Interntionl Journl of Mthemtis Reserh. ISSN 0976-5840 Volume 6, Numer (04), pp. 8-9 Interntionl Reserh Pulition House http://www.irphouse.om Study on the Properties of Rtionl Tringles M. Q. lm, M.R. Hssn nd 3 M.. Hssn eprtment of Mthemtis, Mhil ollege, rhiy, 830, Indi ssoite professor in Mthemtis, S. M. ollege, hglpur-800, Indi 3 Mehnil Engineer, H. M. S. I. T. Tumkur, Krntk, 5704, Indi E-mil: hssnsm@mil.om strt In this mnusript n ttempt hs een mde to ollorte geometry nd rtionl numer system. In support of this some properties of rtionl tringles hve een disussed nd some theorems hve een estlished following the geometril onept of rhme Gupt ( 598 ), het nd Viet et. Keywords: Opposite Prity, Primitive tringle, Rtionl tringle, Juxtposition nd Olique tringle.. Introdution Let us define some key terms to mke the mnusript more understndle. The definitions re s follows: Rtionl tringle: - tringle with rtionl sides nd rtionl re is lled rtionl tringle. Juxtposition: - The proess of omintion of two right tringles with ommon leg is lled Juxtposition. The omined tringle is lled olique tringle. Exmple: - Let & e two right tringles with ommon leg nd Right <ed t, then the omined tringle is the olique tringle.
8 M. Q. lm, M.R. Hssn nd M.. Hssn Olique Tringle: fter juxtposition of two right tringles, the omined tringle is lled olique tringle. Opposite prity: Two reltively prime numers re sid to e of opposite prity if one of them is odd nd the other is even. Ex.: & 3 re of opposite prity, 3 & 4 re of opposite prity. Primitive tringle: right tringle with reltively prime integrl sides is lled primitive tringle. Ex.: Pythgoren triplet (3, 4, 5) represents primitive right tringle s (3, 4, 5) =. Setion: - Erly works out rtionl tringles rhme Gupt (.598-) hose ny three positive rtionl numers,, nd estlished tht,, nd re the side of n olique tringle whose ltitude nd re re rtionl, where the olique tringle is formed y juxtposition of two right tringles & with the ommon height =. + = Olique tringle Juxtposition
Study on the Properties of Rtionl Tringles 83. G. het [] introdued two methods to solve rtionl tringles s follows: Method : In the first method het onsidered right tringle with the sides 0, 8, 6 nd = N suh tht N 8, where is rtionl. From the properties of tringles, < is ute or otuse ording s or i. e. 6N 8 or 6N 8 se : When N 3 / 3 is ute if N 3 / 3 & is otuseif N 3 / 3. ute 0 8 6 N N Let us hoose positive rtionl numer x suh tht 8 8 xn 6x then, N 3 / 3 x x x 0 x ( s x 0) () Ex.: Let x = 5/, then from () N 60 / Thus the sides of the tringles with ute nd ltitude 8 re 0, 6+N, 8 N i.e., 0, 86/ nd 3/ re the rtionl sides. se : When N 3 / 3
84 M. Q. lm, M.R. Hssn nd M.. Hssn N Similr to se () ` N 8 8 xn 6x N 3 / 3 x x x 0 x ( s x 0) Ex: Let x 3 /, then from () N 8 (8 3 N / ) Thus the sides of the tringle with otuse nd ltitude = 8 re 96 96 0,6, 8 i. e. 0,6 / 5,04 / 5 5 5 re rtionl sides. Method : In the seond method het used the method of juxtposition of two rtionl right tringles with ommon height. He onsidered = nd other two numers x nd y suh tht x ny squre nd y ny squre Thus x 35 35 37 & y 6, 6 0 Using (m -n ) + (mn) = (m +n ) Hene one n find the rtionl tringle y juxtposing s 37, 0, 35 + 6 = 5 with ltitude. Similrly y juxtposing the tringles 35 37 & 9 5 or 35 37 & 5 3, We get the rtionl tringles (37, 5, 35+9) or (37, 3, 35+5).
