MathDB
Problems
Contests
National and Regional Contests
Taiwan Contests
TST Round 1
2022 Taiwan TST Round 1
2022 Taiwan TST Round 1
Part of
TST Round 1
Subcontests
(4)
G
1
Hide problems
Geometric inequality in Taiwan TST
Two triangles
A
B
C
ABC
A
BC
and
A
′
B
′
C
′
A'B'C'
A
′
B
′
C
′
are on the plane. It is known that each side length of triangle
A
B
C
ABC
A
BC
is not less than
a
a
a
, and each side length of triangle
A
′
B
′
C
′
A'B'C'
A
′
B
′
C
′
is not less than
a
′
a'
a
′
. Prove that we can always choose two points in the two triangles respectively such that the distance between them is not less than
a
2
+
a
′
2
3
\sqrt{\dfrac{a^2+a'^2}{3}}
3
a
2
+
a
′2
.
5
1
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Chasing After the Tangent
Let
H
H
H
be the orthocenter of a given triangle
A
B
C
ABC
A
BC
. Let
B
H
BH
B
H
and
A
C
AC
A
C
meet at a point
E
E
E
, and
C
H
CH
C
H
and
A
B
AB
A
B
meet at
F
F
F
. Suppose that
X
X
X
is a point on the line
B
C
BC
BC
. Also suppose that the circumcircle of triangle
B
E
X
BEX
BEX
and the line
A
B
AB
A
B
intersect again at
Y
Y
Y
, and the circumcircle of triangle
C
F
X
CFX
CFX
and the line
A
C
AC
A
C
intersect again at
Z
Z
Z
. Show that the circumcircle of triangle
A
Y
Z
AYZ
A
Y
Z
is tangent to the line
A
H
AH
A
H
.Proposed by usjl
C
1
Hide problems
Mingmingsan playing with an equilateral triangle
Let
△
P
1
P
2
P
3
\triangle P_1P_2P_3
△
P
1
P
2
P
3
be an equilateral triangle. For each
n
≥
4
n\ge 4
n
≥
4
, Mingmingsan can set
P
n
P_n
P
n
as the circumcenter or orthocenter of
△
P
n
−
3
P
n
−
2
P
n
−
1
\triangle P_{n-3}P_{n-2}P_{n-1}
△
P
n
−
3
P
n
−
2
P
n
−
1
. Find all positive integer
n
n
n
such that Mingmingsan has a strategy to make
P
n
P_n
P
n
equals to the circumcenter of
△
P
1
P
2
P
3
\triangle P_1P_2P_3
△
P
1
P
2
P
3
. Proposed by Li4 and Untro368.
A
2
Hide problems
Inequality on eventually constant sequence
Let
a
1
,
a
2
,
a
3
,
…
a_1, a_2, a_3, \ldots
a
1
,
a
2
,
a
3
,
…
be a sequence of reals such that there exists
N
∈
N
N\in\mathbb{N}
N
∈
N
so that
a
n
=
1
a_n=1
a
n
=
1
for all
n
≥
N
n\geq N
n
≥
N
, and for all
n
≥
2
n\geq 2
n
≥
2
we have
a
n
≤
a
n
−
1
+
2
−
n
a
2
n
.
a_{n}\leq a_{n-1}+2^{-n}a_{2n}.
a
n
≤
a
n
−
1
+
2
−
n
a
2
n
.
Show that
a
k
>
1
−
2
−
k
a_k>1-2^{-k}
a
k
>
1
−
2
−
k
for all
k
∈
N
k\in\mathbb{N}
k
∈
N
.Proposed by usjl
Functional Equation with Floors
Find all
f
:
Z
→
Z
f:\mathbb{Z}\to\mathbb{Z}
f
:
Z
→
Z
such that
f
(
⌊
f
(
x
)
+
f
(
y
)
2
⌋
)
+
f
(
x
)
=
f
(
f
(
y
)
)
+
⌊
f
(
x
)
+
f
(
y
)
2
⌋
f\left(\left\lfloor\frac{f(x)+f(y)}{2}\right\rfloor\right)+f(x)=f(f(y))+\left\lfloor\frac{f(x)+f(y)}{2}\right\rfloor
f
(
⌊
2
f
(
x
)
+
f
(
y
)
⌋
)
+
f
(
x
)
=
f
(
f
(
y
))
+
⌊
2
f
(
x
)
+
f
(
y
)
⌋
holds for all
x
,
y
∈
Z
x,y\in\mathbb{Z}
x
,
y
∈
Z
.Proposed by usjl