1SKOLIAD No. 114

V�a lav LinekPlease send your solutions to problems in this Skoliad by August 1, 2009.Solutions should be sent to Lily Yen andMogensHansen at the address insidethe ba k over. The Skoliad se tion is in transition and, unfortunately, wehave lost several of the submitted solutions to past ontests. If you have opies of solutions that you sent to past ontests, please send them againso that we an mention any orre t solutions we re eive. (This in ludes any ontest in Skoliad appearing in or after the Mar h 2008 issue of CRUX).Our �rst problem set of the year is the Math Kangaroo Contest Pra ti eSet. The Kangaroo Contest is international in s ope and supported in Canadaby the Canadian Mathemati al So iety and the Institute of Ele tri al andEle troni s Engineers (Northern Se tion).Our thanks go to Valeria Pandelieva, the Canadian representative of theKangaroo Contest, for bringing this ontest to our attention, and for makingus aware of the need for ontests and math-parti ipation in the lower yearsin Canada. For that reason, and also sin e this ontest is straightforwardto administer (see www.mathkangaroocanada.com), we are featuring its entirerange of questions over all grades.Finally, while it is a multiple hoi e test, we ask our readers to send in omplete solutions showing all the steps and details so that we an evaluatethe solutions and give full redit to the solvers.Math Kangaroo ContestPra ti e SetPart A (3 points per question)1. (Grades 3-4) In the addition example, ea h letterrepresents a digit. Equal digits are represented by thesame letter. Di�erent digits are represented by di�er-ent letters. Whi h digit does the letter K represent?

O K+ K OW O W(A) 0 (B) 1 (C) 2 (D) 8 (E) 92. (Grades 5-6) Ten aterpillars, arranged in a row one behind another,walked in the park. The length of ea h aterpillar was equal to 8 m, andthe distan e any two adja ent aterpillars kept for safety reasons was 2 m.What is the total length of their row?(A) 100 m (B) 98 m (C) 82 m (D) 102 m (E) 96 m

23. (Grades 7-8) An ant is running along a rulerof length 10 m with a onstant speed of 1 mper se ond (see the �gure). Any time whenthe ant rea hes one of the ends of the ruler, it ..

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1 2 3 4 5 6 7 8 9turns ba k and runs in the opposite dire tion. It takes the ant exa tly 1 se -ond to make a turn. The ant starts from the left end of the ruler. Nearestwhi h number will it be after 2009 se onds?(A) 1 m (B) 2 m (C) 3 m (D) 4 m (E) 5 m4. (Grades 9-10) Whi h of the numbers 26, 35, 44, 53, 62 is the greatest?(A) 26 (B) 35 (C) 44 (D) 53 (E) 625. (Grades 11-12) A de orator has prepared a mixed paint, in whi h thevolumes of red and yellow olours were in the ratio 2 : 3. The resulting olour seemed too light to him, so he added 2 L of red paint. This way, theratio of the volumes of the red and yellow olours hanged to 3 : 2. Howmany litres of paint did the de orator use?(A) 5 L (B) 6 L (C) 7 L (D) 8 L (E) 9 LPart B (4 points per question)6. (Grades 3-4) Two boys are playing tennis until one of them wins fourtimes. A tennis mat h annot end in a draw. What is the greatest number ofgames they an play?(A) 8 (B) 7 (C) 6 (D) 5 (E) 97. (Grades 5-6) In two years, my son will be twi e as old as he was two yearsago. In three years, my daughter will be three times as old as she was threeyears ago. Whi h of the following best des ribes the ages of the daughter andthe son?(A) The son is older; (B) The daughter is older; (C) They are twins;(D) The son is twi e as old as the daughter;(E) The daughter is twi e as old as the son.8. (Grades 7-8) Some points are marked on a straight line so that all distan es1 m, 2 m, 3 m, 4 m, 5 m, 6 m, 7 m, and 9 m are among the distan esbetween these points. At least how many points are marked on the line?(A) 4 (B) 5 (C) 6 (D) 7 (E) 89. (Grades 9-10) Eva, Betty, Linda, and Cathy went to the inema. Sin e itwas not possible to buy four seats next to ea h other, they bought ti kets forseats number 7 and 8 in the 10th row and ti kets for seats number 3 and 4in the 12th row. How many seating arrangements an they hoose from, ifCathy does not want to sit next to Betty?(A) 24 (B) 20 (C) 16 (D) 12 (E) 8

310. (Grades 11-12) Triangle ABC isisos eles with BC = AC. The segmentsDE, FG, HI, KL, MN , OP , and XYdivide the sides AC and CB into equalparts. Find XY , if AB = 40 m.(A) 38 m (B) 35 m ..

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(C) 33 m (D) 30 m (E) 27 mPart C (5 points per question)11. (Grades 3-4) Matt and Ni k onstru tedtwo buildings, shown in the �gures, using iden-ti al ubes. Matt's building weighs 200 g, andNi k's building weighs 600 g. How many ubesfrom Ni k's building are hidden and annot beseen in the �gure?(A) 1 (B) 2 (C) 3(D) 4 (E) 5

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Matt'sbuilding12. (Grades 5-6) Consider all four-digit numbers divisible by 6 whose digitsare in in reasing order, from left to right. What is the hundreds digit of thelargest su h number?(A) 7 (B) 6 (C) 5 (D) 4 (E) 313. (Grades 7-8) A square of side length 3 is dividedby several segments into polygons as shown in the �g-ure. What per ent of the area of the original square isthe area of the shaded �gure?(A) 30% (B) 3313% (C) 35%(D) 40% (E) 50% ..

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14. (Grades 9-10) A boy always tells the truth on Thursdays and Fridays,always tells lies on Tuesdays, and tells either truth or lies on the rest of thedays of the week. Every day he was asked what his name was and six times ina row he gave the following answers: John, Bob, John, Bob, Pit, Bob. Whatdid he answer on the seventh day?(A) John (B) Bob (C) Pit (D) Kate(E) Not enough information to de ide15. (Grades 11-12) An equilateral triangle and a ir leM are ins ribed in a ir le K, as shown in the �gure.What is the ratio of the area of K to the area ofM?(A) 8 : 1 (B) 10 : 1 (C) 12 : 1(D) 14 : 1 (E) 16 : 1

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4Con ours Math KangarooFeuille d'entra�nement

Partie A (3 points par question)1. (Classes 3-4) Dans l'exemple d'addition i-dessus, haque lettre di� �erente repr �esente un hi�redi� �erent. Quel hi�re la lettre K repr �esente-t-elle ? O K+ K OW O W(A) 0 (B) 1 (C) 2 (D) 8 (E) 92. (Classes 5-6) Dix henilles se promenaient �a la �le indienne dans un par .Chaque henille mesurait 8 m et, pour des raisons de s �e urit �e, elles gar-daient une distan e de 2 m entre ha une d'elles. Quelle �etait la longueurtotale de leur ort �ege ?(A) 100 m (B) 98 m (C) 82 m (D) 102 m (E) 96 m3. (Classes 7-8) Une fourmi ourt le longd'une r �egle de 10 m de longueur, �a la vitesse onstante de 1 m �a la se onde (voir la �gure).Chaque fois qu'elle atteint une extr �emit �e, elle ..

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1 2 3 4 5 6 7 8 9 ourt dans la dire tion oppos �ee et elle met exa tement 1 se onde pour han-ger de dire tion. La fourmi part de l'extr �emit �e gau he de la r �egle. Pr �es dequel hi�re sera-telle apr �es 2009 se ondes ?(A) 1 m (B) 2 m (C) 3 m (D) 4 m (E) 5 m4. (Classes 9-10) Lequel des nombres 26, 35, 44, 53, 62 est-il le plus grand ?(A) 26 (B) 35 (C) 44 (D) 53 (E) 625. (Classes 11-12) Un d �e orateur a pr �epar �e un m�elange de peinture o �u lesvolumes des ouleurs rouge et jaune �etaient dans un rapport de 2 : 3. Trou-vant le m�elange trop lair, il ajouta 2 L de peinture rouge. Le rapport desvolumes des ouleurs rouge et jaune devint alors de 3 : 2. Combien de litresde peinture le d �e orateur a-t-il utilis �e ?(A) 5 L (B) 6 L (C) 7 L (D) 8 L (E) 9 LPartie B (4 points par question)6. (Classes 3-4) Deux gar� ons jouent au tennis jusqu' �a e que l'un d'eux gagnequatre fois. Un mat h de tennis ne peut �nir en un pointage nul. Quel est leplus grand nombre de jeux qu'ils peuvent jouer ?(A) 8 (B) 7 (C) 6 (D) 5 (E) 9

57. (Classes 5-6) Dans deux ans, mon �ls aura deux fois l'age qu'il avait il y adeux ans. Dans trois ans, ma �lle aura trois fois l'age qu'elle avait il y a troisans. Quelle r �eponse d �e rit-elle le mieux l'age de la �lle et du �ls ?(A) Le �ls est plus ag �e ; (B) La �lle est plus ag �ee ;(C) Ils sont des jumeaux ; (D) Le �ls est deux fois plus ag �e que la �lle ;(E) La �lle est deux fois plus ag �ee que le �ls.8. (Classes 7-8) Sur une droite on marque des points de sorte que toutesles distan es de 1 m, 2 m, 3 m, 4 m, 5 m, 6 m, 7 m et 9 m �gurentparmi les distan es entre es points. Combien y a-t-il au minimum de pointsmarqu �es sur ette droite ?(A) 4 (B) 5 (C) 6 (D) 7 (E) 89. (Classes 9-10) Liliane, Ni ole, Katia et Charlotte sont all �ees au in �ema.Comme il n' �etait pas possible d'a heter quatre pla es ensemble, elles onta het �e des billets pour les si �eges num�ero 7 et 8 dans la 10e-rang �ee et d'autrespour les si �eges num�ero 3 et 4 dans la 12e-rang �ee. De ombien de mani �erespeuvent-elles hoisir de s'asseoir, si Charlotte ne veut pas etre assise �a ot �ede Ni ole ?(A) 24 (B) 20 (C) 16 (D) 12 (E) 810. (Classes 11-12) Soit ABC un triangleiso �ele ave BC = AC. Les segmentsDE, FG, HI, KL, MN , OP et XYdivisent les ot �es AC et CB en parties�egales. Trouver XY si AB = 40 m.(A) 38 m (B) 35 m ..

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(C) 33 m (D) 30 m (E) 27 mPartie C (5 points par question)11. (Classes 3-4) En utilisant des ubes iden-tiques, Mathieu et Ni olas ont onstruit deuxbatiments, omme illustr �es dans les �gures. Lebatiment de Mathieu p �ese 200 g et elui de Ni- olas 600 g. Combien de ubes du batiment deNi olas sont-ils a h �es et ne peuvent etre vusdans la �gure ?(A) 1 (B) 2 (C) 3(D) 4 (E) 5

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batimentde Mathieu12. (Classes 5-6) On onsid �ere tous les nombres de quatre hi�res, divisiblespar 6 et dont les hi�res, lus de gau he �a droite, vont en ordre roissant. Quelest le hi�re des entaines dans le plus grand de es nombres ?(A) 7 (B) 6 (C) 5 (D) 4 (E) 3

613. (Classes 7-8) On divise un arr �e de ot �e 3 en po-lygones ave plusieurs segments omme indiqu �e dansla �gure. Quel est le per entage de l'aire de la �gureombr �ee par rapport �a elle du arr �e ?(A) 30% (B) 331

3% (C) 35%(D) 40% (E) 50% ..

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14. (Classes 9-10) Un gar� on dit toujours la v �erit �e les jeudis et vendredis,ment toujours les mardis et, les autres jours de la semaine, soit il dit la v �erit �esoit il ment. On lui demanda son nom haque jour de la semaine et les sixpremi �eres fois, il donna les r �eponses suivantes : Jean, Bernard, Jean, Bernard,Paul, Bernard. Quelle fut sa r �eponse le septi �eme jour ?(A) Jean (B) Bernard (C) Paul (D) Lu (E) Pas possible de d �e ider15. (Classes 11-12) On ins rit un triangle �equilat �eralet un er leM dans un er leK, omme indiqu �e dansla �gure. Quel est le rapport de l'aire de K �a elle deM ? (A) 8 : 1 (B) 10 : 1 (C) 12 : 1(D) 14 : 1 (E) 16 : 1

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M

K

Next we shall give solutions to the Mathemati s Asso iation of Quebe Contest (Se ondary level) February 9, 2006 [2008 : 67-68℄. We apologize toany readers who sent in solutions to this ontest but whose solutions we havelost.1. A parti ular magi square. It is well known that a magi square is obtainedby putting numbers in a square su h that the sum of ea h row, olumn, anddiagonal is the same, as for example,8 1 63 5 74 9 2Imagine now that we de ide to invent a new form of su h squares by repla ingthe sum by a produ t. We ask you to �nd su h a square by repla ing theasterisks, ∗, by natural numbers, not ne essarily distin t or onse utive, inthe following square:∗ 1 ∗4 ∗ ∗∗ ∗ 2

7Solution by the editor.Suppose that the square A below is a magi square. Then the squareB is a magi square for produ ts. For example, by the Law of Exponents, theprodu t along the �rst row of B is xaxbxc = xa+b+c and the produ t alongthe �rst olumn of B is xaxdxg = xa+d+g and these are the same be ausea + b + c = a + d + g. The same is true for the other rows, olumns, anddiagonals of B.

A =

a b cd e fg h i

, B =

xa xb xc

xd xe xf

xg xh xi

.Now, if we subtra t 1 from every entry of the �rst square given in the questionand if we take x = 2, then the square B below is a solution to the problem.

A =7 0 52 4 63 8 1

, B =27 20 25

22 24 26

23 28 21=

128 1 324 16 648 256 2

.2. Clovis' outing. Clovis likes to take an outing in the natural numbers.Ea h day, he starts with a natural number of his hoi e, the biggest possible.Then, during his day, he passes from number to number using the followingrules. Suppose that the sequen e of numbers is urrently at n.(1) If n is divisible by 3 without remainder, then the next number is n/3.(2) If the remainder after dividing n by 3 is 1, then the next number is

2n+ 1.(3) If the remainder after dividing n by 3 is 2, then the next number is2n− 1.(4) If n = 1, then the sequen e stops.Over the years that he has played this game, he noti ed that, whatever thestarting number, the sequen e always ended up with the number 1. How-ever, he wonders if there is a sequen e that in reases inde�nitely, with largerand larger numbers on average, or su h that it ends up in a loop of numbersthat does not ontain 1. Determine if su h a sequen e is possible and givean example, or show that su h a sequen e does not exist by showing that allsequen es using the above rules inevitably end up at the number 1.Here is an example of su h a sequen e: Starting with 55, we get 111,

37, 75, 25, 51, 17, 33, 11, 21, 7, 15, 5, 9, 3 and 1, whi h ends the sequen e.Solution by the editor.Note that Clovis' sequen e starting with 55 has a de reasing subse-quen e that goes to 1, given by the underlined numbers: 55, 111, 37, 75, 25,

851, 17, 33, 11, 21, 7, 15, 5, 9, 3, 1. We will show that for any number a > 1in one of Clovis' sequen es, there is always a number b oming after a in thesequen e su h that a > b. Thus, if Clovis starts with n > 1, then there willbe a subsequen e n,m, p, . . . with n > m > p > · · · and this subsequen emust eventually hit the number 1 (be ause all of the terms in it are positive,it annot de rease forever).If a > 1 and a = 3k, k > 1, then by rule (1) the number b = k omesright after a and a > b.If a > 1 and a = 3k+ 1, k > 0, then by rule (2) the number 2a+ 1 =2(3k+1)+1 = 6k+3 omes right after the number a, and then by rule (1) thenumber (6k+3)/3 = 2k+1 omes after 2a+1. Sin e a = 3k+1 > 2k+1,we see that the number b = 2k + 1 omes after a and a > b.If a > 1 and a = 3k+ 2, k ≥ 0, then by rule (3) the number 2a− 1 =2(3k+2)+1 = 6k+3 omes right after the number a, and then by rule (1) thenumber (6k+3)/3 = 2k+1 omes after 2a−1. Sin e a = 3k+2 > 2k+1,we see that the number b = 2k + 1 omes after a and a > b.Thus, in all ases where a > 1, there is a number b oming after a inthe sequen e su h that a > b, and we are done.3. Eight balls in two urns. We give you two similar urns, four white balls,and four bla k balls. You must separate the balls amongst the two urns(not ne essarily the same number in ea h urn), after whi h both urns willbe made indistinguishable. How should the balls be distributed to maximizethe han es that, if you draw a ball randomly from a randomly hosen urn,you will obtain a white ball?Solution by the editor.Put 1 white ball in one urn and all the other balls in the other urn. Theprobability of hoosing the urn with 1 white ball and then drawing that whiteball from it is 1

2·1 = 1

2and the probability of hoosing the other urn and thendrawing a white ball from it is 1

2· 33+4

= 314. Thus, with this distribution, theoverall probability of ultimately obtaining a white ball is p = 1

2·1+ 1

2· 37

= 57.Now let p1 and p2 be the probabilities of drawing white balls from thetwo urns (p1 = 1 and p2 = 3

7above). The overall probability of ultimatelyobtaining a white ball is then p = 1

2p1+ 1

2p2 =

p1 + p2

2, whi h is the averageof the probabilities p1 and p2. Therefore, p > 5

7implies that p1 > 5

7or

p2 >57. To make an urn with p1 >

57(say) we must have (w, b) = (1, 0),

(2, 0), (3, 0), (3, 1), (4, 0), or (4, 1), where the urn ontains w white ballsand b bla k balls. These are the only distributions that ould yield p > 57.In the ase of (w, b) = (2, 0), (3, 0), or (4, 0) we have p < 5

7, as moving allbut one white ball to the other urn in reases p2 but leaves p1 = 1. Finally,

(w, b) = (3, 1) or (4, 1) yields p = 12or p = 2

5, ea h less than 5

7.Our �rst distribution maximizes our han e of obtaining a white ball.

