How do you prove that the circumference of a circle is proportional to it's radius?
david
it's radius?
What's the modern version and how did the greeks prove it?
/david
Virgil
wrote:
The approximations to that ratio are based on
inscribing/circumscribing regular polygons, each of which may be
partitioned into congruent isosceles triangles.
For any such polygon, the odd(circuferential) side of any of those
isosceles triangles is proportional to the other (radial) sides by
similar triangles.
Thus each approximate circumference is proportional to its radius.
Gerry Myerson
Virgil <vmh...@comcast.net> wrote:
> In article <27Prb.2428$%W3.18852@amstwist00>, david <s...@asf.com>
> wrote:
>
> > How do you prove that the circumference of a circle is proportional to
> > it's radius?
> > What's the modern version and how did the greeks prove it?
>
> The approximations to that ratio are based on
> inscribing/circumscribing regular polygons, each of which may be
> partitioned into congruent isosceles triangles.
>
> For any such polygon, the odd(circuferential) side of any of those
> isosceles triangles is proportional to the other (radial) sides by
> similar triangles.
>
> Thus each approximate circumference is proportional to its radius.
Now if you know/accept that
perimeter of inscribed polygon
< circumference of circle
< perimeter of circumscribed polygon,
then it's easy to complete the proof. Now the first inequality,
perimeter of inscribed polygon < circumference of circle,
is clear; a straight line is the shortest distance between two points.
The other inequality,
circumference of circle < perimeter of circumscribed polygon,
is certainly plausible, but is it actually possible to prove it
using only tools available to the ancients?
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
David C. Ullrich
Well, I don't know what Euclid's _definition_ of the circumference
of a circle was, but if he had one I can't imagine that it was not
something essentially equivalent to what we would call the least
upper bound of the perimeter of an inscribed polygon. Assuming
that, then first we need to show that
(*) perimeter of inscribed polygon
< perimeter of circumscribed polygon;
that shows that the least upper bound mentioned above
_exists_, and also makes the other inequlalities clear.
(Regardless of how much of this they were explicit about,
I do tend to suspect that they would have said the question
was the same as (*), for some reason.)
The circumcribed polygon is the union of triangles with height
equal to the radius of the circle and base one of the sides of
the polygon, so the _area_ of the circumcribed polygon is
equal to r times its circumference.
Otoh, let r' < r. Note that adding points to the inscribed
polygon increases the circumference, by the triangle inequality.
So starting with an inscribed polygon we can find another
with larger circumference, and with area > r' * perimeter.
Since the relation between the areas is clear, it follows
that r' * inscribed perimeter < r * circumscribed perimeter
for all r' < r, hence (*).
>Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
David C. Ullrich
George Baloglou
proportional to the square of the radius (A = K*r^2) and that the
circumference of a circle is proportional to the radius (C = l*r), how do
we establish l = 2*k (especially in case the ancients are listening)?
Here is a 'heuristic' argument of debatable merit (that I recently came up
with, assuming nonetheless that it must have been known for quite a while):
Partition a disk of radius r into a disk of radius r/2 and a ring of width
r/2 ... and express the ring's area as both the difference between the
areas of the two disks and the area of a trapezoid (that is stretched-out
ring) of bases l*r, l*r/2 and height r/2: l = 2*k follows easily from
K*r^2 - k*(r/2)^2 = (1/2)*(l*r + l*r/2)*(r/2).
[In a similar manner one may compute the volume of the torus -- I recall
that on my first Calculus final I asked the students to do that using
Calculus, using Pappus' theorem, and ... using just their imagination :-) ]
baloglouAToswego.edu
Elaine Jackson
there's something about finding out the value of Pi by comparing the number of
lattice-points in a circle with the number in a square. (You get exact values by
letting the size of the shapes in question increase indefinitely.) I'm not sure
the complete details are there, but probably at least enough so the rest can be
filled in with a little effort. I thought this might be interesting to you
because of the historical connection: apparently the idea was initiated by
Gauss.
"david" <s...@asf.com> wrote in message news:27Prb.2428$%W3.18852@amstwist00...
David C. Ullrich
wrote:
>Of some relevance: once we know/accept that the area of a circle is
>proportional to the square of the radius (A = K*r^2) and that the
>circumference of a circle is proportional to the radius (C = l*r), how do
>we establish l = 2*k (especially in case the ancients are listening)?
