Increase your mathematical intelligence [1 ed.] 8073683894

Table of contents :
How to use this book and about the authors
First Semester
Second Semester
Third Semester
Fourth Semester
Hints of Solutions
18
28
38
48

Citation preview

OSTRAVSKA UNIVERZITA PRIRODOVEDECKA FAKULTA

INCREASE YOUR MATHEMATICAL INTELLIGENCE Jaroslav Hanél - Pavel Rucki

Q) UNIVERSITAS OSTRAVIENSIS

Ostrava 2007

Vojtéch J arnik International Mathematical Competition The oldiest mathematical competition for university students at the European Union

Based in 1991 by the student of the student of Vojtéch Jarm’k

Organised by the students of the student of the student of Vojtéch Jarnik

Department of Mathematics

University of Ostrava http://jarnik.osu.cz/

UNIVERSITY OF OSTRAVA FACULTY OF SCIENCE

INCREASE YOUR MATHEMATICAL INTELLIGENCE Jaroslav Hanél - Pavel Rucki

Ostrava 2007

Reviewer: doc. RNDr. Josef 'Ikadlec, CSc. Department of Mathematics, Faculty of Electrical Engineering,

Czech Technical University, Praha

INCREASE YOUR MATHEMATICAL INTELLIGENCE Autors: Publisher Number of Copies: Pages: Edition: Priming:

Doc. RNDr. Jaroslav Hand], CSc., RNDr. Pavel Rucki, PhD. University of Oshawa

500 ks 85 First, 2007 Ediéni stl‘edisko Ostravské univeizity v Ostrave

0 Doc. RNDr. Jaroslav Hancl, CSc, RNDr. Pavel Rucki, PhD.., 2007 0 University of Ostrava

ISBN 978- 80-7368-389—4

Contents How to use this book and about the authors First Semester Second Semester

18

Third Semester

28

Fourth Semester

38

Hiii'tE 6f ’Sél'fitions

48

How to use this book and about the authors This book consists of forty quaternion of various mathematical problems from the difl‘erent mathematical branches. Each quaternion of these problems is suitable for the ninetieth minutes seminar. Ten quaternion of these problems is approximately one semester. It means that one group of the students can use this book for two years. The usual procedure is to put the quatemion of the problems at the end of

the seminar and ask the students to solve this problems at home without any help. Our experiences show that to put more problems for a homework caused

the students that they lost the concentrations on the problems and jump from one to the other without any serious solution. The book is created in the form that it is possible to copy one page consisting of quaternion problems for all students participating in the special seminar. It is good to emphasize that we are interested only in own student’s solutions and not in the solutions which students can found in this or other books even so the solution will not be completely right. The seminar is based on the student’s invention with the minimal knowledge and not on the finding solutions elsewhere. This forced the students to device something. The problems will be solved on the next seminar. The students will do this and if there will not be any student’s idea. how to solve the problem then the head of the seminar will show the solution or start with the hint and ask to students to finish the solution on the next seminar. The solution is good to accompanied with many notes, comments and generalizations. Every quaternion of the problems consists of the different mathematical branches. There are also some simple examples. This is very important for the not strong students in mathematics to solve a. problem from the psychological point of view. If there will be only advanced problems then the great part of the students lost the interest in this seminar. Let us also note that it is suitable to praise the student if he comes with own solution. This is also important for the psycholoy of this student and our experiences are that he will be more interested in the finding the solutions on the next seminar. The main aim of this book is to teach the students to invent something. The book contains many tricks and unusual ways of solutions. Some problems are very hard but after the solution they seemed very simple. This enable the students to look on the problem from the difi'erent point of views. If the

students solve mainly problems of this book then they extend own intelligence and own ability to find something new. The aim is not to extend any special branch of the mathematical area or to collect new mathematical problems. The quatemion of the problems are more simple at the start and more hard

at the end. Jaroslav Hanél led the above special seminar for a long time on the University of Ostrava. He also based the Vojtéch Jarnik International Mathematical Competition which is the oldest mathematical competition of this type in the European Union. The problem seminar was also a good base for training and preparing students on this competition. Pavel Rucki was the Hanél’s student. He took part in the Vojtéch Jarnik international Mathematical Competition several times, participated also in the above special problem seminars and later he helped with this competition and with the problem seminars.

