Proving Triangle Area ≤ $\frac{1}{2}$ in a Square with $(n+1)^2$ Points

In summary, for a square with side length $n$ and $(n+1)^2$ points inside it, we can choose 3 points to determine a triangle with area at most $\frac{1}{2}$. This is proven by dividing the square into smaller squares and using the pigeonhole principle to choose 3 points from the boundary of one of these squares and 3 points from inside it. This approach works for all values of $n$.
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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?
 
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Sure, let's consider the case $n=3$. In this case, we have a square with side length 3 and 16 points inside it. We can divide the square into 9 smaller squares, each with side length 1. By pigeonhole principle, at least one of these squares will contain at least 3 points. Let's call this square $S$. Now, we have 9 points inside $S$ and we need to choose 3 of them to determine a triangle with area at most $\frac{1}{2}$.

One way to approach this is to consider the points on the boundary of $S$. There are 12 points on the boundary, and we can choose any 3 of them to form a triangle with area at most $\frac{1}{2}$. This is because the area of a triangle is given by the formula $\frac{1}{2}bh$, where $b$ is the base and $h$ is the height. Since the base and height are both 1 in this case, the area of the triangle will be at most $\frac{1}{2}$.

So, we have 12 points on the boundary of $S$ that we can choose from. This leaves us with 9 points inside $S$ that we can choose from. Using the same argument as in the case $n=2$, we can divide $S$ into 4 smaller squares, each with side length $\frac{1}{2}$. By pigeonhole principle, at least one of these squares will contain at least 3 points. We can then use the same proof as in the case $n=1$ to show that these 3 points determine a triangle with area at most $\frac{1}{2}$.

So, in total, we have chosen 3 points from the boundary of $S$ and 3 points from inside $S$ to form a triangle with area at most $\frac{1}{2}$. Thus, we have shown that for $n=3$, we can choose 3 points to determine a triangle with area at most $\frac{1}{2}$.

You can follow a similar approach for other values of $n$ as well. The key is to use the pigeonhole principle to divide the square into smaller squares and then use the same proof as in the case $n=1$ to show that the 3 points chosen from
 

FAQ: Proving Triangle Area ≤ $\frac{1}{2}$ in a Square with $(n+1)^2$ Points

1. What is the purpose of proving triangle area ≤ $\frac{1}{2}$ in a square with $(n+1)^2$ points?

The purpose of this proof is to demonstrate that the maximum possible area of a triangle formed by $(n+1)^2$ points within a square is equal to or less than half of the area of the square. This has implications in various fields such as geometry, computer science, and data analysis.

2. How is this proof relevant in computer science?

In computer science, this proof is relevant in the study of algorithms and computational complexity. It can be used to analyze the efficiency of algorithms that involve triangulation of points, such as in computer graphics and data visualization.

3. Can this proof be applied to other shapes besides a square?

Yes, this proof can be applied to any regular polygon with $(n+1)^2$ points, as long as the points are arranged in a grid-like pattern. However, the maximum area of the triangle may vary depending on the shape of the polygon.

4. What are the assumptions made in this proof?

The main assumptions made in this proof are that the points are arranged in a grid-like pattern within the square, and that the triangle formed by these points is a right triangle. Additionally, the proof assumes that the square has a side length of 1 unit, but this can be easily generalized to any square with a side length of s units.

5. How can this proof be used in real-life applications?

This proof has various real-life applications, such as in GPS technology, where triangulation is used to determine the location of a device. It can also be used in surveying and land mapping, as well as in data analysis and pattern recognition.

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