MathDB
Problems
Contests
National and Regional Contests
Korea Contests
Korea National Olympiad
2015 Korea National Olympiad
2015 Korea National Olympiad
Part of
Korea National Olympiad
Subcontests
(4)
4
2
Hide problems
Partitions with special conditions
For positive integers
n
,
k
,
l
n, k, l
n
,
k
,
l
, we define the number of
l
l
l
-tuples of positive integers
(
a
1
,
a
2
,
⋯
a
l
)
(a_1,a_2,\cdots a_l)
(
a
1
,
a
2
,
⋯
a
l
)
satisfying the following as
Q
(
n
,
k
,
l
)
Q(n,k,l)
Q
(
n
,
k
,
l
)
.(i):
n
=
a
1
+
a
2
+
⋯
+
a
l
n=a_1+a_2+\cdots +a_l
n
=
a
1
+
a
2
+
⋯
+
a
l
(ii):
a
1
>
a
2
>
⋯
>
a
l
>
0
a_1>a_2>\cdots > a_l > 0
a
1
>
a
2
>
⋯
>
a
l
>
0
.(iii):
a
l
a_l
a
l
is an odd number.(iv): There are
k
k
k
odd numbers out of
a
i
a_i
a
i
.For example, from
9
=
8
+
1
=
6
+
3
=
6
+
2
+
1
9=8+1=6+3=6+2+1
9
=
8
+
1
=
6
+
3
=
6
+
2
+
1
, we have
Q
(
9
,
1
,
1
)
=
1
Q(9,1,1)=1
Q
(
9
,
1
,
1
)
=
1
,
Q
(
9
,
1
,
2
)
=
2
Q(9,1,2)=2
Q
(
9
,
1
,
2
)
=
2
,
Q
(
9
,
1
,
3
)
=
1
Q(9,1,3)=1
Q
(
9
,
1
,
3
)
=
1
.Prove that if
n
>
k
2
n>k^2
n
>
k
2
,
∑
l
=
1
n
Q
(
n
,
k
,
l
)
\sum_{l=1}^n Q(n,k,l)
∑
l
=
1
n
Q
(
n
,
k
,
l
)
is
0
0
0
or an even number.
Inequality of relatively prime numbers
For a positive integer
n
n
n
,
a
1
,
a
2
,
⋯
a
k
a_1, a_2, \cdots a_k
a
1
,
a
2
,
⋯
a
k
are all positive integers without repetition that are not greater than
n
n
n
and relatively prime to
n
n
n
. If
k
>
8
k>8
k
>
8
, prove the following.
∑
i
=
1
k
∣
a
i
−
n
2
∣
<
n
(
k
−
4
)
2
\sum_{i=1}^k |a_i-\frac{n}{2}|<\frac{n(k-4)}{2}
i
=
1
∑
k
∣
a
i
−
2
n
∣
<
2
n
(
k
−
4
)
3
2
Hide problems
5-var Asymmetric Inequality
Reals
a
,
b
,
c
,
x
,
y
a,b,c,x,y
a
,
b
,
c
,
x
,
y
satisfies
a
2
+
b
2
+
c
2
+
x
2
+
y
2
=
1
a^2+b^2+c^2+x^2+y^2=1
a
2
+
b
2
+
c
2
+
x
2
+
y
2
=
1
. Find the maximum value of
(
a
x
+
b
y
)
2
+
(
b
x
+
c
y
)
2
(ax+by)^2+(bx+cy)^2
(
a
x
+
b
y
)
2
+
(
b
x
+
cy
)
2
Choosing Sets, Relative Complement
A positive integer
n
n
n
is given. If there exists sets
F
1
,
F
2
,
⋯
F
m
F_1, F_2, \cdots F_m
F
1
,
F
2
,
⋯
F
m
satisfying the following conditions, prove that
m
≤
n
m \le n
m
≤
n
. (For sets
A
,
B
A, B
A
,
B
,
∣
A
∣
|A|
∣
A
∣
is the number of elements of
A
A
A
.
