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2006 Paper 3 Q5
D: 1700.0 B: 1516.0

Show that the distinct complex numbers \(\alpha\), \(\beta\) and \(\gamma\) represent the vertices of an equilateral triangle (in clockwise or anti-clockwise order) if and only if \[ \alpha^2 + \beta^2 +\gamma^2 -\beta\gamma - \gamma \alpha -\alpha\beta =0\,. \] Show that the roots of the equation \begin{equation*} z^3 +az^2 +bz +c=0 \tag{\(*\)} \end{equation*} represent the vertices of an equilateral triangle if and only if \(a^2=3b\). Under the transformation \(z=pw+q\), where \(p\) and \(q\) are given complex numbers with \(p\ne0\), the equation (\(*\)) becomes \[ w^3 +Aw^2 +Bw +C=0\,. \tag{\(**\)} \] Show that if the roots of equation \((*)\) represent the vertices of an equilateral triangle, then the roots of equation \((**)\) also represent the vertices of an equilateral triangle.


Solution: The complex numbers represent an equilateral triangle iff \(\gamma\) is a \(\pm 60^\circ\) rotation of \(\beta\) around \(\alpha\), ie \begin{align*} && \gamma - \alpha &= \omega(\beta - \alpha) \\ \Leftrightarrow && \omega &= \frac{\gamma - \alpha}{\beta - \alpha} \\ \Leftrightarrow && -1 &= \left (\frac{\gamma - \alpha}{\beta - \alpha} \right)^3 \\ \Leftrightarrow && -(\beta - \alpha)^3 &=(\gamma - \alpha)^3 \\ \Leftrightarrow && 0 &= (\gamma-\alpha)^3+(\beta-\alpha)^3 \\ &&&= \gamma^3-3\gamma^2\alpha +3\gamma\alpha^2-\alpha^3 +\beta^3-3\beta^2\alpha+3\beta\alpha^2-\alpha^3 \\ &&&= (\beta + \gamma - 2\alpha)(\alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta) \\ \Leftrightarrow && 0 &= \alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta \end{align*} The roots of the equation \(z^3+az^2+bz+c = 0\) represents the vertices of an equilateral triangle iff \(a^2-3b = (\alpha+\beta+\gamma^2) - 3(\alpha\beta+\beta\gamma+\gamma\alpha) = \alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta = 0\) as erquired. Suppose \(a^2 = 3b\), then consider \(z = pw +q\), we must have \begin{align*} && 0 &= (pw+q)^3+a(pw+q)^2 + b(pw+q)+c \\ &&&= p^3w^3 +(3p^2q+ap^2)w^2+(3pq^2+2apq+bp)w+(q^3+aq^2+bq+c) \\ &&&= p^3w^3+p^2(3q+a)w^2+p(3q^2+2aq+b)w+(q^3+aq^2+bq+c) \\ \end{align*} We need to check if \(\left(\frac{3q+a}{p} \right)^2 = 3 \left (\frac{3q^2+2qa+b}{p^2} \right)\). Clearly the denominators match, so consider the numerators \begin{align*} && (3q+a)^2 &= 9q^2+6aq+a^2 \\ &&&= 9q^2+6aq+3b \\ &&&= 3(3q^2+2qa+b) \end{align*} as required

2000 Paper 3 Q5
D: 1700.0 B: 1500.0

Given two non-zero vectors $\mathbf{a}=\begin{pmatrix}a_{1}\\ a_{2} \end{pmatrix}\( and \)\mathbf{b}=\begin{pmatrix}b_{1}\\ b_{2} \end{pmatrix}\( define \)\Delta\!\! \left( \bf a, \bf b \right)\( by \)\Delta\!\! \left( \bf a, \bf b \right) = a_1 b_2 - a_2 b_1$. Let \(A\), \(B\) and \(C\) be points with position vectors \(\bf a\), \(\bf b\) and \(\bf c\), respectively, no two of which are parallel. Let \(P\), \(Q\) and \(R\) be points with position vectors \(\bf p\), \(\bf q\) and \(\bf r\), respectively, none of which are parallel.

  1. Show that there exists a \(2 \times 2\) matrix \(\bf M\) such that \(P\) and \(Q\) are the images of \(A\) and \(B\) under the transformation represented by \(\bf M\).
  2. Show that \( \Delta\!\! \left( \bf a, \bf b \right) \bf c + \Delta\!\! \left( \bf c, \bf a \right) \bf b + \Delta\!\! \left( \bf b, \bf c \right) \bf a = 0. \) Hence, or otherwise, prove that a necessary and sufficient condition for the points \(P\), \(Q\), and \(R\) to be the images of points \(A\), \(B\) and \(C\) under the transformation represented by some \(2 \times 2\) matrix \(\bf M\) is that \[ \Delta\!\! \left( \bf a, \bf b \right) : \Delta\!\! \left( \bf b, \bf c \right) : \Delta\!\! \left( \bf c, \bf a \right) = \Delta\!\! \left( \bf p, \bf q \right) : \Delta\!\! \left( \bf q, \bf r \right) : \Delta\!\! \left( \bf r, \bf p \right). \]


Solution:

