One side of a triangle is one unit, the angles lying on it are acute, and the ratio of their sines is $1: 2$. The area of the triangle is $1 / 4$. What are the other two sides?
Based on the sine theorem, according to the problem statement,

$$
\frac{\sin \alpha}{\sin \beta}=\frac{a}{b}=\frac{1}{2}, \quad \text { from which } \quad b=2 a . \quad \text { (see the labels in the figure.) }
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_84d4db7c90d0e00ab58bg-1.jpg?height=220&width=499&top_left_y=340&top_left_x=802)

From the known area formula of the triangle,

$$
\frac{1}{4}=\frac{a b \sin \gamma}{2}
$$

which means

$$
4 a^{2} \sin \gamma=1
$$

We apply the cosine theorem to the angle $\gamma$:

$$
\begin{gathered}
1^{2}=a^{2}+4 a^{2}-4 a^{2} \cos \gamma \\
\cos \gamma=\frac{5 a^{2}-1}{4 a^{2}}
\end{gathered}
$$

and from the Pythagorean identity,

$$
\sin \gamma=\sqrt{1-\left(\frac{5 a^{2}-1}{4 a^{2}}\right)^{2}}=\frac{1}{4 a^{2}} \sqrt{-9 a^{4}+10 a^{2}-1}
$$

Substituting this into (1),

$$
\sqrt{-9 a^{4}+10 a^{2}-1}=1
$$

Squaring and rearranging,

$$
9 a^{4}-10 a^{2}+2=0
$$

from which

$$
a_{1}=0.9217, \quad a_{2}=0.5115 \quad \text { and } \quad b_{1}=1.8434, \quad b_{2}=1.0230
$$

If we apply the cosine theorem to the angles obtained, in the first case, the angle $\beta$ is greater than $90^{\circ}$. This is therefore not a solution to our problem. However, it is easy to verify that in the second case, all angles of the triangle are acute.

Note. Most solvers calculated the missing sides using Heron's formula. Naturally, they also received 5 points, provided they did not make any calculation errors.