385
SKOLIAD
No. 113
Robert Bilinski
Please send your solutions to the problems in this edition by April 1, . A opy of MATHEMATICAL MAYHEM Vol. 7 will be presented to one pre-university reader who sends in solutions before the deadline. The de ision of the editor is nal. 2009
Our ontest this month is the National Bank of New Zealand Competition 2005. Our thanks go to Warren Palmer, Otago University, Otago, New Zealand for providing us with this ontest and for permission to publish it. We also thank Rolland Gaudet, College universitaire de Saint-Bonifa e, Winnipeg, MB, for translation of this ontest. National Bank of New Zealand Junior Mathemati s Competition 2005 (Years 9 and above) 1 hour allowed 1 (For year 9, Form 3 only). Note: In this question the word \digit" means a positive single-digit whole number, that is, a member of the set {1, . . . , 9}. On a long journey by ar, Mi hael was starting to get bored. To keep him amused, his mother asked him some arithmeti questions. The rst question she asked was \Can you think of ve di erent digits whi h add to a multiple of 5?" Mi hael answered straight away \That's easy Mum. 1, 2, 3, 4, and 5 work be ause 1 + 2 + 3 + 4 + 5 = 15, and 15 is a multiple 5 be ause 15 = 5 × 3." Now answer the other questions whi h Mi hael's mother asked.
(a) Write down a set of ve di erent digits whi h whi h add to 35. (b) Write down a set of three di erent digits whi h add to a multiple of 5, but whi h don't in lude 5 itself or 1. ( ) How many di erent sets of four di erent digits are there whi h add to a multiple of 5, but whi h don't in lude 5 itself or 1? (Note that writing the same set of numbers in a di erent order doesn't ount here.) (d) Is it possible to write down a set of six di erent digits whi h add to a multiple of 5, but whi h don't in lude 5 itself? If it is possible, write down su h a set. If it is not possible, explain brie y why it annot be done.
386 (e) Is it possible to write down a set of seven di erent digits whi h add to a multiple of 5, but whi h don't in lude 5 itself? If it is possible, write down su h a set. If it is not possible, explain brie y why it annot be done. 2. Braille is a ode whi h lets blind people read and write. It was invented by a blind Fren hman, Louis Braille, in 1829. Braille is based on a pattern of dots embossed on a 3 by 2 re tangle. It is read with the ngers moving a ross the top of the dots. Altogether there are 63 possible ways to emboss one to six dots on a 3 by 2 re tangle. (We will not ount zero dots in this question.) Figure 1 shows the pattern for the letter h. Note that if we re e t this pattern (down the middle of the re tangle) the result is the Braille letter j, shown in Figure 2. ................................................................ ..... ... ... .... .... .... ... ... .... .... ... .... ... ... ... .... . ..... ... .. ... .... .... .... .... ... .... .... ... ... ... ................................................................
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Fig. 2: Fig. 3: a simpli ed Fig. 1: the letter j version of Braille. the letter h (a) How many di erent patterns are possible using just one dot? (b) There are 15 di erent ways to emboss two dots on a 3 by 2 re tangle. How many ways are there to emboss four dots on a 3 by 2 re tangle? Brie y explain your answer. ( ) In luding the two patterns shown in Figures 1 and 2, how many possible patterns are there using three dots? (d) A simpli ed version of Braille has been proposed. In this version the dots will be embossed on a 2 by 2 re tangle. An example is shown in Figure 3. How many possible patterns would there be in this simpli ed version (assuming that we will not ount zero dots)? (e) Write down how many possible patterns there will be if we were to develop a more ompli ated version of Braille using a 4 by 2 re tangle. (Again assume that we would not ount zero dots.) 3. Around the year 2000 BC, the Babylonians used a number system based on the number 60. For example, where we would write 0.25 (meaning \one quarter" or 1/4), they would write something like || 15 | (meaning \ fteen sixtieths" or 15/60, whi h does simplify to be ome 1/4 in our number system). The table below shows some numbers and their re ipro als written a
ording to the Babylonian system. (A spe ial feature of a number and its
387 re ipro al is that when you multiply them together, the result is always 1.) In the table below, ea h olumn represents a pla e value of 1/60 of the previous olumn. For example, 7 | 30 means 7/60 + 30/3600. Number 2 3 4 5
|| Re ipro al || 30 | || 20 | || | || |
Number 6 7 8 9
|| Re ipro al || | || | | || 7 | 30 | || | |
|
| ...
(a) Write down in the orre t order the three missing numbers whi h should go beside 4, 5, and 6. (b) If we added our number 2.5 (meaning \two and a half") into the table above, what number would we write beside it to show the Babylonian version of its re ipro al? ( ) The re ipro al of the number 8 is shown in the table as 7 | 30 |. Using the same notation, what is the re ipro al of the number 9? (d) Write down the rst two numbers whi h should go beside 7 in the table above. (e) Write down the Babylonian version for the re ipro al of our number 192. . When we nally landed on Mars, we dis overed that Martians love to play a game alled HitBall. In this game two teams of players try to hit a ball between poles pla ed at ea h end of a eld. The team that s ores the most points within one Martian hour is the winner. There are three ways to s ore points. An Inner s ores 7 points, an Outer s ores 4 points, while a Wide s ores 2 points. The rst mat h report sent ba k to Earth was not very lear be ause of stati , so not all the details are ertain. However, we did hear that the Red Team won. They s ored \something{seven" points altogether (only the last number ould be learly heard). We also learned that they had 16 su
essful s oring shots. 4
(a) Write down in a list (from smallest to largest) the possible numbers of Inners whi h the Red Team ould have s ored a
ording to this rst mat h report. (b) A later report added the information that the Red Team s ored the same number of Inners and Outers. Using this extra information, write down how many points the Red Team s ored altogether. ( ) Explain why your answer to (b) is the only possible solution. As part of your explanation, make sure you in lude how many Inners, Outers, and Wides the Red Team s ored.
388 . During 2004, a Dunedin newspaper held a ompetition to nd a new ag design for the provin e of Otago. Wendy entered the ompetition. Her entry was based on the design shown below (Figure 4). Her ag featured a gold
ross with a blue ba kground. She also pla ed a ir le into her design. The top and bottom of the ir le just tou h the orners of the top and bottom triangles, as shown in Figure 4.
5
40 6 30 ?
40 -
30 6 ? 40
6 30 - ? 40
... ... ... ... ...... ...................................................................................................................................................................................................... ....... . ... ....... ....... ..... . ...... ....... . .... . . . . . . . ....... .. ...... ....... .... ....... ...... .......... ....... ....... ............. ... ....... ....... ....... ....... .. ....... . .. ...... ..... ....... . . . ....... . .... ............. . . . . . . . . . .. ....... ......... ....... ....... ..... ..... .. ....... ....... .......................... ....... .... .... ...... ..... ....... ... ....... .... ... ... ....... ... ...... . . ....... ... ... . . .... . . .. ....... ... ... . . . . . .... . . ... . ........... ..... . . . . . ..... .... . . ....... ... . .... . . . . . . . . . . .... . . . ... . ....... ........ ..... ..... . . . . . . . . . . . . . . . . . . . . . .... .... . . . ....... .... .......... ..... . . . ... . . . . . . . . .. . . . ....... ....... .. ..... . . . . . . .... ............. . . . . . ....... ..... ....... .... . . ......... . . . . . . . . . . . ............... ....... ..... ...... .... . . . . . . . . .. ....... ... ..... ....... ... .... ...... ....... ... ....... .... ....... . ....... . . . . . . . . . . ...... .................................................................................................................................................................................................... ... ... ... ...
6 30 ?
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Fig. 4 (not to s ale)
Fig. 5 (not to s ale)
(a) Wendy designed her ag to be 240 m long and 150 m high. If the edges of the ross are 40 m and 30 m away from ea h of the orners, as shown in Figure 4, what is the radius of the entre ir le? (b) Wendy de ided to remove the ir le from her design (see Figure 5). With the ir le removed, what is the total area of the ross? National Bank of New Zealand Con ours mathematique 2005 Niveau junior (se ondaire 3 et plus) 1 heure au total
. Note : Dans ette question, le mot \ hi re" represente un entier positif a une position de imale, don un entier venant de l'ensemble {1, 2, . . ., 9}. Lors d'un long voyage en voiture, Mi hel s'ennuyait. Comme divertissement, sa mere lui a pose quelques questions d'arithmetique. La premiere question : \Donner 5 hi res di erents dont la somme est multiple de 5." Mi hel repond : \Fa ile, maman ! 1, 2, 3, 4, 5. Car 1+2+3+4+5 = 15, qui est un multiple de 5 ar 15 = 3 × 5." Repondre aux autres questions de la maman de Mi hel. (a) Donner 5 hi res di erents dont la somme est 35. (b) Donner trois hi res di erents, in luant ni 1 ni 5, dont la somme est multiple de 5. ( ) Combien d'ensembles de quatre hi res di erents y a-t-il, in luant ni 1 ni 5, dont la somme est un multiple de 5 ? 1
389 (d) Est-il possible de donner un ensemble de six hi res di erents, n'in luant pas 5, dont la somme est un multiple de 5 ? Si possible, fournir un exemple. Sinon, expliquer pourquoi e n'est pas possible. (e) Est-il possible de donner un ensemble de sept hi res di erents, n'in luant pas 5, dont la somme est un multiple de 5 ? Si possible, fournir un exemple. Sinon, expliquer pourquoi e n'est pas possible. . Le braille est un systeme d'e riture ta tile a l'usage de personnes aveugles ou ave de serieuses de ien es visuelles ; il porte le nom de son inventeur, le Fran ais Louis Braille, en 1829. En braille standard, un ara tere est represent e par la ombinaison de 1 a 6 points en relief, disposes sur un re tangle 3 de haut par 2 de large. Au total il y a 63 fa ons possibles de pla er en relief de 1 a 6 points dans e re tangle. La Figure 1 donne le ara tere braille pour la lettre h. Si on fait la re exion de e ara tere par rapport a une droite verti ale au entre du re tangle, on obtient la Figure 2, donnant le ara tere braille de la lettre j. 2
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(a) (b) ( ) (d)
(e)
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Fig. 3 : une version Fig. 2 : Fig. 1 : simpli ee de braille la lettre j la lettre h Combien de ara teres braille sont possibles, ha un utilisant un seul point ? de hoisir 2 points dans le re tangle braille. Il y a 15 fa ons di erentes Combien de fa ons y a-t-il alors pour hoisir 4 points dans e m^eme re tangle ? Expliquer brievement. In luant les ara teres braille dans les Figures 1 et 2, ombien de ara teres sont representables ave 3 points ? Un systeme braille modi e a et e propose. Dans ette version, on utilise plutot ^ un re tangle de taille 2 par 2. Un exemple se trouve dans la Figure 3. Combien de ara teres sont alors possibles, supposant toujours au moins un point ? Combien de ara teres seraient possibles dans une version de braille sur un re tangle de taille 4 par 2, supposant toujours au moins un point ?
3. En Babylonie aux environs de l'an 2000 av. J.-C., la population utilisait un systeme de nombres en base 60. Par exemple, pour e rire notre 0.25 (en systeme de imal, don 1/4), ils auraient e rit quelque hose omme || 15 | (voulant dire \quinze soixantiemes" ou 15/60, qui reduit a 1/4).
390 Le tableau i-dessous donne ertains nombres et leurs re iproques, e rits dans le systeme babylonien. (Rappel : le produit d'un nombre et sa re iproque donne toujours 1.) Chaque olonne represente 1/60 de la olonne anterieure. Par exemple, 7 | 30 represente 7/60 + 30/3600. Nombre 2 3 4 5
|| Re iproque || 30 | || 20 | || | || |
Nombre 6 7 8 9
|| Re iproque || | | | || || 7 | 30 | || | |
|
| ...
(a) Donner, en ordre, les nombres devant appara^tre a droite de 4, 5 et 6. (b) Si on ajoutait notre nombre de imal 2.5 (don deux et demi) au tableau, quel nombre se retrouverait omme version babylonienne de sa re iproque ? ( ) La re iproque du nombre 8 est donnee dans le tableau omme || 7 | 30 |. Dans la m^eme notation, quelle est la re iproque du nombre 9 ? (d) Donner les deux premiers nombres devant aller a droite de 7 au tableau. (e) Donner la version babylonienne de la re iproque du nombre de imal 192. . En atterrissant sur Mars, on a de ouvert le jeu pref er e des martiens : Ballon-frappe. Dans e jeu, deux equipes tentent tour a tour de frapper un ballon entre deux poteaux pla es a haque bout du terrain de jeu. L'equipe ayant ompte le plus grand nombre de points dans une heure martienne gagne. Il y a trois fa ons de ompter des points : I, omptant 7 points, O omptant 4 points et W omptant 2 points. Les resultats du premier mat h nous ont et e ommuniques, mais des erreurs de transmission font que ertains details ne sont pas onnus. Voi i e qu'on sait. L'equipe Rouge a gagne le mat h ave un s ore total de \quelque
hose, sept" points, le premier hi re n'ayant pas et e entendu lairement. Nous avons aussi appris que l'equipe Rouge a ompte lors de 16 lan ers pre is ement. 4
(a) Donner une liste de toutes les possibilites de lan er I que l'equipe Rouge aurait pu ompter dans e mat h, listes en ordre allant du plus petit au plus grand. (b) Un deuxieme ommunique a ajoute ette information supplementaire : le nombre de lan ers I egale le nombre de lan ers O. l'aide de ette nouvelle information, fournir le nombre de points omptes au total par l'equipe Rouge. ( ) Expliquer pourquoi la reponse en (b) est unique. Fournir le nombre exa t de I, O et W omptes par l'equipe Rouge.
391 . En 2004, un journal de Dunedin a tenu un on ours visant a proposer un nouveau drapeau pour la provin e d'Otago. Wendy a parti ipe a e on ours. Sa soumission etait basee sur le dessin (voir gure 4). Son drapeau omportait une roix de ouleur or, sur un fond bleu. De plus, elle a ajoute un er le au dessin, dont le haut et le bas tou hent tout juste les oins des triangles situes au haut et au bas du drapeau, tel qu'illustre dans la gure 4. 5
40 30 6 ?
40 -
30 6 ? 40
6 30 - ? 40
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.................................................................................................................................................................................................. ....... . ... .... ....... ...... .... .... ....... ....... ....... ....... .... .... ....... ....... . . . . . . . ............ . . ....... .......... ..... . . . ....... . . . . . . . .... .......... ....... .... ... .... . . . . ....... . . ... . . . . . . ....... ...... . ....... ... . . ... . . . . . . . . . . ..... . . . ........ ....... . ..... ....... .... .... ...... ....... ....... .... ... ....... ...... . . . ... ....... . . .... . ..... ...... ... . . . .... . ....... .... . . . . . ..... . . .... . ....... ..... . . . . . . .... . . .... ....... ..... . . . . . . . .... . . ... ....... ............ .... . . . . . . . . . . . . .... ... . . . ....... ....... . ..... . . . . ... ............. . . . . . . . ....... .... ....... ..... . . . . . . . . ............... . . . . . ........... ....... .... . . . . . . ...... . . ... . ....... ...... ....... .... .... ...... ....... ....... .... .... ....... ...... . . . . . ... . . .. . ............................................................................................................................................................................................
6 30 ?
Fig. 5 (pas a l'e helle) Fig. 4 (pas a l'e helle) (a) Wendy a pris pour a quis un drapeau de 240 m de large et de 150 m de haut. Si les bords de la roix se trouvent a 40 m et 30 m des oins, tel qu'illustre a la gure 4, quel est le rayon du er le ? (b) En n, Wendy a de id e d'enlever le er le de son dessin (voir gure 5). Ave le er le enleve, quelle est la surfa e totale de la roix ? We now give solutions to the Montmoren y Contest 2005{06, Se V, given in [2008 : 2{4℄. . Evaluate the following, giving the answer as a fra tion:
1
1 1 1 1− 2 1 − 2 ··· 1 − 2 3 20052
.
Solution by Natalia Desy, student, SMA Xaverius 1, Palembang, Indonesia. We have
= = = =
1 1 1 1− 2 1 − 2 ··· 1 − 2 2 3 2005 1 1 1 1 1 1 1+ 1− 1+ 1− ··· 1 + 1− 2 2 3 3 2005 2005 1 1 1 1 1 1 1+ 1+ ··· 1 + 1− 1− ··· 1 − 2 3 2 3 2005 2005 3 4 5 2005 2006 1 2 3 2003 2004 · · ··· · · · ··· · 2 3 4 2004 2005 2006 1 1003 · = . 2 2005 2005
2
3
4
2004
2005
392 Also solved by ALEX SONG, Elizabeth Ziegler Waterloo, Publi S hool, ON. Natalia Desy also noted that the formula 1 − 212 1 − 312 · · · 1 − n12 = holds for ea h integer n ≥ 2.
∠AOD + ∠BOC = 180◦ .
A .... ................. ........................ ........ ..... . C............................................................................................................D ............
2. In a ir le having entre O , the hords AB and CD are perpendi ular to ea h other and neither hord passes through the entre. Show that
n+2 2n+2
. .. .. . .. .... ... ..... ..... ... ... .. .. ...... .. .. ... .. .. ...... .. ..... ... .. ... ...... ... ... .. .. ... . . ... .. ... .. . . . . .. ... . .. ... ... .. . . ... ... .. ... .... . .... ...... ........ .... ....... .... ....... . ............. . . . . . . . . . .....................
O
B
Solution by Alex Song, Elizabeth Ziegler Publi S hool, Waterloo, ON. We have that ∠AOC + ∠BOD = 2 ∠AEC , where E is the interse tion of AB and CD. Thus, ∠AOC + ∠BOD = 180◦ , from whi h it follows that ∠AOD + ∠BOC = 180◦ . No other solutions were submitted. 3. Show that for real x and y where x 6= 0 and y for the equation (x + y)4 = x4 + y 4 .
