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Problems
Contests
International Contests
International Zhautykov Olympiad
2018 International Zhautykov Olympiad
2018 International Zhautykov Olympiad
Part of
International Zhautykov Olympiad
Subcontests
(6)
6
1
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izho2018 p6 (geometry)
In a circle with a radius
R
R
R
a convex hexagon is inscribed. The diagonals
A
D
AD
A
D
and
B
E
BE
BE
,
B
E
BE
BE
and
C
F
CF
CF
,
C
F
CF
CF
and
A
D
AD
A
D
of the hexagon intersect at the points
M
M
M
,
N
N
N
and
K
K
K
, respectively. Let
r
1
,
r
2
,
r
3
,
r
4
,
r
5
,
r
6
r_1,r_2,r_3,r_4,r_5,r_6
r
1
,
r
2
,
r
3
,
r
4
,
r
5
,
r
6
be the radii of circles inscribed in triangles
A
B
M
,
B
C
N
,
C
D
K
,
D
E
M
,
E
F
N
,
A
F
K
ABM,BCN,CDK,DEM,EFN,AFK
A
BM
,
BCN
,
C
DK
,
D
EM
,
EFN
,
A
F
K
respectively. Prove that.
r
1
+
r
2
+
r
3
+
r
4
+
r
5
+
r
6
≤
R
3
r_1+r_2+r_3+r_4+r_5+r_6\leq R\sqrt{3}
r
1
+
r
2
+
r
3
+
r
4
+
r
5
+
r
6
≤
R
3
.
5
1
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izho2018 P5 (algebra)
Find all real numbers
a
a
a
such that there exist
f
:
R
→
R
f:\mathbb{R} \to \mathbb{R}
f
:
R
→
R
with
f
(
x
−
f
(
y
)
)
=
f
(
x
)
+
a
[
y
]
f(x-f(y))=f(x)+a[y]
f
(
x
−
f
(
y
))
=
f
(
x
)
+
a
[
y
]
for all
x
,
y
∈
R
x,y\in \mathbb{R}
x
,
y
∈
R
4
1
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izho 2018 (P4)Combinatorics
Crocodile chooses
1
1
1
x
4
4
4
tile from
2018
2018
2018
x
2018
2018
2018
square.The bear has tilometer that checks
3
3
3
x
3
3
3
square of
2018
2018
2018
x
2018
2018
2018
is there any of choosen cells by crocodile.Tilometer says "YES" if there is at least one choosen cell among checked
3
3
3
x
3
3
3
square.For what is the smallest number of such questions the Bear can certainly get an affirmative answer?
3
1
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IZHO 2018 P3 number theory
Prove that there exist infinitely pairs
(
m
,
n
)
(m,n)
(
m
,
n
)
such that
m
+
n
m+n
m
+
n
divides
(
m
!
)
n
+
(
n
!
)
m
+
1
(m!)^n+(n!)^m+1
(
m
!
)
n
+
(
n
!
)
m
+
1
2
1
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IZHO 2018 P2(geometry)
Let
N
,
K
,
L
N,K,L
N
,
K
,
L
be points on
A
B
,
B
C
,
C
A
AB,BC,CA
A
B
,
BC
,
C
A
such that
C
N
CN
CN
bisector of angle
∠
A
C
B
\angle ACB
∠
A
CB
and
A
L
=
B
K
AL=BK
A
L
=
B
K
.Let
B
L
∩
A
K
=
P
BL\cap AK=P
B
L
∩
A
K
=
P
.If
I
,
J
I,J
I
,
J
be incenters of triangles
△
B
P
K
\triangle BPK
△
BP
K
and
△
A
L
P
\triangle ALP
△
A
L
P
and
I
J
∩
C
N
=
Q
IJ\cap CN=Q
I
J
∩
CN
=
Q
prove that
I
Q
=
J
P
IQ=JP
I
Q
=
J
P
1
1
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IZHO 2018 P1(inequality)
Let
α
,
β
,
γ
\alpha,\beta,\gamma
α
,
β
,
γ
measures of angles of opposite to the sides of triangle with measures
a
,
b
,
c
a,b,c
a
,
b
,
c
respectively.Prove that
2
(
c
o
s
2
α
+
c
o
s
2
β
+
c
o
s
2
γ
)
≥
a
2
b
2
+
c
2
+
b
2
a
2
+
c
2
+
c
2
a
2
+
b
2
2(cos^2\alpha+cos^2\beta+cos^2\gamma)\geq \frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}
2
(
co
s
2
α
+
co
s
2
β
+
co
s
2
γ
)
≥
b
2
+
c
2
a
2
+
a
2
+
c
2
b
2
+
a
2
+
b
2
c
2