353EDITORIALV�a lav Linek

Many CRUX with Mayhem readers have ommented on the urrentba klog of four-digit problem proposals (for those problems that appear nearthe end of ea h issue). The ba klog of these is due to the fa t that we pub-lish only 100 four-digit problems per year, but we re eive many more than100 problem proposals per year. This situation is daunting, espe ially fornew proposers, but also for established proposers who may have a favouriteproblem languishing in a sta k.To at least partially address this situation, any new proposer may nowsubmit up to two (2) problem proposals and request that they be expedited,meaning that we will do our best to evaluate those proposals as soon aspossible. In addition, anyone who has already proposed problems to CRUXmay also submit two proposals to be expedited. To help us implement thisnew poli y, we advise proposers to refer to the proposal numbers that wegive out whenever a problem is proposed. This makes it mu h easier for usto tra k your problem proposal, whereas without a proposal number we mayhave to sear h literally hundreds of pages of paper in dozens of �le folders.We will also begin streaming the problem proposals that we re eive into ategories, namely: Algebra and Number Theory (a single ategory), Logi ,Cal ulus, Combinatori s, Inequalities (in luding Geometri Inequalities),Geometry, Probability, and (if any of these do not �t!) Mis ellaneous Prob-lems. Currently we have a large store of proposed inequalities, so we areparti ularly interested in re eiving problem proposals in the other ategoriesin order to a hieve a balan e of topi s.In another department, our Contributor Pro�les will be resuming in2010 and we have re eived at least two re ommendations so far this year (anytime is a good time to re ommend someone to be pro�led and we wel omemore re ommendations).Lastly, we ask that all solvers begin ea h new solution on a separatepage, with their name and onta t information at the top of the page. Inparti ular our Mayhem editor, Ian VanderBurgh, makes a spe ial appeal toall Mayhem solvers to follow this pro edure.

354SKOLIAD No. 119

Lily Yen and Mogens HansenPlease send your solutions to problems in this Skoliad by 1 January, 2010. A opy of Crux will be sent to one pre-university reader who sends in solutionsbefore the deadline. The de ision of the editors is �nal.Our ontest this month is the World Youth Mathemati s Inter ityCompetition, Individual Contest, Part I, 2005, whi h was held in Kaosiung,Taiwan. We thank Wen-Hsien Sun, Chiu Chang Mathemati s Edu ationFoundation, Taipei, Taiwan, for providing us with this ontest and for per-mission to publish it. We also thank Rolland Gaudet, University College ofSaint-Bonifa e, Winnipeg, MB, for translation of this ontest.Comp�etition math �ematique mondiale des jeunesCon ours individuel, Partie I, 2005Temps limite : 2 heures1. La somme d'un entier �a 4 hi�res et de ses 4 hi�res est 2005. Quel est et entier �a 4 hi�res ?2. Dans le triangle ABC, on a AB = 10 et AC = 18 ; M est le point milieude BC ; la ligne passant par M et parall �ele �a la bisse tri e de ∠CAB oupe

AC au point D. D �eterminer la longueur de AD.3. Soient x, y et z des nombres positifs tels que x+y+xy=8, y+z+yz=15et z + x + zx = 35. D �eterminer la valeur de x + y + z + xy.4. Au total, le nombre de hampignons ueillis par 11 gar� ons et n �lles est�egal �a n2 +9n−2, o �u haque personne a ueilli exa tement le meme nombrede hampignons. D �eterminer l'entier positif n.5. L'entier positif x est tel que x et x + 99 sont tous les deux des arr �es.D �eterminer la somme de tous les entiers x ayant la meme propri �et �e.6. Les ot �es d'un triangle re tangle sont tous les deux des entiers positifs ;un de es ot �es est de longueur au plus 20. Le rapport du rayon du er le ir ons rit au rayon du er le ins rit de e triangle est de 5 : 2. D �eterminerla valeur maximale du p �erim �etre de e triangle.7. Soit α la plus grande ra ine de (2004x)2 − 2003 · 2005x − 1 = 0 et soitβ la plus petite ra ine de x2 + 2003x − 2004 = 0. D �eterminer la valeur deα − β.

3558. Soit a un nombre positif tel que a2 +

1

a2= 5. D �eterminer la valeur de

a3 +1

a3.

9. Dans la �gure, ABCD est un re tangleave AB = 5 et tel que le demi er le dediam�etre AB oupe CD en deux points.Si la distan e d'un e es points �a A est 4,d �eterminer la surfa e de ABCD. ..................................................................................................................................................................................................

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A B

CD

10. Soit a donn �e par 9(n (10

9

)n

− 1 − 10

9−(

10

9

)2

− · · · −(

10

9

)n−1) o �unest un entier positif. Si a est un entier, d �eterminer la valeur maximale de a.11. Dans un entier �a deux positions d �e imales, le hi�re en position dix estplus grand que le hi�re en position un ; le produit de es deux hi�res estdivisible par leur somme. Quel est et entier �a deux positions d �e imales ?

12. Dans la �gure, PQRS est un re -tangle de surfa e 10. A est un pointsur RS et B est un point sur PS, telsque la surfa e du triangle QAB est 4.D �eterminer la plus petite valeur possiblepour PB + AR. ..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

A

BP

Q R

S

World Youth Mathemati s Inter ity CompetitionIndividual Contest, Part I, 20052 hours allowed1. The sum of a four-digit number and its four digits is 2005. What is thisfour-digit number?2. In triangle ABC, AB = 10 and AC = 18. M is the midpoint of BC,and the line through M parallel to the bise tor of ∠CAB uts AC at D. Findthe length of AD.3. Let x, y, z be positive numbers su h that x+y+xy = 8, y+z+yz = 15,and z + x + zx = 35. Find the value of x + y + z + xy.4. The number of mushrooms gathered by 11 boys and n girls is n2+9n−2,with ea h person gathering exa tly the same number. Determine the positiveinteger n.5. The positive integer x is su h that both x and x + 99 are squares ofintegers. Find the sum of all su h integers x.

3566. The side lengths of a right triangle are all positive integers, and the lengthof one of the legs is at most 20. The ratio of the ir umradius to the inradiusof this triangle is 5 : 2. Determine the maximum value of the perimeter ofthis triangle.7. Let α be the larger root of (2004x)2 − 2003 · 2005x −1 = 0 and β be thesmaller root of x2 + 2003x − 2004 = 0. Determine the value of α − β.8. Let a be a positive number su h that a2 +

1

a2= 5. Determine the valueof a3 +

1

a3.

9. In the �gure, ABCD is a re tanglewith AB = 5 su h that the semi ir le withdiameter AB uts CD at two points. Ifthe distan e from one of them to A is 4,�nd the area of ABCD. ....................................................................................................................................................................................................

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A B

CD

10. Let a be 9

(

n(

10

9

)n

− 1 − 10

9−(

10

9

)2

− · · · −(

10

9

)n−1) where n isa positive integer. If a is an integer, determine the maximum value of a.11. In a two-digit number, the tens digit is greater than the ones digit. Theprodu t of these two digits is divisible by their sum. What is this two-digitnumber?

12. In the �gure, PQRS is a re tangleof area 10. A is a point on RS and B isa point on PS su h that the area of tri-angle QAB is 4. Determine the smallestpossible value of PB + AR. ..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

A

BP

Q R

S

Next follow the solutions to the National Bank of New Zealand JuniorMathemati s Competition 2005 [2008 : 385{391℄.1 (For year 9, Form 3 only). Note: In this question the word \digit" means apositive single-digit whole number, that is, a member of the set {1, . . . , 9}.On a long journey by ar, Mi hael was starting to get bored. To keep himamused, his mother asked him some arithmeti questions. The �rst questionshe asked was \Can you think of �ve di�erent digits whi h add to a multipleof 5?" Mi hael answered straight away \That's easy Mum. 1, 2, 3, 4, and5 work be ause 1 + 2 + 3 + 4 + 5 = 15, and 15 is a multiple of 5 be ause15 = 5 × 3."

357Now answer the other questions whi h Mi hael's mother asked.(a) Write down a set of �ve di�erent digits whi h add to 35.(b) Write down a set of three di�erent digits whi h add to a multiple of 5,but whi h don't in lude 5 itself or 1.( ) How many di�erent sets of four di�erent digits are there whi h add toa multiple of 5, but whi h don't in lude 5 itself or 1? (Note that writingthe same set of numbers in a di�erent order doesn't ount here.)(d) Is it possible to write down a set of six di�erent digits whi h add to amultiple of 5, but whi h don't in lude 5 itself? If it is possible, writedown su h a set. If it is not possible, explain brie y why it annot bedone.(e) Is it possible to write down a set of seven di�erent digits whi h add toa multiple of 5, but whi h don't in lude 5 itself? If it is possible, writedown su h a set. If it is not possible, explain brie y why it annot bedone.Solution by Alex Song, student, Elizabeth Ziegler Publi S hool, Waterloo,ON.(a) The set {5, 6, 7, 8, 9} works.(b) The sets {2, 4, 9}, {2, 6, 7}, {3, 4, 8}, {3, 8, 9}, and {4, 7, 9} all work.( ) Let Dn denote the set of allowed digits that leave remainder n whendivided by 5, 0 < n < 5. Then D1 = {6}, D2 = {2, 7}, D3 = {3, 8}, andD4 = {4, 9}. Now two possibilities exist: either use one digit from ea h ofD1, D2, D3, and D4, or use two digits from ea h of D2 and D3. The �rst ase leads to 1 ·2 ·2 ·2, or eight sets, while the se ond leads to one set. Thusthere are a total of nine sets.(d) Any of the sets {1, 2, 3, 4, 6, 9}, {1, 2, 3, 4, 7, 8}, {1, 2, 3, 7, 8, 9},{1, 2, 4, 6, 8, 9}, {1, 3, 4, 6, 7, 9}, {1, 4, 6, 7, 8, 9}, {2, 3, 4, 6, 7, 8},or {2, 3, 6, 7, 8, 9} will work.(e) The eight allowed digits 1, 2, 3, 4, 6, 7, 8, and 9 add up to 40. If the sumof seven of these were divisible by 5, then the remaining digit would also bedivisible by 5. This is not the ase, so no su h set of seven digits exists.Also solved by JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON; andJIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.In Part (a), Jixuan Wang points out that the largest possible sum of �ve distin t digits is5 + 6 + 7 + 8 + 9 = 35, so {5, 6, 7, 8, 9} is the only solution.In Part (b), �nding a set is suÆ ient; to see that these are the only possible solutions,systemati ally onsider all sets of three digits. You an begin by onsidering sets ontaining thedigit 2, then sets whose smallest member is 3, and so on.In Part ( ), 1 + 2 + 3 + 4 = 10 is divisible by 5, so hoosing one number from ea hset works. Now, put one number ba k into the set it ame from and repla e it by a numberfrom a di�erent set. This auses the sum 10 to hange by ±1, ±2, or ±3, and no new sum isdivisible by 5. Therefore, you annot pi k two numbers from one set and one number from ea hof two di�erent sets. That only leaves the possibility of hoosing the four numbers in D2 ∪D3,D2 ∪ D4, or D3 ∪ D4, and of these possibilities only D2 ∪ D3 yields a solution.

358In Part (d), �nding a set is suÆ ient; to see that these are all of the solutions, note thatthe sum of the eight allowed digits is 40 and that, if six hosen digits have a sum divisible by 5,then the remaining two digits also have a sum divisible by 5. Systemati ally sear hing for allsu h sets of size two yields {1, 4}, {1, 9}, {2, 3}, {2, 8}, {3, 7}, {4, 6}, {6, 9}, and {7, 8}.The omplements of these sets are the sets given in the solution above (in reverse order).2. Braille is a ode whi h lets blind people read and write. It was inventedby a blind Fren hman, Louis Braille, in 1829. Braille is based on a patternof dots embossed on a 3 by 2 re tangle. It is read with the �ngers movinga ross the top of the dots.Altogether there are 63 possible ways to emboss one to six dots on a 3by 2 re tangle. (We will not ount zero dots in this question.)Figure 1 shows the pattern for the letter h. Note that if we re e t thispattern (down the middle of the re tangle) the result is the Braille letter j,shown in Figure 2.

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Fig. 1:the letter h.......................................................................................................................................................................................................................................................................................................................

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Fig. 2:the letter j.........................................................................................................................................................................................................................................................

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Fig. 3: a simpli�edversion of Braille.(a) How many di�erent patterns are possible using just one dot?(b) There are 15 di�erent ways to emboss two dots on a 3 by 2 re tangle.How many ways are there to emboss four dots on a 3 by 2 re tangle?Brie y explain your answer.( ) In luding the two patterns shown in Figures 1 and 2, howmany possiblepatterns are there using three dots?(d) A simpli�ed version of Braille has been proposed. In this version thedots will be embossed on a 2 by 2 re tangle. An example is shown inFigure 3. How many possible patterns would there be in this simpli�edversion (assuming that we will not ount zero dots)?(e) Write down how many possible patterns there will be if we were todevelop a more ompli ated version of Braille using a 4 by 2 re tangle.(Again assume that we would not ount zero dots.)Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON.(a) The single dot must be pla ed in one of the six positions of a two-by-three re tangle, so six patterns exist.(b) There are �fteen ways. For every two-dot pattern, swapping dots forblanks and blanks for dots produ es a unique four-dot pattern. Thereforeequally many two-dot and four-dot patterns exist.

359Alternatively, the number of ways one an hoose four positions fromthe six possible positions is 6C4 =

(64

)

=6!

4! (6 − 4)!= 15. Of ourse (6

2

) isalso 15.( ) If you remove one dot from a four-dot pattern, a three-dot pattern re-mains. Sin e any of the four dots an be removed, ea h four-dot pattern givesrise to four three-dot patterns in this way. However, ea h possible three-dotpattern arises from three di�erent four-dot patterns, namely one four-dotpattern for ea h of the three blanks in the three-dot pattern. Sin e the num-ber of four-dot patterns is �fteen, it follows that the number of three-dotpatterns is 15 · 4

3= 20.Alternatively, the number of ways one an hoose three positions fromthe six possible is 6C3 =

(63

)

=6!

3! (6 − 3)!= 20.(d) Here ea h pattern has four positions. Thus you an arrange one dot infour ways, three dots (so one blank) in four ways, and four dots in one way.By reasoning as in Part ( ), the number of two-dot patterns is 4 · 3/2 = 6.Thus the total number of patterns is 4 + 4 + 1 + 6 = 15.Alternatively, (4

1

)

+(42

)

+(43

)

+(44

)

= 4 + 6 + 4 + 1 = 15 is a di�erentway to ount the number of patterns.(e) Here ea h pattern has eight positions, so the number of patterns is(81

)

+(82

)

+ · · · +(88

)

= 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 255.Also solved by JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON; andALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON.In Parts (b) through (d) van Gaalen ommendably provides solutions from �rst prin iplesin the spirit of the question rather than resorting to general ounting methods. However, ashis alternative solutions show, familiarity with binomial oeÆ ients simpli�es the solutions onsiderably.Song provides even simpler solutions to Parts (d) and (e): Ea h of the four (respe tivelyeight) positions must be either blank or a dot. Thus 24 (respe tively 28) patterns exist. Disal-lowing the all-blank pattern thus leaves �fteen (respe tively 255) patterns.3. Around the year 2000 BC, the Babylonians used a number system basedon the number 60. For example, where we would write 0.25 (meaning \onequarter" or 1/4), they would write something like || 15 | (meaning \�fteensixtieths" or 15/60, whi h does simplify to be ome 1/4 in our number sys-tem).The table below shows some numbers and their re ipro als written a - ording to the Babylonian system. (A spe ial feature of a number and itsre ipro al is that when you multiply them together, the result is always 1.)In the table below, ea h olumn represents a pla e value of 1/60 of the pre-vious olumn. For example, 7 | 30 means 7/60 + 30/3600.Number || Re ipro al Number || Re ipro al2 || 30 | 6 || |3 || 20 | 7 || | | | | . . .4 || | 8 || 7 | 30 |5 || | 9 || | |

360(a) Write down in the orre t order the threemissing numbers whi h shouldgo beside 4, 5, and 6.(b) If we added our number 2.5 (meaning \two and a half") into the tableabove, what number would we write beside it to show the Babylonianversion of its re ipro al?( ) The re ipro al of the number 8 is shown in the table as 7 | 30 |. Usingthe same notation, what is the re ipro al of the number 9?(d) Write down the �rst two numbers whi h should go beside 7 in the tableabove.(e) Write down the Babylonian version for the re ipro al of our number

192.Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON.(a) Sin e 1

4=

15

60, 1

5=

12

60, and 1

6=

10

60; the re ipro al of 4 is ‖15|, there ipro al of 5 is ‖12|, and the re ipro al of 6 is ‖10|.(b) Sin e 1

2.5=

1

5/2=

2

5=

24

60, the re ipro al of 2.5 is ‖24|.( ) Sin e 1

9=

400

602=

360 + 40

602=

6

60+

40

602, the re ipro al of 9 is ‖6|40|.(d) The largest integer, x, su h that x

60<

1

7is 8. Thus, the �rst digit is

x = 8. Now 1

7− 8

60=

1

105, and the largest integer, y, su h that y

602<

1

105is 34. Thus the se ond digit is y = 34. Therefore, the re ipro al of 7 begins‖8|34| . . . .(e) Sin e 1

192=

1125

603=

1080 + 45

603=

18

602+

45

603, the re ipro al of 192is ‖0|18|45|.Also solved by ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; and(in part) JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON.You may enjoy verifying that the re ipro al of 7 (see Part (d)) ontinues the pattern

‖8|34|17|8|34|17| . . . . That is, the Babylonian \digits" 8|34|17| ontinue inde�nitely.4. When we �nally landed on Mars, we dis overed that Martians love toplay a game alled HitBall. In this game two teams of players try to hit a ballbetween poles pla ed at ea h end of a �eld. The team that s ores the mostpoints within one Martian hour is the winner.There are three ways to s ore points. An Inner s ores 7 points, an Outers ores 4 points, while a Wide s ores 2 points.The �rst mat h report sent ba k to Earth was not very lear be ause ofstati , so not all the details are ertain. However, we did hear that the RedTeam won. They s ored \something{seven" points altogether (only the lastnumber ould be learly heard). We also learned that they had 16 su essfuls oring shots.

