The Orthic-of-Intouch and Intouch-of-Orthic Triangles

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1 Forum Geometricorum Volume 6 ( FRUM GEM SSN The rthic-of-ntouch and ntouch-of-rthic Triangles Sándor Kiss bstract arycentric coordinates are used to prove that the othic of intouch and intouch of orthic triangles are homothetic ndeed, both triangles are homothetic to the reference triangle Ratios and centers of homothety are found, and certain collinearities are proved 1 ntroduction We consider a pair of triangles associated with a given triangle the orthic triangle of the intouch triangle, and the intouch triangle of the orthic triangle See Figure 1 lark Kimberling [1, p 74] asks if these two triangles are homothetic We shall show that this is true if the given triangle is acute, and indeed each of them is homothetic to the reference triangle n this paper, we adopt standard notations of triangle geometry, and denote the side lengths of triangle by a, b, c Let denote the incenter, and the incircle (with inradius r touching the sidelines,, at D, E, F respectively, so that DEF is the intouch triangle of Let be the orthocenter of, and let D =, =, =, so that D is the orthic triangle of We shall also denote by the circumcenter of and R the circumradius n this paper we make use of homogeneous barycentric coordinates ere are the coordinates of some basic triangle centers in the notations introduced by John onway =(a b c, = ( 1 S 1 S 1 S =(S S S, =(a S b S c S =(S (S + S S (S + S S (S + S, where and S = b + c a, S = c + a b, S = a + b c, S = S S, S = S S, S = S S Publication Date May 1, 006 ommunicating Editor Paul Yiu The author thanks Paul Yiu for his help in the preparation of this paper

2 17 S Kiss Two pairs of homothetic triangles 1 Perspectivity of a cevian triangle and an anticevian triangle Let P and Q be arbitrary points not on any of the sidelines of triangle t is well known that the cevian triangle of P =(u v w is perspective with the anticevian triangle of Q =(x y z at P/Q = ( x ( x u + y v + z w See, for example, [3, 83] ( x y u y v + z ( x z w u + y v z w The intouch and the excentral triangles The intouch and the excentral triangles are homothetic since their corresponding sides are perpendicular to the respective angle bisectors of triangle The homothetic center is the point P 1 =(a( a(s a+b(s b+c(s c b(a(s a b(s b+c( c(a(s a+b(s b c( =(a(s b( b((s a c(s a(s b ( a = s a b s b c This is the triangle center X 57 in [] b c F P 1 E D a Figure 1

3 The orthic-of-intouch and intouch-of-orthic triangles The orthic and the tangential triangle The orthic triangle and the tangential triangle are also homothetic since their corresponding sides are perpendicular to the respective circumradii of triangle The homothetic center is the point P = ( a S + b S + c S b ( b S + c S + a S c ( c S + a S + b S =(a S b S c S = S b S c S This is the triangle center X 5 in [] P D P D Figure Figure The ratio of homothety is positive or negative according as is acute-angled and obtuse-angled 1 See Figures and When is acute-angled, D, and are the angle bisectors of the orthic triangle, and is the incenter of the orthic triangle f is obtuse-angled, the incenter of the orthic triangle is the obtuse angle vertex 3 The orthic-of-intouch triangle The orthic-of-intouch triangle of is the orthic triangle UV W of the intouch triangle DEF Let h 1 be the homothety with center P 1, swapping D, E, F into U, V, W respectively onsider an altitude DU of DEF This is the image of the altitude a of the excentral triangle under the homothety h 1 n particular, U = h 1 ( See Figure 3 Similarly, the same homothety maps and 1 This ratio of homothety is cos cos cos

4 174 S Kiss into V and W respectively t follows that UVW is the image of under the homothety h 1 Since the circumcircle of UVW is the nine-point circle of DEF, it has radius r t follows that the ratio of homothety is r R b c U E F P 1 V W D a Figure 3 Proposition 1 The vertices of the orthic-of-intouch triangle are ( b + c U =((b + c(s b( b((s a c(s a(s b = s a b s b V =(a(s b( (c + a((s a c(s a(s b = s a c + a s b c W =(a(s b( b((s a (a + b(s a(s b = s a b s b a + b c, Proof The intouch triangle DEF has vertices D =(0s c s b, E =(s c 0s a, F =(s b s a 0 The sidelines of the intouch triangle have equations EF (s ax +(s by +(s cz =0, FD (s ax (s by +(s cz =0, DE (s ax +(s by (z =0,

