Tuyển tập Olympic Toán sinh viên quốc tế 1994 - 2013

Phudinhgioihan Diendantoanhoc.net
TUYỂN TẬP ĐỀ THI OLYMPIC
TOÁN SINH VIÊN QUỐC TẾ
International Mathematics Compe tition for
University Students
1994-2013
Mục lục
IMC 1994 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
IMC 1995 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
IMC 1996 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
IMC 1997 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
IMC 1997 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
IMC 1998 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
IMC 1998 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
IMC 1999 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
IMC 1999 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
IMC 2000 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
IMC 2000 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
IMC 2001 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
IMC 2001 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
IMC 2002 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
IMC 2002 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
IMC 2003 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
IMC 2003 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
IMC 2004 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
IMC 2004 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
IMC 2005 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
IMC 2005 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
IMC 2006 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
IMC 2006 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
IMC 2007 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

IMC 2007 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
IMC 2008 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
IMC 2008 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
IMC 2009 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
IMC 2009 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
IMC 2010 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
IMC 2010 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
IMC 2011 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
IMC 2011 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
IMC 2012 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
IMC 2012 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
IMC 2013 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
IMC 2013 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
1
International Competition in Mathematics for
Universtiy Students
in
Plovdiv, Bulgaria
1994
1
1
PROBLEMS AND SOLUTIONS
First day — July 29, 1994
Problem 1. (13 points)
a) Let A be a n × n, n ≥ 2, symmetric, invertible matrix with real
positive elements. Show that z
n
≤ n
2
− 2n, where z

n
is the number of zero
elements in A
−1
.
b) How many zero elements are there in the inverse of the n × n matrix
A =









1 1 1 1 . . . 1
1 2 2 2 . . . 2
1 2 1 1 . . . 1
1 2 1 2 . . . 2
. . . . . . . . . . . . . . . . . . . .
1 2 1 2 . . . . . .










?
Solution. Denote by a
ij
and b
ij
the elements of A and A
−1
, respectively.
Then for k = m we have
n

i=0
a
ki
b
im
= 0 and from the positivity of a
ij
we
conclude that at least one of {b
im
: i = 1, 2, . . . , n} is positive and at least
one is negative. Hence we have at least two non-zero elements in every
column of A
−1
. This proves part a). For part b) all b
ij
are zero except
b

1,1
= 2, b
n,n
= (−1)
n
, b
i,i+1
= b
i+1,i
= (−1)
i
for i = 1, 2, . . . , n − 1.
Problem 2. (13 points)
Let f ∈ C
1
(a, b), lim
x→a+
f(x) = +∞, lim
x→b−
f(x) = −∞ and
f

(x) + f
2
(x) ≥ −1 for x ∈ (a, b). Prove that b − a ≥ π and give an example
where b − a = π.
Solution. From the inequality we get
d
dx
(arctg f (x) + x) =

f

(x)
1 + f
2
(x)
+ 1 ≥ 0
for x ∈ (a, b). Thus arctg f(x)+x is non-decreasing in the interval and using
the limits we get
π
2
+ a ≤ −
π
2
+ b. Hence b − a ≥ π. One has equality for
f(x) = cotg x, a = 0, b = π.
Problem 3. (13 points)
2
2
Given a set S of 2n − 1, n ∈ N, different irrational numbers. Prove
that there are n different elements x
1
, x
2
, . . . , x
n
∈ S such that for all non-
negative rational numbers a
1
, a

2
, . . . , a
n
with a
1
+ a
2
+ · · · + a
n
> 0 we have
that a
1
x
1
+ a
2
x
2
+ · · · + a
n
x
n
is an irrational number.
Solution. Let I be the set of irrational numbers, Q – the set of rational
numbers, Q
+
= Q ∩[0, ∞). We work by induction. For n = 1 the statement
is trivial. Let it be true for n − 1. We start to prove it for n. From the
induction argument there are n − 1 different elements x
1

, x
2
, . . . , x
n−1
∈ S
such that
(1)
a
1
x
1
+ a
2
x
2
+ · · · + a
n−1
x
n−1
∈ I
for all a
1
, a
2
, . . . , a
n
∈ Q
+
with a
1

+ a
2
+ · · · + a
n−1
> 0.
Denote the other elements of S by x
n
, x
n+1
, . . . , x
2n−1
. Assume the state-
ment is not true for n. Then for k = 0, 1, . . . , n − 1 there are r
k
∈ Q such
that
(2)
n−1

i=1
b
ik
x
i
+ c
k
x
n+k
= r
k

for some b
ik
, c
k
∈ Q
+
,
n−1

i=1
b
ik
+ c
k
> 0.
Also
(3)
n−1

k=0
d
k
x
n+k
= R for some d
k
∈ Q
+
,
n−1


k=0
d
k
> 0, R ∈ Q.
If in (2) c
k
= 0 then (2) contradicts (1). Thus c
k
= 0 and without loss of
generality one may take c
k
= 1. In (2) also
n−1

i=1
b
ik
> 0 in view of x
n+k
∈ I.
Replacing (2) in (3) we get
n−1

k=0
d
k

−
n−1


i=1
b
ik
x
i
+ r
k

= R or
n−1

i=1

n−1

k=0
d
k
b
ik

x
i
∈ Q,
which contradicts (1) because of the conditions on b

s and d

s.

