# Trisecting a Line Segment (With World Record Efficiency!)

### Abstract

The book Proofs without Words is a collection of many short simple proofs, including a construction to trisect a line segment.  This short construction naturally leads one to wonder what the shortest construction is (that is, the construction using the least number of circles and lines).  We find the shortest trisection of a line segment, then the shortest construction using only circles (Mohr-Mascheroni) and using only lines (Poncelet-Steiner).

This article uses Flash animations to illustrate many of the constructions.  You will need a reasonably current version of the Flash Player plug-in installed for your browser in order to view the animations.

## Trisecting a Line

### Introduction

In high school geometry, one of the most useful ruler and compass constructions is bisecting a line segment.  This beautiful construction can be done with only two circles and one line.

Euclid's very first proposition contains the essence of this construction, though he does not explicitly bisect a segment until the tenth proposition.

But what if we wanted to trisect the line segment? Given a line with segment $$AB$$, construct a point $$F$$ on the segment so that $$AF = (1/3) AB$$, using the classical straightedge and compass.

### Trisecting with Two Circles and Four Lines

Scott Coble found a clever construction, reprinted in the wonderful book Proofs without Words [1]. (Here is a proof why it works.)

This construction uses two circles and four additional lines.  Certainly two is the fewest number of circles that is possible, since one needs two circles to construct a point not on the given line through $$AB$$.

Can we do better than two circles and four lines?

### Trisecting with Two Circles and Three Lines

This elegant construction takes two circles and only three additional lines. (Here is a proof why it works.)

### Trisecting with Three Circles and Two Lines

What if one allows a third circle?

Hartshorne in his wonderful textbook Companion to Euclid [3] says the average good geometer can trisect a segment with a "par" of 5 (average of five steps). Here is a version of the classical five step construction.

We have used three circles and two additional lines, which is Hartshorne's "par 5." (Here is a proof.)

### Trisecting with Four Circles

If we wanted to use circles and no additional lines, how many would it take?

Only four circles, one under "par"!

Here is a proof. This construction can be generalized to construct a segment of length  $$1/n$$  by replacing the circle of radius 3 by one of radius $$n$$.

Four circles is in fact the best possible, since three circles or three circles and one line cannot suffice (details). Nor are two circles and two additional lines enough to trisect the segment (details).

By the way, it is impossible to trisect an arbitrary angle with unmarked ruler and compass.

## Proofs of Constructions with Five or Six Steps

### Picturing the Proof of the Coble construction (Two Circles and Four Lines)

The following diagrams outline the geometric reasoning behind the Coble construction.

Begin with the original diagram and add several auxiliary circles and lines.  Note the three congruent rhombuses, $$BHCA$$, $$GBAD$$, and $$JGDI$$, with $$E$$ at the intersection of the diagonals of the middle rhombus $$DGBA$$, which is known to be at the midpoint of each diagonal.  Note the similar triangles $$ACF$$ and $$ICJ$$.
Since $$|CA| = 1$$, $$|CI| = 3$$, and $$|IJ| =1$$, ratios of similar triangles shows that $$|AF| = 1/3$$.

A referee suggested a nice alternative outline of a proof.  The triangles $$ADG$$ and $$ABG$$ are both equilateral triangles.  Thus, the measure of angle $$DAE$$ equals the measure of angle $$BAE$$; by Side-Angle-Side, the triangles $$DAE$$ and $$BAE$$ are congruent.  Then the length of the segments $$DE$$ and $$BE$$ are equal, and also the length of the segments $$CA$$ and $$AD$$ are equal, so the segments $$AB$$ and $$CE$$ are medians of the triangle $$BCD$$.  It is now a well known fact that the intersection $$F$$ of the medians is at the centroid which is one-third of the distance along the median $$AB$$.

