InEuclidean geometry, theFermat point of atriangle, also called theTorricelli point orFermat–Torricelli point, is a point such that the sum of the three distances from each of the three vertices of the triangle to the point is the smallest possible^[1] or, equivalently, thegeometric median of the three vertices. It is so named because this problem was first raised byFermat in a private letter toEvangelista Torricelli, who solved it.

The Fermat point gives a solution to thegeometric median andSteiner tree problems for three points.

The Fermat point of a triangle with largest angle at most 120° is simply itsfirst isogonic center orX(13),^[2] which is constructed as follows:

1. Construct anequilateral triangle on each of two arbitrarily chosen sides of the given triangle.
2. Draw a line from each newvertex to the opposite vertex of the original triangle.
3. The two lines intersect at the Fermat point.

An alternative method is the following:

1. On each of two arbitrarily chosen sides, construct anisosceles triangle, with base the side in question, 30-degree angles at the base, and the third vertex of each isosceles triangle lying outside the original triangle.
2. For each isosceles triangle draw a circle, in each case with center on the new vertex of the isosceles triangle and with radius equal to each of the two new sides of that isosceles triangle.
3. The intersection inside the original triangle between the two circles is the Fermat point.

When a triangle has an angle greater than 120°, the Fermat point is sited at the obtuse-angled vertex.

In what follows "Case 1" means the triangle has an angle exceeding 120°. "Case 2" means no angle of the triangle exceeds 120°.

Fig. 2 shows the equilateral triangles△ARB, △AQC, △CPB attached to the sides of the arbitrary triangle△ABC.Here is a proof using properties ofconcyclic points to show that the three linesRC, BQ, AP in Fig 2 all intersect at the pointF and cut one another at angles of 60°.

The triangles△RAC, △BAQ arecongruent because the second is a 60° rotation of the first aboutA. Hence∠ARF = ∠ABF and∠AQF = ∠ACF. By the converse of theinscribed angle theorem applied to the segmentAF, the pointsARBF areconcyclic (they lie on a circle). Similarly, the pointsAFCQ are concyclic.

∠ARB = 60°, so∠AFB = 120°, using theinscribed angle theorem. Similarly,∠AFC = 120°.

So∠BFC = 120°. Therefore,∠BFC + ∠BPC = 180°. Using theinscribed angle theorem, this implies that the pointsBPCF are concyclic. So, using theinscribed angle theorem applied to the segmentBP,∠BFP = ∠BCP = 60°. Because∠BFP + ∠BFA = 180°, the pointF lies on the line segmentAP. So, the linesRC, BQ, AP areconcurrent (they intersect at a single point).Q.E.D.

This proof applies only in Case 2, since if∠BAC> 120°, pointA lies inside the circumcircle of△BPC which switches the relative positions ofA andF. However it is easily modified to cover Case 1. Then∠AFB = ∠AFC = 60° hence∠BFC = ∠AFB + ∠AFC = 120° which meansBPCF is concyclic so∠BFP = ∠BCP = 60° = ∠BFA. Therefore,A lies onFP.

The lines joining the centers of the circles in Fig. 2 are perpendicular to the line segmentsAP,BQ,CR. For example, the line joining the center of the circle containing△ARB and the center of the circle containing△AQC, is perpendicular to the segmentAP. So, the lines joining the centers of the circles also intersect at 60° angles. Therefore, the centers of the circles form an equilateral triangle. This is known asNapoleon's Theorem.

Given any Euclidean triangle△ABC and an arbitrary pointP let${\displaystyle d(P)=|PA|+|PB|+|PC|.}$ The aim of this section is to identify a pointP₀ such that for all${\displaystyle P\ne P_0.}$ If such a point exists then it will be the Fermat point. In what followsΔ will denote the points inside the triangle and will be taken to include its boundaryΩ.

A key result that will be used is the dogleg rule, which asserts that if a triangle and a polygon have one side in common and the rest of the triangle lies inside the polygon then the triangle has a shorter perimeter than the polygon:

IfAB is the common side, extendAC to cut the polygon at the pointX. Then the polygon's perimeter is, by thetriangle inequality:

${\displaystyle \text{perimeter}>|AB|+|AX|+|XB|=|AB|+|AC|+|CX|+|XB|\ge |AB|+|AC|+|BC|.}$

LetP be any point outsideΔ. Associate each vertex with its remote zone; that is, the half-plane beyond the (extended) opposite side. These 3 zones cover the entire plane except forΔ itself andP clearly lies in either one or two of them. IfP is in two (say theB andC zones’ intersection) then setting${\displaystyle P^{\prime }=A}$ implies by the dogleg rule. Alternatively ifP is in only one zone, say theA-zone, then whereP' is the intersection ofAP andBC. Sofor every pointP outsideΔ there exists a pointP' inΩ such that

Case 1. The triangle has an angle ≥ 120°.

