FORUM GEOMETRICORUM. A Journal on Classical Euclidean Geometry and Related Areas published by

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1 FORUM GEOMETRICORUM A Journl on Clssicl Eucliden Geometry nd Relted Ares published by Deprtment of Mthemticl Sciences Florid Atlntic University FORUM GEOM Volume ISSN

2 Editoril Bord Advisors: John H. Conwy Princeton, New Jersey, USA Julio Gonzlez Cbillon Montevideo, Uruguy Richrd Guy Clgry, Albert, Cnd Clrk Kimberling Evnsville, Indin, USA Kee Yuen Lm Vncouver, British Columbi, Cnd Tsit Yuen Lm Berkeley, Cliforni, USA Fred Richmn Boc Rton, Florid, USA Editor-in-chief: Pul Yiu Editors: Clyton Dodge Rolnd Eddy Jen-Pierre Ehrmnn Lwrence Evns Chris Fisher Rudolf Fritsch Bernrd Gibert Antres P. Htzipolkis Michel Lmbrou Floor vn Lmoen Fred Pui Fi Leung Dniel B. Shpiro Steve Sigur Mn Keung Siu Peter Woo Technicl Editors: Yundn Lin Aron Meyerowitz Xio-Dong Zhng Consultnts: Frederick Hoffmn Stephen Locke Heinrich Niederhusen Boc Rton, Florid, USA Orono, Mine, USA St. John s, Newfoundlnd, Cnd Pris, Frnce L Grnge, Illinois, USA Regin, Ssktchewn, Cnd Munich, Germny St Etiene, Frnce Athens, Greece Crete, Greece Goes, Netherlnds Singpore, Singpore Columbus, Ohio, USA Atlnt, Georgi, USA Hong Kong, Chin L Mird, Cliforni, USA Boc Rton, Florid, USA Boc Rton, Florid, USA Boc Rton, Florid, USA Boc Rton, Floird, USA Boc Rton, Florid, USA Boc Rton, Florid, USA

3 Tble of Contents Nikolos Dergides nd Pul Yiu, Antiprllels nd concurrency of Euler lines, 1 Kurt Hofstetter, Another 5-step division of segment in the golden section, 21 Mirko Rdić, Extremes res of tringles in Poncelet s closure theorem, 23 Hiroshi Okumur nd Msyuki Wtnbe, The Archimeden circles of Schoch nd Woo, 27 Jen-Pierre Ehrmnn, Steiner s theorems on the complete qudrilterl, 35 Atul Dixit nd Drij Grinberg, Orthopoles nd the Pppus theorem, 53 Jun Crlos Slzr, On the res of the intouch nd extouch tringles, 61 Nikolos Dergides, Signed distnces nd the Erdős-Mordell inequlity, 67 Eric Dnneels, A simple construction of the congruent isoscelizers point, 69 K. R. S. Sstry, Tringles with specil isotomic conjugte pirs, 73 Lev Emelynov, On the intercepts of the OI-line, 81 Chrles Ths, On the Schiffler point, 85 Zvonko Čerin, The vertex-midpoint-centroid tringles, 97 Nicole Anghel, Miniml chords in ngulr regions, 111 Li C. Tien, Three pirs of congruent circles in circle, 117 Eric Dnneels, The intouch tringle nd the OI-line, 125 Eric Dnneels nd Nikolos Dergides, A theorem on orthology centers, 135 Roger C. Alperin, A grnd tour of pedls of conics, 143 Minh H Nguyen nd Nikolos Dergides, Grfunkel s inequlity, 153 Pris Pmfilos, On some ctions of D 3 on the tringle, 157 Bernrd Gibert, Generlized Mndrt conics, 177 Nguyen Minh H, Another proof of Fgnno s inequlity, 199 Wlther Jnous, Further inequlities of Erdős-Mordell type, 203 Floor vn Lmoen, Inscribed squres, 207 Victor Oxmn, On the existence of tringles with given lengths of one side nd two djcent ngle bisectors, 215 Jen-Louis Ayme, A purely synthetic proof of the Droz-Frny line theorem, 219 Jen-Pierre Ehrmnn nd Floor vn Lmoen, A projective generliztion of the Droz-Frny line theorem, 225 Hiroshi Okumur nd Msyuki Wtnbe, The twin circles of Archimedes in skewed rbelos, 229 Drij Grinberg nd Alex Mykishev, A generliztion of the Kiepert hyperbol, 253 Author Index, 261

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5 Forum Geometricorum Volume 4 (2004) FORUM GEOM ISSN Antiprllels nd Concurrent Euler Lines Nikolos Dergides nd Pul Yiu Abstrct. We study the condition for concurrency of the Euler lines of the three tringles ech bounded by two sides of reference tringle nd n ntiprllel to the third side. For exmple, if the ntiprllels re concurrent t P nd the three Euler lines re concurrent t Q, then the loci of P nd Q re respectively the tngent to the Jerbek hyperbol t the Lemoine point, nd the line prllel to the Brocrd xis through the inverse of the delongchmps point in the circumcircle. We lso obtin n interesting cubic s the locus of the point P for which the three Euler lines re concurrent when the ntiprllels re constructed through the vertices of the cevin tringle of P. 1. Thébult s theorem on Euler lines We begin with the following theorem of Victor Thébult [8] on the concurrency of three Euler lines. Theorem 1 (Thébult). Let A B C be the orthic tringle of ABC. The Euler lines of the tringles AB C, BC A, CA B re concurrent t the Jerbek center. 1 A O G J B C B H G c O c C A Figure 1. Thébult s theorem on the concurrency of Euler lines We shll mke use of homogeneous brycentric coordintes. With reference to tringle ABC, the vertices of the orthic tringle re the points A =(0:S C : S B ), B =(S C :0:S A ), C =(S B : S A :0). Publiction Dte: Februry 3, Communicting Editor: Antres P. Htzipolkis. 1 Thébult [8] gve n equivlent chrcteriztion of this common point. See lso [7].

6 2 N. Dergides nd P. Yiu These re the trces of the orthocenter H =(S BC : S CA : S AB ). The centroid of AB C is the point (S AA +2S AB +2S AC +3S BC : S A (S C + S A ):S A (S A + S B )). The circumcenter of A BC, being the midpoint of AH, hs coordintes (S CA + S AB +2S BC : S AC : S AB ). It is strightforwrd to verify tht these two points lie on the line S AA(S BS C)(x+y+z) =(S A+S B)(S AB+S BC2S CA)y(S C+S A)(S BC+S CA2S AB)z, (1) which is therefore the Euler line of tringle AB C. Furthermore, the line (1) lso contins the point J =(S A (S B S C ) 2 : S B (S C S A ) 2 : S C (S A S B ) 2 ), which is the center of the Jerbek hyperbol. 2 Similr resoning gives the equtions of the Euler lines of tringles BC A nd A B C, nd shows tht these contin the sme point J. This completes the proof of Thébult s theorem. 2. Tringles intercepted by ntiprllels Since the sides of the orthic tringles re ntiprllel to the respective sides of tringle ABC, we consider the more generl sitution when the residuls of the orthic tringle re replced by tringles intercepted by lines l 1, l 2, l 3 ntiprllel to the sidelines of the reference tringle, with the following intercepts on the sidelines BC CA AB l 1 B C l 2 A b C b l 3 A c B c These lines re prllel to the sidelines of the orthic tringle A B C. We shll ssume tht they re the imges of the lines B C, C A, A B under the homotheties h(a, 1 t 1 ), h(b,1 t 2 ), nd h(c, 1 t 3 ) respectively. The points B, C etc. hve homogeneous brycentric coordintes B =(t 1 S A + S C :0:(1t 1 )S A ), C =(t 1 S A + S B :(1t 1 )S A :0), C b = ((1 t 2 )S B : t 2 S B + S A :0), A b =(0:t 2 S B + S C :(1t 2 )S B ), A c =(0:(1t 3 )S C : t 3 S C + S B ), B c = ((1 t 3 )S C :0:t 3 S C + S A ). 2 The point J ppers s X125 in [4].