### Picturing the Proof of the Two Circles and Three Lines construction

Begin with the original diagram, add several auxiliary circles and lines, then note the similar triangles $$ACF$$ and $$GCH$$.  Since $$|AC| = 1$$, $$|GC| = 3$$, and $$|GH| =1$$, we see that $$|AF| = 1/3$$.

Of course, the picture hides the hardest part of the proof: showing that the circle with center B and the two lines $$CH$$ and $$DJ$$ go through the point $$E$$.  Possibly the easiest way to see this is to use analytic geometry.  We will find the point $$E_2$$ on the intersection of the lines $$CH$$ and $$DJ$$, and then show this point is exactly distance one from $$B$$, that is, the point $$E_2$$ is the same as the point $$E$$.   For convenience, we will make $$B$$ the origin, then one can easily verify the coordinates of these points:

• $$A\quad (-1,0)$$
• $$B\quad (0,0)$$
• $$C\quad (-1/2, \sqrt{3}/2)$$
• $$D\quad (-3/2, -\sqrt{3}/2)$$
• $$H\quad (-1, -\sqrt{3} )$$
• $$J\quad (1,0)$$.

Then the slope of the line $$DJ$$ is $$\sqrt{3}/5$$ and the equation is  $$y = (\sqrt{3}/5) x - (\sqrt{3}/5)$$.  The slope of the line $$CH$$ is  $$3 \sqrt{3}$$ and the equation is  $$y = 3 \sqrt{3} x + 2 \sqrt{3}$$.  Some algebra shows that the point $$E_2$$ where these two lines intersect has $$x = -11/14$$, so $$y = -5 \sqrt{3} / 14$$.  The point  $$E_2$$ $$( -11/14, -5 \sqrt{3}/ 14 )$$ satisfies  $$(11/14)^2 + ( 5 \sqrt{3} / 14 )^2 = ( 121 + 75)/ 14^2 = 1$$,  thus showing it is the point $$E$$ from our construction.

### Picturing the Proof of the Three Circles and Two Lines construction

Begin with the original diagram, add a few auxiliary circles and lines, then note the similar triangles $$ADF$$ and $$GDE$$.  Since $$|AD|=1$$, $$|GD|=3$$, and $$|GE| =1$$, we see that $$|AF| = 1/3$$.

Elegant! Beautiful! Satisfying!

## Proofs of Constructions with Four Steps

### Picturing the Proof of the Four Circles Construction

Begin with the original diagram and add a few auxiliary lines.  Angle $$ADG$$ is a right angle since it is an inscribed angle subtending a diameter.  Now $$GDA$$ and $$DJA$$ are similar right triangles.  By our construction, $$|AG| = 6$$ and $$|AD| = 1$$, and by similar triangles (or trigonometry) $$|AG| / |AD| = |AD| / |AJ|$$.  Thus,  $$|AJ| = 1 / |AG| = 1/6$$.  Since $$DJ$$ is a perpendicular bisector of the chord $$AF$$, we have $$|AF| = 2 |AJ|$$ and so  $$|AF| = 1/3$$.

### Proof that Three Circles and One Additional Line is Not Enough

Beginning with points $$A$$ and $$B$$ and the line through them, we construct the circle with center at $$B$$ and radius to $$A$$.  There are only two ways, up to symmetry, to construct a second circle.

The following outlines all the possibilities for three circles and a line.  We use symmetry whenever possible; for instance, we only need to consider third circles with centers in the "first quadrant" ($$E$$, $$D$$ or $$C$$) for the case of the two congruent circles.  We only draw lines from points in the top half to points in the bottom half since a line through an existing point on the $$x$$-axis does not create a new point on the $$x$$-axis.

