MathDB
Problems
Contests
National and Regional Contests
Belarus Contests
Belarus Team Selection Test
2011 Belarus Team Selection Test
2011 Belarus Team Selection Test
Part of
Belarus Team Selection Test
Subcontests
(4)
4
2
Hide problems
beetles creeping on neighbouring cells on a n x n square
Given a
n
×
n
n \times n
n
×
n
square table. Exactly one beetle sits in each cell of the table. At
12.00
12.00
12.00
all beetles creeps to some neighbouring cell (two cells are neighbouring if they have the common side). Find the greatest number of cells which can become empty (i.e. without beetles) if a)
n
=
8
n=8
n
=
8
b)
n
=
9
n=9
n
=
9
Problem Committee of BMO 2011
(1/a+1/b+1/c)(a+b+c-2) if a+b+c=a^2+b^2+c^2=a^3+b^3+c^3
Given nonzero real numbers a,b,c with
a
+
b
+
c
=
a
2
+
b
2
+
c
2
=
a
3
+
b
3
+
c
3
a+b+c=a^2+b^2+c^2=a^3+b^3+c^3
a
+
b
+
c
=
a
2
+
b
2
+
c
2
=
a
3
+
b
3
+
c
3
. (
∗
*
∗
) a) Find
(
1
a
+
1
b
+
1
b
)
(
a
+
b
+
c
−
2
)
\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\right)(a+b+c-2)
(
a
1
+
b
1
+
b
1
)
(
a
+
b
+
c
−
2
)
b) Do there exist pairwise different nonzero
a
,
b
,
c
a,b,c
a
,
b
,
c
satisfying (
∗
*
∗
)?D. Bazylev
3
3
Hide problems
max value of summands each natural >=3 may be represented as sum
Any natural number
n
,
n
≥
3
n, n\ge 3
n
,
n
≥
3
can be presented in different ways as a sum several summands (not necessarily different). Find the greatest possible value of these summands.Folklore
f(f(x+y))=xf(y)+g(x)
Find all functions
f
:
R
→
R
,
g
:
R
→
R
f: R \to R ,g: R \to R
f
:
R
→
R
,
g
:
R
→
R
satisfying the following equality
f
(
f
(
x
+
y
)
)
=
x
f
(
y
)
+
g
(
x
)
f(f(x+y))=xf(y)+g(x)
f
(
f
(
x
+
y
))
=
x
f
(
y
)
+
g
(
x
)
for all real
x
x
x
and
y
y
y
.I. Gorodnin
f(x-f(x/y))=xf(1-f(1/y))
Find all functions
f
:
R
→
R
f:R\to R
f
:
R
→
R
such that for all real
x
,
y
x,y
x
,
y
with
y
≠
0
y\ne 0
y
=
0
f
(
x
−
f
(
x
/
y
)
)
=
x
f
(
1
−
f
(
1
/
y
)
)
f(x-f(x/y))=xf(1-f(1/y))
f
(
x
−
f
(
x
/
y
))
=
x
f
(
1
−
f
(
1/
y
))
and a)
f
(
1
−
f
(
1
)
)
≠
0
f(1-f(1))\ne 0
f
(
1
−
f
(
1
))
=
0
b)
f
(
1
−
f
(
1
)
)
=
0
f(1-f(1))= 0
f
(
1
−
f
(
1
))
=
0
S. Kuzmich, I.Voronovich
1
8
Show problems
2
6
Show problems