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Adele06

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Consider a square with the side of length n and $(n+1)^2$ points inside it. Show that we can choose 3 of them to determine a triangle (possibly degenerate) of area at most $\frac{1}{2}$.

I think that I know how to solve the problem for the cases $n=1$ and $n=2$:

For $n=1$ we can easily prove that the triangle formed by those 3 points has the area at most $\frac{1}{2}$,like here:https://math.stackexchange.com/q/173333

For $n=2$ we can divide the square into 4 smaller squares (by uniting the midpoints of the opposite sides),each of them with the length of the side equal to 1.By pigeonhole principle we know that at least one of those squares contains at least 3 of the 9 points inside it.From here we can use the same proof from the case $n=1$ to prove that those 3 points determine a triangle of area at most $\frac{1}{2}$

This idea doesn't work for the other cases because we would need $2n^2+1$ points instead of $(n+1)^2$

Can you help me please with the other cases?

I think that I know how to solve the problem for the cases $n=1$ and $n=2$:

For $n=1$ we can easily prove that the triangle formed by those 3 points has the area at most $\frac{1}{2}$,like here:https://math.stackexchange.com/q/173333

For $n=2$ we can divide the square into 4 smaller squares (by uniting the midpoints of the opposite sides),each of them with the length of the side equal to 1.By pigeonhole principle we know that at least one of those squares contains at least 3 of the 9 points inside it.From here we can use the same proof from the case $n=1$ to prove that those 3 points determine a triangle of area at most $\frac{1}{2}$

This idea doesn't work for the other cases because we would need $2n^2+1$ points instead of $(n+1)^2$

Can you help me please with the other cases?

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