Editing Angle trisection (section)

==Other methods==
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''.<ref name ="trisectors">{{Citation|last = Dudley|first = Underwood|author-link = Underwood Dudley|title = The trisectors|publisher = [[Mathematical Association of America]]|year = 1994|isbn = 978-0-88385-514-0}}</ref>

===Approximation by successive bisections===
Trisection can be approximated by repetition of the compass and straightedge method for bisecting an angle. The geometric series {{nowrap|1={{sfrac|1|3}} = [[1/4 + 1/16 + 1/64 + 1/256 + ⋯|{{sfrac|1|4}} + {{sfrac|1|16}} + {{sfrac|1|64}} + {{sfrac|1|256}} + ⋯]]}} or {{nowrap|1={{sfrac|1|3}} = {{sfrac|1|2}} − {{sfrac|1|4}} + {{sfrac|1|8}} − {{sfrac|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.<ref>{{cite web|title=Trisection of an Angle |author=Jim Loy |orig-year=1997 |year=2003 |url=http://www.jimloy.com/geometry/trisect.htm |access-date=30 March 2012 |url-status=dead |archive-url=https://web.archive.org/web/20120225124232/http://www.jimloy.com/geometry/trisect.htm |archive-date=February 25, 2012 }}</ref>

===Using origami===
{{Main|Mathematics of origami#Trisecting an angle}}
Trisection, like many constructions impossible by ruler and compass, can easily be accomplished by the 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===
[[File:Sylvester's Link Fan.svg|275px|thumb|Sylvester's Link Fan]]
There are a number of simple [[Linkage (mechanical)|linkages]] which can be used to make an instrument to trisect angles including Kempe's Trisector and Sylvester's Link Fan or Isoklinostat.<ref>{{cite book|title=The Trisection Problem |url=http://files.eric.ed.gov/fulltext/ED058058.pdf#53 |archive-url=https://ghostarchive.org/archive/20221009/http://files.eric.ed.gov/fulltext/ED058058.pdf#53 |archive-date=2022-10-09 |url-status=live |first=Robert C| last=Yates |pages=39–42 |year=1942 |publisher=The National Council of Teachers of Mathematics}}</ref>

{{Clear}}

===With a right triangular ruler ===
[[File:01-Dreiteilung-des-Winkels-Bieberbach.svg|thumb|upright=1.35|Bieberbach's trisection of an angle (in blue) by means of a right triangular ruler (in red)]]

In 1932, [[Ludwig Bieberbach]] published in ''[[Crelle's Journal|Journal für die reine und angewandte Mathematik]]'' his work ''Zur Lehre von den kubischen Konstruktionen''.<ref name="Ludwig Bieberbach">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 [http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN243919689_0167&DMDID=DMDLOG_0020 online-copie (GDZ)]. Retrieved on June 2, 2017.</ref> 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 construction begins with drawing a [[circle]] passing through the vertex {{mvar|P}} of the angle to be trisected, centered at {{mvar|A}} on an edge of this angle, and having {{mvar|B}} as its second intersection with the edge. A circle centered at {{mvar|P}} and of the same radius intersects the line supporting the edge in {{mvar|A}} and {{mvar|O}}.

Now the ''[[set square|right triangular ruler]]'' is placed on the drawing in the following manner: one [[cathetus|leg]] of its right angle passes through {{mvar|O}}; the vertex of its right angle is placed at a point {{mvar|S}} on the line {{mvar|PC}} in such a way that the second leg of the ruler is tangent at {{mvar|E}} to the circle centered at {{mvar|A}}. It follows that the original angle is trisected by the line {{mvar|PE}}, and the line {{mvar|PD}} perpendicular to {{mvar|SE}} and passing through {{mvar|P}}. This line can be drawn either by using again the right triangular ruler, or by using a traditional [[straightedge and compass construction]]. With a similar construction, one can improve the location of {{mvar|E}}, by using that it is the intersection of the line {{mvar|SE}} and its perpendicular passing through {{mvar|A}}.

