2011-I-12

$\begin{array}{rcl}\mathrm{\angle }PCD+\mathrm{\angle }ABP& =& {180}^{\circ }\text{(同旁內角，}AB//CD\text{)}\\ \mathrm{\angle }PCD+{90}^{\circ }& =& {180}^{\circ }\\ \mathrm{\angle }PCD& =& {90}^{\circ }\end{array}$

$\therefore \mathrm{\angle }ABP=\mathrm{\angle }PCD={90}^{\circ }$。

$\begin{array}{rcll}\mathrm{\angle }BAP& =& {180}^{\circ }–\mathrm{\angle }ABP–\mathrm{\angle }APB& \text{(}\mathrm{\Delta }\text{ 的內角和)}\\ & =& {180}^{\circ }–\mathrm{\angle }APD–\mathrm{\angle }APB& \text{(已知)}\\ & =& \mathrm{\angle }CPD\end{array}$

$\begin{array}{rcll}\mathrm{\angle }APB& =& {180}^{\circ }–\mathrm{\angle }ABP–\mathrm{\angle }BAP& \text{(}\mathrm{\Delta }\text{ 的內角和)}\\ & =& {180}^{\circ }-\mathrm{\angle }PCD–\mathrm{\angle }CPD& \text{(已證)}\\ & =& \mathrm{\angle }PDC& \text{(}\mathrm{\Delta }\text{ 的內角和)}\end{array}$

$\therefore \mathrm{\Delta }ABP\sim \mathrm{\Delta }PCD\text{ (A.A.A.)}$。

$\begin{array}{rcl}{\displaystyle \frac{BP}{CD}}& =& {\displaystyle \frac{AB}{PC}}\\ {\displaystyle \frac{x}{k}}& =& {\displaystyle \frac{3}{11-x}}\\ 11x-x^2& =& 3k\\ x^2-11x+3k& =& 0\end{array}$

$\begin{array}{rcl}\mathrm{\Delta }& \ge & 0\\ (-11{)}^2–4(1)(3k)& \ge & 0\\ -12k& \ge & -121\\ k& \le & 10{\displaystyle \frac{1}{12}}\end{array}$

所以，$k$ 的最大值為 $10$。