We assume that $$B$$ is at the origin and that $$A$$ is at -1 and $$C$$ is at 1 on the $$x$$-axis.  Since we could equally well consider $$A$$ or $$C$$ to be the origin, we only consider the constructible values on the $$x$$-axis modulo 1, hence only need to list values between -0.5 and 0.5. Since we can flip the diagrams, we also can consider the constructible values on the x-axis modulo the +/- sign, hence we only need to list values between 0 and 0.5.  For instance, in the third 3-circle-1-line construction below which adds the circle with radius $$DB$$, the point $$M$$ is actually at  $$x =$$ 2.6180339887  which modulo 1 is  -0.3819660113  which modulo the +/- sign is  0.3819660113.  This happens to be the value of $$K$$ as well.

### Proof that Two Circles and Two Lines are Not Enough

We first consider the two ways we can construct two circles.  The second case does not allow any new lines to be drawn so may be eliminated.  For the case with two congruent circles, by symmetry we only need consider lines through the "first quadrant:": through the point $$C$$, $$D$$ or $$E$$. The only possibilities are the blue lines.  The only new points created are $$G$$, $$H$$, $$I$$, $$J$$.  The only new lines that go through one of these new points and the original points off the $$x$$-axis are the two red lines which are parallel to the $$x$$-axis.  In any case, the only constructible points on the $$x$$-axis are integral or half-integral.

## Mohr-Mascheroni and Poncelet-Steiner Constructions

### The Mohr-Mascheroni Construction

Someone might object that we began with the line through $$A$$ and $$B$$.  What if we do not want to begin with this line?  So our goal is to find a point $$F$$ such that the distance from $$A$$ to $$F$$ is one third that of $$A$$ to $$B$$ without using any line.

Mascheroni (1797) showed that all Euclidean constructions could be done just using a compass without a ruler.  A few decades later, a work by Georg Mohr in 1672 was discovered proving the same result.  (details in [2])

Here is a Mohr-Mascheroni (uses no lines) trisection construction with seven circles.  (Here is a proof that this construction works.  Seven circles is in fact the least number possible: see the details.)

Beautiful!

### The Poncelet-Steiner Construction

I personally love circles, but...  What if we like straight lines and really do not like circles?

In the 1800s, Poncelet and Steiner showed that all Euclidean constructions can be done with a ruler only provided one is given a single circle, its center, and a couple suitable points off the circle.  Milos Tatarevic (emails dated 7 June 2003) found a construction using twelve lines.  Here is his construction; he notes the key point is constructing the parallel line $$A'C'$$ and the midpoint $$E$$.  (We cannot prove this is the shortest such construction.)

Martin [9] points out that when one wishes to construct a rational point, then one need not use the circle.  Following his convention, we begin with the points $$(0,1)$$, $$(1,0)$$, $$(0,2)$$, $$(2,0)$$.  Let the point $$A$$ be $$(1,0)$$ and the point $$B$$ be $$(2,0)$$ so we want to construct the trisecting point $$(4/3, 0)$$.  Of course we easily construct the origin $$(0,0)$$.

Here is a Poncelet-Steiner construction, using Martin's starting convention, of the trisecting point $$(4/3, 0)$$ using eight lines.  Here is a proof that it trisects, illustrated with a slightly more general starting convention.  This is the least number of lines needed: see the details.

For more information on trisections and geometric constructions, see the annotated reference page.

### Picturing the Proof of the Mohr-Mascheroni Construction

Begin with the original diagram and add a few auxiliary lines.  The angle $$CDG$$ is a right angle since this angle subtends a diameter of the circle. Then  $$CDG$$ and $$CHD$$ are similar right triangles.  By our construction, $$|CD| = 2$$ and $$|CG| = 6$$, and by similar triangles (or trigonometry) $$|CH| / |CD| = |CD| / |CG|$$ so $$|CH| = 2/3$$.  Since $$DH$$ is a perpendicular bisector of the chord $$CF$$, we have $$|CF| = 4/3$$.  Since $$|CA| = 1$$, we have $$|AF| = |CF| - |CA| = 1/3$$.

### Proof that Six Circles are Not Enough for Mohr-Mascheroni

We have seen a trisecting construction using seven circles.