361(a) Write down in a list (from smallest to largest) the possible numbersof Inners whi h the Red Team ould have s ored a ording to this �rstmat h report.(b) A later report added the information that the Red Team s ored the samenumber of Inners and Outers. Using this extra information, write downhow many points the Red Team s ored altogether.( ) Explain why your answer to (b) is the only possible solution. As partof your explanation, make sure you in lude how many Inners, Outers,and Wides the Red Team s ored.Solution by Alex Song, student, Elizabeth Ziegler Publi S hool, Waterloo,ON.(a) The Red Team must have s ored an odd number of Inners, sin e theother point s ores are even and the total number of points won by the RedTeam is odd. Thus, the number of Innersmust be one of 1, 3, 5, 7, 9, 11, 13,or 15.Suppose the Red Team s ored x Inners, 0 Outers, and 16 − x Wides.Then the total number of point is 2(16 − x) + 7x = 32 + 5x. Sin e x is odd,

32 + 5x is \something-seven." Thus any of 1, 3, 5, 7, 9, 11, 13, or 15 arepossible.(b), ( ) Again let x be the number of Inners. Then the Red Team s oredx Outers and 16 − 2x Wides, and 7x + 4x + 2(16 − 2x) = 7x + 32 is thetotal number of points. If 7x + 32 leaves remainder 7 when divided by 10,then 7x + 25 is divisible by 10, as is 7x + 5. Therefore 7x must be an oddmultiple of 5, so x must be an odd multiple of 5. Sin e the number ofWides,16 − 2x, must be nonnegative, it follows that x = 5 and that this is the onlysolution. The total number of points is then 7 · 5 + 32 = 67.Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON;and JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON.Our solver's original solution used modular arithmeti : If the integers x and y leavethe same remainder when divided by the natural number n, then we say that x and y areequivalent modulo n, and we write x ≡ y (mod n). In Part (b), 7x + 32 ≡ 7 (mod 10), so7x + 25 ≡ 7x + 5 ≡ 0 (mod 10). Thus 7x ≡ 5 (mod 10), when e x ≡ 5 (mod 10). Thatis, x is an odd multiple of 5.5. During 2004, a Dunedin newspaper held a ompetition to �nd a new agdesign for the provin e of Otago. Wendy entered the ompetition. Her entrywas based on the design shown below (Figure 4). Her ag featured a gold ross with a blue ba kground. She also pla ed a ir le into her design. Thetop and bottom of the ir le just tou h the orners of the top and bottomtriangles, as shown in Figure 4.(a) Wendy designed her ag to be 240 m long and 150 m high. If theedges of the ross are 40 m and 30 m away from ea h of the orners,as shown in Figure 4, what is the radius of the entre ir le?

362(b) Wendy de ided to remove the ir le from her design (see Figure 5). Withthe ir le removed, what is the total area of the ross?

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30

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4040

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Fig. 5 (not to s ale)Solution by Alex Song, student, Elizabeth Ziegler Publi S hool, Waterloo,ON.(a) Label the points as shown.Sin e it is a ross, K, P , M , andC are ollinear. Thus, △QPM ∼△RPK ∼ △ACK, and we haveQM

QP=

RK

RP=

AK

AC=

150 − 30

240 − 40=

3

5.NowRmust be the midpoint ofAJ ,so RK =

150

2− 30 = 45. Thus,

RP =5

3(45) = 75, and moreover

RP + PQ =1

2AD = 120, so that

PQ=45. Hen e QM =3

5QP =27.That is, the radius is 27 m.

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A D

GJ

E

FK

L

C

HI

B

M

O

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(b) Sin e QM + MS =1

2AJ = 75, it follows that MS = 48, when e thearea of △BMC is 160 · 48

2= 3840. Likewise, the area of △HIO is 3840.Moreover, the area of △LPK is 90 · 75

2= 3375, and the area of △ENF isalso 3375. Sin e the entire re tangle ADGJ has area 36000, the area of the ross is 36000 − 2 · 3840 − 2 · 3375 = 21570 m2.Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.Here is an analyti al approa h: Say J = (0,0), A = (0,150), D = (240, 150),and G = (240, 0). Then K = (0,30) and C = (200, 150), and the line CK is given by

y = 35

x + 30. If x = 120, then y = 35

· 120 + 30 = 102, so M = (120, 102). Sin eQ = (120, 75), the radius is 102 − 75 = 27. Similarly, if y = 75, then 75 = 3

5x + 30, so

x = 75, when e P = (75,75). The area of the ross is now al ulated as above.That ompletes another Skoliad. This issue's prize of a opy of CRUXwith Mayhem goes to Alex Song, student, Elizabeth Ziegler Publi S hool,Waterloo, ON. We hope to re eive more solutions from more readers.

363MATHEMATICAL MAYHEMMathemati al Mayhem began in 1988 as a Mathemati al Journal for and byHigh S hool and University Students. It ontinues, with the same emphasis,as an integral part of Crux Mathemati orum with Mathemati al Mayhem.The Mayhem Editor is Ian VanderBurgh (University of Waterloo). Theother sta� members are Monika Khbeis (As ension of Our Lord Se ondaryS hool, Mississauga) and Eri Robert (Leo Hayes High S hool, Frederi ton).

Mayhem ProblemsVeuillez nous transmettre vos solutions aux probl �emes du pr �esent num�eroavant le 15 d �e embre 2009. Les solutions re� ues apr �es ette date ne seront prisesen ompte que s'il nous reste du temps avant la publi ation des solutions.Chaque probl �eme sera publi �e dans les deux langues oÆ ielles du Canada(anglais et fran� ais). Dans les num�eros 1, 3, 5 et 7, l'anglais pr �e �edera le fran� ais,et dans les num�eros 2, 4, 6 et 8, le fran� ais pr �e �edera l'anglais.La r �eda tion souhaite remer ier Jean-Mar Terrier, de l'Universit �e deMontr �eal, d'avoir traduit les probl �emes.M381. Corre tion. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.D �eterminer toutes les solutions de l' �equation

1

x − 1+

2

x − 2+

6

x − 6+

7

x − 7= x2 − 4x − 4 .

M401. Propos �e par l' �Equipe de Mayhem.Graham et Vazz sont en train de projeter une nouvelle pelouse auxQuartiers G �en �eraux de CRUX. Graham dit : \ Si tu hoisis une pelouse 9m�etres plus longue et 8 m�etres plus �etroite, la surfa e sera la meme." Vazzr �epond : \Si tu la hoisis 12 m�etres plus ourte et 16 m�etres plus large, lasurfa e sera aussi la meme." Quelles sont les dimensions de la pelouse pro-jet �ee ?M402. Propos �e par Ne ulai Stan iu, �E ole Te hnique Sup �erieure de SaintMu eni Sava, Ber a, Roumanie.Trouver toutes les paires ordonn �ees (a, b) d'entiers tels queabba + ab + ba = 89 .

364M403. Propos �e par Matthew Babbitt, �etudiant, Albany AreaMath Cir le,Fort Edward, NY, �E-U.Jean a �e rit un programme sur son ordinateur pour tester si un entierplus grand que 1 est un nombre premier. Sa sournoise de soeur, Ali e, a �edit �ele ode de telle sorte que si l'entr �ee est impaire, la probabilit �e que le pro-gramme donne une r �eponse orre te est de 52% et si l'entr �ee est paire, etteprobabilit �e est de 98%. Jean v �eri�e le programme en testant deux entiersplus grands que 1 hoisis au hasard. Quelle est la probabilit �e que les deuxr �eponses soient orre tes ?M404. Propos �e par Bill Sands, Universit �e de Calgary, Calgary, AB.Un magasin vend des opies d'un ertain arti le �a x$ pi �e e, ou y$ pour a opies, ou en ore z$ pour b opies, a et b �etant des entiers tels que 1 < a < bet x, y et z des nombres r �eels positifs. Pour rendre le rabais \y$ pour a opies" int �eressant, y$ devrait etre plus bas que le prix pay �e pour a a hatsau prix de x$, don y < ax. Pour rendre le deuxi �eme rabais \ z$ pour b opies" int �eressant aussi, on pourrait insister sur une des deux onditions :(a) z

b<

y

a; .- �a-d. que le prix moyen pour une opie soit moindre que leprix orrespondant du premier rabais.(b) Chaque fois qu'on peut �e rire b = qa + r ave q et r des entiers nonn �egatifs, alors on a z < qy + rx ; .- �a-d. qu'il en outerait plus herd'a heter b opies en ombinant le premier rabais ave l'a hat individuelde opies suppl �ementaires plutot que d'opter pour le deuxi �eme rabais.Montrer que si la ondition (a) est satisfaite, alors la ondition (b) l'est aussi.Donner aussi un exemple pour montrer que la ondition (b) pourrait etresatisfaite sans que la ondition (a) le soit.M405. Propos �e par George Apostolopoulos, Messolonghi, Gr �e e.Trouver une formule donnant la valeur de la somme

17 + 187 + 1887 + 18887 + · · · + 188 . . . 87 ,o �u le dernier terme ontient exa tement n hi�res 8.M406. Propos �e par Constantino Ligouras, �etudiant, Coll �ege S ienti�queE. Majorana, Putignano, Italie.Le arr �e ABCD est ins rit dans le huiti �eme d'un er le de rayon 1 etde entre O de sorte qu'il ait un sommet sur haque rayon et les sommetsB et C sur l'ar . Le arr �e EFGH est ins rit dans le triangle DOA de sorteque E et H soient sur les rayons et F et G soient sur AD. Dans le probl �emeM295 [2007 : 200, 202 ; solution 2008 : 203-204℄, on a vu que l'aire du arr �eABCD est 2 −

√2

3. Trouver l'aire du arr �e EFGH..................................................................

365M381. Corre tion. Proposed by Mih�aly Ben ze, Brasov, Romania.Determine all of the solutions to the equation

1

x − 1+

2

x − 2+

6

x − 6+

7

x − 7= x2 − 4x − 4 .

M401. Proposed by the Mayhem Sta�.Graham and Vazz were marking out a new lawn at CRUXHeadquarters.Graham said: \If you make the lawn 9 metres longer and 8 metres narrower,the area will be the same". Vazz said: \If you make it 12 metres shorter and16 metres wider, the area will still be the same". What are the dimensionsof the lawn?M402. Proposed by Ne ulai Stan iu, Saint Mu eni Sava Te hnologi alHigh S hool, Ber a, Romania.Determine all ordered pairs (a, b) of positive integers su h that

abba + ab + ba = 89 .M403. Proposed byMatthew Babbitt, student, Albany AreaMath Cir le,Fort Edward, NY, USA.Jason wrote a omputer program that tests if an integer greater than 1is prime. His devious sister Ali e has edited the ode so that if the input isodd, the probability that the program gives the orre t output is 52% and ifthe input is even, the probability that the program gives the orre t output is98%. Jason tests the program by inputting two random integers ea h greaterthan 1. What is the probability that both outputs are orre t?M404. Proposed by Bill Sands, University of Calgary, Calgary, AB.A store sells opies of a ertain item at $x ea h, or at a items for $y, orat b items for $z, where a and b are positive integers satisfying 1 < a < b andx, y, and z are positive real numbers. To make \a items for $y" a sensiblebargain, $y should be less than buying a separate items; in other words weshould have y < ax. To make \b items for $z" also a sensible bargain, we ould insist on one of two onditions:(a) z

b<

y

a; that is, the average pri e of an item under the \b items for $z"deal is less than under the \a items for $y" deal.(b) Whenever we an write b = qa + r for nonnegative integers q and r,then z < qy + rx holds; that is, it should always ost more to buy bitems by buying a ombination of a items plus individual items, thanby hoosing the \b items for $z" deal.Show that if ondition (a) is true, then ondition (b) is also true. Give anexample to show that ondition (b) ould be true while ondition (a) is false.

366M405. Proposed by George Apostolopoulos, Messolonghi, Gree e.Determine a losed form expression for the sum

17 + 187 + 1887 + 18887 + · · · + 188 . . . 87 ,where the last term ontains exa tly n 8's.M406. Proposed by Constantino Ligouras, student, E. Majorana S ien-ti� High S hool, Putignano, Italy.Square ABCD is ins ribed in one-eighth of a ir le of radius 1 and entre O so that there is one vertex on ea h radius and two verti es B andC on the ar . Square EFGH is ins ribed in △DOA so that E and H lieon the radii, and F and G lie on AD. In problem M295 [2007 : 200, 202;solution 2008 : 203-204℄, we saw that the area of square ABCD is 2 −

√2

3.Determine the area of square EFGH.

Mayhem SolutionsM369. Proposed by the Mayhem Sta�.A re tangle has verti es A(0, 0), B(6, 0), C(6, 4), and D(0, 4). A hor-izontal line is drawn through P (4, 3), meeting BC at M and AD at N . Averti al line is drawn through P , meeting AB at Q and CD at R. Prove thatAP , DM , and BR all pass through the same point.Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON, modi�ed by the editor.Sin e the sides of re tangle ABCD are parallel to the axes, any pointon BC has x- oordinate 6 and any point on CD has y- oordinate 4. Thus,M has oordinates (6, 3) and R has oordinates (4, 4).The line through A(0, 0) and P (4, 3) has slope 3 − 0

4 − 0=

3

4and passesthrough the origin, so has equation y =

3

4x.The line through D(0, 4) and M(6, 3) has equation y − 4

x − 0=

4 − 3

0 − 6, orequivalently y = −1

6x + 4.The line through B(6, 0) and R(4, 4) has equation y − 0

x − 6=

0 − 4

6 − 4, orequivalently y = −2x + 12.Next, we �nd the point of interse tion of the lines AP and DM byequating to obtain 3

4x = −1

6x + 4, or 11

12x = 4, or x =

48

11. Sin e y =

3

4x

367on line AP , then y =

3

4

(

48

11

)

=36

11. Therefore, the point of interse tion oflines AP and DM is (48

11,36

11

).Next, we �nd the point of interse tion of lines AP and BR by equatingto obtain 3

4x = −2x + 12, or 11

4x = 12, or x =

48

11. As before, we see thatthe point of interse tion of lines AP and BR is (48

11,36

11

).Sin e line AP interse ts linesDM and BR at the same point, it followsthat all three lines are on urrent, meeting at (48

11,36

11

).Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; JACLYN CHANG, stu-dent, Western Canada High S hool, Calgary, AB; JULIA CLINE, student, Walt Whitman HighS hool, Bethesda, MD, USA; KATHERINE JANELL EYRE, student, Angelo State University, SanAngelo, TX, USA; ANTONIO GODOY TOHARIA, Madrid, Spain; RICHARD I. HESS, Ran hoPalos Verdes, CA, USA; HUGO LUYO S�ANCHEZ, Ponti� ia Universidad Cat �oli a del Peru, Lima,Peru; RICARD PEIR �O, IES \Abastos", Valen ia, Spain; BRUCE SHAWYER, Memorial Universityof Newfoundland, St. John's, NL; and NECULAI STANCIU, Saint Mu eni Sava Te hnologi alHigh S hool, Ber a, Romania.After �nding the �rst point of interse tion, it would suÆ e to show that this point lieson line BR. Stri tly speaking, we have proven that the lines through A and P , D and M, andB and R all pass through the same point; line segments DM and BR pass through this point,but line segment AP does not ontain the point ( 48

11, 3611

) (though its extension does).M370. Proposed by the Mayhem Sta�.(a) Prove that cos(A + B) + cos(A − B) = 2 cos A cos B for all anglesA and B.(b) Prove that cos C + cos D = 2 cos

(

C + D

2

)

cos(

C − D

2

) for all anglesC and D.( ) Determine the exa t value of cos 20◦ + cos 60◦ + cos 100◦ + cos 140◦,without using a al ulator.

Solution by Courtis G. Chryssostomos, Larissa, Gree e.(a) Applying the sum and di�eren e formulae for the osine, we obtaincos(A + B) + cos(A − B)

= (cos A cos B − sin A sin B) + (cos A cos B + sin A sin B)

= 2 cos A cos B .(b) Applying the formula from (a) with A =C + D

2and B =

C − D

2, weobtain A + B = C and A − B = D, and so

cos C + cos D = 2 cos(

C + D

2

)

cos(

C − D

2

) .

