The regular tridecagon is another regular polygon that cannot be constructed using a compass and straightedge. In this post I want to show how the tridecagon can be constructed using the intersection of a circle and a hyperbola.

In my previous posts “The Heptagon, Hyperbola and Lill’s Circle” and “The Nonagon, Hyperbola and Lill’s Method” I showed how to construct a heptagon and a nonagon by solving cubic equations that have 2cos(2pi/7) and cos(2pi/9) as roots. To construct a regular tridecagon inscribed inside the unit circle, it would be useful to find a way to construct cos(2pi/13). In the next sections I’ll discuss how to solve a specific cubic equation that has cos(2pi/13) as one of its roots.

### Cubic Equation

The cubic equation relevant to the construction of the regular tridecagon is x^3+(((3-sqrt(13))/4)-1/2)x^2-0.25x-((3-sqrt(13))/16). A more elegant representation can be seen in the image below.

If we let r=(3-sqrt(13))/2, then the cubic equation can be represented by (8x^3+4(r-1)x^2-2x-r)/8. The polynomial in the parenthesis is divided by 8 to create a monic polynomial, since it’s more convenient to work with monic polynomials when you use Lill’s method.

The cubic equation is relevant to the construction of a regular tridecagon because one of the roots is x_{1} =cos(2pi/13). The other roots are x_{2} = cos(6pi/13) and x_{3} =cos(8pi/13).

The cubic equation (not divided by 8) is mentioned in this stackexchange discussion. The cubic equation was obtained from the sextic equation 64 x^6 + 32 x^5 – 80 x^4 – 32 x^3 + 24 x^2 + 6 x – 1 = 0.

### Hyperbola and Lill’s Circle

The cubic equation is much uglier than the cubic equation from the posts on the heptagon or the nonagon. So I’ll not bother creating the Lill method representation of the cubic equation. However, the equation of the Lill’s circle of the cubic equation is x^2 + y^2 + ((5/16) (3 – Sqrt[13]) – 0.5) x + 0.75 y = 0.25 .

The Lill’s circle of the cubic is intersected with the hyperbola (x + (3 – Sqrt[13])/16) y = -((3 – Sqrt[13])/16). The circle and the hyperbola intersect at the point A(0,-1) and 3 other points B,C and D. From A we draw 3 lines that connect A to B,C and D. The line AB intersects the x-axis at X_{1} ( cos(2pi/13),0). The line AC intersects the x-axis at X_{2} =(cos(6pi/13),0). Finally, the line AD intersects the x-axis at the point X_{3} =(cos(8pi/13),0). Below you can see the circle and the hyperbola.

At the origin we can construct a unit circle. With the use of the point X_{1} we can construct a tridecagon inscribed inside the unit circle. I will leave the construction of the tridecagon to the reader.

### Final Notes

The equation of Lill’s circle and hyperbola can be written in terms of r. The equation of the hyperbola in terms of r is (x+(r/8))y=-r/8. The equation of Lill’s circle of the cubic is x^2+y^2+((r/8)+0.5(r-1))x+0.75y=0.25. So the coefficients of the circle and the hyperbola are connected to the coefficients of the cubic equation.

Is there a more elegant cubic equation that has cos(2pi/13) as a root? A cubic equation with more convenient coefficients can probably also be solved by the intersection of a circle and a hyperbola that also have more convenient coefficients.