[ Coverings ]

It is known that a set $M$ of points on a line can be covered by three segments of length 1.

What is the smallest number of segments of length 1 that can certainly cover the set of midpoints of segments with endpoints in the points of the set $M$?

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If point $A$ is covered by the segment $[a, a+1]$, and point $B$ by the segment $[b, b+1]$, then the midpoint of segment $A B$ is covered by the segment $\left[a+b / 2, a+b / 2+1\right]$.

## Solution

Let two points $A$ and $B$ be covered by one unit segment. Then the midpoint of segment $A B$ is also covered by this segment. If point $A$ is covered by the unit segment $[a, a+1]$, and point $B$ by the segment $[b, b+1]$, then the midpoint of segment $A B$ is covered by the unit segment $\left[a+b / 2, a+b / 2+1\right]$.

Example. Let $[a, a+1],[b, b+1],[c, c+1]$ be three segments covering the given set $M$. As shown above, the set of midpoints of segments with endpoints in the points of set $M$ is covered by six segments $[a, a+1],[b, b+1],[c, c+1],\left[a+b / 2, a+b / 2+1\right],\left[a+c / 2, a+c / 2+1\right],\left[b+c / 2, b+c / 2+1\right]$.

Estimate. Let $M=[0,1] \cup[2,3] \cup[6,7]$. Then the set of midpoints of segments with endpoints in the points of set $M$ is $[0,5] \cup [6,7]$. Therefore, it is impossible to cover with fewer than six segments of length 1.

## Answer

Six segments.