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Contests
National and Regional Contests
Kazakhstan Contests
Kazakhstan National Olympiad
2010 Kazakhstan National Olympiad
2010 Kazakhstan National Olympiad
Part of
Kazakhstan National Olympiad
Subcontests
(6)
2
2
Hide problems
Kazakhstan NMO 2010 grade 9 P 2
Exactly
4
n
4n
4
n
numbers in set
A
=
{
1
,
2
,
3
,
.
.
.
,
6
n
}
A= \{ 1,2,3,...,6n \}
A
=
{
1
,
2
,
3
,
...
,
6
n
}
of natural numbers painted in red, all other in blue. Proved that exist
3
n
3n
3
n
consecutive natural numbers from
A
A
A
, exactly
2
n
2n
2
n
of which numbers is red.
Kazakhstan NMO 2010 10 grade P 5
On sides of convex quadrilateral
A
B
C
D
ABCD
A
BC
D
on external side constructed equilateral triangles
A
B
K
,
B
C
L
,
C
D
M
,
D
A
N
ABK, BCL, CDM, DAN
A
B
K
,
BC
L
,
C
D
M
,
D
A
N
. Let
P
,
Q
P,Q
P
,
Q
- midpoints of
B
L
,
A
N
BL, AN
B
L
,
A
N
respectively and
X
X
X
- circumcenter of
C
M
D
CMD
CM
D
. Prove, that
P
Q
PQ
PQ
perpendicular to
K
X
KX
K
X
1
2
Hide problems
Kazakhstan NMO 2010 grade 9 P 1
Triangle
A
B
C
ABC
A
BC
is given. Circle
ω
\omega
ω
passes through
B
B
B
, touch
A
C
AC
A
C
in
D
D
D
and intersect sides
A
B
AB
A
B
and
B
C
BC
BC
at
P
P
P
and
Q
Q
Q
respectively. Line
P
Q
PQ
PQ
intersect
B
D
BD
B
D
and
A
C
AC
A
C
at
M
M
M
and
N
N
N
respectively. Prove that
ω
\omega
ω
, circumcircle of
D
M
N
DMN
D
MN
and circle, touching
P
Q
PQ
PQ
in
M
M
M
and passes through B, intersects in one point.
Geometry whith ellipses( Kazakhstan NMO 2010 10 grade 1 P)
Triangle
A
B
C
ABC
A
BC
is given. Consider ellipse
Ω
1
\Omega _1
Ω
1
, passes through
C
C
C
with focuses in
A
A
A
and
B
B
B
. Similarly define ellipses
Ω
2
,
Ω
3
\Omega _2 , \Omega _3
Ω
2
,
Ω
3
with focuses
B
,
C
B,C
B
,
C
and
C
,
A
C,A
C
,
A
respectively. Prove, that if all ellipses have common point
D
D
D
then
A
,
B
,
C
,
D
A,B,C,D
A
,
B
,
C
,
D
lies on the circle.Ellipse with focuses
X
,
Y
X,Y
X
,
Y
, passes through
Z
Z
Z
- locus of point
T
T
T
, such that
X
T
+
Y
T
=
X
Z
+
Y
Z
XT+YT=XZ+YZ
XT
+
Y
T
=
XZ
+
Y
Z
3
2
Hide problems
Kazakhstan NMO 2010 grade 9 P 3
Positive real
A
A
A
is given. Find maximum value of
M
M
M
for which inequality
1
x
+
1
y
+
A
x
+
y
≥
M
x
y
\frac{1}{x}+\frac{1}{y}+\frac{A}{x+y} \geq \frac{M}{\sqrt{xy}}
x
1
+
y
1
+
x
+
y
A
≥
x
y
M
holds for all
x
,
y
>
0
x, y>0
x
,
y
>
0
Number of year CNT 6P Kazakhstan NMO 2010 11 grade
Call
A
∈
N
0
A \in \mathbb{N}^0
A
∈
N
0
be
n
u
m
b
e
r
o
f
y
e
a
r
number of year
n
u
mb
ero
f
ye
a
r
if all digits of
A
A
A
equals
0
0
0
,
1
1
1
or
2
2
2
(in decimal representation). Prove that exist infinity
N
∈
N
N \in \mathbb{N}
N
∈
N
, such that
N
N
N
can't presented as
A
2
+
B
A^2+B
A
2
+
B
where
A
∈
N
0
;
B
A \in \mathbb{N}^0 ; B
A
∈
N
0
;
B
-
n
u
m
b
e
r
o
f
y
e
a
r
number of year
n
u
mb
ero
f
ye
a
r
.
