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National and Regional Contests
Turkey Contests
National Olympiad First Round
2012 National Olympiad First Round
2012 National Olympiad First Round
Part of
National Olympiad First Round
Subcontests
(36)
36
1
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Turkish NMO First Round - 2012 Problem - 36 {Combinatorics}
k
k
k
stones are put into
2012
2012
2012
boxes in such a way that each box has at most
20
20
20
stones. We are chosing some of the boxes. We are throwing some of the stones of the chosen boxes. Whatever the first arrangement of the stones inside the boxes is, if we can guarantee that there are equal stones inside the chosen boxes and the sum of them is at least
100
100
100
, then
k
k
k
can be at least?
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349
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296
<span class='latex-bold'>(A)</span>\ 500 \qquad <span class='latex-bold'>(B)</span>\ 450 \qquad <span class='latex-bold'>(C)</span>\ 420 \qquad <span class='latex-bold'>(D)</span>\ 349 \qquad <span class='latex-bold'>(E)</span>\ 296
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349
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296
32
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Turkish NMO First Round - 2012 Problem - 32 {Combinatorics}
How many permutations
(
a
1
,
a
2
,
…
,
a
10
)
(a_1,a_2,\dots,a_{10})
(
a
1
,
a
2
,
…
,
a
10
)
of
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
1,2,3,4,5,6,7,8,9,10
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
satisfy
∣
a
1
−
1
∣
+
∣
a
2
−
2
∣
+
⋯
+
∣
a
10
−
10
∣
=
4
|a_1-1|+|a_2-2|+\dots+|a_{10}-10|=4
∣
a
1
−
1∣
+
∣
a
2
−
2∣
+
⋯
+
∣
a
10
−
10∣
=
4
?
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60
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36
<span class='latex-bold'>(A)</span>\ 60 \qquad <span class='latex-bold'>(B)</span>\ 52 \qquad <span class='latex-bold'>(C)</span>\ 50 \qquad <span class='latex-bold'>(D)</span>\ 44 \qquad <span class='latex-bold'>(E)</span>\ 36
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36
28
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Turkish NMO First Round - 2012 Problem - 28 {Combinatorics}
At the beginning, three boxes contain
m
m
m
,
n
n
n
, and
k
k
k
pieces, respectively. Ayşe and Burak are playing a turn-based game with these pieces. At each turn, the player takes at least one piece from one of the boxes. The player who takes the last piece will win the game. Ayşe will be the first player. They are playing the game once for each
(
m
,
n
,
k
)
=
(
1
,
2012
,
2014
)
(m,n,k)=(1,2012,2014)
(
m
,
n
,
k
)
=
(
1
,
2012
,
2014
)
,
(
2011
,
2011
,
2012
)
(2011,2011,2012)
(
2011
,
2011
,
2012
)
,
(
2011
,
2012
,
2013
)
(2011,2012,2013)
(
2011
,
2012
,
2013
)
,
(
2011
,
2012
,
2014
)
(2011,2012,2014)
(
2011
,
2012
,
2014
)
,
(
2011
,
2013
,
2013
)
(2011,2013,2013)
(
2011
,
2013
,
2013
)
. In how many of them can Ayşe guarantee to win the game?
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<span class='latex-bold'>(A)</span>\ 1 \qquad <span class='latex-bold'>(B)</span>\ 2 \qquad <span class='latex-bold'>(C)</span>\ 3 \qquad <span class='latex-bold'>(D)</span>\ 4 \qquad <span class='latex-bold'>(E)</span>\ 5
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5
24
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Turkish NMO First Round - 2012 Problem - 24 {Combinatorics}
There are
2012
2012
2012
backgammon checkers (stones, pieces) with one side is black and the other side is white. These checkers are arranged into a line such that no two consequtive checkers are in same color. At each move, we are chosing two checkers. And we are turning upside down of the two checkers and all of the checkers between the two. At least how many moves are required to make all checkers same color?
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1006
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2011
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None
<span class='latex-bold'>(A)</span>\ 1006 \qquad <span class='latex-bold'>(B)</span>\ 1204 \qquad <span class='latex-bold'>(C)</span>\ 1340 \qquad <span class='latex-bold'>(D)</span>\ 2011 \qquad <span class='latex-bold'>(E)</span>\ \text{None}
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2011
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None
20
1
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Turkish NMO First Round - 2012 Problem - 20 {Combinatorics}
For each permutation
(
a
1
,
a
2
,
…
,
a
11
)
(a_1,a_2,\dots,a_{11})
(
a
1
,
a
2
,
…
,
a
11
)
of the numbers
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
,
11
1,2,3,4,5,6,7,8,9,10,11
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
,
11
, we can determine at least
k
k
k
of
a
i
a_i
a
i
s when we get
(
a
1
+
a
3
,
a
2
+
a
4
,
a
3
+
a
5
,
…
,
a
8
+
a
10
,
a
9
+
a
11
)
(a_1+a_3, a_2+a_4,a_3+a_5,\dots,a_8+a_{10},a_9+a_{11})
(
a
1
+
a
3
,
a
2
+
a
4
,
a
3
+
a
5
,
…
,
a
8
+
a
10
,
a
9
+
a
11
)
.
k
k
k
can be at most ?
