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Problem
Problem
In triangle
A
B
C
ABC
A
B
C
,
∠
A
>
π
2
\angle A > \frac{\pi}{2}
∠
A
>
2
π
.
A
A
1
AA_1
A
A
1
and
A
A
2
AA_2
A
A
2
are the median and altitude, respectively. If
∠
B
A
A
1
=
∠
A
1
A
A
2
=
∠
A
2
A
C
\angle BAA_1 = \angle A_1AA_2 = \angle A_2AC
∠
B
A
A
1
=
∠
A
1
A
A
2
=
∠
A
2
A
C
, then
sin
3
A
3
⋅
cos
A
3
\sin^3 \frac{A}{3} \cdot \cos \frac{A}{3}
sin
3
3
A
⋅
cos
3
A
is equal to a.
3
a
3
16
b
2
c
\frac{3a^3}{16b^2c}
16
b
2
c
3
a
3
b.
3
a
3
64
b
2
c
\frac{3a^3}{64b^2c}
64
b
2
c
3
a
3
c.
3
a
2
4
b
2
c
\frac{3a^2}{4b^2c}
4
b
2
c
3
a
2
d.
3
a
3
12
b
2
c
\frac{3a^3}{12b^2c}
12
b
2
c
3
a
3
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