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Contests
National and Regional Contests
Turkey Contests
Turkey Team Selection Test
1994 Turkey Team Selection Test
1994 Turkey Team Selection Test
Part of
Turkey Team Selection Test
Subcontests
(3)
3
2
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Turkey TST 1994 - P3
All sides and diagonals of a
25
25
25
-gon are drawn either red or white. Show that at least
500
500
500
triangles, having all three sides are in same color and having all three vertices from the vertices of the
25
25
25
-gon, can be found.
Turkey TST 1994 - P6
Find all integer pairs
(
a
,
b
)
(a,b)
(
a
,
b
)
such that
a
⋅
b
a\cdot b
a
⋅
b
divides
a
2
+
b
2
+
3
a^2+b^2+3
a
2
+
b
2
+
3
.
1
2
Hide problems
Turkey TST 1994 - P1
f
f
f
is a function defined on integers and satisfies
f
(
x
)
+
f
(
x
+
3
)
=
x
2
f(x)+f(x+3)=x^2
f
(
x
)
+
f
(
x
+
3
)
=
x
2
for every integer
x
x
x
. If
f
(
19
)
=
94
f(19)=94
f
(
19
)
=
94
, then calculate
f
(
94
)
f(94)
f
(
94
)
.
Turkey TST 1994 - P4
Let
P
,
Q
,
R
P,Q,R
P
,
Q
,
R
be points on the sides of
△
A
B
C
\triangle ABC
△
A
BC
such that
P
∈
[
A
B
]
,
Q
∈
[
B
C
]
,
R
∈
[
C
A
]
P \in [AB],Q\in[BC],R\in[CA]
P
∈
[
A
B
]
,
Q
∈
[
BC
]
,
R
∈
[
C
A
]
and
∣
A
P
∣
∣
A
B
∣
=
∣
B
Q
∣
∣
B
C
∣
=
∣
C
R
∣
∣
C
A
∣
=
k
<
1
2
\frac{|AP|}{|AB|} = \frac {|BQ|}{|BC|} =\frac{|CR|}{|CA|} =k < \frac 12
∣
A
B
∣
∣
A
P
∣
=
∣
BC
∣
∣
BQ
∣
=
∣
C
A
∣
∣
CR
∣
=
k
<
2
1
If
G
G
G
is the centroid of
△
A
B
C
\triangle ABC
△
A
BC
, find the ratio
A
r
e
a
(
△
P
Q
G
)
A
r
e
a
(
△
P
Q
R
)
\frac{Area(\triangle PQG)}{Area(\triangle PQR)}
A
re
a
(
△
PQR
)
A
re
a
(
△
PQG
)
.
2
2
Hide problems
Turkey TST 1994 - P2
Let
O
O
O
be the center and
[
A
B
]
[AB]
[
A
B
]
be the diameter of a semicircle.
E
E
E
is a point between
O
O
O
and
B
B
B
. The perpendicular to
[
A
B
]
[AB]
[
A
B
]
at
E
E
E
meets the semicircle at
D
D
D
. A circle which is internally tangent to the arc \overarc{BD} is also tangent to
[
D
E
]
[DE]
[
D
E
]
and
[
E
B
]
[EB]
[
EB
]
at
K
K
K
and
C
C
C
, respectively. Prove that
E
D
C
^
=
B
D
C
^
\widehat{EDC}=\widehat{BDC}
E
D
C
=
B
D
C
.
Turkey TST 1994 - P5
Show that positive integers
n
i
,
m
i
n_i,m_i
n
i
,
m
i
(
i
=
1
,
2
,
3
,
⋯
)
(i=1,2,3, \cdots )
(
i
=
1
,
2
,
3
,
⋯
)
can be found such that
lim
i
→
∞
2
n
i
3
m
i
=
1
\mathop{\lim }\limits_{i \to \infty } \frac{2^{n_i}}{3^{m_i }} = 1
i
→
∞
lim
3
m
i
2
n
i
=
1