MathDB
Problems
Contests
National and Regional Contests
Turkey Contests
National Olympiad First Round
2013 National Olympiad First Round
2013 National Olympiad First Round
Part of
National Olympiad First Round
Subcontests
(36)
36
1
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Chess tournament
A chess club consists of at least
10
10
10
and at most
50
50
50
members, where
G
G
G
of them are female, and
B
B
B
of them are male with
G
>
B
G>B
G
>
B
. In a chess tournament, each member plays with any other member exactly one time. At each game, the winner gains
1
1
1
, the loser gains
0
0
0
and both player gains
1
/
2
1/2
1/2
point when a tie occurs. At the tournament, it is observed that each member gained exactly half of his/her points from the games played against male members. How many different values can
B
B
B
take?
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<span class='latex-bold'>(A)</span>\ 5 \qquad<span class='latex-bold'>(B)</span>\ 4 \qquad<span class='latex-bold'>(C)</span>\ 3 \qquad<span class='latex-bold'>(D)</span>\ 2 \qquad<span class='latex-bold'>(E)</span>\ 1
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1
24
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Dividing 77 stones into k groups
77
77
77
stones weighing
1
,
2
,
…
,
77
1,2,\dots, 77
1
,
2
,
…
,
77
grams are divided into
k
k
k
groups such that total weights of each group are different from each other and each group contains less stones than groups with smaller total weights. For how many
k
∈
{
9
,
10
,
11
,
12
}
k\in \{9,10,11,12\}
k
∈
{
9
,
10
,
11
,
12
}
, is such a division possible?
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None of above
<span class='latex-bold'>(A)</span>\ 4 \qquad<span class='latex-bold'>(B)</span>\ 3 \qquad<span class='latex-bold'>(C)</span>\ 2 \qquad<span class='latex-bold'>(D)</span>\ 1 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
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None of above
12
1
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100 students study in groups of two
In the morning,
100
100
100
students study as
50
50
50
groups with two students in each group. In the afternoon, they study again as
50
50
50
groups with two students in each group. No matter how the groups in the morning or groups in the afternoon are established, if it is possible to find
n
n
n
students such that no two of them study together, what is the largest value of
n
n
n
?
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None of above
<span class='latex-bold'>(A)</span>\ 42 \qquad<span class='latex-bold'>(B)</span>\ 38 \qquad<span class='latex-bold'>(C)</span>\ 34 \qquad<span class='latex-bold'>(D)</span>\ 25 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
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None of above
32
1
Hide problems
P32 [Combinatorics] - Turkish NMO 1st Round - 2013
How many
10
10
10
-digit positive integers containing only the numbers
1
,
2
,
3
1,2,3
1
,
2
,
3
can be written such that the first and the last digits are same, and no two consecutive digits are same?
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768
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642
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510
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456
<span class='latex-bold'>(A)</span>\ 768 \qquad<span class='latex-bold'>(B)</span>\ 642 \qquad<span class='latex-bold'>(C)</span>\ 564 \qquad<span class='latex-bold'>(D)</span>\ 510 \qquad<span class='latex-bold'>(E)</span>\ 456
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510
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456
28
1
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P28 [Combinatorics] - Turkish NMO 1st Round - 2013
In the beginning, there is a pair of positive integers
(
m
,
n
)
(m,n)
(
m
,
n
)
written on the board. Alice and Bob are playing a turn-based game with the following move. At each turn, a player erases one of the numbers written on the board, and writes a different positive number not less than the half of the erased one. If a player cannot write a new number at some turn, he/she loses the game. For how many starting pairs
(
m
,
n
)
(m,n)
(
m
,
n
)
from the pairs
(
7
,
79
)
(7,79)
(
7
,
79
)
,
(
17
,
71
)
(17,71)
(
17
,
71
)
,
(
10
,
101
)
(10,101)
(
10
,
101
)
,
(
21
,
251
)
(21,251)
(
21
,
251
)
,
(
50
,
405
)
(50,405)
(
50
,
405
)
, can Alice guarantee to win when she makes the first move?
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None of above
<span class='latex-bold'>(A)</span>\ 4 \qquad<span class='latex-bold'>(B)</span>\ 3 \qquad<span class='latex-bold'>(C)</span>\ 2 \qquad<span class='latex-bold'>(D)</span>\ 1 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
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None of above
16
1
Hide problems
P16 [Combinatorics] - Turkish NMO 1st Round - 2013
16
16
16
white and
4
4
4
red balls that are not identical are distributed randomly into
4
4
4
boxes which contain at most
5
5
5
balls. What is the probability that each box contains exactly
1
1
1
red ball?
