Problem 1. Consider an equilateral triangle with side length 1 and center $O$, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_04_26_012fb179f3572dc85143g-1.jpg?height=317&width=354&top_left_y=247&top_left_x=885)

A ray starts from $O$ and reflects off the three sides, $\overline{A B}, \overline{A C}$, and $\overline{B C}$, (in the given order), until it reaches vertex $A$.

Determine the minimum length of the ray's path.

Note: When the ray reflects off a side, the angles of incidence and reflection are equal.
Solution. Since the ray reflects off the indicated sides, it is sufficient to unfold the path taken by the ray. For this, we unfold the triangle according to the following figure.

![](https://cdn.mathpix.com/cropped/2024_04_26_012fb179f3572dc85143g-1.jpg?height=703&width=488&top_left_y=1085&top_left_x=887)

This figure indicates that there is a unique path from point $O$ to point $A$, reflecting off the sides of the triangle in the indicated order. To calculate the distance traveled by the ray, it is sufficient to consider the triangle $A P O^{\prime}$; it is a right triangle whose hypotenuse $A O^{\prime}$ we need to calculate. We know that $O^{\prime} P$ is equal to $\frac{1}{2}$. The distance $P A$ is $1+1+\frac{1}{3}=\frac{7}{3}$ of the height $h=\frac{\sqrt{3}}{2}$ of the triangle. In this case, we have:

$$
\left(A O^{\prime}\right)^{2}=\left(O^{\prime} P\right)^{2}+(P A)^{2}=\left(\frac{1}{2}\right)^{2}+\left(\frac{7}{3} \frac{\sqrt{3}}{2}\right)^{2}=\frac{1}{4}+\frac{7^{2} \times 3}{3^{2} \times 2^{2}}=\frac{1}{4}+\frac{49}{12}=\frac{13}{3}
$$

Therefore, the distance $A O^{\prime}$ is: $\sqrt{\frac{13}{3}}=\frac{\sqrt{39}}{3}$.