7 Antiprllels nd concurrent Euler lines 3 A B 1 B C C 1 C b B K H B c C A b A c A 1 Figure 2. Tringles intercepted by ntiprllels 2.1. The Euler lines L i, i =1, 2, 3. Denote by T 1 the tringle AB C intercepted by l 1 ; similrly T 2 nd T 3. These re oppositely similr to ABC. We shll study the condition of the concurrency of their Euler lines. Proposition 2. With reference to tringle ABC, the brycentric equtions of the Euler lines of T i, i =1,2,3,re (1 t 1)S AA(S B S C)(x + y + z) =c 2 (S AB + S BC 2S CA)y b 2 (S BC + S CA 2S AB)z, (1 t 2)S BB(S C S A)(x + y + z) = 2 (S BC + S CA 2S AB)z c 2 (S CA + S AB 2S BC)x, (1 t 3)S CC(S A S B)(x + y + z) =b 2 (S CA + S AB 2S BC)x 2 (S AB + S BC 2S CA)y. Proof. It is enough to estblish the eqution of the Euler line L 1 of T 1. This is the imge of the Euler line L 1 of tringle AB C under the homothety h(a, 1 t 1 ).A point (x : y : z) on L 1 corresponds to the point ((1 t 1 )x t 1 (y + z) :y : z) on L 1. The eqution of L 1 cn now by obtined from (1). From the equtions of these Euler lines, we esily obtin the condition for their concurrency. Theorem 3. The three Euler lines L i, i =1, 2, 3, re concurrent if nd only if t 1 2 (S B S C )S AA + t 2 b 2 (S C S A )S BB + t 3 c 2 (S A S B )S CC =0. (2) Proof. From the equtions of L i, i =1, 2, 3, given in Proposition 2, it is cler tht the condition for concurency is (1t 1 ) 2 (S B S C )S AA +(1t 2 )b 2 (S C S A )S BB +(1t 3 )c 2 (S A S B )S CC =0. This simplifies into (2) bove.

8 4 N. Dergides nd P. Yiu 2.2. Antiprllels with given common point of L i, i =1, 2, 3. We shll ssume tringle ABC sclene, i.e., its ngles re unequl nd none of them is right ngle. For such tringles, the Euler lines of the residuls of the orthic tringle nd the corresponding ltitudes intersect t finite points. Theorem 4. Given point Q in the plne of sclene tringle ABC, there is unqiue triple of ntiprllels l i, i =1, 2, 3, for which the Euler lines L i, i =1, 2, 3, re concurrent t Q. Proof. Construct the prllel through Q to the Euler line of AB C to intersect the line AH t O. The circle through A with center O intersects AC nd AB t B nd C respectively. The line B C is prllel to B C. It follows tht its Euler line is prllel to tht of AB C. This is the line O Q. Similr constructions give the other two ntiprllels with corresponding Euler lines pssing through Q. We mke useful observtion here. From the equtions of the Euler lines given in Proposition 2 bove, the intersection of ny two of them hve coordintes expressible in liner functions of t 1, t 2, t 3. It follows tht if t 1, t 2, t 3 re liner functions of prmeter t, nd the three Euler lines re concurrent, then s t vries, the common point trverses stright line. In prticulr, t 1 = t 2 = t 3 = t, the Euler lines re concurrent by Theorem 3. The locus of the intersection of the Euler lines is stright line. Since this intersection is the Jerbek center when t =0 (Thébult s theorem), nd the orthocenter when t = 1, 3 this is the line L c : S AA (S B S C )(S CA + S AB 2S BC )x =0. cyclic We give summry of some of the interesting loci of common points of Euler lines L i, i =1, 2, 3, when the lines l i, i =1, 2, 3, re subjected to some further conditions. In wht follows, T denotes the tringle bounded by the lines l i, i = 1, 2, 3. Line Construction Condition Reference L c HJ T homothetic to orthic tringle t X 25 L q Remrk below l i, i =1, 2, 3, concurrent 3.2 L t KX 74 l i re the ntiprllels 6 of Tucker hexgon L f X 5 X 184 L i intersect on Euler line 7.2 of T L r GX 110 T nd ABC perspective 8.3 Remrk. L q cn be constructed s the line prllel to the Brocrd xis through the intersection of the inverse of the delongchmps point in the circumcircle. 3 For t =1, this intersection is the point X74 on the circumcircle, the isogonl conjugte of the infinite point of the Euler line.

9 Antiprllels nd concurrent Euler lines 5 3. Concurrent ntiprllels In this section we consider the cse when the ntiprllels l 1, l 2, l 3 ll pss through point P =(u : v : w). In this cse, B =((S C + S A )u (S B S C )v :0:(S A + S B )v +(S C + S A )w), C =((S A + S B )u +(S B S C )w :(S A + S B )v +(S C + S A )w :0), C b =((S B + S C )w +(S A + S B )u :(S A + S B )v (S C S A )w :0), A b =(0:(S B + S C )v +(S C S A )u :(S B + S C )w +(S A + S B )u), A c =(0:(S C + S A )u +(S B + S C )v :(S B + S C )w (S A S B )u), B c =((S C + S A )u +(S B + S C )v :0:(S C + S A )w +(S A S B )v). For exmple, when P = K, these re the vertices of the second cosine circle. X 74 A C b Q(K) J B c B K X 184 O C B C A c A b X 110 Figure 3. Q(K) nd the second Lemoine circle Proposition 5. The Euler lines of tringles T i, i =1, 2, 3, re concurrent if nd only if P lies on the line S A (S B S C ) L p : 2 x + S B(S C S A ) b 2 y + S C(S A S B ) c 2 z =0. When P trverses L p, the intersection Q of the Euler lines trverses the line (b 2 c 2 )( 2 (S AA + S BC ) 4S ABC ) L q : 2 x =0. cyclic

10 6 N. Dergides nd P. Yiu For point P on the line L p, we denote by Q(P ) the corresponding point on L q. Proposition 6. For points P 1, P 2, P 3 on L p, Q(P 1 ), Q(P 2 ), Q(P 3 ) re points on L q stisfying Q(P 1 )Q(P 2 ):Q(P 2 )Q(P 3 )=P 1 P 2 : P 2 P The line L p. The line L p contins K nd is the tngent to the Jerbek hyperbol t K. See Figure 4. It lso contins, mong others, the following points. J X 25 X 184 K H O L p X 1495 Figure 4. The line L p (1) X 25 = ( 2 S A : b 2 S B : c2 S C ) which is on the Euler line of ABC, nd is the homothetic center of the orthic nd the tngentil tringles, 4 (2) X 184 =( 4 S A : b 4 S B : c 4 S C ) which is the homothetic center of the orthic tringle nd the medil tngentil tringle, 5 4 See lso For other interesting properties of X184, see [6], where it is nmed the procircumcenter of tringle ABC.

11 Antiprllels nd concurrent Euler lines 7 (3) X 1495 =( 2 (S CA +S AB 2S BC ): : ) which lies on the prllel to the Euler line through the ntipode of the Jerbek center on the nine-point circle The line L q. The line L q is prllel to the Brocrd xis. See Figure 5. It contins the following points. (1) Q(K) =( 2 S A (b 2 c 2 (S BB S BC +S CC )2 2 S ABC ): : ). It cn be constructed s the intersection of the lines joining K to X 74, nd J to X 110. See Figure 3 nd 6 below. The line L q cn therefore be constructed s the prllel through this point to the Brocrd xis. (2) Q(X 1495 )=( 2 S A ( 2 S 2 6S ABC ): : ), which is on the line joining O to X 184 (on L p ). X 74 X 2071 L q Q(K) Q(X 1495) X 184 K O H L p X 1495 Figure 5. The line L q The line L q intersects the Euler line of ABC t the point X 2071 =( 2 ( 2 S AAA + S AA (S BB 3S BC + S CC ) S BBCC ): : ), 6 This is the point X113.