First Semester 1.1. Find all integers x, y satisfying the equation :1: + y = my.

1.2. Can the discriminant of the quadratic equation c1932 + bx + c = 0 with integer coefficients a, b, c be equal to 23? 1.3. Find all real numbers :13, y satisfying the equation

c0322: +log(1+ lyl) = 0. 1.4. Find all real solutions of the following equation

1 m2+2x—3

+

18

_

18

m2+2x+2~x2+2z+1'

2.1. Let a, a1, a2, . . . , (L5 be positive integers such that a1+a2+a3+a4+a5=a.

Prove that

a!

a1! a2! a3! a4! a5! is an integer. 2.2. Let a: and y be positive integers. Find the solution of the following equation

1

1_1

x

y‘m'

2.3. Let a, b and c be positive real numbers. Prove the following inequality

1 1 1 — — -c)_ >9 m+b+® - (a+b+ 2.4. Find all integers n for which the fraction 19n+7 7n+11

is an integer.

3.1. Find all primes m, y and z satisfying the equation my + 1 = z. 3.2. Prove that for each positive integer n > 2, 1 1 1 l ¥+§+E+H'+fio+

Vsina:

13

.

7.1. Find a polynomial P(a:) with integer coefficients such that 7r P (cos E) = 0. 7.2. Find two convex functions f(cc) and g(a:) such that

f(m) — g(x) = cosx. 7.3. Let the polynomial m3 + pa: + q have three distinct real roots. Which conditions must the coefficients 1)

and q fulfill? 7.4. Find the limit

1351;0[(1+x)(1+x)(1+x) I

2

4

for [x] < 1.

14

.....

(1+9: )] 2n

8.1. Let {tn-L921 be a sequence of positive real numbers such that a1 = —2-

and

(1% g Lin—an“.

Prove that for each n E N

1 an no.

i=1 20.4. Find the sum of the series 101

5':

.

3

{L'__l_

g 1 — 3x,- + 3953

27

h

z .=__

W em a“ 101

Third Semester 21.1. Prove that for any positive integer n, there exists a number :1: E N such that :5 consists only of the digits 1 and 2 in the decadic system and a: is divisible by 2". 21.2. Let 171, 172, . . ., 61980 be 1980 vectors in the plane such that they are linearly independent each other. Assume that the sum of any distinct 1979 vectors is linearly dependent with the remaining one. Prove that the sum of all 1980 vectors equals to 6. 21.3. Find all solutions of the equation

x3 — y3 = my + 61 in the set of positive integers. 21.4. Prove that for each positive integer n, 8

Isin 1| + Isin 2| + - - - + Isin(3n —1)|+Isin3n| > 311..

28

22.1. Is it possible to cover the plane with the interiors of the finite number of parabolas? 22.2. Find the limit

.

1—1+2—1+---+n-1

hm ——————.

n—-)oo

1n n

22.3. Let n be a positive integer and let p be a prime such that p = 22" + 1. Prove that it is impossible to express the number p in this way

p=z5-y5 Where 9:, y are positive integers. 22.4. Let a be an irrational number. Let {:3} denote the fractional part of a real number a: defined as follows

{1:} := :1: — [:3].

Prove that the set A = {{na}, 71 e N} is dense on (0,1). In other words, for every interval I C (0, 1), there exists an element from the set A belonging to I.

29

23.1. Let k and n be positive integers. Find all solutions of the following equations

fi+fi+m+fi=1 (1+x1)(1+$2)---(1+xn)=2 in the set of nonnegative real numbers.

23.2. Find the limit '

23.3. Let P(:z:) be a polynomial of the degree n satisfying the equation k PUG—E3:

f0rk—0,1,...,n.

FindP(n+l). 23.4. Let 51, $2, ..., 3,, be n line segments in the plane such that any two line segments have no common point and they do not lie in a line. Is it possible to create a polygonal line containing all these line segments by connecting of their endpoints?

30

24.1. Prove that

3"+1|23"+1

and

3"+2+23"+1.

for each positive integer 11. 24.2. Let f be a continuous function on the interval [0, 21r].

