A unit-sided square can be covered by 9 smaller squares, each with a side length of $1 / 3$. The diagonal of a small square is $\sqrt{2} / 3 < 0.5$, so the distance between any two points within a small square is less than $1 / 2$. Consequently, we can place at most one point in each small square, meaning that the entire unit square can contain at most 9 points.

On the other hand, 8 points can be placed as desired (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_dd3d7085de5edb6fccf2g-1.jpg?height=513&width=525&top_left_y=367&top_left_x=781)

Figure 1

Thus, we can place a maximum of 8 or 9 points in the unit square such that the distance between any two points is greater than $1 / 2$.

Jenő Törőcsik (Bp., Fazekas M. Gyak. Gimn., IV. o. t.)

Note. We will show that the exact upper limit is 8. For this, assume that it is possible to place 9 points in the square according to the requirements—we will derive a contradiction from this.

Consider an arbitrary monotonically decreasing sequence $\{a_n\}$ that converges to 0. For each $n$, draw strips of width $a_n$ along the sides and midlines of the unit square (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_dd3d7085de5edb6fccf2g-1.jpg?height=648&width=650&top_left_y=1392&top_left_x=716)

Figure 2

These strips cut out 9 smaller squares of size $a_n \times a_n$ from the large square. If we can find a sequence $\{a_n\}$ such that for every $n$, one of the 9 points falls into each of the small squares, we are done: since $\lim a_n = 0$, the selected points can only be the vertices, midpoints of the sides, and the center of the square—contradicting the requirement that the distance between any two points is greater than $1 / 2$.

What remains is to provide a suitable sequence $\{a_n\}$. $a_1 = 1 / 3$ is clearly good: it divides the unit square into 9 smaller squares, and we have already seen that exactly one point must fall into each of these.

Assume that for some $a_n \leq 1 / 3$, we know that the points fall into the corresponding regions of Figure 2. Let $ABCD$ and $A'B'C'D'$ be two "adjacent" $a_n \times a_n$ squares (Figure 3), with the distance between them being $CB' = DA' = (1 - 3a_n) / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_02_dd3d7085de5edb6fccf2g-2.jpg?height=607&width=1095&top_left_y=133&top_left_x=496)

Figure 3

One of the 9 points falls into each of the $ABCD$ and $A'B'C'D'$ squares. The distance between these two points is greater than $1 / 2$. Therefore, for example, the point in the $ABCD$ square cannot be closer than $PC = QD = a_n / 10$ to the side $CD$. If we take any point in the rectangle $PCDQ$, its distance from any point in the $A'B'C'D'$ square—thus from the selected point in it—is at most

\[
\begin{gathered}
PD' = \sqrt{PQ^2 + QD'^2} = \\
= \sqrt{a_n^2 + \left(\frac{a_n}{10} + \frac{1 - 3a_n}{2} + a_n\right)^2} = \\
= \sqrt{\frac{1}{4} - \frac{2}{5} a_n (1 - 2.9a_n)}
\end{gathered}
\]

which is less than $1 / 2$ for $0 < a_n \leq 1 / 3$.

If we remove all such "forbidden" rectangles from the squares in Figure 2, the 9 points must fall into the remaining regions (Figure 4).

![](https://cdn.mathpix.com/cropped/2024_05_02_dd3d7085de5edb6fccf2g-2.jpg?height=910&width=900&top_left_y=1510&top_left_x=599)

Figure 4

These regions, however, are covered by the smaller squares of size $a_{n+1} = 0.9a_n$. Therefore, exactly one of the given points must fall into each of these smaller squares.

The sequence defined by $a_1 = 1 / 3$ and $a_{n+1} = 0.9a_n$ is thus suitable, and we have found a suitable sequence $\{a_n\}$. This proves the statement.
In the unit square, we placed points such that the distance between any two is greater than $1 / 2$. Provide an upper bound for the number of points.

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In the unit square, we placed points such that the distance between any two is greater than $1 / 2$. Provide an upper bound for the number of points.