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### Trisecting an Angle

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Date: 06/17/2000 at 18:47:06
From: Martin Vatis
Subject: Trisecting an Angle

that "insist" that an angle cannot be trisected. Nevertheless, I
believe I have the solution to this age-old problem. It seems to work
on any angle except a right angle. It does not involve any of the
problems I've seen described in the literature and is in strict
compliance with the use of only a straightedge and a compass. I
developed the solution when I was in high school (over 50 years ago)
and have worked on it ever since, but have not been able to actually
write a proof. So my question is, are there any accomplished
mathematicians out there who are willing to take some time to work
with me and possibly come up with the mathematical proof? Should I
contact a math professor at a local college or university? Can you
recommend one? (I live in the San Francisco area.) Thank you in
```

```
Date: 06/17/2000 at 21:59:19
From: Doctor Peterson
Subject: Re: Trisecting an Angle

Hi, Martin.

I'd suggest you start out by sending us your construction, carefully
explained of course, so we can check it out to see whether it either
bends the rules in some subtle way, or is not quite precise. We've
done this for several others who made similar claims. I can tell you
ahead of time, of course, that we will find something "wrong," because
it has been definitely proved that it can't be done within the
specified rules; but it may still be an interesting construction.

The fact that you don't yourself have a proof strongly suggests that
it will turn out only to be a very close approximation of a
trisection, as the last trisection claim I dealt with was; any true
trisection must be proved before one can really even make the claim,
since an approximation is not really a trisection at all, as far as
math is concerned, no matter how accurate it looks on paper. But we'll
be happy to look at it, and if somehow it turned out to be valid, we
would have the resources to find a proof.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 06/20/2000 at 15:55:47
From: Mark Gordon
Subject: Re: Trisecting an Angle

Dear Dr. Math:

I am not adept at drawing lines/circles in e-mail, so I will provide
my construction without benefit of a diagram. If you follow the steps
carefully, you will be able to construct the diagram without much
difficulty. Although this works on any angle except a straight angle,
I suggest for STEP 1 you use an angle of about 70 to 90 degrees and

STEP 1: Draw angle BAC.

STEP 2: From point A, draw circle to intersect line BA at point D
and line CA at point E.

STEP 3: Extend line BDA until it intersects circle at point F.

STEP 4: Extend line CEA until it intersects circle at point G.

STEP 5: Bisect arc DE at point H.

STEP 6: Draw lines GD and FH until they intersect at point I.

STEP 7: Draw lines FE and GH until they intersect at point J.

STEP 8: Draw lines AI and AJ.

STEP 9: Angle BAI = angle IAJ = angle JAC!?!

Thank you,
Martin D. Vatis
```

```
Date: 06/21/2000 at 11:51:12
From: Doctor Rick
Subject: Re: Trisecting an Angle

Hi, Mark. I'm not sure Dr. Peterson has seen your response. He
probably has some good things to share with you, but I thought I'd
show you what I've found.

Interactive math tools such as the Geometer's Sketchpad make it a lot
easier to test hypotheses like yours. I drew a figure with Sketchpad,
BAI, IAJ, and JAC, which should be equal if your algorithm works. Here
is the figure:

The angles are pretty close (within a degree) if angle BAC is less
than 90 degrees or so, but the difference increases to 10 degrees when
angle BAC approaches a straight angle. Your algorithm clearly is not
an exact trisection of an angle, but only an approximation.

Let's do some geometry on your figure. If we call angle BAC (= angle
DAE) "angle A", then:

1. Angle DAH = A/2, by construction.
2. Angle AGD = A/2, because it subtends arc DE which is angle A.
3. Angle DFI = A/4, because it subtends arc DH which is angle A/2.
4. Angle ADG = A/2, because ADG + AGD = DAE, an exterior angle of
5. Angle GDA = A/2, because triangle ADG is isosceles.
6. Angle BDI = A/2, because it and ADG are vertical angles.
7. Angle DIF = A/4, because DFI + DIF = BDI (exterior angle of
FDI).
8. Triangle FDI is isosceles, because angle DIF = DFI.
9. Segment DI = DF.

