GAZETA MATEMATICA - A BRIDGE OVER THREE CENTURIES
Cover photo. Cernavoda railway bridge over Danube river was open for traffic on 14 September 1895. The first issue of "Gazeta, Matematica" was published one day later, on 15 September 1895. Both of them continue their existence and mission...
VASILE BERINDE EUGEN PALTANEA (coordinators)
GAZETA MATEMATICA - A BRIDGE OVER THREE CENTURIES
ROMANIAN MATHEMATICAL SOCIETY (SOCIETATEA DE $TIINTE MATEMATICE DIN ROMANIA) 2004
© 2004 SOCIETATEA DE $TIINTE MATEMATICE DIN ROMANIA
Toate drepturile asupra acestei editii apartin Societatii de $tiinte Matematice din Romania. Nici o parte din acest volum nu poate fi copiath, sau transmisa prin alt mijloc, electronic sau mecanic, inclusiv fotocopiere, fg.r5, permisiunea scrissa a SocietAii de Stiinte Matematice din Romania.
© 2004 Societatea de Stiinte Matematice din Romania (Romanian Mathematical Society) All right reserved
Romanian Mathematical Society str. Academiei nr. 14 cod 010014, Bucuresti, Romania E-mail: officeOrms.unibuc.ro
ISBN 973-0-03549-0
Contents Preface
7
1 Problems for the 5th form
13
2 Problems for the 6th form
25
3 Problems for the 7th form
35
4 Problems for the 8th form
47
5 Problems for the annual competition - Secondary level
57
6 Problems for the 9th form
69
7 Problems for the 10th form
91
8 Problems for the llth form
113
9 Problems for the 12th form
141
10 Problems for the annual competition. Forms 9-10
167
11 Problems for the annual competition. Forms 11-12
181
GAZETA MATEMATIC A
-
A BRIDGE OVER THREE
CENTURIES
List of contributors
Eugen Paltanea, "Transilvania University" Brasov Radu Paltanea, "Transilvania University" Brasov Emil Stoica, "Transilvania University" Brasov Marin Marin, "Transilvania University" Brasov Marius Paun, "Transilvania University" Brasov Ovidiu Muntean, "Transilvania University" Brasov Vasile Berinde, North University of Baia Mare Madalina Berinde, "Babes-Bolyai" University Cluj-Napoca Gheorghe Boroica, "Gh. $incai" National College Baia Mare Nicolae Mu§uroia, "Gh. Sinai" National College Baia Mare Cristian Heuberger, "Gh. incai" National College Baia Mare Dana Heuberger, "Gh. *incai" National College Baia Mare
Preface The present book was prepared for the 10thInternational Congress on Mathematical Education, 4 - 11 July 2004, Copenhagen, with the aim to give an idea on what does represent "Gazeta Matematid" for the Romanian mathematical education system. The journal whose name is written quite similar to other languages is the oldest and most respectable Romanian periodical on elementary mathematics and didactics of mathematics, which has been issued continuously as a monthly ever since its foundation in 1895. In the framework of the Romanian mathematics education system, "Gazeta Matematicg," is like an institution, being responsible for the creation, improvement, and keeping up a high interest in practising mathematics in Romania at secondary and high-school levels, and even to primary or higher education levels. The book was designed to offer a selection of problems published during its long life, grouped on the current school levels in Romania. The pre-university Romanian education system can be thought as having three broadly defined levels (kindergarten level is not considered here): elementary school (grades I - IV, corresponding to ages 6/7 - 10/11); middle school or secondary school (grades V - VIII, corresponding to ages 11/12 - 14/15); and high-school (grades IX-XII, corresponding to ages 15/16 18/19). In spite of the fact that "Gazeta Matematid," also includes since the 80th a column of proposed problems for elementary school, we decided to consider in this book only problems addressing to secondary school and high school levels. Chapters 1 - 4 are devoted to problems for secondary school students, grouped on grades in a relative curriculum division, while Chapters 6 - 9 include problems for high school students. In order to illustrate the competition problems column in "Gazeta Matematid" , we also included such kind of problems as follows: in Chapter 5, 7
8
"Gazeta Matematice — a bridge over three centuries
problems for secondary school level, in Chapter 10, problems for high school students in Forms 9 - 10, while Chapter 11 is devoted to Forms 11 - 12. At each chapter, the problems are numbered specifically, for instance, IX.13 denotes Problem 13 in the list of problems addressing to students in 9th form etc. The idea of the selection was to include problems covering the long life of "Gazeta Matematicg.", but as curriculum and also tastes have changed during the XXth century, the vast majority of problems are taken from the last 40 years issues. It is important to stress on the fact that the selection itself is not unitary, due to diversity of personal tastes of those who worked to this book. Although we have tried to impose a unified manner is selecting and solving the problems, the readers will notice the diversity of styles in these respects, however. Therefore, the collection of problems from "Gazeta Matematich'" we are offering is certainly a multi-subjective one. We still hope it will give some sample problems published in "Gazeta Matematicg" during its centennial existence. We initially planned to include some representative mathematical notes and articles published in "Gazeta Matematicr but finally resigned, due to space limiting. However, a few data on the number of articles and notes published in "Gazeta Matematice so far would be interesting by themselves. In the first 54 years of publication (1895-1949) Gazeta MatematicA published 1,037 articles and notes; 114 of them are in arithmetics and number theory; 371 articles are on geometry; 164 articles on algebra, 19 of trigonometry, 59 articles are on applied mathematics (mechanics etc.). In the same period, 6,645 problems were proposed, while only 5,279 were accompanied by complete solutions in the subsequent issues. Since 1950 until nowadays (2004), a number of 2,130 articles were published in "Gazeta Matematicr: 717 in algebra, 681 in geometry, 238 in calculus and the remaining to applied mathematics (astronomy, mechanics) etc. In the same period, for 8,344 problems, "Gazeta Matematia," also published their complete solutions. If we refer to the whole period 1895-2004, about 44,323 problems, and about 13,623 solutions, were published in " Gazeta Matematicr . These numbers include only proposed problems in regular columns. Many other problems related to various mathematical competitions, leaving certificate exams, university entrance exams etc., were usually published in most of issues. Therefore, by its about 280 problems included, the present book is just a drop in an ocean of problems.
Preface
9
It is impossible to acknowledge individually colleagues, friends and collaborators to whom I am indebted for support in preparing this book. I must, however, express my thanks to all who contributed by a significant amount of work. The list includes Radu Paltg,nea, Marin Marin, Emil Stoica, Eugen PaltAnea, Marius Faun and Ovidiu Muntean from "Transilvania" University Bra§ov; Gheorghe Boroica, Nicolae Mu§uroia, Cristian Heuberger and Dana Heuberger from "Gheorghe Sincai" National College Baia Mare and MAdAlina Berinde from "Babe§ Bolyai" University Cluj-Napoca. These ones, together with the undersigned, have worked to the selection of problems, translating their statements and solutions into English and doing the first correction of the type-written version. A sequence of second, third,..., corrections were also done by my colleagues Maria Sanziana Pop, Gabriella Kovacs, Iulian Coroian, Dan Rarbosu, Laszlo Balogh and Huba Sass, from North University of Baia Mare. In order to help readers who are not familiar with some notations, a list of the most important notations in the book was prepared by my colleague Iulian Coroian. Concluding this Preface, I would like to stress on the main merit of this problem book: the very fact that it was written down.
Vasile Berinde
"Gazeta Matematica" - a bridge over three centuries
10 List of symbols
N = {0,1,2,...,n,...} Z = {..., -n,..., -2, ---1,0,1,2,...,n,...} N* = N\{0} R = the set of real numbers R*=R\{0} Q = the set of rational numbers [a, b] = the closed interval (a, b) = the open interval n! = 1,2,3,...,n n! n) k!(n - k)! Card A = the cardinal of the set A df f (x) = — = the derivative of f dx dm f f (n) (x) = — = n - th derivative of f dxn b
f f (X)dX = the Riemann integral on [a, b]
a
f : A --> B = a map from A to B (a function defined on set A with its values in set B) f -1 = the inverse of function f an =n-th term of sequence (an ) or (an)n>i Ntn,m(R) = the set of all matrices with n rows and m columns having real elements Mn(C) = the set of n-th order squares matrices with complex elements det(A) = the determinant of square matrix A lirn an = the limit of sequence (an) as n co n—■ oo P(M) = the family of subsets of set M including 0 and M p
E ai = E i=k
ai = the sum of the numbers ak, ak+i, , ap
k
= the m-th order square matrix with elements aid Jag b 2 if x = a + ib is Ix' = the absolute value of real number x or complex [x] = the integer part of real number x {x} = x - [x] = the fractional part of real number x = the conjugate of complex number z Mp = a multiple of integer p min{a, b} = the minimum between real numbers a and b max{a, b} = the maximum between real numbers a and b
Preface
11
(a, b, c) = the biggest common divisor of integers a, b, c [a, b, = the smallest common multiple of integers a, b, c Im(z) = the imaginary part of complex number z Re(z) = the real part of complex number z R[X] = the set of polynomials with real coefficients Z[X] = the set of polynomials with integer coefficients Q[X] = the set of polynomials with rational coefficients A U B= the union of sets A and B A n B= the intersection of sets A and B = the product of numbers ak, ak+i, • , ap fl ai =
i=k
k
alb = means b = Ma, that is the integer a is a divisor of integer b ma, mb, me = the lengths of the medians of the triangle ABC of sides a, b, c ha, hb, he = the lengths of the altitudes of the triangle ABC of sides a, b, c.
Chapter 1
Problems for the 5th form Problem V.1. Prove that if a = m19 +10, the positive integer a can't be a perfect square or a perfect cube. (We denote by mi9 a random multiple of 19) Solution. We observe that m19 + 10 = mi9 + 19 — 9 = m19 — 9. Let n be a natural number. Then there exists a E {0, 1, ... , 18} such that n = 19k + a. Computing, we deduce that none of the numbers 02 + 9, 12 + 9, ... , 182 + 9 and none of the numbers 03 +9, 13+9, ... ,183+9 is multiple of 19. Then: n2 + 9 = (19k ± a)2± 9 = m19 ± a2+ 9 and because a2 + 9 is not a multiple of 19, we obtain n2+ 9 V Mis i.e. n2 V M19 - 9. With the same reasoning, we deduce that n3V M19 - 9. Note. M19 means the set of the multiples of 19. Problem V.2. Find the digits a and b such that the number 543ab2 is divisible by 42. Solution. We can check that 543ab2 is divisible by 2. It is therefore enough to find the digits a and b such that 543ab2 is divisible by 21. We observe that 542997 is divisible by 21. We have 543ab2 = 542997 + ab0 + 5 = 542997 + ab5, it is consequently sufficient to determine the digits a and b such that ab5 is divisible by 21. Because ab5 is divisible by 5, we obtain that ab5 is an odd multiple of 105 = 5 - 21. Hence, ab5 E {105, 315, 525, 735, 945} and therefore f a=1 I a=3 f a=5 f a=7 { a=9 lb= 0 ' 1 b=1 'lb= 2 ' 1 b= 3 ' b= 4 are the solutions of the problem. 13
14
"Gazeta Matematice — a bridge over three centuries
Problem V.3. If the square of a positive integer ends with 69, prove that the previous or the next perfect square ends with 96. Solution. Let n = xi • • • xkab be the given number, where xi, • • • , xk,a,b, are its digits. The last two digits of n2coincide with the last two digits of ab2. We thus have: ab2 = yi • yt69 , whence we deduce that ab E {13, 37, 63, 87}. Also, (13 + 1)2= 196, (37 — 1)2 = 1296, (63 + 1)2 = 4096, (87 — 1)2 = 7396, therefore the assertion from the problem is true. Problem V.4. Find the non-null digits x, y, z, knowing that 3367 is a divisor of the number 2x03y6z3. Solution. We observe that 3367 • 6246 = 21030282. 336712x03y6z3 <=> 33671 (2x03y6z3 — 21030282) .#• X33671 ((x — 1)• 106+ y • 103+ 10z + 321) <=> <=> 33671(999999(x — + (x
—
1) + 1000y + 10z + 321) .
Because 999999 = 297.3367, we deduce that 33671(1000y + 10z + x + 320) •:=> 33671 (y3zx + 20) . The four digit multiples of 3367 are: 3367 and 6734. Because either digit 3 or the digit 4 is in the hundreds'place in the number y3zx + 20, we conclude that the only solution is y3zx + 20 = 3367. We obtain x = 7, y = 3, z = 4, the unique solution. Problem V.5. Find the least positive integer that has got exactly 42 divisors. Solution. Since 42 = 2 • 3 • 7 = (1 + 1) • (2 + 1) • (6 + 1), the number will be of the form a • b2 C6 , where a, b, c are its (distinct) prime factors. In order to obtain the minimum value, we must have c < b < a and also c, b and a be the smallest possible positive numbers. So, we obtain c = 2, b = 3 and a = 5 and the required number is 2880. Note. The number of divisors of n = aPkk equals (pk +1), where a2, , ak are the distinct prime factors (P1 + 1)(P2 +1) ... of n. . . .
Problem V.6. A group of tourists could go through a certain distance in 10 hours by train and in 12 hours by car. They went however, 6 hours by train, hours by car and finally they walked 36 kilometers. Find
Problems for the 5th form
15
a) The length of the route. b) The speed of the train and the speed of the car. c) How many kilometers did they travel by train and how many by car? 4 6 Solution. The tourist travelled — of the distance by train and — of the 10 12 distance by car. of the total distance. 10 12 15 a) The length of the route is therefore 36 • 15 = 540 km. b) The speed of the train is 54 km/h. The speed of the car is 540 : 12 = 45 km/h. 6 c) They travelled by train — 10• 540 = 324 km and
Consequently they walked 1 —
they travelled by car
• 540 = 180 km. t
Problem V.7. Prove that from all the people that ever lived on earth (including those who live today), the number of those who shaked hands an odd number of times is even. Solution. There are two people involved in every handshake. Therefore, the number of those who shaked hands is even (counting each person n times, if he shaked hands with n other persons). Consequently, the number of the people who shaked hands an odd number of times must be even. Problem V.8. Find the digits indicated by x in the following multiplication
2 x x x x 9 x x 5 x x 5 X XX X X X
Solution. The multiplication is: a 5 c 7 2 e f h 9 i j 5 k t 5 m n p q r
b x d g
s
"Gazeta Matematica" — a bridge over three centuries
16
Because the last digit of c. b is 5, we must have c = 5 or b = 5. Moreover, the digits b and c must be odd. Because a5b•7 = h9ij, we obtain: 7•a+3 = h9 or 7 • a + 4 = h9 and hence a = 8 or a = 5. I. a = 8. From a5b • d = 85b • d = 2e f g, we deduce d = 3. From a5b • c = 85b • c = 5kt5, we deduce c = 7 and b = 5. Finally, the multiplication becomes:
5 6
5 9 6
8 5 5 7 7 3 5 6 5 8 5 5 9 1 5
2 9 8 0
x
II. a = 5. From a5b • d= 55b • d=2efg, we deduce d = 4 or d= 5. From a5b • c = 55b • c = 5kt5, we deduce c = 9 and b = 5. But 555. 9 = 4995 5kt5. Therefore in this case we do not have solution. Problem V.9. Compute:
1+ 1+
2
2 N .1+ 1+ 1 1 1+ 1,5 7 • 5 — g • (1 + 0, 405 : 0, 945) + 1, 25 — 0, (1) • (5 — 0, 32 : 0, 35) : 5 1] [
Solution. We have:
N . 1+
+
3 21+— 2
r5
L
(1+ 405 1000 \ )+ 8 1000 945
1+ 1 2 1+— 3 5 1(5 _ 32 1001 oi 4 9• 100 35 ) 'fi —
17
Problems for the 5th form 1+
= 1+
1+
3
r5 7
[
1
1+
+ 3\
b
7)
4
32)1 35 J
3
7 5 (c 7 10 + 5_ 5+ 32) . 1 + 4 • -8-• Y' — -8- • 74 35l 35 32 = 2. = 1+ 32 35 Problem V.10. Prove that there is no natural number such that, dividing it by 25 we obtain the remainder 7 and dividing it by 20, we obtain the remainder 4. Solution. Suppose such a number, n, exists. Let a and 0 be the quotients of the two divisions. Then: n = 25a + 7 and n = 200 + 4. Therefore, 25a + 7 = 200 + 4 <=> 7 — 4 = 200 — 25a <=> 3 = 5(40 — 5a) The last equality is false, because 3 is not a multiple of 5. Therefore we deduce that such a natural number n does not exist. Problem V.11. Assuming unlimited source of water, how can one cam' 13 liters of water in two pitchers, one pitcher with the capacity of 8 liters and the other with the capacity of 9 liters? Solution. We denote by V8 the 8 liter pitcher and by V9 the 9 liter pitcher. We fill up V9 and we pour the water from V9 into V8, filling up V8. One liter of water remains in V9. We empty V8. We pour the liter of water from V9 into V8. Then, again, we fill up V9 and we pour the water from V9 into V8 filling up. V8. Two liters of water remain into V9.We empty V8. We pour the two liters of water from V9 into V8. We'll continue in the same way, until we'll have 4 liters in V8. Filling V9, we'll have the 13 liters. Problem V.12. Let a, b, c be given digits. Find the digits x and y from the equality abc — cba = 3xy. Solution. abc — cba = (100a + 10b + c) — (100c + 10b + a) = 99(a — Therefore, 99(a — c) = 3xy. The only possibility is that a — c = 4, and in this case, 3xy = 99 •4 = 396. We obtain x = 9, y = 6. Problem V.13. How can one share 8 identical breads to 15 persons, such that none of the breads be divided in 15 equal parts?
18
"Gazeta Matematica" — a bridge over three centuries
8 Solution. Each person will have — of the total quantity of bread. 15 But: --8— = 3 Since there are 15 persons, 3 breads will be 15 15 15 = 5 divided in 5 equal parts and 5 breads will be divided in 3 equal parts. 1 1 Each person will receive 5 — of a bread and — of another bread. 3 Problem V.14. A construction can be finished by 15 workers in 20 days. Because after 8 days of work, some of them left for another site, the working is finished in 28 days. How many workers did leave? 1 Solution. A worker a day finishes the 20 •
th part of the working. In 15 the first 8 days, the 8 th part of the working is realized. The team must p finish the remaining 12 in 20 days. 20 12 12 In each day, they finish the th part of the working. We have 20 • 20 20 • 20 3 1 = = 9 — and hence we deduce that there are still 9 workers in 100 300 100 this site, so 6 workers have left. -
Problem V.15. Radu, Doina, Matei and $erban are four pupils studying for the final exams, from algebra, physics, chemistry and geography textbooks. The book covers have different colors: green, white, yellow and blue. We know that: a) Doina's textbook is green. b) Radu dreams of himself being a well-known geographer. c) The physics textbook's cover isn't white. d) Matei and erban want to become engineers. e) 9erban finished studying physics. f) The future engineers' textbook covers aren't blue. Can you specify from which textbook each pupil is studying and what color have the cover of the respective book? Solution. The condition d) implies that Matei and Serban are studying algebra and physics. From e) we deduce that Serban studies physics and therefore Matei studies algebra. The condition b) implies that Radu studies geography and consequently Doina studies chemistry. Because Doina's textbook isn't green (from a)) and (from f)) Matei and Serban don't study from blue textbooks, we deduce that Radu's textbook is"blue. The condition c) implies that Matei's textbook (algebra) is while. Consequently, Serban's textbook is yellow.
Problems for the 5th form
19
Problem V.16. Find the sets X and Y that verify simultaneously the conditions: (1) X\ {1, 2} = Y U {0, 4, 5, 6} (2) Y\ {1,2,3} = X U {0,5} (9) XUY C {0,1,2,3,4,5,6} Solution. We deduce from (1) that 1,2 Y and 0,4, 5,6 E X. We find from (2) that 5 E Y and 1,2,3 X. Therefore X = {0, 4,5,6}. (1) becomes: {0, 4, 5, 6} = YU{O, 4, 5, 6}, whence we obtain Y C {0, 4, 5, 6} (i) (2) becomes: Y\ {1, 2, 3} = {0, 4, 5, 6} (ii) Using (i), the relation (ii) becomes: Y = 10,4,5,61. Problem V.17. Prove that for any value of the positive integer n, the frac65n + 3 tion is in lowest terms. 39n + 2 Solution. Let d be a common factor of the numerator and of the denominator of the fraction. d1(65n + 3) = d13 • (65n + 3) = di (195n + 9) (i) dk39n + 2)
d15 • (39n + 2)
di(195n + 10) (ii)
From (i) and (ii) we deduce di((195n + 10) - (195n + 9)) <=> del<=> d = 1. Therefore, the fraction is in lowest terms. Problem V.18. Find the prime number p such that the numbers a = 2p + 1, b = 3p+ 2, c = 4p+ 3 and d = 6p+1 are simultaneously prime numbers. Solution. We observe that p = 2 and p = 3 don't verify the conditions; p = 5 implies a = 11, b = 17, c = 23 and d = 31 prime numbers, so p = 5 verifies all the conditions. If p > 7, then p = 6k + 1 or p = 6k + 5 with k natural number, k > 1; - in case of p = 6k + 1 the number a = 2p + 1 = 2(6k + 1) + 1 = 12k + 3 is not prime; - in case of p = 6k + 5, for k = 5a, with a > 1 natural number p = 30a + 5 is not prime k = 5a + 1, with a natural number b = 90a + 35 is not prime k = 5a + 2, with a natural number a = 60a + 35 is not prime k = 5a + 3, with a natural number c = 120a + 95 is not prime k = 5a + 4, with a natural number d = 180a + 175 is not prime. Therefore p = 5 is the only value of p satisfying all requirements. Problem V.19. Find two positive integers, if their sum is 65 and their greatest common factor is 13.
"Gazeta Matematicr
20
—
a bridge over three centuries
Solution. Let x and y be the two numbers. Because their greatest common factor is 13, we find two relative prime integers a and b such that x = 13 • a and y = 13 • b
From x + y = 65 we deduce a + b = 5 and hence: {a=1 b=4
or
a=2 b=3
Therefore, the solutions are: f x = 13 y = 52
and
x = 26 y = 39.
Problem V.20. Find the set of the pairs (x, y) for which: 0, xx(y)
—
0, yy(x) =
1 97 ) • (0 98(1) + 5 ' 900
Solution. The given equality is
xxy xx yyx yy _ 1(981 — 98 ± 97 900 900 5 900 9001 100x + 10x + y — 10x — x 100y + lOy + x — lOy — y 900 900 1 980 aR 5 900 .<4' • (x y) =98 *2. Therefore x y = 2 and the set we were looking for is: —
—
—
A = {(3,1), (4, 2), (5,3)(6,4), (7,5)(8, 6)1 . Problem V.21. The consequent of the sum of three consecutive prime numbers is also a prime number. Find the three prime numbers. Solution. Let pi < p2 < p3 be the three prime numbers. Then pi + P2 + P3 + 1 = p4 and p4 is also a prime number. Because p2,p3 and p4 are odd, we deduce that pi is even. pi being a prime number, we must have pi = 2. We obtain /32 = 3 and p3 = 5 and therefore p4 = 11. Problem V.22. There are drawn 16 smaller rectangles (named cells) in the next rectangle. We will say that two cells are neighbours if they have at least one common vertex or one common side. We colour in red, green, yellow and black the cells from the first row. Stain also in red, green, yellow and black the other cells such that on every row there are all the colours and that two neighbour cells have not got the same colours. Is there an unique way to do this?
Problems for the 5th form
21
Solution. We filled the second row this way: - Only black can be on the second position. - Only red can be on the third position. - Yellow can't be on the last position, so yellow is on the first cell of the row. - Green remains for the last position. We observe that we had a single choice of completing the second row. The same for the other rows. Problem V.23. Prove that three positive integers x, y, z exist, such that x2 + y2 + z2 = 4113 . Solution. We observe that 12 + 22 + 62 = 41. Then, we can choose x= 1 • 416, y = 2 • 416 and z = 6 • 416. We have x 2 + y2 + z2 = 4112 (12 + 22 , n2\= 4113, so a solution is: ( x =1. 416 y = 2 • 416 z = 6 • 416. Problem V.24. Consider the number N = ab(4) + bc(3)+i,(5). a) Prove that N is even if and only if a = 2. b) Find a,b,c for which N = 30. Solution. Because a, b, c are digits of numbers in the bases of the number systems 3,4,5 we must have aE
b E {1,2} and c E {1,2}
a) N= 4a+b+3b+c+5c+a=5a+4b+6c=a+2(2a+2b+3c). N is even <=> a is even a a = 2 (the only even number of {1,2,3}). b) Because N = 30 is an even number, we deduce a = 2. Then N = 5 2 + 4b + 6c •#> 20 = 4b + 6c. The last equality holds for the maximal values of b and c, i.e. b = c = 2. Problem V.25. Let A = {0,1,2,...,11,22,...,121,...,1991, ...} be the set of the natural numbers (written in increasing order) such that reading their digits from left to right or from right to left, we obtain the same number. Find the 2002-th number of the set A.
"Gazeta Matematice - a bridge over three centuries
22
Solution. We can write in the following way one n-digit number (n being a natural number, n > 3) in the set A: 1. If n is odd: We choose at random the first digit a(=the last digit) from the set {1, 2, ..., 9} . There are still n - 2 places of the number to fill in with digits. We choose at random the middle digit b from the set {0, 1, , 9}. Between the first digit, a and the middle digit b, there are n 3 places to 2 fill in. The same between b and the last digit, a. It is therefore enough to fill in the first group, of n 3 places, the other group of digits being 2 symmetrically disposed around the middle digit b. n On the whole, we can fill the 2 3 places with numbers from 00 0 to n 3digits 2
n-3
9 in 10-2-ways. Since have 9 possible values for n and 10 possible
99
n2 3digits
values for b, there are: n-3
n-1
9 • 10 • 10 2=9 • 10 2 n-digit numbers in the set A. 2. If 7/ is even: With the same reasoning, we deduce that there are 2 n-digit numbers in the set A. 9 We conclude: - there are 10 one-digit numbers in the set A; - there are 9 two-digit numbers in the set A; - there are 9.10Y = 90 three-digit numbers in the set A; - there are 9.102 = 90 four-digit numbers in the set A; - there are 9.102= 900 five-digit numbers in the set A; - there are 9.10 = 900 six-digit numbers in the set A; - the seven digit-numbers of the set A are: 1000001, 1001001, 1002001,... Therefore, there are 10+9+2.90+2.900 = 1999 numbers with less than 7 digits in the set A. The number we were looking for is the third seven-digit number from A, i.e. 1002001. 1L -
Problems for the 5th form.
List of authors. Problems for 5th form 1. Atanasiu I., V.14 (G.M. 84(1978), no.1) 2. Banu R., V.16 (G.M. 1/1983) 3. Bee D., V.13 (G.M. 9/1976) 4. Bostan Gh., V.11 (G.M. 5/1972) 5. Buicliu G., V.3 (G.M. VII/1901) 6. Ciorobatca M., V.6 (G.M. 8/1959) 7. Constantinescu D.,V.21 (G.M. 1/1995) 8. Constantinescu L., V.15 (G.M. 7/1981) 9. Craciun M., V.24 (G.M. 9/2002) 10. Ghica R., Ghita I., V.22 (G.M. 5-6/2000) 11. Grecu C., V.23 (G.M. 12/2001) 12. Ionescu I., V.1 (G.M. 1895) 13. Ivanov H., V.4 (G.M. XXII/1915) 14. Lamba S., V.8 (G.M. 2/1963) 15. Mocanu I., V.10 (G.M. 1967) 16. Nan C., V.19 (G.M. 2/1991) 17. Nicolescu N., V.2 (G.M. VI/1900) 18. Ottescu C., V.12 (G.M. 3/1975) 19. Paponiu D., V.18 (G.M. 2/1989) 20. Petre A. A., V.20 (G.M. 9-10-11-12/1992) 21. Sabau St., V.5 (G.M. 7-8/1987) 22. Sandulache C., V.9 (G.M. 3/1965) 23. Stretcu D., V.25 (G.M. 2/2003) 24. Stanica I., V.17 (G.M. 7/1984) 25. Tudor C. N., V.7 (G.M. 12/1962)
23
Chapter 2
Problems for the 6th form Problem VI.1. A number N with three digits is the square of an integer number n. If we interchange the last two digits of N, the square of n +1 will be obtained. Find the integer n. Solution. Let us denote N = abc. Then n2 = 100a+ 10b+ c and (n + 1)2 = 100a + 10c + b. It follows by subtraction that 2n + 1 = 9 (c — b) . On the other hand, N has three digits, therefore 10 < n +1 < 31. As a consequence, we get that 3 < c — b < 6 and, since c — b is odd, c — b = 3 or c — b = 5. Hence there are two possible values for n, namely n = 13 and n = 22, and only n = 13 is a solution to our problem. Problem VI.2. Determine the digits a and b such that the expression
E = 1931a1929 3 ±1929b1931 3 is divisible by 1930. Solution. We divide each number in the expression by 1930 and look for the remainder. Therefore we have the remainders Ei.= (1569 + 35a)3+ (361 + 35b)3 and using the formulas
x3± y3 = (x ± y)(x2 + xy + y2) we need to consider the following cases: 1) x + y = M1930. Obviously, x + y = 1930 + 35(a + b) and since a and b are digits, we conclude that the only solution is a = b = 0. 2) x — y = M1930. Then x — y = 1208 + 35(a — b) that is smaller than 1930 so it is not a solution. 25
"Gazeta Matematica," — a bridge over three centuries
26
3) x2 + y2— xy = M1930. Then
X2 +y2= M1930 + 92 + 175a + 211b + 125(a2 + b2) and xy = M1930 + 829 + 1055a + 875b + 1225ab Therefore we must have 1225(a2 + b2— ab)
—
737 — 880a — 664b = M1930
The division by 2 leads to the conclusion that a and b must be odd digits or a and b must have different parities. But in the expression E1 we can see that a and b have the same parity, consequently a and b must be odd digits. The division by 5 leads to the conclusion that 737 + 664b = M5 so that b = 7 is the only odd solution. The division by 193 leads, for b = 7, to the conclusion that 67a2+ 2a + 72 = M193 that has no solution. 4) x2 + y2+ xy M1930. Then, as in case 3), we conclude that we must have 1225(a2 + b2 +ab) + 921 + 1230a + 108Gb = M1930 Taking into account the expression of E1 the division by 2 leads to the conclusion that a and b must be odd digits. The division by 5 leads to the conclusion that 921 + 108Gb = M5 so that b = 9 is the only odd solution. For b = 9 the division by 193 leads to the conclusion that 67a2+96a-90 = M193 that has no solution. In conclusion, the only solutions are a = b = 0. Problem VI.3. Let p be a given odd prime number. Prove that the product of two integers which have the difference equal to p cannot be a perfect square but for a single value of the factors. Solution. Consider the two positive integers n and n + p. We observe that if p divides n then n(n + p) = p2k(k + 1) and it is not a perfect square. If (p, n) = 1 then (n, n + p) = 1 and n(n + p) is a perfect square if and only if = k2and n +p = 12. Then p = 12 k 2= (1 k)(1+ k). Since p is prime we —
—
(.23
have / — k = 1 or / + k = 1. If / — k = 1 then p = 2k + 1 and n =
1)2.
2
the same if / + k = 1. Problem VI.4. A pool is filled by two taps. The water flows at 15°C through the first tap and at 60°C through the second one. If both taps are opened then the water in the pool is at 30°C. The first tap alone fills the pool in one hour. Find the time to fill the pool if both taps are opened.
Problems for the 6th form
27
Solution. The technique to solve this problem is: From 30°C the first tap gains 15°C and the second tap loses 30°C. Then the water in the first tap flows two times faster than that in the second tap. It means that in a minute 1 1 1 both taps fill — + — = of the pool. Therefore the answer is 40 minutes. 60 120 40 Problem VI.5. Find the three digit number with the property
abc = ab + bc +-Ed. Solution. We observe that a + b = 10 because the last digit of ab + bc+, must be c. From 100a + 10b + c = 10a + b + 10b + c + 10c + a we obtain 11(a + b + c) = 100a + 10b + c. Therefore 11(10 + c) = 90a + 100 + c that means 10 + 10c = 90a, i.e., c + 1 = 9a. Thus we obtain a = 1, b = 9 and c = 8. The number is 198. Problem VI.6. Two points are moving at uniform speed on a circle. They leave a point A on the circle at the same time and go in opposite directions. After they meet at some point B, the first point needs 4 seconds to arrive at A, and the second point, by keeping its direction, needs 9 seconds to arrive at A. How many times would the circle be covered by each of these two points in one minute. Solution. We denote by T the time until the meeting that is the same for both points, and by V1 and V2 their speeds. Because the distance covered by the first point to the meeting place is the same as the distance covered by the second point from the meeting place to the point A, we have V1.T = V2.9. Also, the distance covered by the second point to the meeting place is the same as the distance covered by the first point from the meeting place to the point A. Therefore we have V2.T = V1.4. These two equalities are also written in the following form -T 9 , and V1 =T Vi V2 V2
T 9
=
r2 , =9•4
T = 6.
Thus, the first point covers the circle in 6 + 4 = 10 seconds that means it covers the circle six times in a minute. The second point covers the circle in 6 + 9 = 15 seconds and so it covers the circle four times in a minute. Problem VI.7. In a triangle ABC, the angleBAC is of 60°. Let S be the midpoint of the internal bisector AI, where I E (BC). Suppose that SBA is of 30°. Answer the following questions: i) What is the center of the circle defined by A, B and I? ii) Is AB equal to AI? iii) Are IC and BI equal? iv) Is BS the median of AC in the triangle ABC?
28
"Gazeta Matematica" - a bridge over three centuries
Solution. First we prove that ABC = 90°. Indeed, SAB = SBC = 30° implies that ISB = 60°, and SB = SA= SI. Then, the triangle SIB is equilateral. It follows that ABC = ABS + IBS = 90°. i) Since S is the midpoint of IA and SI = SB = SA, S is the center of the circle. ii) AB is strictly less than Al, because E3 = 90°. In fact, BI =1AI and AB = 2AI. iii) If IC = BI, then AI is a bisector and a median in AABC. Therefore, AC = AB, impossible. iv) The answer is negative. Let R be the intersection of BS and AC. Then BR±AC, because RBA = 30°, and RAB = 60°. If BR were the median of AC, then AB and BC would be equal. Problem VI.8. The ages of the father, son and grandson are prime numbers. Then after five years they become perfect squares natural numbers. What is the age of each of them ? Solution. Let us consider the sequence of the natural numbers that are perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ... If we reduce them by five we obtain the numbers -4, -1, 4, 11, 21, 31, 44, 59, 76, 96, 117, ... The only prime numbers in the last sequence are 11, 31 and 59. These are the ages we were looking for. Problem VI.9. Find all numbers that have three digits, namely abc, and satisfy abc = a + b2+ c3. Solution. Consider a number satisfying abc = a + b2 +c3. Since abc = 100a+10b+c, it results that a, b, c are connected by 99a+10b-b2+c = c3. The left term is greater than 100, therefore c> 5. We would like to study all the cases c E {5, 6, 7, 8, 9} . First, we notice that b (10 - b) E {0, 9,16, 21, 24, 25} and each of these values, but for the value 0, corresponds to two values of b. For instance, b (10 - b) = 9 at b = 1 or b = 9. For c = 5 we get the equation 99a + b (10 - b) = 120. Then a = 1 and b (10 - b) = 21, that is b = 7 or b = 3. Hence 135 and 175 are two solutions. At c = 6 we have 99a + b (10 - b) = 210. Then, taking into account the possible values of b (10 - b), it results a = 2. We obtain no solutions in this case because
-
29
Problems for the 6th form
we cannot have b (10 - b) = 12. Furthermore, as above, for c = 7, a must be 3 and b must satisfy b (10 - b) = 39, impossible. For c = 8 we get a = 5, b (10 - b) = 9 thus b E 11,91 . Other two solutions to our problem are 518, 598. For c = 9 there are no solutions. Our conclusion is that the solutions are {135, 175, 518, 598} . Problem VI.10. Find a three digits number which is divisible by 22, divided to 5 gives the remainder 2, and the hundreds digit is 4 greater than the units digit. Solution. The number is of the form abc; divided to 5 gives the remainder 2, so c E {2, 7}. Since abc is divisible by 22, it is an even number. So c = 2. It results that a = c + 4 = 6. The three digits multiples of 22 hawing the hundreds digit 6 are: 28.22 = 616, 29.22 = 638, 30.22 = 660, 31 . 22 = 682. The unique number with the required properties is 682. Problem VI.11. Prove that there is an infinity of natural numbers xo, YO) zo, to roots of the equation x2 +y3+ z4 = Solution. Let x = 3k, y = 3P, z = 31 and t = 38. Then the equation becomes 32k + 33P + 341 = 3118. If 2k = 3p = 4/ = lls - 1 we obtain a solution of this equation. It means that lls -1 = 12u and this equation has an infinity of roots when s = 12v + 11 and u = llv + 10. Problem VI.12. We consider 1000 nonzero natural numbers with the property "the sum of their inverses is greater than or equal to 10". Then at least two of these numbers are equal. Solution. Suppose that the numbers are all distinct. Then the greatest sum 1 of their inverses is obtained for the numbers 1,2,... 1000. Denote S = 1+ 2+
31
1 1 1 1 1 1 1 1 1 .ButS<1-F 2 + 2 + 4 + 4 + 74 + 71 +•••+ 2 +•-•+ 29 = +•••± 700 =1
= 10.
-1
-1
Problem VI.13. Prove that 1998 divides the number 19971997 +19991999 . Solution. We use the following identities: xn - 1 =(x-1)(xn-l +xn-2 +•••+x+1), nEN x2n-I-1 —
(X
+ 1) (X271— x291-1
X
+ 1)
n E N.
"Gazeta Matematica"
30
-
a bridge over three centuries
Then we have: 19991999- 1 = 1998 (19991998+ • • + 1999 + 1) , 19971997 + 1 = 1998 (19971996 - 19971995+ • • • - 1997 + 1) . We obtain that 1998 divides the number (19971997+ 1) + (19991999 - 1) = 19971997 + 19991999. Problem VI.14. Let p be a prime number greater than 3. Find the remainder of the division of p2 by 12. Solution. Let us write p as p = 12m + n, where m,n E N and 0 < n < 12. Since p is prime, it follows that n E {1, 5, 7,11} . For instance, we cannot have n = 3 because in this case p = 12m + 3 is not prime, being divisible by 3. We clearly have that p2= 12M + n2, where n2 E {1, 25, 49,121} . Now it is sufficient to notice that n2- 1 is divisible by 12, so that the remainder of the division of p2by 12 is always 1. Problem VI.15. Solve in the set of natural numbers the equation:
1 • 2 • 3. • •n + 241 = m2. Solution. Suppose n E {1,2, • • • , 6}. The only solutions are n = 5, m = 19 and n = 6, m = 31. Consider now n > 7. In this case, we have 1 • 2 • 3 • • • 7 • • • n + 241 = M7 + 3. But any square of a natural number has one of the next forms: M7, M7 + 1, M7 + 2, M7 + 4. It follows that the equation has only two solutions. Problem VI.16. Let AOB = 150° and C the point of intersection of the midperpendiculars of OA and OB. Show that ABC is an equilateral triangle. Solution. Since C belongs to the midperpendicular of AO, we have AC = 00. Similarly, OC = BC and hence AC = BC. The midperpendiculars of OA and OB are the bisectors of ACO and OCB, respectively. Therefore
ACB = 2 (21 OCB + ACO) = 2 • 30°= 60° , since the quadrilateral determined by OA, OB and the two midperpendiculars is circumscriptible, that is,
AOB +
1 -
2
OCB +
1 -
2
ACO = 180° .
Problem VI.17. Write 2003 as a sum of successive integers.