Study on the Properties of Rtionl Tringles 85 F. Viet s Method: F. Viet [3] otined rtionl tringle y juxtposing two right tringles s follows: He onsidered rtionl right tringle with right legs nd d nd hypotenuse z nd seond right tringle hving ltitude = (f + d), then the hypotenuse nd se of the seond tringle re & f d f d respetively. Now multiplying the sides of the first tringle y nd the sides of the seond tringle y d then the two right tringles eome of ommon ltitude d. Thus y Juxtposition of two resulting right tringles with ommon ttitude we otined the rtionl olique tringle with sides z., d[( f d ) ], d[( f d ) ]. For Ex, tking = 3, d = 4, z = 5, f = 6, so tht = 60nd ignoring the proportionl ftor 4, we get the rtionl tringle of sides 75, 09, 36 with ltitude 60. Setion (present works out rtionl tringles) In this setion we hve estlished some new theorems pplying the onept of juxtposition of two right tringles nd solved some olique tringles. Theorem () Sttement: In ny tringle with rtionl sides,, nd rtionl re. [( ps qr)( pr qs)] p q r s : : : :, pqrs pq rs where p, q, r, nd s re some rtionl numers nd every pir of sides re in the rtio u v of two numers of the form. uv Proof: p q pq r s rs p q pq r s rs onsidering two right tringles nd with sides
86 M. Q. lm, M.R. Hssn nd M.. Hssn p q p q r s r s, nd,, pq pq rs rs with ommon ltitude =. Now juxtposing these two tringles we get the olique tringle, whose sides re [( ps qr)( pr qs)] p q r s,,, pqrs pq rs where the upper nd lower sign is to e tken ording s the tringles nd do not or do overlp. Moreover we n esily verify tht r s x y : : for x ps qr, y pr qs, rs xy So tht x y ( p q )( r s ) Theorem : Sttement: There exit three tringles with integrl sides nd re hving equl perimeters nd res in the rtio of : :. Proof: Let the three required tringles F, F, F stnding on the sme se nd of ommon ltitude EF= t F t E Tking p, q, r, s F t F t F t F t p q r s p q r s Then E t, E t, E t, E t. p q r s
Study on the Properties of Rtionl Tringles 87 p q q r E E t, E E t p q q r r s nd E E t. r s From the sttement, ll the three tringles hve sme perimeters so F F F F F F p q s r q r () q p, q r () & r q s r (3) & p r s q (4) gin from the sttement, F F F [s height is ommon to ll the tringles] t p q t q r t r s p q q r r s rp p r p r q s s q (5) rp s p r p r q p r (6) s rp (s p r ) q q p, s r (7) From (5) nd (7) we hve r q q q s q r r q r (8)
88 M. Q. lm, M.R. Hssn nd M.. Hssn Eliminting p, q nd s from (), (7) nd (8), we get r r r r r r r 4 d d r de 0 (9) where, d, e Now hoosing integrl vlues of,, so tht the iqudrti in r gives rtionl root, then we n esily see tht p, q & s re rtionls. Thus we otined the required tringles. Exmple: For exmple tking =, = 7, = 5, we get the rtionl root r = 5/3. Thus 5 p, q, s 0 4 54 5 7 4 F t, F t, F t, F t, 40 4 5 4 Tking t = 40, we get F 54, F 55, F 476, F 4, 6, 9, 95 The perimeters of the tringle re sme, whih is equl to 09. Prolem () If the sides nd re of the tringle re integers, then the re is divisile y 6. Proof: If p, q, r, s ll re integers then in ny rtionl tringle, the sides re proportionl to the numers ( ps qr)( pr qs) p q r s,, pqrs pq rs Thus the sides of the rtionl tringles re tken s = (ps+qr)(pr-qs), =rs(p +q ), =pq(r +s ) Let t e the perimeter of the rtionl tringle, then t. ). The re of the tringle is given y t t t t pqrs ps qr pr qs (0) If p, q, r, s re ll odd integers or t lest one of them is even then 0 (mod If one of p, q, r, s is odd then 0 (mod 3). Now suppose tht none of p, q, r, s is divisile y 3, then
