a=1/2(y+l), b=y-1, y!2, y(n and get a

Transcription

1 Internat. J. Math. & Math. Sci. VOL. II NO. 4 (1988) PYTHAGOREAN TRIANGLES OF EQUAL AREAS MALVINA BAICA Department of Mathematics and Computer Science The University of Wisconsin Whitewater, Wisconsin U.S.A. (Received January 8, 1987 and in revised form April 13, 1987) ABSTRACT. The main intent in this paper is to find triples of Rational Pythagorean Triangles (abbr. RPT) having equal areas. A new method of solving a +ab+ b c is to set a=y-1, b=y+l, y6n- {O,11 and get Pell s equation c 3y 1 To solve a ab + b c we set a=1/(y+l), b=y-1, y!, y(n and get a corresponding Pell s equation. The infinite number of solutions in Pell s equation gives rise to an infinity of solutions to a + ab + b c. From this fact the following theorems are proved. Theorem 1 Let c a +ab+ b a+b> c> b> a> O, then the three RPT-s formed by (c,a), (c,b), (a+b,c) have the same area S=abc (b+a) and there are infinitely many such triples of RPT. Theorem Let c=a-ab+b b> c> a> 0 then the three RPT-s formed by (b,c), (c,a), (c,b-a) have the same area S abc (b-a) and there are infinitely many such triples of RPT. KEY WORDS AND PHRASES. Rational Pythagorean Triangles (abbr. RPT), Perimeter of the RPT, Pell s (Euler s) equation, Fibonacci sequence AMS SUBJECT CLASSIFICATION CODE. 10B99 1. DEFINITIONS AND PREVIOUS RESULTS. In one of his papers Bernstein [1] returned to the Grecian classical mathematics, and investigated primitive rational Pythagorean Triangles concerning mainly k-tuples of them having equal perimeters P. In this paper we deal with rational integral right triangles having equal areas and give the following

2 770 M. BAICA DEFINITION A rational triangle with sides a,b,c which are represented by a triple (a,b,c) of natural numbers will be called a Rational Pythagorean Triangle, if and only if there exists (u,v),(u,v) N- {(0,0)I, l u > v, such that { (1.1) a u-v, b uv, c u+v, a,b,c cn t0} 1,,... We abbreviate Rational Pythagorean Triangles by RPT and we write RPT(u,v) for RPT formed by (u,v). We also write S(u,v) for the area and P(u,v) for half perimeter of the RPT(u,v) D=RPT(u,v):S(u,v) 1/ab= uv(u v (1.) P(u,v) 1/(a+b+c) u(u+v) The main intent in this paper is to find triplets of RPT-s having equal areas. The first who asked this question was the great Diophantus [3] and Dickson [] enlarged the topic. Let D be a triangle with integral sides and C 10 one of its angles. Then if c is the side opposite C and a,b the two adjacent ^ b ^ sides of C, we have, by c a+ -ab cos C a + ab + b c a+b> c>b> a; a,b,c N {0} and if C 60 we have a-ab+b= c (1.4) b> c>a> O; a,b,c N [OI. ] The totality of solutions to a ab + b c is given in parameter form that (1.3) by Hasse [6]...T.e new idea in this payer rests in the fact and (1.4) are connecte.d with. the a.reas of th 9 trigpgles. In order to find a formula to derive explicitly the l infinity of RPT-s of equal area, since we cannot use Hasse s [6] parametric form, we will give a new method to prove that the equations a ab + b= c have infinitely many solutions and state some of them explicitly. The new method will bring us to the solution of a Pell s equation. The infinite number of solutions of Pell s equation [43 to an infinity of solutions to a+ab+b=c. will give rise. PELL.S...(EULER S EQUATION: Un 3Vn I; n 0,I, In the sequel we shall permanently have to make use of Pell s equation u n 3v n i, n=0,1, (.1)

