Research & Reviews: Journal of Statistics and Mathematical Sciences

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1 Research & Reviews: Journal of tatistics and Mathematical ciences A Note on the First Fermat-Torricelli Point Naga Vijay Krishna D* Department of Mathematics, Narayana Educational Institutions, Bangalore, India Research Article Received: 08/07/0 Accepted: /08/0 Published: /08/0 ABTRACT The aim of this note is to prove some well-known results related to the Fermat-Torricelli point in a new prominent way. *For Correspondence Department of Mathematics, Narayana Educational Institutions, Bangalore, India. vijay @gmail.com 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 0 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 [-4]. Notations: In this note we will try to establish the very fundamental results related to this point. a+ b+ c s, its area by, its Circumra- Let ABC be a triangle. We denote its side-lengths by a, b, c, its semi perimeter by dius by R abc, In radius by r 4 s Let us define, and as described below: 4 + ( b + c, 4 + (a + c b and 4 + (a + b c + + [4 + (a + b + c ] + [4 + c ], + [4 + a ], + [4 + b ] ome Basic Lemma s: Lemma - If, and are described as mentioned above then ( + + RRJM Mathematics Trends and Multivariate tatistical Analysis-, 0 55

2 Clearly c + 9c + 8ab 9a 9b a + 9a + 8bc 9b 9c b + 9b + 8ac 9a 9c so ( a + b + c + 9( ab + ab + ab a 4 b 4 c 4 ( a b c ( ( a b c ( Hence proved. Lemma - If, and are described as mentioned above then, 4 bcsin(0 + A, 4 acsin(0 + B and 4 absin(0 + C b + c a a We have sin(0+a sin0 cos A + sin A cos 0 + bc R It implies sin(0+a (b + c a bc 4 bc Further simplification gives required conclusions. Theorem- If Triangle ABC is an arbitrary triangle (whose all angles are less than 0 degrees let the triangles A BC, B CA and C AB are equilateral triangles constructed outwardly on the sides BC, CA and AB of triangle ABC then AA, BB and CC are concurrent and the point of concurrence is called as First Fermat Torricelli Point(T or Outer Fermat Torricelli Point (T. Let D, E and F are the point of intersections of the lines AA, BB and CC 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. RRJM Mathematics Trends and Multivariate tatistical Analysis-, 0 5

3 o [ ] [ ] + AB A B + BD ABD A BD ABD A BD A BA DC [ ACD] [ ] + A CD ACD A CD A CA AC A + C.sin(0 c.sin(0 B It implies BD + + AB B DC AC.sin(0 + C b.sin(0 + C And we have sin(0+b 4 bc Hence imilarly BD DC CE EA and AF FB Now by the converse of Ceva s theorem,..sin(0 B. C.sin(0 CE BD AF ince.... EA DC FB The lines AA, BB and CC are concurrent and the point of concurrence is called as First Fermat Point (T. Theorem- Triangles A BC, B CA and C AB are equilateral triangles constructed outwardly on the sides BC, CA and AB of triangle ABC then AA, BB and CC are equal in length. (For the recognition sake let us call the lines AA, BB and CC as Fermat Lines [5]. Clearly from triangle ABA, By cosine rule ( ( ( AA AB + A B. AB. A B.cos ( ABA It implies ( AA c + a accos ( 0 + B It further gives ( AA a + b + c imilarly we can prove that ( BB ( CC Hence Theorem- AA BB CC + + Let D, E and F are the point of intersections of the lines AA, BB and CC with the sides BC, CA, AB respectively and if T is the First Fermat Point then (a AT : T D + : BT : T E + : CT : T F + : (b AT ( + +, BT ( + + and CT ( + + RRJM Mathematics Trends and Multivariate tatistical Analysis-, 0 57

4 ( + + (c AT + BT + CT AA BB CC Clearly we have, BD DC, CE EA and AF FB Now from triangle ABT, the line BT E is acts as transversal so by Menelaus theorem we have: AT AE CB T D EC BD a + AT + It implies TD a imilarly we can prove that BT : T E + : and CT : T F + : Hence the conclusion (a follows: Now from conclusion (a we have AT ( + K, T D K for some constant K It follows that AD( + + K And clearly AD [ ABD] [ ACD] [ ABD] + [ ACD] [ ABC] AA ABA ACA ABA + ACA BCA + [ ABC] AD 4 8 It gives that AA 4 + a + + a K ( o Using the above relation and lemma- we can find the proportionality constant K and by replacing the value of K in AT ( + K we can arrive at the required conclusion (b. Now using (b we can prove the conclusion (c. Theorem-4 Triangles A BC, B CA and C AB are equilateral triangles constructed outwardly on the sides BC, CA and AB of triangle ABC then the circumcircles of the Triangles A BC, B CA and C AB conccur at T. We need to prove that set of the points {A, B, C, T }, {A, B, C, T }, {A, B, C, T } are concyclic. o it is enough to prove that by ptolemy s theorem A T BT +CT, B T AT +CT and C T AT +BT Clearly A T AA AT It implies that A T BT +CT imilarly we can prove the remaining two relations. Corollary: If T is the First Fermat point of triangle ABC then AT BT + BT CT + CT AT 4 (a ( ( RRJM Mathematics Trends and Multivariate tatistical Analysis-, 0 58

5 (b a AT + b BT + c CT + ab AT BT cosc + bc BT CT cosa + ca CT AT cosb 8 For (a, Clearly by Theorem-4 we have angle AT B angle BT C angle CT A 0 0 ATB ATBT sin (AT B ATBT (p BTC BTCT (q and [ CTA] CT AT (r o [ ] imilarly [ ] Now using the fact [ ABC] [ AT B] [ BT C] [ CT A] alternative manner, + + and (p, (q and (r we can prove conclusion (a.in the Using theorem, lemma- 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 B, BT C and CT A we can prove that a BT + CT + BT CT (x b CT + AT + CT AT (y c AT + BT + ATBT (z Now consider ab AT BT cosc + bc BT CT cosa + ca CT AT cosb ( a + b c ( c AT BT abc,, (ab + bc + bc a b c aat aat Equivalently, a AT + b BT + c CT + ab AT BT cosc + bc BT CT cosa + ca CT AT cosb 8 This finishes proof of conclusion (b. RRJM Mathematics Trends and Multivariate tatistical Analysis-, 0 59

6 REMARK When one of the angles of the triangle is 0 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- 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. REFERENCE. Mortici C. A note on the fermat-torricelli point of a class of polygons. Forum Geometricorum. 04;4:7 8.. Vijay Krishna DN. Weitzenbock inequality proofs in a more geometrical way using the idea of lemoine point and Fermat point, 05.. De Villiers M. From the Fermat point to the De Villiers points of a triangle, Proceedings of the 5th Annual AMEA Congress, University of Free tate, Bloemfontein. 4. ándor NK. The metric characterization of the generalized Fermat points. Global Journal of cience Frontier Research Mathematics and Decision ciences. 0;. 5. Yiu P. On the Fermat lines. Forum Geometricorum. 00;:7 8. RRJM Mathematics Trends and Multivariate tatistical Analysis-, 0 0

A NEW PROOF OF PTOLEMY S THEOREM

A NEW PROOF OF PTOLEMY S THEOREM DASARI NAGA VIJAY KRISHNA Abstract In this article we give a new proof of well-known Ptolemy s Theorem of a Cyclic Quadrilaterals 1 Introduction In the Euclidean geometry,