Some solid geometry

1. Oct 31, 2004

symplectic_manifold

Hi!
I'm new here and this is my first post.
I'm not sure whether this thread is where it should be, but the post doesn't deal with homework...I'm working on my own at the moment, and the things I'd like to present here is really a private business, so I decided to post them in the general maths section...I hope they will be of interest to you.

I made up my mind to revise the whole course of Euclid geometry once more...but I'm doing it from a slightly different perspective. The problems I'd like to offer are actually simple...but I solve them using pure axiomatic method, so I pay attention to every detail...i.e. even if something is obvious and clear I try to prove it (except axioms).
Please, check the following solutions for a number of problems and say whether they are correct...or, in addition, whether any other solution is possible.

I'm dealing with solid geometry at the moment...so the whole thing is based on the three main axioms of geometry in 3-dimensional space, which come in addition to those of plane geometry:
Axiom 1. There exist points, which belong to a plane and which do not;
Axiom 2. If two different planes have a common point, then they intersect each other at the line, containing this point.
Axiom 3. If two different lines have a common point, then they set a unique plane, which contains the two lines.

1. Problem:
The points A, B, C all lie in each of two different planes (e.g. alpha and betta). Prove that these points lie on the same line (e.g. "a").
Proof:
(Axiom 1 is subtely involved in each sentence and in what is given, since it allows us to operate with points and planes, so there seems to be no need to mention it anymore)
1) According to Axiom 2, if two different planes have a common point, then they intersect each other at the line, containing this point. It follows that if the points A, B, C lie in the plane alpha then they also lie in some planes, which intersect alpha at the lines, containing the points A, B, C.
2) It is given that the points also lie in betta, then the three planes, containing the points A, B, C are actually the same plane betta (I imagine them merging together to form the plane betta). Hence, the three lines, which are set by the three planes are the intersection line "a" of alpha and betta (the lines merge to the unique line). It finally follows that all three points A, B, C belong to the same line "a".
q.e.d.
I think a proof by contradiction is also possible here.
Suppose the points A, B, C do not lie on the same line a (Axiom 1 of plane geometry), then, since they lie in alpha (this is given), they also lie in some planes, which intersect alpha at some lines, containing these points (Axiom 2), which is impossible, because according to what is given all three points A, B and C must also lie in the plane betta, diffrent from alpha and intersecting it (this is given).

It's pure logic actually. I hope the proofs are consistent. I understand that this all might seem nonsense to you, but I just want to learn how to work with detailed proofs and weird mathematical structures.

If you have an interest in such things there is more to come.

2. Oct 31, 2004

Gokul43201

Staff Emeritus
I'm not convinced that either of your proofs is rigorous.

In Proof #1, the line "It is given that the points also lie in betta, then the three planes, containing the points A, B, C are actually the same plane betta" is dubious. For starters you have not established why there should be 3 planes. Previously, when you said "they also lie in some planes, which intersect alpha at the lines, containing the points A, B, C", do you mean the line containing all three points, or any of the lines containing any of the three points ?

In proof #2, it is not clear why these points may not lie in planes other than beta in addition to lying on beta (and alpha).

Here's how I'd do it, by contradiction :

Let points A and B lie on line L1 (axiom : a line can be drawn through any 2 points). Assume point C lies outside L1.

Let line L1 be the line of intersection of planes alpha and beta; then A and B belong to both these planes (converse of axiom 2, above).

Let C lie on plane alpha (given). Consider the line BC, which by definition, is different from line L1. Let's call BC line L2. Line L2 (or BC) lies on alpha since B and C lie on alpha.

Now consider a third (and different) plane gamma, that intersects plane alpha at L2. There are an infinite number of such planes (axiom : there are an infinite number of planes containing a given line), so we may pick any one and call it gamma.

Let gamma intersect beta along some line L3. Now, irrespective of the choice of gamma, L3 must contain B, since B is on beta (given) and B is on gamma by virtue of its being on L2 (or BC). Now since C lies on beta (given), C must lie on some line on beta passing through B (axiom : the line drawn through two points on a plane lies on the plane). Let the choice of gamma be such that this line is L3.

Now C lies on L3 and L2, so must be the point of intersection of L3 and L2. But we have shown that b is the point of intersection of L3 and L2, and since C is different from B, we have two lines intersecting at two different points, B and C.

This is in violation of the axiom that two lines may intersect, at most, at one point. Hence we have a contradiction.

Last edited: Oct 31, 2004
3. Nov 1, 2004

HallsofIvy

Your "axiom 2" makes no sense:
"Axiom 2. If two different planes have a common point, then they intersect each other at the line, containing this point."

There is no "THE line, containing this point". Given any point there exist an infinite number of lines containing that point.

It is, of course, true that "If two different planes have TWO distinct common points, then they intersect each other at the line, containing those points."
But that's the standard axiom and makes the proof you are seeking trivial.