353
EDITORIAL Va lav Linek
Many CRUX with Mayhem readers have ommented on the urrent ba klog of four-digit problem p...
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353
EDITORIAL Va lav Linek
Many CRUX with Mayhem readers have ommented on the urrent ba klog of four-digit problem proposals (for those problems that appear near the end of ea h issue). The ba klog of these is due to the fa t that we publish only 100 four-digit problems per year, but we re eive many more than 100 problem proposals per year. This situation is daunting, espe ially for new proposers, but also for established proposers who may have a favourite problem languishing in a sta k. To at least partially address this situation, any new proposer may now submit up to two (2) problem proposals and request that they be expedited, meaning that we will do our best to evaluate those proposals as soon as possible. In addition, anyone who has already proposed problems to CRUX may also submit two proposals to be expedited. To help us implement this new poli y, we advise proposers to refer to the proposal numbers that we give out whenever a problem is proposed. This makes it mu h easier for us to tra k your problem proposal, whereas without a proposal number we may have to sear h literally hundreds of pages of paper in dozens of le folders. We will also begin streaming the problem proposals that we re eive into
ategories, namely: Algebra and Number Theory (a single ategory), Logi , Cal ulus, Combinatori s, Inequalities (in luding Geometri Inequalities), Geometry, Probability, and (if any of these do not t!) Mis ellaneous Problems. Currently we have a large store of proposed inequalities, so we are parti ularly interested in re eiving problem proposals in the other ategories in order to a hieve a balan e of topi s. In another department, our Contributor Pro les will be resuming in 2010 and we have re eived at least two re ommendations so far this year (any time is a good time to re ommend someone to be pro led and we wel ome more re ommendations). Lastly, we ask that all solvers begin ea h new solution on a separate page, with their name and onta t information at the top of the page. In parti ular our Mayhem editor, Ian VanderBurgh, makes a spe ial appeal to all Mayhem solvers to follow this pro edure.
354
SKOLIAD
No. 119
Lily Yen and Mogens Hansen
Please send your solutions to problems in this Skoliad by 1 January, 2010. A
opy of Crux will be sent to one pre-university reader who sends in solutions before the deadline. The de ision of the editors is nal. Our ontest this month is the World Youth Mathemati s Inter ity Competition, Individual Contest, Part I, 2005, whi h was held in Kaosiung, Taiwan. We thank Wen-Hsien Sun, Chiu Chang Mathemati s Edu ation Foundation, Taipei, Taiwan, for providing us with this ontest and for permission to publish it. We also thank Rolland Gaudet, University College of Saint-Bonifa e, Winnipeg, MB, for translation of this ontest.
Competition mathematique mondiale des jeunes Con ours individuel, Partie I, 2005 Temps limite : 2 heures
. La somme d'un entier a 4 hi res et de ses 4 hi res est 2005. Quel est
et entier a 4 hi res ? 1
2. Dans le triangle ABC , on a AB = 10 et AC = 18 ; M est le point milieu de BC ; la ligne passant par M et parallele a la bisse tri e de ∠CAB oupe AC au point D . Determiner la longueur de AD.
. Soient x, y et z des nombres positifs tels que x+y+xy = 8, y+z+yz = 15 et z + x + zx = 35. Determiner la valeur de x + y + z + xy.
3
. Au total, le nombre de hampignons ueillis par 11 gar ons et n lles est egal a n2 + 9n − 2, ou haque personne a ueilli exa tement le m^eme nombre de hampignons. Determiner l'entier positif n.
4
. L'entier positif x est tel que x et x + 99 sont tous les deux des arres. Determiner la somme de tous les entiers x ayant la m^eme propriet e.
5
6. Les ot ^ es d'un triangle re tangle sont tous les deux des entiers positifs ; un de es ot ^ es est de longueur au plus 20. Le rapport du rayon du er le
ir ons rit au rayon du er le ins rit de e triangle est de 5 : 2. Determiner la valeur maximale du perim etre de e triangle.
. Soit α la plus grande ra ine de (2004x)2 − 2003 · 2005x − 1 = 0 et soit β la plus petite ra ine de x2 + 2003x − 2004 = 0. Determiner la valeur de α − β.
7
355 . Soit a un nombre positif tel que a2 + 1 a3 + 3 . a 8
1 = 5. a2
. Dans la gure, ABCD est un re tangle ave AB = 5 et tel que le demi er le de diametre AB oupe CD en deux points. Si la distan e d'un e es points a A est 4, determiner la surfa e de ABCD.
D ...................................................................................................................................... C
9
n
Determiner la valeur de
A
... .. ...... . ........ ............. .... ... ........ ... ... .. ......... . ... ... . . . . . . ... ... .. ... .. ......... ... ..... ...... ........ ... ...... ...... ................ ...... ... .......... ...........................................................................................
2
B
n−1
10 10 10 . Soit a donne par 9 n 10 −1− − − ··· − ou n 9 9 9 9 est un entier positif. Si a est un entier, determiner la valeur maximale de a. 10
11. Dans un entier a deux positions de imales, le hi re en position dix est plus grand que le hi re en position un ; le produit de es deux hi res est divisible par leur somme. Quel est et entier a deux positions de imales ?
. Dans la gure, P QRS est un re tangle de surfa e 10. A est un point sur RS et B est un point sur P S , tels que la surfa e du triangle QAB est 4. Determiner la plus petite valeur possible pour P B + AR.
12
P ..............................B S ............................................................
Q
... . ... .... ... .. ................... .. .......... ... ... .......... .... .. .......... .... .. ... . . . ........... .. ... .. ... ............ ..... . . . . . . . . . . . .. ... . ... .......... .. .. ... ............ ... .. ... ............. ............................... ...............................................................................................
A
R
World Youth Mathemati s Inter ity Competition Individual Contest, Part I, 2005 2 hours allowed
. The sum of a four-digit number and its four digits is 2005. What is this four-digit number?
1
. In triangle ABC , AB = 10 and AC = 18. M is the midpoint of BC , and the line through M parallel to the bise tor of ∠CAB uts AC at D. Find the length of AD. 2
3. Let x, y , z be positive numbers su h that x+y +xy = 8, y +z +yz = 15, and z + x + zx = 35. Find the value of x + y + z + xy.
. The number of mushrooms gathered by 11 boys and n girls is n2 + 9n − 2, with ea h person gathering exa tly the same number. Determine the positive integer n. 4
5. The positive integer x is su h that both integers. Find the sum of all su h integers x.
x
and
x + 99
are squares of
356 . The side lengths of a right triangle are all positive integers, and the length of one of the legs is at most 20. The ratio of the ir umradius to the inradius of this triangle is 5 : 2. Determine the maximum value of the perimeter of this triangle. 6
. Let α be the larger root of (2004x)2 − 2003 · 2005x − 1 = 0 and β be the smaller root of x2 + 2003x − 2004 = 0. Determine the value of α − β .
7
. Let a be a positive number su h that a2 + a12 of a3 + a13 . 8
. In the gure, ABCD is a re tangle with AB = 5 su h that the semi ir le with diameter AB uts CD at two points. If the distan e from one of them to A is 4, nd the area of ABCD. 9
. Let a be 9
10
= 5.
Determine the value
D ...................................................................................................................................... C
A
... . .. ... .... ... ........ ......... ............ ..... ... ... .. ........ ... ... . . . . . . . . ... ... .. ... .. ........ ... ..... ...... ........ ... ...... ..... ................ .... ... .......... .............................................................................................
B
n 2 n−1 10 10 10 10 n −1− − − ··· − where n is 9
9
9
9
a positive integer. If a is an integer, determine the maximum value of a.
. In a two-digit number, the tens digit is greater than the ones digit. The produ t of these two digits is divisible by their sum. What is this two-digit number? 11
. In the gure, P QRS is a re tangle of area 10. A is a point on RS and B is a point on P S su h that the area of triangle QAB is 4. Determine the smallest possible value of P B + AR.
P ..............................B S ............................................................
12
Q
... . .. ... .. ................... ... ... .. .......... .. .......... ... .... .......... ..... .. .. .......... ... . . . . . ........... .. ... ... .. ... ............. ..... .. .. ............ . . ... . . . . . . . . . . ... .. ... ........... . ...... ........................ ..................................................................................................
A
R
Next follow the solutions to the National Bank of New Zealand Junior Mathemati s Competition 2005 [2008 : 385{391℄. 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 of 5 be ause 15 = 5 × 3."
357 Now answer the other questions whi h Mi hael's mother asked. (a) Write down a set of ve di erent digits whi h add to 35. (b) Write down a set of three di erent digits whi h add to a multiple of 5, but whi h don't in lude 5 itself or 1. ( ) How many di erent sets of four di erent digits are there whi h add 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. (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. Solution by Alex Song, student, Elizabeth Ziegler Publi S hool, Waterloo, ON. (a) The set {5, 6, 7, 8, 9} works. (b) The sets {2, 4, 9}, {2, 6, 7}, {3, 4, 8}, {3, 8, 9}, and {4, 7, 9} all work. ( ) Let Dn denote the set of allowed digits that leave remainder n when divided by 5, 0 < n < 5. Then D1 = {6}, D2 = {2, 7}, D3 = {3, 8}, and D4 = {4, 9}. Now two possibilities exist: either use one digit from ea h of D1 , D2 , D3 , and D4 , or use two digits from ea h of D2 and D3 . The rst
ase leads to 1 · 2 · 2 · 2, or eight sets, while the se ond leads to one set. Thus there are a total of nine sets. (d) Any of the sets {1, 2, 3, 4, 6, 9}, {1, 2, 3, 4, 7, 8}, {1, 2, 3, 7, 8, 9}, {1, 2, 4, 6, 8, 9}, {1, 3, 4, 6, 7, 9}, {1, 4, 6, 7, 8, 9}, {2, 3, 4, 6, 7, 8}, or {2, 3, 6, 7, 8, 9} will work. (e) The eight allowed digits 1, 2, 3, 4, 6, 7, 8, and 9 add up to 40. If the sum of seven of these were divisible by 5, then the remaining digit would also be divisible by 5. This is not the ase, so no su h set of seven digits exists. Also solved by JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. In Part (a), Jixuan Wang points out that the largest possible sum of ve distin t digits is 5 + 6 + 7 + 8 + 9 = 35, so {5, 6, 7, 8, 9} is the only solution. In Part (b), nding a set is suÆ ient; to see that these are the only possible solutions, systemati ally onsider all sets of three digits. You an begin by onsidering sets ontaining the digit 2, then sets whose smallest member is 3, and so on. In Part ( ), 1 + 2 + 3 + 4 = 10 is divisible by 5, so hoosing one number from ea h set works. Now, put one number ba k into the set it ame from and repla e it by a number from a di erent set. This auses the sum 10 to hange by ±1, ±2, or ±3, and no new sum is divisible by 5. Therefore, you annot pi k two numbers from one set and one number from ea h of two di erent sets. That only leaves the possibility of hoosing the four numbers in D2 ∪ D3 , D2 ∪ D4 , or D3 ∪ D4 , and of these possibilities only D2 ∪ D3 yields a solution.
358 In Part (d), nding a set is suÆ ient; to see that these are all of the solutions, note that the sum of the eight allowed digits is 40 and that, if six hosen digits have a sum divisible by 5, then the remaining two digits also have a sum divisible by 5. Systemati ally sear hing for all su h sets of size two yields {1, 4}, {1, 9}, {2, 3}, {2, 8}, {3, 7}, {4, 6}, {6, 9}, and {7, 8}. The omplements of these sets are the sets given in the solution above (in reverse order).
. 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. 2
...................................................................... ... .... .... .... ... ... .... .... ... .... ... ... ... .... . ..... ... .. ... .... .... .... .... ... .... .... ... ... .... .................................................................
}
} }
................................................................ ... .... ... .... ..... .... ... ... .... .... ... .... ... .... ... ... .... .... ... .... ... ... .... ... .... ... .... .... .... .. ... ................................................................
}
} }
.................................................................... ... ... .... .... ... .... ... .... ... ... ... ... ... .... . ..... .... ... ... ...................................................................
}
}
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.) Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. (a) The single dot must be pla ed in one of the six positions of a two-bythree re tangle, so six patterns exist. (b) There are fteen ways. For every two-dot pattern, swapping dots for blanks and blanks for dots produ es a unique four-dot pattern. Therefore equally many two-dot and four-dot patterns exist.
359 Alternatively, the number of ways one an hoose four positions from the six possible positions is 6 C4 = 64 = 4! (66!− 4)! = 15. Of ourse 62 is also 15. ( ) If you remove one dot from a four-dot pattern, a three-dot pattern remains. Sin e any of the four dots an be removed, ea h four-dot pattern gives rise to four three-dot patterns in this way. However, ea h possible three-dot pattern arises from three di erent four-dot patterns, namely one four-dot pattern for ea h of the three blanks in the three-dot pattern. Sin e the number of four-dot patterns is fteen, it follows that the number of three-dot patterns is 153· 4 = 20. Alternatively, the number of ways one an hoose three positions from the six possible is 6 C3 = 63 = 3! (66!− 3)! = 20. (d) Here ea h pattern has four positions. Thus you an arrange one dot in four ways, three dots (so one blank) in four ways, and four dots in one way. By reasoning as in Part ( ), the number of two-dot patterns is 4 · 3/2 = 6. Thus the total number of patterns is 4 + 4 + 1 + 6 = 15. Alternatively, 41 + 42 + 43 + 44 = 4 + 6 + 4 + 1 = 15 is a di erent way to ount the number of patterns. (e) Here ea h pattern has eight positions, so the number of patterns is 8 8 8 + + · · · + = 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 255. 1 2 8
Also solved by JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON; and ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON. In Parts (b) through (d) van Gaalen ommendably provides solutions from rst prin iples in the spirit of the question rather than resorting to general ounting methods. However, as his alternative solutions show, familiarity with binomial oeÆ ients simpli es the solutions
onsiderably. Song provides even simpler solutions to Parts (d) and (e): Ea h of the four (respe tively eight) positions must be either blank or a dot. Thus 24 (respe tively 28 ) patterns exist. Disallowing the all-blank pattern thus leaves fteen (respe tively 255) patterns. 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 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 | || | |
|
| ...
360 (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. Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. 10 (a) Sin e 14 = 15 , 1 = 12 , and 16 = 60 ; the re ipro al of 4 is k15|, the 60 5 60 re ipro al of 5 is k12|, and the re ipro al of 6 is k10|. 1 1 2 24 , the re ipro al of 2.5 is k24|. (b) Sin e 2.5 = = = 5/2 5 60
360 + 40 6 40 ( ) Sin e 19 = 400 = = + 2 , the re ipro al of 9 is k6|40|. 602 602 60 60 x 1 (d) The largest integer, x, su h that 60 < is 8. Thus, the rst digit is 7 8 1 1 1 = x = 8. Now − , and the largest integer, y, su h that 60y 2 < 105 7 60 105 is 34. Thus the se ond digit is y = 34. Therefore, the re ipro al of 7 begins k8|34| . . . . 1 1125 1080 + 45 18 45 (e) Sin e 192 = = = + , the re ipro al of 192 603 603 602 603 is k0|18|45|.
Also solved by ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; and (in part) JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON. You may enjoy verifying that the re ipro al of 7 (see Part (d)) ontinues the pattern k8|34|17|8|34|17| . . . . That is, the Babylonian \digits" 8|34|17| ontinue inde nitely.
. 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
361 (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. Solution by Alex Song, student, Elizabeth Ziegler Publi S hool, Waterloo, ON. (a) The Red Team must have s ored an odd number of Inners, sin e the other point s ores are even and the total number of points won by the Red Team is odd. Thus, the number of Inners must be one of 1, 3, 5, 7, 9, 11, 13, or 15. Suppose the Red Team s ored x Inners, 0 Outers, and 16 − x Wides. Then the total number of point is 2(16 − x) + 7x = 32 + 5x. Sin e x is odd, 32 + 5x is \something-seven." Thus any of 1, 3, 5, 7, 9, 11, 13, or 15 are possible. (b), ( ) Again let x be the number of Inners. Then the Red Team s ored x Outers and 16 − 2x Wides, and 7x + 4x + 2(16 − 2x) = 7x + 32 is the total number of points. If 7x + 32 leaves remainder 7 when divided by 10, then 7x + 25 is divisible by 10, as is 7x + 5. Therefore 7x must be an odd multiple of 5, so x must be an odd multiple of 5. Sin e the number of Wides, 16 − 2x, must be nonnegative, it follows that x = 5 and that this is the only solution. The total number of points is then 7 · 5 + 32 = 67.
Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON; and JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON. Our solver's original solution used modular arithmeti : If the integers x and y leave the same remainder when divided by the natural number n, then we say that x and y are equivalent modulo n, and we write x ≡ y (mod n). In Part (b), 7x + 32 ≡ 7 (mod 10), so 7x + 25 ≡ 7x + 5 ≡ 0 (mod 10). Thus 7x ≡ 5 (mod 10), when e x ≡ 5 (mod 10). That is, x is an odd multiple of 5.
. 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
(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?
362 (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? 40 30 6 ?
40 -
30 6 ? 40
6 30 - ? 40
... ... ... ... ...... .................................................................................................................................................................................................... ....... ... .. .... ....... ...... . . . . . . .... . . ....... . ....... ....... ... ..... ....... ...... ........ ...... ........... ....... ....... .......... ....... ....... ... ......... ....... ..... ....... ...... . . . . . . . . . . .... .......... . . ....... .... .... . . ....... . . . . . . . . . . .... ..... . . . .. . ....... ..... .......................... ....... .... .... ....... ....... .... ....... ....... .... .... .... .. ....... ...... . . . ... . ... ....... .... .... . ..... ...... ... . . . . . .... ... . . . ......... ... ... . . . . . .... .... . . ....... .. . ..... . . . . . . . . . . .... .... . . ....... ..... .. .... . . . . . . . . . . . . . . . . .... .... . . . .................... ....... ..... . . . . . . . . . . . . . . .... .. . . ....... ..... ........... ..... . . . . . ... . . . . . . . . ....... ..... ....... ..... ... ........... . . . . . . . . . . . ....... ............ . .. .... . . . . . . . . . . ...... .......... ........ . . ....... ..... ....... ... .... ...... ....... ....... .... ... ....... ...... . . . ... . . . . ... . . ... ... ...... .................................................................................................................................................................................................. ... ... ... ... . . . .
