THE EXTREMUM OF A FUNCTION DEFINED ON THE EUCLIDEAN PLANE

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1 INTERNATIONAL JOURNAL OF GEOMETRY Vol. 3 (2014), No. 2, THE EXTREMUM OF A FUNCTION DEFINED ON THE EUCLIDEAN PLANE DORIN ANDRICA Abstract. The main urose of the note is to resent three di erent solutions to a geometric extremal roblem roosed to the 2012 Balkan Mathematical Olymiad. 1. INTRODUCTION There are many well-known geometric situations involving extremal roerties of various functions de ned on the Euclidean lane. Considering a triangle ABC, then the minimum of the function P 7! P A + P B + P C is attained in the famous Fermat-Toricelli oint, and the value of the minimum can be e ectively comuted only in terms of the elements of triangle (see [5]). Another interesting function is given by S q (P ) = x q + y q + z q, where x; y; z denote the distances of the oint P to the sides of triangle ABC, and q is a xed real number. The minimum of S q (P ) is attained at a oint deending on q, and the locus of these oints is the so-called "owers curve" (see [7]). According to notations in the oen encycloedia of the triangle centers [8], the following oints situated on this curve give the above extremal roerty : X 6 (the Lemoine oint of triangle) for S 2, X 365 for S 3, X 31 for S 3=2, X 32 for S 4=3, X 75 for S 1=2, X 76 for S 2=3, X 366 for S 1, etc. (see [7]). The geometric extremal roblems is a very oular toic in many mathematical olymiad and cometitions (see the reference [1]). In this short note we discuss three di erent solutions to a such roblem roosed by the author to the 2012 Balkan Mathematical Olymiad (see [3] and [4]). In this case the geometric extremum is given by a rational function in the distances from a oint P to the vertices of a square. Keywords and hrases: Ptolemy inequality; Ptolemy theorem; olar coordinates (2010)Mathematics Subject Classi cation: 97G40, 51M04. Received: In revised form: Acceted:

2 The extremum of a function de ned on the Euclidean lane MAIN RESULTS We will resent and discuss three di erent solutions to the following extremal roblem roosed by the author in [3] and [4] : Let ABCD be a square situated in the lane P. Find the minimum and the maximum of the function f : P! R de ned by P A + P B P C + P D : a a = 1 2+a Solution 1.(Dorin Andrica [3]) We have f(a) = 2+1 = 2 1, and let us rove that this value is the minimum of function f, i.e. for every P 2 P we have f(p ) 2 1. The last inequality is equivalent to (1) P A + P B ( 2 1)(P C + P D): Figure 1 Alying Ptolemey s inequality for the oints P; A; B; C we have ap A + a 2P B ap C, that is P A + 2P B P C: Alying Ptolemey s inequality for the oints P; A; B; D we have a 2P A+ ap B ap D, that is 2P A + P B P D: Adding the above two inequalities we get ( 2 + 1)(P A + P B) P C + P D; hence we obtain the inequality (1). Using Ptolemey s Theorem it follows that the minimum is attained if and only if the oint P belongs to the arc AB of the circumcircle of the square. To nd the maximum, it is enough to determine the minimum of the function g : P! R, where g(p ) = P C + P D P A + P B :