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None
<span class='latex-bold'>(A)</span>\ 11 \qquad <span class='latex-bold'>(B)</span>\ 6 \qquad <span class='latex-bold'>(C)</span>\ 5 \qquad <span class='latex-bold'>(D)</span>\ 2 \qquad <span class='latex-bold'>(E)</span>\ \text{None}
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16
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Turkish NMO First Round - 2012 Problem - 16 {Combinatorics}
Every cell of
8
×
8
8\times8
8
×
8
chessboard contains either
1
1
1
or
−
1
-1
−
1
. It is known that there are at least four rows such that the sum of numbers inside the cells of those rows is positive. At most how many columns are there such that the sum of numbers inside the cells of those columns is less than
−
3
-3
−
3
?
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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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(
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
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
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s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
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an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
2
12
1
Hide problems
Turkish NMO First Round - 2012 Problem - 12 {Combinatorics}
How many subsets of the set
{
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
}
\{1,2,3,4,5,6,7,8,9,10\}
{
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
}
are there that does not contain 4 consequtive integers?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
596
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
648
<
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
>
679
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
773
<
s
p
a
n
c
l
a
s
s
=
′
l
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t
e
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−
b
o
l
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>
(
E
)
<
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a
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>
812
<span class='latex-bold'>(A)</span>\ 596 \qquad <span class='latex-bold'>(B)</span>\ 648 \qquad <span class='latex-bold'>(C)</span>\ 679 \qquad <span class='latex-bold'>(D)</span>\ 773 \qquad <span class='latex-bold'>(E)</span>\ 812
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
596
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
648
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
679
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
773
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
812
8
1
Hide problems
Turkish NMO First Round - 2012 Problem - 08 {Combinatorics}
In how many different ways can one select two distinct subsets of the set
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
\{1,2,3,4,5,6,7\}
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
, so that one includes the other?
<
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c
l
a
s
s
=
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l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
2059
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2124
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2187
<
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p
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n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
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>
(
D
)
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/
s
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a
n
>
2315
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o
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(
E
)
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>
2316
<span class='latex-bold'>(A)</span>\ 2059 \qquad <span class='latex-bold'>(B)</span>\ 2124 \qquad <span class='latex-bold'>(C)</span>\ 2187 \qquad <span class='latex-bold'>(D)</span>\ 2315 \qquad <span class='latex-bold'>(E)</span>\ 2316
<
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c
l
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a
t
e
x
−
b
o
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d
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>
(
A
)
<
/
s
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an
>
2059
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2124
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
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d
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>
(
C
)
<
/
s
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an
>
2187
<
s
p
an
c
l
a
ss
=
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l
a
t
e
x
−
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d
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>
(
D
)
<
/
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an
>
2315
<
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l
a
ss
=
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x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
2316
4
1
Hide problems
Turkish NMO First Round - 2012 Problem - 04 {Combinatorics}
How many
f
:
A
→
A
f : A \rightarrow A
f
:
A
→
A
are there satisfying
f
(
f
(
a
)
)
=
a
f(f(a)) = a
f
(
f
(
a
))
=
a
for every
a
∈
A
=
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
a \in A=\{1,2,3,4,5,6,7\}
a
∈
A
=
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
?
<
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l
a
s
s
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l
a
t
e
x
−
b
o
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d
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>
(
A
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
106
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
127
<
s
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c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
232
<
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s
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(
E
)
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None
<span class='latex-bold'>(A)</span>\ 1 \qquad <span class='latex-bold'>(B)</span>\ 106 \qquad <span class='latex-bold'>(C)</span>\ 127 \qquad <span class='latex-bold'>(D)</span>\ 232 \qquad <span class='latex-bold'>(E)</span>\ \text{None}
<
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a
ss
=
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t
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x
−
b
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d
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>
(
A
)
<
/
s
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an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
106
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
C
)
<
/
s
p
an
>
127
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
D
)
<
/
s
p
an
>
232
<
s
p
an
c
l
a
ss
=
′
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a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
35
1
Hide problems
Turkish NMO First Round - 2012 Problem - 35 {Algebra}
For every positive real pair
(
x
,
y
)
(x,y)
(
x
,
y
)
satisfying the equation
x
3
+
y
4
=
x
2
y
x^3+y^4 = x^2y
x
3
+
y
4
=
x
2
y
, if the greatest value of
x
x
x
is
A
A
A
, and the greatest value of
y
y
y
is
B
B
B
, then
A
/
B
=
?
A/B = ?