Problems for the 6th form
31
Solution. Since 2003 is an odd number it will be the sum of an odd number of successive integers. If we denote the integer in the middle of these numbers by a, then the successive numbers are a — n, a — n + 1, ..., a — 1, a,a + 1, ..., a + n and their sum is (2n + 1) • a. Therefore we have (2n + 1) • a = 2003 and since 2003 is a prime number the only non trivial solution is obtained for 2n + 1 = 2003 and a = 1, that is a = 1, n = 1001. The successive integer numbers required in the problem are —1000, —999, ... , —1, 0, 1, . , 1002 . Problem VI.18. Let ABCD be a parallelogram. Denote by M the midpoint of [AD] and by P the projection of B on CM. Prove that AP = AB. Solution. B
Let S be the intersection of AB and CM. The triangles AMS and CDM are congruent: AM = MD, ZSMA--=. LCMD and LASM LMCD. To prove that LASM = LMCD, we used the fact that SBIICD. Therefore AS HE CD. In the right angle triangle SBP, PA is the median drawn from P. Hence
1 PA= — SB = AB . 2 Problem VI.19. Let ABC be a triangle with m(g) = 40° and m(C) = 30°. On the segment (BC) we take a point D such that m(BAD) = 60°. Show that [AB] [CD]. Solution. We are given that:
m(ABC) = 40°, m(ACB) = 30°, m(BAD) = 60° , so we find that: m(DAC) = 50°, m(ADB) = 80°.
(1)
"Gazeta Matematicr — a bridge over three centuries
32
We draw AA' and CC' perpendicular on AC at A and C, respectively, such that AA' = CC' = AB, and thus: a)
AABA' is isosceles;
b) ACC'A'
(2)
is a rectangle.
First we are going to prove that ADA'C' is congruent to ADAC. 1. By construction, m(A'AC) = 90° and from (1), m(DAC) = 50°, therefore m(A'AD) = 40°. But m(BAD) = 60°, so rn(BAA') = 20° and, considering the fact that AABA' is isosceles (also by construction, see (2a)), we obtain that
m(ABA') = m(AA'B) = 80° .
(3)
On the other hand, in AABD we have m(BAD) = 60° and m(ABD)= 40°, so that (4) m(BDA) = 80° . From (3) and (4) we obtain that ABA'D is inscribed so
m(ABD) = m(AA'D) = 40° .
(5)
But m(AA'C') = 90°, as AA'C'C is a rectangle, therefore m(DA'C') = 50°. We know that m(DAC)= 50°, so we obtain that:
DA'C' DAC .
(6)
2. In (5) we obtained that m(AA'D) = 40°, and we also know that m(A'AD) = 40°, so AAA'D is isosceles. Consequently,
AD
D
(7)
Problems for the 6th form
33
3. As AA'C'C is a rectangle, we have that
A'C' = AC .
(8)
From (6), (7) and (8) we obtain that the triangles LDA'C' and LDAC are congruent, having two sides and the angle between them respectively congruent. This implies that CD = C'D, so LCC'D is isosceles. But m(ACD) = 30° and we have drawn CC' I AC, so m(DCC') = 60°. Therefore ACC'D is equilateral and, consequently, CC' = CD. As CC' = AB by construction, this implies that
CD = AB .
34
"Gazeta Matematice - a bridge over three centuries
List of authors. Problems for 6th form 1. Acu D., VI.5 (G.M. 8/1966) 2. 13516,ucbi A., VI.15 (G.M. 10/2002) 3. Branzei D., VI.4 (G.M. 8/1965) 4. Bogdan I., VI.16 (G.M. 8/1986) 5. Carbunaru C., VI.7 (G.M. 5/1974) 6. Chirobocea S., VI.14 (G.M. 10/2000) 7. Duinitru A., VI.8 (G.M. 9/1980) 8. Ghitg R. and Ghita I., VI.17 (G.M. 2/2004) 9. Grigorescu N., VI.19 (G.M. 2/2004) 10. Ionescu M., VI.12 (G.M. 8/1991) 11. Ligor I.C., VI.6 (GM 12/1973) 12. Lintel I., VI.2 (G.M. 5/1935) 13. Moraru Gh., VI.9 (G.M. 12/1980) 14. Odobescu A., VI.1 (G.M. 1909) 15. Patra§cu I., VI.18 (G.M. 5-6/2003) 16. Paran L., VI.11 (G.M. 8-9/1990) 17. Safta I., VI.13 (G.M. 3/1998) 18. Sergescu P., VI.3 (G.M. 1936) 19. Tena M., VI.10 (G.M. 5/1971)
Chapter 3
Problems for the 7th form Problem VII.1. On the side AB of an arbitrary triangle ABC we take a mobile point P. Let's have N the midpoint of the segment CP. Through the points A and B we draw parallels to the lines BN and AN, respectively. Find the locus of the point 11/I which is the intersection of the above lines. Solution. The quadrilateral ANBM is a parallelogram and M is the symmetrical of the point N with respect to 0 the middle of the segment AB. The point N describes the middle line [AiBi], Al E AC, B1 E CB, so M will describe [AcBc. ] the symmetrical of [A1 B1] with respect to 0. -
-
Problem VII.2. Find a number of three distinct digits, equal to the sum of all two-digit numbers that can be formed by the digits of the number. Solution. Let's take abc the sought number. The two-digit numbers that can be formed are:
To,
a,
cb.
It results:
abc = lla + 11b+ 11c+ (10a + b) + (10b+ + (10c+ + + (10a + c) + (10b + c) + (10c + = 33(a + b + c). So 33Iabc= 31(a + b + c). Then 9Iabc 9I(a + b + .ci,+b+cE {9,18,27}. It results abc E {33.9, 33.18, 33 • 27} = {297, 594, 891}. Only 594 verifies the hypothesis in the problem. Problem VII.3. Find two prime numbers so that the difference between their squares should equal the square of their arithmetical mean. 35
"Gazeta Matematica," — a bridge over three centuries
36
Solution. Let's denote required a, b the numbers. We have
a2
—
b2 = (a + 2
ab=
a+b b a =3 LEN 2 3
/b,
that is, b = 3k. But b is a prime number so k = 1. It results b = 3, a = 5. Problem VII.4. We have the triangle MAB, C E (MA, with MC = pMA, D E (MB, with MD = qMB, where p and q are strictly positive real numbers. Show that the line that links M to the midpoint of [CD] cuts the line AB at a fixed point whichever the position of M towards A and B. Solution. We'll only analyze the case C E (MA), D E (MB). Let's have AS II CD, S E MB. We denote by N the midpoint of (CD) and {P} = MN fl AB, {R} = MN fl AS. From A/14-CD ti AMAS we have MS MA MS 1 = . or MD MC MD p
In the triangle ASB tutted by the transversal M P R we have by Manelaos theorem: PA MB RS -= PB MS RA —
—
PA MS p But RA = RS so — = = const. PB MB q Then P divides the segment (AB) on a constant ratio, so it is fixed. Analogously we proceed for other positions of the point C and D on the semi-line (MA, respectively (MB. Problem VII.5. In a parallelogram ABCD, the perpendicular drawn on the side AB through its midpoint I, cuts the diagonals BD, CD at M and N. respectively. Let's have OH the perpendicular drawn from 0 - the intersection of the diagonals of the parallelogram on AB. Show that: 1 1 2 . IM IN OH
37
Problems for the 7th form
Solution. From the similarity of the triangles AIN and AHO; BHO and HO AH HO HB and = , respectively. BIM, we have = I IN Summing up the two relations and taking into account that I is the midpoint of (AB), we obtain the required relation. Note. The relation in the problem expresses the fact that OH is the harmonic mean of IM and IN. —
—
/M
/B
Problem VII.6. Let's have D the foot of the interior bisector of the angle A in the triangle ABC; M and N be the projections of D on the altitudes BB' and CC' and P the foot of the simedian AP. Show that:
MC _ MB PB Solution. Denote by Q and R, the projections of P on the sides AC, AB. Knowing that the distances from an arbitrary point of the simedian to the PQ AC adjacent sides are on a ratio with these ones, we have — = PR AB• DC AC From the bisector theorem we have = It results: DB AB
PQ DC PR DB
(1)
But ADNC ti ABRP and ABMD ti APDC. It results:
_
NC PR BP
MB DB and PQ PC
Dividing these relations and using (1) we obtain
NC PC = MB PB
—
Problem VII.7. Demonstrate geometrically, that we have 1
"Gazeta Matematica" — a bridge over three centuries
38
A'B _ AB Let's A'C AC • NB AB have BM the median from B and {N} = BM fl AA'. Then = NM AM NB A'B AB Using the hypothesis, it results NM = 2AC. In AABB', [AA'] being the bisector of the angle A we have Solution. AA' being the bisector of the angle A we have
I B AB IB' AB'
( 1)
and [BB' the bisector of the angle B:
B' A AB B'C CB
B'A AB AC AB + CD
or
(2)
Replacing in (1) the expression of B'A from (2) we obtain the relation in the problem. Problem VII.9. If a, b, c are positive rational numbers, with a c b and = VT + VC, then N/a, are on a ratio with rational numbers. Solution. We have VT) — by squaring it results a + b a+b 2a
—
c
= —
b c Vb +
b
—
—
2.\,/a = c. We obtain: VT, =
— N/T)
N7C,
a+b c 2/a —
b + c = vi:ta — b + c 2a 21ci,
—
We cannot have a + b — c = 0 <=> b c = a <=> together with fa = -VT+ would imply b = 0. —
2a
From
• Analogously
=a
So — =
c
a
—
16 — NiJ =
which
=
a+b—c a—b+c
Problem VII.10. The orthogonal projection of the intersection point of the diagonals of an inscribing quadrilateral on its sides divide these sides in ratios whose product is equal to the unity. Solution. Using the annotations in the figure we have:
Problems for the 7th form
AANO
ADQO
39
AN_ 73
y
DO BP BO ABPO N AAMO =Az„ 1A AO CQ CO Tv= = ACQO ABNO NB BO ADMO ,ACP()
DM DO = CP CO
By multiplying these relations we have:
AN PB CQ DM = 1. DQ MA NB MA
Problem VII.11. Determine a number of the form abba, knowing that in any base of a numerical system it is a perfect cube. Solution. Let's have x the base of a numerical system into which the given number N is written. It results:
N = ax3+ bx2+ bx + a = a(x +1)(x2— x +1) + bx(x +1) = (x + 1) [ax2+ (b — a)x + . As N must be a cube in any base, it results a,x2+ (b (V) x E N, x > 2. We obtain a = 1 and b a = 2. The sought number is N = 1331.
-
a)x + a = (x + 1)2,
-
Problem VII.12. Irrespective of the base of a numerical system, there is no number of n> 2 digits equal to the product of its digits. Solution. Let's have m the base of a numerical system and N = x1x2 • • • xn, the number written in the base m. We must have:
N = ximn—i
x2mn-2 + • • • + xn—im + xn
N
(1)
(2)
= xi • x2 • • • • • Xn
From (1) it results N > xi.m"-1, and from (2) N < xi•m•m • • • m = X1M71-11 which is impossible. Problem VII.13. Show that: 1 3 2 • • 2 4 B = 7. 3 • 5• A =
5 66 7 7 -
899 900 898 899
is smaller than is bigger than
1 —
30 1 — 30
"Gazeta Matematice — a bridge over three centuries
40 Solution.
1 • 2 • 3 • 4 • • • 898 • 8991 = (1) 2 • 3 - 4 • • . 898.899.900 900 1 2 3 4 897 898 1 3 5 897 2 4 898 899 , we have • zi • As < 5, 6 • • • 898 < 5 • g 4 < 5 ' • • • ' 898 < 899' 900 900 that is — A < B. As A < — A it results that 899 899 ' A-B=
A < B. From (1) and (2) we obtain A <
1
and B >
(2) 1
Problem VII.14. On the sides AC and BC of the equilateral triangle ABC a with length side a we consider the points N and P so that AN = BP = 3 . Let's consider M the midpoint of AB. Show that:
a) AMPN is a trapezium; b) AP = 2 MN; area(AMPN)7 = c) area(ABC) 88 ' d) compute the perimeter of the trapezium AMPN Solution.
AN BP = =1-. a) We have AC BC 3 Using the reciprocal to the Thales theorem it results that NP AB. b) Given NN' I AB, N' E AB. Then m (ANN') = 30°. N'
M
B
1 1 a aid a a a It results AN' = 2 AN = NN' = 3CM = ; N' M = –= 6" 6 2 6 3• of In ANN/ NI : NM = VN/N 2+.1\17 M 2 = . 6 2a-1 7In ANN'B : NB = VAT'Ar 2+ N'B2 = . 6 But NB = AP. We obtain AP = 2MN. c) area(AMPN) + NP) • NN' (AM NP) NN' = + area(ABC) AB • MC AB AB MC (1 ± 2) 1 7 2 3 3 18.
_
Problems for the 7th form
41
a alt: 2a a 9 + Ir7a. + + + = 2 3 3 6 Problem VII.15. Show that if the sides of the triangle ABC satisfy the c — b a — c b —a = 0 then AABC is isosceles. relation: + + a b c Solution. We have: c — b a—c b—a = 0 a bc(c — ac(a — c) + ab(b — = 0 .#> a bc2— b2c + a2 c — ac2 + ab2— a2b = 0 4#. b(C2— a2) + ac(a — c) + b2(a — = 0 a (a — c)(b — a)(b — = 0 <=> a s= c or a= b or b= c AABC isosceles. d) AM+11/1P+PN+NA=
Problem VII.16. Suppose ABCD is a rhombus. Through the vertex A we draw an arbitrary line that intersects BC at E, CD at F and the diagonal BD at G. Show that the line GC is tangent at C to the circumcircle of the triangle ECF. Solution. Let's have {T} = AB fl GC. The triangles GBC and GBA are congruent as GB is a common side: (BC) -a (BA) and LABG ZGBC. From AGBC AGBA LGCB LBAG; but LBAG LGFC, so 1 m(GCB) = m(GFC) = — m(EC). 2 Note. The conclusion also holds for G on the prolongation of DB. Problem VII.17. Find the regular polygons that have all the diagonals equal to each other. Solution. The equilateral triangle does not have diagonals. The square has equal diagonals. The regular pentagon A1A2A3A4A5 has all the diagonals equal to each other, as from every vertex we can draw only two diagonals which are equal, as they are synnnetrically placed with the diameter of the circle circumscribed to the triangle that passes through the respective vertex. If we have a regular polygon A1A2 A„ with n > 5, from vertex A we can draw the diagonals AlA3, A1A4, , A1A„_2, that cannot be all equals, as from a point on a circle we can only draw 2 equal cords at the outmost. Thus the required polygon are the square and the regular pentagon.
42
"Gazeta MatematicA" — a bridge over three centuries
Problem VII.18. In the isosceles triangle ABC(AB = AC), the ratio between the lengths of the sides AC and BC is equal to 2.5. Show that the gravity centre of the triangle belongs to the inscribed circle. Solution. Let's take A' the midpoint of BC, G - the gravity center, I - the center of the incircle of A ABC. The triangle ABC being isosceles, the points A, G, I, A' are collinear. Then: SABC
I A' = PABC
a • AA' 5 5 a + —a ± —a 2 2
AA' , 6
where
BC = a ,
AA' AA' = the radius of the and GA' = — . So I E (GA') and IG = IA' = 6 3 incircle of ABC. (PABc,denotes the semiperimeter of ABC). Problem VII.19. Solve in the set of integer numbers the equation: 1 1 1 2x + 3y 4 1 1 1 Solution. From — -1- — = 7 Gy+4x = 3xy = 4(x-2) = 3y(x-2)-8 2x 3y 4 (x — 2)(3y — 4) = 8. We have the cases: {x— 2 = 1 x=3 1) {y=4 3y — 4 = 8 {x — 2 = —1 2)
3)
y=
3y — 4 - = —8
{ x -2 = 4 3y — 4 = 2 { x — 2 = —4
4)
1 x=1
3y — 4 = —2
_
-Th4 V Z
{ lx = 6 y=2 (
x = —2 2
l Y— -3-V
z
So the solutions are: (x = 2, y = 4) and (x = 6, y = 2). Problem VII.20. A rhombus of side 4/ has acute angles of 30°. There are 33 interior points given, any three of them not collinear. Prove that there are three points that form a triangle whose area is smaller than 1.
43
Problems for the 7th form Solution. rhombus of side AB = Let's have ABCD the_ 4.\/ and m(A) = m(C) = 30°. The area of the rhombus will be AD • AB sin 30° = 16. Lets have M, N, P, Q the midpoints of the sides AB,BC,CD,DA and {O} = AC n BD.
Unifying the points M, N, P, Q, 0 we obtain 16 right congruent triangles of area 1. Having 33 interior points, there exists a triangle containing three points (interior or on the sides) because if all the triangles contained 2 points at the outmost there would be in all 16.2 = 32 points, which is a contradiction. It results that these 3 points form a triangle of area smaller than the area of the triangle that contains it, that is smaller than 1. Problem VII.21. There are given the points A, B, C, D so that the lines AC, BD intersect at I. Let's have M, respectively N the midpoints of the segMents (AB), respectively (CD). Show that the points I, M, N are collinear if and only if AB II CD. Solution. The implication AB II CD I, M, N collinear is known: in a trapezium the midpoints of the bases, the intersection point of the diagonals and the intersection point of the sides that are not parallel are four collinear points. To prove the remaining implication: let's have I, M, N collinear. We can discern the following situations: a) MN n AD = {G}. As M, S, G are collinear, from the Menelaus theorem in ,LABD we have: MB GA ID NIA GD IB
ID GD = IB GA .
( 1)
Analogously, from I, N, G collinear in ACAD IC GA ND IC G D — • — — = I, so — IA GD NC IA GA . ID IC From (1) and (2) it results that -1T3= IA
(2)
44
"Gazeta Matematicg" — a bridge over three centuries
As the angles CID FE AID (opposite at the their vertices), it results that LCID AAIB, so BAC F.. DCA, that is AB II CD. b) MN n AD = 0; MN n BC = {G}, analogous to a); c) MN n AD =0 and MN n BC = 0 is obvious. Problem VII.22. Determine the smallest positive integer a so that 71997 -a is divisible by 1000. Solution. As 75 = 16807 = M1000 + 807; 710 = (75) 2 =M1000 + 651249 = M1000 + 249; 720 = (710) 2 = M1000 + 2492 = Mi000 + 62001 = M1000 + 1,
we have: 71997
=
720.99+15+2 = (720) 99 715 72 = (M1000 + 1) 99 • (75) 3 •72 = (M1000 + 1) (Ml000
+ 807) 3 •49 =
(Ml000 + 1) (Mi000 + 943) • 49 = M1000 + 943 • 49 = M1000 + 207. It results a = 207. Problem VII.23. In a cartesian system xOy from the point (a, b) it is permitted the displacement to one of the following four points: (a —1,b —1), (a —1,b +1), (a +1,b —1), (a +1,b +1). Show that from the point (0, 0) it is impossible to get by successive displacements to the point (1997,1998). Solution. Let's notice that a point (a, b) having the sum of its co-ordinates a + b, an even number, can displace itself to a point having the sum of its co-ordinates also even. 1) (a— 1,b— 1):a—l+b— 1=a+b— 2 2) (a— 1,b+1):a-1+b+1 =a+b 3)(a+1,b-1):a+1+b-1=a+b 4) (a+ 1,b+1):a+1+b+1 =a+b+ 2. As the point (0, 0) has the sum of its co-ordinates 0, which is even, it can displace itself successively to points of the same parity of the sum of the co-ordinates, so it cannot displace to the point (1997,1998) whose sum of co-ordinates is an odd number.
45
Problems for the 7th form
Problem VII.24. Determine the triangles circumscribed to the circle of radius 1 that have the altitudes expressed by numbers in the set of positive integers. Solution. Let's have ha < hb < '1,, the altitudes of the triangle of sides a, b, c and r = 1 the radius of the inscribed circle. We have: 1 1 1 a b c a-l-b+c ha +hb + ha aha + bhb + clic = 2S
2p 1 2S — -77. = 1.
Then: (ha, hb, hc) = (3, 3, 3) or (ha, hb, he) = (2, 4, 4) or (ha , hb, he) = (2, 3, 6).
46
"Gazeta Matematica" - a bridge over three centuries
List of authors. Problems for 7th form 1. Acu D., VII.17 (G.M. 1/1977) 2. Abramescu N., VII.4 (G.M. 18/1913) 3. Bogdan I., VII.19 (G.M. 2/1983) 4. Buicliu Gh., VII.2 (G.M. XVI/1910) 5. Bunescu A. D., VII.3 (G.M. XVII/1911) 6. Branzei D., VII.22 (G.M. 9/1994) 7. Calistru C., VII.25 (G.M. 9/2002) 8. Cocea Th. Gh., VII.11 (G.M. 11/1956) 9. Constantinescu L., VII.15 (G.M. 10/1974) 10. Cornea V., VII.23 (G.M. 2/1997) 11. Cristescu V., VII.1 (G.M. 1/1895), VII.6 (G.M. 26/1921) 12. Galbura Gh., VII.10 (G.M. 4/1950) 13. Ionescu I., VII.8 (G.M. XLVII/1941) 14. Manea M., VII.24 (G.M. 9/1999) 15. Manescu L., VII.12 (G.M. 3/1959) 16. Niculescu M., VII.5 (G.M. 25/1920) 17. Stainica, L., VII.20 (G.M. 9/1925) 18. Stoica G., VII.16 (G.M. 12/1976) 19. Teler M., VII.21 (G.M. 2/1990) 20. Tutescu L., VII.18 (G.M. 5/1977) 21. Tiu C. I., VII.14 (G.M. 9/1974) 22. 'Vino 0. M., VII.9 (G.M. 8/1942) 23. Vasiliu Gh., VII.7 (G.M. 7/1935) 24. Voiculescu D., VII.13 (G.M. 6/1965).
Chapter 4
Problems for the 8th form Problem VIII.1. Find all pairs of quadratic numbers with three digits that differ from each other only by the first digit. Solution. Let x and y be two solutions, and let d be the difference of the first digits of x and y, respectively. Then there exist a and b such that a2 = x and b2= y. It follows that 10 < a < 31, 10 < b < 31 and 1 < d < 8. Moreover, 100d = x — y = a2 — b2 = (a — b) (a + b). Then a and b have the same parity. Let us suppose first that a and b are both even, a = 2n and b = 2m. Then n and m must satisfy: 5 < n < 15, 5 < m < 15 and (n — m) (n + m) = 25d. Since 0 < n — m < 10 and 10 < n + m < 30, it results that we have two possibilities: i) 51(n + m) and 51(n — in) , then 51n and 51m . Hence n E {10, 15} and m E {5,10} . It is easy to see that (20, 10) , (30, 20) and (30, 10) are the only solutions (in this case) to our problem. Here, the first number in the parentheses is a and the second is b. ii) 251(n + m) , then n + m = 25. Taking into account the upper bounds for n and m, it follows that we have two more solutions, namely (26, 24) and (28, 22). Suppose now that a and b are both odd, a = 2n+1 and b = 2m +1. Then n and m must satisfy: (n + m + 1) (n — m) = 25d, where 11 < n+m+1 < 31 and 0 < n — m < 10. One can make the same reasoning as above and obtain four more solutions: (25, 15) , (31,19) , (29, 21) and (27, 23) . Al together, we have 9 pairs of solutions. Problem VIII.2. Consider the integers a, b, c so that (a, b, c) = 1. Let ma = [b, c] and da = (a,ma). We define in the same manner db and de. Prove that the product dadbdcis a perfect square. Solution. Let xa = (b, c), xb = (a, c) and x, = (a, b). Then (xa, xb, xc) = 1 and so a = xbxca, b = xa xcl3 and c = xaxb'y and we have (a, 0,7) = 1. 47
48
"Gazeta Matematicr — a bridge over three centuries
Consequently da =(a, Mai = (xbxca, [xexa0, XbXa7]) = XhXc. Then dadbdc = xa2 Problem VIII.3. Consider a quadrilateral formed by the sides of a triangle and by a Simpson line of this triangle. Prove that the line of Gauss for this quadrilateral is perpendicular to the considered Simpson's line and, in the same time, it is a Simpson's line for median triangle of the initial triangle. Solution. It is well known that the line of Gauss for this quadrilateral is perpendicular to the line of orthocenters of the triangles formed by the sides of the quadrilateral. Now, we must prove that this line of the orthocenters of a quadrilateral formed by the sides of a triangle and a Simpson's line of it, is parallel to this Simpson's line. Consider LABC an arbitrary triangle, P a point on the circumcircle of AABC, A1, B1 and C1 the projections of P on the sides BC, CA and AB, respectively, AA' B'C' the median triangle of LABC, H the orthocenter of AABC and Ha, Hb, Hethe orthocenters of the triangles AABiCi, ABC]. Ai , ACA1B1i respectively. We must prove that the lines AiBiCi and HaHbli, are parallel. It is known that the Simpson's line with respect to the previous point P passes through the middle of the side that links point P with orthocenter of AABC. But, AiBiCi is the Simpson's line of P with regard to each of the ABC, ABiCi, BC1A1, CA1B1 triangles. Then it passes through the middle of the segments PH, PHa, P Hb, PH, such that it is parallel to HHa HbHc. Let L, M, N be the midpoints of the segments AA1, BB1, CCII, respectively. The line LMN is the Gauss's line of a quadrilateral formed by AABC and the line AiBiCi. It is obvious that points L, M, N are on B'C', C' A', A'B' of the median triangle A'B'C' and the perpendicular lines in L, M, N on B'C', C'A', A'B' are concurrent at the point P' which is the midpoint of the segment PH. Therefore, LMN is the Simpson's line of the point P' with respect to the median triangle A' B' . Problem VIII.4. If x, y, z are three real numbers satisfying xy+y z+xz > 0, then the following inequality holds: (x + y + z) (x3 +y3 + z3 +xyz) > x4 + y4 + z4. Solution. By direct calculations we have:
(x + y + z) (x3 + y3 + z3 + xyz)= x4 + y4 , _z 4 xy3+ xz3 +x2yz + +yx3 +yz3+ xy2z + zx3 + zy3+ xyz2 =x4 + y4 + z4 + + (xy + yz + zy) (x2 + y2 + z2) > x4 + y4 + z4
49
Problems for the 8th form Problem VIII.5. Prove that if a2 + b2 + c2 = ab + bc + ca then
(a — b + x)(a — b) + (b — c + x)(b — c) + (c — a + x)(c — a) = 0 Vx E R. (a — b)2+ (b — c)2 + (c_ a)2 = 0. Solution. a2 + b2 + c2 = ab + bc + ca The relationship we are asked for becomes E= (a — b)2+ (b — c)2+ (c — a)2+ x (a —b+ b— c+c— a) and so E= 0, V x E R. Problem VIII.6. Let ABCD be a tetrahedron and P some inside point. The straight lines AP, BP, CP, DP intersect the opposite faces at A', B', C' , D'. Prove that PA' PB' PC' PD' + + + =1 AA' BB' CC' DD' Solution. We denote by hA the altitude of the tetrahedron ABCD drawn from A and by hA, the altitude of the tetrahedron PBCD drawn from P. Then we have PA' PB' VolpAcD hA, VolpscD -, . .Similarly we obtain __, = _ , v otABcp , AA' .t5.51 hA VOIABCD PC' Vo/PBcA = Vo/PBAD and PD' = . CC' V ol ABC D DD' V 01 ABC D By summing these relations we obtain the conclusion. Problem VIII.7. Consider a piece that has the shape of a circular right cone with the length of the radius of the base equal to 2 m and with the height equal to 2N/-- m. We cut from this piece an object with the maximum volume with the shape of a cube with one face on the base of the cone. Prove that in this way we use less than a quarter from the material. Solution. Let us denote by x the side of the cube and observe that the diameter of the horizontal section is the diagonal of a face, so it is x/2-. From the similitude of the triangles resulted on an axial section, we obtain x.‘/ 2.1 — x = ,-4 2.V2
x fj . — =1 2
x = N/2.
The volume of the cube is therefore Vi. = x3 = 2N/2 and the volume of the cone is V2 = 87/273. We get VI
— = V2
2.1.3 87r/2-
=
3 47r
<
3 12.56
VI
<
V2 —,-. 4
50
"Gazeta Matematica" — a bridge over three centuries
Problem VIII.8. Let ABCD be a parallelogram, 0 the point of intersection of its diagonals and S a point of the orthogonal line at A on the plane of the parallelogram. We denote by M, N, and P the projections of S on the straight lines BC, CD and DB, respectively. Show that the points M, N, 0 and P belong to the same circle. Solution. If the plane of ABCD is denoted by it, then from SA I 7r, BC C 7 and SM 1 BC, it follows AM 1 BC. Similarly, AP 1 BD and AN 1CD.
N
D
We observe that APDN and AMCN are inscribed quadrilaterals. Thus, m(NPA) = m(NDA) = 90° — m(NAD) and we obtain m(NPO) = 90° + m(NPA) = 180° — m(NAD). Also m(CAB) = rn(ACN) = m(AMN). In the triangle AMC we have m(AMC) = 90° and IA01 = IOC'. It follows that m(OAM) = m(OM A). Therefore m(OMN) = m(BAM). From NA 1 AB and DA 1 AM we find that m(NAD) = m(BAM). Hence it results m(NPO) + m(OMN) = 180°. So the quadrilateral OPNM is an inscribed quadrilateral. Problem VIII.9. Let ABCD be an arbitrary tetrahedron. We denote by E and F the projections of the point A on the bisectors of the angles ABC and ABD, respectively. Prove that the line EF is parallel to the plane BCD. Solution. We denote by M the midpoint of the edge AB. In the rectangular triangle LAEB, ME is median, therefore we have ME = MB. Then AEBM is isosceles, so that MEB = MBE. But, by hypothesis, MBE = EBC and then MEB = EBC. Therefore we have ME II BC and, consequently, ME II (C BD). Similarly, we obtain M F II (BCD), such that we deduce that (MEF) II (BCD). But EF belongs to the plane MEF so that EF II (BCD). Problem VIII.10. Solve in Z the equation:
x(3x + 1) + 2y(y — 2x) — 2 = 0.
51
Problems for the 8t' Solution. The equation is equivalent to: 2(x — y)2 = —(x — 1)(x + 2), x, y E Z.
= Since 2(x —y)2 > 0, we obtain (x —1)(x +2) < 0. Therefore, x E [ 2, {-2, —1,0, 11. The substitution of these values in the equation gives four corresponding equations in y. Solving these equations, we get the following solutions in Z x Z: -
(-2, —2), (-1, —2), (-1,0), (0,-1), (0,1), (1,1). Problem VIII.11. Let M be some point inside the tetrahedron AiA2A3A4The straight lines AiM , A2M , A3M , A4M intersect the opposite face at A'1, A'2, A'3, A'4, respectively. Prove the relationship: MA' 1 MA'2 MA/3 MA'4 4 MAi ± MA2 + MA3 4-MA4 ?. 3 • Solution. Let us consider the tetrahedrons AiA2A3A4 and MA2A3A4. We M h' Vi =— hi = — where h1is the altitude from Al in AiA2A3A4, have V is the altitude from Min MA2A3A4, V and Vi are the volumes of the two MA'1_V1_V—T1 tetrahedrons. Denote by T1 = V — V1. We obtain: MA i T1 MA'1 Using similar notations for the vertices A2, A3, A4, then the ratio M Ai can be expressed as: + T2 + T3 + T4 V —T1 3 T1 1 1 ( T2 T3 T4 + + + -) 1 3 3 Ti Ti
MA'1 M Ai
T1 T1 ±
T2 + T3 + T4 - 371 3T1
MA'14 1 (T2 T3 T4 iviAi =3 +7E — 4. MA/1 4 Ti But — +R> 2 and so E = +7•62—4= MA 3 T3 Ti Then
E
4.
Problem VIII.12. Prove that V3n + 2n + 2 is irrational for any natural number n. Solution. Let us suppose that there exists a natural number n such that p defined as p = V3n2+ 2n + 2 is a rational number. Since p2 =3n2+ 2n + 2,
52
"Gazeta Matematice — a bridge over three centuries
p must be in fact natural. If n is even, then p2 = 4M + 2, therefore p2 is divisible by 2. It follows that 21p too, hence 4 1p2 and 4 l(p2— 4M) . This implies that 412 , impossible. Then n must be odd In this case, p2 is of the form p2 = 4N + 3. As above, we also get that p2— 1 is divisible by 2, so 2 1(p — 1) (p + 1) . But p — 1 and p+ 1 have the same parity, therefore 4 (p2— 1) . We obtain another contradiction, that is 41(4N + 2) . We conclude that there is no such number n. Problem VIII.13. If a, b, c E R, then prove the inequality a2 + b2 + c2— ab — ac — bc > 3(a — b)(b — c).
In what case does the equality hold? Solution. Denoting a — b = x, b— c. = y, we deduce a — c = x+ y, x,y E R. By taking into account the equality a2 + b2 + C2— ab — ac — bc =
1
[(a — b)2 + (b c)2+ (c
a)2] ,
and the above notations, the initial inequality becomes 1
(x2
+ y2 + x +y)2) > 3xy, (
an inequality that reduces to the obvious inequality x2 + y2 > 2xy. As it is well known, we have an equality when x = y. For our initial inequality, the equality appears only in the case x = y, that is, 2b = a + c. Problem VIII.14. Solve in the set of natural numbers the following equation 2 (xy + yz + xz) = 3xyz . Solution. Without loss of generality we may assume that x > y > z > 0. Hence yz < xy, xz < xy and so the equation yields 3xyz < 6xy. Let us analyze the following cases: i) If y = 0, then y = z = 0. Clearly (x, 0,0) is a solution to our problem, for any x > 0. ii) If y > 0, then x > 0 and z < 2. We cannot have z = 0. Hence z E {1, 2} . For z = 1 we obtain the equation 2x + 2y = xy, or otherwise take (x — 2) (y — 2) = 4. The only solutions in natural numbers are (6, 3) and (4,4) . For z = 2 we obtain x + y = xy, that is (x — 1) (y — 1) = 1. This equation has only one solution, (2,2) . Thus, {(x, 0, 0) , (6, 3,1) , (4,4,1) , (2,2,2)1, x E N is the complete set of solutions.
Problems for the 8th form
53
Problem VIII.15. We consider three different numbers a, b, e E R*. Prove that if the equations axe+bx + c = 0, bx2 +cx + a = 0 and cx2+ ax +b = 0 cab have the roots - , - , - , respectively, then they have a common real root. a b c c Solution. Since - is a root of axe + bx + c = 0, it follows that a
c2+ bc + ac = 0 <#. c(a + b + c) = 0. As c 0, it results a + b + c = 0, that is, 1 is a common root of the three given equations.
Problem V111.16. Show that there exists a unique positive integer n such that the number A = 217 + 17212 + Z is a perfect square. Solution. Let a E N be a positive integer such that A = a2. Then 217 + 17 . 212 + 2n = a2 .#>. 2n = a2
212 (17+ 25)
2n =
= a2 - (26• 7)24=> 2n = (a - 26• 7) (a + 26• 7) . The last equality implies there exist p, q E N, p + q = n, q > p such that f a+26 •7 = 2q
a - 26 .7 = 2P, which yields (a + 26 • 7) - (a - 26 • 7) = 24 - 2P, i.e. 2P • (2" - 1) = 27 .7 . The last equality holds if and only if p = 7 and q - p = 3, which gives p = 7, q = 10 and therefore n = 17 is the required value in the problem.
Problem V111.17. Compute the sum 1+
al a2 + 1 - al (1 - al)(1
Solution. We have
1 al - al
=
+ + al 1 +
1 - ai
an, (1 - ai)(1 - a2) • • • (1 - an) •
= -1 +
1 1 - al ;
1 1 a2 - 1 + 1 a2 = + = (1 - ai)(1 - a2) ' • • . 1 - at (1 - ai)(1 - a2) (1 - ai)(1 - a2) an _ + • ' • ' (1 - a1)(11- a2) ... (1 - a„) (1 - a1)(1 - a2) ... (1 - an-1 +
(1 - ai)(1 - a2)... (1 - an)
'
"Gazeta Matematica" - a bridge over three centuries
54
and hence we have a telescoping sum 1 1 + 1 - ai (1 - ai.)(1 - a2) .•• 1 1 (1 - ai)(1 - a2)... (1 - an-1 4- (1 - ai)(1 - a2) - - an)
1-1+
1 1 - ai
which equals [(1 - ai)(1 - a2)... (1 - an)]-1. Problem VIII.18. Construct using the ruler and compass an angle of 37°30'. Solution. We consider the right angle LAOB and construct its bisector OE, and then construct an equilateral triangle FOE like in the figure. We have LFOE = 60°, LAOE = 45° and hence LFOA = 15°. We construct the bisector OG of the angle LADE. Then
0 LAOG = 22°30' and LFOG = 15° + 22°30' = 37°30'.
B
Problems for the 8th form
List of authors. Problems for 8th form 1. Achim Gh., VIII.14 (G.M. 2/1999) 2. Carbunaru C., VIII.? (G.M. 1/1974) 3. Dimca A., VIII.6 (G.M. 5/1970) 4. Dobogan A., VIII.15 (G.M. 9/2003) 5. Giurgiu G., VIII.18 (G.M. 7-8/2003) 6. Haivas M., VIII.16 (G.M. 10/1985) 7. Ionescu Tiu C., VIII.5 (G.M. 9/1967) 8. Lalescu T., VIII.1, VIII.2 (G.M. 6/1900; G.M. 1912) 9. Lascu M., VIII.13 (G.M. 3/1994) 10. Luca T., VIII.17 (G.M. 2/2004) 11. Miculita M., VIII.8, VIII.9 (G.M. 2/1980; G.M. 11/1980) 12. MihAescu St., VIII.3 (G.M. 12/1945) 13. Petre S., VIII.12 (G.M. 8-9/1990) 14. Romilai B., VIII.11 (G.M. 3/1990) 15. Smarandache St., VIII.10 (G.M. 5/1987) 16. Tomescu I., VIII.4 (G.M. 5/1964)
55
Chapter 5
Problems for the annual competition - Secondary level Problem CG.1. Determine the smallest positive integer a, so that for any n E N, n > 2, amongst the numbers: n,n + 1,n + 2, ... , n2 + n + a there exists at least one perfect square. Solution. From n < n2 < n2+ n + a, (V) n E N, n > 2 and a E N, it results that the smallest value is a = 0. Problem CG.2. On the median AL of the arbitrary triangle ABC we take the point P so that m(BPC) = 90°. BP and CP cut the circle circumscribed to the triangle ABC at M, respectively N. Show that AL 1 MN. Solution. Let denote AL n MN = We have: m(MP'L) = — 1 = (m(A1C) + m(CM) + m(AN)) and (PL) (LC) (LB).
1 1 Then m(MP'L) = —m(NB) + — m(CM) = m(M PC) = 90°, so MN d_ AL. 2 2 Problem CG.3. Let's have ABCD a convex quadrilateral and M a point in its plane. a) Find the locus of the point M, for which the sum MA2+ M B2+ MC2+ M D2is constant; 57
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▪
"Gazeta Matematice — a bridge over three centuries
58
b) Find the position of M for which the sum M A2 mB2+mc2 mD2 is minimal. Solution. B
a) Let's have 01 the midpoint of (AC); 02 the midpoint of (BD) and 0 the midpoint of the segment [0102]. Applying the median theorem we obtain:
MA2
MB2
= 2 (2M02 +
MD2 = 2M0? +
MC2
Ac2 2
+ 2M03 +
BD2 2
0103) +2 AC2 +2BD2 = 2
= 4 • MO2 +0103 +
2
(AC2 + BD2).
By the hypothesis, it results that MO = constant, i.e., the locus is either 0, or the point 0, or a circle having its center in 0. b) The sum is minimal for M 0. Problem CG.4. Show that if the real numbers a, b, c satisfy the inequality a + b + c> 3, then a2 + b2 + c2 +ab + bc + ca > 6. Solution. From the obvious inequality (a + b + c)2 < 3(a2 b2 c2) it results, in view of a + b + c > 3, that a2 b2 c2 > 3. But (a + b + c)2 a2 b2 c2 > 9 ±3 a2 b2 -2 +ab+bc+ca>6. Problem CG.5. In the triangle ABC we inscribe the rhombus DAFE, where D E (AB), F E (AC), E E (BC). Prove that the area [DAFE] < AB • AC 4 • Solution. Let's have AD = x; S1 = area of (BDE), S2 = area of (EFC); S = area of (ABC) and S' = area of (ADEF).
