Combinatorial problem based on the Pigeonhole Principle

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Good day,i came across this problem from the section on the Dirichlet principle. And I’m completely stumped, I’d be grateful for any idea

A point $(a_1;a_2)$ in the x — y plane is called a lattice point if both $a_1$ and $a_2$ are integers. Given any set $L_2$ of 5 lattice points in the x — y plane, show that there exist 2 distinct members in $L_2$ whose midpoint is also a lattice point (not necessarily in $L_2$).

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Observe that parity is important. For instance, if we have x-coordinates with different parities, such as (1,0) and (2,0), they will not have a lattice point between them since $1.5 \notin \mathbb{Z}$.

Therefore, we need coordinates whose x-coord and y-coord have the same parity. There are 4 options for these: (odd, odd), (odd, even), (even, odd), and (even, even). By the pigeonhole principle, since we have 5 lattice points, there will be at least 2 with the same option.

WLOG, suppose we have (even, even). Then, we have $(2x_1, 2y_1)$ and $(2x_2, 2y_2)$ for the 2 points. Their midpoint will be $(\frac{2x_1 + 2x_2}{2}, \frac{2y_1 + 2y_2}{2}) = (x_1 + x_2, y_1 + y_2)$. Since these are both integers, we have a lattice point.