55. At the base of the triangular pyramid $S A B C$ lies an isosceles right triangle $A B C$ ( $\widehat{A}=$ $=90^{\circ}$ ). The angles $\widehat{S A B}, S \widehat{C A}, S \widehat{A C}, S \widehat{B A}$ (in the given order) form an arithmetic progression with a non-zero common difference. The areas of the faces $S A B, A B C$ and $S A C$ form a geometric progression. Find the angles that form the progression.
55. Let the angles $\widehat{S A B}, \widehat{S C A}, \widehat{S A C}, \widehat{S B A}$ be $\alpha-2 \varphi, \alpha-\varphi$, $\alpha, \alpha+\varphi$. From the Law of Sines in $\triangle S A B$ we find:

$$
|S A|=|A B| \frac{\sin (\alpha+\varphi)}{\sin (2 \alpha-\varphi)}
$$

and from $\triangle S A C$ we find:

$$
|S A|=|C A| \frac{\sin (\alpha-\varphi)}{\sin (2 \alpha-\varphi)}
$$

But by the condition $|A B|=|A C|$. Therefore, $\sin (\alpha+\varphi)=\sin (\alpha-\varphi)$, from which $\alpha=\pi / 2$. The condition relating the areas of triangles $S A B, A B C$ and $S A C$ leads to the equation $\operatorname{ctg}^{2} \varphi \cos 2 \varphi=1$, from which $\varphi=\frac{1}{2} \arccos (\sqrt{2}-1)$.

Answer: $\frac{\pi}{2}-\arccos (\sqrt{2}-1), \frac{\pi}{2}-\frac{1}{2} \arccos (\sqrt{2}-1)$, $\frac{\pi}{2}, \frac{\pi}{2}+\frac{1}{2} \arccos (\sqrt{2}-1)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-053.jpg?height=304&width=446&top_left_y=120&top_left_x=150)

Fig. 2.