自然科學與教育 Natural Science and Education 2015,第一期第二卷,17-30 2015,1(2),17-30
Collinearity of Points Related to Mixtilinear Excircles
Tze-ming To Department of Mathematics, National Changhua University of Education E-mail:[email protected]
In this paper, we first state results that generalize those on mixtilinear incircles to mixtilinear excircles which can be proved in the same manner. Then we present two concurrences of points related to mixtilinear excircles that have no counterpart for mixtilinear incircles. Brute force barycentric coordinates computation is used in the proof of the concurrences as we are unable to find a synthetic proof .
Key words: Mixtilinear circle, collinearity, concurrence, barycentric coordinates Classiciation:Mathematics
1. Introduction
The geometric configuration contains what now named mixtilinear incircles first appeared on a San Gaku in Iwata prefecture, Japan, dated 1842 (H. Fukagawa, D. Pedoe, 1989). San Gaku are wooden tablets with geometric problems or theorems on them which were placed as offerings at Shinto Shrines or Buddhist temples. The same configuration rediscovered by Leon Bankoff in the American Mathematical Monthly at 1954. Since then mixtilinear incircles have been studied and many of its properties had been found. For a full account on mixtilinear incircles, we refer the readers to P. Yiu (1999) and a series of articles (in french) in J.L. Ayme’s website http://jl.ayme.pagesperso-orange.fr and the reference therein. Let ()O be the circumcircle of ABC . Then a mixtilinear incircle or Longchamp circle as called by J.L.Ayme, is a circle which touches two sides of ABC and ()O internally. In this article, we will consider circles which touch the circumcircle externally instead. We call such a circle a mixtilinear incircle excircle. Results known on a single mixtilinear incircle if replaced by a mixtilinear excircle are still valid with incenter replaced by excenters and their proof essentially the same. We state without proof of the basic properties of mixtilinear excircles which are essentially the same as those for Longchamp circles (see Figure 1 for the situation).
Theorem 1: Let ()La be the circle which touches the circumcircle ()O of ABC externally at A and touches its sides AB, AC at RQaa, respectively. Let A be the mid-point of the arc BAC . Then (i) Suppose BICIaa, intersects ()O again at QRaa, , respectively. Then QAQaa,, (similarly RARaa,, ) are collinear. (ii) ABRI,,, aa are concyclic; similarly, ACQI,,,aa are concyclic.
(iii) The center of A-excircle Ia lies on RQaa, in fact, it is the mid-point of RQaa.
As a result, the radius of ()La is
Collinearity of Points Related to Mixtilinear Excircles 18
rbc rrs sectansec22AAA aa 222 ()sa 2 1 where s a b c2 () is half of the perimeter of ABC , r is the in-radius and ra is the radius of the A-excircle. (iv) A A' , , I a are collinear.
Definition: Call ()La the mixtilinear A-excircle of ABC and A the A-ex- mixtilinear point.
But if all three mixtilinear incircles are involves, some results are still valid but some may be failed for mixtilinear excircles. Here we present collinearity of points that is valid for mixtilinear excircles but we could not think of any counterpart for mixtilinear incircles. We prove this result by brute force computation using barycentric coordinates and we hope in the near future we can find a synthetic proof. We refer to the course note “An introduction to triangle geometry” by Paul Yiu in his website for readers who are unfamiliar with barycentric coordinates .
2. Main Theorem and Its Proof
2.1 Statement of the Main Theorem
Before we state the main theorem, we first state a concurrence result whose proof are analogue to that for mixtilinear incircles.
Theorem 2: Let I Iabc I,, be excenters of ABC . Let ()O be the circumcircle of . Let Abe the mid-point of arcs BAC of and let A be the mid-point of BC of which does not contain A. The points BBCC,,, are defined in the same manner. The mixtilinear A (B, C) - circles of ABC touch ()O at A ( BC, ). The lines AABBCC, , form a triangle with vertices ABC###,,. Then ### (a) AIBICIa, b , c are concurrent. (b) AABBCC### , , are concurrent. (c) AABBCC### , , are concurrent. (see Figure 2 below)
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Figure 1
Figure 2
Now we state the main theorem on collinearity of points. ### # ### # Main Theorem: Suppose A IBabc, IC, I concur at M1 , A AB, , BC C concur at M 2 ### # and AABBCC, , concur at M 3 . Then ## (a) IMM, 12 , are collinear, ## (b) GMMex , 13 , are collinear.
