âVIVYA`S POINTâ - International Journal of Mathematical Sciences
Int. Jr. of Mathematical Sciences and Applications Vol. 5, No. 2, July-December 2015 Copyright Mind Reader Publications ISSN No: 2230-9888 A FEW MINUTES WITH A NEW TRIANGLE CENTRE COINED AS “VIVYA’S POINT” D. N.Vijay Krishna Keshava Reddy Educational Institutions Machilipatnam, Kurnool, Andhra Pradesh (India). E-mail: [email protected] Abstract We have nearly 6102 triangle centers in the literature of ‘geometry of triangles’, “vivya’s point” is one of such triangle center, in this paper let us discuss very few and important properties of this point in a more analytical way. Key Words: Vivya’s point, vivians,maneeal’s identity, cevas theorem,eulers inequality, maneeals inequality, weill’s point INTRODUCTION In our classical “Euclidian Geometry” and in “Modern Geometry” we come across with a lot of different classical centers of triangle. These are all defined by the concurrence of important lines in the triangle. These centres and the cevians that create them are Centroid medians Incenter angle bisectors Circumcenter penpendicular bisectors Orthocenter altitudes Others discovered more recently (about 100 to 150 years ago) are Associated with the incenter Gergonne point. cevians to contact points of incircle Nagel Point, cevians to contract points of excircles Spieker center incircle to medial triangle. 357 VIJAY DASARI Associated with the centroid symmedian point cevians which are reflections of medians respect to angular bisectors weills point centroid of intouch triangle 9 pt center center of 9 pt circle Points that occur in pairs such as the isodynamic and isogonal points. They play vital Role in Discussion and Improvement of our study in the Abstract Geometry. Now in this paper Let us spend very few minutes with a new triangle centre coined as “Vivya’s point” and let us discuss its properties, using this Discussion we will try to prove the famous Inequality named as ‘Maneeal’s Inequality” and also Famous classical Inequality “Euler’s Inequality” and also many more curiosities. 358 Int. Jr. of Mathematical Sciences and Applications Formal notations and some standard formulas :1) We know = s ( s a )(s b )(s c) 2 s ( s a)( s b)( s c) 2 s ( s 3 s 2 (a b c) s (ab bc ca) abc] Where rs and abc 4 R (rs) 2 s ( s 3 s 2 (2s ) s ( ab bc ca) 4Rrs ] r 2 s 2 s 2 [ s 2 2s 2 (ab bc ca) 4 Rr ] r 2 s 2 (ab bc ca) 4 Rr ab bc ca r ² s ² 4 Rr And a2+b2+c2= (a+b+c) 2-2(ab+bc+ca) = (2s) 2-2(r2+s2+4Rr) =2s2-2r2-8Rr a² b² c² 2 s ² r ² 4Rr 2) Let K = a(s-b) (s-c) + b(s-c) (s-a) +c(s-b) (s-a) = s2 (a+b+c)-2s (ab+ac+bc) + 3abc = s2 [2s]-2s [r2+s2+4Rr] + 3 [4Rrs] = 2s [s2-r2-s2-4Rr+6Rr] = 2s [-r2+2Rr] K 2rs 2R r = 2 2 R r 2 R r K 2 K a s b s c b s c s a c s b s a 2rs 2R r 3. Let P= c(s-b) +b(s-c) P=s (b+c) -2bc So, K = a(s-b) (s-c) + b(s-c) (s-a) +c(s-b) (s-a) = a(s-b) (s-c) + (s-a) P = as2-as (b+c) + abc + P(s-a) = as2-a(s (b+c)-2bc) - abc + P(s-a) = as2-aP-abc + P(s-a) = as2+P(s-2a) - abc = as2-abc + P(s-2a) K= as²-abc + P s 2a 359 A few minutes with a new triangle centre coined as “ VIVYA’S POINT” 4. aK +a2bc-4(s-b)2(s-c)2 = a [as2-abc+P(s-2a)] + a2bc-4(s-b) (s-c)2 = a2s2-a2bc+aP(s-2a) + a2bc-(2(s-b) (s-c)]2 = (as) 2- [2(s-b) (s-c)] 2 + P[as-2a2] = [as + 2(s-b) (s-c)] [as- 2(s-b)(s-c)] + P(as-2a2) = [as+2s2-2s (b+c) +2bc] [as-2s2+2s (b+c)-2bc] +P (2s-2a2)] = [s (2s)-(b+c)] +2s2-2s (b+c)+2bc] [s[(2s)-(b+c)]-2s2+2s (b+c)-2bc] + P[as-2a2] = [2s2 -s(b+c)+2s2-2s(b+c)+2bc][ 2s2-s(b+c)-2s2 +2s(b+c)-2bc]+P(as-2a2 ) = [4s2-3s (b+c) +2bc] [s(b+c)-2bc]+P(as-2a2) = [4s2-3s (b+c) +2bc] [P] +P (as-2a2) = P [4s2-3s (b+c) +2bc+as-2a2] = P [(a+b+c) 2-3(a+b+c) (s) +4as+2bc-2a2] a b c abc 2 4a 2bc 2 a ] 2 2 = P [a2+b2+c2+2ab+2bc+2ca-3(a+b+c) 3 2 = P [a2+b2+c2+2ab+2bc+2ca- (a2+b2+c2+2ab+2bc+2ca)+2a2+2ab+2ac+2bc-2a2 ] = P [2a2+2b2+2c2+4ab+4bc+4ca-3a2-3b2-3c2-6ab-6bc-6ca+4a2+4ab+4ac+4bc-4a2] 2 = P [2ab+2bc+2ca-a2-b2-c2] 2 = P [2(ab+bc+ca)-(a2+b2+c2)] 2 = P [2(r2+s2+4Rr)-2(s2-r2-4Rr)] 2 = P (2) [r2+s2+4Rr-s2+r2+4Rr] 2 = P [2r2+8Rr] =2rp[r+4R] aK +a2bc-4(s-b)2(s-c)2=2rP(r+4R) 360 VIJAY DASARI 5. If We is a Weill’s point and X be any point in the plant of le ABC, then 1 sb s c r 2 WeX² = 3 c b AX 9 R r 4 R 2r 3 R Formal Definitions : If ABC is a triangle and D, E, F are the points of contacts of the Incircle of le ABC on the sides BC, CA, AB. Let DP, EQ, FR are the perpendiculars Drawn from the vertices D,E,F to the sides EF, DF, DE of le DEF then the lines formed by joining vertices A,B,C of le ABC and the points P, Q, R are called as “Vivian’s” . We have 3 “Vivians” in the triangle and If we extend the 3 Vivians in the triangle they meet the opposite sides of the le ABC at A1, B1, C1 and these 3 Vivians Intersect at a point called as “Vivya’s point” denoted by ‘Vm’and the triangle formed by joining the 3 points A1, B1, C1 is called as “Vivians Triangle” and for any orbitary triangle the Circumcentre (S), Vivya’s point (Vm) and Incentre (I) are collinear. The line through these 3 points is called as Vivya’s line The following figures demonstrate about the vivians. A A A P F Q B D Fig (1) C B P E F E F R D Fig (2) E R Q C B D Fig (3) C Fig (4) In the above 4 figures we can see how the vivians can be constructed and also we can identify how the new triangle centre “Vivya’s point” denoted by “Vm” formed. Now let us prove some theorems related to this point. 361 Int. Jr. of Mathematical Sciences and Applications Theorem 1 : The three Vivians of triangle are concurrent and the point of concurrence is called as “Vivya’s point” (Vm). Proof: Step - I “From Figure 1” We can observe that le DEF is a contact triangle (or) In touch triangle, so we can find the lengths of AE, AF, BF, BD, CE, CD as follows. Let us consider formal notations Say BC= a units, CA= b units, AB= c units. BDC , BFA are Tangents to the Incircle. We know that “Tangents Drawn from an External Point are equal in length”. So BD=BF=x, CD=CE=y, AF=AE=z (let) Hence BC = a = BD + BC = xy CA = b = CE + EA = y +z AB = c = AF + BF = x + z 2s = a + b +c = 2(x+y+z) s= x + y + z where s is semi perimeter A So BD=BF= x = (x+y+z) – (y+z) = s-b z CD= CE = y =(x+y+z) –(x+z) = s-c x AE = AF = z= (x+y+z) –(x+y) = s-a Now from le BDF, BD=BF and FBD =B F x B C Similarly for le CDE, CDE= CED = 90 ( OCE =C ) 2 A and for le AEF, AEF= AFE= 90 ( EAF =A) 2 o A) DFE = 180 – ( BFD + EFA) B A = 180o – 90 90 2 2 362 E y B So BED= BDE = 90 2 z D C y A few minutes with a new triangle centre coined as “ VIVYA’S POINT” = A B C 90 2 2 2 DEF 90 C ( A+ B+ C =180o) 2 Similarly, FDE 90 A DEF 90 B 2 2 Now from le BDF, By Consine Rule DF2 = BD2 + BF2 – 2BD BF Cos B DF2 = (s-b) 2 +(s-b)2 -2 (s-b) (s-b) Cos B DF2= 2(s-b) 2 -2(s-b) 2 Cos B DF2 = 2(s-b) 2 [1-cos B] =2(s-b)22sin2 B 2 =4(s-b)2.sin2 B 2 =[2(s-b)(sin B 2 )]2 DF = [2(s-b)(sin B 2 )] Similarly, DE= 2(s-c) Sin C 2 EF = 2(s-a) Sin A 2 The sides of le DEF are 2(s-c) sin C/2, 2(s-a) Sin A/2 