Fermat Point Research Articles

A simple proof of Lester’s theorem

Lester’s theorem (1997) states that in any scalene triangle the two Fermat points F and F' (to be defined later), the nine-point centre N, and the circumcentre O, are concyclic, and that the pair of points O,F separates the pair N, F'. (In certain geometrical situations a line is regarded as a circle of infinite radius, so that the word ‘concyclic’ includes ‘collinear’ as a special case, but here ‘concyclic’ means ‘lying on a proper circle of finite radius’.) Previous proofs of Lester’s theorem have involved advanced techniques and/or computer algebra; to quote from Ron Shail’s recent article [1],‘Lester’s original computer-assisted discovery and proof make use of her theory of “complex triangle coordinates” and “complex triangle functions”. ... A proof has also been given by Trott ... using the advanced concept of GrObner bases in the reduction of systems of polynomial equations to “diagonal” form. Trott’s work uses the computer algebra system Mathematica as an essential tool.’

A journey round the triangle Lester’s circle, Kiepert’s hyperbola and a configuration from Morley

In a recent article [1], Ron Shail has given a Cartesian proof of an interesting theorem due to J. A. Lester. This states that, for any triangle, the circumcentre O, the nine-point centre O9 and the two Fermat points F and Fʹ, (which are the points of concurrence of the joins of its vertices to the vertices of equilateral triangles drawn outwards/inwards on the opposite sides), are concyclic. He refers to Lester's own treatment as needing complex coordinates with computer-assisted algebra; his own proof uses an unpromising method, and results in similar problems. Contemplation of the configuration would suggest that the location of the point of intersection of FFʹ with the Euler line OO9 might lead to a simple proof. The theorem is in fact a corollary from the properties of a remarkable configuration originating with Morley [2, p. 209], and shown in Figure 1. He did not deduce Lester’s result, nor label the crucial point J in the diagram, which was drawn without that particular intersection. Also involved in this figure is a rectangular hyperbola known by the name of its describer Kiepert [3]. It is helpful to discuss all three together. What follows is a journey through country nowadays rather unfamiliar, avoiding the computerised motorway and using older tracks via complex numbers, trilinear coordinates and Euclidean methods which reveal much more than is apparent from a Cartesian treatment.

Some features of the general Fermat point

Given a triangle with vertices P1, P2, P3 its Fermat point is defined as that point F which minimises the sum P1F + P2F + P3F. It is known that if each of the angles P1, P2, P3 is less than 120°, then F will be the point inside the triangle at which each pair of P1, P2, P3 subtend an angle of 120°, while if any of the angles at P1, P2, P3 exceed 120° then F will coincide with that vertex – see Figure (1).

A proof of Lester’s Theorem

The object of this note is to draw readers’ attention to a very new theorem in the Euclidean geometry of the triangle and to provide a straightforward Cartesian proof. This remarkable theorem, due to Lester, asserts that in any scalene triangle the two Fermat points, the nine-point centre and the circumcentre are concyclic. Lester’s original computer-assisted discovery and proof make use of her theory of ‘complex triangle coordinates’ and ‘complex triangle functions’ as expounded in, and . A proof has also been given by Trott using the advanced concept of Grobner bases in the reduction of systems of polynomial equations to ‘diagonal’ form. Trott’s work uses the computer algebra system Mathematica as an essential tool and he also provides an animation of the Lester circle as one vertex of the triangle is varied. Further information on the configuration is given in.

