Question: Take a half circle with radius 2r with the center of O. Now take the half points from the radiuses from O forming the border of the half circle and draw two more half circles with radius r. Finally draw a circle touching the original half circle from the inside and the smaller half circles from the outside. What's the radius of this circle?

So let x be the radius in question, the centre of one of the smaller half circles be A, the centre of the circle in question C, then OC is 2r-x, AC is x+r and OA is r. With the Pythagorean theorem, $(r+x)^2=r^2+(2r-x)^2$, $x=2/3r$.

[![enter image description here](https://i.stack.imgur.com/P6xFt.png)](https://i.stack.imgur.com/P6xFt.png)

<http://ggbm.at/v6XZM74R>

Is there a way to prove this without the Pythagorean theorem? Similar triangles? Alternatively, how would you do a compass-and-straightedge construction?

Answer: 
Here's a way to do it:

This element is certainly in the field $K=\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$. Choose a basis for this field as a vector space over $\mathbb{Q}$; I'd probably choose $1,\sqrt[3]{2},\sqrt[3]{3},\sqrt[3]{6},\sqrt[3]{4},\sqrt[3]{9},\sqrt[3]{12},\sqrt[3]{18},\sqrt[3]{24}$. Kind of an extensive list.

Now, multiplication by $\sqrt[3]{2}+\sqrt[3]{3}$ acts on $K$ as a $\mathbb{Q}$-linear endomorphism. With the basis chosen above, it isn't too dificult to write out a matrix for this endomorphism. Taking the characteristic polynomial of it will give you the minimal polynomial (well, a power of the minimal polynomial - of course this field could be a finite extension of the splitting field. It's not but you'd have to show the characteristic polynomial is irreducible - which it is).

If you can use a computer, then a $9\times 9$ determinant isn't too bad.
First, Mathematica is wrong. The minimal polynomial is $-125-87X^3-15X^6+X^9$.

The irreducible equation for $\sqrt[3]2$ over $\Bbb Q$ is $X^3-2$, and the irreducible equation for $\sqrt[3]2+\sqrt[3]3$ over $\Bbb Q(\sqrt[3]3\,)$would apparently be $F(X)=(X-\sqrt[3]3\,)^3-2=X^3-3\sqrt[3]3X^2+3\sqrt[3]9X-5$. Now, to get a $\Bbb Q$-polunomial out of $F$, you need to multiply it by its two conjugates, gotten by (1) replacing $\sqrt[3]3$ by $\omega\sqrt[3]3$, and (2) replacing $\sqrt[3]3$ by $\omega^2\sqrt[3]3$, where $\omega=\frac12(-1+\sqrt{-3}\,)$ is a chosen primitive cube root of $1$. The conjugate polynomials are
\begin{align}
F\_1(X)&=X^3-3\omega\sqrt[3]3X^2+3\omega^2\sqrt[3]9X-5\>,\>\text{and}\\
F\_2(X)&=X^3-3\omega^2\sqrt[3]3X^2+3\omega\sqrt[3]9X-5\,.
\end{align}
I’ll leave it to you to do the multiplication $FF\_1F\_2$. It’s tedious by hand, but far easier than computing a nine-by-nine determinant!