MathDB
Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch Mathematical Olympiad
2007 Dutch Mathematical Olympiad
2007 Dutch Mathematical Olympiad
Part of
Dutch Mathematical Olympiad
Subcontests
(5)
2
1
Hide problems
partition {1, 2, 3, . . . , 32, 33} into 11 subsets of 3, under conditions
Is it possible to partition the set
A
=
{
1
,
2
,
3
,
.
.
.
,
32
,
33
}
A = \{1, 2, 3, ... , 32, 33\}
A
=
{
1
,
2
,
3
,
...
,
32
,
33
}
into eleven subsets that contain three integers each, such that for every one of these eleven subsets, one of the integers is equal to the sum of the other two? If so, give such a partition, if not, prove that such a partition cannot exist.
3
1
Hide problems
444...4443 (all ... are fours) dividible by 13 exists?
Does there exist an integer having the form
444...4443
444...4443
444...4443
(all fours, and ending with a three) that is divisible by
13
13
13
? If so, give an integer having that form that is divisible by
13
13
13
, if not, prove that such an integer cannot exist.
4
1
Hide problems
no of integers a with 1 \le a \le 100 such that a^a is a perfect square
Determine the number of integers
a
a
a
satisfying
1
≤
a
≤
100
1 \le a \le 100
1
≤
a
≤
100
such that
a
a
a^a
a
a
is a perfect square. (And prove that your answer is correct.)
5
1
Hide problems
area of hexagon independent of the position of P inside the triangle
A triangle
A
B
C
ABC
A
BC
and a point
P
P
P
inside this triangle are given. Define
D
,
E
D, E
D
,
E
and
F
F
F
as the midpoints of
A
P
,
B
P
AP, BP
A
P
,
BP
and
C
P
CP
CP
, respectively. Furthermore, let
R
R
R
be the intersection of
A
E
AE
A
E
and
B
D
,
S
BD, S
B
D
,
S
the intersection of
B
F
BF
BF
and
C
E
CE
CE
, and
T
T
T
the intersection of
C
D
CD
C
D
and
A
F
AF
A
F
. Prove that the area of hexagon
D
R
E
S
F
T
DRESFT
D
RESFT
is independent of the position of
P
P
P
inside the triangle.[asy] unitsize(1 cm);pair A, B, C, D, E, F, P, R, S, T;A = (0,0); B = (5,0); C = (1.5,4); P = (2,2); D = (A + P)/2; E = (B + P)/2; F = (C + P)/2; R = extension(A,E,B,D); S = extension(B,F,C,E); T = extension(C,D,A,F);draw(A--B--C--cycle); draw(A--P); draw(B--P); draw(C--P); draw(A--F--B); draw(B--D--C); draw(C--E--A);dot("
A
A
A
", A, SW); dot("
B
B
B
", B, SE); dot("
C
C
C
", C, N); dot("
D
D
D
", D, dir(270)); dot("
E
E
E
", E, dir(270)); dot("
F
F
F
", F, W); dot("
P
P
P
", P, dir(270)); dot("
R
R
R
", R, dir(270)); dot("
S
S
S
", S, SW); dot("
T
T
T
", T, SE); [/asy]
1
1
Hide problems
equilateral inside a circle
Consider the equilateral triangle
A
B
C
ABC
A
BC
with
∣
B
C
∣
=
∣
C
A
∣
=
∣
A
B
∣
=
1
|BC| = |CA| = |AB| = 1
∣
BC
∣
=
∣
C
A
∣
=
∣
A
B
∣
=
1
. On the extension of side
B
C
BC
BC
, we define points
A
1
A_1
A
1
(on the same side as B) and
A
2
A_2
A
2
(on the same side as C) such that
∣
A
1
B
∣
=
∣
B
C
∣
=
∣
C
A
2
∣
=
1
|A_1B| = |BC| = |CA_2| = 1
∣
A
1
B
∣
=
∣
BC
∣
=
∣
C
A
2
∣
=
1
. Similarly, we define
B
1
B_1
B
1
and
B
2
B_2
B
2
on the extension of side
C
A
CA
C
A
such that
∣
B
1
C
∣
=
∣
C
A
∣
=
∣
A
B
2
∣
=
1
|B_1C| = |CA| =|AB_2| = 1
∣
B
1
C
∣
=
∣
C
A
∣
=
∣
A
B
2
∣
=
1
, and
C
1
C_1
C
1
and
C
2
C_2
C
2
on the extension of side
A
B
AB
A
B
such that
∣
C
1
A
∣
=
∣
A
B
∣
=
∣
B
C
2
∣
=
1
|C_1A| = |AB| = |BC_2| = 1
∣
C
1
A
∣
=
∣
A
B
∣
=
∣
B
C
2
∣
=
1
. Now the circumcentre of 4ABC is also the centre of the circle that passes through the points
A
1
,
B
2
,
C
1
,
A
2
,
B
1
A_1,B_2,C_1,A_2,B_1
A
1
,
B
2
,
C
1
,
A
2
,
B
1
and
C
2
C_2
C
2
. Calculate the radius of the circle through
A
1
,
B
2
,
C
1
,
A
2
,
B
1
A_1,B_2,C_1,A_2,B_1
A
1
,
B
2
,
C
1
,
A
2
,
B
1
and
C
2
C_2
C
2
.[asy] unitsize(1.5 cm);pair[] A, B, C;A[0] = (0,0); B[0] = (1,0); C[0] = dir(60); A[1] = B[0] + dir(-60); A[2] = C[0] + dir(120); B[1] = C[0] + dir(60); B[2] = A[0] + dir(240); C[1] = A[0] + (-1,0); C[2] = B[0] + (1,0);draw(A[1]--A[2]); draw(B[1]--B[2]); draw(C[1]--C[2]); draw(circumcircle(A[2],B[1],C[2]));dot("
A
A
A
", A[0], SE); dot("
A
1
A_1
A
1
", A[1], SE); dot("
A
2
A_2
A
2
", A[2], NW); dot("
B
B
B
", B[0], SW); dot("
B
1
B_1
B
1
", B[1], NE); dot("
B
2
B_2
B
2
", B[2], SW); dot("
C
C
C
", C[0], N); dot("
C
1
C_1
C
1
", C[1], W); dot("
C
2
C_2
C
2
", C[2], E); [/asy]