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Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch Mathematical Olympiad
2011 Dutch Mathematical Olympiad
2011 Dutch Mathematical Olympiad
Part of
Dutch Mathematical Olympiad
Subcontests
(5)
3
1
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6 consecutive numbers, scores of a 6-team tournament
In a tournament among six teams, every team plays against each other team exactly once. When a team wins, it receives
3
3
3
points and the losing team receives
0
0
0
points. If the game is a draw, the two teams receive
1
1
1
point each. Can the final scores of the six teams be six consecutive numbers
a
,
a
+
1
,
.
.
.
,
a
+
5
a,a +1,...,a + 5
a
,
a
+
1
,
...
,
a
+
5
? If so, determine all values of
a
a
a
for which this is possible.
4
1
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2x2 system a\sqrt{a}+ b\sqrt{b} = 134 , a\sqrt{b}+ b\sqrt{a} = 126
Determine all pairs of positive real numbers
(
a
,
b
)
(a, b)
(
a
,
b
)
with
a
>
b
a > b
a
>
b
that satisfy the following equations:
a
a
+
b
b
=
134
a\sqrt{a}+ b\sqrt{b} = 134
a
a
+
b
b
=
134
and
a
b
+
b
a
=
126
a\sqrt{b}+ b\sqrt{a} = 126
a
b
+
b
a
=
126
.
5
1
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all numbers coloured bw, if a,b white then a-b and a+b have diff. colours
The number devil has coloured the integer numbers: every integer is coloured either black or white. The number
1
1
1
is coloured white. For every two white numbers
a
a
a
and
b
b
b
(
a
a
a
and
b
b
b
are allowed to be equal) the numbers
a
−
b
a-b
a
−
b
and
a
+
a +
a
+
b have di fferent colours. Prove that
2011
2011
2011
is coloured white.
1
1
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diophantine with factorials, a! + b! = 2^n
Determine all triples of positive integers
(
a
,
b
,
n
)
(a, b, n)
(
a
,
b
,
n
)
that satisfy the following equation:
a
!
+
b
!
=
2
n
a! + b! = 2^n
a
!
+
b
!
=
2
n
2
1
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6 concyclic points, sides' trisectors, lead to equilateral triangle
Let
A
B
C
ABC
A
BC
be a triangle. Points
P
P
P
and
Q
Q
Q
lie on side
B
C
BC
BC
and satisfy
∣
B
P
∣
=
∣
P
Q
∣
=
∣
Q
C
∣
=
1
3
∣
B
C
∣
|BP| =|PQ| = |QC| = \frac13 |BC|
∣
BP
∣
=
∣
PQ
∣
=
∣
QC
∣
=
3
1
∣
BC
∣
. Points
R
R
R
and
S
S
S
lie on side
C
A
CA
C
A
and satisfy
∣
C
R
∣
=
∣
R
S
∣
=
∣
S
A
∣
=
13
∣
C
A
∣
|CR| =|RS| = |SA| = 1 3 |CA|
∣
CR
∣
=
∣
RS
∣
=
∣
S
A
∣
=
13∣
C
A
∣
. Finally, points
T
T
T
and
U
U
U
lie on side
A
B
AB
A
B
and satisfy
∣
A
T
∣
=
∣
T
U
∣
=
∣
U
B
∣
=
1
3
∣
A
B
∣
|AT| = |TU| = |UB| =\frac13 |AB|
∣
A
T
∣
=
∣
T
U
∣
=
∣
U
B
∣
=
3
1
∣
A
B
∣
. Points
P
,
Q
,
R
,
S
,
T
P, Q,R, S, T
P
,
Q
,
R
,
S
,
T
and
U
U
U
turn out to lie on a common circle. Prove that
A
B
C
ABC
A
BC
is an equilateral triangle.