5 The orthic-of-intouch and intouch-of-orthic triangles 175 The point U is the intersection of the lines P 1 and EF See Figure 3 The line P 1 has equation c(s by + b(z =0 Solving this with that of EF, we obtain the coordinates of U given above Those of V and W are computed similarly orollary The equations of the sidelines of the orthic-of-intouch triangle are VW s(s ax +(s b(y +(s b(z =0, WU ((s ax s(s by +(s c(s az =0, UV (s a(s bx +(s a(s by s(z =0 4 The intouch-of-orthic triangle Suppose triangle is acute-angled, so that its orthic triangle D has incenter, and is the image of the tangential triangle under a homothety h with center P onsider the intouch triangle XY Z of D Under the homothety h, the segment is swapped into D X See Figure 4 n particular, h ( =X For the same reason, h ( =Y and h ( =Z Therefore, the intouch-of-orthic triangle XY Z is homothetic to under h X P Z D Y Figure 4 Proposition 3 f is acute angled, the vertices of the intouch-of-orthic triangle are ( b X =((b + c S b S c + c b c S =, S S S Y =(a S (c + a S c S = c + a c, S S S Z =(a S b S (a + b b S = a + b S S S

6 176 S Kiss Proof The orthic triangle D has vertices D =(0S S, =(S 0S, =(S S 0 The sidelines of the orthic triangle have equations S x + S y + S z =0, D S x S y + S z =0, D S x + S y S z =0 The point X is the intersection of the lines P and See Figure 4 The line P has equation c S y + b S z =0 Solving this with that of, we obtain the coordinates of U given above Those of Y and Z are computed similarly orollary 4 f is acute-angled, the equations of the sidelines of the intouchof-orthic triangle are YZ S (S + S + S x + S y + S z =0, ZX S x S (S + S + S y + S z =0, UV S x + S y S (S + S + S z =0 5 omothety of the intouch-of-orthic and orthic-of-intouch triangles Proposition 5 f triangle is acute angled, then its intouch-of-orthic and orthic-of-intouch triangles are homothetic at the point (a(b + c (b + c Q = b(b(c + a (c + a c(c(a + b (a + b (s as (s bs (S Proof The homothetic center is the intersection of the lines UX, VY, and WZ See Figure 5 Making use of the coordinates given in Propositions 1 and 3, we obtain the equations of these lines as follows UX bc(s as (c(s b(s bs x +c(s bs ((b + c (s as (b + cc(s y +b(s (b(b + c(s bs (b + c (s as z =0, VY c(s as (c(c + a(s (c + a (s bs x +ca(s bs (a(s as c(s y +a(s ((c + a (s bs (c + aa(s as z =0, WZ b(s as ((a + b (S (a + bb(s bs x +a(s bs (a(a + b(s as (a + b (S y +ab(s (b(s bs a(s as z =0 t is routine to verify that Q lies on each of these lines Remark Q is the triangle center X 1876 in []

7 The orthic-of-intouch and intouch-of-orthic triangles 177 Q F Z X U Y E W V D D Figure 5 6 ollinearities ecause the circumcenter of XY Z is the orthocenter of, the center of homothety P of and XY Z lies on the Euler line of See Figure 4 We demonstrate a similar property for the point P 1, namely, that this point lies on the Euler line F of DEF, where F is the circumcenter of UVW learly,, F, P 1 are collinear Therefore, it suffices to prove that the points,, P 1 are collinear This follows from cos cos cos (s b( ((s a (s a(s b =0, which is quite easy to check See Figure 1 References [1] Kimberling, Triangle centers and central triangles, ongressus Numerantium, 19 ( [] Kimberling, Encyclopedia of Triangle enters, available at http//facultyevansvilleedu/ck6/encyclopedia/ethtml [3] P Yiu, ntroduction to the Geometry of the Triangle, Florida tlantic University lecture notes, 001 Sándor Kiss Satu Mare, str vram ancu, nr 58, sc, ap 8, Romania address kissandor@clicknetro

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