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Fermat's point from five perspectives
a
Jungeun Park & Alfinio Flores
a
a
Department of Mathematical Sciences, University of Delaware,
Newark, USA
Published online: 18 Nov 2014.
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To cite this article: Jungeun Park & Alfinio Flores (2014): Fermat's point from five
perspectives, International Journal of Mathematical Education in Science and Technology, DOI:
10.1080/0020739X.2014.979894
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International Journal of Mathematical Education in Science and Technology, 2014
http://dx.doi.org/10.1080/0020739X.2014.979894
CLASSROOM NOTE
Fermat’s point from five perspectives
Jungeun Park and Alfinio Flores∗
Department of Mathematical Sciences, University of Delaware, Newark, USA
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(Received 7 March 2014)
The Fermat point of a triangle is the point such that minimizes the sum of the distances
from that point to the three vertices. Five approaches to study the Fermat point of
a triangle are presented in this article. First, students use a mechanical device using
masses, strings and pulleys to study the Fermat point as the one that minimizes the
potential energy of the system. Second, students use soap films between parallel planes
connecting three pegs. The tension on the film will be minimal when the sum of
distances is minimal. Third, students use an empirical approach, measuring distances in
an interactive GeoGebra page. Fourth, students use Euclidean geometry arguments for
two proofs based on the Torricelli configuration, and one using Viviani’s Theorem. And
fifth, the kinematic method is used to gain additional insight on the size of the angles
between the segments joining the Fermat point with the vertices.
Keywords: Fermat point; Torricelli configuration; soap films; minimal sum of distances
1. Introduction
The Fermat point of a triangle ABC is the point P such that the sum of the distances
PA + PB + PC from that point to the three vertices is minimal (Figure 1). This problem
can arise as a purely geometrical question, the way Fermat posed it in 1643, and Torricelli
(1608–1647) solved it,[1] or in contexts where reducing the sum of distances results in an
advantage in terms of effort or cost, for example, locating a power plant serving three cities
(assuming each city receives the same energy) or finding the place to anchor a boat that
will drop the cable to retrieve three treasury chests from the bottom of the sea (assuming
the chests have equal weights).[2]
Solving the problem for three points using a mechanical device goes back to Lamé and
Clapeyron in 1829.[3] The history of minimizing the sum of distances to a set of more
than three points starts with Gergonne in 1811.[1] The letter from 1836 from Carl Friedrich
Gauss to Schumacher about the minimum distance to four points describes Gauss’ idea
of extending Fermat’s problem to more than three points in terms of finding a network
of minimal length connecting the original points rather than finding a single point that
minimizes the sum of distances. Gauss introduces additional points to form such networks,
which later became known as Steiner points.[1]
This paper will deal mainly with triangles whose biggest angle is less than 120◦ . We
address the case where one of the angles is more than 120◦ in the Final Remarks section.
Interested readers may find various methods and proofs in other sources, including weighted
cases,[4] and more than three points.[5,p.194]
∗
Corresponding author. Email: [email protected]
C 2014 Taylor & Francis
2
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Figure 1.
Classroom Note
Sum of distances to three vertices.
In this article, students look at the properties of the point P inside a triangle and
the corresponding segments and angles from five perspectives. First, students explore the
problem using a mechanical device, using string and weights. Second, students use soap
films connecting three pins between parallel plates to represent the problem. Third, students
use an interactive web page made with GeoGebra to experiment and verify the result by
measuring. Fourth, students use classical Euclidean geometry arguments to prove properties
of Fermat’s point and related segments. Finally, students use the kinematic method to gain
additional insight about the angles of the segments joining Fermat’s point with the vertices.
By using different approaches and connections, students can develop a better understanding
of Fermat’s point and the properties of the segments joining this point with the vertices of
the triangle.
2. Minimizing the sum of distances to the vertices of a triangle
with a physical device
To find the point P inside the triangle ABC, such that the sum of the distances AP + BP + CP
is as small as possible, De Finetti [6] suggests using a physical argument. One way to find
this point is by drilling three holes on a rigid horizontal board corresponding to the positions
of the three points A, B and C (Figure 2). We tie three strings together at the movable point
P, pass one string through each one of the holes and hang equal masses from each string.
The strings move freely through the holes. For a computer simulation, see [7].
