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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 $AOB$. |
According to Jamison (see the reference at the top of this page) the construction's main idea comes from an unpublished work by C. R. Lindberg.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 $E'OD$ 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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