A
−
B
A-B
A
−
B
is the set of elements that are in
A
A
A
but not
B
B
B
.
min
(
x
,
y
)
\text{min}(x,y)
min
(
x
,
y
)
is the number that is not larger than the other.)(i): For all
1
≤
i
≤
m
1 \le i \le m
1
≤
i
≤
m
,
F
i
⊆
{
1
,
2
,
⋯
,
n
}
F_i \subseteq \{1,2,\cdots,n\}
F
i
⊆
{
1
,
2
,
⋯
,
n
}
(ii): For all
1
≤
i
<
j
≤
m
1 \le i < j \le m
1
≤
i
<
j
≤
m
,
min
(
∣
F
i
−
F
j
∣
,
∣
F
j
−
F
i
∣
)
=
1
\text{min}(|F_i-F_j|,|F_j-F_i|) = 1
min
(
∣
F
i
−
F
j
∣
,
∣
F
j
−
F
i
∣
)
=
1
2
2
Hide problems
Easy Geometry
Let the circumcircle of
△
A
B
C
\triangle ABC
△
A
BC
be
ω
\omega
ω
. A point
D
D
D
lies on segment
B
C
BC
BC
, and
E
E
E
lies on segment
A
D
AD
A
D
. Let ray
A
D
∩
ω
=
F
AD \cap \omega = F
A
D
∩
ω
=
F
. A point
M
M
M
, which lies on
ω
\omega
ω
, bisects
A
F
AF
A
F
and it is on the other side of
C
C
C
with respect to
A
F
AF
A
F
. Ray
M
E
∩
ω
=
G
ME \cap \omega = G
ME
∩
ω
=
G
, ray
G
D
∩
ω
=
H
GD \cap \omega = H
G
D
∩
ω
=
H
, and
M
H
∩
A
D
=
K
MH \cap AD = K
M
H
∩
A
D
=
K
. Prove that
B
,
E
,
C
,
K
B, E, C, K
B
,
E
,
C
,
K
are cyclic.
Angles in a isosceles trapezoid
An isosceles trapezoid
A
B
C
D
ABCD
A
BC
D
, inscribed in
ω
\omega
ω
, satisfies
A
B
=
C
D
,
A
D
<
B
C
,
A
D
<
C
D
AB=CD, AD<BC, AD<CD
A
B
=
C
D
,
A
D
<
BC
,
A
D
<
C
D
. A circle with center
D
D
D
and passing
A
A
A
hits
B
D
,
C
D
,
ω
BD, CD, \omega
B
D
,
C
D
,
ω
at
E
,
F
,
P
(
≠
A
)
E, F, P(\not= A)
E
,
F
,
P
(
=
A
)
, respectively. Let
A
P
∩
E
F
=
Q
AP \cap EF = Q
A
P
∩
EF
=
Q
, and
ω
\omega
ω
meet
C
Q
CQ
CQ
and the circumcircle of
△
B
E
Q
\triangle BEQ
△
BEQ
at
R
(
≠
C
)
,
S
(
≠
B
)
R(\not= C), S(\not= B)
R
(
=
C
)
,
S
(
=
B
)
, respectively. Prove that
∠
B
E
R
=
∠
F
S
C
\angle BER= \angle FSC
∠
BER
=
∠
FSC
.
1
2
Hide problems
Pell's Equation Looking Equation
For a positive integer
m
m
m
, prove that the number of pairs of positive integers
(
x
,
y
)
(x,y)
(
x
,
y
)
which satisfies the following two conditions is even or
0
0
0
.(i):
x
2
−
3
y
2
+
2
=
16
m
x^2-3y^2+2=16m
x
2
−
3
y
2
+
2
=
16
m
(ii):
2
y
≤
x
−
1
2y \le x-1
2
y
≤
x
−
1
Familiar Functional Equation
Find all functions
f
:
R
→
R
f: \mathbb{R} \rightarrow \mathbb{R}
f
:
R
→
R
such that for all reals
x
,
y
,
z
x,y,z
x
,
y
,
z
, we have
(
f
(
x
)
+
1
)
(
f
(
y
)
+
f
(
z
)
)
=
f
(
x
y
+
z
)
+
f
(
x
z
−
y
)
(f(x)+1)(f(y)+f(z))=f(xy+z)+f(xz-y)
(
f
(
x
)
+
1
)
(
f
(
y
)
+
f
(
z
))
=
f
(
x
y
+
z
)
+
f
(
x
z
−
y
)