A
/
B
=
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
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o
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d
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>
(
A
)
<
/
s
p
a
n
>
2
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
512
729
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
C
)
<
/
s
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a
n
>
729
1024
<
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c
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a
s
s
=
′
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a
t
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x
−
b
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>
(
D
)
<
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>
3
4
<
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E
)
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243
256
<span class='latex-bold'>(A)</span>\ \frac{2}{3} \qquad <span class='latex-bold'>(B)</span>\ \frac{512}{729} \qquad <span class='latex-bold'>(C)</span>\ \frac{729}{1024} \qquad <span class='latex-bold'>(D)</span>\ \frac{3}{4} \qquad <span class='latex-bold'>(E)</span>\ \frac{243}{256}
<
s
p
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c
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a
ss
=
′
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a
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x
−
b
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>
(
A
)
<
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>
3
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
729
512
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
1024
729
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
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an
>
4
3
<
s
p
an
c
l
a
ss
=
′
l
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t
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x
−
b
o
l
d
′
>
(
E
)
<
/
s
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an
>
256
243
31
1
Hide problems
Turkish NMO First Round - 2012 Problem - 31 {Algebra}
f
:
Z
→
Z
f : \mathbb{Z} \rightarrow \mathbb{Z}
f
:
Z
→
Z
satisfies
m
+
f
(
m
+
f
(
n
+
f
(
m
)
)
)
=
n
+
f
(
m
)
m+f(m+f(n+f(m))) = n + f(m)
m
+
f
(
m
+
f
(
n
+
f
(
m
)))
=
n
+
f
(
m
)
for every integers
m
,
n
m,n
m
,
n
. If
f
(
6
)
=
6
f(6) = 6
f
(
6
)
=
6
, then
f
(
2012
)
=
?
f(2012) = ?
f
(
2012
)
=
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
−
2010
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
−
2000
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2000
<
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
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x
−
b
o
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d
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>
(
E
)
<
/
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a
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>
2012
<span class='latex-bold'>(A)</span>\ -2010 \qquad <span class='latex-bold'>(B)</span>\ -2000 \qquad <span class='latex-bold'>(C)</span>\ 2000 \qquad <span class='latex-bold'>(D)</span>\ 2010 \qquad <span class='latex-bold'>(E)</span>\ 2012
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
−
2010
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
−
2000
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2000
<
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
>
2012
27
1
Hide problems
Turkish NMO First Round - 2012 Problem - 27 {Algebra}
What is the least real number
C
C
C
that satisfies
sin
x
cos
x
≤
C
(
sin
6
x
+
cos
6
x
)
\sin x \cos x \leq C(\sin^6x+\cos^6x)
sin
x
cos
x
≤
C
(
sin
6
x
+
cos
6
x
)
for every real number
x
x
x
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
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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
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a
n
>
2
2
<
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
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
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a
n
>
2
<
s
p
a
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c
l
a
s
s
=
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−
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o
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d
′
>
(
E
)
<
/
s
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a
n
>
None
<span class='latex-bold'>(A)</span>\ \sqrt3 \qquad <span class='latex-bold'>(B)</span>\ 2\sqrt2 \qquad <span class='latex-bold'>(C)</span>\ \sqrt 2 \qquad <span class='latex-bold'>(D)</span>\ 2 \qquad <span class='latex-bold'>(E)</span>\ \text{None}
<
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p
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c
l
a
ss
=
′
l
a
t
e
x
−
b
o
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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
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
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
23
1
Hide problems
Turkish NMO First Round - 2012 Problem - 23 {Algebra}
a
,
b
,
c
a,b,c
a
,
b
,
c
are distinct real roots of
x
3
−
3
x
+
1
=
0
x^3-3x+1=0
x
3
−
3
x
+
1
=
0
.
a
8
+
b
8
+
c
8
a^8+b^8+c^8
a
8
+
b
8
+
c
8
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
156
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
171
<
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
>
180
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
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a
n
>
186
<
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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>
201
<span class='latex-bold'>(A)</span>\ 156 \qquad <span class='latex-bold'>(B)</span>\ 171 \qquad <span class='latex-bold'>(C)</span>\ 180 \qquad <span class='latex-bold'>(D)</span>\ 186 \qquad <span class='latex-bold'>(E)</span>\ 201
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
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(
A
)
<
/
s
p
an
>
156
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
171
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
180
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
186
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
201
19
1
Hide problems
Turkish NMO First Round - 2012 Problem - 19 {Algebra}
What is the sum of real roots of the equation
x
4
−
7
x
3
+
14
x
2
−
14
x
+
4
=
0
x^4-7x^3+14x^2-14x+4=0
x
4
−
7
x
3
+
14
x
2
−
14
x
+
4
=
0
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
<
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
>
3
<
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
>
5
