11
th
International Mathematical Competition for University Students
Skopje, 25–26 July 2004
Solutions for problems on Day 2
1. Let A be a real 4 × 2 matrix and B be a real 2 × 4 matrix such that
AB =




1 0 −1 0
0 1 0 −1
−1 0 1 0
0 −1 0 1




.
Find BA. [20 points]
Solution. Let A =

A
1
A
2

and B =

B

2


B
1
B
2

=

A
1
B
1
A
1
B
2
A
2
B
1
A
2
B
2

therefore, A
1
B

−1
2
=
−A
1
. Finally,
BA =

B
1
B
2


A
1
A
2

= B
1
A
1
+ B
2
A
2
= 2I
2
=

a

1 + f(t) dt ≥

b
a

1 + g(t) dt. [20 points]
Solution. Let F (x) =

x
a

f(t) dt and G(x) =

x
a

g(t) dt. The functions F, G are convex, F (a) = 0 =
G(a) and F (b) = G(b) by the hypothesis. We are supposed to show that

b
a

1 +

F

(t)


2
, . . . , p
n
be fixed points in D. Show that there
exists a point p in D such that the sum of the distances of p to each of p
1
, p
2
, . . . , p
n
is greater than or
equal to 1. [20 points]
Solution. considering as vectors, thoose p to be the unit vector which points into the opposite direction as
n

i=1
p
i
. Then, by the triangle inequality,
n

i=1
|p − p
i
| ≥






, λ
2
, . . . , λ
k
, with multiplicities
m
1
, m
2
, . . . , m
k
, respectively. Consider the linear operator L
M
defined by L
M
(X) = MX + XM
T
, for any
complex n × n matrix X. Find its eigenvalues and their multiplicities. (M
T
denotes the transpose of M;
that is, if M = (m
k,l
), then M
T
= (m
l,k
).) [20 points]
Solution. We first solve the problem for the special case when the eigenvalues of M are distinct and all sums
λ

= (Mv
r
)(v
s
)
T
+v
r

Mv
s

T
= λ
r
v
r
(v
s
)
T
+λ
s
v
r
(v
s
)
T
,

2
eigenvalues. We already
found n
2
eigenvalues, so there exists no more and the problem is solved for the special case.
In the general case, matrix M is a limit of matrices M
1
, M
2
, . . . such that each of them belongs to the
special case above. By the continuity of the eigenvalues we obtain that the eigenvalues of L
M
are
• 2λ
r
with multiplicity m
2
r
(r = 1, . . . , k);
• λ
r
+ λ
s
with multiplicity 2m
r
m
s
(1 ≤ r < s ≤ k).
(It can happen that the sums λ
r

(x
−1
− 1)

= −
1
x
2
≤ −
1
x
= | ln x|

, x ∈ (0, 1].
Therefore

1
0

1
0
dx dy
x
−1
+ | ln y| − 1
≤

1
0


u
dx
x

du
| ln u|
=

1
0
| ln u| ·
du
| ln u|
= 1.
Solution 2. Substituting s = x
−1
− 1 and u = s − ln y,

1
0

1
0
dx dy
x
−1
+ | ln y| − 1
=

∞

(s+1)
2
is convex,

u
0
e
s
(s + 1)
2
ds ≤
u
2

e
u
(u + 1)
2
+ 1

so

1
0

1
0
dx dy
x
−1

0
e
−u
du

= 1.
6. For n ≥ 0 define matrices A
n
and B
n
as follows: A
0
= B
0
= (1) and for every n > 0
A
n
=

A
n−1
A
n−1
A
n−1
B
n−1

and B
n

equals
to the number of k-tuples of integers such that every two consecutive integers correspond to the filling of
n × 2 table without 2 × 2 squares filled with 1’s.
Consider binary expansions of integers i and j
i
n
i
n−1
. . . i
1
and j
n
j
n−1
. . . j
1
. There are two cases:
1. If i
n
j
n
= 0 then i and j can be consecutive iff i
n−1
. . . i
1
and j
n−1
. . . j
1
can be consequtive.

