MathDB
Problems
Contests
National and Regional Contests
Taiwan Contests
TST Round 2
2016 Taiwan TST Round 2
2016 Taiwan TST Round 2
Part of
TST Round 2
Subcontests
(4)
6
1
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pentagon and quadrilateral
Let
A
X
Y
Z
B
AXYZB
A
X
Y
ZB
be a convex pentagon inscribed in a semicircle with diameter
A
B
AB
A
B
, and let
K
K
K
be the foot of the altitude from
Y
Y
Y
to
A
B
AB
A
B
. Let
O
O
O
denote the midpoint of
A
B
AB
A
B
and
L
L
L
be the intersection of
X
Z
XZ
XZ
with
Y
O
YO
Y
O
. Select a point
M
M
M
on line
K
L
KL
K
L
with
M
A
=
M
B
MA=MB
M
A
=
MB
, and finally, let
I
I
I
be the reflection of
O
O
O
across
X
Z
XZ
XZ
. Prove that if quadrilateral
X
K
O
Z
XKOZ
X
K
OZ
is cyclic then so is quadrilateral
Y
O
M
I
YOMI
Y
OM
I
.Proposed by Evan Chen
3
1
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Triangle Grids
There is a grid of equilateral triangles with a distance 1 between any two neighboring grid points. An equilateral triangle with side length
n
n
n
lies on the grid so that all of its vertices are grid points, and all of its sides match the grid. Now, let us decompose this equilateral triangle into
n
2
n^2
n
2
smaller triangles (not necessarily equilateral triangles) so that the vertices of all these smaller triangles are all grid points, and all these small triangles have equal areas. Prove that there are at least
n
n
n
equilateral triangles among these smaller triangles.
1
1
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Circles and Radical Axis
Let
O
O
O
be the circumcenter of triangle
A
B
C
ABC
A
BC
, and
ω
\omega
ω
be the circumcircle of triangle
B
O
C
BOC
BOC
. Line
A
O
AO
A
O
intersects with circle
ω
\omega
ω
again at the point
G
G
G
. Let
M
M
M
be the midpoint of side
B
C
BC
BC
, and the perpendicular bisector of
B
C
BC
BC
meets circle
ω
\omega
ω
at the points
O
O
O
and
N
N
N
. Prove that the midpoint of the segment
A
N
AN
A
N
lies on the radical axis of the circumcircle of triangle
O
M
G
OMG
OMG
, and the circle whose diameter is
A
O
AO
A
O
.
2
3
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Number theory and Fibonacci
Let
<
F
n
>
\left< F_n\right>
⟨
F
n
⟩
be the Fibonacci sequence, that is,
F
0
=
0
F_0=0
F
0
=
0
,
F
1
=
1
F_1=1
F
1
=
1
, and
F
n
+
2
=
F
n
+
1
+
F
n
F_{n+2}=F_{n+1}+F_{n}
F
n
+
2
=
F
n
+
1
+
F
n
holds for all nonnegative integers
n
n
n
. Find all pairs
(
a
,
b
)
(a,b)
(
a
,
b
)
of positive integers with
a
<
b
a < b
a
<
b
such that
F
n
−
2
n
a
n
F_n-2na^n
F
n
−
2
n
a
n
is divisible by
b
b
b
for all positive integers
n
n
n
.
An interesting function equation
Find all function
f
:
Z
→
Z
f:\mathbb{Z}\rightarrow\mathbb{Z}
f
:
Z
→
Z
such that
f
(
f
(
x
)
+
f
(
y
)
)
+
f
(
x
)
f
(
y
)
=
f
(
x
+
y
)
f
(
x
−
y
)
f(f(x)+f(y))+f(x)f(y)=f(x+y)f(x-y)
f
(
f
(
x
)
+
f
(
y
))
+
f
(
x
)
f
(
y
)
=
f
(
x
+
y
)
f
(
x
−
y
)
for all integer
x
,
y
x,y
x
,
y
Inequality
Let
x
,
y
x,y
x
,
y
be positive real numbers such that
x
+
y
=
1
x+y=1
x
+
y
=
1
. Prove that
x
x
2
+
y
3
+
y
x
3
+
y
2
≤
2
(
x
x
+
y
2
+
y
x
2
+
y
)
\frac{x}{x^2+y^3}+\frac{y}{x^3+y^2}\leq2(\frac{x}{x+y^2}+\frac{y}{x^2+y})
x
2
+
y
3
x
+
x
3
+
y
2
y
≤
2
(
x
+
y
2
x
+
x
2
+
y
y
)
.