94. The three atta hed barrels. Three big ylindri albarrels, lying parallel to the earth, are atta hed bya steel able at their onta t points, A and B, su hthat they stay �xed in pla e. Knowing that the twosmaller ones ea h have a radius of 4 metres and thebiggest one has a radius of 9 metres, what is thelength of the steel able?

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...................................................................................................................................................q qA B

Solution by the editor.Join the entres of the smaller barrelsand drop a perpendi ular to this segment fromthe entre of the larger barrel, as in the dia-gram at right. Sin e the barrels rest on theearth, the length of CZ is the di�eren e oftheir radii, that is, |CZ| = 9 − 4 = 5. Alsothe length of CY is the sum of the radii, thatis, |CY | = 9 + 4 = 13. By the Pythagorean.................................................................................................................................................................

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A B...........................................................

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C

X YZ........................................

Theorem |ZY | =√

132 − 52 = 12. Thus, |XY | = 24. Finally, sin e triangleCAB is similar to triangle CXY , we have |AB| =

9

13|XY | =

216

13.5. The magi words. An illusionist is sear hing formagi words to a ompany his many magi tri ks.He de ides to onstru t his magi words startingwith the diagram on the right. He takes a paththrough the diagram and jots downs the letters he�nds on it. Ea h magi word must have exa tly 11letters and must start and end with the letter A.Two onse utive letters must never be identi al.How many magi words are there?

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

A

BC

D R

Note: Here are two possible magi words: ABRACADABRA andARADCABARBA.Solution by the editor.LetMk be the number of k-letter words that start withA and end withA and that an be formed by travelling through the bowtie. Let Nk be thenumber of k-letter words starting with A but not ending with the letter Athat an be similarly formed.If k ≥ 2, then we see thatMk = Nk−1, be ause removing the letter Afrom a word ending with A leaves a word not ending in A (but still startingwithA) and the pro ess an be reversed. Similarly, by deleting the last letterof a word of length k that starts with A but does not end in A, we see thatNk = Nk−1 + 4Mk−1, be ause any of the four letters di�erent from A anbe added to a word not ending in A or else there is only way to extend a(k−1)-letter word not ending inA to one that still does not end inA. Sin eMk−1 = Nk−2 the last equation be omes Nk = Nk−1 + 4Nk−2, wherek ≥ 1.

10We now haveN1 = 0,N2 = 4,N3 = N2 +4N1 = 4+4 ·0 = 4, and soforth. The results of al ulating the Ni are summarized in the table below:

N1 N2 N3 N4 N5 N6 N7 N8 N9 N10

0 4 4 20 36 116 260 724 1764 4660FinallyM11 = N10, so there are 4660 magi words altogether.6. All ten digits. Find the smallest positive natural number N su h that, inthe de imal notation, N and 2N together use all ten digits: 0, 1, 2, . . . , 9.Solution by the editor.We have 2(13485) = 26970, and we will prove that if N1 and 2N1together use all ten digits and N1 ≤ N = 13485, then N1 = N .As N1 has �ve digits and N1 ≤ N , then N1 = 1 . . . and 2N1 = 2 . . ..Digits 1 and 2 are now used and 2N1 uses 0 (otherwise N1 uses 0 and 2N1then uses 0 or 1, a ontradi tion). Thus,N1 = 13 . . .. The smallest availabledigit for N1 is now a 4 and N1 ≤ N , hen e N1 = 134 . . . and 2N1 = 26 . . ..The numberN1 uses 5, be ause 2N1 uses 0. IfN1 = 1345x, then x is a digitgreater than 5 and 2N1 = 2691y, a ontradi tion. Thus, N1 = 134x5 ≤ N .Finally, x 6= 7, hen e x = 8.Therefore, N1 = N and N is the smallest positive integer with thegiven property.[Ed.: Rolland Gaudet o�ers the solutionN = 6792 if initial zeroes areallowed, for then 2(6792) = 013584.℄7. The pizza toppings. At the Julio pizzeria, all the pizzas have heese andtomato sau e on them. The hoi e of toppings is limited to bla k olives,an hovies, and sausage. Of the 200 lients Julio had yesterday, 40 took an- hovies, 80 took bla k olives, 120 took sausage, 60 took at the same timebla k olives and sausage, but none took at the same time an hovies andbla k olives or an hovies and sausage. How many lients took none of thethree toppings?Solution by the editor.Let t be the number of ustomers who took at least one topping. Any ustomer who took an hovies took no other topping, so t = 40 + x where xis the number of ustomers who took bla k olives or sausage (or both). Therewere 60 ustomers who took both bla k olives and sausage, so 20 = 80 − 60took just bla k olives and nothing else. Similarly, 60 = 120 − 60 ustomerstook sausage and nothing else. Thus, t = 40+x = 40+(20+60+60) = 180,and the number of ustomers who took no toppings is 200 − t = 20.

11MATHEMATICAL 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 ProblemsPlease send your solutions to the problems in this edition by 1May 2009. Solu-tions re eived after this date will only be onsidered if there is time before publi ationof the solutions.Ea h problem is given in English and Fren h, the oÆ ial languages of Canada.In issues 1, 3, 5, and 7, English will pre ede Fren h, and in issues 2, 4, 6, and 8,Fren h will pre ede English.The editor thanks Jean-Mar Terrier of the University of Montreal for transla-tions of the problems.M376. Proposed by the Mayhem Sta�.Determine the value of x if (102009 + 25

)2 −(102009 − 25

)2= 10x.M377. Proposed by the Mayhem Sta�.An arithmeti sequen e onsists of 9 positive integers. The sum of theterms in the sequen e is greater than 200 and less than 220. If the se ondterm in the sequen e is 12, determine the sequen e.M378. Proposed by the Mayhem Sta�.Points C and D are hosen on the semi- ir le with diameter AB sothat C is loser to A. Segments CB and DA interse t at P ; segments ACand BD extended interse t at Q. Prove that QP extended is perpendi ularto AB.M379. Proposed by John GrantM Loughlin, University of New Brunswi k,Frederi ton, NB.The integers 27+C, 555+C, and 1371+C are all perfe t squares, thesquare roots of whi h form an arithmeti sequen e. Determine all possiblevalues of C.

12M380. Proposed by Bru e Shawyer, Memorial University of Newfound-land, St. John's, NL.Triangle ABC is right-angled at C and has BC = a and CA = b, witha ≥ b. SquaresABDE,BCFG, andCAHI are drawn externally to triangleABC. The lines through FI and EH interse t at P , the lines through FIand DG interse t at Q, and the lines through DG and EH interse t at R. Iftriangle PQR is right-angled, determine the value of b

a.M381. Proposed by Mih�aly Ben ze, Brasov, Romania.Determine all solutions to the equation

1

x− 1+

1

x− 2+

1

x− 6+

1

x− 7= x2 − 4x− 4 .

.................................................................M376. Propos �e par l' �Equipe de Mayhem.D �eterminer la valeur de x si (102009 + 25)2 −

(102009 − 25

)2= 10x.M377. Propos �e par l' �Equipe de Mayhem.D �eterminer la suite arithm�etique form�ee de 9 entiers positifs dont lasomme se situe entre 200 et 220 et dont le se ond terme vaut 12.M378. Propos �e par l' �Equipe de Mayhem.A partir du point A, on hoisit deux points C et D sur un demi- er lede diam�etre AB. Soit P l'interse tion des droites CB et DA, et Q elle desdroites AC et BD. Montrer que la droite PQ est perpendi ulaire �a AB.M379. Propos �e par John Grant M Loughlin, Universit �e du Nouveau-Brunswi k, Frederi ton, NB.Les entiers 27 + C, 555 + C et 1371 + C sont tous des arr �es parfaitsdont les ra ines arr �ees forment une suite arithm�etique. Trouver toutes lesvaleurs possibles de C.M380. Propos �e par Bru e Shawyer, Universit �e Memorial de Terre-Neuve,St. John's, NL.Dans un triangle ABC d'angle droit en C, soit BC = a et CA = b,ave a ≥ b. Ext �erieurement au triangleABC, on onstruit les arr �esABDE,

BCFG etCAHI. Soit respe tivementP ,Q etR les interse tions des droitesFI etEH, FI etDG,DG etEH. D �eterminer la valeur de b

apour que PQRsoit un triangle re tangle.

13M381. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.D �eterminer toutes les solutions de l' �equation

1

x− 1+

1

x− 2+

1

x− 6+

1

x− 7= x2 − 4x− 4 .

Mayhem SolutionsM338. Proposed by the Mayhem Sta�.Two students mis opy the quadrati equation x2 + bx + c = 0 thattheir tea her writes on the board. Jim opies b orre tly but mis opies c;his equation has roots 5 and 4. Vazz opies c orre tly, but mis opies b; hisequation has roots 2 and 4. What are the roots of the original equation?Solution by Taylor Thetford, student, Lakeview High S hool, San Angelo, TX,USA. The roots of the quadrati equation that Jim writes down are 5 and 4.His quadrati equation is thus (x−5)(x−4) = x2 −9x+20 = 0. Sin e Jim opied b orre tly, we an on lude that in the original quadrati equation,b = −9.Similarly, sin e Vazz's roots are 2 and 4, his quadrati equation has theform (x− 2)(x− 4) = x2 − 6x+ 8 = 0. Sin e Vazz opied c orre tly, thenc = 8.Thus, the original equation was x2 − 9x+ 8 = 0. Fa toring, we obtain(x−1)(x−8) = 0. Therefore, the roots of the original equation are 1 and 8.Also solved by CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long,Vietnam; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; PETER CHIEN,student, Central Elgin Collegiate, St. Thomas, ON; IAN JUNE L. GARCES, Ateneo de ManilaUniversity, Quezon City, The Philippines; JOHAN GUNARDI, student, SMPK 4 BPK PENABUR,Jakarta, Indonesia; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIR �O, IES\Abastos", Valen ia, Spain; JOS �E HERN�ANDEZ SANTIAGO, student, Universidad Te nol �ogi ade la Mixte a, Oaxa a, Mexi o; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong,India; BILLY SUANDITO, Palembang, Indonesia; LUYAN ZHONG-QIAO, Columbia Interna-tional College, Hamilton, ON; and TITU ZVONARU, Com�ane�sti, Romania.M339. Proposed by the Mayhem Sta�.(a) Determine the number of integers between 100 and 199, in lusive,whi h ontain exa tly two equal digits.(b) An integer between 1 and 999 is hosen at random, with ea h integerbeing equally likely to be hosen. What is the probability that the in-teger has exa tly two equal digits?

14Solutions by Taylor Thetford, student, Lakeview High S hool, San Angelo,TX, USA.(a) Ea h of the integers in the given range an be written in the form1xy. There are three ases to onsider.Case 1. The �rst and last digits are the same. Here, we are looking forintegers 1x1 where the middle digit an take any value ex ept 1. This yields9 possibilities.Case 2. The �rst and se ond digits are the same. Here, we are looking forintegers 11y where the last digit an take any value ex ept 1. This againyields 9 possibilities.Case 3. The se ond and last digits are the same. Here, we are looking forintegers 1xx where x is not 1. Again, there are 9 possibilities.Adding the results from our three ases, we �nd that there are 27 num-bers between 100 and 199, in lusive, that ontain exa tly two equal digits.(b) We ount the number of integers in the range 1 to 999, in lusive,that have exa tly two equal digits.First, between 1 and 99, there are 9 of these, namely, 11, 22, . . . , 99.Next, between 100 and 199, we have ounted 27 in part (a). Using the sameargument as in (a), we an show that there are 27 numbers between 200 and299, in lusive, and for every other interval of one hundred numbers up tothe range of 900 to 999.There are therefore 9+9 ·27 = 252 numbers between 1 and 999 whi h ontain exa tly two equal digits.The probability that a randomly sele ted integer between 1 and 999has exa tly two equal digits is thus 252

999=

28

111.Also solved by PETER CHIEN, student, Central Elgin Collegiate, St. Thomas, ON; IANJUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines; RICHARDI. HESS, Ran ho Palos Verdes, CA, USA; and LUYAN ZHONG-QIAO, Columbia InternationalCollege, Hamilton, ON. There were 3 in orre t solutions and 1 partial solution submitted.M340. Proposed by the Mayhem Sta�.Let ABC be an isos eles triangle with AB = AC, and let M be themid-point of BC. Let P be any point on BM . A perpendi ular is drawn to

BC at P , meeting BA at K and CA extended at T . Prove that PK + PTis independent of the position of P (that is, the value of PK+PT is alwaysthe same, no matter where P is pla ed).

15Solution by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long,Vietnam.Sin e △ABC is isos eles with sides ABandAC of equal length, we haveMA ⊥ BC.Also, sin e PT ⊥ BC, thenMA||PT .Sin eMA||PK, then △MBA is similarto △PBK sin e ea h is right-angled and theyshare the angle at B. From this, we obtainPK

PB=MA

MB, hen e PK =

PB ·MA

MB.Similarly, sin e MA||PT , then △CPTis similar to △CMA, when e PT

PC=

MA

MCand so PT =PC · MA

MC. ................................................................................................................................................

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A

CB

K

P M

T

Sin eMB = MC = 12BC, we an on lude that

PK + PT =PB ·MA

MB+

PC ·MA

MC=

(PB + PC) ·MA

MB

=BC · MA

MB= 2MA .Thus, PK + PT is independent of the position of P , sin e it dependsonly on the length ofMA.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, ThePhilippines; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIR �O, IES \Abastos",Valen ia, Spain; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; BILLYSUANDITO, Palembang, Indonesia; and TITU ZVONARU, Com�ane�sti, Romania. There was 1in orre t solution submitted.M341. Proposed by the Mayhem Sta�.Let ABC be a right triangle with right angle at B. Sides BA and BCare in the ratio 3 : 2. Altitude BD divides CA into two parts that di�er inlength by 10. What is the length of CA?Solution by Taylor Thetford, student, Lakeview High S hool, San Angelo, TX,USA. Let 2x and 3x be the lengths of

CB and AB, respe tively. Let y andy+ 10 be the lengths of CD and DA,respe tively. Let z be the length ofBD. We wish to �nd 2y + 10, whi his the length of CA.By applying the PythagoreanTheorem in △ABC, we �nd that(2x)2 + (3x)2 = (2y + 10)2 and so13x2 = 4y2 + 40y + 100.

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

C D A

B

z

y y + 10

2x3x

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16Applying the Pythagorean Theorem to △BDC and △BDA, we �ndthat y2 + z2 = 4x2 and z2 + (y + 10)2 = 9x2.Eliminating z in the last two equations gives 4x2−y2 = 9x2−(y+10)2.Therefore, 5x2 = (y + 10)2 − y2 = 20y + 100 or x2 = 4y + 20, and so

13x2 = 52y + 260.Combining this result with 13x2 = 4y2 + 40y + 100, we �nd that52y + 260 = 4y2 + 40y + 100 ;

4y2 − 12y − 160 = 0 ;y2 − 3y − 40 = 0 ;

(y + 5)(y − 8) = 0 .Sin e y > 0, then y = 8, and so CA = 2y + 10 = 26.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; IANJUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines; RICHARDI. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIR �O, IES \Abastos", Valen ia, Spain;KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; BILLY SUANDITO,Palembang, Indonesia; LUYAN ZHONG-QIAO, Columbia International College, Hamilton, ON;and TITU ZVONARU, Com�ane�sti, Romania. There was 1 in orre t solution submitted.M342. Proposed by the Mayhem Sta�.Quin y and Celine have to move 10 small boxes and 10 large boxes. The hart below indi ates the time that ea h person takes to move ea h type ofbox. Celine Quin ysmall box 1 min. 3 min.large box 6 min. 5 min.They start moving the boxes at 9:00 am. What is the earliest time at whi hthey an be �nished moving all of the boxes?Solution by Mayhem Sta�.Let x represent the number of small boxes and y represent the numberof large boxes that Celine moves. Sin e there are 10 small boxes and 10 largeboxes, then Quin y moves 10 − x small boxes and 10 − y large boxes.Given the lengths of time that ea h takes, it takes Celine x+6y minutesand it takes Quin y 3(10 −x)+5(10− y) = 80−3x−5y minutes. If x = 9and y = 4, then Celine takes 33 minutes and Quin y takes 33 minutes. Weshow that it annot be done faster than this.If Quin y and Celine �nish in fewer than 33 minutes, then ea h takesat most 32 minutes, so the total working time is at most 64 minutes, sox+ 6y + (80 − 3x− 5y) = 80 − 2x+ y ≤ 64 or 2x− y ≥ 16.Sin e x and y are nonnegative integers and ea h is less than 10, thenthe possible pairs (x, y) that satisfy this inequality are (8, 0), (9, 0), (9, 1),(9, 2), (10, 0), (10, 1), (10, 2), (10, 3), and (10, 4).