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64
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(
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4
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4
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32
<span class='latex-bold'>(A)</span>\ \dfrac{5}{64} \qquad<span class='latex-bold'>(B)</span>\ \dfrac{1}{8} \qquad<span class='latex-bold'>(C)</span>\ \dfrac{4^4}{\binom{16}{4}} \qquad<span class='latex-bold'>(D)</span>\ \dfrac{5^4}{\binom{20}{4}} \qquad<span class='latex-bold'>(E)</span>\ \dfrac{3}{32}
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32
3
8
1
Hide problems
P08 [Combinatorics] - Turkish NMO 1st Round - 2013
How many kites are there such that all of its four vertices are vertices of a given regular icosagon (
20
20
20
-gon)?
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105
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
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−
b
o
l
d
′
>
(
B
)
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/
s
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a
n
>
100
<
s
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c
l
a
s
s
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′
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t
e
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−
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o
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>
(
C
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>
95
<
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s
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90
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(
E
)
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85
<span class='latex-bold'>(A)</span>\ 105 \qquad<span class='latex-bold'>(B)</span>\ 100 \qquad<span class='latex-bold'>(C)</span>\ 95 \qquad<span class='latex-bold'>(D)</span>\ 90 \qquad<span class='latex-bold'>(E)</span>\ 85
<
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(
A
)
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105
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t
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d
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(
B
)
<
/
s
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100
<
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an
c
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ss
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t
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−
b
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(
C
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>
95
<
s
p
an
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t
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−
b
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(
D
)
<
/
s
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an
>
90
<
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−
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o
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d
′
>
(
E
)
<
/
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>
85
31
1
Hide problems
P31 [Algebra] - Turkish NMO 1st Round - 2013
Let
(
a
n
)
n
=
1
∞
(a_n)_{n=1}^\infty
(
a
n
)
n
=
1
∞
be a real sequence such that
a
n
=
(
n
−
1
)
a
1
+
(
n
−
2
)
a
2
+
⋯
+
2
a
n
−
2
+
a
n
−
1
a_n = (n-1)a_1 + (n-2)a_2 + \dots + 2a_{n-2} + a_{n-1}
a
n
=
(
n
−
1
)
a
1
+
(
n
−
2
)
a
2
+
⋯
+
2
a
n
−
2
+
a
n
−
1
for every
n
≥
3
n\geq 3
n
≥
3
. If
a
2011
=
2011
a_{2011} = 2011
a
2011
=
2011
and
a
2012
=
2012
a_{2012} = 2012
a
2012
=
2012
, what is
a
2013
a_{2013}
a
2013
?
<
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s
s
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(
A
)
<
/
s
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a
n
>
6025
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
5555
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
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a
n
>
4025
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
3456
<
s
p
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n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
2013
<span class='latex-bold'>(A)</span>\ 6025 \qquad<span class='latex-bold'>(B)</span>\ 5555 \qquad<span class='latex-bold'>(C)</span>\ 4025 \qquad<span class='latex-bold'>(D)</span>\ 3456 \qquad<span class='latex-bold'>(E)</span>\ 2013
<
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p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
6025
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
5555
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
4025
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3456
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
2013
27
1
Hide problems
P27 [Algebra] - Turkish NMO 1st Round - 2013
For how many pairs
(
a
,
b
)
(a,b)
(
a
,
b
)
from
(
1
,
2
)
(1,2)
(
1
,
2
)
,
(
3
,
5
)
(3,5)
(
3
,
5
)
,
(
5
,
7
)
(5,7)
(
5
,
7
)
,
(
7
,
11
)
(7,11)
(
7
,
11
)
, the polynomial
P
(
x
)
=
x
5
+
a
x
4
+
b
x
3
+
b
x
2
+
a
x
+
1
P(x)=x^5+ax^4+bx^3+bx^2+ax+1
P
(
x
)
=
x
5
+
a
x
4
+
b
x
3
+
b
x
2
+
a
x
+
1
has exactly one real root?
<
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A
)
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4
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s
s
=
′
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>
(
B
)
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/
s
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>
3
<
s
p
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l
a
s
s
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−
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o
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C
)
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>
2
<
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a
s
s
=
′
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>
(
D
)
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>
1
<
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>
(
E
)
<
/
s
p
a
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>
0
<span class='latex-bold'>(A)</span>\ 4 \qquad<span class='latex-bold'>(B)</span>\ 3 \qquad<span class='latex-bold'>(C)</span>\ 2 \qquad<span class='latex-bold'>(D)</span>\ 1 \qquad<span class='latex-bold'>(E)</span>\ 0
<
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p
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−
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o
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(
A
)
<
/
s
p
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>
4
<
s
p
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a
ss
=
′
l
a
t
e
x
−
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o
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d
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>
(
B
)
<
/
s
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>
3
<
s
p
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c
l
a
ss
=
′
l
a
t
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−
b
o
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d
′
>
(
C
)
<
/
s
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an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
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x
−
b
o
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′
>
(
D
)
<
/
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>
1
<
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−
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′
>
(
E
)
<
/
s
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an
>
0
23
1
Hide problems
P23 [Algebra] - Turkish NMO 1st Round - 2013
If the conditions
f
(
2
x
+
1
)
+
g
(
3
−
x
)
=
x
f
(
(
3
x
+
5
)
/
(
x
+
1
)
)
+
2
g
(
(
2
x
+
1
)
/
(
x
+
1
)
)
=
x
/
(
x
+
1
)
\begin{array}{rcl} f(2x+1)+g(3-x) &=& x \\ f((3x+5)/(x+1))+2g((2x+1)/(x+1)) &=& x/(x+1) \end{array}
f
(
2
x
+
1
)
+
g
(
3
−
x
)
f
((
3
x
+
5
)
/
(
x
+
1
))
+
2
g
((
2
x
+
1
)
/
(
x
+
1
))
=
=
x
x
/
(
x
+
1
)
hold for all real numbers
x
≠
1
x\neq 1
x
=
1
, what is
f
(
2013
)
f(2013)
f
(
2013
)
?
