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2024 Paper 2 Q3
D: 1500.0 B: 1500.0
coordinate geometry unit circle trigonometric identities half-angle formulae geometric transformations rotation tangent half-angle substitution

The unit circle is the circle with radius 1 and centre the origin, \(O\). \(N\) and \(P\) are distinct points on the unit circle. \(N\) has coordinates \((-1, 0)\), and \(P\) has coordinates \((\cos\theta, \sin\theta)\), where \(-\pi < \theta < \pi\). The line \(NP\) intersects the \(y\)-axis at \(Q\), which has coordinates \((0, q)\).

  1. Show that \(q = \tan\frac{1}{2}\theta\).
  2. In this part, \(q \neq 1\).
    1. Let \(\mathrm{f}_1(q) = \dfrac{1+q}{1-q}\). Show that \(\mathrm{f}_1(q) = \tan\frac{1}{2}\!\left(\theta + \frac{1}{2}\pi\right)\).
    2. Let \(Q_1\) be the point with coordinates \((0, \mathrm{f}_1(q))\) and \(P_1\) be the point of intersection (other than \(N\)) of the line \(NQ_1\) and the unit circle. Describe geometrically the relationship between \(P\) and \(P_1\).
    1. \(P_2\) is the image of \(P\) under an anti-clockwise rotation about \(O\) through angle \(\frac{1}{3}\pi\). The line \(NP_2\) intersects the \(y\)-axis at the point \(Q_2\) with co-ordinates \((0, \mathrm{f}_2(q))\). Find \(\mathrm{f}_2(q)\) in terms of \(q\), for \(q \neq \sqrt{3}\).
    2. In this part, \(q \neq -1\). Let \(\mathrm{f}_3(q) = \dfrac{1-q}{1+q}\), let \(Q_3\) be the point with coordinates \((0, \mathrm{f}_3(q))\) and let \(P_3\) be the point of intersection (other than \(N\)) of the line \(NQ_3\) and the unit circle. Describe geometrically the relationship between \(P\) and \(P_3\).
    3. In this part, \(0 < q < 1\). Let \(\mathrm{f}_4(q) = \mathrm{f}_2^{-1}\!\Big(\mathrm{f}_3\!\big(\mathrm{f}_2(q)\big)\Big)\), let \(Q_4\) be the point with coordinates \((0, \mathrm{f}_4(q))\) and let \(P_4\) be the point of intersection (other than \(N\)) of the line \(NQ_4\) and the unit circle. Describe geometrically the relationship between \(P\) and \(P_4\).

Solution:

TikZ diagram
  1. \(\,\) \begin{align*} && \frac{y-0}{x-(-1)} &= \frac{\sin \theta }{\cos \theta + 1} \\ \Rightarrow && y_0 &= \frac{\sin \theta}{\cos \theta + 1} \\ &&&= \frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}+1} \\ &&&= t = \tan \tfrac{\theta}{2} \end{align*} Alternatively, it is straightforward to see from the angles.
    1. \(f_1(q) = \frac{1+q}{1-q}\) so \begin{align*} && f_1(\tan\tfrac12\theta) &= \frac{1+\tan\tfrac12\theta}{1-\tan\tfrac12\theta} \\ &&&= \frac{\cos \tfrac12 \theta + \sin \tfrac12 \theta}{\cos \tfrac12 \theta - \sin \tfrac12 \theta} \\ &&&= \frac{\sin(\tfrac14 \pi + \tfrac12 \theta)}{\cos(\tfrac14 \pi + \tfrac12 \theta)} \\ &&&= \tan \tfrac12(\theta + \tfrac{\pi}{2}) \end{align*}
    2. \(Q_1\) is the point \((0, f_1(q))\) so \(P_1\) will be the point \((\cos (\theta + \tfrac{\pi}{2}), \sin (\theta + \tfrac{\pi}{2}))\) which is a rotation anticlockwise by \(\frac{\pi}{2}\)
    1. \(P_2 = (\cos(\theta + \tfrac{\pi}{3}), \sin( \theta + \tfrac{\pi}{3})\) and so \(f_2(q) = \tan (\tfrac12(\theta + \tfrac{\pi}{3}))\) so \begin{align*} && f_2(q) &= \tan (\tfrac12(\theta + \tfrac{\pi}{3})) \\ &&&= \frac{q + \tan \frac{\pi}{3}}{1 - \tan \frac{\pi}{3} \cdot q} \\ &&&= \frac{q + \frac{1}{\sqrt3}}{1 - \frac{q}{\sqrt{3}}} \\ &&&= \frac{\sqrt3 q + 1}{\sqrt3-q} \end{align*}
    2. Since \(q \to -q\) reflects \((0,q)\) in the \(x\)-axis, \(f_3(q) = f_1(-q)\) so \(P_3\) is the reflection of \(P_1\) so it's rotation by \(\frac{\pi}{2}\) followed by reflection in the \(x\)-axis, which is reflection in \(y=x\). [ie \(\theta \to -\theta + \frac{\pi}{2} \to \frac{\pi}{2}-\theta\)]
    3. We are rotating by \(\frac{\pi}{3}\) then reflecting in \(y=x\) and then rotating by \(-\frac{\pi}{3}\), ie \(\theta \to \theta + \frac{\pi}{3} \to \frac{\pi}{6}-\theta \to -\theta -\frac{\pi}{6} \)