−1
i
1
=0
· · ·

2
n
−1
i
k
=0
a
i
1
i
2
a
i
2
i
3
· · · a
i
k−1
i
k
counts the possible fillings. Therefore F
nk
= S(A

1
n
, ∞) for any integer n > 0. It follows from the inequality that |S
n
| < n. Similarly, if we
define S
−n
= S ∩ (−∞, −
1
n
), then |S
−n
| < n. Any nonzero x ∈ S is an element of some S
n
or S
−n
, because there
exists an n such that x >
1
n
, or x < −
1
n
. Then S ⊂ {0}∪

n∈N
(S
n
∪S
−n

(x) = a, where a > 0, has exactly two distinct real solutions. To this end we
use mathematical induction by n. If n = 1 the assertion follows directly. Assuming that the assertion holds for a
n ∈ N we prove that it must also hold for n + 1. Since P
n+1
(x) = a is equivalent to P
1
(P
n
(x)) = a, we conclude
that P
n
(x) =
√
a + 1 or P
n
(x) = −
√
a + 1. The equation P
n
(x) =
√
a + 1, as
√
a + 1 > 1, has exactly two distinct
real solutions by the inductive hypothesis, while the equation P
n
(x) = −
√
a + 1 has no real solutions (because
−

(x) =
√
2 and P
n
(x) = −
√
2.
By the inductive hypothesis the equation P
n
(x) = 0 has exactly n + 1 distinct real solutions, while the equations
P
n
(x) =
√
2 and P
n
(x) = −
√
2 have two and no distinct real solutions, respectively. Hence, the sets above being
pairwise disjoint, the equation P
n+2
(x) = 0 has exactly n + 3 distinct real solutions. Thus we have proved that,
for each n ∈ N, the equation P
n
(x) = 0 has exactly n + 1 distinct real solutions, so the answer to the question
posed in this problem is 2005.
Problem 3. Let S
n
be the set of all sums
n

n
. [10 points]
Solution. (a) Equivalently, we consider the set
Y = {y = (y
1
, y
2
, , y
n
)| 0 ≤ y
1
, y
2
, , y
n
≤ 1, y
1
+ y
2
+ + y
n
= 1} ⊂ R
n
and the image f(Y ) of Y under
f(y) = arcsin y
1
+ arcsin y
2
+ + arcsin y
n

graph. Hence
2x
π
≤ sin x for all x ∈ [0,
π
2
]
and we can deduce the right-hand side of the claim:
2
π
(x
1
+ x
2
+ + x
n
) ≤ sin x
1
+ sin x
2
+ + sin x
n
= 1.
The value 1 can be reached choosing x
1
=
π
2
and x
2

1
+ x
2
+ + x
n
n
.
Equality holds if x
1
= ··· = x
n
= arcsin
1
n
.
Now we have computed the minimum and maximum of interval S
n
; we can conclude that S
n
= [n arcsin
1
n
,
π
2
].
Thus l
n
=
π

. The given condition becomes

X∈S
f (X) = 0
for any sphere S which passes through at least 4 points of M. For any 3 given points A, B, C in M, denote by
S (A, B, C) the set of all spheres which pass through A, B, C and at least one other point of M and by |S (A, B, C)|
the number of these spheres. Also, denote by

the sum

X∈M
f (X).
We have
0 =

S∈S(A,B,C)

X∈S
f (X) = (|S (A, B, C)| −1) (f (A) + f (B) + f (C)) +

(1)
since the values of A, B, C appear |S (A, B, C)| times each and the other values appear only once.
If there are 3 points A, B, C such that |S (A, B, C)| = 1, the proof is finished.
If |S (A, B, C)| > 1 for any distinct points A, B, C in M, we will prove at first that

