Tạp chí toán học AMM tháng 1 năm 2008

... all n ≥ 1, 1 2F 1/ Fnn+ L 1/ Lnn≤ 2 −Fn +1 F2n.(The Fibonacci and Lucas numbers are given by the recurrence an +1 = an+an 1 , withF0= 0, F 1 = 1, L0= 2andL 1 = 1. ) 11 340. Proposed ... fora = 1orx = 1. If f (x) =nj =1 (1 − xaj) for 0 ≤ x ≤ 1, then0 ≤ f (x) ≤ (1 − xa 1 ) (1 − xa2) ≤ a 1 a2 (1 − x)2.Hence f (x)< ;1 − x as x approaches 1 −. On the other hand, 1 ≥ ... C)=(sin A)2(sin B)2(sin C)2 (1 − cos A) (1 − cos B) (1 − cos C)= (1 − cos2A) (1 − cos2B) (1 − cos2C) (1 − cos A) (1 − cos B) (1 − cos C)= (1 + cos A) (1 +cos B) (1 +cos C) ≤3 + cos A + cos...

Tạp chí toán học AMM tháng 2 năm 2008

... 16 7 12 3456789645978 312 8972 315 64978 312 6452 315 6489745678 912 35648972 31 78 912 3456 312 645978 12 345678978 912 345645678 912 32 315 648978972 315 645648972 31 312 645978978 312 645645978 312 12 34567898972 315 64645978 312 5648972 31 312 64597878 912 3456978 312 64545678 912 32 315 64897 12 3456789978 312 6455648972 31 8972 315 6445678 912 3 312 645978645978 312 2 315 6489778 912 3456Editorial ... squares. 12 345678945678 912 378 912 3456 312 645978645978 312 978 312 6452 315 648975648972 31 8972 315 64 12 34567895648972 31 978 312 645645978 312 78 912 34562 315 648978972 315 64 312 64597845678 912 3February 2008] PROBLEMS AND SOLUTIONS 16 7 12 3456789645978 312 8972 315 64978 312 6452 315 6489745678 912 35648972 31 78 912 3456 312 645978 12 345678978 912 345645678 912 32 315 648978972 315 645648972 31 312 645978978 312 645645978 312 12 34567898972 315 64645978 312 5648972 31 312 64597878 912 3456978 312 64545678 912 32 315 64897 12 3456789978 312 6455648972 31 8972 315 6445678 912 3 312 645978645978 312 2 315 6489778 912 3456Editorial ... − 1) mutuallyorthogonal p2× p2Latin Sudoku squares. 12 345678945678 912 378 912 3456 312 645978645978 312 978 312 6452 315 648975648972 31 8972 315 64 12 34567895648972 31 978 312 645645978 312 78 912 34562 315 648978972 315 64 312 64597845678 912 3February...

Tạp chí toán học AMM tháng 6 2008

... http://www.cut-the-knot.org/Gener-alization/ceva.shtml). Now[A 1 B3B2]= 1 2A 1 B3· A 1 B2sin a 1 = 1 2t3 1 + t3A 1 A2· 1 1 + t2A 1 A3sin a 1 =t3 (1 + t3) (1 + t2)[A 1 A2A3].June–July 2008] PROBLEMS AND SOLUTIONS ... have[A 1 A2B 1 ][A 1 B 1 A3]= t 1 ,[PA2B 1 ][PB 1 A3]= t 1 , so[PA 1 A2][PA3A 1 ]= t 1 ,and similarly for t2, t3.Now writea 1 = x 1 + x2a2= y 1 + y2a3= z 1 + z2dividing ... [A2B 1 B3] and [A3B2B 1 ] we get[B 1 B2B3]=[A 1 A2A3]−[A 1 B3B2]−[A2B 1 B3]−[A3B2B 1 ]=2 (1 + t 1 ) (1 + t2) (1 + t3)[A 1 A2A3].Subtracting areas, we have[A 1 A2B 1 ][A 1 B 1 A3]=...