## Task 3 - 280923

In a triangle $ABC$, let the side lengths and angle measures be denoted as usual by $a, b, c$ and $\alpha, \beta, \gamma$.

The angle bisector of $\angle BAC$ intersects the side $BC$ at a point $D$. Here, $AD=b$.

Furthermore, it is assumed that one of the three angle measures $\alpha, \beta, \gamma$ is the arithmetic mean of the other two.

Determine all possibilities for the angle measures $\alpha, \beta, \gamma$ under these conditions!
I. If the angle sizes $\alpha, \beta, \gamma$ of a triangle $ABC$ satisfy the conditions, then:

Because $AD=b$, it follows from the base angle theorem that $\angle ADC = \angle ACD = \gamma$. Since $AD$ bisects the angle $\angle BAC$, it follows from the interior angle theorem for $\triangle ACD$ that

$$
\frac{\alpha}{2} = 180^{\circ} - 2\gamma \quad ; \quad \alpha = 360^{\circ} - 4\gamma
$$

From the interior angle theorem for $\triangle ABC$, it follows that

$$
\beta = 180^{\circ} - (360^{\circ} - 4\gamma) - \gamma = 3\gamma - 180^{\circ}
$$

For the condition that one of the three angle sizes $\alpha, \beta, \gamma$ is the arithmetic mean of the other two, there are only the following possibilities:

1. Case: $\alpha = \frac{1}{2}(\beta + \gamma)$

This leads, according to (1), (2), to

$$
360^{\circ} - 4\gamma = 2\gamma - 90^{\circ} ; \quad \gamma = 75^{\circ} ; \quad \alpha = 60^{\circ} ; \quad \beta = 45^{\circ}
$$

2. Case: $\beta = \frac{1}{2}(\alpha + \gamma)$

This leads, according to (1), (2), to

$$
3\gamma - 180^{\circ} = 180^{\circ} - \frac{3}{2}\gamma ; \quad \gamma = 80^{\circ} ; \quad \alpha = 40^{\circ} ; \quad \beta = 60^{\circ}
$$

3. Case: $\gamma = \frac{1}{2}(\alpha + \beta)$

This leads, according to (1), (2), to

$$
\gamma = 90^{\circ} - \frac{\gamma}{2} ; \quad \gamma = 60^{\circ} ; \quad \beta = 0^{\circ}
$$

Thus, the 3rd case is excluded.

II. The angle sizes of the 1st and 2nd cases, i.e.,

$$
\alpha = 60^{\circ} ; \quad \beta = 45^{\circ} ; \quad \gamma = 75^{\circ} \quad \text{(3)} \quad \text{or} \quad \alpha = 40^{\circ} ; \quad \beta = 60^{\circ} ; \quad \gamma = 80^{\circ}
$$

are the interior angle sizes of triangles $ABC$ (since they are positive and satisfy $\alpha + \beta + \gamma = 180^{\circ}$). They also satisfy the condition that $AD = b$ for the angle bisector $AD$; because they satisfy the equation $\frac{\alpha}{2} = 180^{\circ} - 2\gamma$, which, according to the interior angle theorem, implies $\angle ADC = \gamma$ and hence, by the converse of the base angle theorem, $AD = AC$.

With I., II., it is shown that (3), (4) contain all the desired possibilities for $\alpha, \beta, \gamma$.