and 2(s-b) sin B/2. The angles of le DEF are 90-C/2, 90 –A/2, 90 –B/2 respectively. . Step - 2: A P E F “From Fig 2 “ Since EQ, DP, FR are r to the sides FD, EF, DE For le FED, EQ is cevian (the line from the vertex to the opposite side Now For le EQF, EQF =90o and QFE = 90-C/2 So FEQ= 180o – ( EQF + QFE) =C/2 Similarly QED = A/2 363 Q B R D C VIJAY DASARI So since EQ is cevian of the le FED meets the opposite side DF in Q then the Ratio it divides the opposite side equal to the ratio of Areas of two triangles formed by the cevian. [le FEQ ] FQ i.e., QB le [ EQD ] ( [ ] represents area) and we know that Area of triangle le = ½ ab sinC So [ le FEQ] = 1 EF .EQ. Sin PEQ 2 1 2 = 2 (s-a) Sin Similarly A C A C .EQ sin = (s-a) Sin .EQ sin 2 2 2 2 1 1 le EQR ED.EQ sin QED = .2 s c sin C .EQ.sin A = (s-c) sin A .EQ sin C 2 2 2 2 2 2 C A FQ s a sin 2 sin 2 EQ s a FQ s a QD s c sin A sin C EQ s c QD s c 2 2 Similarly we can find And DR RE s b s c EP PF s c s b Step - 3 From figure 3 Let FAP A1 , EAP A2 ECR=C1, DCR=C2, DBQ=B1, FBQ=B2 Now consider le BFD, Since BQ is cevian BQ divides the opposite side DF in the ratio of Area of le BQD, le BQF 1 . BD.BQ sin B1 DQ [ le BQD ] 2 i.e. le FQ [ BQF ] 1 .BQ.BFsin B 2 2 364 Int. Jr. of Mathematical Sciences and Applications FQ BF sin B2 DQ BD sin B1 s a s b sin B2 s c s b sin B1 (Using step -2) sin B2 s a sin B1 s c sin B1 s c sin B2 s a similarly we can prove that sin A1 s b sin C1 s a and sin A2 s c sin C2 s b Step -4 “From Fig 4” Now AA1, BB1, CC1 are required Vivian’sand also they are Cevians. We know that Cevian of triangle divides the opposite side in the Ratio of Areas of triangles thus formed. le 1 BA1 B AA So, A1C le 1 C AA 1 BA. AA1.Sin A1 1 BA 2 1 A C 1 CA. AA1.Sin A 2 2 A P C1 BA1 c.Sin A1 1 A C b.Sin A2 BA1 c s b 1 (using step-3) A C b s c BA1 c( s b ) 1 A C b(s c ) A2 A1 B2 Vm Q C2 A1 CB1 a s c AC1 b s a Similarly can prove that B1 A c s a , C1 B a s b Final step AC1 BA1 CB1 b( s a ) c( s b) a ( s b) . . Now 1 . 1 C B A C B1 A a (s b) b(s c ) c ( s a ) 365 R C1 B1 B B1 C A few minutes with a new triangle centre coined as “ VIVYA’S POINT” = abc (s a)( s b)(s c) 1 abc (s a)( s b)(s c) Hence by the converse of ceva’s theorem we can conclude that the threeVivian’s AA1, BB1, CC1 are Concurrent and the point of concurrence is called as “Vivya’s point (Vm)”. COROLLARIES : I. Let AA1, BB1, CC1 are Vivians then BA1 c( s b ) A1C b( s a ) CB1 a ( s c ) AC 1 b (s a ) Similarly 1 and 1 B A c( s a ) C B a ( s b) II. The point of concurrency of 3 Vivian’s is called as “Vivya’s point” and is denoted as Vm’ III. If Vm is the Vivya’s point then the length of the each Vivian in the triangle ABC is given by | 2 AA a 2b 2 c 2 4 abc( s b) 2 ( s c) 2 = [c( s b) b( s c)]2 1 Hence 2 2 2 2 2 AA1 = c( s b) b( s c) a b c 4bc( s b) ( s c) Similarly 2 2 2 2 2 BB1 = a( s c) c( s a) a b c 4ac(s a) ( s c) 1 1 2 2 2 2 2 CC1 = a(s b) b( s c) a b c 4ab( s a) ( s b) Proof:Let AA1 is Vivian We know by using previous Discussion BA1 c( s b) A1C c( s c ) So let BA| = c(s-b) t for some constant t A|C= b(s-c) t 366 VIJAY DASARI BA|+A|C = t [c(s-b) +b(s-c)] BC = t[c(s-b)+b(s-c) ] a = t[c(s-b)+b(s-c)] a t = c ( s b) b( s c ) ca ( s b) BA| = c(s-b) t = c( s b) b( s c) ba( s c) AC| = b(s-c) t = c( s b) b( s c) Since AA| is a vivian as well as Cevian So By STEWART’S THEOREM. BA| . AC 2 CA| .BA2 BA| .CA| BC BC (AA|)2 = | 2 AA ac( s b)b 2 ab(s c)c 2 a 2bc(s b )(s c) = + [c( s b) b( s c)](a) [c( s