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The Fermat Point and GeneralizationsP. Fermat (1601-1665) challenged Evangelista Torricelli(*) (1608-1647), the inventor of barometer with the following question
Torricelli presented several solutions. In one he observed that the circumcircles of the equilateral triangles constructed externally on the sides of a given triangle meet in a point. Many more have been found since. I'll present several solutions and two quite surprising generalizations. Solution 1(Published by Joseph Ehrenfried Hofmann (1900--1973) in 1929. It was independently discovered by Tibor Galai and others [Honsberger, p. 26].) 
In ΔABC, select a point P and connect it with vertices A, B, and C. Rotate ΔABP 60o around B into position C'BP'. By construction, ΔBPP' is equilateral, The result is clearly related to Napoleon's theorem. On the sides of ΔABC construct equilateral triangles ABC', ACB', and BCA'. We know that the Fermat point P that minimizes the sum So the Fermat point is unique and lies at the intersection of three straight lines that connect vertices of the triangle with opposite vertices of Napoleon's triangles. The construction fails if one of the internal angles of ΔABC is 120o or more. In this case, the vertex corresponding to the largest angle of the triangle solves Fermat's problem. (There is a dynamic illustration pertinent to the above proof.)  Solution 2
This solution only works under the assumption that the Fermat point exists and is unique. Let for a given ΔABC, P be that point. Let's move, say, A a little into position A1 on the ray PA. Will P move along into a different position? Assuming it will and denoting its new position as P1, we arrive at contradiction. Indeed, for ΔABC,
Also, by our assumption for ΔA1BC,
Summing up and cancelling, we get
Recollect that we selected A1 on the ray PA so that
which is of course absurd. Therefore, when A slides along the ray PA, point P does not change. Slide then vertices A and B so as to make ΔABC equilateral. This is always possible. For example, start with sliding A so as to make Solution 3([Honsberger, p. 26] ascribes this solution to Torricelli himself and mentions that it was rediscovered almost 300 years later by F. Riesz. In a private communication, Douglas Rogers mentioned that the solution has been also proposed by J. Steiner. Thus the same object that is most frequently referred to as Fermat's point, Torricelli's point and Fermat-Torricelli point, is sometimes also named after J. Steiner, as Steiner's point, especially in the framework of Steiner's networks. [Tikhomirov, p. 31] and [Courant and Robbins, p. 354] even refer to Fermat's problem as Steiner's. [Johnson, p. 221] attributes the present solution to Steiner.) The following solution depends on explicit knowledge of the fact that, for Fermat's point,
At the vertices of ΔABC draw perpendiculars to lines AF, BF, and CF. The new lines form an equilateral triangle RST. Let F1 be a point different from F. Drop perpendiculars F1A1, F1B1 and F1C1 to the sides of ΔRST. It's a nice property (known as Viviani's theorem) of equilateral triangles that in ΔRST,
(The sum of the distances from a point inside an equilateral triangle to the sides of the triangle does not depend on the point. ) On the other hand, obviously,
All that remains is to combine the two. (Douglas Rogers made a delightful observation. ΔRST is the largest equilateral triangle circumscribing ΔABC. According to Viviani's theorem, FA, FB, FC add up to the altitude of ΔRST. For any other equlateral ΔLMN curcumscribing ΔABC, at least one of FA, FB, or FC is not perpendicular to the side and hence is longer than such perpenduclar. It follows that the sum FA + FB + FC is bound to be larger than the alititude of ΔLMN. This implies that the side of ΔLMN is smaller than that of ΔRST. In passing, ΔRST is known as the antipedal triangle of F with respect two ΔABC.) Solution 4(The proof by Lou Talman is found on one of the discussions at the mathforum. Still very simple, it uses a little bit of analytic geometry and calculus. This is probably very close to one of Torricelli's original proofs.) One can see that the Fermat point does minimize Solution 5(The argument is similar to that in Solution 4, but from a different perspective. [F. G.-M., p. 442] credits Lhuilier (1811) with the proof.) Assume FC is constant. Then point F that minimizes
For more detail, draw a tangent to the circle at F and choose any point G on the tangent other than F. Let H be the intersection of HC with the circle. Then, assuming angles AFC and BFC equal,
(The latter inequality warrants extra attention. It follows from the fact the various ellipses with foci at A and B do not intersect.) From here,
The argument can be repeated assuming either FA or FB constant. As a result, at Fermat's point F all three angles AFC, BFC and AFB are found to be equal. Solution 6Fermat's point can be located with the help of Euler's generalization of Ptolemy's Theorem, see [Pedoe, pp. 93-94].
Any triangle has at least two acute angles. Given ΔABC, let angles at B and C be acute. Form an equlateral triangle BCD, with D and A on opposite sides of BC. Let (O) be the circumcircle of ΔBCD. For a point Q, by Ptolemy's inequality
with equality only if Q lies on (O) and such that the quadrilateral BQCD is convex. Note that, by construction,
Therefore
Unless, Q lies on AD, AQ + BQ + CQ > AD. Let P be the intersection of AD with (O) other than D. BPCD is an inscribed convex quadrilateral and P lies on AD. So in this (and only in this) case
Angle BPC is supplementary to If the triangle is acute the same construction applies to the other two sides which brings up the framework of Napoleon's theorem. The three circles intersect at Fermat's point and three lines joining vertices of ΔABC with the opposite vertices of the Napoleon triangles concur at the point P. The latter fact can be used for a more direct proof. Solution 7(This was a part of a solution by Grimbal to a problem posted at the wu::forum.)
