MathDB
Problems
Contests
National and Regional Contests
Korea Contests
Korea Junior Mathematics Olympiad
2016 Korea Junior Math Olympiad
2016 Korea Junior Math Olympiad
Part of
Korea Junior Mathematics Olympiad
Subcontests
(7)
3
1
Hide problems
Split into three or less ranked sets
n
n
n
players participated in a competition. Any two players have played exactly one game, and there was no tie game. For a set of
k
(
≤
n
)
k(\le n)
k
(
≤
n
)
players, if it is able to line the players up so that each player won every player at the back, we call the set ranked. For each player who participated in the competition, the set of players who lost to the player is ranked. Prove that the whole set of players can be split into three or less ranked sets.
8
1
Hide problems
number of paths
One moving point in the coordinate plane can move right or up one position.
N
N
N
is a number of all paths : paths that moving point starts from
(
0
,
0
)
(0, 0)
(
0
,
0
)
, without passing
(
1
,
0
)
,
(
2
,
1
)
,
.
.
.
,
(
n
,
n
−
1
)
(1, 0), (2, 1), . . . , (n, n-1)
(
1
,
0
)
,
(
2
,
1
)
,
...
,
(
n
,
n
−
1
)
and moves
2
n
2n
2
n
times to
(
n
,
n
)
(n, n)
(
n
,
n
)
.
a
k
a_k
a
k
is a number of special paths : paths include in
N
N
N
, but
k
k
k
th moves to the right,
k
+
1
k+1
k
+
1
th moves to the up. find
1
N
(
a
1
+
a
2
+
.
.
.
+
a
2
n
−
1
)
\frac{1}{N} (a_1+a_2+ . . . + a_{2n-1})
N
1
(
a
1
+
a
2
+
...
+
a
2
n
−
1
)
5
1
Hide problems
prove polynomial
n
∈
N
+
n \in \mathbb {N^+}
n
∈
N
+
Prove that the following equation can be expressed as a polynomial about
n
n
n
.
[
2
1
]
+
[
2
2
]
+
[
2
3
]
+
.
.
.
+
[
2
n
2
]
\left[2\sqrt {1}\right]+\left[2\sqrt {2}\right]+\left[2\sqrt {3}\right]+ . . . +\left[2\sqrt {n^2}\right]
[
2
1
]
+
[
2
2
]
+
[
2
3
]
+
...
+
[
2
n
2
]
7
1
Hide problems
maximum of
positive integers
a
1
,
a
2
,
.
.
.
,
a
9
a_1, a_2, . . . , a_9
a
1
,
a
2
,
...
,
a
9
satisfying
a
1
+
a
2
+
.
.
.
+
a
9
=
90
a_1+a_2+ . . . +a_9 =90
a
1
+
a
2
+
...
+
a
9
=
90
find maximum of
1
a
1
⋅
2
a
2
⋅
.
.
.
⋅
9
a
9
a
1
!
⋅
a
2
!
⋅
.
.
.
⋅
a
9
!
\frac{1^{a_1} \cdot 2^{a_2} \cdot . . . \cdot 9^{a_9}}{a_1! \cdot a_2! \cdot . . . \cdot a_9!}
a
1
!
⋅
a
2
!
⋅
...
⋅
a
9
!
1
a
1
⋅
2
a
2
⋅
...
⋅
9
a
9
I was really shocked because there are no inequality problems at KJMO and the test difficulty even more lower...
6
1
Hide problems
nice geometry
circle
O
1
O_1
O
1
is tangent to
A
C
AC
A
C
,
B
C
BC
BC
(side of triangle
A
B
C
ABC
A
BC
) at point
D
,
E
D, E
D
,
E
. circle
O
2
O_2
O
2
include
O
1
O_1
O
1
, is tangent to
B
C
BC
BC
,
A
B
AB
A
B
(side of triangle
A
B
C
ABC
A
BC
) at point
E
,
F
E, F
E
,
F
The tangent of
O
2
O_2
O
2
at
P
(
D
E
∩
O
2
,
P
≠
E
)
P(DE \cap O_2, P \neq E)
P
(
D
E
∩
O
2
,
P
=
E
)
meets
A
B
AB
A
B
at
Q
Q
Q
. A line passing through
O
1
O_1
O
1
(center of
O
1
O_1
O
1
) and parallel to
B
O
2
BO_2
B
O
2
(
O
2
O_2
O
2
is also center of
O
2
O_2
O
2
) meets
B
C
BC
BC
at
G
G
G
,
E
Q
∩
A
C
=
K
,
K
G
∩
E
F
=
L
EQ \cap AC=K, KG \cap EF=L
EQ
∩
A
C
=
K
,
K
G
∩
EF
=
L
,
E
O
2
EO_2
E
O
2
meets circle
O
2
O_2
O
2
at
N
(
≠
E
)
N(\neq E)
N
(
=
E
)
,
L
O
2
∩
F
N
=
M
LO_2 \cap FN=M
L
O
2
∩
FN
=
M
. IF
N
N
N
is a middle point of
F
M
FM
FM
, prove that
B
G
=
2
E
G
BG=2EG
BG
=
2
EG
4
1
Hide problems
find all integer n
find all positive integer
n
n
n
, satisfying
n
(
n
+
2016
)
(
n
+
2
⋅
2016
)
(
n
+
3
⋅
2016
)
.
.
.
(
n
+
2015
⋅
2016
)
1
⋅
2
⋅
3
⋅
.
.
.
.
.
⋅
2016
\frac{n(n+2016)(n+2\cdot 2016)(n+3\cdot 2016) . . . (n+2015\cdot 2016)}{1\cdot 2 \cdot 3 \cdot . . . . . \cdot 2016}
1
⋅
2
⋅
3
⋅
.....
⋅
2016
n
(
n
+
2016
)
(
n
+
2
⋅
2016
)
(
n
+
3
⋅
2016
)
...
(
n
+
2015
⋅
2016
)
is positive integer.
1
1
Hide problems
prove integer
positive reals
a
1
,
a
2
,
.
.
.
a_1, a_2, . . .
a
1
,
a
2
,
...
satisfying (i)
a
n
+
1
=
a
1
2
⋅
a
2
2
⋅
.
.
.
⋅
a
n
2
−
3
a_{n+1}=a_1^2\cdot a_2^2 \cdot . . . \cdot a_n^2-3
a
n
+
1
=
a
1
2
⋅
a
2
2
⋅
...
⋅
a
n
2
−
3
(all positive integers
n
n
n
) (ii)
1
2
(
a
1
+
a
2
−
1
)
\frac{1}{2}(a_1+\sqrt{a_2-1})
2
1
(
a
1
+
a
2
−
1
)
is positive integer. prove that
1
2
(
a
1
⋅
a
2
⋅
.
.
.
⋅
a
n
+
a
n
+
1
−
1
)
\frac{1}{2}(a_1 \cdot a_2 \cdot . . . \cdot a_n + \sqrt{a_{n+1}-1})
2
1
(
a
1
⋅
a
2
⋅
...
⋅
a
n
+
a
n
+
1
−
1
)
is positive integer