b) b( s c)](a) [c( s b) b( s c)]2 AA| = bc a 2 bc( s b)(s c) [b( s b) c( s c)] [c( s b) b( s c)] [c( s b ) b( s c)]2 AA| = bc [b( s b) c( s c)][c( s b ) b( s c)] [a 2 ( s b)( s c ) [c( s b) b ( s c)]2 AA| = bc bc ( s b) 2 b2 ( s b)( s c) c 2 ( s c )( s b) bc( s c )2 a 2 ( s b )( s c) [c( s b) b ( s c)]2 AA| = bc bc s b 2 s c 2 s b s c b² c² a ² [c( s b) b ( s c)]2 AA| = bc bc s b s c 2 2bc s b s c s b s c b² c² a ² [c( s b) b ( s c)]2 2 2 2 2 2 bc 2 2 AA| = [c( s b ) b( s c)] 2 bc 2 s b c s b s c b ² c ² a ² 2bc bc 2 2 2 AA| = [c( s b ) b( s c)] 2 bc a s b s c b c a ² bc 2 AA| = [c( s b ) b( s c)] 2 a ²bc s b s c b c a b c a bc 2 AA| = [c( s b ) b( s c)] 2 a ²bc s b s c 2 s 2c 2 s 2b where 2s= a+b+c | 2 AA a 2b 2 c 2 4bc( s b) 2 ( s c ) 2 = [c (s b ) b( s c)]2 367 Int. Jr. of Mathematical Sciences and Applications 1 2 2 2 2 2 AA1 = c( s b) b (s c) a b c 4bc( s b) (s c) Similarly we can prove that BB1= 1 a (s c) c( s a ) a 2 b 2 c 2 4bc( s a ) 2 ( s c ) 2 1 CC1= a s b b s a a 2 b 2 c 2 4 ab( s a ) 2 ( s b) 2 IV. If Vm is the Vivya’s point of a triangle then the Vivya’s point divides each Vivian in the ratio is given by AVm c( s a )( s b) b (s a )( s c) Vm A1 a ( s b)( s c) BVm c( s b)( s a ) a ( s b)( s c) Vm B1 b( s a )( s c) Similarly and CVm a( s c)( s b) b( s c)( s a ) VmC1 c( s a )( s b ) Proof :Let AA1, CC1 are vivians of le ABC and Vm is “Vivya’s point” Since C1, Vm, C are collinear For le A BA1 and C 1VmC is a transversal Hence By Menelau’s theorem | AC | BC AV m . . We have C | B CA| V A 1 m A1Vm BC1 CA1 So V A C1 A BC m b sc (“-” is removed since it represents only direction but not magnitude) we know that BC1 a( s b ) C 1 A b( s a) and BA1 c( s b ) A1C b( s c) (By adding 1 on both sides ) BA1 c ( s b) 1 1 1 AC b( s c ) 368 A few minutes with a new triangle centre coined as “ VIVYA’S POINT” BA1 A1C c( s b) b ( s c) A1C b( s c ) BC c( s b) b( s c) A1C b( s c ) CA1 b( s c ) BC c( s b) b( s c ) A1Vm BC1 CA since V A C 1 A . BC m a s b s c A1Vm a ( s b) b( s c ) . = Vm A ( s a ) c( s b) b( s c) c s a s b s a s c AVm c( s a )( s b) b ( s a )( s c) . Vm A1 a ( s a )( s c) Similarly we can prove BVm c( s b)( s a ) a ( s b)( s c) Vm B1 b( s a )( s c) CVm And V C| m V. a s c s b b s c s a c s a s b If Vm is Vivya’s points of le ABC and AA1 , BB 1 ,CC 1 are Vivian’s then 2 R r r ² 4Rr 8R ² r ² AVm2 BVm2 CVm2 S 2 2 2R r 2R r Proof: We have by previous theorem : AVm s a s b b s a s c K a s b s c Vm A1 a s b s c a s b s c s a c s b b s c s a p a s b s c a s b s c So AVm K a s b s c t Vm A1 a s b s c t Now AVm Vm A1 t K AA1 tK 369 for some real constant ‘’t’’ VIJAY DASARI t AA1 K 1 AA1 p s a AA So AVm K a s b s c K K Vm A1 a s b s c AA1 K k b s a s b BB b s a s c | BB Siimilarly BVm , Vm B| 1 K K 1 k c s a s b CC1 c s a s b CC 1 CVm , VmC k k and Now AVm2 BVm2 CVm2 AVm2 = p2 s a k AA1 2 p2 s a k 2 2 1 a 2b 2 c 2 4bc s b 2 s c 2 c s b b s c 2 2 2 2 2 p 2 s a a 2b2 c 2 4bc s b s c k2 p² a 2b 2c 2 s a 2 4bc s a 2 s b 2 s c 2 k2 k2 a 2b 2 c 2 2 2 2 4 2 2 2 s a s b s c 2 s a s b s c ab bc ca k2 k 2 2 2 2 a 2b 2c 2 4 s s a s b s c 2 2 2 2 3 s 2 s a b c a b c ab bc ca k2 k2 s2 16 R 2 2 4 4 2 2 2 2 2 3 s 4 s a b c k 2 s 2 ab bc ca k2 = 16 R 2 2 4 4 s 2 2 s 2 2 r 2 8 Rr 2 2 r 2 s 2 4 Rr 2 k k s 4 2 = k2 = abc 4 R and 2s a b c 2 2 2 2 2 2 4 R s 2 r 8 Rr r s 4 Rr rs 2 s 4 2 2 2 4 R s 8 R 2 r 2 32 R 3 r r 2 r 2 rs 2 4 Rr k2 370 Int. Jr. of Mathematical Sciences and Applications 4 2 = 2 4 R 2 s 2 8R 2 r 2 32R 3 