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Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch Mathematical Olympiad
2021 Dutch Mathematical Olympiad
2021 Dutch Mathematical Olympiad
Part of
Dutch Mathematical Olympiad
Subcontests
(5)
4
1
Hide problems
EM_|_AB and similar triangles wanted, <ACB = 90^o
In triangle
A
B
C
ABC
A
BC
we have
∠
A
C
B
=
9
0
o
\angle ACB = 90^o
∠
A
CB
=
9
0
o
. The point
M
M
M
is the midpoint of
A
B
AB
A
B
. The line through
M
M
M
parallel to
B
C
BC
BC
intersects
A
C
AC
A
C
in
D
D
D
. The midpoint of line segment
C
D
CD
C
D
is
E
E
E
. The lines
B
D
BD
B
D
and
C
M
CM
CM
are perpendicular. (a) Prove that triangles
C
M
E
CME
CME
and
A
B
D
ABD
A
B
D
are similar. (b) Prove that
E
M
EM
EM
and
A
B
AB
A
B
are perpendicular.[asy] unitsize(1 cm);pair A, B, C, D, E, M;A = (0,0); B = (4,0); C = (2.6,2); M = (A + B)/2; D = (A + C)/2; E = (C + D)/2;draw(A--B--C--cycle); draw(C--M--D--B);dot("
A
A
A
", A, SW); dot("
B
B
B
", B, SE); dot("
C
C
C
", C, N); dot("
D
D
D
", D, NW); dot("
E
E
E
", E, NW); dot("
M
M
M
", M, S); [/asy]Be aware: the figure is not drawn to scale.
1
1
Hide problems
+1,-1 colours in 4x4 square
Niek has
16
16
16
square cards that are yellow on one side and red on the other. He puts down the cards to form a
4
×
4
4 \times 4
4
×
4
-square. Some of the cards show their yellow side and some show their red side. For a colour pattern he calculates the monochromaticity as follows. For every pair of adjacent cards that share a side he counts
+
1
+1
+
1
or
−
1
-1
−
1
according to the following rule:
+
1
+1
+
1
if the adjacent cards show the same colour, and
−
1
-1
−
1
if the adjacent cards show different colours. Adding this all together gives the monochromaticity (which might be negative). For example, if he lays down the cards as below, there are
15
15
15
pairs of adjacent cards showing the same colour, and
9
9
9
such pairs showing different colours.[asy] unitsize(1 cm);int i;fill(shift((0,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((1,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((2,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((3,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((0,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((1,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red); fill(shift((2,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((3,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((0,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((1,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((2,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((3,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red); fill(shift((0,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red); fill(shift((1,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((2,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow); fill(shift((3,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red);for (i = 0; i <= 4; ++i) { draw((i,0)--(i,4)); draw((0,i)--(4,i)); } [/asy]The monochromaticity of this pattern is thus
15
⋅
(
+
1
)
+
9
⋅
(
−
1
)
=
6
15 \cdot (+1) + 9 \cdot (-1) = 6
15
⋅
(
+
1
)
+
9
⋅
(
−
1
)
=
6
. Niek investigates all possible colour patterns and makes a list of all possible numbers that appear at least once as a value of the monochromaticity. That is, Niek makes a list with all numbers such that there exists a colour pattern that has this number as its monochromaticity.(a) What are the three largest numbers on his list? (Explain your answer. If your answer is, for example,
12
12
12
,
9
9
9
and
6
6
6
, then you have to show that these numbers do in fact appear on the list by giving a colouring for each of these numbers, and furthermore prove that the numbers
7
7
7
,
8
8
8
,
10
10
10
,
11
11
11
and all numbers bigger than
12
12
12
do not appear.)(b) What are the three smallest (most negative) numbers on his list?(c) What is the smallest positive number (so, greater than
0
0
0
) on his list?
5
1
Hide problems
n is prime if for every divisor d of n, d + 1 is a divisor of n + 1
We consider an integer
n
>
1
n > 1
n
>
1
with the following property: for every positive divisor
d
d
d
of
n
n
n
we have that
d
+
1
d + 1
d
+
1
is a divisor of
n
+
1
n + 1
n
+
1
. Prove that
n
n
n
is a prime number.
3
1
Hide problems
a frog jumps around on the grid points in the plane
A frog jumps around on the grid points in the plane, from one grid point to another. The frog starts at the point
(
0
,
0
)
(0, 0)
(
0
,
0
)
. Then it makes, successively, a jump of one step horizontally, a jump of
2
2
2
steps vertically, a jump of
3
3
3
steps horizontally, a jump of
4
4
4
steps vertically, et cetera. Determine all
n
>
0
n > 0
n
>
0
such that the frog can be back in
(
0
,
0
)
(0, 0)
(
0
,
0
)
after
n
n
n
jumps.
2
1
Hide problems
max n for which a balanced tournament with n teams exists
We consider sports tournaments with
n
≥
4
n \ge 4
n
≥
4
participating teams and where every pair of teams plays against one another at most one time. We call such a tournament balanced if any four participating teams play exactly three matches between themselves. So, not all teams play against one another. Determine the largest value of
n
n
n
for which a balanced tournament with
n
n
n
teams exists.