In a certain triangle, the difference between two sides: $b-c$ is twice the distance from the angle bisector $f_{\alpha}$, which starts from the common vertex $A$ of the two sides, to the foot $M$ of the altitude. What is the angle $\alpha$ formed by the two sides?
I. Solution: The labeling is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_02bb75d715f59873d07eg-1.jpg?height=526&width=442&top_left_y=222&top_left_x=820)

Figure 1

Starting from the equation

$$
2 \cdot M N=b-c
$$

we need to transform this equation so that we can calculate $\alpha$ from it, which can only be achieved if all other elements of the triangle except $\alpha$ are eliminated from the transformed equation, leaving a pure trigonometric equation for $\alpha$. Therefore, it will be useful to aim for a situation where only one (and the same) length appears on both sides of the equation, along with the angles of the triangle (or functions of the angles).

Transformation of the left side: From the right triangle $\mathrm{AMN}$, we have $M N=A M \sin \varphi$, where

$$
\varphi=\frac{\alpha}{2}-\left(90^{\circ}-\beta\right)=\frac{\alpha+2 \beta-180^{\circ}}{2}=\frac{\alpha+2 \beta-(\alpha+\beta+\gamma)}{2}=\frac{\beta-\gamma}{2}
$$

From the right triangle $\mathrm{AMC}_{1}$,

$$
A M=\frac{A C_{1}}{\sin \beta}
$$

and from the right triangle $\mathrm{ACC}_{1}$,

$$
A C_{1}=b \cos \alpha
$$

After substituting these values, the left side of (1) transforms into

$$
2 \cdot M N=\frac{b \cos \alpha}{\sin \beta} \cdot 2 \sin \frac{\beta-\gamma}{2}
$$

We also transform the right side so that only the angles and $b$ appear in it. By the sine rule, $c=\frac{b \sin \gamma}{\sin \beta}$, so

$$
b-c=b-\frac{b \sin \gamma}{\sin \beta}=b \frac{\sin \beta-\sin \gamma}{\sin \beta}
$$

Writing the transformed equation (1) and simplifying by $\frac{b}{\sin \beta}$ (since $b \neq 0, \sin \beta \neq 0$),

$$
\cos \alpha \cdot 2 \sin \frac{\beta-\gamma}{2}=\sin \beta-\sin \gamma
$$

We multiply the left side by $\cos \frac{\beta+\gamma}{2}$ and the right side by the equivalent $\sin \frac{\alpha}{2}$,

$$
\cos \alpha\left(2 \cos \frac{\beta+\gamma}{2} \sin \frac{\beta-\gamma}{2}\right)=(\sin \beta-\sin \gamma) \sin \frac{\alpha}{2}
$$

As is known, the product in the parentheses on the left side is identically equal to $(\sin \beta-\sin \gamma)$. If its value is 0, then $\beta=\gamma$, so the triangle is isosceles, for which (1) holds but is trivial. If $\sin \beta-\sin \gamma \neq 0$, then simplifying by $(\sin \beta-\sin \gamma)$, we get the equation

$$
\cos \alpha=\sin \frac{\alpha}{2}
$$

Substituting $\cos \alpha$ with $1-2 \sin ^{2} \frac{\alpha}{2}$ and rearranging,

$$
2 \sin ^{2} \frac{\alpha}{2}+\sin \frac{\alpha}{2}-1=0
$$

from which

$$
\sin \frac{\alpha}{2}=\frac{-1 \pm \sqrt{1+8}}{4}
$$

According to the problem, only the positive root makes sense, so $\sin \frac{\alpha}{2}=\frac{1}{2}$, from which $\left(\frac{\alpha}{2}0\right)$, then

$$
0<\frac{\sqrt{2 k^{2}+1}-1}{2 k}<\frac{\sqrt{2}}{2}, \quad \text { and thus } \quad 0<\frac{\alpha}{2}<\frac{\pi}{4}
$$