Archimedes' Spiral

Problem Statement

The task for this lab was to learn how to use the Geometers Sketchpad and to investigate Archimedes’ approach to squaring the circle, or finding a square with the same area as a given circle. What we are interested in here is finding a triangle with the same area as the circle. Archimedes’ method of doing this involved constructing a spiral out to 360° and a line tangent to the spiral at that point. The details of this can be seen in the figure(s).

Key Mathematical Concepts

  1. The first problem is to be able to make a spiral. Using the concept of polar coordinates, one can make a spiral by plotting, for each angle, a constant multiple of that angle as the radius. If the radius is r, the angle is q , and the constant multiple is a, then r(q ) = aq .
  2. The other problem is to find a line tangent to the spiral at a particular point on the spiral. One way to find this tangent line is by first finding it’s two components. I think that the easiest way to understand this is to think in physical terms: If there is a particle following some curved path, like a circle, then the instantaneous velocity at some point is the velocity that the particle would have if the forces keeping it on the path were suddenly gone. The direction of this velocity is along a line tangent to the curve at that point. Consider a mass attached to a string and swung around in a circle. If the string breaks, the mass goes off in a straight line tangent to the circle at the point the string broke. The tangent line that the mass follows only has one component, and that is in the tangent-to-the-circle direction. Call this the angular component; it’s magnitude is the angular velocity times the radius of the circle, or if the angular velocity was set to one, it would be just r, which is aq . If the string is being let out at some constant rate (say a cm per radian), then the mass follows a spiral and the tangent line gains another component, one perpendicular to the tangent-to-the-circle direction. In our spiral, this constant rate at which the radius is gaining length is simply a (It would actually be a times the angular velocity, which we said was one).

Here is the clip from a textbook (Edwards & Penny, pg.663)

Technology Used

We of course used the most versatile Geometer’s Sketchpad to construct the construction, and here are our basic directions and key concepts.

  1. First, construct a segment and measure it and call that length a.
  2. Next, you need an angle to be q . It has to be a number in radians that has a fixed finite range of 0 to 2p . If you make a circle, and measure the angle of a point as it moves around the circle, you end up with an angle measured in radians that goes from 0 to p and from 0 to -p . This doesn’t work—you need an angle to go from 0 to 2p radians in order to even make a proper spiral. The angle on a circle can be manipulated to give the proper numbers, but here is an interesting alternative: If you measure a segment that is 2p cm long and then play with the units by multiplying by 1 rad, and dividing by 1cm, you get 2p radians. If you measuree the distance from a point that moves on that segment to one of the endpoints, and change the units to radians in the same way, then you have your q .
  3. To make the spiral, you also need a length, aq , to plot against q . Simply multiply a times q and you have it, except that that is not a length, it has units of rad× cm, which isn’t good for anything. Divide this number by 1 rad, and you have your length, aq . (The necessity of modifying the units reflects the fact that a should have had units of cm/rad instead of just cm. This is consistent with the description of a in the math section above, and modifying the units of the original measurement a accordingly might be a better solution).
  4. Construct the spiral. You can plot aq against q on a polar grid, but it is simpler and amounts to the same thing to place a point A somewhere to be the center of the spiral, and translate it by a polar vector of magnitude aq and direction q . (select transform, translate, check ‘by a polar vector’, and click directly on the measurements you have for q and aq ; they will automatically be copied to the proper place in the translate dialogue, because of their units)
  5. Now you have a point on the spiral, you need all the points on the spiral. To construct a locus of that point (it is point B by the way), you need to select three things: The point itself, the point that controls what the measurement q is (changing q moves B around the spiral), and the line or circle that defines this point’s range of motion (so that it knows the limits of the range of q ). With these selected, construct a locus (under ‘construct’) and you have the spiral.
  6. The tangent line is made from the two components, a and aq . Translate the point B by a polar vector as before. Once with magnitude aq and direction q ± p /2 in order to be in the tangent-to-the-circle, or angular direction, and once with magnitude a and direction q . This one is in the radial direction. (I might have made these terms up, I hope they make sense). Then make a rectangle from the two components by constructing a couple of parallel lines, and the tangent is a line through B and the opposite corner of the rectangle.
  7. Now you have everything you need to make the circle and triangle. The circle is simply constructed from the center, A, with radius out to B. To make the triangle, draw the segment from A to B, and then a line perpendicular to this through A (select both the segment and the point A and ‘construct’, ’perpendicular line’). Click on the intersection of this line with the tangent line, and call that point C. And you are done! Measure the areas of the triangle and the circle (you have to ‘construct’ the interiors first) and if q is equal to 2p , the areas will (should? I hope so) be the same.

Principle Finding

Hurrah, the areas are the same. It is also notable that if q isn’t equal to 2p , the areas aren’t the same. Using the method of finding the components for the tangent line, it is easy to prove that the areas are the same when q equals 2p .

Here is a proof. Hopefully it makes sense when referenced to the figure below.

  1. The area of the circle is p r2, but r is aq , which is 2ap . So the area is 4a2 p 3.
  2. The tangent of the inside angle of the triangle at point B (from the components) is aq /a, but q is 2p , so it is just 2p . The tangent of that angle is also, by definition, AC/AB. So AC/AB = 2p , but AB is aq , or 2ap , so AC = 4ap 2. This is the height of the triangle.
  3. The area of the triangle is ½ × base × height, or ½ × aq × 4ap 2, or ½ × 2ap × 4ap 2, or 4a2 p 3.

Summary and Applications

I hope that this tangent line construction makes sense, and that it helps in understanding how Archimedes’ approach works. I also hope that the sketchpad construction might help others in the class get more out of the skethpad, especially concerning how to manipulate the units in order to use measurements in transformations (Assuming my directions made any sense).

The sketchpad file that we made should be here

(In this applet below (if it works) it was hard to get theta to be exactly 360. It turned out that even if theta is exactly 360.000, there is still a little roundoff error in the areas. Also, I think the 'units' are pixels. Drag the red points, and push the up arrow to reset the applet.)

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