Two Interesting Integer Parameters of Integer-sided Triangles

Transcription

1 Forum Geometricorum Volume 17 (2017) FORUM GEOM ISSN Two Interesting Integer Prmeters of Integer-sided Tringles Jose A. De l Cruz nd John F. Goehl, Jr. Abstrct. When tringle is described in terms of the segments into which its sides re divided by n inscribed circle, it permits determintion of ll integer sided tringles for which the re is n integer multiple of the perimeter. It is not possible to hve integer-sided tringles with R/r n integer, where R nd r re the rdii of the circumcircle nd incircle respectively, for right tringles, isosceles tringles, nd tringles whose sides re in rithmetic progression. The exception is the equilterl tringle for which R/r = 2. First consider integer-sided tringles for which the reais n integer multiple m of the perimeter P. A = mp. (1) For the prticulr cse of right tringles (Figure 1), eqution (1) becomes: 1/2b = m(+b+c). (2) c c b Figure 1. θ b Figure 2. Rerrngement gives b 2m(+b) = 2mc. Squring both sides of the eqution, noting tht c 2 = 2 +b 2, gives which cn be simplified to 2 b 2 +4m 2 (+b) 2 4mb(+b) = 4m 2 ( 2 +b 2 ), This lst eqution cn be fctored to give b+8m 2 4m(+b) = 0. ( 4m)(b 4m) = 8m 2. (3) Publiction Dte: November 28, Communicting Editor: Pul Yiu.

2 412 J. A. De l Cruz nd J. F. Goehl The fctors of 8m 2 yield ll right tringles stisfying eqution (2). This solution ws given by Goehl [1]. For generl tringle [2, 3, 4, 5, 6], with b c, eqution (2) is generlized to: 1 bsinθ = m(+b+c). 2 (4) Rerrngement gives: bsinθ 2m(+b) = 2mc. Squring both sides of the eqution, noting tht c 2 = 2 +b 2 2bcosθ, gives 2 b 2 sin 2 θ +4m 2 (+b) 2 4mb(+b)sinθ = 4m 2 ( 2 +b 2 2bcosθ) which cn be simplified to bsin 2 θ +8m 2 4m(+b)sinθ = 8m 2 bcosθ. This lst eqution cn be fctored to give (sinθ 4m)(bsinθ 4m) = 8m 2 (1 cosθ). (5) Note tht, of course, eqution (5) reduces to eqution (3) when θ = 90. In order to determine llowed vlues of θ, consider the circle inscribed in the tringle. Now Since b c,α β γ. β = α+β, b = α+γ, c = β +γ. β r r c γ α α θ 2 r Figure 3. b γ From Heron s formul: A = s(s )(s b)(s c), where s is the semi perimeter ndp = 2s = 2(α+β +γ). The problem A = mp is equivlent to (α+β +γ)(γ)(β)(α( = m 2(α+ β + γ). Simplifiction gives [5] αβγ = 4m 2 (α+β +γ). (6) Clerly there re no equilterl tringles s solutions becuse if α = β = γ, eqution (6) implies tht α 3 = 4m 2 (3α) or α 2 = 12m 2. So α is not n integer. Note tht the rdius of the incircle r is equl to 2m [2, 6]. The combined re of

