Angle trisection

the construction of an angle equal to one third a given angle

Angles may be trisected via a neusis construction using tools beyond an unmarked straightedge and a compass. The example shows trisection of any angle

θ> 3π / 4

by a ruler with length equal to the radius of the circle, giving trisected angle

φ= θ / 3

.

Angle trisection is a classical problem of compass and straightedge constructions of ancient Greek mathematics. It concerns construction of an angle equal to one third of a given arbitrary angle, using only two tools: an unmarked straightedge and a compass.

The problem as stated is impossible to solve for arbitrary angles, as proved by Pierre Wantzel in 1837. However, although there is no way to trisect an angle in general with just a compass and a straightedge, some special angles can be trisected. For example, it is relatively straightforward to trisect a right angle (that is, to construct an angle of measure 30 degrees).

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 of pseudomathematical attempts at solution by naive enthusiasts. These "solutions" often involve mistaken interpretations of the rules, or are simply incorrect.^[1]

4 Other methods
- 4.1 Approximation by successive bisections
- 4.2 Using origami
- 4.3 Using a linkage
- 4.4 With a right triangle ruler
- 4.5 With an auxiliary curve
  - 4.5.1 Application example
  - 4.5.2 Angle trisection
- 4.6 With a marked ruler
- 4.7 With a string
- 4.8 With a "tomahawk"
- 4.9 With interconnected compasses

5 Uses of angle trisection

6 Generalization

7 See also

8 References

9 Further reading

10 External links
- 10.1 Other means of trisection

Background and problem statement[edit]

Bisection of arbitrary angles has long been solved.

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, trisecting the angle, doubling the cube, and squaring the circle. The problem of angle trisection reads:

Construct an angle equal to one-third of a given arbitrary angle (or divide it into three equal angles), using only two tools:

an unmarked straightedge, and

a compass.

Proof of impossibility[edit]

Rulers. The displayed ones are marked — an ideal straightedge is un-marked

compasses

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 abstract algebra of field extensions, a topic now typically combined with Galois theory. However Wantzel published these results earlier than Galois (whose work was published in 1846) and did not use the connection between field extensions and groups that is the subject of Galois theory itself.^[3]

The problem of constructing an angle of a given measure $θ$ is equivalent to constructing two segments such that the ratio of their length is $cos θ$ . From a solution to one of these two problems, one may pass to a solution of the other by a compass and straightedge construction. The triple-angle formula gives an expression relating the cosines of the original angle and its trisection: $cos θ$ = $4cos 3 (θ /3) - 3cos(θ /3)$ . 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 a cubic polynomial. This equivalence reduces the original geometric problem to a purely algebraic problem.

Every rational number is constructible. Every irrational number that is constructible in a single step from some given numbers is a root of a polynomial of degree 2 with coefficients in the field generated by these numbers. Therefore, any number that is constructible by a sequence of steps is a root of a minimal polynomial whose degree is a power of two. Note also that $π/3$ radians (60 degrees, written 60°) is constructible. 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 of rational numbers by $Q$ . If 60° could be trisected, the degree of a minimal polynomial of $cos(20°)$ over $Q$ would be a power of two. Now let $x = cos(20°)$ . Note that $cos(60°)$ = $cos(π/3)$ = $1/2$ . Then by the triple-angle formula, $cos(π/3) = 4 x 3 - 3 x$ and so $4 x 3 - 3 x = 1/2$ . Thus $8 x 3 - 6 x - 1 = 0$ . Define $p (t)$ to be the polynomial $p (t) = 8 t 3 - 6 t - 1$ .

Since $x = cos(20°)$ is a root of $p (t)$ , the minimal polynomial for $cos(20°)$ is a factor of $p (t)$ . Because $p (t)$ has degree 3, if it is reducible over by $Q$ then it has a rational root. By the rational root theorem, this root must be $\pm1, \pm1/2, \pm1/4$ or $\pm1/8$ , but none of these is a root. Therefore, $p (t)$ is irreducible over by $Q$ , and the minimal polynomial for $cos(20°)$ is of degree $3$ .

So an angle of measure $60°$ cannot be trisected.

