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A white regular polygon with area $1$ has a red circle inscribed(the circle touches the edges of the square), with a white regular polygon with the same amount of sides as the first one(the polygon's corners touch the circle) and this pattern continues forever, creating a pattern like this: (https://i.stack.imgur.com/IARUV.jpg) 1. What is the area that is coloured red if the polygon is a square?

  1. If you knew the amount of sides, what would be a formula to find the area coloured red, in terms of $s$ (for sides)
Kyan Cheung
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2 Answers2

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As the part inside the second polygon is a scaled copy of the entire construction, you only need to consider the area between the first and second copy of the polygon.

If it's a square that's easy.

If it's not a square you need to consider whether the question is even well-defined, i.e. whether the ratio of the white/red areas is the same. (And if it is, you should be able to realise quite quickly what the answer must be)

Henrik supports the community
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  • Could you please clarify "whether the ratio of the white/red areas is the same"? I don't exactly understand what you mean. – Kyan Cheung Oct 28 '17 at 08:13
  • I mean: If the polygon is a triangle, a pentagon or perhaps an irregular (you haven't specified in the question that the polygon is regular, you also need to think about your definition of "inscribed" in that case) 35645-gon, can you be sure the areas coloured red will be the same? – Henrik supports the community Oct 28 '17 at 08:20
  • Thanks for clarifying. The polygons are regular and similar. – Kyan Cheung Oct 28 '17 at 08:25
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Consider the larger red area. It is the area of the circle having radius $r=1$ , $C=\pi$ less the square having area $A=\sqrt{2}^2=2$

So its area is $R_1=\pi-2$

Now consider the similitude ratio between the inscribed and the circumscribed squares. The inner square has side $l_1=\sqrt{2}$ and the outer is $L_1=2$

The ratio is $r_1=\frac{\sqrt 2}{2}$ then the ratio for areas is $r_1^2=\frac{1}{2}$

As the red part become smaller, the ratio between them is constant and is $\frac{1}{2}$

Therefore the sum of the red areas is

$$R=R_1+\frac{1}{2}R_1+\frac{1}{2}\frac{1}{2}R_1+\ldots=R_1\sum_{n=0}^{\infty}\frac{1}{2^n}=2R_1$$

so the red area is $R=2(\pi-2)$

Hope this helps

Raffaele
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