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Problems
Contests
National and Regional Contests
Russia Contests
Saint Petersburg Mathematical Olympiad
2011 Saint Petersburg Mathematical Olympiad
2011 Saint Petersburg Mathematical Olympiad
Part of
Saint Petersburg Mathematical Olympiad
Subcontests
(7)
7
3
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Secret Society
There is secret society with
2011
2011
2011
members. Every member has bank account with integer balance ( can be negative). Sometimes some member give one dollar to every his friend. It is known, that after some such moves members can redistribute their money arbitrarily. Prove, that there are exactly
2010
2010
2010
pairs of friends.
Convex quadrilateral
A
B
C
D
ABCD
A
BC
D
- convex quadrilateral.
P
P
P
is such point on
A
C
AC
A
C
and inside
△
A
B
D
\triangle ABD
△
A
B
D
, that
∠
A
C
D
+
∠
B
D
P
=
∠
A
C
B
+
∠
D
B
P
=
90
−
∠
B
A
D
\angle ACD+\angle BDP = \angle ACB+ \angle DBP = 90-\angle BAD
∠
A
C
D
+
∠
B
D
P
=
∠
A
CB
+
∠
D
BP
=
90
−
∠
B
A
D
. Prove that
∠
B
A
D
+
∠
B
C
D
=
90
\angle BAD+ \angle BCD =90
∠
B
A
D
+
∠
BC
D
=
90
or
∠
B
D
A
+
∠
C
A
B
=
90
\angle BDA + \angle CAB = 90
∠
B
D
A
+
∠
C
A
B
=
90
Plays with polygon
Sasha and Serg plays next game with
100
100
100
-angled regular polygon . In the beggining Sasha set natural numbers in every angle. Then they make turn by turn, first turn is made by Serg. Serg turn is to take two opposite angles and add
1
1
1
to its numbers. Sasha turn is to take two neigbour angles and add
1
1
1
to its numbers. Serg want to maximize amount of odd numbers. What maximal number of odd numbers can he get no matter how Sasha plays?
6
3
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Interesting point
A
B
C
D
ABCD
A
BC
D
- convex quadrilateral.
M
M
M
-midpoint
A
C
AC
A
C
and
∠
M
C
B
=
∠
C
M
D
=
∠
M
B
A
=
∠
M
B
C
−
∠
M
D
C
\angle MCB=\angle CMD =\angle MBA=\angle MBC-\angle MDC
∠
MCB
=
∠
CM
D
=
∠
MB
A
=
∠
MBC
−
∠
M
D
C
. Prove, that
A
D
=
D
C
+
A
B
AD=DC+AB
A
D
=
D
C
+
A
B
Lights in garland
We have garland with
n
n
n
lights. Some lights are on, some are off. In one move we can take some turned on light (only turned on) and turn off it and also change state of neigbour lights. We want to turn off all lights after some moves.. For what
n
n
n
is it always possible?
Sequence and divisors
There is infinite sequence of composite numbers
a
1
,
a
2
,
.
.
.
,
a_1,a_2,...,
a
1
,
a
2
,
...
,
where
a
n
+
1
=
a
n
−
p
n
+
a
n
p
n
a_{n+1}=a_n-p_n+\frac{a_n}{p_n}
a
n
+
1
=
a
n
−
p
n
+
p
n
a
n
;
p
n
p_n
p
n
is smallest prime divisor of
a
n
a_n
a
n
. It is known, that
37
∣
a
n
37|a_n
37∣
a
n
for every
n
n
n
. Find possible values of
a
1
a_1
a
1
5
2
Hide problems
Working with divisors
Let
M
(
n
)
M(n)
M
(
n
)
and
m
(
n
)
m(n)
m
(
n
)
are maximal and minimal proper divisors of
n
n
n
Natural number
n
>
1000
n>1000
n
>
1000
is on the board. Every minute we replace our number with
n
+
M
(
n
)
−
m
(
n
)
n+M(n)-m(n)
n
+
M
(
n
)
−
m
(
n
)
. If we get prime, then process is stopped. Prove that after some moves we will get number, that is not divisible by
17
17
17
Area of convex quadrilatera
A
B
C
D
ABCD
A
BC
D
- convex quadrilateral.
