On a Problem Regarding the n-sectors of a Triangle

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1 Forum Geometricorum Volume 5 (2005) FORUM GEOM ISSN O a Problem Regardig the -Sectors of a Triagle Bart De Bruy Abstract. Let be a triagle with vertices A, B, C ad agles α = BAC, β = ÂBC, γ = ÂCB. The 1 lies through A which, together with the lies AB ad AC, divide the agle α i 2 equal parts are called the - sectors of. I this paper we determie all triagles with the property that all three edges ad all 3( 1) -sectors have ratioal legths. We show that such triagles exist oly if {2, 3}. 1. Itroductio Let be a triagle with vertices A, B, C ad agles α = BAC, β = ÂBC, γ = ÂCB. The 1 lies through A which, together with the lies AB ad AC, divide the agle α i 2 equal parts are called the -sectors of. A triagle has 3( 1) -sectors. The 2-sectors ad 3-sectors are also called bisectors ad trisectors. I this paper we study triagles with the property that all three edges ad all 3( 1) -sectors have ratioal legths. We show that such triagles ca exist oly if =2or 3. We also determie all triagles with the property that all edges ad bisectors (trisectors) have ratioal legths. I each of the cases =2 ad =3, there are ifiitely may osimilar triagles havig that property. I umber theory, there are some ope problems of the same type as the abovemetioed problem. (i) Does there exists a perfect cuboid, i.e. a cuboid i which all 12 edges, all 12 face diagoals ad all 4 body diagoals are ratioal? ([3, Problem D18]). (ii) Does there exist a triagle with iteger edges, medias ad area? ( [3, Problem D21]). 2. Some properties A elemetary proof of the followig lemma ca also be foud i [2, p. 443]. Lemma 1. The umber cos π, 2, is ratioal if ad oly if =2or =3. Proof. Suppose that cos π is ratioal. Put ζ 2 =cos 2π 2π + i si 2 2, Publicatio Date: March 29, Commuicatig Editor: J. Chris Fisher.

2 48 B. De Bruy the ζ 2 is a zero of the polyomial X 2 (2 cos π ) X +1 Q[X]. So, the miimal polyomial of ζ 2 over Q is of the first or secod degree. O the other had, we kow that the miimal polyomial of ζ 2 over Q is the 2-th cyclotomic polyomial Φ 2 (x), see [4, Theorem 4.17]. The degree of Φ 2 (x) is φ(2), where φ is the Euler phi fuctio. Wehaveφ(2) =2 p1 1 p 1 p2 1 p 2... pk 1 p k, where p 1,...,p k are the differet prime umbers dividig 2. From φ(2) {1, 2}, it easily follows {2, 3}. Obviously, cos π 2 ad cos π 3 are ratioal. Lemma 2. For every N \{0}, there exist polyomials f (x),g 1 (x) Q[x] such that (i) deg(f ) =, f (x) = 2 1 x + ad cos(x) = f (cos x) for every x R; (ii) deg(g 1 )= 1, g 1 (x) =2 1 x 1 + ad si(x) si x = g 1 (cos x) for every x R \{kπ k Z}. Proof. From cos x =cosx, si x si x =1, cos(k +1)x =cos(kx)cosx si(kx) si x (1 cos2 x), si(k +1)x = si(kx) cos x +cos(kx) si x si x for k 1, it follows that we should make the followig choices for the polyomials: f 1 (x) :=x, g 0 (x) :=1; f k+1 (x) :=f k (x) x g k 1 (x) (1 x 2 ) for every k 1; g k (x) :=g k 1 (x) x + f k (x) for every k 1. Oe easily verifies by iductio that f ad g 1 ( 1) have the claimed properties. Lemma 3. Let N \{0}, q Q + \{0} ad x 1,...,x R. If cos x 1, q si x 1,..., cos x, q si x are ratioal, the so are cos(x x ) ad q si(x x ). Proof. This follows by iductio from the followig equatios (k 1). cos(x x k+1 )=cos(x x k ) cos(x k+1 ) 1 q ( q si(x x k )) ( q si(x k+1 )) ; q si(x1 + + x k+1 )=( q si(x x k )) cos(x k+1 ) +cos(x x k ) ( q si(x k+1 )). Lemma 4. Let be a triagle with vertices A, B ad C. Put a = BC, b = AC, c = AB, α = BAC, β = ÂBC ad γ = BCA. Let N \{0} ad suppose that cos( α ), cos( β ) ad cos( γ ) are ratioal. The the followig are equivalet:

