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Trisecting an Angle Using Compass and Straightedge

Date: 04/29/2004 at 14:17:13
From: Joe 
Subject: Trisect

I've been working on trisecting an arbitrary angle, and I can do it
with no measuring and only a straightedge and a compass.  I can prove
it with "Geometer's Sketchpad."

I divide the arc of the angle and not the angles as you would in 
bisecting.  My reasoning is that one half the circumference of a
circle that is two-thirds the size of another is one third the
circumference of the larger.  I know the chords do not follow but I
have a means of correcting this.  This follows for any segment of the
circle.  It is simple and it checks out with any angle and by double
checking it with The Geometer's Sketchpad or using a known angle and
checking my answer by figuring the chord and comparing.

Where can I submit my effort for confirmation?  It has been very
frustrating to get anyone to take a look.  All I have heard is that
it's impossible.

Date: 04/29/2004 at 14:28:45
From: Doctor Peterson
Subject: Re: Trisect

Hi, Joe.

Sorry--you can't prove ANYTHING with Geometer's Sketchpad!  That just 
draws things and calculates angles and lengths, but only to some level 
of accuracy.  The kind of proof that is required for a geometric 
construction, under the rules of the trisection game, is a logical 
proof that it is EXACTLY correct, which the computer can't do (at 
least not that program).

The following page may help you see what I mean by saying that 
Sketchpad can't prove anything: a proof is a logical derivation, not 
just a demonstration using measurements or calculations.  What you 
have done is not a proof, judging by your description.

  Angle Trisection: Construction vs. Drawing 

But I sort of collect attempted trisections, and I'd be interested in 
seeing how accurate yours is compared to others I've seen; if it
really shows zero error on Sketchpad for all angles, then it must be
pretty interesting!  I know it WILL turn out to be wrong in some way,
since it has been proven to be impossible:

  Impossible Constructions 

But examining it can still be instructive for me, and it will help 
you understand what's going on.

Please send me instructions for the construction (no picture is 
necessary--it's better if you make sure the instructions completely 
describe what you do), and I'll have a look at it and let you know 
what I think.

- Doctor Peterson, The Math Forum 

Date: 05/24/2004 at 12:58:01
From: Joe
Subject: Trisect

Thanks for your interest.  As per your invite, here is my method.

1. Construct an angle AOB. (In the example it will be larger than 90 
degrees for simplicity.)

2. Bisect AOB with ray OC, so that COA = COB.

3. Bisect COB with ray OD, so that COD = DOB; and extend the bisector 
OD beyond O for later use.

4. On OA use the compass to mark off three equal segments OX, XY, YZ 
of (arbitrary) unit length.

5. Using O as center, construct arcs through X, Y, and Z that 
intersect OA and OB; the intersection of the arc through Y will be 
called U.

6. With the compass set at three units (the length of OZ), find the 
point V on the "back" side of line OD that is that distance from U; 
with the compass point at V, construct an arc through U.  Let W be the 
point where this arc crosses ray OC.

7. Construct segment YW.  Copy this length to the arc through Z (with 
radius 3 units), forming a chord ZT of the same length.  OT trisects
AOB, so that BOT = 1/3 AOB.

Date: 05/24/2004 at 23:39:42
From: Doctor Peterson
Subject: Re: Trisect

Hi, Joe.

I took the time tonight to try doing your construction on Sketchpad.
It looks like I have it right, because it is pretty close to a
trisection.  I wonder if you might have Sketchpad set to show angles
only to the nearest degree; I have it set for thousandths of a degree,
and find that for a 90 degree angle, the error in the construction is
about 0.034 degrees (about 2 minutes of angle), and for a 180 degree
angle, the construction is off by about 0.45 degrees (almost 30
minutes).  That's not bad as approximate trisections go, but it
certainly is not a true trisection.

