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Fermat's point from five perspectivesJungeun Parka & Alfinio Floresa

a Department of Mathematical Sciences, University of Delaware,Newark, USAPublished online: 18 Nov 2014.

To cite this article: Jungeun Park & Alfinio Flores (2014): Fermat's point from fiveperspectives, International Journal of Mathematical Education in Science and Technology, DOI:10.1080/0020739X.2014.979894

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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

(Received 7 March 2014)

The Fermat point of a triangle is the point such that minimizes the sum of the distancesfrom that point to the three vertices. Five approaches to study the Fermat point ofa triangle are presented in this article. First, students use a mechanical device usingmasses, strings and pulleys to study the Fermat point as the one that minimizes thepotential energy of the system. Second, students use soap films between parallel planesconnecting three pegs. The tension on the film will be minimal when the sum ofdistances is minimal. Third, students use an empirical approach, measuring distances inan interactive GeoGebra page. Fourth, students use Euclidean geometry arguments fortwo proofs based on the Torricelli configuration, and one using Viviani’s Theorem. Andfifth, the kinematic method is used to gain additional insight on the size of the anglesbetween 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 distancesPA + PB + PC from that point to the three vertices is minimal (Figure 1). This problemcan 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 anadvantage 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 thatwill drop the cable to retrieve three treasury chests from the bottom of the sea (assumingthe chests have equal weights).[2]

Solving the problem for three points using a mechanical device goes back to Lame andClapeyron in 1829.[3] The history of minimizing the sum of distances to a set of morethan three points starts with Gergonne in 1811.[1] The letter from 1836 from Carl FriedrichGauss to Schumacher about the minimum distance to four points describes Gauss’ ideaof extending Fermat’s problem to more than three points in terms of finding a networkof minimal length connecting the original points rather than finding a single point thatminimizes 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◦. Weaddress 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 weightedcases,[4] and more than three points.[5,p.194]

∗Corresponding author. Email: alfinio@udel.edu

C© 2014 Taylor & Francis

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Figure 1. Sum of distances to three vertices.

In this article, students look at the properties of the point P inside a triangle andthe corresponding segments and angles from five perspectives. First, students explore theproblem using a mechanical device, using string and weights. Second, students use soapfilms connecting three pins between parallel plates to represent the problem. Third, studentsuse an interactive web page made with GeoGebra to experiment and verify the result bymeasuring. Fourth, students use classical Euclidean geometry arguments to prove propertiesof Fermat’s point and related segments. Finally, students use the kinematic method to gainadditional 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 understandingof Fermat’s point and the properties of the segments joining this point with the vertices ofthe triangle.

2. Minimizing the sum of distances to the vertices of a trianglewith a physical device

To find the point P inside the triangle ABC, such that the sum of the distances AP + BP + CPis as small as possible, De Finetti [6] suggests using a physical argument. One way to findthis point is by drilling three holes on a rigid horizontal board corresponding to the positionsof the three points A, B and C (Figure 2). We tie three strings together at the movable pointP, 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 thatAP + 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, theresultant of two of them is on the angle bisector between them (Figure 3). And because theresultant of these two forces is in equilibrium with the third force, the line of action of thisone is also the angle bisector. Thus PA is on the angle bisector of PB and PC, PB is on theangle bisector of PA and PC, and PC is on the angle bisector of PA and PB. Therefore, eachpair 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 thesame kind of argument. When the masses are hanged from the pulleys so that the anglebetween the strings connecting the knot with the pulleys is 120◦, the masses will remain inequilibrium (see Figure 5). If one of the masses is moved up or down, the system will tendto go back to the equilibrium position on its own. Because of friction, it may not completelygo back to form exactly angles of 120◦.

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International Journal of Mathematical Education in Science and Technology 3

Figure 2. 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 studyFermat’s point with soap films, two parallel transparent plates are attached with three pinsperpendicular to the plates. The plates are submerged in a soap solution and slowly pulledout from the liquid. A soap film will be formed connecting the three pins (Figure 6). Thetension 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.

