The Problem Corner. Edited by Pat Costello

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1 The Problem Corer Edited by Pat Costello The Problem Corer ivites questios of iterest to udergraduate studets. As a rule, the solutio should ot demad ay tools beyod calculus ad liear algebra. Although ew problems are preferred, old oes of particular iterest or charm are welcome, provided the source is give. Solutios should accompay problems submitted for publicatio. Solutios of the followig ew problems should be submitted o separate sheets before March 15, 018. Solutios received after this will be cosidered up to the time whe copy is prepared for publicatio. The solutios received will be published i the Sprig 018 issue of The Petago. Preferece will be give to correct studet solutios. Affirmatio of studet status ad school should be icluded with solutios. New problems ad solutios to problems i this issue should be set to Pat Costello, Departmet of Mathematics ad Statistics, Easter Ketucky Uiversity, 51 Lacaster Aveue, Richmod, KY ( pat.costello@eku.edu, fax: (859) ) NEW PROBLEMS Problem 798. Proposed by the editor. I 00, Britey Galliva (high school juior) foud a formula for paper foldig ad maaged to do 1 folds of a log sheet of toilet paper. She foud that L t where t represets the thickess of the material to be folded, L is the legth of the paper to be folded ad is the umber of folds desired (i oly oe directio). Suppose you tape together sheets of stadard 8.5 x 11 copier paper (thickess.0035 ) ed to ed, how may sheets would be eeded to be able to fold the log taped sheet 14 times? Problem 799. Proposed by Daiel Sitaru, Theodor Costescu Natioal Ecoomic College, Drobeta Turu Severi, Mehediti, Romaia. Prove that if a,b,c (0,] the b( a a a b c 3 c b c a Problem 800. Proposed by Daiel Sitaru, Theodor Costescu Natioal Ecoomic College, Drobeta Turu Severi, Mehediti, Romaia.

2 Prove that if a R, the a5 a8 a a11 x x x x l(1 e ) dx l(1 e ) dx l(1 e ) dx l(1 e ) dx a3 a6 a a9. Problem 801. Proposed by Jose Luis Diaz-Barrero, Barceloa Tech-UPC, Barceloa, Spai. Compute 1 k lim k k 1 Problem 80. Proposed by Jose Luis Diaz-Barrero, Barceloa Tech-UPC, Barceloa, Spai. Let 1 be a iteger. Prove that 1 F F L k1 k k 1 k k1 k1 where F ad L are the th Fiboacci ad Lucas umbers defied by F1 = F = 1 ad F = F-1 + F- for 3 ad by L1 = 1, L = 3 ad L = L-1 + L- for 3. Problem 803. Proposed by Ovidiu Furdui ad Alia Sitamaria, Techical Uiversity of Cluj-Napoca, Cluj-Napoca, Romaia. Calculate Hm 1 m 1 ( m) where H = 1 + ½ + + 1/ deotes the th harmoic umber. Problem 804. Proposed by D.M. Batietu-Giurgiu, Matei Basarab Natioal College, Bucharest, Romaia ad Neculai Staciu, George Emil Palade School, Buzau, Romaia. Compute the followig limit lim 3! 3!...! 1 ( 1)!!

3 Problem 805. Proposed by D.M. Batietu-Giurgiu, Matei Basarab Natioal College, Bucharest, Romaia ad Neculai Staciu, George Emil Palade School, Buzau, Romaia. Let zk = xk + i yk be a complex umber where k { 1,,, }. Prove that 4 4 xk yk 1 zk k1 k1 Problem 806. Proposed by Marius Draga, Bucharest, Romaia ad Neculai Staciu, George Emil Palade School, Buzau, Romaia. If a1, a,, a > 0 such that k1 a k 1, the prove that 1 1 a a a a / a 11/ a... 11/ a 11/ a 1 Problem 807. Proposed by Titu Zvoaru, Comaesti, Romaia. If A, B, ad C are the agles of a triagle ad = A/, = B/, = C/, prove that 6(1 cos Acos Bcos C) si si si (1 8si si si ) 4cos cos cos SOLUTIONS TO PROBLEMS Problem 780. Proposed by Daiel Sitaru, Colegiul Natioal Ecoomic College Theodor Costescu, Drobeta Turu Severi, Mehediti, Romaia. Prove that if a,b,c [1, ), the ab + bc + ca 3 + l(a b b c c a ). Solutio by Richdad Phuc, Uiversity of Scieces, Haoi, Vietam. We have LHS RHS = b(a l a) + c(b l b) + a(c l c) 3 or LHS RHS = (b/a)a(a l a) + (c/b)b(b l b) + (a/c)c(c l c) - 3 Let f(x) = x(x l x) for x 1. The derivative is f (x) = x l x ad f (x) = /x 0 for all x 1 f (x) f (1) = 0 for all x 1. This meas f(x) is strictly icreasig o [1, ). The f(x) f(1) for all x 1. Hece a(a l a) 1, b(b l b) 1, c(c l c) 1. The LHS RHS (b/a) + (c/b) + (a/c) 3 which is 0 by the AM-GM iequality. Equality holds if a = b = c = 1.

