MathDB
Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch Mathematical Olympiad
1975 Dutch Mathematical Olympiad
1975 Dutch Mathematical Olympiad
Part of
Dutch Mathematical Olympiad
Subcontests
(5)
3
1
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double sum (t_i-t_j)^2x_ix_j <= 0 if sum x_i = 0
Given are the real numbers
x
1
,
x
2
,
.
.
.
,
x
n
x_1,x_2,...,x_n
x
1
,
x
2
,
...
,
x
n
and
t
1
,
t
2
,
.
.
.
,
t
n
t_1,t_2,...,t_n
t
1
,
t
2
,
...
,
t
n
for which holds:
∑
i
=
1
n
x
i
=
0
\sum_{i=1}^n x_i = 0
∑
i
=
1
n
x
i
=
0
. Prove that
∑
i
=
1
n
(
∑
j
=
1
n
(
t
i
−
t
j
)
2
x
i
x
j
)
≤
0.
\sum_{i=1}^n \left( \sum_{j=1}^n (t_i-t_j)^2x_ix_j \right)\le 0.
i
=
1
∑
n
(
j
=
1
∑
n
(
t
i
−
t
j
)
2
x
i
x
j
)
≤
0.
5
1
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convert any triangle into a rectangle with side 1 and area equal to the original
Describe a method to convert any triangle into a rectangle with side 1 and area equal to the original triangle by dividing that triangle into finitely many subtriangles.
4
1
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d(A,BC)/BC is rational if AB=AC and A,B,C are lattice points
Given is a rectangular plane coordinate system.(a) Prove that it is impossible to find an equilateral triangle whose vertices have integer coordinates.(b) In the plane the vertices
A
,
B
A, B
A
,
B
and
C
C
C
lie with integer coordinates in such a way that
A
B
=
A
C
AB = AC
A
B
=
A
C
. Prove that
d
(
A
,
B
C
)
B
C
\frac{d(A,BC)}{BC}
BC
d
(
A
,
BC
)
is rational.
2
1
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gcd of the elements of P, numbers of 2-digits, product of 8 consecutive
Let
T
=
{
n
∈
N
∣
T = \{n \in N|
T
=
{
n
∈
N
∣
n consists of
2
2
2
digits
}
\}
}
and
P
=
{
x
∣
x
=
n
(
n
+
1
)
.
.
.
(
n
+
7
)
;
n
,
n
+
1
,
.
.
.
,
n
+
7
∈
T
}
.
P = \{x|x = n(n + 1)... (n + 7); n,n + 1,..., n + 7 \in T\}.
P
=
{
x
∣
x
=
n
(
n
+
1
)
...
(
n
+
7
)
;
n
,
n
+
1
,
...
,
n
+
7
∈
T
}
.
Determine the gcd of the elements of
P
P
P
.
1
1
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x^7 \in Q \land x^{12} \in Q => x \in Q
Are the following statements true?
x
7
∈
Q
∧
x
12
∈
Q
⇒
x
∈
Q
x^7 \in Q \land x^{12} \in Q \Rightarrow x \in Q
x
7
∈
Q
∧
x
12
∈
Q
⇒
x
∈
Q
, and
x
9
∈
∧
x
12
∈
Q
⇒
x
∈
Q
x^9 \in \land x^{12} \in Q \Rightarrow x \in Q
x
9
∈
∧
x
12
∈
Q
⇒
x
∈
Q
.