Problem 4. (18 points)
Let α ∈ R \ {0} and suppose that F and G are linear maps (operators)
from R
n
into R
n
satisfying F ◦ G − G ◦ F = αF .
a) Show that for all k ∈ N one has F
k
◦ G − G ◦ F
k
= αkF
k
.
b) Show that there exists k ≥ 1 such that F
k
= 0.
3
3
Solution. For a) using the assumptions we have
F
k
◦ G − G ◦ F
k
=
k

i=1

F

k−i+1
◦ G ◦ F
i−1
− F
k−i
◦ G ◦ F
i

=
=
k

i=1
F
k−i
◦ (F ◦ G − G ◦ F ) ◦ F
i−1
=
=
k

i=1
F
k−i
◦ αF ◦ F
i−1
= αkF
k
.
b) Consider the linear operator L(F ) = F ◦G−G◦F acting over all n×n

matrices F . It may have at most n
2
different eigenvalues. Assuming that
F
k
= 0 for every k we get that L has infinitely many different eigenvalues
αk in view of a) – a contradiction.
Problem 5. (18 points)
a) Let f ∈ C[0, b], g ∈ C(R) and let g be periodic with period b. Prove
that

b
0
f(x)g(nx)dx has a limit as n → ∞ and
lim
n→∞

b
0
f(x)g(nx)dx =
1
b

b
0
f(x)dx ·

b
0
g(x)dx.

b) Find
lim
n→∞

π
0
sin x
1 + 3cos
2
nx
dx.
Solution. Set g
1
=

b
0
|g(x)|dx and
ω(f, t) = sup {|f(x) − f(y)| : x, y ∈ [0, b], |x − y| ≤ t} .
In view of the uniform continuity of f we have ω(f, t) → 0 as t → 0. Using
the periodicity of g we get

b
0
f(x)g(nx)dx =
n

k=1

bk/n

b(k−1)/n
f(x)g(nx)dx
=
n

k=1
f(bk/n)

bk/n
b(k−1)/n
g(nx)dx +
n

k=1

bk/n
b(k−1)/n
{f(x) − f(bk/n)}g(nx)dx
=
1
n
n

k=1
f(bk/n)

b
0
g(x)dx + O(ω(f, b/n)g
1

)
4
4
=
1
b
n

k=1

bk/n
b(k−1)/n
f(x)dx

b
0
g(x)dx
+
1
b
n

k=1

b
n
f(bk/n) −

bk/n
b(k−1)/n

f(x)dx


b
0
g(x)dx + O(ω(f, b/n)g
1
)
=
1
b

b
0
f(x)dx

b
0
g(x)dx + O(ω(f, b/n)g
1
).
This proves a). For b) we set b = π, f(x) = sin x, g(x) = (1 + 3cos
2
x)
−1
.
From a) and

π
0

sin xdx = 2,

π
0
(1 + 3cos
2
x)
−1
dx =
π
2
we get
lim
n→∞

π
0
sin x
1 + 3cos
2
nx
dx = 1.
Problem 6. (25 points)
Let f ∈ C
2
[0, N] and |f

(x)| < 1, f

(x) > 0 for every x ∈ [0, N]. Let

0 ≤ m
0
< m
1
< · · · < m
k
≤ N be integers such that n
i
= f(m
i
) are also
integers for i = 0, 1, . . . , k. Denote b
i
= n
i
− n
i−1
and a
i
= m
i
− m
i−1
for
i = 1, 2, . . . , k.
a) Prove that
−1 <
b
1
a

1
<
b
2
a
2
< · · · <
b
k
a
k
< 1.
b) Prove that for every choice of A > 1 there are no more than N/A
indices j such that a
j
> A.
c) Prove that k ≤ 3N
2/3
(i.e. there are no more than 3N
2/3
integer
points on the curve y = f(x), x ∈ [0, N]).
Solution. a) For i = 1, 2, . . . , k we have
b
i
= f (m
i
) − f(m
i−1
) = (m

i
− m
i−1
)f

(x
i
)
for some x
i
∈ (m
i−1
, m
i
). Hence
b
i
a
i
= f

(x
i
) and so −1 <
b
i
a
i
< 1. From the
convexity of f we have that f


is increasing and
b
i
a
i
= f

(x
i
) < f

(x
i+1
) =
b
i+1
a
i+1
because of x
i
< m
i
< x
i+1
.
5
5
b) Set S
A

= {j ∈ {0, 1, . . . , k} : a
j
> A}. Then
N ≥ m
k
− m
0
=
k

i=1
a
i
≥

j∈S
A
a
j
> A|S
A
|
and hence |S
A
| < N/A.
c) All different fractions in (−1, 1) with denominators less or equal A are
no more 2A
2
. Using b) we get k < N/A + 2A
2

. Put A = N
1/3
in the above
estimate and get k < 3N
2/3
.
Second day — July 30, 1994
Problem 1. (14 points)
Let f ∈ C
1
[a, b], f (a) = 0 and suppose that λ ∈ R, λ > 0, is such that
|f

