MathDB
Problems
Contests
National and Regional Contests
Japan Contests
Japan MO Finals
2020 Japan MO Finals
2020 Japan MO Finals
Part of
Japan MO Finals
Subcontests
(5)
4
1
Hide problems
Japan MO Finals 2020-4
Let
n
≥
2
n\geq 2
n
≥
2
be an integer.
3
n
3n
3
n
distinct points are plotted on the circle, where A and B perform the following operation : Firstly,
A
A
A
picks exactly 2 points which haven't been connected yet and connects them by a segment. Secondly, B picks exactly 1 point with no piece and place a piece. Prove that, after consecutive
n
n
n
operations, despite how B acts, A can make the number of segments no less than
n
−
1
6
\displaystyle \frac{n-1}{6}
6
n
−
1
, which connect a point with a piece and a point with no piece.
5
1
Hide problems
Japan MO Finals 2020-5
Find all infinite sequences of positive integers
{
a
n
}
n
≥
1
\{a_{n}\}_{n\geq 1}
{
a
n
}
n
≥
1
satisfying the following condition : there exists a positive constant
c
c
c
such that
gcd
(
a
m
+
n
,
a
n
+
m
)
>
c
(
m
+
n
)
\gcd(a_{m}+n,a_{n}+m)>c(m+n)
g
cd
(
a
m
+
n
,
a
n
+
m
)
>
c
(
m
+
n
)
holds for all pairs of positive integers
(
m
,
n
)
(m,n)
(
m
,
n
)
.
2
1
Hide problems
Japan MO Finals 2020-2
Triangle
A
B
C
ABC
A
BC
satisfies
B
C
<
A
B
BC<AB
BC
<
A
B
and
B
C
<
A
C
BC<AC
BC
<
A
C
. Points
D
,
E
D,E
D
,
E
lie on segments
A
B
,
A
C
AB,AC
A
B
,
A
C
respectively, satisfying
B
D
=
C
E
=
B
C
BD=CE=BC
B
D
=
CE
=
BC
. Lines
B
E
BE
BE
and
C
D
CD
C
D
meet at point
P
P
P
, circumcircles of triangle
A
B
E
ABE
A
BE
and
A
C
D
ACD
A
C
D
meet at point
Q
Q
Q
other than
A
A
A
. Prove that lines
P
Q
PQ
PQ
and
B
C
BC
BC
are perpendicular.
3
1
Hide problems
Japan MO Finals 2020-3
Find all functions
f
:
Z
+
→
Z
+
f:\mathbb{Z}^{+}\to\mathbb{Z}^{+}
f
:
Z
+
→
Z
+
such that
m
2
+
f
(
n
)
2
+
(
m
−
f
(
n
)
)
2
≥
f
(
m
)
2
+
n
2
m^{2}+f(n)^{2}+(m-f(n))^{2}\geq f(m)^{2}+n^{2}
m
2
+
f
(
n
)
2
+
(
m
−
f
(
n
)
)
2
≥
f
(
m
)
2
+
n
2
for all pairs of positive integers
(
m
,
n
)
(m,n)
(
m
,
n
)
.
1
1
Hide problems
Japan MO Finals 2020-1
Find all pairs of positive integers
(
m
,
n
)
(m,n)
(
m
,
n
)
such that
n
2
+
1
2
m
\frac{n^2+1}{2m}
2
m
n
2
+
1
and
2
n
−
1
+
m
+
4
\sqrt{2^{n-1}+m+4}
2
n
−
1
+
m
+
4
are both integers.