In triangle $ABC$, side $AC$ is one unit. The median from vertex $A$ divides the angle at $A$ in the ratio $1:2$. What can the length of side $AB$ be?
I. Solution. Let $F$ be the midpoint of side $BC$, then $AF$ is the median of triangle $ABC$. Let $\angle CAB = \varepsilon + \varphi$, where $\angle CAF = \varepsilon$ and $\angle FAB = \varphi$. Since the areas of triangles $CAF$ and $FAB$ are equal, the following holds:

$$
t = \frac{AF \cdot AC \cdot \sin \varepsilon}{2} = \frac{AB \cdot AF \cdot \sin \varphi}{2}
$$

Given $AC = 1$, it follows that $AB = \frac{\sin \varepsilon}{\sin \varphi}$.

In our case, the median divides the angle $\angle CAB$ in the ratio $1:2$, so two cases are possible:
a) $\varphi = 2 \varepsilon$ or \quad b) $\varepsilon = 2 \varphi$.

In case a), $AB = \frac{\sin \varepsilon}{\sin 2 \varepsilon} = \frac{\sin \varepsilon}{2 \cos \varepsilon \cdot \sin \varepsilon} = \frac{1}{2 \cos \varepsilon}$, and since $0^\circ < \varepsilon < 60^\circ$, it follows that $AB > \frac{1}{2}$.

In case b), $AB = \frac{\sin 2 \varphi}{\sin \varphi} = \frac{2 \sin \varphi \cos \varphi}{\sin \varphi} = 2 \cos \varphi$, and since $0^\circ < \varphi < 60^\circ$, it follows that $1 < AB < 2$.

In case a), $AB_1 = \frac{1}{2 \cos \varepsilon}$, and since $0^\circ < \varepsilon < 60^\circ$, it follows that $AB_1 > \frac{1}{2}$. Since $\angle AB_1C_1 = 180^\circ - 2 \delta > 60^\circ$ (where $\delta < 60^\circ$), and by the triangle inequality, $AB_1 > \frac{1}{2}$.

In case b), $AB_2 = 2 \cos \varphi$, and since $0^\circ < \varphi < 60^\circ$, it follows that $1 < AB_2 < 2$. Since $AB_2 > AC = 1$ by the triangle inequality, and $\angle AC_2B_2 = 180^\circ - 2 \delta > 60^\circ$, it follows that the larger side must be opposite the larger angle, so $AB_2 > AC = 1$, hence $1 < AB_2 < 2$.

Therefore, in the first case, the length of side $AB$ is between $\frac{1}{2}$ and 1, and in the second case, it is between 1 and 2, and it can take any value in these ranges, as a triangle can be constructed for any such value. (Note, $AB \neq 1!$)

Based on the work of Gergő Takács (Tatabánya, Árpád Gymnasium and Pedagogical College, 10th grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_98b98cd0f95f213c67d2g-1.jpg?height=474&width=339&top_left_y=1823&top_left_x=882)

![](https://cdn.mathpix.com/cropped/2024_05_02_98b98cd0f95f213c67d2g-2.jpg?height=588&width=911&top_left_y=178&top_left_x=604)