The axial section of a cone is an equilateral triangle with side $a$. Two planes perpendicular to each other are drawn through the axis of the cone, dividing the cone into four parts. Find the radius of the sphere inscribed in one of these parts.
Let $P$ be the vertex of the cone, $O$ the center of the base, $r$ the radius of the sphere with center $Q$, which is tangent to the perpendicular planes $COP$ and $DOP$, the base plane of the cone at point $M$, and the lateral surface of the cone at point $F$. We can assume that

$$
OC=OD=\frac{a}{2}, OP=\frac{a \sqrt{3}}{2}
$$

Through the height $PO$ of the cone and the line $QM$ parallel to it, we draw a plane. We obtain the axial section of the cone as an equilateral triangle $APB$ with side length $a$. The circle with radius $r$ and center $Q$ is inscribed in the angle $PAO$, so

$AM=QM \operatorname{ctg} \angle MAQ = r \operatorname{ctg} 30^{\circ} = r \sqrt{3}$.

Let the line $OQ$ form an angle $\alpha$ with the base plane of the cone. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_35476b28e4766e84fc81g-43.jpg?height=906&width=869&top_left_y=1868&top_left_x=1152)

$OM=QM \operatorname{ctg} \angle MOQ = r \operatorname{ctg} \alpha$.
$a = r \sqrt{3} + r \operatorname{ctg} \alpha$

Thus, to solve the problem, it is sufficient to find $\operatorname{ctg} \alpha$.

Consider a cube with vertex $L$. Its diagonal, drawn from vertex $L$, forms an angle $\alpha$ with the planes of the three faces containing vertex $L$. If the edge of the cube is $c$, then the diagonal of the face is $c \sqrt{2}$. Therefore,

$$
\operatorname{ctg} \alpha = \frac{c \sqrt{2}}{c} = \sqrt{2}
$$

$\begin{array}{ll}\text { From } & a \\ \text { equation } & 2\end{array}=r \sqrt{3} + r \sqrt{2} \begin{aligned} \text { we find that }\end{aligned} a(\sqrt{3} - \sqrt{2})$.