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
1
B
2
where A
1
, A
2
, B
1
, B
2
are 2 × 2 matrices. Then
1 0 −1 0
0 1 0 −1
−1 0 1 0
0 −1 0 1
=
A
1
A
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
= A
2
B
2
= I
2
and A
1
B
2
= A
2
B
1
= −I
2
. Then B
1
= A
−1
1
, B
2
= −A
−1
1
and A
2
= 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
=
2 0
0 2
2. Let f, g : [a, b] → [0, ∞) be continuous and non-decreasing functions such that for each x ∈ [a, b] we
have
x
a
f(t) dt ≤
x
a
g(t) dt
and
b
a
f(t) dt =
b
a
g(t) dt.
Prove that
b
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
dt ≥
b
a
1 +
G
(t)
2
dt
i.e. The length ot the graph of F is ≥ the length of the graph of G. This is clear since both functions are
convex, their graphs have common ends and the graph of F is below the graph of G — the length of the
graph of F is the least upper bound of the lengths of the graphs of piecewise linear functions whose values
at the points of non-differentiability coincide with the values of F , if a convex polygon P
1
is contained in
a polygon P
2
then the perimeter of P
1
is ≤ the perimeter of P
2
.
3. Let D be the closed unit disk in the plane, and let p
1
, p
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
| ≥
np −
n
i=1
p
i
= n +
n
i=1
p
i
≥ n
4. For n ≥ 1 let M be an n × n complex matrix with distinct eigenvalues λ
1
, 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
r
+ λ
s
are different. Let λ
r
and λ
s
be two eigenvalues of M and v
r
, v
s
eigenvectors associated to them, i.e.
Mv
j
= λv
j
for j = r, s. We have Mv
r
(v
s
)
T
+v
r
(v
s
)
T
M
T
= (Mv
r
)(v
s
)
T
+v
r
Mv
s
T
= λ
r
v
r
(v
s
)
T
+λ
s
v
r
(v
s
)
T
,
so v
r
(v
s
) is an eigenmatrix of L
M
with the eigenvalue λ
r
+ λ
s
.
Notice that if λ
r
= λ
s
then vectors u, w are linearly independent and matrices u(w)
T
and w(u)
T
are
linearly independent, too. This implies that the eigenvalue λ
r
+ λ
s
is double if r = s.
The map L
M
maps n
2
–dimensional linear space into itself, so it has at most n
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
+ λ
s
are not pairwise different; for those multiple values the multiplicities
should be summed up.)
5. Prove that
1
0
1
0
dx dy
x
−1
+ | ln y| − 1
≤ 1. [20 points]
Solution 1. First we use the inequality
x
−1
− 1 ≥ | ln x|, x ∈ (0, 1],
which follows from
(x
−1
− 1)
x=1
= | ln x||
x=1
= 0,
(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
1
0
dx dy
| ln x| + | ln y|
=
1
0
1
0
dx dy
| ln(x · y)|
.
Substituting y = u/x, we obtain
1
0
1
0
dx dy
| ln(x · y)|
=
1
0
1
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
=
∞
0
∞
s
e
s−u
(s + 1)
2
u
duds =
∞
0
u
0
e
s
(s + 1)
2
ds
e
−u
u
dsdu.
Since the function
e
s
(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
+ | ln y| − 1
≤
∞
0
u
2
e
u
(u + 1)
2
+ 1
e
−u
u
du =
1
2
∞
0
du
(u + 1)
2
+
∞
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
=
A
n−1
A
n−1
A
n−1
0
.
Denote the sum of all elements of a matrix M by S(M). Prove that S(A
k−1
n
) = S(A
n−1
k
) for every n, k ≥ 1.
[20 points]
Solution. The quantity S(A
k−1
n
) has a special combinatorical meaning. Consider an n × k table filled with
0’s and 1’s such that no 2 × 2 c ontains only 1’s. Denote the number of such fillings by F
nk
. The filling of
each row of the table corresponds to some integer ranging from 0 to 2
n
− 1 written in base 2. F
nk
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.
2. If i
n
= j
n
= 1 then i and j can be consecutive iff i
n−1
j
n−1
= 0 and i
n−2
. . . i
1
and j
n−2
. . . j
1
can be
consecutive.
