1. In a regular polygon with 67 sides, we draw all segments connecting two vertices, including the sides of the polygon. We choose $n$ of these segments and assign each one of them a color from 10 possible colors. Find the minimum value of $n$ that guarantees, regardless of which $n$ segments are chosen and how the colors are assigned, there will always be a vertex of the polygon that belongs to 7 segments of the same color.

## SOLUTION:

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Let's first see that with $n=2010$ it is not sufficient.

We will say that a segment is of size $r$ if it connects two vertices between which, following the shortest path along the sides of the polygon, there are $r-1$ other vertices. We choose the 2010 segments of size greater than 3. For each $r \in\{1,2, \ldots, 10\}$, we assign the color $r$ to the segments of size $3 r+1, 3 r+2$, and $3 r+3$. It is obvious that each vertex belongs to 6 segments of each color.

Now let's prove that if $n=2011$, there is some vertex that is in 7 segments of the same color. Indeed, in the 2011 segments, 4022 vertices are involved, counting repetitions, so, by the pigeonhole principle, since $4022>60 \times 67$, some vertex is involved in at least 61 segments, of which, again by the pigeonhole principle, at least 7 will be of the same color.