Trouble with using vectors for geometric proofs.

[Mobyseven]

First, the standard disclaimer: This isn't a homework problem.

To cut to the chase, the problem is this - I'm working through the VCE Specialist Maths text book, and I'm up to using vectors for geometric proofs. I keep hitting dead ends, and any general hints on how to approach these proofs would be appreciated.

For an example - using vectors to prove that if the midpoints on each side of a square are connected, the resulting shape is a square.



I'm currently at the point in this proof where I have shown that WX=ZY (not sure how to get an arrow above those to show that they are vectors) and that WZ=XY. I know that, by definition, |OA|=|OC|. If OA=a and OC=c, then I know:

|a|=|c|

WX=ZY=(1/2)a+(1/2)c=(1/2)(a+c)
WZ=XY=(1/2)c-(1/2)a=(1/2)(c-a)

But then I get stuck!

I'm not looking for an answer to this problem in particular, just a hint that will help me with the next stage, and hopefully with using vectors for geometric proof in general.

PS - Again, sorry for the horrible maths notation. I know that there is some sort of code one can use for notation, but I don't know what it is...

[GreedyAlgorithm]

Here's a trick:
<(a+c)/2,(c-a)/2> = <a/2,c/2> + <a/2,-a/2> + <c/2,c/2> + <c/2,-a/2> = 0+(-|a|^2)+|c|^2+0 = 0

Dot product equaling zero is an easy way to prove things are perpendicular.

[Mobyseven]

Thanks for the tip, but I'm afraid I'm not sure how to use that in this situation.

Surely what I want to prove is...wait a minute.

Dang. Thought I had it, then remembered that I got stuck doing just that...

Surely what I want to prove is that |WX|=|WZ|

(1/2)|a+c|=(1/2)|c-a|

If I can do that, then I've shown that |WX|=|WZ|=|XY|=|ZY|. All sides are of equal length, and so the shape is a square. Unless the point of proving that WX and WZ are perpendicular is to show that the sides are of equal length...but I'm not sure I follow that.

[Modified]

Originally Posted by Mobyseven

If I can do that, then I've shown that |WX|=|WZ|=|XY|=|ZY|. All sides are of equal length, and so the shape is a square.

rhombus

[bjb]

...rhombus, which is why to prove a square, you have to show two things:

1) All sides are equal in length (use the Euclidean norm for length)
2) The adjacent sides are perpendicular to each other (use the dot product as suggested by GA)

By the way, WX does not equal WZ ! They do have the same magnitude but they point in different directions. However, you can say that ||WX|| = ||WZ|| because the norms are equal.

Also, |x| is 'absolute value of x'. Use ||x|| to denote 'norm of x. Sorry for being nitpicky but this is how I was taught and this is how I graded tests and homework when I was a TA in grad school. Don't want to miss points on your work just because you used the wrong notation!

[GreedyAlgorithm]

On paper I'll use ||. On Internet ||WX||=||WZ||=||XY||=||ZY|| is too irritating to look at. Or even worse ||A||||B||

[Mobyseven]

Originally Posted by Modified

rhombus

Dang. Yes, a rhombus. Silly, silly me...

Originally Posted by bjb

...rhombus, which is why to prove a square, you have to show two things:

1) All sides are equal in length (use the Euclidean norm for length)
2) The adjacent sides are perpendicular to each other (use the dot product as suggested by GA)

Okay, I'll try and figure that out...

Originally Posted by bjb

By the way, WX does not equal WZ ! They do have the same magnitude but they point in different directions. However, you can say that ||WX|| = ||WZ|| because the norms are equal.

I didn't say that WX=WZ, I said WX=ZY. That ||WX||=||WZ|| is what I'm trying to prove now...

Originally Posted by bjb

Also, |x| is 'absolute value of x'. Use ||x|| to denote 'norm of x. Sorry for being nitpicky but this is how I was taught and this is how I graded tests and homework when I was a TA in grad school. Don't want to miss points on your work just because you used the wrong notation!

Thanks for the hint - |x| for absolute value (where the result is always positive) and ||x|| for Euclidean norm (which is the magnitude of a vector?).

I just checked - the textbook I'm using uses |x| for both, so thanks for the tip. I'll start writing in your way, as GA and wikipedia both agree with you. Better to agree with the world than with a textbook.

Again, thanks for picking up on that - there isn't anyone marking my work, as I'm working through this on my own, but hopefully I'll get into the Arts/Science degree (I want to major in maths in the science component), and then there will be people marking my work. I find out in the next two days whether or not I'm accepted into the course.

[Matteo Martini]

Originally Posted by Mobyseven

First, the standard disclaimer: This isn't a homework problem.

To cut to the chase, the problem is this - I'm working through the VCE Specialist Maths text book, and I'm up to using vectors for geometric proofs. I keep hitting dead ends, and any general hints on how to approach these proofs would be appreciated.