<span class='latex-bold'>(A)</span>\ 1 \qquad <span class='latex-bold'>(B)</span>\ 2 \qquad <span class='latex-bold'>(C)</span>\ 3 \qquad <span class='latex-bold'>(D)</span>\ 4 \qquad <span class='latex-bold'>(E)</span>\ 5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
A
)
<
/
s
p
an
>
1
<
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
>
3
<
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
>
5
15
1
Hide problems
Turkish NMO First Round - 2012 Problem - 15 {Algebra}
If
x
4
+
8
x
3
+
18
x
2
+
8
x
+
a
=
0
x^4+8x^3+18x^2+8x+a = 0
x
4
+
8
x
3
+
18
x
2
+
8
x
+
a
=
0
has four distinct real roots, then the real set of
a
a
a
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
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−
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o
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d
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>
(
A
)
<
/
s
p
a
n
>
(
−
9
,
2
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
(
−
9
,
0
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
[
−
9
,
0
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
[
−
8
,
1
)
<
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
>
(
−
8
,
1
)
<span class='latex-bold'>(A)</span>\ (-9,2) \qquad <span class='latex-bold'>(B)</span>\ (-9,0) \qquad <span class='latex-bold'>(C)</span>\ [-9,0) \qquad <span class='latex-bold'>(D)</span>\ [-8,1) \qquad <span class='latex-bold'>(E)</span>\ (-8,1)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
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d
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(
A
)
<
/
s
p
an
>
(
−
9
,
2
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
(
−
9
,
0
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
[
−
9
,
0
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
[
−
8
,
1
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
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an
>
(
−
8
,
1
)
11
1
Hide problems
Turkish NMO First Round - 2012 Problem - 11 {Algebra}
The number of real quadruples
(
x
,
y
,
z
,
w
)
(x,y,z,w)
(
x
,
y
,
z
,
w
)
satisfying
x
3
+
2
=
3
y
,
y
3
+
2
=
3
z
,
z
3
+
2
=
3
w
,
w
3
+
2
=
3
x
x^3+2=3y, y^3+2=3z, z^3+2=3w, w^3+2=3x
x
3
+
2
=
3
y
,
y
3
+
2
=
3
z
,
z
3
+
2
=
3
w
,
w
3
+
2
=
3
x
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
8
<
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
>
3
<
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>\ 8 \qquad <span class='latex-bold'>(B)</span>\ 5 \qquad <span class='latex-bold'>(C)</span>\ 3 \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
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an
>
8
<
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
>
3
<
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
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an
>
None
7
1
Hide problems
Turkish NMO First Round - 2012 Problem - 07 {Algebra}
How many
f
:
R
→
R
f:\mathbb{R} \rightarrow \mathbb{R}
f
:
R
→
R
are there satisfying
f
(
x
)
f
(
y
)
f
(
z
)
=
12
f
(
x
y
z
)
−
16
x
y
z
f(x)f(y)f(z)=12f(xyz)-16xyz
f
(
x
)
f
(
y
)
f
(
z
)
=
12
f
(
x
yz
)
−
16
x
yz
for every real
x
,
y
,
z
x,y,z
x
,
y
,
z
?
<
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
<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}
<
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
>
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
3
1
Hide problems
Turkish NMO First Round - 2012 Problem - 03 {Algebra}
Which one satisfies the equation
6
+
x
3
+
6
−
x
3
=
3
3
\sqrt[3]{6+\sqrt{x}} + \sqrt[3]{6-\sqrt{x}} = \sqrt[3]{3}
3
6
+
x
+
3
6
−
x
=
3
3
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
27
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
32
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
45
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
52
<
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
)
<
/
s
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a
n
>
63
<span class='latex-bold'>(A)</span>\ 27 \qquad <span class='latex-bold'>(B)</span>\ 32 \qquad <span class='latex-bold'>(C)</span>\ 45 \qquad <span class='latex-bold'>(D)</span>\ 52 \qquad <span class='latex-bold'>(E)</span>\ 63
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
27
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
32
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
45
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
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52
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63
34
1
Hide problems
Turkish NMO First Round - 2012 Problem - 34 {Number Theory}
If
10
10
10
divides the number
1
⋅
2
1
+
2
⋅
2
2
+
3
⋅
2
3
+
⋯
+
n
⋅
2
n
1\cdot2^1+2\cdot2^2+3\cdot2^3+\dots+n\cdot2^n
1
⋅
2
1
+
2
⋅
2
2
+
3
⋅
2
3
+
⋯
+
n
⋅
2
n
, what is the least integer
n
≥
2012
n\geq 2012
n
≥
2012
?
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A
)
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2012
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B
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2013
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a
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s
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−
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C
)
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2014
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a
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x
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(
D
)
<
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>
2015
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2016
<span class='latex-bold'>(A)</span>\ 2012 \qquad <span class='latex-bold'>(B)</span>\ 2013 \qquad <span class='latex-bold'>(C)</span>\ 2014 \qquad <span class='latex-bold'>(D)</span>\ 2015 \qquad <span class='latex-bold'>(E)</span>\ 2016
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A
)
<
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2012
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s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
B
)
<
/
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>
2013
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
C
)
<
/
s
p
an
>
2014
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
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an
>
2015
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
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an
>
2016
30
1
Hide problems
Turkish NMO First Round - 2012 Problem - 30 {Number Theory}
How many integer triples
(
x
,
y
,
z
)
(x,y,z)
(
x
,
y
,
z
)
are there satisfying
x
3
+
y
3
=
x
2
y
z
+
x
y
2
z
+
2
x^3+y^3=x^2yz+xy^2z+2
x
3
+
y
3
=
x
2
yz
+
x
y
2
z
+
2
?
