MathDB
Problems
Contests
National and Regional Contests
Belgium Contests
Flanders Math Olympiad
2022 Flanders Math Olympiad
2022 Flanders Math Olympiad
Part of
Flanders Math Olympiad
Subcontests
(4)
2
1
Hide problems
you can divide a square into n dominoes for all n >= 5 , but not for n=3 or 4
A domino is a rectangle whose length is twice its width. Any square can be divided into seven dominoes, for example as shown in the figure below. https://cdn.artofproblemsolving.com/attachments/7/6/c055d8d2f6b7c24d38ded7305446721e193203.png a) Show that you can divide a square into
n
n
n
dominoes for all
n
≥
5
n \ge 5
n
≥
5
. b) Show that you cannot divide a square into three or four dominoes.
3
1
Hide problems
Arne has 2n + 1 tickets
Arne has
2
n
+
1
2n + 1
2
n
+
1
tickets. Each card has one number on it. One card has the number
0
0
0
on it. The natural numbers
1
,
2
,
.
.
.
,
n
1, 2, . . . , n
1
,
2
,
...
,
n
occur on exactly two cards each. Prove that Arne can arrange cards in a row so that there are exactly
m
m
m
cards between the two cards with the number
m
m
m
, for every
m
∈
{
1
,
2
,
.
.
.
,
n
}
m \in \{1, 2, . . . , n\}
m
∈
{
1
,
2
,
...
,
n
}
.
4
1
Hide problems
kP (k + 1) - (k + 1)P (k) = k^2 + k + 1 for all k \in {1, 2, 3, . . . , 21, 22|
Determine all real polynomials
P
P
P
of degree at most
22
22
22
for which
k
P
(
k
+
1
)
−
(
k
+
1
)
P
(
k
)
=
k
2
+
k
+
1
kP (k + 1) - (k + 1)P (k) = k^2 + k + 1
k
P
(
k
+
1
)
−
(
k
+
1
)
P
(
k
)
=
k
2
+
k
+
1
for all
k
∈
{
1
,
2
,
3
,
.
.
.
,
21
,
22
}
k \in \{1, 2, 3, . . . , 21, 22\}
k
∈
{
1
,
2
,
3
,
...
,
21
,
22
}
.
1
1
Hide problems
equal angles wanted, intersecting chords
The points
A
,
B
,
C
,
D
A, B, C, D
A
,
B
,
C
,
D
lie in that order on a circle. The segments
A
C
AC
A
C
and
B
D
BD
B
D
intersect at the point
P
P
P
. The point
B
′
B'
B
′
lies on the line
A
B
AB
A
B
such that
A
A
A
is between
B
B
B
and
B
′
B'
B
′
and
∣
A
B
′
∣
=
∣
D
P
∣
|AB'| = |DP |
∣
A
B
′
∣
=
∣
D
P
∣
. The point
C
′
C'
C
′
lies on the line
C
D
CD
C
D
such that
D
D
D
is between
C
C
C
and
C
′
C'
C
′
lies and
∣
D
C
′
∣
=
∣
A
P
∣
|DC' | = |AP|
∣
D
C
′
∣
=
∣
A
P
∣
. Prove that
∠
B
′
P
C
′
=
∠
A
B
D
′
\angle B'PC' = \angle ABD'
∠
B
′
P
C
′
=
∠
A
B
D
′
.https://cdn.artofproblemsolving.com/attachments/2/2/7ec65246ff5ecfebc25ca13f3709d1791ceb6c.png =