6 30 ?
................................................................................................................................................................................................ ....... ... .... ... ....... ...... .... ....... .... ....... ....... ....... .... ... ....... ...... . . . . . ........... . . ....... . . .... .... .... . . . ....... . . . . . . . .... .......... . . ....... .... .... ... . . ....... . . . . . . . . .... . . . ....... ...... . ....... ..... . . ............. . . ....... .... ..... . ..... ....... ... .... ...... ....... ....... .... ....... ... ....... . ....... . . . .... .... . ...... ....... . . . .... .... . ....... .... . . . . . . . .... .... . ....... ..... . . . . . . . . .... .... ....... ..... . . . . . . . . . .... ... ....... .............. .... . . . . . . . ... . . . . . ... . . . ....... ....... ... ..... . . . . . . . . . . .... ............ . ....... ..... ....... .... . . . . . . . . . . . .............. . . . .......... ....... ..... . . . . . . . . .... .. ....... ... ....... ... .... ....... ....... ... ....... .... ....... ....... . . . . . . . ..... .. . . .........................................................................................................................................................................................
Fig. 5 (not to s ale) Fig. 4 (not to s ale) Solution by Alex Song, student, Elizabeth Ziegler Publi S hool, Waterloo, ON. (a) Label the points as shown. Sin e it is a ross, K , P , M , and A B C ...................................................................................................................... D C are ollinear. Thus, △QP M ∼ . .. ....... ... ....... ....... △RP K ∼ △ACK , and we have . . ..... . . . . ....... .. .... E . . L .. ...... . . . . . . RK AK 150 − 30 3 QM . . .......M...... ... ....... .... . = = = = . ............... ...... ...... ... QP RP AC 240 − 40 5 . .... . . . . . . . . . . . . . . . . ... ..... ....... .. ... .. .. ...... Now R must be the midpoint of AJ , ..... .. .. . . . . R .... . N . . . . 150 . ... P ..... Q... ......... . .. . ... so RK = 2 − 30 = 45. Thus, . . . ...... ................. ... .......... . . . ....... ... .. ........ 5 . . . . . . . . . . . . ......... . . RP = (45) = 75, and moreover . ........ 3 ....... O ............ .. F . . . K . . . . ....... 1 ... .. . . . . RP + P Q = AD = 120, so that . . . ...................................................................................................................... 2 3 J G P Q = 45. Hen e QM = QP = 27. I H 5 That is, the radius is 27 m. (b) Sin e QM + M S = 12 AJ = 75, it follows that M S = 48, when e the area of △BM C is 1602· 48 = 3840. Likewise, the area of △HIO is 3840. Moreover, the area of △LP K is 90 2· 75 = 3375, and the area of △EN F is also 3375. Sin e the entire re tangle ADGJ has area 36000, the area of the
ross is 36000 − 2 · 3840 − 2 · 3375 = 21570 m2 . .... .... ... .... ... .... . . ... . . ...........................................................
Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. Here is an analyti al approa h: Say J = (0, 0), A = (0, 150), D = (240, 150), and G = (240, 0). Then K = (0, 30) and C = (200, 150), and the line CK is given by y = 3 x + 30. If x = 120, then y = 3 · 120 + 30 = 102, so M = (120, 102). Sin e 5 5 Q = (120, 75), the radius is 102 − 75 = 27. Similarly, if y = 75, then 75 = 3 x + 30, so 5 x = 75, when e P = (75, 75). The area of the ross is now al ulated as above.
That ompletes another Skoliad. This issue's prize of a opy of CRUX goes to Alex Song, student, Elizabeth Ziegler Publi S hool, Waterloo, ON. We hope to re eive more solutions from more readers.
with Mayhem
363
MATHEMATICAL MAYHEM Mathemati al Mayhem began in 1988 as a Mathemati al Journal for and by . 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) and Eri Robert (Leo Hayes High S hool, Frederi ton). High S hool and University Students
Mayhem Problems Veuillez nous transmettre vos solutions aux problemes du present numero avant le . Les solutions re ues apres ette date ne seront prises en ompte que s'il nous reste du temps avant la publi ation des solutions. Chaque probleme sera publie dans les deux langues oÆ ielles du Canada (anglais et fran ais). Dans les numeros 1, 3, 5 et 7, l'anglais pre edera le fran ais, et dans les numeros 2, 4, 6 et 8, le fran ais pre edera l'anglais. La reda tion souhaite remer ier Jean-Mar Terrier, de l'Universite de Montreal, d'avoir traduit les problemes. 15 de embre 2009
. Corre tion. Propose par Mihaly Ben ze, Brasov, Roumanie. Determiner toutes les solutions de l'equation
M381
1 x−1
+
2 x−2
+
6 x−6
+
7 x−7
= x2 − 4x − 4 .
. Propose par l'Equipe de Mayhem. Graham et Vazz sont en train de projeter une nouvelle pelouse aux Quartiers Gen eraux de CRUX. Graham dit : \ Si tu hoisis une pelouse 9 metres plus longue et 8 metres plus etroite, la surfa e sera la m^eme." Vazz repond : \Si tu la hoisis 12 metres plus ourte et 16 metres plus large, la surfa e sera aussi la m^eme." Quelles sont les dimensions de la pelouse projetee ? M401
. Propose par Ne ulai Stan iu, E ole Te hnique Superieure de Saint Mu eni Sava, Ber a, Roumanie. Trouver toutes les paires ordonnees (a, b) d'entiers tels que M402
ab ba + ab + ba = 89 .
364 . Propose par Matthew Babbitt, etudiant, Albany Area Math Cir le, Fort Edward, NY, E-U. Jean a e rit un programme sur son ordinateur pour tester si un entier plus grand que 1 est un nombre premier. Sa sournoise de soeur, Ali e, a edit e le ode de telle sorte que si l'entree est impaire, la probabilite que le programme donne une reponse
orre te est de 52% et si l'entree est paire, ette probabilite est de 98%. Jean veri e le programme en testant deux entiers plus grands que 1 hoisis au hasard. Quelle est la probabilite que les deux reponses soient orre tes ? M403
. Propose par Bill Sands, Universite de Calgary, Calgary, AB. Un magasin vend des opies d'un ertain arti le a x$ pie e, ou y$ pour a
opies, ou en ore z$ pour b opies, a et b etant des entiers tels que 1 < a < b et x, y et z des nombres reels positifs. Pour rendre le rabais \y$ pour a
opies" interessant, y$ devrait e^ tre plus bas que le prix paye pour a a hats rabais \ z$ pour b au prix de x$, don y < ax. Pour rendre le deuxieme
opies" interessant aussi, on pourrait insister sur une des deux onditions : que le prix moyen pour une opie soit moindre que le (a) zb < ay ; .-a-d. prix orrespondant du premier rabais. (b) Chaque fois qu'on peut e rire b = qa + r ave q et r des entiers non negatifs, alors on a z < qy + rx ; .-a-d. qu'il en outerait ^ plus her d'a heter b opies en ombinant le premier rabais ave l'a hat individuel de opies supplementaires plutot ^ que d'opter pour le deuxieme rabais. Montrer que si la ondition (a) est satisfaite, alors la ondition (b) l'est aussi. Donner aussi un exemple pour montrer que la ondition (b) pourrait e^ tre satisfaite sans que la ondition (a) le soit. M404
. Propose par George Apostolopoulos, Messolonghi, Gre e. Trouver une formule donnant la valeur de la somme
M405
17 + 187 + 1887 + 18887 + · · · + 188 . . . 87 ,
ou le dernier terme ontient exa tement n hi res 8.
. Propose par Constantino Ligouras, etudiant, College S ienti que E. Majorana, Putignano, Italie. Le arre ABCD est ins rit dans le huitieme d'un er le de rayon 1 et de entre O de sorte qu'il ait un sommet sur haque rayon et les sommets B et C sur l'ar . Le arre EF GH est ins rit dans le triangle DOA de sorte que E et H soient sur les rayons et F et G soient sur AD. Dans le probleme M295 [2007 : 200, 202 ; solution 2008 : 203-204℄, on a vu que l'aire du
arr e √ 2− 2 ABCD est . Trouver l'aire du arre EF GH . 3 M406
.................................................................
365 . Corre tion. Proposed by Mihaly Ben ze, Brasov, Romania. Determine all of the solutions to the equation
M381
1 2 6 7 + + + = x2 − 4x − 4 . x−1 x−2 x−6 x−7
. Proposed by the Mayhem Sta . Graham and Vazz were marking out a new lawn at CRUX Headquarters. Graham said: \If you make the lawn 9 metres longer and 8 metres narrower, the area will be the same". Vazz said: \If you make it 12 metres shorter and 16 metres wider, the area will still be the same". What are the dimensions of the lawn? M401
M402. Proposed by Ne ulai Stan iu, Saint Mu eni Sava Te hnologi al High S hool, Ber a, Romania. Determine all ordered pairs (a, b) of positive integers su h that
ab ba + ab + ba = 89 .
. Proposed by Matthew Babbitt, student, Albany Area Math Cir le, Fort Edward, NY, USA. Jason wrote a omputer program that tests if an integer greater than 1 is prime. His devious sister Ali e has edited the ode so that if the input is odd, the probability that the program gives the orre t output is 52% and if the input is even, the probability that the program gives the orre t output is 98%. Jason tests the program by inputting two random integers ea h greater than 1. What is the probability that both outputs are orre t? M403
. Proposed by Bill Sands, University of Calgary, Calgary, AB. A store sells opies of a ertain item at $x ea h, or at a items for $y, or at b items for $z , where a and b are positive integers satisfying 1 < a < b and x, y , and z are positive real numbers. To make \a items for $y " a sensible bargain, $y should be less than buying a separate items; in other words we should have y < ax. To make \b items for $z " also a sensible bargain, we
ould insist on one of two onditions: M404
(a)
z y < ; b a
that is, the average pri e of an item under the \b items for $z " deal is less than under the \a items for $y" deal.
(b) Whenever we an write b = qa + r for nonnegative integers q and r, then z < qy + rx holds; that is, it should always ost more to buy b items by buying a ombination of a items plus individual items, than by hoosing the \b items for $z " deal. Show that if ondition (a) is true, then ondition (b) is also true. Give an example to show that ondition (b) ould be true while ondition (a) is false.
366 . Proposed by George Apostolopoulos, Messolonghi, Gree e. Determine a losed form expression for the sum
M405
17 + 187 + 1887 + 18887 + · · · + 188 . . . 87 ,
where the last term ontains exa tly n 8's. . Proposed by Constantino Ligouras, student, E. Majorana S ienti High S hool, Putignano, Italy. Square ABCD is ins ribed in one-eighth of a ir le of radius 1 and
entre O so that there is one vertex on ea h radius and two verti es B and C on the ar . Square EF GH is ins ribed in △DOA so that E and H lie on the radii, and F and G lie on AD. In problem M295 [2007 : 200, 202; √ 2− 2 solution 2008 : 203-204℄, we saw that the area of square ABCD is 3 . Determine the area of square EF GH . M406
Mayhem Solutions . Proposed by the Mayhem Sta . A re tangle has verti es A(0, 0), B(6, 0), C(6, 4), and D(0, 4). A horizontal line is drawn through P (4, 3), meeting BC at M and AD at N . A verti al line is drawn through P , meeting AB at Q and CD at R. Prove that AP , DM , and BR all pass through the same point. M369
Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON, modi ed by the editor. Sin e the sides of re tangle ABCD are parallel to the axes, any point on BC has x- oordinate 6 and any point on CD has y- oordinate 4. Thus, M has oordinates (6, 3) and R has oordinates (4, 4). 0 3 The line through A(0, 0) and P (4, 3) has slope 34 − = and passes −0 4 through the origin, so has equation y =
3 x. 4
The line through D(0, 4) and M (6, 3) has equation equivalently y = − 61 x + 4.
y−4 4−3 = , x−0 0−6
or
0 0−4 The line through B(6, 0) and R(4, 4) has equation xy − = , or −6 6−4 equivalently y = −2x + 12. Next, we nd the point of interse tion of the lines AP and DM by 48 equating to obtain 34 x = − 16 x + 4, or 11 x = 4, or x = . Sin e y = 34 x 12 11
367 on line AP , then y
3
48 36 . = 11 11 48 36 , . 11 11
= 4
Therefore, the point of interse tion of
lines AP and DM is Next, we nd the point of interse tion of lines AP and BR by equating 48 to obtain 34 x = −2x + 12, or 11 . As before, we see that x = 12, or x = 4 11 48 36 the point of interse tion of lines AP and BR is 11 , 11 . Sin e line AP interse ts lines DM and BR at thesame point, it follows 36 that all three lines are on urrent, meeting at 48 . , 11 11
Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; JULIA CLINE, student, Walt Whitman High S hool, Bethesda, MD, USA; KATHERINE JANELL EYRE, student, Angelo State University, San Angelo, TX, USA; ANTONIO GODOY TOHARIA, Madrid, Spain; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; HUGO LUYO SANCHEZ, Ponti ia Universidad Catoli a del Peru, Lima, IES \Abastos", Valen ia, Spain; BRUCE SHAWYER, Memorial University Peru; RICARD PEIRO, of Newfoundland, St. John's, NL; and NECULAI STANCIU, Saint Mu eni Sava Te hnologi al High S hool, Ber a, Romania. After nding the rst point of interse tion, it would suÆ e to show that this point lies on line BR. Stri tly speaking, we have proven that the lines through A and P , D and M , and B and R all pass through the same point; line segments DM and BR pass through this point, , 36 (though its extension does). but line segment AP does not ontain the point 48 11 11
. Proposed by the Mayhem Sta .
M370
(a) Prove that cos(A + B) + cos(A − B) A and B . (b) Prove that cos C + cos D = 2 cos C and D .
for all angles
= 2 cos A cos B
C +D 2
cos
C−D 2
for all angles
( ) Determine the exa t value of cos 20◦ + cos 60◦ + cos 100◦ + cos 140◦ , without using a al ulator. Solution by Courtis G. Chryssostomos, Larissa, Gree e. (a) Applying the sum and di eren e formulae for the osine, we obtain cos(A + B) + cos(A − B) = (cos A cos B − sin A sin B) + (cos A cos B + sin A sin B) = 2 cos A cos B .
(b) Applying the formula from (a) with A = obtain A + B = C and A − B = D, and so cos C + cos D = 2 cos
C +D 2
C +D 2
cos
and B
C−D 2
=
.
C−D , 2
we
368 ( ) We rearrange the given expression and apply the formula from (b): cos 20◦ + cos 60◦ + cos 100◦ + cos 140◦ = cos 140◦ + cos 20◦ + cos 100◦ + cos 60◦ 140◦ + 20◦ 140◦ − 20◦ = 2 cos cos + cos 100◦ + cos 60◦ 2
◦
2
◦
◦
= 2 cos 80 cos 60 + cos 100 + cos 60◦ 1 1 1 = 2 cos 80◦ + cos 100◦ + = cos 80◦ + cos 100◦ + 2 2 2 1 1 . = cos 80◦ + − cos 80◦ + = 2
2
Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JULIA CLINE, student, Walt Whitman High S hool, Bethesda, MD, USA; ANTONIO GODOY TOHARIA, Madrid, Spain; RALPH LOZANO, student, Missouri State University, Missouri, USA; HUGO LUYO SANCHEZ, IES \Abastos", Valen ia, Ponti ia Universidad Catoli a del Peru, Lima, Peru; RICARD PEIRO, Spain; NECULAI STANCIU, Saint Mu eni Sava Te hnologi al High S hool, Ber a, Romania; VASILE TEODOROVICI, Toronto, ON; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. There were two in omplete solutions submitted.
. Proposed by Panagiote Ligouras, Leonardo da Vin i High S hool, No i, Italy. Suppose that the line segment AB has length 3 and C is on AB with AC = 2. Equilateral triangles ACF and CBE are onstru ted on the same side of AB . If K is the midpoint of F C , determine the area of △AKE .
M371
Solution by Antonio Godoy Toharia, Madrid, Spain. Sin e K is the midpoint of segment F 1 F C , we have KC = (F C) = 1 = CE . ........ . 2 ... ...... . Be ause △ACF and △CBE are equilat. ... .. ... E ... ... K eral, ∠F CA = ∠ECB = 60◦ , so we have . . ... . . ....... ... . . ◦ . . . ... ∠KCE = 60 . . ... ... ...... . .... ... ... ... Therefore △KCE is also equilat... .. .... ... ..... . . . . . . ..................................................................... ................................... eral, and so KE = 1. A B C Sin e ∠EKC = ∠KCA, then KE is parallel to AB , and so we an think of △AKE as having base KE = 1 and height equal to the verti al distan e between the lines through KE and AB . √ ◦ The height of △CBE is h = 1 sin 60 = 23 . Thus, the distan e . ... ... ... ... ... ................ .. .. . ... . ... ... ... ... . ... ... ... .. .. . . . . . . . .. .. .. . . . . . . . .. . ... . ... ... .. .. .. ... ... ....... ... .......
between KE and AB is
√
3 . 2
Therefore, the area of △AKE is equal to
1 (1) 2
√ 3 2
=
√
3 . 4
Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG,
369 Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; JULIA CLINE, student, Walt Whitman High S hool, Bethesda, MD, USA; EMILY HENDRYX, student, Angelo State University, San Angelo, TX, USA; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; HUGO LUYO SANCHEZ, Ponti ia IES \Abastos", Valen ia, Spain; Universidad Catoli a del Peru, Lima, Peru; RICARD PEIRO, and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. There were two in orre t solutions submitted.