3 PYTHAGOREAN TRIANGLES OF EQUAL AREAS 771 This could be solved by continued fraction with V + 1 [-], but we use a simpler method since a solution of (.1) is easily found. Neglecting (u0,vo) (1,0) we have (u,v) (,), (.) hence using [5] u n+vnv ( + /)n, n:o,l,... (.3) From (.3) we deduce io n Un (i) n-i3i n=o,1,... (.4) n Vn (i+l) n-i-i3i n=l,,... From (.4) we obtain m Um lo= (i) m-i3i; m=l,,... m-i Vm m (i+i) Pm-l-i3 i m=l,,... (.5) (Uo,Vo) (,o) m Um+l:lO: (I) m+i-ii; m:o,l,... m (.6) m+l Vm+l (i+i) m-i3i m 0,i,... In Um all summands are even but the last which is 3m; in Vm all summands are even, hence F+I; Um Vm G; F,G N J In Um+l all summands are even; in Vm+l last which is 3m; hence Um+l S; (.7) all summands are even but the (.7a) Vm+l T+l; S,T N. We have the initial values (u0,v0) (i,0); (Ul,Vl)= (,1); (Ua,V)= (7,4) (u3,v3) (6,15); (u4,v4)= (97,56); (Us,V5) (36,09 (.8) (u6,v6) (1351,780); (UT,V7) (509,911).

4 77 M. BICA 3. A SOLUTION OF (1.3) a +ab +.b -c Here we shall give infinitely many solutions of (1.3) in explicit form, though they won t constitute all solutions of (1.3). Let We set a +ab+ b c a+b>c>b>a>o, ] a,b,c N- 0}. (3.1) f a=y-l; b:y+l; yen- 0. (3.) Substituting the values of a,b from (3.) in (3.1) we obtain (y-l)+ (y_l) + (y+l) c or c- 3y 1. Now we set c Un, y v n, n=l,,.., and we get Un - 3Vn 1 or u n 3v n 1 which is (.1). The infinite number of solutions un,v n in Pell s equation (. i), give rise to an infinity of solutions to a+ ab + b c. THEOREM I. Let c a b +ab+ a+b> c> b> a> 0 then the three RPT-s fo.rmed by (c,a), (c,b), (a+b,c) have the same area S abc(b+a) and there are infinitely many such triples of RPT-s. Proof. To get the example given by Diophantus [3, p. 17] as a particular case of our formulas we consider (a,b,c)= 1. prefer (a,b,c) l, though this is not absolutely necessary, we have to set v n even in (.1), so that a,b are both odd. If v n Vm we obtain Since we a=vm-l; b=vm+l; c=um; m--l,,... (3.3) Using (1.) and a+ab+b c it is easy to show that S(c,a) S(c,b) S(a+b,c) abc(b+a). D 1 RPT(c,a) RPT(Um,Vm-l); D RPT(c,b) RPT(Um,Vm+l); D 3 RPT(a+b,c) RPT(Vm,Um), and following S abc(b+a) the common area is S UmVm(V l) m That there are infinitely many such triples of RPT-s follow from the infinity of solutions of (.1). Choosing m l, we have from (.8) (u,v) (7,4); u 7, v 4 c 7; a 3; b 5 S and this is exactly the example given by Diophantus. If we choose m, we have

