Extreme Cut-the-Rope

Lemma: there must be a line between any two vertices.

Proof: suppose this is not true, that there are two vertices without a line between them. Then their shared edge would not be cut, contradiction.

Lemma: the number of sections the lines divide the plane must be equal or more than the number of vertices, $2014$ .

Proof: suppose that we have less sections of the plane than vertices. Then, by Pigeonhole Principle, we must have at least $2$ vertices in a section of the graph. But then these two vertices, being part of the same section, cannot have a line between them, so their edge is not cut, contradiction.

Thus, for $n$ straight lines, if the maximum number of sections they divide the plane is less than $2014$ , then it is not possible to cut all edges with $n$ lines.

Now we just have to calculate the smallest $n$ such that the number of sections they create is more than $2014$ .

First off, the formula for the maximum number of regions cut with $n$ lines is $\dfrac{n^2+n+2}{2}$ .

Thus, we want $\dfrac{n^2+n+2}{2}\ge 2014$

$n^2+n+2\ge 4028$

$n^2+n\ge 4026$

We approximate the square root of $4026$ to be $\approx 63\to 64$ , so we check $n$ around this region.

Checking for $n=62$ , we see that $62^2+62=3096 < 4028$

Checking for $n=63$ , we see that $63^2+63=4032 > 4028$ , so the correct answer is $\boxed{n=63}$ .

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Note that we barely made it $n=63$ ; if we, instead, had $2018$ vertices, then $n=64$ is necessary.