257
SKOLIAD
No. 111
Robert Bilinski
Please send your solutions to the problems in this edition by 1 February, 2009. ...
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257
SKOLIAD
No. 111
Robert Bilinski
Please send your solutions to the problems in this edition by 1 February, 2009. A opy of MATHEMATICAL MAYHEM Vol. 2 will be presented to one pre-university reader who sends in solutions before the deadline. The de ision of the editor is nal.
Our problem set this time around omes from the 2007 Christopher Newport University Math Contest. Only a sele tion of the problems is presented here. My thanks go to R. Persky, Christopher Newport University, Newport News, VA. 2007 Christopher Newport University Math Contest (sele ted questions)
1. Find the mid-point of the domain of the fun tion f (x) = (A)
1 4
(B)
3 2
(C)
2 3
(D)
p √ 4 − 2x + 5.
−2 5
2. The sum a+b, the produ t ab and the di eren e a2 −b2 for two positive numbers a and b is the same non-zero number. What is b? (A) 2
(B)
√ 1+ 5 2
(C)
√ 5
3. Let f (x) be a quadrati polynomial with f (3) Find the oeÆ ient of x in f (x). (A) 2
(B) 3
(C) 1
(D) = 15
√ 3− 5 3
and f (−3)
= 9.
(D) −2
4. A pair of fair di e is ast. What is the probability that the sum of the numbers falling uppermost is 7 or 11 if it is known that one of the numbers is a 5? (A)
2 9
(B) p
7 36
(C)
1 9
√
(D)
4 11
√
√
5. The number 24 + 572 an be written in the form a + b, where a and b are whole numbers and b > a. What is the value b − a? (A) 4
(B) −2
(C) 2
(D) 3
258 6. Let x =
1 2+
be the indi ated ontinued fra tion. Whi h one
1 3+
1
2+
1 3+···
of the following is equal to x? (A) 7. If f
r
√
15 + 1 2
1+x 1−x
(B)
= 5x,
√
2+1 3
√
15
−
(D)
√
15 − 3 2
nd f (2). √
(A) −15
(C)
−3 + 2
(B) 15
−1
(C) 3
(D) −4
Universite Christopher Newport 2007 Con ours de maths (questions individuelles)
1. Trouver le point milieu du domaine de la fon tion f (x) = (A)
1 4
(B)
3 2
(C)
2 3
p √ 4 − 2x + 5.
−2 5
(D)
2. La somme a+b, le produit ab et la di eren e a2 −b2 pour deux nombres positifs a et b est le m^eme nombre non-nul. Que vaut b? (A) 2
(B)
√ 1+ 5 2
(C)
√ 5
3. Soit f (x) un polynome ^ quadratique tel que f (3) Trouver le oeÆ ient de x dans f (x). (A) 2
(B) 3
(C) 1
√ 3− 5 3
(D) = 15
et f (−3)
= 9.
(D) −2
4. Une paire de des honn^etes est lan ee. Quelle est la probabilite que la somme des nombres sur le dessus est 7 ou 11 si on sait qu'un des nombres est un 5? (A)
2 9
(B) p
7 36
(C)
1 9
√
(D)
4 11
√
5. Le nombre 24 + 572 peut e^ tre e rit sous la forme a + et b sont des nombres entiers ave b > a. Que vaut b − a? (A) 4
(B) −2
6. Soit la fra tion ontinue x = suivantes est egale a x?
(C) 2 1 2+
ou a
(D) 3 . Laquelle des expressions
1 3+
√ b,
1
1 2+ 3+···
259 (A) 7. Si f
r
√
15 + 1 2
1+x 1−x
(B)
= 5x,
√
2+1 3
−3 + 2
√
15
(D)
−
√
15 − 3 2
trouver f (2). √
(B) 15
(A) −15
(C)
−1
(C) 3
(D) −4
Next we give solutions to the National Bank of New Zealand Junior Mathemati s Competition 2004 run by the University of Otago with the support of The National Bank of New Zealand [2007 : 386-392℄.
1 (For year 9 only).
Linda starts to write down the natural numbers in the square ells of a very large pie e of graph paper. (The graph paper is mu h larger than shown below.) She starts at the bottom left orner and writes down the numbers using the following arrangement: . . . . . . .... .... .... .... .... .... .... .... .... .... .... .... ... ... ... ... ... ... .................................................................................................................................................................. ... ... ... ... ... ... .... .... .... .... .... .... .... ... ... ... ... ... .................................................................................................................................................................. ... ... ... ... ... ... ... ... ... ... ... ... ... .. .. .. .. .. .................................................................................................................................................................. .... .... .... .... .... .... .... .... .... .... .... .... ... .. .. .. .. .. .................................................................................................................................................................. .... .... .... .... .... .... .... .... .... .... .... .... ... .. .. .. .. .. ............................................................................................................................................................... ... ... ... ... ... ... .... .... .... .... .... .... .... ... ... ... ... ... .........................................................................................................................................................
-
6 6
-
6
6 ?-
?
. . . . . . .... .... .... .... .... .... .... .... .... .... .... .... ... ... ... ... ... ... .................................................................................................................................................................. ... ... ... ... ... ... .... .... .... .... .... .... .... ... ... ... ... ... .................................................................................................................................................................. ... ... ... ... ... ... ... ... ... ... ... ... ... .. .. .. .. .. .................................................................................................................................................................. .... .... .... .... .... .... .... .... .... .... .... .... ... .. .. .. .. .. .................................................................................................................................................................. .... .... .... .... .... .... .... .... .... .... .... .... ... .. .. .. .. .. ............................................................................................................................................................... ... ... ... ... ... ... .... .... .... .... .... .... .... ... ... ... ... ... .........................................................................................................................................................
17 · · ·
16 15 14 13 5
6
7 12
4
3
8 11
1
2
9 10
(The arrangement is suggested in the left diagram; some of the numbers are shown in the right diagram.) We identify ea h of the ells using oordinates (a, b), where a is the number of positions to the right, and b is the number of positions up from the bottom. For example, the ell ontaining the number 1 has the oordinates (1, 1), while the ell ontaining the number 8 has the oordinates (3, 2). (a) What are the oordinates of the ell ontaining the number 15? (b) Starting with 1, 9, . . . , every se ond number along the bottom row follows a ertain pattern. In a few words, or using an algebrai expression, des ribe these numbers. ( ) The ell ontaining the number 21 has the oordinates (5, 5). What is the number ontained in the ell with oordinates (6, 6)? As well, nd the number ontained in the ell with oordinates (7, 7). (d) What is the number ontained in the ell with oordinates (20, 20)? (e) What are the oordinates of the ell ontaining the number 2004?
260 Solutions to (a), (b), and (d) by Jo hem van Gaalen, grade 9 student, Medway High S hool, Arva, ON. Solution to ( ) by the editor. OÆ ial solution to (e) modi ed by the editor. (a) The number 15 is pla ed two squares to the right and 4 squares up, so it is pla ed in the ell with oordinates (2, 4). (b) The numbers in these ells are x2 , where x is odd and x is the rst
oordinate of the square we are looking at. ( ) The square with orners (1, 1), (1, k), (k, 1), and (k, k) has k2 ells in it and when traversed ontains all the numbers from 1 to k2 . To rea h the
ell at (k, k) we arrive from one side of the square and we stop at (k, k), leaving un lled the k − 1 ells on the other side of the square. Sin e k − 1
ells of the square are un lled, then k2 − (k − 1) ells are lled. Therefore, 62 − (6 − 1) = 31 is in ell (6, 6) and 72 − (7 − 1) = 43 is in ell (7, 7). (d) By part ( ), the number 202 − (20 − 1) = 381 is in ell (20, 20). (e) By part (b), the number 452 = 2025 is in the ell (45, 1). It was lled going down, so we subtra t from 2025 by the orre t amount to go up in the y - oordinates. Sin e 2025 − 2004 = 21, the number 2004 is in ell (45, 22). A orre t answer to part ( ) was given by JOCHEM VAN GAALEN, grade 9 student, Medway High S hool, Arva, ON, but without an explanation for the formula.
2. The diagram shows a 4× 4 grid ontaining four oins.
....................................................................................................................... ... ... ... ...... ...... ... ... ...... .... .... ....... .... .... ... ... ... ... ....... ................................................................................................................................................ ... ... ................. .... ... ... ..... . ... ... ... . ... . .... ........ ..... .... ...... .... .... ......................... .... ... ................................................................................................................................ ... ... ... ....... ....... ... ... ...... .... .... ....... .... .. .... ... ... ... ... ..... .............................................................................................................................................. .. ................. .. ... .. .. .. .. ...... .... ... . . .... ....... ..... ..... .... ...... .... .......................... .... .... ... ................................................................................................................
Imagine that we have enough oins available to pla e anywhere we like on the grid. However, we would like to pla e oins so that we do not have three pla ed anywhere along a line, either horizontally, or verti ally, or diagonally. (a) Imagine that we add one more oin to the given layout. In how many di erent squares ould we pla e the extra oin so that we would not have three oins pla ed anywhere along a line? (b) Is it possible to add two more oins into the given layout so that we would not have three oins pla ed anywhere along a line? If it is possible, show by drawing a diagram where the two extra oins ould be pla ed. If it is not possible, explain why not. ( ) Imagine now that the grid ontains no oins at all. What is the smallest number of oins whi h ould be pla ed onto the grid so that we would not have three pla ed anywhere along a line, but if we were then to add an extra oin we ould not avoid having three pla ed along a line? Des ribe, perhaps in luding a diagram, where the oins would be pla ed. (d) Imagine again that the grid ontains no oins at all. What is the largest number of oins whi h ould be pla ed onto the grid so that nowhere are there three oins pla ed anywhere along a line? Des ribe, perhaps in luding a diagram, where the oins would be pla ed.
261 Solutions to (a) and ( ) by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. Solutions to (b) and (d) by the editor. (a) We an look at two oins at a time and eliminate N N any horizontal, verti al, or diagonal line that the two
oins may determine. In the squares that are left over N N (see the diagram), we put the letter N to show that it is N N possible to pla e a new oin there without making three
oins that all lie along a line. We see that there are seven N squares where a new oin an be pla ed. (b) For ea h of the N 's in part (a) we pla e a new oin there and then we number all the pla es where a se ond oin ould be added without making three oins along a line provided that the se ond oin is either below the rst oin or to the right of it. We nd the following six grids, for a total of 5 + 2 + 2 + 2 + 2 + 1 = 14 solutions. ............................................................................................................................ ... ... ... ...... ...... ... ... .... .... ....... ............. ....... .... .... .... .... .... ......... ........ .... ... ... ... ................. ... .. ................................................................................................................................... .... .... ............. ......... .... .... .... ....... .............. ...... .... .... .... ... .... .... ... ... ... ......................................................................................................................................... .... .... .... ....... ........ .... .... .... .... ....... ............. ....... .... .... ... ... ... ......... ...... ... .............................................................................................................................................. ... ................ ... .... ... ... .... ... .......... ........ .... .... .... ........ ............. ...... .... .... .... ... ........ ....... ... ... ... .. ....................................................................................................................
.............................................................................................................................. ... ... ... ....................... ... .... .... .... .... ... .... ....... .... .... ... ... ... ........ ...... ... ........................................................................................................................................ ... ... .................. ... ... ... .... .... ... .......... ....... .... .... .... ........ ............ ....... .... .... .... .... ................... .... .... ... ................................................................................................................................. .... .... .... ...................... .... .... .... .... .... ... ... ...... .... .... ... ... ... ........ ...... ... ........................................................................................................................................ ... .................. ... ... ... ... .... ... .......... ....... .... .... .... ........ ............ ....... .... .... .... .... ................... .... .... .... .. ............................................................................................................
............................................................................................................................. ... ... ... ................... ... ... .... .... ........ ... .... ........ .... .... ... ... ... ........ ...... ... .......................................................................................................................................... ... ... .................. ... ... ... .... .... ... .......... ........ .... .... .... ........ ............ ....... .... .... .... .... .................. .... .... ... ................................................................................................................................. .... .... .... ..................... .... .... .... ........ ... ... ....... .... .... ... ... ... ........ ...... ... .......................................................................................................................................... ... ................... ... ... ... ... .... ... .......... ....... .... .... .... ........ ............ ....... .... .... .... .... ................... .... .... .... .. ............................................................................................................
............................................................................................................................. ... ... ... .................... ... ... .... .... .... .... ... .... ....... .... .... ... ... ... ........ ...... ... ........................................................................................................................................ ... ... .................. ... ... ... .... .... ... .......... ....... .... .... .... ........ ............ ....... .... .... .... .... ................... .... .... .. .............................................................................................................................. .... .... .... .......................... .... .... .... .... .... ... ... ...... .... .... ... ... ... ........ ...... ... ........................................................................................................................................ ... .................. ... ... ... ... .... ... .......... ....... .... .... .... ........ ............ ....... .... .... .... .... ................... .... .... .... .. ............................................................................................................
............................................................................................................................. . . ..... ..... ..... ........................... ..... .... .... ....... ............. ....... .... ... .. .. .... ... .. .. ............................................................................................................................................. .... .... ........ ........ .... .... .... .... ........ ............. ........ .... .... ... ... .... ......... ...... ... ... ... ... ............... ... ... ... .......................................................................................................................................... .... .... .... ... .......... ....... .... .... .... ....... ............ ........ .... ... .. .. ..... .... .. .. ............................................................................................................................................. .... ....... ....... .... .... .... .... ........ ............. ........ .... .... .... ... .... ......... ....... .. .. .. ...........................................................................................................................................
............................................................................................................................. . . ..... ..... ..... ............................ ..... .... .... ........ ............. ....... .... ... .. .. ... ... .. .. ............................................................................................................................................. .... .... ........ ........ .... .... .... .... ........ ............. ........ .... .... ... ... .... ......... ...... ... ... ... ... ............... .... ... ... .......................................................................................................................................... .... .... .... ... .......... ........ .... .... .... ........ ............ ....... .... ... .. .. .... .... .. .. ............................................................................................................................................. .... ....... ....... .... .... .... .... ........ ............. ........ .... .... .... ... .... ......... ....... .. .. .. ...........................................................................................................................................
............................................................................................................................. . . ..... ..... ..... ........................... ..... .... .... ....... ............. ....... .... ... .. .. .... ... .. .. ............................................................................................................................................. .... .... ........ ........ .... .... .... .... ........ ............. ........ .... .... ... ... .... ......... ...... ... ... ... ... ............... ... ... .. ......................................................................................................................... . . . .. .. . . ..... ..... ..... ......................... ..... .... .... ....... ............ ........ .... ... .. .. ..... .... .. .. ............................................................................................................................................. .... ....... ....... .... .... .... .... ........ ............. ........ .... .... .... ... .... ......... ....... .. .. .. ...........................................................................................................................................
1
4
N
2
5
3
1
2
N
N
1 2
N 1 2
( ) For ea h square in the diagram, we ount how many lines ( onsisting of 3 or more squares) pass through it. A
oin an be on at most 4 lines, so 3 oins an be on at most 3 · 4 = 12 lines. There are 14 lines in total, so a new oin
an be added to any 3. Four oins pla ed on the 4's in the grid is a smallest solution to the problem. [Ed.: Another solution is to put oins in the four orners.℄ (d) Clearly any row of the grid an ontain at most two
oins. Therefore, at most 4 · 2 = 8 oins an be pla ed on the grid. [Ed.: There are many ways to draw su h patterns. We will present one that wasn't handed in by our two solvers, who both gave the same right answer but laimed that it was unique.℄
1 2
N
N 1
.................................................................................................................... .... .... .... .... .... .... .... .... .... .... ... .. .. .. .. ........................................................................................................................ .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .. ...................................................................................................................... .... .... .... .... .... .... .... .... .... .... ... .. .. .. .. ........................................................................................................................ .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .. ............................................................................................................
3
3
3
3
3
4
4
3
3
4
4
3
3
3
3
3
................................................................................................................................. . .. .. . . .. .. . . .... ......................... .... ......................... .... ....... ............ ....... ....... ............ ....... .... .... ................... .... ... ..... .... ... ... ....................................................................................................................................................... ... ... ... ................... ... ................... ... .... .... ........ ... ... ............ ... ... ........ .... ... ... ... ......... ...... ... ......... ...... ............................................................................................................................................................... .. ................ .. ................ .. .. .. .... .... .......... ........ .... .......... ........ .... .... ....... ............ ........... ............ ........ .... .... .... .................... .... .................... .... .... ... ........................................................................................................................................... ... ... ................... ... ... .................. ... .... ........ .... ... ....... ........ .... ... ....... .... ... ... ........ ..... ... ... ........ ..... ...............................................................................................................................................................
Also solved by RUIQI YU, student, Stephen Lea o k Collegiate Institute, Toronto, ON
(parts (b) and (d)); and JOCHEM VAN GAALEN, grade 9 student, Medway High S hool, Arva, ON.
Both solvers gave solutions to part (b) but neither gave a omplete list of all solutions.
262
3.
The diagram shows an equilateral triangle di3 vided into three smaller triangles. Small ir les have been pla ed on ea h of the verti es, and pos8 itive whole numbers (in this ase 3, 3, 2, and 1) 7 6 have been written inside ea h small ir le. Over1 all the shape forms four regions: three triangles 6 and an outer region. In ea h of these regions the 3 2 sum of the orresponding verti es has been written. For example, the outer region ontains the value 8, be ause 3 + 3 + 2 = 8. In this question we shall be investigating what happens when the numbers in the small ir les are hanged. (Throughout this question, only positive whole numbers will be used.) (a) Imagine that the number in ea h one of the small ir les is 5 . What is the total when the numbers inside all four regions are added together? (b) Find possible numbers inside ea h of the four small ir les so that the sums in the three triangles are 8, 9, and 10, respe tively, while the sum of the outer region is 6. ( ) Find possible numbers inside ea h of the four small ir les so that the sums in the three triangles are 8, 9, and 9, respe tively. (Do not worry about the sum of the outer region in this part of the question.) (d) Is your answer to part ( ) the only possible answer? If it is, explain why no other answer is possible. If it is not, nd another answer. ............. .... ... ..... .... ......
............ ..... .... ..............
........... .............................................................. .......... ........ .. .... ........ ....... ................ ....... ..... . ... .. .... .... . . . . . . .... ... ... ..... .... . . . ... ... .. .... .... ... . . ... ... .. ... .. . . ... ... . .. .. . . ... ... .. . .... . .. . ... . ... .. ... ..... . . ... ... ... . .. . .. . ... ... . .............. . . . ... ... . . ... ... . .. ... . . . .. . . ... . ............................. . . .. . . ... . . . . . . . ........ ... . .. ...... . .. . . . . . . . . . . . . . . ........ .. ... . . ........... . . . . . . . . . ........ ............ .. .. ............. ........ .... ... . ... .. .... ............................................................................................... . .. ...... ................... ................. . .... ... .... ... . .... . . ... ..... ....... ..... ........ ....... ........... ........ ....................................................
............ ..... ..... ..............
