2023-I-19

$\begin{array}{rcl}32& =& {\displaystyle \frac{a+50}{2}}\\ 64& =& a+50\\ a& =& 14\end{array}$

Also,

$\begin{array}{rcl}t& =& {\displaystyle \frac{b+0}{2}}\\ b& =& 2t\end{array}$

Therefore, $R=(14,2t)$.

Since $OP$ is a horizontal line, then $RH$ is a vertical line. Hence, the $x$-coordinate of $H$

$\begin{array}{cl}=& x\text{-coordinate of }R\\ =& 14\end{array}$

Let $H=(14,d)$. Since $OH\perp PR$, then we have

$\begin{array}{rcl}{m}_{OH}\times m_{PR}& =& -1\\ {\displaystyle \frac{d-0}{14-0}}\times {\displaystyle \frac{2t-0}{14-50}}& =& -1\\ dt& =& 252\\ d& =& {\displaystyle \frac{252}{t}}\end{array}$

Therefore, the coordinates of $H$ are $(15,{\displaystyle \frac{252}{t}})$.

Since $OP$ is a horizontal line, then the perpendicular bisector of $OP$ is a vertical line. Hence, the $x$-coordinate of $G$

$\begin{array}{cl}=& {\displaystyle \frac{0+50}{2}}\\ =& 25\end{array}$

Let $G=(25,c)$. Since $Q$ is the mid-point of $PR$, then $GQ$ is the perpendicular bisector of $PR$. Hence, we have

$\begin{array}{rcl}{m}_{GQ}\times m_{PR}& =& -1\\ {\displaystyle \frac{c-t}{25-32}}\times {\displaystyle \frac{2t-0}{14-50}}& =& -1\\ t(c-t)& =& -126\\ c& =& t-{\displaystyle \frac{126}{t}}\\ c& =& {\displaystyle \frac{t^2-126}{t}}\end{array}$

Therefore, the coordinates of $G$ are $(25,{\displaystyle \frac{t^2-126}{t}})$.

1. Since $\mathrm{\angle }PQS=\mathrm{\angle }POQ$, then we have

   $\begin{array}{rcl}\tan \mathrm{\angle }PQS& =& \tan \mathrm{\angle }POQ\\ {\displaystyle \frac{PS}{QS}}& =& {\displaystyle \frac{QS}{SO}}\\ {\displaystyle \frac{50-32}{t-0}}& =& {\displaystyle \frac{t-0}{32-0}}\\ t^2& =& 576\end{array}$

   $\therefore t=24$ or $t=-24$ (rejected).

2. Note that the coordinates of $G$ and $Q$ are $(25,{\displaystyle \frac{75}{4}})$ and $(32,24)$ respectively.

   The slope of $OG$

   $\begin{array}{cl}=& {\displaystyle \frac{\frac{75}{4}-0}{25-0}}\\ =& {\displaystyle \frac{3}{4}}\end{array}$

   The slope of $OQ$

   $\begin{array}{cl}=& {\displaystyle \frac{24-0}{32-0}}\\ =& {\displaystyle \frac{3}{4}}\end{array}$

   Since $m_{OG}=m_{OQ}$ and $O$ is the common point, then $O$, $G$ and $Q$ are collinear.

3. Add the angle bisector of $\mathrm{\angle }OPR$ to the graph. Let $J$ be the foot of perpendicular of $I$ on $OP$.
   

   Since $OQ\perp PQ$ and $IJ\perp OP$, then $QP$ and $JP$ are tangents to the inscribed circle of $\mathrm{\Delta }OPR$ at $Q$ and $J$ respectively.

   $\begin{array}{rcl}QP& =& \sqrt{(32-50{)}^2+(24-0{)}^2}\\ QP& =& 30\end{array}$

   Therefore, the coordinates of $J$ are $(50-30,0)$, i.e. $(20,0)$.

   Since $IJ$ is a vertical line, then the $x$-coordinates of $I$ and $J$ are equal. Let $I=(20,e)$.

   Note that $IJ$ and $IQ$ are radii of the inscribed circle of $\mathrm{\Delta }OPR$.

   $\begin{array}{rcl}IJ& =& IQ\\ e& =& \sqrt{(32-20{)}^2+(24-e{)}^2}\\ e^2& =& 144+576-48e+e^2\\ 48e& =& 720\\ e& =& 15\end{array}$

   Therefore, the coordinates of $I$ are $(20,15)$.

   By the result of (a) and (b)(i), the coordinates of $H$ are $(14,{\displaystyle \frac{21}{2}})$.

   Since $OQ$ is the perpendicular bisector of $PR$, then we have $PQ=RQ$ and $OQ\perp PR$.

   $\begin{array}{cl}& \text{area of }\mathrm{\Delta }GHR:\text{area of }\mathrm{\Delta }IPQ\\ =& {\displaystyle \frac{1}{2}}(GH)(RQ):{\displaystyle \frac{1}{2}}(IQ)(PQ)\\ =& GH:IQ\\ =& \sqrt{(25-14{)}^2+{({\displaystyle \frac{75}{4}}-{\displaystyle \frac{21}{2}})}^2}:\sqrt{(32-20{)}^2+(24-15{)}^2}\\ =& {\displaystyle \frac{55}{4}}:15\\ =& 11:12\end{array}$