MathDB
Problems
Contests
National and Regional Contests
Turkey Contests
Turkey Team Selection Test
2001 Turkey Team Selection Test
2001 Turkey Team Selection Test
Part of
Turkey Team Selection Test
Subcontests
(3)
2
2
Hide problems
Coaxal Circles
A circle touches to diameter
A
B
AB
A
B
of a unit circle with center
O
O
O
at
T
T
T
where
O
T
>
1
OT>1
OT
>
1
. These circles intersect at two different points
C
C
C
and
D
D
D
. The circle through
O
O
O
,
D
D
D
, and
C
C
C
meet the line
A
B
AB
A
B
at
P
P
P
different from
O
O
O
. Show that
∣
P
A
∣
⋅
∣
P
B
∣
=
∣
P
T
∣
2
∣
O
T
∣
2
.
|PA|\cdot |PB| = \dfrac {|PT|^2}{|OT|^2}.
∣
P
A
∣
⋅
∣
PB
∣
=
∣
OT
∣
2
∣
PT
∣
2
.
The line through orthocenter and midpoint
Let
H
H
H
be the intersection of the altitudes of an acute triangle
A
B
C
ABC
A
BC
and
D
D
D
be the midpoint of
[
A
C
]
[AC]
[
A
C
]
. Show that
D
H
DH
DH
passes through one of the intersection point of the circumcircle of
A
B
C
ABC
A
BC
and the circle with diameter
[
B
H
]
[BH]
[
B
H
]
.
1
2
Hide problems
Can 2001 numbers be all positive?
Each one of
2001
2001
2001
children chooses a positive integer and writes down his number and names of some of other
2000
2000
2000
children to his notebook. Let
A
c
A_c
A
c
be the sum of the numbers chosen by the children who appeared in the notebook of the child
c
c
c
. Let
B
c
B_c
B
c
be the sum of the numbers chosen by the children who wrote the name of the child
c
c
c
into their notebooks. The number
N
c
=
A
c
−
B
c
N_c = A_c - B_c
N
c
=
A
c
−
B
c
is assigned to the child
c
c
c
. Determine whether all of the numbers assigned to the children could be positive.
5^x = 1 + 4y + y^4
Find all ordered pairs of integers
(
x
,
y
)
(x,y)
(
x
,
y
)
such that
5
x
=
1
+
4
y
+
y
4
5^x = 1 + 4y + y^4
5
x
=
1
+
4
y
+
y
4
.
3
2
Hide problems
S^{n+k}(a,b,c) = S^{n}(a,b,c) (mod abc)
For all integers
x
,
y
,
z
x,y,z
x
,
y
,
z
, let
S
(
x
,
y
,
z
)
=
(
x
y
−
x
z
,
y
z
−
y
x
,
z
x
−
z
y
)
.
S(x,y,z) = (xy - xz, yz-yx, zx - zy).
S
(
x
,
y
,
z
)
=
(
x
y
−
x
z
,
yz
−
y
x
,
z
x
−
zy
)
.
Prove that for all integers
a
a
a
,
b
b
b
and
c
c
c
with
a
b
c
>
1
abc>1
ab
c
>
1
, and for every integer
n
≥
n
0
n\geq n_0
n
≥
n
0
, there exists integers
n
0
n_0
n
0
and
k
k
k
with
0
<
k
≤
a
b
c
0<k\leq abc
0
<
k
≤
ab
c
such that
S
n
+
k
(
a
,
b
,
c
)
≡
S
n
(
a
,
b
,
c
)
(
m
o
d
a
b
c
)
.
S^{n+k}(a,b,c) \equiv S^n(a,b,c) \pmod {abc}.
S
n
+
k
(
a
,
b
,
c
)
≡
S
n
(
a
,
b
,
c
)
(
mod
ab
c
)
.
(
S
1
=
S
S^1 = S
S
1
=
S
and for every integer
m
≥
1
m\geq 1
m
≥
1
,
S
m
+
1
=
S
∘
S
m
.
S^{m+1} = S \circ S^m.
S
m
+
1
=
S
∘
S
m
.
(
u
1
,
u
2
,
u
3
)
≡
(
v
1
,
v
2
,
v
3
)
(
m
o
d
M
)
⟺
u
i
≡
v
i
(
m
o
d
M
)
(
i
=
1
,
2
,
3
)
.
(u_1, u_2, u_3) \equiv (v_1, v_2, v_3) \pmod M \Longleftrightarrow u_i \equiv v_i \pmod M (i=1,2,3).
(
u
1
,
u
2
,
u
3
)
≡
(
v
1
,
v
2
,
v
3
)
(
mod
M
)
⟺
u
i
≡
v
i
(
mod
M
)
(
i
=
1
,
2
,
3
)
.
)
Continuous f(x-f(x))=x/2 does not exist
Show that there is no continuous function
f
:
R
→
R
f:\mathbb{R}\rightarrow \mathbb{R}
f
:
R
→
R
such that for every real number
x
x
x
f
(
x
−
f
(
x
)
)
=
x
2
.
f(x-f(x)) = \dfrac x2.
f
(
x
−
f
(
x
))
=
2
x
.