- In $\Delta ABP$ and $\Delta PCD$,
$\begin{array}{rcll}
\angle PCD + \angle ABP & = & 180^\circ \text{(int $\angle$s，$AB//CD$)} \\
\angle PCD + 90^\circ & = & 180^\circ \\
\angle PCD & = & 90^\circ
\end{array}$$\therefore \angle ABP = \angle PCD = 90^\circ$.
$\begin{array}{rcll}
\angle BAP & = & 180^\circ – \angle ABP – \angle APB & \text{($\angle$ sum of $\Delta$)} \\
& = & 180^\circ – \angle APD – \angle APB & \text{(given)} \\
& = & \angle CPD
\end{array}$$\begin{array}{rcll}
\angle APB & = & 180^\circ – \angle ABP – \angle BAP & \text{($\angle$ sum of $\Delta$)} \\
& = & 180^\circ- \angle PCD – \angle CPD & \text{(proved)} \\
& = & \angle PDC & \text{($\angle$ sum of $\Delta$)}
\end{array}$$\therefore \Delta ABP \sim \Delta PCD \text{ (A.A.A.)}$.
- Since $\Delta ABP \sim \Delta PCD$, we have
$\begin{array}{rcl}
\dfrac{BP}{CD} & = & \dfrac{AB}{PC} \\
\dfrac{x}{k} & = & \dfrac{3}{11-x} \\
11x-x^2 & = & 3k \\
x^2 -11x+3k & = & 0
\end{array}$ - Note that $x$ is the length of $BP$ and the root of the equation in (b). Therefore, the equation in (b) must has real roots. Hence, we have
$\begin{array}{rcl}
\Delta & \ge & 0 \\
(-11)^2 – 4(1)(3k) & \ge & 0 \\
-12k & \ge & -121 \\
k & \le & 10\dfrac{1}{12}
\end{array}$Therefore, the greatest value of $k$ is $10$.
2011-I-12
Ans: (c) $10$