(x)| ≤ λ|f(x)|
for all x ∈ [a, b]. Is it true that f(x) = 0 for all x ∈ [a, b]?
Solution. Assume that there is y ∈ (a, b] such that f (y) = 0. Without
loss of generality we have f(y) > 0. In view of the continuity of f there exists
c ∈ [a, y) such that f(c) = 0 and f(x) > 0 for x ∈ (c, y]. For x ∈ (c, y] we
have |f

(x)| ≤ λf(x). This implies that the function g(x) = ln f(x) − λx is
not increasing in (c, y] because of g

(x) =
f

(x)
f(x)
−λ ≤ 0. Thus ln f(x)−λx ≥
ln f(y) − λy and f(x) ≥ e

λx−λy
f(y) for x ∈ (c, y]. Thus
0 = f(c) = f(c + 0) ≥ e
λc−λy
f(y) > 0
— a contradiction. Hence one has f(x) = 0 for all x ∈ [a, b].
Problem 2. (14 points)
Let f : R
2
→ R be given by f(x, y) = (x
2
− y
2
)e
−x
2
−y
2
.
a) Prove that f attains its minimum and its maximum.
b) Determine all points (x, y) such that
∂f
∂x
(x, y) =
∂f
∂y
(x, y) = 0 and
determine for which of them f has global or local minimum or maximum.
Solution. We have f(1, 0) = e
−1

, f(0, 1) = −e
−1
and te
−t
≤ 2e
−2
for
t ≥ 2. Therefore |f(x, y)| ≤ (x
2
+ y
2
)e
−x
2
−y
2
≤ 2e
−2
< e
−1
for (x, y) /∈
M = {(u, v) : u
2
+ v
2
≤ 2} and f cannot attain its minimum and its
6
6
maximum outside M. Part a) follows from the compactness of M and the
continuity of f. Let (x, y) be a point from part b). From

∂f
∂x
(x, y) =
2x(1 − x
2
+ y
2
)e
−x
2
−y
2
we get
(1) x(1 − x
2
+ y
2
) = 0.
Similarly
(2) y(1 + x
2
− y
2
) = 0.
All solutions (x, y) of the system (1), (2) are (0, 0), (0, 1), (0, −1), (1, 0)
and (−1, 0). One has f (1, 0) = f(−1, 0) = e
−1
and f has global maximum
at the points (1, 0) and (−1, 0). One has f(0, 1) = f (0, −1) = −e
−1

and
f has global minimum at the points (0, 1) and (0, −1). The point (0, 0)
is not an extrema point because of f(x, 0) = x
2
e
−x
2
> 0 if x = 0 and
f(y, 0) = −y
2
e
−y
2
< 0 if y = 0.
Problem 3. (14 points)
Let f be a real-valued function with n + 1 derivatives at each point of
R. Show that for each pair of real numbers a, b, a < b, such that
ln

f(b) + f

(b) + · · · + f
(n)
(b)
f(a) + f

(a) + · · · + f
(n)
(a)


= b − a
there is a number c in the open interval (a, b) for which
f
(n+1)
(c) = f(c).
Note that ln denotes the natural logarithm.
Solution. Set g(x) =

f(x) + f

(x) + · · · + f
(n)
(x)

e
−x
. From the
assumption one get g(a) = g(b). Then there exists c ∈ (a, b) such that
g

(c) = 0. Replacing in the last equality g

(x) =

f
(n+1)
(x) − f(x)

e
−x

we
finish the proof.
Problem 4. (18 points)
Let A be a n × n diagonal matrix with characteristic polynomial
(x − c
1
)
d
1
(x − c
2
)
d
2
. . . (x − c
k
)
d
k
,
where c
1
, c
2
, . . . , c
k
are distinct (which means that c
1
appears d
1

times on the
diagonal, c
2
appears d
2
times on the diagonal, etc. and d
1
+d
2
+· · ·+d
k
= n).
7
7
Let V be the space of all n × n matrices B such that AB = BA. Prove that
the dimension of V is
d
2
1
+ d
2
2
+ · · · + d
2
k
.
Solution. Set A = (a
ij
)
n

i,j=1
, B = (b
ij
)
n
i,j=1
, AB = (x
ij
)
n
i,j=1
and
BA = (y
ij
)
n
i,j=1
. Then x
ij
= a
ii
b
ij
and y
ij
= a
jj
b
ij
. Thus AB = BA is

equivalent to (a
ii
− a
jj
)b
ij
= 0 for i, j = 1, 2, . . . , n. Therefore b
ij
= 0 if
a
ii
= a
jj
and b
ij
may be arbitrary if a
ii
= a
jj
. The number of indices (i, j)
for which a
ii
= a
jj
= c
m
for some m = 1, 2, . . . , k is d
2
m
. This gives the

desired result.
Problem 5. (18 points)
Let x
1
, x
2
, . . . , x
k
be vectors of m-dimensional Euclidian space, such that
x
1
+x
2
+· · ·+x
k
= 0. Show that there exists a permutation π of the integers
{1, 2, . . . , k} such that





n

i=1
x
π(i)






≤

k

i=1
x
i

2

1/2
for each n = 1, 2, . . . , k.
Note that  ·  denotes the Euclidian norm.
Solution. We define π inductively. Set π(1) = 1. Assume π is defined
for i = 1, 2, . . . , n and also
(1)





n

i=1
x
π(i)






