17. In $\triangle A B C, \angle C=90^{\circ}, \angle B A C<45^{\circ}$ and $A B=7$. $P$ is a point on $A B$ such that $\angle A P C$ $=2 \angle A C P$ and $C P=1$. Find $\frac{A P}{B P}$.
(2 marks)
在 $\triangle A B C$ 中, $\angle C=90^{\circ}, \angle B A C<45^{\circ}$ 且 $A B=7 \circ P$ 是 $A B$ 上的一點, 使得 $\angle A P C=2 \angle A C P$ 及 $C P=1$ 。求 $\frac{A P}{B P}$ 。
17. $6+\sqrt{35}$
17. Let $O$ be the mid-point of $A B$, which is also the circumcentre of $\triangle A B C$. Extend $C P$ to meet the circumcircle at $D$. Note that $P$ is between $O$ and $B$ (if $P$ is between $O$ and $A$ then $\angle A C P$ is less than $45^{\circ}$ while $\angle A P C$ is obtuse, contradicting $\angle A P C=2 \angle A C P$ ).
Let $\angle A C P=\theta$. Then $\angle D P B=\angle A P C=2 \angle A C P=2 \theta$ and $\angle A O D=2 \theta$. It follows that $D O=D P$, both being 3.5 (radius of the circle). Using the power chord theorem $P A \times P B=P C \times P D$, we have $P A \times(7-P A)=1 \times 3.5$. Solving, we get $P A=\frac{7 \pm \sqrt{35}}{2}$. The positive square root is taken as $P$ is between $O$ and $B$. It follows that $P B=\frac{7-\sqrt{35}}{2}$ and so the answer is $\frac{7+\sqrt{35}}{7-\sqrt{35}}=6+\sqrt{35}$.