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Contests
National and Regional Contests
Romania Contests
Romania Team Selection Test
2000 Romania Team Selection Test
2000 Romania Team Selection Test
Part of
Romania Team Selection Test
Subcontests
(4)
4
1
Hide problems
N-gon must be a triangle if always exists a 60 degree angle
Let
P
1
P
2
…
P
n
P_1P_2\ldots P_n
P
1
P
2
…
P
n
be a convex polygon in the plane. We assume that for any arbitrary choice of vertices
P
i
,
P
j
P_i,P_j
P
i
,
P
j
there exists a vertex in the polygon
P
k
P_k
P
k
distinct from
P
i
,
P
j
P_i,P_j
P
i
,
P
j
such that
∠
P
i
P
k
P
j
=
6
0
∘
\angle P_iP_kP_j=60^{\circ}
∠
P
i
P
k
P
j
=
6
0
∘
. Show that
n
=
3
n=3
n
=
3
.Radu Todor
1
3
Hide problems
f mapping {1,2...,n} to {1,2,3,4,5} with condition - Rom TST
Let
n
≥
2
n\ge 2
n
≥
2
be a positive integer. Find the number of functions
f
:
{
1
,
2
,
…
,
n
}
→
{
1
,
2
,
3
,
4
,
5
}
f:\{1,2,\ldots ,n\}\rightarrow\{1,2,3,4,5 \}
f
:
{
1
,
2
,
…
,
n
}
→
{
1
,
2
,
3
,
4
,
5
}
which have the following property:
∣
f
(
k
+
1
)
−
f
(
k
)
∣
≥
3
|f(k+1)-f(k)|\ge 3
∣
f
(
k
+
1
)
−
f
(
k
)
∣
≥
3
, for any
k
=
1
,
2
,
…
n
−
1
k=1,2,\ldots n-1
k
=
1
,
2
,
…
n
−
1
.Vasile Pop
Least n such that 2^2000 divides a^n-1
Let
a
>
1
a>1
a
>
1
be an odd positive integer. Find the least positive integer
n
n
n
such that
2
2000
2^{2000}
2
2000
is a divisor of
a
n
−
1
a^n-1
a
n
−
1
.Mircea Becheanu
There exists a monochromatic isosceles trapezium ABCD
Let
P
1
P_1
P
1
be a regular
n
n
n
-gon, where
n
∈
N
n\in\mathbb{N}
n
∈
N
. We construct
P
2
P_2
P
2
as the regular
n
n
n
-gon whose vertices are the midpoints of the edges of
P
1
P_1
P
1
. Continuing analogously, we obtain regular
n
n
n
-gons
P
3
,
P
4
,
…
,
P
m
P_3,P_4,\ldots ,P_m
P
3
,
P
4
,
…
,
P
m
. For
m
≥
n
2
−
n
+
1
m\ge n^2-n+1
m
≥
n
2
−
n
+
1
, find the maximum number
k
k
k
such that for any colouring of vertices of
P
1
,
…
,
P
m
P_1,\ldots ,P_m
P
1
,
…
,
P
m
in
k
k
k
colours there exists an isosceles trapezium
A
B
C
D
ABCD
A
BC
D
whose vertices
A
,
B
,
C
,
D
A,B,C,D
A
,
B
,
C
,
D
have the same colour.Radu Ignat
3
3
Hide problems
TST-Romania, 2000
Prove that for any positive integers
n
n
n
and
k
k
k
there exist positive integers
a
>
b
>
c
>
d
>
e
>
k
a>b>c>d>e>k
a
>
b
>
c
>
d
>
e
>
k
such that
n
=
(
a
3
)
±
(
b
3
)
±
(
c
3
)
±
(
d
3
)
±
(
e
3
)
n=\binom{a}{3}\pm\binom{b}{3}\pm\binom{c}{3}\pm\binom{d}{3}\pm\binom{e}{3}
n
=
(
3
a
)
±
(
3
b
)
±
(
3
c
)
±
(
3
d
)
±
(
3
e
)
Radu Ignat
Romania 2000 TST
Determine all pairs
(
m
,
n
)
(m,n)
(
m
,
n
)
of positive integers such that a
m
×
n
m\times n
m
×
n
rectangle can be tiled with L-trominoes.
Mapping of interior points of spehere and circle
Let
S
S
S
be the set of interior points of a sphere and
C
C
C
be the set of interior points of a circle. Find, with proof, whether there exists a function
f
:
S
→
C
f:S\rightarrow C
f
:
S
→
C
such that
d
(
A
,
B
)
≤
d
(
f
(
A
)
,
f
(
B
)
)
d(A,B)\le d(f(A),f(B))
d
(
A
,
B
)
≤
d
(
f
(
A
)
,
f
(
B
))
for any two points
A
,
B
∈
S
A,B\in S
A
,
B
∈
S
where
d
(
X
,
Y
)
d(X,Y)
d
(
X
,
Y
)
denotes the distance between the points
X
X
X
and
Y
Y
Y
.Marius Cavachi
2
3
Hide problems
ROMANIA 2000 inequality
Let
n
≥
1
n\ge 1
n
≥
1
be a positive integer and
x
1
,
x
2
…
,
x
n
x_1,x_2\ldots ,x_n
x
1
,
x
2
…
,
x
n
be real numbers such that
∣
x
k
+
1
−
x
k
∣
≤
1
|x_{k+1}-x_k|\le 1
∣
x
k
+
1
−
x
k
∣
≤
1
for
k
=
1
,
2
,
…
,
n
−
1
k=1,2,\ldots ,n-1
k
=
1
,
2
,
…
,
n
−
1
. Prove that
∑
k
=
1
n
∣
x
k
∣
−
∣
∑
k
=
1
n
x
k
∣
≤
n
2
−
1
4
\sum_{k=1}^n|x_k|-\left|\sum_{k=1}^nx_k\right|\le\frac{n^2-1}{4}
k
=
1
∑
n
∣
x
k
∣
−
k
=
1
∑
n
x
k
≤
4
n
2
−
1
Gh. Eckstein
N satisfies ∠ NBA =∠ BAM and ∠ NCA = ∠ CAM
Let
A
B
C
ABC
A
BC
be an acute-angled triangle and
M
M
M
be the midpoint of the side
B
C
BC
BC
. Let
N
N
N
be a point in the interior of the triangle
A
B
C
ABC
A
BC
such that
∠
N
B
A
=
∠
B
A
M
\angle NBA=\angle BAM
∠
NB
A
=
∠
B
A
M
and
∠
N
C
A
=
∠
C
A
M
\angle NCA=\angle CAM
∠
NC
A
=
∠
C
A
M
. Prove that
∠
N
A
B
=
∠
M
A
C
\angle NAB=\angle MAC
∠
N
A
B
=
∠
M
A
C
.Gabriel Nagy
monic polynomials
Let
P
,
Q
P,Q
P
,
Q
be two monic polynomials with complex coefficients such that
P
(
P
(
x
)
)
=
Q
(
Q
(
x
)
)
P(P(x))=Q(Q(x))
P
(
P
(
x
))
=
Q
(
Q
(
x
))
for all
x
x
x
. Prove that
P
=
Q
P=Q
P
=
Q
.Marius Cavachi