12 8 N. Dergides nd P. Yiu which is the inverse of the de Longchmps point in the circumcircle. This corresponds to the ntiprllels through P 2071 =( 4 (( 2 S AAA + S AA (S BB 3S BC + S CC ) S BBCC ): : ) on the line L p. This point cn be constructed by simple ppliction of Theorem 4 or Proposition 6. (See lso Remrk 2 following Theorem 12) The intersection of L p nd L q. The lines L p nd L q intersect t the point M =( 2 S A (S AB + S AC + S BB 4S BC + S CC ): : ). (1) Q(M) is the point on L q with coordintes ( 2 S A (S AA (S BB + S CC )+ 2 S A (S BB 3S BC + S CC )+S BC (S B S C ) 2 ): : ). (2) The point P on L p for which Q(P )=M hs coordintes ( 2 ( 2 (2S AA S BC )+2S A (S BB 3S BC + S CC )) : : ). 4. The tringle T bounded by the ntiprllels We ssume the line l i, i =1, 2, 3, nonconcurrent so tht they bound nondegenerte tringle T = A 1 B 1 C 1. Since these lines hve equtions t 1 S A (x + y + z) =S A x+s B y+s C z, t 2 S B (x + y + z) = S A xs B y+s C z, t 3 S C (x + y + z) = S A x+s B ys C z, the vertices of T re the points A 1 =( 2 (t 2 S B + t 3 S C ):2S CA + t 2 b 2 S B + t 3 S C (S C S A ) :2S AB + t 2 S B (S B S A )+t 3 c 2 S C ), B 1 =(2S BC + t 3 S C (S C S B )+t 1 2 S A : b 2 (t 3 S C + t 1 S A ) :2S AB + t 3 c 2 S C + t 1 S A (S A S B )) C 1 =(2S BC + t 1 2 S A + t 2 S B (S B S C ):2S CA + t 1 S A (S A S C )+t 2 b 2 S B : c 2 (t 1 S A + t 2 S B )) Homothety with the orthic tringle. The tringle T = A 1 B 1 C 1 is homothetic to the orthic tringle A B C. The center of homothety is the point ( t2 S B + t 3 S C P (T) = : t 3S C + t 1 S A : t ) 1S A + t 2 S B, (3) S A S B S C nd the rtio of homothety is 1+ t 1 2 S AA + t 2 b 2 S BB + t 3 c 2 S CC. 2S ABC Proposition 7. If the Euler lines L i, i =1, 2, 3, re concurrent, the homothetic center P (T) of T nd the orthic tringle lies on the line L p.

13 Antiprllels nd concurrent Euler lines 9 Proof. If we write P (T) =(x : y : z). From (3), we obtin t 1 = xs A + ys B + zs C 2S A, t 2 = ys B + zs C + xs A 2S B, t 3 = zs C + xs A + ys B 2S C. Substitution in (2) yields the eqution of the line L p. For exmple, if t 1 = t 2 = t 3 = t, P (T) =X 25 = ( 2 S A : b 2 S B : c2 S C ). 7 If the rtio of homothety is 0, tringle T degenertes into the point X 25 on L p. The intersection of L c nd L q is the point Q(X 25 )=( 2 S A (b 4 S 4 B + c 4 S 4 C + 2 S AAA (S B S C ) 2 S ABC (4 2 S BC +3S A (S B S C ) 2 )) : : ). Remrk. The line L p is lso the locus of the centroid of T for which the Euler lines L i, i =1, 2, 3, concur Common point of L i, i =1, 2, 3, on the Brocrd xis. We consider the cse when the Euler lines L i, i =1, 2, 3, intersect on the Brocrd xis. A typicl point on the Brocrd xis, dividing the segment OK in the rtio t :1t, hs coordintes ( 2 (S A (S A + S B + S C )+(S BC S AA )t) : : ). This point lies on the Euler lines L i, i =1, 2, 3, if nd only if we choose t 1 = (S A + S B + S C )(S 2 S AA )+b 2 c 2 (S B + S C 2S A )t, 2S AA (S A + S B + S C ) t 2 = (S A + S B + S C )(S 2 S BB )+c 2 2 (S C + S A 2S B )t, 2S BB (S A + S B + S C ) t 3 = (S A + S B + S C )(S 2 S CC )+ 2 b 2 (S A + S B 2S C )t. 2S CC (S A + S B + S C ) The corresponding tringle T is homothetic to the orthic tringle t the point ( 2 ((S A + S B + S C) 2 S A + t((2s A + S B + S C)S BC + b 2 S CA + c 2 S AB) : : ), which divides the segment X 184 K in the rtio 2t :12t. The rtio of homothety is 2 b 2 c 2 4S ABC. These tringles re ll directly congruent to the medil tngentil tringle of ABC. We summrize this in the following proposition. Proposition 8. Corresponding to the fmily of tringles directly congruent to the medil tngentil tringle, homothetic to orthic tringle t points on the line L p, the common points of the Euler lines of L i, i =1, 2, 3, ll lie on the Brocrd xis. 7 See lso 3.1(1). The tngentil tringle is T with t =1.

14 10 N. Dergides nd P. Yiu 5. Perspectivity of T with ABC Proposition 9. The tringles T nd ABC re perspective if nd only if (S B S C )(t 1 S AA t 2 t 3 S BC )=0. (4) cyclic Proof. From the coordintes of the vertices of T, it is strightforwrd to check tht T nd ABC re perspective if nd only if t 1 2 S AA + t 2 b 2 S BB + t 3 c 2 S CC +2S ABC =0 or (4) holds. Since the re of tringle T is (t 1 2 S AA + t 2 b 2 S BB + t 3 c 2 S CC +2S ABC ) 2 2 b 2 c 2 S ABC times tht of tringle ABC, we ssume t 1 2 S AA +t 2 b 2 S BB +t 3 c 2 S CC +2S ABC 0 nd (4) is the necessry nd sufficient condition for perspectivity. Theorem 10. If the tringle T is nondegenerte nd is perspective to ABC, then the perspector lies on the Jerbek hyperbol of ABC. Proof. If tringles A 1 B 1 C 1 nd ABC re perspective t P =(x : y : z), then A 1 =(u + x : y : z), B 1 =(x : v + y : z), C 1 =(x : y : w + z) for some u, v, w. Since the line B 1 C 1 is prllel to B C, which hs infinite point (S B S C : (S C + S A ):S A + S B ),wehve S B S C (S C + S A ) S A + S B x y + v z x y z + w =0, nd similrly for the other two lines. These cn be rerrnged s (S C + S A )x (S B S C )y (S B S C )z +(S A + S B )x =S B S C, v w (S A + S B )y (S C S A )z (S C S A )x +(S B + S C )y =S C S A, w u (S B + S C )z (S A S B )x (S A S B )y +(S C + S A )z =S A S B. u v Multiplying these equtions respectively by S A (S B + S C )yz, S B (S C + S A )zx, S C (S A + S B )xy nd dding up, we obtin ( 1+ x u + y v + z ) S A (S BB S CC )yz =0. w cyclic Since the re of tringle T is ( uvw 1+ x u + y v + z ) w

