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Among the many famous discoveries of modern mathematics is that an arbitrary angle cannot be trisected using only a compass and unmarked straightedge. Over the years, many people have found constructions that come close; some are only meant to be approximations, while others are purported to be exact.
Several years ago I came across a Web page, no longer online, devoted to debunking trisection claims. The site used π/3 radians (60°) as a test angle, a common choice since its trisector π/9 is known to be impossible to construct. The author reproduced several constructions, calculated the produced "trisections", and showed that none of them were exact.
One construction in particular caught my attention. It does not work for an arbitrary angle, but there does exist an angle for which it is exact. The author, however, did not know the precise value of that angle. This document will demonstrate the construction, show that it does not trisect π/3, and attempt to find the elusive angle (or possibly multiple angles) that it does trisect.
We begin with an angle ∠AOB with AO=BO. Draw the line AB to create an isosceles triangle. Trisect BO to produce BCDO and draw the line AC. Bisect ∠ABO; this bisection intersects AC at E. Draw the line OE.
Figure 1: "Trisection" of ∠AOB=π/3
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It appears ∠BOE is very close to a trisection of ∠AOB--but is it exact? If we didn't construct π/9, what angle did we construct?
Because ∠AOB=π/3, ∆AOB is actually an equilateral triangle, which aids in our analysis. Fix O at the origin of the Cartesian plane and B at (1,0); then A=(1/2,sqrt(3)/2) and C=(2/3,0). Bisecting ∠ABO=π/3 yields π/6, of which we can evaluate the tangent to produce the slope tan(-π/6)=-sqrt(3)/3.
We now have two lines, one of which is defined by two points and the other by a point and a slope. E lies at their intersection.
AC: (1/2,sqrt(3)/2), (2/3,0) -> y=-3sqrt(3)x+2sqrt(3) BE: (1,0), -sqrt(3)/3 -> y=(-sqrt(3)/3)x+sqrt(3)/3 E=(x,y)=(5/8,sqrt(3)/8)
With E known, we can find the equation for the line OE:
OE: (0,0), (5/8,sqrt(3)/8) -> y=(sqrt(3)/5)x
∠BOE is given by the arctangent of the slope of OE.
arctan(sqrt(3)/5)=0.333473...≠π/9=0.349066...
Thus ∠BOE, while not far off, is not a true trisection of ∠AOB.
If this method produced a valid trisection of every angle, then the proportion ∠BOE/∠AOB would always be 1/3. In reality, this proportion seems to decrease as ∠AOB increases. If we halve our test angle to π/6, ∠BOE more closely resembles a bisection; conversely, doubling to 2π/3 yields an angle almost a fifth as large.
Figure 2: π/6 and 2π/3 are clearly not trisected
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So if the construction doesn't give a general trisection, what does it do?
Consider an angle θ inscribed in the unit circle; we will write this angle as θ=4φ. Then A is located at (cos(4φ),sin(4φ)) and B is located at (1,0) as before. By connecting A and B, we get an isosceles triangle with a vertex angle of θ and two base angles (π-θ)/2. ∠ABO is one such base angle, so bisecting it gives:
(π-θ)/4=π/4-φ
With this information, we can find a formula for the location of E by expressing AC and BE as linear equations and finding their intersection point. We use several trigonometric identities to simplify the result.
AC: (cos(4φ),sin(4φ)), (2/3,0) -> y=(3sin(4φ)/(3cos(4φ)-2))(x-2/3) BE: (1,0), tan(φ-π/4) -> y=tan(φ-π/4)(x-1)
tan(φ-π/4)(x-1)=(3sin(4φ)/(3cos(4φ)-2))(x-2/3)
(-cos(2φ)/(1+sin(2φ)))(x-1)=(sin(4φ)/(3cos(4φ)-2))(3x-2)
(-1/(1+sin(2φ)))(x-1)=(2sin(2φ)/(3cos(4φ)-2))(3x-2)
((3cos(4φ)-2)/2sin(2φ))(x-1)=(-sin(2φ)-1)(3x-2)
((cos(4φ)-4sin^2(2φ))/2sin(2φ))(x-1)=(-sin(2φ)-1)(3x-2)
(cos(4φ)-4sin^2(2φ))x-(cos(4φ)-4sin^2(2φ))=(-6sin^2(2φ)-6sin(2φ))x-(-4sin^2(2φ)-4sin(2φ))
x=(4sin(2φ)+cos(4φ))/(cos(4φ)+2sin^2(2φ)+6sin(2φ))
=(4sin(2φ)+cos(4φ))/(6sin(2φ)+1)
y=(-cos(2φ)/(1+sin(2φ))((4sin(2φ)+cos(4φ))/(6sin(2φ)+1))
=(-cos(2φ)/(1+sin(2φ))((cos(4φ)-2sin(2φ)-1)/(6sin(2φ)+1))
=(cos(2φ)/(1+sin(2φ))((2sin(2φ)+2sin^2(2φ))/(6sin(2φ)+1))
=(2sin(2φ)cos(2φ))/(6sin(2φ)+1)
=sin(4φ)/(6sin(2φ)+1)
E is thus located at:
(x,y)=((4sin(2φ)+cos(4φ))/(6sin(2φ)+1),sin(4φ)/(6sin(2φ)+1))
From here the equation for OE is easy:
y=(sin(4φ)/(4sin(2φ)+cos(4φ)))x
∠BOE is the arctangent of the slope of OE.
