Ruler and Compass Constructions

Ruler and Compass Constructions
Arthur Baragar
UNLV
http://www.nevada.edu/∼baragar/
The Mathematical Interests of Peter Borwein
IRMACS at SFU, Burnaby, Canada, May 16th, 2008
SFU, May 2008 – p. 1/30

No fractals appear in this talk
SFU, May 2008 – p. 2/30

It is impossible to trisect an arbitrary angle
SFU, May 2008 – p. 3/30

It is impossible to trisect an arbitrary angle using a
straightedge and compass.
SFU, May 2008 – p. 4/30

It is impossible to trisect an arbitrary angle using a
straightedge and compass.
But it is possible to trisect an arbitrary angle using a
ruler and compass. This was known to Archimedes.
SFU, May 2008 – p. 5/30

It is impossible to trisect an arbitrary angle using a
straightedge and compass.
But it is possible to trisect an arbitrary angle using a
ruler and compass. This was known to Archimedes.
The ancient Greeks investigated a number of
variations on the classical tools.
SFU, May 2008 – p. 6/30

It is impossible to trisect an arbitrary angle using a
straightedge and compass.
But it is possible to trisect an arbitrary angle using a
ruler and compass. This was known to Archimedes.
The ancient Greeks investigated a number of
variations on the classical tools.
• Solid constructions
• Ruler and compass constructions
SFU, May 2008 – p. 7/30

Solid constructions
The tools include a conic drawing tool.
Using the curve y = x 2 and the circle centered at
(a,b) that goes through (0, 0),
(x − a) 2 + (x 2 − b) 2 = a 2 + b 2
x 4 + (1 − 2b)x 2 y=x
− 2ax = 0
2
(a,b)
0 1
SFU, May 2008 – p. 8/30

Solid constructions
The tools include a conic drawing tool.
Using the curve y = x 2 and the circle centered at
(a,b) that goes through (0, 0),
(x − a) 2 + (x 2 − b) 2 = a 2 + b 2
x 3 y=x
+ (1 − 2b)x − 2a = 0
2
(a,b)
0 1
SFU, May 2008 – p. 9/30

Solid constructions
x 3 + (1 − 2b)x − 2a = 0
• We can extract cube roots 3 √ α:
Set (a,b) = (α/2, 1/2).
SFU, May 2008 – p. 10/30

Solid constructions
x 3 + (1 − 2b)x − 2a = 0
• We can extract cube roots 3 √ α:
Set (a,b) = (α/2, 1/2).
• We can trisect the angle θ:
Identity: cos 3α = 4 cos 3 α − 3 cos α
So solve: 4x 3 − 3x − cosθ = 0.
SFU, May 2008 – p. 11/30

Solid constructions
x 3 + (1 − 2b)x − 2a = 0
• We can extract cube roots 3 √ α:
Set (a,b) = (α/2, 1/2).
• We can trisect the angle θ:
Identity: cos 3α = 4 cos 3 α − 3 cos α
So solve: 4x 3 − 3x − cosθ = 0.
• We can construct anything in a 2-3-tower.
• The points of intersection of two conics lie in a
2-3-tower.
SFU, May 2008 – p. 12/30

The regular 7-gon
• Set ω = e 2πi/7
• ω 7 − 1 = 0
• Set x = ω + ω −1 = 2 cos(2π/7)
• x 3 + x 2 − 2x − 1 = 0
• Complete the cube: Set x = u − 1/3
• u 3 − 7 3 u − 7 27 = 0 SFU, May 2008 – p. 13/30

The regular 7-gon
• u 3 − 7 3 u − 7 27 = 0
• x 3 + (1 − 2b)x − 2a = 0
• So set (a,b) = ( 7
54 , )
5
3
SFU, May 2008 – p. 14/30

The regular 7-gon
(— 5
74 ,5)
—
3
y=x 2 SFU, May 2008 – p. 15/30
0 1

The regular 7-gon
(— 5
74 ,5)
—
3
y=x 2 SFU, May 2008 – p. 16/30
0 1 u + 1 3 –

The regular 7-gon
( 5
7
— —
4 ,5) 3
u
0 – 2
1 u + 1–
3
y=x 2 SFU, May 2008 – p. 17/30

The regular 7-gon
( 5
7
— —
4 ,5) 3
u
0 – 2
1 u + 1–
3
y=x 2 SFU, May 2008 – p. 18/30

The regular 7-gon
( 5
7
— —
4 ,5) 3
u
0 – 2
1 u + 1–
3
y=x 2 SFU, May 2008 – p. 19/30

Ruler and compass
constructions
Our ruler will have two markings on it, a unit distance
apart.
In constructions involving rulers, we are allowed to
place a marking on one constructed object (line or
circle), and the other marking on another constructed
object. We are then allowed to move the ruler,
keeping the markings on the objects, until the ruler
passes through some already constructed point.
SFU, May 2008 – p. 20/30

Archimedes’ trisection
2α
α
2α
α
3α
SFU, May 2008 – p. 21/30

Curve interpretation
The conchoid of Nicomedes
SFU, May 2008 – p. 22/30

Curve interpretation
The Limaçon
SFU, May 2008 – p. 23/30

The conchoid
The line has polar equation
r cos θ = a, so
r = a secθ ± 1
r cos θ − a = ± cosθ
r(x − a) = ±x
(x 2 + y 2 )(x − a) 2 = x 2 SFU, May 2008 – p. 24/30

The generalized limaçon
Algebraic of degree 6.
SFU, May 2008 – p. 25/30

Intersections
Intersecting
• the conchoid with a line gives 4 points.
• the conchoid with a circle gives 8 points, but 2
are at infinity and factor out.
• the generalized limaçon with a line gives 6 points.
• the generalized limaçon with a circle gives 12
points, but 6 are at infinity and factor out.
SFU, May 2008 – p. 26/30

The possibilities
• One can extract cube roots of lengths
(Nicomedes).
• All points in a 2-3-tower over Q are constructible.
• All constructible points lie in a 2-3-5-6-tower
over Q.
SFU, May 2008 – p. 27/30

A provocative example
Let us intersect the conchoid
with a = 2 with the circle
centered at (1, 1) and
through the point (0, 2).
SFU, May 2008 – p. 28/30

A provocative example
Solving, we get
x 6 − 7x 5 + 14x 4 − 2x 3 − 10x 2 − 12x + 18 = 0.
Factoring out x − 3, we get
x 5 − 4x 4 + 2x 3 + 4x 2 + 2x − 6 = 0.
By Eisenstein’s criterion, this is irreducible. The
polynomial has two complex roots, so the Galois
group includes a 2-cycle. Hence, it is all of S 5 . Thus
we can solve for x using a ruler and compass, but
cannot solve it using extractions of roots.
SFU, May 2008 – p. 29/30

Happy Birthday Peter
SFU, May 2008 – p. 30/30