On the Inradius of a Tangential Quadrilateral

Transcription

1 Foum Geometicoum Volume 10 (2010) FORUM GEOM ISSN On the Inadius of a Tangential Quadilateal Matin Josefsson bstact. We give a suvey of known fomulas fo the inadius of a tangential quadilateal, and deive the possibly new fomula (M uvx)(m vxy)(m xyu)(m yuv) =2 uvxy(uv + xy)(ux + vy)(uy + vx) whee u, v, x and y ae the distances fom the incente to the vetices, and M = 1 (uvx + vxy + xyu + yuv) Intoduction tangential quadilateal 1 is a convex quadilateal with an incicle, that is, a cicle which is tangent to all fou sides. Not all quadilateals ae tangential. The most useful chaacteization is that its two pais of opposite sides have equal sums, a + c = b + d, whee a, b, c and d ae the sides in that ode [1, pp ]. 2 h h g g e f e Figue 1. f Klamkin s poblem It is well known that the inadius of the incicle is given by = K s whee K is the aea of the quadilateal and s is the semipeimete 3. The aea of a tangential quadilateal with sides a, b, c and d is accoding to P. Yiu [10] Publication ate: Mach 22, ommunicating Edito: Paul Yiu. 1 Othe names fo these quadilateals ae cicumsciptible quadilateal [10], cicumscibable quadilateal [9] and cicumscibed quadilateal [4]. 2 Thee exists a lot of othe inteesting chaacteizations, see [9] and [7]. 3 This fomula holds fo all polygons with an incicle, whee K in the aea of the polygon.

2 28 M. Josefsson given by 4 K = abcd sin +. 2 Fom the fomulas fo the adius and aea we conclude that the inadius of a tangential quadilateal is not detemined by the sides alone; thee must be at least one angle given, then the opposite angle can be calculated by tigonomety. nothe fomula fo the inadius is efg + fgh+ ghe + hef = e + f + g + h whee e, f, g and h ae the distances fom the fou vetices to the points whee the incicle is tangent to the sides (see Figue 1). This is inteesting, since hee the adius is only a function of fou distances and no angles! The poblem of deiving this fomula was a quickie by M. S. Klamkin, with a solution given in [5]. h 4 h g 3 g e f 1 2 e f Figue 2. Minkus 5 cicles If thee ae fou cicles with adii 1, 2, 3 and 4 inscibed in a tangential quadilateal in such a way, that each of them is tangent to two of the sides and the incicle (see Figue 2), then the adius of the incicle is a oot of the quadatic equation 2 ( ) =0 accoding to J. Minkus in [8, editoial comment]. 5 In [2, p.83] thee ae othe fomulas fo the inadius, whose deivation was only a pat of the solution of a contest poblem fom hina. If the incicle in a tangential quadilateal is tangent to the sides at points W, X, Y and Z, and if E, 4 long synthetic poof can be found in [4]. nothe way of deiving the fomula is to use the fomula K = (s a)(s b)(s c)(s d) abcd cos 2 + fo the aea of a geneal quadilateal, deived in [10, pp ], and the chaacteization a + c = b + d. 2 5 The coesponding poblem fo the tiangle is an old Sangaku poblem, solved in [8], [3, pp. 30, ].

3 On the inadius of a tangential quadilateal 29 F, G and H ae the midpoints of ZW, WX, XY and YZ espectively, then the inadius is given by the fomulas = I IE = I IF = I IG = I IH whee I is the incente (see Figue 3). The deivation is easy. Tiangles IW and IEW ae simila, so I = IE which gives the fist fomula and the othes follow by symmety. Y Z H I G X E F Figue 3. W The poblem fom hina The main pupose of this pape is to deive yet anothe (pehaps new) fomula fo the inadius of a tangential quadilateal. This fomula is also a function of only fou distances, which ae fom the incente I to the fou vetices,, and (see Figue 4). y x I u v Figue 4. The main poblem Theoem 1. If u, v, x and y ae the distances fom the incente to the vetices of a tangential quadilateal, then the inadius is given by the fomula (M uvx)(m vxy)(m xyu)(m yuv) =2 (1) uvxy(uv + xy)(ux + vy)(uy + vx)

