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| [[Image:Fermat Point.svg|thumb|right|300px|Fig 1. Construction of the first isogonic center, X(13). When no angle of the triangle exceeds 120°, this point is the Fermat point.]]
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| In [[geometry]], the '''Fermat point''' of a [[triangle]], also called the '''Torricelli point''', is a point such that the total distance from the three vertices of the triangle to the point is the minimum possible.<ref>[http://www.cut-the-knot.org/Generalization/fermat_point.shtml Cut The Knot - The Fermat Point and Generalizations]</ref> It is so named because this problem is first raised by [[Pierre de Fermat|Fermat]] in a private letter to [[Evangelista Torricelli]], who solved it.
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| The Fermat point gives a solution to the [[geometric median]] and [[Steiner tree problem]]s for three points.
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| == Construction ==
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| The Fermat point of a triangle with largest angle at most 120° is simply its '''first isogonic center''' or '''X(13)''', which is constructed as follows:
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| # Construct an [[equilateral triangle]] on each of two arbitrarily chosen sides of the given triangle.
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| # Draw a line from each new [[Vertex (geometry)|vertex]] to the opposite vertex of the original triangle.
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| # The two lines intersect at the Fermat point.
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| An alternate method is the following:
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| # On each of two arbitrarily chosen sides, construct an isosceles 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.
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| # 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.
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| # The intersection inside the original triangle between the two circles is the Fermat point.
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| When a triangle has an angle greater than 120°, the Fermat point is sited at the obtuse-angled vertex.
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| In what follows "Case 1" means the triangle has an angle exceeding 120°. "Case 2" means no angle of the triangle exceeds 120°.
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| ==Location of X(13)==
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| [[Image:Fermat Point Proof.svg|thumb|right|300px|Fig 2. Geometry of the first isogonic center.]]
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| Here is a proof using properties of [[concyclic points]] to show that the three red lines in Fig 1 are [[Concurrent lines|concurrent]] and cut one another at angles of 60°.
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| The triangles RAC and BAQ are [[Congruence (geometry)|congruent]] because the second is a 60° rotation of the first about A. Hence ∠ARF = ∠ABF and ∠AQF = ∠ACF. By converse of angle in the same segment, ARBF and AFCQ are both concyclic. Thus ∠AFB = ∠AFC = ∠BFC = 120°. Because ∠BFC and ∠BPC add up to 180°, BPCF is also concyclic. Hence ∠BFP = ∠BCP = 60°. Because ∠BFP + ∠BFA = 180°, AFP is a straight line.
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| [[Q.E.D.]]
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| This proof only applies in Case 2 since if ∠BAC > 120° A lies inside the circumcircle of BPC which switches the relative positions of A and F. However it is easily modified to cover Case 1. Then ∠AFB = ∠AFC = 60° hence ∠BFC = ∠AFB = ∠AFC = 120° which means BPCF is concyclic so ∠BFP = ∠BCP = 60° = ∠BFA. Therefore A lies on FP.
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| Clearly any 3 lines perpendicular to the red ones in Fig 1, in particular those joining the centres of the circles, must also cut at angles of 60° and thereby form an equilateral triangle. This curiosity is known as [[Napoleon's Theorem]].
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| ==Location of the Fermat point==
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| ===Traditional geometry===
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| [[Image:Fermat Point Scope.svg|thumb|right|300px|Fig 3. Geometry of the Fermat point]]
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| Given any Euclidean triangle ABC and an arbitrary point P let d(P) = PA+PB+PC. The aim of this section is to identify a point P<sub>0</sub> such that d(P<sub>0</sub>) < d(P) for all P ≠ P<sub>0</sub>. If such a point exists then it will be the Fermat point. In what follows Δ will denote the points inside the triangle and will be taken to include its boundary Ω.
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| 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. <br />[If AB is the common side extend AC to cut the polygon at X. Then by the triangle inequality the polygon perimeter > AB + AX + XB = AB + AC + CX + XB ≥ AB + AC + BC.]
