An angle trisection

Construction by Chris Alberts

Sorry, no pictures. Explanation here.

The construction described in this page is due to Chris Alberts, who sent it to me in an email on March 15, 2011. I have paraphrased and rearranged his construction, but the 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 that the angle $AOB$ is between 0 and 90 degrees. To trisect $AOB$, do:

1. Draw the 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 circle $C''$ (shown in green) centered at $B'$ through the point $O$.
3. Let $E$ be the midpoint of the line segment $OB'$. Draw a line through $E$ parallel to $OA$ and mark its intersections $P$ and $R$ with the circles $C''$ and $C'$, as shown.
4. Let $M$ be the midpoint of the line segment $PR$. Draw the line $B'M$ and extend to the intersection point $N$ with the circle $C'$.
5. Draw a line through $B'$ parallel to $OA$, and select the point $F$ on it so that $NB' = NF$.
6. Extend the line segment $NF$ to intersect the circle $C$ at $G$.
7. Draw the line $GO$ and extend it to the point of intersection $H$ with the circle $C'$. Note: If you look at the diagram closely, you will be able to see that the line segments $GO$ and $GN$ are not collinear.
8. Let $D$ be diagonally opposite the point $B'$ in the circle $C'$. Draw a line through $D$ parallel to $OA$, and select the point $J$ on it so that $HD=HJ$. Note: Although it is impossible to discern visually, the line segments $OH$ and $HJ$ are not collinear.
9. Extend the line segment $HJ$ to intersect the circle $C$ at $K$.
10. Reflect $K$ about the line $OA$ to get the point $T$.

Error analysis

The calculation of the coordinates of all the points that appear in the construction is elementary but the resulting expressions are massively large, therefore I will refrain from putting them on this web page. (I did the calculations in Maple).

Let $\alpha$ and $\beta(\alpha)$ be the sizes of the angles $AOB$ and $AOT$, respectively. Express the trisection error as $e(\alpha) = \frac{\alpha}{3} - \beta(\alpha)$. It turns out that $e(0) = e(\pi/2) = 0$. In the range $0$ to $\pi/2$ the error is the largest near 1.22175 radians = 70.0013 degrees. The maximum error is $2.32\times10^{-18}$ radians = $1.33\times10^{-16}$ degrees.

Expanding $\beta(\alpha)$ in power series we get: \[ \beta(\alpha) = \frac{1}{3} \alpha + \frac{5^9}{2^{13} \cdot 3^{40}} \alpha^{27} + O(\alpha^{29}) = \frac{1}{3} \alpha + \frac{1953125}{99595595440594360737792} \alpha^{27} + O(\alpha^{29}). \]

Alberts' refinement

The extraordinary precision of Chris Alberts' trisection is a result of the application of a refinement technique which I will call Alberts' refinement. The 10 steps of his trisection procedure, described above, consist of three distinct stages:

1. Stage 1, corresponding to the steps 1–5 of the construction, produces the angle $FB'N$ which is roughly one third of the angle $AOB$. The trisection error at this stage may be as large as 0.7 degrees.
2. Stage 2, corresponding to the steps 6–7 of the construction, produces the angle $A'OH$ which is quite close to $\frac13 \alpha$. The worst error is 0.00013791 degrees and occurs when $AOB$ is near 70 degrees.
3. Stage 3, corresponding to the steps 8–10 of the construction, produces the angle $AOT$ which, as noted above, is within $1.33\times10^{-16}$ degrees of the exact trisection.

The lines $L_1$ and $L_2$ intersect at the point $O$. Suppose that the line $OX$ is a crude trisector of the angle between $L_1$ and $L_2$. The rest of the diagram shows a straightedge and compass construction that produces a much finer trisection. Here are the details of the construction:

1. Pick an arbitrary point $A$ (other than $O$) on the line $L_1$.
2. Locate the point $P$ on the line $OX$ so that $AO = AP$.
3. Locate the point $B$ on the line $L_2$ so that $PO = PB$.
4. Draw a circle centered at $A$ and radius equal to three times the length of $OA$. (This circle is shown in magenta.)
5. Extend the line segment $PB$ to a intersect that circle at a point $G$. (This extension is shown in cyan.)
6. Draw a line $AL$ through $A$ and parallel to the line $L_2$.

A quite straightforward calculation, involving an application of the law of sines in the triangle $APG$ leads to the equation: \[ \delta' = \delta - \arcsin\Big(\frac13 \sin3\delta\Big). \] Expanding this into power series in $\delta$, we obtain: \[ \delta' = \frac43 \delta^3 - \frac45 \delta^7 + O(\delta^9). \] This explains the notable efficiency of the refinement. For instance, if the value of $\beta$ has two significant digits after the decimal point, the value of $\beta'$ will have six significant digits after the decimal point.

Remark 1: Move the point $X$ in the diagram and note how insensitive the angle $LAG$ is to the choice of $X$. This indicates that even a crude initial approximation produces an excellent trisection.

Remark 2: If you examine closely Chris Alberts' trisection described earlier in this page, you will find buried in it two instances of Alberts' refinement.

A final comment

Comparing the precision of the trisection described in this page to those of others presented on my website may not seem to be quite fair. After all, any approximate trisection may be applied iteratively to refine its own result. Nevertheless, I am making an exception in this case because in Chris Alberts' trisection, the iterative refinement is an inherent feature of the method.

This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on March 23, 2011.

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