**Reply to Robin, Diminishing Circle Argument**

**Harry Sedinger**

**Subject:** Re: Reply to Robin, Diminishing Circle
Argument

**Author:** Harry Sedinger <hsed@sbu.edu>

**Organization:** St. Bonaventure University

**Date:** Thu, 30 Oct 1997 14:05:08 EST5DST

Dear Jesse,

You wrote:

"Robin - Thanks very much for your response to the note I posted on calculus and geometry. You wrote: ëI'm intrigued, but how does the diameter equal circumference divided by 2? It seems you have eliminated Pi from the equation. Perhaps this is why Pi figures in so importantly when computing curved areas within the Cartesian coordinate system.

Good luck with your work. Send me more info any time.í

I concede that the idea that diameter of a circle = the circumference divided by 2 is counterintuitive. This comes about from what I call the "diminishing circle argument." This argument goes as follows:

Take any circle E and draw two smaller circles inside the larger circle. The two smaller circles should be equal in area, with one edge intersecting the outside edge of the circle E (the point where the diameter intersects the circumference) and the other edge passing through the radius of E (this is easier to draw than to describe). This gives you two smaller circles, A and B, each with a diameter that is half the diameter of the original circle E. This is because the area of E is (pi x r squared), and the area of circle B is (pi x (r/2) squared), (where r refers to the radius of A). This equals pi x r squared/4, or 1/4 the area of the original circle E. Since A and B are equal in size, they have the same area, which is equal to 1/4 the area of E. Hence the areas of A and B together equal 1/2 the area of E, the large circle. This means that the remaining areas, C and D (which are also equal, but not circles) are together each equal in area to the areas of the smaller circles A and B.

The above shows that circle E is divided into four equal areas: the areas of A, B, C, and D. The next step shows that the circumference is B = the Circumference of E/2. The circumference of E = 2 x pi x r. Since the radius of B is 1/2 the radius of E, the radius of B is 2 x pi x r/2 (where r refers to the radius of E). But 2 x pi x r/2 = pi x r, which is 1/2 the circumference of E. So the circumference of B (one of the smaller circles) is half the circumference of E. This means that half way around the circumference of E, which is C/2, equals the entire circumference of circle B. This also means that halfway around the circumference of B plus halfway around the circumference of A equals halfway around the circumference of E (since A and B are equal, their circumferences are equal). So going from one end of the diameter of E to the other via the circumference of E is the same distance as going from one end of the diameter of E to the other via the two circumferences of the smaller circles A and B.

This last result is the main
result I need to get the argument that the diameter of a circle equals the
circumference divided by 2. Now we have circle E divided into four equal areas,
the areas of A, B, C, and D. Now imagine further dividing circles A and B each
into four smaller areas in the same way, by drawing two smaller but equal
circles within A and B, each dividing the areas of A and B into four smaller
areas. Again, going halfway around A is the same distance as traveling along
circumferences of the two smaller circles within A. Continue this process
indefinitely. At the end point, we have the diameter consisting of the smallest
allowable points packed up against one another. But the distance around all
these points still equals the distance around the circumference of E, so the
diameter of circle E will equal half the circumference. The same argument can
be repeated for any circle."

I enjoyed reading this and
thinking about it. You have started with a semicircle, divided it into two
semicircles, divided each of those into two semicircles, etc., while the total
length remained the same. You could have started with an isosceles triangle
instead of a semicircle and performed similar operations. Your argument would
then say that the diameter of the circle is two times the length of a side of
the original triangle. This argument would then assert that the diameter of a
triangle is any number greater than half the circumference.

The problem here is the following.
If you have a sequence of functions f(n) on an interval (whose length is the
length of the diameter) which converge to another function f on the interval
(in this case the zero function), convergence either point wise or uniform, it
does not mean that the lengths of the graphs of the f(n)'s converge to the
length of the graph of f. To see why this is true one need only to look at the
formula for arc length and consider your most interesting example.

Harry

To view original post, go to:

http://forum.swarthmore.edu/epigone/geometry-research/froxpalzhand/A1F63EE155A@acs.sbu.edu

`Jesse L. Yoder`

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