Cyclic Regular Polygon Fractals

Calculus Level 3

The n th n^{\text{th}} figure in the above sequence is constructed by the following procedure:

  1. Draw a blue disc of radius 2016 π \displaystyle\sqrt{\frac{2016}{\pi}}
  2. Remove a regular n n -gon area from the (smallest) disc
  3. Inscribe a blue disc inside the empty n n -gon space
  4. Repeat steps 2-4

Let A n A_n be the total blue area of the n th n^{\text{th}} figure in the sequence.

Compute lim ⁡ n → ∞ A n \displaystyle\lim_{n\to\infty}A_n .


The answer is 1344.

Kerry Soderdahl by Kerry Soderdahl , 23, USA

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1 solution

Kerry Soderdahl
Mar 15, 2016

Relevant wiki: Regular Polygons - Problem Solving - Medium

I'll write a better and more complete solution as soon as I have the time to.

The radius of the outer circle is r 0 = 2016 π r_0=\sqrt{\frac{2016}{\pi}}

The blue area after steps 1 and 2 is π r 0 2 − n r 0 2 cos ⁡ ( π n ) sin ⁡ ( π n ) \pi r_0^2-nr_0^2\cos(\frac{\pi}{n})\sin(\frac{\pi}{n})

This area is repeated through a geometric sequence with a ratio of cos ⁡ 2 ( π n ) \cos^2(\frac{\pi}{n}) to construct the entire figure:

lim ⁡ n → ∞ ∑ k = 0 ∞ ( cos ⁡ 2 k ( π n ) ) ( π r 0 2 − n r 0 2 cos ⁡ ( π n ) sin ⁡ ( π n ) ) \lim_{n\to\infty}\sum_{k=0}^{\infty}(\cos^{2k}(\frac{\pi}{n}))(\pi r_0^2-nr_0^2\cos(\frac{\pi}{n})\sin(\frac{\pi}{n}))

= r 0 2 lim ⁡ n → ∞ π − n cos ⁡ ( π n ) sin ⁡ ( π n ) 1 − cos ⁡ 2 ( π n ) = r 0 2 lim ⁡ n → ∞ π − n cos ⁡ ( π n ) sin ⁡ ( π n ) sin ⁡ 2 ( π n ) =r_0^2\lim_{n\to\infty}\frac{\pi-n\cos(\frac{\pi}{n})\sin(\frac{\pi}{n})}{1-\cos^2(\frac{\pi}{n})}=r_0^2\lim_{n\to\infty}\frac{\pi-n\cos(\frac{\pi}{n})\sin(\frac{\pi}{n})}{\sin^2(\frac{\pi}{n})}

= r 0 2 lim ⁡ n → ∞ π csc ⁡ 2 ( π n ) − n cot ⁡ ( π n ) =r_0^2\lim_{n\to\infty}\pi\csc^2(\frac{\pi}{n})-n\cot(\frac{\pi}{n})

I don't actually know how to evaluate this limit -- I'd really appreciate if someone could post how to do it. According to Wolfram Alpha, lim ⁡ n → ∞ π csc ⁡ 2 ( π n ) − n cot ⁡ ( π n ) = 2 π 3 \lim_{n\to\infty}\pi\csc^2(\frac{\pi}{n})-n\cot(\frac{\pi}{n})=\frac{2\pi}{3} .

r 0 2 ∗ 2 π 3 = 1344 r_0^2*\frac{2\pi}{3}=\boxed{1344}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Easiest way would be use series expansion. Atleast I did it that way. Lhopital is still available on taking common denominator though.

A Former Brilliant Member - 5 years, 2 months ago

Try L'Hospitals' rule. It works with a bit of manipulation. A very nice problem though.

Vignesh S - 5 years, 2 months ago

Hey, @Kerry Soderdahl lim ⁡ n → ∞ A n = lim ⁡ n → ∞ π r 0 2 ( 1 − sin ⁡ 2 π n 2 π n sin ⁡ 2 π n ) \lim_{n\rightarrow\infty}A_n=\lim_{n\rightarrow\infty}\pi r^2_0\left(\frac{1-\frac{\sin{\frac{2\pi}{n}}}{\frac{2\pi}{n}}}{\sin^2{\frac{\pi}{n}}}\right)

Now, using the power series of sin upto third order terms, and simplifying, lim ⁡ n → ∞ A n = 2 π 3 r 0 2 \lim_{n\rightarrow\infty}A_n=\frac{2\pi}{3}r^2_0

Use could also put π n = α \dfrac{\pi}{n}=\alpha so that, α → 0 \alpha\rightarrow0 as n → ∞ n\rightarrow\infty and then use L'Hospitals' rule (but you'll have to use it twice or thrice, which is labourious for the given expression).

A Former Brilliant Member - 5 years, 2 months ago

I got the expression but I do not know how to evaluate the limit

Paul Romero - 8 months, 2 weeks ago

I used a spreadsheet to find the limit as n tends to a big value, nut then to check I found a similar formula for the white areas, and found the ratio of white to blue was 2:1, so the limit is 2/3 the area of the circle.

Ken Garner - 4 months, 1 week ago

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