\centerline{\bf How to Trisect an Angle}
\centerline{without significant physical injury}
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{\sl Stephen Judd 7/1/93
Los Alamos National Laboratory / Northwestern University
judd@sgt-york.lanl.gov}
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{\bf Abstract}
{\it There are three ways to get something done:
\qquad 1. Hire someone else to do it.
\qquad 2. Do it yourself.
\qquad 3. Forbid your children from doing it.}
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In this paper I present an iterative method of
trisecting an angle, where each iteration involves the use of a compass and
a straight-edge. The method has an error of about $10^{-{N\over 2}}$,
where $N$ is the number of iterations.
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{\bf Fi-Fi }
It can be easily shown that
$$\eqalign{ 2^n &= 3z + 1,\qquad {\rm n\ even}\cr
2^n &= 3z - 1,\qquad {\rm n\ odd}\cr} \eqno(1)$$
where $z$ is some integer, naturally different for different integer $n$.
This is
simply an artifact of the fact that any multiple of two is one away from
a multiple of three.
Duplicating angles is a simple thing to do with a compass and a
straight-edge. Simply draw a circle around the angle, with the apex of the
angle as the center of the circle. By drawing a line between the two points
of the intersection of the angle with the circle, a unique isoceles triangle
is formed for that angle. Hence, by using a compass the length of this third
side is known, as is the radius of the circle, hence the angle may be
reproduced.
Everyone knows how to bisect an angle using a compass.
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{\bf The Method}
Let us say that we are given an angle $\alpha$ to trisect. First
bisect the angle. This gives us two angles each of size $\alpha \over 2$.
Next add one of these half-angles to $\alpha$, which gives us an angle of
$3\alpha\over 2$. Bisect this angle, and bisect the bisection. This leaves
us with an angle of size
$$\alpha ^\prime = {3\over 8}\alpha = .375\alpha.$$
Next, add this new angle $\alpha^\prime$ to $\alpha$, and bisect it
twice. This gives a new angle of
$$\alpha^{\prime\prime} = {11\over 32}\alpha = .343\alpha.$$
If we again add and bisect twice we get
$$\alpha^{\prime\prime\prime} = {43\over 128}\alpha = .336\alpha.$$
The next three terms are
$$\alpha^{(4)} = {171\over 512}\alpha = .3339\alpha$$
$$\alpha^{(5)} = {683\over 2048}\alpha = .3334\alpha$$
$$\alpha^{(6)} = {2731\over 8192}\alpha = .33337\alpha.$$
As you can see, $\alpha^{(i)}$ is rapidly converging to ${1\over 3}\alpha$. The
error decreases by about an order of magnitude every two iterations.
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{\bf Analysis}
The iterations generate a sequence of numbers $x_i$, where
$$x_{i+1} = {1\over 4}(x_i + 1)\eqno(2)$$
or
$$x_i = {1 + \sum_{j=1}^i 2^{2j-1}\over 2^{2i+1}}.$$
However, by using equation (1) we can also write the sequence as
$$x_i = {z_1\over 3z_2 - 1}$$
where $z_1$ and $z_2$ are each some integer and
$n$ is some odd integer. In fact, if we look
at the first term in the iteration $({3\over 8}\alpha)$,
we see that $z_2 = z_1$. Let us then
assume that this property holds for all $x_i$, that is,
$$x_i = {z\over 3z-1}\eqno(3)$$
for some integer $z$. From equation (2) we see that
$$\eqalign{x_{i+1} &= {1\over 4}\Bigl({z\over 3z-1} + 1\Bigr)\cr
&= {4z - 1\over 4(3z-1)}\cr
&= {z^\prime \over 3z^\prime - 1}\cr}$$
where
$$z^\prime = 4z - 1.$$
It is then clear that successive iterations preserve the property in
equation (3), so that each term in the sequence is of the form
$$z\over 3z-1.$$
This can be rewritten as
$${1\over 3} {z\over z - {1\over 3}}$$
or, after doing the very important step of making the equation look really
mathematically suave,
$${1\over 3}{z\over z - \epsilon}, \qquad \epsilon < 1 < z.$$
Since z is monotonically increasing, it is clear that this sequence
approaches $1\over 3$ in the limit.
Therefore, successive application of the aforementioned iterative
process yields highly desirable results wherein the angle in question is
thrice reduced.
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{\bf Conclusions}
Hang on...
\bye