368( ) We rearrange the given expression and apply the formula from (b):

cos 20◦ + cos 60◦ + cos 100◦ + cos 140◦

= cos 140◦ + cos 20◦ + cos 100◦ + cos 60◦

= 2 cos

(

140◦ + 20◦

2

)

cos

(

140◦ − 20◦

2

)

+ cos 100◦ + cos 60◦

= 2 cos 80◦ cos 60◦ + cos 100◦ + cos 60◦

= 2 cos 80◦(

1

2

)

+ cos 100◦ +1

2= cos 80◦ + cos 100◦ +

1

2

= cos 80◦ +(

− cos 80◦)

+1

2=

1

2.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo,Bosnia and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINHQUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JULIA CLINE, student, WaltWhitman High S hool, Bethesda, MD, USA; ANTONIO GODOY TOHARIA, Madrid, Spain;RALPH LOZANO, student, Missouri State University, Missouri, USA; HUGO LUYO S�ANCHEZ,Ponti� ia Universidad Cat �oli a del Peru, Lima, Peru; RICARD PEIR �O, IES \Abastos", Valen ia,Spain; NECULAI STANCIU, Saint Mu eni Sava Te hnologi al High S hool, Ber a, Romania;VASILE TEODOROVICI, Toronto, ON; and JIXUAN WANG, student, Don Mills CollegiateInstitute, Toronto, ON. There were two in omplete solutions submitted.M371. Proposed by Panagiote Ligouras, Leonardo da Vin i High S hool,No i, Italy.Suppose that the line segment AB has length 3 and C is on AB with

AC = 2. Equilateral triangles ACF and CBE are onstru ted on the sameside of AB. If K is the midpoint of FC, determine the area of △AKE.Solution by Antonio Godoy Toharia, Madrid, Spain.Sin e K is the midpoint of segmentFC, we have KC =

1

2(FC) = 1 = CE.Be ause △ACF and △CBE are equilat-eral, ∠FCA = ∠ECB = 60◦, so we have

∠KCE = 60◦.Therefore △KCE is also equilat-eral, and so KE = 1.Sin e ∠EKC = ∠KCA, then KEis parallel to AB, and so we an think ....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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A BC

K E

F

of △AKE as having base KE = 1 and height equal to the verti al distan ebetween the lines through KE and AB.The height of △CBE is h = 1 sin(

60◦)

=

√3

2. Thus, the distan ebetween KE and AB is √

3

2.Therefore, the area of △AKE is equal to 1

2(1)

(√3

2

)

=

√3

4.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG,

369Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JACLYN CHANG, student, WesternCanada High S hool, Calgary, AB; JULIA CLINE, student, Walt Whitman High S hool, Bethesda,MD, USA; EMILY HENDRYX, student, Angelo State University, San Angelo, TX, USA; JOS �EHERN�ANDEZ SANTIAGO, student, Universidad Te nol �ogi a de la Mixte a, Oaxa a, Mexi o;RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; HUGO LUYO S�ANCHEZ, Ponti� iaUniversidad Cat �oli a del Peru, Lima, Peru; RICARD PEIR �O, IES \Abastos", Valen ia, Spain;and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. There were twoin orre t solutions submitted.M372. Proposed by the Mayhem Sta�.A real number x satis�es x3 = x + 1. Determine integers a, b, and cso that x7 = ax2 + bx + c.Solution by Paul Bra ken, University of Texas, Edinburg, TX, USA.Sin e x3 = x + 1, we have

x3 · x3 = (x + 1)(x + 1) = x2 + 2x + 1 ,whi h yieldsx7 = x · x6 = x

(

x2 + 2x + 1)

= x3 + 2x2 + x

= (x + 1) + 2x2 + x = 2x2 + 2x + 1 .Thus, if a = 2, b = 2, and c = 1, we have x7 = ax2 + bx + c.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG,Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; ANTONIO GODOY TOHARIA, Madrid,Spain; JOS �E HERN�ANDEZ SANTIAGO, student, Universidad Te nol �ogi a de la Mixte a, Oaxa a,Mexi o; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; HUGO LUYO S�ANCHEZ, Ponti� iaUniversidad Cat �oli a del Peru, Lima, Peru; RICARD PEIR �O, IES \Abastos", Valen ia, Spain;NECULAI STANCIU, Saint Mu eni Sava Te hnologi al High S hool, Ber a, Romania; EDWARDT.H. WANG, Wilfrid Laurier University, Waterloo, ON; and JIXUAN WANG, student, Don MillsCollegiate Institute, Toronto, ON.It an be proven using algebra that a = 2, b = 2, and c = 1 is the only solution.M373. Proposed by Kunal Singh, student, Kendriya Vidyalaya S hool,Shillong, India.The side lengths of a triangle are three onse utive positive integersand the largest angle in the triangle is twi e the smallest one. Determine theside lengths of the triangle.I. Solution by Jixuan Wang, student, DonMills Collegiate Institute, Toronto,ON. Suppose that the side lengths of the triangle are n − 1, n, and n + 1for some positive integer n ≥ 3. (Smaller values of n do not give a triangle.)Suppose that the smallest angle is θ, whi h is opposite the shortest side (oflength n − 1). Thus, the largest angle is 2θ, whi h is opposite the longestside (of length n + 1).By the Law of Sines, n − 1

sin θ=

n + 1

sin 2θand so n − 1

sin θ=

n + 1

2 sin θ cos θ. Sin e

sin θ 6= 0, then cos θ = n+12(n−1)

.

370Now, by the Law of Cosines,

(n − 1)2 = n2 + (n + 1)2 − 2n(n + 1) cos θ ;(n − 1)2 = n2 + (n + 1)2 − 2n(n + 1)

n + 1

2(n − 1);

n2 − 2n + 1 = n2 + n2 + 2n + 1 − n(n + 1)2

n − 1;

n(n + 1)2

n − 1= n2 + 4n ;

n(n2 + 2n + 1) = (n2 + 4n)(n − 1) ;n2 + 2n + 1 = (n + 4)(n − 1) (sin e n 6= 0) ;n2 + 2n + 1 = n2 + 3n − 4 ;

n = 5 .Therefore, n = 5, and the side lengths are 4, 5, and 6.II. Solution by Vasile Teodorovi i, Toronto, ON.Let the triangle be ABC, with side lengths AB = m, AC = m + 1,and BC = m + 2 for some positive integer m ≥ 2. Then the smallest angleis ∠ACB, whi h we label θ, and the largest is ∠BAC, whi h we label 2θ.Let AD be the angle bise tor of∠BAC, with D on BC. Let BD = x andDC = y. Note that BC = AB + 2, sox + y = m + 2.By the Angle Bise tor Theorem,AC

AB=

DC

DB, so m + 1

m=

y

x.Thus, m + 1

m+ 1 =

y

x+ 1 and so

2m + 1

m=

x + y

x=

m + 2

x.

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

A

B CD

m m + 1

x y

This gives x =m(m + 2)

2m + 1and y =

m + 1

mx =

(m + 1)(m + 2)

2m + 1.Next, we note that ∠BAD = ∠BCA = θ so △BAD is similar to

△BCA, when e BD

BA=

BA

BCor m(m + 2)

2m + 1

/

m =m

m + 2, whi h gives theequivalent equations (m+2)2 = m(2m+1), and m2 +4m+4 = 2m2 +m,and 0 = m2 − 3m − 4.Sin e m is a positive integer, then m = 4, and so the side lengths are

4, 5, and 6.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG,Nguyen Binh KhiemHigh S hool, Vinh Long, Vietnam; COURTIS G. CHRYSSOSTOMOS, Larissa,Gree e; ANTONIO GODOY TOHARIA, Madrid, Spain; RICHARD I. HESS, Ran ho Palos Verdes,

371CA, USA; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring�eld, MO, USA;RICARD PEIR �O, IES \Abastos", Valen ia, Spain; NECULAI STANCIU, Saint Mu eni SavaTe hnologi al High S hool, Ber a, Romania; and EDWARD T.H. WANG, Wilfrid LaurierUniversity, Waterloo, ON.M374. Proposed by Mih�aly Ben ze, Brasov, Romania.Suppose that p is a �xed prime number with p ≥ 3. Determine thenumber of solutions to x3 + y3 = x2y + xy2 + p2009, where x and y areintegers.Solution by Missouri State University Problem Solving Group, Spring�eld,MO, USA.The equation an be rewritten as

p2009 = x3 − x2y − xy2 + y3 = x2(x − y) − y2(x − y)

= (x − y)(x2 − y2) = (x − y)2(x + y) .Sin e x and y are integers and the only divisors of p2009 are of the form±pn, then x − y = ±pk for some integer k with 0 ≤ k ≤ 1004. Thus,(x−y)2 = p2k and so x+y = p2009−2k. (The upper bound on k omes fromthe fa t that (x − y)2 is also a divisor of p2009.)This yields

x =1

2

(

(x + y) + (x − y))

=1

2

(

p2009−2k ± pk) ,

y =1

2

(

(x + y) − (x − y))

=1

2

(

p2009−2k ∓ pk) .These are all integers sin e p is odd (in fa t this is still true if p = 2 as longas k > 0). Ea h pair (x, y) is distin t sin e for ea h pair the values of x + yand x − y are di�erent.Sin e there are 1005 possible values for k in the range 0 ≤ k ≤ 1004,there are then 2(1005) = 2010 solutions.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; and EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON. Therewere four in orre t solutions and one in omplete solution submitted.M375. Proposed by Ne ulai Stan iu, Saint Mu eni Sava Te hnologi alHigh S hool, Ber a, Romania.Determine all real solutions to the system of equations

1

x2+

4

y2+

9

z2= 4 ; x2 + y2 + z2 = 9 ; xyz =

9

2.

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON.There are exa tly four solutions for (x, y, z), namely (√6

2,

√3,

3√

2

2

),(√

6

2, −

√3, −3

√2

2

), (−√

6

2,

√3, −3

√2

2

), and (−√

6

2, −

√3,

3√

2

2

).

372Note that if (x, y, z) is a solution, then so are (x, −y, −z), (−x, y, −z),and (−x, −y, z). (No other ombination of minus signs will work, sin e aneven number of minus signs is ne essary to make the sign of xyz orre t.)Hen e, we may assume that x, y, z > 0 (sin e none of x, y, z an equal 0).The Cau hy{S hwarz Inequality says that if a, b, c, A, B, and C arereal numbers, then

(

a2 + b2 + c2)(

A2 + B2 + C2)

≥ (aA + bB + cC)2 ,with equality if and only if (a, b, c) is a s alar multiple of (A, B, C).By the Cau hy{S hwarz Inequality, we have36 =

(

x2 + y2 + z2)

(

1

x2+

4

y2+

9

z2

)

≥(

x

(

1

x

)

+ y

(

2

y

)

+ z

(

3

z

))2

= 36 .Sin e equality holds, then (x, y, z) must be a s alar multiple of ( 1

x,2

y,3

z

).Thus, x =k

x, y =

2k

y, and z =

3k

z, for some k > 0, and so x2 = k, y2 = 2k,and z2 = 3k.Substituting ba k into the original se ond equation, we obtain 6k = 9,and so k =

3

2, when e x =

√

3

2=

√6

2, y =

√3, and z =

√

9

2=

3√

2

2. Thisgives the four solutions above, as laimed.Also solved by GEORGE APOSTOLOPOULOS,Messolonghi, Gree e; CAOMINHQUANG,Nguyen Binh KhiemHigh S hool, Vinh Long, Vietnam; COURTIS G. CHRYSSOSTOMOS, Larissa,Gree e; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; and RICARD PEIR �O, IES \Abastos",Valen ia, Spain. There were two in omplete solutions submitted.

Problem of the MonthIan VanderBurghSometimes we mathemati ians like to use a sledgehammer when per-haps a smaller instrument would be in order.Problem 1 (2009 Gauss Contest, Grade 7) If x, y, and z are positive integerswith xy = 18, xz = 3, and yz = 6, what is the value of x + y + z?(A) 6 (B) 10 (C) 25 (D) 11 (E) 8If we weren't in Grade 7, then here are two ways that we might ta klethis problem:

373Solution 1 Sin e xy = 18, xz = 3, and yz = 6, by multiplying these equa-tions we have (xy)(xz)(yz) = 18(3)(6), or x2y2z2 = 182, or (xyz)2 = 182.Sin e x, y, and z are positive, then xyz > 0, so xyz = 18. We annow ombine this with the original three equations as follows.Sin e xy = 18, we have xyz

xy=

18

18, or z = 1. Sin e xz = 3, we have

xyz

xz=

18

3, or y = 6. Sin e yz = 6, we have xyz

yz=

18

6, or x = 3.Therefore, x + y + z = 3 + 6 + 1 = 10.Solution 2 Sin e xy = 18 and xz = 3, then xy

xz=

18

3, or y

z= 6, or y = 6z.Sin e yz = 6, then (6z)z = 6, or z2 = 1. Sin e z > 0, then z = 1, and so

y = 6z = 6. Sin e xz = 3 and z = 1, then x = 3.Therefore, x + y + z = 3 + 6 + 1 = 10.These are two great solutions using standard te hniques for solving sys-tems of equations. But they're hardly suitable for Grade 7 students. Thisproblem is a perfe tly good Grade 7 problem, though.Solution 3 We are told that x, y, and z are positive integers. Sin e xz = 3and 3 is a prime, then x and z must be 1 and 3, or 3 and 1, respe tively.Let's look at the ase x = 1 and z = 3. Sin e x = 1 and xy = 18and the number that we multiply 1 by to get 18 is 18 itself, then y = 18.Therefore, yz = 18(3) = 54, whi h disagrees with the equation yz = 6.Thus, x = 1 and z = 3 is not the orre t ombination. (For the re ord, Inever hoose the orre t line in the gro ery store either.)So we look at the ase x = 3 and z = 1. Sin e x = 3 and xy = 18 andthe number that we multiply 3 by to get 18 is 6, then y = 6. (The produ tyz does in fa t equal 6, as required). Thus, x = 3, y = 6, and z = 1.Therefore, x + y + z = 3 + 6 + 1 = 10.It's easy to get \trapped" into using the high powered te hniques thatwe know, but sometimes there is a ni er solution that uses less ma hinery.Here is a problem in a similar vein that has developed a following inthe math ontest world.Problem 2 (1988 UK S hools Mathemati al Challenge) Weighing the baby atthe lini was a problem. The baby would not keep still and aused the s alesto wobble. So I held the baby and stood on the s ales while the nurse reado� 78 kg. Then the nurse held the baby while I read o� 69 kg. Finally, I heldthe nurse while the baby read o� 137 kg. What is the ombined weight of allthree (in kg)?(A) 142 (B) 147 (C) 206 (D) 215 (E) 284Again, there is a standard system-of-equations type solution whi h isworth seeing. To save onfusion, we'll give the narrator a name hosen atrandom, say, Tony.

374Solution 1 Let Tony's weight be xkg, let the baby's weight be y kg, and letthe nurse's weight be z kg.The ombined weight of Tony and the baby is 78 kg, so x + y = 78.The ombined weight of the baby and the nurse is 69kg, so y + z = 69. The ombined weight of Tony and the nurse is 137 kg, so x + z = 137.We an show this system of equations ni ely in a visual way:

x + y = 78 ,y + z = 69 ,

x + z = 137 .By the way, laying out the equations in this fashion is a really useful thing todo and gives you a mu h better idea of what to do than by writingx + y = 78 ,y + z = 69 ,x + z = 137 .In any event, the \ni e" way of writing the equations allows us to see thatadding the three equations is a really good idea. When we do this, we obtain

2x + 2y + 2z = 78 + 69 + 137, or 2(x + y + z) = 284, or x + y + z = 142.Noti e that we don't a tually need to determine x, y, and z at all!But, again, we don't need to do anything nearly that fan y. In fa t, we an get away without doing any algebra at all.Solution 2 First, we look at the fa t that the ombined weight of Tony andthe baby is 78 kg and the ombined weight of the nurse and the baby is 69kg.Sin e the baby's weight is in luded in both of these totals, then Tony mustbe 78 − 69 = 9 kg heavier than the nurse (that is, the di�eren e betweenTony's weight and the nurse's weight is 9 kg).But the ombined weight of Tony and the nurse is 137 kg. We want to�nd two numbers that add to 137, one of whi h is 9 greater than the other.To �nd these numbers, we an subtra t 9 from 137 to get 128 and then divideby 2 to get 64. The numbers 64 and 64 + 9 = 73 di�er by 9 and add to 137,so must be the weights, in kg, of the nurse and Tony, respe tively.Sin e the ombined weight of Tony and the baby is 78kg and Tonyweighs 73 kg, then the weight of the baby is 5 kg.Therefore, the ombined weight of all three is 64 + 73 + 5 = 142 kg.So put your sledgehammer ba k in the garage, and think before youleap.

375THE OLYMPIAD CORNERNo. 280

R.E. WoodrowAs a �rst problem set for this number we look to the Grade 12 problemsof the Final Round of the Estonian National Olympiad 2005{2006. My thanksgo to Robert Morewood, Canadian Team Leader to the 2006 IMO in Slovenia,for olle ting them for our use.ESTONIAN NATIONAL OLYMPIAD 2005{2006Final RoundGrade 121. We all a �gure a ship if it is made up of unit squares onne ted by om-mon edges. Prove that if there is an odd number of possible di�erent ships onsisting of n unit squares on a 10 × 10 board, then n is divisible by 4.2. Find the smallest distan e between points P and Q in the xy-plane, if Plies on the line y = x and Q lies on the urve y = 2x.3. Prove or disprove the following statements:(a) For every integern ≥ 3, there existn distin t positive integers su h thatthe produ t of any two of them is divisible by the sum of the remainingn − 2 numbers.(b) For some integer n ≥ 3, there exist n distin t positive integers su hthat the sum of any n − 2 of them is divisible by the produ t of theremaining two numbers.4. The a ute triangle ABC has ir um entre O and triangles BCO, CAO,and ABO have ir um entres A′, B′, and C′, respe tively. Prove that thearea of triangle ABC does not ex eed the area of triangle A′B′C′.5. The Ababi alphabet onsists of the letters A and B, and the words in theAbabi language are pre isely those that an be formed by the following tworules:(a) The letter A is a word.(b) If s is a word, then s⊕s and s⊕ s are words, where s denotes the wordthat is obtained by repla ing all letters A in s with letters B, and vi eversa; and x ⊕ y denotes the on atenation of words x and y.

