# Problem 4.

In-1 The angle at the vertex in the axial section of a cone is $60^{\circ}$. Outside this cone, there are 11 spheres of radius 3, each touching two adjacent spheres, the lateral surface of the cone, and the plane of its base. Find the radius of the base of the cone.
Answer: $\frac{3}{\sin \frac{\pi}{11}}-\sqrt{3}$.

Solution. The general condition: Around (or inside) a cone with an angle at the vertex in the axial section equal to $2 \alpha$, there are $n$ balls of radius $r$, each of which touches two adjacent balls, the lateral surface of the cone, and the plane of its base. Find the radius of the base of the cone.

General answer: $R=r\left(\frac{1}{\sin \frac{\pi}{n}}-\frac{1}{\operatorname{tg}\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)}\right)$ or $R=r\left(\frac{1}{\sin \frac{\pi}{n}}+\frac{1}{\operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}\right)$.

General solution. Let $O$ be the center of the circle at the base of the cone, with radius $R, Q_{1}$ be the center of one of the balls with radius $r, H_{1}$ be the point of tangency of this ball with the plane of the base (the ball is either inside or outside the cone, see Fig.1), $H_{2}$ be the point of tangency of the adjacent ball with the plane of the base of the cone (see Fig.2). Therefore, for external tangency, $A H_{1}=\frac{r}{\operatorname{tg}\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)}$, and for internal tangency, $A H_{1}=\frac{r}{\operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}$ (see Fig.1). Since each ball touches two adjacent balls, the points of tangency of these balls with the plane of the base of the cone are located at the vertices of a regular polygon inscribed in a circle with center at point $O$, radius $O H_{1}$, and side length equal to $2 r$. Therefore, $r=O H_{1} \sin \frac{\pi}{n}$, where $O H_{1}=R+A H_{1}$ or $O H_{1}=R-A H_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_453d03b8571fc3668c57g-4.jpg?height=503&width=617&top_left_y=1533&top_left_x=160)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_453d03b8571fc3668c57g-4.jpg?height=526&width=580&top_left_y=1439&top_left_x=978)

Fig. 2

V-2 The angle at the vertex in the axial section of the cone is $60^{\circ}$. Inside this cone, there are 13 balls of radius 2, each of which touches two adjacent balls, the lateral surface of the cone, and the plane of its base. Find the radius of the base of the cone.

Answer: $\frac{2}{\sin \frac{\pi}{13}}+2 \sqrt{3}$

V-3 The angle at the vertex in the axial section of the cone is $60^{\circ}$. Outside this cone, there are 17 balls of radius 2, each of which touches two adjacent balls, the lateral surface of the cone, and the plane of its base. Find the radius of the base of the cone.

Answer: $\frac{2}{\sin \frac{\pi}{17}}-\frac{2}{\sqrt{3}}$.

V-4 The angle at the vertex in the axial section of the cone is $60^{\circ}$. Inside this cone, there are 19 balls of radius 3, each of which touches two adjacent balls, the lateral surface of the cone, and the plane of its base. Find the radius of the base of the cone.

Answer: $\frac{3}{\sin \frac{\pi}{19}}+3 \sqrt{3}$.

## Lomonosov Olympiad for Schoolchildren in Mathematics

Final stage $2021 / 22$ academic year for 11th grade

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