Department of Mathematics, Narayana Educational Institutions, Bangalore, India

- *Corresponding Author:
- Naga Vijay Krishna D

Department of Mathematics, Narayana Educational Institutions, Bangalore, India.

**E-mail:**vijay9290009015@gmail.com

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

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

First Fermat point, Fermat lines

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 S_{2} , S_{2} and S_{3} as described below:

**Some Basic Lemma’s:**

**Lemma -1**

If S_{1}, S_{2} and S_{3} are described as mentioned above then

**Proof:**

Clearly

Hence proved.

**Lemma -2**

If S_{1}, S_{2} and S_{3} 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 A^{1}BC, B^{1}CA and C^{1}AB are
equilateral triangles constructed outwardly on the sides BC, CA and AB of triangle ABC then AA^{1}, BB^{1} and CC^{1} are concurrent and
the point of concurrence is called as First Fermat Torricelli Point(T_{1}) or Outer Fermat Torricelli Point (T_{1} ) .

**Proof:**

Let D, E and F are the point of intersections of the lines AA^{1}, BB^{1} and CC^{1} 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 AA^{1}, BB^{1} and CC^{1} are concurrent and the point of concurrence is called as First Fermat Point (T_{1}).

**Theorem-2**

Triangles A^{1}BC, B^{1}CA and C^{1}AB are equilateral triangles constructed outwardly on the sides BC, CA and AB of triangle ABC
then AA^{1}, BB^{1} and CC^{1} are equal in length. (For the recognition sake let us call the lines AA^{1}, BB^{1} and CC^{1} as Fermat Lines) [5].

**Proof:**

Clearly from triangle ABA^{1},

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 AA^{1}, BB^{1} and CC^{1} with the sides BC, CA , AB respectively and if T_{1} is
the First Fermat Point then

**Proof:**

Clearly we have,

Now from triangle ABT, the line BT_{1}E 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 AT_{1} = (S_{1}S_{2}+S_{1}S_{3}) K we can arrive at the required conclusion (b).

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

**Theorem-4**

Triangles A^{1}BC, B^{1}CA and C^{1}AB are equilateral triangles constructed outwardly on the sides BC, CA and AB of triangle ABC
then the circumcircles of the Triangles A^{1}BC, B^{1}CA and C^{1}AB conccur at T_{1}.

**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 A^{1}T_{1}= BT_{1}+CT_{1}

Similarly we can prove the remaining two relations.

**Corollary:**

If T_{1} is the First Fermat point of triangle ABC then

**Proof:**

For (a),

Clearly by Theorem-4 we have angle AT_{1}B = angle BT_{1}C = angle CT_{1}A = 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 AT_{1}B, BT_{1}C and CT_{1}A we can prove that

(y)

(z)

Now consider

Equivalently,

This finishes proof of conclusion (b).

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.

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.

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