Renata constructed a trapezoid $P R S T$ with bases $P R$ and $S T$, in which the following conditions are simultaneously satisfied:

- trapezoid $P R S T$ is not a right trapezoid;
- triangle $T R P$ is equilateral;
- triangle $T R S$ is right-angled;
- one of the triangles $T R S, T R P$ has an area of $10 \mathrm{~cm}^{2}$.

Determine the area of the other of these two triangles. Find all possible solutions.

(M. Petrová)
From the second condition, we know that triangle $T R P$ is equilateral, so all its internal angles are $60^{\circ}$. Angles $T R P$ and $S T R$ are alternate, so the size of angle $S T R$ is also $60^{\circ}$.

From the third condition, we know that triangle $T R S$ is a right triangle. From the previous paragraph, we know that the right angle cannot be at vertex $T$, and from the first condition, we know that the right angle cannot be at vertex $S$ either. Therefore, the right angle is at vertex $R$. Further, let $K$ be the midpoint of the hypotenuse $S T$ of triangle $T R S$. Since this triangle is a right triangle, vertex $R$ lies on a circle with center $K$ and radius $|K S|=|K T|$. Therefore,

$$
|K S|=|K T|=|K R| .
$$

Triangle $T R K$ is thus isosceles with base $R T$. Additionally, from the previous paragraph, we know that angle $R T K$ has a size of $60^{\circ}$, so the second angle at the base also has a size of $60^{\circ}$. Triangle $T R K$ is therefore equilateral and congruent to the equilateral triangle $T R P$.

![](https://cdn.mathpix.com/cropped/2024_04_17_a582478f85ae48908ebfg-2.jpg?height=406&width=939&top_left_y=628&top_left_x=564)

Triangles $T R P$ and $T R K$ have the same area because they are congruent. Triangles $T R K$ and $S R K$ have the same area because sides $K T$ and $K S$ are of equal length and the height to these sides is common. This means that triangle $T R S$ has twice the area of triangle $T R P$,

$$
S_{T R S}=2 S_{T R P}
$$

From the fourth condition, we know that the area of one of these two triangles is $10 \mathrm{~cm}^{2}$:

- If $S_{T R S}=10 \mathrm{~cm}^{2}$, then $S_{T R P}=5 \mathrm{~cm}^{2}$.
- If $S_{T R P}=10 \mathrm{~cm}^{2}$, then $S_{T R S}=20 \mathrm{~cm}^{2}$.

Another solution. In the same way as in the previous solution, we determine the internal angles of triangle $T R S$. Point $T$ is reflected in the axis $R S$, and the symmetric point is denoted as $I$. All internal angles of triangle $T I S$ are $60^{\circ}$, so the triangle is equilateral.

![](https://cdn.mathpix.com/cropped/2024_04_17_a582478f85ae48908ebfg-2.jpg?height=722&width=945&top_left_y=1712&top_left_x=561)

Side $T I$ is twice the length of side $T R$, so triangles $T R P$ and $T I S$ are similar with a similarity ratio of $1: 2$. Therefore, their areas are in the ratio $1: 4$,

$$
S_{T I S}=4 S_{T R P}
$$

Triangle $T R S$ forms half of triangle $T I S$, so its area is twice the area of triangle $T R P$. Thus, we arrive at the relationship (1) and conclude the problem in the same way as in the previous solution.

Evaluation. 1 point for determining the sizes of the internal angles $R T S$ and $T R S$; 3 points for justifying the equality (1); 1 point for the final areas of $5 \mathrm{~cm}^{2}$ and $20 \mathrm{~cm}^{2}$.