15 Antiprllels nd concurrent Euler lines 11 times tht of tringle ABC, we must hve 1+ x u + y v + z w S A (S BB S CC )yz =0. cyclic This mens tht P lies on the Jerbek hyperbol. 0. It follows tht We shll identify the locus of the common points of Euler lines in 8.3 below. In the mentime, we give construction for the point Q from the perspector on the Jerbek hypebol. Construction. Given point P on the Jerbek hyperbol, construct prllels to A B nd A C through n rbitrry point A 1 on the line AP. Let M 1 be the intersection of the Euler lines of the tringles formed by these ntiprllels nd the sidelines of ABC. With nother point A 1 obtin point M 2 by the sme construction. Similrly, working with two points B 1 nd B 1 on BP, we construct nother line M 3 M 4. The intersection of M 1 M 2 nd M 3 M 4 is the common point Q of the Euler lines corresponding the ntiprllels tht bound tringle perspective to ABC t P. 6. The Tucker hexgons nd the line L t It is well known tht if the ntiprllels, together with the sidelines of tringle ABC, bound Tucker hexgon, the vertices lie on circle whose center is on the Brocrd xis. If this center divides the segment OK in the rtio t :1t, the ntiprllels pss through the points dividing the symmedins in the sme rtio. The vertices of the Tucker hexgon re B =(S C +(1t)c 2 :0:tc 2 ), C =(S B +(1t)b 2 : tb 2 :0), C b =(t 2 : S A +(1t) 2 :0), A b =(0:S C +(1t)c 2 : tc 2 ), A c =(0:tb 2 : S B +(1t)b 2 ), B c =(t 2 :0:S A +(1t) 2 ). In this cse, 1t 1 = t b 2 c 2 S A(S A + S B + S, C) 1t2 = t c 2 2 S B(S A + S B + S, C) 1t3 = t 2 b 2 S C(S A + S B + S. C) It is cler tht the Euler lines L i, i = 1, 2, 3, re concurrent. As t vries, this common point trverses stright line L t. We show tht this is the line joining K to Q(K). (1) For t =1, this Tucker circle is the second Lemoine circle with center K, the tringle T degenertes into the point K. The common point of the Euler lines is therefore the point Q(K). See 3.2 nd Figure 3. (2) For t = 3 2, the vertices of the Tucker hexgon re B =( 2 + b 2 2c 2 :0:3c 2 ), C =(c b 2 :3b 2 :0), C b =(3 2 : b 2 + c :0), A b =(0: 2 + b 2 2c 2 :3c 2 ), A c =(0:3b 2 : c b 2 ), B c =(3 2 :0:b 2 + c ).

16 12 N. Dergides nd P. Yiu The tringles T i, i = 1, 2, 3, hve common centroid K, which is therefore the common point of their Euler lines. The corresponding Tucker center is the point X 576 (which divides OK in the rtio 3:1). From these, we obtin the eqution of the line L t : b 2 c 2 S A (S B S C )(S CA + S AB 2S BC )x =0. cyclic Remrks. (1) The tringle T is perspective to ABC t K. See, for exmple, [5]. (2) The line L t lso contins X 74 which we my regrd s corresponding to t =0. For more bout Tucker hexgons, see Concurrency of four or more Euler lines 7.1. Common point of L i, i =1, 2, 3, on the Euler line of ABC. We consider the cse when the Euler lines L i, i =1, 2, 3, intersect on the Euler line of ABC. A typicl point on the Euler line xis divides the segment OH in the rtio t :1 t, hs coordintes ( 2 S A (S CA + S AB 2S BC )t) : : ). This lies on the Euler lines L i, i =1, 2, 3, if nd only if we choose t 1 = (S2 S AA )+(S 2 3S AA )t, 2S AA t 2 = (S2 S BB )+(S 2 3S BB )t, 2S BB t 3 = (S2 S CC )+(S 2 3S CC )t. 2S CC Independently of t, the corresponding tringle T is lwys homothetic to the medil tngentil tringle t the point P 2071 on the line L p for which Q(P 2071 )= X 2071, the intersection of L q with the Euler line. See the end of 3.2 bove. The rtio of homothety is 1+t 8S ABC 2 b 2 c t. We summrize this in the following proposition. 2 Proposition 11. Let P 2071 be the point on L p such tht Q(P 2071 )=X The Euler lines L i, i =1, 2, 3, corresponding to the sidelines of tringles homothetic t P 2071 to the medil tngentil tringle intersect on the Euler line of ABC The line L f. The Euler line of tringle T is the line (x + y + z) t 1 2 S AA (S B S C )(S 2 + S BC )(S 2 S AA ) cyclic =2S ABC (S 2 + S CA )(S 2 + S AB )x. (5) cyclic

17 Antiprllels nd concurrent Euler lines 13 Theorem 12. The Euler lines of the four tringles T nd T i, i = 1, 2, 3, re concurrent if nd only if t 1 = 16S2 S ABC + t( 2 b 4 c 4 4S ABC (3S 2 S AA )) 4S AA ( 2 b 2 c 2, +4S ABC ) t 2 = 16S2 S ABC + t( 4 b 2 c 4 4S ABC (3S 2 S BB )) 4S BB ( 2 b 2 c 2, +4S ABC ) t 3 = 16S2 S ABC + t( 4 b 4 c 2 4S ABC (3S 2 S CC )) 4S CC ( 2 b 2 c 2, +4S ABC ) 24 2 b 2 c 2 S ABC with t ( 2 b 2 c 2 8S ABC )(3(S A +S B +S C )S 2 +S ABC. The locus of the common point ) of the four Euler lines is the line L f joining the nine-point center of ABC to X 184, with the intersection with L q deleted. Proof. The eqution of the Euler line L i, i =1, 2, 3, cn be rewritten s t 1S A(S B S C)(x + y + z)+s AA(S B S C)x +(S AB(S B S C) (S AA S BB)S C)y +(S AC(S B S C)+(S AA S CC)S B)z =0, (6) t 2S A(S B S C)(x + y + z)+s BB(S C S A)y +(S BA(S C S A)+(S BB S AA)S C)x +(S BC(S C S A) (S BB S CC)S A)z =0, (7) t 3S C(S A S B)(x + y + z)+s CC(S A S B)z +(S CA(S A S B) (S CC S AA)S B)x +(S CB(S A S B)+(S CC S BB)S A)y =0. (8) Multiplying (4), (5), (6) respectively by 2 S A (S 2 +S BC )(S 2 S AA ), b 2 S B (S 2 +S CA )(S 2 S BB ), c 2 S C (S 2 +S AB )(S 2 S CC ), nd dding, we obtin by Theorem 10 the eqution of the line L f : (S B S C )(S 2 (2S AA S BC )+S ABC S A )x =0 cyclic which contins the common point of the Euler lines of T i, i =1, 2, 3, if it lso lies on the Euler line L of T. The line L f contins the nine-point center X 5 nd X 184 =( 4 S A : b 4 S B : c 4 S C ). Let Q t be the point which divides the segment X 184 X 5 in the rtio t :1t. It hs coordintes ((1 t)4s 2 4 S A + t( 2 b 2 c 2 +4S ABC )(S CA + S AB +2S BC ) :(1t)4S 2 b 4 S B + t( 2 b 2 c 2 +4S ABC )(2S CA + S AB + S BC ) :(1t)4S 2 c 4 S C + t( 2 b 2 c 2 +4S ABC )(S CA +2S AB + S BC )). The point Q t lies on the Euler lines L i, i =1, 2, 3, respectively if we choose t 1, t 2, t 3 given bove.