Is it possible to express the function f in the following

way f(:z:) = g(a:) sina: + h(:v) 0031: where g and h are a. continuous functions on [0, 211']? 24.3. There are given 1978 sets, each containing 40 elements. Every two sets have exactly one element in common. Prove that all, 1978 sets have a common element. 24.4. Let an be real numbers defined as follows a

_

"

1

(n+1)\/H+n\/n+1

Find the sum a1 +a2 + - - - +a99.

31

forlSn_ 0. Write f(z) = h(a:) + sins: and 9(3) = h(.1:) and find the function h(z) so that the above inequalities are valid. 4.2 Use the following identities

a3+b3=(a.+b)(a’—ab+,b2),

for a,b€R

aa — b3 = (a — b)(a2 + ab + b“). 4.3 Prove that

1 + a.- 2 2W for each positive integer 1' = 1, 2, ..., n using the arithmetic-geometric mean inequality. 4.4 Multiply the numerator and the denominator by

Vic + — x/E. 5.1 Multiply the expression by sin(:c/2”) sin(:z/2") and use a formula for sin 22:. 5.2 Use the identity 1

1

l

150:: + 1)(lc + 2) = 2m: + 1) “ 20: + 1)(lc + 2) ' 5.3 The polynomial f(1) can be written in the form n

f(1:) = 201-2"

where a.- e Z.

i=0

49

Then f(15) — f (7) = 4. Use the identity

11." - v" = (u — v)(u.""1 + u""u + - - - + 1411"" + 11"“) to show that the polynomial f(1;) with the required properties cannot exist. 5.4

.211.” sin’ftn/n2 + n) = "IE“ (sin(1rVn’ + n — 1m + 1m))2. REF

nel"D

Then use the identity

sin(1: + y) = sinxcosy +cosz + siny. 6.1

ik3+6k2+llk+5_ 2k3+6k7+11k+6_ i

m

(k + 3)!

(k + 3)!

1 (k+3)'

_ 21— _ i _1_ .. E ‘ H k!

H (H3)! ' 3'

6.2 Express the function f(z) in the following way

f=J—.1=%(£Trfim)6.3 Multiply the numerator and the denominator by (3x)‘ and use the limits sin: lim—=

M

and

a:

lim

e' — 1

2—50

3

=1.

Then use the Taylor expansions of the functions sin a; and e‘

.

a.“

4

.

smx=z-§!-+:(z),

ifs—+0and

2:3

e‘=1+1£!z—+m+3—!+%‘+o($),

ifs—)0

where 0(3‘) is defined as follows

f0?) = 0(z‘) =

$40

6.4

m—m_ m

—.

(l—e—=)z zsina;

’(1-cosz)si.nz _

sinzz

'

Use the same limits as in the previous example. 7.1 Express the function cos 4:1: as a polynomial of 005 z using identifies

cos 2:1: = cos’z -— sinaz, sin2a: = 2sinxcosz,

1 =sinzz+coszz. 7.2 Let f(z) and g(x) be convex functions. Then f”(z) > 0 and g”(:z) > 0. Write f (2:) h(::) + cosz, g(z) = h(z) and find the function 11(3) so that the above inequalitiw are valid. 7.3 Prove that the polynomial p(z) = za+p¢+q must have two extremes, thus the equation p’(a:) = 0 has two distinct solutions 2:; and z; with 2:1 > .22. Rom this we obtain the condition for the coefiicient p. Then prove that

11(21) < 0 and 12(12) > 0. Using these inequalities find the next conditions for the coefficients p and q. 7.4 Multiply the limit by the fraction 1-2: 1—2:

and use the identity (a" —- b")(a’I + b") = a3" — b’“. 8.1 Use the mathematical induction. Let the inequality be valid for k S 1:, thus

a 0 : g’(:co) = 0. 15.4 Let a: and y be positive real numbers such that y e (0, l) and 1: e [0, y]. The geometric series 2:; 3"” is uniformly convergent on the interval [0, y],

flit-n) a“;[M n=1

Putting y = % and using this identity, find the sum of the series. 16.1 Modify the expression

(3 + w/E)" + (3 — V5)" andusethefactthat3-x/5 P(2) = 2a 19(2) = %(P(3) + 11(1)) = P(3) = 3a. Finally, prove that P(z)

-

Using the mathematical induction prove that P(n) = no for every integer 1:. a2: is the zero polynomial.

