A problem in my Galois theory syllabus in the chapter on constructible numbers is as follows:
Check if the small square if constructible from the big square. [Hint: Choose coordinates $0,1\in\mathbf{C}$ as in the picture, and find $z=a+bi$.]
I found two equations from which I can solve $a$ and $b$. Adding up the area's of the four congruent triangles and the small square, we get $1=4\cdot \frac{1}{2}b+(1-a)^2$. Using similar triangles, I can extract a second relation, which is $\frac{1}{\sqrt{b^2+(1-a)^2}}=\frac{\sqrt{b^2+(1-a)^2}}{1-a}$ or $b^2=a(1-a)$.
This gives the following system of equations to solve:
$$\begin{cases} 2b+(1-a)^2=1 \\ b^2=a(1-a) \end{cases}$$
I tried finding the real solution of this system of equations, in vain, which after seeing the solution from Mathematica is not such a surprise.
The minimal polynomial of $z=a+bi$ is, according to Mathematica, $X^3 - 4 X^2 + 6 X - 2$. This polynomial is Eisenstein with $p=2$, so irreducible. By Galois theory, we can then conclude that $z$ is not constructible.
My question is: how can I find either a closed expression for $z$ or it's minimal polynomial without a ton of calculations or use of Mathematica? The author of this problem in my syllabus obviously intended an approach by hand.
We can get the polynomial which $a$ satisfies: $4a(1 - a) = 4b^2 = (1 - (1 - a)^2)^2$ and thus $4(1 - a) = a(2 - a)^2$, equivalently: $a^3 - 4a^2 + 8a - 4 = 0$ which is irreducible $\mathrm{mod}$ $3$ (because it got no root $\mathrm{mod}$ $3$ ). So we can't construct $a$. But $z$ is constructible if and only if both $a, b$ are.