Fermat point

First Fermat point

The first Fermat point and the second Fermat point , named after the French judge and mathematician Pierre de Fermat , are among the special points of a triangle . Both Fermat points are isogonally conjugated to the two isodynamic points . They also lie on the Kiepert hyperbola . In the relevant literature, the much better known first Fermat point is usually referred to as the Fermat point .

First Fermat point

Already in 1647 Bonaventura Cavalieri showed : If all angles of the triangle are smaller than 120 °, then the first Fermat point of the triangle is the point inside the triangle from which all three sides are seen at a 120 ° angle ( Picture 1 and 3); this means ${\displaystyle ABC}$ ${\displaystyle F_{1}}$

(1)  ${\displaystyle \angle AF_{1}B=\angle BF_{1}C=\angle CF_{1}A=120^{\circ }.}$

History

It was probably the year 1646 when Fermat wrote the manuscript "MAXIMA ET MINIMA" by setting the following task to the scholars of his time:

Datis tribus punctis, quartum reperire, a quo si ducantur tres rectæ ad data puncta, summa trium harum rectarum sit minima quantitas

“ There are three points, we are looking for a fourth point, so that the sum of its distances from the three given points becomes a minimum. "

In the same year Evangelista Torricelli found three elementary solutions that Torricelli's pupil Vincenzo Viviani published together with one of his own in 1659.

Torricelli supplied u. a. a geometric solution (Fig. 1) that can be represented with a compass and ruler . The perimeters of the three equilateral triangles, built over the sides of the starting triangle (location triangle), intersect at a point. The Fermat point generated in this way is also called the Torricelli point . Incidentally, his method is suitable for both the first Fermat point and the second Fermat point (Fig. 4–5). ${\displaystyle F_{1}}$ ${\displaystyle F_{2}}$

Finally proved Thomas Simpson , that the three lines extending from each one of the equilateral triangles of the opposite corner of the triangle site, meet in the Torricelli-point . These three lines are therefore also called the Simpson lines .

construction

Over the sides of a given triangle , three equilateral triangles are built , hereinafter referred to as attachment triangles . Now there are two ways to determine the first Fermat point (Fig. 1): ${\displaystyle ABC}$ ${\displaystyle F_{1}}$

A) You connect the newly added points and with the opposite corners of the triangle (i.e. with and ), so these connecting lines intersect at a point This is called the first Fermat point of the triangle. ${\displaystyle A_{1},B_{1}}$ ${\displaystyle C_{1}}$ ${\displaystyle A,B}$ ${\displaystyle C}$ ${\displaystyle F_{1}.}$

B) Determine the circumference of the three attachment triangles. As an intersection, they provide the first Fermat point , as described above in History , also called the Torricelli point . ${\displaystyle F_{1}}$

Figure 1: General triangle, first Fermat point or Torricelli point ${\displaystyle F_{1}}$

Fig. 2: Triangle, angle ≥120 °, first Fermat point or Torricelli point lies on an angle vertex ≥120 ° ${\displaystyle F_{1},}$ ${\displaystyle F_{1}}$

Image 3: Equilateral triangle, first Fermat point or Torricelli point equal to the center of the circle ${\displaystyle F_{1},}$ ${\displaystyle F_{1}}$

properties

• If all angles of the given triangle are smaller than 120 ° (Fig. 1 and 3), then the first Fermat point is the point for which the sum of the distances from the corners of the triangle (i.e. the sum ) assumes the smallest possible value. ${\displaystyle ABC}$ ${\displaystyle F_{1}}$ ${\displaystyle ABC}$ ${\displaystyle {\overline {F_{1}A}}+{\overline {F_{1}B}}+{\overline {F_{1}C}}}$

The proof of this fact comes from the Italian Evangelista Torricelli. This is why the Fermat-Torricelli point is sometimes used .