5 PYTHAGOREAN TRIANGLES OF EQUAL AREAS 773 (u4,v4) (99,56); u 4 9?; v 4 56; c 97; a 55; b 57. S "11 34,058,640. For RPT(97,55) using (1.) we have S(97,55) 97"55( S in (3.6). 4. SOLUTION 0F (i.). 9.- ab + = c., In this chapter we state more triples of triangles RPT having the same area. We shall first prove Let a ab+ b c a,b,c N-0}; then (4.1) b> c> a> O; (a,b,c)! In (4.1) we set arbitrarily b > a; then we have b c a(b-a) > 0; b > c; c a b(b-a) > O; (4 c > a; J ) Thus b> c> a> O, as stated in (4.1). We shall now prove H.EOREM. Let c=a-ab+b, b> c> a> O. Then the three RPT-s formed by (b,c), (c,a), (c,b-a) have the same area S, viz. S abe(b-a) (4.3) and there are infinitely many such triples RPT. Poof. If we set in (1.4) a -->-a, we obtain from equation (1.3), the equation (4.1). But this is only an algebraic formality, since the RPT ormed by (c,-a) makes sense, though from S abc(a+b) we obtain (4.4) by substituting -a for a. We also have c (a-b) +ab, c> a-b, b-a. (4.4) Now D 1 RPT(b,c); S(b,c) I be( b _ c 1/ xy I bca(b-a)= abe(b-a); 1 D RPT(c,a); S(c,a) ca(c-a xy 1 cab(b-a) 1 abe(b-a); c D 3 RPT(c,D-a); S(c,b-a) Re(b-a)[ (b-a) ] =1/ xy c(b-a)ab abe(b-a) S S(D,c) S(c,a) S(c,b-a) abe(b-a). We still have to prove that (4.1) has infinitely many solutions. We shall give two methods to find these solutions, though these may not be all ininitely many solutions of (4.1). This was done by Hasse [6] with algebraic number theory which we shall avoid here, giving simple methods to solve (4.1) in explicit form.

6 774 M. BAICA In (4.1) we set b=a +v, v=b-a> 0; we obtain, setting a b-v, (b-v) (b-v)+ b= c b bv+v c 0 v bv- (c-b) 0. (4.6) From (4.6) we obtain b) 1, v 1/ (b _+ & + (c - v 1/ (b _+ &o"- ). (.r) We have, since setting 4c- 3b=x contradicts our condition (a,b,c)= 1 4c 3b i. (4.8) Thus we have arrived at Pell s equation, setting c Un, b v n. (4.9).Thus u n is even, and we must take u n Um+l b Vm+l; c 1/ Um+l; a b-v=b-1/ (b+l). and from (.7a) we obtain (4.o) Since we obtain two values for a, we may have obtained six RPT-s with 1 equal area. We shall investigate this later. We first take v = (b+l) ba { (b-l) 1/1_(Vm+l-l) I Vm+l; c 4. Um+l; b-a 1/ (Vm+l +I)" J Hence, from (4.3), or forming Ii DI=RPT(b,c), D =RPT(c,a), D3=RPT(c,b-a) ]. 1 S=1/(Vm+l l)-vm+l Um+l (Vm+l+l) s {- Um/Vm/<v/ (4.1) We take, for an example m i, (u3,v3) (6,15) Um+l 6; Vm+l 15 S -i.6.15(15-1) 10,90. The reader should note that since Vm/l S + 1 v m+l-1 O(mod 8). Hence, in (4.1) S is an integer. We now take v from (4. I0) 1/ (b-l) and obtain

7 PYTHAGOREAN TRIANGLES OF EQUAL ARF_JkS 775 i b Vm+l; c Um+l; a b-v 1/(b+l) 1/(Vm+l +i) i b-a (Vm+l I), (4.) and from D4=RPT(b,c), D 5=RPT(c,a), D 6=RPT(c,b-a). Previously, we had for v 1/ (b+l), a (Vm-1) b Vm+l, c Um+l b-a 1/(Vm+l). Comparing (4.11) with (4.13) we see that D D 6 D 3 D 5 D I D 4 and thus get the same triple of RTS-s in both cases. 5 SECOND SOLUTION OF c a ab + b We give a second method of finding infinitely many solutions of (4.1). We set here a 1/ (y+l); b=y-1; y>; y E N. (5.1) Substituting these values in (4.1), we obtain [1/ (y+l)]-1/ (y_l) + (y-l) c y +y+l-y++4y-8y+4 4c y_6y+t 4c }(y-1) + 4 4c 4c 3(y-I) 4. (5.) Since from (.7) y-i has to be even we can cancel (5.) by 4 and we obtain and with c y-i Vm, y Vm+l; (5.3) Um we obtain 3v 1 Um a 1/ (Vm + ), a Vm+l; b Vm, c Um (5.4) S abc(b-a) (Vm + 1)Vm Um(Vm- i) S UmVm(Vl- I). (5.5) Comparing (5.5) with (4.1) we may think about the different expressions for area S; though one is expressed by Um+l, Vm+l, the other by Um,