r r 4 r 2 s 2 4R 4r 3 k 4 2 2 = 2 s 4R 2 r 2 8r 2 8R 2 r 2 4Rr 8R 2 r 2 k 4 2 2 = 2 s 4R2 r 2 r 2 4Rr 8R 2 r ² k 4 ² 4 ² 2 R r 2 s ² 4 R ² r ² r ² 4 Rr 8 R ² r ² 2 R r r ² 4 Rr 8 R² r ² S² 2 2R r 2R r 2 R r r ² 4 Rr 8R ² r ² AVm2 BVm2 CVm2 S 2 2 2 R r 2R r Theorem 2 : MANEEAL’s Identity: Let ABC is a triangle and Vm is it’s “Vivya’s point” and X be any point in the plane of le ABC Then Vm X 2 a (s b)(s c ) b( s a)(s c) c(s a)(s b) 2 Rr 2 (r 4 R ) AX 2 BX 2 CX 2 K K K (2 R r ) 2 where K 2rs 2R r The other form of maneeal’s Identity as Vm X 2 2 RSin 2 A / 2 2 RSin 2 B / 2 2 RSin 2 C / 2 2 Rr 2 (r 4 R ) AX 2 BX 2 CX 2 (2 R r ) 2R r 2R r (2 R r ) 2 Proof:Let R= circum radius of ABC Vm r = In radius of ABC s =semi perimeter Vm = Vivya’s point X = Any point in the plane of ABC 371 A few minutes with a new triangle centre coined as “ VIVYA’S POINT” AB = c, BC=a, CA=b P = c(s-b) +b(s-c) K = a(s-b) (s-c) +b(s-a) (s-c) +(s-a) (s-b) K-a(s-b) (s-c) = (s-a) [b(s-c) +c(s-b)] =(s-a) P We know by previous calculations BA1 c( s b) A1C b( s c) ac( s b ) And BA1 = c( s b) b( s c) ac( s b) p = Similarly A1C= ab( s c) p and Vm is Vivya’s point and we have AVm c( s a )( s b) b( s a )(s c) Vm A1 a ( s b)( s c ) by previous theorems we have c s a s b b s a s c AA| K a s b s c AA| AVm K K and VmA1 = a (s b)( s c) AA1 k Step - 1 : Now from le BXC, Since XA| is a cevian so by STEWARTS THEOREM, we have BA| .CX 2 CA| .BX 2 | BA| . AC BC BC 2 XA| 2 XA| 2 XA| ac s b ab s c a 2bc s b s c CX 2 BX ² Pa Pa P² c s b b s c a 2bc s b s c CX 2 BX ² P P P² Step - 2 : Now From le AXA|, 372 VIJAY DASARI Since VmX is a cevian so by Stewart’s theorem, we have AVm XA| 2 Vm X AA| 2 Vm A| . AX 2 AVm .Vm A| | AA K a s b s c a s b s c AA| K a s b s c AA| a s b s c AA| | 2 2 Vm X 2 XA AX K . AA| K . AA| K2 K a s b s c c s b CX 2 b s c BX 2 a 2bc s b s c Vm X 2 K P P P2 a s b s c K K a s b s c a s b s c AA| AX K2 2 2 s a P c s b CX 2 b s c BX 2 a 2bc s b s c Vm X P P P2 K 2 Vm X 2 a s b s c K 2 AX s a P a s b s c AA| 2 K2 c( s b)( s a)CX 2 b( s c)( s a) BX 2 a( s b)( s c) AX 2 K K K a 2 bc( s a )( s b)( s c ) p p( s a)( s b)( s c) 12 AA K P2 K2 Vm X 2 a( s b)(s c) a ( s a)(s b)(s c) 2 12 AX 2 Kabc P AA K K2P Step - 3 : Now Kabc + P2 AA| 2 a 2 b 2 c 2 4bc( s b) 2 ( s c ) 2 P2 = Kabc+P2 Where AA| = 2 a 2b 2 c 2 4bc( s b) 2 ( s c ) 2 P2 = K abc+a2b2c2 -4bc(s-b)2 (s-c)2 = bc[Ka+a2bc-4(s-b)2(s-c)2] = bc[2rp(r+4R)] [By notation 4] 373 Int. Jr. of Mathematical Sciences and Applications Vm X 2 2 a ( s b)(s c) a( s a )(s b )(s c) AX 2 Kabc P 2 AA1 2 K K P Vm X 2 a( s b)( s c) a ( s a )(s b)( s c) AX 2 2rpbc (r 4 R) K K2 P Vm X 2 a (s b )(s c) abc(s a )( s b)(s c)(r 4 R)(2 r ) AX 2 K K2 Vm X 2 a ( s b)( s c) abc( s a )( s b)( s c )(2 r )(r 4 R ) AX 2 2rs(2R - r) (2rs)2 ( 2 R r ) 2 Vm X 2 a ( s b)( s c) abc(s a )(s b)( s c)(2r )(r 4 R ) AX 2 2rs(2R - r) 4 r 2 s 2 (2 R r ) 2 Step - 4 : Now abc( s a )( s b)( s c )(2r )(r 4 R ) 4r 2 s 2 (2 R r ) 2 = 4 Rrs ( s a)( s b)( s c)(2r )(r 4 R ) 4r 2 s 2 (2 R r ) 2 = 8 Rr 2 s ( s a )( s b)( s c)(r 4 R) 4(rs ) 2 (2 R r ) 2 abc 4 Rrs 8 Rr 2 2 ( r 4 R) rs = 42 (2 R r ) 2 = 2 Rr 2 (r 4 R) (2 R r ) 2 Vm X 2 2 Vm X a ( s b)( s c) 2 Rr 2 (r 4 R) AX 2 2 rs(2 R r ) (2 R r ) 2 a (s b)(s c ) b( s a)(s c) c(s a)(s b) 2 Rr 2 (r 4 R ) AX 2 BX 2 CX 2 K K K (2 R r ) 2 where K 2rs 2R r This is our MANEEAL’s Identity Now a( s b)( s c) abc sin 2 A / 2 4 R sin 2 A / 2 2 R sin 2 A / 2 = 2rs(2 R r ) 2 rs(2 R r ) 2 ( 2 R r ) 2R r Hence