3. Consider the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ and $N \in(C D)$ such that $D N=\frac{1}{3} D C$.
a) Determine the edge of the cube knowing that the area of triangle $A^{\prime} A N$ is $\frac{8 \sqrt{10}}{3} \mathrm{~cm}^{2}$.

b) Determine the cosine of the angle between the planes ( $\left.A^{\prime} A N\right)$ and ( $\left.D^{\prime} D A\right)$.

c) Calculate the distance from $D$ to the plane $\left(A^{\prime} A N\right)$.

Daniela Tilincă and Adriana Mihăilă, Brăila
Solution. a) Let $\mathrm{x}$ be the edge of the cube, from the right-angled triangle AND we calculate

$\mathrm{AN}=\frac{\mathrm{x} \sqrt{10}}{3} \Rightarrow \mathrm{A}_{\mathrm{A}^{\prime} \mathrm{AN}}=\frac{x^{2} \sqrt{10}}{6}=\frac{8 \sqrt{10}}{3} \Rightarrow x^{2}=16 \Rightarrow x=4$

b) The projection of triangle A'AN on the plane ( D'DA ) is triangle A'AD =>

Area[ $\mathrm{A}^{\prime} \mathrm{AD}$ ]=Area $\left[\mathrm{A}^{\prime} \mathrm{AN}\right] \cdot \cos $

$\frac{x \cdot x}{2}=\frac{x^{2} \sqrt{10}}{6} \cdot \text{cosine of}\ \mathrm{DT}=$ distance from $\mathrm{D}$ to the plane (A'AN) $; \mathrm{DT} \perp \mathrm{AN} ; \mathrm{DT} \perp \mathrm{AA}$ ' ;

$\mathrm{AA}^{\prime}, \mathrm{AN} \subset\left(\mathrm{A}^{\prime} \mathrm{AN}\right) \Rightarrow \mathrm{DT} \perp\left(\mathrm{A}^{\prime} \mathrm{AN}\right) ; \mathrm{DT}=\frac{\mathrm{AD} \cdot \mathrm{DN}}{\mathrm{AN}}=\frac{x}{\sqrt{10}}=\frac{4}{\sqrt{10}}=\frac{2 \cdot \sqrt{10}}{5}$