# An angle trisection

#### Construction by Mark Stark

Drag the point $B$ to change the angle $AOB$ (but stay on the upper half of the semicircle). Drag the point “$E$” to change the radius of the circle centered at $A'$. Note how little $G$ is affected by the choice of $E$. Press “r” to reset the diagram to its initial state. The red line ($O$T) is an approximate trisector of the angle $AOB$.

## Construction

This construction, due to Mark Stark, was announced in an message in the `geometry.puzzles` newsgroup on Jun 20, 2002. Scroll to the bottom of that page to view the related discussion thread.

The construction is unusual because one of the steps involves an arbitrary choice. It is interesting that the result is quite insensitive to that choice.

Here I have paraphrased Mark's construction but differences from the original are cosmetic. The error analysis is mine.

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

1. Draw a circle $C'$ centered at $O$ with a radius 1/3 $OA$. Mark $A'$ and $B'$ its intersections with the line segments $OA$ and $OB$, respectively.
2. Draw the line $A'B'$ (shown in green).
3. Draw a circle $C''$ centered at $A'$ with an arbitrary(!) radius. The accuracy of the trisection will be affected by the choice of the radius, albeit only slightly. Best results are obtained when the angle $EA'D$ (see the next step) is close to one third of the angle $AOB$.
4. Let $D$ be the intersection of $C''$ with the line segment $A'B'$.
   Let $E$ be the intersection of $C''$ with the $C'$, as shown.
5. Draw the line $ED$ and extend to the intersection point $G$ with the the circle $C$.
6. Draw the diameter $GOT$.

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

## Error Analysis

In the diagram below, I have duplicated the previous diagram and added the lines $OE$ and $EA'$ which are not needed in the construction, but are needed for the error analysis.

You may zoom and translate the diagram to examine its details. To zoom in, grab the point $B'$ with the mouse and move it away from $O$. To translate, grab $O$ and move it around. As always, type “r” to reset the diagram to its initial state.

To zoom in, grab the point $B'$ with the mouse and move it away from $O$. To translate, grab $O$ and move it around. Type “r” to return to the initial state.

Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $AOT$, respectively. We will show that $\beta \approx \frac{1}{3}\alpha$.

The construction leaves the size of the circle $C''$ (centered at $A'$) unspecified. We parametrize the circle by the position of the point $E$ along the arc $A'B'$, or more precisely, by the value $\gamma$ of the angle angle $B'A'E$. Thus $\gamma=0$ when $E$ coincides with $B'$ and $\gamma=\alpha/2$ (easy to verify) when E coincides with $A'$.

Since the angles $B'A'E$ and $B'OE$ subtend the arc $B'E$ of the circle $C'$, then the angle $B'OE$ is $2\gamma$. Therefore the angle $EOA'$ is $\alpha - 2\gamma$. But $EOA'$ is the vertex angle of the triangle $EOA'$, therefore the base angle $OEA'$ is $\frac{1}{2}(\pi - \alpha + 2\gamma)$.

In the isosceles triangle $DA'E$, the vertex angle is $\gamma$, therefore the base angle $DEA'$ is $\frac{1}{2}(\pi - \gamma)$.

Putting the assertions of the two previous paragraphs together, we calculate the angle $OED$: \[ OED = OEA' - DEA' = \frac{1}{2}(3\gamma - \alpha) \]

In the triangle $GOE$, we have just computed the angle at $E$ (because $OED$ is the same as $OEG$). Let us write $x$ for the angle at $G$. Then $x$ may be computed by applying the law of sines and noting that the ratio of the sides $OG$ to $OE$ is 3. We get: $3\sin x = \sin\frac{1}{2}(3\gamma-\alpha)$.