6= 0,
there is no solution
Solution by Alex Song, Elizabeth Ziegler Publi S hool, Waterloo, ON, modi ed by the editor. We have (x + y)4 = x4 + 4x3 y + 6x2 y 2 + 4xy 3 + y 4 = x4 + y 4 .
Thus, 2x2 + 3yx + 2y2 = 0 (by simplifying and dividing by 2xy 6= 0). Then x =
−3y ±
p √ (3y)2 − 4 · 2 · 2y 2 −3y ± |y| −7 = 2·2 4
.
This means that if y is a real number, then x is not a real number. Therefore, and y annot both be non-zero real numbers.
x
There was one in orre t solution submitted.
. (a) How many positions on average an a knight rea h in one move on a 8 × 8 hessboard? (b) What result do we get for an n × n hessboard? 4
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A YK H * H A H A H j A U
The movement of a knight has the shape of an \L": a shift of two squares horizontally followed by one verti ally, or vi e versa.
Solution by Alex Song, Elizabeth Ziegler Publi S hool, Waterloo, ON. We answer part (b) rst. Let the bottom left square be x1,1 , and let xa,b be the square in the ath row above the bottom and the bth olumn from the left.
393 From any orner square x1,1 , x1,n , xn,1 , or xn,n there are 2 knight moves, and a total of 4 su h squares. From any square x1,c , xc,1 , xn,c , or xc,n , where 3 ≤ c ≤ n − 2, there are 4 knight moves, and a total of 4(n − 4) su h squares. From any square x1,c , xc,1 , xn,c , or xc,n , where c = 2 or c = n − 1, there are 3 knight moves, and a total of 8 su h squares. From any square x2,2 , x2,n−1 , xn−1,2 , or xn−1,n−1 , there are 4 knight moves, and a total of 4 su h squares. From any square x2,c , xc,2 , xn−1,c , or xc,n−1 , where 3 ≤ c ≤ n − 2, there are 6 knight moves, and a total of 4(n − 4) su h squares. From any square xd,c , where 3 ≤ c, d ≤ n − 2, there are 8 knight moves, and a total of (n − 4)2 su h squares. Thus, the average number of squares a knight an rea h on an n × n
hessboard is n2
When n = 8 the answer to part (a) is
` 2 ´ 8n − 24n + 16 21 = . n2 4
8 + 16(n − 4) + 24 + 16 + 24(n − 4) + 8(n − 4)2 8n − 24n + 16 2
=
n2
.
No other solutions were submitted. Alex Song's derivation impli itly assumes that n ≥ 4, but the formula he obtained is also valid for n = 1, n = 2, and n = 3.
. Show that by pla ing 5 points inside a right-triangle with sides 6 m, 8
m, and 10 m, at least two of them have a distan e smaller than 5 m. 5
Solution by Alex Song, Elizabeth Ziegler Publi S hool, Waterloo, ON. We partition the triangle into four parts. Let A = (6, 0), B = (0, 0), be the verti es of the triangle. To save on writing, let 5 s = √ − 0.001. One part is the square with verti es (0, 0), (s, 0), (0, s), 2 and (s, s) interse ted with the triangle. The se ond part is this square moved s units to the right and also bounded by the triangle. The third part is the rst square moved s units up and bounded by the triangle. The last part is the rst square moved 2s units up and bounded by the triangle. [Ed.: One must he k that the four squares ompletely over the triangle, for example, the point (s, s) is above the line AC .℄ By the Pigeonhole Prin iple, there are 2 points that lie in the same part. Sin e ea h part is inside a square of side s, the maximum distan e between these two points is less than 5. and
C = (0, 8)
No other solutions were submitted.
394 . Three runners, Yves, Jules, and Bertrand meet one day at a ir ular tra k. Knowing that they start running at the same moment, that Bertrand, the fastest of the three, does a full revolution of the tra k in 2 minutes and that Yves does it in 8 minutes, how long does it take Jules to do a full revolution, if we further note that all three runners meet at the same pla e on the tra k before Yves, the slowest of the three runners, has done a full revolution? 6
rBertrand Ir Jules r Yves
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Start
Solution by Alex Song, Elizabeth Ziegler Publi S hool, Waterloo, ON, expanded by the editor. Bertrand's speed is 12 a lap (revolution) per minute, and Yves' speed is 1 of a lap per minute. If t is the time in minutes when Bertrand and Yves 8 rst meet, then t t −1 = , 2
8
be ause Bertrand runs one lap before they rst meet. Solving for t gives 8 1 t = minutes and this rst meeting o
urs at a point of the way along the 3 3 tra k. After rst meeting, Bertrand and Yves meet again in t = 83 + 38 = 16 3 minutes at a point 23 of the way along the tra k. The next time they meet Yves will have done a full revolution, so we only need to think about these two meeting points. Jules is slower than Bertrand, so he annot at h Yves the rst time that Bertrand does. Thus, Jules meets them both at the se ond meeting point. When Jules does meet them there, he annot have done less than a lap (otherwise he would be just as slow as Yves) and he annot have done two laps (otherwise he would be just as fast as Bertrand). Therefore, he runs 1 omplete lap before the se ond meeting point and his speed is 2 16 5 1+ / = laps per minute and it takes Jules 16 minutes to run one 3 3 16 5 lap. No other solutions were submitted. 7
7
. How many numbers are there smaller than 10,000 that ontain the digit at least on e?
Solution by Alex Song, Elizabeth Ziegler Publi S hool, Waterloo, ON. Write a whole number less than 10,000 in the form abcd, where a, b, c, and d are digits. We onsider the numbers that do not have a 7 as one of the digits. For these there are 9 hoi es for ea h of a, b, c, and d, resulting in 94 = 6561 numbers. Therefore, the total ount of numbers that have at least one 7 is 10,000 − 6561 = 3439. No other solutions were submitted.
395 . We build a one from a ir ular pie e of ardboard having a radius of 10 m by utting out a se tor from it. Determine the volume of the one we obtain as a fun tion of the angle, in radians, at the enter of the se tor. Re all that Volume = 13 πr2 h, where r is the radius of the base of the one, h is the altitude of the one, and 2π radians = 360◦ . Solution by Alex Song, Elizabeth Ziegler Publi S hool, Waterloo, ON, expanded by the editor. Let θ be the angle of the se tor that is ut out from the one. The ir umferen e of the pie e of ardboard is 2π · 10 = 20π, and the ar along the θ se tor has length 2π · 20π = 10πθ. If r is the radius of the base of the one made from the se tor, then 2πr = 10πθ, hen e r = 5θ. Following from the tip of the one to the entre of the base to the edge of the base along a radius, we have a right triangle of height h, base 5θ, and hypotenuse 10. By the √ Pythagorean Theorem, h2 + (5θ)2 = 102 , hen e h = 100 − 25θ2 and the √ √ volume of the resulting one is 13 π(5θ)2 100 − 25θ2 = 25π θ2 100 − 25θ2 . 3 [Ed.: In the question it appears the se tor is to be dis arded and the material remaining forms the one. However, for θ > π, the material remaining is less than what is dis arded.℄ 8
................................................ .......... . ........ ............ ..... .... ... ........ ... ..... . .. .... . . ... .. ... ... . . . . . ... ... . ... ... ... .. .. ..................................... .. . . ... ...... ...... .... .. ... .. ... ... . ... .. ... .. .... ... ... ... ... . .. . . .. . .. .. ... ... ... ... .... .... .... . . . ..... ...... ....... ....... ......... ........ ............... ..............................
No other solutions were submitted.
That brings us to the end of another issue. This month's winner of a past volume of Mayhem is Alex Song. Congratulations, Alex! Continue sending in your ontests and solutions.
396
MATHEMATICAL MAYHEM Mathemati al Mayhem began in 1988 as a Mathemati al Journal for and by High 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). The other sta members are Monika Khbeis (As ension of Our Lord Se ondary S hool, Mississauga), Eri Robert (Leo Hayes High S hool, Frederi ton), Larry Ri e (University of Waterloo), and Ron Lan aster (University of Toronto).
Mayhem Problems
Please send your solutions to the problems in this edition by 15 January 2009. Solutions re eived after this date will only be onsidered if there is time before publi ation of 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 translations of the problems. M363. Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL. Suppose that A is a six-digit positive integer and B is the positive integer formed by writing the digits of A in reverse order. Prove that A − B is a multiple of 9.
. Proposed by the Mayhem Sta . A semi- ir le of radius 2 is drawn with diameter AB . The square P QRS is drawn with P and Q on the semi- ir le and R and S on AB . Is the area of the square less than or greater than one-half of the area of the semi- ir le? M364
. Proposed by Alexander Gurevi h, student, University of Waterloo, Waterloo, ON. Let D be the family of lines of the form y = nx + n2 , with n ≥ 2 a positive integer. Let H be the family of lines of the form y = m, where m ≥ 2 is a positive integer. Prove that a line from H has a prime y -inter ept if and only if this line does not interse t any line from D at a point with an x- oordinate that is a non-negative integer. M365
397 . Proposed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB. The roots of the equation x3 + bx2 + cx + d = 0 are p, q, and r. Find a quadrati equation with roots (p2 + q2 + r2 ) and (p + q + r). M366
. Proposed by George Tsapakidis, Agrinio, Gree e. For the positive real numbers a, b, and have a + √ c we √ √b + c Determine the maximum possible value of a bc + b ac + c ab. M367
= 6.
. Proposed by J. Walter Lyn h, Athens, GA, USA. An in nite series of positive rational numbers a1 + a2 + a3 + · · · is the fastest onverging in nite series with a sum of 1, a1 = 12 , and ea h ai having numerator 1. (By \fastest onverging", we mean that ea h term an is su
essively hosen to make the sum a1 + a2 + · · · + an as lose to 1 as possible.) Determine a5 and des ribe a re ursive pro edure for nding an . ................................................................. M368
. Propose par Bru e Shawyer, Universite Memorial de Terre-Neuve, St. John's, NL. Soit A un entier positif de six hi res et B l'entier positif forme des
hi res de A e rits dans l'ordre inverse. Montrer que A − B est un multiple de 9. M363
. Propose par l'Equipe de Mayhem. Dans un demi- er le de rayon 2 et de diametre AB , on dessine un arre P QRS ave P et Q sur le demi- er le, et R et S sur AB . L'aire du arre estelle plus petite ou plus grande que la moitie de elle du demi- er le ? M364
. Propose par Alexander Gurevi h, etudiant, Universite de Waterloo, Waterloo, ON. Soit D la famille des droites de la forme y = nx + n2 , ave n ≥ 2, un entier positif. Soit H la famille des droites de la forme y = m, ave m ≥ 2, un entier positif. Montrer que l'ordonnee du point d'interse tion d'une droite de H ave l'axe des y est un nombre premier si et seulement si
ette droite ne oupe au une droite de D en un point d'abs isse entiere non negative.
M365
. Propose par John Grant M Loughlin, Universite du NouveauBrunswi k, Frederi ton, NB. Soit p, q et r les ra ines de l'equation x3 + bx2 + cx + d = 0. Trouver une equation quadratique dont les ra ines sont (p2 + q2 + r2 ) et (p + q + r). M366
398 . Propose par George Tsapakidis, Agrinio, Gre e. Soit a, b et c trois nombres reels positifs que a +√b + √ tels √ Determiner la valeur maximale possible de a bc + b ac + c ab. M367
c = 6.
. Propose par J. Walter Lyn h, Athens, GA, E-U. Une serie in nie de nombres rationnels positifs a1 + a2 + a3 + · · · est la serie la plus rapidement onvergente de somme 1, ave a1 = 12 et
haque ai de numerateur 1. (Par \la plus rapidement onvergente", on entend que haque terme an est tour a tour hoisi de maniere a e que la somme a1 + a2 + · · · + an soit aussi pro he de 1 que possible.) Determiner a5 et de rire une pro edure re ursive pour trouver an . M368
Mayhem Solutions The Editor would like to thank Emily Saltstone, Gravenhurst High S hool, Gravenhurst, ON, for her help in preparing this month's solutions.
M325. Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL.
Let a, b, and c be non-zero digits. A student takes the fra tion ab , ca where ab and ca represent the two-digit integers 10a + b and 10c + a, and applies a (false) an ellation law, an elling the a from the numerator with the a from the denominator. For example, if a = 6, b = 5, and c = 2, the student would obtain 65/26 = 5/2 (by ` an elling' the 6s!). Determine all triples (a, b, c) for whi h this student a tually obtains the
orre t answer. Solution by Samuel Gomez Moreno, Universidad de Jaen, Jaen, Spain, modi ed by the editor. If
10a + b b = , then c(10a + b) = b(10c + a) or 10c(a − b) = b(a − c) 10c + a c
so 2 · 5 · c(a − b) = b(a − c). Therefore, 5 must be a divisor of the right hand side. Sin e 5 is a prime number, then 5 must be a divisor of either b or a − c. Sin e a, b, and c are integers su h that 1 ≤ a, b, c ≤ 9, then either a − c = 0 (whi h would imply that a − b = 0), b = 5, or a − c = 5. Case I: We have a − c = 0 and a − b = 0. Here a = b = c, and we get nine triples (a, a, a), where a ∈ {1, 2, . . . , 9}.
399 Case II: We have b = 5. If b = 5, then the equation be omes 2c(a − 5) = a − c. Solving for c we obtain, c = 2a a− 9 . We have a ≥ 5, otherwise c < 0, a ontradi tion. The value a = 5 yields (5, 5, 5), whi h we found in Case I. Of the remaining possibilities, a = 6 and a = 9 yield two new triples, (6, 5, 2) and (9, 5, 1), respe tively. Case III: We have a − c = 5. If a − c = 5, then 2c(a − b) = b. Substituting a = c + 5, we then obtain c+5 2c(c + 5) = (c + 5) − 2c(c + 5 − b) = b, when e, b = . Now, if c > 4, 2c + 1 2c + 1
c+5 then 2c + 1 > c + 5, and onsequently 0 < 2c < 1, ontradi ting the +1 fa t that b is an integer. Therefore, c ≤ 4, and of these possibilities c = 1 and c = 4 yield the last two triples, (6, 4, 1) and (9, 8, 4). The thirteen triples (a, b, c) are (6, 5, 2), (9, 5, 1), (6, 4, 1), (9, 8, 4), and all triples (a, a, a), where a ∈ {1, 2, . . . , 9}.
Also solved by NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; ~O N GRANDES, student, Universidad de La Rioja, Logrono, MIGUEL MARAN ~ La Rioja, Spain; student, Sarajevo RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; and SALEM MALIKIC, College, Sarajevo, Bosnia and Herzegovina. There were 3 in orre t solutions submitted.
. Proposed by the Mayhem Sta . The notation a89b means the four-digit (base 10) integer whose thousands digit is a, whose hundreds digit is 8, whose tens digit is 9, and whose units digit is b. Determine all pairs of non-zero digits a and b su h that a89b − 5904 = b98a. M326
Solution by Ja lyn Chang, student, Western Canada High S hool, Calgary, AB. Sin e a89b − 5904 = b98a, then the following equations are true: 1000a + 890 + b − 5904 1000a + b 999a
= 1000b + 980 + a , = 1000b + a + 5994 , = 999b + 5994 ,
a = b + 6.
Sin e both a and b represent non-zero digits, they must be whole numbers between 1 and 9 in lusive. This gives the pairs (a, b) = (7, 1), (8, 2), (9, 3). Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; NATALIA DESY, stu \Abastos", Valen ia, Spain; dent, SMA Xaverius 1, Palembang, Indonesia; RICARD PEIRO, IES CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan; GEORGE TSAPAKIDIS, Agrinio, Gree e; VINCENT ZHOU, student, Dr. Norman Bethune Collegiate Institute, Agin ourt, ON; and TITU ZVONARU, Comane sti, Romania. There was 1 in orre t solution submitted.
400 M327. Proposed by Lino Demasi, student, University of Waterloo, Waterloo, ON. Kaitlyn bought a new eraser. Her new eraser is in the shape of a re tangular √ prism.√She al ulates the lengths of the diagonals of the fa es to be 10, 61, and 89. What is the volume of Kaitlyn's eraser?
Solution by Mrinal Singh, student, Kendriya Vidyalaya S hool, Shillong, India. Let the three edge lengths of the prism be l, b, and h. Using the Pythagorean Theorem and the given lengths of the fa e diagonals, we obtain l2 + b2 = 100 and b2 + h2 = 61 and h2 + l2 = 89. Adding these equations, we obtain 2l2 + 2b2 + 2h2 = 250, so that 2 l + b2 + h2 = 125. Therefore, l2 = 125 − (b2 + h2 ) = 125 − 61 = 64, hen e l = 8 sin e l > 0. Sin e l2 + b2 = 100 and l = 8, then b2 = 100 − 64 = 36, hen e b = 6 sin e b > 0. Lastly, sin e b2 + h2 = 61 and b = 6, then h2 = 61 − 36 = 25, hen e h = 5 sin e h > 0. The volume of the re tangular prism equals lbh = 8(6)(5) = 240. Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; PETER CHIEN, student, Central Elgin Collegiate, St. Thomas, ON; NATALIA DESY, stu MEZ MORENO, Universidad de dent, SMA Xaverius 1, Palembang, Indonesia; SAMUEL G O Jaen, Jaen, Spain; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan; GEORGE TSAPAKIDIS, Agrinio, Gree e; JOCHEM VAN GAALEN, grade 9 student, Medway High S hool, Arva, ON; VINCENT ZHOU, student, Dr. Norman Bethune Collegiate Institute, Agin ourt, ON; and TITU ZVONARU, Comane sti, Romania. There was 1 in omplete solution submitted.