. Proposed by the Mayhem Sta . A real number x satis es x3 = x + 1. Determine integers a, b, and c so that x7 = ax2 + bx + c.
M372
Solution by Paul Bra ken, University of Texas, Edinburg, TX, USA. Sin e x3 = x + 1, we have x3 · x3 = (x + 1)(x + 1) = x2 + 2x + 1 , whi h yields x7
= =
x · x6 = x x2 + 2x + 1 = x3 + 2x2 + x
(x + 1) + 2x2 + x = 2x2 + 2x + 1 .
Thus, if a = 2, b = 2, and c = 1, we have x7 = ax2 + bx + c. Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; ANTONIO GODOY TOHARIA, Madrid, Spain; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; HUGO LUYO SANCHEZ, Ponti ia IES \Abastos", Valen ia, Spain; Universidad Catoli a del Peru, Lima, Peru; RICARD PEIRO, NECULAI STANCIU, Saint Mu eni Sava Te hnologi al High S hool, Ber a, Romania; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. It an be proven using algebra that a = 2, b = 2, and c = 1 is the only solution.
. Proposed by Kunal Singh, student, Kendriya Vidyalaya S hool, Shillong, India. The side lengths of a triangle are three onse utive positive integers and the largest angle in the triangle is twi e the smallest one. Determine the side lengths of the triangle. M373
Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. Suppose that the side lengths of the triangle are n − 1, n, and n + 1 for some positive integer n ≥ 3. (Smaller values of n do not give a triangle.) Suppose that the smallest angle is θ, whi h is opposite the shortest side (of length n − 1). Thus, the largest angle is 2θ, whi h is opposite the longest side (of length n + 1). −1 −1 n+1 n+1 By the Law of Sines, nsin = and so nsin = . Sin e θ sin 2θ θ 2 sin θ cos θ n+1 sin θ 6= 0, then cos θ = 2(n−1) . I.
370 Now, by the Law of Cosines, (n − 1)2
=
(n − 1)2
=
n2 − 2n + 1 = n(n + 1)2 n−1 2 n(n + 2n + 1) n2 + 2n + 1 n2 + 2n + 1 n
n2 + (n + 1)2 − 2n(n + 1) cos θ ; n+1 n2 + (n + 1)2 − 2n(n + 1) 2(n − 1) n(n + 1)2 n2 + n2 + 2n + 1 − ; n−1
=
n2 + 4n ;
= = = =
(n2 + 4n)(n − 1) ; (n + 4)(n − 1) (sin e n 6= 0) ; 2 n + 3n − 4 ; 5.
;
Therefore, n = 5, and the side lengths are 4, 5, and 6. Solution by Vasile Teodorovi i, Toronto, ON. Let the triangle be ABC , with side lengths AB = m, AC = m + 1, and BC = m + 2 for some positive integer m ≥ 2. Then the smallest angle is ∠ACB , whi h we label θ, and the largest is ∠BAC , whi h we label 2θ. Let AD be the angle bise tor of ∠BAC , with D on BC . Let BD = x and A DC = y . Note that BC = AB + 2, so ......... . . x + y = m + 2. .. ... ...... ... ... ..... II.
By the Angle Bise tor Theorem, +1 y so mm = . x
AC DC = , AB DB
... ..... ........ ..... m + 1 ... ..... ... . ..... . . . ... ..... ... ..... ... . . ..... . . . ... . ..... . . ... . . . ...............................................................................................................
m .......
y m+1 y x B D +1 = + 1 and so m x 2m + 1 x+y m+2 = = . m x x + 2) +1 (m + 1)(m + 2) This gives x = m(m and y = mm x= . 2m + 1 2m + 1
Thus,
C
Next, we note that ∠BAD = ∠BCA = θ so △BAD is similar to BD BA m(m + 2) m △BCA, when e = or m = , whi h gives the BA BC 2m + 1 m+2 2 2 equivalent equations (m + 2) = m(2m + 1), and m + 4m + 4 = 2m2 + m, and 0 = m2 − 3m − 4. Sin e m is a positive integer, then m = 4, and so the side lengths are 4, 5, and 6. Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; COURTIS G. CHRYSSOSTOMOS, Larissa, Gree e; ANTONIO GODOY TOHARIA, Madrid, Spain; RICHARD I. HESS, Ran ho Palos Verdes,
371 CA, USA; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring eld, MO, USA; IES \Abastos", Valen ia, Spain; NECULAI STANCIU, Saint Mu eni Sava RICARD PEIRO, Te hnologi al High S hool, Ber a, Romania; and EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON.
. Proposed by Mihaly Ben ze, Brasov, Romania. Suppose that p is a xed prime number with p ≥ 3. Determine the number of solutions to x3 + y3 = x2 y + xy2 + p2009 , where x and y are integers. M374
Solution by Missouri State University Problem Solving Group, Spring eld, MO, USA. The equation an be rewritten as p2009
= =
x3 − x2 y − xy 2 + y 3 = x2 (x − y) − y 2 (x − y) (x − y)(x2 − y 2 ) = (x − y)2 (x + y) .
Sin e x and y are integers and the only divisors of p2009 are of the form ±pn , then x − y = ±pk for some integer k with 0 ≤ k ≤ 1004. Thus, (x − y)2 = p2k and so x + y = p2009−2k . (The upper bound on k omes from the fa t that (x − y)2 is also a divisor of p2009 .) This yields x = y
=
1 1 2009−2k (x + y) + (x − y) = p ± pk , 2 2 1 1 2009−2k (x + y) − (x − y) = p ∓ pk . 2 2
These are all integers sin e p is odd (in fa t this is still true if p = 2 as long as k > 0). Ea h pair (x, y) is distin t sin e for ea h pair the values of x + y and x − y are di erent. Sin e there are 1005 possible values for k in the range 0 ≤ k ≤ 1004, there are then 2(1005) = 2010 solutions. Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; and EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON. There were four in orre t solutions and one in omplete solution submitted. M375. Proposed by Ne ulai Stan iu, Saint Mu eni Sava Te hnologi al High S hool, Ber a, Romania. Determine all real solutions to the system of equations
1 4 9 + 2 + 2 = 4; x2 y z
x2 + y 2 + z 2 = 9 ;
xyz =
9 2
.
Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. √
√ 6 √ 3 2 There are exa tly four solutions for (x, y, z), namely 2 , 3, 2 , √ √ √ √ √ √ √ 3√2 6 3 2 6 √ 3 2 6 , − 3, − , − 2 , 3, − 2 , and − 2 , − 3, 2 . 2 2
372 Note that if (x, y, z) is a solution, then so are (x, −y, −z), (−x, y, −z), and (−x, −y, z). (No other ombination of minus signs will work, sin e an even number of minus signs is ne essary to make the sign of xyz orre t.) Hen e, we may assume that x, y, z > 0 (sin e none of x, y, z an equal 0). The Cau hy{S hwarz Inequality says that if a, b, c, A, B , and C are real numbers, then a2 + b2 + c2 A2 + B 2 + C 2 ≥ (aA + bB + cC)2 ,
with equality if and only if (a, b, c) is a s alar multiple of (A, B, C). By the Cau hy{S hwarz Inequality, we have 36
= ≥
x2 + y 2 + z 2
1
+
4
+
9
x2 y2 z2 2 1 2 3 x +y +z = 36 . x y z
Sin e equality holds, then (x, y, z) must be a s alar multiple of
1 2 3 , , x y z
.
Thus, x = xk , y = 2k , and z = 3k , for some k > 0, and so x2 = k, y2 = 2k, y z and z 2 = 3k. Substituting ba k into r the original se ond equation, we r obtain√6k = 9, √ √ 3 3 6 and so k = 2 , when e x = 2 = 2 , y = 3, and z = 92 = 3 2 2 . This gives the four solutions above, as laimed. Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; COURTIS G. CHRYSSOSTOMOS, Larissa, IES \Abastos", Gree e; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; and RICARD PEIRO, Valen ia, Spain. There were two in omplete solutions submitted.
Problem of the Month Ian VanderBurgh
Sometimes we mathemati ians like to use a sledgehammer when perhaps a smaller instrument would be in order. (2009 Gauss Contest, Grade 7) If x, y, and z are positive integers with xy = 18, xz = 3, and yz = 6, what is the value of x + y + z ? (A) 6 (B) 10 (C) 25 (D) 11 (E) 8
Problem 1
If we weren't in Grade 7, then here are two ways that we might ta kle this problem:
373 Solution 1 Sin e xy = 18, xz = 3, and yz = 6, by multiplying these equations we have (xy)(xz)(yz) = 18(3)(6), or x2 y2 z 2 = 182 , or (xyz)2 = 182 . Sin e x, y, and z are positive, then xyz > 0, so xyz = 18. We an now ombine this with the original three equations as follows. 18 Sin e xy = 18, we have xyz = , or z = 1. Sin e xz = 3, we have xy 18
18 xyz = , xz 3
or y = 6. Sin e yz = 6, we have xyz yz Therefore, x + y + z = 3 + 6 + 1 = 10.
=
18 , 6
or x = 3.
18 Sin e xy = 18 and xz = 3, then xy , or yz = 6, or y = 6z . = xz 3 Sin e yz = 6, then (6z)z = 6, or z 2 = 1. Sin e z > 0, then z = 1, and so y = 6z = 6. Sin e xz = 3 and z = 1, then x = 3. Therefore, x + y + z = 3 + 6 + 1 = 10. Solution 2
These are two great solutions using standard te hniques for solving systems of equations. But they're hardly suitable for Grade 7 students. This problem is a perfe tly good Grade 7 problem, though. Solution 3 We are told that x, y , and z are positive integers. Sin e xz = 3 and 3 is a prime, then x and z must be 1 and 3, or 3 and 1, respe tively. Let's look at the ase x = 1 and z = 3. Sin e x = 1 and xy = 18 and the number that we multiply 1 by to get 18 is 18 itself, then y = 18. Therefore, yz = 18(3) = 54, whi h disagrees with the equation yz = 6. Thus, x = 1 and z = 3 is not the orre t ombination. (For the re ord, I never hoose the orre t line in the gro ery store either.) So we look at the ase x = 3 and z = 1. Sin e x = 3 and xy = 18 and the number that we multiply 3 by to get 18 is 6, then y = 6. (The produ t yz does in fa t equal 6, as required). Thus, x = 3, y = 6, and z = 1. Therefore, x + y + z = 3 + 6 + 1 = 10.
It's easy to get \trapped" into using the high powered te hniques that we know, but sometimes there is a ni er solution that uses less ma hinery. Here is a problem in a similar vein that has developed a following in the math ontest world. (1988 UK S hools Mathemati al Challenge) Weighing the baby at the lini was a problem. The baby would not keep still and aused the s ales to wobble. So I held the baby and stood on the s ales while the nurse read o 78 kg. Then the nurse held the baby while I read o 69 kg. Finally, I held the nurse while the baby read o 137 kg. What is the ombined weight of all three (in kg)? (A) 142 (B) 147 (C) 206 (D) 215 (E) 284 Problem 2
Again, there is a standard system-of-equations type solution whi h is worth seeing. To save onfusion, we'll give the narrator a name hosen at random, say, Tony.
374 Solution 1 Let Tony's weight be x kg, let the baby's weight be y kg, and let the nurse's weight be z kg. The ombined weight of Tony and the baby is 78 kg, so x + y = 78. The ombined weight of the baby and the nurse is 69 kg, so y + z = 69. The
ombined weight of Tony and the nurse is 137 kg, so x + z = 137. We an show this system of equations ni ely in a visual way:
x +
y y
x
+ +
z z
= 78 , = 69 , = 137 .
By the way, laying out the equations in this fashion is a really useful thing to do and gives you a mu h better idea of what to do than by writing x + y + x +
y z z
= = =
78 , 69 , 137 .
In any event, the \ni e" way of writing the equations allows us to see that adding the three equations is a really good idea. When we do this, we obtain 2x + 2y + 2z = 78 + 69 + 137, or 2(x + y + z) = 284, or x + y + z = 142. Noti e that we don't a tually need to determine x, y, and z at all! But, again, we don't need to do anything nearly that fan y. In fa t, we
an get away without doing any algebra at all. Solution 2 First, we look at the fa t that the ombined weight of Tony and the baby is 78 kg and the ombined weight of the nurse and the baby is 69 kg. Sin e the baby's weight is in luded in both of these totals, then Tony must be 78 − 69 = 9 kg heavier than the nurse (that is, the di eren e between Tony's weight and the nurse's weight is 9 kg). But the ombined weight of Tony and the nurse is 137 kg. We want to nd two numbers that add to 137, one of whi h is 9 greater than the other. To nd these numbers, we an subtra t 9 from 137 to get 128 and then divide by 2 to get 64. The numbers 64 and 64 + 9 = 73 di er by 9 and add to 137, so must be the weights, in kg, of the nurse and Tony, respe tively. Sin e the ombined weight of Tony and the baby is 78 kg and Tony weighs 73 kg, then the weight of the baby is 5 kg. Therefore, the ombined weight of all three is 64 + 73 + 5 = 142 kg. leap.
So put your sledgehammer ba k in the garage, and think before you
375
THE OLYMPIAD CORNER No. 280
R.E. Woodrow
As a rst problem set for this number we look to the Grade 12 problems of the Final Round of the Estonian National Olympiad 2005{2006. My thanks go to Robert Morewood, Canadian Team Leader to the 2006 IMO in Slovenia, for olle ting them for our use. ESTONIAN NATIONAL OLYMPIAD 2005{2006 Final Round
Grade 12
. We all a gure a ship if it is made up of unit squares onne ted by ommon edges. Prove that if there is an odd number of possible di erent ships
onsisting of n unit squares on a 10 × 10 board, then n is divisible by 4. 1
. Find the smallest distan e between points P and Q in the xy-plane, if P lies on the line y = x and Q lies on the urve y = 2x .
2
. Prove or disprove the following statements:
3
(a) For every integer n ≥ 3, there exist n distin t positive integers su h that the produ t of any two of them is divisible by the sum of the remaining n − 2 numbers. (b) For some integer n ≥ 3, there exist n distin t positive integers su h that the sum of any n − 2 of them is divisible by the produ t of the remaining two numbers. 4. The a ute triangle ABC has ir um entre O and triangles BCO , CAO , and ABO have ir um entres A′ , B ′ , and C ′ , respe tively. Prove that the area of triangle ABC does not ex eed the area of triangle A′ B ′ C ′ .
. The Ababi alphabet onsists of the letters A and B , and the words in the Ababi language are pre isely those that an be formed by the following two rules:
5
(a) The letter A is a word. (b) If s is a word, then s ⊕ s and s ⊕ s¯ are words, where s¯ denotes the word that is obtained by repla ing all letters A in s with letters B , and vi e versa; and x ⊕ y denotes the on atenation of words x and y.
376 The Ululu alphabet also onsists of the letters A and B , and the words in the Ululu language are pre isely those that an be formed by the following two rules: (a) The letter A is a word. (b) If s is a word, then s⊗s and s⊗ s¯ are words, where s¯ is de ned as above and x ⊗ y is the word obtained from words x and y of equal length by writing the letters of x and y alternatingly, starting from the rst letter of x. Prove that the two languages onsist of the same words.
Next we turn to the Kazakhs and the problems of the II International Zhautykov Olympiad in Mathemati s, Almaty, 16{17 January, 2006. Thanks again go to Robert Morewood, Canadian Team Leader to the 2006 IMO in Slovenia, for olle ting them for us. II INTERNATIONAL ZHAUTYKOV OLYMPIAD IN MATHEMATICS Almaty { Day 1 (January 16, 2006)
. Find all positive integers n su h that n = ϕ(n)+402, where ϕ is the Euler phi-fun tion (it is known if p1 , p2 , . . . , pkare the distin t prime divisors that 1 1 1 of n, then ϕ(n) = n 1 − p 1− ··· 1 − ; and ϕ(1) = 1). p p
1
1
k
2
. The points K and L lie on the sides AB and AC , respe tively, of the triangle ABC su h that BK = CL. Let P be the point of interse tion of the segments BL and CK , and let M be an inner point of the segment AC su h that the line M P is parallel to the bise tor of the angle ∠BAC . Prove that CM = AB . 2
3. An m × n table with 4 ≤ m ≤ n is alled good if the numbers 0 and 1 an be written in the unit squares of the table su h that the following onditions hold:
(a) not all numbers are 0's and not all are 1's; (b) the number of 1's in any 3 × 3 subsquare is onstant; ( ) the number of 1's in any 4 × 4 subsquare is also onstant. Find all pairs of positive integers table.
(m, n)
for whi h there is a good
m×n
377 Almaty { Day 2 (January 17, 2006)
. A heap of 100 stones is given. A partition of the heap into k smaller heaps is spe ial provided that (i) no two heaps have the same number of stones, and (ii) any further partition of any of the heaps into two new heaps results in k + 1 heaps for whi h (i) fails (a heap always ontains at least one stone).
4
(a) Find the maximum number k for whi h there is a spe ial partition of the given heap of 100 stones into k heaps. (b) Find the minimum number k for whi h there is a spe ial partition of the given heap of 100 stones into k heaps. . Let a, b, c, and d be real numbers su h that a + b + c + d = 0. Prove that
5
(ab + ac + ad + bc + bd + cd)2 + 12 ≥ 6(abc + abd + acd + bcd) . 6. Let ABCDEF be a onvex hexagon su h that AD = BC + EF , CD EF AB BE = AF + CD , and CF = DE + AB . Prove that = = .
