MathDB
Problems
Contests
National and Regional Contests
China Contests
China Northern MO
2009 China Northern MO
2009 China Northern MO
Part of
China Northern MO
Subcontests
(8)
6
1
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area of polygon AQPBO does not change when points P,C move
Given a minor sector
A
O
B
AOB
A
OB
(Here minor means that
∠
A
O
B
<
90
\angle AOB <90
∠
A
OB
<
90
).
O
O
O
is the centre , chose a point
C
C
C
on arc
A
B
AB
A
B
,Let
P
P
P
be a point on segment
O
C
OC
OC
, join
A
P
AP
A
P
,
B
P
BP
BP
, draw a line through
B
B
B
parallel to
A
P
AP
A
P
, the line meet
O
C
OC
OC
at point
Q
Q
Q
,join
A
Q
AQ
A
Q
. Prove that the area of polygon
A
Q
P
B
O
AQPBO
A
QPBO
does not change when points
P
,
C
P,C
P
,
C
move . https://cdn.artofproblemsolving.com/attachments/3/e/4bdd3a20fe1df3fce0719463b55ef93e8b5d7b.png
2
1
Hide problems
angle bisector and BE+CF=EF wanted, cos B+ cos C=1, <ABF =< ACE = 90^o
In an acute triangle
A
B
C
ABC
A
BC
,
A
B
>
A
C
AB>AC
A
B
>
A
C
,
cos
B
+
cos
C
=
1
\cos B+ \cos C=1
cos
B
+
cos
C
=
1
,
E
,
F
E,F
E
,
F
are on the extend line of
A
B
,
A
C
AB,AC
A
B
,
A
C
such that
∠
A
B
F
=
∠
A
C
E
=
90
\angle ABF = \angle ACE = 90
∠
A
BF
=
∠
A
CE
=
90
. (1) Prove :
B
E
+
C
F
=
E
F
BE+CF=EF
BE
+
CF
=
EF
; (2) Assume the bisector of
∠
E
B
C
\angle EBC
∠
EBC
meet
E
F
EF
EF
at
P
P
P
, prove that
C
P
CP
CP
is the bisector of
∠
B
C
F
\angle BCF
∠
BCF
. https://cdn.artofproblemsolving.com/attachments/a/2/c554c2bc0b4e044c45f88138568f5234d544a8.png
7
1
Hide problems
xy=1 and y irrational >1 if [x [ny] ] =n-1 ,
Let
⌊
m
⌋
\lfloor m \rfloor
⌊
m
⌋
be the largest integer smaller than
m
m
m
. Assume
x
,
y
∈
R
+
x,y \in \mathbb{R+}
x
,
y
∈
R
+
, For all positive integer
n
n
n
,
⌊
x
⌊
n
y
⌋
⌋
=
n
−
1
\lfloor x \lfloor ny \rfloor \rfloor =n-1
⌊
x
⌊
n
y
⌋⌋
=
n
−
1
. Prove :
x
y
=
1
xy=1
x
y
=
1
,
y
y
y
is an irrational number larger than
1
1
1
.
8
1
Hide problems
sum of digits of number N is 209 and 209 | N
Find the smallest positive integer
N
N
N
satisfies : 1 .
209
209
209
│
N
N
N
2 .
S
(
N
)
=
209
S (N) = 209
S
(
N
)
=
209
( # Here
S
(
m
)
S(m)
S
(
m
)
means the sum of digits of number
m
m
m
)
5
1
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sum (x^{2009}-2008(x-1))/(y+z) >= 1/2 (x+y+z), if x,y,z>0, x^2+y^2+z^2 = 3
Assume :
x
,
y
,
z
>
0
x,y,z>0
x
,
y
,
z
>
0
,
x
2
+
y
2
+
z
2
=
3
x^2+y^2+z^2 = 3
x
2
+
y
2
+
z
2
=
3
. Prove the following inequality :
x
2009
−
2008
(
x
−
1
)
y
+
z
+
y
2009
−
2008
(
y
−
1
)
x
+
z
+
z
2009
−
2008
(
z
−
1
)
x
+
y
≥
1
2
(
x
+
y
+
z
)
{\frac{x^{2009}-2008(x-1)}{y+z}+\frac{y^{2009}-2008(y-1)}{x+z}+\frac{z^{2009}-2008(z-1)}{x+y}\ge\frac{1}{2}(x+y+z)}
y
+
z
x
2009
−
2008
(
x
−
1
)
+
x
+
z
y
2009
−
2008
(
y
−
1
)
+
x
+
y
z
2009
−
2008
(
z
−
1
)
≥
2
1
(
x
+
y
+
z
)
4
1
Hide problems
captain and his 3 sailors split 2009 golden coins
The captain and his three sailors get
2009
2009
2009
golden coins with the same value . The four people decided to divide these coins by the following rules : sailor
1
1
1
,sailor
2
2
2
,sailor
3
3
3
everyone write down an integer
b
1
,
b
2
,
b
3
b_1,b_2,b_3
b
1
,
b
2
,
b
3
, satisfies
b
1
≥
b
2
≥
b
3
b_1\ge b_2\ge b_3
b
1
≥
b
2
≥
b
3
, and
b
1
+
b
2
+
b
3
=
2009
{b_1+b_2+b_3=2009}
b
1
+
b
2
+
b
3
=
2009
; the captain dosen't know what the numbers the sailors have written . He divides
2009
2009
2009
coins into
3
3
3
piles , with number of coins:
a
1
,
a
2
,
a
3
a_1,a_2,a_3
a
1
,
a
2
,
a
3
, and
a
1
≥
a
2
≥
a
3
a_1\ge a_2\ge a_3
a
1
≥
a
2
≥
a
3
. For sailor
k
k
k
(
k
=
1
,
2
,
3
k=1,2,3
k
=
1
,
2
,
3
) , if
b
k
<
a
k
b_k<a_k
b
k
<
a
k
, then he can take
b
k
b_k
b
k
coins from the
k
k
k
th pile ; if
b
k
≥
a
k
b_k\ge a_k
b
k
≥
a
k
, then he can't take any coins away . At last , the captain own the rest of the coins .If no matter what the numbers the sailors write , the captain can make sure that he always gets
n
n
n
coins . Find the largest possible value of
n
n
n
and prove your conclusion .
3
1
Hide problems
6 numbers in 26 one of them can be devided by the other 5 numbers .
Given
26
26
26
different positive integers , in any six numbers of the
26
26
26
integers , there are at least two numbers , one can be devided by another. Then prove : There exists six numbers , one of them can be devided by the other five numbers .
1
1
Hide problems
x_n=\sqrt{x_{n-1}^2+x_{n-1}}+x_{n-1}} ,x_1=1
Sequence {
x
n
x_n
x
n
} satisfies:
x
1
=
1
x_1=1
x
1
=
1
,
x
n
=
x
n
−
1
2
+
x
n
−
1
+
x
n
−
1
{x_n=\sqrt{x_{n-1}^2+x_{n-1}}+x_{n-1}}
x
n
=
x
n
−
1
2
+
x
n
−
1
+
x
n
−
1
(
n
>
=
2
{n>=2}
n
>=
2
) Find the general term of {
x
n
x_n
x
n
}