Heronian Triangles of Class K: Congruent Incircles Cevian Perspective

Transcription

1 Foum Geometicoum Volume 5 (05) 5. FORUM GEOM ISSN Heonian Tiangles of lass K: onguent Incicles evian Pespective Fank M. Jackson and Stalislav Takhaev bstact. We elate the popeties of a cevian that divides a efeence tiangle into two sub-tiangles with conguent incicles to the system of inne and oute Soddy cicles of the same efeence tiangle. We show that if constaints ae placed on the efeence tiangle then elationships exist between the Soddy cicles, the incicle of the efeence tiangle and the conguent incicles of the subtiangles. In paticula, we show that a class of Heonian tiangles exists with inadius equal to intege multiples of thei inne and oute Soddy cicle adii.. onguent incicles cevian It has been shown by Yiu [4, pp.7 3] that if a tiangle (with sidelengths a, b, c) is divided by a cevian though into two subtiangles with conguent incicles of adius ρ, then the length of the conguent incicles cevian is s(s a), and ρ = + t b t c = a (s s(s a)), () whee s is the semipeimete and the inadius of tiangle, and t a =tan = s a, t b =tan = s b, t c =tan = s c ae the tangents of the half angles of the tiangle (see Figue ). These numbes satisfy the basic elation t a t b + t b t c + t c t a =. () I I I ρ Figue. Publication ate: Januay, 05. ommunicating Edito: Paul Yiu.

2 6 F. M. Jackson and S. Takhaev Poposition. If θ denotes angle fo the conguent incicle cevian, then cos θ = t b t c = b c t b + t c a, (3) sin θ = t b t c = ()(). (4) t b + t c a Poof. This follows fom the fomula tan θ = tc tb in [4, p.3], and the identities cos θ = t and sin θ = t whee t =tan θ +t +t. Q P I Y Figue. Now conside the tiad of mutually tangent cicles with centes at the vetices,,. These have adii s a,, espectively. Without loss of geneality we may assume b c. If the extenal common tangent of the - and - cicles on the same side of touches these cicles at P and Q espectively, then cos PYQ = (s c) (s b) (s c)+(s b) = b c a (see Figue ). It follows fom (3) that PQ is pependicula to the conguent incicle cevian. This leads to a simple ule and compass constuction of the conguent incicles cevian. Theoem. The conguent incicle cevian is the pependicula though to extenal common tangent of the - and - cicles (of the tiad of mutually tangent cicles with centes at the vetices) on the same side of as vetex.. Radii of Soddy cicles The standad configuation fo the Soddy cicles of a efeence tiangle is shown in Figue 3. It has been shown by egiades [3] that the adii of S( i ) and S ( o ) ae given by the fomulas: i = Δ 4R + +s and o = Δ 4R + s. (5)

3 Heonian tiangles of class K: conguent incicles cevian pespective 7 whee Δ is the aea of the efeence tiangle, R its cicumadius and its inadius. s a s a S S Figue 3. Hee ae two well-known identities associated with the adii of the Soddy cicles: s a =, (6) i s a + + =. (7) o If we wite K := t a + t b + t c, these can be put in the fom = K +, = K. i o Fom these, 3. Soddyian tiangles o = K + i K. (8) The case K =has been consideed by Jackson []. In this case, the oute Soddy cicle has degeneated into a staight line, and the tiangle is called Soddyian. It has the popety that if the sides ae a b c, then s a = +. y multiply though by and conveting to tangent half angles we get: t a =+ t b t c.

4 8 F. M. Jackson and S. Takhaev s a s a Figue 4. ompaing this with the adius of the conguent incicles in (), we obtain the following theoem. Theoem 3. In the tiad of mutually tangent cicles with centes at the vetices of a Soddyian tiangle, the smallest cicle is conguent to the incicles of the subtiangles divided by the conguent incicle cevian though its cente (see Figue 4). We pove anothe inteesting popety of the conguent incicles cevian tiangle of a Soddyian tiangle. Theoem 4. In a Soddyian tiangle with a b c, the conguent incicle cevian is paallel to the Soddy line (joining the incente to the Gegonne point); see Figue 5. Poof. Set up a ectangula coodinate system with as the oigin, and positive x-axis along the line, so the the coodinates of the vetices and the incente ae ( ) =(ccos, csin ), =(0, 0), =(a, 0), I =(s b, ) =,. t b The Gegonne point has homogeneous baycentic coodinates ( ) G e = s a : : =(t a : t b : t c ). Since t a + t b + t c =, this has atesian coodinates

5 Heonian tiangles of class K: conguent incicles cevian pespective 9 G e I Figue 5. G e = ( (t ta c cos + t c a a + t b + t c ) =, t ) ac sin ( ) ( ) ( ) = t a t a + t b t b + t +t c t b b + t c t a t a + t t b b +t, b ( = (t a + t b )( t b )+(t b + t c )( + t b ) ) t b ( + t b ), ta + t b +t b ( ( t = b t b (t b + t c ) + t ) ) b + t c,. t b t b + t c Let ψ be the angle between the Soddy line and the base line. tan ψ = = = = t b +t c ( ) t b t b (t b +t + t b+t c c) t b t b t b (t b + t c ) ( t b )+(t b + t c )(t b + t c ) t b (t b + t c ) ( t b ) t a(t b + t c ) t b (t b + t c ) ( t b ) ( t bt c ) = ( t a) t c t b = (t a ) t b t c.

