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Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch BxMO/EGMO TST
2013 Dutch BxMO/EGMO TST
2013 Dutch BxMO/EGMO TST
Part of
Dutch BxMO/EGMO TST
Subcontests
(5)
3
1
Hide problems
diophantine x^3 + 3x + 14 = 2 p^n , where p is prime
Find all triples
(
x
,
n
,
p
)
(x,n,p)
(
x
,
n
,
p
)
of positive integers
x
x
x
and
n
n
n
and primes
p
p
p
for which the following holds
x
3
+
3
x
+
14
=
2
p
n
x^3 + 3x + 14 = 2 p^n
x
3
+
3
x
+
14
=
2
p
n
2
1
Hide problems
triple (x, y, z) replaced by (y+z- x, z+x- y,x +y -z), a+b+c=2013, 10 steps
Consider a triple
(
a
,
b
,
c
)
(a, b, c)
(
a
,
b
,
c
)
of pairwise distinct positive integers satisfying
a
+
b
+
c
=
2013
a + b + c = 2013
a
+
b
+
c
=
2013
. A step consists of replacing the triple
(
x
,
y
,
z
)
(x, y, z)
(
x
,
y
,
z
)
by the triple
(
y
+
z
−
x
,
z
+
x
−
y
,
x
+
y
−
z
)
(y + z - x,z + x - y,x + y - z)
(
y
+
z
−
x
,
z
+
x
−
y
,
x
+
y
−
z
)
. Prove that, starting from the given triple
(
a
,
b
,
c
)
(a, b,c)
(
a
,
b
,
c
)
, after
10
10
10
steps we obtain a triple containing at least one negative number.
5
1
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(AN)(NC) =(CD)(BN) in a cyclic ABCD with (AD)=(BD)
Let
A
B
C
D
ABCD
A
BC
D
be a cyclic quadrilateral for which
∣
A
D
∣
=
∣
B
D
∣
|AD| =|BD|
∣
A
D
∣
=
∣
B
D
∣
. Let
M
M
M
be the intersection of
A
C
AC
A
C
and
B
D
BD
B
D
. Let
I
I
I
be the incentre of
△
B
C
M
\triangle BCM
△
BCM
. Let
N
N
N
be the second intersection pointof
A
C
AC
A
C
and the circumscribed circle of
△
B
M
I
\triangle BMI
△
BM
I
. Prove that
∣
A
N
∣
⋅
∣
N
C
∣
=
∣
C
D
∣
⋅
∣
B
N
∣
|AN| \cdot |NC| = |CD | \cdot |BN|
∣
A
N
∣
⋅
∣
NC
∣
=
∣
C
D
∣
⋅
∣
BN
∣
.
1
1
Hide problems
equal areas of triangles in ABCD leads to parallel lines
In quadrilateral
A
B
C
D
ABCD
A
BC
D
the sides
A
B
AB
A
B
and
C
D
CD
C
D
are parallel. Let
M
M
M
be the midpoint of diagonal
A
C
AC
A
C
. Suppose that triangles
A
B
M
ABM
A
BM
and
A
C
D
ACD
A
C
D
have equal area. Prove that
D
M
/
/
B
C
DM // BC
D
M
//
BC
.
4
1
Hide problems
Functional equation.
Determine all functions
f
:
R
→
R
f:\mathbb{R}\to\mathbb{R}
f
:
R
→
R
satisfying
f
(
x
+
y
f
(
x
)
)
=
f
(
x
f
(
y
)
)
−
x
+
f
(
y
+
f
(
x
)
)
f(x+yf(x))=f(xf(y))-x+f(y+f(x))
f
(
x
+
y
f
(
x
))
=
f
(
x
f
(
y
))
−
x
+
f
(
y
+
f
(
x
))