MathDB
Problems
Contests
National and Regional Contests
Belarus Contests
Belarus Team Selection Test
2015 Belarus Team Selection Test
2015 Belarus Team Selection Test
Part of
Belarus Team Selection Test
Subcontests
(4)
4
2
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(a+b+c)^5 >= 81 (a^2+b^2+c^2)abc for a,b,c>0
Prove that
(
a
+
b
+
c
)
5
≥
81
(
a
2
+
b
2
+
c
2
)
a
b
c
(a+b+c)^5 \ge 81 (a^2+b^2+c^2)abc
(
a
+
b
+
c
)
5
≥
81
(
a
2
+
b
2
+
c
2
)
ab
c
for any positive real numbers
a
,
b
,
c
a,b,c
a
,
b
,
c
I.Gorodnin
p(x^2)=p(x)q(1-x)+p(1-x)q(x)
Find all pairs of polynomials
p
(
x
)
,
q
(
x
)
∈
R
[
x
]
p(x),q(x)\in R[x]
p
(
x
)
,
q
(
x
)
∈
R
[
x
]
satisfying the equality
p
(
x
2
)
=
p
(
x
)
q
(
1
−
x
)
+
p
(
1
−
x
)
q
(
x
)
p(x^2)=p(x)q(1-x)+p(1-x)q(x)
p
(
x
2
)
=
p
(
x
)
q
(
1
−
x
)
+
p
(
1
−
x
)
q
(
x
)
for all real
x
x
x
.I.Voronovich
3
2
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CX is angle bisector of <ACB if C,X,Y are collinear, touchpoints of incircle
The incircle of the triangle
A
B
C
ABC
A
BC
touches the sides
A
C
AC
A
C
and
B
C
BC
BC
at points
P
P
P
and
Q
Q
Q
respectively.
N
N
N
and
M
M
M
are the midpoints of
A
C
AC
A
C
and
B
C
BC
BC
respectively. Let
X
=
A
M
∩
B
P
,
Y
=
B
N
∩
A
Q
X=AM\cap BP, Y=BN\cap AQ
X
=
A
M
∩
BP
,
Y
=
BN
∩
A
Q
. Given
C
,
X
,
Y
C,X,Y
C
,
X
,
Y
are collinear, prove that
C
X
CX
CX
is the angle bisector of the angle
A
C
B
ACB
A
CB
.I. Gorodnin
cyclic wanted, 3 incircles related
Let the incircle of the triangle
A
B
C
ABC
A
BC
touch the side
A
B
AB
A
B
at point
Q
Q
Q
. The incircles of the triangles
Q
A
C
QAC
Q
A
C
and
Q
B
C
QBC
QBC
touch
A
Q
,
A
C
AQ,AC
A
Q
,
A
C
and
B
Q
,
B
C
BQ,BC
BQ
,
BC
at points
P
,
T
P,T
P
,
T
and
D
,
F
D,F
D
,
F
respectively. Prove that
P
D
F
T
PDFT
P
D
FT
is a cyclic quadrilateral.I.Gorodnin
1
6
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2
4
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