376The Ululu alphabet also onsists of the letters A and B, and the wordsin the Ululu language are pre isely those that an be formed by the followingtwo rules:(a) The letter A is a word.(b) If s is a word, then s⊗s and s⊗ s are words, where s is de�ned as aboveand x ⊗ y is the word obtained from words x and y of equal length bywriting the letters of x and y alternatingly, starting from the �rst letterof x.Prove that the two languages onsist of the same words.Next we turn to the Kazakhs and the problems of the II InternationalZhautykov Olympiad in Mathemati s, Almaty, 16{17 January, 2006. Thanksagain go to Robert Morewood, Canadian Team Leader to the 2006 IMO inSlovenia, for olle ting them for us.II INTERNATIONAL ZHAUTYKOV OLYMPIADIN MATHEMATICSAlmaty { Day 1 (January 16, 2006)

1. Find all positive integers n su h that n = ϕ(n)+402, where ϕ is the Eulerphi-fun tion (it is known that if p1, p2, . . . , pk are the distin t prime divisorsof n, then ϕ(n) = n

(

1 − 1

p1

)(

1 − 1

p2

)

· · ·(

1 − 1

pk

); and ϕ(1) = 1).2. The points K and L lie on the sides AB and AC, respe tively, of thetriangle ABC su h that BK = CL. Let P be the point of interse tion ofthe segments BL and CK, and let M be an inner point of the segment ACsu h that the line MP is parallel to the bise tor of the angle ∠BAC. Provethat CM = AB.3. An m×n table with 4 ≤ m ≤ n is alled good if the numbers 0 and 1 anbe written in the unit squares of the table su h that the following onditionshold:(a) not all numbers are 0's and not all are 1's;(b) the number of 1's in any 3 × 3 subsquare is onstant;( ) the number of 1's in any 4 × 4 subsquare is also onstant.Find all pairs of positive integers (m, n) for whi h there is a good m × ntable.

377Almaty { Day 2 (January 17, 2006)4. A heap of 100 stones is given. A partition of the heap into k smaller heapsis spe ial provided that (i) no two heaps have the same number of stones,and (ii) any further partition of any of the heaps into two new heaps resultsin k + 1 heaps for whi h (i) fails (a heap always ontains at least one stone).(a) Find the maximum number k for whi h there is a spe ial partition ofthe given heap of 100 stones into k heaps.(b) Find the minimum number k for whi h there is a spe ial partition of thegiven heap of 100 stones into k heaps.5. Let a, b, c, and d be real numbers su h that a+ b+ c+d = 0. Prove that

(ab + ac + ad + bc + bd + cd)2 + 12 ≥ 6(abc + abd + acd + bcd) .6. Let ABCDEF be a onvex hexagon su h that AD = BC + EF ,BE = AF + CD, and CF = DE + AB. Prove that AB

DE=

CD

AF=

EF

BC.

Next is the 50th Mathemati al Olympiad of the Republi of Moldova,and problems of the 10th form. We again thank Robert Morewood, CanadianTeam Leader to the 2006 IMO in Slovenia, for olle ting them for us.50th MATHEMATICAL OLYMPIAD OF THEREPUBLIC OF MOLDOVA10th FormMar h 20061. Let a, b, and c be the side lengths of a right triangle with hypotenuse oflength c, and let h be the altitude from the right angle. Find the maximumvalue of c + h

a + b.2. Let n ≥ 2 be an integer and M = {0, 1, 2, . . . , n − 1}. For a nonzerointeger a let fa : M → M where fa(x) is the remainder of ax upon divisionby n. Give a ne essary and suÆ ient ondition for the fun tion fa to bebije tive and prove that an(n−1) −1 is divisible by n2 whenever n is a primenumber.3. The quadrilateralABCD is ins ribed in a ir le. The tangents to the ir leat A and C interse t at a point P not on BD and su h that PA2 = PB ·PD.Prove that BD passes through the midpoint of AC.

3784. Find all values of the real number a for whi h the following equation hasa unique solution:

2x2 − 6ax + 4a2 − 2a − 2 + log2

(

2x2 + 2x − 6ax + 4a2)

= log2

(

x2 + 2x − 3ax + 2a2 + a + 1) .

5. Let n be a positive integer and let x1, x2, . . . , xn be real numbers in theinterval (14,2

3

). Find the minimum value of the expressionlog1.5x1

(

1

2− 1

36x22

)

+ log1.5x2

(

1

2− 1

36x23

)

+ · · · + log1.5xn

(

1

2− 1

36x21

) .When is this value attained?6. Triangle ABC is isos eles with AC = BC and P is a point inside thetriangle su h that ∠PAB = ∠PBC. If M is the midpoint of AB, prove that∠APM + ∠BPC = 180◦.7. The interior angles of a onvex o tagon are all equal and all side lengthsare rational numbers. Prove that the o tagon has a entre of symmetry.8. Let M = {x2 +x | x is a positive integer}. For ea h integer k ≥ 2 provethat there exist a1, a2, . . . , ak, bk in M su h that a1 + a2 + · · · + ak = bk.

We also give the problems of the Se ond and Third Team Sele tion Testsfor the IMO 2006 from Chi�sin�au, Moldova, Mar h 25-26, 2006. Thanks againgo to Robert Morewood, Canadian Team Leader to the 2006 IMO in Slovenia,for olle ting them for us.REPUBLIC OF MOLDOVA INTERNATIONALMATHEMATICAL OLYMPIADSe ond Team Sele tion TestMar h 25, 2006 (Chi�sin�au)1. Let {an}∞n=0 be a sequen e, with a0 = 2, a1 = 1, and an+1 = an +an−1for n ≥ 1. Prove that a59 divides the number (a30)

59 − 1.2. The point P lies in the interior of the ir le C1, whi h in turn lies in theinterior of the ir le C2, and the point Q lies in the exterior of the ir le C2.Ea h line ℓi (i ≥ 3) is di�erent from the line PQ, passes through P , and utsthe ir le C1 at the points Ai and Bi; and the ir um ir le of triangle QAiBi uts the ir le C2 at the points Mi and Ni. Prove that all the lines MiNi are on urrent.

3793. Let a, b, c be the side lengths of a triangle and let s be the semiperimeter.Prove that

a

√

(s − b)(s − c)

bc+ b

√

(s − c)(s − a)

ac+ c

√

(s − a)(s − b)

ab≥ s .

4. Let n be a positive integer and let A = {1, 2, . . . , n}. Determine thenumber of distin t solutions of the equationX ∪ Y ∪ Z = A ,where two solutions are onsidered the same if they di�er only in the orderingof X, Y , and Z. Third Team Sele tion TestMar h 26, 2006 (Chi�sin�au)

5. The pointP is in the interior of triangleABC. The raysAP , BP , andCP ut the ir um ir le of the triangle at the points A1, B1, and C1, respe tively.Prove that the sum of the areas of the triangles A1BC, B1AC, and C1ABdoes not ex eed s(R − r), where s, R, and r are the semiperimeter, the ir umradius, and the inradius of triangle ABC, respe tively.6. Let n ≥ 2 be an integer and let X be a set with n + 1 elements. Thesequen es (a1, a2, . . . , an) and (b1, b2, . . . , bn) of distin t elements from theset X are alled separated if ai = bj for some i 6= j. Determine the maxi-mum number of ordered sequen es of n elements from X su h that any twoof them are separated.7. Let a, b, and c be positive real numbers su h that abc = 1. Prove thata + 3

(a + 1)2+

b + 3

(b + 1)2+

c + 3

(c + 1)2≥ 3 .

8. Let f(n) be the number of permutations (a1, a2, . . . , an) of the set{1, 2, . . . , n} su h that a1 = 1 and |ai −ai+1| ≤ 2 for 1 ≤ i ≤ n−1. Provethat f(2006) is divisible by 3.

Next we give the problems of the Final Round, 2nd Grade High S hool ofthe Youth Mathemati al Olympiad of the Aso ia i �on, Venezolana de Compe-ten iasMatem�ati as, writtenMay 27, 2006. Thanks go to RobertMorewood,Canadian Team Leader to the IMO in Slovenia, for olle ting them.

380YOUTH MATHEMATICAL OLYMPIADAso ia i �on Venezolana de Competen ias Matem�ati asFinal Round, 2nd Grade High S hool (May 27, 2006)1. A positive integer has 223 digits and the produ t of these digits is 3446.What is the sum of the digits?2. Find all solutions of the equation m2 − 3m + 1 = n2 + n − 1, where mand n are positive integers.3. De�ne the sequen e a1, a2, a3, . . . as follows: let a1 = a2 = 1003;

a3 = a2−a1 = 0; a4 = a3−a2 = −1003; and in general an+1 = an −an−1for any n ≥ 2. Compute the sum of the �rst 2006 terms of the sequen e.4. Joseph, Dario, and Henry prepared some labels. On ea h label they wroteone of the numbers 2, 3, 4, 5, 6, 7, or 8. David joined them and stu k onelabel on the forehead of ea h friend. Joseph, Dario, and Henry ould not seethe numbers on their own foreheads, they only saw the numbers of the othertwo. David said, \You do not have distin t numbers on your foreheads, andthe produ t of the three numbers is a perfe t square." Ea h friend then triedto dedu e what number he had on his forehead. Could anyone dis over it?5. Let ABC be an isos eles triangle with ∠B = ∠C = 72◦. Find the valueof BC

AB − BC. Hint: Consider the bise tor CD of ∠ACB.

As a �nal problem set from our 2006 IMO �le we turn to the 42nd Mon-golian Mathemati al Olympiad, 10th Grade, written at Ulaanbaatar in Apriland May of 2006. Thanks go to Robert Morewood, Canadian Team Leaderat the 2006 IMO in Slovenia, for olle ting them for us.Unfortunately, our version of problem 5 below was in omplete, and we ould not remedy this in time. We will print this problem in a future issueas soon as we lo ate it.42nd MONGOLIAN MATHEMATICAL OLYMPIAD10th GradeApril 28-May 4, 2006 (Ulaanbaatar)1. Let a, b, c, d, e, and f be positive integers satisfying the relationab + ac + bc = de + df + ef , and let N = a + b + c + d + e + f.Prove that if N | (abc + def), then N is a omposite number.2. Let m and n be positive integers. In how many di�erent ways an one�ll in an n × n grid with nonnegative integers su h that the sum of every nnumbers from di�erent rows and olumns is m?

3813. Cir les ω, ω0 have distin t entres and ω0 lies inside ω. The four ir lesω1, ω2, ω3, and ω4 ea h tou h ω internally and ω0 externally. Prove that theinterse tion points of the external tangents of ωi and ωj (i 6= j) are ollinear.4. Let p and q be distin t prime numbers and let a and b be distin t positiveintegers. Prove that there exists a prime number r su h that r | (apq − bpq)and r ≡ 1 (mod pq).5. Omitted.6. Let a, b, c be positive real numbers su h that a + b + c = 1. Prove that

(1 + a)

√

1 − a

a+ (1 + b)

√

1 − b

b+ (1 + c)

√

1 − c

c

≥ 3√

3

4· (1 + a)(1 + b)(a + c)√

(1 − a)(1 − b)(1 − c).

Now we give readers' solutions to problems in the September 2008Corner and the X BosnianMathemati al Olympiad given at [2008 : 284{285℄.1. Let H be the ortho entre of an a ute-angled triangle ABC. Prove thatthe midpoints of AB and CH and the interse tion point of the interior bi-se tors of ∠CAH and ∠CBH are ollinear.Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain;George Apostolopoulos, Messolonghi, Gree e; Oliver Geupel, Br �uhl, NRW,Germany; D.J. Smeenk, Zaltbommel, the Netherlands; and Titu Zvonaru,Com�ane�sti, Romania. We give the version of Geupel.Let us denote α = ∠CAB,β = ∠ABC, γ = ∠BCA, and letA′ = AH ∩ BC, B′ = BH ∩ AC.Let M , N , O be the midpoints ofsegments AB, A′B′, CH, respe -tively, let P = CH ∩ A′B′, and letQ be the interse tion point of thebise tors of ∠CAA′ and ∠CBB′,respe tively. We are to show thatthe points M , O, Q are ollinear.It is a well-known fa t that themidpoints of the diagonals of a om-plete quadrilateral are ollinear.

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A B

C

A′

B′

H

M

NO

P

α

α

β

β

Applying this to the omplete quadrilateral generated by the quadrilateralA′CB′H, we see that the points M , N , and O are ollinear. Due to the fa t

382that ∠HA′C = ∠CB′H = 90◦, the quadrilateral A′CB′H is y li with ir um entre O.Hen e, ON ⊥ A′B′, whi h implies that

∠HOM = ∠PON = 90◦ − ∠NPO = 90◦ − ∠A′PC

= 90◦ −(

180◦ − ∠PA′C − ∠A′CP)

= ∠PA′C + ∠BCH − 90◦

= α +(

90◦ − β)

− 90◦ = α − β ;thus,∠AMO = 90◦ − ∠HOM = 90◦ − α + β . (1)Sin e ∠BAA′ = 90◦ − β and ∠A′AC = 90◦ − γ, we have

∠BAQ = ∠BAA′ +1

2∠A′AC

= 90◦ − β +1

2

(

90◦ − γ)

= 135◦ − β − 1

2γ .Similarly, ∠ABQ = 135◦ − α − 1

2γ. Hen e,

∠AQB = 180◦ − ∠BAQ − ∠ABQ

= 180◦ −(

135◦ − β − 1

2γ

)

−(

135◦ − α − 1

2γ

)

= α + β + γ − 90◦ = 90◦ .We dedu e that △AMQ is isos eles with AM = MQ. Thus,∠AMQ = 180◦ − 2∠MAQ = 180◦ − 2

(

135◦ − β − 1

2γ

)

= 90◦ − α + β . (2)From (1) and (2) it is readily seen that ∠AMO = ∠AMQ, that is, the pointsM , O, and Q are ollinear. This ompletes the proof.2. Let a1, a2, and a3 be nonnegative real numbers with a1 + a2 + a3 = 1.Prove that

a1√

a2 + a2√

a3 + a3√

a1 ≤ 1√

3.Solved by George Apostolopoulos,Messolonghi, Gree e; Mi hel Bataille,Rouen, Fran e; George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru, Com�ane�sti,Romania. We give the solution by Bataille.Sin e the fun tion f(x) =

√x is on ave on [0, ∞) and a1+a2+a3 = 1,Jensen's Inequality yields

a1√

a2 + a2√

a3 + a3√

a1 ≤√

a1a2 + a2a3 + a3a1 . (1)

383Now, from the well-known inequality a2

1 + a22 + a2

3 ≥ a1a2 + a2a3 + a3a1,we obtain3(a1a2 + a2a3 + a3a1) ≤ a2

1 + a22 + a2

3 + 2(a1a2 + a2a3 + a3a1)

= (a1 + a2 + a3)2 = 1 ,so that a1a2 + a2a3 + a3a1 ≤ 1

3. From (1), it follows that

a1√

a2 + a2√

a3 + a3√

a1 ≤ 1√

3.

4. Given are a ir le and its diameter PQ. Let t be a tangent to the ir le,tou hing it at T , and let A be the interse tion of the lines t and PQ. Let pand q be the tangents to the ir le at P and Q respe tively, and letPT ∩ q = {N} and QT ∩ p = {M} .Prove that the points A, M , and N are ollinear.Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain;Mi hel Bataille, Rouen, Fran e; D.J. Smeenk, Zaltbommel, the Netherlands;and Titu Zvonaru, Com�ane�sti, Romania. We give the write-up of AmengualCovas.Equivalently, we will prove that: Let T be a point of a ir le with agiven diameter PQ. Let p and q be the tangents to the ir le at P and Q,respe tively, and let

PT ∩ q = {N} , QT ∩ p = {M} , and MN ∩ PQ = {A} .Then AT tou hes the given ir le at T .

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P O Q A

N

M

U

T

V

p q

t

............................................................................

384Let AT ∩ p = {U}. Then, by Ceva's Theorem,

MU

UP· PQ

QA· AN

NM= 1 .

But p ⊥ PQ and q ⊥ PQ, so that p‖q, implying that P Q

QA=

NM

AN. Hen e,

MU

UP= 1 and MU = UP .Let AT ∩ q = {V }. Sin e

NV

MU=

AV

AU=

V Q

UP,we obtain NV = V Q.Thus, U and V are the midpoints of MP and NQ, respe tively.Sin e ∠PTQ is a right angle, so also ∠MTP = 90◦ = ∠NTQ. Hen eboth ir les onstru ted with MP and NQ as diameters pass through T .Sin e the distan e between their entres U and V equals the sum oftheir radii, these ir les are externally tangent to ea h other at T .Let their ommon tangent at T interse t PQ at O. This tangent isperpendi ular to the line UV through the entres, and sin e

PO = OT = OQ ,the point O is the midpoint of PQ, making OT a radius of the given ir le inour problem.Thus, OT ⊥ AT , and the on lusion follows.5. A permutation (a1, a2, . . . , an) of the set {1, 2, . . . , n} satis�es theinequality a2k

ak+1

≤ k + 2 for ea h k = 1, 2, . . . , n − 1. Prove it is theidentity.Solution by Titu Zvonaru, Com�ane�sti, Romania.Suppose that an 6= n. Then an ≤ n − 1, and we havea2

n−1 ≤ (n + 1)an ≤ (n + 1)(n − 1) = n2 − 1 ,hen e, an−1 ≤ n − 1.Now, sin e a2n−2 ≤ nan−1 ≤ n2 − n, we dedu e that an−2 ≤ n − 1.In general, if 1 ≤ k ≤ n − 1 and ak+1 ≤ n − 1, then

a2k ≤ (k + 2)ak+1 ≤ (n + 1)(n − 1) = n2 − 1 ,and we dedu e that ak ≤ n − 1.From an 6= n it then follows that ak < n for ea h index k. However,

n ∈ {a1, a2, . . . , an}, a ontradi tion!Therefore, an = n, and similarly, an−1 = n − 1, . . . , a2 = 2, a1 = 1,whi h means that (a1, a2, . . . , an) is the identity.

