7. Alessandra and Luigi are playing a board game whose pieces are spaceships, like the one shown in gray in the figure on the side. If the dashed squares have sides of length 1, what is the area of a piece?

![](https://cdn.mathpix.com/cropped/2024_04_17_0fac47a360be2b49c478g-03.jpg?height=228&width=619&top_left_y=2079&top_left_x=1164)
(A) $\frac{2}{3}$
(B) $\frac{\sqrt{3}}{6}$
(C) $\frac{1}{6}$
(D) $\frac{\sqrt{2}}{8}$
(E) $\frac{1}{8}+\frac{\sqrt{2}}{12}$
7. The answer is $\mathbf{( C )}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_0fac47a360be2b49c478g-09.jpg?height=437&width=960&top_left_y=238&top_left_x=591)

## FIRST SOLUTION

We assign names to the points involved, as shown in the figure. The quadrilateral $A B C D$ is formed by two triangles symmetric with respect to the line $A C$; its area can thus be calculated as $\frac{1}{2} A C \cdot B D$. On the other hand, $A$ is the center of the square $P_{1} Q_{1} Q_{2} P_{2}$ and $C$ is the midpoint of the segment $Q_{2} P_{2}$, so the length of $A C$ is $\frac{1}{2}$. Note now that the line $A C$ is parallel to the line $Q_{1} Q_{2}$ and consequently the triangles $A B C$ and $Q_{2} B Q_{1}$ are similar; if $h$ is the height of $Q_{2} B Q_{1}$ relative to $Q_{1} Q_{2}$, then $h$ : $\frac{B D}{2}=Q_{1} Q_{2}: A C=1: \frac{1}{2}$, or $B D=h$; since on the other hand $B D+2 h=1$, the length of $B D$ is $\frac{1}{3}$.

We now calculate the area of the triangle $B E Q_{1}$ as the difference between the areas of the triangles $Q_{1} Q_{2} B$ and $Q_{1} Q_{2} E$, that is, as $\frac{1}{2} Q_{1} Q_{2} \cdot\left(h-h^{\prime}\right)$, where $h^{\prime}$ is the height of $Q_{1} Q_{2} E$ relative to $Q_{1} Q_{2}$. The intersection point $F$ between the lines $Q_{1} P_{4}$ and $P_{1} Q_{4}$ is the center of the square $Q_{2} Q_{3} P_{3} P_{2}$; the line $A F$ is parallel to the line $Q_{1} Q_{2}$ and thus the triangles $Q_{1} Q_{2} E$ and $F A E$ are similar. On the other hand, $A F=1=Q_{1} Q_{2}$, and therefore they are actually congruent triangles, from which $h^{\prime}=\frac{1}{4}$. The area of $B E Q_{1}$ is thus $\frac{1}{2}\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{24}$.

The area of the tile, in conclusion, is $2 \cdot \frac{1}{24}+\frac{1}{12}=\frac{1}{6}$.

## SECOND SOLUTION

Consider the coordinates of the points in a Cartesian plane centered at $Q_{1}$, with $P_{1}$ on the $y$-axis and $Q_{2}$ on the $x$-axis. It is easy to verify that $A$ has coordinates $\left(\frac{1}{2}, \frac{1}{2}\right)$, $C$ has coordinates $\left(1, \frac{1}{2}\right)$, and $F$ has coordinates $\left(\frac{3}{2}, \frac{1}{2}\right)$. The point $B$ is the intersection of the line $Q_{1} P_{3}$ (with equation $y=\frac{1}{2} x$) and the line $P_{1} Q_{2}$ (with equation $x+y=1$), so it has coordinates $\left(\frac{2}{3}, \frac{1}{3}\right)$; $E$, as the intersection of the line with equation $y=\frac{1}{3} x$ and the line with equation $x+y=1$, has coordinates $\left(\frac{3}{4}, \frac{1}{4}\right)$. It is now straightforward to calculate the area of the tile, which results in $\frac{1}{6}$.