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## Various geometry problems

1. Suppose we have a polygon $P$ in plane, not necessarily convex such that its area is strictly less than $1$. Prove that we can translate the polygon in such a way that there are no points with integer coordinates inside or on its edges.

Solution: Imagine we break the lattice into unit squares and overlap all of them on the square $[0,1]\times [0,1]$. Then, because the area of our polygon is strictly smaller than 1, there exists one point $A \in [0,1]\times [0,1]$ such that $A$ is not covered. Consider all the points $\mathcal{P}=\{ (a,b)+\overrightarrow{OA}: a,b \in \Bbb{Z}\}$ in the plane; by the property of $A$, these points are all outside $P$. Consider the translation of vector $\overrightarrow{OA}$, where $O$ is the origin of the lattice. Then all the lattice points are translated in the set $\mathcal{P}$, which does not intersect $P$. This means that a translation of $P$ with vector $-\overrightarrow{OA}$ leaves all lattice points outside $P$.

2. We have two sets of points in plane $A$ which has $2n$ elements and $B$ which has $2m$ elements, with $m,n$ positive integers such that no three points from their union are collinear. Prove that there exists a line which splits the plane in two regions each of which contain $n$ elements of $A$ and $m$ elements of $B$.

3. Inside the unit square consider $n^2$ points. Prove that there exists a polygonal line joining these points which has length smaller than $3n$.

4. A plane convex figure has the property that any segment which divides it in two figures of equal areas has length at most $1$. Prove that the figure has area at most $2$.

These problems are solvable, but I don’t have any solutions momentarily. If you solve them, you can post a hint in comments. Thanks

Can you prove this? Still I don’t think that this is enough. If $n^2+1$ points always give a polygonal line with length greater than 1 then this does not tell us anything about the length of a polygonal line of length $n^2$.