MathDB
Problems
Contests
National and Regional Contests
Ukraine Contests
Official Ukraine Selection Cycle
Ukraine Team Selection Test
2016 Ukraine Team Selection Test
2016 Ukraine Team Selection Test
Part of
Ukraine Team Selection Test
Subcontests
(4)
5
1
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Grasshoppers on triangle lattice
Let
A
B
C
ABC
A
BC
be an equilateral triangle of side
1
1
1
. There are three grasshoppers sitting in
A
A
A
,
B
B
B
,
C
C
C
. At any point of time for any two grasshoppers separated by a distance
d
d
d
one of them can jump over other one so that distance between them becomes
2
k
d
2kd
2
k
d
,
k
,
d
k,d
k
,
d
are nonfixed positive integers. Let
M
M
M
,
N
N
N
be points on rays
A
B
AB
A
B
,
A
C
AC
A
C
such that
A
M
=
A
N
=
l
AM=AN=l
A
M
=
A
N
=
l
,
l
l
l
is fixed positive integer. In a finite number of jumps all of grasshoppers end up sitting inside the triangle
A
M
N
AMN
A
MN
. Find, in terms of
l
l
l
, the number of final positions of the grasshoppers. (Grasshoppers can leave the triangle
A
M
N
AMN
A
MN
during their jumps.)
1
1
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(6n+3)-game
Consider a regular polygon
A
1
A
2
…
A
6
n
+
3
A_1A_2\ldots A_{6n+3}
A
1
A
2
…
A
6
n
+
3
. The vertices
A
2
n
+
1
,
A
4
n
+
2
,
A
6
n
+
3
A_{2n+1}, A_{4n+2}, A_{6n+3}
A
2
n
+
1
,
A
4
n
+
2
,
A
6
n
+
3
are called holes. Initially there are three pebbles in some vertices of the polygon, which are also vertices of equilateral triangle. Players
A
A
A
and
B
B
B
take moves in turn. In each move, starting from
A
A
A
, the player chooses pebble and puts it to the next vertex clockwise (for example,
A
2
→
A
3
A_2\rightarrow A_3
A
2
→
A
3
,
A
6
n
+
3
→
A
1
A_{6n+3}\rightarrow A_1
A
6
n
+
3
→
A
1
). Player
A
A
A
wins if at least two pebbles lie in holes after someone's move. Does player
A
A
A
always have winning strategy?Proposed by Bohdan Rublov
2
1
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Function
Find all functions from positive integers to itself such that
f
(
a
+
b
)
=
f
(
a
)
+
f
(
b
)
+
f
(
c
)
+
f
(
d
)
f(a+b)=f(a)+f(b)+f(c)+f(d)
f
(
a
+
b
)
=
f
(
a
)
+
f
(
b
)
+
f
(
c
)
+
f
(
d
)
for all
c
2
+
d
2
=
2
a
b
c^2+d^2=2ab
c
2
+
d
2
=
2
ab
10
1
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Root in (0; 2016)
Let
a
1
,
…
,
a
n
a_1,\ldots, a_n
a
1
,
…
,
a
n
be real numbers. Define polynomials
f
,
g
f,g
f
,
g
by
f
(
x
)
=
∑
k
=
1
n
a
k
x
k
,
g
(
x
)
=
∑
k
=
1
n
a
k
2
k
−
1
x
k
.
f(x)=\sum_{k=1}^n a_kx^k,\ g(x)=\sum_{k=1}^n \frac{a_k}{2^k-1}x^k.
f
(
x
)
=
k
=
1
∑
n
a
k
x
k
,
g
(
x
)
=
k
=
1
∑
n
2
k
−
1
a
k
x
k
.
Assume that
g
(
2016
)
=
0
g(2016)=0
g
(
2016
)
=
0
. Prove that
f
(
x
)
f(x)
f
(
x
)
has a root in
(
0
;
2016
)
(0;2016)
(
0
;
2016
)
.