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International Contests
Romanian Masters of Mathematics Collection
2019 Romanian Master of Mathematics Shortlist
2019 Romanian Master of Mathematics Shortlist
Part of
Romanian Masters of Mathematics Collection
Subcontests
(14)
original P6
1
Hide problems
Small degree implies linear factorisation
Let
P
(
x
)
P(x)
P
(
x
)
be a nonconstant complex coefficient polynomial and let
Q
(
x
,
y
)
=
P
(
x
)
−
P
(
y
)
.
Q(x,y)=P(x)-P(y).
Q
(
x
,
y
)
=
P
(
x
)
−
P
(
y
)
.
Suppose that polynomial
Q
(
x
,
y
)
Q(x,y)
Q
(
x
,
y
)
has exactly
k
k
k
linear factors unproportional two by tow (without counting repetitons). Let
R
(
x
,
y
)
R(x,y)
R
(
x
,
y
)
be factor of
Q
(
x
,
y
)
Q(x,y)
Q
(
x
,
y
)
of degree strictly smaller than
k
k
k
. Prove that
R
(
x
,
y
)
R(x,y)
R
(
x
,
y
)
is a product of linear polynomials.Note: The degree of nontrivial polynomial
∑
m
∑
n
c
m
,
n
x
m
y
n
\sum_{m}\sum_{n}c_{m,n}x^{m}y^{n}
∑
m
∑
n
c
m
,
n
x
m
y
n
is the maximum of
m
+
n
m+n
m
+
n
along all nonzero coefficients
c
m
,
n
.
c_{m,n}.
c
m
,
n
.
Two polynomials are proportional if one of them is the other times a complex constant.Proposed by Navid Safaie
original P4
1
Hide problems
Uniquely determined side
Let there be an equilateral triangle
A
B
C
ABC
A
BC
and a point
P
P
P
in its plane such that
A
P
<
B
P
<
C
P
.
AP<BP<CP.
A
P
<
BP
<
CP
.
Suppose that the lengths of segments
A
P
,
B
P
AP,BP
A
P
,
BP
and
C
P
CP
CP
uniquely determine the side of
A
B
C
ABC
A
BC
. Prove that
P
P
P
lies on the circumcircle of triangle
A
B
C
.
ABC.
A
BC
.
G5
1
Hide problems
concurrency wanted, starting with a tangential ABCD
A quadrilateral
A
B
C
D
ABCD
A
BC
D
is circumscribed about a circle with center
I
I
I
. A point
P
≠
I
P \ne I
P
=
I
is chosen inside
A
B
C
D
ABCD
A
BC
D
so that the triangles
P
A
B
,
P
B
C
,
P
C
D
,
PAB, PBC, PCD,
P
A
B
,
PBC
,
PC
D
,
and
P
D
A
PDA
P
D
A
have equal perimeters. A circle
Γ
\Gamma
Γ
centered at
P
P
P
meets the rays
P
A
,
P
B
,
P
C
PA, PB, PC
P
A
,
PB
,
PC
, and
P
D
PD
P
D
at
A
1
,
B
1
,
C
1
A_1, B_1, C_1
A
1
,
B
1
,
C
1
, and
D
1
D_1
D
1
, respectively. Prove that the lines
P
I
,
A
1
C
1
PI, A_1C_1
P
I
,
A
1
C
1
, and
B
1
D
1
B_1D_1
B
1
D
1
are concurrent.Ankan Bhattacharya, USA
C3
1
Hide problems
p_i +p_{i+1} +...+p_j is divisible by n
Fix an odd integer
n
>
1
n > 1
n
>
1
. For a permutation
p
p
p
of the set
{
1
,
2
,
.
.
.
,
n
}
\{1,2,...,n\}
{
1
,
2
,
...
,
n
}
, let S be the number of pairs of indices
(
i
,
j
)
(i, j)
(
i
,
j
)
,
1
≤
i
≤
j
≤
n
1 \le i \le j \le n
1
≤
i
≤
j
≤
n
, for which
p
i
+
p
i
+
1
+
.
.
.
+
p
j
p_i +p_{i+1} +...+p_j
p
i
+
p
i
+
1
+
...
