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Circle inscribed in a square inscribed in a circle question

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by pufftissue » Wed Aug 13, 2008 7:22 am
Hi, please help:

An inner circle is inscribed in a square, which is inscribed in another circle (outer circle).

If "a" is the difference between the radius of the two circles, then what is the side length of the square in terms of "a"?

Thank you!
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by santa_dem » Wed Aug 13, 2008 7:34 am
let's say x is the side of the square.

The radius of the inner circle is x/2

The radius of the outer circle is x*sqrt(2)/2.

The difference of the radius is x*sqrt(2)/2-x/2=a


x/2*(sqrt(2)-1)=a

x=2a/(sqrt(2)-1)
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by pepeprepa » Wed Aug 13, 2008 8:38 am
Let's say
R=radius of the big circle
r=radius of the small circle
a=R-r according to the question

Look at the draw and take half of the square, you have a rectangular triangle and we can use pyhtagore theorem. You can see that 2r is one side of the square and that 2R is the hypothenuse of the square.

(2R)^2 = (2r)^2 + (2r)^2
R=r*sqrt(2)

Now let's use R-r=a
a+r=r*sqrt(2)
a/(sqrt(2) -1)=r

We want one side, which is 2r so:
2r= 2a/(sqrt(2) -1)
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