Problem 3. At the vertex $\mathrm{D}$ of the rectangle $\mathrm{ABCD}$, a perpendicular is drawn to the plane of the rectangle, and a point M is taken on this perpendicular. Let $\mathrm{AC} \cap B D=\{O\}$ and $[O N$ be the bisector of $\angle M O C, \mathrm{~N} \in(\mathrm{MC})$. Consider a point $\mathrm{P} \in(\mathrm{MA})$ such that the projection of $\mathrm{P}$ onto the line $\mathrm{ON}$ is the point $\mathrm{O}$.

If $\mathrm{AB}=4 \mathrm{~cm}, \mathrm{BC}=3 \mathrm{~cm}$ and $\mathrm{d}(\mathrm{M}, \mathrm{AC})=\frac{12}{5} \sqrt{2}$, determine:

a) the measure of the angle between PN and the plane (ABC);

b) the distance from point D to the plane (MAC).

![](https://cdn.mathpix.com/cropped/2024_06_07_e3d15057ad6f2c4f2a36g-3.jpg?height=811&width=848&top_left_y=733&top_left_x=671)
Solution:
a) $m(\angle A O P)=180^{\circ}-90^{\circ}-m(\angle N O C)=90^{\circ}-m(\angle N O M)=m(\angle P O M) \Rightarrow[O P$ bisector of $\angle M O A$....

$1 p$

Applying the angle bisector theorem in triangles $M O C$ and $M O A$ we get:

![](https://cdn.mathpix.com/cropped/2024_06_07_e3d15057ad6f2c4f2a36g-3.jpg?height=118&width=1541&top_left_y=1925&top_left_x=308)

But $O C=O A \Rightarrow \frac{M N}{N C}=\frac{M P}{P A} \xrightarrow{\text { T.R.Th }} P N \| A C$

Since $P N\|A C, A C \subset(ABC) \Rightarrow P N\|(ABC) \Rightarrow m \angle[P N,(ABC)]=0^{\circ}$
b) Let $D Q \perp A C$. Since $M D \perp(ABC), D Q \cap A C=\{Q\} \xrightarrow{T_{3 \downarrow}} M Q \perp A C$ $.1 p$

We have $D Q \perp A C, M Q \perp A C$ and let $D S \perp M Q, M Q \cap A C=\{Q\} \xrightarrow{T . R . T_{3} \perp} D S \perp(M A C)$

It follows that $d[D,(M A C)]=D S$

In $\triangle A D C$ using $T . P$ and $T . h_{2}$ we get: $A C=5$ and $D Q=\frac{12}{5}$

In $\triangle M D Q \Rightarrow \cos Q=\frac{D Q}{M Q} \Rightarrow \cos Q=\frac{\sqrt{2}}{2} \Rightarrow m \angle(M Q D)=45^{\circ}$

$.0,5 p$

$\triangle M D Q$ is a right isosceles triangle at $\mathrm{D}$ and since $[D S]$ is the altitude $\Rightarrow[D S]$ is also the median.

Thus $D S=\frac{M Q}{2}=\frac{6 \sqrt{2}}{5}$

$.0,5 p$