13.265. A material point moves straight-line within the interior of a $60^{\circ}$ angle. After leaving the vertex of this angle, it found itself at a distance $a$ from one side of the angle and at a distance $b$ from the other side after some time. Then it changed its direction of motion and fell onto the side to which it was closer by the shortest path. Find the length of the path traveled by the point if $a < b$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1178.jpg?height=474&width=572&top_left_y=449&top_left_x=687)

Fig. 13.8
## Solution.

The length of the path traveled by the point is equal to the sum of the lengths of segments $O A$ and $A B=a$ (Fig. 13.8). We obtain

$$
\begin{aligned}
& O A=\frac{a}{\sin \alpha} \text { or } O A=\frac{b}{\sin \left(60^{\circ}-\alpha\right)} \Rightarrow \frac{\sin \left(60^{\circ}-\alpha\right)}{\sin \alpha}=\frac{b}{a} \Leftrightarrow \\
& \Leftrightarrow \frac{\sqrt{3}}{2} \operatorname{ctg} \alpha-\frac{1}{2}=\frac{b}{a} \Leftrightarrow \operatorname{ctg} \alpha=\frac{2 b+a}{a \sqrt{3}}
\end{aligned}
$$

Since $\sin ^{2} \alpha=\frac{1}{1+\operatorname{ctg}^{2} \alpha}$, then $\sin \alpha=\frac{\sqrt{3}}{2} \cdot \frac{a}{\sqrt{a^{2}+a b+b^{2}}}$.

Thus, the desired length is

$$
a+\frac{a}{\sin \alpha}=a+2 \sqrt{\frac{a^{2}+a b+b^{2}}{3}}
$$

Answer: $a+2 \sqrt{\frac{a^{2}+a b+b^{2}}{3}}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1179.jpg?height=414&width=638&top_left_y=84&top_left_x=93)

Fig. 13.9