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>
(
A
)
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>
5
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4
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s
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C
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3
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a
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D
)
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2
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E
)
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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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5
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(
B
)
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4
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3
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(
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)
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2
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(
E
)
<
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1
26
1
Hide problems
Turkish NMO First Round - 2012 Problem - 26 {Number Theory}
How many prime numbers less than
100
100
100
can be represented as sum of squares of consequtive positive integers?
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A
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3
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s
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=
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x
−
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>
(
B
)
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a
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>
4
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l
a
s
s
=
′
l
a
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−
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o
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C
)
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5
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−
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D
)
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/
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>
6
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E
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7
<span class='latex-bold'>(A)</span>\ 3 \qquad <span class='latex-bold'>(B)</span>\ 4 \qquad <span class='latex-bold'>(C)</span>\ 5 \qquad <span class='latex-bold'>(D)</span>\ 6 \qquad <span class='latex-bold'>(E)</span>\ 7
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A
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3
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(
B
)
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4
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(
C
)
<
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5
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a
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(
D
)
<
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6
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(
E
)
<
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7
22
1
Hide problems
Turkish NMO First Round - 2012 Problem - 22 {Number Theory}
How many integer pairs
(
m
,
n
)
(m,n)
(
m
,
n
)
are there satisfying
4
m
n
(
m
+
n
−
1
)
=
(
m
2
+
1
)
(
n
2
+
1
)
4mn(m+n-1)=(m^2+1)(n^2+1)
4
mn
(
m
+
n
−
1
)
=
(
m
2
+
1
)
(
n
2
+
1
)
?
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(
A
)
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>
5
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s
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−
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o
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d
′
>
(
B
)
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>
4
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c
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a
s
s
=
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−
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>
(
C
)
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3
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a
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−
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>
(
D
)
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2
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−
b
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>
(
E
)
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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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5
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(
B
)
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4
<
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a
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=
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(
C
)
<
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3
<
s
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a
ss
=
′
l
a
t
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x
−
b
o
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d
′
>
(
D
)
<
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an
>
2
<
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p
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a
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=
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>
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E
)
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1
18
1
Hide problems
Turkish NMO First Round - 2012 Problem - 18 {Number Theory}
If the representation of a positive number as a product of powers of distinct prime numbers contains no even powers other than
0
0
0
s, we will call the number singular. At most how many consequtive singular numbers are there?
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6
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)
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7
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8
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)
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9
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None
<span class='latex-bold'>(A)</span>\ 6 \qquad <span class='latex-bold'>(B)</span>\ 7 \qquad <span class='latex-bold'>(C)</span>\ 8 \qquad <span class='latex-bold'>(D)</span>\ 9 \qquad <span class='latex-bold'>(E)</span>\ \text{None}
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6
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B
)
<
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7
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(
C
)
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8
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D
)
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9
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None
14
1
Hide problems
Turkish NMO First Round - 2012 Problem - 14 {Number Theory}
What is the sum of distinct remainders when
(
2
n
−
1
)
502
+
(
2
n
+
1
)
502
+
(
2
n
+
3
)
502
(2n-1)^{502}+(2n+1)^{502}+(2n+3)^{502}
(
2
n
−
1
)
502
+
(
2
n
+
1
)
502
+
(
2
n
+
3
)
502
is divided by
2012
2012
2012
where
n
n
n
is positive integer?
<
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s
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a
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−
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>
(
A
)
<
/
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a
n
>
3
<
s
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c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1510
<
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s
=
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e
x
−
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o
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>
(
C
)
<
/
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a
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>
1511
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c
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a
s
s
=
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x
−
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o
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(
D
)
<
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>
1514
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E
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None
<span class='latex-bold'>(A)</span>\ 3 \qquad <span class='latex-bold'>(B)</span>\ 1510 \qquad <span class='latex-bold'>(C)</span>\ 1511 \qquad <span class='latex-bold'>(D)</span>\ 1514 \qquad <span class='latex-bold'>(E)</span>\ \text{None}
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A
)
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3
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=
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x
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>
(
B
)
<
/
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1510
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a
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−
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>
(
C
)
<
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1511
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a
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=
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−
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(
D
)
<
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1514
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E
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None
10
1
Hide problems
Turkish NMO First Round - 2012 Problem - 10 {Number Theory}
How many positive integers
n
n
n
are there such that there are
20
20
20
positive integers that are less than
n
n
n
and relatively prime with
n
n
n
?