<
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s
s
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o
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(
A
)
<
/
s
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>
1007
<
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c
l
a
s
s
=
′
l
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t
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x
−
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o
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>
(
B
)
<
/
s
p
a
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>
4021
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
C
)
<
/
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a
n
>
6037
7
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
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a
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>
4029
5
<
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s
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−
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o
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(
E
)
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a
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>
None of above
<span class='latex-bold'>(A)</span>\ 1007 \qquad<span class='latex-bold'>(B)</span>\ \dfrac {4021}{3} \qquad<span class='latex-bold'>(C)</span>\ \dfrac {6037}7 \qquad<span class='latex-bold'>(D)</span>\ \dfrac {4029}{5} \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
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p
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a
ss
=
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l
a
t
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x
−
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o
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d
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>
(
A
)
<
/
s
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an
>
1007
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
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o
l
d
′
>
(
B
)
<
/
s
p
an
>
3
4021
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
7
6037
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
5
4029
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
19
1
Hide problems
P19 [Algebra] - Turkish NMO 1st Round - 2013
What is the minimum value of
x
2
−
4
x
+
7
−
2
2
+
x
2
−
8
x
+
27
−
6
2
\sqrt {x^2 - 4x + 7 - 2\sqrt 2} + \sqrt {x^2 - 8x + 27 - 6\sqrt 2}
x
2
−
4
x
+
7
−
2
2
+
x
2
−
8
x
+
27
−
6
2
where
x
x
x
is a real number?
<
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s
s
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−
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o
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>
(
A
)
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a
n
>
2
<
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p
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c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
B
)
<
/
s
p
a
n
>
3
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
1
+
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
2
2
<
s
p
a
n
c
l
a
s
s
=
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a
t
e
x
−
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o
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′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 2 \qquad<span class='latex-bold'>(B)</span>\ 3\sqrt 2 \qquad<span class='latex-bold'>(C)</span>\ 1 + \sqrt 2 \qquad<span class='latex-bold'>(D)</span>\ 2\sqrt 2 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
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(
A
)
<
/
s
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an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
3
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
1
+
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
20
1
Hide problems
Two-pan balance showing the difference
The numbers
1
,
2
,
…
,
2013
1,2,\dots, 2013
1
,
2
,
…
,
2013
are written on
2013
2013
2013
stones weighing
1
,
2
,
…
,
2013
1,2,\dots, 2013
1
,
2
,
…
,
2013
grams such that each number is used exactly once. We have a two-pan balance that shows the difference between the weights at the left and the right pans. No matter how the numbers are written, if it is possible to determine in
k
k
k
weighings whether the weight of each stone is equal to the number that is written on the stone, what is the least possible value of
k
k
k
?
<
s
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n
c
l
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s
s
=
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a
t
e
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−
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o
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d
′
>
(
A
)
<
/
s
p
a
n
>
15
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
12
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
10
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
7
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 15 \qquad<span class='latex-bold'>(B)</span>\ 12 \qquad<span class='latex-bold'>(C)</span>\ 10 \qquad<span class='latex-bold'>(D)</span>\ 7 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
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c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
15
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
12
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
10
<
s
p
an
c
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7
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None of above
4
1
Hide problems
1,2,..,49 written into chessboard
The numbers
1
,
2
,
…
,
49
1,2,\dots, 49
1
,
2
,
…
,
49
are written on unit squares of a
7
×
7
7\times 7
7
×
7
chessboard such that consequtive numbers are on unit squares sharing a common edge. At most how many prime numbers can a row have?