We prove six circles is not enough by using a Maple program that creates all points that can be constructed with a given number of circles (Maple file, pdf). Essentially, we construct all four-circle constructions (there are 14 up to symmetry) and note that none of these go through the point $$(1/3, 0)$$ or $$(-1/3, 0)$$. If a six circle construction exists, the next two circles added must go through the desired trisecting point. It is then easy to verify that no fifth circle goes through the desired $$(1/3,0)$$ or $$(-1/3,0)$$ point, hence one requires at least seven circles.

We summarize by giving the number of points generated by N circles by our Mohr-Mascheroni construction.

For convenience, we assume we begin with the two points, $$(1,0)$$ and $$(-1,0)$$. By symmetry, we need only list those points in the first quadrant. Here are the number of new points in the first quadrant generated by $$N$$ circles:

• $$N =$$ 2 circles: 2 points
• $$N =$$ 3 circles: 2 new points
• $$N =$$ 4 circles: 11 new points
• $$N =$$ 5 circles: 300 new points

A list of all points that can be constructed by 2, 3, 4, or 5 circles is in this Maple file (pdf).

### Picturing the Proof of Martin's Poncelet-Steiner Construction

In order to see the essence of the proof, we draw a more general picture: the key assumptions are that $$|OA| = |AB|$$ and that $$|OG| = |GH|$$.

### Proof that Seven Lines are Not Enough For Poncelet-Steiner

Beginning with Martin's four points, we have seen a trisecting construction using eight lines.

The proof that seven lines is not enough is contained in this Maple file (pdf). This worksheet generates all points that can be constructed by line-only constructions up to ten lines. A list of all points that can be constructed by two through ten lines is given in this Maple file (pdf). Note that seven lines can generate the point $$(2/3, 0)$$ so seven lines does trisect a line segment, though not the segment $$AB$$.

We summarize by giving the number of points generated by $$N$$ lines with Martin's starting configuration.

We begin with Martin's four points $$(1,0)$$, $$(2,0)$$, $$(0,1)$$ and $$(0,2)$$. If we use $$N$$ lines, we have $$M$$ possible new points where

• $$N=$$ 2 lines: $$M =$$ 2 new points
• $$N=$$ 3, 4, 5 lines: no additional points
• $$N=$$ 6 lines: $$M =$$ 2 new points
• $$N =$$ 7: $$M =$$ 6
• $$N=$$ 8: $$M =$$ 11
• $$N=$$ 9: $$M =$$ 70
• $$N =$$ 10: $$M =$$ 309

## Appendix: Trisecting Angles

A very famous problem is trisecting arbitrary angles.  This problem occupied mathematicians for thousands of years, and although in 1837 Pierre Wantzel proved that angles cannot be trisected by Euclidean methods, people still keep trying (see references, especially Dudley's Budget of Trisections [1]).

If you change the rules and allow something other than a straightedge and compass, you can often trisect angles.  The most famous method is Archimedes' who used a straightedge with two marked points on it; see the Geometry Forum's  http://www.geom.uiuc.edu/docs/forum/angtri/.  Origami paper folding is another elegant way to trisect an angle.

Here are a few of many web pages discussing angle trisection:

MacTutor History of Mathematics
http://www-history.mcs.st-and.ac.uk/history/HistTopics/Trisecting_an_angle.html

Math Forum's Ask Dr. Math FAQ has a link to an outstanding web page by Jim Loy:
http://www.jimloy.com/geometry/trisect.htm

The Mathematical Atlas
http://www.math.niu.edu/~rusin/known-math/index/51M15.html

One of the earliest methods, the Quadratrix of Hippias
http://www.perseus.tufts.edu/GreekScience/Students/Tim/Trisection.page.html

## Annotated References

[1] Dudley, Underwood, A Budget of Trisections, Springer Verlag, 1987.

