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Problems
Contests
National and Regional Contests
Moldova Contests
Moldova Team Selection Test
2006 Moldova Team Selection Test
2006 Moldova Team Selection Test
Part of
Moldova Team Selection Test
Subcontests
(4)
1
3
Hide problems
Divisors and finding n
Determine all even numbers
n
n
n
,
n
∈
N
n \in \mathbb N
n
∈
N
such that
1
d
1
+
1
d
2
+
⋯
+
1
d
k
=
1620
1003
,
{ \frac{1}{d_{1}}+\frac{1}{d_{2}}+ \cdots +\frac{1}{d_{k}}=\frac{1620}{1003}},
d
1
1
+
d
2
1
+
⋯
+
d
k
1
=
1003
1620
,
where
d
1
,
d
2
,
…
,
d
k
d_1, d_2, \ldots, d_k
d
1
,
d
2
,
…
,
d
k
are all different divisors of
n
n
n
.
Lucas divisibility
Let
(
a
n
)
(a_n)
(
a
n
)
be the Lucas sequence:
a
0
=
2
,
a
1
=
1
,
a
n
+
1
=
a
n
+
a
n
−
1
a_0=2,a_1=1, a_{n+1}=a_n+a_{n-1}
a
0
=
2
,
a
1
=
1
,
a
n
+
1
=
a
n
+
a
n
−
1
for
n
≥
1
n\geq 1
n
≥
1
. Show that
a
59
a_{59}
a
59
divides
(
a
30
)
59
−
1
(a_{30})^{59}-1
(
a
30
)
59
−
1
.
Modlova 3rd tst, problem 1
Let the point
P
P
P
in the interior of the triangle
A
B
C
ABC
A
BC
.
(
A
P
,
(
B
P
,
(
C
P
(AP, (BP, (CP
(
A
P
,
(
BP
,
(
CP
intersect the circumcircle of
A
B
C
ABC
A
BC
at
A
1
,
B
1
,
C
1
A_{1}, B_{1}, C_{1}
A
1
,
B
1
,
C
1
. Prove that the maximal value of the sum of the areas
A
1
B
C
A_{1}BC
A
1
BC
,
B
1
A
C
B_{1}AC
B
1
A
C
,
C
1
A
B
C_{1}AB
C
1
A
B
is
p
(
R
−
r
)
p(R-r)
p
(
R
−
r
)
, where
p
,
r
,
R
p, r, R
p
,
r
,
R
are the usual notations for the triangle
A
B
C
ABC
A
BC
.
3
3
Hide problems
Cyclic ineq with triangle sides
Let
a
,
b
,
c
a,b,c
a
,
b
,
c
be sides of the triangle. Prove that
a
2
(
b
c
−
1
)
+
b
2
(
c
a
−
1
)
+
c
2
(
a
b
−
1
)
≥
0.
a^2\left(\frac{b}{c}-1\right)+b^2\left(\frac{c}{a}-1\right)+c^2\left(\frac{a}{b}-1\right)\geq 0 .
a
2
(
c
b
−
1
)
+
b
2
(
a
c
−
1
)
+
c
2
(
b
a
−
1
)
≥
0.
Sum a sin a/2>=p
Let
a
,
b
,
c
a,b,c
a
,
b
,
c
be sides of a triangle and
p
p
p
its semiperimeter. Show that
a
(
p
−
b
)
(
p
−
c
)
b
c
+
b
(
p
−
c
)
(
p
−
a
)
a
c
+
c
(
p
−
a
)
(
p
−
b
)
a
b
≥
p
a\sqrt{\frac{(p-b)(p-c)}{bc}}+b \sqrt{\frac{(p-c)(p-a)}{ac}}+c\sqrt{\frac{(p-a)(p-b)}{ab}}\geq p
a
b
c
(
p
−
b
)
(
p
−
c
)
+
b
a
c
(
p
−
c
)
(
p
−
a
)
+
c
ab
(
p
−
a
)
(
p
−
b
)
≥
p
Moldova tst iii, problem 3
Positive real numbers
a
,
b
,
c
a,b,c
a
,
b
,
c
satisfy the relation
a
b
c
=
1
abc=1
ab
c
=
1
. Prove the inequality:
a
+
3
(
a
+
1
)
2
+
b
+
3
(
b
+
1
)
2
+
c
+
3
(
c
+
1
)
2
≥
3
\frac{a+3}{(a+1)^{2}}+\frac{b+3}{(b+1)^{2}}+\frac{c+3}{(c+1)^{2}}\geq3
(
a
+
1
)
2
a
+
3
+
(
b
+
1
)
2
b
+
3
+
(
c
+
1
)
2
c
+
3
≥
3
.
2
3
Hide problems
Fixed point
Let
C
1
C_1
C
1
be a circle inside the circle
C
2
C_2
C
2
and let
P
P
P
in the interior of
C
1
C_1
C
1
,
Q
Q
Q
in the exterior of
C
2
C_2
C
2
. One draws variable lines
l
i
l_i
l
i
through
P
P
P
, not passing through
Q
Q
Q
. Let
l
i
l_i
l
i
intersect
C
1
C_1
C
1
in
A
i
,
B
i
A_i,B_i
A
i
,
B
i
, and let the circumcircle of
Q
A
i
B
i
QA_iB_i
Q
A
i
B
i
intersect
C
2
C_2
C
2
in
M
i
,
N
i
M_i,N_i
M
i
,
N
i
. Show that all lines
M
i
,
N
i
M_i,N_i
M
i
,
N
i
are concurrent.
