INEQUALITY AND IDENTITY (M, N ) DamVanNhi Pedagogical University Ha Noi 136 Xuan Thuy Road, Hanoi, Vietnam Email: damvannhi@yahoo.com 2010 Mathematics subject classification: 26D05,26D15,51M16 Abstract In this paper we introduce the inequalty and identity (M, N ), of which Hayashi is a spcial case: Inequality, Identity (M, N ) And after then we will present some interesting applications Keywords Hayashi’s inequality, point, triangle, polygon Introduction In Euclidian geometry, the Hayashi’s Inequality in R2 states: Suppose given a triangle ABC of the lengths of sides a, b, c Then, with any point M, we have an inequality aM B.M C + bM C.M A + cM A.M B abc (see [3, pp 297, 311]) In this paper we propose to give a new inequality and it’s identity, which is a generalization of the above inequality, and after then we want to give some interesting applications about triangle Inequality and Identity (M, N ) Now we prove an inequality, of which Hayashi is a special case Use this result we can give some new interesting inequalities We give the following: Theorem 2.1 Let A1 A2 An be a polygon, s be an integer, s < n, and arbitrary points N1 , N2 , , Ns , M in euclidean plane R2 we have the following inequality s s M Nj j=1 M Ai i=1 Ak Nj n j=1 n k=1 Ak Ai M Ak i=k 16 , (M, N ) (i) If s = 0, we have Hayashi inequality (ii) If n = 3, s = 1, and A, B, C, N belong to the circle with the centere M we have the inequality aAN + bBN + cCN 4SABC Proof Suppose that Ak have affixe ak , M has affixe z and Nh affixe zh Using the s s (ak − zj ) n j=1 (z − zj ) = Lagrange interpolation formula, we have j=1 (ak − ) k=1 (z − ) and i=k i=k s s |z − zj | deducing |z − | |ak − zj | n j=1 j=1 |ak − ||z − ak | k=1 i=1 From this, we deduce geometric inequal- i=k ity s s M Nj Ak Nj n j=1 n j=1 M Ai k=1 i=1 (i) When s = we have Ak Ai M Ak i=k M Nj = = Hayashi inequality for the polygon Ak Nj and the inequality (M, N ) becomes the n A A M Ak M Ai k=1 i=k k i n i=1 (ii) When n = 3, s = and A, B, C, N belong to the circle with the centere M we have abc the inequality aAN + bBN + cCN or aAN + bBN + cCN 4SABC R Remark 2.2 Denote N as the center of circumcircle From the iequality aAN +bBN + cCN 4SABC by the inequality (M,N) (ii) we deduce R(a + b + c) 2r(a + b + c) or R 2r [Euler] Corollary 2.3 Suppose that O, I and G are respectively the centre of circumcirle and incircle of ABC Denote the radii of circumcircles of the triangles GBC, GCA, GAB by R1 , R2 , R3 , respectivly Let , rb , rc be the the radius of circumcircle of the triangles IBC, ICA, IAB, and let R1 , R2 , R3 be the radii of circumcircles of the triangles OBC, OCA, OAB, respectivly We will have (i) R2 abc a+b+c (ii) R1 + R2 + R3 3R (see [1]) (iii) rb rc + + hb hc R with , hb , hc are the lengths of altitudes of ∆ABC r (iv) R1 x R2 y R3 z + + R when ∆ABC is not obtituse and x, y, z are the distances hb hc from O to the sides 17 Proof (i) Applying the inequality (M,N) (ii) we obtain aOB.OC+bOC.OA+cOA.OB