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National and Regional Contests
Moldova Contests
Moldova Team Selection Test
2017 Moldova Team Selection Test
2017 Moldova Team Selection Test
Part of
Moldova Team Selection Test
Subcontests
(12)
12
1
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MDA 2017 TST 2nd combinatorics
There are
75
75
75
points in the plane, no three collinear. Prove that the number of acute triangles is no more than
70
%
70\%
70%
from the total number of triangles with vertices in these points.
11
1
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Surprising Moldova 17 TST B11 problem
Find all ordered pairs of nonnegative integers
(
x
,
y
)
(x,y)
(
x
,
y
)
such that
x
4
−
x
2
y
2
+
y
4
+
2
x
3
y
−
2
x
y
3
=
1.
x^4-x^2y^2+y^4+2x^3y-2xy^3=1.
x
4
−
x
2
y
2
+
y
4
+
2
x
3
y
−
2
x
y
3
=
1.
10
1
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MDA TST B10, floor function problem
Let
p
p
p
be an odd prime. Prove that the number
⌊
(
5
+
2
)
p
−
2
p
+
1
⌋
\left\lfloor \left(\sqrt{5}+2\right)^{p}-2^{p+1}\right\rfloor
⌊
(
5
+
2
)
p
−
2
p
+
1
⌋
is divisible by
20
p
20p
20
p
.
9
1
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P^2(1) \geq 2b_{n-1} [Moldova TST 2017, D3, P1]
Let
P
(
X
)
=
a
0
X
n
+
a
1
X
n
−
1
+
⋯
+
a
n
P(X)=a_{0}X^{n}+a_{1}X^{n-1}+\cdots+a_{n}
P
(
X
)
=
a
0
X
n
+
a
1
X
n
−
1
+
⋯
+
a
n
be a polynomial with real coefficients such that
a
0
>
0
a_{0}>0
a
0
>
0
and
a
n
≥
a
i
≥
0
,
a_{n}\geq a_{i}\geq 0,
a
n
≥
a
i
≥
0
,
for all
i
=
0
,
1
,
2
,
…
,
n
−
1.
i=0,1,2,\ldots,n-1.
i
=
0
,
1
,
2
,
…
,
n
−
1.
Prove that if
P
2
(
X
)
=
b
0
X
2
n
+
b
1
X
2
n
−
1
+
⋯
+
b
n
−
1
X
n
+
1
+
⋯
+
b
2
n
,
P^{2}(X)=b_{0}X^{2n}+b_{1}X^{2n-1}+\cdots+b_{n-1}X^{n+1}+\cdots+b_{2n},
P
2
(
X
)
=
b
0
X
2
n
+
b
1
X
2
n
−
1
+
⋯
+
b
n
−
1
X
n
+
1
+
⋯
+
b
2
n
,
then
P
2
(
1
)
≥
2
b
n
−
1
.
P^2(1)\geq 2b_{n-1}.
P
2
(
1
)
≥
2
b
n
−
1
.
8
1
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There are at least 120 students [2017 Moldova TST, D2, P4]
At a summer school there are
7
7
7
courses. Each participant was a student in at least one course, and each course was taken by exactly
40
40
40
students. It is known that for each
2
2
2
courses there were at most
9
9
9
students who took them both. Prove that at least
120
120
120
students participated at this summer school.
7
1
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Cute, but easy geometric inequality [2017 Moldova TST, Day 2, Problem 3]
Let
A
B
C
ABC
A
BC
be an acute triangle, and
H
H
H
its orthocenter. The distance from
H
H
H
to rays
B
C
BC
BC
,
C
A
CA
C
A
, and
A
B
AB
A
B
is denoted by
d
a
d_a
d
a
,
d
b
d_b
d
b
, and
d
c
d_c
d
c
, respectively. Let
R
R
R
be the radius of circumcenter of
△
A
B
C
\triangle ABC
△
A
BC
and
r
r
r
be the radius of incenter of
△
A
B
C
\triangle ABC
△
A
BC
. Prove the following inequality:
d
a
+
d
b
+
d
c
≤
3
R
2
4
r
d_a+d_b+d_c \le \frac{3R^2}{4r}
d
a
+
d
b
+
d
c
≤
4
r
3
R
2
.
6
1
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Basic inequality involving a condition [Moldova TST 2017, D2, P2]
Let
a
,
b
,
c
a,b,c
a
,
b
,
c
be positive real numbers that satisfy
a
+
b
+
c
=
a
b
c
a+b+c=abc
a
+
b
+
c
=
ab
c
. Prove that
(
1
+
a
2
)
(
1
+
b
2
)
+
(
1
+
b
2
)
(
1
+
c
2
)
+
(
1
+
a
2
)
(
1
+
c
2
)
−
(
1
+
a
2
)
(
1
+
b
2
)
(
1
+
c
2
)
≥
4.