Angles which can be trisected[edit]

However, some angles can be trisected. For example, for any constructible angle $θ$ , an angle of measure $3 θ$ can be trivially trisected by ignoring the given angle and directly constructing an angle of measure $θ$ . There are angles that are not constructible but are trisectible (despite the one-third angle itself being non-constructible). For example, $ 3 π ⁄ 7$ is such an angle: five angles of measure $ 3 π ⁄ 7$ combine to make an angle of measure $ 15 π ⁄ 7$ , which is a full circle plus the desired $ π ⁄ 7$ .

For a positive integer $N$ , an angle of measure $ 2 π ⁄ N$ is trisectible if and only if $3$ does not divide $N$ .^[4]^[5] In contrast, $ 2 π ⁄ N$ is constructible if and only if $N$ is a power of $2$ or the product of a power of $2$ with the product of one or more distinct Fermat primes.

Algebraic characterization[edit]

Again, denote the set of rational numbers by $Q$ .

Theorem: An angle of measure $θ$ may be trisected if and only if $q (t) = 4 t 3 - 3 t - cos(θ)$ is reducible over the field extension $Q (cos(θ))$ .

The proof is a relatively straightforward generalization of the proof given above that a $60°$ angle is not trisectible.^[6]

Other methods[edit]

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 mathematician Underwood Dudley has detailed some of these failed attempts in his book The Trisectors.^[1]

Approximation by successive bisections[edit]

Trisection can be approximated by repetition of the compass and straightedge method for bisecting an angle. The geometric series 1/3 = 1/4 + 1/16 + 1/64 + 1/256 + ⋯ or 1/3 = 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]

Using origami[edit]

Trisection, like many constructions impossible by ruler and compass, can easily be accomplished by the more powerful operations of paper folding, or origami. 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).

Using a linkage[edit]

There are a number of simple linkages which can be used to make an instrument to trisect angles including Kempe's Trisector and Sylvester's Link Fan or Isoklinostat.^[8]

Sylvester's Link Fan

With a right triangle ruler[edit]

Trisection of the angle by means of the Rechtwinkelhaken according to Ludwig Bieberbach, with continuation of the construction, animation 1 min 35 s, of which break at the end 30 s.

In 1932, Ludwig Bieberbach published in Journal für die reine und angewandte Mathematik his work Zur Lehre von den kubischen Konstruktionen.^[9] He states therein (free translation):

"As is known ... every cubic construction can be traced back to the trisection of the angle and to the multiplication of the cube, that is, the extraction of the third root. I need only to show how these two classical tasks can be solved by means of the right angle hook."

The following description of the adjacent construction (animation) contains their continuation up to the complete angle trisection.

It begins with the first unit circle around its center $A$ , the first angle limb $displaystyle overline BP$ , and the second unit circle around $P$ following it. Now the diameter $displaystyle overline BP$ from $P$ is extended to the circle line of this unit circle, the intersection point $O$ being created. Following the circle arc around $P$ with the radius $displaystyle overline BP$ and the drawing of the second angle limb from the angle $delta$ , the point $C$ results. Now the so-called additional construction mean is used, in the illustrated example it is the Geodreieck. This geometry triangle, as it is also called, is now placed on the drawing in the following manner: The vertex of the right angle determines the point $S$ on the angle leg $displaystyle overline PC$ , a cathetus of the triangle passes through the point $O$ and the other affects the unit circle $A$ .
After connecting the point $O$ to $S$ and drawing the tangent from $S$ to the unit circle around $A$ , the above-mentioned right angle hook respectively Rechtwinkelhaken is shown. The angle enclosed by the segments $displaystyle overline OS$ and $displaystyle overline PS$ is thus exactly $displaystyle frac delta 3$ . It goes on with the parallel to $displaystyle overline OS$ from $P$ , the alternate angle $displaystyle frac delta 3$ and the point $D$ are being created. A further parallel to $displaystyle overline OS$ from $A$ determines the point of contact $E$ from the tangent with the unit circle about $A$ .Finally, draw a straight line from $P$ through $E$ until it intersects the unit circle in $F$ . Thus the angle $delta$ has exactly three parts.