................ . ..... ................
(e) We have been using positive whole numbers throughout this question. In a few words, or using an algebrai expression, give a general des ription of the total when the numbers inside all four regions are added together. Explain your reasoning for your des ription. Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. (a) Let a, b, c, and d be the numbers in the small ir les (with b in the
entre) and let A, B , C , and D be the numbers in the regions (with D in the outer region). Then we have A B C D
= = = =
a a b a
+ + + +
b b c c
+ + + +
c, d, d, d.
If a = b = c = d = 5, then A = B = C = D = 15 and their sum is 60. (b) and (e) We are given that A = 8, B = 9, C = 10, and D = 6. From A + B + C + D = 3(a + b + c + d) = 33
263 we have S = a + b + c + d = 11. Then a = S − C = 11 − 10 = 1, and similarly b = 5, c = 2, and d = 3. In parti ular, the sum of the numbers in the regions is three times the sum of the numbers in the ir les. ( ) and (d) Let C = 8 and A = B = 9. Then A − C = a − d = 1 and B − C = a − c = 1, hen e, c = d. We an have a solution only for c, d in {1, 2, 3}, otherwise c, d ≥ 4 and then b ≤ 0, a ontradi tion. Therefore, for (a, b, c, d) we have the three solutions (2, 6, 1, 1), (3, 4, 2, 2), and (4, 2, 3, 3). Also solved by Jo hem van Gaalen, grade 9 student, Medway High S hool, Arva, ON.
4.
A lass of students votes to sele t one andidate as their representative on the s hool oun il. Their tea her de ides on the following voting system: \You have to rank the three andidates in order: rst, se ond, and third. Your rst hoi e will re eive one point, your se ond hoi e will re eive two points, and your third hoi e will re eive four points. The winner will be the student with the smallest total." After the voting has been ompleted, the tea her dis overs that there is a problem with this voting system. She explains the problem to the prin ipal: \The student with the smallest s ore is Diane, who re eived 44 points. However, only four people voted for her as their rst hoi e. Next was Belinda with 45 points. She re eived more rst hoi e votes than anyone else. Colin was in last pla e with 51 points, and he had more people voting for him as their third hoi e than voted for the other andidates. It looks as though I will have to announ e to the lass that Diane is the winner, even though she had the smallest number of people voting for her as rst hoi e." (a) Show that 20 students took part in the voting. (b) How many people voted for Belinda as their rst hoi e? ( ) Explain why your answer to (b) is the only possible answer whi h ts the tea her's des ription of how the votes were ast. (d) How many people voted for Belinda as their se ond hoi e, and how many people voted for her as their third hoi e? Solution to (a) by Jo hem van Gaalen, grade 9 student, Medway High S hool, Arva, ON. OÆ ial solution to (b) and ( ). Solution to (d) by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. (a) The total number of points re eived is 44 + 45 + 51 = 140. A single ballot gives out 7 points (1, 2, and 4 points for the rst, se ond, and third
hoi es, respe tively) Therefore, 140 = 20 people took part in the voting. 7
(b) and ( ) Four people voted " rst" for Diane, leaving 16 " rst" votes for Belinda and Colin between them. Belinda and Colin re eived an odd number of points, so both must have had an odd number of " rst" votes. Be ause Belinda had the highest number of " rst" votes, and be ause both of them
264 re eived more than Diane's four votes, the only possibilities are Belinda: 11, Colin: 5 or Belinda: 9, Colin: 7. If Belinda re eived 11 " rst" votes, then to obtain a total 45 she must have re eived 11(1) + 1(2) + 8(4) = 45. Then Colin would have re eived 5(1)+7(2)+8(4) = 51. However this gives Colin the same number of "third" votes as Belinda, whi h ontradi ts one of the statements in the question. If Belinda re eived 9 " rst" votes and Colin re eived 7 " rst" votes, then the totals for the andidates are Belinda: 9(1) + 4(2) + 7(4) = 45, Colin: 7(1) + 4(2) + 9(4) = 51, and Diane: 4(1) + 12(2) + 4(4) = 44. (Diane appears to be the " ompromise" andidate, not really supported but not disliked either.) In on lusion, Belinda re eiving nine " rst" votes is the only solution. (d) By parts (a) and (b), Belinda has 11 votes other than " rst". If all of these are "se ond" votes, then this makes up 22 points, leaving a shortfall of 14 points to rea h her total. Divide the 14 points by 2 to get 7. So 7 "third" votes and 4 "se ond" votes make up her remaining 36 points. Therefore, Belinda got 4 se ond hoi e votes and 7 third hoi e votes. Also solved by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON
(parts (a), (b), and ( )); and Jo hem van Gaalen, grade 9 student, Medway High S hool, Arva, ON (parts (b), ( ), and (d)).
5. Ari has ut some regular pentagons out of ardboard and is joining them
together to make a ring (see Figure 1). He has ut them using a template so that they are all the same size. ............................. .. .. ... ... .. ... ... .. . ........ . . . ..... . . . . .... .......... ...... . . . . . . . . . . . . ....... ....... ... .... ................................... .. .. .. ... ... ... .. .. .. ... .. .. . . ...... . ....... ............................... ....... ............ .............
............................. ... ... .. ... .. .. ... ... ... . ..... .. ......... . . . . . . . ....... ........ .. . . . . . ........... ... . . . . .. ..... ... ...... ............ ....... ...... ....... ...... ....... . ........... . . . . . . . . . . . . . .. ........... .. ... .. .. .. .. .. .. ... .. ... ........................................... ...........
Figure 1 Figure 2 ◦ (a) The external angle of a regular pentagon is 72 . Explain how this value is al ulated. (b) When the ring is omplete, how many pentagons will there be? Next Ari de ides to join his pentagons with squares whi h have the same side length (see Figure 2). He would like to ombine them all together to make a new ring with alternating squares and pentagons. ( ) Is it possible for Ari to onstru t a ring in this way? If it is possible, explain why. If it is not possible, explain why not. (d) Ari nally de ides to onstru t a ring using regular hexagons (six sides) joined together. (This is not shown in any diagram.) If the hexagons have side length of exa tly one unit, what is the area of the shape en losed inside the ring?
265 Solutions to (a), (b), and ( ) by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON (part ( ) solution modi ed by the editor). Solution to (d) by the editor. (a) The sum of all the external angles of a regular pentagon is 360◦ and they = 72◦ . are all equal, hen e, the external angle of a regular pentagon is 360 5 (b) If a ring an be onstru ted, then the internal gure of the ring is a regular polygon, and the internal angle of the polygon is 72◦ +72◦ = 144◦ . If n is the number of sides of the polygon, then we have (n−2)·180◦ = n·144◦ , and hen e, n = 10. Ea h regular pentagon in the ring shares exa tly one side with the inner polygon, so there are 10 pentagons in the ring. ( ) Label 5 onse utive points along what would be the inner side of the ring, as shown E in the diagram at right. We have that ◦
D
∠ABC = 360◦ − 90◦ − 108◦ = 162◦ ,
and that △ABC is iso eles with AB = BC , hen e, ∠CAB = ∠BCA = 9◦ . By symmetry we also have ∠DCE = 9◦ . It follows that ∠ACE = 360◦ − 198◦ − 18◦ = 144◦ .
............................................... .. ... .. .. .. .. .. ... .. . . ... . .. ... .. .. . ........ . ...... . . . . ..... . ... ............ . . . . . . . ....... ... . . . . . . . . . . . . ....... ..... . ............. .... .. .... ... ... .... ................. . . . . . . . . . ....... . .... . . . . . . . . . . . . ....... . ....... ....... ...... ....... ........ ........ .. ............. .... .. ............ . . . . . . . . . . . . . . . ..... .. .. ... .. .. .. ... .. .. .. .. .. .. .. .. . . . .. .. ................................................ .. ............ ......................... .
C
B
A
By the result of part (b) we an state that the ring will be ompleted by alternating 10 squares and 10 pentagons. (d) If regular hexagons of the same size are used to onstru t a ring, then the shape in luded in the ring is an identi al regular hexagon. Thus, the area of the inner shape is 6 times the area of an equilateral triangle of side 1, or Area
= 6
1·
√ 3 2
2
!
√ 3 3 = 2
.
There was one in orre t solution submitted.
That brings us to the end of another issue. This month's winner of a past Volume of Mayhem is Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. Congratulations Ruiqi! Continue sending in your
ontests and solutions.
266
MATHEMATICAL MAYHEM Mathemati al Mayhem began in 1988 as a Mathemati al Journal for and by High S hool and University Students. It ontinues, with the same emphasis, as an integral part of Crux Mathemati orum with Mathemati al Mayhem. The Mayhem Editor is Ian VanderBurgh (University of Waterloo). The other sta members are Monika Khbeis (As ension of Our Lord Se ondary S hool, Mississauga), Eri Robert (Leo Hayes High S hool, Frederi ton), Larry Ri e (University of Waterloo), and Ron Lan aster (University of Toronto).
Mayhem Problems Please send your solutions to the problems in this edition by 15 November . Solutions re eived after this date will only be onsidered if there is time before publi ation of the solutions. Ea h problem is given in English and Fren h, the oÆ ial languages of Canada. In issues 1, 3, 5, and 7, English will pre ede Fren h, and in issues 2, 4, 6, and 8, Fren h will pre ede English. The editor thanks Jean-Mar Terrier of the University of Montreal for translations of the problems. 2008
M350.
Proposed by the Mayhem Sta . Dean rides his bi y le from Coe Hill to Apsley. By distan e, one-third of the route is uphill, one-third of the route is downhill, and the rest of the route is on at ground. Dean rides uphill at an average speed of 16 km/h and on at ground at an average speed of 24 km/h. If his average speed over the whole trip is 24 km/h, then what is his average speed while riding downhill?
M351.
Proposed by Kunal Singh, student, Kendriya Vidyalaya S hool, Shillong, India. Let C be a point on a ir le with entre O and radius r. The hord AB is of length r and is parallel to OC . The line AO uts the ir le again at E and it uts the tangent to the ir le at C at the point F . The hord BE uts OC at L and AL uts CF at M . Determine the ratio CF : CM .
M352.
Proposed by the Mayhem Sta . Consider the numbers 37, 44, 51, . . . , 177, whi h form an arithmeti sequen e. A number n is the sum of ve distin t numbers from this sequen e. How many possible values of n are there?
267
M353.
Proposed by Mihaly Ben ze, Brasov, Romania. Determine all pairs (x, y) of real numbers for whi h xy +
1 x
+
1 y
=
1 xy
+ x + y.
M354.
Proposed by the Mayhem Sta . Without using a al ulating devi e, determine the prime fa torization of 320 + 319 − 12.
M355.
Proposed by the Mayhem Sta . A right ir ular one with vertex C has a base with radius 8 and a slant height of 24. Points A and B are diametri ally opposite points on the ir umferen e of the base. Point P lies on CB . (a) If CP = 18, determine the shortest path from A through P and ba k to A that travels ompletely around the one. (b) Determine the position of P on CB that minimizes the length of the shortest path in part (a).
M356.
Proposed by Mihaly Ben ze, Brasov, Romania. Determine all pairs (k, n) of positive integers for whi h k(k + 1)(k + 2)(k + 3) = n(n + 1) .
................................................................. M350. Propose par l'Equipe de Mayhem.
Daniel se rend a bi y lette de Montignez a Bon ourt. Un tiers du hemin est a la montee, un tiers a la des ente et le reste au plat. Il grimpe a une vitesse moyenne de 16 km/h et fait du 24 km/h au plat. Si sa vitesse moyenne pour tout le trajet est 24 km/h, quelle est-elle pour la portion des endante ?
M351. Propose par Kunal Singh, etudiant, Kendriya Vidyalaya S hool,
Shillong, Inde. Soit C un point sur un er le de entre O et de rayon r. Soit AB une
orde de longueur r parallele a OC . La droite AO oupe de nouveau le er le en E et elle oupe la tangente au er le par C au point F . La orde BE oupe OC en L et AL oupe CF en M . Determiner le rapport CF : CM . M352. Propose par l'Equipe de Mayhem.
On onsidere la suite arithmetique 37, 44, 51, . . . , 177. Si n designe la somme de 5 nombres distin ts de ette suite, ombien de valeurs le nombre n peut-il prendre ?
268
M353. Propose par Mihaly Ben ze, Brasov, Roumanie.
Trouver toutes les paires (x, y) de nombres reels pour lesquelles xy +
1 1 1 + = + x + y. x y xy
M354. Propose par l'Equipe de Mayhem.
Sans l'aide d'une al ulatri e, trouver la de omposition en fa teurs premiers de 320 + 319 − 12.
M355. Propose par l'Equipe de Mayhem.
Un one ^ de revolution de sommet C a une base de rayon 8 et une apotheme mesurant 24. Soit A et B deux points du er le de base situes sur un m^eme diametre, et P un point sur CB . (a) Si CP = 18, trouver le plus ourt hemin autour du one ^ partant de A et passant par P pour nir en A. (b) Trouver la position de P sur CB minimisant la longueur du plus ourt
hemin omme mentionne en (a).
M356. Propose par Mihaly Ben ze, Brasov, Roumanie.
Trouver toutes les paires (k, n) d'entiers positifs pour lesquelles k(k + 1)(k + 2)(k + 3) = n(n + 1) .
Mayhem Solutions M301. Proposed by D.E. Prithwijit, University College Cork, Republi of
Ireland. The general term of a sequen e is tn = n2 + 20, for n ≥ 1. Show that for all n ≥ 1, the greatest ommon divisor of tn and tn+1 must be a divisor of 81. Solution by Geo rey A. Kandall, Hamden, CT, USA Let d be any ommon divisor of tn and tn+1 . Sin e tn+1 − tn
=
(n + 1)2 + 20 − (n2 + 20)
= (n2 + 2n + 21) − (n2 + 20) = 2n + 1 ,
we nd that d | (2n + 1) be ause d | tn and d | tn+1 .
269 Also, sin e 4tn − (2n + 1)(2n − 1) = 81, we nd that d | 81 (be ause and d | (2n + 1)). Thus, any ommon divisor of tn and tn+1 (in parti ular the greatest ommon divisor) must divide 81.
d | tn
MEZ MORENO, Universidad de Jaen, Also solved by SAMUEL G O Jaen, Spain; RICHARD student, Sarajevo College, Sarajevo, I. HESS, Ran ho Palos Verdes, CA, USA; SALEM MALIKIC, \Abastos", Valen ia, Spain; KUNAL SINGH, Bosnia and Herzegovina; RICARD PEIRO, IES student, Kendriya Vidyalaya S hool, Shillong, India; J. SUCK, Essen, Germany; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and VINCENT ZHOU, student, Dr. Norman Bethune Collegiate Institute, Agin ourt, ON.
M302. Proposed by Babis Stergiou, Chalkida, Gree e.
A triangle ABC has ∠ABC = ∠ACB = 40◦ . If P is a point in the interior of the triangle su h that ∠P BC = 20◦ and ∠P CB = 30◦ , prove that BP = BA. Solution by D.J. Smeenk, Zaltbommel, the Netherlands. Without loss of generality, we may assume that AB = AC = 1. Sin e triangle ABC is isos eles, we have that BC = 2 cos(40◦ ). Also, ∠BP C = 180◦ − ∠P BC − ∠P CB , hen e, ∠BP C = 130◦ . 20◦ Considering triangle BP C and 20◦ using the Law of Sines, we obtain the B following equivalent ratios:
A
........ ..... ........ .... .... .... .... ................ .... . . . . ........ .... .... ........ .... .... . .... . . . . . . . . ........ ..... ... . . . . . . . ..................... ........ . . . .... . . . . . . . .. . . . . . . . ........ ....... . . ... . . . . ........ ........ . . . . . . . ....... ........ ..... . . ... . . . ◦ . . . . . . . . ........ .... ....... . . ... . . . . . ........ .... . . . . . . ........ .... .... .......... ... .. .... ........... ........... .... ◦................................ ......................... . . . ........ ................. . . . . .........................................................................................................................................................................................................................
BP sin(30◦ ) BP 1/2
= =
P
130
30
C
BC , sin(130◦ ) 2 cos(40◦ ) . sin(130◦ )
We also know that sin(130◦ ) = sin(90◦ + 40◦ ) = cos(40◦ ), so the last equation simpli es to BP = 1. Therefore, BP = BA. Also solved by PAUL BRACKEN and N. NADEAU, University of Texas, Edinburg, TX, USA; MEZ MORENO, Universidad de COURTIS G. CHRYSSOSTOMOS, Larissa, Gree e; SAMUEL G O Jaen, Jaen, Spain; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; GEOFFREY A. KANDALL, student, Sarajevo College, Sarajevo, Bosnia and Hamden, CT, USA; SALEM MALIKIC, \Abastos", Valen ia, Spain; CAO MINH QUANG, Herzegovina; RICARD PEIRO, IES Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and VINCENT ZHOU, student, Dr. Norman Bethune Collegiate Institute, Agin ourt, ON. There were 5 in orre t or in omplete solutions submitted.
M303. Proposed by Neven Juri , Zagreb, Croatia.
A urious relation among squares states that the sum of n + 1 onse utive squares, beginning with the square of n(2n + 1), is equal to the sum of the squares of the next n onse utive integers. (For example, when n = 1 we have 32 + 42 = 52 , and when n = 2 we have 102 + 112 + 122 = 132 + 142 .) Show that this property holds for any n ≥ 1.
270 I. Solution by Gustavo Krimker, Universidad CAECE, Buenos Aires, Argentina. We must prove the equality 2 2nX +2n
j
2
2 2nX +3n
=
j=2n2 +n
whi h is equivalent to n+1 X k=1
j2 ,
j=2n2 +2n+1
(2n2 + 2n + 1 − k)2 =
n X
(2n2 + 2n + k)2 .
k=1
For any n ≥ 1 the required equality follows from the al ulation below, where the se ond equality arises by fa toring a di eren e of squares, and we use the fa t that the sum of the rst n odd positive integers is n2 : n X
= =
2
2
(2n + 2n + k) −
k=1 n X
n X
(2n2 + 2n + 1 − k)2 − (2n2 + n)2
k=1
(2n2 + 2n + k)2 − (2n2 + 2n + 1 − k)2 − (2n2 + n)2
k=1 n X
!
2
(2k − 1)(4n + 4n + 1)
k=1 2
= (4n + 4n + 1)
n X
!
(2k − 1)
k=1
− (2n2 + n)2 − (2n2 + n)2
= (2n + 1)2 n2 − (2n2 + n)2 = 0 .
II. Solution submitted independently by Carl Libis, University of Rhode Island, Kingston, RI, USA; and Salem Maliki , student, Sarajevo College, Sarajevo, Bosnia and Herzegovina. Let S be the sum of the n + 1 onse utive squares beginning with the n P n(n + 1)(2n + 1) square of n(2n + 1). Sin e i2 = , we nd that 6 i=1
2
S
=
2nX +2n
i=2n2 +n
=
=
=
2 2nX +2n
i2 =
i=1
2n2X +n−1
i2 −
i=1
i2
1 (2n2 + 2n)(2n2 + 2n + 1)(4n2 + 4n + 1) 6 − (2n2 + n − 1)(2n2 + n)(4n2 + 2n − 1) 1 (16n6 + 48n5 + 60n4 + 40n3 + 14n2 + 2n) 6 − (16n6 + 24n5 − 10n3 − n2 + n)
4n5 + 10n4 +
25 3 n 3
+ 52 n2 + 16 n .
