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Sorry in advance for not attaching a picture to help explain what I'm talking about, I don't have the reputation necessary to do so yet.

Consider a square with side length $a$ with a circle inside it, with the circumference just touching the sides of the square (i.e. $2r=a$). 4 smaller circles are located in each corner of the square such that their circumferences touch each side of the square and the larger circle.

All the circles are shaded black and the remaining area of the square is shaded red. Determine the percentage area of the square that's shaded red.

I know that to find the remaining percentage area of a single circle inscribed in a square you'd simply use $A_{\text{%}}=100(1-\frac {A_{circle}} {A_{square}})=100(1-\frac{\pi r^2}{a^2})=100(1-\frac{\pi r^2}{4r^2})=100(1-\frac{\pi}{4}) \approx 21.4602 \text{%}$

However, the 4 smaller circles in the corners have really stumped me with this problem. Any help is greatly appreciated.

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  • $\begingroup$ Hint: (assuming I have the picture right in my head). You need to compute $r$, the radius of a little circle. Draw the line segment from the center of the big circle to a corner of the square. We know its length. Convince yourself that this line passes through the center of a little circle. That makes its length $\frac a2+ r +x$ where $x$ is the distance from the center of the small circle to the corner. Work with the Pythagorean Theorem to express $x$ as a simple function of $r$. That's enough to compute $r$. $\endgroup$ – lulu Nov 16 '19 at 1:58
  • $\begingroup$ @lulu Thanks for you reply. I may be miss interpreting what you said, or maybe I just worded the original question poorly and it's thrown everything off, but I'm a but confused by the length being $\frac {a} {2} + r + x$. Shouldn't it be $\frac {a} {2} + 2r + x$, or am I even more lost than I thought? $\endgroup$ – Aussie Mathematician Nov 16 '19 at 2:26
  • $\begingroup$ Yes, that's right. I dropped a factor of $2$. $\endgroup$ – lulu Nov 16 '19 at 10:54
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Let $b$ the radius of the four small circles. Apply the Pythagorean formula to the isosceles right triangle whose hypotenuse connect the centers of a small and the large circle, i.e.

$$(r+b)^2 = (r-b)^2 + (r-b)^2$$

which yields,

$$b = (3-2\sqrt2)r$$

Then, the total area of the four small circles is $\pi(17-12\sqrt2)a^2$.

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