+
p
j
is divisible by
n
n
n
. Determine the maximum possible value of
S
S
S
.Croatia
C2
1
Hide problems
largest no of moves the fairy chess piece leopard could have made
Fix an integer
n
≥
2
n \ge 2
n
≥
2
. A fairy chess piece leopard may move one cell up, or one cell to the right, or one cell diagonally down-left. A leopard is placed onto some cell of a
3
n
×
3
n
3n \times 3n
3
n
×
3
n
chequer board. The leopard makes several moves, never visiting a cell twice, and comes back to the starting cell. Determine the largest possible number of moves the leopard could have made. Dmitry Khramtsov, Russia
C1
1
Hide problems
I see equally many knights to my left and to my right, knights always tell truth
Let
k
k
k
and
N
N
N
be integers such that
k
>
1
k > 1
k
>
1
and
N
>
2
k
+
1
N > 2k + 1
N
>
2
k
+
1
. A number of
N
N
N
persons sit around the Round Table, equally spaced. Each person is either a knight (always telling the truth) or a liar (who always lies). Each person sees the nearest k persons clockwise, and the nearest
k
k
k
persons anticlockwise. Each person says: ''I see equally many knights to my left and to my right." Establish, in terms of
k
k
k
and
N
N
N
, whether the persons around the Table are necessarily all knights.Sergey Berlov, Russia
N1
1
Hide problems
set A contains at least (p - 1)(q + 1)/8 pairs whose entries are both even
Let
p
p
p
and
q
q
q
be relatively prime positive odd integers such that
1
<
p
<
q
1 < p < q
1
<
p
<
q
. Let
A
A
A
be a set of pairs of integers
(
a
,
b
)
(a, b)
(
a
,
b
)
, where
0
≤
a
≤
p
−
1
,
0
≤
b
≤
q
−
1
0 \le a \le p - 1, 0 \le b \le q - 1
0
≤
a
≤
p
−
1
,
0
≤
b
≤
q
−
1
, containing exactly one pair from each of the sets
{
(
a
,
b
)
,
(
a
+
1
,
b
+
1
)
}
,
{
(
a
,
q
−
1
)
,
(
a
+
1
,
0
)
}
,
{
(
p
−
1
,
b
)
,
(
0
,
b
+
1
)
}
\{(a, b),(a + 1, b + 1)\}, \{(a, q - 1), (a + 1, 0)\}, \{(p - 1,b),(0, b + 1)\}
{(
a
,
b
)
,
(
a
+
1
,
b
+
1
)}
,
{(
a
,
q
−
1
)
,
(
a
+
1
,
0
)}
,
{(
p
−
1
,
b
)
,
(
0
,
b
+
1
)}
whenever
0
≤
a
≤
p
−
2
0 \le a \le p - 2
0
≤
a
≤
p
−
2
and
0
≤
b
≤
q
−
2
0 \le b \le q - 2
0
≤
b
≤
q
−
2
. Show that
A
A
A
contains at least
(
p
−
1
)
(
q
+
1
)
/
8
(p - 1)(q + 1)/8
(
p
−
1
)
(
q
+
1
)
/8
pairs whose entries are both even.Agnijo Banerjee and Joe Benton, United Kingdom
G4 ver.II
1
Hide problems
< KCE = < LCP , 4 circles related, hard version
Let
Ω
\Omega
Ω
be the circumcircle of an acute-angled triangle
A
B
C
ABC
A
BC
. A point
D
D
D
is chosen on the internal bisector of
∠
A
C
B
\angle ACB
∠
A
CB
so that the points
D
D
D
and
C
C
C
are separated by
A
B
AB
A
B
. A circle
ω
\omega
ω
centered at
D
D
D
is tangent to the segment
A
B
AB
A
B
at
E
E
E
. The tangents to
ω
\omega
ω
through
C
C
C
meet the segment
A
B
AB
A
B
at
K
K
K
and
L
L
L
, where
K
K
K
lies on the segment
A
L
AL
A
L
. A circle
Ω
1
\Omega_1
Ω
1
is tangent to the segments
A
L
,
C
L
AL, CL
A
L
,
C
L
, and also to
Ω
\Omega
Ω
at point
M
M
M
. Similarly, a circle
Ω
2
\Omega_2
Ω
2
is tangent to the segments
B
K
,
C
K
BK, CK
B
K
,
C
K
, and also to
Ω
\Omega
Ω
at point
N
N
N
. The lines
L
M
LM
L
M
and
K
N
KN
K
N
meet at
P
P
P
. Prove that
∠
K
C
E
=
∠
L
C
P
\angle KCE = \angle LCP
∠
K
CE
=
∠
L
CP
.Poland
G4 ver.I
1
Hide problems
< KCE = < LCP , 4 circles related, easy version
Let
Ω
\Omega
Ω
be the circumcircle of an acute-angled triangle
A
B
C
ABC
A
BC
. Let
D
D
D
be the midpoint of the minor arc