Here Gex is the point of concurrence of the lines IAIBICa,, b c .
19 Collinearity of Points Related to Mixtilinear Excircles 20
Remark 1: Because A B,, C are mid-points of I Ibccaab I I I,, I , respectively, Gex is the centroid of I Ia Ib c .
In this paper, we prove the theorem by using barycentric coordinates. So we first give a definition and notation for this notion. We will use ( :x : y) z to denote barycentric coordinates of a point P if xyzsBCPsCAPsABP::():():() , where s X() Y Z is the signed area of X Y Z . When x y z 1, instead of , we use ( ,x , )y z ; and it is called the absolute barycentric coordinates of P. We divide the proof of the main theorem in parts.
2.2 Coorindates of A B ,, C
Here we use Theorem 1 to compute the coorindates of A and we divide the computation in parts. Some parts will be used in later calculation.
1. (i) As the altitudes from Ia to the sides of ABC are A-exradii, the barycentric coordinates of Ia is ( : : )abc. Similarly, I ab b c( : : ) and I ac b c( : : ) . The absolute coordinates of the excenters are abcabcabc IIIabc ,,,,,,,,. sasasasbsbsbscscsc Hence 111111111 Gabcex :: sbscsascsasbsasbsc
(ii) It is known that A is the mid-point of IIbc. Hence aabbcc A :: sbscsbscsbsc (asbcb (2) :() bcc : ()) bc (:()ab2 : bcc ()) bc and similarly Ba cabc( ()::()),((): caCa 22 a ():). bb a bc
2. It is known that A lies on AI a and the circumcircle has equation (in barycentric coordinates) is ayzbxzcxy222 0 . The equation of is bcacab xyz 0, bb() c ()()() cb ca cb22 cabb c which is simplified to bcb()()()0 cx acs cy abs bz or a[ c ()() sc yb ] sb z x . bc() bc Substitute this to , we obtain c()() s c y b s b z a 2 yz a (b22 z c y ) 0 . bc() b c The denominator of the coefficient of the yz term is
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abcbcbcsccbsb()()() 22 bcabcbsccsb[()()()] bcbcsa()(). So the equation becomes cscybc3232()()()()0 bcsayzbsbz . It has discriminant b2 c 2223() bcsab ()4()() 3 c sb sc bc22 [()bcbcabc ()4()()] 22 acb abc 4 bc22 {()bcbcabc ()4[()2222 ]} abc 4 bc22 {()bcbcabcbca [()4]4}222 4 bc22 {()bcbcbca [()422222 ()]4} bcabca 4 bc22 ()bc4 2a ()()[()4] bc bcabcbc222 4 bc22 ()2bca ()()() bc4222 bca bc 4 b2 cbca 222[()()] bc . 4
Therefore ybc b2( c s a )( bc ) b c [( a ) b2 c ( )] zc s c 43 ( ) b [ (b c )() b ( c a ) b ( c2 )] a b c 4c2 ( s c ) b [(b 2 ) c 22 ( a b ) c ( b c )] a b c 4c2 ( s c ) bb (2 c22 2 bc 2 acb ) or ( bc 2 ab 2 2 ) 4c22 ( s cc )4 s ( c ) b b2 () s b or . cc s c2 () The first solution is clearly associated with A so the second solution is associated with A .
y b Remark 2: Alternatively, because one of the solution of is (corresponding to the z c barycentric coordinates of A ), the other solution is
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ybcbcsab()() zcscc 3 () b [()()()]bcsacsc csc2 () b {()[()()]}bsacsasc csc2 () b [()]bsabc csc2 () bbsb2 () ().cas csccsc22()()
Substitute the second solution back to the equation of AI a , we obtain xab 2 ()sb csc() b ()sb z bc() bc c 2 ()sc asb() asbsc()() ()bc . cbc2 () csc2 () Thus Aasbscbsbcsc (()():():()) 22 and similarly Basab (() scsacsc22 :()() :()),
Casabsbc (()22 :() sasb:()()).
Remark 3: It follows that AABBCC,, concur at a point M and its barycentric coordinates is (():():())asabsbcsc222 .
