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Contests
National and Regional Contests
Romania Contests
Romania Team Selection Test
1979 Romania Team Selection Tests
1979 Romania Team Selection Tests
Part of
Romania Team Selection Test
Subcontests
(6)
6.
1
Hide problems
Max max of product of binomial coefficients
If
n
>
2
n>2
n
>
2
is a positive integer, compute
max
1
⩽
k
⩽
n
max
n
1
+
.
.
.
+
n
k
=
n
(
n
1
2
)
(
n
2
2
)
…
(
n
k
2
)
.
\max_{1\leqslant k\leqslant n}\max_{n_1+...+n_k=n} \binom{n_1}{2}\binom{n_2}{2}\ldots\binom{n_k}{2}.
1
⩽
k
⩽
n
max
n
1
+
...
+
n
k
=
n
max
(
2
n
1
)
(
2
n
2
)
…
(
2
n
k
)
.
Ioan Tomescu
5.
2
Hide problems
Counting grid configurations with \pm 1
In how many ways can we fill the cells of a
m
×
n
m\times n
m
×
n
board with
+
1
+1
+
1
and
−
1
-1
−
1
such that the product of numbers on each line and on each column are all equal to
−
1
-1
−
1
?
(Not) covering an unit square with (rectangle or) triangles
a) Are there rectangles
1
×
1
2
1\times \dfrac12
1
×
2
1
rectangles lying strictly inside the interior of a unit square? b) Find the minimum number of equilateral triangles of unit side which can cover completely a unit square.Laurențiu Panaitopol
4.
2
Hide problems
Quadratic polynomial inequality
Give an example of a second degree polynomial
P
∈
R
[
x
]
P\in \mathbb{R}[x]
P
∈
R
[
x
]
such that
∀
x
∈
R
with
∣
x
∣
⩽
1
:
∣
P
(
x
)
+
1
x
−
4
∣
⩽
0.01.
\forall x\in \mathbb{R}\text{ with } |x|\leqslant 1: \; \left|P(x)+\frac{1}{x-4}\right| \leqslant 0.01.
∀
x
∈
R
with
∣
x
∣
⩽
1
:
P
(
x
)
+
x
−
4
1
⩽
0.01.
Are there linear polynomials with this property?Octavian Stănășilă
Surface area on sphere
Let
A
1
A
2
A
3
A
4
A_1A_2A_3A_4
A
1
A
2
A
3
A
4
be a tetrahedron. Consider the sphere centered at
A
1
A_1
A
1
which is tangent to the face
A
2
A
3
A
4
A_2A_3A_4
A
2
A
3
A
4
of the tetrahedron. Show that the surface area of the part of the sphere which is inside the tetrahedron is less than the area of the triangle
A
2
A
3
A
4
A_2A_3A_4
A
2
A
3
A
4
.Sorin Rădulescu
3.
2
Hide problems
An old problem with permutations by mavropnevma
Let
M
n
M_n
M
n
be the set of permutations
σ
∈
S
n
\sigma\in S_n
σ
∈
S
n
for which there exists
τ
∈
S
n
\tau\in S_n
τ
∈
S
n
such that the numbers
σ
(
1
)
+
τ
(
1
)
,
σ
(
2
)
+
τ
(
2
)
,
…
,
σ
(
n
)
+
τ
(
n
)
,
\sigma (1)+\tau(1),\, \sigma(2)+\tau(2),\ldots,\sigma(n)+\tau(n),
σ
(
1
)
+
τ
(
1
)
,
σ
(
2
)
+
τ
(
2
)
,
…
,
σ
(
n
)
+
τ
(
n
)
,
are consecutive. Show that
(
M
n
≠
∅
⇔
n
is odd
)
(M_n\neq \emptyset\Leftrightarrow n\text{ is odd})
(
M
n
=
∅
⇔
n
is odd
)
and in this case for each
σ
1
,
σ
2
∈
M
n
\sigma_1,\sigma_2\in M_n
σ
1
,
σ
2
∈
M
n
the following equality holds:
∑
k
=
1
n
k
σ
1
(
k
)
=
∑
k
=
1
n
k
σ
2
(
k
)
.
\sum_{k=1}^n k\sigma_1(k)=\sum_{k=1}^n k\sigma_2(k).
k
=
1
∑
n
k
σ
1
(
k
)
=
k
=
1
∑
n
k
σ
2
(
k
)
.