. Proposed by Hugo Cuellar, Columbia Aprendiendo, Zipaquira, Coloumbia. Prove that, if from any positive integer we subtra t the sum of ea h of its digits raised to any odd power (not ne essarily the same), then the result is always a multiple of 3. M328
Solution by Daniel Tsai, student, Taipei Ameri an S hool, Taipei, Taiwan, modi ed by the editor. n−1 P
Let dk 10k be a positive integer, where 0 ≤ dk < 10 is an integer k=0 for ea h integer 0 ≤ k < n and dn−1 6= 0. Let ek be a positive odd integer for ea h integer 0 ≤ k < n. We need to prove that n−1 X k=0
dk 10k −
n−1 X k=0
dkek ≡ 0 (mod 3)
,
(1)
401 whi h is equivalent to proving that n−1 X k=0
dk 10k ≡
n−1 X
dkek (mod 3)
.
(2)
k=0
To prove (2), it suÆ es to prove that dk 10k ≡ dke (mod 3) for ea h k. Sin e 10k ≡ 1 (mod 3) for any non-negative integer k, we have k dk 10 ≡ dk (mod 3), so it suÆ es to prove that dk ≡ dke (mod 3) for ea h integer 0 ≤ k < n. This follows from the lemma below. k
k
If d and e are integers and e > 0 is odd, then d ≡ de (mod 3). Proof: Without loss of generality, we may assume that d = 0, 1, or 2. If d = 0 or d = 1, then d = de , so the result follows. If d = 2, then we have de ≡ 2e ≡ (−1)e ≡ −1 ≡ 2 (mod 3), sin e e is odd. This ompletes our solution. Lemma
Also solved by LAURIN M. COLEMAN, student, Auburn University Montgomery, MEZ MORENO, Universidad de Jaen, Montgomery, Alabama, USA; SAMUEL G O Jaen, Spain; \Abastos", Valen ia, Spain; JOSE HERNANDEZ RICARD PEIRO, IES SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; VINCENT ZHOU, student, Dr. Norman Bethune Collegiate Institute, Agin ourt, ON; and TITU ZVONARU, Comane sti, Romania. There were 3 in omplete solutions submitted.
. Proposed by the Mayhem Sta . Determine the value of
M329
cos2 1◦ + cos2 2◦ + cos2 3◦ + · · · + cos2 89◦ + cos2 90◦ .
Solution by Kunal Singh, student, Kendriya Vidyalaya S hool, Shillong, India. We have cos2 1◦ + cos2 2◦ + cos2 3◦ + · · · + cos2 89◦ + cos2 90◦ = (cos2 1◦ + cos2 89◦ ) + (cos2 2◦ + cos2 88◦ ) + (cos2 3◦ + cos2 87◦ ) + · · · + (cos2 44◦ + cos2 46◦ ) + cos2 45◦ + cos2 90◦ = cos2 1◦ + cos2 (90◦ − 1◦ ) + cos2 2◦ + cos2 (90◦ − 2◦ ) 2 1 + · · · + cos2 44◦ + cos2 (90◦ − 44◦ ) + √ +0 2
= (cos2 1◦ + sin2 1◦ ) + (cos2 2◦ + sin2 2◦ ) + (cos2 3◦ + sin2 3◦ ) 1 2
+ · · · + (cos2 44◦ + sin2 44◦ ) + 1
= 1 + 1 + ··· + 1 + | {z } 2 44 times
= 44 +
1 2
=
89 2
.
402 Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; MIHALY BENCZE, Brasov, Romania; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; DENISE CORNWELL, student, Angelo State University, San Angelo, TX, USA; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; SAMUEL MEZ MORENO, Universidad de Jaen, GO Jaen, Spain; RICHARD I. HESS, Ran ho Palos Verdes, \Abastos", Valen ia, Spain; CAO MINH QUANG, Nguyen Binh CA, USA; RICARD PEIRO, IES Khiem High S hool, Vinh Long, Vietnam; DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan; GEORGE TSAPAKIDIS, Agrinio, Gree e; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and VINCENT ZHOU, student, Dr. Norman Bethune Collegiate Institute, Agin ourt, ON. Hernandez Santiago and Zvonaru both noted that M329 is the same as M170 from 2004.
. Proposed by the Mayhem Sta . If n is a positive integer, the nth triangular number is de ned as Tn = 1 + 2 + · · · + (n − 1) + n = 12 n(n + 1). Determine all pairs of triangular numbers whose di eren e is 2008. M330
Solution by Titu Zvonaru, Comane sti, Romania. We must solve the equation 12 n(n + 1) − 12 m(m + 1) = 2008, whi h is equivalent to n2 − m2 + n − m = 2 · 2008, or (n − m)(n + m + 1) = 24 · 251. Note that n + m + 1 > n − m. Also, (n − m) and (n + m + 1) have di erent parity, be ause their sum is 2n + 1, whi h is odd. Therefore, n − m = 1 and n + m + 1 = 4016, or n − m = 16 and n + m + 1 = 251. (Sin e 251 is a prime number, these are the only two ways to fa tor 24 · 251 as the produ t of one even and one odd positive integer.) In the rst ase, adding the two equations gives 2n + 1 = 4017 when e n = 2008 and the equation n − m = 1 gives m = 2007. Repeating this in the se ond ase, we obtain n = 133 and m = 117. The two solutions are the pairs (T2008 , T2007 ) and (T133 , T117 ).
Also solved by MIHALY BENCZE, Brasov, Romania; NATALIA DESY, student, SMA MEZ MORENO, Universidad de Jaen, Xaverius 1, Palembang, Indonesia; SAMUEL G O Jaen, \Abastos", Spain; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIRO, IES Valen ia, Spain; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan; and EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON. There were 3 in omplete solutions and 1 in orre t solution submitted.
. Proposed by the Mayhem Sta . In trapezoid ABCD, side AB is parallel to DC , and diagonals AC and BD interse t at P . (a) If the area of △AP B is 4 and the area of △DP C is 9, (i) prove that AP : P C = 2 : 3, (ii) explain why the ratio of the area of △BP C to the area of △BP A equals 3 : 2, and (iii) determine the area of trapezoid ABCD. (b) If the area of △AP B is x and the area of △DP C is y, determine the area of trapezoid ABCD in terms of x and y.
M331
403 Solution by Denise Cornwell, student, Angelo State University, San Angelo, TX, USA, modi ed slightly by the editor. Suppose that AP = q, CP = r, BP = s, and DP = t. We use the A B notation [△XY Z] to denote the area of △XY Z . s q (a) Sin e AB k CD and diagonals AC and BD interse t at P , then ∠AP B = ∠CP D , ∠ABP = ∠CDP , and ∠BAP = ∠DCP , so △AP B ∼ △CP D , when e s q = . r t
.................................................................................................. .. ................ .... ..... .. ........ .... ..... ........ .. .... ..... ........ .. .... ........ .......... ... .... ........... . . .... . ... ............ . .... . . ... . ........ .... .... . . . . . . . . . ........ .... .... . .. . . . . . .... . . . . ........ . .... .... . . . . . . . . ... ........ .... ... . . . . . . . . . . . .... ........ . .... . . . .... . . . . . ... ........ .... . . . . . . . . . . . ........ ....... .. . ........ .... ... .......... ........ ... . ..... . ........ .... ........... .. .... .. . .. ..... ...................................................................................................................................................................................................................................
P
t
r
D
C
(i) Now, [△AP B] = 12 qs sin ∠AP B and [△CP D] = 12 rt sin ∠CP D. Sin e ∠AP B = ∠CP D, the ratio of the areas of these triangles is 4 qs q s 4 q2 2 q = = gives = 2 , or = . . Using the fa t that 9 rt r t 9 r 3 r Therefore, AP : P C = 2 : 3. (ii) Now, [△CP B] = 12 rs sin ∠CP B and [△AP B] = 12 qs sin ∠AP B . Also, ∠CP B = π − ∠AP B , and sin e sin θ = sin(π − θ) we then have sin ∠CP B = sin ∠AP B . Hen e, the ratio of the area of the triangles is [△CP B] [△AP B]
=
1 rs sin ∠CP B 2 1 qs sin ∠AP B 2
=
r q
=
3 2
.
(iii) Now we an nd the area of the trapezoid by nding the areas of △CP B and △DP A. [△CP B] =
r
q
[△AP B] =
3 2
(4) = 6 .
By a similar argument, we an show that [△DP A] = 6. That is, we an swit h the pairs of labels A and B , C and D, q and s, and r and t, and then arry out same arguments that we did before. Altogether, [ABCD] = 4 + 6 + 6 + 9 = 25. (b) If [△AP B] = x and [△DP C] =√y, then using the same steps as above 2 we see that xy = qr 2 , hen e qr = √xy . As above, [△BP C] = [△DP A] =
√ r y √ [△AP B] = √ (x) = xy ; q x √ r y √ [△AP B] = √ (x) = xy . q x
404 Thus, the area of the trapezoid ABCD is [ABCD] = x +
√ √ √ √ √ xy + xy + y = x + 2 xy + y = ( x + y)2 .
Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; RICHARD \Abastos", Valen ia, Spain; I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIRO, IES CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan; GEORGE TSAPAKIDIS, Agrinio, Gree e; VINCENT ZHOU, student, Dr. Norman Bethune Collegiate Institute, Agin ourt, ON; and TITU ZVONARU, Comane sti, Romania. There were 2 in omplete solutions and 1 in orre t solution submitted.
Problem of the Month Ian VanderBurgh
It's time for a road trip! Problem (1997 UK Junior Mathemati al Challenge) Seven towns P , Q, R, S , T , U , V lie (in that order) along a road. The table on the right is meant to give all the distan es between pairs of towns, in km; for example, the distan e from P to S is 23 km. Unfortunately, fteen of the distan es are missing. How many of the missing distan es an be al ulated from the given information? (A) 0 (B) 1 (C) 6 (D) 12 (E) 15
P ? Q ? ? R 23 ? ? S ? 30 ? ? T 58 ? 40 ? ? U ? 68 ? 53 ? ? V
You have probably seen this type of hart before on the ba k of a roadmap. We should make sure rst that we know how to interpret the
hart. From the hart, the distan e from Q to T is 30 km, but we don't know the distan e from Q to U . Q My next in lination P R S T U V s s s s s s s is to draw a s hemati diagram. I say the diagram is \s hemati " sin e it is not to s ale. Next, we should indi ate the information we are given on the diagram. I'm going to do this not by marking the a tual lengths, but by indi ating under the diagram the lengths that we know using dashed line segments. Now we should try to nd some new lengths. Can you see how to use the given information to nd the distan e from P to V ? See if you an gure this out! .............................................................................................................................................................................................................................................................................................. ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ......
...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... .. ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ......
...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... .. ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ......
...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ......
405 From the information, the distan e from P to S is 23 km. We'll write P S = 23 as shorthand. Also, SV = 53. Therefore, P V = P S + SV = 23 + 53 = 76. Can you see what this is saying physi ally? It says that the distan e along the line from P to V equals the distan e from P to S plus the distan e from S to V . If you're not onvin ed, try going for a drive! (A
question to think about: Why is it important that the points lie along a line?) So we've got at least 1 of the 15 missing pie es of information sorted out. We an ontinue by trial and error to nd more pie es of information. But is there a strategy that we an use to do this systemati ally? Another question: Are some segments more important than others? Put another way, are there segments whose lengths give us the most information about other lengths? The answer is yes { it's the \basi " segments: P Q, QR, RS , ST , T U , and U V whi h are ru ial. All of the other segments are made up from these, so let's try to gure out the lengths of these basi segments. At the very least, this gives us a way to organize our work. Solution
We have the partial hart above and we have already gured out
P V . Is there a basi segment we an dedu e in one step? Yes, we an gure out U V . Sin e P V = 76 and P U = 58, then U V = P V − U V = 18.
Let's ontinue to work from right to left. We don't have enough information yet to determine T U in one step, but sin e we know U V , then if we knew T V , we ould nd T U . But T V = QV − QT = 68 − 30 = 38. Thus, T U = T V − U V = 38 − 18 = 20. Next, we determine ST . We have SV = 53 and T V = 38. Therefore, ST = 53 − 38 = 15. Also, SU = ST + T U = 15 + 20 = 35. Now we gure out RS . We know that RU = 40 and SU = 35, so RS = RU − SU = 5. Thus, RT = RS + ST = 5 + 15 = 20 and RV = RS + SV = 5 + 53 = 58. We're getting loser: QR is next. We know QT = 30 and RT = 20, so QR = QT − RT = 10. Thus, QS = QR + RS = 10 + 5 = 15 and QU = QR + RU = 10 + 40 = 50. Lastly, we dedu e P Q. We know that P V = 76 and QV = 68. Thus, P Q = P V − QV = 8. Also, P R = P Q + QR = 8 + 10 = 18 and P T = P Q + QT = 8 + 30 = 38. Our hart is now omplete and we deterP mined all 15 pie es of missing information! 8 Q I always nd these kinds of problems 18 10 R surprising. It's amazing how mu h information we an get from so little. But wait { is it so 23 15 5 S surprising? While there are 21 distan es in to38 30 20 15 T tal that we need, how many of these are truly 58 50 40 35 20 U \independent"? A tually, only 6 distan es are (essentially, these 6 basi segments that we 76 68 58 53 38 18 V looked at earlier). Do you think that it's a
oin iden e that we were given 6 pie es of information to begin? Do you think that we ould do this if we were given only 5 pie es of information to start?
406 As a posts ript, I would suggest that you not approa h your lo al Department of Transportation to see if they want to save on printing osts by removing most of the distan es on their lo al hart...
Adding Up Bru e Shawyer
A daily exer ise is o ered in the British Newspaper The Times and is there alled Add Up. The stated rules are: The number in ea h ir le is the sum of the two below it. Work out the top number. Try it in your head, if you an. A typi al example is .................. ... .... . ..... .. ... ...................................... . . . . . . . . . . . . . .... .... .... ... . . . . ... ..... .... .. .. ... ... ... ... ........................................................................................... . . . . . ... ... ... ... ... ..... . . . . ... ..... ..... ... .. ... .. ... .. .......... .............. .............. ......... ....... ............................ ............................ ............................. ......... . ...... ...... ...... ... ..... .... .... .... .. ... ... ... ... ... ... ... ... .................. .......................... .......................... ......................... .................. ..... .................. .................. .................. ................... ....... . ..... . ..... ...... . . . .... . . .. .. .. .. ..... ..... ... ... .................... ...................... ...................... ...................... ......................
11
10
4
3
1
With a little bit of work, we an see that the number that must be in the top position is 78. There are two ways (at least) to solve this problem. There is the obvious straight forward way of lling in every number on the olle tion. In the given example, we see the missing numbers in the bottom row: .................... ... ... ..... .. ... ... ...................................................... . . . . ... ... ... ..... . . ... ...... ... ... .... .... .... ................................................................................................ . . . .. .. ....... .. . ..... . . . . .. ...... ....... ... ............ ................. ................. ............ ...... ........................... ........................... ........................... ......... . . ....... ..... ...... ... ..... . . .. .. ... ... ... .. ... .. .......... ............... .............. ............. ......... ....... ............................. ............................. .............................. .............................. ......... ...... ...... ...... ... ..... ..... ... .. .. .. .. . .... .. ... .. ... .. ... .. ... .. ................. ..................... .................... .................... ....................
11
7
10
4
3
9
1
then ll in the se ond row, and ea h subsequent row, to get ............ ..... ....... ..... . .. ... ... . . . . . . . . . . . . . . . ........................... ..... ...... ...... ...... . ..... ... .... .. ... ... ... ......... .............. ......... ....... .............................. .............................. .......... ..... ...... . .. ... ... ...... ...... ... ... ... ... ... ... ... ................................................................................................................................ . . . ... .. ... ... ... ... ... . ..... ...... .... ..... ... .. .. .. .. ... .. ... .. ............. .................. .................. .................. ............. ...... .......................... .......................... .......................... .......................... ........ . ...... ...... ...... ..... .. ..... ..... ..... ..... ...... ... ... .................... ...................... ...................... ...................... ......................
78 37 41 18 19 22 11 7 12 10 7 4 3 9 1
Copyright
c 2008
Canadian Mathemati al So iety
Crux Mathemati orum with Mathemati al Mayhem, Volume 34, Issue 7
407 However, those with a little mathemati al imagination an work out an interesting algebrai formula for the top number, based on the bottom row. Suppose that the bottom row is a, b, c, d, e. Then the next row is a + b, b + c, c + d, d + e. The following row is (a + b) + (b + c) , (b + c) + (c + d) , (c + d) + (d + e) ; that is, a + 2b + c , b + 2c + d , c + 2d + e . The next row is (a + 2b + c) + (b + 2c + d) , (b + 2c + d) + (c + 2d + e) ; that is, a + 3b + 3c + d , b + 3c + 3d + e . Has the pattern be ome lear? The top number is a + 4b + 6c + 4d + e . The binomial oeÆ ients are the key to the solution. And, we an see, that if we take a bottom row of n + 1 numbers a0 , a1 , . . . , an , we end up with a n P top number of ak n . k k=0
I will on lude with three hallenges to the reader. Consider an Add Up diagram with n + 1 ir les in the bottom row. 1. Take the bottom row to onsist entirely of 1's. Where in the diagrams do powers of 2 arise? 2. Take the bottom row to be the Fibona
i numbers {1, 1, 2, 3, 5, 8, . . . }. What is the top number? 3. The top number is always
n P
k=0
ak
n k
. Can you prove that?