271 Similarly, let T be the sum of the n onse utive squares, beginning with the square of 2n2 + 2n + 1. Then T
2 2nX +3n
=
i=2n2 +2n+1
=
=
1 6
2 2nX +3n
i2 =
i=1
2 2nX +2n
i2 −
i=1
i2
(2n2 + 3n)(2n2 + 3n + 1)(4n2 + 6n + 1)
− (2n2 + 2n)(2n2 + 2n + 1)(4n2 + 4n + 1) 1 (16n6 + 72n5 + 120n4 + 90n3 + 29n2 + 3n) 6 − (16n6 + 48n5 + 60n4 + 40n3 + 14n2 + 2n)
= 4n5 + 10n4 +
25 3 n 3
+ 52 n2 + 16 n .
We then have S = T for any n ≥ 1, whi h is the required equality.
MEZ MORENO, Universidad de Jaen, Also solved by SAMUEL G O Jaen, Spain; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; GEOFFREY A. KANDALL, Hamden, CT, USA; DEREK MERRELL and SUE YANG, students, California State University, Fresno, CA, USA; J. SUCK, Essen, Germany; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and VINCENT ZHOU, student, Dr. Norman Bethune Collegiate Institute, Agin ourt, ON.
M304. Corre ted. Proposed by Mihaly Ben ze, Brasov, Romania.
Let a, b, and c be real numbers su h that both a + b+ c and ab+ bc + ca are rational numbers, and a + b + c 6= 0. Show that a4 + b4 + c4 is a rational number if and only if the produ t abc is a rational number. Solution by Miguel Maran~ on Grandes, student, Universidad de La Rioja, Logrono, ~ La Rioja, Spain. Let a + b + c = pq and ab + bc + ca = rs , where p, q, r, and s are integers with q 6= 0 and s 6= 0. We have a2 + b2 + c2 + 2(ab + bc + ca) = a2 + b2 + c2
=
(a + b + c)2 = p2 q2
p2
;
q2 r p1 − 2 = s q1
,
where p1 and q1 are integers, hen e, a2 + b2 + c2 is a rational number. Setting A = a2 b2 + b2 c2 + c2 a2 , we have a4 + b4 + c4 + 2 a2 b2 + b2 c2 + c2 a2
a4 + b4 + c4
= =
a2 + b2 + c2 p21 q12
− 2A .
2
=
p21 q12
;
This means that a4 + b4 + c4 is a rational number if and only if A is a rational number.
272 Finally we obtain a2 b2 + b2 c2 + c2 a2 + 2abc(a + b + c) = r2 p A = 2 − 2abc . s q
(ab + bc + ca)2 =
r2 s2
;
Sin e pq = a + b + c 6= 0, then A is a rational number if and only if abc is a rational number. Thus, a4 + b4 + c4 is a rational number if and only if abc is a rational number, with a + b + c 6= 0, as we wanted to show.
MEZ MORENO, Universidad de Jaen, Also solved by SAMUEL G O Jaen, Spain; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; D. KIPP JOHNSON, Beaverton, OR, USA; GEOFFREY A. KANDALL, Hamden, CT, USA; CARL LIBIS, University of Rhode Island, Kingston, RI, USA; student, THANOS MAGKOS, 3rd High S hool of Kozani, Kozani, Gree e; SALEM MALIKIC, Sarajevo College, Sarajevo, Bosnia and Herzegovina; MISSOURI STATE UNIVERSITY \Abastos", PROBLEM SOLVING GROUP, Spring eld, MO, USA; RICARD PEIRO, IES Valen ia, Spain; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; J. SUCK, Essen, Germany; and EDWARD T.H. WANG and KAIMING ZHAO, Wilfrid Laurier University, Waterloo, ON. There were 7 in omplete or in orre t solutions submitted. Our apologies for any awed solutions due to the original in orre t√version of the prob√ lem. A ounterexample to the original problem statement is a = b = 2 and c = −2√ 2. 4 4 4 These give a + b + c = 0, ab + ac + bc = −6, and a + b + c = 72, but abc = −4 2.
M305. Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain. Find all real solutions to the following system of equations: √ √ √ x+ y+ z √ √ √ x x+y y+z z √ √ √ x2 x + y 2 y + z 2 z
= = =
3, 3, 3.
Solution by J. Su k, Essen, Germany. Adding the rst and third equations and subtra ting twi e the se ond equation, we obtain √ (x2 + 1 − 2x) x + +
or
(x − 1)2
√ (y 2 + 1 − 2y) y √ (z 2 + 1 − 2z) z = 3 + 3 − 2(3) ,
√ √ √ x + (y − 1)2 y + (z − 1)2 z = 0 .
Ea h of the nonnegative, so √ three terms2is √ ea h must equal zero; that is, √ x = (y − 1) y = (z − 1)2 z = 0. Therefore, x, y , and z ea h take the √ value 0 or√1. Sin e x + √y + z = 3, we must have x = y = z = 1, whi h learly satis es all three equations. (x − 1)2
273 Also solved by PAUL BRACKEN, University of Texas, Edinburg, TX, USA; SAMUEL MEZ MORENO, Universidad de Jaen, GO Jaen, Spain; RICHARD I. HESS, Ran ho Palos Verdes, student, Sarajevo CA, USA; GEOFFREY A. KANDALL, Hamden, CT, USA; SALEM MALIKIC, \Abastos", Valen ia, Spain; College, Sarajevo, Bosnia and Herzegovina; RICARD PEIRO, IES CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; URZICA SORIN, Grigore Cobal es u High s hool, Moinesti, Romania; GEORGE TSAPAKIDIS, Agrinio, Gree e; EDWARD T.H. WANG and KAIMING ZHAO, Wilfrid Laurier University, Waterloo, ON; VINCENT ZHOU, student, Dr. Norman Bethune Collegiate Institute, Agin ourt, ON (2 solutions); and TITU ZVONARU, Comane sti, Romania. There were 3 solutions submitted that were in omplete or in orre t.
M306. Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL. Find all solutions to the following addition problem, in whi h ea h letter represents a distin t digit:
+
T
E
N
T
E
N
N
I
N
E
E
I
G H
T
T
H R
E
E
.................................................................................................................................................................................
F
O
R
T
Y
Solution by Candi e Arredondo, student, California State University, Fresno, CA, USA. The solutions are 417 329 308 718 701 713 631
+ + + + + + +
417 329 308 718 701 713 631
+ + + + + + +
7871 9492 8780 8281 1410 3431 1013
+ + + + + + +
18324 24863 07643 12347 04267 14207 30756
+ + + + + + +
42511 36022 34200 74011 76500 70611 65233
= = = = = = =
624 + 624 + 4042 + 20756 + 65322 148 + 148 + 8284 + 42391 + 19644 047 + 047 + 7174 + 41690 + 09344
= = =
69540 , 71035 , 51239 , 96075 , 83579 , 89675 , 98264 ,
91368 , 70615 , 58302 .
Also solved by RICHARD I. HESS, Ran ho Palos Verdes, CA, USA. There was 1 in orre t solution submitted. Ms. Arredondo des ribed the exhaustive trial and error approa h that she took to nd all the solutions. Is there a \ leaner" approa h? Hess also obtained the same solutions as above, but without the solutions ontaining leading zeros.
274
M307. Proposed by Neven Juri , Zagreb, Croatia.
Two 4×4 magi squares have the property that all four of their rows, all four of their olumns, and their two diagonals all sum to the same value N . Consider the sum of the four orner elements of ea h square. Can these sums be di erent, or must they be the same? (In other words, does the orner sum depend on the square itself, or only on the magi sum N ?) Either determine the onstant sum, or show that these sums an di er. Almost all submitted solutions were the same. Let aij denote the entry in the ith row and j th olumn. The sum of the four orner elements of a 4 × 4 magi square is then a11 + a14 + a41 + a44 . Using the properties of the magi square, we obtain the following =
= =
2(a11 + a14 + a41 + a44 ) (a11 + a12 + a13 + a14 ) + (a41 + a42 + a43 + a44 ) + (a11 + a22 + a33 + a44 ) + (a14 + a23 + a32 + a41 ) − (a12 + a22 + a32 + a42 ) − (a13 + a23 + a33 + a43 ) N +N +N +N −N −N 2N .
Therefore, the four orner elements of a 4 × 4 magi square satisfying the given properties have a onstant sum of N . Solved by RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; CARL LIBIS, University of student, Sarajevo College, Sarajevo, Bosnia Rhode Island, Kingston, RI, USA; SALEM MALIKIC, and Herzegovina; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring eld, MO, USA; and JUSTIN YANG, student, Lord Byng Se ondary S hool, Van ouver, BC. There was 1 in orre t solution submitted.
M308. Proposed by Babis Stergiou, Chalkida, Gree e.
Let ABC be a right triangle with A = 90◦ , and let M be the mid-point of side AB . If D is the foot of the perpendi ular from A to CM and N is the mid-point of DC , prove that BD ⊥ AN . Solution by Missouri State University Problem Solving Group, Spring eld, MO, USA. We assign oordinates. Let A, B , and C C have oordinates (0, 0), (b, 0), and (0, c), respe tively. Then the oordinates of M are b −2c N ( , 0). Also, the slope of CM is and the 2 b D equation of CM is y = − 2c x + c. Sin e AD b and CM are perpendi ular, the slope of AD is the negative re ipro al of −2c , therefore, A M B b .......... ............... .... ............... ... ... ........ ... .... ....... .. ... ... ... ............ ... ... ....... . ....... .... ..... ....... .... ... ...... ....... ... ....... .. .... ... ....... .... .... ... ....... . ... . . ....... ... ... ... . ....... . ... ... ... ....... . . ....... .......... .... ... . . . . ....... . . ... .. ... .... . ....... . . . . . . ... .. ... . ....... ... ... ... ............. ....... ... ... .. ........ ....... . ... .... .................. . ...................................................................................................................................................
the equation of AD is y =
b x. 2c
275 Sin e D is the point of interse tion of CM and AD, we an nd its oordinates using the equations of CM and AD. Equating values of y, we obtain 2c b 2bc2 − x+c = x whi h upon solving for x yields x = 2 . Therefore, b 2c b + 4c2 y=
b b2 c x= 2 2c b + 4c2
and D has oordinates
The point N has oordinates point of DC . The slope of AN is is
b2 c
2bc2 b2 c , 2 2 2 b + 4c b + 4c2
b2 c + 2c3 bc2 , 2 2 2 b + 4c b + 4c2 2 3 b c + 2c b2 + 2c2 = 2 bc bc
.
, sin e it is the mid-
and the slope of BD
b2 c bc b2 + 4c2 = = − 2 2 2 2 2 2bc 2bc − b(b + 4c ) b + 2c2 − b b2 + 4c2
.
The produ t of the slopes is −1, hen e, AN and BD are perpendi ular.
Also solved by RICARDO BARROSO CAMPOS, University of Seville, Seville, Spain; COURTIS G. CHRYSSOSTOMOS, Larissa, Gree e; RICHARD I. HESS, Ran ho Palos Verdes, CA, student, Sarajevo College, USA; GEOFFREY A. KANDALL, Hamden, CT, USA; SALEM MALIKIC, Sarajevo, Bosnia and Herzegovina; ANDREA MUNARO, student, University of Trento, Trento, \Abastos", Valen ia, Spain; CAO MINH QUANG, Nguyen Binh Italy; RICARD PEIRO, IES Khiem High S hool, Vinh Long, Vietnam; DANIEL REISZ, Auxerre, Fran e; and D.J. SMEENK, Zaltbommel, the Netherlands.
M309. Proposed by Mihaly Ben ze, Brasov, Romania. Determine all possible non-negative integers x, y, z , and t su h that is a perfe t ube.
3x + 3y + 3z + 3t
Partial solution independently by Ja lyn Chang, student, Western Canada High S hool, Calgary, AB; and the proposer. For any non-negative integers a and b, (3a + 3b )3
=
(3a )3 + 3(3a )2 (3b ) + 3(3a )(3b )2 + (3b )3
=
33a + 32a+b+1 + 3a+2b+1 + 33b .
Thus, x = 3a, y = 2a + b + 1, z = a + 2b + 1, and t = 3b, and their permutations are solutions for any non-negative integers a and b. No other solutions were submitted. The partial answer gives an in nite family of solutions. Are there any more solutions? Can any readers prove that there are not any more solutions? As it turns out, we had intended to ask for an in nite family of solutions rather than all solutions but somehow asked for all solutions { our apologies for this.
276
M310. Proposed by J. Walter Lyn h, Athens, GA, USA.
Four ongruent re tangles are arranged in a square pattern so that they en lose a smaller square. Let S be the area of the outer square and Q the area of the inner square. If the area of the outer square is 9 times the area of the inner square, determine the ratio of the sides of the re tangles.
...................................................................................................................................... ... ... ... ... ... ... .... .... .... .... .... .... .... .... .... .... .... ... .... .... ... . .. .... . ................................................................................. .... . . . .... ..... ... .... .... .... .... .... .... .. ... ... .................................................................................. .... ... .... .... .. ... . ... ... .... .... ... . . .... . . ..... ..... .... .... ... ... .... .... ... .................................................................................................................................
Almost all submitted solutions were the same. Let x and y be the side lengths of one of the four re tangles. Without loss of generality, we may assume that x > y. Therefore, the outer square has side length x + y and the inner square has side length x − y. From the given information S = 9Q, we obtain su
essively (x + y)2 x + 2xy + y 2 8x2 − 20xy + 8y 2 2x2 − 5xy + 2y 2 (2x − y)(x − 2y) 2
= = = = =
9(x − y)2 , 9x2 − 18xy + 9y 2 , 0, 0, 0.
Sin e x > y, we have x = 2y and the required ratio is
x = 2. y
Solved by JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; GEOFFREY A. KANDALL, Hamden, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; CT, USA; SALEM MALIKIC, MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring eld, MO, USA; ANDREA MUNARO, student, University of Trento, Trento, Italy; RICARD PEIRO, IES \Abastos", Valen ia, Spain; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; DANIEL REISZ, Auxerre, Fran e; and KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India. There was 1 in orre t solution submitted.
M311. Proposed by Mihaly Ben ze, Brasov, Romania.
Let a, b, and c be positive real numbers, and let m ∈ at least one of the following equations has real roots: ax2 + bx + cm = bx2 + cx + am = cx2 + ax + bm =
0, 0, 0.
0,
1 4
. Show that
Solution submitted independently by Geo rey A. Kandall, Hamden, CT, USA; and the proposer. Assume that none of the three equations has real roots. Thus, ea h of the quadrati polynomials has a negative dis riminant; that is, b2 −4acm < 0
277 and
c2 − 4abm < 0 and a2 − 4bcm < 0. 4abm > c2 and 4bcm > a2 .
Equivalently,
4acm > b2
and
Sin e all quantities are positive, multiplying a ross these inequalities 1 , or m > 14 . This is yields 64a2 b2 c2 m3 > a2 b2 c2 . This implies that m3 > 64 a ontradi tion, sin e 0 < m < 14 . Therefore, at least one of the equations has real roots. Also solved by COURTIS G. CHRYSSOSTOMOS, Larissa, Gree e; RICHARD I. HESS, student, Sarajevo College, Sarajevo, Bosnia Ran ho Palos Verdes, CA, USA; SALEM MALIKIC, and Herzegovina; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring eld, MO, USA; ANDREA MUNARO, student, University of Trento, Trento, Italy; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; and JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o.
M312. Proposed by G.P. Henderson, Garden Hill, Campbell roft, ON.
John is negotiating the terms of a mortgage with his bank manager. They have agreed that the loan will be for L dollars and that the annual interest rate will be i. John says, \I will make payments of P dollars at the end of ea h year for the next n years. This is more than enough to pay the interest. The ex ess will redu e the prin ipal outstanding for the next year. At the end of n years, I will arrange a new mortgage for the remaining prin ipal." The manager responds, \I would like more frequent payments. I suggest payments of P/4 ea h quarter-year with interest rate i/4 applied to the previous quarter's balan e." John obje ts, \But then the e e tive annual interest rate will be greater than i!" The manager replies, \Yes, but the amount outstanding at time n will be less!" John nds this hard to believe. Is it true?
Solution by Ri hard I. Hess, Ran ho Palos Verdes, CA, USA, modi ed by the Mayhem Sta . We will write the interest rate as a de imal rather than as a per entage (for example, we would write 0.05 rather than 5%). Let Jn be John's amount outstanding after n years under his arrangement and let Bn be the amount outstanding under the banker's arrangement. Under John's arrangement, the amount outstanding after one year is J1 = L(1 + i) − P , sin e interest a
umulates at a rate of i over the year on the initial loan and then John pays P dollars at the end of the year. Similarly, for ea h k ≥ 1, the amount outstanding at the end of year k + 1 will be Jk+1 = Jk (1 + i) − P . Using these equations, we determine that Jn = L(1 + i)n − P
(1 + i)n − 1 i
.
[Ed.: See the note after the solution for a more detailed explanation.℄
(1)
278 Similarly, under the banker's arrangement, we determine that i 4n 4n 1 + − 1 P i 4 − = L 1+ i 4 4
Bn
4
i 4n = L 1+ −
P i
4
1+
i 4
4n
−1
,
whi h we an obtain by using the formula in (1) for Jn , but with 4n periods (quarter-years) instead of n periods (years). Now we have Jn − Bn
(1 + i)n − 1 n = L(1 + i) − P i P i 4n i 4n − L 1+ − 1+ −1 4
i
4
i 4n P i 4n = L (1 + i)n − 1 + − (1 + i)n − 1 + 4
=
1+
i 4
4n
i
− (1 + i)n
4
P −L i
.
(2)
In order for the loan to be paid o , we need P > iL (otherwise the payment ea h year will be less than the total interest). Therefore, Pi − L > 0. Also, (1 +
i i 4 ) = 1 + 4( ) + · · · , 4 4
the remaining terms being positive,
hen e, (1 + 4i )4n > (1 + i)n . (Sin e (1 + 4i )4 > 1 + i, the annual interest rate is greater than i.) Therefore, Jn − Bn > 0, as the two fa tors on the right side of (2) are positive. Thus, the bank manager's statement is true. [Ed.: Given that J1 = L(1 + i) − P and Jk+1 = Jk (1 + i) − P for k ≥ 1, we an prove the formula in (1) for Jn either by using Mathemati al Indu tion, or more informally by writing out J1 J2 J3
= = =
.. . Jn
=
L(1 + i) − P , J1 (1 + i) − P = L(1 + i)2 − (1 + i)P − P 3
2
,
J2 (1 + i) − P = L(1 + i) − (1 + i) P − (1 + i)P − P
,
L(1 + i)n − (1 + i)n−1 + (1 + i)n−2 + · · · + (1 + i) + 1 P
and using the formula for the sum of a geometri series. There was 1 in orre t solution submitted.