59
Problems for the annual competition - Secondary level 1 ( x ) S , = From DE if AC, EF ll AB we have: -, A)
2
82 _ (
2 x
' S — AB
and
1) x . BD AB — x i x (1 . BA = i BA x AB + AC) — 1 AC BA 1 1 1 2 2X2 2 ( 1 1 + = x2 ' X AB2 AC2 ) • (1) ,LI B ± AC/ - 1 AB • AC As (1) 1 2x2 ) _ S' = S— (Si + S2). S — x 2S ( 1 +•) = S (1 1+ AC2 AB2 AB • AC AB • AC • sin A 2x2 2x2 = x2 • sin 2;1-< x2 = = S = AB • AC AB • AC 2 _ AB2• AC2 1 = (1 1 ) 2 (AB + AC)2 ' AB +AC and so the relation
AC2• AB2 AC • AB < implies the conclusion of the (AC + AB)2 4 —
problem. Problem CG.6. For m E R, solve in R the inequality 3x3 +m(x2 +x+1)-3>0. Solution. We have: 3x3 +m(x2 +x+1)-3 > 0 .#> (x2 +x+1)(3x-3+m) > (m — 3 0<=>3x-3+m>0.#>xE 3 7 , since x2 + x + I > 0, V x E R. Problem CG.7. Let's have SABC a triangular pyramid and M, N, P the ends of the bisectors for BSC, CSA, ASB on BC, AC and AB, respectively. Show that AM, BN, CP are concurrent. Solution. S
From the bisector theorem we obtain: MC SC NA SA PBS.13. MB SB' NC SC' PA SA MC PB NA It results that MB PA NC= SC SB SA — = 1 and from the reSB SA SC ciprocal to the Menelaus theorem, the conclusion to the problem follows. -
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60
Problem CG.8. If a = .....„,...—• 11...1, then 9a2 + 2a = ...„...., 11 ... 1 . k digits
2k digits
Solution.
9a2 + 2a = a(9a + 2) = a(99...9+2) = a(1.....„—• 00...0+1) = ...„._.• k digits
k digits
= a • 1•-.„....., 00...0+a=....„.....„_., 11...1 00...0+11...1=11...1 ....„..• ....„...., . k digits
k digits
k digits
k digits
2k digits
Problem CG.9. Let's x, y, z E (0, oo) be so that xyz = 1. Prove that: y2 X2 Z2 y2 + yz + z2 + X2 +XZ + Z2 + x2 + xy +y2 ? 1.
Solution. Using the inequality a • b < x2
Y
a 2+b2 2 , (V) a, b E R, we have:
2
Z2
y 2 + yz + z2 + X2 + XZ + Z2 + x2 + xy +y2 > y2 >2 Z2 x2 3 ( y2 ± z2 + z2 + x2 + x2 + y2) .2( a + b 4. c ) > 2 3 1)-1--c c+a a+b) - 3
3 2
Note. To prove the last inequality involved in the solution, see Problem CG.18. Problem CG.10. Prove that there exists an infinity of numbers x E Q so that: Vx + 1997 E Q and Vx + 1998 E Q. 1 1._ 2 Solution. Take for example x = - ( t - 1997, with t E Q*. 4 t Problem CG.11. Solve in the set of integer numbers the equation:
x(x + 3)(x + 4)(x + 7) -F 25 = y2. Solution. We have:
x(x + 7)(x + 3)(x + 4) = y2— 25 <=> (x2 +7x)(x2 + 7x + 12) = = y2— 25 <=> (a - 6)(a + 6) = y2- 25 <=> a2 - 11 = y2<=> <=> (a - y)(a ± y) = 11 <=> .#> (a - y, a + y) E {(-11, -1), (-1,-11),(1,11),(11,1)1 <=> <=> (a, y) E {(-6, 5), (-6, -5), (6, 5), (6, -5)1 ,
Problems for the annual competition - Secondary level
61
where a = x2 +7x + 6. The solutions to the equation are
S = {(-3, 5), (-4, 5), (-3, —5), (-4, —5), (-7,5), (0,5), (-7, —5), (0, —5)} . Problem CG.12. If a > 0 and x,y,z E [0,4 prove that:
x(a — z) + y(a — x) + z(a — y) < a2. Solution. Given LABC equilateral with side length =a and M, N, P points situated respectively on the sides [AB], [AC], [CB] to that BM = x, AN =- y, CP = z.
Denote by Si, S2 and S3 the areas of triangles BMP, AMN and NPC, respectively. We have:
.15 0. 2 0 1+ y(a — x).71- + z(a — y)-7 1 < a 7. Si + S2 + S3 < S <=> x(a — z)-7.4=> x(a — z)+ y(a — x) + z(a — y) < a2. Problem CG.13. Let's consider the following polynomial equality: (3x2 — 3x + 1)2002 )N2002 = a2004x2004 + ... + a2x2 + aix + a0.
Show that: a) ao + a2 + a4 + • • • + a2004 is divisible by 25; b) al + a3 + a5 + • • • + a2003 is divisible by 24. Solution. Denote f (x) = 3x2— 3x + 1. We have: 2002 + 1 7 = f(1)+ i(-1) a) ao + a2 + • • • + a2004
(50 — wool + 1 2
=
=
M50
2
=
2
=
2
=
= M25.
b) al + a3 + • • • + a2003 w — 1-L M48 (48 + wool
2
2
401001 ± 1
=
f( 1)
2 = 24•
— f(-1) _ 72002 _ 1 = 401001 — 1 = 2
2
2
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62
Problem CG.14. If a, b, c E Z are such that 71a3 + b3 + c3. show that 141(a3+ b3)(b3 + c3)(c3 + a3). Solution. If x E Z, then we have: x = Nt7 x3= M7; x = M7 + i E {1, 2, 4} x3= M7 + 1; x = M7 + j, E {3,5,6} x3= M7 — 1. The following cases are to be analyzed: a) a3= M7, b3= M7, c3 = M7 = a3+ b3 = M7 and at least one of the numbers a3+ b3; b3+ C3; C3 + a3is even, so 141(a3+ b3)(b3 + c3)(c3 + a3). b) a3= M7 (a = M7); b3= M7 + 1; c3= M7 — 1 = 7c' — 1 and, similarly to the previous case, we obtain the conclusion. Problem CG.15. If n E N, find the integer part of the real number: a = Vn2
_n in2 + 2n ± vn2 3 V 3
n.
1 ,., 2n n 1 Solution. As n < Vn2 + — < n + — n < n,h + — < n + — and n < 3 6' 3 3 Vn2 +n
nEN,wehavea>n+n+n=3nanda<3n+1. 2' Therefore [a] = 3n. Problem CG.16. Let's consider a triangle ABC in which 2AB = AC. The bisector of the angle BAC intersects the circle circumscribed to the triangle in D, and the perpendicular from D on BC cuts the circle a second time in E. Show that 2m(AB) = m(EC) <#. [AC] is a diameter <=> ABCE is an orthodiagonal quadrilateral. Solution.
Let's have m(AB) = x, m(BD) = m(DC) = y, m(CE) = z, m(EA) = t. From DE 1 BC y+z=180°=x+y+t.
(1)
I. "2x = z [AC] diameter"; From z = 2x m(ABC) =
(i)
t = x and hence
t+z 3x = — = m(ACB). 2 2
(2)
Problems for the annual competition - Secondary level
63
Let's take M the midpoint of [AC]. We have: (AM) E-_- (AB) = AMB a-ABM; m(AMB) = m(MCB) + m(MBC) m(ABM) = m(MBC)+ m(MCB) (W m(ABM) = 2m(ACB), so AC m(MBC) = m(MCB) = (MB) = (MC) BM = [AC] diameter; II. "[AC] diameter = ABCE orthodiagonal quadrilateral"; [AC] diameter m(ACB) = 30°, m(BAC) = GO° = y = x = 60°. We have
m(BFC) =
2y t x+y+t 2 2
—
m(BGE) = 90°
BF I AC;
III. ABCE orthodiagonal quadrilateral = [AC] diameter and IV. [AC] diameter 2x = z, are similarly obtained. Problem CG.17. If x, y E [0, 1], show that:
Vx 2 + (1
+ V(1
x)2 + y2 ?_ V2(x2 +y2).
When the equality holds? Solution. On the cartesian coordinates xOy system let's have the points A(1, 0), B(0, 1) and C(1, 1). We have to demonstrate that for M(x, y) belonging to the surface OACB we have MA+ MB > f • MO. For M on the surface OACB intersected with the disc of center 0 and radius 1 D(0; 1) we have: MA + MB > AB = 12. > v4-0 with equality for
M E [AB] n D(0;1) = {A, B} . For M on the surface OACB, M [OB] U [OA] we have:
MB • OA + MA • OB > AB • OM (Ptolemy's inequality) <=> <=> MB +MA> 12-• MO. We have equality in the case in which OAMB is an inscribable quadrilateral, i.e. M C. So the inequality holds with equality for M E {A, B, C}.
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64
Problem CG.18. Show that for all positive real numbers x, y, z the inequality
z x 1 Y <— + + 13x+y+z x+13y+z x+y+13z -5 '
holds.
Solution. Denote 13x + y + z = a; x + 13y + z = b; x + y + 13z = c. a+b+c If follows x + y + z = and then 15
x=
a 12
a+b+c b Y= ; 12 12 • 15
x x 1 = = 13x + y + z a 12 Consequently
But
x 13x+y+z
= <
=
a+b+c c • z= 12 • 15 ' 12
a+b+c 12 • 15
a+b+ c and the analogous for y and z. 12 . 15a
3 1 12 12 • 15 1 1 4 4.15 1 1 4 4.15
(bcacab) 3 + T1, + T 1+ ++ + < c b b 1 (2+2+2)= 12 . 15 1 1 2-15 5'
that proves the inequality. We have equality for a b a = b. x = y = z, since - + - = <>o b a 2 a, Problem CG.19. If x
E
R, prove that
(x + 1)(x + 2) . ... • (x + 1999)(x + 2000) < (x +
2001 2000 2 ) 2
2 001 Solution. For k = 1,1000 we have (x+ k)(x +2001 - k) < ( x + 2) <=> ( k
2
2001) > 0, obviously true. Multiplying these inequalities for 2 k = 1, 1000, we obtain the required inequality. Problem CG.20. Show that, for all non zero real numbers al, a2 ... an, the equation
ix + aii + • • • + lx + ani + lx has no solution.
—
ail + • • • + Ix
—
an' = lai + • • • + anl
65
Problems for the annual competition - Secondary level Solution. Using la + bI < lal + Ibl, (V) (a, b) E R, we have:
lad
lai + a2 + • • • + an l < i=1
and
E (Ix + ail + lx — ail) E +
— x + ail = 2E lail .
i=i If the equality in the problem would be satisfied, we'll obtain 2
E lad <
E lai l<=> i=i E Iai l< 0- which is a contradiction. Problem CG.21. Let's have the set A = {8888 • n In E N*}. Determine the smallest element of the set A that has all the digits distinct one from the other. Solution. The required number is 391072 = 44 • 8888. For the complete solution, we refer to [V. Berinde, Exploring, Investigating and Discovering in Mathematics, Birkhauser, 2004], where a similar problem is solved in Chapter 8. Problem CG.22. Let's consider ABC D a square with the side length 1 and the points M E (AB), N E (BC), P E (CD), Q E (DA). Show that: \ISMBN -F\ISNPC -F \ISPDQ+ \ISQAM
Solution. C
D
Let's consider AM = x, BN = y, CP = z and DQ = t.We'll have 1 1y+1—X and VSMBN = -Y(1- X) 2 v2 2 the other similar.
N y A
x M
B
By summing up these inequalities, we obtain the requested inequality. Problem CG.23. Let's consider V ABC D a regular quadrilateral pyramid with the property that SvAB = SVAC• Show that the size of the dihedral angle formed by two adjoint faces is 120° and m(,ZAV B) > 60°.
"Gazeta Matematice - a bridge over three centuries
66 Solution.
a) Sv Ac = 2SV AO 2SV AB • cos a =
1
Sv AB cos a = 2
= 60°, where
a = in ((VAB), (VAC)). So, m ((V-AB), (VAD)) = 120°. D'
b) Let's take AB = a, V 0 = h, D' E AB, (AD') (BD'). From the hypothesis it results: a2 1 2= AC • VO = - AB • V 0' col -• h = a h2 + — 2 4 2h 2
h2
= h=
-2-
and
V D' =
T a2 +11,2
< aNid 2
m(AV B) > 60°. 2 Problem CG.24. Determine the function f : R -> R which verifies the relation: 3f (x) + f (3 x) = x + 3f (3), for any x E R. -
Solution. We replace x by 3 - x in the relation from the hypothesis, solve x +6 the system with respect to (x) and obtain f (x) = 2 Problem CG.25. How many regular polygons have the size of their angles equal to n° degrees, where n E N? Solution. If a is the measure of an angle of a regular polygon with m sides, 180°(m - 2) 360° m > 3, then: a = <#. a = 180° m in 360° Wehaven= 180° -71 ENmED360\{1,2}andmElmtakes 22 values. It follows that there are 22 regular polygons fulfilling the condition. Problem CG.26. Let a, b, c be real numbers such that 0 < a < 1, 0 < b < 1, 0 < c < 1 and ab + be + ca = 1. Show that a2 b2 c2 < 2. Solution. Since a, b, c E [0, 1] we have a2 < a, b2 < b and c2 < c, hence a2 b2+ c e2
-
1)(b
-
1)(c
-
1) <0.#>a+b+c<1-abc+ab+bc+ca.
Problems for the annual competition - Secondary level
List of authors. Problems for the annual competition Secondary level 1. Achim Gh., C.G.11 (G.M. 5-6/1999) 2. Apostol C., C.G.5 (G.M. 3/1997) 3. Batinetu-Giurgiu D. M., C.G.8 (G.M. 7/1995) 4. Batranetu D., C.G.25 (G.M. 11/1996) 5. Becheanu M., G.M. CG.26 (G.M. 4/2003) 6. Birsan M., C.G.3 (G.M. 9/1987) 7. Blaga A., C.G.1 (G.M. 2/1986) 8. Carjan F., C.G.12 (G.M. 5-6/2002) 9. Cindrea A., C.G.2 (G.M. 8/1986) 10. Cojocaru I., C.G.6 (G.M. 1/2002) 11. Cristea C., C.G.19 (G.M. 3/2001) 12. Eckstein A., C.G.24 (G.M. 10/2002) 13. Florea C. S., C.G.13 (G.M. 7-8/2001) 14. Heuberger C., C.G.16 (G.M. 9/1994) 15. Ilie R., C.G.4 (G.M. 2/1990) 16. Marinescu D., C.G.18 (G.M. 1/1991) 17. Popa E. C., C.G.21 (G.M. 7-8/1987) 18. Popescu D., C.G.10 (G.M. 2/1998) 19. Prajea M., C.G.22 (G.M. 7-8/2000) 20. Rotaru F., C.G.7 (G.M. 6/1991) 21. Safta I., C.G.15 (G.M. 5-6/2002) 22. Szollosy Gh., C.G.17 (G.M. 9-10/2001)
67
68
"Gazeta Matematica" — a bridge over three centuries
23. erb'anescu D., C.G.23 (G.M. 5-6/1998) 24. Ursarescu M., C.G.9 (G.M. 3/1998) 25. Vlaicu L., C.G.20 (G.M. 9/1989) 26. Zanoschi A., C.G.14 (G.M. 9/2002).
Chapter 6
Problems for the 9th form Problem IX.1. Let f :R R be a polynomial function of degree 2. We call xo to be a fixed point of f , if x0is a solution of the equation f (x) = x. Prove that if f and f o f have the same value of extremum then f has two fixed points. The reciprocal statement holds true? Solution. Consider f : R —> R, f (x) = ax2+ bx + c, a, b, cER,a have to study the cases a > 0 and a < 0. A I. a > 0. f has a minimum value equal to – — for x = 4a A = b2 – 4ac. –
0. We
—
where
2a
In order to have min(f o f) =min f we must have – 2 E Imf = –
4a'
oo ,
b A hence – — > – — i.e. A – 2b > 0. 2a – 4a f has two fixed points if the equation f (x) x = 0 has distinct real solutions. The discriminant of this equation is b2– 4ac – 2b + 1 and from above, we have b2 – 4ac 2b + 1 > 0. II. For a < 0 we proceed in the same manner and we obtain the same conclusion. As for the reciprocal statement we prove that this does not hold true 1 by an example. The function f(x) = x2 + x has two fixed points but –
– —
8
min f (x) < min( f o f)(x). xER
xEIR
Problem IX.2. If 0 < x < 2 and 0 < y < 2 then Solution. From x, y E (0, 2) we obtain –1 < x 69
–
x—y < 1. x + y — xy
1 < 1 and –1 < y
–
1 < 1,
"Gazeta Matematica." — a bridge over three centuries
70
i.e., Ix — 11 < 1 and ly — 11 < 1. Therefore —
11 • ly — 11 < 1 •<=> 1(x — 1)(y — 1)1 < 1 •<=>. —1 < (x —1)(y — 1) < 1.
From (x 1)(y — 1) < 1 we obtain x y — xy > 0, which shows that the fraction in our problem is well defined. Since x —y x —y < 1 •<=> 1 < <1 x 1-- y — xy x + y — xy —
-
and using the fact that x+ y—xy > 0, the previous double inequality reduces to x(y — 2) < 0 and y(2 — x) > 0, which are obvious. Problem IX.3. Solve the following system a2x2 + a3x3 + + anxn — 1= aixi + a3x3 + + anxn— 1 al a2 —1 a1x1 + a2x2 + + = \M. + 4 + + an Prove also, that if al
+ an = 0, then
a2
aixi + a2x2 + + anxn =
n—1
Solution. From the properties of the sequence of equal fractions, the value of each fraction is equal to the the fraction having as numerator, the sum of the numerators and as denominator, the sum of the denominators of the fractions, i.e., (n — 1)(aixi +a2x2 + + anxn ) — n (1) a1 + a2 + + an Denote S = al + a2 + ...an. Using the same properties, the value of each of the fractions is equal to the fractions obtained by subtracting from the numerator and the denominator of fraction (1), the numerator and the denominator of each fraction respectively, multiplied by it — 1. We obtain the equalities: (n — 1)aixi — 1
S
—
(n — 1)ai
=
(n— 1)a2x2 — 1
S
—
(n — 1)a2
= Vx? + 4 + ... + x?, = t.
(n — 1)anxn— 1 = S (n — 1)a„ —
71
Problems for the 9th form Then we have xi
S
=
- 1)ai S - 1)a2
X2 =
xn
=
1 t+ 1
t
1
(n - 1)ai 1 (n -1)a2
(2)
S [(n 1)an 1 ] t + (n -11)an;
In order to obtain t we square the equalities (2) and add them. By choosing 1 1 1 1 1 1 , we have + ... + bi = — + — + • • . + 7 c n, b2 = --2- + an al al a2 a2
t2 — r
S2
[(n 1)2
b2
2S
n 1 bi + nt ] 2 +2 +
1 S b2 1.(n -1) 2
b1 ] t+ n 1
1 b2, (n - 1)2
and then [ S2 b2 [ (n - 1)2
2S bi +n 1] t2 + 2 [ S n -1 (n - 1)2
b2
b1
n -1
I t+
b2 = O. (n -1 1)2 From this equation one obtains the values of t, and from (2) one obtains the solutions of the system (two solutions). 1 ' and consequently If S = al + a2 + • • + an = 0, then xi = -t + (n - 1)ai +
n
E_ E aixi = -t E ai + i=i n- 1 n- 1 i=i
Problem IX.4. Prove that in an arbitrary .triangle the midpoint of the external simedians are three co-linear points. Solution. The external simedians of any LABC are the tangent lines ATa, BTb, CT, at the points A, B, C to the circumscribed circle limited by the opposite sides. According to Pascal's theorem, the points Ta, Tb, T, are co-linear so that ABCTaTbTC is a complete quadrilateral in which the diagonals are the external simedians of the quadrilateral. Applying the Gauss's theorem to this quadrilateral, we have that the midpoints of the simedians are co-linear.
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72
Problem IX.5. In a triangle ABC the small angles made by the medians BB' and CC' with the line BC are denoted by and 7, respectively. Prove that the following relationship holds sin (3 — y) sin (,3 + y)
1 sin (B C) 3 sin (B + C) • —
Solution. Let G be the point where CC' and BB' intersect each other. sin + -y) sin 3 sin 7 In ABCG we have • We can rewrite this formula BC GC BG sin G(3 + y) sin (f3 — = Certainly, this last term equals to as BC GC cos ry — BG cos /3 sin G(3 — 7) 3 By projecting CC' and BC' onto BC, one obtains 2 CC' cos 7 — BB' cos 13 1 that BC = CC' cosy + AB cos B. Similarly, BC = BB' cos 13 + 2 AC cos C. — 2
1 By substracting these two formulas we get CC' cos 7 BB' cos 3 = — AC cos C2 1 — AB cos B. —
2
AB AC BC = = , and plug the obsin B sin (B + C) sin C sin (3 + 7) tamed result into this formula for BC sin (/3 — ry) 1 sin (B — C) The result is precisely sin (3 ± ry) 3 sin (B C) • Finally, let us use that
Problem IX.6. Let ABC be a triangle and M E Int (ABC). The parallels through M to the sides AB and AC intersect the sides AC and AB at the points P and N, respectively. Find the geometrical locus of M if AP • AB + AN • AC = constant.
Solution. Let us consider the expression
E=
AP • AB AN AC AC sin A + sin A = o-(ABP) + o-(AC N) = constant 2
73
Problems for the 9th form
By the relations MP AB and MN II AC one obtains u(APB) = a(AMB) and a(ANC) = o-(AMC)). So, a(AMB) + a(AMC) = constant. But a(BMC) = o- (ABC)— (cr(AMB)+o-(AMC)) = constant. That means M belongs to a line segment which is parallel to the side BC, bounded by AB and AC. Problem IX.7. The numbers xi (1 < i < n), satisfy the conditions a < xi < 13 and xi E R, a,0 E R. Find the minimum of the expression n-1
E=
XiXi —
a+ n E xi. 2 i=i
(n 1) •
1
Solution. First we find the minimum of the expression n-1
E Yiyi, where Yi E R, and In < k, k > 0. j>i
We have n-1
1 [( n
) 2 n
— E yi2
i=i j>i 2
> Yi must be minimum and the term E Yi2 i=1 i=1
It follows that the term
must be maximum. If n is even, we take all Y such that lYil = k, half of them being equal to +k and the other half of them being equal to —k. From this we have n
EY
2
= 0, and E Yi2 = nk2, so that i=1
i=1
(n-1
n = — — • k`.
E YYj
2
min
If n is odd we have for [— 21 of the terms
Y
[n the value k, for -. , the
value —k and for one of them the value Y, which we shall choose. (n-1
= [172
E YiYj j>i
21 min
y2
(n _ 1)k2 ] ,
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74
hence Y can be arbitrary, and by unifying the results we get (n-1
E YiYj
] k2
-L 2
i=1
J>i
•
min
The relation: a < xi < ,i3 can be rewritten in the form: O-a a+/ 2 - 2
xi
By taking into account the previous results we have: [9/-1
E (xi /=1.
3)1
a 2 Q1
[ 2 ] (a-4 /3)2 ,
/
a 13) (xi
2 )
j>i
min
or n-1
Exixj -(n 1)• a +2 En xi.
(a + 0)2n(n - 1) 4
2
n(n - 1 2
(a + 0)2
_ min in] L2J
(a 0)2 4
Hence Emin =
4
in] L2J
(a - 0)2. 4
For n even, we have:
Emir, =
[n(a2 + 02) + 2(n - 2 )a0].
For n odd, we have: Emin =
n-1, 8 t(Th + 1)(a2 + /32 ) + 2(b - 1)ai3J.
Problem IX.8. Prove that the sum of the distances from the middle point of the edges of a triangle to the tangent lines to the circumscribed circle drawn a 2 b2 e2 through the opposite vertices is 2S abc
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Problems for the 9th form
Solution. Let A', B', C' be the midpoints of edges BC, CA, AB and A", B", C" their projections on the considered tangents, D, E the projections of B and C on the tangent at A, and F the projection of A on BC. Then BDEC is trapezoid, A'A" is its middle line so A'A" = BD -CE
AAHC , ABDA having a right angle and DAB = ACH. From this H AC BH we obtain: ' 731) BA and BD = ACor in classical notations ha =2 _ 2Sc2 2Sb2 i -F•Q, — BD = cha ba a — abc ' Same for CE, we have CE = -7-b -
and the problem is solved. A'A" = 512---12 a c -
'
Problem IX.9. Show that if a, b, c are positive numbers such that a+b+c=1, then the following inequality is true:
8 1 1 1 > - + + - 1. 27abc a b c -
Solution. The inequality to be proved is: 8 > 27E ab 27abc. By taking into account the relation a + b + c = 1, the inequality can be given in the form 8 (E a) 3 > 27 (E ab) (E a) - 27abc. -
After computations, this inequality becomes: 8 (E a3) - 3 (Da2 b + ab2)) - 6abc > O. We have 2 (E a3) - (Da2b + ab2)) = (a + b)(a
-
b)2? O.
Also
E(a2b± ab2)
-
6abc = E a(b
-
c) 2 > 0.
Therefore 8 (E a3) - 3 E(a2 b + ab2)) - 6abc = + 4 [2 (E a3) - (>(a2 b + ab2))] + + [(a2 b + ab2) - 6abc] > 0. Problem IX.10. Is it possible that two rectangular triangles that have five equal elements (sides, angles) may exist and still be not equal ?
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76
Solution. The answer is yes ! We built two triangles AABC and AAB'C' with all angles equal, B = 90'; B' = 90°, but the triangles are different. If the hypotenuse AC' of the small triangle is equal to the big cathetus AB of the big triangle and the small cathetus BC of the big triangle is equal to the big cathetus AB' of the small triangle, the problem is solved because although the triangles have five equal elements (three angles and two sides), they are not equal. Indeed, since AB = AC' and B = 90° we deduce cos x =tgx and then cost x = sin x
sine x + sin x
-
1=0
sin x =
-1 ± 162
The only convenient solution is sin x =
-1 + 2
< 1.
Problem IX.11. Consider the convex pentagon OABCD, and E, F, G, M the midpoints of the sides AB, DC, EF, OG respectively. We denote by A1, B1, C1, M1 the projections of A, B, C, M, on the line AD respectively. Prove that AA1+ BB]. + CC1 = 8M Solution. Our solution uses vectorial calculus. We have that MM1 = 0M1 - OM, where
1 1 -4 1-4 1 OG = 71 (0E + OF) = (0A+0B)+ g (0C + OD) 8 1 ---> ---> --+ ---+ --> --> = g (0A1 - AA1 OB1 - BB1 OC1 CC' + OD) .
OM =
Thus, the following formula holds for MM1:
8MM1 = OM1 + AA1 + BB1 + CC].
-
°Ai
-
OB1
-
OCi
OD.
-
Now, let V> = 8MM1 - AA1 - B131 - CCi and Vt) = 0/1/1-1 - 0A1 - OCi - OD. The above formula says that Ti = Vu but, on the other hand -7_4/lie . Hence v = w = 0. The formula AA]. + BB1 + CC]. = 8MM1 is a consequence of 7> = 0. Problem IX.12. Let a, b, c E R be such that 1) b2— 4ac < 0; 2) lax2 +bx + <1, V x E [-1, 1]. Show that Icx2+ bx + al < 1, V x E [-1, 1]. Solution. Condition 1) shows that the graph of the quadratic function f (x) = ax 2+ bx + c has one of the forms (and the others similar): and hence the maximum value of I f (x)I is obtained on the boundary of [-1, 1].
Problems for the 9th form
77
1 I
Therefore la — b + cl <1 and la + b + cl < 1 and using the same arguments for g(x) = cx2+ bx + a, we get
Ig(x)1 < max {la — b+ cl, la+ b+ el} 5_ 1, since the discriminant of g is b2— 4ac < 0, too. Problem IX.13. Consider an acute-angled triangle ABC with the side lengths a, b, c and area S. If P E Int(ABC) such that alP AI + bIPBI± cIPCI = 4S, then P is the orthocenter of the triangle ABC. Solution. Let A'B'C' be the complementary triangle of ABC. A Al B' C'
N14
A' We have S = SABC' = SBCA' = SCAB' and SA,B,c,= 4S. On the other hand, we have 4S = SpB/C -FSPC'Al -FSPA'B'• But SpBIC,= 11B1Ci l • IPAll = alPAll < a IPAI, where PA1 1 B'C'. The equality holds if and only if PA 1 C' B' . Similarly, Spc/A1 < b IPBI, SpAIB/
Therefore, the equality a IPAI + b IPBI + cIPCI = 4S takes place if and only if PA 1 BC, PB I CA and PC 1 AB. It follows that P is the orthocenter of the triangle ABC. Problem IX.14. Find the function f :NT N that satisfies
f ( f (n)) ± f (n,) = 2n + 6, V n E N,
(1)
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78
Solution. First of all we prove that f is injective. Let m, n E N such that 1(m) = 1(n). We have
1(m) =
f(n) = 1(1(m)) = f(f(n))1(1(m))+1(m) = i(i(n))+ Am) 2m+6=2n+6<=>m=n,
hence f is injective. Denote 1(0) = a and let's compute it. Taking n = 0 in (1) 1(1(0)) + f (0) = 6 1 (a) + a = 6 1(a) = 6 - a and since f (a) E N, it follows 0 < a < 6; f (1 (a)) + f (a) = 2a + 6 -1(6 - a) = 3a; Taking n= a in (1) Taking n = 6-a in (1) = f (1(6-a))+ f (6-a) = 18-2a = f (3a) = 18-5a and hence 18 - 5a > 0 which gives a < 3; Taking n = 3a in (1) f(18 - 5a) =f (1 (3a)) + f (3a) = 6a + 6 11a - 12 E N which gives a > 2. Taking 71 = 18 - 5a in (1) f(f(18 - 5a)) + f(18 - 5a) = 6a + 6 f (11a - 12) = 54 - 21a E N which gives a < 2. Therefore a = 2, i.e., f (0) = 2. We can now prove that f (2n) = 2n + 2, for all n E N, using induction. Since f is injective, it follows that f(2n + 1) is an odd natural number, for each n E Let's now compute 1(1) = Q. Taking n = 1 in (1) we get f(f(1))+ f(1) = 8 f(0) = 8 - E N and so 0 < < 8. Since /3 is odd number, we can have E {1, 3, 5, 7}. We shall prove that 13 = 3. If Q = 1, then 1(1) = 1 and hence 1(1) = 7, a contradiction. If = 5, then 1(1) = 5 and hence 1(5) = 3 and by (1) we get -
-
1(1(5)) + 1(5) = 16
1(3) + 3 = 16 -<=> f(3) = 13.
For n = 3 in (1) we obtain
1(1(3)) + 1(3) = 12 f(13) + 13 = 12 = f(13) = -1' N, a contradiction. If Q = 7, i.e., f(1) = 7, then 1(7) = 1 and hence by (1) we get 1(1(7)) + 1(7) = 20 1(1) = 19, a contradiction. Therefore 13 = 3, that is 1(1) = 3. We now prove by induction that f (2n + 1) = 2n + 3, and so f (n) =71+2, VnEN is the unique function that fulfills the given requirements.
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Problems for the 9th form
Problem IX.15. Find the natural numbers x and y for which VX2 + y + 1+ y2 + x + 4 is natural.
V
Solution. We must have ✓x 2+ y + 1 E N and ✓y2+ x + 4 E N. Since x, y > 0 it follows that x2 +y+1 > (x+1)2 and y2 +x+4 > (y+1)2, that is y > 2x and x + 3 > 2y. These two inequalities imply x + y < 3. By a direct verification we find the solutions { xi = 0 =0
and
{ x2 = 1 y2 = 2.
Problem IX.16. The inequality 2R+r > 2V holds in any triangle, where R and r are the radius of circumcircle and incircle respectively, and S is the area of the triangle. Solution. It is well-known the relation R > 2r (Euler's inequality). Hence 2R + r > R + 3r > 2.V3Rr. From the sinus theorem, we obtain R=
sin A + sin B + sin C
(where p is the semiperimeter). The above relations give us: 3pr
2R-pr>2
E sin A
—2
3
E sin A S > 2\ig.
Problem IX.17. Find all finite sets X C R with the property: x + jx1 E X, V x E X. Solution. Consider X = {x1, x2, • • • , xp}, with x1 < x2 < • • • xp. If xp > 0, then 2x2 > xp and 2x2 E X; contradiction. Thus x1 < x2 < • • • xp < 0. We obtain 0 = xi — xi = xi + lxi I E X. So xp = 0 and we have xi + = 0 = xp, V i E {1, 2, • • • ,p}. It follows that X is any finite subset of (—oo, 0], which contains 0. Problem IX.18. Let f : R —> R be a function for which there exists n E N, n> 2 and a E IR, a 0, such that (f of o...o f)(x) = x + a, (V)x E n On
Show that f can be written as a sum of a periodic function and of a polynomial function of degree one.
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80
Solution. Using the substitution x > f (x) we get —
(f of o ... 0 f)(x) = f(x) + a, n+lori
and applying the function f to the initial relation we obtain (f o f o ... 0 f)(x) = f(x + a). n+1 on Hence 1(x) + a = f(x + a) = f(x + a) — (x + a) = f(x) — x. Define g(x) = f(x) x. It follows that g has the period a and f (x) = g(x) + x. —
Problem IX.19. Find the functions f, g, h : R —> R with the property
- x+2y+3z, (V)x,y,zER. (hogof)(x+y+z)+(goi)(Y+z)+f(z) = Solution. Taking x = 0 and y = 0 in the relation in the statement of the problem we obtain
(h o g 0 f)(z) =
—
(g o f)(z)
—
f (z) + 3z, (V)z E R.
(1)
Taking z —* x + y + z in (1) one obtains
(hogof)(x+y+z)=—(g0f)(x+y+z)—Ex+y+z)+3(x+y+z), which together with the hypothesis relation yields:
(9(1)(x+Y+z) — f(x+y+z)+2y+z+(gof)(Y+z)+f(z) = 0, (V)x,y, z E R. (2) If in (2) we take x = 0 it follows —f (y + z) + f (z) + y = 0, (V)y, z E R. If in the last relation we take z = 0 it follows 1(y) = f (0) + y, (V)y E R, that is —
f (x) = a + x, (V)x E R, where a = f (0). By replacing in (2) the expression of f it follows —
g(x+y+z+a)
—
x
—
y
—
z
—
a+2x+Y+g(Y+z+a)+z+a= 0
i.e. g(x+y+z+a)+g(y+z+a)+x=0, (V)x, y, z E R, from which, taking y = 0 and z = —a we deduce: —
g(x) = x + b, (V)x E R
Problems for the 9th form
81
where b = g(0). Replacing the expression of f and g in (1) we get, after short computations that h(z + a + b) = z — 2a — b, (V)z E R, in which we make the substitution z ---4 x — a — b and we obtain h(x) = x — 3a — 2b, (V)x E R. Therefore the functions f, g, h are f (x) = a + x. g(x) = x + b, h(x) = x — 3a — 2b. It is immediately that these functions verify the relation given in the problem. Problem IX.20. Let MON be a proper angle and P E Int(MON). We consider a fixed point A El OM, a mobile point B El ON and D the midpoint of I AB I. a) Prove that the parallel through D to OM intersects I OP at a point C. b) Find the locus of the baricenter of AABC if B describes I ON and D VI OP Solution. We consider the parallel to ON by Al. This line intersects OP at E and ON at H. The parallel to OM through D intersects AE at F. Then ACBF is parallelogram (CD=DF, BD=DA) BCI I AE. Then the midpoint of BC describes the middle line from 0 at OEH. If A2 is the midpoint of BC then A2 describes a parallel to the middle line from 0 in OEH. Problem IX.21. Let a, b, c be non-negative real numbers with sum equal to 1. Prove the inequality: 1 ab be ca + < —. + c +1 a +1 b+ 1 — 4 When does the equality hold? Solution. We define a + b = S and ab = P. We have c = 1 — S and S E [0, 1], P E [0, S2 /4]. Then we can rewrite the relation to be proved as: P (1 — S)(S2 — 2P + S) 1 + 2— S 1+ S + P 4.
(1)
After computations, we obtain the next equivalent inequality: 4 P2— (8 S2— 29 S + 14) P — (2 S — 1)2(S +1) (2 — S) 5_ 0. For a fixed number S E [0,1], we consider the polynomial function f : R —* R, defined by f(x) = 4x2— (8 S2 — 29S + 14) x — (2 S — 1)2(S + 1) (2 — S), V x E R. We have f (0) = —(2 S — 1) 2(S + 1) (2 — S)
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82
1 (3S — 2)2(1 — S)(2 8). We see that f(0) < 0 and 4 f(S2 /4) 5 0. Then f (x) < 0, V x E [0, S2 /4]. Thus the inequality (1) is proved. The equality holds in the next three situations:
and f(S2 /4) =
1 1 1 1 1 1. S=- P=0 <=> a = 2, b = 0, c = — or a = 0, b = c = 2; 2' 2 2' s2 <=> 4
S=
1' 1 a.— b=— c=0. 2 2"
1 2 52 3. s= 5 , P=-71- <=> a=b=c= 5 . Problem IX.22. Prove the following inequality
(x + y + z){
(2x + y + z)(y + z) +
(2z + x
z)(z + x)+ 9 > where x, y, z > O. y)(x y)] I (2y
x
Solution. Because of the homogeneity we can consider only the case where E x = 1. Then the inequality can be rewritten in the form Ex/(1- x2) > 9/8. Using the symmetry we can consider that x < y < z. It follows 1/(1— x2) < 1/(1— y2) < z(1— z2). Now, applying the Chebyshev inequality 1 1 1 and the inequality (a + b ± c) ( — + b— + — > 9, we obtain c— Ca
v
(v,
1 v■ 1 - X2 3 L--d x .c-a 1—x2 = 6 L-'1 —x + r—d1+x) = 1 1 (E(i - x)E 1 -x Di +x)E 1 +x + 4 6 2 ± ,
9 8.
) Problem IX.23. If f :R --> R is a strictly monotonous function, determine the function g :R --> R such that
(f o g)(x + 1985) > f(x) > f (g(x) + 1985) , Vx E TR. Solution. Without loss of generality, we assume that f is a strictly increasing function (similarly the case when f is a strictly decreasing function can be analyzed). By hypothesis, we deduce the following relations
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Problems for the 9th form
(1) g(x + 1985) ? x _>. g(x) +1985, Vx E R <=> g(x +1985) > x, Vx E R. (2) x ? g(x) + 1985, Vx E R or g(x) < x — 1985, Vx E R. Substituting in (1) x by x — 1985 we obtain (3) g(x) > x — 1985, Vx E R. From (2) and (3) we deduce g(x) = x — 1985, Vx E IR. Problem IX.24. a) Prove that there is an infinity of positive integers such that their squares can be written as a sum of three cubes of positive integers. b) Prove that there is an infinity of positive integers such that their cubes can be written as a sum of three cubes of positive integers. Solution. a) It is easy to verify the following identity: (71,4 + 2n)3+ (2)13 +1)3 + (3712)3 = (7.7,6 + 7n3+ 1)2
,
n E N.
Since the set of positive integers n6 + 7n3 +1, n E N, is infinite, the assertion is proved. b) Similarly, we have: (3n2+5n-5)3+(4n2 -4n+6)3+ (5n2-5n-3)3= (67/2-4n+4)3, n E N, n > 2. Problem IX.25. Prove that 2 ,— nvn + 1 < 1 + .■ /+ V5+ ... + vn < (n + 1)vn, n E N. —
3
Solution. The inequality in the problem follows by adding for 1 < k < n the following inequalities 2r ,— 2 r ikITd- 1 — (k — 1)fi c < v k < — [(k 3 3
1).Vic — kVk — 11 .