Bru e Shawyer Department of Mathemati s and Statisti s Memorial University of Newfoundland St. John's, NL, A1C 5S7 Canada
[email protected]
408
THE OLYMPIAD CORNER No. 273 R.E. Woodrow
We start this number of the Corner with the problems of both rounds of the 2005/6 British Mathemati al Olympiad. Thanks go to Robert Morewood, 2006 Canadian Team Leader to the 47th IMO in Slovenia, for olle ting them for our use. 2005/6 BRITISH MATHEMATICAL OLYMPIAD Round 1
. Let n be an integer greater than 6. Prove that if n − 1 and n + 1 are both prime, then n2 (n2 + 16) is divisible by 720. Is the onverse true?
1
2. Adrian tea hes a lass of six pairs of twins. He wishes to set up teams for a quiz, but wants to avoid putting any pair of twins into the same team. Subje t to this ondition:
(i) In how many ways an he split them into two teams of six? (ii) In how many ways an he split them into three teams of four? . In the y li quadrilateral ABCD, the diagonal AC bise ts the angle DAB . The side AD is extended beyond D to a point E . Show that CE = CA if and only if DE = AB . 3
. The equilateral triangle ABC has sides of integer length N . The triangle is ompletely divided (by drawing lines parallel to the sides of the triangle) into equilateral triangular ells of side length 1. A ontinuous route is hosen, starting inside the ell with vertex A and always rossing from one ell to another through an edge shared by the two
ells. No ell is visited more than on e. Find, with proof, the greatest number of ells whi h an be visited. 4
. Let G be a onvex quadrilateral. Show that there is a point X in the plane of G with the property that every straight line through X divides G into two regions of equal area if and only if G is a parallelogram.
5
. Let T be a set of 2005 oplanar points with no three ollinear. Show that, for any of the 2005 points, the number of triangles it lies stri tly within, whose verti es are points in T , is even.
6
409 Round 2
. Find the minimum possible value of x2 + y2 given that x and y are real numbers satisfying xy(x2 − y 2 ) = x2 + y 2 and x 6= 0 .
1
. Let x and y be positive integers with no prime fa tors larger than 5. Find all su h x and y whi h satisfy
2
x2 − y 2 = 2k
for some non-negative integer k.
. Let ABC be a triangle with AC > AB . The point X lies on the side BA extended through A, and the point Y lies on the side CA in su h a way that BX = CA and CY = BA. The line XY meets the perpendi ular bise tor of side BC at P . Show that ∠BP C + ∠BAC = 180◦ . 3
. An exam onsisting of six questions is sat by 2006 hildren. Ea h question is marked either right or wrong. Any three hildren have right answers to at least ve of the six questions between them. Let N be the total number of right answers a hieved by all the hildren (that is, the total number of questions solved by hild 1 + the total solved by hild 2 + · · · + the total solved by hild 2006). Find the least possible value of N . 4
Next we give the problems of the Bulgarian National Olympiad, National Round, May 20{21, 2006. Thanks again go to Robert Morewood, 2006 Canadian Team Leader to the 47th IMO in Slovenia, for olle ting them for us. BULGARIAN NATIONAL OLYMPIAD National Round May 20{21, 2006
. (Aleksandar Ivanov) Consider the set A = {1, 2, 3, . . . , 2n }, n ≥ 2. Find the number of subsets B of A, su h that if the sum of two elements of A is a power of 2 then exa tly one of them belongs to B . 1
. (Oleg Mushkarov, Nikolai Nikolov) Let R+ be the set of all positive real numbers and f : R+ → R+ be a fun tion su h that p f (x + y) − f (x − y) = 4 f (x)f (y) . for all x > y > 0. (a) Prove that f (2x) = 4f (x) for all x ∈ R+ . (b) Find all su h fun tions. 2
410 3. (Aleksandar Ivanov, Emil Kolev) Consider the in nite sequen e of digits obtained by writing all the positive integers one after another in in reasing order. Find the least positive integer k su h that among the rst k digits of the above sequen e every two non-zero digits appear a di erent number of times.
. (Aleksandar Ivanov) Let p be a prime number su h that p2 is a divisor of 2 − 1. Prove that for any positive integer n, the integer (p − 1)(p! + 2n ) has at least three distin t prime divisors.
4
p−1
. (Emil Kolev) Let △ABC be su h that ∠BAC = 30◦ and ∠ABC = 45◦ . −→ Consider all pairs of points X and Y su h that X is on the ray AC , Y is on − − → the ray BC , and OX = BY , where O is the ir um enter of △ABC . Prove that the perpendi ular bise tors of the segments XY pass through a xed point. 5
6. (Nikolai Nikolov and Slavomir Dinev) Given a point O in the plane, nd all sets S in the same plane, ontaining at least two points, su h that for any point A ∈ S , A 6= O, the ir le with diameter OA is ontained in S .
Next we give the problems used in the Sele tion of the Indian Team 2006. Thanks again go to Robert Morewood, 2006 Canadian Team Leader to the 47th IMO in Slovenia, for olle ting them for our use. INDIAN MATHEMATICAL OLYMPIAD 2006 Team Sele tion Problems 1. Let n be a positive integer divisible by 4. Find the number of permutations σ of (1, 2, 3, . . . , n) whi h satisfy the ondition σ(j) + σ−1 (j) = n + 1 for all j ∈ {1, 2, 3, . . . , n}.
. (Short list, IMO 2005) Let ABCD be a parallelogram. A line l through the point A interse ts the rays BC and DC at X and Y respe tively. Let K and L be the entres of the ex ir les of triangles ABX and ADY , tou hing the sides BX and DY respe tively. Prove that the size of ∠KCL does not depend on the hoi e of the line l. 2
3. (Short list, IMO 2005) There are n markers, ea h with one side white and the other side bla k, aligned in a row with their white sides up. In ea h step (if possible) we pi k a marker with the white side up that is not an outermost marker, remove it, and turn over the losest marker to the left and the losest marker to the right of it. Prove that one an rea h a terminal state of exa tly two markers if and only if (n − 1) is not divisible by 3.
411 4. Let ABC be a triangle and let P be a point in the plane of ABC that is inside the region of the angle BAC but outside triangle ABC . (a) Prove that any two of the following statements imply the third:
−→
(i) The ir um entre of triangle P BC lies on the ray P A. − − → (ii) The ir um entre of triangle CP A lies on the ray P B . −→ (iii) The ir um entre of triangle AP B lies on the ray P C . (b) Prove that if the onditions in (a) hold, then the ir um entres of triangles BP C , CP A, and AP B lie on the ir um ir le of triangle ABC . 5. Let p be a prime number and let X be a nite set ontaining at least p elements. A olle tion of pairwise mutually disjoint p-element subsets of X is alled a p-family. (In parti ular, the empty olle tion is a p-family.) Let A (respe tively, B ) denote the number of p-families having an even (respe tively, odd) number of p-element subsets of X . Prove that A and B di er by a multiple of p.
. Let ABC be an equilateral triangle, and let D, E , and F be points on BC , and AB respe tively. Let ∠BAD = α, ∠CBE = β , and ∠ACF = γ . Prove that if α + β + γ ≥ 120◦ , then the union of the triangular regions BAD , CBE , and ACF overs the triangle ABC . 6
CA,
. Let ABC be a triangle with inradius r, ir umradius R, a = BC , b = AC , and c = AB . Prove that
7
R ≥ 2r 8
and with sides
64a2 b2 c2 4a2 − (b − c)2 4b2 − (c − a)2 4c2 − (a − b)2
!2
.
. The positive divisors d1 , d2 , . . . , dl of a positive integer n are ordered 1 = d1 < d2 < · · · < dl = m .
Suppose it is known that d27 + d215 = d216 . Find all possible values of d17 . . Let A1 , A2 , . . . , An be arithmeti progressions of integers, ea h of k terms, su h that any two of these arithmeti progressions have at least two
ommon elements. Suppose b of these arithmeti progressions have ommon di eren e d1 and the remaining arithmeti progressions have ommon di eren e d2 , where 0 < b < n. Prove that 9
b ≤ 2 k−
d2 gcd(d1 , d2 )
− 1.
. Find all triples (a, b, c) su h that a, b, and c are integers in the {2000, 2001, . . . , 3000} satisfying a2 + b2 = c2 and gcd(a, b, c) = 1.
10
set
412 . Let ujk be a real number for ea h j = 1, 2, 3 and ea h k = 1, 2; and let be an integer su h that
11
N
max
1≤k≤2
3 X
j=1
|ujk | ≤ N
.
Let M and l be positive integers su h that l2 < (M + 1)3 . Prove that there exist integers ξ1 , ξ2 , and ξ3 , not all zero, su h that max ξj ≤ M
1≤j≤3
and
X 3 MN u ξ jk k ≤ l j=1
,
for k = 1 ,2 .
12. Let A1 , A2 , . . . , An be subsets of a nite set S su h that |Aj | = 8 for ea h j . For a subset B of S , let F (B) = {j : 1 ≤ j ≤ n and Aj ⊂ B}. Suppose for ea h subset B of S , at least one of the following onditions holds:
(a)
|B| > 25,
(b)
F (B) = ∅, T Aj 6= ∅.
( )
j∈F (B)
Prove that A1 ∩ A2 ∩ · · · ∩ An 6= ∅. Another set of problems for your pleasure are those of the South Afri an Mathemati al Olympiad, Third Round 2004, Senior Division. Thanks go to Robert Morewood, Canadian Team Leader to the 47th IMO in Slovenia, for
olle ting them for our use. THE SOUTH AFRICAN MATHEMATICAL OLYMPIAD 2004 Third Round Senior Division (Grades 10-12)
. Let a = 1111 · · · 1111 and b = 1111 · · · 1111, where a has forty ones and has twelve ones. Determine the greatest ommon divisor of a and b.
1
b
2. Fifty points are hosen inside a onvex polygon having eighty sides su h that no three of the fty points lie on the same straight line. The polygon is ut into triangles su h that the verti es of the triangles are just the fty points and the eighty verti es of the polygon. How many triangles are there?
413 3. Find all real numbers x su h that x⌈x⌈x⌈x⌉⌉⌉ = means: \the least integer whi h is not less than x".
88.
The notation ⌈x⌉
. Let ABC be an isos eles triangle with CA = CB and ∠C > 60◦ . Let A1 and B1 be two points on AB su h that ∠A1 CB1 = ∠BAC . A ir le externally tangent to the ir um ir le of triangle A1 B1 C is tangent also to the rays CA and CB at points A2 and B2 respe tively. Prove A2 B2 = 2AB . 4
5. Let n ≥ 2 be an integer. Find the number of integers x with 0 ≤ x < n and su h that x2 leaves a remainder of 1 when divided by n. 6
. Let a1 , a2 , and a3 be distin t positive integers su h that a1 a2 a3
is a divisor of is a divisor of is a divisor of
a2 + a3 + a2 a3 , a3 + a1 + a3 a1 , a1 + a2 + a1 a2 .
and
Prove that a1 , a2 , and a3 annot all be prime. As a nal set of problems for this number, we give the problems of the two days of the 2006 Vietnamese Olympiad. Thanks go to Robert Morewood, Canadian Team Leader to the 47th IMO in Slovenia, for olle ting them for us. 2006 VIETNAMESE MATHEMATICAL OLYMPIAD Days 1 and 2 February 23-24, 2006
. Find all real solutions of the system of equations
1
p x2 − 2x + 6 · log3 (6 − y) = x , p y 2 − 2y + 6 · log3 (6 − z) = y , p z 2 − 2z + 6 · log3 (6 − x) = z .
2. Let ABCD be a given onvex quadrilateral. A point M moves on the line AB but does not oin ide with A or B . Let N be the se ond point of interse tion (distin t from M ) of the ir les (M AC) and (M BD), where (XY Z) denotes the ir le passing through the points X , Y , and Z . Prove that
(a)
N
moves on a xed ir le,
(b) The line M N passes through a xed point.
414 . A re tangular m × n board is given, where m and n are integers ea h greater than 3. At ea h step, one puts 4 marbles into 4 ells of the board (one marble per ell) so that these four ells form one of the following shapes:
3
....................................................... ... . . ... ..... ..... ... ... ... ........................................................................................................ ... .... .... ... ... ... ... ... .... .....................................................
................................................... ..... ... ... ... .... .... .... .... .... . . . . ..................................................................................................... ... ... .. ... . ... . ... ... ..... .. ... .................................................
............................. ... ... ..... ... ... ............................... ... .... ... .... . .... ... ..................................................... ... .... ... ... ... ... .............................. ... .... ... ... ... .... . ...........................
.......................... .... ... ... .... ... .... .............................. ... .... ... .... ... .... . . . ..................................................... ... .. .. ... ... .. . ... . ............................ ..... ... ... ... ... .... ... .........................
After a nite number of steps is it possible for ea h ell to ontain exa tly the same (positive) number of marbles if (a)
m = 2004
and n = 2006?
(b)
m = 2005
and n = 2006?
(At ea h step any of the four hosen ells may or may not have marbles in them.) . Consider the fun tion
4
f (x) = −x +
p (x + a)(x + b)
where a and b are distin t positive real numbers. Prove that for every real number s in the interval (0, 1), there exists a unique positive real number α su h that s a + bs 1/s f (α) = . 2
. Find all polynomials P (x) with real oeÆ ients satisfying
5
P (x2 ) + x 3P (x) + P (−x) = P (x)2 + 2x2 ,
for all real numbers x.
. A set of integers T is alled sum-free if for every two (not ne essarily distin t) elements u and v in T , their sum u + v does not belong to T . Prove that 6
(a) a sum-free subset of S = {1, 2, . . . , 2006} has at most 1003 elements,
(b) any set S onsisting of 2006 positive integers has a sum-free subset
onsisting of 669 elements. Now we turn to the le of readers' solutions to problems given in the De ember 2007 number of the Corner beginning with a problem of the 2004 Chinese Mathemati al Olympiad, given at [2007 : 470℄.
415 . Given any positive integer n ≥ 2, let ai (i = 1, 2, . . . , n) be positive n P 1 integers satisfying a1 < a2 < · · · < an and ≤ 1. Prove that for any a i=1 i real number x, 5
n X
i=1
1 a2i + x2
!2
1 1 · 2 a1 (a1 − 1) + x2
≤
.
Solution by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain. We assume rst that x2 ≥ a1 (a1 − 1). Then x is non-zero and from 1 1 ≤ (ai −|x|)2 ≥ 0, we have a2i +x2 ≥ 2ai |x| for ea h i. Thus, 2 , a i + x2 2ai |x| and we have n X
i=1
1 a2i + x2
!2
≤
n X
i=1
1 2ai |x|
!2
=
1 4x2
n X 1
k=1
ai
!2
1 1 1 ≤ · . 2 4x 2 a1 (a1 − 1) + x2 − u = √1a1 , √1a2 , . . . , √1an and Now we suppose that x2 < a1 (a1 − 1). Let → √ √ √ a a an → − v = a2 +x1 2 , a2 +x2 2 , . . . , a2 +x . We apply the Cau hy-S hwarz Inequality 2 n 1 2 → − → − to u and v to obtain ≤
n X
i=1
1 2 ai + x2
!2
≤
n X 1 a i=1 i
!
n X
i=1
ai a2i + x2
2
!
≤
n X
i=1
ai a2i + x2
2
.
1 For ea h i we have a2i ≥ 16 and a2i + x2 2 ≥ a2i + x2 + 14 2 − a2i . Furthermore, from a1 < a2 < · · · < an , it follows that ai+1 ≥ ai + 1 and 1 1 ai+1 − ≥ ai + . Therefore, 2 2
ai a2i + x2
2
≤ =
= ≤
ai a2i + x2 +
1 2 4
− a2i
1 2ai 2 1 1 2 2 2 2 ai − + x a + + x i 2 2 1 2
ai −
1 2
ai −
1 1 2 2
1 1 2 2
+ x2 + x2
− −
ai +
1 1 2 2
+ x2
!
1
ai+1 −
1 2 2
+ x2
!
.
416 Finally, n X
i=1
ai a2i + x2
2
≤ ≤
n 1X 2 i=1
1
1 1 2
− 2 ai − 2 + x2 ai+1 − 12 + x2 1 1 1 1 ≤ · · , 2 1 2 2 a1 (a1 − 1) + x2 a1 − 2 + x2
!
and the proof is omplete.
Next we look at solutions to problems of the Singapore Mathemati al Olympiad 2004, Open Se tion, Spe ial Round, given at [2007 : 470℄. 3. Let AD be the ommon hord of two ir les Γ1 and Γ2 . A line through D interse ts Γ1 at B and Γ2 at C . Let E be a point on the segment AD di erent from A and D. The line CE interse ts Γ1 at P and Q. The line BE interse ts Γ2 at M and N . (i) Prove that P , Q, M , and N lie on the ir umferen e of a ir le Γ3 . (ii) If the entre of Γ3 is O, prove that OD is perpendi ular to BC . Solution by Mi hel Bataille, Rouen, Fran e. (i) We have EP · EQ = EA · ED, sin e the hords P Q and AD of Γ1 interse t at E . Similarly, we obtain EM · EN = EA · ED . It follows that EP · EQ = EM · EN and so P , Q, M , and N are on y li . (ii) Sin e M N and CD interse t at B , we have BD · BC = BM · BN . But BM · BN is the power of B with respe t to Γ3 , so BM · BN = BO2 − ρ2 where ρ is the radius of Γ3 . Similarly, CD · CB = CP · CQ = CO 2 − ρ2 ,
.................................................. ............ ......... ........ ....... ....... ...... ...... ..... . . . . .... .. . . .... . ... .... . . ... ... . . ... .. . ... . ... .. .. ... .. . ....... . .. . ....... . ... . ....... .. ... ....... . . ....... ... ... ....... .. ... ... ....... .. . .. ......... ......................................... . . .......... ............... . . . . . . . . . . . . . . . . . . . . ....... . . . ...... .... ............. ..... .. ..... ... ........ ....... .. .. .... ....... ..... .. .......... ... ....... .... ... ............................................... .... ................... ... . . ... . . . . . . . . . . . . . . . . . . . . . . ............. ....... .... ........ . . . .. . . . . . . . . . . . . . . . . . ....... . . .... ....... ........ .... ... .......... .... ...... ..... ... ......... . .. .. ................... ........ ... ...... ... . . .. ....... .... .. .......... ... .............. .. ......... ....... .. ............ ........................................ ................... ... ..... ................................................... .. ... . . . . . . . . . . . ...... .... ...... ... ........... ... . . . . . . . . ... . ....... ... .. .. ... ...... . . . .. . . . . . . . . . . . . ....... .. ... ....... .. .. ....... . . . . . . . . . . . ... . . . . . . ....... ... .. .. . ...... . . . . . . . . . ... . . . . . . . . ....... .. ... ... .. . ............ . . .. .......................................................................................................................................................................................................................................................................... . ........ ....... ...... ............. .... .............. .......... ....... ....... ....................................... ..... . . . . . ........ . ....... ....... ........... ....... ...........................................
