18. The administrators of the Instagram page for the Math Olympics noticed that the icon corresponding to the "likes" received is heart-shaped, composed of a square and two semicircles constructed on consecutive sides. The figure can be inscribed in a rectangle with sides parallel to the diagonals of the square, as shown alongside. Knowing that the square has a side length of $1 \mathrm{~cm}$, how many $\mathrm{cm}^{2}$ is the area of the rectangle?

![](https://cdn.mathpix.com/cropped/2024_04_17_594e72200cb405ca2f14g-2.jpg?height=257&width=280&top_left_y=647&top_left_x=2463)
(A) $3 \sqrt{2}-2$
(D) $\frac{3+2 \sqrt{2}}{4}$
(B) $\frac{3+\sqrt{2}}{2}$
(C) $\frac{7}{8}(1+\sqrt{2})$
(18) The correct answer is $(E)$.

The tangency between the sides of the rectangle and the semicircles occurs at points $B, C, F, E$ indicated in the figure, where the radii are parallel to one of the diagonals of the square.

Given that, in cm, $\overline{A D}=\frac{\sqrt{2}}{2}$ and $\overline{A B}=\overline{D E}=\frac{1}{2}$, the horizontal side of the rectangle thus measures $2 \overline{A B}+\overline{A D}=\frac{2+\sqrt{2}}{2}$.

The vertical side of the rectangle is instead equal to $3 / 4$ of the diagonal of the square plus $\overline{A C}$ (which is $\frac{1}{2}$), that is: $\frac{1}{2}+\frac{3}{4} \cdot \sqrt{2}=\frac{2+3 \sqrt{2}}{4}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c43da7de8bb5572fa51ag-4.jpg?height=363&width=423&top_left_y=1209&top_left_x=1469)

The area of the rectangle in $\mathrm{cm}^{2}$ is therefore $\frac{2+\sqrt{2}}{2} \cdot \frac{2+3 \sqrt{2}}{4}=\frac{5}{4}+\sqrt{2}$.

Question proposed by Andrea Gallese.