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Contests
National and Regional Contests
Taiwan Contests
Taiwan APMO Prelininary
2019 Taiwan APMO Preliminary Test
2019 Taiwan APMO Preliminary Test
Part of
Taiwan APMO Prelininary
Subcontests
(7)
P7
1
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2019 Taiwan APMO Preliminary P7
Let positive integer
k
k
k
satisfies
1
<
k
<
100
1<k<100
1
<
k
<
100
. For the permutation of
1
,
2
,
.
.
.
,
100
1,2,...,100
1
,
2
,
...
,
100
be
a
1
,
a
2
,
.
.
.
,
a
100
a_1,a_2,...,a_{100}
a
1
,
a
2
,
...
,
a
100
, take the minimum
m
>
k
m>k
m
>
k
such that
a
m
a_m
a
m
is at least less than
(
k
−
1
)
(k-1)
(
k
−
1
)
numbers of
a
1
,
a
2
,
.
.
.
,
a
k
a_1,a_2,...,a_k
a
1
,
a
2
,
...
,
a
k
. We know that the number of sequences satisfies
a
m
=
1
a_m=1
a
m
=
1
is
100
!
4
\frac{100!}{4}
4
100
!
. Find the all possible values of
k
k
k
.
P6
1
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2019 Taiwan APMO Preliminary P6
Let
N
\mathbb{N}
N
denote the set of all positive integers.Function
f
:
N
∪
0
→
N
∪
0
f:\mathbb{N}\cup{0}\rightarrow\mathbb{N}\cup{0}
f
:
N
∪
0
→
N
∪
0
satisfies :for any two distinct positive integer
a
,
b
a,b
a
,
b
, we have
f
(
a
)
+
f
(
b
)
−
f
(
a
+
b
)
=
2019
f(a)+f(b)-f(a+b)=2019
f
(
a
)
+
f
(
b
)
−
f
(
a
+
b
)
=
2019
(1)Find
f
(
0
)
f(0)
f
(
0
)
(2)Let
a
1
,
a
2
,
.
.
.
,
a
100
a_1,a_2,...,a_{100}
a
1
,
a
2
,
...
,
a
100
be 100 positive integers (they are pairwise distinct), find
f
(
a
1
)
+
f
(
a
2
)
+
.
.
.
+
f
(
a
100
)
−
f
(
a
1
+
a
2
+
.
.
.
+
a
100
)
f(a_1)+f(a_2)+...+f(a_{100})-f(a_1+a_2+...+a_{100})
f
(
a
1
)
+
f
(
a
2
)
+
...
+
f
(
a
100
)
−
f
(
a
1
+
a
2
+
...
+
a
100
)
P5
1
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2019 Taiwan APMO Preliminary P5
Find the minimum positive integer
n
n
n
such that for any set
A
A
A
with
n
n
n
positive intergers has
15
15
15
elements which sum is divisible by
15
15
15
.
P4
1
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2019 Taiwan APMO Preliminary P4
We define a sequence
a
n
{a_n}
a
n
:
a
1
=
1
,
a
n
+
1
=
a
n
+
n
2
,
n
=
1
,
2
,
.
.
.
a_1=1,a_{n+1}=\sqrt{a_n+n^2},n=1,2,...
a
1
=
1
,
a
n
+
1
=
a
n
+
n
2
,
n
=
1
,
2
,
...
(1)Find
⌊
a
2019
⌋
\lfloor a_{2019}\rfloor
⌊
a
2019
⌋
(2)Find
⌊
a
1
2
⌋
+
⌊
a
2
2
⌋
+
.
.
.
+
⌊
a
20
2
⌋
\lfloor a_{1}^2\rfloor+\lfloor a_{2}^2\rfloor+...+\lfloor a_{20}^2\rfloor
⌊
a
1
2
⌋
+
⌊
a
2
2
⌋
+
...
+
⌊
a
20
2
⌋
P3
1
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2019 Taiwan APMO Preliminary P3
Let
△
A
B
C
\triangle ABC
△
A
BC
be an acute triangle,
H
H
H
is its orthocenter.
A
H
→
,
B
H
→
,
C
H
→
\overrightarrow{AH},\overrightarrow{BH},\overrightarrow{CH}
A
H
,
B
H
,
C
H
intersect
△
A
B
C
\triangle ABC
△
A
BC
's circumcircle at
A
′
,
B
′
,
C
′
A',B',C'
A
′
,
B
′
,
C
′
respectively. Find the range (minimum value and the maximum upper bound) of
A
H
A
A
′
+
B
H
B
B
′
+
C
H
C
C
′
\dfrac{AH}{AA'}+\dfrac{BH}{BB'}+\dfrac{CH}{CC'}
A
A
′
A
H
+
B
B
′
B
H
+
C
C
′
C
H
P2
1
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2019 Taiwan APMO Preliminary P2
Put
1
,
2
,
.
.
.
.
,
2018
1,2,....,2018
1
,
2
,
....
,
2018
(2018 numbers) in a row randomly and call this number
A
A
A
. Find the remainder of
A
A
A
divided by
3
3
3
.
P1
1
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2019 Taiwan APMO Preliminary Problem 1
In
△
A
B
C
\triangle ABC
△
A
BC
,
∠
B
=
9
0
∘
\angle B=90^\circ
∠
B
=
9
0
∘
, segment
A
B
>
B
C
AB>BC
A
B
>
BC
. Now we have a
△
A
i
B
C
(
i
=
1
,
2
,
.
.
.
,
n
)
\triangle A_iBC(i=1,2,...,n)
△
A
i
BC
(
i
=
1
,
2
,
...
,
n
)
which is similiar to
△
A
B
C
\triangle ABC
△
A
BC
(the vertexs of them might not correspond). Find the maximum value of
n
+
2018
n+2018
n
+
2018
.