Tài liệu Đề thi Olympic sinh viên thế giới năm 2003 pdf

10
th
International Mathematical Competition for University Students
Cluj-Napoca, July 2003
Day 1
1. (a) Let a
1
, a
2
, . . . be a sequence of real numbers such that a
1
= 1 and a
n+1
>
3
2
a
n
for all n.
Prove that the sequence
a
n

3
2

n−1
has a finite limit or tends to infinity. (10 points)
(b) Prove that for all α > 1 there exists a sequence a
1
, a

2
, . . . with the same properties such
that
lim
a
n

3
2

n−1
= α.
(10 points)
Solution. (a) Let b
n
=
a
n

3
2

n−1
. Then a
n+1
>
3
2
a
n

is equivalent to b
n+1
> b
n
, thus the sequence
(b
n
) is strictly increasing. Each increasing sequence has a finite limit or tends to infinity.
(b) For all α > 1 there exists a sequence 1 = b
1
< b
2
< . . . which converges to α. Choos ing
a
n
=

3
2

n−1
b
n
, we obtain the required sequence (a
n
).
2. Let a
1
, a
2

. . . , a
51
be non-zero elements of a field. We simultaneously replace each element with
the sum of the 50 remaining ones. In this way we get a sequence b
1
. . . , b
51
. If this new sequence is
a permutation of the original one, what can be the characteristic of the field? (The characteristic
of a field is p, if p is the smallest positive integer such that x + x + . . . + x

 
p
= 0 for any element x
of the field. If there exists no such p, the characteristic is 0.) (20 points)
Solution. Let S = a
1
+ a
2
+ · · · + a
51
. Then b
1
+ b
2
+ · · · + b
51
= 50S. Since b
1
, b

2
, · · · , b
51
is a
permutation of a
1
, a
2
, · · · , a
51
, we get 50S = S, so 49S = 0. Assume that the characteristic of the
field is not equal to 7. Then 49S = 0 implies that S = 0. Therefore b
i
= −a
i
for i = 1, 2, · · · , 51.
On the other hand, b
i
= a
ϕ(i)
, where ϕ ∈ S
51
. Therefore, if the characteristic is not 2, the sequence
a
1
, a
2
, · · · , a
51
can be partitioned into pairs {a

i
, a
ϕ(i)
} of additive inverses. But this is impossible,
since 51 is an odd number. It follows that the characteristic of the field is 7 or 2.
The characteristic can be either 2 or 7. For the case of 7, x
1
= . . . = x
51
= 1 is a possible
choice. For the case of 2, any elements can be chosen such that S = 0, since then b
i
= −a
i
= a
i
.
3. Let A be an n × n real matrix such that 3A
3
= A
2
+ A + I (I is the identity matrix). Show
that the sequence A
k
converges to an idempotent matrix. (A matrix B is called idempotent if
B
2
= B.) (20 points)
Solution. The minimal polynomial of A is a divisor of 3x
3

− x
2
− x −1. This polynomial has three
different roots. This implies that A is diagonalizable: A = C
−1
DC where D is a diagonal matrix.
The eigenvalues of the matrices A and D are all roots of polynomial 3x
3
− x
2
− x − 1. One of the
three roots is 1, the remaining two roots have smaller absolute value than 1. Hence, the diagonal
elements of D
k
, which are the kth powers of the eigenvalues, tend to either 0 or 1 and the limit
M = lim D
k
is idempotent. Then lim A
k
= C
−1
MC is idempotent as well.
4. Determine the set of all pairs (a, b) of positive integers for which the set of positive integers
can be decomposed into two sets A and B such that a · A = b · B. (20 points)
Solution. Clearly a and b must be different since A and B are disjoint.
1
10
th
International Mathematical Competition for University Students
Cluj-Napoca, July 2003

Day 2
1. Let A and B be n × n real matrices such that AB + A + B = 0. Prove that AB = BA.
Solution. Since (A + I)(B + I) = AB + A + B + I = I (I is the identity matrix), matrices
A + I and B + I are inverses of each other. Then (A + I)(B + I) = (B + I)(A + I) and
AB + BA.
2. Evaluate the limit
lim
x→0+