6 0 F. M. Jackson and S. Takhaev Howeve, fom Poposition, we have tan θ = t b t c = (t a ). t b t c t b t c This shows that the Soddy line is paallel to the conguent incicles cevian. 4. Heon tiangles fom Soddy cicles Soddyian tiangles with intege sides ae always Heonian [, 4]. We shall say that a tiangle has class K if the sum of the tangents of its half angles is equal to K. Thus, Soddyian tiangles have class. Heonian tiangles of class ae constucted in []. Let K be a positive intege. n intege tiangle of class K is Heonian if and only if the tangents of its half angles ae ational. Let θ be the angle fo the conguent incicle cevian. We have t b t c = (t b + t c )cosθ. Togethe with t a + t b + t c = K and the basic elation (), we have t a = K( + cos θ)+ε K 3 cos θ 3+cos, θ t b = ( + cos θ)(k ε K 3 cos θ) 3+cos, (9) θ t c = ( cos θ)(k ε K 3 cos θ) 3+cos θ fo ε = ±. lealy, t a, t b, t c ae ational if and only if K 3 cos θ = v fo a ational numbe v, i.e., K 3 is a sum of two squaes of ational numbes. Equivalently, K 3 is a sum of squaes of two integes. Lemma 5. n intege is a sum of two squaes of ational numbes if and only if it is a sum of squaes of two integes. Poof. We need only pove the necessity pat, fo squae-fee integes. Let n = u + v fo two ational numbes. Witing u = h q and v = k q fo integes h, k, q, wehavenq = h + k fo integes h, k, q. Hee, h and k must be elatively pime, since any common diviso must be pime to q, and so its squae must divide n, contay to the assumption that n is squae-fee. Let p be a pime diviso of n. Modulo p, h + k 0. Since at least one of h and k, say, k, is nonzeo modulo p, wehave is a quadatic esidue modulo p, and p (mod4). Thus, p is a sum of two squaes of integes. This being tue fo evey pime diviso of n, the numbe n is itself a sum of two squaes of integes. Theoem 6. Let K> be a positive intege. Heonian tiangles of class K exists if and only if K 3 is a sum of two squaes of integes. The necessity pat follows fom Lemma 5 above. We shall constuct Heonian tiangles of class 4 in the next section. The constuction clealy applies to class K with K 3 equal to a sum of two squaes of integes.

7 Heonian tiangles of class K: conguent incicles cevian pespective 5. Heonian tiangles of class 4 The atio of the adii of the Soddy cicles of a tiangle is given by (8). Fo intege values of K := t a + t b + t c, this atio is an intege only when K =3, 4, 6, and is equal to 5, 3, espectively. y Theoem 6 above, thee is no Heonian tiangle of class K =3, 6. We constuct Heonian tiangles with K =4. Without loss of geneality, assume a b c. The paametes t a, t b, t c ae given in (9) with K =4. Hee, K 3= 3, and we equie cos θ and v := 3 cos θ to be ational numbes. Since 3 = 3 +, we ewite v =3 cos θ as (3 cos θ)(3 + cos θ) =(v )(v +). Since all factos involved ae ational, we assume 3 cos θ = w(v +)fo a ational numbe w. It follows that w(3 + cos θ) =v. Solving these fo cos θ and v,wehave 3 4w 3w +6w w cos θ = +w, v = +w. (0) Note that t b = t c if and only if cos θ =0. In this case, w cannot be ational. We shall assume b>c, so that θ is an acute angle, and 0 < cos θ<. Fo this, 3 <w< 3 3. Substitution of cos θ and v = 3 cos θ given in (0) into (9) (with K =4), we obtain, fo ε =, t a = 3w +w + w +3w +3, t b = w w + w +3w +3, t c = w +w w +3w +3, () and, fo ε =, t a = w w +3 3w 3w +, t b = w w + 3w 3w +, t c = w +w 3w 3w +. () ( 3 ) In the latte case, t a cannot be geate than both t b and t c fo w, 3 3. Theefoe, Heonian tiangles of class 4 ae given by (). Witing w = m n fo elatively pime integes m and n, and using s a : : = t a : t b : t c, we may take s a = (m +mn n )( m mn +n ), = (m +mn n )(3m +mn +n ), = ( m mn +n )(3m +mn +n ). This gives a = (m + n )(3m +mn +n ), b = ( m mn +n )(5m +4mn +0n ), c = (m +mn n )(m +0mn +3n ). Fo integes m, n 5 giving w in the ange, we obtain pimitive Heonian tiangles of class 4 by dividing a, b, c by thei geatest common denominato, as pesented in the table below. n example is shown in Figue 6.

8 F. M. Jackson and S. Takhaev m n a b c s Δ R S o S i θ Figue 6. Refeences [] N. egiades, The Soddy cicles, Foum Geom., 7 (007) [] F. M. Jackson, Soddyian tiangles, Foum Geom., 3 (03) 6. [3]. Kimbeling, Encyclopedia of Tiangle entes, available at [4] P. Yiu, Notes on Euclidean Geomety, Floida tlantic Univesity, 998; available at Fank M. Jackson: ldebaan, Mixbuy, Nothamptonshie NN3 5RR United Kingdom addess: fjackson@matix-logic.co.uk Stalislav Takhaev: Pybezhnay Steet, 4 KV.33, 9076 Saint-Petesbug, Russia addess: stalislavt@mail.u