A square is inscribed in a unit circle, a circle is inscribed in the square, a regular octagon is inscribed in the second circle, a circle is inscribed in the octagon, a regular 16-gon is inscribed in the 3rd circle and so on … , 2ⁿ⁺ ¹ -gon is inscribed in the n-th circle and (n+1)th circle is inscribed in the 2ⁿ⁺ ¹ -gon. If R(n) is the radius of the n-th circle, find lim[n–>∞] R(n) !
Update:
… , a regular 2ⁿ⁺ ¹ -gon is inscribed in the n-th circle and …
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The answer is 2/π. The proof uses the following well-known and easily to prove statements:
(1) If r is inradius, R circumradius for a regular n-gon, then r/R = cos(π/n);
(2) sin 2x = 2 cos x * sin x;
(3) lim[x→0] (sin(ax)/x) = a
Let R_{1}, R_{2}, . . , R_{n}, . . /n = 1, 2, . . / are the radiuses of the 1st, 2nd, . . , nth circle, then according (1) we have:
R_{2}/R_{1} = cos(π/4); R_{3}/R_{2} = cos(π/8); R_{4}/R_{3} = cos(π/16); . . R_{n}/R_{n-1} = cos(π/2ⁿ)
Multiplying them (using R_{1} = 1) we get
R_{n} = cos(π/4) cos(π/8) cos(π/16) . . cos(π/2ⁿ), /n = 2, 3, . . /
Using (2) several times, we get
1 = sin(π/2) = 2 cos(π/4) sin(π/4) = 4 cos(π/4) cos(π/8) sin(π/8) =
= 8 cos(π/4) cos(π/8) cos(π/16) sin(π/16) = . . .
= 2ⁿ⁻¹ cos(π/4) cos(π/8) cos(π/16) . . cos(π/2ⁿ) * sin(π/2ⁿ), hence
R_{n} = cos(π/4) cos(π/8) cos(π/16) . . cos(π/2ⁿ) =
= 1/(2ⁿ⁻¹ *sin(π/2ⁿ)) = 2 * (1/2ⁿ) / sin(π/2ⁿ)
Now according (3) lim[n→∞] (1/2ⁿ) / sin(π/2ⁿ) = 1/π since 1/2ⁿ → 0 and finally
lim[n→∞] R_{n} = 2 * lim[n→∞] (1/2ⁿ) / sin(π/2ⁿ) = 2/π
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