Art. XXX.—On the Trisection of an Angle. By H. W. Segar, M.A., Professor of Mathematics, Auckland University College. [Read before the Auckland Institute, 17th August, 1908.] Three problems of an apparently simple character are famous as having engaged the attention of the ancients, who sought in vain for solutions by means of the straight-edge and compass. These were (1) the trisection of an arbitrary angle; (2) the quadrature of the circle; (3) the duplication of the cube. These have generally ceased now to be objects of attack by mathematicians, because it is now known that it is impossible to solve any one of theses problems by means of the straight-edge and compass. The proofs of this impossibility are available to the English reader in the translation of Klein's “Famous Problems in Elementary Geometry,” published by Ginn and Company. Occasionally, however, one or other of these problems takes possession of some person unaware of these investigations, and with only a slight knowledge of mathematics. I was recently approached by Mr. Viggo Hansen, of this city, who submitted to me what he considered was a solution of the first problem—the trisection of an angle On testing his drawing I could discover no inequality in the three parts into which his construction had divided the original angle Thinking that the construction happened to suit the particular angle chosen, I carried it out for three other angles very different in magnitude from one another and from Mr. Hansen's, with exactly the same result. It was evident then that Mr. Hansen had obtained an approximate solution of the problem of exceptional interest, especially as the construction is very simple, and depends on the bisection of angles. The construction was as follows:—

Let AOB be the given angle. With O as centre, describe a circle meeting OA, OB, in C, D, and these lines produced through O in E, F, respectively. Let G, J, be the middle points of the arcs CD, EF; and Fig. 1. H, I, those of the arcs CG, GD, respectively. Now draw JH, EG, meeting in K; and JI, FG, meeting in L. Then OK, OL, are the trisectors as given by this construction. We shall now proceed to investigate the degree of approximation. We shall use the methods of co-ordinate geometry. Take OK as the axis of x, and a line through O perpendicular to OK as the axis of y. Let the angle GOH be a. Then we easily obtain the following system of co-ordinates:— Point. x. y. H a cos a a sin a G a cos(a - θ/4) a sin(a - θ/4) J –a cos(a - θ/4) –a sin(a - θ/4) E –a cos(a - θ/4) –a sin(a - θ/4) where θ is the original angle AOB. The equation of JH is x - a cos (a/y - a sin(a = cos a + cos(a θ/4)/sin a + sin (a + θ/4) which reduces to x - a cos(a/y - a sin(a = cot(a + θ/8)…(1).

The equation of EG is x - a cos(a - θ/4)/y - a sin (a - θ/4) = cos(a - θ/4) + cos(a - θ/4)/sin(a - θ/4) + sin(a + θ/4) which reduces to x - a cos(a - θ/4)/y - a sin (a - θ/4) = cot a…(2) These meet on OK at K. Hence put y = 0 and equate the resulting values of x. We get cos a - sin a cot (a - θ/8) = cos (a - θ/4) - sin (a - θ/4) cot a which becomes - sin θ/8/sin(a - θ/8) = sin θ/4/sin a = 2 sin θ/8 cos θ/8/sina ∴ sin a = 2 sin(θ/8 - a) cos θ/8 = sin (θ/4 - sin a ∴ 2 sin a = sin(θ/4 - a)…(3) Now, OK being almost coincident with a true trisector, we have a nearly equal to (θ/3 - θ/4)—i.e., to θ/12 This may also be seen from (3). Hence put a = θ/12 - ∊ where ∊ is very small. Then (3) gives 2 sin (θ/12 - ∊) = sin (θ/6 + ∊) Expanding and retaining only the first power of ∊, we get 2 sin θ/12 - 2 ∊ cos θ/12 = sin θ/6 + ∊ cos θ/6 and therefore ∊ = 2 sin θ/12 - sin θ/6/2 cos θ/12 + cos thetas/6 Here θ/6 and θ/12 are proper fractions. Expanding in powers of θ, and retaining only the lowest power of θ that remains, we get ∊ = θs/5184 This is the angle between a trisector as given by the construction and a real trisector. The accuracy of the construction increases rapidly as the angle diminishes, the error being approximately proportional to the cube of the angle. If the given angle be a right angle the error is 0.67 of a degree, for an angle of 45° it is 0.008 of a degree, and for one of 30° it is only 0.002 of a degree. In conclusion, we shall indicate briefly how the trisection of any angle between 0° and 180° can be derived from that of an angle not greater

than 45°, and therefore be effected by the construction of this article with an error less than one-hundredth of a degree. Let the angle AOB lie between 45° and 90° (fig. 2). Draw OC perpendicular to OA, and on the same side as OB. The angle BOC will be less than 45°. Let OD, OE, be the trisectors, OD being the nearer to OC. The trisectors OK, OL, of AOB will make angles of 60° and 30° with OB, OE, respectively; and the angles 60° and 30° are easily constructed with straight-edge-and compass. Fig. 2. Fig. 3. Fig. 4. Let the angle AOB lie between 90° and 135° (fig. 3). Draw OC as before. The angle BOC will again be less than 45°. Let OD, OE, be the trisectors, OD being nearer to OC. The trisectors OK, OL, of AOB will make angles of 60° and 30° with OD, OE, respectively. Lastly, let the angle AOB lie between 135° and 180° (fig. 4). Produce AO to C. The angle BOC will be less than 45°. Let OD, OE, be the tusectors, OD being nearer to OC. The trisectors OK, OL, of AOB will make angles of 120° and 60° with OD, OE, respectively.