2.3 Coorindates of ABC###,,
We first determine the barycentric coordinates of A which is the point at where AI meets the circumcircle again. The incenter Iabc (::) and the equation of AI is cybz 0 or ybzc / . Substitute this to ayzbxzcxy222 0 , we obtain a22 z c( b c ) xz 0 . The solution z 0 , which is corresponding to A, is excluded and xzac::() bc 2 . Therefore Aab bcc(: ():2 bc ()) and, similarly, Ba cabc( () ::ca ()), 2
Ca ab( () b : abc () :). 2
The equation of AA is b2 () s b c2()s c a ()()() s b s c c 2 s c a ())(s b() s cs b 2 b x y z 0. bbccbc()()()( a2 cbc a2 bcb ) But
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bsbc2 () 2 ()sc b()() sbc sc bc()()[ bcbc ()()] bcb sbc sc b()() bcc bc 11 bc()()[()]()()(), bcbcsbcbc bcbcsa
ac()())sbscsc 2 ( sbcsbc ac()() scac sc ac2 bc () abcabs b sb c ac() scac s ( cbcsa)()(), sc b
ab()()()sbscsb 2 ab()()(). sasbbc ab2 bc () Thus the equation of A A is bcbcxacscyabsbz()()()0. Similarly, BBbcscxaccayabsaz :()()()0,
CCbcsbxacsayababz :()()()0. Since A# is the intersection point of BB and CC , it has barycentric coordinates ac()()()()()() a cab s aab s abc s cbc s cac a c :: . ac()()()()()() s aab a bab a bbc s bbc s bac s a Now
ac()() a cab s a a c s a a222bca bcs bs cs a [(2)(2 ) () ] ac()() s aab a b s a a b a222 bc[3] s 2s () b c abc a a222 bc[34 ss ()] s abc a 222 a bsc[], 4as bc a ab()() s a bc s c s a s c ab22cabcsasb[( sc )( sc ) ( )(2 )] ab()() a b bc s b a b s b abc2 [( s 22 )3] sab ab cs c 2 22 ab c(3), sab4csc bcsc()() acac scac abc2 ab 2c(3 s 2 4 bs b 2 ac ). bcsb()() acsa sbs a
Note that a22bc 2() sa a (). bc a a b cab c b ca Set ()abbcca . Since is a symmetric polynomial in a, b, c, we have a2bc 222 as b 2 ca bs c 2 ab c s . Then A# (((( a s2 2(s s a ) ): b s22 2( s s c ) ): c s 2( s s b ) )), and, similarly,
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Ba# sss(((( 222s ()) scbs :2 ()) sbcs :2 sa ()) 2 ),
C# (((()as2222s ()) sbbs : sacs s ss2 ()) : 2 ()) c .
### # 2.4 Concurrence of A I Iabc C,,B I and Coordinates of point of concurrence M1 Now we show that are concurrent and find barycentric coordinates of the # # point of concurrence M1 . The equation of the line AIa is b((ss222s ())2 sc ()) cs sb x bc cs((sas sbs222 sa () )2 ()) y ca a(ss222(s sabs))2 sc ()) ( z 0. a b We have b((ss222 s ( s c ) ) c 2 s ( s b ) ) bc ss222s ( s c ) 2 s ( s b ) bc 2 bcs ( b c ), 11
c((s222 s ( s b ) ) a s 2 s ( s a ) ) ca ss222s ( s b ) 2 s ( s a ) ca 2 ca( sc ), 11
a(( s222s ( s a ) ) b s 2 s ( s c ) ) ab (s2 2ss ( a ) )(s2 2s ( s c ) ) ab 2ab ( bs) . 11 # Thus A Ibca :()( bc sxaccs)()0 yabbs z , and similarly B# Ic:()(bccs a()) s y x ab 0, caas z b # C Ibcac : b(()bs)()0 x ca as y ab s z . Note that bcbcsac()((()( cs))) ab bs bcs cs bs bc(()(()(cs))) accasab as a2 b 2 c 2 cs cas as bc((()bs)) ac as ab a b)(s () bs as a b s and all column (row) sums of the matrix on the right hand side are zero. Therefore the ### # determinant has value 0 and the three lines AIBICIa,, b c concur. Denote M1 the point of concurrence. Then