the other form of MANEEAL’s identity is 374 A few minutes with a new triangle centre coined as “ VIVYA’S POINT” Vm X 2 2 R sin 2 A / 2 2 R sin 2 B / 2 2 R sin 2 C / 2 2 Rr 2 (r 4 R) AX 2 BX 2 CX 2 2R r 2R r 2R r (2 R r )2 Corollaries : I. If Vivya’s point (Vm) and circumcentre (S) of a triangle then 2R r VmS² = R R 2r 2 R r 2 Proof 1. If we put instead of X as S (Circumcentre) in VmX using the previous theorem then we can find the distance between our ‘Vivya’s point and Circumcentre (S). Vm S 2 a ( s b)( s c ) b( s a)( s c) 2 c( s a )( s b) 2 2 Rr 2 (r 4 R) AS 2 BS CS 2rs (2 R r ) 2 rs( 2 R r ) 2rs (2 R r ) (2 R r ) 2 and we have AS=BS=CS=R= Circum Radius 2 Hence Vm S a ( s b)( s c) 2 b (s a )(s c) 2 c( s a )( s b) 2 2 Rr 2 (r 4 R) R R Rr K K K (2 R r ) 2 2 Rr 2 (r 4 R ) R2 Vm S a( s a)(s c) b( s a)( s c ) c (s a)( s b)- (2 R r ) 2 K 2 Vm S 2 R2 2 Rr 2 (r 4 R) (k ) K ( 2R r ) 2 Vm S 2 R 2 2 Rr 2 (r 4 R) (2 R r ) 2 R ( 2 R r ) 2 2r 2 ( r 4 R) Vm S 2 R (2 R r ) 2 Vm S 2 R R[4 R 2 r 2 4 Rr ] 2r 2 (r 4 R) (2 R r ) 2 Vm S 2 R 4 R 3 Rr 2 4 R 2 r 2r 3 8 Rr 2 (2 R r ) 2 Vm S 2 R 4 R 3 4 R 2 r 7 Rr 2 2r 3 (2 R r ) 2 Vm S ² = R 2R r R 2R r 2 2 4 R ³ 4 R ²r 7 Rr ² 2r ³ R 2r 2R r 2 R R 2r 2 R r 2R r 375 2 VIJAY DASARI 2R r VmS² = R R 2r 2 R r II. 2 If Vivya’s point (Vm) and incentre (I) of a triangle then VmI² = 4 Rr 2 R 2r 2R r 2 Proof : If we put instead of X in VmX2 as I (In centre) we can get a relation to find the distance between our “Vivya’s point and Incentre. VmI = 2 2 R sin 2 A / 2 2R r AI 2 2 R sin 2 B / 2 2R r BI 2 2 R sin 2 C / 2 2R r 2 Rr 2 (r 4 R ) 2R r 2 And we know that AI r r r , BI , CI r 2 AI 2 sin 2 A BI 2 sin B CI 2 sin 2 C 2 2 2 sin A / 2 sin B / 2 sin C / 2 2 Rr 2 2 Rr 2 2 Rr 2 2 Rr 2 (r 4 R) Vm I = 2R r 2 R r 2 R r (2 R r ) 2 2 Vm I 2 = 6 Rr 2 2 Rr 2 (r 4 R) 2R r (2 R r ) 2 Vm I 2 = 2 Rr 2 [3(2 R r ) (r 4 R )] 2R r 2 Rr 2 [ 2 R 4r ] Vm I = 2( R r ) 2 2 2 Vm I 4 Rr 2 R 2r 2R r 2 Vm S 2 IS 2 2 R III. The relation between the VmS, VmI, IS is Vm I 2 r 2 2 Proof : We know that SI =R -2Rr=R[R-2r] where S, I are circum centre, In centre. By previous theorems we have 2 4 Rr 2 R 2r 2 2R r V I 2 VmS² = R R 2r & m 2R r 2 R r 2R r (2 R r ) 2 [Vm S 2 SI 2 ] = Now R R 2 R 4 R 3 4 R 2 r 7 Rr 2 2 r 3 R R 2r 2 2 R r = [4R3-4R2r-7R2-2r3-(R-2r)(2R-r)2] = 4R3-4R2r-7Rr2-2r3-(R-2r)(4R2+r2-4Rr) 376 Int. Jr. of Mathematical Sciences and Applications = 4R3-4R2r-7Rr2-2r3-[4R3+Rr2-4R2r-8R2r-2r3+8Rr2] = 8R2r-16Rr2 = 8Rr(R-2r) 8Rr 2 R 2 r 2 R r 2 2 2R r r 2 4 Rr 2 R 2r 2 2 R r = (V I2) = 2R r m r 2 (2R-r)2 r 2 (2 R r ) 2 [VmS2-IS2]= (2R-r)2 (VmI2) r R r(VmS2 – IS 2) =2R(VmI2) r Vm S 2 IS 2 2 R Vm I 2 IV. Vm S 2 IS 2 2 R Vm I 2 r If VmS, VmI are the distances from Vivya’s point (Vm) to circumcentre (S), Incentre V S m (I) then V I m 2R r 2r Proof : We know by previous theorem 2 2 2 R r V I 2 4 Rr R 2r 2 2 R R 2 r m 2 VmS² = , 2R r & SI =R -2Rr=R[R-2r] 2 R r Vm S 2 So clearly we have R 2R r 2 Vm I 2 4 Rr 2 Vm S V I m 2 R r 2r Vm S 2 R r Vm I 2r V. r If We is the Weill’s point then WeI²= 1 3R 2 R² 2Rr Proof : If We is a Weill’s point and X is any point in the plane of le , then we havee 377 A few minutes with a new triangle centre coined as “ VIVYA’S POINT” We X 2 1 s b s c r AX 2 r 4R 2r 3R 3 c b 9R Put X = S where AS BS CS R We S2 1 s b s c 2 r AS r 4 R 2r 3R 3 c b 4R 1 s b s c r R ² r 4R 2r 3R 3 c b 9R 1 s b s c s a s b s c s a r R² r 4R 2r 3R 3 c b b a a c 9 R 1 r R ² 1 1 1 r 4R 2r 3R 3 9R R² r r 4R 2r 3R 9R 