<
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s
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a
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>
(
A
)
<
/
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a
n
>
1
<
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p
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n
c
l
a
s
s
=
′
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a
t
e
x
−
b
o
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d
′
>
(
B
)
<
/
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a
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>
2
<
s
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c
l
a
s
s
=
′
l
a
t
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−
b
o
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′
>
(
C
)
<
/
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<span class='latex-bold'>(A)</span>\ 1 \qquad <span class='latex-bold'>(B)</span>\ 2 \qquad <span class='latex-bold'>(C)</span>\ 3 \qquad <span class='latex-bold'>(D)</span>\ 4 \qquad <span class='latex-bold'>(E)</span>\ \text{None}
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None
6
1
Hide problems
Turkish NMO First Round - 2012 Problem - 06 {Number Theory}
Which one statisfies
n
29
≡
7
(
m
o
d
65
)
n^{29} \equiv 7 \pmod {65}
n
29
≡
7
(
mod
65
)
?
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37
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39
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43
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46
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55
<span class='latex-bold'>(A)</span>\ 37 \qquad <span class='latex-bold'>(B)</span>\ 39 \qquad <span class='latex-bold'>(C)</span>\ 43 \qquad <span class='latex-bold'>(D)</span>\ 46 \qquad <span class='latex-bold'>(E)</span>\ 55
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55
2
1
Hide problems
Turkish NMO First Round - 2012 Problem - 02 {Number Theory}
Find the sum of distinct residues of the number
201
2
n
+
m
2
2012^n+m^2
201
2
n
+
m
2
on
m
o
d
11
\mod 11
mod
11
where
m
m
m
and
n
n
n
are positive integers.
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43
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39
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37
<span class='latex-bold'>(A)</span>\ 55 \qquad <span class='latex-bold'>(B)</span>\ 46 \qquad <span class='latex-bold'>(C)</span>\ 43 \qquad <span class='latex-bold'>(D)</span>\ 39 \qquad <span class='latex-bold'>(E)</span>\ 37
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37
33
1
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Turkish NMO First Round - 2012 Problem - 33 {Geometry}
Let
A
B
C
D
A
′
B
′
C
′
D
′
ABCDA'B'C'D'
A
BC
D
A
′
B
′
C
′
D
′
be a rectangular prism with
∣
A
B
∣
=
2
∣
B
C
∣
|AB|=2|BC|
∣
A
B
∣
=
2∣
BC
∣
.
E
E
E
is a point on the edge
[
B
B
′
]
[BB']
[
B
B
′
]
satisfying
∣
E
B
′
∣
=
6
∣
E
B
∣
|EB'|=6|EB|
∣
E
B
′
∣
=
6∣
EB
∣
. Let
F
F
F
and
F
′
F'
F
′
be the feet of the perpendiculars from
E
E
E
at
△
A
E
C
\triangle AEC
△
A
EC
and
△
A
′
E
C
′
\triangle A'EC'
△
A
′
E
C
′
, respectively. If
m
(
F
E
F
′
^
)
=
6
0
∘
m(\widehat{FEF'})=60^{\circ}
m
(
FE
F
′
)
=
6
0
∘
, then
∣
B
C
∣
/
∣
B
E
∣
=
?
|BC|/|BE| = ?
∣
BC
∣/∣
BE
∣
=
?
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2
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15
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3
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<span class='latex-bold'>(A)</span>\ \sqrt\frac53 \qquad <span class='latex-bold'>(B)</span>\ \sqrt\frac{15}2 \qquad <span class='latex-bold'>(C)</span>\ \frac32\sqrt{15} \qquad <span class='latex-bold'>(D)</span>\ 5\sqrt\frac53 \qquad <span class='latex-bold'>(E)</span>\ \text{None}
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15
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15
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5
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None
29
1
Hide problems
Turkish NMO First Round - 2012 Problem - 29 {Geometry}
Let
D
D
D
and
E
E
E
be points on
[
B
C
]
[BC]
[
BC
]
and
[
A
C
]
[AC]
[
A
C
]
of acute
△
A
B
C
\triangle ABC
△
A
BC
, respectively.
A
D
AD
A
D
and
B
E
BE
BE
meet at
F
F
F
. If
∣
A
F
∣
=
∣
C
D
∣
=
2
∣
B
F
∣
=
2
∣
C
E
∣
|AF|=|CD|=2|BF|=2|CE|
∣
A
F
∣
=
∣
C
D
∣
=
2∣
BF
∣
=
2∣
CE
∣
, and
A
r
e
a
(
△
A
B
F
)
=
A
r
e
a
(
△
D
E
C
)
Area(\triangle ABF) = Area(\triangle DEC)
A
re
a
(
△
A
BF
)
=
A
re
a
(
△
D
EC
)
, then
A
r
e
a
(
△
A
F
C
)
/
A
r
e
a
(
△
B
F
C
)
=
?
Area(\triangle AFC)/Area(\triangle BFC) = ?
A
re
a
(
△
A
FC
)
/
A
re
a
(
△
BFC
)
=
?
