The side of the base of a regular hexagonal pyramid is $a$, and the lateral edge is $b$. Find the radius of the sphere that touches the base plane and the lateral edges of the pyramid.
Let $PM$ be the height of the regular hexagonal prism $PABCDEF$, and $r$ be the desired radius. Since the pyramid is regular, the center $Q$ of the sphere that touches the base plane and the lateral edges of the pyramid lies on the line $PM$, and the point of tangency of the sphere with the base plane coincides with the point $M$. If $L$ is the point of tangency of the sphere with the lateral edge $AP$, then

$QL \perp AP, \quad AM=a, \quad AL=AM=a, \quad PL=AP-AL=b-a$.

From the right triangle $AMR$, we find that
$PM = \sqrt{AP^2 - AM^2} = \sqrt{b^2 - a^2}$.

From the similarity of the right triangles $PLQ$ and $PMA$, it follows that $QL : PL = AM : PM$.

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_35476b28e4766e84fc81g-04.jpg?height=1104&width=1170&top_left_y=0&top_left_x=884)

$r = QL = PL \cdot \begin{aligned} & AM \\ & PM \end{aligned} = (b-a) \cdot \begin{aligned} & a \\ & \sqrt{b^2 - a^2} \end{aligned} = \begin{aligned} & a(b-a) \\ & \sqrt{b^2 - a^2} \end{aligned}$.

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