Tài liệu Đề thi Olympic sinh viên thế giới năm 2004 pdf
Tài liệu Đề thi Olympic sinh viên thế giới năm 2004 pdf
... 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 ... ≥
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 ... u,...
Tài liệu Đề thi Olympic sinh viên thế giới năm 2003 pdf
... 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 ... 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 diago...
Tài liệu Đề thi Olympic sinh viên thế giới năm 1994 pptx
... 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
= ... −
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...
Tài liệu Đề thi Olympic sinh viên thế giới năm 1995 docx
... −
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, ... 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...
Tài liệu Đề thi Olympic sinh viên thế giới năm 1996 doc
... consider k ≥ 2.
For any m we have
(2)
cosh θ = cosh ((m + 1)θ − mθ) =
= cosh (m + 1)θ.cosh mθ − sinh (m + 1)θ .sinh mθ
= cosh (m + 1)θ.cosh mθ −
cosh
2
(m + 1)θ − 1.
√
cosh
2
mθ − 1
Set cosh kθ = a, ... the first and the third integral tend to −
1
f
(0)
as n → ∞, hence
so does the second.
Also n
1
A
(f(x))
n
dx ≤ n(f(A))
n
−→
n→∞
0 (f (A) < 1). We get L = −
1
f
(0)
in this case....
Tài liệu Đề thi Olympic sinh viên thế giới năm 1997 ngày 1 pptx
... suffices
to show that this sequence is strictly decreasing. Now,
p
k
− q
k
− (p
k−1
− q
k−1
) = n
k
p
k−1
− (n
k
+ 1)q
k−1
− p
k−1
+ q
k−1
= (n
k
− 1)p
k−1
− n
k
q
k−1
and this is negative because
p
k−1
q
k−1
= ... is the rational number
p
q
. Our aim is to show
that for some m,
θ
m−1
=
n
m
n
m
− 1
.
Suppose this is not the case, so that for every m,
(3) θ
m−1
<
n
m
n
m
− 1
.
4
For each k we...
Tài liệu Đề thi Olympic sinh viên thế giới năm 1997 ngày 2 docx
... then there
are axes making k
2π
n
angle). If A is infinite then we can think that A = Z
and f (m) = m + 1 for every m ∈ Z. In this case we define g
1
as a symmetry
relative to
1
2
, g
2
as a symmetry ... It is enough to prove the theorem for every such set. Let A = T (x).
If A is finite, then we can think that A is the set of all vertices of a regular
n polygon and that f is rotation by
2π
n
.
Tài liệu Đề thi Olympic sinh viên thế giới năm 1998 ppt
... contains
only powers of ¯y = yK, i.e., S
4
/K is cyclic. It is easy to see that this factor-group is not comutative
(something more this group is not isomorphic to S
3
).
3) n = 5
a) If x = (12), then for ... vector space.
Solution First choose a basis {v
1
, v
2
, v
3
} of U
1
. It is possible to extend this basis with vectors v
4
,v
5
and
v
6
to get a basis of U
2
. In the same way we can e...
Tài liệu Đề thi Olympic sinh viên thế giới năm 1999 doc
... (1)
From this recursion we can compute the probabilities for small values of n and can conjecture that p
(r)
n
=
1
5
+
4
5·6
n
if n ≡ r (mod )5 and p
(r)
n
=
1
5
−
1
5·6
n
otherwise. From (1), this ... opposite inequality holds ∀m 1. This
contradiction shows that g is a constant, i.e. f(x) = Cx, C > 0.
Conversely, it is easy to check that the functions of this type verify the conditions...
Tài liệu Đề thi Olympic sinh viên thế giới năm 2000 doc
... e, this yields e + 2ef = 0, and similarly f + 2ef = 0,
so that f = −2ef = e, hence e = f = g by symmetry. Hence, finaly, 3e = e + f + g = 0,
i.e. e = f = g = 0.
For part (i) just omit some of this.
Problem ... into n smaller cubes may be refined to
give a dissection into n + (a
d
− 1) cubes, for any a ≥ 1. This refinement is achieved by
picking an arbitrary cube in the dissection, and cutting it...