Fermat Point is the point that Pierre de Fermat, the 17th-century French mathematician, posed as a challenge to his compatriot Evangelista Torricelli to geometrically determine, is named in Fermat’s honour as the solution that would minimize the total combined distance from each vertex of a triangular figure to any single internal locus. Torricelli solved the problem, therefore other than Fermat Point, it is also known as the Fermat Point or the Torricelli Point or Fermat Torricelli Point.

Although various methods exist to locate the Fermat point, connecting the vertices of the original triangle to those of the equilateral triangles constructed on each of its sides furnishes a straightforward technique. The intersection of these segments is the Fermat point.

The Fermat point gives a solution to the geometric median and Steiner tree problems for three points.

What is Fermat Point or Torricelli Point?

This point is related to the concept of Geometry. It is the point in inside a triangle from which the sum of the distance between vertices is minimum. the Fermat point can be found by constructing an equilateral triangle on the sides of the original triangle.

It is also known as the Torricelli point or the Fermat–Torricelli point. It is named after Pierre de Fermat and Evangelista Torricelli, who independently solved or worked on the problem of finding the Fermat point in the 17th century.

This question is proposed by Fermat, as a challenge in front of Italian mathematician Evangelista Torricelli. And He solved the problem in a similar way to Fermat, but the difference is that he used a circumcircle instead of three equilateral triangles as solved by Fermat.

Fermat Point Definition

In a Triangle, the sum of the distances to the three vertices is smallest from point which is known as Fermat Point. It is also known as Torricelli point or Fermat Torricelli Point. It is named after a French Mathematician, Pierre de Fermat. And it is named as Torricelli Point after a name of Italian Mathematician, Evangelista Torricelli.

Terminologies related to Fermat Point

Some of the common terms used to explain Fermat point and its construction are listed below:

• First Isogonic center: The first isogonic center of a triangle is a point in the plane of the triangle at which the three sides subtend each an angle of 120°.

• Second Isogonic center: The second isogonic center is that point at which two sides subtend each at an angle of 60° and the third side an angle of 120°.

• First Isodynamic center: The first isodynamic point is the intersection of three circles through the pairs of points AB, AC, and BC, where each of these circles intersects the circumcircle of △ABC to form a lens with an apex angle of 2π/3.

• Second Isodynamic Point: The Second Isodynamic Point is the intersection of three circles that intersect the circumcircle to form lenses with an apex angle of π/3.

• Euler Line: The Euler line of a triangle is a line going through several important triangle centers, including the orthocenter, circumcenter, centroid, and center of the nine-point circle.

• Euler Infinity Point: The Euler infinity point is the intersection of the Euler line and line at infinity. Since it lies on the line at infinity, it is a point at infinity.

How to Find Fermat Point in Triangle?

The Fermat point of a triangle with the largest angle at most 120° is simply its first isogonic center, which is constructed as follows:

Step 1: First, we need a triangle for finding a Fermat point. So, Draw a triangle with vertices A, B, and C.

Step 2: Then, draw an equilateral triangle on side AB of side length equal to side AB. And named it Triangle ABD.
 

Step 3: Similarly, let’s draw equilateral triangle’s on left sides AC and BC of their respective side lengths, and named them triangle BAE and triangle BCF.

Step 4: Now, Draw line DB.

Step 5: Similarly, draw a line between EC and FA.

Step 6: The point where these 3 lines meet is known as Fermat Line.

Note: There are different ways to find Fermat Point, but this one is popular one.

Different Cases of Fermat Point

For different triangles, Fermat Point can be inside or on the boundary of the triangle under consideration. Let’s discuss the following two cases of Fermat Point.

• Case 1: Triangle has an Angle ≥ 120°
• Case 2: Triangle has no Angle ≥ 120°

Case 1: 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 a triangle (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 triangle ABC, by the dogleg rule the length of this path exceeds |AC| + |AF| = d(A).

Therefore, d(A) < d(P) for all [Tex]P\in \Delta ,P\neq A. [/Tex]

Now allow P to range outside the triangle. From above a point [Tex] P’ \in \Omega   [/Tex] exists such that d(P’)<d(P) and as d(A) ≤ d(P’) it follows that d(A) < d(P) for all P outside triangle. Thus d(A) < d(P) for all [Tex]P\neq A   [/Tex] which means that A is the Fermat point of the triangle. In other words, the Fermat point lies at the obtuse-angled vertex.