This delightful book begins with a historical overview of angle trisections, using devices other than a straightedge and compass, such as Archimedes' angle trisection using a compass and a straightedge with two marks. He then describes the personalities of some would-be angle trisectors, then details dozens of angle trisection attempts.

[2] Eves, Howard, A Survey of Geometry, Allyn and Bacon, 1963.

This encyclopedic text is chock full of fascinating tidbits.  In particular, Section 4.4 (pp. 198-204) discuss the Mohr-Mascheroni construction theorem which states that all Euclidean constructions could be carried out with a compass alone and no straightedge.   Section 4.5 (pp.204-210) details the Poncelet-Steiner theorem, which shows that a straightedge along with one given circle and its center is sufficient to carry out any desired Euclidean construction.  Section 4.6 (pp. 210-217) discusses other construction results, and mentions Lemoine's 1907 geometrography, a counting scheme for determining the complexity of a construction. Lemoine generally counts three operations where Hartshorne's construction counts one "step."   We have followed Hartshorne's simpler counting method.

[3] Hartshorne, Robin, Companion to Euclid, American Mathematical Society, Berkeley Mathematics Lecture Notes, Volume 9, 1997
[4] Hartshorne, Robin,  Geometry: Euclid and Beyond, Springer, Undergraduate Texts in Mathematics, 2000.

The 2000 title is an update of the 1997 version.  Pages 20-22 in the newer version discuss the number of steps Euclid used for a proof versus the number needed for a mere construction. The homework following this section contains several constructions with an average or "par" estimate of how many steps an experienced geometer might use. Page 25, problem 2.14, asks for both of the trisection points of a segment and rates it "par 6".  Of course, we are only asking for the left trisection point, so our equivalent is "par 5."  Fascinatingly, in the 1997 version, the same problem 2.14, only now on page 23, says par=9.  Either experienced geometers improved a lot in three years, or the original textbook had a typo.

[5] Heath, Sir Thomas, The Thirteen Books of Euclid's Elements with introduction and commentary, Dover Publications, 1956.

The bisection construction pictured is very close to that in Euclid's first proposition.  Euclid actually does not prove the bisection construction until Proposition 10.  (The relevant Propositions 1 and 10 are pages 241 and 267-268.)  The bisection construction we have pictured is said to be due to Appolonius.  Euclid clearly is more interested in the logical development of the proofs rather than in the shortest constructions.

[6] Hull, Thomas,   http://kahuna.merrimack.edu/~thull/origamimath.html

Tom Hull has a fascinating set of origami geometric constructions, including how to trisect an angle.  He has numerous books on origami and math, such as Project Origami: Activities for Exploring Mathematics.

[7] Lang, Robert J.,  Origami and Geometric Constructions, pdf file available at  http://www.langorigami.com/science/hha/origami_constructions.pdf

Robert Lang is a physicist who is an expert on mathematics and origami.  In particular, he summarizes the best set of mathematical axioms  for paper-folding, and discusses the most efficient ways to trisect a line segment.  His web site http://www.langorigami.com/ has a beautiful collection of origami objects he has folded.

[8] Lemoine, Emile, Geometrographie, ou Art des Constructions Geometriques, Scientia, Phys.-Math. no. 18, Paris, February 1902.

Lemoine invented a method to measure the complexity of geometric constructions.   His method has four parameters; one gives the required number of lines, a second the number of circles, while two others count the moves needed to place the ruler and the compass.  Trisection is not explicitly mentioned in this monograph, although pages 34-36 give a more general construction, a corollary of which is essentially our third trisection method (see Reusch and Ringenberg below).

[9] Martin, George, Geometric Constructions, Springer, 1998.

This nice undergraduate geometry textbook explicitly develops the Mohr-Mascheroni and the Poncelet-Steiner constructions.  We use his convention for "ruler points", pages 69-82, for the Poncelet-Steiner type of constructions.

[10] Nelson, Roger, Proofs Without Words, Mathematical Association of America, 1993.

Our first trisection is taken from page 13, attributed to Scott Coble.

