From Euclid to Spigotry #6
Pentagon Construction
Last modified: September 15, 2004
The Greeks knew how to construct a regular pentagon with ruler and compass, and so 
soon will you.  But they failed to construct a regular heptagon.  There is a reason: 
constructing a pentagon is equivalent to solving a quadratic equation, while constructing a 
heptagon is equivalent to solving an irreducible cubic equation. 

References:
Bold, Chapter III, pages 23-27
Bold, Chapter V, pages 49-61
Lecture topic 1: Quadratic equation for a pentagon
By considering the complex "fifth roots of unity", show that cos (2p/5) can be computed by solving a quadratic equation and hence can be constructed with ruler and compass.
Notes:
The ancient Greeks knew how to construct a regular pentagon, but their approach does not generalize.  Here is an approach that is similar to the one that Gauss used for the regular heptadecagon. It is well explained in Bold. The secret is to note that the complex fifth roots of unity, when graphed in the complex plane, form a regular pentagon.

Number the roots R1 through R5, as
suggested by the diagram.  
R1 = cos 2p/5 + i sin 2p/5
R2 = cos 4p/5 + i sin 4p/5
All of the roots satisfy R5 - 1 = 0,
which can be factored to
(R-1)(R4 + R3 + R2 + R + 1) = 0.

Now let 
y1 = R1 + R4 = 2 cos 2p/5
y2 = R2 + R3 = 2 cos 4p/5

Multiply out
(y - y1) (y- y2)
using the fact that 
R1 + R2 + R3 + R4 + 1 = 0,
and you discover that y1 and y2
are the solutions of
y2 + y - 1 = 0. Details of the algebra
are on the next page.

The five fifth roots of unity, numbered counterclockwise as shown in the diagram, are R, R2, R3, R4, R5.  Now we can follow the approach on page 57 of Bold.  Since R5 = 1,

In terms of the complex quantity R, we can express trig functions of the angles 2p/5 and 4p/5:

So 
Remarkably, 
Therefore and  are roots of the quadratic equation

and can be constructed with compass and straightedge.
The quadratic formula gives
and 
Check

Lecture topic 2: The quadratic equation for a pentagon by multiple angle formulas

We know that cos 2q = 2 cos2 q - 1
and that cos 3q = cos3 q - 3 cos q sin2 q = cos3 q - 3 cos q(1 - cos2 q)

By looking at a diagram of the regular pentagon, we can see that the
three angles for which cos 2q = cos 3q are 0, 2p/5, and 4p/5.

So set  cos 2q = cos 3q  to get a cubic equation for cos q.

One factor of this equation must be (cos q - 1).

The other factor turns out to be 4 cos2 q + 2 cos q -1.

If you set y = 2 cos q this is again the same quadratic equation
y2 + y - 1 = 0.

Lecture topic 3: Constructing a pentagon - quadratic method 1 
Show how to construct a regular pentagon by solving the equation  graphically,
using the technique from page 3 of Bold with a = -1, b = -1.
Constructing a pentagon by solving  graphically 

A is the point x = 0, y = 1.
B is the point x = -1/2, y = 0
To solve the equation, construct a circle with center at B that passes through A
(review Ruler and Compass to see why this is correct - the diameter must join
the points (0, 1) and (a, b) = (-1, -1) ).
The circle cuts the x axis at the roots C and D of 
We know that these roots are 2 cos 2p/5 and 2 cos 4p/5.
Bisect segment OC to get cos 2p/5, then construct the perpendicular bisector
of OC to locate vertices V1 and V4

Construct the perpendicular bisector of OD to locate vertices V2 and V3.
This construction is on the course Web site.

Lecture topic 4: Constructing a pentagon - quadratic method 2 
Show how to construct a regular pentagon by solving the equation  graphically,
using the technique from the last topic of "Ruler and Compass".
Constructing a pentagon by solving  graphically, using the technique that works
with one positive root and one negative root. 

Let z1 = cos 2p/5,  z2 = cos 4p/5, so that z1 = y1/2 and z2= y2/2
Then z1 + z2 = - 1/2    while z1 z2 = 1/4
Now use the technique from the last topic of "Ruler and Compass".
The average of z1 and z2 is the center of the circle, 1/4.
The square root of the product z1 z2  , 1/2, is where the circle cuts the y axis.
Simply construct a circle centered at ( -1/4, 0) that passes through ( 0, 1/2).
This circle intersects the x axis at z1 and z2
Construct vertical lines through these two points to locate the vertices of the pentagon.
This is essentially the same as the previous construction, but the circle is only half as large.
This construction is also on the course Web site.

Lecture topic 5: Constructing a pentagon - Richmond's method
This clever way of solving the quadratic equation above that leads to a simple construction for the regular pentagon was invented by Richmond in the late 19th century.

Let and so that 
Then 
In order for  and  to be the solutions of 
we require that  and that  = -1.
Thus  and  
This leads to a simple construction:
1. Construct an acute angle whose tangent is 2: the angle OAB in the diagram below.
2. Bisect it to get the angle f : the angle OAD in the diagram.
3. Construct DE parallel to OC.
4. Thus cos b = 1/2 tan f 
Now consider the quadratic equation
 (y - tan f) (y + cot f) = + (cot f - tan f) y - cot f  tan f = 0
                                    + 2 cot 2f  y - 1 =  = 0
We know that the roots of this equation are y = 2 cos (2p/5) and y = 2 cos (4p/5).
So 1/2 tan f = cos (2p/5) and b = 2p/5 as required.
By constructing the perpendicular to AD at A and finding where it intersects the y axis,
you can easily find the two vertices below the x axis
This construction is on the course Web site, showing the steps to locate all five vertices.

Lecture topic 6: Euclid's construction - preliminary.
This approach to constructing the pentagon seems more complicated by modern standards, but it
is purely geometric.  While the details are all in Bold, in it interesting to go to the Euclid Web site
and study Book II, proposition 11. Here is the diagram from Euclid, with the addition of the two circles that need to be drawn to make EF equal to EB and then AH equal to AF.

Euclid puts the proof in purely geometric terms, but you should just show that if the square ABCD
has side 1 and the square AFGH has side x, then x(x+1) = 1 or x2 + x - 1 = 0.

Euclid expresses this by saying that the rectangle FGKC has the same area as the square ABCD
and that A divides CF in "extreme and mean proportion," which means that 1/x = 1+x:
the larger part is to the smaller as the whole is to the larger part.

This construction is on the course Web site.

Lecture topic 7: Euclid's construction
Look at Book IV, proposition 10.
The key point is that in a 360 - 720 -720 isosceles triangle,
the bisector of either base angle divides the opposite side in extreme and mean proportion, and conversely.  There is a nice diagram for this in Bold.

Euclid's proof is unnecessarily complicated, because he has not yet developed the theory of
similar triangles in Book IV.  Here is an easier approach.

Suppose that the angle at A is 360
and that the base angles at B and C
are both 720.
If BD bisects the angle ABC into two 
360 angles, then triangles BCD and
ABC are similar.  Furthermore, triangle
ADB is isosceles.

It follows that 1/x = 1+x.

So once Euclid has a line divided in extreme 
and mean proportion, like ADC, he can
construct a 360 - 720 -720 triangle.

Book IV, proposition 11 finishes the construction.
Circumscribe a circle about this triangle.
Then A, B, and C are three of the vertices
of a regular pentagon, and the bisectors
of the angles at B and C intersect the
circle in the other two vertices.
This construction is on the course Web site.