Solution. Let $V$ be the midpoint of $[B C]$. [TV] is the midline in triangle $A B C, m(A C B)=90^{\circ}$, so $T V \perp B C$.

Since $[D T]$ is the midline in triangle $D A B, D A=D B$, it follows that $D T \perp B C$. We also have $(A B C) \perp(A B D)$, so $D T \perp(A B C)$. By the theorem of the three perpendiculars, it follows that $D V \perp B C$, hence $\operatorname{tg}((A B C),(D B C))=\operatorname{tg}(T V, D V)=\frac{D T}{T V}$.

Let $C M=A N=x, B M=C N=2 x, A B=3 x \sqrt{2}, C T=\frac{3 x \sqrt{2}}{2} \Rightarrow S D=\frac{5 x \sqrt{2}}{2}$.

Applying Menelaus in $\triangle A B C, P-N-M$:
$\frac{P A}{P B} \cdot \frac{M B}{M C} \cdot \frac{N C}{N A}=1 \Rightarrow \frac{P A}{P A+3 \sqrt{2} x} \cdot \frac{2 x}{x} \cdot \frac{2 x}{x}=1 \Rightarrow P A=x \sqrt{2}$.

Applying Menelaus in $\triangle A T D, P-S-G$:

$\frac{P A}{P T} \cdot \frac{G T}{G D} \cdot \frac{S D}{S A}=1 \Rightarrow \frac{x \sqrt{2}}{x \sqrt{2}+\frac{3 \sqrt{2} x}{2}} \cdot \frac{1}{2} \cdot \frac{\frac{5 x \sqrt{2}}{2}}{S A}=1 \Rightarrow S A=\frac{x \sqrt{2}}{2}$.

Applying the Pythagorean theorem in $\triangle A T D$, we have $\left(\frac{x \sqrt{2}}{2}+\frac{5 x \sqrt{2}}{2}\right)^{2}=\left(\frac{3 x \sqrt{2}}{2}\right)^{2}+D T^{2} \Rightarrow D T=\frac{3 x \sqrt{6}}{2}$,

thus $\operatorname{tg}(T V, D V)=\frac{\frac{3 x \sqrt{6}}{2}}{\frac{3 x}{2}}=\sqrt{6}$.
4. Consider triangle $ABC$, where $AC = BC$, $m(ACB) = 90^{\circ}$, and triangle $DAB$, where $DA = DB$, located in perpendicular planes. Let $\quad M \in (BC), \quad BM = 2CM, \quad N \in (AC)$, $AC = 3AN, P \in MN \cap AB$, $T$ be the midpoint of segment $[AB]$, and $G$ be the centroid of triangle $DAB$. Calculate the tangent of the plane angle corresponding to the dihedral angle determined by the planes $(ABC)$ and $(DBC)$, given that $3SD = 5CT$, where $S \in PG \cap AD$.

Narcis Gabriel Turcu, Brăila