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National and Regional Contests
China Contests
China Team Selection Test
1986 China Team Selection Test
1986 China Team Selection Test
Part of
China Team Selection Test
Subcontests
(4)
1
1
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China TST 1986 geometry problem on square perimeter
Given a square
A
B
C
D
ABCD
A
BC
D
whose side length is
1
1
1
,
P
P
P
and
Q
Q
Q
are points on the sides
A
B
AB
A
B
and
A
D
AD
A
D
. If the perimeter of
A
P
Q
APQ
A
PQ
is
2
2
2
find the angle
P
C
Q
PCQ
PCQ
.
4
2
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sum is over the consecutive square of the segments
Given a triangle
A
B
C
ABC
A
BC
for which
C
=
90
C=90
C
=
90
degrees, prove that given
n
n
n
points inside it, we can name them
P
1
,
P
2
,
…
,
P
n
P_1, P_2 , \ldots , P_n
P
1
,
P
2
,
…
,
P
n
in some way such that:
∑
k
=
1
n
−
1
(
P
K
P
k
+
1
)
2
≤
A
B
2
\sum^{n-1}_{k=1} \left( P_K P_{k+1} \right)^2 \leq AB^2
∑
k
=
1
n
−
1
(
P
K
P
k
+
1
)
2
≤
A
B
2
(the sum is over the consecutive square of the segments from
1
1
1
up to
n
−
1
n-1
n
−
1
). Edited by orl.
China TST 1986 4k circle markers
Mark
4
⋅
k
4 \cdot k
4
⋅
k
points in a circle and number them arbitrarily with numbers from
1
1
1
to
4
⋅
k
4 \cdot k
4
⋅
k
. The chords cannot share common endpoints, also, the endpoints of these chords should be among the
4
⋅
k
4 \cdot k
4
⋅
k
points. i. Prove that
2
⋅
k
2 \cdot k
2
⋅
k
pairwisely non-intersecting chords can be drawn for each of whom its endpoints differ in at most
3
⋅
k
−
1
3 \cdot k - 1
3
⋅
k
−
1
. ii. Prove that the
3
⋅
k
−
1
3 \cdot k - 1
3
⋅
k
−
1
cannot be improved.
3
2
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China TST 1986 construction of sequence
Given a positive integer
A
A
A
written in decimal expansion:
(
a
n
,
a
n
−
1
,
…
,
a
0
)
(a_{n},a_{n-1}, \ldots, a_{0})
(
a
n
,
a
n
−
1
,
…
,
a
0
)
and let
f
(
A
)
f(A)
f
(
A
)
denote
∑
k
=
0
n
2
n
−
k
⋅
a
k
\sum^{n}_{k=0} 2^{n-k}\cdot a_k
∑
k
=
0
n
2
n
−
k
⋅
a
k
. Define
A
1
=
f
(
A
)
,
A
2
=
f
(
A
1
)
A_1=f(A), A_2=f(A_1)
A
1
=
f
(
A
)
,
A
2
=
f
(
A
1
)
. Prove that:I. There exists positive integer
k
k
k
for which
A
k
+
1
=
A
k
A_{k+1}=A_k
A
k
+
1
=
A
k
. II. Find such
A
k
A_k
A
k
for
1
9
86
.
19^{86}.
1
9
86
.
China TST 1986 symmetric sum
Let
x
i
,
x_i,
x
i
,
1
≤
i
≤
n
1 \leq i \leq n
1
≤
i
≤
n
be real numbers with
n
≥
3.
n \geq 3.
n
≥
3.
Let
p
p
p
and
q
q
q
be their symmetric sum of degree
1
1
1
and
2
2
2
respectively. Prove that: i)
p
2
⋅
n
−
1
n
−
2
q
≥
0
p^2 \cdot \frac{n-1}{n}-2q \geq 0
p
2
⋅
n
n
−
1
−
2
q
≥
0
ii)
∣
x
i
−
p
n
∣
≤
p
2
−
2
n
q
n
−
1
⋅
n
−
1
n
\left|x_i - \frac{p}{n}\right| \leq \sqrt{p^2 - \frac{2nq}{n-1}} \cdot \frac{n-1}{n}
x
i
−
n
p
≤
p
2
−
n
−
1
2
n
q
⋅
n
n
−
1
for every meaningful
i
i
i
.
2
2
Hide problems
China TST 1986 equivalence of two sum inequalities
Let
a
1
a_1
a
1
,
a
2
a_2
a
2
, ...,
a
n
a_n
a
n
and
b
1
b_1
b
1
,
b
2
b_2
b
2
, ...,
b
n
b_n
b
n
be
2
⋅
n
2 \cdot n
2
⋅
n
real numbers. Prove that the following two statements are equivalent: i) For any
n
n
n
real numbers
x
1
x_1
x
1
,
x
2
x_2
x
2
, ...,
x
n
x_n
x
n
satisfying
x
1
≤
x
2
≤
…
≤
x
n
x_1 \leq x_2 \leq \ldots \leq x_ n
x
1
≤
x
2
≤
…
≤
x
n
, we have \sum^{n}_{k \equal{} 1} a_k \cdot x_k \leq \sum^{n}_{k \equal{} 1} b_k \cdot x_k, ii) We have \sum^{s}_{k \equal{} 1} a_k \leq \sum^{s}_{k \equal{} 1} b_k for every s\in\left\{1,2,...,n\minus{}1\right\} and \sum^{n}_{k \equal{} 1} a_k \equal{} \sum^{n}_{k \equal{} 1} b_k.
China TST 1986 tetrahedron perimeter max inequality
Given a tetrahedron
A
B
C
D
ABCD
A
BC
D
,
E
E
E
,
F
F
F
,
G
G
G
, are on the respectively on the segments
A
B
AB
A
B
,
A
C
AC
A
C
and
A
D
AD
A
D
. Prove that: i) area
E
F
G
≤
EFG \leq
EFG
≤
max{area
A
B
C
ABC
A
BC
,area
A
B
D
ABD
A
B
D
,area
A
C
D
ACD
A
C
D
,area
B
C
D
BCD
BC
D
}. ii) The same as above replacing "area" for "perimeter".