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Problems
Contests
International Contests
Czech-Polish-Slovak Match
1999 Czech and Slovak Match
1999 Czech and Slovak Match
Part of
Czech-Polish-Slovak Match
Subcontests
(6)
6
1
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inequality with least common multiple of n numbers
Prove that for any integer
n
≥
3
n \ge 3
n
≥
3
, the least common multiple of the numbers
1
,
2
,
.
.
.
,
n
1,2, ... ,n
1
,
2
,
...
,
n
is greater than
2
n
−
1
2^{n-1}
2
n
−
1
.
3
1
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exactly k pairwise non-congruent triangles of sides
Find all natural numbers
k
k
k
for which there exists a set
M
M
M
of ten real numbers such that there are exactly
k
k
k
pairwise non-congruent triangles whose side lengths are three (not necessarily distinct) elements of
M
M
M
.
2
1
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circumcircle of triangle passes through segment's midpoint
The altitudes through the vertices
A
,
B
,
C
A,B,C
A
,
B
,
C
of an acute-angled triangle
A
B
C
ABC
A
BC
meet the opposite sides at
D
,
E
,
F
,
D,E,F,
D
,
E
,
F
,
respectively. The line through
D
D
D
parallel to
E
F
EF
EF
meets the lines
A
C
AC
A
C
and
A
B
AB
A
B
at
Q
Q
Q
and
R
R
R
, respectively. The line
E
F
EF
EF
meets
B
C
BC
BC
at
P
P
P
. Prove that the circumcircle of the triangle
P
Q
R
PQR
PQR
passes through the midpoint of
B
C
BC
BC
.
1
1
Hide problems
Poland Inequalities
Leta,b,c are postive real numbers,proof that \frac{a}{b\plus{}2c}\plus{}\frac{b}{c\plus{}2a}\plus{}\frac{c}{a\plus{}2b}\geq1
4
1
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Find All k
Find all positive integers
k
k
k
for which the following assertion holds: If
F
(
x
)
F(x)
F
(
x
)
is polynomial with integer coefficients ehich satisfies
F
(
c
)
≤
k
F(c) \leq k
F
(
c
)
≤
k
for all
c
∈
{
0
,
1
,
⋯
,
k
+
1
}
c \in \{0,1, \cdots,k+1 \}
c
∈
{
0
,
1
,
⋯
,
k
+
1
}
, then
F
(
0
)
=
F
(
1
)
=
⋯
=
F
(
k
+
1
)
.
F(0)= F(1) = \cdots =F(k+1).
F
(
0
)
=
F
(
1
)
=
⋯
=
F
(
k
+
1
)
.
5
1
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F from(1,infinity) to r
Find all functions
f
:
(
1
,
∞
)
to R
f: (1,\infty)\text{to R}
f
:
(
1
,
∞
)
to R
satisfying
f
(
x
)
−
f
(
y
)
=
(
y
−
x
)
f
(
x
y
)
f(x)-f(y)=(y-x)f(xy)
f
(
x
)
−
f
(
y
)
=
(
y
−
x
)
f
(
x
y
)
for all
x
,
y
>
1
x,y>1
x
,
y
>
1
. [hide="hint"]you may try to find
f
(
x
5
)
f(x^5)
f
(
x
5
)
by two ways and then continue the solution. I have also solved by using this method.By finding
f
(
x
5
)
f(x^5)
f
(
x
5
)
in two ways I found that
f
(
x
)
=
x
f
(
x
2
)
f(x)=xf(x^2)
f
(
x
)
=
x
f
(
x
2
)
for all
x
>
1
x>1
x
>
1
.