9.12. (New York, 76). Find at least one right triangle with integer sides, each angle of which can be divided into three equal parts using a compass and straightedge.
9.12. Let's choose an angle $\alpha$ that satisfies the conditions $6 \alpha < 90^{\circ}$ and $\operatorname{tg} \alpha \in Q$ (for example, the value $\operatorname{tg} \alpha = 1 / 4$ is suitable). Then each of the numbers

$$
\begin{aligned}
& \operatorname{tg} 2 \alpha = \frac{2 \operatorname{tg} \alpha}{1 - \operatorname{tg}^{2} \alpha}, \operatorname{tg} 3 \alpha = \frac{\operatorname{tg} 2 \alpha + \operatorname{tg} \alpha}{1 - \operatorname{tg} 2 \alpha \operatorname{tg} \alpha} \\
& \cos 2 \alpha = \frac{1 - \operatorname{tg}^{2} \alpha}{1 + \operatorname{tg}^{2} \alpha}, \sin 2 \alpha = \frac{2 \operatorname{tg} \alpha}{1 + \operatorname{tg}^{2} \alpha} \\
& \cos 6 \alpha = \frac{1 - \operatorname{tg}^{2} 3 \alpha}{1 + \operatorname{tg}^{2} 3 \alpha}, \quad \sin 6 \alpha = \frac{2 \operatorname{tg} 3 \alpha}{1 + \operatorname{tg}^{2} 3 \alpha}
\end{aligned}
$$

is rational. Therefore, for the right triangle $A_{1} B_{1} C_{1}$, satisfying the conditions

$$
\angle C_{1} = 90^{\circ}, \angle A_{1} = 6 \alpha, A_{1} B_{i} = 1, A_{1} C_{1} = \cos 6 \alpha, B_{1} C_{1} = \sin 6 \alpha
$$

the lengths of all sides are rational, hence it is similar to some triangle $A B C$ with integer sides (for example, with $\operatorname{tg} \alpha = 1 / 4$, the sides of triangle $A B C$ can be $A B = 4913$, $A C = 495$, $B C = 4888$). On the other hand, for the right triangle $A_{2} B_{2} C_{2}$, satisfying the conditions

$$
\angle C_{2} = 90^{\circ}, \angle A_{2} = 2 \alpha, A_{2} B_{2} = 1, A_{2} C_{2} = \cos 2 \alpha, B_{2} C_{2} = \sin 2 \alpha
$$

the lengths of all sides are also rational. Therefore, it is possible to construct the angle $2 \alpha = (1 / 3) \angle A_{1}$ using a compass and a ruler, i.e., to divide the angle $A$ of triangle $A B C$ into three equal parts. Since an angle of $30^{\circ}$ can also be constructed and

$$
\frac{1}{3} \angle B = \frac{1}{3}(\angle C - \angle A) = 30^{\circ} - 2 \alpha
$$

each angle of triangle $A B C$ can be divided into three equal parts using a compass and a ruler. Thus, triangle $A B C$ satisfies the requirements of the problem.