The right above inequality is equivalent to the immediate inequality 4k2(k — 1)2 < k(2k — 1)2. The left inequality is analogous. Problem IX.26. Solve the equation {x} =
x+ [x] + (x) 10 ,
where [x] is the integer part of x, {x} = x — [x] and (x) is the closest integer to x. 1 [x] Solution. We define (x) = i i 1 + [x]
if if
1 0 < {x} < -1
< {x} < 1
84
"Gazeta Matematica" - a bridge over three centuries
1 I. Suppose 0 < {x} < -. Then x = [x] + {x} and (x) =- [x] then the 2 3 equation becomes 9{x} = 3[x]. From here we have 0 < [x] < 2. Then [x] = 0 which implies {x} = 0 and we have the solution x1 = 0, or 1 4 [x] = 1 which implies {x} = .5and x2 = d • 1 II. For the case - < {x} < 1 we have (x) = [x] + 1 and the equation 27 8 becomes 9{x} - 1 = 3[x]. After some calculations we obtain - [x] < . 6< 3 7 Then [x] = 2 and {x} = - and x3 = 9 9
L5 .
Problem IX.27. Let M be a finite set of positive real numbers and v) x E M. f:M-4M a bijective function such that f (x) • f-4( x) ______ x2, ( Prove that f = 1 m , Solution. Suppose there exists an element x1 E M such that f(xi) xi. Define recursively xn2+1 = f(x,„), for m > 1. Since M is finite there are two indices k < n such that f (xk) = f (xn). Let k be the smallest index for which there exists n > k with f (xk) = f (xn). Since f is one-to-one function it follows that we must have k = 1. Therefore we have x2 = f(x1), x3 = f(x2),... xn = f(xn_i and xl = f(xn). Take xn+1 = xi. We have the equivalence f (xi) = < > xi = f -1(xi+i), j = 1,2, ... ,n. From the relation in the problem we obtain the system: 2 2 2 2 Xi = X2Xn , X2 = X1X3, • • • Xn -i = Xn-2Xn) Xn = Xn-1X1•
Denote these equations by (1), (2), ... (n). Suppose that x1 < x2. Then we have the implications: (i) x1 <x2 xn < xi and (2) (3) Xi < X2 =. X2 < X3 X3 < X4
(n-1) .. •
Xn-1 < Xn-
Hence x1 < x„, a contradiction. In an analogous way one obtains a contradiction in the case xl > x2. Consequently we must have x2 = xl. But this contradicts the assumption f (xi) xi. It follows that f = liw • Problem IX.28. Let f = ax2+ bx + c be a polynomial with integer coefficients. If for any positive integer n there is an integer cnsuch that n divides f (ch), then f has rational roots. Solution. For a = 0 the conclusion is immediate.
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Problems for the 9th form
Suppose a 0 0. From the hypothesis, for any positive integer n, there is an integer k„ such that the equation with integer coefficients ax2+bx+c—knn = 0 has an integer root cn. It follows that the discriminant D = b2 -4a(c— knn) of the equation is the square of an integer un. We will show that A = —4ac is also the square of an integer. From the above relations, we can write: VnEN\ {0} 3 k„, un E Z : A + 4ak„n = un2 .
(1)
Consider I A I = 1. Since 4m-1, m E Z, cannot be the square of any integer, we exclude the case A = —1. Then A = 1 = 12. Consider now IAI > 1. Let epr • • • phi be the factorization of A, with E E {-1, 1}, pi prime number and qi positive integer, i = 1, 2, • • • , j. Suppose, ad absurdum, that in the above decomposition there is an odd exponent q3, with s E {1, 2, • • • , j}. In this case, using the relation (1) for n = As+1, we get that 71' divides u2n . Since ps is a prime number, it results that p•i divides un2 . Therefore, there is an integer v such that pss Olios pr 4apsk„) = pr 1 v2. It follows that psdivides {hos p"; contradiction. So, all the exponents qi, q2, • • • , qi are even positive integers. Hence A = ed2, d E Z, E E {-1, 1}. Observe that, for n = d2, we cannot find k„ E Z such that —d2 + 4aknn = d2 (4akn— 1) should be the square of an integer. Therefore A = d2. But A is the discriminant of the polynomial f. We conclude that f has rational roots. Problem IX.29. Solve and discuss the system:
I
1
axi — bx3 = c ax — 2 — bx3 = c axn_i — bxn2 = c ax„— bx? = c
where a, b, c > 0. Solution. By subtracting the second equation from the first one, the third from the second one and so on, we obtain a(xl — x2) = a(x2 — x3) =
b(x2 — x3)(x2 + x3) b(x3 — x4)(x3 + x4)
an) = b(xn— xi)(xn + xi) a(xn — xi) = b(x i — X2)(X1 + X2)
a xn _i (
—
Obviously, from ax1 = c + bxi > 0 we obtain x1 > 0. In a similar way we get x2, , x„ > 0. Suppose that xi > x2. From the first equation of
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86
the last system we obtain x2 > x3 and then using the other equations we obtain the sequence of inequalities: xi > x2 > x3 > • • • > xn—i > xn > xi. Consequently x1 = x2 = = xn. Such a solution exists if we have a2— 4bc > 0. Problem IX.30. Let ABC be a triangle with the circumcircle C. A straight line that contains the point A intersects BC at Al and C at A2. Prove that 1 1 4 > — and find the conditions when the equality holds. AiA2 — BC 1 Solution. Appling the inequalities of means, we find:
2 1 1 + AA1 A1A2
< /AA1• A1A2
BA1+ CA1 and VBA1 • AiC < 2
But, in the circle C, we known the property AAi• AiA2 = BA1-CA1. Hence it follows that 1 1 2 2 4 + AA1 AiA2 \/AA1 • AiA2 VBAi • CAI BA1 + CAI
4 BC •
Equality holds if and only if AA]. = AiA2 and BA1= CAi, i.e. AB A2C is a parallelogram. Problem IX.31. Prove that a trapezoid is isosceles if and only if the difference between the lengths of its diagonals equals the difference between the lengths of the two unparallel sides. Solution. Let ABCD, AB II CD be a certain trapezoid. We want to prove that AD = BC if and only if AC — BD = AD — BC. Obviously, if AD = BC then AC = BD, so the equality holds. Conversely, if AD—BC = AC—BD it follows that AD2+BC2-2AD•BC = BD2 AC2— 2BD • AC. We now use the fact that in every trapezoid, AC2 + BD2 = AD2 + BC2+ 2AB • CD holds and obtain that BD • AC = AD • BC + AB • CD. Here we have equality in the known inequality of Ptolemeu. The equality holds if and only if the trapezoid is inscribed in a circle. It follows that the trapezoid is in fact isosceles. Problem IX.32. Let f : R —* l be a function having the property: f (f (x)) = x f (x), Vx E R. Compute f (0) and prove that if f (x) function.
0, Vx E R*, then f is an one-to-one
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87
Solution. In the equality from hypothesis, we substitute x by f (x) and obtain f (1(1(x))) = f (xf(x))= f(x)f (f(x)) = f(x).x.f(x), that is f (f (1(x))) = xf 2(x). For x = 0 we obtain ( 1) 1(f (Po))) = O. But f (f (0)) = 0 such that from (1) we deduce f (0) = 0. To prove the second statement, we take xi, x2 E R such that f(xi) = f(x2). Then we obtain f (f (f (xi))) = f (f (f (x2))) .4=> xif (xi) = x2f(x2). If x1 0 0 x2 then, according to the hypothesis, we have f (xi) = f (x2) 0 such that from the above equality, we deduce xi f(xi) — x2f(x2) = 0 or (xi — x2)f(xi) = 0 whence xi = x2. In the case x1 0 0 = x2 we have f(x2) = 0 and f(xi) 0 0, according to the hypothesis, therefore, f(xi) f(x2). In conclusion, f is an one-to-one function. Problem IX.33. Prove that the points M in the plane that correspond to complex numbers z satisfying the relationship 1z2— a21 = 12az + 1)1, where a, b E R and Ibl < 2a2, belong to two orthogonal circles. Solution. The given relationship is equivalent to (z2 a2) (z2 a2)
= (2az + b)(2a-z- b).
We define t = p2 = z z > 0, where p = 1z1. The relation becomes t2— a2[(z + z)2 — 2t} a4 = 4a2t + 2ab(z + + b2, that is (t — a2)2 (a(z + 7z") + b)2. From this we deduce z • -f a2 = a(z + 727,) + b. The last relation can be rewriten equivalently in the following forms (z — a)( — a) = 2a2 +b or (z + a) (z+ a) = 2a2— b. Since 1b1 < 2a2, it follows 2a2 +b > 0 and 2a2— b > 0. Hence lz — al = .V2a2 +b or Iz + = N/2a2— b. These equations represent two circles with the centers Ci(a, 0) and C2(—a, 0) and the radii R1 = V2a2 +b and R2 = V2a2— b. But 11C1C2112 = 4a2 = R? + that implies that the circles are orthogonals. —
Problem IX.34. Find all functions f any quadratic function g.
R that satisfy fog = g o f, for
88
"Gazeta Matematica" - a bridge over three centuries
Solution. Consider any point xo E R. We want to prove that f (xo) = xo. Then the problem possesses only one solution, the identical function on IR. To this end, let g : R -4 R, g (x) = x2+ (1 — 2xo) x + 4 be a quadratic function. The function g was chosen such that it satisfies: g (u) = u, where u E R, if and only if u = xo. Indeed, g (x0) = xo, and if g (u) = u, then u2— 2xou + x8 = 0, that is (u — x0)2 = 0. Now let us return to the property of f, which in our context is g (f (x0)) = f (g (x0)) = f (xo) . It follows that f (xo) is a fixed point for g, thus we may infer that f (x0) = xo.
Problems for the 9th form
List of authors. Problems for 9th form 1. Abramescu N., IX.3 (G.M. 1903) 2. Andreescu T., IX.13 (G.M. 2/1980) 3. Atanasiu I., IX.2 (GM. 3/1988) 4. Bandilh V., IX.19 (G.M. 5/1983) 5. Watinetu-Giurgiu D.M., IX.23 (G.M. 12/1985) 6. Bencze M., IX.24 (G.M. 3/1987) 7. Bi§boaca N., IX.26 (G.M. 1/1987) 8. Brinzainescu V., IX.6, (G.M. 12/1964) 9. Burdu§el C., IX.30 (G.M. 3/1989) 10. Caragea C., G.M. 9/1979 11. Cavachi M., IX.28 (G.M. 1/1988) 12. Chirita M., IX.15 (G.M. 11/1981) 13. Constantinescu Laura, IX.16 (G.M. 5/1982) 14. Cognita C., IX.4 (G.M. 5/1935) 15. Crainic M., IX.31 (G.M. 4/1989) 16. Cri§an I., IX.29 (G.M. 8/1988) 17. Cucurezeanu I., IX.25, IX.32 (G.M. 7-8/1987, G.M. 2/1991) 18. Gologan R., IX.9 (G. M. 11/1970) 19. Milan N.N., IX.12 (G. M. 1/1987) 20. Ionescu Tiu C., IX.11 (G.M. 10/1975) 21. Manolache N., IX.7 (GM. 12/1967) 22. Metier M., IX.8 (G.M. 4/1968) 23. Mihet D., IX.17 (G.M. 11/1982) 24. Mortici C., IX.14 (G.M. 6/1987)
89
90
"Gazeta Matematica" - a bridge over three centuries
25. Nedelcu I., IX.27 (G.M. 9/1987) 26. Nicula V., G.M. IX.33 (G.M. 2-3/1992) 27. Olteanu M., IX.35 (G.M. 11/2001) 28. Paltgnea E., IX.20 (G.M. 6/1984) 29. Par§an L., IX.10 (G.M. 5/1973) 30. Popa V. M., IX.22 (G.M. 6/1985) 31. Popa V., IX.34 (G.M. 3/1994) 32. Tino 0. N., IX.5 (G.M. 1940) 33. Vijiitu N., Zaharescu A., IX.21 (G.M. 3/1985) 34. Vulpescu Jalea F., IX.18 (G.M. 1/1983).
Chapter 7
Problems for the 10th form Problem X.1. Prove that for n E N* the number 42n+1— 1 does not admit other divisor having the form 221 + 1 except for 3. Solution. We have 429+1— 1 =
4 • 42n — 1 = 4 (22n + 1) (22n — 1) + 3 = = 4 (22n + 1) (229-1 + 1) (220 + 1) (220 + 1) + 3 =
229-1 + i) = 3 + 4 (22n +1) (
(220 + i)
In the second term we have all the numbers having the form 22P + 1 with p < n. Any one of these numbers, in order to divide the given number, should also divide 3. It results that the only number having the form 22P +1 with p < n, divisor of the given number, is 3. We prove that we do not have divisors of the same form for p > 71. We notice that 42n +1-1 < 229+2 +1, which means that we cannot have p > n+2, so p n +1. It's enough, then, to prove that 22n+1 +1 doesn't divide 429+1— 1. This immediately follows if we put the given number under the form 4 (22n+ +1) — 5 and if we take into account the fact that for n E N*, the number 22n+1+ 1 is not divisible by 5. Problem X.2. Let's denote by a a primitive root of the equation xn —1 = 0 ax + b and let's denote f (x) , S = f (a) ± f (a2) + • • • + f (an). Show that cx d we have: Sa be — ad dn+1 — 71 = + c do — (—c)n • 91
"Gazeta Matematice - a bridge over three centuries
92
Solution. Let's have x a root of the equation x" - 1 = 0 and y = We obtain x =
ax +b cx + d
dy - b . As x" - 1 = 0, we have a - cy
(dy - b)n - (a - cy)n= 0 .4=> <#. [dn
Cr] yn
-
n [be-1 + a(-c)n-1] y'l + =0 (1)
-
On the other hand, a being a primitive root of the equation xn -1 = 0, the n successive powers of a give all the roots of this equation. The numbers f (a), f (a2) , , f (a") will all be solutions to the equation (1), corresponding to the roots of the binomial equation x" - 1 = 0. We'll have: b ±a f(ai) = =n do ( c)n i=i i=i
E
E
-
-
that is ad" - a • (-c)n S = n
+ bdn-1
C
ad" -
do
-
( Cr -
S a be - ad do 1 <=> = ± Ti c c dn - (- c)' • -
C
Problem X.3. Solve, in the set of integer numbers the equation x3 +y3 + z3 = 42 + x2 + y2 + z2, x, y, z being successive terms of an arithmetical progression. Solution. Let's consider x = y + u, z = y - u (x, y, z E Z, u E Z+). It results
(y + u)3+ y3+ (y - u)3 = 16 ± (y + u)2 + y2 + (y - u)2<=> <=> 3y3 + 6u2y = 16 + 3y2 + 2u2.#> 2(3y - 1)u2 = 16 - 3y2(y - 1). If y < 0 then the two numbers bear different signs, so the equation has no solution. If y is a positive integer number, that is y > 1, then the number in the left-hand side is positive, while the number in the right-hand side is positive only if y = 1 or y = 2. For y = 1 we have u2 4 < u = 2, hence x = -1, y = 1 and z = 3. u>o For y = 2 we obtain 10u2 = u u Z. The solution is (-1, 1, 3) or (3, 1, -1). Problem X.4. Solve, in the set of integer numbers, the following equation: x3 +24 = 2x.
93
Problems for the 10th form
Solution. For x < 0 we have x3+ 24 E Z and 2x V Z so the equation has no solution. For x > 10, through mathematical induction we can prove that ,10}, from > x3+24, so the equation has no solution. For x E 10,1,2, + 24 = 2' it results that x is even (x = 0 is not a solution) and it follows x3 that only x = 10 is a solution to the given equation. Problem X.5. A sphere is cut by two parallel planes; one passes through the center 0 of the sphere, the second so that it cuts the sphere on a circle having the radius equal to half the radius of the sphere. Compute the total area of the frustum of the cone that has as bases the sections made by to two planes. Solution. N
Let's consider an axial section (see drawing) in which we denote OA = R, R MD = r = — . Hence it results 2 OE = AE = R.
From ,Z:OMD OM =
2
. The generatrix of the frustum of the cone
is AD = R and the total surface required is S =
Problem X.6. Prove that:
(m, + 72 — 1
)
1171R2 4 .
< mn and
n ) . (+13-1) m < ()(n m+P
P -1 ) P
p )'
where m,n,p E N, m > I; n > m + p. Solution. We have
m+n n But
m+i1
+ n I)! (m n!•(m I)! < m for i E {1,2,
results that we have
(m + n 1)
n
1)(m
n — 2) • ... • (m + 1)m n!
, n 1} because (m 1)(i — I) > 0. It < ma with an equality if n = 1. ,
"Gazeta Matematia" - a bridge over three centuries
94 We have:
n )(m+p-1) = p (n)(n (m+p) p m m+ p p-1 and as p < (m + p), the inequality in the problem follows.
Problem X.7. If in a triangle the square of the length of the sides are in arithmetical progression in the order a2, b2, c2, then the measure of the angle B is at most 60°. Solution. We have 2b2 = a2 + c2 and b2 = a2 + c2 - 2ac cos B. We obtain a2 + c2 cos B = 4ac We have to show that 2 C2 > 1 60° •#• cos B > cos 60° aa+ •#. (a - c)2 > 0 , 4ac
which is true. We have equality for a = c.
Problem X.8. Prove that if for the sides a, b, c of a given triangle ABC there exists the relation b2+c2= (b+c).a, then the radiuses of the exinscribed circles, ra, raand r, are in geometrical progression. A Solution. We have ra = ptg - rb = ptg - §i = ptg 2' 2 2 A B c If tg2 9= tg -9 tg 9-then also 7-2 = rb • rc . The relation in the hypothesis will be-writter: b2 + c2 = (b + c) • a <1.> (b + c)(b + c - a) = 2bc. The last relation can be written -4=> (sin B + sin C)(sin B + sin C - sin A) = 2 sin B • sin C 4=> B C A B-C B+C <=> 2 sin cos 4 sin - sin - • cos 2 = 2 2 2 2 = 2 • 2 sin -12 cos/ • 2 sin e cos C-4=> 2 2 2 2 2A B-C B C > cos - • cos = cos - • cos - .4* 2 •2 2 2 1 1B C <=> <=> tg• •tg ,A B C 1 ± tgh — 1+tg2tg2 2
Problems for the 10th form
95
Problem X.9. Prove the inequality: sin A + isTr17.B + sinC < where A, B, C are the angles of an arbitrary triangle. Solution. For any triangle ABC we have: .N/ 3 sin A + sin B + sin C < 3 2 and cos A + cos B + cos C < 2 with equality for A = B = C =
7r
1 Let's have a triangle A1B1C1having the angles Al = -(7r - A), B1 = 2 1 = -(7 - C). Writing for these angles the above inequalities -(7r - B), 2 2 we obtain:
C 31d cosA - + cos + cos - < — and sin 24- + sin- + sin- 23 < . 2 2 2 2 2 2 2 2 From the Cauchy-Buniakovski-Schwarz inequality we have: 2
A B C C (Vsin - cos 21+ \isin -cos + \isin - cos -) < 2 2 2 2 2 2 B —
< (sinA-+ sin LEI + sin C cos A-+ cos + cos -62 <9V5 2 2 2 4 2 2 B —
or
e<3
sin i41- cos A + Vsin Bcos + 2 2 2 2
C cos 2 2 3 4.
B —
<=> Vsin A + sinB + Nis7).0 < 3 •
•4#•
Problem X.10. Prove the following identity: 2 (E 2k • (22nk)) k=0
2
n-1
2
2k k=0
) -1 (k2 1
Solution. We consider the binomial formula (1 +
2n
2n
(271)
2k k=0
E 2k (2n) k=0
n-1
v2 •
2k ( k=0
2n 2k +1) •
"Gazeta Matematice - a bridge over three centuries
96 Similarly,
\ 2n
(1 - v2)
?I k (2n
=E2 •
k=0
2k
)-
71-1
E 2k • (
4- •
k=0
2n
2k + 1) •
Multiplying the two relations side by side, it results
(1 - 2)2n =
2 2 (t 2k On)) _ (v2- . nt1 2k ( 2n ))
k=0
<=>
2k)
2k 1
4=>
2 C-1k=1.° 2n x-, 72 k (2n 2 22 - 2k)) -2 E 2-1 . (k=0 k=0 (2k+1))
Problem X.11. Solve, in the set of integer numbers, the following equation: x2 +24x + 44 = 5Y . Solution. The given equation can be written: (x + 2)(x + 22) = 5Y We can discern two cases: a) x + 2 = 512 , x + 22 = 5u, u, V E Z and u + v = y. By substracting the two relations we obtain:
5u (5v-u _ 1) = 5 • 4 u = 1 and v=2= y = 3 and
x = 3.
b) x + 2 = -5u, x + 22 = -5u = u = 1, -5' = 4 (false). x=3 It follows that the unique solution is Y=3 Problem X.12. Consider a polynomial P = Xn+1- aX" - aX +1 E R[X], n E N*. Show that if lal < 1 then all the roots of the polynomial P have the module equal to 1. Solution. Let's have z a root of the polynomial. It results:
a =
Zn+1 +1 (1 + rn+1cos(n + 1)t) + i •• rn+1- sin(n + 1)t = r" cos nt + r cos t + i (rn sin nt + r sin t) z'1 + z 1 + 2ru+1cos(n + 1)t + r2n+2
lal 2 = r2n + 20+1cos(n - 1)t + r2 •
Problems for the 10th form
97
We have: r2" + 2r"±1cos(n — 1)t + r2 > r2" — 2rn+1 + r2 ..._...., (rn, — r)2 > 0 and
+ 2Tn+1cos(n — 1)t + r2•4=> lal2< 1 <=> 1 + 2rn+1cos(n + 1)t + r2n+2 < T 2n 4=> (1 - r2 )(1 - r2n) < 4rn+1Sill nt sin t.
(1)
The condition a E R .' [1 + r"+1cos(n + 1)t] (r" sin nt + r sin t) —
(2)
— e+1sin(n + 1)t [r" cos nt + r cos t] = 0 •#;• 44. (1 — r2) [(1 + r2 + r4 ± ... + r2n-2 ) Sill t ± Tn-1sin nt] = O. Let's denote A = (1+ r2 + r4 ± .. . + r2n-2 ) sin td-rn-1- sin nt. We successively have: 4r2A sin t = (1 + r2 + r4 ± ... + r2n-2)4r2 sin2 t + 4r"4-1sin nt sin t > > (1 ± r2 .1_ r4 ± ... + r2n-2) 4r2 sin2 t + + (1 —r2)2 (1 +r2 ± r4 + ... ± r2n-2) = = (1 ± r2 + r4 + .. . + r2n-2) (1 _4_ r4 _ 2r -2 cos 2t) > >
(1 + r2 + r 4 + ...+ r2n-2) (1 - r2)2 > 0,
so 4r2• A • sin t > 0
A 0 0. From the relation (2)
r = 1, q.e.d.
Problem X.13. Construct a triangle T, knowing the feet of the altitudes of the triangle T', which has as vertices the midpoints of the sides of the triangle T. Solution. Let's consider T" the triangle obtained by linking the feet of the altitudes of the triangle r . The bisectors of the triangle T", which is orthic to T', are altitudes of T'. From the given T", from each vertex we draw perpendiculars on the corresponding bisector. We obtain the triangle T'. The parallels from the vertices of T' to the opposite sides are the sides of the triangle T. Problem X.14. Find a positive integer x so that 2x +1 is a perfect square and such that among the numbers 2x + 2, 2x + 3, ... , 3x + 2 there aren't any perfect squares.
"Gazeta Matematica" - a bridge over three centuries
98
Solution. We must have 2x + 1 = n2 n2 fi x= 2 n2 •<#. x = 2 n2 fi x = 2 n2 ta x = 2
n2 < 3x + 2 < (n + 1)2•<=> 3n22 +1 < n2+ 2n + 1 .#• and n2 < 2
and 1 1
and 2n2 < 3n2 + 1 < 2n2 + 4n + 2 <=> 1 1
and
n E (2 — V-5,2 +
<=> nEN*
and n E {1, 2, 3, 4} <#. n = 3 and
x = 4.
The only possible solution to the problem is x = 4; 2x + 1 = 9 = 32; 3x + 2 = 14 and among the numbers 10, 11, 12, 13 and 14 there is no a perfect square. Problem X.15. If in triangle ABC we have A > B > C then a(C - B) + c(B - A) + b(A - C) > 0 where the angles A, B, C are expressed in radians and a, b, c are the lengths of the sides of the triangle. Solution. On the graphic of the concave function f : (0,7) -4 R, f(x) = sin x, we consider the points M(C, sin C), N(B, sin B) and P(A, sin A). Between the areas of the subsequently formed trapezium there exists the relation: SMNBC SNBAP > SMPAC '#>
sin A +sin B sin B+sin C sin A + sin C > (A C) +(A B) 2 2 2 •4=> (C - B) sin A + (B - A) sin C + (A - C) sin B > 0 <=> <=> (B C)
<=> (C - B)2R sin A + (B - A)2R sin C + (A - C)2R sin B > 0 .4=> <=> a(C - B) + c(B - A) + b(A - C) > 0. Problem X.16. Let's consider the arithmetical progression (an)n>1 and n,p E N*, with n > 2. Prove the identity: n ap - ( ) • a 1+ • • • + 1 P+
1.)n • () • ap+n =O.
(-
Solution. Let's denote by r the ratio of the given progression. We have ap+k k • r. The relation in the hypothesis can be written:
aP
[1.
1) ± (n2) (
+ (-1)n (71)1
=0. -r (ni- 2 (1 + 3 (1 - • • • + (-1)n±ln • (a 2 3 )
Problems for the 10th form As 1 — (11) (n) — 2/
99
+ (-1)n (n) = (1 — 1)n = 0 and
71) Th = ) + • • + (-1)7741 n ( n 3 3 ( (n1) — 2 (2) +
E(1)k+1 k (n) = E(_nic+1 k=1
k=1
k — 1)
= 0,
it results that the given relation is true. Problem X.17. Let's have x1, x2, x3 the roots of the equation x3— x2— 2x + 4 = 0; lxi I >1x21> ix31• Determine the polynomial f E Z[X] of minimal degree which has the number Xi5 + X23 + x 23as its root. Solution. Let's have a = x7+4+4. As xi, i = 1,3 are roots of the given equation, we have: 4 - 4 - 2xi + 4 = 0, i = 1, 3. We have: x i = X? • X? = 4(X? + 2X1 - 4) = x1 • 4 +2x1- 4x7 = xi(x?+2xi -4) + 2(x7 + 2x" — 4) — 4xT = x? + 2x1 — 12; X2 = 4 + 2X2 - 4 and 3
a
2
2x1 — 12 +4+2x2 — 4+4=E4+2Exi— 16 = i=i 1 — 2(-2) + 2(1 — x3) — 16 = —2x3 — 9.
We'll thus look for the polynomial f E Z[X] of minimal degree and which has as its root a = —2x3 — 9. The polynomial in the problem, g = X 3 'X 2— 2X + 4, with roots x1, x2, x3 is of degree 3 and not have any roots in Q, so it is irreducible in Q[X]. The polynomial h E Q[X] of degree 3, which has its roots a = —2x3 — 9, = —2x2 — 9, -y = -2xi - 9 will also be irreducible in Q[X], because is obtained from h through a linear transformation of the unknown, namely: X —2X — 9. Being irreducible, h will be a polynomial of minimal degree from Q[X] which admits a as a root.
"Gazeta Mateinatica." - a bridge over three centuries
100
As g(X) = h(-2X - 9) it results g(X) = g
(X + 9)3 (X +9)2 +2 X +9
- X+9 ( 2
-g (X'
4
2
+4=
29X2 ±271X + 463) .
To obtain a polynomial from Z[X] we'll have:
f = -8h = X 3 +29X2 + 271X + 463. Problem X.18. Show that any nonconstant polynomial with integer coefficients can be written as the sum of polynomials with integer coefficients, irreducible in Q[X]. Solution. Let's have f = a0 + a1X + • • • + n E N*, the polynomial in the problem. We search for two polynomials g,1r E Z[X], degree g = degree h = n, g = bo + biX + • • • + b,,X", h = co + ciX + • • • + cn Xn, g, h irreducible in Z[X] (so also in Q[X]) and so that f = g + h. Taking, for example: bo = 28(ao - 5) + 2; bi = 10ai i = 1, n co = -27(ao - 5) + 3;
=
b„ = 10(a„ - 2) + 1; i = 1,n - 1; c,, = -9(a„ - 2) + 1,
we have bi+c,, = ai, i = 0,n (f = g +h). Moreover, taking p = 2, we realize that the polynomial g verifies the hypotheses of the Eisenstein criterion, and if we take p = 3, the polynomial h verifies the hypotheses of the same criterion, hence g, h are irreducible in Z[X] (so also in Q[X]). Problem X.19. Let's have et, b, c E (0, 1), 0 < a < b < c < 1. Compare the numbers:
A = logb a + log, b + loga c and B = loga b + logy, c + log, a . Solution. Denoting a = lg a, /3 = lg b, y = lg c, we'll have a < < y < 0 and A = a + + = cc2,.y ± 02a +720 a a07 a 027 +1,2a +a20 + 7 += 7 a a07 7
B
Then A- B = (a -)(a-077)(7 - a)< 0, asa-13 <0, 13-7 < 0, 7- a > 0 and a13-y < 0. Therefore, B > A.
Problems for the 10th form
101
Problem X.20. Given the arbitrary triangle ABC and a point M on the circumcircle of this triangle, the centers of the Euler circles of the triangles M AB, MBC and MCA form a triangle identical to the triangle ABC. Solution. Consider an axis system with coordinates having the origin in the center 0 of the circle circumscribed to the triangle ABC. For a point P in the plane we'll denote by p its affix with respect to the given system. Let's have G the gravity center and H the orthocenter of the triangle ABC and w the center of the Euler circle of the triangle ABC. We denote by w1, w2, c.,./3 the centers of the Euler circles of the triangles IvIAB, MBC, MCA and by z„,„ zw2 , zw, the affixes of these centers. a+b c a+bc We have: g ,h=a+b+c,w= 3 2 +a+b c +a m+b Fc ,z,2= and zw„ = zu„ = m 2 2 2 The distances between the points col, w2, w3 are: -
1 = la — cl 1AC. 2 2 lb — al AB. w2w3 = Izw2 zw3 l = 2 = 1 2 lc —bl 1 — zw, I = = BC. wawl = 2 2 — zw2
w1w2
—
So,
W1W2
AC
(-4)2W3 W3W 1
AB
BC
1
2 Awic,)2w3
ACAB.
Problem X.21. Find the polynomials f E R[X] which satisfy
f (x") = [f (x)r , V x E C,
where n E
n > 2.
Solution. Obviously, the zero polynomial verifies the condition. We are interested to find f 0 that satisfies the given condition. For such a polynomial we have
f (zn2) = f [(z")"1= [f (z")]n= [f (z)]"2 , Vn EN and by induction we find that
f (z&') = [f(z)]"P , Vz E cC, Vp E N. We choose p E N such that k = nP > degree (f). Let xo be an arbitrary root of f. We shall prove that xo = 0. Indeed, assume xo 0. Then, the equation 2:k = X0 has in C k distinct roots:
"Gazeta Matematic0 - a bridge over three centuries
102 xi, z2, • • • 5 zk •
Due to
[f (zi)]k = ff (zt)rr = f (zr) = f (4) = f (x0) = 0, it follows that f (zi) = 0, for all / = 1, 2, . , k. Therefore f has k > degree (f) distinct roots, that is, a contradiction. This implies xo= 0, hence all roots of f are equal to zero. Let m = degree (f). Then f is of the form f = axm, a E
.
From f (xn) = [f (x)]n, V x E C, it follows axmn = anxmn Vx E C and since a 5L 0, we get an-1= 1. Therefore the polynomials that satisfy the condition in our problem are f -a 0; f = xm, f = -xm for n = odd and f 0; f = xm for n = even number, and m E N, arbitrary. Problem X.22. Determine the polynomials f E C[X] having the property: f (xn) = f(x)n, (V) x E C
where n E N and n > 2.
Solution. We assume the problem is solved and let's have the polynomials P, Q E C[X], P(X) = f (Xn), Q (X) = (f (X)n). We'll have P = Q as P - Q has infinity many roots. We can write f (z") = (f (z))nfor any z E C. Hence, f (z"2 ) = f ((zn)') = (f(zn))n = ((f(z))n)n = (f(z))n2 and through mathematical induction, f (znP ) = (f(z))nP, (V) p E N. This allows us to choose p E N, so that nP > degree f, and to denote nP = g E N*, taking into account the property g > degree f. Let's take xo E C a root of f and z1, z2, , z9the complex roots of order that will be distinct if xo # 0. Applying the property of polynomials f in the problem, we'll write (f(zk))g = f (z) = f (x0) = 0, which shows that Azk) = 0. Accordingly, the polynomial f has g > degree f distinct roots, which constitutes a contradiction. So, f has zero roots. If m = degree f, then f = aX", a E C. From f(xn) = (f(x))n, (V) x EC = a• xn•m = an • xn'n, (V) x E C a = 0 or an-i 1. We obtain f = 0 or f = aXm with a E C, an-1 = 1 and m E N*, arbitrary.
Problems for the 10th form
103
Problem X.23. Demonstrate the inequality
(ax + by)3< (1 + a3)(1 + b3)(x3 + y3) for any a,b,x,y E (0, oo). 1 Solution. After denoting b = – and t = –, the given inequality can be written: (a • t - y + – 1y)3 < (1+ a3) c
(1 c3
c3
g, g - y"(1 + t" ) <=>
(1)
<=> (act + 1)3< (1 + a3)(1 + c3)(1 + t3) <=>
.@> act +1< -01+a3)(1 + c3)(1 + t3). On the other hand, according to the inequality of means we have:
act + a3)(1 + c3)(1 + t3) < 3
(
a3 c3 t3 1+a3 + 1 + C3 + 1 ± t3
and 1
1 1 1+c33)
1( 1
Summing the previous inequalities we obtain (1). The equality holds for a = c = t .#› ab = 1 and – = a. y
Problem X.24. Solve in the set of strictly positive numbers the equation:
xx = ax-Fa2
where a > 1 is fixed.
Solution. We apply logein both sides to obtain
x logex = x + a2 which can be written a2 logex = 1+ — . x The left hand side has an increasing function, while in the right hand side is a decreasing function. Therefore, the equation has at most one solution. As x = a2verifies the equation, it follows that it is the only solution.
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Problem X.25. Solve in R3the system:
xV1 + y2 + y07F x2
1
yV1 + x2 + z\/1 + y2 = 2 + X2 + +z2 = 3 Solution. We denote x = sh a, y = sh b, z = sh c, where sh t = et + e-t cht = tE R.
et - e-t 2
2 We have: ch2t = 1 + sh2t and sh(u + v) = sh u • ch v sh v ch u. The system will be sh(a + b) = 1 1 a + b = arc shl b-Fc = arcsh2 , sh(b+c) = 2 4 sh(c + a) = 3 l c + a = arc sh3 where arc sht = ln (t + Vt2 + 1), t E Then 1 a -(arc shl + arc sh3 - arc sh2) = 2 1 - ln (2 + = [ln (1 + 4 + ln (3 + 1 2
ln
(1+ V-2-)
=
(3+
2+
and sh a = 1 2
(1 + 4(3+ 10) 2+f
2 + Nig
1=
(1 + A (3 +VT .0)
1 1+3V2+VT)+\16 = „ '.\/(2 + A (1 + N12-) (3+ V1C) 1 1 (1 + 3.\/+ 10+ V(2 + /6) (1 + N/2) (3 + 10). Similarly we obtain y and z.
Problem X.26. Let's have x, y,zEC*,x0y0z0x, so that lxj = ly1= Izi = 1 and
(x+y+z)5 =(y+z-x)5 +(x+z-y)5 +(x+y-z)5 Compute lx y + zi.
Problems for the 10th form
105
Solution. Consider the polynomial:
P(X,Y, Z) = (X +Y + Z)5— (Y + Z — X)5— (Z + X — Y)5 — (X + Y Z)5 . As P(0, Y, Z) = P(X, 0, Z) = P(X, Y, 0) = 0, it results that:
P(X, Y, Z) = XY Z • Q(X, Y, Z). As P is a symmetrical and homogenous polynomial of degree 5, with integer coefficients, it results that Q is a symmetrical and homogenous polynomial of degree 2, with integer coefficients. We search for Q having the form:
a(X +Y + Z)2+ b(XY + XZ +Y Z),
where a,b E Z.
We have: P(1, 1, 1) = 35— 3 = 240 = 9a + 3b P(1, 1, —1) = 3 — 35= —240 = —a + b, whence a= 80, b = —160. It results that P(X, Y, Z) = 80XY Z (X 2 +Y2 + Z2). According to the relation in the problem, we have x2 + y2 z2 = 0. Through conjugation it results: 1 1 1 x2 +y2 + z2 0 X2y2
y2z2
z2x2 0 <4.
.#› (xy + yz + zx)2 = 2xyz(x + y + z) (1) On the other hand, from x2 + y2 + z2 = 0 we obtain
(x + y + z)2 = 2(xy + yz + zx)
(2)
From (1) and (2) we obtain (x+y+z)4 = 8xyz(x+y+z), whence x+y+z = 0 or (x + y + z)3= 8xyz. In the second case we have:
Ix + y + zI3= 81x1 lyl Izi = 8
ix + y + zl = 2
So, Ix + y+ zl E {0, 2}. Problem X.27. Show that in any triangle we have Ma
>
la —
b2
c2
2bc •
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106
2(b2 +c2) — a2 2bc A and la = cos — 4 b+c 2 a+ b +c . Vbcp(p a), where p = 2 b+c 0 + c)2 _ a2 = be [(b + 02 _ az] 4 the inequality in Since l?, = be ' 4 (b + c)2 (b + c)2 the problem is equivalent to b2 + c2 <=> n'ta > 421 2bc 2(b2 ±c2) _ az be [(b+ c)2— a2] b2 + c2 > <=> <=> 2bc 4 (b + c) 2 <#. 0 + 02 [2(b2 +ez) _ a21i 2(b2 + c2) • kb + c)2— •4=> a2(b 4_ c)2 > c2) 2a2(b2 c2) 44 + c)2(b2 .#› 2(b + c)2(b2 c2) — Solution. We know that ma =
,
2(b2 + c) ? (b + c)2<#. (b — c)2 > O. The equality holds for b = c, that is, for isosceles triangles. Problem X.28. Find the relationship between the angles of a triangle whose center of gravity lies on the circumference of its incircle. Solution. Given by G. Titeica (Tzitzeica) (1873-1939). The condition in the problem expresses the fact that the distance between the center of gravity and incenter equals the radius r of the incircle. Denote G and I the circumcentre and incenter of the triangle ABC, respectively and let M and D be the midpoints of BC and AG, respectively. Let also E and F be the contact points of the incircle to BC and AC. Then we have AD = DG = GM. On the other hand, in the triangle IDM we have I D2 + 1m2 MG = ID2 + 1E2 ME2 AM2 2 /G2 2 2 9 ID2 +1E2+ (MC — CE)2 AM2 2 9 • In the triangle IAG we have
2 /G2 + /A2 ID = 2
GD2 = IG
2+ IA2 AM2 2 9 •
Therefore
2 IG=
/G2 + IA2 AM2 4
18
7,2
a a+b—c) 2 ' 2 2 ) 2
AM 2 9
Problems for the 10th form
107
which yields
3IG2= IA2
2AM 2 (b +2r2 + 3 2
But IA2 = /F2 AF2
r2
(b + c— 4
a)2
and AM2 =
2 b2 +e2 a 7 2
and so after all computations we find 5 (a2 +b2 + c2) — — 1b e + ca + ab). 3/G2 = 3r2+ c • — 12 2( Using now IG = r, we get 5(a2 + b2 + c2) = 6(ab+ be + ca) a homogenous relation with respect to a, b and c. Using now the sine theorem, we obtain 5(sin2 A + sin2 B sin2 C) = 6(sin B sin C + sin C sin A + sin A sin B) which is the required relationship between the angles of the triangle ABC. Problem X.29. Prove that between the elements of an arbitrary triangle we have following identities a sin B sin(C — a) — c sin C sin a = b sin C sin(A — a) — a sin A sin a = = c sin A sin(B — a) — b sin B sin a; a sin C sin(B — a) — b sin B sin a = b sin A sin(C — a) — c sin C sin a = = c sin B sin(A — a) — a sin A sin a,
where a is an arbitrary angle. Solution. Let us consider the first group of identities. If we expand sin(A — a), . . . and replace a, b and c by 2R sin A, 2R sin B and 2R sin C, respectively, we obtain sin A sin B sin C cos a — sin A cos B sin C sin a — sin2 Bsin a = = sin A sin B sin C cos a — cos A sin B sin C sin a — sin2A sin a = = sin A sin B sin C cos a — sin A sin B cos C sin a — sin2 Csin a.