3856. Let a, b, and c be integers su h that

a

b+

b

c+

c

a= 3 .Prove that abc is a perfe t ube.Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille,Rouen, Fran e; George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru, Com-�ane�sti, Romania. We give the generalization of Bataille.We generalize and show that if a

b+

b

c+

c

a= n for some integer n,then abc is a perfe t ube. Assume that a

b+

b

c+

c

a= n. Certainly abc isa perfe t ube if |abc| = 1. From now on, we suppose that |abc| > 1. ItsuÆ es to prove that for any prime p dividing abc, the exponent of p in theprime fa torization of abc is a multiple of 3.Let p be su h a prime. There exist nonnegative integers r, s, and t with

(r, s, t) 6= (0, 0, 0) and su h that a = pra1, b = psb1, and c = ptc1, wherea1, b1, and c1 are integers not divisible by p. We have to show that r + s + tis divisible by 3. From the hypothesis, we obtain

pr−sa21c1 + ps−tb2

1a1 + pt−rc21b1 = na1b1c1 . (1)Let N be the number of negative integers among r − s, s − t, t − r, and onsider the possibilities for N .First, note that (r − s) + (s − t) + (t − r) = 0, so N = 3 annothold. Also, N = 1 annot hold either: if we had, say, r − s, s − t ≥ 0 and

t − r < 0, then the relation (1) would implyc21b1 = pr−t

(

na1b1c1 − pr−sa21c1 − ps−tb2

1a1

) , ontradi ting the fa t that p does not divide c21b1.If N = 0, then r − s, s − t, and t − r are nonnegative and sum to zero,hen e r = s = t and r + s + t is a multiple of 3.If N = 2, then suppose (for example) that s−t, t−r < 0 and r−s > 0.Then (1) yields

pr−sa21c1 +

b21a1

pt−s+

c21b1

pr−t= na1b1c1 .If, in addition, we had t − s 6= r − t, then by multiplying by pmax(t−s,r−t)we would dedu e that p divides b2

1a1 (if t − s > r − t) or p divides c21b1(if r−t > t−s), thus arriving at a ontradi tion in either ase. Consequently,

t − s = r − t and r + s + t = 3t, a multiple of 3.This last ase ompletes the analysis.

386We next look at solutions from our readers to the problems of theI elandi Mathemati al Contest 2004{2005 (Final Round) given in the Cornerat [2008 : 285℄.1. How many subsets with three elements an be formed from the set

{1, 2, . . . , 20} so that 4 is a fa tor of the produ t of the three numbersin the subset?Solved by George Apostolopoulos, Messolonghi, Gree e; and John GrantM Loughlin, University of New Brunswi k, Frederi ton, NB. We give thesolution of M Loughlin.There are (203

)

= 1140 subsets with 3 elements.All of these subsets have elements that when multiplied will have 4 asa fa tor, ex ept in two ases:• all elements are odd, of whi h there are (10

3

)

= 120 subsets;• two elements are odd and the third element is even but not a multipleof 4, of whi h there are (10

2

)(51

)

= 225 subsets.Therefore, 1140 − (120 + 225) = 795 subsets meet the requirements.2. Triangle ABC is equilateral, D is a point inside the triangle su h thatDA = DB, and E is a point that satis�es the onditions ∠DBE = ∠DBCand BE = AB. How large is the angle ∠DEB?Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; and TituZvonaru, Com�ane�sti, Romania. We give Zvonaru's answer.Sin e DA = DB, the point D lieson the perpendi ular bise tor of AB,hen e ∠DCB = 30◦. We have

BE = BC ;∠DBE = ∠DBC ;

BD = BD .It follows that △BDE and △BDCare ongruent (side-angle-side), hen e∠DEB = DCB = 30◦ .

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A

B C

D

E

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3. Find a three-digit number n that is equal to the sum of all the two-digitnumbers that an be formed by using only the digits of the number n. (Notethat if a is one of the digits of the number n, then aa is one of the two-digitnumbers that an be formed.)

387Solution by Titu Zvonaru, Com�ane�sti, Romania.Let a, b, and c denote distin t digits. The following ases are ruled out:(a) n = aaa = aa ⇒ 111a = 11a ⇒ 100a = 0; no solution in this ase.(b) n = aab = aa+ bb+ab+ ba ⇒ 110a+ b = 22a+22b ⇒ 88a = 21b.There is no solution here, sin e 11 divides 88a but does not divide 21b.( ) n = aba = aa + bb + ab + ba ⇒ 79a = 12b; no solution as b is notdivisible by 79.(d) n = baa = aa + bb + ab + ba ⇒ 78b = 11a; no solution as 78b is notdivisible by 11.Thus, n = abc = aa + bb + cc + ab + ba + bc + cb + ac + ca and wehave abc = 33(a + b + c).It follows that 3 divides abc, hen e 3 divides a+b+c and then 9 divides33(a + b + c). Now 9 divides abc, hen e 9 divides a + b + c and n an onlybe one of

33 · 9 = 297 ; 33 · 18 = 594 ; 33 · 27 = 891 .Therefore, 594 = 55 + 99 + 44 + 59 + 95 + 94 + 49 + 54 + 45 is theonly solution.4. Transportation between six ities is su h that between any two itiesthere is either a bus or a train but not both of these. Show that among thesesix ities there are three ities that are linked either only by buses or only bytrains.Solved by George Apostolopoulos, Messolonghi, Gree e; and John GrantM Loughlin, University of New Brunswi k, Frederi ton, NB. We give theresponse of M Loughlin.Consider six ities A, B, C, D, E, and F . There must be either atleast three train links to ity A or at least three bus links to ity A. We willassume three train links without loss of generality. Let us suppose that A is onne ted to ea h of B, C, and F by train, as indi ated in the diagram.Assume that no three ities anbe linked by train or bus. Then CR,BC, and BF annot be train links, asthat would result in linking three itiesby train.Using dashed segments for buslinks, it now happens thatB, C, and Fare all onne ted by a bus link. Hen e,using proof by ontradi tion, it has ..

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A B

CF

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T T

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been shown that there are three ities linked only by buses or only by trains.

388Comment. This is a tually the game of SIM in disguise. The game requiresplayers to alternately (using one olour ea h) join two verti es with a seg-ment. The player who �rst forms a triangle with all segments in their olourloses. The game annot end in a tie as shown in this problem.5. Determine whether the fra tion 1

2005 an be written as a sum of 2005di�erent unit fra tions. (A unit fra tion is a fra tion of the form 1

n, where nis a natural number.)Solution by John Grant M Loughlin, University of New Brunswi k, Freder-i ton, NB.Yes, 1

2005 an be written as the sum of 2005 distin t unit fra tions.For all positive integers m, 1

m=

1

m + 1+

1

m(m + 1). If m ≥ 2, this an be employed to in rease the number of unit fra tions in a sum withoutlimit:

1

2005=

1

2006+

1

(2005)(2006)

=1

2006+

1

(2005)(2006) + 1+

1

(2005)(2006)[

(2005)(2006) + 1]

and so on!6. Let h be the altitude from A in an a ute triangle ABC. Prove that(b + c)2 ≥ a2 + 4h2 ,where a, b, and c are the lengths of the sides opposite A, B, and C respe -tively.Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; GeorgeApostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; OliverGeupel, Br �uhl, NRW, Germany; and Titu Zvonaru, Com�ane�sti, Romania. Wegive the solution of Amengual-Covas.The triangle ABC need not be a ute, as the proof will show.Let wa be the length of the internal bise tor of angle A.From the known relations

wa =2bc

b + ccos

A

2and 4bc ≤ (b + c)2 ,we obtain

w2a = bc · 4bc

(b + c)2cos2

A

2≤ bc · cos2

A

2.

389However, h ≤ wa, so we have

h2 ≤ w2a ≤ bc · cos2

A

2= bc · 1 + cos a

2

= bc ·1 +

(

b2 + c2 − a2

2bc

)

2=

(b + c)2 − a2

4and (b + c)2 ≥ a2 + 4h2 follows, as desired.Equality holds only if b = c.To omplete this number of the Corner we begin to look at solutionsfrom our readers to problems given in the O tober 2008 issue, starting withthe Olimpiada Matem�ati a Espa ~nola 2005 given at [2008 : 341{342℄.3. Let r, s, u, and v be real numbers. Prove that

min{r − s2, s − u2, u − v2, v − r2} ≤ 1

4.Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille,Rouen, Fran e; and Titu Zvonaru, Com�ane�sti, Romania. We give Bataille'swrite-up.Suppose that min

{

r − s2, s − u2, u − v2, v − r2}

>1

4. Then ea h ofthe numbers r − s2, s − u2, u − v2, and v − r2 is greater than 1

4, hen e

(r + u) −(

s2 + v2)

>1

2(1)and

(s + v) −(

r2 + u2)

>1

2. (2)Sin e (s + v)2 ≤ 2

(

s2 + v2), inequality (1) yields

2(r + u) − (s + v)2 > 1and similarly from inequality (2) we obtain2(s + v) − (r + u)2 > 1 .Let a = r + u. Then,

2a > 1 +

(

1 + a2

2

)2 ,whi h is equivalent to the inequality a4 +2a2 −8a+5 < 0, a ontradi tion,sin e a4 + 2a2 − 8a + 5 = (a − 1)2(

a2 + 2a + 5) is nonnegative for all realnumbers a. The result follows.

That ompletes the Corner for this issue. Send me your ni e solutionsand generalizations!

390BOOK REVIEWSAmar SodhiSink or Float? Thought Problems in Math and Physi sBy Keith Kendig, Mathemati al Asso iation of Ameri a, 2008ISBN 978-0-88385-339-9, hard over, 375+xiii pages, US$59.95Reviewed by Nora Franzova, Langara College, Van ouver, BCMay the title not onfuse you! Only a small part of the book is gearedtowards Ar himedes enthusiasts. A tually there are ten di�erent ategoriesof problems spanning geometry, numbers, probability, lassi al me hani s,astronomy, linear algebra, ele tri ity and magnetism, the leaking tank, heatandwave phenomena, and Ar himedes' Prin iple. These ategories are neatlyseparated into groups, where transitioning between math and physi s prob-lems happens without any a tual transition. Like walking on a M�obius strip,there is only one side to it: the hallenge and joy of solving problems.The book opens with \A Sampler" of problems that immediately pullyou in and, as their diÆ ulty in reases, you realize that this is a treasure tobe savoured slowly, sin e ea h problem, no matter how short and sweet itseems, an lead to surprising truths.Ea h of the ten se tions of the book starts with an introdu tion thatbrings up interesting histori al information and lays down notation, basi de�nitions, and o asional theorems that are needed in the se tion. Shortexamples are in luded. These are not aimed to tea h the material, just tobring the reader to the starting point from whi h the author is observingthat parti ular subje t area.Problems are not numbered, but ea h has a title, a name that is alreadya part of the story. Ea h title is followed by a short des ription, and thena list of possible answers is given in a multiple hoi e format with most ofthe answers being oh-so-tempting. In between individual problems, addi-tional histori al and theoreti al information is provided in learly separateddarkened frames, whi h seem very inviting. Kendig revisits several questionsmore than on e, so we en ounter not only a problem alled \Friday the 13th",but also \Friday the 13th Again" and \Friday the 13th OneMore Time". Noneinvolves any bla k ats.The author states: \I hope that this book will serve as a springboard {either individually or in lassroom { for mathemati al thought, ase testing,dis ussions, arguments." The �rst step an be taken by following his advi efrom the prefa e. The author re ommends one read the problem, think aboutit, then lose the book and think about it some more. Only then should oneturn the pages to the designated pla e where the omplete solution an befound. The solutions are more than omplete; many also in lude short sidedis ussions about related questions. Very helpful are the graphi al imagesthat are part of many of the problems and their solutions. These images talkto the reader through questions in framed \bubbles".

391The book brought forth an immediate urry of interest and ommentslike \every problem is a gem" or \this is so neat," from family and guestsafter it had be ome a staple on our o�ee table. One might say that a \mathbook" is not o�ee table material, but the problems in the book are attra -tively presented, and fairly simple to understand, so a lay person (let's saya \mathemati al muggle⋆") an be tempted to approa h a problem like thefollowing: \Is it possible to stand at just the right pla e on the earth so thatthe moon's res ent looks like a smiley fa e instead of being more verti al?"And a mathemati al wizard (just to sti k with Harry Potter terminology) willhumbly turn the pages, pretending only mild interest while still trying tosolve that probability problem from the beginning. A math wizard knowsthat these questions only look inno ent. In reality they hide a lot of deeptruths, ready to be un overed again and again.Even though the book is from the Dol iani Mathemati al Expositionsseries, it de�nitely makes an ex ellent o�ee table book. Many of the prob-lems an lead to lively dis ussions and arguments for any group of uriousindividuals. For a similar reason the book would make an ex ellent prize tobe given to winners of math ontests or s ien e fairs. And, sin e it is not to beread qui kly, professors, graduate and undergraduate students, or even highs hool students and their tea hers, will be oming ba k to it to rejuvenatetheir uriosity and wonder about nature's mysteries.We all know that solving a good problem brings us not only satisfa tionbut also the joy of understanding. And that an only be repeated by �ndingand solving another interesting problem and then another ... .At this point, all that is left to the thoughtful person is to open KeithKendig's book and start solving problems.⋆Muggle is the word used in the Harry Potter series of books by J.K. Rowlingto refer to a person who la ks any sort of magi al ability.

Number Theory Through InquiryBy David C. Marshall, Edward Odell, and Mi hael Starbird, Mathemati alAsso iation of Ameri a, Washington, DC, 2007ISBN 978-0-88385-751-9, hard over, 140+x pages, US$51.00Reviewed by Jeff Hooper, A adia University, Wolfville, NSAlthough we an learn mathemati s by reading texts or by sitting in lasses, mathemati s is not a spe tator sport: most of us learn new math-emati al ideas by a tively working through examples and problems. Onere ent, and very laudable, trend in the tea hing of mathemati s has been to-wards the adoption of tea hing methods whi h push students towardsbe oming more independent in their mathemati al thinking. An importantexample of this is the method known as Inquiry-Based Learning (or IBL forshort).The basi notion behind IBL-related methods is to have the students'ideas and questions drive the development of the material, with the tea her

392a ting as fa ilitator. In other words, in its purest form, students would workthrough examples, use them to suggest onne tions among ideas, formulatestatements of onje tures, and then prove these results, with the instru torthere only to help push things along when needed. In prin iple, be ausestudents must work out mu h of the details and proofs themselves, theygain a deeper understanding of the material, and more importantly developa mu h higher level of mathemati al skill. Of ourse, this pure form of IBL,as I've des ribed it, would be extremely taxing for the students involved,and far too time- onsuming, so in pra ti e the approa h gets modi�ed sothat students are at least provided, as the ourse moves along, with pie esof an outline: examples to onsider and statements of results whi h must beproved. This skeletal outline ensures that the inquiry moves in appropriatedire tions.In mathemati s, a version of this approa h was hampioned in the 1950sand 60s by R.L. Moore of the University of Texas at Austin. Moore used thismethod to tea h sele ted lasses of top students, many of whom went onto earn Ph.D. degrees and be ome well-established resear hers. Moore'smethod was very pure, and students often had to on o t their own de�ni-tions of ne essary on epts.The authors have been involved in a `modi�ed-Moore' approa h totea hing mathemati s, whi h is aimed at a wider audien e, and this book isan outgrowth of their ourses. The text has two main goals in mind: tea h-ing the fundamental material overed in a standard undergraduate numbertheory ourse, and tea hing students to be ome independent mathemati althinkers.The book overs a fairly standard sele tion of topi s: divisibility andgreatest ommon divisors, primes and their distribution, the arithmeti of ongruen es, Fermat's Little Theorem and Euler's generalization, appli a-tions to publi -key ryptography and RSA, higher-order ongruen es andprimitive roots, quadrati re ipro ity, Pythagorean triples, sums of squares,Pell's equation and introdu tory Diophantine approximation, and primalitytesting. So the outline is mu h like what one would �nd in many popularelementary number theory texts.What makes this text more unusual is the style. The authors use thismaterial for introdu ing students to proofs and abstra t mathemati s, so the�rst ouple of hapters are more leisurely than the rest, and there are many omments to help the reader along. More importantly, be ause the entirepoint of this book is its use in an IBL-style number theory ourse, the nar-rative in ludes de�nitions, exer ises for students to work through, as wellas statements of lemmas, propositions, theorems, and orollaries. But noproofs or solutions are in luded. The intent is that the reader will workout the proofs of these ideas for herself. The authors also in lude numerousquestions for students to ponder, many of them open-ended. These generallyrequire students to think more deeply about the material just en ountered,and often ask students to formulate onje tures and prove them. Many se -tions on lude with a `Blank-Paper Exer ise'. These ask students to return,

393after a day or two away from the material, and try to outline the ideas in asmu h detail as possible without looking at the text.Other number theory texts have been written from a problems point-of-view. Three whi h an be found in many libraries are: R.P. Burn's A Pathwayinto Number Theory, Joe Roberts' Elementary Number Theory: A ProblemOriented Approa h, and Sierpi �nski's 250 Problems in Elementary NumberTheory.So who would bene�t from this text? On the ba k over the book sug-gests its use as a text for a proof-transition or introdu tory abstra t mathe-mati s ourse, or for independent study. But I must add a aveat to this lastsuggestion: be ause there are no proofs or solutions, I'm not so sure of itsappropriateness for independent work. It would ertainly be useful in thisregard, but likely would require that the student or reader have some a essto help when they get stu k. I'm not omplaining here; it's lear that provid-ing proofs and solutions would detra t onsiderably from its other purpose.However, a good high-s hool student ould work through a text like this onlyif she an depend on some a ess to tea her support.As a professor, I found myself disappointed in one respe t. I've beenusing a related approa h in some of my senior-level undergraduate ourseswith some su ess, and I was really looking forward to reading about theauthors' approa h. How do they use this material in their ourses? Whatadvi e would they have for an instru tor hoping to use this material? Mosttexts treat these sorts of questions in the prefa e. Unfortunately, for somereason the authors in lude no prefa e, and their introdu tory hapter, whiledes ribing the text and speaking to its goals, doesn't address these questionsor explain how the material ould be used. Reading through the text fairly arefully, the `Questions' and `Blank-Paper Exer ises' give some hint at theapproa h the authors use, but I would have wished for more here.My small disappointment aside, this is a good text, and should er-tainly be onsidered by anyone looking to experiment with an inquiry-basedapproa h to the tea hing and learning of mathemati s.Geometri OrigamiBy Robert Gerets hl�ager, Arbelos, 2008ISBN 978-0-9555477-1-3, soft over, 198+xiii pages, US$24.99Reviewed by Georg Gunther, Sir Wilfred Grenfell College, Corner Brook,NL Origami is the traditional Japanese art of paper folding whi h has beenpra tised in that ountry for over 400 years. Not surprisingly, this art formwith its stri t dependen e on a sequen e of pre ise folds, has long been of in-terest to mathemati ians. In 1989, the �rst International Meeting of OrigamiS ien e and Te hnology was held; sin e that time there have been numerous onferen es devoted to this burgeoning �eld, and today the resear h litera-ture is extensive and ever growing.

