<s>[INST] sangaku - a geometrical puzzle 
Find the radius of the circles if the size of the larger square is 1x1.
Enjoy!  
(read about the origin of sangaku)
 [/INST] Let $r$ be the length the radius of the circles, and let $\theta$ be the measure of the (smaller) angle made at the corner of the big square.

The width of the square is equal to two radii and the projection of a double diameter (a quadruple-radius), so that
$(1)\hspace{1.0in}4r\cos\theta=1-2r$
Looking at the four right triangles, we see that the center circle's diameter is equal to the difference in the lengths of the legs; since the hypotenuse has length $1$, we have
$(2)\hspace{1.0in}2r = \cos\theta - \sin\theta$
From here, we simply need to eliminate $\theta$.
Multiplying (2) through by $4r$ and substituting in from (1) ...
$$8 r^2 = 4r\cos\theta - 4r \sin\theta = 1 - 2r - 4r \sin\theta$$
$$4r \sin\theta = 1 - 2r - 8 r^2$$
Therefore,
$$\begin{eqnarray}16r^2 &=& (4r \cos\theta)^2 + (4 r \sin\theta)^2 \\\ 
&=& ( 1 - 2r )^2 + ( 1 - 2r - 8 r^2 )^2 \\\
&=& 2 - 8 r - 8 r^2 + 32r^3 + 64 r^4 \end{eqnarray}$$
so that
$$0 = 32 r^4 + 16 r^3 - 12 r^2 - 4 r + 1 = (2r+1)(2r-1)(8 r^2 + 4 r - 1)$$ 
The roots of the polynomial are $\pm1/2$ and $(-1\pm\sqrt{3})/4$. We can eliminate three of them from consideration to conclude that $r = (-1+\sqrt{3})/4$.
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