Christmas Streak 82/88: Limits on Geometry #3

Calculus Level 5

As shown on the right, there is a circle C 1 \rm C_1 with radius 2 2 . 2\sqrt{2}.

For every natural number n ≤ 2 N , n\le 2N, conduct these:

1. Inscribe a square in C 2 n − 1 \rm C_{2n-1} and call it A n . \rm A_{n}. S 2 n − 1 S_{2n-1} is the difference of the area of C 2 n − 1 \rm C_{2n-1} and A n . \rm A_{n}.

2. Inscribe a circle in A n \rm A_{n} and call it C 2 n . \rm C_{2n}.

3. Inscribe an equilateral triangle in C 2 n \rm C_{2n} and call it B n . \rm B_{n}. S 2 n S_{2n} is the difference of the area of C 2 n \rm C_{2n} and B n . \rm B_{n}.

lim ⁡ N → ∞ ∑ n = 1 2 N S n = p q ( a π + b − c ) \displaystyle \lim_{N\to\infty}\sum_{n=1}^{2N}S_n=\dfrac{p}{q}(a\pi+b-\sqrt{c}) where p , q p,~q are coprime positive integers, a , b , c a,~b,~c are integers, and p q \dfrac{p}{q} is as large as possible.

Find p + q + a + b + c . p+q+a+b+c.

This problem is a part of <Christmas Streak 2017> series .


The answer is 38.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Tapas Mazumdar
May 18, 2018

Using simple trigonometry and geometry, we can create a table as follows (here R = 2 2 R = 2\sqrt{2} is the radius of outermost circle C 1 C_1 ):

Value of n C 2 n − 1 C 2 n A n B n n = 1 π R 2 − 2 R 2 − − π 2 R 2 − 3 3 8 R 2 n = 2 π 8 R 2 − 1 4 R 2 − − π 16 R 2 − 3 3 64 R 2 n = 3 π 64 R 2 − 1 32 R 2 − − π 128 R 2 − 3 3 512 R 2 ⋮ ⋮ ⋮ ⋮ ⋮ n = N π 8 N − 1 R 2 − 2 8 N − 1 R 2 − − 4 π 8 N R 2 − 3 3 8 N R 2 \begin{array}{c}\\ & \boxed{\text{Value of } n} & & \boxed{C_{2n-1}} & & \boxed{C_{2n}} & & \boxed{A_n} & & \boxed{B_n} \\ \\ & n=1 & & \pi R^2 & & - & & 2R^2 & & - \\ & & & - & & \dfrac{\pi}{2} R^2 & & - & & \dfrac{3 \sqrt{3}}{8} R^2 \\ & n=2 & & \dfrac{\pi}{8} R^2 & & - & & \dfrac{1}{4} R^2 & & - \\ & & & - & & \dfrac{\pi}{16} R^2 & & - & & \dfrac{3 \sqrt{3}}{64} R^2 \\ & n=3 & & \dfrac{\pi}{64} R^2 & & - & & \dfrac{1}{32} R^2 & & - \\ & & & - & & \dfrac{\pi}{128} R^2 & & - & & \dfrac{3 \sqrt{3}}{512} R^2 \\ \\ & \vdots & & \vdots & & \vdots & & \vdots & & \vdots \\ \\ & n=N & & \dfrac{\pi}{8^{N-1}} R^2 & & - & & \dfrac{2}{8^{N-1}} R^2 & & - \\ & & & - & & \dfrac{4\pi}{8^N} R^2 & & - & & \dfrac{3 \sqrt{3}}{8^N} R^2 \\ \end{array}

Thus

lim ⁡ N → ∞ ∑ n = 1 2 N S n = lim ⁡ N → ∞ ( ∑ n = 1 N S 2 n − 1 + ∑ n = 1 N S 2 n ) = lim ⁡ N → ∞ ( ∑ n = 1 N ( C 2 n − 1 − A n ) + ∑ n = 1 N ( C 2 n − B n ) ) = lim ⁡ N → ∞ ( ∑ n = 1 N ( π − 2 ) R 2 8 n − 1 + ∑ n = 1 N ( 4 π − 3 3 ) R 2 8 n ) = ∑ n = 1 ∞ ( π − 2 ) R 2 8 n − 1 + ∑ n = 1 ∞ ( 4 π − 3 3 ) R 2 8 n = 8 7 ( π − 2 ) R 2 + 1 7 ( 4 π − 3 3 ) R 2 = 64 7 ( π − 2 ) + 8 7 ( 4 π − 3 3 ) = 8 7 ( 12 π − 16 − 27 ) \begin{aligned} \displaystyle \lim_{N \to \infty} \sum_{n=1}^{2N} S_n &= \lim_{N \to \infty} \left( \sum_{n=1}^{N} S_{2n-1} + \sum_{n=1}^{N} S_{2n} \right) \\ &= \displaystyle \lim_{N \to \infty} \left( \sum_{n=1}^{N} ( C_{2n-1} - A_n ) + \sum_{n=1}^{N} ( C_{2n} - B_n ) \right) \\ &= \displaystyle \lim_{N \to \infty} \left( \sum_{n=1}^{N} \dfrac{(\pi -2)R^2}{8^{n-1}} + \sum_{n=1}^{N} \dfrac{(4\pi -3\sqrt{3})R^2}{8^n} \right) \\ &= \sum_{n=1}^{\infty} \dfrac{(\pi -2)R^2}{8^{n-1}} + \sum_{n=1}^{\infty} \dfrac{(4\pi -3\sqrt{3})R^2}{8^n} \\ &= \dfrac{8}{7} (\pi -2) R^2 + \dfrac{1}{7} (4\pi -3\sqrt{3}) R^2 \\ &= \dfrac{64}{7} (\pi -2) + \dfrac{8}{7} (4\pi -3\sqrt{3}) \\ &= \dfrac{8}{7} \left( 12 \pi - 16 - \sqrt{27} \right) \end{aligned}

So p + q + a + b + c = 8 + 7 + 12 + ( − 16 ) + 27 = 38 p+q+a+b+c = 8+7+12+(-16)+27 = \boxed{38} .

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