394The book Geometri Origami provides a ni e balan e between theoryand pra ti al appli ation. If your interest is to fold a regular triskaide agon(otherwise known as a 13-gon), you will �nd detailed instru tions, a om-panied by lear drawings, to guide you on your way. On the other hand, forthose whose interests are theoreti al, they will �nd a lear exposition of themathemati s underlying this onstru tion.This is a book whi h yields wonderful nuggets, regardless whether itis read arefully from beginning to end, or whether it is skimmed super�- ially. Part I, whi h develops the underlying mathemati al theory, providesa lear and areful dis ussion of the axioms of origami onstru tions, andshows that these are more powerful than the orresponding axioms govern-ing Eu lidean onstru tions. Spe i� ally, origami onstru tions are powerfulenough to provide geometri solutions to the problem of solving the general ubi equation (and, indeed, the general quarti as well), something that isbeyond the s ope of Eu lidean onstru tions. Hen e, as the author shows,origami onstru tions enable one to solve two of the lassi al onstru tionproblems: dupli ating the ube and trise ting a general angle.In Part II, the author gives lear step-by-step origami onstru tionsfor a fas inating lass of problems. All of these results are bolstered by athorough dis ussion of the underlying theory. Mu h of this se tion fo useson origami onstru tions of regular polygons, and Gerets hl�ager ta kles thisfrom several dire tions. First, he explores the general question of whi h

n-gons are onstru tible (for example, he shows that if n = 5 · 2k · 3ℓ, thenthe orresponding n-gon an be folded). Subsequently, the author devotesone hapter to questions around the theory and onstru tion of the regularpentagon, followed by a hapter exploring similar questions for the regularheptagon. The book on ludes with a �nal hapter des ribing how to fold theregular nonagon, triskaide agon, 17-gon, and 19-gon.This book is a delightful addition to a wonderful orner of mathemati swhere art and geometry meet in su h a fruitful union. Anyone with a workingknowledge of elementary geometry, algebra, and the geometry of omplexnumbers will have little trouble following the theoreti al dis ussions. It willbe of value to a wide range of audien es; new omers to the �eld of mathemat-i al paper-folding will �nd, easily a essible, all the tools they need, whileexperts will enjoy a well-written, beautifully illustrated referen e work.

395PROBLEMS

Toutes solutions aux probl �emes dans e num�ero doivent nous parvenir au plustard le 1er avril 2010. Une �etoile (⋆) apr �es le num�ero indique que le probl �eme a �et �esoumis sans solution.Chaque probl �eme sera publi �e dans les deux langues oÆ ielles du Canada(anglais et fran� ais). Dans les num�eros 1, 3, 5 et 7, l'anglais pr �e �edera le fran� ais,et dans les num�eros 2, 4, 6 et 8, le fran� ais pr �e �edera l'anglais. Dans la se tion dessolutions, le probl �eme sera publi �e dans la langue de la prin ipale solution pr �esent �ee.La r �eda tion souhaite remer ier Jean-Mar Terrier, de l'Universit �e deMontr �eal, d'avoir traduit les probl �emes.3439. Rempla ement. Propos �e par D.J. Smeenk, Zaltbommel, Pays-Bas.Soit Γ un er le de entre O et de rayon R. La droite t est tangente �a Γau point A, et P est un point sur t distin t de A. La droite ℓ, distin te de t,passe par P et oupe Γ aux points B et C. Le point K est sur la droite ACet tel que PK‖AB, et le point L est sur la droite AB et tel que PL‖AC.Montrer que KL ⊥ OP .3463. Propos �e par Mi hel Bataille, Rouen, Fran e.Soit Γ un er le de entre O et de rayon r, et soit P un point tel queOP > r. Soit L l'ensemble de toutes les droites ℓ telles que P 6∈ ℓ et ℓ oupeΓ aux points A, B de sorte que PA · PB = OP 2 − r2. Montrer que L estun fais eau de droites on ourantes.3464. Propos �e par Mi hel Bataille, Rouen, Fran e.Soit ABC un triangle ave ∠A = 90◦ et H le pied de la hauteurabaiss �ee de A. Soit J le point sur l'hypot �enuse BC tel que CJ = HB etsoit K, L les proje tions respe tives de J sur AB, AC. Montrer que

M(

−2

3; AK, AL

)

≤ M (−2; AB, AC) ,o �u M (α; x, y) =

(

xα + yα

2

)1/α.3465. Propos �e par Xavier Ros ( �etudiant) and Jos �e Luis D��az-Barrero, Uni-versit �e Polyte hnique de Catalogne, Bar elone, Espagne.Montrer que ∞∑

i=1

∞∑

j=1

1

ij(i + j)<

π2

6+

1

2+

3

4log 2.

3963466. Propos �e par Tuan Le, �etudiant, Fairmont High S hool, Anaheim,CA, �E-U.Soit x, y et z des nombres r �eels positifs tels que xyz ≥ 10 + 6

√3.Montrer que

y

x + y3 + z2+

z

y + z3 + x2+

x

z + x3 + y2≤ 1

2.

3467. Propos �e par Tuan Le, �etudiant, Fairmont High S hool, Anaheim,CA, �E-U.Soit x, y et z des nombres r �eels positifs. Montrer que3

√

x3 + y3 + z3

xyz+

√

xy + yz + xz

x2 + y2 + z2≥ 3

√3 + 1 .

3468. Propos �e par Joseph DeVin entis, Salem, MA, �E-U ; BernardoRe am�an, Institut Alberto Merani, Bogota, Colombie ; Peter Saltzman,Berkeley, CA, �E-U ; et Stan Wagon, Ma alester College, St. Paul, MN, �E-U.Trouver tous les entiers positifs n pour lesquels on peut faire orres-pondre �a haque nombre pair de {2, 4, . . . , 2n} exa tement un nombreimpair de {1, 3, . . . , 2n − 1} de telle sorte que les sommes des n paires,prises deux �a deux, soient des nombres relativement premiers. Par exemple,si n = 3, on a une solution, 2 + 3 = 5, 4 + 5 = 9, 6 + 1 = 7.3469. Propos �e par Mihaela Blanariu, Coll �ege Columbia Chi ago, Chi ago,IL, �E-U.Soit p ≥ 2 un nombre r �eel. Trouver la limitelim

n→∞

1 +√

2(2!)p + 3√

3(3!)p + · · · + n√

n(n!)p

(n!)p.

3470. Propos �e par Mihaela Blanariu, Coll �ege Columbia Chi ago, Chi ago,IL, �E-U.Soit p ≥ 2 un nombre r �eel. Trouver la limitelim

n→∞

1 +(√

2!)p

+(

3√

3!)p

+ · · · +(

n√

n!)p

np+1.3471. Propos �e par C�at�alin Barbu, Ba �au, Roumanie.Soit ABC un triangle a utangle et M , N , P les points milieu respe tifsdes petits ar s BC, CA, AB. Si [XY Z] d �esigne l'aire du triangle XY Z,montrer que [MBC] + [NCA] + [PAB] ≥ s(3r − R), o �u s, r et R sontrespe tivement le demi-p �erim �etre, le rayon du er le ins rit et le rayon du er le ir ons rit du triangle ABC.

3973472. Propos �e par Xavier Ros, �etudiant, Universit �e Polyte hnique deCatalogne, Bar elone, Espagne.Soit x, y et z des nombres r �eels positifs tels que x ≤ y ≤ z. Montrerquelog x · log

(

1 + y

1 + z

)

+ log y · log

(

1 + z

1 + x

)

+ log z · log

(

1 + x

1 + y

)

≥ 0 .3473. Propos �e par Walther Janous, Ursulinengymnasium, Innsbru k,Autri he, �a la m�emoire de Jim Totten.Soit c un nombre r �eel donn �e et tel que 0 < c ≤ 1.(a) Montrer que pour tout nombre r �eel positif µ, on a

2√

cµ + 1+

µ√

c + µ2≤ 3

√c + 1

.(b) ⋆. Trouver tous les nombres r �eels positifs λ tels que la relation

λ√

cµ + 1+

µ√

c + µ2≤ λ + 1

√c + 1soit v �eri� �ee pour tous les nombres r �eels positifs µ.3474. Propos �e par Silouanos Brazitikos et Christos Patilas, Trikala, Gr �e e.Soit x, y et z des nombres r �eels positifs tels que xyz = 1. Montrer sioui ou non, on a l'in �egalit �e 2

√

3(

xy + yz + zx)

+ x + y + z ≥ 9.[Ed. : Les proposeurs font remarquer que l'in �egalit �e est appuy �ee pardes al uls sur ordinateur.℄.................................................................3439. Repla ement. Proposed by D.J. Smeenk, Zaltbommel, the Nether-lands.Let Γ be a ir le with entre O and radius R. Line t is tangent to Γ atthe point A, and P is a point on t distin t from A. The line ℓ distin t from tpasses through P and interse ts Γ at the points B and C. The point K is onthe line AC and su h that PK‖AB, and the point L is on the line AB andsu h that PL‖AC. Prove that KL ⊥ OP .3463. Proposed by Mi hel Bataille, Rouen, Fran e.Let Γ be a ir le with entre O and radius r, and let P be a point withOP > r. Let L be the set of all lines ℓ su h that P 6∈ ℓ and ℓ interse ts Γat points A, B su h that PA · PB = OP 2 − r2. Show that L is a pen il of on urrent lines.

3983464. Proposed by Mi hel Bataille, Rouen, Fran e.Let ABC be a triangle with ∠A = 90◦ and H be the foot of the altitudefrom A. let J be the point on the hypotenuse BC su h that CJ = HB andlet K, L be the proje tions of J onto AB, AC, respe tively. Prove that

M(

−2

3; AK, AL

)

≤ M (−2; AB, AC) ,where M (α; x, y) =

(

xα + yα

2

)1/α.3465. Proposed by Xavier Ros (student) and Jos �e Luis D��az-Barrero, Uni-versitat Polit �e ni a de Catalunya, Bar elona, Spain.Prove that ∞∑

i=1

∞∑

j=1

1

ij(i + j)<

π2

6+

1

2+

3

4log 2.

3466. Proposed by Tuan Le, student, Fairmont High S hool, Anaheim,CA, USA.Let x, y, and z be positive real numbers su h that xyz ≥ 10 + 6√

3.Prove thaty

x + y3 + z2+

z

y + z3 + x2+

x

z + x3 + y2≤ 1

2.

3467. Proposed by Tuan Le, student, Fairmont High S hool, Anaheim,CA, USA.Let x, y, and z be positive real numbers. Prove that3

√

x3 + y3 + z3

xyz+

√

xy + yz + xz

x2 + y2 + z2≥ 3

√3 + 1 .

3468. Proposed by Joseph DeVin entis, Salem, MA, USA; BernardoRe am�an, Instituto Alberto Merani, Bogot �a, Colombia; Peter Saltzman,Berkeley, CA, USA; and Stan Wagon, Ma alester College, St. Paul, MN, USA.Find all positive integers n for whi h one an mat h ea h even numberfrom {2, 4, . . . , 2n} with exa tly one odd number from {1, 3, . . . , 2n − 1}so that the sums of the resulting n pairs are pairwise relatively prime. Forexample, if n = 3 a solution is 2 + 3 = 5, 4 + 5 = 9, 6 + 1 = 7.3469. Proposed byMihaela Blanariu, Columbia College Chi ago, Chi ago,IL, USA.Let p ≥ 2 be a real number. Find the limitlim

n→∞

1 +√

2(2!)p + 3√

3(3!)p + · · · + n√

n(n!)p

(n!)p.

3993470. Proposed byMihaela Blanariu, Columbia College Chi ago, Chi ago,IL, USA.Let p ≥ 2 be a real number. Find the limit

limn→∞

1 +(√

2!)p

+(

3√

3!)p

+ · · · +(

n√

n!)p

np+1.

3471. Proposed by C�at�alin Barbu, Ba �au, Romania.Let ABC be an a ute triangle and M , N , P be the midpoints of theminor ar sBC, CA, AB; respe tively. If [XY Z] denotes the area of triangleXY Z, prove that [MBC]+[NCA]+[PAB] ≥ s(3r−R), where s, r, and Rare the semiperimeter, the inradius, and the ir umradius of triangle ABC,respe tively.3472. Proposed by Xavier Ros, student, Universitat Polit �e ni a de Catalunya,Bar elona, Spain.Let x, y, and z be positive real numbers with x ≤ y ≤ z. Prove thatlog x · log

(

1 + y

1 + z

)

+ log y · log

(

1 + z

1 + x

)

+ log z · log

(

1 + x

1 + y

)

≥ 0 .3473. Proposed by Walther Janous, Ursulinengymnasium, Innsbru k,Austria, in memory of Jim Totten.Let c be a �xed real number su h that 0 < c ≤ 1.(a) For all positive real numbers µ prove that

2√

cµ + 1+

µ√

c + µ2≤ 3

√c + 1

.(b)⋆. Determine all positive real numbers λ su h that

λ√

cµ + 1+

µ√

c + µ2≤ λ + 1

√c + 1holds for all positive real numbers µ.3474⋆. Proposed Silouanos Brazitikos and Christos Patilas, Trikala,Gree e.Let x, y, and z be positive real numbers with xyz = 1. Prove ordisprove that 2

√

3(

xy + yz + zx)

+ x + y + z ≥ 9.[Ed.: The proposers indi ate that the inequality is supported by om-puter omputations.℄

400SOLUTIONSAu un probl �eme n'est immuable. L' �editeur est toujours heureux d'en-visager la publi ation de nouvelles solutions ou de nouvelles perspe tivesportant sur des probl �emes ant �erieurs.

3363. [2008 : 362, 364℄ Proposed by Toshio Seimiya, Kawasaki, Japan.Let ABC be a triangle with ∠ACB = 90◦ + 12∠ABC. Let M be themidpoint of BC. Prove that ∠AMC < 60◦.Composite of similar solutions by Ri ardo Barroso Campos, University ofSeville, Seville, Spain; Ri hard I. Hess, Ran ho Palos Verdes, CA, USA; MattiLehtinen, National Defen e College, Helsinki, Finland; Madhav R. Modak,formerly of Sir Parashurambhau College, Pune, India; and Peter Y. Woo,Biola University, La Mirada, CA, USA.In △ABC let 4β = ∠B and let W be the foot of the bise tor of ∠A.We are given that ∠C = 90◦ + 2β, when e ∠A = 90◦ − 6β. We dedu e that

6β < 90◦, so that β < 15◦. Moreover, be ause ∠AWC is an external angleof △ABW , it satis�es∠AWC = 4β +

(

45◦ − 3β)

= 45◦ + β < 60◦ .It remains to show that ∠AMC < ∠AWC. Be ause we are given that ∠Cis obtuse, we have AB > AC, when e the foot W of the angle bise torsatis�es BW > WC. That is, W lies between M and C; thus ∠AWC is anexternal angle of △AMW and therefore satis�es ∠AMC < ∠AWC < 60◦,as desired.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; �SEFKETARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA,Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS,Missouri Southern State University, Joplin, MO, USA; GERALD EDGECOMB and JULIESTEELE, students, California State University, Fresno, CA, USA; OLIVER GEUPEL, Br �uhl, NRW,Germany; JOHN G. HEUVER, Grande Prairie, AB; WALTHER JANOUS, Ursulinengymnasium,Innsbru k, Austria; V �ACLAV KONE �CN �Y, Big Rapids, MI, USA; KEE-WAI LAU, Hong Kong,China; TAICHI MAEKAWA, Takatsuki City, Osaka, Japan; THANOS MAGKOS, 3rd High S hoolof Kozani, Kozani, Gree e; SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia andHerzegovina; SKIDMORE COLLEGE PROBLEM SOLVING GROUP, Skidmore College, SaratogaSprings, NY, USA; GEORGE TSAPAKIDIS, Agrinio, Gree e; TITU ZVONARU, Com�ane�sti,Romania; and the proposer.3364. [2008 : 362, 364℄ Proposed by Toshio Seimiya, Kawasaki, Japan.Let ABC be a triangle with ∠BAC = 120◦ and AB > AC. Let M bethe midpoint of BC. Prove that ∠MAC > 2 ∠ACB.

