minimum perimeter is 21 and triangle has side lengths 3, 8, 10. SMO 2013, open section, 2nd round, q3

Let the sides be $a,b,c$. From the sine rule, we have \[\frac{a}{b}=\frac{\sin 3B}{\sin B}=4\cos^2 B-1\]\[\frac{c}{b}=\frac{\sin C}{\sin B}=\frac{\sin 4B}{\sin B}=8\cos^3 B-4\cos B\]Thus, \[2\cos B=\frac{a^2+c^2-b^2}{ac}\in \mathbb{Q}.\]Hence, there exist coprime positive integers $p,q$ such that $2\cos B=\frac{p}{q}$. Hence, \[\frac{a}{b}=\frac{p^2}{q^2}-1 \quad\Leftrightarrow\quad \frac{a}{p^2-q^2}=\frac{b}{q^2}.\]\[\frac{c}{b}=\frac{p^3}{q^3}-\frac{2p}{q} \quad\Leftrightarrow\quad \frac{c}{p^3-2pq^2}=\frac{b}{q^3}.\]Thus, \[\frac{a}{(p^2-q^2)q}=\frac{b}{q^3}=\frac{c}{p^3-2pq^2}=\frac{e}{f},\qquad \gcd(e,f)=1.\]Since the perimeter is minimum, $\gcd(a,b,c)=1$. From $\gcd(e,f)=1$, we have $f|q^3$ and $f|p^3-2pq^2$. We shall prove that $f=1$. If $f>1$, then it has a prime divisor $f'>1$ such that $f'|q^3$ and $f'|p^3-2pq^2$. Thus $f'|q$ and $f'|p$, contradicting $\gcd(p,q)=1$. Thus, $f=1$. From $\gcd(a,b,c)=1$, we conclude that $e=1$. Thus, \[a=(p^2-q^2)q,\quad b=q^3,\quad c=p^3-2pq^2.\]From $0^\circ < \angle A+\angle B=4\angle B<180^\circ$, we get $0^\circ < \angle B < 45^\circ$ and hence $\sqrt{2}<2\cos B<2$ implying that $\sqrt{q}<p<2q$. The smallest positive integers satisfying this inequality are $p=3$ and $q=2$. Since $a+b+c=p^2q+p(p^2-2q^2)$ and $p^2-2q^2=1$, we see that the minimum perimeter is achieved when $p=3$ and $q=2$, and the value is 21.