🚀 go-pugleaf

Shimpei Yamashita (shimpei@leland.Stanford.EDU) wrote:
: In article <33mbhc$gug@nermal.cs.uoguelph.ca>,
: Douglas Harder <dharder@sgi2.mathstat.uoguelph.ca> wrote:
: :One is the trisection of an angle.
: :
: :I believe another is finding a focus of any conic which is
: :not a circle.
: Hmm. I think they were trisection of the angle, doubling a cube, and
: squaring a circle. The one you mention may be equivalent to these; I can't
: say offhand.

You're right.  The second one I mentioned is something I tried to
do in high school, and one of my teachers told me it was 
impossible.  Much to my teacher's annoyance, I tried anyway and
found the best I could do was get successively better approximations
(if I recall correctly).

: The reason for this is because two circles indeed don't intersect each 
: other if the sum of their radii is less than the distance between their
: centers. Of course, this is not the case here. That such a point (OK, two
: such points) exists follows almost trivially from Pythagorian theorem.
: (Draw perpendicular bisector, etc.). Then drawing the circles *must*
: yield two intersections because you are taking the intersection of two
: loci which, as you just proved, consists of exactly two points. There may
: be a way to prove it without using the fifth axiom (the infamous one about
: a unique parallel line through a point, which is required by Pythagorian
: Theorem), but I can't think of one off the top of my head....
: -- 

It can't be proved from Euclid's axioms (that given a line segment, compass,
and rule that you can create an equalateral triangle with the line segment
as one of the sides.)

Nowhere does it say in Euclid's axioms that if you have two circles
of equal radii, and with centres closer together than twice the
radius, that the two circles must intersect.  Neither is there any
guarentee (in the axioms) that if any point on a line is closer to the 
centre of a circle than the radius of the circle, that the line must 
intersect the circle at two points.

Take for example the rational plane (all points with rational coordinates).
It can be shown that all five of Euclid's axioms hold on the rational
plane, but you cannot create an equalateral triange triangle out of
the (rational) line segment connecting (0,0) and (1,0).  This is since
the vertex will be at the point (0.5, sqrt(3)/2), which is not a point
on the rational plane.  (for ease, let us ignore the second vertex.:)
Thus, the equation of the line connecting the two is y = sqrt(3)x, 
where x only takes rational values.  But it follows then that the only
y value that is rational (given x rational) is y = 0 when x = 0.
Thus we find that no points in the rational plane connect the points
(0,0) and (0.5, sqrt(3)/2), and thus, no line exists (since a line
is a connection between two points).

What this shows is that there are some assumptions which Euclid made,
but did not bother writing down in his axioms.  That
you _can_ make an equalateral triangle is what Euclid wanted,
but he didn't say enough in his axiom to ensure this.

It was not until the 19th century that Hilbert and others created
a set of axioms which satisfied what Euclid wanted.  Hilbert's axioms
should be found in any book on Modern Geometry (eg, Cederberg).  I
think that he had to have around 16 independent axioms.

I personally think the above arguements are initially rather
subtle, and interesting.  I hope I presented them in a reasonable
and convincing fashion.  (It took me about a month to get used
to the above when after the prof initially presented it).

Douglas.
-IPU(pbuh)-----------------------------------------
GM/S/(CS) -d+ H-->++ s+: g+ p1 au a24 w+ v>+ C+>++ UL+ P?
L++>+++ 3 E--- N+ K+ W--- M-- V- -po+ Y(+) t+ 5 j(-) R
G++>+++ !tv b+ D B e+>++++ u++(---) h-- f(?) r- n+>++ y+

Good evening, could you tell MacLean that Roger Styles
from the Foreign Office is here to see him?