Extreme Values for the Area of Rectangles with Vertices on Concentrical Circles

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1 olumbus State University SU eress Faculty ibliography 2007 Etreme Values for the rea of Rectangles with Vertices on oncentrical ircles Eugen J. Ionascu Follow this and additional works at: art of the Mathematics ommons Recommended itation Ionascu, Eugen J., "Etreme Values for the rea of Rectangles with Vertices on oncentrical ircles" (2007). Faculty ibliography. aper This rticle is brought to you for free and open access by SU eress. It has been accepted for inclusion in Faculty ibliography by an authoried administrator of SU eress.

2 Elem. Math. 62 (2007) /07/ c Swiss Mathematical Society, 2007 Elemente der Mathematik Etreme values for the area of rectangles with vertices on concentrical circles Eugen J. Ionascu and antelimon Stanica Eugen J. Ionascu obtained his h.d. from the University of Teas &M in He is now associate professor at olumbus State University, Georgia, US. His interests are in areas of analysis such as ergodic theory, wavelet analysis, comple variables, or operator theory. antelimon Stanica obtained his h.d. from the State University of New York at uffalo in He is currently a distinguished research associate professor at uburn University Montgomery, labama, US. His main interests are in oolean functions in cryptography, combinatorics, discrete mathematics, and number theory. 1 Introduction In [3], the following unusual geometrical problem was proposed: roblem 1. onsider three positive numbers, y and such that there eists a rectangle D with the property that for some point in the interior of the rectangle, =, = y and =. Find the maimum possible area of a rectangle with this property in terms of, y,. We are going to consider also the problem of finding the minimum area of such a rectangle, where is not necessarily in the interior of the rectangle. For eample, if = 7, y = 5 Im nachfolgenden eitrag bestätigen die utoren unächst die ehauptung, dass die edingung y 2 + t 2 = an die Radien, y,, t von vier konentrischen Kreisen notwendig und hinreichend für die Eisten eines Rechtecks mit je einer Ecke auf je einem der vier Kreise ist. bgesehen von entarteten Fällen stellen die utoren überdies fest, dass es in der Regel jeweils unendlich viele Lösungsmöglichkeiten gibt, und sie stellen sich die ufgabe, das Rechteck mit dem maimalen/minimalen Flächeninhalt u finden. Es gelingt ihnen, dieses Optimierungsproblem mit Hilfe von Lagrange-Multiplikatoren epliit u lösen. Es sei darauf hingewiesen, dass eine verwandte ufgabenstellung kürlich im merican Mathematical Monthly Erwähnung fand (siehe [3])..

3 Etreme values for the area of rectangles with vertices on concentrical circles 31 and = 1, the maimum area is 32 and the minimum area is 18 ( not necessarily in the interior). We find a surprisingly simple formula for the maimum and minimum area using various tools that range from geometry and calculus to elementary number theory. Further, we show that the eample given is one of the infinitely many instances when the etreme values for the area and the data are integers. D O u v y Fig. 1 Let us observe that there are some restrictions that one would have to impose on, y and in orderfor a rectangle with the required propertiesto eist. Let O be the intersection of the diagonals D and. Using the formula relating the median to the side lengths of thetriangleweget4o 2 = 2( ) 2 = 2(t 2 + y 2 ) D 2, where t = D. Thus, t = y 2, so we need to have y < (1) This condition is also sufficient for the eistence of a rectangle D if the point is not required to be necessarily in the interior of the rectangle with =, = y and =, as we shall show in the net section. Hence, we may consider the slightly modified and in some respect more general problem: roblem 2. Given four concentrical circles (, ), (, y), (, ) and (, t) of center and radii, y,, t, resp., where = y 2 + t 2, consider four points, (, ), (, y), (, ) and D (, t), such that the quadrilateral D is a rectangle (in general, there may be two such rectangles for each particular position of and see Fig. 2). Find the etreme values for the area of such a rectangle. roblem 1 follows from roblem 2 since the maimum is attained in a position where is in the interior of the rectangle as shown in Section 2. Let us observe that roblem 2 is equivalent to the following: roblem 3. Given three positive numbers, y, such that y < 2 + 2, determine the etreme values of the area of a right triangle ( = π/2) with (, ), (, y), (, ). n advantage of formulating the problem this way is that one can easily see that the lengths of the rectangle s sides are bounded quantities. Therefore, the maimum and minimum

