Re: Quite complex problem from the euclidean geometry
- From: Angus Rodgers <twirlip@xxxxxxxxxxx>
- Date: Tue, 02 Oct 2007 15:43:43 +0100
On Sat, 29 Sep 2007 14:50:40 +0200, "Philippe 92"
<nospam@xxxxxxxxxxxx> wrote:
Angus Rodgers wrote :
On Fri, 28 Sep 2007 12:37:30 +0200, "Philippe 92"
<nospam@xxxxxxxxxxxx> wrote:
[Simson line]
The classic proof of this uses _oriented line angles_ to deal
with "different locations of points" and answer to Angus' fears.
To answer Angus' "how to deal with ASCII", figure at
http://i22.tinypic.com/35cla4z.gif
(Neat. What software did you use to generate this?
I've been making do with pencil, paper and a rather
inflexible pair of compasses - because my best pair
of compasses seems to be permanently out on loan.)
There is a lot of available software to do this.
The one I would recommand is Compass-and-Straightedge
(Zirkel-und-Lineal)
<http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/>
Thanks for the link. I've installed Z.u.L. (C.a.R.,
in English), but I haven't played with it much yet.
(Perhaps after the exam - two weeks from today, and
no revision done, so I haven't really got any time
to think about geometry at the moment! - but this is
interesting, so I don't want to let it drop completely.)
The proper rules in these cases is to use the oriented line
angles geometry. (oriented angles, not oriented lines !)
That is considering two lines, this angle is defined
modulo 180° (pi), in an oriented plane.
Given two lines AB and CD, and noting oriented angle
AB to CD : (AB,CD),
we have (AB,CD) = (BA,CD) = -(CD,AB) mod pi.
Then we have 'Chasles relations' between any three lines
AB, CD, DE :
(AB,CD) + (CD,DE) = (AB,DE) mod pi
That is, if the angle (l,m) is measured anticlockwise from
(unoriented) line l to (unoriented) line m, taking values
in R/(pi*Z), then (l,m) + (m,n) + (n,l) = 0, for any three
lines l,m,n.
Then the "A,B,C in line" condition is just :
(AB, AC) = 0 mod pi, this is true whatever the order of
points on the line.
and A,B,C,D 'cocyclic or in line' iff
(AC,AD) = (BC,BD) mod pi (cocyclic if =/= 0)
This is very interresting as the 'usual' angles
CAD and CBD may not be equal ! They are 'equal or suplementary'
This is the reason of the 'modulo pi'.
And finally AB parallel or coincident to DE iff (AB,DE)=0 mod pi
From Chasle relations, this is equivallent to(AB,CD) = (DE,CD) mod pi for any line CD.
This is an equivallent formulation of the "alternate angle
criterion for parallel lines".
Now we are ready for reasoning only once, whatever the figure
looks like. [...]
This is indeed very interesting. I get the impression
that it's not just an isolated trick. But do you know
of a systematic presentation of Euclidean geometry in
which angles are treated in this way? It's new to me.
I see no mention of it in, for instance, Hartshorne's
"Geometry: Euclid and Beyond", which I was thinking of
using as a textbook - along with Heath's edition of
Euclid, Hilbert's "Foundations of Geometry", and Ian
Mueller's "Philosophy of Mathematics and Deductive
Structure in Euclid's Elements" - when I do get around
to studying Euclidean geometry properly. (At school, I
learned to do the exercises, but I didn't really think
about what I was doing, and my subsequent learning of
a modicum of Bourbaki-style mathematics - arguably also
in parrot fashion - didn't provide any opportunity to
sharpen my understanding.)
For instance, although I don't really have time to go
into this, I can't help noticing that the condition for
distinct points A,B,C,D to be cocyclic can be expressed
in 4! = 24 superficially different ways, and even after
allowance has been made for the obvious equivalences
between the four conditions (AC,CD) = (BC,BD), (BC,BD) =
(AC,AD), (AD,AC) = (BD,BC), and (BD,BC) = (AD,AC), this
still leaves six apparently different conditions, which
are all in fact equivalent. I guess this equivalence
has been written out in a tidy fashion somewhere?
Do you find that there are parts of Euclidean geometry
which can't be handled in this way, so that you have to
distinguish awkwardly between different cases according
to the ordering of points (as I would have been obliged
to do if I had developed my partial solution to this
problem, without this neat notion of the oriented angle
between an ordered pair of two unoriented lines)?
So to be prooved that AY is parallel to PQR.
There are only the following specific cases :
When XR is tangent to circumcircle, then X=Y.
When X is in A, B or C.
I'll discuss only the general case (X != Y, A, B, C) :
X,Y,A,C cyclic (by definition) => (YA,YX) = (CA,CX) mod pi
Also (RC,RX) = (QC,QX) = pi/2 mod pi => R,X,C,Q cyclic
=> (CQ,CX) = (RQ,RX) mod pi
But CA is line CQ, and RX is line YX.
Hence (YA,YX) = (CA,CX) = (CQ,CX) = (RQ,RX) = (RQ,YX) mod pi
That is (YA,YX) = (RQ,YX) mod pi
=> YA and RQ are parallel (or coincident), QED.
Very nice. (I think the problem is well and truly done now.)
--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril
.
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