17.5.1. Midpoint of the Line Joining Points

Let us start with question number 8. You are given here arbitrary distinct points in 2D planes. Each point will have an x-coordinate and y-coordinate, and the points have integer coordinates. So they are arbitrary points except that they are distinct. I am denoting the points and their respective coordinates as (x1, y1), (x2, y2), (x3, y3), (x4, y4), (x5, y5). Our goal is to show that irrespective of the way these 5 points are chosen arbitrarily, always exists a pair of points such that if you consider the midpoint of the line joining those 2 points, it has integer coordinates.

This section introduces a problem involving 5 distinct points on a 2D plane. Each point has integer coordinates, meaning the x and y values are whole numbers. The objective is to prove that it is always possible to find a pair of these points where their midpoint also has integer coordinates. The midpoint formula is where you add the x-coordinates of the two points together, divide by 2, and do the same for the y-coordinates.

Imagine you are at a park with 5 different trees (each represented by a point). The coordinates represent their positions in the park. No two trees are exactly at the same spot (they are distinct), and our task is to see if between any pair of trees, we can find a spot (midpoint) that is also on an integer coordinate, such as (2,2) or (3,4), rather than a fractional coordinate like (2.5, 3.5).

So, we want to apply here the pigeonhole principle. For the pigeonhole principle, we have to identify the set of pigeons and the set of holes here and then the mapping which relates the pigeons and the holes. Consider the set of 5 arbitrary points... Depending upon whether the x-coordinate is even or odd, or the y-coordinate is odd or even, I have 4 possible combinations.

In this chunk, the focus is on applying the pigeonhole principle to the problem at hand. The idea is to categorize the points based on the parity (even/odd nature) of their coordinates. There are 4 combinations: (even, even), (even, odd), (odd, even), and (odd, odd). With 5 points to classify and only 4 combinations (holes), the pigeonhole principle guarantees that at least two points must fall into the same category.

Think of a classroom where you have 5 students (pigeons) and only 4 different color shirts available (holes). If the students randomly choose shirts, at least two of them will have to choose the same color shirt. Similarly, in our problem, we are guaranteed that at least two points will share the same parity combination.

Now we want to inspect what happens to the midpoint of the line joining these 2 points. As per the formula, the midpoint will be (x1 + x2)/2, (y1 + y2)/2. Since both points are mapped to the same ordered pair; they could be both odd or both even.

Once we have identified that at least one pair of points shares the same parity for both coordinates, we apply the midpoint formula. The critical point is that the sum of two odd numbers is even, and the sum of two even numbers is also even. This means that in both cases, dividing by 2 will yield an integer. Thus, the x and y coordinates of the midpoint will also be integers.

Imagine two friends are at two spots in the park. If both spots are either an even distance from a point or both odd distances, when they take the average of their positions (the midpoint), they will land on a spot that is also an integer location. Let's say friend one is at (2, 4) and friend two is at (6, 8); their midpoint (4, 6) is also an integer point.

In the same way, we cannot have the case where one coordinate is even and the other is odd because that is not the nature of the points in this chosen pair. Hence, this shows that the midpoint is an integer.

This final chunk highlights that since we established the nature of coordinates shared between points, the proof is complete. The midpoint will always yield integer coordinates based on our previous reasoning about the sums of odd and even integers. Therefore, the proof validates our initial claim using logical deductions and the pigeonhole principle.

Returning to our tree analogy, by ensuring that we only consider trees that are either both tall or both short, we can confidently say that the average height (midpoint) is also a whole number, rather than a fractional value. This reinforces that our conclusion holds true universally for any distinct set of points.

--