\section*{Problem 1 - 181011}

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1402.jpg?height=320&width=328&top_left_y=477&top_left_x=864)

The figure shows four parallel lines \(g_{1}, g_{2}, g_{3}, g_{4}\), where a uniform length \(a\) for each \(i=1,2,3,4\) is the distance between \(g_{i}\) and \(g_{i+1}\), and four other lines \(h_{1}, h_{2}, h_{3}, h_{4}\) perpendicular to the \(g_{i}\), where \(a\) for each \(i=1,2,3,4\) is also the distance between \(h_{i}\) and \(h_{i+1}\).

Furthermore, the figure shows a numbering of the resulting intersection points.

a) Determine the number of all those squares that have only numbered points as vertices and only segments lying on lines \(g_{i}\) or \(h_{i}\) as sides.

b) Investigate whether it is possible to color all 16 numbered points using the colors Red, Blue, Green, and Yellow (each numbered point with exactly one of these colors) such that for the vertices of each square mentioned in a), all four colors appear!
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a) The mentioned squares are exactly the following:

9 squares of side length a,

4 squares of side length 2a,

1 square of side length 3a,

thus exactly 14 squares in total.

b) Suppose there were such a coloring. In it,

A denotes the color of point 1,

\(B\) denotes the color of point 3,

\(C\) denotes the color of point 11,

\(D\) denotes the color of point 9,

so \(A, B, C, D\) in some order are all four colors Red, Blue, Green, Yellow. Furthermore, it follows that:

Point 6 cannot have the color \(A\), because otherwise this color would appear for two corners of the square 1/2/6/5, so that one of the other colors would be missing at its corners.

Similarly, by considering the squares 2/3/7/6, 6/7/11/10, 5/6/10/9, it is derived that point 6 cannot have the colors \(B, C, D\) either.

This contradiction proves that there is no coloring of the kind mentioned in (b).