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Cyprus Team Selection Test
2018 Cyprus IMO TST
2018 Cyprus IMO TST
Part of
Cyprus Team Selection Test
Subcontests
(5)
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1
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Cyprus IMO TST 2018
[url=https://artofproblemsolving.com/community/c677808]Cyprus IMO TST 2018[url=https://artofproblemsolving.com/community/c6h1666662p10591751]Problem 1. Determine all integers
n
≥
2
n \geq 2
n
≥
2
for which the number
11111
11111
11111
in base
n
n
n
is a perfect square. [url=https://artofproblemsolving.com/community/c6h1666663p10591753]Problem 2. Consider a trapezium
A
B
Γ
Δ
AB \Gamma \Delta
A
B
ΓΔ
, where
A
Δ
∥
B
Γ
A\Delta \parallel B\Gamma
A
Δ
∥
B
Γ
and
∡
A
=
12
0
∘
\measuredangle A = 120^{\circ}
∡
A
=
12
0
∘
. Let
E
E
E
be the midpoint of
A
B
AB
A
B
and let
O
1
O_1
O
1
and
O
2
O_2
O
2
be the circumcenters of triangles
A
E
Δ
AE \Delta
A
E
Δ
and
B
E
Γ
BE\Gamma
BE
Γ
, respectively. Prove that the area of the trapezium is equal to six time the area of the triangle
O
1
E
O
2
O_1 E O_2
O
1
E
O
2
. [url=https://artofproblemsolving.com/community/c6h1666660p10591747]Problem 3. Find all triples
(
α
,
β
,
γ
)
(\alpha, \beta, \gamma)
(
α
,
β
,
γ
)
of positive real numbers for which the expression
K
=
α
+
3
γ
α
+
2
β
+
γ
+
4
β
α
+
β
+
2
γ
−
8
γ
α
+
β
+
3
γ
K = \frac{\alpha+3 \gamma}{\alpha + 2\beta + \gamma} + \frac{4\beta}{\alpha+\beta+2\gamma} - \frac{8 \gamma}{\alpha+ \beta + 3\gamma}
K
=
α
+
2
β
+
γ
α
+
3
γ
+
α
+
β
+
2
γ
4
β
−
α
+
β
+
3
γ
8
γ
obtains its minimum value.[url=https://artofproblemsolving.com/community/c6h1666661p10591749]Problem 4. Let
Λ
=
{
1
,
2
,
…
,
2
v
−
1
,
2
v
}
\Lambda= \{1, 2, \ldots, 2v-1,2v\}
Λ
=
{
1
,
2
,
…
,
2
v
−
1
,
2
v
}
and
P
=
{
α
1
,
α
2
,
…
,
α
2
v
−
1
,
α
2
v
}
P=\{\alpha_1, \alpha_2, \ldots, \alpha_{2v-1}, \alpha_{2v}\}
P
=
{
α
1
,
α
2
,
…
,
α
2
v
−
1
,
α
2
v
}
be a permutation of the elements of
Λ
\Lambda
Λ
.(a) Prove that
∑
i
=
1
v
α
2
i
−
1
α
2
i
≤
∑
i
=
1
v
(
2
i
−
1
)
2
i
.
\sum_{i=1}^v \alpha_{2i-1}\alpha_{2i} \leq \sum_{i=1}^v (2i-1)2i.
i
=
1
∑
v
α
2
i
−
1
α
2
i
≤
i
=
1
∑
v
(
2
i
−
1
)
2
i
.
(b) Determine the largest positive integer
m
m
m
such that we can partition the
m
×
m
m\times m
m
×
m
square into
7
7
7
rectangles for which every pair of them has no common interior points and their lengths and widths form the following sequence:
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
,
11
,
12
,
13
,
14.
1,2,3,4,5,6,7,8,9,10,11,12,13,14.
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
,
11
,
12
,
13
,
14.
2
1
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Area of trapezium is six times that of triangle [Cyprus IMO TST 2018]
Consider a trapezium
A
B
Γ
Δ
AB \Gamma \Delta
A
B
ΓΔ
, where
A
Δ
∥
B
Γ
A\Delta \parallel B\Gamma
A
Δ
∥
B
Γ
and
∡
A
=
12
0
∘
\measuredangle A = 120^{\circ}
∡
A
=
12
0
∘
. Let
E
E
E
be the midpoint of
A
B
AB
A
B
and let
O
1
O_1
O
1
and
O
2
O_2
O
2
be the circumcenters of triangles
A
E
Δ
AE \Delta
A
E
Δ
and
B
E
Γ
BE\Gamma
BE
Γ
, respectively. Prove that the area of the trapezium is equal to six time the area of the triangle
O
1
E
O
2
O_1 E O_2
O
1
E
O
2
.
1
1
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All n s.t. 11111 is a perfect square in base n [Cyprus IMO TST 2018]
Determine all integers
n
≥
2
n \geq 2
n
≥
2
for which the number
11111
11111
11111
in base
n
n
n
is a perfect square.
4
1
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Inequality on 2v elements [Cyprus IMO TST 2018]
Let
Λ
=
{
1
,
2
,
…
,
2
v
−
1
,
2
v
}
\Lambda= \{1, 2, \ldots, 2v-1,2v\}
Λ
=
{
1
,
2
,
…
,
2
v
−
1
,
2
v
}
and
P
=
{
α
1
,
α
2
,
…
,
α
2
v
−
1
,
α
2
v
}
P=\{\alpha_1, \alpha_2, \ldots, \alpha_{2v-1}, \alpha_{2v}\}
P
=
{
α
1
,
α
2
,
…
,
α
2
v
−
1
,
α
2
v
}
be a permutation of the elements of
Λ
\Lambda
Λ
.(a) Prove that
∑
i
=
1
v
α
2
i
−
1
α
2
i
≤
∑
i
=
1
v
(
2
i
−
1
)
2
i
.
\sum_{i=1}^v \alpha_{2i-1}\alpha_{2i} \leq \sum_{i=1}^v (2i-1)2i.
i
=
1
∑
v
α
2
i
−
1
α
2
i
≤
i
=
1
∑
v
(
2
i
−
1
)
2
i
.
(b) Determine the largest positive integer
m
m
m
such that we can partition the
m
×
m
m\times m
m
×
m
square into
7
7
7
rectangles for which every pair of them has no common interior points and their lengths and widths form the following sequence:
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
,
11
,
12
,
13
,
14.
1,2,3,4,5,6,7,8,9,10,11,12,13,14.
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
,
11
,
12
,
13
,
14.
3
1
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Weird sum obtains min value [Cyprus IMO TST 2018]
Find all triples
(
α
,
β
,
γ
)
(\alpha, \beta, \gamma)
(
α
,
β
,
γ
)
of positive real numbers for which the expression
K
=
α
+
3
γ
α
+
2
β
+
γ
+
4
β
α
+
β
+
2
γ
−
8
γ
α
+
β
+
3
γ
K = \frac{\alpha+3 \gamma}{\alpha + 2\beta + \gamma} + \frac{4\beta}{\alpha+\beta+2\gamma} - \frac{8 \gamma}{\alpha+ \beta + 3\gamma}
K
=
α
+
2
β
+
γ
α
+
3
γ
+
α
+
β
+
2
γ
4
β
−
α
+
β
+
3
γ
8
γ
obtains its minimum value.