The dihedral angles of a trihedron are equal; let their size be denoted by $\alpha$. Let the measure of the face angles be $x$. What is the relationship between $\cos \alpha$ and $\cos x$?

What are $\alpha$ and $x$ if they are complementary angles?
I. Solution Let the vertex of the trihedron be the center of a sphere. The trihedron defines a spherical triangle on the sphere, whose sides are the dihedral angles of the trihedron, and whose angles are the face angles of the trihedron. Let $\alpha, \beta, \gamma$ be the sides of the spherical triangle and $x$ the angle opposite to $\alpha$; then

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

If $\beta = \gamma = \alpha$, then $\cos \alpha = \cos^2 \alpha + \sin^2 \alpha \cos x$, and thus

$$
\cos \alpha = \cos^2 \alpha + (1 - \cos^2 \alpha) \cos x
$$

and from this

$$
\cos x = \frac{\cos \alpha (1 - \cos \alpha)}{1 - \cos^2 \alpha}
$$

We exclude the case $\alpha = 0, \cos \alpha = 1$, and thus

$$
\cos x = \frac{\cos \alpha}{1 + \cos \alpha}
$$

For $\alpha < \frac{\pi}{2}, \cos x < \cos \alpha$.

If $\alpha = \frac{\pi}{2}, \cos \alpha = 0$ and $\cos x = 0$, so $x = \frac{\pi}{2}$: we have a right-angled trihedron.

For $\alpha > \frac{\pi}{2}, \cos \alpha < 0$ and thus $x > \alpha$. (Both $\alpha$ and $x$ are obtuse angles!)

If $\alpha + x = \frac{\pi}{2}$, then $\cos x = \sin \alpha$. Thus,

$$
\sin \alpha (1 + \cos \alpha) = \cos \alpha \quad \text{or} \quad \sin \alpha \cos \alpha = \cos \alpha - \sin \alpha
$$

Squaring both sides: $(\sin \alpha \cos \alpha)^2 = \cos^2 \alpha + \sin^2 \alpha - 2 \cos \alpha \sin \alpha$

which simplifies to

$$
\sin^2 2\alpha + 4 \sin 2\alpha - 4 = 0
$$

From this,

$$
\sin 2\alpha = 2(-1 + \sqrt{2})
$$

The negative root of the equation, in absolute value, is greater than 1 and cannot be considered.

From the tables: $2\alpha = 55^\circ 56' 15''$, $\alpha = 27^\circ 58' 7.5''$ and

$$
x = 90^\circ - \alpha = 62^\circ 1' 52.5''
$$

The same sine value applies to the complementary angle of $55^\circ 56' 15''$; however, this cannot be valid because then $2\alpha > \frac{\pi}{2}$, $\alpha > \frac{\pi}{4}$, and $x = \frac{\pi}{2} - \alpha < \alpha$!

George Bizám (Bólyai g. VI. o. Bp. V.)

II. Solution. On the edge of the trihedron, take a point $A$ and construct a plane perpendicular to the edge $SA$; this plane intersects the other two edges at points $B$ and $C$. Since $SAB$ and $SAC$ are right-angled and congruent triangles at $A$, $SB = SC$ and $AB = AC$. From the triangle $SAB$:

$$
AB = SB \sin \alpha
$$

From the isosceles triangle $BSC$:

$$
BC = 2 SB \sin \frac{\alpha}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_a7f64ad404dd2b61594fg-1.jpg?height=377&width=382&top_left_y=1904&top_left_x=861)

In the isosceles triangle $BAC$, $\angle BAC = x$, so

$$
BC = 2 AB \sin \frac{x}{2} = 2 SB \sin \alpha \sin \frac{x}{2}
$$

Thus, $2 SB \sin \frac{\alpha}{2} = 2 SB \sin \alpha \sin \frac{x}{2}$,

$$
\sin \frac{\alpha}{2} = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \sin \frac{x}{2} \quad \text{or} \quad 2 \cos \frac{\alpha}{2} \sin \frac{x}{2} = 1
$$

Squaring both sides:

$$
4 \cos^2 \frac{\alpha}{2} \sin^2 \frac{x}{2} = 1
$$

However, $2 \cos^2 \frac{\alpha}{2} = 1 + \cos \alpha$, $2 \sin^2 \frac{x}{2} = 1 - \cos x$,

and thus

$$
(1 + \cos \alpha)(1 - \cos x) = 1, \quad \text{or} \quad \cos x = \frac{\cos \alpha}{1 + \cos \alpha}
$$

[^0]:    ${ }^{1}$ One side and the two angles on it are equal!

    ${ }^{2}$ From this, it can be seen that $\sin \frac{\alpha}{2} \leqq \sin \frac{x}{2}$, so $\alpha \leqq x \cdot\left(\sin \alpha > 0, \sin \frac{\alpha}{2} > 0, \sin \frac{x}{2} > 0.\right)$