9 R³ r 2r ² 3Rr 8Rr 12R ² 9R 9 R³ 2r ³ 11Rr ² 12R r² 9R R² 2Rr 9R ² r ² 6Rr 9 R² 3R r 3 R 2 3R r 3 R 2 R² 2Rr R² 2Rr 2 3R r SI ² 3 R 2 3R r r SI 1 SI 3R 3R r WeS² 1 SI 3R 2 2 r WeS 1 SI 3 R 378 VIJAY DASARI VI. If We is the Weill's point then We I 2 r 2 R 2r 9R Proof : If We is a Weill’s point and X is any point in the plane of le , then we havee We X 2 1 s b s c r AX 2 r 4R 2r 3R 3 c b 9R r Put X=I and AI = Sin A r ² s a 2 r BI = Sin B 2 r ² s b 2 2 r CI = Sin C r ² s c 2 2 1 sb s c 1 sb s c r 2 2 So We I 3 c b AI 9 R r 4 R 2r 3R 2 r = 3 c b r ² s a 9 R r 4R 2r 3R 1 sb sc sa sb sa s c = 3 r ² c b b a c a 2 2 2 2 c b c a 2 2 b a s a s b s a s c s b s a s b s c s c s a s c s b r r 4 R 2r 3R 9R 1 1 s a s b 1 s b s c r ² 3 s a s b s b s c 3 3 c 3 a 1 s c s a r s c s a r 4R 2r 3R 3 b 9R 1 r r ² s a s b s b s c s c s a r 4R 2r 3R 3 9R 1 r r 2 3S² 2S a b c ab bc ca r 4R 2r 3R 3 9R 379 Int. Jr. of Mathematical Sciences and Applications 1 r r ² 3s ² 2s a b c s ² r ² Rr 4 r R4 r2 R3 3 9R 1 r r ² r ² 4Rr r 4R 2r 3R 3 9R 4r ² 4Rr r r 4 R 2r 3R 3 9R 12r ²R 12R ²r 2r ³ 11 r R ² 12 R r² 9R r ²R 2r ³ r ² R 2r r ² R ² 2Rr 9R 9R 9 R² We I ² We I VII. r ² R ² 2Rr 9 R² 2 r r SI ² SI 3R 3R 2 r SI 3R 2 If We is the Weill's point and V m is vivya's point then VmWe r 2 ( r 4 R)2 ( R 2 2 Rr ) 9 R 2 (2 R r )2 Proof : we have by theorem 2 Vm X 2 a( s b)( s c ) 2 Rr 2 (r 4 R) AX 2 K (2 R r )2 Put X= We and AWe 2 1 s a s c 2 1 s a s b 2 r b c r 4 R 2r 3R 3 c a 3 b a 9R BWe 2 1 s b s c 2 1 s b s a 2 r a c r 4R 2r 3R 3 c b 3 a b 9R CWe 2 1 s c s a 2 1 s c s b 2 r b a r 4R 2r 3R 3 a c 3 b c 9R so VmWe 2 a( s b)( s c ) 2 Rr 2 (r 4 R) AWe 2 K (2 R r )2 2 Rr 2 (r 4 R) a( s b)( s c ) 1 s c s a 2 1 s c s b 2 r b a r 4 R 2 r 3 R (2R r )2 3 a K c 3 b c 9R 2 2 1 s b s c c 2 s c s b b2 ac s a s b s c ab s a s b s c 3K 380 A few minutes with a new triangle centre coined as “ VIVYA’S POINT” = 1 1 s b s c c 2 s c b 2 s b s a s b s c ab ac 3K 3K r 2 Rr 2 ( r 4 R) r 4 R 2r 3 R a s b s c 9 RK (2 R r )2 r 2 Rr 2 (r 4 R) r 4 R 2r 3 R s2 ( a b c ) 2 s(ab bc ca) 3abc 9 RK (2 R r )2 1 2 s b s c c 2 s c b 2 s b s a s b s c ab bc ac 3K 3K r 2 Rr 2 ( r 4 R) r 4 R 2r 3 R s2 (2 s) 2 s(r 2 s 2 4 Rr ) 34 Rrs 9 RK (2 R r )2 Where K = 2rs (2R-r) by brute force algebraic manipulation we get 2 2 VmWe r 2 r 4 R R2 2 Rr 9 R2 2 R r 2 2 r 2 r 4 R SI 2 = 9 R2 2 R r 2 r r 4 R SI = 3R 2R r 2 (where SI is the distance between the circumcenter and incenter) r r 4 R SI hence VmWe 3R 2 R r Theorem 3 : Prove that Vivya’s point (Vm), circumcentre (S), Incentre (I) are collinear. The line through these points are called as Vivya’s line. Proof: Method - I By previous theorems 2 We have Vm s R R 2r R R 2r 2 2 2r 2R r and Vm I 2 2 2R r 2R r 2 So clearly Vm s 2 , Vm I 2 We have Vm S 2 R r Vm S 2 IS 2 2R and 2 Vm I 2r Vm I r Vm S 1 2 R 1 Vm S 2 IS 2 2 R and 2 Vm I r Vm I 2 r 2 381 VIJAY DASARI Vm S 1 Vm S 2 IS 2 1 Vm I 2 Vm I 2 2 2Vm I .Vm S Vm S 2 IS 2 Vm I 2 2 IS 2 Vm S Vm I Vm S Vm I IS Method - II We know that Vm S R R 2r 2R r 2R r Vm I R R 2r 2r 2R r IS R R 2r 2 R r 2r 2R r Vm I IS R R 2r R R 2r = Vm s 2R r 2 R r The Vivya’s point (Vm), circumcentre (S), Incentre (I) are collinear. and Vm I 2r IS 2R r Incentre (I) Divides the line joining of Vm , S in the ratio 2r : 2R-r Theorem : 4 Prove that weill's point also lies on the vivya's line. Proof : we have r r 4 R SI VmWe 3R 2 R r r WeS 1 SI 3 R Vm I 2r SI 2R r Vm S 2R r SI 2R r SI ( R 2 2 Rr ) consider 382 Int. Jr. of Mathematical Sciences and Applications VmWe We I r r 4R 2R r SI 3R 2R r 2r SI 2R r = Vm I VmWe We I Vm I ............... 