O
P
A
q
N
B
M
E
Q
D
q
C
and therefore BO 2 − CO 2
From BO2 − CO2 BC .
= BD · BC − CD · CB = BC(BD − DC) = (BD + DC)(BD − DC) = BD 2 − DC 2 .
= BD 2 − DC 2 ,
it follows that OD is perpendi ular to
. If 0 < x1 , x2 , . . . , xn ≤ 1, where n ≥ 1, show that
4
x1 x2 xn + + ··· + ≤ 1. 1 + (n − 1)x1 1 + (n − 1)x2 1 + (n − 1)xn
417 Solved by Mi hel Bataille, Rouen, Fran e; and Pavlos Maragoudakis, Pireas, Gree e. We give the write-up of Maragoudakis. It is enough to prove that 1 + (nx− 1)x ≤ n1 for every x ∈ (0, 1]. However, this is equivalent to nx ≤ 1 + (n − 1)x, whi h is equivalent to x ≤ 1. Next we turn to solutions to problems of the 18th Nordi Mathemati al Contest 2004 given at [2007: 471℄. 2. Let f1 = 0, f2 = 1, and fn+2 = fn+1 + fn for n = 1, 2, . . . , be the sequen e of Fibona
i numbers. Show that there exists a stri tly in reasing in nite arithmeti sequen e of integers whi h has no numbers in ommon with the Fibona
i sequen e. Solution by Salem Maliki , student, Sarajevo College, Sarajevo, Bosnia and Herzegovina. We onsider the Fibona
i sequen e modulo 8. The rst few terms (mod 8) are: 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, . . . | {z }
,
so we see that the two-term subsequen e "1, 1" o
urs again, hen e the Fibona
i sequen e is y li modulo 8. We on lude that the Fibona
i sequen e does not ontain any terms
ongruent to 4 or 6 (mod 8). Thus, the arithmeti sequen e A, with A1 = 4 (or 6) and ommon di eren e An+1 − An = 8, has no numbers in ommon with the Fibona
i sequen e. . Let a, b, c, and R be the side lengths and the ir umradius of a triangle. Show that 1 1 1 1 . + + ≥ 2
4
ab
bc
ca
R
Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; Mi hel Bataille, Rouen, Fran e; Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain; Salem Maliki , student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; Pavlos Maragoudakis, Pireas, Gree e; and D.J. Smeenk, Zaltbommel, the Netherlands. We rst give the write-up of Amengual Covas. Let s and r be the semiperimeter and the inradius of the given triangle. We have 1 1 1 a+b+c 2s 1 1 + + = = = ≥ ab bc ca abc 4Rrs 2Rr R2
,
418 sin e R ≥ 2r (Euler's Inequality). Equality o
urs only if and only if the triangle is equilateral. We next give the solution by Maliki . Let s, r, and P be the semiperimeter, the inradius, and the area of the triangle. It is well known that 1 R2
= = =
4P 2 16P 2 = abc (abc)2 16(s(s − a)(s − b)(s − c) (abc)2 (a + b + c)(b + c − a)(c + a − b)(a + b − c) (abc)2
,
so it is enough to prove 1 ab
+
1 bc
+
1 ca
= ≥
a+b+c abc (a + b + c)(b + c − a)(c + a − b)(a + b − c) (abc)2
,
whi h is equivalent to (a + b − c)(c + a − b)(b + c − a) ≤ abc .
There are several ways to prove this last inequality. One may, for instan e, expand the left-hand side and redu e the inequality to S hur's Inequality. We give another proof, where we use the fa t √ b that a + b − c > 0, b + c − a > 0, c + a − b > 0, and ab ≤ a + : 2 (a + b − c)(b + c − a)(a + c − b) Y p = (a + b − c)(b + c − a)
y li
=
Y (a + b − c) + (b + c − a)
y li
2
=
Y
b = abc .
y li
Equality holds if and only if a = b = c, in whi h ase the triangle is equilateral. Next we look at readers' solutions to problems proposed, but not used, at the 2004 IMO in Athens given at [2007: 471{475℄.
419 . Let a, b, c > 0 and ab + bc + ca = 1. Prove the inequality
A3
r 3
1 + 6b + a
r 3
1 + 6c + b
r 3
1 1 + 6a ≤ c abc
.
Solution by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain. From the identity (x + y + z)2 = x2 + y2 + z 2 + 2(xy + yz + zx) and the inequality x2 + y2 + z 2 ≥ xy + yz + zx, we get (x + y + z)2 ≥ 3(xy + yz + zx) .
Setting x = ab, y = bc and c = ca into the pre eding inequality yields 1 = (ab + bc + ca)2 ≥ 3(ab2 c + bc2 a + ca2 b) = 3abc(a + b + c)
from whi h we have
3(a + b + c) ≤
1 abc
.
(1)
On the other hand, applying the AM{GM Inequality, we have 1 = ab + bc + ca ≥ 3
hen e,
(abc)2 ≤
1 27
p 3 (abc)2 ,
.
(2)
Using power mean inequalities, and taking into a
ount that ab+bc+ca = 1, (1), and (2), we obtain r 3
1 + 6b + a 3 = √ 3 3
r 3
r 3
1 + 6c + b
r
1 + 6a c
≤
1 + 6(a + b + c) abc
≤
3
r 3 3 1 1 1 + + + 6(a + b + c) √ 3 b c 3 a r r 3 3 3 1 1 3 = 3 ≤ . √ 3 abc abc abc 3
Equality holds when a = b = c = 13 . . The fun tion ψ from the set N of positive integers into itself is de ned by the equality n X ψ(n) = (k, n) , for n ∈ N,
N2
k=1
where (k, n) denotes the greatest ommon divisor of k and n. (a) Prove that ψ(mn) = ψ(m)ψ(n) whenever m, n ∈ N are relatively prime. (b) Prove that, for ea h a ∈ N, the equation ψ(x) = ax has a solution.
( ) Find all a ∈ N su h that the equation ψ(x) = ax has a unique solution.
420 Solution by Daniel Tsai, student, Taipei Ameri an S hool, Taipei, Taiwan. First we prove a lemma. P Lemma 1. ψ(n) = )d for n ∈ N, where the sum is over the positive φ( n d d|n
divisors d of n and φ is the Euler Totient Fun tion (that is, φ(m) is the number of positive integers relatively prime to m and not ex eeding m). n P Proof. Sin e ea h term of ψ(n) = (k, n) is a positive divisor of n, k=1
n X
(k, n) =
k=1
X d|n
|{1 ≤ k ≤ n : (k, n) = d}| · d =
X n φ d d d|n
and the lemma is proved. Let m, n ∈ N be relatively prime, then by the Lemma, ψ(mn)
= = = =
X X mn mn φ d = φ de d ef d|mn e|m f |n X X m n φ e · φ f e f e|m f |n X m X n φ e φ f e f e|m f |n X
ψ(m)ψ(n) .
This proves the statement in part (a). Lemma 2. Let p be a prime number and α ≥ 0 be an integer. Then
ψ(pα ) = (α + 1)pα − αpα−1 =
Proof. By Lemma 1, we have ψ(pα ) =
X
d|pα
=
α−1 X
φ
pα d
d =
α X
=
k=0
=
p
α + 1 pα
φ(pα−k )pk
k=0
φ(pα−k )pk + pα =
k=0
α−1 X
p−1
α−1 X k=0
(pα−k − pα−k−1 )pk + pα
(pα − pα−1 ) + pα = α(pα − pα−1 ) + pα
(α + 1)pα − αpα−1 =
and the proof is omplete.
p−1 p
α + 1 pα .
421 . Let n ∈ N be odd, then ψ(n) is odd. Proof. In view of part (a), it suÆ es to prove that if p is an odd prime number and α ≥ 0 is an integer, then ψ(pα ) is odd. Clearly ψ(p0 ) = 1 is odd. If α ≥ 1, then by Lemma 2, ψ(pα ) = (α + 1)pα − αpα−1 is the di eren e of two integers of opposite parity, and therefore it is odd. For any integer α ≥ 0, we have by Lemma 2 Corollary
α
ψ(2 ) =
α + 1 2α . 2
(1)
For ea h a ∈ N take α = 2(a − 1) and x = 2α , then the above equation yields ψ(x) = ax. This proves the statement in part (b). To settle part ( ), we will prove that the only integers a ∈ N su h that ψ(x) = ax has a unique solution are the non-negative powers of 2. For any integer α ≥ 0, we have by Lemma 2 α
ψ(3 ) =
2α 3
+ 1 3α .
(2)
Let a ∈ N be of the form 2m n, where m ≥ 0 is an integer and n > 1 is an odd number. By equation (1), the number x = 22(2 −1) is a solution to ψ(x) = 2m x and by equation (2), the number x = 3 (n−1) is a solution to ψ(x) = nx. Sin e 22(2 −1) and 3 (n−1) are relatively prime, it follows from part (a) that x = 22(2 −1) 3 (n−1) is a solution to ψ(x) = ax. However, by equation (1), the number x = 22(a−1) is also a solution to ψ(x) = ax that is di erent from the solution just obtained. Finally, let a = 2k , where k ≥ 0 is an integer, and let x = 2m n be a solution to ψ(x) = ax, where m ≥ 0 is an integer and n ∈ N is odd. By + 1)2m ψ(n) = 2k+m n, part (a) and the equation in (1), we have ψ(x) = ( m 2 hen e (m + 2)ψ(n) = 2k+1 n . (3) Now 2k+1 divides the left-hand side of (3), so by the Corollary, it divides m + 2 and m + 2 ≥ 2k+1 . If n > 1, then ψ(n) ≥ n + 1 > 1, whi h implies that the left-hand side of (3) is greater than the right-hand side of (3), a ontradi tion! Hen e, n = 1 and by (3) we have m = 2(2k − 1). m
3 2
m
m
3 2
3 2
That ompletes the material for this number of the Corner. There are lots of opportunities to send in your ni e solutions and generalizations to problems.
422
BOOK REVIEWS John Grant M Loughlin
Impossible? Surprising Solutions to Counterintuitive Conundrums By Julian Havil, Prin eton University Press, Prin eton & Oxford, 2008 ISBN 978-0-691-13131-3, loth, 264 pages, US$27.95 Reviewed by Edward Barbeau, University of Toronto, Toronto, ON Julian Havil, a master at Win hester College in England, is a type of mathemati s tea her that is all too rare, a well-read onnoisseur of his subje t who is eager to explore its byways and share his dis overies with his pupils and the publi at large. (Another su h was F.J. Budden, whose ne book Fas ination of Groups, published by Cambridge, dates ba k to 1972.) This is Havil's third book, after Nonplussed! Mathemati al Proof of Implausible Ideas (2007) and Gamma: Exploring Euler's Constant (2003), both published by Prin eton. There is a lot to savour, and Havil has both the understanding and te hni al pro ien y to onvey it to his readers. This book annot be read asually. It is espe ially suitable for mathemati s undergraduates, but only for the most apable se ondary students and tea hers. Professional mathemati ians will be familiar with many of the topi s, but will undoubtedly nd something they have not yet en ountered, parti ularly the puzzles that have re ently ome onstream. While most of the topi s date from the last entury, he does rea h ba k in time for some of the dis ussion. The in nitely long \trumpet" of Torri elli, with its in nite volume and nite surfa e area provides the o
asion for a dis ussion of the divergen e of the harmoni series and the onvergen e and evaluation by Euler of the sum of the series of square re ipro als. The study of the length of a run of heads in a sequen e of oin tosses in ludes an a
ount of the work of DeMoivre, an early investigator of the problem. A dis ussion of the frequen y of poker hands is extended to in lude the e e t of wild ards. There are three enjoyable hapters for those with an aÆnity for numbers, espe ially large ones. A treatment of the o
urren e of sequen es of digits in powers of 2 leads smoothly into Benford's Law, to wit that in a \natural" set of numeri al data, the numbers start with the smaller digits more frequently than the larger. The hapter titled Goodstein Sequen es pa ks a surprise. We start the sequen e with a number written in the omplete base 2 representation; we start with the standard base 2 representation and (re ursively) write all exponents that o
ur in base 2. Thus, 2+1
2136 = 22
+2+1
2
+ 22
+2
2
+ 22
+ 22+1 .
The Goodstein sequen e starting with n is de ned initially for r = 2 by G2 (n) = n (written in omplete base 2). For r ≥ 3, Gr (n) is obtained by taking Gr−1 (n), a number written in omplete base r − 1, hanging all the
423 o
urren es of r − 1 to r, subtra ting 1 and then adjusting the result to get a omplete base r representation. Thus, 3+1
G3 (2136) = 33
+3+1
3
+ 33
+3
+ 33 + 2 · 33 + 2 · 32 + 2 · 3 + 2 . 3
With this ontinual bumping up of the base, one might expe t the terms of the sequen e to keep in reasing rapidly, whi h indeed they do to begin with. But eventually, they rash to 0 in nite time. This result turns out to be unprovable in ordinary arithmeti , and, indeed, Goodstein established it in 1944 by appealing to the well-ordering of the trans nite ordinals. Geometry has a small but interesting ni he. A brief look at Eu lid's axioms leads to a dis ussion of Cantor ardinality and situations where the whole need not be greater than the part, su h as the equipollen e of the
losed unit interval and n−dimensional Eu lidean spa e. A hapter treats in detail the solution of the Kakeya problem where a needle an be reversed in dire tion within an arbitrarily small area. The parti ularly ne last hapter
onveys the essen e of the argument behind the Bana h-Tarski Paradox. Other paradoxes make an appearan e. Within a dis ussion of omplex numbers, we learn about the reality of ii and determine logarithms of negative numbers. Simpson's paradox is analyzed, although the insight o ered by the observation that a mediant of two positive rationals (obtained by adding the numerators and the denominators) depends on their parti ular representations and an lie anywhere in the interval between them, would have helped. Probably less familiar to the reader would be Braess' paradox that the onstru tion of an additional road in a town may a tually worsen the
ongestion. Problems of re ent notoriety are treated in detail: ar and goats (Monty Hall); pla ement of oloured hats; intera tions between two individuals to identify a number pair; and the han e of getting an elevator in your dire tion. These raise deli ate issues in logi , probability and oding theory that are ompetently handled. The reader will want to he k out two ard tri ks that exemplify the Kruskal and Gilbreath prin iples, the latter based on the stru ture preserved in a de k of ards by an imperfe t rie shue. The author analyzes the phenomena in some depth without being tedious. The book is lightened by ane dotes and histori al asides; it is well referen ed, so that the reader an go to the literature. An interesting feature of the book is the set of illustrations that a
ompany ea h of the eighteen
hapter headings. There are seventeen di erent symmetry groups for wallpaper patterns; the basi pattern is presented with Chapter 1 and ea h of the remaining hapters gives a fundamental region for one of these groups. While the author dedi ates the book to Martin Gardner, the treatment is more overtly mathemati al than what Gardner would provide. A better
omparator is Sherman Stein's How the Other Half Thinks: Adventures in Mathemati al Reasoning (M Graw-Hill, 2001) or Ross Honsberger's essays. In any ase, this book makes a pleasant and absorbing read.
424
Twin Problems on Non-Periodi Fun tions Eugen J. Ionas u
1
Introduction
Two very similar problems were proposed in the Ameri an Mathemati al Monthly by P.P. Dalyay. . ([3℄) Let f and g be non onstant, ontinuous periodi fun R into R. Is it possible that the fun tion h on R given by xg(x) is periodi ?
Problem 11111
tions mapping h(x) = f
Problem 11174. ([4℄) Let f and g be non onstant, ontinuous fun tions mapping R into R and satisfying the following onditions: 1. f is periodi .
g(xn ) 2. There is a sequen e hxn in≥1 su h that n→∞ lim xn = ∞ and lim = n→∞ xn ∞. 3. f ◦ g is not onstant on R. Determine whether h = f ◦ g an be periodi .