2
≤
n

i=1
x
π(i)

2
.
Note (1) is true for n = 1. We choose π(n + 1) in a way that (1) is fulfilled
with n + 1 instead of n. Set y =
n

i=1
x
π(i)
and A = {1, 2, . . . , k} \ {π(i) : i =
1, 2, . . . , n}. Assume that (y, x
r
) > 0 for all r ∈ A. Then

y,

r∈A
x
r


> 0
and in view of y +

r∈A
x
r
= 0 one gets −(y, y) > 0, which is impossible.
Therefore there is r ∈ A such that
(2) (y, x
r
) ≤ 0.
Put π(n + 1) = r. Then using (2) and (1) we have





n+1

i=1
x
π(i)





2
= y + x

r

2
= y
2
+ 2(y, x
r
) + x
r

2
≤ y
2
+ x
r

2
≤
8
8
≤
n

i=1
x
π(i)

2
+ x
r


2
=
n+1

i=1
x
π(i)

2
,
which verifies (1) for n + 1. Thus we define π for every n = 1, 2, . . . , k.
Finally from (1) we get





n

i=1
x
π(i)





2
≤

n

i=1
x
π(i)

2
≤
k

i=1
x
i

2
.
Problem 6. (22 points)
Find lim
N→∞
ln
2
N
N
N−2

k=2
1
ln k · ln(N − k)
. Note that ln denotes the natural
logarithm.

Solution. Obviously
(1) A
N
=
ln
2
N
N
N−2

k=2
1
ln k · ln(N − k)
≥
ln
2
N
N
·
N − 3
ln
2
N
= 1 −
3
N
.
Take M, 2 ≤ M < N/2. Then using that
1
ln k · ln(N − k)

is decreasing in
[2, N/2] and the symmetry with respect to N/2 one get
A
N
=
ln
2
N
N



M

k=2
+
N−M −1

k=M +1
+
N−2

k=N −M



1
ln k · ln(N − k)
≤
≤

ln
2
N
N

2
M − 1
ln 2 · ln(N − 2)
+
N − 2M − 1
ln M · ln(N − M)

≤
≤
2
ln 2
·
M ln N
N
+

1 −
2M
N

ln N
ln M
+ O

1

ln N

.
Choose M =

N
ln
2
N

+ 1 to get
(2) A
N
≤

1 −
2
N ln
2
N

ln N
ln N − 2 ln ln N
+O

1
ln N

≤ 1+O


ln ln N
ln N

.
Estimates (1) and (2) give
lim
N→∞
ln
2
N
N
N−2

k=2
1
ln k · ln(N − k)
= 1.
9
International Competition in Mathematics for
Universtiy Students
in
Plovdiv, Bulgaria
1995
10
1
PROBLEMS AND SOLUTIONS
First day
Problem 1. (10 points)
Let X be a nonsingular matrix with columns X
1

, X
2
, . . . , X
n
. Let Y be a
matrix with columns X
2
, X
3
, . . . , X
n
, 0. Show that the matrices A = Y X
−1
and B = X
−1
Y have rank n − 1 and have only 0’s for eigenvalues.
Solution. Let J = (a
ij
) be the n × n matrix where a
ij
= 1 if i = j + 1
and a
ij
= 0 otherwise. The rank of J is n − 1 and its only eigenvalues are
0

s. Moreover Y = XJ and A = Y X
−1
= XJX
−1

, B = X
−1
Y = J. It
follows that both A and B have rank n −1 with only 0

s for eigenvalues.
Problem 2. (15 points)
Let f be a continuous function on [0, 1] such that for every x ∈ [0, 1] we
have

1
x
f(t)dt ≥
1 − x
2
2
. Show that

1
0
f
2
(t)dt ≥
1
3
.
Solution. From the inequality
0 ≤

1

0
(f(x) − x)
2
dx =

1
0
f
2
(x)dx − 2

1
0
xf(x)dx +

1
0
x
2
dx
we get

1
0
f
2
(x)dx ≥ 2

1
0

xf(x)dx −

1
0
x
2
dx = 2

1
0
xf(x)dx −
1
3
.
From the hypotheses we have

1
0

1
x
f(t)dtdx ≥

1
0
1 − x
2
2
dx or


1
0
tf(t)dt ≥
1
3
. This completes the proof.
Problem 3. (15 points)
Let f be twice continuously differentiable on (0, +∞) such that
lim
x→0+
f

(x) = −∞ and lim
x→0+
f

(x) = +∞. Show that
lim
x→0+
f(x)
f

(x)
= 0.
11
2
Solution. Since f

tends to −∞ and f


tends to +∞ as x tends to
0+, there exists an interval (0, r) such that f

(x) < 0 and f

(x) > 0 for all
x ∈ (0, r). Hence f is decreasing and f

is increasing on (0, r). By the mean
value theorem for every 0 < x < x
0
< r we obtain
f(x) − f (x
0
) = f

(ξ)(x − x
0
) > 0,
for some ξ ∈ (x, x
0
). Taking into account that f

is increasing, f

(x) <
f

(ξ) < 0, we get
x − x

0
<
f

(ξ)
f

(x)
(x − x
0
) =
f(x) − f(x
0
)
f

(x)
< 0.
Taking limits as x tends to 0+ we obtain
−x
0
≤ lim inf
x→0+
f(x)
f

(x)
≤ lim sup
x→0+
f(x)

f

(x)
≤ 0.
Since this happens for all x
0
∈ (0, r) we deduce that lim
x→0+
f(x)
f

(x)
exists and
lim
x→0+
f(x)
f

(x)
= 0.
Problem 4. (15 points)
Let F : (1, ∞) → R be the function defined by
F (x) :=

x
2
x
dt
ln t
.