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108
To prove the previous relations it suffices to show that sin A cos B cos C
sin2 B = cos A sin B sin C = sin A sin B cos C
sin2 A = sin2 C.
Indeed, we have sin2A — sin2 B = sin(A
B) sin(A — B) = sin C sin(A — B),
or expanding sin(A — b) and grouping: sin2A + cos A sin B sin C = sin2 B + sin A cos B sin C. Similarly, sin2 A + cos A sin B sin C = sin2C + sin A sin B cos C. The second group of formulae can be reduced to the first one. Problem X.30. Let F(x, y) be a rational function of variables x and y that has real coefficients. Prove that we can find three polynomials P(x, y), Q(x, y), R(x, y) with real coefficients symmetrical with regard to x and y, such that F(x,y) = P(x, Y) + (x — y)Q(x, Y) . R(x , y) Solution. We have F (x, y)
0+F(y, x)
coje(x ( x : Yy) ,where f
,
E 111[x , y]. Then
f (x,Y)f (y, x) f (x,Y)(p(Y, x) + f (Y, x)(p(x, y) (1) = yo(x,y) co(y, x) 49(x OP(Y x)
and
F (x, y) — F (y, x) = f (x, y) (P(x,Y)
x) V(Y,x)
f (x, Y)(00(Y , x) — f (y, x)co(x,Y) . (2) go(x, Y)(10(Y x)
We take R(x , y) = cp(x, y)cp(y,, x), which is a symmetrical polynomial. Then we can write f (x, y)co(y, x) + f (y, x)cp(x, y) = 2P (x , y), where P (x, y) is a symmetrical polynomial. Equation (1) becomes , y)
F(x,y) + F (y , x) = 2 P (x R(x,y)
(3)
The polynomial f (x, y)cp(y,, x) — f (y, x)cp(x , y) is antisyrnmetrical and it is divisible by x — y (since for x = y it vanishes). Then we can write f (x,y)(p(Y,x) — f (Y, x)co(x, y) = 2(x — y)Q(x, y).
(4)
Problems for the 10th form
109
Since the polynomial from the left hand side is antisymmetrical and x y is alike, we obtain from equation (4) that Q is symmetrical. Using (2) and (4) we deduce 2(x F(x, y) + F(y, x) = (5) —
—
y)Q(x,y)
R(x, y)
From (3) and (5) we get the equality in the statement of the problem.
110
"Gazeta Matematice - a bridge over three centuries
List of authors. Problems for 10th form 1. Albu T., X.9 (G.M. 3/1963) 2. Angheluta Th., X.6 (G.M. XLIV/1938) 3. Berinde V., X.16 (G.M. 7/1991) 4. Callan F., X.15 (G.M. 1/1977) 5. Chite§ C., X.20 (G.M. 7/1991) 6. Cristescu V., (G.M. 1/1895) 7. Diaconu I., X.19 (G.M. 4/2001) 8. Dumitrescu Gh., X.3 (G.M. XXIII/1917) 9. Georgescu-Roegen N., X.4 (G.M. 27/1921) 10. Gheorghiu M., X.7 (G.M. 3/1958) 11. Gologan R. N., X.14 (G.M. 10/1971) 12. Halanay A., X.12 (G.M. 3/1972) 13. Ioachhnescu A. G., X.2 (G.M. X/1905) 14. Ionescu I., X.13 (G.M. 1/1896) 15. Istrate G., X.18 (G.M. 3/1990) 16. Lalescu T., X.1 (G.M. VI/1901) 17. Mu§uroia N., X.23 (G.M. 7-8/1996) 18. Niculescu L., X.22 (G.M. 9/1994) 19. Neac§u M., X.11 (G.M. 11/1969) 20. Nedeianu D., X.25 (G.M. 11/2002) 21. Panaitopol L., X.17 (G.IVI. 5/1989) 22. Par§an L., X.24 (G.M. 7-8/2001) 23. Simionescu G. D., X.5 (G.M. 9/1937)
Problems for the 10th form 24. Tknaian T., X.26 (G.M. 1/2003) 25. Timofte V., X.21 (G.M. 5/1991) 26. Tomescu I., X.10 (G.M. 11/1965) 27. Titeica G., (G.M 2/1896) 28. Vocia V. G., X.8 (G.M. 1/1962)
111
Chapter 8
Problems for the 11th form Problem XI.1. Consider Sn
=
1
1 1 + + • • • +— 07— zvn. .1, V1 V2
Prove that: 1) Snis negative and increases in absolute value as n is increasing. 2) if n tends to infinity, then Sntends to a finite limit between 2 and —1. —
Solution. 1) Let f : (0, oo) (0, oo) be the function defined by f(x) = 2 \5. For a positive integer k, using the Lagrange's theorem on the interval [k, k 1], we find a number ck E (k, k + 1) so that
t(ck) =
1 <=> — A
(k + 1) — f (k) (k + 1) — k
.
2 \511-1. — 2\5.
The inequality k < ck < k + 1 implies 1 \,/k + 1 <2 So, we obtain S„+1 — Sn
1 1— 2\5< f— rz' k = 1,2,--• 1
(1)
2-VTI + 2\ n < 0, for any positive +1 integer it. Then (Sn)n>1 is a decreasing sequence. But S1 = —1. It follows that Sn < 1, V n > 2. Therefore, (1Sni)n>1 is an increasing sequence. 2) To prove the convergence of the sequence (SS)n>1, it is enough to find a lower bound. The sum of the right inequalities of the relations (1), for k = 1, 2, • • • , n, gives us the next classical inequality 27 \711-1 — 2 <
1
+
113
1
1 VF?..
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114
We obtain Sn >2\/n + 1 -2071, -2 > -2, V n > 1. Therefore, the sequence is bounded and hence (Sn)n>i converges to a finite limit s E (-1, -2). The value of the limit s of the sequence (Sn) is s = -1.455345. Note. In the Romanian literature (Sn) is known as the Ioachimescu's sequence. Problem XI.2. Find lim anfor the sequence of real numbers (an)nEN, n—∎oo an — n+ 1 Solution. We give here a solution based on an ideea of M. 'ena presented by 0. Carja in the journal Recreatii matematice, Iasi, an V, nr. 2/2003, as the following proposition: If (an)nEN is a strictly positive real sequence satisfying the conditions: ti an an+i liin — = a E (0, oo) and lim = Q E (0, oo), then n—*oo n n—)ao a„ lim (an+i - a,1) = a ln n—,co 1 an By applying this proposition to a„ = rni we have lim — = Ern ( c/In)11 = lim an 1 71-■ 00
n V(n + 11
( n+
n---■co k
c/T 1,!
■n--.00 n n
= iiM
(n + 1)n+1
lira
". , 11.
e
= e.
1 1 Then lim a„ = -In e = -. n—+co e Note. In the Romanian literature, (an)nis known as Lalescu's sequence. Problem XI.3. Find the sum of the following series 1+
1 2•3
1 1 1 + 3 • 5 5 • 8 8 • 13+
,
where the denominator of any fraction in the series is the product of two factors each of them being the sum of the factors (on the same position) in the denominators of the previous two fractions. the Solution. The series can be written in the form: 00
(_1)-
1+ ,z=1
1 where an on
(1)
a() = 2, bo = 3, al = 3, bi = 5, and an =an-9 ± bn-2, bn = an-1 bn-1, (V) n > 2. From these recurrence relations an+2 = an+1 + an and bn = an+i, (V) nENU {0}.
(2)
Problems for the 11th form
115
one can easily verify by induction that an > n, (V) n E N and hence the absolute value of the general term of the series satisfies (-1)n ailbn
1 anb„
1 anan+i
1 1 < —. n(n -I- 1) n2
Consequently the series is absolutely convergent. We prove by induction the equality a„an+2 - an2+1 = (-1)n, (V) it E N U {0}. ( 3)
For n = 0, n = 1 the proof is easy. If we suppose that this equality is true up to '71, we have: an+ian+3 - an+2 = (a„_i a„)(a„±i + an+2) - (a„ an+1)2 = nan+ i + an_ian+2 - 2anan+i = - an n + anan+i - an+1 a = = Hir-1 (_i )n ani an-2 - anan+1 = an- (an + - anan+i = = anan-i + an-ian+i - an(an + an-1) = an_ian+i - an = (-1)11-1 = Hir+1, that is the statement is true for n 1, too. Using equation (3) we can rewrite the general term in the form: (_i)k
1 akbk
akak+2 - a/2,±1 = ak+2 ak+i akak+1 ak+.1 ak
(4)
In (4) we choose k = 0, 1, 2, ... , 71. and we obtain 1 aobo 1 al bi 1 a2b2 (-1)n
1-1,
a2 _ al al ao a3 a2 _ a2 al a4_a3 a3 a2 an+2
an+1
an+1
an
00 1 Adding these inequalities we get the partial sum of the series > (-1)" : anan 71=0 Sn =
an+2 3 an+i 2
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116
Since (Sn) is a convergent sequence, it follows that
(an+2)
an+1
is also a conver-
gent sequence. Using (2) we have an+2
= an+i + an = 1 +
an+1
.
If we define 1 = hm
a n+2
an+i
an+1 a,1
, by taking the limit in the last equality we obtain
CO an+
/=1
1 and 1 =
±1
2
Finally, by taking into account equation (1), the sum of the initial series is 1 + Ern Sn = + n—■ co
■ ,/g + 1 3
2
2
= -\/2
Problem XI.4. Compute the determinant: (11)
(P-N
P )
(p•-I2-1)
(712)
(2p —1) P )
Ap =
(2p-1) P ) (73,24134 ) •
( 2pp —12 ) ( 2pp— 1) (2p-3 ) (2p-2 ) (2pp-1) i) ) (p+1)
(2p-2) (2p-1)
Solution. Substracting each line from the next one and taking into account the formula (n+1) n 1) = (ti) we obtain: k+1 — (k+ (2p-2) kp-1
(2p-1) p )
(2p-2) (2p-1) p+1 ) p
Ap = p )
• (3p-53p-4 ) 2p-2) k2p-3) (
(2p-2) 2p-1) ) ( p+1 ) \
(3p-4) (3p-3) k2p-2) k2p-1)
) ( p-1)
117
Problems for the 11th form
We repeat this operation for the lines having the index greater than or equal to 2, 3, ... ,p and we obtain (2p-2 \ p-1 /
(2p-1) P
2p-2\ P
(2p-1 \ p+1 /
(
• • •
Ap = (22p p-32)
k 2p- 2/) (2p-1 (22p p-- 22 ) (2p-1 k2p- 1 )
Now we make the same sequence of operations but with regard to the columns and we obtain (p7) 1)
(77) )
(p)
0
0
0
0
0
Op =
Bicause (P) = 1 and signum (
1
2 p 1— 1
P
1
= ( 1) -
p(2 1)
2
,
we get
= (-1) P(P 2-1) Problem XI.5. Let Ox, Oy be two orthogonal lines and A some fixed point on Oy. Consider AA' parallel to Ox and some point M that moves on AA'. Let B be the midpoint of AM. Consider the equilateral hyperbola (H) which passes through A and M and is tangent to Ox at 0. Then find i) the locus where (H) intersects the parallel to OB through M; ii) the locus of the projection of D on the parallel to OB through M where D is the center of (H) . Solution. Assume that A (0, a), B (k, a) , and M (2k, a) . Then AM has the equation y — a = 0, OA has the equation x = 0 and OM: ax 2ky = 0. Certainly, the tangent in 0 to (H) is y = 0. Then, any conic passing through A and M and tangent to Ox at 0 corresponds to a value of the parameter m in the equation y (y — a) + mx (ax 2ky) = 0. This equation might be written as max 2 2mkxy + y2— ay = 0. This is a hyperbola equation if —
—
—
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118
00 I . Moreover, (2' the hyperbola must be equilateral, i.e. the two asymptotic directions are orthogonal. Hence ma = –1. We obtained that (H) has the equation axe – 2kxy – ay2 + a2y – 0. Now we are in position to find the two loci. i) The parallel to OB through M has the equation ax – ky – ak = 0. Then the locus must satisfy (y – a) (x2 + y2+ ay) = 0. Obviously y – a = 0 is the locus of M, therefore the locus is given by x2 + y2 + ay = 0, the a\ a2 and radius — equation of a circle of center 4• a2k a3 ii) It can be easily found that the center of (H) is D ( 2a2 + 2k2' 2a2 + 2k2) and the locus is the circle 2x2 + 2y2 + ay = a2.
and only if ma :– k2m2 < 0, that is m E ( 00, 0) U -
CO, --2
Problem XI.6. Prove that the sequence 1 1 1 – + log sin – , n= 1, 2, • • • Gn =sin 1 + sin – + • - • + sin n
n
2
is convergent as n –+ oo. Solution. Define x„ = sin 4., n = 1, 2, • • • . Clearly we have x„ > x_ n+i for all positive integers n. From the well-known inequality 2
x+y
<
log x – log y < x–y
1
, V x, y > 0, x 0 y,
we get Xn – Xn+1 2(xn – xn+i) < log xn log xn+1 < , n= 1,2,• – . xn + xn+i Vxnx,/±1
Also we have
x3 x – — < sin x < x, V x > O. 6
(1)
(2)
So, we obtain 1 1 __ n 6n3 n + 1
2 (1 2(Xn – Xn+1) Xn + xn+1
>
1 1 1 7-t 6n3± n + 1
1 1 – 1)(4n +1) = (n > xn+1) V n > 1. + 12n3 + 6n2 –n– 1 n+ 1 > n+1
Problems for the 11th form
119
Using the above inequality and the left inequality of (1) we find xn+1 < log xn - log xn+1. Therefore Gn -Gn+i = log xn - log xn+i-xn+i > 0, 72 = 1,2, ... and (Gn)n>i is a decreasing sequence. To prove the convergence of (Gn)n>1it suffices to find a lower bound of the sequence. From (2) we have , V n > 2. (3) > \i Vxnxn+i > \/ (n 6n3 ) n + 1 6(n + 1)3) n(n + 2) sin x is decreasing on (0, oo). x 1 n 1 xn+i > V n > 1. Then, using It follows that f (n +1) > f (n), or n+ l' xn (3), we obtain The function 1(x) =
Xn — Xn+1
= 1
Xn+1 Xn
Xn
<1
n = n+1
1 1 . < Vn(n + 2) < Vxnxn+i, V n > 2. n+1 The above inequality and the right inequality of (1) imply log xk - log xk+i < xk, k = 2, 3, • • • , n - 1. The summation of these inequalities gives us: log x2 - log x„ < x2 + x3 + • • • + xn-1, n > 3. 1 Hence, it follows that Gn >sin 1 + log sin ., for n > 3, i.e. (G„) is bounded. Therefore the convergence of the sequence (Gn )n >1 is proved. Problem XI.7. Compute the derivative of the expression
y=11x+Vx4-...+ \5, that contains n radicals. Solution. Using the notations
Y1 = \5, Y2 =
1/X
+ fi, Y3 = VX + Vx + Nii, and so on,
we obtain dy dx
1 + 2y1 + 22y1y2 + 23y1y2y3+ ... + 2n-1Y1Y2•••Yn-1
2"Y1Y2••• Yn
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120
Problem XL& Consider the function f(x) = aoxm + aien-1 + + am, and An, Gnthe arithmetic mean and the geometric mean of the numbers f (1), f(2), f(n). Prove that:
ai > 0, i = 0,1, ... , m
lim
n /Gn = em/(m + 1).
n—>oo A
Solution. We have f (x) > 0, x = 1, 2, ..., n. So the geometric mean is well defined. Let Sn,k = 1 + 2k + 3k+ . . . + nk . Then lim Sn'k/nk4-1= 1/(k +1) n--■co by Stolz-Cesaro's theorem. We have
An= (1(1) + f (2) + + f (n))1n = aoSn,m /n + aiSn,m-1/n
+ am .
Consequently, lim An/um = ao/(m + 1). It is known that if there exists n—∎ oo lim un = u, then we have lim (ui u2 + + u„)/n = u. n—000 72,--∎o0 Choose un = log[f (n)/nm]. Then lim un = log ao, and noo
(log [f (1)111 + log [f (2)/21
log[f (nm) nni ]) = log[Gn / (nrin].
We get lim log[Gn/(n!)nin] = log ao or lim Gn/(nrin = ao. 72-±00 n—,00 Hence ipm = 1/(m + 1). nlirrl oo(An /Gn) • ( VrIt is known that if there exists lim un+i /un, then there exists lim Or r, n—>oo and these two limits are equal. Taking un = n!/nn, we have lim un+i/u„ = lim (n/(n +1))n = n--*oo
n--000
and hence lim It follows
= Ern i'/Ti!ln = 1/e. lien An1Gn = em/(m + 1).
71-00
Problem XI.9. Prove that in a triangle ABC the following relation holds: sin A sin B sin C sin 2A sin 2B sin 2C sin 3A sin 3B sin 3C
(a + b c)2(a - b)(a -c)(b - c) 8R5
•
Solution. Denoting A our determinant, we have: sin A sin B sin C A = 2 sin A cos A 2 sin B cos B 2 sin C cos C sin A(4 cos2 A - 1) sin B(4 cos2 B -1) sin C(4 cos2 C - 1)
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121
1 1 1 cos C = 2 sin A sin B sin C cos A cos B 4 cos2A— 1 4 cos2 B 1 4 cos2 C — 1 —
1 1 1 = 8 sin A sin B sin C cos A cos B cos C cost A cos2 B cos2 C = 8 sin A sin B sin C(cos B — cos A)(cosC — cos A) (cos C — cos B), because the last determinant is a Vanderrnonde determinant. It is easy to show that: C cos — 2(sin A sin B), cos B — cos A = sin — 2 and the analogous relations. Thus we have: —
A A = 64 cost— cost 2
cos2
— B
(sin A — sin B)(sin A — sin C)(sin B
— C
2
—
sin C).
Finally, we use the relation
a+b+c H cos —A = 2 8R and the sinus theorem to obtain:
b + c)2(a A = (a +
b)(a c)(b c) 8R5 Problem XI.10. Let xOy be a Cartesian reference system and A(a, 0) a fixed point. A moving straight line through the origin intersects the straight line x = a at B. On the x-axis we consider a point C so that OBC = BOC. The parallel through A to OB intersects BC at P. Find the locus of the projection of the point P on OB. —
—
—
Solution. We try a sinthetical solution. Consider a moving straight line through the origin of the frame, a parallel through the point A(a, 0) to the yaxis and a circle centered on x-axis and tangent to Oy and to the considered parallel. The mobile line intersects the parallel at S and the circle at 0 and T. We consider also a point M on [OS so that OM = TS. Then M describes a cisoide. For our problem OB intersects the circle at D, the parallel to OB intersects the circle at E and Oy at F. Then OBAP is an isoscel trapezoid and OBAF is parallelogram. Then AFOP is isoscel and this equalities hold FE = EP = OQ = DB. It results that Q describes a cisoide. The analytical solution is quite immediate.
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"Ga.zeta Matematica" - a bridge over three centuries
Problem XI.11. Let f : [a, b] -> [a, b] be a function such that f (f (x)) = x and f(x) 0 0 for any x E [a,b]. Prove that f has an infinite number of points of discontinuity. Solution. If f (xi) = f (x 2), then f (f (xi)) = f (f (x2)) and from hypothesis xi = x2. Therefore the function f is one-to-one. Also, for any x E [a, b] there is t E [a, b], such that f(t) = x, namely t = f (x). Therefore f is a bijection. Suppose ad absurdum, that f has a finite number (or zero) number of points of discontinuity. Let a = xo < xi < < xr, = b be the points of discontinuity, together with the endpoints of the interval. Set h = xi) and Ji = f (Ii,), for i = 1, , n. Then f is continuous on each interval I. From the Darboux property it follows that Ji are also intervals. Since f is one-to-one, f is a one-to-one correspondence between h and Ji for each i. It is known that any continuous bijection between two intervals has a continuous inverse. Note, that from the condition f (f (s)) = x, for all x E i = 1, n, it follows that the inverse function of the restriction of f to h is the restriction of f to Ji. We conclude that f is continuous on each interval Hence Jin{xo , xi, ... xn,} = 0. Since Ji is an interval it follows that there is an index k E {1, 2, ... n}, such that Ji C I. Denote this index k by co(i), for i E {1, 2, ... , n}. We obtain a function yo : {1, 2, ... , n} -> {1, 2, ... , n}, such that f(h) c ' p(i), for all i E {1, 2, ... , n}. On the other hand, since f is surjective we have ,
[a, b] =
U
gii)U {f (x0), f(x1), • • • , f(xn)}.
This implies that the map yo must be surjective (and consequently a bijection) on the set {1, 2, ... , n}, and also we must have f (10 = ' phi), for all i E {1, 2, ... , n}.
(1)
Otherwise, since the set {f(xo), f (xi), • • • , f (xn.)} is finite, it would follow that there is i E {1, 2, ... , n} and a subinterval (a, f3) C h such that (a, [3) ,Z f ([a, b]). It follows that f ({1, 2, • • • , n}) = {f(xo), f (xi), • • • ,.f(x.)}. Let us denote by b : {0, 1, , n} -> {0, 1, , n} the bijection defined by the equality f (xi) = zoo), j E {0, 1, . , n} . (2) From the condition f(x) x, (V) x E [a, b] we have OW j (V) j E {0, 1, , n}. This implies that n must be odd. But from the same condition of the hypothesis we can also derive that cp(i) i
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123
(V) i E {1, 2, ... , n}, which implies that n must be even. So the entire proof will follow. In order to prove this, let us suppose that there is i E {1, 2, ... , n}, such that co(i) = i, that is WO = I. Since f is continuous and injective on the interval h, it follows that f is strictly monotonous on this interval. If f is strictly decreasing, since the function f(x) x changes the sign on /i, from the Darboux property we must have a root of the equation f(x) = x in the interval I. But this is impossible. Consider now that f is strictly increasing on the interval I. From the hypothesis and from the Darboux property it follows that one of the following two conditions is satisfied: i) or ii) f (s) > x, (V) x E Ii. Choose e E Ii , arbitrarily. f (x) < x, (V) x E If the condition i) is satisfied then we have x* = f (f (x*)) < f (e) < x*; contradiction. If the condition ii) is satisfied then we have e = f (f (e)) > f (x*) > ; contradiction. Because we obtain contradictions in all cases, the assumption that there is i E {1, 2, ... , n} such that y)(i) = i, is false. —
Problem XI.12. In a given circle we inscribe a regular hexagon. In this hexagon we inscribe a circle. In the second circle we inscribe the second hexagon and then we inscribe another circle and so on. Demonstrate that the sum of the areas of all hexagons cannot be greater than four times the area of the first hexagon. Solution. Denote by Ri the radius of all inscribed circles (R1 = R is the radius of the first circle), ai and Si the apothems and the areas of the inscribed hexagones, respectively. We have = R, R2=
R.4 2
3R 3-13-R R4 =4 8 •••
RO-
3R 30.R 9R 2 ' a2 = 4 ' a3 — 8 'a4 = 16 ••• 6R.a1 3R2.4 , 9R2.13- , 27R2 V303 = Si = , 02 2 2 8 32 ••• So, we obtain al =
-
sn =
3R2 -4 2
+
3 (3)2 + + 4 4
+ (7 33n 2 n
3R2 0-1 — 2
'
(1_ (1 — ( ) 1 =6/R2 = 4 1— 4
Taking the limit, we obtain lim sn = 6\R2. 00
=
).
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"Gazeta Matematice – a bridge over three centuries
Then 4S1 = 4
3R2' = 6R2 V-3-
and, consequently, 4S1 > sn• Problem X1.13. Let A be a quadratic matrix with real entries such that the sum of the entries on the main diagonal is less than the sum of the entries on the secondary diagonal. Does there exist any matrix X such that X • Xt = A, where Xt is the transpose of X ? Solution. Let us suppose that the equality A = X • Xt holds for a matrix X. If (xii) i=1n; j=1m and (aik)i,k=i7i denote the entries of X and A, respectively, then aik = Eixii • xki, for any i, k = 1, n. The sum on the main diagonal is Ei aii = Ei (xii)2and the sum on the secondary diagonal is Ei Ei ixiixn_i,j. We apply the Cauchy-Schwartz inequality and get
Exiix,i_i
■ Since Eid Ei 4 j, it follows that Ei = that is Ei < Ei But the matrix A satisfies exactly the opposite inequality. Consequently, such a matrix X does not exist. Problem XI.14. Let ai bi E (0, 1) U (0, 00), for i E {1, 2, ... , n}, such that ,
aibf + a2t, + +ang > al + a2 + . • • + an•
Prove that b711 2i2• ... • bc,tn = 1. Solution. Consider the function f : IR —> IR, f(x) = al bf. + + a ng. Note that 1(0) = al + an, and the relationship in the problem becomes f(x) > f(0), that is 0 is a minimum point for f . From Fermat's theorem we have 1(0) = 0. But ,1(0) = al In b1 + .. • + b,1In bn= In b71• ... • ban" and In u = 0 u = 1. Problem XI.15. Prove that any function f :R R satisfying:
f (x)EQ if andonlyif f (x+1)0Q has at least one point of discontinuity.
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125
Solution. Let us suppose that f is continuous on R. Then the function g : R R, defined by g (x) = f (x + 1) f (x) is also continuous on R. Now, notice that g (x) E R\Q, for any x E R. Indeed, f (x + 1) and f (x) are never both rational, or both irrational. Now, since g is continuous, g (IR) is an interval, and this clearly implies that g (R) is in fact a single irrational point. Hence f (x + 1) = f (x) + a, for any x E R, where a E \Q. Consider x such that f (x) E Q. Then, f (x + 1) E R \Q and f (x + 2) E Q. Indeed, if f (x + 2) were irrational, then f (x + 1) would be rational, impossible. For the proof, notice that f (x + 2) = f (x + 1) + a = f (x) + 2a implies that a is rational, contradiction. —
Problem XI.16. Prove that for all positive distinct real numbers a and b, the following inequalities hold:
2 ( al < e. a + b bb ) Solution. Observe that the symmetrical identity ib 2 (0)g 2 (bb ) a + b bb b + a aa < 2<
holds. Thus, we can suppose a < b. We have: 2 (b a +1) a a 2 (aa aq7, 1b — a)b-. ) ( with b — a s >0. a + b bb ) a ) b—a +2 a Consider the differentiable function f : (0, oo) ----> R, defined by —
f (x)
2 (+ x 1) +2(1 + x) x , V x > O.
The derivative of f is f'(x)
2 (1 + x)1+1 ( 2x
x2(x + 2)
+2
log(1 + x)) , x > 0.
2x x2 2 log(l+x), x > O. We have g'(x). <0 x+ (x + 1)(x + 2)2 , V x > 0. Since g(0) = 0, we get g(x) < 0, V x > 0. So, f has a negative derivative on (0, oo). Thus f is a decreasing function on (0, oo) and we obtain: Define g(x),
e= lim (x) > f (x) > lim f (x) = 2. x.00 s\o
"Gazeta Matematice — a bridge over three centuries
126 2 ( as) a-17 a+ bb is proved. But
f
(b a) — a
. Therefore the statement of the problem
Problem XI.17. Let f : [0,1] > R be a continuous function. Prove that co the series E f(xn)/2n is convergent for any x E [0, 1]. Find the function f —
such that
n=1
00
f (x) =
E f (x")12n , (V) x E [0, 1]. n=1
Solution. Since f is continuous, from Weierstrass theorem, it is bounded and it attains its bounds. So there exists a constant M > 0 such that I f(x)I < M, (V) x E [0,1]. The partial sums Sn of the series of absolute values verifies n
n
Sn=
E k=1
Then the series
f (x)I12k < M
E 112k= M(1 — 1/2n ) < M. k=1
00
E 11(x) I /2nis absolutely convergent and so convergent.
n=1 00 Let now a function with the property f (x) = E f (xn)12n ,(V) x E [0,1]. 71=1 inf f (x). From WeierLet a E (0,1). Put Ma := sup f(x) and ma sE[0,a] sE [0,a] strass theorem (3) xa E [0, a] such that f(xa) = Ma and (3) ya E [0, a] such 00 00 that f (ya) = ma. We have Ma = f (xa) = > f((xa)n)/2n < Ma E 1/2n = n=1 n=1 Ma. It follows that we must have f ((xa)n) = Ma, for any n E N. But
the sequence ((xa)n )netsr tends to zero, and since f is continuous, it follows 1(0) = lim f((xa)n) = Ma. 00 00 In a similar way ma = f(Ya) = E f ((ya)n)/ ma E 1/ 2n = ma , n=1
71=1
and as above we conclude that f(0) = ma. Consequently f is a constant on the interval [0, a]. Because a E (0,1) was arbitrarily, by making a —p 1, we obtain that f is constant on the interval [0,1]. Conversely, every constant function on [0,1] satisfies the condition in the problem. Therefore the functions satisfying this condition are the constant functions. Problem XI.18. Determine how many nonzero terms appear in the expansion of a determinant of order n in which only three entries are zero. Solution. We will consider four different cases.
127
Problems for the 11th form
1) The three non-zero entries are on the same row (or column). Then the expansion of the determinant contains 3(n— 1)! zero terms and (n —1)!(n— 3) non-zero terms. 2) Two of the three entries are on the same row (or column) and the third is on the same column (or row) with one of the other two. In this case we have (n — 2)!(n — 2)2non-zero terms. 3) Two of the three entries are on the same row (or column) and the third is not on the same row (or column) with none of the other two. In this case we have (n — 2)!(n2— 4n + 5) non-zero terms. 4) Any two of the three entries are not situated on the same row or on the same column. In this case the number of non-zero terms is (n — 3)!(n3— Gn + 14n — 13). Problem XI.19. Prove that for any polynomial P(x) with simple roots we have inxi) 0 i=i P(xi) Solution. The following relationship is easy to be obtained: n
P'(x) = P(x) • E
,xER\fxl X2, • • • Xi}. ,
j=1 X Xi
(1)
By differentiating this relation we obtain P"(x) = .13/(x) E = P(x)
2 (x — x )2 =
1 x — x2
("
E
1
i=1 x - xi ) 2
P(x) E
1
(x -
xi )2
and then we get P"(x)
P(x)
=
n
1 )2 11 =2 x — xii - il (x — Xi) 2 — Xj)(X — Xi) 1
(2)
Multiplying (2) by x — xk we obtain P"(x) • (x — xk) P(x)
n
—
1
x — Xk + 2 1
2E i0k
(3)
i,j0k
In equation (3), valid for x E R C {xl, x2, ... , xn}, we take t --4 xk and since urn X-xk
x — Xk X — Xk 1 r.,, , = lim = ./---x) :c—xk P(x) — P(xk) P'(xk)'
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128
P"(x) — 2 t 1 . P'(x) i=1 xk— xi
we obtain
n pH (xi) = 2 t Therefore
P'(xi)
1
i=i i=i xk — xi i0k
Problem XI.20. Let A, B be square matrices of order k. Prove that if det(A + nB) = det(nA + B) for at least k +1 distinct values of the natural number n > 1 then det A = det B. R, f (x) = det(A + Solution. Consider the polynomial function f : R xB) det(xA + B) and the polynom P(X) E R[X] attached to f . Then deg P < k and P is vanishing at k + 1 distinct points that means P is identically zero so f is the zero constant function. From f (0) = 0 it results det A = det B. —
Problem XI.21. If Ix' < 7r, x
0, show the inequality
lx1 sinx <— 1——< 7r C11— x 2 Solution.
Since the function
inequality for x E (0,7r].
sin x x
7
is even it is sufficient to prove the
x2 First consider the function 9.2(x) = sin x — x + — 7r , x E [0,7r]. We have 2x „, 2 sin x + — 7r and (r(x) = — cos x. Since So(x) = 7r 7r 7r cp'"(x) < 0, for 0 < x < — and cpw(x) ? 0, for — < x < 7r it follows that the 2 2 7r 7r function e(x) is decreasing on the interval 0, [ — and increasing on —2'7r . 2 (7r Moreover, we have e(0)> 0, (p" — < 0 and cou(r) > 0. Consequently, co'(x) = cos x — 1
2
7r 7 there are two points xi E 0, ( — and x2 E —2'7r , such that (i.9''(c) > 0, 2 for x E [0, xi] U [x2,7r] and ce(x) < 0, for x E [xi, x2]. It follows that the function (p'(x) is increasing on the intervals [0, xi], [x2, 7r] and is decreasing on the interval [xi, x2]. Since (p'(0) = yd(7r) = 0 it follows that there is a point y E (xi, x2) C (0,7r) such that cp'(x) < 0, for x E [0, y] and cci(x) > 0, for x E [y, 7r]. Therefore the function cp(x) is increasing on the interval [0, y]
Problems for the 11th form
129
and decreasing on the interval [y, ir]. Since cp(0) = 0 = co(r), we obtain that (p(x) > 0, for x E [0, 7r]. This implies the left inequality in the problem. In order to prove the right inequality in the problem it is sufficient to 7r 2 show that 0(x) < 0, for x E [0,7r], where 7b (s) sin x — — . x + 2. We have 2 7r zp'(x) = cos — — + x, (x) = —sins +1. Since 0"(x) > 0, for x E [0, 7r], it 2 follows that the function 1P'(x) is increasing on the interval [0.7r]. We have Y(0) < 0 and V(7r) > 0. Then there is a point z E (0.7r), such zli(x) < 0, for x E [0, .2] and 111(x) > 0. for x E [0, y] and E [y, 7r] and hence the function 0(x) is decreasing on the interval [0, y] and increasing on the interval [y, 7r]. Finally, since '(0) = (ir), we conclude that OW < 0, for x E [0, 7r]. Problem XI.22. Let x„ E [0,1] be, such that xn+i.clne(1 — x„) > 1, (V)n E N. Prove that {x„}„E N is increasing and compute the limit (e is the base of the natural logarithm). Solution.
Using Lagrange's theorem for the function f(x) = Ins, 1 1 x E [n, n +1], n E N we obtain n+ 1 < ln(n +1) —ln n < — , or equivalently, n (
1)n
1 7L+1
1 + — < e < (1 — n
(1)
Take some value for n. Consider the function g(x) = xnne(1 — x) — 1, x E [0, 1]. From its derivative and the right inequality in (1), we obtain max g(x) = g (
xE [0,1j
n n +1 ) ( e l) n+1 1+— n
1 < 0.
Therefore xn, Vne(1 — x„) < 1. It follows x„ < xn+1 for any n E N. Consequently the sequence {x„}„E N is convergent. Define 1 = lim xn. n—∎oo If we suppose that 1 < 1, then by using Cauchy's criterion we get lien xn+i /ne(1 — x„) = 1. By taking into account the hypothesis, this n—* oo leads to a contradiction. Hence 1 = 1. Finally, note that there exist sequences satisfying the conditions in the n — 1. problem, for example, the sequence x„ = This follows from the left inequality in (1). Problem XI.23. Let f : (0, oo) properties:
(1, oo) be, a function with the following
"Gazeta Matematica" - a bridge over three centuries
130
i) liln f (x) = a, a > 0, ii) > 0 such that f(xy) = A(f2(x) + f 2 (y)), (V)x, y E (0, oo) \ {1 } . One claims: 1 a) Show that A < n-1
b) Compute lien H f (x 22) , x > 0. Solution. a) Suppose that A > 2. It follows that A(f2(x) + f 2(y)) > f (x) f (y) and consequently f (xy) > f (x) f (y), for all x, y E (0, DO) \ {1}. In particular we have f (x2) > f 2(x), x E (0, oo) \ {1}. Let show that a = 1. Indeed, since f (x) > 1, for x > 0 we obtain a > 1. Also, we obtain lim f(x 2) > slim f 2(x), that is, a > a2. Then, since a > 1 x01 we have a = 1. Let x > 1 be fixed and let 1 < t < x be arbitrarily taken. We have 1 2 (x) + 1) < A(f2(x) + 1) < A (12(x) + f 2 ()) = (t). Passing to limit t
1, we obtain a > 2 (f 2(x) + 1) > 1. The contradiction 1 shows that we must have A < -. 2 n-1 (
b) Let x > 0, x 0 1 be fixed. Denote c.,,, =
i=o relation ii) we obtain f(t2) = 2Af2(t), for t > 0, t have, for any i > 0:
f (x7z.1-) =
f (x2Z) = 2Af 2 (x FI")
f xF), n > 0. From 1. Consequently we
/f (x 22) ) 2 2A
From this relation, by induction with regard to i we obtain
1-1,
1
f (x*) = (f(x))7
(V)i ?. 0.
Indeed, for i = 0 this relation is obvious, and if we suppose them true for i, then we obtain 1
f (x7'7 1- =
( (f(X ))*(1;017 2A
2=
1 ( (f (x)) TrT F— 2A
2e+'
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131
It follows that n-1
n-1 (
)
cn = (f(x)) i=°
E
1-2-2)
-T1+1 (
.f ((x))2-2
`
1 91,-2-F2-n+i
1 Since 0 < A < - we obtain firn cn = 00. 2' Problem XI.24. Let A, B E Mn(R) be two matrices with the property that there is A E R \ {0} such that AAB + A + B = 0 (0 is the zero matrix of M„(R)). Prove that AB = BA. Solution. Denote by U the unit matrix of Mn(R). From the hypothesis we obtain A2AB + AA + AB + = U. The factorization of the left side of the identity gives (AA+ U)(AB +U) = U. It follows that the matrix AA + U is invertible, with the inverse matrix AB + U. In this case, we have also (AB + U)(AA + U) = U. Thus, A2AB = -A(A + B) = A2BA. Since A 0 0, one obtains AB = BA. Problem XI.25. Let f : (0, oo) R be a continuous function with the property f (2x) = f (3x), V x E R. Prove that f is a constant function. Solution. Let y be a positive real number, y (Yn)nEN by yn
1. Define the sequence
1212)n
y k log 3 ,
n = 0,1,2, • • • .
log 2
We have
ynlog 3
Yn+i /
f (Yn) = f
log yn
and, from the hypothesis, \
)= f
log 3
/ log 2 \ f (y7';)g 3 ) =
\ = f(yn+i), V n > 0.
log 2 E (0,1) implies yn log 3 The continuity of the function f gives us urn f (yn) = f(1). n--■oo We obtain f (y) = f (1), V y > 0. Therefore f(yn) = f(y), V n
E
N. But
y° = 1.
Problem XI.26. Let (an)„>1 be a sequence of real numbers such that the sequence (2an+i + sin a„)„>1 is convergent. Show that the sequence (a„)„>1 is convergent.
