Problem 3. Inside the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, consider the regular quadrilateral pyramid $S A B C D$ with base $A B C D$, such that $\angle\left(\left(S A^{\prime} B^{\prime}\right),\left(S C^{\prime} D^{\prime}\right)\right)=30^{\circ}$. Let point $M$ be on the edge $A^{\prime} D^{\prime}$ such that $\angle A^{\prime} B^{\prime} M=30^{\circ}$.

a) Find the angle between the apothem of the pyramid $S A B C D$ and the plane $(A B C)$.

b) Determine the tangent of the angle between the planes $(MAB')$ and $(S A B)$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_06_07_61d9b6d719f0f01da53ag-3.jpg?height=662&width=808&top_left_y=1859&top_left_x=669)
a) The intersection between the surface of the cube and the plane parallel to $\left(A A^{\prime} D^{\prime} D\right)$ that contains point $S$ is the square $U U^{\prime} V^{\prime} V$, where $U \in A B$ and $V \in C D$. Clearly, $\Varangle\left(\left(S A^{\prime} B^{\prime}\right),\left(S C^{\prime} D^{\prime}\right)\right)=$ $\Varangle\left(U^{\prime} S, V^{\prime} S\right)$, consequently $\Varangle U^{\prime} S V^{\prime}=30^{\circ}$ or $\Varangle U^{\prime} S V^{\prime}=150^{\circ}$. Point $S$ is inside the square $U U^{\prime} V^{\prime} V$, so $\Varangle U^{\prime} S V^{\prime}$ is inside the circle circumscribed to the square. Therefore, we get $\Varangle U^{\prime} S V^{\prime}>\frac{1}{2} \overparen{U^{\prime} V^{\prime}}=45^{\circ}>30^{\circ}$, which implies that $\Varangle U^{\prime} S V^{\prime}=150^{\circ}$.

$1 p$

It is easy to show that triangle $U S V$ is equilateral (we choose point $S^{\prime}$ inside the square $U U^{\prime} V^{\prime} V$ such that triangle $U V S^{\prime}$ is equilateral, and we show that points $S$ and $S^{\prime}$ coincide), hence $\Varangle(S U,(A B C D))=\Varangle S U V=60^{\circ}$. $1 p$

b) Since triangles $M A^{\prime} A$ and $M A^{\prime} B^{\prime}$ are congruent (S.S.), we deduce that $\Varangle M A A^{\prime}=30^{\circ}$, so $\Varangle M A D=60^{\circ}$. Therefore, $\Varangle(M A,(A B C D))=\Varangle(S U,(A B C D))=60^{\circ}$.

It follows that $S U \| M A$, so $S U \|\left(M A B^{\prime}\right)$. 

$1 p$ Therefore, the line of intersection of the planes $(MAB')$ and $(SAB)$ is parallel to $S U$ and passes through $A$, i.e., $\left(M A B^{\prime}\right) \cap(S A B)=M A$. $1 p$

Let $T \in A M$ such that $B^{\prime} T \perp A M$.

Since $A B \subset(S A B)$ and $A B \perp A M$, it follows that $\Varangle\left(\left(M A B^{\prime}\right),(S A B)\right)=\Varangle\left(B^{\prime} T, A B\right)$. Since $A^{\prime} B^{\prime} \| A B$, we have $\Varangle\left(\left(M A B^{\prime}\right),(S A B)\right)=\Varangle\left(B^{\prime} T, A^{\prime} B^{\prime}\right)=\Varangle A^{\prime} B^{\prime} T$.

$1 p$

By equating the area of the isosceles triangle $M A B^{\prime}$ written in terms of two bases, we get $B^{\prime} T=\frac{a \sqrt{5}}{2}$, where $a$ is the side length of the cube. From the right triangle $T A^{\prime} B^{\prime}$, with $\Varangle T A^{\prime} B^{\prime}=90^{\circ}$, we deduce $A^{\prime} T=\frac{a}{2}$, and $\operatorname{tg}\left(\Varangle A^{\prime} B^{\prime} T\right)=\frac{A^{\prime} T}{A^{\prime} B^{\prime}}=\frac{1}{2}$.