The system of masses will be in equilibrium when the potential energy is minimal,
that is, when the sum of distances from the masses to the floor is minimal. Therefore,
the sum of the distances from the vertices to the masses is maximal. This implies that
AP + BP + CP should be minimum, because the total length of the strings is constant.
Because the magnitudes of the three forces acting along AP, BP and CP are equal, the
resultant of two of them is on the angle bisector between them (Figure 3). And because the
resultant of these two forces is in equilibrium with the third force, the line of action of this
one is also the angle bisector. Thus PA is on the angle bisector of PB and PC, PB is on the
angle bisector of PA and PC, and PC is on the angle bisector of PA and PB. Therefore, each
pair of segments form an angle of 120◦ at P (Figure 4).
Polya [8,p.148] uses a triangle on a vertical plane, and pulleys instead of holes for the
same kind of argument. When the masses are hanged from the pulleys so that the angle
between the strings connecting the knot with the pulleys is 120◦ , the masses will remain in
equilibrium (see Figure 5). If one of the masses is moved up or down, the system will tend
to go back to the equilibrium position on its own. Because of friction, it may not completely
go back to form exactly angles of 120◦ .
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International Journal of Mathematical Education in Science and Technology
Figure 2.
3
Equal masses suspended from the vertices.
3. Using soap films to minimize sum of distances
Using soap films is a great way for students to visualize minimal surfaces.[9,10] To study
Fermat’s point with soap films, two parallel transparent plates are attached with three pins
perpendicular to the plates. The plates are submerged in a soap solution and slowly pulled
out from the liquid. A soap film will be formed connecting the three pins (Figure 6). The
tension along the soap film will be minimal when the total area of the film is minimal.
Figure 3.
Two vectors with equal magnitudes and their resultant.
4
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Figure 4.
Classroom Note
Sum of distances to the three vertices is minimal.
Because the width of the film is constant, this is equivalent to have a sum of distances that
is minimal. When seen from above, you can see that the films join at a point inside the
triangle and that the angle between the soap films is 120◦ (see Figure 7).
4. Interactive website to minimize sum of distances
Students can also experiment using GeoGebra and find the approximate position of the
point inside the triangle that minimizes the sum of the distances to the vertices (Figure 8).
Figure 5.
Equal masses hanging from three pulleys.
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International Journal of Mathematical Education in Science and Technology
Figure 6.
5
Soap films minimize sum of lengths.
Students will realize that when using experimental methods it is not always possible to get
exactly three angles of 120◦ ; however, Guven, Cekmez and Karatas [11] have pointed to
the importance of the use of dynamical geometry programs to facilitate the transition from
empirical data to deductive proof. An interactive web page where students can experiment
is available at http://www.geogebratube.org/student/m60133.
Figure 7.
Soap films viewed from above.
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6
Figure 8.
Classroom Note
Empirical verification at interactive website.
5. Using Euclidean geometry arguments to locate Fermat’s point
We will find the position of the point that minimizes the sum of distances to the three
vertices in two steps. We will first show that the point has to be on the line that connects C
with the outside vertex F of an equilateral triangle constructed on side AB (Figure 9). Let P
be a point in the inside of triangle ABC, and triangle BAF an equilateral triangle. Construct
the segment PB and let BPE be an equilateral triangle on this segment. Angle PBA is
congruent to angle EBF because side BE is BP rotated 60◦ and side BF is BA also rotated
60◦ . Because PB is congruent to EB (sides of an equilateral triangle) and BA is congruent
to BF, we have that triangles BPA and BEF are congruent (because of side-angle-side).
Therefore, segment PA is congruent to segment EF. The sum of distances PA + PB + PC is
equal to the sum of the lengths of the segments FE + EP + PC. This sum will be minimal
if points C, P, E and F are on the straight line CF. In the same way, the position of P that
minimizes the sum of distances is on the line that connects A with the outside vertex of
an equilateral triangle on BC (Figure 10). The intersection of these two lines is the desired
position.
