$\left.\begin{array}{l}{\left[\begin{array}{l}\text { Cone }\end{array}\right]} \\ {[\underline{\text { Scrap (other) }}]}\end{array}\right]$

The axial section of the cone is an equilateral triangle with a side length of 1. Find the radius of the sphere that touches the axis of the cone, its base, and the lateral surface.
Let a sphere of radius $r$ with center $Q$ touch the base plane of a cone with vertex $A$ at point $M$, the axis $A O-$ at point $P$ ($O$ is the center of the base of the cone), and the lateral surface at point $K$ (Fig.1). Consider the section of the cone and the sphere by a plane passing through the parallel lines $Q M$ and $A O$ (Fig.2). We obtain an equilateral triangle $A B C$ with side length 1 and a circle of radius $r$ with center $Q$, touching side $B C$ at point $M$, height $A O$ at point $P$, and side $A B$ at point $K$. Since $r$ is the radius of the circle inscribed in the right triangle $A O B$ with legs $A O=\frac{\sqrt{3}}{2}, O B=\frac{1}{2}$ and hypotenuse $A B=1$, then

$$
r=\frac{1}{2}(A O+O B-A B)=\frac{1}{2}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}-1\right)=\frac{\sqrt{3}-1}{4}
$$

## Answer

$\frac{\sqrt{3}-1}{4}$