,
279
Problem of the Month Ian VanderBurgh
(2004 Hypatia Contest) green stones, 4 yellow stones, and 5 red stones are pla ed in a bag. Two stones of di erent olours are sele ted at random. These two stones are then removed and repla ed with one stone of the third olour. (Enough extra stones of ea h olour are kept to the side for this purpose.) This pro ess
ontinues until there is only one stone left in the bag, or all of the stones are the same olour. What is the olour of the stone or stones that remain at the end? Problem 1
3
This is the rst of three problems that we'll look at this month, all of whi h deal with piles of stones. However, the problems all require di erent approa hes. To me, this is one of the fas inating things about mathemati s { problems that appear to be very similar often require quite di erent approa hes in their solutions. From the wording of Problem 1, we an dedu e that the olour of the stone or stones left at the very end is always the same, regardless of the order in whi h the stones are removed. Thus, it shouldn't be too diÆ ult to determine that olour. (Just trying one parti ular order will do it!) The tri ky part will be to try to justify why the olour is always the same, regardless of what we do. Let's try a parti ular order to see what answer we get. We'll make a table to keep tra k of the olours: Colours Removed Colour Added # Green # Yellow # Red | | 3 4 5 Green, Yellow Red 2 3 6 Yellow, Red Green 3 2 5 Green, Red Yellow 2 3 4 Yellow, Red Green 3 2 3 Green, Red Yellow 2 3 2 Green, Red Yellow 1 4 1 Green, Red Yellow 0 5 0 So in this parti ular ordering of hoi es, we are left with only yellow stones. Thus, we must always be left with yellow stones. But why is the olour always yellow at the end? Try looking for a pattern among the numbers in the hart. Need a hint? Try looking at the parity of the numbers (that is, whether they are even or odd). Solution to Problem 1: For simpli ity, we'll all ea h \sele tion and repla ement" a turn. Also, we'll use G to represent the number of green stones, Y to represent the number of yellow stones, and R to represent the number of red stones.
280 At the beginning, G and R are ea h odd, and Y is even. Ea h turn auses ea h of G, Y , and R to either in rease by 1 or de rease by 1. This means that after ea h turn, the parity of ea h of G, Y , and R has
hanged. Therefore, after the rst turn, the parities of G, R, and Y are even, odd, and even, respe tively, regardless of whi h olours are in reased or de reased. Similarly, after the se ond turn, the respe tive parities will be odd, even, and odd. Continuing in this way, we an see that the parities will always be odd, even, and odd; or even, odd, and even. We an also see that the parities of G and R are always the same, and the parity of Y is di erent from the parity of these two. Suppose that we rea h the state where there are only stones of one
olour left. Thus, two of G, Y , and R equal 0. For this to be true, then we must be in the \even, odd, and even" ase (be ause 0 is even). Therefore, G and R are both even (and so equal to 0) and Y is odd. So the nal stone or stones in the bag are always yellow. Parity arguments rop up in unexpe ted pla es. This type of argument
an often be useful when looking at a problem in whi h there seem to be a lot of ases to onsider. See Problem 3 at the end of this olumn for a problem requiring a di erent, but related, argument. Here's a similar problem that needs another neat argument. Problem 2
Za h has three piles of stones, ontaining 5, 49, and 51 stones. He an
ombine any two piles together into one pile and he an also divide a pile
ontaining an even number of stones into two piles of equal size. Can he ever a hieve 105 piles, ea h with 1 stone?
Unless you have a ro k garden, this one might be a tad more diÆ ult to simulate than the rst one, given that you need 105 stones! Let's do some investigation. Sin e none of the piles ontains an even number of stones, the rules tell us that initially we have to ombine piles rather than separate piles. If we ombine the rst two piles, we will have two piles remaining with 5 + 49 = 54 and 51 stones. If we ombine the rst and last piles, we will have piles with 56 and 49 stones. If we ombine the last two piles, we will have piles with 5 and 100 stones. Hopefully, it is lear that we don't want to ombine again at this stage, otherwise we'd end up with one big pile of 105 stones in ea h ase; sin e 105 is odd, we'd be unable to go any further. This of ourse doesn't yet mean that we an't a hieve the desired result { it simply means that we've gone down the wrong path. Let's look at our rst option where we had piles with 54 and 51 stones. The only useful move here is to divide the pile ontaining 54 stones into two piles of 12 (54) = 27 stones. So we now have piles of 27, 27, and 51 stones. So we ould re ombine a pair of piles again. Can you see why this isn't going
281 to lead us anywhere? Let's look at the se ond option to see if anything be omes learer. Starting with 56 and 49, we ould split the 56 into two piles of 28, giving piles of 28, 28, and 49 stones. Here we have more hoi es { ombining an odd and an even pile, splitting one or both of the even piles, and so on. At this point, you ould spend quite a lot of time trying di erent approa hes to get down to 105 piles of one stone ea h without su
ess. This, of
ourse, would make you very frustrated with me and probably onvin e you that it seems as if the desired out ome is impossible. (But as every intrepid problem solver knows, just be ause you've tried it for 47 hours, doesn't ne essarily mean it won't work!) If you're stu k at this stage, read on! Solution to Problem 2: On his rst move, Za h must ombine two piles and obtain piles of 54 and 51 stones, or 56 and 49 stones, or 5 and 100 stones. In the rst ase, the number of stones in the pile have a ommon fa tor of 3, in the se ond ase, a ommon fa tor of 7, and in the third ase, a ommon fa tor of 5. Suppose that we are starting now from piles of 54 and 51 stones. The two possible moves are to ombine two piles or to divide one pile in half. If two integers are both multiples of 3 and are added together, their sum is also a multiple of 3; if an even integer is a multiple of 3 and is divided in half, ea h of the two resulting integers will also be a multiple of 3. (Can you see why?) Thus, no matter how many moves we make starting with piles of 54 and 51 stones, ea h of the piles remaining will always ontain a number of stones that is a multiple of 3. Thus, we an never rea h 105 piles of 1, as with this starting move, ea h pile will always ontain a number of stones that is a multiple of 3, and 1 is not a multiple of 3. The argument that we made above about ombining or dividing piles whose sizes are multiples of 3 also works for multiples of 5 or 7. (In fa t, it works for multiples of any odd number greater than 1.) Thus, starting with 56 and 49, ea h of the piles will always ontain a number of stones that is a multiple of 7, and starting with 5 and 100, ea h of the piles will always
ontain a number of stones that is a multiple of 5. In summary, no matter what rst move is made, it is impossible to
reate 105 piles of 1 stone ea h, as a ommon fa tor is always introdu ed at the rst step that an never be made to disappear. What do you think? I found this problem quite neat when I saw it for the rst time. This argument is in some ways similar and in some ways di erent to the argument needed for Problem 1. Here is a hallenge problem for you to onsider. We'll brie y look at the solution next month. Problem 3 (2004 Hypatia Contest) 3 green stones, 4 yellow stones, and 5 red stones are pla ed in a bag. This time, two stones of di erent olours are sele ted at random, removed and repla ed with two stones of the third olour. Show that it is impossible for all of the remaining stones to be the same olour, no matter how many times this pro ess is repeated.
282
THE OLYMPIAD CORNER No. 271 R.E. Woodrow
Wel ome ba k from the break. We start this number of the Corner with the 20 problems of the 19th Lithuanian Team Contest in Mathemati s. Thanks go to Felix Re io, Canadian Team Leader to the IMO in Mexi o for
olle ting them.
19th LITHUANIAN TEAM CONTEST IN MATHEMATICS O tober 2, 2004
1.
Twelve numbers { four 1's, four 5's, and four 6's { are written in some order around a ir le. Does there always exist a three-digit number omprised of three neighbouring numbers (its digits an be taken lo kwise or
ounter lo kwise) that is divisible by 3?
2. Solve the equation 2 cos(2πx) + cos(3πx) = 0. 3. Solve the equation 3x⌊x⌋ = 13, where ⌊x⌋ denotes the integer part of the
number x.
4. Let 0 ≤ a ≤ 1 and 0 ≤ b ≤ 1. Prove the inequality √
a 2b2 + 5
b 2 + √ ≤ √ 2a2 + 5 7
.
5. If a, b, and c are nonzero real numbers, what values an be taken by the expression
a2 − b2 a2 + b2
+
b2 − c2
b2 + c2
+
c2 − a2
c2 + a2
?
6. Determine all pairs of real numbers (x, y) su h that x6 y6
= =
y 4 + 18 , x4 + 18 .
7. Find all triples (m, n, r) of positive integers su h that 2001m + 4003n = 2002r .
283
8. Assume that m and n are positive integers. Prove that, if mn − 23 is divisible by 24, then m3 + n3 is divisible by 72. 9. Is it possible that, for some a, both expressions 1 − 2a a
2
are integers?
√
35
and a +
√ 35
10.
Prove that among any six onse utive positive integers it is always possible to nd a number that is relatively prime to the produ t of the other ve integers.
11. What is the greatest value that a produ t of positive integers an take if their sum is equal to 2004?
12. Positive integers a, b, c, u, v, and w satisfy the system of equations a+u b+v c+w
= 21 , = 31 , = 667 .
Can abc be equal to uvw?
13. Let u be the real root of the equation x3 − 3x2 + 5x − 17 = 0, and let
v
be the real root of the equation x3 − 3x2 + 5x + 11 = 0. Find u + v.
14.
Is it possible to write all the integers from 1 to 10 000 in the ells of a table so that the number in ea h ell is either smaller or greater than all the numbers written in all the ells having a ommon side with that
ell?
100 × 100
15. Does there exist a polynomial, P (x), with integer oeÆ ients su h that 4 9 2 for all x in the interval [ 10 , 10 ] the inequality |P (x) − 3 | < 10−10 is valid?
16. Does there exist a positive number a0 su h that all the members of the in nite sequen e a0 , a1 , a2 , . . . , de ned by the re urren e formula p an = an−1 + 1, n ≥ 1, are rational numbers? 17. Let a, b, and c be the sides of a triangle and let x, y, and numbers su h that x + y + z = 0. Prove that
z
be real
a2 yz + b2 zx + c2 xy ≤ 0 .
18.
Points M and N are on the sides AB and BC of the triangle ABC , AM BN respe tively. It is given that M = = 2 and ∠ACB = 2∠M N B . B NC Prove that ABC is an isos eles triangle.
284
19. The two diagonals of a trapezoid divide it into four triangles. The areas
of three of them are 1, 2, and 4 square units. What values an the area of the fourth triangle have?
20. The ratio of the lengths of the diagonals of a rhombus is a : b. Find the ratio of the area of the rhombus to the area of an ins ribed ir le.
Next we give the problems of the X Bosnian Mathemati al Olympiad. Thanks again go to Felix Re io for olle ting them for use in the Corner.
X BOSNIAN MATHEMATICAL OLYMPIAD First Day
1.
Let H be the ortho enter of an a ute-angled triangle ABC . Prove that the mid-points of AB and CH and the interse tion point of the interior bise tors of ∠CAH and ∠CBH are ollinear.
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
3.
Let n ≥ 2 be an integer, let x1 , x2 , . . . , xn be positive integers, and let Si = x1 + · · · + xi−1 + xi+1 + · · · + xn for i = 1, 2, . . . , n. Find the maximum of the fun tion f (x1 , x2 , . . . , xn ) =
gcd(x1 , S1 ) + gcd(x2 , S2 ) + · · · + gcd(xn , Sn ) x1 + x2 + · · · + xn
if (x1 , x2 , . . . , xn ) runs through the set of all n-tuples of distin t positive integers. Se ond Day
4.
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.
5.
A permutation 2 inequality aak ≤ k+1 identity.
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
285
6. Let a, b, and c be integers su h that a b c + + = 3. b c a
Prove that abc is a perfe t ube.
The I elandi Mathemati al Contest 2004{2005 (Final Round) is the last set of problems we give for this number. Thanks again to Felix Re io for
olle ting them for our use.
ICELANDIC MATHEMATICAL CONTEST 2004{2005 Final Round Mar h 12, 2005
1.
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, 2, . . . , 20}
2.
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 ?
3. 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.) 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. 1 5. 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.) 6. Let h be the altitude from A to in an a ute triangle ABC . Prove that (b + c)2 ≥ a2 + 4h2 ,
where a, b, and c are the lengths of the sides opposite A, B , and C respe tively.
286 Two solutions arrived to problems we dis ussed in the April and May numbers of the Corner. A solution to problem 1 of the Thai Mathemati al Olympiad [2007: 277; 2008: 155-156℄ was re eived from Miguel Amengual Covas, Cala Figuera, Mallor a, Spain. He also sent a solution to problem 2 of the 25th Albanian Mathemati al Olympiad [2007: 278; 2008: 220-221℄.
Next we present a \missed ase" in the featured solution to problem 2 of the XX Olimpiadi Italiane Della Matemati a, Cesenati o, given at [2007: 149-150; 2008: 83℄.
2. Let r and s be two parallel lines in the plane, and P and Q two points su h that P ∈ r and Q ∈ s. Consider ir les CP and CQ su h that CP is tangent to r at P , CQ is tangent to s at Q, and CP and CQ are tangent externally to ea h other at some point, say T . Find the lo us of T when (CP , CQ ) varies over all pairs of ir les with the given properties. Comment and solution by Jan Verster, Kwantlen University College, BC. The solution given is missing the ase where one of the ir les is outside the lines. To handle this ase, let the ir les be oriented as shown in the diagram, where O1 and O2 are the entres of ir les CP and CQ , respe tively. Let H and K be the points where the tangent line ommon to the two ir les rosses the lines r and s, respe tively. Let M be the inO1 terse tion of the lines KO2 and O1 H . Then △KT M ∼ = △KQM (as KO2 bise ts the angle at K ) r T and so we have M T = M Q H P and ∠KT M = ∠KQM . SimiO2 larly △M P O1 ∼ = △M T O1 , and thus, we have M T = M P and also ∠M P O1 = ∠M T O1 . By M subtra ting right angles we obtain ∠M P H = ∠M T K = ∠KQM , s hen e, P , M , and Q are ollinear K Q and M is the mid-point of P Q. Thus, T is on a ir le with diameter P Q. In summary, the lo us of points T is the line segment P Q, together with the part of the ir le with diameter P Q whi h is outside the two lines (and if one allows degenerate ir les, the four points where the ir le tou hes the lines).
............. .............. ...................... ........ ..... ..... ... .... .... ... .... .... ... . .... ... ... .. . . . . . .. ... .. .. .. .. .... ... .. ..... ...... . ... .................................. ................................. .... . . . . . . . . . . ......... . .... ... ....... .. ... ......... ..................... .. ...... .................... .. .. ....... .. . . ..... . . ... ........ .. ... ........ . .... . . . ....... ... ... ... ... .... . . . . .... ........ .... ... ... . . .... . ... . . . . ... ... ...... ..... ... .... . . . . . . . . . ...... . ........ ... ...... .......... ..... ............................. ... .................................................................................................................................................................................................................................................................... .. ....... .... ........ .. . . . .... .. .. .. ....... .. ....... .. ... ...... ... ...... . ...... .. . .. . ...... .. .. ........ ... . . . . ...... .. . .... ... ... . . . . . ... . . ... .... ... ...... . . .. . . . . . . . . . . . . .... .. ... .. ..... .... . .. . . . . . . . . . . . . . ...... .. .. . ...... . ... . . . . . . . . . . . . . . . ....... .. . . ...... . ... . . . . . . . . . . . . . . ............ .. .. .. . .. ..... .. ... .. ...................... .. .... ... ........ .... .... ... .... ... ... ........... . . ... . . . . . . . . . . . . . . . ... . .... . ... ... .... ......... .... ..... ... ... ........ .... .... ... ........ .... .... .... .... .... ......... ... ..... .... .... .... ......... . ... . . . . . . . . . . . . . . . . . . . . ....... .... .. .. ... ........ ........ ....... .... ... .......... ... ........ ........ ... .. .... ... .......... ........... .......................................................................................................................................................................................................................................................................................... . . . .. ....
287 Next we turn to solutions from our readers to problems given in the O tober 2007 number of the Corner. We begin with a solution to Problem 1 of the 2003 Kurs h ak Competition [2007: 336℄.
1.
Let EF be a diameter of the ir le Γ, and let e be the tangent line to Γ at E . Let A and B be any two points of e su h that E is an interior point of the segment AB , and AE · EB is a xed onstant. Let AF and BF meet Γ at A′ and B ′ , respe tively. Prove that all su h segments A′ B ′ pass through a ommon point.
Solved by Geo rey A. Kandall, Hamden, CT, USA; and Titu Zvonaru, Comane sti, Romania. We give Kandall's solution, modi ed by the editor. Let G be the point of interse tion of A′ B ′ and EF . Let F A′ = p, ′ F B = q , EA′ = r , EB ′ = s, AE = t, and EB = u. It is given that tu = k, where k is a xed onstant. F We have that ∠EAF and ∠AEF are right angles. The pair of angles ∠F EA′ and ∠A′ F E are q supplementary, as are the pair p ′ ′ of angles ∠F EA and ∠A EA. q Therefore, ∠A′ F E = ∠A′ EA, and B′ Gq hen e, △EF A′ ∼ △AEA′ . Cons sequently, pt = r · EF , and by A′ r e analogy, qu = s · EF . Therefore, u t pqk = rs · EF 2 . A E B ′ ′ ′ ′ Now ∠A F B and ∠A EB are supplementary angles, hen e, ................. ............
............................. ............ ............. ..................... . .. .. ....... ......... ...... ....... ... ... ..... .... ..... . . . ............ ...... .... ... .. ..... .... . .... . . ... .. . .. ... . . . . . ... .... ... .. .. . . . . .... .. .... .. .. . .. . .... .. .. .. . . .... . .... .. .... .. . .. . .... .... .... .. .. . . ... .... ... .. ... . . .... ..... .. ... ..... . . .... ... . .. .. . . . . . .. ..... . .. .. .... ..................... .. .. ... ..................................... ............... ...... ... .... ... . .... ... ............. ..................... ....... ..... .... . . ............................................... .... . . . . .. .. .... ... ................. ....... ...... .... ..... ....... . ... ......................... .... ...... .......... . ... . ....... ........ . .. . .... . . . . ........ .......... .. ........ .......... .... .. ............ ........ ... ...... .......... . ..................................................................................................................................................................................................................................................................
FG GE
=
[A′ F B ′ ] [A′ EB ′ ]
=
pq rs
=
EF 2 k
,
whi h is a xed ratio. Thus, all su h segments A′ B ′ pass through G. Next we move to solutions for problems of the Helleni Mathemati al Competitions 2004 given at [2007: 336-337℄.
