Angle trisection is the construction of anangle equal to one third of a given arbitrary angle, using only two tools: an unmarkedstraightedge and acompass. It is a classical problem ofstraightedge and compass construction of ancientGreek mathematics.
In 1837,Pierre Wantzel proved that the problem, as stated, isimpossible to solve for arbitrary angles. However, some special angles can be trisected: for example, it is trivial to trisect aright angle.
It is possible to trisect an arbitrary angle by using tools other than straightedge and compass. For example,neusis construction, also known to ancient Greeks, involves simultaneous sliding and rotation of a marked straightedge, which cannot be achieved with the original tools. Other techniques were developed by mathematicians over the centuries.
Because it is defined in simple terms, but complex to prove unsolvable, the problem of angle trisection is a frequent subject ofpseudomathematical attempts at solution by naive enthusiasts. These "solutions" often involve mistaken interpretations of the rules, or are simply incorrect.[1]
Using only an unmarkedstraightedge and a compass,Greek mathematicians found means to divide aline into an arbitrary set of equal segments, to drawparallel lines, tobisect angles, to construct manypolygons, and to constructsquares of equal or twice the area of a given polygon.
Three problems proved elusive, specifically, trisecting the angle,doubling the cube, andsquaring the circle. The problem of angle trisection reads:
Construct anangle equal to one-third of a given arbitrary angle (or divide it into three equal angles), using only two tools:
Pierre Wantzel published a proof of the impossibility of classically trisecting an arbitrary angle in 1837.[2] Wantzel's proof, restated in modern terminology, uses the concept offield extensions, a topic now typically combined withGalois theory. However, Wantzel published these results earlier thanÉvariste Galois (whose work, written in 1830, was published only in 1846) and did not use the concepts introduced by Galois.[3]
The problem of constructing an angle of a given measureθ is equivalent to constructing two segments such that the ratio of their length iscos θ. From a solution to one of these two problems, one may pass to a solution of the other by a compass and straightedge construction. Thetriple-angle formula gives an expression relating the cosines of the original angle and its trisection:cos θ = 4 cos3θ/3 − 3 cosθ/3.
It follows that, given a segment that is defined to have unit length, the problem of angle trisection is equivalent to constructing a segment whose length is the root of acubic polynomial. This equivalence reduces the original geometric problem to a purely algebraic problem.
Every rational number is constructible. Everyirrational number that isconstructible in a single step from some given numbers is a root of apolynomial of degree 2 with coefficients in thefield generated by these numbers. Therefore, any number that is constructible by a sequence of steps is a root of aminimal polynomial whose degree is apower of two. The angleπ/3radians (60degrees, written 60°) isconstructible. The argument below shows that it is impossible to construct a 20° angle. This implies that a 60° angle cannot be trisected, and thus that an arbitrary angle cannot be trisected.
Denote the set ofrational numbers byQ. If 60° could be trisected, the degree of a minimal polynomial ofcos 20° overQ would be a power of two. Now letx = cos 20°. Note thatcos 60° =cosπ/3 =1/2. Then by the triple-angle formula,cosπ/3 = 4x3 − 3x and so4x3 − 3x =1/2. Thus8x3 − 6x − 1 = 0. Definep(t) to be the polynomialp(t) = 8t3 − 6t − 1.
Sincex = cos 20° is a root ofp(t), the minimal polynomial forcos 20° is a factor ofp(t). Becausep(t) has degree 3, if it is reducible over byQ then it has arational root. By therational root theorem, this root must be±1, ±1/2, ±1/4 or±1/8, but none of these is a root. Therefore,p(t) isirreducible over byQ, and the minimal polynomial forcos 20° is of degree 3.
So an angle of measure60° cannot be trisected.
However, some angles can be trisected. For example, for anyconstructible angleθ, an angle of measure3θ can be trivially trisected by ignoring the given angle and directly constructing an angle of measureθ. There are angles that are not constructible but are trisectible (despite the one-third angle itself being non-constructible). For example,3π/7 is such an angle: five angles of measure3π/7 combine to make an angle of measure15π/7, which is a full circle plus the desiredπ/7.
For apositive integerN, an angle of measure2π/N istrisectible if and only if3 does not divideN.[4][5] In contrast,2π/N isconstructible if and only ifN is a power of2 or the product of a power of2 with the product of one or more distinctFermat primes.
Again, denote the set ofrational numbers byQ.
Theorem: An angle of measureθ may be trisectedif and only ifq(t) = 4t3 − 3t − cos(θ) is reducible over thefield extensionQ(cos(θ)).
Theproof is a relatively straightforward generalization of the proof given above that a60° angle is not trisectible.[6]
For any nonzero integerN, an angle of measure2π⁄N radians can be divided inton equal parts with straightedge and compass if and only ifn is either a power of2 or is a power of2 multiplied by the product of one or more distinct Fermat primes, none of which dividesN. In the case of trisection (n = 3, which is a Fermat prime), this condition becomes the above-mentioned requirement thatN not be divisible by3.[5]
The general problem of angle trisection is solvable by using additional tools, and thus going outside of the original Greek framework of compass and straightedge.
