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National and Regional Contests
Turkey Contests
Turkey Team Selection Test
1992 Turkey Team Selection Test
1992 Turkey Team Selection Test
Part of
Turkey Team Selection Test
Subcontests
(3)
2
2
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Turkey TST 1992 - P5: m balls n boxes
There are
n
n
n
boxes which is numbere from
1
1
1
to
n
n
n
. The box with number
1
1
1
is open, and the others are closed. There are
m
m
m
identical balls (
m
≥
n
m\geq n
m
≥
n
). One of the balls is put into the open box, then we open the box with number
2
2
2
. Now, we put another ball to one of two open boxes, then we open the box with number
3
3
3
. Go on until the last box will be open. After that the remaining balls will be randomly put into the boxes. In how many ways this arrangement can be done?
Turkey TST 1992 - P2: Altitude meets circumcircle
The line passing through
B
B
B
is perpendicular to the side
A
C
AC
A
C
at
E
E
E
. This line meets the circumcircle of
△
A
B
C
\triangle ABC
△
A
BC
at
D
D
D
. The foot of the perpendicular from
D
D
D
to the side
B
C
BC
BC
is
F
F
F
. If
O
O
O
is the center of the circumcircle of
△
A
B
C
\triangle ABC
△
A
BC
, prove that
B
O
BO
BO
is perpendicular to
E
F
EF
EF
.
3
2
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Turkey TST 1992 - P3: x_1.x_2...x_(n+1) >= n^(n+1)
x
1
,
x
2
,
⋯
,
x
n
+
1
x_1, x_2,\cdots,x_{n+1}
x
1
,
x
2
,
⋯
,
x
n
+
1
are posive real numbers satisfying the equation
1
1
+
x
1
+
1
1
+
x
2
+
⋯
+
1
1
+
x
n
+
1
=
1
\frac{1}{1+x_1} + \frac{1}{1+x_2} + \cdots + \frac{1}{1+x_{n+1}} =1
1
+
x
1
1
+
1
+
x
2
1
+
⋯
+
1
+
x
n
+
1
1
=
1
Prove that
x
1
x
2
⋯
x
n
+
1
≥
n
n
+
1
x_1x_2 \cdots x_{n+1} \geq n^{n+1}
x
1
x
2
⋯
x
n
+
1
≥
n
n
+
1
.
Turkey TST 1992 - P6: Find circle with 11 pts out of 251 pts
A circle with radius
4
4
4
and
251
251
251
distinct points inside the circle are given. Show that it is possible to draw a circle with radius
1
1
1
and containing at least
11
11
11
of these points.
1
2
Hide problems
Turkey TST 1992 - P1: 14 conseq. divisible by first 5 primes
Is there
14
14
14
consecutive positive integers such that each of these numbers is divisible by one of the prime numbers
p
p
p
where
2
≤
p
≤
11
2\leq p \leq 11
2
≤
p
≤
11
.
Turkey TST 1992 - P4: PQ+RS=QR+SP in cyclic quadrilateral
The feet of perpendiculars from the intersection point of the diagonals of cyclic quadrilateral
A
B
C
D
ABCD
A
BC
D
to the sides
A
B
,
B
C
,
C
D
,
D
A
AB,BC,CD,DA
A
B
,
BC
,
C
D
,
D
A
are
P
,
Q
,
R
,
S
P,Q,R,S
P
,
Q
,
R
,
S
, respectively. Prove
P
Q
+
R
S
=
Q
R
+
S
P
PQ+RS=QR+SP
PQ
+
RS
=
QR
+
SP
.