## Task B-3.5.

A sphere with a radius of 1 is inscribed in a right circular cone. The slant height of the cone forms a $60^{\circ}$ angle with the base of the cone. The cone is intersected by a plane that is parallel to the base of the cone and tangent to the sphere. What is the area of the intersection of this plane and the cone?
## First Solution.

The axial section of the given cone and the inscribed sphere is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_c317f9bfd22736a3ecebg-25.jpg?height=765&width=511&top_left_y=1379&top_left_x=698)

From the right triangle \(A N S\) with the given labels, we get

\[
\operatorname{tg} \frac{\alpha}{2} = \frac{r}{R} \Rightarrow R = \frac{r}{\operatorname{tg} \frac{\alpha}{2}} = \frac{1}{\operatorname{tg} 30^{\circ}} = \sqrt{3}
\]

The trapezoid \(A B C D\) is tangential, so it holds that \(2 R + 2 r_{1} = 2 y \Rightarrow y = R + r_{1} = \sqrt{3} + r_{1}\).

Furthermore, from the right triangle \(E B C\) we get \(\sin \alpha = \frac{2 r}{y}\) so

\[
y = \frac{2 r}{\sin \alpha} = \frac{2}{\sin 60^{\circ}} = \frac{4 \sqrt{3}}{3}
\]

Since \(r_{1} = y - R\), we finally get \(r_{1} = \frac{4 \sqrt{3}}{3} - \sqrt{3} = \frac{\sqrt{3}}{3}\).

Thus, the area of the intersection of the plane and the cone is \(P = r_{1}^{2} \cdot \pi = \frac{\pi}{3}\).