∠
A
+
∠
D
=
150
,
∠
B
<
150
,
∠
C
<
150
\angle A+ \angle D=150, \angle B<150, \angle C<150
∠
A
+
∠
D
=
150
,
∠
B
<
150
,
∠
C
<
150
Prove, that area
A
B
C
D
ABCD
A
BC
D
is greater than
1
4
(
A
B
∗
C
D
+
A
B
∗
B
C
+
B
C
∗
C
D
)
\frac{1}{4}(AB*CD+AB*BC+BC*CD)
4
1
(
A
B
∗
C
D
+
A
B
∗
BC
+
BC
∗
C
D
)
4
2
Hide problems
Far from squares and cubes.
Call integer number
x
x
x
as far from squares and cubes, if for every integer
k
k
k
it is true :
∣
x
−
k
2
∣
>
1
0
6
,
∣
x
−
k
3
∣
>
1
0
6
|x-k^2|>10^6,|x-k^3|>10^6
∣
x
−
k
2
∣
>
1
0
6
,
∣
x
−
k
3
∣
>
1
0
6
. Prove, that there are infinitely many far from squares and cubes degrees of
2
2
2
Friends in city
In some city there are
2000000
2000000
2000000
citizens. In every group of
2000
2000
2000
citizens there are
3
3
3
pairwise friends. Prove, that there are
4
4
4
pairwise friends in city.
3
2
Hide problems
Building from bricks
Can we build parallelepiped
6
×
7
×
7
6 \times 7 \times 7
6
×
7
×
7
from
1
×
1
×
2
1 \times 1 \times 2
1
×
1
×
2
bricks, such that we have same amount bricks of one of
3
3
3
directions ?
Point in triangle
Point
D
D
D
is inside
△
A
B
C
\triangle ABC
△
A
BC
and
A
D
=
D
C
AD=DC
A
D
=
D
C
.
B
D
BD
B
D
intersect
A
C
AC
A
C
in
E
E
E
.
B
D
B
E
=
A
E
E
C
\frac{BD}{BE}=\frac{AE}{EC}
BE
B
D
=
EC
A
E
. Prove, that
B
E
=
B
C
BE=BC
BE
=
BC
2
3
Hide problems
Some geometry
A
B
C
ABC
A
BC
-triangle with circumcenter
O
O
O
and
∠
B
=
30
\angle B=30
∠
B
=
30
.
B
O
BO
BO
intersect
A
C
AC
A
C
at
K
K
K
.
L
L
L
- midpoint of arc
O
C
OC
OC
of circumcircle
K
O
C
KOC
K
OC
, that does not contains
K
K
K
. Prove, that
A
,
B
,
L
,
K
A,B,L,K
A
,
B
,
L
,
K
are concyclic.
Sum of divisors
n
n
n
- some natural. We write on the board all such numbers
d
d
d
, that
d
≤
1000
d\leq 1000
d
≤
1000
and
d
∣
n
+
k
d|n+k
d
∣
n
+
k
for some
1
≤
k
≤
1000
1\leq k \leq 1000
1
≤
k
≤
1000
. Let
S
(
n
)
S(n)
S
(
n
)
-sum of all written numbers. Prove , that
S
(
n
)
<
1
0
6
S(n)<10^6
S
(
n
)
<
1
0
6
and
S
(
n
)
>
1
0
6
S(n)>10^6
S
(
n
)
>
1
0
6
has infinitely many solutions.
GCD and LCM
a
,
b
a,b
a
,
b
are naturals and
a
×
G
C
D
(
a
,
b
)
+
b
×
L
C
M
(
a
,
b
)
<
2.5
a
b
a \times GCD(a,b)+b \times LCM(a,b)<2.5 ab
a
×
GC
D
(
a
,
b
)
+
b
×
L
CM
(
a
,
b
)
<
2.5
ab
. Prove that
b
∣
a
b|a
b
∣
a
1
1
Hide problems
Easy problem
f
(
x
)
,
g
(
x
)
f(x),g(x)
f
(
x
)
,
g
(
x
)
- two square trinomials and
a
,
b
,
c
,
d
a,b,c,d
a
,
b
,
c
,
d
- some reals.
f
(
a
)
=
2
,
f
(
b
)
=
3
,
f
(
c
)
=
7
,
f
(
d
)
=
10
f(a)=2,f(b)=3,f(c)=7,f(d)=10
f
(
a
)
=
2
,
f
(
b
)
=
3
,
f
(
c
)
=
7
,
f
(
d
)
=
10
and
g
(
a
)
=
16
,
g
(
b
)
=
15
,
g
(
c
)
=
11
g(a)=16,g(b)=15,g(c)=11
g
(
a
)
=
16
,
g
(
b
)
=
15
,
g
(
c
)
=
11
Find
g
(
d
)
g(d)
g
(
d
)