نقطة فيرما

إنشاء نقطة فيرما.

في الهندسة الرياضية، يطلق اسم نقطة فيرما على حل مسألة إيجاد نقطة F داخل مثلث ABC بحيث أن المسافة الكلية من هذه النقطة إلى رؤوس المثلث الثلاثة تكون أصغرية. سميت هذه النقطة على اسم بيير فيرما الذي كان أول من قام بإنشاءها.

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 إنشاء نقطة فيرما

تتبع الخطوات التالية لإنشاء نقطة فيرما كما هي موضحة بالشكل:

1. أنشىء ثلاث مثلثات متساوية الأضلاع من كل ضلع من أضلاع المثلث المفترض
2. من كل رأس جديد في المثلث المتساوي الأضلاع ارسم مستقيم يصل هذا الرأس بالرأس المقابل للمثلث الأصلي
3. تتقاطع المستقيمات الثلاثة في نقطة هي نقطة فيرما.

 موضع X(13)

Fig 2. Geometry of the first isogonic center.

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

The triangles △RAC, △BAQ are congruent because the second is a 60° rotation of the first about A. Hence ∠ARF = ∠ABF and ∠AQF = ∠ACF. By the converse of the inscribed angle theorem applied to the segment AF, the points ARBF are concyclic (they lie on a circle). Similarly, the points AFCQ are concyclic.

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

So ∠BFC = 120°. Therefore, ∠BFC + ∠BPC = 180°. Using the inscribed angle theorem, this implies that the points BPCF are concyclic. So, using the inscribed angle theorem applied to the segment BP, ∠BFP = ∠BCP = 60°. Because ∠BFP + ∠BFA = 180°, the point F lies on the line segment AP. So, the lines RC, BQ, AP are concurrent (they intersect at a single point). Q.E.D.

This proof applies only in Case 2, since if ∠BAC > 120°, point 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.

The lines joining the centers of the circles in Fig. 2 are perpendicular to the line segments AP, 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 segment AP. 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 as Napoleon's Theorem.

 موضع نقطة فيرما

 الهندسة التقليدية

Fig 3. Geometry of the Fermat point

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 P₀ such that $d(P_{0})<d(P)$ for all $P\neq P_{0}.$ 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.

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:

If AB is the common side, extend AC to cut the polygon at the point X. Then the polygon's perimeter is, by the triangle inequality:

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

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 $d(P')<d(P).$

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

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 $P\in \Delta ,P\neq A.$ Now allow P to range outside Δ. From above a point $P'\in \Omega$ exists such that $d(P')<d(P)$ and as $d(A)\leq d(P')$ it follows that $d(A)<d(P)$ for all P outside Δ. Thus $d(A)<d(P)$ for all $P\neq A$ which means that A is the Fermat point of Δ. 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, 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 P₀ be the point where AD and CF intersect. This point is commonly called the first isogonic center. Carry out the same exercise with P₀ as you did with P, and find the point Q₀. By the angular restriction P₀ lies inside △ABC. Moreover, △BCF is a 60° rotation of △BDA about B, so Q₀ must lie somewhere on AD. Since ∠CDB = 60° it follows that Q₀ lies between P₀ and D which means AP₀Q₀D is a straight line so $d(P_{0}=|AD|.$ Moreover, if $P\neq P_{0}$ then either P or Q won't lie on AD which means $d(P_{0})=|AD|<d(P).$ Now allow P to range outside Δ. From above a point $P'\in \Omega$ exists such that $d(P')<d(P)$ and as $d(P_{0})\leq d(P')$ it follows that $d(P_{0})<d(P)$ for all P outside Δ. That means P₀ is the Fermat point of Δ. In other words, the Fermat point is coincident with the first isogonic center.

 تحليل المتجهات

Let O, A, B, C, X be any five points in a plane. Denote the vectors ${\overrightarrow {OA}},\ {\overrightarrow {OB}},\ {\overrightarrow {OC}},\ {\overrightarrow {OX}}$ by a, b, c, x respectively, and let i, j, k be the unit vectors from O along a, b, c.

${\begin{aligned}|\mathbf {a} |&=\mathbf {a\cdot i} =(\mathbf {a} -\mathbf {x} )\mathbf {\,\cdot \,i} +\mathbf {x\cdot i} \leq |\mathbf {a} -\mathbf {x} |+\mathbf {x\cdot i} ,\\|\mathbf {b} |&=\mathbf {b\cdot j} =(\mathbf {b} -\mathbf {x} )\mathbf {\,\cdot \,j} +\mathbf {x\cdot j} \leq |\mathbf {b} -\mathbf {x} |+\mathbf {x\cdot j} ,\\|\mathbf {c} |&=\mathbf {c\cdot k} =(\mathbf {c} -\mathbf {x} )\mathbf {\,\cdot \,k} +\mathbf {x\cdot k} \leq |\mathbf {c} -\mathbf {x} |+\mathbf {x\cdot k} .\end{aligned}}$

Adding a, b, c gives

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

If a, b, c meet at O at angles of 120° then i + j + k = 0, so

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

for all x. In other words,

$|OA|+|OB|+|OC|\leq |XA|+|XB|+|XC|$

and hence O is the Fermat point of △ABC.