The solutions to Problem 11111 and Problem 11174 appeared in [5℄ and [8℄. In this note we are going to onsider a new question whi h is similar to but more general than ea h of these problems. The proofs here are based on a parti ular ase of the well-known Stolz-Cesaro Lemma and on the fa t that a ontinuous periodi fun tion on R is uniformly ontinuous. The use of the latter idea is not new as it was used in the published solutions of these problems. On the other hand, the use of the Stolz-Cesaro Lemma is a good example of where an old tool from analysis appears unexpe tedly (see [1℄, [6℄, [7℄, and [10℄). L'Hospital's rule, whi h is well known to al ulus students, is its \di erentiable" ounterpart. For ea h version of L'Hospital's rule, there is an analogous version of the Stolz-Cesaro Lemma. For example, one version of L'Hospital's Rule states that if f and g are two di erentiable fun tions on (a, ∞) su h that f ′ (x) f (x) lim f (x) = lim g(x) = ∞ and lim ′ = L, then lim = L. x→∞ x→∞ x→∞ g (x) x→∞ g(x) The Stolz-Cesaro analog of this version of L'Hospital's Rule is: For two y − yn sequen es hxn in≥1 and hyn in≥1 , if n→∞ lim xn = ∞ and lim n+1 = L, n→∞ x −x y then n→∞ lim n x
n+1
=L
n
Copyright
c 2008
Canadian Mathemati al So iety
Crux Mathemati orum with Mathemati al Mayhem, Volume 34, Issue 7
n
425 Indeed, one may think of the derivative y′ (a) as a spe ial quotient y(a + δ) − y(a) y ′ (a) = , where δ takes its smallest possible in nitesimal δ value. In the dis rete ase we write yn = y(n) for n ∈ N, and we let δ take its smallest value, namely δ = 1, to obtain the dis rete derivative y(n + 1) − y(n) y − yn y ′ (n) = = yn+1 −yn . Thus, lim n+1 = L indi ates n→∞ xn+1 − xn 1 (in this interpretation) that a ratio of dis rete derivatives tends to L. The following problem may be solved by applying both the Stolz-Cesaro Lemma and L'Hospital's Rule. We in lude it for the interested reader. Let x0 ∈ (0, π) and de ne the sequen e√hxn in≥0 by the re ursion √ xn+1 = sin xn , n ≥ 0. Show that lim xn n = 3. n→∞
Problem
2
1/xn Solution The required limit is equivalent to n→∞ lim n Cesaro Lemma, it suÆ es to show that 1 x2 n+1
lim
−
1 x2 n
(n + 1) − n
n→∞
or equivalently lim
n→∞
x2n − sin2 xn x2n sin2 xn
=
!
1 3
=
=
1 . 3
By the Stolz-
,
1 3
.
Sin e 0 < sin x < x whenever x ∈ (0, π), the sequen e hxn i is de reasing and bounded below by 0 and so it onverges to a limit ℓ ∈ [0, π). This limit ℓ satis es sin ℓ = ℓ, hen e ℓ = 0. 1 x2 − sin2 x Thus, it suÆ es to show that x→0 = , whi h an be done lim 3 x2 sin2 x by applying L'Hospital's Rule: lim
x→0
x2 − sin2 x x2
2
sin x
= = = =
lim
(x − sin x)(x + sin x)
x2 sin2 x x − sin x sin x x 2 lim lim 1 + lim x→0 x→0 x→0 sin x x3 x 1 − cos x lim · (1 + 1) · 12 x→0 3x2 sin x 1 2 lim = . 3 x→0 6x x→0
We will use a variant of the Stolz-Cesaro Lemma to prove our main theorem, where we weaken the onditions in Problem 11174 but obtain the same on lusion.
426 Let f and g be non onstant, ontinuous fun tions from R into R that satisfy the following onditions: Theorem 1
(i) The fun tion f is periodi . (ii) There exist sequen es hxn in≥1 and hyn in≥1 su h that inf |xn − yn | > 0 n
and
g(xn ) − g(yn ) = ∞. lim n→∞ x −y n
n
Then the fun tion h = f ◦ g is not periodi .
2
Some Facts from Real Analysis
The variant of the Stolz-Cesaro Lemma that we will use is stated next. Let han i and hbn i be two sequen es su h that hbn i is in reasing and a a − an lim bn = ∞. If lim n = ∞, then lim sup n+1 = ∞. n→∞ n→∞ b b −b
Lemma 1
n
n→∞
n+1
n
For ompleteness we in lude a proof of Lemma 1 along lassi al lines. an Assume to the ontrary that γ = lim sup abn+1 − < ∞. Then there n+1 − bn n→∞ exists n0 ∈ N su h that for n ≥ n0 we have
or equivalently
an+1 − an ≤ γ + 1, bn+1 − bn an+1 − an ≤ (bn+1 − bn )(γ + 1) ,
(1) for n ≥ n0 . Adding up the inequalities in (1) for n = k, k + 1, . . . , l, where n0 ≤ k < l, we obtain al+1 − ak ≤ (bl+1 − bk )(γ + 1) .
For suÆ iently large l the term bl+1 is positive and we may divide the last inequality by bl+1 and then let l → ∞. Using the hypothesis we then obtain ∞ ≤ γ + 1, a ontradi tion. We now re all that a (real- or omplex-valued) fun tion with domain D ⊂ R is uniformly ontinuous on D if for ea h ǫ > 0 there exists a δ > 0 su h that |f (x) − f (y)| < ǫ whenever x, y ∈ D and |x − y| < δ . The following basi fa t about ontinuous fun tions on a losed interval is all we need, though it an be generalized onsiderably (see [9℄, Theorem 4.19). Every ontinuous, real-valued fun tion whose domain is a losed interval D = [a, b] is uniformly ontinuous on D.
Theorem 2
As an easy onsequen e of this theorem, we have
427 . Every ontinuous, periodi fun tion f on R is uniformly ontin-
Corollary 1
uous.
This an be seen by applying Theorem 2 to the restri tion of f to the
losed interval [0, 2T ], where T > 0 is a period of f . That is, given ǫ > 0, by Theorem 2 there is a δ su h |f (x) − f (y)| < ǫ whenever x, y ∈ [0, 2T ] and |x − y| < δ . Thus, if x, y ∈ R and |x − y| < δ , then there exist integers n and m su h that both x1 = x − nT and y1 = y − mT are in [0, 2T ] and |x1 − y1 | < δ , so that |f (x) − f (y)| = |f (x1 ) − f (y1 )| < ǫ. The idea of our proofs is to show that the fun tion h is not uniformly
ontinuous. It then follows from Corollary 1 that h is not periodi . Let us see how Problem 11111 an be solved using Theorem 1. Sin e g is not onstant there exist a and b su h that g(a) − g(b) 6= 0. Let T > 0 be a period of g, and de ne the sequen es hxn in≥1 and hyn in≥1 by xn = a + nT and yn = b + nT . Then |xn − yn | = |a − b| > 0 and lim
n→∞
|xn g(xn ) − yn g(yn )| |xn − yn |
= |a − b|−1 lim ag(a) − bg(b) + nT g(a) − g(b) = ∞ , n→∞
whi h says that f and the fun tion g1 on R given by g1 (x) = xg(x) both satisfy the onditions (i) and (ii) in Theorem 1, hen e, h = f ◦ g1 is not periodi . We have solved Problem 11111. To show that Problem 11174 an be solved using Theorem 1, we need the weaker version of the Stolz-Cesaro Lemma given in Lemma 1. Let us assume that f , g, and hxn in≥1 satisfy the three onditions in Problem 11174. There is a subsequen e hxn ik≥1 of hxn in≥1 , su h that g(xn ) = ∞ or xn − xn ≥ 1 for all k, and for whi h either lim x k
k
k+1
k
k→∞
g(xn k ) = −∞. xn k
nk
Without loss of generality we may suppose the former, be ause the latter ase follows from the former ase by repla ing g with −g and f with f1 (x) = f (−x), x ∈ R. By Lemma 1 we have lim
k→∞
lim sup k→∞
g(xnk+1 ) − g(xnk ) xnk+1 − xnk
= ∞,
whi h proves the existen e of the two sequen es in the hypothesis (ii) of Theorem 1. Hen e, Theorem 1 an be applied to f and g and we dedu e that h = f ◦ g is not periodi . We have solved Problem 11174.
3
Proof of Theorem 1
Let f and g satisfy the hypotheses of Theorem 1. Sin e g is ontinuous and satis es ondition (ii), the interval In = g [xn , yn ] (or In = g [yn , xn ] )
428 has length greater than the period T of f for suÆ iently large n. Hen e, f and h = f ◦ g have the same range. Sin e f is not onstant, h is not
onstant. Therefore, there exist α and β su h that f g(α) 6= f g(β) , and we let ǫ0 = |f g(α) − f g(β) | > 0. As we said in the introdu tion, the key idea is to prove that h is not uniformly ontinuous. In fa t, we will show that the de nition of uniform ontinuity is not satis ed for this ǫ0 . We x n ∈ N large enough so that |g(xn ) − g(yn )| > 2T , and we denote by ♯ g(α) the number of integers k for whi h g(α) + kT is in In . It is then easy to see that |g(xn ) − g(yn )| ♯ g(α) > − 1 > 1. T Similarly, we denote by ♯ g(β) the number of integers k for whi h g(β)+kT is in In . Similarly, we have ♯(g(β)) > 1. It is lear that the values g(α) + kT , k ∈ Z, interla e with those of g(β) + kT , k ∈ Z. Using again the fa t that g is ontinuous and by repeated
appli ation of the Intermediate Value Theorem, we an nd two nite sequen es huk i and hvk i in the interval [xn , yn ] (or [yn , xn ]) both in reasing and interla ing and su h that g(uk ) = g(α) + lk T and g(bk ) = g(β) + sk T , with lk , sk ∈ Z. The number of intervals of the form [uk , vk ) (or [vk , uk )) is at least n o M = min 2 ♯ (g(α) − 1 , 2 ♯ (g(β) − 1 ≥ 2.
These intervals form a partition of a subinterval of Jn = [xn , yn ] (or of Jn = [yn , xn ]) of length |xn − yn |. Then one of these intervals has length at − yn | most |xn M . We denote su h an interval by [ζn , ηn ] and noti e that |ζn − ηn | ≤
|xn − yn | M
<
|xn − yn | |g(xn ) − g(yn )| 2 −4 T
1 → 0 as n → ∞ , (2) 2 |g(xn ) − g(yn )| 4 − T |xn − yn | |xn − yn | and f g(ζn ) − f g(ηn ) = ǫ0 . Given δ > 0, we may hoose n so large that |ζn − ηn | < δ . This an be done be ause of (2). For su h an n we still have |h(ζn ) − h(ηn )| ≥ ǫ0 , whi h proves that h is not uniformly ontinuous. =
4
Conclusion
We would like to leave the reader with a natural question: Can Theorem 1 be generalized to almost periodi fun tions? There are various on epts of almost periodi ity, but here we will only give Bohr's de nition:
429 A ontinuous fun tion F : R → R is said to be almost periodi if for all ǫ > 0, there is an L > 0 su h that every interval of length L ontains an ǫ-period, that is, a number T su h that |F (x + T ) − F (x)| < ǫ for all x ∈ R. What is interesting and related to the question above is the fa t that every almost periodi fun tion is uniformly ontinuous (see [2℄).
Acknowledgment We thank Professor Albert VanCleave, who gave us helpful suggestions after reading an earlier version of this arti le. Referen es
[1℄ D.M. Batinetu-Giurgiu, Siruri (Sequen es), Editura Albatros, Bu uresti 1979. [2℄ C. Corduneanu, Almost Periodi Fun tions. New York: Wiley Inters ien e, 1961. [3℄ P.P. Dalyay, Problem 11111, Amer. Math. Monthly, Vol. 111, 2004, p. 822.
no. 9,
[4℄ P.P. Dalyay, Problem 11174, Amer. Math. Monthly, Vol. 112, 2005, p. 749.
no. 8,
[5℄ GCHQ Problem Solving Group, Amer. Math. Monthly, Vol. 113, no. 5, 2006, p. 467. [6℄ A. Mannino, Some lassi Stolz-Cesaro theorems. (Italian), Atti A
ad. S i. Lett. Arti Palermo Parte I (4) 37 (1977/78), pp. 409{414 (1980). [7℄ G. Nagy, The Stolz-Cesaro Theorem http://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf
[8℄ National Se urity Agen y Problems Group, Amer. Math. Monthly, Vol. 114, no. 9, 2007, p. 836. [9℄ W. Rudin, Prin iples of Mathemati al Analysis, Third Edition, M GrawHill, In . 1964. [10℄ Gh. Siret hi, The Toeplitz Theorem and some of its onsequen es (Romanian), Gaz. Mat. (Bu harest) 90 (1985), no. 3, pp. 65{70. Eugen J. Ionas u Department of Mathemati s Columbus State University 4225 University Avenue Columbus, GA, 31907 USA ionascu
[email protected]
430
PROBLEMS
Solutions to problems in this issue should arrive no later than 1 May 2009. An asterisk (⋆) after a number indi ates that a problem was proposed without a solution. 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 in the language of the primary featured solution. The editor thanks Jean-Mar Terrier of the University of Montreal for translations of the problems. . Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. The verti es of quadrilateral ABCD lie on a ir le. Let K , L, M , and N be the in entres of △ABC , △BCD , △CDA, and △DAB , respe tively. Show that quadrilateral KLM N is a re tangle. 3376
. Proposed by Toshio Seimiya, Kawasaki, Japan. Let ABC be a triangle with ∠B = 2∠C . The interior bise tor of ∠BAC meets BC at D . Let M and N be the mid-points of AC and BD , respe tively. Suppose that A, M , D, and N are on y li . Prove that ∠BAC = 72◦ . 3377
. Proposed by Mihaly Ben ze, Brasov, Romania. Let x, y, and z be positive real numbers. Prove that
3378
X
y li
X xy x ≤ xy + x + y 2x + z
y li
.
. Proposed by Mihaly Ben ze, Brasov, Romania. Let a1 , a2 , . . . , an be positive real numbers. Prove that
3379
n X
i=1
ai ≥ 1, ai + (n − 1)ai+1
where the subs ripts are taken modulo n. . Proposed by Mihaly Ben ze, Brasov, Romania. Let a, b, c, x, y, and z be real numbers. Show that
3380
(a2 + x2 )(b2 + y 2 ) (b2 + y 2 )(c2 + z 2 ) (c2 + z 2 )(a2 + x2 ) + + (c2 + z 2 )(a − b)2 (a2 + x2 )(b − c)2 (b2 + y 2 )(c − a)2 2 2 2 a +x b + y2 c2 + z 2 + + ≥ |(a − b)(a − c)| |(b − a)(b − c)| |(c − a)(c − b)|
.
431
⋆.
Proposed by Shi Changwei, Xian City, Shaan Xi Provin e, China. Let n be a positive integer. Prove that
3381
(a)
1 1 1 4 1− 1 − 2 ··· 1 − n > ; 6
6
6
5
(b) Let an = a1 q , where 0 < a1 < 1 and 0 < q < 1. Evaluate n
lim
n→∞
n Y
(1 − ai ) .
i=1
. Proposed by Jose Luis Daz-Barrero and Josep Rubio-Masseg u, Universitat Polite ni a de Catalunya, Bar elona, Spain. Let n be a positive integer. Prove that 3382
sin
Pn+2 4Pn Pn+1
+ cos
Pn+2 4Pn Pn+1
<
3 2
sec
3Pn + 2Pn−1 4Pn Pn+1
where Pn is the nth Pell number, whi h is de ned by P0 Pn = 2Pn−1 + Pn−2 for n ≥ 2.
,
= 0, P1 = 1
and
. Proposed by Mi hel Bataille, Rouen, Fran e. Let ABC be a triangle with ∠BAC 6= 90◦ , let O be its ir um entre and let M be the mid-point of BC . Let P be a point on the ray M A su h that ∠BP C = 180◦ − ∠BAC . Let BP meet AC at E and let CP meet AB and F . If D is the proje tion of the mid-point of EF onto BC , show that 3383
is a symmedian of △ABC ;
(a)
AD
(b)
O , P , and
the ortho entre of △EDF are ollinear.
. Proposed by Mi hel Bataille, Rouen, Fran e. Show that, for any real number x,
3384
n−1 1 X n−k−1 ⌊x⌋ + {x}2 lim 2 k· x+ = n→∞ n n 2 k=1
,
where ⌊a⌋ is the integer part of the real number a and {a} = a − ⌊a⌋. . Proposed by Mi hel Bataille, Rouen, Fran e. Let p1 , p2 , . . . , p6 be primes with pk+1 = 2pk + 1 for k = 1, 2, . . . , 5. Show that X 3385
pi pj
1≤i<j≤6
is divisible by 15.
432 . Proposed by Ovidiu Furdui, University of Toledo, Toledo, OH, USA. Evaluate the integral
3386
Z
∞
0
e
−x
Z
x 0
e−t − 1 t
!
ln x dx .
. Proposed by Ovidiu Furdui, University of Toledo, Toledo, OH, USA. Let k > l ≥ 0 be xed integers. Find
3387
lim 2x ζ(x + k)ζ(x+k) − ζ(x + l)ζ(x+l)
x→∞
,
where ζ is the Riemann Zeta fun tion.
. Proposed by Paul Bra ken, University of Texas, Edinburg, TX, USA, in memory of Murray S. Klamkin. For all real x ≥ 1, show that 3388
(x − 1)2 1√ x2 x−1 + √ < √ √ √ 2 x−1+ x+1 x + x+2
.
................................................................. . Propose par D.J. Smeenk, Zaltbommel, Pays-Bas. Les sommets d'un quadrilatere ABCD sont situes sur un er le. Soit respe tivement K , L, M et N les entres des er les ins rits des triangles ABC , BCD , CDA et DAB . Montrer que le quadrilatere KLM N est un re tangle.
3376
. Propose par Toshio Seimiya, Kawasaki, Japon. Soit ABC un triangle ave ∠B = 2∠C . La bisse tri e interieure de l'angle BAC oupe BC en D. Soit M et N les points milieux respe tifs de AC et BD . Supposons que A, M , D et N soient o y liques. Montrer que l'angle BAC = 72◦ . 3377
. Propose par Mihaly Ben ze, Brasov, Roumanie. Soit x, y et z trois nombres reels positifs. Montrer que
3378
X
y lique
xy ≤ xy + x + y
X
y lique
x 2x + z
.
