51. What is the greatest number of squares with side 1 that can be placed ${ }^{1}$ ) next to a given unit square $K$ so that no two of them intersect?



$^{1}$) The superscript "1" is kept as is, since it might refer to a footnote or additional information in the original text.
51. First solution. Let $O$ be the center of the main square $K$, and $O_{1}$ and $O_{2}$ be the centers of the non-overlapping squares $K_{1}$ and $K_{2}$ attached to it (Fig. $135, a$). Since the smallest distance from the center of a unit square to its boundary is $\frac{1}{2}$, the segments $O O_{1}$, $O O_{2}$, and $O_{1} O_{2}$ are not less than 1. Furthermore, since the squares $K$ and $K_{1}$ touch at some point $A$, we have $O O_{1} \leqslant O A + O_{1} A \leqslant \sqrt{2}$ (since the greatest distance from the center of a unit square to its boundary is $\frac{\sqrt{2}}{2}$). Therefore, the points $O_{1}$ and $O_{2}$ lie within the annulus centered at $O$ and bounded by circles of radii 1 and $\sqrt{2}$; and since $O_{1} O_{2} \geqslant 1$, the angle $O_{1} O O_{2}$, by the result of problem 46, is not less than $2 \arcsin \frac{1}{2 \sqrt{2}} \approx 41.4^{\circ}$. From this,

![](https://cdn.mathpix.com/cropped/2024_05_21_962b75a91b7185035ab9g-166.jpg?height=390&width=396&top_left_y=454&top_left_x=209)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_962b75a91b7185035ab9g-166.jpg?height=239&width=234&top_left_y=601&top_left_x=739)

b)

Fig. 135.

it follows that more than eight non-overlapping unit squares cannot be attached to the square $K$, since $9 \times 41.4^{\circ} > 360^{\circ}$. That eight squares can be attached is evident from Fig. 135, b.

Note. This problem can also be solved without using trigonometric tables. $2 \arcsin \frac{1}{2 \sqrt{2}}$ is the angle at the vertex of an isosceles triangle with legs $\sqrt{2}$ and base 1. If we attach nine such triangles to each other as shown in Fig. 136, the sum of the lengths of the bases of all these triangles will exceed the length of the circumference of radius $\sqrt{2}$; indeed, $2 \pi \sqrt{2} < 360^{\circ}$, which is what we needed to show.

Second solution. Let $A B C D$ be the original square $K$; we will take the length of its side as the unit. Consider the square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with side 2, having the same center as the square $K$ and the same orientation of sides (Fig. $137, a$); the perimeter of the square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is obviously 8. We will show that the length $l$ of the part of the perimeter of the square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ cut by the side of one attached square cannot be less than 1; from this it will follow that no more than eight non-overlapping squares can be attached to $K$. Consider all possible positions of the attached square.

![](https://cdn.mathpix.com/cropped/2024_05_21_962b75a91b7185035ab9g-167.jpg?height=432&width=438&top_left_y=574&top_left_x=225)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_962b75a91b7185035ab9g-167.jpg?height=496&width=478&top_left_y=574&top_left_x=623)

b)

Fig. 137.

$1^{\circ}$ Suppose no vertex of the square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ falls inside the attached square (cases of squares $E F G H$ and $E_{1} F_{1} G_{1} H_{1}$ in Fig. 137, a). If in this case only one vertex of the attached square falls inside the square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ (as in the case of square $E F G H$), then, denoting the angle between the sides $B C$ and $E F$ by $\alpha$, we have

$$
\begin{aligned}
& l = M K + K N = E K \operatorname{tg} \alpha + E K \operatorname{ctg} \alpha = \frac{1}{2} \left( \operatorname{tg} \alpha + \frac{1}{\operatorname{tg} \alpha} \right) = \\
& = 1 + \frac{1}{2} \left( \sqrt{\operatorname{tg} \alpha} - \frac{1}{\sqrt{\operatorname{tg} \alpha}} \right)^2 \geqslant 1
\end{aligned}
$$

If two vertices of the attached square fall inside the square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ (as in the case of square $E_{1} F_{1} G_{1} H_{1}$), then obviously $l = M_{1} N_{1} \geqslant N_{1} L_{1} = 1$.
$2^{\circ}$ Suppose one vertex of the square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ falls inside the attached square (squares $A F_{2} C_{2} H_{2}$, $E_{3} F_{3} C_{3} H_{3}$, and $E_{4} F_{4} G_{4} H_{4}$ in Fig. 137, b). If the attached square touches the square $A B C D$ at a point on the boundary other than a vertex (case of square $E_{3} F_{3} G_{3} H_{3}$), then

$$
\begin{aligned}
l = M B^{\prime} + B^{\prime} N & = \left( P B^{\prime} + B^{\prime} Q \right) + (Q N - P M) = \\
& = 1 + (Q N - P M).
\end{aligned}
$$

But (Fig. $137, \sigma$)

$P M = E_{3} P \operatorname{tg} \alpha, \quad Q N = R Q \operatorname{tg} \alpha, \quad R Q - E_{3} P = R B - E_{3} B \geqslant 0$, since $\alpha \leqslant 45^{\circ}$. Therefore, $Q N > P M$ and $l > 1$. These same arguments apply in the case where the attached square touches the square $A B C D$ at a vertex (case of square $A F_{2} G_{2} H_{2}$); in this case only $P^{\prime} M^{\prime} = Q^{\prime} N^{\prime}$ and, therefore, $l = M^{\prime} A^{\prime} + A^{\prime} N^{\prime} = 1$. Finally, if the attached square touches the vertex of the square $A B C D$ with a side (case of square $E_{4} F_{4} G_{4} H_{4}$), then

$$
l = M_{1} D^{\prime} + D^{\prime} N_{1} > M_{1} N_{1} \geqslant G_{4} F_{4} = 1
$$

Thus, in all cases, the length of the part $M N$ of the perimeter of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ indeed cannot be less than 1.