Show that F is one-to-one (i.e. injective) and find the range (i.e. set of
values) of F .
Solution. From the definition we have
F

(x) =
x − 1
ln x
, x > 1.
Therefore F

(x) > 0 for x ∈ (1, ∞). Thus F is strictly increasing and hence
one-to-one. Since
F (x) ≥ (x
2
− x) min

1
ln t
: x ≤ t ≤ x
2

=
x
2
− x
ln x
2
→ ∞
12

3
as x → ∞, it follows that the range of F is (F (1+), ∞). In order to determine
F (1+) we substitute t = e
v
in the definition of F and we get
F (x) =

2 ln x
ln x
e
v
v
dv.
Hence
F (x) < e
2 ln x

2 ln x
ln x
1
v
dv = x
2
ln 2
and similarly F (x) > x ln 2. Thus F (1+) = ln 2.
Problem 5. (20 points)
Let A and B be real n × n matrices. Assume that there exist n + 1
different real numbers t
1
, t

2
, . . . , t
n+1
such that the matrices
C
i
= A + t
i
B, i = 1, 2, . . . , n + 1,
are nilpotent (i.e. C
n
i
= 0).
Show that both A and B are nilpotent.
Solution. We have that
(A + tB)
n
= A
n
+ tP
1
+ t
2
P
2
+ ··· + t
n−1
P
n−1
+ t

n
B
n
for some matrices P
1
, P
2
, . . . , P
n−1
not depending on t.
Assume that a, p
1
, p
2
, . . . , p
n−1
, b are the (i, j)-th entries of the corre-
sponding matrices A
n
, P
1
, P
2
, . . . , P
n−1
, B
n
. Then the polynomial
bt
n

+ p
n−1
t
n−1
+ ··· + p
2
t
2
+ p
1
t + a
has at least n + 1 roots t
1
, t
2
, . . . , t
n+1
. Hence all its coefficients vanish.
Therefore A
n
= 0, B
n
= 0, P
i
= 0; and A and B are nilpotent.
Problem 6. (25 points)
Let p > 1. Show that there exists a constant K
p
> 0 such that for every
x, y ∈ R satisfying |x|

p
+ |y|
p
= 2, we have
(x − y)
2
≤ K
p

4 − (x + y)
2

.
13
4
Solution. Let 0 < δ < 1. First we show that there exists K
p,δ
> 0 such
that
f(x, y) =
(x − y)
2
4 − (x + y)
2
≤ K
p,δ
for every (x, y) ∈ D
δ
= {(x, y) : |x −y| ≥ δ, |x|
p

+ |y|
p
= 2}.
Since D
δ
is compact it is enough to show that f is continuous on D
δ
.
For this we show that the denominator of f is different from zero. Assume
the contrary. Then |x + y| = 2, and




x + y
2




p
= 1. Since p > 1, the function
g(t) = |t|
p
is strictly convex, in other words




x + y

2




p
<
|x|
p
+ |y|
p
2
whenever
x = y. So for some (x, y) ∈ D
δ
we have




x + y
2




p
<
|x|
p

+ |y|
p
2
= 1 =




x + y
2




p
. We get a contradiction.
If x and y have different signs then (x, y) ∈ D
δ
for all 0 < δ < 1 because
then |x−y| ≥ max{|x|, |y|} ≥ 1 > δ. So we may further assume without loss
of generality that x > 0, y > 0 and x
p
+ y
p
= 2. Set x = 1 + t. Then
y = (2 − x
p
)
1/p
= (2 − (1 + t)

p
)
1/p
=

2 − (1 + pt +
p(p−1)
2
t
2
+ o(t
2
))

1/p
=

1 − pt −
p(p − 1)
2
t
2
+ o(t
2
)

1/p
= 1 +
1
p


−pt −
p(p − 1)
2
t
2
+ o(t
2
)

+
1
2p

1
p
− 1

(−pt + o(t))
2
+ o(t
2
)
= 1 −t −
p − 1
2
t
2
+ o(t
2

) −
p −1
2
t
2
+ o(t
2
)
= 1 −t −(p −1)t
2
+ o(t
2
).
We have
(x − y)
2
= (2t + o(t))
2
= 4t
2
+ o(t
2
)
and
4−(x+y)
2
=4−(2−(p−1)t
2
+o(t
2

))
2
=4−4+4(p−1)t
2
+o(t
2
)=4(p−1)t
2
+o(t
2
).
So there exists δ
p
> 0 such that if |t| < δ
p
we have (x−y)
2
< 5t
2
, 4−(x+y)
2
>
3(p − 1)t
2
. Then
(∗) (x − y)
2
< 5t
2
=