The external angle $EOT$ of the triangle $GOE$ equals the sum of the remaining internal angles, that is: \[ EOT = x + \frac{1}{2}(3\gamma-\alpha). \] On the other hand, \[ EOT = EOB' - TOB' = EOB' - (A'OB' - A'OT) = 2\gamma - (\alpha - \beta). \] We see then $x + \frac{1}{2}(3\gamma-\alpha) = 2\gamma - (\alpha - \beta)$, whence $x = \beta + \gamma/2 - \beta/2$. This leads to the equation: \[ 3\sin\big(\beta + \frac{1}{2}\gamma - \frac{1}{2}\beta\big) = \sin\frac{1}{2}(3\gamma-\alpha), \] which we may solve for $\beta$: \[ \beta = \frac{1}{3}\alpha + \frac{1}{6}(\alpha-3\gamma) - \arcsin\bigg[ \frac{1}{3}\sin\Big(\frac{\alpha-3\gamma}{2} \Big) \bigg] \]

As expected, the constructed angle, $\beta$, depends on the original angle $\alpha$ we well as the choice of $\gamma$. Let us express this dependence as $\beta = \tau(\alpha,\gamma)$. Expanding $\tau$ in power series we get: \[ \beta = \tau(\alpha,\gamma) = \frac{1}{3}\alpha + \frac{4}{3}\Big(\frac{\alpha-3\gamma}{6} \Big)^3 - \frac{4}{5}\Big(\frac{\alpha-3\gamma}{6} \Big)^7 + O\bigg( \Big(\frac{\alpha-3\gamma}{6}\Big)^9 \bigg). \] The term with exponent 5 is absent in the series expansion; that's not a typo.

### On the choice of $\gamma$

We see that $\tau(\alpha,\alpha/3) = \alpha/3$, that is, the construction produces an exact trisection with the choice $\gamma=\alpha/3$. Of course, constructing such a $\gamma$ is equivalent to solving the original trisection problem, therefore that is not an option. On the other hand, a constructible $\gamma$ that comes close to $\alpha/3$ will serve just fine. The function $\tau$ is not very sensitive to the variations of $\gamma$ as is evident from: \[ \frac{\partial \tau(\alpha,\gamma)}{\partial \gamma} = \frac{1}{2} \bigg( \frac{3\cos 3x}{\sqrt{9 - \sin^2 3x}} -1 \bigg), \] where I have let $x=(\alpha-3\gamma)/6$ to simplify the notation. As noted above, best results are achieved when $\gamma$ is close to $\alpha/3$. Even with a not-so-optimal choice of $\gamma=\alpha/4$ we get $x=\alpha/24$. With such a choice, the value of partial derivative in the range $0 \le \alpha \le \pi/2$ does not exceed 0.01, indicating that the value of the function is essentially independent of $\gamma$ on that range.

An excellent choice for $\gamma$ is obtained as follows. In Step 3 of the construction, first select the point $D$ on the line segment $A'B'$ such that $A'D = \frac{1}{3} A'B'$. Then draw the circle $C''$ with center $A'$ passing through $D$. One may verify that this results in an angle $\gamma$ given by: \[ \hat{\gamma} = \frac{1}{2} \alpha - \arcsin\Big( \frac{1}{3} \sin\frac{\alpha}{2} \Big) = \frac{1}{3} \alpha + \frac{1}{2\cdot3^4} \alpha^3 + O(\alpha^7). \] Then, the constructed angle is: \[ \beta = \tau(\alpha,\hat{\gamma}) = \tau\bigg(\alpha, \frac{1}{2} \alpha - \arcsin\Big( \frac{1}{3} \sin\frac{\alpha}{2} \Big) \bigg) = \frac{1}{3} \alpha - \frac{1}{2^4\cdot3^{13}} \alpha^9 + O(\alpha^{13}). \] The construction error, $e(\alpha) = \frac{1}{3}\alpha-\beta$, is monotone increasing. Since $e(\alpha) = O(\alpha^9)$, we expect it to be very small. Indeed, the worst error on the interval $0 \le \alpha \le \pi/2$ is the incredibly small $e(\pi/2)$ = 0.00000226 radians = 0.00013 degrees. The worst error on the interval $0 \le \alpha \le \pi$ is $e(\pi)$ = 0.00103 radians = 0.0592 degrees.

Despite its extraordinary accuracy, this is not among my favorite trisection methods because the points $D$ and $E$ are too close to each other for locating the point $G$ reliably. For practical purposes, should there be such a need, I would much rather use a more robust, albeit less accurate, method.

This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on July 26, 2002.
The error analysis was thoroughly revised and extensive cosmetic changes were made on June 7, 2010.

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