''Proof:'' One has to prove the angle equalities <math>\widehat{EPD}= \widehat{DPS}</math> and <math>\widehat{BPE} = \widehat{EPD}.</math> The three lines {{mvar|OS}}, {{mvar|PD}}, and {{mvar|AE}} are parallel. As the [[line segment]]s {{mvar|OP}} and {{mvar|PA}} are equal, these three parallel lines delimit two equal segments on every other secant line, and in particular on their common perpendicular {{mvar|SE}}. Thus {{math|1=''SD{{'}}'' = ''D{{'}}E''}}, where {{mvar|D'}} is the intersection of the lines {{mvar|PD}} and {{mvar|SE}}. It follows that the [[right triangle]]s {{mvar|PD{{'}}S}} and {{mvar|PD{{'}}E}} are congruent, and thus that <math>\widehat{EPD}= \widehat{DPS},</math> the first desired equality. On the other hand, the triangle {{mvar|PAE}} is [[isosceles triangle|isosceles]], since all [[radius]]es of a circle are equal; this implies that <math>\widehat{APE}=\widehat{AEP}.</math> One has also <math>\widehat{AEP}=\widehat{EPD},</math> since these two angles are [[alternate angles]] of a transversal to two parallel lines. This proves the second desired equality, and thus the correctness of the construction.

===With an auxiliary curve===
<gallery heights="320" widths="320">
File:Archimedean spiral trisection.svg|Trisection using the Archimedean spiral
File:01-Angel Trisection.svg|Trisection using the Maclaurin trisectrix
</gallery>There are certain curves called [[trisectrix|trisectrices]] which, if drawn on the plane using other methods, can be used to trisect arbitrary angles.<ref>Jim Loy {{cite web|url=http://www.jimloy.com/geometry/trisect.htm |title=Trisection of an Angle |access-date=2013-11-04 |url-status=dead |archive-url=https://web.archive.org/web/20131104113041/http://www.jimloy.com/geometry/trisect.htm |archive-date=November 4, 2013 }}</ref>  Examples include the [[Trisectrix of Maclaurin|trisectrix of Colin Maclaurin]], given in [[Cartesian coordinate system|Cartesian coordinates]] by the [[Implicit curve|implicit equation]]
:<math>2x(x^2+y^2)=a(3x^2-y^2),</math>
and the [[Archimedean spiral]].  The spiral can, in fact, be used to divide an angle into ''any'' number of equal parts.
Archimedes described how to trisect an angle using the Archimedean spiral in [[On Spirals#Trisecting an angle|On Spirals]] around 225 BC.

===With a marked ruler===
[[File:Trisecting angles three.svg|thumb|355px|Trisection of the angle using a marked ruler]]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 [[Pons asinorum|meet the third side at the same angle]].
{{Clear}}

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

[[Hypothesis]]: Given {{mvar|AD}} is a straight line, and {{mvar|AB}}, {{mvar|BC}}, and {{mvar|CD}} all have equal length,

[[logical consequence|Conclusion]]: angle {{math|''b'' {{=}} {{sfrac|''a''|3}}}}.

[[Mathematical proof|Proof]]:
# From Fact 1) above, <math> e + c = 180</math>°.
# Looking at triangle ''BCD'', from Fact 2) <math> e + 2b = 180</math>°.
# From the last two equations, <math> c = 2b</math>.
# Therefore, <math>a=c+b=2b+b=3b</math>.

and the [[theorem]] is proved.

Again, this construction stepped outside the [[Greek mathematics|framework]] of [[compass and straightedge constructions|allowed constructions]] by using a marked straightedge.

===With a string===
Thomas Hutcheson published an article in the ''[[Mathematics Teacher]]''<ref>{{cite journal|journal=Mathematics Teacher|volume=94 |issue=5 |date=May 2001 |pages=400–405 |last=Hutcheson |first=Thomas W. |title=Dividing Any Angle into Any Number of Equal Parts|doi=10.5951/MT.94.5.0400 }}</ref> 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"===
{{main|Tomahawk (geometry)}}
[[File:Tomahawk2.svg|thumb|right|A tomahawk trisecting an angle. The tomahawk is formed by the thick lines and the shaded semicircle.]]

A "[[Tomahawk (geometry)|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 a [[set square]], it can be also used for trisection angles by the method described in {{slink||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 ([[Steel square#Carpenter's square|Carpenter's Square]]).

===With interconnected compasses===

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