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2018 Paper 2 Q7
D: 1600.0 B: 1500.0
vectors geometric proof triangle position vectors ratio parallel lines intersection point cevians

The points \(O\), \(A\) and \(B\) are the vertices of an acute-angled triangle. The points \(M\) and \(N\) lie on the sides \(OA\) and \(OB\) respectively, and the lines \(AN\) and \(BM\) intersect at \(Q\). The position vector of \(A\) with respect to \(O\) is \(\bf a\), and the position vectors of the other points are labelled similarly. Given that \(\vert MQ \vert = \mu \vert QB\vert \), and that \(\vert NQ \vert = \nu \vert QA\vert \), where \(\mu\) and \(\nu\) are positive and \(\mu \nu <1\), show that \[ {\bf m} = \frac {(1+\mu)\nu}{1+\nu} \, {\bf a} \,. \] The point \(L\) lies on the side \(OB\), and \(\vert OL \vert = \lambda \vert OB \vert \,\). Given that \(ML\) is parallel to \(AN\), express \(\lambda\) in terms of \(\mu\) and \(\nu\). What is the geometrical significance of the condition \(\mu\nu<1\,\)?

Solution:

TikZ diagram
The line \(AN\) is \(\mathbf{a} + \alpha (\mathbf{n}-\mathbf{a})\) \\ The line \(BM\) is \(\mathbf{b} + \beta (\mathbf{m} - \mathbf{b})\) The point \(OQ = OB + BQ = \mathbf{b} + \frac{1}{\mu+1} (\mathbf{m}-\mathbf{b})\) It is also \(OQ = OA + AQ = \mathbf{a} + \frac{1}{\nu+1} ( \mathbf{n} - \mathbf{a})\) \begin{align*} && \mathbf{q} &= \mathbf{a} + \frac{1}{\nu+1} ( n\mathbf{b} - \mathbf{a}) \\ && \mathbf{q} &= \mathbf{b} + \frac{1}{\mu+1} ( m\mathbf{a} - \mathbf{b}) \\ \Rightarrow && \frac{\nu}{\nu+1} &= \frac{m}{\mu+1} \\ \Rightarrow && m &= \frac{(1+\mu)\nu}{1+\nu} \\ \Rightarrow && \mathbf{m} &= \frac{(1+\mu)\nu}{1+\nu} \mathbf{a} \end{align*} By similar triangles (\(\triangle OAN \sim \triangle OML\), we can observe that \(\lambda = \mu \nu\). The significance of \(\mu \nu < 1\) \(L\) lies on the side \(OB\) and both \(M\) and \(N\) lie between \(O\) and \(A\) and \(B\) respectively.

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2007 Paper 2 Q8
D: 1600.0 B: 1529.3
vectors locus ratio collinearity Cevians Menelaus-Ceva geometric proof position vectors triangle

The points \(B\) and \(C\) have position vectors \(\mathbf{b}\) and \(\mathbf{c}\), respectively, relative to the origin \(A\), and \(A\), \(B\) and \(C\) are not collinear.

  1. The point \(X\) has position vector \(s \mathbf{b}+t\mathbf{c}\). Describe the locus of \(X\) when \(s+t=1\).
  2. The point \(P\) has position vector \(\beta \mathbf{b}+\gamma\mathbf{c}\), where \(\beta\) and \(\gamma\) are non-zero, and \(\beta+\gamma\ne1\). The line \(AP\) cuts the line \(BC\) at \(D\). Show that \(BD:DC=\gamma:\beta\).
  3. The line \(BP\) cuts the line \(CA\) at \(E\), and the line \(CP\) cuts the line \(AB\) at \(F\). Show that \[ \frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA}=1\,. \]

Solution:

  1. \(X\) lies on the line including \(B\) and \(C\).
  2. points on the line \(AP\) have the form \(\lambda(\beta \mathbf{b}+\gamma\mathbf{c})\), and the point \(D\) will be the point where \(\lambda\beta + \lambda \gamma = 1\). \begin{align*} && \frac{|BD|}{|DC|} &= \frac{|\mathbf{b} -\lambda(\beta \mathbf{b}+\gamma\mathbf{c})| }{|\lambda(\beta \mathbf{b}+\gamma\mathbf{c})- \mathbf{c}|} \\ &&&= \frac{|(1-\lambda \beta)\mathbf{b} - \lambda \gamma \mathbf{c}|}{|\lambda \beta \mathbf{b}+(\lambda \gamma -1)\mathbf{c}|}\\ &&&= \frac{|\lambda \gamma\mathbf{b} - \lambda \gamma \mathbf{c}|}{|\lambda \beta \mathbf{b}-(\lambda \beta)\mathbf{c}|} \\ &&&= \frac{\gamma}{\beta} \end{align*}
  3. The line \(BP\) is \(\mathbf{b} + \mu(\beta \mathbf{b}+\gamma\mathbf{c})\) and will meet \(CA\) when \(1+\mu\beta = 0\), ie \(\mu = -\frac{1}{\beta}\), therefore \(E\) is \(-\frac{\gamma}{\beta}\mathbf{c}\), and so \(\frac{|CE|}{|EA|} = \frac{1+\gamma/\beta}{\gamma/\beta} = \frac{\beta+\gamma}{\gamma}\). Similarly, \(F\) is \(-\frac{\beta}{\gamma}\mathbf{b}\) and \(\frac{|AF|}{|FB|} = \frac{\beta/\gamma}{1+\frac{\beta}{\gamma}} = \frac{\beta}{\gamma+\beta}\), and so \[\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = \frac{\beta}{\gamma+\beta} \frac{\gamma}{\beta} \frac{\beta+\gamma}{\gamma} = 1 \]

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