Task B-4.5. (20 points) The cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is intersected by a plane that contains the spatial diagonal of the cube $\overline{B D^{\prime}}$ and has the smallest possible area of intersection with the cube. What is the cosine of the angle between that plane and the plane $A B C D$?
Solution. Let $a$ be the length of the edge of the cube. The intersection of the cube and the plane is a parallelogram $B E D^{\prime} F$ whose one diagonal is the spatial diagonal $\overline{B D^{\prime}}$ of the cube. The area of the parallelogram is $P=\left|D^{\prime} E\right| \cdot|B E| \cdot \sin \alpha$, where $\alpha=\angle D^{\prime} E B$.

![](https://cdn.mathpix.com/cropped/2024_05_30_978576e245d98ac92b08g-21.jpg?height=607&width=607&top_left_y=1599&top_left_x=609)

Let $x=\left|C^{\prime} E\right|$. Then $\left|D^{\prime} E\right|=\sqrt{a^{2}+x^{2}},|B E|=\sqrt{a^{2}+(a-x)^{2}}$.

Using the cosine rule for triangle $D^{\prime} B E$, we have

$$
\left|B D^{\prime}\right|^{2}=\left|D^{\prime} E\right|^{2}+|B E|^{2}-2 \cdot\left|D^{\prime} E\right| \cdot|B E| \cos \alpha
$$

Substituting $\left|B D^{\prime}\right|=a \sqrt{3}$ and the expressions for $\left|D^{\prime} E\right|,|B E|$, after simplification, we get

$$
\cos \alpha=\frac{x^{2}-a x}{\sqrt{\left(a^{2}+x^{2}\right)\left(a^{2}+(a-x)^{2}\right)}}
$$

From $\sin \alpha=\sqrt{1-\cos ^{2} \alpha}$, we obtain

$$
\sin \alpha=\frac{a \sqrt{2 x^{2}-2 a x+2 a^{2}}}{\sqrt{\left(a^{2}+x^{2}\right)\left(a^{2}+(a-x)^{2}\right)}}
$$

Substituting into the formula for the area of the parallelogram, we get $P=a \sqrt{2 x^{2}-2 a x+2 a^{2}}$.

Since the expression under the square root is a quadratic function, it achieves its minimum at $x=\frac{-(-2 a)}{2 \cdot 2}=\frac{a}{2}$

and the minimum value is $P_{\min }=\frac{a^{2} \sqrt{6}}{2}$.

The desired angle is $\angle D B D^{\prime}$. The lengths of the sides of the right triangle $D B D^{\prime}$ are $\left|D D^{\prime}\right|=a$, $|B D|=a \sqrt{2}$, and $\left|B D^{\prime}\right|=a \sqrt{3}$, so

$$
\cos \varphi=\frac{|B D|}{\left|B D^{\prime}\right|}=\frac{\sqrt{6}}{3}
$$

Note. $\cos \varphi$ can also be calculated using the formula

$$
\cos \varphi=\frac{P_{A B C D}}{P_{\min }}=\frac{a^{2}}{\frac{a^{2} \sqrt{6}}{2}}=\frac{\sqrt{6}}{3}
$$