1. Find the greatest possible value of the positive real number M su h that, for all x, y, z ∈ R,
x4 + y 4 + z 4 + xyz(x + y + z) ≥ M (xy + yz + zx)2 .
Solved by Mi hel Bataille, Rouen, Fran e; Andrea Munaro, student, University of Trento, Trento, Italy; Pavlos Maragoudakis, Pireas, Gree e; and Panos E. Tsaoussoglou, Athens, Gree e. We give Bataille's write-up.
288 The greatest possible value of M is 23 . First assume that (1) for all real x, y, and z . In parti ular, (1) must hold for x = y = z = 1, hen e, 6 ≥ M · 9 and M ≤ 23 . To omplete the proof, it remains to prove that (1) holds for all x, y, and z on e 23 is substituted for M , or equivalently that 3(x4 + y 4 + z 4 ) ≥ 2(x2 y 2 + y 2 z 2 + z 2 x2 ) + x2 yz + xy 2 z + xyz 2 . (2) Now, from the well-known inequality a2 + b2 + c2 ≥ ab+ bc + ca, we dedu e that 2(x4 + y 4 + z 4 ) ≥ 2(x2 y 2 + y 2 z 2 + z 2 x2 ) , (3) and also x4 + y 4 + z 4 ≥ (xy)2 + (yz)2 + (zx)2 ≥ xy · yz + yz · zx + zx · xy , that is, x4 + y 4 + z 4 ≥ x2 yz + xy 2 z + xyz 2 . (4) Inequality (2) now follows by addition of (3) and (4). x4 + y 4 + z 4 + xyz(x + y + z) ≥ M (xy + yz + zx)2
3. A ir le (O, r) and a point A outside the ir le are given. From A we draw a straight line ε, di erent from the line AO, whi h interse ts the ir le at B and Γ, with B between A and Γ. Next we draw the symmetri line of ε with respe t to the axis AO, whi h interse ts the ir le at E and ∆, with E between A and ∆. Prove that the diagonals of the quadrilateral BΓ∆E pass through a xed point; that is, they always interse t at the same point, independent of the position of the line ε. Solved by Geo rey A. Kandall, Hamden, CT, USA; Andrea Munaro, student, University of Trento, Trento, Italy; Pavlos Maragoudakis, Pireas, Gree e; and Titu Zvonaru, Comane sti, Romania. We give Maragoudakis' solution. Let AO ∩ B∆ = {P }. By ǫ Γ symmetry, ∠AΓP = ∠B∆A. Sin e 1 B ∠B∆A = 2 ∠BOE = ∠BOA, we have ∠AΓP = ∠BOA. Thus, triangles AΓP and AOB are similar, so A p AP AB
=
AΓ AO
, or
AP =
AB · AΓ AO
...... ............ ............ ............. .................. . . . . . . . . . . . .. ............. ...... ............ ... ............ .............. .... ... ............. .......................... . . . . . . . . . . . . . . ... ... ......... ....... .. . . . . . . . . . . . . . . . . . . . .... ... ..... ............ ............ ... .... ............. ............ .. .. .. ..... .............. ........................................................................................................................................................................................................................................................... ............... .... ....... . . ... ............. . ............. ......... . . ............. ....... ....... .... ............. ........ .... ............ .... ............. .... .................. .... ............ .... ............. ............. ............. ....... .................. ............ ............. ............. ...........
P
.
O
E
∆ Sin e AB · AΓ = AO2 − r2 , we have 2 2 AO − r , whi h is independent of the line ǫ. By symmetry, EΓ and AP = AO AO also interse t at the point P , and the proof is omplete.
289 We nish this number with solutions provided by our readers to some problems of the Vietnamese Mathemati al Olympiad 2004 [2007: 337-338℄.
1. Solve the system of equations = 2, = 30 , = 16 .
x3 + x(y − z)2 y 3 + y(z − x)2 z 3 + z(x − y)2
Solved by Arkady Alt, San Jose, CA, USA; Pavlos Maragoudakis, Pireas, Gree e; Panos E. Tsaoussoglou, Athens, Gree e; and Titu Zvonaru, Comane sti, Romania. We give the write-up of Maragoudakis. We note that x, y, and z are positive. We rewrite the equations as x2 + (y − z)2
=
y 2 + (z − x)2
=
z 2 + (x − y)2
=
2 , x 30 , y 16 . z
We subtra t the se ond equation from the rst to obtain 2 30 (x − y)(x + y) + (y − z) − (z − x) (y − z) + (z − x) = − x y
.
Isolating x − y in the above yields the rst equation below; the other two equations below follow similarly: x−y
=
y−z
=
y − 15x , xyz 15z − 8y xyz 8x − z
z−x =
xyz
,
.
The left sides of the above sum to zero, hen e, the right sides sum to zero and we obtain (y − 15x) + (15z − 8y) + (8x − z) = 0, or z = 12 (x + y). Substituting this into the very rst equation, we obtain x3 +
hen e, (x − y)2 =
8 − 4x3 . x
put in the form (x − y)2 =
x(x − y)2 4
= 2,
The third equation in the original system an be 16 − z 3 , z
hen e,
8 − 4x3 16 − z 3 = x z
.
290 2 It follows that 8(z −2x)+xz(z −4x2 ) = 0, whi h upon fa toring be omes (z − 2x) 8 + xz(z + 2x) = 0. Sin e x, y , and z are positive, we dedu e that z = 2x, and hen e, y = 2z − x = 3x. Now the very rst equation redu es to x3 + x(3x − 2x)2 = 2, or x = 1. Therefore, (x, y, z) = (1, 3, 2) is the unique solution to the system.
2. Solve the system of equations x3 + 3xy 2 x2 − 8xy + y 2
= −49 , = 8y − 17x .
Solved by Arkady Alt, San Jose, CA, USA, Pavlos Maragoudakis, Pireas, Gree e; Andrea Munaro, student, University of Trento, Trento, Italy; and Titu Zvonaru, Comane sti, Romania. We give the write-up of Zvonaru. 3
From the rst equation we have x 6= 0, and y2 = −x 3x− 49 . Beginning with the se ond equation we dedu e a su
ession of equations as follows: x2 − 8xy + y 2 x2 + 17x + y 2 3 x + 49 x2 + 17x − 3x 3 2x + 51x2 − 49 (x + 1)(2x2 + 49x − 49)
= =
8y − 17x ; 8y(x + 1) ;
=
8y(x + 1) ;
= =
24xy(x + 1) ; 24xy(x + 1) .
We now make two ases. . We have x = −1. Then y2 only two solutions. Case 1
= 16,
hen e, (x, y) = (−1, ±4) are the
. We have x 6= −1. Using the relations obtained so far we dedu e a su
ession of equations as follows:
Case 2
2x2 + 49x − 49 (2x2 + 49x − 49)2 (2x2 + 49x − 49)2 196x4 + 196x3 + 2205x2 + 4606x + 2401 49(x + 1)2 (4x2 − 4x + 49)
Sin e x 6= −1, we have 4x
2
− 4x + 49 = 0,
= 24xy ; = 576x2 y 2 ; −x3 − 49 = 576x2 3x = 0; = 0.
so that
1 ± 4i x= 2
√
3
.
;
291 We obtain y as follows: y
= = = = =
2x2 + 49x − 49
24x 2x2 − 2x − 49 + 51x 24x − 49 − 49 + 51x 2 17 8 17 8
− −
24x 49
16x 49
= √ 8(1 ± 4i 3)
√ 4±i 3 2
1 ± 4i 2
The solutions for (x, y) are (−1, ±4) and a onsistent hoi e of signs is taken in the last pair.
√
.
3 4±i , 2
√ 3
, where
3. Let ABC be a triangle in a plane. The internal angle bise tor of ∠ACB
uts the side AB at D. Consider an arbitrary ir le Γ1 passing through C and D so that the lines BC and CA are not its tangents. This ir le uts the lines BC and CA again at M and N , respe tively. (a) Prove that there exists a ir le tou hing the line DN at N .
Γ2
tou hing the line
DM
at
M
and
(b) The ir le Γ2 from part (a) uts the lines BC and CA again at P and Q, respe tively. Prove that the measures of the segments M P and N Q are onstant as Γ1 varies. Solved by Andrea Munaro, student, University of Trento, Trento, Italy; and Titu Zvonaru, Comane sti, Romania. We give Munaro's solution. (a) Let γ = 12 ∠ACB . Consider the perpendi ulars to DM at M and to DN at N . Clearly they interse t in a point R on Γ1 , where DR is a diameter of the ir le. Sin e ∠N RD = γ = ∠M RD, then N R = M R. Thus, Γ2 exists, namely Γ2 is the ir le with entre at R and passing through the points M and N . (b) Using the fa ts established in part (a), the Law of Sines, and the fa t that DM RC is y li with a right angle at M , we have PC sin ∠P RC
= =
CR MR = sin ∠CP R sin ∠RCM MR MR = . sin ∠RDM cos γ
292 Sin e ∠RM C = ∠RN C , then triangles M RP and N RQ are ongruent and it is easy to show that ∠P RC = ∠CDN . Then PC =
MR · sin ∠CDN cos γ
.
On the other hand, CM sin ∠CDM
=
MD sin γ
,
from whi h we obtain CM
= =
MD
· sin ∠CDM sin γ M R tan γ MR · sin ∠CDM = · sin ∠CDM . sin γ cos γ
Hen e, P C + CM
However,
MR
· (sin ∠CDN + sin ∠CDM ) ∠N DM ∠CDN − ∠CDM = · sin · cos cos γ 2 2 ∠CDN − ∠CDM = 2M R · cos . 2 =
cos γ 2M R
∠CDN − ∠CDM = ∠RDN − (90◦ − ∠CM D) − (∠RDM + 90◦ − ∠CM D) = 2∠CM D − 180◦ .
Then
P C + CM
= = =
2M R · sin ∠CM D MR 2CD · sin γ · MD 2CD · cos γ ,
whi h is a onstant. The same holds for N Q, sin e M P
= N Q.
4. Given an a ute triangle ABC ins ribed in a ir le Γ in a plane, let H be
its ortho entre. On the ar BC of Γ not ontaining A, take a point P distin t − − → −→ from B and C . Let D be the point su h that AD = P C . Let K be the ortho entre of triangle ACD, and let E and F be the orthogonal proje tions of K onto the lines BC and AB , respe tively. Prove that the line EF passes through the mid-point of HK .
293 Solution and note by Mi hel Bataille, Rouen, Fran e. − − →
−→
Sin e AD = P C , the quadrilateral ADCP is a parallelogram and so AP kCD and CP kAD . Sin e we have AK ⊥ CD , we also have AK ⊥ AP . It follows that the ir le with diameter KP passes through A. Similarly, this
ir le passes through C . Finally, this
ir le is Γ and in parti ular, K is on Γ. As a result, the line EF is just the Simson line of K relative to △ABC and it is a well-known result that this line bise ts the segment joining K to the ortho entre H . Note. The latter property of the Simson line an be found, for example, in R. Honsberger's book Episodes in 19th and 20th Century Eu lidean Geometry, MAA, 1995, pp. 43-6.
D ...... ........ .. ........ ..... ........ ... ........ . . . . . .... . . . ... ........ ........ . . . .... . . . . ..... ... . . . . . . . ..... .... . . . . . . . ... .............. ............... . .... .. .... ................. . ... ...... .. ............. ... . ............................................................... . ... . . ............... ... ................ . ... . . . . . . . . . . . . . . . . . . . . . ........... ... ............. ...... ... ... ..... .... . . .... . . . . . . . ....... . .... ...... ... . ...... ... .... ..... . . . . . . . . ..... . . . . . . ..... .... ... ... ... .. . ... . .... ... . . . ... . . . . . . . . . . . ...... .. .. .. . .. .... . .... .... ... . . . . . . . . . . . . . . . . .... .. .. .. .... ... . .. . ... . . . . . . . . . . . . . . ... .... .. .. .... ... . .. ... .... . . . . . . . . . . . . . . ... .... ... ...... . ... . .. . .. . . . . . . . . . . . . ... ... .... .. .... ......... .. .. .. . ... . . . . . . ... ... .... . . .. ... ... ....... .. . . . . . . . . ..... .. .... ... . . . ......... ..... ... .. .... ... ... ... .. ..... .. .... ... .... .... ... ... .. . ..... . ..... .. .... .. .. .... . . . . . ... ... .... .. .... ... .. ... .... ......... . . ... . . . .... ... .. ... .. ..... .. .. ... . ... . . . . . .... ... ... .. ... ... .. ....... .. . . . . .... . . ... .. .... ... ... .. . ... . . . . . ... . . . . ... .. .... .. .. .. .. .. ... ... . . . . . . . . . . .... .. .... .... ... . ... ... .. ... . . . . . ... .... ... ..... . .. ... . . . . . . . . . . . .. .. ....... ........ .... .. .. .. . .................................................................................................................................................................................................................................. ....... .. .......... . . .. ... . . ..... .. ... .... ... ........ .... ... ..... .. ........ .. ... ..... ........ . .. . . . . . . . . . . . ... .... .. ... .... ........ ... .. .... ... ........ . .... . . . . . . . . . . . .... . ... ... .. ........ .... ....... ... .. ........ ..... . .. ..... ........ ..... ....... .... ........ ... ....... ... ....... . ........ .... ... ............... . . . . . ... ......... .... .. ......... ......... ....................... ........... ................ .............................................
F
K
A
H
B
O r
E
C
P
5. Consider the sequen e of real numbers {xn }∞ n=1 de ned by x1 = 1 and xn+1 =
(2 + cos 2α)xn + cos2 α (2 − 2 cos 2α)xn + 2 − cos 2α
for every n = 1, 2, . . . , where α is a real parameter. For ea h n = 1, 2, . . . , n P 1 let yn = . Determine all values of α so that the sequen e k=1 2xk + 1 {yn }∞ n=1 has a nite limit. Find this limit in these ases. Solved by Mi hel Bataille, Rouen, Fran e; and Andrea Munaro, student, University of Trento, Trento, Italy. We give Munaro's solution. First we prove by indu tion that 1 3(3n−1 − 1) sin2 α + 1 = 2xn + 1 3n
.
(1)
For n = 1 it is learly true. On the other hand, we have xn+1 =
(1 + 2 cos2 α)xn + cos2 α 4(1 − cos 2α)xn + 3 − 2 cos2 α
=
cos2 α(2xn + 1) + xn 4xn + 3 − 2 cos2 α(2xn + 1)
.
294 Hen e, 2xn+1 + 1 =
3(2xn + 1) 4xn + 3 − 2 cos2 α(2xn + 1)
1
=
2xn+1 + 1
= = =
and
4xn + 3 − 2 cos2 α(2xn + 1) 3(2xn + 1) 2(2xn + 1) + 1 2 − cos2 α + 3 3(2xn + 1) 2 2 1 − cos2 α + + 3 3 3(2xn + 1) 2 1 sin2 α + . 3 3(2xn + 1)
Suppose that (1) holds, then 1 2xn+1 + 1
2
=
3
sin2 α+
3(3n−1 − 1) sin2 α + 1 3n+1
=
3(3n − 1) sin2 α + 1 3n+1
,
and the indu tion proof is omplete. Now n X
k=1
1 2xk + 1
= =
n X 3(3k−1 − 1) sin2 α + 1
k=1 n X
1
=
3k
+ sin2 α
n X 1−
1
3k 3k−1 k=1 1 1 3 1 1 − n + sin2 α n − 1− n 2 3 2 3 1 1 3 1− n − sin2 α + n sin2 α . 3 2 2 k=1
=
!
If sin2 α > 0, then yn
1 1 3 2 = 1− n − sin α ≥ −1 + n sin2 α , 3 2 2
and hen e, yn→ ∞ as n → ∞. In the other ase, sin2 α = 0, and then we have yn = 12 1 − 31n → 12 as n → ∞. Hen e, {yn }∞ n=1 has a nite limit if and only if α = kπ, k ∈ Z, for whi h the orresponding limit is 12 .
6. Find the least value and the greatest value of the expression P =
x4 + y 4 + z 4 (x + y + z)4
,
where x, y, and z are positive real numbers satisfying the ondition (x + y + z)3 = 32xyz .
295 Solution by Arkady Alt, San Jose, CA, USA. Sin e P is homogeneous, we an assume that x + y + z 1 subje t to onditions x + y + z = 1 and xyz = 32 we have P
= 1.
=
x4 + y 4 + z 4
=
1 − 4(xy + yz + zx) + 2(xy + yz + zx)2 + 4xyz
=
2(xy + yz + zx)2 − 4(xy + yz + zx) + 1 +
=
2(1 − xy − yz − zx)2 −
Sin e xy + yz + zx ≤
1 (x 3
min P
=
max P
=
+ y + z)2 =
1 3
7 8
Then
1 8
.
, then 1 − xy − yz − zx > 0, so
2 7 2 1 − max(xy + yz + zx) − , 8 2 7 2 1 − min(xy + yz + zx) − . 8
1 +z(1−z), sin e x+y = 1−z and xy = Moreover, xy +yz +zx = 32z 1 Setting h(z) = 32z + z(1 − z), we have that
min P
=
max P
=
1 . 32z
2 7 2 1 − max h(z) − , 8 2 7 2 1 − min h(z) − , 8
where z is onstrained by the solvability of the Viete System x+y
=
xy
=
1−z, 1 32z
,
in positive real numbers. That is, z ∈ (0, 1) and z must additionally satisfy 1 the inequality (1 − z)2 − 4 · 32z ≥ 0. We have √ √ 1 1 1 3− 5 3+ 5 (1 − z) − 4 · = z− z− z− , 32z 2 4 4 z √ √ and 0 < 3 −4 5 < 12 < 3 +4 5 , thus, for z ∈ (0, 1) the above expression is √ non-negative for z ∈ 3 −4 5 , 12 , and we must nd min h(z) and max h(z) 2
on this interval. We have
√ √ 32z 2 − 64z 3 − 1 2 1 1− 5 1+ 5 h (z) = = − 2 z− z− z− , 32z 2 z 4 8 8 √ hen e, z = 14 and z = 1 +8 5 are the only roots of h′ in the interval of 5 interest. By dire t al ulation we have h 14 = h 21 = 16 and also that ′
296 h
√ √ √ 3− 5 1+ 5 5 5−1 = h = , 4 8 32
so the minimum and maximum
values of h(z) in the interval of interest are Finally, the extreme values of P are min P max P
5 16
and
√ 5 5−1 , 32
respe tively.
2 √ √ 5 5−1 7 383 − 165 5 = 2 1− − = 32
8
5 2 7 = 2 1− − = 16
8
9 128
256
,
.