5
3(p − 1)
· 3(p − 1)t
2
<
5
3(p − 1)
(4 − (x + y)
2
)
14
5
if |x − 1| < δ
p
. From the symmetry we have that (∗) also holds when
|y − 1| < δ
p
.
To finish the proof it is enough to show that |x − y| ≥ 2δ
p
whenever
|x −1| ≥ δ
p
, |y −1| ≥ δ
p
and x
p
+ y
p
= 2. Indeed, since x

p
+ y
p
= 2 we have
that max{x, y} ≥ 1. So let x − 1 ≥ δ
p
. Since

x + y
2

p
≤
x
p
+ y
p
2
= 1 we
get x + y ≤ 2. Then x − y ≥ 2(x − 1) ≥ 2δ
p
.
Second day
Problem 1. (10 points)
Let A be 3×3 real matrix such that the vectors Au and u are orthogonal
for each column vector u ∈ R
3
. Prove that:
a) A


= −A, where A

denotes the transpose of the matrix A;
b) there exists a vector v ∈ R
3
such that Au = v × u for every u ∈ R
3
,
where v × u denotes the vector product in R
3
.
Solution. a) Set A = (a
ij
), u = (u
1
, u
2
, u
3
)

. If we use the orthogonal-
ity condition
(1) (Au, u) = 0
with u
i
= δ
ik
we get a
kk

= 0. If we use (1) with u
i
= δ
ik
+ δ
im
we get
a
kk
+ a
km
+ a
mk
+ a
mm
= 0
and hence a
km
= −a
mk
.
b) Set v
1
= −a
23
, v
2
= a
13
, v

3
= −a
12
. Then
Au = (v
2
u
3
− v
3
u
2
, v
3
u
1
− v
1
u
3
, v
1
u
2
− v
2
u
1
)


= v ×u.
Problem 2. (15 points)
Let {b
n
}
∞
n=0
be a sequence of positive real numbers such that b
0
= 1,
b
n
= 2 +

b
n−1
− 2

1 +

b
n−1
. Calculate
∞

n=1
b
n
2
n

.
15
6
Solution. Put a
n
= 1 +
√
b
n
for n ≥ 0. Then a
n
> 1, a
0
= 2 and
a
n
= 1 +

1 + a
n−1
−2
√
a
n−1
=
√
a
n−1
,
so a

n
= 2
2
−n
. Then
N

n=1
b
n
2
n
=
N

n=1
(a
n
− 1)
2
2
n
=
N

n=1
[a
2
n
2

n
− a
n
2
n+1
+ 2
n
]
=
N

n=1
[(a
n−1
− 1)2
n
− (a
n
− 1)2
n+1
]
= (a
0
− 1)2
1
− (a
N
− 1)2
N+1
= 2 − 2

2
2
−N
− 1
2
−N
.
Put x = 2
−N
. Then x → 0 as N → ∞ and so
∞

n=1
b
n
2
N
= lim
N→∞

2 − 2
2
2
−N
− 1
2
−N

= lim
x→0


2 − 2
2
x
−1
x

= 2 − 2 ln 2.
Problem 3. (15 points)
Let all roots of an n-th degree polynomial P (z) with complex coefficients
lie on the unit circle in the complex plane. Prove that all roots of the
polynomial
2zP

(z) − nP (z)
lie on the same circle.
Solution. It is enough to consider only polynomials with leading coef-
ficient 1. Let P (z) = (z − α
1
)(z −α
2
) . . . (z − α
n
) with |α
j
| = 1, where the
complex numbers α
1
, α
2

, . . . , α
n
may coincide.
We have

P (z) ≡ 2zP

(z) − nP (z) = (z + α
1
)(z − α
2
) . . . (z −α
n
) +
+(z − α
1
)(z + α
2
) . . . (z −α
n
) + ··· + (z −α
1
)(z − α
2
) . . . (z + α
n
).
Hence,

P (z)

P (z)
=
n

k=1
z + α
k
z − α
k
. Since Re
z + α
z − α
=
|z|
2
−|α|
2
|z − α|
2
for all complex z,
α, z = α, we deduce that in our case Re

P (z)
P (z)
=
n

k=1
|z|
2

− 1
|z − α
k
|
2
. From |z| = 1
it follows that Re

P (z)
P (z)
= 0. Hence

P (z) = 0 implies |z| = 1.
16
7
Problem 4. (15 points)
a) Prove that for every ε > 0 there is a positive integer n and real
numbers λ
1
, . . . , λ
n
such that
max
x∈[−1,1]





x −

n

k=1
λ
k
x
2k+1





< ε.
b) Prove that for every odd continuous function f on [−1, 1] and for every
ε > 0 there is a positive integer n and real numbers µ
1
, . . . , µ
n
such that
max
x∈[−1,1]





f(x) −
n

k=1

µ
k
x
2k+1





< ε.
Recall that f is odd means that f (x) = −f (−x) for all x ∈ [−1, 1].
Solution. a) Let n be such that (1 − ε
2
)
n
≤ ε. Then |x(1 − x
2
)
n
| < ε
for every x ∈ [−1, 1]. Thus one can set λ
k
= (−1)
k+1

n
k

because then
x −

n

k=1
λ
k
x
2k+1
=
n

k=0
(−1)
k

n
k

x
2k+1
= x(1 − x
2
)
n
.
b) From the Weierstrass theorem there is a polynomial, say p ∈ Π
m
, such
that
max
x∈[−1,1]

|f(x) − p(x)| <
ε
2
.
Set q(x) =
1
2
{p(x) − p(−x)}. Then
f(x) − q(x) =
1
2
{f(x) − p(x)} −
1
2
{f(−x) − p(−x)}
and
(1) max
|x|≤1
|f(x) −q(x)| ≤
1
2
max
|x|≤1
|f(x) −p(x)|+
1
2
max
|x|≤1
|f(−x) −p(−x)| <
ε