\sqrt{(1+a^2)(1+b^2)}+\sqrt{(1+b^2)(1+c^2)}+\sqrt{(1+a^2)(1+c^2)}-\sqrt{(1+a^2)(1+b^2)(1+c^2)} \ge 4.
(
1
+
a
2
)
(
1
+
b
2
)
+
(
1
+
b
2
)
(
1
+
c
2
)
+
(
1
+
a
2
)
(
1
+
c
2
)
−
(
1
+
a
2
)
(
1
+
b
2
)
(
1
+
c
2
)
≥
4.
5
1
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Very easy functional equation
Find all continuous functions
f
:
R
→
R
f : R \rightarrow R
f
:
R
→
R
such, that
f
(
x
y
)
=
f
(
x
2
+
y
2
2
)
+
(
x
−
y
)
2
f(xy)= f\left(\frac{x^2+y^2}{2}\right)+(x-y)^2
f
(
x
y
)
=
f
(
2
x
2
+
y
2
)
+
(
x
−
y
)
2
for any real numbers
x
x
x
and
y
y
y
4
1
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n of the form...
Determine all natural numbers
n
n
n
of the form
n
=
[
a
,
b
]
+
[
b
,
c
]
+
[
c
,
a
]
n=[a,b]+[b,c]+[c,a]
n
=
[
a
,
b
]
+
[
b
,
c
]
+
[
c
,
a
]
where
a
,
b
,
c
a,b,c
a
,
b
,
c
are positive integers and
[
u
,
v
]
[u,v]
[
u
,
v
]
is the least common multiple of the integers
u
u
u
and
v
v
v
.
3
1
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P on altitude and concurrent lines
Let
ω
\omega
ω
be the circumcircle of the acute nonisosceles triangle
Δ
A
B
C
\Delta ABC
Δ
A
BC
. Point
P
P
P
lies on the altitude from
A
A
A
. Let
E
E
E
and
F
F
F
be the feet of the altitudes from P to
C
A
CA
C
A
,
B
A
BA
B
A
respectively. Circumcircle of triangle
Δ
A
E
F
\Delta AEF
Δ
A
EF
intersects the circle
ω
\omega
ω
in
G
G
G
, different from
A
A
A
. Prove that the lines
G
P
GP
GP
,
B
E
BE
BE
and
C
F
CF
CF
are concurrent.
2
1
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Complex root of a polynomial [Moldova TST 2017, D1, P2]
Let
f
(
X
)
=
a
n
X
n
+
a
n
−
1
X
n
−
1
+
⋯
+
a
1
X
+
a
0
f(X)=a_{n}X^{n}+a_{n-1}X^{n-1}+\cdots +a_{1}X+a_{0}
f
(
X
)
=
a
n
X
n
+
a
n
−
1
X
n
−
1
+
⋯
+
a
1
X
+
a
0
be a polynomial with real coefficients which satisfies
a
n
≥
a
n
−
1
≥
⋯
≥
a
1
≥
a
0
>
0.
a_{n}\geq a_{n-1}\geq \cdots \geq a_{1}\geq a_{0}>0.
a
n
≥
a
n
−
1
≥
⋯
≥
a
1
≥
a
0
>
0.
Prove that for every complex root
z
z
z
of this polynomial, we have
∣
z
∣
≤
1
|z|\leq 1
∣
z
∣
≤
1
.
1
1
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Inequality in Moldova TST 2017, D1, P1
Let the sequence
(
a
n
)
n
⩾
1
(a_{n})_{n\geqslant 1}
(
a
n
)
n
⩾
1
be defined as:
a
n
=
A
n
+
2
1
A
n
+
3
2
A
n
+
4
3
A
n
+
5
4
5
4
3
,
a_{n}=\sqrt{A_{n+2}^{1}\sqrt[3]{A_{n+3}^{2}\sqrt[4]{A_{n+4}^{3}\sqrt[5]{A_{n+5}^{4}}}}},
a
n
=
A
n
+
2
1
3
A
n
+
3
2
4
A
n
+
4
3
5
A
n
+
5
4
,
where
A
m
k
A_{m}^{k}
A
m
k
are defined by
A
m
k
=
(
m
k
)
⋅
k
!
.
A_{m}^{k}=\binom{m}{k}\cdot k!.
A
m
k
=
(
k
m
)
⋅
k
!
.
Prove that
a
n
<
119
120
⋅
n
+
7
3
.
a_{n}<\frac{119}{120}\cdot n+\frac{7}{3}.
a
n
<
120
119
⋅
n
+
3
7
.