255. A finite set $K_{0}$ is given on a plane (or in space). To it are added all points that can be obtained by symmetric reflection of one point of this set relative to another. The resulting set is denoted as $K_{1}$. Similarly, from the set $K_{1}$, $K_{2}$ is obtained, from $K_{2}$ - $K_{3}$, and so on.

a) Let the set $K_{0}$ consist of two points $A$ and $B$ at a distance of 1. For what smallest $n$ will the set $K_{n}$ contain a point at a distance of 1000 from point $A$?

b) Let $K_{0}$ consist of the three vertices of an equilateral triangle with an area of 1. Find the area of the smallest convex polygon containing the set $K_{n}$ (for $n=1,2, \ldots$).

In the following parts, $K_{0}$ is a set of four points that are the vertices of a regular tetrahedron with a unit volume.

c) Consider the smallest convex polyhedron containing all points of the set $K_{1}$. How many and what types of faces does this polyhedron have?

d) What is the volume of this polyhedron?

e)* Find the volume of the smallest convex polyhedron containing the sets $K_{n}$ (for $n=2,3, \ldots$).
255. a) Answer: $n=7$.

The set $K_{n}$ consists of points on the line $A B$, removed from $A$ and $B$ by integer distances, and the extreme points of this set are removed from the midpoint of the segment $A B$ by a distance of $3^{n} / 2$ $(n=1,2,3, \ldots)$, i.e., from $A$ - by distances $\left(3^{n}-1\right) / 2$ and $\left(3^{n}+1\right) / 2$. Since $\left(3^{6}+1\right) / 2=365$ $>1000$, the required set is $-K_{7}$.

b) The polygons that are the convex hulls (see P17) of the sets $K_{1}, K_{2}, \ldots$ are hexagons (Fig. 104, a). The vertices of the convex hull $K_{n}$ are obtained as follows: take each pair of vertices of the triangle $A B C$ and perform $n$ operations of "symmetric reflection" - as in part a). To prove that all points of $K_{n}$ lie within the convex hull $H_{n}$ of the six constructed "extreme points," it is convenient to consider $H_{n}$ as the intersection of three strips, the edges of which run along the opposite sides of the hexagon $H_{n}$ ( $n=1,2,3, \ldots$ ). The initial triangle $A B C$ can also be considered as the intersection of three strips; one strip has one edge along the line $A B$ and the other passes through the point $C$, and two analogous ones. All symmetric reflections of a strip relative to any of its points belong to a strip with the same axis but three times wider. Using this, it can be proven that $K_{n}(n=1,2,3, \ldots)$ belongs to each of the three strips, the intersection of which gives $H_{n}$.

The area of $H_{n}$ can be calculated as a combination of the areas of triangles homothetic to $A B C$. It is equal to $\left(3^{2 n+1}-1\right) / 2$.
![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-227.jpg?height=434&width=926&top_left_y=396&top_left_x=162)

Fig. 104

c) The set $K_{1}$ contains, in addition to the points of $K_{0}$, 12 more points obtained by reflecting one of the vertices of the tetrahedron $A B C D$ relative to another. It is convenient to represent these 12 "extreme" points of $K_{1}$ by constructing a cube $L_{0}$, whose points $A, B, C, D$ are non-adjacent vertices, and a cube $L_{1}$ homothetic to $L_{0}$ relative to the center with a coefficient of 3, with corresponding vertices $A_{1}, B_{18}$ $C_{1}, D_{1}: 12$ points of $K_{1}$ lie one by one on the edges of $L_{1}$ and divide them, counting from the vertices $A_{1}, B_{1}, C_{1}, D_{1}$, in the ratio $1: 2$. If from each vertex of $L_{1}$ a plane passing through 3 "extreme" points on the edges emanating from this vertex is used to cut off a pyramid, the remaining polyhedron $M_{1}$ - the convex hull of $K_{1}$ - will have 14 faces: 6 rectangles $a \times 2 a$ (where $a$ is the length of the edge of the tetrahedron $A B C D$) and 8 equilateral triangles - four with side $a$ and 4 with side $2 a$ (Fig. 104, b).

d) The volume of $M_{1}$ can be calculated by subtracting the volume of the cut-off pyramids $\left(4 \cdot \frac{1}{2}+4 \cdot 4=18\right)$ from the volume of the cube (81). It is equal to 63.

e) The convex hull $M_{n}$ of the set $K_{n}$ will be a convex polyhedron with vertices at the 12 "extreme" points constructed for each pair of vertices of the tetrahedron $A B C D$. The vertices of $M_{n}$ lie one by one on the edges of the cube $L_{n}$, obtained from $L_{0}$ by central homothety with a coefficient of $3^{n}$, and divide them in the ratios $\left(3^{n}-1\right):\left(3^{n}+1\right)$.

In the proof, it is convenient to consider 7 strips, the intersection of which gives $M_{n}$: each of them is obtained by expanding in $3^{n}$ times from a strip whose edges contain all four vertices of the tetrahedron $A B C D$. The volume of $M_{n}$ is equal to $\left(5 \cdot 3^{3 n}-3^{n+1}\right) / 2$.