18 14 N. Dergides nd P. Yiu If Q lies on L q, then Q t = Q(P ) for some point P on L p. 8 In this cse, the tringle T degenertes into the point P Q nd its Euler line is not defined. It should be excluded from L f. The corresponding vlue of t is s given in the sttement bove. Here re some interesting points on L f. (1) For t =0, T is perspective with ABC t X 74, nd the common point of the four Euler lines is X 184. The ntiprllels re drwn through the intercepts ( of the triliner polrs of X 186 = 2 S A (S 2 3S AA ) : : ), the inversive imge of the orthocenter in the circumcircle. (2) For t =1, this common point is the nine-point of tringle ABC. The tringle T is homothetic to the orthic tringle t X 51 nd to the medil tngentil tringle t the point P 2071 in 3.2. (3) t = 2 b 2 c 2 4S ABC gives X 156, the nine-point center of the tngentil tringle. In these two cses, we hve the concurrency of five Euler lines. (4) The line L f intersects the Brocrd xis t X 569. This corresponds to t = 2 2 b 2 c b 2 c 2 +4S ABC. Proposition 13. The tringle T is perspective with ABC nd its Euler line contins the common point of the Euler lines of T i, i =1, 2, 3 precisely in the following three cses. (1) t =0, with perspector X 74 nd common point of Euler line X (2) t = 2 b 2 c 2 S ABC 4 b 4 c b 2 c 2 S ABC 16(S ABC ), with perspector K. 2 Remrks. (1) In the first cse, t 1 = k, t 2 = k, t 3 = k S AA S BB S CC 4S2 S ABC for k = 2 b 2 c 2 +4S ABC. The ntiprllels pss through the intercepts of the triliner polr of X 186, the inversive imge of H in the circumcircle. (2) In the second cse, the ntiprllels bound Tucker hexgon. The center of the Tucker circle divides OK in the rtio t :1t, where t = S2 (S A + S B + S C )( 2 b 2 c 2 16S ABC ) 4 b 4 c b 2 c 2 S ABC 16(S ABC ) 2. It follows tht the common point of the Euler lines is the intersection of the lines L f = X 5 X 184 nd L t. 8. Common points of L i, i =1, 2, 3, when T is perspective If the Euler lines L i, i =1, 2, 3, re concurrent, then, ccording to (2) we my put t 1 = k(λ + S A) 2, t 2 = k(λ + S B) S AA b 2, t 3 = k(λ + S C) S BB c 2 S CC 8 This point is the intersection of Lp with the line joining the Jerbek center J to X 323, the reflection in X 110 of the inversive imge of the centroid in the circumcircle.

19 Antiprllels nd concurrent Euler lines 15 for some λ nd k. If, lso, the T is perspective, (4) gives k(kλ + S ABC )(λ + S A + S B + S C )(k(3λ + S A + S B + S C )+2S ABC )=0. If k =0, T is the orthic tringle. We consider the remining three cses below The cse k(s A + S B + S C +3λ)+2S ABC =0. In this cse, t 1 = 2S ABC + k(s B + S C 2S A ) 3 2, S AA t 2 = 2S ABC + k(s C + S A 2S B ) 3b 2, S BB t 3 = 2S ABC + k(s A + S B 2S C ) 3c 2. S CC The ntiprllels re concurrent The cse kλ + S ABC =0. In this cse, t 1 = k S BC 2, t 2 = k S CA S A b 2, t 3 = k S AB S B c 2. S C In this cse, the perspector is the Lemoine point K. The ntiprllels bound Tucker hexgon. The locus of the common point of Euler lines is the line L t. Here re some more interesting points on this line. (1) For k =0,wehve S BC S CA S AB t 1 = S A (S B + S C ), t 2 = S B (S C + S A ), t 3 = S C (S A + S B ). This gives the Tucker hexgon with vertices B =(S CC :0:S 2 ), C =(S BB : S 2 :0), C b =(S 2 : S AA :0), A b =(0:S CC : S 2 ), A c =(0:S 2 : S BB ), B c =(S 2 :0:S AA ). These re the pedls of A, B, C on the sidelines. The Tucker circle is the Tylor circle. The tringle T is the medil tringle of the orthic tringle. The corresponding Euler lines intersect t X 974, which is the intersection of L t = KX 74 with X 5 X 125. See [2]. (2) For k = S ABC S A +S B +S C,wehve S BC t 1 = S A(S A + S B + S, C) t2 = S CA S B(S A + S B + S, C) t3 = S AB S C(S A + S B + S. C) The Tucker circle is the second Lemoine circle, considered in 6. (3) The line L t intersects the Euler line t ( 2 (S 2 ) +3S AA ) X 378 = : :. S A The corresponding Tucker circle hs center

20 16 N. Dergides nd P. Yiu (S 2 (S B + S C )(S C S A )(S A S B )+3(S A + S B )(S B + S C )(S C + S A )S BC : : ) which is the intersection of the Brocrd xis nd the line joining the orthocenter to X 110. X 74 A K O H X 378 B C X 110 Figure 6. Intersection of 4 Euler lines t X The cse λ = (S A + S B + S C ). In this cse, we hve t 1 = k, t 2 = k, t 3 = k. S AA S BB S CC In this cse, the perspector is the point ( ) 1 2S ABC S A k(b 2 c 2 : : 2S BC ) on the Jerbek hyperbol. If the point on the Jerbek hyperbol is the isogonl conjugte of the point which divides OH in the rtio t :1 t, then k = 4tS 2 S ABC 2 b 2 c 2 (1 + t)+4t S ABC. The locus of the intersection of the Euler lines L i, i =1, 2, 3, is clerly line. Since this intersection is the Jerbek center for k =0(Thébult s theorem) nd the

21 Antiprllels nd concurrent Euler lines 17 centroid for k = S2 3, this is the line L r : (S B S C )(S BC S AA )x =0. cyclic This line lso contins, mong other points, X 110 nd X 184. We summrize the generl sitution in the following theorem. Theorem 14. Let P be point on the Euler line other thn the centroid G. The ntiprllels through the intercepts of the triliner polr of P bound tringle perspective with ABC (t point on the Jerbek hyperbol). The Euler lines of the tringles T i, i =1, 2, 3, re concurrent (t point Q on the line L r joining the centroid G to X 110 ). Here re some interesting exmples with P esily constructed on the Euler line. P Perspector Q H H X 125 O X 64 = X20 X 110 X 30 X2071 G X 186 X 74 X 184 X 403 X 265 = X186 X 1899 X 23 X 1177 = X858 X 182 X 858 X 1352 X 1316 X 98 Remrks. (1) X 186 is the inversive imge of H in the circumcircle. (2) X 403 is the midpoint between H nd X 186. (3) X 23 is the inversive imge of G in the circumcircle. (4) X 858 is the inferior of X 23. (5) X 182 is the midpoint of OK, the center of the Brocrd circle. (6) X 1352 is the reflection of K in the nine-point center. (7) X 1316 is the intersection of the Euler line nd the Brocrd circle prt from O. 9. Two loci: line nd cubic We conclude this pper with brief discussion on two locus problems Antiprllels through the vertices of pedl tringle. Suppose the ntiprllels l i, i =1, 2, 3, re constructed through the vertices of the pedl tringle of finite point P. Then the Euler lines L i, i =1, 2, 3, re concurrent if nd only if P lies on the line S A (S B S C )(S AA S BC )x =0. cyclic This is the line contining H nd the Trry point X 98.ForP = H, the common point of the Euler line is X 185 =( 2 S A (S A (S BB + S CC )+ 2 S BC ): : ).

