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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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1989 Paper 1 Q3
D: 1516.0 B: 1516.0
vectors scalar product geometric proof triangle perpendicular inequality optimisation intersection point

In the triangle \(OAB,\) \(\overrightarrow{OA}=\mathbf{a},\) \(\overrightarrow{OB}=\mathbf{b}\) and \(OA=OB=1\). Points \(C\) and \(D\) trisect \(AB\) (i.e. \(AC=CD=DB=\frac{1}{3}AB\)). \(X\) and \(Y\) lie on the line-segments \(OA\) and \(OB\) respectively, in such a way that \(CY\) and \(DX\) are perpendicular, and \(OX+OY=1\). Denoting \(OX\) by \(x\), obtain a condition relating \(x\) and \(\mathbf{a\cdot b}\), and prove that \[ \frac{8}{17}\leqslant\mathbf{a\cdot b}\leqslant1. \] If the angle \(AOB\) is as large as possible, determine the distance \(OE,\) where \(E\) is the point of intersection of \(CY\) and \(DX\).

Solution:

TikZ diagram
Denoting \(\overrightarrow{OY}\) by \(\mathbf{y}\) and \(\overrightarrow{OC}\) by \(\mathbf{c}\) etc, we have: \begin{align*} \mathbf{c} &= \frac23 \mathbf{a} + \frac13 \mathbf{b} \\ \mathbf{d} &= \frac13 \mathbf{a} + \frac23 \mathbf{b} \\ \mathbf{x} &= \lambda \mathbf{a} \\ \mathbf{y} &= (1-\lambda) \mathbf{b} \\ 0 &= (\mathbf{d}-\mathbf{x}) \cdot (\mathbf{c} - \mathbf{y}) \\ &=((\frac13 -\lambda)\mathbf{a} + \frac23 \mathbf{b})\cdot(\frac23 \mathbf{a} + (\frac13-1+\lambda) \mathbf{b} ) \\ &= \frac{2}{3} \cdot (\frac13-\lambda) +\frac23 \cdot(\lambda - \frac23)+(\frac{4}{9}+(\frac13-\lambda)(-\frac23+\lambda))\mathbf{a}\cdot\mathbf{b} \\ &= -\frac{2}{9} + (\frac{4}{9} - \frac{2}{9}+\lambda-\lambda^2)\mathbf{a}\cdot \mathbf{b} \\ &= - \frac{2}{9} + (\frac{2}{9} + \lambda - \lambda^2)\mathbf{a}\cdot \mathbf{b} \\ \mathbf{a}\cdot \mathbf{b} &= \frac{2/9}{2/9+\lambda - \lambda^2} \end{align*} Since \(0 \leq \lambda - \lambda^2 \leq \frac14\), \(\frac{\frac29}{\frac29+\frac14} = \frac8{17} \leq \mathbf{a}\cdot\mathbf{b} \leq 1\) If \(\angle AOB\) is as large as possible, \(\mathbf{a}\cdot\mathbf{b}\) is as small as possible, ie \(\lambda = \frac12\) and \(\mathbf{a}\cdot \mathbf{b} = \frac{8}{17}\) First notice that the length \(OM\) to the midpoint of \(AB\) is \(\sqrt{\frac14 (\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})} = \sqrt{\frac14 (2 + 2\mathbf{a}\cdot \mathbf{b})} = \sqrt{\frac12 + \frac4{17}} = \sqrt{\frac{25}{34}} = \frac{5}{\sqrt{34}}\) Notice that \(XYE\) and \(DCE\) are similar triangles, and so the heights satisfy \(\frac{h_1}{h_2} = \frac{\frac12}{\frac13} = \frac32\). Therefore the length \(OE\) is \(\frac12 \frac{5}{\sqrt{34}} + \frac{3}{5} \frac12 \frac{5}{\sqrt{34}} = \frac{8}{10} \frac{5}{\sqrt{34}} = \frac{4}{\sqrt{34}}\)

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