Problem 4. In the plane $\alpha$, consider the segment $[A B]$ with a length of $12 \, \text{cm}$ and two mobile points $M$ and $N$, symmetric with respect to point $O$, the midpoint of segment $[A B]$, such that $[M N] \equiv [A B]$. On the perpendicular to the plane $\alpha$ at point $A$, consider point $P$, which is $2 \sqrt{7} \, \text{cm}$ away from point $A$.

a) Find the perimeter of triangle $PMN$ with the maximum area;

b) If point $S$ is the symmetric point of point $P$ with respect to point $O$, find the distance from point $A$ to the plane of quadrilateral $PMSN$, which has the maximum area.
## Solution. 1p by default

a) $A_{P M N}=\frac{M N \cdot d(P, M N)}{2}$ is maximum if $\mathrm{d}(\mathrm{P}, \mathrm{MN})$ is maximum.

It follows that d(A,MN) is maximum.

Let $A I \perp M N, I \in(M N)$.

From $\mathrm{AB}=\mathrm{MN}$ and $\mathrm{N}=\mathrm{S}_{\mathrm{o}}(\mathrm{M}) \mathrm{with} O \in(A B) ; A O=O B$ it follows that points

$A ; B ; M$ and $N$ are concyclic and d (A; MN)

is maximum when $\mathrm{AI}=\mathrm{AO}=6 \mathrm{~cm}$.

Thus $A O \perp M N$ $\qquad$

![](https://cdn.mathpix.com/cropped/2024_06_07_b85f7de9e1f92d54ac64g-5.jpg?height=517&width=534&top_left_y=958&top_left_x=1403)

(1p)

From $P A \perp \alpha, A O \perp M N$ it follows by $T_{3 \perp} c a ̆ ~ P O \perp M N(\mathbf{0}, \mathbf{5 p})$

$\mathrm{PO}=\sqrt{P A^{2}+A O^{2}}=\sqrt{(2 \sqrt{7})^{2}+6^{2}}=\sqrt{64}=8$

In $\triangle P O M$, applying $T . P$ we get $P M=10(\mathbf{0 , 5 p})$

In $\triangle P M N, P O \perp M N$ and $M O=O N$ we get $\triangle P M N$ is isosceles

$P_{M N P}=2 \cdot 10+12=32(\mathrm{~cm})(\mathbf{0 , 5 p})$

b)PMSN is a rhombus because the diagonals bisect each other and are perpendicular, having the maximum area for APMN to be maximum. (1p)

We have $A O \perp M N, P O \perp M N$ and let $A T \perp P S$

$$
\Rightarrow \text{by} T R T_{3 \perp} c a ̆ a T \perp(P M N)(\mathbf{1} \boldsymbol{p})
$$

In $\triangle A P D$ we get by $T h_{2}: A T=\frac{A O \cdot A P}{P O} \Rightarrow A T=\frac{6 \cdot 2 \sqrt{7}}{8}=\frac{3 \sqrt{7}}{2}(\mathrm{~cm})$ (1p)

![](https://cdn.mathpix.com/cropped/2024_06_07_b85f7de9e1f92d54ac64g-5.jpg?height=777&width=674&top_left_y=1710&top_left_x=1291)