In the tetrahedron $O A B C$, the edges meeting at vertex $O$ are pairwise perpendicular to each other. Let the projection of vertex $O$ onto the plane $A B C$ be $H$. The perpendicular planes raised at points $A$, $B$, and $C$ to the edges $O A$, $O B$, and $O C$ respectively intersect $O H$ at points $K, L, M$. If it is already known that

$$
\frac{1}{O K}+\frac{1}{O L}+\frac{1}{O M}=\frac{1}{l},
$$

where $l$ is a given length, what surface is enclosed by the planes $A B C$? [This problem is related to problem 1259 solved in issue 4. (1936/12.118. p. - the editor.)]
$\mathrm{Az} O A K, O B L, O C M$ triangles are right-angled at $A, B, C$ respectively. If $O H \perp[A B C]$, then $O H$ is perpendicular to $A H, B H, C H$, meaning $A H, B H, C H$ are the altitudes corresponding to the hypotenuses $O K, O L, O M$ in the right-angled triangles $O A K, O B L, O C M$ respectively, and therefore

$$
\overline{O H} \cdot \overline{O K}=\overline{O A}^{2}, \quad \overline{O H} \cdot \overline{O L}=\overline{O B}^{2}, \quad \overline{O H} \cdot \overline{O M}=\overline{O C}^{2}
$$

or

$$
O K=\frac{\overline{O A}^{2}}{O H}, \quad O L=\frac{\overline{O B}^{2}}{O H}, \quad O M=\frac{\overline{O C}^{2}}{O H} \ldots
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_236ee3c8f093661bd26bg-1.jpg?height=404&width=461&top_left_y=606&top_left_x=821)

If we substitute the values of $O K, O L, O M$ into the equation

$$
\frac{1}{O K}+\frac{1}{O L}+\frac{1}{O M}=\frac{1}{l} \ldots
$$

we get:

$$
O H\left(\frac{1}{O A^{2}}+\frac{1}{O B^{2}}+\frac{1}{O C^{2}}\right)=\frac{1}{l} \ldots
$$

In problem 1259, we saw that

$$
\frac{1}{O A^{2}}+\frac{1}{O B^{2}}+\frac{1}{O C^{2}}=\frac{1}{O H^{2}}
$$

Considering this, we have: $\quad O H=l$,

which means that the planes $A B C$ satisfying condition 2) are tangent planes to a sphere with center $O$ and radius $l$. This sphere is the envelope of the planes $A B C$.

Pálos Peregrin (Pope Gregory VIII,)

(Note: The name "Pálos Peregrin" is a Hungarian name, and "Pope Gregory VIII" is the English equivalent for the historical figure.)