4. In the regular tetrahedron \(ABCD\), consider point \(M\) - the midpoint of \([AB]\) and \(N\) - the midpoint of \([AC]\). If the edge of the tetrahedron is \(8 \, \text{cm}\), determine:

(3 p) a) The perimeter of quadrilateral \(BCNM\)

(2 p) b) The measure of the angle formed by line \(MN\) with \(AB\)

(2 p) c) The distance from point \(C\) to the plane \((ABD)\).

Note: All subtasks are mandatory,

Time allowed: 3 hours

Each subtask is graded from 0 to 7 points.

## GRADING AND SCORING CRITERIA FOR THE MATHEMATICS OLYMPIAD LOCAL PHASE
4. Desen

![](https://cdn.mathpix.com/cropped/2024_06_07_41eee465e45fe0070e7bg-4.jpg?height=674&width=605&top_left_y=1630&top_left_x=520)

a) $\triangle A B C, M N$ - midline $\Rightarrow M N \| B C, M N=\frac{B C}{2}=4 \text{ cm}$ $B C N M$ - isosceles trapezoid $P_{B C N M}=B C+C N+N M+M B=8 \text{ cm}+3 \cdot 4 \text{ cm}=$ $20 \text{ cm}$

b) $M N \| B C, A B$ secant $m(\angle(M N, A B))=60^{\circ}$

Let $P$ be the midpoint of $[B C] \Rightarrow A P \perp B C$

Since $D O \perp(A B C) \Rightarrow D O \perp B C$, it follows that $\left\{\begin{array}{l}B C \perp(D P O) \\ A D \subset(D P O)\end{array} \Rightarrow B C \perp A D\right.$.

From $M N \| B C$ and $A D \perp M N \Rightarrow m(\angle(M N, A D))=90^{\circ}$

c) $\left\{\begin{array}{l}A B \perp C M \\ A B \perp D M\end{array} \Rightarrow\left\{\begin{array}{l}A B \perp(C M D) \\ A B \subset(A B D)\end{array} \Rightarrow(A B D) \perp(C D M)\right.\right.$

We construct $C E \perp D M, C E \subset(C D M) \Rightarrow C E \perp(A D B) \Rightarrow d(C,(A B D))=C E$

In $\triangle C D M, D M=\frac{l \sqrt{3}}{2}=4 \sqrt{3}$

Applying the Pythagorean theorem, we get $D O=\frac{8 \sqrt{6}}{3}$

$A_{\triangle C D M}=\frac{C M \cdot D O}{2}=16 \sqrt{2}$

$A_{\triangle C D M}=\frac{D M \cdot C E}{2}=\frac{4 \sqrt{3} \cdot C E}{2}=2 \sqrt{3} \cdot C E$. We obtain $C E=\frac{8 \sqrt{6}}{3} d(C,(A B D))=\frac{8 \sqrt{6}}{3} \text{ cm}$