3 O a problem regardig the -sectors of a triagle 49 (i) b a ad c a are ratioal umbers. (ii) si β si α ad si γ si α are ratioal umbers. Proof. We have b a = si β si α = si β si β si β si β By Lemma 2, si α si α ratioal. A similar remark holds for the fractio c a. 3. Necessary ad sufficiet coditios si α si α si β si α. Q+ \{0}. So, b a is ratioal if ad oly if si β si α Theorem 5. Let 2 ad 0 < α,β,γ < π with α + β + γ = π. There exists a triagle with agles α, β ad γ all whose edges ad -sectors have ratioal legths if ad oly if the followig coditios hold: (1) cos π Q, (2) cot π 2 ta α 2 Q, (3) cot π 2 ta β 2 Q. Proof. (a) Let be a triagle with the property that all edges ad all -sectors have ratioal legths. Let A, B ad C be the vertices of. Put α = BAC, β = ÂBC ad γ = ÂCB. Let A 0,...,A be the vertices o the edge BC such that A 0 = B, A = C ad A i 1 AA i = α for all i {1,...,}. Put a i = A i 1 A i for every i {1,...,}. For every i {1,..., 1}, the lie AA i is a bisector of the a triagle with vertices A i 1, A ad A i+1. Hece, i a i+1 = AA i 1 AA i+1 Q. Together with a a = BC Q, it follows that a i Q for every i {1,...,}. The cosie rule i the triagle with vertices A, A 0 ad A 1 gives cos α = AA2 0 + AA 1 2 a 2 1 Q. 2 AA 0 AA 1 I a similar way oe shows that cos β, cos γ Q. Put q := (1 cos2 α ) 1.By Lemma 4, q si α, q si β ad q si γ are ratioal. From Lemma 3, it follows that cos π Q ad q si π Q. Hece, cot π 2 ta α 2 = 1+cosπ q si π Similarly, cot π 2 ta β π 2 Q ad cot 2 ta γ 2 Q. q si α 1+cos α Q. (b) Coversely, suppose that cos π Q, cot π 2 ta α 2 Q ad cot π 2 ta β 2 Q. Put q := si 2 π =1 cos2 π Q. From q cot π 2 = q 1+cos π si π Q, it follows that q ta α 2 Q, q ta β 2 Q, cos α = 1 ta2 α 2 1+ta 2 α Q, 2 cos β Q, q si α = 2 q ta α 2 1+ta 2 α Q, q si β Q. By Lemma 3, also cos γ, 2 q si γ Q. Now, choose a triagle with agles α, β ad γ such that the edge is

4 50 B. De Bruy opposite the agle α has ratioal legth. By Lemma 4, it the follows that also the edges opposite to β ad γ have ratioal legths. Let A, B ad C be the vertices of such that BAC = α, ÂBC = β ad ÂCB = γ. As before, let A 0,...,A be vertices o the edge BC such that the +1lies AA i, i {0,...,}, divide the agle α i equal parts. By the sie rule, AA i = AB si β si( iα + β). Now, si( iα + β) si β = si iα si α q si α q si β si β si β cos β +cosiα. By Lemma 2, this umber is ratioal. Hece AA i Q. By a similar reasoig it follows that the legths of all other -sectors are ratioal as well. By Lemma 1 ad Theorem 5 (1), we kow that the problem ca oly have a solutio i the case of bisectors or trisectors. 4. The case of bisectors The bisector case has already bee solved completely, see e.g. [1] or [5]. Here we preset a complete solutio based o Theorem 5. Without loss of geerality, we may suppose that α β γ. These coditios are equivalet with 0 <α π 3, (1) α β π 2 α 2. (2) By Theorem 5, q α := ta α 4 ad q β := ta β 4 are ratioal. Equatio (1) implies 0 <q α ta π 12 ad equatio (2) implies q α q β x, where x := ta( π 8 α 8 ). 2x Now, =ta( π 1 x 2 4 α 1 qα 4 )= 1+q α ad hece x = we have the followig restrictios for q α Q ad q β Q: 2+2q 2 α 1 q α 1 q α. Summarizig, 0 <q α ta π 12, q α q β 2+2q 2 α 1 q α 1 q α. I Figure 1 we depict the area G correspodig with these iequalities. Every poit i G with ratioal coordiates i G will give rise to a triagle all whose edges ad bisectors have ratioal legths. Two differet poits i G with ratioal coefficiets correspod with osimilar triagles.

5 O a problem regardig the -sectors of a triagle 51 q β (0, ta π 8 ) G (ta π 12, ta π 12 ) q α Figure 1 5. The case of trisectors A ifiite but icomplete class of solutios for the trisector case did also occur i the solutio booklet of a mathematical competitio i the Netherlads (uiversitaire wiskude competitie, 1995). Here we preset a complete solutio based o Theorem 5. Agai we may assume that α β γ; so, equatios (1) ad (2) remai valid here. By Theorem 5, q α := 3 ta α 6 ad q β := 3 ta β 6 are ratioal. As before, oe ca calculate the iequalities that eed to be satisfied by q α Q ad q β Q: 0 <q α 3 ta π 18, q α q β q 2 α 3 q α 1 q α. q β (0, ta π 8 ) G ( 3 ta π 18, 3 ta π 18 ) q α Figure 2 I Figure 2 we depict the area G correspodig with these iequalities. Every poit i G with ratioal coordiates will give rise to a triagle all whose edges ad

6 52 B. De Bruy trisectors have ratioal legths. Two differet poits i G with ratioal coefficiets correspod with osimilar triagles. Refereces [1] W. E. Buker ad E. P. Starke. Problem E418, Amer. Math. Mothly, 47 (1940) 240; solutio, 48 (1941) [2] H. S. M. Coxeter. Itroductio to Geometry. 2d editio, Joh Wiley & Sos, New York, [3] R. K. Guy. Usolved problems i umber theory. Problem books i Mathematics. Spriger Verlag, New York, [4] N. Jacobso. Basic Algebra I. Freema, New York, [5] D. L. Mackay ad E. P. Starke. Problem E331, Amer. Math. Mothly, 45 (1938) 249; solutio, 46 (1939) 172. Bart De Bruy: Departmet of Pure Mathematics ad Computer Algebra, Ghet Uiversity, Galglaa 2, B-9000 Get, Belgium address: bdb@cage.uget.be

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