4 32 E.J. Ionascu and. Stanica D D' ' Fig. 2 area should be positive numbers (it is easy to see that the minimum is ero if and only if y = or y = ). roblem 1 is unusual because of its surprisingly simple answer in spite of our rather laborious solution. Theorem 1.1. The maimum area of the rectangle in roblem 1 is equal to + yt and the minimum area is yt. 2 Solution and commentaries Our strategy to solve roblem 1 is a classical one. First of all we are mainly concerned with roblem 3. Referring to Fig. 1, we introduce parameters u and v which are oriented angles u = m( ) and v = m( ) (angles rotated clockwise are considered negative and angles rotated counterclockwise are considered positive) and so u,v ( π,π]). ll other angles are considered as in elementary Euclidean geometry: angles between 0 and π. We write the problem as an optimiation of a function of two variables with a constraint (since our problem has essentially one degree of freedom). We apply the Lagrange multipliers method to find the associated critical points, and finally, we identify the points that indeed give the maimum or the minimum for our objective function. roposition 2.1 (Objective and onstraint Function). The area of the rectangle D is twice the area of the triangle and as a function of u and v it is given by the formula rea(d) = sin u + y sin v sin(u + v), (2) where is not necessarily in the interior of the rectangle. The constraint equation is cos u + y cos v cos(u + v) y 2 = 0. (3) roof. We use vector calculus to derive rea(d) = 2rea( ) = 2 = ( ) ( ). Further,rea(D) = 2 +

5 Etreme values for the area of rectangles with vertices on concentrical circles 33 = sin u + y sin v sin(u + v). For the constraint equation we employ the notion of inner product of two vectors as opposed to the cross product that appeared in the derivation of (2). The constraint is given by the fact that is a right triangle at. This can be equivalently written as = 2. This can be written as ( ) 2 + ( ) 2 = ( ) 2. fter simplifications this can be reduced to (3). Let us reduce the number of given data by introducing a = y and b = y. The inequality (1) is equivalent to a 2 b 2 < a 2 + b 2. So, roblem 3 is rephrased as Maimie/Minimie f (u,v) = a sin u + b sin v absin(u + v) under the constraint g(u,v):= a cos u + b cos v cos(u + v) ab, (u,v) ( π,π] ( π,π]. This analysis problem is interesting in itself because of the symmetry of the two functions involved and of its relatively simple answer. We would like to make more precise the domain D of u which is determined by the eistence of a right triangle, with (, ), (, y), (, ) having = u and a negative oriented angle = π/2. We call this the positive orientation of. In the parts (b) and (c) of the net proposition, y is assumed to satisfy (1). roposition 2.2. (a) ondition (1) is necessary and sufficient for the eistence of. (b) If y, then D is ( π,π]. (c) If y >, then D =[arccos( y2 2 +t ), arccos( y2 2 t )]. roof. The necessity of (1) was already shown, and the sufficiency is a consequence of (b) and (c). First, if y, thecircle(, y) is inside of or equal to (, ) and so, the perpendicular on at is going to intersect the circle (, ) at one point (if y = ), or at two points. We choose to be one of these points in such a way that the orientation of the triangle is positive for every u D. Hence, since this is possible for every particular position of (with fied), (b) follows. We assume now that y >. Thus, (, ) is in the interior of (, y), andsotheperpendicular on at is not going to intersect the circle (, ) unless the angle u is in the right range. This range is given by the fact that must be limited in length in the following way. pplying the triangle inequality in D (see Fig. 3), we see that t + t. Equivalently, using the law of cosines in, we obtain (t ) y 2 2cos(u) ( + t) 2. Therefore, y2 2 t (4) cos(u) y2 2 +t. Thus, u D or u D. Every angle u D has to be ecluded because it makes positive. Now, if u D, then we construct D with (, ), D (, t), andd parallel and of the same length as. This is possible because u D implies that a triangle