22 18 N. Dergides nd P. Yiu 9.2. Antiprllels through the vertices of cevin tringle. If, insted, the ntiprllels l i, i =1, 2, 3, re constructed through the vertices of the cevin tringle of P, then the locus of P for which the Euler lines L i, i =1, 2, 3, re concurrent is the cubic K : S A + S B + S C xyz + ( x SA + S B y 2 S ) C + S A z 2 =0. S ABC S A (S B S C ) S C S B cyclic This cn lso be written in the form ( (S B + S C )yz)( S A (S B S C )(S B + S C S A )x) cyclic cyclic =( S A (S B S C )x)( S A (S B + S C )yz). cyclic cyclic From this, we obtin the following points on K: the orthocenter H (s the intersection of the Euler line nd the line cyclic S A(S B S C )(S B + S C S A )x) =0), the Euler reflection point X 110 (s the fourth intersection of the circumcircle nd the circumconic cyclic S A(S B + S C )yz =0with center K), the intersections of the Euler line with the circumcircle, the points X 1113 nd X Corresponding to P = X 110, the Euler lines L i, i =1, 2, 3, intersect t the circumcenter O. On the other hnd, X 1113 nd X 1114 re the points ( 2 S A + λ(s CA + S AB 2S BC ): : ) bc for λ = nd λ = bc respectively. The ntiprllels 2 b 2 c 2 8S ABC 2 b 2 c 2 8S ABC through the trces of ech of these points correspond to t 1 = t 2 = t 3 = λ 1 λ +1. This mens tht the corresponding intersections of Euler lines lie on the line L c = HJ in The cubic K. The infinite points of the cubic K cn be found by rewriting the eqution of K in the form ( S A (S B S C )(S B + S C )yz)( (S B + S C )x) cyclic cyclic =(x + y + z)( (S B + S C )(S B S C )(S A (S A + S B + S C ) S BC )yz) cyclic They re the infinite points of the Jerbek hyperbol nd the line (S B + S C )x + (S C + S A )y +(S A + S B )z =0. The ltter is X 523 =(S B S C : S C S A : S A S B ). The symptotes of K re

23 Antiprllels nd concurrent Euler lines 19 the prllels to the symptotes of Jerbek hyperbol through the ntipode the Jerbek center on the nine-point circle, i.e., X 113 =((S CA + S AB 2S BC )(b 2 S BB + c 2 S CC 2 S AA 2S ABC ): : ), the perpendiculr to the Euler line (of ABC) t the circumcenter O, intersecting K gin t ( Z = S CA + S AB 2S BC b 2 S BB + c 2 S CC 2 S AA 2S ABC : : which lso lies on the line joining H to X 110. See Figure 7. 9 ), Jerbek hyperbol A J=X125 X1114 B X1313 O X113 N K H X1312 X1113 C Γ Z X110 Figure 7. The cubic K 9 We thnk Bernrd Gibert for providing the sketch of K in Figure 7.

24 20 N. Dergides nd P. Yiu Remrk. The symptotes of K nd the Jerbek hyperbol bound rectngle inscribed in the nine-point circle. Two of the vertices nd J = X 125 nd its ntipode X 113. The other two re the points X 1312 nd X 1313 on the Euler line. References [1] N. Dergides, Hycinthos 7777, September 4, [2] J.-P. Ehrmnn, Hycinthos 3693, September 1, [3] D. Grinberg, Hycinthos 7781, September 4, [4] C. Kimberling, Encyclopedi of Tringle Centers, vilble t [5] F. M. vn Lmoen, Concurrencies from Tucker hexgons, Forum Geom., 2 (2002) [6] A. Mykishev, On the procircumcenter nd relted points, Forum Geom., 3 (2003) [7] V. Thébult, Concerning the Euler line of tringle, Amer. Mth. Monthly, 54 (1947) [8] V. Thébult, O. J. Rmmler, nd R. Goormghtigh, Problem 4328, Amer. Mth. Monthly, 56 (1949) 39; solution, (1951) 45. Nikolos Dergides: I. Znn 27, Thessloniki 54643, Greece E-mil ddress: ndergides@yhoo.gr Pul Yiu: Deprtment of Mthemticl Sciences, Florid Atlntic University, Boc Rton, Florid, , USA E-mil ddress: yiu@fu.edu

25 Forum Geometricorum Volume 4 (2004) FORUM GEOM ISSN Another 5-step Division of Segment in the Golden Section Kurt Hofstetter Abstrct. We give one more 5-step division of segment into golden section, using ruler nd compss only. Insmuch s we hve given in [1, 2] 5-step constructions of the golden section we present here nother very simple method using ruler nd compss only. It is fscinting to discover how simple the golden section ppers. For two points P nd Q, we denote by P (Q) the circle with P s center nd P Q s rdius. C 3 D C 1 C 2 F E A G B C Figure 1 Construction. Given segment AB, construct (1) C 1 = A(B), (2) C 2 = B(A), intersecting C 1 t C nd D, (3) the line AB to intersect C 1 t E (prt from B), (4) C 3 = E(B) to intersect C 2 t F (so tht C nd F re on opposite sides of AB), (5) the segment CF to intersect AB t G. The point G divides the segment AB in the golden section. Publiction Dte: Februry 10, Communicting Editor: Pul Yiu.

26 22 K. Hofstetter Proof. Suppose AB hs unit length. It is enough to show tht AG = 1 2 ( 5 1). Extend BA to intersect C 3 t H. Let CD intersect AB t I, nd let J be the orthogonl projection of F on AB. In the right tringle HFB, BH =4, BF =1. Since BF 2 = BJ BH, BJ = 1 4. Therefore, IJ = 1 4. It lso follows tht JF = D F G H G E A I J B C Now, IG GJ = IC JF = = 2 Figure 2 5. It follows tht IG = IJ = 52 2, nd AG = IG = This shows tht G divides AB in the golden section. Remrk. If FD is extended to intersect AH t G, then G is such tht G A : AB = 1 2 ( 5+1):1. After the publiction of [2], Dick Klingens nd Mrcello Trquini hve kindly written to point out tht the sme construction hd ppered in [3, p.51] nd [4, S.37] lmost one century go. References [1] K. Hofstetter, A simple construction of the golden section, Forum Geom., 2 (2002) [2] K. Hofstetter, A 5-step division of segment in the golden section, Forum Geom., 3 (2003) [3] E. Lemoine, Géométrogrphie ou Art des Constructions Géométriques, C. Nud, Pris, [4] J. Reusch, Plnimetricsche Konstruktionen in Geometrogrphischer Ausführung, Teubner, Leipzig, Kurt Hofstetter: Object Hofstetter, Medi Art Studio, Lngegsse 42/8c, A-1080 Vienn, Austri E-mil ddress: pendel@sunpendulum.t

27 Forum Geometricorum Volume 4 (2004) FORUM GEOM ISSN Extreme Ares of Tringles in Poncelet s Closure Theorem Mirko Rdić Abstrct. Among the tringles with the sme incircle nd circumcircle, we determine the ones with mximum nd miniumum res. These re lso the ones with mximum nd minimum perimeters nd sums of ltitudes. Given two circles C 1 nd C 2 of rdii r nd R whose centers re t distnce d prt stisfying Euler s reltion R 2 d 2 =2Rr, (1) by Poncelet s closure theorem, for every point A 1 on the circle C 2, there is tringle A 1 A 2 A 3 with incircle C 1 nd circumcircle C 2. In this rticle we determine those tringles with extreme res, perimeters, nd sum of ltitudes. A t M C 1 C 2 C r I d O R t m I O C 2 C 1 B Figure 1 Figure 1b Denote by t m nd t M respectively the lengths of the shortest nd longest tngents tht cn be drwn from C 2 to C 1. These re given by t m = (R d) 2 r 2, t M = (R + d) 2 r 2. (2) We shll use the following result given in [2, Theorem 2.2]. Let t 1 be ny given length stisfying t m t 1 t M, (3) Publiction Dte: Februry 23, Communicting Editor: Pul Yiu.

