The Trisection Triangle

This was a problem where I was primarily interested in the construction of a model, so the fact of the matter is there's probably a better way to solve this problem rather than jumping through so many mental hula hoops.

Let's say this is the complex plane and lets label the green line $z$ . Since $z$ has length of 1, $z^{2}$ is the blue line, and $z^{3}$ is the red line. (Due to Euler's Formula)

The area of the triangle is $\frac{1}{2}*a*b*sin(\theta)$ or one half of the cross product of the vector form of $a$ and $b$

In complex numbers, we can hobble something together something that acts like the cross product: $Im(\frac{a}{b})|b|^{2}$ (compare complex number multiplication with the calculation of a 2x2 matrix determinant to see why.)

Now, there are three sides to the triangle, $z^{3}-z$ , $z^{2}-z$ , $z^{3}-z^{2}$ , but for use in this operation I will use $z^{3}-z^{2}$ for $a$ and $z^{2}-z$ for $b$ , for reasons that will become apparent.

Let's simplify the first part of the "cross product" expression.

$Im(\frac{z^{3}-z^{2}}{z^{2}-z})$

$Im(\frac{z^{2}-z}{z-1})$

$Im(z)$

$Im(cos(\theta)+i*sin(\theta))$

$sin(\theta)$

Now let's simplify the second part of the "cross product" expression.

$|z^{2}-z|^{2}$

$|e^{2\theta}-e^{\theta}|^{2}$

$|e^{1.5\theta+0.5\theta}-e^{1.5\theta-0.5\theta}|^{2}$

$|e^{1.5\theta}|^{2}|e^{0.5\theta}-e^{-0.5\theta}|^{2}$

$|e^{0.5\theta}-e^{-0.5\theta}|^{2}$

By definition of sine

$4|sin(0.5\theta)|^{2}$

Put them together and this is what you get:

$\frac{1}{2}*sin(\theta)*4|sin(0.5\theta)|^{2}$

$2*sin(\theta)*|sin(0.5\theta)|^{2}$ = area of triangle.

A surprisingly simple result!

So, this is where people might get mad, because from here it's essentially just checking off the possible solutions.

We know that since the area is nice recognizable irrational number, it probably came from a nice recognizable angle. Not only that, but this angle has to be 'nice' for sine even when it is halved. Setting our upper boundary of search at pi, we see this applies for only pi/2 and pi/3. Since pi/2 obviously has an area of 1, we can hazard a guess that the correct angle is pi/3, or 60 degrees.

Let the three angles described be α,2α and 3α. Then the vertices of the triangle are (cosα,sinα),(cos2α,sin2α) and (cos3α,sin3α). The area of the triangle can easily be calculated. It is sinα|1-cosα| which is given as $\dfrac{√3}{4}$ . Solving this equation, we get α=60 degrees.