A
B
AB
A
B
of
Ω
\Omega
Ω
. A circle
ω
\omega
ω
centered at
D
D
D
is tangent to
A
B
AB
A
B
at
E
E
E
. The tangents to
ω
\omega
ω
through
C
C
C
meet the segment
A
B
AB
A
B
at
K
K
K
and
L
L
L
, where
K
K
K
lies on the segment
A
L
AL
A
L
. A circle
Ω
1
\Omega_1
Ω
1
is tangent to the segments
A
L
,
C
L
AL, CL
A
L
,
C
L
, and also to
Ω
\Omega
Ω
at point
M
M
M
. Similarly, a circle
Ω
2
\Omega_2
Ω
2
is tangent to the segments
B
K
,
C
K
BK, CK
B
K
,
C
K
, and also to
Ω
\Omega
Ω
at point
N
N
N
. The lines
L
M
LM
L
M
and
K
N
KN
K
N
meet at
P
P
P
. Prove that
∠
K
C
E
=
∠
L
C
P
\angle KCE = \angle LCP
∠
K
CE
=
∠
L
CP
.Poland
G3
1
Hide problems
the concurrency point of 3 lines is the orthocenter, incenter, circumcenter
Let
A
B
C
ABC
A
BC
be an acute-angled triangle with
A
B
≠
A
C
AB \ne AC
A
B
=
A
C
, and let
I
I
I
and
O
O
O
be its incenter and circumcenter, respectively. Let the incircle touch
B
C
,
C
A
BC, CA
BC
,
C
A
and
A
B
AB
A
B
at
D
,
E
D, E
D
,
E
and
F
F
F
, respectively. Assume that the line through
I
I
I
parallel to
E
F
EF
EF
, the line through
D
D
D
parallel to
A
O
AO
A
O
, and the altitude from
A
A
A
are concurrent. Prove that the concurrency point is the orthocenter of the triangle
A
B
C
ABC
A
BC
. Petar Nizic-Nikolac, Croatia
G2
1
Hide problems
MC = MB + BH wanted, KH // AC, perpendiculars, external angle bisector
Let
A
B
C
ABC
A
BC
be an acute-angled triangle. The line through
C
C
C
perpendicular to
A
C
AC
A
C
meets the external angle bisector of
∠
A
B
C
\angle ABC
∠
A
BC
at
D
D
D
. Let
H
H
H
be the foot of the perpendicular from
D
D
D
onto
B
C
BC
BC
. The point
K
K
K
is chosen on
A
B
AB
A
B
so that
K
H
∥
A
C
KH \parallel AC
KH
∥
A
C
. Let
M
M
M
be the midpoint of
A
K
AK
A
K
. Prove that
M
C
=
M
B
+
B
H
MC = MB + BH
MC
=
MB
+
B
H
.Giorgi Arabidze, Georgia,
G1
1
Hide problems
circumcenter of BJK lies on line AC, median, right angle, circumcircle related
Let
B
M
BM
BM
be a median in an acute-angled triangle
A
B
C
ABC
A
BC
. A point
K
K
K
is chosen on the line through
C
C
C
tangent to the circumcircle of
△
B
M
C
\vartriangle BMC
△
BMC
so that
∠
K
B
C
=
9
0
∘
\angle KBC = 90^\circ
∠
K
BC
=
9
0
∘
. The segments
A
K
AK
A
K
and
B
M
BM
BM
meet at
J
J
J
. Prove that the circumcenter of
△
B
J
K
\triangle BJK
△
B
J
K
lies on the line
A
C
AC
A
C
.Aleksandr Kuznetsov, Russia
A1
1
Hide problems
f(a)=a,-a
Determine all the functions
f
:
R
↦
R
f:\mathbb R\mapsto\mathbb R
f
:
R
↦
R
satisfies the equation
f
(
a
2
+
a
b
+
f
(
b
2
)
)
=
a
f
(
b
)
+
b
2
+
f
(
a
2
)
∀
a
,
b
∈
R
f(a^2 +ab+ f(b^2))=af(b)+b^2+ f(a^2)\,\forall a,b\in\mathbb R
f
(
a
2
+
ab
+
f
(
b
2
))
=
a
f
(
b
)
+
b
2
+
f
(
a
2
)
∀
a
,
b
∈
R
A2
1
Hide problems
Inequality with permutations
Given a positive integer
n
n
n
, determine the maximal constant
C
n
C_n
C
n
satisfying the following condition: for any partition of the set
{
1
,
2
,
…
,
2
n
}
\{1,2,\ldots,2n \}
{
1
,
2
,
…
,
2
n
}
into two
n
n
n
-element subsets
A
A
A
and
B
B
B
, there exist labellings
a
1
,
a
2
,
…
,
a
n
a_1,a_2,\ldots,a_n
a
1
,
a
2
,
…
,
a
n
and
b
1
,
b
2
,
…
,
b
n
b_1,b_2,\ldots,b_n
b
1
,
b
2
,
…
,
b
n
of
A
A
A
and
B
B
B
, respectively, such that
(
a
1
−
b
1
)
2
+
(
a
2
−
b
2
)
2
+
…
+
(
a
n
−
b
n
)
2
≥
C
n
.
(a_1-b_1)^2+(a_2-b_2)^2+\ldots+(a_n-b_n)^2\ge C_n.
(
a
1
−
b
1
)
2
+
(
a
2
−
b
2
)
2
+
…
+
(
a
n
−
b
n
)
2
≥
C
n
.
(B. Serankou, M. Karpuk)