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### Trisecting Angles

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Date: 03/10/98 at 04:47:49
From: Melanie
Subject: trisecting angles not divisible by 3

Hi. I have been faced with this rather annoying problem:

"The measure in degrees of a given angle is 180/n, where n is a
positive integer not divisible by 3. Prove that the angle CAN be
trisected using Euclidean means."

There isn't any 'rule' for trisecting angles. The only way I can think
of is to construct a 60 degree angle to trisect a straight angle, and
then to bisect this to obtain a right angle. How do you sugest I go
stuff? (I suppose this relates to 60, which is a multiple of 3?)

Thank you,
Melanie.
```

```
Date: 03/12/98 at 14:56:00
From: Doctor Wilkinson
Subject: Re: trisecting angles not divisible by 3

This is a very nice little problem. At first glance, it looks very
difficult or even impossible, but it turns out to be quite easy.

Since n is not divisible by 3, it must be of the form

3q + 1
or
3q - 1,

where q is a positive integer. Suppose that

n = 3q + 1

Divide both sides by 3n, and you get

1/3 = q/n + 1/(3n)

or

1/(3n) = 1/3 - q/n

and, multiplying by 180,

180/(3n) = 180/3 - q(180/n)

Now, you're given the angle 180/n. Draw a circle with its center at
the vertex of the angle. Suppose the angle is AOB, where O is the
center of the circle and going the short way from A to B around the
circle is counterclockwise. Now set the compass to the radius of the
circle, put the point of the compass at A and mark a point C on the
circle, again going counterclockwise. This makes AOC 60 degrees, or
180/3. Now set the compass point on A and the other arm at B, and step
clockwise from C for q steps. You will now be at a point X, with AOX
equal to 180/3 - q(180/n) = 180/(3n), so you have trisected the given
angle.

I'll let you work out how to do the case when n = 3q - 1.

-Doctor Wilkinson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry

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