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Contests
International Contests
Romanian Masters of Mathematics Collection
2021 Romanian Master of Mathematics
2021 Romanian Master of Mathematics
Part of
Romanian Masters of Mathematics Collection
Subcontests
(6)
5
1
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Social Distancing Kingdom
Let
n
n
n
be a positive integer. The kingdom of Zoomtopia is a convex polygon with integer sides, perimeter
6
n
6n
6
n
, and
6
0
∘
60^\circ
6
0
∘
rotational symmetry (that is, there is a point
O
O
O
such that a
6
0
∘
60^\circ
6
0
∘
rotation about
O
O
O
maps the polygon to itself). In light of the pandemic, the government of Zoomtopia would like to relocate its
3
n
2
+
3
n
+
1
3n^2+3n+1
3
n
2
+
3
n
+
1
citizens at
3
n
2
+
3
n
+
1
3n^2+3n+1
3
n
2
+
3
n
+
1
points in the kingdom so that every two citizens have a distance of at least
1
1
1
for proper social distancing. Prove that this is possible. (The kingdom is assumed to contain its boundary.)Proposed by Ankan Bhattacharya, USA
4
1
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Numbers on a Board
Consider an integer
n
≥
2
n \ge 2
n
≥
2
and write the numbers
1
,
2
,
…
,
n
1, 2, \ldots, n
1
,
2
,
…
,
n
down on a board. A move consists in erasing any two numbers
a
a
a
and
b
b
b
, then writing down the numbers
a
+
b
a+b
a
+
b
and
∣
a
−
b
∣
\vert a-b \vert
∣
a
−
b
∣
on the board, and then removing repetitions (e.g., if the board contained the numbers
2
,
5
,
7
,
8
2, 5, 7, 8
2
,
5
,
7
,
8
, then one could choose the numbers
a
=
5
a = 5
a
=
5
and
b
=
7
b = 7
b
=
7
, obtaining the board with numbers
2
,
8
,
12
2, 8, 12
2
,
8
,
12
). For all integers
n
≥
2
n \ge 2
n
≥
2
, determine whether it is possible to be left with exactly two numbers on the board after a finite number of moves.Proposed by China
6
1
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(A,B)-polynomial on a board
Initially, a non-constant polynomial
S
(
x
)
S(x)
S
(
x
)
with real coefficients is written down on a board. Whenever the board contains a polynomial
P
(
x
)
P(x)
P
(
x
)
, not necessarily alone, one can write down on the board any polynomial of the form
P
(
C
+
x
)
P(C + x)
P
(
C
+
x
)
or
C
+
P
(
x
)
C + P(x)
C
+
P
(
x
)
where
C
C
C
is a real constant. Moreover, if the board contains two (not necessarily distinct) polynomials
P
(
x
)
P(x)
P
(
x
)
and
Q
(
x
)
Q(x)
Q
(
x
)
, one can write
P
(
Q
(
x
)
)
P(Q(x))
P
(
Q
(
x
))
and
P
(
x
)
+
Q
(
x
)
P(x) + Q(x)
P
(
x
)
+
Q
(
x
)
down on the board. No polynomial is ever erased from the board. Given two sets of real numbers,
A
=
{
a
1
,
a
2
,
…
,
a
n
}
A = \{ a_1, a_2, \dots, a_n \}
A
=
{
a
1
,
a
2
,
…
,
a
n
}
and
B
=
{
b
1
,
…
,
b
n
}
B = \{ b_1, \dots, b_n \}
B
=
{
b
1
,
…
,
b
n
}
, a polynomial
f
(
x
)
f(x)
f
(
x
)
with real coefficients is
(
A
,
B
)
(A,B)
(
A
,
B
)
-nice if
f
(
A
)
=
B
f(A) = B
f
(
A
)
=
B
, where
f
(
A
)
=
{
f
(
a
i
)
:
i
=
1
,
2
,
…
,