4
3
Hide problems
Kazakhstan NMO 2010 grade 9 P 4
Let
x
x
x
- minimal root of equation
x
2
−
4
x
+
2
=
0
x^2-4x+2=0
x
2
−
4
x
+
2
=
0
. Find two first digits of number
{
x
+
x
2
+
.
.
.
.
+
x
20
}
\{x+x^2+....+x^{20} \}
{
x
+
x
2
+
....
+
x
20
}
after
0
0
0
, where
{
a
}
\{a\}
{
a
}
- fractional part of
a
a
a
.
Kazakhstan NMO 2010 11 grade 1P
It is given that for some
n
∈
N
n \in \mathbb{N}
n
∈
N
there exists a natural number
a
a
a
, such that
a
n
−
1
≡
1
(
m
o
d
n
)
a^{n-1} \equiv 1 \pmod{n}
a
n
−
1
≡
1
(
mod
n
)
and that for any prime divisor
p
p
p
of
n
−
1
n-1
n
−
1
we have
a
n
−
1
p
≢
1
(
m
o
d
n
)
a^{\frac{n-1}{p}} \not \equiv 1 \pmod{n}
a
p
n
−
1
≡
1
(
mod
n
)
. Prove that
n
n
n
is a prime.
Kazakhstan NMO 2010 11 grade 4P
For
x
;
y
≥
0
x;y \geq 0
x
;
y
≥
0
prove the inequality:
x
2
−
x
+
1
y
2
−
y
+
1
+
x
2
+
x
+
1
y
2
+
y
+
1
≥
2
(
x
+
y
)
\sqrt{x^2-x+1} \sqrt{y^2-y+1}+ \sqrt{x^2+x+1} \sqrt{y^2+y+1} \geq 2(x+y)
x
2
−
x
+
1
y
2
−
y
+
1
+
x
2
+
x
+
1
y
2
+
y
+
1
≥
2
(
x
+
y
)
6
2
Hide problems
Kazakhstan NMO 2010 grade 9 P 6
Let numbers
1
,
2
,
3
,
.
.
.
,
2010
1,2,3,...,2010
1
,
2
,
3
,
...
