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Problems
Contests
National and Regional Contests
Ukraine Contests
Official Ukraine Selection Cycle
Ukraine Team Selection Test
2019 Ukraine Team Selection Test
2019 Ukraine Team Selection Test
Part of
Ukraine Team Selection Test
Subcontests
(3)
2
2
Hide problems
x divides p(x) if p(n+1)p(n+2)...p(n+k) / p(1)p(2)...p(k) is an integer
Polynomial
p
(
x
)
p(x)
p
(
x
)
with real coefficients, which is different from the constant, has the following property: for any naturals
n
n
n
and
k
k
k
the
p
(
n
+
1
)
p
(
n
+
2
)
.
.
.
p
(
n
+
k
)
p
(
1
)
p
(
2
)
.
.
.
p
(
k
)
\frac{p(n+1)p(n+2)...p(n+k)}{p(1)p(2)...p(k)}
p
(
1
)
p
(
2
)
...
p
(
k
)
p
(
n
+
1
)
p
(
n
+
2
)
...
p
(
n
+
k
)
is an integer. Prove that this polynomial is divisible by
x
x
x
.
2n rooks in a regular hexagon split into 6n^2 equilateral triangles
There is a regular hexagon that is cut direct to
6
n
2
6n^2
6
n
2
equilateral triangles (Fig.). There are arranged
2
n
2n
2
n
rooks, neither of which beats each other (the rooks hit in directions parallel to sides of the hexagon). Prove that if we consider chess coloring all
6
n
2
6n^2
6
n
2
equilateral triangles, then the number of rooks that stand on black triangles will be equal to the number of rooks standing on white triangles. https://cdn.artofproblemsolving.com/attachments/d/0/43ce6c5c966f60a8ec893d5d8cd31e33c43fc0.png [hide=original wording] Є правильний шестикутник, що розрізаний прямими на 6n^2 правильних трикутників (рис. 2). У них розставлені 2n тур, ніякі дві з яких не б'ють одна одну (тура б'є в напрямках, що паралельні до сторін шестикутника). Доведіть, що якщо розглянути шахове розфарбування всіх 6n^2 правильних трикутників, то тоді кількість тур, що стоять на чорних трикутниках, буде рівна кількості тур, що стоять на білих трикутниках.
3
1
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tangent circumcircles if BA_1 = 3A_1C, altitudes, orthocenter related
Given an acute triangle
A
B
C
ABC
A
BC
. It's altitudes
A
A
1
,
B
B
1
AA_1 , BB_1
A
A
1
,
B
B
1
and
C
C
1
CC_1
C
C
1
intersect at a point
H
H
H
, the orthocenter of
△
A
B
C
\vartriangle ABC
△
A
BC
. Let the lines
B
1
C
1
B_1C_1
B
1
C
1
and
A
A
1
AA_1
A
A
1
intersect at a point
K
K
K
, point
M
M
M
be the midpoint of the segment
A
H
AH
A
H
. Prove that the circumscribed circle of
△
M
K
B
1
\vartriangle MKB_1
△
M
K
B
1
touches the circumscribed circle of
△
A
B
C
\vartriangle ABC
△
A
BC
if and only if
B
A
1
=
3
A
1
C
BA1 = 3A1C
B
A
1
=
3
A
1
C
.(Bondarenko Mykhailo)
1
1
Hide problems
incenter lies on P_1T_1, <ABC=60^o, PI //BC, TI //AB, AP_1 = B , CT_1 = BT
In a triangle
A
B
C
ABC
A
BC
,
∠
A
B
C
=
6
0
o
\angle ABC= 60^o
∠
A
BC
=
6
0
o
, point
I
I
I
is the incenter. Let the points
P
P
P
and
T
T
T
on the sides
A
B
AB
A
B
and
B
C
BC
BC
respectively such that
P
I
∥
B
C
PI \parallel BC
P
I
∥
BC
and
T
I
∥
A
B
TI \parallel AB
T
I
∥
A
B
, and points
P
1
P_1
P
1
and
T
1
T_1
T
1
on the sides
A
B
AB
A
B
and
B
C
BC
BC
respectively such that
A
P
1
=
B
P
AP_1 = BP
A
P
1
=
BP
and
C
T
1
=
B
T
CT_1 = BT
C
T
1
=
BT
. Prove that point
I
I
I
lies on segment
P
1
T
1
P_1T_1
P
1
T
1
.(Anton Trygub)