PROBLEM 3. Consider the square $ABCD$ and the point $S$ outside its plane such that $SA=SB=SC=SD$. If $AB=12 \text{ cm}, AP \perp CS, P \in (SC)$ and $AP=SO$, where $O$ is the center of the square, the following are required:

a) The measure of the angle formed by the lines $SC$ and $BD$;

b) The distance from point $A$ to the plane $(BPD)$;

c) The distance from point $B$ to the line of intersection of the planes $(BPD)$ and $(ADS)$.
## PROBLEM 3.

![](https://cdn.mathpix.com/cropped/2024_06_07_f2c3e35ea523bc113df1g-3.jpg?height=571&width=579&top_left_y=157&top_left_x=750)

(2p) a) $S O \perp(A B C) \Rightarrow S O \perp B D, B D \perp A C \Rightarrow B D \perp(S O C) \Rightarrow B D \perp S C \Rightarrow$ $m(\widehat{B D, S C})=90^{\circ}$

(1p) b) Triangles $\triangle S A C$ and $\triangle O P C$ are equilateral

$(1 p) \quad d(A,(B P D))=d(C,(B P D))=d(C, O P)=3 \sqrt{6} \text{ cm}$

$(3 p) \quad$ c) $O P$ is the midline in $\triangle S A C \Rightarrow O P\|A S \Rightarrow O P\|(A D S)$;

$$
\begin{aligned}
& O P \subset(B P D) \Rightarrow O P \|(A D S) \cap(B P D) \\
& B D \perp O P \Rightarrow B D \perp(A D S) \cap(B P D) \Rightarrow d(B,(A D S) \cap(B P D))=B D=12 \sqrt{2} \text{ cm}
\end{aligned}
$$