# Archimedes Trisection of Angle

Standard

The theorem that the general angle cannot be trisected with ruler and compass alone is true only when the ruler is regarded as an instrument for drawing a straight line through any two given points and nothing else.

By permitting other uses of the ruler the totality of possible constructions may be greatly extended. The following method for trisecting the angle, found in the works of Archimedes, is a good example.

METHOD:

Figure 1 : Archimedes’ trisection of an angle
[pp. 138, What is Mathematics?, © Oxford University Press Inc. ]

STEP – 1 Let an arbitrary angle x be given, as in Fig. 1.

STEP – 2 Extend the base of the angle to the left, and swing a semicircle with O as center and arbitrary radius r.

STEP – 3 Mark two points A and B on the edge of the ruler such that AB = r. (use compass to measure, thus using ruler to measure!!)

STEP – 4 Keeping the point B on the semicircle, slide the ruler into the position where A lies on the extended base of the angle x, while the edge of the ruler passes through the intersection of the terminal side of the angle x with the semicircle about O.

STEP – 5 With the ruler in this position draw a straight line, making an angle y with the extended base of the original angle x.

RESULT: This construction actually yields$y=x/3$

PROOF:

Figure 2: Labeling Archimedes’ trisection of an angle
[pp. 138, What is Mathematics?, © Oxford University Press Inc. ]

As in Fig 2,

$\Delta ABO$$\angle BAO = \angle BOA = y$ and $\angle ABO = \pi - z$

Similarly in , $\Delta BDO$$\angle BDO = \angle DBO = z$ and $\angle BOD = \pi - y - x$

From $\Delta ABO$,  we have, $2y-\pi - z = \pi \Rightarrow 2y = z \ldots (i)$

Similarly, from$\Delta BDO$, we have, $2z = y+x \ldots (ii)$

Now using $(ii)$ in$(i)$ , we can eliminate $z$ to get: $3y=x$