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Problems
Contests
National and Regional Contests
China Contests
China Northern MO
2012 China Northern MO
2012 China Northern MO
Part of
China Northern MO
Subcontests
(8)
4
1
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congruent pairs of lines = 2012 more than not congruent
There are
n
n
n
(
n
≥
4
n \ge 4
n
≥
4
) straight lines on the plane. For two straight lines
a
a
a
and
b
b
b
, if there are at least two straight lines among the remaining
n
−
2
n-2
n
−
2
lines that intersect both straight lines
a
a
a
and
b
b
b
, then
a
a
a
and
b
b
b
are called a congruent pair of staight lines, otherwise it is called a separated pair of straight lines. If the number of congruent pairs of straight line among
n
n
n
straight lines is
2012
2012
2012
more than the number of separated pairs of straight line , find the smallest possible value of
n
n
n
(the order of the two straight lines in a pair is not counted).
8
1
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p!|(a +1) if $p!|(a^p + 1)
Assume
p
p
p
is a prime number. If there is a positive integer
a
a
a
such that
p
!
∣
(
a
p
+
1
)
p!|(a^p + 1)
p
!
∣
(
a
p
+
1
)
, prove that :(1)
(
a
+
1
,
a
p
+
1
a
+
1
)
=
p
(a+1, \frac{a^p+1}{a+1}) = p
(
a
+
1
,
a
+
1
a
p
+
1
)
=
p
(2)
a
p
+
1
a
+
1
\frac{a^p+1}{a+1}
a
+
1
a
p
+
1
has no prime factors less than
p
p
p
.(3)
p
!
∣
(
a
+
1
)
p!|(a +1)
p
!
∣
(
a
+
1
)
.
3
1
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x=a^2+ab+b^2 with a,b \in Z
Suppose
S
=
{
x
∣
x
=
a
2
+
a
b
+
b
2
,
a
,
b
∈
Z
}
S= \{x|x=a^2+ab+b^2,a,b \in Z\}
S
=
{
x
∣
x
=
a
2
+
ab
+
b
2
,
a
,
b
∈
Z
}
. Prove that:(1) If
m
∈
S
m \in S
m
∈
S
,
3
∣
m
3|m
3∣
m
, then
m
3
∈
S
\frac{m}{3} \in S
3
m
∈
S
(2) If
m
,
n
∈
S
m,n \in S
m
,
n
∈
S
, then
m
n
∈
S
mn\in S
mn
∈
S
.
2
1
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max sum x_i/n if sum x_i^2=111
Positive integers
x
1
,
x
2
,
.
.
.
,
x
n
x_1,x_2,...,x_n
x
1
,
x
2
,
...
,
x
n
(
n
∈
N
+
n \in N_+
n
∈
N
+
) satisfy
x
1
2
+
x
2
2
+
.
.
.
+
x
n
2
=
111
x_1^2 +x_2^2+...+x_n^2=111
x
1
2
+
x
2
2
+
...
+
x
n
2
=
111
, find the maximum possible value of
S
=
x
1
+
x
2
+
.
.
.
+
x
n
n
S =\frac{x_1 +x_2+...+x_n}{n}
S
=
n
x
1
+
x
2
+
...
+
x
n
.
5
1
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a_n=\frac{1}{a_{n-1}-2}, b_n=\frac{2b_{n-1}+1} {b_{n-1}}
Let
{
a
n
}
\{a_n\}
{
a
n
}
be the sequance with
a
0
=
0
a_0=0
a
0
=
0
,
a
n
=
1
a
n
−
1
−
2
a_n=\frac{1}{a_{n-1}-2}
a
n
=
a
n
−
1
−
2
1
(
n
∈
N
+
n\in N_+
n
∈
N
+
). Select an arbitrary term
a
k
a_k
a
k
in the sequence
{
a
n
}
\{a_n\}
{
a
n
}
and construct the sequence
{
b
n
}
\{b_n\}
{
b
n
}
:
b
0
=
a
k
b_0=a_k
b
0
=
a
k
,
b
n
=
2
b
n
−
1
+
1
b
n
−
1
b_n=\frac{2b_{n-1}+1} {b_{n-1}}
b
n
=
b
n
−
1
2
b
n
−
1
+
1
(
n
∈
N
+
n\in N_+
n
∈
N
+
) . Determine whether the sequence
{
b
n
}
\{b_n\}
{
b
n
}
is a finite sequence or an infinite sequence and give proof.
7
1
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AC bisects BE, BC = DE, CD//BE, <BAC = <DAE, AB/BD=AE/ED
As shown in figure , in the pentagon
A
B
C
D
E
ABCDE
A
BC
D
E
,
B
C
=
D
E
BC = DE
BC
=
D
E
,
C
D
∥
B
E
CD \parallel BE
C
D
∥
BE
,
A
B
>
A
E
AB>AE
A
B
>
A
E
. If
∠
B
A
C
=
∠
D
A
E
\angle BAC = \angle DAE
∠
B
A
C
=
∠
D
A
E
and
A
B
B
D
=
A
E
E
D
\frac{AB}{BD}=\frac{AE}{ED}
B
D
A
B
=
E
D
A
E
. Prove that
A
C
AC
A
C
bisects the line segment
B
E
BE
BE
. https://cdn.artofproblemsolving.com/attachments/3/2/5ce44f1e091786b865ae4319bda56c3ddbb8d7.png
1
1
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DF _|_ AI wanted, incenter of a right triangle
As shown in figure, given right
△
A
B
C
\vartriangle ABC
△
A
BC
with
∠
C
=
9
0
o
\angle C=90^o
∠
C
=
9
0
o
.
I
I
I
is the incenter. The line
B
I
BI
B
I
intersects segment
A
C
AC
A
C
at the point
D
D
D
. The line passing through
D
D
D
parallel to
A
I
AI
A
I
intersects
B
C
BC
BC
at point
E
E
E
. The line
E
I
EI
E
I
intersects segment
A
B
AB
A
B
at point
F
F
F
. Prove that
D
F
⊥
A
I
DF \perp AI
D
F
⊥
A
I
. https://cdn.artofproblemsolving.com/attachments/2/4/6fc94adb4ce12c3bf07948b8c57170ca01b256.png
6
1
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China Northern Mathematical Olympiad 2012 , Problem 6
Prove that
(
1
+
1
3
)
(
1
+
1
3
2
)
⋯
(
1
+
1
3
n
)
<
2.
(1+\frac{1}{3})(1+\frac{1}{3^2})\cdots(1+\frac{1}{3^n})< 2.
(
1
+
3
1
)
(
1
+
3
2
1
)
⋯
(
1
+
3
n
1
)
<
2.