6 34 E.J. Ionascu and. Stanica = D y t D Fig. 3 congruent to D eists and then one rotates until the side Dbecomes parallel to. Hence, the quadrilateral D is a parallelogram. The fact that D is a rectangle follows from the equality = y 2 + t 2 since this implies that the diagonals of D are equal using an argument as the one used to derive condition (1). The orientation of is positivebecause u D and (, ) is in the interior of (, y). Remark. In case (c) we consider both solutions for.thismeansthatv is going to have two solutions in general (ecept for the endpoints of the interval) for each value of u D. We do not lose any generality when looking for the etreme values because for every value of u D the corresponding rectangles (right triangles) are congruent to the ones corresponding to u D by symmetry. In case (b), due to the periodicity of the functions involved in (4), the maimum and minimum are going to be given by critical points. We want to show that those critical points are in the range given in (c), as well. roof of Theorem 1.1. y Lagrange multipliers theorem (see [1, p ], Theorem 3.1) the critical points are given by the system of equations in u, v and λ, f g u (u,v) = λ u (u,v) a cos u cos(u + v) = λ( a sin u + sin(u + v)) f g v (u,v) = λ v (u,v) b cos v cos(u + v) = λ( b sin v + sin(u + v)) g(u,v)= 0 a cos u + b cos v cos(u + v) ab= 0. (5) One can check that g g u (u,v) = v (u,v) = g(u,v) = 0leadstonosolutioninu and v if a and b are not 1. So, in order to apply Theorem 3.1 in [1, p. 136] we are going to work under this assumption and deal with this particular (simple) case separately. Eliminating λ,we get the system: ab cos u sin v + a cos u sin(u + v) + b cos(u + v) sin v + ab sin u cos v a sin u cos(u + v) b sin(u + v) cos v = 0 a cos u + b cos v cos(u + v) ab= 0. (6)

7 Etreme values for the area of rectangles with vertices on concentrical circles 35 Eliminating v from these equations, we get sin u (a 2 1)[a 2 b 4 4 b 3 a 2 cos u 5 b 2 a 2 sin 2 u + 6 a 2 b a 2 b sin 2 u cos u 4 ba 2 cos u + a 2 a 2 sin 2 u b 4 sin 2 u + 2 b 3 sin 2 u cos u b 2 sin 2 u]=0. (7) We then get a family of critical points from the equation sin u = 0. These turn out not to be relative etrema for the function f. Then other critical points may be given by the remaining factor of (7). fter substituting sin 2 u = 1 cos 2 u and cos u = ω, the equation in question becomes ( 2 ω b b 2 )(a 2 b 2 b 2 + b 2 ω 2 2 a 2 ω b + a 2 ω 2 ) = 0. (8) One solution of this equation is cos u = ω = b2 +1 2b. learly ω>1, by our assumption, and so this leads to no critical value of u. The other critical values are cos u 1,2 = a2 ± b a 2 + b 2 a 2 b 2 b 2 + a 2 = ± t (9) We are going to show that the global maimum, respectively, minimum are given by t cos u M = 2 + 2, respectively, cos u + t m = 2 + 2, (10) for a corresponding v m, v M determined obviously by u m, respectively, u M.Letusfirst show that these two values of cos u M and cos u m are indeed in the interval [ 1, 1]. Let α, β [0,π/2] be defined by cos α =,sinα = 2 +,cosβ = y and t t sin β =. With this notation, the fractions in (10) become = cos α cos β sin α sin β = cos(α + β) and similarly, +t = cos(α β) ase I: y. Thus, (b) of roposition 2.2 holds. In this case, if we construct with the positive orientation for every u D, rea(d) in (2) as a function of u is continuous and periodic. Hence, the maimum and the minimum should appear as critical points. The values of u that give the maimum and minimum, respectively, are: u M = α + β, u m = α β. (11) Indeed, using the law of cosines in the triangle, we obtain = t+y 2 + 2,and again the law of cosines for the side gives 2 cos( ) = y 2 = 22 t 2 +2t > 0. This implies, in particular, that m( )<π/2. fter more t simplifications we obtain cos( ) = = cos(π/2 β), which means that m( ) = π/2 β. Hence m( ) = π/2 α and m( ) = α. Therefore, cos( ) = cos(α) = Now, we apply the law of cosines in to obtain 2 = y ( ) 2 y cos( ), or after completing the square ( y cos( )) 2 = t