3 22 Dorin Andrica Taking in to account the rst art of the solution, this is 2 1 and the function g takes this value if and only if P belongs to the arc CD of the circumcircle of the square. Finally, we have min P 2P f = 2 1 and max P 2P f = 2 + 1: Solution 2.(Daniel Lasaosa [4]) Let P be any oint in the lane such that f(p ) is maximum, and let, for this oint P; P A+P B = u and P C+P D = v. The resective loci of oints in P such that P A+P B = u and P C +P D = v are two ellises with foci A; B and foci C; D, hence both symmetric around the common erendicular bisector of AB and CD. Assume that there is one oint Q on the ellise with foci A; B which is inside the ellise with foci C; D. Then, the ellise with foci C; D through Q is such that QC + QD < P C + P D, while QA + QB = P A + P B, or f(p ) is not maximum because f(q) > f(p ). It follows that the ellise with foci A; B cannot have any oint inside the ellise with foci C; D, or since P is on both ellises, they must touch externally at P, which is clearly on the common symmetry axis of both ellises. Similarly, if a oint on the ellise with foci A; B is outside the ellise with foci C; D; f(p ) cannot be minimum, or again any oint P such that f(p ) is minimum must be the oint where the ellise with foci A; B is internally tangent to the ellise with foci C; D, hence on the common erendicular bisector of AB and CD. Figure 2 Since the roblem is invariant under scalings, consider a coordinate system such that A( 1; 1); B( 1; 1); C(1; 1) and D(1; 1). By the revious argument, it su ces to nd the maximum and minimum of f over the horizontal axis, for which P A 2 = P B 2 = (x + 1) 2 + 1, and P C 2 = P D 2 = (x 1) 2 + 1, or for both extrema, we would have g(x) = (f(p )) 2 = P A2 P C 2 = (x + 1)2 + 1 (x 1) ; and it su ces to nd the extrema of this function. Note rst that the limit when x! 1 is 1, that the exression is continuous and di erentiable, or it su ces to check the values of g(x) when its rst derivative is zero, which clearly satisfy 2(x + 1)((x 1) 2 + 1) = 2(x 1)((x + 1) 2 + 1); x 2 = 2. Taking

4 x = 2, we nd while for x = The extremum of a function de ned on the Euclidean lane 23 q g( 2) = 2, we have q g( 2) = s ( s ) ( 2 1) = = > 1; 2 1 s ( s ) ( 2 1) = = 2 1 < 1: It follows that the maximum and minimum of f(p ) are resectively 2+1 and 2 1, and occur for oints P + ; P, which are the intersections of the circumcircle of the square and the common erendicular bisector of AB; CD, such that P + is the one closest to side AB, and P is the one closest to side CD. Solution 3. (by the Jury of 2012 Balkan Mathematical Olymiad [3]) Place the square in the Cartesian lane with the vertices at (1; 0); (0; 1); ( 1; 0); (0; 1): Use the olar coordinates to obtain r + 1 2r cos + r r sin r r cos + r r sin : The change of variables u = reduces the roblem to showing that 2r cos 2r sin ; v = 1 + r2 1 + r u + 1 v 1 + u v for u 2 + v 2 1. Note that this exression is decreasing in u for constant v, and decreasing in v for constant u. On the unit circle u 2 + v 2 = 1, the tangent half-angle substitutions u = 2t 1 + t 2 ; v = 1 t2 1 + t 2 give 1 u + 1 v = j1 tj + 2jtj 1 + u v j1 + tj + 2 which reduces to 2 1 in the rst quadrand (that is, for 0 t 1) and to in the third quadrand (that is, for t 1) nishing the roof.

5 24 Dorin Andrica References [1] Andreescu, T., Mushkarov, O. and Stoyanov, L., Geometric Problems on Maxima and Minima, Birkauser, Boston-Basel-Berlin, [2] Andrica, D., Varga, Cs. and Vacaretu, D., Selected Toics and Problems in Geometry (Romanian), Ed. Plus, Bucharest, [3] Andrica, D., Problem 4, Shortlist of 29th Balkan Mathematical Olymiad, Antalya, Turkey, [4] Andrica, D., Problem O288, Mathematical Re ections, 5(2013). [5] Barbu, C., Fundamental Theorems of Triangle Geometry (Romanian), Ed.Unique, Bacau, [6] Coxeter, H.S.M. and Greitzer, S.L., Geometry Revisited, The Math. Assoc. of America, [7] Kimberling, C., Trilinear distance inequalities for the symmedian oint, the centroid, and other triangle centers, Forum Geometricorum, 10(2010), [8] Kimberling, C., Encycloedia of triangle centers, htt://faculty.evansville.edu/ck6/encycloedia/etc.html,2008. "BABEŞ-BOLYAI" UNIVERSITY FACULTY OF MATHEMATICS AND COMPUTER SCIENCE KOGALNICEANU 1, CLUJ-NAPOCA, ROMANIA address: