1.4. The spiral of Archimedes 11 Proof. Suppose that the angle ∠AOB is given, with O being the origin. See Figure 1.4.2. Let P be the point where the ray OB meets the spiral, S. If ∠AOB has α radians, then the equation of the spiral, r = θ, implies Figure 1.4.2 that |OP| = α. Now use Proposition 1.3.1 to trisect the line segment OP . Denote the equidistant points on OP as R and Q. See Figure 1.4.3. So Figure 1.4.3 |OR| = |RQ| = |QP| = 1 3 α. As in Figure 1.4.3, draw the circle with center O and radius |OR| let T denote the point where this circle meets the spiral. Consider ∠AOT . Because of the nature of the spiral, ∠AOT has 1 3 α radians. Therefore we have trisected ∠AOB. Archimedes was aware of this. Don’t worry. He was not under the illusion that he had solved the ancient trisection problem of the Greeks. Archimedes was a truly great mathematician and no such misunderstanding was possible.
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