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It is well known that the angle trisection or doubling the cube is impossible by using only ruler and compass. The usual algebraic proof uses the fact that if a real number $\xi$ is constructible using only ruler and compass then there is a tower of fields $$\mathbb Q= F_0 \subset F_1 \subset \cdots \subset    F_n$$ such that if $n \geq 1$, then $F_i = F_{i-1}(u_i)$ where $u_i\notin F_{i-1}$ but $u_i^2\in F_{i-1}$ and  $\xi \in F_n$. Hence $2^m = [F_n : \mathbb Q]$ for some $m \in \mathbb N$. This shows that $$2^m = [F_n : \mathbb Q]= [F_n : \mathbb Q(\xi)][\mathbb Q(\xi) : \mathbb Q]$$ So, if the minimal polynomial of $\xi$ in  $\mathbb Q(x)$ is of odd degree, then $\xi$ is not constructible by only ruler and compass. Moreover, the minimal polynomial of $\cos 20^\circ$ has degree $3$. So it is impossible to trisect $60^\circ$ by using only ruler and compass.

However, this proof requires the fundamentals of vector spaces. In this talk, we will tweak the last part of the proof to avoid vector spaces.