Problem 12.4. The point $K$ on the edge $A B$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ is such that the angle between the line $A_{1} B$ and the plane $\left(B_{1} C K\right)$ is equal to $60^{\circ}$. Find $\tan \alpha$, where $\alpha$ is the angle between the planes $\left(B_{1} C K\right)$ and $(A B C)$.

Oleg Mushkarov
12.4. We may assume that the edges of the cube have length 1 . Denote $M=A_{1} B \cap K B_{1}$ and set $K B=x$. Since $\triangle K B M \sim \triangle A_{1} M B_{1}$ it follows that $\frac{M B}{A_{1} M}=x$. Now using the identity $A_{1} M+M B=\sqrt{2}$, we get $M B=\frac{x \sqrt{2}}{x+1}$. Denote by $V$ the volume of the tetrahedron $K B C B_{1}$. Then

$$
V=\frac{1}{3} B B_{1} \cdot S_{K B C}=\frac{x}{6}
$$

On the other hand, the altitude of $K B C B_{1}$

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-155.jpg?height=630&width=596&top_left_y=141&top_left_x=1115)

This and (1) imply that

$$
\frac{x}{6}=\frac{x \sqrt{6\left(2 x^{2}+1\right)}}{12(x+1)}
$$

and we obtain easily that $x=\frac{1}{2}$.

Denote by $L$ the foot of the perpendicular from $B$ to $K C$. Then $K C \perp B L$ and $K C \perp B B_{1}$ which shows that $K C \perp B_{1} L$. Hence $\alpha=\Varangle B_{1} L B$. We get from $\triangle K B C$ that $B L=\frac{1}{\sqrt{5}}$ and therefore $\tan \alpha=\sqrt{5}$.