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# A Note on the first Fermat-Torricelli Point

Naga Vijay Krishna D*

Department of Mathematics, Narayana Educational Institutions, Bangalore, India

*Corresponding Author:
Naga Vijay Krishna D
Department of Mathematics, Narayana Educational Institutions, Bangalore, India.
E-mail: [email protected]

Received date: 08/07/2016; Accepted date: 22/08/2016; Published date: 26/08/2016

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## Abstract

The aim of this note is to prove some well-known results related to the Fermat-Torricelli point in a new prominent way.

#### Keywords

First Fermat point, Fermat lines

#### Introduction

The Fermat point is named for the point which is the solution to a geometric challenge that Pierre Fermat posed for Evangelista Torricelli, who was briefly an associate of the aged Galileo. Fermat challenged Torricelli to find the point P in an acute triangle ABC which would minimize the sum of the distances to the vertices A, B, and C. The triangle need not actually be acute, but if the largest angle reaches 120 degrees or more, then the vertex at the largest angle is the solution. For a general solution, one approach is to construct equilateral triangles on each side of the triangle (actually only two are needed) and draw the segments connecting the opposite vertices of the original triangle and the newly created equilateral vertices. They intersect in a point which is the solution. The point is called the Fermat point. The more details about this point and its generalizations is in [1-4].

In this note we will try to establish the very fundamental results related to this point.

Notations:

Let ABC be a triangle. We denote its side-lengths by a, b, c, its semi perimeter by its area by Δ , its Circumradius by In radius by Let us define S2 , S2 and S3 as described below: Some Basic Lemma’s:

Lemma -1

If S1, S2 and S3 are described as mentioned above then Proof:

Clearly   Hence proved.

Lemma -2

If S1, S2 and S3 are described as mentioned above then, Proof:

We have sin(60+A) = sin60 cos A + sin A cos 60 = It implies sin(60+A) Further simplification gives required conclusions.

Theorem-1

If Triangle ABC is an arbitrary triangle (whose all angles are less than 120 degrees) let the triangles A1BC, B1CA and C1AB are equilateral triangles constructed outwardly on the sides BC, CA and AB of triangle ABC then AA1, BB1 and CC1 are concurrent and the point of concurrence is called as First Fermat Torricelli Point(T1) or Outer Fermat Torricelli Point (T1 ) .

Proof: Let D, E and F are the point of intersections of the lines AA1, BB1 and CC1 with the sides BC, CA and AB.

Now clearly by angle chasing and using the fact “cevian divides the triangle into two triangles whose ratio between the areas is equal to the ratio between the corresponding bases.” It implies And we have Hence Similarly Now by the converse of Ceva’s theorem,

Since The lines AA1, BB1 and CC1 are concurrent and the point of concurrence is called as First Fermat Point (T1).

Theorem-2

Triangles A1BC, B1CA and C1AB are equilateral triangles constructed outwardly on the sides BC, CA and AB of triangle ABC then AA1, BB1 and CC1 are equal in length. (For the recognition sake let us call the lines AA1, BB1 and CC1 as Fermat Lines) .

Proof:

Clearly from triangle ABA1,

By cosine rule It implies It further gives Similarly we can prove that Hence Theorem-3

Let D, E and F are the point of intersections of the lines AA1, BB1 and CC1 with the sides BC, CA , AB respectively and if T1 is the First Fermat Point then   Proof:

Clearly we have, Now from triangle ABT, the line BT1E is acts as transversal so by Menelaus theorem we have: It implies Similarly we can prove that Hence the conclusion (a) follows:

Now from conclusion (a) we have for some constant K

It follows that And clearly It gives that So Using the above relation and lemma-1 we can find the proportionality constant K and by replacing the value of K in AT1 = (S1S2+S1S3) K we can arrive at the required conclusion (b).

Now using (b) we can prove the conclusion (c).

Theorem-4

Triangles A1BC, B1CA and C1AB are equilateral triangles constructed outwardly on the sides BC, CA and AB of triangle ABC then the circumcircles of the Triangles A1BC, B1CA and C1AB conccur at T1.

Proof:

We need to prove that set of the points are concyclic.

So it is enough to prove that by ptolemy’s theorem Clearly It implies that A1T1= BT1+CT1

Similarly we can prove the remaining two relations.

Corollary:

If T1 is the First Fermat point of triangle ABC then   Proof:

For (a),

Clearly by Theorem-4 we have angle AT1B = angle BT1C = angle CT1A = 1200

So (p)

Similarly (r)

Now using the fact and (p), (q) and (r) we can prove conclusion (a).In the alternative manner,

Using theorem – 3, lemma-1 and by little algebra we can prove the conclusion (a).

Now for (b),

Clearly by Theorem – 4 and by applying cosine rule for the triangles AT1B, BT1C and CT1A we can prove that (x) (y) (z)

Now consider Equivalently, This finishes proof of conclusion (b).

#### Remark

When one of the angles of the triangle is 120° or greater, then the Fermat point (which still exists) is no longer the point that minimizes the sum of the distances to the vertices, but the minimal point is located at the vertex of the obtuse angle. Clearly Theorem-3 derives this fact.

#### Acknowledgement

The author is grateful to the creators of the free Geogebra software, without which this work would have been impossible and the author is would like to thank an anonymous referee for his/her kind comments and suggestions, which lead to a better presentation of this paper.