2x
x
sin
m
t
t
n
dt (m, n ∈ N).
Solution. We use the fact that
sin t
t
is decreasing in the interval (0, π) and lim
t→0+0
sin t
t
= 1.
For all x ∈ (0,
π
2
) and t ∈ [x, 2x] we have
sin 2x

2
x <
sin t
t
< 1, thus

sin 2x
2x

m

2x
x
t
m
t
n
<

2x
x
sin
m
t
t
n
dt <

2x
x

t
m
t
n
dt,

2x
x
t
m
t
n
dt = x
m−n+1

2
1
u
m−n
du.
The factor

sin 2x
2x

m
tends to 1. If m − n + 1 < 0, the limit of x
m−n+1
is infinity; if
m − n + 1 > 0 then 0. If m − n + 1 = 0 then x

m−n+1

2
1
u
m−n
du = ln 2. Hence,
lim
x→0+0
2x

x
sin
m
t
t
n
dt =





0, m ≥ n
ln 2, n − m = 1
+∞, n − m > 1.
.
3. Let A be a closed subset of R
n
and let B b e the set of all those points b ∈ R

n
for which
there exists exactly one point a
0
∈ A such that
|a
0
− b| = inf
a∈A
|a − b|.
Prove that B is dense in R
n
; that is, the closure of B is R
n
.
Solution. Let b
0
/∈ A (otherwise b
0
∈ A ⊂ B),  = inf
a∈A
|a−b
0
|. The intersection of the ball
of radius  + 1 with centre b
0
with set A is compact and there exists a
0
∈ A: |a
0

− b
0
| = .
1
Denote by B
r
(a) = {x ∈ R
n
: |x − a| ≤ r} and ∂B
r
(a) = {x ∈ R
n
: |x − a| = r} the
ball and the sphere of center a and radius r, respectively.
If a
0
is not the unique nearest point then for any point a on the open line segment (a
0
, b
0
)
we have B
|a−a
0
|
(a) ⊂ B

(b
0
) and ∂B

|a−a
0
|
(a)

∂B

(b
0
) = {a
0
}, therefore (a
0
, b
0
) ⊂ B and
b
0
is an accumulation point of set B.
4. Find all positive integers n for which there exists a family F of three-element subsets
of S = {1, 2, . . . , n} satisfying the following two conditions:
(i) for any two different eleme nts a, b ∈ S, there exists exactly one A ∈ F containing
both a, b;
(ii) if a, b, c, x, y, z are elements of S such that if {a, b, x}, {a, c, y}, {b, c, z} ∈ F, then
{x, y, z} ∈ F.
Solution. The condition (i) of the problem allows us to define a (well-defined) operation
∗ on the set S given by
a ∗ b = c if and only if {a, b, c} ∈ F, where a = b.
We note that this operation is still not defined completely (we need to define a ∗ a), but
nevertheless let us investigate its features. At first, due to (i), for a = b the operation

obviously satisfies the following three conditions:
(a) a = a ∗ b = b;
(b) a ∗ b = b ∗ a;
(c) a ∗ (a ∗ b) = b.
What does the condition (ii) give? It claims that
(e’) x ∗ (a ∗ c) = x ∗ y = z = b ∗ c = (x ∗ a) ∗ c
for any three different x, a, c, i.e. that the operation is asso c iative if the arguments are
different. Now we can complete the definition of ∗. In order to save associativity for non-
different arguments, i.e. to make b = a ∗ (a ∗ b) = (a ∗ a) ∗ b hold, we will add to S an extra
element, call it 0, and define
(d) a ∗ a = 0 and a ∗ 0 = 0 ∗ a = a.
Now it is easy to check that, for any a, b, c ∈ S ∪ {0}, (a),(b),(c) and (d), still hold, and
(e) a ∗ b ∗ c := (a ∗ b) ∗ c = a ∗ (b ∗ c).
We have thus obtained that (S ∪ {0}, ∗) has the structure of a finite Abelian group,
whose elements are all of order two. Since the order of every such group is a power of 2,
we conclude that |S ∪ {0}| = n + 1 = 2
m
and n = 2
m
− 1 for some integer m ≥ 1.
Given n = 2
m
−1, according to what we have proven till now, we will construct a f amily
of three-element subsets of S satisfying (i) and (ii). Let us define the operation ∗ in the
following manner:
if a = a
0
+ 2a
1
+ . . . + 2

m−1
a
m−1
and b = b
0
+ 2b
1
+ . . . + 2
m−1
b
m−1
, where a
i
, b
i
are either 0 or 1, we put a ∗ b = |a
0
− b
0
| + 2|a
1
− b
1
| + . . . + 2
m−1
|a
m−1
− b
m−1
|.

