MathDB
Problems
Contests
National and Regional Contests
China Contests
(China) National High School Mathematics League
2006 China Second Round Olympiad
2006 China Second Round Olympiad
Part of
(China) National High School Mathematics League
Subcontests
(15)
15
1
Hide problems
2006 China Second Round Olympiad Test 1 #15
Suppose
f
(
x
)
=
x
2
+
a
f(x)=x^2+a
f
(
x
)
=
x
2
+
a
. Define
f
1
(
x
)
=
f
(
x
)
f^1(x)=f(x)
f
1
(
x
)
=
f
(
x
)
,
f
n
(
x
)
=
f
(
f
n
−
1
(
x
)
)
f^n(x)=f(f^{n-1}(x))
f
n
(
x
)
=
f
(
f
n
−
1
(
x
))
,
n
=
2
,
3
,
⋯
n=2, 3, \cdots
n
=
2
,
3
,
⋯
, and let
M
=
{
a
∈
R
∣
∣
f
n
(
0
)
∣
≤
2
,
for any
n
∈
N
}
M=\{a\in\mathbb{R}| |f^n(0)|\le 2, \text{for any } n\in\mathbb{N}\}
M
=
{
a
∈
R
∣∣
f
n
(
0
)
∣
≤
2
,
for any
n
∈
N
}
. Prove that
M
=
[
−
2
,
1
4
]
M=[-2, \frac{1}{4}]
M
=
[
−
2
,
4
1
]
.
14
1
Hide problems
2006 China Second Round Olympiad Test 1 #14
Let
2006
2006
2006
be expressed as the sum of five positive integers
x
1
,
x
2
,
x
3
,
x
4
,
x
5
x_1, x_2, x_3, x_4, x_5
x
1
,
x
2
,
x
3
,
x
4
,
x
5
, and
S
=
∑
1
≤
i
<
j
≤
5
x
i
x
j
S=\sum_{1\le i<j\le 5}x_ix_j
S
=
∑
1
≤
i
<
j
≤
5
x
i
x
j
.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
<span class='latex-bold'>(A)</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
What value of
x
1
,
x
2
,
x
3
,
x
4
,
x
5
x_1, x_2, x_3, x_4, x_5
x
1
,
x
2
,
x
3
,
x
4
,
x
5
maximizes
S
S
S
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
<span class='latex-bold'>(A)</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
Find, with proof, the value of
x
1
,
x
2
,
x
3
,
x
4
,
x
5
x_1, x_2, x_3, x_4, x_5
x
1
,
x
2
,
x
3
,
x
4
,
x
5
which minimizes of
S
S
S
if
∣
x
i
−
x
j
∣
≤
2
|x_i-x_j|\le 2
∣
x
i
−
x
j
∣
≤
2
for any
1
≤
i
1\le i
1
≤
i
,
j
≤
5
j\le 5
j
≤
5
.
13
1
Hide problems
2006 China Second Round Olympiad Test 1 #13
Given an integer
n
≥
2
n\ge 2
n
≥
2
, define
M
0
(
x
0
,
y
0
)
M_0 (x_0, y_0)
M
0
(
x
0
,
y
0
)
to be an intersection point of the parabola
y
2
=
n
x
−
1
y^2=nx-1
y
2
=
n
x
−
1
and the line
y
=
x
y=x
y
=
x
. Prove that for any positive integer
m
m
m
, there exists an integer
k
≥
2
k\ge 2
k
≥
2
such that
(
x
0
m
,
y
0
m
)
(x^m_0, y^m_0)
(
x
0
m
,
y
0
m
)
is an intersection point of
y
2
=
m
x
−
1
y^2=mx-1
y
2
=
m
x
−
1
and the line
y
=
x
y=x
y
=
x
.
12
1
Hide problems
2006 China Second Round Olympiad Test 1 #12
Suppose there are 8 white balls and 2 red balls in a packet. Each time one ball is drawn and replaced by a white one. Find the probability that the last red ball is drawn in the fourth draw.
