From point $S$ in space, three non-coplanar rays originate: $a, b$, and $c$. The plane determined by $c$ and $b$ is perpendicular to the plane determined by $c$ and $a$. Given the acute angles $\alpha$ between $b$ and $c$, and $\beta$ between $a$ and $c$, calculate the angle $\gamma$ formed by $a$ and $b$.
I. solution. Let $A$ be a point on the ray $a$ different from $S$. Raise a perpendicular to $a$ at point $A$ in the plane of $a$ and $c$. Since $\alpha$ is an acute angle, this intersects the ray $c$ at a point $C$. Similarly, raising a perpendicular to $c$ at point $C$ in the plane of $c$ and $b$, this intersects $b$ at a point $B$ (1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_b67259815a1610c6e8e2g-1.jpg?height=453&width=585&top_left_y=294&top_left_x=759)

The plane $\Sigma$ passing through $A, B, C$ is perpendicular to $a$, indeed, the line $A C$ is perpendicular to $a$ by construction, and the line $C B$ is in the plane $S A C$ and perpendicular to the intersection line $S C$ of the planes $S A C$ and $S C B$, so $C B$ is perpendicular to the plane $S A C$, and thus to $a$. Therefore, $A C$ and $C B$ are two non-parallel lines in the plane $\Sigma$ that are perpendicular to $a$. Thus, $\Sigma$ is perpendicular to $a$, and the angle $S A B$ is a right angle, as is the angle $A C B$, because $A C$ lies in the plane $S A C$, and $C B$ is perpendicular to this plane. The angle $S C B$ is also a right angle by construction. According to the problem, $\angle B S C = \alpha$, $\angle C S A = \beta$, and we need to calculate the angle $\angle A S B = \gamma$.

Now, from the triangles $S B C$ and $S C A$, we have $S C = S B \cos \alpha$, $S A = S C \cos \beta$, and thus from the triangle $S B A$,

$$
\cos \gamma = \cos \angle A S B = \frac{S A}{S B} = \frac{S C}{S B} \cos \beta = \cos \alpha \cos \beta.
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_b67259815a1610c6e8e2g-1.jpg?height=520&width=450&top_left_y=1296&top_left_x=770)

2nd figure

Remark. The relationships used can be made clearer with geometric considerations. Consider the plane of $a, c$ as the first image plane, and the plane of $b, c$ as the second image plane, then $c$ lies on the axis of the image plane system. Take the half-planes of the image planes containing the rays $a$ and $b$ as positive. The plane determined by the rays $a$ and $b$ can be rotated around the line $a$ (as the first trace line of the plane) into the first image plane, resulting in the 2nd figure, where all three right-angled triangles used are shown in their true size. (The point $B$ can be determined from the perpendicular to $a$ through $C$ by the method of the difference triangle, based on the equality $S B = S(B)$, since the segment $S B$ is visible in its true size both in the second image plane and in the rotation.

Most contestants performed the calculation using more trigonometric knowledge but with less discussion. The following is an example:

![](https://cdn.mathpix.com/cropped/2024_05_02_b67259815a1610c6e8e2g-1.jpg?height=423&width=420&top_left_y=2279&top_left_x=777)

II. solution. Let $D$ be a point on the ray $c$ different from $S$, and raise a perpendicular to $c$ in both the planes $a, c$ and $b, c$. Since $\alpha$ and $\beta$ are acute angles, these intersect $a$ at $E$ and $b$ at $F$ (3rd figure). The angle $E D F$ is the dihedral angle between the planes $S D E$ and $S D F$, so in the triangle $D E F$, the angle at $D$ is a right angle, and the triangles $S E D$ and $S F D$ are also right-angled. The sides of the triangle $S E F$ containing the angle $\gamma$ can be expressed in terms of $S D$ and the known angles:

$$
S E = \frac{S D}{\cos \alpha}, \quad S F = \frac{S D}{\cos \beta}
$$

and $D E = S D \tan \beta$, $D F = S D \tan \alpha$, and thus

$$
E F = \sqrt{D E^2 + D F^2} = S D \sqrt{\tan^2 \alpha + \tan^2 \beta}
$$

With these, by the cosine rule,

$$
\cos \gamma = \cos \angle E S F = \frac{S E^2 + S F^2 - E F^2}{2 S E \cdot S F}.
$$

The numerator can be transformed using well-known identities:

$$
\begin{aligned}
S E^2 & + S F^2 - E F^2 = S D^2 \left( \frac{1}{\cos^2 \alpha} + \frac{1}{\cos^2 \beta} - \tan^2 \alpha - \tan^2 \beta \right) = \\
& = S D^2 \left( \frac{\sin^2 \alpha + \cos^2 \alpha}{\cos^2 \alpha} - \tan^2 \alpha + \frac{\sin^2 \beta + \cos^2 \beta}{\cos^2 \beta} - \tan^2 \beta \right) = 2 S D^2
\end{aligned}
$$

thus

$$
\cos \gamma = 2 S D^2 : \frac{2 S D^2}{\cos \alpha \cos \beta} = \cos \alpha \cos \beta
$$

Remark. Several contestants pointed out that $\gamma$ can be similarly calculated even if the angle between the planes determined by the ray pairs $a, c$ and $b, c$ is an arbitrary angle $\varepsilon$. The angle $E D F$ measures this dihedral angle, so

$$
E F^2 = D E^2 + D F^2 - 2 D E \cdot D F \cos \varepsilon
$$

and thus, considering $D E / S E = \sin \beta$, $D F / S F = \sin \alpha$,

$$
\cos \gamma = \cos \alpha \cos \beta + \sin \alpha \sin \beta \cos \varepsilon
$$

This relationship is often referred to as the cosine rule of the spherical triangle; we will not further discuss its validity conditions here.