433 . Propose par Mihaly Ben ze, Brasov, Roumanie. Soit a1 , a2 , . . . , an nombres reels positifs. Montrer que
3379
n X
i=1
ai
≥ 1,
ai + (n − 1)ai+1
ou les indi es sont al ules modulo n. . Propose par Mihaly Ben ze, Brasov, Roumanie. Soit a, b, c, x, y et z six nombres reels positifs. Montrer que
3380
(a2 + x2 )(b2 + y 2 ) (c2 + z 2 )(a − b)2 ≥
+
(b2 + y 2 )(c2 + z 2 )
+
(c2 + z 2 )(a2 + x2 )
(a2 + x2 )(b − c)2 (b2 + y 2 )(c − a)2 2 2 b + y2 c2 + z 2 a +x + + |(a − b)(a − c)| |(b − a)(b − c)| |(c − a)(c − b)| 2
.
⋆
3381 . Propose par Shi Changwei, Xian City, Provin e de Shaan Xi, Chine. Soit n un entier positif. Montrer que (a) 1 − 16 1 − 612 · · · 1 − 61n > 45 ; (b) Soit an = a1 qn , ou 0 < a1 < 1 et 0 < q < 1. Trouver la valeur de
lim
n→∞
n Y
(1 − ai ) .
i=1
. Propose par Jose Luis Daz-Barrero et Josep Rubio-Masseg u, Universite Polyte hnique de Catalogne, Bar elone, Espagne. Soit n un entier positif. Montrer que
3382
sin
Pn+2 4Pn Pn+1
+ cos
Pn+2 4Pn Pn+1
ou Pn est le n-ieme nombre de Pn = 2Pn−1 + Pn−2 pour n ≥ 2.
<
3 2
sec
Pell, de ni par
3Pn + 2Pn−1 4Pn Pn+1
,
P0 = 0, P1 = 1
et
. Propose par Mi hel Bataille, Rouen, Fran e. Soit ABC un triangle, d'angle en A di erent de 90◦ , O le entre de son
er le ir ons rit et M le point milieu de BC . Soit P un point sur le rayon M A tel que l'angle en P du triangle BP C soit le supplementaire de elui en A. Soit E l'interse tion de BP ave AC et F elle de CP ave AB . Si D est la proje tion sur BC du point milieu de EF , montrer que (a) AD est une symediane du triangle ABC ; (b) O, P et l'ortho entre du triangle EDF sont olineaires. 3383
434 . Propose par Mi hel Bataille, Rouen, Fran e. Pour un nombre reel quel onque x, montrer que
3384
lim
n→∞
n−1 1 X n−k−1 ⌊x⌋ + {x}2 k · x + = n2 k=1 n 2
,
ou ⌊a⌋ est la partie entiere du nombre reel a et {a} = a − ⌊a⌋. . Propose par Mi hel Bataille, Rouen, Fran e. Soit p1 , p2 , . . . , p6 nombres premiers tels que pk+1 k = 1, 2, . . . , 5. Montrer que
3385
X
= 2pk + 1
ave
pi pj
1≤i<j≤6
est divisible par 15. . Propose par Ovidiu Furdui, Universite de Toledo, Toledo, OH, E-U. Evaluer l'integrale
3386
Z
∞
0
e
−x
Z
x 0
e−t − 1 t
!
ln x dx .
. Propose par Ovidiu Furdui, Universite de Toledo, Toledo, OH, E-U. Pour des entiers donnes k et l, ave k > l ≥ 0, al uler
3387
lim 2x ζ(x + k)ζ(x+k) − ζ(x + l)ζ(x+l)
x→∞
,
ou ζ designe la fon tion zeta de Riemann.
. Propose par Paul Bra ken, Universite du Texas, Edinburg, TX, USA, a la memoire de Murray S. Klamkin. Pour tous les nombres reels x ≥ 1, montrer que
3388
1√ (x − 1)2 x2 x−1 + √ < √ √ √ 2 x−1+ x+1 x + x+2
.
435
SOLUTIONS No problem is ever permanently losed. The editor is always pleased to onsider for publi ation new solutions or new insights on past problems. . [2007 : 428, 430℄ Proposed by Neven Juri , Zagreb, Croatia. A sequen e {an }∞ n=0 of positive real numbers satis es the re urren e relation an+3 = an+1 + an for n ≥ 0. Simplify
3276
q
a2n+5 + a2n+4 + a2n+3 − a2n+2 + a2n+1 − a2n .
Solution submitted independently by Arkady Alt, San Jose, CA, USA; Mi hel Bataille, Rouen, Fran e; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Thanos Magkos, 3rd High S hool of Kozani, Kozani, Gree e. Clearly, all the terms of the given sequen e are positive. Set A =
q a2n+5 + a2n+4 + a2n+3 − a2n+2 + a2n+1 − a2n .
By the re urren e relation, an+3 = an+1 + an , an+4 = an+2 + an+1 , an+5 = an+3 + an+2 = an+2 + an+1 + an , so that
and
a2n+5 + a2n+4 + a2n+3 = =
(an+2 + an+1 + an )2 + (an+2 + an+1 )2 + (an+1 + an )2 2a2n+2 + 3a2n+1 + 2a2n + 4an+2 an+1 + 2an+2 an + 4an+1 an .
Thus, A2
=
a2n+2 + 4a2n+1 + a2n + 4an+2 an+1 + 2an+2 an + 4an+1 an
=
(an+2 + 2an+1 + an )2 .
Therefore, A
= = = =
an+2 + 2an+1 + an (an+2 + an+1 ) + (an+1 + an ) an+4 + an+3 an+6 .
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; DIONNE ARSLANAGIC, BAILEY, ELSIE CAMPBELL, CHARLES DIMINNIE, KARL HAVLAK and PAULA KOCA, Angelo State University, San Angelo, TX, USA; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; OLIVER GEUPEL, Bruhl, NRW, Germany; RICHARD I. HESS,
436 Ran ho Palos Verdes, CA, USA; JOE HOWARD, Portales, NM, USA; CARL LIBIS, University student, Sarajevo College, Sarajevo, of Rhode Island, Kingston, RI, USA; SALEM MALIKIC, Bosnia and Herzegovina; ANDREA MUNARO, student, University of Trento, Trento, Italy; JOEL SCHLOSBERG, Bayside, NY, USA; GEORGE TSAPAKIDIS, Agrinio, Gree e; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.
. [2007 : 428, 430℄ Proposed by Ovidiu Furdui, student, Western Mi higan University, Kalamazoo, MI, USA. The Lu as numbers Ln satisfy the re urren e relation L0 = 2, L1 = 1, and Ln+2 = Ln+1 + Ln for n ≥ 0. Let k be an even positive integer. Find 3277
lim
p p p k Ln − k Ln−k + k Ln−2k ,
n→∞
where {x} is the fra tional part of x (that is, {x} the integer part of x).
= x − ⌊x⌋,
where ⌊x⌋ is
Comment: All three submissions laimed that the limit is 0, but no satisfa tory argument was provided that the limit exists for every positive even integer k. Problem 3277, therefore, remains open. . [2007 : 428, 431℄ Proposed by Virgil Ni ula, Bu harest, Romania. Let P be a point in the plane of △ABC su h that P C = P B and P A = AB . Let x be the measure of ∠P BC . Prove that 3278
sin(B − C) = 2 sin C cos(B + 2εx) ,
where ε = otherwise.
1
if the line
BC
separates the points
P
and
A,
and
ε = −1
Solution by Mi hel Bataille, Rouen, Fran e. Let AB = c, BC = a, and CA = b. First, suppose that BC separates P and A (see the gure on the left). ... . ′ .... .... ............ . .. ...... .... . ... .... .. ... ... ..... .. ..... ... ... .. ... .. . ... .. ... ... ... .... .. ... .. ... .. . ... .. . .... .. ... .. .. . . . ... .. .. ........ . . ... . ................... ...... . ... . . ... .... ... .... . . .. . . . . . .... .... .... ........ ... . . . .. . . . . . .... ... ..... ...... ... . . .. . . . . . . . .... ... . ...... .... ... ... ... ... ... ....... .... .. ... ... .. ... ....... .... ... ... ..... .... .. ............. . . ...... . .. ... .. ............. ..... . . . . . ............................................................................................................................................................... ........ . . . . .... . . . . . ........ ...... . . . . . ........ ...... ..... ... . .......... ....... ....... ....... ........ .... ........ ........ ........ ........ ............... ........ .... ........ ............ ... .. ..
A
A
B
P
q
... . ... ... ............... . . . ...... ... .... . . ... .. .. .. ... . .. .. .... ... ..... .... ... .... .. .... ... .... ... . . . ... .. ... .... ....... ... ... .. ................ . . . ... .. . ..... ... ................... ... . .. . ....... ... ... . .. ... ... ............. ... . ......... ... .. ... . . .. . . . ... .. ..... ............ ... ... ... ....... .. ... ... ... ....... ... ..... . .. . . ....... . ....... .... ′ ........ .... ....... ... ... .. . . ....... ... . ....... .. ..... ....... ......... ... .... ............................................................................................................................................... ... .. ..... ... .... ... .
qP
qP
q
C
B
P
C
437 a/2 and cos(x + B) = cos(∠P BA) = P B/2 , so that PB c a sin A A 2 cos x cos(x+B) = = . This yields cos(2x+B)+cos B = 2sin 2c 2 sin C sin C
Then, cos x
=
and, sin e sin A = sin(B + C) = sin B cos C + sin C cos B , we obtain
2 sin C cos(B + 2x) = sin B cos C − sin C cos B = sin(B − C) ,
as desired. a/2 and If P and A are on the same side of BC , then we have cos x = P B ∠P BA = x − B or ∠P BA = B − x, depending on the lo ation of point P . [Ed.: The ases are (a) P outside △ABC and A inside △P BC ( gure on the left, P ′ repla ing P ), (b) P outside △ABC and A outside △P BC ( gure on the right), and ( ) P inside △ABC ( gure on the right, P ′ repla ing P ).℄ In any ase, P B/2 = cos(x − B), and, in the same way as above, we obtain c a 2 cos x cos(x − B) = , whi h leads to 2 sin C cos(B − 2x) = sin(B − C). 2c Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; ROY BARBARA, Lebanese University, Fanar, Lebanon; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; ANDREA MUNARO, student, University of Trento, Trento, Italy; JOEL SCHLOSBERG, Bayside, NY, USA; D.J. SMEENK, Zaltbommel, the Netherlands; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.
. [2007 : 428, 431℄ Proposed by Virgil Ni ula, Bu harest, Romania. Let O, I , R, and r be the ir um entre, in entre, ir umradius, and inradius of △ABC , and let a, b, and c be the lengths of the sides of △ABC opposite the angles A, B , and C , respe tively. Let IO meet the lines AB and AC at M and N , respe tively. Prove that the points B , C , N , and M are on y li if and only if ha = R + r (where ha is the altitude to the side 1 1 1 BC ), and, in this ase, we also have = + . MN a b+c 3279
A omposite of similar solutions by Tai hi Maekawa, Takatsuki City, Osaka, Japan and D.J. Smeenk, Zaltbommel, the Netherlands. For an arbitrary triangle ABC let D be the foot of the altitude from A to BC and let P be the foot of the perpendi ular from I to AD ; thus P D = r . Moreover, it is always the ase that AI bise ts ∠DAO ; that is, ∠DAI = ∠IAO . Be ause ha = AP + P D = AP + r , we therefore have ha = R + r
⇐⇒ ⇐⇒ ⇐⇒
AO = △AP I ∠AOI
AP ∼ = △AOI = 90◦ .
The rst part of the problem therefore redu es to proving that B , C , N , and M
lie on a ir le
⇐⇒
IO ⊥ AO .
438 This is not quite orre t, however: the four y li points must be distin t to for e a nontrivial ondition. To that end we will assume the equivalent
ondition that no two angles of △ABC are equal. Under this assumption, we have that (be ause M ∈ AB and N ∈ AC ) B , C , N , and M lie on a ir le if and only if ∠AN M = ∠ABC , if and only if ∠AN M equals the angle between the hord AC and the tangent to the
ir um ir le at A (on the side that ontains the ar AC opposite B ), if and only if M N is parallel to that tangent, if and only if AO ⊥ M N . Sin e M N is the same line as IO, the proof of the rst part is omplete. For the laim on erning M1N , we assume that B , C , N , and M lie on a ir le, in whi h ase ∠AN M = ∠B and ∠AM N = ∠C . Let E and F be the feet of the perpendi ulars from I to AC and AB , respe tively. Then in the right triangles IN E and IM F we have NI =
Hen e, M N
sin B
and
IM =
r r + , sin C sin B
= M I + IN =
MN
Thus,
r
r sin C
.
and the Law of Sines gives us
2R 2R b+c = r + = 2rR sin A c b bc sin A ar(b + c) ar(b + c) = = . 2 Area(ABC) r(a + b + c) 1 MN
=
a+b+c a(b + c)
=
1
+
a
1 b+c
,
as laimed. Also solved by RICARDO BARROSO CAMPOS, University of Seville, Seville, Spain; MICHEL BATAILLE, Rouen, Fran e; OLIVER GEUPEL, Bruhl, NRW, Germany; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. Both Geupel and the proposer avoid appealing to the Law of Sines to prove the nal
laim as follows: The triangles ABC and ANM are assumed to be similar. Sin e the angle bise tor AI is ommon to both triangles, if W is the point where AI meets BC we have AI MN = IW CB
.
AI b+c = (see, for example, Nathan Altshiller IW a+b+c Court, College Geometry, page 75, Theorem 121), it follows that
Sin e CB = a and it is known that
MN =
whi h is the re ipro al of the desired equality.
a(b + c) a+b+c
,
439 . [2007 : 429, 431℄ Proposed by Virgil Ni ula, Bu harest, Romania. Let O and R be the ir um entre and ir umradius, respe tively, of △ABC . Let E and F be points on AB and AC , respe tively, su h that O is the mid-point of segment EF . Let A′ be the point where the line AO meets the ir um ir le Γ of △ABC a se ond time, and let P be the point on the line EF su h that A′ P ⊥ EF . Prove that the lines EF , BC , and the tangent line to Γ at A′ are on urrent, and that ∠BP A′ = ∠CP A′ . 3280
A omposite of solutions by Salem Maliki , student, Sarajevo College, Sarajevo, Bosnia and Herzegovina and by Andrea Munaro, student, University of Trento, Trento, Italy. One sees that E is uniquely de ned as the point where AB interse ts the image of the line AC under the halfturn about O, while F is the interse tion of AC with the image of AB under that halfturn. More relevant for us, however, is that be ause A and A′ are inter hanged by that halfturn it follows that A′ E||AF and A′ F ||AE . Be ause A′ P ⊥ P F (given) and A′ C ⊥ F C (be ause A′ A is a diameter of the ir um ir le Γ of △ABC ), the quadrilateral A′ P F C is y li , when e the dire ted angles satisfy ∠A′ P C = ∠A′ F C . The latter angle equals ∠BAC (be ause their sides are parallel), so that ∠A′ P C = ∠A .
Analogously, A′ BEP is y li and ∠BP A′ = ∠BEA′ = ∠A .
Consequently,
∠BP A′ = ∠A′ P C = ∠A ,
whi h is the se ond laim that we were to prove. For the on urren y laim, we will prove that EF , BC , and the tangent to Γ at A′ all ontain the image of P under the inversion de ned by Γ. We note that ∠BOC = 2∠A, and we have just seen that ∠BP C = 2∠A; we therefore on lude that the points B, P, O, and C lie on a ir le. Inversion in Γ xes the line EF (be ause that line passes through the entre O of Γ), it takes line BC to ir le BOC (whi h, we have seen, ontains P ), and it takes the tangent (to Γ) at A′ to the ir le on diameter OA′ (whi h ontains P be ause A′ P ⊥ P O ). We therefore see that the images (under inversion) of the three lines ontain P , so that these three lines themselves must on ur at the inverse of P , as laimed. [Editor's omment. Maliki ontributed the ni e treatment of angles; it was Munaro's idea to invert the gure.℄ Also solved by MICHEL BATAILLE, Rouen, Fran e; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. There was one in orre t submission.
440 3281. [2007 : 429, 431; Corre tion, 2008: 44, 46℄ Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain and Pantelimon George Popes u, Bu harest, Romania. Let a1 , a2 , . . . , an be positive real numbers. Prove that
X n
n+1
ak 2
k=1
n
n n X Y
≤
k=1
akj
j=1
.
I. Solution by Mi hel Bataille, Rouen, Fran e. n P
Let Ak =
j=1
akj
and let P
n Q
=
Ak
k=1
denote the right side of the given
inequality. Then P 2 = (A1 An )(A2 An−1 ) · · · (An A1 ). For ea h k we have, by the Cau hy-S hwarz Inequality, that Ak An+1−k
n n n+1−k 2 X X k 2 aj2 aj 2
=
j=1
≥
Hen e, P
2
n P
≥
j=1
j=1
aj
2
n X
aj aj
!2n
, from whi h the result follows.
n+1−k 2
k 2
j=1
n+1 2
2
=
n X
n+1 2
aj
j=1
.
II. Solution by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam. Note rst that 1 ≤ k, ℓ ≤ n.
n(n + 1) = 1 + 2 + · · · + n, 2
and let bkℓ = 1/n
n+1 2
q n aℓk
for ea h
For ea h k we have ak = a1+2+···+n = bk1 bk2 · · · bkn . k Therefore, by the generalized Holder Inequality, we have n X
n+1
ak 2
=
k=1
n X
k=1
≤ =
bk1 bk2 · · · bkn
n X
bn k1
k=1 n X
ak
k=1
=
n Y
k=1
!1/n
!1/n
n X
j=1
n X
k=1 n X
k=1 1/n
akj
bn k2
from whi h the given inequality follows.
