The angles of a certain triangle form an arithmetic progression. The smallest side is half of the largest one. The angles need to be calculated!
I. Solution. Let $a<b<c$ and thus $\alpha<\beta<\gamma$, so that

$$
\alpha=\beta-x, \quad \gamma=\beta+x, \quad \alpha+\beta+\gamma=3 \beta=180^{\circ}, \quad \beta=60^{\circ}.
$$

Accordingly, $\quad \alpha=60^{\circ}-x, \quad \gamma=60^{\circ}+x=30^{\circ}$. and $a: c=1: 2$.

Using the tangent theorem:

$$
\operatorname{tg} \frac{\gamma+\alpha}{2}: \operatorname{tg} \frac{\gamma-\alpha}{2}=(2+1):(2-1) \quad \text { that is } \quad \operatorname{tg} 60^{\circ}: \operatorname{tg} x=3: 1
$$

Accordingly,

$$
\operatorname{tg} x=\frac{\operatorname{tg} 60^{\circ}}{3}=\frac{\sqrt{3}}{3}, \quad \text { so } \quad x=30^{\circ}.
$$

The angles of the triangle are $30^{\circ}, 60^{\circ}, 90^{\circ}$. (In this case $c=2a$.)

László Grosz (Balassi Bálint g. VI. o. Balassagyarmat).

II. Solution. In the $\Delta ABC$, as we have seen, $\beta=60^{\circ}$. If $AB=2BC$ and $M$ bisects $AB$, then in $\Delta BCM$, $BM=BC$ and the angle between these two sides is $60^{\circ}$; it must be that the angles opposite the $CM$ side are each $60^{\circ}$, so $\Delta BCM$ is equilateral. Since $CM \angle=120^{\circ}$ and $CM=BM=AM$, in $\angle ACM$ the angles opposite the $AC$ side are equal, each $30^{\circ}$.

Accordingly,

$$
\alpha=30^{\circ}, \quad \gamma=60^{\circ}+30^{\circ}=90^{\circ}
$$

Róbert Frisch (Szent István g. V. o. Bp. XIV.)