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2
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<span class='latex-bold'>(A)</span>\ 4 \qquad <span class='latex-bold'>(B)</span>\ 2\sqrt2 \qquad <span class='latex-bold'>(C)</span>\ 2 \qquad <span class='latex-bold'>(D)</span>\ \sqrt2 \qquad <span class='latex-bold'>(E)</span>\ 1
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)
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2
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25
1
Hide problems
Turkish NMO First Round - 2012 Problem - 25 {Geometry}
The midpoint
M
M
M
of
[
A
C
]
[AC]
[
A
C
]
of a triangle
△
A
B
C
\triangle ABC
△
A
BC
is between
C
C
C
and the feet
H
H
H
of the altitude from
B
B
B
. If
m
(
A
B
H
^
)
=
m
(
M
B
C
^
)
m(\widehat{ABH}) = m(\widehat{MBC})
m
(
A
B
H
)
=
m
(
MBC
)
,
m
(
A
C
B
^
)
=
1
5
∘
m(\widehat{ACB}) = 15^{\circ}
m
(
A
CB
)
=
1
5
∘
, and
∣
H
M
∣
=
2
3
|HM|=2\sqrt{3}
∣
H
M
∣
=
2
3
, then
∣
A
C
∣
=
?
|AC|=?
∣
A
C
∣
=
?
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6
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(
B
)
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2
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c
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(
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8
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(
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16
3
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10
<span class='latex-bold'>(A)</span>\ 6 \qquad <span class='latex-bold'>(B)</span>\ 5 \sqrt 2 \qquad <span class='latex-bold'>(C)</span>\ 8 \qquad <span class='latex-bold'>(D)</span>\ \frac{16}{\sqrt3} \qquad <span class='latex-bold'>(E)</span>\ 10
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(
B
)
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2
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(
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)
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8
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(
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)
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3
16
<
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10
21
1
Hide problems
Turkish NMO First Round - 2012 Problem - 21 {Geometry}
The angle bisector of vertex
A
A
A
of
△
A
B
C
\triangle ABC
△
A
BC
cuts
[
B
C
]
[BC]
[
BC
]
at
D
D
D
. The circle passing through
A
A
A
and touching to
B
C
BC
BC
at
D
D
D
meets
[
A
B
]
[AB]
[
A
B
]
and
[
A
C
]
[AC]
[
A
C
]
at
P
P
P
and
Q
Q
Q
, respectively.
A
D
AD
A
D
and
P
Q
PQ
PQ
meet at
T
T
T
. If
∣
A
B
∣
=
5
,
∣
B
C
∣
=
6
,
∣
C
A
∣
=
7
|AB|=5, |BC|=6, |CA|=7
∣
A
B
∣
=
5
,
∣
BC
∣
=
6
,
∣
C
A
∣
=
7
, then
∣
A
T
∣
∣
T
D
∣
=
?
\frac{|AT|}{|TD|}=?
∣
T
D
∣
∣
A
T
∣
=
?
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5
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(
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2
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<span class='latex-bold'>(A)</span>\ \frac75 \qquad <span class='latex-bold'>(B)</span>\ 2 \qquad <span class='latex-bold'>(C)</span>\ 3 \qquad <span class='latex-bold'>(D)</span>\ \frac72 \qquad <span class='latex-bold'>(E)</span>\ 4
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7
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B
)
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7
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4
17
1
Hide problems
Turkish NMO First Round - 2012 Problem - 17 {Geometry}
Let
D
D
D
be a point inside
△
A
B
C
\triangle ABC
△
A
BC
such that
m
(
B
A
D
^
)
=
2
0
∘
m(\widehat{BAD})=20^{\circ}
m
(
B
A
D
)
=
2
0
∘
,
m
(
D
A
C
^
)
=
8
0
∘
m(\widehat{DAC})=80^{\circ}
m
(
D
A
C
)
=
8
0
∘
,
m
(
A
C
D
^
)
=
2
0
∘
m(\widehat{ACD})=20^{\circ}
m
(
A
C
D
)
=
2
0
∘
, and
m
(
D
C
B
^
)
=
2
0
∘
m(\widehat{DCB})=20^{\circ}
m
(
D
CB
)
=
2
0
∘
.
m
(
A
B
D
^
)
=
?
m(\widehat{ABD})= ?
m
(
A
B
D
)
=
?
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A
)
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∘
<
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a
s
s
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x
−
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o
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>
(
B
)
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1
0
∘
<
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p
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l
a
s
s
=
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−
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5
∘
<
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p
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c
l
a
s
s
=
′
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a
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x
−
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o
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d
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>
(
D
)
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>
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0
∘
<
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a
s
s
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x
−
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(
E
)
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5
∘
<span class='latex-bold'>(A)</span>\ 5^{\circ} \qquad <span class='latex-bold'>(B)</span>\ 10^{\circ} \qquad <span class='latex-bold'>(C)</span>\ 15^{\circ} \qquad <span class='latex-bold'>(D)</span>\ 20^{\circ} \qquad <span class='latex-bold'>(E)</span>\ 25^{\circ}
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∘
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s
p
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a
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l
a
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x
−
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d
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(
B
)
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1
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
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1
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
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an
>
2
0
∘
<
s
p
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c
l
a
ss
=
′
l
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x
−
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o
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d
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>
(
E
)
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5
∘
13
1
Hide problems
Turkish NMO First Round - 2012 Problem - 13 {Geometry}
20
20
20
points with no three collinear are given. How many obtuse triangles can be formed by these points?