2
It is simple to check that this ∗ satisfies (a),(b),(c) and (e’). Therefore, if we include in
F all possible triples a, b, a ∗ b, the condition (i) follows from (a),(b) and (c), whereas the
condition (ii) follows from (e’)
The answer is: n = 2
m
− 1.
5. (a) Show that for each function f : Q × Q → R there exists a function g : Q → R such
that f(x, y) ≤ g(x) + g(y) for all x, y ∈ Q.
(b) Find a function f : R × R → R for which there is no function g : R → R such that
f(x, y) ≤ g(x) + g(y) for all x, y ∈ R.
Solution. a) Let ϕ : Q → N be a bijection. Define g(x) = max{|f(s, t)| : s, t ∈ Q, ϕ(s) ≤
ϕ(x), ϕ(t) ≤ ϕ(x)}. We have f(x, y) ≤ max{g(x), g(y)} ≤ g(x) + g (y).
b) We shall show that the function defined by f(x, y) =
1
|x−y|
for x = y and f(x, x) = 0
satisfies the problem. If, by contradiction there exists a function g as above, it results, that
g(y) ≥
1
|x−y|
− f(x) for x, y ∈ R, x = y; one obtains that for each x ∈ R, lim
y→x
g(y) = ∞.
We show, that there exists no function g having an infinite limit at each point of a bounded
and closed interval [a, b].
For each k ∈ N
+
denote A
k

= {x ∈ [a, b] : |g(x)| ≤ k}.
We have obviously [a, b] = ∪
∞
k=1
A
k
. The set [a, b] is uncountable, so at least one of the
sets A
k
is infinite (in fact uncountable). This set A
k
being infinite, there exists a sequence
in A
k
having distinct terms. This sequence will contain a convergent subsequence (x
n
)
n∈N
convergent to a point x ∈ [a, b]. But lim
y→x
g(y) = ∞ implies that g(x
n
) → ∞, a contradiction
because |g(x
n
)| ≤ k, ∀n ∈ N.
Second solution for part (b). Let S be the set of all sequences of real numbers. The
cardinality of S is |S| = |R|
ℵ
0

= 2
ℵ
2
0
= 2
ℵ
0
= |R|. Thus, there exists a bijection h : R → S.
Now define the function f in the following way. For any real x and positive integer n,
let f(x, n) be the nth element of sequence h(x). If y is not a positive integer then let
f(x, y) = 0. We prove that this function has the required property.
Let g be an arbitrary R → R function. We show that there exist real numbers x, y
such that f(x, y) > g (x) + g(y). Consider the sequence (n + g(n))
∞
n=1
. This sequence is an
element of S, thus (n + g(n))
∞
n=1
= h(x) for a certain real x. Then for an arbitrary positive
integer n, f(x, n) is the nth element, f(x, n) = n + g(n). Choosing n such that n > g(x),
we obtain f(x, n) = n + g(n) > g(x) + g(n).
6. Let (a
n
)
n∈N
be the sequence define d by
a
0
= 1, a

n+1
=
1
n + 1
n

k=0
a
k
n − k + 2
.
Find the limit
lim
n→∞
n

k=0
a
k
2
k
,
3
if it exists.
Solution. Consider the generating function f(x) =

∞
n=0
a
n

x
n
. By induction 0 < a
n
≤ 1,
thus this series is absolutely convergent for |x| < 1, f(0) = 1 and the function is positive
in the interval [0, 1). The goal is to compute f(
1
2
).
By the recurrence formula,
f