4 30 M. Josefsson whee uvx + vxy + xyu + yuv M =. 2 Remak. It is notewothy that fomula (1) is somewhat simila to Paameshvaas fomula fo the cicumadius R of a cyclic quadilateal, 6 R = 1 (ab + cd)(ac + bd)(ad + bc) 4 (s a)(s b)(s c)(s d) whee s is the semipeimete, which is deived in [6, p.17]. 2. Peliminay esults about tiangles The poof of fomula (1) uses two equations that holds fo all tiangles. These ae two cubic equations, and one of them is a sot of coespondence to fomula (1). The fact is, that while it is possible to give as a function of the distances I, I, I and I in a tangential quadilateal, the same poblem of giving as a function of I, I and I in a tiangle is not so easy to solve, since it gives a cubic equation. The second cubic equation is found when solving the poblem, in a tiangle, of finding an exadius as a function of the distances fom the coesponding excente to the vetices. Lemma 2. If x, y and z ae the distances fom the incente to the vetices of a tiangle, then the inadius is a oot of the cubic equation 2xyz 3 +(x 2 y 2 + y 2 z 2 + z 2 x 2 ) 2 x 2 y 2 z 2 =0. (2) z γ x I α β y Figue 5. The incicle 6 quadilateal with a cicumcicle.

5 On the inadius of a tangential quadilateal 31 Poof. If α, β and γ ae the angles between these distances and the inadius (see Figue 5), we have α + β + γ = π, socos (α + β) =cos(π γ) and it follows that cos α cos β sin α sin β = cos γ. Using the fomulas cos α = x, cos β = y and sin 2 α +cos 2 α =1, we get x y 1 2 x y 2 = z o 2 xy + (x z = 2 2 )(y 2 2 ). xy Multiplying both sides with xyz, educing common factos and squaing, we get (z 2 + xy) 2 = z 2 (x 2 2 )(y 2 2 ) which afte expansion and simplification educes to (2). Lemma 3. If u, v and z ae the distances fom an excente to the vetices of a tiangle, then the coesponding exadius c is a oot of the cubic equation 2uvz 3 c (u 2 v 2 + v 2 z 2 + z 2 u 2 ) 2 c + u 2 v 2 z 2 =0. (3) γ γ α α z u v β β c I c x y E Figue 6. Excicle to tiangle and incicle to E Poof. efine angles α, β and γ to be between u, v, z and the sides of the tiangle o thei extensions (see Figue 6). Then 2α + = π, 2β + = π and 2γ =. Fom the sum of angles in a tiangle, + + = π, this simplifies to α+β = π 2 +γ. Hence cos (α + β) =cos(π 2 + γ) and it follows that cos α cos β

6 32 M. Josefsson sin α sin β = sin γ. Fo the exadius c,wehavesin α = c c u, sin β = v, sin γ = c z, and so 1 2 c u c v 2 c u c v = c z. This can, in the same way as in the poof of Lemma 2, be ewitten as z 2 (u 2 c 2 )(v 2 c 2 )=(zc 2 uv c ) 2 which afte expansion and simplification educes to (3). 3. Poof of the theoem Given a tangential quadilateal E whee the distances fom the incente to the vetices ae u, v, x and y, we see that if we extend the two sides and E to meet at, then the incicle in E is both an incicle in tiangle E and an excicle to tiangle (see Figue 6). The incicle and the excicle theefoe have the same adius, and fom (2) and (3) we get that 2xyz 3 +(x 2 y 2 + y 2 z 2 + z 2 x 2 ) 2 x 2 y 2 z 2 = 0, (4) 2uvz 3 (u 2 v 2 + v 2 z 2 + z 2 u 2 ) 2 + u 2 v 2 z 2 = 0. (5) We shall use these two equations to eliminate the common vaiable z. To do so, equation (4) is multiplied by uv and equation (5) by xy, giving 2uvxyz 3 + uv(x 2 y 2 + y 2 z 2 + z 2 x 2 ) 2 uvx 2 y 2 z 2 = 0, 2uvxyz 3 xy(u 2 v 2 + v 2 z 2 + z 2 u 2 ) 2 + xyu 2 v 2 z 2 = 0. Subtacting the second of these fom the fist gives ( uv(x 2 y 2 + y 2 z 2 + z 2 x 2 )+xy(u 2 v 2 + v 2 z 2 + z 2 u 2 ) ) 2 uvx 2 y 2 z 2 xyu 2 v 2 z 2 =0 fom which it follows uvxy 2 (xy+uv)+z 2 ( (uvy 2 + uvx 2 + xyv 2 + xyu 2 ) 2 uvxy(xy + uv) ) =0. Solving fo z 2, z 2 uvxy(uv + xy) 2 = uvxy(uv + xy) 2 (uvy 2 + uvx 2 + xyv 2 + xyu 2 ). (6) Now we multiply (4) by u 2 v 2 and (5) by x 2 y 2, which gives 2u 2 v 2 xyz 3 + u 2 v 2 (x 2 y 2 + y 2 z 2 + z 2 x 2 ) 2 u 2 v 2 x 2 y 2 z 2 = 0, 2x 2 y 2 uvz 3 x 2 y 2 (u 2 v 2 + v 2 z 2 + z 2 u 2 ) 2 + u 2 v 2 x 2 y 2 z 2 = 0. dding these we get 2uvxyz(uv + xy) 3 +(u 2 v 2 y 2 + u 2 v 2 x 2 x 2 y 2 v 2 x 2 y 2 u 2 )z 2 2 =0 and since z 2 0this educes to 2uvxy(uv + xy) +(u 2 v 2 y 2 + u 2 v 2 x 2 x 2 y 2 v 2 x 2 y 2 u 2 )z =0.