7. Find all triples of positive integers (x, y, z) satisfying the ondition (x + y)(1 + xy) = 2z .
Solved by Mi hel Bataille, Rouen, Fran e; Pavlos Maragoudakis, Pireas, Gree e; and Panos E. Tsaoussoglou, Athens, Gree e. We give the solution of Bataille. The solutions are the triples (1, 2j − 1, 2j), (2j − 1, 1, 2j), where j is a positive integer and (2k − 1, 2k + 1, 3k + 1), (2k + 1, 2k − 1, 3k + 1), where k is an integer with k ≥ 2. It is readily he ked that these triples are solutions. Conversely, suppose (x, y, z) is a solution. Then x + y = 2a and 1 + xy = 2b for some positive integers a and b. It follows that both x and y are odd. Note that (y, x, z) is also a solution, so we may suppose that x ≤ y , and we have that b ≥ a, sin e 1 + xy − (x + y) = (1 − x)(1 − y) ≥ 0. If x = 1, then (1 + y)2 = 2z so that z = 2j , 1 + y = 2j for some positive integer j and (x, y, z) = (1, 2j − 1, 2j). Now, suppose 3 ≤ x ≤ y, in whi h ase a ≥ 3 and b ≥ 4. Let x = 2m + 1 and y = 2n + 1. From x + y = 2a , 1 + xy = 2b , we dedu e that m and n are of opposite parity and mn (m + 1)(n + 1)
= 2a−2 (2b−a − 1) ,
= 2a−2 (2b−a + 1) .
Thus, either one or the other of the following holds: (m, n) = (2a−2 , 2b−a − 1) , (m + 1, n + 1) = (2b−a + 1, 2a−2 ) ; (m, n) = (2b−a − 1, 2a−2 ) , (m + 1, n + 1) = (2a−2 , 2b−a + 1) . In any ase, b − a = a − 2, so x + y = 2a and 1 + xy = 22a−2 . As a result, the quadrati polynomial X 2 −2a X +(22a−2 −1) has x, y as roots. We re all that x ≤ y and set k = a − 1 to obtain (x, y, z) = (2k − 1, 2k + 1, 3k + 1). This ompletes the proof. That ompletes the Corner for this month. Send me your ni e solutions and generalizations!
297
BOOK REVIEWS John Grant M Loughlin
Digital Di e By Paul J. Nahin, Prin eton University Press, 2008 ISBN 978-0-691-12698-2, hard over, 263+xi pages, US$27.95 Reviewed by Amar Sodhi, Sir Wilfred Grenfell College, Corner Brook, NL A olle tion of twenty-one problems in probability whi h the reader is invited to solve using Monte Carlo methods surely annot pique the interest of a person who does not relish omputer programming? Not if the book is Digital Di e! This book ontains more than enough material to interest anyone who enjoys the re reational side of mathemati s. The problems sele ted by Nahin are intrinsi ally interesting. Determining an ideal pla ement of poli e patrol ars on a stret h of highway, omputing the expe ted number of stops that an elevator makes on its way up in an oÆ e tower, and deriving an optimal dating strategy are three examples that spring to mind. There are also problems involving oin ipping, voting pro edures, and random walks as well as a ouple of paradoxes whi h are bound to please. All problems are well referen ed, enabling one to pursue a parti ular topi in more detail. Of parti ular interest is Parrondo's Paradox. Dis overing that one an onstru t a winning game by randomly swit hing between two losing games is mind boggling to say the least! Rather than use either Monte Carlo methods or pen and paper analysis to investigate this problem, I simply enjoyed myself by using 100-sided "digital-di e" to play Parrondo's game many times and found that I did indeed ome out ahead. The material is very well presented and an be followed by a reader with just a modi um of knowledge of basi probability. The lengthy introdu tion provided by the author arefully shows how both exa t methods and Monte Carlo methods an be used to solve a variety of problems. One su h problem, a famous one in geometri probability, involves omputing the han e that three points randomly hosen inside a re tangle are the verti es of an obtuse triangle. Nahin has a warm and witty style of writing whi h makes his book a pleasure to read. By s attering interesting ane dotes, histori al fa ts, and mathemati al insights throughout the book he does ater to an audien e who have no desire to write omputer ode to solve a probability problem. However, even the most pure mathemati ian might see the worthiness of using Monte Carlo methods for some of the problems. This is an enjoyable book for anyone who likes probability problems and/or solving problems using methods of simulation.
298
PROBLEMS Solutions to problems in this issue should arrive no later than 1 Mar h 2009. An asterisk (⋆) after a number indi ates that a problem was proposed without a solution. Ea h problem is given in English and Fren h, the oÆ ial languages of Canada. In issues 1, 3, 5, and 7, English will pre ede Fren h, and in issues 2, 4, 6, and 8, Fren h will pre ede English. In the solutions' se tion, the problem will be stated in the language of the primary featured solution. The editor thanks Jean-Mar Terrier of the University of Montreal for translations of the problems. 3351.
Proposed by Toshio Seimiya, Kawasaki, Japan. Let ABC be a triangle with AB > AC . Let P be a point on the line AB beyond A su h that AP + P C = AB . Let M be the mid-point of BC , and let Q be the point on the side AB su h that CQ ⊥ AM . Prove that BQ = 2AP .
3352.
Proposed by Toshio Seimiya, Kawasaki, Japan. Let ABC be a right-angled triangle with right angle at A. Let I be the in entre of △ABC , and let D and E be the interse tions of BI and CI with AC and AB , respe tively. Prove that BI · ID CI · IE
=
AB AC
.
3353.
Proposed by Mihaly Ben ze, Brasov, Romania. Let ABC be a triangle all of whose side lengths are positive integers.
(a) Determine all su h triangles where one angle has twi e the measure of a se ond angle. (b) Determine all su h triangles where two medians are perpendi ular.
3354.
Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain. Evaluate lim
n→∞
n X
k=1
ln
n2 + k2 n2
k3 /n4
.
299
3355.
Proposed by Todor Yalamov, So a University, So a, Bulgaria. For the triangle ABC let (x, y)ABC denote the line whi h interse ts the union of the segments AB and BC in X and the segment AC in Y su h that g AY x · AB + y · BC AX = = , AB + BC
AC
(x + y)(AB + BC)
g is either the length of the segment AX if X lies between A and where AX B , or the sum of the lengths of the segments AB and BX if X lies between B and C . Prove that the three lines (x, y)ABC , (x, y)BCA , and (x, y)CBA interse t in a point dividing the segment N I in the ratio x : y, where N is the Nagel point and I the in entre of △ABC .
3356.
Proposed by Cristinel Morti i, Valahia University of Targoviste, Romania. Let f : [0, ∞) → R be integrable on [0, 1] and have period 1 (that is, f (x + 1) = f (x) for all x ∈ [0, ∞)). If {xn }∞ n=1 is any stri tly in reasing, unbounded sequen e with x0 = 0 for whi h (xn+1 − xn ) → 0, denote r(n) = max{k ∈ N | xk ≤ n} .
(a) Prove that lim
n→∞
r(n) 1 X
n
k=1
(xk − xk−1 )f (xk ) =
(b) Prove that 1
lim
n→∞
ln n
n X f (ln k)
k=1
k
=
Z
1
0
Z
1
0
f (x) dx .
f (x) dx .
3357. Z
Proposed by Ovidiu Furdui, University of Toledo, Toledo, OH, USA. Let a be a real number su h that −1 < a ≤ 1. Prove that
1
0
x+a x2 + 2ax + 1
ln(1 − x) dx =
1 2 θ ln 2 sin 2 2
+
θ2 θπ π2 − + 8 4 24
,
where θ is the unique solution in (0, π] of the equation cos θ = −a.
3358.
Proposed by Toshio Seimiya, Kawasaki, Japan. The interior bise tor of ∠BAC of triangle ABC meets BC at D. Suppose that 1 1 2 + = . 2 2 2 Prove that ∠BAC
BD
◦
= 90
.
CD
AD
300
3359. Proposed by Ray Killgrove, Vista, CA, USA and David Koster, University of Wis onsin, La Crosse, WI, USA. 2 Consider the sequen e {an }∞ n=1 de ned by an = n + n + 1. Find ∞ a subsequen e {bn }n=1 su h that b1 = a1 , b2 = a2 , b3 > a3 , every pair of terms from this subsequen e are relatively prime, and there are primes whi h divide no term of the subsequen e. 3360.
Proposed by Mi hel Bataille, Rouen, Fran e. For omplex numbers a, b, and c, not all zero, let N (a, b, c) denote the number of solutions (z1 , z2 , z3 ) ∈ C3 to the system: z1 z3 z1 z2 + z2 z3 z12 + z22 + z32
= a, = b, = c.
For whi h a, b, and c does N (a, b, c) attain its minimal value?
3361.
Proposed by Mi hel Bataille, Rouen, Fran e. Let the in ir le of triangle ABC meet the sides CA and AB at E and F , respe tively. For whi h points P of the line segment EF do the areas of △EBC , △P BC , and △F BC form an arithmeti progression?
3362.
Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain. Prove that Z
1
0
r 3
ln(1 + x) dx x
Z
1 0
r 3
ln2 (1 + x) π2 dx < x2 12
.
.................................................................
3351.
Propose par Toshio Seimiya, Kawasaki, Japon. Soit ABC un triangle ave AB > AC . Soit P un point sur la droite AB , au-dela de A et tel que AP + P C = AB . Soit M le point milieu de BC , et soit Q le point sur le ot ^ e AB tel que CQ ⊥ AM . Montrer que BQ = 2AP .
3352.
Propose par Toshio Seimiya, Kawasaki, Japon. Soit ABC un triangle re tangle ave son angle droit en A. Soit I le
entre du er le ins rit du triangle ABC , et soit D et E les interse tions respe tives de BI et CI ave AC et AB . Montrer que BI · ID AB = CI · IE AC
.
301
3353.
Propose par Mihaly Ben ze, Brasov, Roumanie. Soit ABC un triangle dont la longueur de ha un de ses ot ^ es est un entier positif. (a) Trouver tous les triangles qui, en plus, ont un angle de mesure double d'un autre angle. (b) Trouver tous les triangles qui, en plus, ont deux medianes perpendi ulaires.
3354.
Propose par Jose Luis Daz-Barrero, Universite Polyte hnique de Catalogne, Bar elone, Espagne. Evaluer lim
n→∞
n X
ln
k=1
3355.
n2 + k2 n2
k3 /n4
.
Propose par Todor Yalamov, Universite de So a, So a, Bulgarie. par (x, y)ABC la droite qui oupe la Dans un triangle ABC designons reunion des segments AB et BC en X et le segment AC en Y de sorte que g AX
AB + BC
=
AY AC
=
x · AB + y · BC
(x + y)(AB + BC)
,
g est soit la longueur du segment AX si X est entre A et B , ou la ou AX somme des longueurs des segments AB et BX si X est entre B et C . Montrer que les trois droites (x, y)ABC , (x, y)BCA et (x, y)CBA se oupent en un point qui divise le segment N I dans le rapport x : y, ou N est le point de Nagel et I le entre du er le ins rit du triangle ABC .
3356.
Propose par Cristinel Morti i, Valahia Universite de Targoviste, Roumanie. Soit f : [0, ∞) → R une fon tion integrable sur [0, 1] et de periode 1 ( -a-d., f (x+1) = f (x) pour tout x ∈ [0, ∞)). Pour toute suite non bornee, stri tement roissante {xn }∞ n=1 , ave x0 = 0 et telle que (xn+1 − xn ) → 0, on pose r(n) = max{k ∈ N | xk ≤ n} . (a) Montrer que lim
n→∞
(b) Montrer que
r(n) 1 X
n
k=1
lim
n→∞
(xk − xk−1 )f (xk ) =
1 ln n
n X f (ln k)
k=1
k
=
Z
1
0
Z
1
0
f (x) dx .
f (x) dx .
302
3357. Z
Propose par Ovidiu Furdui, Universite de Toledo, Toledo, OH, E-U. Soit a un nombre reel tel que −1 < a ≤ 1. Montrer que
1
0
x+a x2
+ 2ax + 1
1 2 θ ln 2 sin 2 2
ln(1 − x) dx =
+
θ2 θπ π2 − + 8 4 24
,
ou θ est la seule solution de l'equation cos θ = −a dans (0, π].
3358.
Propose par Toshio Seimiya, Kawasaki, Japon. de l'angle BAC oupe Dans le triangle ABC , la bisse tri e interieure BC en D . On suppose que 1 2 1 + = 2 2 BD CD AD 2
Montrer que l'angle BAC
.
= 90◦ .
3359.
et David Koster, UniverPropose par Ray Killgrove, Vista, CA, E-U site de Wis onsin, La Crosse, WI, E-U eral an = n2 + n + 1. Trouver une Soit la suite {an }∞ n=1 de terme gen ∞ sous-suite {bn }n=1 telle que b1 = a1 , b2 = a2 , b3 > a3 , haque paire de termes de ette sous-suite soient relativement premiers, et qu'il existe des nombres premiers qui ne divisent au un terme de ette sous-suite.
3360.
Propose par Mi hel Bataille, Rouen, Fran e. Pour des nombres omplexes (a, b, c), non tous nuls, notons N (a, b, c) le nombre de solutions (z1 , z2 , z3 ) ∈ C3 du systeme : z1 z3 = a , z1 z2 + z2 z3 = b , z12 + z22 + z32 = c . Trouver les nombres a, b et c pour lesquels N (a, b, c) atteint sa valeur minimale.
3361.
Propose par Mi hel Bataille, Rouen, Fran e. Le er le ins rit du triangle ABC est tangent aux ot ^ es CA et AB en E et F respe tivement. Pour quels points P du segment de droite EF les aires des triangles EBC , P BC et F BC forment-elles un progression arithmetique ?
3362.
Propose par Jose Luis Daz-Barrero, Universite Polyte hnique de Catalogne, Bar elone, Espagne. Montrer que Z
1
0
r 3
ln(1 + x) dx x
Z
1 0
r 3
ln2 (1 + x) π2 dx < x2 12
.
303
SOLUTIONS No problem is ever permanently losed. The editor is always pleased to onsider for publi ation new solutions or new insights on past problems. We have re eived late solutions to problem 3221 from the following solvers: Ateneo Problem Solving Group; John G. Heuver; Thanos Magkos; and Pavlos Maragoudakis. Our apologies to Chip Curtis, Missouri Southern State University, Joplin, MO, USA, for a lost bat h of orre t solutions to problems 3239, 3241, and 3243{3248.
3251. [2006 : 297,299℄ Proposed by Mi hel Bataille, Rouen, Fran e. Let u1 , u2 , and u3 be any real numbers. Prove that
3 1 X cos2 (ui − ui+1 ) + cos2 (ui + ui+1 ) 6 i=1
≥ (cos u1 cos u2 cos u3 )2 + (sin u1 sin u2 sin u3 )2 ,
where the subs ripts in the summation are taken modulo 3.
Solution by Daniel Tsai, student, Taipei Ameri an S hool, Taipei, Taiwan. Sin e cos(x + y) = cos x cos y − sin x sin y for real numbers x and y, it follows that for ea h i we have cos2 (ui − ui+1 ) + cos2 (ui + ui+1 ) = 2(cos2 ui cos2 ui+1 + sin2 ui sin2 ui+1 ) ,
where subs ripts are taken modulo 3. Sin e 0 ≤ sin2 ui ≤ 1 for ea h i, we have
0 ≤ cos2 ui ≤ 1
and also
3 1X cos2 (ui − ui+1 ) + cos2 (ui + ui+1 ) 6 i=1
=
= ≥
3 1X
6 1 3 1 3
i=1 3 X i=1 3 X
2(cos2 ui cos2 ui+1 + sin2 ui sin2 ui+1 ) (cos2 ui cos2 ui+1 + sin2 ui sin2 ui+1 ) (cos2 ui cos2 ui+1 cos2 ui+2 + sin2 ui sin2 ui+1 sin2 ui+2 )
i=1
1 = · 3(cos2 u1 cos2 u2 cos2 u3 + sin2 u1 cos2 u2 sin2 u3 ) 3 = (cos u1 cos u2 cos u3 )2 + (sin u1 cos u2 sin u3 )2 .
304 University of Sarajevo, Sarajevo, Bosnia Also solved by SEFKET ARSLANAGIC, and Herzegovina; IAN JUNE L. GARCES and WINFER C. TABARES, Ateneo de Manila University, Quezon City, The Philippines; ATENEO PROBLEM SOLVING GROUP, Ateneo de Manila University, Quezon City, The Philippines; DIONNE BAILEY, ELSIE CAMPBELL, and CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; ROY BARBARA, SCAR CIAURRI, EMILIO Lebanese University, Fanar, Lebanon; MANUEL BENITO, O 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; APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; TOM LEONG, Brooklyn, NY, USA; SALEM MALIKIC, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; PAVLOS MARAGOUDAKIS, Pireas, Gree e; ANDREA MUNARO, student, University of Trento, Trento, Italy; JOSE H. NIETO, Universidad del Zulia, Mara aibo, Venezuela; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. Ros proved two generalizations of this inequality. For n > 2 and u1 , u2 , . . . , un real numbers,
and
n 1 X [cos2 (ui − ui+1 ) + cos2 (ui + ui+1 )] ≥ 2n i=1
1 n(n − 1)
X
1≤i<j ≤n
n Y
n Y
cos ui
!2
i=1
where subs ripts in the summation are taken modulo n. π 2
+
i=1
[cos2 (ui − uj ) + cos2 (ui + uj )] ≥
u1 ≡ · · · ≡ un ≡ 0 (mod π), or u1 ≡ · · · ≡ un ≡
!2
cos ui
n Y
sin ui
i=1
+
n Y
!2
sin ui
i=1
!2
,
Equality holds if and only if
(mod π).
Benito, Ciaurri, Fernandez, and Ron al proved another generalization. For n and ui
(1 ≤ i ≤ n) as above,
n 1 X [cos2 (ui − ui+1 ) + cos2 (ui + ui+1 )] ≥ 2n i=1
n Y
i=1
where subs ripts in the summation are taken modulo n.
cos ui
!4/n
+
n Y
sin ui
i=1
!4/n
3252. [2007 : 297,300℄ Proposed by Mi hel Bataille, Rouen, Fran e.
Let S be a set of omplex 2 × 2 matri es su h that, for all A, B , C we have ABCAB = C . (a) Show that (ABC)n = matri es A, B , C ∈ S .
An B n C n
for all positive integers
n
,
∈ S,
and all
(b) Give an example of su h a set S ontaining at least three matri es with two of them non- ommuting. Solution by Mi hael Parmenter, Memorial University of Newfoundland, St. John's, NL. (a) The ondition ABCAB = C implies ABCABC = C 2 , whi h, up to symmetry, is the same as CABCAB = B 2 . Also, ABCAB = C implies
305 CABCAB = C 2 , so that B 2 = C 2 for any two matri es B ,C ∈ S . Now, B 2 = C 2 implies B 2 C = C 3 and CB 2 = C 3 , so that B 2 ommutes with C for any two matri es B ,C ∈ S . Applying the ondition ABCAB = C for A = B = C , we obtain A5 = A for every A ∈ S . Let k be any positive integer. Using the above observations, we have k (ABC)2k = (ABC)2 = (C 2 )k = C 2k
and
A2k B 2k C 2k
= (A2 )k (B 2 )k (C 2 )k = (C 2 )3k = C 6k = C k C 5k = C k (C 5 )k = C k C k = C 2k .