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A
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7
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(
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6
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3
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3
<span class='latex-bold'>(A)</span>\ 7 \qquad<span class='latex-bold'>(B)</span>\ 6 \qquad<span class='latex-bold'>(C)</span>\ 5 \qquad<span class='latex-bold'>(D)</span>\ 3 \qquad<span class='latex-bold'>(E)</span>\ 3
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7
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3
35
1
Hide problems
f(x)=x+1+[x]
What is the least positive integer
n
n
n
such that
f
(
f
(
…
f
⏞
21
times
(
n
)
)
)
=
2013
\overbrace{f(f(\dots f}^{21 \text{ times}}(n)))=2013
f
(
f
(
…
f
21
times
(
n
)))
=
2013
where
f
(
x
)
=
x
+
1
+
⌊
x
⌋
f(x)=x+1+\lfloor \sqrt x \rfloor
f
(
x
)
=
x
+
1
+
⌊
x
⌋
? (
⌊
a
⌋
\lfloor a \rfloor
⌊
a
⌋
denotes the greatest integer not exceeding the real number
a
a
a
.)
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1214
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1202
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1186
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1178
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None of above
<span class='latex-bold'>(A)</span>\ 1214 \qquad<span class='latex-bold'>(B)</span>\ 1202 \qquad<span class='latex-bold'>(C)</span>\ 1186 \qquad<span class='latex-bold'>(D)</span>\ 1178 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
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1186
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1178
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None of above
15
1
Hide problems
Scalene Polygon with sides 1<= S_i <= 2013
No matter how
n
n
n
real numbers on the interval
[
1
,
2013
]
[1,2013]
[
1
,
2013
]
are selected, if it is possible to find a scalene polygon such that its sides are equal to some of the numbers selected, what is the least possible value of
n
n
n
?
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14
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13
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10
<span class='latex-bold'>(A)</span>\ 14 \qquad<span class='latex-bold'>(B)</span>\ 13 \qquad<span class='latex-bold'>(C)</span>\ 12 \qquad<span class='latex-bold'>(D)</span>\ 11 \qquad<span class='latex-bold'>(E)</span>\ 10
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13
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10
11
1
Hide problems
P11 [Algebra] - Turkish NMO 1st Round - 2013
How many pairs of real numbers
(
x
,
y
)
(x,y)
(
x
,
y
)
are there such that
x
4
+
y
4
+
2
x
2
y
+
2
x
y
2
+
2
=
x
2
+
y
2
+
2
x
+
2
y
x^4+y^4 + 2x^2y + 2xy^2+ 2 = x^2 + y^2 + 2x + 2y
x
4
+
y
4
+
2
x
2
y
+
2
x
y
2
+
2
=
x
2
+
y
2
+
2
x
+
2
y
?
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6
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4
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2
<span class='latex-bold'>(A)</span>\ 6 \qquad<span class='latex-bold'>(B)</span>\ 5 \qquad<span class='latex-bold'>(C)</span>\ 4 \qquad<span class='latex-bold'>(D)</span>\ 3 \qquad<span class='latex-bold'>(E)</span>\ 2
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4
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(
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3
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(
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2
7
1
Hide problems
P07 [Algebra] - Turkish NMO 1st Round - 2013
What is the sum of real roots of the equation
x
4
−
8
x
3
+
13
x
2
−
24
x
+
9
=
0
x^4-8x^3+13x^2 -24x + 9 = 0
x
4
−
8
x
3
+
13
x
2
−
24
x
+
9
=
0
?
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8
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7
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6
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E
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4
<span class='latex-bold'>(A)</span>\ 8 \qquad<span class='latex-bold'>(B)</span>\ 7 \qquad<span class='latex-bold'>(C)</span>\ 6 \qquad<span class='latex-bold'>(D)</span>\ 5 \qquad<span class='latex-bold'>(E)</span>\ 4
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8
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6
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E
)
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4
3
1
Hide problems
P03 [Algebra] - Turkish NMO 1st Round - 2013
If the remainder is
2013
2013
2013
when a polynomial with coefficients from the set
{
0
,
1
,
2
,
3
,
4
,
5
}
\{0,1,2,3,4,5\}
{
0
,
1
,
2
,
3
,
4
,
5
}
is divided by
x
−
6
x-6
x
−
6
, what is the least possible value of the coefficient of
x
x
x
in this polynomial?
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1
<span class='latex-bold'>(A)</span>\ 5 \qquad<span class='latex-bold'>(B)</span>\ 4 \qquad<span class='latex-bold'>(C)</span>\ 3 \qquad<span class='latex-bold'>(D)</span>\ 2 \qquad<span class='latex-bold'>(E)</span>\ 1
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(
B
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(
E
)
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1
34
1
Hide problems
P34 [Number Theory] - Turkish NMO 1st Round - 2013
How many triples of positive integers
(
a
,
b
,
c
)
(a,b,c)
(
a
,
b
,
c
)
are there such that
a
!
+
b
3
=
18
+
c
3
a!+b^3 = 18+c^3
a
!
+
b
3
=
18
+
c
3
?
