[ Linear dependence of vectors ] [ Angles between lines and planes $]$

The side of the base of a regular triangular pyramid is equal to $a$. The lateral edge forms an angle of $60^{\circ}$ with the base plane. Find the radius of the sphere inscribed in the pyramid.
Let $ABCP$ be a given regular triangular pyramid with vertex $P, AB=BC=AC=a, M$ - the center of the equilateral triangle $ABC, L$ - the midpoint of $BC, \angle PAM=\angle PBM=\angle PCM=60^{\circ}$. Since the pyramid is regular, $PM$ is its height. From the right triangle $PAM$, we find that

$$
PM=AM \operatorname{tg} \angle PAM=\frac{2}{3} \cdot \frac{a \sqrt{3}}{2} \cdot \operatorname{tg} 60^{\circ}=\frac{a \sqrt{3}}{3} \cdot \sqrt{3}=a
$$

Since $PL \perp BC$ and $ML \perp BC$, the angle $PLM$ is the linear angle of the dihedral angle between the plane of the lateral face $BCP$ and the plane of the base $ABC$. From the right triangle $PLM$, we find that

$$
\operatorname{tg} \beta=\operatorname{tg} \angle PLM=\frac{PM}{ML}=\frac{\frac{a}{a}}{2 \sqrt{3}}=2 \sqrt{3}
$$

The center $O$ of the sphere inscribed in the regular pyramid $ABC$ lies on its height $PM$, and since this sphere is inscribed in the dihedral angle between the planes of the faces $BCP$ and $ABC$, the point $O$ lies in the bisector plane of this angle. Consider the section of the pyramid by the plane $APL$. We get the triangle $APL$ and a circle tangent to its two sides $LP$ and $LA$, and the side $KC$ - at point $M$. The radius $r$ of this circle is equal to the radius of the sphere inscribed in the pyramid, the center $O$ lies on the height $PM$, and $\angle OLM=\frac{9}{2}$. From the right triangle $OML$, we find that

$$
r=OM=LM \operatorname{tg} \frac{\frac{g}{2}}{\frac{a \sqrt{3}}{6}} \cdot \operatorname{tg} \frac{\frac{3}{2}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_db48a70f246fb4a097d0g-04.jpg?height=120&width=783&top_left_y=543&top_left_x=0)

$$
2 \sqrt{3}=\frac{2 \operatorname{tg}^{\frac{3}{2}}}{1-\operatorname{tg}^2 \frac{3}{2}}
$$

from which we find that

$$
\operatorname{tg} \frac{9}{2}=\frac{\sqrt{13}-1}{2 \sqrt{3}}
$$

Therefore,

$$
r=\frac{a \sqrt{3}}{6} \cdot \operatorname{tg} \frac{\frac{9}{2}}{2}=\frac{a \sqrt{3}}{6} \cdot \frac{\sqrt{13}-1}{2 \sqrt{3}}=\frac{a(\sqrt{13}-1)}{12}
$$

## Answer

$\frac{a(\sqrt{13}-1)}{12}$.