We will now prove that the angle between segments AG and CF is 120◦ using a rotation
of 60◦ .[12] Rotate triangle ABG 60◦ counterclockwise around B. Segment BG goes to BC,
AB goes to FB and GA goes to CF. Therefore, the angle between GA and CF is 60◦ , and angle
APC is equal to 120◦ . In the same way, we can prove that angle APB is 120◦ . This means
that quadrilateral APBF is an inscribed quadrilateral. The same is true for quadrilaterals
APCH and BPCG (Figure 11). This means that the circumcircles of the equilateral triangles
on the sides of the original triangle intersect at Fermat’s point (Figure 12). The configuration
formed by the original triangle together with the three equilateral triangles on its sides is
known as Torricelli’s configuration. An interactive Torricelli configuration is available at
http://www.geogebratube.org/student/m66431.
We will now look at two alternative proofs. First, in Figure 13 angle EFB is congruent
to angle DAB (corresponding angles of congruent triangles BEF and BDA). Both cover
segment DB. Therefore, F and A are on the same circle, so that ADBF is a cyclical
quadrilateral. Thus angle ADF is congruent to angle ABF and angle FDB is congruent to
angle BAF. Therefore angle ADB is 120◦ .
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International Journal of Mathematical Education in Science and Technology
Figure 9.
7
Sum of distances is equal to length of path CDEF.
Similar proofs can be designed with dynamic geometry activities, which guide students through the process of exploring, making conjectures and proving their conjectures.[5,p.108–114] For example, De Villiers [5] created such activities with Geometer’s
Sketchpad, through which students make conjectures about the intersection of three segments connecting AG, CF and BF in Figure 11, and their lengths, and prove their conjectures
that the segments are equal length using the property of congruent triangles, and that the
segments intersect at one point using the property of cyclic quadrilaterals, inscribed in three
circles in Figure 12.[5,p.108–114]
Second, we can also use Viviani’s Theorem, which states that the sum of distances of
a point inside an equilateral triangle to its sides is equal to its height, to prove that the
point P that minimizes AP + BP + CP is the point where angle APB = angle BPC = angle
CPA.[13] In Figure 14, P is the point where angle APB = angle BPC = angle CPA = 120◦ .
Construct a perpendicular line to segment PA, and repeat the procedure for PB and PC.
Then, triangle EFG formed from this procedure is equilateral because each angle is 60◦
(e.g. the sum of three angles in quadrilateral CPBE is angle ECP + angle EBP + angle BPC
= 90◦ + 90◦ + 120◦ . Therefore, angle BEC = 60◦ ).[13]
Then, to complete the proof, we will pick another point P and show AP + BP + CP <
AP + BP + CP . Draw a line passing through P that is perpendicular to side FG, and
label the intersection point A . Similarly, construct B and C on GE and EF, respectively
(Figure 15). Then, because A P is the shortest distance between side FG and point P , A P ≤
AP , and similarly, B P ≤ BP and C P ≤ CP . Because P is different from P, not all three
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8
Figure 10.
Classroom Note
Location of point that minimizes the sum of distances.
equalities hold. Therefore, A P + B P + C P < AP + BP + CP . Viviani’s Theorem gives
AP + BP + CP = A P + B P + C P , and thus AP + BP + CP < AP + BP + CP .[13]
6. Using the kinematic method
The last approach is using the kinematic method. This method uses a constant relation
between two vector functions to find a similar relation between the velocities of the corresponding endpoints of the vectors. And vice versa, if we know that a relation holds for
the velocities of the endpoints for all times, we know that the same relation will hold
for the vector functions (except possibly by having to add a constant vector). This is similar
to the case of real functions. For example, if two functions f and g are related by f (x) =
kg(x) for all x, where k is a constant, then f (x) = kg (x) for all x; vice versa if f (x) = kg
(x) for all x, then f (x) = kg(x) + c, where c is a constant. We will use the case where a vector
function r1 (t) is identical to the rotation of another vector function r2 (t) by a constant angle
α. We will express this relation by r1 = rot(α)r2 . Let v1 (t) be the derivative of r1 (t), that
is
v1 (t) = lim
h→0
r1 (t + h) − r1 (t)
h
(1)
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International Journal of Mathematical Education in Science and Technology
Figure 11.
9
Cyclic quadrilaterals.
We will use the following two theorems.
Theorem 1: If r1 = rot(α)r2 for all times, then v1 = rot(α)v2 for all times (Figure 16).