If P is a circumscribed polygon then as noted above the area of P is
(r/2) times the perimeter of P.
(If we don't define the circumference of a circle as the sup of the
perimeter of inscribed polygons then I don't know how we _do_ define
it. If we do define it that way then it follows easily that it's the
inf of the perimeter of circumscribed polygons, qed.)
>Here is a 'heuristic' argument of debatable merit (that I recently came up
>with, assuming nonetheless that it must have been known for quite a while):
>
>Partition a disk of radius r into a disk of radius r/2 and a ring of width
>r/2 ... and express the ring's area as both the difference between the
>areas of the two disks and the area of a trapezoid (that is stretched-out
>ring) of bases l*r, l*r/2 and height r/2: l = 2*k follows easily from
>K*r^2 - k*(r/2)^2 = (1/2)*(l*r + l*r/2)*(r/2).
>
>[In a similar manner one may compute the volume of the torus -- I recall
>that on my first Calculus final I asked the students to do that using
>Calculus, using Pappus' theorem, and ... using just their imagination :-) ]
>
> baloglouAToswego.edu
David C. Ullrich
Bill Kleinhans
>On Tue, 11 Nov 2003 09:40:09 +1100, Gerry Myerson
><ge...@maths.mq.edi.ai.i2u4email> wrote:
>
>>In article <vmhjr2-23EEA1....@news.newsguy.com>,
>> Virgil <vmh...@comcast.net> wrote:
>>
>>> In article <27Prb.2428$%W3.18852@amstwist00>, david <s...@asf.com>
>>> wrote:
>>>
>>> > What's the modern version and how did the greeks prove it?
>>>
Euclid did not discuss the length of curved lines, or the area
of curved surfaces, apparently because he had no satisfactory definition of such things. I believe Archimedes was the first.
He proposed several postulates, from which he could conclude
that any convex curve is longer than an inscribed polygon,
and shorter than a circumscribing polygon. Effectively, he had a
definition for the length of convex curves, and similarly for
convex surfaces. To these he applied the method of exhaustion.
Bill Kleinhans
Eur Ing Panagiotis Stefanides
>In article <27Prb.2428$%W3.18852@amstwist00>, david <s...@asf.com>
>wrote:
>
>> How do you prove that the circumference of a circle is proportional to
>> it's radius?
Please refer to:
http://mathforum.org/discuss/sci.math/a/m/107260/107288
and
http://mathforum.org/discuss/sci.math/a/m/107260/107289
Regards,
Panagiotis Stefanides
http://www.stefanides.gr
Ken Pledger
> How do you prove that the circumference of a circle is proportional to
> its radius? .... and how did the Greeks prove it?
You've asked about history. You shall have history!
I wrote a fairly detailed answer to a similar question in sci.math
on the 8th June 2001, so here's a copy of that.
> In article <eb16264a.01060...@posting.google.com>,
> satya_...@hotmail.com (satya) wrote:
>
> > Is there a straightforward way to prove that the ratio of
> > circumference to the diameter of any arbitrary circle is always a
> > constant? It feels like this should be easy but I have not seen such a
> > proof in any book so far.
>
> That's good mathematical thinking! You've seen that rough ideas
> about similarity aren't enough. It was the ancient Greeks who realized
> that it's not straightforward at all, and invented the first technique of
> limits to handle just such problems. You've already had limit arguments
> outlined by DWIII and Peter M. Jack, so I'll just mention some of the
> history.
>
> The area was tackled before the circumference, perhaps because it's
> easier. Euclid XII.2 (probably due to Eudoxus) proved that "Circles are
> to one another as the squares on the diameters." A modern statement of
> that might say
>
> (area of first circle)/(area of second circle)
> = (diameter of first circle)^2/(diameter of second circle)^2.
>
> This shows that our modern area formula (pi)(r^2) or (pi)(d^2)/4
> really does have the same constant pi for all circles. Euclid's proof
> used inscribed and circumscribed polygons of more and more sides, and then
> a rigorous limit argument entirely in Greek geometrical terms. It's not
> easy!
