5. Inside a square of side length 1 , four quarter-circle arcs are traced with the edges of the square serving as the radii. It is known that these arcs intersect pairwise at four distinct points, which in fact are the vertices of a smaller square. Suppose this process is repeated for the smaller square, and so on and so forth. What is the sum of the areas of all squares formed in this manner?
Answer: $\frac{1+\sqrt{3}}{2}$.
Solution: By similarity, we note that the areas of the squares are in geometric progression. Hence, we need only to find out the area of the first smaller square. Note that the diagonal of the smaller square is the overlap of two line segments of length $\frac{\sqrt{3}}{2}$, with a total length of 1 . This is going to be $\sqrt{3}-1$, and so the area of the smaller square is $\frac{1}{2}(\sqrt{3}-1)^{2}=2-\sqrt{3}$.
Thus, the sum of the areas of all squares is given by $\frac{1}{1-(2-\sqrt{3})}=\frac{1+\sqrt{3}}{2}$.