The Torricelli-Fermat point (TF-point) of a triangle is that point which minimises the sum of its distances from the vertices. I generalise this definition, replacing the triangle by a set of M+1 points in E^N. Using the theory of convex functions, I show that the TF-point is unique and find explicit conditions to to determine whether it coincides with any of the given points. If it does not, it may be found by solving a set of ordinary differential equations

Synge, J. L.

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DIAS-STP-87-38The Torricelli-Fermat Point GeneralisedJ. L. SyngeDublin Institute for tdvanced StudiesAbstract: The Torricelli—Fermat point (TF-point) of atriangle is that point which minimises the sum of itsdistances from the vertices. I generalise thisdefinition, replacing the triangle by a set of M÷l pointsin EN. Using the theory of convex functions, I showthat the TF-point is unique and find explicit conditionsto determine whether it coincides with any of the givenpoints. If it does not, it may be found by solving aset of ordinary differential equations.1. Introduction. In the geometry of the triangle thereare certain familiar points — centroid, circumcentre,etc. The point discussed in this note is much lessfamiliar: it is that point which minimises the sum ofits distances from the vertices of a given triangle. Itis strange that this point should be so little known:one can think of obvious applications, such as thelocation of a centre to supply three outposts with aminimum of distance travelled.The problem is a very old one, having been stated byFermat (1601—1665) and solved by Torricelli (1608—1647),but only for acute—angled triangles. Coxeter1 describesg2a proof due to Hofmann in 1929 and remarks that therestriction to acute—angled triangles was removed byPedoe In 1957. In correspondence with me Coxeter hassuggested that the point should be called Torricelli—Fermat (briefly TF-point), and I adopt that name.In the present paper I generalise the problem: 12.S(P) = PA + PA + . . . + PA , (1.1)0 1 MM , 2, N 2For the classical problem M = N = 2 and the threepoints form an undegenerate triangle.2. Notation. Vectors in EN are indicated by heavytype. A. (1= 0,1 M) are the position vectors of- 1given points relative to an arbitrary origin 0. Scalarproducts are indicated by dots. If P is the positionvector of an arbitrary point, (1.1) may be writtenMS(P)=. (P—A.)j1”2 (2.1)If we give an arbitrary infinitesimal displacementto P, we havedS(P) =— dP.Q, (2.2)where Q is a sum of unit vectors3Q = I., I. = (A.—P)/PA . (2.3)•r1 - IThese unit vectors are drawn from p in the directions ofthe A—points, and are well defined unless P coincideswith an A-point, in which case the corresponding I-vectordoes not exist.3•Proof: Since S(P) as in (2.1) is positive, there is atleast one point P at which it has an absolute minimum.Thus at least one TF—point exists, To prove uniqueness,one appeals to the theory of convex functions.2 Afunction f(x) is convex if it satisfies÷ (l—ex2] f(x1) ÷ (l—if(x2) (3.1)for every pair of distinct values of x1, x2 and for allin the open range (0,1). This means that the graph off(x) from x1 to x2, excluding end—points, lies below oron (but not above) the straight line joining the endpoints of the graph. For a stri convex function thesign of equality in (3.1) is deleted; the graph of f(x)lies below the straight line joining the end points ofthe graph.It is easy to see the sum of convex functions isItself convex, and a set of functions of which some are4convex and some strictly convex is itself strictlyconvex.Suppose now that there are two TF-pointS. Let L bethe infinite straight line through them and x a measureof length on it, so that, if P lies on L, we may writefr’S(P) = ‘f1(x). If A. is not on L, a simplecalculation shows that f(x) is positive, and thisimplies strict convexity. If A. lies n L it is easy tosee that f.(x) is convex. Since we have assumed that nothree A—points are collinear, the sum S(P) contains atleast one strictly convex function, and so S(P) on L is astrictly convex function of x, and it is known that astrictly convex function has at most one minimum. Thusthe assumption of two TF—points is false, and uniquenessis proved.Theorem II: If S(P) has a local minimum or a stationaryvalue for some point P, then P is the TF-point.Proof: Let T be the TF—point. Suppose that P is not T.Draw an infinite straight line L through P and T, with xa measure of distance on L. Then S = f(x) on L, andthis function is strictly convex; this is inconsistentwith the assumption that p is not T. Therefore P is T,and the theorem is proved.Theorem III: A point P which is not one of the givenpoints is the TP—point iffQ = I ÷ I ÷ . . . + I = 0, (3.2)A.0 t.