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Problems
Contests
National and Regional Contests
Turkey Contests
JBMO TST - Turkey
2024 JBMO TST - Turkey
2024 JBMO TST - Turkey
Part of
JBMO TST - Turkey
Subcontests
(8)
8
1
Hide problems
Colors and *sigh* 207 boxes
There is
207
207
207
boxes on the table which numbered
1
,
2
,
…
,
207
1,2, \dots , 207
1
,
2
,
…
,
207
respectively. Firstly Aslı puts a red ball in each of the
100
100
100
boxes that she chooses and puts a white ball in each of the remaining ones. After that Zehra, writes a pair
(
i
,
j
)
(i,j)
(
i
,
j
)
on the blackboard such that
1
≤
i
≤
j
≤
207
1\leq i \leq j \leq 207
1
≤
i
≤
j
≤
207
. Finally, Aslı tells Zehra that for every pair; whether the color of the balls which is inside the box which numbered by these numbers are the same or not. Find the least possible value of
N
N
N
such that Zehra can guarantee finding all colors that has been painted to balls in each of the boxes with writing
N
N
N
pairs on the blackboard.
7
1
Hide problems
Circumcircle geometry
Let
A
B
C
D
ABCD
A
BC
D
be circumscribed quadrilateral such that the midpoints of
A
B
AB
A
B
,
B
C
BC
BC
,
C
D
CD
C
D
and
D
A
DA
D
A
are
K
K
K
,
L
L
L
,
M
M
M
,
N
N
N
respectively. Let the reflections of the point
M
M
M
wrt the lines
A
D
AD
A
D
and
B
C
BC
BC
be
P
P
P
and
Q
Q
Q
respectively. Let the circumcenter of the triangle
K
P
Q
KPQ
K
PQ
be
R
R
R
. Prove that
R
N
=
R
L
RN=RL
RN
=
R
L
6
1
Hide problems
Interesting sequence for Juniors
Let
(
a
n
)
n
=
0
∞
{(a_n)}_{n=0}^{\infty}
(
a
n
)
n
=
0
∞
and
(
b
n
)
n
=
0
∞
{(b_n)}_{n=0}^{\infty}
(
b
n
)
n
=
0
∞
be real squences such that
a
0
=
40
a_0=40
a
0
=
40
,
b
0
=
41
b_0=41
b
0
=
41
and for all
n
≥
0
n\geq 0
n
≥
0
the given equalities hold.
a
n
+
1
=
a
n
+
1
b
n
and
b
n
+
1
=
b
n
+
1
a
n
a_{n+1}=a_n+\frac{1}{b_n} \hspace{0.5 cm} \text{and} \hspace{0.5 cm} b_{n+1}=b_n+\frac{1}{a_n}
a
n
+
1
=
a
n
+
b
n
1
and
b
n
+
1
=
b
n
+
a
n
1
Find the least possible positive integer value of
k
k
k
such that the value of
a
k
a_k
a
k
is strictly bigger than
80
80
80
.
5
1
Hide problems
\frac{2^{n!}-1}{2^n-1} be a square
Find all positive integer values of
n
n
n
such that the value of the
2
n
!
−
1
2
n
−
1
\frac{2^{n!}-1}{2^n-1}
2
n
−
1
2
n
!
−
1
is a square of an integer.
4
1
Hide problems
Nice nt with remainder set
Let
n
n
n
be a positive integer and
d
(
n
)
d(n)
d
(
n
)
is the number of positive integer divisors of
n
n
n
. For every two positive integer divisor
x
,
y
x,y
x
,
y
of
n
n
n
, the remainders when
x
,
y
x,y
x
,
y
divided by
d
(
n
)
+
1
d(n)+1
d
(
n
)
+
1
are pairwise distinct. Show that either
d
(
n
)
+
1
d(n)+1
d
(
n
)
+
1
is equal to prime or
4
4
4
.
3
1
Hide problems
Strange question with positive reals
Find all
x
,
y
,
z
∈
R
+
x,y,z \in R^+
x
,
y
,
z
∈
R
+
such that the sets
(
23
x
+
24
y
+
25
z
,
23
y
+
24
z
+
25
x
,
23
z
+
24
x
+
25
y
)
(23x+24y+25z,23y+24z+25x,23z+24x+25y)
(
23
x
+
24
y
+
25
z
,
23
y
+
24
z
+
25
x
,
23
z
+
24
x
+
25
y
)
and
(
x
5
+
y
5
,
y
5
+
z
5
,
z
5
+
x
5
)
(x^5+y^5,y^5+z^5,z^5+x^5)
(
x
5
+
y
5
,
y
5
+
z
5
,
z
5
+
x
5
)
are same
2
1
Hide problems
Maximum sum for column
A real number is written on each square of a
2024
×
2024
2024 \times 2024
2024
×
2024
chessboard. It is given that the sum of all real numbers on the board is
2024
2024
2024
. Then, the board is covered by
1
×
2
1 \times 2
1
×
2
or
2
×
1
2\times 1
2
×
1
dominos such that there isn't any square that is covered by two different dominoes. For each domino, Aslı deletes
2
2
2
numbers covered by it and writes
0
0
0
on one of the squares and the sum of the two numbers on the other square. Find the maximum number
k
k
k
such that after Aslı finishes her moves, there exists a column or row where the sum of all the numbers on it is at least
k
k
k
regardless of how dominos were replaced and the real numbers were written initially.
1
1
Hide problems
Prove the perpendicularity
In the acute-angled triangle
A
B
C
ABC
A
BC
,
P
P
P
is the midpoint of segment
B
C
BC
BC
and the point
K
K
K
is the foot of the altitude from
A
A
A
.
D
D
D
is a point on segment
A
P
AP
A
P
such that
∠
B
D
C
=
90
\angle BDC=90
∠
B
D
C
=
90
. Let
(
A
D
K
)
∩
B
C
=
E
(ADK) \cap BC=E
(
A
DK
)
∩
BC
=
E
and
(
A
B
C
)
∩
A
E
=
F
(ABC) \cap AE=F
(
A
BC
)
∩
A
E
=
F
. Prove that
∠
A
F
D
=
90
\angle AFD=90
∠
A
F
D
=
90
.