STRAIGHT-LINE AND COMPASS CONSTRUCTIONS.

Problem 16

16. In honour of this being the 17th question: construct a regular 17-sided polygon.

Solution

1. This is possible to do iff we can construct an angle of 2pi / 17.
2. This is possible to do iff we can (given an arbitrary unit length) construct the value of cos(2pi / 17).
3. Assuming that such a construction is possible, we can draw a circle of arbitrary unit radius and proceed to construct the polygon as shown in the diagram.

It now remains to show how we can construct cos (2pi/17).

Finding the roots of an order-17 polynomial

Remember cos a + i.sin a = e ^i.a ? That's the trick. Because we have:

• cos a + i.sin a = e ^i.a
• cos 17a + i.sin 17a = e ^17i.a

If we set a = 2pi / 17, and remember that e^xy = (e^x)^y, we get:

• cos 2pi + i.sin 2pi = e ^17i.a
• 1 + 0i = (e ^i.a)¹⁷
• 1 + 0i = (cos a + i.sin a)¹⁷

Now, we set c=cos a and s=sin a (and later use c² + s² = 1), and can expand the resulting equation:

• 1 + 0i = (cos a + i.sin a)¹⁷

You can see what's coming next: expand the large polynomial above, equate real and imaginary parts of the resulting equation. Replacing through with 1-c² for s², we wind up with a large polynomial in c:

• 1 + 0i = (c + i.s)¹⁷
• 1 + 0i = c¹⁷ + 17c¹⁶i¹s¹ + 136c¹⁵i²s² + 680c¹⁴i³s³ + 2380c¹³i⁴s⁴ + 6188c¹²i⁵s⁵ + 12376c¹¹i⁶s⁶ + 19448c¹⁰i⁷s⁷ + 24310c⁹i⁸s⁸ + 24310c⁸i⁹s⁹ + 19448c⁷i¹⁰s¹⁰ + 12376c⁶i¹¹s¹¹ + 6188c⁵i¹²s¹² + 2380c⁴i¹³s¹³ + 680c³i¹⁴s¹⁴ + 136c²i¹⁵s¹⁵ + 17c¹i¹⁶s¹⁶ + i¹⁷s¹⁷

We equate the real parts of the above equation, and perform the substitution s²=1-c²:

• 1 = c¹⁷
  - 136c¹⁵(1 - c²)
  + 2380c¹³(1 - c²)²
  - 12376c¹¹(1 - c²)³
  + 24310c⁹(1 - c²)⁴
  - 19448c⁷(1 - c²)⁵
  + 6188c⁵(1 - c²)⁶
  - 680c³(1 - c²)⁷
  + 17c(1 - c²)⁸
• 1 = c¹⁷
  - 136c¹⁵(1 - c²)
  + 2380c¹³(1 - 2c² + c⁴)
  - 12376c¹¹(1 - 3c² + 3c⁴ - c⁶)
  + 24310c⁹(1 - 4c² + 6c⁴ - 4c⁶ + c⁸)
  - 19448c⁷(1 - 5c² + 10c⁴ - 10c⁶ + 5c⁸ - c¹⁰)
  + 6188c⁵(1 - 6c² + 15c⁴ - 20c⁶ + 15c⁸ - 6c¹⁰ + c¹²)
  - 680c³(1 - 7c² + 21c⁴ - 35c⁶ + 35c⁸ - 21c¹⁰ + 7c¹² - c¹⁴)
  + 17c(1 - 8c² + 28c⁴ - 56c⁶ + 70c⁸ - 56c¹⁰ + 28c¹² - 8c¹⁴ + c¹⁶)
• 1 = c¹⁷ (1 + 136 + 2380 + 12376 + 24310 + 19448 + 6188 + 680 + 17)
  + c¹⁵ (-136 - 2.2380 - 3.12376 - 4.24310 - 5.19448 - 6.6188 - 7.680 - 8.17)
  + c¹³ (2380 + 3.12376 + 6.24310 + 10.19448 + 15.6188 + 21.680 + 28.17)
  + c¹¹ (-12376 - 4.24310 - 10.19448 - 20.6188 - 35.680 - 56.17)
  + c⁹ (24310 + 5.19448 + 15.6188 + 35.680 + 70.17)
  + c⁷ (-19448 - 6.6188 - 21.680 - 56.17)
  + c⁵ (6188 + 7.680 + 28.17)
  + c³ (-680 - 8.17)
  + c (17)
  
• 1 = 65536c¹⁷ - 278528c¹⁵ + 487424c¹³ - 452608c¹¹ + 239560c⁹ - 71808c⁷ + 11424c⁵ - 816c³ + 17c

We can now consider the equation:

• 0 = 65536x¹⁷ - 278528x¹⁵ + 487424x¹³ - 452608x¹¹ + 239560x⁹ - 71808x⁷ + 11424x⁵ - 816x³ + 17x - 1

Roots of this equation, values of x which satisfy this, will be the x-coordinates of points around a unit circle centred on the origin at intervals of 2pi/17.

Clearly, then, (x - 1) is a factor of the RHS (we can demonstrate this by substitution). One of the other roots we're after will be x=c, and so on.

From previous results, if a root of this belongs to the closure of the integers under the operations of addition, subtraction, multiplication, division, and taking the square root, then we can construct a 17-sided polygon. And in fact, the algebraic construction of the root encodes the geometric operations required to produce such a ruler-and-compass construction.

We can factor out (x - 1).

At this point I'm going to stop until more browsers support MathML; if anyone's really interested then I can send them the LaTeX source. However, it's traditional at this stage to whip out a long algebraic monster and say something like, "we note that x is a root of this equation" :-)

x in this case is -1/16 + 1/16 sqrt(17) + 1/16 sqrt[34 - 2sqrt(17)] + 1/8 sqrt[17 + 3sqrt(17) - sqrt(34-2sqrt(17)) - 2sqrt(34+2sqrt(17)]