1 and similarly consider We I IS r r SI SI 1 SI 3R 3R WeS We I IS WeS ................... 2 From (1), Vm , We and I are collinear and We lies in between Vm and I From (2), We, I, S are collinear and I lies in between We and S i.e., Vm, We, I, S are collinear and they lies on line Join of Vm, I, S. So The point We is also lies on the Vivya's line Now. VmWe : WeI : IS r r 4 R 3R 2 R r SI : r SI : SI 3R r r 4 R : r 2 R r : 3 R i.e., the points Vm, We, I, S are collinear and VmWe : WeI : IS = r(r+4R) : r(2R-r) : 3R r r 4R Vm 3R We Vivya's line r 2R r I S In the similar argument what ever we adopted by using 'Maneeal's Identity', we can also prove that the Vivya's point lies on the line join of different triangle centres. Such as Vivya's point also lies on the line join of spieker centre and schiffler point. (I am omitting proof of this statement) etc., 383 A few minutes with a new triangle centre coined as “ VIVYA’S POINT” Proving of famous Inequalities using Vivya’s point : I. If R, r are circumradii, inradii then prove that Proof : 2 We already proved that Vm I Since Vm I 2 o R 2 [Euler’s inequality] r 4 Rr 2 ( R 2r ) (2 R r )2 4 Rr 2 ( R 2r ) o (2 R r )2 4 Rr 2 R 2r o o 2 2( R r ) R 2r R 2 r Which is a famous Euler’s Inequality. II. If R, r are circumradii, inradii then prove that 4 R 3 4 R 2 r 7 Rr 2 2r 3 [Maneeals inequality] Proof : We already proved that Vm S 2 R 2R r 2 4 R 3 4 R 2 r 7 Rr 2 2r 3 Since Vm S 2 o R 2R r 2 4 R 3 4 R 2 r 7 Rr 2 2 r 3 o R o 2 4 R 4 R r 7 Rr 2r o 2 R r 3 2 2 3 4 R 3 4 R 2 r 7 Rr 2 2 r 3 Which is a famous maneeal’s inequality III. Prove that Euler’s inequality using Maneeals inequality. Proof : Consider 4 R 3 4 R 2 r 7 Rr 2 2 r 3 384 VIJAY DASARI 4 R 3 4 R 2 r 7 Rr 2 2 r 3 o 4 R 3 4 R 2 r Rr 2 8 R 2 r 8 Rr 2 2r 3 o R[4 R 2 4 Rr r 2 ] 2 r[4 R 2 4 Rr r 2 ] o R 2r 4 R 2 4 Rr r 2 o 2 R 2r 2 R r o R 2r o (2 R r ) 2 o R 2r R 2 r Which is a famous Euler’s inequality IV. Prove that if VmS, VmI, IS are the distances from Vivya’s point to Circumcentre, Incentre and Circumcentre to Incentre respectively. Then prove that Vm S 2 4Vm I 2 IS 2 . Proof : We already proved that r (Vm s 2 Is 2 ) 2 R Vm I v But So Vm s 2 IS 2 2 R Vm I 2 r R 2R 2 4 r r Vm S 2 IS 2 4 Vm I 2 Vm S 2 IS 2 4Vm I 2 Vm S 2 4Vm I 2 IS 2 These are the very few properties for the Vivya’s point. 385 Int. Jr. of Mathematical Sciences and Applications ACKNOWLEDGMENT The author is grateful to the creators of the free Geogebra software, without which this work would have been impossible and the author is would like to thank an anonymous refree for his/her kind comments and suggestions, which lead to a better presentation of this paper. References : 1. Applications of Stewart's theorem in Geometric Proofs. — WILLIAM CHAU 2. College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle (Dover Books on Mathematics) by Nathan Altshiller-Court 3. en.wikipedia.org/wiki/encyclopedia-of-triangle-centre 4. en.wikipedia.org/wiki/stewart's theorem 5. forum geometricorum 6. H.S.M Coxeter, Introduction to Geometry, John Wiley 8 Sons, NY, 1961 7. H.S.M Coxeter and S.L.Greitzer, Geometry Revisited, MAA, 1967 8. MODERN GEOMETRY OF A TRIANGLE by WILLIAM GALLATLY 9. Ross Honsberger: Episodes in Nineteenth and Twentieth Century Euclidean Geometry, USA 1995. 387
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