The tangents of the angles of a triangle are in the ratio $1: 2: 3$. How do the sides of the triangle relate to each other?
I. Solution: Let $\operatorname{tg} \alpha=\lambda, \operatorname{tg} \beta=2 \lambda, \operatorname{tg} \gamma=3 \lambda$.

$$
\operatorname{tg} \gamma=\operatorname{tg}\left[180^{\circ}-(\alpha+\beta)\right]=-\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}
$$

thus

$$
3 \lambda=-\frac{\lambda+2 \lambda}{1-\lambda \cdot 2 \lambda}=-\frac{3 \lambda}{1-2 \lambda^{2}}
$$

from which

$$
1-2 \lambda^{2}=-1
$$

hence

$$
\lambda^{2}=1
$$

Since $\lambda$ cannot be negative (otherwise all angles of the triangle would be greater than a right angle), therefore $\lambda=1$, and thus

$$
\operatorname{tg} \alpha=1, \quad \operatorname{tg} \beta=2, \quad \operatorname{tg} \gamma=3
$$

Therefore

$$
\begin{gathered}
a: b: c=\sin \alpha: \sin \beta: \sin \gamma=\frac{\operatorname{tg} \alpha}{\sqrt{1+\operatorname{tg}^{2} \alpha}}: \frac{\operatorname{tg} \beta}{\sqrt{1+\operatorname{tg}^{2} \beta}}: \frac{\operatorname{tg} \gamma}{\sqrt{1+\operatorname{tg}^{2} \gamma}}= \\
=\frac{1}{\sqrt{2}}: \frac{1}{\sqrt{5}}: \frac{1}{\sqrt{10}}=\sqrt{5}: 2 \sqrt{2}: 3
\end{gathered}
$$

Zawadowski Alfréd (Bp., I. Petőfi g. III. o. t.)

II. Solution: Based on the first solution, $\operatorname{tg} \alpha=1, \operatorname{tg} \beta=2, \operatorname{tg} \gamma=3$.

We construct a triangle where the height corresponding to side $b$ is $B B_{1}=m_{b}=3$ (see the first figure) and the two segments on side $b$ are $A B_{1}=b_{1}=3$ and $B_{1} C=b_{2}=1$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9e1edda02492b7fa9ac9g-1.jpg?height=336&width=393&top_left_y=1315&top_left_x=844)

In this triangle, indeed $\operatorname{tg} \alpha=1, \operatorname{tg} \gamma=3$ and $\operatorname{tg} \beta=-\frac{1+3}{1-3}=2$.

The sides of this triangle are

$$
\begin{aligned}
& a=\sqrt{m_{b}^{2}+b_{2}^{2}}=\sqrt{9+1}=\sqrt{10} \\
& b=b_{1}+b_{2}=4
\end{aligned}
$$

and

$$
c=\sqrt{m_{b}^{2}+b_{1}^{2}}=\sqrt{9+9}=\sqrt{18}
$$

Therefore

$$
a: b: c=\sqrt{10}: \sqrt{16}: \sqrt{18}=\sqrt{5}: \sqrt{8}: \sqrt{9}=\sqrt{5}: 2 \sqrt{2}: 3
$$

Lázár László (Mezőkövesd, I. László g. III. o. t.)