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Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch BxMO/EGMO TST
2015 Dutch BxMO/EGMO TST
2015 Dutch BxMO/EGMO TST
Part of
Dutch BxMO/EGMO TST
Subcontests
(5)
5
1
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(x^2 + y^2)f(xy) = f(x)f(y)f(x^2 + y^2)
Find all functions
f
:
R
→
R
f : R \to R
f
:
R
→
R
satisfying
(
x
2
+
y
2
)
f
(
x
y
)
=
f
(
x
)
f
(
y
)
f
(
x
2
+
y
2
)
(x^2 + y^2)f(xy) = f(x)f(y)f(x^2 + y^2)
(
x
2
+
y
2
)
f
(
x
y
)
=
f
(
x
)
f
(
y
)
f
(
x
2
+
y
2
)
for all real numbers
x
x
x
and
y
y
y
.
3
1
Hide problems
dominoes on a red-blue nxn boad, colourful dominoes with no overlaps
Let
n
≥
2
n \ge 2
n
≥
2
be a positive integer. Each square of an n\times n board is coloured red or blue. We put dominoes on the board, each covering two squares of the board. A domino is called even if it lies on two red or two blue squares and colourful if it lies on a red and a blue square. Find the largest positive integer
k
k
k
having the following property: regardless of how the red/blue-colouring of the board is done, it is always possible to put
k
k
k
non-overlapping dominoes on the board that are either all even or all colourful.
2
1
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a_1 <= a_2 <= ..., exists r : r/ a_r= k + 1 => exists t : t/ a_t = k
Given are positive integers
r
r
r
and
k
k
k
and an infi nite sequence of positive integers
a
1
≤
a
2
≤
.
.
.
a_1 \le a_2 \le ...
a
1
≤
a
2
≤
...
such that
r
a
r
=
k
+
1
\frac{r}{a_r}= k + 1
a
r
r
=
k
+
1
. Prove that there is a
t
t
t
satisfying
t
a
t
=
k
\frac{t}{a_t}=k
a
t
t
=
k
.
1
1
Hide problems
(5m+ n) / (5n +m) => m / n
Let
m
m
m
and
n
n
n
be positive integers such that
5
m
+
n
5m+ n
5
m
+
n
is a divisor of
5
n
+
m
5n +m
5
n
+
m
. Prove that
m
m
m
is a divisor of
n
n
n
.
4
1
Hide problems
dutch angle chasing for TST, 2 circles and angle bisector related
In a triangle
A
B
C
ABC
A
BC
the point
D
D
D
is the intersection of the interior angle bisector of
∠
B
A
C
\angle BAC
∠
B
A
C
and side
B
C
BC
BC
. Let
P
P
P
be the second intersection point of the exterior angle bisector of
∠
B
A
C
\angle BAC
∠
B
A
C
with the circumcircle of
∠
A
B
C
\angle ABC
∠
A
BC
. A circle through
A
A
A
and
P
P
P
intersects line segment
B
P
BP
BP
internally in
E
E
E
and line segment
C
P
CP
CP
internally in
F
F
F
. Prove that
∠
D
E
P
=
∠
D
F
P
\angle DEP = \angle DFP
∠
D
EP
=
∠
D
FP
.