STEP II 2007 question 8 solution - The Student Room
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STEP II 2007 question 8 solution

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i) Line through points B and C.

ii) Line AP = t(\textbf{p} - \textbf{a}) = t\beta\textbf{b} + t\gamma\textbf{c}

Line BC = \textbf{b} + \mu(\textbf{c} - \textbf{b})

They intersect when (1 - \mu)\textbf{b} + \mu\textbf{c} = t\beta\textbf{b} + t\gamma\textbf{c}

t\beta = 1 - \mu, \; t\gamma = \mu \Rightarrow t(\beta + \gamma) = 1 \Rightarrow t = \frac{1}{\beta + \gamma}

So the point of intersection is \frac{\beta}{\beta + \gamma}\textbf{b} + \frac{\gamma}{\beta + \gamma}\textbf{c}

So, as required, BD : DC = \gamma : \beta

iii) It is first necessary to find the position vectors of F and E.

F is at the point where AB and CP intersect.

Line AB = \mu\textbf{b}

Line CP = \textbf{c} + t(\textbf{p} - \textbf{c}) = t\beta\textbf{b} + (1 - t + t\gamma)\textbf{c}

AB = CP

\mu = t\beta

1 - t + t\gamma = 0 \Rightarrow t = \frac{1}{1-\gamma} \Rightarrow \mu = \frac{\beta}{1-\gamma}

Line segment AF = \textbf{f} = \frac{\beta}{1-\gamma}\textbf{b}

Line segment FB = \textbf{b} - \textbf{f} = \frac{1 - \beta - \gamma}{1 - \gamma}\textbf{b}

AF : FB = \beta : 1 - \beta - \gamma

Similarly, E is the point of intersection of the lines CA and PB.

Line CA = -t\textbf{c}

Line PB = \textbf{b} + \mu(\textbf{b} - \textbf{p}) = (1 + \mu - \mu\beta)\textbf{b} - \mu\gamma\textbf{c}

\mu = \frac{1}{\beta - 1}

t = \mu\gamma = \frac{\gamma}{\beta - 1}

Line segment EA = -\textbf{e} = \frac{\gamma}{\beta - 1}\textbf{c}

Line segment CE = \textbf{e} - \textbf{c} = \frac{1 - \beta - \gamma}{\beta - 1}\textbf{c}

CE : EA = 1 - \beta - \gamma : \gamma

So \frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = \frac{\beta}{1 - \beta - \gamma} \times \frac{\gamma}{\beta} \times \frac{1 - \beta - \gamma}{\gamma} = 1

As required.

Solution by Dystopia.

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