STEP II 2007 question 8 solution

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STEP II 2007 question 8 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 2007 question 8 solution

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