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Problems
Contests
International Contests
Tuymaada Olympiad
2002 Tuymaada Olympiad
2002 Tuymaada Olympiad
Part of
Tuymaada Olympiad
Subcontests
(7)
8
1
Hide problems
area of a triangle is 1/4 area of another triangle iff <A=60^o
The circle with the center of
O
O
O
touches the sides of the angle
A
A
A
at the points of
K
K
K
and
M
M
M
. The tangent to the circle intersects the segments
A
K
AK
A
K
and
A
M
AM
A
M
at points
B
B
B
and
C
C
C
respectively, and the line
K
M
KM
K
M
intersects the segments
O
B
OB
OB
and
O
C
OC
OC
at the points
D
D
D
and
E
E
E
. Prove that the area of the triangle
O
D
E
ODE
O
D
E
is equal to a quarter of the area of a triangle
B
O
C
BOC
BOC
if and only if the angle
A
A
A
is
6
0
∘
60^\circ
6
0
∘
.
6
1
Hide problems
changing numbers in a 100x100 table
In the cells of the table
100
×
100
100 \times100
100
×
100
are placed in pairs different numbers. Every minute each of the numbers changes to the largest of the numbers in the adjacent cells on the side. Can after
4
4
4
hours all the numbers in the table be the same?
5
1
Hide problems
5(x^2+ y^2) ^2 \leq 4 + (x +y) ^4 for x,y in [0,1]
Prove that for all
x
,
y
∈
x, y \in
x
,
y
∈
0
,
1
0, 1
0
,
1
the inequality
5
(
x
2
+
y
2
)
2
≤
4
+
(
x
+
y
)
4
5 (x^2+ y^2) ^2 \leq 4 + (x +y) ^4
5
(
x
2
+
y
2
)
2
≤
4
+
(
x
+
y
)
4
holds.
3
3
Hide problems
Circle + triangle = hexagon ==> two parallel lines.
A circle having common centre with the circumcircle of triangle
A
B
C
ABC
A
BC
meets the sides of the triangle at six points forming convex hexagon
A
1
A
2
B
1
B
2
C
1
C
2
A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}
A
1
A
2
B
1
B
2
C
1
C
2
(
A
1
A_{1}
A
1
and
A
2
A_{2}
A
2
lie on
B
C
BC
BC
,
B
1
B_{1}
B
1
and
B
2
B_{2}
B
2
lie on
A
C
AC
A
C
,
C
1
C_{1}
C
1
and
C
2
C_{2}
C
2
lie on
A
B
AB
A
B
). If
A
1
B
1
A_{1}B_{1}
A
1
B
1
is parallel to the bisector of angle
B
B
B
, prove that
A
2
C
2
A_{2}C_{2}
A
2
C
2
is parallel to the bisector of angle
C
C
C
.Proposed by S. Berlov
trinomial with integer coefficients, and powers of two as natural values
Is there a quadratic trinomial with integer coefficients, such that all values which are natural to be natural powers of two?
Acute triangle, circumcircle, isoceles triangle, altitudes.
The points
D
D
D
and
E
E
E
on the circumcircle of an acute triangle
A
B
C
ABC
A
BC
are such that
A
D
=
A
E
=
B
C
AD=AE = BC
A
D
=
A
E
=
BC
. Let
H
H
H
be the common point of the altitudes of triangle
A
B
C
ABC
A
BC
. It is known that
A
H
2
=
B
H
2
+
C
H
2
AH^{2}=BH^{2}+CH^{2}
A
H
2
=
B
H
2
+
C
H
2
. Prove that
H
H
H
lies on the segment
D
E
DE
D
E
.Proposed by D. Shiryaev
4
2
Hide problems
1x2 tiling of a 2001x2002 table.
A rectangular table with 2001 rows and 2002 columns is partitioned into
1
×
2
1\times 2
1
×
2
rectangles. It is known that any other partition of the table into
1
×
2
1\times 2
1
×
2
rectangles contains a rectangle belonging to the original partition. Prove that the original partition contains two successive columns covered by 2001 horizontal rectangles.Proposed by S. Volchenkov
Tuymaada 2002
A real number
a
a
a
is given. The sequence
n
1
<
n
2
<
n
3
<
.
.