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| Let P be any point outside Δ. 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 for Δ itself and P clearly lies in either one or two of them. If P is in two (say the B and C zones intersection) then setting P' = A implies d(P') = d(A) < d(P) by the dogleg rule. Alternatively if P is in only one zone, say the A-zone, then d(P') < d(P) where P' is the intersection of AP and BC. So '''for every point P outside Δ there exists a point P' in Ω such that d(P') < d(P)'''.
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| '''Case 1. The triangle has an angle ≥ 120°.'''
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| Without loss of generality suppose that the angle at A is ≥ 120°. Construct the equilateral triangle AFB and for any point P in Δ (except A itself) construct Q so that the triangle AQP is equilateral and has the orientation shown. Then the triangle ABP is a 60° rotation of the triangle AFQ about A so these two triangles are congruent and it follows that d(P) = CP+PQ+QF which is simply the length of the path CPQF. As P is constrained to lie within ABC, by the dogleg rule the length of this path exceeds AC+AF = d(A). Therefore d(A) < d(P) for all P є Δ, P ≠ A. Now allow P to range outside Δ. From above a point P' є Ω exists such that d(P') < d(P) and as d(A) ≤ d (P') it follows that d(A) < d(P) for all P outside Δ. Thus d(A) < d(P) for all P ≠ A which means that A is the Fermat point of Δ. In other words '''the Fermat point lies at the obtuse angled vertex'''.
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| '''Case 2. The triangle has no angle ≥ 120°.'''
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| Let P be any point inside Δ and construct the equilateral triangle CPQ. Then CQD is a 60° rotation of CPB about C so d(P) = PA+PB+PC = AP+PQ+QD which is simply the length of the path APQD. Let P<sub>0</sub> be the point where AD and CF intersect. This point is commonly called the first isogonic center. By the angular restriction P<sub>0</sub> lies inside Δ moreover BCF is a 60° rotation of BDA about B so Q<sub>0</sub> must lie somewhere on AD. Since CDB = 60° it follows that Q<sub>0</sub> lies between P<sub>0</sub> and D which means AP<sub>0</sub>Q<sub>0</sub>D is a straight line so d(P<sub>0</sub>) = AD. Moreover if P ≠ P<sub>0</sub> then either P or Q won't lie on AD which means d(P<sub>0</sub>) = AD < d(P). Now allow P to range outside Δ. From above a point P' є Ω exists such that d(P') < d(P) and as d(P<sub>0</sub>) ≤ d(P') it follows that d(P<sub>0</sub>) < d(P) for all P outside Δ. That means P<sub>0</sub> is the Fermat point of Δ. In other words '''the Fermat point is coincident with the first isogonic center'''.
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| ===Vector analysis===
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| Let ''O'', ''A'', ''B'', ''C'', ''X'' be any five points in a plane. Denote the vectors <math>\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}}, \overrightarrow{\mathrm{OC}}, \overrightarrow{\mathrm{OX}}</math> by '''a''', '''b''', '''c''', '''x''' respectively, and let '''i''', '''j''', '''k''' be the unit vectors from ''O'' along '''a''', '''b''', '''c'''. <br />
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| Now |'''a'''| = '''a⋅i''' = ('''a''' − '''x''')'''⋅i''' + '''x⋅i''' ≤ |'''a''' − '''x'''| + '''x⋅i''' and similarly |'''b'''| ≤ |'''b''' − '''x'''| + '''x⋅j''' and |'''c'''| ≤ |'''c''' − '''x'''| + '''x⋅k'''.<br />
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| Adding gives |'''a'''| + |'''b'''| + |'''c'''| ≤ |'''a''' − '''x'''| + |'''b''' − '''x'''| + |'''c''' − '''x'''| + '''x⋅'''('''i''' + '''j''' + '''k''').<br />
