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 BIaa, CI 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 r rsec22AAA s tan sec aa 2 2 2 ()sa 2 1 where s2 () a b c is half of the perimeter of ABC , r is the in-radius and ra is the radius of the A-excircle. (iv) AAI', , 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 IIIa,, b c 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) 19 Tze-ming To Figure 1 ### AIBICIa, b, c AABBCC###, , AABBCC###, , Figure 2 Now we state the main theorem on collinearity of points. # # Main Theorem: Suppose concur at M1 , concur at M 2 # and 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 ABC,, are mid-points of IIIIIIb c,, c a a b , respectively, Gex is the centroid of IIIa b 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 x::():():() y z s BCP s CAP s ABP , where s() XYZ is the signed area of XYZ . 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 ABC,, 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, Ib (::) a b c and Ic (::) a b c . The absolute coordinates of the excenters are a b c a b c a b c IIIa,,,,,,,,. b c sasasa sbsbsb scscsc Hence 1 1 1 1 1 1 1 1 1 Gex a :: b c sbscsa scsasb sasbsc (ii) It is known that A is the mid-point of IIbc. Hence a a b b c c A :: sbscsbscsbsc ((2a s b c ): b ( b c ):( c b c )) (a2 : b ( b c ) : c ( b c )) and similarly B(( acab )::(22 ccaC )), (( aabbabc ):( ):). 2. It is known that A lies on AI a and the circumcircle has equation (in barycentric coordinates) is a2 yz b 2 xz c 2 xy 0 . The equation of is b c a c a b x y z 0, bbccbc()()()() a22 cbc a bbc which is simplified to bcb( cx ) acs ( cy ) abs ( bz ) 0 or a[()()] c s c y b s b z x . bc() b c 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 21 Tze-ming To abcb()()() c bcs22 c cbs b bcab[ ( c ) bs ( c ) cs ( b )] bc( b c )( s a ). So the equation becomes cs3( cy ) 2 bcb ( cs )( ayz ) bs 3 ( bz ) 2 0. It has discriminant bcbcsa2 2( ) 2 ( ) 2 4 bcsbsc 3 3 ( )( ) bc22 [(b c )(22 b c a )4( bc a c b )( a b c )] 4 bc22 {(b c )(2 b c a )4[ 2 bc a 2 ( b c )]} 2 4 bc22 {(b c )2 [( b c a ) 2 4 bc ] 4 bca 2 } 4 bc22 {(b c )[(2 b c )4 2 bc 2( a b c ) a 2 ]4 bca 2 } 4 bc22 ()bc4 2a ( b c )( b c )2 a 2 [( b c ) 2 4 bc ] 4 bc22 (b c )4 2 a ( b c )( b c ) 2 a 2 ( b c ) 2 4 b2 c 2[( b c ) 2 a ( b c )] 2 . 4 Therefore y2( bc b c )( s a ) bc [( b c )2 a ( b c )] z4 c3 ( s c ) b [ (b c )( b c a ) ( b c )2 a ( b c )] 4c2 ( s c ) b [ b2 c 2 a ( b c ) ( b c ) 2 a ( b c )] 4c2 ( s c ) bb (2c22 2 bc 2 ac ) or ( 2 b 2 bc 2 ab ) 4c22 ( s c ) 4 c ( s c ) b b2 () s b or . c c2 () s c 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 21 Collinearity of Points Related to Mixtilinear Excircles 22 y bc( b c )( s a ) b z c3 () s c c b [ (b c )( s a ) c ( s c )] c2 () s c b { b ( s a ) c [( s a ) ( s c )]} c2 () s c b [()] b s a bc c2 () s c b2 b() s b ().c a s c22()() s c c s c Substitute the second solution back to the equation of AI a , we obtain xab 2 ()sb c() s c b ()sb z bc() b c c 2 ()sc a() s b a( s b )( s c ) ()bc . c2 () b c c2 () s c Thus A (( asbscbsbcsc )( ):(22 ):( )) and similarly B (( asa22 ): bscsacsc ( )( ):( )), C (( asabsb22 ):( ): csasb ( )( )). Remark 3: It follows that AA,, BB CC concur at a point M and its barycentric coordinates is (a2 ( s a ): b 2 ( s b ): c 2 ( s c )) . 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 I (::) a b c and the equation of AI is cy bz 0 or y bz/ c .