Case 2: Triangle has no Angle ≥ 120°

• Construct the equilateral triangle BCD, let P be any point inside the triangle, and construct the equilateral triangle CPQ. Then triangle CQD is a 60° rotation of triangle 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 triangle ABC. Moreover, triangle BCF is a 60° rotation of triangle BDA about B, so Q₀ must lie somewhere on AD. Since angle CBD = 60°, it follows that Q₀ lies between P₀ and D which means AP₀Q₀D is a straight line so d(P₀) = |AD|.
• Moreover, if P ≠ P₀ then either P or Q won’t lie on AD which means d(P₀) = |AD| < d(P)
• Now allow P to range outside Δ. From above a point [Tex]\displaystyle P’\in \Omega   [/Tex] exists such that d(P’) < d(P) and as d(P₀) ≤ d(P) it follows that d(P₀) < d(P) for all P outside triangle BDA.
• That means P₀ is the Fermat point of the triangle. In other words, the Fermat point is coincident with the first isogonic center.

Properties of Fermat Point

There are various properties associated with Fermat Point, some of these properties are:

• If the triangle is equilateral, then the Fermat point will coincide with one of the vertex. Because in equilateral distance between any point within it is always equal.
• In a triangle whose all three interior angles are less than 120°, the Fermat point lies inside a triangle.
• When the largest angle of the triangle is not larger than 120°, the first isogonic center is the Fermat point.
• It lies on the circumcircle of the triangle.
• It is unique or we can say that in a triangle, there will be one and only one Fermat point exists.
• If We make a reflection of any of the triangle’s vertices across the corresponding side then the reflected vertex will lie on the line connecting the other two vertices. In other words, we can say that the reflected vertex will be collinear with the original triangle’s vertices and the Fermat point.
• And If one of the interior angles is greater than 120°, then the Fermat point lies outside a triangle.
• The circumcircles of the three constructed equilateral triangles are concurrent at Fermat Point.
• Trilinear coordinates for the first isogonic center:

[Tex]\displaystyle {\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}} [/Tex]

• Trilinear coordinates for the second isogonic center:

[Tex]\displaystyle {\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}} [/Tex]

• Trilinear coordinates for the Fermat point:

[Tex]\displaystyle 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) [/Tex]

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

• The isogonal conjugate of the first isogonic center is the first isodynamic point:

[Tex]\displaystyle \sin \left(A+{\tfrac {\pi }{3}}\right):\sin \left(B+{\tfrac {\pi }{3}}\right):\sin \left(C+{\tfrac {\pi }{3}}\right). [/Tex]

• The isogonal conjugate of the second isogonic center is the second isodynamic point:

[Tex]\displaystyle \sin \left(A-{\tfrac {\pi }{3}}\right):\sin \left(B-{\tfrac {\pi }{3}}\right):\sin \left(C-{\tfrac {\pi }{3}}\right). [/Tex]

• The lines joining the first isogonic center and first isodynamic point, and the second isogonic center and second isodynamic point are parallel to the Euler line. The three lines meet at the Euler infinity point.

These properties highlight the geometrical significance of the Fermat Point and its interesting characteristics of it, making it an interesting concept in the field of geometry.

Important Points about Fermat Point

Some important points about Fermat Point are:

• Fermat Point mostly be inside a triangle, and it is equidistant from the three medians of a triangle.
• It is the center of the smallest circle which can be inscribed in a triangle.

Solved Problems on Fermat point

Problem 1: Find the Fermat Point of an equilateral triangle with vertices A(0,0), B(2,0), and C[Tex](1, \sqrt{3})[/Tex].

Solution:

In an equilateral triangle, the Fermat Point is the centroid.

The centroid coordinates are: [Tex]\left(\frac{0+2+1}{3}, \frac{0+0+\sqrt{3}}{3}\right) [/Tex]

[Tex]= \left(1, \frac{\sqrt{3}}{3}\right)[/Tex].

Problem 2: Find the Fermat Point of an isosceles triangle with vertices A(0,0), B(1,1), and C(2,0).

Solution:

For an isosceles triangle with base angles less than 120 degrees, the Fermat Point is the centroid. [Tex]\left(\frac{0+1+2}{3}, \frac{0+1+0}{3}\right) [/Tex]

[Tex]= \left(1, \frac{1}{3}\right).[/Tex]

Problem 3: Given an acute triangle △ABC with vertices A(0,0), B(1,0), and C(0,1), find the Fermat Point.

Solution:

For an acute triangle, the Fermat Point is the point where the angles between PA, PB, and PC are each 120 degrees. In practice, iterative methods or geometric constructions are used.

Problem 4: Find the Fermat Point of an obtuse triangle with vertices A(0,0), B(1,0), and C(0.5,2).

Solution:

For an obtuse triangle, the Fermat Point is the vertex where the largest angle is located. In this case, it is C(0.5,2)C(0.5,2)C(0.5,2).

Problem 5: Find the Fermat Point of an isosceles triangle with sides a = 5, b = 5, and base c=8.

Solution:

Since no angle is greater than 120 degrees, the Fermat Point will be inside the triangle, where the angles between PA, PB, and PC are each 120 degrees. This involves construction or numerical methods.