7 On the inadius of a tangential quadilateal 33 Solving fo z, we get 2uvxy(uv + xy) z = u 2 v 2 y 2 + u 2 v 2 x 2 x 2 y 2 v 2 x 2 y 2 u 2. Squaing and substituting z 2 fom (6), we get the equality uvxy(uv + xy) 2 uvxy(uv + xy) 2 (uvy 2 + uvx 2 + xyv 2 + xyu 2 ) 4(uvxy(uv + xy)) 2 2 = (u 2 v 2 y 2 + u 2 v 2 x 2 x 2 y 2 v 2 x 2 y 2 u 2 ) 2, which, since uvxy(uv + xy) 2 0, ewites as 4uvxy(uv + xy)(ux(vx + uy)+vy(uy + vx)) 2 =(2uvxy(uv + xy)) 2 (u 2 v 2 y 2 + u 2 v 2 x 2 x 2 y 2 v 2 x 2 y 2 u 2 ) 2. What is left is to facto this equation. Using the basic algebaic identities a 2 b 2 = (a + b)(a b), a 2 +2ab + b 2 =(a + b) 2 and a 2 2ab + b 2 =(a b) 2 we get 4uvxy(uv + xy)(uy + vx)(ux + vy) 2 = ( 2uvxy(uv + xy)+(uvy) 2 +(uvx) 2 (xyv) 2 (xyu) 2) (2uvxy(uv + xy) (uvy) 2 (uvx) 2 +(xyv) 2 +(xyu) 2) = ( (uvy + uvx) 2 (xyv xyu) 2)( (xyv + xyu) 2 (uvy uvx) 2) =((uvy + uvx + xyv xyu)(uvy + uvx xyv + xyu)) ((xyv + xyu + uvy uvx)(xyv + xyu uvy + uvx)). (7) Now using M = 1 2 (uvx + vxy + xyu + yuv) we get uvy + uvx + xyv xyu =(uvx + vxy + xyu + yuv) 2xyu =2(M xyu) and in the same way Thus, (7) is equivalent to uvy + uvx xyv + xyu =2(M vxy), xyv + xyu + uvy uvx =2(M uvx), xyv + xyu uvy + uvx =2(M uvy). 4uvxy(uv + xy)(uy + vx)(ux + vy) 2 =2(M uxy) 2(M vxy) 2(M uvx) 2(M uvy). Hence 2 4(M uxy)(m vxy)(m uvx)(m uvy) =. uvxy(uv + xy)(uy + vx)(ux + vy) Extacting the squae oot of both sides finishes the deivation of fomula (1).

8 34 M. Josefsson Refeences [1] T. ndeescu and. Enescu, Mathematical Olympiad Teasues, ikhäuse, oston, [2] X. in and L. P. Yee (editos), Mathematical Olympiad in hina, East hina Nomal Univesity Pess, [3] H. Fukagawa and. Pedoe, Japanese Temple Geomety Poblems San Gaku, The hales abbage Reseach ente, Winnipeg, anada, [4]. Ginbeg, icumscibed quadilateals evisited, 2008, available at ginbeg/icumrev.pdf [5] M. S. Klamkin, Five Klamkin Quickies, ux Math., 23 (1997) [6] Z.. Melzak, Invitation to Geomety, John Wiley & Sons, New Yok, 1983; new edition 2008 by ove Publications. [7] N. Minculete, haacteizations of a Tangential Quadilateal, Foum Geom., 9 (2009) [8]. Soland, Poblem 11046, me. Math. Monthly, 110 (2003) 844; solution, ibid., 113 (2006) [9]. Woall, jouney with cicumscibable quadilateals, Mathematics Teache, 98 (2004) [10] P. Yiu, Euclidean Geomety Notes, 1998 p. 157, Floida tlantic Univesity Lectue Notes, available at Matin Josefsson: Västegatan 25d, Makayd, Sweden addess: matin.makayd@hotmail.com