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Problems
Contests
International Contests
Romanian Masters of Mathematics Collection
2018 Romanian Master of Mathematics Shortlist
2018 Romanian Master of Mathematics Shortlist
Part of
Romanian Masters of Mathematics Collection
Subcontests
(8)
G1
1
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Equal angles with a point on altitude
Let
A
B
C
ABC
A
BC
be a triangle and let
H
H
H
be the orthogonal projection of
A
A
A
on the line
B
C
BC
BC
. Let
K
K
K
be a point on the segment
A
H
AH
A
H
such that
A
H
=
3
K
H
AH = 3 KH
A
H
=
3
KH
. Let
O
O
O
be the circumcenter of triangle
A
B
C
ABC
A
BC
and let
M
M
M
and
N
N
N
be the midpoints of sides
A
C
AC
A
C
and
A
B
AB
A
B
respectively. The lines
K
O
KO
K
O
and
M
N
MN
MN
meet at a point
Z
Z
Z
and the perpendicular at
Z
Z
Z
to
O
K
OK
O
K
meets lines
A
B
,
A
C
AB, AC
A
B
,
A
C
at
X
X
X
and
Y
Y
Y
respectively. Show that
∠
X
K
Y
=
∠
C
K
B
\angle XKY = \angle CKB
∠
X
K
Y
=
∠
C
K
B
.Italy
G2
1
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Two tangents meet on BC
Let
△
A
B
C
\triangle ABC
△
A
BC
be a triangle, and let
S
S
S
and
T
T
T
be the midpoints of the sides
B
C
BC
BC
and
C
A
CA
C
A
, respectively. Suppose
M
M
M
is the midpoint of the segment
S
T
ST
ST
and the circle
ω
\omega
ω
through
A
,
M
A, M
A
,
M
and
T
T
T
meets the line
A
B
AB
A
B
again at
N
N
N
. The tangents of
ω
\omega
ω
at
M
M
M
and
N
N
N
meet at
P
P
P
. Prove that
P
P
P
lies on
B
C
BC
BC
if and only if the triangle
A
B
C
ABC
A
BC
is isosceles with apex at
A
A
A
.Proposed by Reza Kumara, Indonesia
N1
1
Hide problems
Polynomial f in Z[x] satisfies f(p)|2^p-2
Determine all polynomials
f
f
f
with integer coefficients such that
f
(
p
)
f(p)
f
(
p
)
is a divisor of
2
p
−
2
2^p-2
2
p
−
2
for every odd prime
p
p
p
.[I]Proposed by Italy
N2
1
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b-ary Fibonacci Numbers
Prove that for each positive integer
k
k
k
there exists a number base
b
b
b
along with
k
k
k
triples of Fibonacci numbers
(
F
u
,
F
v
,
F
w
)
(F_u,F_v,F_w)
(
F
u
,
F
v
,
F
w
)
such that when they are written in base
b
b
b
, their concatenation is also a Fibonacci number written in base
b
b
b
. (Fibonacci numbers are defined by
F
1
=
F
2
=
1
F_1 = F_2 = 1
F
1
=
F
2
=
1
and
F
n
+
2
=
F
n
+
1
+
F
n
F_{n+2} = F_{n+1} + F_n
F
n
+
2
=
F
n
+
1
+
F
n
for all positive integers
n
n
n
.) Proposed by Serbia
C3
1
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Outcome of all the matches is unique
N
N
N
teams take part in a league. Every team plays every other team exactly once during the league, and receives 2 points for each win, 1 point for each draw, and 0 points for each loss. At the end of the league, the sequence of total points in descending order
A
=
(
a
1
≥
a
2
≥
⋯
≥
a
N
)
\mathcal{A} = (a_1 \ge a_2 \ge \cdots \ge a_N )
A
=
(
a
1
≥
a
2
≥
⋯
≥
a
N
)
is known, as well as which team obtained which score. Find the number of sequences
A
\mathcal{A}
A
such that the outcome of all matches is uniquely determined by this information.[I]Proposed by Dominic Yeo, United Kingdom.
C2
1
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Least number of integer valued coins
Fix integers
n
≥
k
≥
2
n\ge k\ge 2
n
≥
k
≥
2
. We call a collection of integral valued coins
n
−
d
i
v
e
r
s
e
n-diverse
n
−
d
i
v
erse
if no value occurs in it more than
n
n
n
times. Given such a collection, a number
S
S
S
is
n
−
r
e
a
c
h
a
b
l
e
n-reachable
n
−
re
a
c
hab
l
e
if that collection contains
n
n
n
coins whose sum of values equals
S
S
S
. Find the least positive integer
D
D
D
such that for any
n
n
n
-diverse collection of
D
D
D
coins there are at least
k
k
k
numbers that are
n
n
n
-reachable.[I]Proposed by Alexandar Ivanov, Bulgaria.
C1
1
Hide problems
Operation on lattice points
Call a point in the Cartesian plane with integer coordinates a
l
a
t
t
i
c
e
lattice
l
a
tt
i
ce
p
o
i
n
t
point
p
o
in
t
. Given a finite set
S
\mathcal{S}
S
of lattice points we repeatedly perform the following operation: given two distinct lattice points
A
,
B
A, B
A
,
B
in
S
\mathcal{S}
S
and two distinct lattice points
C
,
D
C, D
C
,
D
not in
S
\mathcal{S}
S
such that
A
C
B
D
ACBD
A
CB
D
is a parallelogram with
A
B
>
C
D
AB > CD
A
B
>
C
D
, we replace
A
,
B
A, B
A
,
B
by
C
,
D
C, D
C
,
D
. Show that only finitely many such operations can be performed.[I]Proposed by Joe Benton, United Kingdom.
A1
1
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2018 RMM Shortlist A1
Let
m
m
m
and
n
n
n
be integers greater than
2
2
2
, and let
A
A
A
and
B
B
B
be non-constant polynomials with complex coefficients, at least one of which has a degree greater than
1
1
1
. Prove that if the degree of the polynomial
A
m
−
B
n
A^m-B^n
A
m
−
B
n
is less than
min
(
m
,
n
)
\min(m,n)
min
(
m
,
n
)
, then
A
m
=
B
n
A^m=B^n
A
m
=
B
n
. Proposed by Tobi Moektijono, Indonesia