2
.
But q is an odd polynomial in Π
m
and it can be written as
q(x) =
m

k=0
b
k
x
2k+1
= b
0
x +
m

k=1
b
k
x
2k+1
.
17
8
If b
0
= 0 then (1) proves b). If b
0

= 0 then one applies a) with
ε
2|b
0
|
instead
of ε to get
(2) max
|x|≤1





b
0
x −
n

k=1
b
0
λ
k
x
2k+1






<
ε
2
for appropriate n and λ
1
, λ
2
, . . . , λ
n
. Now b) follows from (1) and (2) with
max{n, m} instead of n.
Problem 5. (10+15 points)
a) Prove that every function of the form
f(x) =
a
0
2
+ cos x +
N

n=2
a
n
cos (nx)
with |a
0
| < 1, has positive as well as negative values in the period [0, 2π).
b) Prove that the function
F (x) =

100

n=1
cos (n
3
2
x)
has at least 40 zeros in the interval (0, 1000).
Solution. a) Let us consider the integral

2π
0
f(x)(1 ± cos x)dx = π(a
0
± 1).
The assumption that f (x) ≥ 0 implies a
0
≥ 1. Similarly, if f (x) ≤ 0 then
a
0
≤ −1. In both cases we have a contradiction with the hypothesis of the
problem.
b) We shall prove that for each integer N and for each real number h ≥ 24
and each real number y the function
F
N
(x) =
N

n=1

cos (xn
3
2
)
changes sign in the interval (y, y + h). The assertion will follow immediately
from here.
18
9
Consider the integrals
I
1
=

y+h
y
F
N
(x)dx, I
2
=

y+h
y
F
N
(x)cos x dx.
If F
N
(x) does not change sign in (y, y + h) then we have
|I

2
| ≤

y+h
y
|F
N
(x)|dx =






y+h
y
F
N
(x)dx





= |I
1
|.
Hence, it is enough to prove that
|I
2

| > |I
1
|.
Obviously, for each α = 0 we have






y+h
y
cos (αx)dx





≤
2
|α|
.
Hence
(1) |I
1
| =






N

n=1

y+h
y
cos (xn
3
2
)dx





≤ 2
N

n=1
1
n
3
2
< 2

1 +

∞
1

dt
t
3
2

= 6.
On the other hand we have
I
2
=
N

n=1

y+h
y
cos xcos (xn
3
2
)dx
=
1
2

y+h
y
(1 + cos (2x))dx +
+
1
2

N

n=2

y+h
y

cos

x(n
3
2
− 1)

+ cos

x(n
3
2
+ 1)

dx
=
1
2
h + ∆,
where
|∆| ≤
1
2


1 + 2
N

n=2

1
n
3
2
− 1
+
1
n
3
2
+ 1

≤
1
2
+ 2
N

n=2
1
n
3
2
− 1

.
19
10
We use that n
3
2
− 1 ≥
2
3
n
3
2
for n ≥ 3 and we get
|∆| ≤
1
2
+
2
2
3
2
− 1
+ 3
N

n=3
1
n
3
2

<
1
2
+
2
2
√
2 − 1
+ 3

∞
2
dt
t
3
2
< 6.
Hence
(2) |I
2
| >
1
2
h − 6.
We use that h ≥ 24 and inequalities (1), (2) and we obtain |I
2
| > |I
1
|. The
proof is completed.

Problem 6. (20 points)
Suppose that {f
n
}
∞
n=1
is a sequence of continuous functions on the inter-
val [0, 1] such that

1
0
f
m
(x)f
n
(x)dx =

1 if n = m
0 if n = m
and
sup{|f
n
(x)| : x ∈ [0, 1] and n = 1, 2, . . .} < +∞.
Show that there exists no subsequence {f
n
k
} of {f
n
} such that lim
k→∞

f
n
k
(x)
exists for all x ∈ [0, 1].
Solution. It is clear that one can add some functions, say {g
m
}, which
satisfy the hypothesis of the problem and the closure of the finite linear
combinations of {f
n
}∪{g
m
} is L
2
[0, 1]. Therefore without loss of generality
we assume that {f
n
} generates L
2
[0, 1].
Let us suppose that there is a subsequence {n
k
} and a function f such
that
f
n
k
(x) −→
k→∞

f(x) for every x ∈ [0, 1].
Fix m ∈ N. From Lebesgue’s theorem we have
0 =

1
0
f
m
(x)f
n
k
(x)dx −→
k→∞

1
0
f
m
(x)f(x)dx.
Hence

1
0
f
m
(x)f(x)dx = 0 for every m ∈ N, which implies f(x) = 0 almost
everywhere. Using once more Lebesgue’s theorem we get
1 =