I also wanted to further explain why what you have given is not what 
you need to do to solve the trisection problem.  Let's compare it to 
a bisection of an angle.  If you gave me the standard construction 
for bisecting an angle, and said that you proved it by applying it 
to many angles and seeing that it was always correct, I would have 
to tell you that you had not yet demonstrated that you really had a 
bisection, because, although the construction was correct, you had 
not PROVED it.  To prove it, you have to make a logical demonstration 
that the construction MUST ALWAYS work.  Your proof might look 
something like this:


  Given angle AOB.

  Construct arc AB with center O, with A and B on the rays of the 
  given angle.
  Construct a circle with center A passing through B, and one with
  center B passing through A.
  Construct the ray OC, where C is one of the points of intersection
  of the two circles just drawn.
  It is claimed that OC bisects angle AOB, that is, that angles AOC
  and BOC are congruent.


  Consider triangles AOC and BOC.  By the construction of points A
  and B, corresponding sides AO and BO are congruent.  By the
  construction of the circles with centers at A and B, sides AC and
  BC are congruent, since they are radii of two circles with the
  same radius.  Trivially, sides OC in both triangles are congruent.
  Therefore, by the SSS congruence theorem, triangles AOC and BOC
  are congruent, and so angles AOC and BOC are congruent.

That is what you have to present in order to submit a trisection: 
not a measurement that makes it look right, but a proof that it must 
be exactly right, no matter what angle you start with.  And that is 
what has been proved to be impossible.  So no construction you do 
will meet the requirements of the "game".

Your construction is better than many I've seen, especially for 
acute angles; but "close" doesn't count here.

- Doctor Peterson, The Math Forum 

Date: 05/25/2004 at 12:38:42
From: Joe
Subject: Thank you (Trisect)

Thanks so much for the evaluation of my trisection method.  I think I
used 1/100th accuracy for my Sketchpad drawing.  If I go backwards
from a known angle and chord and calculate it comes out as accurate as pi.

After reading your proof comment I can see that I have a lot to learn.
I was not aware that construction was not enough to satisfy the
requirement.  I hate to accept anything as impossible so I may
continue.  It keeps my mind alert.  Thanks again.

Date: 05/25/2004 at 22:21:32
From: Doctor Peterson
Subject: Re: Thank you (Trisect)

Hi, Joe.

I think I referred you earlier to this page:

  Impossible constructions 

As that explains, it has been proved to be impossible to exactly 
trisect any arbitrary angle using only compass and unmarked 
straightedge; and as I pointed out, "exact" to a mathematician means 
something entirely different from "as close as you can measure"; it 
means "provably exact".

"Impossible" is a common thing in math.  For example, it is impossible
to find two odd numbers whose sum is odd; it is easy to prove that the
sum of two odd numbers is always even, so we just accept that.  I've
never heard of people who, when told this fact, spend their lives
trying to find two odd numbers whose sum is odd; but there have been
thousands of people who have wasted a lot of time trying to find a way
to trisect an angle.  Most of them probably have no idea that a valid
trisection is not just a drawing but a proof!  They just hear that it
can't be done, don't understand how that could be proved, and decide
to show that it doesn't apply to them.  Since it's not as obvious as
adding odd numbers, they never realize how silly what they are doing is.

To keep your mind alert, I would suggest something more "constructive"
than trying to do what has been proved impossible.  Take some time to
learn the basics of geometric construction and proof, then work
through the exercises in a good book on the subject, which will ask
you to do constructions that are hard but possible.  That way you are
giving yourself a major mental challenge, but one that you can really
master with effort.  It's a lot more rewarding!

- Doctor Peterson, The Math Forum 

Date: 05/26/2004 at 10:40:36
From: Joe
Subject: Thank you (Trisect)

Thank you, Doctor, for the lesson.  I really appreciate it.  I came
close but no prize.  Thanks again.  I will put your note about proof
under the glass on my desk as a reminder to leave the trisection
impossibility alone!  Have a nice summer.  Sincerely, Joe
Associated Topics:
High School Constructions

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