Hence i and j can be consecutive iff (i + 1, j + 1)-th entry of A
n
is 1. Denoting this entry by a
i,j
, the sum
S(A
k−1
n
) =
2
n
−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
k−1
n
).
The the obvious statement F
nk
= F
kn
completes the proof.
11
th
International Mathematical Competition for University Students
Skopje, 25–26 July 2004
Solutions for problems on Day 1
Problem 1. Let S be an infinite set of real numbers such that |s
1
+ s
2
+ ··· + s
k
| < 1 for every finite subset
{s
1
, s
2
, . . . , s
k
} ⊂ S. Show that S is countable. [20 points]
Solution. Let S
n
= S ∩(
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
), S is a countable union of finite sets, and
hence countable.
Problem 2. Let P (x) = x
2
− 1. How many distinct real solutions does the following equation have:
P (P (. . . (P
2004
(x)) . . . )) = 0 ? [20 points]
Solution. Put P
n
(x) = P (P ( (P
n
(x)) )). As P
1
(x) ≥ −1, for each x ∈ R, it must be that P
n+1
(x) = P
1
(P
n
(x)) ≥
−1, for each n ∈ N and each x ∈ R . Therefore the equation P
n
(x) = a, where a < −1 has no real solutions.
Let us prove that the equation P
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
−
√
a + 1 < −1). Hence the equation P
n+1
(x) = a, has exactly two distinct real solutions.
Let us prove now that the e quation P
n
(x) = 0 has exactly n + 1 distinct real solutions. Again we use
mathematical induction. If n = 1 the solutions are x = ±1, and if n = 2 the solutions are x = 0 and x = ±
√
2,
so in both cases the number of solutions is equal to n + 1. Suppose that the assertion holds for some n ∈ N.
Note that P
n+2
(x) = P
2
(P
n
(x)) = P
2
n
(x)(P
2
n
(x) − 2), so the set of all real solutions of the equation P
n+2
= 0 is
exactly the union of the sets of all real solutions of the equations P
n
(x) = 0, P
n
(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
k=1
x
k
, where n ≥ 2, 0 ≤ x
1
, x
2
, . . . , x
n
≤
π
2
and
n
k=1
sin x
k
= 1 .
a) Show that S
n
is an interval. [10 points]
b) Let l
n
be the length of S
n
. Find lim
n→∞
l
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
.
Note that f(Y ) = S
n
. Since Y is a connected subspace of R
n
and f is a continuous function, the image f(Y ) is
also connected, and we know that the only connected subspaces of R are intervals. Thus S
n
is an interval.
(b) We prove that
n arcsin
1
n
≤ x
1
+ x
2
+ + x
n
≤
π
2
.
Since the graph of sin x is concave down for x ∈ [0,
π
2
], the chord joining the points (0, 0) and (
π
2
, 1) lies below the
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
= ··· = x
n
= 0.
The left-hand side follows immediately from Jensen’s inequality, since sin x is concave down for x ∈ [0,
π
2
] and
0 ≤
x
1
+x
2
+ +x
n
n
<
π
2
1
n
=
sin x
1
+ sin x
2
+ + sin x
n
n
≤ sin
x
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
2
− n arcsin
1
n
and
lim
n→∞
l
n
=
π
2
− lim
n→∞
arcsin(1/n)
1/n
=
π
2
− 1.
Problem 4. Suppose n ≥ 4 and let M be a finite set of n points in R
3
, no four of which lie in a plane. Assume
that the p oints can be coloured black or white so that any sphere which intersects M in at least four points has
the property that exactly half of the points in the intersection of M and the sphere are white. Prove that all of
the points in M lie on one sphere. [20 points]
Solution. Define f : M → {−1, 1}, f (X) =
−1, if X is white
1, if X is black
. 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
n
3
possible
choices of (A, B, C) we obtain that
n
3
< 0, which means
< 0 (contradicts the starting assumption). The
same reasoning is applied when assuming
< 0.
Now, from
= 0 and (1), it follows that f (A) + f (B) + f (C) = 0 for any distinct points A, B, C in M .
