3. In the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, point $M$ is the midpoint of [CC'], $S$ is the midpoint of [BM], $T$ is the midpoint of [A S] and $\{P\}=A^{\prime} T \cap(A B C)$. If $A B=4 \text{ cm}$, then determine the length of the segment [P T].

Daniela and Nicolae Stănică, Brăila
## Solution:

$E$ is the midpoint of $[B' C']$ and $F$ is the midpoint of segment $[B C] \Rightarrow A P \cap B C=\{F\}$

If $A' T \cap E F=\{G\} \Rightarrow \frac{A' P}{P G}=\frac{A P}{F P}=\frac{A A'}{F G}$ and $\frac{A A'}{S G}=\frac{A' T}{T G}=\frac{A T}{T S}=1$

$2 \mathbf{p}$

Thus $A' T=T G,[S F]$ is the midline in $\triangle B C M$ and $A' A=S G=4 \mathrm{~cm}$

$2 p$

$\Rightarrow F G=3 \mathrm{~cm}$

From the Pythagorean theorem in triangle $A' E G$ we get $A' G=\sqrt{69} \mathrm{~cm}$.

$1 \mathbf{1 p}$

From $\frac{A' P}{P G}=\frac{A A'}{F G}=\frac{4}{3} \Rightarrow \frac{\sqrt{69}}{P G}=\frac{7}{3} \Rightarrow P G=\frac{3 \sqrt{69}}{7} \mathrm{~cm}$.

$2 \mathbf{p}$

We obtain $P T=\frac{\sqrt{69}}{14} \mathrm{~cm}$.