8 776 M. BAICA Vm, it is so because we are dealing here with different areas, depending on the values of a,b,c, which are different in each case. For an example for (5.5) we choose m, (u4,v4) (97,56), If we set S.97" 56-(56-1) 34,058,640. a=m-l, b=vm, c=u n, b-a=vm+l 3v (5.6) Um then a ab + c S= UmVm(Vm-1). But we obtain nothing new b since RPT(c,a), RPT(c,b-a) are interchanged with RPT(c,b-a), RPT(c,a). The author is asking whether two successive Fibonacci numbers could be solutions a,b of a + ab + b c The Fibonacci sequence, see n=l,,.., goes i, l,, 3, 5, 8, [8] with F l=f=1, Fn+=Fn+Fn+l, 13, 1, 34, 55, 89, 144, Now (F4,F5) (3,5) is a solution of the Diophantine example 3+3 " and (Fs,F6) (5,8) is a solution of 5 5" Here S abc(b-a) whether there are more pairs of adjacent Fibonacci numbers serving as solutions of a ab + b c could not be decided generally here. 6. PERIMETERS AND AREAS. As known, P, half the perimeter of a RPT is a divisor of its area S, since P u(u+v), S u-v(u-v) (6.1) s" P v(u-v) where (u,v) forms the RPT. The question arises whether there are other connections between these two elements. Here we could only prove THEOREM 3. Let a ab+b c b> c> a>o, a,b,c N- {0}. Then the sum of the three perimeters formed by a,b,c, viz. RPT(b,c), RPT(c,a), RPT(c,b-a) is the sum of two squares. Proof. We have D 1 RPT(b c): PI Xl+Yl +zl P 1 b (b+c) b c +bc+ +c b D RPT(c,a) P x + Y + z c_ a + ca + c + a c(c+a) D 3 RPT(c,b-a) P3 x3 + Y3 + z3 c (b-a) + c(b-a) + (b-a) + c P1 +P +P3 =c + c(b-a) c(c+b-a). b(b+c) +c(c+a) +c(c+b)-ac c+bc+b (b+c)+c

9 PYTHAGOREAN TRIANGLES OF EQUAL AREAS 7. ENTIRELY NEW RPT-s. In a very little known paper Hillyer [ 7] has given a most surprising infinity of triples of Rational Pythagorean triangles in an explicit form, having equal areas. These are formed by D 1 RPT(u,v) RPT(a + ab + b b -a D RPT(u,v) =RPT(a+ab+b ab+a D 3 RPT(u,v) =RPT(ab+b, a+ab+b) (7.1) b > a > 0; b,a Q+ As we see, Hillyer was not concerned about RPT-s; his Pythagorean triangles had just to have rational sides. We shall operate with RPT-s only and set a,b N- {0; b>a>0. (.) We first investigate, whether the condition u > v > 0 (7.3) is fulfilled for D l, D, D 3. We have Dl: u-v a + ab + b (b a) b + ab > 0; u > v > 0. D: u-v a + ab + b (ab + a) b ab b(b-a) > 0, since b > a hence u > v > 0 D3: u-v ab + b (a +ab+ b ab a a(b-a) > 0, since b-a > 0, u > v > 0. We shall now find the areas of D1, D, D3, and have D 1 :S 1 (a+ab+b) (b-a)[ (a+ab+b)-(b-a) ] (a+ab+b) (b-a) (a+ab+b+b_a). (a+ab+b-b+a) (a+ab+b) (b-a) (ab+b) (ab+a) S 1 ab(a+ab+b) (b-a) (a+b) (b+a). (7.4) D:S (a+ab+b) (ab+a)[ (a+ab+b)-(ab+a)] (a+ab+b)a(a+b) (a+ab+b+ab+a) (a+ab+b-ab-a) (a+ab+b)a(a+b) (a+3ab+b). (b-ab). Now hence D3:S3 Now a+3ab+b S (a+b)(a+b), ab(a + ab + b) (b-a) (a+b) (b+a) (ab+b) (am+ab+b)[ (ab+br)m-(ar+ab+b) ] b(a+b) (a+ab+b) (b+3ab+b) (ab-a). b+3ab+b (b-a)(b+a).