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2021 Paper 3 Q7
D: 1500.0 B: 1500.0
complex numbers unit circle cotangent modulus and argument perpendicularity orthocentre geometric transformations

  1. Let \[ z = \frac{e^{i\theta} + e^{i\phi}}{e^{i\theta} - e^{i\phi}}, \] where \(\theta\) and \(\phi\) are real, and \(\theta - \phi \neq 2n\pi\) for any integer \(n\). Show that \[ z = i\cot\!\bigl(\tfrac{1}{2}(\phi - \theta)\bigr) \] and give expressions for the modulus and argument of \(z\).
  2. The distinct points \(A\) and \(B\) lie on a circle with radius \(1\) and centre \(O\). In the complex plane, \(A\) and \(B\) are represented by the complex numbers \(a\) and \(b\), and \(O\) is at the origin. The point \(X\) is represented by the complex number \(x\), where \(x = a + b\) and \(a + b \neq 0\). Show that \(OX\) is perpendicular to \(AB\). If the distinct points \(A\), \(B\) and \(C\) in the complex plane, which are represented by the complex numbers \(a\), \(b\) and \(c\), lie on a circle with radius \(1\) and centre \(O\), and \(h = a + b + c\) represents the point \(H\), then \(H\) is said to be the orthocentre of the triangle \(ABC\).
  3. The distinct points \(A\), \(B\) and \(C\) lie on a circle with radius \(1\) and centre \(O\). In the complex plane, \(A\), \(B\) and \(C\) are represented by the complex numbers \(a\), \(b\) and \(c\), and \(O\) is at the origin. Show that, if the point \(H\), represented by the complex number \(h\), is the orthocentre of the triangle \(ABC\), then either \(h = a\) or \(AH\) is perpendicular to \(BC\).
  4. The distinct points \(A\), \(B\), \(C\) and \(D\) (in that order, anticlockwise) all lie on a circle with radius \(1\) and centre \(O\). The points \(P\), \(Q\), \(R\) and \(S\) are the orthocentres of the triangles \(ABC\), \(BCD\), \(CDA\) and \(DAB\), respectively. By considering the midpoint of \(AQ\), show that there is a single transformation which maps the quadrilateral \(ABCD\) onto the quadrilateral \(QRSP\) and describe this transformation fully.

Solution:

  1. \(\,\) \begin{align*} && z &= \frac{e^{i \theta} + e^{i \phi}}{e^{i \theta} - e^{i \phi}} \\ &&&= \frac{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})}{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})} \\ &&&= \frac{(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})/2}{i(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})/2i} \\ &&&= \frac{\cos \frac12(\theta-\phi)}{i \sin \frac12(\theta-\phi)} \\ &&&= -i \cot \tfrac12(\theta-\phi) \end{align*} Therefore \(|z| = \cot \tfrac12(\theta-\phi)\) and \(\arg z = \frac{\pi}{2}\)
  2. Since \(a,b\) lie on the unit circle, wlog \(a = e^{i \theta}, b = e^{i\phi}\). Not that the line \(OX\) has vector \(a+b\) and \(AB\) has vector \(b-a\) and not their ratio has argument \(\frac{\pi}{2}\) and hence they are perpendicular.
  3. \(AH\) has vector \(h - a = (a+b+c) - a = b+c\) which we've already established is perpendicular to \(c-b\) which is the vector for \(BC\) (unless \(b+c = 0\) in which case \(h = a\)).
  4. \(p = a +b+c, q = b+c+d\) etc. The midpoint of \(AQ = \frac12(a+b+c+d)\) which is the same as the midpoint of \(BR\), \(CS\) and \(DP\). Therefore we could say the transformation is reflection in the point \(\frac12(a+b+c+d)\)

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2020 Paper 3 Q3
D: 1500.0 B: 1500.0
complex numbers rotation centroid equilateral triangle parallelogram roots of unity geometric transformations

Given distinct points \(A\) and \(B\) in the complex plane, the point \(G_{AB}\) is defined to be the centroid of the triangle \(ABK\), where the point \(K\) is the image of \(B\) under rotation about \(A\) through a clockwise angle of \(\frac{1}{3}\pi\). Note: if the points \(P\), \(Q\) and \(R\) are represented in the complex plane by \(p\), \(q\) and \(r\), the centroid of triangle \(PQR\) is defined to be the point represented by \(\frac{1}{3}(p+q+r)\).

