# An angle trisection

#### Free Jamison, Trisection Approximation, American Mathematical Monthly, vol. 61, no. 5, May 1954, pp. 334–336.

Drag the point $B$ to change the angle $AOB$ (but stay on the right half of the circle). Press “r” to reset the diagram to its initial state. The red line $OE$ is an approximate trisector of the angle $AOB$.

## The construction

This construction, due to Free Jamison (see the reference at the top of this page) is a more accurate variant of the construction described in a simpler construction.

Consider the circular arc $AB$ centered at $O$, shown in the diagram above. Assume the angle $AOB$ is between 0 and 180 degrees. To trisect $AOB$, do:

1. Pick the points $F$ and $D$ on the arc $BA$ such that arc $BF$ = 2/8 of the arc $BA$ and arc $BD$ = 3/8 of the arc $BA$.
2. Extend $FO$ to intersect the circle at a point $C$.
3. Draw the line $CD$ and extend it to a point $E$ such that $DE$ equals the circle's diameter.

The line $OE$ is an approximate trisector of the angle $AOB$.

## Error Analysis

Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $EOB$, respectively. The angle $FOD$ equals $\alpha/8$ by the construction, therefore the angle $FCD$, which is half the central angle $FOD$, is equal to $\alpha/16$. The triangle $DOC$ is isosceles, therefore the angle $ODC$ also equals $\alpha/16$.

In the triangle $OED$, let $x$ and $y$ be the sizes of the angles $OED$ and $EOD$, respectively. Since the sum $x+y$ of the triangle's internal angles equals the triangle's external angle $ODC$, we have $x+y = \alpha/16$. Let us note, however, that the angle $y$ equals $DOB$ minus $EOB$. Thus $y = 3\alpha/8 - \beta$, whence $x = \beta - 5\alpha/16$.

In the triangle $OED$, the side $DE$ is twice the side $OD$ by the construction, therefore the law of sines gives $\sin y = 2 \sin x$. Consequently, $\sin(3\alpha/8 - \beta) = 2 \sin(\beta - 5\alpha/16)$. Solving this for $\beta$ we arrive at: \[ \beta = \frac{5}{16} \alpha + \arctan \frac{\sin(a/16)}{2+\cos(a/16)} = \frac{1}{3} \alpha - \frac{1}{2^{12}\cdot3^4} \alpha^3 + O(\alpha^5) = \frac{1}{3} \alpha - \frac{1}{331776} \alpha^3 + O(\alpha^5). \]

We see that the trisection error $e(\alpha) = \alpha/3 - \beta$ is given by: \[ e(\alpha) = \frac{1}{48}\alpha - \arctan \frac{\sin(a/16)}{2+\cos(a/16)}. \] (This formula is also given in Jamison's article.) The function $e(a)$ is monotonically increasing in $\alpha$. The worst error on the interval $0 \le \alpha \le \pi/2$ is $e(\pi/2)$ = 0.0000117 radians = 0.00067 degrees. The worst error on the interval $0 \le \alpha \le \pi$ is $e(\pi)$ = 0.000093756 radians = 0.00537 degrees. Quite impressive!

This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on July 22, 2002.
Cosmetic revisions on June 7, 2010.

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