17.1 Let A" be a set with n elements and h(A,.) be the number of different couples of disjunctive subsets of A... Let 2: g! A" and An“ = A" U {z}. The set of all difl'erent couples of disjunctive subsets of A...” contains all difl‘erent couples of disjunctive subsets of A“. The number of these couples is h(A,.). We obtain the next couples by adding of the element a: into either the first or the second subset of any couple of disjunctive subsets of A... The number of these couples is 2h(A,.). However, if the element :5 is added into the couple

{93,0} in this way then the same couples {1, 0} and {0,x} arise. This implies that

h(An+1) = MA") + 2h(An) - 1 = 3h(An) - 1Using this recurrence equation find h(A,.+1).

59

17.2 Prove that

2W5+2~V522-2-‘gfiu’d’i22.2?"E using the arithmetic-geometric mean inequality.

17.3 Prove that

23 f(:z:) = 2 arcgz+arcsxn1+zz t ' —

is a constant function and put :1: = 1. 17.4 Prove that

1) mwmgm and 2)Ifi+m2[m. ad 1) It is suficient to prove that \fii + V71 + 1 3 \/1n + 2. ad 2) Suppose the contrary, [J71+\/n + 1| < [V411 + 2]. It implies that there eadsts a natural number a such that

W + «m < a. s m. Horn this and the fact that a and n are natural numbers, we obtain that

(a2 — 2n- 1)’-4n’—4n= 1. This is equivalent to

a2(a2 — 4n — 2) = 0. Prove that a aé 0 and a2 96 4n + 2, if n is a positive integer. 18.1 Write n! as a product 5‘ r where r is not divisible by 5. Prove that the number of nulls occurring at the end of n! equals to s.

18.2 Let 100! = fl"-=1 pfi‘, where {Pi}?=r is an increasing sequence of primes, be the prime decox'nposition of 100!. Each factor of 100! can be written as follows n

24100! => 2:11p?

forOSti.

i=1

The sum of all factors of the number 100! equals to n

I]

J2

lg

n

n

phi.) _ 1

Z HP?=ZZ“'ZHP?‘=H'—:T-

315:5; i=1

k1=0 k2=0

kn=0 i=1

60

5:]

p‘

18.3 Use pigeon hole theorem. 18.4

LetX = {a}; i: 1,2,...,4000} beasetof4000 points. Let Lbeasetofall lines passing through the arbitrary two points from X. Let p0 be a line not parallel to any

line from L that divides a plane into two half planes such that all 4000 points are situated

exactly in one of them. Construct a line p1 parallel to the line pa such that there are even 4 points in the strip between the lines 110 and

, ”a

p1 .

p]. In the same manner construct a line 1;;

p 2

such that there are even 4 points from the remaining 3996 once in the strip between the lines 111 and 11;. Repeat this procedure for lines pa, 1)., .. . , mm. 19.1 Let f(3,34) be a. real function. Use the following identity

6 d5 m, b)- m. a) — /.b 5mg) and tini's Theorem. 19.2 Prove that z...” = $n$n_1 . . .21(a:1 — 1) + 1. Use the fact that 2_=Z$n_1.. 11(11—1)+1—1=

(11—1).a:1

“Ix”

.,.$

=2l(11;, — 1):c1.. .zn_1

2 (2:1 — 1)1zl..

to find the sum of the series. 19.3 Let P(:c) be a polynomial, thus P(:z) = aux" + a,._1x"‘1 + ‘ - - + an: + no. Compare the coefficients of the polynomials P(a:2) and (P(:r:))’.

19.4 Let {MEL-1 be a sequence of integers such that b;:=a.-a.-+1

fori=1,2,...,n—1and

bu := anal.

