MathDB
Problems
Contests
National and Regional Contests
Ukraine Contests
Official Ukraine Selection Cycle
Kyiv City MO
2022 Kyiv City MO
2022 Kyiv City MO Round 1
2022 Kyiv City MO Round 1
Part of
2022 Kyiv City MO
Subcontests
(5)
Problem 5
4
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Problem 4
5
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Problem 3
3
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Subsets with sum equal to one
You are given
n
n
n
not necessarily distinct real numbers
a
1
,
a
2
,
…
,
a
n
a_1, a_2, \ldots, a_n
a
1
,
a
2
,
…
,
a
n
. Let's consider all
2
n
−
1
2^n-1
2
n
−
1
ways to select some nonempty subset of these numbers, and for each such subset calculate the sum of the selected numbers. What largest possible number of them could have been equal to
1
1
1
?For example, if
a
=
[
−
1
,
2
,
2
]
a = [-1, 2, 2]
a
=
[
−
1
,
2
,
2
]
, then we got
3
3
3
once,
4
4
4
once,
2
2
2
twice,
−
1
-1
−
1
once,
1
1
1
twice, so the total number of ones here is
2
2
2
.(Proposed by Anton Trygub)
Geometry from legendary author
In triangle
A
B
C
ABC
A
BC
∠
B
>
9
0
∘
\angle B > 90^\circ
∠
B
>
9
0
∘
. Tangents to this circle in points
A
A
A
and
B
B
B
meet at point
P
P
P
, and the line passing through
B
B
B
perpendicular to
B
C
BC
BC
meets the line
A
C
AC
A
C
at point
K
K
K
. Prove that
P
A
=
P
K
PA = PK
P
A
=
P
K
.(Proposed by Danylo Khilko)
Bisector and circles
Let
A
L
AL
A
L
be the inner bisector of triangle
A
B
C
ABC
A
BC
. The circle centered at
B
B
B
with radius
B
L
BL
B
L
meets the ray
A
L
AL
A
L
at points
L
L
L
and
E
E
E
, and the circle centered at
C
C
C
with radius
C
L
CL
C
L
meets the ray
A
L
AL
A
L
at points
L
L
L
and
D
D
D
. Show that
A
L
2
=
A
E
×
A
D
AL^2 = AE\times AD
A
L
2
=
A
E
×
A
D
.(Proposed by Mykola Moroz)
Problem 2
3
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Sticks forming non-degenerate triangles
There are
n
n
n
sticks which have distinct integer length. Suppose that it's possible to form a non-degenerate triangle from any
3
3
3
distinct sticks among them. It's also known that there are sticks of lengths
5
5
5
and
12
12
12
among them. What's the largest possible value of
n
n
n
under such conditions?(Proposed by Bogdan Rublov)
Hardest algebra ever
For any reals
x
,
y
x, y
x
,
y
, show the following inequality:
(
x
+
4
)
2
+
(
y
+
2
)
2
+
(
x
−
5
)
2
+
(
y
+
4
)
2
≤
(
x
−
2
)
2
+
(
y
−
6
)
2
+
(
x
−
5
)
2
+
(
y
−
6
)
2
+
20
\sqrt{(x+4)^2 + (y+2)^2} + \sqrt{(x-5)^2 + (y+4)^2} \le \sqrt{(x-2)^2 + (y-6)^2} + \sqrt{(x-5)^2 + (y-6)^2} + 20
(
x
+
4
)
2
+
(
y
+
2
)
2
+
(
x
−
5
)
2
+
(
y
+
4
)
2
≤
(
x
−
2
)
2
+
(
y
−
6
)
2
+
(
x
−
5
)
2
+
(
y
−
6
)
2
+
20
(Proposed by Bogdan Rublov)
Selecting pairs with composite sums
You are given
2
n
2n
2
n
distinct integers. What's the largest integer
C
C
C
such that you can always form at least
C
C
C
pairs from them, so that no integer is in more than one pair, and the sum of integers in each pair is a composite number?(Proposed by Anton Trygub)
Problem 1
5
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