MathDB
Problems
Contests
National and Regional Contests
Korea Contests
Korea National Olympiad
2019 Korea National Olympiad
2019 Korea National Olympiad
Part of
Korea National Olympiad
Subcontests
(8)
2
1
Hide problems
Circle centered on A-excenter passing A
Triangle
A
B
C
ABC
A
BC
is an scalene triangle. Let
I
I
I
the incenter,
Ω
\Omega
Ω
the circumcircle,
E
E
E
the
A
A
A
-excenter of triangle
A
B
C
ABC
A
BC
. Let
Γ
\Gamma
Γ
the circle centered at
E
E
E
and passes
A
A
A
.
Γ
\Gamma
Γ
and
Ω
\Omega
Ω
intersect at point
D
(
≠
A
)
D(\neq A)
D
(
=
A
)
, and the perpendicular line of
B
C
BC
BC
which passes
A
A
A
meets
Γ
\Gamma
Γ
at point
K
(
≠
A
)
K(\neq A)
K
(
=
A
)
.
L
L
L
is the perpendicular foot from
I
I
I
to
A
C
AC
A
C
. Now if
A
E
AE
A
E
and
D
K
DK
DK
intersects at
F
F
F
, prove that
B
E
⋅
C
I
=
2
⋅
C
F
⋅
C
L
BE\cdot CI=2\cdot CF\cdot CL
BE
⋅
C
I
=
2
⋅
CF
⋅
C
L
.
1
1
Hide problems
Inequality in sequence
The sequence
a
1
,
a
2
,
.
.
.
,
a
2019
{a_1, a_2, ..., a_{2019}}
a
1
,
a
2
,
...
,
a
2019
satisfies the following condition.
a
1
=
1
,
a
n
+
1
=
2019
a
n
+
1
a_1=1, a_{n+1}=2019a_{n}+1
a
1
=
1
,
a
n
+
1
=
2019
a
n
+
1
Now let
x
1
,
x
2
,
.
.
.
,
x
2019
x_1, x_2, ..., x_{2019}
x
1
,
x
2
,
...
,
x
2019
real numbers such that
x
1
=
a
2019
,
x
2019
=
a
1
x_1=a_{2019}, x_{2019}=a_1
x
1
=
a
2019
,
x
2019
=
a
1
(The others are arbitary.) Prove that
∑
k
=
1
2018
(
x
k
+
1
−
2019
x
k
−
1
)
2
≥
∑
k
=
1
2018
(
a
2019
−
k
−
2019
a
2020
−
k
−
1
)
2
\sum_{k=1}^{2018} (x_{k+1}-2019x_k-1)^2 \ge \sum_{k=1}^{2018} (a_{2019-k}-2019a_{2020-k}-1)^2
∑
k
=
1
2018
(
x
k
+
1
−
2019
x
k
−
1
)
2
≥
∑
k
=
1
2018
(
a
2019
−
k
−
2019
a
2020
−
k
−
1
)
2
4
1
Hide problems
Three three-dimensional vectors modulo three!
Let
(
x
1
,
y
1
,
z
1
)
,
(
x
2
,
y
2
,
z
2
)
,
⋯
,
(
x
19
,
y
19
,
z
19
)
(x_1, y_1, z_1), (x_2, y_2, z_2), \cdots, (x_{19}, y_{19}, z_{19})
(
x
1
,
y
1
,
z
1
)
,
(
x
2
,
y
2
,
z
2
)
,
⋯
,
(
x
19
,
y
19
,
z
19
)
be integers. Prove that there exist pairwise distinct subscripts
i
,
j
,
k
i, j, k
i
,
j
,
k
such that
x
i
+
x
j
+
x
k
x_i+x_j+x_k
x
i
+
x
j
+
x
k
,
y
i
+
y
j
+
y
k
y_i+y_j+y_k
y
i
+
y
j
+
y
k
,
z
i
+
z
j
+
z
k
z_i+z_j+z_k
z
i
+
z
j
+
z
k
are all multiples of
3
3
3
.
8
1
Hide problems
There are even number of Hamiltonian cycles
There are two countries
A
A
A
and
B
B
B
, where each countries have
n
(
≥
2
)
n(\ge 2)
n
(
≥
2
)
airports. There are some two-way flights among airports of
A
A
A
and
B
B
B
, so that each airport has exactly
3
3
3
flights. There might be multiple flights among two airports; and there are no flights among airports of the same country. A travel agency wants to plan an exotic traveling course which travels through all
2
n
2n
2
n
airports exactly once, and returns to the initial airport. If
N
N
N
denotes the number of all exotic traveling courses, then prove that
N
4
n
\frac{N}{4n}
4
n
N
is an even integer.(Here, note that two exotic traveling courses are different if their starting place are different.)
6
1
Hide problems
Mansion point and incenter
In acute triangle
A
B
C
ABC
A
BC
,
A
B
>
A
C
AB>AC
A
B
>
A
C
. Let
I
I
I
the incenter,
Ω
\Omega
Ω
the circumcircle of triangle
A
B
C
ABC
A
BC
, and
D
D
D
the foot of perpendicular from
A
A
A
to
B
C
BC
BC
.
A
I
AI
A
I
intersects
Ω
\Omega
Ω
at point
M
(
≠
A
)
M(\neq A)
M
(
=
A
)
, and the line which passes
M
M
M
and perpendicular to
A
M
AM
A
M
intersects
A
D
AD
A
D
at point
E
E
E
. Now let
F
F
F
the foot of perpendicular from
I
I
I
to
A
D
AD
A
D
. Prove that
I
D
⋅
A
M
=
I
E
⋅
A
F
ID\cdot AM=IE\cdot AF
I
D
⋅
A
M
=
I
E
⋅
A
F
.
7
1
Hide problems
Existance of such m
For prime
p
≡
1
(
m
o
d
7
)
p\equiv 1\pmod{7}
p
≡
1
(
mod
7
)
, prove that there exists some positive integer
m
m
m
such that
m
3
+
m
2
−
2
m
−
1
m^3+m^2-2m-1
m
3
+
m
2
−
2
m
−
1
is a multiple of
p
p
p
.
5
1
Hide problems
Functional Equation
Find all functions
f
f
f
such that
f
:
R
→
R
f:\mathbb{R}\rightarrow \mathbb{R}
f
:
R
→
R
and
f
(
f
(
x
)
−
x
+
y
2
)
=
y
f
(
y
)
f(f(x)-x+y^2)=yf(y)
f
(
f
(
x
)
−
x
+
y
2
)
=
y
f
(
y
)
3
1
Hide problems
Residues
Suppose that positive integers
m
,
n
,
k
m,n,k
m
,
n
,
k
satisfy the equations
m
2
+
1
=
2
n
2
,
2
m
2
+
1
=
11
k
2
.
m^2+1=2n^2, 2m^2+1=11k^2.
m
2
+
1
=
2
n
2
,
2
m
2
+
1
=
11
k
2
.
Find the residue when
n
n
n
is divided by
17
17
17
.