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| If '''a''', '''b''', '''c''' meet at ''O'' at angles of 120° then '''i''' + '''j''' + '''k''' = '''0''' so |'''a'''| + |'''b'''| + |'''c'''| ≤ |'''a''' − '''x'''| + |'''b''' − '''x'''| + |'''c''' − '''x'''| for all '''x'''.<br />
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| In other words ''OA'' + ''OB'' + ''OC'' ≤ ''XA'' + ''XB'' + ''XC'' and hence ''O'' is the Fermat point of ''ABC''. <br />
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| This argument fails when the triangle has an angle ''∠C'' > 120° because there is no point ''O'' where '''a''', '''b''', '''c''' meet at angles of 120°. Nevertheless it is easily fixed by redefining '''k''' = − ('''i''' + '''j''') and placing ''O'' at ''C'' so that '''c''' = '''0'''. Note that |'''k'''| ≤ 1 because the angle between the unit vectors '''i''' and '''j''' is ''∠C'' which exceeds 120°. Since |'''0'''| ≤ |'''0''' − '''x'''| + '''x⋅k''' the third inequality still holds, the other two inequalities are unchanged. The proof now continues as above (adding the three inequalities and using '''i''' + '''j''' + '''k''' = '''0''') to reach the same conclusion that ''O'' (or in this case ''C'') must be the Fermat point of ''ABC''.<br />
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| ===Lagrangian multipliers===
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| Another approach to find a point within the triangle, from where sum of the distances to the [[vertex (geometry)|vertices]] of triangle is minimum, is to use one of the [[optimization (mathematics)]] methods. In particular, method of the [[Lagrange multipliers]] and the [[law of cosines]].
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| We draw lines from the point within the triangle to its vertices and call them '''X''', '''Y''' and '''Z'''. Also, let the lengths of these lines be x, y, and z, respectively. Let the angle between '''X''' and '''Y''' be α, '''Y''' and '''Z''' be β. Then the angle between '''X''' and '''Z''' is (2π − α − β). Using the method of Lagrange multipiers we have to find the minimum of the Lagrangian ''L'', which is expressed as:
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| : ''L'' = ''x'' + ''y'' + ''z'' + ''λ''<sub>1</sub> (''x''<sup>2</sup> + ''y''<sup>2</sup> − 2''xy'' cos(''α'') − ''a''<sup>2</sup>) + ''λ''<sub>2</sub> (''y''<sup>2</sup> + ''z''<sup>2</sup> − 2''yz'' cos(β) − ''b''<sup>2</sup>) + ''λ''<sub>3</sub> (''z''<sup>2</sup> + ''x''<sup>2</sup> − 2''zx'' cos(''α'' + ''β'') − ''c''<sup>2</sup>)
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| where ''a'', ''b'' and ''c'' are the lengths of the sides of the triangle.
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| Equating each of the five partial derivatives δ''L''/δx, δ''L''/δy, δ''L''/δz, δ''L''/δα, δ''L''/δβ to zero and eliminating ''λ''<sub>1</sub> , ''λ''<sub>2</sub> , ''λ''<sub>3</sub> eventually gives sin(α) = sin(β) and sin(α + β) = − sin(β) so α = β = 120°. However the elimination is a long tedious business and the end result only covers Case 2. <br />
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| == Properties ==
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| [[File:Isogonic centres and vesicae piscis.png|thumb|300px|The two isogonic centers are the intersections of three [[vesica piscis|vesicae piscis]] whose paired vertices are the vertices of the triangle]]
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| * When the largest angle of the triangle is not larger than 120°, ''X''(13) is the Fermat point.
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| * The angles subtended by the sides of the triangle at ''X''(13) are all equal to 120° (Case 2), or 60°, 60°, 120° (Case 1).
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| * The [[circumcircle]]s of the three constructed equilateral triangles are concurrent at ''X''(13).