28 24 M. Rdić nd let t 2 nd t 3 be given by t 2 = 2Rrt 1 + D r 2 + t 2 1, t 3 = 2Rrt 1 D r 2 + t 2, (4) 1 where D =4R 2 r 2 t 2 1 r 2 (r 2 + t 2 1)(4Rr + r 2 + t 2 1). Then there is tringle A 1 A 2 A 3 with incircle C 1 nd circumcircle C 2 with side lengths i = A i A i+1 = t i + t i+1, i =1, 2, 3. (5) Here, the indices re tken modulo 3. It is esy to check tht (t 1 + t 2 + t 3 )r 2 =t 1 t 2 t 3, t 1 t 2 + t 2 t 3 + t 3 t 1 =4Rr + r 2, nd tht these re necessry nd sufficient for C 1 nd C 2 to be the incircle nd circumcircle of tringle A 1 A 2 A 3. Denote by J(t 1 ) the re of tringle A 1 A 2 A 3. Thus, J(t 1 )=r(t 1 + t 2 + t 3 ). (6) Note tht D =0when t 1 = t m or t 1 = t M. In these cses, 2Rrt m, if t r 2 +t 2 1 = t m, m t 2 = t 3 =, if t 1 = t M. For convenience, we shll write t m = 2Rrt m r 2 + t 2 m 2Rrt M r 2 +t 2 M nd t M = 2Rrt M r 2 + t 2. (7) M Theorem 1. J(t 1 ) is mximum when t 1 = t M nd minimum when t 1 = t m. In other words, J(t m ) J(t 1 ) J(t M ) for t m t 1 t M. Proof. It follows from (6) nd (4) tht ( J(t 1 )=r t 1 + 4Rrt ) 1 r 2 + t 2. 1 From d dt 1 J(t 1 )=0, we obtin the eqution t 4 1 2(2Rr r 2 )t Rr 3 + r 4 =0, nd t 2 1 =2Rr r2 ± 2r R 2 2Rr =2Rr r 2 ± 2rd. Since 4R 2 r 2 =(R 2 d 2 ) 2,wehve

29 Extreme res of tringles in Poncelet s closure theorem 25 2Rr r 2 +2rd t m 2 =2Rr r 2 +2rd (R + d)2 ((R d) 2 r 2 ) (R d) 2 = (R d)2 (2Rr r 2 +2rd) (R + d) 2 ((R d) 2 r 2 ) (R d) 2 = ((R + d)2 (R d) 2 )r 2 +2r(R + d)(r d) 2 (R 2 d 2 ) 2 (R d) 2 = 4Rdr2 +2r(R d)(2rr) (2Rr) 2 (R d) 2 =0. Similrly, 2Rr r 2 2rd t M 2 =0. It follows tht d dt 1 J(t 1 )=0for t 1 = t m, t M. The mximum of J occurs t t 1 = t M nd t M while the minimum occurs t t 1 = t m nd t m. J(t 1) J(t M ) J(t m) t 1 t m t M t m t M Figure 2 The tringle determined by t m (respectively t M ) is exctly the one determined by t m (respectively t M ). We conclude with n interesting corollry. Let h 1, h 2, h 3 be the ltitudes of the tringle A 1 A 2 A 3. Since

30 26 M. Rdić 2R(h 1 + h 2 + h 3 )= =(t 1 + t 2 + t 3 ) 2 +4Rr + r 2, the following re equivlent: the tringle hs mximum (respectively minimum) re, the tringle hs mximum (respectively minimum) perimeter, the tringle hs mximum (respectively minimum) sum of ltitudes. It follows tht these re precisely the two tringles determined by t M nd t m. t M t M t m I O t m I O Figure 3 Figure 3b References [1] H. Dörrie, 100 Gret Problems of Elementry Mthemtics, Dover, [2] M. Rdić, Some reltions concerning tringles nd bicentric qudrilterls in connection with Poncelet s closure theorem, Mth. Mced. 1 (2003) Mirko Rdić: Deprtment of Mthemtics, Fculty of Philosophy, University of Rijek, Rijek, Omldinsk 14, Croti E-mil ddress: mrdic@pefri.hr

31 Forum Geometricorum Volume 4 (2004) FORUM GEOM ISSN The Archimeden Circles of Schoch nd Woo Hiroshi Okumur nd Msyuki Wtnbe Abstrct. We generlize the Archimeden circles in n rbelos (shoemker s knife) given by Thoms Schoch nd Peter Woo. 1. Introduction Let three semicircles α, β nd γ form n rbelos, where α nd β touch externlly t the origin O. More specificlly, α nd β hve rdii, b>0nd centers (, 0) nd (b, 0) respectively, nd re erected in the upper hlf plne y 0. The y-xis divides the rbelos into two curviliner tringles. By fmous theorem of Archimedes, the inscribed circles of these two curviliner tringles re congruent nd hve rdii r = b +b. See Figure 1. These re clled the twin circles of Archimedes. Following [2], we cll circles congruent to these twin circles Archimeden circles. β(2b) U 2 α(2) γ γ β α β α O Figure 1 O Figure 2 2r For rel number n, denote by α(n) the semicircle in the upper hlf-plne with center (n, 0), touching α t O. Similrly, let β(n) be the semicircle with center (n, 0), touching β t O. In prticulr, α() =α nd β(b) =β. T. Schoch hs found tht (1) the distnce from the intersection of α(2) nd γ to the y-xis is 2r, nd (2) the circle U 2 touching γ internlly nd ech of α(2), β(2b) externlly is Archimeden. See Figure 2. P. Woo considered the Schoch line L s through the center of U 2 prllel to the y-xis, nd showed tht for every nonnegtive rel number n, the circle U n with center on L s touching α(n) nd β(nb) externlly is lso Archimeden. See Figure 3. In this pper we give generliztion of Schoch s circle U 2 nd Woo s circles U n. Publiction Dte: Mrch 3, Communicting Editor: Pul Yiu.

32 28 H. Okumur nd M. Wtnbe β(nb) L s β(2b) U n α(n) U 2 α(2) Figure 3 O 2. A generliztion of Schoch s circle U 2 Let nd b be rel numbers. Consider the semicircles α( ) nd β(b ). Note tht α( ) touches α internlly or externlly ccording s > 0 or < 0; similrly for β(b ) nd β. We ssume tht the imge of α( ) lies on the right side of the imge of β(b ) when these semicircles re inverted in circle with center O. Denote by C(,b ) the circle touching γ internlly nd ech of α( ) nd β(b ) t point different from O. Theorem 1. The circle C(,b ) hs rdius b( +b ) +bb + b. α( ) α( ) β(b ) β(b ) b b O b b O Figure 4 Figure 4b Proof. Let x be the rdius of the circle touching γ internlly nd lso touching α( ) nd β(b ) ech t point different from O. There re two cses in which this circle touches both α( ) nd β(b ) externlly (see Figure 4) or one internlly nd the other externly (see Figure 4b). In ny cse, we hve