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a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
0
<span class='latex-bold'>(A)</span>\ 4 \qquad<span class='latex-bold'>(B)</span>\ 3 \qquad<span class='latex-bold'>(C)</span>\ 2 \qquad<span class='latex-bold'>(D)</span>\ 1 \qquad<span class='latex-bold'>(E)</span>\ 0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
0
30
1
Hide problems
P30 [Number Theory] - Turkish NMO 1st Round - 2013
For how many postive integers
n
n
n
less than
2013
2013
2013
, does
p
2
+
p
+
1
p^2+p+1
p
2
+
p
+
1
divide
n
n
n
where
p
p
p
is the least prime divisor of
n
n
n
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
212
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
206
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
191
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
185
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
173
<span class='latex-bold'>(A)</span>\ 212 \qquad<span class='latex-bold'>(B)</span>\ 206 \qquad<span class='latex-bold'>(C)</span>\ 191 \qquad<span class='latex-bold'>(D)</span>\ 185 \qquad<span class='latex-bold'>(E)</span>\ 173
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
212
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
206
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
191
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
185
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
173
26
1
Hide problems
P26 [Number Theory] - Turkish NMO 1st Round - 2013
What is the maximum number of primes that divide both the numbers
n
3
+
2
n^3+2
n
3
+
2
and
(
n
+
1
)
3
+
2
(n+1)^3+2
(
n
+
1
)
3
+
2
where
n
n
n
is a positive integer?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 3 \qquad<span class='latex-bold'>(B)</span>\ 2 \qquad<span class='latex-bold'>(C)</span>\ 1 \qquad<span class='latex-bold'>(D)</span>\ 0 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
22
1
Hide problems
P22 [Number Theory] - Turkish NMO 1st Round - 2013
For how many integers
0
≤
n
<
2013
0\leq n < 2013
0
≤
n
<
2013
, is
n
4
+
2
n
3
−
20
n
2
+
2
n
−
21
n^4+2n^3-20n^2+2n-21
n
4
+
2
n
3
−
20
n
2
+
2
n
−
21
divisible by
2013
2013
2013
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
8
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
12
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
16
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 6 \qquad<span class='latex-bold'>(B)</span>\ 8 \qquad<span class='latex-bold'>(C)</span>\ 12 \qquad<span class='latex-bold'>(D)</span>\ 16 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
8
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
12
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
16
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
18
1
Hide problems
P18 [Number Theory] - Turkish NMO 1st Round - 2013
What is remainder when the sum
(
2013
1
)
+
2013
(
2013
3
)
+
201
3
2
(
2013
5
)
+
⋯
+
201
3
1006
(
2013
2013
)
\binom{2013}{1}+2013\binom{2013}{3} + 2013^2\binom{2013}{5} + \dots + 2013^{1006}\binom{2013}{2013}
(
1
2013
)
+
2013
(
3
2013
)
+
201
3
2
(
5
2013
)
+
⋯
+
201
3
1006
(
2013
2013
)
is divided by
41
41
41
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
20
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
14
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
7
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None
<span class='latex-bold'>(A)</span>\ 20 \qquad<span class='latex-bold'>(B)</span>\ 14 \qquad<span class='latex-bold'>(C)</span>\ 7 \qquad<span class='latex-bold'>(D)</span>\ 1 \qquad<span class='latex-bold'>(E)</span>\ \text{None}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
20
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
14
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
7
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
14
1
Hide problems
P14 [Number Theory] - Turkish NMO 1st Round - 2013
Let
d
(
n
)
d(n)
d
(
n
)
be the number of positive integers that divide the integer
n
n
n
. For all positive integral divisors
k
k
k
of
64800
64800
64800
, what is the sum of numbers
d
(
k
)
d(k)
d
(
k
)
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1440
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1650
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
1890
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
2010
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 1440 \qquad<span class='latex-bold'>(B)</span>\ 1650 \qquad<span class='latex-bold'>(C)</span>\ 1890 \qquad<span class='latex-bold'>(D)</span>\ 2010 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
1440
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
1650
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
1890
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2010
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
10
1
Hide problems
P10 [Number Theory] - Turkish NMO 1st Round - 2013
How many positive integers
n
n
n
are there such that there are exactly
20
20
20
positive odd integers that are less than
n
n
n
and relatively prime with
n
n
n
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 5 \qquad<span class='latex-bold'>(B)</span>\ 4 \qquad<span class='latex-bold'>(C)</span>\ 3 \qquad<span class='latex-bold'>(D)</span>\ 2 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
6
1
Hide problems
P06 [Number Theory] - Turkish NMO 1st Round - 2013
What is the
11
1
st
111^{\text{st}}
11
1
st
smallest positive integer which does not have
3
3
3
and
4
4
4
in its base-
5
5
5
representation?