11
1
Hide problems
2006 China Second Round Olympiad Test 1 #11
Find the number of real solutions to the equation
(
x
2006
+
1
)
(
1
+
x
2
+
x
4
+
…
+
x
2004
)
=
2006
x
2005
(x^{2006}+1)(1+x^2+x^4+\ldots +x^{2004})=2006x^{2005}
(
x
2006
+
1
)
(
1
+
x
2
+
x
4
+
…
+
x
2004
)
=
2006
x
2005
10
1
Hide problems
2006 China Second Round Olympiad Test 1 #10
Suppose four solid iron balls are placed in a cylinder with the radius of 1 cm, such that every two of the four balls are tangent to each other, and the two balls in the lower layer are tangent to the cylinder base. Now put water into the cylinder. Find, in
cm
2
\text{cm}^2
cm
2
, the volume of water needed to submerge all the balls.
9
1
Hide problems
2006 China Second Round Olympiad Test 1 #9
Suppose points
F
1
,
F
2
F_1, F_2
F
1
,
F
2
are the left and right foci of the ellipse
x
2
16
+
y
2
4
=
1
\frac{x^2}{16}+\frac{y^2}{4}=1
16
x
2
+
4
y
2
=
1
respectively, and point
P
P
P
is on line
l
:
l:
l
:
,
x
−
3
y
+
8
+
2
3
=
0
x-\sqrt{3} y+8+2\sqrt{3}=0
x
−
3
y
+
8
+
2
3
=
0
. Find the value of ratio
∣
P
F
1
∣
∣
P
F
2
∣
\frac{|PF_1|}{|PF_2|}
∣
P
F
2
∣
∣
P
F
1
∣
when
∠
F
1
P
F
2
\angle F_1PF_2
∠
F
1
P
F
2
reaches its maximum value.
8
1
Hide problems
2006 China Second Round Olympiad Test 1 #8
Let complex number
z
=
(
a
+
cos
θ
)
+
(
2
a
−
sin
θ
)
i
z = (a+\cos\theta)+(2a-\sin \theta)i
z
=
(
a
+
cos
θ
)
+
(
2
a
−
sin
θ
)
i
. Find the range of real number
a
a
a
if
∣
z
∣
≥
2
|z|\ge 2
∣
z
∣
≥
2
for any
θ
∈
R
\theta\in \mathbb{R}
θ
∈
R
.
7
1
Hide problems
2006 China Second Round Olympiad Test 1 #7
Let
f
(
x
)
=
sin
4
x
−
sin
x
cos
x
+
c
o
s
4
x
f(x)=\sin^4x-\sin x\cos x+cos^4 x
f
(
x
)
=
sin
4
x
−
sin
x
cos
x
+
co
s
4
x
. Find the range of
f
(
x
)
f(x)
f
(
x
)
.
6
1
Hide problems
2006 China Second Round Olympiad Test 1 #6
Let
S
S
S
be the set of all those 2007 place decimal integers
2
s
1
a
2
a
3
…
a
2006
‾
\overline{2s_1a_2a_3 \ldots a_{2006}}
2
s
1
a
2
a
3
…
a
2006
which contain odd number of digit
9
9
9
in each sequence
a
1
,
a
2
,
a
3
,
…
,
a
2006
a_1, a_2, a_3, \ldots, a_{2006}
a
1
,
a
2
,
a
3
,
…
,
a
2006
. The cardinal number of
S
S
S
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
2
(
1
0
2006
+
8
2006
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1
2
(
1
0
2006
−
8
2006
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
1
0
2006
+
8
2006
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
1
0
2006
−
8
2006
{ <span class='latex-bold'>(A)</span>\ \frac{1}{2}(10^{2006}+8^{2006})\qquad<span class='latex-bold'>(B)</span>\ \frac{1}{2}(10^{2006}-8^{2006})\qquad<span class='latex-bold'>(C)</span>\ 10^{2006}+8^{2006}\qquad <span class='latex-bold'>(D)</span>}\ 10^{2006}-8^{2006}\qquad
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2
1
(
1
0
2006
+
8
2006
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
1
(
1
0
2006
−
8
2006
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
1
0
2006
+
8
2006
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
1
0
2006
−
8
2006
5
1
Hide problems
2006 China Second Round Olympiad Test 1 #5
Suppose
f
(
x
)
=
x
3
+
log
2
(
x
+
x
2
+
1
)
f(x) = x^3 + \log_2(x + \sqrt{x^2+1})
f
(
x
)
=
x
3
+
lo
g
2
(
x
+
x
2
+
1
)
. For any
a
,
b
∈
R
a,b \in \mathbb{R}
a
,
b
∈
R
, to satisfy
f
(
a
)
+
f
(
b
)
≥
0
f(a) + f(b) \ge 0
f
(
a
)
+
f
(
b
)
≥
0
, the condition
a
+
b
≥
0
a + b \ge 0
a
+
b
≥
0
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
necessary and sufficient
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
not necessary but sufficient
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
necessary but not sufficient
