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Eureka!-Archimedes and π

I had wanted to talk about something other than my current classes today; however, I spent the entire day thinking about Archimedes and how best to explain his ideas to my students in my math history course.  Therefore, it really is all I can think about at the moment.

It’s not a bad topic to be stuck on.  Archimedes was quite the character according to the stories.  From the story of shouting Eureka as he jumped naked out of a bath tub and ran across town as he was overly excited about realizing how submersing an object leads to displacement, to making a death ray to destroy Roman ships, and ultimately being killed by a Roman soldier when Syracus fell, there are many great stories to be had.  Whether or not any or all of them are true isn’t really the important part right now, I’ll leave that for my discussion in class today.  However, the stories are exciting enough that they are worth sharing again and again as they get people to pay attention to subjects they otherwise may not.

Here, I wanted to talk about the problem I’ve been contemplating all day.  That is, how did Archimedes show that the ratio of the perimeter of a circle to its diameter is the same as the ratio of the area to the radius squared.  That is, how can we show A=πr2 and C=2πr, where π is the same both the same number in both situations and constant (this is not the notation that would have been used, however, it is more clear to the reader at this point).  Eudoxus had already shown that the ratio of the area of a circle to the square of the radius is the same for all circle (this appears in Euclid’s Elements). So, how does Archimedes proceed?   Well, instead of dealing with a circle, he looks at regular polygons that are inscribed in and circumscribe the circle in question.  Below we have such an example using squares.

squares

From here, we work under the assumption that we already have a regular polygon with n edges drawn and use this to recursively find the area and perimeter of a regular polygon with 2n edges.  In the picture below, CB is the edge of the n-gon, while CE and EB are edges of the 2n-gon.

inner.png

By comparing congruent triangles, we see that all the angles at point F are right angles.  Therefore, the area of ΔAEC=(AE)*CF/2.  If we let ln be the side length of the regular n-gon inscribed in the circle, and r the radius of the circle, we have that this area is then r*ln/4.  We then note that the regular 2n-gon is made up of 2n triangles congruent to this one, so, if an is the area of the regular n-gon and pn is the perimeter of the n-gon, we have that

innerform.png

We will now look out the polygons that circumscribe the circle.  Again, we will start with a regular n-gon and look at the regular 2n-gon.  Below we have the FG is the side of the regular 2n-gon, whereas, CE and BE are half the sides of the regular n-gon.

outer1.png

Note that the line FG is the tangent to circle at the point D, so the angles at point D are all right angles.  Therefore, the area of ΔAFD=(FD)(AD)/2.  Noting that there are 4n such congruent pieces in the regular 2n-gon, and letting An be the area of the regular n-gon, Pn is the perimeter of the n-gon and Ln the side of the n-gon, we have that

outerform.png

We now note that since the circle is larger than all the inscribed polygon and smaller than all the circumscribed polygon, we have that

comparison.png

From here, Archimedes assumed that the values were distinct, then used the method of exhaustion to give a contradiction.  However, I will instead use the ideas of limit and state that we note that the inscribed polygons and the circumscribed polygons get arbitrarily close to each other as n gets large.  That is, the area of the circumscribed and inscribed are the same as n goes to infinity, therefore, the values of P/2r and A/r2  must be equal to the limit of these ratios and therefore to each other.

We have therefore shown that A=πr2 and C=2πr where π is the same number in both cases.  Now, how do we find π?  While π is a well known number now and many people can recite many digits, we do need to have some way of figuring out what the numerical approximation would be.  Here, again, Archimedes was able to help.  If we look back at the picture of inscribed polygon, we can use this to find a recursive definition of the side lengths of the polygons.

inner.png

Since we know that π is the same for all circles, we will proceed with the assumption that the radius of the circle is 1 in order to simplify the calculations.  Using the repeated application of the Pythagorean theorem, we find thatperimeter.png

To approximate π, you would then need a starting polygon, and you could find each successive value using your previous approximation.  Archimedes started with n=3, then continued until he reached a 96-gon to approximate π.  Now, this would give a lower bound since we are using inscribed polygons.  If we wanted to find an upper bound, we could then do this again with circumscribed polygons.  Again, Archimedes did this and found that 223/71<π<22/7.  If you’d like to repeat the process for yourself, you can start with n=4, then the side length is √2.  Using a spreadsheet, you can then find your own estimate of π with an accuracy up to the level of accuracy of the spreadsheet you are using.

I hope you enjoyed the exploration into π, if you did, be sure to follow the page to get updates.

References

Boyer, C. B., & Merzbach, U. C. (2011). A History of mathematics. Hoboken, NJ: Wiley.

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