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The side length of the square is $a$. Two different quadrants are inscribed in it as follows. A small red circle is also inscribed between them.

Find the radius of the smallest circle (red one) in terms of '$a$' only.

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1 Answer 1

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$r=\frac{3^2-2^2\sqrt{2}}{7^2} a$.

Take $a=1$ to simplify. To find $r$, find the solution to the following that has $(x,y) \in [0,1]^2$:

$$x^2+y^2 = (1+r)^2\\ (x-1)^2+(y-1)^2 = (\sqrt{2}-1+r)^2\\ x = 1-r .$$ $x,y$ represent the center of the smaller circle, $r$ is the radius. There are two solutions, only one has $(x,y) \in [0,1]^2$. Since I took $a=1$ to simplify, I need to scale the answer by $a$, which gives the answer above.

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Do you mean $r=\frac{9-4\sqrt{2}}{7^2} a$? Cheers. –  user26872 Jul 8 '12 at 3:28
    
Yes @oen, good catch, I meant to write $2^2$. Thanks! –  copper.hat Jul 8 '12 at 3:57
    
Glad to help. (+1) –  user26872 Jul 8 '12 at 4:00

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