8 36 E.J. Ionascu and. Stanica ecause of reasons of orientation, and also because we want to maimie the area of D, should be equal to y cos( ) + t = 2 + +t.thisgivesthearea (still need to check it is maimal) Maarea(, y, ) = = (For easy writing we use Maarea for Maarea(, y, ).) t + y + t = + yt Now, consider the case of sin u = 0. That means that the point is on the line determined by the side, which implies that = + y (or, = y,when is not on ). In any case, one obtains = 2 y 2 = t 2 2. Furthermore, the area becomes = ( + y) 2 y 2 = = y 2 y 2 ( + y) t yt = Maarea, or 2 y 2 + y 2 y 2 + yt = Maarea. To show that Maarea is indeed a global maimum we have to compare it with the other values of rea(d) obtained by choosing the other possible cases for u given by (9). Due to symmetry, the value of rea(d) (from (2)) for u = u M can be computed as in the case u = u M. One gets rea(d)( u M ) = (t+y)(t ) Maarea. Then, 2 + we just need to compare Maarea with the values of rea(d) 2 at u =±u m. Now, if u m = α β, using the law of cosines in we obtain = t y It is easy to see that t y = 0 is equivalent to y =. Ify =, theny = and therefore 0 is indeed the minimum area. So, we may assume that y =. s before, similar calculations show that 2 y cos( ) = y = 22 y 2 2t 2 + 2, or t y cos( ) = (y t), which in turn gives cos( ) =± = ± sin α = cos(π/2 ± α). Thisimpliesthatm( ) = π/2 ± β. nalying the two different situations y > and y <, we see from Fig. 4 that m( ) = π α if y >,orm( ) = 2 π (π/2 + α + π/2) = π α if y <. y > y y < y Fig. 4

9 Etreme values for the area of rectangles with vertices on concentrical circles 37 So the measure of the angle does not depend on the two cases y < or y >. Hence, from we get 2 = y y cos( ). ut cos( ) = cos α =, and so after completing the square, satisfies the equation ( / ) 2 = 2 t 2 /( ).Thisgives = (t )/ Wehaveto point out that t > because this is equivalent to > y, which is our assumption. Then rea(d) = = t y (t )/( ) = yt :=Minarea. For u = u m one obtains with a similar analysis rea(d)( u m ) = t y (t + )/( ). Since rea(d)( u m ) and rea(d)( u M ) are between Minarea and Maarea, we have found the etreme values of the function rea(d) as a function of u. Let us notice that we have solved in particular roblem of [3, p. 64], since in the case of maimum area, is in the interior of the rectangle (all the angles,, and are acute angles, by similar calculations). Now, we deal with the case of a = 1 and/or b = 1. It follows that y = and = t. This means that is on the perpendicular bisector of and D determined in Fig. 5 by midpoints E and F. E D y F Fig. 5 Hence,theareaofD is 2 rea(ef) = 2 2 rea(e) = 4 rea() = 4 sin u 2 = ( + yt) sin u Maarea. In this situation, Minarea = 0. ase II: y >. So, D =[arccos( y2 2 +t ), arccos( y2 2 t )] =[u 1, u 2 ]. In this case, as we mentioned before, we consider for each u D the two right triangles 1 and 2, which are both positively oriented and the area of the corresponding rectangles f 1 (u) := 2 rea( 1 ) and f 2 (u) := 2 rea( 2 ) ( 2 > 1 ). We want to show that the maimum is attained on the branch f 2 and the minimum is attained on the branch f 1. The difference between these two functions compared to (2) is that in f 1 the angle v is such that 2 is acute and in f 2 the angle v is such that 1 is obtuse.

