Verify Archimedes’ construction for angle trisection

Verify Archimedes’ construction for trisecting an angle.

We are allowed to construct points using a straightedge and compass. Moreover, our straightedge has an (arbitrary) unit distance marked on it. We are permitted to ‘slide’ the marked straightedge on the plane.

Suppose we have an angle \angle zxy of measure \theta, where 0 < \theta < \pi/2. Using our unit ruler, we can say (without loss of generality) that \overline{xy} has length 1. Construct the circle with center x and containing y; this circle intersects the opposite ray of our angle at a point with we call (again without loss of generality) z. So xz = 1. Now extend the segment \overline{xy} to a line.

Now slide the straightedge so that one mark is on the line \overline{xy} and the other is on the circle centered at x. Finally, slide the straightedge (keeping the marks on the line and circle) so that the edge intersects point z. Construct this segment, calling the points where the edge intersects the line and circle w and u, respectively, as in the following diagram.

Archimedes' Construction for angle trisection

In addition, label the angles \alpha, \beta, and \gamma as shown.

Note that \triangle xzu is isocoles, so that \beta = \gamma. Moreover, \triangle uwx is isocoles. Now \angle uxz has measure \pi - \theta - \alpha = \pi - \gamma - \beta, whence we have \theta = 2\beta - \alpha. Now \angle wux has measure \pi - 2\alpha, and moreover \beta + \pi - 2\alpha = \pi, so that \beta = 2\alpha. Thus \theta = 3\alpha.

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