Problem 6: Find the Fermat Point of a triangle △ABC with angles 130 degrees, 25 degrees, and 25 degrees.

Solution:

Fermat Point is the vertex where the 130-degree angle is located, as it is greater than 120 degrees. If [Tex]\angle A = 130^\circ[/Tex], then the Fermat Point is A.

Applications of Fermat Point

Fermat points have interesting properties and various applications in various fields such as engineering, physics, and computer graphics. Its have a geometric significance that lies in optimizing. It’s some applications are:-

• The total distance travelled from a point to a given three locations has practical implications in situations like determining the optimal location for a facility or analyzing the efficiency of transportation routes.
• It also helps a lot in the field of Physics. For example:- It used in finding the center of Gravity of an Object.
• There are various applications in the field of Computer Science such it is used in many cryptographic algorithms.

Worksheet: Fermat Point

Problem 1: Given an acute triangle △ABC with vertices A(0,0), B(4,0), and C(2,3), find the Fermat Point.

Problem 2: Determine the Fermat Point of an isosceles triangle with vertices A(0,0), B(5,0), and C(2.5,4).

Problem 3: Find the Fermat Point of a right triangle with vertices A(0,0), B(6,0), and C(0,8).

Problem 4: Find the Fermat Point of an obtuse triangle with vertices A(0,0), B(4,0), and C(1,5).

Problem 5: For a scalene triangle with vertices A(1,1), B(4,1), and C(2,5), find the coordinates of the Fermat Point

Problem 6: Three towns form a triangle with coordinates A(2,3), B(5,1), and C(6,4). Determine the optimal location for a new facility (Fermat Point) such that the total distance to the three towns is minimized.

Problem 7: A triangular plot of land has vertices A(0,0), B(10,0), and C(4,7). Find the Fermat Point to determine the optimal location for a water source to minimize the distance to all three vertices.

Problem 8: Find the Fermat Point of a triangle with angles 130 degrees, 30 degrees, and 20 degrees. Verify if it coincides with the vertex of the largest angle.

Problem 9: Use a geometry software (e.g., GeoGebra) to construct the Fermat Point for a triangle with vertices A(1,2), B(4,6), and C(7,3).

Problem 10: For a triangle with vertices A(1,1), B(5,1), and C(4,5), find the Fermat Point using coordinates.

Read More:

• Congruence of Triangles
• Triangle Inequality Theorem
• Construction of Triangles
• Area of a Triangle in Coordinate Geometry

FAQs of Fermat Point

Define Fermat Point.

The Fermat point of a triangle is the point that minimizes the sum of the distances from each of the three vertices of the triangle to the point. It is also known as the Torricelli point or the Fermat–Torricelli point.

How to Find Fermat Point in Triangle?

To find Fermat Point in Triangle, we need to do a geometrical construction that is known as Fermat’s problem or the Steiner problem, and the construction of this is discussed in the article in good detail.

Are there any Other Names for the Fermat point?

Yes, the Fermat point is also known as the Fermat-Torricelli point, Torricelli point, or Steiner point.

Where does the Fermat Point is Located for an Obtuse Triangle?

The Fermat point in the obtuse triangle is the obtuse-angled vertex.

When will the Fermat Point and the Centroid be the Same for a Triangle?

In a Equilateral triangle, Fermat Point and the centroid coincide.

Does every Triangle have a Fermat point?

Yes, every triangle has a Fermat point. It may lie either within the triangle or on its boundary.

What are Some Properties of the Fermat Point?

Some properties of Fermat Points are:

• Fermat point is always inside the triangle, unless one of the angles of the triangle is 120 degrees or more. In that case, the Fermat point is the vertex with the largest angle.
• Fermat point is the geometric median of the three vertices of the triangle. This means that it is the point that divides each of the medians (the line segments that connect a vertex to the midpoint of the opposite side) in the ratio 2:1.
• Fermat point is also the circumcenter of the triangle if and only if the triangle is equilateral.

What is the geometric construction approach to finding the Fermat Point?

The geometric construction approach involves:

1. Constructing an equilateral triangle on one side of the given triangle.
2. Connecting the new vertex of the equilateral triangle to the opposite vertex of the original triangle.
3. Repeating the process for the other two sides.
4. The intersection point of these three lines is the Fermat Point.

How is the Fermat Point related to other triangle centers like the centroid or circumcenter?

Fermat Point is different from other triangle centers:

• Centroid: The centroid is the average of the vertices’ coordinates, while the Fermat Point minimizes total distance.
• Circumcenter: The circumcenter is the center of the circle passing through all vertices, whereas the Fermat Point minimizes total distance to vertices.

What are some real-life applications of the Fermat Point?

Real-life applications include:

• Optimal location planning: Minimizing travel distances in logistics and network design.
• Urban planning: Determining the best location for facilities such as fire stations, hospitals, and distribution centers to minimize response time.