1

0
f
2
n
k
(x)dx −→
k→∞

1
0
f
2
(x)dx = 0.
The contradiction proves the statement.
20
International Competition in Mathematics for
Universtiy Students
in
Plovdiv, Bulgaria
1996
21
1
PROBLEMS AND SOLUTIONS
First day — August 2, 1996
Problem 1. (10 points)
Let for j = 0, . . . , n, a
j
= a
0
+ jd, where a

0
, d are fixed real numbers.
Put
A =







a
0
a
1
a
2
. . . a
n
a
1
a
0
a
1
. . . a
n−1
a
2
a

1
a
0
. . . a
n−2
. . . . . . . . . . . . . . . . . . . . . . . . . . .
a
n
a
n−1
a
n−2
. . . a
0







.
Calculate det(A), where det(A) denotes the determinant of A.
Solution. Adding the first column of A to the last column we get that
det(A) = (a
0
+ a
n
) det








a
0
a
1
a
2
. . . 1
a
1
a
0
a
1
. . . 1
a
2
a
1
a
0
. . . 1
. . . . . . . . . . . . . . . . . . . . . . .
a
n

a
n−1
a
n−2
. . . 1







.
Subtracting the n-th row of the above matrix from the (n+1)-st one, (n−1)-
st from n-th, . . . , first from second we obtain that
det(A) = (a
0
+ a
n
) det







a
0
a

1
a
2
. . . 1
d −d −d . . . 0
d d −d . . . 0
. . . . . . . . . . . . . . . . . . . .
d d d . . . 0







.
Hence,
det(A) = (−1)
n
(a
0
+ a
n
) det








d −d −d . . . −d
d d −d . . . −d
d d d . . . −d
. . . . . . . . . . . . . . . . . . . .
d d d . . . d







.
22
2
Adding the last row of the above matrix to the other rows we have
det(A) = (−1)
n
(a
0
+a
n
) det








2d 0 0 . . . 0
2d 2d 0 . . . 0
2d 2d 2d . . . 0
. . . . . . . . . . . . . . . . . . .
d d d . . . d







= (−1)
n
(a
0
+a
n
)2
n−1
d
n
.
Problem 2. (10 points)
Evaluate the definite integral

π
−π
sin nx

(1 + 2
x
)sin x
dx,
where n is a natural number.
Solution. We have
I
n
=

π
−π
sin nx
(1 + 2
x
)sin x
dx
=

π
0
sin nx
(1 + 2
x
)sin x
dx +

0
−π
sin nx

(1 + 2
x
)sin x
dx.
In the second integral we make the change of variable x = −x and obtain
I
n
=

π
0
sin nx
(1 + 2
x
)sin x
dx +

π
0
sin nx
(1 + 2
−x
)sin x
dx
=

π
0
(1 + 2
x

)sin nx
(1 + 2
x
)sin x
dx
=

π
0
sin nx
sin x
dx.
For n ≥ 2 we have
I
n
− I
n−2
=

π
0
sin nx − sin (n − 2)x
sin x
dx
= 2

π
0
cos (n − 1)xdx = 0.
The answer

I
n
=

0 if n is even,
π if n is odd
23
3
follows from the above formula and I
0
= 0, I
1
= π.
Problem 3. (15 points)
The linear operator A on the vector space V is called an involution if
A
2
= E where E is the identity operator on V . Let dim V = n < ∞.
(i) Prove that for every involution A on V there exists a basis of V
consisting of eigenvectors of A.
(ii) Find the maximal number of distinct pairwise commuting involutions
on V .
Solution.
(i) Let B =
1
2
(A + E). Then
B
2
=

1
4
(A
2
+ 2AE + E) =
1
4
(2AE + 2E) =
1
2
(A + E) = B.
Hence B is a projection. Thus there exists a basis of eigenvectors for B, and
the matrix of B in this basis is of the form diag(1, . . . , 1, 0, . . . , 0).
Since A = 2B − E the eigenvalues of A are ±1 only.
(ii) Let {A
i
: i ∈ I} be a set of commuting diagonalizable operators
on V , and let A
1
be one of these operators. Choose an eigenvalue λ of A
1
and denote V
λ
= {v ∈ V : A
1
v = λv}. Then V
λ
is a subspace of V , and
since A
1

A
i
= A
i
A
1
for each i ∈ I we obtain that V
λ
is invariant under each
A
i
. If V
λ
= V then A
1
is either E or −E, and we can start with another
operator A
i
. If V
λ
= V we proceed by induction on dim V in order to find
a common eigenvector for all A
i
. Therefore {A
i
: i ∈ I} are simultaneously
diagonalizable.
If they are involutions then |I| ≤ 2
n
since the diagonal entries may equal

1 or -1 only.
Problem 4. (15 points)
Let a
1
= 1, a
n
=
1
n
n−1

k=1
a
k
a
n−k
for n ≥ 2. Show that
(i) lim sup
n→∞
|a
n
|
1/n
< 2
−1/2
;
(ii) lim sup
n→∞
|a
n

|
1/n
≥ 2/3.
Solution.
(i) We show by induction that
(∗) a
n
≤ q
n
for n ≥ 3,
24