With an auxiliary curve[edit]

There are certain curves called trisectrices which, if drawn on the plane using other methods, can be used to trisect arbitrary angles.^[10]

Application example[edit]

Angle trisection (exact) with Maclaurin's trisectrix curve as additional auxiliary mean

The known Trisectrix of Colin Maclaurin from the year 1742 is used.

In Cartesian coordinates this curve is described with the equation

displaystyle 2x(x^2+y^2)=a(3x^2-y^2),

or, in implicit form,

displaystyle 2x^3+2xy^2-a3x^2+ay^2=0.

Angle trisection[edit]

First, the diameter $overline AB$ with its center $M$ is determined. This is followed by the semicircle $displaystyle MBA$ with the subsequent generation of the trisectrix as the implicit curve.^[11] Thus, the basic construction for the angle trisection of angles $displaystyle 0^circ <alpha leq 180^circ$ is completed. Now, the second angle limb $displaystyle overline MC$ is drawn in such a way that it encloses with the first angle limb $displaystyle overline MB$ the angle $alpha$ to be divided. The angle limb $displaystyle overline MC$ intersects the trisectrix in $D$ . Next, a straight line from $A$ is drawn through $D$ to the semicircle, resulting in the intersection $E$ . The angle $beta$ generated by $displaystyle overline BA$ and $displaystyle overline AE$ is the angle $displaystyle frac 13alpha$ sought.

With a marked ruler[edit]

Basic triangle's angle properties

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 un-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):

Any full set of angles on a straight line add to 180°,

The sum of angles of any triangle is 180°, and,

Any two equal sides of an isosceles triangle will meet the third in the same angle.

Trisection of the angle using marked ruler

Let $l$ be the horizontal line in the adjacent diagram. Angle $a$ (left of point $B$ ) is the subject of trisection. First, a point $A$ is drawn at an angle's ray, one unit apart from $B$ . A circle of radius $AB$ is drawn. Then, the markedness of the ruler comes into play: one mark of the ruler is placed at $A$ and the other at $B$ . While keeping the ruler (but not the mark) touching $A$ , the ruler is slid and rotated until one mark is on the circle and the other is on the line $l$ . The mark on the circle is labeled $C$ and the mark on the line is labeled $D$ . This ensures that $CD = AB$ . A radius $BC$ is drawn to make it obvious that line segments $AB$ , $BC$ , and $CD$ all have equal length. Now, triangles $ABC$ and $BCD$ are isosceles, thus (by Fact 3 above) each has two equal angles.

Hypothesis: Given $AD$ is a straight line, and $AB$ , $BC$ , and $CD$ all have equal length,

Conclusion: angle $b = a / 3$ .

Proof:

From Fact 1) above, $e+c=180$ °.

Looking at triangle BCD, from Fact 2) $e+2b=180$ °.

From the last two equations, $c=2b$ .

From Fact 2), $d+2c=180$ °, thus $d=180$ ° $-2c$ , so from last, $d=180$ ° $-4b$ .

From Fact 1) above, $a+d+b=180$ °, thus $a+(180$ ° $-4b)+b=180$ °.

Clearing, $a - 3 b = 0$ , or $a = 3 b$ , and the theorem is proved.

Again, this construction stepped outside the framework of allowed constructions by using a marked straightedge.

With a string[edit]

Thomas Hutcheson published an article in the Mathematics Teacher^[12] 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.

With a "tomahawk"[edit]

A tomahawk trisecting an angle. The handle forms one trisector and the blue line shown forms the other.

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.

Note that 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.

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).

With interconnected compasses[edit]

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.^[13]

Uses of angle trisection[edit]

An animation of a neusis construction of a heptagon with radius of circumcircle

displaystyle overline OA=6

, based on Andrew M. Gleason, using angle trisection by means of the tomahawk^[14]^{:p. 186}

A cubic equation with real coefficients can be solved geometrically with compass, straightedge, and an angle trisector if and only if it has three real roots.^[14]^{:Thm. 1}