"Gazeta MatematicA" — a bridge over three centuries
132
Solution. Denote bn, = + sin an, n > 1. From the hypothesis, the sequence (bn)n>1 converges to a finite limit b. Consider the function f R R, f (x) = 2x + sin x. Since 2x - 1 < f (x) < 2x + 1, V x E R, we obtain lien f(x) = oo and lien f(x) = oo. Also, we have f' (x) x ,00 x, 00 2 + cos x > 0, V x E R, and we get that f is an increasing function on R. Therefore, f has a continuous inverse function f -1: R -> R. By definition, bn = f (an), n > 1. It follows an = f -1(bn,), n > 1. The convergence of the sequence (bn)n>1and the continuity of f-1impliy the convergence of the sequence (an)n>1, with nfirlan = f -1(b). -
-
-
Problem XI.27. Determine the triangle LABC if a = A, b = B and c = C, where a, b, c are the lengths of the sides and A, B, C are the measures expressed in radians of the opposite angles in the triangle. Solution. From the hypothesis and sinus theorem, we obtain
A sin A Consider the function f : (0, 7r)
sin B sin C R defined by
—7-f( x) = sin x
(x ) =
sin x — x cos x sing x
We have f' (x) > 0, Vx E (0,7r), because -x cos x > 0, for x E (7r/2, 7r). Then sin x x cos x > 0. For x E (0, 7/2) we have cos(tgx - x) > 0 because tgx > x. Since the derivative is positive, we deduce that f is increasing on the interval (0, ir), therefore f is an one-to-one function. Therefore, the equality A sin A sin B implies A = B. Similarly, we deduce B = C. In conclusion, the only triangle satisfying the hypothesis is the equilateral triangle with A = B = C = a = b = c = 7r/3. -
Problem XI.28. Let A E Mn ( R) be a matrix that has an inverse and A — A-1= In. Show that
Solution. Let B E M„(R) with A =
2
follows A2 = A + In , that is 3 1,/
2
B2 =
g2-F
<
(15-- 1) n < \ 2 j
1+
2
B. Since A
+
-
A-1= In it
(1)
Problems for the 11th form
133
Let M = max{ max {Ibij1}, 1}, where B = (bij). We prove by induction 1
with respect to k, the proposition P(k) : Ib I < M, (V)i, j = 1,n. From the definition of M it follows that P(0) and P(1) are true. Assuming that propositions P(i), for 0 < i < k, are true and taking into account the relation (1) we obtain 3+2
=
1+ 2
i3
+ 071
1+ 2 -
Ibt I+
fib
M.
I<3 2
It follows I det Bk I <
E
... • IbLe,,,i< n! 11//7, (V)k E N.
aES,i
Since k is arbitrary, it follows that I det BI < 1. Then I det Ai <
2
In order to prove the left inequality consider the matrix C +1 Then C - C-1 = In. From above we obtain I det CI < 2 • Hence I det = 1/I det CI > 1/ Remark: For A = become equalities.
-
A-1.
(1/5- + 1) n(Vg — 1 2
2 ) • +l Nig 1 Inthe two inequalities in and A = 2 2
Problem XI.29. Find the continuous functions f : R f(0) = 0 and for all x E R, we have
R, such that
f (2x) > x + f (x), f(3x) <2x + f(x). Solution. Define the function g(x) = f(x)- x, x E R. We have g(2x) > g(x), and g(3x) < g(x), (V)x E R. Let a > 0 arbitrarily chosen. Define m = min g(x) and M = max g(x). xE[—a,a]
xE[—a,a]
From the Weierstrass theorem there is xo E [—a, a] such that g(xo) = M. Consider the sequence (xn)n>o, defined by xn = 3-7xo, fm E N. From the relation g(3x) < g(x), (V)x E R., we obtain g(xn) < g(xn+i), (V)n E N. Hence g(xn) > g(xo) = M, (V)n E N. Since xnE [—a, a], it follows g(xn) < M. Therefore we have g(xn) = M, (V)n E N. Using the continuity of the
"Gazeta Matematicr – a bridge over three centuries
134
function g we obtain 0 = g(0) = n—oo lirn g(xn) = M. Consequently we have g(x) 5_ 0, (V)x E [—a, a]. In a similar mode, using the relation g(2x) > g(x), (V)x E R, one obtains that g(x) > 0, (V)x E [—a, a]. Then we have g(x) = 0, (V)x E [–a, a], and since a was taken arbitrarily, we obtain g(x) = 0, x E R. It follows f (x) = x, x E R. Problem XI.30. Consider a sequence (xn)nEN,satisfying the inequalities a2 0 < xn < a and (a– xn)xn+i > (V)n E N; a > 0. Prove that the sequence 4 is convergent and find its limit lien xn• nEoo –
Solution. xn+1
a – ,it follows If xn < 2
a Xn+1 < — 2
a and if xn > – it follows 2,
a a] n E N or xn E [2, a Therefore we have xn, E [0, – > – 2 2•
n E N.
Consider that xn E [0, a] – n E N. Write xn = –a – Yn7 Yn E 0, i . 2 2 ' a a a2 The condition in the problem implies – + Yn – Yn+1 > — • 2 4 2
–
–
Consequently, Yn+1 5_ yr, . We conclude that the sequence (Y 1 gent and then (xn)nE N is also convergent. Let 1 = lim xn.
is conver-
nEoo
a2
a We obtain (a – /)/ > — and hence 1 = 2. 4 Problem XI. 31. Let A E 31/2(C) be a square matrix such that Tr(A) = Tr(A2) = 0, where Tr(A) is the trace of A. Prove that det(A2) = 0. Is the converse true? Solution. If A E M2(C), it follows that we can write A = ( ac
db )
, where
a, b, c, d E C. It is known that A, verifies the equality A2– (Tr(A))A + (det A)I2 = 02, where /2 is the unitar matrix and 02 is the null matrix. Since Tr(A) = 0, this relation becomes A2 + (det A)/2 = 02. From this last equality we deduce Tr(A2+ (det A)12) = Tr(02) = Tr(A2) + 2 det A = 0.
135
Problems for the 11th form Since Tr(A2) = 0 we obtain det A = 0. Then det(A2) = (det A)2 = 0. 1 0 10 then A2 = The converse is not true: if A = ( 0 0 / 0 0 )• Tr(A) = Tr(A2) = 1. We have det(A2) = 0 and Problem XI.32. Let A, ai, bi,ci, (i = 1, 2, 3) be real numbers such that al A2 + bi A + = a2A2+ b2A + c2 = a3A2+ b3A + c3 Prove that 1 al Cl det ( 1 a2 c2 1 a3 C3
2
( 1 b1 Cl = det 1 b2 C2 1 b3 C3
(
det
1 al b1 1 a2 b2 1 a3 b3
)
Solution. We introduce the following notations: = A2, A2 = A and A3 = — (a1A2 +10 + ci) • Then the following system: + b1A2 + A3 = a2A1 + b2A2 + A3 = —C2 1 a3A1 + b3A2 + A3 = —C3 has at least the solution (A1, A2, A3) defined above. ( al bi 1 If A = det a2 b2 1 = 0, the system is compatible, but it has a3 b3 1 more than a single solution. The theorem of Kroneker-Capelli says that ( al Cl 1 det a2 C2 1 = 0, so the equation is verified. Suppose now that the a3 C3 1 determinant of the system is not zero, i.e., the system has a unique solution. We apply Cramer's formulas and obtain: AA1 2— — 612 , where Al =,A A 2 — (
Dal
= det
b1 1'\ ( al , 02 = det a2 — C2 b2 1 —C3 b3 1 a3
—Cl
—Cl —
•11
C2
—C3 1
For the proof of the formula, recall that Al = A2 therefore (& \2 )2 = A • AA, . ,
136
"Gazeta Matematica" - a bridge over three centuries
Problem XI.33. For any n E N, consider the equation x"+1 - (n 2)x n - 1 = 0. Show that this equation has a unique positive real root an and find Bin an. 7L—∎ CO
Solution. Consider the function f (x) = xn+1 - (n + 2)x - n - 1, x E From the derivative of f it follows that x = 1 is the unique point of local extremum, namely a minimum point. Since f (0) < 0 and f (1) < 0, it follows that the equation f (x) = 0 has a unique positive root an and that an > 1, for each n. We have 0 =
n+i f (an) = (an - (E an - (n + 2))- 2(n + 1) = k=0 n+1 k-1
n+1
E E a.;„ - 2(n + 1) =
(an - 1)(an- 1) - 2(n + 1) = (an - 1)2
k=1 j=0
k=0 n+1 k-1
> (an —
1)2 E E - 2(n + 1) = (n + [(an - 1)2n
2 2. ,
k=1 j=0
It follows (an - 1)2 1. Problem XI.34. Let p E N be a fixed positive integer and consider the sequences (xn)n>1, (Yn)n>i, (Zn)n>l, (tn)n>1 defined by the equality below
(
p 1
(Xn
1 0
zn
yn o t
(v) n E N.
Xn Xn Compute the limits: lien — and lim n—■ oo yn
n—∎ co to
Solution. We have (x71+1 Yn+1) 2:7/1- 1
tn+1
(-P
1) 71+1
(Xn Zn
0
yn) (p to
1) = (PXn
1 0
PZ72,
Y71,
tn
Zn Xn)
we get the following recurrence relations Xn +1 = PXn yn; Zn+1 = pZn
Yn+1 = Xn;
to ; tn+1 = Z.
Therefore (xn )n>1 satisfies xn+1 — pxn — Xn-1 = 0
(1)
137
Problems for the 11th form and xi = p, x2 = p2 +1. The characteristic equation of (1) is t2— pt —1 = 0 with the roots
p+
+4
Vp2
A, T with A =
>1.
2
We know that (xn) is given by 1n x„ = ciA"± c2 • (--) ,
n>1
where ci,c2 are constants that can be found using the known values for xi and x2. Therefore )
n-1
,n>1
= c1 n-1 + C2 • (--
and so 1 n
+ C2 . Xn 11M —=
n—oo yn
A lim + C2
(
1 — A2
A
)n-1
.
Xn
To find urn — we follow the same way. n—•00 to
Problem XI.35. Let ai,a2,• • • ,anbe positive numbers with the property aia2••-a„ = 1. Prove the inequality: al + a2 + • • • + +
1 1+a1
+
1 + 1 + a2
1 3n -- --• 1+a„ — 2
Solution. Consider the convex function f(x) = x + Using Jensen's inequality we find
1 , x E (0, oo). 1±x
( n ak 7
E f (ak)
nf
k=1
n
But, from the inequality of means and from our hypothesis, we get iL
ak k=1
7 ak -> 11 k=1
= 1.
"Gazeta Matematica" - a bridge over three centuries
138
3 To get the conclusion, it is enough to prove that f (x) > - , V x > 1, which -2 is equivalent to
f( x)
3 (In - 1)(2m + 1) >0, V x > 1. 2= 2(m+ 1)
Problems for the 11th form
List of authors. Problems for llth form 1. Alexe St., XI.18 (G.M. 7/1980) 2. B5.lima M., XI.17 (G.M. 7/1980) 3. Bi§boach N., Bottesch M., XI.27 (G.M. 1/1992) 4. Buicliu Gh., XI.5 (G.M. 1930) 5. Bu§neag D., XI.20 (G.M. 4/1982) 6. Cavachi M., XI.25 (G.M. 1/1988) 7. Ghermanescu M., XI.6, XI.7 (G.M. 3/1935) 8. Haivas M., XI.30 (G.M. 1/1994) 9. Ioachimescu A.G, XI.1 (G.M. 2/1895) 10. Ionescu-Tiu C., XI.12 (G.M. 9/1973) 11. Ilie R., XI.23 (G. M. 11/1982) 12. Istrate G., XI.28 (G.M. 5-6/1993), 13. Lalescu T., XI.2 (G.M. 1901) 14. Lohanel G., XI.10 (G.M. 2/1969) 15. Lupa§ A., XI.21 (G. M. 7/1982) 16. Onofra§ E., XI.16 (G.M. 2/1980) 17. Otet A., XI.19 (G.M. 9/1981) 18. Panaitopol L., Singer B., XI.15 (G.M. 6/1979) 19. Panaitopol L., XI.22 (G.M. 7/1982) 20. Pantazi A., XI.4 (G.M. 1913) 21. Parsan L., XI.34 (G.M. 7-8/2002) 22. Popoviciu T., XI.8 (G.M. 1937) 23. Popescu Gh., XI.9 (G.M. 6/1965)
139
140
"Gazeta Matematica" — a bridge over three centuries
24. Radu D., XI.32 (G.M. 7-8/1996) 25. Radulescu S., XI.11 (G.M. 3/1973) 26. Rotaru F., XI.29 (G.M. 9/1993) 27. Secleman D., XI.14 (G.M. 4/1978) 28. Szollosy Gh., XI.33 (G.M. 9/1997) 29. Urlea I., XI.13 (G.M. 9/1974) 30. Ursarescu M., G.M. 12/2002 31. Valcovici V., XI.3 (G.M. 1906) 32. Vijitu V., XI.26 (G.M. 8/1991) 33. Vijdeliuc M., XI.31 (G.M. 2/1995) 34. Vlaicu L., XI.24 (G.M. 10/1986)
Chapter 9
Problems for the 12th form Problem XII.1. Compute
1
d"y for x = 0 and y = (arctan x)2. • n dxn
1•2
Solution. The function f : R R, f(x) = y is even, because is the square of the function "arctan", which is an odd function. d"y Conclusion: if n is odd, the n-th derivative — (0) = 0, and if n is even dx" (n = 2k, with k a positive integer), the requested number is the coefficient of x 2k from the Mac Laurin expansion of the function f . We have: 1
=
x2 + x4
x6 +
+ i)k x2k + 1)k-1-1
1 + x2
t2k+2
1 + t2,
for any x E (-1, 1), where t is between 0 and x. Therefore, x t2k+2 x2k+1 x 3 x 5 x7 k arctan x = x – + ( 1)k+1 dt. + 5– + • • • +(-1) 3 f + t2 7 2k + 1
Because the following inequality holds
1 < 1 it results that 1 + t2 , +2k+2
dt < t2 0
x2k+3
2k + 3 .
Hence, x i2k+3 <
r j2k+2
(-1)k+1
0
1 t2
dt < i "
1
2k+3 2k+3'
141
142
"Gazeta Matematica" - a bridge over three centuries
and the desired expansion is: arctanx = x-
x2k-1
x3 x 5 x7
(arctan x)2 = + • • • (+D k-1 ' 2k - 1 + 3 5 7 1 1 4. .x2k+... = • • •±2(-1)k-1 + + [1 • (2k 1- 1) 3 - (2k - 3) 5 • (2k - 5) ' • Because
+
-
_ 1 (1 1 ) 1 i(2k - i) - 2k i + 2k - i) '
the requested coefficient is: + 1 + 1 ) 5 2k - 3 2k- 1
(-1 i )k-i )k-1 1 . (1 + 1 + 1 +
3 and
1
0, 1 d"y (0\ ) 71 ,!dxn \ ' 7= ) (-0-12 (1 + 1+ 1+ n 3 5 l
for n odd +
1 n-1
, for
n even
Problem XII.2. If p is a prime number and a is a positive integer divisible by p, prove that p_i
x = 1 +E(a + i)P-1 i=i
is also divisible by p. Solution. Let us notice that, due to the fact that p is a prime number, we have gcd(a + k, p) = 1, for any k E {1,2, ... , p - 1}. Using Fermat's theorem we obtain: (a + 1)P-1
-_ 1 (modp)
(a + 2)P-1
-_- 1 (modp)
(a + p - 1)P-1 .- 1 (mod p). Summing the previous relations we deduce: p_i
p--1 (a + i)P-1E_-- (p - 1) (mod p),
i=i
that is 1 +
(a + i)P-1.7, -0 (mod p).
i=i.
Problems for the 12th form
143
Problem XII.3. Consider a mobile line that pass as through the vertex of a parabola. Find the locus of the center of gravity of the area bounded by the parabola and the mobile line. Solution. Consider a Cartesian coordinates system (0,T, j). Let the equation of the parabola be: y2= 2px (p > 0), and the equation of the mobile line: y = Ax (A E R*). We denote by M the second point where the straight line intersects the parabola (the first one being the origin, 0). Solving the system given by the equations of the two curves, we obtain the 2p 2p coordinates of the point 1k!: x m = — , y m = — . Let G(a, (3) be the center A2 A of gravity of the given area. Thus,
xm f x (OW - Ax) dx
a
o
1
f (IT/Tx - Ax) dx
xm f (2px - A2x2)dx 1 0 Q= 2 xm f (V2Tx - Ax) dx
1
if A > 0,
0
and
fm x (Ax - (- OW)) dx
x A,1 f (A2x2- 2px)dx
xr(Ax_(_.3x))
, if A < 0. , =2 xm ° f (Ax - (- -V2px)) dx
1
a= 0
0
4p In both cases, after calculations, we deduce: a =5A2 — , (3 = A- . 5 Hence, we obtain the equation (32 = -p • a, whence the locus of the point G 4 is also a parabola, with the same vertex and the same axis of symmetry. Problem XII.4. Consider the derivable functions f, co : R R, such that for any real number x, f(x) (p(x), and the quadratic equations:
Y 2+ Y • f (x) + (x) = 0
(1)
y2+ y • ca(x) + cd(x) = 0,
(2)
and where the unknown is y. 1) Find the functions f and co such that the roots of the equation (2) are the inverses of the roots of the equation (1). 2) Prove that the following relation holds: f(x) - (p(x) = c • f (x) • (p(x), x E R, where c is a real constant.
"Gazeta Matematice — a bridge over three centuries
144
Solution. For x E R, let yi, y2 be the roots of the equation (1) and y3, y4 be the roots of the equation (2). Using Viete's relations, we obtain:
1 yi + t
1
1 y3 y4 1 Y2 = y3 y4 =
• (p' = (10
<=>
(3)
f' • co' = 1
The roots of the equation (2) being the inverses of the roots of the equation (1), we must have yl • y2 = f'(x) 0 and y3 • y4 = cd(x) 0 0, for any x E R. The functions f' and (p' have the Darboux property, therefore f' and co' have constant sign on R, whence using f' • (p' = 1, we deduce that the functions f and yo are both strictly increasing or strictly decreasing on R. Suppose xo E R exists, such that f (x0) = 0 (we could find only one such point, due to the strictly monotonicity of f). From (3) it follows cp(x0) = 0, whence (f co)(4) = 0, in contradiction with the assumption we made. So, for every x E R, f (x) 0. Then: —
( f — (py
(f' — (20')f (do — ( f — (P)(1' (do + f co')
f'
v) • (do — •
f 2(p2
f (p
f' • (p • co — f • (p
f 2,p2
f co' (3)
f 2,p2
(f' • (p — f) • co (3) (f' • f • (P' — f) • co —
f 2(p2
(3)
(f • 1 — 1) • (do =0,
f 2,702
whence there exists c E R* such that (4)
1-99=c•f•
We have: f — yo (1j f f • co' = f • (1 co') and using (4) we deduce f • (1— cp') = -
—
= c•f•co f49 1 So' = c•cp, i.e. 1—co'(x) = c•co(x), x E R <=> co' (x)-Fc•co(x) = 1, x E R. Multiplying the last relation by e" 0 0, we obtain: (co(x) • e")' = ex, eox 1 x E R, whence 99(x) • e" = — k with k E R co(x) = k • e c*s + , c c —
—
x E R. Using again (4), we find f(x) =
1
1
k
— — , x E R. c
Problem XII.5. Consider the functions f, g : I = [a, b] —4 R, such that: 1. a • b > 0 and f(I), g(I) c I;
Problems for the 12th form
145
2. f and g are strictly monotonous and have derivatives on I and g' 0 0; f 0 g ft 3. < — for x E (a, b). g o f gl Find the biggest of the following numbers: g(b) Ii
=f
f(b)
f(x)dx, 12 =
g(a)
f
g(x)dx.
f(a)
Solution. The first condition implies that f 0 and g # 0 on I. Moreover, fog f and g have the same sign on I, therefore > 0 on I, and hence, using go f f' 3, we obtain .9 —/ > 0 on I. (*) Whence, f' 0 0 on I and the functions f and g have the same monotonicity on I. The functions f : I f (I) and g : I g(I) are bijections. x = gt) ( For computing /1, we put dx = g'(t)dt. Then x = g(a) t = a and x = g(b) t = b. Thus, /1 =
a
f (g(t)) • g' (t)dt = f (F (g (OW dt = (F o g)(b) — (F o g)(a), where a
F is a primitive of the function f. With the same reasoning, we get /2 = (G o f)(b) — (G o f)(a), with G a primitive of g. (F o g)(x) Consider the function h : I —> h(x) = , continuously deri(G o f)(x) vable on I. Using the Cauchy's theorem, we find c E (a, b) such that (F o g)(b) — (F o g)(a) (G of)(b) — (G o f)(a)
o g)/(c) <#. (G o f)/(c) (f o g)(c) g'(c) (3),(*) <=> < 1. = (g 0 f)(c) f(c) Iz
1) If a < b < 0, then f < 0 and g < 0 on I. When f and g are strictly increasing on I, we have f (a) < f(b) and g(a) < g(b), whence we deduce /1 < 0, /2 < 0. In this case, from (**) we obtain 1/11 < I/21, i.e. /1 > When f and g are strictly decreasing on I, we have f(a) > f(b) and g(a) > g(b), whence we deduce /1 > 0, /2 > 0. From (**) we obtain /1 <12. 2) If 0 < a < b, with the same reasoning we deduce that if f and g are strictly increasing on /, we have /1 < /2 and if f and g are strictly decreasing on /, we have > /2.
146
"Gazeta Matematica" - a bridge over three centuries
i 7 • 97r Problem XII.6. a) Prove that f sin10 x dx = 29 ; o i 3 • 7 • 117r b) Prove that f sin12 x dx = 211 " 0
_IL 2
Ir.
2
c) Prove that lim f sin2n x dx = liraf cos2" x dx = 0. n --K>o 0
n-400 0
7r
Solution. Denote
2
I„ = f sine x dx, for all integer n > 0.
o Applying the formula of integration by parts, we obtain:
7r
2
2
In = f sine x • sinn-2x dx = / (1 - cos2 x) sin"-2x dx = o
o .
M 2
2
/
x (sin"-1 x) = i„ _ 2 — f cos2 x • sinn-2x dx = In_2 - f cos dx = n-1 0 0 i f sin"x 1 sin"-1 x 1 i ) = 1n-2 <=> x dx •#› In (1+ cos 7 1-2 = 1. in-1 n - 1 lo n -1 0 n-1 /„_2 , for all integers n > 2. (1) •#> /n, = n n 4-- n - 1
/n_1 =
n-2 n-1
/n _3,
for all integer
n > 3.
(2)
Multiplying the relations (1) and (2), we obtain In • /71-1 =
n -2
1,1-2 • 1n-3 1
n
for all integer
n>3
(3)
From (3), we deduce:
In • In 1 = -
n-2 7/
r
r
in-2 • in-3 =
n-4 n
( 1 i - • ./). • io, if n is odd n 2 1 7 71 • 1-2 • li, if n is even
/n-4 • -in-5 = • • •
(4)
Problems for the 12th form
147
/„±i — In =f sin" x • (sin x — 1)dx < 0, for every integer n > 0, because the [Ir integrand is a continuous and negative function on 0, — .
2
Therefore, the sequence (in)n>0 is strictly decreasing. 7r Because sin" x E [0, 1], for all x E [0, — 2] , we obtain in E [0, -2-1, for any integer n > 0, hence (In )n>0 is convergent. Denote L = lim In. Then, lira I2n = L = urn /2n4-1. n—>oo
n--400
Using (4) we deduce that L2 =0, then L = 0. It is well known that, for b
b
f : [a, b]
1m continuous, f f (x)dx = f f (a + b — x)dx . Thus, a
a
2
lim sine" x dx = n—Kx) 0
0 = fim 12n n--o co
lr 2
ir 2
7r = lim f sine" (0 + — x) dx = lim f cos2n x dx . n—,00
n—■ oo
0
0
Using (1) we obtain: ,
/2n=
2n-1 2n-3 , (2n-1)(2n-3) • .... 3 . 1 2n — 1 , 1 In. 2n-2= /2n-4= = 2n 2n 2n-2 (2n)(2n-2) .... • 4 . 2
Therefore 9 • 7 • 5 •3 . 1 7 97r /0= and 29 10 • 8 • 6 •4 2 11. 9 .7. 5. 3 • 1 3 • 7.117r 2" 1.12= 12 •10.8.6 • 4.21.0=
ho=
Problem XII.7. Consider the function f :
R, which admits primitives, and the sequence (a„)„>0, such that lim an =0, and f (x a„) = f (x), for any positive integer n, and any real number x. Prove that f is constant on R.
Solution. Let F be a primitive of the function f. Then f(x + an) = f(x) q (F(x + an))' = Fi(x), for any integer n > 0 and any real number x. Therefore, we find the sequence of real numbers (cn)n>0 such that
F(x + an) = F(x) + en
(1)
"Gazeta MatematicA" — a bridge over three centuries
148
for any integer n > 0 and any real number x. x = 0 in (1) cn = F(a72) F(0) and hence —
F(x F an) = F(x)+F(an) -
—
F(0) <=> F(x+a„)
—
F(x) = F(an)
—
F(0). (2)
We obtain:
(x) = (x) = lirn
n—*oo
F(x + an) — F(x) (2)F(a . n)— F(0) = 1(0), n—>oo an — x+ x —
for any x E R. Problem XII.8. Consider the odd integer n > 3, r E R+, the complex numbers al , a2, , an-1and the equation: Xn
ai x n-1 + a2 • x n-2 + • • • ± an_i • x
—
1 = O.
The roots of the equation, x1,x2,...,x„ are such that for any k E {1,2,...,n}, ixkl = r. Prove that: a) Im(xi • x2 • xj—i • xj±i • • • Xn + xi) = 0, for any j =1,n. b) Im(ak) = Irn(an_k), for any k = 1, n —1. Solution. The integer n being odd, from Viete's relations we obtain: xi • X2 • ... • Xn = (-1) • (-1)n = 1 We therefore find ti, t2, xi = cost•3• + i • sin ti • a) We have:
X2 • . .. •
IX ni = rn= 1 •4=>
r = 1.
, t„ E R such that for any j E {1, 2, ... , n},
X1 • X2 • ... • X j-1 • X j+1 • • • • • Xn = Xi •
IX11 1x21 • • •
1 Xi
1 xi
Xj_i • Xj+i • • • • • Xn Xj = —
Moreover, xi • x2 • . .. • xj_i • xj+i • ... • xn b) For any k =1, n, we denote Sk =
Xj = Xj
Xj E R.
= 2 cos ti E xi, • xi2 • ... • xik .
If {ii,i2,• • • ik}
Oh :721 • • • jn-k} =
Xii• Xj2 • . .• Xin_k •Xi •Xj2 • . . xik
=1
{1, 2, . , n}, then xii aj2 .. . .•xin_ k =
1 xil•
Xj2 • . . . • Xik
Problems for the 12th form
149
Therefore, Sk Sn _k =
+
Xil• Xi2 • . . . • Xik +
x j, • xj2• ... • xj„_, •;=› Sk Sn-k =
E 1<ji<j2<•••<jn _k
(xi, • Xi2 .
1
. . X.
xi l •
1
Xj2 • ... • Xi k
) <#.
2 cos (til.+ ti, + • • • + tik )
Sk Sn-k =
1
The integer
rt
and
Cl„ k = (-1)71-k • an-k• -
being odd, we have Sn _k = -(-1) k •an_k and consequently:
Sk Sn-k = (-1)k ' (ak - an-k) •#• ak
-
an k = ( 1)k • (Sk + Sn _k) E R, -
-
Thus, we deduce that Im(ak) = Im(a„_k), for any k = 1, n - 1. Problem XII.9. Consider S = {z E C : Izi = 1}. Prove that we can't find any polynomial P E R[X], with deg (P) > 1 and such that P(S) C R, where P(S) = {P(z) : z E S}. Solution. Suppose a polynomial P E R[X], P(X) = an Xn + • • • + aiX exists, with an 0 and with the previous properties.
P(z) E R, for any z E C, Izi = 1
ao
P(z) = P(z) PE4X1P{z) =
1 1 P(z) = P (-1) •<=> a„ • z” + an_i • 7-21-1+ • • • + al • z + ao =
= P (T) I
1 1 1 = an • — + an_i kn-1+ • + al • - + ao = 0 •;=> Zn Z an • Z2n + an— 1 • Z 2n.-1 + • • *H-al*Z n+l — ai•Zn-1-a2.zn 2 -• • --an_i-z-an = 0, -
for any z E C, Izi = 1. Consequently, the polynomial Q(x) =anx2n an_ix2„-i + • • • +aiX11+1 _
_ _ an_i x _an
has an infinity of roots, therefore Q 0, i.e. an = an_i = • • • = al = 0, false. We conclude that the assertion of the problem is true.
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Problem XII.10. Let p = 4k + 3 (k positive integer) be a prime number. Prove that for any integer n > 0, the polynomial f (X) = (X 2 +1)n p is irreducible in the ring 7L[X] of the polynomials with integer coefficients. Solution. Suppose that we find a positive integer n 0, such that the polynomial f is reducible in Z[X]. That is, we find the polynomials g, h E Z[X], non-invertibles, i.e. g, h { 1,1}, g(X) = atX t+• • • +aiX+ao, h(X) = bsxs + • • • + bi x + bo, with t, s positive integers, such that -
f = g • h,
(1)
(by -1 and 1 we denoted the constant polynomials a (X) = Writing (1) in the ring Zp[X], we obtain:
-
1, /3(X) = 1).
= 74 • T,
(X2 +
(2)
where g, IT, E Zp[X] are
g(x) = at xt +•• • + aix + ao Ti(x). 6s xs + ••• + bi x +bo. ,
It is well known that if p = 4k + 3 (k positive integer) is a prime number, then for any 1. E zp, we have "i -1, i.e., the polynomial a = X2 -1-1 E Zp[X] has no roots in the field Z. Therefore, having the second degree, a is irreducible in Zp[X]. From (2) results that there exists an k E {0,1, , n
n-k
1}, such that g = (x 2 +i)k and ti = ( x2 + . Therefore, we find two polynomials gi, hi in the ring Z[X], such that:
g(X) = (X 2 +1) k p•gi (X) and h(X) = (X2 + 1)71-k p • hi(X). (3) Then
g(X) • h(X) = (X 2 +1)71 p • [(X2 +1)n k • gi(X)+ + (X2 + 1)k • hi(X)] + p2• gi(X) hi(X)
(4)
Comparing (1) and (4) we conclude that we must have: (X2
-k 1)n91(X) + (X2 i) k • h1(X) = 1
(5
)
and
gi (X) • hi(X) =- 0. If gi (X) = 0, we obtain from (5) that k = 0 and from (3) we find g(X) = 1, false. If h1(X) = 0, we obtain from (5) that 11 - k = 0 and from (3) we find h(X) = 1, false. Therefore, the equality (1) is impossible, so f is irreducible in the ring Z[X].
Problems for the 12th form
151
Problem XII.11. Consider the abelian group (G,-) with the identity element e, and the endomorphism f : G ---> G, such that 1(x) 0 x, for any x E G, x 0 e. Prove that the function g : G —> G, g(x) = x-1• f(x) is bijective if and only if for any y E G, there exists x E G such that (f o f)(x) = f(x • y). Solution. We'll first prove that g is injective. We have g(e) = e-1• f(e) = e • e = e. Suppose that we find x E GVel, g(x) = e <=. x-1- 1(x) = e <=> f (x) = x 4=> x = e. Therefore, if x E G, x 0 e, then
g(x) 0 e.
(1)
Moreover, because f is an endomorphism, we have f (x-1) = (f (x))-1, for any x E G. Suppose that we find x, y E G, g(x) = g(y). Then
g(x • y-1) = (x • y-1)-1• f (x • y-1) = y • x-1• f (x) - f (y-1) = = g(x) • (g(y))-1= e 4 x - y-1= e<= . x = y, which proves that g is injective. " " Suppose that g is a surjection. We'll prove that for any y E G, there exists x E G such that (2)
(f 0 f)(x) = f (x • Y)• Since g is surjective, for any y E G, there exists x E G such that
g(x) = y .4=> x-1• f(x) = y <=> f(x) = x • y
f(f(x)) = f(x • y),
i.e., the relation (2). "." Suppose that the assertion (2) is true. f being an endomorphism,
(2) <=> f(f(x)) = f(x) • f(Y) <=> .4=> f (y) = (f (x))-1• f (f (x)) <=> f (y) = g(f (x))•
(3)
We'll prove that there exists z E G, such that g(z) = y. g ( f (x) . y .1 )
( 1 (x) . y ..1. ) 1 . f (f (x) . y ) ..... ( f (X))-1 . y . f (f (x)) . f (y.. . . 1) = g(f(x)) • y • (f (0-1 (3)Ay) • (f (y))-1• y = y.
Consequently, for any y E G, there exists z = f(x) • y-1 E G, such that f (z) = y, whence f is surjective, therefore f is bijective.
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"Gazeta Matematica" - a bridge over three centuries
Problem XII.12. Consider the group (G,.) for which there exists an integer such that the functions f, g,h : G -4 G, f (x) = x"+1, g(x) = x2n+1, h(s) = x4n-1 are endomorphisms. Prove that (G,-) is an abelian group.
71
Solution. The function f being an endomorphism, we have: (x• y)n+1= xn+1• yn+14.> (y • x)n = xn • yn , for any
x, y E G.
(1)
Using the fact that g and h are also endomorphisms of the group G, we find: y2n
(y 5)2n = 52n
for any
(x • y)4n-1 =y4n-1
x, y E G,
for any
(2)
x, y E G.
(3)
(1)
x 2n y2n = for any x,y E G, ((y x )71)2 = x71 y71 . x11 . yn. = xn • yn • xn • yn•:=> Xn ' Yn =yn• Xn. (4)
(4)
(y x)n = yn . acn. <> (5 yr-1 = yn-1 5n-1, for any x,y E G. (5
(2) <=> (y 5)2n = x2n y2n = y2n 52n <=> ( x
•y
= (x2)n (y2)n = (y2)n (52)n =
)2n-1 = y2n-1 52n-1 <=> (5 y)n-1 (x y)n =
= yn_ 1 . y71 xn-1 xn = yn-1 yn
)
xn-1 x n
yn-1 xn-1 ( x
xn-1 yn
x71 = yn
<=> xn-1 y7/ = y71 xn-1
= xn-1 xn <=>
for any
x, y E G. (6)
yn (/) x.yn. 5n-1 <=> (4) > yn • Xn = Xn • yn = X • Xn-1 •
•4=> y" • x = x • yn, for any
(y 5)4n
(x291 y2n )2 = 52n y2n .52n y 2n
54n y4n
(y5)4n = 54n • y4n (1) y4n x 4n,
(3) <#, (x y )4n
= x y 54n-1
y4n-1
54n
x, y E G. (7)
for any x, y E G. (8)
y4n = x •y
54n-1 • y4n-1 <#,
•y=x•y•x 4n-1(7) y X4n = x y x4n-1 •4=>y•x=x• y, for any x, y E G and therefore (G,.) is an abelian group. 54n
Problems for the 12th form
153
Problem XII.13. Consider a E R* and the functions f, g, h : (0, oo) —> R, so that g(x) = (ax + x3) • f (x4) and h(x) (—ax2+ x5) • f (x6), for all x > 0. Prove that f has primitives on (0, oo) •:=> g and h have primitives on (0, oo). Solution. The following theorem is well-known: Theorem. Let I C JR be an interval. If the functions f, g : I —> JR are such as f has primitives on I, g is continuously derivable on I, then the function f • g has primitives on I. " Assume that f has the primitive F in (0, oo). That means: F'(x) = f(x), x > 0. We have
f(x 4) = Fi (x4)= 4x3• F'(x4)
4x3
= (F(x4))
'
1 — for x > 0. 4x3
(1)
1 R, t(x) = — x > 0 is derivable, t' is continuous 4x3 on (0, oo) and (F(x4))' has the primitive F(x4) on (0, oo). Using (1) and the Theorem, we find that the function a : (0, oo) R, a(x) = f (x4), has primitives on (0, oo). Because g(x) = (ax + x3) • f (x4), x > 0, the function 3: (0, oo) R, /3(x) = ax + x3is derivable, 3' is continuous on (0, oo) and a has primitives on (0, oo), we obtain, using the theorem, that g has primitives on (0, oo). Using the same method, we find that h has also primitives in (0, oo). Supposing that the functions g and h have the primitives G and H on (0, oo), we have to prove that the function f has also primitives on (0, co). The function u.: (0, co) (0, oo), u(x) = -'/Y is bijective. We obtain thus g ( 45) = (a`./T + 4 x3 • f (x), for all x E (0, co). It follows: The function t : (0, oo)
.
g ()-x 1 -- ) =
+ 1) • f (x),
for all
x E (0, co).
(2)
The same as above, we find: h (x5)
( ——
+ 1) • f (x),
for all
x E (0, oo).
(3)
Summing the relations (2) and (3) we deduce: 4 (G ( 4 Y))/ +6 (H (Y7))' = 2.1(x), for all x E (0, oo), which means that f has the primitive F : (0, oo) F(x) = 2G + 3H (sIX).
R,
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"Gazeta Matematica" - a bridge over three centuries
(1 Problem XII.14. Prove that there exists only one point c E —2 , 1 I , so C
that f ez dx = (1 - c) • ec2 0
Solution. Let us define the function g : [0,1]
2 R, g(x) = (x - 1) f et dt.
The function g satisfies Rolle's theorem, because g is continuous on [0,1] and has the derivative g' (x) = f et2 dt + (x - 1) • eX2 , for any x E (0, 1). Hence, we find c E (0, 1), such that g'(c) = 0. It follows: g"(x) = eX2 ex2 + (x - 1) • eX2 • 2x = eX 2(2x2 -2x + 2) > 0, for any x E (0, 1). Therefore, the function g' increases in the interval (0, 1), and consequently, c is unique. Hence,
f
2 (c - 1) • ec = 0, for one number c E (0, 1).
et2dt
(1)
0
1 We have to prove that c E (2, 1 I . The function f (x) = eX2is continuous and increases on [0, c], and using the first mean-value theorem for the integrals, we find an unique number a E (0, c) so that
f eX2 dx = c • 02 .
(2)
From (1) and (2), we deduce that: (1 - c) • ec2
2 2 2 c • ec' •:=> eC -0t -
(3)
1 -c
We have 0
c2 - a2 > 0
eC2—a2 > 1
(3) 1 cc
> 1 •#›
1 2c -- 1 >04cE (-- 1). 1 c 2'
Problem XII.15. Let f : (0, oo) -> R be a function with the properties: a) f is continuous at x = 1; b) f is integrable on [1, 2];
155
Problems for the 12th form
c) f (x2) < x • f (x), for any x E (0, oo). Prove that the following inequality holds: 2 1 f (x)dx < 3 • f (1). , cri in the inequality
Solution. Replacing successively x by c), we obtain: f (x2) < x • f (x) < x • VX • f ( N5) < • • • < x •
•
. . . • 2'1\5 • f ( 21Nri) ,
for any x > 0. Therefore, f ( 2nfi)
f (x) < x2+1+-141 r • f ( 2'./Y) <#> f(x) < x1-
for
x > 0.
Hence, f(x) = lim f(x) < lim x l— itt • lim f ( c5) n-400
n--)co
n--400
f (x) x • f (1), for any
4
x > 0.
Using b), we find 2
2
2
J f(X)dX < f x • f (1)dx <=> 2 f f(x)dx < 3 • f(1).
