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Problems
Contests
National and Regional Contests
Greece Contests
Greece National Olympiad
2024 Greece National Olympiad
2024 Greece National Olympiad
Part of
Greece National Olympiad
Subcontests
(4)
4
1
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Diophantine equation has too many solutions for some n
Prove that there exists an integer
n
≥
1
n \geq 1
n
≥
1
, such that number of all pairs
(
a
,
b
)
(a, b)
(
a
,
b
)
of positive integers, satisfying
1
a
−
b
−
1
a
+
1
b
=
1
n
\frac{1}{a-b}-\frac{1}{a}+\frac{1}{b}=\frac{1}{n}
a
−
b
1
−
a
1
+
b
1
=
n
1
exceeds
2024.
2024.
2024.
3
1
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Sets, differences and counting
Let
n
≥
2
n \geq 2
n
≥
2
be a positive integer and let
A
,
B
A, B
A
,
B
be two finite sets of integers such that
∣
A
∣
≤
n
|A| \leq n
∣
A
∣
≤
n
. Let
C
C
C
be a subset of the set
{
(
a
,
b
)
∣
a
∈
A
,
b
∈
B
}
\{(a, b) | a \in A, b \in B\}
{(
a
,
b
)
∣
a
∈
A
,
b
∈
B
}
. Achilles writes on a board all possible distinct differences
a
−
b
a-b
a
−
b
for
(
a
,
b
)
∈
C
(a, b) \in C
(
a
,
b
)
∈
C
and suppose that their count is
d
d
d
. He writes on another board all triplets
(
k
,
l
,
m
)
(k, l, m)
(
k
,
l
,
m
)
, where
(
k
,
l
)
,
(
k
,
m
)
∈
C
(k, l), (k, m) \in C
(
k
,
l
)
,
(
k
,
m
)
∈
C
and suppose that their count is
p
p
p
. Show that
n
p
≥
d
2
.
np \geq d^2.
n
p
≥
d
2
.
2
1
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Geometry with circle centered at arc midpoint
Let
A
B
C
ABC
A
BC
be a triangle with
A
B
<
A
C
<
B
C
AB<AC<BC
A
B
<
A
C
<
BC
with circumcircle
Γ
1
\Gamma_1
Γ
1
. The circle
Γ
2
\Gamma_2
Γ
2
has center
D
D
D
lying on
Γ
1
\Gamma_1
Γ
1
and touches
B
C
BC
BC
at
E
E
E
and the extension of
A
B
AB
A
B
at
F
F
F
. Let
Γ
1
\Gamma_1
Γ
1
and
Γ
2
\Gamma_2
Γ
2
meet at
K
,
G
K, G
K
,
G
and the line
K
G
KG
K
G
meets
E
F
EF
EF
and
C
D
CD
C
D
at
M
,
N
M, N
M
,
N
. Show that
B
C
N
M
BCNM
BCNM
is cyclic.
1
1
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Algebra with quadratic
Let
a
,
b
,
c
a, b, c
a
,
b
,
c
be reals such that some two of them have difference greater than
1
2
2
\frac{1}{2 \sqrt{2}}
2
2
1
. Prove that there exists an integer
x
x
x
, such that
x
2
−
4
(
a
+
b
+
c
)
x
+
12
(
a
b
+
b
c
+
c
a
)
<
0.
x^2-4(a+b+c)x+12(ab+bc+ca)<0.
x
2
−
4
(
a
+
b
+
c
)
x
+
12
(
ab
+
b
c
+
c
a
)
<
0.