Construct Napoleon's equilateral triangles ABC', AB'C', A'BC externally on the sides of ΔABC. Let P be the intersection of AA' and BB'. AA' is CC' rotated clockwise 60° about B, and BB' is CC' rotated counterclockwise 60° about A, it follows that angle APB' is 60°. Let X be the point on BB' making triangle APX equilateral. Now, if BB' is rotated clockwise 60° about A, then X goes to P, B' to C, and B to C'. Hence P, C, and C' are collinear; and so AA', BB', and CC' are concurrent, and any two of these lines make an angle of 120°. Searching for the Fermat point we discovered a nice property of Napoleon's triangles. I found quite surprising generalizations of those properties in a column by David Gale in The Mathematical Intelligencer which was kindly pointed out to me by Professor McWorter. 
Two lines passing through a vertex of a triangle are called isogonal with respect to that vertex if they form equal angles with its (internal) angle bisector. Following is the first generalization. Theorem 1As in the diagram, assume lines AB' and AC' are isogonal as are pairs CB', CA' and BA', BC'. Then three lines AA', BB', and CC' are concurrent, i.e., meet at a common point.
Napoleon's theorem is obtained when all three angles involved are equal to 30o. The common point is then known as the First Napoleon point. (The Second Napoleon Point is obtained when the equlateral triangles are formed internally to the given triangle. Similarly, there are two Fermat's points, which are also known as the first and second isogonic centers.) If the base angles are just equal between themselves, the theorem bears the name of Ludwig Kiepert who replaced equilateral triangles with isosceles ones. An interesting specification of the theorem is obtained when the three angles add up to 180o. But the theorem also admits further generalization. In Theorem 1, pairs of (isogonal) lines were related to angle bisectors. As is well known from elementary geometry, angle bisectors meet at a single point (incidentally, the incenter of the triangle.) 
Two lines AB' and AC' through a vertex A of a triangle are said to be isotomic if they intersect the opposite side BC in points equidistant from its midpoint Ma. Theorem 1 remains valid for isotomic lines as well. And, in a certain sense, for any three concurrent lines. A second generalization that I am about to formulate and then prove belongs to the realm of Projective Geometry. Theorem 2Let p, q, r be concurrent lines through the vertices A, B, and C, respectively, of ΔABC. Let PA be the pencil of lines at A and let TA be the (unique) projective mapping on PA which
Define PA, PB, and TA, TB similarly. For any line a in PA, let a' = TA(a). Similarly, for b in PB and c in PC, let b' = TB(b) and c' = TC(c). Let C' = a' Then AA', BB', and CC' are concurrent. Note that (an apparently more general) Theorem 2 follows from Theorem 1 with a projective mapping that leaves vertices A,B,C fixed but takes the incenter into any point P (the intersection of lines 
To prove Theorem 2, perform the projective mapping that carries C to the origin, A to the point at infinity at y-axis, B to the point at infinity at x-axis, and P to the point Lines p, q, and r are carried onto the lines
Transformation TA preserves the vertical direction and, therefore, is in reality a 1-dimensional projective transformation. So it's a Möbius transformation which, in general, has the form f(x) = (ax + b)/(cx + d). Transformations that interchange x = 0 with the point at infinity are given by f(x) = a + b/x (c = 1 and d = 0.) Among those, there is a single one that leaves x = 1 fixed: f(x) = 1/x. So TA maps x = a onto x = 1/a. To simplify the notations, we will denote the vertical line with x = a by a, the horizontal line y = b by b, and the line y = cx through the origin, by c. Then with similar definitions for TB and TC we have
Further
Thus AA' has equation x = 1/(bc), BB' has equation y = a/c, and CC' has equation Looking back at the two theorems, the fact that Theorem 2 is a theorem of Projective Geometry implies that Theorem 1 belongs to Absolute Geometry. As such, besides Euclid's, it also holds in geometries of Riemann and Lobachevsky. (In the later, existence of points A', B', C' must be stipulated.) Reference
E. Torricelli is sometimes said to be a student of Galileo [Honsberger, p. 24], which is not quite correct. Torricelli was not directly a student of Galileo, but rather of Galileo's student Bendetto Castelli. Castelli did introduce Torrecelli's work to Galileo, who then invited Torricelli to come to work with him in Florence. Unfortunately, Torricelli could not come until 1641, and Galileo died in early January, 1642, so their cooperation was rather short. Upon Galileo's death, Torricelli was named the successor of Galileo as the court mathematician to the Grand Duke of Tuscany. He lived and died in the ducal palace in Florence.
Copyright © 1996-2008 Alexander Bogomolny |
CC'
BPC' = 60o.
b, B' = c'
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