Problem XII.16. Find all derivable functions f : properties: f(0) = 0;
Tr i
) --> R with the
Ir b) f (x) • sin2 x + (x) • cos2x = ex, for any x E {0, — ). 2 Solution. We choose the function g : [0, 2R, g(x) = f(x) • e'. It is obvious that g is derivable on [0, 2) and g(0) = 0. Replacing f (x) = g(x) • ex in b), it results: g(x) • sine x +
(x) • cos2x + g(x) • cos2x = 1 •:#, •#;• (x) • cos2= 1— g(x),
It is impossible that, for all x E [0,
2)
for
, we have g(x) = 1.
x E [0,
(1)
"Gazeta Matematica" - a bridge over three centuries
156
7r Indeed, supposing the contrary, we obtain f (x) = ex, for any x E {0, 2 ), r 7r \ therefore f (0) = 1, in contradiction with a). Hence, we find an xo E 2 ), so that g(xo) 0 1. 7r If g(xo) < 1, the function g being continuous on [0 ) a whole interval 2
I C [0,
2
(which includes x0), so that g(x) < 1 for any x E I. )
g'(x) 1 . , for any x E I. The function g 1 - g(x) cos2 x being derivable on I, we obtain: Then, from (1) we get
(x)
1
J 1 g(x) dx = f cost x
dx <=> ln(1
-
g(x)) = - tan x + c,
c E R. (2)
-
— g (x )
g(0) = o
e — tan x
.4=> g (x)
X E I •#. 1 _ e-tanx,
for any
x E I. (3)
If g(xo) > 1, the proof is the same, and we obtain again the relation (3). 7T
We choose the function c : [0' 2 )
R, c(x) = (1 - g(x)).etan x, derivable
7r 7r 0 -). Therefore, g(x) = 1 c(x) • e- tan x , for x E [0, — 2 2 Hence, for the interval I so that g(x) 0 1 for any x E I we have c(x) = 1. 7r Replacing g in (1), it results d(x). e- tan x • cos2 x = 0, for any x E [0 — ) '2 (x) = 0 = c(x) = 1, for any x E [0, 2 ) g(x) = 1 — e-tan s for any on
-
[
E [0 7r- ). Therefore, f (x) = ex ex—tan x , for any x E [0, 2 Problem XII.17. Consider the function f : [0,1] (0,1n 2), continuous. a) Prove that the following inequality holds: X
1 < I e-f (s)dx • f ef (x)dx < 2.
e-f(x)dx b) Prove that
= 0, where [•] denote the integer part.
1 ef (x) dx 0
Problems for the 12th form
157
Solution. a) Using the Cauchy-Buniakovski-Schwarz inequality, we obtain: 1 ( _ f(2x) )2
e
f 0
2
1 f(x)
2
dx • f (e) dx >_
.
1.
0
We have 0 < 1(x) < In 2, hence we obtain: ef(x) E (1, 2) and e-f(s) E (
2 '
I ) •
i 1 1 It follows that Ii. = f ef(x)dx E (I, 2), /2 = f e-f(x)dx E ( 1 , there0 o i 1 fore f e-f(x)dx • f ef(x)dx < 2. o o 1 ) /1 (1 /1 1) and hence [ d = O. —E b) Ii E (1, 2), 12 E ( 1 4' 2' /2 12 2 ,
Problem XII.18. Consider the field (K,+,•), K = {0,1, a, b}. Prove that there exists x E K, such that x3m+2— x3r+1+ 1 = 0, for any integers m, r. Solution. For any ring (A, +, •), with 1 0 0, char(A) = p, where p is the smallest integer, p > 0, such that 1 + 1 + • • • + 1 = 0, and char(A) = 0, if P for any positive integer p, 1 + 1 + • • • +10 0. P It is well known that if (K,+,.) is a field and char(K) = p > 0, then p is a prime number. Moreover, if K is a finite field, we find a positive integer n, such that K has p" elements. In our case IKI = 4, therefore char (K) = 2. We also know that for any field (K, +, •), if a, /3 E K, 0 0 0 and a .0 = 0, then a = 0. Therefore, for the field K = {0, 1, a, b} we have: 1 + 1 = 0, a + a = 0, b + b = 0 and the tables of the two operations are the following:
+ 0 1 a b
Olab 0 1 a b 1 0 b a a b 0 1 b a 1 0
1 a b
1 ab 1 a b a b 1 b 1 a
+ 1 = a2 - a+ 1, We have: a2 =b a3 = a = 1. Hence a = a3m-1-2 -a3r+1-tfor any positive integers m, r. Further, 0 = a3- 1 = (a - 1) • a, and because a - 1 0 0 and K is a field, we obtain a = 0.
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Problem XII.19. Consider a > 0 and the functions f, g : [-a, a] -4 R with the properties: a) f is continuously derivable on [-a, a]. b) g is an odd and continuous function on [-a, a] and g(x) 0 0, for any x E (0, a) . Prove that there exists at least one value c E ( - a , a) such that
a
a f (x)g(x)dx = 2 • f' (c) • f x • g(x)dx . —a 0
f
Solution. The following second mean-value theorem for integrals is wellknown: Theorem. Consider a, b E R, a < b and the functions f, g : [a, b] -> R with the properties: a) f is continuous on [a, b]. b) g is integrable on [a,b] and g ? 0 on [a,b]. Then there exists at least one value e E (a, b) such that b
b
f f (x)g(x)dx = f (e) • f g(x)dx. a I = f f (x)g(x)dx = —a 0 a a odd = f f (x)g(x)dx + f f (x)g(x)dx g . f g(x)[f (x) - f (-x)]dx. —a 0 0 ( f(x) - f(-x) , x 00 We choose the function h : [-a, a] -4 R , h(x) = 1 f,(0), 2x x=0 a Obviously, h is continuous on [0, a]. Hence, I = f 2x • g(x) • h(x)dx and o using the mentioned theorem, there exists at least one value e E (0, a), such that I = h(e) • 12x • g(x)dx. f(e)
f (-0 and using the Lagrange's theorem for the 2e function f, continuously derivable on [-c, l;], we find at least one value Then, h(e) =
c E (---e, e) such that
,
f (6 — A-0= f'(c). - (-0
Problems for the 12th form
159 a
We conclude that I = 2 • ft(c) 1.2x • g(x)dx. Problem XII.20. Consider (A, +, •) a commutative ring, with the properties: a) 1 +1 and 1 +1 +1 are invertibles (1 being the multiplicative identity of the ring). b) if x, y E A such that x3 =y3, then x = y. Prove that if a,b,c E A such that a2 + b2 + c2 =ab+bc+ ca, then a = b = c. Solution. We have: a2 + b2 +C2 = ab + be + ca •<=> (a — b)2+ (b — c)2+ (c — a)2 = 0. (1) We denote a — b= x, b — c= y, c — a = z. Replacing in (1), we obtain: x 2 +y2 ± Z2 = x+y+z = 0
(2)
a2 + b2 + c2 =- ab + bc + ca <4. a(a — b) -I- b(b — c) + c(c — a) = 0 4=>
<=>a•x+b•y+c•z=0.
(3)
But x+y+z=01•aa•x+a•y+a•z=0
(4)
From (3) and (4) we deduce:
b•yd- c•z=a-y+a•z•#>(c — a)•z=(a — b)•y•#-z2 =x - y. Replacing in (2), we find: x2 +y2 +x•y=01•(x—y)x3 —y3 =0<#.x3 =y3 4x=y<#.
•<=>•a — b=b — c42-a+c=2b•#>a+b+c=3b. Using the same reasoning, we find: a + b + c = 3a. We obtain: 3a = 3b <=> (1 + 1 + 1)(a — b) = 0 4 a — b = 0 <=> a = b. Similarly, we find b = c, whence a = b = c. We must notice that the condition "1 + 1 is invertible" was not necessary. Problem XII.21. Consider the function f : [0, 1] > R, having a derivative of the second order in [0, 1], with f"(x) > 0 for any x E [0, 1]. If p is a positive integer, p > 2, prove that the following inequality holds: —
1 f f (x)dx < P p2 —
f(1)± (P— 1 2 I f (x)dx . )
0
"Gazeta Matematice - a bridge over three centuries
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Solution. f" (x) > 0 for any x E [0, 1] implies the convexity of the graph of the function f. That means: f ((1 - t) • xi t • x2) < (1- t) • f (xi) +t • f (x2), for any x1, x2, t E [0, 1]. 1 1 Replacing t = - < - in the last inequality, we obtain: P 2 1 f(xi)d- - • f(x2), for any xi, x2 E [0, 1].
p-1
1 - x2 P
f (P
P
1 We choose the partition A = (- = xo < x1 < • • • < xn = 1 of the interval P ,1) with xk = 1 + ic(1) -1) ,for k = 0,n, and the points 6,6, • • • ,n, P Pn P such that EL fr P - IL Xk_i, Xkj, for k =1,n. sn, = -k pWe have ii•Aii = pn 1 and we deduce: 1
(
-
1 z n f (Lc p -1 1 f f(x)dx = lim P1 + — •1 ) 00 Dm 72—■ n p p k=1 1 P
1
< lim
n--°° Pn 1 1 •4=> f f (x)dx 5_. P 2 P 1
n io (Z' k=1
1 /k\ 1 f )+— P n P
f( 1 ) ) <=>
1• N-. f (LC ) — f (1) + (1) p 1) 211111 00 n, L--/ 71,--■ k=1
=
n)
P
1
p- 1 (p -1 ) 2 • f f(x)dx. = 2 f (1) + P p 12
Problem XII.22. Compute: f 12
Solution. We put
I
12
I=
-f-r2
x = -y dx = -dy
tan x • tan 3x 1
dx.
eX
. Then
tan x • tan 3x dx = 1+ ex
12
ex • tan x • tan 3x dx, 1+ ex
Problems for the 12th form 12 (1 ±
whence 2/ = f
161
12 ex) tan x • tan 3x = f tan x • tan 3x dx. 1 + ex -12
'71"
7i
The function f : [— — -> IR, f (x) = tan x•tan 3x being even, we obtain: 12 ' 12 12
21 = 2 • f tan x • tan 3x dx <=> M 12
12 f
<=> / = f tan x • tan 3x dx = 0 0
tan x • (3 tan x - tan3 x) dx 1 - 3 tan2 x
18 8 dx = <=> / = f (-tan2 x 9 9 3 tan21x - 1 ) 3 0 12
= l• tan x 3
0
fir COS2 X = f We denote Ii dx, /2 = 0 3 sin2 x - cos2 x r
1
7r
tan 12
17r 91 • 12
12
0
cos2 X dx. 3 sine x - cos2 x
sine x dx and thus 3 sing x - cos2 x ' 8 9
•
We have: 312 — 1-1 =xI17-'2 = 0
1
12
(1)
12 •
m 12
12
1 1 1 dx = f dx = f dx = Ii + 12 = f 2(1 - cos 2x) - 1 1 - 2 cos 2x 4 sine x - 1 o o M ii 3i 12 1 + tan2 x f 1 + tan2 X , dx = 1 dx = = o 1 2 - 2 tan2 x dx '>1-1 4-12=io 3 tan2 x - i 0 tang x - 1 + tan2 x 3
f
1 = - In 3
1 tan x - — tan x +
1
12
7r 0 • tan 12 -1 = 1 In 3
7r 0 • tan 1 2 +1
162
"Gazeta MatematicA"
—
a bridge over three centuries
7r Because tan 12 = 2 — -13-, we deduce —3/1 — 3/2 = ln Using (1), we obtain: Id— 1 7r —4/1 =ln a /1 = 2 12 whence / =
2—/ 3
—
2 •
7r 1 .4 — 1 + In 48 4 2 '
2 —1 / —1 in . 12 9 2 'Jr
Problem XII.23. Let H be a subgroup of the group (R, +). Prove that one and only one of the following properties holds: a) H = (a). {n •aln integer}, where a is a given real number (i.e., H is cyclic). b) For any x,y E R, x < y, we find z E H such that x < z < y (H is a dense set in R). Solution. We must observe that certain by there are subgroups of (R, +) with the a) property - the group (Z, +) of the integers, for instance - and others with the b) property - an example being (R, +). We consider H a subgroup which doesn't have the a) property. We will prove that H has the b) property. Because H doesn't have the a) property, we certainly find a, b E H, a > b, such that for every 0 E H,{a,b} V (0) .
(1)
(H being a subgroup, we may choose a, b E (0, oo)). The following assertion is known: For any a, b E (0, oo) with a > b, we find an unique positive integer, n 0, and an unique r E [0, b) such as a, = b • n + r.
(2)
(indeed, n= [—a] — b • {1). b ' r— b For a, b, n, r used in (1) and (2), knowing that a, b E H, we deduce that r E H. If r = 0, we obtain a, b E (b), a contradiction with (1). Therefore, for a, b E H used in (1), we have 11 > 0, integer and r E (0, b), such that a = b • n + r. (2') Using the same reasoning, for a <— b, b 4-r, we obtain the strictly positive integer n1 and the real number r1 E (0,r), such that b = r • ni + Ti. (If r1 = 0, we obtain b E (r) a E (r), in contradiction with (1)). Using the mathematical induction, we obtain the strictly decreasing sequence of real numbers (rt)t,>o, and the sequence of strictly positive integers
Problems for the 12th form
163
(nt)t>o, such that: ro = r < b, no =n rt > 0, for any positive integer t, rt-i = rt • nt±i + rt+1, for any t > 1. (3) The sequence (rt )t>0is strictly decreasing and inferior bounded by 0, so, using the Weierstrass theorem, we deduce that there exists lim rt =a E R+. t-.00 Suppose that a > 0. Using (3), we obtain: lim rt_i = lim rt • lim nt+i + lim rt+1 •<=> lim nt+i = 0, t.00 t-00 t-,00
t---goo
impossible, because a sequence of strictly positive integers can't have 0 for limit. Therefore, we found a sequence (rt)t>o,with rt E H, for every positive integer, and lim rt = 0. t--.D0 Let x, y be real numbers, x < y. We'll find z E H, x < z < y., Denoting Q = y - x > 0, because the sequence of positive numbers (rt)t>0 has 0 for limit, we find a positive integer t such that 0 < rt < 0. Consider the sequence (z„)n>o, z„ = n • rt E H, for 71, > 0. It is impossible that all terms of the sequence are in the exterior of (x, y), because the length of (x, y) is Q and zn+1 - zn= rt E (0,(3), for any integer n > 0. Therefore, we find a value for n, such that x < zn < y. Problem XII.24. We consider the ring (A, +, •), with 1 0 0 and such that
x3 -y3 = (x - y)(x2 xy + y2),
for x, y E A.
a) Prove that (A,+,.) is a commutative ring. b) Find whether the ring is also commutative, knowing that there exists no unit for the operation ".", but for every x E A, x2 = 0 implies x = 0. Solution. a) Consider x, y E A. X3 -y3 = (x - y)(x2
xy + y2) x2y xy2 = yx2
yxy
.
( 1)
y4-- y + 1 W x2(y + 1) + x(y2+ 2y + 1) = (y + 1)x2+ (y + 1)x(y + 1) <=> <#. (x2y + xy2) x2+ 2xy x = (yx2 + yxy) + x2 + yx +xy+x<=> (x2y xy2) + xy = yyx /_.__2 + yxy) + yx xy = yx, x,y E A,
"Gazeta Matematica" — a bridge over three centuries
164
and thus (A, +, •) is a commutative ring.
+ y W x2(x + y) + x(x2 + + +y2) = = (x + y)x2 + (x y)x(x + y) <#. (x2y xy2) + 2x3 + x2y + xyx =
b)
y
x 2y yx2
= (yx2+ yxy) + 2x3
xyx = yx2,
( x y, y x yxy = xy2,
for any
for any
x, y E A (2)
x, y E A (3)
(xy—yx)2=--xyxy—(xy2)x—(yx2)y+yxyx (3): 2)xyxy—yxyx—xyxy+yxyx=O. From (xy — yx)2= 0 and the hypothesis we obtain xy — yx = 0, for x, y E A, therefore (A, +, •) is a commutative ring. Problem XII.25. Find the real numbers a, b, c, such that the set of the roots of the equation x3+ axe+ bx + c = 0 form a group together with the x+ y operation x * y = 1— xy • Solution. Let G = {e, x1, x2} be the set of the roots of the given equation. If (G, *) is a group, let e E G be its unit. Hence,
x+e = x, for any x E G <=> e(1 + x2) = 0, for any x E G. (1) 1 — xe If x1= i E C is an element of the set G, because the equation has real coefficients, x2 = —i must also be a root of the equation. In this case xl*X2 does not exist, because the denominator is 0, false. Therefore, for x E G, 1+ x2 00, and (1) <#. e = 0. Thus, e = 0 being a root of the equation, we obtain c = 0. Then, xi and x2 are the roots of the equation: x2 +ax+b=0.
(2)
There exists only one type of groups with 3 elements, as in the next table:
* 0 X1 X2
Xi * X2 = 0 <=>
X1 + X2
1— xix2 2xi
xi * xi = X2 •4=>
0 0
x1 x2 Xi
X2
X1 X2 0 X2
0
= 0 <=>
X1
a=0
= X2 <=> 2X1 = X2 - (Xi • X2) • Xi <=> 1 - X21 2x1 =x2 —
b•xi <=> (2 + b) • xi = x2. (3)
Problems for the 12th form
165
From xi * xi = x2, we obtain (2 + b) • x2 = x1 and multiplying this equality by (3), we deduce: (2 + b) •xi •x2 = xi. x2 <=. (2 + b)2 b= b .#> b(b2+ 4b + 3) = O.
(4)
If b = 0, then G = {0}, and (G, *) is a group. Therefore, a = b = c = 0 is a solution. If b 0, then (4) <=> b E {-1, —3}. If b = —1, we obtain x1 = 1, x2 = —1, but xl*x1 does not exist, because the denominator is 0, false. If b = —3, we find G = 01, and it is easy to prove that in that case, (G, *) is a group. Therefore, the second solution of the problem is: a = c = 0, b = —3.
166
"Gazeta Matematice - a bridge over three centuries
List of authors. Problems for 12th form 1. Andreescu T., XII.12 (G.M. 9/1987) 2. BAnicg. A., XII.5 (G.M. 10/1972) 3. Batinetu-Giurgiu D.M., XII.11 (G.M. 7/1986) 4. Berinde V., XII.15 (G.M. 1/1991) 5. Berinde *t., XII.16 (G.M. 9/1991) 6. Burdu§el C., XII.21 (G.M. 4-5/1997) 7. Cerchez M., XII.4 (G.M. XLV, 1939) 8. Cocea C-tin, XII.9 (G.M. 5/1980) 9. Constantinescu G., XII.2 (G.M. VIII, 1902) 10. Cornea V., Marinescu D.$t., XII.22 (G.M. 5-6/2000) 11. Dobo§an A., XII.25 (G.M. 5-6/2003) 12. Ionescu-Tiu C-tin, XII.6 (G.M. 8/1974) 13. Ionescu I., XII.3 (G.M. XIV, 1908) 14. Ionescu M., Popescu C., XII.18 (G.M. 8/1994) 15. Macarie Gr. N., XII.1 (G.M. III, 1897) 16. Mihalca D., XII.13 (G.M. 4/1988) 17. Millet D., XII.7 (G.M. 8/1976) 18. Mocuta C., XII.8 (G.M. 11/1978) 19. Mortici C., XII.20, (G.M. 2/1996) 20. Nicula V., XII.19 (G.M. 9/1995) 21. Pan. F., XII.14 (G.M. 10/1989) 22. Popescu N., XII.10 (G.M. 2/1984) 23. Tetiva M., XII.24 (G.M. 1/2002) 24. Tolo§i M., XII.23 (G.M. 12/2001) 25. Tudor I., XII.17 (G.M. 9-12/1992).
Chapter 10
Problems for the annual competition. Forms 9-10 Problem C IX-X.1. Consider a trapezoid with the property that the sum of the square of a diagonal and the square of a nonparallel edge equals the sum of the square of the other diagonal and the square of the other nonparallel edge. Prove that it is isosceles. Solution. Denote by d1, d2 the diagonals, b, c the nonparallel edges, a and d the bases and a, 0 the angles between b and a, respectively c and a. The cosinus theorem gives us d? = b2 + a2— 2ab cos a and d3 = e2 + a2— 2ac cos 0. Subtracting these equalities and using the hypothesis we obtain b cos a = c cos 0. If h is the altitude of the trapezoid we have (by Pythagora's theorem) b2 = h2+ (b cos a)2 = h2 + (c cosi3)2 = c2. Therefore the trapezoid is isosceles.
Problem C IX-X.2. Find the values of the real parameter a so that the equation Ix — a2 + aI + lx 3 + aI = 20 has two and only two solutions. —
Solution. Consider the equation lx — al + Ix — 01= y. In order to have two and only two solutions a, 0 and y must verify the relations f a + 13 — 7 5 min(a, 13) c t - F )3 + 7 _> max(a, 3). Therefore we must solve the system { a2—a — a + 3 — 20 < min(a2— a, 3 — a) a2— a — a + 3 + 20 > max(a2— a, 3 — a) 1) If a E (oo, —V-31 then the first inequality implies a2— 2a — 17 < 3 — a with the solution a E [-4, —4. 167
"Gazeta Matematica" — a bridge over three centuries
168
2 For a E (-0,4 we have a2 - 2a - 17 < a2 - a with the solution a E ( - 0, 3) a E (0, oo) gives the solution a E (0,5]. Hence, the first inequality has the solution a E [-4, 5] In the same way we obtain from the second inequality that a E (—co, 23]. Intersecting the two sets we have that for a E [-4, 5], the equation has two and only two solutions.
4.
Problem C IX-X.3. Let xi, i = 1, 2, ..., n, be n positive numbers with the 71
property
>2 xi = 1.
xk + > n + 1
Then there exists at least one k E {1, 2, • • • , n}, so that
.
Solution. Assuming xi > n-1, V i E {1,2,• • • , n}, we find
>2 xi > 1;
i=i contradiction. Therefore, there is k E {1, 2, • , n} so that xk E (0,n-1). But the function f : (0, co) -+ IR, defined by f (x) = x + x-1, is decreasing on the interval (0,1]. It follows f(xk) > f (71-1). So we obtain the conclusion.
Problem C IX-X.4. a) Prove that for any positive integer n, the following inequality holds: 1 1 1 10n + > + 1 5n+1 5n+2 + + 10n 15n
b) Prove that for all positive integers p and n, the following inequality holds: 1 1 1 2pn + + 2pn >- 3pn . ±1 pn + 1 pn + 2 Solution. Clearly, we have to prove the next statement: 1 1 1 2k k > k + 1 k + 2 ++ 2k 3k + 1'
2, 3,••• .
Let k > 2 be an integer number. The identity of Catalan-Botez gives us: 1 Ek+i = i=i E 2i - 1 2i = t=i 2i(2i - 1) •
(1)
2k k ( 2i 2i 2(i - 1) _ kk 2 -j 3k + 1 = j=1 3k+1 + 1 3(i - 1) + 1) E j=1(3i - 2)(3i + 1) •
(2)
(
Also we have:
169
Problems for the annual competition. Forms 9-10 2 (3i — 2)(3i + 1)
1 2i(2i — 1) Then,
But
(i — 1)(i +2) 2i(2i — 1)(3i — 2)(3i + 1)•
2 1 > i =1,2,...,k. 2i(2i — 1) (3i — 2)(3i + 1) ' The equality holds only for i = 1. The summation of these inequalities, gives (by (1) and (2)): k 1 2k 3k+1. 1 k 2 2k = log 2. Also Remark. It is well-known that lien > k.0.0 i=ik + i 3k +1 3 2 Hence we get the inequality log 2 > 3 Problem C IX X.5. Let ABC be an acute triangle with the altitudes IAAli, IBA], ICC]] and the orthocenter H. If: 1 area(HA1B) + area(HB1C) + area(HC1A) = — area(ABC), 2 then the triangle ABC is isosceles. -
Solution. Since ABC is an acute triangle, we have
area(HA1 B) + area(HB1C) + area(HC1A) + +area(H AiC) + area(HB1A) + area(HC1B) = area(ABC). Then, from the hypothesis, we get
area(HA1B) + area(B1C) + area(HC1A) = = area(HA1C) + area(HB1A) + area(HC1B). Using the sinus theorem, we obtain BA1 = 2R sin C cos B and HA1 = 2R cos B cos C. Thus, 1 area(HA1B) = BAi HA1 = 2R2 sin C cos C cost B = 1 R2 sin 2C(l+cos 2B). —
—
2
2
In a similar way, we find the following relations: 1 area(HA1B) = R2sin 2A(1 + cos 2C), —
2
1
area(HC1A) = R2sin 2B(1 + cos 2A), —
2
1
area(HA1C) = R2sin 2B(1 + cos 2C), —
2
area(HB1 A) = 2 R2sin 2C(1 + cos 2A)
"Gazeta Matematice — a bridge over three centuries
170 and
1 2 area(HC1B) = — R sin2A(1+ cos2B). 2
Therefore: sin 2C(1 + cos 2B) + sin 2A(1 + cos 2C) + sin 2B(1 + cos 2A) = sin 2B(1 + cos 2C) + sin 2C(1 + cos 2A) + sin 2A(1 + cos 2B). Using the relation A + B + C = 7 and some usual trigonometrical identities in the above equality, we obtain the next factorization: 4 sin(A — B) sin(B — C) sin(C — A) = 0. Hence sin(A—B) = 0 or sin(B—C) = 0 or sin(C—A) = 0. So, we conclude that ABC is an isosceles triangle. Problem C IX-X.6. Let b > 0 and ai,a2,..., a, E R \ {0}, (n > 2) such that a2 + + an = b, and
R+
+...+
p
(p, q E N, 7/ odd and p < q). Show that at least one of the numbers al, a2, . , an is not positive. Solution. Suppose that ai > 0, for all the indices 1 < i < n. Define the numbers xi = V.-l a• 1 < i < n. b Then we obtain the relations
xq +
= 1, and 4 + 4+ • • • + 4= 1.
It follows that xi E (0, 1) and consequently xl < xP, (V)i, which is a contradiction. Problem C IX X.7. Two points M1 and M2 are moving clockwise (MO and counterclockwise (M2) by constant speeds on a circle. At a moment t they are at opposite positions Al and A2, then when they meet for the first time: m(Al OMI) = m(Ai0M2) = a is true, with a known. Determine the angle 13 = m(A10M1) = m(Ai0M2) when they meet the second time. -
Solution. We can consider a the angular speed of M1 and 7 - a the angular speed of M2. They meet again after 2a, that means Q = 3a.
Problems for the annual competition. Forms 9-10
171
Problem C IX X.8. For a,b E Z, prove that (a + n,b + n) = 1 Vn E N if and only if a — b = ±1. -
Solution. Suppose that a — b = k, k ±1 then a, and b are congruent mod k. a = sk +p, b = lk + p and for n k — p we have (a + n,b+ n) 0 1 Conversely, a + n and b + n are consecutive numbers so (a + n, b + n) = 1 V E N. Problem C IX X.9. Prove that in any triangle we have: -
R b c —>--. r c b Solution. We know that A B C r = 4R sin — sin — sin — . 2 2 2 From the sinus theorem, we can write:
b c sin B sin C - + = + c b sin C sin B • R b c Thus, the inequality — > -+ is equivalent to the following one: r c -
b
1 sin B sin C > + A B C sin C sin B • 4 sin — sin — sin — 2 2 2 Using the relation A+ B +C = 7t and some known trigonometrical identities, we obtain the next equivalent inequality:
cos
B+C B—C + cos 2 2 > 2 cos B
2
C (1 (2 cos2
B C
2
2 B— C 1)(2 cos 2
1)l. (1)
B—CB+C 7 < 2 < — and cosinus is a decreasing function on 2 2 B -I- C B C we obtain 0 < cos cos < 1. 2 < 2 B+C Denote m = cos E (0, 1) and consider the function f : R R, 2 defined by: Since 0 <
—
f (x) = 4m(2m2— 1)x2+ x + m — 4m3, x E R.
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172
Remark that the inequality (1) holds if and only if f (x) > 0, V x E (m, 1]. We have f (m) = 2m(2m2 - 1)2 >0 and f (1) = (m + 1)(2m2 - 1)2 > 0 and, regarding the sign of the coefficient of x2, we must analyse the following three situations: 1. m E (0,
; in this case we have 4m(2m2 - 1) < 0; since
f (m), f (1) ?_ 0 we get f(x) > 0, V x E (m,1]; 2.. m = 3. m E
1 we have f (x) = x - — > 0, V x E
1
1 1/2
1
, 1].
in this case we have 4m(2m2 - 1) > 0 and f has a
1 ; thus f is an increasing function 842t2 on [m, 1] C [xo, oo); therefore f (x) > f (m) > 0, V x E (m, 1]. negative minimum point xo =
Problem C IX-X.10. Let fi(X), f2(X) be polynomials in Q[X] of degree at least one. Prove that there exists h(X) E Q[X] so that fi(X) h(f2(X)). Solution. We solve this problem by considering that we can write the equality fi(X) • g(X) = h(f2(X)). The polynomials fi,g,h, f2 have the degrees p,k,l,q, respectively, and we know that p and q> 1. The following equalities hold: p + k = lq and k +1+ 1 +1 = p + k + 1. Then 1 = p - 1 and k = (p - 1)q - p. The polynomials g and h can now be found by identification. If p = q = 1 then k = 0 and 1 = 1 and h is determined by identification. Problem C IX-X.11. Prove that, in any acute triangle ABC, min{ma, Mb) Mc} < max{ha, hb, he},
where ma, mb, meare the lengths of the medians and ha, hb, It, are the lengths of the altitudes. Solution. We can suppose that A < B < C. Then a < b < c and we obtain the following orderings:
V 2(a2 and
2) - b2 V2(b2 b 2) - < \ /2(c2 a 4 4 2S 2S 2S <—< c b a
In this case, me < mb < ma and h„ <
hb
< hc.
c2) _ a2 4
Problems for the annual competition. Forms 9-10
173
Therefore, we have to prove that 7n, < ha. Using the sinus theorem, we obtain the equivalent trigonometrical inequality: 2 sin2 A + 2 sin2 B - sin2C < 4 sin2 Bsin2 C. But 0
Problem C IX-X.12. If ai,a2, n E N, then
,an E [1, oo) or al, a2, • • • ,an E (0) 1),
(1 +(11(22 . . an) n > (aia2 ... an) 2 (1 + al)(1 a2)
(1 ± an).
Solution. We consider only the case ai E (0, 1), since the other case can be reduced to this one by dividing to (aia2...a n )n and using the substitution 1 — = xi. Consider the function ai
n-1
F(ai,a2,... , an) = (1 + aia2...an)n - (aia2...an) 2 (1 + ai) ... (1 + an). Using the symmetry we can assume that al < a2. It follows, for 0 < A < 1, a2 F (Aai — a3, • • • , an)< F(ai, a2, • • • 'an). ' ' After a short calculation this is equivalent to the inequality (1 + Aai) (1 + a2 -A-)
(1 + ai)(1 a2),
which becomes the obvious inequality (1- A)(a2 - Aai) > 0, since A < 1 and al < a2. Choose A = a2. It follows F(ai, a2, , an ) > F(ala2, 1, a3, I an). Now, if we consider, using the symmetry, that al < a2 < < a,„ then, by repeating the argument above, we get F(ai, a2,
, an) > F(aia2, 1, a3, . , an) >
> F(aia2 ... an, 1, . , 1).
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Let x = aia2 ... an. We have
F(x,1,... ,1) = (1 + x)n - xY • (1 + x) - 2n-1 >0, 1+x > 1,5. 2 Problem C IX X.13. Find all polynomials P E Z[X] such that P(a) divides a, VaEN\101.
since
-
Solution. Let P E Z[X] be a polynomial with the property in the problem. We have IP(a)I < a, VaEN\ 101. If the degree of P is greater than 1, then there is a positive integer a such that IP(a)1 > a, contradiction. It follows that P = mX + n, with m, n E Z. 1. Suppose that m, n 0 0. Since POO divides Ink, m + 1 divides 1, if n > 0, or m - 1 divides 1, if n < O. We obtain m = -2 and n > 0 or m = 2 and n < 0. In the first case, P(1) divides 1 implies n E {1, 3}; but we see that P(4) doesn't divide 4. In the second case, P(1) divides 1 implies n E { -1, -3} and also we observe the same situation as above. 2. If n = 0, then ma divides a, VaEN\ {0} if and only if m E 1-1, 11. 3. If m = 0, then we obtain clearly P = 1 or P = -1. Therefore, there are only four polynomials that verify the property in the problem: Pi = XI P2 = - X , P3 = 1, P4 = -1. Problem C IX X.14. We consider the function f : R -+ JR , f(x) = -x + 4k for x E (2k - 1, 2k + 1], Vk E Z. a) prove that V y E R, 3 an unique x E R so that f (x) = y; b) find the values of x so that f(f(x)) = x. -
Solution. For each k, the function fl(2k-1,2k+11 that is strictly decreasing maps the interval (2k - 1, 2k + 1] in the interval [2k - 1,2k + 1). We observe that f (2k) = 2k so f (f (2k)) = 2k for all k. Problem C IX X.15. Let A C R be a finite non-empty set such that -
An{-1,1}=0andxeA
x 3 EA. x +1
a) Show that the number of elements of the set A is a multiple of 3. b) Find the set A, if all elements are integers. x Solution. a) Consider the function f : R \ {-1, 1} -4 R, 1 (x) = x
3 . One +1 obtain immediately that f : R \ {-1, 1} -4 JR \ {-1, 1} is a bijection. Also we can derive easily the relations:
(f o f o f)(x) = x, and f(x) 0 x, (V)x E R \ 1-1,11.
175
Problems for the annual competition. Forms 9-10
From these relations we obtain that, for any x E R \ {-1, 1}, the element: x, f(x), (f o f)(x) are distinct. Define 215 = {x, 1(x), (f o f)(x)}, for any x E A. We have A= U Ax. xEA Also, if Ax n Au 0 0, for some x, y E A, then it follows A5 0 Ay. Therefore the family {Ax :x E A} is a partition of the set A. It follows that the cardinal number of the set A is multiple of 3. b) Let x E A. Since f(x) is integer, we obtain x + 1I( 4) x E {0, -2,1, -3,3, -5}. Since A U1-1,11 = 0 and f(A) C {0, -2,1,-3,3, -5} we obtain x ¢ {1, -2, -5}. Finally we get A = 10, -3,31. -
Problem C IX X.16. Let a, b E (0, 00) be such that ab = a + b +1. Solve the equation: -
ax
a2x bs — 2as - bs
1 1 + a2x
1 asbx 1 •
Solution. After an elementary computation, the equation can be written in the equivalent form: (ax + bx)2 = (axbx
-
1)2.
If x > 0, this equation is equivalent to ax + bx = axbx — 1, i.e., 1 x 1x (Tx) + (I))
1x =1. -a7)
Since the function f (x) = (1/a)5+ (1/b)5+ (1/ (ab))x is strictly decreasing and f (1) = 1, it follows that x = 1 is the unique non negative root. If x < 0, we take x = y, y > 0. We derive (a-y b—y)2 = (a —yb—y 1)2 and then, multiplying by a2Yb2✓, we obtain (ay + bY)2 = (OP - 1)2. From the above result we obtain y = 1. In conclusion the roots of the given equation are x = 1 and x = 1. -
-
Problem C IX-X.17. Let (Fn)n>1be the Fibonacci sequence: F1 = F2 = 1 and F,,,,±1 = Fn + Fn-1, for all n > 2. Put
Sk =
k
> F2j _ 1, for all k> 2. j=1
Prove that: E(_i)k • (n) • Sk = (-I)" • Fn. k=1
"Gazeta Matematica"
176
—
a bridge over three centuries
Solution. The characteristic equation of the Fibonacci sequence (Fn)n>1, is X2— x 1= 0 and has the roots: a = (1 + 4/2 and 0 = (1 — -4)/2. We obtain that 1 1 n> 1. —
The following relations are immediate: 1 — a = 0, 1 — = a, 1 + a = a2, 1 + = 02, a = —1. So we have: k
Sk
—
(a23-1
/32j-1) —
0
a 1 _a2k .16
1 — 02k
1— a2 16 1 — 02 —
/3 1 _ 02k 2k 2k a02 = Tea —13 ).
a 1 _ a2k 0a2 Consequently:
E(-1)k. (k)
.
sk
=
1 E(_,) n k ()(a2k k=1
k=1 1 n
k=0
=
_ 02k) =
k (n)(a2k _ 02k) = 1 [(1 _ 2n — (1 _ )132)nj k
(Or (an_ on) = (-1)n Fn.
[0na2n — an/2n1 =
Problem C IX-X.18. The interior, respectively the exterior bisectors of the angles B and C of triangle ABC intersect the tangent at A to the circumcircle at D and E, respectively at M and N. Prove that 1 11 1 = AD AE AM AN Solution. We use the sinus theorem to compute the length of the segments AD, AE, AM and AN. In .LABD : AD
AB
sin —) 2
sin (C —
2
But AB = 2R sin C. It results: 2R sin C sin AD= sin (C —
2
—B2 =
R sin B sin C cos —sin C 2
-
2
Problems for the annual competition. Forms 9-10
177
Similarly, in AAEC we have :
R sin B sin C
c
AE .--
cos --sin (B — 9 • 2 2 Then
1 1 + AD AE
C Cl 2 — B) + cos — sin (B — 2 —2 2 = R sin B sin C sin C + sin B = 1 1 = R sin B sin C AC + AB' cos
B
sin (C
—
=
In a similar way, we prove that 1 11 1 + = + AM AN AC AB So we obtain the conclusion. Problem C IX X.19. Prove that in any triangle the following inequality holds: -
ra rb ra r he h e rc + + + e rb + + > ha + hb+ha + hb + + . rb ra r c r a re rb hb ha he ha he hb
—
—
—
—
—
—
—
—
—
—
—
—
—
Solution. The radii of the excircles have the following lengths: ra = A B C p tan —, rb = p tan —, rc = p tan —, where p is the semi-perimeter of the 2 2 2 S triangle. The altitudes of the triangle can be expressed as ha = R sin A' S S hb = hc = where S is the area of the triangle and R is R sin B' R sin C length of the radius of the circumcircle. So, we obtain: (ra + 72) _ (ha + hb) k rb ra ) 111) ha
(sin
(tan A — tan — 13 2 2 B A tan tan .0
=
A — 2 —
sin
= sin
B — 2
cos2
B — 2 —
A cos2 2
AB AB sin — cos —cos — 2 2 2
— 2
(sin B sin A
sin A ) sin B
"Gazeta Matematice — a bridge over three centuries
178
But sin is a positive increasing function on the interval (0,
and cos is a
AB positive decreasing function on the same interval. Since - - E (0, 72r) one 2' 2 obtains: 2B 9B A A sin- - sin -) (cos- - cos- -) 2 2 2 > 0. AB A B sin - sin cos cos 2 2 2 Therefore ra rb ha hb rb ra hb ha Adding up all inequalities such as the above inequality, we get the conclusion. (
Problem C IX-X.20. Let a E R be such that sin a + cos a E Q. Prove that sine a + cos" a E Q, for each positive integer n, n > 2. Solution. Denote s = sin a + cos a E Q. Then sine a + cos2 a = (sin a + cos a)2- 2 sin a cos a = 1 and hence p = sin a cos a =
s2
-
2
1 EQ.
Consider now the quadratic equation t2 -st +p = 0. Its roots are t1 = sin a, t2 = cos a and by the Vieta's formulae we get sin2 s sin a + p = 0 • sine-2 a cos2 a -s cos a + q = 0
I • cos' -2 a
sin" a- s sinn-1 a+ p sinn-2 a= 0 cosh a- s cosn-1 a+p cos"-2 a= 0
n > 0.
This yields sin?' a+cosn a = s (sin'l a + cos"1 a) p (sinn-2a + cosn-2 a), and so, by induction, we obtain the conclusion. -
2 . 6 10 ... (4n - 2) + 1 is a (fl + 5)(n + 6) ... 2n positive integer and a perfect square, for each n E N, n > 5.
Problem C IX-X.21. Prove that the number
Solution. We have 2 6 • 10 ... (4n - 2) 2" • [1 . 3 . 5 ... (2n - 1)] (n + 5)(n + 6) ... 2n (n + 5)(n + 6) ... 2n 2" . (2n)! (n + 1)(n +2) ... 2n 2" • nqn + 5)(n + 6) ... 2n (n + 5)(n + 6) ... 2n = (n + 1)(n + 2)(n + 3)(n + 4).
Problems for the annual competition. Forms 9-10
179
Moreover 2 • 6 • 10 ... (4n — 2) + 1 = (n + 1)(n + 2)(n + 3)(n + 4) + 1 = (n + 5)(n + 6) ...2n = 1(n + 1)(n + 4)] • [(n + 2)(n + 3)] + 1= =(n2 +5n+4)(n2 +5n+6)+1 = (n2 + 5n + 5)2 , q.e.d.
