Join the discussion
Write your take first — we'll ask for email only when you're ready to publish.
- Hacker News
- Futility closet is fantastic!by emil-lp
- In the proof, it claims that "At least two of these must fall on the same side of P′, the perpendicular projection of P on ℓ.".
However, this is not true as it is possible that P'=B. However it seems the proof still goes through (at least as depicted in the image, haven't thought hard about the general case).
by frotaur - I dont think theres an issue, the point P' simply belongs to both sides of itself. With this convention in place it is still true that there are two points in the same side, and the proof goes through verbatim.
Its mostly a question of whether you count the line defining a half-plane as belonging to the half plane or not, and clearly they do here
by cirpis - >Every finite set of points in the Euclidean plane that is not collinear has a line that passes through exactly two of the points.
Isn't this a tautology?
The problem definition states that the set of points is in Euclidean space, which from Euclid's Axioms means we can draw a line between any two points. The set of points is defined to be not collinear, thus we cannot draw a line passing through more than two of them. This is just simple logic.
by stackghost - "The set is not collinear" here means "there is no straight line passing through all the points simultaneously", not "there is no straight line passing through some three points".
- Say you've got a set of 5 points that are not all collinear (e.g. everytime you draw a line, you never get all 5 of them). Now every time you try to draw a line between two points, you always end up with a third one. Can this happen?
This is what is stated here: there always exists at least a pair where this does not happen.
by Mithriil - That is not what was meant. Here is a better rephrasing:
Let X be a set of points not all of which are collinear. Then, there are two points a, b in X such that the line l passing through X only passes through a and b.
by agnishom - > The set of points is defined to be not collinear, thus we cannot draw a line passing through more than two of them
This is wrong and you are misunderstanding what collinearity means. You could have a set of points where all but one are on the same line, and the set of a whole will be not collinear, while we can obviously draw a line that passes through more than two of them.
A different way of stating the theorem is that any finite set of points has either a line passing through all points (i.e. the set is collinear) or there exists a line that passes through exactly two points. This dichotomy (why two and not three? Why can't we construct a set where any line passes through at least three points?) is not immediately obvious.
by fxwin - > According to a strengthening of the theorem, every finite point set (not all on one line) has at least a linear number of ordinary lines. An algorithm can find an ordinary line in a set of n points in time O(n log n). [1]
There are many such lines (think convex hull) and they are easy to find. This makes it hard to appreciate the theorem. You keep thinking oh whats the big deal.
[1] https://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theor...
- The boundary lines of the convex hull can easily have 3 points each. So not as obvious as all that.by tirutiru
- I might be too stupid to understand why this is interesting and useful. If it helps I am a working physicist, and a lot of pure math is lost on me. I think I followed this, but I don't know why one would care or this would be interesting.by Nail2680
- Also, see the title of the website.by Mithriil
- The Appolonius circle is a theorem in pure geometry. It is used in the 'method of images' in physics. And not just for ancient textbook problems either. I saw a paper in experimental fluid mechanics that builds up the intuition with images (2D incompressible flow). Appolonius live and kicking in the 21st century.
The Kochen-Specker theorem is interesting (if not useful). The proofs have a similar flavour to Sylvester. One tries to make a set of projections 'compatible' and it turns out to be impossible. Maybe there's a deeper connection.
by tirutiru - It is interesting to a pure mathematician. Since it is obviously true, one's intuition is that it should have a simple proof. In particular, the obvious induction ought to work.
The base case is n=2. The line joining them passes through exactly two points because that is all you have.
Now we attempt the induction step. We have n+1 points. Leave one, p, out. We know that the theorem applies to the n points by the induction hypothesis. So we have points q and r that have a line going through them. And the point of the theorem is that the line goes through only q and r, exactly two points of the n. All we have to do is add in p, not on that line, and we are done.
But we are also stuck. Point p is not one of the n points participating in the induction hypothesis. Nothing tells us that p is not on the line joining q and r.
So how do we prove it? It is a good, intriguing puzzle, but in proof theory, not geometry.
by alan-crowe - > Every finite set of points in the Euclidean plane that is not collinear has a line that passes through exactly two of the points.
I can't make out the point here (no pun). Of course a line can pass through any two points. It could pass through three if those points were collinear but the statement says they're not. So what is the new fact?
by hyperhello - It’s that the line passes through exactly two points, which if you think about it is not exactly obvious.by chenb4425
- > So what is the new fact?
For all arbitrarily sized (but finite) sets of not collinear points, there's always a line that passes through exactly two points in the set.
by math_loser - It can help to think about theorems like this by restating them as a puzzle asking for a counterexample.
Given N points, N > 2, can you arrange them in a Euclidean plane so that (1) they are not all on the same line, and (2) every line that goes through two of the points must also go through at least one more of the points?
The theorem says that you cannot do this.
by tzs - Maybe this'll help: Every bunch of points on a plane has a lonely bridge. A lonely bridge is a line connecting only two points, no more.
Obvious caveat: The points can't all lie in the same line (the collinear condition).
Not-so-obvious caveat: There can't be an infinite number of points.
by ky3 - Try to come up with a set non-colinear points where NO line passes through two and ONLY TWO points and you'll see the value of the statement.
You may think "I'm sure I can arrange these points in a way where EVERY line will cross three or more points" but you will fail if you try unless ALL points are colinear.
by glimshe - > It could pass through three if those points were collinear but the statement says they're not
The theorem doesn't presume that no three points are collinear, it presumes that the set as a whole isn't collinear, which is a much weaker statement.
by fxwin - I think what's going on here is that you've misunderstood the theorem's hypothesis. The hypothesis isn't that no three of the points are collinear; rather, it's the weaker statement that there isn't any one single line that all the points lie on. It's true that with your version of the hypothesis the theorem would be trivial; but with the actual hypothesis it is is nontrivial.by Sniffnoy