10 778 M. BAICA Hence S 3 ab(ar+ab+b)(br-ar)(a+b)(b+a). (7.6) Thus S 1 S S 3, ab(a + ab + DR) (br-ar)(a+rb) (b+ra). (7.7) We now return to the equation of Diophantus from (i. 3) a + ab + b c ] a+b > c > b > a > 0 (1.3) a,b,c N- 0}. We found that the three triples RPT(c,b), RPT(c,a), RPT(a+b,c) have equal areas, S viz. abc(b+a). If in (7.7) we demand that a b c + ab + solvable in natural number a+b > c > b > a > 0 we obtain abcr(b-ar)(a+rb)(b+ra), (7.8) and from (7.8), and S abc(a+b), S c(b-a)(a+b)(b+ra). (7.9) Now the quotient S is a natural number, and many authors have asked and solved the question of the ratio of the areas of two RPT-s. 8. THE MAIN RESULT. We now form three RPT-s, having equal areas. They are entirely new and unknown. We investigate D b b a 1 RPT(u v) RPT(a ab + ); D RPT(u,v) RPT(a-ab+b, ab-b); D 3 RPT(u,v) RPT(ab-a, a-ab+b) a ) b ) a > 0; b,a + We first check for DI, D, D3, (8.1) u > v > O, u-v > O. The reader note the condition a>b. Later when we shall operate with RPT-s and the equation a ab + b= c we shall see that solutions of this equation are possible with a > b. We have, (u,v) g Q+) DI: u-v a ab + b (b a a ab a(a-b) > 0 v b a > O.

11 PYTHAGOREAN TRIANGLES OF EQUAL AREAS 779 D : u-v a-ab+b- (ab-b) a- 3ab + b (b-a)(b-a) > 0 v ab-b=b(a-b) > 0. D3: u-v ab-a- (a -ab/b) =3ab-a-b (a-b)(b-a) > O. u-- ab-a=a(b-a) > 0 v a-ab+b (a-b)+ab>o. We shall now find the areas formed by DI, D, D3, and have ml:s1 (a-ab+br)(b-a)[ (a-ab+b)_(b_a) (a-ab+b) (b-a) (a-ab+b-b+a). (a-ab+b+b-a) (am-ab+b) (br-a (ar-ab) br-ab (a-ab+b) (br-ar)a (a-b) (b-a). S 1 ab(ar-ab+b) (b-a) (a-b) (b-a). DR: S (a-ab+b) (ab-b)[ (a-ab-br)-(rab-a) ] (a-ab+b)b(a-b) (a-ab+b-ab+b) (ab+a) (a-ab+br)b(a-b) (ar-3ab+rbr)a(b+a) (ar-ab+br)b (a-b) (b-a) (b-a)a (b+a) ab (a-ab+b (b-a a-b b-a ). ab(ar-ab+b)(b-ar)(ra-b)(b-a). S D3: S 3 (ab-a)(am-ab+b)[ (ab-a)r-(ar-ab+b) ] (ab-a) (a-ab+b) (ab-a-a+ab-b) (ab-a+a-ab+b) a b-a (ar-ab+b) (3ab-am-b)- (ab+b ab(ar-ab+br)(rb-a)(ra-b)(b-a) (a+b) ab (a-ab+b) (b-a) (a-b) (b-a). S 3 ab (ar-ab+b) (b-a) (a-b) (b-a). Thus we have obtained the wanted result S 1 S S 3. We now return to the equation a-ab+b c, b>a>0 b>c>a>o; b,c,a g N- 0} and recall from (5.4) that there is a solution with b a=+l; a> b as we needed. With equation (4.1), takes the form (8.) (8.3) (8.5)