61

Then 23:1 bi = 0. Using this prove that 2 | n. Next prove that H}; b,- = 1 and this impliw that 4 | n. 20.1 Express the polynomial as a product of two polynomials in this way

kHz—as: (23—9)(.1:+4) =1»? where p is a prime. There are six possibilities how to write 1:” as a product of two integers. Thus,

2:0—9=d:l 2:1:—9=:l:p2

and and

an+4=ip2 $+4=:|:1

22-9=:l:p

and

z+4=:|:p.

or or

Find the solution of the equations. 20.2 Use Cauchy’s Theorem for the functions

I

a:

20.3 Use Stolz’s Theorem for the sequences {zn};'|"=, and {31”}?=1 defined as follows 2,, = i 0.13:. i=1

lln = ¢lvn+13fl+1 foremhn=1, 2,

20.4 Let {123.42}, be a. sequence of positive rational numbers such that a;

-_$?___.3?__ —1—3z.~+3$?—(1—z.')3+22

for each i = o, 1, ..., 101. Prove. that a.- + am-.- = 1. Then 50

S = 20154-01014) = 5].. (:0

21.1 Prove this using the mathematical induction. Let X.- e {1, 2} for i = 1, 2, ..., k andlet A = X1X2 . . . X;

62

be a positive integer expressed in the decadic system where X; are digits. Let A = 2*3 for s e N. Prove that 2"+1 | 1X1...X2, if s is odd and 2k+1|2X1.. .Xk, ifs is even. 21.2 Let 1‘13, 172,

, 1719“ be vectors. Then there exist two constants 1:1 and

In; such that

61+---+171979=k1‘l7mo 172+---+171m=k2171 Thisimpliesthath:

9 fil+"'+171980=(k1+1)fi193° => 171+---+1719w=(k2+1)61980-

=—1and171+172~-+51930=6-

21.3 It is obvious that z > y. Put a substitution 1: = y + k where k is a positive integer and prove that k3 < 61. This inequality is satisfied only by the values k = 1, k = 2, k = 3. Set each value into the substitution 1: = y+k and find all solutions of the equation.

21.4 Let

m) = Isinz|+ Isin(z+1)|+ lsin(z+2)l-

The function f(z) is periodic with the period 1r. Find the global minimum

of f(a:) and prove that f(a:) > 8/5. 22.1

Construct a line p not parallel to the axis of any parabola. Specify all positions between the line p and a parabola that can arise. Using the fact that there are only the finite number of parabolas we can easy get the answer. 22.2 Use Stolz’s Theorem.

22.3 Suppose the contrary that there exist two positive integers 2: and y such that

p = z“ — y‘ = (a; — y)(1:‘ + may + zzyz +:I:y3 +y‘). Prove that :c — y = 1. Then

p=z5—(z—l)’=5.1:‘—101:3+10:1:2—5z+1 => 2“" =51' where 1' is a positive integer. This is the contradiction. 63

22.4 Let I be an interval such that I C (0,1) and the length of interval I is e. Let a be an irrational number. Prove that

{ka} = {la} e» k=l. This implies that the set A is infinite. Then there exist two positive integers n; and n: such that n; > M and

{ma} — {ma} < a. Prove that {11.10:} — {ma} = {(111 - 1:2)a} and show that there exist at least one positive integer s such that {3(n1 - 11.2) a} e I. 23.1 It is easy to find one solution 3i=11 $1='”=zi—l=$l'+l="'=3n=o

for i = 1, 2, . . . , n. Prove that if there exist other solutions then the necessary

condition is z; e [0, 1) for 1 S 2' S n. From the second equation we obtain that

2: (l+$1)(1+$2)...(1+1n)=1+(£1+$2+"'+1n)

+ z Ii1j+"'+$[12...zn 2 1+(xf+1:',‘+~-+zf.) igiqsn +

Z

1513 = 2+

19‘a

2

353%.

15i 0. It means that there do not exist any other solutions. 23.2Wehave

2:12-32...(2n—1)’= 22-42...(2n)2 _l_-§3_5 (2n—1)(2n+1)_

‘22

42'”

(2n)2

Henceitfollows liman=0. 11—500

64

1

2n+1




a=e.

, 2;; be complex roots of the equation 2" = 1. Find the

sum using the following two equalities, 2 2n— e“+e’+-~+e -2: fi+fi+m+4 m N

n=0

i n $1" i=1

=

0, 1:,

k f 71 k: | n.

26.2 Split the unit square into 25 squares with side length g Using pigeon hole theorem we prove that there exists at least one square that contains at

66

least three points of the given 50 ones.