Dan Schwarz
Inequality in system of equations
Let
a
,
b
,
c
∈
R
a,b,c\in \mathbb{R}
a
,
b
,
c
∈
R
with
a
2
+
b
2
+
c
2
=
1
a^2+b^2+c^2=1
a
2
+
b
2
+
c
2
=
1
and
λ
∈
R
>
0
∖
{
1
}
\lambda\in \mathbb{R}_{>0}\setminus\{1\}
λ
∈
R
>
0
∖
{
1
}
. Then for each solution
(
x
,
y
,
z
)
(x,y,z)
(
x
,
y
,
z
)
of the system of equations:
{
x
−
λ
y
=
a
,
y
−
λ
z
=
b
,
z
−
λ
x
=
c
.
\begin{cases} x-\lambda y=a,\\ y-\lambda z=b,\\ z-\lambda x=c. \end{cases}
⎩
⎨
⎧
x
−
λ
y
=
a
,
y
−
λ
z
=
b
,
z
−
λ
x
=
c
.
we have
x
2
+
y
2
+
z
2
⩽
1
(
λ
−
1
)
2
\displaystyle x^2+y^2+z^2\leqslant \frac1{(\lambda-1)^2}
x
2
+
y
2
+
z
2
⩽
(
λ
−
1
)
2
1
.Radu Gologan
2.
2
Hide problems
Plane cuts pyramid in polyhedra of equal volumes
Let
V
A
1
A
2
A
3
A
4
VA_1A_2A_3A_4
V
A
1
A
2
A
3
A
4
be a pyramid with the vertex at
V
V
V
. Let
M
,
N
,
P
M,\, N,\, P
M
,
N
,
P
be the midpoints of the segments
V
A
1
VA_1
V
A
1
,
V
A
3
VA_3
V
A
3
, and
A
2
A
4
A_2A_4
A
2
A
4
. Show that the plane
(
M
N
P
)
(MNP)
(
MNP
)
cuts the pyramid into two parts with the same volume.Radu Gologan
Algebra with sequences related to sqrt n
For each
n
∈
Z
>
0
n\in \mathbb{Z}_{>0}
n
∈
Z
>
0
let
a
n
a_n
a
n
be the closest integer to
n
\sqrt{n}
n
. Compute the general term of the sequence:
(
b
n
)
n
⩾
1
(b_n)_{n\geqslant 1}
(
b
n
)
n
⩾
1
with
b
n
=
∑
k
=
1
n
2
a
k
.
b_n=\sum_{k=1}^{n^2} a_k.
b
n
=
k
=
1
∑
n
2
a
k
.
Pall Dalyay
1.
2
Hide problems
Locus geometry from old Romanian TST
Let
△
A
B
C
\triangle ABC
△
A
BC
be a triangle with
∠
B
A
C
=
6
0
∘
\angle BAC=60^\circ
∠
B
A
C
=
6
0
∘
,
M
M
M
be a point in its interior and
A
′
,
B
′
,
C
′
A',\, B',\, C'
A
′
,
B
′
,
C
′
be the orthogonal projections of
M
M
M
on the sides
B
C
,
C
A
,
A
B
BC,\, CA,\, AB
BC
,
C
A
,
A
B
. Determine the locus of
M
M
M
when the sum
A
′
B
+
B
′
C
+
C
′
A
A'B+B'C+C'A
A
′
B
+
B
′
C
+
C
′
A
is constant.Horea Călin Pop
Polynomial equation with rational numbers
Determine the polynomial
P
∈
R
[
x
]
P\in \mathbb{R}[x]
P
∈
R
[
x
]
for which there exists
n
∈
Z
>
0
n\in \mathbb{Z}_{>0}
n
∈
Z
>
0
such that for all
x
∈
Q
x\in \mathbb{Q}
x
∈
Q
we have:
P
(
x
+
1
n
)
+
P
(
x
−
1
n
)
=
2
P
(
x
)
.
P\left(x+\frac1n\right)+P\left(x-\frac1n\right)=2P(x).
P
(
x
+
n
1
)
+
P
(
x
−
n
1
)
=
2
P
(
x
)
.
Dumitru Bușneag