11.5. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with edge length a. The ends of a segment forming an angle of $60^{\circ}$ with the plane of the face $A B C D$ lie on the lines $A B_{1}$ and $B C_{1}$. What is the minimum length that this segment can have?

## § 2. Area and Volume
11.5. Let's introduce a coordinate system, directing the axes $O x, O_{y}$, and $O_{z}$ along the rays $B C, B A$, and $B B_{\mathbf{i}}$ respectively. Suppose point $M$ on the line $B C_{1}$ has coordinates $(x, 0, x)$, and point $N$ on the line $B_{1} A$ has coordinates $(0, y, a-y)$. Then the square of the length of segment $M N$ is $x^{2}+y^{2}+(a-x-y)^{2}$, and the square of the length of its projection $M_{1} N_{i}$ on the plane of face $A B C D$ is $x^{2}+y^{2}$. Since the angle between the lines $M N$ and $M_{1} N_{\mathrm{i}}$ is $60^{\circ}$, we have $M N=2 M_{1} N_{1}$, i.e., $(a-x-y)^{2}=3\left(x^{2}+y^{2}\right)$.

Let $u^{2}=x^{2}+y^{2}$ and $v=x+y$. Then $M N=2 M_{1} N_{i} = 2 u$. Furthermore, $(a-v)^{2}=3 u^{2}$ by the condition and $2 u^{2} \geqslant v^{2}$. Therefore, $(a-v)^{2} \geqslant 3 v^{2} / 2$, which means $v \leqslant a(\sqrt{6}-2)$. Hence, $u^{2}=(a-t)^{2} / 3 \geqslant a^{2}(3-\sqrt{6})^{2 / 3}=a^{2}(\sqrt{3}-\sqrt{2})^{2}$, i.e., $M N \geqslant 2 a(\sqrt{3}-\sqrt{2})$. Equality is achieved when $x = y = a(\sqrt{6}-2) / 2$.