10.32 Given a finite set $K_{0}$ in the plane (or in space), add to $K_{0}$ all points obtained by reflecting a point in $K_{0}$ about another point in $K_{0}$, to get the set $K_{1}$. Similarly, we can obtain $K_{2}$ from $K_{1}$, $K_{3}$ from $K_{2}$, and so on.
(1) Suppose the set $K_{0}$ consists of two points $A$ and $B$ that are 1 unit apart. What is the smallest value of $n$ such that there is a point in $K_{n}$ that is 1000 units away from $A$?
(2) Suppose the set $K_{0}$ consists of the three vertices of an equilateral triangle with area 1. Find the area of the smallest convex polygon, i.e., the convex hull, that contains $K_{n}$, for $n=1,2, \cdots$.
In the following sub-questions, assume that $K_{0}$ consists of the four vertices of a unit volume tetrahedron.
(3) How many faces does the convex hull of $K_{1}$ have, and what are the shapes of these faces?
(4) What is the volume of the convex hull of $K_{1}$?
(5) Find the volume of the convex hull of $K_{n}$, for $n=1,2, \cdots$.
[Solution] (1) It is easy to see that the set $K_{n}$ consists of all points on the line $AB$ that have an integer distance from $A$ and whose distance from the midpoint of the segment $AB$ does not exceed $-\frac{1}{2} \times 3^{n}$, for $n=1,2, \cdots$. Since $\frac{1}{2}\left(3^{6}+1\right)=3651000$, the smallest value of $n$ is 7.
(2) The convex hull of $K_{n}$ is a hexagon. Taking any two points in $K_{0}$ and performing $n$ symmetric reflections as in (1), we can obtain these six vertices.

To prove that all points of $K_{n}$ are within the convex hull $H_{n}$ formed by the six "support points," we can view $H_{n}$ as the intersection of three strips, each defined by a pair of opposite sides of the hexagon $(n=1,2, \cdots)$. Clearly, the initial equilateral triangle is the intersection of three strips, each of which has a side of the triangle and a parallel line through the opposite vertex. After each symmetric reflection, the three strips remain as three strips, with the center line unchanged but the width increasing by a factor of 3. Thus, the intersection of the three strips, i.e., $K_{n}$, remains a hexagon.
The area of $H_{n}$ can be calculated using the areas of triangles that are similar to $\triangle ABC$ in the same direction and in the opposite direction, and it equals $\frac{1}{2}\left(3^{2 n+1}-1\right)$.
(3) In addition to the points of $K_{0}$, $K_{1}$ also includes 12 points that are images of symmetric reflections. It can be seen that these 12 points are the support points of the convex hull of $K_{1}$. We construct a cube $L_{0}$ such that the four vertices of the given tetrahedron $ABCD$ are exactly one set of four non-adjacent vertices of the cube. By performing a homothety with a coefficient of 3 about the center of $L_{0}$, we obtain the cube $L_{1}$ and its corresponding non-adjacent vertices $A_{1}, B_{1}, C_{1}, D_{1}$. At this point, the 12 support points of the convex hull of $K_{1}$ are precisely on the 12 edges of the cube $L_{1}$, and each point divides its edge in the ratio $1:2$, with the shorter segment having one of $A_{1}, B_{1}, C_{1}, D_{1}$ as an endpoint (see the figure above). If a plane is used to cut off a cone from each vertex of $L_{1}$, the remaining polyhedron $M_{1}$ is the convex hull of $K_{1}$, where the planes pass through one support point on each of the three edges emanating from the vertex. In this way, it is easy to see that $M_{1}$ has 14 faces, of which 6 are $a \times 2a$ rectangles (where $a$ is the edge length of the tetrahedron $ABCD$), and 8 are equilateral triangles, 4 with side length $a$ and the other 4 with side length $2a$.
(4) The volume of $M_{1}$ can be calculated by subtracting the volumes of the 8 truncated cones $\left(4 \times \frac{1}{2}+4 \times 4=18\right)$ from the volume of the cube $L_{1}$ (which is 81), so the volume of $M_{1}$ is 63.
(5) The convex hull $M_{n}$ of the set $K_{n}$ is similar to $M_{1}$, also being a convex polyhedron with 12 support points as vertices, which are obtained by performing $n$ symmetric reflections on the vertices of the tetrahedron $ABCD$. The vertices of $M_{n}$ are on the edges of the cube $L_{n}$, dividing each edge in the ratio $\left(3^{n}-1\right):\left(3^{n}+1\right)$, where $L_{n}$ is the cube obtained by performing a homothety with a coefficient of $3^{n}$ about the center of $L_{0}$. Similar to (4), we can obtain that the volume of $M_{n}$ is $\frac{1}{2}\left(5 \times 3^{3 n}-3^{n+1}\right)$.