17Sin e we want ea h of Celine's time and Quin y's time to be at most

32 minutes, then we need x + 6y ≤ 32 and 80 − 3x − 5y ≤ 32. The�rst inequality eliminates the pair (10, 4) from the list of possible pairs. These ond inequality simpli�es to 3x + 5y ≥ 48; none of the remaining pairssatisfy this inequality.Thus, none of these possibilities take any less time than 33 minutes.Therefore, the earliest possible �nishing time is 9:33 a.m.There were 4 in orre t and 3 in omplete solutions submitted.An expanded treatment of a similar problem appeared in the Problem of the Month olumn in CRUX with MAYHEM, volume 34, number 2.M343. Proposed by the Mayhem Sta�.The Fibona i numbers are de�ned by f1 = f2 = 1 and, for n ≥ 2,by fn+1 = fn + fn−1. The �rst few Fibona i numbers are 1, 1, 2, 3, 5, 8,13, 21, 34, 55, 89, 144, . . . . Find the sum of the �rst 100 even Fibona inumbers.Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON.Sin e f1 = f2 = 1, f3 = 2, and fn = fn−1 + fn−2, then fm is evenif and only if m is a multiple of 3. (This is be ause the parities of the termswill form the pattern Odd, Odd, Even, Odd, Odd, Even, and so on.)If Sn =

n∑

k=1

f3k, thenSn =

1

2

n∑

k=1

(f3k+f3k) =1

2

n∑

k=1

((f3k−2+f3k−1)+f3k

)=

1

2

3n∑

k=1

fk . (1)Next, we have f1 = f3 − f2, and f2 = f4 − f3, and also f3 = f5 − f4,and so on until fr−1 = fr+1 − fr and fr = fr+2 − fr+1.Sin e the right side of the sum of the n equations above \teles opes",it follows thatr∑

k=1

fk = fr+2 − f2 = fr+2 − 1 . (2)From (1) and (2), we �nd that Sn = 12(f3n+2 − 1). In our parti ular ase, S100 = 1

2(f302 − 1). Maple omputes the value of S100 to be exa tly290905784918002003245752779317049533129517076702883498623284700.For the re ord, by Binet's formula for Fibona i numbers we have that

fm =1√5

(αm − βm), where α =1

2(1 +

√5) and β =

1

2(1 +

√5). Hen ethe required sum is also given by S100 =

1

2√

5

(α302 − β302

)− 1

2.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam;IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines; RICHARDI. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIR �O, IES \Abastos", Valen ia, Spain;DIVYANSHU RANJAN, Delhi, India; JOS �E HERN�ANDEZ SANTIAGO, student, UniversidadTe nol �ogi a de la Mixte a, Oaxa a, Mexi o; and TITU ZVONARU, Com�ane�sti, Romania.

18Problem of the Month

Ian VanderBurghApproximation is one of the most important on epts in mathemati s.Problem (2006 Canadian Open Mathemati s Challenge) Determine, with jus-ti� ation, the fra tion p

q, where p and q are positive integers and q < 100,that is losest to, but not equal to, 3

7.While it is tempting to get out your al ulator, it an initially only helpso mu h. If you al ulate 3

7, you'll obtain 0.428571 . . .. This doesn't help inany obvious way to answer the question.A �rst approa h after the al ulator might be to go for the fra tion withthe largest possible denominator. This makes a lot of sense in some ways,as the fra tions with the largest denominators will be losest together and sowould seem to have the best han e of being losest to 3

7. In our ase, thelargest possible denominator is q = 99. The given fra tion, 3

7, is between

42

99= 0.424242 . . . and 43

99= 0.434343 . . .. After a qui k look, we an tellthat 3

7is loser to 42

99. From the de imal approximations, 3

7and 42

99di�er byabout 0.004. Is this the losest of all possible fra tions?Another idea is to try to onvert 3

7into the equivalent fra tion withthe largest possible denominator and then adjust from there. Multiplyingnumerator and denominator by 14, we obtain 42

98. We ould then add 1 or

−1 to the numerator to obtain 41

98or 43

98, whi h di�er from 3

7by 1

98. But thismeans that the di�eren e is bigger than 0.01, whi h is worse than before, sothis approa h doesn't give a loser fra tion.Can we do better than 42

99? It is possible that, even though fra tionswith smaller denominators are further apart, they an be between some ofthe other fra tions that we've looked at, for example between 42

99and 3

7orbetween 43

99and 3

7.Solution We want to use the fra tion p

qto approximate 3

7. Let's al ulatetheir di�eren e, whi h is what we want to minimize:

∣∣∣∣

p

q− 3

7

∣∣∣∣

=

∣∣∣∣

7p− 3q

7q

∣∣∣∣

=|7p− 3q|

7q.

19What an we do to make this as small as possible? Two approa heswould be to make the numerator of the di�eren e as small as possible or tomake the denominator of the di�eren e as large as possible.Let's fo us initially on the numerator. The numerator annot equal 0be ause the fra tions p

qand 3

7are not equal. Thus, the smallest possiblevalue for the numerator is 1, be ause p and q are integers. So let's try to�nd values of p and q for whi h the numerator equals 1. In this ase, thedi�eren e equals 1

7qwhi h is minimized when q is largest.For the numerator to equal 1, we need 7p − 3q = ±1. Sin e we alsowant to maximize q, we rewrite this as 7p = 3q±1 and work from the largestpossible integer values of q to see when we also get an integer value for p.If q = 99, the equation be omes 7p = 3(99) ± 1 = 297 ± 1. Neitherpossibility is a multiple of 7.If q = 98, the equation be omes 7p = 3(98) ± 1 = 294 ± 1. Neitherpossibility is a multiple of 7.If q = 97, the equation be omes 7p = 3(97) ± 1 = 291 ± 1. Neitherpossibility is a multiple of 7.If q = 96, the equation be omes 7p = 3(96) ± 1 = 288 ± 1. Sin e

287 is a multiple of 7, then taking q = 96 and p = 41 gives a di�eren e withnumerator 1.So we have ∣∣∣4196

− 3

7

∣∣∣ =

1

7 · 96=

1

672and this is the smallest possibledi�eren e with the numerator equal to 1.If the numerator equalled 2 or something larger, then the smallest pos-sible di�eren e o urs when the numerator is as small as possible and thedenominator is as large as possible, so is 2

7 · 99=

2

693. This is the smallestpossible di�eren e with numerator at least 2.Combining the ases, the smallest possible di�eren e is indeed 1

672,and so the losest fra tion to 3

7of all of the fra tions under onsideration is

p

q=

41

96.The approximation of fun tions with polynomials is often seen in �rst-year university al ulus ourses. As part of these investigations, we learnhow to estimate the amount of error when approximating, for example, sin xwith x − 1

6x3 +

1

120x5. The te hniques used to estimate this type of errorare not dissimilar to what we have seen above, and are very useful in manytypes of al ulations.

20THE OLYMPIAD CORNERNo. 275R.E. WoodrowThe year has own by, and it has brought many hanges to Crux andto the Corner. Of ourse it has been overshadowed by the sudden and un-timely loss of a great friend and a devoted olleague, Jim Totten, mid-waythrough the transition to a new Editor-in-Chief. I think Vazz Linek has donea wonderful job of stepping in and keeping the journal on tra k with only anunderstandable slowing of the produ tion pa e in the interim.Readers will have noti ed the announ ement at the end of the De em-ber Corner that Joanne Canape, who has transformed my s ribbles into ahigh quality tex �le for many years, has de ided that two de ades is enough.As I ustomarily begin the year by thanking all those who ontributed to theCorner in the last year, I would be very remiss not to lead o� with sin erethanks to Joanne.It is also appropriate to thank those who submitted problem sets for ouruse as well as a spe ial thanks to the dedi ated readers who furnish their ni esolutions whi h we use. Hoping, as always, that I've not missed someone,here is the list for the 2008 members of the Corner.Arkady AltMiguel Amengual CovasJean-Claude AndrieuxHouda AnounRi ardo Barroso CamposMi hel BatailleJos �e Luis D��az-BarreroJ. Chris FisherKipp JohnsonGeo�rey A. KandallIoannis KatsikisR. LaumenSalem Maliki Pavlos Maragoudakis

Robert MorewoodAndrea MunaroVedula N. MurtyFelix Re ioXavier RosD.J. SmeenkBabis StergiouDaniel TsaiPanos E. TsaoussoglouGeorge TsapakidisJan VersterEdward T.H. WangLuyan Zhong-QiaoLi ZhouTitu ZvonaruOur apologies to Svetoslav Sav hev for the misspelling of his name inthe De ember 2008 Olympiad.For your problem solving pleasure in the new year we start o� with theproblems of the German Mathemati al Olympiad, Final Round, 2006. Mythanks go to Robert Morewood, Canadian Team Leader to the 47th IMO inSlovenia 2006, for olle ting them for our use.

21German Mathemati al OlympiadFinal Round, Grades 12{13Muni h, April 29 { May 2, 2006

First Day1. Determine all positive integers n for whi h the numberzn = 101 · · · 101

︸ ︷︷ ︸

2n+1 digitsis a prime.2. Five points are on the surfa e of a sphere of radius 1. Let amin denotethe smallest distan e (measured along a straight line in spa e) between anytwo of these points. What is the maximum value for amin, taken over allarrangements of the �ve points?3. Find all positive integers n for whi h the numbers 1, 2, 3, . . . , 2n an be oloured with n olours in su h a way that every olour appears twi e andevery number 1, 2, 3, . . . , n appears exa tly on e as the di�eren e of twonumbers with the same olor. Se ond Day4. Let D be a point inside the triangle ABC su h that AC − AD ≥ 1 andBC −BD ≥ 1. Prove that EC −ED ≥ 1 for any point E on the side AB.5. Let x be a nonzero real number satisfying the equation ax2 + bx+ c = 0.Furthermore, let a, b, and c be integers satisfying |a| + |b| + |c| > 1. Provethat

|x| ≥ 1

|a| + |b| + |c| − 1.

6. Let a ir le through B and C of a triangle ABC interse t AB and AC inY and Z, respe tively. Let P be the interse tion of BZ and CY , and let Xbe the interse tion of AP and BC. LetM be the point that is distin t fromX and on the interse tion of the ir um ir le of the triangleXY Z with BC.Prove thatM is the midpoint of BC

Our se ond problem set for this number is a set of sele ted problemsfrom the Thai Mathemati al Olympiad Examinations 2005. Again, thanksgo to Robert Morewood, team leader to the 47th IMO in Slovenia 2006, for olle ting them for the Corner.

22Thai Mathemati al Olympiad Examinations 2005Sele ted Problems1. Let P (x), Q(x), and R(x) be polynomials satisfying

2xP(x3)

+ Q(−x− x2

)=(1 + x+ x2

)R(x) .Show that x− 1 is a fa tor of P (x) −Q(x).2. Find all fun tions f : R → R su h that

f(x+ y + f(xy)

)= f

(f(x+ y)

)+ xyfor all x, y ∈ R.3. Let a, b, and c be positive real numbers. Prove that

1 +3

ab+ bc+ ca≥ 6

a+ b+ c.

4. Let n be a positive integer. Prove that n(n + 1)(n + 2) is not a perfe tsquare.5. Find the least positive integer n su h that 2549|(n2545 − 2541

).6. Do there exist positive integers x, y, and z su h that2548x + (−2005)y = (−543)z ?7. Show that there exist positive integersm and n su h that gcd(m,n) = 1and 2549| ((25 · 49)m + 25n − 2 · 49n).8. The median AM of a triangle ABC interse ts its in ir le ω at K and L.The lines through K and L parallel to BC interse t ω again at X and Y ,respe tively. The lines AX and AY interse t BC at P and Q. Prove that

BC = CQ. (Shortlist 2005)9. Let ABC be an a ute-angled triangle with AB 6= AC, let H be itsortho entre and M the midpoint of BC. Points D on AB and E on ACare su h that AE = AD and D, H, and E are ollinear. Prove that HM isorthogonal to the ommon hord of the ir um ir les of triangles ABC andADE. (Shortlist 2005)10. Assume ABC is an isos eles triangle with AB = AC. Suppose thatP is a point on the extension of side BC. X and Y are points on lines ABand AC su h that PX‖AC and PY ‖AB. Let T be the midpoint of ar BC.Prove that PT ⊥ XY . (Iran 2004)

23As a third set of problems we give the 46th Ukrainian Mathemati alOlympiad Final Round 2006 - 11th form problems. Again, thanks go to RobertMorewood, team leader to the 47th IMO in Slovenia 2006, for olle ting themfor our use.46th Ukrainian Mathemati al Olympiad 2006Final Round11th Form1. (V.V. Plakhotnyk) Prove that for any rational numbers a and b the graphof the fun tion

f(x) = x3 − 6abx− 2a3 − 4b3 , x ∈ Rhas exa tly one point in ommon with the x-axis.2. (O.A. Sarana) A ir le is divided into 2006 equal ar s by 2006 points.Baron Mun hausen laims that he an onstru t a losed polygonal urvewith the set of verti es onsisting of these 2006 points su h that amongst its2006 edges there are no two whi h are parallel to ea h other. Is his laimtrue or false?3. (T.M. Mitelman)(a) Prove that for any rational number α ∈ (0, 1) there exists an in�niteset of real numbers that satisfy the equation {x⌊x{x}

⌋}= α and anytwo of them have the same fra tional part. (The fra tional part of a realnumber a is given by {a} = a− ⌊a⌋, where ⌊a⌋ is its integer part, thatis, the greatest integer that does not ex eed a.)(b) Prove that for any rational number α ∈ (0, 1) there exists an in�niteset of real numbers that satisfy the equation {x⌊x{x}

⌋}= α and anytwo of them have di�erent fra tional parts.4. (V.A. Yasinskiy) Two ir les ω1 and ω2 interse t ea h other at two distin tpoints A and B. The tangent line of the ir le ω1 at the point A and thetangent line of the ir le ω2 at the point B meet at point C. The �rst ofthese two lines interse ts the ir le ω2 for the se ond time at point T 6= A.The point X (distin t from A and B) is on the ir le ω1, and the line XAinterse ts the ir le ω2 for the se ond time at point Y (distin t from A). Thelines Y B and XC meet at point Z. Prove that TZ is parallel to XY .5. (O.O. Kur henko) Prove that for any real numbers x and y

| cosx| + | cos y| + | cos(x+ y)| ≥ 1 .6. (T.M. Mitelman) Find all fun tions f : R → R su h thatf(x3 + y3

)= x2f(x) + yf(y2)for all real numbers x and y.

247. (V.A. Yasinskiy) A pointM lies inside a ube ABCDA1B1C1D1. PointsA′, B′, C′, D′, A′

1, B′1, C′

1, and D′1 belong to the rays MA, MB, MC,

MD, MA1, MB1, BC1, and MD1 respe tively. Prove that if the poly-hedron A′B′C′D′A′1B

′1C

′1D

′1 is a parallelepiped (that is, all of its fa es areparallelograms), then it is a ube.8. (V.A. Yasinskiy) There are n ≥ 3 soldiers in aptain Petrenko's squad,no two of the same height. The aptain orders them to stand single-�le (notne essarily sorted by height). A \wave" is any subsequen e of (not ne essar-ily next to ea h other) soldiers in this line su h that the �rst (leftmost) soldierin the wave is taller than the se ond soldier in it, but the se ond soldier in itis shorter than the third one, who is in turn taller than the fourth one, andso on. (For example, if n = 9, the soldiers are enumerated by height, andthe aptain lines them up as 9, 3, 5, 7, 1, 2, 6, 4, 8 then a longest wave forthis line-up is 9, 3, 7, 1, 6, 4, 8. However, if the aptain lines them up as

1, 2, 3, 4, 5, 6, 7, 8, 9, then a longest wave onsists of (any) one soldier.)For every n, onsider the number of possible lines with the longest waves ofeven lengths and the number of possible lines with the longest waves of oddlengths. Whi h of these numbers is bigger?Continuing with problems for readers to solve we give the Cze h-Polish-Slovak Mathemati s Competition written on June 26-28, 2006 at �Zilina,Slovakia. Thanks again go to Robert Morewood, Canadian team leader tothe 47th IMO in Slovenia 2006, for olle ting them for our use.Cze h-Polish-Slovak Mathemati s Competition 2006

1. Five distin t points A, B, C, D, and E lie in this order on a ir le ofradius r and satisfy AC = BD = CE = r. Prove that the ortho entresof the triangles ACD, BCD, and BCE are the verti es of a right-angledtriangle.2. There are n hildren sitting at a round table. Erika is the oldest amongthem and she has n andies. No other hild has any andy. Erika distributesthe andies as follows. In every round, all the hildren with at least two andies show their hands. Erika hooses one of them and he/she gives one andy to ea h of the hildren sitting next to him/her. (So in the �rst roundErika must hoose herself to begin the distribution.) For whi h n ≥ 3 is itpossible to redistribute the andies so that ea h hild has exa tly one andy?3. The sum of four real numbers is 9 and the sum of their squares is 21.Prove that these four numbers an be labelled as a, b, c, and d so that theinequality ab− cd ≥ 2 holds.

254. Prove that for every positive integer k there is a positive integer n su hthat the de imal representation of 2n has a blo k of exa tly k onse utivezeros, that is, 2n = · · · a00 · · · 0b · · · , where a and b are nonzero digits withk zeros between them.5. Find the number of integer sequen es (an)∞

n=1 su h that an 6= −1 andan+2 =

an + 2006

an+1 + 1for every positive integer n.6. Is there a onvex pentagon A1A2 . . . A5 su h that for ea h i the linesAiAi+3 and Ai+1Ai+2 interse t in Bi and the points B1, B2, . . . , B5 are ollinear? (By onvention A6 = A1, A7 = A2, and A8 = A3.)