Taking another point D ∈ M , the following equalities take place
f (A) + f (B) + f (C) = 0
f (A) + f (B) + f (D) = 0
f (A) + f (C) + f (D) = 0
f (B) + f (C) + f (D) = 0
which easily leads to f (A) = f (B) = f (C) = f (D) = 0, which contradicts the definition of f.
Problem 5. Let X be a set of
2k−4
k−2
+ 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
|x
i+1
− x
1
| ≥ 2|x
i
− x
1
| ∀i = 2, . . . , k − 1,
or X contains a decreasing sequence {x
i
}
l
i=1
⊆ X of length l and
|x
i+1
− x
1
| ≥ 2|x
i
− x
1
| ∀i = 2, . . . , l − 1.
Proof of the lemma. We use induction on k + l. In case k = 2 or l = 2 the lemma is obviously true.
Now let us make the induction step. Let m be the minimal element of X, M be its maximal element. Let
X
m
= {x ∈ X : x ≤
m + M
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
+ 1, or |X
M
| ≥
k + (l − 1) − 4
k − 2
+ 1.
In the first case we apply the inductive assumption to X
m
and either obtain a decreasing sequence of length l
with the required properties (in this case the inductive step is made), or obtain an increasing sequence {x
i
}
k−1
i=1
⊆
X
m
of length k −1. Then we note that the sequence {x
1
, x
2
, . . . , x
k−1
, M} ⊆ X has length k and all the required
properties.
In the case |X
M
| ≥
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
n
−4
, and the branches of the logarithms do not matter because all branches are taken.
It is easy to check that the convergence is locally uniform on C \{0, 1}; therefore, f is a holomorphic function on
the complex plane, except possibly for isolated singularities at 0 and 1. (We omit the detailed estimates here.)
The function log has its only (simple) zero at z = 1, so f has a quadruple pole at z = 1.
Now we investigate the behavior near infinity. We have Re(log(z)) = log |z|, hence (with c := log |z|)
|
(log z)
−4
| ≤
|log z|
−4
=
(log |z| + 2πin)
−4
+ O(1)
=
∞
−∞
(c + 2πix)
−4
dx + O(1)
= c
−4
∞
−∞
(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
a, b, c.
There are a number of ways to compute the coefficients a, b, c, which turn out to be a = c = 1/6, b = 2/3.
Since f (z) = f(1/z), it follows easily that a = c. Moreover, the fact lim
z→1
(z − 1)
4
f(z) = 1 implies a + b + c = 1
(this fact follows from the observation that at z = 1, all summands cancel pairwise, except the principal branch
which contributes a quadruple pole). Finally, we can calculate
f(−1) = π
−4
nodd
n
−4
= 2π
−4
n≥1odd
n
−4
= 2π
−4
n≥1
n
−4
−
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
can be computed from lim
z→1
(z −1)
2
g(z) = 1; it is d = 1. Then the exponent 2 in the denominator can be increased
by taking derivatives (see Solution 2). Similarly, one can start with exponent 3 directly.
A more straightforward, though tedious way to find the constants is computing the first four terms of the
Laurent series of f around z = 1. For that branch of the logarithm which vanishes at 1, for all |w| <
1
2
we have
log(1 + w) = w −
w
2
2
+
w
3
3
−
w
4
4
+ O(|w|
5
);
after some computation, one can obtain
1
log(1 + w)
4
= 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
(the remainder vanishes) and
f(z) =
1 + 2(z −1) +
7
6
(z −1)
2
+
1
6
(z −1)
3
(z −1)
4
=
z(z
2
+ 4z + 1)
6(z −1)
4
.
Solution 2. ¿From the well-known series for the cotangent function,
lim
N→∞
N
k=−N
1
w + 2πi ·k
=
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·
i log z
2
− 1
=
1
2
+
1
z −1
.
Taking derivatives we obtain
1
(log z)
2
= −z ·
1
2
+
1
z −1
=
z
(z −1)
2
,
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
=
z(z
2
+ 4z + 1)
2(z −1)
4
.
. 2π
−4
n≥1
n
−4
−
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. +
G
(t)
2
dt
i.e. The length ot the graph of F is ≥ the length of the graph of G. This is clear since both functions are
convex, their graphs have common ends