14. [11] In triangle $A B C, \angle A=2 \angle C$. Suppose that $A C=6, B C=8$, and $A B=\sqrt{a}-b$, where $a$ and $b$ are positive integers. Compute $100 a+b$.
Answer:
7303

Solution: Let $x=A B$, and $\angle C=\theta$, then $\angle A=2 \theta$ and $\angle B=180-3 \theta$.
Extend ray $B A$ to $D$ so that $A D=A C$. We know that $\angle C A D=180-2 \theta$, and since $\triangle A D C$ is isosceles, it follows that $\angle A D C=\angle A C D=\theta$, and so $\angle D C B=2 \theta=\angle B A C$, meaning that $\triangle B A C \sim \triangle B C D$. Therefore, we have
$$
\frac{x+6}{8}=\frac{8}{x} \Longrightarrow x(x+6)=8^{2}
$$

Since $x>0$, we have $x=-3+\sqrt{73}$. So $100 a+b=7303$.