1. A cube with edge $a=\sqrt{2+\sqrt{3}}$ is illuminated by a cylindrical beam of light with radius $\rho=\sqrt{2}$, directed along the main diagonal of the cube (the axis of the beam contains the main diagonal). Find the area of the illuminated part of the cube's surface.
Answer: $\sqrt{3}(\pi+3)$.

Solution. Using the scalar product, it is easy to show that the cosine of the angle between the main diagonal (the axis of the beam) and any edge of the cube is $1 / \sqrt{3}$. The same value is the cosine of the angle between the axis of the beam and the normal to any face of the cube.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd5278a2a8e19ad9915eg-4.jpg?height=825&width=817&top_left_y=796&top_left_x=1807)

Let's draw the projection of the cube onto a plane perpendicular to the axis of the beam. The area of the projection on this plane of any planar figure located on any face of the cube is equal to the area of the figure itself, multiplied by the cosine of the angle between the face and the plane of projection, i.e., by $1 / \sqrt{3}$. Therefore, by calculating the area of the projection of the illuminated part of the cube and multiplying it by $\sqrt{3}$, we obtain the required answer.

The length of the projection of any edge is the product of the edge length $a$ by the cosine of the angle between the edge and the plane of projection or by the sine of the angle between the edge and the axis of the beam, i.e., $a \sqrt{2 / 3}$.

In the case where the radius of the beam $\rho$ does not exceed the radius $r$ of the circle inscribed in the depicted regular hexagon, the projection of the illuminated part has an area equal to $\pi \rho^{2}$. The radius of the inscribed circle in the hexagon is $r=a \sqrt{2 / 3} \cdot \sqrt{3} / 2=a / \sqrt{2}$. Thus, when $\rho \leqslant a / \sqrt{2}$, the area of the illuminated part is $\pi \rho^{2} \sqrt{3}$. If the radius $\rho$ is greater than or equal to the radius $R$ of the circle circumscribed around the hexagon, then three faces of the cube are fully illuminated, i.e., when $\rho \geqslant R=a \sqrt{2 / 3}$, the area of the illuminated part is $3 a^{2}$.

Consider the case $a / \sqrt{2}<\rho<a \sqrt{2 / 3}$. The area of the projection of the illuminated part is obtained by subtracting from $\pi \rho^{2}$ the areas of six segments that extend beyond the hexagon. The area of each such segment is the difference between the areas of the corresponding sector and triangle. The angle of the sector is $2 \arccos a / \rho \sqrt{2}$. Therefore, the area of the segment is

$$
\rho^{2} \arccos \frac{a}{\rho \sqrt{2}}-\frac{1}{2} \cdot 2 \sqrt{\rho^{2}-\frac{a^{2}}{2}} \cdot \frac{a}{\sqrt{2}}=\rho^{2}\left(\arccos q-q \sqrt{1-q^{2}}\right)
$$

where $q=a / \rho \sqrt{2} \in(\sqrt{3} / 2 ; 1)$, and the area of the illuminated part is

$$
S=\rho^{2} \sqrt{3}\left(\pi-6 \arccos q+6 q \sqrt{1-q^{2}}\right)
$$

Note that the relation $q=a / \rho \sqrt{2} \in(\sqrt{3} / 2 ; 1)$ is equivalent to the inequality $a / \sqrt{2}<\rho<a \sqrt{2 / 3}$, which defines the third case considered.

According to the condition, we have

$$
\begin{gathered}
q=\frac{a}{\rho \sqrt{2}}=\sqrt{\frac{2+\sqrt{3}}{2 \cdot 2}}=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}=\cos \frac{\pi}{12} \in\left(\frac{\sqrt{3}}{2} ; 1\right) \\
q \sqrt{1-q^{2}}=\sqrt{\frac{2+\sqrt{3}}{4}} \cdot \sqrt{1-\frac{2+\sqrt{3}}{4}}=\frac{1}{4} \\
S=2 \sqrt{3}\left(\pi-\frac{\pi}{2}+\frac{6}{4}\right)=\sqrt{3}(\pi+3)
\end{gathered}
$$