10 38 E.J. Ionascu and. Stanica Maimum 1 Minimum u M u m 2 Fig. 6 In this case u M = α + β and u m = α β. Let us show that these angles are in D. Since cos(α + β) = cos α cos β sin α sin β = t 2 + 2, we need to show that y 2 2 t t y2 2 + t. These two inequalities are in fact equivalent to the auchy-schwart inequality ( t) 2 ( )(y 2 + t 2 ). Since f 1 [respectively f 2 ] agrees with (2) [for the appropriate choice of v] we obtain the same epressions for Maarea and Minarea. t the endpoints of the interval D, the two functions have the same values because 1 = 2 := (v 1 = v 2 ), and becomes tangent to the circle (, y). Moreover, = y 2 2 and = t + or = t. Hence, f 2 (u 1 ) = = y 2 2 t y 2 2 (t+) = f 2 (u 2 ) = 2 t 2 + y 2 2 t + yt = Maarea, and f 1 (u 2 ) = = y 2 2 (t + ) y 2 2 t = f 1 (u 1 ) = y 2 2 t yt =Minarea. The last inequality can be shown to be true by analying the two cases y and y >. The same analysis appliesin this case to show thatfor the maimumarea, the point is in the interior of the rectangle and one can show that the minimum area is happening for a position of in the eterior of the rectangle. 3 Further results and proposed problem Obviously, the maimal area is an integer if, y,, t are integers. The following theorem provides a practical way to construct eamples of such values of, y, (and t). Theorem 3.1. Let m, n, p, q be arbitrary positive integers. If, y,, t are given by = mn + pq, = mq np, y= mn pq and t = mq + np, then Maarea and Minarea are integers.

11 Etreme values for the area of rectangles with vertices on concentrical circles 39 roof. Using [2, p. 15], we get that all integer solutions of the Diophantine equation = y 2 + t 2 are of the mentioned forms. Since, y,, t are therefore integers, it follows that Maarea(, y, ) = + yt is also an integer. Similarly, for Minarea. re these the only cases when Maarea is an integer? ertainly not, since one can multiply the values of, y, and t given by Theorem 3.1 by l (l positive integer which is not a perfect square), for instance, and still obtain an integer as a result. lso, one may ask similar questions of finding the etreme values in various situations like having the circles in roblem 3 not necessarily concentrical or on different curves. The following is a problem we propose for further study. roblem 4. Determine the etreme values for the area of a right triangle ( = π/2) with (, ), (Q, y), (R, ) where circle (Q, y) is contained in the interior of one of the other two circles. References [1] Lang, S.: alculus of Several Variables. (3rd ed.), Springer-Verlag, [2] Mordell, L.J.: Diophantine Equations. cademic ress, [3] Zahlreich roblems Group, roblem mer. Math. Monthly 111 (2004), 64. Eugen J. Ionascu olumbus State University 4225 University venue olumbus, G 31907, U.S.. ionascu eugen@colstate.edu antelimon Stanica uburn University Montgomery Mathematics Department Montgomery, L 36124, U.S.. pstanica@mail.aum.edu