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# acababbc(((()()(csbsbscs)))) bc sbc bc sac M1 ,,. ac() ca saba (() asasc))) bbcbc((( scs ) ac ca s Now acab((csbscsbs))() a2bc ac()( ca sab asca)()() sas ()() asba s a2222 bca bcasab [( )()()] ca s , ()()ca sas ab(()bsbsbc)()bc bc s s ab2c abbc((ascsascs))() bsbc() s ab2c ab222 cbsbcab[()()()] s , (ba )sbs bc()( bc sac csbc)()() scs abc2 bcc((cscsca)()acs s a) abc222[(. csbc)()() cs a ] Therefore # 22 Ma1 ( [()()() asa ] b ca s :bbsb [()()() c ] a22 b s :)()()ccsb[( ] c ca22 s )
2.5 Concurrence of AC###ABBC,, Now we show that are concurrent and find barycentric coordinates of the # # point of concurrence M 2 . The equation of the line AA is bs((ss s22 ccs2 ())2 s ())b x b()() b cc b c cs((sa s22 b2 s ())2 ()) s s a y c() b ca 2 a(( ss222s ())2 s abs s c ()) z 0. a2 b() b c We have
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bs((ss sccs222 sb ())2 ()) b()() bcc bc ss222s ()2 scs () sb bc() bc 11 2bc()[, bc s 22ssbcbc(2)]2()() bc sas
cs((sas sbs222 sa ())2 ()) c() bca 2 ss222s ()2 sbs () sa ca b ca s2 2s()2 sbssab (2)2 ca bcabc s2 2s () sb cs 2ca bcsc s2 ss(2 2bb cs) 2 2ca bcsc s22 ass 2ca bcsc as s2 2ca ()sbsc 2[()(()]ca sb sa 2 ) ss c , and, by interchanging b and c in the second equality, we obtain ass(( 222s ())2 sabs ()) sc ab2 bc () 2[(ab ss c)( 2 )()]. assb Set s2 . Then Aca# A: s bc ()( bas b cas s c xa ) [()() ][()y ab s c sb()]0s z and similarly BBbcsa# :[( )bs ( s c )] x ca (c a )( bs ) y abscbssaz [( ) ( )] 0,
CC# :bcsa[( ) cssbxcasb ( )] [( ) cssayababcsz ( )] ( )( )0 . The lines A###AB,, BC C are concurrent if and only bc(b c )()( asa )] ca[( s b )( as )][( s cab ) s c s s b bc[( s aab )[( sbs )(( c s cca )]( bs)] c s 0 a )(a bs ) bc[(sa )( cs )][( s bca s b ) cs( s aab)]( a b )() cs if and only if (b c )( as )()()()s b as s c s c as ( s b ) ()s abs(s c ) ( c a )( bs ) ( s c ) bs ( s a ) 0. ()()()s a cs s b s b cs( s a ) ( a b )( cs )
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Note that the sum of the entries of first row of the last matrix is 2a ( ) s b c , and similarly, the sums of the entries of the second and third rows are 2b ( ) s c a and 2c ( )s a b , respectively. It follows that the sum of all entries is 0. Moreover, the sum of the entries of the first column is ()b c ( asbs)(sc)(cs sb )()()()()bcassbcas2 , and similarly, the sum of the entries of the second column is (c a )( bs) . Thus ()()()b casas s b ()()()s bas s cs c ()()()s as cbsbs()()() s cc abs sa ()()()s acs s bs b cs()()() s aa bcs ()()()b casa b c ()()s bas s c 2()ss a bs()()()() s cc absb c a ()()bc as (c a)( bs)0 ()()s bas s c a() b c 2s ()()b cas ()()()c absb c a ()()(b casa b c) ()(ca bs) ()sa bs()() s cb c a ()()s bas s c a() b c 2s ()()b c c a () as bsb asa ( bs) (sa) bs()() s cb c a The value of the determinant is then zero as ()()()()s b as s ca()() b c s b as s b a b c bs b b as() s b a()() b c b s b b as() s b a()() b c b s b b
as()(s b b2 ac 2) bs bs ab b as() s b bs ab sb2 a() bs bs (bs)( s2 ab ) sb2 and
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asaa ()()sasa bs()()()() scb cabs sab ca a bs()()() sab caa sa a bs()(2) saab 2 casasab sa2 bs() saasab sa2 (assab)(). 2 basas() Therefore A###A B,, B C C are concurrent.