∠BOE=arctan(sin(4φ)/(4sin(2φ)+cos(4φ)))
Figure 2 led us to conjecture that the proportion ∠BOE/∠AOB and ∠AOB are inversely related. Now that we can express ∠BOE in exact terms, we see our conjecture is true: the derivative of the proportion is strictly negative in the open interval (0,π), so the proportion is monotonically decreasing in the same interval. Here ∠BOE/∠AOB is continuous and has range (0,1), so by the intermediate value theorem of calculus, there must be some angle for which:
∠BOE/∠AOB=1/3
In the introduction we mentioned that several angles might be trisected by the construction. Since the proportion decreases monotonically, we now know this is false: there is exactly one angle that is trisected. Indeed, given any real number 0<r<1, this construction produces exactly one angle θ_r such that ∠BOE/∠AOB=r.
Figure 3: ∠BOE/∠AOB decreases monotonically from 0 to π
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We now know there exists one angle that can be trisected with this construction, but how can we find it?
We'll continue to analyze our generic angle, which we'll limit to the range 0<θ<π. When we've found an exact trisection, ∠BOE will be one third of ∠AOB:
θ=3arctan(sin(4φ)/(4sin(2φ)+cos(4φ)))
We'll define another quantity 2φ=3ψ and substitute. (Note that θ=6ψ.)
tan(2ψ)=sin(6ψ)/(4sin(3ψ)+cos(6ψ))
4sin(3ψ)sin(2ψ)+cos(6ψ)sin(2ψ)=sin(6ψ)cos(2ψ)
2cos(5ψ)+sin(4ψ)-2cos(ψ)=0
This is a trigonometric polynomial of degree 5. Trigonometric polynomials have infinitely many complex zeroes because they are periodic functions, but we can consider their principal zeroes in the complex strip -π<Re(c)≤π. Inside this strip we can transform a trigonometric polynomial into an algebraic polynomial with analogous roots. This is done by expanding each term using the following substitutions:
t=tan(ψ/2)
cos(ψ)=(1-t^2)/(1+t^2)
sin(ψ)=2t/(1+t^2)
cos(nψ)=sum(((1-t^2)/(1+t^2))^k(2t/(1+t^2))^(n-k)cos((n-k)π/2),k,0,n)
sin(nψ)=sum(((1-t^2)/(1+t^2))^k(2t/(1+t^2))^(n-k)sin((n-k)π/2),k,0,n)
After clearing fractions and simplifying, we are left with the following sextic equation in t:
0=p(t)=t^6-12t^5-5t^4+40t^3-5t^2-12t+1
p(t) is irreducible over the rationals but is solvable in radicals, which is a property not all sextic polynomials have. The field of mathematics that proves this property is Galois theory--the same area of study that showed the impossibility of general angle trisection to begin with. The zeroes of p(t), as found by the GAP software's "Radiroot" package, are as follows, where the radicals are evaluated at each possible complex value.
t=2+(1/3)(ζ_3-(ζ_3)^2)ω_2-(9/98)(ω_2)^2+(1/294)(ζ_3-(ζ_3)^2)ω_1(ω_2)^2-ω_3 ζ_3=1^(1/3) ω_1=sqrt(443) ω_2=(18(ζ_3-(ζ_3)^2)-2ω_1)^(1/3) ω_3=sqrt(37/3+(37/21)(ζ_3-(ζ_3)^2)ω_2+(1/21)ω_1(ω_2)-(103/147)(ω_2)^2+(2/147)(ζ_3-(ζ_3)^2)ω_1(ω_2)^2)
As cube roots are involved in the expressions of these zeroes, none of their analogues in θ are constructible.