For an example - using vectors to prove that if the midpoints on each side of a square are connected, the resulting shape is a square.

http://forums.randi.org/imagehosting...88b5d5edeb.bmp

I'm currently at the point in this proof where I have shown that WX=ZY (not sure how to get an arrow above those to show that they are vectors) and that WZ=XY. I know that, by definition, |OA|=|OC|. If OA=a and OC=c, then I know:

|a|=|c|

WX=ZY=(1/2)a+(1/2)c=(1/2)(a+c)
WZ=XY=(1/2)c-(1/2)a=(1/2)(c-a)

But then I get stuck!

I'm not looking for an answer to this problem in particular, just a hint that will help me with the next stage, and hopefully with using vectors for geometric proof in general.

PS - Again, sorry for the horrible maths notation. I know that there is some sort of code one can use for notation, but I don't know what it is...

^
y|
|
|
|
|
|
-------------> x

a=(0,1)
c=(1,0)

|a|=squareroot(0^2+1^2)=1
|c|=squareroot(1^2+0^2)=1

WX=ZY=(1/2,1/2)
WZ=XY=(1/2,-1/2)

|WX|=|ZY|=squareroot((1/2)^2+(1/2)^2)=squareroot(1/2)=1/squareroot(2)
|WZ|=|XY|=squareroot((1/2)^2+(-1/2)^2)=squareroot(1/2)=1/squareroot(2)

[DavidS]

Originally Posted by Modified

Originally Posted by Mobyseven

If I can do that, then I've shown that |WX|=|WZ|=|XY|=|ZY|. All sides are of equal length, and so the shape is a square.

rhombus

Alas: only that the four vectors are of equal length. As noted, squaring the vertices requires a little more work.

Originally Posted by bjb

...rhombus, which is why to prove a square, you have to show two things:

1) All sides are equal in length (use the Euclidean norm for length)
2) The adjacent sides are perpendicular to each other (use the dot product as suggested by GA)

Alas: only that the four vectors are perpendicular. You still need to show:

3) The figure is closed
4) The four vectors are coplanar
5) The figure is convex

As others have noted, zero dot products will demonstrate perpendicularity, but without "external knowledge" (eyeballing the figure) they don't differentiate between closed polygon and open zigzag.

You can demonstrate that the figure is closed by summing the vectors to yield the zero vector: <sigh... I tried LaTEX, but I'm no hand at it... below BOLD are vectors>

3) V_wx+V_xy+V_yz+V_zw = 0

I suppose the identity of the "w" in "wx" and "zw" reveals the same thing, depending on how formal you want to be.

4) Cross products will also reveal that all the angles between successive vectors around the square are all right angles; the magnitude of each cross-product equals the product of the magnitudes of the vector.

V_wx X V_xy = P_x
V_xy X V_yz = P_y
V_yz X V_zw = P_z
V_zw X V_wx = P_w

sin(|pi/2|) ||V_wx|| ||V_xy|| = ||P_x||
sin(|pi/2|) ||V_xy|| ||V_yz|| = ||P_y||
sin(|pi/2|) ||V_yz|| ||V_zw|| = ||P_z||
sin(|pi/2|) ||V_zw|| ||V_wx|| = ||P_w||

The cross product also yields vectors that can be used to reveal that all the sides are coplanar and the figure is convex, because

P_i . P_j = cos(0) ||P_i|| ||P_j||

For all i and j in w,x,y,z. The cross-product of two vectors yields a vector perpendicular to the plane containing those vectors; these equalities reveal that all the vector pairs lie in parallel planes. Along with the closure of the path, this means they're all coplanar.

5) The figure is convex if all the vertex angles are of the same hand and sum to +/- 2*pi (full circle) and all less than straight angles (that is, all same sign and all less than pi in absolute value). The norms of the cross product vectors reveals that all the angles are the same size -- right angles. The common sign of all the dot products of each of the cross products reveals that all the angles are of the same hand -- right or "left" angles.

Now that's a square: A closed, convex polygon with four (coplanar) sides of equal length, with right angles included at all four vertices.

I realize that one could appeal to the problem formulation for evidence of closure (the inscribed square vectors were chosen for closure), convexity (a polygon inscribed in a convex polygon will be convex), and coplanarity (the circumscribed square, and therefore all points thereon, is coplanar). Isn't the point of the exercise, though, to wring that sort of geometrical information from the vectors themselves?

EDITED: Fixed innocent typo with sin(pi/2) on wrong side in (4). Struck out the booger-eating-moron convexity criteria with something more possibly correct (at least less easily refutable by triangle and pentagon counterexamples).

[DavidS]

Actually, you also need one more criteria but I'm not certain what to call it:

6) The absolute value of the sum of the vertex angles must be (v-2)*pi, where v is the number of vertices; that is, this included angles of this 4-cornered figure must sum to |2pi|

This excludes figures that go round and round the square multiple times, which could pass the other five criteria and still not be a square.

ETA: Of course, four vectors passing (1-5) can't fail (6)

[Mobyseven]

Thanks for all your help everyone. I'm going to print out this thread so I can actually have it sitting next to me while I try to work out the problem.

I'm pretty sure DavidS has brought in some stuff that wasn't in the textbook, but I might as well take it and run with it. A friend is coming around later to help me out with some of the stuff, so if I still don't have it by then perhaps having someone assist me in person will help me 'click'.

Thanks again!

M7