0},A_{3})).$
(2.9)

By squaring both parts of (2.6), (2.7),(2.8) and taking into account (2.5), we
obtain (2.1), (2.2) and (2.3), respectively.

By squaring both parts of (2.9)and by substituting (2.5) in the derived
equation and taking into account (2.1), (2.2) and (2.3), we get (2.4).
Therefore, we derive that $\alpha_{102}=\alpha_{304},$
$\alpha_{203}=\alpha_{104}$ and $\alpha_{103}=\alpha_{204}.$

∎

######  Theorem 1.

The three bisecting lines having as a common vertex the Fermat-Torricelli
point $A_{0}$ and formed by each pair of equal angles, which are seen by the
opposite edges of $A_{1}A_{2}A_{3}A_{4}$ meet perpendicularly at $A_{0}.$

###### Proof.

We express the unit vectors $\vec{u}(A_{0},A_{i})$ for $i=1,2,3,4,$ using
spherical coordinates:

$\vec{u}(A_{0},A_{1})=(1,0,0),$ (2.10)

$\vec{u}(A_{0},A_{2})=(\cos\alpha_{102},\sin\alpha_{102},0),$ (2.11)

$\vec{u}(A_{0},A_{3})=(\cos a_{3,102}\cos\omega_{3,102},\cos
a_{3,102}\sin\omega_{3,102},\sin a_{3,102}),$ (2.12)

$\vec{u}(A_{0},A_{4})=(\cos a_{4,102}\cos\omega_{4,102},\cos
a_{4,102}\sin\omega_{4,102},\sin a_{4,102}).$ (2.13)

The unit vector $\vec{\delta}_{i0j}$ of the angle bisector that corresponds to
the angle $\alpha_{i0j}$ is given by:

$\vec{\delta}_{i0j}=\vec{u}(A_{0},A_{i})+\vec{u}(A_{0},A_{j})$ (2.14)

for $i,j=1,2,3,4,i\neq j.$

By replacing (2.10), (2.11), (2.12), (2.13) in (2.14), we get:

$\vec{\delta}_{102}=(1+\cos\alpha_{102},\sin\alpha_{102},0)$ (2.15)

$\vec{\delta}_{103}=(1+\cos a_{3,102}\cos\omega_{3,102},\cos
a_{3,102}\sin\omega_{3,102},\sin a_{3,102})$ (2.16)

$\vec{\delta}_{104}=(1+\cos a_{4,102}\cos\omega_{4,102},\cos
a_{4,102}\sin\omega_{4,102},\sin a_{4,102})$ (2.17)

$\vec{\delta}_{203}=(\cos\alpha_{102}+\cos
a_{3,102}\cos\omega_{3,102},\sin\alpha_{102}+\cos
a_{3,102}\sin\omega_{3,102},\sin a_{3,102})$ (2.18)

$\vec{\delta}_{204}=(\cos\alpha_{102}+\cos
a_{4,102}\cos\omega_{4,102},\sin\alpha_{102}+\cos
a_{4,102}\sin\omega_{4,102},\sin a_{4,102})$ (2.19)

$\displaystyle\vec{\delta}_{304}=(\cos a_{3,102}\cos\omega_{3,102}+\cos
a_{4,102}\cos\omega_{4,102},\cos a_{3,102}\sin\omega_{3,102}+$
$\displaystyle\cos a_{4,102}\sin\omega_{4,102},\sin a_{3,102}+\sin a_{4,102})$
(2.20)

Taking into account (2.15), (2.16), (2.17), (2.18), (2.19), (2), we obtain
that:

$\vec{\delta}_{102}\cdot\vec{\delta}_{203}=1+\cos\alpha_{102}+\cos\alpha_{103}+\cos\alpha_{203}.$
(2.21)

$\vec{\delta}_{102}\cdot\vec{\delta}_{103}=1+\cos\alpha_{102}+\cos\alpha_{103}+\cos\alpha_{203}.$
(2.22)

By applying Lemma 1 in (2.21), (2.22), we derive that:

$\vec{\delta}_{102}\cdot\vec{\delta}_{203}=\vec{\delta}_{102}\cdot\vec{\delta}_{103}=0,$

which yields that
$\vec{\delta}_{102}\perp\vec{\delta}_{203}\perp\vec{\delta}_{103}.$ Therefore,
$\vec{\delta}_{102},\vec{\delta}_{203},\vec{\delta}_{103}$ is an orthonormal
system of unit vectors.

Taking into account (2.15), (2.16), (2.17), (2.18), (2.19), (2), we obtain
that:

$\frac{\vec{\delta}_{102}}{|\vec{\delta}_{102}|}\cdot\frac{\vec{\delta}_{304}}{|\vec{\delta}_{304}|}=\frac{1}{\sqrt{2(1+\cos\alpha_{102})}}\frac{1}{\sqrt{2(1+\cos\alpha_{304})}}(-2(1+\cos\alpha_{102})).$
(2.23)

By replacing $\alpha_{304}=\alpha_{102}$ (Lemma 1) in (2.23), we derive that:

$\frac{\vec{\delta}_{102}}{|\vec{\delta}_{102}|}\cdot\frac{\vec{\delta}_{304}}{|\vec{\delta}_{304}|}=-1.$

Hence, the angle bisectors of the angles $\alpha_{102}$ and $\alpha_{304}$
belong to the same line.

