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Math Prize For Girls Problems
2023 Math Prize for Girls Problems
20
20
Part of
2023 Math Prize for Girls Problems
Problems
(1)
Math Prize 2023 Problem 20
Source:
10/12/2023
Let
f
1
(
x
)
=
2
π
sin
(
x
)
f_1(x) = 2\pi \sin (x)
f
1
(
x
)
=
2
π
sin
(
x
)
. For
n
>
1
n > 1
n
>
1
, define
f
n
(
x
)
f_n(x)
f
n
(
x
)
recursively by
f
n
(
x
)
=
2
π
sin
(
f
n
−
1
(
x
)
)
.
f_n(x) = 2\pi \sin(f_{n-1}(x)).
f
n
(
x
)
=
2
π
sin
(
f
n
−
1
(
x
))
.
How many intervals
[
a
,
b
]
[a, b]
[
a
,
b
]
are there such that \bullet \
0
≤
a
<
b
≤
2
π
0 \le a < b \le 2\pi
0
≤
a
<
b
≤
2
π
, \bullet \
f
6
(
a
)
=
−
2
π
f_6(a) = -2\pi
f
6
(
a
)
=
−
2
π
, \bullet \
f
6
(
b
)
=
2
π
f_6(b)=2\pi
f
6
(
b
)
=
2
π
, \bullet \ and
f
6
f_6
f
6
is increasing on
[
a
,
b
]
[a, b]
[
a
,
b
]
?