.
n_{1}< n_{2}< n_{3}< ...
n
1
<
n
2
<
n
3
<
...
consists of all the positive integral
n
n
n
such that
{
n
a
}
<
1
10
\{na\}< \frac{1}{10}
{
na
}
<
10
1
. Prove that there are at most three different numbers among the numbers
n
2
−
n
1
n_{2}-n_{1}
n
2
−
n
1
,
n
3
−
n
2
n_{3}-n_{2}
n
3
−
n
2
,
n
4
−
n
3
n_{4}-n_{3}
n
4
−
n
3
,
…
\ldots
…
.A corollary of a theorem from ergodic theory
1
2
Hide problems
One hexagon and 18 segments.
Each of the points
G
G
G
and
H
H
H
lying from different sides of the plane of hexagon
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
is connected with all vertices of the hexagon. Is it possible to mark 18 segments thus formed by the numbers
1
,
2
,
3
,
…
,
18
1, 2, 3, \ldots, 18
1
,
2
,
3
,
…
,
18
and arrange some real numbers at points
A
,
B
,
C
,
D
,
E
,
F
,
G
,
H
A, B, C, D, E, F, G, H
A
,
B
,
C
,
D
,
E
,
F
,
G
,
H
so that each segment is marked with the difference of the numbers at its ends?Proposed by A. Golovanov
sequence of primes build recursively
A positive integer
c
c
c
is given. The sequence
{
p
k
}
\{p_{k}\}
{
p
k
}
is constructed by the following rule:
p
1
p_{1}
p
1
is arbitrary prime and for
k
≥
1
k\geq 1
k
≥
1
the number
p
k
+
1
p_{k+1}
p
k
+
1
is any prime divisor of
p
k
+
c
p_{k}+c
p
k
+
c
not present among the numbers
p
1
p_{1}
p
1
,
p
2
p_{2}
p
2
,
…
\dots
…
,
p
k
p_{k}
p
k
. Prove that the sequence
{
p
k
}
\{p_{k}\}
{
p
k
}
cannot be infinite.Proposed by A. Golovanov
2
3
Hide problems
Four variables plus a restraint
Let
a
,
b
,
c
,
d
a,b,c,d
a
,
b
,
c
,
d
be positive real numbers such that
a
b
c
d
=
1
abcd=1
ab
c
d
=
1
. Prove that
1
+
a
b
1
+
a
+
1
+
b
c
1
+
b
+
1
+
c
d
1
+
c
+
1
+
d
a
1
+
d
≥
4.
\frac{1+ab}{1+a} + \frac{1+bc}{1+b} + \frac{1+cd}{1+c} + \frac{1+da}{1+d} \geq 4 .
1
+
a
1
+
ab
+
1
+
b
1
+
b
c
+
1
+
c
1
+
c
d
+
1
+
d
1
+
d
a
≥
4.
Proposed by A. Khrabrov
Double functional inequation : f(3x-2) <= f(x) <= f(2x-1).
Find all the functions
f
(
x
)
,
f(x),
f
(
x
)
,
continuous on the whole real axis, such that for every real
x
x
x
f
(
3
x
−
2
)
≤
f
(
x
)
≤
f
(
2
x
−
1
)
.
f(3x-2)\leq f(x)\leq f(2x-1).
f
(
3
x
−
2
)
≤
f
(
x
)
≤
f
(
2
x
−
1
)
.
Proposed by A. Golovanov
if the inscribed one is equilateral then the other is also equilateral
Points on the sides
B
C
BC
BC
,
C
A
CA
C
A
and
A
B
AB
A
B
of the triangle
A
B
C
ABC
A
BC
are respectively
A
1
A_1
A
1
,
B
1
B_1
B
1
and
C
1
C_1
C
1
such that
A
C
1
:
C
1
B
=
B
A
1
:
A
1
C
=
C
B
1
:
B
1
A
=
2
:
1
AC_1: C_1B = BA_1: A_1C = CB_1: B_1A = 2: 1
A
C
1
:
C
1
B
=
B
A
1
:
A
1
C
=
C
B
1
:
B
1
A
=
2
:
1
. Prove that if triangle
A
1
B
1
C
1
A_1B_1C_1
A
1
B
1
C
1
is equilateral, then triangle
A
B
C
ABC
A
BC
is also equilateral.