33 The Archimeden Circles of Schoch nd Woo 29 ( b + b ) 2 +( + b x) 2 (b + x) 2 2( b + b )( + b x) = ( ( b)) 2 +( + b x) 2 ( + x) 2 2(, ( b))( + b x) by the lw of cosines. Solving the eqution, we obtin the rdius given bove. b Note tht the rdius r = +b of the Archimeden circles cn be obtined by letting = nd b,or nd b = b. Let P ( ) be the externl center of similitude of the circles γ nd α( ) if > 0, nd the internl one if < 0, regrding the two s complete circles. Define P (b ) similrly. Theorem 2. The two centers of similitude P ( ) nd P (b ) coincide if nd only if + b b =1. (1) Proof. If the two centers of similitude coincide t the point (t, 0), then by similrity, : t = + b : t ( b) =b : t + b. Eliminting t, we obtin (1). The converse is obvious by the uniqueness of the figure. From Theorems 1 nd 2, we obtin the following result. Theorem 3. The circle C(,b ) is n Archimeden circle if nd only if P ( ) nd P (b ) coincide. When both nd b re positive, the two centers of similitude P ( ) nd P (b ) coincide if nd only if the three semicircles α( ), β(b ) nd γ shre common externl tngent. Hence, in this cse, the circle C(,b ) is Archimeden if nd only if α( ), β(b ) nd γ hve common externl tngent. Since α(2) nd β(2b) stisfy the condition of the theorem, their externl common tngent lso touches γ. See Figure 5. In fct, it touches γ t its intersection with the y-xis, which is the midpoint of the tngent. The originl twin circles of Archimedes re obtined in the limiting cse when the externl common tngent touches γ t one of the intersections with the x-xis, in which cse, one of α( ) nd β(b ) degenertes into the y-xis, nd the remining one coincides with the corresponding α or β of the rbelos. Corollry 4. Let m nd n be nonzero rel numbers. Archimeden if nd only if 1 m + 1 n =1. The circle C(m, nb) is

34 30 H. Okumur nd M. Wtnbe O Figure 5 3. Another chrcterizton of Woo s circles The center of the Woo circle U n is the point ( b r, 2r n + b + r + b ). (2) Denote by L the hlf line x =2r, y 0. This intersects the circle α(n) t the point ( 2r, 2 ) r(n r). (3) In wht follows we consider β s the complete circle with center (b, 0) pssing through O. Theorem 5. If T is point on the line L, then the circle touching the tngents of β through T with center on the Schoch line L s is n Archimeden circle. L s L T O Figure 6

35 The Archimeden Circles of Schoch nd Woo 31 Proof. Let x be the rdius of this circle. By similrity (see Figure 6), b +2r : b =2r b b + r : x. From this, x = r. The set of Woo circles is proper subset of the set of circles determined in Theorem 5 bove. The externl center of similitude of U n nd β hs y-coordinte 2 n + r + b. When U n is the circle touching the tngents of β through point T on L, we shll sy tht it is determined by T. The y-coordinte of the intersection of α nd L is 2 r +b. Therefore we obtin the following theorem (see Figure 7). Theorem 6. U 0 is determined by the intersection of α nd the line L : x =2r. L s L T O Figure 7 As stted in [2] s the property of the circle lbeled s W 11, the externl tngent of α nd β lso touches U 0 nd the point of tngency t α coincides with the intersection of α nd L. Woo s circles re chrcterized s the circles determined by the points on L with y-coordintes greter thn or equl to 2 r +b. 4. Woo s circles U n with n<0 Woo considered the circles U n for nonnegtive numbers n, with U 0 pssing through O. We cn, however, construct more Archimeden circles pssing through points on the y-xis below O using points on L lying below the intersection with α. The expression (2) suggests the existence of U n for r n<0. (4) + b

36 32 H. Okumur nd M. Wtnbe In this section we show tht it is possible to define such circles using α(n) nd β(nb) with negtive n stisfying (4). Theorem 7. For n stisfying (4), the circle with center on the Schoch line touching α(n) nd β(nb) internlly is n Archimeden circle. Proof. Let x be the rdius of the circle with center given by (2) nd touching α(n) nd β(nb) internlly, where n stisfies (4). Since the centers of α(n) nd β(nb) re (n, 0) nd (nb, 0) respectively, we hve ( b b + r n ) 2 ( +4r 2 n + r + b ) =(x + n) 2, nd ( ) b 2 ( b + r + nb +4r 2 n + r ) =(x + nb) 2. + b Since both equtions give the sme solution x = r, the proof is complete. 5. A generliztion of U 0 We conclude this pper by dding n infinite set of Archimeden circles pssing through O. Let x be the distnce from O to the externl tngents of α nd β. By similrity, b : b + = x :. This implies x =2r. Hence, the circle with center O nd rdius 2r touches the tngents nd the lines x = ±2r. We denote this circle by E. Since U 0 touches the externl tngents nd psses through O, the circles U 0, E nd the tngent touch t the sme point. We esily see from (2) tht the distnce between the center of U n nd O is 4n +1r. Therefore, U 2 lso touches E externlly, nd the smllest circle touching U 2 nd pssing through O, which is the Archimeden circle W 27 in [2] found by Schoch, nd U 2 touches E t the sme point. All the Archimeden circles pss through O lso touch E. In prticulr, Bnkoff s third circle [1] touches E t point on the y-xis. Theorem 8. Let C 1 be circle with center O, pssing through point P on the x-xis, nd C 2 circle with center on the x-xis pssing through O. IfC 2 nd the verticl line through P intersect, then the tngents of C 2 t the intersection lso touches C 1. x r 2 x r 2 O P O P Figure 8 Figure 8b

37 The Archimeden Circles of Schoch nd Woo 33 Proof. Let d be the distnce between O nd the intersection of the tngent of C 2 nd the x-xis, nd let x be the distnce between the tngent nd O. We my ssume r 1 r 2 for the rdii r 1 nd r 2 of the circles C 1 nd C 2.Ifr 1 <r 2, then r 2 r 1 : r 2 = r 2 + d = x : d. See Figure 8. If r 1 >r 2, then r 1 r 2 : r 2 = r 2 : d r 2 = x : d. See Figure 8b. In ech cse, x = r 1. Let t n be the tngent of α(n) t its intersection with the line L. This is well defined if n b +b. By Theorem 8, t n lso touches E. This implies tht the smllest circle touching t n nd pssing through O is n Archimeden circle, which we denote by A(n). Similrlry, nother Archimeden circle A (n) cn be constructed, s the smllest circle through O touching the tngent t n of β(nb) t its intersection with the line L : x = 2r. See Figure 9 for A(2) nd A (2). Bnkoff s circle is A ( ) 2r = A ( ) 2r b, since it touches E t (0, 2r). On the other hnd, U 0 = A(1) = A (1) by Theorem 6. L L s L β(2b) U 2 α(2) γ β α E O Figure 9 Theorem 9. Let m nd n be positive numbers. The Archimeden circles A(m) nd A (n) coincide if nd only if m nd n stisfy 1 m + 1 nb = 1 r = b. (5) Proof. By (3) the equtions of the tngents t m nd t n re (m +(m2)b)x +2 b(m +(m 1)b)y =2mb, (nb +(n2))x +2 (nb +(n 1))y =2nb. These two tngents coincide if nd only if (5) holds.

38 34 H. Okumur nd M. Wtnbe The line t 2 hs eqution x + b(2 + b)y =2b. (6) It clerly psses through (2b, 0), the point of tngency of γ nd β (see Figure 9). Note tht the point ( 2r ) + b, 2r b(2 + b) + b lies on E nd the tngent of E is lso expressed by (6). Hence, t 2 touches E t this point. The point lso lies on β. This mens tht A(2) touches t 2 t the intersection of β nd t 2. Similrly, A (2) touches t 2 t the intersection of α nd t 2. The Archimeden circles A(2) nd A (2) intersect t the point ( b b + r, r + b ( ( +2b)+ ) b(2 + b)) on the Schoch line. References [1] L. Bnkoff, Are the twin circles of Archimedes relly twin?, Mth. Mg., 47 (1974) [2] C. W. Dodge, T. Schoch, P. Y. Woo nd P. Yiu, Those ubiquitous Archimeden circles, Mth. Mg., 72 (1999) Hiroshi Okumur: Deprtment of Informtion Engineering, Mebshi Institute of Technology, Kmisdori Mebshi Gunm , Jpn E-mil ddress: okumur@mebshi-it.c.jp Msyuki Wtnbe: Deprtment of Informtion Engineering, Mebshi Institute of Technology, Kmisdori Mebshi Gunm , Jpn E-mil ddress: wtnbe@mebshi-it.c.jp