(x) =
∞

n=0
(n + 1)a
n+1
x
n
=
∞

n=0
n

k=0
a
k

n − k + 2
x
n
=
=
∞

k=0
a
k
x
k
∞

n=k
x
n−k
n − k + 2
= f (x)
∞

m=0
x
m
m + 2
.
Then
ln f(x) = ln f(x) − ln f(0) =

x

0
f

f
=
∞

m=0
x
m+1
(m + 1)(m + 2)
=
=
∞

m=0

x
m+1
(m + 1)
−
x
m+1
(m + 2)

= 1 +

1 −
1
x


∞

m=0
x
m+1
(m + 1)
= 1 +

1 −
1
x

ln
1
1 − x
,
ln f

1
2

= 1 − ln 2,
and thus f(
1
2
) =
e
2
.

4
Let {a, b} be a solution and consider the sets A, B such that a · A = b · B. Denoting d = (a, b)
the greatest common divisor of a and b, we have a = d·a
1
, b = d·b
1
, (a
1
, b
1
) = 1 and a
1
·A = b
1
·B.
Thus {a
1
, b
1
} is a solution and it is enough to determine the solutions {a, b} with (a, b) = 1.
If 1 ∈ A then a ∈ a · A = b · B, thus b must be a divisor of a. Similarly, if 1 ∈ B, then a is a
divisor of b. Therefore, in all solutions, one of numbers a,b is a divisor of the other one.
Now we prove that if n ≥ 2, then (1, n) is a solution. For each positive integer k, let f(k)
be the largest non-negative integer for which n
f(k)
|k. Then let A = {k : f(k) is odd} and
B = {k : f (k) is even}. This is a decomposition of all positive integers such that A = n · B.
5. Let g : [0, 1] → R be a continuous function and let f
n
: [0, 1] → R be a sequence of functions

defined by f
0
(x) = g(x) and
f
n+1
(x) =
1
x

x
0
f
n
(t)dt (x ∈ (0, 1], n = 0, 1, 2, . . .).
Determine lim
n→∞
f
n
(x) for every x ∈ (0, 1]. (20 points)
B. We shall prove in two different ways that lim
n→∞
f
n
(x) = g(0) for every x ∈ (0, 1]. (The
second one is more lengthy but it tells us how to calculate f
n
directly from g.)
Proof I. First we prove our claim for non-decreasing g. In this case, by induction, one can
easily see that
1. each f

n
is non-decrasing as well, and
2. g(x) = f
0
(x) ≥ f
1
(x) ≥ f
2
(x) ≥ . . . ≥ g(0) (x ∈ (0, 1]).
Then (2) implies that there exists
h(x) = lim
n→∞
f
n
(x) (x ∈ (0, 1]).
Clearly h is non-decreasing and g(0) ≤ h(x) ≤ f
n
(x) for any x ∈ (0, 1], n = 0, 1, 2, . . Therefore
to show that h(x) = g(0) for any x ∈ (0, 1], it is enough to prove that h(1) cannot be greater than
g(0).
Supp ose that h(1) > g(0). Then there exists a 0 < δ < 1 such that h(1) > g(δ). Using the
definition, (2) and (1) we get
f
n+1
(1) =

1
0
f
n

(t)dt ≤

δ
0
g(t)dt +

1
δ
f
n
(t)dt ≤ δg(δ) + (1 − δ)f
n
(1).
Hence
f
n
(1) − f
n+1
(1) ≥ δ(f
n
(1) − g(δ)) ≥ δ(h(1) − g(δ)) > 0,
so f
n
(1) → −∞, which is a contradiction.
Similarly, we can prove our claim for non-increasing continuous functions as well.
Now suppose that g is an arbitrary continuous function on [0, 1]. Let
M(x) = sup
t∈[0,x]
g(t), m(x) = inf
t∈[0,x]

g(t) (x ∈ [0, 1])
Then on [0, 1] m is non-increasing, M is non-decreasing, both are continuous, m(x) ≤ g(x) ≤ M(x)
and M(0) = m(0) = g(0). Define the sequences of functions M
n
(x) and m
n
(x) in the same way
as f
n
is defined but starting with M
0
= M and m
0
= m.
Then one can easily see by induction that m
n
(x) ≤ f
n
(x) ≤ M
n
(x). By the first part of the
proof, lim
n
m
n
(x) = m(0) = g(0) = M(0) = lim
n
M
n
(x) for any x ∈ (0, 1]. Therefore we must

