3B. Given a parallelogram $A B C D$. The bisector of $\angle D A B$ intersects side $D C$ at point $L$, and diagonal $B D$ at point $K$, such that $\overline{D K}: \overline{K B}=3: 4$. Calculate the length of segment $L C$, if the perimeter of the parallelogram is 28.
Solution. AL is the bisector of $\measuredangle D A B$, so it holds that

$$
\measuredangle D A L=\measuredangle L A B=\angle A L D
$$

(the last two angles are equal as alternate interior angles when the transversal is $A L$), from which $\triangle A L D$ is isosceles. Let's denote $\overline{D L}=\overline{A D}=b$

![](https://cdn.mathpix.com/cropped/2024_06_05_a11ff5dd9a0a1c832502g-13.jpg?height=358&width=623&top_left_y=489&top_left_x=842)
and $\overline{A B}=\overline{D C}=a$. For the areas of the triangles in the diagram, we have

$$
P_{\triangle A K D}: P_{\triangle A B K}=\frac{\overline{A D} \cdot \overline{M K}}{2}: \frac{\overline{A B} \cdot \overline{N K}}{2}=b: a,
$$

where $M K, N K$ are the corresponding heights dropped to the sides, and for them, $\overline{M K}=\overline{N K}$ because point $K$, as a point on the bisector of $\measuredangle D A B$, is equidistant from the sides of the angle. On the other hand,

$$
P_{\triangle A K D}: P_{\triangle A B K}=\frac{\overline{D K} \cdot \overline{A H}}{2}: \frac{\overline{B K} \cdot \overline{A H}}{2}=\overline{D K}: \overline{B K}=3: 4
$$

Now, equating the ratios of the areas, we get that $b: a=3: 4$, and from the perimeter of the parallelogram $2(a+b)=28$, by substitution, we have $b=6, a=8$. Finally, $\overline{L C}=a-b=2$.

4AAB. Determine the maximum value of the expression

$$
y=\operatorname{tg}\left(x+\frac{2 \pi}{3}\right)-\operatorname{tg}\left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right), x \in\left[-\frac{5 \pi}{12},-\frac{\pi}{3}\right]
$$