Lemma. Among any five nodes of a grid of equilateral triangles, there will be two such that the midpoint of the segment between them is also a grid node.

![](https://cdn.mathpix.com/cropped/2024_05_06_ef1f6e5b136aa760f4dfg-25.jpg?height=368&width=537&top_left_y=1734&top_left_x=766)

Proof of the lemma. Introduce the origin at one of the grid nodes and denote by $\$ \mid v e c\{a\} \$$ and $\$ \mid v e c\{b\} \$$ the radius vectors to the two nearest nodes, as shown in the figure. Then the grid nodes are points of the form $\$ m \mid v e c\{a\}+ n \mid v e c\{b\} \$$ for integers \$m $\$$ and \$n $\$$. By the pigeonhole principle, among five points, there will be two points $\$m_1 \mid v e c\{a\}+ n_1 \mid v e c\{b\} \$$ and $\$m_2 \mid v e c\{a\}+ n_2 \mid v e c\{b\} \$$, for which the parity of \$m_1 and \$ \$m_2\$, and the parity of \$n_1 and \$n_2\$, coincide simultaneously. The midpoint of the segment connecting these two points is the point $\$ \backslash frac \left\{m \_1+m \_2\right\} 2 \backslash m k e r n 2 m u \operatorname{vec}\{a\}+\backslash f r a c\left\{n \_1+n \_2\right\} 2 \backslash m k e r n 2 m u \operatorname{vec}\{b\} \$$$. It is a grid node, since the numbers $\$ \backslash frac \left\{m \_1+m \_2\right\} 2 \$$ and $\$ \backslash frac \left\{n \_1+n \_2\right\} 2 \$$ are integers due to the same parity of \$ \mathrm{~m} \_1 \$$ and \$m_2\$, and \$n_1\$, and \$n_2\.

Solution. In the figure on the left, you can see an example of the placement of 8 grid nodes, among which there are no two such that the midpoint of the segment between them is a grid node. We will prove that nine nodes are sufficient. Note that the hexagonal grid is divided into the union of two triangular grids (see the figure on the right). By the pigeonhole principle, among any nine nodes, at least five will be in one of these two triangular grids. By the lemma, among these five nodes, there will be two of the required ones.

## Answer

9.
Kuuyggin A.K.

Among any five nodes of a regular square grid, there will always be two nodes such that the midpoint of the segment between them is also a node of the grid. What is the minimum number of nodes of a regular hexagonal grid that must be taken so that among them there will always be two nodes such that the midpoint of the segment between them is also a node of this grid?