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Figure 4. 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 thatis minimal. When seen from above, you can see that the films join at a point inside thetriangle 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 thepoint 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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Figure 6. Soap films minimize sum of lengths.

Students will realize that when using experimental methods it is not always possible to getexactly three angles of 120◦; however, Guven, Cekmez and Karatas [11] have pointed tothe importance of the use of dynamical geometry programs to facilitate the transition fromempirical data to deductive proof. An interactive web page where students can experimentis available at http://www.geogebratube.org/student/m60133.

Figure 7. Soap films viewed from above.

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Figure 8. 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 threevertices in two steps. We will first show that the point has to be on the line that connects Cwith the outside vertex F of an equilateral triangle constructed on side AB (Figure 9). Let Pbe a point in the inside of triangle ABC, and triangle BAF an equilateral triangle. Constructthe segment PB and let BPE be an equilateral triangle on this segment. Angle PBA iscongruent to angle EBF because side BE is BP rotated 60◦ and side BF is BA also rotated60◦. Because PB is congruent to EB (sides of an equilateral triangle) and BA is congruentto 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 isequal to the sum of the lengths of the segments FE + EP + PC. This sum will be minimalif points C, P, E and F are on the straight line CF. In the same way, the position of P thatminimizes the sum of distances is on the line that connects A with the outside vertex ofan equilateral triangle on BC (Figure 10). The intersection of these two lines is the desiredposition.

We will now prove that the angle between segments AG and CF is 120◦ using a rotationof 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 angleAPC is equal to 120◦. In the same way, we can prove that angle APB is 120◦. This meansthat quadrilateral APBF is an inscribed quadrilateral. The same is true for quadrilateralsAPCH and BPCG (Figure 11). This means that the circumcircles of the equilateral triangleson the sides of the original triangle intersect at Fermat’s point (Figure 12). The configurationformed by the original triangle together with the three equilateral triangles on its sides isknown as Torricelli’s configuration. An interactive Torricelli configuration is available athttp://www.geogebratube.org/student/m66431.

We will now look at two alternative proofs. First, in Figure 13 angle EFB is congruentto angle DAB (corresponding angles of congruent triangles BEF and BDA). Both coversegment DB. Therefore, F and A are on the same circle, so that ADBF is a cyclicalquadrilateral. Thus angle ADF is congruent to angle ABF and angle FDB is congruent toangle BAF. Therefore angle ADB is 120◦.

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Figure 9. Sum of distances is equal to length of path CDEF.

Similar proofs can be designed with dynamic geometry activities, which guide stu-dents through the process of exploring, making conjectures and proving their conjec-tures.[5,p.108–114] For example, De Villiers [5] created such activities with Geometer’sSketchpad, through which students make conjectures about the intersection of three seg-ments connecting AG, CF and BF in Figure 11, and their lengths, and prove their conjecturesthat the segments are equal length using the property of congruent triangles, and that thesegments intersect at one point using the property of cyclic quadrilaterals, inscribed in threecircles in Figure 12.[5,p.108–114]

Second, we can also use Viviani’s Theorem, which states that the sum of distances ofa point inside an equilateral triangle to its sides is equal to its height, to prove that thepoint P that minimizes AP + BP + CP is the point where angle APB = angle BPC = angleCPA.[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, andlabel 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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Figure 10. Location of point that minimizes the sum of distances.

equalities hold. Therefore, A′P′ + B′P′ + C′P′ < AP′ + BP′ + CP′. Viviani’s Theorem givesAP + 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 relationbetween two vector functions to find a similar relation between the velocities of the cor-responding endpoints of the vectors. And vice versa, if we know that a relation holds forthe velocities of the endpoints for all times, we know that the same relation will holdfor the vector functions (except possibly by having to add a constant vector). This is similarto 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 vectorfunction r1(t) is identical to the rotation of another vector function r2(t) by a constant angleα. We will express this relation by r1 = rot(α)r2. Let v1(t) be the derivative of r1(t), thatis

v1 (t) = limh→0

r1 (t + h) − r1 (t)

h(1)