Our �nal problem set for this issue is the XXI Olimpiadi Italiano dellaMatemati a, Cesenati o, written 5 May 2006. Thanks again go to RobertMorewood, Canadian team leader to the 47th IMO in Slovenia, for olle tingthem for our use.XXI Olimpiadi Italiano della Matemati aCesenati oMay 5, 20061. Rose and Savino play a game with a de k of traditional Neapolitan playing ards whi h onsists of 40 ards of four di�erent suits, numbered 1 to 10. Atthe start ea h player has 20 ards. Taking turns, one shows a ard on thetable. Whenever some ards on the table add to exa tly 15, these are thenremoved from the game (if the sum 15 an be obtained in more than one way,the player who last moved de ides whi h ards adding to 15 to remove). Atthe end of the game only one ard, a 9, is left on the table. Savino holds two ards numbered 3 and 5, and Rose holds one ard. What is the number ofRose's ard?2. Find all values ofm, n, and p su h thatpn + 144 = m2 ,wherem and n are positive integers and p is a prime number.3. Let A and B be two points on a ir le Γ su h that AB is not a diameter.Let P be a point on Γ di�erent from A and B, and letH be the ortho entreof the triangle ABP . Find the lo us of H as P varies over all points of Γdi�erent from A and B.

264. On an in�nite hessboard all thepositive integers are written in as- ending order along a spiral, startingfrom 1 and pro eeding anti lo kwise;a portion of the hessboard is shownin the �gure.A "right half-line" of the hess-board is the set of squares given bya square C and by all squares in thesame row as C and to the right of C.

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(a) Prove that there exists a right half-line none of whose squares ontainsa multiple of 3.(b) Determine if there exist in�nitely many pairwise disjoint right half-linesnone of whose squares ontains a multiple of 3.5. Consider the inequality(x1 + · · · + xn)2 ≥ 4(x1x2 + x2x3 + · · · + xnx1) .(a) Determine for whi h n ≥ 3 the inequality holds true for all possible hoi es of positive real numbers x1, x2, . . . , xn.(b) Determine for whi h n ≥ 3 the inequality holds true for all possible hoi es of any real numbers x1, x2, . . . , xn.6. Albert and Barbara play a game. At the start there are some piles of oinson the table, not all ne essarily with the same number of oins. The playersmove in turn and Albert starts. At ea h turn a player may either take a oinfrom a pile or divide a pile into two piles with ea h pile ontaining at leastone oin (a player may exer ise only one of these options).The one who takes the last oin wins the game. In terms of the numberof piles and the number of oins in ea h pile at the start, determine whi h ofthe players has a winning strategy.

Now we turn to our �le of solutions from the readers to problems fromthe Mar h 2008 number of the Corner and the Estonian IMO Sele tion Con-test 2004-2005, given at [2008 : 79-80℄.3. Find all pairs (x, y) of positive integers satisfying (x+ y)x = xy.Solution by Konstantine Zelator, University of Toledo, Toledo, OH, USA.We show that there are exa tly two su h pairs, (x, y) = (2, 6), (3, 6).We will make use of two basi fa ts from elementary number theory.

27(a) If a, b ∈ Z and gcd(a, b) = 1 , then gcd(am, bn) = 1 for any positiveintegersm and n.(b) If a|b, a is a positive integer, b ∈ Z, and gcd(a, b) = 1, then a = 1.Let x and y be positive integers that satisfy (x + y)x = xy. Write

d = gcd(x, y) and let x = dx1, y = dy1, where x1 and y1 are positiveintegers that are relatively prime. Substituting for x and y yieldsdx1d · (x1 + y1)

x1d = xy1d1 · dy1d . (1)We make ases by omparing the sizes of x1 and y1.Case 1. Suppose that x1 = y1. Sin e gcd(x1, y1) = 1, we have x1 = y1 = 1.Thus, equation (1) be omes dd · 2d = dd, whi h is impossible sin e d ≥ 1.Case 2. Suppose that y1 < x1. Then x1 − y1 is a positive integer and fromequation (1) we obtain

dd(x1−y1) · (x1 + y1)x1d = xy1d

1 . (2)Sin e gcd(x1, y1) = 1 it follows that gcd(x1 + y1, x1) = 1. By (a) above, wehave gcd((x1 + y1)

x1d, xy1d1

)= 1. However, by equation (2), the positiveinteger (x1+y1)

x1d is a divisor of xy1d1 . Sin e these two integers are relativelyprime, it follows by (b) that (x1 + y1)

x1d = 1, whi h is impossible sin ex1 + y1 ≥ 2 and x1 · d ≥ 1.Case 3. Suppose that x1 < y1. From (1) we obtain

(x1 + y1)x1d = xy1d

1 · dd(y1−x1) . (3)Sin e gcd(x1, y1) = 1 we have gcd((x1 + y1)

x1d, xy1d1

)= 1 and from equa-tion (3) we see that xy1d

1 is a divisor of (x1 + y1)x1d, whi h implies that

xy1d1 = 1. Sin e y1d is a positive integer this means that x1 = 1. Going ba kto equation (3) we see that (1 + y1)

d = dd(y1−1), hen e1 + y1 = dy1−1 . (4)Note that d 6= 1; otherwise equation (4) be omes 1+y1 = 1, ontrary to thefa t that y1 is a positive integer. Thus, d ≥ 2. Sin e y1 = 1 does not satisfyequation (4), we also have y1 ≥ 2. Setting k = y1 − 1 equation (4) thenbe omes k + 2 = dk. By Indu tion (or the Binomial Theorem) we obtain

2k > k+ 2 for all integers k ≥ 3. Sin e dk ≥ 2k, it follows from k+ 2 = dkthat k = 1 or k = 2.For k = 2 we have 4 = d2, hen e d = 2. From 2 = k = y1 −1 we thenhave y1 = 3. Re all that x1 = 1. Going ba k, we have x = x1d = 1 · 2 = 2and y = y1d = 3 · 2 = 6. This is the solution (x, y) = (2, 6).Similarly, for k = 1 we have d = 3. Then y1 = k + 1 = 2 and sin ex1 = 1 we obtain x = dx1 = 3 · 1 = 3 and y = dy1 = 3 · 2 = 6. This is theother solution (x, y) = (3, 6).

284. Find all pairs (a, b) of real numbers su h that all roots of the polynomials6x2 − 24x− 4a and x3 + ax2 + bx− 8 are non-negative real numbers.Solution by Titu Zvonaru, Com�ane�sti, Romania.Let β1, β2, and β3 be the roots of the polynomial x3 +ax2 +bx−8, sothat x3 + ax2 + bx− 8 = (x−β1)(x−β2)(x−β3). Comparing oeÆ ientsyields β1 + β2 + β3 = −a and β1β2β3 = 8. Sin e β1, β2, and β3 arenonnegative real numbers, by the AM-GM Inequality we have

β1 + β2 + β3 ≥ 3 3

√

β1β2β3 ,hen e −a ≥ 6 or a ≤ −6. The equation 6x2 − 24x − 4a = 0 has real rootsif and only if 242 − 4 · (−4a) · 6 ≥ 0, whi h implies 242 + 24 · 4a ≥ 0 andhen e a ≥ −6. Therefore, a = −6.Now we have β1 + β2 + β3 = 6 = 3 3√β1β2β3, from whi h it followsthat β1 = β2 = β3 = 2 and b = 12. Thus, the only pair satisfying the ondition is (a, b) = (−6, 12).

Next we turn to a solution to a problem of the Trenti �eme OlympiadMath �ematique BelgeMaxi Finale,Mer redi 20 avril 2005 given at [2008 : 80℄.3. Dans le triangle ABC, les droites AE et CD sont les bisse tri esint �erieures des angles ∠BAC et ∠ACB respe tivement ; E appartient �aBCet D appartient �a AB. Pour quelles amplitudes de l'angle ∠ABC a-t-on ertainement(a) |AD| + |EC| = |AC|? (b) |AD| + |EC| > |AC|?( ) |AD| + |EC| < |AC|?Solution by Titu Zvonaru, Com�ane�sti, RomaniaAs usual let a = BC, b = CA,and c = AB. The �rst equation belowis the Angle Bise tor Theorem; the fol-lowing equations are equivalent to it:BE

EC=

AB

AC;

BE

EC=

c

b;

BE + EC

EC=

b+ c

b;

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EB

D

A

CSimilarly, AD =bc

a + b.

29We have

|AD| + |EC| − |AC| =bc

a+ b+

ab

b+ c− b

= b

(c(b+ c) + a(a+ b) − (a+ b)(b+ c)

(a+ b)(b+ c)

) .By the Law of Cosines a2 + c2 − b2 = 2ac cosB, so the equation above anbe rewritten as|AD| + |EC| − |AC| =

2abc(cos(B) − 1

2

)

(a+ b)(b+ c).Hen e,

|AD| + |EC| = |AC| ⇐⇒ ∠ABC = 60◦ ,|AD| + |EC| > |AC| ⇐⇒ ∠ABC < 60◦ ,|AD| + |EC| < |AC| ⇐⇒ ∠ABC > 60◦ .

Next we turn to solutions from our readers to problems of the 2005Vietnam Mathemati al Olympiad given at [2008 : 81℄.1. Find the smallest and largest values of the expression P = x+ y, wherex and y are real numbers satisfying x− 3

√x+ 1 = 3

√y + 2 − y.Solved by Arkady Alt, San Jose, CA, USA; andMi hel Bataille, Rouen, Fran e.We give the solution of Bataille.We show that Pmin = 1

2

(9 + 3

√21) and Pmax = 9 + 3

√15.First, x = −1 and y = 1

2

(11 + 3

√21) satisfy the onstraint equation

x−3√x+ 1 = 3

√y + 2−y (easily he ked using 10+2

√21 =

(√3+

√7)2

)and P = 12

(9+3

√21). Similarly, for x = 1

2

(10+3

√15), y = 1

2

(8+3

√15),we have P = 9 + 3

√15 and the onstraint is satis�ed. Now, let x and ysatisfy the onstraint equation. Then P = 3

√x+ 1 + 3

√y + 2, so that

P 2 = 9(

P + 3 + 2√

(x+ 1)(y + 2)) . (1)It follows that P ≥ 0 and P 2 − 9P − 27 ≥ 0. Thus, P is not less thanthe positive solution of the quadrati x2 − 9x− 27 = 0 and we dedu e that

P ≥ 12

(9 + 3

√21). From the AM-GM Inequality and (1), we obtain

P 2 ≤ 9(P + 3 + x+ 1 + y + 2) = 9(2P + 6) = 18P + 54 ,or P 2 − 18P − 54 ≤ 0, whi h implies that P ≤ 9 + 3√

15. The proof is omplete.

304. Find all real-valued fun tions f de�ned on R that satisfy the identityf(f(x− y)

)= f(x)f(y) − f(x) + f(y) − xy.Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille,Rouen, Fran e; and Daniel Tsai, student, Taipei Ameri an S hool, Taipei,Taiwan. We give the write up of Bataille.It is readily he ked that the fun tion f(x) = −x for all x is a solution.We show that there are no other solutions. Let f satisfyf(f(x− y)

)= f(x)f(y) − f(x) + f(y) − xy (1)for all x, y ∈ R and let a = f(0). Taking x = y = 0 in (1) gives f(a) = a2and then taking y = x shows thatf(x)2 = x2 + a2for all real numbers x. In parti ular f(a)2 = 2a2, that is, a4 = 2a2, hen e

a ∈ {0, √2, −

√2}. Assume that a =

√2. Then for any given x we have

f(x) =√x2 + 2 or f(x) = −

√x2 + 2 and taking y = 0 in (1), we obtain

f(f(x)

)=(√

2 − 1)f(x) +

√2 . (2)Now, f(√2

)= f(a) = a2 = 2 so that f(f(√2

))= f(2) = ±

√6, but thentaking x =

√2 in (2) yields a ontradi tion sin e 6 6=

(3√

2 − 2)2. Similarly,the assumption a = −

√2 leads to a ontradi tion. It follows that a = 0,hen e f(x) = x or f(x) = −x for any given x. However, if f(x0) = x0 forsome nonzero real number x0, then f(f(x0)

)= f(x0) = x0, while from (1)we have f(f(x0)

)= f

(f(x0 − 0)

)= −f(x0) = −x0. This is impossiblesin e x0 6= 0, hen e f(x) = −x for all real x.5. Find all triples of non-negative integers (x, y, n) su h that x! + y!

n!= 3n(with the onvention 0! = 1).Solved by Daniel Tsai, student, Taipei Ameri an S hool, Taipei, Taiwan; andKonstantine Zelator, University of Toledo, Toledo, OH, USA. We give thesolution of Tsai.Let S be the set of all ordered triples (x, y, n) of nonnegative integerssu h that x! + y!

n!= 3n, or equivalently x! + y! = 3nn!. For n = 0, it isseen at on e that there is no orresponding (x, y, n) ∈ S. For n = 1, let

(x, y, n) ∈ S; then if x ≥ 3 or y ≥ 3 we have x! + y! ≥ 7 > 3 = 311!,thus x, y < 3 and simple he king yields that {(0, 2, 1), (1, 2, 1), (2, 0, 1),(2, 1, 1)} ⊂ S.Lemma Let (x, y, n) ∈ S with n ≥ 2. Then x, y ≥ n and x > n or y > n.Proof: If x, y ≤ n, then x! + y! ≤ 2n! < 3nn!, so x > n or y > n. Ifx > n and y < n, then x! + y!

n!=

x!

n!+y!

n!= 3n, a ontradi tion sin e x!

n!is

31an integer but y!

n!is not. Thus, if x > n, then y ≥ n. Similarly, in the ase

y > n we have x ≥ n by symmetry.We shall prove that for n ≥ 2 there is no orresponding (x, y, n) ∈ Sby onsidering ases on n modulo 3.Case 1. n ≡ 0 (mod 3). Let (x, y, n) ∈ S and assume without loss ofgenerality that x ≤ y. By the Lemma, n ≤ x ≤ y and one of these twoinequalities is stri t. If x > n, then from x! + y!

n!= 3n it follows that

(n+1)|3n. However, n+1 has a prime divisor other than 3, a ontradi tion.Therefore, n = x < y, and onsequentlyx! + y!

n!=

x!

n!+

y!

n!= 1 + (n+ 1)(n+ 2) · · · y = 3n .Thus, 3 divides 1+(n+1)(n+2) · · · y and (n+1)(n+2) · · · y ≡ 2 (mod 3),whi h implies that y = n+ 2. However, 1 + (n+ 1)(n+ 2) < 3n for n ≥ 3(by indu tion) and 1 + (2 + 1)(2 + 2) 6= 32, ontradi ting the fa t that

(x, y, n) ∈ S.Case 2. n ≡ 1 (mod 3). Let (x, y, n) ∈ S and assume without loss ofgenerality that x ≤ y. By reasoning similar to that of Case 1 it follows thatx = n and y = n + 1. However, 1 + (n + 1) < 3n for ea h integer n ≥ 2, ontradi ting the fa t that (x, y, n) ∈ S.Case 3. n ≡ 2 (mod 3). Let (x, y, n) ∈ S and assume without loss ofgenerality that x ≤ y. By the Lemma, n ≤ x ≤ y and one of these twoinequalities is stri t. If x ≥ n + 2, then from x! + y!

n!= 3n it follows that

(n+2)|3n. However, n+2 has a prime divisor other than 3, a ontradi tion,hen e n ≤ x < n+ 2.If x = n, then n = x < y andx! + y!

n!=

x!

n!+

y!

n!= 1 + (n+ 1)(n+ 2) · · · y = 3n , ontradi ting n+ 1 ≡ 0 (mod 3).If x = n+ 1, then

x! + y!

n!=

x!

n!+

y!

n!= (n+ 1) + (n+ 1)(n+ 2) · · · y = 3n .If furthermore y = n+ 1, then

2(n+ 1) = (n+ 1) + (n+ 1)(n+ 2) · · · y = 3nis even, a ontradi tion. Thus, y > n+1 and (n+1)(1+(n+2) . . . y

)= 3n.It follows that 1+(n+2) · · · y ≡ 0 (mod 3) and (n+2) · · · y ≡ 2 (mod 3),whi h implies that y = n+ 3. However, (n+ 1)

(1 + (n+ 2)(n+ 3)

)6= 3n

32for 1 ≤ n ≤ 5 and by indu tion (n+1)

(1+(n+2)(n+3)

)< 3n for n ≥ 6, ontradi ting the fa t that (x, y, n) ∈ S.6. Let the sequen e x1, x2, x3, . . . , be de�ned by x1 = a, where a is a realnumber, and the re ursion xn+1 = 3x3

n − 7x2n + 5xn for n ≥ 1.Find all values of a for whi h the sequen e has a �nite limit as n tendsto in�nity, and �nd this limit.Solved by Arkady Alt, San Jose, CA, USA; Mi hel Bataille, Rouen, Fran e;and Daniel Tsai, student, Taipei Ameri an S hool, Taipei, Taiwan. We giveBataille's write-up.Let f(x) = 3x3 − 7x2 + 5x and g(x) = f(x) − x = x(x− 1)(3x− 4).The sequen e {xn}, whi h satis�es xn+1 = f(xn) for all positive integers n, an only onverge to a root of g(x) = 0. Thus, the only possible �nite limitsof {xn} are 0, 1, and 4

3. We show that the sequen e is onvergent if and onlyif 0 ≤ a ≤ 4

3, in whi h ase the limit is 1 ex ept if a = 0 and lim

n→∞xn = 0 orif a = 4

3and lim

n→∞xn = 4

3.Suppose �rst a < 0. Sin e g(x) < 0 when x < 0, it follows that

xn < x1 = a < 0 for all positive integers n. If {xn} had a �nite limit, ℓ,we would have ℓ ≤ a, ontradi ting the fa t that ℓ ∈ {0, 1, 43}. Thus, {xn}is divergent when a < 0. Using the fa t that g(x) > 0 for x > 4

3, similarreasoning shows that {xn} is divergent when a > 4

3.If a ∈ {0, 1, 4

3}, then the sequen e {xn} is onstant.If a ∈ (1, 4

3), then using f(x) − 1 = (x − 1)2(3x − 1) an easy indu -tion shows that 1 < xn+1 < xn for all positive integers n. Thus, {xn} isde reasing and bounded, hen e onvergent. Its limit ℓ satis�es ℓ ≥ 1 and

ℓ ∈ {0, 1, 43}, that is, ℓ = 1.If a ∈ [13, 1) then x2 = f(a) ≥ 1 and x2 <

43, as the maximum of f on

[0, 1] is f(59

)= 275

243< 4

3. From the previous ase, we see that lim

n→∞xn = 1.It remains to study the ase a ∈

(0, 1

3

). Then, 1

3m+1≤ a <

1

3mforsome unique positive integerm. If any of the numbers x2, x3, . . . , xm is notless than 1

3, let xk be the one with the smallest index. Then 1

3≤ xk <

43andby the previous ases {xn}n≥k onverges to 1 and lim

n→∞xn = 1. Otherwise,noting that f(x)−3x = x(x−2)(3x−1) is positive for x ∈

(0, 1

3

), we havex2 = f(x1) > 3x1 = 3a ≥ 1

3m,

x3 = f(x2) > 3x2 ≥ 1

3m−1,

· · ·xm = f(xm−1) > 3xm−1 ≥ 1

32,and �nally xm+1 >

13. So {xn}n≥m+1 onverges to 1 and again lim

n→∞xn =1.