401I. Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA.Let β = ∠CBA and γ = ∠BCA; then (be ause ∠BAC = 120◦)β + γ = 60◦. Let D be the point of segment BC for whi h ∠BAD = 2β;then ∠DAC = 2γ. The Law of Sines tells us that

BD =AD sin 2β

sin β= 2AD cos β ;also,

CD =AD sin 2

(

60◦ − β)

sin(

60◦ − β) = 2AD cos

(

60◦ − β) .But β+γ = 60◦ and γ > β; hen e, 60◦−β > β so that cos

(

60◦−β)

< cos β.Consequently, CD < BD, when e M lies between B and D. It follows that∠MAC > ∠DAC = 2∠ACB, as desired.II. Solution by Matti Lehtinen, National Defen e College, Helsinki, Finland.Let O be the entre of the ir um ir le Γ of △ABC, and D be the mid-point of the ar CAB of Γ. Note that A is on the smaller ar DC. Sin e120◦ = ∠BAC = ∠BDC, DB = DC = DO. Draw the ir le Γ′ with en-tre D and radius DB (passing through B, O, and C). Be ause M is the om-mon midpoint of BC and DO, ir les Γ and Γ′ are symmetri with respe t toM . Let AM meet Γ′ at E (so that M is also the midpoint of AE); note that∠EDC = ∠AOB = 2∠ACB. Extend AE to meet Γ again at F and extendDE to meet Γ again at G. Be ause the line FA separates D from G andC, G is on the ar FC opposite D and A; onsequently, ∠FAC > ∠GDC.But, ∠FAC = ∠MAC, and ∠GDC = ∠EDC = ∠AOB = 2∠ACB, so weare done.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; �SEFKETARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA,Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS,Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br �uhl, NRW,Germany; V �ACLAV KONE �CN �Y, Big Rapids, MI, USA; KEE-WAI LAU, Hong Kong, China;THANOS MAGKOS, 3rd High S hool of Kozani, Kozani, Gree e; SALEM MALIKI �C, student,Sarajevo College, Sarajevo, Bosnia and Herzegovina; GEORGE TSAPAKIDIS, Agrinio, Gree e;TITU ZVONARU, Com�ane�sti, Romania; and the proposer.3365. [2008 : 362, 364℄ Proposed by Toshio Seimiya, Kawasaki, Japan.A square ABCD is ins ribed in a ir le Γ. Let P be a point on theminor ar AD of Γ, and let E and F be the interse tions of AD with PBand PC, respe tively. Prove that

AE · DF = 2(

[PAE] + [PDF ]) ,where [KLM ] denotes the area of triangle KLM .

402Solution by John G. Heuver, Grande Prairie, AB.The altitudes from pointP to segmentsAB andCD lead us to on ludethat

[PAB] + [PDC] =1

2AB2 ,Furthermore,

1

2AB2 = [PAE] + [AEB] + [PDF ] + [DFC] ,Sin e

[AEB] + [DFC] =1

2AB(AE + DF ) ,we have

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A

B C

DE F

P

q q

b

b

[PAE] + [PDF ] =1

2AB2 −

(

[AEB] + [DFC])

=1

2AB2 − 1

2AB(AE + DF )

=1

2AB(AB − AE − DF ) =

1

2AB · EF .Now, ∠BPC = ∠CPD = ∠45◦ as angles subtended by ar s equal toa quarter of the ir le. Hen e PF is the bise tor of ∠EPD, and therefore,

EF

FD=

PE

PD.Also, ∠ABP = ∠PDA as angles subtended by the ar AP , so that triangles

PDE and ABE are similar, and therefore,PE

PD=

AE

AB.Consequently,

EF

FD=

AE

AB,or EF · AB = AE · DF , and then

[PAE] + [PDF ] =1

2AB · EF =

1

2AE · DF ,whi h ompletes the proof.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; �SEFKETARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA,Lebanese University, Fanar, Lebanon; RICARDO BARROSO CAMPOS, University of Seville,Seville, Spain; MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern StateUniversity, Joplin, MO, USA; OLIVER GEUPEL, Br �uhl, NRW, Germany; WALTHER JANOUS,Ursulinengymnasium, Innsbru k, Austria; V �ACLAV KONE �CN �Y, Big Rapids, MI, USA; SALEM

403MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; MADHAV R. MODAK,formerly of Sir Parashurambhau College, Pune, India; JOEL SCHLOSBERG, Bayside, NY, USA;D.J. SMEENK, Zaltbommel, the Netherlands; PETER Y. WOO, Biola University, La Mirada,CA, USA; TITU ZVONARU, Com�ane�sti, Romania; and the proposer. There was one in orre t omment submitted.Kone� n �y and the proposer ea h began by letting Z be the foot of the perpendi ular fromP to AD, then dedu ing that △DPZ ∼ △BEA and △APZ ∼ △CFD, whi h then enablesthe al ulation of AE · DF .3366. [2008 : 362, 365℄ Proposed by Ovidiu Furdui, Campia Turzii, Cluj,Romania.Let {x} denote the fra tional part of the real number x; that is,{x} = x − ⌊x⌋, where ⌊x⌋ is the greatest integer not ex eeding x.Evaluate

∫ 1

0

{

1

x

}4

dx .Solution by Chip Curtis, Missouri Southern State University, Joplin, MO,USA. Let I be the integral to be evaluated. ThenI = lim

N→∞

N∑

n=1

∫ 1n

1n+1

{

1

x

}4

dx = limN→∞

N∑

n=1

∫ n+1

n

{y}4

y2dy

= limN→∞

N∑

n=1

∫ n+1

n

(y − n)4

y2dy = lim

N→∞

N∑

n=1

In ,where

In =

∫ n+1

n

(y − n)4

y2dy .We have

In =

∫ n+1

n

(

y2 − 4ny + 6n2 − 4n3

y+

n4

y2

)

dy

=

(

y3

3− 2ny2 + 6n2y − 4n3 ln y − n4

y

)

∣

∣

∣

∣

∣

n+1

n

= 3n2 − n +1

3− n4

n + 1+ n3 − 4n3 ln

(

1 +1

n

)

= 3n2 − n +1

3+

(

n2 − n − 1

n + 1+ 1

)

− 4n3 ln

(

1 +1

n

)

= 4n2 − 2n +4

3− 1

n + 1− 4n3 ln

(

1 +1

n

) .

404Substituting the series expansions

x

x + 1=

∞∑

k=0

(−1)kxk+1 and ln(1 + x) =∞∑

k=1

(−1)k+1xk

k,

with x =1

n(n ≥ 2) into the last expression for In yields

In = 4n2 − 2n +4

3−

∞∑

k=0

(−1)k

nk+1− 4n3

∞∑

k=1

(−1)k+1

knk

= 4n2 − 2n +4

3− 4n3

(

1

n− 1

2n2+

1

3n3

)

−∞∑

k=0

(−1)k

nk+1− 4n3

∞∑

k=4

(−1)k+1

knk

= −∞∑

k=0

(−1)k

nk+1− 4n3

∞∑

k=4

(−1)k+1

knk

=

∞∑

k=1

(−1)k

nk− 4

∞∑

k=4

(−1)k+1

knk−3

=∞∑

k=1

(−1)k

nk− 4

∞∑

k=1

(−1)k

(k + 3)nk

=∞∑

k=2

(

− 1

n

)k

− 4∞∑

k=2

(−1)k

(k + 3)nk

=1

n(n + 1)− 4

∞∑

k=2

(−1)k

(k + 3)nk,and the last formula is valid for I1 as well. Thus,

I =

∞∑

n=1

1

n(n + 1)− 4

∞∑

n=1

∞∑

k=2

1

k + 3

(

− 1

n

)k

= 1 − 4∞∑

k=2

(−1)k

k + 3

∞∑

n=1

1

nk

= 1 − 4

∞∑

k=2

(−1)k

k + 3ζ(k) .Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; RICHARDI. HESS, Ran ho Palos Verdes, CA, USA; OLIVER GEUPEL, Br �uhl, NRW, Germany; MADHAVR. MODAK, formerly of Sir Parashurambhau College, Pune, India; PAOLO PERFETTI,Dipartimento di Matemati a, Universit �a degli studi di Tor Vergata Roma, Rome, Italy; andthe proposer. There were two in orre t solutions submitted.The proposer obtained the answer −1

3− γ + 2 ln 2π − 12 ln A + 12 ln B, where γ isEuler's onstant and where

A = limn→∞

1122 · · · nn

nn2

2+ n

2+ 1

12 e−n2

4

; B = limn→∞

112222

· · · nn2

nn3

3+ n2

2+ n

6 e−n3

9+ n

12

405are the Glaisher{Kinkelin onstants of order 1 and order 2, respe tively.Geupel obtained I = 24

∞∑

k=2

ζ(k) − 1

k(k + 1)(k + 2)(k + 3)= 0.14553289 . . . , whi h heremarked onverges more rapidly than the series 1−

∞∑

k=2

(−1)kζ(k)

k + 3and is thus better suitedfor numeri al approximation.Janous reports the generalization

∫ 1

0

{

1

x

}N

dx =

N −2∑

j=0

(−1)j(N

j

) 1

N − j − 1

(

2N −j−1 − 1)

− (−1)N

(

N ln 2 −1

2

)

+∞∑

p=2

(−1)p(p − 1)

p + N − 1

(

ζ(p) − 1) .Keith Ekblaw, Walla Walla, WA, USA, obtained the estimate I ≈ 0.146 by aMonte Carloapproa h, whi h is orre t to 3 de imal pla es after rounding.

3367. [2008 : 362, 365℄ Proposed by Li Zhou, Polk Community College,Winter Haven, FL, USA.Let p(x) = anxn +an−1xn−1 + · · ·+a1x be a polynomial with integer oeÆ ients, where an > 0 and n∑

k=1

ak = 1. Prove or disprove that there arein�nitely many pairs of positive integers (k, ℓ) su h that p(k +1) − p(k) andp(ℓ + 1) − p(ℓ) are relatively prime.Solution by Cristinel Morti i, Valahia University of Targoviste, Romania,modi�ed slightly by the editor.Let Q(x) = p(x + 1) − p(x). Then we have

Q(0) = p(1) − p(0) =

n∑

k=1

ak − 0 = 1 .Hen e, Q(x) = xq(x) + 1 for some polynomial q(x) of degree n − 1.We need to �nd in�nitely many positive integers k and l su h thatgcd

(

Q(k), Q(l))

= 1.Sin e the leading term in Q(x) is nanxn−1 we have Q(k) > 0 forsuÆ iently large positive integers k. For su h k let l = Q(k) = kq(k) + 1,so that l is a positive integer. If, on the ontrary, gcd(Q(k), Q(l))

6= 1, thenthere is a prime number p su h that p | Q(k) and p | Q(l). Then, sin eQ(l) = lq(l) + 1 = Q(k)q(l) + 1, we on lude that p | 1, a ontradi tion.Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; OLIVER GEUPEL,Br �uhl, NRW, Germany; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; and theproposer.Morti i remarked that the on lusion in fa t holds for any polynomial Q(k) whose on-stant term is either 1 or −1. This is lear from the proof featured above. Barbara proved thestronger results that (1) if an 6= 0 and n

∑

k=1

ak = 1, then there are in�nitely many positive

406integers k su h that p(k + 1) − p(k) are all pairwise relatively prime, and (2) if f(x) is apolynomial with integer oeÆ ients and positive degree su h that f(0) = 1, then there arein�nitely many integers k1 < k2 < k3 < · · · su h that the f(ki)'s are all pairwise relativelyprime.3368. [2008 : 363, 365℄ Proposed by Neven Juri� , Zagreb, Croatia.Let m be an integer, m ≥ 2, and let A = [Aij ] be a blo k matrix ofdimension 2m × 2m with Aij ∈ M4,4(N) for 1 ≤ i, j ≤ 2m−2, de�ned byAij = 2mBij + Cij , where

Bij =

2m − 4i + 4 4i − 4 4i − 4 2m − 4i + 44i − 3 2m − 4i + 3 2m − 4i + 3 4i − 34i − 2 2m − 4i + 2 2m − 4i + 2 4i − 2

2m − 4i + 1 4i − 1 4i − 1 2m − 4i + 1

,and Cij =

4 − 4j 4j − 2 4j − 1 1 − 4j4j − 3 3 − 4j 2 − 4j 4j4j − 3 3 − 4j 2 − 4j 4j4 − 4j 4j − 2 4j − 1 1 − 4j

.Show that matrix A is a magi square of order 2m.Solution by Oliver Geupel, Br �uhl, NRW, Germany.For onvenien e, write Bij = B(i) =

(

b(i)kℓ

) and Cij = C(j) =(

c(j)kℓ

),and note that these 4 × 4 matri es have integer entries.Claim 1 For ea h entry a of A, 1 ≤ a ≤ 22m.Proof: We have a = 2mb(i)kℓ + c

(j)kℓ for some i, j, k, and ℓ. If we also have

(k, ℓ) ∈ {(1, 2), (1, 3)}, then b(i)kℓ ≥ 0 and c

(j)kℓ ≥ 1; hen e a ≥ 1. Otherwise,

b(i)kℓ ≥ 1 and c

(j)kℓ ≥ 1−2m; hen e a ≥ 2m+1−2m = 1. This proves the lowerbound. For the upper bound, we observe that for (k, ℓ) ∈ {(1, 1), (1, 4)} wehave b

(i)kℓ ≤ 2m and c

(j)kℓ ≤ 0; hen e a ≤ 2m · 2m = 22m. In any other ase

b(i)kℓ ≤ 2m − 1 and c

(j)kℓ ≤ 2m; so again a ≤ 2m

(

2m − 1)

+ 2m = 22m.Claim 2 The entries of A are distin t.Proof: If two entries of A are equal then 2mb(i)kℓ + c

(j)kℓ = 2mb

(i′)kℓ + c

(j′)kℓ forsome j, j′, k, k′, ℓ, and ℓ′. Let b = b

(i)kℓ , b′ = b

(i′)k′ℓ′ , c = c

(j)kℓ , and c′ = c

(j′)k′ℓ′.We then have 1 ≤ |c − c′| ≤ 4 · 2m−2 −

(

1− 4 ·2m−2)

= 2m+1 − 1 < 2m+1.Without loss of generality, suppose that c < c′. Then c′ − c = 2m andb = b′ + 1, hen e c ≡ c′ (mod 4) and b ≡ b′ + 1 (mod 4). There are four ases for c and c′:Case 1 (k, ℓ) ∈ {(1, 1), (4, 1)} and (k′, ℓ′) ∈ {(2, 4), (3, 4)};Case 2 (k, ℓ) ∈ {(1, 4), (4, 4)} and (k′, ℓ′) ∈ {(2, 1), (3, 1)};

407Case 3 (k, ℓ) ∈ {(2, 3), (3, 3)} and (k′, ℓ′) ∈ {(1, 2), (4, 2)};Case 4 (k, ℓ) ∈ {(2, 2), (3, 2)} and (k′, ℓ′) ∈ {(1, 3), (4, 3)};ea h of them in ompatible with the ondition b ≡ b′ + 1 (mod 4).Claim 3 The sum of the entries along any horizontal, verti al, or main diagonalline of A is 23m−1 + 2m−1.Proof: The sum along any given horizontal or verti al line of B(i) and C(j) isSB = 2m+1 and SC = 2, respe tively. Therefore, the sum of the entries inea h line ofA is 2m−2

(

2mSB+SC

)

= 2m−2(

2m·2m+1+2)

= 23m−1+2m−1.The main diagonal sums of B(i) and C(j) are b(i) = 2m+2 − 16i + 10 andc(j) = 10 − 16j. The entries in ea h main diagonal of A then also have thesum 2m−2

∑

i=1

2mb(i) +2m−2∑

j=1

c(j) = 23m−1 + 2m−1.Also solved by the proposer. One in omplete solution was re eived that veri�ed the row, olumn, and diagonal sums, but did not show that the entries of the magi square onsisted of1, 2, . . . , 22m .3369. [2008 : 363, 365℄ Proposed by George Tsintsifas, Thessaloniki,Gree e.Let A1A2A3A4 be a tetrahedron whi h ontains the entre O of its ir umsphere as an interior point. Let ρi be the distan e from O to the fa eopposite vertex Ai. If R is the radius of the ir umsphere, prove that

4

3R ≥

4∑

i=1

ρi .Solution by Oliver Geupel, Br �uhl, NRW, Germany.The laim is false; the following ounterexample is adapted from [1℄.First, onsider the degenerate tetrahedron A′

1A′2A′

3A′4, where

A′1 = A′

2 = (−1, 1, 0) ,A′

3 = (1, 1, 0) ,A′

4 = (−1, −1, 0) ,O = (0, 0, 0), and R =

√2. We have ρ1 = ρ2 = 0, and ρ3 = ρ4 = 1, sothat

4R4∑

i=1

ρi

=4√

2

2= 2

√2 < 3 .

408Now, let

A1 =(

−1 + ε, 1 − ε,√

2ε(2 − ε)) ,

A2 =(

−1 + ε, 1 − ε, −√

2ε(2 − ε)) ,

A3 =(√

2 cos(

π

4− ε

)

,√

2 sin(

π

4− ε

)

, 0) ,

A4 =(√

2 cos(

−3π

4+ ε

)

,√

2 sin(

−3π

4+ ε

)

, 0) .It is easy to verify that pointO is in the interior of the tetrahedronA1A2A3A4,and that Ai → A′

i for ea h i as ε → 0, as well as4R4∑

i=1

ρi

−→ 2√

2 .The bound 2

√2 is the best possible. This is a orollary from the fol-lowing:Theorem [1℄. LetA1A2A3A4 be a nondegenerate tetrahedron whose ir um- entreO is not an exterior point. LetP be a point not exterior toA1A2A3A4.Let the distan es from P to the verti es and to the fa es of A1A2A3A4 bedenoted by Ri and ρi, respe tively. Then

4∑

i=1

Ri

4∑

i=1

ρi

> 2√

2 ,and 2

√2 is the greatest lower bound.Referen es[1℄ Ni holas D. Kazarino�, \D.K. Kazarino�'s inequality for tetrahedra",Mi higan Mathemati al Journal, Vol. 4, No. 2 (1957), pp. 99-104.Counterexample also given by Peter Y. Woo, Biola University, La Mirada, CA, USA.George Apostolopoulos, Messolonghi, Gree e, ited the book by Kazarino�, Geometri Inequalities, Yale University Press, 1961, p. 116.No omplete solutions or other omments were submitted.

