Rationality and triangles 
Consider a triangle with angles $\alpha, 5\alpha, 180-6\alpha$. What is the minimum perimeter of that triangle, if it has integer sides and $5\alpha<90$?.

Let's call the sides that face each angle $x,y,z$ respectively, and set $\cos \alpha =t$. Using the law of sines , $x=2R\sin\alpha, y=2R\sin5\alpha, z=2R\sin6\alpha$. Setting $2R\sin\alpha=k$, we have the parametrized forms
$$x=k$$
$$y=k(16t^4-12t^2+1)$$
$$z=k(32t^5-32t^3+6t)$$
I got stuck here. If $t$ is rational, I think  we could make an argument for concluding that $t=20/21$(Final update: Nope, I was wrong). But that would depend on this lemma:

If $32t^5-32t^3+6t$ and $16t^4-12t^2+1$ are rational, then so is $t$.

Is this assertion correct? If not, is there an easier way to solve the original question?

Edit:
I managed to prove that my later proposition is false when $t=\pm\frac{\sqrt{3}}{2}, t=\frac{1\pm\sqrt5}4$, but true otherwise. $16t^4-12t^2+1$ is rational iff $16t^4-12t^2$ is, and then we set:
$$16t^4-12t^2=p$$
$$32t^5-32t^3+6t=q$$
$$\implies 2tp-\frac{p}{2t}=q$$
Solving for $t$, we get
$$t=\frac{q\pm\sqrt{4p^2+q^2}}{4p}$$
Substituting in the first equation with the positive case(it is analogous in the negative), we get
$$\frac{q^4}{2 p^4}+\frac{q^2}{2 p^2}-\frac{q\sqrt{4 p^2+q^2}}{2 p^2}+\frac{q^3\sqrt{4 p^2+q^2}}{2 p^4}-2=p$$
If $4p^2+q^2$ is not a perfect square, then $p=\pm q$ or $q=0$. Wolframalpha sledgehammering both cases (it is only necessary to check one case, since the other one corresponds to inverting  the sign of $t$), we see that that  all solutions yield rational $p,q$.
Going back to the original problem, we can forget those solutions since we wanted $0<5\alpha<90$ 

Second update: Substituting $2t=a/b$ with coprime $a,b$, we see that we have to minimize $x+y+z$ subject to $x,y,z$ being integers and $2\cos18<a/b<2$. Therefore, we want to minimize:
$$k\left(\left(\frac ab\right)^5+\left(\frac ab\right)^4-4\left(\frac ab\right)^3-3\left(\frac ab\right)^2+3\left(\frac ab\right)+2\right)$$
It is easy to see that when reducing to the common denominator $b^5$. the numerator will be coprime due to the leading coefficint being $1$. Therefore, $k$ must be $b^5$ in order to have integers(and not another multiple of $b^5$ since we want to minimize the perimeter). So the problem is reduced to:

Minimize $$a^5+a^4b-4a^3b^2-3a^2b^3+3ab^4+2b^5$$
  Subject to $a,b$ positive integers, $2\cos18<a/b<2$


Finally solved the problem! for shortness, let's set $2\cos18=x$. From the inequalities $a\le2b$ and $2\cos(18)b\le a$, we see the trivial inequalities:
$$a^5\ge(xb)^5,a^4\ge(xb)^4, -a^3\ge-8b^3,-a^2\ge-4b^2,a\ge xb$$
So
\begin{align*}
P(a,b)&=a^5+a^4b-4a^3b^2-3a^2b^3+3ab^4+2b^5\\
&\ge b^5(x^5+x^4-32-12+3x+2)\\
&\approx 1.69549b^5
\end{align*}
If $b\ge15, P>1287512.71875\implies P\ge1287513$. We now have an upper bound on the minimum $P$. Since we want $2\cos18<\frac ab\le \frac{2b-1}{b}<2$, we get that $b>(2-2\cos18)^{-1}>10$, so $b\ge 11$. Finally, we are left to try the pairs $a,b$ that meet the requirements and compare the $P$ that corresponds to each pair. So we calculate 
$$P(21,11)=1224640$$
$$P(23,12)=1972355$$
$$P(25,13)=3046536$$
$$P(27,14)=4543825$$
So the minimum perimter of the proposed triangle is $1224640$, with $t=21/22$ and not $20/21$ as initially hypothesised.
