MathDB
Problems
Contests
National and Regional Contests
Belgium Contests
Flanders Math Olympiad
1996 Flanders Math Olympiad
1996 Flanders Math Olympiad
Part of
Flanders Math Olympiad
Subcontests
(4)
4
1
Hide problems
polynomial
Consider a real poylnomial
p
(
x
)
=
a
n
x
n
+
.
.
.
+
a
1
x
+
a
0
p(x)=a_nx^n+...+a_1x+a_0
p
(
x
)
=
a
n
x
n
+
...
+
a
1
x
+
a
0
. (a) If
deg
(
p
(
x
)
)
>
2
\deg(p(x))>2
de
g
(
p
(
x
))
>
2
prove that
deg
(
p
(
x
)
)
=
2
+
d
e
g
(
p
(
x
+
1
)
+
p
(
x
−
1
)
−
2
p
(
x
)
)
\deg(p(x)) = 2 + deg(p(x+1)+p(x-1)-2p(x))
de
g
(
p
(
x
))
=
2
+
d
e
g
(
p
(
x
+
1
)
+
p
(
x
−
1
)
−
2
p
(
x
))
. (b) Let
p
(
x
)
p(x)
p
(
x
)
a polynomial for which there are real constants
r
,
s
r,s
r
,
s
so that for all real
x
x
x
we have
p
(
x
+
1
)
+
p
(
x
−
1
)
−
r
p
(
x
)
−
s
=
0
p(x+1)+p(x-1)-rp(x)-s=0
p
(
x
+
1
)
+
p
(
x
−
1
)
−
r
p
(
x
)
−
s
=
0
Prove
deg
(
p
(
x
)
)
≤
2
\deg(p(x))\le 2
de
g
(
p
(
x
))
≤
2
. (c) Show, in (b) that
s
=
0
s=0
s
=
0
implies
a
2
=
0
a_2=0
a
2
=
0
.
1
1
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Geometry (flanders '96)
In triangle
Δ
A
D
C
\Delta ADC
Δ
A
D
C
we got
A
D
=
D
C
AD=DC
A
D
=
D
C
and
D
=
10
0
∘
D=100^\circ
D
=
10
0
∘
. In triangle
Δ
C
A
B
\Delta CAB
Δ
C
A
B
we got
C
A
=
A
B
CA=AB
C
A
=
A
B
and
A
=
2
0
∘
A=20^\circ
A
=
2
0
∘
. Prove that
A
B
=
B
C
+
C
D
AB=BC+CD
A
B
=
BC
+
C
D
.
2
1
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greatest common divisor
Determine the gcd of all numbers of the form
p
8
−
1
p^8-1
p
8
−
1
, with p a prime above 5.
3
1
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[solved] - Real Axis
Consider the points
1
,
1
2
,
1
3
,
.
.
.
1,\frac12,\frac13,...
1
,
2
1
,
3
1
,
...
on the real axis. Find the smallest value
k
∈
N
0
k \in \mathbb{N}_0
k
∈
N
0
for which all points above can be covered with 5 closed intervals of length
1
k
\frac1k
k
1
.