DE
AF
BC
Next is the 50th Mathemati al Olympiad of the Republi of Moldova, and problems of the 10th form. We again thank Robert Morewood, Canadian Team Leader to the 2006 IMO in Slovenia, for olle ting them for us. 50
th
MATHEMATICAL OLYMPIAD OF THE REPUBLIC OF MOLDOVA th 10
Form
Mar h 2006
. Let a, b, and c be the side lengths of a right triangle with hypotenuse of length c, and let h be the altitude from the right angle. Find the maximum h value of ca + . +b 1
. Let n ≥ 2 be an integer and M = {0, 1, 2, . . . , n − 1}. For a nonzero integer a let fa : M → M where fa (x) is the remainder of ax upon division by n. Give a ne essary and suÆ ient ondition for the fun tion fa to be bije tive and prove that an(n−1) − 1 is divisible by n2 whenever n is a prime number. 2
. The quadrilateral ABCD is ins ribed in a ir le. The tangents to the ir le at A and C interse t at a point P not on BD and su h that P A2 = P B · P D. Prove that BD passes through the midpoint of AC .
3
378 4. Find all values of the real number a for whi h the following equation has a unique solution:
2x2 − 6ax + 4a2 − 2a − 2 + log2 2x2 + 2x − 6ax + 4a2
= log2 x2 + 2x − 3ax + 2a2 + a + 1
.
5. Let n be a positive integer and let x1 , x2 , . . . , xn be real numbers in the interval 14 , 23 . Find the minimum value of the expression
log1.5x1
1 1 − 2 36x22
+ log1.5x2
1 1 − 2 36x23
+ · · · + log1.5xn
1 1 − 2 36x21
.
When is this value attained? . Triangle ABC is isos eles with AC = BC and P is a point inside the triangle su h that ∠P AB = ∠P BC . If M is the midpoint of AB , prove that ∠AP M + ∠BP C = 180◦ .
6
. The interior angles of a onvex o tagon are all equal and all side lengths are rational numbers. Prove that the o tagon has a entre of symmetry.
7
. Let M = {x2 + x | x is a positive integer}. For ea h integer k ≥ 2 prove that there exist a1 , a2 , . . . , ak , bk in M su h that a1 + a2 + · · · + ak = bk . 8
We also give the problems of the Se ond and Third Team Sele tion Tests for the IMO 2006 from Chisinau, Moldova, Mar h 25-26, 2006. Thanks again go to Robert Morewood, Canadian Team Leader to the 2006 IMO in Slovenia, for olle ting them for us. REPUBLIC OF MOLDOVA INTERNATIONAL MATHEMATICAL OLYMPIAD Se ond Team Sele tion Test
Mar h 25, 2006 (Chisin au)
. Let {an }∞ n=0 be a sequen e, with a0 = 2, a1 = 1, and an+1 = an + an−1 for n ≥ 1. Prove that a59 divides the number (a30 )59 − 1.
1
2. The point P lies in the interior of the ir le C1 , whi h in turn lies in the interior of the ir le C2 , and the point Q lies in the exterior of the ir le C2 . Ea h line ℓi (i ≥ 3) is di erent from the line P Q, passes through P , and uts the ir le C1 at the points Ai and Bi ; and the ir um ir le of triangle QAi Bi
uts the ir le C2 at the points Mi and Ni . Prove that all the lines Mi Ni are
on urrent.
379 . Let a, b, c be the side lengths of a triangle and let s be the semiperimeter. Prove that 3
s
a
(s − b)(s − c) bc
s
+ b
s
(s − c)(s − a)
+ c
ac
4. Let n be a positive integer and let A = number of distin t solutions of the equation
(s − a)(s − b) ab
{1, 2, . . . , n}.
≥ s.
Determine the
X ∪ Y ∪ Z = A,
where two solutions are onsidered the same if they di er only in the ordering of X , Y , and Z . Third Team Sele tion Test
Mar h 26, 2006 (Chisin au)
. The point P is in the interior of triangle ABC . The rays AP , BP , and CP
ut the ir um ir le of the triangle at the points A1 , B1 , and C1 , respe tively. Prove that the sum of the areas of the triangles A1 BC , B1 AC , and C1 AB does not ex eed s(R − r), where s, R, and r are the semiperimeter, the
ir umradius, and the inradius of triangle ABC , respe tively. 5
. Let n ≥ 2 be an integer and let X be a set with n + 1 elements. The sequen es (a1 , a2 , . . . , an ) and (b1 , b2 , . . . , bn ) of distin t elements from the set X are alled separated if ai = bj for some i 6= j . Determine the maximum number of ordered sequen es of n elements from X su h that any two of them are separated. 6
7
. Let a, b, and c be positive real numbers su h that abc = 1. Prove that a+3 (a +
1)2
+
b+3 (b +
1)2
+
c+3 (c + 1)2
≥ 3.
. Let f (n) be the number of permutations (a1 , a2 , . . . , an ) of the set {1, 2, . . . , n} su h that a1 = 1 and |ai − ai+1 | ≤ 2 for 1 ≤ i ≤ n − 1. Prove that f (2006) is divisible by 3.
8
Next we give the problems of the Final Round, 2nd Grade High S hool of the Youth Mathemati al Olympiad of the Aso ia ion, Venezolana de Competen ias Matemati as, written May 27, 2006. Thanks go to Robert Morewood, Canadian Team Leader to the IMO in Slovenia, for olle ting them.
380 YOUTH MATHEMATICAL OLYMPIAD Aso ia ion Venezolana de Competen ias Matemati as
nd Grade High S hool (May 27, 2006)
Final Round, 2
. A positive integer has 223 digits and the produ t of these digits is 3446 . What is the sum of the digits?
1
. Find all solutions of the equation m2 − 3m + 1 = n2 + n − 1, where m and n are positive integers.
2
. De ne the sequen e a1 , a2 , a3 , . . . as follows: let a1 = a2 = 1003; a3 = a2 − a1 = 0; a4 = a3 − a2 = −1003; and in general an+1 = an − an−1 for any n ≥ 2. Compute the sum of the rst 2006 terms of the sequen e. 3
. Joseph, Dario, and Henry prepared some labels. On ea h label they wrote one of the numbers 2, 3, 4, 5, 6, 7, or 8. David joined them and stu k one label on the forehead of ea h friend. Joseph, Dario, and Henry ould not see the numbers on their own foreheads, they only saw the numbers of the other two. David said, \You do not have distin t numbers on your foreheads, and the produ t of the three numbers is a perfe t square." Ea h friend then tried to dedu e what number he had on his forehead. Could anyone dis over it? 4
. Let ABC be an isos eles triangle with ∠B = ∠C = 72◦ . Find the value of ABBC . Hint: Consider the bise tor CD of ∠ACB . − BC 5
As a nal problem set from our 2006 IMO le we turn to the 42nd Mongolian Mathemati al Olympiad, 10th Grade, written at Ulaanbaatar in April and May of 2006. Thanks go to Robert Morewood, Canadian Team Leader at the 2006 IMO in Slovenia, for olle ting them for us. Unfortunately, our version of problem 5 below was in omplete, and we
ould not remedy this in time. We will print this problem in a future issue as soon as we lo ate it. nd 42 MONGOLIAN MATHEMATICAL OLYMPIAD th 10
Grade
April 28-May 4, 2006 (Ulaanbaatar)
. Let a, b, c, d, e, and f be positive integers satisfying the relation ab + ac + bc = de + df + ef , and let N = a + b + c + d + e + f . Prove that if N | (abc + def ), then N is a omposite number.
1
. Let m and n be positive integers. In how many di erent ways an one ll in an n × n grid with nonnegative integers su h that the sum of every n numbers from di erent rows and olumns is m? 2
381 3. Cir les ω , ω0 have distin t entres and ω0 lies inside ω . The four ir les ω1 , ω2 , ω3 , and ω4 ea h tou h ω internally and ω0 externally. Prove that the interse tion points of the external tangents of ωi and ωj (i 6= j ) are ollinear.
. Let p and q be distin t prime numbers and let a and b be distin t positive integers. Prove that there exists a prime number r su h that r | (apq − bpq ) and r ≡ 1 (mod pq). 4
. Omitted.
5
. Let a, b, c be positive real numbers su h that a + b + c = 1. Prove that
6
r
(1 + a)
r r 1−a 1−b 1−c + (1 + b) + (1 + c) a b c √ 3 3 (1 + a)(1 + b)(a + c) ≥ ·p 4 (1 − a)(1 − b)(1 − c)
.
Now we give readers' solutions to problems in the September 2008 Corner and the X Bosnian Mathemati al Olympiad given at [2008 : 284{285℄. . Let H be the ortho entre of an a ute-angled triangle ABC . Prove that the midpoints of AB and CH and the interse tion point of the interior bise tors of ∠CAH and ∠CBH are ollinear.
1
Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; George Apostolopoulos, Messolonghi, Gree e; Oliver Geupel, Bruhl, NRW, Germany; D.J. Smeenk, Zaltbommel, the Netherlands; and Titu Zvonaru, Comane sti, Romania. We give the version of Geupel. Let us denote α = ∠CAB , β = ∠ABC , γ = ∠BCA, and let C ..... A′ = AH ∩ BC , B ′ = BH ∩ AC . ............ .... ..... ....... . .... Let M , N , O be the midpoints of . .... ... ... ..... segments AB , A′ B ′ , CH , respe β ..... ...... .... .... A′ . ..... O . ′ ′ .. α . ..... tively, let P = CH ∩ A B , and let . ... . ...................... . . . . . . ... ...N . . . ... . . ..... . . . . . . . . . . . . Q be the interse tion point of the .... . P ............ ... .... ..................... ..... ........... .... . ′ ′ . . . ′ .. ......... .. .... .. bise tors of ∠CAA and ∠CBB , ..... B .. .................. ... .... . . .... . ... respe tively. We are to show that ..... .... ... H .... . . . .... ... the points M , O, Q are ollinear. ... ..... ... . .... . . ... .. β It is a well-known fa t that the ........α ................................................................................................................................ midpoints of the diagonals of a omA B M plete quadrilateral are ollinear. Applying this to the omplete quadrilateral generated by the quadrilateral A′ CB ′ H , we see that the points M , N , and O are ollinear. Due to the fa t ... .. .. .. .. ... .. .
382 that ∠HA′ C = ∠CB ′ H = 90◦ , the quadrilateral
ir um entre O. Hen e, ON ⊥ A′ B ′ , whi h implies that ∠HOM
= = = =
thus,
A′ CB ′ H
is y li with
∠P ON = 90◦ − ∠N P O = 90◦ − ∠A′ P C 90◦ − 180◦ − ∠P A′ C − ∠A′ CP ∠P A′ C + ∠BCH − 90◦ α + 90◦ − β − 90◦ = α − β ;
∠AM O = 90◦ − ∠HOM = 90◦ − α + β .
Sin e ∠BAA′ = 90◦ − β and ∠A′ AC ∠BAQ =
∠BAA′ +
=
90◦ − β +
1 2 1 2
(1)
= 90◦ − γ , we have
∠A′ AC 90◦ − γ
= 135◦ − β −
1 2
γ.
Similarly, ∠ABQ = 135◦ − α − 12 γ . Hen e, ∠AQB
= 180◦ − ∠BAQ − ∠ABQ 1 1 = 180◦ − 135◦ − β − γ − 135◦ − α − γ 2 2 ◦ ◦ = α + β + γ − 90 = 90 .
We dedu e that △AM Q is isos eles with AM ∠AM Q = =
= M Q. Thus, 1 ◦ ◦ ◦ 180 − 2∠M AQ = 180 − 2 135 − β − γ 2 90◦ − α + β .
(2)
From (1) and (2) it is readily seen that ∠AM O = ∠AM Q, that is, the points M , O , and Q are ollinear. This ompletes the proof. 2. Let a1 , a2 , and a3 be nonnegative real numbers with a1 + a2 + a3 Prove that 1 √ √ √ a1 a2 + a2 a3 + a3 a1 ≤ √ .
= 1.
3
Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru, Comane sti, Romania. We give the solution by Bataille. √ Sin e the fun tion f (x) = x is on ave on [0, ∞) and a1 +a2 +a3 = 1, Jensen's Inequality yields a1
√ √ √ √ a2 + a2 a3 + a3 a1 ≤ a1 a2 + a2 a3 + a3 a1 .
(1)
383 Now, from the well-known inequality a21 + a22 + a23 ≥ a1 a2 + a2 a3 + a3 a1 , we obtain 3(a1 a2 + a2 a3 + a3 a1 ) ≤ a21 + a22 + a23 + 2(a1 a2 + a2 a3 + a3 a1 ) = (a1 + a2 + a3 )2 = 1 ,
so that a1 a2 + a2 a3 + a3 a1 ≤ 13 . From (1), it follows that a1
1 √ √ √ a2 + a2 a3 + a3 a1 ≤ √ 3
.
. Given are a ir le and its diameter P Q. Let t be a tangent to the ir le, tou hing it at T , and let A be the interse tion of the lines t and P Q. Let p and q be the tangents to the ir le at P and Q respe tively, and let P T ∩ q = {N } and QT ∩ p = {M } . Prove that the points A, M , and N are ollinear. 4
Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; Mi hel Bataille, Rouen, Fran e; D.J. Smeenk, Zaltbommel, the Netherlands; and Titu Zvonaru, Comane sti, Romania. We give the write-up of Amengual Covas. Equivalently, we will prove that: Let T be a point of a ir le with a given diameter P Q. Let p and q be the tangents to the ir le at P and Q, respe tively, and let P T ∩ q = {N } , QT ∩ p = {M } , and M N ∩ P Q = {A} . Then AT tou hes the given ir le at T . p
q
... ... ... .... ... .... ... ... ... . . ... ................................................................... . . . . . . . . . . . . ... . ........... ... ........... ....... . . . . . . . . . . .... . . ........ ... .... ...... ...... . . . . . . . . . . . .... . . . ....... .. .... ..... .... . . . . . . . . . ... . ..... ... ...... ... .... . . . . . . .... . . . . ..... ..... ... .. ... . . . . . . .... . . . ... ..... .... ... ... . . . . . . .... . . ..... ... .... .. .... . . . . . . . .... ..... ... ... . ... ... . . . . . ... . . ..... ... ... ... .. . . .... . . ... ..... ... ... .. ... . . ... . ..... .. ... ... . . . . . .... ... ..... .. ... .... . . . . ... .... . . ........ ... ... ...... ... . ... ... ... . ... . . . . . . .... . ... . ... ... .. ....... .. ... ... ... .... . .. ........ ... .... .. . . . . . ... .. ...... ..... ... ..... ... . . . . . ... ..... ... ... . ... .... .................. . . . . . . .... ... .......... ... ... ..... .......... ... ... ..................................... .. ... .......... .... .......... ........ ... ...... .......... ... ... ... .......... ................ ............ .... ......... .......... . . .. .. . . . . . . . .......... . ..... ... . ... .. .......... ...... ........................ ..... ... .. ..... ... . ..... ... .... .. ... .......................................................................... ..... .. .. ... ........... . ....... ... .... . ............................................ ........ . . ...... . . . . . . ... . . ..... . . . . .......... ..... .......... ..... ........................ .... . .... . ... ... . . . . . . . . ... ... ..... .............. ... .... ....... ... .. . . . . .... . . . . . ... . . . . ... ......... ... .... .. ........... ..... ... .... ... ..... . . .... . . . ..... . . . . . . . . . ... . . . .......... ... .... ... . ... .... .. ... .... .... . ..... . . . . . . . . . . . . . . . . . . . . . . . . .......... ..... . ..... ... .. ... .... .. ... .... . . . . . . . . .... .... . ..... . ..... . . . . . . . ... .. . ... . .................. .. ... .... . ..... . . . . . . . . . . . . . . . . . . . . .......... ......... ... .... .... ....... . .... .. ..... . . . . . . .......... ..... ......... . . . . . . ........ . . . . . . . . . . . ..... ... ..... .......... .... . .. ..... ......... . ..... ........... . . . . . . . . . . . . .......... ..... . . . . ... .. ........ . ........... .. .. ....... .... . .. .. . ........ ................. ........... ................................................................................................................................................................................................................................................................................................................................................... .. .. .... ... . . .. .. .. .. ... .. .. ... .. ... . . ... .. .... ... .... .... ..... .... ....... ..... . . . ........ . . . . ........... ..... .....................................................
M
U
N
t
T
.... ......... ..... ..........
P
O
V
Q
A
384 Let AT
∩ p = {U }.
Then, by Ceva's Theorem, M U P Q AN · · = 1. U P QA N M
But p ⊥ P Q and q MU =1 UP
Let
and M U
⊥ P Q,
so that pkq, implying that
PQ NM = . QA AN
Hen e,
= UP . AT ∩ q = {V }. Sin e NV AV VQ = = MU AU UP
,
we obtain N V = V Q. Thus, U and V are the midpoints of M P and N Q, respe tively. Sin e ∠P T Q is a right angle, so also ∠M T P = 90◦ = ∠N T Q. Hen e both ir les onstru ted with M P and N Q as diameters pass through T . Sin e the distan e between their entres U and V equals the sum of their radii, these ir les are externally tangent to ea h other at T . Let their ommon tangent at T interse t P Q at O. This tangent is perpendi ular to the line U V through the entres, and sin e P O = OT = OQ ,
the point O is the midpoint of P Q, making OT a radius of the given ir le in our problem. Thus, OT ⊥ AT , and the on lusion follows. . A permutation 2 inequality aak ≤ k+1 identity. 5
of the set {1, 2, . . . , n} satis es the for ea h k = 1, 2, . . . , n − 1. Prove it is the
(a1 , a2 , . . . , an ) k+2
Solution by Titu Zvonaru, Comane sti, Romania. Suppose that an 6= n. Then an ≤ n − 1, and we have
a2n−1 ≤ (n + 1)an ≤ (n + 1)(n − 1) = n2 − 1 ,
hen e, an−1 ≤ n − 1. Now, sin e a2n−2 ≤ nan−1 ≤ n2 − n, we dedu e that an−2 In general, if 1 ≤ k ≤ n − 1 and ak+1 ≤ n − 1, then a2k ≤ (k + 2)ak+1 ≤ (n + 1)(n − 1) = n2 − 1 ,
≤ n − 1.
and we dedu e that ak ≤ n − 1. From an 6= n it then follows that ak < n for ea h index k. However, n ∈ {a1 , a2 , . . . , an }, a ontradi tion! Therefore, an = n, and similarly, an−1 = n − 1, . . . , a2 = 2, a1 = 1, whi h means that (a1 , a2 , . . . , an ) is the identity.