If we call the radius of the circle (DA) 1 unit, then DI = 2 units.
Now let's draw a line AK, making an angle of A/3 with AD, and
intersecting DI at K. If your algorithm trisected the angle A, then K
would be coincident with I. We can find the length of DK and see if
this is true.

Angle AKD = A/6 since angle DAK (A/3) + angle AKD = angle BDI (A/2).
We know the angles and one side of triangle ADK, so we can use
trigonometry (the Law of Sines) to find side DK, which I'll call s.
The ratio of the sine of any angle to the opposite side is the same,
so:

sin(A/3)/s = sin(A/6)/1

Therefore

s = sin(A/3)/sin(A/6)

Using the double-angle formula,

sin(A/3) = 2*sin(A/6)*cos(A/6)

s = 2*cos(A/6)

That's less than 2, unless angle A = 0 (a rather uninteresting angle).
If angle A = 90 degrees, s = 2*cos(15) = 1.93, less than a 4% error,
so your trisection is not too bad an approximation. In fact, you get A
= 30.36 degrees instead of 30 degrees.

If angle A = 180 degrees, s = 2*cos(30) = 1.732, not so close to 2.

As you can see, your construction is not what we call a trisection,
only an approximation, which explains why you couldn't prove it to be
a trisection. Approximations are possible (even closer approximations
than yours), but an exact trisection has been proved to be impossible
with compass and unmarked straightedge alone.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 06/21/2000 at 12:01:38
From: Doctor Peterson
Subject: Re: Trisecting an Angle

A careful explanation is actually better than a picture in this sort
of situation, since it avoids the temptation to make assumptions from
the picture; and you did a great job of describing it. Here's a
which allows me to try it out easily for different angles, and even
measure the resulting angles:

You'll see that the angles you produce are close to a trisection, but
are off by 1/6 and 1/3 degree when I start with a 70-degree angle. It
gets worse for larger angles. So as I expected, you have a
construction that gives what looks to the eye like a reasonably
accurate trisection, but is not exact, and therefore does not qualify
as a real trisection.

You may be interested in this similar, but considerably more complex
and accurate, construction someone else sent in last year:

http://mathforum.org/dr.math/problems/kovach.6.15.99.html

In that case, I was able to see a modification to the construction
(which unfortunately broke the rules) that made it give a perfect
trisection, and that explained why it was so close; in your case, I
haven't been able to find a similar modification, so I can't say why
it almost works. (If you think it's hard to prove something is true,
try proving it's "almost true"!) But I can use trigonometry to see
what the angle actually is. By introducing a coordinate system with
origin at A and x axis along AB, with AD = 1, the coordinates of E are
(cos(a/2), sin(a/2), where a = angle BAC. Point I can be obtained by
simply dilating by a factor of 2 around point F (since triangles FAH
and FDI are similar), so the coordinates of I turn out to be:

(2 cos(a/2) + 1, 2 sin(a/2)

and angle BAI is:

2 sin(a/2)             2 sin(a) - sqrt(2 - 2 cos(a))
arctan(--------------) = arctan(-----------------------------)
2 cos(a/2) + 1                    1 + 2 cos(a)

If you graph this, you'll find that it's close to a/3 for a wide range
of angles, but not equal to it. For a = 180 degrees, the error reaches
3.435 degrees.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 06/24/2000 at 11:53:49
From: Mark Gordon
Subject: Re: Trisecting an Angle

Dr. Math:

Thank you for all the time and effort you put into evaluating my
"trisection." It is of course disappointing that it doesn't work. I
may continue to work on a solution, but am becoming convinced that an
angle cannot be trisected. I would someday like to see the proof that
it cannot be trisected.

Again, thanks.
Martin Vatis
```
Associated Topics:
High School Constructions
High School Euclidean/Plane Geometry
High School Geometry

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