Pigeonholing mod 4 points on plane.

Question

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.

Answer 1

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.