<
s
p
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n
c
l
a
s
s
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′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
760
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
756
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
755
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
752
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
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a
n
>
750
<span class='latex-bold'>(A)</span>\ 760 \qquad<span class='latex-bold'>(B)</span>\ 756 \qquad<span class='latex-bold'>(C)</span>\ 755 \qquad<span class='latex-bold'>(D)</span>\ 752 \qquad<span class='latex-bold'>(E)</span>\ 750
<
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a
t
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x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
760
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
756
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
755
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
752
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
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an
>
750
2
1
Hide problems
P02 [Number Theory] - Turkish NMO 1st Round - 2013
How many triples
(
p
,
q
,
n
)
(p,q,n)
(
p
,
q
,
n
)
are there such that
1
/
p
+
2013
/
q
=
n
/
5
1/p+2013/q = n/5
1/
p
+
2013/
q
=
n
/5
where
p
p
p
,
q
q
q
are prime numbers and
n
n
n
is a positive integer?
<
s
p
a
n
c
l
a
s
s
=
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a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
7
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
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a
n
>
4
<span class='latex-bold'>(A)</span>\ 7 \qquad<span class='latex-bold'>(B)</span>\ 6 \qquad<span class='latex-bold'>(C)</span>\ 5 \qquad<span class='latex-bold'>(D)</span>\ 4 \qquad<span class='latex-bold'>(E)</span>\ 4
<
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p
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c
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a
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=
′
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a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
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>
7
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
4
33
1
Hide problems
P33 [Geometry] - Turkish NMO 1st Round - 2013
Let
D
D
D
be a point on side
[
B
C
]
[BC]
[
BC
]
of triangle
A
B
C
ABC
A
BC
such that
[
A
D
]
[AD]
[
A
D
]
is an angle bisector,
∣
B
D
∣
=
4
|BD|=4
∣
B
D
∣
=
4
, and
∣
D
C
∣
=
3
|DC|=3
∣
D
C
∣
=
3
. Let
E
E
E
be a point on side
[
A
B
]
[AB]
[
A
B
]
and different than
A
A
A
such that
m
(
B
E
D
^
)
=
m
(
D
E
C
^
)
m(\widehat{BED})=m(\widehat{DEC})
m
(
BE
D
)
=
m
(
D
EC
)
. If the perpendicular bisector of segment
[
A
E
]
[AE]
[
A
E
]
meets the line
B
C
BC
BC
at
M
M
M
, what is
∣
C
M
∣
|CM|
∣
CM
∣
?
<
s
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a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
12
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
9
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
7
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 12 \qquad<span class='latex-bold'>(B)</span>\ 9 \qquad<span class='latex-bold'>(C)</span>\ 7 \qquad<span class='latex-bold'>(D)</span>\ 5 \qquad<span class='latex-bold'>(E)</span>\ \text { None of above}
<
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p
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c
l
a
ss
=
′
l
a
t
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x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
12
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
9
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
7
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
29
1
Hide problems
P29 [Geometry] - Turkish NMO 1st Round - 2013
Let
O
O
O
be the circumcenter of triangle
A
B
C
ABC
A
BC
with
∣
A
B
∣
=
5
|AB|=5
∣
A
B
∣
=
5
,
∣
B
C
∣
=
6
|BC|=6
∣
BC
∣
=
6
,
∣
A
C
∣
=
7
|AC|=7
∣
A
C
∣
=
7
. Let
A
1
A_1
A
1
,
B
1
B_1
B
1
,
C
1
C_1
C
1
be the reflections of
O
O
O
over the lines
B
C
BC
BC
,
A
C
AC
A
C
,
A
B
AB
A
B
, respectively. What is the distance between
A
A
A
and the circumcenter of triangle
A
1
B
1
C
1
A_1B_1C_1
A
1
B
1
C
1
?