• If, on the other hand, one of the angles of the triangle is greater than or equal to 120 ° (Fig. 2), then the solution is precisely the point at which this angle is located , i.e. H. the first Fermat point coincides with the vertex of the 120 ° angle. ${\displaystyle ABC}$ ${\displaystyle F_{1}}$
• In the case of an equilateral triangle (Fig. 3), the first Fermat point corresponds to the inscribed center . ${\displaystyle ABC}$ ${\displaystyle F_{1}}$

application

The first Fermat point is used in business mathematics , especially in location planning . Suppose three companies want to build a central warehouse in such a way that the transport costs to this central warehouse are minimal. The central warehouse would have to be built at the location of the Fermat point , if one imagines the location of the three companies as a triangle, since the sum of the distances to the corners of the triangle is minimal for the Fermat point (with all angles in the triangle smaller than 120 °). ${\displaystyle F_{1}}$ ${\displaystyle F_{1}}$ ${\displaystyle F_{1}}$

Second Fermat point

For the second Fermat point (vertex) the following applies, regardless of which interior angle the triangle has ${\displaystyle F_{2}}$ ${\displaystyle ABC}$

${\displaystyle \angle AF_{2}B=\angle BF_{2}C=60^{\circ }}$ and ${\displaystyle \angle AF_{2}C=120^{\circ }.}$

construction

The second Fermat point of a triangle results from the same construction as the first Fermat point, only the three top triangles do not have to be erected “outwards” over the sides of the triangle, but “inwards”. ${\displaystyle F_{2}}$ ${\displaystyle F_{1},}$

Figure 4: General triangle, second Fermat point or Torricelli point ${\displaystyle F_{2}}$

Figure 5: Triangle, angle 60 °, second Fermat point or Torricelli point lies on the vertex of the angle 60 ° ${\displaystyle F_{2},}$ ${\displaystyle F_{2}}$

properties

• In contrast to the first Fermat point , the second Fermat point generally does not have the minimum property. It only fulfills them if it coincides with one of the corner points of the starting triangle (Fig. 5). ${\displaystyle F_{2}}$ ${\displaystyle F_{1}}$ ${\displaystyle \triangle {ABC}}$
• If the triangle has a 60 ° angle (Fig. 5), then the second Fermat point corresponds to the vertex of the 60 ° angle. ${\displaystyle ABC}$ ${\displaystyle F_{2}}$
• If the triangle is equilateral, it is congruent to the three (equilateral) attachment triangles , i.e. H. the four triangles are on top of each other, so and correspond ${\displaystyle ABC}$ ${\displaystyle A_{1}=A,}$ ${\displaystyle B_{1}=B,}$ ${\displaystyle C_{1}=C.}$

Consequently, each of three vertices or less a second Fermat point be. ${\displaystyle A,B}$ ${\displaystyle C}$ ${\displaystyle F_{2}}$

proofs

In Lemma 1 and Lemma 2 we use the properties of vectors and their scalar product in the Euclidean plane .

Lemma 1

For all vectors is ${\displaystyle {\overrightarrow {a}},{\overrightarrow {b}},{\overrightarrow {c}}\neq {\overrightarrow {0}},}$

${\displaystyle {\frac {\overrightarrow {a}}{|{\overrightarrow {a}}|}}+{\frac {\overrightarrow {b}}{|{\overrightarrow {b}}|}}+{\frac {\overrightarrow {c}}{|{\overrightarrow {c}}|}}={\overrightarrow {0}}}$

equivalent to saying that

${\displaystyle {\frac {\overrightarrow {a}}{|{\overrightarrow {a}}|}},{\frac {\overrightarrow {b}}{|{\overrightarrow {b}}|}},{\frac {\overrightarrow {c}}{|{\overrightarrow {c}}|}}}$ each have an angle of 120 ° to each other.