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Figure 11. 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

limh→0

r1 (t + h) − r1 (t)

h=

limh→0

rot (α) r2 (t + h) − rot (α) r2 (t)

h=

limh→0

rot (α)r2 (t + h) − r2 (t)

h=

rot (α) limh→0

r2 (t + h) − r2 (t)

h= rot (α) v2

Thus v1 = rot(α)v2.

Theorem 2: Assume that the functions r1 and r2 are such that for all times their cor-responding velocities satisfy v1 = rot(α)v2, then r1 = rot(α)r2 + k where k is a constantvector (Figure 17).

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Figure 12. Intersection of circumcircles.

Proof: We will use the fact that if the derivative of a vector function is constant equal tozero, then the function is constant. Consider the function r1 − rot(α)r2. Its derivative isv1 − 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 methodin 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 usetheorem 1 to say that the velocity of F has the same magnitude as the velocity of C but isrotated 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 rotated60◦ 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 seethat 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 inthe 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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Figure 13. Congruent triangles.

Figure 14. Triangle EFG formed from the perpendicular lines of PA, PB, and PC (recreated from[13]).

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Figure 15. 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 calledthe Torricelli point of a triangle.[1] If the largest angle in the triangle is 120◦ or less, theTorricelli point coincides with the Fermat point. If the biggest angle in the original triangleis more than 120◦, the point that minimizes the sum of distances to the three vertices willbe the vertex of this angle (e.g. C in triangle ABC in Figure 20). To prove this, we willshow that for any point X other than C, AC + BC < AX + BX + CX.[13] Let angle ACXbe 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 theintersection 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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Figure 17. Relation of velocities and relation of vectors.

CXE is a right triangle, CE = CXcosα and CF = CXcosβ. Because AC = AE + CE andBC = 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 + β

2 cos α −β2 ). Be-

cause (α + β) = angle ACB > 120◦, α + β2 > 60◦, cos α +β

2 < 12 and 0 < cos α−β

2 ≤ 1. Thus,(CE + CF) < CX. Therefore, AC + BC < AE + BF + CX < AX + BX + CX (note thatAE < 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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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 problembefore 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 observationthat 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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Figure 22. 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 anodd number of points are given on a straight line, the point where the sum of the distancesfrom all the original points is minimal is the middle point (median), whereas when an evennumber of points are given, the point that minimizes the sum of distances can be locatedanywhere between the two middle points among the original points.[5]

Students can also explore finding minimal configurations of lines in squares, pentagonsor other geometrical figures. During the explorations, students can make conjectures basedon their investigation on triangles, where there is only one Fermat point such that thesum of the distance between the point and three vertices are minimal. In the case of aconcave quadrilateral, students may find the case similar to the triangle whose biggestangle is more than 120◦ because such a point in a concave quadrilateral is the vertex of thereflex angle.[5,p.194)] In the case of a convex quadrilateral, students may realize a needfor multiple additional points to find such an optimal path,[2,p.3] and choose between theoptimal path and having one point that connects all vertices based on the real life situationand limited resources that they may have.

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[2] Dicken B. Sunken treasure. In: Gould H, Murray D, Sanfratello A, editors. Mathematicalmodeling handbook. Bedford, MA: The Consortium for Mathematics and its ApplicationsCOMAP; 2012. p. 99–106.

[3] Franksen OI, Grattan-Guinness I. The earliest contribution to location theory. Spacio-economicequilibrium with Lame 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: Birkhauser; 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] Lyubich YuI, Shor LA. Metodo cinematico en problemas geometricos [Kinematic method in

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