MathDB
Problems
Contests
National and Regional Contests
China Contests
(China) National High School Mathematics League
2016 China Second Round Olympiad
2016 China Second Round Olympiad
Part of
(China) National High School Mathematics League
Subcontests
(6)
2
1
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Geometry
Let
X
,
Y
X,Y
X
,
Y
be two points which lies on the line
B
C
BC
BC
of
△
A
B
C
(
X
,
B
,
C
,
Y
lies in sequence
)
\triangle ABC(X,B,C,Y\text{lies in sequence})
△
A
BC
(
X
,
B
,
C
,
Y
lies in sequence
)
such that
B
X
⋅
A
C
=
C
Y
⋅
A
B
BX\cdot AC=CY\cdot AB
BX
⋅
A
C
=
C
Y
⋅
A
B
,
O
1
,
O
2
O_1,O_2
O
1
,
O
2
are the circumcenters of
△
A
C
X
,
△
A
B
Y
\triangle ACX,\triangle ABY
△
A
CX
,
△
A
B
Y
,
O
1
O
2
∩
A
B
=
U
,
O
1
O
2
∩
A
C
=
V
O_1O_2\cap AB=U,O_1O_2\cap AC=V
O
1
O
2
∩
A
B
=
U
,
O
1
O
2
∩
A
C
=
V
. Prove that
△
A
U
V
\triangle AUV
△
A
U
V
is a isosceles triangle.
1
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Number Theory
Let
p
p
p
and
p
+
2
p+2
p
+
2
be primes,
p
>
3
p>3
p
>
3
. Sequence
{
a
n
}
:
a
1
=
2
,
a
n
=
a
n
−
1
+
⌈
p
a
n
−
1
n
⌉
\{a_n\}:a_1=2,a_n=a_{n-1}+\left\lceil{\frac{pa_{n-1}}{n}}\right\rceil
{
a
n
}
:
a
1
=
2
,
a
n
=
a
n
−
1
+
⌈
n
p
a
n
−
1
⌉
. Prove that
n
∣
p
a
n
−
1
+
1
n\mid pa_{n-1}+1
n
∣
p
a
n
−
1
+
1
for all
n
=
3
,
4
,
…
,
p
−
1
n=3,4,\dots,p-1
n
=
3
,
4
,
…
,
p
−
1
.
3
1
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Maximum number of the lines
Given
10
10
10
points in the space such that each
4
4
4
points are not lie on a plane. Connect some points with some segments such that there are no triangles or quadrangles. Find the maximum number of the segments.
4
1
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nice problem
Let
p
>
3
p>3
p
>
3
and
p
+
2
p+2
p
+
2
are prime numbers,and define sequence
a
1
=
2
,
a
n
=
a
n
−
1
+
⌊
p
a
n
−
1
n
⌋
a_{1}=2,a_{n}=a_{n-1}+\lfloor \dfrac{pa_{n-1}}{n}\rfloor
a
1
=
2
,
a
n
=
a
n
−
1
+
⌊
n
p
a
n
−
1
⌋
show that:for any
n
=
3
,
4
,
⋯
,
p
−
1
n=3,4,\cdots,p-1
n
=
3
,
4
,
⋯
,
p
−
1
have
n
∣
p
a
n
−
1
+
1
n|pa_{n-1}+1
n
∣
p
a
n
−
1
+
1
Q10
1
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China Second Round Olympiad 2016 Test 1 Q10
Let
f
(
x
)
f(x)
f
(
x
)
is an odd function on
R
R
R
,
f
(
1
)
=
1
f(1)=1
f
(
1
)
=
1
and
f
(
x
x
−
1
)
=
x
f
(
x
)
f(\frac{x}{x-1})=xf(x)
f
(
x
−
1
x
)
=
x
f
(
x
)
(
∀
x
<
0
)
(\forall x<0)
(
∀
x
<
0
)
. Find the value of
f
(
1
)
f
(
1
100
)
+
f
(
1
2
)
f
(
1
99
)
+
f
(
1
3
)
f
(
1
98
)
+
⋯
+
f
(
1
50
)
f
(
1
51
)
.
f(1)f(\frac{1}{100})+f(\frac{1}{2})f(\frac{1}{99})+f(\frac{1}{3})f(\frac{1}{98})+\cdots +f(\frac{1}{50})f(\frac{1}{51}).
f
(
1
)
f
(
100
1
)
+
f
(
2
1
)
f
(
99
1
)
+
f
(
3
1
)
f
(
98
1
)
+
⋯
+
f
(
50
1
)
f
(
51
1
)
.
1
1
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China Second Round Olympiad 2016 Test 2 Q1
Let
a
1
,
a
2
,
…
,
a
2016
a_1, a_2, \ldots, a_{2016}
a
1
,
a
2
,
…
,
a
2016
be real numbers such that
9
a
i
≥
11
a
i
+
1
2
9a_i\ge 11a^2_{i+1}
9
a
i
≥
11
a
i
+
1
2
(
i
=
,
2
,
⋯
,
2015
)
(i=,2,\cdots,2015)
(
i
=
,
2
,
⋯
,
2015
)
. Find the maximum value of
(
a
1
−
a
2
2
)
(
a
2
−
a
3
2
)
⋯
(
a
2015
−
a
2016
2
)
(
a
2016
−
a
1
2
)
.
(a_1-a^2_2)(a_2-a^2_3)\cdots (a_{2015}-a^2_{2016})(a_{2016}-a^2_{1}).
(
a
1
−
a
2
2
)
(
a
2
−
a
3
2
)
⋯
(
a
2015
−
a
2016
2
)
(
a
2016
−
a
1
2
)
.