MathDB
Problems
Contests
National and Regional Contests
China Contests
China National Olympiad
2006 China National Olympiad
2006 China National Olympiad
Part of
China National Olympiad
Subcontests
(6)
2
1
Hide problems
2006 positive integers and 2006 pairwise distinct fractions
For positive integers
a
1
,
a
2
,
…
,
a
2006
a_1,a_2 ,\ldots,a_{2006}
a
1
,
a
2
,
…
,
a
2006
such that
a
1
a
2
,
a
2
a
3
,
…
,
a
2005
a
2006
\frac{a_1}{a_2},\frac{a_2}{a_3},\ldots,\frac{a_{2005}}{a_{2006}}
a
2
a
1
,
a
3
a
2
,
…
,
a
2006
a
2005
are pairwise distinct, find the minimum possible amount of distinct positive integers in the set
{
a
1
,
a
2
,
.
.
.
,
a
2006
}
\{a_1,a_2,...,a_{2006}\}
{
a
1
,
a
2
,
...
,
a
2006
}
.
6
1
Hide problems
A set with 56 elements, and 15 subsets of his
Suppose
X
X
X
is a set with
∣
X
∣
=
56
|X| = 56
∣
X
∣
=
56
. Find the minimum value of
n
n
n
, so that for any 15 subsets of
X
X
X
, if the cardinality of the union of any 7 of them is greater or equal to
n
n
n
, then there exists 3 of them whose intersection is nonempty.
4
1
Hide problems
Right angle
In a right angled-triangle
A
B
C
ABC
A
BC
,
∠
A
C
B
=
9
0
o
\angle{ACB} = 90^o
∠
A
CB
=
9
0
o
. Its incircle
O
O
O
meets
B
C
BC
BC
,
A
C
AC
A
C
,
A
B
AB
A
B
at
D
D
D
,
E
E
E
,
F
F
F
respectively.
A
D
AD
A
D
cuts
O
O
O
at
P
P
P
. If
∠
B
P
C
=
9
0
o
\angle{BPC} = 90^o
∠
BPC
=
9
0
o
, prove
A
E
+
A
P
=
P
D
AE + AP = PD
A
E
+
A
P
=
P
D
.
5
1
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2006cmo problem5
Let
{
a
n
}
\{a_n\}
{
a
n
}
be a sequence such that:
a
1
=
1
2
a_1 = \frac{1}{2}
a
1
=
2
1
,
a
k
+
1
=
−
a
k
+
1
2
−
a
k
a_{k+1}=-a_k+\frac{1}{2-a_k}
a
k
+
1
=
−
a
k
+
2
−
a
k
1
for all
k
=
1
,
2
,
…
k = 1, 2,\ldots
k
=
1
,
2
,
…
. Prove that
(
n
2
(
a
1
+
a
2
+
⋯
+
a
n
)
−
1
)
n
≤
(
a
1
+
a
2
+
⋯
+
a
n
n
)
n
(
1
a
1
−
1
)
(
1
a
2
−
1
)
⋯
(
1
a
n
−
1
)
.
\left(\frac{n}{2(a_1+a_2+\cdots+a_n)}-1\right)^n \leq \left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^n\left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)\cdots \left(\frac{1}{a_n}-1\right).
(
2
(
a
1
+
a
2
+
⋯
+
a
n
)
n
−
1
)
n
≤
(
n
a
1
+
a
2
+
⋯
+
a
n
)
n
(
a
1
1
−
1
)
(
a
2
1
−
1
)
⋯
(
a
n
1
−
1
)
.
3
1
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Equation with mn=k^2+k+3
Positive integers
k
,
m
,
n
k, m, n
k
,
m
,
n
satisfy
m
n
=
k
2
+
k
+
3
mn=k^2+k+3
mn
=
k
2
+
k
+
3
, prove that at least one of the equations
x
2
+
11
y
2
=
4
m
x^2+11y^2=4m
x
2
+
11
y
2
=
4
m
and
x
2
+
11
y
2
=
4
n
x^2+11y^2=4n
x
2
+
11
y
2
=
4
n
has an odd solution.
1
1
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2006 chinese mathematical olympiad 1
Let
a
1
,
a
2
,
…
,
a
k
a_1,a_2,\ldots,a_k
a
1
,
a
2
,
…
,
a
k
be real numbers and
a
1
+
a
2
+
…
+
a
k
=
0
a_1+a_2+\ldots+a_k=0
a
1
+
a
2
+
…
+
a
k
=
0
. Prove that
max
1
≤
i
≤
k
a
i
2
≤
k
3
(
(
a
1
−
a
2
)
2
+
(
a
2
−
a
3
)
2
+
⋯
+
(
a
k
−
1
−
a
k
)
2
)
.
\max_{1\leq i \leq k} a_i^2 \leq \frac{k}{3} \left( (a_1-a_2)^2+(a_2-a_3)^2+\cdots +(a_{k-1}-a_k)^2\right).
1
≤
i
≤
k
max
a
i
2
≤
3
k
(
(
a
1
−
a
2
)
2
+
(
a
2
−
a
3
)
2
+
⋯
+
(
a
k
−
1
−
a
k
)
2
)
.