33To �nish this number of the Corner we give solutions from the readersto problems of the 2005 GermanMathemati al Olympiad, given at [2008 : 82℄.1. Determine all pairs (x, y) of reals, whi h satisfy the system of equations

x3 + 1 − xy2 − y2 = 0 ,y3 + 1 − x2y − x2 = 0 .

Solved by George Apostolopoulos, Messolonghi, Gree e; Konstantine Zela-tor, University of Toledo, Toledo, OH, USA; and Titu Zvonaru, Com�ane�sti,Romania. We give the write-up of Apostolopoulos.We subtra t the two equations of the system to obtain(x3 − y3

)+ xy(x− y) + x2 − y2 = 0 ,whi h upon fa toring be omes

(x− y)(x+ y)(x+ y + 1) = 0 .Thus, y = x or y = −x or y = −x− 1.If y = x, then x = ±1, hen e (x, y) = (1, 1) or (x, y) = (−1,−1).If y = −x, then again x = ±1, hen e (x, y) = (1,−1) or (x, y) = (−1, 1).If y = −x− 1 we substitute into the �rst equation to obtainx3 + 1 − x(x+ 1)2 − (x+ 1)2 = −3x(x+ 1) = 0 ,hen e x = 0 or x = −1 and (x, y) = (0,−1) or (x, y) = (−1, 0).2. Let A, B, and C be three distin t points on the ir le k. Let the lines hand g ea h be perpendi ular to BC with h passing through B and g passingthrough C. The perpendi ular bise tor of AB meets h in F and the perpen-di ular bise tor of AC meets g in G. Prove that the produ t |BF | · |CG| isindependent of the hoi e of A, whenever B and C are �xed.Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; Mi helBataille, Rouen, Fran e; Konstantine Zelator, University of Toledo, Toledo,OH, USA; and Titu Zvonaru, Com�ane�sti, Romania. We give the solution ofAmengual Covas.Let A′ be the point diametri ally opposite to A. Let M and N bethe midpoints of the segments AB and AC, respe tively. Let B′ and C′be the se ond points of interse tion of the lines g and h with the ir le k,respe tively.

34

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k

gh

Sin e C′BC and BCB′ are right angles, the segments BB′ and CC′are diameters of the ir le k. Thus, the two quadrilaterals AC′A′C andABA′B′ are re tangles, and hen e parallelograms, so we have

|C′A′| = |CA| and |A′B′| = |AB| .Sin e the right triangles BMF and A′C′A are similar, as are the right tri-angles CNG and A′B′A, we haveBF

AA′ =BM

C′A′ =12AB

CAandCG

AA′ =CN

A′B′ =12CA

AB,from whi h we obtain

BF · CG =1

4|AA′|2as the square of the radius of k, whi h is independent of the hoi e of A.3. A lamp is pla ed at ea h latti e point (x, y) in the plane (that is, x and

y are both integers). At time t = 0 exa tly one lamp is swit hed on. Atany integer time t ≥ 1, exa tly those lamps are swit hed on whi h are at adistan e of 2005 from some lamp whi h is already swit hed on. Prove thatevery lamp will be swit hed on at some time.Solution by Titu Zvonaru, Com�ane�sti, Romania.Assume that at time t = 0 the lamp at O(0, 0) is swit hed on. Sin e2005 =

√13572 + 14762 then at some time the lamps at the following latti e

35points will be swit hed on:

A1(1357, 1476) , O1(2 · 1357, 0) ,A2(3 · 1357, 1476) , O2(4 · 1357, 0) ,... ...

Ak

((2k − 1) · 1357, 1476

) , Ok(2k · 1357, 0) ;and then the lamps at these latti e points will be swit hed on:B1(2k · 1357 − 2005, 0) ,B2(2k · 1357 − 2 · 2005, 0) ,B3(2k · 1357 − 3 · 2005, 0) ,...Bt(2k · 1357 − t · 2005, 0) .The equation 2k · 1357−2005t = 1 is the same as 2714k−2005t = 1,whi h has a solution in positive integers k, t be ause gcd(2714, 2005) = 1,for example, 2714 · 1134 − 2005 · 1535 = 1. Thus the lamp at (1, 0) will beswit hed on at some time. It follows (by symmetry) that every lamp will beswit hed on at some time.4. Let Q(n) denote the sum of the digits of the positive integer n. Provethat Q(Q(Q(20052005)

))

= 7.Solution by Titu Zvonaru, Com�ane�sti, Romania.It is well known thatQ(n) ≡ n (mod 9). Let k = Q(

Q(Q(20052005)

)),then k ≡ 20052005 (mod 9) and we have20052005 ≡ (−2)2005 = −2 · 22004 = −2 ·

(23)668

≡ −2(−1)668 = −2 ≡ 7 (mod 9) ,so that k ≡ 7 (mod 9).The number 20052005 has at most 4 · 2005 = 8020 digits. Hen e,Q(20052005

) is at most 9 · 8020 = 72180. This implies thatQ(

Q(20052005

))

≤ 5 · 9 = 45and hen e k = Q(

Q(Q(20052005)

)) is at most Q(39) = 12.Altogether, k is an integer satisfying 0 ≤ k ≤ 12 and k ≡ 7 (mod 9),hen e k = 7, as desired.That ompletes the Corner for this issue. Send me your ni e solutions,generalizations, and Olympiad problem sets.

36BOOK REVIEWSAmar SodhiPutnam and BeyondBy R�azvan Gel a and Titu Andrees u, Springer S ien e+ BusinessMedia LLC,New York, 2007ISBN-13: 978-0-387-25765-5, soft over, 798+xvi pages, US$69.95e-ISBN-13: 978-0-387-68445-1Reviewed by Jeff Hooper, A adia University, Wolfville, NSOne of my favourite problem books is the one by Andrees u and Gel a,Mathemati al Olympiad Challenges (Birkh�auser, 2000), in whi h the authors olle t numerous problems entredmainly around past Olympiads, and groupthem together by similar topi . In Putnam and Beyond, the authors again ombine to deliver a similar problem book, based this time around on thePutnam Competition.Stru turally, the book onsists of six major hapters: Methods of Proof,Algebra, Real Analysis, Geometry and Trigonometry, Number Theory, andCombinatori s and Probability. Ea h hapter is divided into numerous se -tions and subse tions, ea h of whi h fo uses on an important problem idea.For instan e, the initial se tion of the Algebra hapter is Identities andInequalities, and here we �nd the topi divided into a number of impor-tant ideas: Algebrai Identities, the equation x2 ≥ 0, the Cau hy-S hwartzInequality, the Triangle Inequality, the AM{GM Inequality, and so on. Ea hse tion ontains explanations of the key ideas and several worked problems,along with a number of problems to solve. Full solutions are provided atthe ba k of the book, along with sour es and/or helpful referen es wherene essary. The initial hapter is a parti ularly ni e introdu tion for students.There is some overlap, of ourse, with the topi s overed in the authors'previous book, and I was expe ting there to be mu h in ommon. To their redit, the authors have avoided that sort of mild heating: for the topi s in ommon with their �rst book, the examples and problems they o�er are new.This new book is also a mu h larger and far more extensive e�ort. In additionto all of the examples provided, the book ontains more than 900 problems.And these are Putnam-level problems, so they are mainly a level up fromthe earlier book. In fa t the topi s in the book go very deep, and over mostof the major ideas found in the undergraduate mathemati s urri ulum. Sothere's lots here.The book is not error-free, however, and a tea her or oa h who usesthis book should be a little areful. I ertainly haven't worked through everyproblem in the book, but did dip into them in a number of se tions. Justto give an example, Problem 33 asks the solver, "Given 50 distin t positiveintegers stri tly less than 100, prove that some two of them sum to 99." Alittle thought shows that one an take the numbers 50, 51, . . . , 99 and thestatement fails.

37But these problems are of the minor variety. This wonderful book is anex ellent problems resour e and should be ome a part of any serious libraryfor problem solving. By olle ting together problems by topi , the authorsprovide readers the han e to study ea h of these important problem-solvingte hniques and ideas in isolation, and help them begin to see the inherentpatterns. This should be one of the �rst books onsidered as a resour e byanyone oa hing groups of problem solvers.

Mathemati al Conne tions: A Companion for Tea hers and OthersBy Al Cuo o, Mathemati al Asso iation of Ameri a, 2005ISBN 978-0-88385-739-7, hard over, 239+xix pages, US$54.95Reviewed by Peter S. Brouwer, State University of New York, Potsdam,NY, USAAl Cuo o is the Dire tor of the Center for Mathemati s Edu ation atthe Edu ation Development Center in Newton, MA, where he works in theareas of urri ulum development and professional development of tea hers.This book joins a number of re ent others in addressing se ondary s hoolmathemati s ontent topi s from an advan ed (or deeper) perspe tive. Theprimary audien e is high s hool mathemati s tea hers, and the book is basedon the assumption that providing a more advan ed treatment of some of themathemati al topi s taught at that level is a valuable form of professional de-velopment. The author's emphasis is on making onne tions between topi sand developing mathemati al habits of mind.The hoi e of topi s is somewhat idiosyn rati , and re e ts the inter-related topi s that Cuo o is interested in exploring. The hapter titles are:1. Di�eren e Tables and Polynomial Fits, 2. Form and Fun tion: The Alge-bra of Polynomials, 3. Complex Numbers, ComplexMaps, and Trigonometry,4. Combinations and Lo ks, and 5. Sums of Powers.The strength of this book is that it is essentially a problems book (onthe above topi s). There are many problems, in luding 90 in the �rst hap-ter alone, and the reader is asked to work them sequentially while readingthrough the text. These are grouped by themes, whi h aids oheren e. Inaddition, there are many problems given as exer ises. The author in ludeshelpful notes on sele ted problems at the end of ea h hapter.Many of the problems in this book are quite hallenging, but its in re-mental and themati approa h helps. As polynomial algebra (and patterns inpolynomial oeÆ ients) appear in every hapter, the reader must be omfort-able manipulating rather ompli ated algebrai expressions. I would re om-mend this book for serious, mathemati s-based professional developmentprograms as well as for experien ed independent readers who would enjoypursuing a fruitful intelle tual journey through sele ted advan ed se ondarymathemati s topi s.

38Velo ity Analysis: an Approa h to SolvingGeometry Problems

Peng YuChen1 IntroductionWe introdu e velo ity analysis for solving otherwise omplex geometry prob-lems, then we give two examples of the use of this method: (1) to give a verybrief proof of the opti al property of the ellipse, and (2) to �nd the length ofa logarithmi spiral. At the end of this note we leave some problems for thereader; the solutions will be very brief if velo ity analysis is used.2 Velocity AnalysisIn analyti geometry, a urve is a set of points whose oordinates (x, y) sat-isfy a ertain formula. For example, the set of points {(x, y) : x2+y2 = R2}is a ir le or radius R.As we all know, the tra e of a moving point is a urve. In the analyti notion of a urve above, we mainly pay attention to the position of a movingpoint, rather than its velo ity. In fa t, either the position or the velo ity ofa moving point an des ribe the ourse of its motion, and in some ases itis more onvenient to study the velo ity rather than the position. If insteadof studying the oordinates of a moving point we study its velo ity, thenit is very simple for us to dedu e ertain geometri properties of the tra ed urve without engaging in omplex mathemati s, espe ially when seeking thetangent or ar length. (As we know, the velo ity ve tor of a moving pointgives a tangent to the urve tra ed by the point.)Nowwe introdu e the basi idea of velo ity analysis in solving geometryproblems.A point P is moving under a ertain restri tion on its velo ity −→

V . Forexample, if −→V ·−→

OP ≡ 0 (here −→OP is the position ve tor of P ), then obviouslythe point P tra es out a ir le whose entre is O. That is to say, we �nd anequivalent way of de�ning the ir le by studying the velo ity rather than the oordinates of a moving point.When studying velo ity, usually it is more onvenient to study om-ponents. In our example of the ir le, we break the velo ity of P into twodire tions: along −→

OP and orthogonal to −→OP . Name the two omponents −→

V1Copyright c© 2009 Canadian Mathemati al So ietyCrux Mathemati orum with Mathemati al Mayhem, Volume 35, Issue 1

39and −→

V2 respe tively; if the omponent −→V1 along −→

OP satis�es −→V1 ≡ 0, thenthe point P tra es out a ir le.This is the basi idea of velo ity analysis: if we de ompose the velo ityof a moving point into two appropriate dire tions, giving ertain restri tionsto the omponents of the velo ity, the tra e of the moving point will be omea ertain urve. In this setting it is very easy to analyse the tangent to the urve, for we already know the restri tion on the velo ity of the point.Example 1 The opti al property of the ellipse. To show the power of the ve -tor analysis approa h, we use it to solve this lassi al geometry problem. Inan ient times people found that oni se tions have very spe ial and beauti-ful opti al properties. One example is this: if a ray of light leaves one fo usof an ellipse and strikes the ellipse, it will be re e ted to the other fo us ofthe ellipse (see Figure 1).

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r r

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Figure 1.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................r r

F1 F2

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β

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L1

Figure 2Using al ulus to prove this property would be ompli ated, so we usethe approa h of analysing the velo ity of a moving point to give a mu hshorter proof. As in Figure 2, to prove the opti al property of the ellipse,we need to prove that α = β, whi h is enough to satisfy the Law of Re e -tion. Here L1 is tangent to the ellipse.We usually de�ne an ellipse likethis: two points F1 and F2 in the planeare �xed. A point P moves so that|−−→PF1|+|−−→PF2| is always a onstant. We all the tra e of P an ellipse. In a or-dan e with the idea of velo ity analy-sis, we des ribe an ellipse as follows:De�nition. Two points F1, F2 in theplane are �xed (see Figure 3). A mov-ing point, P , has velo ity ve tor −→

V .Let −→V1 and −→

V2 be the omponents of −→Vtowards F1 and away from F2, respe -tively. If |−→V1| ≡ |−→V2|, then the tra e ofP is an ellipse.

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..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................r r

r

F1 F2

P

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−→V2−→

V

−→V1

Figure 3

40It is obvious that the two de�nitions are equivalent. For when P movestowards F1 a short distan e, it must then move away from F2 by that samedistan e to ensure that |−−→PF1| + |−−→PF2| is always a onstant.Now−→

V is a tangent ve tor of the ellipse. Sin e |−→V1| ≡ |−→V2|, thenα = γ.Sin e β = γ, we also have α = β. That is all.Example 2 The length of the logarithmi spiral. As another example, we �ndthe length of the logarithmi spiral ρ = ρ0eaθ between ρ = ρ1 and ρ = ρ2.Usually we annot �gure out this problem without using a lot of al ulus, sowe introdu e a physi al model.As in Figure 4, three pointsA,B, and C are ea h at a vertex of an equi-lateral triangle of side √

3ρ0. Point A moves towards B, B moves towardsC, and C moves towards A. The speed of ea h point is s. Where will thethree points meet and how far will they travel before meeting?

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Figure 4 ................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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. . . . . . . . . . . . . . . . . . .. .. .. .. .. .. .. .. .. ..−→

V1

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Figure 5The solution of this problem is easy: obviously the three points willmeet at the point O, whi h is the entre of the triangle. We de ompose thevelo ity −→V of A at ea h instant into two omponents: one towards the point

O and the other perpendi ular to −→AO, and we all these two omponents −→

V1and −→V2, respe tively. We have that |−→V1| = |−→V | cos 30◦ =

√3

2s (see Figure 5),so the time until they meet is t =

ρ0

|−→V1|=

ρ0

(√

3/2)sand the distan e traveledis d = st =

2ρ0√3.Now we onsider the urve tra ed by A. We use a polar oordinatesystem with the origin at O and with the ve tor −→

V2 pointing in the dire tionof in reasing θ. At ea h instant, |−→V2| = |−→V | sin 30◦ = |−→V1| tan 30◦. Sin e|−→V2| =

ρdθ

dtand |−→V1| = −dρ

dt, we have dρ = aρdθ, the di�erential equationof the spiral ρ = ρ0e

aθ with a = cot 150◦ = −√

3. Thus, the length ofthe urve from ρ = ρ1 to ρ = ρ2 is |ρ1 − ρ2||−→V1|

· s =2√3|ρ1 − ρ2|. We leave

41the problem of �nding a suitable physi al model for other values of a to thereader.We leave three more problems for the reader tosolve. They ould be solved by using analyti ge-ometry and al ulus, but are more onvenientlysolved using the approa h of velo ity analysis.1. The pursuant traje tory problem. As in Figure6, suppose that an obje t A starts from the point(0, 1), and moves with a onstant speed s in thedire tion of the positive y{axis. At the same timeanother obje t B starts from the point (−1, 0),and moves with speed 2s and always in the dire -tion of the obje t A. When will the obje ts meet? ..........................................................................................................................................................................................................................................................