<span class='latex-bold'>(A)</span>\ \text{necessary and sufficient}\qquad<span class='latex-bold'>(B)</span>\ \text{not necessary but sufficient}\qquad<span class='latex-bold'>(C)</span>\ \text{necessary but not sufficient}\qquad
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
necessary and sufficient
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
not necessary but sufficient
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
necessary but not sufficient
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
neither necessary nor sufficient
<span class='latex-bold'>(D)</span>\ \text{neither necessary nor sufficient}\qquad
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
neither necessary nor sufficient
4
1
Hide problems
2006 China Second Round Olympiad Test 1 #4
Given a right triangular prism
A
1
B
1
C
1
−
A
B
C
A_1B_1C_1 - ABC
A
1
B
1
C
1
−
A
BC
with
∠
B
A
C
=
π
2
\angle BAC = \frac{\pi}{2}
∠
B
A
C
=
2
π
and
A
B
=
A
C
=
A
A
1
AB = AC = AA_1
A
B
=
A
C
=
A
A
1
, let
G
G
G
,
E
E
E
be the midpoints of
A
1
B
1
A_1B_1
A
1
B
1
,
C
C
1
CC_1
C
C
1
respectively, and
D
D
D
,
F
F
F
be variable points lying on segments
A
C
AC
A
C
,
A
B
AB
A
B
(not including endpoints) respectively. If
G
D
⊥
E
F
GD \bot EF
G
D
⊥
EF
, the range of the length of
D
F
DF
D
F
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
[
1
5
,
1
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
[
1
5
,
2
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
[
1
,
2
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
[
1
2
,
2
)
{ <span class='latex-bold'>(A)</span>\ [\frac{1}{\sqrt{5}}, 1)\qquad<span class='latex-bold'>(B)</span>\ [\frac{1}{5}, 2)\qquad<span class='latex-bold'>(C)</span>\ [1, \sqrt{2})\qquad<span class='latex-bold'>(D)</span>} [\frac{1}{\sqrt{2}}, \sqrt{2})\qquad
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
[
5
1
,
1
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
[
5
1
,
2
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
[
1
,
2
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
[
2
1
,
2
)
3
2
Hide problems
solve the system of equations
Solve the system of equations in real numbers:
{
x
−
y
+
z
−
w
=
2
x
2
−
y
2
+
z
2
−
w
2
=
6
x
3
−
y
3
+
z
3
−
w
3
=
20
x
4
−
y
4
+
z
4
−
w
4
=
66
\begin{cases} x-y+z-w=2 \\ x^2-y^2+z^2-w^2=6 \\ x^3-y^3+z^3-w^3=20 \\ x^4-y^4+z^4-w^4=66 \end{cases}
⎩
⎨
⎧
x
−
y
+
z
−
w
=
2
x
2
−
y
2
+
z
2
−
w
2
=
6
x
3
−
y
3
+
z
3
−
w
3
=
20
x
4
−
y
4
+
z
4
−
w
4
=
66
2006 China Second Round Olympiad Test 1 #3
Suppose
A
=
x
∣
5
x
−
a
≤
0
A = {x|5x-a \le 0}
A
=
x
∣5
x
−
a
≤
0
,
B
=
x
∣
6
x
−
b
>
0
B = {x|6x-b > 0}
B
=
x
∣6
x
−
b
>
0
,
a
,
b
∈
N
a,b \in \mathbb{N}
a
,
b
∈
N
, and
A
∩
B
∩
N
=
2
,
3
,
4
A \cap B \cap \mathbb{N} = {2,3,4}
A
∩
B
∩
N
=
2
,
3
,
4
. The number of such pairs
(
a
,
b
)
(a,b)
(
a
,
b
)
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
20
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
25
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
30
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
42
{ <span class='latex-bold'>(A)</span>\ 20\qquad<span class='latex-bold'>(B)</span>\ 25\qquad<span class='latex-bold'>(C)</span>\ 30\qquad<span class='latex-bold'>(D)</span>} 42\qquad
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
20
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
25
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
30
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
42
2
2
Hide problems
a recursive sequence concerning Fibonacci
Let
x
,
y
x,y
x
,
y
be real numbers. Define a sequence
{
a
n
}
\{a_n \}
{
a
n
}
through the recursive formula
a
0
=
x
,
a
1
=
y
,
a
n
+
1
=
a
n
a
n
−
1
+
1
a
n
+
a
n
−
1
,
a_0=x,a_1=y,a_{n+1}=\frac{a_na_{n-1}+1}{a_n+a_{n-1}},
a
0
=
x
,
a
1
=
y
,
a
n
+
1
=
a
n
+
a
n
−
1
a
n
a
n
−
1
+
1
,
Find
a
n
a_n
a
n
.
