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Problems
Contests
National and Regional Contests
Kazakhstan Contests
Kazakhstan National Olympiad
2024 Kazakhstan National Olympiad
2024 Kazakhstan National Olympiad
Part of
Kazakhstan National Olympiad
Subcontests
(6)
6
2
Hide problems
Sequence of positive integers, are they all equal 2?
An integer
m
≥
3
m\ge 3
m
≥
3
and an infinite sequence of positive integers
(
a
n
)
n
≥
1
(a_n)_{n\ge 1}
(
a
n
)
n
≥
1
satisfies the equation
a
n
+
2
=
2
a
n
+
1
m
−
1
+
a
n
m
−
1
m
−
a
n
+
1
.
a_{n+2} = 2\sqrt[m]{a_{n+1}^{m-1} + a_n^{m-1}} - a_{n+1}.
a
n
+
2
=
2
m
a
n
+
1
m
−
1
+
a
n
m
−
1
−
a
n
+
1
.
for all
n
≥
1
n\ge 1
n
≥
1
. Prove that
a
1
<
2
m
a_1 < 2^m
a
1
<
2
m
.
Perpendiculars to the harmonic lines are also harmonic lines
The circle
ω
\omega
ω
with center at point
I
I
I
inscribed in an triangle
A
B
C
ABC
A
BC
(
A
B
≠
A
C
AB\neq AC
A
B
=
A
C
) touches the sides
B
C
BC
BC
,
C
A
CA
C
A
and
A
B
AB
A
B
at points
D
D
D
,
E
E
E
and
F
F
F
, respectively. The circumcircles of triangles
A
B
C
ABC
A
BC
and
A
E
F
AEF
A
EF
intersect secondary at point
K
.
K.
K
.
The lines
E
F
EF
EF
and
A
K
AK
A
K
intersect at point
X
X
X
and intersects the line
B
C
BC
BC
at points
Y
Y
Y
and
Z
Z
Z
, respectively. The tangent lines to
ω
\omega
ω
, other than
B
C
BC
BC
, passing through points
Y
Y
Y
and
Z
Z
Z
touch
ω
\omega
ω
at points
P
P
P
and
Q
Q
Q
, respectively. Let the lines
A
P
AP
A
P
and
K
Q
KQ
K
Q
intersect at the point
R
R
R
. Prove that if
M
M
M
is a midpoint of segment
Y
Z
,
YZ,
Y
Z
,
then
I
R
⊥
X
M
IR\perp XM
I
R
⊥
XM
.
5
1
Hide problems
Center on BC
In triangle
A
B
C
ABC
A
BC
(
A
B
≠
A
C
AB\ne AC
A
B
=
A
C
), where all angles are greater than
4
5
∘
45^\circ
4
5
∘
, the altitude
A
D
AD
A
D
is drawn. Let
ω
1
\omega_1
ω
1
and
ω
2
\omega_2
ω
2
be-- circles with diameters
A
C
AC
A
C
and
A
B
AB
A
B
, respectively. The angle bisector of
∠
A
D
B
\angle ADB
∠
A
D
B
secondarily intersects
ω
1
\omega_1
ω
1
at point
P
P
P
, and the angle bisector of
∠
A
D
C
\angle ADC
∠
A
D
C
secondarily intersects
ω
2
\omega_2
ω
2
at point
Q
Q
Q
. The line
A
P
AP
A
P
intersects
ω
2
\omega_2
ω
2
at the point
R
R
R
. Prove that the circumcenter of triangle
P
Q
R
PQR
PQR
lies on line
B
C
BC
BC
.
4
2
Hide problems
Classic number theory, look at the differences
Prove that for any positive integers
a
a
a
,
b
b
b
,
c
c
c
, at least one of the numbers
a
3
b
+
1
a^3b+1
a
3
b
+
1
,
b
3
c
+
1
b^3c+1
b
3
c
+
1
,
c
3
a
+
1
c^3a+1
c
3
a
+
1
is not divisible by
a
2
+
b
2
+
c
2
a^2+b^2+c^2
a
2
+
b
2
+
c
2
.
Finite game on coordinate plane, nut's game
Players
A
A
A
and
B
B
B
play the following game on the coordinate plane. Player
A
A
A
hides a nut at one of the points with integer coordinates, and player
B
B
B
tries to find this hidden nut. In one move
B
B
B
can choose three different points with integer coordinates, then
A
A
A
tells whether these three points together with the nut's point lie on the same circle or not. Can
B
B
B
be guaranteed to find the nut in a finite number of moves?
1
2
Hide problems
Two equations with result a = bc.
Positive integers
a
,
b
,
c
a,b,c
a
,
b
,
c
satisfy the equations
a
2
=
b
3
+
a
b
a^2=b^3+ab
a
2
=
b
3
+
ab
and
c
3
=
a
+
b
+
c
c^3=a+b+c
c
3
=
a
+
b
+
c
. Prove that
a
=
b
c
a=bc
a
=
b
c
.
