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Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch BxMO/EGMO TST
2019 Dutch BxMO TST
2019 Dutch BxMO TST
Part of
Dutch BxMO/EGMO TST
Subcontests
(5)
5
1
Hide problems
2018 cities, some of which are connected by roads
In a country, there are
2018
2018
2018
cities, some of which are connected by roads. Each city is connected to at least three other cities. It is possible to travel from any city to any other city using one or more roads. For each pair of cities, consider the shortest route between these two cities. What is the greatest number of roads that can be on such a shortest route?
4
1
Hide problems
a_n = gcd(a_{n+k}, a_{n+k+1}) for all n, sequence a_n
Do there exist a positive integer
k
k
k
and a non-constant sequence
a
1
,
a
2
,
a
3
,
.
.
.
a_1, a_2, a_3, ...
a
1
,
a
2
,
a
3
,
...
of positive integers such that
a
n
=
g
c
d
(
a
n
+
k
,
a
n
+
k
+
1
)
a_n = gcd(a_{n+k}, a_{n+k+1})
a
n
=
g
c
d
(
a
n
+
k
,
a
n
+
k
+
1
)
for all positive integers
n
n
n
?
3
1
Hide problems
(a)x^3 - y^3 >= 4x => x^2 > 2y (b) x^5 - y^3 >= 2x => $x^3 >= 2y
Let
x
x
x
and
y
y
y
be positive real numbers. 1. Prove: if
x
3
−
y
3
≥
4
x
x^3 - y^3 \ge 4x
x
3
−
y
3
≥
4
x
, then
x
2
>
2
y
x^2 > 2y
x
2
>
2
y
. 2. Prove: if
x
5
−
y
3
≥
2
x
x^5 - y^3 \ge 2x
x
5
−
y
3
≥
2
x
, then
x
3
≥
2
y
x^3 \ge 2y
x
3
≥
2
y
.
1
1
Hide problems
a^2 + b = n, a+b = 2^k for each positive integer k
Prove that for each positive integer
n
n
n
there are at most two pairs
(
a
,
b
)
(a, b)
(
a
,
b
)
of positive integers with following two properties: (i)
a
2
+
b
=
n
a^2 + b = n
a
2
+
b
=
n
, (ii)
a
+
b
a+b
a
+
b
is a power of two, i.e. there is an integer
k
≥
0
k \ge 0
k
≥
0
such that
a
+
b
=
2
k
a+b = 2^k
a
+
b
=
2
k
.
2
1
Hide problems
Angle Bisector at intersection of circumcircles
Let
Δ
A
B
C
\Delta ABC
Δ
A
BC
be a triangle with an inscribed circle centered at
I
I
I
. The line perpendicular to
A
I
AI
A
I
at
I
I
I
intersects
⊙
(
A
B
C
)
\odot (ABC)
⊙
(
A
BC
)
at
P
,
Q
P,Q
P
,
Q
such that,
P
P
P
lies closer to
B
B
B
than
C
C
C
. Let
⊙
(
B
I
P
)
∩
⊙
(
C
I
Q
)
=
S
\odot (BIP) \cap \odot (CIQ) =S
⊙
(
B
I
P
)
∩
⊙
(
C
I
Q
)
=
S
. Prove that,
S
I
SI
S
I
is the angle bisector of
∠
P
S
Q
\angle PSQ
∠
PSQ