In the Egyptian triangle, whose sides are 3, 4, and 5 units long, inscribe a rectangle whose vertices lie on the sides of the triangle and the ratio of its sides is $1: 3$. Determine the lengths of the sides of the rectangle.
Solution. The given rectangular shape can be placed in the triangle in several ways. The simplest case is when the sides of the rectangle are parallel to the legs of the triangle:
![](https://cdn.mathpix.com/cropped/2024_05_02_30539c7daf111229d35fg-1.jpg?height=648&width=378&top_left_y=266&top_left_x=861)

The vertices of the triangle are $A, B$, and $C$. $AC=4, BC=3, AB=5$ units. The vertices of the rectangle are $C, D, E$, and $F$ as shown in the figure. From the similarity of the right triangles $ABC$ and $AED$, in the first case $\frac{4-3x}{4}=\frac{x}{3}$, and from this $12-9x=4x$, $x=\frac{12}{13}$ and $3x=\frac{36}{13}$.

In the second case, $\frac{4-x}{4}=\frac{3x}{3}; x=\frac{4}{5}$ and $3x=\frac{12}{5}$.

Another possibility is when one side of the rectangle is perpendicular to the hypotenuse:
![](https://cdn.mathpix.com/cropped/2024_05_02_30539c7daf111229d35fg-1.jpg?height=646&width=374&top_left_y=1300&top_left_x=862)

The right triangles formed are all similar because their corresponding angles are equal. In the first case: $\frac{v}{x}=\frac{5}{3}$, $v=\frac{5}{3}x; \quad \frac{4-v}{3x}=\frac{4}{5}, 4-v=\frac{12}{5}x;$ therefore

$$
\frac{5}{3}x+\frac{12}{5}x=4, \quad \text{so} \quad x=\frac{60}{61}, \quad 3x=\frac{180}{61}
$$

In the second case: $\frac{v}{3x}=\frac{5}{3}, v=5x; \frac{4-v}{x}=\frac{4}{5}, 4-v=\frac{4}{5}x$; therefore

$$
5x+\frac{4}{5}x=4, \quad \text{so} \quad x=\frac{20}{29}, \quad 3x=\frac{60}{29}.
$$