,
a2k
!1/n
!1/n
···
···
n X
bn kn
k=1 n X
k=1
an k
!1/n
!1/n
441 Equality holds if and only if a1 = a2 = · · · = an .
Also solved by MOHAMMED AASSILA, Strasbourg, Fran e; ARKADY ALT, San Jose, CA, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; CHIP USA; SEFKET ARSLANAGIC, CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW, student, Sarajevo College, Germany; JOE HOWARD, Portales, NM, USA; SALEM MALIKIC, Sarajevo, Bosnia and Herzegovina, DUNG NGUYEN MANH, High S hool of HUS, Hanoi, Vietnam; PAOLO PERFETTI, Dipartimento di Matemati a, Universita degli studi di Tor Vergata Roma, Rome, Italy; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; PETER Y. WOO, Biola University, La Mirada, CA, USA; BINGJIE WU, student, High S hool AÆliated to Fudan University, Shanghai, China; TITU ZVONARU, Comane sti, Romania; and the proposers. ! Wu proved a generalization:
m
n m P Q
aij
i=1 j =1
≤
m Q
j =1
n P
i=1
am ij
whenever aij is posi-
tive for 1 ≤ i ≤ n and 1 ≤ j ≤ m. The proposed inequality is the spe ial ase when m = n and aij = aji /n .
. [2007 : 429, 432℄ Proposed by M.N. Deshpande, Nagpur, India. Of the n! permutations σ of (1, 2, . . . , n), for how many is σ3 the identity permutation?
3283
A omposite of solutions by Mi hel Bataille, Rouen, Fran e and Ri hard I. Hess, Ran ho Palos Verdes, CA, USA. The permutation σ3 is the identity if and only if σ is itself the identity or a produ t of disjoint 3- y les. We will investigate the ase n = 8 expli itly and then generalize. Cy le type
Total number of permutations
(1)(2)(3) · · · (8) (1, 2, 3)(4)(5)(6)(7)(8) 1 2
(1, 2, 3)(4, 5, 6)(7)(8)
1 8 2 = 112 3 2 83 2 53 = 1120
Thus, for n = 8 the desired number is 1 + 112 + 1120 = 1233. In general, to nd the number of permutations of n elements into n−3 k 3- y les, we hoose the rst 3- y le in n ways, the next in ways, 3 3 and so on until the kth 3- y le, hosen in n−3(k−1) ways. The produ t of 3 these binomial oeÆ ients is n n−3 3
3
···
n − 3(k − 1) 3
=
n! (n − 3k)!(3!)k
.
There are two possible y li permutations of ea h triple (su h as (1, 2, 3) and (1, 3, 2) for the triple {1, 2, 3}), whi h means that we must multiply the above produ t by 2k ; moreover, these k 3- y les an be hosen in k! orders, so that we must divide the produ t by k!. We therefore have
442 Cy le type
Total number of permutations
no
3- y le
20 n! 1 n! = 0 = 1 0! (n − 0)!(3!)0 3 0! (n − 0)!
1
3- y le
21 n! 1 n! = 1 1! (n − 3)!(3!)1 3 1! (n − 3)!
2
3- y les
k
.. . 3- y les
22 n! 1 n! = 2 2! (n − 6)!(3!)2 3 2! (n − 6)!
.. .
2k n! 1 n! = k k! (n − 3k)!(3!)k 3 k! (n − 3k)!
Of ourse, k an be any number from 0 to ⌊ n3 ⌋. It follows that the total number of permutations for whi h σ3 is the identity is ⌊n 3⌋
n!
k=0
3k k!(n − 3k)!
X
.
The rst eight values are 1, 1, 3, 9, 21, 81, 351, 1233. Also solved by MOHAMMED AASSILA, Strasbourg, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; and JOEL SCHLOSBERG, Bayside, NY, USA. There were three in orre t submissions. If we denote by P (n) the number of permutations for whi h σ3 is the identity, then alternative expressions obtained by our orrespondents are ⌊n ⌋ 3
P (n)
=
X n (3k)!
k=0
=
3k 3k k!
⌊n ⌋ 3
= 1+
X n(n − 1) · · · (n − 3k + 1) 3k k! k=1
P (n − 1) + (n − 1)(n − 2)P (n − 3) .
Using an argument analogous to our featured solution, Geupel proved that for any positive integer n and any prime p, the number P (n, p) of permutations σ of (1, 2, . . . , n) su h that σp is the identity is ⌋ ⌊n p
P (n, p) =
X n (kp)! kp pk k! k=0
.
For a dis ussion of the ase p = 2 he refers to Arthur Engel, Problem-solving Strategies (Springer, 1998), Chapter 5, Problem 15, pages 101 and 107. Our sequen e P (n) an be found in the on-line en y lopedia of integer sequen es: http://www.research.att.com/~njas/sequences/ (enter 1, 1, 3, 9, 21, 81, 351). That web page provides several referen es, one of whi h gives (without proof) the formula of Chowla, Herstein, and S ott (from 1952) for the number P (n, m) of permutations for whi h σm is the identity, where m is any given integer: If d0 = 1, d1 , d2 , . . . , dl = m are the divisors of m, then P (n, m)/(n!) is the oeÆ ient of xn in the Taylor expansion of 1 x d1
ex e d1
1 x d2
e d2
1 x dl
· · · e dl
.
In parti ular, for our problem m = 3 and P (n, 3) = P (n) equals n! times the oeÆ ient of 1 3 xn in the Taylor expansion of ex e 3 x .
443 3284. [2007 : 429, 432℄ Proposed by K.S. Bhanu and M.N. Deshpande, Institute of S ien es, Nagpur, India. Let x, y, and z be positive real numbers whi h satisfy x2 + y2 = z 2 . Constru t a line segment AC with length z . Let B be any point su h that BC = x and 90◦ < ∠ABC < 180◦ . Let M be a point on AC su h that ∠M AB = ∠M BC . Let D be the point on line BM on the opposite side of AC from B su h that AD = y . Show that ∠ADM = ∠DCM .
Solution by Oliver Geupel, Bruhl, NRW, Germany; and Mi hael Parmenter, Memorial University of Newfoundland, St. John's, NL. Sin e the triangles CAB and CBM are similar, we have su
essively MC BC
=
MC x
=
MC
=
CB CA x , z x2 z
D
,
y
A
.......... ........................ ............ .................. ...... ... .................. ...... ................. .... . ...... . . . . . . . . . . . . . . . . . . . ...... ... ... ...... .................. ..... .... ................. . . . . . . . . . . . . . . . . . . . . ...................................................................................................................................................................................................................................... ......... ..... . . . ... . . . . ......... . . . . .. ....... . ......... . . . . . . . . . . . ......... .... .. ......... .......... .... ......... .......... .... .......... ......... .... .......... ......... .... ................... ......... . . ......... .................. ..........
z
θ
M
θ
.
C
x
B
Then we have AM = z − AM y = , z y
x2 z 2 − x2 y2 = = z z z
or equivalently AD = AC AM D and ADC are similar. Therefore, Thus,
AM , AD
.
whi h implies that triangles
∠ADM = ∠ACD = ∠DCM ,
as laimed.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE, ARSLANAGIC, Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; PAUL DEIERMANN, Southeast Missouri State University, Cape Girardeau, MO, USA; SALEM student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; PETER Y. WOO, Biola MALIKIC, University, La Mirada, CA, USA; and the proposers.
3285. [2007 : 429, 432℄ Proposed by Gregory Akulov, student, University of Regina, Regina, SK. Solve the following for x:
r ! r q p 3 2 2 x 3 − 2x + 5(1 − x ) + = 2
3
.
444 Solution by Mi hel Bataille, Rouen, Fran e. The equation an be rewritten as q p √ x 6 3 − 2x + 5(1 − x2 ) = 2 − 3x ,
so that the given equation is equivalent to 0 < x <
2 3
and
p 6x2 3 − 2x + 5 (1 − x2 ) = (2 − 3x)2 .
Sin e (3 − 2x)2 − (2 − 3x)2 = 5 equation as 6x2
=
1 − x2
, we su
essively rewrite the above
p (2 − 3x)2 3 − 2x − 5 (1 − x2 ) (3 − 2x)2 − 5 (1 − x2 )
p 6x2 + 2x − 3 = − 5 (1 − x2 ) . 2 α = arccos and x = cos θ where θ ∈ (α, π ),
Setting 3 is equivalent to ea h of
−
Hen e,
√ 5 3
2
√ − 5 sin θ sin θ −
2
3
cos θ
,
the last equation
= 6 cos2 θ + 2 cos θ − 3 , = 2 cos2 θ − 1 = cos(2θ) ,
cos(θ − α) = cos(π − 2θ) . ` ´ π + arccos 23 π+α θ= and x = cos . 3
3
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; DIONNE ARSLANAGIC, BAILEY, ELSIE CAMPBELL, CHARLES DIMINNIE, KARL HAVLAK and PAULA KOCA, Angelo State University, San Angelo, TX, USA; ROY BARBARA, Lebanese University, Fanar, Lebanon; PAUL DEIERMANN, Southeast Missouri State University, Cape Girardeau, MO, USA; KEE-WAI LAU, Hong Kong, China; JOEL SCHLOSBERG, Bayside, NY, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. There were two in omplete and two in orre t solutions submitted.
. [2007 : 430,432℄ Proposed by Neven Juri , Zagreb, Croatia. Is it possible to nd a fun tion f : [0, 1] → R su h that
3286
f (x) = 1 + x
Z
1 0
f (t) dt + x2
Z
1
0
2 f (t) dt ?
445 All submitted solutions were similar to those of Mi hel Bataille, Rouen, Fran e and Ri hard I. Hess, Ran ho Palos Verdes, CA, USA. R R Let a = 01 f (t) dt and b = 01 [f (x)]2 dt. Thus, f (t) = 1 + at + bt2 and we have a =
Z
1
1 + at + bt
0
2
hen e a = 2 + 2b . Also 3 b =
Z
1
0
= = =
Hen e,
1 bt3 a b at2 + dt = t + = 1+ + 2 3 2 3
2
1 + 2at + 2b + a 2
t + at +
,
0
2b + a2 t3 3
2
3
2 4
t + 2abt + b t +
abt4 2
+
dt
! 1 5
b2 t5
0
2b + a2 ab b2 1 + a + + + 3 2 5 2b 1 8b 4b2 b2 b2 1 + 2 + + 2b + 4 + + + b + + 3 3 3 9 3 5 92b2 135
+
20b 9
+
13 3
.
= 0,
and the two roots of the equation 92b2 + 300b + 585 = 0 are b =
√ −75 ± 3i 870 46
.
Thus, su h afun tion not exist. However, the omplex valued fun tion √ f does √ 21 ± i 870 −75 ± 3i 870 f (z) = 1+ z+ z 2 does satisfy the equation 23 46 of the problem. Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; ROY BARBARA, Lebanese University, Fanar, Lebanon; DIONNE BAILEY, ELSIE CAMPBELL, CHARLES DIMINNIE, KARL HAVLAK and PAULA KOCA, Angelo State University, San Angelo, TX, USA; PAUL BRACKEN, University of Texas, Edinburg, TX, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; JOSE LUIS D I AZ-BARRERO, Universitat Polite ni a de Catalunya, Bar elona, Spain; OLIVER GEUPEL, Bruhl, NRW, Germany; KEE-WAI LAU, Hong Kong, China; THANOS MAGKOS, 3rd High S hool of Kozani, Kozani, Gree e; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; ROBERT P. SEALY, Mount Allison University, Sa kville, NB; DIGBY SMITH, Mount Royal College, Calgary, AB; DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan; GEORGE TSAPAKIDIS, Agrinio, Gree e; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.
446 . [2007 : 430, 432℄ Proposed by Virgil Ni ula, Bu harest, Romania. Let x, y, and z be positive real numbers satisfying
3287
xy + yz + zx + xyz = 4 .
Prove that (a)
(x+2)(y +2)+(y +2)(z +2)+(z +2)(x+2) = (x+2)(y +2)(z +2);
(b) there is a triangle whose sides have lengths (x+2)(y+2), (y+2)(z+2), and (z + 2)(x + 2). Solution by Joe Howard, Portales, NM, USA. Let a = x + 2, b = y + 2, and c = z + 2. A simple al ulation shows that 4 = = =
xy + yz + zx + xyz (a − 2)(b − 2) + (b − 2)(c − 2) + (c − 2)(a − 2) + (a − 2)(b − 2)(c − 2) abc − (ab + bc + ca) + 4 .
Hen e, abc = ab + bc + ca, and the equation in part (a) holds. Clearly a, b, and c are positive. By part (a) ab + bc − ca
= ab + bc + ca − 2ca = abc − 2ca = ac(b − 2) = acy > 0 .
Similarly, bc + ca − ab > 0 and ab + ca − bc > 0, and the result in part (b) follows. Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS, University of Sarajevo, Sarajevo, Bosnia and Messolonghi, Gree e; SEFKET ARSLANAGIC, Herzegovina; DIONNE BAILEY, ELSIE CAMPBELL, and CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; JOSE LUIS D I AZ-BARRERO, Universitat Polite ni a de Catalunya, Bar elona, Spain; OLIVER GEUPEL, Bruhl, NRW, Germany; KARL HAVLAK, Angelo State University, San Angelo, TX, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; student, THANOS MAGKOS, 3rd High S hool of Kozani, Kozani, Gree e; SALEM MALIKIC, Sarajevo College, Sarajevo, Bosnia and Herzegovina; DUNG NGUYEN MANH, High S hool of HUS, Hanoi, Vietnam; ANDREA MUNARO, student, University of Trento, Trento, Italy; MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's, NL; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; JOEL SCHLOSBERG, Bayside, NY, USA; BOB SERKEY, Leonia, NJ, USA; D.J. SMEENK, Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Gree e; GEORGE TSAPAKIDIS, Agrinio, Gree e; PETER Y. WOO, Biola University, La Mirada, CA, USA; BINGJIE WU, student, High S hool AÆliated to Fudan University, Shanghai, China; TITU ZVONARU, Comane sti, Romania; and the proposer.
447 . [2007 : 430, 432℄ Proposed by Ovidiu Furdui, student, Western Mi higan University, Kalamazoo, MI, USA. Let n be a positive integer. Evaluate the sum: 3288
⌊ n−1 2 ⌋ X n − i − 1 2n−2i−1 i n − 2i i=0
,
where ⌊x⌋ is the integer part of x.
Solution by Mi hel Bataille, Rouen, Fran e. Let Sn denote the sum to be evaluated. Then Sn =
1 2
Z
2
⌊ n−1 2 ⌋
X
0
i=0
n−i−1 i
xn−2i−1 dx .
From a known result (for example, see the solution to 3217 in [2008: 112-3℄) we have ⌊ n−1 2 ⌋
X
n−i−1 i
i=0
(1)
Crux
problem
xi = Fn (x)
for non-negative x, where 1
1+
Fn (x) = √ 1 + 4x Cal ulating xn−1 Fn 12
!n √ 1 + 4x 2
−
1−
!n ! √ 1 + 4x 2
.
(2)
for x > 0 using (1) and then (2) yields
x
⌊ n−1 2 ⌋
X
i=0
n − i − 1 n−2i−1 x i n n p p 1 x + x2 + 4 − x − x2 + 4 = √ 2n x2 + 4
, (3)
1 a formula that still holds if x = 0 [Ed.: if x = 0, n is odd, and i = n − , 2 0 then the last term of the sum in (3) is 0 , whi h we interpret as 1℄. It then follows from (1) and (3) that
Sn =
Substituting x integral yields
1 2n+1
Z
n n √ √ x + x2 + 4 − x − x2 + 4 dx . √ x2 + 4
2
0
= 2 sinh t = et − e−t Sn =
Z
ln(1+
0
√ 2)
and
dx = 2 cosh t dt
ent + (−1)n−1 e−nt 2
dt .
in the above
448 Thus, if n is odd, Z
Sn =
ln(1+
√
2)
cosh(nt) dt
=
0
=
and if n is even, Sn =
Z
ln(1+
√ 2)
sinh(nt) dt
=
0
=
√ n 1 sinh ln 1 + 2 n √ n √ n 1 2+1 − 2−1 , 2n
√ n 1 1 cosh ln 1 + 2 − cosh(0) n n √ n √ n 1 1 2+1 + 2−1 − 2n n
.
Also solved by the proposer, whose proof used Fibona
i polynomials and similar arguments as given in the featured solution above.
The ATOM Series (whi h is an a ronym for A Taste Of Mathemati s, in Fren h: Aime-T-On les Mathematiques) is a series of 64 page booklets published by the CMS, designed as enri hment materials for high s hool students with an interest and aptitude for mathemati s. Some booklets in the series will over materials useful for mathemati al ompetitions. Eight volumes have been published so far (see the CMS website for details). We are always on the lookout for interesting proposals from authors in any part of the world. Proposals should be sent to the Editor-in-Chief, Bru e Shawyer at
[email protected]
Crux Mathemati orum with Mathemati al Mayhem Former Editors / An iens Reda teurs: Bru e L.R. Shawyer, James E. Totten
Crux Mathemati orum Founding Editors / Reda teurs-fondateurs: Leopold Sauve & Frederi k G.B. Maskell Former Editors / An iens Reda teurs: G.W. Sands, R.E. Woodrow, Bru e L.R. Shawyer
Mathemati al Mayhem Founding Editors / Reda teurs-fondateurs: Patri k Surry & Ravi Vakil Former Editors / An iens Reda teurs: Philip Jong, Je Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia, Shawn Godin, Je Hooper