<
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c
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s
s
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l
a
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x
−
b
o
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d
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>
(
A
)
<
/
s
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a
n
>
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
29
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
19
2
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
35
4
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
35
3
<span class='latex-bold'>(A)</span>\ 6 \qquad<span class='latex-bold'>(B)</span>\ \sqrt {29} \qquad<span class='latex-bold'>(C)</span>\ \dfrac {19}{2\sqrt 6} \qquad<span class='latex-bold'>(D)</span>\ \dfrac {35}{4\sqrt 6} \qquad<span class='latex-bold'>(E)</span>\ \sqrt {\dfrac {35}3}
<
s
p
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c
l
a
ss
=
′
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a
t
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x
−
b
o
l
d
′
>
(
A
)
<
/
s
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an
>
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
29
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
6
19
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
4
6
35
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
3
35
25
1
Hide problems
P25 [Geometry] - Turkish NMO 1st Round - 2013
Let
D
D
D
be a point on side
[
A
B
]
[AB]
[
A
B
]
of triangle
A
B
C
ABC
A
BC
with
∣
A
B
∣
=
∣
A
C
∣
|AB|=|AC|
∣
A
B
∣
=
∣
A
C
∣
such that
[
C
D
]
[CD]
[
C
D
]
is an angle bisector and
m
(
A
B
C
^
)
=
4
0
∘
m(\widehat{ABC})=40^\circ
m
(
A
BC
)
=
4
0
∘
. Let
F
F
F
be a point on the extension of
[
A
B
]
[AB]
[
A
B
]
after
B
B
B
such that
∣
B
C
∣
=
∣
A
F
∣
|BC|=|AF|
∣
BC
∣
=
∣
A
F
∣
. Let
E
E
E
be the midpoint of
[
C
F
]
[CF]
[
CF
]
. If
G
G
G
is the intersection of lines
E
D
ED
E
D
and
A
C
AC
A
C
, what is
m
(
F
B
G
^
)
m(\widehat{FBG})
m
(
FBG
)
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
15
0
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
13
5
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
12
0
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
10
5
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 150^\circ \qquad<span class='latex-bold'>(B)</span>\ 135^\circ \qquad<span class='latex-bold'>(C)</span>\ 120^\circ \qquad<span class='latex-bold'>(D)</span>\ 105^\circ \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
15
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
13
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
12
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
10
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
21
1
Hide problems
P21 [Geometry] - Turkish NMO 1st Round - 2013
Let
D
D
D
and
E
E
E
be points on side
[
A
B
]
[AB]
[
A
B
]
of a right triangle with
m
(
C
^
)
=
9
0
∘
m(\widehat{C})=90^\circ
m
(
C
)
=
9
0
∘
such that
∣
A
D
∣
=
∣
A
C
∣
|AD|=|AC|
∣
A
D
∣
=
∣
A
C
∣
and
∣
B
E
∣
=
∣
B
C
∣
|BE|=|BC|
∣
BE
∣
=
∣
BC
∣
. Let
F
F
F
be the second intersection point of the circumcircles of triangles
A
E
C
AEC
A
EC
and
B
D
C
BDC
B
D
C
. If
∣
C
F
∣
=
2
|CF|=2
∣
CF
∣
=
2
, what is
∣
E
D
∣
|ED|
∣
E
D
∣
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
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p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
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/
s
p
a
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>
1
+
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
2
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
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x
−
b
o
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d
′
>
(
E
)
<
/
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a
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>
None of above
<span class='latex-bold'>(A)</span>\ \sqrt 2 \qquad<span class='latex-bold'>(B)</span>\ 1+\sqrt 2 \qquad<span class='latex-bold'>(C)</span>\ 2 \qquad<span class='latex-bold'>(D)</span>\ 2\sqrt 2 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
1
+
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
17
1
Hide problems
P17 [Geometry] - Turkish NMO 1st Round - 2013
Let
A
B
C
ABC
A
BC
be an equilateral triangle with side length
10
10
10
and
P
P
P
be a point inside the triangle such that
∣
P
A
∣
2
+
∣
P
B
∣
2
+
∣
P
C
∣
2
=
128
|PA|^2+ |PB|^2 + |PC|^2 = 128
∣
P
A
∣
2
+
∣
PB
∣
2
+
∣
PC
∣
2
=
128
. What is the area of a triangle with side lengths
∣
P
A
∣
,
∣
P
B
∣
,
∣
P
C
∣
|PA|,|PB|,|PC|
∣
P
A
∣
,
∣
PB
∣
,
∣
PC
∣
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
6
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
7
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
8
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
9
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
10
3
<span class='latex-bold'>(A)</span>\ 6\sqrt 3 \qquad<span class='latex-bold'>(B)</span>\ 7 \sqrt 3 \qquad<span class='latex-bold'>(C)</span>\ 8 \sqrt 3 \qquad<span class='latex-bold'>(D)</span>\ 9 \sqrt 3 \qquad<span class='latex-bold'>(E)</span>\ 10 \sqrt 3
<
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p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
6
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
7
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
8
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
9
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
10
3
13
1
Hide problems
P13 [Geometry] - Turkish NMO 1st Round - 2013
Let
D
D
D
and
E
E
E
be points on side
[
B
C
]
[BC]
[
BC
]
of a triangle
A
B
C
ABC
A
BC
with circumcenter
O
O
O
such that
D
D
D
is between
B
B
B
and
E
E
E
,
∣
A
D
∣
=
∣
D
B
∣
=
6
|AD|=|DB|=6
∣
A
D
∣
=
∣
D
B
∣
=
6
, and
∣
A
E
∣
=
∣
E
C
∣
=
8
|AE|=|EC|=8
∣
A
E
∣
=
∣
EC
∣
=
8
. If
I
I
I
is the incenter of triangle
A
D
E
ADE
A
D
E
and
∣
A
I
∣
=
5
|AI|=5
∣
A
I
∣
=
5
, then what is
∣
I
O
∣
|IO|
∣
I
O
∣
?
