4. What values can the angle at the vertex of an isosceles triangle take if it is known that exactly three lines can be drawn that divide both the area and the perimeter of this triangle in half.
Answer: $2 \arcsin (\sqrt{2}-1)\sqrt{2}-1$, one - if $\sin \frac{\alpha}{2}=\sqrt{2}-1$ and none if $\sin \frac{\alpha}{2}<\frac{1}{2}$, then the roots of equation (1)

$$
x_{1,2}=\frac{2 b(1+t) \pm 2 b \sqrt{t^{2}+2 t-1}}{4 b}=\frac{1+t \pm \sqrt{t^{2}+2 t-1}}{2}
$$

have the property:

$x_{2}=\frac{1+t+\sqrt{t^{2}+2 t-1}}{2}>1 \Leftrightarrow \sqrt{t^{2}+2 t-1}>1-t \Leftrightarrow t>\frac{1}{2}$. The formula for the smaller root $x_{1}=\frac{1+t-\sqrt{t^{2}+2 t-1}}{2}$ implies that $y=\frac{1}{2 x}=\frac{1}{1+t-\sqrt{t^{2}+2 t-1}}=\frac{1+t+\sqrt{t^{2}+2 t-1}}{2}>1$, i.e., there are no lines to consider in this case when $t>\frac{1}{2}$, while for $\sqrt{2}-1<\sin \frac{\alpha}{2}<\frac{1}{2}$, there are $2 \arcsin (\sqrt{2}-1)$ lines.