1.3. Some possible constructions 9 two triangles are similar. By Proposition 1.2.2, |OP| |OA| = |OR| |OC| . But |OC| = r and |OA| = 3r. Doing some algebra shows that |OP| = 3|OR|, so that the point R trisects the segment OP . While we will show that not every angle can be trisected, some angles can be. An easy example is an angle with 3π/2 radians. In fact, to do this we merely construct a line perpendicular to one of the sides of the 3π/2 angle. The next result is a little more interesting. 1.3.2. Proposition. The angle of π 2 radians can be trisected. Proof. Let ∠AOB be a right angle. Choose any radius r for the compass and draw a circle with center O and radius r let P be the point where this circle meets the ray OB. Now using P as the center, draw another circle of radius r let Q be the point where this second circle meets the first. See Figure 1.3.3. Draw the segments OQ and PQ. Note that |OP| = |PQ| = Figure 1.3.3 |OQ| = r. So ΔOPQ is an equilateral triangle and all its interior angles are π 3 . Hence ∠AOQ = π 6 , trisecting the right angle. Exercises 1. Show that any line segment can be cut into 5 equal pieces using straight- edge and compass alone. 2. Is it possible to cut a segment into as many equal parts as desired using straightedge and compass alone? 3. Show that an angle of π 4 radians can be trisected.
Previous Page Next Page