Proof of Lemma 1

We define unit vectors by ${\displaystyle {\overrightarrow {e_{i}}}\ (i=0,1,2)}$

${\displaystyle {\overrightarrow {e_{0}}}={\frac {\overrightarrow {a}}{|{\overrightarrow {a}}|}},{\overrightarrow {e_{1}}}={\frac {\overrightarrow {b}}{|{\overrightarrow {b}}|}},{\overrightarrow {e_{2}}}={\frac {\overrightarrow {c}}{|{\overrightarrow {c}}|}}.}$

and denote by the angle between the two unit vectors . ${\displaystyle \theta _{ij}}$ ${\displaystyle {\overrightarrow {e_{i}}},{\overrightarrow {e_{j}}}}$

Then we have for example

${\displaystyle 1=|-{\overrightarrow {e_{2}}}|^{2}=|{\overrightarrow {e_{0}}}+{\overrightarrow {e_{1}}}|^{2}=|{\overrightarrow {e_{0}}}|^{2}+2{\overrightarrow {e_{0}}}\cdot {\overrightarrow {e_{1}}}+|{\overrightarrow {e_{1}}}|^{2}=2+2{\overrightarrow {e_{0}}}\cdot {\overrightarrow {e_{1}}}}$,

So , the same for the other pairs of points. ${\displaystyle {\overrightarrow {e_{0}}}\cdot {\overrightarrow {e_{1}}}=-{\tfrac {1}{2}}}$

So we get and the values ​​of the inner product as ${\displaystyle \theta _{ij}=\theta _{ji}}$

${\displaystyle {\overrightarrow {e_{i}}}\cdot {\overrightarrow {e_{j}}}=\cos \theta _{ij}={\begin{cases}1&(i=j)\\-{\frac {1}{2}}&(i\neq j).\end{cases}}}$

With that we get ${\displaystyle \theta _{ij}=120^{\circ }\ (i\neq j).}$

Conversely, if unit vectors have an angle of 120 ° to one another, one obtains ${\displaystyle {\overrightarrow {e_{i}}}\ (i=0,1,2)}$

${\displaystyle |{\overrightarrow {e_{0}}}+{\overrightarrow {e_{1}}}+{\overrightarrow {e_{2}}}|^{2}=\sum _{i=j}^{}{\overrightarrow {e_{i}}}\cdot {\overrightarrow {e_{j}}}+\sum _{i\neq j}^{}{\overrightarrow {e_{i}}}\cdot {\overrightarrow {e_{j}}}=3\times 1+6\times \left(-{\frac {1}{2}}\right)=0.}$

That's why we get

${\displaystyle {\overrightarrow {e_{0}}}+{\overrightarrow {e_{1}}}+{\overrightarrow {e_{2}}}={\overrightarrow {0}}.}$ Qed

Lemma 2

For all vectors and applies ${\displaystyle {\overrightarrow {a}}\neq {\overrightarrow {0}}}$ ${\displaystyle {\overrightarrow {x}}}$

${\displaystyle |{\overrightarrow {a}}-{\overrightarrow {x}}|\geq |{\overrightarrow {a}}|-{\frac {\overrightarrow {a}}{|{\overrightarrow {a}}|}}\cdot {\overrightarrow {x}}.}$

Proof of Lemma 2

This follows from the inequality that applies to all vectors by inserting Qed ${\displaystyle {\overrightarrow {u}},{\overrightarrow {v}}}$ ${\displaystyle |{\overrightarrow {u}}||{\overrightarrow {v}}|\geq {\overrightarrow {u}}\cdot {\overrightarrow {v}}}$ ${\displaystyle {\overrightarrow {u}}={\frac {\overrightarrow {a}}{|{\overrightarrow {a}}|}},{\overrightarrow {v}}={\overrightarrow {a}}-{\overrightarrow {x}}.}$

If all interior angles in the triangle are smaller than 120 °, we can construct the Fermat point inside the triangle . Then we set ${\displaystyle ABC}$ ${\displaystyle F_{1}}$ ${\displaystyle ABC}$ ${\displaystyle {\overrightarrow {a}}={\overrightarrow {F_{1}A}},{\overrightarrow {b}}={\overrightarrow {F_{1}B}},{\overrightarrow {c}}={\overrightarrow {F_{1}C}},{\overrightarrow {x}}={\overrightarrow {F_{1}X}}.}$

If the Fermat point is, then by definition so we get the equation from Lemma 1. ${\displaystyle F_{1}}$ ${\displaystyle \angle AF_{1}B=\angle BF_{1}C=\angle CF_{1}A=120^{\circ },}$

