## Task 10/76

We are looking for all triangles $ABC$ with sides $a, b, c$ and angles $\alpha, \beta, \gamma$ (usual notation), that satisfy the following conditions:

1. It is $a=9$ length units, $b$ and $c$ are integer multiples of the length unit.
2. It is $\beta=2 \alpha$
From $\beta=2 \alpha$ it follows that $\gamma=\pi-3 \alpha$, so $0<\alpha<\frac{\pi}{3}$. Since $a=3 \mathrm{LE}$, we have $b=6 \sin \alpha$ and $c=6 \cos \alpha \cdot 3-3=18 \cos \alpha-3$. Since $0<\alpha<\frac{\pi}{3}$, we have $0<\cos \alpha<\frac{1}{2}$, and thus $-3<18 \cos \alpha-3<6$, so $c>0$. For $c$ to be an integer, it is necessary and sufficient that $6 \cos \alpha$ is an integer, i.e., $\cos \alpha=\frac{p}{6}$ with $3<p<6$, $p$ an integer. Then $c=\left(p^{2}-9\right)$ units of length (LE), and also $b=3 p$ becomes an integer. Therefore, there are exactly 2 solutions:

$$
\begin{array}{ccc}
p_{1}=4 ; & \cos \alpha_{1}=\frac{4}{6} ; \quad b_{1}=12 \mathrm{LE} ; & c_{1}=7 \mathrm{LE} \\
p_{2}=5 ; & \cos \alpha_{2}=\frac{5}{6} ; \quad b_{2}=15 \mathrm{LE} ; \quad c_{2}=16 \mathrm{LE}
\end{array}
$$