385 6
. Let a, b, and c be integers su h that b c a + + = 3. b c a
Prove that abc is a perfe t ube. Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru, Comane sti, Romania. We give the generalization of Bataille. a b c + + = n for b c a that ab + bc + ac = n.
We generalize and show that if
some integer n,
then abc is a perfe t ube. Assume Certainly abc is a perfe t ube if |abc| = 1. From now on, we suppose that |abc| > 1. It suÆ es to prove that for any prime p dividing abc, the exponent of p in the prime fa torization of abc is a multiple of 3. Let p be su h a prime. There exist nonnegative integers r, s, and t with (r, s, t) 6= (0, 0, 0) and su h that a = pr a1 , b = ps b1 , and c = pt c1 , where a1 , b1 , and c1 are integers not divisible by p. We have to show that r + s + t is divisible by 3. From the hypothesis, we obtain pr−s a21 c1 + ps−t b21 a1 + pt−r c21 b1 = na1 b1 c1 .
(1)
Let N be the number of negative integers among r − s, s − t, t − r, and
onsider the possibilities for N . First, note that (r − s) + (s − t) + (t − r) = 0, so N = 3 annot hold. Also, N = 1 annot hold either: if we had, say, r − s, s − t ≥ 0 and t − r < 0, then the relation (1) would imply c21 b1 = pr−t na1 b1 c1 − pr−s a21 c1 − ps−t b21 a1
,
ontradi ting the fa t that p does not divide c21 b1 . If N = 0, then r − s, s − t, and t − r are nonnegative and sum to zero, hen e r = s = t and r + s + t is a multiple of 3. If N = 2, then suppose (for example) that s−t, t−r < 0 and r−s > 0. Then (1) yields pr−s a21 c1 +
b21 a1 pt−s
+
c21 b1 pr−t
= na1 b1 c1 .
If, in addition, we had t − s 6= r − t, then by multiplying by pmax(t−s,r−t) we would dedu e that p divides b21 a1 (if t − s > r − t) or p divides c21 b1 (if r−t > t−s), thus arriving at a ontradi tion in either ase. Consequently, t − s = r − t and r + s + t = 3t, a multiple of 3. This last ase ompletes the analysis.
386 We next look at solutions from our readers to the problems of the I elandi Mathemati al Contest 2004{2005 (Final Round) given in the Corner at [2008 : 285℄. . How many subsets with three elements an be formed from the set so that 4 is a fa tor of the produ t of the three numbers in the subset?
1
{1, 2, . . . , 20}
Solved by George Apostolopoulos, Messolonghi, Gree e; and John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB. We give the solution of M Loughlin. = 1140 subsets with 3 elements. There are 20 3 All of these subsets have elements that when multiplied will have 4 as a fa tor, ex ept in two ases: • •
all elements are odd, of whi h there are
10 3
= 120
subsets;
two elements are odd and the 5third element is even but not a multiple of 4, of whi h there are 10 = 225 subsets. 2 1 Therefore, 1140 − (120 + 225) = 795 subsets meet the requirements.
. Triangle ABC is equilateral, D is a point inside the triangle su h that DA = DB , and E is a point that satis es the onditions ∠DBE = ∠DBC and BE = AB . How large is the angle ∠DEB ? 2
Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; and Titu Zvonaru, Comane sti, Romania. We give Zvonaru's answer. Sin e DA = DB , the point D lies on the perpendi ular bise tor of AB , hen e ∠DCB = 30◦ . We have = = =
BC ; ∠DBC ; BD .
E
..... ... ....... ......... ..... .....
BE ∠DBE BD
A ........ ... ....... ... .. ... ... .... ..... . .. .. ... ... ... .. .. ... ... ... .. ... .. .... . . ... ... .. . . .. ... .. ... . ... . . .. . ... .... . ... ... ................. ... . .. ...... . . . . . ... .................. .... ............... . . . . . .. ...... .... ........ ... ... ........ ........ .. ........ ....... .. ... ... ........ ............. .... ... ... ..... .......... ... .. .... ... . . . . . . . . . . . . ........ ... .... ... ........ .... ... ....... ... ........ ..... ... ........ ... ............ ... ........ . . . . ... . . . . . . . . . . . . . ........ ......... ..... . . ... . . . . . ... . . . . . . . . . . . . . . . . . ........ ... ........ ..... . . . . . . . . . . .. . . . . . . . . . . . . . ........ .. . ............. ......... . . . . . . . . . . . . . . . . . . . ........ .... . ................................................ ........ ... . ... . ....................... .. . . . .......................................................................................................................................................................................
It follows that △BDE and △BDC are ongruent (side-angle-side), hen e ∠DEB = DCB = 30◦ .
D
B
C
. Find a three-digit number n that is equal to the sum of all the two-digit numbers that an be formed by using only the digits of the number n. (Note that if a is one of the digits of the number n, then aa is one of the two-digit numbers that an be formed.)
3
387 Solution by Titu Zvonaru, Comane sti, Romania. Let a, b, and c denote distin t digits. The following ases are ruled out: (a)
n = aaa = aa ⇒ 111a = 11a ⇒ 100a = 0;
(b)
n = aab = aa + bb + ab + ba ⇒ 110a + b = 22a + 22b ⇒ 88a = 21b. There is no solution here, sin e 11 divides 88a but does not divide 21b.
( )
n = aba = aa + bb + ab + ba ⇒ 79a = 12b; divisible by 79.
(d)
no solution in this ase.
n = baa = aa + bb + ab + ba ⇒ 78b = 11a;
divisible by 11.
no solution as b is not
no solution as 78b is not
Thus, n = abc = aa + bb + cc + ab + ba + bc + cb + ac + ca and we have abc = 33(a + b + c). It follows that 3 divides abc, hen e 3 divides a+b+c and then 9 divides 33(a + b + c). Now 9 divides abc, hen e 9 divides a + b + c and n an only be one of 33 · 9 = 297 ;
33 · 18 = 594 ;
33 · 27 = 891 .
Therefore, 594 = 55 + 99 + 44 + 59 + 95 + 94 + 49 + 54 + 45 is the only solution. 4. Transportation between six ities is su h that between any two ities there is either a bus or a train but not both of these. Show that among these six ities there are three ities that are linked either only by buses or only by trains. Solved by George Apostolopoulos, Messolonghi, Gree e; and John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB. We give the response of M Loughlin. Consider six ities A, B , C , D, E , and F . There must be either at least three train links to ity A or at least three bus links to ity A. We will assume three train links without loss of generality. Let us suppose that A is
onne ted to ea h of B , C , and F by train, as indi ated in the diagram. Assume that no three ities an be linked by train or bus. Then CR, A ................................T B ..................................... .. ............. BC , and BF annot be train links, as . . . ........ .. ........ ........ ... that would result in linking three ities ........ .... ........ T . ........ by train. T... ........ . ........ . . ........ Using dashed segments for bus . . ........ . . ........ . . ...... links, it now happens that B , C , and F .. C are all onne ted by a bus link. Hen e, F using proof by ontradi tion, it has been shown that there are three ities linked only by buses or only by trains. . ....... ...... .... ...... .. . . . . .. .... .. ...... .. ...... .. . . . . . .. .... . . .. . ... . .. . . . .. ..... . . .. ... . . . .. ... . . .. . . .. .... . . . .. ... . . . . ... ...................................................................................................................
388 Comment. This is a tually the game of SIM in disguise. The game requires players to alternately (using one olour ea h) join two verti es with a segment. The player who rst forms a triangle with all segments in their olour loses. The game annot end in a tie as shown in this problem. 1 . Determine whether the fra tion 2005
an be written as a sum of 2005 di erent unit fra tions. (A unit fra tion is a fra tion of the form n1 , where n is a natural number.) Solution by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB.
5
Yes,
1 2005
an be written as the sum of 2005 distin t unit fra tions.
1 1 1 For all positive integers m, m . If m ≥ 2, this = + m+1 m(m + 1)
an be employed to in rease the number of unit fra tions in a sum without limit:
1 2005
= =
and so on!
1 1 + 2006 (2005)(2006) 1 1 1 + + 2006 (2005)(2006) + 1 (2005)(2006) (2005)(2006) + 1
. Let h be the altitude from A in an a ute triangle ABC . Prove that
6
(b + c)2 ≥ a2 + 4h2 ,
where a, b, and c are the lengths of the sides opposite A, B , and C respe tively. Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; Oliver Geupel, Bruhl, NRW, Germany; and Titu Zvonaru, Comane sti, Romania. We give the solution of Amengual-Covas. The triangle ABC need not be a ute, as the proof will show. Let wa be the length of the internal bise tor of angle A. From the known relations wa =
2bc A cos b+c 2
we obtain wa2 = bc ·
and
4bc (b +
c)2
cos2
4bc ≤ (b + c)2 ,
A 2
≤ bc · cos2
A 2
.
389 However, h ≤ wa , so we have h2
≤
wa2 ≤ bc · cos2
A 2
= bc ·
1 + cos a 2
1+ = bc ·
b2 + c2 − a 2 2bc
2
(b + c)2 − a2 4
=
and (b + c) ≥ a + 4h follows, as desired. Equality holds only if b = c. 2
2
2
To omplete this number of the Corner we begin to look at solutions from our readers to problems given in the O tober 2008 issue, starting with the Olimpiada Matemati a Espanola ~ 2005 given at [2008 : 341{342℄. . Let r, s, u, and v be real numbers. Prove that
3
min{r − s2 , s − u2 , u − v 2 , v − r 2 } ≤
1 4
.
Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; and Titu Zvonaru, Comane sti, Romania. We give Bataille's write-up. Suppose that min r − s2 , s − u2 , u − v2 , v − r2 > 14 . Then ea h of the numbers r − s2 , s − u2 , u − v2 , and v − r2 is greater than 14 , hen e and Sin e
(r + u) − s2 + v 2 (s + v) − r 2 + u2 (s + v)2 ≤ 2 s2 + v 2 , inequality (1)
>
>
1 2 1 2
(1) .
(2)
yields
2(r + u) − (s + v)2 > 1
and similarly from inequality (2) we obtain Let a = r + u. Then,
2(s + v) − (r + u)2 > 1 . 2a > 1 +
1 + a2 2
2
,
whi h is equivalent to the inequality a4 + 2a2 − 8a + 5 < 0, a ontradi tion, sin e a4 + 2a2 − 8a + 5 = (a − 1)2 a2 + 2a + 5 is nonnegative for all real numbers a. The result follows. That ompletes the Corner for this issue. Send me your ni e solutions and generalizations!
390
BOOK REVIEWS Amar Sodhi
Sink or Float? Thought Problems in Math and Physi s By Keith Kendig, Mathemati al Asso iation of Ameri a, 2008 ISBN 978-0-88385-339-9, hard over, 375+xiii pages, US$59.95 Reviewed by Nora Franzova, Langara College, Van ouver, BC May the title not onfuse you! Only a small part of the book is geared towards Ar himedes enthusiasts. A tually there are ten di erent ategories of problems spanning geometry, numbers, probability, lassi al me hani s, astronomy, linear algebra, ele tri ity and magnetism, the leaking tank, heat and wave phenomena, and Ar himedes' Prin iple. These ategories are neatly separated into groups, where transitioning between math and physi s problems happens without any a tual transition. Like walking on a Mobius strip, there is only one side to it: the hallenge and joy of solving problems. The book opens with \A Sampler" of problems that immediately pull you in and, as their diÆ ulty in reases, you realize that this is a treasure to be savoured slowly, sin e ea h problem, no matter how short and sweet it seems, an lead to surprising truths. Ea h of the ten se tions of the book starts with an introdu tion that brings up interesting histori al information and lays down notation, basi de nitions, and o
asional theorems that are needed in the se tion. Short examples are in luded. These are not aimed to tea h the material, just to bring the reader to the starting point from whi h the author is observing that parti ular subje t area. Problems are not numbered, but ea h has a title, a name that is already a part of the story. Ea h title is followed by a short des ription, and then a list of possible answers is given in a multiple hoi e format with most of the answers being oh-so-tempting. In between individual problems, additional histori al and theoreti al information is provided in learly separated darkened frames, whi h seem very inviting. Kendig revisits several questions more than on e, so we en ounter not only a problem alled \Friday the 13th ", but also \Friday the 13th Again" and \Friday the 13th One More Time". None involves any bla k ats. The author states: \I hope that this book will serve as a springboard { either individually or in lassroom { for mathemati al thought, ase testing, dis ussions, arguments." The rst step an be taken by following his advi e from the prefa e. The author re ommends one read the problem, think about it, then lose the book and think about it some more. Only then should one turn the pages to the designated pla e where the omplete solution an be found. The solutions are more than omplete; many also in lude short side dis ussions about related questions. Very helpful are the graphi al images that are part of many of the problems and their solutions. These images talk to the reader through questions in framed \bubbles".
391 The book brought forth an immediate urry of interest and omments like \every problem is a gem" or \this is so neat," from family and guests after it had be ome a staple on our o ee table. One might say that a \math book" is not o ee table material, but the problems in the book are attra tively presented, and fairly simple to understand, so a lay person (let's say a \mathemati al muggle⋆ ") an be tempted to approa h a problem like the following: \Is it possible to stand at just the right pla e on the earth so that the moon's res ent looks like a smiley fa e instead of being more verti al?" And a mathemati al wizard (just to sti k with Harry Potter terminology) will humbly turn the pages, pretending only mild interest while still trying to solve that probability problem from the beginning. A math wizard knows that these questions only look inno ent. In reality they hide a lot of deep truths, ready to be un overed again and again. Even though the book is from the Dol iani Mathemati al Expositions series, it de nitely makes an ex ellent o ee table book. Many of the problems an lead to lively dis ussions and arguments for any group of urious individuals. For a similar reason the book would make an ex ellent prize to be given to winners of math ontests or s ien e fairs. And, sin e it is not to be read qui kly, professors, graduate and undergraduate students, or even high s hool students and their tea hers, will be oming ba k to it to rejuvenate their uriosity and wonder about nature's mysteries. We all know that solving a good problem brings us not only satisfa tion but also the joy of understanding. And that an only be repeated by nding and solving another interesting problem and then another ... . At this point, all that is left to the thoughtful person is to open Keith Kendig's book and start solving problems. ⋆
Muggle is the word used in the Harry Potter series of books by J.K. Rowling
to refer to a person who la ks any sort of magi al ability.
Number Theory Through Inquiry By David C. Marshall, Edward Odell, and Mi hael Starbird, Mathemati al Asso iation of Ameri a, Washington, DC, 2007 ISBN 978-0-88385-751-9, hard over, 140+x pages, US$51.00 Reviewed by Jeff Hooper, A adia University, Wolfville, NS Although we an learn mathemati s by reading texts or by sitting in
lasses, mathemati s is not a spe tator sport: most of us learn new mathemati al ideas by a tively working through examples and problems. One re ent, and very laudable, trend in the tea hing of mathemati s has been towards the adoption of tea hing methods whi h push students towards be oming more independent in their mathemati al thinking. An important example of this is the method known as Inquiry-Based Learning (or IBL for short). The basi notion behind IBL-related methods is to have the students' ideas and questions drive the development of the material, with the tea her
392 a ting as fa ilitator. In other words, in its purest form, students would work through examples, use them to suggest onne tions among ideas, formulate statements of onje tures, and then prove these results, with the instru tor there only to help push things along when needed. In prin iple, be ause students must work out mu h of the details and proofs themselves, they gain a deeper understanding of the material, and more importantly develop a mu h higher level of mathemati al skill. Of ourse, this pure form of IBL, as I've des ribed it, would be extremely taxing for the students involved, and far too time- onsuming, so in pra ti e the approa h gets modi ed so that students are at least provided, as the ourse moves along, with pie es of an outline: examples to onsider and statements of results whi h must be proved. This skeletal outline ensures that the inquiry moves in appropriate dire tions. In mathemati s, a version of this approa h was hampioned in the 1950s and 60s by R.L. Moore of the University of Texas at Austin. Moore used this method to tea h sele ted lasses of top students, many of whom went on to earn Ph.D. degrees and be ome well-established resear hers. Moore's method was very pure, and students often had to on o t their own de nitions of ne essary on epts. The authors have been involved in a `modi ed-Moore' approa h to tea hing mathemati s, whi h is aimed at a wider audien e, and this book is an outgrowth of their ourses. The text has two main goals in mind: tea hing the fundamental material overed in a standard undergraduate number theory ourse, and tea hing students to be ome independent mathemati al thinkers. The book overs a fairly standard sele tion of topi s: divisibility and greatest ommon divisors, primes and their distribution, the arithmeti of
ongruen es, Fermat's Little Theorem and Euler's generalization, appli ations to publi -key ryptography and RSA, higher-order ongruen es and primitive roots, quadrati re ipro ity, Pythagorean triples, sums of squares, Pell's equation and introdu tory Diophantine approximation, and primality testing. So the outline is mu h like what one would nd in many popular elementary number theory texts. What makes this text more unusual is the style. The authors use this material for introdu ing students to proofs and abstra t mathemati s, so the rst ouple of hapters are more leisurely than the rest, and there are many
omments to help the reader along. More importantly, be ause the entire point of this book is its use in an IBL-style number theory ourse, the narrative in ludes de nitions, exer ises for students to work through, as well as statements of lemmas, propositions, theorems, and orollaries. But no proofs or solutions are in luded. The intent is that the reader will work out the proofs of these ideas for herself. The authors also in lude numerous questions for students to ponder, many of them open-ended. These generally require students to think more deeply about the material just en ountered, and often ask students to formulate onje tures and prove them. Many se tions on lude with a `Blank-Paper Exer ise'. These ask students to return,
393 after a day or two away from the material, and try to outline the ideas in as mu h detail as possible without looking at the text. Other number theory texts have been written from a problems point-ofview. Three whi h an be found in many libraries are: R.P. Burn's A Pathway into Number Theory, Joe Roberts' Elementary Number Theory: A Problem Oriented Approa h, and Sierpinski's 250 Problems in Elementary Number Theory. So who would bene t from this text? On the ba k over the book suggests its use as a text for a proof-transition or introdu tory abstra t mathemati s ourse, or for independent study. But I must add a aveat to this last suggestion: be ause there are no proofs or solutions, I'm not so sure of its appropriateness for independent work. It would ertainly be useful in this regard, but likely would require that the student or reader have some a
ess to help when they get stu k. I'm not omplaining here; it's lear that providing proofs and solutions would detra t onsiderably from its other purpose. However, a good high-s hool student ould work through a text like this only if she an depend on some a
ess to tea her support. As a professor, I found myself disappointed in one respe t. I've been using a related approa h in some of my senior-level undergraduate ourses with some su
ess, and I was really looking forward to reading about the authors' approa h. How do they use this material in their ourses? What advi e would they have for an instru tor hoping to use this material? Most texts treat these sorts of questions in the prefa e. Unfortunately, for some reason the authors in lude no prefa e, and their introdu tory hapter, while des ribing the text and speaking to its goals, doesn't address these questions or explain how the material ould be used. Reading through the text fairly
arefully, the `Questions' and `Blank-Paper Exer ises' give some hint at the approa h the authors use, but I would have wished for more here. My small disappointment aside, this is a good text, and should ertainly be onsidered by anyone looking to experiment with an inquiry-based approa h to the tea hing and learning of mathemati s. Geometri Origami By Robert Gerets hlager, Arbelos, 2008 ISBN 978-0-9555477-1-3, soft over, 198+xiii pages, US$24.99 Reviewed by Georg Gunther, Sir Wilfred Grenfell College, Corner Brook, NL Origami is the traditional Japanese art of paper folding whi h has been pra tised in that ountry for over 400 years. Not surprisingly, this art form with its stri t dependen e on a sequen e of pre ise folds, has long been of interest to mathemati ians. In 1989, the rst International Meeting of Origami S ien e and Te hnology was held; sin e that time there have been numerous
onferen es devoted to this burgeoning eld, and today the resear h literature is extensive and ever growing.