  1. If \(A\), \(B\) and \(G_{AB}\) are represented in the complex plane by \(a\), \(b\) and \(g_{ab}\), show that \[ g_{ab} = \frac{1}{\sqrt{3}}(\omega a + \omega^* b), \] where \(\omega = \mathrm{e}^{\frac{\mathrm{i}\pi}{6}}\).
  2. The quadrilateral \(Q_1\) has vertices \(A\), \(B\), \(C\) and \(D\), in that order, and the quadrilateral \(Q_2\) has vertices \(G_{AB}\), \(G_{BC}\), \(G_{CD}\) and \(G_{DA}\), in that order. Using the result in part (i), show that \(Q_1\) is a parallelogram if and only if \(Q_2\) is a parallelogram.
  3. The triangle \(T_1\) has vertices \(A\), \(B\) and \(C\) and the triangle \(T_2\) has vertices \(G_{AB}\), \(G_{BC}\) and \(G_{CA}\). Using the result in part (i), show that \(T_2\) is always an equilateral triangle.

Solution:

  1. Note that the vector \(\overrightarrow{AB}\) is \(b-a\), and if we rotate this by \(\frac13\pi\) we get \(e^{-i\pi/3}(b-a)\) after rotating it. Therefore the point \(K\) is represented by \(a + e^{-i\pi/3}(b-a)\) and so \(G_{AB}\) is \begin{align*} && g_{ab} &= \tfrac13(a + b + a + e^{-i\pi/3}(b-a)) \\ &&&= \tfrac13((1+ e^{-i\pi/3})b+(2-e^{-i\pi/3})a)\\ &&&= \tfrac13((1+\tfrac12 - \tfrac{\sqrt3}{2}i)b + ((2-\tfrac12+\tfrac{\sqrt3}{2}i)a) \\ &&&= \tfrac13((\tfrac32 - \tfrac{\sqrt3}{2}i)b + ((\tfrac32+\tfrac{\sqrt3}{2}i)a) \\ &&&= \tfrac1{\sqrt3}((\tfrac{\sqrt3}2 - \tfrac{1}{2}i)b + ((\tfrac{\sqrt3}2+\tfrac{1}{2}i)a) \\ &&&= \frac{1}{\sqrt3}(\omega^* b + \omega a) \end{align*}
  2. First note that \(Q_1\) is a parallelogram iff \(c - a = (b-a) + (d-a)\) ie \(a + c = b+d\) (indeed this is true for all quadrilaterals), so. \begin{align*} && Q_1 &\text{ is a parallelogram} \\ \Longleftrightarrow && a + b &= c + d \\ \Longleftrightarrow && \frac{1}{\sqrt{3}}(\omega - \omega^*)(a + c) &= \frac{1}{\sqrt{3}}(\omega -\omega^*)(b + d) \\ \Longleftrightarrow && \frac{1}{\sqrt{3}}(\omega a + \omega^*b)+\frac{1}{\sqrt{3}}(\omega c + \omega^*d) &=\frac{1}{\sqrt{3}}(\omega b + \omega^*c)+\frac{1}{\sqrt{3}}(\omega d + \omega^*a) \\ \Longleftrightarrow && g_{ab}+g_{cd} &=g_{bc}+g_{da} \\ \Longleftrightarrow && Q_2 &\text{ is a parallelogram} \\ \end{align*}
  3. We consider \(\frac{g_{ab}-g_{bc}}{g_{ca}-g_{bc}}\) so \begin{align*} && \frac{g_{ab}-g_{bc}}{g_{ca}-g_{bc}} &= \frac{(\omega a + \omega^*b)-(\omega b + \omega^* c)}{(\omega c + \omega^*a)-(\omega b + \omega^* c)} \\ &&&= \frac{\omega a- \omega^* c -(\omega- \omega^*)b }{\omega^*a-\omega b -(\omega^* -\omega )c} \\ &&&= \frac{\omega^2 a- c -(\omega^2- 1)b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&=\omega^2\frac{ a- \omega^4 c -(1- \omega^4)b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&=\omega^2\frac{ a- (1-\omega^2) c -\omega^2b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&= \omega^2 \end{align*} Therefore the triangle is equilateral.

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