have lim
n
f
n
(x) = g(0).
2
Proof II. To make the notation clearer we shall denote the variable of f
j
by x
j
. By definition
(and Fubini theorem) we get that
f
n+1
(x
n+1
) =
1
x
n+1

x
n+1
0
1
x
n

x
n

0
1
x
n−1

x
n−1
0
. . .

x
2
0
1
x
1

x
1
0
g(x
0
)dx
0
dx
1
. . . dx
n
=
1

x
n+1

0≤x
0
≤x
1
≤ ≤x
n
≤x
n+1
g(x
0
)
dx
0
dx
1
. . . dx
n
x
1
. . . x
n
=
1
x
n+1

x

n+1
0
g(x
0
)


x
0
≤x
1
≤ ≤x
n
≤x
n+1
dx
1
. . . dx
n
x
1
. . . x
n

dx
0
.
Therefore with the notation
h
n

(a, b) =

a≤x
1
≤ ≤x
n
≤b
dx
1
. . . dx
n
x
1
. . . x
n
and x = x
n+1
, t = x
0
we have
f
n+1
(x) =
1
x

x
0
g(t)h
n

(t, x)dt.
Using that h
n
(a, b) is the same for any permutation of x
1
, . . . , x
n
and the fact that the integral
is 0 on any hyperplanes (x
i
= x
j
) we get that
n! h
n
(a, b) =

a≤x
1
, ,x
n
≤b
dx
1
. . . dx
n
x
1
. . . x
n

=

b
a
. . .

b
a
dx
1
. . . dx
n
x
1
. . . x
n
=


b
a
dx
x

n
= (log(b/a))
n
.
Therefore
f

n+1
(x) =
1
x

x
0
g(t)
(log(x/t))
n
n!
dt.
Note that if g is constant then the definition gives f
n
= g. This implies on one hand that we
must have
1
x

x
0
(log(x/t))
n
n!
dt = 1
and on the other hand that, by replacing g by g − g(0), we can suppose that g(0) = 0.
Let x ∈ (0, 1] and ε > 0 be fixed. By continuity there exists a 0 < δ < x and an M such that
|g(t)| < ε on [0, δ] and |g(t)| ≤ M on [0, 1] . Since
lim
n→∞

(log(x/δ))
n
n!
= 0
there exists an n
0
sucht that log(x/δ))
n
/n! < ε whenever n ≥ n
0
. Then, for any n ≥ n
0
, we have
|f
n+1
(x)| ≤
1
x

x
0
|g(t)|
(log(x/t))
n
n!
dt
≤
1
x


δ
0
ε
(log(x/t))
n
n!
dt +
1
x

x
δ
|g(t)|
(log(x/δ))
n
n!
dt
≤
1
x

x
0
ε
(log(x/t))
n
n!
dt +
1
x


x
δ
Mεdt
≤ ε + Mε.
Therefore lim
n
f(x) = 0 = g(0).
3
6. Let f(z) = a
n
z
n
+ a
n−1
z
n−1
+ . . . + a
1
z + a
0
be a polynomial with real coefficients. Prove that
if all roots of f lie in the left half-plane {z ∈ C : Re z < 0} then
a
k
a
k+3
< a
k+1
a

k+2
holds for every k = 0, 1, . . . , n − 3. (20 points)
Solution. The polynomial f is a product of linear and quadratic factors, f(z) =

i
(k
i
z + l
i
) ·

j
(p
j
z
2
+q
j
z +r
j
), with k
i
, l
i
, p
j
, q
j
, r
j

∈ R. Since all roots are in the left half-plane, for each i, k
i
and l
i
are of the same sign, and for each j, p
j
, q
j
, r
j
are of the same sign, too. Hence, multiplying
f by −1 if necessary, the roots of f don’t change and f becomes the polynomial with all positive
coefficients.
For the simplicity, we extend the sequence of coefficients by a
n+1
= a
n+2
= . . . = 0 and
a
−1
= a
−2
= . . . = 0 and prove the same statement for −1 ≤ k ≤ n − 2 by induction.
For n ≤ 2 the statement is obvious: a
k+1
and a
k+2
are positive and at least one of a
k−1
and