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, j is ∠C which exceeds 120°. Since

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

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.

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 Lagrange multipliers

Another approach to finding the point within a triangle, from which the sum of the distances to the vertices of the triangle is minimal, is to use one of the mathematical optimization methods; specifically, the method of Lagrange multipliers and the law of cosines.

We draw lines from the point within the triangle to its vertices and call them X, Y, Z. Also, let the lengths of these lines be x, y, z respectively. Let the angle between X and Y be α, Y and Z be β. Then the angle between X and Z is π − α − β. Using the method of Lagrange multipliers we have to find the minimum of the Lagrangian L, which is expressed as:

$L=x+y+z+\lambda _{1}(x^{2}+y^{2}-2xy\cos(\alpha )-a^{2})+\lambda _{2}(y^{2}+z^{2}-2yz\cos(\beta )-b^{2})+\lambda _{3}(z^{2}+x^{2}-2zx\cos(\alpha +\beta )-c^{2})$

where a, b, c are the lengths of the sides of the triangle.

Equating each of the five partial derivatives ${\tfrac {\partial L}{\partial x}},{\tfrac {\partial L}{\partial y}},{\tfrac {\partial L}{\partial z}},{\tfrac {\partial L}{\partial \alpha }},{\tfrac {\partial L}{\partial \beta }}$ to zero and eliminating λ₁, λ₂, λ₃ eventually gives sin α = sin β and sin(α + β) = − sin β so α = β = 120°. However the elimination is a long and tedious business, and the end result covers only Case 2.

 الخصائص

The two isogonic centers are the intersections of three vesicae piscis whose paired vertices are the vertices of the triangle

• 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 at X(13) are all equal to 120° (Case 2), or 60°, 60°, 120° (Case 1).
• The circumcircles of the three constructed equilateral triangles are concurrent at X(13).
• Trilinear coordinates for the first isogonic center, X(13):

${\begin{aligned}&\csc \left(A+{\tfrac {\pi }{3}}\right):\csc \left(B+{\tfrac {\pi }{3}}\right):\csc \left(C+{\tfrac {\pi }{3}}\right)\\&=\sec \left(A-{\tfrac {\pi }{6}}\right):\sec \left(B-{\tfrac {\pi }{6}}\right):\sec \left(C-{\tfrac {\pi }{6}}\right).\end{aligned}}$

• Trilinear coordinates for the second isogonic center, X(14):

${\begin{aligned}&\csc \left(A-{\tfrac {\pi }{3}}\right):\csc \left(B-{\tfrac {\pi }{3}}\right):\csc \left(C-{\tfrac {\pi }{3}}\right)\\&=\sec \left(A+{\tfrac {\pi }{6}}\right):\sec \left(B+{\tfrac {\pi }{6}}\right):\sec \left(C+{\tfrac {\pi }{6}}\right).\end{aligned}}$

• Trilinear coordinates for the Fermat point:

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

where u, v, w respectively denote the Boolean variables (A < 120°), (B < 120°), (C < 120°).

• The isogonal conjugate of X(13) is the first isodynamic point, X(15):

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

• The isogonal conjugate of X(14) is the second isodynamic point, X(16):

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

• The following triangles are equilateral:
  • antipedal triangle of X(13)
  • Antipedal triangle of X(14)
  • Pedal triangle of X(15)
  • Pedal triangle of X(16)
  • Circumcevian triangle of X(15)
  • Circumcevian triangle of X(16)
• 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).
• The points X(13), X(14), the circumcenter, and the nine-point center lie on a Lester circle.
• The line X(13)X(14) meets the Euler line at midpoint of X(2) and X(4).
• The Fermat point lies in the open orthocentroidal disk punctured at its own center, and could be any point therein.

 أسماء بديلة

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.

 انظر أيضاً

• Geometric median or Fermat–Weber point, the point minimizing the sum of distances to more than three given points.
• مبرهنة لستر
• Triangle center
• نقاط ناپليون
• Weber problem

 وصلات خارجية

• Hazewinkel, Michiel, ed. (2001), "Fermat-Torricelli problem", Encyclopaedia of Mathematics, Kluwer Academic Publishers, ISBN 978-1556080104 
• Fermat Point by Chris Boucher, The Wolfram Demonstrations Project.
• Fermat-Torricelli generalization at Dynamic Geometry Sketches Interactive sketch generalizes the Fermat-Torricelli point.
• A practical example of the Fermat point
• iOS Interactive sketch