B2. Two angles of a triangle are in the ratio 1:2. The third interior angle is equal to the arithmetic mean of the first two. Determine the sizes of the interior angles of this triangle.
B2. From the ratio $\alpha: \beta=1$ : 2 we express $\alpha=t$ and $\beta=2 t$. We consider that the third angle is equal to the arithmetic mean: $\gamma=\frac{1}{2}(\alpha+\beta)=\frac{3 t}{2}$. Since the sum of the interior angles of a triangle is $180^{\circ}$, we write the equation $t+2 t+\frac{3 t}{2}=180^{\circ}$. The solution to the equation is $t=40^{\circ}$. Thus, $\alpha=40^{\circ}, \beta=80^{\circ}$ and $\gamma=60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_bce9c6e2a6f4f88bb0deg-12.jpg?height=58&width=1628&top_left_y=1927&top_left_x=277)

Considering the arithmetic mean and writing $\gamma=\frac{1}{2}(\alpha+\beta)=\frac{3 t}{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots$ point

Considering the sum of the interior angles of the triangle $t+2 t+\frac{3 t}{2}=180^{\circ} \ldots \ldots \ldots \ldots \ldots \ldots . .1$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_bce9c6e2a6f4f88bb0deg-12.jpg?height=51&width=1639&top_left_y=2079&top_left_x=274)

Solutions $\alpha=40^{\circ}, \beta=80^{\circ}, \gamma=60^{\circ} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$