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Problem
Problem
Let
Δ
k
\Delta_k
Δ
k
be the area of triangle
A
P
k
B
AP_k B
A
P
k
B
which is inscribed in a circle of radius 2 units. If
A
B
AB
A
B
diameter of circle,
∠
A
B
P
k
=
k
π
2
n
and
∑
k
=
1
n
+
1
Δ
k
=
4
cot
π
32
,
then
n
2
is equal to
\angle ABP_k = \frac{k\pi}{2n} \text{ and } \sum_{k=1}^{n+1} \Delta_k = 4 \cot \frac{\pi}{32}, \text{ then } \frac{n}{2} \text{ is equal to}
∠
A
B
P
k
=
2
n
k
π
and
k
=
1
∑
n
+
1
Δ
k
=
4
cot
32
π
,
then
2
n
is equal to
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