Without loss of generality, suppose that the angle atA is ≥ 120°. Construct the equilateral triangle△AFB and for any pointP inΔ (exceptA itself) constructQ so that the triangle△AQP is equilateral and has the orientation shown. Then the triangle△ABP is a 60° rotation of the triangle△AFQ aboutA so these two triangles are congruent and it follows that${\displaystyle d(P)=|CP|+|PQ|+|QF|}$ which is simply the length of the pathCPQF. AsP is constrained to lie within△ABC, by the dogleg rule the length of this path exceeds${\displaystyle |AC|+|AF|=d(A).}$ Therefore, for all${\displaystyle P\in \mathrm{\Delta },P\ne A.}$ Now allowP to range outsideΔ. From above a point${\displaystyle P^{\prime }\in \mathrm{\Omega }}$ exists such that and as${\displaystyle d(A)\le d(P^{\prime })}$ it follows that for allP outsideΔ. Thus for all${\displaystyle P\ne A}$ which means thatA is the Fermat point ofΔ. In other words,the Fermat point lies at the obtuse-angled vertex.

Case 2. The triangle has no angle ≥ 120°.

Construct the equilateral triangle△BCD, letP be any point insideΔ, and construct the equilateral triangle△CPQ. Then△CQD is a 60° rotation of△CPB aboutC so

${\displaystyle d(P)=|PA|+|PB|+|PC|=|AP|+|PQ|+|QD|}$

which shows that the sum of the distances sought is just the length of the pathAPQD from A to D along a piecewise linear line. Now we show that ifP is chosen to be the isogonic center of△ABC the pathAPQD lies on a straight line - and thus it is minimal. To do this, construct the equilateral triangle△ABF. LetP₀ be the point whereAD andCF intersect. By construction, this point is the first isogonic center (see above) of△ABC. Carry out the same exercise withP₀ as you did withP, and find the pointQ₀. By the angular restrictionP₀ lies inside△ABC. SinceP₀ is the isogonic center,∠AP₀C = 120°; by construction∠CP₀Q₀ = 60°, thereforeA,P₀ andQ₀ are aligned on the line fromA toD. (Also,△BCF is a 60° rotation of△BDA aboutB, soQ₀ must lie somewhere onAD). Since∠CDB = 60° it follows thatQ₀ lies betweenP₀ andD. Since the pathAP₀Q₀D now lies on a straight line,${\displaystyle d(P_0)=|AD|.}$ Moreover, if${\displaystyle P\ne P_0}$ then eitherP orQ won't lie onAD which means Now allowP to range outsideΔ. From above a point${\displaystyle P^{\prime }\in \mathrm{\Omega }}$ exists such that and as${\displaystyle d(P_0)\le d(P^{\prime })}$ it follows that for allP outsideΔ. That meansP₀ is the Fermat point ofΔ. In other words,the Fermat point is coincident with the first isogonic center.

LetO, A, B, C, X be any five points in a plane. Denote the vectors${\displaystyle \overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC},\overrightarrow{OX}}$ bya,b,c,x respectively, and leti,j,k be the unit vectors fromO alonga,b,c.

Addinga,b,c gives

${\displaystyle |\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}|\le |\mathbf{a}-\mathbf{x}|+|\mathbf{b}-\mathbf{x}|+|\mathbf{c}-\mathbf{x}|+\mathbf{x}\cdot (\mathbf{i}+\mathbf{j}+\mathbf{k}).}$

Ifa,b,c meet atO at angles of 120° theni +j +k =0, so

${\displaystyle |\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}|\le |\mathbf{a}-\mathbf{x}|+|\mathbf{b}-\mathbf{x}|+|\mathbf{c}-\mathbf{x}|}$

for allx. In other words,

${\displaystyle |OA|+|OB|+|OC|\le |XA|+|XB|+|XC|}$

and henceO is the Fermat point of△ABC.

This argument fails when the triangle has an angle∠C > 120° because there is no pointO wherea,b,c meet at angles of 120°. Nevertheless, it is easily fixed by redefiningk = − (i +j) and placingO atC so thatc =0. Note that|k| ≤ 1 because the angle between the unit vectorsi,j is∠C which exceeds 120°. Since

${\displaystyle |\mathbf{0}|\le |\mathbf{0}-\mathbf{x}|+\mathbf{x}\cdot \mathbf{k},}$

the third inequality still holds, the other two inequalities are unchanged. The proof now continues as above (adding the three inequalities and usingi +j +k =0) to reach the same conclusion thatO (or in this caseC) must be the Fermat point of△ABC.