4 Also solved by Agel Plaza, Uiversidad de Las Palmas de Gra Caaria, Spai; Hery Ricardo, New York Math Circle, NY; ad the proposer. Problem 781. Proposed by Daiel Sitaru, Colegiul Natioal Ecoomic College Theodor Costescu, Drobeta Turu Severi, Mehediti, Romaia. Prove that if a,b,c (0, ), the 4 4 a ( b c ) / a ( b c ) b ( a c ) c ( a b ) 3 abc. Solutio by the proposer. We prove that if x,y (0, ), the x y x y xy (*). x y We deote u which meas u = x + y ad let v xy v With these otatios, we have u + v = x +xy + y = (x+y) We ca rewrite (*) as x + y v u or (x + y) (u + v) u + v (u + v) u + v u v uv 0 (u v) 0 Now replace x with x/y ad y with y/x i (*) to get x y ( x / y) ( y / x) x y y x y x xy x y 1 x y xy xy x y x y xy For x = a ad y = b ad multiplyig by c we have 4 4 a b a c b c c abc Aalogously, 4 4 b c b a c a a abc 4 4 c a ad c b a b b abc Addig the last three iequalities gives the desired result.

5 Also solved by Titu Zvoaru, Comaesti, Romaia ad Neculai Staciu, George Emil Palade School, Buzau, Romaia; Ioa Viorel Codreau, Satulug, Maramures, Romaia; Soumitra Moukherjee, Scottish Church College, Chadar Nagore, Idia; Ravi Prakash, Oxford Uiversity Press, New Delhi, Idia. Problem 78. Proposed by Jose Luis Diaz-Barrero, Barceloa Tech-UPC, Barceloa, Spai. Let a, b, c be the legths of the sides of triagle ABC ad ma, mb, ad mc the legths of its medias. Prove that a b c ma mb mc 1. Solutio by Titu Zvoaru, Comaesti, Romaia ad Neculai Staciu, George Emil Palade School, Buzau, Romaia. a b Sice mc, by the AM-GM iequality we have ab a b a b m c Writig the other two similar iequalities ad addig all three gives the desired result. Also solved by Madiso Estabrook, Missouri State Uiversity, Sprigfield, MO; Rovse Pirkulyev, Baku State Uiversity, Sumgait, Azerbaidjia; ad the proposer. Problem 783. Proposed by Jose Luis Diaz-Barrero, Barceloa Tech-UPC, Barceloa, Spai. Fid all real solutios of the followig system of equatios x 3 + x + y = 9 + 3x 3y + 6y + z = 1 + 9y 5z z + x = z. Solutio by the proposer. We ca rewrite the system as 3 y = x 3-3x + x 6 3 z = 3(y 3 3y + y - 6) 3 x = 5(z 3 3z +z 6) Sice t 3 3t + t - 6 = (t 3)(t +), we have 3 y = (x 3)(x +) 3 z = 3(y 3)(y +) 3 x =5(z 3)(z +). Multiplyig these together gives -(x-3)(y-3)(z-3) = 15(x-3)(y-3)(z-3)(x +)(y +)(z +). From this we get 0 = (x-3)(y-3)(z-3)(15(x +)(y +)(z +) + 1)

6 Sice the last factor above is positive, either x = 3, y = 3 or z = 3. If we assume that x = 3, the first equatio says y = 3. Substitutig this ito the secod equatio implies that z=3. The same occurs for startig with y=3 or z=3. So x = y = z = 3 is the oly real solutio. Also solved by Soumava Chakraborty, Softweb Techologies, Kolkota, Idia. Problem 784. Proposed by D.M. Batietu-Giurgiu, Matei Basarab Natioal College, Bucharest, Romaia, Neculai Staciu, George Emil Palade School, Buzau, Romaia. Prove that i ay triagle ABC with BC = a, CA = b, AB = c ad area F, the followig iequalities are true. A B C ( b c )si ( c a )si ( a b )si 4 3 F, C A B ab bc ca F (1 si ) (1 si ) (1 si ) 4 3. Solutio by Ioa Viorel Codreau, Satulug, Maramures. Romaia. A A bcsi A F A We have ( b c )si bc si F sec A A cos cos B B C C Similarly, ( c a )si F sec ad ( a b )si F sec A A The ( b c )si F sec. Usig Jese s Iequality ad that f(x) = sec x o (0, π/) is a covex fuctio, we get A A sec 3sec A 3. Thus ( b c )si 4 3 F. 6 C C absi C C Next ab(1 si ) absi F sec. C cos A A B B Similarly, bc(1 si ) F sec ad ca(1 si ) F sec. C A The ab(1 si ) F sec 4 3 F. Also solved by Soumava Chakraborty, Softweb Techologies, Kolkota, Idia; Ravi Prakash, Oxford Uiversity Press, New Delhi, Idia; Soumitra Moukherjee, Scottish Church College, Chadar Nagore, Idia; ad the proposer.

