# An angle trisection

#### R. L. Durham, A simple construction for the approximate trisection of an angle, American Mathematical Monthly, vol. 51, no. 4, April 1944, pp. 217–218.

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Drag the point $B$ to change the angle $AOB$. Press “r” to reset the diagram to its initial state. The red line, $OT$, is an approximate trisector of the angle $AOB$.

## Construction

We wish to trisect the given angle $AOB$ represented by the circular arc $AB$ centered at $O$, as shown in the diagram above.

1. Draw the bisector $OC$ of the angle $AOB$.
2. Draw the chord $AC$ and trisect it at point $F$ so that $CF=\frac{1}{3}CA$.
3. Locate point $G$ on the extension of the chord $AC$ so that $GC=CF$.
4. Draw a circle (shown in green) centered at $G$ and through point $F$.
5. Let $T$ on the green circle be such that $TC$ is perpendicular to $OC$.

The line $OT$ (shown in red) is an approximate trisector of the angle $AOB$.

R. L. Durham (see the reference at the top of this page) goes one step further. Using the line $OT$ as a starting point, he produces a substantially better approximation $OT'$. The construction for this second stage is complex and not particularly pretty, so I won't go into that here.

## Error Analysis

Let $\alpha$ and $\beta $ be the sizes of the angles $AOB$ and $AOT$, respectively. It is possible to show that \[ \beta = \frac{1}{2}{\alpha} - \arctan\bigg( \frac{4}{3}\sin\Big( \frac{\alpha}{4} - \arcsin\big(\frac{1}{2}\sin\frac{\alpha}{4}\big) \Big)\bigg) = \frac{1}{3}\alpha + \frac{7}{2^7\cdot3^4} \alpha^3 + O(\alpha^5) = \frac{1}{3}\alpha + \frac{7}{10368} \alpha^3 + O(\alpha^5). \]

The error $\ds e(\alpha) = \beta - \frac{\alpha}{3}$ is monotonically increasing in $\alpha$. The worst error on the interval $0 \le \alpha \le \pi/2$ is $e(\pi/2) =$ 0.00265 radians = 0.152 degrees. The worst error on the interval $0 \le \alpha \le \pi$ is $e(\pi)$ = 0.0218 radians = 1.252 degrees.

## Calculation details

The derivation of the formula for $\beta$ shown above is a straightforward application of trigonometry. Here are the details.

The angles $AOC$ and $ACT$ subtend the arc $AB$ of the circle centered at $O$. Since $AOC = \alpha/2$, then $ACT=\alpha/4$. Consequently, the angle $GCT$ is $\pi - \alpha/4$.

We apply the law of sines in the triangle $GCT$. (The edge $GT$ is not shown in the diagram to reduce clutter.) Let us write $\theta$ for the angle $CGT$. The the angle $CTG$ is $\alpha/4 - \theta$. The law of sines is: \[ \frac{\sin\theta}{CT} = \frac{\sin(\alpha/4-\theta)}{CG} = \frac{\sin(\pi-\alpha/4)}{GT}. \] But $GT=2CG$, therefore $2\sin(\alpha/4-\theta) = \sin(\pi-\alpha/4) = \sin(\alpha/4)$, whence \[ \theta = \frac{\alpha}{4} - \arcsin\big(\frac{1}{2}\sin\frac{\alpha}{4}\big). \] Going back to the equation of law of since, we compute $CT$: \[ CT = \frac{\sin\theta}{\sin(\alpha/4)} GT. \] But $GT = GF = 2CF = \frac{2}{3} AC = \frac{2}{3} \big(2OA\sin\frac{\alpha}{4}\big) = \frac{4}{3} OA\sin\frac{\alpha}{4}$. We conclude that $CT = \frac{4}{3} OA\sin\theta$. Then $COT = \arctan \frac{CT}{OC} = \arctan \frac{4}{3} \sin\theta$. Finally, the angle $\beta$, which equals $COA - COT$, is given by: \[ \beta = \frac{\alpha}{2} - \arctan \frac{4}{3} \sin\theta. \] Substituting the expression for $\theta$ calculated above, we arrive at the desired expression for $\beta$.

## Aside

In 1990, the well-known logician Willard Van Orman Quine wrote an expository article in the Mathematics Magazine where he proves that some angles cannot be trisected by ruler and compass. The proof is elementary (but not short) and requires nothing but simple algebra. Here is the full reference:

#### W. V. Quine, Elementary proof that some angles cannot be trisected by ruler and compass, Mathematics Magazine, vol. 63, no. 2, April 1990, pp. 95–105.

In a prefatory note he refers to Durham, the author of the trisection described in this web page. He writes:

> This purely expository paper dates from April 1946. Robert Lee Durham, president emeritus of Southern Seminary Junior and College, had sent me a hundred dollars and asked me to make it clear to him why an angle cannot in general be trisected by ruler and compass. He had himself presented a way of almost trisecting any angle by ruler and compass, to an accuracy for acute angles of 1/720 of a degree. [Here he refers to Durham's 1944 article cited at the top of this web page.]

> I welcomed the money and the occasion to familiarize myself with the famous proof. I was guided in large part by L. E. Dickson, Why it is impossible trisect to an angle or construct a regular polygon of 7 or 9 sides by ruler and compass, Mathematics Teacher, vol. 14 (1921), 217–223.

> Mr. Durham expressed satisfaction with my report and proposed paying for publishing it as a pamphlet. With his approval I submitted instead to a mathematics journal. After waiting nineteen months for a decision from the journal, I recalled the paper and dropped the matter.

Quine goes on to explain how this article was eventually published more than 50 years after it was written.

This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on June 13, 2010.

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