From Euclid to Spigotry #17
Heptadecagon Construction
Last modified: September 14, 2004
Reference: Mathematica notebook heptadecagon.nb (attached)

Lecture topic 1: Algebra of
the 17th roots of unity.

Call the root numbered 1 on
the diagram R. Then the powers of
the complex number R are all the
other roots, and R17=1.

To construct the heptadecagon,
we need to know cos (2p/17), etc.

So define z1= 2 cos (2p/17) = R +R16
z2= 2 cos (4p/17) = R2 +R15
z3= 2 cos (6p/17) = R3 +R14
z4= 2 cos (8p/17) = R4 +R13
z5= 2 cos (10p/17) = R5 +R12
z6= 2 cos (12p/17) = R6 +R11
z7= 2 cos (14p/17) = R7 +R10
z8= 2 cos (16p/17) = R8 +R9
Gauss found the following clever way to group these:
y1 = z1 + z4
y2 = z2 + z8
y3 = z3 + z5
y4 = z6 + z7
Then Gauss combined the y's as follows:
x1 = y1 + y2
x2 = y3 + y4
The construction of the heptadecagon relies on the following algebra:
Easily done by hand: x1 + x2 = -1
z3 z5 = y2
Perhaps worth doing by hand: y1 y2 = -1
Best left to Mathematica: y3 y4 = -1
x1 x2 = -4

Lecture topic 2: Proof that the heptadecagon is constructible

If we know the sum and product of two quantities, we can determine them by solving a quadratic
equation, and this can be accomplished with compass and straightedge.
Since x1 + x2 = -1 and x1 x2 = -4 we can determine x1 and x2 by solving
x2 + x - 4 = 0 using the quadratic formula
Now we know x1 and x2.
Since y1 + y2 = x1 and y1 y2 = -1 we can determine y1 and y2 by solving
y2 - x1 y - 1 = 0 using the quadratic formula. This is best left to Mathematica.
Since y3 + y4 = x2 and y3 y4 = -1 we can determine y3 and y4 by solving
y2 - x2 y - 1 = 0 using the quadratic formula. This is also best left to Mathematica.
Since z3 + z5 = y3 and z3 z5 = y2 we can determine z3 and z5 by solving
z2 - y3 z + y2 = 0 using the quadratic formula. This leads to very messy formulas for z3 and z5
but Mathematica does it instantly.
Once we know z3 and z5, we have cos (6p/17) and cos (10p/17). In principle we can use these
to construct two vertices of the heptadecagon and the rest will be straightforward.
Lecture topic 3: Richmond's clever trigonometry
Richmond's idea was to construct an angle f, chosen so that all the eqations above for 
y1, y2, y3, y4, x1, and x2 are automatically satisfied.  
The angle must be chosen so that 4 cot (4f) = 1.

Now 4 cot (4f) = 2 cot (2f) - 2 tan(2f). If we set
x1 = 2 tan (2f) and x2 = -2 cot (2f) then  -1 and x1 x2 = -4. 
x1 + x2 = 2 tan (2f)  - 2 cot (2f) = - 4 cot (4f) = -1.
So x1 and x2 have the correct sum and product.
Next, 2 cot (2f) = cot f -  tan f. If we set
y3 = tan f and y4 = - cot f  then   y3 + y4 = - 2 cot (2f) = x2 and y3 y4 = -1. 
So y3 and y4 have the correct sum and product.
The last part involves an unfamiliar identity. Using tan (a+b) = 
you can show that y1  = tan (f+p/4) = 
and that y2  =tan (f-p/4) = .
It follows that y1 y2 = -1 and that y1 + y2 =  = 2 tan(2f) = x1
So y1 and y2 also have the correct sum and product.

Lecture topic 4: Richmond's construction of y2 and y3
The diagram is on the Web site, but everyone should replicate it step by step.
Since 4 cot (4f) = 1, cot (4f) = 1/4.  Richmond constructs 4f by making a right
triangle whose sides are 1 and 1/4.  He then bisects the angle twice to get f itself.
The x-coordinate of the point E where the second bisector cuts the x axis is
1/4 tan f, which is equal to y3/4.

Next Richmond constructs a line at an angle of p/4 to the line IE. 
The x-coordinate of the point F where this line cuts the x axis is
1/4 tan (f-p/4), which is equal to y2/4.
Lecture topic 5: Richmond's construction of z3/2 and z5/2
We already know that z3 + z5 = y3 and z3 z5 = y2 
so the quantities c3 = cos (6p/17) = z3/2 and c5 = cos (10p/17) = z5/2
satisfy c3 + c5 = y3/2 and c3 c5 = y2/4.
Since c3 is positive and c5 is negative, we can construct them both by a variant of Bold's
square-root construction.

The center of the circle
needs to be midway between
c3 and c5 ,at y3/4, which has 
already been constructed.
The circle must cut the y axis
at the point K whose vertical
coordinate y satisfies 
y2 =  - c3 c5 = y2/4.
But since the point y2/4 has
also been constructed, a 
circle whose diameter extends
from x =y2/4 (negative) to x = 1
also cuts the y axis
at the point K whose vertical
coordinate y satisfies 
y2 =   y2/4.