Error estimate for Mark Stark's approximate angle trisection

Please refer to An angle trisection page for the description of the problem the corresponding diagram.

Let T be the measure of the angle to be trisected, that is BOA = T. Our purpose is to calculate the size of the construction error, that is, the difference between T/3 and the measure of the constructed angle E'OA.

The error depends on T as well as the choice of the point E. We specify the point E in terms the measure `a' of the angle BAE. Note that a varies between 0 (when E coincides with B) and T/2 (when E coincides with A). We write the error as e = e(T,a).

Remark 1: When E corresponds to an exact trisection, the points E and E' coincide, the angle EOA is T/3, therefore the arc BE is 2T/3 hence the a = T/3.

We will show that for any T and a:

(1)

e(T,a) = T/6 - a/2 - arcsin((1/3)sin(T/2 - 3a/2)).

Note, in particular, that e(T,T/3) = 0, which confirms the statement made in Remark 1 above.

The key idea in the derivation of (1) is the observation that in the triangle OEG, the side OG is three time the side OE, therefore by the law of sines, the angle at vertex E is approximately 3 times as large as the angle at vertex G. (These angles are approximately equal to their sines, since they are pretty small.)

Note: I have not drawn the line segment OE in the diagram to avoid clutter. Because of this, the triangle OEG does not really look like a closed polygon at all!) You just have to imagine that there is a line segment joining O and E.

Note: A similar observation (about angles being propotional to the side lengths) was made in an unrelated trisection construction described by Free Jamison in Trisection Approximation, American Mathematical Monthly, vol. 61, no. 5, May 1954, pp. 334-336.

Before describing the derivation of equation (1), let's look at some of its implications.

Implication 1: The series expansion of e(T,a) about a=T/3 for a fixed T, is:

e(T,a) = (1/6) (T/3 - a)^3 - (1/160) (T/3 - a)^7 + O(T/3 - a)^9

which indicates that the construction error is related to the cube of the error in the choice of the point E.

Implication 2: If we select E such that a = T/4 (which can be constructed by bisecting T twice) we get:

e(T,T/4) = T/24 - arcsin((1/3)sin(T/8))

This is an increasing function of T. The worst error occurs at T = Pi and equals e(Pi,Pi/4) = 0.00299 radians. The corresponding relative error is 0.00095176, that is, 0.095%.

If we restrict the angle T to the range [0, Pi/2], then the worst error occurs at T = Pi/2 and equals e(Pi/2,Pi/8) = 0.00037382 radians. The corresponding relative error is 0.0002379797, that is, approximately 238 parts in a million.

The derivation of the expression (1)

We proceed by determining several angles in the diagram in terms of T and a.

1. The central angle BOE is twice BAE, therefore BOE = 2a.
2. Therefore EOA = T - 2a.
3. The triangle EOA is isosceles, therefore the base angle OEA = (Pi - EOA)/2 = Pi/2 - T/2 + a.
4. The triangle EAD is isosceles and its vertex angle is a, therefore the base angle DEA = (Pi - a)/2.
5. We now can calculate the angle DEO = DEA - OEA = T/2 - 3a/2.
6. Consider the triangle OEG. Its angle at vertex E is GEO = DEO = T/2 - 3a/2.
7. Since the side OG of the triangle OEG is 3 times as long as the side OE, the law of sines implies that sin(EGO) / sin(GEO) = 1/3, therefore EGO = arcsin((1/3)sin(GEO)) = arcsin((1/3)sin(T/2 - 3a/2).
8. The external angle EOE' of the triangle OEG is the sum of the other two internal angles, therefore EOE' is approximately T/2 - 3a/2 + arcsin((1/3)sin(T/2 - 3a/2).
9. Therefore the angle E'OA = BOA - ( BOE + EOE') is approximately T - [ 2a + T/2 - 3a/2 + arcsin((1/3)sin(T/2 - 3a/2) ]. The error e(T,a) is the difference between this expression and T/3. This simplifies to give (1).