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)
  • $\begingroup$ Interesting, but could you tell us what is your work on the subject ? $\endgroup$ – Jean Marie Oct 28 '17 at 8:07
  • $\begingroup$ Seems to boil down to a geometric series, but I am not sure. $\endgroup$ – Peter Oct 28 '17 at 8:18

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)

  • $\begingroup$ Could you please clarify "whether the ratio of the white/red areas is the same"? I don't exactly understand what you mean. $\endgroup$ – Kyky Oct 28 '17 at 8:13
  • $\begingroup$ 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? $\endgroup$ – Henrik Oct 28 '17 at 8:20
  • $\begingroup$ Thanks for clarifying. The polygons are regular and similar. $\endgroup$ – Kyky Oct 28 '17 at 8:25

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


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

Hope this helps


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.