Download background on constructible angles

Math 4000: Modern Algebra and Geometry I
Spring 2014, Dr. Klipper
Extra Examples for Week 15
Practice and Results with Constructible Angles
In class, we defined an angle θ to be constructible when it is possible to construct two
intersecting lines that form an angle of θ. (We allow ourselves to add or subtract multiples
of 360◦ or 2π.) This isn’t the same definition as a constructible number ; indeed, π is not a
constructible number, but π radians is clearly a constructible angle, since it’s formed by a
straight line! Let’s say T is the set of constructible angles. We’ve seen that:
• T is closed under sums and differences (as well as integer multiples): if θ ∈ T , then
nθ ∈ T for n ∈ Z.
• We can bisect angles in T : if θ ∈ T , then θ/2 ∈ T .
• Trisection isn’t always possible: for instance, 60◦ ∈ T but 60◦ /3 6∈ T .
• A regular n-gon is constructible iff θ = 2π/n (or 360◦ /n) is constructible.
In this document, we’ll explore these ideas more, show a difficult example, and discover
a surprising result about which degree values are constructible!
Relationship to Sines and Cosines
Let’s say F is the set of constructible numbers. We mentioned in class that a point (x, y)
is constructible iff x, y ∈ F . When we apply that to points on the unit circle, where the
angle θ yields the point (cos(θ), sin(θ)), we find that
θ ∈ T if and only if cos(θ) and sin(θ) are constructible numbers.
In fact, you only need to check one of these: cos(θ) ∈ F automatically implies sin(θ) ∈ F
and vice versa. (Figure out why!)
This allows us to make some nifty algebraic arguments for constructibility, as opposed
to using geometric arguments!
Theorem. If θ and φ are constructible angles (i.e. θ, φ ∈ T ), then so are θ + φ and nθ for
any n ∈ N.
Alternate proof. We will just show cos(θ + φ) and cos(nθ) are in F . Using the angle-addition
formula for cosine,
cos(θ + φ) = cos(θ) cos(φ) − sin(θ) sin(φ)
By assumption, all four of these trigonometric values are in F , so their products and difference
are in F . Hence, cos(θ + φ) ∈ F . (A similar identity shows that sin(θ + φ) ∈ F as well.)
To get an identity for cos(nθ), we use DeMoivre’s Theorem and the Binomial Theorem
to state that
n X
n
n
(cos(nθ) + i sin(nθ)) = (cos(θ) + i sin(θ)) =
cos(θ)k sin(θ)n−k in−k
k
k=0
1
By taking the real part of that sum, we obtain a formula for cos(nθ) using a sum whose terms
have products of integers, cos(θ), and sin(θ). All those factors are constructible numbers, so
cos(nθ) ∈ F . (Similarly, the imaginary part shows that sin(nθ) ∈ F too.)
Proving an Angle is Non-Constructible
Questions like “is it possible to trisect an angle with compass and straightedge?” were
very old mathematical questions, many of these questions dating back to the ancient Greeks
and Euclid’s famous textbook Elements. With the advent of abstract algebra, a convenient
technique finally emerged for showing something was non-constructible. We’ve seen in class
that all elements of F have degrees over Q which are powers of 2. Thus, to show an angle θ
is non-constructible, we should show that the degree of either cos(θ) or sin(θ) is not a power
of 2.
The textbook (and class) covers one example. We’ll provide another demonstration. The
main technique, like in the proof on the previous page, is to relate cos(nθ) to cos(θ) by using
an identity obtained from DeMoivre’s Theorem.
Cutting arccos(5/7) into five parts:
Let φ = arccos(5/7) and θ = φ/5. Certainly φ is constructible, because cos(φ) = 5/7 ∈ Q,
and rational numbers are constructible. However, we will prove θ is non-constructible by
showing that cos(θ) has degree 5 over Q.
By using DeMoivre’s Theorem with n = 5, and a little algebra, you can eventually find
that
cos(5θ) = cos5 (θ) − 10 cos3 (θ) sin2 (θ) + 5 cos(θ) sin4 (θ)
By using the identity sin2 (θ) = 1 − cos2 (θ) to substitute for sin2 (θ) (and sin4 (θ) becomes
(1 − cos2 (θ))2 ), we expand and eventually obtain
cos(5θ) = 16 cos5 (θ) − 20 cos3 (θ) + 5 cos(θ)
Let’s say x = cos(θ) for clarity. Since 5θ = φ, cos(5θ) = cos(φ) = 5/7 as shown earlier.
Therefore,
5
= 16x5 − 20x3 + 5x
7
Multiply the equation by 7, and then subtract 5, to obtain
112x5 − 140x3 + 35x − 5 = 0
This shows that cos(θ) is a root of a fifth-degree polynomial. This polynomial is irreducible
thanks to Eisenstein’s Criterion with p = 5: note 112 is not divisible by 5, but the other
coefficients are divisible by 5, and 5 is not a multiple of 52 . This proves cos(θ) has degree 5.
Whole Degree Amounts
Now, we turn to an interesting question: which degree amounts are constructible? Certainly, 1◦ cannot be constructible, or else all of its multiples would be constructible, and that
would contradict the class proof that 20◦ is not constructible. In fact, no integer divisor of
20◦ could be constructible, so 2◦ and 4◦ are also non-constructible.
However, it turns out that 3◦ is constructible. Here’s an outline of how we show this:
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1. First, a regular hexagon is constructible, using six angles of 60◦ . In fact, we know 60◦
is constructible because its cosine is 1/2.
2. A regular pentagon is also constructible! One way to show this is to get an explicit
formula for sin(2π/5) or cos(2π/5). (It’s a tricky task; see Exercise 5.2 #20 and its
associated hints.) There are also some very elegant geometric constructions. I found
a very nice applet at http://www.mathopenref.com/constinpentagon.html, and it
seems that site has plenty of other geometric constructions as well. (However, the site
does not prove why their final construction really produces a regular pentagon.)
As a result, 360◦ /5 = 72◦ is constructible.
3. Since 72◦ and 60◦ are constructible, so is their difference 12◦ .
4. If we bisect 12◦ twice, we obtain 3◦ , so 3◦ is constructible.
Therefore, the smallest n ∈ N for which n◦ is constructible is n = 3. This leads to the
following theorem:
Theorem. Let n ∈ N. The angle n◦ is constructible if and only if 3 | n.
Proof. Divide n by 3 to get n = 3q + r with q, r ∈ Z and 0 ≤ r ≤ 2. Since 3◦ is constructible,
so is its integer multiple (3q)◦ . Therefore, n◦ is constructible iff r◦ is constructible. However,
0◦ is constructible, whereas we showed earlier that 1◦ and 2◦ are not.
Note, however, that this result only applies to integer degree amounts. Thus, it only helps
us analyze polygons where the number of sides is a divisor of 360. We can’t, for example,
use this result to study heptagons with 7 sides, because 360/7 isn’t a natural number.
The textbook has some results that say Gauss figured out exactly which regular n-gons
were constructible. Some of the results are surprising. For instance, it turns out that the
regular 17-gon is constructible, whereas the regular 18-gon is not!
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