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Problems
Contests
International Contests
Tuymaada Olympiad
2013 Tuymaada Olympiad
2013 Tuymaada Olympiad
Part of
Tuymaada Olympiad
Subcontests
(8)
7
1
Hide problems
stereometric inequality
Points
A
1
A_1
A
1
,
A
2
A_2
A
2
,
A
3
A_3
A
3
,
A
4
A_4
A
4
are the vertices of a regular tetrahedron of edge length
1
1
1
. The points
B
1
B_1
B
1
and
B
2
B_2
B
2
lie inside the figure bounded by the plane
A
1
A
2
A
3
A_1A_2A_3
A
1
A
2
A
3
and the spheres of radius
1
1
1
and centres
A
1
A_1
A
1
,
A
2
A_2
A
2
,
A
3
A_3
A
3
. Prove that
B
1
B
2
<
max
{
B
1
A
1
,
B
1
A
2
,
B
1
A
3
,
B
1
A
4
}
B_1B_2 < \max\{B_1A_1, B_1A_2, B_1A_3, B_1A_4\}
B
1
B
2
<
max
{
B
1
A
1
,
B
1
A
2
,
B
1
A
3
,
B
1
A
4
}
. A. Kupavsky
8
2
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divisibility the number of ways
Cards numbered from 1 to
2
n
2^n
2
n
are distributed among
k
k
k
children,
1
≤
k
≤
2
n
1\leq k\leq 2^n
1
≤
k
≤
2
n
, so that each child gets at least one card. Prove that the number of ways to do that is divisible by
2
k
−
1
2^{k-1}
2
k
−
1
but not by
2
k
2^k
2
k
. M. Ivanov
inequality with areas
The point
A
1
A_1
A
1
on the perimeter of a convex quadrilateral
A
B
C
D
ABCD
A
BC
D
is such that the line
A
A
1
AA_1
A
A
1
divides the quadrilateral into two parts of equal area. The points
B
1
B_1
B
1
,
C
1
C_1
C
1
,
D
1
D_1
D
1
are defined similarly. Prove that the area of the quadrilateral
A
1
B
1
C
1
D
1
A_1B_1C_1D_1
A
1
B
1
C
1
D
1
is greater than a quarter of the area of
A
B
C
D
ABCD
A
BC
D
. L. Emelyanov
4
1
Hide problems
Three positive variables with product 1
Prove that if
x
x
x
,
y
y
y
,
z
z
z
are positive real numbers and
x
y
z
=
1
xyz = 1
x
yz
=
1
then
x
3
x
2
+
y
+
y
3
y
2
+
z
+
z
3
z
2
+
x
≥
3
2
.
\frac{x^3}{x^2+y}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+x}\geq \dfrac {3} {2}.
x
2
+
y
x
3
+
y
2
+
z
y
3
+
z
2
+
x
z
3
≥
2
3
.
A. Golovanov
5
2
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Representation of a degree 4 polynomial
Prove that every polynomial of fourth degree can be represented in the form
P
(
Q
(
x
)
)
+
R
(
S
(
x
)
)
P(Q(x))+R(S(x))
P
(
Q
(
x
))
+
R
(
S
(
x
))
, where
P
,
Q
,
R
,
S
P,Q,R,S
P
,
Q
,
R
,
S
are quadratic trinomials.A. GolovanovEDIT. It is confirmed that assuming the coefficients to be real, while solving the problem, earned a maximum score.
Colouring a cube of side 7
Each face of a
7
×
7
×
7
7 \times 7 \times 7
7
×
7
×
7
cube is divided into unit squares. What is the maximum number of squares that can be chosen so that no two chosen squares have a common point?A. Chukhnov
6
2
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A Diophantine equation in prime numbers
Solve the equation
p
2
−
p
q
−
q
3
=
1
p^2-pq-q^3=1
p
2
−
pq
−
q
3
=
1
in prime numbers.A. Golovanov
Quadratic trinomials in an array
Quadratic trinomials with positive leading coefficients are arranged in the squares of a
6
×
6
6 \times 6
6
×
6
table. Their
108
108
108
coefficients are all integers from
−
60
-60
−
60
to
47
47
47
(each number is used once). Prove that at least in one column the sum of all trinomials has a real root.K. Kokhas & F. Petrov
2
2
Hide problems
Equal angles in a rhombus
Points
X
X
X
and
Y
Y
Y
inside the rhombus
A
B
C
D
ABCD
A
BC
D
are such that
Y
Y
Y
is inside the convex quadrilateral
B
X
D
C
BXDC
BX
D
C
and
2
∠
X
B
Y
=
2
∠
X
D
Y
=
∠
A
B
C
2\angle XBY = 2\angle XDY = \angle ABC
2∠
XB
Y
=
2∠
X
D
Y
=
∠
A
BC
. Prove that the lines
A
X
AX
A
X
and
C
Y
CY
C
Y
are parallel.S. Berlov
Metric relation in a hexagon
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
is a convex hexagon, such that in it
A
C
∥
D
F
AC \parallel DF
A
C
∥
D
F
,
B
D
∥
A
E
BD \parallel AE
B
D
∥
A
E
and
C
E
∥
B
F
CE \parallel BF
CE
∥
BF
. Prove that
A
B
2
+
C
D
2
+
E
F
2
=
B
C
2
+
D
E
2
+
A
F
2
.
AB^2+CD^2+EF^2=BC^2+DE^2+AF^2.
A
B
2
+
C
D
2
+
E
F
2
=
B
C
2
+
D
E
2
+
A
F
2
.
N. Sedrakyan
1
1
Hide problems
100 heaps of stones
100
100
100
heaps of stones lie on a table. Two players make moves in turn. At each move, a player can remove any non-zero number of stones from the table, so that at least one heap is left untouched. The player that cannot move loses. Determine, for each initial position, which of the players, the first or the second, has a winning strategy.K. KokhasEDIT. It is indeed confirmed by the sender that empty heaps are still heaps, so the third post contains the right guess of an interpretation.
3
2
Hide problems
Connected, not n-colourable graph
The vertices of a connected graph cannot be coloured with less than
n
+
1
n+1
n
+
1
colours (so that adjacent vertices have different colours). Prove that
n
(
n
−
1
)
2
\dfrac{n(n-1)}{2}
2
n
(
n
−
1
)
edges can be removed from the graph so that it remains connected.V. DolnikovEDIT. It is confirmed by the official solution that the graph is tacitly assumed to be finite.
Inequality involving GM, QM, HM
For every positive real numbers
a
a
a
and
b
b
b
prove the inequality
a
b
≤
1
3
a
2
+
b
2
2
+
2
3
2
1
a
+
1
b
.
\displaystyle \sqrt{ab} \leq \dfrac{1}{3} \sqrt{\dfrac{a^2+b^2}{2}}+\dfrac{2}{3} \dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}.
ab
≤
3
1
2
a
2
+
b
2
+
3
2
a
1
+
b
1
2
.
A. Khabrov