2014
Problem - 109
Suppose that the angles of $\triangle ABC$ satisfy $cos(3A)+cos(3B)+cos(3C)=1.$ Two sides of the triangle have lengths $10$ and $13$. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}$. Find $m$.
As above, we can see that $\cos3A+\cos3B-\cos(3A+3B)=1$
Expanding, we get
$\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1$
$\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B$
$(\cos3A-1)(\cos3B-1)=\sin3A\sin3B$
$\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1$
$\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1$
Note that $\tan{x}=\frac{1}{\tan(90-x)}$, or $\tan{x}\tan(90-x)=1$
Thus $\frac{3A}{2}+\frac{3B}{2}=90$, or $A+B=60$.
Now we know that $C=120$, so we can just use Law or Cosines to get $\boxed{399}$