By following the same process and by applying Lemma 1, we get:

$\frac{\vec{\delta}_{203}}{|\vec{\delta}_{203}|}\cdot\frac{\vec{\delta}_{104}}{|\vec{\delta}_{104}|}=-1$

and

$\frac{\vec{\delta}_{103}}{|\vec{\delta}_{103}|}\cdot\frac{\vec{\delta}_{204}}{|\vec{\delta}_{204}|}=-1,$

which yields that the angle bisectors of the angles $\alpha_{203}$ and
$\alpha_{104}$ belong to the same line and the angle bisectors of the angles
$\alpha_{103}$ and $\alpha_{204}$ belong to the same line, respectively. ∎

######  Remark 1.

We should not confuse the fundamental property of the Fermat-Torricelli tree
solution for tetrahedra with the Steiner tree solution for tetrahedra having
two nodes (Fermat-Torricelli points) in $\mathbb{R}^{3}.$ A specific case was
proved in [13, Theorem 3.1] for the Steiner tree problem, which states that:
if $A_{1}A_{2}A_{3}A_{4}$ is a centralized symmetric tetrahedron, then its
three Simpson lines meet perpendicularly at its center $A_{0}.$ Therefore, the
fundamental property of the Fermat-Torricelli tree for any tetrahedron
$A_{1}A_{2}A_{3}A_{4},$ where three bisecting lines meet perpendicularly is
much stronger than the one proved for Steiner trees for centralized symmetric
tetrahedra in $\mathbb{R}^{3}.$

##  3\. Unsolvability of the Fermat-Torricelli problem for tetrahedra in
$\mathbb{R}^{3}$ by Compass and Ruler

We need the following lemma, which gives the position of four rays, which meet
at a common point, in order to prove the unsolvability of the Fermat-
Torricelli problem for $A_{1}A_{2}A_{3}A_{4}$ in $\mathbb{R}^{3}$ by compass
and ruler.

######  Lemma 2.

[19, Proposition 1] The position of four line segments $A_{i}A_{0}$ which meet
at $A_{0}$ for $i=1,2,3,4$ depend on exactly five given angles $\alpha_{102},$
$\alpha_{103},$ $\alpha_{104},$ $\alpha_{203}$ and $\alpha_{204}.$

The sixth angle $\alpha_{304}$ is calculated by the following formula:

$\displaystyle\cos\alpha_{304}=\frac{1}{4}[4\cos\alpha_{103}(\cos\alpha_{104}-\cos\alpha_{102}\cos\alpha_{204})+$
$\displaystyle+2\left(b+2\cos\alpha_{203}\left(-\cos\alpha_{102}\cos\alpha_{104}+\cos\alpha_{204}\right)\right)]\csc{}^{2}\alpha_{102}$

where

$\displaystyle
b\equiv\sqrt{\prod_{i=3}^{4}\left(1+\cos\left(2\alpha_{102}\right)+\cos\left(2\alpha_{10i}\right)+\cos\left(2\alpha_{20i}\right)-4\cos\alpha_{102}\cos\alpha_{10i}\cos\alpha_{20i}\right)}$
. (3.2)

for $i,k,m=1,2,3,4,$ and $i\neq k\neq m.$

######  Theorem 2.

The Fermat-Torricelli problem for four non-collinear and non-coplanar points
forming a tetrahedron in $\mathbb{R}^{3}$ is not in general solvable by
Euclidean constructions.

###### Proof.

By replacing $\alpha_{304}=\alpha_{102},$ $\alpha_{104}=\alpha_{203}$ and
$\alpha_{103}=\alpha_{204}$ in (2) taken from Lemma 2, we derive the implicit
function $a_{102}=g(\alpha_{102},\alpha_{203}).$ Therefore, we cannot derive
in general from this implicit expression an explicit function of
$\alpha_{102}$ with respect to $\alpha_{203}.$ Hence, this functional
dependence is responsible for the unsolvability of the Fermat-Torricelli
problem for tetrahedra in $\mathbb{R}^{3}.$ ∎

##  4\. Concluding Remarks

By enriching Synge’s construction with the fundamental property of the Fermat-
Torricelli point for tetrahedra in $\mathbb{R}^{3},$ we may obtain the Fermat-
Torricelli tree sausage, in order to develop some models for the determination
of 3-D minimum-energy configurations for macromolecular structures such as
proteins and DNA, instead of working with the method of Steiner minimal trees
established by Stanton and J. Mc Gregor Smith and W. Smith ([15]),[14]).

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