3 Two interesting integer prmeters of integer-sided tringles 413 the 6 smll right tringles in the figure tht hve r s one side is equl to the re of the tringle: A = 1 2 (rα+rα+rβ +rβ +rγ +rγ) = r(α+β +γ) = rp 2 = mp, sor = 2m. From eqution (6), it follows tht α 2 βγ = αr 2 (α + β + γ) nd fctoriztion yields (αβ r 2 )(αγ r 2 ) = r 2 (α 2 +r 2 ). (7) This result lso follows from identifying the ngle in eqution (5). From the bove figure, sin θ 2 = r nd cos θ α 2 +r 2 2 =. Now, sinθ = 2 α α 2 +r 2sin θ 2 cos θ 2 = 2rα nd cosθ = cos 2 θ α 2 +r 2 2 sin2 θ 2 = α2 r 2. Substituting these α 2 +r 2 vlues into eqution (5) yields: ( ( ) )( ( ) ) 2rα 2rα α 2 +r 2 2r b α 2 +r 2 4m = 2r (1 2 α2 r 2 ) α 2 +r 2. Simplifying: (α (α 2 +r 2 ))(bα (α 2 +r 2 )) = r 2 (α 2 +r 2 ). Substituting = α+β, ndb = α+γ yields: (αβ r 2 )(αγ r 2 ) = r 2 (α 2 +r 2 ). Sincecis the lrgest side,θ is the lrgest ngle. The lrgest vlue of the smllest prmeter, α, results when α = β = γ nd θ = 60. So ( tn θ ) 2 min = r 3 = 1 3 nd r α 1 3. Therefore llowed vlues forαre given by1 α < 3r = 2 3m. From the figure,tn θ 2 = r α ndtnθ = 2tn θ 2 1 tn 2 θ 2 = 2rα α 2 r 2. Of course,m = 1 corresponds to re=perimeter. For m = 1, r = 2m = 2. So 1 α < , nd the vlues for α re 1, 2, nd3. Eqution (7) becomes (αβ 4)(αγ 4) = 4(α 2 +4). For α = 1, tnθ = = 4 3, θ = , nd (β 4)(γ 4) = 4(1 + 4). The three fctoriztions of 20, nmely, (4)(5), (2)(10), nd (1)(20), ll yield tringles: 1, 8, 9 α, β, γ = 1, 6, 14 1, 5,24 9, 10, 17 nd, b, c = 7, 15, 20 6, 25,29 For α = 2, θ = 90, nd (β 2)(γ 2) = Both fctoriztions of 8, (2)(4) nd (1)(8), yield tringles: { { 2, 4, 6 6, 8, 10 α, β, γ = nd, b, c =. 2, 3, 10 6, 12,13.

4 414 J. A. De l Cruz nd J. F. Goehl For α = 3, tnθ = = 12 5, θ = , nd (3β 4)(3γ 4) = 4(9 + 4). The three fctoriztions of 52, nmely, (4)(13), (2)(26), nd (1)(52), yield no new solutions becuse α < β. The results form = 1 ndm = 2 re shown in Tble 1. A β γ α B c +b+c m θ Tble 1. An interesting specil cse is tht of isosceles tringles. For isosceles tringles, let β = γ, then αβ 2 = 4m 2 (α+2β). The result is the qudrtic eqution: αβ 2 8m 2 β 4m 2 α = 0. The solutions re: β = 2m α (2m± 4m 2 +α 2 ). (8) The positive vlue is chosen. The restriction,α β, cnnot be ssumed. For m = 1, β = 2 α (2+ ) 4+α 2. To get integer solutions, 4 + α 2 must be the squre of n integer,i 2. SoI 2 α 2 = 4, or(i+α)(i α) = 4. The two possible fctoriztions,(2)(2) nd (4)(1), result in no solutions. The first fctoriztion gives α = 0 while the second givesα = 2.5.

5 Two interesting integer prmeters of integer-sided tringles 415 For m = 2, β = 4 α (4+ ) 16+α 2. To get integer solutions, 16 + α 2 must be the squre of n integer,i 2. So I 2 α 2 = 16, or (I + α)(i α) = 16. The three possible fctoriztions, (4)(4),(8)(2), nd(16)(1), result in one solution. These fctoriztions give vlues for α of 0, 3, nd 7.5. The one integer vlue for α lso gives n integer vlue for β:12. The results for vlues of m from 2 through 6 re shown in Tble 2. Note tht there re no solutions form = 1 ndm = 7. m α β γ b c Tble 2. Tringles with sides in rithmetic progression Another interesting specil cse is tht of tringles with sides in rithmetic progression. The problem hs been studied before for consecutive integers [7]. Now consider the generl cse for n rithmetic progression α = β δ nd γ = β +δ. The sides become: = α+β = 2β δ,b = α+γ = 2β, ndc = β+γ = 2β+δ. Now eqution (6) becomes (β δ)β(β +δ) = 4m 2 (β δ +β +β +δ) or (β δ)(β +δ) = 12m 2. To find ll tringles with sides in rithmetic progression, find ll fctoriztions of12m 2 tht result in integer vlues ofβ ndδ. For exmple, for m = 1, the possible fctoriztions re (12)(1), (6)(2), nd (4)(3). Only the second fctoriztion gives integer vlues: β = 4 nd δ = 2. The resulting tringle hs sides = 6,b = 8, ndc = 10. The results for vlues ofmfrom1through4re shown in Tble 3.