a
k+3
is 0; hence, a
k+1
a
k+2
> a
k
a
k+3
= 0.
Now assume that n ≥ 3 and the statement is true for all smaller values of n . Take a divisor of
f(z) which has the form z
2
+ pz + q where p and q are positive real numbers. (Such a divisor can
be obtained from a conjugate pair of roots or two real roots.) Then we can write
f(z) = (z
2
+ pz + q)(b
n−2
z
n−2
+ . . . + b
1
z + b
0
) = (z
2
+ pz + q)g(x). (1)
The roots polynomial g(z) are in the left half-plane, so we have b

k+1
b
k+2
< b
k
b
k+3
for all −1 ≤
k ≤ n − 4. Defining b
n−1
= b
n
= . . . = 0 and b
−1
= b
−2
= . . . = 0 as well, we also have
b
k+1
b
k+2
≤ b
k
b
k+3
for all integer k.
Now we prove a
k+1
a
k+2

> a
k
a
k+3
. If k = −1 or k = n − 2 then this is obvious since a
k+1
a
k+2
is positive and a
k
a
k+3
= 0. Thus, assume 0 ≤ k ≤ n − 3. By an easy computation,
a
k+1
a
k+2
− a
k
a
k+3
=
= (qb
k+1
+ pb
k
+ b
k−1
)(qb
k+2

+ pb
k+1
+ b
k
) − (qb
k
+ pb
k−1
+ b
k−2
)(qb
k+3
+ pb
k+2
+ b
k+1
) =
= (b
k−1
b
k
− b
k−2
b
k+1
) + p(b
2
k
− b
k−2

b
k+2
) + q(b
k−1
b
k+2
− b
k−2
b
k+3
)+
+p
2
(b
k
b
k+1
− b
k−1
b
k+2
) + q
2
(b
k+1
b
k+2
− b
k
b

k+3
) + pq(b
2
k+1
− b
k−1
b
k+3
).
We prove that all the six terms are non-negative and at leas t one is positive. Term p
2
(b
k
b
k+1
−
b
k−1
b
k+2
) is positive since 0 ≤ k ≤ n−3. Also terms b
k−1
b
k
−b
k−2
b
k+1
and q
2

(b
k+1
b
k+2
−b
k
b
k+3
)
are non-negative by the induction hypothesis.
To check the sign of p(b
2
k
− b
k−2
b
k+2
) consider
b
k−1
(b
2
k
− b
k−2
b
k+2
) = b
k−2
(b

k
b
k+1
− b
k−1
b
k+2
) + b
k
(b
k−1
b
k
− b
k−2
b
k+1
) ≥ 0.
If b
k−1
> 0 we can divide by it to obtain b
2
k
−b
k−2
b
k+2
≥ 0. Otherwise, if b
k−1
= 0, either b

k−2
= 0
or b
k+2
= 0 and thus b
2
k
− b
k−2
b
k+2
= b
2
k
≥ 0. Therefore, p(b
2
k
− b
k−2
b
k+2
) ≥ 0 for all k . Similarly,
pq(b
2
k+1
− b
k−1
b
k+3
) ≥ 0.

The sign of q(b
k−1
b
k+2
− b
k−2
b
k+3
) can be checked in a similar way. Consider
b
k+1
(b
k−1
b
k+2
− b
k−2
b
k+3
) = b
k−1
(b
k+1
b
k+2
− b
k
b
k+3
) + b

k+3
(b
k−1
b
k
− b
k−2
b
k+1
) ≥ 0.
If b
k+1
> 0, we can divide by it. Otherwise either b
k−2
= 0 or b
k+3
= 0. In all cases, we obtain
b
k−1
b
k+2
− b
k−2
b
k+3
≥ 0.
Now the signs of all terms are checked and the proof is complete.
4