4. Consider a regular quadrilateral pyramid $S M N P Q$, with vertex $S$ and the center of the base $O$, having all edges of length $l$. Let $T$ be the symmetric point of $M$ with respect to the line $N P$.

a) Calculate the distance between the lines $S O$ and $P T$.

b) Calculate the distance from the point $M$ to the plane (SNP).

c) Find the tangent of the angle formed by the planes $(S M T)$ and (MPS).

d) An ant starts from $S$, moves on the face $S M N$, touches the edge $[M N]$ at a point $A$, and then moves on the base $M N P Q$ until it reaches the point $F$, the midpoint of the edge $[N P]$. Calculate $N A$ knowing that the path traveled by the ant from $S$ to $F$ is the shortest possible.

Dorina Bocu
## Solution.

a) $\triangle P M N$ and $\triangle P N T$ are right-angled isosceles triangles. It follows that $O P \perp P T$. But $S O \perp(M N P)$, so $O P \perp S O$. Then $O P$ is the common perpendicular of the lines $S O$ and $P T$. Therefore, $d(S O, P T)=O P=\frac{l \sqrt{2}}{2}$. (1p)

b) Let $E$ and $F$ be the midpoints of the edges $[M Q]$ and $[N P]$. Since $M Q \| N P$ and $N P \subset(S N P)$, it follows that $M Q \|(S N P)$. Therefore, $d(M,(S N P))=d(E,(S N P))$.

We construct $E H \perp S F, H \in S F$. Then, from $S F \perp N P, E F \perp N P$ and $S F, N P \in(S N P)$, it follows that $E H \perp(S N P)$, so $d(E,(S N P))=E H$. From $\triangle S O M$ we obtain $S O=l / \sqrt{2}$.

$$
\mathcal{A}_{S E F}=\frac{E F \cdot S O}{2}=\frac{S F \cdot E H}{2}, \text{ hence } E H=\frac{E F \cdot S O}{S F}=l \cdot \frac{l \sqrt{2}}{2} \cdot \frac{2}{l \sqrt{3}}=\frac{l \sqrt{6}}{3} \cdot(\mathbf{2} \mathbf{p})
$$

c) $P T \perp M P$ and $P T \perp S O$, so $P T \perp(S M P)$. $\triangle S M P \equiv \triangle N M P$ (S.S.S.), so $m(\overline{M S P})=90^{\circ}$. By the theorem of the three perpendiculars, $P T \perp(S M P), P S \perp M S$ and $S P, M S \subset (S M P)$ imply $T S \perp M S$. $(S M T) \cap(M S P)=M S, S T \perp M S, S T \subset(S T M), S P \perp M S$

![](https://cdn.mathpix.com/cropped/2024_06_07_6e5542fea3c43ef4ac06g-3.jpg?height=72&width=1033&top_left_y=1803&top_left_x=243)

Triangle $\triangle S P T$ is right-angled at $P$, so $\operatorname{tg}(\widehat{P S T})=\frac{P T}{P S}=\frac{l \sqrt{2}}{l}=\sqrt{2}$. $(2 \mathbf{p})$

d) Let $S^{\prime} \in(M N P)$, such that $\triangle S^{\prime} M N$ is equilateral, and $S^{\prime}$ and $F$ are in opposite half-planes with respect to $M N$. The length of the shortest path traveled by the ant is the length of the segment $\left[S^{\prime} F\right]$. Thus, $\{A\}=\left[S^{\prime} F\right] \cap[M N]$. Let $B$ be the midpoint of $[M N]$. We have $S^{\prime} B=\frac{l \sqrt{3}}{2}$. $S B \| N F$ implies $\triangle N A F \sim \triangle B A S$, so $\frac{N A}{B A}=\frac{N F}{S^{\prime} B}=\frac{1}{\sqrt{3}}$. Then $\frac{N A}{N B}=\frac{1^{2}}{1+\sqrt{3}}$. We obtain $N A=\frac{l(\sqrt{3}-1)}{4}$. (2p)