RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS


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1 RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is shown in the following figure  As lredy shown by Euclid in his Elements, there re n infinite number of such right ngle tringles whose sides hve ll integer vlues. They re referred to s tringle integer triplets nd designted by [,b,c]. The first few of these triplets where <b<c cn be found by simple numericl evlution nd re [,b,c][3,4,5],[5,,3],[6,8,0],[7,4,5],[8,5,7],[9,,5],[9,40,4],[0,4,6], [,60,6],[,6,0],[,35,37],[3,84,85],[4,48,50],[5,,3],[6,63,65], [7,44,45],[8,80,8],[9,80,8],[0,99,0],[0,,9] One notices t once tht some of the triplets re just n erlier triplet multiplied by n integer nd thus form similr tringles. Thus [6,8,0], [9,,5],[,6,0] re ll similr to the [3,4,5] bse triplet for the group. Likewise [5,,3] is the bse triplet for the group [0,4,6],[5,36,39], etc. Also no rel new informtion is obtined by interchnging nd b. Keeping only the bse triplets leves one with [3,4,5],[5,,3],[7,4,5],[8,5,7],[9,40,4],[,60,6],[,35,37],[3,84,85],[4,48,50] [5,,3],[6,63,65],[7,44,45],[8,80,8],[9,80,8],[0,99,0],[0,,9],.. You will notice in this collection tht the numbers b nd c differ by one if is n odd integer nd by two if is n even integer for most cses with few exceptions such
2 s [0,,9]. This mens we my generte subclss of the infinite number of bse triplets by the formuls n, b ( n 4) / 4, nd c ( n + 4) / 4 nd for even n, b ( n ) /, nd c ( n + ) / for odd with n,, 3, As exmples, for 3 we find b550 nd c55 nd for 44 we hve b44943 nd c Both triplets stisfy the Pythgoren Theorem for right tringles. Note tht in the bove list the lst triplet [0,,9] does not stisfy the just stted formul for even nor is it obtinble from one of the lower number bse triplets. The difference between c nd b is 8 suggesting the modifiction n, b [ n ]/, nd c [ n + ]/ where (cb)k nd k,, 3, for both even nd odd. We cn lso multiply this lst triplet by to obtin the ll integer triplets n, b n, c n + with n > This lst expression represents the clssicl form for triplets lredy known to Euclid nd referred to in the literture s the Pythgoren Triplets. One exmple of such triplet obtined for 99 nd n is [, b, c] [3958, 4840, 444] Note tht this Pythgoren Triplet is not bse triplet since ll components cn be divided by 4 to yield the bse triplet [0,99,0]. Among the bove listed bse triplets we find [0,,9] is generted by nd n5. Also [0,99,0] is generted by nd n0. To determine if triplet generted by the Euclid formul is bse triplet one needs to simply see if the elements divide by n integer. Thus [ 96,80,04] > 4[4,45,5] > [8,5,7] shows tht the bse triplet in this cse is [8,5,7] nd is chrcterized by n4,. One cn tke the originl Pythgoren Theorem nd divide it by c. Then letting x/c nd yb/c, one hs the formul for unit rdius circle
3 x + y n n + n + n + nd cn red off the points on this circle corresponding to bse triplet by the pproprite choice of n nd. Keeping n even number, one hs for the first few bse triplets [n, ] [,],[3,],[4,3],[4,],[6,5],[6,],[8,7],[8,],[9,8],[9,],[0,9],.. The corresponding vlues of x nd y re [ x, y] [, ],[, ],[, ],[, ],[, ],[, ], [, ],[, ],[, ],[, ],[, ], Noting tht x cn be interchnged with y, we hve the bse triplets [, b, c] [3,4,5],[5,,3],[7,4,5],[8,5,7],[,60,6],[[,35,37], [5,,3],[6,63,65],[7,44,45],[8,80,8],[9,80,8],... All points [x,y] nd their complements[y,x] lie on circle of rdius one s shown
4 The two cute ngles of the right tringles defined by bse triplet [,b,c] re Brctn(b/) nd Arctn(/b). Thus we hve tht Also, it follows tht b π rctn( ) + rctn( ) b b sin( A) sin( B) nd cos( A) b, cos( B) + b + b For the [3,4,5] right tringle one hs Arcos(4/5) deg nd Brcos(3/5) deg with C90deg. An interesting clcultion involving right tringles is the determintion of the rdius of the lrgest circle which my be inscribed in n oblique tringle. The derivtion involves the Heron formul for the re of such tringle nd the bisection of the three ngles A,B, C of the tringle surrounding circle of rdius R. We hve the following geometry
5 One sees tht the re of the tringle ABC must equl the re of the six right tringles shown. The height of ech of the tringles is equl to the inscribed circle rdius R. From the geometry one sees tht the hlf circumference is s ( + b+ c) ( + + b + b + c + c ) nd lso it is observed from the picture tht, b c c b, Thus one hs fter little mnipultion tht, c s so tht s + b + c s b b s c, which mens tht the tringle re is Are + b + c ) R sr (,
6 Equting to the known Heron result, one hs tht the rdius of the lrgest inscribed circle in ny oblique tringle will be R ( s )( s b)( s c) s The lrgest circle fitting into [3,4,5] right tringle will thus be R nd for n equilterl tringle of sidelength the rdius will be R/[sqrt(3)]. The rdius of the lrgest circle inscribed in the right tringle defined by the triplet [0,99,0] will be R9. An interesting observtion is tht R is lwys n integer vlue s long s the triplet [,b,c] consists of integers nd defines right tringle. This condition is violted for non right tringles such s isosceles nd equilterl tringles or right tringles where one of the sides is noninteger such s for [,,sqrt()]. Just to confirm this sttement consider the bse triple [060808, 70985, 5443] generted by n997 nd 53. Here we find the lrgest inscribed circle hs integer rdius R withsemiperimeters 5443 The frction of the tringle re covered by the circle will be f circlere tringlere π ( s )( s b)( s c) 3/ s Finlly, we look for the lrgest right ngle tringle, designted by the triple[,b,c], which just fits into circle. The solution is stright forwrd when one reclls from elementry geometry tht two lines drwn from opposite ends of circle s dimeter lwys intersect t right ngle when they meet somewhere on the circle. Thus the rdius of such circle will be Rc/ nd the frction of the circle covered by the [,b,c] tringle will be f tringlere circlere b π ( + b ) Thus [3,4,5] right tringle will hve n f This problem becomes more complicted when the tringle becomes oblique. Now the rdil distnce from the circle center is R nd the circle touches ll three corners of the [,b,c] tringle. Clling the three ngles subtended t the circle center s θ A, θ B,θ C we hve from the lw of cosines tht b c cos( θ A), cos( θ ), cos( ) B θ C R R R
7 Denoting these cosines by α, β nd γ, we hve π rccos( α) + rccos( β ) + rccos( γ ) Combining these three rccosine terms two t time nd then tking the cos of the result leds to the equlity ( γ αβ) ( α )( β ) Which yields the implicit result for the circle rdius R for ny inscribed tringle. Tke the cse of n equilterl tringle of side length bc. Here And leds to the cubic eqution α β γ R γ 3 3γ + 0 with solutions γ/,, nd.the first of these is of interest to us nd leds to R/sqrt(3). A plot of this inscribed equilterl tringle follows
8 Its cover frction equls f3sqrt(3)/(4π) nd is the lrgest possible for n inscribed tringle within circle. September 00
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