<
s
p
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n
c
l
a
s
s
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′
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a
t
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x
−
b
o
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d
′
>
(
A
)
<
/
s
p
a
n
>
29
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
23
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
21
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
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a
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>
None of above
<span class='latex-bold'>(A)</span>\ \dfrac {29}{5} \qquad<span class='latex-bold'>(B)</span>\ 5 \qquad<span class='latex-bold'>(C)</span>\ \dfrac {23}{5} \qquad<span class='latex-bold'>(D)</span>\ \dfrac {21}{5} \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
5
29
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
5
23
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
5
21
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
9
1
Hide problems
P09 [Geometry] - Turkish NMO 1st Round - 2013
Let
A
B
C
ABC
A
BC
be a triangle with
∣
A
B
∣
=
18
|AB|=18
∣
A
B
∣
=
18
,
∣
A
C
∣
=
24
|AC|=24
∣
A
C
∣
=
24
, and
m
(
B
A
C
^
)
=
15
0
∘
m(\widehat{BAC}) = 150^\circ
m
(
B
A
C
)
=
15
0
∘
. Let
D
D
D
,
E
E
E
,
F
F
F
be points on sides
[
A
B
]
[AB]
[
A
B
]
,
[
A
C
]
[AC]
[
A
C
]
,
[
B
C
]
[BC]
[
BC
]
, respectively, such that
∣
B
D
∣
=
6
|BD|=6
∣
B
D
∣
=
6
,
∣
C
E
∣
=
8
|CE|=8
∣
CE
∣
=
8
, and
∣
C
F
∣
=
2
∣
B
F
∣
|CF|=2|BF|
∣
CF
∣
=
2∣
BF
∣
. Let
H
1
H_1
H
1
,
H
2
H_2
H
2
,
H
3
H_3
H
3
be the reflections of the orthocenter of triangle
A
B
C
ABC
A
BC
over the points
D
D
D
,
E
E
E
,
F
F
F
, respectively. What is the area of triangle
H
1
H
2
H
3
H_1H_2H_3
H
1
H
2
H
3
?
<
s
p
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n
c
l
a
s
s
=
′
l
a
t
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x
−
b
o
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d
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>
(
A
)
<
/
s
p
a
n
>
70
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
72
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
84
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
96
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
108
<span class='latex-bold'>(A)</span>\ 70 \qquad<span class='latex-bold'>(B)</span>\ 72 \qquad<span class='latex-bold'>(C)</span>\ 84 \qquad<span class='latex-bold'>(D)</span>\ 96 \qquad<span class='latex-bold'>(E)</span>\ 108
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
70
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
72
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
84
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
96
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
108
5
1
Hide problems
P05 [Geometry] - Turkish NMO 1st Round - 2013
Let
D
D
D
be a point on side
[
B
C
]
[BC]
[
BC
]
of triangle
A
B
C
ABC
A
BC
where
∣
B
C
∣
=
11
|BC|=11
∣
BC
∣
=
11
and
∣
B
D
∣
=
8
|BD|=8
∣
B
D
∣
=
8
. The circle passing through the points
C
C
C
and
D
D
D
touches
A
B
AB
A
B
at
E
E
E
. Let
P
P
P
be a point on the line which is passing through
B
B
B
and is perpendicular to
D
E
DE
D
E
. If
∣
P
E
∣
=
7
|PE|=7
∣
PE
∣
=
7
, then what is
∣
D
P
∣
|DP|
∣
D
P
∣
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 5 \qquad<span class='latex-bold'>(B)</span>\ 4 \qquad<span class='latex-bold'>(C)</span>\ 3 \qquad<span class='latex-bold'>(D)</span>\ 2 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
1
1
Hide problems
P01 [Geometry] - Turkish NMO 1st Round - 2013
Let
A
B
C
ABC
A
BC
be a triangle with incenter
I
I
I
, centroid
G
G
G
, and
∣
A
C
∣
>
∣
A
B
∣
|AC|>|AB|
∣
A
C
∣
>
∣
A
B
∣
. If
I
G
∥
B
C
IG\parallel BC
I
G
∥
BC
,
∣
B
C
∣
=
2
|BC|=2
∣
BC
∣
=
2
, and
A
r
e
a
(
A
B
C
)
=
3
5
/
8
Area(ABC)=3\sqrt 5 / 8
A
re
a
(
A
BC
)
=
3
5
/8
, then what is
∣
A
B
∣
|AB|
∣
A
B
∣
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
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9
8
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11
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13
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15
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17
8
<span class='latex-bold'>(A)</span>\ \dfrac 98 \qquad<span class='latex-bold'>(B)</span>\ \dfrac {11}8 \qquad<span class='latex-bold'>(C)</span>\ \dfrac {13}8 \qquad<span class='latex-bold'>(D)</span>\ \dfrac {15}8 \qquad<span class='latex-bold'>(E)</span>\ \dfrac {17}8
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11
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15
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8
17