In the trihedron $O A B C$, the dihedral angles are right angles. On two edges, we mark $O A=O B=a$, while on the third one, the distance $O C=x$. The value of $x$ is to be determined so that the volume of the tetrahedron $O A B C$ is equal to $\frac{F m}{3}$, where $F$ is the surface area of the tetrahedron and $m$ is a given length. For which values of $m$ is there a solution?
The volume of the tetrahedron: $\frac{1}{3}(A O B) \cdot O C=\frac{a^{2} x}{6}$.

The surface area of the tetrahedron $F=(A O B)+2(A O C)+(A B C)=\frac{a^{2}}{2}+a x+(A B C)$.

Draw a perpendicular from vertex $C$ to $A B$, denoted as $C D$.

![](https://cdn.mathpix.com/cropped/2024_05_02_31bd771bbafd3f43e554g-1.jpg?height=309&width=325&top_left_y=360&top_left_x=889)

Clearly, $O D=\frac{A B}{2}=\frac{a \sqrt{2}}{2}$.

$$
(A B C)=\frac{1}{2} \overline{A B} \cdot \overline{C D}=\frac{1}{2} a \sqrt{2} \sqrt{x^{2}+\left(\frac{a \sqrt{2}}{2}\right)^{2}}=\frac{a}{2} \sqrt{2 x^{2}+a^{2}}
$$

Thus, $F=\frac{a^{2}}{2}+a x+\frac{a}{2} \sqrt{2 x^{2}+a^{2}}$.

The problem requirement is:

$$
\frac{a^{2} x}{6}=\frac{m}{3}\left(\frac{a^{2}}{2}+a x+\frac{a}{2} \sqrt{2 x^{2}+a^{2}}\right)
$$

Each term in the equation is divisible by $a$. Removing the fractions and performing the indicated operations, we get:

$$
(a-2 m) x-a m=m \sqrt{2 x^{2}+a^{2}} \ldots
$$

Squaring both sides and simplifying:

$$
\left[(a-2 m)^{2}-2 m^{2}\right] x^{2}-2 a m(a-2 m) x=0 \ldots
$$

One root of (2) is: $x=0$. This does not satisfy (1), because substituting $x=0$ results in $-a m=a m$, a contradiction if $m \neq 0$.

The other root of (2) is:

$$
x=\frac{2 a m(a-2 m)}{(a-2 m)^{2}-2 m^{2}} \cdots
$$

This is valid if it is positive and makes the left side of (1) positive. However, this condition cannot be met if $a-2 m$ is negative (or zero). Therefore, we need $a-2 m>0$ and $m>0$, so that:

$$
(a-2 m)^{2}-2 m^{2}=(a-2 m+m \sqrt{2})(a-2 m-m \sqrt{2}) \geq 0
$$

which means:

$$
a-2 m+m \sqrt{2}>0
$$

$$
a-2 m-m \sqrt{2} \leq 0, \quad \text { i.e., } \quad m \leq \frac{a}{2+\sqrt{2}}=\frac{a}{2}(2-\sqrt{2}) \ldots
$$

If $x$ is substituted into the left side of (1) according to (3):

$$
\frac{2 a m(a-2 m)^{2}}{(a-2 m)^{2}-2 m^{2}}-a m=\frac{a m\left[(a-2 m)^{2}+2 m^{2}\right]}{(a-2 m)^{2}-2 m^{2}}>0, \quad \text { if } \quad m \leq \frac{a}{2}(2-\sqrt{2})
$$

Thus, the maximum value of $m$ is $m'=\frac{a}{2}(2-\sqrt{2})$. This is the radius of the circle inscribed in the $\triangle A O B$. On the other hand, $m$ is equal to the radius of the sphere inscribed in the tetrahedron $A O B C$. Clearly, $m < m'$. If $m = m'$, then $x = \infty$; the tetrahedron becomes a triangular prism bounded by $\triangle A O B$. The sphere that touches the faces of this prism has its largest circle coinciding with the circle inscribed in $\triangle A O B$.[^0]

[^0]:    ${ }^{1}$ When $x=0$, $F=a^{2}$, which is twice the area of $\triangle A O B$. In this case, the volume of the tetrahedron $=0$, so $m=0$.

    ${ }^{2} \varrho=\frac{2 t}{2 s}=\frac{a^{2}}{2 a+a \sqrt{2}}=\frac{a}{2+\sqrt{2}}$