14. A square $A B C D$ with a side length of $2 \mathrm{~cm}$ is inscribed in a circle. In the circular segments delimited by the sides of the square, 4 rectangles are inscribed. How many cm must each side of these rectangles measure, so that they are squares?
(A) $\sqrt{2} / 4$
(B) $1 / 3$
(C) $3 / 8$
(D) $2 / 5$
(E) $1 / 4$

![](https://cdn.mathpix.com/cropped/2024_04_17_e21ecd5de9c42efa6d1eg-2.jpg?height=284&width=279&top_left_y=1406&top_left_x=1095)
(14) The correct answer is $(D)$.

First, observe that the radius of the circle, which is half the diagonal of the square $ABCD$, has a length (in $\mathrm{cm}$) equal to $\sqrt{2}$. Let $O$ be the center of the circle. Suppose that $EFGH$ is a square inscribed in the circular segment delimited by $AB$, with vertices $E, F$ on the side $AB$ and vertices $G, H$ on the circle. Let $M$ be the midpoint of $AB$, which is also the midpoint of $EF$. Denote by $x$ the measure (always in $\mathrm{cm}$) of the side of the square $EFGH$. By the Pythagorean theorem, we have

$$
(\overline{OM} + x)^{2} + (x / 2)^{2} = \overline{OG}^{2}
$$

that is, $(1 + x)^{2} + (x / 2)^{2} = (\sqrt{2})^{2}$, which simplifies to $5x^{2} + 8x - 4 = 0$. The solutions to the equation are $2 / 5$ and -2, so it must be that $x = 2 / 5$.

Problem proposed by Matteo Rossi.