<
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2
(
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3
)
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3
(
10
3
)
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E
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(
20
3
)
<span class='latex-bold'>(A)</span>\ 6 \qquad <span class='latex-bold'>(B)</span>\ 20 \qquad <span class='latex-bold'>(C)</span>\ 2{{10}\choose{3}} \qquad <span class='latex-bold'>(D)</span>\ 3{{10}\choose{3}} \qquad <span class='latex-bold'>(E)</span>\ {{20}\choose{3}}
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−
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)
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20
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(
C
)
<
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2
(
3
10
)
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a
t
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x
−
b
o
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d
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(
D
)
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3
(
3
10
)
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p
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c
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a
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a
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x
−
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(
E
)
<
/
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(
3
20
)
9
1
Hide problems
Turkish NMO First Round - 2012 Problem - 09 {Geometry}
The chord
[
C
D
]
[CD]
[
C
D
]
of the circle with diameter
[
A
B
]
[AB]
[
A
B
]
is perpendicular to
[
A
B
]
[AB]
[
A
B
]
. Let
M
M
M
and
N
N
N
be the midpoints of
[
B
C
]
[BC]
[
BC
]
and
[
A
D
]
[AD]
[
A
D
]
, respectively. If
∣
B
C
∣
=
6
|BC|=6
∣
BC
∣
=
6
and
∣
A
D
∣
=
2
3
|AD|=2\sqrt{3}
∣
A
D
∣
=
2
3
, then
∣
M
N
∣
=
?
|MN|=?
∣
MN
∣
=
?
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A
)
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(
B
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2
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None
<span class='latex-bold'>(A)</span>\ 4 \qquad <span class='latex-bold'>(B)</span>\ 3 \sqrt 2 \qquad <span class='latex-bold'>(C)</span>\ \sqrt{21} \qquad <span class='latex-bold'>(D)</span>\ 5 \qquad <span class='latex-bold'>(E)</span>\ \text{None}
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a
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a
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x
−
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d
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>
(
B
)
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2
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x
−
b
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C
)
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c
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a
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a
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−
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d
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(
D
)
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None
5
1
Hide problems
Turkish NMO First Round - 2012 Problem - 05 {Geometry}
△
A
B
C
\triangle ABC
△
A
BC
is given with
∣
A
B
∣
=
7
,
∣
B
C
∣
=
12
|AB|=7, |BC|=12
∣
A
B
∣
=
7
,
∣
BC
∣
=
12
, and
∣
C
A
∣
=
13
|CA|=13
∣
C
A
∣
=
13
. Let
D
D
D
be a point on
[
B
C
]
[BC]
[
BC
]
such that
∣
B
D
∣
=
5
|BD|=5
∣
B
D
∣
=
5
. Let
r
1
r_1
r
1
and
r
2
r_2
r
2
be the inradii of
△
A
B
D
\triangle ABD
△
A
B
D
and
△
A
C
D
\triangle ACD
△
A
C
D
, respectively. What is
r
1
/
r
2
r_1/r_2
r
1
/
r
2
?
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(
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)
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13
12
<
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c
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a
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s
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x
−
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o
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C
)
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7
5
<
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c
l
a
s
s
=
′
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a
t
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x
−
b
o
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d
′
>
(
D
)
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3
2
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E
)
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None
<span class='latex-bold'>(A)</span>\ 1 \qquad <span class='latex-bold'>(B)</span>\ \frac{13}{12} \qquad <span class='latex-bold'>(C)</span>\ \frac{7}{5} \qquad <span class='latex-bold'>(D)</span>\ \frac{3}{2} \qquad <span class='latex-bold'>(E)</span>\ \text{None}
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(
B
)
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13
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a
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a
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x
−
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C
)
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7
<
s
p
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c
l
a
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=
′
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a
t
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x
−
b
o
l
d
′
>
(
D
)
<
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2
3
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(
E
)
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None
1
1
Hide problems
Turkish NMO First Round - 2012 Problem - 01 {Geometry}
Find the perimeter of a triangle whose altitudes are
3
,
4
,
3,4,
3
,
4
,
and
6
6
6
.
<
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(
A
)
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12
3
5
<
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c
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a
s
s
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′
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x
−
b
o
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>
(
B
)
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16
3
5
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c
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a
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s
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(
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)
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3
5
<
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n
c
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a
s
s
=
′
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a
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x
−
b
o
l
d
′
>
(
D
)
<
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24
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5
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<span class='latex-bold'>(A)</span>\ 12\sqrt\frac35 \qquad <span class='latex-bold'>(B)</span>\ 16\sqrt\frac35 \qquad <span class='latex-bold'>(C)</span>\ 20\sqrt\frac35 \qquad <span class='latex-bold'>(D)</span>\ 24\sqrt\frac35 \qquad <span class='latex-bold'>(E)</span>\ \text{None}
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12
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16
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20
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D
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24
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None