3370. [2008 : 363, 365℄ Proposed by George Tsintsifas, Thessaloniki,Gree e.Let ai and bi be positive real numbers for 1 ≤ i ≤ k, and let n be apositive integer. Prove that(

k∑

i=1

a1n

i

)n

≤(

k∑

i=1

ai

bi

)(

k∑

i=1

b1

n−1

i

)n−1.

409Composite of similar or identi al solutions submitted by all the solvers whosenames appear below (ex ept the two solvers identi�ed by a \*" before theirnames).Let p = n and q =

n

n − 1. Then p > 1, q > 1 and 1

p+

1

q= 1.[Ed: learly, n > 1 for the given inequality to make sense.℄Let xi =

(

ai

bi

)1n and yi = b

1n

i . Then xi and yi are positive for ea h i.ByH�older's Inequality, we have k∑

i=1

xiyi ≤(

k∑

i=1

xpi

)

1p(

k∑

i=1

yqi

)

1q whi h

be omes k∑

i=1

a1n

i ≤(

k∑

i=1

ai

bi

)

1n(

k∑

i=1

bn−1i

)

n−1n .The result follows by raising both sides to the nth power.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; �SEFKETARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE,Rouen, Fran e; MANUEL BENITO, �OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL,Logro ~no, Spain; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam;CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br �uhl,NRW, Germany; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinengymnas-ium, Innsbru k, Austria; SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia andHerzegovina; DUNG NGUYEN MANH, High S hool of HUS, Hanoi, Vietnam; *CRISTINELMORTICI, Valahia University of Targoviste, Romania; PAOLO PERFETTI, Dipartimento diMatemati a, Universit �a degli studi di Tor Vergata Roma, Rome, Italy; JOEL SCHLOSBERG,Bayside, NY, USA; *PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU,Com�ane�sti, Romania; and the proposer.

3371. [2008 : 363, 368℄ Proposed by George Tsintsifas, Thessaloniki,Gree e.Let ABC be a triangle with a, b, and c the lengths of the sides oppositethe verti es A, B, and C, respe tively, and let M be an interior point of△ABC. The lines AM , BM , and CM interse t the opposite sides at thepoints A1, B1, and C1, respe tively. Lines through M perpendi ular to thesides of △ABC interse t BC, CA, and AB at A2, B2, and C2, respe tively.Let p1, p2, and p3 be the distan es from M to the sides BC, CA, and AB,respe tively. Prove that

[A1B1C1]

[A2B2C2]=

(ap1 + bp2)(bp2 + cp3)(cp3 + ap1)

8a2b2c2

(

a

p1

+b

p2

+c

p3

) ,where [KLM ] denotes the area of triangle KLM .Solution by Joel S hlosberg, Bayside, NY, USA.By the basi formula for the area of a triangle, [ACM ] = 12bp2 and

[CMB] = 12ap1. It is well known that if segments Y Z and Y ′Z′ lie on the

410same line and X is any point, then [XY Z] : [XY ′Z′] = Y Z : Y ′Z′, so that

[ACC1]

[ACM ]=

[CC1B]

[CMB]=

CC1

CM;

[ACC1]

[CC1B]=

[ACM ]

[CMB]=

AC1

C1B=

bp2

ap1

.By similar reasoning, BA1

A1C=

cp3

bp2

and CB1

B1A=

ap1

cp3

. A known formula (seeEri W. Weisstein, \Routh's Theorem," at http://mathworld.wolfram.com/RouthsTheorem.html) states that if A′, B′, and C′ are points on the sidesBC, CA, and AB of △ABC, respe tively, then

[A′B′C′] =

AC ′

C ′B· BA′

A′C· CB′

B′A+ 1

(

AC ′

C ′B+ 1

)(

BA′

A′C+ 1

)(

CB′

B′A+ 1

) [ABC] .Therefore,

[A1B1C1] =

bp2

ap1

· cp3

bp2

· ap1

cp3

+ 1(

bp2

ap1

+ 1

)(

cp3

bp2

+ 1

)(

ap1

cp3

+ 1

) [ABC]

=2abcp1p2p3

(ap1 + bp2)(bp2 + cp3)(cp3 + ap1)[ABC] .Sin e ∠MA2C and ∠MB2C are right angles,

∠A2MB2 = 360◦ − ∠MA2C − ∠MB2C − ∠A2CB2

= 180◦ − ∠ACB ,so sin ∠A2MB2 = sin C, and sin e all four angles of quadrilateral A2MB2Care less than 180◦, A2MB2C is onvex. The well-known area formula for atriangle, [XY Z] = 12XY · XZ sin ∠Y XZ, yields

[MA2B2] =1

2MA2 · MB2 sin ∠A2MB2

=p1p2

ab

(

1

2ab sin C

)

=p1p2

ab[ABC] .By similar reasoning, [MB2C2] =

p2p3

bc[ABC] and [MC2A2] =

p3p1

ca[ABC],and B2MC2A and C2MA2B are onvex. Sin e A2MB2C, B2MC2A, and

C2MA2B are onvex, M is outside of △A2B2C, △B2C2A, and △C2A2B,and sin e M is in the interior of △ABC, M must be in the interior of△A2B2C2. Therefore,

[A2B2C2] = [MA2B2] + [MB2C2] + [MC2A2] ,

411and so

[A2B2C2] =

(

p1p2

ab+

p2p3

bc+

p3p1

ca

)

[ABC]

=p1p2p3

abc

(

a

p1

+b

p2

+c

p3

)

[ABC] .Finally,[A2B2C2]

[A1B1C1]=

p1p2p3

abc

(

a

p1

+b

p2

+c

p3

)

[ABC]

2abcp1p2p3

(ap1 + bp2)(bp2 + cp3)(cp3 + ap1)[ABC]

=(ap1 + bp2)(bp2 + cp3)(cp3 + ap1)

2a2b2c2

(

a

p1

+b

p2

+c

p3

) .Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; ARKADY ALT, SanJose, CA, USA; MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern StateUniversity, Joplin, MO, USA; OLIVER GEUPEL, Br �uhl, NRW, Germany; WALTHER JANOUS,Ursulinengymnasium, Innsbru k, Austria; PETER Y. WOO, Biola University, La Mirada, CA,USA; and TITU ZVONARU, Com�ane�sti, Romania. There was one in orre t solution submitted.Both Bataille and Geupel al ulated with bary entri oordinates. Geupel also used theformula for the area of the pedal triangle of M, to wit [A2B2C2] = |R2 −OM2|/4R2, whereO and R are the ir um entre and ir umradius of triangle ABC, respe tively.3372. [2008 : 364, 366℄ Proposed by Vo Quo Ba Can, Can Tho Universityof Medi ine and Pharma y, Can Tho, Vietnam.If x, y, z ≥ 0 and xy + yz + zx = 1, prove that(a) 1

√

2x2 + 3yz+

1√

2y2 + 3zx+

1√

2z2 + 3xy≥ 2

√6

3;

(b)⋆ 1√

x2 + yz+

1√

y2 + zx+

1√

z2 + xy≥ 2

√2 .Solution by George Apostolopoulos, Messolonghi, Gree e.(a) Sin e 2x2 ≤ 3x2 we have 2x2 + 3yz ≤ 3x2 + 3yz, thus

1√

2x2 + 3yz≥ 1

√

3x2 + 3yz=

1√

3√

x2 + yz.Similar inequalities hold for the other two terms on the left side of the desiredinequality, and we now have

1√

2x2 + 3yz+

1√

2y2 + 3zx+

1√

2z2 + 3xy

≥ 1√

3

(

1√

x2 + yz+

1√

y2 + zx+

1√

z2 + xy

) .

412The desired inequality follows now from (b), whi h is proven below.(b)⋆ More generally, we will prove that if x, y, z ≥ 0 and xy+yz+zx > 0,then

1√

x2 + yz+

1√

y2 + zx+

1√

z2 + xy≥ 2

√2

√xy + yz + zx

. (1)Sin e this is symmetri in x, y, z, we may assume that x ≥ y ≥ z. Noti ethat1

√

y2 + zx+

1√

z2 + xy≥ 2

√2

√

y2 + z2 + xy + zx.

[Ed: we have 1p

y2 + zx+

1p

z2 + xy≥ 2

√

1p

y2 + zx

1p

z2 + xyfrom theAM{GM Inequality, and also 1

p

y2 + zxp

z2 + xy≥ 2

y2 + zx + z2 + xy; theinequality above now follows from these two.℄So it suÆ es to prove that

1√

x2 + yz+

2√

2√

y2 + z2 + xy + zx≥ 2

√2

√xy + yz + zx

.Let K = xy + yz + zx and L = y2 + z2 + xy + zx. Then2√

2√

K− 2

√2

√L

=2√

2(√

L −√

K)

√KL

=2√

2(

y2 − yz + z2)

√KL

(√L +

√K) .

It is lear that L ≥ K, L ≥ 2(

y2 − yz + z2), and K ≥ y

√

x2 + yz.[Ed: Sin e x ≥ y ≥ z, we have L = y2 + z2 + xy + zx ≥ y2 + z2

+ y2 + z2 ≥ 2(y2 − yz + z2) and K = xy + yz + zx ≥ xy + yz + yz

= y√

x2 + 4xz + 4z2 ≥ y√

x2 + 4yz + 4z2 ≥ y√

x2 + yz.℄Thus,2√

2(

y2 − yz + z2)

√KL(

√L +

√K)

≤ 2√

2(

y2 − yz + z2)

√KL(2

√K)

=

√2(

y2 − yz + z2)

K√

L

≤√

2(

y2 − yz + z2)

y√

x2 + yz√

2(

y2 − yz + z2)

=

√

y2 − yz + z2

y√

x2 + yz≤

√

y2

y√

x2 + yz

=1

√

x2 + yz,

413and we have proved (1).Sin e xy + yz + zx = 1, inequality (b) now follows from (1).Part (a) also solved by Oliver Geupel, Br �uhl, NRW, Germany; and the proposer.Geupel notes that the problem is Problem 1.89 on page 74.f with solution in the e-papertoanhocmuonmaumain.pdf written by the proposer in Vietnamese, available on the MathLinksforum website http://www.mathlinks.ro/viewtopic.php?t=197674.3373. Proposed by Vo Quo Ba Can, Can Tho University of Medi ine andPharma y, Can Tho, Vietnam.Let x, y, z, and t be positive real numbers. Prove that(x + y)(x + z)(x + t)(y + z)(y + t)(z + t) ≥ 4xyzt(x + y + z + t)2 .Solution by Cristinel Morti i, Valahia University of Targoviste, Romania.Dividing by xyzt(x + z)(y + t)(x + t)(y + z) the inequality be omes

x + y

xy· z + t

zt≥ 2(x + y + z + t)

(x + z)(y + t)· 2(x + y + z + t)

(x + t)(y + z).Thus we need to show:

(

2

x + z+

2

y + t

)(

2

x + t+

2

y + z

)

≤(

1

x+

1

y

)(

1

z+

1

t

) .By the AM{GM Inequality we have(

2

x + z+

2

y + t

)(

2

x + t+

2

y + z

)

≤(

1√

xz+

1√

yt

)(

1√

xt+

1√

yz

) ,while by the Cau hy-S hwartz Inequality we have1

√xz

+1

√yt

≤√

(

1

x+

1

y

)(

1

z+

1

t

)

and1

√xt

+1

√yz

≤√

(

1

x+

1

y

)(

1

z+

1

t

) .Combining the last three inequalities we obtain the desired inequality.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; �SEFKETARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE,Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVERGEUPEL, Br �uhl, NRW, Germany; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS,Ursulinengymnasium, Innsbru k, Austria; SALEM MALIKI �C, student, Sarajevo College,

414Sarajevo, Bosnia and Herzegovina; KHANH BAO NGUYEN, High S hool for Gifted Stu-dents, Hanoi University of Edu ation, Hanoi, Vietnam; PAOLO PERFETTI, Dipartimento diMatemati a, Universit �a degli studi di Tor Vergata Roma, Rome, Italy; PETER Y. WOO, BiolaUniversity, La Mirada, CA, USA; TITU ZVONARU, Com�ane�sti, Romania; and the proposer.Howard notes that this problem is similar to CRUX 2393(a) [1999 : 447℄. In fa t, Maliki � observes that by using 2393(a) for (a, b, c, d) = (x, y, z, t), (a, b, c, d) = (x, z, t, y), and(a, b, c, d) = (x, t, y, z) (that is, all the y li permutations of y, z, t) and multiplying theresulting inequalities together we get exa tly the desired inequality.Geupel notes that the problem is Problem 1.73 on page 64.f with solution in the e-papertoanhocmuonmaumain.pdf written by the proposer in Vietnamese, available on the MathLinksforum website http://www.mathlinks.ro/viewtopic.php?t=197674.3374. [2008 : 364, 366℄ Proposed by Pham Huu Du , Ballajura, Australia.Let a, b, and c be positive real numbers. Prove that

a2

a2 + bc+

b2

b2 + ca+

c2

c2 + ab≤ a + b + c

23√

abc.Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA.By homogeneity, we may take abc = 1 and a ≤ b ≤ c. Also, if a = 1,then a = b = c = 1 and there is nothing to prove, so we take a < 1.For onvenien e, write x = a3, y = b3, and z = c3; let f(x) =

x

x + 1,and let the left and right sides of the inequality be L = f(x) + f(y) + f(z)and R =

1

2

(

x1/3 + y1/3 + z1/3), respe tively.Setting m =

√yz and r =

√

z

y, we have that m > 1 (be ause a < 1implies that yz > 1), y =

m

r, and z = mr. We then �nd that

2f(m) − f(y) − f(z) =2m

m + 1− m

m + r− mr

mr + 1

=m(m − 1)(r − 1)2

(m + 1)(m + r)(mr + 1)≥ 0 ,hen e L annot de rease if ea h of y and z are repla ed by their geometri mean. On the other hand, from √

bc ≤ b + c

2, we see that R annot in reaseif y and z are ea h repla ed by their geometri mean.Therefore, it suÆ es to prove the inequality under the additional as-sumption that b = c and a =

1

b2. This new relation yields

R − L =(1 + 2y)

(

y3 + y2 + y + 1)

− 4b11 − 6b5 − 2b2

2b2(

y3 + y2 + y + 1) ,where the denominator is positive and (after some omputation) the numer-ator be omes

(b − 1)2(

(b − 1)(

2b9 + 2b8 − b6 − b5)

+ 5b4 + 3b3 + b2 + 2b + 1) ,

415whi h is also a positive quantity sin e b > 1.Thus, R − L > 0 in this last ase, and the proof is omplete.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CHIP CURTIS,Missouri Southern State University, Joplin, MO, USA (2 solutions); WALTHER JANOUS,Ursulinengymnasium, Innsbru k, Austria; KEE-WAI LAU, Hong Kong, China; and the proposer.There were two in orre t solutions submitted.3375. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.Let p be a non-negative integer and x any real number. Find the sum

∞∑

n=1

(−1)n

(

ex − 1 − x

1!− x2

2!− · · · − xn+p

(n + p)!

) .Solution by Cristinel Morti i, Valahia University of Targoviste, Romania.If r > 0 and f is a fun tion with

f(x) =

∞∑

n=0

anxn , x ∈ (−r, r) ,then we will show that

∞∑

n=1

(−1)n(

f(x) − a0 − a1x − · · · − an+pxn+p)

=

p/2∑

i=0

a2ix2i − f(x) + f(−x)

2, p even ,

(p−1)/2∑

i=0

a2i+1x2i+1 − f(x) − f(−x)

2, p odd . (1)

Taking f(x) = ex, we obtain the answer∞∑

n=1

(−1)n

(

f(x) − 1 − x

1!− · · · − xn+p

(n + p)!

)

=

p/2∑

i=0

x2i

(2i)!− ex + e−x

2, p even,

(p−1)/2∑

i=0

x2i+1

(2i + 1)!− ex − e−x

2, p odd.

To prove (1), note that the general terman = (−1)n

(

f(x) − a0 − a1x − · · · − an+pxn+p)

416 onverges to zero, so it suÆ es to �nd the limit of the sequen e {s2n}, where

sn =n∑

k=1

ak ,be ause s2n+1 = s2n + a2n+1 and a2n+1 → 0. We haves2n =

n∑

k=1

(a2k−1 + a2k) =n∑

k=1

(

−a2k+px2k+p) .If p is even, then

s2n =

p/2∑

i=0

a2ix2i −

n+p/2∑

i=0

a2ix2i ;while if p is odd, then

s2n =

(p−1)/2∑

i=0

a2i+1x2i+1 −n+(p−1)/2∑

i=0

a2i+1x2i+1 .Now the relation (1) follows from the equations∞∑

i=0

a2ix2i =

f(x) + f(−x)

2; ∞

∑

i=0

a2i+1x2i+1 =f(x) − f(−x)

2.Also solved byMICHEL BATAILLE, Rouen, Fran e; MANUEL BENITO, �OSCAR CIAURRI,EMILIO FERNANDEZ, and LUZ RONCAL, Logro ~no, Spain; CHIP CURTIS, Missouri SouthernState University, Joplin, MO, USA; OLIVER GEUPEL, Br �uhl, NRW, Germany; RICHARD I. HESS,Ran ho Palos Verdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria;and the proposer. There were three in orre t solutions submitted.

Crux Mathemati orumwith Mathemati al MayhemFormer Editors / An iens R �eda teurs: Bru e L.R. Shawyer, James E. TottenCrux Mathemati orumFounding Editors / R �eda teurs-fondateurs: L �eopold Sauv �e & Frederi k G.B. MaskellFormer Editors / An iens R �eda teurs: G.W. Sands, R.E. Woodrow, Bru e L.R. ShawyerMathemati al MayhemFounding Editors / R �eda teurs-fondateurs: Patri k Surry & Ravi VakilFormer Editors / An iens R �eda teurs: Philip Jong, Je� Higham, J.P. Grossman,Andre Chang, Naoki Sato, Cyrus Hsia, Shawn Godin, Je� Hooper