A regular polygon with n sides can be constructed with ruler, compass, and angle trisector if and only if $displaystyle n=2^r3^sp_1p_2cdots p_k,$ where r, s, k ≥ 0 and where the p_i are distinct primes greater than 3 of the form $displaystyle 2^t3^u+1$ (i.e. Pierpont primes greater than 3).^[14]^{:Thm. 2}

Generalization[edit]

For any nonzero integer $N$ , an angle of measure $ 2 π ⁄ N$ radians can be divided into $n$ equal parts with straightedge and compass if and only if $n$ is either a power of $2$ or is a power of $2$ multiplied by the product of one or more distinct Fermat primes, none of which divides $N$ . In the case of trisection ( $n = 3$ , which is a Fermat prime), this condition becomes the above-mentioned requirement that $N$ not be divisible by $3$ .^[5]

References[edit]

^ ^a^b Dudley, Underwood (1994), The trisectors, Mathematical Association of America, ISBN 0-88385-514-3

^ Wantzel, P M L (1837). "Recherches sur les moyens de reconnaître si un problème de Géométrie peut se résoudre avec la règle et le compas" (PDF). Journal de Mathématiques Pures et Appliquées. 1. 2: 366–372. Retrieved 3 March 2014.

^ For the historical basis of Wantzel's proof in the earlier work of Ruffini and Abel, and its timing vis-a-vis Galois, see Smorynski, Craig (2007), History of Mathematics: A Supplement, Springer, p. 130, ISBN 9780387754802 .

^ MacHale, Desmond. "Constructing integer angles", Mathematical Gazette 66, June 1982, 144–145.

^ ^a^b McLean, K. Robin, "Trisecting angles with ruler and compasses", Mathematical Gazette 92, July 2008, 320–323. See also Feedback on this article in vol. 93, March 2009, p. 156.

^ Stewart, Ian (1989). Galois Theory. Chapman and Hall Mathematics. pp. g. 58. ISBN 0-412-34550-1.

^ Jim Loy (2003) [1997]. "Trisection of an Angle". Archived from the original on February 25, 2012. Retrieved 30 March 2012.

^ Yates, Robert C (1942). The Trisection Problem (PDF). The National Council of Teachers of Mathematics. pp. 39–42.

^ Ludwig Bieberbach (1932) Zur Lehre von den kubischen Konstruktionen, Journal für die reine und angewandte Mathematik, H. Hasse und L. Schlesinger, Band 167 Berlin, p. 142–146 online-copie (GDZ). Retrieved on June 2, 2017.

^ Jim Loy "Archived copy". Archived from the original on November 4, 2013. Retrieved 2013-11-04.

^ Dörte Haftendorn: "Kurven erkunden und verstehen" (German) Kapitel 3.3: S. 62, ISBN 9783658147488, Spektrum, Akademischer Verlag, Springer, 2016, "Trisektrix Konstruktion" (German). Retrieved on January 12, 2017.

^ Hutcheson, Thomas W. (May 2001). "Dividing Any Angle into Any Number of Equal Parts". Mathematics Teacher. 94 (5): 400–405.

^ Isaac, Rufus, "Two mathematical papers without words", Mathematics Magazine 48, 1975, p. 198. Reprinted in Mathematics Magazine 78, April 2005, p. 111.

^ ^a^b^c Gleason, Andrew Mattei (March 1988). "Angle trisection, the heptagon, and the triskaidecagon" (PDF). The American Mathematical Monthly. 95 (3): 185–194. doi:10.2307/2323624. Archived from the original (PDF) on November 5, 2014.

External links[edit]

MathWorld site

Geometric problems of antiquity, including angle trisection

Some history

One link of marked ruler construction

Another, mentioning Archimedes

A long article with many approximations & means going outside the Greek framework

Geometry site

Other means of trisection[edit]

Approximate angle trisection as an animation, max. error of the angle ≈ ±4E-8°

Trisecting via (Archived 2009-10-25) the limacon of Pascal; see also Trisectrix

Trisecting via an Archimedean Spiral

Trisecting via the Conchoid of Nicomedes

sciencenews.org site on using origami

Hyperbolic trisection and the spectrum of regular polygons

[trisectors-1] Dudley, Underwood (1994), The trisectors, Mathematical Association of America, ISBN 0-88385-514-3

[2] Wantzel, P M L (1837). "Recherches sur les moyens de reconnaître si un problème de Géométrie peut se résoudre avec la règle et le compas" (PDF). Journal de Mathématiques Pures et Appliquées. 1. 2: 366–372. Retrieved 3 March 2014.

