17. In $\triangle A B C, A B=2 A C$ and $\angle B A C=112^{\circ}$. $P$ and $Q$ are points on $B C$ such that $A B^{2}+B C \cdot C P=B C^{2}$ and $3 A C^{2}+2 B C \cdot C Q=B C^{2}$. Find $\angle P A Q$.
在 $\triangle A B C$ 中, $A B=2 A C$ 且 $\angle B A C=112^{\circ} \circ P$ 和 $Q$ 是 $B C$ 上的點, 使得 $A B^{2}+B C \cdot C P=B C^{2}$ 及 $3 A C^{2}+2 B C \cdot C Q=B C^{2}$ 。求 $\angle P A Q$ 。
17. $22^{\circ}$
17. By scaling we may assume $A C=1$ and $A B=2$. Then the cosine law asserts that $B C=\sqrt{5+4 \cos 68^{\circ}}$. Since $C P=\frac{B C^{2}-A B^{2}}{B C}$ and $C Q=\frac{B C^{2}-3 A C^{2}}{2 B C}$, it can be shown that $P$ is between $B$ and $Q$. Since $A B^{2}=B C(B C-C P)=B C \times B P$, we have $\frac{A B}{B C}=\frac{P B}{B A}$.
Together with the common $\angle B$, we have $\triangle A B C \sim$ $\triangle P B A$ and so $\angle B A P=\angle B C A$. It follows that
$$
\angle A P Q=\angle A B P+\angle B A P=\angle A B P+\angle B C A=180^{\circ}-112^{\circ}=68^{\circ} .
$$

On the other hand, we have $\angle A Q P=90^{\circ}$ since
$$
A B^{2}-B Q^{2}=(2 A C)^{2}-(B C-C Q)^{2}=4 A C^{2}-B C^{2}+2 B C \cdot C Q-C Q^{2}=A C^{2}-C Q^{2} \text {. }
$$

Therefore $\angle P A Q=180^{\circ}-68^{\circ}-90^{\circ}=22^{\circ}$.