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Figure 6

2. The opti al property of a hyperbola. If a light ray leaves one fo us of ahyperbola and strikes the hyperbola, then the (reverse) extended line of itsre e tion will pass through the other fo us of the hyperbola (see Figure 7).3. The opti al property of a parabola. If a light ray leaving the fo us is re- e ted in the parabola, then its re e tion is parallel to the axis of symmetry(see Figure 8).

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

Figure 8..................................................................................................................................................................................................................................................................................................................................................................................................................

..................................................................................................................................................................................................................................................................................................................................................................................................................Referen es[1℄ Cai Suilin, Ordinary Di�erential Equations, ZheJiang University Press,1988, pp. 46-48.[2℄ R.T. Co�man and C.S. Ogilvy, The \Re e tion Property" of the Coni s,Mathemati s Magazine, Vol. 36, No. 1 (Jan., 1963), pp. 11-12.Peng YuChenLantian-1 4083Zi Jingang Campus, ZheJiang University388 Yu Hangtang Road, Hangzhou, China

007-sunboy@163.com

42PROBLEMSSolutions to problems in this issue should arrive no later than 1 August 2009.An asterisk (⋆) after a number indi ates that a problem was proposed without asolution.Ea h problem is given in English and Fren h, the oÆ ial languages of Canada.In issues 1, 3, 5, and 7, English will pre ede Fren h, and in issues 2, 4, 6, and 8,Fren h will pre ede English. In the solutions' se tion, the problem will be stated inthe language of the primary featured solution.The editor thanks Rolland Gaudet of the University College of Saint Bonifa eand Jean-Mar Terrier of the University of Montreal for translations of the problems.

3401. Proposed by Tigran Sloyan, Basi Gymnasium of SEUA, Yerevan,Armenia.Let ABCDE be a onvex pentagon su h that ∠BAC = ∠EAD and∠BCA = ∠EDA, and let the lines CB and DE interse t in the point F .Prove that the midpoints of CD, BE, and AF are ollinear.3402. Proposed by Mih�aly Ben ze, Brasov, Romania.Let D and E be the midpoints of the sides AB and AC in triangleABC, respe tively. Prove that CD is perpendi ular to BE if and only if

5BC2 = AC2 +AB2 .3403. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.The ir les Γ1 and Γ2 interse t at P andQ. A line ℓ through P interse tsΓ1 and Γ2 for the se ond time at A and B, respe tively. The tangents toΓ1 and Γ2 at A and B interse t at C. If O is the ir um entre of △ABCdetermine the lo us of O when ℓ rotates about P .3404. Proposed by Mi hel Bataille, Rouen, Fran e.Let Q be a y li quadrilateral. The perpendi ulars to ea h diagonalthrough its endpoints form a parallelogram, P . Chara terize the entre of Pand show that opposite sides of Q interse t on a diagonal of P .3405. Proposed by Mi hel Bataille, Rouen, Fran e.Find the minimum value of| cosα| + | cosβ| + | cos γ| + | cos(α− β)| + | cos(β − γ)| + | cos(γ − α)| ,where α, β, and γ are real numbers.

433406. Proposed by Jos �e Luis D��az-Barrero and Miquel Grau-S �an hez,Universitat Polit �e ni a de Catalunya, Bar elona, Spain.Find

limn→∞

ln

[

1

2n

n∏

k=1

(

2 +k

n2

)] .3407. Proposed by Roy Barbara, Lebanese University, Fanar, Lebanon.Let S be a set of positive integers ontaining the integer 2007 and su hthat(a) If x, y ∈ S and x 6= y, then |x− y| ∈ S, and(b) If x ∈ S, then (x3 − 1007x+ 3007

)∈ S.Prove that S is the set of all positive integers.3408. Proposed by Slavko Simi , Mathemati al Institute SANU, Belgrade,Serbia.Let {ci}∞

i=1 be a sequen e of distin t positive integers, and let |q| < 1.Prove that the inequality∞∑

i=1

ciqci

1 +

∞∑

i=1

qci

≤ q

1 − q

holds for all su h sequen es {ci}∞i=1 if and only if q ∈

[0, 1

2

].3409. Proposed by Jos �e Luis D��az-Barrero, Universitat Polit �e ni a deCatalunya, Bar elona, Spain.Let a, b, c, and d be positive real numbers. Prove thatab+ bc + ca

a3 + b3 + c3+ab+ bd+ da

a3 + b3 + d3+ac + cd+ da

a3 + c3 + d3+bc + cd+ db

b3 + c3 + d3

≤ min

{a2 + b2

(ab)3/2+c2 + d2

(cd)3/2, a2 + c2

(ac)3/2+b2 + d2

(bd)3/2, a2 + d2

(ad)3/2+b2 + c2

(bc)3/2

}.3410. Proposed by Joe Howard, Portales, NM, USA.Let a, b, and c be the sides of triangle ABC, let R be its ir umradius,and let F be its area. Prove that

∑ y li bc sin2A/2

b+ c≥ F

2R.

443411. Proposed by Mih�aly Ben ze, Brasov, Romania.Let a, b, and c be positive real numbers su h that

a6 + b6 + c6 <32

33

(a3 + b3 + c3

)2 .Prove that at least one of the quadrati s ax2 + bx + c, bx2 + cx + a, orcx2 + ax+ b has no real roots.3412. Proposed by Cao Minh Quang, Nguyen Binh Khiem High S hool,Vinh Long, Vietnam.Let a, b, and c be positive real numbers su h that abc = 1. Prove that

∑ y li 1√a3 + 2b3 + 6

≤ 1 .3413. Proposed by Vo Quo Ba Can, Can Tho University of Medi ine andPharma y, Can Tho, Vietnam.Let a, b, c, and d be real numbers in the interval [1, 2]. Prove that

a+ b

c+ d+

c+ d

a+ b− a+ c

b+ d≤ 3

2.

.................................................................3401. Propos �e par Tigran Sloyan, Ly �ee de la SEUA, Erevan, Arm�enie.Soit ABCDE un pentagone onvexe tel que ∠BAC = ∠EAD et∠BCA = ∠EDA. Soit F le point d'interse tion des droites CB et DE.Montrer que les milieux des CD, BE et AF sont olin �eaires.3402. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.Soit le triangle ABC, o �u D et E sont les milieux des ot �es AB et AC,respe tivement. Montrer que CD est perpendi ulaire �a BE si et seulementsi

5BC2 = AC2 +AB2 .3403. Propos �e par D.J. Smeenk, Zaltbommel, Pays-Bas.Les er les Γ1 et Γ2 interse tent �a P et Q. Une ligne ℓ passant par Pinterse te une se onde fois Γ1 et Γ2 �aA etB, respe tivement. Les tangentesde Γ1 et Γ2 �a A et B interse tent �a C. Si O est le entre du er le ir ons ritdu △ABC d �eterminer le lieu de O lorsque ℓ tourne autour de P .

453404. Propos �e par Mi hel Bataille, Rouen, Fran e.Soit Q un quadrilat �ere y lique. Les perpendi ulaires �a haque diago-nale issues de ses sommets forment un parall �elogramme P . Cara t �eriser le entre de P et montrer que les ot �es oppos �es de Q se oupent sur une dia-gonale de P .3405. Propos �e par Mi hel Bataille, Rouen, Fran e.D �eterminer la valeur minimum de| cosα| + | cosβ| + | cos γ| + | cos(α− β)| + | cos(β − γ)| + | cos(γ − α)| ,o �u α, β et γ sont des nombres r �eels.3406. Propos �e par Jos �e Luis D��az-Barrero et Miquel Grau-S �an hez, Uni-versit �e Polyte hnique de Catalogne, Bar elone, Espagne.Cal uler

limn→∞

ln

[

1

2n

n∏

k=1

(

2 +k

n2

)] .3407. Propos �e par Roy Barbara, Universit �e Libanaise, Fanar, Liban.Soit S un ensemble d'entiers ontenant l'entier 2007 et tel que(a) Si x, y ∈ S et x 6= y, alors |x− y| ∈ S, et(b) Si x ∈ S, alors (x3 − 1007x + 3007

)∈ S.Montrer que S est l'ensemble de tous les entiers positifs.3408. Propos �e par Slavko Simi , Institut de Math �ematiques SANU,Belgrade, Serbie.Soit {ci}∞

i=1 une suite d'entiers positifs distin ts, et soit |q| < 1. Mon-trer que l'in �egalit �e∞∑

i=1

ciqci

1 +∞∑

i=1

qci

≤ q

1 − q

vaut pour toutes les suites de e type si et seulement si q ∈[0, 1

2

].3409. Propos �e par Jos �e Luis D��az-Barrero, Universit �e Polyte hnique deCatalogne, Bar elone, Espagne.Soit a, b, c et d des nombres r �eels positifs. Montrer queab+ bc + ca

a3 + b3 + c3+ab+ bd+ da

a3 + b3 + d3+ac + cd+ da

a3 + c3 + d3+bc + cd+ db

b3 + c3 + d3

≤ min

{a2 + b2

(ab)3/2+c2 + d2

(cd)3/2, a2 + c2

(ac)3/2+b2 + d2

(bd)3/2, a2 + d2

(ad)3/2+b2 + c2

(bc)3/2

}.

463410. Propos �e par Joe Howard, Portales, NM, �E-U.Soit a, b et c les ot �es du triangle ABC, soit R le rayon de son er le ir ons rit et F son aire. Montrer que

∑ y lique bc sin2A/2

b+ c≥ F

2R.

3411. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.Soit a, b et c des nombres r �eels positifs tels quea6 + b6 + c6 <

32

33

(a3 + b3 + c3

)2 .Montrer que au moins une des quadratiques ax2 + bx+ c, bx2 + cx+ a etcx2 + ax+ b n'a au une ra ine r �eelle.3412. Propos �e par Cao Minh Quang, Coll �ege Nguyen Binh Khiem, VinhLong, Vietnam.Soit a, b et c trois nombres r �eels positifs tels que abc = 1. Montrerque

∑ y lique 1

a3 + 2b3 + 6≤ 1 .

3413. Propos �e par Vo Quo Ba Can, Universit �e de M�ede ine et Pharma iede Can Tho, Can Tho, Vietnam.Soit a, b, c et d quatre nombres r �eels dans l'intervalle [1, 2]. Montrerquea+ b

c+ d+

c+ d

a+ b− a+ c

b+ d≤ 3

2.

47SOLUTIONSNo problem is ever permanently losed. The editor is always pleasedto onsider for publi ation new solutions or new insights on past problems.Last year we re eived a bat h of orre t solutions from Steven Karp,student, University ofWaterloo,Waterloo, ON, to problems 3289, 3292, 3294,3296, 3297, 3298, and 3300, whi h did not make it into the De ember issuedue to being mis�led. Our apologies for this oversight.

3301. [2008 : 44, 46℄ Proposed by Ovidiu Furdui, University of Toledo,Toledo, OH, USA.Prove that∞∑

n=1

ln 2 −(

1

n + 1+

1

n + 2+ · · · +

1

2n

)

n=

∞∑

n=1

1 +1

2+ · · · +

1

n

(2n+ 1)(2n + 2).

What is this ommon value ?Solution by Manuel Benito, �Os ar Ciaurri, and Emilio Fern�andez, Logro ~no,Spain, expanded by the editor.Let A and B denote the summations on the left side and the right sideof the proposed equality, respe tively. Also, letHn = 1+1

2+ · · ·+ 1

n. Then

1

n+ 1+

1

n+ 2+ · · · +

1

2n= H2n −Hn

= H2n − 2

(1

2+

1

4+ · · · +

1

2n

)

=2n∑

j=1

(−1)j−1

j.

Sin e it is well known that ∞∑

j=1

(−1)j−1

j= ln 2, by hanging the orderof the double summation we have

A =∞∑

n=1

1

n

∞∑

j=2n+1

(−1)j−1

j

=∞∑

j=3

(−1)j−1

j

⌊ j−1

2⌋

∑

n=1

1

n

=∞∑

j=3

(−1)j−1

jH⌊ j−1

2⌋ =

∞∑

k=1

Hk

(1

2k + 1− 1

2k + 2

)

=∞∑

k=1

Hk

(2k + 1)(2k + 2)=

∞∑

n=1

1 + 12

+ · · · + 1n

(2n+ 1)(2n+ 2)= B .

48To �nd the ommon value of the two absolutely onvergent series, let

f(x) =∞∑

j=3

(−1)j−1

jH⌊ j−1

2⌋x

j ,where the power series for f onverges for all x ∈ (−1, 1). Then

f ′(x) =

∞∑

j=3

(−1)j−1H⌊ j−1

2⌋x

j−1 =

∞∑

j=2

(−1)jH⌊ j2

⌋xj

=∞∑

n=1

(Hnx2n −Hnx

2n+1) = (1 − x)∞∑

n=1

Hnx2n . (1)

Now, it is well known that1

1 − xln

(1

1 − x

)

=∞∑

n=1

Hnxn . (2)

[Ed : Multiply 1

1 − x= 1+x+x2+· · · with− ln(1−x) = x+

1

2x2+

1

3x3+· · ·and observe that the oeÆ ient of xn is 1 +

1

2+ · · · +

1

n.℄From (2) we have (1 − x)

∞∑

n=1Hnx

n = − ln(1 − x).Thus, (1 − x2)∞∑

n=1Hnx

2n = − ln(1 − x2), and it follows that(1 − x)

∞∑

n=1

Hnx2n = − 1

1 + xln(1 − x2) . (3)

From (1) and (3) we obtainf ′(x) = − 1

1 + xln(1 − x2) .Sin e f(0) = 0 and the last improper integral below is onvergent, byapplying Abel's Continuity Theorem for power series we have

A = limx→1−

f(x) =

∫ 1

0

f ′(x) dx = −∫ 1

0

ln(1 − x2

)

1 + xdx . (4)It remains to evaluate the last integral in (4).With the hange of variable x = 2u− 1, we have

∫ 1

0

ln(1 − x2

)

1 + xdx =

∫ 1

1/2

ln(4u(1 − u)

)

udu = I1 + I2 , (5)

49where

I1 =

∫ 1

1/2

ln(4u)

udu =

1

2ln2(4u)

∣∣∣∣

1

1/2

=1

2

((ln 4)2 − (ln 2)2

)

=1

2(ln 4 + ln 2)(ln 4 − ln 2) =

1

2(ln 8)(ln 2) =

3

2(ln 2)2 . (6)On the other hand, using integration by parts and then making the hange of variable u = 1 − t, we have

I2 =

∫ 1

1/2

ln(1 − u)

udu = (lnu)

(ln(1 − u)

)

∣∣∣∣∣

1

1/2

+

∫ 1

1/2

lnu

1 − udu

= −(ln 2)2 +

∫ 1/2

0

ln(1 − t)

tdt = −(ln 2)2 +

∫ 1

0

ln(1 − t)

tdt − I2 ,from whi h we obtain

2I2 = −(ln 2)2 +

∫ 1

0

ln(1 − t)

tdt . (7)[Ed : Sin e the integrals in the omputations above are improper, are mustbe taken ; e.g., the evaluation of (lnu)

(ln(1 −u)

) at u = 1 must be done by omputing limu→1−

(lnu)(ln(1 − u)

) using L'Hospital's Rule.℄Finally,∫ 1

0

ln(1 − t)

tdt = −

∫ 1

0

1

tln

(1

1 − t

)

dt

= −∫ 1

0

1

t

( ∞∑

n=1

tn

n

)

dt = −∞∑

n=1

∫ 1

0

tn−1

ndt

= −∞∑

n=1

(

tn

n2

∣∣∣∣

1

0

)

= −∞∑

n=1

1

n2= −π

2

6. (8)

From (4) { (8), we on lude that A = B =π2

12− (ln 2)2.Also solved by PAOLO PERFETTI, Dipartimento di Matemati a, Universit �a degli studi diTor Vergata Roma, Rome, Italy ; and the proposer. There were also three in omplete solutions,all of whi h only demonstrated that the two given summations are equal.

3302. [2008 : 44, 47℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Let s, r, and R denote the semiperimeter, the inradius, and the ir umradius of a triangle ABC, respe tively. Show that(s2 + r2 + 4Rr)(s2 + r2 + 2Rr) ≥ 4Rr(5s2 + r2 + 4Rr) ,and determine when equality holds.

