4B. The lengths of the smaller diagonal, the side, and the larger diagonal of the rhombus form a geometric progression. Determine the angles of the rhombus!
Solution. Let $ABCD$ be a rhombus in which the diagonal $d_{1}=\overline{BD}$, the side $a=\overline{AB}$, and the diagonal $d_{2}=\overline{AC}$ are consecutive terms of a geometric progression. Thus, there exist numbers $b$ and $q$ such that $d_{1}=b$, $a=bq$, and $d_{2}=bq^{2}$.

From the right triangle $AOB$, by the Pythagorean theorem, we have

$$
\left(\frac{d_{1}}{2}\right)^{2}+\left(\frac{d_{2}}{2}\right)^{2}=a^{2}, \quad \frac{b^{2}}{4}+\frac{b^{2}}{4} q^{4}=b^{2} q^{2}, \quad q^{4}-4 q^{2}+1=0
$$

If we introduce the substitution $q^{2}=t$, we get the quadratic equation $t^{2}-4 t+1=0$ whose solutions are $t_{1/2}=2 \pm \sqrt{3}$.

For the solution $t=2+\sqrt{3}$, we get

$$
\begin{aligned}
& q=\sqrt{2+\sqrt{3}}, d_{1}=b \\
& a=b \sqrt{2+\sqrt{3}}, d_{2}=b(2+\sqrt{3})
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_a6c2bf676fcc39861820g-12.jpg?height=293&width=495&top_left_y=1435&top_left_x=973)

From the right triangle $AOB$, we get

$$
\operatorname{tg} \frac{\alpha}{2}=\frac{\overline{OB}}{OA}=\frac{\frac{d_{1}}{2}}{\frac{d_{2}}{2}}=\frac{d_{1}}{d_{2}}=\frac{b}{b(2+\sqrt{3})}=2-\sqrt{3}
$$

Finally, the angles of the rhombus $ABCD$ are

$$
\alpha=2 \operatorname{arctg}(2-\sqrt{3}) \text { and } \beta=\pi-2 \operatorname{arctg}(2-\sqrt{3}) \text {. }
$$

The same procedure is followed for the solution $t=2-\sqrt{3}$.