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| * [[Trilinear coordinates]] for the first isogonic center, ''X''(13):
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| :csc(''A'' + π/3) : csc(''B'' + π/3) : csc(''C'' + π/3), or, equivalently,
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| :sec(''A'' − π/6) : sec(''B'' − π/6) : sec(''C'' − π/6).<ref>Entry X(13) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers]</ref>
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| * [[Trilinear coordinates]] for the second isogonic center, ''X''(14):
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| :csc(''A'' − π/3) : csc(''B'' − π/3) : csc(''C'' − π/3), or, equivalently,
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| :sec(''A'' + π/6) : sec(''B'' + π/6) : sec(''C'' + π/6).<ref>Entry X(14) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers]</ref>
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| * [[Trilinear coordinates]] for the Fermat point:
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| :1 − ''u'' + ''uvw'' sec(''A'' − π/6) : 1 − ''v'' + ''uvw'' sec(''B'' − π/6) : 1 − ''w'' + ''uvw'' sec(''C'' − π/6)
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| : where ''u, v, w'' respectively denote the [[Boolean domain|Boolean variables]] (''A''<120°), (''B''<120°), (''C''<120°).
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| * The isogonal conjugate of ''X''(13) is the [[isodynamic point|first isodynamic point]], ''X''(15):
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| :sin(''A'' + π/3) : sin(''B'' + π/3) : sin(''C'' + π/3).<ref>Entry X(15) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers]</ref>
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| * The isogonal conjugate of ''X''(14) is the [[isodynamic point|second isodynamic point]], ''X''(16):
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| :sin(''A'' − π/3) : sin(''B'' − π/3) : sin(''C'' − π/3).<ref>Entry X(16) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers]</ref>
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| * The following triangles are equilateral:
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| :[[pedal triangle|antipedal triangle]] of ''X''(13)
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| :antipedal triangle of ''X''(14)
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| :[[pedal triangle]] of ''X''(15)
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| :pedal triangle of ''X''(16)
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| :circumcevian triangle of ''X''(15)
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| :circumcevian triangle of ''X''(16)
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| * The lines ''X''(13)''X''(15) and ''X''(14)''X''(16) are parallel to the [[Euler line]]. The three lines meet at the Euler infinity point, ''X''(30).
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| * The points ''X''(13), ''X''(14), the [[circumcircle|circumcenter]], and the [[nine-point circle|nine-point center]] lie on a [[Lester's theorem|Lester circle]].
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| == Aliases ==
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| The '''isogonic centers''' ''X''(13) and ''X''(14) are also known as the '''first Fermat point''' and the '''second Fermat point''' respectively. Alternatives are the '''positive Fermat point''' and the '''negative 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 the '''Fermat point''' and the '''first Fermat point''' whereas it is only in Case 2 above that they are actually the same.
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| == History ==
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| This question was proposed by Fermat, as a challenge to [[Evangelista Torricelli]]. He solved the problem in a similar way to Fermat's, albeit using intersection of the circumcircles of the three regular triangles instead. His pupil, Viviani, published the solution in 1659.<ref>{{MathWorld|urlname=FermatPoints |title=Fermat Points}}</ref>
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| ==See also==
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| *[[Geometric median]] or Fermat–Weber point, the point minimizing the sum of distances to more than three given points.
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| *[[Lester's theorem]]
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| *[[Triangle center]]
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| *[[Napoleon points]]
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| *[[Weber problem]]
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| ==References==
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| {{reflist}}
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| == External links ==
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| * {{springer|title=Fermat-Torricelli problem|id=p/f130050}}
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| * ''[http://demonstrations.wolfram.com/FermatPoint/ Fermat Point]'' by Chris Boucher, [[The Wolfram Demonstrations Project]].
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| * [http://dynamicmathematicslearning.com/fermat-general.html Fermat-Torricelli generalization] at [http://dynamicmathematicslearning.com/JavaGSPLinks.htm Dynamic Geometry Sketches] Interactive sketch generalizes the Fermat-Torricelli point.
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| * [http://www.matifutbol.com/en/triangle.eng.html A practical example of the Fermat point]
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| [[Category:Triangle centers]]
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| [[Category:Articles containing proofs]]
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