= 0.
Assume that

> 0. From (1) it follows that f (A) + f (B) + f (C) < 0 and summing by all


+ 1 real numbers, k ≥ 2. Prove that there exists a monotone sequence
{x
i
}
k
i=1
⊆ X such that
|x
i+1
− x
1
| ≥ 2|x
i
− x
1
|
for all i = 2, . . . , k − 1. [20 points]
Solution. We prove a more general statement:
Lemma. Let k, l ≥ 2, let X be a set of

k+l−4
k−2

+1 real numbers. Then either X contains an increasing sequence
{x
i
}
k
i=1
⊆ X of length k and

2
}, X
M
= {x ∈ X : x >
m + M
2
}.
Since

k+l−4
k−2

=

k+(l−1)−4
k−2

+

(k−1)+l−4
(k−1)−2

, we can see that either
|X
m
| ≥

(k − 1) + l − 4
(k − 1) −2



k+(l−1)−4
k−2

+ 1 the inductive step is made in a similar way. Thus the lemma is proved.
The reader may check that the number

k+l−4
k−2

+ 1 cannot be smaller in the lemma.
Problem 6. For every complex numb e r z /∈ {0, 1} define
f(z) :=

(log z)
−4
,
where the sum is over all branches of the complex logarithm.
a) Show that there are two polynomials P and Q such that f(z) = P (z)/Q(z) for all z ∈ C \ {0, 1}. [10
points]
b) Show that for all z ∈ C \ {0, 1}
f(z) = z
z
2
+ 4z + 1
6(z −1)
4
. [10 points]
Solution 1. It is clear that the left hand side is well defined and independent of the order of summation, because
we have a sum of the type


∞
−∞
(1 + 2πix/c)
−4
dx + O(1)
= c
−3

∞
−∞
(1 + 2πit)
−4
dt + O(1)
≤ α(log |z|)
−3
for a universal constant α. Therefore, the infinite sum tends to 0 as |z| → ∞ . In particular, the isolated singularity
at ∞ is not essential, but rather has (at least a single) zero at ∞.
The remaining singularity is at z = 0. It is readily verified that f(1/z) = f(z) (because log(1/z) = −log(z));
this implies that f has a zero at z = 0.
We conclude that the infinite sum is holomorphic on C with at most one pole and without an essential singularity
at ∞, so it is a rational function, i.e. we can write f(z) = P (z)/Q(z) for some polynomials P and Q which we
may as well assume coprime. This solves the first part.
Since f has a quadruple pole at z = 1 and no other poles, we have Q(z) = (z − 1)
4
up to a constant factor
which we can as well set equal to 1, and this determines P uniquely. Since f(z) → 0 as z → ∞, the degree of P
is at most 3, and since P(0) = 0, it follows that P(z) = z(az
2
+ bz + c) for yet undetermined complex constants


n≥1even
n
−4


=
1
48
.
This implies a −b + c = −1/3. These three equations easily yield a, b, c.
Moreover, the function f satisfies f(z) + f(−z) = 16f(z
2
): this follows because the branches of log(z
2
) =
log((−z)
2
) are the numbers 2 log(z) and 2 log(−z). This observation supplies the two equations b = 4a and a = c,
which can be used instead of some of the considerations above.
Another way is to compute g(z) =

1
(log z)
2
first. In the same way, g(z) =
dz
(z−1)
2
. The unknown coefficient d

= w
−4
+ 2w
−2
+
7
6
w
−2
+
1
6
w
−1
+ O(1).
The remaining branches of logarithm give a bounded function. So
f(1 + w) = w
−4
+ 2w
−2
+
7
6
w
−2
+
1
6
w
−1

i
2
cot
iw
2
and
lim
N→∞
N

k=−N
1
log z + 2πi ·k
=
i
2
cot
i log z
2
=
i
2
· i
e
2i·
i log z
2
+ 1
e
2i·

1
(log z)
3
= −
z
2
·

z
(z −1)
2


=
z(z + 1)
2(z −1)
3
and

1
(log z)
4
= −
z
3
·

z(z + 1)
2(z −1)
3