3. One diagonal of a rhombus is the geometric mean of the side length and the other diagonal. Determine the angles of the rhombus.
3. Let $|A C|=d_{1}, \quad|B D|=d_{2}$. Since $d_{2}=\sqrt{a d_{1}}$, we have $d_{2}^{2}=a d_{1}$. From this, we get

$$
\frac{d_{1}}{d_{2}}=\frac{d_{2}}{a}=k, \quad \text { i.e. } \quad d_{2}=a k, \quad d_{1}=a k^{2} . \quad 5 \text { points }
$$

![](https://cdn.mathpix.com/cropped/2024_06_03_843fcc95a1719fe2319dg-06.jpg?height=480&width=574&top_left_y=180&top_left_x=570)

Since

$a^{2}=\left(\frac{d_{1}}{2}\right)^{2}+\left(\frac{d_{2}}{2}\right)^{2}, \quad \text{i.e.}$

$a^{2}=\left(\frac{a k^{2}}{2}\right)^{2}+\left(\frac{a k}{2}\right)^{2} \quad$ we obtain the equation

$k^{4}+k^{2}-4=0$.

The only positive solution is $k^{2}=\frac{1}{2}(\sqrt{17}-1)$.

10 points

Let's calculate the angle $\epsilon$ :

$$
\cos \epsilon=\frac{\frac{d_{1}}{2}}{a}=\frac{k^{2}}{2}=\frac{1}{4}(\sqrt{17}-1)=0.780776
$$

$\epsilon=\arccos \frac{1}{4}(\sqrt{17}-1)$,

$\alpha=2 \epsilon, \quad \beta=180^{\circ}-2 \epsilon$.

10 points