5. Find all values of the angle $\alpha$ from the interval $\left[0^{\circ} ; 360^{\circ}\right]$ such that the system

$$
\left\{\begin{array}{l}
y^{3} \geqslant 3 x^{2} y \\
(x-\cos \alpha)^{2}+(y-\sin \alpha)^{2}=\frac{2-\sqrt{3}}{4}
\end{array}\right.
$$

has a unique solution.

(R. Alishev)
# Answer: $\alpha \in\left\{15^{\circ}, 45^{\circ}, 135^{\circ}, 165^{\circ}, 255^{\circ}, 285^{\circ}\right\}$  Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_815d06088b341653082cg-4.jpg?height=1005&width=1228&top_left_y=748&top_left_x=471)

Note that $\sin ^{2} 15^{\circ}=\frac{1-\cos 30^{\circ}}{2}=\frac{2-\sqrt{3}}{4}$. Rewrite our system as

$$
\left\{\begin{array}{l}
y(y-\sqrt{3} x)(y+\sqrt{3} x) \geqslant 0 \\
(x-\cos \alpha)^{2}+(y-\sin \alpha)^{2}=\sin ^{2} 15^{\circ}
\end{array}\right.
$$

The shaded area satisfies the first inequality. Therefore, if we consider an arbitrary point $(\cos \alpha ; \sin \alpha)$ and draw a circle with this point as the center and radius $\sin 15^{\circ}$, the circle must have exactly one common point with the shaded area $\left((x-\cos \alpha)^{2}+(y-\sin \alpha)^{2}-\right.$ the distance from the point $(x, y)$ to the point $(\cos \alpha ; \sin \alpha)$ ).

As can be seen from the graph, this is possible if and only if the circle is tangent to one of the lines $y=0$, $y=\sqrt{3} x$ or $y=-\sqrt{3} x$ and lies entirely outside the shaded area. Since the radius of all considered circles is $\sin 15^{\circ}$, the tangency will occur only for $\alpha \in\left\{15^{\circ}, 45^{\circ}, 135^{\circ}, 165^{\circ}, 255^{\circ}, 285^{\circ}\right\}$.