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Contests
National and Regional Contests
Romania Contests
JBMO TST - Romania
1999 Junior Balkan Team Selection Tests - Romania
1999 Junior Balkan Team Selection Tests - Romania
Part of
JBMO TST - Romania
Subcontests
(4)
4
2
Hide problems
Inequality involving area of intersected discs
Let be three discs
D
1
,
D
2
,
D
3
.
D_1,D_2,D_3.
D
1
,
D
2
,
D
3
.
For each
i
,
j
∈
{
1
,
2
,
3
}
,
i,j\in\{1,2,3\} ,
i
,
j
∈
{
1
,
2
,
3
}
,
denote
a
i
j
a_{ij}
a
ij
as being the area of
D
i
∩
D
j
.
D_i\cap D_j.
D
i
∩
D
j
.
If
x
1
,
x
2
,
x
3
∈
R
x_1,x_2,x_3\in\mathbb{R}
x
1
,
x
2
,
x
3
∈
R
such that
x
1
x
2
x
3
≠
0
,
x_1x_2x_3\neq 0,
x
1
x
2
x
3
=
0
,
then
a
11
x
1
2
+
a
22
x
2
2
+
a
33
x
3
2
+
2
a
12
x
1
x
2
+
2
a
23
x
2
x
3
+
2
a
31
x
3
x
1
>
0.
a_{11} x_1^2+a_{22} x_2^2+a_{33} x_3^2+2a_{12} x_1x_2+2a_{23 }x_2x_3+2a_{31} x_3x_1>0.
a
11
x
1
2
+
a
22
x
2
2
+
a
33
x
3
2
+
2
a
12
x
1
x
2
+
2
a
23
x
2
x
3
+
2
a
31
x
3
x
1
>
0.
Vasile Pop
Constructive geometry 1999
Let be a convex quadrilateral
A
B
C
D
.
ABCD.
A
BC
D
.
On the semi-straight line extension of
A
B
AB
A
B
in the direction of
B
,
B,
B
,
put
A
1
A_1
A
1
such that
A
B
=
B
A
1
.
AB=BA_1.
A
B
=
B
A
1
.
Similarly, define
B
1
,
C
1
,
D
1
,
B_1,C_1,D_1,
B
1
,
C
1
,
D
1
,
for the other three sides.a) If
E
,
E
1
,
F
,
F
1
E,E_1,F,F_1
E
,
E
1
,
F
,
F
1
are the midpoints of
B
C
,
A
1
B
1
,
A
D
BC,A_1B_1,AD
BC
,
A
1
B
1
,
A
D
respectively,
C
1
,
D
1
,
C_1,D_1,
C
1
,
D
1
,
show that
E
E
1
=
F
F
1
.
EE_1=FF_1.
E
E
1
=
F
F
1
.
b) Delete everything, but
A
1
,
B
1
,
C
1
,
D
1
.
A_1,B_1,C_1,D_1.
A
1
,
B
1
,
C
1
,
D
1
.
Now, find a way to construct the initial quadrilateral. Vasile Pop
3
2
Hide problems
x in A implies (x\2 in A and 1\(1+x) in A)
Let be a subset of the interval
(
0
,
1
)
(0,1)
(
0
,
1
)
that contains
1
/
2
1/2
1/2
and has the property that if a number is in this subset, then, both its half and its successor's inverse are in the same subset. Prove that this subset contains all the rational numbers of the interval
(
0
,
1
)
.
(0,1).
(
0
,
1
)
.
gcd(2n+3m+13,3n+5m+1,6n+6m-1)
Consider the set
M
=
{
gcd
(
2
n
+
3
m
+
13
,
3
n
+
5
m
+
1
,
6
n
+
6
m
−
1
)
∣
m
,
n
∈
N
}
.
\mathcal{M}=\left\{ \gcd(2n+3m+13,3n+5m+1,6n+6m-1) | m,n\in\mathbb{N} \right\} .
M
=
{
g
cd
(
2
n
+
3
m
+
13
,
3
n
+
5
m
+
1
,
6
n
+
6
m
−
1
)
∣
m
,
n
∈
N
}
.
Show that there is a natural
k
k
k
such that the set of its positive divisors is
M
.
\mathcal{M} .
M
.
Dan Brânzei
2
2
Hide problems
locus of center of masses
Consider, on a plane, the triangle
A
B
C
,
ABC,
A
BC
,
vectors
x
⃗
,
y
⃗
,
z
⃗
,
\vec x,\vec y,\vec z,
x
,
y
,
z
,
real variable
λ
>
0
\lambda >0
λ
>
0
and
M
,
N
,
P
M,N,P
M
,
N
,
P
such that
{
A
M
→
=
λ
⋅
x
⃗
A
N
→
=
λ
⋅
y
⃗
A
P
→
=
λ
⋅
z
⃗
.
\left\{\begin{matrix} \overrightarrow{AM}=\lambda\cdot\vec x\\\overrightarrow{AN}=\lambda\cdot\vec y \\\overrightarrow{AP}=\lambda\cdot\vec z \end{matrix}\right. .
⎩
⎨
⎧
A
M
=
λ
⋅
x
A
N
=
λ
⋅
y
A
P
=
λ
⋅
z
.
Find the locus of the center of mass of
M
N
P
.
MNP.
MNP
.
Dan Brânzei and Gheorghe Iurea
number of perfect squares of five digits whose last two digits are equal
Find the number of perfect squares of five digits whose last two digits are equal. Gheorghe Iurea
1
2
Hide problems
Polynomial in two variables at JBMOTST
Let be a natural number
n
.
n.
n
.
Prove that there is a polynomial
P
∈
Z
[
X
,
Y
]
P\in\mathbb{Z} [X,Y]
P
∈
Z
[
X
,
Y
]
such that
a
+
b
+
c
=
0
a+b+c=0
a
+
b
+
c
=
0
implies
a
2
n
+
1
+
b
2
n
+
1
+
c
2
n
+
1
=
a
b
c
(
P
(
a
,
b
)
+
P
(
b
,
c
)
+
P
(
c
,
a
)
)
a^{2n+1}+b^{2n+1}+c^{2n+1}=abc\left( P(a,b)+P(b,c)+P(c,a)\right)
a
2
n
+
1
+
b
2
n
+
1
+
c
2
n
+
1
=
ab
c
(
P
(
a
,
b
)
+
P
(
b
,
c
)
+
P
(
c
,
a
)
)
Dan Brânzei
Condition for a triangle to be separated in two isosceles ones
Find a relation between the angles of a triangle such that this could be separated in two isosceles triangles by a line. Dan Brânzei