26.3 Use Stolz’s lemma. 26.4 Use Jensen’s inequality.

27.1 Let S be the center of the unit circle and let A1, A2, ..., A, be n vertices of a polygon. Split the polygon into n triangles by connecting the vertices A; (i = 1, 2, n) with the center S. Let 0:. denote the angle AAgSAHl for each [C = 1, 2, ..., n - 1 and an = AAHSAl. The area of the

triangle AAkAkHS equals to l

,

1 ,

AAAsAufiS = EIAkSHAkHSI 51110:: = 5 man where IAgSI resp. [AHIS] is the distance between the points A, and S resp. A.“ and S. The area of the polygon equals to the sum of the area of all n triangles, thus 1|

7|

i=1

k=l

1 . AMI = E :AAAhAI-+1S = 5 2 :ma‘k-

The function sin a: is a concave function on the interval [0,1r]. Using Jensen’s inequality we obtain that Sina1+~~+an

>

sin a1 + - - - + sin an n .

n Ftom this and the fact that 22:10:). =21r we obtain that 1 n

n

27r

— ' 2. 31.3 Without low of generality suppose that a 2 b 2 c. Then “Zn—(He) bib—(n+2) ck- (n+b) ___ an—caa—bbb—abb—ccc-acc—b =

=(%)“‘(%)“"(2)”2172

Multiplying both sides by ambflcf‘l‘”, we get the required inequality. 31.4 It is obvious that b,. _>_ 4 for n = 1, 2, ...Prove by induction that a,l > bfhl. Let n = 3 then 113 = 3“ > 4‘ = (21. Suppose that an > b2_, for n=3,4, ..., k. Then

«n+1 = 3°~ > 3"i—x > 2°14 > 24““ = (40“)2 = bfi. 32.1 Let 2:, y and 2 be three roots of the polynomial

P(u) = 1.3 — 9n“ + 271; — 27. Using Vieta’s formulas, we obtain that

zy+zz+yz=27. This is exactly the system of equations which we solve now. Using the binomial theorem, we can find out that

P(u) = (u. — 3)3. Thisimpliwthatz=y=z=3. 32.2 Multiply the inequality by n! we obtain that

1.. n

1

> 1 +

n(n+l)...(n+k)‘n+1

1

(n+1)(n+2)

+...

'“+(n+1)...(n+k)

$1 n(n+1)...(n+k) 1

—

1

n+1

11

—

1

—.a-

(n+1)(n+2) 1

(n+1)...(n+k)' Denote the right side of the inequality by the symbol S. Then using thefact that

1 1 n(n+1)...(n+m—1)_(n+1)(n+2)...(n+m) m

=11(n+1)...('n+m) 2 n(n+1)...(n+m) 73

1

5:1n

n+1

_

1

(n+1)(n+2)

1

_...

(n+1)...(n+k)

1

1

=n'(n+1)‘(n+1)(n+2) """ (n+1)...(n+k) .

1

(n+1)...(n+k)

1‘ 1 Z 'n(n+1)(n+2)(n+3) — (n+1)(n+2)(n+3)(n+4) 1

”‘(n+1')...‘(n+'k) Z"'3_n(n+i)...'(n"+' k)' 32.3 Let a: = (J5+‘/§)%. Then 1/: = (fi— fi)*.1fz+1/z is a rational number then

1

1

2

x2+—2= (1+—) —2

m z is also rational. Using the mathematical induction, we obtain that z" + 1/2:" is rational. On the other side

1

a: n + _ z“

=

2V5 .

Therefore a: + 1/2; is irrational. 32.4 Let Izl < 1. We express the function f(:5) as a power series. f(:v)

=

2a:

—--—-————-—=]_

1

1+:1:+::;2

222—21:

_—

+ 1—23

=1+(2z’—2z)(1+z3+a:°+...)= =1—2z+222-2z‘+2z5—2z7+2m‘-.... Thus no

f(2)=1+2a.z'

where a,=

5:1

Use this formula to find f1°°(0). 74

—2, 2,

ifs=3k~2 ifs=3k-—l

0

ifs=3kfork€N.

,

33.1 For fixed m, the following inequality is valid. '

nmamlan

—

211m

'