,
2010
stand in a row at random. Consider row, obtain by next rule: For any number we sum it and it's number in a row (For example for row
(
2
,
7
,
4
)
( 2,7,4)
(
2
,
7
,
4
)
we consider a row
(
2
+
1
;
7
+
2
;
4
+
3
)
=
(
3
;
9
;
7
)
(2+1;7+2;4+3)=(3;9;7)
(
2
+
1
;
7
+
2
;
4
+
3
)
=
(
3
;
9
;
7
)
); Proved, that in resulting row we can found two equals numbers, or two numbers, which is differ by
2010
2010
2010
Kazakhstan NMO 2010 11 grade P3
Let
A
B
C
D
ABCD
A
BC
D
be convex quadrilateral, such that exist
M
,
N
M,N
M
,
N
inside
A
B
C
D
ABCD
A
BC
D
for which
∠
N
A
D
=
∠
M
A
B
;
∠
N
B
C
=
∠
M
B
A
;
∠
M
C
B
=
∠
N
C
D
;
∠
N
D
A
=
∠
M
D
C
\angle NAD= \angle MAB; \angle NBC= \angle MBA; \angle MCB=\angle NCD; \angle NDA=\angle MDC
∠
N
A
D
=
∠
M
A
B
;
∠
NBC
=
∠
MB
A
;
∠
MCB
=
∠
NC
D
;
∠
N
D
A
=
∠
M
D
C
Prove, that
S
A
B
M
+
S
A
B
N
+
S
C
D
M
+
S
C
D
N
=
S
B
C
M
+
S
B
C
N
+
S
A
D
M
+
S
A
D
N
S_{ABM}+S_{ABN}+S_{CDM}+S_{CDN}=S_{BCM}+S_{BCN}+S_{ADM}+S_{ADN}
S
A
BM
+
S
A
BN
+
S
C
D
M
+
S
C
D
N
=
S
BCM
+
S
BCN
+
S
A
D
M
+
S
A
D
N
, where
S
X
Y
Z
S_{XYZ}
S
X
Y
Z
-area of triangle
X
Y
Z
XYZ
X
Y
Z
5
3
Hide problems
Kazakhstan NMO 2010 grade 9 P 5
Arbitrary triangle
A
B
C
ABC
A
BC
is given (with
A
B
<
B
C
AB<BC
A
B
<
BC
). Let
M
M
M
- midpoint of
A
C
AC
A
C
,
N
N
N
- midpoint of arc
A
C
AC
A
C
of circumcircle
A
B
C
ABC
A
BC
, which is contains point
B
B
B
. Let
I
I
I
- in-center of
A
B
C
ABC
A
BC
. Proved, that
∠
I
M
A
=
∠
I
N
B
\angle IMA = \angle INB
∠
I
M
A
=
∠
I
NB
Adhesion Operation - Kazakhstan NMO 2010 P2
Let
n
≥
2
n \geq 2
n
≥
2
be an integer. Define
x
i
=
1
x_i =1
x
i
=
1
or
−
1
-1
−
1
for every
i
=
1
,
2
,
3
,
⋯
,
n
i=1,2,3,\cdots, n
i
=
1
,
2
,
3
,
⋯
,
n
.Call an operation adhesion, if it changes the string
(
x
1
,
x
2
,
⋯
,
x
n
)
(x_1,x_2,\cdots,x_n)
(
x
1
,
x
2
,
⋯
,
x
n
)
to
(
x
1
x
2
,
x
2
x
3
,
⋯
,
x
n
−
1
x
n
,
x
n
x
1
)
(x_1x_2, x_2x_3, \cdots ,x_{n-1}x_n, x_nx_1)
(
x
1
x
2
,
x
2
x
3
,
⋯
,
x
n
−
1
x
n
,
x
n
x
1
)
. Find all integers
n
≥
2
n \geq 2
n
≥
2
such that the string
(
x
1
,
x
2
,
⋯
,
x
n
)
(x_1,x_2,\cdots, x_n)
(
x
1
,
x
2
,
⋯
,
x
n
)
changes to
(
1
,
1
,
⋯
,
1
)
(1,1,\cdots,1)
(
1
,
1
,
⋯
,
1
)
after finitely adhesion operations.
Kazakhstan NMO 2010 11 grade 5P
Let
O
O
O
be the circumcircle of acute triangle
A
B
C
ABC
A
BC
,
A
D
AD
A
D
-altitude of
A
B
C
ABC
A
BC
(
D
∈
B
C
D \in BC
D
∈
BC
),
A
D
∩
C
O
=
E
AD \cap CO =E
A
D
∩
CO
=
E
,
M
M
M
-midpoint of
A
E
AE
A
E
,
F
F
F
-feet of perpendicular from
C
C
C
to
A
O
AO
A
O
. Proved that point of intersection
O
M
OM
OM
and
B
C
BC
BC
lies on circumcircle of triangle
B
O
F
BOF
BOF