Right-angled triangle and maximal area
Consider a right-angled triangle
A
B
C
ABC
A
BC
with the hypothenuse
A
B
=
1
AB=1
A
B
=
1
. The bisector of
∠
A
C
B
\angle{ACB}
∠
A
CB
cuts the medians
B
E
BE
BE
and
A
F
AF
A
F
at
P
P
P
and
M
M
M
, respectively. If
A
F
∩
B
E
=
{
P
}
{AF}\cap{BE}=\{P\}
A
F
∩
BE
=
{
P
}
, determine the maximum value of the area of
△
M
N
P
\triangle{MNP}
△
MNP
.
Modlova 3rd tst, problem 2
Let
n
∈
N
n\in N
n
∈
N
n
≥
2
n\geq2
n
≥
2
and the set
X
X
X
with
n
+
1
n+1
n
+
1
elements. The ordered sequences
(
a
1
,
a
2
,
…
,
a
n
)
(a_{1}, a_{2},\ldots,a_{n})
(
a
1
,
a
2
,
…
,
a
n
)
and
(
b
1
,
b
2
,
…
b
n
)
(b_{1},b_{2},\ldots b_{n})
(
b
1
,
b
2
,
…
b
n
)
of distinct elements of
X
X
X
are said to be
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
s
e
p
a
r
a
t
e
d
<
/
s
p
a
n
>
<span class='latex-italic'>separated</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
se
p
a
r
a
t
e
d
<
/
s
p
an
>
if there exists
i
≠
j
i\neq j
i
=
j
such that
a
i
=
b
j
a_{i}=b_{j}
a
i
=
b
j
. Determine the maximal number of ordered sequences of
n
n
n
elements from
X
X
X
such that any two of them are
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
s
e
p
a
r
a
t
e
d
<
/
s
p
a
n
>
<span class='latex-italic'>separated</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
se
p
a
r
a
t
e
d
<
/
s
p
an
>
. Note: ordered means that, for example
(
1
,
2
,
3
)
≠
(
2
,
3
,
1
)
(1,2,3)\neq(2,3,1)
(
1
,
2
,
3
)
=
(
2
,
3
,
1
)
.
4
3
Hide problems
Counting numbers on circle
Let
m
m
m
circles intersect in points
A
A
A
and
B
B
B
. We write numbers using the following algorithm: we write
1
1
1
in points
A
A
A
and
B
B
B
, in every midpoint of the open arc
A
B
AB
A
B
we write
2
2
2
, then between every two numbers written in the midpoint we write their sum and so on repeating
n
n
n
times. Let
r
(
n
,
m
)
r(n,m)
r
(
n
,
m
)
be the number of appearances of the number
n
n
n
writing all of them on our
m
m
m
circles. a) Determine
r
(
n
,
m
)
r(n,m)
r
(
n
,
m
)
; b) For
n
=
2006
n=2006
n
=
2006
, find the smallest
m
m
m
for which
r
(
n
,
m
)
r(n,m)
r
(
n
,
m
)
is a perfect square. Example for half arc:
1
−
1
1-1
1
−
1
;
1
−
2
−
1
1-2-1
1
−
2
−
1
;
1
−
3
−
2
−
3
−
1
1-3-2-3-1
1
−
3
−
2
−
3
−
1
;
1
−
4
−
3
−
5
−
2
−
5
−
3
−
4
−
1
1-4-3-5-2-5-3-4-1
1
−
4
−
3
−
5
−
2
−
5
−
3
−
4
−
1
;
1
−
5
−
4
−
7
−
3
−
8
−
5
−
7
−
2
−
7
−
5
−
8
−
3
−
7
−
4
−
5
−
1
1-5-4-7-3-8-5-7-2-7-5-8-3-7-4-5-1
1
−
5
−
4
−
7
−
3
−
8
−
5
−
7
−
2
−
7
−
5
−
8
−
3
−
7
−
4
−
5
−
1
...
Number of distinct solutions of x,y,z=a
Let
A
=
{
1
,
2
,
…
,
n
}
A=\{1,2,\ldots,n\}
A
=
{
1
,
2
,
…
,
n
}
. Find the number of unordered triples
(
X
,
Y
,
Z
)
(X,Y,Z)
(
X
,
Y
,
Z
)
that satisfy
X
⋃
Y
⋃
Z
=
A
X\bigcup Y \bigcup Z=A
X
⋃
Y
⋃
Z
=
A
Modlova 3rd tst, problem 4
Let
f
(
n
)
f(n)
f
(
n
)
denote the number of permutations
(
a
1
,
a
2
,
…
,
a
n
)
(a_{1}, a_{2}, \ldots ,a_{n})
(
a
1
,
a
2
,
…
,
a
n
)
of the set
{
1
,
2
,
…
,
n
}
\{1,2,\ldots,n\}
{
1
,
2
,
…
,
n
}
, which satisfy the conditions:
a
1
=
1
a_{1}=1
a
1
=
1
and
∣
a
i
−
a
i
+
1
∣
≤
2
|a_{i}-a_{i+1}|\leq2
∣
a
i
−
a
i
+
1
∣
≤
2
, for any
i
=
1
,
2
,
…
,
n
−
1
i=1,2,\ldots,n-1
i
=
1
,
2
,
…
,
n
−
1
. Prove that
f
(
2006
)
f(2006)
f
(
2006
)
is divisible by 3.