From Lemma 2 we see that

${\displaystyle |{\overrightarrow {XA}}|\geq |{\overrightarrow {F_{1}A}}|-{\frac {\overrightarrow {a}}{|{\overrightarrow {a}}|}}\cdot {\overrightarrow {x}},}$

${\displaystyle |{\overrightarrow {XB}}|\geq |{\overrightarrow {F_{1}B}}|-{\frac {\overrightarrow {b}}{|{\overrightarrow {b}}|}}\cdot {\overrightarrow {x}},}$

${\displaystyle |{\overrightarrow {XC}}|\geq |{\overrightarrow {F_{1}C}}|-{\frac {\overrightarrow {c}}{|{\overrightarrow {c}}|}}\cdot {\overrightarrow {x}}.}$

From these three inequalities and the equation of Lemma 1 it follows

${\displaystyle |{\overrightarrow {XA}}|+|{\overrightarrow {XB}}|+|{\overrightarrow {XC}}|\geq |{\overrightarrow {F_{1}A}}|+|{\overrightarrow {F_{1}B}}|+|{\overrightarrow {F_{1}C}}|.}$

This applies to every point X in the Euclidean plane. We have shown that if X = , then the value becomes minimal. Qed ${\displaystyle F_{1}}$ ${\displaystyle |{\overrightarrow {XA}}|+|{\overrightarrow {XB}}|+|{\overrightarrow {XC}}|}$

Hofmann proof

The following proof for triangles with inner angles smaller than comes from Joseph Ehrenfried Hofmann , from his 1929 article Elementary solution of a minimum problem in the journal for mathematical and scientific teaching . Even if less known than the classic, analytical evidence, it convinces with its simplicity and comprehensibility. ${\displaystyle 120^{\circ }}$

The approach is the rotation ( coordinate transformation ) of any point within a triangle with the center (coordinate origin). The definition of the function in the plane applies ${\displaystyle P}$ ${\displaystyle ABC}$ ${\displaystyle M.}$

${\displaystyle f_{M}\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\;f_{M}\left(P\right)=P'}$

If the point of the triangle is the center and the angle of rotation , then (Fig. 6) ${\displaystyle A}$ ${\displaystyle ABC}$ ${\displaystyle 60^{\circ }}$

${\displaystyle f_{A}\left(P\right)=P'}$ and

${\displaystyle f_{A}\left(C\right)=B_{1}.}$

Figure 6: Hofmann proof of the first Fermat point, broken line

Fig. 7: Hofmann proof of the first Fermat point, shortest possible length , see animation

One can deduce from the equilateral triangles and and the congruent triangles and ( second congruence theorem SWS ) ${\displaystyle APP'}$ ${\displaystyle ACB_{1}}$ ${\displaystyle APC}$ ${\displaystyle AP'B_{1}}$

${\displaystyle d=d(P,A)+d(P,B)+d(P,C)=|PA|+|PB|+|PC|=|PP'|+|PB|+|P'B_{1}|.}$

In words: stands for the sum of the three distances , i.e. H. for the length of the broken line It is equal to the sum of the connecting lines from to the corners of the triangle ${\displaystyle d}$ ${\displaystyle B_{1}P'PB.}$ ${\displaystyle P}$ ${\displaystyle ABC.}$

The shortest possible length of is achieved when the points and lie on a common straight line, because this results in (Figure 7) ${\displaystyle d}$ ${\displaystyle B_{1},P',P}$ ${\displaystyle B}$

${\displaystyle \angle {B_{1}P'P}=180^{\circ }}$ and ${\displaystyle \angle {P'PB}=180^{\circ },}$

because of

${\displaystyle \angle {AP'P}=60^{\circ },}$ or because of the circumference of the ( circle angle set with center angle ) ${\displaystyle P}$ ${\displaystyle \triangle ACB_{1}}$ ${\displaystyle 120^{\circ }}$

follows

${\displaystyle \angle {B_{1}P'A}=\angle {CPA}=120^{\circ };}$

consequently is also

${\displaystyle \angle {APB}=\angle {BPC}=\angle {CPA}=120^{\circ }.}$

If one sets for the same (Fig. 1), this is proven ${\displaystyle P}$ ${\displaystyle F_{1}}$