Note. It is easy to verify that the first part of the problem remains true if we consider the fraction 2 • 6 • 10 ... (4n — 2) (n + p)(n + p + 1) ... 2n ' where 0 < p < 2n is a positive integer.
180
"Gazeta Matematica" - a bridge over three centuries
List of authors. Problems for the annual competition. 9th - 10th forms 1. Alexandru M., C IX-X.1 (G.M. 3/1981) 2. Bandila V., C IX-X.9 (G.M. 2/1985) 3. Beju I., C IX-X.7 (G.M. 4/1983) 4. Blaga A., Pop 0., C IX-X.20 (G.M. 6/1986) 5. Burcea N., C IX-X.3 (G.M. 12/1981) 6. Caragea C., Bordea G., C IX-X.12 (G.M. 5-6/1988) 7. Chirciu M., C X-IX.16 (G.M. 11/2001) 8. Chirita M., C IX-X.2 (G.M. 6/1981) 9. Cocea C., C IX-X.5 (G.M. 9-10/1982) 10. Constantinescu Laura, C IX-X.18 (G.M. 7-8/2002) 11. Ghioca A. P., C IX-X.14 (G.M. 8-9/1990) 12. Mihet D., C IX-X.11 (G.M. 11-12/1986) 13. Niculescu Liliana, C IX-X.4 (G.M. 2-3/1982) 14. Nita, C., C IX-X.6 (G. M. 4/1983) 15. Olteanu M., C IX-X.19 (G.M. 1/2004) 16. Panaitopol L., C IX-X.8 (G.M. 6/1984) 17. Papacu N., C IX-X.15 (G.M. 3/1999) 18. Popescu N., C IX-X.10 (G.M. 4/1986) 19. Rotaru Fl., C IX-X.21 (G.M. 9/1985) 20. Ursarescu M., C IX-X.17 (G. M. 1/2002) 21. Zidaru V., C IX-X.13 (G.M. 3/1989)
Chapter 11
Problems for the annual competition. Forms 11-12 Problem C XI XII.1. Denote by S„ the set of all permutations of order n. If a E S„ denote by m(o-), the number of the inversions of the permutation a. Prove that (n m(0-) = n! — •2 ). 2 0-esn -
Solution. Define M := {(i, j E {1,2, ...,n}, i < j}. For any finite set A, denote by Card A, the cardinal of the set A. We have successively:
E
m(a)
=E
Card {(i,
E M I cf(i) > a(j)}
,Esn
of7 E Sn
E creSn
Card lo- E Sn I cr(i) > u(i)} =
E Card
(
crES,i
E E
U lo- E Sn lo-(i) = a(j)=k} 1
Card lo- E Sn I o-(i) = /, o-(j) = k} =
o- ES,,1
E E
n! n (n — 2)! = (2) (n 2 ) (n — 2)! = 2 • (2).
crES„ 1
Problem C XI XII.2. Consider the matrices -
(
A=
B=( a (' 3 ') C= 181
as ba ay by bi3 aS 1)5 ca da c-y dy cl3 dd c(5 d8)
aQ
"Gazeta Matematicr - a bridge over three centuries
182
where a, b, c, d, a, 0,7, 6 is not singular, too.
E
R. Prove that if A and B are not singular then C
Solution. Consider the equation AX B = D, where X
(m n =( x y and D = P z t )' q )
It follows
X = A-1DB-1•
(1)
From (1) it follows that the matrix X is uniquely determined and hence x, y, z, t are uniquely defined. On the other hand, the matrix equation can be written in the form
n (a b ) ( x 6) = ( 77; q z t) ('Y '3 c d
)
or
(ax + bz ay + bt ) (a /3 ) = (m n cx + dz cy + dt -y 5 Pq or
( a,ax + baz + ayy + Int a0x + b/3z + a,0y + bOt) _ ( m n ) cax + daz + c-yy + dyt c/3x + d/3z + cOz + dot ) Pq Or
+ baz + ayy + byt = m I aax a0x + b0z + aSy + bot = n + daz + c-yy + d-yt = p I cax c/3x ( + d/3z + cSz + dot = q
(2)
But the system (2) is equivalent to the matrix equation (1) and then it has an unique solution. This implies that the determinant of the system (2) is non zero. Hence det C 0 0, i.e., C is not singular.
Problem C XI-XII.3. Let p E NU {0} . Consider the sequence given by n (lnk)P (ln n)P+1 Xn = k p +1 • k=1
(Xn)n>1 .1
(1)
Show that this sequence is bounded and beginning with certain rank, is monotone, and consequently it is convergent, (A generalization of the sequence which has as limit, the Euler's constant; this one is obtained for p = 0.)
Problems for the annual competition. Forms 11-12
183
Solution. Consider the function f (x) = (lnx)P+1/(p + 1), x > 0. We have f' (x) = (In x)P/x and f"(x) = (In x)P-1(p - In x)/x2. We can write
E f(k) - f (n + 1).
Xn = k=1
Define no = [eP+1], where [•] denote the integer part. We have f"(x) < 0, for x > no and then the function f' is decreasing on the interval [no, oo). Let k > no. From the Lagrange's theorem we obtain f (k + 1) - f (k) = f' (6k) with 6k E (k, k + 1). Therefore we have
(k + 1) < f + 1) - f (k) < (k), (V)k > no. If n > no, from (1) we obtain x„+1 - x„ =
(n + 1) - f (n + 1) - f (n) < 0.
Also, for n > no, we have no
Xn =
E
(k) +
k=1 no
> E
k=1 no
E
(k) - f (n + 1) >
k=n0-1-1
(k) +
E
+ 1) =
(f (k + 1) - f (k)) -
k=n0-1-1
= E f'(k) - f(no+ 1):= m. k=1
Hence we have for any n > 1 ,x„,„m} < x„ <
xmo}.
Problem C XI-XII.4. Compute the limit 1 lint n2 [(1 +n
n-4o0
+1
)n
n+1
- 1 + n
.
Solution. Consider the function
f (x) = (x + 1) In
x+2 x ln x+1
+1 , x > O.
We have for x > 0
(x)
=
1 + 1)(x + 2)
+ In(1
1 , 3x + 4 and in (x) = x(x (x +1) 2 ) 2 + 3x + 2) 2.
"Gazeta MatematicA" - a bridge over three centuries
184
Using the Heine theorem and the L'Hopital's rule we obtain: lim n2 [(1 + 1 )71+1 - (1 + -1)711 = n +1 re(n+1)1n n+i 1 • bin n2 [ef (n) 1] = e n In n+-J- = lim e n In + = lim n2 n-+oo n-∎co L n-400 1 f (n) 1(x) lim = e lhn lim e = e lim n-.00 Th2 x-K:o x-2 = n-*00 f (n) (3x + 4)x3 e f (x) f"(x) lim 6x-4 = e lim = e x.co lim -2x-3 = e x--400 =. x-,c0 6(x2 + 3x + 2)2 2 n-'00
7
Problem C XI XII.5. Let f : [a, b] ---4 IR be a function such that f is n +1 times differentiable on (a, b), n > 2 and xo E (a, b) with f(x0) = c(x0) = r(x0) = 0, f ("±1) (x0) 0 0. Prove that -
lim
f(x) if "(x) + 1 1 _ 1 f'(x) [ f'(x) x - xoi .
Solution. Let p be the unique integer for which f (x0) (xo) = • • • = f (P) (xo) = 0 and f (P+1) (x0) 0 0. From the hypothesis it follows that p > 2. By induction and L'Hospital's rule, any two functions cp and V) that are k times differentiable on an interval I, k > 1 and for which there exists a point xo E I such that (p(x0) = co)(xo) = • • • = 42(k-1)(xo) = 0, 0(xo) = 0)(4) = = 0(k-1)(x0) = 0 and 0(k)(x0) 0 0, satisfy lim ca(x)/1/)(x) (19 (k) (4)10k) (x0). Hence we obtain
X-*
.f (i)f x\
) (x - xo)P+1-3 .
f (13+1) fx° ) , + 1 - j)!
j = 0, 1, 2.
Consequently it follows
1
lim f (x) P ell(x) + 1 = lim f (x)(f"(x)(x x°) f'(x)) (X)) x-4x°.f'(x) fi(x) x — x0] x—xo (f (x))2(x — x0) f ll (x) f (x) [ f'(x) 1 (x )73+1 - x0 )P-1 (x - xo) Pi (x - x0 = lilil x-∎ xo f'(x) I 2 [(X
1 + + 1)! [(p -1)!
- X0)P
1
1
=1.
Problems for the annual competition. Forms 11-12
185
IR, f (a) = 0, f (b) = 1 is a convex Problem C XI XII.6. If f : [a, b] increasing differentiable function on the interval [a, b], prove that -
fab
2 fb f 2(x) dx < — f (x) dx. 3 a Solution. Let c = sup{x E [a, b] I f (x) = 0}. From the continuity of the function f we find f(c) = 0. Let show that f is strictly increasing on the interval [c, b]. Indeed, assume that there exist c < x1 < x2 < b such that f (xi) = f (x2). We have f (x2) > 0 and then by the convexity of f we have
f (xi) <
X2 — x1 X2 — C
xl
AC)
—C
X2 — C
X 1 —C
f (X2) =
X2 — C
f (X2) < f (X2)
a contradiction. Therefore we have f(x) > 0, for x E (c, b]. f (x) We have lim = 0. Indeed, if x E (c, b], from Lagrange's theorem x—.c+ 1' (x) there is G E (c, x), such that f (x) = (G)(x — c), and since f is convex, we f (( x) obtain < x — c. f'x ) Let 0 < e < 1 be arbitrarily chosen. Then we can find a point d E (c, b) such that: f (d) d—c<e, f (d) < E, and < E. (1) f'(d) Define q = f (d). Since f : [d, b] [q,1] is a continuous bijection, it has an inverse g : [q,1] [d, b] and we have g' (f (x)) = 1/ f' (x), for x E [d, Since f' is increasing it follows that g' is decreasing and also it is bounded: 0 < g' (t) < (q), t E [q,1]. Consequently g' is integrable. From relation (1) we obtain q g' (q) < e. (2) By applying the change of variable x = g(t), we obtain for j = 1, 2
fdb 1
7 (x) dx = f 1 f (g(t))g' (t) dt =
f1
t3 (t) dt.
(3)
From the integrability of the functions t- (t), t E [q, 1], we can choose a partition q = to < t1 < . . . < tn, = 1, such that ti— < q and 1
t3g/(t) dt — E(ti)3g/(ti)(ti— ti-1) i=i Define
n Aj=
xi g
i=1
<E, j = 1,
(ti)f u3 du, j = 1,2. xi-
2.
"Gazeta Matematica" - a bridge over three centuries
186 We have for j = 1, 2:
n
Ai -E(ti)-1g'(4)(ti - ti-1)
<
i=1
ti Eg'(ti) f ((ti)j - uj) du < ✓ ti-i t=1 nfti
—
ti-1)
E
j=1 ti- i
5 9(q)
(ti + ti_oi-1 du <
• q[2(1 - 03-1(1 - q) < 23-1E. (4)
Therefore it follows
j = 1,2.
t3 (t) dt - Ai < (1
(5)
Also we have: n
A3
k(i)
= E
n-1
± E(g1(t k
i=1
ti )
-
9(4+1)1 f
k=i
1
n-1 k
t,,
n-1
du =
i-lui t.
= 9/(1) f ui du + E E(g'(tk) - gl(tk44) f u3du = ti _ i q k=1 i=1 tk
= 9/(1) f u3du + E q k=1
- gi (tk+i) f u3du. q
Now, for 1 < k < n we have tk
U2 du
1 = ((tk)3 -
q3) 5.
-
3
Since 9' (tk) relation (1):
A2 <
-
2
3
(tk)3 =
2 ftk 3
2 u du +
q
q —
3
g'(tk+l) > 0 and 0 < g/(1), we obtain from above and from
to
• gi (1) f
q
n-1 2 tk u du + E (g/(4) - g'(tk+i))k- f u du + 3 o k=1
2q2 n-1
5 .Al + -3- E(d(tk) - g'(tk+i)) < k=1 ,2
3•A1+
2
5• Ai. + E.
q
• Problems for the annual competition. Forms 11-12
187
Taking into account equations (3) and (5) we obtain
fab
f 2(x) dx
E3 + I t2 (t) dt < A2 ± 4E < <
2 -3- • Al + 5e < Ll tg'(t) dt + 6E = fdb (x) dx + GE <
•
—2 3
f(x) dx + 6e.
'
a
Since 0 < e < 1 was arbitrary, we obtain the inequality in the problem. Problem C XI XII.7. Prove that any ring with 23- q elements, where p and q are two prime numbers, is abelian. -
Solution. Consider Z (A) = {x E A : xy = yx, for any y E A}. It is known that Z (A) is an abelian subring of A. In particular, Lagrange's theorem implies that card (Z (A)) I pq. Clearly, {0,1} C Z (A), thereby card (Z (A)) E {p, q, pq} . We want to prove that card (Z (A)) = pq. Then it follows that Z (A) = A i.e., A is abelian. To this end, suppose that card (Z (A)) = p. Select a E A\Z (A) and consider the subset B = {a0 + aia + + and' : n E N, ak E Z (A)}. One can verify that in fact B is an abelian subring of A and a E B, so that Z (A) is strictly included in B. Lagrange's theorem implies that card (B) pq, and card (Z (A)) I car dB . We conclude that card (B) = pq, impossible. The case card (Z (A)) = q can be analyzed in a similar way. Problem C XI XII.8. Let K and L be two finite fields such that there exists a morphism of fields f : K L satisfying the property: for any irreducible polynomial ao+aiX+... +a„X" in K [X], the polynomial f (a0)+f (ai.) X+ + f (an) xn is irreducible in L [X] . Prove that K and L are isomorphic. -
Solution. It is known that any morphism of fields is injective, therefore it is sufficient to prove that f is surjective. Suppose there exists an element a E L\Imf. Notice that a 0 0, because 0 E Imf. If q denotes the number of elements of L, then aq-1= 1. Hence a is a root of the polynomial Xq-1— 1, which has coefficients in I in f Consider P E Imf [X], the minimal polynomial of a over Imf. Then P is irreducible over Imf, and it is of degree n > 2. We now use the fact that K is isomorphic Imf, which implies that P is of the form P = f (ao) + f (ai) X +... + f (an) X' . Hence ao + al X + + anXn
188
"Gazeta Matematica" — a bridge over three centuries
is irreducible in K [X] . According to the hypothesis, P is now irreducible in L [X]! This contradicts the fact that a is a root of P. Consequently, f : K L is an isomorphism of fields.
Problem C XI XII.9. Let (G,.) be a group with unit e, and a E G be fixed. Suppose that x2k = (ax)k holds for any x E G, where k E N* is a fixed prime number. Prove that xk =e, for any x E G. -
Solution. For x = e we find that ak =e. Replace x with ax; then (a2x) k = (ax)2k = ((ax)k)2
x4k.
By induction it can be found that (anx)k = x2n•k.
For n = k we get xk = x2k , therefore we proved that x(2k-1).k = e, for any x E G. Let us now return to the hypothesis: x2k = (ax)k , for any x E G. Since x commutes with x2k, we may infer that x commutes with (ax)k .It follows that (ax)k = (xa)k ,for any x E G. This relation holds if x is replaced by k a —l x, and we get xk = (a —l xa) = a —l xka. Hence axk = Xk a thus, if x is replaced by xk in the hypothesis, it results (xk) 2k = =ak = x L.2 . that x2k2 = (aXk) k ( x k) We proved that xk2 =e, for any x E G. Now it suffices to prove that k = (k2, (2k — 1) k) , because if this is true, we will have that ord (x) = k. We have that 1 = (k, 2k — 1) . Indeed, k is a prime number, so that by Fermat's first theorem 2k-1 a- 1 mod(k). Hence (2k— 1) -a: 1 mod(k) that is to say (k, 2k — 1) = 1. "k
Problem C XI XII.10. Let (G,.) be a group with unit e and with an element a E G\ {e} that belongs to any proper subgroup of G. Prove that if H and K are two subgroups of G, then either H C K or K C H. -
Solution. Let us consider L a finite subgroup of G. The group (L,.) is abelian, thereby isomorphic with the direct product of some cyclic groups, i.e., (L,.) ^ (z 1 x X Z p r,+) where pi are prime numbers (not necesPi sarily distinct) and cti E N*. The elements b = ( i, 6„ , and c = (6, 1 ...6) in Z .1 x x Z Pr satisfy bn = cm if and only if bn = = (6, ..., 6) . It Pi follows that r = 1, because otherwise a cannot belong to both < b > and . We proved that any finite subgroup of G is cyclic, isomorphic with (Zr ., +) , for p a prime number and a E N*. In fact, p is the same for all the subgroups, since ord (a) is a divisor of their orders. The subgroups of (Zp. , +) are {0} c< >c c< p >c< 1 >= Zpa. Consider now x, y E G. First, notice that (Z, +) does not have the property in the hypothesis, thus any cyclic subgroup is finite, in particular < x > ,
189
Problems for the annual competition. Forms 11-12
and < y > are finite. Second, the subgroup < x > • < y > is also finite and includes both < x > and < y >, therefore either < x >C< y > or < y >c< x > . Indeed, recall that we pointed out that the subgroups of Z have this property. Summarizing, we proved that in the group (G,.) any x and y satisfy either < x >c< y > or < y >C< x > . In particular, if x and y have the same order, then < x >=< y > . It results that if G is finite, then it is cyclic, therefore any subgroups H and K satisfy either H C K or K C H. If G is infinite, consider M = {a E N* : there is xa in G of order pi. Certainly, G = UaEM < x,„>, and < xa >C< Xo >, provided that a < 0. Then M is infinite, as a matter of fact, M = N*. One can see that G = Un < xn > and G has no proper infinite subgroups. oo such that If H is such a subgroup, then there is a subsequence nk Uk < Xnk ›C H. But the left term is still G, therefore G = H. We have proved that G has only finite subgroups, so that the result holds even if G is infinite. Problem C XI-XII.11. Let f be a polynomial of degree > 2 with rational coefficients. Let us consider the set A = {x E R \Q : 1(x) E Q}. If ep : R R, co(x) = infteA Ix - tl, prove that cp(x) = 0, V x E R. Solution. Assume f = a(an Xn + an_iXn-1+ • • • 4-aiX + ao), where n > 2 is an integer, a E \ {0}, and ak E Z, = 0,1,• • • ,n, with an 0 O. Let x and E. > 0 be two real numbers. n We can find a prime number p so that p > max {-, Ian ' . Define u = E
e)p] + 1. From the inequalities (x - E)p < u < (x - E)p + 1 and n + 2 2n < — < 2E, we obtain P - p uu+1 u+n+1 x e<< < < < x + E. P p p We denote by g the polynomial with rational coefficients of degree n - 1, 1 defined by g = f ( X + -) - f (X). In the set { u ± k, k = 0,1, . . , n} P P [(x —
-
•
there is a number q so that g q 0 0, i.e. f -(- of q+1 . But the P P P P associated real function of the polynomial f has the Darboux property. Then, for any A E (0,1) there is cA E ( q ±1)so that f(cA) = f ( q) ' 13' qP P b ( g+ 1) (g +1) f (q) f ()} q . We can write f A{f p = a pn-k., with P P P
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1 k E {0, 1, • • • , n} and b E Z, so that (b, p) = 1. We choose A = k 1 and P + q q + 1) mp + b we get f (c) = a pn+1 , where c = ci/pk+i E and m E Z. , P P u, v We prove that c ¢ Q. Let us suppose that c = — Z, (u, v) = 1. We have f (— ) = a anon
0, u, v E
nilv,
, E Z. In this case, v vn pn±l (ctriu"-Frniv) = vn (mp+ b). It follows that p divides v, i.e., there is v1 E Z so that v = pvi. Now we obtain p(anun + miv) = vli(mp + b); thus p divides v1, i.e., v1 = pv2, with v2 E Z. An elementary reasoning gives us the conclusion that the prime number p divides anvil. But p > Ian' > 0, (u, v) = 1 and ply, a contradiction. So c E (x — E, x + E) \ (2 and f (c) E Q. Therefore cE An (x e, x + E). Because x and E > 0 are arbitrary real numbers, it results co(x) = inftE A lx tl = 0, V E R. -
Problem C XI-XII.12. Compute 1
lim
n-4 co n + 1
1 n+2
1 (n
n
1 ( 1)n 2n+1
n \in
Solution. Let n be a fixed positive integer. For i E {1, 2, • • • , n}, we define (-1)k
Si— k=0
2n
1 k 1—
For i E {1,•, 71 — 1}, we have i+1
Si±i = E(-1)k •
1
i+1
- (2n — i — 1) E(-1)k
k=0 i+1
k=0
EHlyc-1
1 k 2n + k —
1 (i 1 )+ k 2n + k — i
i+1 k
—(2n — i — 1)St+ 1 + (i + 1)Si, k=1. and we find the following reccurent relations: =
1 1 2n 2n + 1
1 2n(2n + ;
i+1 2n—
i'
= 1) 2' • • •
Therefore: Sn =
2 • 3. • • n (n +1)(n + 2) • • • (2n — 1)
=
(n!)2 (2n + 1) !.
1'
191
Problems for the annual competition. Forms 11-12
Using 1
.
Sn+1
firrl n--■ oo
n+1
1
= Inn = n— 2(2n ± 3) 4' ) 00
we get (by Cauchy-d'Alembert criterion):
lim
1
1 ( n , n -I- 2
n—'eo {11 ±
( 1.)11
2n 1+ 1 ( nn,
=lim VT ?, =
4.
Problem C XI-XII.13. Consider a, b E R \ {0} and the real sequences (xn)nE N, (yn)nEN that verify the recurrence relationship Xn+2 = aXn+1+bxn, respectively yn+2 = —ayn±i + byn, (V)n E N. Prove the identity n) ak bn-k (xk k=o k (
Solution. Define zn = xn obtain
y = x2n + y2n, (V)n E N.
( -1)nyn, n E N. From the hypothesis we
zn4.2 = azn-Fi + bzn, (V)n E N. Denote by t1, t2 the complex roots of the equation t2 = at + b. It follows that there exist two complex constants a, 0, such that Zn =
a(ti)n 4-3(t2)n , (V)n E N.
We have to prove b zk = En (n) akn-k
Z2n•
k=0
It is sufficient to show, for j = 1,2, that n N"`
akbn—k(tok =
) , (V)n E N.
k=0
We use the induction. For n = 1, it is obvious. If we suppose true for n,
"Gazeta Matematica" - a bridge over three centuries
192 then we obtain: (t j)2n+2
,,,
n
= qv-, (lakbn-k(ti)k = k k=0 n
= (ate ±b)
t( i
=
k =0 n+1 k-.=0
n k bn-k (ti)k r____ E ()ak k=0
(n) a k+lbn-k(ti)k+1 + (k)ak bn-1-1-k (ti)k) k n
\\
t j) k = k/ + ( k :1)) akbrt+i-k(
n+1 (
k)ak bn+i-k (ti)k. En
=
k=0
Note. Here (-1 n = 0 and (\ n-1n 1/ = 0. Problem C XI XII.14. Let m,n E N, n > 3, m > 1, m < 71. and A E Mn (C) be a complex matrix. Prove that the rank of A is m if and only if there exist two matrices B E Mn fin (C) and C E Mm,n(C) of rank m, such that A = B • C. )
-
Solution. Let us prove that if A = BC as above, then rank (A) = m. First, notice that rank (A) = rank (BC) < rank (B) = in. It remains to prove that there is a nonzero minor in A of order m. Suppose that bit, big are m rows in B which form a nonzero minor. These rows must exist, for rank (B) = m. Suppose further that eh, cam are m columns in C that form a nonzero minor. The minor det (airjs)r,s=1,m is the product of the two nonzero minors, therefore is also nonzero. Hence rank (A) = m. Let us prove now the converse statement. We will suppose, without loss of generality, that det (ajj)ii=on 0 0. Take B with columns the first m columns in A, and C with its first m columns al, .., am, the columns of the unit matrix _rm. Denote by crn+i, c,2 the last n-m columns of C. Certainly, rank (B) = rank (C) = m. We have A = B • C if and only if (cm+k ) k=i,n_rn satisfy Acm+k = am+k. For a fixed k, this system has n equations and m variables. It is compatible if and only if rank (A) = rank (A) , where A is obtained adding to A the column arn+k. Obviously, this condition is fulfilled. ...,
Problem C XI XII.15. Let n, k be positive integers and A E Mk (R) such that An+1 =Ok. Prove that -
det
+
1
1 2
1
+ y A + ...± wAn ) =1.
Problems for the annual competition. Forms 11-12
193
1 1 —X + — Xn. It 1! n! is known the fact that if A has the eigenvalues A1, Ak E C, then P (A) has the eigenvalues P (Ai) , P (Ak) . In our context, the matrix A has the eigenvalues Al = 0, ..., Ak = 0. Indeed, a more general result asserts that if Q (A) = 0 for some polynomial Q, then any eigenvalue of A is a root of Q. It follows that P (A) has the eigenvalues P (Ai) = 1, P (4) = 1. We conclude that det P (A) = P (Ai) • .... • P (Ak) = 1. Problem C XI-XII.16. Let A be a ring such that the equality x2 +y2 +z2 Solution.
Consider the polynomial P (X) = 1
0 holds in A if and only if x = y = z = 0. Prove that the ring A is infinite. Solution. First, our goal is to prove that there exist x, y, z E 4, which are not all zero, such that x2 + y2 + z2 = 0, provided that p is prime. The polynomial P (X) = XP-1 1 has p - 1 roots in Z [X] . We decompose P as P (X) = (X 2 — 1) (X 2 +1) ; each of these two polynomials has
p
1
roots. 2 Consider now the case p 1(mod 4). There is a E 7Z such that ar e = p1 -1. Let us denote x = a 4 it results that x2 = -1. Conversely, if the
exactly
,
p-1
p1
equation x2 = -1 has a solution, then 1 = (x2) = (-1)r21 , leading to p 1(mod 4). Take y = 1 and z = 0, it follows that x2 +y2+z2= 0. The other possible case is p 3(mod 4). Consider the polynomials P1(X) = X 2 — X p and P2 (X) = X 2 + X. Each polynomial has 2 1 roots in ZP , one root p -1 is 0 and the other are the roots of the polynomials considered in the 2 previous case. Then any x E Zp*is a root of either P1 or P2. Similar to above, it follows that either x or -x is a square in Z. Suppose that -2 is a square in Zp*, i.e., there exists a E such that a2 = -2. Then x = a, y = 1, z = 1 satisfy x2 +y2 + z2 = 0. If -2 is not a square in 4, then 2 is. We continue assuming that -3 is a square in Z1,. If this is true, we have found again x, y, z such that x2 +y2 + z2 = 0, because -3 + 2 + 1 = 0. If -3 is not a square, then 3 is. Notice that p - 1 is not a square in 4, because p is not -a- 1(mod 4). In particular, our reasoning ends to x, y, z such that x2 ±y2 + z2 = 0. Let us return to the ring A, and suppose that A is finite. Then (< 1 >, +, -) is a subring of (A, +, •) isomorphic with (Z„, +, •) , were n is the characteristic of A. Consider p a prime divisor of n and a E (< 1 >, +) an element of order p. There exist k , 1, m E N, not all 0 (mod p), such that k2 + /2 + m2 r= 0
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(mod p) . Then ka, la, ma are not all zero and (ka)2 + (la)2 + (ma)2 = 0, because pa = 0. Problem C XI-XII.17. Prove that for any a E [0, oo) U {oo} there exists
a sequence of positive integers (kn)n>i such that 1 1 + — + +••+ n-4 00 V 2 3
=a.
Solution. Let a E (1, oo) be a real number. Let us consider the positive integer a = [2a11] + 1. We observe that a > 2, and hence for any positive integer k, k > a, we have ka-1 > 2. It follows that [e] > k, V k > a. We define the recurrent sequence (kn)n>1 of positive integers as k1= a, and kn±i = [kg] , n = 1,2, • . Clearly, (kn)n>i is an increasing sequence, with lim kn = 00.
n—oo
Denote xn =
V
1+
1 1
1
+ • • • +
n-)
1 1 = 1 •
-3-+ • • •
1
and
1 1 n 2 converges to the Euler's constant C E (0, 1). We have yn = ckn+ log kn and xn = VT-4. From the recurrent definition of kn+i, the inequality x — 1 < [x] < x, V x E R, and the property cn E (0,1), V it > 1, we deduce: cn = 1 + — + • • • + — — log n, n = 1,2, • , the classical sequence which
ayn— a + log (1 _ But yn > log kn and kn .
hm
Yn+1
n— oo yn
1)
oo. Therefore yn
(1)
oo. From (1), we obtain
= a. It results lim xn= lim Yri+i = a (by Cauchy-d'Alembert n•—'oo
∎
n—>co yn
criterion). For a = 1 we define kn= 1 for any positive integer n and we have xn = 1, V 71 > 1. Suppose now a = oo. We consider the sequence (kn )n>i of positive integers defined by k1 = 2 and kn+.1 = kg. In that case, Yn+1 yn Yn+I.
log knn + ck,„+1 log kn +ckn
n log kn V n > 1. n log kn + 1
We obtain lim — = oo, so that lim xn =00. n•—■ oo yn
Problems for the annual competition. Forms 11-12
195
Problem C XI XII.18. Prove that there are exactly (n + 2)m functions -
f : Mn(Ill) —4 9 ({1,2, ...,m}) such that f(X • Y) C f (X) fl f(Y) , for all X, Y E Mn(IR) . (Here P(A) denotes the family of the subsets of a set A). Solution. We have: f(X) = f (X • In) f (X) fl f (In) C f (In) , for any X E Mn (R) . If, in addition, Y E Mn (R) is invertible, then gin) = f (YY -1) C f (Y) n f (Y-1) C f (Y), thereby f (Y) = f (I„), for any invertible Y. Moreover, if A is invertible and X is given, then f (AX) C f (X) and f (X) = f (A-1AX) f (AX), that is to say f (AX) = f (X). There exist R, Ur, S, with R, S invertible and U,. a diagonal matrix having the first r entries on the diagonal equal to 1, and the rest equal to 0, such that X = RUTS. Notice that f (X) = f (Ur ), where the sequence { f (Ur )}r=0,12is ascending. The last assertion is a consequence of Ur Ur+i = Ur. In fact, such a sequence determines the function f. Indeed, let (Ai)i=0,9, be an ascending sequence of subsets of {1, .., m}. Define f by f (X) = Ar, if rank X = r. Since rank (XY) < min {rank X, rank Y} it results that f satisfies f(XY)C f (X) nf(Y),VX,Y E Mn (R) . Let us determine the number of such ascending sequences (Ai)i=01, Ai E P({1, 2, m,}). Let k0 := card (AO ; we have (km0) ways to single out Ao. For such a choice, the elements of Ai\ Aoare selected from the m — ko elements left. If k1 := card (A1) — card (A0) , k1 > 0, then we have (mk—k°) ways to select Ai \ Ao. By induction, the total number of such sequences is: ko
N :=
(m)(m—k0)
0
(m—(k0+•••+k7,1
len
It can be proved, simply using Newton's binomial, that N = (n + 2)772 . Problem C XI XII.19. Let (G,.) be a finite group with 2n elements. Find the numbers of the endomorphisms of G such that for any elements a, b E G\ {e} , there is an endomorphism f with f (a) = b. (e denotes the unit of the group G). -
Solution. The Cauchy theorem applied to G states that there exists an element a E G such that ord (a) = 2. Consider now an arbitrary element x E G\ {e} . There is f E End (G) such that f (a) = x. Then x2 = f (a)2 = f (a2) = f (e) = e, thereby we have
"Gazeta Matematice — a bridge over three centuries
196
proved that any element in G satisfies x2 =e. It is a fact that such a group is of order 2m, for m E N*. In our context, m = 1 + log2 n. Another fact is that the group G is isomorphic with (Z2 x ... x Z2 , +) , where the direct product is taken m times. Hence, without loss of generality from now on (G, .) = (V, +) , where V = Z2 x ... x Z2. This notation is good for the fact that V is a Z2— vector field of dimension m. Moreover, one can verify that End (V, +) = Mm(Z2) , the set of matrices of dimension m with entries in Z2. Now we will prove that Mm(Z2) satisfies the property in the hypothesis, namely for any v, w E V, the system Av = w has at least one solution A E Mn(Z2) . If (vi)i_ im, and (w ) j=1in are the components of v and w in the canonical basis of V, and A = (ao)ii=1,77, , then the system is Ei aovj = wi. If wi = 0, for some i, then regardless of vj we may take ao = 0, for any j. If wi = 1, since v 0, v has at least one nonzero component, say vj0 0. Then A may be defined on the row i by aik = 5. Clearly, the matrix A will satisfy Av = w. Now let us count the endomorphisms of (V, +) . A matrix A is determined by its m2entries, therefore we have 2m2endomorphisms, where m = 1 + loge n. Problem C XI XII.20. Let a < b be real numbers and let f : [a, b] —4 IR be a continuous function, differentiable on (a, b) and with f (a) = f (b). Prove that for any integer n > 1 one can find n distinct numbers ci, c2, • • • , E (a, b) such that: f (ci) + (c2) + • • • + (c) = 0. -
Solution. If f is a constant function, then the conclusion is clear. Let f be a nonconstant function. From Rolle's theorem, there is .a point c E (a, b), chosen as an extremum point of f, such that f'(c) = 0 (by Fermat's theorem). Without loss of generality, we can suppose that c is an absolute maximum point of f on [a, b] (i.e. f (x) < f (c), V x E [a, b]), such that f (a) = f (b) < f (c). We prove the next assertion: For any real numbers a and /3, with a 1) O. a— c —c Using Lagrange's theorem, on [a, c] and [c, )3] we find a' E (a, c) and f (a) c (c) /3' E (c, /3), respectively, such that (a') = and n0') = —
f(0) — f(c) .It follows f (o‘')+ f (13') = 0. If f (cr') = f (c), then f (x) LS f (a/), —c
Problems for the annual competition. Forms 11-12
197
V x E [a, b] , and we get f'(a') = 0; contradiction. So, f (a' < f (c) . Similarly f( 8') < f(c)• f (a) f (c) f (0) — f (c) 2) 0 < < a—c —c• f (x) — f (c) Consider the continuous function cp : [c, —4 R, (p(x) = x—c f (a) f (c) if x E (c, 0], and co(c) = 0. But, from the hypothesis, a—c f (3) — f (0) — (co(c),co(0)). Using the Darboux property, there is /3 —c
CO,
f(131) f (c) je(a) f (6. Hence, we are in a— c —c the situation 1), with /3 replaced by /31. 3) f(a) f (c) > f (0) — f (c) > 0 —c a— c This case can be discussed in a similar way as 2). Using the above result, we obtain the conclusion for all even integers n = 2k, after a recurrent reasoning. For odd integers 71 = 2k + 1, we add the point c to the 2k chosen points. E (c, 0) so that y,(01) =
Problem C XI XII.21. Let A1, ..., Anbe second order matrices with real entries. Prove that there is a choice Ei E { — 1, I.} of the signs + and — such that: -
det (EiAi + e2A2 + + En A„) > det Al + det A2
det An
ei
and there is another choice such that the opposite inequality holds. Solution. We use the following result: If A and B are second order real matrices then the following identity holds: det (A + B) + det (A — B) = 2 (det A + det B) . For the proof, notice that the polynomial P (X) = det (A + X B) is of second order, P (X) = ((let B) X 2+ aX + det A, where a E R. The left term in the identity is precisely P (1) + P (-1) . On the other hand, P (1) + P (-1) = det B + a + det A + det B — a + det A =the right term. Using this identity one can see that exactly one of the determinants det (A + B) and det (A — B) is greater than det A + det B, and the other is less than det A + det B. This solves the problem for n = 2. Let us extend the identity for n matrices. Notice that we have det (Ai + + An_1+ An) + det (Ai + + = 2 (det (Al + ••• + An-i) det An)
— An)
"Gazeta Matematica" — a bridge over three centuries
198 or, otherwise choose
E det (Ai + ... + An-1 + enAn) = 2 (det (A1 + .. + An—i) + det An) , en
where the sum is taken over the possible options of signs +, —. We use the above identity for n — 1 matrices to prove that
E det (A1 + ... + An-2 + En•-1An-1 + EnAn) = 2 E (det (Ai + .. + An-2 + En—iAn—i) + det An) en -1,en
en-1
= 22 (det (Ai + ... + An-2) + det A n_i + det An) •
By induction it can be proved that
E
det (A1+ E2A2 + .. + EnAn) = 2'1(det Al + ... + det An) .
627••len
Since the sum in the left term is of 71— 1 determinants, at least one choice of signs e will produce a determinant greater than det Al + ... + det An, and another choice for the opposite inequality. Problem C XI XII.22. Let f : [0,1] --4 [0,1] be a function continuous at 0 and 1 whose lateral limits at x verifies f(x — 0) < f(x) < f(x + 0), for each x E (0,1). Show that there exists xo E [0,1] such that f (x0) = xo. -
Solution. Denote A = {x E [0,1] : f (x) > x}. A is nonempty since f(0) > 0. Since A C [0, 1], there exists xo = sup A E [0,1]. We consider the following three cases: i) xo = 0. Let xn, E (0,1], x n---+ 0 (as n —+ oo). As f(xn) < xn and f is continuous at 0, it results f(0) < 0, that is f(0) = 0. ii) x0 = 1. Let xn E A, xn—> 1 (as n —+ oo). As f(xn) > xn and f is continuous at 1, it results f (1) > 1. Since f(1) < 1, we get f (1) = 1. iii) xo E (0,1). Let xn, E (x0,1], xn—4 xo (as n —÷ oo). Since f (xn) < xn, it follows f(xo + 0) < xo and hence f (x0) < xo. Assume f(xo) < xo. Then xoV A and since xo = sup A, there exists yn E A, yn - xo. By yn < xo and f (Yn) > y we deduce f (x0 — 0) > x. Therefore f (xo) > xo, a contradiction. This shows that f(x0) = 0, q.e.d.
Problems for the annual competition. Forms 11-12
199
List of authors. Problems for the annual competition. Forms 11 - 12 1. Alexe St., Simion S., C XI-XII.12 (G.M. 3/1989) 2. Andrica D., Buzeteanu
5., C XI-XII.3 (G.M. 6/1983)
3. Andronache M., C XI-XII.19 (G.M. 2/2003) 4. Batinetu-Giurgiu D. M., C XI-XII.13 (G.M. 1/1991) 5. Gramatovici R., C XI-XII.9 (G.M. 5-6/1988) 6. Ignat R. R., C XI-XII. 14 (G.M. 10/1996) 7. Isac D., C XI-XII.10 (G.M. 10/1988) 8. Lupa§ A., C XI-XII.6 (G.M. 4-5/1984) 9. Manea M., C XI-XII.21 (G.M. 4/2003) 10. Mortici C., C
C XI-XII.17 (G.M. 2/1997, G.M. 4/1998)
11. Munteanu 0., C XI-XII.18 (G.M. 12/2002) 12. Ngslasescu C., C XI-XII.1, C XI-XII.7 (G.M. 6/1987, G.M. 1/1982) 13. Panaitopol L., C XI-XII.16 (G.M. 4-5/1997) 14. Pir§an L., C XI-XII.2 (G.M. 5/1982) (C:214) 15. Piticari M., C XI-XII.22 (G.M. 3/2003) 16. Savu I., C XI-XII.11 (G.M. 3/1989) 17. Stoica Gh., C XI-XII.4 (G.M. 8/1983) 18. Tena M., C XI-XII.8 (G.M. 7-8/1987) 19. Vi§an A., C XI-XII.20 (G.M. 4/2003) 20. Vulpescu-Jalea F., C XI-XII.5 (G.M. 12/1983)
Technical editor: Marta Gae Processed: Department of Mathematics and Computer Science North University of Baia Mare Victoriei 76, 430122 Baia Mare ROMANIA