Another approach to finding the point within a triangle, from which the sum of the distances to thevertices of the triangle is minimal, is to use one of themathematical optimization methods; specifically, the method ofLagrange multipliers and thelaw of cosines.

We draw lines from the point within the triangle to its vertices and call themX,Y,Z. Also, let the lengths of these lines bex, y, z respectively. Let the angle betweenX andY beα,Y andZ beβ. Then the angle betweenX andZ isπ −α −β. Using the method of Lagrange multipliers we have to find the minimum of the LagrangianL, which is expressed as:

${\displaystyle L=x+y+z+{\lambda }_1(x^2+y^2-2xy\cos (\alpha )-a^2)+{\lambda }_2(y^2+z^2-2yz\cos (\beta )-b^2)+{\lambda }_3(z^2+x^2-2zx\cos (\alpha +\beta )-c^2)}$

wherea, b, c are the lengths of the sides of the triangle.

Equating each of the five partial derivatives${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }x},\frac{\mathrm{\partial }L}{\mathrm{\partial }y},\frac{\mathrm{\partial }L}{\mathrm{\partial }z},\frac{\mathrm{\partial }L}{\mathrm{\partial }\alpha },\frac{\mathrm{\partial }L}{\mathrm{\partial }\beta }}$ to zero and eliminatingλ₁,λ₂,λ₃ eventually givessinα = sinβ andsin(α +β) = − sinβ soα =β = 120°. However the elimination is a long and tedious business, and the end result covers only Case 2.

• When the largest angle of the triangle is not larger than 120°,X(13) is the Fermat point.
• The angles subtended by the sides of the triangle atX(13) are all equal to 120° (Case 2), or 60°, 60°, 120° (Case 1).
• Thecircumcircles of the three constructed equilateral triangles are concurrent atX(13).
• Trilinear coordinates for the first isogonic center,X(13):^[3]

• Trilinear coordinates for the second isogonic center,X(14):^[4]

• Trilinear coordinates for the Fermat point:

${\displaystyle 1-u+uvw\sec \left(A-\frac{\pi }{6}\right):1-v+uvw\sec \left(B-\frac{\pi }{6}\right):1-w+uvw\sec \left(C-\frac{\pi }{6}\right)}$

whereu, v, w respectively denote theBoolean variables(A < 120°), (B < 120°), (C < 120°).

• The isogonal conjugate ofX(13) is thefirst isodynamic point,X(15):^[5]

${\displaystyle \sin \left(A+\frac{\pi }{3}\right):\sin \left(B+\frac{\pi }{3}\right):\sin \left(C+\frac{\pi }{3}\right).}$

• The isogonal conjugate ofX(14) is thesecond isodynamic point,X(16):^[6]

${\displaystyle \sin \left(A-\frac{\pi }{3}\right):\sin \left(B-\frac{\pi }{3}\right):\sin \left(C-\frac{\pi }{3}\right).}$

• The following triangles are equilateral:
  • antipedal triangle ofX(13)
  • Antipedal triangle ofX(14)
  • Pedal triangle ofX(15)
  • Pedal triangle ofX(16)
  • Circumcevian triangle ofX(15)
  • Circumcevian triangle ofX(16)
• The linesX(13)X(15) andX(14)X(16) are parallel to theEuler line. The three lines meet at the Euler infinity point,X(30).
• The pointsX(13),X(14), thecircumcenter, and thenine-point center lie on aLester circle.
• The lineX(13)X(14) meets the Euler line at midpoint ofX(2) andX(4).^[7]
• The Fermat point lies in the openorthocentroidal disk punctured at its own center, and could be any point therein.^[8]

Theisogonic centersX(13) andX(14) are also known as thefirst Fermat point and thesecond Fermat point respectively. Alternatives are thepositive Fermat point and thenegative Fermat point. However these different names can be confusing and are perhaps best avoided. The problem is that much of the literature blurs the distinction between theFermat point and thefirst Fermat point whereas it is only in Case 2 above that they are actually the same.

This question was proposed by Fermat, as a challenge toEvangelista Torricelli. He solved the problem in a similar way to Fermat's, albeit using the intersection of the circumcircles of the three regular triangles instead. His pupil, Viviani, published the solution in 1659.^[9]

• Geometric median or Fermat–Weber point, the point minimizing the sum of distances to more than three given points.
• Lester's theorem
• Triangle center
• Napoleon points
• Weber problem

• "Fermat-Torricelli problem",Encyclopedia of Mathematics,EMS Press, 2001 [1994]
• Fermat Point by Chris Boucher,The Wolfram Demonstrations Project.
• Fermat-Torricelli generalization atDynamic Geometry Sketches Interactive sketch generalizes the Fermat-Torricelli point.
• A practical example of the Fermat point

• Triangle centers

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