50Solution by Mi hel Bataille, Rouen, Fran e.First, using the two known formulas s2 + r2 +4Rr = ab+ bc+ ca andabc = 4Rrs, where a, b, and c are the sides of the triangle, we dedu e that

(a+ b)(b+ c)(c+ a) = (a+ b+ c)(ab+ bc+ ca) − abc

= 2s(s2 + r2 + 4Rr

)− 4Rrs

= 2s(s2 + r2 + 2Rr

) .For onvenien e, let e1 = a + b + c, e2 = ab + bc + ca, and e3 = abc.It follows that the required inequality is su essively equivalent to(a+ b)(b+ c)(c+ a)e2 ≥ 8Rrs(5s2 + r2 + 4Rr) ,(a+ b)(b+ c)(c+ a)e2 ≥ 2e3

(e21 + e2

) ,(ab2 + a2b+ bc2 + b2c+ ca2 + c2a

)e2 ≥ 2e3e

21 ,

a2b3 + a3b2 + b2c3 + b3c2 + c2a3 + c3a2 ≥ 2a2b2c+ 2a2bc2 + 2ab2c2 ,and �nally toa2(b− c)2(b+ c) + b2(c− a)2(c+ a) + c2(a− b)2(a+ b) ≥ 0 .The last inequality is obviously true, whi h ompletes the proof. Equalityholds if and only if a = b = c, that is, if and only if the triangle ABC isequilateral.Also solved by ARKADY ALT, San Jose, CA, USA ; GEORGE APOSTOLOPOULOS,Messolonghi, Gree e ; �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina ; ROY BARBARA, Lebanese University, Fanar, Lebanon ; CHIP CURTIS, MissouriSouthern State University, Joplin, MO, USA ; CHARLES R. DIMINNIE, Angelo StateUniversity, San Angelo, TX, USA ; OLIVER GEUPEL, Br �uhl, NRW, Germany ; JOE HOWARD,Portales, NM, USA ; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria ; KEE-WAILAU, Hong Kong, China ; THANOSMAGKOS, 3rd High S hool of Kozani, Kozani, Gree e ; SALEMMALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina ; ANDREA MUNARO,student, University of Trento, Trento, Italy ; PANOS E. TSAOUSSOGLOU, Athens, Gree e ;GEORGE TSAPAKIDIS, Agrinio, Gree e ; PETER Y. WOO, Biola University, La Mirada, CA, USA ;TITU ZVONARU, Com�ane�sti, Romania ; and the proposer.

3303. [2008 : 44, 47℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Let a, b, and c be positive real numbers. Show that∏ y li (2(a+ b)3

)≥

∏ y li ((a+ s1)(bc+ s2)) ,

where s1 = a+ b+ c and s2 = ab+ bc+ ca.

51Composite of similar solutions by Manuel Benito, �Os ar Ciaurri, EmilioFern�andez, and Luz Ron al, Logro ~no, Spain ; and Chip Curtis, MissouriSouthern State University, Joplin, MO, USA.By straightforward omputations, we have

2(a+ b)(b+ c)(c+ a) − (a+ s1)(bc+ s2)

= 2(a+ b)(b+ c)(c+ a) − (2a+ b+ c)(ab+ 2bc+ ca)

= 2(a2b+ ab2 + b2c+ bc2 + c2a+ ca2 + 2abc)

− (2a2b+ ab2 + 2b2c+ 2bc2 + c2a+ 2ca2 + 6abc)

= ab2 + ac2 − 2abc = a(b− c)2 ≥ 0 .Hen e,2(a+ b)(b+ c)(c+ a) ≥ (a+ s1)(bc+ s2) .Similarly, we have2(a+ b)(b+ c)(c+ a) ≥ (b+ s1)(ca+ s2) ;2(a+ b)(b+ c)(c+ a) ≥ (c+ s1)(ab+ s2) .The result now follows by multiplying a ross the last three inequalities.Also solved by ARKADY ALT, San Jose, CA, USA ; GEORGE APOSTOLOPOULOS,Messolonghi, Gree e ; �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Her-zegovina ; DIONNE BAILEY, ELSIE CAMPBELL, and CHARLES R. DIMINNIE, Angelo State Uni-versity, San Angelo, TX, USA ;MICHEL BATAILLE, Rouen, Fran e ; CAOMINHQUANG, NguyenBinh Khiem High S hool, Vinh Long, Vietnam ; OLIVER GEUPEL, Br �uhl, NRW, Germany ;SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina ; ANDREAMUNARO, student, University of Trento, Trento, Italy ; NGUYEN MANH DUNG, High s hoolstudent, HUS, Hanoi, Vietnam ; PAOLO PERFETTI, Dipartimento di Matemati a, Universit �adegli studi di Tor Vergata Roma, Rome, Italy ; DANIEL TSAI, student, Taipei Ameri an S hool,Taipei, Taiwan ; PANOS E. TSAOUSSOGLOU, Athens, Gree e ; GEORGE TSAPAKIDIS, Agrinio,Gree e ; PETER Y. WOO, Biola University, La Mirada, CA, USA ; TITU ZVONARU, Com�ane�sti,Romania ; and the proposer.

3304. [2008 : 45, 47℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Let a1, a2, . . . , an be positive real numbers and identify an+1 with a1.Prove thatn∑

k=1

a3k ≥

n∑

k=1

aka2k+1 .Similar solutions by Mi hel Bataille, Rouen, Fran e ; Walther Janous, Ur-sulinengymnasium, Innsbru k, Austria ; Steven Karp, student, University ofWaterloo, Waterloo, ON ; Xavier Ros, student, Universitat Polit �e ni a de Ca-talunya, Bar elona, Spain ; and Daniel Tsai, student, Taipei Ameri an S hool,Taipei, Taiwan.Reorder the numbers a1, a2, . . . , an from smallest to largest and re-name them x1, x2, . . . , xn. Then 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn and if yi = x2

i

52for ea h i then we also have 0 ≤ y1 ≤ y2 ≤ · · · ≤ yn. The Rearrange-ment Inequality states that n∑

i=1

xiyi ≤n∑

i=1

xiyσ(i) for any permutation σ of{1, 2, . . . , n}. Sin e n∑

k=1

a3k =

n∑

i=1

xiyi and n∑

k=1

aka2k+1 =

n∑

i=1

xiyσ(i) for anappropriate permutation σ, the result follows.Also solved by MOHAMMED AASSILA, Strasbourg, Fran e ; ARKADY ALT, San Jose,CA, USA ; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e ; �SEFKET ARSLANAGI �C, Uni-versity of Sarajevo, Sarajevo, Bosnia and Herzegovina ; ROY BARBARA, Lebanese University,Fanar, Lebanon ; MANUEL BENITO, �OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZRONCAL, Logro ~no, Spain ; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long,Vietnam ; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA ; OVIDIUFURDUI, University of Toledo, Toledo, OH, USA ; OLIVER GEUPEL, Br �uhl, NRW, Germany ;JOE HOWARD, Portales, NM, USA ; SALEMMALIKI �C, student, Sarajevo College, Sarajevo, Bos-nia and Herzegovina ; DUNG NGUYEN MANH, High S hool of HUS, Hanoi, Vietnam ; PAOLOPERFETTI, Dipartimento di Matemati a, Universit �a degli studi di Tor Vergata Roma, Rome,Italy ; HENRY RICARDO, Medgar Evers College (CUNY ), Brooklyn, NY, USA ; PANOSE. TSAOUSSOGLOU, Athens, Gree e ; GEORGE TSAPAKIDIS, Agrinio, Gree e ; PETER Y. WOO,Biola University, La Mirada, CA, USA ; TITU ZVONARU, Com�ane�sti, Romania ; and the propo-ser. Ri ardo omments that this problem appears as Problem 11.7 on p. 148 of ElementaryInequalities by D.S. Mitrinovi� (P. Nordho�, 1964), but that no solution is provided there.3305. [2008 : 45, 47℄ Proposed by Stanley Rabinowitz, MathPro Press,Chelmsford, MA, USA.Prove that

tan2π

13+ 4 sin

6π

13= tan

4π

13+ 4 sin

π

13

= tan5π

13+ 4 sin

2π

13=

√

13 + 2√

13 .Solution by Manuel Benito, �Os ar Ciaurri, Emilio Fern�andez, and Luz Ron al,Logro ~no, Spain, in memory of Jim Totten.We will prove thattan

2π

13+ 4 sin

6π

13= tan

5π

13+ 4 sin

2π

13

= tan6π

13− 4 sin

5π

13=

√

13 + 2√

13 (1)andtan

4π

13+ 4 sin

π

13= − tan

π

13+ 4 sin

3π

13

= − tan3π

13+ 4 sin

4π

13=

√

13 − 2√

13 . (2)We will make use of two elegant results due to K.F. Gauss and in luded inthe Se tio VII of the Disquisitiones Arithmeti � (DA).

53Lemma (DA, art. 362, II). Let n > 1 be an odd number and ω =

2kπ

n, where

k is any of the numbers 1, 2, . . . , n− 1. Then,tanω = 2

[sin(2ω) − sin(4ω) + sin(6ω) + · · · ∓ sin

((n− 1)ω

)] .Theorem (DA, art. 356). Let n > 1 be an odd prime number, R be the set ofthe (positive and less than n) quadrati residues modulo n, and N be the setof the (positive and less than n) quadrati non-residues modulo n. Then,

∑

r∈R

cos2πr

n−∑

m∈N

cos2πm

n=

{ √n if n ≡ 1 (mod 4) ,0 if n ≡ 3 (mod 4) ,and

∑

r∈R

sin2πr

n−∑

m∈N

sin2πm

n=

{0 if n ≡ 1 (mod 4) ,√n if n ≡ 3 (mod 4) .

For n = 13 with ω =2kπ

nand 1 ≤ k ≤ 12 the Lemma yields

tanω = 2[sin(2ω) − sin(4ω) + sin(6ω) − sin(8ω) + sin(10ω) − sin(12ω)

] .We ompute with di�erent values of k in this identity as follows.If k = 1 and ω =2π

13, then the Lemma yields

tan2π

13= 2

(

sin4π

13− sin

5π

13+ sin

π

13+ sin

3π

13− sin

6π

13+ sin

2π

13

) . (3)If k = 3 and ω =6π

13, then the Lemma yields

tan6π

13= 2

(

sinπ

13+ sin

2π

13+ sin

3π

13+ sin

4π

13+ sin

5π

13+ sin

6π

13

) . (4)If k = 4 and ω =8π

13, then tan

5π

13= − tan

8π

13and the Lemma yields

tan5π

13= 2

(

sin3π

13+ sin

6π

13+ sin

4π

13+ sin

π

13− sin

2π

13− sin

5π

13

) . (5)By omparing the equations (3), (4), and (5) we see that the �rst threeexpressions in equation (1) are equal.If k = 2 and ω =4π

13, then the Lemma yields

tan4π

13= 2

(

sin5π

13+ sin

3π

13− sin

2π

13− sin

6π

13− sin

π

13+ sin

4π

13

) . (6)If k = 5 and ω =10π

13, then tan

3π

13= − tan

10π

13and the Lemma yields

tan3π

13= 2

(

sin6π

13− sin

π

13− sin

5π

13+ sin

2π

13+ sin

4π

13− sin

3π

13

) . (7)

54If k = 6 and ω =

12π

13, then tan

π

13= − tan

12π

13and the Lemma yields

tanπ

13= 2

(

sin2π

13− sin

4π

13+ sin

6π

13− sin

5π

13+ sin

3π

13− sin

π

13

) . (8)By omparing the equations (6), (7), and (8) we see that the �rst threeexpressions in equation (2) are equal.Now we take A = tan2π

13+ 4 sin

6π

13and B = tan

4π

13+ 4 sin

π

13.Clearly A and B are positive numbers. From (3) and (6) it follows that

A+B = 4(

sinπ

13+ sin

3π

13+ sin

4π

13

)

andA−B = 4

(

sin2π

13− sin

5π

13+ sin

6π

13

) .Then,A2 −B2 = 16

(

sinπ

13+ sin

3π

13+ sin

4π

13

) (

sin2π

13− sin

5π

13+ sin

6π

13

) .Applying the identity 2 sin a sin b = cos(a− b) − cos(a+ b), we haveA2 −B2 = 8

(

cosπ

13+ cos

2π

13+ cos

3π

13− cos

4π

13− cos

5π

13+ cos

6π

13

) .However, for n = 13 ≡ 1 (mod 4), the sets R and N in the Theorem areR = {1,4,9,3,12,10} and N = {2,8,6,11,5,7} ; thus, by the Theorem

2(

cos2π

13+ cos

6π

13− cos

5π

13+ cos

π

13+ cos

3π

13− cos

4π

13

)

=√

13 ,and thereforeA2 −B2 = 4

√13 .Similarly, using the identity 2 sin2 a = 1 − cos(2a) we dedu e that

AB = 4(

sinπ

13+ sin

2π

13+ sin

3π

13+ sin

4π

13− sin

5π

13+ sin

6π

13

)

×(

sinπ

13− sin

2π

13+ sin

3π

13+ sin

4π

13+ sin

5π

13− sin

6π

13

)

= 6(

cosπ

13+ cos

2π

13+ cos

3π

13− cos

4π

13− cos

5π

13+ cos

6π

13

)

=6√

13

2= 3

√13 .For positive real numbers A and B with A > B, the solutions of the equa-tions A2 −B2 = 4

√13 and AB = 3

√13 are

A =

√

13 + 2√

13 and B =

√

13 − 2√

13 .

55This ompletes the proof of the identities (1) and (2). The following similaridentities an be dedu ed when n = 11 :

tanπ

11+ 4 sin

3π

11= − tan

2π

11+ 4 sin

5π

11

= tan3π

11+ 4 sin

2π

11= tan

4π

11+ 4 sin

π

11

= tan5π

11− 4 sin

4π

11

=√

11 .Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Her-zegovina ; MICHEL BATAILLE, Rouen, Fran e ; APOSTOLIS K. DEMIS, Varvakeio High S hool,Athens, Gree e (2 solutions) ; JOHN HAWKINS and DAVID R. STONE, Georgia Southern Uni-versity, Statesboro, GA, USA ; and PETER Y. WOO, Biola University, La Mirada, CA, USA.All solvers noted that tan 4π13

+4 sin π13

6=√

13 + 2√

13, as did George Apostolopoulos,Messolonghi, Gree e ; and Luyan Zhong-Qiao, Columbia International College, Hamilton, ON.Wagon used Mathemati a to he k that the �rst and third identities are orre t and the se ondis in orre t. The proposer o�ered a partially orre t solution.Woo wondered if similar results hold for π5, π

7, π

11or π

17. For the ase of π

11Benito et al.answered (above) in the aÆrmative. The interested reader may want to investigate the other ases. Woo also hallenges the readers to �nd geometri proofs for the equalities in (1) and (2).

3306. [2008 : 45, 47℄ Proposed by Stanley Rabinowitz, MathPro Press,Chelmsford, MA, USA.Find a real number t and polynomials f(x), g(x), and h(x) with integer oeÆ ients, su h thatf(t) =

√2 , g(t) =

√3 , and h(t) =

√7 .Solution by Roy Barbara, Lebanese University, Fanar, Lebanon.Set θ =

√2+

√3+

√7 and ψ =

√2√

3√

7. Computing θ3, θ5, and θ7,we obtain√

2 +√

3 +√

7 = θ ,16

√2 + 15

√3 + 11

√7 + 3ψ =

1

2θ3 ,

281√

2 + 241√

3 + 161√

7 + 60ψ =1

4θ5 ,

4796√

2 + 3975√

3 + 2611√

7 + 1043ψ =1

8θ7 .This is a linear system for √

2, √3, √

7, and ψ. Solving for √2, √

3, and √7

56yields

√2 =

59

20θ − 1

2θ3 +

1

80θ5 ,

√3 =

313

80θ − 297

160θ3 +

67

320θ5 − 3

640θ7 ,

√7 = −469

80θ +

377

160θ3 − 71

320θ5 +

3

640θ7 .

Finally, setting t =θ

80=

√2 +

√3 +

√7

80we obtain √

2 = f(t), √3 = g(t),and √

7 = h(t), where the polynomials f(x), g(x), and h(x) have integer oeÆ ients :f(x) = 236x − 1

2(80)3x3 + (80)4x5 ,

g(x) = 313x − 1

2(80)2 · 297x3 +

1

4(80)4 · 67x5 − 3

8(80)6x7 ,

h(x) = −469x +1

2(80)2 · 377x3 − 1

4(80)4 · 71x5 +

3

8(80)6x7 .Also solved by MOHAMMED AASSILA, Strasbourg, Fran e ; BRIAN D. BEASLEY,Presbyterian College, Clinton, SC, USA ; MANUEL BENITO, �OSCAR CIAURRI, EMILIOFERNANDEZ, and LUZ RONCAL, Logro ~no, Spain ; CHIP CURTIS, Missouri Southern StateUniversity, Joplin, MO, USA ; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria ;and the proposer.