Proof: We have
lim
h→0
lim
h→0
r1 (t + h) − r1 (t)
=
h
rot (α) r2 (t + h) − rot (α) r2 (t)
=
h
lim rot (α)
h→0
r2 (t + h) − r2 (t)
=
h
r2 (t + h) − r2 (t)
= rot (α) v2
h→0
h
rot (α) lim
Thus v1 = rot(α)v2 .
Theorem 2: Assume that the functions r1 and r2 are such that for all times their corresponding velocities satisfy v1 = rot(α)v2, then r1 = rot(α)r2 + k where k is a constant
vector (Figure 17).
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10
Figure 12.
Classroom Note
Intersection of circumcircles.
Proof: We will use the fact that if the derivative of a vector function is constant equal to
zero, then the function is constant. Consider the function r1 − rot(α)r2 . Its derivative is
v1 − rot(α)v2 . This is constant equal to zero, thus r1 − rot(α)r2 = k and r1 = rot(α)r2 + k.
(For other theorems for the kinematic method see [14] or [15].)
Students can use an interactive web page that illustrates the use of the kinematic method
in relation to Fermat’s point. It is available at http://www.geogebratube.org/student/m60129.
Triangles ACF and BCD are equilateral triangles on two sides of arbitrary triangle ABC
(Figure 18).
Vector AF is congruent with AC, and is rotated 60◦ counterclockwise. We can use
theorem 1 to say that the velocity of F has the same magnitude as the velocity of C but is
rotated 60◦ to the left. Vector BD is congruent to BC and is rotated 60◦ clockwise. Therefore,
by theorem 1 the velocity of D has the same magnitude as the velocity of C, but is rotated
60◦ to the right. Therefore, the velocity of D has the same magnitude as the velocity of F,
rotated 120◦ . We can now use theorem 2 to conclude that BF = rot(120◦ )AD + k. To see
that k = 0, we can start with triangle ABC as an equilateral triangle (Figure 19). In this case,
it is clear that BF and AD are medians/altitudes/perpendicular bisectors/angle bisectors in
the bigger equilateral triangle DFE. Segments AD and BF are thus congruent. Therefore,
the angle between segments BF and AD is also 120◦ .
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International Journal of Mathematical Education in Science and Technology
11
Figure 13.
Congruent triangles.
Figure 14.
[13]).
Triangle EFG formed from the perpendicular lines of PA, PB, and PC (recreated from
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12
Figure 15.
Classroom Note
Verification of Fermat point using Viviani’s Theorem (recreated from [13]).
7. Final remarks
The point of intersection of the three circumcircles of the equilateral triangles is called
the Torricelli point of a triangle.[1] If the largest angle in the triangle is 120◦ or less, the
Torricelli point coincides with the Fermat point. If the biggest angle in the original triangle
is more than 120◦ , the point that minimizes the sum of distances to the three vertices will
be the vertex of this angle (e.g. C in triangle ABC in Figure 20). To prove this, we will
show that for any point X other than C, AC + BC < AX + BX + CX.[13] Let angle ACX
be α, and angle BCX be β. Then, α + β = angle ACB > 120◦ (Figure 20).
Now draw a line passing through point X that is perpendicular to side AC and label the
intersection point as E, and similarly construct F on side AB (Figure 21). Then, because
Figure 16.
Relation of vectors and relation of velocities.
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International Journal of Mathematical Education in Science and Technology
Figure 17.
13
Relation of velocities and relation of vectors.
CXE is a right triangle, CE = CXcosα and CF = CXcosβ. Because AC = AE + CE and
BC = BF + CF, AC + BC = (AE + BF) + (CE + CF).[13] The second term (CE + CF)
is equal to CX cos α + CX cos β = CX(cos α + cos β) = 2CX(cos α +2 β cos α −2 β ). Because (α + β) = angle ACB > 120◦ , α +2 β > 60◦ , cos α +2 β < 12 and 0 < cos α−β
≤ 1. Thus,
2
(CE + CF) < CX. Therefore, AC + BC < AE + BF + CX < AX + BX + CX (note that
AE < AX because AX is the hypotenuse of the right triangle AEX). Similarly, BF <
BX.[13]
Note that if the biggest angle in the original triangle is more than 120◦ , the three
circumcircles of the equilateral triangles will no longer intersect at the minimal point.
However, the segments connecting the point of intersection J of the circumcircles of the
Figure 18.
Rotated velocities with the same magnitude.
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14
Classroom Note
Figure 19.