7 Problem 785. Proposed by Iuliaa Trasca, Scoricesti, Romaia. Show that x, y, z > 0, the x z y x z y x y z 3xyz. x y z Solutio by Soumava Chakraborty, Softweb Techologies, Kolkota, Idia. The iequality is equivalet to (x 6 z 3 + y 6 x 3 +z 6 y 3 ) x 5 y z +y 5 z x +z 5 x y + 3x 3 y 3 z 3. Usig the AM-GM iequality, we have x 6 z 3 + x 6 z 3 + x 3 y 6 3x 5 y z y 6 x 3 + y 6 x 3 + y 3 z 6 3y 5 z x z 6 y 3 + z 6 y 3 + z 3 x 6 3z 5 x y Addig these gives 3(x 6 z 3 + y 6 x 3 + z 6 y 3 ) 3(x 5 y z + y 5 z x + z 5 x y ). Dividig by 3 says x 6 z 3 + y 6 x 3 + z 6 y 3 x 5 y z +y 5 z x + z 5 x y The AM-GM iequality also says x 6 z 3 + y 6 x 3 + z 6 y 3 3x 3 y 3 z 3. Summig the previous two iequalities gives the iequality that is equivalet to the oe of the problem. Also solved by Soumitra Moukherjee, Scottish Church College, Chadar Nagore, Idia; Ioa Viorel Codreau, Satulug, Maramures, Romaia; Titu Zvoaru, Comaesti, Romaia; ad the proposer. Problem 786. Proposed by Thomas Chu, Macomb, Illiois. Prove that if x,y,z > 1, the ( x y z )( x y z) x y z 4xy 4xz 4yz. Solutio by Agel Plaza, Uiversidad de Las Palmas de Gra Caaria, Spai. By chagig variables x = 1+a, y = 1+b ad z = 1+c, the problem reads as: Prove that if a,b,c > 0, the 3 (1 a) 3 a (1 a) 4(1 a)(1 b) 4(1 b)(1 c) 4(1 c)(1 a) Expadig the right-had side ad left-had sides, we get LHS = 1 + 1(a+b+c) + 4(ab+ac+bc) + 8 a a b a RHS = 1 + 8(a+b+c) + + 4(ab+ac+bc) We ca clearly see that the LHS > RHS whe a,b,c > 0. 3 Also solved by Aas Adlay (studet), Omar Be Abdelaziz Uiversity, El Jadida, Morroco; Myagmarsure Yadamsure, Ulabataar Uiversity, Ulabataar, Mogolia; Soumava Chakraborty, Softweb Techologies, Kolkota, Idia; ad the proposer.

8 Problem 787. Proposed by the editor. Mike buys some pats ad shorts at the Great Pats Store. Mike buys shorts that cost $11 each ad pats that cost $14 each. His total before taxes is $83. How may shorts ad how may pats did Mike buy? Solutio by Robert Bailey (former KME atioal Presidet ), Niagara Uiversity, NY. Let x = umber of shorts ad y = umber of pats. We have 11x + 14y = 83 which is a liear Diophatie equatio i two variables. The 14y = 83 11x which is equivalet to 14y 83 (mod 11) or 3y 8 (mod 11) or 3y -3 (mod 11). Sice 3 is relatively prime to 11, we get y -1 (mod 11). This meas y = 10, 1, 3,. The oly value for y that causes x to be positive i the equatio 11x + 14y = 83 is y = 10 i which case x = 13. Also solved by Michael Bhujel, Bobbie Legg, Katie Tyso (studets), ad Bill Yakosky, North Carolia Wesleya College, Rocky Mout, NC; Jeremiah Bartz, Uiversity of North Dakota, Grad Forks, ND; ad the proposer. Problem 788. Proposed by George Heiema, Worcester Polytechic Istitute, Worcester, MA. A Sujike puzzle has a triagular grid of cells cotaiig digits from 1 to 9. You must place a digit i each of the empty cells with the costrait that o digit ca repeat i ay row, colum, or diagoal. Additioally, o digit ca repeat i the 3x3 large squares with thick borders or the three triagular regios with thick borders. The puzzle below is of itermediate difficulty.

Jeremiah Bartz, Uiversity of North Dakota, Grad Forks, ND; Katie Tyso (studet),

9 Solutio Solved by Jamie Farrar, Destiee Fisher, Nicole Kettle, Courtey Lush (studets), Ed Wilso (retired faculty), Easter Ketucky Uiversity, Richmod, KY; Jeremiah Bartz, Uiversity of North Dakota, Grad Forks, ND; Katie Tyso (studet), Gail Stafford, Carol Lawrece, Bill Yakosky, North Carolia Wesleya College, Rocky Mout, NC.

ELEMENTARY PROBLEMS AND SOLUTIONS

ELEMENTARY PROBLEMS AND SOLUTIONS EDITED BY HARRIS KWONG Please submit solutios ad problem proposals to Dr. Harris Kwog Departmet of Mathematical Scieces SUNY Fredoia Fredoia NY 14063 or by email at wog@fredoia.edu.