>
> Later, Archimedes used the same limit technique in his "Measurement
> of the Circle". He first proved that "The area of any circle is equal to
> a right-angled triangle in which one of the sides about the right angle is
> equal to the radius, and the other to the circumference of the circle." A
> modern version is area = (radius)(circumference)/2. Since each
> circle's area = (pi)(r^2) with constant pi, it follows that its
> circumference = 2(pi)r with the same constant pi. Then Archimedes went
> on to calculate bounds for that constant, and proved that 3 1/7 < pi < 3
> 10/71. Those bounds aren't very tight, but nothing like them had ever
> before been rigorously proved.
>
> So your unease about the constancy of pi was well justified. The
> question needed some very great mathematical minds to handle.
>
> Ken Pledger.
huffy
"The police and so forth only exist insofar as they can demonstrate
their authority. They say they're here to preserve order, but in fact
they'd go absolutely mad if all the criminals of the world went on
strike for only a month. They'd be on their knees waiting for a
crime. That's the only existence they have."
William S. Burroughs (American writer)
Guardian, 1966
huffy
Luis A. Rodriguez
Very simple. It's enough to accept that equal cords substends equal
arcs.
If you have a circumference and an inscrit polygon, its n sides
(cords) are proportional to radius. If you now have a bigger polygon,
the new n arcs must increase proportional to radius. Otherwise the new
cords cannot substend equal arcs. L.Rodriguez
Eur Ing Panagiotis Stefanides
>In article <27Prb.2428$%W3.18852@amstwist00>, david <s...@asf.com> wrote:
>
>
>
> You've asked about history. You shall have history!
>
> I wrote a fairly detailed answer to a similar question in sci.math
>on the 8th June 2001, so here's a copy of that.
>
>
>> In article <eb16264a.01060...@posting.google.com>,
>> satya_...@hotmail.com (satya) wrote:
>>
>> > Is there a straightforward way to prove that the ratio of
>> > circumference to the diameter of any arbitrary circle is always a
>> > constant? It feels like this should be easy but I have not seen such a
>> > proof in any book so far.
>>
>> That's good mathematical thinking! You've seen that rough ideas
>> about similarity aren't enough. It was the ancient Greeks who realized
>> that it's not straightforward at all, and invented the first technique of
>> limits to handle just such problems. You've already had limit arguments
>> outlined by DWIII and Peter M. Jack, so I'll just mention some of the
>> history.
>>
>> The area was tackled before the circumference, perhaps because it's
>> easier. Euclid XII.2 (probably due to Eudoxus) proved that "Circles are
>> to one another as the squares on the diameters." A modern statement of
>> that might say
>>
>> (area of first circle)/(area of second circle)
>> = (diameter of first circle)^2/(diameter of second circle)^2.
>>
>> This shows that our modern area formula (pi)(r^2) or (pi)(d^2)/4
Ken ,
Is this really a modern formula for circle's area?
If You see it is equal to the product of [pi*d/4]*[d].
pi*d/4=quarter circle circumference .
Thus the area comes after one has completed the required details
and relationships of circumference and diameter.
One has to start finding circumference [quadrature] relationships
with square sides [Perimeter].
This is at least a logical procedure.
[ Further analysis ,may be found below:
http://www.stefanides.gr/quadcirc.htm
http://www.stefanides.gr/piquad.htm ]
I, believe that the Greek Philosophers started first with the
circumference relationships,before going to the area ones.
Regards,
Panagiotis Stefanides
httm://www.stefanides.gr
Eur Ing Panagiotis Stefanides
>david <s...@asf.com> wrote in message news:<27Prb.2428$%W3.18852@amstwist00>...
>> What's the modern version and how did the greeks prove it?
>>
>> /david
>
>Very simple. It's enough to accept that equal cords substends equal
>arcs.
How does one accept this ,though !
Regards,
Panagiotis Stefanides
http://www.stefanides.gr
>If you have a circumference and an inscrit polygon, its n sides
Ken Pledger
pana...@otenet.gr (Eur Ing Panagiotis Stefanides) wrote:
> On 12 Nov 2003, Ken Pledger wrote:
> >> ....
> >> This shows that our modern area formula (pi)(r^2) or (pi)(d^2)/4 ....
>
> Ken ,
> Is this really a modern formula for circle's area?....
What I meant was modern _notation_ for the formula. I'm sorry if
that wasn't clear.
Ken Pledger.