In the figure, $ABCD$ is a square, whose area measures $7 / 32$ of the area of triangle $XYZ$. What is the ratio between $XA$ and $XY$?

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-020.jpg?height=223&width=437&top_left_y=2290&top_left_x=1392)
Solution 1: Let $S_{1}, S_{2}$, and $S_{3}$ be the areas of the triangles $\triangle X A B, \triangle A Y D$, and $\triangle B C Z$, and let $S_{q}$ be the area of the square $A B C D$, as indicated in the figure. If $S$ denotes the area of the triangle $\triangle X Y Z$, then $S=S_{1}+S_{2}+S_{3}+S_{q}$ and, by hypothesis, $S_{q}=(7 / 32) S$, so we establish that

$$
\begin{aligned}
\frac{S_{1}+S_{2}+S_{3}}{S} & =\frac{S-S_{q}}{S}=1-\frac{S_{q}}{S} \\
& =1-\frac{7}{32}=\frac{25}{32}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-078.jpg?height=363&width=791&top_left_y=367&top_left_x=1018)

Since $\triangle X A B$ and $\triangle X Y Z$ are similar triangles, the ratio of their areas is equal to the square of the similarity ratio, that is,

$$
\frac{S_{1}}{S}=\left(\frac{X A}{X Y}\right)^{2}
$$

By horizontally translating the triangle $\triangle B C Z$ to juxtapose it with the triangle $\triangle A D Y$, we obtain a triangle similar to $\triangle X Y Z$, but with area $S_{2}+S_{3}$. Thus,

$$
\frac{S_{2}+S_{3}}{S}=\left(\frac{A Y}{X Y}\right)^{2}
$$

Adding these two ratios and using the established proportion, we conclude that

$$
\begin{aligned}
\frac{25}{32} & =\frac{S_{1}+S_{2}+S_{3}}{S}=\frac{(X A)^{2}+(A Y)^{2}}{(X Y)^{2}}=\frac{(X A)^{2}+(X Y-X A)^{2}}{(X Y)^{2}} \\
& =\frac{(X Y)^{2}+2(X A)^{2}-2(X Y)(X A)}{(X Y)^{2}}=1+2\left(\frac{X A}{X Y}\right)^{2}-2\left(\frac{X A}{X Y}\right)
\end{aligned}
$$

In other words, the ratio between $X A$ and $X Y$ we are looking for satisfies the quadratic equation

$$
2\left(\frac{X A}{X Y}\right)^{2}-2\left(\frac{X A}{X Y}\right)+\frac{7}{32}=0
$$

Using the quadratic formula, we obtain two possible values for this ratio, namely, $\frac{X A}{X Y}=\frac{7}{8}$ and $\frac{X A}{X Y}=\frac{1}{8}$.

Solution 2: Let the side length of the square $A B C D$ be $l$, the height of the triangle $\triangle X Y Z$ be $H$, the height of the triangle $\triangle X A B$ be $h$, and the length of the side $Y Z$ be $b$. The area of the square is $l^{2}$ and the area of the triangle $\triangle X Y Z$ is $\frac{1}{2} b H$. Since the triangles $X Y Z$ and $X A B$ are similar, we have

$$
\frac{l}{b}=\frac{h}{H}=\frac{X A}{X Y}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-078.jpg?height=351&width=788&top_left_y=2280&top_left_x=1022)

Therefore, $b=\frac{H l}{h}$, the area of the triangle $\triangle X Y Z$ is $\frac{1}{2} b H=\frac{H^{2} l}{2 h}=\frac{(h+l)^{2} l}{2 h}$, and the desired ratio is given by

$$
\frac{X A}{X Y}=\frac{h}{H}=\frac{h}{h+l}=\frac{1}{1+\frac{l}{h}}
$$

Thus, it remains to calculate the ratio $l / h$.

Since the ratio between the area $\frac{(h+l)^{2} l}{2 h}$ of the triangle $\triangle X Y Z$ and the area $l^{2}$ of the square is $32 / 7$, we obtain $7(h+l)^{2}=64 h l$. Expanding and dividing by $h^{2}$, we get the quadratic equation

$$
7\left(\frac{l}{h}\right)^{2}-50\left(\frac{l}{h}\right)+7=0
$$

which has solutions

$$
\frac{l}{h}=\frac{50 \pm \sqrt{50^{2}-4 \times 49}}{14}=\frac{25 \pm \sqrt{25^{2}-7^{2}}}{7}=\frac{25 \pm 24}{7}
$$

Thus, $l / h$ has two possible values, $1 / 7$ and 7, and therefore, $\frac{X A}{X Y}$ has two possible values, $\frac{X A}{X Y}=\frac{7}{8}$ and $\frac{X A}{X Y}=\frac{1}{8}$.