(1) ${\displaystyle \angle AF_{1}B=\angle BF_{1}C=\angle CF_{1}A=120^{\circ }.}$

Coordinates

Fermat points ( and ) ${\displaystyle X_{13}}$ ${\displaystyle X_{14}}$
Trilinear coordinates	${\displaystyle \csc \left(\alpha \pm {\frac {\pi }{3}}\right)\,:\,\csc \left(\beta \pm {\frac {\pi }{3}}\right)\,:\,\csc \left(\gamma \pm {\frac {\pi }{3}}\right)}$
Barycentric coordinates	${\displaystyle a\cdot \csc \left(\alpha \pm {\frac {\pi }{3}}\right)\,:\,b\cdot \csc \left(\beta \pm {\frac {\pi }{3}}\right)\,:\,c\cdot \csc \left(\gamma \pm {\frac {\pi }{3}}\right)}$

See also

• Excellent points in the triangle

literature

• Joseph Ehrenfried Hofmann : Elementary solution to a minimum problem. In: Journal for mathematical and scientific teaching , 60, 1929, pp. 22–23.
• Harold Scott MacDonald Coxeter: Immortal Geometry. (= Science and Culture , Volume 17). Birkhäuser, Basel / Stuttgart 1963, pp. 39–39.
• Hans Schupp: Elementarge Geometry (= university pocket books. 669 Mathematics ). Schöningh, Paderborn 1977, ISBN 3-506-99189-2 , pp. 79-82.
• Hans Schupp: Figures and illustrations (= study and teaching of mathematics ). Franzbecker, Hildesheim 1998, ISBN 3-88120-288-9 , pp. 54-55.

Web links

• Eric W. Weisstein : Fermat Points . In: MathWorld (English).
• Fermat point - a Java visualization with the help of GeoGebra .
• A practical example of the Fermat point (English)

Individual evidence

1. ↑ ^{a b c} Ulrich Eckhardt: Shortest routes and optimal locations - From industrial sites, bombs and soap bubbles. (PDF) 2.1 The task of Fermat. University of Hamburg Department of Mathematics, April 11, 2008, p. 15 , accessed on September 8, 2019 . 
2. ↑ Pierre de Fermat: Œvres de Fermat. Tom Premier. Œvres math ́ematiques diverse. - Observations on Diophante. (PDF) MAXIMA ET MINIMA. University of Michigan Library Digital Collections, 1841, p. 153 , accessed September 8, 2019 . 
3. ↑ ^{a b} Ulrich Eckhardt: Shortest routes and optimal locations - From industrial sites, bombs and soap bubbles. (PDF) 2.1 The task of Fermat. University of Hamburg Department of Mathematics, April 11, 2008, p. 13 , accessed on September 8, 2019 . 
4. ↑ Ulrich Eckhardt: Shortest routes and optimal locations - From industrial sites, bombs and soap bubbles. (PDF) 2.1 The task of Fermat. University of Hamburg Department of Mathematics, April 11, 2008, p. 14 , accessed on September 8, 2019 . 
5. ↑ Tasja Werner: The Fermat point - an extension of the school geometry. (PDF) 7.4 The 2nd Fermat point. University of Bremen, Department of Mathematics / Computer Science, April 21, 2008, p. 13 , accessed on September 8, 2019 . 
6. ↑ Peter Andree: The Point of Fermat. (PDF) 9.1 Fermat's assignment to Torricelli. Headquarters for teaching media on the Internet e. V., January 15, 2004, p. 1 , accessed on September 15, 2019 . 
7. ↑ ^{a b} H.S. Coxeter: Immortal geometry, triangles . Springer Basel, 1981, ISBN 978-3-0348-5152-7 , p. 38 Google
8. ↑ Peter Andree: The Point of Fermat. (PDF) 9.3 The Hofmann proof. The rotation and the point of Fermat. Headquarters for teaching media on the Internet e. V., January 15, 2004, p. 4 , accessed on September 15, 2019 .