Thus, (ABC)2k = A2k B 2k C 2k . Using this result and the established ommutativity, we obtain (ABC)2k+1
(ABC)2k ABC = A2k B 2k C 2k ABC A2k AB 2k BC 2k C = A2k+1 B 2k+1 C 2k+1 .
= =
Therefore, (ABC)n = An B n C n for all positive integers n. (b) We rst give a general approa h to nding su h an example. Let A be any matrix with A4 = I and A2 6= I , where I is the identity matrix. Let B = A3 and let C be any matrix su h that C 2 = A2 and AC 6= CA. Then AB = BA = A4 = I . It is easy to show that the ondition XY ZXY = Z holds for any permutation (X, Y, Z) of the matri es A, B , and C : ICI = C , ICI = C , ACIC = AC 2 = A3 = B , CICA = C 2 A = A3 = B ,
ABCAB BACBA ACBAC CABCA
= = = =
BCABC CBACB
= BCIC = BC 2 = A3 A2 = A , = CICB = C 2 B = A2 A3 = A .
One parti ular example is A=
i 0
0 −i
,
3
B=A =
−i 0 0 i
, and
C=
where x 6= 0. Clearly, A and C do not ommute, be ause AC − CA =
0 0
2xi 0
i x 0 −i
,
.
Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; JOSE H. NIETO, Universidad del Zulia, Mara aibo, Venezuela; PETER Y. WOO, Biola University, La Mirada, CA, USA (part (a) only); and the proposer. Both Barbara and Parmenter noted that the result in part (a) is true in any semigroup.
306
3253. [2007 : 297, 300℄ Proposed by Mihaly Ben ze, Brasov, Romania. Prove that
loge (eπ − 1) loge (eπ + 1) + logπ (π e − 1) logπ (π e + 1) < e2 + π 2 .
I. Essentially similar solutions by Sefket Arslanagi , University of Sarajevo, Sarajevo, Bosnia and Herzegovina; Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; Tom Leong, Brooklyn, NY, USA; Andrea Munaro, student, University of Trento, Trento, Italy; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; and the proposer. By the AM-GM Inequality, we have π
π
loge (e − 1) loge (e + 1) ≤ = <
loge (eπ − 1) + loge (eπ + 1)
2 2 2 1 loge (eπ − 1)(eπ + 1) = loge (e2π − 1) 4 4 1 2 loge (e2π ) = π2 . 4 1
2
(1)
Similarly, we have
logπ (π e − 1) logπ (π e + 1) < e2 .
(2)
The result follows by adding (1) and (2). s ar Ciaurri, and Emilio Fernandez, Logrono, II. Solution by Manuel Benito, O ~ Spain. For a > 1 and b > 0, we have loga (ab − 1) loga (ab + 1) = b + loga (1 − a−b ) b + loga (1 + a−b )
= b2 + b loga (1 − a−2b ) + loga (1 − a−b ) loga (1 + a−b ) < b2 ,
sin e b and 1 − a−2b are positive, 1 − a−b < 1, and 1 + a−b > 1. From the inequality above, we on lude that n X
i=1
logai (abi i − 1) logai (abi i + 1) ≤
n X
b2i ,
i=1
where ai > 1 and bi > 0 for i = 1, 2, . . . , n. The proposed inequality is the spe ial ase when n = 2, a1 and a2 = b1 = π.
= b2 = e
Also solved by DIONNE BAILEY, ELSIE CAMPBELL, CHARLES DIMINNIE, KARL HAVLAK, PAULA KOCA, and ANDREW SIEFKER, Angelo State University, San Angelo, TX, USA; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e;
307 RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium, student, Sarajevo Innsbru k, Austria; KEE-WAI LAU, Hong Kong, China; SALEM MALIKIC, College, Sarajevo, Bosnia and Herzegovina; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring eld, MO, USA; JOSE H. NIETO, Universidad del Zulia, Mara aibo, Venezuela; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; JOEL SCHLOSBERG, Bayside, NY, USA; ASHLEY TANGEMAN and PETRUS MARTINS, students, California State University, Fresno, CA, USA; and PETER Y. WOO, Biola University, La Mirada, CA, USA.
3254. [2007 : 298, 300℄
Proposed by G. Tsintsifas, Thessaloniki, Gree e. Let C be a onvex gure in the plane. A diametri al hord AB of C − parallel to the dire tion ve tor → v is a hord of C of maximal length parallel → − to the dire tion ve tor v . Prove that if every diametri al hord of C bise ts the area en losed by C , then C must be entro-symmetri . Solution by P.C. Hammer and T. Je erson Smith from 1964, adapted by the editor. What follows is a simpli ed version of the proof of Theorem 2.4 in [3℄. In that work the authors prove that any onvex planar body is entrally symmetri provided that ea h line bise ting the area is a diametral line. (Hammer and Smith use the words diametral and entrally symmetri rather than the equivalent but less ommon diametri al and entro-symmetri .) Be ause in every dire tion there is exa tly one line that bise ts the given area, our assumption that every diametral hord is area bise ting implies that there is a unique diametral hord in every dire tion. The Hammer and Smith result is therefore stronger sin e it applies also to entrally symmetri regions whose boundary ontains line segments (for whi h points of parallel sides are joined by parallel diametral hords, one of whi h bise ts the area). In order to avoid a page of te hni al arguments, we will further restri t our result by assuming that the boundary of the onvex region, denoted by C , is a di erentiable urve. For su h boundaries our assumption that an areabise ting hord is diametral implies that the tangent lines at its ends are parallel. We treat the plane as a ve tor spa e and let u(θ) = (cos θ, sin θ) be a unit ve tor fun tion; then u′ (θ) = (− sin θ, cos θ) = u(θ + 12 π). Let m(θ) be the unique diametral line parallel to the dire tion of u(θ). Then there exists a unique real number p(θ) su h that m(θ) is representable as {x : x · u′ (θ) = p(θ)} = {x : x = p(θ)u′ (θ) + tu(θ), t ∈ R} .
(1)
x(θ) = p(θ)u′ (θ) + f (θ)u(θ) ,
(2)
Note that p(θ)u′ (θ) is the foot of the perpendi ular from the origin to m(θ). Be ause we assume that C is di erentiable, it follows easily that so is p(θ). We now represent C by a fun tion x(θ) in the following way. Ea h line m(θ) interse ts the boundary in two points, one of whi h is given by
308 where f (θ) is assigned a value that makes it ontinuous. With that assignment then, sin e p(θ + π) = −p(θ) while u(θ + π) = −u(θ), the other point of m(θ) = m(θ + π) on the boundary would be x(θ + π) = p(θ)u′ (θ) − f (θ + π)u(θ) .
(3)
f (θ) + f (θ + π) = |x(θ) − x(θ + π)|
(4)
|x′ (θ + π)| · x′ (θ) = −|x′ (θ)| · x′ (θ + π) .
(5)
Subtra ting (3) from (2) gives us x(θ) − x(θ + π) = f (θ) + f (θ + π) u(θ). Therefore, we hoose the fun tion f (θ) so that f (θ) + f (θ + π) > 0, when e
for this hoi e. With this notation the ondition that the tangents at the ends of a diametral hord are parallel be omes We must prove that x′ (θ + π) = −x′ (θ). This will require two simple observations about neighbouring diametral hords. First, in the limit, they interse t at s(θ) = p(θ)u′ (θ) − p′ (θ)u(θ) . (6) More pre isely, the diametral line m(θ) (whose points, a
ording to (1) with x = (x1 , x2 ), satis es the equation x1 sin θ − x2 cos θ + p(θ) = 0) interse ts the diametral line m(θ + h) in the point
p(θ) cos(θ + h) − p(θ + h) cos θ p(θ) sin(θ + h) − p(θ + h) sin θ , sin h sin h
.
The limit of these interse tion points on m(θ) as h → 0 is s(θ) in (6). Alternatively, one an avoid su h al ulations by referring to the diagram below, ....... ............ ............ ............ ............ ............ ....... S ............ ...... ................. . .............................................................................................. ............. . . . . . . . . .... ............ ...................................... . ...... . . .................... . ....... . . . .... . .................. R ....... ..... . . . . ............ . .... . ....... ..... . . ............. ....... . ... m(θ + dθ) ... .......dp ..... ....................................... . . . . . ............ ....... ... ... .... .. ... . . ............ ... ... . . . ....... ... .. ............ ... ... ... . . . . . . . . . ....... ............ ... ... .. ... ... ..... ... ... ... ... ... ... .. . . . . .. . . . . ............ . . . . . . . . . . . . . . . ... ... .. ...... . .. . .. ............ .......... ....... ............ ... ... .. ... .... .... . ........... ....... ... .. ... ... .. . . . . . . . . ......................... ........ . ... . Q . .. . . . . . . . .................................. .. . ... ... ... .. . . . ........................... P . . . ... ........ . ..... . . . ... p dθ ............................... .. ..... .... .. .... .. .. .. .... .. .... . . ... .... .... ................. m(θ) .. . . .. ... . . . . . . ......... . . .. . .. .. .. ........ .. .. ..... .... .. ... .. .. ....... .. ... .. ... .... .. .. ........ .... .. .. .... .. ... .. ........ .. .... . . . ... ....... . ... . .... .. ... .. . .. . .. . .. . . . . . .. . . .. . .. . . . . ..... ... . .. . . . . . ... . ... . . ... ... . .. .. . .. . . . . . . .. .. ..... .. .. .... .. p(θ) ...... ... .... .... .. ... .... ... .. .. ... .. dθ...... .. .. . . . ... . ..... .... .... .. .... .. ... .... .... ... .. .... .... .... .... ... .... .... ... . . . . . . . . . . ..... . . . .... ..... .... ... ...... ..... ...... .... ... .... ...... ....... ....... ........ .... ... ..... ....... . . . ......... . . . . . . . . . . . ........... .... .......... .............. ........... .......................................................................
r
r
r
r
r
r
O
where m(θ + h) = SR interse ts m(θ) = SP at S , while R and P are the feet of the perpendi ulars from the origin O. The laim to be veri ed
309 is that the distan e P S from the foot of the perpendi ular P to the point of interse tion S approa hes p′ (θ): Be ause of the right angles at R and P , these points lie on the ir le whose diameter is OS . The ir le with entre O and radius p = p(θ) = OP (whi h is tangent to SP at P ) interse ts OR at Q (as in the gure). Thus, for small values of h = dθ we see that P Q is approximately p dθ, and QR = dp. As h → 0, RP approa hes the tangent SP to ir le OP S at P , so that ∠RP S → ∠P OR. Consequently, pdp → , dθ p so that the limit satis es SP =
dp = p′ (θ) , dθ
as laimed. The se ond required observation is that as h → 0, the point where m(θ + h) interse ts m(θ) approa hes the midpoint 12 x(θ) + x(θ + π) of the hord of C along m(θ). This is an immediate onsequen e of the fa t that two area bise ting hords, namely m(θ + h) and m(θ), divide the region bounded by C into four se tors su h that opposite se tors have the same area. [Ed.: The interse ting lines are be oming the sides of isos eles triangles with equal areas and vertex angles: If the area bise ting hords AB and A′ B ′ interse ts at X , then ∠A′ XA = ∠B ′ XB while XA′ → XA as A′ → A and XB ′ → XB as B ′ → B .℄ This observation and equations (2), (3), and (5) yield s(θ) =
1 2
x(θ) + x(θ + π)
from whi h we on lude that
= p(θ)u′ (θ) +
1
f (θ) − f (θ + π) u(θ)
2 = p(θ)u′ (θ) − p′ (θ)u(θ) , −p′ (θ) = 12 f (θ) − f (θ + π) , or
p′ (θ) + f (θ) = f (θ + π) − p′ (θ) .
(7)
By taking the derivatives of the expressions in equations (2) and (3), we note that the oeÆ ient of u′ (θ) in x′ (θ) is p′ (θ) + f (θ), while in x′ (θ + π) it is − f (θ + π) − p′ (θ) . Setting the oeÆ ients of u′ (θ) equal in equation (5) therefore gives us p′ (θ) + f (θ) |x′ (θ + π)| =
f (θ + π) − p′ (θ) |x′ (θ)| .
Be ause p′ (θ) + f (θ) is stri tly positive ( ompare formula (2) with (6) and re all that the midpoint of ea h hord is interior to C ), this last equation together with the equality in (7) implies that |x′ (θ + π)| = |x′ (θ)| . ′ ′ Equation (5) now says that x (θ) = −x (θ + π). On integration we nd that 1 x(θ) + x(θ + π) is a onstant, namely the midpoint of every diametral 2
hord, whi h ompletes the proof.
310 Also solved by PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. The editors are grateful for the help and the referen es provided by Paul Goodey, University of Oklahoma, Norman, OK; and Horst Martini University of Te hnology, Chemnitz, Germany. By oin iden e, ea h has re ently published a paper that dis usses the laim in our problem, see [1℄ and [2℄ where related theorems are proved and further referen es provided. In [3℄ the authors prove that C will also be entrally symmetri if ea h diametral line bise ts the ir umferen e. They tra e their theorems ba k to a 1921 work of Konrad Zindler [4℄. Referen es
[1℄ G. Averkov, E. Makai, Jr., and H. Martini, Chara terizations of entral symmetry for onvex bodies in Minkowski spa es. Studia S i. Math. Hungar. To appear. [2℄ Paul Goodey, Area and perimeter bise tors of planar onvex sets. Integral Geometry and Convexity, Pro eedings of the International Conferen e, Wuhan, China, World S ienti , New Jersey (2006) 29{35. [3℄ P.C. Hammer and T. Je erson Smith, Conditions equivalent to entral symmetry of onvex
urves. Pro . Camb. Phil. So . 60 (1964) 779{785. [4℄ Konrad Zindler, Uber konvexe Gebilde. Monatsh. Math.-Phys. 31 (1921) 25{56.
3255. [2007 : 298, 300℄
Proposed by G. Tsintsifas, Thessaloniki, Gree e. Prove that, as the points A, B , C move over the surfa e of an ellipsoid
entred at O while the lines OA, OB , OC stay mutually perpendi ular, the plane ABC remains tangent to a xed sphere. Solution by Apostolis K. Demis, Varvakeio High S hool, Athens, Gree e. 2
2
2
Let xa2 + yb2 + zc2 = 1 be the equation of the given ellipsoid in an orthonormal oordinate system; that is, the entre of the ellipsoid is O(0, 0, 0) and its axes are along the x-, y-, and z -axes. If the ve tors from the origin to −→ −→ −→ −→ the moving points are OA = |OA|(a1 , a2 , a3 ), OB = |OB|(b1 , b2 , b3 ), and −→ −→ OC = |OC|(c1 , c2 , c3 ), then the ondition that the lines OA, OB , and OC stay mutually perpendi ular implies the equations 3 X
i=1 3 X
a2i
ai bi
=
=
i=1
3 X
i=1 3 X
b2i
=
bi ci =
i=1
3 X
c2i
i=1 3 X
= 1,
ci ai = 0 .
(1)
i=1
The points A, B , and C lie on the ellipse, hen e, their oordinates satisfy the equation of the ellipse: 1 −→ 2 |OA| 1 −→ 2 |OB| 1 −→ 2 |OC|
= = =
a21 a2
+
a22 b2
+
a23 c2
,
b21 b2 b2 + 22 + 32 2 a b c
,
c21
.
a2
+
c22 b2
+
c23 c2
(2)
311 If we de ne the matrix
a1 M = b1 c1
a2 b2 c2
a3 b3 c3
,
then the rst set of equations in (1) imply that M M t = I . That is, M is an orthogonal matrix. This means that we also have M t M = I , so that a21 + b21 + c21 = a22 + b22 + c22 = a23 + b23 + c23 = 1 .
(3)
Combining the equations in (2) and (3), we dedu e that 1 1 1 −→ 2 + −→ 2 + −→ 2 |OA| |OB| |OC| a21 + b21 + c21 a22 + b22 + c22 a23 + b23 + c23 = + + a2 b2 c2 1 1 1 = + 2 + 2. a2 b c
Letting d denote the distan e from O to the plane ABC , we re ognize the left side of the pre eding set of equations to be d12 . [Ed.: Although Demis in ludes a short proof of this laim, it an be found in standard referen es: the 1 distan e from the origin to the plane px + qy + rz = 1 is √p +q , where +r 2
1 1 , , p q
and
1 r
2
2
are the distan es from the origin to the points where the oordi-
nate axes interse t the plane.℄ The equation tells us that is a onstant. In other words, abc d= √ 2 2 a b + b2 c2 + c2 a2
and we on lude that the plane
entre O and radius d.
ABC
1 1 1 1 = 2+ 2+ 2 d2 a b c
,
remains tangent to the sphere with
SCAR CIAURRI, Also solved by MICHEL BATAILLE, Rouen, Fran e; MANUEL BENITO, O EMILIO FERNANDEZ, and LUZ RONCAL, Logrono, ~ Spain; JOEL SCHLOSBERG, Bayside, NY, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. Although all the submitted solutions were essentially the same, only the treatment of Benito, Ciaurri, Fernandez, and Ron al expli itly dealt with the n-dimensional version of the problem { the result (and the proof) is independent of dimension; it applies to an n-dimensional ellipsoid. The n-dimensional version of our problem an be found as exer ise 15.7.20 in Mar el Berger's Geometry, volume 2 (Springer 1987). His solution, whi h is also similar to our featured solution, appears in the a
ompanying volume by Berger et al., Problems in Geometry (Springer Problem Books in Mathemati s, 1984), problem 15.4, pages 227-229. After their proof the authors investigate the dual problem: the polarity with respe t to the xed sphere in our problem takes A, B, C to mutually orthogonal planes that interse t in the pole of the plane ABC (whi h is the point where the plane is tangent to the sphere). Be ause A, B, C lie on an ellipsoid, their polar planes are tangent to an ellipsoid, the polar body of the ellipsoid we started with. If, instead, we start with that new ellipsoid, we get the dual result:
312 The lo us of those points ommon to three mutually orthogonal planes that are tangent to an ellipsoid is a sphere, on entri with the ellipsoid. The dual result seems to be better known, and it omes with a variety of proofs. The sphere so obtained is variously known as the orthopti sphere, Monge's sphere, or the dire tor sphere.
3256.
USA.
