16. $A, B, C, M, N$ are points on the circumference of a circle with $M N$ as a diameter. $A$, $B$ are on the same side of $M N$ and $C$ is on the other side, with $A$ being the midpoint of arc $M N$. $C A$ and $C B$ meet $M N$ at $P$ and $Q$ respectively. If $M N=1$ and $M B=\frac{12}{13}$, find the greatest length of $P Q$.
(2 marks)
$A 、 B 、 C 、 M 、 N$ 為某個圓的圓周上的點, 而 $M N$ 為該圓的直徑。 $A 、 B$ 位於 $M N$ 的同一方, 而 $C$ 則位於 $M N$ 的另一方, 且 $A$ 為弧 $M N$ 的中點 $\circ C A$ 和 $C B$分別與 $M N$ 交於 $P$ 和 $Q$ 。若 $M N=1$ 而 $M B=\frac{12}{13}$, 求 $P Q$ 的長度的最大值。
16. $\frac{17-4 \sqrt{15}}{7}$
16. Let $\frac{C M}{C N}=r$ and $d$ be the length of $P Q$. Then we have $\frac{M Q}{N Q}=\frac{[B M C]}{[B N C]}=\frac{B M \times C M}{B N \times C N}=\frac{12 r}{5}$. Since $M N=1$, we have $N Q=\frac{5}{12 r+5}$. Similarly, $\frac{M P}{N P}=\frac{[A M C]}{[A N C]}=\frac{A M \times C M}{A N \times C N}=r$ and this gives $N P=\frac{1}{r+1}$. It follows that
$$
d=P Q=\frac{1}{r+1}-\frac{5}{12 r+5} .
$$

To find the maximum value of $d$, we rewrite the above as a quadratic equation in $r$, namely, $12 d r^{2}+(17 d ?) r+5 d=0$. Since $r$ is a real number, the discriminant must be non-negative, i.e. $(17 d-7)^{2}-4(12 d)(5 d) \geq 0$. Bearing in mind that $d<1$, solving the inequality gives $d \leq \frac{17-4 \sqrt{15}}{7}$ and it is easy to check that such maximum is indeed attainable (by choosing the corresponding value of $r$ which determines the position of $C$ ).