## Solution:

The set of all vertices of the five unit squares is an example of a set of 12 points whose pairwise distances are all at least equal to 1. Consequently, $n \geq 13$. We now show that two of the 13 points always have a distance $<1$. To do this, divide the Swiss cross into 12 sub-areas as shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_04_28_a20106fbf267e8f87defg-8.jpg?height=451&width=397&top_left_y=820&top_left_x=858)

Figure 2: The 12 areas

The eight areas of the shape $a$ are rectangles of size $0.5 \times 0.8$. If two points $P, Q$ lie in or on the boundary of such a rectangle, then their distance is, according to Pythagoras' theorem, at most

$$
|P Q| \leq \sqrt{0.5^{2}+0.8^{2}}=\sqrt{0.89}<1
$$

The four areas of the form $b$ are each part of a square of side length 0.7 . Two points $P, Q$ in or on the edge of such an area then have a distance of at most

$$
|P Q| \leq \sqrt{0.7^{2}+0.7^{2}}=\sqrt{0.98}<1
$$

According to the drawer principle, two of the 13 points now lie in or on the edge of the same area, i.e. they are $<1$ apart.
5. a Swiss cross consists of five unit squares, one in the center and four on the sides. Determine the smallest natural number $n$ with the following property: For every $n$ points inside or on the edge of a Swiss cross, there are always two whose distance is less than 1.