Sorry, no pictures. Explanation here.
According to Jamison (see the reference at the top of this page) the construction's main idea comes from an unpublished work by C. R. Lindberg.
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:
The line $OE$ is an approximate trisector of the angle $AOB$.
Here is a heuristic explanation for why the construction works, as explained by Jamison. The key lies in the observation that (i) in the triangle $ODE$ the angles $O$ and $E$ are “small”, and (ii) the side $ED$ is twice as long as the side $OD$. Therefore from the law of sines we have $\sin(O)/\sin(E) = ED/OD = 2$, which implies that the angle $O$ is approximately twice the angle $E$ in the triangle $ODE$.
Let the measure of the angle $OED$ be $x$. Then the triangle's external angle at $D$, that is the angle $ODC$, is the sum of the internal angles $O$ and $E$, therefore it is approximately $3x$. Therefore the angle $OCD$ is $3x$. Therefore the angle $BOD$ is $6x$. Since the angle $EOD$ is $2x$, we conclude that the angle $BOE$ is $4x$. Furthermore, since the angle $BOD$ is $6x$, the angle $BOA$ is $12x$. This shows that $BOA$ is 3 times $BOE$, as asserted.
Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $EOB$, respectively. The angle $DOB$ is half of $AOB$ by the construction, therefore it is equal to $\alpha/2$. Consequently the angle $DCB$, which is half the central angle $DOB$, equals $\alpha/4$. The triangle $DOC$ is isosceles, therefore the angle $ODC$ also equals $\alpha/4$.
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 $x+y = \alpha/4$. Let us note, however, that the angle $y$ equals $DOB$ minus $EOB$. Thus $y = \alpha/2 - \beta$, whence $x = \beta - \alpha/4$.
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, $\sin(\alpha/2 - \beta) = 2 \sin(\beta - \alpha/4)$. Solving this for $\beta$ we arrive at: \[ \beta = \frac{1}{4} \alpha + \arctan \frac{\sin(a/4)}{2+\cos(a/4)} = \frac{1}{3} \alpha - \frac{1}{2^6\cdot3^4} \alpha^3 + O(\alpha^5) = \frac{1}{3} \alpha - \frac{1}{5184} \alpha^3 + O(\alpha^5). \]
We see that the trisection error $e(\alpha) = \alpha/3 - \beta$ is given by: \[ e(\alpha) = \frac{1}{12}\alpha - \arctan \frac{\sin(a/4)}{2+\cos(a/4)}. \] (This formula is also given in Jamison's article.) The function $e(a)$ is monotonically increasing. The worst error on the interval $0 \le \alpha \le \pi/2$ is $e(\pi/2)$ = 0.000757 radians = 0.0434 degrees. The worst error on the interval $0 \le \alpha \le \pi$ is $e(\pi)$ = 0.0063 radians = 0.361 degrees. That's quite good for such a simple construction.
The angle $\beta$ constructed by this method coincides exactly with that of Pllana's construction, where $\beta$ is given as: \[ \beta = \arctan \frac {\sin\frac{\alpha}{2} + 2\sin\frac{\alpha}{4}} {\cos\frac{\alpha}{2} + 2\cos\frac{\alpha}{4}}. \] One way to verify that the seemingly different expressions for $\beta$ are in fact identical, is to compare their derivatives. In both cases we have: \[ \frac{d\beta}{d\alpha} = \frac{3(1 + \cos\frac{\alpha}{4})}{2(5 + 4\cos\frac{\alpha}{4})}. \]
Although this construction is quite good as is, Jamison proceeds to give an extension of Lindberg's method which requires a bit more work but is substantially more accurate.
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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