The problem of trisecting the angle is a classic problem of compass and straightedge constructions of ancient Greek mathematics. Two tools are allowed
Problem: construct an angle onethird a given arbitrary angle.
With such tools, it is generally impossible, as shown by Pierre Wantzel (1837). This requires taking a cube root, impossible with the given tools. The fact that there is no way to trisect an angle in general with just a compass and a straightedge does not mean that it is impossible to trisect all angles so. Also, it is possible to trisect any angle analytically.
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Using only an unmarked straightedge and a compass, Greek mathematicians found means to divide a line into an arbitrary set of equal segments, to draw parallel lines, to bisect angles, to construct many polygons, and to construct squares of equal or twice the area of a given polygon.
Three problems proved elusive, specifically:
The geometric problem of angle trisection can be related to algebra – specifically, the roots of a cubic polynomial – since by the tripleangle formula, cos(3θ) = 4cos^{3}(θ) − 3cos(θ).
Denote the rational numbers .
Note that a number constructible in one step from a field K is a solution of a secondorder polynomial. Note also that π / 3 radians (60 degrees, written 60°) is constructible.
However, the angle of π / 3 radians (60 degrees) cannot be trisected. Note .
If 60° could be trisected, the minimal polynomial of over would be of second order. Note the trigonometric identity cos(3α) = 4cos^{3}(α) − 3cos(α). Now let .
By the above identity, . So 4y^{3} − 3y − 1 / 2 = 0. Multiplying by two yields 8y^{3} − 6y − 1 = 0, or (2y)^{3} − 3(2y) − 1 = 0. Now substitute x = 2y, so that x^{3} − 3x − 1 = 0. Let p(x) = x^{3} − 3x − 1.
The minimal polynomial for x (hence ) is a factor of p(x). If p(x) has a rational root, by the rational root theorem, it must be 1 or −1, both clearly not roots. Therefore p(x) is irreducible over , and the minimal polynomial for is of degree 3.
So an angle of radians cannot be trisected.
However, some angles may be trisected. Given angle θ, angle 3θ trivially trisects to θ. More notably, 2π / 5 radians (72°) may be constructed, and may be trisected.^{[1]} Also there are angles, while nonconstructable, but (if somehow given) are trisectable, for example 3π / 7.^{[2]}
Again, denote the rational numbers :
Theorem: The angle θ may be trisected if and only if q(t) = 4t^{3} − 3t − cos(θ) is reducible over the field extension .
Proof. The proof would take us afield, but it may be derived from the above trigonometric identity.^{[3]}
Trisection, like many constructions impossible by ruler and compass, can easily be accomplished by the more powerful (but physically easy) operations of paper folding, or origami. Huzita's axioms (types of folding operations) can construct cubic extensions (cube roots) of given lengths, whereas rulerandcompass can construct only quadratic extensions (square roots). See mathematics of paper folding.
There are certain curves called trisectrices which, if drawn on the plane using other methods, can be used to trisect arbitrary angles.^{[4]}
Another means to trisect an arbitrary angle by a "small" step outside the Greek framework is via a ruler with two marks a set distance apart. The next construction is originally due to Archimedes, called a Neusis construction, i.e., that uses tools other than an unmarked straightedge.
This requires three facts from geometry (at right):
Look to the diagram at right; note angle a left of point B. We trisect angle a.
First, a ruler has two marks distance AB apart. Extend the lines of the angle and draw a circle of radius AB.
"Anchor" the ruler at point A, and move it until one mark is at point C, one at point D, i.e., CD = AB. A radius BC is drawn as obvious. Triangle BCD has two equal sides, thus is isosceles.
That is to say, line segments AB, BC, and CD all have equal length. Segment AC is irrelevant.
Now: Triangles ABC and BCD are isosceles, thus by Fact 3 each has two equal angles. Now redraw the diagram, and label all angles:
Hypothesis: Given AD is a straight line, and AB, BC, and CD are all equal length,
Conclusion: angle b = (1 / 3)a.
Steps:
Clearing, a − 3b = 0, or a = 3b, and the theorem is proved.
Again: this construction stepped outside the framework of allowed constructions by using a marked straightedge. There is an unavoidable element of inaccuracy in placing the straightedge.
Hutcheson published an article in Mathematics Teacher, vol. 94, No. 5, May, 2001 that used a string instead of a compass and straight edge. A string can be used as either a straight edge (by stretching it) or a compass (by fixing one point and identifying another), but can also wrap around a cylinder, the key to Hutcheson's solution.
Hutcheson constructed a cylinder from the angle to be trisected by drawing an arc across the angle, completing it as a circle, and constructing from that circle a cylinder on which a, say, equilateral triangle was inscribed (a 360degree angle divided in three). This was then "mapped" onto the angle to be trisected, with a simple proof of similar triangles.
For the detailed proof and its generalization, see the article cited: Mathematics Teacher, vol. 94, No. 5, May, 2001, pp. 400405.
There are other constructions (references).