  1. First notice that there is a matrix taking \((1,0)\) and \((0,1)\) to \(P\) and \(Q\). Notice there is also a matrix taking \((1,0)\) and \((0,1)\) to \(A\) and \(B\). Since \(A\) and \(B\) are not parallel, this map is invertible. Then we must be able to compose this inverse with the second map to obtain a matrix \(\mathbf{M}\) satisfying our conditions.
  2. \(\,\) \begin{align*} && LHS &= \Delta\!\! \left( \bf a, \bf b \right) \bf c + \Delta\!\! \left( \bf c, \bf a \right) \bf b + \Delta\!\! \left( \bf b, \bf c \right) \bf a \\ &&&= (a_1b_2-a_2b_1) \binom{c_1}{c_2} + (c_1a_2-c_2a_1)\binom{b_1}{b_2} + (b_1c_2-b_2c_1)\binom{a_1}{a_2} \\ &&&= \binom{a_1b_2c_1-a_2b_1c_1+c_1a_2b_1-c_2a_1b_1+b_1c_2a_1-b_2c_1a_1}{a_1b_2c_2-a_2b_1c_2+c_1a_2b_2-c_2a_1b_2+b_1c_1a_2-b_2c_1a_2} \\ &&&= \binom{0}{0} \\ &&&= \mathbf{0} \end{align*} First note that the matrix taking \(P\), \(Q\) to \(A\), \(B\) is unique. (\(\Rightarrow\)) Suppose \(\mathbf{Ma} = \mathbf{p}\) and \(\mathbf{Mb} = \mathbf{q}\) and \(\mathbf{Mc} = \mathbf{r}\). Then notice that \begin{align*} && \mathbf{0} &= \mathbf{M0} \\ &&&= \mathbf{M}\left ( \Delta\!\! \left( \bf a, \bf b \right) \bf c + \Delta\!\! \left( \bf c, \bf a \right) \bf b + \Delta\!\! \left( \bf b, \bf c \right) \bf a\right) \\ &&&= \Delta\!\! \left( \bf a, \bf b \right)\mathbf{M} \bf c + \Delta\!\! \left( \bf c, \bf a \right) \mathbf{M}\bf b + \Delta\!\! \left( \bf b, \bf c \right) \mathbf{M}\bf a\\ &&&= \Delta\!\! \left( \bf a, \bf b \right)\bf r + \Delta\!\! \left( \bf c, \bf a \right)\bf q + \Delta\!\! \left( \bf b, \bf c \right) \bf p\\ \end{align*} However, since \(\mathbf{p}, \mathbf{q}, \mathbf{r}\) are not parallel, then these coefficients must be a scalar multiples of \(\Delta(\mathbf{p}, \mathbf{q}), \cdots\) as required. \((\Leftarrow)\) Suppose we have this relationship, and \(\mathbf{Ma} = \mathbf{p}\) and \(\mathbf{Mb} = \mathbf{q}\), then \begin{align*} && \mathbf{0} &= \mathbf{M0} \\ &&&= \mathbf{M}\left ( \Delta\!\! \left( \bf a, \bf b \right) \bf c + \Delta\!\! \left( \bf c, \bf a \right) \bf b + \Delta\!\! \left( \bf b, \bf c \right) \bf a\right) \\ &&&= \Delta\!\! \left( \bf a, \bf b \right)\mathbf{M} \bf c + \Delta\!\! \left( \bf c, \bf a \right) \mathbf{M}\bf b + \Delta\!\! \left( \bf b, \bf c \right) \mathbf{M}\bf a\\ &&&= \Delta\!\! \left( \bf a, \bf b \right)\mathbf{Mc} + \Delta\!\! \left( \bf c, \bf a \right)\bf q + \Delta\!\! \left( \bf b, \bf c \right) \bf p\\ \end{align*} Since these are scalar multiples of \(\Delta(\mathbf{p}, \mathbf{q}), \cdots\) and we write this as \begin{align*} && \mathbf{0} &= \Delta(\mathbf{p}, \mathbf{q})\mathbf{Mc} + \Delta(\mathbf{r}, \mathbf{p})\mathbf{q} + \Delta (\mathbf{q}, \mathbf{r})\mathbf{p} \end{align*} But since \(\mathbf{p}, \mathbf{q}, \mathbf{r}\) are not parallel, this means that \(\mathbf{Mc}\) is uniquely defined to be \(\mathbf{r}\) as required.

1991 Paper 3 Q6
D: 1700.0 B: 1516.0

The transformation \(T\) from \(\binom{x}{y}\) to \(\binom{x'}{y'}\) in two-dimensional space is given by \[ \begin{pmatrix}x'\\ y' \end{pmatrix}=\begin{pmatrix}\cosh u & \sinh u\\ \sinh u & \cosh u \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix}, \] where \(u\) is a positive real constant. Show that the curve with equation \(x^{2}-y^{2}=1\) is transformed into itself. Find the equations of two straight lines through the origin which transform into themselves. A line, not necessary through the origin, which has gradient \(\tanh v\) transforms under \(T\) into a line with gradient \(\tanh v'\). Show that \(v'=v+u\). The lines \(\ell_{1}\) and \(\ell_{2}\) with gradients \(\tanh v_{1}\) and \(\tanh v_{2}\) transform under \(T\) into lines with gradients \(\tanh v_{1}'\) and \(\tanh v_{2}'\) respectively. Find the relation satisfied by \(v_{1}\) and \(v_{2}\) that is the necessary and sufficient for \(\ell_{1}\) and \(\ell_{2}\) to intersect at the same angle as their transforms. In the case when \(\ell_{1}\) and \(\ell_{2}\) meet at the origin, illustrate in a diagram the relation between \(\ell_{1}\), \(\ell_{2}\) and their transforms.