2006-I-16 – Solving Master

Ans: (b) (i) $x^2+y^2+2x-10y-24=0$ (ii) $(0,2)$

1. Since $H$ is the orthocentre of $\Delta ABC$, then $AO\perp BC$. Hence, we have $\angle AOC = 90^\circ$.
  Since $BS$ is a diameter of the circle, then we have
  
  $\begin{array}{ll}
  \angle BCS = 90^\circ & \text{($\angle$ in semi-circle)}
  \end{array}$
  
  Consider $AH$ and $SC$,
  
  $\angle AOC + \angle BCS$
  
  $\begin{array}{cl}
  = & 90^\circ + 90^\circ \\
  = & 180^\circ
  \end{array}$
  
  $\therefore AH\text{//} SC$ (int. $\angle$s supp.).
  
  Now, we extend $CH$ to $P$ on $AB$. Since $H$ is the orthocentre of $\Delta ABC$, then $CP \perp AB$. Hence, we have $\angle CPA=90^\circ$.
  
  Since $BS$ is a diameter of the circle, then we have
  
  $\begin{array}{ll}
  \angle BAS = 90^\circ & \text{($\angle$ in semi-circle)}
  \end{array}$
  
  Consider $CH$ and $SA$,
  
  $\angle CPA + \angle BAS$
  
  $\begin{array}{cl}
  = & 90^\circ + 90^\circ \\
  = & 180^\circ
  \end{array}$
  
  $\therefore CH\text{//} SA$ (int. $\angle$s supp.).
  
  Since $AH\text{//} SC$ and $CH\text{//} SA$, then $AHCS$ is a parallelogram.
2. Given that $G$ is the centre and $GR \perp BC$. Hence, we have
  $\begin{array}{ll}
  BR = RC & \text{($\perp$ line from centre to chord bisects chord)}
  \end{array}$
  
  In $\Delta BCS$,
  
  $\begin{array}{ll}
  BG = GS & \text{(radii)} \\
  BR = RC & \text{(proved)}
  \end{array}$
  
  $\therefore SC = 2RG$ (mid-pt theorem).
  
  By the result of (a)(i), we have
  
  $\begin{array}{ll}
  AH = SC & \text{(opp. sides of //gram)}
  \end{array}$
  
  Hence, we have $AH=2GR$.
1. Let $x^2 +y^2 +Dx +Ey +F=0$ be the equation of the required circle. Since the circle passes through $A$, $B$ and $C$, then we can substitute the coordinates of $A$, $B$ and $C$ into the equation of the circle. Hence, we have
  $\left\{ \begin{array}{l}
  (0)^2 + (12)^2 +D(0) + E(12) +F =0 \\
  (-6)^2 + (0)^2 +D(-6) + E(0) +F =0 \\
  (4)^2 + (0)^2 +D(4) +E(0) +F =0
  \end{array}\right.$
  
  $\left\{ \begin{array}{ll}
  12E + F = -144 & \ldots \unicode{x2460} \\
  -6D + F = -36 & \ldots \unicode{x2461} \\
  4D + F = -16 & \ldots \unicode{x2462}
  \end{array}\right.$
  
  $\unicode{x2462} – \unicode{x2461}$, we have
  
  $\begin{array}{rcl}
  10D & = & 20 \\
  D & = & 2
  \end{array}$
  
  Sub. $D=2$ into $\unicode{x2462}$, we have
  
  $\begin{array}{rcl}
  4(2) + F & = & -16 \\
  F & = & -24
  \end{array}$
  
  Sub. $F=-24$ into $\unicode{x2460}$, we have
  
  $\begin{array}{rcl}
  12E + (-24) & = & -144 \\
  12E & = & -120 \\
  E & = & -10
  \end{array}$
  
  Therefore, the required equation is $x^2 +y^2 +2x – 10y -24=0$.
2. Note that $H$ lies on the positive $y$-axis, let $H=(0,y)$. Since $H$ is the orthocentre of $\Delta ABC$, we have $AB\perp CH$. Hence, we have
  $\begin{array}{rcl}
  m_{AB} \times m_{CH} & = & -1 \\
  \dfrac{12 – 0}{0 – (-6)} \times \dfrac{y-0}{0-4} & = & -1 \\
  \dfrac{y}{-2} & = & -1 \\
  y & = & 2
  \end{array}$
  
  Therefore, $H=(0,2)$.
3. By the result of (b)(i), $G=(-1,5)$. Consider
  $m_{GH} \times m_{BG}$
  
  $\begin{array}{cl}
  = & \dfrac{5-2}{-1-0} \times \dfrac{5-0}{-1-(-6)} \\
  = & -3 \\
  \neq & -1
  \end{array}$
  
  Therefore $GH\not \perp BG$. Hence, we have $\angle BGH \neq 90^\circ$.
  
  Since $H$ is the orthocentre of $\Delta ABC$, then $HO \perp BC$.
  
  Hence, we have $\angle HOB = 90^\circ$. Therefore, $\angle BGH + \angle HOB \neq 180^\circ$.
  
  Hence, $B$, $O$, $H$ and $G$ are no concyclic.

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