In an equilateral triangle $ABC$, the intersection points of a line $f$ passing through the center $O$ with the side lines are denoted by $A_{1}, B_{1}, C_{1}$. What angle does $f$ form with the side lines if the product $O A_{1} \cdot O B_{1} \cdot O C_{1}$ is minimal?

What are the angles if the product is twice the minimum value?
On the one hand, the regular triangle, and on the other hand, the symmetry of the questioned product with a line $f$ results in the same value of the product for its symmetric images. We can avoid repetitions with the following agreement: let the points on the perimeter of the triangle on $f$ be $B_{1}$ and $C_{1}$, and let $A_{1}$ be on the extension of the side $BC$ beyond $B$, or possibly at $B$ itself.

![](https://cdn.mathpix.com/cropped/2024_05_02_62e5bc7453504e5147b1g-1.jpg?height=597&width=979&top_left_y=344&top_left_x=562)

Let the angle $\angle O A_{1} C$ be denoted by $\varphi$ (where $0 < \varphi \leq 30^{\circ}$). Applying the exterior angle theorem to the triangle $B_{1} A_{1} C$, we get $\angle O B_{1} A = \varphi + 60^{\circ}$, $\angle O C_{1} B = \varphi + 120^{\circ}$, and $\angle O C_{1} A = 60^{\circ} - \varphi$. In the right triangle $O A_{1} A_{0}$, the leg $O A_{0}$ opposite to $\varphi$ is $\varrho$, the radius of the inscribed circle. Thus, and similarly,

$$
O A_{1} \cdot O B_{1} \cdot O C_{1} = \frac{\varrho^{3}}{\sin \varphi \sin \left(60^{\circ} + \varphi\right) \sin \left(60^{\circ} - \varphi\right)}
$$

Therefore, the product is minimal for the value of $\varphi$ that maximizes the product in the denominator. Now,

$$
\begin{aligned}
\sin \varphi \{ & \left\{\sin \left(60^{\circ} + \varphi\right) \cdot \sin \left(60^{\circ} - \varphi\right)\right\} = \frac{\sin \varphi}{2} \left\{\cos \left(60^{\circ} + \varphi - (60^{\circ} - \varphi)\right) - \right. \\
\left. - \cos \left(60^{\circ} + \varphi + 60^{\circ} - \varphi\right)\right\} = & \frac{\sin \varphi}{2} \left(\cos 2 \varphi - \cos 120^{\circ}\right) = \\
= & \frac{\sin \varphi}{2} \left(1 - 2 \sin^2 \varphi + \frac{1}{2}\right) = \left(3 \sin \varphi - 4 \sin^3 \varphi\right) = \frac{1}{4} \sin 3 \varphi
\end{aligned}
$$

This is clearly maximal when $\sin 3 \varphi = 1$, $3 \varphi = 90^{\circ}$, $\varphi = 30^{\circ}$. In this case, $A_{1}$ coincides with $B$, $f$ coincides with the symmetry axis passing through here, and the angles are $30^{\circ}$, $30^{\circ}$, and $90^{\circ}$.

The product is twice the minimal value when the denominator is half as large, i.e., $\sin 3 \varphi = \frac{1}{2}$, $\varphi = 10^{\circ}$, and the other two angles are $70^{\circ}$ and $50^{\circ}$.