Galperin G.A.

What is the maximum number of points that can be placed a) on a plane; b)* in space so that no triangle with vertices at these points is obtuse?

(Of course, it is implied in the condition that no three points should lie on the same line - without this restriction, any number of points can be placed.)
Let us prove that a larger number of points cannot be placed.

First of all (this applies to both the planar and spatial problems), if $A_{i}$ and $A_{j}$ are any two of the $n$ points $A_{1}, A_{2}, \ldots, A_{n}$ satisfying the problem's condition, then all these points must belong to the set of points $M$ for which simultaneously $\angle M A_{i} A_{j} \leq 90^{\circ}$ and $\angle M A_{j} A_{i} \leq 90^{\circ}$. This set is a strip between lines (in space - planes) passing through points $A_{i}$ and $A_{j}$ and perpendicular to the segment $A_{i} A_{j}$ (Fig.3); we will denote this strip by $\Pi_{i j}$ below.

From this observation, it follows that if we consider the convex hull $V$ of the given $n$ points - the smallest convex polygon (polyhedron) containing all these points - then all points $A_{1}, A_{2}, \ldots, A_{n}$ must lie on the boundary of the set $V$; none of them can be inside $V$: after all, $V$ is contained in each of the strips $\Pi_{i j}$.

The solution to part (a) is now in two words. In this case, $V$ is a convex $n$-gon with vertices $A_{1}, A_{2}, \ldots, A_{n}$. The sum of its angles is $180^{\circ}(n-2)$, and if each of the angles is no more than $90^{\circ}$, then

$$
180^{\circ}(n-2) \leq 90^{\circ} n
$$

from which $n \leq 4$.

The solution to the planar problem was submitted by many readers. The spatial problem is much more difficult, and we did not receive a complete proof from any of the readers. We will present a solution that was first found by the well-known geometers L. Danzer and B. Grünbaum (1962).

Consider, in addition to the polyhedron $V$ (here $V$ can be understood as either the convex hull of the given $n$ points $A_{1}, A_{2}, \ldots, A_{n}$ or the intersection - the common part - of all $n(n-1) / 2$ strips $\Pi_{i j}$) the following polyhedra: $V_{i}$,

obtained from $V$ by translation by the vector $\overrightarrow{A_{1} A_{i}}\left(i=1,2, \ldots, n ; V_{1}\right.$, of course, coincides with $V$), and $V V^{\prime}$, obtained from $V$ by stretching (homothety) with a coefficient of 2 and center at point $A_{1}$.

We will prove the following three statements.

$1^{o}$. The polyhedron obtained from $V$ by translation by the vector $\overrightarrow{A_{i} A_{j}}$ does not have common interior points with $V$ (i.e., it can intersect $V$ only on the boundary).

$2^{o}$. No two of the polyhedra $V_{1}, V_{2}, \ldots, V_{n}$ have common interior points.

$3^{0}$. All $V_{i}$ are contained in $V^{\prime}$.

Let us prove $1^{o}$. When translated by $\overrightarrow{A_{i} A_{j}}$, the strip $\Pi_{i j}$ transforms into a new strip, which does not have common interior points with it - these two strips have only a common boundary plane. But $V$ is contained in $\Pi_{i j}$, so $1^{0}$ is even more true.

$2^{o}$ immediately follows from $1^{\circ}$ if we note that $V_{j}$ is obtained from $V_{i}$ by translation by the vector $\overrightarrow{A_{i} A_{j}}$ (Fig.4). $3^{\circ}$ follows from a more general fact: if $A B C$ are three points of a convex polyhedron, $W$ and $W^{\prime}$ is the polyhedron obtained from $W$ by stretching by a factor of 2 with center at point $A$, then the fourth vertex $D$ of the parallelogram $A B D C$ (i.e., the point obtained from $C$ by translation by the vector $\overrightarrow{A B}$) belongs to $W^{\prime}$. This fact is easily proven: the point $K$, from which $D$ is obtained by stretching, is the midpoint of segment $A D$ - it is also the midpoint of segment $B C$, so it belongs to $W$ (since the polyhedron $W$ is convex), so $D$ belongs to $W^{\prime}$ (Fig.5).

Thus, statements $2^{o}$ and $3^{o}$ are proven. Let $v$ be the volume of the polyhedron $V$. Then the volume of each of the $V_{i}$ is also equal to $v$, and the volume of $V^{\prime}$ is equal to $8 v$. From $2^{o}$ and $3^{\circ}$, it obviously follows that $n v \leq 8 v$, from which $n \leq 8$.

Note that in passing, we have solved a problem proposed at the XIII International Olympiad (see "Kvant" #12 for 1971, p.54, problem 2). Apparently, the full problem M130 b) was considered too difficult for the Olympiad by the international jury and was excluded, leaving only the first half.

In connection with the proven statement, a whole series of questions arise, which cannot be solved by the beautiful but rather artificial method we have described. For example, what is the maximum number of points that can be placed in space so that all angles of triangles with vertices at these points are acute? From our solution, it is clear that 8 points cannot be placed (convince yourself of this). It is not difficult to construct an example with 5 points. Can there be 6? A more general and, probably, very difficult question - what is the maximum number of points that can be placed so that all angles do not exceed a given $\alpha$?

If any of the readers manage to make progress in solving these questions, we will return to them again.

## Answer

Answer: a) 4 points, b) 8 points. Examples of such placement of points: a) at the vertices of a square; b) at the vertices of a cube.