Pigeonholing mod 4 points on plane. I have the following problem as homework. Suppose there are 13 points in the plane, all with integer coordinates. Prove at least one quadrilateral with vertices on those points has a barycentre with integer coordinates.
 A: Each point on the plane is of one of the following forms. $(o, o), (e, e), (o, e), (e, o)$ where $o$ and $e$ stands for odd and even integers. Since there are $13$ points on the plane - by the pigeonhole principle - at least $4$ of them are of the same form. That is  there are $4$ points on the plane $A, B, C, D$ such that each of their first coordinates are pairwise congruent modulo $2$ and each of their second coordinates are pairwise congruent modulo $2$. 
Now think the vectors representing the points $A, B, C, D$ are $ \vec a, \vec b, \vec c, \vec d$. where the $\vec i, \vec j$ components of the vectors are pairwise congruent to each other modulo 2. 
According to @John above, the medians of the bisecting points of the lines joining the vectors will be $\frac {\vec a + \vec b}{2}, \frac{  \vec b + \vec c} {2}, \frac {  \vec c  + \vec d }{2}, \frac {  \vec d + \vec a}{2}$. The intersection of the lines $M_{AB}M_{CD}$ and $M_{AD}M_{BC}$ is the solution to the following two equations. 
$$\vec r = t \left[{ \frac {  \vec c  + \vec d }{2} - \frac {\vec a + \vec b}{2}   }\right] + \frac {\vec a + \vec b}{2} -------(1)$$
$$ r = s \left[{  \frac{  \vec b + \vec c} {2} -  \frac {  \vec d + \vec a}{2}  }\right] +  \frac {  \vec d + \vec a}{2}-------(2) $$
The above two vectors will be equal  when the parameters $t = s= \frac 1 2$ and hence I believe this is the point of intersection of $M_{AB}M_{CD}$ and $M_{AD}M_{BC}$ - (correct me if I'm wrong). 
Now the vector representing the barycentre is $$\vec r = \frac { \vec a + \vec b + \vec c + \vec d}{4}$$ 
Now, like I said since the components of each each of the vectors $ \vec a, \vec b, \vec c, \vec d$ is congruent modulo 2 to each other's components, it can be seen that the components of $\vec r$ are integers. 
I'm not a 100% sure about this solution. But this problem enthuses me and hope this would help you to come up with an answer. 
A: The solution of the problem relies on this classical problem, given $5$ points on the plane there are $2$ of them which have a midpoint with integer coordinates. This problem is easy since there are $4$ possibilities for the parities of an ordered pair $(x,y)$. Thus there are two which have the same parity $(x,y)$ and $(x_1,y_1)$. Thus the $x_1+x_2$ and $y_1+y_2$ are even, which makes the midpoint have integer coefficients.
To solve the problem here we use this result to extract a pair of points with integer coordinates, we have $11$ remaining points, do it again, we now have $9$ points. Do this after you take out $5$ pairs of points and there are $3$ points remaining, forget about those $3$ extra points.
You know have $5$ line segments with midpoints with integer coordinates. So $5$ midpoints, so there are $2$ of these midpoints that have a midpoint with integer coordinates. Take these midpoints let $M$ be the midpoint of $AB$ and $L$ the midpoint of $CD$. Then the midpoint of $ML$ is the barycentre of $ABCD$ and we are done.