6 416 J. A. De l Cruz nd J. F. Goehl m δ α β γ b c Tble 3. A second interesting prmeter ssocited with tringle is the rtio of the rdii of its circumcircle nd incircle. McLeod [8] hs shown tht, for tringles with sides,b, ndc, this rtio is given by: N = 2bc (+b c)(+c b)(b+c ). (9) He points out tht tringles for which this rtio is n integer re reltively rre nd finds some of them. In fct, it will be shown tht, with the exception of the equilterl tringle, no right tringles, isosceles tringles, or tringles whose sides re in rithmetic progression hve n integer rtio. First consider right tringles. Only primitive tringles need be considered. For such tringles, the sides cn be represented by = 2mn, b = m 2 n 2, nd c = m 2 + n 2, where m nd n re reltively prime nd of opposite prity. Then, from eqution (9): N = 4mn(m 2 n 2 )(m 2 +n 2 ) (2n(m n))(2n(m+n))(2m(m n)). (10) Simplifiction yields: 2n(m+n)N = m 2 +n 2. (11) This implies tht n must divide m. The only possibility is n = 1. But then the left side of (11) would be even nd the right side odd. Therefore there re no right tringles with integer N. Next consider isosceles tringles. Letting b = c in eqution (9) gives N = 2b 2 (2b ). This my be written s qudrtic eqution in:n2 2Nb+2b 2 = 0. This eqution hs solutions = b(n±i) N where I 2 = (N 1) 2 1. Thus N = 2 ndi = 0, nd the only solution is = b = c, the equilterl tringle. Lstly consider tringles with sides in rithmetic progression. Letting = b + d nd c = b d in eqution (9) gives N = 2b(b2 d 2 ) b 2 4d 2 or b2 d 2 = 2(2N 1) N 2. Thus 2(2N 1)(N 2) = J 2 for some integer J. This my be rewritten s (4N 5) 2 (2J) 2 = (4N 5+2J)(4N 5 2J) = 9. One fctoriztion yieldsj = 2

7 Two interesting integer prmeters of integer-sided tringles 417 nd the non-integer N = 5 2. The only other possible fctoriztion yields J = 0 ndn = 2, once gin the equilterl tringle. Conclusion Insight is gined from the geometric pproch to the solution of the problem of A = mp. Eqution (5), in terms of the generl ngle, θ, leds immeditely to eqution (3) for right tringles nd to eqution (7) for the generl cse. When tringle is described in terms of the segments into which its sides re divided by n inscribed circle, eqution (7) permits determintion of ll integer sided tringles for which the re is n integer multiple of the perimeter. The ngle θ, opposite to the gretest side c, is shown to hve reltively smll number of llowed vlues. The rtio of the rdii of circumcircle to incircle is considered. The specil cses of right tringles, isosceles tringles, nd tringles with sides in rithmetic progression re solved generlly for the two prmeters. References [1] J. Goehl, Jr., Are = k(perimeter), Mth. Techer, 76 (1985) [2] J. Li, Finding Heronin tringles whose res re integer multiples of their perimeters, Deprtment of Mthemtics, Michign Technologicl University, Houghton, MI, USA. [3] L. P. Mrkov, Pythgoren triples nd the problema = mp for tringles, Mth. Mg., 79(2006) [4] L. P. Mrkov, Heronin tringles whose res re integer multiples of their perimeters, Forum Geom., 7 (2007) [5] T. Leong, D. T. Biley, E. M. Cmpbell, C. R. Diminnie, nd P. K. Swets, Another pproch to solving A = mp for tringles, Mth. Mg., 80 (2007) [6] A. M. Lmb. The ProblemA = mp for tringles, 4/30/07 [7] J. A. McDougll, Heron Tringles With Sides in Arithmetic Progression, School of Mthemticl nd Physicl Sciences, University of Newcstle, NSW, Austrli [8] A. J. McLeod, Integer tringles with R/r = N, Forum Geom., 10 (2010) Jose A. De l Cruz: Brry University, NE Second Avenue, Mimi Shores, Florid, USA E-mil ddress: jdelcruz@brry.edu John F. Goehl, Jr.: Brry University, NE Second Avenue, Mimi Shores, Florid, USA E-mil ddress: jgoehl@brry.edu