n
}
f(A) = \{ f(a_i) : i = 1, 2, \dots, n \}
f
(
A
)
=
{
f
(
a
i
)
:
i
=
1
,
2
,
…
,
n
}
. Determine all polynomials
S
(
x
)
S(x)
S
(
x
)
that can initially be written down on the board such that, for any two finite sets
A
A
A
and
B
B
B
of real numbers, with
∣
A
∣
=
∣
B
∣
|A| = |B|
∣
A
∣
=
∣
B
∣
, one can produce an
(
A
,
B
)
(A,B)
(
A
,
B
)
-nice polynomial in a finite number of steps. Proposed by Navid Safaei, Iran
2
1
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Xenia and Sergey play an NT game
Xenia and Sergey play the following game. Xenia thinks of a positive integer
N
N
N
not exceeding
5000
5000
5000
. Then she fixes
20
20
20
distinct positive integers
a
1
,
a
2
,
⋯
,
a
20
a_1, a_2, \cdots, a_{20}
a
1
,
a
2
,
⋯
,
a
20
such that, for each
k
=
1
,
2
,
⋯
,
20
k = 1,2,\cdots,20
k
=
1
,
2
,
⋯
,
20
, the numbers
N
N
N
and
a
k
a_k
a
k
are congruent modulo
k
k
k
. By a move, Sergey tells Xenia a set
S
S
S
of positive integers not exceeding
20
20
20
, and she tells him back the set
{
a
k
:
k
∈
S
}
\{a_k : k \in S\}
{
a
k
:
k
∈
S
}
without spelling out which number corresponds to which index. How many moves does Sergey need to determine for sure the number Xenia thought of?Sergey Kudrya, Russia
3
1
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RMM 2021 Problem 3
A number of
17
17
17
workers stand in a row. Every contiguous group of at least
2
2
2
workers is a
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
b
r
i
g
a
d
e
<
/
s
p
a
n
>
<span class='latex-italic'>brigade</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
b
r
i
g
a
d
e
<
/
s
p
an
>
. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker’s number of assignments is divisible by
4
4
4
. Prove that the number of such ways to assign the leaders is divisible by
17
17
17
.Mikhail Antipov, Russia
1
1
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How many points is too many
Let
T
1
,
T
2
,
T
3
,
T
4
T_1, T_2, T_3, T_4
T
1
,
T
2
,
T
3
,
T
4
be pairwise distinct collinear points such that
T
2
T_2
T
2
lies between
T
1
T_1
T
1
and
T
3
T_3
T
3
, and
T
3
T_3
T
3
lies between
T
2
T_2
T
2
and
T
4
T_4
T
4
. Let
ω
1
\omega_1
ω
1
be a circle through
T
1
T_1
T
1
and
T
4
T_4
T
4
; let
ω
2
\omega_2
ω
2
be the circle through
T
2
T_2
T
2
and internally tangent to
ω
1
\omega_1
ω
1
at
T
1
T_1
T
1
; let
ω
3
\omega_3
ω
3
be the circle through
T
3
T_3
T
3
and externally tangent to
ω
2
\omega_2
ω
2
at
T
2
T_2
T
2
; and let
ω
4
\omega_4
ω
4
be the circle through
T
4
T_4
T
4
and externally tangent to
ω
3
\omega_3
ω
3
at
T
3
T_3
T
3
. A line crosses
ω
1
\omega_1
ω
1
at
P
P
P
and
W
W
W
,
ω
2
\omega_2
ω
2
at
Q
Q
Q
and
R
R
R
,
ω
3
\omega_3
ω
3
at
S
S
S
and
T
T
T
, and
ω
4
\omega_4
ω
4
at
U
U
U
and
V
V
V
, the order of these points along the line being
P
,
Q
,
R
,
S
,
T
,
U
,
V
,
W
P,Q,R,S,T,U,V,W
P
,
Q
,
R
,
S
,
T
,
U
,
V
,
W
. Prove that
P
Q
+
T
U
=
R
S
+
V
W
PQ + TU = RS + VW
PQ
+
T
U
=
RS
+
VW
Geza Kos, Hungary