Study on the Properties of Rtionl Tringles 89 ( ps qr)( pr qs) rs( p q ) pq( r s ) rs pq mod (3) ( ps qr)( ps qr) 0(mod 3) euse the squre of ny integer whih is not divisile y 3, is ongruent to (mod 3). Hene it follows tht pqrs ( ps qr)( ps qr) 0 (mod 3). Then, 0mod 6. Theorem (3) Sttement: There exit tringles whose sides re onseutive integers. Proof: Let x, x, x + e the sides of rtionl tringles, then the re is given y where t t t t t 3x hlf of the perimeter of the tringle. 3 x x 4 4 Sine the re of the tringle is rtionl hene we must hve x 3y 4 () Here x nd y oth nnot e odd simultneously sine 4 3 (mod 8) is impossile. lso x nd y nnot e of opposite prity sine 0 4 3 (mod 4), 4 3 (mod 4) eh of whih is impossile hene x nd y oth must e even. Let us write x = u & y = v then from () we get u 3v () Whih is Pell s eqution nd whose fundmentl solution is u =, v =. Hene ll solutions of Pell s eqution () re given y 3 3 r u v, r,,3,... Thus ( u, v) (,),(7, 4)... nd the required tringles re (3, 4, 5), (3, 4, 5) Prolem () The tringle with the ltitude nd sides 3, 4 5 is the only one tringle in whih the ltitude nd sides re onseutive integer. Proof: Let e the ltitude nd +, +, + 3 e the three sides of tringles.
90 M. Q. lm, M.R. Hssn nd M.. Hssn x 3 If the perpendiulr is drwn from to the side + 3 nd x is the se of one of the right tringle formed fter drwing perpendiulr, then (I.) x, 3 x For other perpendiulrs, we hve (II) x, x 3 (III) x 3, x Sutrting the result of (I.), we get + 3 0 (mod ( - 3)) whih is impossile Similrly (II) gives 5 0 (mod( )) whih is impossile only when =. Now sustituting the two equtions of se (III), we get x ( ) 4 x 6 (Negtive x is indmissile) Thus from the first Eqution, we get 3 3 Hene the theorem follows. Theorem (4) Sttement: If in ny primitive rtionl tringle, two sides re odd, the lst side > nd the differene etween the sum of two smller sides nd the lrger sides is not unity. Proof: Let,, e the integrl sides of tringle with (,, ) = nd let < <. (I.) Let us first prove two of,, re odd. The re of the tringle is given y t t t t, where t
Study on the Properties of Rtionl Tringles 9 6 ( ) ( ) (3) 6 0 0 (mod 4) if, nd re even. 6 0 0 (mod 4) if, even nd odd. 6 0 0 (mod 4) if, even nd odd. Hene two of,, must e odd. (II) Now let us prove tht > We hve & s M GM In one of the ove two inequlity ( ) Hene / Similrly from the seond inequlity / Thus (dding two results) 4 5 If =, the lest vlue of = nd lest vlue of = 3 nd so whih is 4 impossile. 7 If =, the lest vlue of = 3 nd lest vlue of = 4 nd whih is 4 impossile. 9 If = 3, the lest vlue of = 4 nd 3 whih is impossile. 4 Hene, >. (III) Here to prove tht. If possile, suppose + =, then from (3) 6 6 I (4) The reltion (4) does not hold euse left side of (4) is even wheres right side is odd. Thus + = is impossile. Hene, +. onlusion: ll the propositions nd theorems re the tools for numer system nd for the theory of numers. With the help of these propositions nd theorems mny unsolved prolems of rtionl tringles n e solved.
9 M. Q. lm, M.R. Hssn nd M.. Hssn Referenes [] rhme-sphut-sidhnt, h., lger with rith nd Mensurtion, from the Snskrit of rhme, Gupt nd hsr, trnslted y H.T. olerooke, LONON, (87) p; 306 [] L.E. ikson: History of the theory of numers, helse Pulishing ompny, New York (95) [3] H. renport: Higher rithmeti, Hithinson, LONON (95)