[3] For the historical basis of Wantzel's proof in the earlier work of Ruffini and Abel, and its timing vis-a-vis Galois, see Smorynski, Craig (2007), History of Mathematics: A Supplement, Springer, p. 130, ISBN 9780387754802 .

[4] MacHale, Desmond. "Constructing integer angles", Mathematical Gazette 66, June 1982, 144–145.

[McLean-5] McLean, K. Robin, "Trisecting angles with ruler and compasses", Mathematical Gazette 92, July 2008, 320–323. See also Feedback on this article in vol. 93, March 2009, p. 156.

[Stewart-6] Stewart, Ian (1989). Galois Theory. Chapman and Hall Mathematics. pp. g. 58. ISBN 0-412-34550-1.

[7] Jim Loy (2003) [1997]. "Trisection of an Angle". Archived from the original on February 25, 2012. Retrieved 30 March 2012.

[8] Yates, Robert C (1942). The Trisection Problem (PDF). The National Council of Teachers of Mathematics. pp. 39–42.

[Ludwig_Bieberbach-9] Ludwig Bieberbach (1932) Zur Lehre von den kubischen Konstruktionen, Journal für die reine und angewandte Mathematik, H. Hasse und L. Schlesinger, Band 167 Berlin, p. 142–146 online-copie (GDZ). Retrieved on June 2, 2017.

[10] Jim Loy "Archived copy". Archived from the original on November 4, 2013. Retrieved 2013-11-04.

[Dörte_Haftendorn-11] Dörte Haftendorn: "Kurven erkunden und verstehen" (German) Kapitel 3.3: S. 62, ISBN 9783658147488, Spektrum, Akademischer Verlag, Springer, 2016, "Trisektrix Konstruktion" (German). Retrieved on January 12, 2017.

[12] Hutcheson, Thomas W. (May 2001). "Dividing Any Angle into Any Number of Equal Parts". Mathematics Teacher. 94 (5): 400–405.

[13] Isaac, Rufus, "Two mathematical papers without words", Mathematics Magazine 48, 1975, p. 198. Reprinted in Mathematics Magazine 78, April 2005, p. 111.

[Gleason-14] Gleason, Andrew Mattei (March 1988). "Angle trisection, the heptagon, and the triskaidecagon" (PDF). The American Mathematical Monthly. 95 (3): 185–194. doi:10.2307/2323624. Archived from the original (PDF) on November 5, 2014.

v t e Ancient Greek mathematics
Mathematicians	Anaxagoras Anthemius Archytas Aristaeus the Elder Aristarchus Apollonius Archimedes Autolycus Bion Bryson Callippus Carpus Chrysippus Cleomedes Conon Ctesibius Democritus Dicaearchus Diocles Diophantus Dinostratus Dionysodorus Domninus Eratosthenes Eudemus Euclid Eudoxus Eutocius Geminus Heron Hipparchus Hippasus Hippias Hippocrates Hypatia Hypsicles Isidore of Miletus Leon Marinus Menaechmus Menelaus Metrodorus Nicomachus Nicomedes Nicoteles Oenopides Pappus Perseus Philolaus Philon Porphyry Posidonius Proclus Ptolemy Pythagoras Serenus Simplicius Sosigenes Sporus Thales Theaetetus Theano Theodorus Theodosius Theon of Alexandria Theon of Smyrna Thymaridas Xenocrates Zeno of Elea Zeno of Sidon Zenodorus
Treatises	Almagest Archimedes Palimpsest Arithmetica Conics (Apollonius) Elements (Euclid) On the Sizes and Distances (Aristarchus) On Sizes and Distances (Hipparchus) On the Moving Sphere (Autolycus) The Sand Reckoner
Problems	Problem of Apollonius Squaring the circle Doubling the cube Angle trisection
Centers	Cyrene Library of Alexandria Platonic Academy
Timeline of Ancient Greek mathematicians