394 The book Geometri Origami provides a ni e balan e between theory and pra ti al appli ation. If your interest is to fold a regular triskaide agon (otherwise known as a 13-gon), you will nd detailed instru tions, a
ompanied by lear drawings, to guide you on your way. On the other hand, for those whose interests are theoreti al, they will nd a lear exposition of the mathemati s underlying this onstru tion. This is a book whi h yields wonderful nuggets, regardless whether it is read arefully from beginning to end, or whether it is skimmed super ially. Part I, whi h develops the underlying mathemati al theory, provides a lear and areful dis ussion of the axioms of origami onstru tions, and shows that these are more powerful than the orresponding axioms governing Eu lidean onstru tions. Spe i ally, origami onstru tions are powerful enough to provide geometri solutions to the problem of solving the general
ubi equation (and, indeed, the general quarti as well), something that is beyond the s ope of Eu lidean onstru tions. Hen e, as the author shows, origami onstru tions enable one to solve two of the lassi al onstru tion problems: dupli ating the ube and trise ting a general angle. In Part II, the author gives lear step-by-step origami onstru tions for a fas inating lass of problems. All of these results are bolstered by a thorough dis ussion of the underlying theory. Mu h of this se tion fo uses on origami onstru tions of regular polygons, and Gerets hlager ta kles this from several dire tions. First, he explores the general question of whi h n-gons are onstru tible (for example, he shows that if n = 5 · 2k · 3ℓ , then the orresponding n-gon an be folded). Subsequently, the author devotes one hapter to questions around the theory and onstru tion of the regular pentagon, followed by a hapter exploring similar questions for the regular heptagon. The book on ludes with a nal hapter des ribing how to fold the regular nonagon, triskaide agon, 17-gon, and 19-gon. This book is a delightful addition to a wonderful orner of mathemati s where art and geometry meet in su h a fruitful union. Anyone with a working knowledge of elementary geometry, algebra, and the geometry of omplex numbers will have little trouble following the theoreti al dis ussions. It will be of value to a wide range of audien es; new omers to the eld of mathemati al paper-folding will nd, easily a
essible, all the tools they need, while experts will enjoy a well-written, beautifully illustrated referen e work.
395
PROBLEMS Toutes solutions aux problemes dans e numero doivent nous parvenir au plus tard le . Une e toile (⋆) apres le numero indique que le probleme a e te soumis sans solution. Chaque probleme sera publie dans les deux langues oÆ ielles du Canada (anglais et fran ais). Dans les numeros 1, 3, 5 et 7, l'anglais pre edera le fran ais, et dans les numeros 2, 4, 6 et 8, le fran ais pre edera l'anglais. Dans la se tion des solutions, le probleme sera publie dans la langue de la prin ipale solution presentee. La reda tion souhaite remer ier Jean-Mar Terrier, de l'Universite de Montreal, d'avoir traduit les problemes. 1er avril 2010
. Rempla ement. Propose par D.J. Smeenk, Zaltbommel, Pays-Bas. Soit Γ un er le de entre O et de rayon R. La droite t est tangente a Γ au point A, et P est un point sur t distin t de A. La droite ℓ, distin te de t, passe par P et oupe Γ aux points B et C . Le point K est sur la droite AC et tel que P KkAB , et le point L est sur la droite AB et tel que P LkAC . Montrer que KL ⊥ OP . 3439
. Propose par Mi hel Bataille, Rouen, Fran e. Soit Γ un er le de entre O et de rayon r, et soit P un point tel que OP > r . Soit L l'ensemble de toutes les droites ℓ telles que P 6∈ ℓ et ℓ oupe Γ aux points A, B de sorte que P A · P B = OP 2 − r 2 . Montrer que L est un fais eau de droites on ourantes. 3463
. Propose par Mi hel Bataille, Rouen, Fran e. Soit ABC un triangle ave ∠A = 90◦ et H le pied de la hauteur abaissee de A. Soit J le point sur l'hypotenuse BC tel que CJ = HB et soit K , L les proje tions respe tives de J sur AB , AC . Montrer que
3464
2 M − ; AK, AL ≤ M (−2; AB, AC) 3
ou M (α; x, y) =
xα + y α 2
1/α
,
.
. Propose par Xavier Ros (etudiant) and Jose Luis Daz-Barrero, Universite Polyte hnique de Catalogne, Bar elone, Espagne.
3465
Montrer que
∞ X ∞ X
i=1 j=1
1 ij(i + j)
<
π2 6
+
1 2
+
3 4
log 2.
396 . Propose par Tuan Le, etudiant, Fairmont High S hool, Anaheim, CA, E-U. √ Soit x, y et z des nombres reels positifs tels que xyz ≥ 10 + 6 3. Montrer que 3466
z x 1 y + + ≤ 3 2 3 2 3 2 x+y +z y+z +x z+x +y 2
.
. Propose par Tuan Le, etudiant, Fairmont High S hool, Anaheim, CA, E-U. Soit x, y et z des nombres reels positifs. Montrer que 3467
r 3
x3 + y 3 + z 3 + xyz
r
√ xy + yz + xz 3 ≥ 3 + 1. 2 2 2 x +y +z
; Bernardo . Propose par Joseph DeVin entis, Salem, MA, E-U Re aman, Institut Alberto Merani, Bogota, Colombie ; Peter Saltzman, ; et Stan Wagon, Ma alester College, St. Paul, MN, E-U. Berkeley, CA, E-U Trouver tous les entiers positifs n pour lesquels on peut faire orrespondre a haque nombre pair de {2, 4, . . . , 2n} exa tement un nombre impair de {1, 3, . . . , 2n − 1} de telle sorte que les sommes des n paires, prises deux a deux, soient des nombres relativement premiers. Par exemple, si n = 3, on a une solution, 2 + 3 = 5, 4 + 5 = 9, 6 + 1 = 7. 3468
. Propose par Mihaela Blanariu, College Columbia Chi ago, Chi ago, IL, E-U. Soit p ≥ 2 un nombre reel. Trouver la limite 3469
lim
1+
√
2(2!)p +
√ √ 3 3(3!)p + · · · + n n(n!)p (n!)p
n→∞
.
. Propose par Mihaela Blanariu, College Columbia Chi ago, Chi ago, IL, E-U. Soit p ≥ 2 un nombre reel. Trouver la limite 3470
lim
n→∞
1+
√ p √ √ p p 3 n 2! + 3! + · · · + n! np+1
.
. Propose par Catalin Barbu, Ba au, Roumanie. Soit ABC un triangle a utangle et M , N , P les points milieu respe tifs des petits ar s BC , CA, AB . Si [XY Z] designe l'aire du triangle XY Z , montrer que [M BC] + [N CA] + [P AB] ≥ s(3r − R), ou s, r et R sont respe tivement le demi-perim etre, le rayon du er le ins rit et le rayon du
er le ir ons rit du triangle ABC . 3471
397 . Propose par Xavier Ros, etudiant, Universite Polyte hnique de Catalogne, Bar elone, Espagne. Soit x, y et z des nombres reels positifs tels que x ≤ y ≤ z . Montrer que 3472
log x · log
1+y 1+z
+ log y · log
1+z 1+x
+ log z · log
1+x 1+y
≥ 0.
. Propose par Walther Janous, Ursulinengymnasium, Innsbru k, Autri he, a la memoire de Jim Totten. Soit c un nombre reel donne et tel que 0 < c ≤ 1. (a) Montrer que pour tout nombre reel positif µ, on a
3473
(b)
2 µ 3 + p ≤ √ √ 2 cµ + 1 c +1 c+µ
.
⋆. Trouver tous les nombres reels positifs λ tels que la relation λ µ λ+1 + p ≤ √ √ 2 cµ + 1 c+1 c+µ
soit veri ee pour tous les nombres reels positifs µ.
. Propose par Silouanos Brazitikos et Christos Patilas, Trikala, Gre e. Soit x, y et z des nombres positifs tels que xyz = 1. Montrer si q reels oui ou non, on a l'inegalit e 2 3 xy + yz + z x + x + y + z ≥ 9. [Ed. : Les proposeurs font remarquer que l'inegalit e est appuyee par des al uls sur ordinateur.℄ ................................................................. 3474
3439. Repla ement. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. Let Γ be a ir le with entre O and radius R. Line t is tangent to Γ at the point A, and P is a point on t distin t from A. The line ℓ distin t from t passes through P and interse ts Γ at the points B and C . The point K is on the line AC and su h that P KkAB , and the point L is on the line AB and su h that P LkAC . Prove that KL ⊥ OP .
. Proposed by Mi hel Bataille, Rouen, Fran e. Let Γ be a ir le with entre O and radius r, and let P be a point with OP > r . Let L be the set of all lines ℓ su h that P 6∈ ℓ and ℓ interse ts Γ at points A, B su h that P A · P B = OP 2 − r2 . Show that L is a pen il of
on urrent lines. 3463
398 . Proposed by Mi hel Bataille, Rouen, Fran e. Let ABC be a triangle with ∠A = 90◦ and H be the foot of the altitude from A. let J be the point on the hypotenuse BC su h that CJ = HB and let K , L be the proje tions of J onto AB , AC , respe tively. Prove that 3464
where
2 M − ; AK, AL ≤ M (−2; AB, AC) 3 α x + y α 1/α M (α; x, y) = .
,
2
. Proposed by Xavier Ros (student) and Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain. 3465
Prove that
∞ X ∞ X
i=1 j=1
1 ij(i + j)
<
π2 6
+
1 2
+
3 4
log 2.
. Proposed by Tuan Le, student, Fairmont High S hool, Anaheim, CA, USA. √ Let x, y, and z be positive real numbers su h that xyz ≥ 10 + 6 3. Prove that 3466
y x+
y3
+
z2
+
z y+
z3
+
x2
+
x z+
x3
+
y2
≤
1 2
.
. Proposed by Tuan Le, student, Fairmont High S hool, Anaheim, CA, USA. Let x, y, and z be positive real numbers. Prove that 3467
s 3
x3 + y 3 + z 3 + xyz
s
√ xy + yz + xz 3 ≥ 3 + 1. 2 2 2 x +y +z
. Proposed by Joseph DeVin entis, Salem, MA, USA; Bernardo Re aman, Instituto Alberto Merani, Bogota, Colombia; Peter Saltzman, Berkeley, CA, USA; and Stan Wagon, Ma alester College, St. Paul, MN, USA. Find all positive integers n for whi h one an mat h ea h even number from {2, 4, . . . , 2n} with exa tly one odd number from {1, 3, . . . , 2n − 1} so that the sums of the resulting n pairs are pairwise relatively prime. For example, if n = 3 a solution is 2 + 3 = 5, 4 + 5 = 9, 6 + 1 = 7. 3468
. Proposed by Mihaela Blanariu, Columbia College Chi ago, Chi ago, IL, USA. Let p ≥ 2 be a real number. Find the limit 3469
lim
n→∞
1+
√
2(2!)p +
√ √ 3 3(3!)p + · · · + n n(n!)p (n!)p
.
399 . Proposed by Mihaela Blanariu, Columbia College Chi ago, Chi ago, IL, USA. Let p ≥ 2 be a real number. Find the limit 3470
lim
n→∞
1+
√ √ √ p p p 3 n 2! + 3! + · · · + n! np+1
.
. Proposed by Catalin Barbu, Ba au, Romania. Let ABC be an a ute triangle and M , N , P be the midpoints of the minor ar s BC , CA, AB ; respe tively. If [XY Z] denotes the area of triangle XY Z , prove that [M BC]+[N CA]+[P AB] ≥ s(3r−R), where s, r , and R are the semiperimeter, the inradius, and the ir umradius of triangle ABC , respe tively. 3471
. Proposed by Xavier Ros, student, Universitat Polite ni a de Catalunya, Bar elona, Spain. Let x, y, and z be positive real numbers with x ≤ y ≤ z . Prove that
3472
log x · log
1+y 1+z
+ log y · log
1+z 1+x
+ log z · log
1+x 1+y
≥ 0.
. Proposed by Walther Janous, Ursulinengymnasium, Innsbru k, Austria, in memory of Jim Totten. Let c be a xed real number su h that 0 < c ≤ 1. (a) For all positive real numbers µ prove that
3473
(b)
2 µ 3 + p ≤ √ √ 2 cµ + 1 c+1 c+µ
.
⋆. Determine all positive real numbers λ su h that
λ µ λ+1 + p ≤ √ √ cµ + 1 c+1 c + µ2
holds for all positive real numbers µ.
⋆
. Proposed Silouanos Brazitikos and Christos Patilas, Trikala, Gree e. Let x, y, q and z be positive real numbers with xyz = 1. Prove or disprove that 2 3 xy + yz + z x + x + y + z ≥ 9. [Ed.: The proposers indi ate that the inequality is supported by omputer omputations.℄ 3474
400
SOLUTIONS Au un probleme n'est immuable. L'editeur est toujours heureux d'envisager la publi ation de nouvelles solutions ou de nouvelles perspe tives portant sur des problemes anterieurs. . [2008 : 362, 364℄ Proposed by Toshio Seimiya, Kawasaki, Japan. Let ABC be a triangle with ∠ACB = 90◦ + 12 ∠ABC . Let M be the midpoint of BC . Prove that ∠AM C < 60◦ . 3363
Composite of similar solutions by Ri ardo Barroso Campos, University of Seville, Seville, Spain; Ri hard I. Hess, Ran ho Palos Verdes, CA, USA; Matti Lehtinen, National Defen e College, Helsinki, Finland; Madhav R. Modak, formerly of Sir Parashurambhau College, Pune, India; and Peter Y. Woo, Biola University, La Mirada, CA, USA. In △ABC let 4β = ∠B and let W be the foot of the bise tor of ∠A. We are given that ∠C = 90◦ + 2β , when e ∠A = 90◦ − 6β . We dedu e that 6β < 90◦ , so that β < 15◦ . Moreover, be ause ∠AW C is an external angle of △ABW , it satis es ∠AW C = 4β + 45◦ − 3β
= 45◦ + β < 60◦ .
It remains to show that ∠AM C < ∠AW C . Be ause we are given that ∠C is obtuse, we have AB > AC , when e the foot W of the angle bise tor satis es BW > W C . That is, W lies between M and C ; thus ∠AW C is an external angle of △AM W and therefore satis es ∠AM C < ∠AW C < 60◦ , as desired.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA, ARSLANAGIC, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; GERALD EDGECOMB and JULIE STEELE, students, California State University, Fresno, CA, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; JOHN G. HEUVER, Grande Prairie, AB; WALTHER JANOUS, Ursulinengymnasium, Y, Big Rapids, MI, USA; KEE-WAI LAU, Hong Kong, Innsbru k, Austria; V ACLAV KONECN China; TAICHI MAEKAWA, Takatsuki City, Osaka, Japan; THANOS MAGKOS, 3rd High S hool student, Sarajevo College, Sarajevo, Bosnia and of Kozani, Kozani, Gree e; SALEM MALIKIC, Herzegovina; SKIDMORE COLLEGE PROBLEM SOLVING GROUP, Skidmore College, Saratoga Springs, NY, USA; GEORGE TSAPAKIDIS, Agrinio, Gree e; TITU ZVONARU, Comane sti, Romania; and the proposer.
. [2008 : 362, 364℄ Proposed by Toshio Seimiya, Kawasaki, Japan. Let ABC be a triangle with ∠BAC = 120◦ and AB > AC . Let M be the midpoint of BC . Prove that ∠M AC > 2 ∠ACB .