[2007 : 298, 300℄ Proposed by Va lav Kone ny, Big Rapids, MI,
A bi entri quadrilateral (also alled a hord-tangent quadrilateral) is a quadrilateral that is simultaneously ins ribed in one ir le and ir ums ribed about another. Let ABCD be a bi entri quadrilateral in whi h there are no parallel sides. Suppose that the ir ums ribed and ins ribed ir les of ABCD have
entres O and I , respe tively. Let AC meet BD at E . Join the points of tangen y on the opposite sides of the quadrilateral, thus obtaining two lines whi h interse t at a point T . Prove that O, E , T , and I are ollinear. When do the points E and T
oin ide? (Compare 2978 [2004 : 429, 432; 2005 : 470{472℄.) Solution by Mi hel Bataille, Rouen, Fran e. First we show the following: If the points A, B , C , and D are arranged so that the lines AB , BC , CD , and DA are tangent to a ir le [Ed.: or, more generally, to a oni ℄ γ at P , Q, R, and S , respe tively, then the point of interse tion T of P R and QS is also the point of interse tion of AC and BD . This result is just a spe ial ase of Brian hon's Theorem. [Ed.: See the omments after the list of solvers.℄ However, the following proof seems more appropriate here. Let AB and CD meet at U , and let AD and BC meet at V . Then P R is the polar of U with respe t to γ , and QS is the polar of V . It follows that U V is the polar of T . But X = P S ∩QR and W = RS ∩P Q are
onjugates of T , hen e, the polar U V of T passes through W and X . Moreover, the polar RS of D and the polar P Q of B pass through W , hen e, the polar of W is BD, and so BD passes through T . Similarly, the polar of X is AC and passes through T . The result follows. Note that the argument is proje tive, so that it is easily adapted if any of U , V , X , or W is at in nity; in other words, the proof remains valid should any or all of the sides of the quadrilaterals ABCD and P QRS be parallel. We now suppose that γ is a ir le with entre I and that A, B , C , and D are, in addition, on a ir le Γ with entre O . Sin e AC and BD meet at T , the polar of T with respe t to Γ passes through U and V , hen e, is the line U V . As a result, U V is the polar of T with respe t to Γ as well as to γ . Be ause lines that are onjugate with respe t to a ir le are perpendi ular, we on lude that E (whi h, as we have seen, oin ides with T ), O, and I are on the perpendi ular to U V through T .
313 University of Sarajevo, Sarajevo, Bosnia Also solved by SEFKET ARSLANAGIC, and Herzegovina; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e; JOHN G. HEUVER, Grande student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; Prairie, AB; SALEM MALIKIC, ANDREA MUNARO, student, University of Trento, Trento, Italy; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. Leon Anne served as rep etiteur at the College Louis-le-Grand. His proof that E = T appeared in Les Nouvelles Annales 1842, page 186; 1844, pages 28 and 485; it was reprodu ed in paragraph 1274 (pages 563-564) in [4℄. Our problem appears in paragraph 1275 (page 565) in [4℄ under the heading "Newton's Theorem," with a proof similar to that of our rst featured solution. No expli it referen e to Newton's work appears there. The result appeared before in Crux Mathemati orum [1989 : 226℄ as an unused I.M.O. proposal. Other referen es, su h as [3℄, page 2, seem to refer to the (restri ted) assertion that E = T for a ir le as Newton's theorem. Proofs (of this restri ted theorem) in [1℄, page 79; and [5℄, page 100, apply Brian hon's theorem (If a hexagon is ir ums ribed about a oni , the three diagonals are on urrent) to the degenerate hexagons AP BCRD and ABQCDS to on lude that AC, P R, BD, and QA all pass through the same point E = T . In Problem 39 of [2℄, pages 188-191, there is a dis ussion of properties of hordtangent quadrilaterals. We thank Curtis, Demis, Heuver (he refers to problem 40 of [6℄), Maliki , and Kone ny for the helpful referen es. Referen es
[1℄ H.S.M. Coxeter and S.L. Greitzer, Geometry Revisited. Mathemati al Asso iation of Ameri a (1967). [2℄ Heinri h Dorrie, 100 Great Problems of Elementary Mathemati s: Their History and Solution. Dover (1965). [3℄ Kin Y. Li, Famous geometry theorems, Mathemati al Ex alibur, 10:3 (2005). (This is an ele troni journal: http://www.math.ust.hk/excalibur/) [4℄ F. G.-M., Exer i es de Geom etrie| omprenant l'expose des methodes geom etriques et 2000 questions resolues, quatrieme edition, Maison A. Mame & Fils, Tours (1907). [5℄ Ross Honsberger, In Polya's Footsteps: Mis ellaneous Problems and Essays. Mathemati al Asso iation of Ameri a (1997). [6℄ I.M. Yaglom, Geometri Transformations III. Mathemati al Asso iation of Ameri a (1973).
3257. [2007 : 298, 300℄ Proposed by Bill Sands, University of Calgary, Calgary, AB. Find the number of ordered pairs (A, B) of subsets of {1, 2, . . . , 13} su h that |A ∪ B| is even.
Essentially similar solutions by Sefket Arslanagi , University of Sarajevo, Sarajevo, Bosnia and Herzegovina; Mi hel Bataille, Rouen, Fran e; Jose H. Nieto, Universidad del Zulia, Mara aibo, Venezuela; and Xavier Ros, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; modi ed by the editor. The solution provided is for S = {1, 2, . . . , n}, the given problem being the spe ial ase n = 13. If A is a subset of S with |A| = k, 0 ≤ k ≤ n, and if B is a subset of S su h that A ∪ B = S, then B is the union of S − A and an arbitrary subset C of A. There are nk su h subsets A and there are
314 2k
possibilities for the subset C , hen e, the number of su h pairs (A, B) is n X n
k=0
k
2k = 3n .
It is lear that the number P of ordered pairs (A, B) of subsets of S su h that n k |A ∪ B| is even, is En = 3 . To evaluate En observe that k k even
n k 3 k k=0 n X n k (−1)k 3 k k=0 n X
= (1 + 3)n = 4n ,
(1)
= (1 − 3)n = (−2)n .
(2)
Adding (1) and (2) yields 2En = 4n +(−2)n , hen e, En = 22n−1 +(−1)n 2n−1 . The required number of pairs is E13 = 225 − 212 = 33550336. Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; MANUEL SCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logrono, BENITO, O ~ Spain; WHITNEY BULLOCK and DEBORAH SALAS-SMITH, students, California State University, Fresno CA, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; JOHN HAWKINS and DAVID R. STONE, Georgia Southern University, Statesboro, GA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; TOM LEONG, Brooklyn, NY, USA; KATHLEEN E. LEWIS, SUNY Oswego, Oswego, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; NY, USA; SALEM MALIKIC, MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring eld, MO, USA; EDMUND SWYLAN, Riga, Latvia; BINGJIE WU, student, High S hool AÆliated to Fudan University, Shanghai, China; and the proposer. Geupel generalized the result by showing that the number of q-tuples (B1 , . . . , B q ) of subsets of {1, . . . , N} su h that |B1 ∪ · · · ∪ Bq | is even is 12 2qN + (−1)N (2 − 2q )N . The Missouri State University Problem Solving Group showed that if |B1 ∪ · · · ∪ Bq | is restri ted to be a multiple of d, then the number of q-tuples is
1 d
primitive dth root of unity.
d−1 P t=0
1 + (2q − 1)ωt
N
, where ω is a
3258⋆. [2007 : 298, 300℄ Proposed by Alper Cay, Uzman Private S hool,
Kayseri, Turkey. Let ABC be a triangle with ∠ABC bise tor of ∠ABC with D on AC . If AD using a purely geometri argument.
= 80◦ . Let BD be the angle = DB + BC , determine ∠A,
Solution. The Problem Editor of Math Horizons, Andy Liu, informed us that this problem appeared re ently as problem 201 in the April 2006 issue of Math Horizons, page 32. A solution (showing that ∠A = 20◦ ) by David Rhee and Jerry Lo was published in the November 2006 issue, pages 41-42. SCAR CIAURRI, EMILIO FERNANDEZ, and LUZ Also solved by MANUEL BENITO, O RONCAL, Logrono, ~ Spain; APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e; Y, Big Rapids, MI, USA; JOSE H. NIETO, Universidad del Zulia, Mara aibo, V ACLAV KONECN Venezuela; and PETER Y. WOO, Biola University, La Mirada, CA, USA.
315
3259. [2007 : 298, 301℄
Proposed by Neven Juri , Zagreb, Croatia. Is it possible to nd a ubi polynomial P su h that, for any positive n integer n, the polynomial P {z· · · ◦ P} has exa tly 3 distin t real roots? | ◦P ◦ n times
Find one, if possible, or show that none exists.
I. Solution by Jose H. Nieto, Universidad del Zulia, Mara aibo, Venezuela, modi ed slightly by the editor. 3 − 3x. Then P has three distin t real roots, namely 0 Let √ P (x) = x ′ and ± 3. Sin e P (x) = 3x2 − 3 = 3(x + 1)(x − 1) we see that P is stri tly in reasing on (−∞, −1] and [1, ∞), and stri tly de reasing on [−1, 1]. Note that P (−2) = −2, P (−1) = 2, P (1) = −2, and P (2) = 2, so it follows that P ([−2, −1]) = P ([−1, 1]) = P ([1, 2]) = [−2, 2]. Now we prove by indu tion that P n = P ◦ P ◦ · · · ◦ P (the n-fold
omposite) has exa tly 3n distin t real roots, all of whi h lie in the interval (−2, 2). This is learly true for n = 1. Suppose the laim holds for some n ≥ 1. Note that for x ∈ (−2, −1), the polynomial P takes ea h value in (−2, 2) exa tly on e. The same is true for x ∈ (−1, 1) and for x ∈ (1, 2). Therefore, P n+1 (x) = P n P (x) has exa tly 3n distin t real roots in ea h of the intervals (−2, −1), (−1, 1), and (1, 2), hen e, P n+1 has 3n+1 distin t real roots in the interval (−2 ,2). These are all the roots of P n+1 , be ause P n+1 is of degree 3n+1 . The indu tion is omplete.
. Solution by the Missouri State University Problem Solving Group. More generally, we show that for any positive integer d, there exists a polynomial P of degree d su h that for any positive integer n, the n-fold
omposite P n = P ◦ P ◦ · · · ◦ P has exa tly dn distin t real roots. Let P (x) = Td (x) = cos(d cos−1 x) be the Chebyshev Polynomial of degree d of the rst kind. It follows from the de nition (and is well known) (2k + 1)π that Td has the d distin t real roots cos , k = 0, 1, . . . , d − 1, 2d and that Ta ◦ Tb = Tab . Therefore, Td = Td ◦ Td ◦ · · · ◦ Td has dn distin t real roots. In parti ular, if we take d = 3, then II
n
P (x) = =
π 5π π x − cos x − cos T3 (x) = x − cos 6 2 6 ! ! √ √ 3 3 1 x x− x+ = (4x3 − 3x) 2 2 4
is a solution to the proposed problem. Also solved by CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; XAVIER SALEM MALIKIC, ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.
316
3260. [2007 : 298, 301℄
Proposed by Virgil Ni ula, Bu harest, Romania. Let a, b be distin t positive real numbers su h that (a − 1)(b − 1) ≥ 0. Prove that ab + ba ≥ 1 + ab + (1 − a)(1 − b) · min{1, ab} .
Solution by Tom Leong, Brooklyn, NY, USA. The numbers a and b need not be distin t, as the proof will show. First suppose that 0 < a < 1. Then we also have 0 < b < 1. Put a = 1 − r and b = 1 − s, where 0 < r < 1 and 0 < s < 1. Sin e in this ase min{1, ab} = ab, the right-hand side of the inequality is 1 + (1 − r)(1 − s) + rs(1 − r)(1 − s) = 1 + (1 − r)(1 − s)(1 + rs) .
We have (1 − r)s ≤ 1 − rs and (1 − s)r Sin e 1 −1 rs > 1 + rs, we obtain ab + ba
=
1−r
(1 −
r)s
≤ 1 − rs by Bernoulli's Inequality.
+
1−s
(1 − s)r
≥
1−r 1−s + 1 − rs 1 − rs
=
1+
>
1 + (1 − r)(1 − s)(1 + rs)
=
(1 − r)(1 − s) 1 − rs
1 + ab + (1 − a)(1 − b)ab .
Next suppose that a ≥ 1. Then we also have b ≥ 1. Put a = 1 + r and where r ≥ 0 and s ≥ 0. Sin e in this ase min{1, ab} = 1, the right-hand side of the inequality is
b = 1 + s,
1 + (1 + r)(1 + s) + rs = 2rs + r + s + 2 .
We have (1+r)1+s ≥ 1+r(1+s) and (1+s)1+r ≥ 1+s(1+r) by Bernoulli's Inequality. Adding the two inequalities, we have ab + ba ≥ 2rs + r + s + 2. Equality holds for a = b = 1. University of Sarajevo, Sarajevo, Bosnia and Also solved by SEFKET ARSLANAGIC, Herzegovina; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; KEE-WAI LAU, Hong Kong, China; THANOS MAGKOS, 3rd High S hool of Kozani, student, Sarajevo College, Sarajevo, Bosnia and Kozani, Gree e; SALEM MALIKIC, Herzegovina; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; and the proposer.
317
3261. [2007 : 299, 301℄
Proposed by Ovidiu Furdui, University of Toledo, Toledo, OH, USA. The Fibona
i numbers Fn and Lu as numbers Ln are de ned by the following re urren es: F0 = 0 , F1 = 1 , L0 = 2 , L1 = 1 ,
and and
Fn+1 = Fn + Fn−1 , Ln+1 = Ln + Ln−1 ,
for n ≥ 1; for n ≥ 1.
Prove that ∞ arctan X
n=1
where β =
1 2
1 L2n
arctan
1
arctan L2n+2 1 F2n+1
≤
4 1 , arctan(β) arctan(β) + 3 π
√ 5−1 .
s ar Ciaurri, Emilio Fernandez, and Luz Ron al, Solution by Manuel Benito, O Logrono, ~ Spain; and Chip Curtis, Missouri Southern State University, Joplin, MO, USA. The following relations between the Fibona
i and Lu as numbers L2n + L2n+2 = 5F2n+1
and
2 L2n L2n+2 − 1 = 5F2n+1 ,
are well known and easy to he k. From these we have 1 1 + L2n + L2n+2 L2n L2n+2 = = 1 1 L2n L2n+2 − 1 1− L2n L2n+2
1 F2n+1
,
so that arctan
1 F2n+1
arctan
1 L2n
1
+
1
L2n L2n+2 = arctan 1 1 1− L2n L2n+2 1 1 = arctan + arctan L2n L2n+2
Applying the inequality xy ≤
arctan
1 (x 4
+ y)2 , we 1
L2n+2
≤
.
obtain 1 4
arctan
1 F2n+1
2
,
318 therefore, ∞ arctan X
1 arctan ∞ 1 X 1 L2n+2 ≤ arctan 1 4 n=1 F2n+1 arctan F2n+1
n=1
1 L2n
.
Using the relation F2n+2 − F2n = F2n+1 and the well known and easy 2 to he k formula F2n F2n+2 + 1 = F2n+1 , we have 1 F2n+1
1 1 − F2n+2 − F2n F2n F2n+2 = = 1 1 F2n F2n+2 + 1 1+ F2n F2n+2
and then arctan
1 F2n+1
Hen e, ∞ 1 X
4
n=1
arctan
1 F2n+1
1
1
−
,
F2n F2n+2 = arctan 1 1 1+ F2n F2n+2 1 1 = arctan − arctan F2n F2n+2
= =
∞ 1 X
4 1 4
arctan
n=1
arctan
1 F2
1 F2n =
1 4
Thus, the sum of the given series does not ex eed proves the proposed upper bound, be ause
− arctan
arctan 1 = π 16
1 F2n+2
π 16
≈ 0.196,
.
.
whi h im-
1 arctan(β) arctan(β) + ≈ 0.625 . π 3 Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; and the proposer. Janous also improved the proposed upper bound. 4
3262. [2007 : 299, 301℄ Proposed by Ovidiu Furdui, University of Toledo, Toledo, OH, USA. Let m be an integer, m ≥ 2, and let a1 , a2 , . . . , am be positive real numbers. Evaluate the limit Lm =
lim
n→∞
1 nm
Z
m e Y
1 k=1
ln 1 + ak xn dx .
319 s ar Ciaurri, Emilio Fernandez, and Luz Ron al, Solution by Manuel Benito, O Logrono, ~ Spain, modi ed by the editor. For ea h integer m ≥ 1 we will show that Lm = (−1)m+1 m! + e
m X
(−1)k
k=0
m! (m − k)!
.
(1)
First note that for x ≥ 1, we have 1/n
xak
≤ (1 + ak xn )1/n ≤ x(1 + ak )1/n .
(2)
Sin e a1/n and (1 + ak )1/n ea h onverge to 1 as n → ∞, it follows from k the above that (1 + ak xn )1/n onverges to x as n → ∞, thus, lim
n→∞
ln(1 + ak xn ) = n
lim ln(1 + ak xn )1/n = ln x .
n→∞
(3)
Taking logarithms a ross the last inequality in (2), we obtain ln(1 + ak xn ) ln(1 + ak ) ≤ ln x + ≤ ln x + ln(1 + ak ) , n n
from whi h it follows that m Y ln(1 + ak xn )
k=1
n
n Y
≤
k=1
ln x + ln(1 + ak ) .
By Lebesgue's Dominated Convergen e Theorem, we may bring the limit inside the integral; then we apply (3) as follows Lm
= = =
Z
n Y ln(1 + ak xn )
e
lim
1 n→∞ k=1 Z e Y n
lim
1 k=1 n→∞ e m
Z
1
(ln x)
n
dx
ln(1 + ak xn ) dx n
dx .
(4)
Next we integrate by parts to derive the re urren e relation Lm = e − mLm−1 .
(5)
Finally, we use indu tion on m to show that (with the appropriate initial ondition) the solution to the re
uren e in (5) is given by (1). The ase when R m = 1 is lear, sin e the right side of (1) is 1 and from (4) we have L1 = 1e ln x dx = 1.
320 Suppose (1) holds for some m ≥ 1. Then using (5) we have Lm
(
m−1 X
=
(m − 1)! e − m (−1)m (m − 1)! + e (−1)k (m − 1 − k)! k=0
=
e + (−1)m+1 m! + e
m−1 X
(−1)k+1
k=0
=
(−1)m+1 m! + e + e
m X
(−1)k
k=1
=
(−1)m+1 m! + e
m X
(−1)k
k=0
and our proof is omplete.
)
m! (m − 1 − k)! m!
(m − k)!
m! (m − k)!
,
Also solved by MICHEL BATAILLE, Rouen, Fran e; PAUL BRACKEN and N. NADEAU, University of Texas, Edinburg, TX, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. There was 1 in orre t solution submitted. Janous notes the interesting fa t that Lm an be expressed in terms of Dm , the number of derangements of 1, 2, . . . , m. (A permutation σ of 1, 2, . . . , m is alled a derangement if σ(i) 6= i for all i = 1, 2, . . . , m.) Sin e it is well known that Dm = m!
m P
(−1)k
k=0
1 , we see k!
that Lm = (−1)m+1 m! + (−1)m eDm . The proposer remarked that his proposal was a generalization of the following problem whi h appeared in the Romanian journal Gazeta in 2000: Z e 1 Compute lim 2 ln(1 + xn ) ln(1 + 2xn ) dx. n→∞
n
1
Both he and Bra ken and Nadeau pointed out the interesting fa t that the answer is
ompletely independent of the ak 's given.
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