Similiarity of triangles
Let
A
B
C
ABC
A
BC
be an acute triangle with an altitude
A
D
AD
A
D
. Let
H
H
H
be the orthocenter of triangle
A
B
C
ABC
A
BC
. The circle
Ω
\Omega
Ω
passes through the points
A
A
A
and
B
B
B
, and touches the line
A
C
AC
A
C
. Let
B
E
BE
BE
be the diameter of
Ω
\Omega
Ω
. The lines~
B
H
BH
B
H
and
A
H
AH
A
H
intersect
Ω
\Omega
Ω
for the second time at points
K
K
K
and
L
L
L
, respectively. The lines
E
K
EK
E
K
and
A
B
AB
A
B
intersect at the point~
T
T
T
. Prove that
∠
B
D
K
=
∠
B
L
T
\angle BDK=\angle BLT
∠
B
DK
=
∠
B
L
T
.
2
2
Hide problems
Two similiar positions of Figure in chessboard.
Given an integer
n
>
1
n>1
n
>
1
. The board
n
×
n
n\times n
n
×
n
is colored white and black in a chess-like manner. We call any non-empty set of different cells of the board as a figure. We call figures
F
1
F_1
F
1
and
F
2
F_2
F
2
similar, if
F
1
F_1
F
1
can be obtained from
F
2
F_2
F
2
by a rotation with respect to the center of the board by an angle multiple of
9
0
∘
90^\circ
9
0
∘
and a parallel transfer. (Any figure is similar to itself.) We call a figure
F
F
F
connected if for any cells
a
,
b
∈
F
a,b\in F
a
,
b
∈
F
there is a sequence of cells
c
1
,
…
,
c
m
∈
F
c_1,\ldots,c_m\in F
c
1
,
…
,
c
m
∈
F
such that
c
1
=
a
c_1 = a
c
1
=
a
,
c
m
=
b
c_m = b
c
m
=
b
, and also
c
i
c_i
c
i
and
c
i
+
1
c_{i+1}
c
i
+
1
have a common side for each
1
≤
i
≤
m
−
1
1\le i\le m - 1
1
≤
i
≤
m
−
1
. Find the largest possible value of
k
k
k
such that for any connected figure
F
F
F
consisting of
k
k
k
cells, there are figures
F
1
,
F
2
F_1,F_2
F
1
,
F
2
similar to
F
F
F
such that
F
1
F_1
F
1
has more white cells than black cells and
F
2
F_2
F
2
has more black cells than white cells in it.
Euler function, degree manipulations.
Given a prime number
p
≥
3
,
p\ge 3,
p
≥
3
,
and an integer
d
≥
1
d \ge 1
d
≥
1
. Prove that there exists an integer
n
≥
1
,
n\ge 1,
n
≥
1
,
such that
gcd
(
n
,
d
)
=
1
,
\gcd(n,d) = 1,
g
cd
(
n
,
d
)
=
1
,
and the product
P
=
∏
1
≤
i
<
j
<
p
(
i
n
+
j
−
j
n
+
i
)
is not divisible by
p
n
.
P=\prod\limits_{1 \le i < j < p} {({i^{n + j}} - {j^{n + i}})} \text{ is not divisible by } p^n.
P
=
1
≤
i
<
j
<
p
∏
(
i
n
+
j
−
j
n
+
i
)
is not divisible by
p
n
.
3
2
Hide problems
Dumpty point problem
An acute triangle
A
B
C
ABC
A
BC
(
A
B
≠
A
C
AB\ne AC
A
B
=
A
C
) is inscribed in the circle
ω
\omega
ω
with center at
O
O
O
. The point
M
M
M
is the midpoint of the side
B
C
BC
BC
. The tangent line to
ω
\omega
ω
at point
A
A
A
intersects the line
B
C
BC
BC
at point
D
D
D
. A circle with center at point
M
M
M
with radius
M
A
MA
M
A
intersects the extensions of sides
A
B
AB
A
B
and
A
C
AC
A
C
at points
K
K
K
and
L
L
L
, respectively. Let
X
X
X
be such a point that
B
X
∥
K
M
BX\parallel KM
BX
∥
K
M
and
C
X
∥
L
M
CX\parallel LM
CX
∥
L
M
. Prove that the points
X
X
X
,
D
D
D
,
O
O
O
are collinear.
Wrapped, hard functional equation.
Find all functions
f
:
R
+
→
R
+
f: \mathbb R^+ \to \mathbb R^+
f
:
R
+
→
R
+
such that
f
(
x
+
f
(
x
y
)
x
)
=
f
(
x
y
)
f
(
y
+
1
y
)
f \left( x+\frac{f(xy)}{x} \right) = f(xy) f \left( y + \frac 1y \right)
f
(
x
+
x
f
(
x
y
)
)
=
f
(
x
y
)
f
(
y
+
y
1
)
holds for all
x
,
y
∈
R
+
.
x,y\in\mathbb R^+.
x
,
y
∈
R
+
.