The nonagon is another polygon that cannot be constructed with ruler and the compass (see OEIS sequence A004169). However, the nonagon can be constructed using conics (see OEIS sequence A051913). In this post I want to show how we can use the intersection of the Lill circle of the polynomial x3 – 0.75x + 0.125 and a hyperbola to construct nonagon.
The polynomial x3 – 0.75x + 0.125 has the same roots as the polynomial 8x3 – 6x + 1, but when dealing with Lill’s method it’s more convenient to deal with monic polynomials. The polynomial x3 – 0.75x + 0.125 has the roots x1=cos(2π/9), x2= cos(4π/9) and x3=cos
(8π/9). If we can construct a segment equal to cos(2π/9), then we can easily construct a nonagon inscribed inside a unit circle.
The method used in this post is similar to the method used in “The Heptagon, Hyperbola and Lill’s Circle” and “Trisection Hyperbolas and Lill’s Circle”.
Geometric Solution
The Lill’s method representation of the polynomial x3 – 0.75x + 0.125 can be seen below. I included the Lill circle of the polynomial.
If you are not familiar with Lill’s method, it’s probably more convenient to use the equation x^(2) + y^(2) – 0.125x + 0.25y = 0.75 for Lill’s circle.
The hyperbola that let’s us solve the polynomial equation is the hyperbola that passes through point P0 (0,-1) and has the x-axis and the line x=0.125 as asymptote lines. the equation of the hyperbola is -8xy+y=-1.
The hyperbola intersects Lill’s circle at point P0 (0,-1), X‘1,X‘2 and X‘3. The line passing through P0 and X‘1 intersects the x-axis at the point X1 (cos(2π/9),0). The line passing through P0 and X‘2 intersects the x-axis at the point X2 (cos(4π/9),0). Finally, the line passing through P0 and X‘3 intersects the x-axis at the point X3 (cos
(8π/9),0).
A nonagon inscribed inside a unit circle can be constructed using the point X1 (cos(2π/9),0).
Final Notes
Another relevant polynomial equation for constructing a nonagon is x3 – 3x + 1. One of the roots of x3 – 3x + 1 is 2cos(2π/9). In the future I may write a post showing how to solve this polynomial in a similar manner.
A third method of constructing the nonagon involves doing angle trisection 2 times . I discussed how to use Lill’s method and hyperbolas to trisect an angle in my “Trisection Hyperbolas and Lill’s Circle”. Doing a double angle trisection using my method is a bit cumbersome, so I’ll probably not dedicate a post on the topic.
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