Many incorrect methods of trisecting the general angle have been proposed. Some of these methods provide reasonable approximations; others (some of which are mentioned below) involve tools not permitted in the classical problem. The mathematicianUnderwood Dudley has detailed some of these failed attempts in his bookThe Trisectors.[1]
Trisection can be approximated by repetition of the compass and straightedge method for bisecting an angle. The geometric series1/3 =1/4 +1/16 +1/64 +1/256 + ⋯ or1/3 =1/2 −1/4 +1/8 −1/16 + ⋯ can be used as a basis for the bisections. An approximation to any degree of accuracy can be obtained in a finite number of steps.[7]
Trisection, like many constructions impossible by ruler and compass, can easily be accomplished by the operations of paper folding, ororigami.Huzita's axioms (types of folding operations) can construct cubic extensions (cube roots) of given lengths, whereas ruler-and-compass can construct only quadratic extensions (square roots).
There are a number of simplelinkages which can be used to make an instrument to trisect angles including Kempe's Trisector and Sylvester's Link Fan or Isoklinostat.[8]
In 1932,Ludwig Bieberbach published inJournal für die reine und angewandte Mathematik his workZur Lehre von den kubischen Konstruktionen.[9] He states therein (free translation):
The construction begins with drawing acircle passing through the vertexP of the angle to be trisected, centered atA on an edge of this angle, and havingB as its second intersection with the edge. A circle centered atP and of the same radius intersects the line supporting the edge inA andO.
Now theright triangular ruler is placed on the drawing in the following manner: oneleg of its right angle passes throughO; the vertex of its right angle is placed at a pointS on the linePC in such a way that the second leg of the ruler is tangent atE to the circle centered atA. It follows that the original angle is trisected by the linePE, and the linePD perpendicular toSE and passing throughP. This line can be drawn either by using again the right triangular ruler, or by using a traditionalstraightedge and compass construction. With a similar construction, one can improve the location ofE, by using that it is the intersection of the lineSE and its perpendicular passing throughA.
Proof: One has to prove the angle equalities and The three linesOS,PD, andAE are parallel. As theline segmentsOP andPA are equal, these three parallel lines delimit two equal segments on every other secant line, and in particular on their common perpendicularSE. ThusSD' =D'E, whereD' is the intersection of the linesPD andSE. It follows that theright trianglesPD'S andPD'E are congruent, and thus that the first desired equality. On the other hand, the trianglePAE isisosceles, since allradiuses of a circle are equal; this implies that One has also since these two angles arealternate angles of a transversal to two parallel lines. This proves the second desired equality, and thus the correctness of the construction.
There are certain curves calledtrisectrices which, if drawn on the plane using other methods, can be used to trisect arbitrary angles.[10] Examples include thetrisectrix of Colin Maclaurin, given inCartesian coordinates by theimplicit equation
and theArchimedean spiral. The spiral can, in fact, be used to divide an angle intoany number of equal parts.Archimedes described how to trisect an angle using the Archimedean spiral inOn Spirals around 225 BC.
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 toArchimedes, called aNeusis construction, i.e., that uses tools other than anun-marked straightedge. The diagrams we use show this construction for an acute angle, but it indeed works for any angle up to 180 degrees.
This requires three facts from geometry (at right):
Letl be the horizontal line in the adjacent diagram. Anglea (left of pointB) is the subject of trisection. First, a pointA is drawn at an angle'sray, one unit apart fromB. A circle ofradiusAB is drawn. Then, the markedness of the ruler comes into play: one mark of the ruler is placed atA and the other atB. While keeping the ruler (but not the mark) touchingA, the ruler is slid and rotated until one mark is on the circle and the other is on the linel. The mark on the circle is labeledC and the mark on the line is labeledD. This ensures thatCD =AB. A radiusBC is drawn to make it obvious that line segmentsAB,BC, andCD all have equal length. Now, trianglesABC andBCD areisosceles, thus (by Fact 3 above) each has two equal angles.
Hypothesis: GivenAD is a straight line, andAB,BC, andCD all have equal length,
Conclusion: angleb =a/3.
and thetheorem is proved.
Again, this construction stepped outside theframework ofallowed constructions by using a marked straightedge.
Thomas Hutcheson published an article in theMathematics Teacher[11] 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 360-degree angle divided in three). This was then "mapped" onto the angle to be trisected, with a simple proof of similar triangles.
A "tomahawk" is a geometric shape consisting of a semicircle and two orthogonal line segments, such that the length of the shorter segment is equal to the circle radius. Trisection is executed by leaning the end of the tomahawk's shorter segment on one ray, the circle's edge on the other, so that the "handle" (longer segment) crosses the angle's vertex; the trisection line runs between the vertex and the center of the semicircle.
While a tomahawk is constructible with compass and straightedge, it is not generally possible to construct a tomahawk in any desired position. Thus, the above construction does not contradict the nontrisectibility of angles with ruler and compass alone.
As a tomahawk can be used as aset square, it can be also used for trisection angles by the method described in§ With a right triangular ruler.
The tomahawk produces the same geometric effect as the paper-folding method: the distance between circle center and the tip of the shorter segment is twice the distance of the radius, which is guaranteed to contact the angle. It is also equivalent to the use of an architects L-Ruler (Carpenter's Square).
An angle can be trisected with a device that is essentially a four-pronged version of a compass, with linkages between the prongs designed to keep the three angles between adjacent prongs equal.[12]
Acubic equation with real coefficients can be solved geometrically with compass, straightedge, and an angle trisector if and only if it has threerealroots.[13]: Thm. 1
Aregular polygon withn sides can be constructed with ruler, compass, and angle trisector if and only if wherer, s, k ≥ 0 and where thepi are distinct primes greater than 3 of the form (i.e.Pierpont primes greater than 3).[13]: Thm. 2
See also Feedback on this article in vol. 93, March 2009, p. 156.