[11] Reusch, J., Planimetrische Konstruktionen in Geometrographischer Ausfuhrung, Druck and Verlag von B. G. Teubner, Leipzig and Berlin, 1904.

Reusch expands on Lemoine's monograph with many diagrams and even more explicit analysis of the basic geometric constructions.  Pages 17-20 explicitly deal with trisections; he gives three different constructions.  The one he calls classical takes 4 lines and 6 circles; his second is the our third one a la Hartshorne that we are calling classical.  Reusch's third construction uses four circles and a line (another "par" 5 construction).  Here are scans of the relevant pages: V, VI, 17, 18, 19 ,20

[12] Ringenberg, Lawrence, Informal Geometry, John Wiley and Sons, 1967.

This standard text contains the typical method of trisecting a segment.   Here is his version of the classical construction, page 139.  In our third construction, we use two circles to construct his point C and draw the line
A C and then a third circle constructs C2.  We do not need to draw C3 nor B C3 since the geometry is such that our line C2 D is already parallel to B C3.  Thus, this slick version of the classical construction takes only three circles and two additional lines.

Our second and fourth trisection constructions do not seem to appear on the web or in standard modern texts, nor are our Mohr-Mascheroni and the Poncelet-Steiner trisection constructions explicitly shown anywhere.  (The 12 line Poncelet-Steiner construction illustrated is due to Milos Tatarevic, who sent it in a couple emails dated June 7, 2003, and is used by permission.)

The third construction given, using three circles and two lines, is well known.   Here is the typical diagram that is used to trisect the segment AB (due to Dr. Math at the Math Forum.)    In our third trisection, the two initial circles construct the point C in Dr. Math's and Ringenberg's classical construction.

             /          D o.           /  .          /     .       C o.       .        /  .       .       /     .       .    A o--------+---------+--------o B                 .       .     /                   .       .  /                     .       o E                       .     /                         .  /                           o F                           /

Here are some trisection constructions that appeared in a Google search.
http://forum.swarthmore.edu/dr.math/problems/cbr.8.13.99.html
http://www.math.niu.edu/~rusin/known-math/98/trisect
http://www.cut-the-knot.com/arithmetic/rational.html#const

Using origami for geometric constructions can delight one for hours:  Robert Lang has a nice discussion showing a trisection using four paper folds.

These emails from Milos Tatrevich discuss the general Poncelet-Steiner constructions: he has generalized the 1/3 construction for 1/n  with beautiful combinatorics and conjectures how many lines are needed, for instance, to construct 1/7 probably takes 14 lines.

I received an email explaining how carpenters use a square and a ruler to trisect line segments:

Subject:
Date:
Sun, 23 Feb 2003 20:36:11 -0500
From: "T. Wilson"

Enjoyed your methods of trisecting a line segment. Now if the criteria include the use of only a compass and an unmarked straightedge, then the method I was taught as an apprentice carpenter would qualify, wouldn't it?

Take line AB. Extend two parallel lines perpendicular to A and B. Take the straightedge and mark off four equidistant points along one edge. Place the straightedge at an angle between the parallel lines so that the first and fourth points coincide with the parallel lines drawn from A and B. With the compass determine the perpendicular distance of each of the three middle points from either parallel line and transfer that distance to the line AB. Voila.

Simple strokes for simple minds (mine, of course, not yours!).

Best wishes,  T. Wilson,  Richmond, VA

Note that carpenters use a marked ruler, which opens a fascinating world of constructions going beyond Euclidean, in particular, one can use a marked ruler to trisect angles.

I received another email with a beautifully symmetric construction using two circles and four lines:

Subject: Trisecting a line segment
Date: Sun, 3 Aug 2003 12:19:14 -0500
From: "Fred Barnes"
To:

I just stumbled across your wonderful web page on trisecting a line segment. I've included a method, not on your page, using four circles and two lines. See attachment.

Fred Barnes