Initial position shows k = 0.
Figure 20.
Triangle whose largest angle >120◦ .
three equilateral triangles with the two other vertices A and B will form an angle of 120◦
(Figure 22).
Of course, there are other situations where the idea of Fermat point can be applied.
Cases where the given points are on a straight line can serve as a good introduction problem
before students explore several triangle cases described above. Similar to triangle cases,
real life contexts such as building a bus station or sunken treasures can motivate students,
and both dynamic explorations and traditional proof can be used. Based on the observation
that the sum of the distances from a point in an interval to the endpoints is constant, no
Figure 21.
Two segments drawn from X perpendicular to AC and AB.
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International Journal of Mathematical Education in Science and Technology
Figure 22.
15
Three circumcircles intersecting at point J in case of largest angle >120◦ .
matter where that point is located,[2,p.101–103; 6,p.194] students may realize that when an
odd number of points are given on a straight line, the point where the sum of the distances
from all the original points is minimal is the middle point (median), whereas when an even
number of points are given, the point that minimizes the sum of distances can be located
anywhere between the two middle points among the original points.[5]
Students can also explore finding minimal configurations of lines in squares, pentagons
or other geometrical figures. During the explorations, students can make conjectures based
on their investigation on triangles, where there is only one Fermat point such that the
sum of the distance between the point and three vertices are minimal. In the case of a
concave quadrilateral, students may find the case similar to the triangle whose biggest
angle is more than 120◦ because such a point in a concave quadrilateral is the vertex of the
reflex angle.[5,p.194)] In the case of a convex quadrilateral, students may realize a need
for multiple additional points to find such an optimal path,[2,p.3] and choose between the
optimal path and having one point that connects all vertices based on the real life situation
and limited resources that they may have.
References
[1] Brazil M, Graham RL, Thomas DA, Zachariasen M. On the history of the Euclidean Steiner
tree problem. Arch Hist Exact Sci. 2014;68:327–354.
Downloaded by [216.243.243.238] at 16:31 20 November 2014
16
Classroom Note
[2] Dicken B. Sunken treasure. In: Gould H, Murray D, Sanfratello A, editors. Mathematical
modeling handbook. Bedford, MA: The Consortium for Mathematics and its Applications
COMAP; 2012. p. 99–106.
[3] Franksen OI, Grattan-Guinness I. The earliest contribution to location theory. Spacio-economic
equilibrium with Lamé and Clapeyron (1829). Math Comput Simulation. 1989;31:195–220.
[4] De Villiers M. Weighted airport problem [Internet]. Dyn Math Learn. 2005. Available from:
http://dynamicmathematicslearning.com/weighted-airport.html
[5] De Villiers M. Rethinking proof with Sketchpad. Emeryville, CA: Key Curriculum Press; 2003.
[6] De Finetti B. Die Kunst des Sehens in der Mathematik [The art of seeing in mathematics].
Basel: Birkhäuser; 1974.
[7] Kabal S, Gevay G. Polya’s mechanical model for the Fermat point [Internet]. 2008. Available
from: http://demonstrations.wolfram.com/PolyasMechanicalModelForTheFermatPoint/
[8] Polya G. Mathematics and plausible reasoning Vol. 1. Induction and analogy in mathematics.
Princeton: Princeton University Press; 1954.
[9] Courant R. Soap film experiments with minimal surfaces. Am Math Monthly. 1940;47:167–
174.
[10] Hoffman DT. Smart soap bubbles can do calculus. Math Teach. 1979;72:377–385, 389.
[11] Guven B, Cekmez E, Karatas I. Using empirical evidence in the process of proving: the case
of dynamic geometry. Teach Math Appl. 2010;29:193–207.
[12] Coxeter HSM, Greitzer SL. Geometry revisited. New York, NY: Random House; 1967.
[13] Woltermann M. Fermat’s problem for Torricelli [Internet]. 2014. Available from: http://www2.
washjeff.edu/users/mwoltermann/Dorrie/91.pdf.
[14] Lyúbich YuI, Shor LA. Método cinemático en problemas geométricos [Kinematic method in
geometrical problems]. Moscow: Mir; 1978.
[15] Flores A. The kinematic method and the Geometer’s Sketchpad in geometrical problems. Int J
Comput Math Learn. 1998;3:1–12.