3364
401 Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA. Let β = ∠CBA and γ = ∠BCA; then (be ause ∠BAC = 120◦ ) β + γ = 60◦ . Let D be the point of segment BC for whi h ∠BAD = 2β ; then ∠DAC = 2γ . The Law of Sines tells us that I.
BD =
also,
AD sin 2β sin β
AD sin 2 60◦ − β CD = sin 60◦ − β
= 2AD cos β ;
= 2AD cos 60◦ − β
.
But β+γ = 60◦ and γ > β ; hen e, 60◦ −β > β so that cos 60◦ −β < cos β . Consequently, CD < BD, when e M lies between B and D. It follows that ∠M AC > ∠DAC = 2∠ACB , as desired.
Solution by Matti Lehtinen, National Defen e College, Helsinki, Finland. Let O be the entre of the ir um ir le Γ of △ABC , and D be the midpoint of the ar CAB of Γ. Note that A is on the smaller ar DC . Sin e 120◦ = ∠BAC = ∠BDC , DB = DC = DO . Draw the ir le Γ′ with entre D and radius DB (passing through B , O, and C ). Be ause M is the ommon midpoint of BC and DO, ir les Γ and Γ′ are symmetri with respe t to M . Let AM meet Γ′ at E (so that M is also the midpoint of AE ); note that ∠EDC = ∠AOB = 2∠ACB . Extend AE to meet Γ again at F and extend DE to meet Γ again at G. Be ause the line F A separates D from G and C , G is on the ar F C opposite D and A; onsequently, ∠F AC > ∠GDC . But, ∠F AC = ∠M AC , and ∠GDC = ∠EDC = ∠AOB = 2∠ACB , so we are done. II.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA, ARSLANAGIC, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW, Y, Big Rapids, MI, USA; KEE-WAI LAU, Hong Kong, China; Germany; V ACLAV KONECN student, THANOS MAGKOS, 3rd High S hool of Kozani, Kozani, Gree e; SALEM MALIKIC, Sarajevo College, Sarajevo, Bosnia and Herzegovina; GEORGE TSAPAKIDIS, Agrinio, Gree e; TITU ZVONARU, Comane sti, Romania; and the proposer.
. [2008 : 362, 364℄ Proposed by Toshio Seimiya, Kawasaki, Japan. A square ABCD is ins ribed in a ir le Γ. Let P be a point on the minor ar AD of Γ, and let E and F be the interse tions of AD with P B and P C , respe tively. Prove that 3365
AE · DF = 2 [P AE] + [P DF ]
where [KLM ] denotes the area of triangle KLM .
,
402 Solution by John G. Heuver, Grande Prairie, AB. The altitudes from point P to segments AB and CD lead us to on lude that [P AB] + [P DC] =
P
1 AB 2 , 2
q q
......................................................... ..................... .... ....................... .................. ............ ....... .................. .... .... ............ ...... ...................... . . . .. ... . . ............................................................................................................................................ . ........ ... . . . . . ..... ...... . ... ... .... . . .... ..... . . ... .. ... ..... . . ... ... ... . ... .. . . ... . . . ... . ... .. .. .. ..... . ... .. .. .. ..... .... ... ... .. ... ... .. . ... . ... . . ... . .. .... . ... .. . . ... . . ... .. . .... ..... . . ... .. .. . .. ... ... . . . ... .. .. ... .. . ... . . . ... . .. . ... .. ... . . . . . ... . .. . . . . ..... ... . .. . ... . . .. ... ..... .... ... .. ... ... ... ... ... .... .... ... ... .... .... .... ... ... ... .... .... ..... . .. ... ......... ................................................................................................................................ ..... ..... ....... ....... ........ . . . . . . .. ......... ......... .............. ............................................
A
Furthermore,
E
1 AB 2 = [P AE] + [AEB] + [P DF ] + [DF C] , 2
Sin e 1 [AEB] + [DF C] = AB(AE + DF ) , 2
we have [P AE] + [P DF ] = = =
F
b
D
b
B
C
1 AB 2 − [AEB] + [DF C] 2 1 1 AB 2 − AB(AE + DF ) 2 2 1 1 AB(AB − AE − DF ) = AB · EF 2 2
.
Now, ∠BP C = ∠CP D = ∠45◦ as angles subtended by ar s equal to a quarter of the ir le. Hen e P F is the bise tor of ∠EP D, and therefore, EF PE = FD PD
.
Also, ∠ABP = ∠P DA as angles subtended by the ar AP , so that triangles P DE and ABE are similar, and therefore, PE PD
Consequently, or EF
· AB = AE · DF ,
=
AE AB
EF AE = FD AB
. ,
and then
[P AE] + [P DF ] =
whi h ompletes the proof.
1 1 AB · EF = AE · DF 2 2
,
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA, ARSLANAGIC, Lebanese University, Fanar, Lebanon; RICARDO BARROSO CAMPOS, University of Seville, Seville, Spain; MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; WALTHER JANOUS, Y, Big Rapids, MI, USA; SALEM Ursulinengymnasium, Innsbru k, Austria; V ACLAV KONECN
403 student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; MADHAV R. MODAK, MALIKIC, formerly of Sir Parashurambhau College, Pune, India; JOEL SCHLOSBERG, Bayside, NY, USA; D.J. SMEENK, Zaltbommel, the Netherlands; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane sti, Romania; and the proposer. There was one in orre t
omment submitted. Kone ny and the proposer ea h began by letting Z be the foot of the perpendi ular from P to AD, then dedu ing that △DP Z ∼ △BEA and △AP Z ∼ △CF D, whi h then enables the al ulation of AE · DF .
. [2008 : 362, 365℄ Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. Let {x} denote the fra tional part of the real number x; that is, {x} = x − ⌊x⌋, where ⌊x⌋ is the greatest integer not ex eeding x. Evaluate Z 1 4 1 dx . 3366
x
0
Solution by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. Let I be the integral to be evaluated. Then I
=
lim
N →∞
=
lim
N →∞
where
N Z X
n=1
N Z X
n=1
1 n 1 n+1
4 1
n+1
n
In =
x
In
= = = = =
n+1
lim
N →∞
(y − n)4 dy = y2
Z
n+1
n
We have
dx =
N Z X
n=1
lim
N →∞
n+1
{y}4 y2
n
N X
dy
In ,
n=1
(y − n)4 dy . y2
n4 4n3 2 2 + 2 dy y − 4ny + 6n − y y n n+1 3 4 y n − 2ny 2 + 6n2 y − 4n3 ln y − 3 y n 4 1 n 1 2 3 3 3n − n + − + n − 4n ln 1 + 3 n+1 n 1 1 1 2 2 3 3n − n + + n − n − + 1 − 4n ln 1 + 3 n+1 n 4 1 1 4n2 − 2n + − − 4n3 ln 1 + . 3 n+1 n Z
404 Substituting the series expansions x x+1
with x =
=
∞ X
ln(1 + x) =
k=0
4
∞ X (−1)k
3
= 4n2 − 2n +
4 − 4n3 3
∞ X (−1)k
k=1 ∞ X
=
(−1)k
− 4n3
∞ X (−1)k+1
k=1
knk
1 1 1 − + 2 n 2n 3n3 ∞ X (−1)k+1 − 4n3 knk k=4
∞ X (−1)k+1
knk
k=4
∞
− 4
nk
−
k=2
1 n
k
1
=
,
X (−1)k+1 (−1)k − 4 nk knk−3 k=4
k=1 ∞ X
=
nk+1
k=0
− 4n3
nk+1
k=0 ∞ X
−
∞ X (−1)k nk+1 k=0
−
=
k
into the last expression for In yields
= 4n2 − 2n +
= −
∞ X (−1)k+1 xk
k=1
1 (n ≥ 2) n
In
and
(−1)k xk+1
n(n + 1)
∞ X
(k + 3)nk
k=1 ∞ X
− 4
− 4
(−1)k
k=2 ∞ X
k=2
(−1)k (k + 3)nk (−1)k
(k + 3)nk
,
and the last formula is valid for I1 as well. Thus, I
=
∞ X
n=1
= =
∞ X ∞ X 1 1 1 k − 4 − n(n + 1) k+3 n n=1 k=2
1 − 4 1 − 4
∞ ∞ X (−1)k X 1
k=2 ∞ X k=2
k+3
n=1
nk
(−1)k ζ(k) . k+3
Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India; PAOLO PERFETTI, Dipartimento di Matemati a, Universita degli studi di Tor Vergata Roma, Rome, Italy; and the proposer. There were two in orre t solutions submitted. The proposer obtained the answer − 31 − γ + 2 ln 2π − 12 ln A + 12 ln B, where γ is Euler's onstant and where A = lim
11 22 · · · nn
n2 n 1 n→∞ n 2 + 2 + 12
−n2 e 4
2
;
B = lim
2
11 22 · · · nn
n3 n2 n n→∞ n 3 + 2 +6
2
−n3 n e 9 + 12
405 are the Glaisher{Kinkelin onstants of order 1 and order 2, respe tively.
ζ(k) − 1 = 0.14553289 . . . , whi h he k(k + 1)(k + 2)(k + 3) ∞ (−1)k ζ(k) P and is thus better suited remarked onverges more rapidly than the series 1 − k+3 k=2
Geupel obtained I
∞ P
= 24
k=2
for numeri al approximation. Janous reports the generalization Z
1 0
1 x
N
dx =
N −2 X
j
(−1)
j =0
+
N j
1 1 N −j −1 2 − 1 − (−1)N N ln 2 − N −j −1 2
∞ X (−1)p (p − 1) ζ(p) − 1 . p + N − 1 p=2
Keith Ekblaw, Walla Walla, WA, USA, obtained the estimate I approa h, whi h is orre t to 3 de imal pla es after rounding.
≈ 0.146 by a Monte Carlo
. [2008 : 362, 365℄ Proposed by Li Zhou, Polk Community College, Winter Haven, FL, USA. Let p(x) = an xn + an−1 xn−1 + · · · + a1 x be a polynomial with integer n P ak = 1. Prove or disprove that there are
oeÆ ients, where an > 0 and k=1 in nitely many pairs of positive integers (k, ℓ) su h that p(k + 1) − p(k) and p(ℓ + 1) − p(ℓ) are relatively prime. 3367
Solution by Cristinel Morti i, Valahia University of Targoviste, Romania, modi ed slightly by the editor. Let Q(x) = p(x + 1) − p(x). Then we have Q(0) = p(1) − p(0) =
n X
k=1
ak − 0 = 1 .
Hen e, Q(x) = xq(x) + 1 for some polynomial q(x) of degree n − 1. We need to nd in nitely many positive integers k and l su h that gcd Q(k), Q(l) = 1. Sin e the leading term in Q(x) is nan xn−1 we have Q(k) > 0 for suÆ iently large positive integers k. For su h k let l = Q(k) = kq(k) + 1, so that l is a positive integer. If, on the ontrary, gcd Q(k), Q(l) 6= 1, then there is a prime number p su h that p | Q(k) and p | Q(l). Then, sin e Q(l) = lq(l) + 1 = Q(k)q(l) + 1, we on lude that p | 1, a ontradi tion.
Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; OLIVER GEUPEL, Bruhl, NRW, Germany; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; and the proposer. Morti i remarked that the on lusion in fa t holds for any polynomial Q(k) whose onstant term is either 1 or −1. This is lear from the proof featured above. Barbara proved the n P stronger results that (1) if an 6= 0 and ak = 1, then there are in nitely many positive k=1
406 integers k su h that p(k + 1) − p(k) are all pairwise relatively prime, and (2) if f(x) is a polynomial with integer oeÆ ients and positive degree su h that f(0) = 1, then there are in nitely many integers k1 < k2 < k3 < · · · su h that the f(ki )'s are all pairwise relatively prime.
. [2008 : 363, 365℄ Proposed by Neven Juri , Zagreb, Croatia. Let m be an integer, m ≥ 2, and let A = [Aij ] be a blo k matrix of dimension 2m × 2m with Aij ∈ M4,4 (N) for 1 ≤ i, j ≤ 2m−2 , de ned by Aij = 2m Bij + Cij , where 3368
and
Bij
=
Cij
=
2m − 4i + 4 4i − 4 4i − 4 m m 4i − 3 2 − 4i + 3 2 − 4i + 3 4i − 2 2m − 4i + 2 2m − 4i + 2 2m − 4i + 1 4i − 1 4i − 1 4 − 4j 4j − 2 4j − 1 1 − 4j 4j − 3 3 − 4j 2 − 4j 4j . 4j − 3 3 − 4j 2 − 4j 4j 4 − 4j 4j − 2 4j − 1 1 − 4j
2m − 4i + 4 4i − 3 4i − 2 2m − 4i + 1
,
Show that matrix A is a magi square of order 2m . Solution by Oliver Geupel, Bruhl, NRW, Germany.
For onvenien e, write Bij = B (i) = b(i) and Cij kℓ and note that these 4 × 4 matri es have integer entries.
(j) = C (j) = ckℓ ,
For ea h entry a of A, 1 ≤ a ≤ 22m . (j) Proof: We have a = 2m b(i) kℓ + ckℓ for some i, j , k, and ℓ. If we also have (i) (j) (k, ℓ) ∈ {(1, 2), (1, 3)}, then bkℓ ≥ 0 and ckℓ ≥ 1; hen e a ≥ 1. Otherwise, (i) (j) bkℓ ≥ 1 and ckℓ ≥ 1−2m ; hen e a ≥ 2m +1−2m = 1. This proves the lower bound. For the upper bound, we observe that for (k, ℓ) ∈ {(1, 1), (1, 4)} we m m have b(i) and c(j) · 2m = 22m . In any other ase kℓ ≤ 2 kℓ ≤ 0; hen e a ≤ 2 (i) (j) m m m bkℓ ≤ 2 − 1 and ckℓ ≤ 2 ; so again a ≤ 2 2m − 1 + 2m = 22m . Claim 1
The entries of A are distin t. (j) (j ) m (i ) Proof: If two entries of A are equal then 2m b(i) kℓ + ckℓ = 2 bkℓ + ckℓ for (i ) (j) (j ) ′ ′ some j , j ′ , k, k′ , ℓ, and ℓ′ . Let b = b(i) kℓ , b = bk ℓ , c = ckℓ , and c = ck ℓ . We then have 1 ≤ |c − c′ | ≤ 4 · 2m−2 − 1 − 4 · 2m−2 = 2m+1 − 1 < 2m+1 . Without loss of generality, suppose that c < c′ . Then c′ − c = 2m and b = b′ + 1, hen e c ≡ c′ (mod 4) and b ≡ b′ + 1 (mod 4). There are four
ases for c and c′ : ′ ′ Case 1 (k, ℓ) ∈ {(1, 1), (4, 1)} and (k , ℓ ) ∈ {(2, 4), (3, 4)}; ′ ′ Case 2 (k, ℓ) ∈ {(1, 4), (4, 4)} and (k , ℓ ) ∈ {(2, 1), (3, 1)}; Claim 2
′
′
′
′
′ ′
′ ′
407 and (k′ , ℓ′ ) ∈ {(1, 2), (4, 2)}; Case 4 and (k′ , ℓ′ ) ∈ {(1, 3), (4, 3)}; ea h of them in ompatible with the ondition b ≡ b′ + 1 (mod Case 3
(k, ℓ) ∈ {(2, 3), (3, 3)} (k, ℓ) ∈ {(2, 2), (3, 2)}
4).
The sum of the entries along any horizontal, verti al, or main diagonal line of A is 23m−1 + 2m−1 . Proof: The sum along any given horizontal or verti al line of B (i) and C (j) is SB = 2m+1 and SC = 2, respe tively. Therefore, the sum of the entries in ea h line of A is 2m−2 2m SB +SC = 2m−2 2m ·2m+1 +2 = 23m−1 +2m−1 . The main diagonal sums of B (i) and C (j) are b(i) = 2m+2 − 16i + 10 and c(j) = 10 − 16j . The entries in ea h main diagonal of A then also have the Claim 3
sum
2m−2 P
2m b(i) +
i=1
2m−2 P j=1
c(j) = 23m−1 + 2m−1 .
Also solved by the proposer. One in omplete solution was re eived that veri ed the row,
olumn, and diagonal sums, but did not show that the entries of the magi square onsisted of 1, 2, . . . , 22m .
. [2008 : 363, 365℄ Proposed by George Tsintsifas, Thessaloniki, Gree e. Let A1 A2 A3 A4 be a tetrahedron whi h ontains the entre O of its
ir umsphere as an interior point. Let ρi be the distan e from O to the fa e opposite vertex Ai . If R is the radius of the ir umsphere, prove that 3369
4 X 4 R ≥ ρi . 3 i=1
Solution by Oliver Geupel, Bruhl, NRW, Germany. The laim is false; the following ounterexample is adapted from [1℄. First, onsider the degenerate tetrahedron A′1 A′2 A′3 A′4 , where A′1 = A′2 A′3 A′4 O = (0, 0, 0),
that
and R
=
√ 2.
= = =
(−1, 1, 0) , (1, 1, 0) , (−1, −1, 0) ,
We have ρ1
= ρ2 = 0,
√ √ 4R 4 2 = = 2 2 < 3. 4 P 2 ρi
i=1
and ρ3
= ρ4 = 1,
so
408 Now, let A1
=
A2
=
A3
=
A4
=
p −1 + ε, 1 − ε, 2ε(2 − ε) , p −1 + ε, 1 − ε, − 2ε(2 − ε) , √ √ π π 2 cos − ε , 2 sin −ε , 0 , 4 √ 4 √ 3π 3π 2 cos − + ε , 2 sin − +ε , 0 4
4
.
