My Methods

 

First solutionLong Line

Like the Archimedes and Tomahawk methods, mine also impose the need to cheat. What sets them apart is that they require only the compass and unmarked ruler that are usually demanded by the terms of the challenge. They are, needless to say, mathematically exact methods for trisecting any arbitrary angle. This is the method I described in the March 2007 issue of the Mathematical Association of America’s College Mathematics Journal.

Angle ABC is the angle to trisect. D bisects BC. An arc is drawn with centre C and radius CD. DE is perpendicular to AB. The ruler is positioned so that its short left edge passes through B while its left upper corner meets DE and its upper edge is tangential to the arc. Angle ABF is exactly one third of angle ABC.

Although my method’s greatest strength is that it uses only a compass and ruler and requires no measurements along the ruler’s edge, it also has a neatness and simplicity that perhaps surpasses the elegance of Archimedes' method in that it places the trisecting line directly where it should be: inside the original angle. Also, without using any tools other than the two traditionally specified, it achieves exactly what the Tomahawk and other such contrived methods do. On the face of it, it meets the exact terms of the challenge as it is commonly, although too simply, expressed by teachers of geometry:

 

Other two-edge solutionsLong Line

Here are a few of my many other methods of unmarked-ruler non-Euclidean trisection using a second edge of the ruler rather than marking it. I may add further methods from time to time.


1. Perhaps the simplest method of all
This construction, like most of those that follow, uses the width of the ruler.

Angle ABC is the angle to trisect. D bisects BC, whose length is twice the width of the ruler. DE is perpendicular to AB. The ruler is positioned so that its corner F lies on DE, its corner G on the arc, and its edge passes through C. Angle ABF is exactly one third of angle ABC.

 

2. Also using the width of the ruler
But the construction relies on somewhat different yet related principles.

Angle ABC is the angle to trisect. The distance from B to D is twice the width of the ruler. E bisects BD.  EF is drawn parallel to AB, and an arc is drawn, with centre E, through B and D. The ruler is positioned so that its corner G lies on the arc, its corner H on EF, and its edge passes through D. Angle ABG is exactly one third of angle ABC.

 

3. Using a circle the width of the ruler
The construction is very similar to the previous two but less elegant.

Angle ABC is the angle to trisect. The distance from B to D is twice the width of the ruler. E bisects BD.  EF is drawn perpendicular to AB and EG parallel to AB.  A circle is drawn with centre D and a radius half the width of the ruler. The ruler is positioned so that its corner H lies on FE, its corner J on EG, and its edges are tangential to the circle. Angle ABH is exactly one third of angle ABC.

 

4. A related but rather different approach
Aspects of my pending one-edge solution are used in this construction.

Angle ABC is the angle to trisect. XXX is at XX degrees to XX, its length twice the width of the ruler. A circle with centre X is drawn through X. XX is perpendicular to XX and XX to XX. An arc is drawn, its centre at X, through the midpoint of XX. The ruler is positioned so that its corner X lies on XX, its long edge passes through X, and its short edge is tangential to the arc. Angle XXX is exactly one third of angle ABC.

 

5. A very different simple construction
It still relies on the width of the ruler but the principles are different.

Angle ABC is the angle to trisect. The radius of arc DE is equal to the width of the ruler. BF bisects angle ABC. The ruler is positioned so that its corner G lies on the arc DE, its corner H on BF, and its edge passes through E. Angle ABG is exactly one third of angle ABC.

 

6. A similar but more complex approach
Here, as in my main construction, the ruler’s width is irrelevant.

Angle ABC is the angle to trisect. The arc DE can have any radius. FG would be perpendicular to a bisector of angle ABC. DH is perpendicular to FG. FH is the same length as DF. The arc drawn through B has its centre at E. The ruler is positioned so that its corner K lies on FG, its short edge passes through H, and its long edge is tangential to the arc that passes through B. Angle ABJ is exactly one third of angle ABC.

 

7. The first of three wholly different trisections
Only one construction line and one circle are needed.

Angle ABC is the angle to trisect. DE is drawn parallel to a bisector of angle ABC that is not shown. The circle, whose centre is at B, has a diameter equal to the width of the ruler. The ruler is positioned so that its top corners lie on AB and DE and its edges are tangential to the circle. Angle ABE is exactly one third of angle ABC.

 

8. The second of three wholly different trisections
Here two construction lines and one circle are needed.

Angle ABC is the angle to trisect. DE is drawn parallel to AB and is tangential to a circle whose centre is at B and whose diameter is equal to the width of the ruler. The ruler is positioned so that its top corners lie on DE and BF and its edges are tangential to the circle. The angle between AB and the ruler is exactly one third of angle ABC.

 

9. The third of three wholly different trisections
Again only one construction line and one circle are needed.

Angle ABC is the angle to trisect. DE is drawn parallel to BC and is tangential to a circle whose centre is at B and whose diameter is equal to the width of the ruler. The ruler is positioned so that its top corners lie on AB and DE and its edges are tangential to the circle. The angle between AB and the ruler is exactly one third of angle ABC.
 

All these trisections have relied on using more than just one edge of the ruler. Now look at my New Method to read about a construction, my latest and most challenging, that will use only one edge — yet will still work without having to rely on any marks made on the ruler.

 

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