-i5where these are the unit vectors drawn from P towards theA—points, that is= (A1—P)/AP (3.3)Proof: This follows immediately from Theorem II, thevariation dS being given by (2.2).4• I!_iYII_IzQ2!_I A 1ffcos ij (1—M)/2 , (4.1)i and is the anglebetween the vectors A.—A and A.-A1 0—J 0Er.91: Take the origin at A. The position vector ofany point P may then be written sI where I is a unitvector and s is the distance PA. Giving all directionsto I and letting s take all positive values, we cover thewhole of EN except the origin where s = 0. Then the sumS as in (2.1) isS(P) = s ÷ [(sI-Ai). (sI-A1)J2 (4.2)Differentiating with respect to s and letting s tend tozero, we get(dS/ds) = (4.3)where6= i ÷r ÷ = A./(A.1 .2 ..i_iI’s being drawn from AA-points.Rotating the unit vector I in all direexpression (4.3) is always positive iff theR is less than unity or equivalentlyButR.R < 1.(4.4)towards theR.R = M + 2cos p..,1J(4.6)where the summationThus we have a localby continuity. ThisIn the classical2. Then the formulaat a vertex iff cos çìtetrahedron in E3, wethe TF—point iffcos ÷ cos01 o2these being the anglesand the angles Ø. . areiiminimum, the equalitycompletes the proof.case of a triangle,(4.1) tells us that— 1/2, i.e. 1 12have M = N/o3of theas in (4.1).sign following5. The TF-conruence. Given the points A. (i=O,1,.in EN and seeking the TF-point, the systematic plan isfirst to test whether it lies at one of the A—points.,M)This is done by investigating the inequality (4.1).RtheseotherI.Munit vectorsCt ionsmagni ofthetude(4.5)we have M = N =the TF—point is00 . For a= 3 and the vertex A is0+ cosat A0- 1, (4.7)faces containing A0.7Suppose that the result is negative: then we must seekthe TF—point elsewhere.By Theorem II we know that we need only apply astationary condition. Now by (2.2)dS(P) =- dP.Q, Q II.,.= (A.-P)/PA.. (5.1)The stationary points are such that Q=O. That conditionis not easy to apply, but if we choosedP = (5.2)where ds is an element of distance, we havedS(P)/d(L-Q.Q . (5.3)This differeential equation defines a congruence ofcurves in EN, and if we proceed in the correct sensealong any one of these curves, S(P) steadily decreases.Since we have ruled out the A-points as possible TFpoints, this congruence of curves must lead us to the TF—point, no matter where we start. Note thatQ.Q = M + 1 + cos (5.4)where in the summation i = 0,1,.. .M and j < 1.6. The tetrahedron. The tetrahedron in E3 stands nextin simplicity to the triangle. In (4.1) we have the8conditions that the TF—point should be at a vertex. Ifit is not there, it is to satisfy (3.2), which it isconvenient to writeQ = I + 3 + K + 0, (6.1)where these are unit vectors drawn from the TF-pointtowards the vertices A, B, C, D.If we transfer L to the other side and square, wegetJ.K + K.I + I.J =- 1, (6.2)p-a result of apparently little interest. But if wetransfer both K and L to the other side and square, wegetI.J = K.L. (6.3)%-Thus at the TF—point the sides AB and CD subtend the sameangle. Obviously this is true for all the three pairsof opposite sides of the tetrahedron.This suggests a construction for the TF—point.With AB as chord, describe a circular arc containing anangle 8 and rotate this arc around AB, forming a spindle.If 0 changes continuously from ii’ to zero, the growingspindle covers all space. If we do the same with CD,using an angle %, we shall get a second system ofspindles. But if we make % = 0 and let their common1•i4e decrease from fl”, there will be a state in which1:9the two spindles touch, and this will be the TF-point ofthe tetrahedron. Since this point is unique, we seethat there is a unique point (the TF-point) at which i-frreach pair of opposite edges, the two edges subtend thesame angle.7. Conclusion. I thank my colleague Professor J. T.Lewis for discussions, and in particular for suggestingthe use of convexity to establish uniqueness. I alsothank Professor H. S. M. Coxeter for correspondence.References.1. H. S. M. Coxeter, Introduction to Geometry, Wiley1961, p. 21.2. R. T. Rockafellar, Convex analysis, Princeton Univ.Press 1972.

The Torricelli-Fermat Point Generalised

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The Torricelli-Fermat Point Generalised

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