3307. [2008 : 45, 47℄ Proposed by D.E. Prithwijit, University College Cork,Republi of Ireland.Eliminate θ from the systemλ cos(2θ) = cos(θ + α) ,λ sin(2θ) = 2 sin(θ + α) .Similar solutions by Manuel Benito, �Os ar Ciaurri, Emilio Fern�andez, andLuz Ron al, Logro ~no, Spain ; Joe Howard, Portales, NM, USA ; and GeorgeTsapakidis, Agrinio, Gree e.The given system an be rewritten as a linear system in cosα and sinα :

cos θ cosα− sin θ sinα = λ(cos2 θ − sin2 θ) ,sin θ cosα+ cos θ sinα = λ sin θ cos θ .Its solution is thencosα = λ cos3 θ , and sinα = λ sin3 θ ;

57when e,

(cosα)2/3 + (sinα)2/3 = λ2/3 . (1)Comments from the Spanish team. Note that the original system has a solu-tion if and only if α and λ satisfy (1). In parti ular, letting x = cosα andy = sinα, we see that for 1 ≤ |λ| ≤ 2 the solutions an be representedby the interse tion points of the unit ir le x2 + y2 = 1 with the astroidx2/3 + y2/3 = λ2/3. Thus for ea h λ with absolute value between 1 and 2,(1) will be satis�ed for eight values of α ; for λ ∈ {±1, ±2}, it will be sa-tis�ed by four values of α. There an be no real solutions for other values ofλ. Also solved by ARKADY ALT, San Jose, CA, USA ; GEORGE APOSTOLOPOULOS,Messolonghi, Gree e ; �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina ; MICHEL BATAILLE, Rouen, Fran e ; CHIP CURTIS, Missouri Southern StateUniversity, Joplin, MO, USA ; APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e ;JOS �E LUIS D�IAZ-BARRERO, Universitat Polit �e ni a de Catalunya, Bar elona, Spain ; OLIVERGEUPEL, Br �uhl, NRW, Germany ; JOHN HAWKINS and DAVID R. STONE, Georgia SouthernUniversity, Statesboro, GA, USA ; WALTHER JANOUS, Ursulinengymnasium, Innsbru k,Austria ; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA ; ANDREA MUNARO,student, University of Trento, Trento, Italy ; PAOLO PERFETTI, Dipartimento di Matemati a,Universit �a degli studi di Tor Vergata Roma, Rome, Italy ; XAVIER ROS, student, Universitat Po-lit �e ni a de Catalunya, Bar elona, Spain ; BOB SERKEY, Leonia, NJ, USA ; PANOS E. TSAOUS-SOGLOU, Athens, Gree e ; and the proposer. There was one in orre t submission.Our readers produ ed solutions in a variety of formats. Here are a few of the ni est.Instead of (1), Alt, Bataille, and the proposer independently obtained the equivalent equation

sin2(2α) =4(λ2 − 1

)3

27λ2.Geupel found that in terms of a real parameter t, the solutions of the given system satisfy

θ = arctan t + mπ , α = arctan(t3) + nπ , and λ = (−1)m+n

√

(1 + t2)3

1 + t6,for integers m and n. In addition, there were several impli it solutions where the solver simplypresented an equation for θ in terms of α or λ ; for example, θ = arctan 3

√tan α + kπ amefrom Arslanagi� and from Ros.

3308. [2008 : 45, 48℄ Proposed by D.E. Prithwijit, University College Cork,Republi of Ireland.Given △ABC, let AD be the altitude to BC. If AB : AC = 1 :√

3,prove that AD ≤√

32BC. When does equality hold ?I. Solution by Joe Howard, Portales, NM, USA.It suÆ es to take AB = 1 and AC =

√3 ; onsequently, AD = sinB.Writing a = BC, we therefore must show that

sinB ≤√

3

2a .

58We start with the inequality

1 ≥ sin(60◦ +A) = sin 60◦ cosA + sinA cos 60◦

=

√3

2cosA +

1

2sinA ,whi h is equivalent to

2 −√

3 cosA ≥ sinA . (1)By the osine law, a2 = 4 − 2√

3 cosA, so the inequality (1) is equivalent toa2 ≥ 2 sinA . (2)By the sine law, sinA

a=

sinB√3

, ora =

√3 sinA

sinB,when e (2) is equivalent to

a

√3 sinA

sinB≥ 2 sinA ,whi h redu es immediately to what we were to show. Equality o urs when

60◦ +A = 90◦, in whi h aseA = 30◦ and a2 = 4−2√

3 cos 30◦ = 1. Thus,equality holds if and only if △ABC is isos eles with A = C = 30◦.II. Solution by Mi hel Bataille, Rouen, Fran e.We are given that vertex A belongs to the lo us of those points P forwhi h PB

PC=

1√3. We re ognize this lo us to be a ir le Γ alled the ir le ofApollonius ; it interse ts symmetri ally the line joining B to C in points thatdivide the segment BC internally and externally in the ratio 1 :

√3. As aneasy onsequen e, the entre K of Γ satis�es −−→

BK = −1

2

−−→BC, and its radiusis √

3BC

2. Let NN ′ be the diameter of Γ that is perpendi ular to BC. Then

N and N ′ are the points of Γ that are farthest from the line BC, hen eAD = d(A,BC) ≤ d(N,BC) =

√3BC

2.Equality holds if and only if A is situated at N or N ', in whi h ase △ABCis isos eles with BA = BC and ∠ABC = 120◦.Also solved by MOHAMMED AASSILA, Strasbourg, Fran e ; ARKADY ALT, SanJose, CA, USA ; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e ; �SEFKET ARSLANAGI �C,University of Sarajevo, Sarajevo, Bosnia and Herzegovina ; ROY BARBARA, LebaneseUniversity, Fanar, Lebanon ; MANUEL BENITO, �OSCAR CIAURRI, EMILIO FERNANDEZ, andLUZ RONCAL, Logro ~no, Spain ; CHIP CURTIS, Missouri Southern State University, Joplin,

59MO, USA ; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA ; IANJUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines ; FRANCISCOJAVIER GARC�IA CAPIT �AN, IES �Alvarez Cubero, Priego de C �ordoba, Spain ; OLIVER GEUPEL,Br �uhl, NRW, Germany ; STEVEN KARP, student, University of Waterloo, Waterloo, ON ;V �ACLAV KONE �CN �Y, Big Rapids, MI, USA ; KEE-WAI LAU, Hong Kong, China ; THANOSMAGKOS, 3rd High S hool of Kozani, Kozani, Gree e ; SALEM MALIKI �C, student, SarajevoCollege, Sarajevo, Bosnia and Herzegovina ; ANDREA MUNARO, student, University ofTrento, Trento, Italy ; JOEL SCHLOSBERG, Bayside, NY, USA ; PANOS E. TSAOUSSOGLOU,Athens, Gree e ; GEORGE TSAPAKIDIS, Agrinio, Gree e ; DANIEL TSAI, student, TaipeiAmeri an S hool, Taipei, Taiwan ; PETER Y. WOO, Biola University, La Mirada, CA, USA ; TITUZVONARU, Com�ane�sti, Romania ; and the proposer.More generally, Kone� n �y proved that AD ≤ q

q2 − 1BC if AC : AB = q with q > 1.His argument was mu h like that of solution II above.

3309. [2008 : 45, 48℄ Proposed by Virgil Ni ula, Bu harest, Romania.Let α, β, and γ be �xed non-zero real numbers. Show that the systemαx+ βy + γz = 1 ,xy + yz + zx = 1 ,has a unique solution for (x, y, z) if and only if

α2 + β2 + γ2 + 1 = 2(αβ + βγ + γα) ,and, in this ase �nd that unique solution.Solution by George Tsapakidis, Agrinio, Gree e, modi�ed by the editor.Substituting αx = 1 − βy − γz into α(xy + yz + zx) = α we obtain(y+z)(1−βy−γz)+αyz = α, whi h upon simplifying yields the followingquadrati equation in y

βy2 −[1 + (α− β − γ)z

]y + γz2 − z + α = 0 . (1)Equation (1) has a unique solution in y if and only if

[1 + (α− β − γ)z]2 − 4β(γz2 − z + α) = 0 ,whi h an be written as the following quadrati equation in z

[(α− β − γ)2 − 4βγ

]z2 + 2(α+ β − γ)z + 1 − 4αβ = 0 . (2)Equation (2) has a unique solution in z if and only if

(α+ β − γ)2 −[(α− β − γ)2 − 4βγ

](1 − 4αβ) = 0 ,whi h, by straightforward omputations, redu es to

α2 + β2 + γ2 + 1 = 2(αβ + βγ + γα) . (3)

60Now, suppose (3) holds. Then we have

(α− β − γ)2 − 4βγ = α2 + β2 + γ2 − 2αβ − 2βγ − 2γα = −1 .Thus, (2) redu es to −z2 + 2(α+ β − γ)z + 1 − 4αβ = 0, the uniquesolution of whi h is given by z = α+ β − γ.Using this and (3), we �nd thaty =

1 + (α− β − γ)(α+ β − γ)

2β=

1 + (α− γ)2 − β2

2β

=1 + α2 − β2 + γ2 − 2αγ

2β=

−2β2 + 2αβ + 2βγ

2β= γ + α− β .

Finally, using (3) we haveαx = 1 − βy − γz = 1 − β(γ + α− β) − γ(α+ β − γ)

= −α2 + αβ + γα ,from whi h it follows that x = β + γ − α.Therefore, the unique solution to the given system is(x, y, z) = (β + γ − α, γ + α− β, α+ β − γ) .Also solved by ARKADY ALT, San Jose, CA, USA ; GEORGE APOSTOLOPOULOS, Mes-solonghi, Gree e ; ROY BARBARA, Lebanese University, Fanar, Lebanon ; MICHEL BATAILLE,Rouen, Fran e ; MANUEL BENITO, �OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL,Logro ~no, Spain ; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA ; OLIVERGEUPEL, Br �uhl, NRW, Germany ; STEVEN KARP, student, University of Waterloo, Waterloo,ON ; PAOLO PERFETTI, Dipartimento di Matemati a, Universit �a degli studi di Tor VergataRoma, Rome, Italy ; DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan ; PETERY. WOO, Biola University, La Mirada, CA, USA ; TITU ZVONARU, Com�ane�sti, Romania ; andthe proposer. There was also one partially in orre t solution submitted.

3310. [2008 : 46, 48℄ Proposed by Virgil Ni ula, Bu harest, Romania.Let a, b, and c denote, as usual, the lengths of the sides BC, CA, andAB, respe tively, in △ABC. Let s be the semiperimeter of △ABC, r theinradius, ha the altitude to side BC, and ra, rb, and rc the exradii to A, B,and C, respe tively.(a) Show that for x > 0, we have ha =

2s(s − a)x

x2 + s(s − a)if and only if x = rbor x = rc.(b) Show that for x > 0, we have ha =

2(s − b)(s − c)x

|x2 − (s − b)(s − c)| if and only ifx = r or x = ra.

61Similar solutions by Mi hel Bataille, Rouen, Fran e ; Oliver Geupel, Br �uhl,NRW, Germany ; Titu Zvonaru, Com�ane�sti, Romania ; and the proposer.It is well known that the area F of △ABC an be variously expressedas 1

2aha, rs, ra(s−a), rb(s− b), rc(s− c), or√s(s− a)(s− b)(s− c). Wehave

rb + rc =F

s− b+

F

s− c=

F (2s− b− c)

(s− b)(s− c)

=a · F · (s− a)

s(s− a)(s− b)(s− c)=

as(s− a)

F

=2as(s− a)

aha

=2s(s− a)

ha

(1)andrbrc =

F 2

(s− b)(s− c)=

s(s− a)(s− b)(s− c)

(s− b)(s− c)= s(s− a) . (2)It follows from (1) and (2) that x = rb and x = rc are the solutions of

hax2 − 2s(s− a)x + has(s− a) = 0 ,hen e these are the solutions of the equation in (a).We also have

ra − r =F

s− a− F

s=

aF

s(s− a)

=a · F · (s− b)(s− c)

F 2=

2(s− b)(s− c)

ha

(3)andrar =

F 2

s(s− a)= (s− b)(s− c) . (4)By (3) and (4) it follows that the equation

hax2 − 2(s− b)(s− c)x − (s− b)(s− a)ha = 0has the solutions x = ra and x = −r and that the equation

hax2 + 2(s− b)(s− c)x − (s− b)(s− a)ha = 0has the solutions x = −ra and x = r. Thus, the only positive solutions ofthe pre eding two equations are r and ra, and it follows that these are theonly positive solutions to the equation in (b).Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e ; MANUEL BENITO,�OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logro ~no, Spain ; CHIP CURTIS,Missouri Southern State University, Joplin, MO, USA ; GEORGE TSAPAKIDIS, Agrinio, Gree e ;and PETER Y. WOO, Biola University, La Mirada, CA, USA.

623311. [2008 : 46, 48℄ Proposed by Mi hel Bataille, Rouen, Fran e.Let n be an integer with n ≥ 2. Suppose that for k = 0, 1, . . . , n − 2we have (

n− 2

k

)

≡ (−1)k(k + 1) (mod n) .Show that n is a prime.Solution by Oliver Geupel, Br �uhl, NRW, Germany.It is suÆ ient to prove that if n is a omposite integer with n ≥ 2, thenthere exists an integer k with 0 ≤ k ≤ n− 2 su h that(n− 2

k

)

6≡ (−1)k(k + 1) (mod n) . (1)Toward that end, let p be the least prime fa tor of n. If (1) holds for somek < p, then we are done. Otherwise the given ongruen e holds in parti ularfor k = p− 2, and we have

(p− 1)

(n− 2

p

)

=

(n− 2

p− 2

)

· n− p

p· (n− p− 1)

≡ (−1)p−2(p− 1) ·(n

p− 1

)

· (−p− 1)

≡ (p− 1) · (−1)p

(

p+ 1 − n

p

)

(mod n) .Be ause p − 1 is oprime to the modulus n, we an divide both sides by itand on lude that (1) holds with k = p :(n− 2

p

)

≡ (−1)p

(

p+ 1 − n

p

)

6≡ (−1)p(p+ 1) (mod n) .Also solved by MANUEL BENITO, �OSCAR CIAURRI, EMILIO FERNANDEZ, andLUZ RONCAL, Logro ~no, Spain ; JOHN HAWKINS and DAVID R. STONE, Georgia SouthernUniversity, Statesboro, GA, USA ; STEVEN KARP, student, University of Waterloo, Waterloo,ON ; SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina ; DAVIDE. MANES, SUNY at Oneonta, Oneonta, NY, USA ; JOEL SCHLOSBERG, Bayside, NY, USA ; andthe proposer.The onverse is a known result : If p is a prime, then (p−2

k

)≡ (−1)k(k + 1) (mod p)for k = 0, 1, . . . , p − 2. Karp provided a simple proof (by indu tion on k) ; Bataille provideda referen e : E. Lu as, Th �eorie des nombres, A. Blan hard (1961), p. 420.

3312. [2008 : 46, 48℄ Proposed by Mi hel Bataille, Rouen, Fran e.Let n be a positive integer ongruent to 1 modulo 6. Show that 3/n anbe expressed as1

a1

+1

a2

+ · · · +1

akfor some distin t positive integers a1, a2, . . . , ak, and �nd the minimal valueof k.

63Solution by Steven Karp, student, University of Waterloo, Waterloo, ON.We solve the problem for all positive integers n. If n ≡ 0 or 3 (mod 6),then 3

n=

1

n/3. If n ≡ −1 (mod 6), then

3

n=

1n+1

3

+1

n(n+1)3

.If n ≡ ±2 (mod 6), then3

n=

1n2

+1

n.If n ≡ 1 (mod 6) and n has a fa tor c su h that c ≡ −1 (mod 6), then

3

n=

1n+c

3

+1

n(n+c)3c

.We see that k is minimal in all of these ases, sin e 3

n=

1

a1

for some positiveinteger a1 only if n ≡ 0 (mod 3). Now, if n > 1, n ≡ 1 (mod 6) and alldivisors of n are ongruent to 1 modulo 6, then3

n=

1n+1

2

+1

n+

1n(n+1)

2

,and we laim that k = 3 is minimal in this ase. To prove this suppose forthe sake of ontradi tion that3

n=

1

a+

1

bfor distin t positive integers a and b. Thenn =

b(3a− n)

a,and we have a|b(3a − n). Therefore, a = pq where p and q are positiveintegers su h that p|b and q|(3a − n). Then b

p

∣∣n, when e b

p≡ 1 (mod 6).Sin e q|a and q|(3a−n), we also have q|n, so q ≡ 1 (mod 6). We now have

1 ≡ qn =

(b

p

)

(3a− n) ≡ 3a− n ≡ −1 or 2 (mod 6) ,a ontradi tion.Finally, for n = 1 we have3 = 1 +

1

2+

1

3+

1

4+

1

5+

1

6+

1

7+

1

8+

1

9+

1

10+

1

15+

1

230+

1

57960.A omputer sear h shows that k = 13 is minimal in this ase.

64Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon ; MANUELBENITO, �OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logro ~no, Spain ; CHIPCURTIS, Missouri Southern State University, Joplin, MO, USA ; JOHN HAWKINS and DAVIDR. STONE, Georgia Southern University, Statesboro, GA, USA ; WALTHER JANOUS, Ursulinen-gymnasium, Innsbru k, Austria ; DAVID E.MANES, SUNY at Oneonta, Oneonta, NY, USA ; JOELSCHLOSBERG, Bayside, NY, USA ; PETER Y. WOO, Biola University, La Mirada, CA, USA ; TITUZVONARU, Com�ane�sti, Romania ; and the proposer.Benito et al. refer to the note by Thomas R. Hagedorn, A proof of a onje ture on Egyptianfra tions, Amer. Math. Monthly, 107 (2000) 62-63, where it is proved that for ea h odd integer

n ≥ 3 not divisible by 3 there exist distin t odd, positive integers a, b, and c su h that3

n=

1

a+

1

b+

1

c.The result had been onje tured by R. Hardin and N. Sloane. When n = 6p+1 Hagedorn givesthe de omposition

3

n=

3

6p + 1=

1

2p + 1+

1

(2p + 1)(4p + 1)+

1

(4p + 1)(6p + 1).Janous refers to a paper by Andrzej S hinzel, Sur quelques propri �et �es des nombres 3/net 4/n, o �u n est un nombre impair, Mathesis 65 (1956) 219-222, for a treatment of our problem.Sin e the improper fra tion 3 = 3/1 arises, he asks if any CRUX readers know the minimumnumber ℓ(n) of distin t Egyptian fra tions needed to represent the positive integer n.

AHappy New Year to all CRUX withMAYHEM readers. The Jim Tottenspe ial issue was slated to be ompleted inMay of this year, but due to delayswe are now going to release the spe ial issue in September 2009 instead.This year we plan on improving our database of names and aÆliationsof you, the readers. If your name does not look quite right, for example, ifthe a ents are not quite right or your family name is in orre t, et ., thenplease let us know and we will update our �les. Many international readerssubs ribe to CRUX with MAYHEM and we want to get these (fas inating !)details right. V �a lav Linek.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