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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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2004 Paper 1 Q6
D: 1484.0 B: 1500.0
coordinate geometry centroid medians perpendicularity orthocentre triangle concurrent lines midpoint

The three points \(A\), \(B\) and \(C\) have coordinates \(\l p_1 \, , \; q_1 \r\), \(\l p_2 \, , \; q_2 \r\) and \(\l p_3 \, , \; q_3 \r\,\), respectively. Find the point of intersection of the line joining \(A\) to the midpoint of \(BC\), and the line joining~\(B\) to the midpoint of \(AC\). Verify that this point lies on the line joining \(C\) to the midpoint of~\(AB\). The point \(H\) has coordinates \(\l p_1 + p_2 + p_3 \, , \; q_1 + q_2 + q_3 \r\,\). Show that if the line \(AH\) intersects the line \(BC\) at right angles, then \(p_2^2 + q_2^2 = p_3^2 + q_3^2\,\), and write down a similar result if the line \(BH\) intersects the line \(AC\) at right angles. Deduce that if \(AH\) is perpendicular to \(BC\) and also \(BH\) is perpendicular to \(AC\), then \(CH\) is perpendicular to \(AB\).

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