Suppose ten points are placed at random inside a triangle whose sides are all of length three. Show that there must be at least two points no more than a distance one apart. Be sure to consider all possible cases.
So what i did was...
dividing the triangle into 4 smaller equilateral triangles like this:
A
AVA
(hope you understand what i was trying to draw out) and each length has length 1.
And I used the pigeohole priciplae. which says if there are n+1 pigeons in n pigeon holes, at least one hall must contain 2 or more pigeon.
so isn't it obvious that if there are 10 points placed in the triangle, obviously there will be at least 2 points which has a length less than one??
Am I being tricked here, and what does it mean by "be consider to show all cases"???
Presumably you meant to divide the triangle into 9 smaller triangles:
AVA
AAVA
AVAVA
It is clear that none of the smaller triangles can contain more than one of the ten points in its interior, but as undefined points out there is nothing to stop one of the smaller triangles containg a point at more than one of its vertices. That is where you need to consider all possible cases carefully.
Edit. Looking at the question more carefully, I don't think that there is anything wrong with simply using the pigeonhole principle. The question does not ask for two points to have distance less than 1, but for two points with distance no more than 1. If two points are in the same smaller triangle then their distance apart certainly cannot be more than 1.
Well I think it's easy to accidentally require the points to be in the interiors of the smaller triangles, thus missing some cases; I wasn't suggesting that a simple application of the pidgeonhole principle did not apply.
A simple solution is to arbitrarily assign each horizontal edge to the triangle above it, every other edge to the triangle to its left, and the middle vertex to the northwest triangle connected to it.
Sorry to bring this topic up again... Just looking through the question again...
but i did not quite understand undefined's solution:
"
A simple solution is to arbitrarily assign each horizontal edge to the triangle above it, every other edge to the triangle to its left, and the middle vertex to the northwest triangle connected to it."
Does that mean, you are placing the "points" on each of the horizontal edge to the traingle above it? So, if i do so, I have used up 6 points, hence 4 points left, thus, there will be at least 2 points no more than a distance one apart.
But I don't get this case "the middle vertex to the northwest triangle connected to it?"
Also, how many possible cases will there there be? Just 3 i pressume?
I was just referring to a partition of the original triangle into subtriangles. A partition divides a set into parts whose intersection is empty and whose union is the original set. We don't really need a partition here, but I thought it easiest to define a partition rather than dealing with cases of points that belong to more than one subtriangle. So I was describing a scheme of assigning edges/vertices to certain subtriangles, so that for any point within the original triangle, we can identify which subtriangle it belongs to. See what I mean?