We are only considering the interval 0<θ<π, which limits our interval for t to 0<t<tan(π/6)=sqrt(3)/3. p(t) has exactly one zero t_0 in this interval, given by the following values:
ζ_3=(-1-sqrt(-3))/2 ω_1 positive ω_2=((-1+sqrt(-3))/2)(18(ζ_3-(ζ_3)^2)-2ω_1)^(1/3) ω_3 positive
Substituting into our expression for t yields:
t_0=2+((3+sqrt(-3))/6)(-2sqrt(443)-18sqrt(-3))^(1/3)+((27-3sqrt(443)+(27+sqrt(443))sqrt(-3))/588)(-2sqrt(443)-18sqrt(-3))^(2/3)-sqrt(5439/121+((111-sqrt(443)+(37+sqrt(443))sqrt(-3))/42)(-2sqrt(443)-18sqrt(-3))+((103-6sqrt(443)+(103+2sqrt(443))sqrt(-3))/294)(-2sqrt(443)-18sqrt(-3))^(2/3)) ≈0.08234631084075932307...
Taking the arctangent of t_0 gives us ψ/2, from which we recover θ:
θ=6ψ=12arctan(t_0)
≈0.98593124023785294531...
≈56.48969895572781875522...°
We finish with some trivia. The proportion ∠BOE/∠AOB monotonically decreases over (0,π), but ∠BOE does not--it increases from 0 to a local maximum, then decreases to 0 again. The maximum occurs at 4arctan(t_1), where t_1 is one of two real zeroes of another sextic polynomial q(t):
t^6+3t^4-32t^3+3t^2+1
Like p(t), q(t) is solvable in radicals but its zeroes are not constructible. The exact form for t_1 is:
t_1=((3+sqrt(-3))/6)(-12sqrt(-3))^(1/3)-sqrt(((1+sqrt(-3))/12)(-12sqrt(-3))^(2/3)-1) ≈0.35458229002881921833...
The maximum value of ∠BOE thus occurs at:
θ≈1.36300476336282708191... ≈78.09442039691748465276...
Upon fixing ∠BOE=∠AOB/3, we can find loci of various intersection points between AC, BE and OE. Below are plotted three loci: the intersection of AC and OE as a dotted (red) curve, the intersection of BE and AC as a solid (green) curve, and the intersection of OE and BE as a dash-dotted (blue) curve. The black point is the location of E that trisects ∠AOB.
Figure 4: Loci of trisector intersection points
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Coincidentally, the angle for which ∠BOE is maximized is very close to the angle quadrisected by the construction--they are less than one degree away from each other. The quadrisected angle is given by:
θ=8arctan((-4-sqrt(19)+2sqrt(11+2sqrt(19)))/3) ≈1.37832202979787466497... ≈78.97203511732310743831...°
Unlike the previous angles we've found, this angle is constructible. A crude compass-and-straightedge method, created using the GeoGebra software, is exhibited below.
Figure 5: The black angle is quadrisected by the construction
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The question of finding a trisectable angle turned out to be much more complicated than I expected at the outset. The construction at the heart of this analysis is almost trivial to create, but the angle trisected by it is not trivial in any way. To find it was not a simple process, either: several trigonometric manipulations were required, culminating in finding the zeroes of a sextic equation. Not all sextics are solvable in radicals, so we might have had to settle for a numerical approximation. Fortunately for us, the answer was expressible in radicals, although square roots of negative numbers are unavoidable in its exact form.
Mathematics tends to evolve under an impetus similar to this problem. Simple questions can, and often do, lead to highly intricate or convoluted solutions. Sometimes an entirely new mathematical field needs to be invented just to solve a problem. Calculus was invented to study celestial mechanics; probability was invented to study gambling; the modularity theorem was proved in a special case to study Fermat's Last Theorem. These things were all thought to be straightforward phenomena when the first questions were asked about them, but a definitive characterization led to insights that were previously unimaginable.
To me, this is the joy of mathematics. Every problem is a chance to discover depths of theory I've never seen before. An expanded horizon of knowledge, a new understanding of the mechanics of the universe, is always just around the corner.
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[Last updated: 2022-01-14]