abc abc or R2 a+b+c (ii) Applying the inequality (M, N ) we obtain aGB.GC + bGC.GA + cGA.GB abc SABC abc abc abc Since aGB.GC = 4R1 SGBC = 4R1 = 4R1 = R1 , bGC.GA = R2 and 3.4R 3R 3R abc abc abc abc therefore R1 + R2 + R3 abc or R1 + R2 + R3 3R cGA.GB = R2 3R 3R 3R 3R (iii) Applying the inequality (M, N ) we have aIB.IC + bIC.IA + cIA.IB abc Since rabc rabc rb rabc aIB.IC = 4ra SIBC = 2ra = = , bIC.IA = and cIA.IB = 4R R hb R rc rabc rabc rabc rabc rb rc R there is + + abc or + + hc R R R R hb hc r (iv) Applying the inequality (M, N ) we have aOB.OC +bOC.OA+cOA.OB abc Since x abc x abc y abc = R1 , bOC.OA = R2 and aOB.OC = 4R1 SOBC = 2R1 xa = 4R1 4R R hb R z abc R x abc R2 y abc R3 z abc Rx Ry Rz cOA.OB = R3 we have + + abc or + + hc R R hb R hc R hb hc R Proposition 2.4 Suppose given a triangle ABC with the lengths of sides a, b, c respectively and R is the radius of circumcircle of ABC Let’s I, Ja , Jb , Jc are the centres of incircle and escribed circles of ∆ABC, respectively Then, with any point M, we have abcM I M A.M B.M C √ MI a + b + c (ii) M A.M B.M C (i) (iii) (iv) aAI bBI cCI + + MA MB MC √ √ √ b+c−a c+a−b a+b−c √ + √ + √ caM B bcM A abM C M Ja + M Jb + M Jc M A.M B.M C AJa + AJb + AJc BJa + BJb + BJc CJa + CJb + CJc + + bcM A caM B abM C M Ja M Jb + M Jb M Jc + M Jc M Ja AJa AJb + AJb AJc + AJc AJa M A.M B.M C bcM A BJa BJb + BJb BJc + BJc BJa CJa CJb + CJb CJc + CJc CJa + + caM B abM C Proof (i) Applying the inequality (M, N ) we have MI M A.M B.M C AI BI CI + + bcM A caM B abM C ca(c + a − b) ab(a + b − c) bc(b + c − a) , IB = , IC = therefore (ii) Since IA2 = a +√b + c a+b+c √ √a + b + c √ MI a + b + c b+c−a c+a−b a+b−c √ + √ + √ M A.M B.M C caM B bcM A abM C 18 (iii) Applying the inequality (M, N ) to n = 3, s = 1, we have the three inequalities M Ja M A.M B.M C M Ja M A.M B.M C M Jc M A.M B.M C AJa BJa CJa + + bcM A caM B abM C AJb BJb CJb + + bcM A caM B abM C AJc BJc CJc + + bcM A caM B abM C M Ja + M Jb + M Jc On adding the three inequalities, we find the inequality M A.M B.M C AJa + AJb + AJc BJa + BJb + BJc CJa + CJb + CJc + + bcM A caM B abM C (iii) Applying the inequality (M, N ) to n = 3, s = 1, we have the three inequalities M Ja M Jb M A.M B.M C M Jb M Jc M A.M B.M C M Jc M Ja M A.M B.M C AJa AJb BJa BJb CJa CJc + + bcM A caM B abM C AJb AJc BJb BJc CJb CJc + + bcM A caM B abM C AJc AJa BJc BJa CJc CJa + + bcM A caM B abM C M Ja M Jb + M Jb M Jc + M Jc M Ja On adding the three inequalities, we find the inequality M A.M B.M C AJa AJb + AJb AJc + AJc AJa BJa BJb + BJb BJc + BJc BJa + + bcM A caM B CJa CJb + CJb CJc + CJc CJa abM C Corollary 2.5 Let be the triangle ABC with the lengths of sides a, b, c and R is the radius of circumcircle of ABC Denote O, H the center of circumcircle and the orthocenter of ABC Then, with any point M, we have the inequality: aAH bBH c CH abcM O.M H + + RM A.M B.M C MA MB MC if M belongs to the circle with the centre O and the radius R, we obtain the inequality √ √ √ abcM H a 4R2 − a2 b 4R2 − b2 c 4R2 − c2 + + M A.M B.M C MA MB MC Proof Applying the inequality (M, N ) to the case n = 3, s = we have the inequality: M O.M H M A.M B.M C AO.AH BO.BH CO.CH + + bcM A caM B abM C abcM O.M H aAH bBH c CH + + Since AH = Thus, we obtain the inequality RM A.M B.M C MA MB MC √ √ √ abcM H 4R2 − a2 , BH = 4R2 − b2 and CH = 4R2 − c2 we obtain M A.M B.M C √ √ √ a 4R2 − a2 b 4R2 − b2 c 4R2 − c2 + + MA MB MC 19 Corollary 2.6 Suppose given the triangle ABC with the lengths of sides a, b, c, respectively I, G, H the center of inscribed circle and the barycenter and the orthocenter of ∆ABC Then, with any point M, we always have the inequality (i) abcM I M A.M B.M C aAI bBI c CI + + MA MB MC (ii) abcM G2 M A.M B.M C aAG2 bBG2 c CG2 + + MA MB MC (iii) abcM H M A.M B.M C a(4R2 − a2 ) b(4R2 − b2 ) c(4R2 − c2 ) + + MA MB MC Proof Applying the inequality (M,N) to the case n = 3, s = 2, N1 ≡ N2 ≡ N, we have MN2 M A.M B.M C AN BN CN + + bcM A caM B abM C aAI bBI c CI abcM G2 + + and MA MB MC M A.M B.M C abcM H aAG2 bBG2 c CG2 + + Then we have (i) and (ii) If N ≡ H we have (iii): MA MB MC M A.M B.M C a(4R2 − a2 ) b(4R2 − b2 ) c(4R2 − c2 ) + + MA MB MC Therefore, we obtain the inequality abcM I M A.M B.M C Proposition 2.7 Suppose the polygon A1 A2 An is inscribed in the circle with the center O and radius R Then, with any s < n points N1 , , Ns in the plane A1 A2 An , s s Ak Ni n ONi i=1 n we always have the inequality i=1 k=1 Ak Ai Rn−1 When R = we obtain i=1,i=k s n Ak Ni i=1 n k=1 s ONi Ak Ai i=1 i=1,i=k When n = 3, s = and a1 = A2 A3 , a2 = A3 A1 , a3 = A1 A2 we obtain the inequality ON a1 A1 N + a2 A2 N + a3 A3 N 4SA1 A2 A3 R s s n Proof The inequality Ak Ni ONi i=1 n k=1 i=1 Rn−1 Ak Ai follows from the inequality (M, N ) i=1,i=k s n when M ≡ O With R = we have Ak Mi i=1 n k=1 OMi Ak Ai i=1,i=k 20 s i=1 Example Giving the triangle ABC with the lengths of sides a, b, c and R is the radius of circumscribed circle; r1 , r2 , r3 are the radius of escribed circles Let’s da , db , dc the distances from the center of circumscribed circle to the center of escribed circles Then, with any point D belong to the circumscribed circle of ∆ABC we always have the inequality: √ (i) (ii) + da db dc a+b+c R3 √ √ bc ca ab DJa DJb DJc √ √ + √ + √ + x b + c − a y c + a − b z a + b − c xyz a + b + c √ √ √ (R + 2r1 )(R + 2r2 )(R + 2r3 ) bc ca ab √ +√ +√ R (a + b + c) b+c−a c+a−b a+b−c √ DJa DJb DJc √ xyz a + b + c Proof (i) We consider M ≡ O bc(a + b + c) ca(a + b − c) ab(a − b + c) Ja A2 = , Ja B = , Ja C = b+c−a b+c−a b+c−a ca(a + b + c) ab(b + c − a) bc(b + a − c) 2 Since Jb A = there, Jb B = , Jb C = c + a − b c + a − b c + a − b bc(c + a − b) ca(c + b − a) ab(a + b + c) Jc A2 = , Jc B = , Jc C = a+b−c a+b−c a+b−c da db dc √ √ √ √ bc ca ab DJa DJb DJc a+b+c √ √ fore we obtain + √ + √ + R x b + c − a y c + a − b z a + b − c xyz a + b + c (ii) Since d2a = R2 + 2Rr1 , d2b = R2 + 2Rr2 , d2c = R2 + 2Rr3 therefore √ √ √ (R + 2r1 )(R + 2r2 )(R + 2r3 ) bc ca ab √ √ √ + + + R3 (a + b + c) b+c−a c+a−b a+b−c DJa DJb DJc √ xyz a + b + c Now, we illustrate the advantage of this identity by addressing several important problems of elementary Geometry Firstly, we use the functions sin and cosin to create the identity under the form of trigonometry Without generality, we can assume that the radius R of the circle C equal to Suppose that every point Ak has affixe ak = cos αk + i sin αk , and M has affixe z = cos u + i sin u and every Nh has affixe zh = cos uh + i sin uh Following the interpolation formula 21 s s (z − zj ) Lagrange, we have j=1 n (ak − zj ) n j=1 = (z − at ) (z − ak ) k=1 u − uj 2i sin e j=1 s (ak − at ) t=k t=1 or u − αt 2i sin e t=1 n i(u + uj ) i(u + αt ) i(αk + uj ) αk − uj 2i sin e j=1 i(u + αk ) i(αk + αt ) αk − αt u − αk 2 e 2i sin e 2i sin 2 t=k s n = k=1 iuj iαt We reduce all the factors 2i, e and e And we receive the relationship u − uj j=1 = n u − αt sin t=1 s sin αk − uj i(s + − n)(αk − u) j=1 αk − αt e u − αk sin sin t=k s n k=1 sin From this relationship, we deduce identities below: u − uj j=1 = n u − αt sin t=1 s sin αk − uj (s + − n)(αk − u) j=1 u − αk αk − αt cos sin sin t=k s n k=1 sin αk − uj (s + − n)(αk − u) j=1 and sin = With this result, we have u − α α − α k k t k=1 sin sin t=k just built the identities under the form of trigonometry and geometry for the inequality (M, N ) as below: s n sin Proposition 2.8 Assume that the polygon A1 A2 An is inscribed in the circle with the center O and radius R Taking s + points N1 , , Ns and M also belonging to this circle C Assuming that the coordinate Ak (cos αk ; sin αk ), k = 1, 2, , n; the coordinate Nj (cos uj ; sin uj ), j = 1, 2, , s and the coordinate M (cos u; sin u) Then, we will have these identities s s u − uj αk − uj sin sin n 2 (s + − n)(αk − u) j=1 j=1 (i) n = cos u − α α − α u − αt k k t k=1 sin sin sin 2 t=k t=1 22 αk − uj n (s + − n)(αk − u) j=1 (ii) sin = α − α u − α k t k k=1 sin sin t=k s sin αk − u1 (iii) α − αt = when n = 3, s = k k=1 sin t=k sin n−1 u − uj αk − uj sin n 2 j=1 j=1 (iv) n = u − αk αk − αt when s = n − u − αt k=1 sin sin sin t=k 2 t=1 n−1 sin n−2 n−2 uj αk − uj αk − uj sin sin n j=1 n j=1 2 αk j=1 = cot (v) n and αk − αt αk − αt = when s = n − αt k=1 k=1 sin sin sin 2 t=k t=k t=1 and u = n−2 sin Remark 2.9 If the quadrilateral ABCD is inscribed in the circle we have DB aDA2 cDC bDB by (iii) or + = Using the relationship ca DA DC DB DA DC + = bc ab DA2 DB.DC.a + DC DA.DB.c = DB DC.DA.b Hence DA2 SDBC + DC SDAB = DB SDCA [Feuerbach] Proposition 2.10 Suppose the polygon A1 A2 An is inscribed in the circle with the radius R = Taking s + points N1 , , Ns and M also belonging to this circle C Assuming that the coordinate Ak (cos αk ; sin αk ), k = 1, 2, , n; the coordinate Nj (cos uj ; sin uj ), j = 1, 2, , s and the coordinate M (cos u; sin u) Then, with the proper choices of + or − we will have the identities s s M Nj (i) j=1 n ± n = k=1 M At Ak Nj j=1 M Ak Ak At cos (s + − n)(αk − u) , (M, N ) t=k t=1 s n (ii) k=1 ± Ak Nj j=1 M Ak Ak At sin (s + − n)(αk − u) = t=k 23 n−2 n−2 M Nj (iii) j=1 n n ± j=1 = M At t=1 n−2 Ak Nj Ak At k=1 αk cot and n ± Ak Aj j=1 Ak At k=1 t=k = t=k Corollary 2.11 Asuming that the points A1 , A2 , , An , M in order belong to the circle C with the center O We convent n + := Then, we have the identities n (i) (−1)r r=1 cos(n − 1)∠M Ar+1 Ar = Ar Ak M Ar k=r n (ii) (−1)r r=1 n M Ak k=1 sin(n − 1)∠M Ar+1 Ar = M Ar Ar Ak k=r Proof These identities follow from the identity (M, N ) with s = Corollary 2.12 Let the quadrilateral ABCD be inscribed in the circle C with the center O Let a = BC, b = CA, c = AB Then, we have identities: (i) a cos(OD, OA) b cos(OD, OB) c cos(OD, OC) abc − + =− DA DB DC DA.DB.DC a sin(OD, OA) c sin(OD, OC) b sin(OD, OB) + = DA DC DB Proof These identities follow from the identity (M, N ) with n = 3, s = (ii) Conjecture Despite of not having been proven yet, these following results are still hoped to be true: Open Problem 3.1 Suppose given a triangle ABC with the lengths of sides a, b, c respectively and R is the radius of circumcircle of ABC Let’s I, Ja , Jb , Jc are the centres of incircle and escribed circles of ∆ABC, respectively Then, with any point M, we have M Ja M Jb M Jc AJa AJb AJc BJa BJb BJc CJa CJb CJc + + (i) M A.M B.M C bcM A caM B abM C √ √ √ M Ja M Jb M Jc bc ca ab √ √ √ √ b+c−a c+a−b a+b−c a+b+c (ii) + + M A.M B.M C MA MB MC Open Problem 3.2 Giving the triangle ABC with the lengths of sides a, b, c and R is the radius of circumscribed circle; r1 , r2 , r3 are the radius of escribed circles Let’s da , db , dc the distances from the center of circumscribed circle to the center of escribed circles Then, with any point D belong to the circumscribed circle of ∆ABC we always have the inequality: 24 √ (i) (ii) da db dc a+b+c R3 √ √ √ bc ca ab √ + √ + √ x b+c−a y c+a−b z a+b−c √ √ √ (R + 2r1 )(R + 2r2 )(R + 2r3 ) bc ca ab √ +√ +√ R(a + b + c) b+c−a c+a−b a+b−c References [1] T Andreescu and D Andrica, Proving some geometric inequalities by using complex numbers, Eduatia Mathematica Vol Nr (2005),19-26 [2] T Hayashi, Two theorems on complex numbers, Thoku Math J., 4(1913/14), pp 68-70 [3] D.S Mitrinovic, J.E Pecaric and V Volenec, Recent Advances in Geometric Inequalities, Acad Publ., Dordrecht, Boston, London 1989 25 ... y c + a − b z a + b − c xyz a + b + c (ii) Since d2a = R2 + 2Rr1 , d2b = R2 + 2Rr2 , d2c = R2 + 2Rr3 therefore √ √ √ (R + 2r1 )(R + 2r2 )(R + 2r3 ) bc ca ab √ √ √ + + + R3 (a + b + c) b+c−a c+a−b... A.M B.M C MA MB MC √ √ √ abcM H 4R2 − a2 , BH = 4R2 − b2 and CH = 4R2 − c2 we obtain M A.M B.M C √ √ √ a 4R2 − a2 b 4R2 − b2 c 4R2 − c2 + + MA MB MC 19 Corollary 2. 6 Suppose given the triangle... G2 M A.M B.M C aAG2 bBG2 c CG2 + + MA MB MC (iii) abcM H M A.M B.M C a(4R2 − a2 ) b(4R2 − b2 ) c(4R2 − c2 ) + + MA MB MC Proof Applying the inequality (M,N) to the case n = 3, s = 2, N1 ≡ N2