The Fermat Point of a Triangle

The Fermat Point of a Triangle

The Generalized Fermat's Point

In the 17th century, Fermat proposed and Torricelli solved the problem of determining a point P in a given acute triangle ABC such that the sum PA + PB + PC of the distances from P to the vertices of the triangle is the minimum. The first proof was published in 1659, and the point P is called Fermat 's point in the literature. There is a very natural generalization of Fermat's problem. Let 1, m, n be three positive real numbers. Find a point P in a given triangle ABC such that 1 PA + m a PB + n * PC is a minimum. This generalization was considered in two articles in this journal [2], [3]. The method in [2] is complicated, while the method in [3] does not use synthetic geometry. In this note, using the idea in the proof of the existence of Fermat's point [1], we give a proof for the generalized case. This proof is just a slight modification of the old one. Let us recall the proof of the existence of Fermat's point. On the sides of AB, BC, CA of triangle ABC, erect externally the equilateral triangles ABC', A'BC, and AB'C. Draw lines CC' and AA' to intersect at P. Then the point P is Fermat's point. The proof is based on the fact that a line segment is the shortest path between two points. Here is the outline of the proof in [1].

The Generalized Fermat's Point

The Generalized Fermat's Point

79.21 A Note on the Fermat Point of a Tetrahedron

79.21 A Note on the Fermat Point of a Tetrahedron

Convex hull properties in location theory

We present new characterizations of inner product spaces which bring into play a property of a family of optimization problems related to the norm of the space. This property concerns the existence of a solution .to some optimization problems which belongs to the convex hull of some set. We thus obtain a generalization of results of V. Klee and A. Garkavi about the Chebychev centers and also of more recent results of the author about Fermat points. Intermediate propositions concerning unicity in some optimization problems, a geometric characterization of finite dimensional inner product spaces and monotone norms seem to have their own interest.

77.15 Tetrahedra – Fermat points and centroids

77.15 Tetrahedra – Fermat points and centroids

Converses of Napoleon's Theorem

Interesting converse results elementary geometry can often be found by taking certain parts of a figure as given in position and investigating the extent to which various other parts of the figure are determined. In this article we use this tactic to obtain some apparently new converses of the well-known theorem of Napoleon. Geometry is more a point of view than a methodology, and we employ a variety of different arguments (synthetic, coordinate, transformational, complex analytic) to establish our results. To set the stage, we begin with an overview of Napoleon's theorem and a glimpse of its long history. 1. NAPOLEON'S THEOREM AND TORRICELLI'S CONFIGURATION. The familiar but curious theorem attributed to Bonaparte asserts that the centers L, M, N of the three equilateral triangles ABXC, A CYA, AAZB built outwards on the sides BC, CA, AB of an arbitrary triangle AABC are the vertices of an equilateral triangle, and the same is true of the centers L', M', N' of the three inward equilateral triangles ACX'B, AAY'C, ABZA. The formed by a triangle, the equilateral triangles on its sides, the Napoleon triangles, and various connecting lines and circles (commonly called Torricelli's configuration a century ago), has many elegant and unexpected properties. The outward case. Suppose (FIGURE 1) that AABC is a positively oriented triangle (so that A -> B -* C -> A is counterclockwise). The outer triangle ALMN is also positively oriented, and its center coincides with the centroid G of AABC. Lines AXZ,T Xare concurrent at a point F, called the outward Fermat point of AABC, and F lies on the circumcircle of each outward equilateral triangle ABXC, A CYA, AAZB and also on the circumcircle of the inner triangle AL'M'N'. Lines 'CZ make acute angles of 600 with each other at F, and AX = BY = CZ = +AF + BF + CF, a minus sign being taken if the angle of AABC at that vertex exceeds 1200. The vertices A, B, C are symmetric to F the sidelines MN,O, NL, of the outer triangle ALMN. Lines AL, are concurrent. When AABC has a 1200 angle, F is the vertex of that angle; when AABC has an angle larger than 1200, F lies the angle vertical to that angle; and when every angle of AABC is smaller than 1200, F lies inside AABC and is the point P that solves the problem Fermat posed to Torricelli: minimize f(P) = PA + PB + PC. When the largest angle of AABC exceeds 1200, the solution of Fermat's problem is the vertex of that largest angle. The inward case. Analogous properties hold for the inward case. Suppose (FIGURE 2) that AABC is a positively oriented scalene triangle. The inner triangle AL'M'N' is negatively oriented, and its centroid coincides with the centroid G of AABC. Lines A*X, Y, CZ' are concurrent at a point F', called