### # 2.6 Concurrence of A A B,, B C C and Coordinates of point of concurrence M 3
Now we show that are concurrent and find the barycentric coordinates of # the point of concurrence M 3 . The equations of are A#22 Ac:(2 ss ())(2 sbyb ())0 ss scz B#22 Bc:(2 ss ())(2 saxa ())0 ss scz C#22 Cb:(2 ss ())(2 saxa ())0 ss sby Divide the equations by bccaab,, , respectively, we see that concur at the point
# abc M3 222 ,, . ss s2 ass ()2 s ()2 bss () s c
## ## 2.7 Proof of the main theorem: collinearity of IMM, , 12 and GMMex , , 13
Since Iabc (::) and # 22 Ma1 ( [() asa ()() b] c a s :bbsb [() c ()() a22 ] b s , :)ccsb[( ()() c ]c22 a s ) # the equation of IM1 is # 22 2 IM1 bc( b c )(( axbcc ca )(()) c a by ab a baa c )(0 b)z as (cs))2 (b c )( c a ) s 2 ( bs 2 ( a b )( b c ) s 2 s( b c)(2 (b c) s ) ( b c )( b c 2as ) 2 2s ( b c )( as) 2 s ( b c )() a2 bc It suffices then to show that and
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B# Bbcsaabscbs :()]()()0,[()[()()] sazbs scxca cabs y
C# Ccs :()]()( sayabbcsacs[()()][ abcs sbxcasb z ( ) )0. are concurrent. But it follows immediately as ()(()(()(bcacababc 222bccaab))) [()()()sascbs sabs()] scca c ()() abs 0 ()()[()s acs sbb s cs()]()( saabc s)
## because the sum of entries in each row in the determinant is 0. So IMM, 12 , are collinear. Recall that
# abc M3 222 ,, ss sass2 ()2 sbss ()2 sc () a(2 ss2 ())(2 sbs ()),, 2 s sc # 22 Ma1 ( [()()()] asabca s :[()()()]bbsbcab s22 :)()()]ccsbcca[( s22) 111111 111 Gabcex : : sbscsascsasbsasbsc ascsasbsasbsc[()() ()( ) ()(: ):] ## That GMMex , 13 , are collinear follows from (as )(2 abcas )( ) 2 ( s 2 2( ssb ))( s 2 2( ssc )) ( as )2 ( a b )( c a ) s 2 ( s 2 ) 2 2s ( s 2 ) [( s b ) ( s c )] 4 s2 ( s b)( s c ) (as )2 ( a b )( c a ) s 2 ( s 2 ) 2 2as ( s 2 ) 4 s2 ( s b )( s c ) aa2ss2 4 2s 2 2s 3 ( a b )( c a ) s 2 4s2 ( s b )( s c ) s2 ( a22 s 2 2 as ( a b )(4 c a) ( s b )( s c )) sa22[ s222as 2( ab bc ca ) ( ab ca bc a ) 4(s2 (b c ) s bc )] s2[s 2 2 as 3( ab ca ) bc 4 ( s 2 (2 s a )s bc )] sa2[2s 2 s 3 a (2 s a) bc 4 (s 2 (2 s a ) s bc )] 3s2(sb 2a 2 c) and ()()sc ()()s as ()() b s as b s c a( s as)[2 ()]bcs bc
a() s a [s2 (2)s a s bc] sa22 bc
Remark 4: We have not written down the barycentric coordinates of the point of concurrence # M 2 because we cannot find a nice and simple form of it. Nevertheless, our proof does not use its explicit coordinates.
Reference
29 Collinearity of Points Related to Mixtilinear Excircles 30
1. Ayme, J.L., A new mixtilinear incircle adventure I, II, III, Geometry Géométrie Geometria, web: http://jl.ayme.pagesperso-orange.fr 2. Fukagawa, H., Pedoe, D. (1989), Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989 3. Yiu, P. (1999), Mixtilinear Circles, Am Math Monthly, v. 106, n. 10, 952-955 4. Yiu, P. (2001), An introduction to triangle geometry, version 13.0411, web: http:// http://math.fau.edu/yiu/Geometry.html.