12 780 M. BAICA a> b> c> a; a,b,c N-10. Now with a ab + b c and (4.3), viz. S abc b abc(b-a) a )(a-b)(b-a) and obtain thus the quotient I s c (b+a) (a-b) (b-a). As an example we shall take a-ab+b=c (a,b,c) (8,15,13). Here a=16> 15 b; we obtain (8.6) , ,831,760 S abc(b-a) 8-15"13" ,90. S ll ,831,760 10,90 6,578 ACKNOWLEDGEMENT. Dedicated to the memory of Leon Bernstein. Dr. Bernstein was the author s professor and advisor at Illinois Institute of Technology, Chicago. REFERENCES [I] L. BERNSTEIN: Primitive Pythagorean triples, The Fibonacci _Quarterly 0, No. 3 (198) [] L. E. DICKSON:. History of Number Theory 3 vols., New York, Chelsea (1919) Vol. II, [3] DIOPHANTUS: Collected works [4] U. DUDLEY: Elementary Number Theory, W. H. Freeman and Co. (1969). [5] A. GIOIA: The Theo.r of Numbers,3An Introduction, Markham Publishing Co. 1970)Jl1-11 [6] H. HASSE: Ein Analogon zu den ganzzahligen pythagorischen Dreiechken, Elemente der Mathematic, Basel 3, No. 1 (1977). [7] C. E. HILLYER: Mathematical Questions and Solutions, from the "Educational Times", vol. 7 (1900) (Hillyer s sblution for R. FJ. Davis problem 14087). [8] V. E. HOGGATT, JR.: Fibonacci and Lucas numbers. Boston: Houghton-Mifflin Co. (1969) (rpt. The Fibonacci Association, 1980).

13 Mathematical Problems in Engineering Special Issue on Modeling Experimental Nonlinear Dynamics and Chaotic Scenarios Call for Papers Thinking about nonlinearity in engineering areas, up to the 70s, was focused on intentionally built nonlinear parts in order to improve the operational characteristics of a device or system. Keying, saturation, hysteretic phenomena, and dead zones were added to existing devices increasing their behavior diversity and precision. In this context, an intrinsic nonlinearity was treated just as a linear approximation, around equilibrium points. Inspired on the rediscovering of the richness of nonlinear and chaotic phenomena, engineers started using analytical tools from Qualitative Theory of Differential Equations, allowing more precise analysis and synthesis, in order to produce new vital products and services. Bifurcation theory, dynamical systems and chaos started to be part of the mandatory set of tools for design engineers. This proposed special edition of the Mathematical Problems in Engineering aims to provide a picture of the importance of the bifurcation theory, relating it with nonlinear and chaotic dynamics for natural and engineered systems. Ideas of how this dynamics can be captured through precisely tailored real and numerical experiments and understanding by the combination of specific tools that associate dynamical system theory and geometric tools in a very clever, sophisticated, and at the same time simple and unique analytical environment are the subject of this issue, allowing new methods to design high-precision devices and equipment. Authors should follow the Mathematical Problems in Engineering manuscript format described at Prospective authors should submit an electronic copy of their complete manuscript through the journal Manuscript Tracking System at mts.hindawi.com/ according to the following timetable: Guest Editors José Roberto Castilho Piqueira, Telecommunication and Control Engineering Department, Polytechnic School, The University of São Paulo, São Paulo, Brazil; piqueira@lac.usp.br Elbert E. Neher Macau, Laboratório Associado de Matemática Aplicada e Computação (LAC), Instituto Nacional de Pesquisas Espaciais (INPE), São Josè dos Campos, São Paulo, Brazil ; elbert@lac.inpe.br Celso Grebogi, Department of Physics, King s College, University of Aberdeen, Aberdeen AB4 3UE, UK; grebogi@abdn.ac.uk Manuscript Due February 1, 009 First Round of Reviews May 1, 009 Publication Date August 1, 009 Hindawi Publishing Corporation