2006 China Second Round Olympiad Test 1 #2
Suppose
l
o
g
x
(
2
x
2
+
x
−
1
)
>
l
o
g
x
2
−
1
log_x (2x^2+x-1)>log_x 2-1
l
o
g
x
(
2
x
2
+
x
−
1
)
>
l
o
g
x
2
−
1
. Then the range of
x
x
x
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
2
<
x
<
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
x
>
1
2
and
x
≠
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
x
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
0
<
x
<
1
{ <span class='latex-bold'>(A)</span>\ \frac{1}{2}<x<1\qquad<span class='latex-bold'>(B)</span>\ x>\frac{1}{2} \text{and} x \not= 1\qquad<span class='latex-bold'>(C)</span>\ x>1\qquad<span class='latex-bold'>(D)</span>}\ 0<x<1\qquad
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2
1
<
x
<
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
x
>
2
1
and
x
=
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
x
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
0
<
x
<
1
1
2
Hide problems
Prove four points are concyclic in an ellipse
An ellipse with foci
B
0
,
B
1
B_0,B_1
B
0
,
B
1
intersects
A
B
i
AB_i
A
B
i
at
C
i
C_i
C
i
(
i
=
0
,
1
)
(i=0,1)
(
i
=
0
,
1
)
. Let
P
0
P_0
P
0
be a point on ray
A
B
0
AB_0
A
B
0
.
Q
0
Q_0
Q
0
is a point on ray
C
1
B
0
C_1B_0
C
1
B
0
such that
B
0
P
0
=
B
0
Q
0
B_0P_0=B_0Q_0
B
0
P
0
=
B
0
Q
0
;
P
1
P_1
P
1
is on ray
B
1
A
B_1A
B
1
A
such that
C
1
Q
0
=
C
1
P
1
C_1Q_0=C_1P_1
C
1
Q
0
=
C
1
P
1
;
Q
1
Q_1
Q
1
is on ray
B
1
C
0
B_1C_0
B
1
C
0
such that
B
1
P
1
=
B
1
Q
1
B_1P_1=B_1Q_1
B
1
P
1
=
B
1
Q
1
;
P
2
P_2
P
2
is on ray
A
B
0
AB_0
A
B
0
such that
C
0
Q
1
=
C
0
Q
2
C_0Q_1=C_0Q_2
C
0
Q
1
=
C
0
Q
2
. Prove that
P
0
=
P
2
P_0=P_2
P
0
=
P
2
and that the four points
P
0
,
Q
0
,
Q
1
,
P
1
P_0,Q_0,Q_1,P_1
P
0
,
Q
0
,
Q
1
,
P
1
are concyclic.
2006 China Second Round Olympiad Test 1 #1
Let
△
A
B
C
\triangle ABC
△
A
BC
be a given triangle. If
∣
B
A
−
t
B
C
∣
≥
∣
A
C
∣
|BA-tBC| \ge |AC|
∣
B
A
−
tBC
∣
≥
∣
A
C
∣
for any
t
∈
R
t \in \mathbb{R}
t
∈
R
, then
△
A
B
C
\triangle ABC
△
A
BC
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
an acute triangle
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
an obtuse triangle
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
a right triangle
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
not known
<span class='latex-bold'>(A)</span>\ \text{an acute triangle}\qquad<span class='latex-bold'>(B)</span>\ \text{an obtuse triangle}\qquad<span class='latex-bold'>(C)</span>\ \text{a right triangle}\qquad<span class='latex-bold'>(D)</span>\ \text{not known}\qquad
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
an acute triangle
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
an obtuse triangle
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
a right triangle
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
not known