It is easy to verify that point O is in the interior of the tetrahedron A1 A2 A3 A4 , and that Ai → A′i for ea h i as ε → 0, as well as 4R
√ 2
The bound 2 lowing:
4 P
ρi
√ −→ 2 2 .
i=1
is the best possible. This is a orollary from the fol-
[1℄. Let A1 A2 A3 A4 be a nondegenerate tetrahedron whose ir um entre O is not an exterior point. Let P be a point not exterior to A1 A2 A3 A4 . Let the distan es from P to the verti es and to the fa es of A1 A2 A3 A4 be denoted by Ri and ρi , respe tively. Then Theorem
4 P
√
and 2
i=1 4 P
Ri
√ > 2 2,
ρi
i=1
2 is the
greatest lower bound.
Referen es
[1℄ Ni holas D. Kazarino , \D.K. Kazarino 's inequality for tetrahedra", Mi higan Mathemati al Journal, Vol. 4, No. 2 (1957), pp. 99-104. Counterexample also given by Peter Y. Woo, Biola University, La Mirada, CA, USA. George Apostolopoulos, Messolonghi, Gree e, ited the book by Kazarino , Geometri Inequalities, Yale University Press, 1961, p. 116. No omplete solutions or other omments were submitted.
. [2008 : 363, 365℄ Proposed by George Tsintsifas, Thessaloniki, Gree e. Let ai and bi be positive real numbers for 1 ≤ i ≤ k, and let n be a positive integer. Prove that 3370
k X
i=1
1 n
ai
!n
≤
k X ai
i=1
bi
!
k X
i=1
1 n−1
bi
!n−1
.
409 Composite of similar or identi al solutions submitted by all the solvers whose names appear below (ex ept the two solvers identi ed by a \*" before their names). Let p = n and q =
n . n−1
Then p > 1, q > 1 and
1 1 + = 1. p q
[Ed: learly, n > 1 for the given inequality to make sense.℄ Let xi = ab i and yi = bi . Then xi and yi are positive for ea h i. 1 n
1 n
i
By Holder's Inequality, we have be omes
k P
i=1
1
ain ≤
k P ai
i=1
bi
1 n
k P
i=1
k P
i=1
bn−1 i
xi yi ≤
n−1 n
.
k P
i=1
xpi
p1
k P
i=1
yiq
q1
whi h
The result follows by raising both sides to the nth power.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE, ARSLANAGIC, Rouen, Fran e; MANUEL BENITO, OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logrono, ~ Spain; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinengymnas student, Sarajevo College, Sarajevo, Bosnia and ium, Innsbru k, Austria; SALEM MALIKIC, Herzegovina; DUNG NGUYEN MANH, High S hool of HUS, Hanoi, Vietnam; *CRISTINEL MORTICI, Valahia University of Targoviste, Romania; PAOLO PERFETTI, Dipartimento di Matemati a, Universita degli studi di Tor Vergata Roma, Rome, Italy; JOEL SCHLOSBERG, Bayside, NY, USA; *PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane sti, Romania; and the proposer.
. [2008 : 363, 368℄ Proposed by George Tsintsifas, Thessaloniki, Gree e. Let ABC be a triangle with a, b, and c the lengths of the sides opposite the verti es A, B , and C , respe tively, and let M be an interior point of △ABC . The lines AM , BM , and CM interse t the opposite sides at the points A1 , B1 , and C1 , respe tively. Lines through M perpendi ular to the sides of △ABC interse t BC , CA, and AB at A2 , B2 , and C2 , respe tively. Let p1 , p2 , and p3 be the distan es from M to the sides BC , CA, and AB , respe tively. Prove that 3371
[A1 B1 C1 ] [A2 B2 C2 ]
=
(ap1 + bp2 )(bp2 + cp3 )(cp3 + ap1 ) 8a2 b2 c2
a p1
+
b p2
+
c p3
,
where [KLM ] denotes the area of triangle KLM . Solution by Joel S hlosberg, Bayside, NY, USA. By the basi formula for the area of a triangle, [ACM ] = 12 bp2 and [CM B] = 12 ap1 . It is well known that if segments Y Z and Y ′ Z ′ lie on the
410 same line and X is any point, then [XY Z] : [XY ′ Z ′ ] = Y Z [ACC1 ] [ACM ] [ACC1 ] [CC1 B]
= =
[CC1 B] [CM B]
=
CC1 CM
: Y ′ Z ′ , so that
;
[ACM ] AC1 bp2 = = [CM B] C1 B ap1
.
cp3 ap1 1 1 By similar reasoning, BA = and CB = . A known formula (see A1 C bp2 B1 A cp3 Eri W. Weisstein, \Routh's Theorem," at http://mathworld.wolfram.com/ RouthsTheorem.html) states that if A′ , B ′ , and C ′ are points on the sides BC , CA, and AB of △ABC , respe tively, then AC ′ BA′ CB ′ · · + 1 C ′ B A′ C B ′ A [ABC] . [A′ B ′ C ′ ] = ′ ′ AC BA CB ′ +1 +1 +1 C′B A′ C B′A
Therefore,
[A1 B1 C1 ] =
=
bp2 cp3 ap1 · · +1 ap1 bp2 cp3 [ABC] bp2 cp3 ap1 +1 +1 +1 ap1 bp2 cp3
2abcp1 p2 p3 (ap1 + bp2 )(bp2 + cp3 )(cp3 + ap1 )
[ABC] .
Sin e ∠M A2 C and ∠M B2 C are right angles, ∠A2 M B2
= 360◦ − ∠M A2 C − ∠M B2 C − ∠A2 CB2 = 180◦ − ∠ACB ,
so sin ∠A2 M B2 = sin C , and sin e all four angles of quadrilateral A2 M B2 C are less than 180◦ , A2 M B2 C is onvex. The well-known area formula for a triangle, [XY Z] = 21 XY · XZ sin ∠Y XZ , yields [M A2 B2 ] = =
1 2
M A2 · M B2 sin ∠A2 M B2
p1 p2 ab
1 ab sin C 2
=
p1 p2 [ABC] . ab
p p 2 p3 By similar reasoning, [M B2 C2 ] = pbc [ABC] and [M C2 A2 ] = 3 1 [ABC], ca and B2 M C2 A and C2 M A2 B are onvex. Sin e A2 M B2 C , B2 M C2 A, and C2 M A2 B are onvex, M is outside of △A2 B2 C , △B2 C2 A, and △C2 A2 B , and sin e M is in the interior of △ABC , M must be in the interior of △A2 B2 C2 . Therefore,
[A2 B2 C2 ] = [M A2 B2 ] + [M B2 C2 ] + [M C2 A2 ] ,
411 and so [A2 B2 C2 ]
= =
Finally, [A2 B2 C2 ] [A1 B1 C1 ]
p1 p2
p2 p3
p3 p1
+ + [ABC] ab bc ca p1 p2 p3 a b c [ABC] . + + abc p1 p2 p3
p1 p2 p3 abc
=
a b c [ABC] + + p1 p2 p3 2abcp1 p2 p3 [ABC] (ap1 + bp2 )(bp2 + cp3 )(cp3 + ap1 )
=
(ap1 + bp2 )(bp2 + cp3 )(cp3 + ap1 ) 2a2 b2 c2
a b c + + p1 p2 p3
.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; ARKADY ALT, San Jose, CA, USA; MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; PETER Y. WOO, Biola University, La Mirada, CA, USA; and TITU ZVONARU, Comane sti, Romania. There was one in orre t solution submitted. Both Bataille and Geupel al ulated with bary entri oordinates. Geupel also used the formula for the area of the pedal triangle of M , to wit [A2 B2 C2 ] = |R2 −OM 2 |/4R2 , where O and R are the ir um entre and ir umradius of triangle ABC, respe tively. 3372. [2008 : 364, 366℄ Proposed by Vo Quo Ba Can, Can Tho University of Medi ine and Pharma y, Can Tho, Vietnam. If x, y, z ≥ 0 and xy + yz + zx = 1, prove that
(a)
1
1
1
p + p + p ≥ 2x2 + 3yz 2y 2 + 3zx 2z 2 + 3xy
(b)⋆
√ 1 1 1 p + p + p ≥ 2 2 2 2 2 x + yz y + zx z + xy
√ 2 6 3
;
.
Solution by George Apostolopoulos, Messolonghi, Gree e. (a) Sin e 2x2 ≤ 3x2 we have 2x2 + 3yz ≤ 3x2 + 3yz , thus
1 1 1 p ≥ p = √ p 2 2 2x + 3yz 3x + 3yz 3 x2 + yz
.
Similar inequalities hold for the other two terms on the left side of the desired inequality, and we now have 1 1 1 p + p + p 2 2 2 2x + 3yz 2y + 3zx 2z + 3xy 1 ≥ √ 3
1
1
1
p + p + p x2 + yz y 2 + zx z 2 + xy
!
.
412 The desired inequality follows now from (b), whi h is proven below. (b)⋆ More generally, we will prove that if x, y, z ≥ 0 and xy+yz+zx > 0, then √ 2 2 p + p + p ≥ √ xy + yz + zx x2 + yz y 2 + zx z 2 + xy 1
1
1
Sin e this is symmetri in x, y, z , we may assume that x that √ [Ed: we
≥ y ≥ z.
1 1 2 2 p + p ≥ p 2 2 2 y + zx z + xy y + z 2 + xy + zx r 1 1 1 p + p ≥ 2 p have p 1 y 2 + zx
z 2 + xy
AM{GM Inequality, and also
.
(1)
Noti e
.
y 2 + zx
from the
z 2 + xy 1 2 p p ≥ 2 ; y + zx + z 2 + xy y 2 + zx z 2 + xy
the
inequality above now follows from these two.℄ So it suÆ es to prove that √ 2 2
√ 2 2 p + p ≥ √ xy + yz + zx x2 + yz y 2 + z 2 + xy + zx 1
.
= xy + yz + zx and L = y 2 + z 2 + xy + zx. Then √ √ √ √ √ √ 2 2 2 2 2 2 L− K 2 2 y 2 − yz + z 2 − = = √ √ . √ √ √ √ K L KL KL L + K p It is lear that L ≥ K , L ≥ 2 y2 − yz + z 2 , and K ≥ y x2 + yz . [Ed: Sin e x ≥ y ≥ z , we have L = y2 + z 2 + xy + zx ≥ y2 + z 2 + y 2 + z 2 ≥ 2(y 2 − yz +pz 2 ) and K = xy + yz p + zx ≥ xy + yz + yz √ = y x2 + 4xz + 4z 2 ≥ y x2 + 4yz + 4z 2 ≥ y x2 + yz .℄
Let K
Thus,
√ 2 2 y 2 − yz + z 2 √ √ √ KL( L + K)
≤ ≤
= =
√ √ 2 2 y 2 − yz + z 2 2 y 2 − yz + z 2 = √ √ √ KL(2 K) K L √ 2 y 2 − yz + z 2 q p y x2 + yz 2 y 2 − yz + z 2
p
p y 2 − yz + z 2 y2 p ≤ p y x2 + yz y x2 + yz 1
p
x2
+ yz
,
413 and we have proved (1). Sin e xy + yz + zx = 1, inequality (b) now follows from (1). Part (a) also solved by Oliver Geupel, Bruhl, NRW, Germany; and the proposer. Geupel notes that the problem is Problem 1.89 on page 74.f with solution in the e-paper toanhocmuonmaumain.pdf written by the proposer in Vietnamese, available on the MathLinks forum website http://www.mathlinks.ro/viewtopic.php?t=197674.
3373. Proposed by Vo Quo Ba Can, Can Tho University of Medi ine and Pharma y, Can Tho, Vietnam. Let x, y, z , and t be positive real numbers. Prove that
(x + y)(x + z)(x + t)(y + z)(y + t)(z + t) ≥ 4xyzt(x + y + z + t)2 .
Solution by Cristinel Morti i, Valahia University of Targoviste, Romania. Dividing by xyzt(x + z)(y + t)(x + t)(y + z) the inequality be omes x+y xy
·
z+t zt
≥
2(x + y + z + t)
·
(x + z)(y + t)
2(x + y + z + t) (x + t)(y + z)
.
Thus we need to show:
2 2 + x+z y+t
2 2 + x+t y+z
1 1 + x y
1 1 + z t
1 1 +√ √ xz yt
1 1 √ +√ yz xt
≤
.
By the AM{GM Inequality we have
2 2 + x+z y+t
2 2 + x+t y+z
≤
,
while by the Cau hy-S hwartz Inequality we have 1 1 +√ ≤ √ xz yt
s
1 1 ≤ √ +√ yz xt
s
and
1 1 + x y
1 1 + x y
1 1 + z t
1 1 + z t
.
Combining the last three inequalities we obtain the desired inequality.
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; OLIVER GEUPEL, Bruhl, NRW, Germany; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, student, Sarajevo College, Ursulinengymnasium, Innsbru k, Austria; SALEM MALIKIC,
414 Sarajevo, Bosnia and Herzegovina; KHANH BAO NGUYEN, High S hool for Gifted Students, Hanoi University of Edu ation, Hanoi, Vietnam; PAOLO PERFETTI, Dipartimento di Matemati a, Universita degli studi di Tor Vergata Roma, Rome, Italy; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane sti, Romania; and the proposer. Howard notes that this problem is similar to CRUX 2393(a) [1999 : 447℄. In fa t, Maliki observes that by using 2393(a) for (a, b, c, d) = (x, y, z, t), (a, b, c, d) = (x, z, t, y), and (a, b, c, d) = (x, t, y, z) (that is, all the y li permutations of y, z, t) and multiplying the resulting inequalities together we get exa tly the desired inequality. Geupel notes that the problem is Problem 1.73 on page 64.f with solution in the e-paper toanhocmuonmaumain.pdf written by the proposer in Vietnamese, available on the MathLinks forum website http://www.mathlinks.ro/viewtopic.php?t=197674.
. [2008 : 364, 366℄ Proposed by Pham Huu Du , Ballajura, Australia. Let a, b, and c be positive real numbers. Prove that
3374
a2 a2
+ bc
+
b2 b2
+ ca
+
c2 c2
+ ab
≤
a+b+c √ 3 2 abc
.
Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA. By homogeneity, we may take abc = 1 and a ≤ b ≤ c. Also, if a = 1, then a = b = c = 1 and there is nothing to prove, so we take a < 1. x For onvenien e, write x = a3 , y = b3 , and z = c3 ; let f (x) = x + 1 , and let the left and right sides of the inequality be L = f (x) + f (y) + f (z) and R = 12 x1/3 + y1/3 + z 1/3 , respe tively. Setting
√ m = yz
and r
implies that yz > 1), y =
m , r
2f (m) − f (y) − f (z)
=
r
z , y
we have that m
>1
(be ause a
<1
and z = mr. We then nd that = =
2m m mr − − m+1 m+r mr + 1 m(m − 1)(r − 1)2
(m + 1)(m + r)(mr + 1)
≥ 0,
hen e L annot de rease if ea h of y and z are repla ed by their geometri √ c mean. On the other hand, from bc ≤ b + , we see that R annot in rease 2 if y and z are ea h repla ed by their geometri mean. Therefore, it suÆ es to prove the inequality under the additional assumption that b = c and a = b12 . This new relation yields (1 + 2y) y 3 + y 2 + y + 1 − 4b11 − 6b5 − 2b2 R−L = 2b2 y 3 + y 2 + y + 1
,
where the denominator is positive and (after some omputation) the numerator be omes (b − 1)2 (b − 1) 2b9 + 2b8 − b6 − b5 + 5b4 + 3b3 + b2 + 2b + 1 ,
415 whi h is also a positive quantity sin e b > 1. Thus, R − L > 0 in this last ase, and the proof is omplete.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA (2 solutions); WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; KEE-WAI LAU, Hong Kong, China; and the proposer. There were two in orre t solutions submitted.
. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. Let p be a non-negative integer and x any real number. Find the sum
3375
∞ X
x x2 xn+p e −1− − − ··· − 1! 2! (n + p)!
n
x
(−1)
n=1
!
.
Solution by Cristinel Morti i, Valahia University of Targoviste, Romania. If r > 0 and f is a fun tion with ∞ X
f (x) =
n=0
then we will show that ∞ X
an xn , x ∈ (−r, r) ,
(−1)n f (x) − a0 − a1 x − · · · − an+p xn+p
n=1
=
p/2 X
a2i x2i −
i=0 (p−1)/2
X
i=0
f (x) + f (−x) 2
a2i+1 x2i+1 −
,
p
f (x) − f (−x) 2
even , (1)
,
p
odd .
Taking f (x) = ex , we obtain the answer ∞ X
n
(−1)
n=1
=
x xn+p f (x) − 1 − − ··· − 1! (n + p)! p/2 X x2i
(2i)!
i=0 (p−1)/2
X
i=0
−
ex + e−x
x2i+1
(2i + 1)!
2 −
,
ex − e−x 2
,
! p
even,
p
odd.
To prove (1), note that the general term an = (−1)n f (x) − a0 − a1 x − · · · − an+p xn+p
416
onverges to zero, so it suÆ es to nd the limit of the sequen e {s2n }, where sn =
n X
ak ,
k=1
be ause s2n+1 = s2n + a2n+1 and a2n+1 → 0. We have s2n =
n X
(a2k−1 + a2k ) =
k=1
If p is even, then
k=1
s2n =
p/2 X
i=0
while if p is odd, then X
i=0
−a2k+p x2k+p
n+p/2
a2i x2i −
(p−1)/2
s2n =
n X
X
.
a2i x2i ;
i=0
n+(p−1)/2
a2i+1 x2i+1 −
X
a2i+1 x2i+1 .
i=0
Now the relation (1) follows from the equations ∞ X
i=0
a2i x2i =
f (x) + f (−x) 2
;
∞ X
a2i+1 x2i+1 =
f (x) − f (−x)
i=0
2
.
Also solved by MICHEL BATAILLE, Rouen, Fran e; MANUEL BENITO, OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logrono, ~ Spain; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; and the proposer. There were three in orre t solutions submitted.
Crux Mathemati 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
